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**Unformatted text preview: **Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 1. (a) The point ( 1, 2 ) is on the graph of f , so f ( 1)= 2 .
(b) When x=2 , y is about 2.8 , so f (2) 2.8 .
(c) f (x)=2 is equivalent to y=2 . When y=2 , we have x= 3 and x=1 .
(d) Reasonable estimates for x when y=0 are x= 2.5 and x=0.3 .
(e) The domain of f consists of all x values on the graph of f . For this function, the domain is
3 x 3 , or 3,3 . The range of f consists of all y values on the graph of f . For this function,
the range is 2 y 3 , or 2,3 .
(f) As x increases from 1 to 3 , y increases from 2 to 3 . Thus, f is increasing on the interval 1,3
.
2. (a) The point ( 4, 2 ) is on the graph of f , so f ( 4)= 2 . The point ( 3,4 ) is on the graph of g , so
g(3)=4 .
(b) We are looking for the values of x for which the y values are equal. The y values for f and g
are equal at the points ( 2,1 ) and ( 2,2 ) , so the desired values of x are 2 and 2 .
(c) f (x)= 1 is equivalent to y= 1 . When y= 1 , we have x= 3 and x=4 .
(d) As x increases from 0 to 4 , y decreases from 3 to 1 . Thus, f is decreasing on the interval 0,4
.
(e) The domain of f consists of all x values on the graph of f . For this function, the domain is
4 x 4 , or 4,4 . The range of f consists of all y values on the graph of f . For this function,
the range is 2 y 3 , or 2,3 .
(f) The domain of g is 4,3 and the range is 0.5,4 .
3. From Figure 1 in the text, the lowest point occurs at about ( t,a ) = ( 12, 85) . The highest point
occurs at about ( 17,115) . Thus, the range of the vertical ground acceleration is 85 a 115 . In
Figure 11, the range of the north south acceleration is approximately 325 a 485 . In Figure 12,
the range of the east west acceleration is approximately 210 a 200 .
4. Example 1: A car is driven at 60 mi / h for 2 hours. The distance d traveled by the car is a function
of the time t . The domain of the function is {t |0 t 2} , where t is measured in hours. The range of
the function is { d |0 d 120} , where d is measured in miles. Example 2: At a certain university, the number of students N on campus at any time on a particular
day is a function of the time t after midnight. The domain of the function is {t |0 t 24} , where t is
measured in hours. The range of the function is { N |0 N k} , where N is an integer and k is the
largest number of students on campus at once. 1 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function Example 3: A certain employee is paid $8.00 per hour and works a maximum of 30 hours per week.
The number of hours worked is rounded down to the nearest quarter of an hour. This employee’s
gross weekly pay P is a function of the number of hours worked h . The domain of the function is
0,30 and the range of the function is { 0,2.00,4.00,... ,238.00,240.00} . 5. No, the curve is not the graph of a function because a vertical line intersects the curve more than
once. Hence, the curve fails the Vertical Line Test.
6. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is
[ 2,2] and the range is [ 1,2] .
7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is
[ 3,2] and the range is 3, 2 ) [ 1,3] .
8. No, the curve is not the graph of a function since for x=0 ,
points on the curve. 1 , and 2 , there are infinitely many 9. The person’s weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years.
The person’s weight dropped to about 120 pounds for the next 5 years, then increased rapidly to
about 170 pounds. The next 30 years saw a gradual increase to 190 pounds. Possible reasons for the
drop in weight at 30 years of age: diet, exercise, health problems.
10. The salesman travels away from home from 8 to 9 A.M. and is then stationary until 10 : 00 . The
salesman travels farther away from 10 until noon. There is no change in his distance from home until
1 : 00 , at which time the distance from home decreases until 3 : 00 . Then the distance starts
increasing again, reaching the maximum distance away from home at 5 : 00 . There is no change from
5 until 6 , and then the distance decreases rapidly until 7 : 00 P.M., at which time the salesman
reaches home.
11. The water will cool down almost to freezing as the ice melts. Then, when the ice has melted, the
water will slowly warm up to room temperature.
2 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 12. The summer solstice (the longest day of the year) is around June 21, and the winter solstice (the
shortest day) is around December 22. 13. Of course, this graph depends strongly on the geographical location! 14. The temperature of the pie would increase rapidly, level off to oven temperature, decrease rapidly,
and then level off to room temperature. 15. 16. (a) (b)
(c)
3 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function (d) 17. (a)
(b) From the graph, we estimate the number of cell phone subscribers in Malaysia to be about 540 in 1994 and 1450 in 1996. 18. (a)
(b) From the graph in part (a), we estimate the temperature at 11:00 A.M. to be about 84.5 C.
2 19. f (x)=3x x+2.
2 f (2)=3(2) 2+2=12 2+2=12.
2 f ( 2)=3( 2) ( 2)+2=12+2+2=16.
4 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 2 f (a)=3a a+2.
2 2 f ( a)=3( a) ( a)+2=3a +a+2.
2 2 2 2 f (a+1)=3(a+1) (a+1)+2=3(a +2a+1) a 1+2=3a +6a+3 a+1=3a +5a+4.
2 2 2 f (a)=2 f (a)=2(3a a+2)=6a 2a+4.
2 2 2 f (2a)=3(2a) (2a)+2=3(4a ) 2a+2=12a 2a+2.
2 22 2 4 2 4 2 f (a )=3(a ) (a )+2=3(a ) a +2=3a a +2.
f (a) 2 2 2 ( 2 = 3a a+2 = 3a a+2
4 3 2 3 ) ( 3a2 a+2) 2 2 4 3 2 =9a 3a +6a 3a +a 2a+6a 2a+4=9a 6a +13a 4a+4.
2 2 2 2 2 f (a+h)=3(a+h) (a+h)+2=3(a +2ah+h ) a h+2=3a +6ah+3h a h+2.
4
3 4
3
2
r +3r +3r+1 . We
( r+1 ) =
3
3
wish to find the amount of air needed to inflate the balloon from a radius of r to r+1 . Hence, we need
4
4 3 4
3
2
2
to find the difference V ( r+1 ) V ( r ) =
r +3r +3r+1
r =
3r +3r+1 .
3
3
3 ( 20. A spherical balloon with radius r+1 has volume V ( r+1 ) = ( 2 ) 2 ( 2 2 21. f (x)=x x , so f (2+h)=2+h (2+h) =2+h (4+4h+h )=2+h 4 4h h =
2 2 2 2 ) ) ( h2+3h+2) , 2 f (x+h)=x+h ( x+h ) =x+h (x +2xh+h )=x+h x 2xh h , and
2 2 2 2 f (x+h) f (x) x+h x 2xh h x+x
h 2xh h
h(1 2x h)
=
=
=
=1 2x h .
h
h
h
h
x
2+h
2+h
x+h
, so f (2+h)=
=
, f (x+h)=
, and
x+1
2+h+1 3+h
x+h+1
x+h
x
f (x+h) f (x) x+h+1 x+1
1
( x+h ) ( x+1 ) x ( x+h+1 )
=
=
=
.
h
h
h ( x+h+1 ) ( x+1 )
( x+h+1 ) ( x+1 ) 22. f (x)= 23. f (x)=x/(3x 1) is defined for all x except when 0=3x 1 { x R| x 1
3 } = /( ,
2 1
3 1
,
3 ) x= 1
, so the domain is
3 .
2 24. f (x)=(5x+4) x +3x+2 is defined for all x except when 0=x +3x+2
1 , so the domain is { x R| x 2, 1} =(
, 2) ( 2, 1) ( 1, ) . 0=(x+2)(x+1) x= 2 or 25.
5 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 3 f (t)= t + t is defined when t 0 . These values of t give real number results for t , whereas any value of t gives a real number result for 3 t . The domain is 0, 26. g(u)= u + 4 u is defined when u 0 and 4 u 0 / 4 ). u 4 . Thus, the domain is 0 u 4= 0,4 . 2 2 2 27. h(x)=1
x 5x is defined when x 5x>0 x(x 5)>0 . Note that x 5x 0 since that would
result in division by zero. The expression x(x 5) is positive if x<0 or x>5 . (See Appendix A for
,0) (5, ) .
methods for solving inequalities.) Thus, the domain is (
2 2 28. h(x)= 4 x . Now y= 4 x 2 2 y =4 x 2 2 x +y =4 , so the graph is the top half of a circle of radius 2 with center at the origin. The domain is
From the graph, the range is 0 y 2 , or 0,2 . { x|4 x2 0} = { x|4 29. f (x)=5 is defined for all real numbers, so the domain is R , or (
horizontal line with y intercept 5 . , 1
(x+3) is defined for all real numbers, so the domain is R , or (
2
3
a line with x intercept 3 and y intercept
.
2 2 x } =[ 2,2] . ) . The graph of f is a 30. F(x)= 31. f (t)=t 6t is defined for all real numbers, so the domain is R , or ( } ={ x|2 2 x , , ) . The graph of F is ) . The graph of f is a 2 parabola opening upward since the coefficient of t is positive. To find the t intercepts, let y=0 and
2 solve for t . 0=t 6t=t(t 6) t=0 and t=6 . The t coordinate of the vertex is halfway between the t
6 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 2 intercepts, that is, at t=3 . Since f (3)=3 6 3= 9 , the vertex is (3, 9) . 2 4 t
(2+t)(2 t)
32. H(t)=
, so for t 2 , H(t)=2+t . The domain is {t |t 2} . So the graph of H is
=
2 t
2 t
the same as the graph of the function f (t)=t+2 (a line) except for the hole at ( 2,4 ) . 33. g(x)= x 5 is defined when x 5 0 or x 5 , so the domain is 5,
2 y =x 5 ) . Since y= x 5 2 x=y +5 , we see that g is the top half of a parabola. 34.
F(x)= 2x+1 = = {
{ 2x+1
(2x+1) if 2x+1 0
if 2x+1<1
1
2
1
if x<
2 if x 2x+1
2x 1 The domain is R , or ( , ). 35.
7 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function G(x)= 3x+ x
x . Since x = G(x)= { { x if x 0
x if x<0 3x+x
x
3x x
x , we have if x>0
=
if x<0 { 4x
x
2x
x Note that G is not defined for x=0 . The domain is (
36. g(x)= x
2 x . Since x = { g(x)= x if x 0
x if x<0 { x x
2 =
if x<0 ,0) (0, { 4 if x>0
2 if x<0 ). , we have x
2 if x>0 if x>0
=
if x<0 x Note that g is not defined for x=0 . The domain is ( { 1
x
1
x ,0) (0, if x>0
if x<0 ). { x
if x 0
x+1 if x>0
Domain is R , or (
, ).
37. f (x)= 8 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 38. f (x)= { 39. f (x)= { 2x+3 if x< 1
3 x if x 1
Domain is R , or (
, ). x+2 if x 1
2
if x> 1
x
2 Note that for x= 1 , both x+2 and x are equal to 1. Domain is R . 40. f (x)= { 1
if x 1
3x+2 if 1<x<1
7 2x if x 1 Domain is R . ( y y 41. Recall that the slope m of a line between the two points x ,y 1 1 2 ) and ( x ,y ) is m= x
2 2 2 1 x and an 1 equation of the line connecting those two points is y y =m(x x ) . The slope of this line segment is
1 1 6 1
7
7
7
4
=
, so an equation is y 1=
(x+2) . The function is f (x)=
x
, 2 x 4.
4 ( 2)
6
6
6
3
42. The slope of this line segment is
f (x)= 3 ( 2) 5
5
= , so an equation is y+2= (x+3) . The function is
6 ( 3) 9
9 5
1
x
, 3 x 6.
9
3 43. We need to solve the given equation for y .
9 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 2 2 x+ ( y 1 ) =0 ( y 1 ) = x y 1=
x y=1
x . The expression with the positive radical
represents the top half of the parabola, and the one with the negative radical represents the bottom
x . Note that the domain is x 0 .
half. Hence, we want f (x)=1
2 2 44. ( x 1 ) +y =1
0 x 2. 2 y= 1 (x 1) = 2 2 2x x . The top half is given by the function f (x)= 2x x , 45. For 1 x 2 , the graph is the line with slope 1 and y intercept 1 , that is, the line y=x+1 . For
3
3
3
2<x 4 , the graph is the line with slope
and x intercept 4 , so y 0=
(x 4)=
x+6 . So the
2
2
2
x+1
if 1 x 2
3
function is f (x)=
x+6 if 2<x 4
2 { 46. For x 0 , the graph is the line y=2 . For 0<x 1 , the graph is the line with slope 2 and y
intercept 2 , that is, the line y= 2x+2 . For x>1 , the graph is the line with slope 1 and x intercept 1 ,
2
if x 0
that is, the line y=1( x 1 ) =x 1 . So the function is f (x)=
2x+2 if 0<x 1 .
x 1
if 1<x { 47. Let the length and width of the rectangle be L and W . Then the perimeter is 2L+2W =20 and the
20 2L
area is A=LW . Solving the first equation for W in terms of L gives W =
=10 L . Thus,
2
2 A(L)=L(10 L)=10L L . Since lengths are positive, the domain of A is 0<L<10 . If we further restrict
L to be larger than W , then 5<L<10 would be the domain.
48. Let the length and width of the rectangle be L and W . Then the area is LW =16 , so that W =16/L .
The perimeter is P=2L+2W , so P(L)=2L+2(16/L)=2L+32/L , and the domain of P is L>0 , since
lengths must be positive quantities. If we further restrict L to be larger than W , then L>4 would be
the domain.
49. Let the length of a side of the equilateral triangle be x . Then by the Pythagorean Theorem, the
2 2
3
1
2
2 2 1 2 3 2
x =x , so that y =x
x = x and y=
x . Using the
height y of the triangle satisfies y +
2
4
4
2
3
3 2
1
1
formula for the area A of a triangle, A= ( base ) ( height ) , we obtain A(x)= (x)
x =
x ,
2
2
2
4
with domain x>0 .
3 3 50. Let the volume of the cube be V and the length of an edge be L . Then V =L so L= V , and the
10 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function surface area is S(V )=6 ( 3 V ) 2=6V 2/3 , with domain V >0 . 51. Let each side of the base of the box have length x , and let the height of the box be h . Since the
2 2 2 volume is 2 , we know that 2=hx , so that h=2/x , and the surface area is S=x +4xh . Thus,
2 2 2 S(x)=x +4x(2/x )=x +(8/x) , with domain x>0 .
1
52. The area of the window is A=xh+
2 2 2 x
, where h is the height of the
=xh+
8
1
1
rectangular portion of the window. The perimeter is P=2h+x+
x=30 2h=30 x
x
2
2
1
h= ( 60 2x x ) . Thus,
4
60 2x
A(x)=x
4 x 1
x
2 2 x
1 2
4 2
2
2
2
2
+
=15x
x
x + x =15x
x
x =15x x
8
2
4
8
8
8 +4
8 Since the lengths x and h must be positive quantities, we have x>0 and h>0 . For h>0 , we have 2h>0
1
60
60
30 x
x>0 60>2x+ x x<
. Hence, the domain of A is 0<x<
.
2
2+
2+
53. The height of the box is x and the length and width are L=20 2x , W =12 2x . Then V =LWx and
so ( 2 ) 3 2 V (x)= ( 20 2x ) ( 12 2x ) (x)=4(10 x)(6 x)(x)=4x 60 16x+x =4x 64x +240x .
The sides L , W , and x must be positive. Thus, L>0 20 2x>0 x<10 ; W >0 12 2x>0
x>0 . Combining these restrictions gives us the domain 0<x<6 .
54. C(x)= { $2.00
2.20
2.40
2.60
2.80
3.00
3.20
3.40
3.60
3.80
4.00 x<6 ; and if 0.0<x 1.0
if 1.0<x 1.1
if 1.1<x 1.2
if 1.2<x 1.3
if 1.3<x 1.4
if 1.4<x 1.5
if 1.5<x 1.6
if 1.6<x 1.7
if 1.7<x 1.8
if 1.8<x 1.9
if 1.9<x<2.0 11 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 55. (a)
(b) On $14 , 000 , tax is assessed on $4000 , and 10%($4000)=$400 .
On $26 , 000 , tax is assessed on $16 , 000 , and 10%($10 , 000)+15%($6000)=$1000+$900=$1900 .
(c) As in part (b), there is $1000 tax assessed on $20,000 of income, so the graph of T is a line
segment from (10 , 000,0) to (20 , 000,1000) . The tax on $30,000 is $2500, so the graph of T for
x>20 , 000 is the ray with initial point (20 , 000,1000) that passes through (30 , 000,2500) . 56. One example is the amount paid for cable or telephone system repair in the home, usually
measured to the nearest quarter hour. Another example is the amount paid by a student in tuition fees,
if the fees vary according to the number of credits for which the student has registered.
57. f is an odd function because its graph is symmetric about the origin. g is an even function
because its graph is symmetric with respect to the y axis.
58. f is not an even function since it is not symmetric with respect to the y axis. f is not an odd
function since it is not symmetric about the origin. Hence, f is neither even nor odd. g is an even
function because its graph is symmetric with respect to the y axis.
59. (a) Because an even function is symmetric with respect to the y axis, and the point ( 5,3) is on
the graph of this even function, the point ( 5,3) must also be on its graph.
(b) Because an odd function is symmetric with respect to the origin, and the point ( 5,3) is on the
graph of this odd function, the point ( 5, 3) must also be on its graph.
60. (a) If f is even, we get the rest of the graph by reflecting about the y axis.
12 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function (b) If f is odd, we get the rest of the graph by rotating 180 about the origin. 61. f (x)=x 2 .
1 2 f ( x) = ( x ) = ( x) 2 1 = 2 x 2 =x = f (x)
So f is an even function. 62. f (x)=x 3 .
3 f ( x) = ( x ) =
= 1
x 3 = 1 ( x) 3 ( x 3) = = 1
x 3 f (x) So f is odd. 2 2 2 63. f (x)=x +x , so f ( x)= ( x ) + ( x ) =x x . Since this is neither f (x) nor f (x) , the function f is
13 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function neither even nor odd.
4 2 64. f (x)=x 4x .
f ( x) = ( x ) 4 4 ( x ) 2
4 2 =x 4x = f (x)
So f is even. 3 65. f (x)=x x .
f ( x) = ( x ) 3 ( x ) = x3+x
= ( x3 x ) = f (x) So f is odd. 3 2 3 2 3 2 66. f (x)=3x +2x +1 , so f ( x)=3( x) +2( x) +1= 3x +2x +1 . Since this is neither f (x) nor f (x) ,
the function f is neither even nor odd. 14 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions 5 1. (a) f (x)= x is a root function with n=5 .
2 (b) g(x)= 1 x is an algebraic function because it is a root of a polynomial.
9 4 (c) h(x)=x +x is a polynomial of degree 9 .
2 (d) r(x)= x +1
3 is a rational function because it is a ratio of polynomials. x +x
(e) s(x)=tan 2x is a trigonometric function.
(f) t(x)=log x is a logarithmic function.
10 2. (a) y= ( x 6 ) / ( x+6 ) is a rational function because it is a ratio of polynomials.
2 / (b) y=x+x
x 1 is an algebraic function because it involves polynomials and roots of
polynomials.
x (c) y=10 is an exponential function (notice that x is the exponent ).
10 (d) y=x 6 is a power function (notice that x is the base ).
4 (e) y=2t +t
is a polynomial of degree 6 .
(f) y=cos +sin is a trigonometric function.
3. We notice from the figure that g and h are even functions (symmetric with respect to the y axis)
and that f is an odd function (symmetric with respect to the origin). So (b) y=x
g is flatter than h near the origin, we must have (c) y=x
with h . 8 5 must be f . Since
2 matched with g and (a) y=x matched 4. (a) The graph of y=3x is a line (choice G ).
x (b) y=3 is an exponential function (choice f ).
3 (c) y=x is an odd polynomial function or power function (choice F ).
3 1/3 (d) y= x =x is a root function (choice g ). 5. (a) An equation for the family of linear functions with slope 2 is y= f (x)=2x+b , where b is the y
intercept. 1 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions (b) f (2)=1 means that the point ( 2,1 ) is on the graph of f . We can use the point slope form of a line
to obtain an equation for the family of linear functions through the point ( 2,1 ) . y 1=m ( x 2 ) , which
is equivalent to y=mx+ ( 1 2m ) in slope intercept form. (c) To belong to both families, an equation must have slope m=2 , so the equation in part (b),
y=mx+ ( 1 2m ) , becomes y=2x 3 . It is the only function that belongs to both families.
6. All members of the family of linear functions f (x)=1+m(x+3) have graphs that are lines passing
through the point ( 3,1) . 7. All members of the family of linear functions f (x)=c x have graphs that are lines with slope 1 .
The y intercept is c . 8. (a)
2 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions (b) The slope of 4 means that for each increase of 1 dollar for a rental space, the number of spaces
rented decreases by 4 . The y intercept of 200 is the number of spaces that would be occupied if
there were no charge for each space. The x intercept of 50 is the smallest rental fee that results in no
spaces rented. 9. (a)
9
9
means that F increases
degrees for each increase of 1 C. (Equivalently, F
5
5
increases by 9 when C increases by 5 and F decreases by 9 when C decreases by 5 .) The F
intercept of 32 is the Fahrenheit temperature corresponding to a Celsius temperature of 0 .
(b) The slope of 10. (a) Let d= distance traveled (in miles) and t= time elapsed (in hours). At t=0 , d=0 and at t=50
1 5
5
40 0
minutes =50
= h, d=40 . Thus we have two points: ( 0,0 ) and
,40 , so m=
=48 and
60 6
6
5
0
6
so d=48t . (b)
(c) The slope is 48 and represents the car’s speed in mi / h.
11. (a) Using N in place of x and T in place of y , we find the slope to be
T T
1
1
173
80 70
10 1
2 1
=
=
= . So a linear equation is T 80= ( N 173) T 80= N
6
6
6
173 113 60 6
N N
2 1 3 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions 307
=51.16 .
6
1
(b) The slope of
means that the temperature in Fahrenheit degrees increases one sixth as rapidly
6
as the number of cricket chirps per minute. Said differently, each increase of 6 cricket chirps per T= 1
307
N+
6
6 minute corresponds to an increase of 1 F.
(c) When N=150 , the temperature is given approximately by T = 1
307
=76.16 F
( 150 ) +
6
6 76 F. 12. (a) Let x denote the number of chairs produced in one day and y the associated cost. Using the
4800 2200 2600
points ( 100,2200 ) and ( 300,4800 ) we get the slope
=
=13 . So y 2200=13 ( x 100 )
300 100
200
y=13x+900 . (b) The slope of the line in part (a) is 13 and it represents the cost (in dollars) of producing each
additional chair.
(c) The y intercept is 900 and it represents the fixed daily costs of operating the factory. change in pressure
4.34
=
=0.434 . Using P for pressure and d for
10 feet change in depth
10
depth with the point ( d,P ) = ( 0,15) , we have the slope intercept form of the line, P=0.434d+15 .
85
(b) When P=100 , then 100=0.434d+15 0.434d=85 d=
195.85 feet. Thus, the pressure is
0.434
13. (a) We are given 2 100 lb/in at a depth of approximately 196 feet.
14. (a) Using d in place of x and C in place of y , we find the slope to be
C C
2 d d
2 1 1 = 460 380 80 1
=
=
800 480 320 4
4 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions So a linear equation is C 460= 1
( d 800 )
4 (b) Letting d=1500 we get C= 1
( 1500 ) +260=635 . The cost of driving 1500 miles is $635.
4 C 460= 1
1
d 200 C= d+260 .
4
4 (c)
The slope of the line represents the cost per mile, $0.25 .
(d) The y intercept represents the fixed cost, $260 .
(e) A linear function gives a suitable model in this situation because you have fixed monthly costs
such as insurance and car payments, as well as costs that increase as you drive, such as gasoline, oil,
and tires, and the cost of these for each additional mile driven is a constant.
15. (a) The data appear to be periodic and a sine or cosine function would make the best model. A
model of the form f (x)=a cos (bx)+c seems appropriate.
(b) The data appear to be decreasing in a linear fashion. A model of the form f (x)=mx+b seems
appropriate.
x 16. (a) The data appear to be increasing exponentially. A model of the form f (x)=a b or
x f (x)=a b +c seems appropriate.
(b) The data appear to be decreasing similarly to the values of the reciprocal function. A model of the
form f (x)=a/x seems appropriate.
17. Some values are given to many decimal places. These are the results given by several computer
algebra systems rounding is left to the reader. (a)
A linear model does seem appropriate.
(b) Using the points ( 4000,14.1 ) and ( 60,000,8.2 ) , we obtain y 14.1=
equivalently, y 8.2 14.1
( x 4000 ) or,
60,000 4000 0.000105357x+14.521429 .
5 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions (c) Using a computing device, we obtain the least squares regression line
y= 0.0000997855x+13.950764 .
The following commands and screens illustrate how to find the least squares regression line on a TI
83 Plus. Enter the data into list one (L1) and list two (L2). Press
to enter the editor. Find the regession line and store it in Y 1 . Press . Note from the last figure that the regression line has been stored in Y 1 and that Plot1 has been turned
on (Plot1 is highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and
or by pressing
.
pressing Now press
to produce a graph of the data and the regression line. Note that choice 9 of the
ZOOM menu automatically selects a window that displays all of the data. (d) When x=25 , 000 , y 11.456 ; or about 11.5 per 100 population.
(e) When x=80 , 000 , y 5.968 ; or about a 6% chance.
(f) When x=200 , 000 , y is negative, so the model does not apply.
18. (a)
6 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions (b)
Using a computing device, we obtain the least squares regression line y=4.856x 220.96 .
(c) When x=100 F, y=264.7 265 chirps / min. 19. (a)
A linear model does seem appropriate. (b)
Using a computing device, we obtain the least squares regression line y=0.089119747x 158.2403249
, where x is the year and y is the height in feet.
(c) When x=2000 , the model gives y 20.00 ft. Note that the actual winning height for the 2000
Olympics is less than the winning height for 1996 so much for that prediction.
(d) When x=2100 , y 28.91 ft. This would be an increase of 9.49 ft from 1996 to 2100. Even though
there was an increase of 8.59 ft from 1900 to 1996, it is unlikely that a similar increase will occur
over the next 100 years.
20. By looking at the scatter plot of the data, we rule out the linear and logarithmic models. 7 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions We try various models:
2 Quadratic: y=0.496x 62.2893x+1970.639 Cubic: y=0.0201243201x 3.88037296x +247.6754468x 5163.935198 Quartic: y=0.0002951049x 0.0654560995x +5.27525641x 180.2266511x+2203.210956 3 2 4 3 2 Exponential: y=2.41422994 ( 1.054516914 ) x
Power: 3 y=0.000022854971x .616078251 After examining the graphs of these models, we see that the cubic and quartic models are clearly the
best. 21.
3 2 Using a computing device, we obtain the cubic function y=ax +bx +cx+d with a=0.0012937 ,
b= 7.06142 , c=12 , 823 , and d= 7 , 743 , 770 . When x=1925 , y 1914 (million).
22. (a) T =1.000396048d 1.499661718 (b) The power model in part (a) is approximately T =d
2 1.5 2 3 . Squaring both sides gives us T =d , so the 3 model matches Kepler’s Third Law, T =kd . 8 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions 1. (a) If the graph of f is shifted 3 units upward, its equation becomes y= f (x)+3 .
(b) If the graph of f is shifted 3 units downward, its equation becomes y= f (x) 3 .
(c) If the graph of f is shifted 3 units to the right, its equation becomes y= f (x 3) .
(d) If the graph of f is shifted 3 units to the left, its equation becomes y= f (x+3) .
(e) If the graph of f is reflected about the x axis, its equation becomes y= f (x) .
(f) If the graph of f is reflected about the y axis, its equation becomes y= f ( x) .
(g) If the graph of f is stretched vertically by a factor of 3 , its equation becomes y=3 f (x) .
1
(h) If the graph of f is shrunk vertically by a factor of 3 , its equation becomes y= f (x) .
3
2. (a) To obtain the graph of y=5 f (x) from the graph of y= f (x) , stretch the graph vertically by a
factor of 5 .
(b) To obtain the graph of y= f (x 5) from the graph of y= f (x) , shift the graph 5 units to the right.
(c) To obtain the graph of y= f (x) from the graph of y= f (x) , reflect the graph about the x axis.
(d) To obtain the graph of y= 5 f (x) from the graph of y= f (x) , stretch the graph vertically by a factor
of 5 and reflect it about the x axis.
(e) To obtain the graph of y= f (5x) from the graph of y= f (x) , shrink the graph horizontally by a
factor of 5 .
(f) To obtain the graph of y=5 f (x) 3 from the graph of y= f (x) , stretch the graph vertically by a
factor of 5 and shift it 3 units downward.
3. (a) (graph 3) The graph of f is shifted 4 units to the right and has equation y= f (x 4) .
(b) (graph 1) The graph of f is shifted 3 units upward and has equation y= f (x)+3 .
1
(c) (graph 4) The graph of f is shrunk vertically by a factor of 3 and has equation y= f (x) .
3
(d) (graph 5) The graph of f is shifted 4 units to the left and reflected about the x axis. Its equation
is y= f (x+4) .
(e) (graph 2) The graph of f is shifted 6 units to the left and stretched vertically by a factor of 2 . Its
equation is y=2 f (x+6) .
4. (a) To graph y= f (x+4) we shift the graph of f , 4 units to the left. The point ( 2,1 ) on the graph of f corresponds to the point
( 2 4,1 ) = ( 2,1 ) .
(b) To graph y= f (x)+4 we shift the graph of f , 4 units upward. 1 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions The point ( 2,1 ) on the graph of f corresponds to the point ( 2,1+4 ) = ( 2,5) .
(c) To graph y=2 f (x) we stretch the graph of f vertically by a factor of 2 . The point ( 2,1 ) on the graph of f corresponds to the point ( 2,2 1 ) = ( 2,2 ) .
1
(d) To graph y=
f (x)+3 , we shrink the graph of f vertically by a factor of 2 , then reflect the
2
resulting graph about the x axis, then shift the resulting graph 3 units upward. The point ( 2,1 ) on the graph of f corresponds to the point 2, 1
1+3 = ( 2,2.5) .
2 5. (a) To graph y= f (2x) we shrink the graph of f horizontally by a factor of 2 . The point ( 4, 1 ) on the graph of f corresponds to the point
(b) To graph y= f 1
x
2 1
4, 1 = ( 2, 1 ) .
2 we stretch the graph of f horizontally by a factor of 2 . The point ( 4, 1 ) on the graph of f corresponds to the point ( 2 4, 1 ) = ( 8, 1 ) .
(c) To graph y= f ( x) we reflect the graph of f about the y axis. 2 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions The point ( 4, 1 ) on the graph of f corresponds to the point ( 1 4, 1 ) = ( 4, 1 ) .
(d) To graph y= f ( x) we reflect the graph of f about the y axis, then about the x axis. The point ( 4, 1 ) on the graph of f corresponds to the point ( 1 4, 1 1 ) = ( 4,1 ) .
2 6. The graph of y= f (x)= 3x x has been shifted 2 units to the right and stretched vertically by a
factor of 2 . Thus, a function describing the graph is
y = 2 f (x 2)=2 3(x 2) (x 2)2
= 2 3x 6 (x2 4x+4) =2 x2+7x 10 2 7. The graph of y= f (x)= 3x x has been shifted 4 units to the left, reflected about the x axis, and
shifted downward 1 unit. Thus, a function describing the graph is
1
reflect
about
x axis y= f x+4 1 shift
4units
left shift
1unit
down This function can be written as
y = f (x+4) 1=
= 2 3(x+4) (x+4) 1= 3x+12 ( x2+8x+16) 1 2 x 5x 4 1 8. (a) The graph of y=2sin x can be obtained from the graph of y=sin x by stretching it vertically by
a factor of 2 . 3 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions (b) The graph of y=1+ x can be obtained from the graph of y= x by shifting it upward 1 unit. 3 3 9. y= x : Start with the graph of y=x and reflect about the x axis. Note: Reflecting about the y
3 3 axis gives the same result since substituting x for x gives us y=( x) = x . 2 2 2 10. y=1 x = x +1 : Start with the graph of y=x , reflect about the x axis, and then shift 1 unit
upward. 2 2 11. y=(x+1) : Start with the graph of y=x and shift 1 unit to the left. 12.
4 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions 2 2 2 2 y=x 4x+3=(x 4x+4) 1=(x 2) 1 : Start with the graph of y=x , shift 2 units to the right, and then
shift 1 unit downward. 13. y=1+2cos x : Start with the graph of y=cos x , stretch vertically by a factor of 2 , and then shift 1
unit upward. 14. y=4sin 3x : Start with the graph of y=sin x , compress horizontally by a factor of 3 , and then
stretch vertically by a factor of 4 . 15. y=sin (x/2) : Start with the graph of y=sin x and stretch horizontally by a factor of 2 . 16. y=1/(x 4) : Start with the graph of y=1/x and shift 4 units to the right. 5 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions 17. y= x+3 : Start with the graph of y= x and shift 3 units to the left. 4 4 18. y=(x+2) +3 : Start with the graph of y=x , shift 2 units to the left, and then shift 3 units upward. 1 2
1 2
1
2
2
(x +8x)= (x +8x+16) 8= (x+4) 8 : Start with the graph of y=x , compress vertically by
2
2
2
a factor of 2 , shift 4 units to the left, and then shift 8 units downward.
19. y= 3 3 20. y=1+ x 1 : Start with the graph of y= x , shift 1 unit to the right, and then shift 1 unit upward. 21. y=2/(x+1) : Start with the graph of y=1/x , shift 1 unit to the left, and then stretch vertically by a
factor of 2. 6 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions 1
tan (x
) : Start with the graph of y=tan x , shift
units to the right, and then compress
4
4
4
vertically by a factor of 4 .
22. y= 23. y= sin x : Start with the graph of y=sin x and reflect all the parts of the graph below the x axis
about the x axis. 2 2 2 2 24. y= x 2x = x 2x+1 1 = (x 1) 1 : Start with the graph of y=x , shift 1 unit right, shift 1 unit
downward, and reflect the portion of the graph below the x axis about the x axis. 25. This is just like the solution to Example 4 except the amplitude of the curve (the 30 N curve in
2
Figure 9 on June 21) is 14 12=2 . So the function is L(t)=12+2sin
(t 80) . March 31 is the
365
90 th day of the year, so the model gives L(90) 12.34 h. The daylight time (5:51 A.M. to 6:18 P.M.)
is 12 hours and 27 minutes, or 12.45 h. The model value differs from the actual value by
12.45 12.34
0.009 , less than 1% .
12.45
26. Using a sine function to model the brightness of Delta Cephei as a function of time, we take its
period to be 5.4 days, its amplitude to be 0.35 (on the scale of magnitude), and its average magnitude
to be 4.0 . If we take t=0 at a time of average brightness, then the magnitude (brightness) as a
2
t .
function of time t in days can be modeled by the formula M(t)=4.0+0.35sin
5.4
7 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions 27. (a) To obtain y= f ( x ) , the portion of the graph of y= f (x) to the right of the y axis is reflected
about the y axis.
(b) y=sin x (c) y= x 28. The most important features of the given graph are the x intercepts and the maximum and
minimum points. The graph of y=1 / f (x) has vertical asymptotes at the x values where there are x
intercepts on the graph of y= f (x) . The maximum of 1 on the graph of y= f (x) corresponds to a
minimum of 1/1=1 on y=1 / f (x) . Similarly, the minimum on the graph of y= f (x) corresponds to a
maximum on the graph of y=1 / f (x) . As the values of y get large (positively or negatively) on the
graph of y= f (x) , the values of y get close to zero on the graph of y=1 / f (x) . 29. Assuming that successive horizontal and vertical gridlines are a unit apart, we can make a table of
approximate values as follows.
x
f (x)
g(x)
g(x)+g(x) 0
2
2
4 1
1.7
2.7
4.4 2
1.3
3
4.3 3
1.0
2.8
3.8 4
0.7
2.4
3.1 5
0.3
1.7
2.0 6
0
0
0 8 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions Connecting the points ( x,f(x)+g(x) ) with a smooth curve gives an approximation to the graph of f +g .
Extra points can be plotted between those listed above if necessary.
30. First note that the domain of f +g is the intersection of the domains of f and g ; that is, f +g is
only defined where both f and g are defined. Taking the horizontal and vertical units of length to be
the distances between successive vertical and horizontal gridlines, we can make a table of
approximate values as follows:
1
2 2.5 3
2 1 0
x
1 2.2 2.0 2.4 2.7 2.7 2.3
f (x)
1
1.3 1.2 0.6 0.3 0.5 0.7
g(x)
f (x)+g(x) 0 0.9 0.8 1.8 3.0 3.2 3.0 Extra values of x (like the value 2.5 in the table above) can be added as needed.
3 2 2 31. f (x)=x +2x ; g(x)=3x 1 . D=R for both f and g .
3 2 2 3 3 2 2 3 2 ( f +g)(x)=(x +2x )+(3x 1)=x +5x 1 , D=R .
2 ( f g)(x)=(x +2x ) (3x 1)=x x +1 , D=R .
3 2 2 5 4 3 2 ( fg)(x)=(x +2x )(3x 1)=3x +6x x 2x , D=R .
f
g 3 (x)= 2 x +2x
2 3x 1 , D= { x| x 1
3 } 2 since 3x 1 0 . 32. f (x)= 1+x , D=[ 1, ) ; g(x)= 1 x , D=(
,1] .
( f +g) (x)= 1+x + 1 x , D= ( ,1] [ 1, ) = 1,1 .
( f g) (x)= 1+x 1 x , D= 1,1 .
2 ( fg) (x)= 1+x 1 x = 1 x , D= 1,1 .
9 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions f
g 1+x (x)= 1 x , D=[ 1,1) . We must exclude x=1 since it would make f
undefined.
g 33. f (x)=x , g(x)=1/x 3 2 34. f (x)=x , g(x)= x 2 35. f (x)=2x x ; g(x)=3x+2 . D=R for both f and g , and hence for their composites.
2 2 2 ( f g)(x)= f (g(x))= f (3x+2)=2(3x+2) (3x+2)=2(9x +12x+4) 3x 2=18x +21x+6 .
2 2 2 2 2 (g f )(x)=g( f (x))=g(2x x)=3(2x x)+2=6x 3x+2 .
2 2 4 3 2 2 4 3 ( f f )(x)= f ( f (x))= f (2x x)=2(2x x) (2x x)=2(4x 4x +x ) 2x +x=8x 8x +x .
(g g)(x)=g(g(x))=g(3x+2)=3(3x+2)+2=9x+6+2=9x+8 .
3 36. f (x)=1 x , D=R ; g(x)=1/x , D={x| x 0} .
3 3 ( f g) (x)= f (g(x))= f ( 1/x ) =1 (1/x) =1 1/x , D={x| x 0} .
(g
(f
(g ( 3 ) 3 f ) (x)=g( f (x))=g 1 x =1/(1 x ) , D={x|1 x
3 33 9 6 3 0}={x| x 1}.
3 f ) (x)= f ( f (x))= f (1 x )=1 (1 x ) [ =x 3x +3x ] , D=R .
g ) (x)=g(g(x))=g ( 1/x ) =1/ ( 1/x ) =x , D={x| x 0} because 0 is not in the domain of g . 37. f (x)=sin x , D=R ; g(x)=1 x , D=[0, ) .
( f g) (x)= f (g(x))= f (1 x )=sin ( 1 x ) , D=[0, ] .
( g f ) (x)=g( f (x))=g ( sin x ) =1 sin x . For sin x to be defined, we must have sin x 0
x 0,
2 ,3
2 ,
4 ,5
4 , 3
... , so
D= { x| x 2n , +2n ,where n is an integer} .
10 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions ( f f ) (x)= f ( f (x))= f ( sin x ) =sin ( sin x ) , D=R .
( g g) (x)=g(g(x))=g ( 1 x ) =1 1 x ,
D={ x 0|1 0} ={ x 0|1 x x } ={ x 0| x 1} =[0,1] . 2 38. f (x)=1 3x , D=R ; g(x)=5x +3x+2 , D=R . ( f g) (x) = f ( g(x) ) = f (5x2+3x+2)=1 3(5x2+3x+2)
2 2 =1 15x 9x 6= 15x 9x 5 , D=R . ( g f ) (x) =g ( f (x) ) =g ( 1 3x ) =5(1 3x)2+3(1 3x)+2=5(1 6x+9x2)+3 9x+2
2 2 =5 30x+45x 9x+5=45x 39x+10 , D=R .
( f f ) (x)= f ( f (x) ) = f ( 1 3x ) =1 3(1 3x)=1 3+9x=9x 2, D=R . ( g g) (x) =g ( g(x) ) =g(5x2+3x+2)=5(5x2+3x+2)2+3(5x2+3x+2)+2
4 3 2 2 =5(25x +30x +29x +12x+4)+15x +9x+6+2
4 3 2 4 3 2 2 =125x +150x +145x +60x+20+15x +9x+8
=125x +150x +160x +69x+28, D=R .
1
x+1
, D= { x| x 0} ; g(x)=
, D= { x| x 2} .
x
x+2
x+1
x+1
1
x+1 x+2
= f (g(x))= f
=
+
=
+
( f g)(x)
x+2
x+2 x+1 x+2 x+1
x+2 39. f (x)=x+ ( ) ( 2 ) 2 2 (x+1)(x+1)+(x+2)(x+2)
x +2x+1 + x +4x+4
2x +6x+5
=
=
=
(x+2)(x+1)
(x+2)(x+1)
(x+2)(x+1)
Since g(x) is not defined for x= 2 and f (g(x)) is not defined for x= 2 and x= 1 , the domain of (
f g)(x) is D= { x| x 2, 1} . (g f )(x)=g( f (x))=g 1
x+
x = 1
x+
x
1
x+
x 2 +1
= x +1+x
x
2 2 = x +x+1
2 2 = x +x+1
2 . x +1+2x x +2x+1 (x+1)
x
Since f (x) is not defined for x=0 and g( f (x)) is not defined for x= 1 , the domain of (g f )(x) is
D= { x| x 1,0} .
+2 11 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions ( f f )(x) = f ( f (x))= f = ( 1
x x+ ) ( 2 = x+ 2 1
x ) x(x) x +1 +1 x +1 +x(x)
2 1 + x+
4 = = 1
x 2 2 1
+
x 1
2 =x+ x +1
x 1
+
x x
2 x +1 2 x +x +x +1+x x(x +1)
4 =x+ 2 x(x +1) 2 x +3x +1
2 , D= { x| x 0} . x(x +1)
x+1
x+1+1(x+2)
+1
x+1
x+2
x+2
x+1+x+2 2x+3
(g g)(x)=g(g(x))=g
=
=
=
=
. Since g(x) is not
x+2
x+1
x+1+2(x+2) x+1+2x+4 3x+5
+2
x+2
x+2
5
defined for x= 2 and g(g(x)) is not defined for x=
, the domain of ( g g ) (x) is
3
5
D= x| x 2,
.
3 { } 40. f (x)= 2x+3 , D=
2 { x| x 3
2 } 2 ; g(x)=x +1 , D=R . 2 2 ( f g)(x)= f (x +1)= 2(x +1)+3 = 2x +5 , D=R .
2 (g f )(x)=g ( 2x+3 ) = ( 2x+3 ) +1=(2x+3)+1=2x+4 , D=
( f f )(x)= f ( {
{ 2x+3 ) = 2 ( 2x+3 ) +3 = 2 2x+3 +3 , D=
2 2 2 4 2 4 x| x
x| x 3
2
3
2 }
} .
. 2 (g g)(x)=g(x +1)=(x +1) +1=(x +2x +1)+1=x +2x +2 , D=R .
41.
( f g h)(x) = f (g(h(x)))= f (g(x 1))= f (2(x 1))
=2(x 1)+1=2x 1
42.
( f g h)(x) = f (g(h(x)))= f (g(1 x))= f ((1 x)2)
2 2 =2(1 x) 1=2x 4x+1
43.
12 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions ( f g h)(x) = f (g(h(x)))= f (g(x+3))= f ((x+3)2+2) ( x2+6x+11) 2 = f (x +6x+11)= 44. ( f g h)(x)= f (g(h(x)))= f ( g ( x+3 ) ) = f ( cos
2 10 45. Let g(x)=x +1 and f (x)=x 2 1 = x +6x+10
x+3 ) = 2
x+3 +1 cos
2 10 . Then ( f g)(x)= f (g(x))=(x +1) =F(x) . 46. Let g(x)= x and f (x)=sin x . Then ( f g)(x)= f (g(x))=sin ( x ) =F(x) . 2 x
x
. Then ( f g)(x)= f (g(x))= 2 =G(x) .
47. Let g(x)=x and f (x)=
x+4
x +4
2 48. Let g(x)=x+3 and f (x)=1/x . Then ( f g)(x)= f (g(x))=1/(x+3)=G(x) .
49. Let g(t)=cos t and f (t)= t . Then ( f g)(t)= f (g(t))= cos t =u(t) .
50. Let g(t)=tan t and f (t)=
2 t
tan t
. Then ( f g)(t)= f (g(t))=
=u(t) .
1+t
1+tan t
2 x x 51. Let h(x)=x , g(x)=3 , and f (x)=1 x . Then ( f g h)(x)=1 3 =H(x) .
3 52. Let h(x)= x , g(x)=x 1 , and f (x)= x . Then ( f g h)(x)=
4 3 x 1 =H(x) . 53. Let h(x)= x , g(x)=sec x , and f (x)=x . Then ( f g h)(x)= ( sec x ) 4 =sec 4 ( x ) =H(x) . 54. (a) f (g(1))= f (6)=5
(b) g( f (1))=g(3)=2
(c) f ( f (1))= f (3)=4
(d) g(g(1))=g(6)=3
(e) (g f )(3)=g( f (3))=g(4)=1
(f) ( f g)(6)= f (g(6))= f (3)=4
55. (a) g(2)=5 , because the point ( 2,5) is on the graph of g . Thus, f (g(2))= f (5)=4 , because the
point ( 5,4 ) is on the graph of f .
(b) g( f (0))=g(0)=3
13 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions (c) ( f g)(0)= f (g(0))= f (3)=0
(d) (g f )(6)=g( f (6))=g(6) . This value is not defined, because there is no point on the graph of g that
has x coordinate 6 .
(e) (g g)( 2)=g(g( 2))=g(1)=4
(f) ( f f )(4)= f ( f (4))= f (2)= 2
56. To find a particular value of f (g(x)) , say for x=0 , we note from the graph that g(0) 2.8 and
f (2.8) 0.5 . Thus, f (g ( 0 ) ) f (2.8) 0.5 . The other values listed in the table were obtained in a
similar fashion.
x
5
4
3
2
1 g(x)
0.2
1.2
2.2
2.8
3 f (g(x))
4
3.3
1.7
0.5
0.2 x g(x)
2.8
2.2
1.2
0.2
1.9
4.1 0
1
2
3
4
5 f (g(x))
0.5
1.7
3.3
4
2.2
1.9 57. (a) Using the relationship distance = rate
r(t)=60t .
(b) A= r 2 2 time with the radius r as the distance, we have
2 (A r)(t)=A(r(t))= (60t) =3600 t . This formula gives us the extent of the rippled area 2 (in cm ) at any time t .
58. (a) d=rt d(t)=350t
(b) There is a Pythagorean relationship involving the legs with lengths d and 1 and the hypotenuse
2 2 2 2 with length s : d +1 =s . Thus, s(d)= d +1 .
2 (c) (s d)(t)=s(d(t))=s(350t)= (350t) +1
59. (a) 14 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions { 0 if t<0
1 if t 0 { 0
if t<0
120 if t 0 H(t)=
(b) V (t)= so V (t)=120H(t) . (c)
Starting with the formula in part (b), we replace 120 with 240 to reflect the different voltage. Also,
because we are starting 5 units to the right of t=0 , we replace t with t 5 Thus, the formula is
V (t)=240H(t 5) .
60. (a)
R(t) =tH(t)
= { 0 if t<0
t if t 0 { 0 if t<0
2t if 0 t
so V (t)=2tH(t) , t 60 .
(b) V (t)= (c) V (t)= { 0
4 ( t 7) 60 if t<7
if 7 t 32 so V (t)=4(t 7)H(t 7) , t 32 . 61. (a) By examining the variable terms in g and h , we deduce that we must square g to get the terms
2 2 2 2 4x and 4x in h . If we let f (x)=x +c , then ( f g)(x)= f (g(x))= f (2x+1)=(2x+1) +c=4x +4x+ ( 1+c ) .
15 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.3 New Functions from Old Functions 2 2 Since h(x)=4x +4x+7 , we must have 1+c=7 . So c=6 and f (x)=x +6 .
(b) We need a function g so that f (g(x))=3(g(x))+5=h(x) . But
2 2 2 2 h(x)=3x +3x+2=3(x +x)+2=3(x +x 1)+5 , so we see that g(x)=x +x 1 .
62. We need a function g so that g( f (x))=g(x+4)=h(x)=4x 1=4(x+4) 17 . So we see that the function
g must be g(x)=4x 17 .
63. We need to examine h( x) .
h( x)=( f g)( x)= f (g( x))= f (g(x)) [because g is even] =h(x)
Because h( x)=h(x) , h is an even function.
64. h( x)= f (g( x))= f ( g(x)) . At this point, we can’t simplify the expression, so we might try to find
2 a counterexample to show that h is not an odd function. Let g(x)=x , an odd function, and f (x)=x +x .
2 Then h(x)=x +x, which is neither even nor odd.
Now suppose f is an odd function. Then f ( g(x))= f (g(x))= h(x) . Hence, h( x)= h(x) , and so h is
odd if both f and g are odd.
Now suppose f is an even function. Then f ( g(x))= f (g(x))=h(x) . Hence, h( x)=h(x) , and so h is
even if g is odd and f is even. 16 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers 4 1. f (x)=x +2
(a) 2,2 by 2,2 (b) 0,4 by 0,4 (c) 4,4 by 4,4 (d) 8,8 by 4,40 (e) 40,40 by 80,800 The most appropriate graph is produced in viewing rectangle (d).
2 2. f (x)=x +7x+6
(a) 5,5 by 5,5 1 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers (b) 0,10 by 20,100 (c) 15,8 by 20,100 (d) 10,3 by 100,20 The most appropriate graph is produced in viewing rectangle (c).
3 3. f (x)=10+25x x
(a) 4,4 by 4,4 (b) 10,10 by 10,10
2 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers (c) 20,20 by (d) 100,100 100,100 by 200,200 The most appropriate graph is produced in viewing rectangle (c) because the maximum and minimum
points are fairly easy to see and estimate.
2 4. f (x)= 8x x
(a) 4,4 by 4,4 (b) 5,5 by 0,100 3 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers (c) 10,10 by (d) 10,40 2,10 by 2,6 The most appropriate graph is produced in viewing rectangle (d).
2 5. Since the graph of f (x)=5+20x x is a parabola opening downward, an appropriate viewing
rectangle should include the maximum point. 3 2 6. An appropriate viewing rectangle for f (x)=x +30x +200x should include the high and low points. 3 2 7. f (x)=0.01x x +5 . Graphing f in a standard viewing rectangle, 10,10 by 10,10 , shows us
what appears to be a parabola. But since this is a cubic polynomial, we know that a larger viewing
rectangle will reveal a minimum point as well as the maximum point. After some trial and error, we
choose the viewing rectangle 50,150 by 2000,2000 . 4 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers 8. f (x)=x(x+6)(x 9) 4 9. f (x)=
4 81 x 0
is 0,3 . 4 81 x is defined when
x 4 4 81 x 3 , so the domain of f is 10. f (x)= 0.1x+20 is defined when 0.1x+20 0 3,3 . Also 0 x 81 x 4 4 81 =3 , so the range 200 , so the domain of f is 200, ). 2 11. The graph of f (x)=x +(100/x) has a vertical asymptote of x=0 . As you zoom out, the graph of f
2 looks more and more like that of y=x . 5 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers 2 12. The graph of f (x)=x/(x +100) is symmetric with respect to the origin. 13. f (x)=cos ( 100x ) 14. f (x)=3sin (120x) 15. f (x)=sin ( x/40 ) 16. f (x)=tan ( 25x ) 17.
6 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers 2 cos (x ) y=3 2 18. y=x +0.02sin (50x) 19. We must solve the given equation for y to obtain equations for the upper and lower halves of the
ellipse.
2 2 4x +2y =1 2 2 2y =1 4x 2 1 4x
y=
2
2 2 y= 1 4x
2 7 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers 2 2 20. y 9x =1 2 2 y =1+9x 2 y= 3 1+9x 2 21. From the graph of f (x)=x 9x 4 , we see that there is one solution of the equation f (x)=0 and it
is slightly larger than 9. By zooming in or using a root or zero feature, we obtain x 9.05. 3 22. We see that the graphs of f (x)=x and g(x)=4x 1 intersect three times. The x coordinates of
these points (which are the solutions of the equation) are approximately 2.11,0.25, and 1.86 .
3 Alternatively, we could find these values by finding the zeros of h(x)=x 4x+1 . 2 23. We see that the graphs of f (x)=x and g(x)=sin x intersect twice. One solution is x=0. The other
solution of f =g is the x coordinate of the point of intersection in the first quadrant. Using an
intersect feature or zooming in, we find this value to be approximately 0.88. Alternatively, we could
2 find that value by finding the positive zero of h(x)=x sin x . 8 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers Note : After producing the graph on a TI 83 Plus, we can find the approximate value 0.88 by using
the following keystrokes:
. The ‘‘1’’ is just a guess for 0.88. 24. (a)
The x coordinates of the three points of intersection are x 3.29 , 2.36 and 1.20 .
(b) Using trial and error, we find that m 0.3365 . Note that m could also be negative.
3 2 25. g(x)=x /10 is larger than f (x)=10x whenever x>100 . 4 3 3 26. f (x)=x 100x is larger than g(x)=x whenever x>101 . 27. We see from the graphs of y= sin x x and y=0.1 that there are two solutions to the equation
9 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers sin x x =0.1 : x
these two values. 0.85 and x 0.85 . The condition sin x x <0.1 holds for any x lying between 28.
5 3 5 P(x)=3x 5x +2x , Q(x)=3x . These graphs are significantly different only in the region close to the
origin. The larger a viewing rectangle one chooses, the more similar the two graphs look.
4 6 29. (a) The root functions y= x , y= x and y= x 3 5 (b) The root functions y=x , y= x and y= x 3 4 5 (c) The root functions y= x , y= x , y= x and y= x (d)
For any n , the n th root of 0 is 0 and the n th root of 1 is 1 ; that is, all n th root functions pass
through the points ( 0,0 ) and ( 1,1 ) .
For odd n , the domain of the n th root function is R , while for even n , it is { x R| x 0} .
Graphs of even root functions look similar to that of x , while those of odd root functions
resemble that of 3 x .
10 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers As n increases, the graph of n x becomes steeper near 0 and flatter for x>1 . 30. (a) The functions y=1/x and y=1/x 2 (b) The functions y=1/x and y=1/x 2 3 4 3 (c) The functions y=1/x , y=1/x , y=1/x and y=1/x 4 (d)
n The graphs of all functions of the form y=1/x pass through the point ( 1,1 ) .
If n is even, the graph of the function is entirely above the x axis. The graphs of 1/x
for n even are similar to one another.
If n is odd, the function is positive for positive x and negative for negative x . The n n graphs of 1/x for n odd are similar to one another.
n As n increases, the graphs of 1/x approach 0 faster as x
4 . 2 31. f (x)=x +cx +x . If c<0 , there are three humps: two minimum points and a
maximum point. These humps get flatter as c increases, until at c=0 two of the humps
disappear and there is only one minimum point. This single hump then moves to the
right and approaches the origin as c increases. 11 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers 2 / / c ,1
c , and
32. f (x)= 1+cx . If c<0 , the function is only defined on 1
its graph is the top half of an ellipse. If c=0 , the graph is the line y=1 . If c>0 , the
graph is the top half of a hyperbola. As c approaches 0 , these curves become flatter
and approach the line y=1 . n x 33. y=x 2 . As n increases, the maximum of the function moves further from the
origin, and gets larger. Note, however, that regardless of n , the function approaches 0
as x
. 34. y= x . The ‘‘bullet’’ becomes broader as c increases.
2 c x 2 3 2 35. y =cx +x
If c<0 , the loop is to the right of the origin, and if c is positive, it is to the left. In both
cases, the closer c is to 0 , the larger the loop is.
12 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers (In the limiting case, c=0 , the loop is ‘‘infinite’’, that is, it doesn’t close.) Also, the
larger c is, the steeper the slope is on the loopless side of the origin. 36. (a) y=sin ( x )
This function is not periodic; it oscillates less frequently as x increases. ( 2) (b) y=sin x
This function oscillates more frequently as x increases. Note also that this function
is even, whereas sin x is odd. 37. The graphing window is 95 pixels wide and we want to start with x=0 and end
with x=2 . Since there are 94 ‘‘gaps’’ between pixels, the distance between pixels is
2 0
2
. Thus, the x values that the calculator actually plots are x=0+
n , where
94
94
n=0 , 1 , 2 , ... , 93 , 94 . For y=sin 2x , the actual points plotted by the calculator are
2
2
sin 2
n
for n=0 , 1 , ... , 94. For y=sin 96x , the points plotted
94
94
2
2
are
sin
96
n
for n=0 , 1 , ... , 94. But
94
94
2
2
2
2
sin
96
n =sin
94
n+2
n =sin 2 n+2
n
94
94
94
94 13 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.4 Graphing Calculators and Computers =sin 2 2
n
94 [ by periodicity of sine ], n=0 , 1 , ... , 94 So the y values, and hence the points, plotted for y=sin 96x are identical to those
plotted for y=sin 2x .
Note: Try graphing y=sin 94x . Can you see why all the y values are zero?
38. As in Exercise 37, we know that the points being plotted for y=sin 45x are
2
2
sin
45
n
for n=0 , 1 , ... , 94 . But
94
94
2
2
2
2
sin
45
n =sin
47
n 2
n =sin
n 2
n
94
94
94
94
2
2
=sin ( n ) cos 2
n cos ( n ) sin 2
n [ Subtraction
94
94
formula for the sine]
2
2
2
=0 cos 2
n ( 1)sin 2
n = sin 2
n , n=0 ,
94
94
94
1 , ... , 94
So the y values, and hence the points, plotted for y=sin 45x lie on either y=sin 2x or
y= sin 2x . 14 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.5 Exponential Functions x 1. (a) f (x)=a , a>0
(b) R
(c) ( 0, )
(d) See Figures (c), (b), and (a), respectively.
2. (a) The number e is the value of a such that the slope of the tangent line at x=0 on the graph of
x y=a is exactly 1 .
(b) e 2.71828
(c) f (x)=e x 3. All of these graphs approach 0 as x
, all of them pass through the point ( 0,1 ) , and all of
them are increasing and approach
as x
. The larger the base, the faster the function increases
for x>0 , and the faster it approaches 0 as x
. 4. The graph of e x x x is the reflection of the graph of e about the y axis, and the graph of 8 is the
x x x reflection of that of 8 about the y axis. The graph of 8 increases more quickly than that of e for
x>0 , and approaches 0 faster as x
. x x 5. The functions with bases greater than 1 ( 3 and 10 ) are increasing, while those with bases less
x
1 x
1
1 x
x
than 1
and
are decreasing. The graph of
is the reflection of that of 3
3
10
3
x
1
x
about the y axis, and the graph of
is the reflection of that of 10 about the y axis. The
10
x x graph of 10 increases more quickly than that of 3 for x>0 , and approaches 0 faster as x . 1 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.5 Exponential Functions 6. Each of the graphs approaches
as x
base, the faster the function grows as x , and each approaches 0 as x
. The smaller the
, and the faster it approaches 0 as x
. x 7. We start with the graph of y=4 (Figure 3) and then shift 3 units downward. This shift doesn’t
x affect the domain, but the range of y=4 3 is ( 3, x ) . There is a horizontal asymptote of y= 3. x y=4 y=4 3
x 8. We start with the graph of y=4 (Figure 3) and then shift 3 units to the right. There is a horizontal
asymptote of y=0 . x y=4 x 3 y=4 2 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.5 Exponential Functions x 9. We start with the graph of y=2 (Figure 2), reflect it about the y axis, and then about the x axis
x (or just rotate 180 to handle both reflections) to obtain the graph of y= 2 . In each graph, y=0 is the
horizontal asymptote. x y=2 y=2 x y= 2 x x 10. We start with the graph of y=e (Figure 13), vertically stretch by a factor of 2, and then shift 1 unit
upward. There is a horizontal asymptote of y=1. x x y=2e y=1+2e
x 11. We start with the graph of y=e (Figure 13), reflect it about the x axis, and then shift 3 units
upward. Note the horizontal asymptote of y=3 . y= e x y=3 e x x 12. We start with the graph of y=e (Figure 13), reflect it about the y axis, and then about the x axis
3 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.5 Exponential Functions x (or just rotate 180 to handle both reflections) to obtain the graph of y= e . Now shift this graph 1
unit upward, vertically stretch by a factor of 5 , and then shift 2 units upward. y= e x x y=2+5(1 e )
x 13. (a) To find the equation of the graph that results from shifting the graph of y=e 2 units downward,
x we subtract 2 from the original function to get y=e 2 .
x (b) To find the equation of the graph that results from shifting the graph of y=e 2 units to the right,
( x 2)
we replace x with x 2 in the original function to get y=e
.
x (c) To find the equation of the graph that results from reflecting the graph of y=e about the x axis,
x we multiply the original function by 1 to get y= e .
x (d) To find the equation of the graph that results from reflecting the graph of y=e about the y axis,
we replace x with x in the original function to get y=e x .
x (e) To find the equation of the graph that results from reflecting the graph of y=e about the x axis
x and then about the y axis, we first multiply the original function by 1 (to get y= e ) and then
replace x with x in this equation to get y= e x . 14. (a) This reflection consists of first reflecting the graph about the x axis (giving the graph with
x x equation y= e ) and then shifting this graph 2 4=8 units upward. So the equation is y= e +8 .
(b) This reflection consists of first reflecting the graph about the y axis (giving the graph with
x
( x 4)
equation y=e ) and then shifting this graph 2 2=4 units to the right. So the equation is y=e
.
x x x 15. (a) The denominator 1+e is never equal to zero because e >0 , so the domain of f (x)=1/(1+e ) is
R.
4 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.5 Exponential Functions x x (b) 1 e =0 e =1 x x=0 , so the domain of f (x)=1/(1 e ) is ( ,0) (0, ).
t 16. (a) The sine and exponential functions have domain R , so g(t)=sin (e ) also has domain R .
t t (b) The function g(t)= 1 2 has domain {t |1 2 t 0}={t |2
1 x 17. Use y=Ca with the points ( 1,6 ) and ( 3,24 ) . 6=Ca
2 4=a a=2 [ since a>0 ] and C= 1}={t |t C= 6
a 3 24= 6
a a 3 6
x
=3 . The function is f (x)=3 2 .
2
x x 1
a= [ since a>0 ]. The function is f (x)=2
3
x+h x x h x 1
3
x 2, 2
9 gives us 2
2
=2a
9 x or f (x)=2(3) .
x ( h ) f (x+h) f (x) 5
5 55 5 5 5 1
x
19. If f (x)=5 , then
=
=
=
=5
h
h
h
h
x ,0] . and 24=Ca 18. Given the y intercept ( 0,2 ) , we have y=Ca =2a . Using the point
1 2
=a
9 0}=( h 5 1
h . 28 1 27 20. Suppose the month is February. Your payment on the 28th day would be 2 =2 =134 , 217 ,
728 cents, or $1 , 342 , 177 . 28 . Clearly, the second method of payment results in a larger amount
for any month.
2 24 21. 2 ft =24 in, f (24)=24 in =576 in =48 ft. g(24)=2 24 in =2 /(12 5280) mi
x 265 mi 5 22. We see from the graphs that for x less than about 1.8 , g(x)=5 > f (x)=x , and then near the point
( 1.8,17.1 ) the curves intersect. Then f (x)>g(x) from x 1.8 until x=5 . At ( 5,3125) there is another
point of intersection, and for x>5 we see that g(x)> f (x) . In fact, g increases much more rapidly than
f beyond that point. 5 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.5 Exponential Functions 23. The graph of g finally surpasses that of f at x 35.8 . x x 9 24. We graph y=e and y=1 , 000 , 000 , 000 and determine where e =1 10 . This seems to be true at
x 20.723 , so
6 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.5 Exponential Functions x 9 e >1 10 for x>20.723 . 25. (a) Fifteen hours represents 5 doubling periods (one doubling period is three hours).
5 100 2 =3200
(b) In t hours, there will be t/3 doubling periods. The initial population is 100 , so the population y at
t/3 time t is y=100 2
(c) t=20 .
20/3 10 , 159 y=100 2 x/3 (d) We graph y =100 2
1 and y =50 , 000 . The two curves intersect at x 26.9 , so the population
2 reaches 50 , 000 in about 26.9 hours. 1 4 1
= g
2
8
(b) In t hours, there will be t/15 half life periods. The initial mass is 2 g, so the mass y at time t is
1 t/15
y=2
.
2
1 96/15
0.024 g
(c) 4 days =4 24=96 hours. t=96 y=2
2
(d) y=0.01 t 114.7 hours
26. (a) Sixty hours represents 4 half life periods. 2 27. An exponential model is
7 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.5 Exponential Functions t y=ab , where a=3.154832569 10
and y(2010) 7417 million. t 12 and b=1.017764706 . This model gives y(1993) 5498 million 9 28. An exponential model is y=ab , where a=1.9976760197589 10 and b=1.0129334321697 . This
model gives y(1925) 111 million, y(2010) 330 million, and y(2020) 375 million. 8 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms 1. (a) See Definition 1.
(b) It must pass the Horizontal Line Test.
1 2. (a) f (y)=x f (x)=y for any y in B . The domain of f
(b) See the steps in (5).
(c) Reflect the graph of f about the line y=x . 1 is B and the range of f 1 is A . 3. f is not one to one because 2 6 , but f (2)=2.0= f (6) .
4. f is one to one since for any two different domain values, there are different range values.
5. No horizontal line intersects the graph of f more than once. Thus, by the Horizontal Line Test, f is
one to one.
6. The horizontal line y=0 (the x axis) intersects the graph of f in more than one point. Thus, by the
Horizontal Line Test, f is not one to one.
7. The horizontal line y=0 (the x axis) intersects the graph of f in more than one point. Thus, by the
Horizontal Line Test, f is not one to one.
8. No horizontal line intersects the graph of f more than once. Thus, by the Horizontal Line Test, f is
one to one.
9. The graph of f (x)= 1
1
(x+5) is a line with slope
. It passes the Horizontal Line Test, so f is one
2
2 to one.
Algebraic solution : If x 1 x , then x +5 x +5
2 1 2 1
x +5
2 1 ( ) 1
x +5
2 2 ( ) ( ) f ( x ) , so f is f x 1 2 one to one.
b
4
=
=2 . Pick any x
2a
2( 1)
values equidistant from 2 to find two equal function values. For example, f (1)=4 and f (3)=4 , so f
is not 1 1 .
2 10. The graph of f (x)=1+4x x is a parabola with axis of symmetry x= 11. g(x)= x g( 1)=1=g(1) , so g is not one to one. 12. x x 1 x 2 1 x 2 ( ) g ( x ) , so g is 1 g x 1 2 1. 13. A football will attain every height h up to its maximum height twice: once on the way up, and
1 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms ( ) may equal f ( t ) , so f is not 1 again on the way down. Thus, even if t does not equal t , f t
1 2 1 2 1 .
14. f is not 1 1 because eventually we all stop growing and therefore, there are two times at which
we have the same height.
15. f does not pass the Horizontal Line Test, so f is not 1 1. 16. f passes the Horizontal Line Test,
so f is 1 1 . 17. Since f (2)=9 and f is 1 1 1 , we know that f (9)=2 . Remember, if the point ( 2,9 ) is on the graph of f , then the point ( 9,2 ) is on the graph of f 1 . 18. (a) First, we must determine x such that f (x)=3 . By inspection, we see that if x=0 , then f (x)=3 .
Since f is 1 1 1 ( f is an increasing function), it has an inverse, and f (3)=0 . ( 1 ) (b) By the second cancellation equation in (4), we have f f (5) =5 .
19. First, we must determine x such that g(x)=4 . By inspection, we see that if x=0 , then g(x)=4 .
Since g is 1
20. (a) f is 1 1 1 ( g is an increasing function), it has an inverse, and g (4)=0 .
1 because it passes the Horizontal Line Test. (b) Domain of f = 3,3 = Range of f 1 . Range of f = 2,2 = Domain of f 1 . 1 (c) Since f ( 2)=1 , f (1)= 2 .
21. We solve
2 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms 5
9
9
(F 32) for F : C=F 32 F= C+32 . This gives us a formula for the inverse function, that
9
5
5
is, the Fahrenheit temperature F as a function of the Celsius temperature C . F 459.67
9
9
C+32 459.67
C 491.67 C 273.15 , the domain of the inverse function.
5
5 C= 0 22. m= 2 2 m v 1 2 2 c 1 v /c 2 v 0
2 = 2 2 m
m c 2 0
2 =1 2 2 m 2 v =c 2 m m 0
2 1 m v=c 1 0
2 . This m m 1 formula gives us the speed v of the particle in terms of its mass m , that is, v= f (m) .
23. f (x)= 10 3x
: y= y= 10 3x (y 0) 2 2 y =10 3x 3x=10 y 1 2 10
1 2 10
1
x+
. So f (x)=
x+
. Note that the domain of f
3
3
3
3 24. f (x)= 4x 1
2x+3 y= 4x 1
2x+3 y(2x+3)=4x 1 2xy+3y=4x 1 1 2 10
. Interchange x and y
y+
3
3 x=
1 is x 0 . 3y+1=4x 2xy 3y+1=(4 2y)x 3y+1
3x+1
. Interchange x and y : y=
.
4 2y
4 2x
3x+1
1
So f (x)=
.
4 2x
x= 25. f (x)=e x 1 3 y=e x 3 ln y=x 3 3 3 x= ln y . Interchange x and y : y= ln x . 3 So f (x)= ln x .
3 26. y= f (x)=2x +3 Interchange x and y : y= 27. y=ln ( x+3) 28. y= 1+e
1 e x x x+3=e
x 3 y y ye =1+e 3 y 3
.
2 x 3
1
3
. So f (x)=
2 y 3=2x y 3 3
=x
2 x 3
.
2 3 x= y x 1 x x=e 3 . Interchange x and y : y=e 3 . So f (x)=e 3 .
x x y 1=ye +e x x y 1=e ( y+1 ) 3 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms x e= y 1
y+1 y 1
y+1 x=ln Note that the domain of f
29. y= f (x)=1 2
2 1 y= x
1 . So f (x)= 1 2 x= 2 x= 2 x x 1
x+1 . 2
1 y x= 2
, since x>0 . Interchange x and y : y=
1 y 2
1 x 2
.
1 x 2 2 1 . So f (x)=ln is x >1 . 30. y= f (x)= x +2x , x>0
2 x 1
x+1 . Interchange x and y : y=ln 2 4 1
2 1 ( y2)
2 2 2 y>0 and y =x +2x 2 2 x +2x y =0 . Now we use the quadratic formula: 2 1+y . But x>0 , so the negative root is inadmissible. Interchange x = 1
1 2 and y : y= 1+ 1+x . So f (x)= 1+ 1+x , x>0 . 31. The function f is one to one, so its inverse exists and the graph of its inverse can be obtained by
reflecting the graph of f about the line y=x . 4 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms 32. The function f is one to one, so its inverse exists and the graph of its inverse can be obtained by
reflecting the graph of f about the line y=x . For the graph of 1/ f , the y coordinates are simply the
1
reciprocals of f . For example, if f (5)=9, then 1/ f (5)= . If we draw the horizontal line y=1 , we see
9
that the only place where the graphs intersect is on that line. 33. (a) It is defined as the inverse of the exponential function with base a , that is, log x=y
a y a =x . (b) ( 0, )
(c)
(d) See Figure .
34. (a) The natural logarithm is the logarithm with base e , denoted ln x .
(b) The common logarithm is the logarithm with base 10 , denoted log x .
(c) See Figure .
6 35. (a) log 64=6 since 2 =64 .
2 1
2 1
= 2 since 6 =
.
(b) log
6 36
36
36. (a) log 2=
8 (b) ln e 2 1
1/3
since 8 =2 .
3 = 2 37. (a) log 1.25+log 80=log
10 10 2 ( 1.25 80 ) =log 10100=log 1010 =2
10
3 (b) log 10+log 20 3log 2=log ( 10 20 ) log 2 =log
5
5
5
5
5
5 200
2
=log 25=log 5 =2
5
5
8
5 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms ( log 23+log 25) =2log 215=15 [ Or: 2 ( log 23+log 25) =2log 23 2log 25=3 5=15 ] 38. (a) 2
(b) e 3ln 2 3 =e ln (2 ) =e ln 8 =8 [ Or: e 3ln 2 ( ln 2) 3=23=8 ] = e 2 39. 2ln 4 ln 2=ln 4 ln 2=ln 16 ln 2=ln
a 16
=ln 8
2 b a b a b 40. ln x+aln y bln z=ln x+ln y ln z =ln (x y ) ln z =ln (xy /z )
2 (1+x ) x
1
2
1/2
2
41. ln (1+x )+ ln x ln sin x=ln (1+x )+ln x ln sin x=ln [(1+x ) x ] ln sin x=ln
2
sin x
2 ln 10
0.926628
12
ln 12
ln 8.4
(b) log 8.4=
3.070389
2
ln 2
42. (a) log 10= 43. To graph these functions, we use log 1.5 x= ln x
ln x
and log x=
. These graphs all approach
50
ln 1.5
ln 50 + as x 0 , and they all pass through the point ( 1,0 ) . Also, they are all increasing, and all
approach
as x
. The functions with larger bases increase extremely slowly, and the ones with
smaller bases do so somewhat more quickly. The functions with large bases approach the y axis
more closely as x + 0 . x 44. We see that the graph of ln x is the reflection of the graph of e about the line y=x , and that the
graph of log x 10 x x is the reflection of the graph of 10 about the same line. The graph of 10 increases
x more quickly than that of e . Also note that log 10 x as x more slowly than ln x . 6 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms 45. 3 ft =36 in, so we need x such that log x=36
2 68 , 719 , 476 , 736 in 1ft
1mi
12in 5280ft 36 x=2 =68 , 719 , 476 , 736 . In miles, this is 1 , 084 , 587.7 mi. 46. 0.1 From the graphs, we see that f (x)=x >g(x)=ln x for approximately 0<x<3.06 , and then g(x)> f (x)
15 for 3.06<x<3.43 10 (approximately). At that point, the graph of f finally surpasses the graph of g
for good.
47. (a) Shift the graph of y=log 10 x five units to the left to obtain the graph of y=log (x+5) . Note the
10 vertical asymptote of x= 5 .
y=log x
10 7 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms y=log (x+5)
10 (b) Reflect the graph of y=ln x about the x axis to obtain the graph of y= ln x .
y=ln x y= ln x 48. (a) Reflect the graph of y=ln x about the y axis to obtain the graph of y=ln ( x ) .
y=ln x y=ln ( x) 8 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms (b) Reflect the portion of the graph of y=ln x to
the right of the y axis about the y axis. The
graph of y=ln x is that reflection in addition
to the original portion.
y=ln x y=ln x 49. (a) 2ln x=1
x (b) e =5 ln x= x=ln 5 2x+3 50. (a) e x 5 x 5 Or: 2 =3 =3 1/2 x=e = e x= ln 5
2x+3 7=0 e (b) ln (5 2x)= 3
51. (a) 2 1
2 =7 5 2x=e 3 2x+3=ln 7
2x=5 e 2x=ln 7 3 3 x= log 3=x 5 (x 5)ln 2=ln 3 1
(ln 7 3)
2 1
3
(5 e )
2 x=5+log 3 . ( ) =ln 3 x= 2
x 5 ln 2 2 1 x 5= ln 3
ln 2 x=5+ ln 3
ln 2 2 (b) ln x+ln (x 1)=ln (x(x 1))=1 x(x 1)=e x x e=0 . The quadratic formula (with a=1 , b= 1 ,
1
and c= e ) gives x= ( 1 1+4e ) , but we reject the negative root since the natural logarithm is not
2
1
defined for x<0 . So x= ( 1+ 1+4e ) .
2
9 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms e 52. (a) ln (ln x)=1
ax (b) e =Ce
ln C
x=
a b bx ax x x e 2 3x ln x 1 >e
2 e <e ln e 2x ln x <e e ln x e 9 2x x ( 1/e,
2 e <x<e bx ax=ln C+bx 9 2 (a b)x=ln C ( e2,e9) x 3x>ln 4 2
2x 0 1
(ln 4 2)
3 x< e 3 2x ln 3 x x , 1
(2 ln 4)
3 1
ln 3 .
2 1
ln 3] .
2
2 2x 2x y =3 e 2 e =3 y 2 2x=ln (3 y ) x= 1
1
2
1
2
ln (3 x ) . So f (x)= ln (3 x ) . For the domain of f
2
2 3 x >0 x <3 |x|< 3
equals the range of f . 3 <x< 3 1
2
ln (3 y ) .
2
1 , we must have 0 x< 3 since x 0 . Note that the domain of f 56. (a) For f (x)=ln (2+ln x) , we must have 2+ln x>0
2 ax bx=ln C ) 2x [ note that y 0 ] Interchange x and y : y= x=e ,ln 10) , we must have 3 e
, e =e x ( 2 3x>ln 4 Thus, the domain of f is (
(b) y= f (x)= 3 e 1 x>e >ln 4 55. (a) For f (x)= 3 e 2 x<ln 10 2 3x >4 1 ln x=e =e bx ln e <ln 10 54. (a) 2<ln x<9
(b) e 1 =e ln e =ln [C(e )] ax=ln C+bx+ln e 53. (a) e <10
(b) ln x> 1 ln (ln x) ln x> 2 1 , [0, 3 ) , 2 x>e . Thus, the domain of f is ). (e , (b) y= f (x)=ln (2+ln x) y e =2+ln x x 1 e 2 f (x)=e . The domain of f 1 y y e 2 ln x=e 2 x=e x . Interchange x and y : y=e , as well as the range of f , is
3 2 e 2 . So .
3 2 57. We see that the graph of y= f (x)= x +x +x+1 is increasing, so f is 1 1 . Enter x= y +y +y+1
and use your CAS to solve the equation for y . Using Derive, we get two (irrelevant) solutions
involving imaginary expressions, as well as one which can be simplified to the following:
1 y= f (x)= 3 4
6 ( 3 2 D 27x +20 3 2 3 D+27x 20 + 2 )
10 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms 4 2 where D=3 3 27x 40x +16 . Maple and Mathematica each give two complex expressions and one
real expression, and the real expression is equivalent to that given by Derive. For example, Maple’s
1 M
expression simplifies to
6 2/3 8 2M 2M 1/3 2 2 , where M=108x +12 1/3 6 48 120x +81x 4 80 . 4 58. (a) If we use Derive, then solving x=y +y for y gives us six solutions of the form y= { A
A
+
, 2cos
+
3 3
3 6
3
6 4
B 1 with B=2sin
inverse for y=x +x ( x 0 ) is y=
3
4
4
is 0,
, this expression is only valid for x
0,
27
27
, where B 2sin A
,2sin
3 } and A=sin A
+
3 3 1 27x 2
2 3
3 B 1 . The , but because the domain of A .
6 4 Happily, Maple gives us the rest of the solution! We solve x=y +y for y to get the two real solutions
6
6 C 1/3 (C 2/3 2C 1/3+4)
C 1/3 6 4 , where C=108x+12 3 x ( 27x 4 ) , and the inverse for y=x +x ( 4
,
.
27
Mathematica also gives two real solutions, equivalent to those of Maple. The positive one is
6 3
1/3 3
1/3
4 D +2 2 D
2 , where D= 2+27x+3 3 x 27x 4 . Although this expression also has
6
4
domain
,
, Mathematica is mysteriously able to plot the solution for all x 0 .
27
x 0 ) is the positive solution, whose domain is ( ) (b) 11 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms n
n
t
n
t/3
=2
log
=
t=3log
. Using formula ( ), we can
2
2
100
100
3
100
ln (n/100)
write this as t=3
. This function tells us how long it will take to obtain n bacteria (given
ln 2
the number n ).
50,000
ln 500
(b) n=50 , 000 t=3log
=3log 500=3
26.9 hours
2 100
2
ln 2
t/3 59. (a) n=100 2 60. (a) Q=Q ( 1 e t/a)
0 Q
=1 e
Q t/a e Q
Q t/a =1 0 0 t
=ln
a Q
Q 1 t= aln (1 Q/Q ) . This
0 0 gives us the time t necessary to obtain a given charge Q .
= 2ln 0.1 4.6 seconds.
(b) Q=0.9Q and a=2 t= 2ln 1 0.9 Q /Q ( 0 ( 0 0 )) 61. (a) To find the equation of the graph that results from shifting the graph of y=ln x3 units upward,
we add 3 to the original function to get y=ln x+3 .
(b) To find the equation of the graph that results from shifting the graph of y=ln x3 units to the left,
we replace x with x+3 in the original function to get y=ln ( x+3) .
(c) To find the equation of the graph that results from reflecting the graph of y=ln x about the x axis,
we multiply the original equation by 1 to get y= ln x .
(d) To find the equation of the graph that results from reflecting the graph of y=ln x about the y axis,
we replace x with x in the original equation to get y=ln ( x) .
(e) To find the equation of the graph that results from reflecting the graph of y=ln x about the line y=x
x , we interchange x and y in the original equation to get x=ln y y=e .
(f) To find the equation of the graph that results from reflecting the graph of y=ln x about the x axis
and then about the line y=x , we first multiply the original equation by 1 and then interchange x and
x y in this equation to get x= ln y ln y= x y=e .
(g) To find the equation of the graph that results from reflecting the graph of y=ln x about the y axis
and then about the line y=x , we first replace x with x in the original equation and then interchange x
x x y=e y= e .
and y to get x=ln ( y)
(h) To find the equation of the graph that results from shifting the graph of y=ln x3 units to the left
and then reflecting it about the line y=x , we first replace x with x+3 in the original equation and then
12 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms y+3=e interchange x and y in this equation to get x=ln (y+3) x x y=e 3 . 62. (a) If the point ( x,y ) is on the graph of y= f (x) , then the point ( x c,y ) is that point shifted c units
1 1 , the point ( y,x ) is on the graph of y= f (x) and the point corresponding to to the left. Since f is 1 1 ( x c,y ) on the graph of f is ( y,x c ) on the graph of f . Thus, the curve’s reflection is shifted down
the same number of units as the curve itself is shifted to the left. So an expression for the inverse
1 1 function is g (x)= f (x) c .
(b) If we compress (or stretch) a curve horizontally, the curve’s reflection in the line y=x is
compressed (or stretched) vertically by the same factor. Using this geometric principle, we see that
1 1 the inverse of h(x)= f (cx) can be expressed as h (x)= ( 1/c ) f (x) .
63. (a) sin 3
2 1 = since sin 3 3 1 (b) cos ( 1)= since cos = 1 and
64. (a) arctan( 1)=
1 (b) csc 2= 65. (a) tan 3= 3 6 since tan 2= 1 (b) arcsin1= 2 = since sec 4 4 since sin 3 is in 0, 6 = 3 and since sin 2 is in 4 0, 3
= 4 = 2 and =1 and 2 is in 4 ,
2 2 . . = 1 and 4 =2 and 1
2 (b) arcsin 66. (a) sec since tan since csc 6 1 4 3
and
is in
2
3 = 4 is in
, 2 is in ,
2 2
3
2 ,
2 2 1
and
2
is in 0, ,
2 2 4 . . .
is in ,
2 2
, 2 3
2 . . . 1 67. (a) sin (sin 0.7)=0.7 since 0.7 is in 1,1 .
4
1
1
(b) tan
tan
=tan
3=
since
is in
3
3
3
68. (a) Let =arctan2 , so tan =2
.
(b) Let sec 2 =1+tan 2 ,
2 2
=1+4=5 .
sec = 5 sec (arctan2)=sec = 5 13 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms =sin 1 5
5
. Then sin =
, so cos
13
13
1 69. Let y=sin x . Then 2 y 2sin 1 5
13 =cos 2 =1 2sin 2 =1 2 1 2 5
13 2 = 119
.
169 2 2 cos y 0 , so cos (sin x)=cos y= 1 sin y = 1 x 1 1 70. Let y=sin x . Then sin y=x , so from the triangle we see that tan (sin x)=tan y= x .
2 1 x 1 1 71. Let y=tan x . Then tan y=x , so from the triangle we see that sin (tan x)=sin y= x .
2 1+x 72.
1 Let y=cos x . Then cos y=x
1 2 sin y= 1 x since 0 y sin (2cos x)=sin 2y=2sin ycos y=2x
73. . So 2 1 x . 1 The graph of sin x is the reflection of the graph of sin x about the line y=x .
74.
14 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.6 Inverse Functions and Logarithms 1 The graph of tan x is the reflection of the graph of tan x about the line y=x .
75.
1 g(x)=sin (3x+1) .
Domain ( g ) = { x| 1 3x+1 1} = { x| 2 3x 0} =
Range ( g ) = { y| 2 ( y
1 76. (a) f (x)=sin sin x 2 } = ,
2 2 { x| 2
3 } x 0 2
,0
3 = . . ) Since one function undoes what the other one does, we get the identity function, y=x , on the
restricted domain 1 x 1 .
(b) g(x)=sin 1 ( sin x ) This is similar to part (a), but with domain R . Equations for g on intervals of the form
n n+1 , for any integer n , can be found using g ( x ) =( 1) x+( 1) n . The sine
2
2
function is monotonic on each of these intervals, and hence, so is g (but in a linear fashion).
+ n, + n 15 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems 1. (a) Using P ( 15,250 ) , we construct the following table:
t Q
5 ( 5,694 ) 10 ( 10,444 ) 20 ( 20,111 ) 25 ( 25,28 ) 30 ( 30,0 ) slope=m PQ 694 250
444
=
= 44.4
5 15
10
444 250
194
=
= 38.8
10 15
5
111 250
139
=
= 27.8
20 15
5
28 250
222
=
= 22.2
25 15
10
0 250
250
=
= 16.6
30 15
15 (b) Using the values of t that correspond to the points closest to P ( t=10 and t=20 ), we have
38.8+ ( 27.8 )
= 33.3
2
(c) From the graph, we can estimate the slope of the tangent line at P to be 300
= 33.3 .
9 2.
2948
42
2948
(c) Slope =
42
(a) Slope = 2530 418
2948
=
69.67 (b) Slope =
36
6
42
2806 142
3080
(d) Slope =
=
=71
40
2
44 2661 287
=
=71.75
38
4
2948 132
=
=66
42
2 From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeats / minute after
42 minutes. After being stable for a while, the patient’s heart rate is dropping.
1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems 3. (a) For the curve y=x/ ( 1+x ) and the point P 1,
x
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii) m Q 0.5
0.9
0.99
0.999
1.1
1.5
1.01
1.001 1
2 PQ ( 0.5,0.333333)
( 0.9,0.473684 )
( 0.99,0.497487)
( 0.999,0.499750 )
( 1.5,06 )
( 1.1,0.523810 )
( 1.01,0.502488 )
( 1.001,0.500250 ) 0.333333
0.263158
0.251256
0.250125
0.2
0.238095
0.248756
0.249875 1
.
4
1 1
1
1
(c) y
= (x 1) or y= x+ .
2 4
4
4
(b) The slope appears to be 4. For the curve y=ln x and the point P(2,ln2) :
(a)
x
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii) 1.5
1.9
1.99
1.999
2.5
2.1
2.01
2.001 PQ ( 1.5,0.405465)
( 1.9,0.641854 )
( 1.99,0.688135)
( 1.999,0.692647)
( 2.5,0.916291 )
( 2.1,0.741937)
( 2.01,0.698135)
( 2.001,0.693647) (b) The slope appears to be
(c) y ln 2= m Q 0.575364
0.512933
0.501254
0.500125
0.446287
0.487902
0.498754
0.499875 1
.
2 1
1
(x 2) or y= x 1+ln 2
2
2 (d) 2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems 2 2 5. (a) y=y(t)=40t 16t . At t=2 , y=40(2) 16(2) =16 . The average velocity between times 2 and 2+h
2 2 y(2+h) y(2)
40(2+h) 16(2+h) 16
24h 16h
is v =
=
=
= 24 16h , if h 0 .
ave
(2+h) 2
h
h
(i) [2,2.5] : h=0.5 , v = 32 ft / s
(ii) [2,2.1] : h=0.1 , v = 25.6 ft / s
ave ave (iii) [2,2.05] : h=0.05 , vave= 24.8 ft / s (iv) [2,2.01] : h=0.01 , vave= 24.16 ft / s
(b) The instantaneous velocity when t=2 ( h approaches 0 ) is 24 ft / s.
6. The average velocity between t and t+h seconds is
2 ( 2 ) 2 58(t+h) 0.83(t+h) 58t 0.83t
58h 1.66th 0.83h
=
=58 1.66t 0.83h if h 0 .
h
h
(a) Here t=1 , so the average velocity is 58 1.66 0.83h=56.34 0.83h .
(i)
(ii) 1,1.5 : h=0.5 , 55.925 m / s
1,2 : h=1,55.51 m / s
(iii) 1,1.1 : h=0.1 , 56.257 m / s
(iv) 1,1.01 : h=0.01 , 56.3317 m / s
(v)
1,1.001 : h=0.001 , 56.33917 m / s
(b) The instantaneous velocity after 1 second is 56.34 m / s.
7. (a)
(i)
(iii) 13
(ii)
ft / s
ave 6
19
1,1.5 : h=0.5 , v =
ft / s (iv)
ave 24
1,3 : h=2 , v = 7
ft / s
ave 6
331
1,1.1 : h=0.1 , v =
ft / s
ave 600
1,2 : h=1 , v = (b) As h approaches 0 , the velocity approaches
3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems 3 1
= ft / s.
6 2 (c) (d)
8. Average velocity between times t=2 and t=2+h is given by
(a)
(i)
(ii)
(iii) h=3
h=2
h=1 s(5)
av
5
s(4)
v =
av
4
s(3)
v =
av
3 v = s(2+h) s(2)
.
h s(2) 178 32 146
=
=
48.7 ft / s
2
3
3
s(2) 119 32 87
=
=
=43.5 ft / s
2
2
2
s(2) 70 32
=
=38 ft / s
2
1 (b) Using the points ( 0.8,0 ) and ( 5,118 ) from the approximate tangent line, the instantaneous
118 0
28 ft / s.
velocity at t=2 is about
5 0.8 9. For the curve y=sin (10 /x) and the point P ( 1,0 ) :
(a)
m
x
Q
PQ 2 ( 2,0 ) 0
4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems 1.5
1.4
1.3
1.2
1.1
x
0.5
0.6
0.7
0.8
0.9 ( 1.5,0.8660 )
( 1.4, 0.4339 )
( 1.3, 0.8230 )
( 1.2,0.8660 )
( 1.1, 0.2817)
Q ( 0.5,0 )
( 0.6,0.8660 )
( 0.7,0.7818 )
( 0.8,1 )
( 0.9, 0.3420 ) 1.7321
1.0847
2.7433
4.3301
2.8173
m PQ 0
2.1651
2.6061
5
3.4202 As x approaches 1 , the slopes do not appear to be approaching any particular value. (b)
We see that problems with estimation are caused by the frequent oscillations of the graph. The
tangent is so steep at P that we need to take x values much closer to 1 in order to get accurate
estimates of its slope.
(c) If we choose x=1.001 , then the point Q is ( 1.001, 0.0314 ) and m
PQ 31.3794 . If x=0.999 , then Q is ( 0.999,0.0314 ) and m = 31.4422 . The average of these slopes is 31.4108 . So we estimate
PQ
that the slope of the tangent line at P is about 31.4 . 5 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function 1. As x approaches 2 , f (x) approaches 5 . [Or, the values of f (x) can be made as close to 5 as we like
by taking x sufficiently close to 2 (but x 2 ).] Yes, the graph could have a hole at ( 2,5) and be
defined such that f (2)=3 .
2. As x approaches 1 from the left, f (x) approaches 3 ; and as x approaches 1 from the right, f (x)
approaches 7 . No, the limit does not exist because the left and right hand limits are different.
3. (a) lim f (x)=
x means that the values of f (x) can be made arbitrarily large (as large as we please) 3 by taking x sufficiently close to 3 (but not equal to 3 ).
means that the values of f (x) can be made arbitrarily large negative by taking x
(b) lim f (x)=
x + 4 sufficiently close to 4 through values larger than 4 .
4. (a) lim f (x)=3
x 0 (b) lim f (x)=4
x 3 (c) lim f (x)=2
x + 3 (d) lim f (x) does not exist because the limits in part (b) and part (c) are not equal.
x 3 (e) f (3)=3
5. (a) f (x) approaches 2 as x approaches 1 from the left, so lim f (x)=2 .
x 1 (b) f (x) approaches 3 as x approaches 1 from the right, so lim f (x)=3 .
x + 1 (c) lim f (x) does not exist because the limits in part (a) and part (b) are not equal.
x 1 (d) f (x) approaches 4 as x approaches 5 from the left and from the right, so lim f (x)=4 .
x 5 (e) f (5) is not defined, so it doesn’t exist.
6. (a) lim g(x)= 1
x 2 (b) lim g(x)=1
+ x 2 (c) lim g(x) doesn’t exist
x 2 (d) g( 2)=1
(e) lim g(x)=1
x 2
1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function (f) lim g(x)=2
+ x 2 (g) lim g(x) doesn’t exist
x 2 (h) g(2)=2
(i) lim g(x) doesn’t exist
+ x 4 (j) lim g(x)=2
x 4 (k) g(0) doesn’t exist
(l) lim g(x)=0
x 0 7. (a) lim g(t)= 1
t 0 (b) lim g(t)= 2
t + 0 (c) lim g(t) does not exist because the limits in part (a) and part (b) are not equal.
t 0 (d) lim g(t)=2
t 2 (e) lim g(t)=0
+ t 2 (f) lim g(t) does not exist because the limits in part (d) and part (e) are not equal.
t 2 (g) g(2)=1
(h) lim g(t)=3
t 4 8. (a) lim R(x)=
x 2 (b) lim R(x)=
x 5 (c) lim R(x)=
x 3 (d) lim R(x)=
+ x 3 (e) The equations of the vertical asymptotes are x= 3 , x=2 , and x=5 .
9. (a) lim f (x)=
x 7 (b) lim f (x)=
x 3 (c)
2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function lim f (x)=
x 0 (d) lim f (x)=
x 6 (e) lim f (x)=
x + 6 (f) The equations of the vertical asymptotes are x= 7 , x= 3 , x=0 , and x=6 .
10. lim f (t)=150 mg and lim f (t)=300 mg. These limits show that there is an abrupt change in the
t + 12 t 12 amount of drug in the patient’s bloodstream at t=12 h. The left hand limit represents the amount of
the drug just before the fourth injection. The right hand limit represents the amount of the drug just
after the fourth injection.
11. (a) lim f (x)=1
x 0 (b) lim f (x)=0
x + 0 (c) lim f (x) does not exist because the limits in part (a) and part (b) are not equal.
x 0 12. lim f (x) exists for all a except a= 1 .
x a 13. lim f (x)=4 , lim f (x)=2 ,
x + 3 x 3
3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function lim f (x)=2 , f (3)=3 , f ( 2)=1
x 2 14. lim f (x)=1 , lim f (x)= 1 , lim f (x)=0 , lim f (x)=1 , f (2)=1 , f (0) is undefined
x 0 x + 0 x 2 x + 2 2 15. For f (x)= x 2x : 2 x x 2
x f (x) 2.5 0.714286 2.1 0.677419 2.05 0.672131 2.01 0.667774 2.005 0.667221
2.001 0.666778
x f (x) 1.9 0.655172 1.95 0.661017 1.99 0.665552 1.995 0.666110
1.999 0.666556
2 It appears that lim
x 2 x 2x
2 x x 2 =0.6= 2
.
3 4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function 2 16. For f (x)= x 2x : 2 x x 2
x
0 f (x)
0 0.5 1 0.9 9 0.95 19 0.99 99 0.999 999
x f (x) 2 2 1.5 3 1.1 11 1.01 101 1.001 1001
2 x 2x It appears that lim
x 1 2 does not exist since f (x) as x 1 and f (x) as x + 1 . x x 2 x 17. For f (x)= e 1 x
2 : x
x f (x) 1 0.718282 0.5 0.594885
0.1 0.517092
0.05 0.508439
0.01 0.501671
x
1 f (x)
0.367879
5 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function 0.5 0.426123
0.1 0.483742
0.05 0.491770
0.01 0.498337
x e 1 x It appears that lim
x 2 0 x ( 2 18. For f (x)=xln x+x
x
f (x)
1 =0.5= 1
.
2 ): 0.693147 0.5 0.143841 0.1 0.220727 0.05 0.147347 0.01 0.045952 0.005 0.026467
0.001 0.006907 ( 2 ) It appears that lim xln x+x =0 .
+ x 19. For f (x)=
x x+4 2
:
x f (x) 1 0 0.236068 0.5 0.242641
0.1 0.248457
0.05 0.249224
0.01 0.249844
x
1 f (x)
0.267949
6 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function 0.5 0.258343
0.1 0.251582
0.05 0.250786
0.01 0.250156
It appears that lim
x 20. For f (x)=
x 0 x+4 2
1
=0.25= .
x
4 tan 3x
:
tan 5x f (x)
0.2 0.439279 0.1 0.566236 0.05 0.591893 0.01 0.599680 0.001 0.599997
It appears that lim
x 0 tan 3x
3
=0.6= .
tan 5x
5 6 21. For f (x)= x 1
10 x
x 0.985337 0.9 0.719397 0.95 0.660186 0.99 1 f (x) 0.5 : 0.612018 0.999 0.601200
x f (x) 1.5 0.183369 1.1 0.484119 1.05 0.540783 1.01 0.588022
7 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function 1.001 0.598800
6 x 1 It appears that lim 10 x 1 x =0.6= 3
.
5 x x 1 9 5
22. For f (x)=
:
x
x
f (x)
0.5 1.527864 0.1 0.711120 0.05 0.646496 0.01 0.599082 0.001 0.588906
x f (x) 0.5 0.227761 0.1 0.485984 0.05 0.534447 0.01 0.576706 0.001 0.586669
x x 9 5
=0.59 . Later we will be able to show that the exact value is ln (9/5) .
It appears that lim
x
x 0
23. lim
x + 5 24. lim
x 5 25. lim
x 1 6
=
x 5 since ( x 5) 6
=
x 5
2 x
2 since ( x 5) = (x 1)
positive values as x 0 as x 0 as x + 5 and 5 and 6
>0 for x>5 .
x 5
6
<0 for x<5 .
x 5 since the numerator is positive and the denominator approaches 0 through
1. 26.
8 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function x 1 lim
x 0 2 = 2 since x 0 as x x (x+2) + x 2 = 2 since ( x+2 ) x x <0 for 0<x<1 and for 2<x<0 .
x 1 + 0 as x 2 and x (x+2) 2 <0 for 2<x<0 . x (x+2) 28. lim csc x=lim ( 1/sin x ) = 29. 2 x (x+2) x 1
27. lim x 1 0 and 0 as x since sin x and sin x>0 for 0<x< . x lim sec x= lim
( 1/cos x ) =
x ( /2 )
( /2 ) 30. lim ln (x 5)=
x since x 5 since cos x + 0 as x ( /2) and cos x<0 for <x< /2 . + 0 as x 5 . + 5 3 31. (a) f (x)=1/(x 1)
x
f (x)
0.5 1.14 0.9 3.69 0.99 33.7 0.999 333.7 0.9999 3333.7 0.99999 33 , 333.7
x f (x) 1.5 0.42 1.1 3.02 1.01 33.0 1.001 333.0 1.0001 3333.0 1.00001 33 , 333.3
From these calculations, it seems that lim f (x)=
x 1 and lim f (x)=
x . + 1 (b) If x is slightly smaller than 1 , then
9 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function 3 3 x 1 will be a negative number close to 0 , and the reciprocal of x 1 , that is, f (x) , will be a negative
.
number with large absolute value. So lim f (x)=
x 1 3 If x is slightly larger than 1 , then x 1 will be a small positive number, and its reciprocal, f (x) , will
be a large positive number. So lim f (x)= .
x + 1 (c) It appears from the graph of f that lim f (x)=
x 32. (a) y= x
2 = x x 2 and lim f (x)= 1 x x
. Therefore, as x
(x 2)(x+1) y>0 for x< 1 and for x>2 , so lim y=lim y=
x + 1 x + 1 or x . Also, as x 1 + 2 , the denominator approaches 0 , and
1 or x 2 , the denominator + 2 approaches 0 and y<0 for 1<x<2 , so lim y=lim y=
x . + 1 x . 2 (b)
33. (a) Let h(x)= ( 1+x )
x
h(x)
0.001 . 2.71964 0.0001 1/x 2.71842 0.00001 2.71830
0.000001 2.71828
0.000001 2.71828
0.00001 2.71827
10 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function 0.0001 2.71815 0.001 2.71692 It appears that lim ( 1+x )
x 1/x 2.71828 , which is approximately e . In Section 7.4 we will see that the 0 value of the limit is exactly e . (b) ( x) : x 34. For the curve y=2 and the points P ( 0,1 ) and Q x,2
x m Q PQ ( 0.1,1.0717735)
0.01
( 0.01,1.0069556 )
0.001 ( 0.001,1.0006934 )
0.0001 ( 0.0001,1.0000693) 0.1 0.71773
0.69556
0.69339 0.69317
The slope appears to be about 0.693 .
35. (a)
x
f (x)
1 0.998000 0.8 0.638259
0.6 0.358484
0.4 0.158680
0.2 0.038851
0.1 0.008928
0.05 0.001465
It appears that lim f (x)=0 .
x (b)
x
0.04 0 f (x)
0.000572
11 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function 0.02 0.000614 0.01 0.000907 0.005 0.000978
0.003 0.000993
0.001 0.001000
It appears that lim f (x)= 0.001 .
x 36. h(x)= tan x x
x (a)
x 0 3 h(x) 1.0 0.55740773 0.5 0.37041992 0.1 0.33467209 0.05 0.33366700
0.01 0.33334667
0.005 0.33333667 (b) It seems that lim h(x)=
x (c) 0 1
.
3 x h(x) 0.001 0.33333350 0.0005 0.33333344 0.0001 0.33333000 0.00005 0.33333600
0.00001 0.33300000
0.000001 0.00000000
Here the values will vary from one calculator to another. Every calculator will eventually give
false values .
(d) As in part (c), when we take a small enough viewing rectangle we get incorrect output.
12 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function 37. No matter how many times we zoom in toward the origin, the graphs of f (x)=sin ( /x) appear to
consist of almost vertical lines. This indicates more and more frequent oscillations as x 0 . m
38. lim m=lim
v c v c 0
2 1 v /c . As v c , 2 1 v /c 2 + 0 , and m . 2 39. There appear to be vertical asymptotes of the curve y=tan (2sin x) at x
0.90 and x
2.24 . To
find the exact equations of these asymptotes, we note that the graph of the tangent function has
13 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.2 The Limit of a Function vertical asymptotes at x=
sin x=
to x + 2 + n . Thus, we must have 2sin x= n . Since 1 sin x 1 , we must have sin x= 4 2
0.90 ). Just as 150 is the reference angle for 30 ,
x= sin 1 3 1 4 + n , or equivalently,
and so x= sin 1 4 is the reference angle for sin (corresponding 1 are also equations of the vertical asymptotes (corresponding to x 4 ( sin 4 2 4 . So
2.24 ). ) 40. (a) Let y= x 1 / ( x 1 ) .
x
y
0.99 5.92531 0.999 5.99250 0.9999 5.99925
1.01 6.07531 1.001 6.00750
1.0001 6.00075 From the table and the graph, we guess that the limit of y as x approaches 1 is 6 .
3 x 1
(b) We need to have 5.5<
<6.5 . From the graph we obtain the approximate points of
x 1
intersection P ( 0.9313853,5.5) and Q ( 1.0649004,6.5) . Now 1 0.9313853 0.0686 and
1.0649004 1 0.0649 , so by requiring that x be within 0.0649 of 1 , we ensure that y is within 0.5 of
6. 14 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 1. (a)
lim f (x)+h(x) =lim f (x)+lim h(x)
x a x a x a = 3+8=5
2 (b) lim
x 2 a x x
3 (c) lim x a a
3 h(x) = 3 lim h(x) = 8 =2
x a (d) lim 2 f (x) = lim f (x) =( 3) =9 a 1
1
1
1
=
= =
f (x) lim f (x)
3
3
x a lim f (x)
f (x) x a
3
3
(e) lim
=
= =
lim h(x) 8
8
x a h(x)
x a lim g(x)
(f) lim
x a g(x) x a
0
=
= =0
f (x) lim f (x)
3
x a (g) The limit does not exist, since lim g ( x ) =0 but lim f (x) 0 .
x a x a 2lim f (x)
2 f (x)
x a
2 ( 3)
6
(h) lim
=
=
=
11
x a h(x) f (x) lim h(x) lim f (x) 8 ( 3 )
x 2. (a) lim
x a x a f (x)+g(x) =lim f (x)+lim g(x)=2+0=2 2 x 2 x 2 (b) lim g(x) does not exist since its left and right hand limits are not equal, so the given limit does
x 1 not exist.
(c) lim f (x)g(x) =lim f (x) lim g(x)=0 1.3=0
x 0 x 0 x 0 (d) Since lim g(x)=0 and g is in the denominator, but lim f (x)= 1 0 , the given limit does not
x 1 x 1 exist.
3 (e) lim x f (x)= lim x
x 2 (f) lim
x 1 x 3+ f (x) = 3 3 lim f (x) =2 2=16 2 x 2 3+lim f (x) = 3+1 =2
x 1 3.
1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 4 2 4 2 lim (3x +2x x+1) =lim 3x +lim 2x lim x+lim 1 [Limit Laws 1 and 2]
x 2 x 2 x 2 x 4 2 x 2 2 =3lim x +2lim x lim x+lim 1 [3]
x 2 x 2 4 x 2 x 2 2 [9, 8, and 7] =3( 2) +2( 2) ( 2)+(1)
=48+8+2+1=59
4.
2x +1 lim
x 2 ( 2x2+1) lim 2 x = 2 x +6x 4 2 lim
x [Limit Law 5] ( x2+6x 4) 2
2 2lim x +lim 1
x 2
2 = x 2 [2, 1, and 3] lim x +6lim x lim 4
x 2 x 2 x 2 = 2(2) +1 9 3
[9, 7, and 8]
=
12 4 = 2 2 (2) +6(2) 4
5.
2 3 2 3 lim (x 4)(x +5x 1) =lim (x 4) lim (x +5x 1)
x 3 x 3 x
2 3 = lim x lim 4
x 3 x 2 [Limit Law 4] 3 lim x +5lim x lim 1 3 x 3 x 3 3 x [2, 1, and 3] 3 [7, 8, and 9] =(3 4) (3 +5 3 1)
=5 41=205
6.
2 3 5 2 3 5 lim (t +1) (t+3) =lim (t +1) lim (t+3)
t 1 t 1 t
2 = lim (t +1)
t 1 [Limit Law 4] 1
3 lim (t+3)
t 5 [6] 1 2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 3 2 = lim t +lim 1
t 1 t 1 2 3 = t lim = ( 1) +1 lim t+lim 3
1 t 5 [1] 1 5 [9, 7, and 8] 1+3 =8 32=256 7.
x 3 1+3x lim
1 2 1+4x +3x 4 x 3 1+3x
2 1 1+4x +3x [6] 4
3 lim (1+3x)
x = 1
2 [5] 4 lim (1+4x +3x )
x 1
3 lim 1+3lim x
x = 1 x 1
2 lim 1+4lim x +3lim x
x 1 x 1 x 2 1 3 1+3(1) = = 4 [2, 1, and 3] 4 1+4(1) +3(1) 4
8 3 = 1
2 3 = 1
8 [7, 8, and 9] 8.
4 u +3u+6 = lim
u ( u4+3u+6) lim
u 2 [11] 2
4 = lim u +3lim u+lim 6 [1, 2, and 3]
u 2 u 2 u 4 ( 2 ) +3 ( 2 ) +6 = 2 [9, 8, and 7] = 16 6+6 = 16 =4
9.
lim
x 2 16 x = lim
x 4 = ( 16 x2) [11] 4
2 lim 16 lim x
x 4 x [2] 4 3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 2 [7 and 9] = 16 ( 4 ) =0 10. (a) The left hand side of the equation is not defined for x=2 , but the right hand side is.
(b) Since the equation holds for all x 2 , it follows that both sides of the equation approach the same
limit as x 2 , just as in Example 3. Remember that in finding lim f (x) , we never consider x=a .
x a 2 x +x 6
(x+3)(x 2)
11. lim
=lim
=lim (x+3)=2+3=5
x 2
x 2 x 2
x 2
x 2
2 x +5x+4 12. lim
x 4 2 = lim
x x +3x 4 (x+4)(x+1)
x+1
4+1
3 3
= lim
=
= =
4 1
5 5
4 (x+4)(x 1)
x
4 x 1 2 x x+6
13. lim
does not exist since x 2
x 2
x 2
2 x 4x 14. lim
x 4 =lim 2 x x 3x 4 4 2 t 9 15. lim
t 3 =lim 2 t 2t +7t+3 2 0 but x x+6 8 as x 2. x(x 4)
x
4
4
=lim
=
=
(x 4)(x+1) x 4 x+1 4+1 5
(t+3)(t 3)
t 3
3 3
6 6
=lim
=
=
=
5 5
3 (2t+1)(t+3) t
3 2t+1 2( 3)+1 2 x 4x 16. lim
x 1 2 2 does not exist since x 3x 4 2 0 but x 4x 5 as x 1. x 3x 4
2 2 2 (4+h) 16
(16+8h+h ) 16
8h+h
h(8+h)
17. lim
=lim
=lim
=lim
=lim (8+h)=8+0=8
h
h
h
h
h 0
h 0
h 0
h 0
h 0
3 18. lim
x 1 x 1
2 x 1 ( x 1 ) ( x +x+1 )
2 =lim
x 1 (x 1)(x+1) 2 2 x +x+1 1 +1+1 3
=lim
=
=
1+1
2
x 1 x+1 19. 4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws ( 1+4h+6h2+4h3+h4) 4 ( 1+h ) 1 lim
h h 0 =lim
h 1 h 0 2 3 2 3 2 3 4h+6h +4h +h
=lim
h
h 0 4 h(4+6h+4h +h )
2 3
=lim
=lim 4+6h+4h +h =4+0+0+0=4
h
h 0
h 0 ( ) 20. ( 3 (2+h) 8
8+12h+6h +h
lim
=lim
h
h
h 0
h 0 21. lim
t 9 2 8 12h+6h +h
=lim
h
h 0 3 ( 12+6h+h2) =12+0+0=12 =lim
h ) 0 ( 3+ t ) ( 3 9 t
=lim
3 t t 9 3 t t ) =lim ( 3+ t ) =3+ 9 =6
t 9 22.
lim
h 0 1+h 1
= lim
h
h 0 1+h 1
h 1+h +1 = lim
h 0 ( 1+h ) 1
h ( 1+h +1 ) = lim
h 0 h( h
1+h +1 ) 1
1
1
=
=
1+h +1
1 +1 2 = lim
h 1+h +1 0 23.
lim
x 7 x+2 3
=lim
x 7
x 7
=lim
x 7 x+2 3
x 7 x+2 +3
x+2 +3 =lim
x x 7
=lim
( x 7) ( x+2 +3) x 7 7 ( x+2 ) 9
( x 7) ( x+2 +3)
1
1
1
=
=
x+2 +3
9 +3 6 24. ( 4 2 ) x 16
2
2
( x+2 ) ( x 2 ) x +4
lim
=lim
=lim ( x+2 ) x +4 =lim ( x+2 ) lim x +4
x 2
x 2 x 2
x 2
x 2
x 2
x 2 ( 2 ( ) ( ) ) = ( 2+2 ) 2 +4 =32
25.
5 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 1 1
x+4
+
4 x
4x
x+4
1
1
1
=lim
=lim
=lim
=
=
4+x x 4 4+x x 4 4x(4+x) x 4 4x 4( 4)
16 lim
x 4 1
t 26. lim
t 0 1 ( t2+t ) t =lim t2 =lim 1 = 1 =1
2
t t(t+1) t 0 t+1 0+1
t ( t +t ) t 0 =lim 2 t t +t 0 27.
2
( x 3)( x +3) ( x+9 )
( x 9 ) ( x+9 )
x 81
=lim
=lim
lim
x 3
x 3
x 9
x 3 x 9
x 9 factor x 9 as a
difference of squares =lim ( x +3) ( x+9 ) = ( 9 +3) ( 9+9 ) =6 18=108
x 9 28.
1 (3+h) 3
lim
h
h 0 1
1
3+h 3
3 ( 3+h )
h
=lim
=lim
h
h 0 h(3+h)3 h 0 h(3+h)3 1 =lim
h 0 1
3(3+h) =lim
h 0 1
1
1
=
=
lim 3(3+h)
3(3+0)
9 = h 0 29.
1
t 1+t lim
t 0 1
t 1 =lim
t t 1+t =lim (1 0 0 t 1
1+t ( 1+ 1+t 0 =lim
t 1+t ) ( 1+ 1+t )
=lim
t+1 ( 1+ 1+t )
t 0
1+t = t ) 1
1+0 ( 1+ 1+0 ) = t
t 1+t ( 1+ 1+t ) 1
2 30.
2 lim
x 1 x x
1 x =lim
x 1 =lim
x 1 ( x 1 x
1 3/2 x ) =lim
x x (1 ) ( 1+ 1 1 x x +x ) [difference of cubes] x x ( 1+ x +x ) =lim 1(1+1+1) =3
x 1 Another method: We ’’add and subtract’’ 1 in the numerator, and then split up the fraction: 6 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 2 lim
x 1 x x
1 x =lim
x ( 1 1 =lim
x ( 2 x 1) + 1 x 1+ (1 x ) =lim
x x ) ( 1+
1 1 1+ 1 (1 x)(1+x)
1 x x ) ( 1+x ) = 1+ ( 1+ 1 ) ( 1+1 ) =3 x 31. (a)
lim
x 0 x
1+3x 1 (b)
x 2
3 f (x) 0.001 0.6661663 0.0001 0.6666167 0.00001 0.6666617
0.000001 0.6666662
0.000001 0.6666672
0.00001 0.6666717
0.0001 0.6667167 0.001 0.6671663 The limit appears to be 2
.
3 (c)
lim
x 0 x
1+3x 1 1+3x +1
1+3x +1 =lim
x 0 x ( 1+3x +1 )
x ( 1+3x +1 )
=lim
3x
( 1+3x ) 1 x 0 1
lim ( 1+3x +1 )
3 x 0
1
=
lim (1+3x) +lim 1
3
x 0
x 0
= = 1
3 lim 1+3lim x +1
x 0 x [Limit Law 3]
[1 and 11]
[1, 3, and 7] 0 7 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 1
( 1+3 0 +1 )
3
1
2
= (1+1)=
3
3 [7 and 8] = 32. (a)
lim
x 0 3+x
x (b)
x 3 0.29 f (x) 0.001 0.2886992 0.0001 0.2886775 0.00001 0.2886754
0.000001 0.2886752
0.000001 0.2886751
0.00001 0.2886749
0.0001 0.2886727 0.001
0.2886511
The limit appears to be approximately 0.2887 .
(c)
(3+x) 3
3+x 3
3+x + 3
=lim
=lim
lim
x
3+x + 3
x 0 x ( 3+x + 3 ) x 0
x 0
lim 1
= x lim
x = 3+x +lim 0 x 3 [Limit Laws 5 and 1] 0 1
lim ( 3+x ) + 3
x = 0 1
3+x + 3 [7 and 11] 0 1
3+0 + 3 [1, 7, and 8] 8 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws =
2 1
2 3 2 2 2 2 2 x x cos 20 x x
33. Let f (x)= x , g(x)=x cos 20 x and h(x)=x . Then 1 cos 20 x 1
f (x) g(x) h(x) . So since lim f (x)=lim h(x)=0 , by the Squeeze Theorem we have lim g(x)=0 .
x 3 34. Let f (x)=
3 2 x +x 0 2 x 3 2 x 2 3 0 3 0 2 x +x , g(x)= x +x sin ( /x) , and h(x)= x +x . Then 1 sin ( /x) 1
3 2 x +x sin ( /x) f ( x) x +x g(x) h(x) . So since lim f (x)=lim h(x)=0 , by the
x 0 x 0 Squeeze Theorem we have lim g(x)=0 .
x 0 2 35. 1 f (x) x +2x+2 for all x . Now lim 1=1 and
x ( x +2x+2) =lim
2 lim
x 1 x 1 2 2 x +2lim x+lim 2= ( 1 ) +2( 1)+2=1 . Therefore, by the Squeeze Theorem,
1 x 1 x 1 lim f (x)=1 .
x 1 36. 3x 3 f (x) x +2 for 0 x 2 . Now lim 3x=3 and lim
x 1 x 1 ( x3+2) =lim x3+lim 2=13+2=3 . Therefore,
x 1 x 1 by the Squeeze Theorem, lim f (x)=3 .
x 37. 1 cos (2/x) 1 x 4 1
4 4 x cos (2/x) x . Since lim
x 0 ( x4) =0 and lim x4=0 , we have
x 0
9 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 4 lim
x x cos (2/x) =0 by the Squeeze Theorem. 0 38. 1 sin ( /x) 1 e 1 sin ( /x) 1 e sin ( /x) x /e e xe x ( lim
x x e ) =0 , we have lim + 0 x ( sin xe /x ) ( x e . Since lim x /e ) =0 and + 0 =0 by the Squeeze Theorem. + 0 x+4 = lim ( x+4 ) = 4+4=0 . 39. If x> 4 , then x+4 =x+4 , so lim
+ x 4 If x< 4 , then x+4 = ( x+4 ) , so lim
x + x 4 ( x+4 ) = ( 4+4 ) =0 . x+4 = lim
4 x 4 Since the right and left limits are equal, lim x+4 =0 .
x 4 40. If x< 4 , then x+4 = ( x+4 ) , so lim
x 41. If x>2 , then x 2 =x 2 , so lim
x lim
x 2 x 2
=lim
x 2
x 2 ( x 2)
x 2 + 2 4 x+4
= lim
x+4
x ( x+4 )
x+4 4 = lim ( 1 ) = 1 .
x 4 x 2
x 2
=lim
=lim 1=1 . If x<2 , then x 2 = ( x 2 ) , so
x 2
+ x 2
+
x 2 x 2 =lim 1= 1 . The right and left limits are different, so lim
x x 2 2 x 2
x 2 does not exist.
3
42. If x> , then 2x 3 =2x 3 , so lim
2
x x< 3
, then 2x 3 =3 2x , so lim
2
x 2 + 1.5
2 2x 3x
= lim
2x 3
x 2x 3x
= lim
2x 3 1.5 x 1.5 2 + 1.5
2 2x 3x
= lim
2x 3
x 2x 3x
= lim
( 2x 3)
x 1.5 + 1.5 x ( 2x 3)
= lim x=1.5 . If
2x 3
+
x x ( 2x 3)
= lim
( 2x 3)
x 1.5 x= 1.5 . The 1.5 2 2x 3x
does not exist.
1.5 2x 3 right and left limits are different, so lim
x 43. Since x = x for x<0 , we have lim
x 0 1
x 1
x =lim
x 0 1
x 1
x =lim
x 0 2
, which does not
x exist since the denominator approaches 0 and the numerator does not. 10 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 1
x 44. Since x =x for x>0 , we have lim
+ x 0 1
x =lim
x + 0 1
x 1
x =lim 0=0 .
x + 0 45. (a)
(b)
(i) Since sgnx=1 for x>0 , lim sgnx=lim 1=1 .
+ x (ii) 0 + x 0 Since sgnx= 1 for x<0 , lim sgn x=lim 1= 1 .
x 0 x 0 (iii) Since lim sgnx lim sgnx , lim sgnx does not exist.
x 0 + x x 0 0 (iv) Since sgnx =1 for x 0 , lim sgnx =lim 1=1 .
x 0 x 0 46. (a) ( 4 x2) =lim 4 lim x2 lim f (x) =lim
x 2 x 2 x 2 x 2 =4 4=0
lim f (x) =lim ( x 1 ) =lim x lim 1
x + 2 x + 2 x + 2 x + 2 =2 1=1 (b) No, lim f (x) does not exist since lim f (x) lim f (x) .
x 2 x 2 x + 2 (c) 47. (a)
11 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws 2 (i) lim
x + 1 2 x 1
x 1
=lim
=lim ( x+1 ) =2
x 1
+ x 1
+
x 1 x 2 x 2 x 1
=lim
x 1 (ii) lim
1 x 1 x 1
=lim
( x 1) 1 x ( x+1 ) = 2 1 (b) No, lim F(x) does not exist since lim F(x) lim F(x) .
x 1 x + 1 x 1 (c) 48. (a)
2 2
(i)
lim h(x)=lim x =0 =0
x (ii) 0 x x 0
2 2 2 0 1 x 2 1 lim h(x)=lim x =2 =4
x (v) + 0 lim h(x)=lim x =1 =1
x (iv) x lim h(x)=lim x=0 , so lim h(x)=0 .
x (iii) + 0 2 x 2 lim h(x)=lim ( 8 x ) =8 2=6
x + 2 x + 2 (vi) Since lim h(x) lim h(x) , lim h(x) does not exist.
x 2 x + 2 x 2 (b)
49. (a)
12 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws [ x]= 2 for 2 x< 1 , so lim [ x]= lim ( 2 ) = 2 (i) + x 2 + x 2 (ii) [ x]= 3 for 3 x< 2 , so lim [ x]= lim ( 3) = 3 .
x 2 x 2 The right and left limits are different, so lim [ x] does not exist.
x 2 (iii) [ x]= 3 for 3 x< 2 , so lim [ x]= lim ( 3) = 3 .
x 2.4 x 2.4 (b)
(i) [ x]=n 1 for n 1 x<n , so lim [ x]=lim ( n 1 ) =n 1 .
x n x n (ii) [ x]=n for n x<n+1 , so lim [ x]=lim n=n .
+ x (c) lim [ x] exists
x n x + n a is not an integer. a 50. (a)
(b)
(i) lim f (x)=lim ( x [ x]) =lim x ( n 1 ) =n ( n 1 ) =1
x n x n x n (ii) lim f (x)=lim ( x [ x]) =lim ( x n ) =n n=0
+ x n x + n (c) lim f (x) exists
x + x n a is not an integer. a 51. The graph of f (x)=[ x]+[ x] is the same as the graph of g(x)= 1 with holes at each integer, since
f (a)=0 for any integer a . Thus, lim f (x)= 1 and lim f ( x ) = 1 , so lim f (x)= 1 . However,
x 2 f (2)=[2]+[ 2]=2+( 2)=0 , so lim f (x)
x x + 2 x 2 f (2) . 2 2 52. lim
v c L 0 1 v
c 2 =L 0 1 1 =0 . As the velocity approaches the speed of light, the length
13 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws approaches 0 .
A left hand limit is necessary since L is not defined for v>c .
2 n 53. Since p(x) is a polynomial, p(x)=a +a x+a x +
0 2 lim p(x) = lim
x a
x a +a x+a x +
0 a 1 1
n x 1 n x n a n = a +a a+a a +
0 +a lim x a 2 n n 2 a 2 +a x 2 = a0+a1lim x+a2lim x +
x +a x . Thus, by the Limit Laws, +a a = p(a) 2 n Thus, for any polynomial p , lim p(x)= p(a) .
x a p(x)
where p(x) and q(x) are any polynomials, and suppose that q(a) 0 . Thus,
q(x)
lim p(x)
p(x) x a
p(a)
lim r(x)=lim
=
[Limit Law 5] =
[ Exercise 53 ] =r(a) .
q(a)
x a
x a q(x) lim q ( x ) 54. Let r(x)= x a 2 55. Observe that 0 2 f (x) x for all x , and lim 0=0=lim x . So, by the Squeeze Theorem,
x 0 x 0 lim f (x)=0 .
x 0 56. Let f (x)=[ x] and g(x)= [ x] . Then lim f (x) and lim g(x) do not exist (Example 10) but
x
x 3 x 3 f (x)+g(x) =lim ( [ x] [ x]) =lim 0=0 . lim
3 x 3 x 3 57. Let f (x)=H(x) and g(x)=1 H(x) , where H is the Heaviside function defined in Exercise 1.3.59.
Thus, either f or g is 0 for any value of x . Then lim f (x) and lim g(x) do not exist, but
x lim
x 0 x 0 f (x)g(x) =lim 0=0 . 0 x 0 58.
lim
x 2 6 x 2
3 x 1 =lim
x 2 6 x 2 6 x +2 3 x +1 3 x 1 6 x +2 3 x +1 14 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws (
( =lim
x 2 =lim
x 2 3 2 ) 22
2 2
x) 1 6 x 3 x +1
6 x +2 x 2 x 12 2a+a+3=0 3( 2) +a( 2)+a+3=0
2 3x +15x+18 x 2 2 =lim
x x +x 2 6 x +2 2 , the limit will exist only if the numerator also ( 3x2+ax+a+3) =0 2 . In order for this to happen, we need lim 2 lim 3 x +1 6 x 4
3 x 1 3 x +1 1
( 2 x ) ( 3 x +1 )
=lim
=
( 2 x ) ( 6 x +2 ) x 2 6 x +2 2 59. Since the denominator approaches 0 as x
approaches 0 as x =lim 2 2 a=15 . With a=15 , the limit becomes 3(x+2)(x+3)
3(x+3) 3( 2+3) 3
=lim
=
= = 1.
(x 1)(x+2) x 2 x 1
2 1
3 60. Solution 1: First, we find the coordinates of P and Q as functions of r . Then we can find the
equation of the line determined by these two points, and thus find the x intercept (the point R ), and
take the limit as r 0 .
2 2 2 The coordinates of P are ( 0,r ) . The point Q is the point of intersection of the two circles x +y =r
1 2
2 2
2 2
2
2
and (x 1) +y =1 . Eliminating y from these equations, we get r x =1 (x 1) r =1+2x 1 x= r .
2
Substituting back into the equation of the shrinking circle to find the y coordinate, we get
1 2 2 2 2
1 2
1 2
2 2
r
+y =r
y =r 1
r
y=r 1
r (the positive y value). So the coordinates of
2
4
4
Q are
r
y r= 1 2
r ,r
2
1 1 1 2
r r
4 1 2
r 0
2 1 2
r
4 ( x 0 ) . We set y=0 in order to find the x intercept, and get 1 2
r
2 x= r
r . The equation of the line joining P and Q is thus = 1 2
r
2 1 2
r +1
4 1 2
1
r 1
4 1 2
1
r 1
4 Now we take the limit as r 1 + 0 : lim x=lim 2
r + 0 r + 0 =2 1 1 1 2
r +1
4 1 2
r +1
4 . =lim 2 ( 1 +1 ) =4 .
r + 0 So the limiting position of R is the point ( 4,0 ) .
15 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.3 Calculating Limits Using the Limit Laws Solution 2: We add a few lines to the diagram, as shown. Note that
diameter PS ). PQS=90 (subtended by So SQR=90 = OQT (subtended by diameter OT ). It follows that OQS= TQR . Also
PSQ=90
SPQ= ORP . Since QOS is isosceles, so is QTR , implying that QT =TR . As the
circle C shrinks, the point Q plainly approaches the origin, so the point R must approach a point
2 twice as far from the origin as T , that is, the point ( 4,0 ) , as above. 16 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit 1. (a) To have 5x+3 within a distance of 0.1 of 13 , we must have 12.9 5x+3 13.1
9.9 5x 10.1 1.98 x 2.02 . Thus, x must be within 0.02 units of 2 so that 5x+3 is within 0.1 of
13 .
(b) Use 0.01 in place of 0.1 in part (a) to obtain 0.002 .
2. (a) To have 6x 1 within a distance of 0.01 of 29 , we must have 28.99 6x 1 29.01
29.99 6x 30.01 4.9983 x 5.0016 . Thus, x must be within 0.0016 units of 5 so that 6x 1 is
within 0.01 of 29 .
(b) As in part (a) with 0.001 in place of 0.01 , we obtain 0.00016 .
(c) As in part (a) with 0.0001 in place of 0.01 , we obtain 0.000016 .
3. On the left side of x=2 , we need x 2 < 10
4
2 = . On the right side, we need
7
7 10
4
2 = . For both of these conditions to be satisfied at once, we need the more
3
3
4
4
restrictive of the two to hold, that is, x 2 < . So we can choose = , or any smaller positive
7
7
number.
x 2 < 4. On the left side, we need x 5 < 4 5 =1 . On the right side, we need x 5 < 5.7 5 =0.7 . For both
conditions to be satisfied at once, we need the more restrictive condition to hold; that is, x 5 <0.7 .
So we can choose =0.7 , or any smaller positive number.
5. The leftmost question mark is the solution of x =1.6 and the rightmost, x =2.4 . So the values are
2 2 1.6 =2.56 and 2.4 =5.76 . On the left side, we need x 4 < 2.56 4 =1.44 . On the right side, we need
x 4 < 5.76 4 =1.76 . To satisfy both conditions, we need the more restrictive condition to hold
namely, x 4 <1.44 . Thus, we can choose =1.44 , or any smaller positive number.
2 6. The left hand question mark is the positive solution of x =
2 question mark is the positive solution of x = 3
, that is, x=
2 1
1
, that is, x=
, and the right hand
2
2
3
. On the left side, we need
2 1
1
0.292 (rounding down to be safe). On the right side, we need
2
3
x 1 <
1
0.224 . The more restrictive of these two conditions must apply, so we choose
2
=0.224 (or any smaller positive number).
x 1 < 7. 4x+1 3 <0.5 2.5< 4x+1 <3.5 . We plot the three parts of this inequality on the same screen
1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit and identify the x coordinates of the points of intersection using the cursor. It appears that the
inequality holds for 1.3125 x 2.8125 . Since 2 1.3125 =0.6875 and 2 2.8125 =0.8125 , we
choose 0< <min { 0.6875,0.8125} =0.6875 . 1
<0.1 0.4<sin x<0.6 . From the graph, we see that for this inequality to hold, we
2
need 0.42 x 0.64 . So since 0.5 0.42 =0.08 and 0.5 0.64 =0.14 , we choose
0< min { 0.08,0.14} =0.08 .
8. sin x 9. For =1 , the definition of a limit requires that we find such that 3 ( 4+x 3x3) 2 <1 3 1<4+x 3x <3 whenever 0< x 1 < . If we plot the graphs of y=1 , y=4+x 3x and y=3 on the same
screen, we see that we need 0.86 x 1.11 . So since 1 0.86 =0.14 and 1 1.11 =0.11 , we choose
=0.11 (or any smaller positive number). For =0.1 , we must find such that ( 4+x 3x3) 2 <0.1 3 1.9<4+x 3x <2.1 whenever 0< x 1 < . From the graph, we see that we need 0.988 x 1.012 .
So since 1 0.988 =0.012 and 1 1.012 =0.012 , we choose =0.012 (or any smaller positive
number) for the inequality to hold. 2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit x 10. For =0.5 , the definition of a limit requires that we find
x such that e 1
1
x <0.5 x e 1
e 1
0.5<
<1.5 whenever 0< x 0 < . If we plot the graphs of y=0.5 , y=
, and y=1.5 on the
x
x
same screen, we see that we need 1.59 x 0.76 . So since 0 ( 1.59 ) =1.59 and 0 0.76 =0.76 ,
we choose =0.76 (or any smaller positive number). For =0.1 , we must find such that
x x e 1
e 1
1 <0.1 0.9<
<1.1 whenever 0< x 0 < . From the graph, we see that we need
x
x
0.21 x 0.18 . So since 0 ( 0.21 ) =0.21 and 0 0.18 =0.18 , we choose =0.18 (or any smaller
positive number) for the inequality to hold. 11. From the graph, we see that x ( x +1) ( x 1 )
2 2 >100 whenever 0.93 x 1.07 . So since 1 0.93 =0.07 and 1 1.07 =0.07 , we can take =0.07 (or any smaller positive number). 12. For M=100 , we need 0.0997<x<0 or 0<x<0.0997 . Thus, we choose =0.0997 (or any smaller
2 positive number) so that if 0< x < , then cot x>100 . 3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit For M=1000 , we need 0.0316<x<0 or 0<x<0.0316 . Thus, we choose =0.0316 (or any smaller
2 positive number) so that if 0< x < , then cot x>1000 . 2 13. (a) A= r and A=1000 cm
r=
(b) 1000
A 1000
995
r
1005 r>0
5 2 2 r =1000 2 r = 1000 17.8412 cm.
5
1005 1000 2 r 1000 5
17.7966 r 1000 5
17.8858 . r 2 1000+5
1000 995 0.04466 and 0.04455 . So if the machinist gets the radius within 0.0445 cm of 17.8412 ,
2 the area will be within 5 cm of 1000 .
(c) x is the radius, f (x) is the area, a is the target radius given in part (a), L is the target area (1000) ,
is the tolerance in the area ( 5 ), and is the tolerance in the radius given in part (b).
2 14. (a) T =0.1w +2.155w+20 and T =200
from the graph] w 33.0 watts ( w>0 ) 2 0.1w +2.155w+20=200 [ by the quadratic formula or 4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit (b) From the graph, 199 T 201 32.89<w<33.11 .
(c) x is the input power, f (x) is the temperature, a is the target input power given in part (a), L is the
target temperature ( 200 ), is the tolerance in the temperature ( 1 ), and is the tolerance in the
power input in watts indicated in part (b) ( 0.11 watts).
15. Given >0 , we need >0 such that if 0< x 1 < , then (2x+3) 5 < . But (2x+3) 5 <
2x 2 <
2 x 1 <
x 1 < /2 . So if we choose = /2 , then 0< x 1 <
(2x+3) 5 < . Thus,
lim (2x+3)=5 by the definition of a limit.
x 1 16. Given >0 , we need >0 such that if 0< x ( 2) < , then
( 1
x+3) 2 <
2 0< x ( 2) < 1
x+1 <
2
( 1
x+2 <
2 ( 1
x+3) 2 < . But
2 x ( 2) <2 . So if we choose =2 , then 1
1
x+3) 2 < . Thus, lim ( x+3)=2 by the definition of a limit.
2
x
2 2 17. Given >0 , we need >0 such that if 0< x ( 3) < , then (1 4x) 13 < . But (1 4x) 13 <
4x 12 <
4 x+3 <
x ( 3) < /4 . So if we choose = /4 , then 0< x ( 3) <
(1 4x) 13 < . Thus, lim (1 4x)=13 by the definition of a limit.
x 3 5 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit 18. Given >0 , we need >0 such that if 0< x 4 < , then (7 3x) ( 5) < . But (7 3x) ( 5) <
3x+12 <
3 x 4 <
x 4 < /3 . So if we choose = /3 , then 0< x 4 <
(7 3x) ( 5) < . Thus, lim (7 3x)= 5 by the definition of a limit.
x 4 19. Given >0 , we need >0 such that if 0< x 3 < , then
So choose =5 . Then 0< x 3 <
limit, lim
x 3 x 3 <5 x 3
<
5 x
5 3
1
<
x 3<
x 3 <5 .
5
5
x 3
< . By the definition of a
5 5 x 3
= .
5 5 20. Given >0 , we need >0 such that if 0< x 6 < , then x
+3
4 9
2 1
x 6 <
x 6 <4 . So choose =4 . Then 0< x 6 <
x 6 <4
4
x 6
x
9
<
+3
< . By the definition of a limit, lim
4 4
4
2
x 6
21. Given >0 , we need >0 such that if 0< x ( 5) < , then 4 3
x
5 < x
4 3
2 < x 6
<
4
x
9
+3 = .
4
2
7 < 6 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit 3
3
5
5
x 3 <
x+5 <
x ( 5) <
. So choose =
. Then x ( 5) <
5
5
3
3
3
3
4
x 7 < . Thus, lim 4
x =7 by the definition of a limit.
5
5
x
5
2 x +x 12
7 < . Notice that if
x 3 22. Given >0 , we need >0 such that if 0< x 3 < , then
2 x +x 12 (x+4)(x 3)
=
=x+4 . Thus, when 0< x 3 , we have
0< x 3 , then x 3 , so
x 3
x 3
2 x +x 12
7 <
x 3 (x+4) 7 < x 3 < . We take = and see that 0< x 3 < 2 2 x +x 12
x +x 12
7 < . By the definition of a limit, lim
=7 .
x 3
x 3
x 3
23. Given >0 , we need >0 such that if 0< x a < , then x a < . So = will work.
24. Given >0 , we need >0 such that if 0< x a < , then c c < . But c c =0 , so this will be
true no matter what we pick.
2 2 25. Given >0 , we need >0 such that if 0< x 0 < , then x 0 <
. Then 0< x 0 < 2 x< 0
3 3 26. Given >0 , we need >0 such that if 0< x 0 < , then x 0 <
3 3 . Then 0< x 0 < 3 x < x 4 . Take 0 x = x . So this is 0 28. Given >0 , we need >0 such that if 9
<x<9 . So take = 3 3 27. Given >0 , we need >0 such that if 0< x 0 < , then x 0 < . But
true if we pick = . Thus, lim x =0 by the definition of a limit. 4 x < x 0 < = . Thus, lim x =0 by the definition of a limit.
x 9 . Take = x 0 < . Thus, lim x =0 by the definition of a limit.
x = x < 2 . Then 9 <x<9 <x<9 , then
4 4 4 9 x 0 < 9 x 0 < . Thus, lim
x 4 9 x< 9 x< 4 9 x =0 by the definition of a 9 limit.
29. Given >0 , we need >0 such that if 0< x 2 < , then ( x2 4x+5) 1 < 2 x 4x+4 <
7 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit 2 (x 2) < . So take = . Then 0< x 2 < 2 (x 2) < . Thus, lim x 2 < x ( x2 4x+5) =1 2 by the definition of a limit. ( ) 2 2 x +x 12 <
30. Given >0 , we need >0 such that if 0< x 3 < , then x +x 4 8 <
(x 3)(x+4) < . Notice that if x 3 <1 , then 1<x 3<1 6<x+4<8
x+4 <8 . So take
=min { 1, /8} . Then 0< x 3 < (x 3)(x+4) 8(x 3) =8 x 3 <8 . Thus, lim
x ( x2+x 4) =8 3 by the definition of a limit. ( x2 1) 31. Given >0 , we need >0 such that if 0< x ( 2) < , then 3 < or upon simplifying we 2 need x 4 < whenever 0< x+2 < . Notice that if x+2 <1 , then 1<x+2<1
5<x 2< 3
x 2 <5 . So take =min { /5,1} . Then 0< x+2 <
x 2 <5 and x+2 < /5 , so ( x2 1) ( x2 1) =3 3 = (x+2)(x 2) = x+2 x 2 <( /5)(5)= . Thus, by the definition of a limit, lim
x 2 .
3 32. Given >0 , we need >0 such that if 0< x 2 < , then x 8 < . Now ( 3 2 x 8 = (x 2) x +2x+4
3 x 8 = x 2
3 x 8 = x 2 ) 2 2 . If x 2 <1 , that is, 1<x<3 , then x +2x+4<3 +2(3)+4=19 and so ( x2+2x+4) <19 x 2 . So if we take =min { 1, 19 } , then 0< x 2 <
( x2+2x+4) < 19 19= . Thus, by the definition of a limit, lim x3=8 .
x 33. Given >0 , we let =min
x+3 <8 . Also x 3 < { }
2, 8 8 are =3 1 9 4<x+3<8 8 = . Thus, lim x =9 . and x = 9+ 2 3 3 . The largest possible choice is the minimum value of { , }; that is, =min{ , }= = 9+
1 2 2<x 3<2
2 2 , so x 9 = x+3 x 3 <8 34. From the figure, our choices for
for . If 0< x 3 < , then x 3 <2 2 1 2 2 3. 8 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit 35. (a) The points of intersection in the graph are (x ,2.6) and (x ,3.4) with x
1 Thus, we can take 2 1 0.891 and x 2 1.093 . to be the smaller of 1 x and x 1 . So =x 1 0.093 .
1 2 2 3 (b) Solving x +x+1=3+ gives us two nonreal complex roots and one real root, which is
x( )= 2/3 ( 216+108 +12 336+324 +81 ) 12 . Thus,
6 ( 216+108 +12 336+324 +81 )
2 2 1/3 =x( ) 1 . (c) If =0.4 , then x( ) 1.093272342 and =x( ) 1 0.093 , which agrees with our answer in part
(a).
36. 1. Guessing a value for Let >0 be given. We have to find a number >0 such that
1 1
1 1
2 x
x 2
< whenever 0< x 2 < . But
=
=
< . We find a positive
x 2
x 2
2x
2x
1
x 2
constant C such that
<C
<C x 2 and we can make C x 2 < by taking x 2 < =
2x
2x
C
1 1
1 1 1
1
1
. We restrict x to lie in the interval x 2 <1 1<x<3 so 1> >
<
<
< . So
x 3
6 2x 2
2x
2
1
C= is suitable. Thus, we should choose =min { 1,2 } .
2
2. Showing that works Given >0 we let =min { 1,2 } . If 0< x 2 < , then x 2 <1 1<x<3
1
1
1 1
x 2
1
< (as in part 1). Also x 2 <2 , so
=
< 2 = . This shows that
2x
2
x 2
2x
2
9 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit lim (1/x)=
x 2 1
.
2 37. 1. Guessing a value for Given
x a
0< x a < . But x a =
x+ a
x a
such that x + a >C then
<
x+ a >0 , we must find >0 such that x a < whenever < (from the hint). Now if we can find a positive constant C x a
< , and we take x a <C . We can find this number by
C
1
1
1
1
3
restricting x to lie in some interval centered at a . If x a < a , then
a<x a< a
a<x< a
2
2
2
2
2
1
1
x+ a>
a + a , and so C=
a + a is a suitable choice for the constant. So
2
2
x a < 1
a+ a
2 2. Showing that . This suggests that we let =min works Given >0 , we let =min { { 1
a,
2 1
a,
2 1
a+ a
2 1
a+ a
2 } } . . If 0< x a < , 1
1
1
a
x+ a>
a + a (as in part 1). Also x a <
a+ a
, so
2
2
2
( a/2 + a )
x a
x a =
<
= . Therefore, lim x = a by the definition of a limit.
x+ a
( a/2 + a )
x a
1
1
38. Suppose that lim H(t)=L . Given = , there exists >0 such that 0< t <
H(t) L <
2
2
t 0
1
1
1
1
1
L
<H(t)<L+ . For 0<t< , H(t)=1 , so 1<L+
L> . For <t<0 , H(t)=0 , so L
<0
2
2
2
2
2
1
1
L< . This contradicts L> . Therefore, lim H(t) does not exist.
2
2
t 0
1
1
39. Suppose that lim f (x)=L . Given = , there exists >0 such that 0< x <
f (x) L < .
2
2
x 0
1
1
Take any rational number r with 0< r < . Then f (r)=0 , so 0 L < , so L L < . Now take
2
2
1
1
1
any irrational number s with 0< s < . Then f (s)=1 , so 1 L < . Hence, 1 L< , so L> . This
2
2
2
1
contradicts L< , so lim f (x) does not exist.
2
x 0
then x a < 40. First suppose that lim f (x)=L . Then, given >0 there exists >0 so that 0< x a <
x f (x) L < . Then a a <x<a 0< x a < so f (x) L < . Thus,
10 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit lim f (x)=L . Also a<x<a+
x 0< x a < so f (x) L < . Hence, lim f (x)=L . a + x a Now suppose lim f (x)=L=lim f (x) . Let >0 be given. Since lim f (x)=L , there exists
x a <x<a a + x a x f (x) L < . Since lim f (x)=L , there exists 1 x be the smaller of 1 and a lim f (x)=L . So we have proved that lim f (x)=L
x a x 1 41. 4 4 >10 , 000 (x+3) < (x+3) >0 so that a<x<a+ f (x) L < . Let 2 a . Then 0< x a < 2 a 2 + >0 so that 1 a <x<a or a<x<a+ 1 lim f (x)=L=lim f (x) .
x 1
10,000 x+3 < so f (x) L < . Hence, 2 a 1
4 + x a x ( 3) < 10,000 42. Given M>0 , we need >0 such that 0< x+3 < 1
10
1 4 1/(x+3) >M . Now 4 >M 4 (x+3) < (x+3)
x+3 < 1
4 . So take = M 1
4 . Then 0< x+3 < = M 1
4 1
4 M x (x+3) 43. Given M<0 we need >0 so that ln x<M whenever 0<x< ; that is, x=e
M This suggests that we take =e . If 0<x<e
lim ln x=
.
x M 1 >M , so lim ln x 4 3 = 1
M
. (x+3) M <e whenever 0<x< . M , then ln x<ln e =M . By the definition of a limit, + 0 44. (a) Let M be given. Since lim f (x)=
x Since lim g(x)=c , there exists
x a smaller of 1 and 2 , there exists a >0 such that 0< x a < 2 . Then 0< x a < g(x)>c/2 . Since lim f (x)= { , } 1 2 a , there exists a . Then 0< x a < f (x) g(x)> g(x)<c/2 . Since lim f (x)=
x a a , there exists 2 f (x)+g(x) = . 1 g(x) c <c/2 f (x)>2M/c . Let >0 such that 0< x a < 1 >0 such that 0< x a < 2 2 2M c
=M , so lim f (x) g(x)=
c 2
x a (c) Let N<0 be given. Since lim g(x)=c<0 , there exists
x >0 such that 0< x a < >0 such that 0< x a < be the a 1 2 f (x)>M+1 c . g(x)>c 1 . Let
x x =min g(x) c <1 2 1 f (x)+g(x)>(M+1 c)+(c 1)=M . Thus, lim (b) Let M>0 be given. Since lim g(x)=c>0 , there exists
x >0 such that 0< x a < 1 .
1 g(x) c < c/2 f (x)>2N/c . (Note that
11 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.4 The Precise Definition of a Limit c<0 and N<0
f (x) g(x)< 2N/c>0 .) Let =min { 2N c
=N , so lim f (x) g(x)=
c 2
x a , } 1 2 . Then 0< x a < f (x)>2N/c . 12 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity 1. From Definition 1, lim f (x)= f (4) .
x 4 2. The graph of f has no hole, jump, or vertical asymptote.
3. (a) The following are the numbers at which f is discontinuous and the type of discontinuity at that
number: 4 (removable), 2 ( jump), 2 ( jump), 4 (infinite).
(b) f is continuous from the left at 2 since lim f (x)= f ( 2) . f is continuous from the right at 2 and
x 2 4 since lim f (x)= f (2) and lim f (x)= f (4) . It is continuous from neither side at 4 since f ( 4) is
x + 2 x + 4 undefined.
4. g is continuous on 4, 2 ) , ( 2,2 ) , 2,4 ) , ( 4,6 ) , and ( 6,8 ) . 5. The graph of y= f (x) must have a discontinuity at x=3 and must show that lim f (x)= f (3) .
x 3 6. 7. (a)
(b) There are discontinuities at times t=1 , 2 , 3 , and 4 . A person parking in the lot would want to
keep in mind that the charge will jump at the beginning of each hour.
8. (a) Continuous; at the location in question, the temperature changes smoothly as time passes,
without any instantaneous jumps from one temperature to another.
1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity (b) Continuous; the temperature at a specific time changes smoothly as the distance due west from
New York City increases, without any instantaneous jumps.
(c) Discontinuous; as the distance due west from New York City increases, the altitude above sea
level may jump from one height to another without going through all of the intermediate values
at
a cliff, for example.
(d) Discontinuous; as the distance traveled increases, the cost of the ride jumps in small increments.
(e) Discontinuous; when the lights are switched on (or off ), the current suddenly changes between 0
and some nonzero value, without passing through all of the intermediate values. This is debatable,
though, depending on your definition of current.
9. Since f and g are continuous functions,
lim 2 f (x) g(x) = 2lim f (x) lim g(x) [by Limit Laws 2 and 3]
x 3 x 3 x 3 [by continuity of f and g at x=3 ]
= 2 f (3) g(3)
= 2 5 g(3)=10 g(3)
Since it is given that lim 2 f (x) g(x) =4 , we have 10 g(3)=4 , so g(3)=6 .
x ( x2+ 10. lim f (x)=lim
x 4 x 3 ) 2 2 7 x =lim x + 4 x lim 7 lim x =4 + 7 4 =16+ 3 = f (4) . 4 x 4 x 4 By the definition of continuity, f is continuous at a=4 . ( x+2x3) 4= 11. lim f (x)=lim
x 1 x 1 lim x+2lim x
x 1 x 4 3 3 4 = 1+2( 1) 4 =( 3) =81= f ( 1) . 1 By the definition of continuity, f is continuous at a= 1 .
lim x+lim 1
12. lim g(x)=lim
x 4 x 4 x+1
2 = 2x 1 x 4 x 4 = 2 2lim x lim 1
x 4 x 4+1
2 2(4) 1 = 5
=g(4) . So g is continuous at 4 .
31 4 lim (2x+3)
2lim x+lim 3
2x+3 x a
x a
x a
13. For a>2 , we have lim f (x)=lim
=
[ Limit Law 5] =
lim (x 2)
lim x lim 2
x a
x a x 2
x a x a x 2a+3
[ 7 and 8] = f (a) . Thus, f is continuous at x=a for every a in (2,
a 2
continuous on (2, ) .
[1, 2, and 3] = 14. For a<3 , we have lim g(x)=lim 2 3 x =2lim
x a x a x a 3 x [ Limit Law 3] =2 a ) ; that is, f is lim (3 x) [ 11]
x a
2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity =2 lim 3 lim x [ 2] =2 3 a [ 7 and 8] =g(a) , so g is continuous at x=a for every a in (
x a x Also, lim g(x)=0=g(3) , so g is continuous from the left at 3 . Thus, g is continuous on (
x ,3) . a ,3]. 3 15. f (x)=ln x 2 is discontinuous at 2 since f (2)=ln 0 is not defined. 16. f (x)= 17. f (x)= { 1/(x 1)
2 { e x
2 x if x 1
if x=1 is discontinuous at 1 because lim f (x) does not exist.
x 1 ifx<0
ifx 0
x The left hand limit of f at a=0 is lim f (x)=lim e =1 . The right hand limit of f at a=0 is
x 0 x 0 2 lim f (x)=lim x =0 . Since these limits are not equal, lim f (x) does not exist and f is discontinuous
x + 0 x + 0 x 0 at 0. 18. 3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity f (x)= { 2 x x if x 1
if x=1 2 x 1
1
2 x(x 1)
lim f (x) =lim x x =lim
x 1
2
x 1 x 1 x 1 (x+1)(x 1)
x
1
=lim
= ,
x 1 x+1 2
but f (1)=1 , so f is discontinous at 1. { 2 x x 12 if x 3
=
19. f (x)=
x+3
if x= 3
5
So lim f (x)=lim (x 4)= 7 and f ( 3)= 5 .
x 3 x Since lim f (x)
x 20. f (x)= { x 4 if x 3
5 if x= 3 3 f ( 3) , f is discontinuous at 3 . 3 { 2 1+x
4 x 2 if x<1
if x 1
2 lim f (x)=lim (1+x )=1+1 =2 and
x 1 x 1 4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity lim f (x)=lim (4 x)=4 1=3 .
x + 1 + x 1 Thus, f is discontinuous at 1 because lim f (x) does not exist.
x x 21. F(x)= 2 1 is a rational function. So by Theorem 5 (or Theorem 7), F is continuous at every x +5x+6 { } 2 number in its domain, x| x +5x+6 0 = { x|(x+3)(x+2) 0} = { x| x
( , 3) ( 3, 2 ) ( 2, ) .
22. By Theorem 7, the root function 3 3, 2} or 3 x and the polynomial function 1+x are continuous on R . By
3 ( part 4 of Theorem 4, the product G(x)= x 1+x 3 ) is continuous on its domain, R . 2 , ) . By Theorem 7, the root
23. By Theorem 5, the polynomials x and 2x 1 are continuous on (
function x is continuous on [0, )] . By Theorem 9, the composite function 2x 1 is continuous on
1
1
2
its domain, [ , )] . By part 1 of Theorem 4, the sum R(x)=x + 2x 1 is continuous on [ , )] .
2
2
24. By Theorem 7, the trigonometric function sin x and the polynomial function x+1 are continuous
sin x
on R . By part 5 of Theorem 4, h ( x ) =
is continuous on its domain, { x| x 1} .
x+1
25. By Theorem 5, the polynomial 5x is continuous on (
continuous on ( ) . By Theorems 9 and 7, sin 5x is
x
, ) . By Theorem 7, e is continuous on (
, ) . By part 4 of Theorem 4, the
, x product of e and sin 5x is continuous at all numbers which are in both of their domains, that is, on
( , ).
2 26. By Theorem 5, the polynomial x 1 is continuous on (
continuous on its domain, { x| 2 } { 1,1 . By Theorem 9, sin
2 1 x 1 1 = x|0 x } 2 ={ x| x 2}= 1 , ) . By Theorem 7, sin 1 is ( x2 1) is continuous on its domain, which is
2, 2 .
5 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity 4 27. By Theorem 5, the polynomial t 1 is continuous on ( , ) . By Theorem 7, ln x is continuous ) . By Theorem 9, ln ( t 1 ) is continuous on its domain, which is
{ t |t 4 1>0} = { t |t 4>1} ={t | t >1} = ( , 1 ) ( 1, ) .
4 on its domain, ( 0, 28. By Theorem 7, x is continuous on 0, ) . By Theorems 7 and 9, e ( ) is continuous on Also by Theorems 7 and 9, cos e
1 29. The function y= 1/x x 0, x is continuous on 0, ). ). is discontinuous at x=0 because the left and right hand limits at x=0 are 1+e
different. 2 30. The function y=tan x is discontinuous at x= ( 2 ) 2 2 + k , where k is any integer. The function ( 2 ) y=ln tan x is also discontinuous where tan x is 0 , that is, at x= k . So y=ln tan x is
discontinuous at x= 2 n , n any integer. 31. Because we are dealing with root functions, 5+ x is continuous on 0, ) , x+5 is continuous
5+ x
is continuous on 0, ) . Since f is continuous at x=4 ,
on 5, ) , so the quotient f (x)=
5+x
7
lim f (x)= f (4)= .
3
x 4
32. Because x is continuous on R , sin x is continuous on R , and x+sin x is continuous on R , the
6 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity composite function f (x)=sin (x+sin x) is continuous on R , so lim f (x)= f ( )=sin ( +sin )=sin =0 .
x
2 2 33. Because x x is continuous on R , the composite function f (x)=e
1 1 is continuous on R , so 0 =e =1 . lim f (x)= f (1)=e
x x x 1 34. Because arctan is a continuous function, we can apply Theorem 8.
2 x 4 lim arctan
x 2 2 35. f (x)= =arctan lim
x 3x 6x { 2 2 (x+2)(x 2)
3x(x 2) =arctan lim
x x 2 2 lim f (x)=lim x =1 and lim f (x)=lim
1 x =arctan 0.588 1 x + 1 x ,1) , f is continuous on (
,1) . By
), f is continuous on (1, ) . At x=1 , x =1 . Thus, lim f (x) exists and equals 1 . Also, f (1)= 1 =1
+ 1 x 1 . Thus, f is continuous at x=1 . We conclude that f is continuous on ( , ). { sin x if x< /4
cos x if x
/4
By Theorem 7, the trigonometric functions are continuous. Since f (x)=sin x on (
f (x)=cos x on ( /4, ) , f is continuous on (
, /4) ( /4, ).
36. f (x)= 2
3 if x<1
if x 1 x By Theorem 5, since f (x) equals the polynomial x on (
Theorem 7, since f (x) equals the root function x on (1,
x 2 x+2
3x , /4) and lim f (x)= lim sin x=sin =1/ 2 since the sine function is continuous at /4. Similarly,
4
x ( /4 )
x ( /4 )
lim f (x)= lim cos x=1/ 2 by continuity of the cosine function at /4 . Thus, lim f (x)
+
+
x ( /4 )
x ( /4 )
x ( /4 )
exists and equals 1/ 2 , which agrees with the value f ( /4) . Therefore, f is continuous at /4 , so f
is continuous on (
, ). 37. f (x)= { 2 1+x
2 x 2 (x 2)
f is continuous on ( if x 0
if 0<x 2
if x>2
,0) , (0,2) , and (2, ) since it is a polynomial on each of these intervals. Now
7 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity 2 lim f (x)=lim (1+x )=1 and
x 0 x 0 lim f (x)=lim (2 x)=2, so f is discontinuous at 0 . Since f (0)=1 , f is continuous from the left at 0.
x + 0 x + 0 2 Also, lim f (x)=lim (2 x)=0,lim f (x)=lim (x 2) =0 , and f (2)=0 , so f is continuous at 2. The only
x 2 x 2 x + 2 x + 2 number at which f is discontinuous is 0 . { x+1
if x 1
1/x
if 1<x<3
x 3 if x 3
f is continuous on (
,1) , (1,3) , and (3, ) , where it is a polynomial, a rational function, and a
composite of a root function with a polynomial, respectively. Now lim f (x)=lim (x+1)=2 and
38. f (x)= x 1 x 1 lim f (x)=lim ( 1/x ) =1 , so f is discontinuous at 1 .
x + 1 x + 1 Since f (1)=2 , f is continuous from the left at 1 . Also, lim f (x)=lim ( 1/x ) =1/3 , and
x lim f (x)=lim
x + 3 x 3 x 3 x 3 =0= f (3) , so f is discontinuous at 3 , but it is continuous from the right at 3.
+ 3 { x+2 if x<0
x
39. f (x)=
if 0 x 1
e
2 x if x>1
f is continuous on (
,0 ) and ( 1, ) since on each of these intervals it is a polynomial; it is
continuous on ( 0,1 ) since it is an exponential. Now
8 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity x lim f (x)=lim (x+2)=2 and lim f (x)=lim e =1 , so f is discontinuous at 0 . Since f (0)=1 , f is
x 0 x x 0 + 0 x + 0 x continuous from the right at 0 . Also lim f (x)=lim e =e and lim f (x)=lim (2 x)=1 , so f is
x 1 x 1 + x 1 x + 1 discontinuous at 1 . Since f (1)=e , f is continuous from the left at 1 . 40. By Theorem 5, each piece of F is continuous on its domain. We need to check for continuity at
r=R .
GMr GM
GM GM
GM
GM
lim F(r)=lim
=
and lim F(r)=lim
=
, so lim F(r)=
. Since F(R)=
,
3
2
2
2
2
2
+
+
r R
R
R
R
R
r R
r R R
r R
r R r
F is continuous at R . Therefore, F is a continuous function of r .
41. f is continuous on ( ,3) and ( 3, ) . Now lim f (x)=lim ( cx+1 ) =3c+1 and
x ( cx 1) =9c 1 . So f is continuous
2 lim f (x)=lim
+ x 3 x + 3 continuous on ( , ) , c= 2 3 x 3 3c+1=9c 1 6c=2 c= 1
. Thus, for f to be
3 1
.
3 2 42. The functions x c and cx+20 , considered on the intervals (
,4 ) and 4, ) respectively, are
continuous for any value of c . So the only possible discontinuity is at x=4 . For the function to be
continuous at x=4 , the left hand and right hand limits must be the same. Now ( x2 c2) =16 c2 and lim g(x)=lim (cx+20)=4c+20=g(4) . Thus, 16 c2=4c+20 lim g(x)=lim
x 4
2 x c +4c+4=0 4 x + 4 x + 4 c= 2 .
2 x 2x 8 (x 4)(x+2)
=
43. (a) f (x)=
has a removable discontinuity at 2 because g(x)=x 4 is
x+2
x+2
continuous on R and f (x)=g(x) for x 2 . [The discontinuity is removed by defining f ( 2)= 6 .]
(b)
9 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity f (x)= x 7
x 7 lim f (x)= 1 and lim f (x)=1 . Thus, lim f (x) does not exist, so the discontinuity is
x 7 + x 7 x 7 not removable. (It is a jump discontinuity.)
3 2 x +64 (x+4)(x 4x+16)
2
has a removable discontinuity at 4 because g(x)=x 4x+16 is
=
(c) f (x)=
x+4
x+4
continuous on R and f (x)=g(x) for x 4 .[The discontinuity is removed by defining f ( 4)=48 .]
3 x
3 x
1
(d) f (x)=
has a removable discontinuity at 9 because g(x)=
is
=
9 x
3+ x
( 3 x ) ( 3+ x )
1
continuous on R and f (x)=g(x) for x 9 . [The discontinuity is removed by defining f (9)= .]
6
44. f does not satisfy the conclusion of the Intermediate Value Theorem. f does satisfy the conclusion of the Intermediate Value Theorem.
3 2 45. f (x)=x x +x is continuous on the interval 2,3 , f (2)=6 , and f (3)=21 . Since 6<10<21 , there is
a number c in ( 2,3) such that f (c)=10 by the Intermediate Value Theorem.
2 46. f (x)=x is continuous on the interval 1,2 , f (1)=1 , and f (2)=4 . Since 1<2<4 , there is a
2 number c in ( 1,2 ) such that f (c)=c =2 by the Intermediate Value Theorem.
4 47. f (x)=x +x 3 is continuous on the interval [1,2], f (1)= 1 , and f (2)=15 . Since 1<0<15 , there is a
number c in (1,2) such that f (c)=0 by the Intermediate Value Theorem. Thus, there is a root of the
4 equation x +x 3=0 in the interval (1,2).
3 48. f (x)= x +x 1 is continuous on the interval [0,1], f (0)= 1 , and f (1)=1 . Since 1<0<1 , there is a
number c in (0,1) such that f (c)=0 by the Intermediate Value Theorem. Thus, there is a root of the
10 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity equation 3 x +x 1=0 , or 3 x =1 x , in the interval (0,1). 49. f (x)=cos x x is continuous on the interval 0,1 , f (0)=1 , and f (1)=cos 1 1 0.46 . Since
0.46<0<1 , there is a number c in ( 0,1 ) such that f (c)=0 by the Intermediate Value Theorem. Thus,
there is a root of the equation cos x x=0 , or cos x=x , in the interval ( 0,1 ) .
x 1 2 50. f (x)=ln x e is continuous on the interval 1,2 , f (1)= e
0.37 , and f (2)=ln 2 e
0.56 .
Since 0.37<0<0.56 , there is a number c in ( 1,2 ) such that f (c)=0 by the Intermediate Value
x x Theorem. Thus, there is a root of the equation ln x e =0 , or ln x=e , in the interval ( 1,2 ) .
x 51. (a) f (x)=e +x 2 is continuous on the interval 0,1 , f (0)= 1<0 , and f (1)=e 1 1.72>0 . Since
1<0<1.72 , there is a number c in ( 0,1 ) such that f (c)=0 by the Intermediate Value Theorem. Thus,
x x there is a root of the equation e +x 2=0 , or e =2 x , in the interval ( 0,1 ) .
(b) f (0.44) 0.007<0 and f (0.45) 0.018>0 , so there is a root between 0.44 and 0.45 .
52. (a) f (x)=sin x 2+x is continuous on 0,2 , f (0)= 2 , and f (2)=sin 2 0.91 . Since 2<0<0.91 ,
there is a number c in ( 0,2 ) such that f (c)=0 by the Intermediate Value Theorem. Thus, there is a root
of the equation sin x 2+x=0 , or sin x=2 x , in the interval ( 0,2 ) .
(b) f (1.10) 0.009<0 and f (1.11) 0.006>0 , so there is a root between 1.10 and 1.11 .
5 2 5 2 5 2 53. (a) Let f (x)=x x 4 . Then f (1)=1 1 4= 4<0 and f (2)=2 2 4=24>0 . So by the Intermediate
5 2 Value Theorem, there is a number c in ( 1,2 ) such that f (c)=c c 4=0 .
(b) We can see from the graphs that, correct to three decimal places, the root is x 1.434 . 54. (a) Let f (x)= x 5 1
1
8
. Then f (5)=
<0 and f (6)= >0 , and f is continuous on 5,
x+3
8
9 ) . So
11 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity by the Intermediate Value Theorem, there is a number c in ( 5,6 ) such that f (c)=0 . This implies that
1
= c 5 .
c+3
(b) Using the intersect feature of the graphing device, we find that the root of the equation is x=5.016
, correct to three decimal places. 55. (
) If f is continuous at a , then by Theorem 8 with g(h)=a+h , we have
lim f (a+h)= f lim ( a+h ) = f (a) .
h ( ( 0 h ) 0 ) Let >0 . Since lim f (a+h)= f (a) , there exists >0 such that 0< h <
h f (a+h) f (a) < . So 0 if 0< x a < , then f (x) f (a) = f (a+(x a)) f (a) < . Thus, lim f (x)= f (a) and so f is continuous
x a at a .
56.
lim sin (a+h) =lim ( sin acos h+cos asin h ) =lim ( sin acos h ) +lim ( cos asin h )
h 0 h ( 0 h = lim sin a
h 0 0 h 0 ) ( lim cos h) +( lim cos a) ( lim sin h)
h 0 h 0 h 0 =(sin a)(1)+(cos a)(0)=sin a
57. As in the previous exercise, we must show that lim cos (a+h)=cos a to prove that the cosine
h 0 function is continuous.
lim cos (a+h) =lim ( cos acos h sin asin h )
h 0 h 0 =lim ( cos acos h ) lim ( sin asin h )
h ( 0 h = lim cos a
h 0 0 ) ( lim cos h) ( lim sin a) ( lim sin h)
h 0 h 0 h 0 =(cos a)(1) (sin a)(0)=cos a
58. (a) Since f is continuous at a , lim f (x)= f (a) .Thus, using the Constant Multiple Law of Limits,
x a
12 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity we have lim (cf )(x)=lim cf (x)=clim f (x)=cf (a)=(cf )(a) . Therefore, cf is continuous at a .
x a x a x a (b) Since f and g are continuous at a , lim f (x)= f (a) and lim g(x)=g(a) . Since g(a) 0 , we can use
x f
g the Quotient Law of Limits: lim
x a a x a lim f (x)
f (x) x a
f (a)
(x)=lim
=
=
=
lim g(x) g(a)
x a g(x)
x f
g (a) . Thus, f
is
g a continuous at a . {( 0 if x is rational
is continuous nowhere. For, given any number a and any >0 ,
1 if x is irrational
the interval a ,a+ ) contains both infinitely many rational and infinitely many irrational numbers.
Since f (a)=0 or 1 , there are infinitely many numbers x with 0< x a < and f (x) f (a) =1 . Thus,
lim f (x) f (a) . [In fact lim f (x) does not even exist.]
59. f (x)= x a x a { 0 if x is rational
is continuous at 0 . To see why, note that x g(x) x , so by
x if x is irrational
the Squeeze Theorem lim g(x)=0=g(0) . But g is continuous nowhere else. For if a 0 and >0 , the 60. g(x)= x 0 interval ( a ,a+ ) contains both infinitely many rational and infinitely many irrational numbers.
Since g(a)=0 or a , there are infinitely many numbers x with 0< x a < and g(x) g(a) > a /2 .
Thus, lim g(x) g(a) .
x a
3 3 61. If there is such a number, it satisfies the equation x +1=x x x+1=0 . Let the left hand side of
this equation be called f (x) . Now f ( 2)= 5<0 , and f ( 1)=1>0 . Note also that f (x) is a polynomial,
and thus continuous. So by the Intermediate Value Theorem, there is a number c between 2 and 1
3 such that f (c)=0 , so that c=c +1 .
62. (a) lim F(x)=0 and lim F(x)=0 , so lim F(x)=0 , which is F(0) , and hence F is continuous at
x + 0 x x 0 0 x=a if a=0 . For a>0 , lim F(x)=lim x=a=F(a) . For a<0 , lim F(x)=lim ( x)= a=F(a) . Thus, F is
x a x a x a x continuous at x=a ; that is, continuous everywhere.
(b) Assume that f is continuous on the interval I . Then for a I , lim
x a a f (x) = lim f (x) = f (a) by
x a Theorem 8. (If a is an endpoint of I , use the appropriate one sided limit.) So f is continuous on I .
1 if x 0
(c) No, the converse is false. For example, the function f (x)=
is not continuous at
1 if x<0
x=0 , but f (x) =1 is continuous on R . { 13 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.5 Continuity 63. Define u(t) to be the monk’s distance from the monastery, as a function of time, on the first day,
and define d(t) to be his distance from the monastery, as a function of time, on the second day. Let D
be the distance from the monastery to the top of the mountain. From the given information we know
that u(0)=0 , u(12)=D , d(0)=D and d(12)=0 . Now consider the function u d , which is clearly
continuous. We calculate that (u d)(0)= D and (u d)(12)=D . So by the Intermediate Value Theorem,
there must be some time t between 0 and 12 such that (u d)(t )=0 u(t )=d(t ) . So at time t after
0 0 0 0 0 7:00 A.M., the monk will be at the same place on both days. 14 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 1. (a) As x becomes large, the values of f (x) approach 5 .
(b) As x becomes large negative, the values of f (x) approach 3 .
2. (a) The graph of a function can intersect a vertical asymptote in the sense that it can meet but not
cross it. The graph of a function can intersect a horizontal asymptote. It can even intersect its horizontal
asymptote an infinite number of times. (b) The graph of a function can have 0 , 1 , or 2 horizontal asymptotes. Representative examples are
shown. No horizontal
asymptote One horizontal
asymptote Two horizontal
asymptotes 3. (a) lim f (x)=
x 2 (b) lim f (x)=
x 1 1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes (c) lim f (x)=
+ x 1 (d) lim f (x)=1
x (e) lim f (x)=2
x (f) Vertical: x= 1 , x=2 ; Horizontal: y=1 , y=2
4. (a) lim g(x)=2
x (b) lim g(x)= 2
x (c) lim g(x)=
x 3 (d) lim g(x)=
x 0 (e) lim g(x)=
x + 2 (f) Vertical: x= 2 , x=0 , x=3 ; Horizontal: y= 2 , y=2
5. f (0)=0 , f (1)=1 , lim f (x)=0 ,
x f is odd 2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 6. lim f (x)=
x , lim f (x)= + 0 x , 0 lim f (x)=1 , lim f (x)=1
x x 7. lim f (x)=
x , lim f (x)= 2 , x lim f (x)=0 , lim f (x)=
x x , + 0 lim f (x)=
x 0 8. lim f (x)=
x , lim f (x)=3 , 2 x lim f (x)= 3
x 2 x 9. If f (x)=x /2 , then a calculator gives f (0)=0 , f (1)=0.5 , f (2)=1 , f (3)=1.125 , f (4)=1 ,
f (5)=0.78125 , f (6)=0.5625 , f (7)=0.3828125 , f (8)=0.25 , f (9)=0.158203125 , f (10)=0.09765625 ,
f (20) 0.00038147 , f (50) 2.2204 10
It appears that lim 12 , f (100) 7.8886 10 27 . ( x /2 ) =0 .
2 x x 3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes x 10. (a) From a graph of f (x)= ( 1 2/x ) in a window of 0,10,000 by 0,0.2 , we estimate that
lim f (x)=0.14 (to two decimal places.)
x (b)
x f (x) 10,000 0.135308 100,000 0.135333 1,000,000 0.135335 From the table, we estimate that lim f (x)=0.1353 (to four decimal places.)
x 11.
2 lim
x 3x x+4
2 2x +5x 8 2 [ divide both the numerator and 2 2 (3x x+4)/x =lim 2 x denominator by x
(the highest power of x that
appears in the denominator)] 2 (2x +5x 8)/x
2 lim (3 1/x+4/x )
= x [ Limit Law 5] 2 lim (2+5/x 8/x )
x
2 lim 3 lim (1/x)+lim (4/x )
= x x x
2 [ Limit Laws 1 and 2] lim 2+lim (5/x) lim (8/x )
x x x
2 3 lim (1/x)+4lim (1/x )
= x x
2 [ Limit Laws 7 and 3] 2+5lim (1/x) 8lim (1/x )
x 3 0+4(0)
2+5(0) 8(0)
3
=
2
= x [Theorem 5 of Section 2.5] 12.
4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 3 lim
x 12x 5x+2
2 1+4x +3x 3 3 = 12x 5x+2 lim 2 x 1+4x +3x [ Limit Law 11] 3 2 = 12 5/x +2/x lim 3
3 [ divide by x ] 3 x 1/x +4/x+3
2 3 lim (12 5/x +2/x )
= x [ Limit Law 5] 3 lim (1/x +4/x+3)
x
2 3 lim 12 lim (5/x )+lim (2/x )
= x x x [ Limit Laws 1 and 2] 3 lim (1/x )+lim (4/x)+lim 3
x x x 2 3 12 5lim (1/x )+2lim (1/x )
x = x [ Limit Laws 7 and 3] 3 lim (1/x )+4lim (1/x)+3
x = 12 5(0)+2(0)
0+4(0)+3 = x 12
= 4 =2
3 [Theorem 5 of Section 2.5] lim (1/x)
lim (1/x)
1
1/x
x
x
0
0
13. lim
=lim
=
=
=
= =0
2x+3 x
(2x+3)/x lim (2+3/x) lim 2+3lim (1/x) 2+3(0) 2
x
x x x lim 3+5lim
14. lim
x 1
x 3x+5
(3x+5)/x
3+5/x x
x
3+5(0)
=lim
=lim
=
=
=3
x 4 x
(x 4)/x x
1 4/x
1
1 4(0)
lim 1 4lim
x
x
x
5 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 15.
2 2 2 1 x x lim = lim 2 x (1 x x )/x
2 x 2x 7 lim (1/x 1/x 1) 2 = 2 x
2 lim (2 7/x ) (2x 7)/x x
2 lim (1/x ) lim (1/x) lim 1
= x x x
2 = lim 2 7 lim (1/x )
x 0 0 1
1
=
2 7(0)
2 x
2 2 2 3y 16. lim 2 y 2 (2 3y )/y =lim 2 y 5y +4y 2 2 lim (2/y 3) 2 y = lim (5+4/y) (5y +4y)/y 2lim (1/y ) lim 3
= y y y lim 5+4lim (1/y)
y = 2(0) 3
3
=
5+4(0)
5 y 3 17. Divide both the numerator and denominator by x (the highest power of x that occurs in the
denominator).
3 x +5x
3 x +5x lim
x 3 2 2x x +4 x =lim 3 3 x 1+
=lim 2 2x x +4
x 3 lim 1+5lim
x x =
lim 2 lim
x 2 18. lim
t t +2
3 2 t +t 1 = lim
t x x 5 lim 2 x 1 4
2
+
3
x
x 5 1+ x 2 x =
lim
x 2 1 4
+
3
x
x 1
2 x 1
+4lim
x x 1
x ( t2+2) /t 3 = lim
( t 3+t2 1) /t 3 t = 1+5(0)
1
=
2 0+4(0) 2 3 1/t+2/t 3 1+1/t 1/t 3 = 0+0
=0
1+0 0 19. First, multiply the factors in the denominator. Then divide both the numerator and denominator by
4 u . 6 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 4 4u +5
4 4 4u +5 lim
u 2 4u +5 =lim 2 4 u (u 2)(2u 1) u =lim 2 2u 5u +2 u u =
lim 5 2 u =
x+2 20. lim u u u u = 2 + lim 2 5lim 4 ( x+2 ) /x u u 2 2 2 + u 4 1
u 1
u 1+2/x =lim 2 x 9x +1 2 u 4 lim 4+5lim 4 5 2 4 4 1 +2lim
u u = 4+5(0)
2 5(0)+2(0) 4 4
=2
2 =lim 2 x 5 4+ u 2u 5u +2 5
u =lim 2 u
lim 4 4 u 4+ 2 9x +1 / x =
2 x 9+1/x 1+0
1
=
9+0 3 21.
6 lim
x 9x x
3 x +1 6 =lim
x 9x x /x
3 (x +1)/x = 3 6 lim 3 (9x x)/x 6 x 3 6 [ since x = x for x>0 ] 3 lim (1+1/x )
x lim 9 1/x 5 5 lim 9 lim (1/x )
x x = = 3 1+0 lim 1+lim (1/x )
x x x = 9 0 =3
22.
6 lim
x 9x x
3 x +1 6 = lim
x 9x x /x
3 (x +1)/x 3 lim 3 = 6 (9x x)/x x 6
3 3 [ since x = 6 x for x<0 ] lim (1+1/x )
x 7 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes lim 9 1/x 5 5 lim 9 lim (1/x )
x x = 3 x = 1+0 lim 1+ lim (1/x )
x x = 9 0= 3 23.
lim ( 2 9x +x 3x ) =lim x ( (
9x +x +3x )
=lim 3x ) ( 2 9x +x 2 x =lim 2 9x 2 x x 9x +x +3x
x/x =lim 2 2 x x =lim 2 2 ) 9x /x +x/x +3x/x 2 ( 3x ) 2 2 x 9x +x +3x ( 9x2+x) 9x +x )
2 2 9x +x +3x
1/x
1/x 2 9x +x +3x
1
1
1
1
=lim
=
=
=
9+1/x +3
9 +3 3+3 6
x 24.
lim ( x+ 2 x +2x x ) = lim ( x+ x +2x x x
2x = lim
x x = lim 2 x x +2x 2 x x +2x
2 2 = lim x ( x2+2x)
2 x
x +2x
x x +2x
2
2
=
= 1
1+ 1+2/x 1+ 1+2 ( 0 ) 2 x . Note: In dividing numerator and denominator by x , we used the fact that for x<0 , x=
25.
lim
x ( 2 x +ax 2 x +bx ) =lim ( 2 2 x +ax x +bx
2 x x 2 2 x +ax + x +bx ) 2 x +ax + x +bx
2 =lim )( 2 (x +ax) (x +bx)
2 2 x +ax + x +bx =lim
x [(a b)x]/x ( 2 2 ) 2 x +ax + x +bx / x 8 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes a b
a b
a b
=
=
2
1+a/x + 1+b/x
1+0 + 1+0 =lim
x 26. lim cos x does not exist because as x increases cos x does not approach any one value, but
x oscillates between 1 and 1 .
27. x is large when x is large, so lim x= . x
3 28. x is large negative when x is large negative, so lim 3 . x= x 29. lim (x x ) =lim x x ( x 1) = since x 3 30. lim x 2x+3
2 x 3 x . =lim 2 x 5 2x 2 (x 2x+3)/x [divide by the highest power of x in the denominator] 2 (5 2x )/x 2 =lim as x and x 1 x x 2/x+3/x
2 = 2 2 because x 2/x+3/x and 5/x 2 2 as x . 5/x 2
4 5 5 31. lim (x +x )= lim x (
x x 32. lim tan 1 because x 5 and 1/x+1 1 as x . ( x2 x4) =lim tan 1 ( x2 ( 1 x2) ) . If we let t=x2 ( 1 x2) , we know that t x x 1
+1)=
x as x
2 2 , since x and 1 x . So lim tan 1 ( x2 ( 1 x2) ) = lim x t 1 tan t= 2 . 33.
3 lim
x 5 2 4 x+x +x
1 x +x 3 =lim
x 5 4 2 4 4 (x+x +x )/x
(1 x +x )/x [ divide by the highest power of x in the denominator ] 3 =lim
x
3 because (1/x +1/x+x) 1/x +1/x+x
4 2 = 1/x 1/x +1
4 2 and (1/x 1/x +1) 34. If we let t=tan x , then as x + ( /2) , t 1 as x . . Thus,
9 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes tan x lim e
+ x t = lim e =0 .
t ( /2) 35. (a)
2 From the graph of f (x)= x +x+1 +x , we estimate the value of lim f (x) to be 0.5 .
x (b)
x f (x) 10,000 0.4999625 100,000 0.4999962 1,000,000 0.4999996 From the table, we estimate the limit to be 0.5 .
(c)
lim ( x 2 ) x +x+1 +x = lim ( 2 x +x+1 +x 2 ) x +x+1 x
2 x x +x+1 x = lim
x = (x+1)(1/x) ( 2 ) = lim x +x+1 x (1/x)
1+0
1
=
2
1+0+0 1 x = lim ( x2+x+1) 2 x 2 x x +x+1 x 1+(1/x) ( 2) 1+(1/x)+ 1/x 1 2 Note that for x<0 , we have x = x = x , so when we divide the radical by x , with x<0 , we get
1
2
2
2
1
x +x+1 =
x +x+1 = 1+(1/x)+ 1/x .
x
2
x ( ) 36. (a)
10 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 2 2 From the graph of f (x)= 3x +8x+6
lim f (x) to be 1.4 . 3x +3x+1 , we estimate (to one decimal place) the value of x (b)
x f (x) 10,000 1.44339 100,000 1.44338 1,000,000 1.44338 From the table, we estimate (to four decimal places) the limit to be 1.4434 .
(c)
lim f (x) =lim
x ( 2 3x +8x+6 x 37. lim
x lim
x + 4 2 2 2 3x +8x+6 + 3x +3x+1
(5x+5)(1/x) ( 2 ) (1/x) 2 3x +8x+6 + 3x +3x+1
5+5/x
2 =
2 3+8/x+6/x + 3+3/x+1/x x
= lim
x+4 x x
=
x+4 ) ( 3x2+8x+6) ( 3x2+3x+1) =lim
x 2 3x +8x+6 + 3x +3x+1 x =lim 2 3x +8x+6 + 3x +3x+1 2 x =lim )( 2 3x +3x+1 5 3
5
5
=
=
6
3+ 3 2 3 1.443376 1
1
x
=
=1 , so y=1 is a horizontal asymptote. lim
=
1+4/x 1+0
x+4
x and 4 , so x= 4 is a vertical asymptote. The graph confirms these calculations. 11 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 2 38. Since x 1 0 as x 2 x +4 lim
x 2 1 1 and y<0 for 1<x<1 and y>0 for x< 1 and x>1 , we have
2 x +4 , lim = x 1 + x 1 2
2 x +4 , lim = x 1 x
2 vertical asymptotes. Also lim 2 39. lim x +3x 10
x lim
x 2 3 x +3x 10
x , lim 2 x 1+(3/x)
3 2 = lim 1 1/x ( 10/x2) 2 + = , so x=1 and x= 1 are x 1 1+0
=1 , so y=1 is a horizontal asymptote.
1 0 , so there is no horizontal asymptote. x x , and lim = 1 2 x
, since
>0 for x>2 . Similarly, lim
(x+5)(x 2) 3 = 2 x = + 3 x
=lim
=
+ (x+5)(x 2)
x 1+4/x x = lim 2 x + 3 x
2 x 1
The graph confirms these calculations. x , and lim = x 1 1 x +4 x 2 2 x +4 2 2 x
2 3 = x +3x 10 3 = , so x=2 and x= 5 are vertical asymptotes. The x +3x 10
x
5 x +3x 10
graph confirms these calculations.
x 5 40.
12 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 3 x +1 lim 3 x 1+1/x = lim 2 x x +x 3 3 =1 , so y=1 is a horizontal asymptote. Since y=
3 x +1 x>0 and y<0 for 1<x<0 , lim 41. lim
x x
4 1/x 4 4 + x +1 1/ x 0 3 x = x =
1 1+ x
lim
x x 1/x 4 4 4 x +1 1/ x 4 x 4 = 2
x ( x +1 ) >0 for =
1 1+ 3 , so x=0 is a vertical asymptote. x +x 0 1
4 = =1 and 1+0 4 1 = lim x +1 and lim 1
4 3 x +1 3 x +x =lim
4 3 x +x 1+1/x x x +1 x 1
4 = 1 , so y= 1 are horizontal asymptotes. 1+0 4 There is no vertical asymptote. 42. lim
x x 9
2 =lim
x 4x +3x+2 Using the fact that 1 9/x ( 2) = 4+ ( 3/x ) + 2/x 1 0
1
= .
4+0+0 2 2 x = x = x for x<0 , we divide the numerator by x and the denominator by 2 x .
Thus, lim
x x 9
2 = lim
x 1+9/x ( 2) = 1+0
1
=
.
2
4+0+0 4+ ( 3/x ) + 2/x
1
2
The horizontal asymptotes are y=
. The polynomial 4x +3x+2 is positive for all x , so the
2
denominator never approaches zero, and thus there is no vertical asymptote.
4x +3x+2 13 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 43. Let’s look for a rational function.
(1)
lim f (x)=0
degree of numerator < degree of denominator
x (2) lim f (x)=
x 2 there is a factor of x in the denominator (not just x , since that would 0 produce a sign change at x=0 ), and the function is negative near x=0 .
(3) lim f (x)= and lim f (x)=
vertical asymptote at x=3 ; there is a factor of ( x 3) in the
x 3 x + 3 denominator.
(4) f (2)=0
2 is an x intercept; there is at least one factor of ( x 2 ) in the numerator.
Combining all of this information and putting in a negative sign to give us the desired left and right
2 x
hand limits gives us f (x)= 2
as one possibility.
x (x 3)
44. Since the function has vertical asymptotes x=1 and x=3 , the denominator of the rational function
we are looking for must have factors ( x 1 ) and ( x 3) . Because the horizontal asymptote is y=1 , the
degree of the numerator must equal the degree of the denominator, and the ratio of the leading
2 x
coefficients must be 1 . One possibility is f (x)=
.
(x 1)(x 3)
2 45. y= f (x)=x (x 2)(1 x) . The y intercept is f (0)=0 , and the x intercepts occur when y=0 x=0 , 1 2 , and 2 . Notice that, since x is always positive, the graph does not cross the x axis at 0 , but does
2 cross the x axis at 1 and 2 . lim x (x 2)(1 x)= , since the first two factors are large positive and x the third large negative when x is large positive.
14 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 2 lim x (x 2)(1 x)= because the first and third factors are large positive and the second large x negative as x . 3 46. y=(2+x) (1 x)(3 x) . As x
, the first factor is large positive, and the second and third factors
, the first factor is large negative, and the
are large negative. Therefore, lim f (x)= . As x
x second and third factors are large positive. Therefore, lim f (x)= . Now the y intercept is x
3 f (0)=(2) (1)(3)=24 and the x intercepts are the solutions to f (x)=0
crosses the x axis at all of these points. 5 4 5 x= 2 , 1 and 3 , and the graph 4 47. y= f (x)=(x+4) (x 3) . The y intercept is f (0)=4 ( 3) =82 , 944 . The x intercepts occur when
y=0 4 x= 4 , 3 . Notice that the graph does not cross the x axis at 3 because (x 3) is always
5 4 positive, but does cross the x axis at 4 . lim (x+4) (x 3) =
5 x
4 when x is large positive. lim (x+4) (x 3) = since both factors are large positive since the first factor is large negative and the second x factor is large positive when x is large negative. 15 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 2 2 48. y=(1 x)(x 3) (x 5) . As x
, the first factor approaches
while the second and third factors
. Therefore, lim (x)=
. As x
, the factors all approach
. Therefore,
approach
x lim (x)= 2 2 . Now the y intercept is f (0)=(1)( 3) ( 5) =225 and the x intercepts are the solutions x to f (x)=0
2 x=1 , 3 , and 5 . Notice that f (x) does not change sign at x=3 or x=5 because the factors
2 (x 3) and (x 5) are always positive, so the graph does not cross the x axis at x=3 or x=5 , but does
cross the x axis at x=1 . 1
for x>0 . As x
, 1/x 0 and 1/x 0 , so by
x
sin x
the Squeeze Theorem, (sin x)/x 0 . Thus, lim
=0 .
x
x
(b) From part (a), the horizontal asymptote is y=0 . The function y=(sin x)/x crosses the horizontal
asymptote whenever sin x=0 ; that is, at x= n for every integer n . Thus, the graph crosses the
asymptote an infinite number of times.
49. (a) Since 1 sin x 1 for all x, 1
x sin x
x 50. (a) In both viewing rectangles, lim P(x)=lim Q(x)=
x and lim P(x)= lim Q(x)= x x . In the x larger viewing rectangle, P and Q become less distinguishable. 16 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 5 3 P(x)
3x 5x +2x
=lim
=lim
(b) lim
5
Q(x) x
x
x
3x
P and Q have the same end behavior. 1 5
3 1
2 + x 2
3 1
x 4 =1 5
2
(0)+ (0)=1
3
3 51. Divide the numerator and the denominator by the highest power of x in Q(x) .
(a) If deg P<deg Q , then the numerator 0 but the denominator doesn’t. So lim P(x)/Q(x) =0 .
x (b) If deg P>deg Q , then the numerator
but the denominator doesn’t, so
lim P(x)/Q(x) =
(depending on the ratio of the leading coefficients of P and Q ).
x 52. (i) n=0 (ii) n>0 (n odd)
n (a) lim x =
x + 0 n (b) lim x =
x 0
n (c) lim x =
x
n { {
{ (d) lim x =
x { (iii) n>0 (n even) (iv) n<0 (n odd) (v) n<0 (n even) 1
0 if n=0
if n>0
if n<0
1 if n=0
0 if n>0
if n<0, n odd
if n<0, n even
1 if n=0
if n>0
0 if n<0
1 if n=0
if n>0, n odd
if n>0, n even
0 if n<0 53.
17 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 4x 1
lim
=lim
x
x
x 2 1
4
x 4x +3x =4 , and lim 2 x =lim
x x 4+ 3
x =4 . Therefore, by the Squeeze Theorem, lim f (x)=4 .
x 54. (a) After t minutes, 25t liters of brine with 30 g of salt per liter has been pumped into the tank, so
it contains ( 5000+25t ) liters of water and 25t 30=750t grams of salt. Therefore, the salt
750t
30t g
=
.
concentration at time t will be C(t)=
5000+25t 200+t L
30t
30t/t
30
(b) lim C(t)=lim
=lim
=
=30 . So the salt concentration approaches that of the
200+t t
200/t+t/t 0+1
t
t
brine being pumped into the tank.
55. (a) lim v(t)=lim v
t * ( 1 e ) =v (1 0)=v
gt/v * * * t
9.8t * (b) We graph v(t)=1 e
and v(t)=0.99v , or in this case, v(t)=0.99 . Using an intersect feature or
zooming in on the point of intersection, we find that t 0.47 s. 56. (a) y=e x/10 If x>x , then e
1 (b) e and y=0.1 intersect at x 1 23.03 . x/10 <0.1 . x/10 <0.1
x/10<ln 0.1
1
1
x> 10ln
= 10ln 10 =10ln 10
10 18 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 2 57. 6x +5x 3
2 2x 1 2 3 <0.2 2.8< 6x +5x 3
2 <3.2 . So we graph the three parts of this inequality on 2x 1
2 the same screen, and find that the curve y= 6x +5x 3
2 seems to lie between the lines y=2.8 and y=3.2 2x 1
whenever x>12.8 . So we can choose N=13 (or any larger number) so that the inequality holds
whenever x N . 2 58. For =0.5 , we must find N such that whenever x N , we have 4x +1
2
x+1 <0.5 2 4x +1
1.5<
<2.5 . We graph the three parts of this inequality on the same screen, and find that it
x+1
holds whenever x 3 . So we choose N=3 (or any larger number). For =0.1 , we must have
2 4x +1
1.9<
<2.1 , and the graphs show that this holds whenever x 19 . So we choose N=19 (or
x+1
any larger number). 19 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 2 2 4x +1
4x +1
59. For =0.5 , we need to find N such that
( 2) <0.5 2.5<
< 1.5
x+1
x+1
whenever x N . We graph the three parts of this inequality on the same screen, and see that the
inequality holds for x 6. So we choose N= 6 (or any smaller number).
2 4x +1
< 1.9 whenever x N. From the graph, it seems that this
For =0.1 , we need 2.1<
x+1
inequality holds for x 22 . So we choose N= 22 (or any smaller number). 2x+1
>100 whenever x N. From the graph, we see that this inequality
x+1
holds for x 2500 . So we choose N=2500 (or any larger number).
60. We need N such that 2 61. (a) 1/x <0.0001 2 x >1/0.0001=10 , 000 x>100 ( x>0 )
20 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes 2 2 x >1/
(b) If >0 is given, then 1/x <
. Let N=1/
x>1/
1
1
1
1
Then x>N x>
0 = < , so lim
=0 .
2
2
2
x
x
x
x
4 62. (a) 1/ x <0.0001 x >1/0.0001=10 . 8 x>10 2 (b) If >0 is given, then 1/ x <
x >1/
x>1/ . Let N=1/
1
1
1
1
Then x>N x>
0 =
< , so lim
=0 .
2
x
x
x
x 2 . 63. For x<0 , 1/x 0 = 1/x . If >0 is given, then 1/x<
x< 1/ .
Take N= 1/ . Then x<N x< 1/
( 1/x ) 0 = 1/x< , so lim ( 1/x ) =0 .
x 64. Given M>0 , we need N>0 such that x>N
3 3 x>N= M 3 x >M , so lim x = 3 3 x x x >M . Now x >M 3 3 x> M , so take N= M . Then . x 65. Given M>0 , we need N>0 such that x>N e >M . Now e >M x>ln M , so take N=max ( 1,ln M ) . (This ensures that N>0 .) Then x>N=max ( 1,ln M )
x lim e = x e >max ( e,M ) M , so . x 66. Definition Let f be a function defined on some interval ( ,a ) . Then lim f (x)= means x that for every negative number M there is a corresponding negative number N such that f (x)<M ( 1+x3) = whenever x<N . Now we use the definition to prove that lim . Given a negative x number M , we need a negative number N such that x<N
3 3 x< M 1 . Thus, we take N= M 1 and find that x<N
lim ( 1+x3) = 3 3 1+x <M . Now 1+x <M 3 x <M 1 3 1+x <M . This proves that . x 67. Suppose that lim f (x)=L . Then for every >0 there is a corresponding positive number N such
x that f (x) L < whenever x>N . If t=1/x , then x>N 0<1/x<1/N 0<t<1/N . Thus, for every >0
there is a corresponding >0 (namely 1/N ) such that f (1/t) L < whenever 0<t< . This proves
that
21 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.6 Limits at Infinity; Horizontal Asymptotes lim f (1/t)=L=lim f (x) .
t + x 0 Now suppose that lim f (x)=L . Then for every >0 there is a corresponding negative number N
x such that f (x) L < whenever x<N . If t=1/x , then x<N 1/N<1/x<0 1/N<t<0 . Thus, for every
>0 there is a corresponding >0 (namely 1/N ) such that f (1/t) L < whenever <t<0 . This
proves that lim f (1/t)=L= lim f (x) .
t 0 x 22 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change 1. (a) This is just the slope of the line through two points: m = PQ y f (x) f (3)
=
.
x
x 3 (b) This is the limit of the slope of the secant line PQ as Q approaches P : m=lim
x 3 f (x) f (3)
.
x 3 s f (a+h) f (a) f (a+h) f (a)
=
=
t
(a+h) a
h
f (a+h) f (a)
(b) Instantaneous velocity =lim
h
h 0
2. (a) Average velocity = 3. The slope at D is the largest positive slope, followed by the positive slope at E . The slope at C is
zero. The slope at B is steeper than at A (both are negative). In decreasing order, we have the slopes
at: D , E , C , A , and B .
4. The curve looks more like a line as the viewing rectangle gets smaller. 5. (a)
(i) Using Definition 1, ( 2 ) f (x) f (a)
f (x) f ( 3)
x +2x (3)
(x+3)(x 1)
m =lim
lim
=lim
=lim
x a
x ( 3)
x ( 3)
x+3
x a
x
3
x
3
x
3
=lim (x 1)= 4
x 3
1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change (ii) Using Equation 2,
2 f (a+h) f (a)
f ( 3+h) f ( 3)
( 3+h) +2( 3+h) (3)
m =lim
=lim
=lim
h
h
h
h 0
h 0
h 0
2 9 6h+h 6+2h 3
h(h 4)
=lim
=lim
=lim (h 4)= 4
h
h
h 0
h 0
h 0
(b) Using the point slope form of the equation of a line, an equation of the tangent line is
y 3= 4(x+3) . Solving for y gives us y= 4x 9 , which is the slope intercept form of the equation of
the tangent line. (c)
6. (a)
(i) 2 x ) ( x2 x+1) =3 =lim
(ii) ( 3 f (x) f ( 1)
x ( 1)
( x+1 ) x x+1
m =lim
=lim
=lim
x ( 1)
x+1
x
1
x
1 x+1
x
1
1
3 3 2 f ( 1+h) f ( 1)
( 1+h) ( 1)
h 3h +3h 1+1
m =lim
=lim
=lim
h
h
h
h 0
h 0
h 0
=lim
h ( h2 3h+3) =3 0 (b) y ( 1)=3 x ( 1) y+1=3x+3 y=3x+2 (c) 3 7. Using (2) with f (x)=1+2x x and P(1,2) ,
2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change 3 f (a+h) f (a)
f (1+h) f (1)
1+2(1+h) (1+h)
m =lim
=lim
=lim
h
h
h
h 0
h 0
h 0
2 3 3 2 2 1+2+2h (1+3h+3h +h ) 2
h 3h h
=lim
=lim
h
h
h 0
h 0
2 h( h 3h 1)
2
=lim
=lim ( h 3h 1)= 1
h
h 0
h 0
Tangent line: y 2= 1(x 1) y 2= x+1 y= x+3 8. Using (1),
2x+1 m=lim
x 4 2(4)+1
=lim
x 4
x 4 2x+1 3
x 4 2x+1 +3
2x+1 +3
2 (2x+1) 3
2(x 4)
=lim
=lim
x 4 (x 4)( 2x+1 +3) x 4 (x 4)( 2x+1 +3)
=lim
x Tangent line: y 3= 1
(x 4)
3 4 2
2
1
=
= .
( 2x+1 +3) 3+3 3 y 3= 1
4
x
3
3 y= 1
5
x+
3
3 x 1
and P ( 3,2 ) ,
x 2
x 1
x 1 2(x 2)
2
f (x) f (a)
x 2
x 2
3 x
m
=lim
=lim
=lim
=lim
x a
x 3
x 3
x a
x 3
x 3
x 3 (x 2)(x 3) 9. Using (1) with f (x)= =lim
x 3 1
1
=
= 1.
x 2 1 Tangent line: y 2= 1(x 3) y 2= x+3
2x
0
2
(x+1)
10. Using (1), m=lim
=lim
x 0
x 0
x 0
Tangent line: y 0=2(x 0) y= x+5
2x
2 x(x+1) =lim
x 0 2
2 (x+1) = 2
2 =2 . 1 y=2x
3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change 11. (a)
f (x) f (a)
2/(x+3) 2/(a+3)
2(a+3) 2(x+3)
=lim
=lim
x a
x a
x a
x a (x a)(x+3)(a+3) m =lim
x a 2(a x)
2
=lim
=
(x a)(x+3)(a+3) x a (x+3)(a+3) =lim
x (b)
(i) a a= 1 (ii)
(iii) a=0
a=1 m= 2
2 ( 1+3)
2
m=
=
2
(0+3)
2
m=
=
2
(1+3) = 2
2 (a+3) 1
2 2
9
1
8 12. (a) Using (1),
m = lim
x x a a = lim
x ( 1+x+x2) ( 1+a+a2) =lim a x a 2 2 x+x a a
x a+(x a)(x+a)
=lim
x a
x a
x a (x a)(1+x+a)
=lim ( 1+x+a ) =1+2a
x a
x a (b)
(i) x= 1 m=1+2( 1)= 1
1
1
(ii) x=
m=1+2
=0
2
2
(iii) x=1 m=1+2(1)=3 (c)
13. (a) Using (1),
4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change m = lim
x ( x3 4x+1) ( a3 4a+1) =lim ( x3 a3)
x a a ( 2 x
2 (x a) x +ax+a
= lim
x a
x a ) 4(x a) 4(x a)
x a a =lim
x ( x2+ax+a2 4) =3a2 4 a 2 (b) At ( 1, 2 ) : m=3(1) 4= 1 , so an equation of the tangent line is y ( 2)= 1(x 1) y= x 1 . At 2 ( 2,1 ) : m=3(2) 4=8 , so an equation of the tangent line is y 1=8(x 2) y=8x 15 . (c)
14. (a) Using (1),
a x
1
1
x
a
ax
( a x ) ( a+ x )
m=lim
=lim
=lim
x a
x a
x a
x a
x a ax ( x a ) ( a + x )
a x
1
1
1
=lim
=lim
=
=
or
3/2
2
x a ax ( x a ) ( a + x ) x a ax ( a + x )
2a
a (2 a ) 1
a
2 3/2 1
1
1
3
, so an equation of the tangent line is y 1=
(x 1) y=
x+ .
2
2
2
2
1
3
1
1
1
: m=
, so an equation of the tangent line is y
=
(x 4) y=
x+ .
16
2
16
16
4 (b) At ( 1,1 ) : m=
At 4, 1
2 (c)
15. (a) Since the slope of the tangent at t=0 is 0 , the car’s initial velocity was 0 .
(b) The slope of the tangent is greater at C than at B , so the car was going faster at C .
(c) Near A , the tangent lines are becoming steeper as x increases, so the velocity was increasing, so
the car was speeding up. Near B , the tangent lines are becoming less steep, so the car was slowing
down. The steepest tangent near C is the one at C , so at C the car had just finished speeding up, and
was about to start slowing down.
5 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change (d) Between D and E , the slope of the tangent is 0 , so the car did not move during that time.
16. Let a denote the distance traveled from 1:00 to 1:02 ,b from 1:28 to 1:30, and c from 3:30 to 3:33,
where all the times are relative to t=0 at the beginning of the trip. 2 17. Let s(t)=40t 16t .
s(t) s(2)
v(2) = lim
=lim
t 2
t 2
t 2
= lim
t 2 ( 40t 16t2) 16 t 2 2 =lim
t ( 2 16t +40t 16
8 2t 5t+2
=lim
t 2
t 2
t 2 2 ) 8(t 2)(2t 1)
= 8lim (2t 1)= 8(3)= 24
t 2
t 2 Thus, the instantaneous velocity when t=2 is 24 ft / s.
18. (a)
v(1) =lim
h 0 =lim
h H(1+h) H(1)
h ( 58+58h 0.83 1.66h 0.83h2) 57.17 h 0 =lim (56.34 0.83h)=56.34 m / s
h 0 (b)
v(a) =lim
h 0 =lim
h 0 H(a+h) H(a)
h ( 58a+58h 0.83a2 1.66ah 0.83h2) ( 58a 0.83a2)
h =lim ( 58 1.66a 0.83h ) =58 1.66a m / s
h 0 (c) The arrow strikes the moon when the height is 0 , that is,
6 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change 58
69.9 s (since t can’t be 0 ).
0.83
58
58
(d) Using the time from part (c), v
=58 1.66
= 58 m / s. Thus, the arrow will
0.83
0.83
have a velocity of 58 m / s.
2 t(58 0.83t)=0 t= 58t 0.83t =0 19.
3 s(a+h) s(a)
4(a+h) +6(a+h)+2
v(a) =lim
=lim
h
h
h 0
h 0
3 2 2 3 ( 4a3+6a+2) 3 4a +12a h+12ah +4h +6a+6h+2 4a 6a 2
=lim
h
h 0
2 2 3 12a h+12ah +4h +6h
2
2
2
=lim
=lim 12a +12ah+4h +6 = 12a +6 m / s
h
h 0
h 0 ( 2 ) ( 2 ) 2 So v(1)=12(1) +6=18 m / s, v(2)=12(2) +6=54 m / s, and v(3)=12(3) +6=114 m / s.
20. (a) The average velocity between times t and t+h is ( s(t+h) s(t) (t+h)2 8(t+h)+18 t 2 8t+18
( t+h ) t =
h
2 2 2 )
2 t +2th+h 8t 8h+18 t +8t 18 2th+h 8h
=
=
h
h
= ( 2t+h 8 ) m/s (i)[3,4] : t=3 , h=4 3=1 , so the average
(ii)[3.5,4] : t=3.5 , h=0.5 , so the average velocity
velocity is 2(3)+1 8= 1 m / s.
is 2(3.5)+0.5 8= 0.5 m / s.
(iii)[4,5] : t=4 , h=1 , so the average velocity is (iv)[4,4.5] : t=4 , h=0.5 , so the average velocity is
2(4)+1 8=1 m / s.
2(4)+0.5 8=0.5 m / s. (b) v(t)=lim
h 0 s(t+h) s(t)
=lim ( 2t+h 8 ) =2t 8 , so v ( 4 ) =0 .
h
h 0 (c) 7 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change 21. The sketch shows the graph for a room temperature of 72 and a refrigerator temperature of 38 .
The initial rate of change is greater in magnitude than the rate of change after an hour. 22. The slope of the tangent (that is, the rate of change of temperature with respect to time) at t=1 h
75 168
seems to be about
0.7 F / min.
132 0 23. (a) (i) 20,23 : 7.9 11.5
= 1.2 C / h
23 20 (ii) 20,22 : 9.0 11.5
= 1.25 C / h
22 20 (iii) 20,21 : 10.2 11.5
= 1.3 C / h
21 20 (b) In the figure, we estimate A to be ( 18,15.5) and B as ( 23,6 ) . So the slope is 6 15.5
= 1.9 C / h
23 18 at 8:00 P.M. 8 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change 24. (a)
(i)
(ii)
(iii) P(1996)
1996
P(1996)
1994,1996 :
1996
P(1998)
1996,1998 :
1998
1992,1996 : P(1992) 10,152 10,036 116
=
=
=29 thousand / year
1992
4
4
P(1994) 10,152 10,109 43
=
=
=21.5 thousand / year
1994
2
2
P(1996) 10,175 10,152 23
=
=
=11.5 thousand / year
1996
2
2 21.5+11.5
=16.5 thousand / year.
2
(c) Estimating A as (1994,10 , 125) and B as ( 1998,10,182 ) , the slope at 1996 is
10,182 10,125 57
=
=14.25 thousand / year.
1998 1994
4
(b) Using the values from (ii) and (iii), we have 25. (a)
N(1997)
1997
N(1996)
(ii) 1995,1996 :
1996
N(1995)
(iii) 1994,1995 :
1995
(i) 1995,1997 : N(1995) 2461 873 1588
=
=
=794 thousand / year
1995
2
2
N(1995) 1513 873
=
=640 thousand / year
1995
1
N(1994) 873 572
=
=301 thousand / year
1994
1 (b) Using the values from (ii) and (iii), we have
9 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change 640+301 941
=
=470.5 thousand / year.
2
2
(c) A as ( 1994,420 ) and B as ( 1996,1275) , the slope at 1995 is 1275 420 855
=
=427.5 thousand
1996 1994
2 / year 26. (a)
N(1998)
1998
N(1998)
(ii) 1997,1998 :
1998
N(1999)
(iii) 1998,1999 :
1999
(i) 1996,1998 : N(1996) 1886 1015 871
=
=
=435.5 locations / year
1996
2
2
N(1997) 1886 1412
=
=474 locations / year
1997
1
N(1998) 2135 1886
=
=249 locations / year
1998
1 474+249 723
=
=361.5 362 locations / year.
2
2
(c) Estimating A as ( 1997,1525) and B as ( 1999,2250 ) , the slope at 1998 is
2250 1525 725
=
=362.5 locations / year.
1999 1997
2
(b) Using the values from (ii) and (iii), we have 27. (a)
10 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.7 Tangents, Velocities, and Other Rates of Change (i)
(ii) C C(105) C(100) 6601.25 6500
=
=
=$20.25/ unit.
x
105 100
5
C C(101) C(100) 6520.05 6500
=
=
=$20.05/ unit.
x
101 100
1 (b)
2
2
C(100+h) C(100)
5000+10(100+h)+0.05(100+h) 6500 20h+0.05h
=
=
h
h
h
=20+0.05h , h 0
C(100+h) C(100)
So the instantaneous rate of change is lim
=lim (20+0.05h)=$20/ unit.
h
h 0
h 0 28.
V =V (t+h) V (t)=100 , 000 1 t+h
60
2 2 100 , 000 1 V by h and then letting h 2 2 t+h ( t+h )
t
t
=100 , 000
1
+
1
+
30
3600
30 3600
100,000
250
=
h ( 120+2t+h ) =
h ( 120+2t+h )
3600
9
Dividing t
60 2 =100 , 000 h
2th
h
+
+
30 3600 3600 0 , we see that the instantaneous rate of change is 500
( t 60 )
9 gal / min.
t
0 Flow rate (gal/min) Water remaining V (t) (gal)
3333.3 100,000 10 2777.7 69,444.4 20 2222.2 44,444.4 30 1666.6 25,000 40 1111.1 11,111.1 50 555.5 2,777.7 60 0
0
The magnitude of the flow rate is greatest at the beginning and gradually decreases to 0. 11 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives 1. The line from P(2, f (2)) to Q(2+h, f (2+h)) is the line that has slope f (2+h) f (2)
h 2. As h decreases, the line PQ becomes steeper, so its slope increases. So
f (4) f (2) f (3) f (2)
f (x) f (2)
1
/
. Thus, 0< [ f (4) f (2)]< f (3) f (2)< f (2) .
0<
<
<lim
4 2
3 2
x 2
2
x 2
/ 3. g (0) is the only negative value. The slope at x=4 is smaller than the slope at x=2 and both are
/ / / / smaller than the slope at x= 2 . Thus, g (0)<0<g (4)<g (2)<g ( 2) .
4. Since (4,3) is on y= f (x) , f (4)=3 . The slope of the tangent line between (0,2) and (4,3) is
/ f (4)= 1
, so
4 1
.
4 5.
We begin by drawing a curve through the origin at a slope of
/ / 3 to satisfy f (0)=0 and f (0)=3 . Since f (1)=0 , we will
round off our figure so that there is a horizontal tangent
directly over x=1 . Lastly, we make sure that the curve has a
slope of 1 as we pass over x=2 . Two of the many
possibilities are shown.
6. 7. Using Definition 2 with
1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives 2 f (x)=3x 5x and the point (2,2) , we have
2 f (2+h) f (2)
[3(2+h) 5(2+h)] 2
=lim
f (2) =lim
h
h
h 0
h 0
/ 2 2 (12+12h+3h 10 5h) 2
3h +7h
=lim
=lim
=lim (3h+7)=7 .
h
h
h 0
h 0
h 0
So an equation of the tangent line at (2,2) is y 2=7(x 2) or y=7x 12 .
3 8. Using Definition 2 with g(x)=1 x and the point (0,1) , we have
3 3 g(0+h) g(0)
[1 (0+h) ] 1
(1 h ) 1
2
g (0)=lim
=lim
=lim
=lim ( h )=0 .
h
h
h
h 0
h 0
h 0
h 0
So an equation of the tangent line is y 1=0(x 0) or y=1 .
/ 3 9. (a) Using Definition 2 with F(x)=x 5x+1 and the point (1, 3) , we have
3 F(1+h) F(1)
[(1+h) 5(1+h)+1] ( 3)
=lim
F (1) =lim
h
h
h 0
h 0
/ 2 3 3 2 (1+3h+3h +h 5 5h+1)+3
h +3h 2h
=lim
=lim
h
h
h 0
h 0
2 h(h +3h 2)
2
=lim
=lim (h +3h 2)= 2
h
h 0
h 0
So an equation of the tangent line at (1, 3) is y ( 3)= 2(x 1) y= 2x 1 . (b)
10. (a)
a+h
/
G (a) =lim G(a+h) G(a) =lim 1+2(a+h)
h
h
h 0
h 0 a
1+2a
2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives 2 2 a+2a +h+2ah a 2a 2ah
1
=lim
=lim
=(1+2a)
h(1+2a+2h)(1+2a)
h 0
h 0 (1+2a+2h)(1+2a)
1 1
1
,
) is m= 1+2(
)
4 2
4 So the slope of the tangent at the point (
y+ 2 2 =4 , and thus an equation is 1
1
1
=4(x+ ) or y=4x+ .
2
4
2 (b)
1+h 1 f (1+h) f (1)
3
3
11. (a) f (1)=lim
.
=lim
h
h
h 0
h 0
/ 1+h So let F(h)= 3 3 . We calculate: h h F(h) h F(h) 0.1 3.484 0.1 3.121 0.01 3.314 0.01 3.278 0.001 3.298 0.001 3.294 0.0001 3.296 0.0001 3.296 / We estimate that f (1) 3.296 . (b)
From the graph, we estimate that the slope of the tangent is about 3.2 2.8
0.4
=
1.06 0.94 0.12 3.3 . 12. (a)
3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives g( / g ( 4 ) =lim
h 0 tan(
So let G(h)= 4 +h) g( 4 h ) tan(
=lim
h 4 +h) tan( 4 ) h 0 . +h) 1
4
. We calculate:
h
h G(h) h G(h) 0.1 2.2305 0.1 1.8237 0.01 2.0203 0.01 1.9803 0.001 2.0020 0.001 1.9980 0.0001 2.0002 0.0001 1.9998 (b)
From the graph, we estimate that the slope of the tangent is about 1.07 0.91 0.16
=
=2 .
0.82 0.74 0.08 2 13. Use Definition 2 with f (x)=3 2x+4x .
2 2 f (a+h) f (a)
[3 2(a+h)+4(a+h) ] (3 2a+4a )
=lim
f (a) =lim
h
h
h 0
h 0
/ 2 2 2 2 [3 2a 2h+4a +8ah+4h ] (3 2a+4a )
2h+8ah+4h
=lim
=lim
h
h
h 0
h 0
=lim
h 0 h( 2+8a+4h)
=lim ( 2+8a+4h)= 2+8a
h
h 0 14.
4 4 f (a+h) f (a)
[(a+h) 5(a+h)] (a 5a)
=lim
f (a) =lim
h
h
h 0
h 0
/ 4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives 4 3 2 2 3 4 4 (a +4a h+6a h +4ah +h 5a 5h) (a 5a)
=lim
h
h 0
3 2 2 3 4 3 2 2 3 4a h+6a h +4ah +h 5h
h(4a +6a h+4ah +h 5)
=lim
=lim
h
h
h 0
h 0
3 2 2 3 3 =lim (4a +6a h+4ah +h 5)=4a 5
h 0 15. Use Definition 2 with f (t)=(2t+1)/(t+3) .
2(a+h)+1 2a+1
/
a+3
f (a) =lim f (a+h) f (a) =lim (a+h)+3
h
h
h 0
h 0
(2a+2h+1)(a+3) (2a+1)(a+h+3)
h(a+h+3)(a+3)
0 =lim
h 2 2 (2a +6a+2ah+6h+a+3) (2a +2ah+6a+a+h+3)
=lim
h(a+h+3)(a+3)
h 0
=lim
h 0 5h
5
=lim
=
h(a+h+3)(a+3) h 0 (a+h+3)(a+3) 5
2 (a+3) 16.
2 2 (a+h) +1
/
f (a+h) f (a)
(a+h) 2
f (a)
=lim
=lim
h
h
h 0
h 0
2 2 a +1
a 2 2 (a +2ah+h +1))(a 2) (a +1)(a+h 2)
=lim
h(a+h 2)(a 2)
h 0
3 2 2 2 2 3 2 2 (a 2a +2a h 4ah+ah 2h +a 2) (a +a h 2a +a+h 2)
=lim
h(a+h 2)(a 2)
h 0
2 2 2 2 a h 4ah+ah 2h h
h(a 4a+ah 2h 1)
=lim
=lim
h(a+h 2)(a 2)
h 0
h 0 h(a+h 2)(a 2)
2 2 a 4a+ah 2h 1 a 4a 1
=lim
=
2
h 0 (a+h 2)(a 2)
(a 2)
5 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives 17. Use Definition 2 with f (x)=1/ x+2 .
1
/
f (a) =lim f (a+h) f (a) =lim (a+h)+2
h
h
h 0
h 0
a+2
h 0 =lim
h 0 =lim
h 0 a+2 a+h+2 a+2 + a+h+2 h a+h+2 a+2 a+2 + a+h+2 (a+2) (a+h+2)
h a+h+2 a+2 ( a+2 + a+h+2 )
h
h a+h+2 a+2 ( a+2 + a+h+2 )
1
a+h+2 a+2 ( a+2 + a+h+2 ) =lim
h a+h+2 a+h+2 a+2
=lim
h
h 0 =lim 1
a+2 0 1 = 1 = 2 ( a+2 ) (2 a+2 ) 3/2 2(a+2) 18.
/
f (a) =lim
h 0 =lim
h 3(a+h)+1
f (a+h) f (a)
=lim
h
h
h 0
( 3a+3h+1 h 0 (3a+3h+1) (3a+1)
3h
=lim
h( 3a+3h+1 + 3a+1 ) h 0 h( 3a+3h+1 + 3a+1 )
3
3
=
3a+3h+1 + 3a+1 2 3a+1 =lim
h 3a+1 )( 3a+3h+1 + 3a+1 ) h( 3a+3h+1 + 3a+1 ) 0 =lim 0 10 (1+h)
19. By Definition 2, lim
h
h 0
10 (1+h)
lim
h
h 0 3a+1 1 / 1 / 10 = f (1) , where f (x)=x
10 = f (0) , where f (x)=(1+x) and a=1 . Or: By Definition 2, and a=0 . 20. By Definition 2,
6 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives 4 lim
h 0 16+h 2
/
4
= f (16) , where f (x)= x and a=16 . Or : By Definition 2, lim
h
h 0 4 16+h 2
/
= f (0) ,
h 4 where f (x)= 16+x and a=0 .
x 2 32
/
x
= f (5) , where f (x)=2 and a=5 .
21. By Equation 3, lim
x 5 x 5
tan x 1
/
= f ( /4) , where f (x)=tan x and a= /4 .
x /4 22. By Equation 3, lim
x /4 23. By Definition 2, lim
h lim
h 0 0 cos ( +h)+1
/
= f ( ) , where f (x)=cos x and a= . Or : By Definition 2,
h cos ( +h)+1
/
= f (0) , where f (x)=cos ( +x) and a=0 .
h
4 t +t 2
/
4
24. By Equation 3, lim
= f (1) , where f (t)=t +t and a=1 .
t 1 t 1
25.
2 2 f (2+h) f (2)
[(2+h) 6(2+h) 5] [2 6(2) 5]
v(2) = f (2)=lim
=lim
h
h
h 0
h 0
/ 2 2 (4+4h+h 12 6h 5) ( 13)
h 2h
=lim
=lim
=lim (h 2)= 2m/s
h
h
h 0
h 0
h 0
26.
/ v(2) = f (2)=lim
h 0 f (2+h) f (2)
h
3 3 [2(2+h) (2+h)+1] [2(2) 2+1]
=lim
h
h 0
3 2 (2h +12h +24h+16 2 h+1) 15
=lim
h
h 0 7 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives 3 2 2h +12h +23h
2
=lim
=lim (2h +12h+23)=23m/s
h
h 0
h 0
/ 27. (a) f (x) is the rate of change of the production cost with respect to the number of ounces of gold
produced. Its units are dollars per ounce.
(b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing
is $17/ ounce. So the cost of producing the 800 th (or 801 st) ounce is about $17 .
/ (c) In the short term, the values of f (x) will decrease because more efficient use is made of start up
/ costs as x increases. But eventually f (x) might increase due to large scale operations.
/ 28. (a) f (5) is the rate of growth of the bacteria population when t=5 hours. Its units are bacteria
per hour.
/ / / (b) With unlimited space and nutrients, f should increase as t increases; so f (5)< f (10) . If the
supply of nutrients is limited, the growth rate slows down at some point in time, and the opposite may
be true.
/ 29. (a) f (v) is the rate at which the fuel consumption is changing with respect to the speed. Its units
are (gal / h) / (mi / h) .
(b) The fuel consumption is decreasing by 0.05( gal / h) / (mi / h) as the car’s speed reaches 20 mi / h
. So if you increase your speed to 21 mi / h , you could expect to decrease your fuel consumption by
about 0.05( gal / h) / (mi / h) .
/ 30. (a) f (8) is the rate of change of the quantity of coffee sold with respect to the price per pound
/ when the price is $8 per pound. The units for f (8) are pounds /( dollars / pound ) .
/ (b) f (8) is negative since the quantity of coffee sold will decrease as the price charged for it
increases. People are generally less willing to buy a product when its price increases.
/ 31. T (10) is the rate at which the temperature is changing at 10:00 A.M. To estimate the value of
/ T (10) , we will average the difference quotients obtained using the times t=8 and t=12 . Let
T (8) T (10) 72 81
T (12) T (10) 88 81
A=
=
=4.5 and B=
=
=3.5 . Then
8 10
2
12 10
2
T (t) T (10) A+B 4.5+3.5
/
T (10)= lim
=
=4 F / h .
t 10
2
2
t 10
32. For 1910: We will average the difference quotients obtained using the years 1900 and 1920.
8 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives E(1900) E(1910) 48.3 51.1
=
=0.28 and
1900 1910
10
E(1920) E(1910) 55.2 51.1
B=
=
=0.41 . Then
1920 1910
10
E(t) E(1910) A+B
/
E (1910)= lim
=0.345 . This means that life expectancy at birth was
t 1910
2
t 1910
increasing at about 0.345 year / year in 1910.
For 1950: Using data for 1940 and 1960 in a similar fashion, we obtain
Let A= / E (1950) [0.31+0.10]/2=0.205 . So life expectancy at birth was increasing at about 0.205 year /
year in 1950.
/ 33. (a) S (T ) is the rate at which the oxygen solubility changes with respect to the water temperature.
Its units are ( mg / L )/ C .
(b) For T =16 C , it appears that the tangent line to the curve goes through the points (0,14) and (32,6)
6 14
8
/
. So S (16)
=
= 0.25( mg / L )/ C . This means that as the temperature increases past
32 0
32
16 C , the oxygen solubility is decreasing at a rate of 0.25( mg / L )/ C .
/ 34. (a) S (T ) is the rate of change of the maximum sustainable speed of Coho salmon with respect to
the temperature. Its units are ( cm / s )/ C .
(b) For T =15 C, it appears the tangent line to the curve goes through the points (10,25) and (20,32) .
32 25
/
So S (15)
=0.7( cm / s )/ C . This tells us that at T =15 C, the maximum sustainable speed
20 10
of Coho salmon is changing at a rate of 0.7 ( cm / s )/ C . In a similar fashion for T =25 C, we can
25 35
/
= 2 (cm / s) / C . As it gets warmer than
use the points (20,35) and (25,25) to obtain S (25)
25 20
20 C, the maximum sustainable speed decreases rapidly.
35. Since f (x)=xsin (1/x) when x 0 and f (0)=0 , we have
f (0+h) f (0)
hsin (1/h) 0
/
f (0)=lim
=lim
=lim (sin(1/h)) . This limit does not exist since sin(1/h)
h
h
h 0
h 0
h 0
takes the values 1 and 1 on any interval containing 0 . (Compare with Example 4 in Section 2.2.)
2 36. Since f (x)=x sin(1/x) when x 0 and f (0)=0 , we have
2 f (0+h) f (0)
h sin (1/h) 0
1
f (0)=lim
=lim
=lim (h sin(1/h)) . Since 1 sin
h
h
h
h 0
h 0
h 0
/ 1 , we have 9 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.8 Derivatives h h sin lim (h sin
h 0 1
h h h hsin 1
h h . Because lim ( h )=0 and lim h =0 , we know that
h 0 h 0 1
/
)=0 by the Squeeze Theorem. Thus, f (0)=0 .
h 10 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 1. Note: Your answers may vary depending on your estimates. By estimating the slopes of tangent
lines on the graph of f , it appears that
/ (a) f (1) (c) f (3) / / 2 (b) f (2) 0.8 1 (d) f (4) / 0.5 2. Note: Your answers may vary depending on your estimates. By estimating the slopes of tangent
lines on the graph of f , it appears that
/ / (a) f (0) 3 (b) f (1) 0 (c) f (2) 1.5 (d) f (3) 2 (e) f (4) 0 (f) f (5) /
/ / / 3. It appears that f is an odd function, so f
/ (a) f ( 3) 1.5 (c) f ( 1) (e) f (1) (g) 1.2 / / / will be an even function ??? that f ( a)= f (a)
is, / f (3) 1.5 /
/ 0
0 (b) f ( 2) 1 (d) f (0) (f) f (2) / / 4
1 / 1 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function / (a) = II, since from left to right, the slopes of the tangents to graph (a) start out negative,
4. (a)
become 0 , then positive, then 0 , then negative again. The actual function values in graph II follow
the same pattern.
/ (b) = IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed
(b)
positive quantity, then suddenly become negative, then positive again. The discontinuities in graph IV
indicate sudden changes in the slopes of the tangents.
/ (c)
(c) = I, since the slopes of the tangents to graph (c) are negative for x<0 and positive for x>0
, as are the function values of graph I.
/ (d)
(d) = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0 ,
then negative, then 0 , then positive, then 0 , then negative again, and the function values in graph III
follow the same pattern. 5. 6.
2 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 7. 8. 9.
10. 3 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 11. 12.
13. 4 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 14. The slopes of the tangent lines on the graph of y=P(t) are always positive, so the y values of
/ y=P (t) are always positive. These values start out relatively small and keep increasing, reaching a
/ maximum at about t=6 . Then the y values of y=P (t) decrease and get close to zero. The graph of
/ P tells us that the yeast culture grows most rapidly after 6 hours and then the growth rate declines. 15. It appears that there are horizontal tangents on the graph of M for t=1963 and t=1971 . Thus, there
/ are zeros for those values of t on the graph of M . The derivative is negative for the years 1963 to
1971. 16. See Figure 1 in Section 3.4.
17. 5 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function The slope at 0 appears to be 1 and the slope at 1 appears to be 2.7 . As x decreases, the slope gets
/ x closer to 0 . Since the graphs are so similar, we might guess that f (x)=e .
18. / / As x increases toward 1 , f (x) decreases from very large numbers to 1 . As x becomes large, f (x)
/ 2 / gets closer to 0 . As a guess, f (x)=1/x or f (x)=1/x make sense.
/ 19. (a)
(b) By zooming in, we estimate that f (0)=0 , f
/ / By symmetry, f ( x)= f (x) . So f
/ / 1
2 / 1
2 / / =1 , f (1)=2 , and f (2)=4 . = 1, / f ( 1)= 2 , and f ( 2)= 4 .
(c)
(d) / / It appears that f (x) is twice the value of x , so we guess that f (x)=2x .
6 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 2 2 f (x+h) f (x)
(x+h) x
f (x)=lim
=lim
h
h
h 0
h 0
/ 2 2 2 2 (x +2hx+h ) x
2hx+h
h(2x+h)
=lim
=lim
=lim
=lim (2x+h)=2x
h
h
h
h 0
h 0
h 0
h 0 / 20. (a) By zooming in, we estimate that f (0)=0 , f / 1
2 / / 0.75 , f (1) 3 , f (2) 12 , and / f (3) 27 .
(b) / / By symmetry, f ( x)= f (x) . So f 1
2 / / / 0.75 , f ( 1) 3 , f ( 2) 12 , and / f ( 3) 27 . (c)
(d) / Since f (0)=0 , it appears that f
/ / / 2 / may have the form f (x)=ax . Using f (1)=3 , we have 2 a=3 , so f (x)=3x .
3 3 3 2 2 3 f (x+h) f (x)
(x+h) x
(x +3x h+3xh +h ) x
f (x)=lim
=lim
=lim
h
h
h
h 0
h 0
h 0
/ (e) 2 2 3 2 3 2 3x h+3xh +h
h(3x +3xh+h )
2
2
2
=lim
=lim
=lim (3x +3xh+h )=3x
h
h
h 0
h 0
h 0
/ 21. f (x)=lim
h 0 f (x+h) f (x)
37 37
0
=lim
=lim =lim 0=0
h
h
h 0
h 0 h h 0
/ Domain of f = domain of f =R .
7 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 22.
/ f (x) =lim
h 0 =lim
h 0 f (x+h) f (x)
[12+7(x+h)] (12+7x)
=lim
h
h
h 0
12+7x+7h 12 7x
7h
=lim
=lim 7=7
h
h 0 h h 0
/ Domain of f = domain of f =R .
23.
2 2 f (x+h) f (x)
[1 3(x+h) ] (1 3x )
=lim
f (x) =lim
h
h
h 0
h 0
/ 2 2 2 2 2 2 [1 3(x +2xh+h )] (1 3x )
1 3x 6xh 3h 1+3x
=lim
=lim
h
h
h 0
h 0
2 6xh 3h
h( 6x 3h)
=lim
=lim
=lim ( 6x 3h)= 6x
h
h
h 0
h 0
h 0
/ Domain of f = domain of f =R .
24.
2 2 f (x+h) f (x)
[5(x+h) +3(x+h) 2] (5x +3x 2)
=lim
f (x) =lim
h
h
h 0
h 0
/ 2 2 2 2 5x +10xh+5h +3x+3h 2 5x 3x+2
10xh+5h +3h
=lim
=lim
h
h
h 0
h 0
=lim
h 0 h(10x+5h+3)
=lim (10x+5h+3)=10x+3
h
h 0
/ Domain of f = domain of f =R .
25.
3 3 f (x+h) f (x)
[(x+h) 3(x+h)+5] (x 3x+5)
=lim
f (x) =lim
h
h
h 0
h 0
/ 8 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 3 2 2 3 3 (x +3x h+3xh +h 3x 3h+5) (x 3x+5)
=lim
h
h 0
2 2 3 2 2 3x h+3xh +h 3h
h(3x +3xh+h 3)
=lim
=lim
h
h
h 0
h 0
2 2 2 =lim (3x +3xh+h 3)=3x 3
h 0
/ Domain of f = domain of f =R .
26.
/
f (x) =lim
h 0 (x+h+ x+h ) (x+ x )
f (x+h) f (x)
=lim
h
h
h 0
x+h
h
+
h
h =lim
h 0 =lim
h 1+ 0 Domain of f = 0, x x+h + x =lim x+h + x 1
x+h + x =1+ h 1+ 0 (x+h) x
h( x+h + x ) 1
1
=1+
x+ x
2 x / ) , domain of f = ( 0, ) . 27.
/
g (x) =lim
h 0 =lim
h 0 1+2(x+h)
g(x+h) g(x)
=lim
h
h
h 0 1+2x 1+2(x+h) + 1+2x
1+2(x+h) + 1+2x (1+2x+2h) (1+2x)
2
2
1
=lim
=
=
h[ 1+2(x+h) + 1+2x ] h 0 1+2x+2h + 1+2x 2 1+2x
1+2x Domain of g= 1
,
2 / , domain of g = 1
,
2 . 28.
3+(x+h)
f (x) =lim f (x+h) f (x) =lim 1 3(x+h)
h
h
h 0
h 0
/ 3+x
1 3x 9 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function (3+x+h)(1 3x) (3+x)(1 3x 3h)
h(1 3x 3h)(1 3x)
0 =lim
h 2 2 (3 9x+x 3x +h 3hx) (3 9x 9h+x 3x 3hx)
=lim
h(1 3x 3h)(1 3x)
h 0
=lim
h 0 10h
10
=lim
=
h(1 3x 3h)(1 3x) h 0 (1 3x 3h)(1 3x)
/ Domain of f = domain of f = , 1
3 1
,
3 10
2 (1 3x) . 29.
4(t+h)
/
G (t) =lim G(t+h) G(t) =lim (t+h)+1
h
h
h 0
h 0
2 4t
t+1 4(t+h)(t+1) 4t(t+h+1)
(t+h+1)(t+1)
=lim
h
h 0 2 (4t +4ht+4t+4h) (4t +4ht+4t)
=lim
h(t+h+1)(t+1)
h 0
=lim
h 0 4h
4
=lim
=
h(t+h+1)(t+1) h 0 (t+h+1)(t+1)
/ Domain of G= domain of G =( , 1) ( 1, 4
2 (t+1) ). 30.
1 1
2 / g (x) =lim g(x+h) g(x) =lim (x+h)
h
h
h 0
h 0
2 2 2 x (x +2xh+h )
=lim
=lim
h
h 0
h 0 2 x 2 =lim
h 0 2 2xh h 2 2 h(x+h) x =lim
h 2 x (x+h) 0 2 2 h(x+h) x
2x h 2 2 (x+h) x = 2x
x 4 = 2x 3 / Domain of g= domain of g = { x| x 0} .
31. 10 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 4 4 4 3 2 2 3 4 f (x+h) f (x)
(x+h) x
(x +4x h+6x h +4xh +h ) x
=lim
=lim
f (x) =lim
h
h
h
h 0
h 0
h 0 4 / 3 2 2 3 4 4x h+6x h +4xh +h
3
2
2 3
3
=lim
=lim (4x +6x h+4xh +h )=4x
h
h 0
h 0
/ Domain of f = domain of f =R . 32. (a) (b) Note that the third graph in part (a) has small negative values for its slope, f / ; but as x 6 , / f
.
See the graph in part (d).
(c)
/
f (x) =lim
h 0 =lim
h 0 6 (x+h)
f (x+h) f (x)
=lim
h
h
h 0 6 (x+h) + 6 x
6 (x+h) + 6 x [6 (x+h)] (6 x)
h
=lim
h[ 6 (x+h) + 6 x ] h 0 h( 6 x h + 6 x ) =lim
h 6 x 0 Domain of f = ( 1
1
=
6 x h+ 6 x 2 6 x
,6 , domain of
11 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function / f =( ,6 ) . (d)
33. (a)
f (x) =lim f (x+h) f (x) =lim
h
h 0
h 0
h
=lim
h 0 =lim
h 0 1+ =lim
h =1+ x 2
x h 2
2
+
(x+h) x
h
2
x(x+h) 2
x+h x+h / 1+ 0 2x+2(x+h)
hx(x+h) =lim
h 1+ 0 2h
hx(x+h) 2
2 x / / (b) Notice that when f has steep tangent lines, f (x) is very large. When f is flatter, f (x) is smaller. 34. (a)
6 6
2 /
f (t) =lim f (t+h) f (t) =lim 1+(t+h)
h
h
h 0
h 0
2 =lim
h 0 12th 6h
2 2 h[1+(t+h) ](1+t ) =lim
h 0 1+t 2 2 =lim
h 0 12t 6h
2 2 [1+(t+h) ](1+t ) 2 6+6t 6 6(t+h)
2 2 h[1+(t+h) ](1+t ) = 12t
22 (1+t ) / (b) Notice that f has a horizontal tangent when t=0 . This corresponds to f (0)=0 . f
when f is increasing and negative when f is decreasing. / is positive
12 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function / 35. (a) U (t) is the rate at which the unemployment rate is changing with respect to time. Its units are
percent per year.
U (t+h) U (t) U (t+h) U (t)
/
for small values of h .
(b) To find U (t) , we use lim
h
h
h 0
/ For 1991: U ( 1991 ) = U (1992) U (1991) 7.5 6.8
=
=0.70
1992 1991
1
/ For 1992: We estimate U (1992) by using h= 1 and h=1 , and then average the two results to obtain
a final estimate.
U (1991) U (1992) 6.8 7.5
/
h= 1 U (1992)
=
=0.70 ;
1991 1992
1
U (1993) U (1992) 6.9 7.5
/
h=1 U (1992)
=
= 0.60 .
1993 1992
1
1
/
So we estimate that U ( 1992 )
[0.70+( 0.60)]=0.05 .
2
t 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 /
U (t) 0.70 0.05 0.70 0.65 0.35 0.35 0.45 0.35 0.25 0.20 / 36. (a) P (t) is the rate at which the percentage of Americans under the age of 18 is changing with
respect to time. Its units are percent per year (%/yr).
P(t+h) P(t) P(t+h) P(t)
/
(b) To find P (t) , we use lim
for small values of h .
h
h
h 0
P(1960) P(1950) 35.7 31.1
/
For 1950: P (1950)=
=
=0.46
1960 1950
10
/ For 1960: We estimate P (1960) by using h= 10 and h=10 , and then average the two results to
obtain a final estimate.
P(1950) P(1960) 31.1 35.7
/
h= 10 P (1960)
=
=0.46
1950 1960
10
13 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function P(1970) P(1960) 34.0 35.7
=
= 0.17
1970 1960
10
1
/
So we estimate that P (1960)
[0.46+( 0.17)]=0.145 .
2
h=10 / P (1960) t 1950 1960 1970 /
P (t) 0.460 0.145 0.385 1980
0.415 1990 2000 0.115 0.000 (c)
/ (d) We could get more accurate values for P (t) by obtaining data for the mid decade years 1955,
1965, 1975, 1985, and 1995.
37. f is not differentiable at x= 1 or at x=11 because the graph has vertical tangents at those points;
at x=4 , because there is a discontinuity there; and at x=8 , because the graph has a corner there.
38. (a) g is discontinuous at x= 2 (a removable discontinuity), at x=0 ( g is not defined there), and at
x=5 (a jump discontinuity).
(b) g is not differentiable at the points mentioned in part (a) (by Theorem 4), nor is it differentiable at
x= 1 (corner), x=2 (vertical tangent), or x=4 (vertical tangent).
39. As we zoom in toward ( 1,0) , the curve appears more and more like a straight line, so
f (x)=x+ x is differentiable at x= 1 . But no matter how much we zoom in toward the origin, the
curve doesn’t straighten out we can’t eliminate the sharp point (a cusp). So f is not differentiable
at x=0 . 14 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 40. As we zoom in toward (0,1) , the curve appears more and more like a straight line, so f is
differentiable at x=0 . But no matter how much we zoom in toward (1,0) or ( 1,0) , the curve doesn’t
straighten out we can’t eliminate the sharp point (a cusp). So f is not differentiable at x= 1 . 41. (a) Note that we have factored x a as the difference of two cubes in the third step.
1/3 1/3 1/3 f (x) f (a)
x a
=lim
f (a) =lim
x a
x a
x a
x a
/ =lim
x a 1
2/3 1/3 1/3 2/3 x +x a +a =lim 2/3 a or 1
a
3 3a f (0+h) f (0)
(b) f (0)=lim
=lim
h
h 0
h 0
/ 1/3 x 1 = 3 1/3 x
(x 1/3 a 2/3 1/3 1/3 2/3 a )(x +x a +a ) 2/3 h 0
1
=lim
. This function increases without bound, so the
h
h 0 h2/3 / limit does not exist, and therefore f (0) does not exist.
/
1
(c) lim f (x) =lim
= and f is continuous at x=0 (root function), so f has a vertical tangent
x 0
x 0 3x2/3
at x=0 .
2/3 g(x) g(0)
x 0
1
42. (a) g (0)=lim
, which does not exist.
=lim
=lim
x 0
x
x 0
x 0
x 0 x1/3
(b)
/ 2/3 2/3 1/3 g(x) g(a)
x a
=lim
g (a) =lim
x a
x a
x a
x a
/ 1/3 =lim
x a 2/3 (c) g(x)=x 1/3 2/3 1/3 1/3 =lim
x a 1/3 x +a 2/3 x +x a +a = (x 2a 2/3 3a 1/3 (x 2 = 1/3 3a or 1/3 1/3 2/3 2
a
3 1/3 1/3 2/3 1/3 / 0 1/3 a )(x +x a +a ) is continuous at x=0 and lim g (x) =lim
x 1/3 a )(x +a ) x 0 2
3 x 1/3 = . This shows that g has a vertical tangent line at x=0 .
15 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function (d)
43. f (x)=|x 6| =
lim
+ x 6 { x(x6 6) f (x) f (6)
= lim
x 6
+
x { if x<6
= 6 x if x<6
if x 6
x 6 if x 6
x 6 0
x 6
= lim
= lim 1=1 .
x 6
+ x 6
+ 6 x 6 x 6 But
lim
x 6 f (x) f (6)
= lim
x 6
x 6 x 6 0
6 x
= lim
x 6
x 6
x 6 = lim ( 1)= 1
x 6 / / So, f (6) does not exist. However, f (x)= { 11 if x<6
Another way of writing the answer is
if x 6 x 6
.
|x 6| / f (x) = 44. f (x)=[[x]] is not continuous at any integer n , so f is not differentiable at n by the contrapositive
/ of Theorem 4. If a is not an integer, then f is constant on an open interval containing a , so f (a)=0 .
/ Thus, f (x)=0 , x not an integer. 45. (a) f (x)=x x = { 2 x 2 x if x 0
if x<0 16 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 2 / (b) Since f (x)=x for x 0 , we have f (x)=2x for x>0 . [See Exercise 2.9.19(d).] Similarly, since
2 / f (x)= x for x<0 , we have f (x)= 2x for x<0 . At x=0 , we have
f (x) f (0)
x x
/
f (0)=lim
=lim
=lim x =0.
x 0
x 0
x 0 x
x 0
So f is differentiable at 0 . Thus, f is differentiable for all x .
/
2x if x 0
(c) From part (b), we have f (x)=
=2 x .
2x if x<0 { / 46. (a) f (4)= lim
h 0 } f (4+h) f (4)
5 (4+h) 1
h
= lim
= lim
= 1 and
h
h
h
h 0 h 0 1
1
f (4+h) f (4)
5 (4+h)
1 (1 h)
1
/
f (4)= lim
= lim
= lim
= lim
=1 .
+
h
h
+
+
+ h(1 h)
+ 1 h
h 0 h 0 h 0 h 0 (b) { 0
if x 0
(c) f (x)=
5 x
if 0<x<4 These expressions show that f is continuous on the intervals
1/(5 x) if x 4
(
,0) , (0,4) , (4,5) and (5, ) . Since lim f (x)= lim (5 x)=5 0= lim f (x) , lim f (x) does not exist,
+ x 0 + x 0 x 0 x 0 so f is discontinuous (and therefore not differentiable) at 0 .
1
=1 , so lim f (x)=1= f (4) and f is
At 4 we have lim f (x)= lim (5 x)=1 and lim f (x)= lim
+
+ 5 x
x 4
x 4 x 4 x 4 x 4 continuous at 4 . Since f (5) is not defined, f is discontinuous at 5 .
/ (d) From (a), f is not differentiable at 4 since f (4) / f (4) , and from (c), f is not differentiable at 0
+ or 5 .
17 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 47. (a) If f is even, then
f ( x+h) f ( x)
f [ (x h)] f ( x)
f (x h) f (x)
/
=lim
=lim
f ( x) = lim
h
h
h
h 0
h 0
h 0
lim = h 0 f (x h) f (x)
f (x+ x) f (x)
/
[let x= h] = lim
= f (x)
h
x
x 0 / Therefore, f is odd.
(b) If f is odd, then
f ( x+h) f ( x)
f [ (x h)] f ( x)
f (x h)+ f (x)
/
=lim
=lim
= lim
f ( x)
h
h
h
h 0
h 0
h 0
= lim
h Therefore, f 0
/ f (x h) f (x)
f (x+ x) f (x)
/
[let x= h] = lim
= f (x)
h
x
x 0 is even. 48. (a)
(b) The initial temperature of the water is close to room temperature because of the water that was in
the pipes. When the water from the hot water tank starts coming out, dT /dt is large and positive as T
increases to the temperature of the water in the tank. In the next phase, dT /dt=0 as the water comes
out at a constant, high temperature. After some time, dT /dt becomes small and negative as the
contents of the hot water tank are exhausted. Finally, when the hot water has run out, dT /dt is once
again 0 as the water maintains its (cold) temperature. (c) 49.
In the right triangle in the diagram, let y be the side opposite angle and x the side adjacent
angle . Then the slope of the tangent line is m= y/ x=tan . Note that
18 Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.9 The Derivative as a Function 0< < 2 2 tangent to the curve at the point (1,1) is 2 . Thus,
; that is, / . We know (see Exercise 19) that the derivative of f (x)=x is f (x)=2x . So the slope of the
is the angle between 0 and 2 whose tangent is 2 1 =tan 2 63 . 19 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions h e 1
=1 .
1. (a) e is the number such that lim
h 0 h
(b)
x x x (2.7 1)/x x (2.8 1)/x 0.001 0.9928 0.001 1.0291 0.0001 0.9932 0.0001 1.0296 0.001 0.9937 0.001 1.0301 0.0001 0.9933 0.0001 1.0297
h h 2.7 1
2.8 1
From the tables (to two decimal places), lim
=0.99 and lim
=1.03 . Since
h
h
h 0
h 0
0.99<1<1.03 , 2.7<e<2.8 . 2. (a)
The function value at x=0 is 1 and the slope at x=0 is 1 .
x e (b) f (x)=e is an exponential function and g(x)=x is a power function. d x x
d e
e
(e )=e and
(x )=ex
dx
dx 1 .
x e (c) f (x)=e grows more rapidly than g(x)=x when x is large.
/ 3. f (x)=186.5 is a constant function, so its derivative is 0 , that is, f (x)=0 .
/ 4. f (x)= 30 is a constant function, so its derivative is 0 , that is, f (x)=0 .
5. f (x)=5x 1
10 6. F(x)= 4x
2 7. f (x)=x +3x 4 / f (x)=5 0=5
/ 10 1 F (x)= 4(10x
/ )= 40x 9 2 1 f (x)=2x +3 0=2x+3
1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions 8 5 8 / 5 11. y=x 13. V (r)= 15. Y (t)=6t 7 = 10 x 5x 7/5 x 3 (
t
5 2 3/5) 1 10 = 3t 8/5 10 = 54t / 8 R (x)= 7 10 x = 7 10
x 1/3 18. y= x =x 1
x
2 21. g(x)=x + )+0=40x 10x 2 = 2 / )+0=40x 10x 7/5 (3r )=4 r Y (t)=6( 9)t 7 20. f (t)= t 4
3 / x 2 = R (t)=5 9 ) 2(5x 2
x
5 2/5) 1 / ) 2(5x 1 5
3
5
3
(6t ) 3(4t )+1=3t 12t +1
2 x 1/2 17. G(x)= x 2e =x 19. F(x)= 4 V (r)= 3/5 x 3 7 2 (
x
5 / 10 16. R(x)= 5 1 y =5(e )+0=5e 4 3
r
3 14. R(t)=5t 8 1 / / x 4 f (t)= y = 12. y=5e +3 7 g (x)=5(8x 1 6 4
t 3t +t
2 2/5 5 1 g (x)=5(8x 9. g(x)=5x 2x +6
10. f (t)= 8 1 / 8. g(x)=5x 2x +6 / y = 2e 1
x
3 5 1
2 = 1 1/2
=t t
t
1
2 x 2 =x +x 2 / x G (x)= 1
x
2 1/2 8 1 x 2e = 2 x 2e x 1 2/3 = 2/3 3x
5 5 x= 1/2 1 5
x
32
/ f (t)=
/ 1
t
2 / F (x)= 1
5 4
4
(5x )=
x
32
32 1
t
2 1/2 3 g (x)=2x+( 2)x =2x 3/2 = 1
2 t + 1
2t t 2
x 3 2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions 22. y= x (x 1)=x 3/2 3 1/2 1
x
x
2
2 / 1/2 y = x 1
x
2 1/2 = 1/2 (3x 1) [ factor out 1
x
2 1/2 3x 1
.
2 x / or y = 2 x +4x+3 3/2
1/2
1/2
23. y=
=x +4x +3x
x
1
1
/ 3 1/2
1/2
y = x +4
x +3
2
2
2
3/2 2/2 x = 1/2 2 x 2 x
24. y=
=x 2x
x
2 1/2 / 2 29. v=t 4 t
3 2 3/4 =t t A 31. z= 3 2/3 y 10 +Be =Ay 10 / 2 3
4 +Be y 2 v 3 b 2cv =ae / u = 2
t
3 3 7/4 t =2t+ 4t
1/3 +2 / z = 10Ay 11 3
2
y +Be = y
32. y=e 1/2 = 2 + 3 /(2 u ) 2c 2 v / 3/2 v y =ae bv 3 2 1
u
2 / g (u)= 2 (1)+ 3 v =2t t +2 t =t +2t 30. u= =1+1/(x x ) y =2ax+b b c
v
1
+ =ae +bv +cv
2
v
v
1 2 3/2 / 27. y=ax +bx+c
v x is a constant. 26. g(u)= 2 u+ 3u = 2 u+ 3 u 28. y=ae + 1
2 / y =1 2 y =0 since 4 2 3
2x x x =x x note that x =x 25. y=4 3
2
x+
2
x 3/2 t 7/4 = 2 3 3 =2t+ 4 4t 1/2 v t 3 +3 t 3 3 t
10A
11 +Be y y x+1 x 1 x +1=e e +1=e e +1
x 33. f (x)=e 5x / / x y =e e =e x+1 x f (x)=e 5 .
3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions / Notice that f (x)=0 when f has a horizontal tangent, f
negative when f is decreasing.
5 3 / 34. f (x)=3x 20x +50x 4 / is positive when f is increasing, and f 2 / 15 3 / 5x +3 14 f (x)=45x / / is an even function while f is an 2 / 1 / 15x . Notice that f (x)=0 when f has a horizontal tangent, f
negative when f is decreasing.
36. f (x)=x+1/x=x+x is f (x)=15x 60x +50 . Notice that f (x)=0 when f has a horizontal tangent and that f
odd function.
35. f (x)=3x / 2 / is positive when f is increasing, and f / is 2 f (x)=1 x =1 1/x . Notice that f (x)=0 when f has a horizontal tangent, f / is positive when f is increasing, and f / is
4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions negative when f is decreasing.
/ 2 3 37. To graphically estimate the value of f (1) for f (x)=3x x , we’ll graph f in the viewing
rectangle [1 0.1,1+0.1] by [ f (0.9),f(1.1)] , as shown in the figure. If we have sufficiently zoomed in
on the graph of f , we should obtain a graph that looks like a diagonal line; if not, graph again with
1 0.01 and 1+0.01 , etc.
2.299 1.701 0.589
/
=
=2.99 .
Estimated value: f (1)
1.1 0.9
0.2
2 Exact value: f (x)=3x x 3 / 2 / f (x)=6x 3x , so f (1)=6 3=3 . 38. See the previous exercise. Since f is a decreasing function, assign Y (3.9) to Y
1 Y min and Y (4.1) to
1 .
0.49386 0.50637
0.01251
=
= 0.06255 .
4.1 3.9
0.2
1 3/2
1 3/2
1
/
/
f (x)=
x
, so f (4)=
(4 )=
2
2
2 / Estimated value: f (4)
Exact value: f (x)=x
4 39. y=x +2e
y=2x+2. x / 1/2 3 x 2 2 3 = 1
= 0.0625 .
16 / y =4x +2e . At (0, 2) , y =2 and an equation of the tangent line is y 2=2(x 0) or 2 / 40. y=(1+2x) =1+4x+4x
y 9=12(x 1) or y=12x 3.
41. y=3x x
y=3x 1 . 1
8 / / y =4+8x. At (1, 9), y =12 and an equation of the tangent line is 2 / y =6x 3x . At (1, 2) , y =6 3=3 , so an equation of the tangent line is y 2=3(x 1) , or 5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions 3 1/2
/ 3
x . At (4, 8) , y = (2)=3, so an equation of the tangent line is
2
2
y 8=3(x 4), or y=3x 4 . 42. y=x x =x 3/2 / y = 43. (a) (b)
From the graph in part (a), it appears that f
negative (so f / (x , x ) and (x ,
1 2 is zero at x 1 1.25 , x 2 0.5 , and x , x ) and (x , x ). The slopes are positive (so f
1 2 3 3 . The slopes are 3
/ is positive) on ). 3 4 is negative) on ( / 3 2 (c) f (x)=x 3x 6x +7x+30 / 3 2 f (x)=4x 9x 12x+7 44. (a) 6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions (b)
From the graph in part (a), it appears that f
(so f / is positive) on (
x 2 (c) g(x)=e 3x , x ) and (x ,
1 / / is zero at x 1 0.2 and x 2.8 . The slopes are positive 2 ) . The slopes are negative (so f 2 / is negative) on (x , x ) .
1 2 x g (x)=e 6x 3 2 / 2 45. The curve y=2x +3x 12x+1 has a horizontal tangent when y =6x +6x 12=0
2 6(x +x 2)=0
and (1, 6) .
3 6(x+2)(x 1)=0 x= 2 or x=1 . The points on the curve are ( 2, 21) 2 / 2 46. f (x)=x +3x +x+3 has a horizontal tangent when f (x)=3x +6x+1=0
6 36 12
1
6 .
x=
= 1
6
3
3 47. y=6x +5x 3 / 2 2 m=y =18x +5 , but x
x 0 for all x , so m 5 for all x .
/ x 48. The slope of y=1+2e 3x is given by m=y =2e 3 .
The slope of 3x y=5 y=3x 5 is 3 .
m=3 x 2e 3=3 x e =3 x=ln 3 . This occurs at the point (ln 3, 7 3ln 3) (1.1, 3.7) .
7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions 49. 2 Let (a, a ) be a point on the parabola at which the tangent line passes through the point (0, 4). The
tangent line has slope 2a and equation y ( 4)=2a(x 0)
2 2 y=2ax 4 . Since (a, a ) also lies on the line, 2 a =2a(a) 4, or a =4 . So a= 2 and the points are (2, 4) and ( 2, 4).
2 / 2 50. If y=x +x , then y =2x+1 . If the point at which a tangent meets the parabola is (a, a +a), then
the slope of the tangent is 2a+1 . But since it passes through (2, 3), the slope must also be
2 y a +a+3
=
.
x
a 2
2 a +a+3
2
2
Therefore, 2a+1=
. Solving this equation for a we get a +a+3=2a 3a 2
a 2
2 a 4a 5=(a 5)(a+1)=0 a=5 or 1 . If a= 1 , the point is ( 1, 0) and the slope is 1, so the equation
is y 0=( 1)(x+1) or y= x 1. If a=5, the point is (5, 30) and the slope is 11, so the equation is
y 30=11(x 5) or y=11x 25.
2 / / f (x)= 2x, so the tangent line at (2, 3) has slope f (2)= 4. The normal line has
51. y= f (x)=1 x
1 1
1
1
7
slope
= and equation y+3= (x 2) or y= x
.
4 4
4
4
2 2 52. y= f (x)=x x / / f (x)=1 2x. So f (1)= 1, and the slope of the normal line is the negative
8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions reciprocal of that of the tangent line, that is, 1/( 1)=1. So the equation of the normal line at (1, 0) is
2 y 0=1(x 1) y=x 1. Substituting this into the equation of the parabola, we obtain x 1=x x
x= 1. The solution x= 1 is the one we require. Substituting x= 1 into the equation of the parabola
to find the y coordinate, we have y= 2. So the point of intersection is ( 1, 2), as shown in the
sketch. 53.
1
1
x+h x
x (x+h)
=lim
h
h 0 hx(x+h) / f (x) =lim f (x+h) f (x) =lim
h
h 0
h 0
=lim
h 0 h
1
1
=lim
=
2
hx(x+h) h 0 x(x+h)
x
2 54. Substituting x=1 and y=1 into y=ax +bx gives us a+b=1 (1) . The slope of the tangent line
/ / y=3x 2 is 3 and the slope of the tangent to the parabola at (x, y) is y =2ax+b . At x=1 , y =3
3=2a+b (2) . Subtracting (1) from (2) gives us 2=a and it follows that b= 1 . The parabola has
2 equation y=2x x .
2 55. f (x)=2 x if x 1 and f (x)=x 2x+2 if x>1 . Now we compute the right and left hand
derivatives defined in Exercise :
f (1+h) f (1)
2 (1+h) 1
h
/
f (1)=lim
=lim
=lim
=lim 1= 1 and
h
h
h
h 0 h 0 h 0 h 0 2 2 f (1+h) f (1)
(1+h) 2(1+h)+2 1
h
f (1)=lim
=lim
=lim
=lim h=0 .
+
h
h
+
+
+ h
+
/ h 0 / h 0 h / Thus, f (1) does not exist since f (1) / 0 h 0 / f (1) , so f is not differentiable at 1 . But f (x)= 1 for x<1
+ / and f (x)=2x 2 if x>1 . 9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions 56. g(x)=
lim
h 0 { 1 2x if x< 1
2
if 1 x 1
x
if x>1
x g( 1+h) g( 1)
=lim
h
h 0 1 2( 1+h) 1
=lim
h
h 2h
=lim ( 2 ) = 2 and
h 0 h 2 lim
h + 0 0 2 g( 1+h) g( 1)
( 1+h) 1
2h+h
=lim
=lim
=lim ( 2+h)= 2,
h
h
h
+
+
+
h 0 h 0 h 0 / so g is differentiable at 1 and g ( 1)= 2.
2 2 g(1) g(1)
(1) 1
2h+h
lim
=lim
=lim
=lim (2+h)=2 and
h
h
h h 0 h 0 h 0 h 0 g(1) g(1)
(1) 1
h
/
lim
=lim
=lim
=lim 1=1 , so g (1) does not exist.
h
h
+
+
+ h
+ h 0 h 0 h 0 h 0 / Thus, g is differentiable except when x=1 , and g (x)= 2 2 57. (a) Note that x 9<0 for x <9
f (x) = { 2 x 9
2 x +9
2 x 9 if x 3 if 3<x<3
ifx 3 | x|<3
/ { 2 if x< 1
2x if 1 x<1
1 if x>1 3<x<3 . So f (x) = { 2x
if x< 3
2x if 3<x<3
2x
if x>3 = { 2x
if | x|>3
2x if | x|<3 To show that
10 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions / f (3) does not exist we investigate lim
h 0 f (3+h) f (3)
by computing the left and right hand
h derivatives defined in Exercise .
2 f (3+h) f (3)
[ (3+h) +9] 0
f (3)=lim
=lim
=lim ( 6 h)= 6 and
h
h
/ h 0 h 0 h 0 2 2 f (3+h) f (3)
(3+h) 9 0
6h+h
f (3)=lim
=lim
=lim
=lim (6+h)=6 .
+
h
h
h
+
+
+
+
/ h 0 h 0 h Since the left and right limits are different, lim
h 0 0 h 0 f (3+h) f (3)
/
does not exist, that is, f (3) does not
h / exist. Similarly, f ( 3) does not exist. Therefore, f is not differentiable at 3 or at 3 . (b)
58. If x 1 , then h(x)=| x 1|+| x+2|=x 1+x+2=2x+1.
If 2<x<1 , then h(x)= (x 1)+x+2=3.
If x 2, then h(x)= (x 1) (x+2)= 2x 1. Therefore, h(x)= / To see that h (1)=lim
x 1 { 2x 1 if x 2
3
if 2<x<1
2x+1 if x 1 / h (x)= { 2 if x< 2
0 if 2<x<1
2 if x>1 h(x) h(1)
h(x) h(1)
3 3
does not exist, observe that lim
=lim
=0 but
x 1
x 1
3 1
x 1 x 1 h(x) h(1)
2x 2
/
lim
=lim
=2 . Similarly, h ( 2) does not exist.
x 1
+
+ x 1
x 1 x 1 11 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions 2 / 59. y= f (x)=ax f (x)=2ax. So the slope of the tangent to the parabola at x=2 is m=2a(2)=4a. The
1
slope of the given line, 2x+y=b y= 2x+b, is seen to be 2, so we must have 4a= 2 a= . So
2
1 2
when x=2, the point in question has y coordinate
2 = 2. Now we simply require that the given
2
line, whose equation is 2x+y=b, pass through the point (2, 2): 2(2)+( 2)=b b=2. So we must have
1
a=
and b=2.
2
/ / 60. f is clearly differentiable for x<2 and for x>2. For x<2, f (x)=2x, so f (2)=4. For x>2,
/ / / / f (x)=m, so f (2)=m. For f to be differentiable at x=2, we need 4= f (2)= f (2)=m. So
+ + f (x)=4x+b. We must also have continuity at x=2, so 4= f (2)=lim f (x)=lim (4x+b)=8+b. Hence,
x + 2 x + 2 b= 4.
3 2 / 2 61. y= f (x)=ax +bx +cx+d f (x)=3ax +2bx+c. The point ( 2, 6) is on f , so f ( 2)=6
8a+4b 2c+d=6 (1) . The point (2, 0) is on f , so f (2)=0 8a+4b+2c+d=0 (2) . Since there are
/ / horizontal tangents at ( 2, 6) and (2, 0), f ( 2)=0. f ( 2)=0 12a 4b+c=0 (3) and / f (2)=0 12a+4b+c=0 (4) . Subtracting equation (3) from (4) gives 8b=0 b=0. Adding (1) and (2)
gives 8b+2d=6, so d=3 since b=0. From (3) we have c= 12a, so (2) becomes 8a+4(0)+2( 12a)+3=0
3
3
9
3=16a a=
. Now c= 12a= 12
=
and the desired cubic function is
16
16
4
3 3 9
y=
x
x+3.
16
4
62. (a) xy=c c
c
y= . Let P= a,
x
a / . The slope of the tangent line at x=a is y (a)= c
2 . Its a
c
2c
2c
c
c
equation is y
=
(x a) or y=
x+ . so its y intercept is
. Setting y=0 gives x=2a, so
2
2
a
a
a
a
a
2c
the x intercept is 2a. The midpoint of the line segment joining 0,
and (2a, 0) is
a
c
a,
=P.
a
(b) We know the x and y intercepts of the tangent line from part (a), so the area of the triangle
1
1
1
bounded by the axes and the tangent is (base)(height)= xy= (2a)(2c/a)=2c, a constant.
2
2
2
12 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.1 Derivatives of Polynomials and Exponential Functions 1000 63. Solution 1: Let f (x)=x . Then, by the definition of a derivative,
1000 / f (1)=lim
x 1 f (x) f (1)
x
1
. But this is just the limit we want to find, and we know (from
=lim
x 1
x 1
x 1
/ the Power Rule) that f (x)=1000x 999 / 1000 999 , so f (1)=1000(1) =1000 . So lim
x 1000 Solution 2: Note that (x
1000 lim
x 1 x x 1 1 = lim
x = (x 1)(x 999 +x 998 999 +x 998 +x 999 +x 998 +x 997 997 + 997 + x 1 1 =1000 . 2 +x +x+1). So 2 +x +
x 1 1 lim (x
x 1)=(x 1)(x 1 x +x +x+1)
2 +x +x+1) = 1+1+1+...+1+1+1 1 1000 ones = 1000, as above. 64.
In order for the two tangents to intersect on the y axis, the points of tangency must be at equal
2 distances from the y axis, since the parabola y=x is symmetric about the y axis. Say the points of
2 2 2 tangency are (a, a ) and ( a, a ), for some a>0. Then since the derivative of y=x is dy/dx=2x, the
2 2 left hand tangent has slope 2a and equation y a = 2a(x+a), or y= 2ax a , and similarly the right
hand tangent line has equation
2 2 2 y a =2a(x a), or y=2ax a . So the two lines intersect at (0, a ). Now if the lines are perpendicular,
1
2 1
then the product of their slopes is 1, so ( 2a)(2a)= 1 a =
a= . So the lines intersect at
4
2
1
0,
.
4 13 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates 1. V =x dV dV dx
2 dx
=
=3x
dt
dx dt
dt 3 2. (a) A= r dA dA dr
dr
=
=2 r
dt dr dt
dt 2 dA
dr
2
=2 r
=2 (30m)(1m/s)=60 m / s
dt
dt (b) dy dy dx
dy
2
2
=
=(3x +2)(5)=5(3x +2) . When x=2 ,
=5(14)=70 .
dt dx dt
dt 3 3. y=x +2x
2 2 4. x +y =25 2x
2 dx
dy
+2y
=0
dt
dt 2 When y=4 , x +4 =25
2 2 dz
dx
dy
dz 1
dx
dy
2 2
2
=2x
+2y
=
x
+y
. When x=5 and y=12 , z =5 +12
dt
dt
dt
dt z
dt
dt
dx
dy
dz
1
46
z= 13 . For
=2 and
=3, =
(5 2+12 3)=
.
dt
dt
dt
13
13
2 5. z =x +y
2 z =169 dx
dy
dx
y dy
= y
=
.
dt
dt
dt
x dt
dy
dx
4
x= 3 . For
=6 ,
=
(6)= 8 .
dt
dt
3
x 6. y= 1+x 3 we have 4= 2z dy dy dx 1
3
=
= (1+x )
dt dx dt 2
3(4) dx
2(3) dt dx
(3x ) =
dt 1/2 2 2 3x 2 1+x 3 dx
dy
. With
=4 when x=2 and y=3 ,
dt
dt dx
=2 cm / s.
dt 7. (a) Given: a plane flying horizontally at an altitude of 1 mi and a speed of 500 mi / h passes
directly over a radar station. If we let t be time (in hours) and x be the horizontal distance traveled by
the plane (in mi), then we are given that dx/dt=500 mi / h.
(b) Unknown: the rate at which the distance from the plane to the station is increasing when it is 2 mi
from the station. If we let y be the distance from the plane to the station, then we want to find dy/dt
when y=2 mi.
(c)
2 2 (d) By the Pythagorean Theorem, y =x +1 2y(dy/dt)=2x(dx/dt) .
3
dy x dx x
dy
2 2
(e)
=
= (500) . Since y =x +1 , when y=2 , x= 3 , so
=
(500)=250 3 433 mi / h.
dt y dt y
dt
2
8. (a) Given: the rate of decrease of the surface area is 1 cm
1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates 2 2 / min. If we let t be time (in minutes) and S be the surface area (in cm ), then we are given that
2 dS/dt= 1 cm / s.
(b) Unknown: the rate of decrease of the diameter when the diameter is 10 cm. If we let x be the
diameter, then we want to find dx/dt when x=10 cm. (c)
(d) If the radius is r and the diameter x=2r , then r= 1
2
x and S=4 r =4
2 dS dS dx
dx
=
=2 x
.
dt dx dt
dt
dS
dx
dx
1
dx
1
(e) 1=
=2 x
=
. When x=10 ,
=
dt
dt
dt
2 x
dt
20
min. 1
x
2 2 2 = x . So the rate of decrease is 1
20 cm / 9. (a) Given: a man 6 ft tall walks away from a street light mounted on a 15 ft tall pole at a rate of 5
ft / s. If we let t be time (in s) and x be the distance from the pole to the man (in ft), then we are given
that dx/dt=5 ft / s.
(b) Unknown: the rate at which the tip of his shadow is moving when he is 40 ft from the pole. If we
let y be the distance from the man to the tip of
d
his shadow (in ft), then we want to find
(x+y) when x=40 ft.
dt (c)
(d) By similar triangles, 15 x+y
=
6
y 15y=6x+6y (e) The tip of the shadow moves at a rate of 9y=6x d
d
(x+y)=
dt
dt 2
x.
3
2
5 dx 5
25
x+ x =
= (5)=
ft / s.
3
3 dt 3
3
y= 10. (a) Given: at noon, ship A is 150 km west of ship B; ship A is sailing east at 35 km / h, and ship B
is sailing north at 25 km / h. If we let t be time (in hours), x be the distance traveled by ship A (in
km), and y be the distance traveled by ship B (in km), then we are given that dx/dt=35 km / h and
dy/dt=25 km / h.
(b) Unknown: the rate at which the distance between the ships is changing at 4:00 P.M. If we let z be
the distance between the ships, then we want to find dz/dt when t=4 h.
2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates (c)
2 2 2 (d) z =(150 x) +y 2z dz
=2(150 x)
dt dx
dt +2y dy
dt
2 2 (e) At 4:00 P.M., x=4(35)=140 and y=4(25)=100 z= (150 140) +100 = 10,100 . So
dz 1
dx
dy
10(35)+100(25)
215
=
(x 150) +y
=
=
21.4 km / h.
dt z
dt
dt
10,100
101 11.
dx
dy
2 2 2
=60 mi / h and
=25 mi / h. z =x +y
dt
dt
dz
dx
dy
dz 1
dx
dy
z =x
+y
=
x
+y
.
dt
dt
dt
dt z
dt
dt We are given that 2 2z dz
dx
dy
=2x
+2y
dt
dt
dt 2 After 2 hours, x=2(60)=120 and y=2(25)=50 z= 120 +50 =130 , so
dz 1
dx
dy
120(60)+50(25)
=
x
+y
=
=65 mi / h.
dt z
dt
dt
130 12.
We are given that
When x=8 , dx
y 2
=1.6 m / s. By similar triangles,
=
dt
12 x y= 24
x dy
24 dx
24
=
=
(1.6) .
2 dt
2
dt
x
x dy
24(1.6)
=
= 0.6 m / s, so the shadow is decreasing at a rate of 0.6 m / s.
dt
64 13.
We are given that dx
dy
2
2
2
=4 ft / s and
=5 ft / s. z =(x+y) +500
dt
dt 2z dz
=2(x+y)
dt dx dy
+
dt dt . 15
3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates minutes after the woman starts, we have x=(4ft/s)(20min)(60s/min)=4800 ft and y=5 15 60=4500
2 2 z= (4800+4500) +500 = 86,740,000 , so
dz x+y
dx dy
4800+4500
837
=
+
=
(4+5)=
dt
z
dt dt
86,740,000
8674
14. We are given that 8.99 ft / s . dx
=24 ft / s.
dt (a)
dx
.
dt
dy 90 x
2
2
When x=45 , y= 45 +90 =45 5 , so
=
dt
y
2 2 2 y =(90 x) +90 2y dy
=2(90 x)
dt dx
dt = so the distance from second base is decreasing at a rate of 24
5 45
24
( 24)=
,
45 5
5
10.7 ft / s. (b) Due to the symmetric nature of the problem in part (a), we expect to get the same answer and we
dz
dx
dz
45
24
2 2
2
. When x=45 , z=45 5 , so
=
(24)=
10.7 ft / s.
do. z =x +90 2z =2x
dt
dt
dt 45 5
5
1
dh
dA
bh , where b is the base and h is the altitude. We are given that
=1 cm / min and
=2
2
dt
dt
dh
db
dA 1
2
=
b
+h
cm / min. Using the Product Rule, we have
. When h=10 and A=100 , we
dt 2
dt
dt
1
1
1
db
db
have 100= b(10)
b=10 b=20 , so 2=
20 1+10
4=20+10
2
2
2
dt
dt
db 4 20
=
= 1.6 cm / min.
dt
10 15. A= 16.
dy
dx
2 2
= 1 m / s, find
when x=8 m. y =x +1
dt
dt
, y= 65 , so
Given 2y dy
dx
=2x
dt
dt dx y dy
y
=
=
. When x=8
dt x dt
x 4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates dx
=
dt 65
. Thus, the boat approaches the dock at
8 65
8 1.01 m / s. 17.
We are given that dx
dy
2
2
2
=35 km / h and
=25 km / h. z = ( x+y ) +100
dt
dt
2 2z dz
=2(x+y)
dt dx dy
+
dt dt . 2 At 4:00 P.M., x=4(35)=140 and y=4(25)=100 z= (140+100) +100 = 67,600 =260 , so
dz x+y
dx dy
140+100
720
=
+
=
(35+25)=
55.4 km / h.
dt
z
dt dt
260
13
18. Let D denote the distance from the origin (0,0) to the point on the curve y= x .
dD 1 2
dx
2x+1 dx
2
2
2
2
2
1/2
D= (x 0) +(y 0) = x +( x ) = x +x
= (x +x) (2x+1) =
. With
dt 2
dt
dt
2
2 x +x
dx
dD
9
27
=3 when x=4 ,
=
(3)=
3.02 cm / s.
dt
dt 2 20
4 5 19.
dV
1 2
=C 10 , 000 , where V =
r h is the volume at
dt
3
2
r h
1
1
1
dV
3
2 dh
time t . By similar triangles, =
r= h V =
h h=
h
= h
.
2 6
3
3
3
27
dt 9
dt
dh
800,000
2
When h=200cm ,
=20cm/min , so C 10 , 000= (200) (20) C=10 , 000+
289 ,
dt
9
9
If C= the rate at which water is pumped in, then 3 253 cm / min. 20.
By similar triangles,
5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates 3 b
1
2
= , so b=3h . The trough has volume V = bh(10)=5(3h)h=15h
1 h
2
1 dh
2
4
. When h= ,
=
= ft / min.
2 dt
1 5
5
2 12= dV
dh
=30h
dt
dt dh 2
=
dt 5h 21.
1
( base + base )( height ) , and the
1
2
2
volume V of the 10 meter long trough is 10A . Thus, the volume of the trapezoid with height h is
1
a 0.25 1
V =(10) [0.3+(0.3+2a)]h . By similar triangles, =
= , so
2
h 0.5 2
dV dV dh
dh
dh
0.2
2
2a=h V =5(0.6+h)h=3h+5h . Now
=
0.2=(3+10h)
=
. When h=0.3
dt
dh dt
dt
dt 3+10h
dh
0.2
0.2
1
10
,
=
=
m / min =
m / min or
cm / min.
dt 3+10(0.3) 6
30
3
The figure is labeled in meters. The area A of a trapezoid is 22.
1
(b+12)h(20)=10(b+12)h and, from similar triangles,
2
x 6
y 16 8
8h
11h
= and =
= , so b=x+12+y=h+12+
=12+
. Thus,
h 6
h 6 3
3
3 The figure is drawn without the top 3 feet. V = 2 11h
110h
dV
220
V =10 24+
h=240h+
and so 0.8=
= 240+
h
3
3
dt
3
dh
0.8
3
=
=
0.00132 ft / min.
dt 240+5(220/3) 2275 dh
. When h=5 ,
dt 23.
dV
1 2 1
3
We are given that
=30 ft / min. V =
r h=
dt
3
3 h
2 2 3 h
h=
12 6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates dV dV dh
=
dt
dh dt 2 h dh
30=
4 dt dh 120
dh 120
6
=
. When h=10 ft,
= 2 =
2
dt
dt
5
10
h 0.38 ft / min. 24.
x
We are given dx/dt=8 ft / s. cot =
100
100 1
When y=200 , sin =
=
200 2
1
rad / s.
50 x=100cot
2 d
(1/2)
=
dt
100 8= 2 dx
2 d
= 100csc
dt
dt d
sin
=
dt
100 8. 1
rad / s. The angle is decreasing at a rate of
50 25.
1
h
1
bh , but b=5 m and sin =
h=4sin , so A= (5)(4sin )=10sin . We are given
2
4
2
d
dA dA d
=0.06 rad / s, so
=
=(10cos )(0.06)=0.6cos . When =
,
dt
dt d dt
3
dA
1
2
=0.6 cos
=(0.6)
=0.3 m / s.
dt
3
2 A= 26.
We are given d /dt=2 / min =
2 2 2 90 rad / min. By the Law of Cosines, x =12 +15 2(12)(15)cos =369 360cos 2x =60 , x= 369 360cos 60 = 189 =3 21 , so dx
=360sin
dt d
dt dx 180sin 60
=
dt
3 21 dx 180sin
=
dt
x
90 = 3
3 21 = 7
21 d
. When
dt
0.396 m / min.
27. Differentiating both sides of PV =C with respect to t and using the Product Rule gives us
dV
dP
dV
V dP
dP
P
+V
=0
=
. When V =600 , P=150 and
=20 , so we have
dt
dt
dt
P dt
dt
7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates dV
600
3
=
(20)= 80 . Thus, the volume is decreasing at a rate of 80 cm / min.
dt
150
28. PV 1.4 =C P 1.4V 0.4 dV
=
dt dV
1.4 dP
+V
=0
dt
dt V 1.4
0.4 dP
V dP
=
. When V =400 ,
dt
1.4P dt P 1.4V
dP
dV
400
250
P=80 and
= 10 , so we have
=
( 10)=
. Thus, the volume is increasing at a
dt
dt
1.4(80)
7
250
3
rate of
36 cm / min.
7
29. With R =80 and R =100 ,
1 2 1 1
1
1
180
9
400
1
=
+
=
+
=
=
, so R=
. Differentiating
R R
9
R 80 100 8000 400
1 2 1 1
1
=
+
with respect to t , we have
R R
R
1 2 dR 2
=R
dt 1 dR 1 1 dt 2 R + 1
2 R 1 1
2 R dR 2 dt 2 dR 2 dt . When R =80 and R =100 ,
1 2 2 2 dR 400
=
2
dt
9 dR
1
1 dR
1
=
2 dt
2 dt
R
R 1
2 (0.3)+ 80 1
2 100 (0.2) = 107
810 0.132 / s. dB
2/3
2.53
when L=18 using B=0.007W and W =0.12L
.
dt
dB
dB dW dL
2
20 15
1/3
1.53
=
= 0.007 W
(0.12 2.53 L )
dt
dW dL dt
3
10,000,000
2
2.53 1/3
1.53
8
5
(0.12 2.53 18 )
1.045 10 g/yr
= 0.007 3 (0.12 18 )
7
10 30. We want to find 31.
We are given that
2=10cos d
4 dt dx
x
x=10sin
=2 ft / s. sin =
dt
10
2
d
2
=
=
rad / s.
dt 10(1/ 2 )
5 dx
=10cos
dt d
. When =
,
dt
4 8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates 32.
Using Q for the origin, we are given dx
dy
= 2 ft / s and need to find
when x= 5 . Using the
dt
dt
2 Pythagorean Theorem twice, we have 2 2 2 x +12 + y +12 =39 , the total length of the rope.
dx
dy
x
y
+
=0 , so
Differentiating with respect to t , we get
2
2 dt
2
2 dt
x +12
y +12
dy
x
=
dt
y 2 2 y +12
2 2 x +12
2 dx
2
2
2
2
2
2
. Now when x= 5 , 39= ( 5) +12 + y +12 =13+ y +12
dt 2 , and y= 26 12 = 532 . So when x= 5 , dy
=
dt ( 5)(26)
( 2)=
532 (13) 10
133 2 2 y +12 =26 0.87 ft / s. So cart B is moving towards Q at about 0.87 ft / s. 33. (a)
2 2 2 By the Pythagorean Theorem, 4000 +y = . Differentiating with respect to t , we obtain
dy
d
dy
2y =2
. We know that
=600 ft / s, so when y=3000 ft,
dt
dt
dt
d
y dy 3000
1800
2
2
=
=
(600)=
=360 ft / s.
= 4000 +3000 = 25,000,000 =5000 ft and
dt
dt 5000
5
y
(b) Here tan =
4000 d
d
(tan )=
dt
dt y
4000 sec 2 d
1 dy
=
dt 4000 dt 2 d
cos
dy
=
.
dt
4000 dt
2 dy
4000 4000 4
d
(4/5)
When y=3000 ft,
=600 ft / s, =5000 and cos =
=
= , so
=
(600)=0.096
dt
5000 5
dt 4000
rad / s. 34.
9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates We are given that
sec 2 =1+ 1
3 d
=4(2 )=8 rad / min. x=3tan
dt
2 10
dx
10
80
=
and
=3
(8 )=
9
dt
9
3 dx
1
2 d
=3sec
. When x=1 , tan = , so
dt
dt
3
83.8 km / min. 35.
dx
=300 km / h. By the Law of Cosines,
dt
1
dy
dx dx
dy 2x+1 dx
2 2 2
2
2
y =x +1 2(1)(x)cos 120 =x +1 2x
=x +x+1 , so 2y =2x
+
=
. After
2
dt
dt dt
dt
2y dt
1 minute,
300
dy 2(5)+1
1650
2
x=
=5 km y= 5 +5+1 = 31 km
=
(300)=
296 km / h.
60
dt 2 31
31
We are given that 36.
dx
dy
=3 mi / h and
=2 mi / h. By the Law of Cosines,
dt
dt
dz
dx
dy
dy
dx
2 2 2
2 2
z =x +y 2xycos 45 =x +y 2 xy 2z =2x
+2y
2x
2y
. After 15 minutes
dt
dt
dt
dt
dt
1
= h ,
4
13 6 2
3
2 1
3 2
2 2
3
2
2
we have x= and y= =
z=
and
z=
+
2
4
4
4 2
4
4
4
4
13 6 2
2
3
1
3
1
2
dz
=
2
3+2
2 2
2 2
3 =
= 13 6 2
4
2
4
2
2
dt
13 6 2
13 6 2
2.125mi/h.
We are given that 37.
Let the distance between the runner and the friend be . Then by the Law of Cosines,
10 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.10 Related Rates 2 2 2 =200 +100 2 200 100 cos =50,000 40,000cos (*). Differentiating implicitly with respect to t
d
d
, we obtain 2
= 40,000( sin )
. Now if D is the distance run when
dt
dt
the angle is radians, then by the formula for the length of an arc on a circle, s=r , we have
1
d
1 dD
7
d
D=100 , so =
D
=
=
. To substitute into the expression for
, we must
100
dt 100 dt 100
dt
1
2
know sin at the time when =200 , which we find from (*): 200 =50,000 40,000cos
cos =
4
sin = 1 1
4 2 = 15
15
d
. Substituting, we get 2(200) =40,000
4
dt
4 7
100 7 15
6.78 m / s. Whether the distance between them is increasing or decreasing depends on
4
the direction in which the runner is running. d /dt= 38.
2
= rad / h.
12 6
The minute hand goes around once an hour, or at the rate of 2 rad / h. So the angle between them
(measuring clockwise from the minute hand to the hour hand) is changing at the rate of
11
d /dt=
2 =
rad / h. Now, to relate to , we use the Law of Cosines:
6
6
The hour hand of a clock goes around once every 12 hours or, in radians per hour, 2 2 2 =4 +8 2 4 8 cos =80 64cos (*). d
d
= 64( sin )
. At 1:00, the angle
dt
dt
2
between the two hands is one twelfth of the circle, that is,
= radians. We use (*) to find at
12 6
d
11
1:00: = 80 64cos
= 80 32 3 . Substituting, we get 2
=64sin
6
dt
6
6
1
11
64
d
2
6
88
=
=
18.6 . So at 1:00, the distance between the tips of the
dt
2 80 32 3
3 80 32 3
hands is decreasing at a rate of 18.6 mm / h 0.005 mm / s.
Differentiating implicitly with respect to t , we get 2 11 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials 1. As in Example 1, T (0)=185 , T (10)=172 , T (20)=160 , and
T (10) T (20) 172 160
/
/
T (20)
=
= 1.2 F / min. T (30) T (20)+T (20)(30 20) 160 1.2(10)=148
10 20
10
F.
We would expect the temperature of the turkey to get closer to 75 F
as time increases. Since the temperature decreased 13 F in the first 10 minutes and 12 F in the
second 10 minutes, we can assume that the slopes of the tangent line are increasing through negative
values: 1.3, 1.2,... . Hence, the tangent lines are under the curve and 148 F is an underestimate. From the figure, we estimate the slope of the tangent line at t=20 to be
184 147
37
=
.
0 30
30
37
2
/
Then the linear approximation becomes T (30) T (20)+T (20) 10 160
(10)=147
147.7 .
30
3
/ 2. P (2) P(1) P(2) 87.1 74.9
=
= 12.2 kilopascals / km.
1 2
1
/ P(3) P(2)+P (2)(3 2) 74.9 12.2(1)=62.7 kPa.
From the figure, we estimate the slope of the tangent line at h=2 to be
/ approximation becomes P(3) P(2)+P (2) 1 74.9 35
3 98 63
35
=
. Then the linear
0 3
3 63.23 kPa. 3. Extend the tangent line at the point (2030,21) to the t axis. Answers will vary based on this
approximation we’ll use t=1900 as our t intercept. The linearization is then
/
P(t)
P(2030)+P (2030)(t 2030)
1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials 21
(t 2030)
130
21
P(2040)=21+
(2040 2030) 22.6%
130
21
P(2050)=21+
(2050 2030) 24.2%
130
21+ These predictions are probably too high since the tangent line lies above the graph at t=2030.
N(1980) N(1985) 15.0 17.0
N(1990) N(1985) 19.3 17.0
=
=0.4 and B=
=
=0.46 . Then
1980 1985
5
1990 1985
5
N(t) N(1985) A+B
/
N (1985)= lim
=0.43 million / year. So
t 1985
2
t 1985 4. Let A= / N(1984) N(1985)+N (1985)(1984 1985) 17.0+0.43( 1)=16.57 million.
N(1995) N(2000) 22.0 24.9
/
N (2000)
=
=0.58 million / year.
1995 2000
5
/ N(2006) N(2000)+N (2000)(2006 2000) 24.9+0.58(6)=28.38 million.
5. f (x)=x 3 / 2 / / f (x)=3x , so f (1)=1 and f (1)=3 . With a=1 , L(x)= f (a)+ f (a)(x a) becomes
/ L(x)= f (1)+ f (1)(x 1)=1+3(x 1)=3x 2 .
/ / 7. f (x)=cos x
L(x)= f f (x)= sin x , so f
+f 2
3 / / f (x)=1/x , so f (1)=0 and f (1)=1 . Thus, L(x)= f (1)+ f (1)(x 1)=0+1(x 1)=x 1. 6. f (x)=ln x 1/3 8. f (x)= x =x / / x 2
/ f (x)= 1
x
3 L(x)= f ( 8)+ f ( 8)(x+8)= 2+ 2
2/3 2
=0 1 =0 and f
x 2 / = 1 . Thus, 2
= x+
/ 2 , so f ( 8)= 2 and f ( 8)= .
1
. Thus,
12 1
1
4
.
(x+8)=
x
12
12
3 9. f (x)= 1 x
2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials 1
1
/
, so f (0)=1 and f (0)=
. Therefore,
2
2 1 x
1 x = f (x) f (0)+ f /(0)(x 0)
1
1
= 1+
(x 0)=1
x
2
2
1
1
So 0.9 = 1 0.1 1
(0.1)=0.95 and 0.99 = 1 0.01 1
(0.01)=0.995 .
2
2
/ f (x)= 1/3 3 10. g(x)= 1+x =(1+x) / g (x)= 1
(1+x)
3 2/3 , 1
1
/
3
. Therefore, 1+x =g(x) g(0)+g (0)(x 0)=1+ x . So
3
3
1
1
3
3
3
0.95= 1+ ( 0.05) 1+ ( 0.05) =0.983 , and 1.1 = 1+0.1 1+ (0.1)=1.03 .
3
3
/ so g(0)=1 and g (0)=
3 1/3 3 11. f (x)= 1 x =(1 x)
f (x) / f (0)+ f (0)(x 0)=1 / f (x)= 1
(1 x)
3 1
x . We need
3 3 2/3 / , so f (0)=1 and f (0)= 1 x 0.1<1 1
. Thus,
3 1 3
x< 1 x +0.1 , which is true when
3 1.204<x<0.706 . / 2 / / 12. f (x)=tan x f (x)=sec x , so f (0)=0 and f (0)=1 . Thus, f (x) f (0)+ f (0)(x 0)=0+1(x 0)=x .
We need tan x 0.1<x<tan x+0.1 , which is true when 0.63<x<0.63 . 3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials 1 13. f (x)= 4 =(1+2x) 4 / 5 f (x)= 4(1+2x) (2)= (1+2x)
f (x) 8 / 5 , so f (0)=1 and f (0)= 8 . Thus, (1+2x) / f (0)+ f (0)(x 0)=1+( 8)(x 0)=1 8x .
4 4 We need 1/(1+2x) 0.1<1 8x<1/(1+2x) +0.1 , which is true when 0.045<x<0.055 . 14. f (x)=e x / x / f (x)=e , so f (0)=1 and f (0)=1 . Thus, f (x) x / f (0)+ f (0)(x 0)=1+1(x 0)=1+x . x We need e 0.1<1+x<e +0.1 , which is true when 0.483<x<0.416. / 4 15. If y= f (x) , then the differential dy is equal to f (x)dx . y=x +5x
16. y=cos x
17. y=xln x dy= x 18. y= 1+t 19. y= dy= sin x 2 dy= u+1
u 1 dy= dx= 4 sin xdx 1
+ln x 1 dx=(1+ln x)dx
x 1
2
(1+t )
2 1/2 (2t)dt= t dt 1+t (u 1)(1) (u+1)(1)
2 (u 1) 20. y=(1+2r) 3 dy=(4x +5)dx . dy= 4(1+2r) du= 2 2
2 du (u 1)
5 5 2dr= 8(1+2r) dr 2 21. (a) y=x +2x dy=(2x+2)dx
(b) When x=3 and
4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials dx= 1
, dy=[2(3)+2]
2 22. (a) y=e x/4 dy= 1
2 =4 . 1 x/4
e dx
4
1 0
e
4 (b) When x=0 and dx=0.1 , dy= 5
dx
2 4+5x
5
5 1
1
(b) When x=0 and dx=0.04 , dy=
(0.04)=
=
=0.05 .
4 25 20
2 4
23. (a) y= 4+5x dy= 24. (a) y=1/(x+1) 1
(4+5x)
2 (0.1)=0.025 . dy= 1
2 1/2 5dx= dx (x+1) (b) When x=1 and dx= 0.01 , dy= 1
2 ( 0.01)= 2
25. (a) y=tan x 1 1
1
=
=0.0025 .
4 100 400 2 dy=sec xdx
2 2 (b) When x= /4 and dx= 0.1 , dy=[sec ( /4)] ( 0.1)=( 2 ) ( 0.1)= 0.2 .
26. (a) y=cos x dy= sin xdx
(b) When x= /3 and dx=0.05 , dy= sin ( /3)(0.05)= 0.5 3 (0.05)= 0.025 3
2 27. y=x , x=1 , x=0.5 28. y= x , x=1 , x=1
1
1
dy=
dx= (1)=0.5
2
2 x 2 0.043 . 2 y=(1.5) 1 =1.25 . dy=2xdx=2(1)(0.5)=1 y= 2 1 = 2 1 0.414 5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials 2 2 29. y=6 x , x= 2 , x=0.4
dy= 2xdx= 2( 2)(0.4)=1.6 30. y= 16
, x=4 ,
x 31. y= f (x)=x 5 x= 1 y= 16
3 16 4
= . dy=
4 3 4 16
2 dx= x 16
2 ( 1)=1 4 4 dy=5x dx . When x=2 and dx=0.001 , dy=5(2) (0.001)=0.08 , so 5 (2.001) = f (2.001)
32. y= f (x)= x
99.8 = f (99.8)
2/3 33. y= f (x)=x
2/3 2 y=(6 ( 1.6) ) (6 ( 2) )=1.44 (8.06) = f (8.06) f (2)+dy=32+0.08=32.08 . dy= 1 dx . When x=100 and dx= 0.2 , dy= 1
( 0.2)= 0.01 , so
2 100 2 x
f (100)+dy=10 0.01=9.99 .
2
2
dy= 3 dx . When x=8 and dx=0.06 , dy= 3 (0.06)=0.02 , so
3 x
3 8
f (8)+dy=4+0.02=4.02 .
2 2 34. y= f (x)=1/x dy=( 1/x )dx . When x=1000 and dx=2,dy=[ 1/(1000) ](2)= 0.000002 , so
1/1002= f (1002) f (1000)+dy=1/1000 0.000002=0.000998
6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials 2 dy=sec xdx . When x=45 and dx= 1 , 35. y= f (x)=tan x
2 2 dy=sec 45 ( /180)=( 2 ) ( /180)= /90 , so tan 44 = f (44 ) f (45 )+dy=1 /90 0.965 . 1
1
dx . When x=1 and dx=0.07 , dy= (0.07)=0.07 , so
x
1
f (1)+dy=0+0.07=0.07. 36. y= f (x)=ln x dy= ln 1.07= f (1.07) / 37. y= f (x)=sec x / f (x)=sec xtan x , so f (0)=1 and f (0)=1 0=0 . The linear approximation of f at / 0 is f (0)+ f (0)(x 0)=1+0(x)=1 . Since 0.08 is close to 0 , approximating sec 0.08 with 1 is
reasonable.
6 / 5 38. If y=x , y =6x and the tangent line approximation at (1,1) has slope 6 . If the change in x is
6 0.01 , the change in y on the tangent line is 0.06 , and approximating (1.01) with 1.06 is reasonable.
/ 39. y= f (x)=ln x / f (x)=1/x , so f (1)=0 and f (1)=1 . The linear approximation of f at 1 is / f (1)+ f (1)(x 1)=0+1(x 1)=x 1 . Now f (1.05)=ln 1.05 1.05 1=0.05 , so the approximation is
reasonable.
2 / 40. (a) f (x)=(x 1) / f (x)=2(x 1) , so f (0)=1 and f (0)= 2 .
/ Thus, f (x) L (x)= f (0)+ f (0)(x 0)=1 2x .
g(x)=e 2x f
/ g (x)= 2e 2x / , so g(0)=1 and g (0)= 2 .
/ Thus, g(x) L (x)=g(0)+g (0)(x 0)=1 2x .
g h(x)=1+ln (1 2x) / h (x)= 2
/
, so h(0)=1 and h (0)= 2 .
1 2x / Thus, h(x) L (x)=h(0)+h (0)(x 0)=1 2x .
h Notice that L =L =L . This happens because f , g , and h have the same function values and the
f g h same derivative values at a=0 . (b)
The linear approximation appears to be the best for the function f since it is closer to f for a larger
domain than it is to g and h . The approximation looks worst for h since h moves away from L faster
7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials than f and g do.
41. (a) If x is the edge length, then V =x 3 2 2 dV =3x dx . When x=30 and dx=0.1 , dV =3(30) (0.1)=270
3 , so the maximum possible error in computing the volume of the cube is about 270 cm . The relative
error is calculated by dividing the change in V , V , by V . We approximate V with dV .
2 V
Relative error =
V dV 3x dx
dx
0.1
=
=3 =3
=0.01 .
3
V
x
30
x
Percentage error = relative error 100%=0.01 100%=1% .
2 (b) S=6x dS=12xdx . When x=30 and dx=0.1 , dS=12(30)(0.1)=36 , so the maximum possible
2 error in computing the surface area of the cube is about 36 cm .
S dS 12xdx
dx
0.1
=
=2 =2
=0.006 .
Relative error =
2
S
S
x
30
6x
Percentage error = relative error 100%=0.006 100%=0.6% .
42. (a) A= r 2 dA=2 r dr . When r=24 and dr=0.2 , dA=2 (24)(0.2)=9.6 , so the maximum
2 possible error in the calculated area of the disk is about 9.6
30 cm .
A dA 2 rdr 2dr 2(0.2) 0.2 1
(b) Relative error =
=
=
=
=
=
=0.016 .
2
A
A
r
24
12 60
r
Percentage error = relative error 100%=0.016 100%=1.6% .
2 43. (a) For a sphere of radius r , the circumference is C=2 r and the surface area is S=4 r , so
2
84
2
2
r=C/(2 ) S=4 (C/2 ) =C /
dS=(2/ )C dC . When C=84 and dC=0.5 , dS= (84)(0.5)=
,
84 so the maximum error is about
4 3 4
(b) V =
r =
3
3
1 dV =
2 2 2 (84) (0.5)= C
2
1764
2 3 C = 6 2 27 cm . Relative error 3 1 dV = 2 2 2
2 ) = 1
56 0.012 2 2 C dC . When C=84 and dC=0.5 , , so the maximum error is about dV
1764/
=
approximately
3
V
(84) /(6 dS 84/
1
= 2 =
S
84
84 / 1764
2 3 179 cm . The relative error is 0.018 . 44. For a hemispherical dome,
8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials 1
(50)=25 m and
2
5
5
2
dr=0.05 cm =0.0005 m, dV =2 (25) (0.0005)=
, so the amount of paint needed is about
8
8 V= 2 3
r
3 2 dV =2 r dr . When r= 2 3 m .
2 V 45. (a) V = r h
(b) The error is dV =2 rhdr=2 rh r V dV =[ (r+ r)2h 2 2 2 r h] 2 rh r= r h+2 rh r+ ( r) h 2 r h 2 rh r 2 = ( r) h
3 dF 4kR dR
dR
46. F=kR dF=4kR dR
=
=4
. Thus, the relative change in F is about 4
4
F
R
kR
times the relative change in R . So a 5% increase in the radius corresponds to a 20% increase in blood
flow.
4 3 dc
dx=0dx=0
dx
d
du
(b) d(cu)= (cu)dx=c
dx=cdu
dx
dx
d
du dv
(c) d(u+v)= (u+v)dx=
+
dx
dx dx
d
dv
du
(d) d(uv)= (uv)dx= u
+v
dx
dx
dx
du
v
u
u
d
u
dx
(e) d
=
dx=
2
v
dx
v
v
d n
n
n 1
(f) d(x )= (x )dx=nx dx
dx
47. (a) dc= 48. (a) f (x)=sin x
f (x)
(b) / du
dv
dx+
dx=du+dv
dx
dx
dv
du
dx=u
dx+v
dx=udv+vdu
dx
dx
dv
du
dv
v
dx u
dx
dx
dx
dx
vdu udv
dx=
=
2
2
v
v
dx= / f (x)=cos x , so f (0)=0 and f (0)=1 . Thus, / f (0)+ f (0)(x 0)=0+1(x 0)=x . 9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials We want to know the values of x for which y=x approximates y=sin x with less than a 2% difference;
that is, the values of x for which
x sin x
x sin x
<0.02
0.02<
<0.02
sin x
sin x { 0.02sin x<x sin x<0.02sin x
0.02sin x>x sin x>0.02sin x if sin x>0
if sin x<0 { 0.98sin x<x<1.02sin x
1.02sin x<x<0.98sin x if sin x>0
if sin x<0 In the first figure, we see that the graphs are very close to each other near x=0 . Changing the viewing
rectangle and using an intersect feature (see the second figure) we find that y=x intersects y=1.02sin x
at x 0.344 . By symmetry, they also intersect at x 0.344 (see the third figure.). Converting 0.344
180 radians to degrees, we get 0.344
/ 19.7 20 , which verifies the statement.
/ 49. (a) The graph shows that f (1)=2 , so L(x)= f (1)+ f (1)(x 1)=5+2(x 1)=2x+3 .
f (0.9) L(0.9)=4.8 and f (1.1) L(1.1)=5.2 .
/ (b) From the graph, we see that f (x) is positive and decreasing. This means that the slopes of the
tangent lines are positive, but the tangents are becoming less steep. So the tangent lines lie above the
curve. Thus, the estimates in part (a) are too large.
10 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials / 2 50. (a) g (x)= x +5 / / g (2)= 9 =3 . g(1.95) g(2)+g (2)(1.95 2)= 4+3( 0.05)= 4.15 . / g(2.05) g(2)+g (2)(2.05 2)= 4+3(0.05)= 3.85 .
/ 2 / (b) The formula g (x)= x +5 shows that g (x) is positive and increasing. This means that the slopes
of the tangent lines are positive and the tangents are getting steeper. So the tangent lines lie below the
graph of g . Hence, the estimates in part (a) are too small. 11 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules 1. Product Rule:
2 3 y=(x +1)(x +1)
/ 2 2 3 4 2 4 4 2 y =(x +1)(3x )+(x +1)(2x)=3x +3x +2x +2x=5x +3x +2x .
2 3 5 3 2 / 2. Quotient Rule: F(x)=
1/2 /
F (x) = x x 3x x
x 9 1/2
x
2 1 = x 3x 4 2 y =5x +3x +2x (equivalent). Multiplying first: y=(x +1)(x +1)=x +x +x +1
3/2 1/2 x 1
x
2 ( x 3x3/2)
( x1/2) 2 1/2 9
1 1/2 3
1 1/2
x
x + x
x 3x
=
2
2
2
2
1 1/2
=
= x
3
x
x
2
x 3x x
1 1/2
1/2
/
= x 3x=x 3x F (x)= x
Simplifying first: F(x)=
3 (equivalent).
2
x
For this problem, simplifying first seems to be the better method.
1/2 x f (x)=x d x x d 2 2 x x
x
(e )+e
(x )=x e +e (2x)=xe (x+2) .
dx
dx 1/2 x / / 2 x 3. By the Product Rule, f (x)=x e x 2 4. By the Product Rule, g(x)= x e =x e 5. By the Quotient Rule, y= 2 x e 2 x / y = d x x d 2
(e ) e
(x )
dx
dx
22 x e
6. By the Quotient Rule, y=
1+x
7. g(x)= 8. f (t)= 3x 1
2x+1 QR 2t
4+t 2 1
x
2 x / 2 x = y = x x 2 x = = x e +xe e
2 1
x
2 x x (1+x)e e (1) 1/2 x (e ) e (2x) (x ) x QR 1/2 x g (x)=x (e )+e x = 4 1/2 x e (2x+1) . x = xe (x 2)
x xe 4 x = e (x 2)
x 3 x
2 . (1+x)
(x+1)
(x+1)
/
(2x+1)(3) (3x 1)(2) 6x+3 6x+2
5
g (x)=
=
=
2
2
2
(2x+1)
(2x+1)
(2x+1)
/ f (t)= 2 (4+t )(2) (2t)(2t)
22 (4+t ) 2 = 8+2t 4t
22 (4+t ) 2 = 8 2t 2 22 (4+t ) 1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules 3 PR 4 9. V (x)=(2x +3)(x 2x)
/ 3 3 4 2 6 3 6 3 6 3 V (x)=(2x +3)(4x 2)+(x 2x)(6x )=(8x +8x 6)+(6x 12x )=14x 4x 6
2 3 5 PR 2 10. Y (u)=(u +u )(u 2u )
/ 2 3 4 5 2 3 Y (u) =(u +u )(5u 4u)+(u 2u )( 2u
2 1 2 2 4 3u )
1 2 2 =(5u 4u +5u 4u )+( 2u 3u+4u +6u )=3u +2u+2u
1 11. F(y)= 3 2 4 y ( 3 2 (y+5y )= y 4 3y ) ( y+5y3) 2 PR y ( 2 3y 4) ( 1+15y2) + ( y+5y3) ( 2y 3+12y 5)
2
4
2
2
4
2
= ( y +15 3y 45y ) +( 2y +12y 10+60y ) / F (y) = y 2 4 2 4 =5+14y +9y or 5+14/y +9/y
t 12. R(t)=(t+e )(3 t )=
1 1/2
t
t
/
R (t) =(t+e )( 2 t )+(3 t )(1+e )
1 1/2 1 1/2 t
t
=
t
t e +(3+3e t
2
2
t 13. y= 2 t t t e )=3+3e 3
t
2 t t t e e /(2 t ) QR 2 3t 2t+1
2 / y = 2 (3t 2t+1)(2t) t (6t 2)
2 2 2 = 2t[3t 2t+1 t(3t 1)]
2 (3t 2t+1)
2 = 2 2t (3t 2t+1 3t +t)
2 2 2t(1 t) = 2 (3t 2t+1)
3 14. y= t +t
4 QR / 2 (3t 2t+1)
4 y = 2 (3t 2t+1) 2 3 3 (t 2)(3t +1) (t +t)(4t )
4 t 2 2 6 = (t 2)
6 = 4 2 t 3t 6t 2
4 2 (t 2) 2 6 4 4 2 (t 2)
6 = 4 (3t +t 6t 2) (4t +4t ) 4 2 4 2 t +3t +6t +2
(t 2) 2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules 2 r / 2 r r r 2 r 2 15. y=(r 2r)e = y =(r 2r)(e )+e (2r 2)=e (r 2r+2r 2)=e (r 2) 16. y= 1 s s+ke s s (s+ke )(0) (1)(1+ke ) / =y = s2 1+ke = s s2 (s+ke ) (s+ke ) 3 v 2v v 2
2
1/2
17. y=
=v 2 v =v 2v
v 1
2 / y =2v 2 We can change the form of the answer as follows: 2v v
3/2 w 5/2 18. z=w (w+ce )=w +cw 19. y= 1
4 4 / y = 2 2 4 1/2 2v v 1 2v 1
1
=
=
v
v
v 3/2 .
3/2 =2v w e +e w 3 1/2
5 3/2 1 1/2 w
w
= w + cw e (2w+3)
2
2
2 2 = 2x(2x +1)
4 2 2 (x +x +1)
( x 1) 1
2 x = 2 1
1
+
2 2 x 1
1
+
2 2 x / f (x)= 2 (cx+d)(a) (ax+b)(c)
(cx+d) / y = (x+1)(2) (2x)(1)
2 = = acx+ad acx bc
2 (cx+d)
2 / 2 (x+1) 2 . At (1,1) , y = = x ( x +1) 2 2 2 1 = ( x +1) x 2c/x
(x+c/x)(1) x(1 c/x ) x+c/x x+c/x
=
f (x)=
=
2
2
2
2
c 2
(x +c)
x +c
x+
x
2
x
x
/ (x+1)
1
1
1
is y 1= (x 1) , or y= x+ .
2
2
2
24. y= 1/2 =2v v ( x +1) ax+b
cx+d 2x
x+1 2 2 x / y = x
21. f (x)=
x+c/x 23. y= 2 1 ( x +1) x +1 w 3 (x +x +1)(0) 1(4x +2x)
(x +x +1) x 1 22. f (x)= z = e x +x +1 20. y= 5 3/2
w +c
2 / 3/2 w 1/2 v 2 x = 2cx
2 2 (x +c) ad bc
2 (cx+d) 1
, and an equation of the tangent line
2 x
x+1
3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules 1 (x+1) x (1) 2 x / y = = 2 (x+1) (2x) 1 x = 2 / . At (4,0.4) , y = 2 (x+1)
2 x (x+1) 2 x (x+1)
equation of the tangent line is y 0.4= 0.03(x 4) , or y= 0.03x+0.52 .
x / x x x / 3
= 0.03 , and an
100 0 25. y=2xe
y =2(x e +e 1)=2e (x+1) . At (0, 0) , y =2e (0+1)=2 1 1=2 , and an equation of the
tangent line is y 0=2(x 0) , or y=2x .
x e
26. y=
x x / y = x x e e 1
2 x = e (x 1)
2 x / . At (1, e) , y =0 , and an equation of the tangent line is x y e=0(x 1) , or y=e .
1 27. (a) y= f (x)= 2 / f (x)= 2 (1+x )(0) 1(2x)
22 1+x point 2x = 22 . So the slope of the tangent line at the (1+x )
(1+x )
1
1 1
1
/
2
is f ( 1)= = and its equation is y
= (x+1) or y= x+1 .
2 2
2 2
2
2 1
1,
2 (b)
x 28. (a) y= f (x)= 2 / f (x)= 2 (1+x )1 x(2x)
22 1+x 2 = (1+x ) / point (3,0.3) is f (3)= 1 x 22 . So the slope of the tangent line at the (1+x ) 8
and its equation is y 0.3= 0.08(x 3) or y= 0.08x+0.54 .
100 (b) 29. (a) f (x)= e
x x
3 / f (x)= 3 x x 2 x (e ) e (3x )
32 (x ) 2 x = x e (x 3)
x 6 x = e (x 3)
x 4 4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules (b)
/ f =0 when f has a horizontal tangent line, f
when f is increasing.
x 30. f (x)= 2 (x 1)1 x(2x) / f (x)= 2 2 x 1 2 / is negative when f is decreasing, and f 2 2 (x 1) is positive 2 x 1 = / = 2 (x 1) x +1
2 2 Notice that the slopes of all tangents to (x 1) / f are negative and f (x)<0 always. / / 31. We are given that f (5)=1 , f (5)=6 , g(5)= 3 , and g (5)=2 .
/ / / (a) ( fg) (5)= f (5)g (5)+g(5) f (5)=(1)(2)+( 3)(6)=2 18= 16
(b) (c) f
g
g
f / / / (5)= / g(5) f (5) f (5)g (5)
2 = ( 3)(6) (1)(2)
2 [g(5)] ( 3) / (5)= = / f (5)g (5) g(5) f (5)
2 = (1)(2) ( 3)(6)
2 [ f (5)] 20
9 =20 (1)
/ / 32. We are given that f (3)=4 , g(3)=2 , f (3)= 6 , and g (3)=5 .
/ / / (a) ( f +g ) (3)= f (3)+g (3)= 6+5= 1
/ / / (b) ( fg ) (3)= f (3)g (3)+g(3) f (3)=(4)(5)+(2)( 6)=20 12=8
(c) f
g / / (3)= / g(3) f (3) f (3)g (3)
2 [g(3)] = (2)( 6) (4)(5)
2 (2) = 32
= 8
4 (d) 5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules / f
f g / (3) = / 2 [ f (3) g(3)]
(4 2)( 6) 4( 6 5)
12+44
=
=
=8
2
2
(4 2)
2
x / 33. f (x)=e g(x)
/ / [ f (3) g(3)] f (3) f (3)[ f (3) g (3)] 0 x / x x / f (x)=e g (x)+g(x)e =e g (x)+g(x) . / f (0)=e g (0)+g(0) =1(5+2)=7
d
34.
dx / h(x)
x d
dx xh (x) h(x) 1 = 2 x h(x)
x / = 2h (2) h(2)
2 x=2 2 = 2( 3) (4)
10
=
= 2.5
4
4 35. (a) From the graphs of f and g , we obtain the following values: f (1)=2 since the point (1,2) is on
/ the graph of f ; g(1)=1 since the point (1,1) is on the graph of g ; f (1)=2 since the slope of the line
4 0
/
segment between (0,0) and (2,4) is
=2 ; g (1)= 1 since the slope of the line segment between
2 0
0 4
/
/
/
( 2,4) and (2,0) is
= 1 . Now u(x)= f (x)g(x) , so u (1)= f (1)g (1)+g(1) f (1)=2 ( 1)+1 2=0 .
2 ( 2)
8
1
2
/
/
2
3
3
2
/
g(5) f (5) f (5)g (5)
3
3
(b) v(x)= f (x)/g(x) , so v (5)=
=
=
=
2
2
4
3
[g(5)]
2
/ / / 2
3
+2 0= .
4
2
1
2
1
5
4
3 36. (a) P(x)=F(x)G(x) , so P (2)=F(2)G (2)+G(2)F (2)=3
/ / (b) Q(x)=F(x)/G(x) , so Q (7)= / G(7)F (7) F(7)G (7)
2 [G(7)]
/ 37. (a) y=xg(x)
x
(b) y=
g(x)
g(x)
(c) y=
x / = 2 1 = 1 10 43
+
=
4 3 12 / y =xg (x)+g(x) 1=xg (x)+g(x)
/ y = / g(x) 1 xg (x)
2 / = g(x) xg (x)
2 [g(x)]
/ y = [g(x)] xg (x) g(x) 1 xg (x) g(x) / 2 (x) / = 2 x 38. (a)
6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules 2 / y=x f (x)
(b) y= 2 / / x f (x) f (x)(2x) y =x f (x)+ f (x)(2x) f (x) 2 y = 2 / 22 x 2 2 / y = x 3 / f (x)(2x) x f (x)
2 [ f (x)] 1+xf (x)
x
/ x[xf (x)+ f (x)] [1+xf (x)] / = (x ) x
(c) y=
f (x)
(d) y= / xf (x) 2 f (x) y = 1
2 x 2 ( x)
x 3/2 / 1/2 f (x)+x 1
x
2 f (x) = 1/2 1 1/2
x f (x)
1/2
2
2x x 1/2 2 = / xf (x)+2x f (x) 1 2x 2x 3/2 39. If P(t) denotes the population at time t and A(t) the average annual income, then T (t)=P(t)A(t) is
/ / / the total personal income. The rate at which T (t) is rising is given by T (t)=P(t)A (t)+A(t)P (t)
/ / / T (1999) =P(1999)A (1999)+A(1999)P (1999)=( 961,400 )( $1400 / yr )+( $30,593 )( 9200 / yr )
=$ 1,345,960,000 / yr +$ 281,455,600 / yr =$ 1,627,415,600 / yr
So the total personal income was rising by about $ 1.627 billion per year in 1999.
/ The term P(t)A (t) $ 1.346 billion represents the portion of the rate of change of total income due to
/ the existing population’s increasing income. The term A(t)P (t) $ 281 million represents the portion
of the rate of change of total income due to increasing population.
40. (a) f (20)=10 , 000 means that when the price of the fabric is $20/ yard, 10 , 000 yards will be
sold.
/ f (20)= 350 means that as the price of the fabric increases past $20/ yard, the amount of fabric
which will be sold is decreasing at a rate of 350 yards per (dollar per yard).
/ / / / (b) R( p)= pf ( p) R ( p)= pf ( p)+ f ( p) 1 R (20)=20 f (20)+ f (20) 1=20( 350)+10,000=3000.
This means that as the price of the fabric increases past $20/ yard, the total revenue is increasing at
$3000/($/yard). Note that the Product Rule indicates that we will lose $7000/($/yard) due to selling
less fabric, but that that loss is more than made up for by the additional revenue due to the increase in
price.
7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules x
/
(x+1)(1) x(1)
1
, then f (x)=
=
. When x=a , the equation of the tangent
2
2
x+1
(x+1)
(x+1)
a
a
1
1
line is y
=
(x a) . This line passes through (1,2) when 2
=
(1 a)
2
2
a+1
a+1
(a+1)
(a+1)
41. If y= f (x)= 2 2 2(a+1) a(a+1)=1 a 2 2a +4a+2 a a 1+a=0 2 a +4a+1=0 . The quadratic formula gives the roots of this equation as a= 2 4 12
4 4(1)(1)
=
= 2
2(1)
2 4 3 , so there are two such tangent lines. Since
f( 2 2 3) = 2 3
3 +1 = 2 3 1 3 1 3 1 3 2 2 3
3 3
1 3 1 3
=
=
,
1 3
2
2
1 3
the lines touch the curve at A 2+ 3,
( 0.27, 0.37) and
2
1+ 3
B 2 3,
( 3.73,1.37) .
2
x 1
/ (x+1)(1) (x 1)(1)
2
42. y=
y =
=
. If the tangent intersects the curve when x=a ,
2
2
x+1
(x+1)
(x+1)
1
2
then its slope is 2/(a+1) . But if the tangent is parallel to x 2y=2 , that is, y= x 1 , then its slope is
2
1
1
2
2
. Thus,
=
(a+1) =4 a+1= 2 a=1 or 3 . When a=1 , y=0 and the equation of the
2 2
2
(a+1)
1
1
1
tangent is y 0= (x 1) or y= x
.
2
2
2
1
1
7
When a= 3 , y=2 and the equation of the tangent is y 2= (x+3) or y= x+ .
2
2
2
= / / / / / / / / / 43. (a) ( fgh) =[( fg)h] =( fg) h+( fg)h =( f g+ fg )h+( fg)h = f gh+ fg h+ fgh /
8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.2 The Product and Quotient Rules (b) Putting f =g=h in part (a) , we have
d
3
/
/
/
/
/
2 /
[ f (x)] = ( fff ) = f ff + ff f + fff =3 fff =3[ f (x)] f (x).
dx (c) d 3x d x 3
x2 x
2x x
3x
(e )= (e ) =3(e ) e =3e e =3e
dx
dx 44. (a)
d
dx d
d
(1) 1
[g(x)]
dx
dx g(x) 1
g(x) = [Quotient Rule] 2 [g(x)]
/ = g(x) 0 1 g (x)
2 / = [g(x)] (b) y= 1
4 2 2 y = 2 2 (x +x +1)
1
x n = 2 [g(x)]
2 4x +2x
4 / g (x) [g(x)]
3 / x +x +1
d
d
n
(x )=
(c)
dx
dx 0 g (x) or 2x(2x +1)
4 2 2 (x +x +1) n / = (x ) n2 (x ) [by the Reciprocal Rule] = nx n 1 2n = nx n 1 2n = nx n 1 x 9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences 2 / 1. (a) s= f (t)=t 10t+12 v(t)= f (t)=2t 10
(b) v(3)=2(3) 10= 4 ft / s
(c) The particle is at rest when v(t)=0 2t 10=0 t=5 s.
(d) The particle is moving in the positive direction when v(t)>0 2t 10>0 2t>10 t>5 .
(e) Since the particle is moving in the positive direction and in the negative direction, we need to
calculate the distance traveled in the intervals [0,5] and [5,8] separately. | f (5) f (0)|=| 13 12|=25 ft
and | f (8) f (5)|=|{ 4 ( 13)|=9 ft. The total distance traveled during the first 8 s is 25+9=34 ft. (f)
3 2 2. (a) s= f (t)=t 9t +15t+10 / 2 v(t)= f (t)=3t 18t+15=3(t 1)(t 5) (b) v(3)=3(2)( 2)= 12 ft / s
(c) v(t)=0 t=1 s or 5 s
(d) v(t)>0 0 t<1 or t>5
(e) | f (1) f (0)|=|17 10|=7 , | f (5) f (1)|=| 15 17|=32 , and | f (8) f (5)|=|66 ( 15)|=81 . Total distance
=7+32+81=120 ft. (f)
3 2 2 / 3. (a) s= f (t)=t 12t +36t v(t)= f (t)=3t 24t+36
(b) v(3)=27 72+36= 9 ft / s
2 (c) The particle is at rest when v(t)=0 . 3t 24t+36=0 3(t 2)(t 6)=0 t=2 s or 6 s.
(d) The particle is moving in the positive direction when v(t)>0 . 3(t 2)(t 6)>0 0 t<2 or t>6 .
(e) Since the particle is moving in the positive direction and in the negative direction, we need to
calculate the distance traveled in the intervals (0,2) , (2,6) , and [6,8] separately.
| f (2) f (0)|=|32 0|=32 .
| f (6) f (2)|=|0 32|=32 .
| f (8) f (6)|=|32 0|=32 .
The total distance is 32+32+32=96 ft.
(f) 1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences 4 / 4. (a) s= f (t)=t 4t+1 3 v(t)= f (t)=4t 4 3 (b) v(3)=4(3) 4=104 ft / s
3 2 (c) It is at rest when v(t)=4(t 1)=4(t 1)(t +t+1)=0 t=1 s. 3 (d) It moves in the positive direction when 4(t 1)>0 t>1 .
(e) Distance in positive direction =| f (8) f (1)|=|4065 ( 2)|=4067 ft
Distance in negative direction =| f (1) f (0)|=| 2 1|=3 ft
Total distance traveled =4067+3=4070 ft (f)
t 5. (a) s= / 2 v(t)=s (t)= 2 (t +1)(1) t(2t)
2 t +1 (t +1)
2 (b) v(3)= 1 (3)
2 2 2 (3 +1) = 1 9
2 10 (c) It is at rest when v=0 = 1 t = 2 2
2 (t +1) 8
2
=
ft / s
100
25
2 1 t =0 t=1 s [ t (d) It moves in the positive direction when v>0 1 since t
2 0 ].
2 1 t >0 t <1 0 t<1 .
1
1
(e) Distance in positive direction =|s(1) s(0)|=
0 = ft
2
2
8 1
49
Distance in negative direction =|s(8) s(1)|=
=
ft
65 2
130
1 49 57
Total distance traveled = +
=
ft
2 130 65 (f)
2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences 2 5/2 3/2 1/2 35t +90t
6. (a) s= t (3t 35t+90)=3t
15 3/2 105 1/2
15
/
1/2 15 1/2 2
v(t)=s (t)=
t
t +45t =
t (t 7t+6)=
(t 1)(t 6)
2
2
2
2 t
15
(b) v(3)=
(2)( 3)= 15 3 ft / s
2 3
(c) It is at rest when v=0 t=1 s or 6 s.
(d) It moves in the positive direction when v>0 (t 1)(t 6)>0 0 t<1 or t>6.
(e)
Distance in positive direction =|s(1) s(0)|+|s(8) s(6)|=|58 0|+|4 2 ( 12 6 )
=58+4 2 +12 6 93.05 ft
Distance in negative direction =|s(6) s(1)|=| 12 6 58|=58+12 6 87.39 ft
Total distance traveled =58+4 2 +12 6 +58+12 6 =116+4 2 +24 6 180.44 ft (f)
3 2 / 2 2 7. s(t)=t 4.5t 7t v(t)=s (t)=3t 9t 7=5 3t 9t 12=0
, the particle reaches a velocity of 5 m / s at t=4 s.
8. (a) s=5t+3t
(b) v(t)=35 3(t 4)(t+1)=0 t=4 or 1 . Since t 0 ds
=5+6t , so v(2)=5+6(2)=17 m / s.
dt
5+6t=35 6t=30 t=5 s. 2 v(t)= dh
=10 1.66t , so v(3)=10 1.66(3)=5.02 m / s.
dt
10 17
2
2
(b) h=25 10t 0.83t =25 0.83t 10t+25=0 t=
3.54 or 8.51 .
1.66
The value t =(10 17 )/1.66 corresponds to the time it takes for the stone to rise 25 m and
9. (a) h=10t 0.83t 2 v(t)= 1 t =(10+ 17 )/1.66 corresponds to the time when the stone is 25 m high on the way down. Thus,
2 v(t )=10 1.66[(10
1 17 )/1.66]= 17 4.12 m / s.
/ 10. (a) At maximum height the velocity of the ball is 0 ft / s. v(t)=s (t)=80 32t=0 32t=80 t= 5
.
2 So the maximum height is
3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences 5
2 s 5
2 =80 16 5
2 2 2 =200 100=100 ft. 2 2 (b) s(t)=80t 16t =96 16t 80t+96=0 16(t 5t+6)=0 16(t 3)(t 2)=0 .
So the ball has a height of 96 ft on the way up at t=2 and on the way down at t=3 . At these times the
velocities are v(2)=80 32(2)=16 ft / s and v(3)=80 32(3)= 16 ft / s, respectively.
2 / / 2 11. (a) A(x)=x
A (x)=2x . A (15)=30 mm / mm is the rate at which the area is increasing with
respect to the side length as x reaches 15 mm.
1
1
(4x)= P(x) . The figure suggests that if x is small,
2
2
then the change in the area of the square is approximately half of its perimeter ( 2 of the 4 sides)
/ (b) The perimeter is P(x)=4x , so A (x)=2x=
times x . From the figure,
A/ x 2x . 2 A=2x( x)+( x) . If dV
dV
2
=3x .
dx
dx
increasing as x increases past 3 mm.
12. (a) V (x)=x 3 x is small, then 2 A 2x( x) and so 3 =3(3) =27 mm / mm is the rate at which the volume is x=3 1
2 1
(6x )= S(x) . The figure suggests that if x is
2
2
small, then the change in the volume of the cube is approximately half of its surface area (the area of
2 / 2 (b) The surface area is S(x)=6x , so V (x)=3x =
3 of the 6 faces) times
V 2 3x ( x) and so x . From the figure, 2 2 3 V =3x ( x)+3x( x) +( x) . If x is small, then 2 V / x 3x . 4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences 13. (a)
(i)
A(3) A(2) 9 4
=
=5
3 2
1
(ii) A(2.5) A(2) 6.25 4
=
=4.5
2.5 2
0.5
(iii) A(2.1) A(2) 4.41 4
=
=4.1
2.1 2
0.1
(b) A(r)= r 2 / / A (r)=2 r , so A (2)=4 .
/ (c) The circumference is C(r)=2 r=A (r) . The figure suggests that if r is small, then the change in
the area of the circle (a ring around the outside) is approximately equal to its circumference times r
. Straightening out this ring gives us a shape that is approximately rectangular with length 2 r and
2 2 2 width r , so A 2 r( r) . Algebraically, A=A(r+ r) A(r)= (r+ r) r =2 r( r)+ ( r) .
So we see that if r is small, then A 2 r( r) and therefore, A/ r 2 r . / 2 14. (a) A (1)=7200 cm / s
/ 2 / 2 (b) A (3)=21 , 600 cm / s
(c) A (5)=36 , 000 cm / s
/ 2 15. (a) S (1)=8 ft / ft
/ 2 / 2 (b) S (2)=16 ft / ft
(c) S (3)=24 ft / ft
5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences 16. (a)
(a)
4
4
(512)
(125)
V (8) V (5) 3
3
=
=172
8 5
3
(b) (c) (d) (e) (f) 4
4
(512)
(125)
V (8) V (5) 3
3
=
=172
8 5
3 3 m / m 3 m / m 4
4
(216)
(125)
V (6) V (5) 3
3
=
=121.3
6 5
1
4
4
(216)
(125)
V (6) V (5) 3
3
=
=121.3
6 5
1 3 m / m 4
3 4
3
(5.1)
(5)
V (5.1) V (5) 3
3
=
=102.013
5.1 5
0.1
4
3 4
3
(5.1)
(5)
V (5.1) V (5) 3
3
=
=102.013
5.1 5
0.1
/ 2 / 3 m / m 3 m / m 3 m / m 3 (b) V (r)=4 r , so V (5)=100
m / m.
4 3
/
2
(c) V (r)=
V (r)=4 r =S(r) . By analogy with Exercise 13(c) , we can say that the change in
r
3
the volume of the spherical shell, V , is approximately equal to its thickness, r , times the surface
area of the inner sphere. Thus, V 2 4 r ( r) and so 2 V/ r 4 r . 17. (a) (1)=6 kg / m
(b) (2)=12 kg / m
(c) (3)=18 kg / m
5
= 218.75 gal / min
40
10
/
(b) V (10)= 250 1
= 187.5 gal / min
40
20
/
(c) V (20)= 250 1
= 125 gal / min
40
(d)
/ 18. (a) V (5)= 250 1 6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences / V (40)= 250 1 / 40
40 =0 gal / min
2 19. (a) Q (0.5)=3(0.5) 4(0.5)+6=4.75 A
/ 2 (b) Q (1)=3(1) 4(1)+6=5 A
dF
3
2GmM
= 2(GmM)r =
, which is the rate of change of the
2
3
dr
r
r
force with respect to the distance between the bodies. The minus sign indicates that as the distance r
between the bodies increases, the magnitude of the force F exerted by the body of mass m on the
body of mass M is decreasing.
/
/
3
2GmM
GmM=20 , 000 .
(b) Given F (20 , 000)= 2 , find F (10 , 000) . 2=
3
20,000
20. (a) F= / GmM F (10,000)= =(GmM)r ( 3 2 20,000 3 2 ) = 2 23= 16 N / km 10,000 21. (a) To find the rate of change of volume with respect to pressure, we first solve for V in terms of
P.
C
dV
C
PV =C V =
=
.
2
P
dP
P
(b) From the formula for dV /dP in part (a), we see that as P increases, the absolute value of dV /dP
decreases. Thus, the volume is decreasing more rapidly at the beginning.
1 dV
1
C
C
1
C
(c) =
=
=
=
=
2
V dP
V
(PV )P CP P
P
22. (a) . C(6) C(2)
0.0295 0.0570
=
6 2
4
(i) . = 0.006875 ( moles/L ) / min
C(4) C(2)
0.0408 0.0570
=
4 2
2
= 0.008 ( moles/L ) / min 7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences . C(2) C(0)
0.0570 0.0800
=
2 0
2 (iii) = 0.0115 ( moles/L ) / min
C
t (b) Slope = 0.077
7.8 0.01 ( moles/L ) / min 1860 1750 110
2070 1860 210
=
=11 , m =
=
=21 ,
1 1920 1910
2 1930 1920
10
10
(m +m )2=(11+21)/2=16 million / year
23. (a) 1920: m =
1 2 4450 3710 740
5280 4450 830
=
=74 , m =
=
=83 ,
1 1980 1970
2 1990 1980
10
10
(m +m )2=(74+83)/2=78.5 million / year
1980: m =
1 2 3 2 3 2 (b) P(t)=at +bt +ct+d (in millions of people), where a 0.0012937063 , b
c 12,822.97902 , and d 7,743,770.396 .
/ (c) P(t)=at +bt +ct+d
(d) 7.061421911 , 2 P (t)=3at +2bt+c (in millions of people per year) / 2 P (1920) =3(0.0012937063)(1920) +2( 7.061421911)(1920)+12,822.97902
14.48 million / year
/ 75.29 million / year (smaller, but close) P (1980)
/ (e) P (1985) 81.62 million / year, so the rate of growth in 1985 was about 81.62 million / year.
4 3 2 6 24. (a) A(t)=at +bt +ct +dt+e , where a= 5.8275058275396 10 , b=0.0460458430461 ,
c= 136.43277039706 , d=179 , 661.02676871 , and e= 88 , 717 , 597.060767 .
4 3 2 (b) A(t)=at +bt +ct +dt+e
(c) / 3 2 A (t)=4at +3bt +2ct+d
8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences / A (1990) 0.0833 years of age per year (d)
2 a kt
25. (a) C =
akt+1 rate of reaction
2 2 2 2 d C
(akt+1)(a k) (a kt)(ak) a k(akt+1 akt)
ak
=
=
=
=
2
2
2
dt
(akt+1)
(akt+1)
(akt+1)
2 2 2 a kt
a kt+a a kt
a
(b) If x= C , then a x=a
=
=
.
akt+1
akt+1
akt+1
2 So k(a x) =k a
akt+1 2 = 2 ak
2 (akt+1) = d C
dt = dx
.
dt
2 26. (a) After an hour the population is n(1)=3 500 ; after two hours it is n(2)=3(3 500)=3 500 ; after
2 3 4 three hours, n(3)=3(3 500)=3 500 ; after four hours, n(4)=3 500 . From this pattern, we see that the
t t population after t hours is n(t)=3 500=500 3 .
d x
t
x
(b) From (5) in Section 3.1, we have
(3 ) (1.10)3 . Thus, for n(t)=500 3 ,
dx
dn
d t
dn
t
6
=500 (3 ) 500(1.10)3
500(1.10)3 400 , 950 bacteria / hour.
dt
dt
dt t=6
27. (a) Using v= P
2 2
(R r with R=0.01 , l=3 , P=3000 , and =0.027 , we have v as a function of r :
4 l 3000
2 2
(0.01 r ) . v(0)=0.925 cm / s, v(0.005)=0.694 cm / s, v(0.01)=0 .
4(0.027)3
P
P
Pr
2 2
/
(R r ) v (r)=
( 2r)=
. When l=3 , P=3000 , and =0.027 , we have
(b) v(r)=
4 l
4 l
2 l v(r)= 9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences 3000r
/
/
/
. v (0)=0 , v (0.005)= 92.592(cm/s)/ cm, and v (0.01)= 185.185(cm/s)/ cm.
2(0.027)3
(c) The velocity is greatest where r=0 (at the center) and the velocity is changing most where
r=R=0.01 cm (at the edge).
/ v (r)= 28. (a)
(a) 1
2L T 1
2L T f= 1
2L T f= 1
2L T f= 1
2L T 1
2L T f=
(b)
f=
(c) (d) (e) (f)
f= (b)
(i) = = 1
2 T 1
2 T L L = 1
2L T = 1
2L T = T
2L
T
2L 1/2 1
2 T df
=
dL 1 1/2 = df
=
dL 1 1
2 T T 1 2 L = 2 2L T 1 2 L = 2 2L 1/2 df 1
=
dT 2 1
2L T 1/2 df 1
=
dT 2 1
2L 1/2 T 1
4L T 1/2 1
4L T = = df
1
=
d
2 T
2L 3/2 df
1
=
d
2 T
2L 3/2 T =
4L T =
4L df
<0 and L is decreasing
dL f is increasing df
>0 and T is increasing
dT f is increasing higher note (iii) df
<0 and
d f is decreasing 3/2 higher note (ii) 3/2 lower note is increasing
2 29. (a) C(x)=2000+3x+0.01x +0.0002x
/ 3 / 2 C (x)=3+0.02x+0.0006x
/ (b) C (100)=3+0.02(100)+0.0006(10 , 000)=3+2+6=$11/ pair. C (100) is the rate at which the cost
is increasing as the 100 th pair of jeans is produced. It predicts the cost of the 101 st pair.
(c) The cost of manufacturing the 101 st pair of jeans is
10 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences C(101) C(100) = (2000+303+102.01+206.0602) (2000+300+100+200)
= 11.0702 $11.07
2 3 / 2 30. (a) C(x)=84+0.16x 0.0006x +0.000003x C (x)=0.16 0.0012x+0.000009x
This is the rate at which the cost is increasing as the 100 th item is produced.
(b) C(101) C(100)=97.13030299 97 $0.13 .
p(x)
31. (a) A(x)=
x / / A (x)= xp (x) p(x) 1
2 / = xp (x) p(x)
2 / . A (x)>0 / C (100)=0.13 . A(x) is increasing; that is, x
x
the average productivity increases as the size of the workforce increases.
/ (b) p (x) is greater than the average productivity
/ xp (x)> p(x) / / xp (x) p(x)>0 xp (x) p(x)
2 >0 / p (x)>A(x) / p (x)> p(x)
x / A (x)>0 . x
32. (a) ( 1+4x
dR
=
S=
dx
= 9.6x 0.4 0.6 ) ( 9.6x 0.6) ( 40+24x0.4) ( 1.6x 0.6)
( 1+4x0.4) 2 +38.4x 0.2 64x 0.6 0.4 2 ( 1+4x ) 38.4x 0.2 = 54.4x 0.6 ( 1+4x0.4) 2 (b)
At low levels of brightness, R is quite large and is quickly decreasing, that is, S is negative with large
absolute value. This is to be expected: at low levels of brightness, the eye is more sensitive to slight
changes than it is at higher levels of brightness.
PV
PV
1
=
=
(PV ) . Using the Product Rule, we have
nR (10)(0.0821) 0.821
dT
1
1
/
/
=
[P(t)V (t)+V (t)P (t)]=
[(8)( 0.15)+(10)(0.10)] 0.2436 K / min.
dt 0.821
0.821 33. PV =nRT T= 11 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.3 Rates of Change in the Natural and Social Sciences 34. (a) If dP/dt=0 , the population is stable (it is constant).
dP
P
P
P
(b)
=0
P=r
1
P
=1
=1
0
dt
P
r
P
P
r
c 0 c c P=P 0 (c) If =0.05 , then P=10 , 000 1 5
5 r 0 If P =10 , 000 , r =5%=0.05 , and =4%=0.04 , then P=10 , 000
c 1 c 1 4
5 .
0 =2000 . =0 . There is no stable population. 35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is,
dC
dW
=0 and
=0 .
dt
dt
(b) ‘‘The caribou go extinct’’ means that the population is zero, or mathematically, C=0 .
dC
dW
(c) We have the equations
=aC bCW and
= cW +dCW . Let dC/dt=dW /dt=0 , a=0.05 ,
dt
dt
b=0.001 , c=0.05 , and d=0.0001 to obtain 0.05C 0.001CW =0 (1) and 0 .05W +0.0001CW =0 (2) .
Adding 10 times (2) to (1) eliminates the CW terms and gives us 0.05C 0.5W =0 C=10W .
Substituting C=10W into (1) results in
2 2 0.05(10W ) 0.001(10W )W =0 0.5W 0.01W =0 50W W =0 W (50 W )=0 W =0 or 50 . Since
C=10W , C=0 or 500 . Thus, the population pairs (C,W) that lead to stable populations are (0,0) and
(500,50) . So it is possible for the two species to live in harmony. 12 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions / f (x)=1 3cos x 1. f (x)=x 3sin x
/ 2. f (x)=xsin x f (x)=x cos x+(sin x) 1=xcos x+sin x
/ 3. y=sin x+10tan x / 4. y=2 x+5cos x
3 5. g(t)=t cos t 2 y =cos x+10sec x
y = 2 xcot x 5sin x
/ 3 2 2 3 2 g (t)=t ( sin t)+(cos t) 3t =3t cos t t sin t or t (3cos t tsin t)
/ 6. g(t)=4sec t+tan t 2 g (t)=4sec ttan t+sec t 7. h( )=csc +e cot
/ h ( )= csc cot +e ( csc
u / 8. y=e (cos u+cu)
9. y= x
cos x y = )+(cot )e = csc cot +e (cot u u 2 ) u y =e ( sin u+c)+(cos u+cu)e =e (cos u sin u+cu+c) / y = (cos x)(1) (x)( sin x)
2 = cos x+xsin x
2 cos x (cos x) (x+cos x)(cos x) (1+sin x)(1 sin x)
2 (x+cos x)
2 = csc 1+sin x
x+cos x 10. y=
/ 2 2 (x+cos x) = 2 2 (x+cos x) 2 xcos x+cos x (cos x) 2 xcos x+cos x (1 sin x) = xcos x
2 (x+cos x) sec
1+sec
(1+sec )(sec tan ) (sec )(sec tan ) 11. f ( )=
/ f ( ) 12. y= = 2 (1+sec ) = (sec tan )[(1+sec ) sec ]
2 (1+sec ) = sec tan
2 (1+sec ) tan x 1
sec x
1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions ( 2 2 ) 2 dy sec xsec x (tan x 1)sec xtan x sec x sec x tan x+tan x
1+tan x
=
=
=
2
2
dx
sec x
sec x
sec x
/ y =cos x+sin x . Another method: Simplify y first: y=sin x cos x 13. y= sin x 2 / y = 2 x cos x (sin x)(2x) ( x2) 2 x = x(xcos x 2sin x)
x 4 = xcos x 2sin x
x 3 14. y= ( +cot )
/ 2 2 y = (1 csc )+( +cot )(
= ( cot
= ( 2 cot cot
2cot cot
2 )= {1+cot 2 2 cot cot 2 ) 2 =csc } cot ( +2cot ) y =sec (sec Using the identity 1+tan
sec (1+2tan ) / 15. y=sec tan
2 2 cot )= (1 csc 2 =sec ) or sec (2sec 2 2 )+tan (sec tan )=sec (sec 2 +tan 2 ) , we can write alternative forms of the answer as
1) / / / / 16. Recall that if y= fgh , then y = f gh+ fg h+ fgh . y=xsin xcos x
dy
2
2
=sin xcos x+xcos xcos x+xsin x( sin x)=sin xcos x+xcos x xsin x
dx
1
sin x = d
d
( sec x ) =
dx
dx 1
cos x = d
d
19.
( cot x ) =
dx
dx cos x
sin x 17. 18. d
d
( csc(x) ) =
dx
dx (sin x)(0) 1(cos x)
2 = sin x 2 = 1
cos x
= xcot x
sin x sin x = 1
sin x
=sec xtan x
cos x cos x sin x (cos x)(0) 1( sin x)
2 cos x = cos x = sin x
2 cos x (sin x)( sin x) (cos x)(cos x)
2 sin x 2 = 2 sin x+cos x
2 = sin x 1
2 = 2 x sin x 20. f (x)=cos x
f (x+h) f (x)
cos (x+h) cos x
cos xcos h sin xsin h cos x
/
= lim
= lim
= lim
f (x)
h
h
h
h 0
h 0
h 0
2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions = lim
h cos x 0 cos h 1
sin h
sin x
h
h =cos x lim
h 0 cos h 1
sin h
sin x lim
h
h 0 h =(cos x)(0) (sin x)(1)= sin x
21. y=tan x / 2 y =sec x the slope of the tangent line at equation of the tangent line is y 1=2
x 22. y=e cos x / x x 4 x or y=2x+1 x y =e ( sin x)+(cos x)e =e (cos x sin x) 4
2 ,1 is sec 2 2 4 =( 2 ) =2 and an . the slope of the tangent line at (0, 1) is 0 e (cos 0 sin 0)=1(1 0)=1 and an equation is y 1=1(x 0) or y=x+1 .
23. y=x+cos x
y=x+1 .
24. y= / / y =1 sin x . At ( 0,1 ) , y =1 , and an equation of the tangent line is y 1=1(x 0) , or 1
sin x+cos x cos x sin x / y = / 2 [ Reciprocal Rule]. At ( 0,1 ) , y = (sin x+cos x)
equation of the tangent line is y 1= 1(x 0) , or y= x+1 . 1 0
2 = 1 , and an (0+1) / 25. (a) y=xcos x y =x( sin x)+cos x(1)=cos x xsin x . So the slope of the tangent at the point
sin = 1 (0)= 1 , and an equation is y+ = (x ) or y= x .
( , ) is cos (b)
/ 26. (a) y=sec x 2cos x y =sec xtan x+2sin x the slope of the tangent line at
equation is y 1=3 3 x 3 3 ,1 is sec or y=3 3 x+1 3 tan 3 +2sin 3 =2 3 +2 3
=3 3 and an
2 3 . (b) 3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions 27. (a) f (x)=2x+cot x / 2 f (x)=2 csc x (b)
/ Notice that f (x)=0 when f has a horizontal tangent.
/ f is positive when f is increasing and f
negative when the graph of f is steep.
28. (a) f (x)= x sin x / / is negative when f is decreasing. Also, f (x) is large / f (x)= x cos x+(sin x) 1
x
2 1/2 = x cos x+ sin x
2 x (b)
/ Notice that f (x)=0 when f has a horizontal tangent.
f / is positive when f is increasing and f / is negative when f is decreasing.
/ 29. f (x)=x+2sin x has a horizontal tangent when f (x)=0
x= 1+2cos x=0 cos x= 2
4
4
2
+2 n or
+2 n , where n is an integer. Note that
and
are
3
3
3
3 3 This allows us to write the solutions in the more compact equivalent form (2n+1) 1
2
units from
3 . , n an integer.
cos x
30. y=
2+sin x / y = (2+sin x)( sin x) cos xcos x
2 (2+sin x) 2 = 2 2sin x sin x cos x
2 (2+sin x) = 2sin x 1
2 =0 when (2+sin x) 4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions 11
7
+2 n or x=
+2 n , n an integer. So y=
6
6
11
1
the points on the curve with horizontal tangents are:
+2 n,
,
6
3
n an integer.
2sin x 1=0 sin x= 1
2 31. (a) x(t)=8sin t v(t)=x
2
(b) The mass at time t=
3
2
2
v
=8cos
=8
3
3
32. (a) s(t)=2cos t+3sin t (b)
(c) s=0 t 2 t 2.55 s.
(d) v=0 t 1 x= / 1
or y=
3
7
+2 n,
6 1
and
3
1
,
3 (t)=8cos t has position x
1
2 2
3 = 4 . Since v =8sin
2
3 2
=8
3 3
2 =4 3 and velocity <0 , the particle is moving to the left. v(t)= 2sin t+3cos t 2.55 . So the mass passes through the equilibrium position for the first time when
0.98 , s(t ) 3.61 cm. So the mass travels a maximum of about 3.6 cm (upward and
1 downward) from its equilibrium position.
(e) The speed |v| is greatest when s=0 ; that is, when t=t +n , n a positive integer.
2 33.
From the diagram we can see that sin =x/10 x=10sin . We want to find the rate of change of x
with respect to ; that is, dx/d . Taking the derivative of the above expression, dx/d =10(cos ) .
dx
1
So when =
,
=10cos
=10
=5 ft/rad.
3 d
3
2
34. (a) F= W
sin +cos dF ( sin +cos )(0) W ( cos
=
2
d
( sin +cos ) sin ) = W (sin cos )
2 ( sin +cos ) (b)
5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions dF
=0
d W (sin cos )=0 sin = cos tan = =tan 1 (c)
From the graph of F= 0.6(50)
0.6sin +cos for 0 1 , we see that dF
=0
d 0.54 . Checking this 1 with part (b) and =0.6 , we calculate =tan 0.6 0.54 . So the value from the graph is consistent
with the value in part (b).
35.
sin 3x
3sin 3x
=lim
[ multiply numerator and denominator by 3 ]
x
3x
0
x 0 lim
x sin 3x
[ as x
0 3x =3 lim
3x =3 lim sin 0 , 3x 0] [ let =3x ] 0 =3(1)
=3 [ Equation 2] 36.
sin 4x
=lim
0 sin 6x x 0 sin 4x
x
x
sin 6x lim
x =lim
x 0 4sin 4x
6x
lim
4x
x 0 6sin 6x sin 4x 1
6x
1
2
lim
=4(1) (1)=
6 x 0 sin 6x
6
3
0 4x =4lim
x 37.
tan 6t
=lim
0 sin 2t t 0 sin 6t
1
t
t
cos 6t sin 2t lim
t =lim
t 0 6sin 6t
1
2t
lim
lim
6t
t 0 cos 6t t 0 2sin 2t sin 6t
1
1
2t
1 1
lim
lim
=6(1)
(1)=3
1 2
0 6t
t 0 cos 6t 2 t 0 sin 2t =6lim
t 38. 6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions cos
cos 1
= lim
sin
0 lim
0 1 cos lim
= sin 1 0 = sin lim 0
=0
1 0 sin
sin (cos )
39. lim
=
sec
0 ( lim (cos )) = sin 1 =sin 1
0 lim (sec ) 1 0 40.
2 lim
t sin 3t 0 t 2 sin 3t sin 3t
t
t =lim
t = 0 sin 3t
t
0 lim
t sin 3t
sin 3t
lim
t
t
0
t 0 =lim
t 2 sin 3t
0 3t = 3lim
t 2 2 =(3 1) =9 41.
cot 2x
cos2x sinx
lim
=lim
=lim cos 2x
x
sin2x
x 0
x 0
x 0
=1 lim [(sin x)/x]
(sin x)/x
(sin 2x)/x =lim
x cos2x 0 x 0 2 lim [(sin 2x)/2x]
x 0 1
1
=
2 1 2 42.
sin x cos x
sin x cos x
sin x cos x
= lim
= lim
2
2
cos 2x
x
/4 cos x sin x x
/4 (cos x+sin x)(cos x sin x)
/4
1
1
1
= lim
=
=
2
x
/4 cos x+sin x cos
+sin
4
4 lim
x 43. Divide numerator and denominator by . ( sin( ) also works.)
sin
sin
lim
sin
0
1
1
lim
= lim
=
=
=
1
sin
1
1+1 1 2
0 +tan
0 1+ sin
1+ lim
lim
cos
0
0 cos
44.
7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions lim
x 1 sin (x 1)
2 sin (x 1)
1
sin (x 1) 1
1
=lim
lim
= 1=
x 1
3
3
1 (x+2)(x 1) x 1 x+2 x 1 =lim
x x +x 2 d
d sin x
45. (a)
tan x=
dx
dx cos x
d
d
1
sec x=
(b)
dx
dx cos x
(c) 2 sec x= sec xtan x= cos xcos x sin x( sin x)
2 cos x
(cos x)(0) 1( sin x)
2 2 = 2 cos x+sin x
2 cos x
. So sec xtan x= cos x 1 2 . So sec x= 2 . cos x
sin x
2 . cos x d
d 1+cot x
(sin x+cos x)=
dx
dx
x
2 cos x sin x = x( csc x) (1+cot x)( xcot x)
2 2 = x[ csc x+(1+cot x)cot x]
2 csc x
2 csc x 2 csc x+cot x+cot x
1+cot x
=
=
x
x
So cos x sin x= cot x 1
.
x 46. Let PR =x . Then we get the following formulas for r and h in terms of and x : sin 2 = r
x h
1 2
1
r=xsin
and cos =
h=xcos
. Now A ( ) =
r and B ( ) = (2r)h=rh . So
2
2 x
2
2
2
1 2
A( )
r
2
1
r 1
xsin ( /2 )
lim
=
lim =
lim
+ B ( ) = lim
rh
2
2
0
+
+ h
+ xcos ( /2 )
0 = 1
2 0 0 lim tan ( /2 ) =0 .
+ 0 47. By the definition of radian measure, s=r , where r is the radius of the circle.
8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.4 Derivatives of Trigonometric Functions By drawing the bisector of the angle , we can see that sin = d/2
r 2
s
r
2 ( /2 )
/2
So lim = lim
= lim
= lim
=1 .
+ d
+ 2rsin ( /2 )
+ 2sin ( /2 )
0 sin ( /2 )
0 0 d=2rsin 2 . 0 9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule 1. Let u=g(x)=4x and y= f (u)=sin u . Then 2. Let u=g(x)=4+3x and y= f (u)= u =u
2 1/2 dy dy du
=
=(cos u)(4)=4cos 4x .
dx du dx
dy dy du 1
=
= u
dx du dx 2 . Then 10 3. Let u=g(x)=1 x and y= f (u)=u . Then 1/2 (3)= 3
3
=
.
2 u 2 4+3x dy dy du
9
29
=
=(10u )( 2x)= 20x(1 x ) .
dx du dx 4. Let u=g(x)=sin x and y= f (u)=tan u . Then dy dy du
2
2
=
=(sec u)(cos x)=(sec u)(sin x) cos x , or
dx du dx 2 equivalently, sec (sin x) cos x .
5. Let u=g(x)=sin x and y= f (u)= u . Then
x 6. Let u=g(x)=e and y= f (u)=sin u . Then
3 7 / 7. F(x)=(x +4x)
2 3 6 3 / 2 4 2 3 2 6 2 1 2 (2+3x ) 3 3/4 4(1+2x+x ) 2+3x = 4(1+2x+x ) 4 4 33 (1+2x+x ) 2
4
f (x)= (1+x )
3 4 2/3 / 10. f (x)=(1+x ) 4 (t +1) 2 2 3 3/4 3 6 [ or 7x (x +4) (3x +4) ] 3 1/4 2+3x 4 cos x
cos x
=
.
2 u 2 sin x 2 1+2x+x =(1+2x+x )
1
3
3 3/4 d
/
(1+2x+x )=
= (1+2x+x )
F (x) 4
dx 11. g(t)= cos x= F (x)=3(x x+1) (2x 1) 9. F(x)= 1 1/2 dy dy du
x
x
x
=
=(cos u)(e )=e cos e .
dx du dx F (x)=7(x +4x) (3x +4) 8. F(x)=(x x+1) = dy dy du 1
=
= u
dx du dx 2 =(t +1) 3 / 1/3 8x 3 (4x )= 3 1+x 3
4 4 3 3 4 3 4 4 g (t)= 3(t +1) (4t )= 12t (t +1) = 12t
4 3
4 (t +1) 12.
1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule 1
f (t)= (1+tan t)
3 1/3 3 / f (t)= 1+tan t =(1+tan t) 3 3 / 13. y=cos (a +x )
3 3 mx 3 2 2 sec t 2 sec t= 3 2 3 (1+tan t) 3 2 3 3 y = sin (a +x ) 3x [ a is just a constant] = 3x sin (a +x )
/ 14. y=a +cos x
15. y=e 3 2/3 2 3 2 y =3(cos x) ( sin x) [ a is just a constant] = 3sin xcos x
/ d
( mx)=e
dx mx y =e mx mx ( m)= me / 16. y=4sec 5x y =4sec 5xtan 5x(5)=20sec 5xtan 5x
5 28 17. g(x)=(1+4x) (3+x x )
/ 5 27 28 4 g (x) =(1+4x) 8(3+x x ) (1 2x)+(3+x x ) 5(1+4x) 4
4 27 2 4 27 2 4 27 2 =4(1+4x) (3+x x ) [2(1+4x)(1 2x)+5(3+x x )]
2 =4(1+4x) (3+x x ) [(2+4x 16x )+(15+5x 5x )]
=4(1+4x) (3+x x ) (17+9x 21x )
4 3 3 4 18. h(t)=(t 1) (t +1)
/ 4 3 3 3 2 3 4 4 2 3 h (t) =(t 1) 4(t +1) (3t )+(t +1) 3(t 1) (4t )
2 4 2 3 3 4 3 2 4 2 3 3 4 =12t (t 1) (t +1) [(t 1)+t(t +1)]=12t (t 1) (t +1) (2t +t 1)
4 2 3 19. y=(2x 5) (8x 5)
/ 3 2 3 4 2 4 y =4(2x 5) (2)(8x 5) +(2x 5) ( 3)(8x 5) (16x)
3 2 =8(2x 5) (8x 5)
2 2 3 1/3 / 20. y=(x +1)(x +2)
2 21. y=xe x / 4 2 4 1/3 2 48x(2x 5) (8x 5)
2 y =2x(x +2) +(x +1)
2 x y =xe ( 2x)+e 2 x 2 1=e x 2 1
3 ( 2x +1)=e 2 (x +2) 2/3 2 1/3 (2x)=2x(x +2) 2 1+ x +1
2 3(x +2)
2 x 2 (1 2x ) 2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule 22. y=e
23. y=e 5x / cos 3x
/ xcos x 5x y =e y =e 5x ( 3sin 3x)+(cos 3x)( 5e 2 1
=
F (z) 2 1/2 1 x y =10 d
2
1
(ln 10)
(1 x )= 2x(ln 10)10
dx 2 x . 1/2 z 1
z+1 z 1
z+1 / 2 / 24. Using Formula 5 and the Chain Rule, y=10
z 1
=
z+1 (3sin 3x+5cos 3x) d
xcos x
xcos x
(xcos x)=e
x( sin x)+(cos x) 1 =e
(cos x xsin x)
dx xcos x 1 x 25. F(z)= 5x )= e d
dz z 1
z+1 1
=
2 1/2 1/2 z+1
z 1 (z+1)(1) (z 1)(1)
2 (z+1) 1/2 1 (z+1)
z+1 z+1 1 (z+1)
2
1
=
=
=
1/2
2
1/2
2
1/2
3/2
2
2
(z 1)
(z+1)
(z 1)
(z+1) (z 1) ( z+1 )
4 (y 1) 26. G(y)= 2 5 (y +2y)
2 / G (y) = 5 3 4 2 4 (y +2y) 4(y 1) 1 (y 1) 5(y +2y) (2y+2)
2 52 [(y +2y) ]
2 = 4 3 2 2(y +2y) (y 1) [2(y +2y) 5(y 1)(y+1)]
2 10 (y +2y)
3 = 2 2 2(y 1) [(2y +4y)+( 5y +5)]
2 6 3 = 2 (y +2y)
27. y= 2 2(y 1) ( 3y +4y+5)
6 (y +2y) r
2 r +1 y 2 r +1 (1) r / = ( 1 2
(r +1)
2
r +1 )
2 2 r 2 1/2 r +1 (2r)
= r +1 )
2 2 2 r +1 2 r +1 ( 2 2 r +1 r 2 2 = r +1 ( r +1 )
2 2 3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule 2 (r +1) r = ( 2
3 r +1 )
2 1 = 2 2 or (r +1) 3/2 3/2 (r +1) 2 Another solution: Write y as a product and make use of the Product Rule. y=r (r +1)
1 2
3/2
2
1/2
/
1
y =r 2 (r +1) (2r)+r +1)
2 3/2 =(r +1) 2 2 1 2 [ r +(r +1) ]=(r +1) 3/2 2 3/2 (1)=(r +1) 2 The step that students usually have trouble with is factoring out (r +1)
2 2 1/2 3/2 . But this is no different 5 than factoring out x from x +x ; that is, we are just factoring out a factor with the smallest exponent
3
1
that appears on it. In this case,
is smaller than
.
2
2
2u e 28. y= u u e +e
u / y = u 2u (e +e )(e 2u u u 2u 2) e (e e ) u e (2e +2e u u u 2u e +e ) u u e (e +3e ) =
=
u
u2
u
u2
u
u2
(e +e )
(e +e )
(e +e )
Another solution: Eliminate negative exponents by first changing the form of y .
2u u e y= u e +e e
u u y = = 3u 2u e +1 e 2u / e 3u 3u 2 2u 3u 2u (e +1)(3e ) e (2e ) = (e +1) 2u 2u e (3e +3 2e )
2u 2 (e +1)
/ 3u = 2u e (e +3)
2u 2 (e +1) 2 2 y =sec (cos x) ( sin x)= sin x sec (cos x) 29. y=tan (cos x)
2 sin x
30. y=
cos x
2 / y = cos x(2sin x cos x) sin x( sin x)
2 cos x 2 = 2 sin x(2cos x+sin x)
2 cos x 2 = sin x(1+cos x)
2 cos x 2 =sin x(1+sec x)
Another method: y=tan x sin x
4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule / 2 2 y =sec x sin x+tan x cos x=sec x sin x+sin x
sin x 31. Using Formula 5 and the Chain Rule, y=2
d
/ sin x
sin x
y =2
(ln 2)
(sin x)=2
(ln 2) cos x
dx
2 2 6 34. y=xsin 1
x 2 y =2(tan 3 ) / 33. y=(1+cos x) 2 ( ln 2)cos x d
(tan 3 )=2tan 3
d / 2 32. y=tan (3 )=(tan 3 ) sin x =2 2 sec 3 2 3=6tan 3 sec 3 5 2 5 y =6(1+cos x) 2cos x( sin x)= 12cos xsin x(1+cos x)
/ y =sin 1
1
+xcos
x
x 2 2 1
2 =sin x 1
x 1
1
cos
x
x 2 35. y=sec x+tan x=(sec x) +(tan x)
/ 2 2 2 2 y =2(sec x) (sec xtan x)+2(tan x)(sec x)=2sec x tan x+2sec x tan x=4sec x tan x
ktan 36. y=e x ktan / y =e 2 x ktan
d
(ktan x )=e
dx x ksec 2 1
x
x
2 1/2 = ksec 2 2 x x ktan e x 2 37. y=cot (sin )=[cot (sin )]
d
/
2
2
y =2[cot (sin )]
[cot (sin )]=2cot (sin ) [ csc (sin ) cos ]= 2cos cot (sin ) csc (sin )
d
d
/
38. y=sin (sin (sin x)) y =cos (sin (sin x)) (sin (sin x))=cos (sin (sin x)) cos (sin x) cos x
dx
39. y= x+ x / y = 40. y= x+ x+ x 1
(x+ x )
2
/ y = 1/2 1+ 1
x
2 1
x+ x+ x
2 ( 1/2 ) 1/2 = 1 2 x+ x
1
1+ (x+ x )
2 1+
1/2 1
2 x
1
1+ x
2 1/2 41. y=sin (tan sin x )
d
d
2
1/2
/
y =cos (tan sin x ) dx (tan sin x )=cos (tan sin x ) sec sin x dx (sin x)
1
2
1/2
=cos (tan sin x )sec sin x
(sin x)
cos x
2
5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule ( =cos (tan sin x ) sec
2
x 2 2
x 1
2 sin x ) sin x 2
x ( )
2 (cos x) 2 d
x
3
x
3 =2 (ln 2)3 (ln 3)(2x)
y =2 (ln 2)
dx 3 / 42. y=2 3 10 / 9 9 / 9 y =10(1+2x) 2=20(1+2x) . At (0,1) , y =20(1+0) =20 , and an equation of the
43. y=(1+2x)
tangent line is y 1=20(x 0) , or y=20x+1 .
2 / / 44. y=sin x+sin x y =cos x+2sin xcos x . At (0,0) , y =1 , and an equation of the tangent line is
y 0=1(x 0) , or y=x .
/ / 45. y=sin (sin x) y =cos (sin x) cos x . At ( ,0) , y =cos (sin ) cos =cos (0) ( 1)=1( 1)= 1 ,
and an equation of the tangent line is y 0= 1(x ) , or y= x+ .
2 46. y=x e x / 2 x x y =x ( e )+e (2x)=2xe 47. (a) y= 1+e
/ At (0, 1) , y = x / y = x 2 xe x . At 1, 1
e / , y =2e 1 1 e = 1
. So an equation of
e 1 1
1
= (x 1) or y= x .
e e
e the tangent line is y
2 x x (1+e )(0) 2( e )
x2 (1+e )
0 2e 02 (1+e ) = 2(1)
2 (1+1) = 2
2 2 = = 2e x
x2 . (1+e )
1
.
2 So an equation of the tangent line is y 1= 1
1
(x 0) or y= x+1 .
2
2 (b)
48. (a) For x>0 , |x|=x , and y= f (x)= x
2 2 x 6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule 1
2 2 2 x (1) x / f (x) = (
2 = 2 x )
2 2 (2 x )+x 2 3/2 ( 2x) 2 1/2 (2 x ) 2 2 1/2 (2 x ) 2 = (2 x ) 2 1/2 (2 x ) 2 3/2 (2 x ) / So at (1,1) , the slope of the tangent line is f (1)=2 and its equation is y 1=2(x 1) or y=2x 1 . (b)
2 49. (a) f (x)= 1
(1 x
2 x / 1 x
x f (x) = 1/2 2 ( 2x) 1 x (1) 2 2 x
2 = 2 x (1 x )
2 x 2 1 x 2 1 x
1 x 1 =
2 x 2 1 x (b)
/ Notice that all tangents to the graph of f have negative slopes and f (x)<0 always.
50. (a) 7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule From the graph of f , we see that there are 5 horizontal tangents, so there must be 5 zeros on the
graph of f / . From the symmetry of the graph of f , we must have the graph of f it is low at x= . The intervals of increase and decrease as well as the signs of f
figure. / / as high at x=0 as are indicated in the (b)
f (x) =sin (x+sin 2x)
d
/
f (x) =cos (x+sin 2x) dx (x+sin 2x)
=cos (x+sin 2x)(1+2cos 2x) / 2 / 51. For the tangent line to be horizontal, f (x)=0 . f (x)=2sin x+sin x f (x)=2cos x+2sin xcos x=0
3
2cos x ( 1+sin x ) =0 cos x=0 or sin x= 1 , so x= +2n or
+2n , where n is any integer.
2
2
3
Now f
=3 and f
= 1 , so the points on the curve with a horizontal tangent are
2
2
3
+2n ,3 and
+2n , 1 , where n is any integer.
2
2
52. f (x)=sin 2x 2sin x
8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule / 2 2 f (x)=2cos 2x 2cos x=4cos x 2cos x 2 , and 4cos x 2cos x 2=0
1
cos x=1 or cos x=
. So x=2n or (2n+1)
, n any integer.
2
3 / / / / 53. F(x)= f (g(x)) / (cos x 1)(4cos x+2)=0 / F (x)= f (g(x)) g (x) , / / / so F (3)= f (g(3)) g (3)= f (6) g (3)=7 4=28 . Notice that we did not use f (3)=2 .
54. w=u v
/ / / / / / w ( x ) =u (v(x)) v (x) , so w(x)=u(v(x))
/ / / w (0)=u (v(0)) v (0)=u (2) v (0)=4 5=20 . The other pieces of information, u(0)=1 , u (0)=3 ,
/ and v (2)=6 , were not needed.
/ 55. (a) h(x)= f (g(x))
(b) H(x)=g( f (x)) / / / / / / / / / / / / H (x)=g ( f (x)) f (x) , so H (1)=g ( f (1)) f (1)=g (3) 4=9 4=36 .
/ 56. (a) F(x)= f ( f (x))
(b) G(x)=g(g(x)) / h (x)= f (g(x)) g (x) , so h (1)= f (g(1)) g (1)= f (2) 6=5 6=30 . / / / / / / F (x)= f ( f (x)) f (x) , so F (2)= f ( f (2)) f (2)= f (1) 5=4 5=20 .
/ / / / / / / G (x)=g (g(x)) g (x) , so G (3)=g (g(3)) g (3)=g (2) 9=7 9=63 .
/ / / / / / / / / u (x)= f (g(x))g (x) . So u (1)= f (g(1))g (1)= f (3)g (1) . To find f (3) ,
3 4
1
/
note that f is linear from (2,4) to (6,3) , so its slope is
=
. To find g (1) , note that g is linear
6 2
4
0 6
1
3
/
/
from (0,6) to (2,0) , so its slope is
= 3 . Thus, f (3)g (1)=
( 3)= .
2 0
4
4
57. (a) u(x)= f (g(x)) (b) v(x)=g( f (x)) / / / / / / / / / / / v (x)=g ( f (x)) f (x) . So v (1)=g ( f (1)) f (1)=g (2) f (1) , which does not / exist since g (2) does not exist.
/ (c) w(x)=g(g(x)) / / / / / w (x)=g (g(x))g (x) . So w (1)=g (g(1))g (1)=g (3)g (1) . To find g (3) ,
2 0 2
2
/
/
note that g is linear from (2,0) to (5,2) , so its slope is
= . Thus, g (3) g (1)=
( 3)= 2
5 2 3
3
.
/ / / / / h (x)= f ( f (x)) f (x) . 58. (a) h(x)= f ( f (x))
/ / / So h (2)= f ( f (2)) f (2)= f (1) f (2) ( 1)( 1)=1 .
d 2
2
/
/ 2
/
/
/ 2
/ 2
(b) g(x)= f (x ) g (x)= f (x )
(x )= f (x )(2x) . So g (2)= f (2 )(2 2)=4 f (4) 4(1.5)=6 .
dx
9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule / / / / / / / / 59. h(x)= f (g(x)) h (x)= f (g(x))g (x) . So h (0.5)= f (g(0.5))g (0.5)= f (0.1)g (0.5) . We can
estimate the derivatives by taking the average of two secant slopes.
m +m
1
2 22+36
14.8 12.6
18.4 14.8
/
/
For f ( 0.1 ) : m =
=22 , m =
=36 . So f (0.1)
=
=29 .
1
2
2
0.1 0
0.2 0.1
2
m +m
1
2
0.10 0.17
0.05 0.10
/
/
= 0.6 .
For g (0.5) : m =
= 0.7 , m =
= 0.5 . So g (0.5)
1
2
2
0.5 0.4
0.6 0.5
/ / / Hence, h (0.5)= f (0.1)g (0.5) (29)( 0.6)= 17.4 .
/ 60. g(x)= f ( f (x)) / / / / / / / g (x)= f ( f (x)) f (x) . So g (1)= f ( f (1)) f (1)= f (2) f (1) .
m +m
1
2
3.1 2.4
4.4 3.1
/
/
For f (2) : m =
=1.4 , m =
=2.6 . So f (2)
=2 .
1 2.0 1.5
2 2.5 2.0
2
m +m
1
2
2.0 1.8
2.4 2.0
/
/
For f (1) : m =
=0.4 , m =
=0.8 . So f (1)
=0.6 .
1 1.0 0.5
2 1.5 1.0
2
/ / / Hence, g (1)= f (2) f (1) (2)(0.6)=1.2 .
/ x (b) G(x)=e / f (x) G (x)=e f (x) d x
/ x x
(e )= f (e )e
dx / F (x)= f (x )
/ (b) G(x)= f (x) x d
f (x) /
f (x)=e f (x)
dx / 62. (a) F(x)= f (x ) G (x)=
4 d
/
(x )= f (x ) x
dx
1 f (x) / 63. (a) f (x)=L(x ) / 4 1 / f (x) 3 4 3 f (x)=L (x ) 4x =(1/x ) 4x =4/x for x>0 .
/ (b) g(x)=L(4x) / F (x)= f (e ) 61. (a) F(x)= f (e ) / g (x)=L (4x) 4=(1/(4x)) 4=1/x for x>0 .
4 / 3 / 3 3 (c) F(x)=[L(x)] F (x)=4[L(x)] L (x)=4[L(x)] (1/x)=4[L(x)] /x (d) G(x)=L(1/x) G (x)=L (1/x) ( 1/x )=(1/(1/x)) ( 1/x )=x ( 1/x )= 1/x for x>0 . 64. r(x)= f (g(h(x)))
/ / / / / 2 / 2 / 2 / r (x)= f (g(h(x))) g (h(x)) h (x) , so
/ / / / / r (1)= f (g(h(1))) g (h(1)) h (1)= f (g(2)) g (2) 4= f (3) 5 4=6 5 4=120
65. s(t)=10+ 1
sin (10 t)
4 the velocity after t seconds is
10 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule / v(t)=s (t)= 1
5
cos (10 t)(10 )=
cos (10 t) cm / s.
4
2
/ 66. (a) s=Acos ( t+ ) velocity =s =
/ (b) If A 0 and 0 , then s =0 2 t
5.4
dB 7
2
(b) At t=1 ,
=
cos
dt 54
5.4 sin ( t+ )=0 t+ =n dB
2 t
= 0.35cos
dt
5.4 67. (a) B(t)=4.0+0.35sin t= 2
5.4 n = , n an integer. 0.7
2 t 7
2 t
cos
=
cos
5.4
5.4 54
5.4 0.16 . 2
(t 80)
365 68. L(t)=12+2.8sin Asin ( t+ ) . 2
(t 80)
365 / L (t)=2.8cos 2
365 / .
/ On March 21, t=80 , and L (80) 0.0482 hours per day. On May 21, t=141 , and L (141) 0.02398 ,
/ which is approximately one half of L (80) .
69. s(t)=2e
/ 1.5t sin 2 t v(t)=s (t)=2 e 1.5t (cos 2 t)(2 )+(sin 2 t)e 70. (a) lim p(t)=lim
t t
kt 1 (b) p(t)=(1+ae ) 1
1+ae kt = 1.5t ( 1.5) =2e 1
=1 , since k>0
1+a 0 dp
kt 2
kt
= (1+ae ) ( kae )=
dt kae 1.5t kt (2 cos 2 t 1.5sin 2 t) e kt 0. kt
kt 2 (1+ae ) (c) 11 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule 0.5t 1 ) , it seems that p(t)=0.8 (indicating that 80% of the population
From the graph of p(t)=(1+10e
has heard the rumor) when t 7.4 hours.
t 71. (a) Using a calculator or CAS, we obtain the model Q=ab with a=100.0124369 and
t t lnb b=0.000045145933 . We can change this model to one with base e and exponent ln b [ b =e
tln b precalculus mathematics or from Section 7.3]: Q=ae
/ =100.012437e from 10.005531t . t x (b) Use Q (t)=ab ln b or the calculator command nDeriv(Y , X, .04) with Y =ab to get
1 1 / Q (0.04) 670.63 A. The result of Example 2 in Section 2.1 was 670 A.
t 20 72. (a) P=ab with a=4.502714 10 and b=1.029953851 ,
where P is measured in thousands of people. The fit appears to be very good. (b) For 1800: m =
1 5308 3929
7240 5308
=137.9 , m =
=193.2 .
2 1810 1800
1800 1790 / So P (1800) (m +m )/2=165.55 thousand people / year.
1 For 1850: m =
1 2 23,192 17,063
31,443 23,192
=612.9 , m =
=825.1 .
2
1850 1840
1860 1850 / So P (1850) (m +m )/2=719 thousand people / year.
1 2 x (c) Use the calculator command nDeriv(Y , X, .04) with Y =ab to get
1 / 1 / P (1800) 156.85 and P (1850) 686.07. These estimates are somewhat less than the ones in part
(b).
(d) P(1870) 41,946.56. The difference of 3.4 million people is most likely due to the Civil War
(1861 1865). 12 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule 8 / 73. (a) Derive gives g (t)= 45(t 2) without simplifying. With either Maple or Mathematica, we first 10 (2t+1)
8 9 (t 2) / get g (t)=9 9 (t 2) 18 (2t+1) 10 , and the simplification command results in the above expression. (2t+1) / 3 3 4 3 2 (b) Derive gives y =2(x x+1) (2x+1) (17x +6x 9x+3) without simplifying.
/ 4 3 4 5 3 3 2 With either Maple or Mathematica, we first get y =10(2x+1) (x x+1) +4(2x+1) (x x+1) (3x 1) .
If we use Mathematica’s Factor or Simplify , or Maple’s factor , we get the above expression, but
Maple’s simplify gives the polynomial expansion instead. For locating horizontal tangents, the
factored form is the most helpful. 4 4 74. (a) f (x)= x x+1 (3x 1) 1/2 4 x +x+1 / . Derive gives f (x)= 4 4 x x+1 x +x+1 4 whereas either Maple or 4 (x +x+1)(x x+1)
4 3x 1 / Mathematica give f ( x ) =
4 x x+1
4 after simplification.
4 2 (x +x+1) x +x+1
/ (b) f (x)=0
/ 4 3x 1=0 x= 4 1
3 0.7598 . (c) f (x)=0 where f has horizontal tangents. f
inflection points. / has two maxima and one minimum where f has 75. (a) If f is even, then f (x)= f ( x) . Using the Chain Rule to differentiate this equation, we get
d
/
/
/
/
/
/
f (x)= f ( x)
( x)= f ( x) . Thus, f ( x)= f (x) , so f is odd.
dx
/ / / (b) If f is odd, then f (x)= f ( x) . Differentiating this equation, we get f (x)= f ( x)( 1)= f ( x) ,
so f / is even. 13 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule 76.
f (x)
g(x) / = { f (x) g(x) 1} /= f /(x) g(x)
/ / f (x)
=
g(x) f (x)g (x)
2 g(x) 1 / = 2 +( 1) g(x) / g (x) f (x) / f (x)g(x) f (x)g (x)
2 g(x) 77. (a)
d
n
sin xcos nx =nsin n 1 xcos xcos nx+sin nx( nsin nx) [Product Rule]
dx ( ) =nsin
=nsin
=nsin n 1
n 1
n 1 n 1 x(cos nxcos x sin nxsin x) [factor out nsin x] xcos (nx+x) [Addition Formula for cosine] xcos (n+1)x [factor out x ] (b)
d
n
cos xcos nx =ncos n 1 x( sin x)cos nx+cos nx( nsin nx) [Product Rule]
dx ( ) = ncos
= ncos
= ncos n 1
n 1
n 1 n 1 x(cos nxsin x+sin nxcos x) [factor out ncos x] xsin (nx+x) [Addition Formula for sine] xsin (n+1)x [factor out x ] 5 78. ‘‘The rate of change of y with respect to x is eighty times the rate of change of y with respect to
d 5
dy
dy
dy
4
4 dy
5y =80 (Note that
x ’’
y =80
5y
=80
0 since the curve never has a
dx
dx
dx
dx
dx
4 y =16 horizontal tangent)
79. Since
d
d y=2 (since y>0 for all x ) = ( sin ) rad, we have
180
d
=
sin
=
cos
=
cos
d
180
180
180
180
2 2 1/2 80. (a) f (x)=|x|= x =(x ) / f (x)= 1 2
(x )
2 1/2 (2x)= x . =
2 x x
for x 0 .
|x| f is not differentiable at x=0 .
14 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.5 The Chain Rule 2 (b) f (x)=|sin x|= sin x
1
sin x
/
2 1/2
f (x)= (sin x)
2sin x cos x=
cos x=
2
|sin x| { cos x if sin x>0
cos x if sin x<0 f is not differentiable when x=n , n an integer.
x
x
/
2
(c) g(x)=sin |x|=sin x
g (x)=cos |x|
= cos x=
|x| |x| { cos x if x>0
cos x if x<0 g is not differentiable at 0 .
81. First note that products and differences of polynomials are polynomials and that the derivative of
a polynomial is also a polynomial. When n=1,
A (x)
/ Q(x)P /(x) P(x)Q /(x)
P(x)
(1)
/
/
1
f (x)=
=
=
, where A (x)=Q(x)P (x) P(x)Q (x).
1
2
1+1
Q(x)
[Q(x)]
[Q(x)]
A (x)
(k)
k
Suppose the result is true for n=k, where k 1. Then f (x)=
, so
k+1
[Q(x)]
f (k+1) k (x) = k+1 / A (x) [Q(x)] / k+1 2 [Q(x)]
k+1 [Q(x)]
= / k k = k+1 k A (x) A (x) (k+1)[Q(x)] Q (x)
{[Q(x)] } / k / A (x) (k+1)A (x)[Q(x)] Q (x)
k k 2k+2 [Q(x)]
k1 / / / [Q(x)] { A (x) (k+1)A (x)Q (x)}
k k = k k+2 [Q(x)] [Q(x)]
=A k+2 (x)/[Q(x)] k+1 k k = k+2 / Q(x)A (x) (k+1)A (x)Q (x)
[Q(x)] , where A / / (x)=Q(x)A (x) (k+1)A (x)Q (x) . k+1 k k We have shown that the formula holds for n=1, and that when it holds for n=k it also holds for
n=k+1. Thus, by mathematical induction, the formula holds for all positive integers n. 15 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation d
d
2
(xy+2x+3x )= (4)
dx
dx
y+2
/
y = 6
.
x / 1. (a) / (x y +y 1)+2+6x=0 xy = y 2 6x / y = y 2 6x
or
x 2 4 2x 3x
4
/
4
(b) xy+2x+3x =4 xy=4 2x 3x
y=
=
2 3x , so y =
3.
2
x
x
x
y 2 6x
(4/x 2 3x) 2 6x
4/x 3x
/
4
(c) From part (a), y =
=
=
=
3.
2
x
x
x
x
2 2 d
d
2
2
(4x +9y )= (36)
dx
dx 2. (a) 2 2 2 2 1
2
(9 x )
3 2 / y= 1/2 2 1 1
+
x y 1 1
+ =1
x y (c) y = y 2 = y =4
/ (c) y = 5. 2 x x = y=(4
4 x
x d 2 2 d
(x +y )=
(1)
dx
dx 1 y =0 / y = 2 1
2 / y = 2 y 2 y
y
x
x
x
/ (x 1)(1) (x)(1)
1
y=
, so y =
=
.
2
2
x 1
(x 1)
(x 1) 2 2 1
2 x 2 1 = x (x 1) + 2 (x 1)
1 / 2 y x ) =16 8 x +x
= 3 9 x / 2 = 2 2 .
2 9 x
1 2x = 2 1 x x x
y 2
3 9 2 d
d
( x + y )=
(4)
dx
dx 4. (a) 4x d
= (1)
dx [x/(x 1)] x (b) 4x
=
9y 1
1 x 1
=1
=
y
x
x 2 / y= 3 9 x
/ (b) 8x
4x
=
18y
9y
2
2
9 x , so
3 2x ( 2x)= (c) From part (a), y = d
3. (a)
dx / y = 4
2
(9 x )
9 2 2 9y =36 4x (b) 4x +9y =36
y = / 8x+18y y =0 y =0
/ y = y / y = x 4
+1
x 4
+1
x
/ 2x+2yy =0 / 2yy = 2x / y = x
y
1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 6. d 2 2 d
(x y )=
(1)
dx
dx 7. d 3 2
d
2
(x +x y+4y )=
(6)
dx
dx
2 / / 2x 2yy =0 2 2 2 y = 2 d 2
d
3
(x 2xy+y )=
(c)
dx
dx
/ 2x 2y
y =
2
2x 3y 2 2 2 / / 2 2 x +8y
/ 2 / / / 2x 2y=2xy 2 2 (x y +y 2x)+(x 2yy +y 1)=3 / 2 3y y / / / 2 2x 2y=y (2x 3y ) / 2 x y +2xyy =3 2xy y 2 / y (x +2xy)=3 2xy y 2 x y +8yy = 3x 2xy x(3x+2y) = 2x 2(xy +y 1)+3y y =0 d 2
d
2
(x y+xy )=
(3x)
dx
dx x
y / 2 3x +2xy 8. / y = / x +8y 9. / / 3x +(x y +y 2x)+8yy =0 / (x +8y) y = 3x 2xy 2x=2yy y = 3 2xy y
2 x +2xy
10. d 5 2 3 d
4
(y +x y )= (1+x y)
dx
dx / 4 2 2 4 3 y (5y +3x y x )=4x y 2xy 4 / 2 2 / 3 3 3 4x y 2xy / y = d 2 2
d
(x y +xsin y)=
(4)
dx
dx 2 / / 2 / 4 2 2 2 / 3 3 5y +3x y x
11. 4 5y y +x 3y y +y 2x=0+x y +y 4x 4 2 / x 2yy +y 2x+xcos y y +sin y 1=0 2x yy +xcos y y = 2xy sin y 2 / 2 (2x y+xcos y)y = 2xy sin y / y = 2 2xy sin y
2 2x y+xcos y
12. d
d
2
(1+x)= [sin (xy )]
dx
dx
2 2 2 1 y cos (xy )=2xycos (xy )y 2 / 2 1=[cos (xy )](x 2yy +y 1)
/ / y = 2 2 / 2 2 1=2xycos (xy )y +y cos (xy ) 2 1 y cos (xy )
2 2xycos (xy )
13. d
d
(4cos xsin y)= (1)
dx
dx / 4[cos x cos y y +sin y ( sin x)]=0 2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation / / y (4cos xcos y)=4sin xsin y 14. y = d
d
2
2
[ysin (x )]= [xsin (y )]
dx
dx / 2 2 4sin xsin y
=tan xtan y
4cos xcos y
2 2 / 2 / 2 ycos (x ) 2x+sin (x ) y =xcos (y ) 2yy +sin (y ) 1 2 2 2 / y [sin (x ) 2xycos (y )]=sin (y ) 2xycos (x ) y = 2 2 2 sin (y ) 2xycos (x )
sin (x ) 2xycos (y ) ( )
2 d
x
15.
e
dx y 2 d
=
(x+y)
dx x y 2 / e (x y +y 2x)=1+y 2 / 2 x y xe 2 / y +2xye x y / =1+y 2 2 2 x y xe y / 2 / y =1 2xye x y 2 / 2 2 x y y (x e 1)=1 2xye x y / y = 1 2xye
2 2 x y xe
16. d
d
2 2
( x+y )=
(1+x y )
dx
dx 1
(x+y)
2 1/2 2 / 1 2 (1+y )=x 2yy +y 2x / 1
y
2
/
2
+
=2x yy +2xy
2 x+y 2 x+y / x y / 2 / 2 1+y =4x y x+y y +4xy x+y
2 y / 2 / 2 4x y x+y y =4xy / x+y 1 2 2 y (1 4x y x+y )=4xy / x+y 1 y = 4xy x+y 1
2 1 4x y x+y
1
(xy)
2 2 17. xy =1+x y
y 1/2 / x
y
2
x =2xy
2 xy
2 xy / y 18. tan (x y)= 2 2 y x 2x / xy = 2 xy 2 2 (1+x )tan (x y)=y 2 x
y
/
2 /
y +
=x y +2xy
2 xy
2 xy / (xy +y 1)=0+x y +y 2x 4xy xy y / y = 2 xy 2 4xy xy y
2 x 2x
/ xy (1+x )sec (x y) (1 y )+tan (x y) 2x=y / 1+x
2 2 2 2 2 / 2 (1+x )sec (x y) (1+x )sec (x y) y +2xtan (x y)=y
2 2 (1+x )sec (x y)+2xtan (x y)=[1+(1+x )]sec (x y) y /
/ / y = 2 2 (1+x )sec (x y)+2xtan (x y)
2 2 1+(1+x )sec (x y)
19. xy=cot (xy) / 2 / y+xy = csc (xy)(y+xy ) / 2 (y+xy )[1+csc (xy)]=0 / y+xy =0 / y = y/x 20. sin x+cos y=sin xcos y
3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation / / / cos x sin y y =sin x( sin y y )+cos ycos x
/ cos x(cos y 1)
y =
sin y(sin x 1)
21. d
d
2
3
{1+ f (x)+x [ f (x)] }= (0)
dx
dx / 2 2 / / 2 d
d 2
= (x )
dx dx 2 / 3 f (x)+x 3[ f (x)] f (x)+[ f (x)] 2x=0 . If x=1 , we have 3 / f (1)+1 3[ f (1)] f (1)+[ f (1)] 2(1)=0
16
/
/
13 f (1)= 16 f (1)=
.
13
22. (sin xsin y sin y) y =cos xcos y cos x / 2 / 3 / f (1)+1 3 2 f (1)+2 2=0 / f (1)+12 f (1)= 16 / g (x)+xcos g(x) g (x)+sin g(x) 1=2x . If x=1 , we have / / / g (1)+1cos g(1) g (1)+sin g(1)=2(1) / / / dx 4
+x 1 =1
dy 2xy g (1)+cos 0 g (1)+sin 0=2 g (1)+g (1)=2 / 2g (1)=2 / g (1)=1 .
4 2 2 3 4 2 22 2 2 24. (x +y ) =ax y 2 2 25. x +xy+y =3 2 3 / / / 2x+xy +y 1+2yy =0 2 4 2 / xy +2yy = 2x y 2 2 dx ax 4y(x +y )
=
2 2
dy
4x(x +y ) 2axy dx
dx
2
2x
+2y =2ayx
+ax
dy
dy 2(x +y ) / + y 4x dx 1 4y 2x y x
=
2
3
dy
2xy +4x y 2 x=1 and y=1 , we have y = dx
dy
3 dx
3 dx
3
2
4
+4x y
=1 4y 2x y x
dy
dy 2 2 4y + x 2y+y 2x 23. y +x y +yx =y+1 / y (x+2y)= 2x y / y = 2x y
. When
x+2y 2 1
3
= = 1 , so an equation of the tangent line is y 1= 1(x 1) or
1+2 1 3 y= x+2 .
2 2 / / / / / 26. x +2xy y +x=2 2x+2(xy +y 1) 2yy +1=0 2xy 2yy = 2x 2y 1 y (2x 2y)= 2x 2y 1
2x 2y 1
2 4 1
7 7
/
/
y =
. When x=1 and y=2 , we have y =
=
= , so an equation of the tangent
2x 2y
2 4
2 2
7
7
3
line is y 2= (x 1) or y= x
.
2
2
2
2 2 2 2 2 27. x +y =(2x +2y x) / 2 2 2x+2yy =2(2x +2y x)(4x+4yy / 1) . When x=0 and y= 1
, we have
2
4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 1
2 / 0+y =2
y=x+ (2y / / 1) y =2y / / y =1 , so an equation of the tangent line is y 1 1
=1(x 0) or
2 1
.
2 2/3 2
x
3 2/3 28. x +y =4 1/3 y =0 1/3 ( 3 3) = = 3 3 ( 3 3) 1 / 2/3 1 / have y = 2
+ y
3 1/3 3 + x y
3 / =0 3 / y = 3 y y . When x= 3 3 and y=1 , we x 3
1
, so an equation of the tangent line is
=
3 3
3 1
1
(x+3 3 ) or y=
x+4 .
3
3 y 1= 2 22 / 2 2 2 2 29. 2(x +y ) =25(x y ) 2 / / 4(x +y )(2x+2yy )=25(2x 2yy ) 2 / / 4(x+yy )(x +y )=25(x yy ) 2 2 / 2 2 4yy (x +y )+25yy =25x 4x(x +y ) / y = 2 2 2 2 25x 4x(x +y ) . When 25y+4y(x +y )
75 120
45
9
=
=
, so an equation of the tangent line is
25+40
65
13
9
9
40
y 1=
(x 3) or y=
x+
.
13
13
13
/ x=3 and y=1 , we have y = 2 2 2 2 30. y (y 4)=x (x 5)
/ / / 32y +16y =0
4 2 4 16y =0
2 31. (a) y =5x x 2 4 2 y 4y =x 5x 3 4y y / / 3 8yy =4x 10x . When x=0 and y= 2 , we have / y =0 , so an equation of the tangent line is y+2=0(x 0) or y= 2 .
3 / 2yy =5(4x ) 2x 3 10x x
y =
. So at the point (1,2) we have
y
/ 3 10(1) 1 9
9
9
5
y =
= , and an equation of the tangent line is y 2= (x 1) or y= x
.
2
2
2
2
2
/ (b)
2 3 2 32. (a) y =x +3x / 2 2yy =3x +3(2x) 2 3x +6x
y =
. So at the point (1, 2) we have
2y
/ 5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 2 3(1) +6(1)
9
9
9
1
y =
=
, and an equation of the tangent line is y+2=
(x 1) or y=
x+ .
2( 2)
4
4
4
4
/ / (b) The curve has a horizontal tangent where y =0 2 3x +6x=0 3x(x+2)=0 x=0 or x= 2 . But note 2 2 3 that at x=0 , y=0 also, so the derivative does not exist. At x= 2 , y =( 2) +3( 2) = 8+12=4 , so y= 2
. So the two points at which the curve has a horizontal tangent are ( 2, 2) and ( 2,2) . (c) 33. (a)
There are eight points with horizontal tangents: four at x 1.57735 and four at x 0.42265 .
2 3x 6x+2 / (b) y = 3 2 / / y = 1 at (0, 1) and y = 2(2y 3y y+1)
Equations of the tangent lines are y= x+1 and y=
/ (c) y =0 2 3x 6x+2=0 x=1 1
at (0, 2)
3 1
x+2 .
3 1
3
3 (d)
By multiplying the right side of the equation by x 3 , we obtain the
first graph.
By modifying the equation in other ways, we can generate the other
graphs. 2 y(y 1)(y 2)=x(x 1)(x 2)(x 3)
6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 2 y(y 4)(y 2)=x(x 1)(x 2) 2 y(y+1)(y 1)(y 2)=(x 1)(x 2) 2 (y+1)(y 1)(y 2)=x(x 1)(x 2) 2 x(y+1)(y 1)(y 2)=y(x 1)(x 2) 2 2 y(y +1)(y 2)=x(x 1)(x 2) 2 2 y(y+1)(y 2)=x(x 1)(x 2)
34. (a)
7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 1
, and 3 at x=1 . The three
2
horizontal tangents along the top of the wagon are hard to find, but by limiting the y range of the
graph (to [1.6,1.7] , for example) they are distinguishable.
(b) There are 9 points with horizontal tangents: 3 at x=0 , 3 at x= / 2 2 35. From Exercise 29, a tangent to the lemniscate will be horizontal if y =0 25x 4x(x +y )=0
2 2
2 2 25
x[25 4(x +y )]=0 x +y =
( 1 ). (Note that when x is 0 , y is also 0 , and there is no horizontal
4
25
2 2
tangent at the origin.) Substituting
for x +y in the equation of the lemniscate,
4
2 22
2 2
2 2 25
2 75
2 25
2(x +y ) =25(x y ) , we get x y =
( 2 ). Solving ( 1 ) and ( 2 ), we have x =
and y =
, so
8
16
16
5 3
5
the four points are
,
.
4
4
2 x 36. 2 2 y + 2 a 2x =1 2 b / 2yy + =0 2 a 2 / y = b x b 2 b x
y y=
0 0 x x
0 the ellipse, we have 2 2 37. 2 2 + a 2yy 2 b 2 a 0 2 x
= b 2x =1 2 y y a
y y y=
0 0 2 a y 2 b / =0 0
2 0
2 2 x x
0 = 2 x
+ a b 0
2 . Since (x ,y ) lies on
0 0 a 2 + a 0
2 =1 . b
/ y = 2 b x
2 an equation of the tangent line at (x ,y ) is
0 0 a y (x x ) . Multiplying both sides by 0 2 b y y y
0 0 gives 2 y y 2 b x y y b 0 2 0 (x x ) . Multiplying both sides by a y x 0 0 y
0 2 an equation of the tangent line at (x ,y ) is 2 a y 0
2 b gives 0 2 b 2 x x y 0
2 b = 0 2 a 2 x 0
2 a . Since (x ,y ) lies on
0 0 the hyperbola, we have
8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation x x 2 y y 0 x 0 2 = 2 a b 2 y 0
2 0
2 a =1 . b / 1 y
+
=0
2 x 2 y 38. x + y = c
y
0 x an equation of the tangent line at (x ,y ) is
0 0 x
y 0 y y= y / y = (x x ) . Now x=0 0 y=y 0 0 x 0 ( x )=y + x
0 0 y+ x
0 y . And y=0 0 0 is x + x
0 y+ x
0 y=
0 x 0 0 y 0 y , so the y intercept is 0 y
(x x ) x x= 0 0 0 x 0 x=x + x
0 y 0 y , so the x intercept 0 0 0 y . The sum of the intercepts is 0 0 y 0 0 + x+ x
0 =x +2 x y 0 0 0 y +y = 0 0 0 2 2 39. If the circle has radius r , its equation is x +y =r
x
tangent line at P(x ,y ) is
0 0 0 x + 2 0 2 2 =( c ) =c . y 0 / / 2x+2yy =0 . The negative reciprocal of that slope is y x
, so the slope of the
y
y
1
0
, which is the
=
x /y
x y = 0 0 0 0 slope of OP , so the tangent line at P is perpendicular to the radius OP .
q 40. y =x p q 1 qy / y = px p 1 / y = px p 1 q 1 = qy
41. y=tan 1 x / y = 1
2 1+( x )
1 px p 1 y q = px qy p 1 p/q x qx d
1
( x )=
dx
1+x 1
x
2 1/2 = p = p ( p/q )
x
q 1 1
2 x (1+x) 1 1/2 42. y= tan x =(tan x)
/ 1
1 1/2 d
1
y = (tan x)
(tan x)=
2
dx 1 1
1 2 tan x 2 1+x = 1
1 2 2 tan x (1+x ) 1 43. y=sin (2x+1)
9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation d
(2x+1)=
dx 1 / y = 2 1 (2x+1)
2 1
2 2 / 2 1 2 H (x)=(1+x ) 2 x x 1
2
(1 x )
2 +arcsinx 1 x
45. H(x)=(1+x )arctanx 2 4x 4x 1 2 h (x)= 1 x 1 = 2 1 (4x +4x+1) / 44. h(x)= 1 x arcsinx 2 2= 1/2 ( 2x) =1 xarcsinx
2 1 x +(arctanx)(2x)=1+2xarctanx 1+x
46. y=tan 1 (x 2 x +1 ) 1 / y = ( ) 2 1+ x x +1 x 1 2 = 2 x +1 2 ( 2 2 2 1+x x x +1 2 ( ) 2 = 2 x +1 x +1 x 1 2 x +1 x +1 x
2 2 x +1 (1+x ) x(x +1) 1 = ) 2 2 2 2 (1+x ) 2 2 x +1 x = 2 1+x 2x x +1 +x +1 2 x +1 x = 2 x +1 x 1 2 2(1+x ) 1 47. h(t)=cot (t)+cot (1/t)
/ h (t)= 1 1
1+t 2 2 1+(1/t) d 1
=
dt t t 1
1+t 1 2 1 x / 1 y =cos x 1 2 2 t +1 Note that this makes sense because h(t)= 48. y=xcos x 2 2 t 2 1+t for t>0 and h(t)= x +
2 x 49. y=cos (e ) / y = 1
2x 2 1 (e ) 2 + 2 1
2 =0 . t +1
for t<0 . 1 =cos x
2 1 x
1 2x 1 = 1 x d 2x
(e )=
dx 2e 2x 1 e 4x 50. y=arctan(cos )
10 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 1 / y = sin ( sin )= 2 1+cos 1+(cos )
x 2 51. f (x)=e x arctanx
x
2
1
/
=e
x
2
f (x)
1+x
=e x 2 +(arctanx)(2x) 2 x 2 2xarctanx 1+x This is reasonable because the graphs show that f is increasing when f
when f has a minimum. / is positive, and f / is zero 2 52. f (x)=xarcsin(1 x )
2x
=x
/
f (x)
22
1 (1 x ) 2 +arcsin(1 x ) 1
2 2x 2 =arcsin(1 x ) 2 2x x 4 This is reasonable because the graphs show that f is increasing when f
inflection point when f / 1 1
2 is positive, and that f has an changes from increasing to decreasing. 53. Let y=cos x . Then cos y=x and 0 y
dy
1
=
=
dx
sin y / 1 cos y = 1 sin y dy
=1
dx . (Note that sin y 0 for 0 y ). 2 1 x 11 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 1 54. (a) Let y=sec x . Then sec y=x and y
dy
dx sec ytan y dy
1
=
=
dx sec ytan y =1 0, , 2
1 = 2 . Differentiate with respect to x:
1 2 2 . Note that tan y=sec y 1 2 sec y 1 x x 1
3
2
tan y= sec y 1 since tan y>0 when 0<y<
or <y<
.
2
2
dy
dy
1
1
2
2
2
(b) y=sec x sec y=x sec ytan y =1
=
. Now tan y=sec y 1=x 1 , so
dx
dx sec ytan y
2 tan y=
dy
=
dx x 1 . For y
1
2 x 0, 1 = 2 ( 2 2 , x 1 , so sec y=x=| x| and tan y 0 2 . For y x 1 | x| x 1
dy
1
1
=
=
dx sec ytan y
2
x
x 1 ) sec y 3
2 , 2
1 =
( x) 2 2 ,x 1 = x 1 2 2 2 2 intersection: (1, 1) . 2x +y =3 . / 3
or 1 , but
2 3
is
2 y= 1 , so there are two points of / 2 / / y = 2x/y , and x=y 1=2yy
y =1/(2y) . At (1,1)
1
, the slopes are m = 2(1)/1= 2 and m =1/(2 1)= , so the curves are orthogonal (since m and m
1
2
1
2
2
are negative reciprocals of each other). By symmetry, the curves are also orthogonal at (1, 1) .
2 2 2 4x+2yy =0 x= (2x+3)(x 1)=0 extraneous since x=y is nonnegative. When x=1 , 1=y
2 x 1 | x| x 1 55. 2x +y =3 and x=y intersect when 2x +x 3=0 2 2 1, so | x|= x and tan y= 2 2 2 56. x y =5 and 4x +9y =72 intersect when 4x +9(x 5)=72
2 2 points of intersection: ( 3, 2) . x y =5 / / 2 x= 3 , so there are four 13x =117
2 2 / 2x 2yy =0 y =x/y , and 4x +9y =72 8x+18yy =0
3
2
/
y = 4x/9y . At (3,2) , the slopes are m = and m =
, so the curves are orthogonal. By symmetry,
1 2
2
3
the curves are also orthogonal at (3, 2) , ( 3,2) and ( 3, 2) . 57.
12 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 58. The orthogonal family represents the direction of the wind. 2 2 2 2 2 59. x +y =r is a circle with center O and ax+by=0 is a line through O . x +y =r 2 / 2x+2yy =0 / y = x/y , so the slope of the tangent line at P (x ,y ) is x /y . The slope of the line OP is y /x ,
0 0 0 0 0 0 0 0 which is the negative reciprocal of x /y . Hence, the curves are orthogonal, and the families
0 0 of curves are orthogonal trajectories of each other. 2 2 2 2 60. The circles x +y =ax and x +y =by intersect at the origin where the tangents are vertical and
2 2 0 2 0 2 0 0 horizontal. If (x ,y ) is the other point of intersection, then x +y =ax ( 1 ) and x +y =by ( 2 ). Now
0 0 2 2 x +y =ax / 2x+2yy =a / y = are orthogonal at
a 2x
b 2y
0
0
=
(x ,y )
0 0
2y
2x
0 2 61. y=cx a 2x
2 2
and x +y =by
2y
2 2 0 0 2ax 4x =4y 2by
0 0 / 2x+2yy =by 0
/ 2 2x
. Thus, the curves
b 2y 2 0 / y = 0 0 ax +by =2(x +y ) , which is true by ( 1 ) and ( 2 ).
0 0 0 / 2 2 y =2cx and x +2y =k / 2x+4yy =0 / 2yy = x / y = x
=
2(y) x
2 2(cx ) = 1
, so the
2cx curves are orthogonal.
13 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 62. y=ax 3 / 2 2 2 / / y =3ax and x +3y =b / 3yy = x 2x+6yy =0 y = x
=
3(y) x
3 3(ax ) = 1
2 , so 3ax the curves are orthogonal. 2 2 63. To find the points at which the ellipse x xy+y =3 crosses the x axis, let y=0 and solve for x .
2 y=0 2 x x(0)+0 =3 3 . So the graph of the ellipse crosses the x axis at the points ( x= / Using implicit differentiation to find y , we get 2x xy
0 2 3 / So y at ( 3,0) is 2(0) / =2 and y at ( 3 3,0) is / / y+2yy =0 0+2 3
2(0)+ 3 / y (2y x)=y 2x 3,0) .
y 2x
/
y =
.
2y x =2 . Thus, the tangent lines at these points are parallel.
y 2x
as in Exercise 49. The slope of the tangent
2y x
1 2( 1) 3
1
line at ( 1,1) is m=
= =1 , so the slope of the normal line is
= 1 , and its equation is
2(1) ( 1) 3
m
/ 64. (a) We use implicit differentiation to find y = y 1= 1(x+1) 2 2 y= x . Substituting x for y in the equation of the ellipse, we get x x( x)+( x) =3 2 3x =3 x= 1 . So the normal line must intersect the ellipse again at x=1 , and since the equation of
the line is y= x , the other point of intersection must be (1, 1) . (b)
2 2 65. x y +xy=2 2 / 2 / x 2yy +y 2x+x y +y 1=0 / 2 2 y (2x y+x)= 2xy y
14 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 2 2xy +y / y = 2 2 . So 2x y+x 2xy +y
2 2 = 1 2 2xy +y=2x y+x y(2xy+1)=x(2xy+1) 2x y+x y(2xy+1) x(2xy+1)=0 (2xy+1)(y x)=0
2 2 4 must have x=y . Then x y +xy=2 2 1
1
or y=x . But xy=
2
2 xy= x +x =2 4 2 x +x 2=0 2 2 2 x y +xy= 2 1
4 1
2 2 , so we 2 (x +2)(x 1)=0 . So x = 2 , which is 2 impossible, or x =1 x= 1 . Since x=y , the points on the curve where the tangent line has a slope of
1 are ( 1, 1) and (1,1) .
x
2
2
. Let (a,b) be a point on x +4y =36 whose tangent line
4y
a
a
passes through (12,3) . The tangent line is then y 3=
(x 12) , so b 3=
(a 12) . Multiplying
4b
4b
2 2 / 66. x +4y =36 2x+8yy =0 2 / y = 2 2 2 2 2 both sides by 4b gives 4b 12b= a +12a , so 4b +a =12(a+b) . But 4b +a =36 , so 36=12(a+b)
2 a+b=3 2 2 2 2 2 b=3 a . Substituting 3 a for b into a +4b =36 gives a +4(3 a) =36 a +36 24a+4a =36
24
24
24
9
2
5a 24a=0 a(5a 24)=0 , so a=0 or a=
. If a=0 , b=3 0=3 , and if a=
, b=3
=
. So
5
5
5
5
24 9
the two points on the ellipse are (0,3) and
,
.
5
5
a
Using y 3=
(x 12) with (a,b)=(0,3) gives us the tangent line y 3=0 or y=3 . With
4b
24 9
24/5
2
2
(a,b)=
,
, we have y 3=
(x 12) y 3= (x 12) y= x 5 . A graph of the
5
5
4( 9/5)
3
3
ellipse and the tangent lines confirms our results. 1 67. (a) If y= f (x) , then f (y)=x . Differentiating implicitly with respect to x and remembering that y
dy
dy
1
1
/
1 /
is a function of x, we get f (y) =1, so
= /
( f ) (x)= / 1
.
dx
dx
f (y)
f ( f (x))
2
3
1
1 /
/
1
/
(b) f (4)=5 f (5)=4. By part (a), ( f ) (5)=1/ f ( f (5))=1/ f (4)=1/
= .
3
2
68. (a) f (x)=2x+cos x
one to one. / f (x)=2 sin x>0 for all x . Thus, f is increasing for all x and is therefore
15 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.6 Implicit Differentiation 1 (b) Since f is one to one, f (1)=k f (k)=1 . By inspection, we see that f (0)=2(0)+cos 0=1 , so 1 k= f (1)=0 .
1 / / 1 / (c) ( f ) (1)=1/ f ( f (1))=1/ f (0)=1/(2 sin 0)= 1
2 x
. Now let h be the height of the lamp, and let (a,b) be the
4y
point of tangency of the line passing through the points (3,h) and ( 5,0) . This line has slope
h 0
1
a
/
= h . But the slope of the tangent line through the point (a,b) can be expressed as y =
,
3 ( 5) 8
4b
b 0
b
a
b
2
2
2
2
2
2
or as
=
, so
=
4b = a 5a a +4b = 5a . But a +4b =5 , so 5= 5a a= 1 .
a ( 5) a+5
4b a+5
2 2 69. x +4y =5 2 / 2x+4(2yy )=0 / y = 2 Then 4b = a 5a= 1 5( 1)=4 b=1 , since the point is on the top half of the ellipse. So
h
b
1
1
=
=
=
h=2 . So the lamp is located 2 units above the x axis.
8 a+5
1+5 4 16 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives / 1. a= f , b= f // , c= f . We can see this because where a has a horizontal tangent, b=0 , and where b has a horizontal tangent, c=0 . We can immediately see that c can be neither f nor f
points where c has a horizontal tangent, neither a nor b is equal to 0 . / , since at the / 2. Where d has horizontal tangents, only c is 0 , so d =c . c has negative tangents for x<0 and b is the
/ only graph that is negative for x<0 , so c =b . b has positive tangents on R (except at x=0 ), and the
/ only graph that is positive on the same domain is a , so b =a . We conclude that d= f , c= f
and a= f /// / , b= f // . 3. We can immediately see that a is the graph of the acceleration function, since at the points where a
has a horizontal tangent, neither c nor b is equal to 0 . Next, we note that a=0 at the point where b has
/ a horizontal tangent, so b must be the graph of the velocity function, and hence, b =a . We conclude
that c is the graph of the position function.
4. a must be the jerk since none of the graphs are 0 at its high and low points. a is 0 where b has a
/ / maximum, so b =a . b is 0 where c has a maximum, so c =b . We conclude that d is the position
function, c is the velocity, b is the acceleration, and a is the jerk.
5 2 / 5. f (x)=x +6x 7x
8 6 6. f (t)=t 7t +2t 4 4 / 7 5 f (t)=8t 42t +8t / y = 2sin 2 8. y= sin y = cos +sin // / 3 6 4 f (t)=56t 210t +24t 2 y = 4cos 2 / 6 3 f (x)=20x +12 // 7. y=cos 2 9. F(t)=(1 7t)
2x+1
10. g(x)=
x 1 // f (x)=5x +12x 7 // y = ( sin )+cos
5 5 1+cos =2cos
// sin 4 4 F (t)=6(1 7t) ( 7)= 42(1 7t)
F (t)= 42 5(1 7t) ( 7)=1470(1 7t)
/
2
(x 1)(2) (2x+1)(1) 2x 2 2x 1
3
g (x)=
=
=
or 3(x 1)
2
2
2
(x 1)
(x 1)
(x 1)
//
3
3
6
g (x)= 3( 2)(x 1) =6(x 1) or
3
(x 1) 11. h(u)=
// 1 4u
1+3u / h (u)= (1+3u)( 4) (1 4u)(3)
2 (1+3u)
3 = 4 12u 3+12u
2 (1+3u) = 7
2 or 7(1+3u) 2 (1+3u) 3 h (u)= 7( 2)(1+3u) (3)=42(1+3u) or
1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives 42
3 (1+3u)
12. b
1/2
1/2
=as +bs
s
1 1/2
1 3/2
1
1/2 1
3/2
/
s
= as
bs
H (s) =a 2 s +b
2
2
2
1
1 3/2
1
3 5/2
1
3/2 3
//
s
b
s
=
as + bs
H (s) = 2 a
2
2
2
4
4
=a s + H(s) 2 h (x)= 1 2
(x +1)
2 2 x +1 1 x // h (x)= (
cx / 14. y=xe // x +1
cx 2/3 1
3 2 y =2x cx c+e y =
3 (x +1) 4/3 x 1/2 (2x)= 2 x +1 1/2 (2x) 2 (x +1) = 2 1/2 2 2 [(x +1) x ] 2 1 cx // 1=e (cx+1)
1/3 2 2 2 3 3 (3x )+(x +1) 1/3 2 3/2 (x +1) cx cx y =e (c)+(cx+1)e (3x )=2x (x +1) 1 = (x +1) 2 3
(x +1)
3 / 15. y=(x +1) ) 2 y =x e 3 1 2
(x +1)
2 / 13. h(x)= x +1 5/2 cx cx c=ce (1+cx+1)=ce (cx+2) 1/3
3 (4x)=4x(x +1) 1/3 4 3 2x (x +1) 4/3 16.
= y y / 4x
x+1
x+1 4 4x = 1
(x+1)
2 1/2 2 ( x+1 )
3/2 y // (x+1) 2 (2x+4) = = 4 x+1 2x/ x+1 4(x+1) 2x
2x+4
=
=
3/2
3/2
x+1
(x+1)
(x+1) 3
1/2
(x+1)
2 3/2 2 [(x+1) ]
2x+2 3x 6
x 4
=
=
5/2
5/2
(x+1)
(x+1) 1/2 = (x+1) [2(x+1) 3(x+2)]
3 (x+1) 17. H(t)=tan 3t
2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives / // 2 d
2
(sec 3t)=6sec 3t (3sec 3t tan 3t)=18sec 3t tan 3t
dt / 2 H (t)=2 3sec 3t H (t)=3sec 3t
2 18. g(s)=s cos s g (s)=2s cos s s sin s // 2 2
g (s)=2cos s 2s sin s 2s sin s s cos s=(2 s )cos s 4s sin s
3 5t / 19. g(t)=t e
// 3 5t 5t 2 2 5t g (t)=t e 5+e 3t =t e (5t+3) 5t 2 5t 2 5t g (t) =(2t)e (5t+3)+t (5e )(5t+3)+t e (5)
=te 5t 5t 2 2(5t+3)+5t(5t+3)+5t =te (25t +30t+6) 20. h(x)=tan 1 1 / 2 h (x)= (x ) 22 2x= 1+(x )
2 21. (a) f (x)=2cos x+sin x 2x
1+x // 4 h (x)= 4 3 (1+x )(2) (2x)(4x )
42 = (1+x ) 2 6x 4 42 (1+x ) / f (x)=2( sin x)+2sin x(cos x)=sin 2x 2sin x // f (x)=2cos 2x 2cos x=2(cos 2x cos x) (b)
/ We can see that our answers are plausible, since f has horizontal tangents where f (x)=0 , and f / // has horizontal tangents where f (x)=0 .
x 22. (a) f (x)=e x 3 / x 2 f (x)=e 3x // x f (x)=e 6x (b)
The graphs seem reasonable since f has horizontal tangents where f
is increasing, and f / / is zero, f / is positive where f is negative where f is decreasing; and the same relationships exist between f / // and f .
3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives y /// = 3
(2x+3)
2 24. y= x
2x 1 5/2 1
(2x+3)
2 / 1/2 23. y= 2x+3 =(2x+3) y = (2x 1)(1) x(2) / 2 3 y = 1( 2)(2x 1) (2)=4(2x 1)
y /// /// /// f or 1(2x 1) 2 3/2 2= (2x+3) 3/2 2 3
4 4 / f (t)=t( sin t)+cos t 1 (x)= /// // f (t)=t( cos t) sin t 1 sin t
/// (t)=tsin t cos t 1 cos t cos t=tsin t 3cos t , so f
/ g (x)= 3
(5 2x)
2 5/2 ( 2)= 3(5 2x)
2 1/2 ( 2)= (5 2x) 5/2 1/2 // g (x)= 2 2 3/2 ( 2)= (5 2x) 3/2 = 3.
2 cot )=2csc 2 cot )cot +2csc ( csc )= 2csc (2cot
2 1
(5 2x)
2 5/2 (2)= 3(1) f ( )= 2csc ( csc
2 ( )=2( 2csc /// , so g (0)=0 3= 3 . // f ( )= csc
2 /// 1
(5 2x)
2 / 27. f ( )=cot
f 1
(2x+3)
2 // y = (2x 1) 4 26. g(x)= 5 2x
g 1/2 =4( 3)(2x 1) (2)= 24(2x 1) or 24/(2x 1) 25. f (t)=tcos t
f 1 = (2x 1) // 2=(2x+3) 5/2 2=3(2x+3) y = 1/2 2 cot 2 +csc ) 2 = 2(2) [2( 3 ) +(2) ]= 80 6 / 28. g(x)=sec x g (x)=sec x tan x // 2 3 2 3 2 3 g (x)=sec x sec x+tan x(sec x tan x)=sec x+sec x tan x=sec x+sec x(sec x 1)=2sec x sec x
g /// 2 2 (x)=6sec x(sec x tan x) sec x tan x=sec x tan x(6sec x 1)
2 2 / 29. 9x +y =9
// y = 9 18x+2yy =0 y 1 x y / = 9 2 y / 2yy = 18x
y x( 9x/y)
2 2 4 = 2 (1)(6 2 1)=11 2 / 2 = 9 2 y +9x
y // /// y = 9x/y y
2 g 3 = 9 9
y 3 [ since x and y must satisfy the 3 original equation, 9x +y =9] . Thus, y = 81/y .
30. 4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives y
+
=0
2 x 2 y x + y =1 y 1 x // y (2 y ) = / 1 / 1 y (2 x ) y / y = x
y 1
y x
= 1
x y x
2x y 1+ x = 2x
x
x+ y
1
=
=
since x and y must satisfy the original equation, x + y =1 .
2x x
2x x
3 3 2 31. x +y =1 2 / 2 x / 3x +3y y =0 y = 2 y
2 2 2 y (2x) x 2yy // y = 2 2xy 2x y / 2 4 y = 22 x 2 = 4 (y ) 4 2xy +2x y
6 y 3 = 6 y
3 3 2xy(y +x ) 2x = y y , 5 3 since x and y must satisfy the original equation, x +y =1 .
4 4 32. x +y =a 4 3 3 3 / 2 3 2 y = / 4y y = 4x 3 / x
y 2 2 = 3x y 32 34. f (x)=
f /// n / n 1 1
=(5x 1)
5x 1 1 f /// / f (x)= 1(5x 1)
4 / 2 2x 2x 3 f (x)= 1
2 1
2 3 2x 4 3 5 f ( n) = 3x y // f (x)=( 1)( 2)(5x 1)
n n 4
7 4 2 = (x)=n(n 1)(n 2) f (x)=( 1) n!5 (5x 1)
2x a 2 y y n 2 2 4 3 3a x
y 7 2 1x n n =n! 2 5 (n+1) 2 2x (n) n 2x f (x)=2 e
/ f (x)=
x = 3x 4 y +x f (x)=2 2e =2 e (x)=2 2e =2 e
1/2 2 (n) 5 // f (x)=2e 36. f (x)= x =x
// // f (x)=n(n 1)x (x)=( 1)( 2)( 3)(5x 1)
2x 3 y f (x)=nx 35. f (x)=e 3 6 (y )
33. f (x)=x 3 3 y = x /y y x / y 3x x 3y y // 3 4x +4y y =0 3/2 f 1
x
2
/// 1/2 (x)= 1
2 1
2 3
2 x 5/2 5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives f
f
f ( 4)
( 5)
( n) 1
2 (x)= 3
2 5
2 1
2 3
2 5
2 1
2 (x)= 1
2 1
2 (x)= 1
2 3
2 3 37. f (x)=1/(3x )=
f
f /// (x)= ( n) 1
x
3 3 = 1 3 5
4 7/2 2
7
2 1
( 3)x
3 / x 9/2 x = 1 3 5 7
5 9/2 x 2
x n
( 2n 1 ) /2
=( 1) 1 1 3 5 (2n 3) x n ( 2n 1 ) /2 2 4 // f (x)= 1
( 3)( 4)x
3 5 6
n 1
(x)= ( 3)( 4)
3 n (n+2) 2 ( 1) (n+2)!
=
n+3
2
6x ( n+3 ) ( 1) 3 4 5
[ (n+2)]x
=
n+3
3x
2 D sin x= sin x 38. Dsin x=cos x 7/2 1
n+1
2 f (x)= 1
( 3)( 4)( 5)x
3 x 3 4 D sin x= cos x D sin x=sin x . The derivatives of sin x occur in 74 2 2 2 a cycle of four. Since 74=4(18)+2 , we have D sin x=D sin x= sin x .
/ // 3 3 39. Let f (x)=cos x . Then Df (2x)=2 f (2x) , D f (2x)=2 f (2x) , D f (2x)=2 f
(n) /// (2x) , ... , n (n) D f (2x)=2 f (2x) . Since the derivatives of cos x occur in a cycle of four, and since 103=4(25)+3 ,
we have f (103) 40. f (x)=xe
f /// (3) 103 (x)= f (x)=sin x and D x / x f (x)=x( e )+e =(1 x)e
x x (x)=(x 2)( e )+e =(3 x)e (n) n f (x)=( 1) (x n)e
1000 So D x x x x (4) 103 (103) 103 // x cos 2x=2
x f (2x)=2 sin 2x .
x f (x)=(1 x)( e )+e ( 1)=(x 2)e
x x f (x)=(3 x)( e )+e ( 1)=(x 4)e x x .
x xe =(x 1000)e . 41. By measuring the slope of the graph of s= f (t) at t=0 , 1 , 2 , 3 , 4 , and 5 , and using the method
/ of Example 1 in Section 3.2, we plot the graph of the velocity function v= f (t) in the first figure. The
// / acceleration when t=2 s is a= f (2) , the slope of the tangent line to the graph of f when t=2 . We
27
//
/
2
=9 ft / s . Similar measurements
estimate the slope of this tangent line to be a(2)= f (2)=v (2)
3
enable us to graph the acceleration function in the second figure. 6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives . 42. (a) Since we estimate the velocity to be a maximum at t=10 , the acceleration is 0 at t=10 .
2 (b) Drawing a tangent line at t=10 on the graph of a , a appears to decrease by 10 ft / s over a / 2 3 period of 20 s. So at t=10 s, the jerk is approximately 10/20= 0.5 (ft/s ) s or ft / s .
3 2 / 43. (a) s=2t 15t +36t+2 2 v(t)=s (t)=6t 30t+36 (b) a(1)=12 1 30= 18 m / s / a(t)=v (t)=12t 30 2 2 2 2 (c) v(t)=6(t 5t+6)=6(t 2)(t 3)=0 when t=2 or 3 and a(2)=24 30= 6 m / s , a(3)=36 30=6 m / s .
3 2 44. (a) s=2t 3t 12t / 2 / v(t)=s (t)=6t 6t 12 a(t)=v (t)=12t 6 2 (b) a(1)=12 1 6=6 m / s
2 (c) v(t)=6(t t 2)=6(t+1)(t 2)=0 when t= 1 or 2 . Since t 0,t 1 and 2 a(2)=24 6=18 m / s .
45. (a) s=sin t +cos 6 / v(t)=s (t)=cos t t ,0 t sin 6 6 t 2.
6 = cos 6 6 sin 6 t 6 cos 6 t 6 2 (b) a(1)= 6 t sin 6 t 2
/ a(t)=v (t)= 6 6 36 =
2 sin (c) v(t)=0 for 0 t 6
2 1 +cos cos 6 6 t =sin 1 =
6 36 36 sin 3
1
+
2
2 6 t +cos 6 t 2 = 72 (1+ 3 ) 0.3745 m / s 2 t
7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives sin 6 1=
cos
tan 6
6 t
t
1 t =1 6 2 3
a( )=
2
36 3
6 2 sin
3 t=tan 1 2 6
4 = 3
=1.5 s. Thus,
2
2 3
6 2 +cos
/ 46. (a) s=2t 7t +4t+1 = 2 36 = 36 2 2 0.3877 m / s . a(t)=v (t)=12t 14 2 ) 2 2 2
2
+
2
2
/ v(t)=s (t)=6t 14t+4 (b) a(1)=12 14= 2 m / s ( t= (c) v(t)=2 3t 7t+2 =2(3t 1)(t 2)=0 when t= 1
or 2 and a
3 1
3 =12 1
3 2 14= 10 m / s , 2 a(2)=12(2) 14=10 m / s .
4 3 / 3 2 / 2 a(t)=v (t)=12t 24t=12t(t 2)=0 when t=0 or 2 .
47. (a) s(t)=t 4t +2 v(t)=s (t)=4t 12t
(b) s(0)=2 m, v(0)=0 m / s, s(2)= 14 m, v(2)= 16 m / s
3 2 / 2 / a(t)=v ( t ) =12t 18=0 when t=1.5 . v(t)=s (t)=6t 18t
48. (a) s(t)=2t 9t
(b) s(1.5)= 13.5 m, v(1.5)= 13.5 m / s
3 2 49. (a) s= f (t)=t 12t +36t , t 0 / 2 v(t)= f (t)=3t 24t+36 . / 2 a(t)=v (t)=6t 24 . a(3)=6(3) 24= 6 ( m/s ) / s or m / s . (b)
(c) The particle is speeding up when v and a have the same sign. This occurs when 2<t<4 and when
t>6 . It is slowing down when v and a have opposite signs; that is, when 0 t<2 and when 4<t<6 . 50. (a) x(t)= t
1+t / a(t)=v (t)= / 2 v(t)=x (t)= 23 22 (1+t ) 2 2t (t 3) 2 (1+t )(1) t(2t) . a(t)=0 2 2t (t 3)=0 = 1 t 2 22 . (1+t )
t=0 or 3 (1+t )
(b) 8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives (c) v and a have the same sign and the particle is speeding up when 1<t< 3 . The particle is slowing
down and v and a have opposite signs when 0<t<1 and when t> 3 .
51. (a) y(t)=Asin / v(t)=y (t)=A cos t 2 2 / t 2 a(t)=v (t)= A sin t 2 (b) a(t)= A sin t=
(Asin t)=
y(t) , so a(t) is proportional to y(t) .
(c) speed = v(t) =A cos t is a maximum when cos t= 1 . But when cos
sin t=0 , and a(t)= A 2 sin t= 1 , we have 2 t= A (0)=0 . dv dv ds dv
dv
=
=
v(t)=v(t)
. The derivative dv/dt is the rate of change
dt ds dt ds
ds
of the velocity with respect to time (in other words, the acceleration) whereas the derivative dv/ds is
the rate of change of the velocity with respect to the displacement. 52. By the Chain Rule, a(t)= 2 / // // 53. Let P(x)=ax +bx+c . Then P (x)=2ax+b and P (x)=2a . P (2)=2
/ P (2)=3
P(2)=5 2(1)(2)+b=3 4+b=3 2 1(2) +( 1)(2)+c=5
3 2a=2 a=1 . b= 1 .
2 2+c=5 2 c=3 . So P(x)=x x+3 .
/ 2 // 54. Let Q(x)=ax +bx +cx+d . Then Q (x)=3ax +2bx+c , Q (x)=6ax+2b and Q
/ // Q ( 1 ) =a+b+c+d=1 , Q ( 1 ) =3a+2b+c=3 , Q (1)=6a+2b=6 and Q /// /// (x)=6a . Thus, (1)=6a=12 . Solving these four
3 2 equations in four unknowns a , b , c and d we get a=2 , b= 3 , c=3 and d= 1 , so Q(x)=2x 3x +3x 1
.
/ // // / 55. y=Asin x+Bcos x y =Acos x Bsin x y = Asin x Bcos x . Substituting into y +y 2y=sin x
gives us ( 3A B)sin x+(A 3B)cos x=1sin x , so we must have 3A B=1 and A 3B=0 . Solving for A
1
3
and B , we add the first equation to three times the second to get B=
and A=
.
10
10
2 56. y=Ax +Bx+C
// y +y / / y =2Ax+B // y =2A . We substitute these expressions into the equation 2 2y=x to get 2
2
(2A)+(2Ax+B) 2(Ax +Bx+C) = x
2 2A+2Ax+B 2Ax 2Bx 2C = x2
9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives 2 2 ( 2A)x +(2A 2B)x+(2A+B 2C) = (1)x +(0)x+(0)
2 The coefficients of x on each side must be equal, so 2A=1
1
and 2A+B 2C=0
2 A=B= rx / rx x / // 1 1
2C=0
2 2 rx // 57. y=e
y =re
y =r e , so y +5y
(r+6)(r 1)=0 r=1 or 6.
x y = e
58. y=e
1
5
=
, since e
2
2 59. f (x)=xg(x )
// / // 2 x / 2 2 rx rx 6y=r e +5re / y = e . Thus, y+y =y
x 3
.
4 C=
/ 1
. Similarly, 2A 2B=0
2 A= // x rx rx 2 rx 6e =e (r +5r 6)=e (r+6)(r 1)=0 x 2 e + e = e x x e ( 2 1)=0 0. / 2 2 2 / 2 3 // f (x)=x g (x ) 2x+g(x ) 1=g(x )+2x g (x ) 2 2 // 2 / 2 / 2 2 f (x)=g (x ) 2x+2x g (x ) 2x+g (x ) 4x=6xg (x )+4x g (x )
g(x)
60. f (x)=
x
2 // f (x)= / / f (x)= xg (x) g(x)
2 x / // / / x [g (x)+xg (x) g (x)] 2x[xg (x) g(x)]
x 4 2 = // / x g (x) 2xg (x)+2g(x)
x 3 61.
/ g ( x)
f (x) =g( x ) f (x)=g ( x ) 1 x 1/2=
2
2 x
1 1/2 /
1 1/2
//
2 x g ( x)
x
g ( x) 2
x
//
x
2
2
f (x) =
=
2
(2 x )
/ // = / 1/2 // / [ x g ( x ) g ( x )]
4x / xg ( x) g ( x)
4x x 62.
10 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives 5 3 / f (x)=3x 10x +5 4 2 // f (x)=15x 30x 3 2 f (x)=60x 60x=60x(x 1)=60x(x+1)(x 1) // So f (x)>0 when 1<x<0 or x>1 , and on these intervals the graph of f lies above its tangent lines;
// and f (x)<0 when x< 1 or 0<x<1 , and on these intervals the graph of f lies below its tangent lines.
63. (a)
f (x) 1 = (2x+1) / f (x)= 2 2 x +x
2 // f (x) = 2 (x +x)
2 2 (x +x) ( 2)+(2x+1)(2)(x +x)(2x+1)
2 4 2 = 2(3x +3x+1)
2 (x +x)
2 f /// (x) = 3 3 (x +x)
2 2 2 (x +x) (2)(6x+3) 2(3x +3x+1)(3)(x +x) (2x+1)
2 6 (x +x)
3 2 6(4x +6x +4x+1) = 2 4 (x +x)
2 (4) f (x) = 4 2 3 2 2 3 (x +x) ( 6)(12x +12x+4)+6(4x +6x +4x+1)(4)(x +x) (2x+1)
2 8 (x +x)
4 = 3 2 24(5x +10x +10x +5x+1)
2 5 (x +x)
(5) f (x) =?
(b) f (x)=
f /// 1
1
=
x(x+1) x 1
x+1 / 4 (x)=( 3)(2)x +(3)(2)(x+1)
7x+17 64. (a) For f (x)= 2 f (x)= x +(x+1)
4 f ( n) , a CAS gives us f 2 2 // f (x)=2x
n (x)=( 1) n! x
/// 3 ( n+1 )
4 (x)= 7x+17
2 = 2x 7x 4
(x)= 144 30
4 4 (2x+1) 3 ( n+1 )
2 2 4 (2x 7x 4) (b) Using a CAS we get f (x)=
/// (x+1) 3 6(56x +544x 2184x +6184x 6139) 2x 7x 4 f 2(x+1) 3
5
+
. Now we differentiate three times to obtain
2x+1 x 4 . (x 4) 65.
2 x / 2 x x x 2 For f (x)=x e , f (x)=x e +e (2x) =e (x +2x). Similarly, we have
11 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives // x 2 x 2 (4) x 2 (5) x 2 f (x) =e (x +4x+2)
/// f (x) =e (x +6x+6) f (x) =e (x +8x+12)
f (x) =e (x +10x+20)
It appears that the coefficient of x in the quadratic term increases by 2 with each differentiation. The
pattern for the constant terms seems to be 0=1 0, 2=2 1, 6=3 2, 12=4 3, 20=5 4. So a reasonable
(n) guess is that f (x)=e x 2 x +2nx+n(n 1) .
(n) Proof: Let S be the statement that f (x)=e x 2 x +2nx+n(n 1) . 1. S is true because n / x 1 2 f (x)=e (x +2x).
(k) 2. Assume that S is true; that is, f (x)=e x k f d
(x) = dx This shows that S k+1 (k) (k+1) =e x 2 x +2kx+k(k 1) . Then
x 2 f (x) =e (2x+2k)+ x +2kx+k(k 1) e
2 2 x +(2k+2)x+(k +k) =e x x 2 x +2(k+1)x+(k+1)k is true.
(n) 3. Therefore, by mathematical induction, S is true for all n; that is, f (x)=e
n x 2 x +2nx+n(n 1) for every positive integer n.
/ // F =( f // / / / / // // g+ f g )+( f g + fg )= f
/// /// // / // / / / / g+2 f g + fg
// / / F = f g+ fg 66. (a) Use the Product Rule repeatedly: F= fg // // / .
/// /// // / / // /// / ( n 1) (b) F = f g+ f g +2( f g + f g )+ f g + fg = f g+3 f g +3 f g + fg
/// /
/// /
// //
// //
/ ///
/ ///
( 4) ( 4)
( 4)
F = f g+ f g +3( f g + f g )+3( f g + f g )+ f g + fg
/// /
// //
/ ///
( 4)
( 4)
= f g+4 f g +6 f g +4 f g + fg
(c) By analogy with the Binomial Theorem, we make the guess:
F ( n) =f ( n) g+nf ( n 1) /
( n 2 ) //
g + n f
g +
2 n!
n(n 1)(n 2)
where n =
=
k!(n k)!
k!
k (n k+1) (n k) (k)
+ n f
g +
k +nf g + fg ( n) . 67. The Chain Rule says that
12 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.7 Higher Derivatives dy dy du
=
, so
dx du dx
2 d y
2 = dx d
dx dy
dx d
du = d
dx =
dy
du dy du
du dx d y
2 2 = dx
3 2 = du 2 2 du
du
dx d y
du 2 du
dx
du
dx d y 3 = du 3 2 2 2 3 dy d u
+
2
du
dx 2 2 d
+
dx 2 d
dx +
du
dx dy d u
2
du
dx
du
dx
2 2 2 2 d y
du
2 du d u d y
+2
+
2
2
dx
dx du 2 2 +
d
du d
dx // /// 2 dy
du
dy
du d u
2 + dx 2 du
dx 2 d
dx d u
2 dx dy
du d u
2 dx
3 d u dy
+
3 du
dx 3 du d u d y dy d u
+3
+
2
2 du
3
dx
dx du
dx 69. We will show that for each positive integer n , the n th derivative f
/ [Product Rule] 2 2 du
dx du
dx dy d u
+
2
du
dx 2 du
dx d y d y d
du = du 2 d
dx 2 2 du
dx d y 2 d d y d
=
=
3
2 dx
dx
dx
dx du dy d
+
dx du dx du dy d u d y
+
=
2
2
dx du
dx
du 2 d y dy
du 2 du
dx 68. From Exercise 65, d
dx = ( p 1) (n) ( p) exists and equals one of f ,
/ // /// ( p 1) f ,f ,f
. Since f = f , the first p derivatives of f are f , f , f
, ... , f
, ... , f
,
and f. In particular, our statement is true for n=1 . Suppose that k is an integer, k 1 , for which f is k
times differentiable with f
S={ f , f
so f
/// (k+1) / ,f // , ... , f (k) in the set ( p 1) } . Since f is p times differentiable, every member of S is differentiable, exists and equals the derivative of some member of S . Thus, f
( p) (k+1) is in the set { f / ,f // , ( p) , ... , f } , which equals S since f = f . We have shown that the statement is true for n=1 and
f
that its truth for n=k implies its truth for n=k+1 . By mathematical induction, the statement is true for
all positive integers n. 13 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions 1. The differentiation formula for logarithmic functions, d
1
(log x)=
, is simplest when a=e
a
dx
xln a because ln e=1.
/ d 2
2x
(x +10)= 2
2
dx
x +10
x +10 / 2 1 1
cos 2. f (x)=ln (x +10) f (x)= 3. f ( )=ln (cos ) f ( )= f (x)= sin (ln x) f (x)= 2 10 x
x 1 =log
1/5 5 5 10 f (x)= 1
(ln x)
5
/ f (x)=
1
x / 9. f (x)= x ln x f (x)= x f (x)= 10 1
ln x
5 1/5 x =ln x = / x log (x 1) / 7. f (x)= ln x =(ln x) 8. f (x)=ln 1
d
3
3
(1 3x)=
or
(1 3x)ln 2 dx
(1 3x)ln 2
(3x 1)ln 2 / 5. f (x)=log (1 3x) = tan 1
sin (ln x)
=
x
x / 4. f (x)=cos (ln x) 6. f (x)=log
d
sin
(cos )=
d
cos 1+ln t
1 ln t
(1 ln t)(1/t) (1+ln t)( 1/t) 4/5 1
xln 10 d
(ln x)=
dx 1
or
(x 1)ln 10
1
=
x 1
4/5 5(ln x) 1
x(x 1)ln 10 1
5 5x 4 (ln x) 1 1 1
=
5 x 5x +(ln x) 1
2 x = 1
ln x 2+ln x
+
=
x 2 x
2 x 10. f (t)=
/ f (t)= 2 = (1/t) (1 ln t)+(1+ln t)
2 (1 ln t)
3 11. F(t)=ln (2t+1) 4 (1 ln t)
3 2 = 2 t(1 ln t) 4 =ln (2t+1) ln (3t 1) =3ln (2t+1) 4ln (3t 1) (3t 1)
1
1
6
/
F (t)=3
2 4
3=
2t+1
3t 1
2t+1 12
6(t+3)
, or combined,
.
3t 1
(2t+1)(3t 1) 12.
1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions ( 2 h(x)=ln x+ x 1 ) 1 / h (x)= x 1+ 2 a x
=ln (a x) ln (a+x)
a+x
1
1
(a+x) (a x)
/
g (x)=
( 1)
=
=
a x
a+x
(a x)(a+x) = 2 x+ x 1 2 x 1 +x 1
2 x 1 2 x+ x 1 1 = 2 x 1 x 1 13. g(x)=ln y 1 / 14. F(y)=yln (1+e ) F (y)=y f (u) =
= 2 y ye y e +ln (1+e ) 1= y 1+e y +ln (1+e ) y ln u
1+ln (2u)
1+ln (2u) / 2 a x
y 1+e
15. f (u)= 2a 1
1
ln u
2
u
2u 1+ln (2u)
1+(ln 2+ln u) ln u
2 u 1+ln (2u)
4 2 2 = 1
1+ln (2u) ln u
u
1+ln (2u) 2 1+ln 2 = u 1+ln (2u) 4 2 2 2 17. y=ln |2 x 5x | 1 / y = 2 2 3u+2 1
=
ln (3u+2) ln (3u 2)
3u 2 2
x x x 10x+1 or 2 5x +x 2 2 x 5x 2 x 5x
18. G(u)=ln y =4 10x 1 ( 1 10x)= 1
1
4
+2
cos x= +2cot x
x
sin x
x / 16. y=ln (x sin x)=ln x +ln (sin x) =4ln x+2ln sin x / G (u)= 1
2 x 19. y=ln (e +xe )=ln (e (1+x))=ln (e )+ln (1+x)= x+ln (1+x) x 2 20. y=[ln (1+e )] / x y =2[ln (1+e )] 1
1+e 21. y=xln x / y =x(1/x)+(ln x) 1=1+ln x x x e= x 3
3u+2
/ 3
3u 2 y = 1+ = 6
2 9u 4 1
1 x+1
x
=
=
1+x
1+x
1+x x 2e ln (1+e )
1+e x // y =1/x
2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions 22. y= ln x 2 x (1/x) (ln x)(2x) / y = 2 2 2 (x ) x
// y = 3 2 x ( 2/x) (1 2ln x)(3x )
32 x 10 x = / 1
1
=
xln 10 ln 10 = 1
x 3 6ln x 5 // y = 4 1
ln 10 1
2 2 x ln 10 x 2 sec xtan x+sec x
y =
=sec x
sec x+tan x 1 = // y =sec xtan x x
1 ln (x 1) 25. f (x)= 1
x 1 [1 ln (x 1)] 1 x f (x) =
= 1 2ln x x / 24. y=ln (sec x+tan x) / 6 = x 2 x y = 4 x ( 2 3+6ln x) (x )
23. y=log x(1 2ln x) = 2 = (x 1)[1 ln (x 1)]+x
x 1 = 2 [1 ln (x 1)]
2x 1 (x 1)ln (x 1) x 1 (x 1)ln (x 1)+x [1 ln (x 1)] (x 1) 2 2 (x 1)[1 ln (x 1)] Dom( f ) ={ x| x 1>0 and 1 ln (x 1) 0}={ x| x>1 and ln (x 1) 1}
1 ={ x| x>1 and x 1 e }={ x| x>1 and x 1+e}=(1, 1+e) (1+e,
26. f (x) = 1
1+ln x / f (x) = 1/x
2 (1+ln x)
{ x| x>0 and x 1/e} = (0, 1/e) (1/e, ).
2 2 27. f (x) = x ln (1 x ) / 1 = 2 f (x) = 2xln (1 x )+ 2 ) . Dom( f ) = { x| x>0 and ln x 1} = x(1+ln x) 2 x ( 2x)
2 2 = 2xln (1 x ) 1 x 2x 3
2 . 1 x 2 Dom( f ) = { x|1 x >0} = { x | | x|<1} = ( 1, 1).
1
1 1
/
28. f (x) = ln ln ln x f (x) =
.
ln ln x ln x x
Dom( f ) = { x|ln ln x>0} = { x|ln x>1} = { x| x>e} = (e, ). 29. f (x) =
3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions x
ln x ln x x(1/x) / f (x) = 2 ln x 1 = / f (e) = 2 (ln x) (ln x) 1
x = 2xln x+x 1
ln x 1
x f (e) = / f (x) = / / f (1) = 2ln 1+1 = 1
1
, so an equation of the tangent line at
e 1
1
(x e), or y = x 1, or x ey = e.
e
e (e, 0) is y 0 = / 3 32. y=ln (x 7) y = 1 2 / 3x 3 y (2)= x 7
y 0=12(x 2) or y=12x 24. 12
=12, so an equation of a tangent line at (2, 0) is
8 7 / 33. f (x)=sin x+ln x 34. y= =0 1 2 / 31. y = f (x) = ln ln x / 2 f (x) = 2xln x+x 2 30. f (x) = x ln x when f 1 1 f (x)=cos x+1/x. This is reasonable, because the graph shows that f increases
/ is positive, and f (x)=0 when f has a horizontal tangent. ln x
x / y = x(1/x) ln x
2 = 1 ln x
2 / . y (1)= 1 0
2 / =1 and y (e)= x
x
1
lines are y 0=1(x 1) or y=x 1 and y 1/e=0(x e) or y=1/e. 5 4 6 5 35. y=(2x+1) (x 3) 4 y =y 3 10
24x
+ 4
2x+1
x 3 5 4 6 =(2x+1) (x 3) 2 =0 equations of tangent e 6 ln y=ln ((2x+1) (x 3) )
1 /
1
4
ln y=5ln (2x+1)+6ln (x 3)
y =5
2+6
y
2x+1
/ 1 1 1 4x 4 3 x 3
3 10
24x
+ 4
2x+1
x 3 .
4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions 2 x 2 2 10 x 36. y= x e (x +1) 1 / 1 1
y =
+2x+10
y
2 x
2 37. y= 1 2x 2 2 2 / 2 1
2
2
ln x+x +10ln (x +1)
2
1
20x
+2x+ 2
2x
x +1 10 x 10 ln y= y = x e (x +1) x +1 4 sin xtan x 2 ln y=ln x +ln e +ln (x +1) 2 4 2 2 ln y=ln (sin xtan x) ln (x +1) 2 (x +1)
2 4 2 2 2 ln y=ln (sin x) +ln (tan x) ln (x +1)
ln y=2ln |sin x|+4ln |tan x| 2ln (x +1)
1 /
1
1
1
2
y =2
cos x+4
sec x 2 2
2x
y
sin x
tan x
x +1
/ y = 2 2 4 4sec x
2cot x+
tan x sin xtan x
2 2 (x +1) 2 38. y= x +1 4 ln y= 2 x 1
2 x +1 / 4 y = 2 x 1
39. y=x x ln y=ln x 40. y=x sin x 2 2 x 43. y=(ln x) 2 4 x +1 2x 2 4 x 1 / 1
x 1 +(ln x) 1 2 1
4 2x 2 x +1 1 x 1
x 1
x +1 4 2 4 x 1 / y =y(1+ln x)
/ 1/x y =x x 2x 2 2 x = x 1 y /y=x(1/x)+(ln x) 1 y
1
=
y x / x y =x (1+ln x) 1 ln x
2 x / ln y=sin xln x sin x
+ln xcos x
x 42. y=(sin x) 1 / 1
y =
y
4 / ln y=xln x ln y=ln x x 1
=
2 x 1 sin x 41. y=x
/ x x +1
x 2 x +1 1
1
2
2
ln (x +1)
ln (x 1)
4
4 x 1
ln y= ln x
x 1/x y =y 1
2 4x / sin x y =x y
1
=(sin x) +(ln x)(cos x)
y
x sin x
+ln xcos x
x / ln y=xln (sin x) x ln y=ln (ln x) y
1
=x
cos x+ ln (sin x) 1
y
sin x / x y =(sin x) xcot x+ln (sin x) / ln y=xln ln x y
1 1
=x
+(ln ln x) 1
y
ln x x
5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions x
+ln ln x
xln x / y =y 44. y=x 45. y=x ln x e / ln y=ln xln x=(ln x) 46. y=(ln x) / 2 d 2 2
(x +y )
dx 1 / y = 2 2 x +y
/ 2 / x e x y =x e ln x+ 1
x y
1 1
=cos x
+(ln ln x)( sin x)
y
ln x x ln y=cos xln (ln x) 47. y=ln (x +y )
2 / 2ln x
x ln x cos x
sin xln ln x
xln x cos x y =(ln x) x y +y y / y =x / cos x 2 1
x y
x 1
x
=e
+(ln x) e
y
x x ln y=e ln x / / y
=2ln x
y 2 x 1
+ln ln x
ln x x y =(ln x) / 2 2yy =2x 2 / y = / / (x +y 2y) y =2x 2x+2yy y = 2 2 / 2 / x y +y y =2x+2yy 2 2 / / 2 x +y
2x
x +y 2y y 48. x =y x yln x=xln y
/ f (x)=1/(x 1)=(x 1) 49. f (x)=ln (x 1)
f /// (x)=2(x 1)
n 1 (n) f (x)=( 1) 3 1
1
/
/
+(ln x) y =x
y +ln y
x
y y (4) f (x)= 2 3(x 1) 1 / y ln x // f (x)= (x 1) x /
y
y =ln y
y
x / y = ln y y/x
ln x x/y 2 4
n n 1 2 3 4 ... (n 1)(x 1) =( 1) (n 1)!
n (x 1)
8 9 8 / 8 7 7 7 50. y=x ln x , so D y=D y =D (8x ln x+x ). But the eighth derivative of x is 0 , so we now have
8 7 7 6 6 7 6 D (8x ln x) = D (8 7x ln x+8x )=D (8 7x ln x)
= D6(8 7 6x5ln x)=...=D(8! x0ln x)=8!/x.
1
/
, so f (0)=1. Thus,
1+x
ln (1+x)
f (x)
f (x) f (0)
/
lim
=lim
=lim
= f (0)=1.
x
x 0
x 0
x 0 x
x 0
/ 51. If f (x)=ln (1+x), then f (x)= 6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.8 Derivatives of Logarithmic Functions 52. Let m=n/x. Then n=xm, and as n
,m
. Therefore,
x n
1 mx
1 m x x
lim 1+
= lim
1+
= lim
1+
=e by Equation 6.
n
m
m
n
m
m 7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions 1 0 0
(e e )=0
2
1 0 0 1
(b) cosh 0= (e +e )= (1+1)=1
2
2
1. (a) sinh 0= 0 2. (a) tanh 0= 0 0 0 (e e )/2 =0 (e +e )/2
1 (b) tanh 1= 1 1 e +e 3. (a) sinh (ln 2)=
(b) sinh 2= 2 1 e e = e 1 0.76159 2 e +1 e ln 2 e
2 ln 2 = e ln 2 (e
2 ln 2 1 ) 1 = 2 2
=
2 2 1
2 3
=
2
4 1 2 2
(e e ) 3.62686
2 1 3 3
(e +e ) 10.06766
2
1
ln 3
ln 3 3+
e +e
3 5
(b) cosh (ln 3)=
=
=
2
2
3
4. (a) cosh 3= 5. (a) sech 0= 1
1
= =1
cosh 0 1 1 (b) cosh 1=0 because cosh 0=1 .
6. (a) sinh 1= 1 1 1
(e e ) 1.17520
2
1 ( 2 ) (b) Using Equation 3, we have sinh 1=ln 1+ 1 +1 =ln (1+ 2 ) 0.88137 .
7. sinh ( x)= 1
e
2 8. cosh ( x)= 1
x
e +e
2 9. cosh x+sinh x= x e ( x) ( x) = 1
(e
2 = 1
x x 1
x
x
(e +e )= (e +e )=cosh x
2
2 x x e )= 1 x x
(e e )= sinh x
2 1 x x 1 x x 1
x
x
(e +e )+ (e e )= (2e )=e
2
2
2
1 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions 10. cosh x sinh x= 1 x x 1 x x 1
x
(e +e )
(e e )= (2e )=e
2
2
2
1 x x
(e e )
2 11. sinh xcosh y+cosh xsinh y=
1
x+y x
(e +e
4
1
x+y
= (2e
2e
4
= y e x+y x y )= e x y )+(e 1 x+y
e
e
2 x+y e (x+y) x y +e x+y 1 y y
(e +e ) +
2 1 x x
(e +e )
2 1 y y
(e e )
2 1 x x
(e e )
2 1 y y
(e e )
2 x y e ) =sinh (x+y) 1 x x
(e +e )
2 12. cosh xcosh y+sinh xsinh y= x 1 y y
(e +e ) +
2 1
x+y x y
x+y
x y
x+y x y
x+y
x y
(e +e +e +e )+(e
e
e +e )
4
1
x+y
x y 1
x+y
(x+y)
= (2e +2e )=
e +e
=cosh (x+y)
4
2
= 2 2 2 13. Divide both sides of the identity cosh x sinh x=1 by sinh x :
2 cosh x
2 sinh x
14. 2 sinh x
2 sinh x = 1 2 2 2 coth x 1=csch x. sinh x sinh xcosh y cosh xsinh y
+
sinh (x+y) sinh xcosh y+cosh xsinh y
cosh xcosh y cosh xcosh y
tanh (x+y) =
=
=
cosh (x+y) cosh xcosh y+sinh xsinh y cosh xcosh y sinh xsinh y
+
cosh xcosh y cosh xcosh y
tanh x+tanh y
=
1+tanh xtanh y
15. Putting y=x in the result from Exercise 11, we have
sinh 2x=sinh (x+x)=sinh xcosh x+cosh xsinh x=2sinh xcosh x.
16. Putting y=x in the result from Exercise 12, we have
2 2 cosh 2x=cosh (x+x)=cosh xcosh x+sinh xsinh x=cosh x+sinh x .
17.
ln x sinh (ln x) (e
e
= ln x
tanh (ln x) =
cosh (ln x)
(e +e ln x )/2 ln x )/2 = x (e ln x 1 x+(e ) ln x 1 ) = x x
x+x 1
1 2 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions 2 2 x 1/x (x 1)/x x 1
=
= 2
= 2
x+1/x
(x +1)/x x +1
18.
1 x x 1 x x
(e +e )+ (e e )
1+tanh x
1+(sinh x)/cosh x cosh x+sinh x 2
2
=
=
=
1 tanh x
1 (sinh x)/cosh x cosh x sinh x 1 x x 1 x x
(e +e )
(e e )
2
2
x = x x e +e +e e
x e +e x x e +e x
x = 2e x 2e 2x x =e x cosh x+sinh x e
2x
Or: Using the results of Exercises 9 and 10,
=
=e
x
cosh x sinh x
e
n xn nx 19. By Exercise 9, (cosh x+sinh x) =(e ) =e =cosh nx+sinh nx .
4
9
25
5
2
2
. cosh x=sinh x+1=
+1=
cosh x= (since cosh x>0 ).
3
16
16
4
4
3/4 3
5
coth x=1/cosh x= , tanh x=sinh x/cosh x=
= , and cothx=1/tanh x= .
5
5/4 5
3 20. sinh x= 3
4 csch x=1/sinh x= 4
5
4 2 9
3
2
2
>0 , so x>0 . cothx=1/tanh x= , sech x=1 tanh x=1
=
sech x=
5
4
5
25
5
5
4 5 4
3
(since sech x>0 ), cosh x=1/sech x= , sinh x=tanh xcosh x=
= , and csch x=1/sinh x= .
3
5 3 3
4
21. tanh x= 22. 23. (a) 3 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions x lim tanh x=lim
x x e e
x x e x e +e e x
x x x x x 1+e
e x e +e
x x x e e (b) lim tanh x= lim =lim 2x 1 e e 1 0
=1
1+0 2x x
x 2x = = lim
x e 1 2x = e +1 0 1
= 1
0+1 x e e
=
(c) lim sinh x=lim
2
x
x
x x e e
=
(d) lim sinh x= lim
2
x
x
2
(e) lim sech x=lim
=0
x
x
x
x
e +e
x (f) lim cothx=lim
x x e +e
x x e x x
x =lim 1+e 2x
2x = x
e e
e
1 e
cosh x
(g) lim cothx=lim
= , since sinh x
+
+ sinh x
x 0 x 0 x cosh x
=
sinh x 0 2 (i) lim csch x= lim
x x x 24. (a) 0 through positive values and cosh x 1. 0 (h) lim cothx=lim
x 1+0
=1
1 0 e e d
d
cosh x=
dx
dx d
d
tanh x=
(b)
dx
dx
(c) 0 through negative values and cosh x 1 x x
1 x x
(e +e ) = (e e )=sinh x
2
2
= cosh xcosh x sinh xsinh x
2 2 = 2 cosh x sinh x
2 cosh x
cosh x
1
cosh x
1
cosh x
=
=
= csch xcoth x
2
sinh x
sinh x sinh x
sinh x
1
sinh x
1
sinh x
=
=
= sech xtanh x
2
cosh x
cosh x cosh x
cosh x d
d
sech x=
dx
dx 1. =0 sinh x
cosh x d
d
csch x=
dx
dx (d) x , since sinh x 1 = 2 2 =sech x cosh x (e)
d
d
coth x =
dx
dx cosh x
sinh x = sinh xsinh x cosh xcosh x
2 sinh x 2 = 2 sinh x cosh x
2 sinh x = 1
2 sinh x 2 = csch x
4 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions 1 2 2 25. Let y=sinh x . Then sinh y=x and, by Example 1(a), cosh y sinh y=1
2 y 2 [ with cosh y>0 ] ( 2 cosh y= 1+sinh y = 1+x . So by Exercise 9, e =sinh y+cosh y=x+ 1+x
2 1 2 y=ln x+ 1+x ). 2 26. Let y=cosh x . Then cosh y=x and y 0 , so sinh y= cosh y 1 = x 1 . So, by Exercise 9,
y ( 2 2 ) y=ln x+ x 1 .
1 y y
Another method: Write x=cosh y= (e +e ) and solve a quadratic, as in Example 3.
2
e =cosh y+sinh y=x+ x 1 y y 2y y sinh y (e e )/2 e
e 1
27. (a) Let y=tanh x . Then x=tanh y=
= y y
= 2y
y
cosh y
(e +e )/2 e e +1
1 2y 2y xe +x=e
2y 1+x
e =
1 x 2y 1 2y 1+x=e xe
1+x
2y=ln
1 x 2y 1+x=e (1 x)
1
1+x
y= ln
2
1 x . 1 (b) Let y=tanh x . Then x=tanh y , so from Exercise 18 we have
1+x
1
1+x
2y 1+tanh y 1+x
e =
=
2y=ln
y= ln
.
1 tanh y 1 x
1 x
2
1 x
28. (a)
(i) 1 y=csch x csch y=x ( x 0 ) (ii) We sketch the graph of csch
y=x. 1 by reflecting the graph of csch (see Exercise 22) about the line
2 1 (iii) Let y=csch x . Then x=csch y= y e e 2 y y y xe xe =2 1 csch x=ln 1
+
x y x(e ) 2e x=0 1+ x +1
y 1
But e >0 , so for x>0 , e =
and for x<0 , e =
x
y y y2 y e= 1 2 x +1
.
x 2 x +1
. Thus,
x 2 x +1
| x| .
5 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions (b)
(i) 1 sech y=x and y>0. y=sech x 1 (ii) We sketch the graph of coth by reflecting the graph of coth (see Exercise 22) about the line
y=x.
2 1 (iii) Let y=coth x , so x=sech y= y y e +e But y>0 y xe +xe =2 y y2 y x(e ) 2e +x=0 y e >1 . This rules out the minus sign because 1 y e= 2 1 1 x
.
x 2 1 x
>1
x 1 2 1 x >x
2 2 2 1 x> 1 x 2 1 2x+x >1 x 1+ 1 x
x>1 , but x=sech y 1 . Thus, e =
x
y 2 x >x 2 1+ 1 x
x 1 sech x=ln . (c)
(i) 1 y=coth x coth y=x
1 (ii) We sketch the graph of coth by reflecting the graph of coth (see Exercise 22) about the line
y=x.
1 y Let y=coth x . Then x=coth y=
(iii)
2y=ln x+1
x 1 e +e
y e e
1
x+1
1
coth x= ln
2
x 1 y
y y y y xe xe =e +e y y (x 1)e =(x+1)e y 2y e = x+1
x 1 6 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions 1 29. (a) Let y=cosh x . Then cosh y=x and y 0
dy
1
=
=
dx sinh y 1 = 2 cosh y 1 1 sinh y dy
=1
dx (since sinh y 0 for y 0 ). Or: Use Formula 4. 2 x 1
2 1 (b) Let y=tanh x . Then tanh y=x sech y dy
=1
dx dy
=
dx 1
2 1 = sech y 2 1 tanh y = 1
2 . 1 x Or: Use Formula 5.
1 (c) Let y=csch x. Then csch y=x
2 coth y= csch ycoth y dy
=1
dx dy
1
=
. By Exercise 13,
dx
csch ycoth y 2 csch y+1 = 2 x +1 . If x>0 , then coth y>0 , so coth y= x +1 . If x<0 , then coth y<0 , so
dy
1
1
2
coth y= x +1 . In either case we have
=
=
.
dx
csch ycoth y
2
| x| x +1
dy
1
(d) Let y=sech x. Then sech y=x
sech ytanh y
=1
dx
dy
1
1
1
=
=
=
. (Note that y>0 and so tanh y>0 .)
dx
sech ytanh y
2
2
sech y 1 sech y
x 1 x
dy
1
1
1
2 dy
1
(e) Let y=coth x. Then coth y=x
csch y
=1
=
=
=
by Exercise
2
2
2
dx
dx
csch y 1 coth y 1 x
13.
/ 2 30. f (x)=tanh 4x f (x)=4sech 4x 31. f (x)=xcosh x f (x)=x(cosh x) +(cosh x)(x) =xsinh x+cosh x 2 / / / / 32. g(x)=sinh x g (x)=2sinh xcosh x
2 33. h(x)=sinh (x ) / 2 2 h (x)=cosh (x ) 2x=2xcosh (x ) 34. F(x)=sinh xtanh x / 2 F (x)=sinh xsech x+tanh xcosh x 1 cosh x
1+cosh x
(1+cosh x)( sinh x) (1 cosh x)(sinh x) 35. G(x)=
/ G (x) = 2 (1+cosh x) 7 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions sinh x sinh xcosh x sinh x+sinh xcosh x = 2 (1+cosh x)
t / 36. f (t)=e sech t 37. h(t)=coth 1+t t t / 2 2 / t t 1
2
(1+t )
2 2 1/2 (2t)= tcsch 1+t 2 t t t 2 t / cosh 3x 42. y=x sinh (2x) y =x / 1 sinh 3x 3=3e cosh 3x sinh 3x 1 2 x 1 2+sinh (2x) 2x=2x
2 2 1 ( x)
1 2 / 1 x + 2 1 x
1 1
2 1 1 1/2 1
2
ln (1 x )
2 = 1
2 x (1 x) 1 2 ( 2x)=tanh x 1 x
2 45. y=xsinh (x/3) 9+x
x
/
1
1/3
y =sinh
+x
3
2
1+(x/3)
46. y=sech 1
x
2 1 x =xtanh x+ 44. y=xtanh x+ln 1+4x 1 1 / y = x 1 +sinh (2x)
2 1+(2x) y =tanh x+ 2 y =cosh (cosh x) sinh x y =e 1 2 1
cosh t=coth t
sinh t / / 43. y=tanh 1+t H (t)=sech (e ) e =e sech (e ) 40. y=sinh (cosh x) 2 2 1+t h (t)= csch f (t)= 39. H(t)=tanh (e ) cosh 3x 2 (1+cosh x) f (t)=e ( sech ttanh t)+(sech t)e =e sech t (1 tanh t) 38. f (t)=ln (sinh t) 41. y=e 2sinh x = 2x
2 =sinh
2 9+x 1 x
3 + x x
2 9+x =sinh
2 9+x 1 x
3 2 1 x 8 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions 1 / y = 2 1 x
47. y=coth 1 2x
2 2 1 (1 x ) 2
2 2 (1 x )| x| 1 x
1 / x +1 x = y = 2x 2 2 1 (x +1) 2 1 = x +1 x 2 x +1 48.
For y=acosh (x/a) with a>0 , we have the y intercept equal to a. As a increases, the graph flattens. / x=7, we have y (7)=sinh
(b) If 1
=sinh (x/20) . Since the right pole is positioned at
20 / 49. (a) y=20cosh (x/20) 15 y =20sinh (x/20) 7
20 0.3572 . is the angle between the tangent line and the x axis, then tan = slope of the line =sinh , so =tan
=90 1 7
20 sinh 0.343 rad 19.66 . Thus, the angle between the line and the pole is 70.34 . 50. We differentiate the function twice, then substitute into the differential equation: y=
dy T
=
sinh
dx
g gx
T 2 g
gx
=sinh
T
T d y
2 dx We evaluate the two sides separately: LHS= d y
2 dx
g
T 1+ dy
dx 2 = g
T 51. (a) y=Asinh mx+Bcosh mx
// 2 2 1+sinh gx
T =cosh 2 RHS= 7
20 2 = T
gx
cosh
g
T g
g
gx
=
cosh
.
T
T
T g
gx
cosh
,
T
T gx
g gx
=
by the identity proved in Example 1(a).
T
T T / y =mAcosh mx+mBsinh mx
2 2 y =m Asinh mx+m Bcosh mx=m (Asinh mx+Bcosh mx)=m y
// (b) From part (a), a solution of y =9y is y(x)=Asinh 3x+Bcosh 3x. So 4=y(0)=Asinh 0+Bcosh 0=B,
9 Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.9 Hyperbolic Functions / so B= 4. Now y (x)=3Acosh 3x 12sinh 3x
sinh x 52. lim
x e x x e e =lim
x 2e x x 1 e
=lim
2
x 2x = / 6=y (0)=3A A=2, so y=2sinh 3x 4cosh 3x . 1 0 1
=
2
2
/ 1 53. The tangent to y=cosh x has slope 1 when y =sinh x=1 x=sinh 1=ln (1+ 2 ) , by Equation 3. 2 Since sinh x=1 and y=cosh x= 1+sinh x , we have cosh x= 2. The point is (ln (1+ 2 ), 2 ).
54.
1 ln (sec +tan ) ln (sec +tan )
e
+e
2
1
1
sec tan
sec +tan +
=
sec +tan +
sec +tan
2
(sec +tan )(sec
sec tan
1
sec +tan +
= (sec +tan +sec tan )=sec
2
2
2
sec
tan cosh x =cosh [ln (sec +tan )]=
1
2
1
=
2
= x tan ) x 55. If ae +be = cosh (x+ ) [ or sinh (x+ ) ], then
x x ae +be =
x of e and e 2
x (e x+ e x )= , we have a= 2 x 2 (e e x e e )= e (1) and b= 2 2
e e e x 2 e (2) . We need to find e x . Comparing coefficients and . Dividing equation (1) by
a
a
1
a
2
= e
(*)2 =ln (
)
= ln
. Solving equations (1) and
b
b
2
b
2a
2a
2
and e =
, so
= 4ab
=2
ab .
(2) for e gives us e =
=
2b
2b
a
a
(*) If >0 , we use the + sign and obtain a cosh function, whereas if <0 , we use the sign and
b
b
obtain a sinh function.
1
a
x
x
In summary, if a and b have the same sign, we have ae +be =2 ab cosh x+ ln
, whereas, if
2
b
1
a
x
x
a and b have the opposite sign, then ae +be =2 ab sinh x+ ln
.
2
b
equation (2) gives us 10 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values 1. A function f has an absolute minimum at x=c if f (c) is the smallest function value on the entire
domain of f , whereas f has a local minimum at c if f (c) is the smallest function value when x is
near c .
2. (a) The Extreme Value Theorem
(b) See the Closed Interval Method.
3. Absolute maximum at b ; absolute minimum at d ; local maxima at b and e ; local minima at d and
s;
neither a maximum nor a minimum at a , c , r , and t .
4. Absolute maximum at e ; absolute minimum at t ; local maxima at c , e , and s ; local minima at b ,
c , d , and r ;
neither a maximum nor a minimum at a .
5. Absolute maximum value is f (4)=4 ; absolute minimum value is f (7)=0 ; local maximum values
are f (4)=4 and f (6)=3 ; local minimum values are f (2)=1 and f (5)=2 .
6. Absolute maximum value is f (8)=5 ; absolute minimum value is f (2)=0 ; local maximum values
are f (1)=2 , f (4)=4 , and f (6)=3 ; local minimum values are f (2)=0 , f (5)=2 , and f (7)=1 .
7. Absolute minimum at 2, absolute maximum at 3, local minimum at 4 8. Absolute minimum at 1, absolute maximum at 5, local maximum at 2, local minimum at 4 9. Absolute maximum at 5, absolute minimum at 2, local maximum at 3, local minima at 2 and 4 1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values 10. f has no local maximum or minimum, but 2 and 4 are critical numbers 11. (a) (b) (c)
12. (a) Note that a local maximum cannot occur at an endpoint. 2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values (b) Note: By the Extreme Value Theorem, f must not be continuous.
13. (a) Note: By the Extreme Value Theorem, f must not be continuous; because if it were, it would
attain an absolute minimum. (b)
14. (a)
3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values (b)
15. f (x)=8 3x , x 1 . Absolute maximum f (1)=5 ; no local maximum. No absolute or local
minimum. 16. f (x)=3 2x , x 5 . Absolute minimum f (5)= 7 ; no local minimum. No absolute or local
maximum. 2 17. f (x)=x , 0<x<2 . No absolute or local maximum or minimum value. 4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values 2 18. f (x)=x , 0<x 2 . Absolute maximum f (2)=4 ; no local maximum. No absolute or local
minimum. 2 19. f (x)=x , 0 x<2 . Absolute minimum f (0)=0 ; no local minimum. No absolute or local
maximum. 2 20. f (x)=x , 0 x 2 . Absolute maximum f (2)=4 . Absolute minimum f (0)=0 . No local maximum
or minimum. 2 21. f (x)=x , 3 x 2 . Absolute maximum f ( 3)=9 . No local maximum. Absolute and local
minimum f (0)=0 . 5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values 2 22. f (x)=1+(x+1) , 2 x<5 . No absolute or local maximum. Absolute and local minimum f ( 1)=1
. 23. f (t)=1/t , 0<t<1 . No maximum or minimum. 24. f (t)=1/t , 0<t
maximum. 25. f ( )=sin
local minima f 1 . Absolute minimum f (1)=1 ; no local minimum. No local or absolute , 2 2 . Absolute and local maxima f
2 =f 3
2 3
2 =f 2 =1 . Absolute and = 1. 6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values 26. f ( )=tan , 4 < 2 . Absolute minimum f 4 = 1 ; no local minimum. No absolute or local maximum. 27. f (x)=1 x . Absolute maximum f (0)=1 ; no local maximum. No absolute or local minimum. x 28. f (x)=e . No absolute or local maximum or minimum value. 29. f (x)= { 30. f (x)= { 1 x if 0 x<2
2x 4 if 2 x 3
Absolute maximum f (3)=2 ; no local maximum. No absolute or local minimum. 2 x if 1 x<0
2 2 x if0 x 1
7 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values Absolute and local maximum f (0)=2 .
No absolute or local minimum. 2 / 31. f (x)=5x +4x
3 / f (x)=10x+4 . f (x)=0 2 / 2 x= / f (x)=3x +2x 1 . f (x)=0 32. f (x)=x +x x 2
, so
5 2
is the only critical number.
5
x= 1 , (x+1)(3x 1)=0 1
. These are the only
3 critical numbers.
3 2 / 33. f (x)=x +3x 24x
/ f (x)=0 ( 2 ) 2 f (x)=3x +6x 24=3 x +2x 8 . 3(x+4)(x 2)=0 x= 4 , 2 . These are the only critical numbers.
2 4 12
3 2
/
2
/
2
34. f (x)=x +x +x
. Neither of these is
f (x)=3x +2x+1 . f (x)=0 3x +2x+1=0 x=
6
a real number. Thus, there are no critical numbers.
4 3 35. s(t)=3t +4t 6t 2 / 3 2 (2 / s (t)=12t +12t 12t . s (t)=0 12t t +t 1 quadratic formula to solve the latter equation gives us t=
1.618 . The three critical numbers are 0 , 36. f (z)= z+1 f 2 z +z+1 / 1 5
2 ) 2 t=0 or t +t 1=0 . Using the 2 1 1 4(1)( 1)
1 5
=
2(1)
2 0.618 , . ( z2+z+1)1 (z+1)(2z+1) = z2 2z =0
(z)=
( z2+z+1) 2
( z2+z+1) 2 z(z+2)=0 z=0 , 2 are the 2 critical numbers. (Note that z +z+1 0 since the discriminant <0 .) 37. g(x)= 2x+3 =
/ { 2x+3
if 2x+3 0
(2x+3) if 2x+3<0 / g (x)= { 3
2
3
2 if x<
2 2 if x> / g (x) is never 0 , but g (x) does not exist for
8 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values x= 3
, so
2 3
is the only critical number.
2 1/3 38. g(x)=x x / / 2/3 g (x)= 1
x
3 2/3 + 2
x
3 5/3 = 1
x
3 x+2 5/3 (x+2)= 3x 5/3 . / g ( 2)=0 and g (0) does not exist, but 0 is not in the domain of g , so the only critical number is 2 .
2/3 39. g(t)=5t +t
/ g (t)= 5
t
3 / 5/3 g (t)= 10
t
3 1/3 + 5 2/3
/
t . g (0) does not exist, so t=0 is a critical number.
3 1/3 (2+t)=0 t= 2 , so t= 2 is also a critical number. 40. g(t)= t (1 t)=t
1 3t
0=g (t)=
2 t
/ 4/5 1/2 t 3/2 / g (t)= 1 3
/
t . g (0) does not exist, so t=0 is a critical number.
2 2 t
1
1
t= , so t= is also a critical number.
3
3
2 41. F(x)=x (x 4) 4 1/5 1 1/5
x = x (x 4) 5 x 2+(x 4) 4
5
5
8
/
(x 4)(14x 16) 2(x 4)(7x 8)
=
=
=0 when x=4 ,
; and F (0) does not exist.
1/5
1/5
7
5x
5x /
F (x) =x 4/5 2 2(x 4)+(x 4) 8
,4.
7 Critical numbers are 0 ,
3 42. G(x)= 2 x x
/ x=0 or 1 . G (x)=0 2/3
1 2
/
2
x x
(2x 1) . G (x) does not exist when x x=0 , that is, when
3
1
1
2x 1=0 x= . So the critical numbers are x=0 ,
,1.
2
2 ( / G (x)= 2 ) / / 43. f ( )=2cos +sin
f ( )= 2sin +2sin cos . f ( )=0 2sin (cos 1)=0 sin =0 or
cos =1
=n ( n an integer) or =2n . The solutions =n include the solutions =2n , so the
critical numbers are =n .
44. g( )=4 tan / 2 g ( )=4 sec / . g ( )=0 2 sec =4 sec = 2 cos = 1
2 = 3 +2n ,
9 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values 5
2
4
+2n ,
+2n , and
+2n are critical numbers.
3
3
3
Note: The values of / that make g ( ) undefined are not in the domain of g . / / 45. f (x)=xln x f (x)=x(1/x)+(ln x) 1=ln x+1 . f (x)=0
critical number is x=1/e .
2x / 46. f (x)=xe
2x+1=0 2x 2x 2x 2x f (x)=x(2e )+e =e (2x+1) . Since e
1
1
x=
. So
is the only critical number.
2
2
2 ln x= 1 1 x=e =1/e . Therefore, the only / is never 0 , we have f (x)=0 only when / 47. f (x)=3x 12x+5 , 0,3 . f (x)=6x 12=0 x=2 . Applying the Closed Interval Method, we find
that f (0)=5 , f (2)= 7 , and f (3)= 4 . So f (0)=5 is the absolute maximum value and f (2)= 7 is the
absolute minimum value.
3 / 2 48. f (x)=x 3x+1 , 0,3 . f (x)=3x 3=0 x= 1 , but 1 is not in 0,3 . f (0)=1 , f (1)= 1 , and
f (3)=19 . So f (3)=19 is the absolute maximum value and f (1)= 1 is the absolute minimum value.
3 2 / ( 2 ) 2 49. f (x)=2x 3x 12x+1 , [ 2,3] . f (x)=6x 6x 12=6 x x 2 =6 ( x 2 ) ( x+1 ) =0 x=2, 1 .
f ( 2)= 3 , f ( 1)=8 , f (2)= 19 , and f (3)= 8 . So f ( 1)=8 is the absolute maximum value and
f (2)= 19 is the absolute minimum value.
3 2 / 4 2 / )3 / ( 2 ) 2 50. f (x)=x 6x +9x+2 , [ 1,4] . f (x)=3x 12x+9=3 x 4x+3 =3(x 1)(x 3)=0 x=1,3 . f ( 1)= 14 ,
f (1)=6 , f (3)=2 , and f (4)=6 . So f (1)= f (4)=6 is the absolute maximum value and f ( 1)= 14 is the
absolute minimum value. ( 3 2 ) 51. f (x)=x 2x +3 , [ 2,3]. f (x)=4x 4x=4x x 1 =4x(x+1)(x 1)=0 x= 1 , 0 , 1 . f ( 2)=11 ,
f ( 1)=2 , f (0)=3 , f (1)=2 , f (3)=66 . So f (3)=66 is the absolute maximum value and f ( 1)=2 is the
absolute minimum value. ( 2 ( 2 )2 2 2 52. f (x)= x 1 , [ 1,2] . f (x)=3 x 1 ( 2x ) =6x ( x+1 ) ( x 1 ) =0 x= 1,0,1 . f ( 1)=0 , f (0)= 1
, and f (2)=27 . So f (2)=27 is the absolute maximum value and f (0)= 1 is the absolute minimum
value. 53. f (x)= x
2 x +1
, f (1)= , 0,2 . f / ( x2+1) x(2x) = 1 x2 =0
(x)=
( x2+1) 2 ( x2+1) 2 x= 1 , but 1 is not in 0,2 . f (0)=0 1
2
1
, f (2)= . So f (1)= is the absolute maximum value and f (0)=0 is the absolute
2
5
2
10 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values minimum value.
2 x 4 54. f (x)= , [ 4,4] . f 2 / x +4
f (0)= 1 . So f ( 4)=
2 55. f (t)=t 4 t , 1
2
f (t)=t
4 t
2 ( / 2 ) 2 4 2t =0 t =2 ( x2+4) (2x) ( x2 4) (2x) = 16x =0
(x)=
( x2+4) 2
( x2+4) 2 x=0 . f ( 4)= 12 3
= and
20 5 3
is the absolute maximum value and f (0)= 1 is the absolute minimum value.
5 1,2 .
1/2 ( 2t ) + ( 4 t 2 1/2 ) 1= t 2 2 + 4 t = 4 t
t= 2 , but t= 2 ( t + 4 t 2 4 t 2 )= 4 2t 2 2 is not in the given interval, 2 4 t / . f (t)=0
2 / 1,2 . f (t) does not exist if 2 4 t =0 t= 2 , but 2 is not in the given interval. f ( 1)= 3 , f ( 2 ) =2 , and f (2)=0 . So
f ( 2 ) =2 is the absolute maximum value and f ( 1)= 3 is the absolute minimum value.
3 56. f (t)= t ( 8 t ) , [0,8] . f (t)=8t 1/3 4/3 / t f (t)= / . f (t) does not exist if t=0 . f (0)=0 , f (2)=6
So f (2)=6 3 3 8
t
3 4 1/3 4
t = t
3
3 2/3 2/3 (2 t)= 3 t 0, . f (0)=1 , f / . f (x)=cos x sin x=0 sin x=cos x 3 = 2 1.41 , f 58. f (x)=x 2cos x ,
5
6 t=2 2 2 7.56 , and f (8)=0 . 3 +1
1.37 . So f
4
4
3
2
maximum value and f (0)=1 is the absolute minimum value. f 3 / . f (t)=0 2 is the absolute maximum value and f (0)= f (8)=0 is the absolute minimum value. 57. f (x)=sin x+cos x ,
x= 4(2 t) 5
6 = 3 / . f (x)=1+2sin x=0 , 0.886 , f / = 6 absolute maximum value and f
x = = 6
x x 6
x 6 sin x=
3 2 f (0)=0 , f (1)=e =1/e 0.37 , f (2)=2/e
f (0)=0 is the absolute minimum value. x= 5
,
6 tan x=1 = 2 is the absolute 4 6 . f( )=2 1.14 , 2.26 , f ( )= +2 5.14 . So f ( )= +2 is the 3 is the absolute minimum value. 59. f (x)=xe , 0,2 . f (x)=x( e )+e =e (1 x)=0
1 1
2 sin x
=1
cos x x=1 . 0.27 . So f (1)=1/e is the absolute maximum value and
11 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values ln x
/
x(1/x) ln x 1 ln x
, 1,3 . f (x)=
=
=0 1 ln x=0 ln x=1 x=e . f (1)=0/1=0 ,
2
2
x
x
x
f (e)=1/e 0.368 , f (3)=(ln 3)/3 0.366 . So f (e)=1/e is the absolute maximum value and f (1)=0 is
the absolute minimum value. 60. f (x)= 3 x 3
/
=
=0 x=3 . f does not exist for x=0 , but 0 is not in the
x
x
domain of f . f (1)=1 , f (3)=3 3ln 3 0.296 , f (4)=4 3ln 4 0.159 . So f (1)=1 is the absolute
maximum value and f (3)=3 3ln 3 0.296 is the absolute minimum value.
/ 61. f (x)=x 3ln x , 1,4 . f (x)=1 x 62. f (x)=e 2x e / x 2 2x , 0,1 . f (x)=e ( 1) e ( 2)= 1 2x x = 2 e x =0 2x x e =2 x=ln 2 0.69 . f (0)=0 e
e
e
1 1 1
1
1
2
, f (ln 2)=e
e
= e
e
=2 2 =
= , f (1)=e e
0.233 . So f (ln 2)=
2 4 4
4
is the absolute maximum value and f (0)=0 is the absolute minimum value.
ln 2 a 1 ( ) ( ) 2ln 2 ln 2 ln 2 2 1 2 b 63. f (x)=x (1 x) , 0 x 1 , a>0 , b>0 .
/ a b 1 f (x) =x b(1 x)
=x a 1 b ( 1)+(1 x) ax a 1 =x a 1 b 1 (1 x) x b( 1)+(1 x) a b 1 (1 x) (a ax bx)
/ At the endpoints, we have f (0)= f (1)=0 [ the minimum value of f ]. In the interval (0,1) , f (x)=0
a
x=
.
a+b
f So f a
a+b = a
a+b a a
a+b a
1
a+b b = a a a (a+b) a+b a
a+b b = b a a b b a a (a+b) b (a+b) = ab ( a+b ) a+b . a b = ab ( a+b ) a+b is the absolute maximum value. 64. / / We see that f (x)=0 at about x=0.0 and 2.0 , and that f (x) does not exist at about x= 0.7 , 1.0 , and
2.7 , so the critical numbers of f are about 0.7 , 0.0 , 1.0 , 2.0 , and 2.7 .
12 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values 65. (a)
From the graph, it appears that the absolute maximum value is about f ( 1.63)=9.71 , and the absolute
minimum value is about f (1.63)= 7.71 . These values make sense because the graph is symmetric
3 about the point ( 0,1 ) . ( y=x 8x is symmetric about the origin.)
3 / (b) f (x)=x 8x+1
8
3 f / 3 8
3 =
= 2 f (x)=3x 8 . So f (x)=0
8 8
3 +1= 32 6
8
16
+1=1
or
9
3
3 16
3 8
.
3 x=
8
3 8
3 8
+1
3 8 32 6
8
+1=1+
9
3 (From the graph, we see that the extreme values do not occur at the endpoints.) 66. (a)
From the graph, it appears that the absolute maximum value is about f ( 0.58)=1.47 , and the absolute
minimum value is about f ( 1)= f (0)=1.00 ; that is, at both endpoints.
(b) f (x)=e
and f ( x 3 x 1/3 ) =e / f (x)=e x 3 x 3 /9+ 3 /3 ( 3x2 1) . So f /(x)=0 on
2 3 /9 =e 1,0 x= 1/3 . f ( 1)= f (0)=1 (minima) (maximum). 67. (a)
From the graph, it appears that the absolute maximum value is about f (0.75)=0.32 , and the absolute
minimum value is f (0)= f (1)=0 ; that is, at both endpoints.
13 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values 1 2x / 2 f (x)=x (b) f (x)=x x x 2 + x x ( x 2x2) + ( 2x 2x2) =
= 2 2 x(3 4x)=0 3x 4x =0
3
4 and f 2 x x
3
x=0 or
. f (0)= f (1)=0 ,
4 = 3
4 3
4 2 3
4 = 2 2 3x 4x / . So f (x)=0
2 2 x x 2 x x 3 3
.
16 68. (a)
From the graph, it appears that the absolute maximum value is about f (5.76)=0.58 , and the absolute
minimum value is about f (3.67)= 0.58 .
(b) f (x)= cos x
2+sin x / So f (x)=0
and f 11
6 / f (x)= (2+sin x)( sin x) (cos x)(cos x) ( 2+sin x ) 1
7
11
x=
or
2
6
6
3 /2
1
=
=
.
3/2
3 sin x= 2 = 1 2sin x ( 2+sin x )
7
6 . Now f = 3 /2
=
3/2 2 . 1
,
3 mass
1000
3
=
(in g / cm ). But a critical point of will also be a
volume V (T )
d
2 dV
critical point of V since [
= 1000V
and V is never 0 ], and V is easier to differentiate than
dT
dT
.
69. The density is defined as = V (T )=999.87 0.06426T +0.0085043T
/ 2 0.0000679T 3 2 V (T )= 0.06426+0.0170086T 0.0002037T . Setting this equal to 0 and using the quadratic formula
to
find T , we get T = 0.0170086 2 0.0170086 4 0.0002037 0.06426
2( 0.0002037) Since we are only interested in the region 0 C T
1000
1.00013 ; (30)
and at 3.9665 C: (0)
999.87 3.9665 C or 79.5318 C. 30 C, we check the density
1000
0.99625 ;
1003.7628 at the endpoints 14 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values (3.9665) 1000
999.7447 70. F= W
sin +cos If tan = , then sin 1.000255 . So water has its maximum density at about 3.9665 C. dF ( sin +cos )(0) W ( cos sin )
W ( cos sin )
=
=
.
2
2
d
( sin +cos )
( sin +cos )
dF
sin
So
=0
cos sin =0
=
=tan . Substituting tan for in F gives us
d
cos
(tan )W
W tan
W tan cos
W sin
F=
=
=
=
=W sin .
2
2
2
(tan )sin +cos
1
sin
sin +cos
+cos
cos
= (see the figure), so F= 2 +1 +1 value of F at the endpoints: F(0)= W and F 2 , we have that =W . Now because 2 2 1 and +1 W 2 +1 W . We compare this with the 2 +1 is less than or equal to each of F(0) and F
value of F( ) , and it occurs when tan 2 . Hence, 2 W is the absolute minimum +1 = . 71. We apply the Closed Interval Method to the continuous function
5 4 3 2 I(t)=0.00009045t +0.001438t 0.06561t +0.4598t 0.6270t+99.33 on 0,10 . Its derivative is
/ 4 3 2 / I (t)=0.00045225t +0.005752t 0.19683t +0.9196t 0.6270 . Since I exists for all t , the only
/ critical numbers of I occur when I (t)=0 . We use a root finder on a computer algebra system (or a
/ graphing device) to find that I (t)=0 when t 29.7186 , 0.8231 , 5.1309 , or 11.0459 , but only the
second and third roots lie in the interval 0,10 . The values of I at these critical numbers are
I(0.8231) 99.09 and I(5.1309) 100.67 . The values of I at the endpoints of the interval are
I(0)=99.33 and I(10) 96.86 . Comparing these four numbers, we see that food was most expensive
at t 5.1309 (corresponding roughly to August, 1989) and cheapest at t=10 (midyear 1994).
15 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values 72. (a)
The equation of the graph in the figure is
3 2 v(t)=0.00146t 0.11553t +24.98169t 21.26872 .
/ 2 / / (b) a(t)=v (t)=0.00438t 0.23106t+24.98169 a (t)=0.00876t 0.23106 . a (t)=0
0.23106
t =
26.4 . a(0) 24.98 , a(t ) 21.93 , and a(125) 64.54 . The maximum acceleration is
1 0.00876
1
2 2 about 64.5 ft / s and the minimum acceleration is about 21.93 ft / s .
2 2 73. (a) v(r)=k(r r)r =kr r kr
0 3 / 2 / v (r)=2kr r 3kr . v (r)=0 0 0 kr(2r 3r)=0
0 r=0 or 2
r (but 0
3 0 1
2
1
1 3
r , r , and r , we get v
r = kr ,
0
2 0 3 0
2 0
8 0
2
4 3
4 1
2
v
r =
kr , and v(r )=0 . Since
> , v attains its maximum value at r= r . This
0
3 0
27 0
27 8
3 0
supports the statement in the text.
4 3
kr .
(b) From part (a), the maximum value of v is
27 0
is not in the interval). Evaluating v at (c)
3 / 2 / 74. g(x)=2+(x 5) g (x)=3(x 5) g (5)=0 , so 5 is a critical number. But g(5)=2 and g takes on
values >2 and values <2 in any open interval containing 5 , so g does not have a local maximum or
minimum at 5 .
101 51 / 100 50 / 75. f (x)=x +x +x+1 f (x)=101x +51x +1 1 for all x , so f (x)=0 has no solution. Thus,
f (x) has no critical number, so f (x) can have no local maximum or minimum.
76. Suppose that f has a minimum value at c , so f (x) f (c) for all x near c . Then
g(x)= f (x)
f (c)=g(c) for all x near c , so g(x) has a maximum value at c .
/ 77. If f has a local minimum at c , then g(x)= f (x) has a local maximum at c , so g (c)=0 by the case
16 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values / / of Fermat’s Theorem proved in the text. Thus, f (c)= g (c)=0 .
78.
(a) 3 2 / 2 f (x)=ax +bx +cx+d , a 0 . So f (x)=3ax +2bx+c is a quadratic and hence has either 2 , 1 ,
or 0 real roots, so f (x) has either 2 , 1 or 0 critical numbers.
Case (i) (2 critical numbers):
3 f (x)=x 3x
/ 2 f (x)=3x 3 , so x= 1,1
are critical numbers. Case (ii) (1 critical
number): Case (iii) (no critical number): f(x)=x f (x)=x +3x 3 / /
2
f (x)=3x,so x=0
f (x)=3x +3 ,
is the only critical number. so there are no real roots. (b) Since there are at most two critical numbers, it can have at most two local extreme values and
by (i) this can occur. By (iii) it can have no local extreme value. However, if there is only one
critical number, then there is no local extreme value. 17 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives 2+1 1+1 2 x
x
3
2
F(x)=6
8
+3x+C=2x 4x +3x+C
2+1
1+1 / 2 1. f (x)=6x 8x+3 2 Check: F (x)=2 3x 4 2x+3+0=6x 8x+3= f (x)
2 3 2. f (x)=4+x 5x F(x)=4x+ 1 3 5 4
x
x +C
3
4
3+1 5+1 7+1 x
x
x
1 4 5 6 3 8
3. f (x)=1 x +5x 3x
F(x)=x
+5
3
+C=x
x+ x
x +C
3+1
5+1
7+1
4
6
8
1 21 4 11
20
10
4. f (x)=x +4x +8 F(x)=
x +
x +8x+C
21
11
3 5 7 1/4+1 1/4 5. f (x)=5x 7x 1.7 2 6. f (x)=2x+3x x =6x 1/2+1 8. f (x)= 9. f (x)= 3 10 4 { 9 1/6
3/2 3/4 x =x +x
9 =10x 7/6 8 8 x
3/2 6 7/6
+C=4x
x +C
7/6
7 7/4 5
4x 8 ,0 ) ( 0, +C ) , so if x<0 1 +C 2 7/3 x
x
4 7/4 3 7/3
F(x)=
+
+C= x + x +C
7/4 7/3
7
7 4/3 has domain ( 10x
+C =
1
8
5 7/4 3 2.7
2 10 2.7
x +C=x +
x +C
2.7
9 x
x
+C=6
1
3/2
+1
6 x + x F(x)= 3 5/4 x 1/6+1 x
F(x)=6
1
+1
2
4 F(x)=x +
1/2 6 7. f (x)=6 x 3/4+1 x
x
x
x
5/4
7/4
F(x)=5
7
+C=5
7
+C=4x 4x +C
1
3
5/4
7/4
+1
+1
4
4 3/4 4x
See Example 1 for a similar problem. if x>0 10.
1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives 3 g(x)= 5 4x +2x
x G(x)= 6 6 { 5 x 5 1
5 2 2 + 2 u +3 u
u 3 1 2 x 5 + 2 u 4
2 x if x>0 + 3u 1/2 u 2 =u +3u 2 3/2 1/2 2 12. f (x)=3e +7sec x 14. h( )= 5sin sin
cos = 2 ( 2 6
+C
u x F(x)=3e +7tan x+C on the interval
n G( )=sin 1
cos 15. f (x)=2x+5 1 x if x<0 1 2 3/2+1 13. g( )=cos ) , so +2x+C 2 u
u
1 3 u
1 3
F(u)= +3
+C= u +3
+C= u
3
3/2+1
3
1/2
3
x ,0 ) ( 0, +2x+C u = 1 +2x+C = x 4 11. f (u)= 3 4x +2 has domain (
2 x 4 5 x 6 =5x sin
cos ) 1/2=2x+ n 2 ,n + 2 . 5( cos )+C=sin +5cos +C =sec 5 tan H( )=sec +C on the interval
n 2 n 2 ,n + 2 . 1 F(x)=x +5sin x+C
2 1 x
2 16. f (x)= x +x+1
1
=x+1+
x
x 4 17. f (x)=5x 2x 5 F(x)= 5 {
6 1 2
x +x+ln x +C if x<0
1
2
1 2
x +x+ln x +C if x>0
2
2 x
x
5 1 6
F(x)=5
2
+C=x
x +C . F(0)=4
5
6
3 5 0 1 6
0 +C=4
3 C=4 , so 1 6
x +4 . The graph confirms our answer since f (x)=0 when F has a local maximum, f is
3
positive when F is increasing, and f is negative when F is decreasing. F(x)=x 5 2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives ( 2 18. f (x)=4 3 1+x ) 1=4 3 1 F(x)=4x 3tan x+C . F(1)=0 2 4 3 1+x +C=0 4 C= 3
4 4 , so 3
4 . Note that f is positive and F is increasing on R . Also, f has smaller
4
values where the slopes of the tangent lines of F are smaller.
1 F(x)=4x 3tan x+ 19. f / / / (x)=6x+12x
3 2 3 x
x
2
3
f (x)=6
+12
+C=3x +4x +C
2
3 2 4 x
x
3 4
f (x)=3
+4
+Cx+D=x +x +Cx+D
3
4
20. f 21. f
22. f
23. f 24. f
f (t)= / / / / / / 3 (x)=2+x +x (x)=1+x / / / (t)=60t (t)=t f (x)=2x+
/ 4/5 (x)=cos x / / / / 6 2 t f (x)=x+ 1 4 1 7
x + x +C
4
7 5 9/5
x +C
9 / f (x)=sin x+C
f f / / 3 (t)=20t +C / / (t)= 2 f (x)=x + f (x)= 1 5 1 8
x+
x +Cx+D
20
56 1 2 5 5 14/5
1 2 25 14/5
x+
x +Cx+D= x +
x +Cx+D
2
9 14
2
126 f (x)= cos x+Cx+D
/ 4 f (t)=5t +Ct+D 1 2 2 3/2
t
t +C
2
3 / f (t)= 5 f (t)=t + 1 2
Ct +Dt+E
2 1 3 4 5/2
t
t +Ct+D
6
15 1 4 8 7/2 1 2
t
t + Ct +Dt+E
24
105
2
/ 25. f (x)=1 6x 2 f (x)=x 3x +C . f (0)=C and f (0)=8 2 C=8 , so f (x)=x 3x +8 .
3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives / 3 4 26. f (x)=8x +12x+3 4 11+C=6 2 C= 5 , so f (x)=2x +6x +3x 5 . / 1/2 27. f (x)= x (6+5x)=6x +5x / 4 4 2 3 3 C=1 , so f (x)=x +1/x +1 . 2 f (t)=2sin t+tan t+C because 29. f (t)=2cos t+sec t =2 ( 3 /2 ) + 3 +C=2 3 +C and f / 30. f (x)=3x 2 f (x)= f ( 1)=3+C =0 { 3 3/x+C
3/x+C C=4 2 3 , so f (t)=2sin t+tan t+4 2 3 . =4 if x<0 f (1)= 3+C =0 2 2 /2<t< /2 . if x>0 1 C = 3 . So f (x)= 2 5/2 f (x)=x +x +C because we’re given that x>0 .
2 f (1)=2+C and f (1)=3 3 5/2 f (x)=4x +2x +C . C=4 , so f (x)=4x +2x +4 . 28. f (x)=2x 3/x =2x 3x / 3/2 3/2 3/2 f (1)=6+C and f (1)=10 f 2 f (x)=2x +6x +3x+C . f (1)=11+C and f (1)=6 { 1 C =3 ,
1 3/x+3 if x>0
3/x 3 if x<0 / 31. f (x)=2/x f (x)=2ln x +C=2ln ( x)+C (since x<0 ). Now f ( 1)=2ln 1+C=2(0)+C=7 C=7 .
Therefore, f (x)=2ln ( x)+7 , x<0 .
1
=4sin
2
2
2
1
C=1
, so f (x)=4sin x+1
.
3
3 / 1 2 f (x)=4sin x+C . f 32. f (x)=4/ 1 x
2
+C=1
3
33. f / / 2 / 3 2 1
2 +C=4 / 6 +C and f 1
2 =1 / f (x)=8x +x +10x+C . f (1)=8+1+10+C and f (1)= 3 19+C= 3
1
/
3 2
4 1 3
2
C= 22 , so f (x)=8x +x +10x 22 and hence, f (x)=2x + x +5x 22x+D . f (1)=2+ +5 22+D and
3
3
7 59
59
4 1 3
2
f (1)=5 D=22
=
, so f (x)=2x + x +5x 22x+
.
3 3
3
3
34. f / / (x)=24x +2x+10 1 (x)=4 6x 40x / 2 3 4 / 2 4 / / f (x)=4x 3x 10x +C . f (0)=C and f (0)=1
2 3 5 C=1 , so f (x)=4x 3x 10x +1 and hence, f (x)=2x x 2x +x+D . f (0)=D and f (0)=2
2 3 D=2 , so 5 f (x)=2x x 2x +x+2 .
35.
4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives f / / / ( )=sin +cos / / f ( )= cos +sin +5 and hence, f ( )= sin
f ( )= sin cos +5 +4 .
36. f / / (t)=3/ t =3t hence, f (t)=4t 3/2 / f ( )= cos +sin +C . f (0)= 1+C and f (0)=4 1/2 / f (t)=6t 1/2 cos +5 +D. f (0)= 1+D and f (0)=3 / / / 2 2 C= 5 , so f (t)=6t D=8 , so f (t)=4t 3/2 1/2 5 and 5t+8 . 3 37. f (x)=2 12x f (x)=2x 6x +C f (x)=x 2x +Cx+D .
f (0)=D and f (0)=9 D=9 . f (2)=4 16+2C+9=2C 3 and f (2)=15
2 D=4 , so / +C . f (4)=12+C and f (4)=7 5t+D . f (4)=32 20+D and f (4)=20 / / C=5 , so 2C=18 C=9 , so 3 f (x)=x 2x +9x+9 .
38. f / / 3 2 / (x)=20x +12x +4 4 3 f / / = 2 / (x)=2+cos x
2 /4+ 2 2 f (x)=x cos x
40. f / / 2 f (x)=x cos x+Cx+D . f (0)= 1+D and f (0)= 1
2 2 C= /4 C= 2 D=0 . , so x. 2
t / (t)=2e +3sin t t t f (t)=2e 3cos t+C . f ( )=2e + C 2 and f ( )=0
/ / 4 C= 7 , so f (x)=x +x +2x 7x+8 . =0 2 2 2 f (x)=2x+sin x+C
C and f 4 f (x)=x +x +2x +Cx+D . f (0)=D and f (0)=8
5 D=8 . f (1)=1+1+2+C+8=C+12 and f (1)=5
39. f 5 f (x)=5x +4x +4x+C 2 f (t)=2e 3sin t+Ct+D . f (0)=2+D and f (0)=0 C=2 2e C= 2 2e t , so f (t)=2e 3sin t+ 2 2e D= 2 t 2. / 41. f (x)=x , x>0 f (x)= 1/x+C f (x)= ln x +Cx+D= ln x+Cx+D
(since x>0 ). f (1)=0 C+D=0 and f (2)=0
ln 2+2C+D=0
ln 2+2C C=0 [ since D= C ]
ln 2+C=0 C=ln 2 and D= ln 2 .
So f (x)= ln x+(ln 2)x ln 2 .
42. f
f / / / / / (x)=sin x (x)= cos x+2 1= f (0)=1+E / / f (x)= cos x+C 1= f / / / (0)= 1+C / f (x)= sin x+2x+D 1= f (0)=D C=2 , so
/ f (x)= sin x+2x+1 2 f (x)=cos x+x +x+E 2 E=0 , so f (x)=cos x+x +x . / 2 43. Given f (x)=2x+1 , we have f (x)=x +x+C . Since f passes through ( 1,6 ) , f (1)=6
2 2 1 +1+C=6 2 C=4 . Therefore, f (x)=x +x+4 and f (2)=2 +2+4=10 .
5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives / 44. f (x)=x 3 f (x)= 1 4
x +C . x+y=0
4 y= x / m= 1 . Now m= f (x) 1=x 3 x= 1 the equation of the tangent line), so ( 1,1 ) is a point on the graph of f . From f , 1=
C= y=1 (from 1
4
( 1) +C
4 3
1 4 3
. Therefore, the function is f (x)= x + .
4
4
4 45. b is the antiderivative of f . For small x , f is negative, so the graph of its antiderivative must be
decreasing. But both a and c are increasing for small x , so only b can be f ’s antiderivative. Also, f
is positive where b is increasing, which supports our conclusion.
46. We know right away that c cannot be f ’s antiderivative, since the slope of c is not zero at the x
value where f =0 . Now f is positive when a is increasing and negative when a is decreasing, so a is
the antiderivative of f .
/ 47. The graph of F will have a minimum at 0 and a maximum at 2 , since f =F goes from negative
to positive at x=0 , and from positive to negative at x=2 . 48. The position function is the antiderivative of the velocity function, so its graph has to be
horizontal where the velocity function is equal to 0 . 49. 6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives { { 2 if0 x<1
2x+C if0 x<1
f (x)=
1 if1<x<2
f (x)=
x+D if1<x<2
f (0)= 1 2(0)+C= 1 C= 1 .
1 if2<x 3
x+E if2<x 3
Starting at the point ( 0, 1 ) and moving to the right on a line with slope 2 gets us to the point ( 1,1 ) .
The slope for 1<x<2 is 1 , so we get to the point ( 2,2 ) . Here we have used the fact that f is
continuous. We can include the point x=1 on either the first or the second part of f . The line
connecting ( 1,1 ) to ( 2,2 ) is y=x , so D=0 . The slope for 2<x 3 is 1 , so we get to ( 3,1 ) . f (3)=1
3+E=1 E=4 . Thus,
/ f (x)= { 2x 1 if 0 x 1
x
if 1<x<2
x+4 if 2 x 3 / Note that f (x) does not exist at x=1 or at x=2 . 50. (a)
(b) Since F(0)=1 , we can start our graph at ( 0,1 ) . f has a minimum at about x=0.5 , so its derivative
is zero there. f is decreasing on ( 0,0.5) , so its derivative is negative and hence, F is CD on ( 0,0.5)
/ and has an IP at x 0.5 . On ( 0.5,2.2 ) , f is negative and increasing ( f is positive), so F is
decreasing and CU. On ( 2.2, ) , f is positive and increasing, so F is increasing and CU. (c) f (x)=2x 3 x 2 F(x)=x 3 2 3/2
x +C . F(0)=C and F(0)=1
3 2 3/2 C=1 , so F(x)=x 2x +1 . (d)
7 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives 2 51. f (x)=sin (x ) , 0 x 4 4 52. f (x)=1/(x +1) 53. 54.
8 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives 55.
x
0
0.5
1.0
1.5
2.0
2.5
3.0 f (x)
1
0.959
0.841
0.665
0.455
0.239
0.047 x
3.5
4.0
4.5
5.0
5.5
6.0 f (x)
0.100
0.189
0.217
0.192
0.128
0.047 We compute slopes [values of f (x)=(sin x) / x for 0 < x < 2 ] as in the table lim f (x)=1
x and + 0 draw a direction field as in Example 6. Then we use the direction field to graph F starting at ( 0,0 ) 56.
x
0
0.2
0.4
0.6
0.8
1.0
1.2 f (x)
0
0.041
0.169
0.410
0.824
1.557
3.087
9 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives 1.4
1.5 8.117
21.152 We compute slopes [values of f (x) = x tan x for
/2 < x
/ 2 ] as in the table and draw a
direction field as in Example 6. Then we use the direction field to graph F starting at ( 0,0 ) and
extending in both directions. Note that if f is an even function, then the antiderivative F that passes
through the origin is an odd function.] 57. Remember that the given table values of f are the slopes of F at any x . For example, at x=1.4 , the
slope of F is f (1.4)=0 . 58. (a) (b) The general antiderivative of f (x)=x 2 { 1/x+C 1 if x<0 since f (x) is not defined
if x>0
2
at x=0 . The graph of the general antiderivatives of f (x) looks like the graph in part (a), as expected.
/ 59. v(t)=s (t)=sin t cos t
.
/ 60. v(t)=s (t)=1.5 t is F(x)= 1/x+C s(t)= cos t sin t+C . s(0)= 1+C and s(0)=0 s(t)=t 3/2 +C . s(4)=8+C and s(4)=10 C=1 , so s(t)= cos t sin t+1 C=2 , so s(t)=t 3/2 +2 .
10 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives / 61. a(t)=v (t)=t 2
s(t)= v(t)= 1 2
t 2t+C . v(0)=C and v(0)=3
2 1 3 2
t t +3t+D . s(0)=D and s(0)=1
6 D=1 , and s(t)= C=3 , so v(t)= 1 2
t 2t+3 and
2 1 3 2
t t +3t+1 .
6 / 62. a(t)=v (t)=cos t+sin t v(t)=sin t cos t+C 5=v(0)= 1+C C=6 , so v(t)=sin t cos t+6
s(t)= cos t sin t+6t+D 0=s(0)= 1+D D=1 , so s(t)= cos t sin t+6t+1 .
/ 63. a(t)=v (t)=10sin t+3cos t
and s(2 )= 3+2 C+D=12
/ 64. a(t)=v (t)=10+3t 3t 2 v(t)= 10cos t+3sin t+C s(t)= 10sin t 3cos t+Ct+D . s(0)= 3+D=0
6
6
D=3 and C=
. Thus, s(t)= 10sin t 3cos t+ t+3 . v(t)=10t+ 3 2 3
t t +C
2
2 10=s(2)=20+4 4+2C C= 5 , so s(t)= 5t+5t + 2 s(t)=5t + 1 3 1 4
t
t +Ct+D
2
4 0=s(0)=D and 1 3 1 4
t
t .
2
4 65. (a) We first observe that since the stone is dropped 450 m above the ground, v(0)=0 and s(0)=450
.
/ v (t)=a(t)= 9.8
D=450 v(t)= 9.8t+C . Now v(0)=0 C=0 , so v(t)= 9.8t 2 s(t)= 4.9t +D . Last, s(0)=450 2 s(t)=450 4.9t .
2 2 (b) The stone reaches the ground when s(t)=0 . 450 4.9t =0 t =450/4.9 t = 450/4.9 9.58 s.
1 ( ) (c) The velocity with which the stone strikes the ground is v t = 9.8 450/4.9 93.9 m / s. 1 (d) This is just reworking parts (a) and (b) with v(0)= 5 . Using v(t)= 9.8t+C , v(0)= 5
2 0+C= 5 2 v(t)= 9.8t 5 . So s(t)= 4.9t 5t+D and s(0)=450 D=450 s(t)= 4.9t 5t+450 . Solving s(t)=0 by
using the quadratic formula gives us t= ( 5 8845 ) / ( 9.8 ) t 9.09 s.
1 / 66. v (t)=a(t)=a
s(t)= v(t)=at+C and v =v(0)=C
0 1 2
at +v t+D
0
2 s =s(0)=D
0 s(t)= v(t)=at+v 0 1 2
at +v t+s
0
0
2
2 / 67. By Exercise with a= 9.8 , s(t)= 4.9t +v t+s and v(t)=s ( t ) = 9.8t+v . So
0
0
0
2 v (t ) = 2 22 ( 9.8t+v ) = ( 9.8) t
0 2 2 0 0 2 2 19.6v t+v =v +96.04t 19.6v t=v 19.6
0 0 2 0 2 4.9t +v t
0 2 is just s(t) without the s term; that is, s(t) s . Thus, v ( t ) =v 19.6 s ( t ) s
0 0 0 0 2 . But 4.9t +v t
0 .
11 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives 2 68. For the first ball, s (t)= 16t +48t+432 from Example 8. For the second ball, a(t)= 32
1 v(t)= 32t+C , but v(1)= 32(1)+C=24
2 s(1)= 16(1) +56(1)+D=432
s (t)=s (t)
1 2 C=56 , so v(t)= 32t+56 2 s(t)= 16t +56t+D , but 2 D=392 , and s (t)= 16t +56t+392 . The balls pass each other when
2 2 2 8t=40 t=5 s. 16t +48t+432= 16t +56t+392 2 2 Another solution: From Exercise , we have s (t)= 16t +48t+432 and s (t)= 16t +24t+432 . We now
1 want to solve s (t)=s (t 1)
1 2 2 2 2 16t +48t+432= 16(t 1) +24(t 1)+432 48t=32t 16+24t 24 40=8t t=5 s.
69. Using Exercise with a= 32 , v =0 , and s =h (the height of the cliff ), we know that the height at
0 2 0 / time t is s(t)= 16t +h . v(t)=s (t)= 32t and v(t)= 120
2 0=s(3.75)= 16(3.75) +h
70. (a) EIy / / =mg(L x)+ 32t= 120 t=3.75 , so 2 h=16(3.75) =225 ft.
1
2
g(L x)
2 / EIy = 1
2 1
3
mg(L x)
g(L x) +C
2
6 1
3 1
4
/
mg(L x) +
g(L x) +Cx+D . Since the left end of the board is fixed, we must have y=y =0
6
24
1
1
2 1
3
3 1
4
when x=0 . Thus, 0=
mgL
gL +C and 0= mgL +
gL +D . It follows that
2
6
6
24
1
1
1
3 1
4
2 1
3
3 1
4
EIy= mg(L x) +
g(L x) +
mgL +
gL x
mgL +
gL
and
6
24
2
6
6
24
1
1
1
1
3 1
4
2 1
3
3 1
4
f (x)=y=
mg(L x) +
g(L x) +
mgL +
gL x
mgL +
gL
EI
6
24
2
6
6
24
(b) f (L)<0 , so the end of the board is a distance approximately f (L) below the horizontal. From
our result in (a), we calculate
EIy= f (L) = 1
EI 1
=
EI 1
3 1
4 1
3 1
4
mgL +
gL
mgL
gL
2
6
6
24
3 1
gL
3 1
4
mgL +
gL =
3
8
EI m
L
+
3
8 Note: This is positive because g is negative.
/ 71. Marginal cost =1.92 0.002x=C (x) 2 C(x)=1.92x 0.001x +K . But C(1)=1.92 0.001+K=562
2 K=560.081 . Therefore, C(x)=1.92x 0.001x +560.081 C(100)=742.081 , so the cost of producing
12 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives 100 items is $742.08 .
dm
1/2
1/2
m(x)=2x +C
=x
dx
and m(0)=C=0 , so m(x)=2 x . Thus, the mass of the 100 centimeter rod is m(100)=2 100 =20 g.
72. Let the mass, measured from one end, be m(x) . Then m(0)=0 and = 73. Taking the upward direction to be positive we have that for 0 t
refer to 0 t / 10 ), a (t)= ( 9 0.9t ) =v (t)
1
1
2 2 v (t)= 9t+0.45t +v , but v (0)=v = 10
1 0 1 0 9 2
3
s (t)=
t +0.15t 10t+s . But s (0)=500=s
1
0
1
0
2 / v (t)= 9t+0.45t 10=s (t)
1 10 (using the subscript 1 to 1 9 2
3
t +0.15t 10t+500 . s (10)= 450+150 100+500=100 , so it takes more than 10 seconds for
1
2 s (t)=
1 / the raindrop to fall. Now for t>10 , a(t)=0=v (t) 2 v(t)= constant =v (10)= 9(10)+0.45(10) 10= 55
1 v(t)= 55 . At 55 ft / s, it will take 100/55 1.8 s to fall the last 100 ft. Hence, the total time is
100 130
10+
=
11.8 s.
55
11
50 5280 220
220
=
ft / s, so v(t)= 22t+
. The
3600
3
3
220 10
2 220
car stops when v(t)=0 t=
=
. Since s(t)= 11t +
t , the distance covered is
3 22 3
3
10
10 2 220 10 1100
s
= 11
+
=
=122.2 ft.
3
3
3
3
9
/ 74. v (t)=a(t)= 22 . The initial velocity is 50 mi / h = 75. a(t)=k , the initial velocity is 30 mi / h =30 5280
=44 ft / s, and the final velocity (after 5
3600 5280 220
=
ft / s . So v(t)=kt+C and v(0)=44 C=44 . Thus, v(t)=kt+44
3600
3
220
220
88
88
2
v(5)=5k+44 . But v(5)=
, so 5k+44=
5k=
k=
5.87 ft / s .
3
3
3
15 seconds) is 50 mi / h =50 76. a(t)= 16 v(t)= 16t+v where v is the car’s speed (in ft / s) when the brakes were applied. The
0 0 1
1
2
2
v . Now s(t)= ( 16)t +v t= 8t +v t . The car travels 200 ft in the
0
0
0
16 0
2
2
1
1
1
1 2
time that it takes to stop, so s
v =200 200= 8
v
+v
v =
v
0
16 0
16 0
16 0
32 0
car stops when 16t+v =0 2 v =32 200=6400
0 t= v =80 ft / s ( 54.54 mi / h).
0 13 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives 2 77. Let the acceleration be a(t)=k km / h . We have v(0)=100 km / h and we can take the initial
position s(0) to be 0 . We want the time t for which v(t)=0 to satisfy s(t)<0.08 km. In general,
f / / v (t)=a(t)=k , so v(t)=kt+C , where C=v(0)=100 . Now s (t)=v(t)=kt+100 , so s(t)= 1 2
kt +100t+D ,
2 1 2
kt +100t . Since v(t )=0 , we have kt +100=0 or t = 100/k , so
f
f
f
2
2
1
100
100
1 1
5,000
s(t )= k
+100
=10 , 000
=
. The condition s(t ) must
f
f
2
k
k
2k k
k
5,000
5,000
2
satisfy is
<0.08
>k [ k is negative] k< 62 , 500 km / h , or equivalently,
k
0.08
3125
2
k<
4.82 m / s .
648
where D=s(0)=0 . Thus, s(t)= 78. (a) For 0 t 3 we have a(t)=60t 2 v(t)=30t +C 2 3 v(t)=30t , so s(t)=10t +C v(0)=0=C 3 s(0)=0=C s(t)=10t . Note that v(3)=270 and s(3)=270 .
For 3<t 17 : a(t)= g= 32 ft / s v(t)= 32(t 3)+C v(3)=270=C
2 v(t)= 32(t 3)+270 2 s(t)= 16(t 3) +270(t 3)+C s(3)=270=C s(t)= 16(t 3) +270(t 3)+270 . Note that v(17)= 178 and
s(17)=914 .
For 17<t 22 : The velocity increases linearly from 178 ft / s to 18 ft / s during this period, so
v
18 ( 178 ) 160
2
=
=
=32 . Thus, v(t)=32(t 17) 178 s(t)=16(t 17) 178(t 17)+914 and
t
22 17
5
s(22)=424 ft.
For t>22 : v(t)= 18 s(t)= 18(t 22)+C . But s(22)=424=C s(t)= 18(t 22)+424 .
Therefore, until the rocket lands, we have v(t)=
and s(t)= { {
10t 2 30t
32 ( t 3) +270
32(t 17) 178
18 3 if
if
if
if 0 t 3
3<t 17
17<t 22
t>22 if 0 t
2 16(t 3) +270(t 3)+270
2 16(t 17) 178 ( t 17) +914
18(t 22)+424 3 if 3<t 17 if 17<t 22
if t>22 14 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.10 Antiderivatives (b) To find the maximum height, set v(t) on 3<t ( ) ( ) 2 17 equal to 0 . 32(t 3)+270=0 t =11.4375 s and
1 ( ) the maximum height is s t = 16 t 3 +270 t 3 +270=1409.0625 ft.
1 1 1 (c) To find the time to land, set s(t)= 18(t 22)+424=0 . Then t 22= 424
=23.5 , so t 45.6 s.
18 5280
2
ft / s =132 ft / s. Then a(t)=4 ft / s
v(t)=4t+C , but
3600
132
C=0 . Now 4t=132 when t=
=33 s, so it takes 33 s to reach 132 ft / s. Therefore, taking
4 79. (a) First note that 90 mi / h =90
v(0)=0 2 s(0)=0 , we have s(t)=2t , 0 t 33 . So s(33)=2178 ft. 15 minutes =15(60)=900 s, so for 33<t 933
we have v(t)=132 ft / s s(933)=132(900)+2178=120 , 978 ft =22.9125 mi.
(b) As in part (a), the train accelerates for 33 s and travels 2178 ft while doing so. Similarly, it
decelerates for 33 s and travels 2178 ft at the end of its trip. During the remaining 900 66=834 s it
travels at 132 ft / s, so the distance traveled is 132 834=110 , 088 ft. Thus, the total distance is
2178+110 , 088+2178=114 , 444 ft =21.675 mi.
(c) 45 mi =45(5280)=237 , 600 ft. Subtract 2(2178) to take care of the speeding up and slowing
down, and we have 233 , 244 ft at 132 ft / s for a trip of 233 , 244/132=1767 s at 90 mi / h. The total
time is 1767+2(33)=1833s=30 min 33s=30.55 min.
(d) 37.5(60)=2250 s. 2250 2(33)=2184 s at maximum speed. 2184(132)+2(2178)=292 , 644 total
feet or 292 , 644/5280=55.425 mi. 15 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.2 The Mean Value Theorem 2 1. f (x)=x 4x+1 , 0,4 . Since f is a polynomial, it is continuous and differentiable on R , so it is
/ continuous on 0,4 and differentiable on ( 0,4 ) . Also, f (0)=1= f (4) . f (c)=0 2c 4=0
which is in the open interval ( 0,4 ) , so c=2 satisfies the conclusion of Rolle’s Theorem.
3 c=2 , 2 2. f (x)=x 3x +2x+5 , 0,2 . f is continuous on 0,2 and differentiable on ( 0,2 ) . Also,
6 36 24
1
/
2
3 , both in ( 0,2 ) .
f (0)=5= f (2) . f (c)=0 3c 6c+2=0 c=
=1
6
3
3. f (x)=sin 2 x , 1,1 . f , being the composite of the sine function and the polynomial 2 x , is
continuous and differentiable on R , so it is continuous on 1,1 and differentiable on ( 1,1 ) . Also,
1
/
f ( 1)=0= f (1) . f (c)=0 2 cos 2 c=0 cos 2 c=0 2 c=
+2 n c=
+n . If n=0 or 1 ,
2
4
1
3
then c=
,
is in ( 1,1 ) .
4
4 ) , and differentiable on ( 6, ) , so it 4. f (x)=x x+6 , 6,0 . f is continuous on its domain, is continuous on 6,0 and differentiable on ( 6,0 ) . Also, f ( 6)=0= f (0) . f (c)=0 6, / 3c+12
=0
2 c+6 c= 4 , which is in ( 6,0 ) .
2/3 5. f (x)=1 x 2/3 / . f ( 1)=1 ( 1) =1 1=0= f (1) . f (x)= 2
x
3 1/3 / , so f (c)=0 has no solution. This / does not contradict Rolle’s Theorem, since f (0) does not exist, and so f is not differentiable on
( 1,1 ) .
2 2 2 / 6. f (x)=(x 1) . f (0)=(0 1) =1=(2 1) = f (2) . f (x)= 2(x 1) 3 / f (x) is never 0 . This does not / contradict Rolle’s Theorem since f (1) does not exist.
f (8) f (0) 6 4 1
1
/
=
= . The values of c which satisfy f (c)= seem to be about c=0.8 , 3.2 , 4.4 ,
8 0
8
4
4
and 6.1 .
7. f (7) f (1) 2 5
1
1
/
=
=
. The values of c which satisfy f (c)=
seem to be about c=1.1 , 2.8 ,
7 1
6
2
2
4.6 , and 5.8 . 8. 1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.2 The Mean Value Theorem 9.
(a),
8.5 5
(b) The equation of the secant line is y 5= 8 1 (x 1) (c) / y= 1
9
x+ .
2
2 2 f (x)=x+4/x f (x)=1 4/x .
1
4
/
2
So f (c)=
=3 2 . Thus, an equation of the tangent
c =8 c=2 2 , and f (c)=2 2 +
2
2 2
1
1
line is y 3 2 = ( x 2 2 ) y= x+2 2 .
2
2 10. (a)
It seems that the tangent lines are parallel to the secant at x 1.2 .
3 / 2 (b) The slope of the secant line is 2 , and its equation is y=2x . f (x)=x 2x f (x)=3x 2 ,
2 3
/
2
so we solve f (c)=2 3c =4 c=
1.155 . Our estimates were off by about 0.045 in each
3
case.
2 1,1 and differentiable on ( 1,1 ) since
f (b) f (a)
/
polynomials are continuous and differentiable on R . f (c)=
b a
11. f (x)=3x +2x+5 , 1,1 . f is continuous on 2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.2 The Mean Value Theorem 6c+2= f (1) f ( 1) 10 6
=
=2
1 ( 1)
2 c=0 , which is in ( 1,1 ) . 6c=0 / 3 12. f (x)=x +x 1 , 0,2 . f is continuous on 0,2 and differentiable on ( 0,2 ) . f (c)=
2 3c +1= 9 ( 1)
2 13. f (x)=e 2x 2 2 c= 3c =5 1 4
3 c= 2
2
, but only
is in ( 0,2 ) .
3
3 , 0,3 . f is continuous and differentiable on R , so it is continuous on 0,3 and f (b) f (a)
differentiable on ( 0,3) . f (c)=
b a
/ 1
c=
ln
2 f (2) f (0)
2 0 1 e
6 2e e 6 0 e
=
3 0 2c e 1 e
=
6 2c 6 2c=ln 1 e
6 6 6 0.897 , which is in ( 0,3) . x
f (b) f (a)
/
, 1,4 . f is continuous on 1,4 and differentiable on ( 1,4 ) . f (c)=
x+2
b a
2 1
3 3
2
2
=
(c+2) =18 c= 2 3 2 . 2+3 2 2.24 is in ( 1,4 ) .
2
4 1
(c+2) 14. f (x)= 15. f (x)= x 1 . f (3) f (0)= 3 1 / / 0 1 =1 . Since f (c)= 1 if c<1 and f (c)=1 if c>1 , / f (c)(3 0)= 3 and so is never equal to 1 . This does not contradict the Mean Value Theorem since
/ f (1) does not exist.
x+1
/
/
2
1(x 1) 1(x+1)
=
. f (2) f (0)=3 ( 1)=4 . f (x)=
. Since f (x)<0 for all x
16. f (x)=
2
2
x 1
(x 1)
(x 1)
/ (except x=1 ), f (c)(2 0) is always <0 and hence cannot equal 4 . This does not contradict the Mean
Value Theorem since f is not continuous at x=1 .
3 5 17. Let f (x)=1+2x+x +4x . Then f ( 1)= 6<0 and f (0)=1>0. Since f is a polynomial, it is
continuous, so the Intermediate Value Theorem says that there is a number c between 1 and 0 such
that f (c)=0. Thus, the given equation has a real root. Suppose the equation has distinct real roots a
and b with a<b . Then f (a)= f (b)=0 . Since f is a polynomial, it is differentiable on ( a,b) and
/ continuous on a,b . By Rolle’s Theorem, there is a number r in ( a,b) such that f (r)=0 . But
/ 2 4 / f (x)=2+3x +20x 2 for all x , so f (x) can never be 0 . This contradiction shows that the equation
can’t have two distinct real roots. Hence, it has exactly one real root.
3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.2 The Mean Value Theorem 18. Let f (x)=2x 1 sin x . Then f (0)= 1<0 and f ( /2)= 2>0 . f is the sum of the polynomial 2x 1
and the scalar multiple ( 1 ) sin x of the trigonometric function sin x , so f is continuous (and
differentiable) for all x . By the Intermediate Value Theorem, there is a number c in ( 0, /2 ) such that
f (c)=0 . Thus, the given equation has at least one real root. If the equation has distinct real roots a
and b with a<b , then f (a)= f (b)=0 . Since f is continuous on a,b and differentiable on ( a,b) ,
/ / Rolle’s Theorem implies that there is a number r in ( a,b) such that f (r)=0 . But f (r)=2 cos r >0
since cos r 1 . This contradiction shows that the given equation can’t have two distinct real roots,
so it has exactly one real root.
3 19. Let f (x)=x 15x+c for x in 2,2 . If f has two real roots a and b in 2,2 , with a<b , then
f (a)= f (b)=0 . Since the polynomial f is continuous on a,b and differentiable on ( a,b) , Rolle’s
/ / 2 Theorem implies that there is a number r in ( a,b) such that f (r)=0 . Now f (r)=3r 15 . Since r is
in ( a,b) , which is contained in 2 2,2 , we have r <2 , so r <4 . It follows that 2 / 3r 15<3 4 15= 3<0 . This contradicts f (r)=0 , so the given equation can’t have two real roots in
2,2 . Hence, it has at most one real root in 2,2 .
4 20. f (x)=x +4x+c . Suppose that f (x)=0 has three distinct real roots a , b , d where a<b<d . Then
f (a)= f (b)= f (d)=0 . By Rolle’s Theorem there are numbers c and c with a<c <b and b<c <d and
1 0= f / / ( c ) = f ( c ) , so f
1 / / 2 ( 3 1 2 (x)=0 must have at least two real solutions. However ) 3 2 ( ) 2 0= f (x)=4x +4=4 x +1 =4 ( x+1 ) x x+1 has as its only real solution x= 1 . Thus, f (x) can have
at most two real roots.
21. (a) Suppose that a cubic polynomial P(x) has roots a <a <a <a , so
1 2 3 4 ( ) ( ) ( ) ( ) . By Rolle’s Theorem there are numbers c , c , c with a <c <a
a <c <a and a <c <a and P ( c ) =P ( c ) =P ( c ) =0 . Thus, the second degree polynomial P a =P a =P a =P a
1 2 3 4 1 / 2 2 3 3 3 / 3 1 1 2 , / 1 4 2 2 3 / P (x) has three distinct real roots, which is impossible.
(b) We prove by induction that a polynomial of degree n has at most n real roots. This is certainly
true for n=1 . Suppose that the result is true for all polynomials of degree n and let P(x) be a
polynomial of degree n+1 . Suppose that P(x) has more than n+1 real roots, say
a <a <a <
<a <a . Then P a =P a =
=P a
=0 . By Rolle’s Theorem
1 2 3 n+1 there are real numbers c ,
1 P / ( ) c =
1 =P / ( ) ( ) n+2 1 ,c n+1 ( ) 2 with a <c <a ,
1 1 2 n+2 ,a <c n+1 ( c ) =0 Thus, the n th degree polynomial P
n+1 <a n+1
/ n+2 and (x) has at least n+1 roots. This contradiction shows that P(x) has at most n+1 real roots.
4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.2 The Mean Value Theorem 22. (a) Suppose that f (a)= f (b)=0 where a<b . By Rolle’s Theorem applied to f on a,b there is a
/ number c such that a<c<b and f (c)=0 .
(b) Suppose that f (a)= f (b)= f (c)=0 where a<b<c . By Rolle’s Theorem applied to f (x) on a,b and
/ / b,c there are numbers a<d<b and b<e<c with f (d)=0 and f (e)=0 . By Rolle’s Theorem applied
/ to f (x) on d,e there is a number g with d<g<e such that f / / (g)=0 . (c) Suppose that f is n times differentiable on R and has n+1 distinct real roots. Then f
least one real root. ( n) has at / 23. By the Mean Value Theorem, f (4) f (1)= f (c)(4 1) for some c ( 1,4 ) . But for every c ( 1,4 )
/ / we have f (c) 2 . Putting f (c) 2 into the above equation and substituting f (1)=10 , we get
/ / f (4)= f (1)+ f (c)(4 1)=10+3 f (c) 10+3 2=16 . So the smallest possible value of f (4) is 16 .
/ / 24. If 3 f (x) 5 for all x , then by the Mean Value Theorem, f (8) f (2)= f (c) ( 8 2 ) for some c
in 2,8 .
( f is differentiable for all x , so, in particular, f is differentiable on ( 2,8 ) and continuous on 2,8 .
/ Thus, the hypotheses of the Mean Value Theorem are satisfied.) Since f (8) f (2)=6 f (c) and
3 / / f (c) 5 , it follows that 6 3 6 f (c) 6 5 18 f (8) f (2) 30. 25. Suppose that such a function f exists. By the Mean Value Theorem there is a number 0<c<2 with
f (2) f (0) 5
5
/
/
f (c)=
= . But this is impossible since f (x) 2< for all x , so no such function can
2 0
2
2
exist.
26. Let h= f g . Then since f and g are continuous on a,b and differentiable on ( a,b) , so is h , and
thus h satisfies the assumptions of the Mean Value Theorem. Therefore, there is a number c with
/ / / a<c<b such that h(b)=h(b) h(a)=h (c)(b a) . Since h (c)<0 , h (c)(b a)<0 , so f (b) g(b)=h(b)<0
and hence f (b)<g(b) .
1
x , and a=0 . Notice that f (0)=1=g(0) and
2
1
1
1
/
/
f (x)=
< =g (x) for x>0 . So by Exercise 26, f (b)<g(b)
1+b <1+ b for b>0 .
2
2 1+x 2
1
Another method: Apply the Mean Value Theorem directly to either f (x)=1+ x 1+x or g(x)= 1+x
2
on 0,b . 27. We use Exercise 26 with f (x)= 1+x , g(x)=1+ 5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.2 The Mean Value Theorem 28. f satisfies the conditions for the Mean Value Theorem, so we use this theorem on the interval
f (b) f ( b)
/
b,b :
= f (c) for some c ( b,b) . But since f is odd, f ( b)= f (b) . Substituting this
b ( b)
f (b)+ f (b)
f (b)
/
/
into the above equation, we get
= f (c)
= f (c) .
2b
b
29. Let f (x)=sin x and let b<a . Then f (x) is continuous on b,a and differentiable on ( b,a ) . By the
Mean Value Theorem, there is a number c ( b,a ) with
/ cos c b a
a b . If
sin a sin b= f (a) f (b)= f (c)(a b)=(cos c)(a b) . Thus, sin a sin b
a<b , then sin a sin b = sin b sin a
b a = a b . If a=b , both sides of the inequality are 0 .
/ / 30. Suppose that f (x)=c . Let g(x)=cx , so g (x)=c . Then, by Corollary 7, f (x)=g(x)+d , where d is
a constant, so f (x)=cx+d .
/ / / / 2 / / 2 31. For x>0 , f (x)=g(x) , so f (x)=g (x) . For x<0 , f (x)=(1/x) = 1/x and g (x)=(1+1/x) = 1/x ,
/ / so again f (x)=g (x) . However, the domain of g(x) is not an interval so we cannot conclude that
f g is constant (in fact it is not).
1 1 ( 2 ) 32. Let f (x)=2sin x cos
1 2x . Then
/
/
2
4x
2
4x
f (x)=
=
=0 (since x 0 ). Thus, f (x)=0 for
2
2
2
2 2
1 x
1 x
2x 1 x
1 1 2x
all x ( 0,1 ) . Thus, f (x)=C on ( 0,1 ) . To find C , let x=0.5 . Thus, ( 1 ) 1 2sin (0.5) cos (0.5)=2 6 3 =0=C . We conclude that f (x)=0 for x in ( 0,1 ) . By continuity
1 of f , f (x)=0 on 0,1 . Therefore, we see that f (x)=2sin x cos
1 2sin x=cos 1 x 1
x+1 . Note that the domain of f is 0, ) . Thus,
2
2
1
1
1
=
=0 . Then
1+x 2 x
x (x+1)
x (x+1) 2arctan x + ( x+1 ) ( x 1 ) 1 / 2 1
f (x)=C on ( 0, ( 1 2x2) =0 ( 1 2x2) . 33. Let f (x)=arcsin
f (x)= 1 x 1 2
( x+1 )
x+1
) by Theorem 5. By continuity of f , f (x)=C on 0, arcsin( 1) 2arctan(0)+ 2 =C 2 0+ 2 =0=C . Thus, f (x)=0 ) . To find C , we let x=0 arcsin x 1
x+1 =2arctan x 2 . 34. Let v(t) be the velocity of the car t hours after 2:00 P.M. Then
6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.2 The Mean Value Theorem v(1/6) v(0) 50 30
1
=
=120 . By the Mean Value Theorem, there is a number c such that 0<c< with
1/6 0
1/6
6
/ / v (c)=120 . Since v (t) is the acceleration at time t , the acceleration c hours after 2:00 P.M. is
2 exactly 120 mi / h .
35. Let g(t) and h(t) be the position functions of the two runners and let f (t)=g(t) h(t) . By
hypothesis, f (0)=g(0) h(0)=0 and f (b)=g(b) h(b)=0 , where b is the finishing time. Then by the
f (b) f (0)
/
. But f (b)= f (0)=0
Mean Value Theorem, there is a time c , with 0<c<b , such that f (c)=
b 0
/ / / / / / , so f (c)=0 . Since f (c)=g (c) h (c)=0 , we have g (c)=h (c) . So at time c , both runners have
/ / the same speed g (c)=h (c) .
/ 36. Assume that f is differentiable (and hence continuous) on R and that f (x) 1 for all x . Suppose
f has more than one fixed point. Then there are numbers a and b such that a<b , f (a)=a , and f (b)=b
. Applying the Mean Value Theorem to the function f on a,b , we find that there is a number c in
f (b) f (a)
b a
/
/
. But then f (c)=
=1 , contradicting our assumption that
( a,b) such that f (c)=
b a
b a
/ f (x) 1 for every real number x . This shows that our supposition was wrong, that is, that f cannot
have more than one fixed point. 7 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph 1. (a) f is increasing on ( 0,6 ) and ( 8,9 ) .
(b) f is decreasing on ( 6,8 ) .
(c) f is concave upward on ( 2,4 ) and ( 7,9 ) .
(d) f is concave downward on ( 0,2 ) and ( 4,7) .
(e) The points of inflection are ( 2,3) , ( 4,4.5) and ( 7,4 ) (where the concavity changes).
2. (a) f is increasing on ( 1, 3.8 ) and ( 5, 6.5) .
(b) f is decreasing on ( 0,1 ) , ( 3.8,5) , ( 6.5,8 ) , and ( 8,9 ) .
(c) f is concave upward on ( 0,3) and ( 8,9 ) .
(d) f is concave downward on ( 3,5) and ( 5,8 ) .
(e) The point of inflection is ( 3, 1.8 ) (where the concavity changes).
3. (a) Use the Increasing/Decreasing (I/D) Test.
(b) Use the Concavity Test.
(c) At any value of x where the concavity changes, we have an inflection point at ( x,f(x) ) .
4. (a) See the First Derivative Test.
(b) See the Second Derivative Test and the note that precedes Example 7.
/ / 5. (a) Since f (x)>0 on ( 1,5) , f is increasing on this interval. Since f (x)<0 on ( 0,1 ) and ( 5,6 ) , f
is decreasing on these intervals.
/ (b) Since f (x)=0 at x=1 and f / changes from negative to positive there, f changes from decreasing
/ / to increasing and has a local minimum at x=1 . Since f (x)=0 at x=5 and f changes from positive
to negative there, f changes from increasing to decreasing and has a local maximum at x=5 .
/ / 6. (a) f (x)>0 and f is increasing on ( 0,1 ) and ( 3,5) . f (x)<0 and f is decreasing on ( 1,3) and
( 5,6 ) .
/ (b) Since f (x)=0 at x=1 and x=5 and f / changes from positive to negative at both values, f
/ changes from increasing to decreasing and has local maxima at x=1 and x=5 . Since f (x)=0 at x=3
/ and f changes from negative to positive there, f changes from decreasing to increasing and has a
local minimum at x=3 .
/ / 7. There is an inflection point at x=1 because f (x) changes from negative to positive there, and so
the graph of f changes from concave downward to concave upward. There is an inflection point at
/ / x=7 because f (x) changes from positive to negative there, and so the graph of f changes from
concave upward to concave downward.
8. (a) f is increasing on the intervals where
1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph / f (x)>0 , namely, ( 2,4 ) and ( 6,9 ) .
(b) f has a local maximum where it changes from increasing to decreasing, that is, where f
changes from positive to negative (at x=4 ). Similarly, where f
has a local minimum (at x=2 and at x=6 ).
(c) When f / is increasing, its derivative f / / / / changes from negative to positive, f is positive and hence, f is concave upward. This
/ happens on ( 1,3) , ( 5,7) , and ( 8,9 ) . Similarly, f is concave downward when f is decreasing
that is, on ( 0,1 ) , ( 3,5) , and ( 7,8 ) .
(d) f has inflection points at x=1 , 3 , 5 , 7 , and 8 , since the direction of concavity changes at each
of these values.
9. The function must be always decreasing and concave downward. 10. (a) The rate of increase of the population is initially very small, then gets larger until it reaches a
maximum at about t=8 hours, and decreases toward 0 as the population begins to level off.
(b) The rate of increase has its maximum value at t=8 hours.
(c) The population function is concave upward on ( 0,8 ) and concave downward on ( 8,18 ) .
(d) At t=8 , the population is about 350 , so the inflection point is about ( 8,350 ) .
3 11. (a) f (x)=x 12x+1 / 2 f (x)=3x 12=3(x+2)(x 2) .
/ We don’t need to include "3" in the chart to determine the sign of f (x) .
/ Interval x+2 x 2 x< 2
2<x<2
x>2 f (x)
+ +
+ + + So f is increasing on ( , 2 ) and ( 2, f
increasing on (
, 2)
decreasing on ( 2,2)
increasing on (2, ) ) and f is decreasing on ( 2,2 ) . (b) f changes from increasing to decreasing at x= 2 and from decreasing to increasing at x=2 . Thus,
f ( 2)=17 is a local maximum value and f (2)= 15 is a local minimum value.
(c) f / / (x)=6x . f / / (x)>0 x>0 and f / / (x)<0 x<0 . Thus, f is concave upward on ( 0, ) and
2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph concave downward on (
( 0,f(0) ) = ( 0,1 ) .
2 ,0 ) . There is an inflection point where the concavity changes, at 3 / 2 / / f (x)= 6x+3x =3x(x 2) . Thus, f (x)>0 x<0 or x>2 and f (x)<0
12. (a) f (x)=5 3x +x
0<x<2 . So f is increasing on (
,0 ) and ( 2, ) and f is decreasing on ( 0,2 ) .
(b) f changes from increasing to decreasing at x=0 and from decreasing to increasing at x=2 . Thus,
f (0)=5 is a local maximum value and f (2)=1 is a local minimum value.
/ / / / / / (c) f (x)= 6+6x=6(x 1) . f (x)>0 x>1 and f (x)<0 x<1 . Thus, f is concave upward on
( 1, ) and concave downward on ( ,1 ) . There is an inflection point at ( 1,3) .
4 2 / Interval x+1 x< 1
1<x<0
0<x<1
x>1 ( 3 ) 2 f (x)=4x 4x=4x x 1 =4x ( x+1 ) ( x 1 ) . 13. (a) f (x)=x 2x +3 x +
+
+ / x 1 f
decreasing on (
, 1)
increasing on ( 1,0)
decreasing on (0,1)
increasing on (1, ) f (x)
+ +
+ + + So f is increasing on ( 1,0) and (1, ) and f is decreasing on (
, 1) and (0,1) .
(b) f changes from increasing to decreasing at x=0 and from decreasing to increasing at x= 1 and
x=1 . Thus, f (0)=3 is a local maximum value and f ( 1)=2 are local minimum values.
/ /
2
2 1
/ /
(c) f (x)=12x 4=12 x
=12 ( x+1/ 3 ) ( x 1/ 3 ) . f (x)>0 x< 1/ 3 or x>1/ 3 and
3
f / / (x)<0 1/ 3 <x<1/ 3 . Thus, f is concave upward on concave downward on ( 2 14. (a) f (x)= x f 2 x +3 ( , 3 /3) and 3 /3,
22
3 /3,
.
9 3 /3, 3 /3) . There are inflection points at / ( x2+3) (2x) x2(2x) = 6x
(x)=
( x2+3) 2
( x2+3) 2 / / ( ) and . The denominator is positive so the sign
/ of f (x) is determined by the sign of x . Thus, f (x)>0 x>0 and f (x)<0 x<0 . So f is increasing
on ( 0, ) and f is decreasing on (
,0 ) .
(b) f changes from decreasing to increasing at x=0 . Thus, f (0)=0 is a local minimum value.
(c)
2 f / / (x) = 2 ( 2 ) x +3 (6) 6x 2 x +3 (2x) ( x +3)
2 2 2 = ( 2 6 x +3 ) 2 2 x +3 4x ( x2+3) 4
3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph = f / / ( 2) = 18(x+1)(x 1)
( x2+3) 3 ( x2+3) 3 6 3 3x (x)>0 1<x<1 and f concave downward on ( / / . x< 1 or x>1 . Thus, f is concave upward on ( 1,1 ) and
1
, 1 ) and ( 1, ) . There are inflection points at
1,
.
4
(x)<0 15. (a) f (x)=x 2sin x on ( 0,3 ) / / f (x)=1 2cos x . f (x)>0 1 2cos x>0 cos x< 1
2 3 <x< 5
3 7
1
5
7
5
/
<x<3 . f (x)<0 cos x>
0<x<
or
<x<
. So f is increasing on
,
3
2
3
3
3
3 3
7
5 7
and
,3
, and f is decreasing on 0,
and
,
.
3
3
3 3
5
(b) f changes from increasing to decreasing at x=
, and from decreasing to increasing at x=
3
3
7
5
5
and at x=
. Thus, f
=
+ 3 6.97 is a local maximum value and
3
3
3
7
7
f
=
3
0.68 and f
=
3 5.60 are local minimum values.
3
3
3
3
or / / / / (c) f (x)=2sin x >0 0<x< and 2 <x<3 , f (x)<0
<x<2 . Thus, f is concave upward on
( 0, ) and ( 2 ,3 ) , and f is concave downward on ( ,2 ) . There are inflection points at ( , ) and
( 2 ,2 ) .
2 / 16. (a) f (x)=cos x 2sin x , 0 x 2 . f (x)= 2cos x sin x 2cos x= 2cos x ( 1+sin x ) . Note that
1+sin x 0 [ since sin x 1 ], with equality sin x= 1 x=3 /2 [ since 0 x 2 ] cos x=0 .
/ / /2<x<3 /2 and f (x)<0 cos x>0 0<x< /2 or 3 /2<x<2 . Thus, f
Thus, f (x)>0 cos x <0
is increasing on ( /2,3 /2 ) and f is decreasing on ( 0, /2 ) and ( 3 /2,2 ) .
(b) f changes from decreasing to increasing at x= /2 and from increasing to decreasing at x=3 /2 .
Thus, f ( /2)= 2 is a local minimum value and f (3 /2)=2 is a local maximum value.
(c)
f / / 2 ( 2 2 (x) =2sin x ( 1+sin x ) 2cos x=2sin x+2sin x 2 1 sin x ) 2 =4sin x+2sin x 2=2(2sin x 1)(sin x+1)
so f / / (x)>0 sin x > 1
2 <x< 5
, and f
6 / / (x)<0 1
and sin x 1 0<x<
or
2
6
5
,
and concave downward on
6 6 sin x < 6
5
3
3
<x<
or
<x<2 . Thus, f is concave upward on
6
2
2 4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph 0, 5 3
,
6 2 , 6 5
1
,
6
4 3
,2
2 , and . There are inflection points at 6 , 1
4 and .
x / x x x / 17. (a) y= f (x)=xe
f (x)=xe +e =e (x+1) . So f (x)>0
( 1, ) and decreasing on ( , 1 ) . x+1>0 x> 1 . Thus, f is increasing on (b) f changes from decreasing to increasing at its only critical number, x= 1 . Thus, f ( 1)= e
local minimum value.
/ x / / x x x 1 is a / / (c) f (x)=e (x+1) f (x)=e (1)+(x+1)e =e (x+2) . So f (x)>0 x+2>0 x> 2 . Thus, f is
concave upward on ( 2, ) and concave downward on (
, 2 ) . Since the concavity changes
direction at x= 2 , the point
2 x ( / 2, 2e
2 x 2 ) is an inflection point.
x x / f (x)=x e +2xe =x(x+2)e . So f (x)>0 x(x+2)>0 either x< 2 or x>0 .
18. (a) y= f (x)=x e
Therefore f is increasing on (
, 2 ) and ( 0, ) , and decreasing on ( 2,0 ) .
2 (b) f changes from increasing to decreasing at x= 2 , so f ( 2)=4e is a local maximum value. f
changes from decreasing to increasing at x=0 , so f (0)=0 is a local minimum value.
/ ( 2 ) (c) f (x)= x +2x e x f / / ( 2 ) x x (x)= x +2x e +e (2x+2)=e x ( x2+4x+2) . f / /(x)=0 2 x +4x+2=0 / / x= 2 2 . f (x)<0
2 2 <x< 2+ 2 , so f is concave downward on ( 2 2 , 2+ 2 ) and
concave upward on (
, 2 2 ) and ( 2+ 2 , ) . There are inflection points at
( 2 2 ,f ( 2 2 ) ) ( 3.41,0.38 ) and ( 2+ 2 ,f ( 2+ 2 ) ) ( 0.59,0.19 ) .
ln x
. (Note that f is only defined for x>0 .)
x
1
ln x
1 1/2
x (1/x) ln x
x
x 2 x 2 x 2 ln x
2
/
f (x)=
=
=
>0
3/2
x
x
2 x
2x 19. (a) y= f (x)= 2 ln x>0 ( 2) and decreasing on ( e2, ) .
2
2
2
2
ln e
f changes from increasing to decreasing at x=e , so f ( e ) =
=
e ln x<2 2 x<e . Therefore f is increasing on 0,e
(b) is a local maximum 2 e value.
(c) 5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph ( 3/2 / / f (x) = ( 2x3/2) 2
x ( 2+3ln x 6)
4x f / / ln x= (x)=0 3 = 8/3 x=e 1/2 1/2 2x +3x (ln x 2)
3 3ln x 8
4x 8
3 )= 4x 1/2 = 1/2 2x ( 1/x) (2 ln x) 3x .f 5/2 / / 8/3 (x)>0 x>e 8/3 , so f is concave upward on (e ,
8/3 8/3 downward on (0,e ) . There is an inflection point at e , 8
e
3 4/3 ) and concave ( 14.39,0.70 ) . 20. (a) y= f (x)=xln x . (Note that f is only defined for x>0 .)
/ / 1 f (x)=x(1/x)+ln x=1+ln x . f (x)>0 ln x+1>0 ln x> 1 x>e . Therefore f is increasing on
( 1/e, ) and decreasing on ( 0,1/e ) .
(b) f changes from decreasing to increasing at x=1/e , so f (1/e)= 1/e is a local minimum value.
(c) f / / (x)=1/x>0 for x>0 . So f is concave upward on its entire domain, and has no inflection point.
5 / 21. f (x)=x 5x+3 ( 4 2 ) f (x)=5x 5=5 x +1 (x+1)(x 1) .
/ / 1<x<1 and f (x)>0 First Derivative Test: f (x)<0 x>1 or x< 1 . Since f positive to negative at x= 1 , f ( 1)=7 is a local maximum value; and since f
to positive at x=1 , f (1)= 1 is a local minimum value.
Second Derivative Test: f / / 3 / (x)=20x . f (x)=0 x= 1 . f / / ( 1)= 20<0 / / changes from changes from negative f ( 1)=7 is a local / / maximum value. f (1)=20>0 f (1)= 1 is a local minimum value.
Preference: For this function, the two tests are equally easy. 22. f (x)= x
2 f / ( x2+4) 1 x(2x) =
(x)=
2 x +4 x +4 2 2 4 x
()
2 2 = (2+x)(2 x) ( x +4)
2 2 . x +4
/ First Derivative Test: f (x)>0 / / 2<x<2 and f (x)<0 x>2 or x< 2 . Since f changes from
1
/
positive to negative at x=2 , f (2)= is a local maximum value; and since f changes from negative
4
1
to positive at x= 2 , f ( 2)=
is a local minimum value. Second Derivative Test:
4 6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph 2 ( 4 x2) 2 ( x2+4) (2x)
2
( x2+4) 2
2
2
2
2
2x ( x +4) ( x +4) +2 ( 4 x )
2x ( 12 x )
=
4
2
( x +4)
( x2+4) 3
2 / / f x +4 ( 2x) (x) = =
/ / / x= 2 . f f (x)=0 ( 2)= 1
>0
16 f ( 2)= 1
is a local minimum value.
4 1
1
<0 f (2)= is a local maximum value.
16
4
Preference: Since calculating the second derivative is fairly difficult, the First Derivative Test is
easier to use for this function.
f / / (2)= 1
. Note that f is defined for 1 x 0 ; that is,
2 1 x
1
1
3
/
/
for x 1 . f (x)=0 2 1 x =1
1 x=
1 x=
x= . f does not exist at x=1 , but we
2
4
4
can’t have a local maximum or minimum at an endpoint.
3
3
/
/
/
First Derivative Test: f (x)>0 x< and f (x)<0
<x<1 . Since f changes from positive to
4
4
3
3
5
negative at x= , f
= is a local maximum value.
4
4
4
1
1
3
/ /
3/2
/ /
1
Second Derivative Test: f (x)=
(1 x) ( 1)=
.f
= 2<0
3
2
2
4
4( 1 x)
3
5
f
= is a local maximum value.
4
4
Preference: The First Derivative Test may be slightly easier to apply in this case.
/ 23. f (x)=x+ 1 x f (x)=1+ 4 3 24. (a) f (x)=x (x 1) / 1
(1 x)
2 4 1/2 ( 1)=1 2 3 3 3 2 3 2 f (x)=x 3(x 1) +(x 1) 4x =x (x 1) 3x+4(x 1) =x (x 1) (7x 4)
4
The critical numbers are 0 , 1 , and
.
7
(b)
f / / 2 2 3 3 2 (x) =3x (x 1) (7x 4)+x 2(x 1)(7x 4)+x (x 1) 7
2 =x (x 1) 3(x 1)(7x 4)+2x(7x 4)+7x(x 1)
Now f / / (0)= f / / (1)=0 , so the Second Derivative Test gives no information for x=0 or x=1 . 7 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph f 4
7 / / = 2 4
7 4
1
7 0+0+7 there is a local minimum at x=
(c) f / is positive on ( 4
7 4
1
7 = 4
7 2 3
7 3
7 (4) >0 , so 4
.
7 ,0 ) , negative on 0, 4
7 4
,1
7 , positive on , and positive on ( 1, ). 4
, and no local maximum or minimum at
7 So f has a local maximum at x=0 , a local minimum at x=
x=1 .
/ 25. (a) By the Second Derivative Test, if f (2)=0 and f
. / / (2)= 5<0 , f has a local maximum at x=2 / / / (b) If f (6)=0 , we know that f has a horizontal tangent at x=6 . Knowing that f (6)=0 does not
provide any additional information since the Second Derivative Test fails. For example, the first and
4 4 3 second derivatives of y= ( x 6 ) , y= ( x 6 ) , and y= ( x 6 ) all equal zero for x=6 , but the first has a
local minimum at x=6 , the second has a local maximum at x=6 , and the third has an inflection point
at x=6 .
/ 26. f (x)>0 for all x 1 with vertical asymptote x=1 , so f is increasing on (
/ / / / f (x)>0 if x<1 or x>3 , and f (x)<0 if 1<x<3 , so f is concave upward on (
and concave downward on ( 1,3) . There is an inflection point when x=3 . / / / 27. f (0)= f (2)= f (4)=0
increasing on (
/ / / ,1 ) and ( 1, ,1 ) and ( 3, horizontal tangents at x=0 , 2 , 4 . f (x)>0 if x<0 or 2<x<4
/ ,0 ) and ( 2,4 ) . f (x)<0 if 0<x<2 or x>4 ).
), f is f is decreasing on ( 0,2 ) and ( 4, ). / / f (x)>0 if 1<x<3 f is concave upward on ( 1,3) . f (x)<0 if x<1 or x>3 f is concave
downward on (
,1 ) and ( 3, ) . There are inflection points when x=1 and 3 . 8 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph 28. / / / horizontal tangents at x= 1 . f (x)<0 if x <1 f (1)= f ( 1)=0
/ f is increasing on ( 2, 1 ) and ( 1,2 ) . f (x)= 1 if x >2 f (x)>0 if 1< x <2 , 2 ) and ( 2,
constant slope 1 on (
. Inflection point ( 0,1 ) .
/ 29. f (x)>0 if x <2
/ and ( 2, f is decreasing on ( 1,1 ) . / ) . f ( 2)=0 ).f / / (x)<0 if 2<x<0 f is concave downward on ( 2,0 ) / f is increasing on ( 2,2 ) . f (x)<0 if x >2
horizontal tangent at x= 2 . lim
x asymptote or vertical tangent (cusp) at x=2 . f
( 2, ) . / / the graph of f has f is decreasing on ( / f (x) = , 2) there is a vertical 2 (x)>0 if x 2 f is concave upward on ( ,2 ) and 30. / / f is increasing on ( 2,2 ) . f (x)<0 if x >2 f (x)>0 if x <2 f is decreasing on ( , 2 ) and / ( 2, ) . f (2)=0 , so f has a horizontal tangent (and local maximum) at x=2 . lim f (x)=1 y=1 is a
x horizontal asymptote. f ( x)= f (x)
Finally, f / / (x)<0 if 0<x<3 and f 31. (a) f is increasing where f
where f / / f is an odd function (its graph is symmetric about the origin).
/ / (x)>0 if x>3 , so f is CD on ( 0,3) and CU on ( 3, is positive, that is, on ( 0,2 ) , ( 4,6 ) , and ( 8, ). ) ; and decreasing is negative, that is, on ( 2,4 ) and ( 6,8 ) . (b) f has local maxima where f
minima where f / / changes from positive to negative, at x=2 and at x=6 , and local changes from negative to positive, at x=4 and at x=8 .
9 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph (c) f is concave upward (CU) where f / is increasing, that is, on ( 3,6 ) and ( 6, ) , and concave / downward (CD) where f is decreasing, that is, on ( 0,3) .
(d) There is a point of in.ection where f changes from being CD to being CU, that is, at x = 3. (e)
/ 32. (a) f is increasing where f
negative, on ( 0,1 ) and ( 6,8 ) . is positive, on ( 1,6 ) and ( 8, (b) f has a local maximum where f
where f / / / ) , and decreasing where f is changes from positive to negative, at x=6 , and local minima changes from negative to positive, at x=1 and at x=8 . (c) f is concave upward where f / is increasing, that is, on ( 0 2 ) , ( 3, 5) ; and ( 7, ) and / concave downward where f is decreasing, that is, on ( 2 3) , ( 5, 7) .
(d) There are points of inflection where f changes its direction of concavity, at x=2 , x=3 , x=5 and
x=7 . (e)
3 2 33. (a) f (x)=2x 3x 12x / 2 ( 2 ) / f (x)=6x 6x 12=6 x x 2 =6(x 2)(x+1) . f (x)>0 x< 1 or x>2 / and f (x)<0 1<x<2 . So f is increasing on (
, 1 ) and ( 2, ) , and f is decreasing on ( 1,2 ) .
(b) Since f changes from increasing to decreasing at x= 1 , f ( 1)=7 is a local maximum value. Since
f changes from decreasing to increasing at x=2 , f (2)= 20 is a local minimum value.
1
1
/ /
/ /
/ /
(c) f (x)=6(2x 1) f (x)>0 on
,
and f (x)<0 on
,
. So f is concave
2
2
1
1
upward on
,
and concave downward on
,
. There is a change in concavity at
2
2
1
1 13
x= , and we have an inflection point at
,
.
2
2 2
(d)
10 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph 3 / ( 2 ) 2 / / f (x)=3 3x = 3 x 1 = 3(x+1)(x 1) . f (x)>0 1<x<1 and f (x)<0
34. (a) f (x)=2+3x x
x< 1 or x>1 . So f is increasing on ( 1,1 ) and f is decreasing on (
, 1 ) and ( 1, ) .
(b) f ( 1)=0 is a local minimum value and f (1)=4 is a local maximum value.
/ /
/ /
/ /
(c)
,0 ) and f (x)<0 on ( 0, ) . So f is concave upward on
f (x)= 6x f (x)>0 on (
( ,0 ) and concave downward on ( 0, ) . There is an inflection point at ( 0,2 ) .
(d) 4 2 35. (a) f (x)=x 6x
Interval
| x< / ( 3 ) 2 f (x)=4x 12x=4x x 3 =0 when x=0 ,
2 4x / x 3
+ 3 f (x)
+ + x> 3 + + (b) Local minimum values f
/ / 2 ( 2 ( f ( , 3)
increasing on ( 3,0 )
decreasing on ( 0, 3 )
increasing on ( 3, )
decreasing on 3 <x<0
0<x< 3 3 . + 3 ) = 9 , local maximum value f (0)=0 ) 2 (c) f (x)=12x 12=12 x 1 >0 x >1
x >1
and CD on ( 1,1 ) . Inflection points at ( 1, 5) x>1 or x< 1 , so f is CU on ( , 1 ) , ( 1, ) (d) 11 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph 3 36. (a) g(x)=200+8x +x 4 / 2 3 2 / g (x)=24x +4x =4x (6+x)=0 when x= 6 and when x=0 . g (x)>0 / ( x 0 ) and g (x)<0 x< 6 , so g is decreasing on (
, 6 ) and g is increasing on ( 6,
horizontal tangent at x=0 .
(b) g( 6)= 232 is a local minimum value. There is no local maximum value.
(c) g / / 2 (x)=48x+12x =12x(4+x)=0 when x= 4 and when x=0 . g / / (x)<0
4<x<0 , so g is CU on (
( 4, 56 ) and ( 0,200 ) g , 4 ) and ( 0, / / (x)>0 x> 6 ) , with a x< 4 or x>0 and ) , and g is CD on ( 4,0 ) . Inflection points at (d)
5 3 / 37. (a) h(x)=3x 5x +3
/ 4 2 2 h (x)=15x 15x =15x ( x2 1) =0 when x=0 , 2 nonnegative, h (x)>0 x >1
x >1 x>1 or x< 1 , so h is increasing on (
decreasing on ( 1,1 ) , with a horizontal tangent at x=0 .
(b) Local maximum value h( 1)=5 , local minimum value h(1)=1
(c)
h h / / / / ( 3 2 2 1 . Since 15x is
, 1 ) and ( 1, ) and ) (x) =60x 30x=30x 2x 1
1
1
=60x x+
x
2
2
1
1
1
or
<x<0 , so h is CU on
,0
2
2
2
1
and 0,
. Inflection points at ( 0,3) and
2 (x)>0 when x>
, 1
2 1
,
2
1
7
,3
2
8
2 and and CD on
. (d) 12 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph ( )3 2 ( / 2 h (x)=6x x 1
38. (a) h(x)= x 1
decreasing on (
,0 ) .
(b) h(0)= 1 is a local minimum value.
(c) h / / ( )2 2 2 (x)=6 x 1 +24x )2 x>0 ( x 1 ), so h is increasing on ( 0, 0 ( x2 1) =6 ( x2 1) ( 5x2 1) . The roots ) and 1
divide R into five
5 1 and intervals.
Interval x
+ x< 1 2 2 5x
+ 1
5 1<x< From the table, we see that h is CU on (
1
5 and 1
,1
5 h
+ / / ( x) + upward upward
downward +
+ , 1) , Concavity downward + 1
1
<x<
5
5
1
<x<1
5
x>1
+ 1, 1 + 1
1
,
5
5 . Inflection points at ( 1,0 ) and upward and ( 1, ) , and CD on 1
64
,
5 125 (d)
39. (a) A(x)=x x+3
The domain of A is / A (x)=x
3, / 1
(x+3)
2 1/2 + x+3 1= x
x+2(x+3)
3x+6
+ x+3 =
=
.
2 x+3
2 x+3
2 x+3 / ) . A (x)>0 for x> 2 and A (x)<0 for 3<x< 2 , so A is increasing on
13 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph ( 2, ) and decreasing on ( 3, 2 ) .
(b) A( 2)= 2 is a local minimum value.
1
2 x+3 3 (3x+6)
/ /
x+3
6(x+3) (3x+6)
3x+12
3(x+4)
(c) A (x)=
=
=
=
A
3/2
3/2
3/2
2
4(x+3)
4(x+3)
4(x+3)
( 2 x+3 )
x> 3 , so A is concave upward on ( 3, ) . There is no inflection point. / / (x)>0 for all (d)
2/3 40. (a) B(x)=3x / B (x)=2x x 1/3 1= 2
3 1= 3 2 x 3 / / . B (x)>0 if 0<x<8 and B (x)<0 if x<0 or x
x
x>8 , so B is decreasing on (
,0 ) and ( 8, ) , and B is increasing on ( 0,8 ) .
(b) B(0)=0 is a local minimum value.
B(8)=4 is a local maximum value.
(c)
2 4/3
/ /
/ /
2
B (x)=
x =
, so B (x)<0 for all x 0 . B is concave downward on (
4/3
3
3x
( 0, ) . There is no inflection point. ,0 ) and (d) 1/3 4/3 1/3 / 41. (a) C(x)=x (x+4)=x +4x C (x)= 4 1/3 4
x + x
3
3 / 2/3 = 4
x
3 2/3 1<x<0 or x>0 and C (x)<0 for x< 1 , so C is increasing on ( 1, (x+1)= 4(x+1)
3 3 / . C (x)>0 if 2 x ) and C is decreasing on ( , 1) .
(b) C( 1)= 3 is a local minimum value.
4 2/3 8 5/3 4 5/3
/ /
/ /
/ /
4(x 2)
(c) C (x)= x
x = x ( x 2) =
. C (x)<0 for 0<x<2 and C (x)>0 for x<0
3 5
9
9
9
9 x
and x>2 , so C is concave downward on ( 0,2 ) and concave upward on (
,0 ) and ( 2, ) . There are ( 3 inflection points at ( 0,0 ) and 2,6 2
(d) ) ( 2,7.56 ) .
14 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph ( 4 42. (a) f (x)=ln x +27 ) / f (x)= 4x 3
/ / . f (x)>0 if x>0 and f (x)<0 if x<0 , so f is increasing on 4 x +27
0, ) and f is decreasing on (
,0 ) .
(
(b) f (0)=ln 27 3.3 is a local minimum value.
(c)
/ / f ( x4+27) ( 12x2) 4x3 ( 4x3) = 4x2 3( x4+27) 4x4
(x) =
( x4+27) 2
( x4+27) 2
2
4
2 2
4x ( 81 x )
4x ( x +9 ) (x+3)(x 3)
=
=
2
4
( x +27)
( x4+27) 2 / / / / (x)>0 if 3<x<0 and 0<x<3 , and f (x)<0 if x< 3 or x>3 . Thus, f is concave upward on
( 3,0 ) and ( 0,3) and f is concave downward on ( , 3) and ( 3, ) . There are inflection points at
( 3,ln 108 ) ( 3,4.68 ) .
f (d)
43. (a) f ( )=2cos cos 2 , 0 2 . / f ( )= 2sin +2sin 2 = 2sin +2(2sin cos )=2sin (2cos
Interval sin 2cos 0< < + + 3 3 < < + 1 / 1) . f ( ) f + increasing on (0,
decreasing on ( 3 3 ) , ) 15 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph 5
3 < < + 5
< <2
3
(b) f increasing on ( , + = 3 3
and f
2 decreasing on ( 5
3 = 5
)
3 5
,2 )
3 3
are local maximum values and f ( )= 3 is a local minimum
2 value.
/ (c) f ( )= 2sin +2sin 2
2 / / 1)
( ) = 2cos +4cos 2 = 2cos +4(2cos
2
=2(4cos
cos 2)
1 33
1 33
1+ 33
/ /
1
1
f ( )=0 cos =
=cos
=cos
0.5678 ,
8
8
8
1+ 33
1 33
1 33
1
1
1
2 cos
5.7154 , cos
2.2057 , or 2 cos
4.0775 .
8
8
8
Denote these four values of by
,
,
, and
, respectively. Then f is CU on 0,
, CD
f 1 ( on , 1 CU on ( 2 4 2 ( ) 3 1 ),
, 2 3 ) , CD on ( we have f ( )=2cos
1 1 , 3 4 ) , and CU on ( cos 2 =2cos
1 1 2cos ,2 4
2 1 ) . To find the exact y
1 =2 1+ 33
8 2 coordinate for =
1+ 33
8 1 , 2 +1 1 1
1
1
33
3
3
3
+
33
33
+1=
+
33 =
( 1+ 33 ) =y1 1.26 . Similarly,
4 4
32 16
32
16 16
16
3
f ( )=
( 1 33 ) =y2 0.89 . So f has inflection points at 1,y1 , 2,y2 , 3,y2 , and
2 16
,y . = ( ( 4 1 ) ( ) ( ) ) (d)
44. (a) f (t)=t+cos t , 2
2 t 2 / f (t)=1 sin t / 0 for all t and f (t)=0 when sin t=1 t= 3
or
2 , so f is increasing on ( 2 ,2 ) .
16 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph (b) No maximum or minimum
(c) f / / (t)= cos t > 0 CD 0n 3
2 2 , 3
,
2 3
2 3
,
2 t
, 2 and 2 , 2
3
, 2
2 , and 2 , 2 , 3
2 , so f is CU on these intervals and . Points of inflection at 2 (d) 2 x 45. f (x)= 2 x 1 2 x
=
has domain (
(x+1)(x 1)
2 (a) lim = = since x x 2 x x 2 1 2 x + 1 2 1 1/x
+ 0 as x = ). 1
=1 , so y=1 is a HA.
1 0
1 , so x= 1 is a VA. x 1
x lim 1 = lim ( x2 1) /x2 x
2
2
since x 1 and ( x 1 ) x lim 2 x /x f (x)= lim , 1 ) ( 1,1 ) ( 1, 2 2 ( 2 1 and x 1 ) + 0 as x + 1 , so x=1 is a VA. x 1
2 (b) f (x)= x
2 f / ( x2 1) (2x) x2(2x) = 2x ( x2 1) x2 = 2x
(x)=
( x2 1) 2
( x2 1) 2
( x2 1) 2 ( 2 . Since x 1 ) 2 is x 1
positive for all x in the domain of f , the sign of the derivative is determined by the sign of 2x .
/ / Thus, f (x)>0 if x<0 ( x 1 ) and f (x)<0 if x>0 ( x 1 ) . So f is increasing on (
and f is decreasing on ( 0,1 ) and ( 1, ) .
/
(c)
f (x)=0 x=0 and f (0)=0 is a local maximum value. , 1 ) and ( 1,0 ) , 17 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph (d) ( x2 1) 2( 2) ( 2x) 2 ( x2 1) (2x)
/ /
f (x) =
( ( x2 1) 2) 2
2
2
2
2
2 ( x 1 ) ( ( x 1 ) +4x ) 2 ( 3x +1 )
=
=
4
2
( x 1)
( x2 1) 3
/ / The sign of f . / / (x) is determined by the denominator; that is, f / / f (x)<0 if x <1 . Thus, f is CU on (
no inflection points. , 1 ) and ( 1, (x)>0 if x >1 and ) , and f is CD on ( 1,1 ) . There are (e) 2 46. f (x)= x has domain ( 2 ,2 ) ( 2, ). (x 2) 2 (a) lim
x 2 x = lim 2 ( x2 4x+4) /x2 x x 4x+4 2 x /x 2 x so y=1 is a HA. lim
x
2 (b) f (x)= x + 2 2 2 2 2 x 1 4/x+4/x
2 since x = 4 and (x 2) + 0 as x 1
=1 ,
1 0+0
+ 2 , so x=2 is a VA. (x 2)
/ (x 2) = 1 = lim f (x)= 2 2 ( x 2 ) (2x) x 2(x 2) 2x(x 2)[(x 2) x]
( x 2) / 2 2 = 4 (x 2) = 4x / 3 . f (x)>0 if 0<x<2 (x 2) and f (x)<0 if x<0 or x>2 , so f is increasing on ( 0,2 ) and f is decreasing on (
(c) ,0 ) and ( 2, ). f (0)=0 is a local minimum value. 18 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph (d) 3 f / / (x) = 2 (x 2) ( 4) ( 4x) 3(x 2)
3 2 (x 2)
2 = 4(x 2) [ (x 2)+3x]
6 = (x 2)
f / / 8(x+1)
4 (x 2)
/ / (x)<0 if x< 1 . Thus, f is CU on ( 1,2 ) and ( 2,
1
, 1 ) . There is an inflection point at
1,
.
9 (x)>0 if x> 1 ( x 2 ) and f is CD on ( ) , and f (e) 47. (a) lim ( 2 ) x +1 x = and x lim ( ) 2 x +1 x =lim x ( 2 x +1 x ) x
2 (b) f (x)= x +1 x 2 x +1 +x
2 x +1 +x
/ f (x)= x
2 x +1 1 =lim 1 . Since 2 x =0 , so y=0 is a HA. x +1 +x
x
2 / <1 for all x , f (x)<0 , so f is decreasing x +1 on R .
(c) No minimum or maximum 19 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph (d)
/ / f (x) = ( x2+1) 1/2 (1) x 1 ( x2+1) 1/2 (2x)
2 (
( x +1)
2 =
= ( x +1) x +1 ) 2 2 1/2 x ( x2+1) 1/2 = ( x2+1) x2
2
x +1
( x2+1) 3/2 1
2 2 3/2 >0 , so f is CU on R . No IP (e) 48. (a) lim xtan x=
x and lim /2 x f (x)=xtan x , 2 (b) <x< xtan x=
/2
/ 2 , so x= + 2 2 . f (x)=xsec x+tan x>0 and x= 0<x< 2 2 are VA. , so f increases on 0, 2 and decreases on
(c)
(d) ,0 .
2
f (0)=0 is a local minimum value.
f / / 2 2 (x)=2sec x+2xtan xsec x>0 for 2 <x< 2 , so f is CU on ,
2 2 . No IP (e) 49. f (x)=ln(1 ln x) is defined when x>0 (so that ln\:\italic{x} is defined) and 1 ln x>0 [so that
ln(1 ln x) is defined]. The second condition is equivalent to 1>ln x
x<e , so f has domain (0,e) .
(a) As x + 0 ,ln x , so 1 ln x and f (x) . As x e ,ln x 1 , so 1 ln x + 0 and
20 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph f (x) . Thus, x=0 and x=e are vertical asymptotes. There is no horizontal asymptote.
1
1
1
/
(b) f (x)=
=
<0 on ( 0,e ) . Thus, f is decreasing on its domain, ( 0,e ) .
1 ln x
x
x(1 ln x)
/ (c) f (x) 0 on ( 0,e ) , so f has no local maximum or minimum value.
(d)
f / / / x(1 ln x) (x) = x(1 ln x)
ln x = 2 2 = x( 1/x)+(1 ln x)
2 2 x (1 ln x) 2 x (1 ln x)
/ / so f (x)>0
at ( 1,0 ) . ln x<0 0<x<1. Thus, f is CU on ( 0,1 ) and CD on ( 1,e ) . There is an inflection point (e) 50. f (x)= e x 1+e x has domain R .
x e /e (a) lim f (x)=lim
x x (b) f ( 1+ex) /ex x lim f (x)= lim
x
/ x e x 1+e x = =lim
x 1
x e +1 = 1
=1 , so y=1 is a HA.
0+1 0
=0 , so y=0 is a HA. No VA.
1+0 ( 1+ex) ex ex ex = ex >0 for all x . Thus, f is increasing on R .
(x)=
( 1+ex) 2
( 1+ex) 2 (c) There is no local maximum or minimum. 21 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph (d) ( 1+ex) 2ex ex 2 ( 1+ex) ex
/ /
f (x) =
2
( 1+ex) 2
x
x
x
x
x
x
e ( 1+e ) ( 1+e ) 2e
e (1 e )
=
=
x 4
( 1+e )
( 1+ex) 3
f / / (x)>0 point at x x<0 , so f is CU on ( 1 e >0
1
0,
.
2 ,0 ) and CD on ( 0, ) . There is an inflection (e) 51. (a) lim e 1/ ( x+1 ) =1 since 1/(x+1) 0 , so y=1 is a HA. lim e x lim e
x x 1/ ( x+1 ) = since 1/(x+1) 1/ ( x+1 ) =0 since 1/(x+1) , + 1 , so x= 1 is a VA. 1 (b) f (x)=e 1/(x+1) / f (x)=e 1/(x+1) ( 1) 1
2 [Reciprocal Rule] =e (x+1)
for all x except 1 , so f is increasing on (
, 1 ) and ( 1,
(c) No local maximum or minimum
(d)
2
1/(x+1)
2 1/(x+1)
1/(x+1) e
2(x+1)
(x+1) e
/ /
f (x) =
2 2
(x+1)
= e 1/(x+1) 1 (2x+2)
4 (x+1)
f / / (x)>0 1
,
2 = e x< 2 /(x+1) / f ( x ) >0 ). 1/(x+1) (2x+1)
4 (x+1) 1
, so f is CU on (
2
1 2
. f has an IP at
,e
.
2
2x+1<0 1/(x+1) , 1 ) and 1, 1
2 , and CD on (e) 22 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph 52. (a) f is periodic with period ( x 2 ) lim ln tan x =
( /2 ) ( 2 (b) f (x)=ln tan x ( , and
x ) , so we consider only / f (x)= 2 2tan xsec x
2 0, and decreasing on
2
(c) No maximum or minimum
2
4
/
(d) f (x)=
=
sin xcos x sin 2x and 0, 4 4 <x< 4 2
f ,0
/ / 2 , so x=0 , x= ( 2 ) . lim ln tan x =
x 2 , 0 are VA. 2 sec x
=2
>0
tan x tan x>0 0<x< , so f is increasing on 2 . (x)= 8cos 2x
2 <0 sin 2x , so f is CD on , and CU on ) lim ln tan x =
+
( /2 )
tan x cos 2x>0 2 2 <x< 2 , 4 4
and ,0
,
4 2 . IP are 4 ,0 . (e) 53. (a)
From the graph, we get an estimate of f (1) 1.41 as a local maximum value, and no local minimum
value.
x+1
/
1 x
f (x)=
f (x)=
.
3/2
2
2
x +1
x +1 ( ) 23 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph / x=1 . f (1)= f (x)=0 2
= 2 is the exact value.
2 (b) From the graph in part (a), f increases most rapidly somewhere between x=
find the exact value, we need to find the maximum value of f
/ critical numbers of f minimum value of f / .f / / 2 2x 3x 1 ( x) = ( x2+1) 5/2 =0 . The maximum value of f x=
/ 3 17
4 is at / 1
1
and x=
. To
2
4 , which we can do by finding the . x= 3 17
4 3+ 17
corresponds to the
4
, 17
6 7
6 ( 0.28,0.69 ) . 54. (a)
Tracing the graph gives us estimates of f (0)=0 for a local minimum value and f (2)=0.54 for a local
maximum value.
2 f (x)=x e x / x / f (x)=xe (2 x) . f (x)=0 x=0 or 2 . f (0)=0 and f (2)=4e (b) From the graph in part (a), f increases most rapidly around x=
need to find the maximum value of f
f / / (x)=e x ( x2 4x+2) =0 maximum value of f
55. f (x)=cos x+ / x=2 ( is at 2 1
cos 2x
2 / / 2 are the exact values. 3
. To find the exact value, we
4 , which we can do by finding the critical numbers of f 2 . x=2+ 2 corresponds to the minimum value of f
2 ,( 2 2 2) e 2+ 2 f (x)= sin x sin 2x f ) / / . . The ( 0.59,0.19 ) .
/ / (x)= cos x 2cos 2x (a)
From the graph of f , it seems that f is CD on ( 0,1 ) , CU on ( 1,2.5) , CD on ( 2.5,3.7) , CU on
( 3.7,5.3) , and CD on ( 5.3,2 ) . The points of inflection appear to be at ( 1,0.4 ) , ( 2.5, 0.6 ) ,
24 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph ( 3.7, 0.6 ) , and ( 5.3,0.4 ) . (b)
/ / From the graph of f
(and zooming in near the zeros), it seems that f is CD on ( 0,0.94 ) , CU on
( 0.94,2.57) , CD on ( 2.57,3.71 ) , CU on ( 3.71,5.35) , and CD on ( 5.35,2 ) . Refined estimates of
the inflection points are ( 0.94,0.44 ) , ( 2.57, 0.63) , ( 3.71, 0.63) , and ( 5.35,0.44 ) .
3 4 / 56. f (x)=x (x 2)
f / / 3 3 4 2 2 3 2 3 f (x)=x 4(x 2) +(x 2) 3x =x (x 2) [4x+3(x 2)]=x (x 2) (7x 6)
3 2 2 2 3 (x) =(2x)(x 2) (7x 6)+x 3(x 2) (7x 6)+x (x 2) (7)
2 =x(x 2) [2(x 2)(7x 6)+3x(7x 6)+7x(x 2)]
2 2 2 ( 2 =x(x 2) 42x 72x+24 =6x(x 2) 7x 12x+4 ) (a)
From the graph of f , it seems that f is CD on (
,0 ) , CU on ( 0,0.5) , CD on ( 0.5,1.3) , and CU on
( 1.3, ) . The points of inflection appear to be at ( 0,0 ) , ( 0.5,0.5) , and ( 1.3,0.6 ) . (b)
/ / From the graph of f
(and zooming in near the zeros), it seems that f is CD on (
,0 ) , CU on
( 0,0.45) , CD on ( 0.45,1.26 ) , and CU on ( 1.26, ) . Refined estimates of the inflection points are
( 0,0 ) , ( 0.45,0.53) , and ( 1.26,0.60 ) .
57. In Maple, we define f and then use the command plot (diff (diff (f, x), x), x= 3..3);. In
Mathematica, we define f and then use Plot [Dt [Dt [f, x], x], {x, 3, 3}]. We see that f / / >0 for
25 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph x>0.1 and f / / <0 for x<0.1 . So f is concave up on ( 0.1, / / ) and concave down on ( 58. It appears that f
is positive (and thus f is concave up) on ( 1.8,0.3) and ( 1.5,
(so f is concave down) on (
, 1.8 ) and ( 0.3,1.5) . ,0.1 ) . ) and negative 59. Most students learn more in the third hour of studying than in the eighth hour, so K(3) K(2) is
larger than K(8) K(7) . In other words, as you begin studying for a test, the rate of knowledge gain is
/ large and then starts to taper off, so K (t) decreases and the graph of K is concave downward.
60. At first the depth increases slowly because the base of the mug is wide. But as the mug narrows,
the coffee rises more quickly. Thus, the depth d increases at an increasing rate and its graph is
concave upward. The rate of increase of d has a maximum where the mug is narrowest; that is, when
the mug is half full. It is there that the inflection point (IP) occurs. Then the rate of increase of d starts
to decrease as the mug widens and the graph becomes concave down. 26 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph 61. From the graph, we estimate that the most rapid increase in the percentage of households in the
United States with at least one VCR occurs at about t=8. To maximize the first derivative, we need to
a
determine the values for which the second derivative is 0. We’ll use V (t)=
, and substitute
ct
1+be
a=85 , b=53 , and c= 0.5 later. ) [ by the Reciprocal Rule] and
( 1+bect ) 2
( 1+bect ) 2 cect ect 2 ( 1+bect ) bcect
/ /
V (t) = abc
2
( 1+bect ) 2
ct
ct
ct
ct
2 ct
ct
abc ce ( 1+be ) ( 1+be ) 2be
abc e ( 1 be )
=
=
ct 4
( 1+be )
( 1+bect ) 3
/ V (t)= So V / / ( ct a bce ct (t)=0 ct 1=be e =1/b . Now graph y=e 0.5t and y= 1
. These graphs intersect at t 7.94
53 years, which corresponds to roughly midyear 1988.
2 ( 2) t , t= x / 2 62. (a) As x , and e 0 . The HA is y=0 . Since t takes on its maximum
2 t value at x=0 , so does e . Showing this result using derivatives, we have f (x)=e
/ 2 ( 2) ( x / 2 2 ) / ( 2) x / 2 / x/
f (x)=e
. f (x)=0 x=0 . Because f changes from positive to negative at x=0 ,
f (0)=1 is a local maximum. For inflection points, we find
2
2
2
2
2
2
/ /
x / (2 )
x / (2 )
2
2 2
1
1 x / (2 )
x/
1 x /
f (x)=
e
1+xe
=
e
. ( 2 f ( / / (x)=0 2 x= 2 x= , ) and ( , ) . IP at
(b) Since we have IP at x= .f ( / / (x)<0
1/2 ) 2 x< 2 2 ( ) <x< . So f is CD on ( , ) and CU on ) ,e
.
, the inflection points move away from the y axis as increases.
27 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph (c)
From the graph, we see that as increases, the graph tends to spread out and there is more area
between the curve and the x axis.
3 2 / 63. f (x)=ax +bx +cx+d 2 f (x)=3ax +2bx+c . We are given that f (1)=0 and f ( 2)=3 , so
/ / f (1)=a+b+c+d=0 and f ( 2)= 8a+4b 2c+d=3 . Also f (1)=3a+2b+c=0 and f ( 2)=12a 4b+c=0 by
2
1
4
7
, d= , so the function
Fermat’s Theorem. Solving these four equations, we get a= , b= , c=
9
3
3
9
1
3
2
is f (x)=
2x +3x 12x+7 .
9 ( ) 2 64. f (x)=axe bx . 2 / f (x)=a xe
/ bx we must have f (2)=0 . f (2)=1
now 1=2ae 1/2 a= 2 2bx+e bx 1=2ae 4b 2 1 =ae bx ( 2bx2+1) . For f (2)=1 to be a maximum value, / and f (2)=0 0=(8b+1)ae 4b . So 8b+1=0 b= 1
and
8 / e 2.
/ 65. Suppose that f is differentiable on an interval I and f (x)>0 for all x in I except x=c . To show
that f is increasing on I , let x , x be two numbers in I with x <x .
1 2 1 2 Case x <x <c . Let J be the interval { x I | x<c} . By applying the Increasing/Decreasing Test to f
1 2
2
on J , we see that f is increasing on J , so f (x )< f (x ) .
1 2 Case c<x <x . Apply the Increasing/Decreasing Test to f on K= { x I | x>c} .
1 2
2
Case x <x =c . Apply the proof of the Increasing/Decreasing Test, using the Mean Value Theorem
1 2
3
(MVT) on the interval x ,x and noting that the MVT does not require f to be
1 2 differentiable at the endpoints of x ,x 1 2 .
28 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph Case c=x <x . Same proof as in Case 3.
1 2
4
Case x <c<x . By Cases 3 and 4, f is increasing on x ,c and on c,x , so f (x )< f (c)< f (x ) .
1
2
1
2
1
2
5
In all cases, we have shown that f (x )< f (x ) . Since x , x were any numbers in I with x <x
1 2 1 2 1 2 , we have shown that f is increasing on I .
66. (a) We will make use of the converse of the Concavity Test (along with the stated assumptions);
/ / that is, if f is concave upward on I , then f
g / / >0 on I , so ( f +g ) / / =f / / +g / / >0 on I . If f and g are CU on I , then f >0 on I / / / / / / =2 f f +2 ff ( / ) 2+2 ff / />0 =2 f / / 2 ( fg) / / =f / / / 0, f
/ / / / >0 , g>0 , g / g+2 f g + fg / / / / f / / / / / /
/ and g we still have 2 f g
on I as in part (a). / 0 on I . Thus, ( fg) / / >0 on I
/ / =f / / / / / / g+2 f g + fg 2 never equal to 0 , we / fg is CU on I . / (c) Suppose f is increasing and g is decreasing . Then f
the argument in parts (a) and (b) fails.
Example 1.
3 / / >0 on I . Then ( fg) = f g+ fg
/ (b) In part (a), if f and g are both decreasing instead of increasing, then f
/ g =2 ff g is CU on I . 0,g g+ fg / >0 on I . So g ( x ) =[ f ( x ) ] 67. (a) Since f and g are positive, increasing, and CU on I with f
have f >0 , f >0 and f +g is CU on I . (b) Since f is positive and CU on I , f >0 and f
g / / 0 and g f
/ / / / 0 and g g+ fg / / / >0 on I
/ 0 on I , so 2 f g I= ( 0, ) , f (x)=x , g(x)=1/x . Then ( fg)(x)=x , so ( fg) (x)=2x and ( fg)
CU on I . / / 0 on I , so / fg is CU
0 on I and (x)=2>0 on I . Thus, fg is Example 2.
/ I= ( 0, ) , f (x)=4x x , g(x)=1/x . Then ( fg)(x)=4 x , so ( fg) (x)=2/ x and ( fg)
on I . Thus, fg is CD on I . / / 3 (x)= 1/ x <0 Example 3.
I= ( 0, 2 ) , f (x)=x , g(x)=1/x . Thus, ( fg)(x)=x , so fg is linear on I . 68. Since f and g are CU on (
h(x)= f (g(x)) / / , ),f / / >0 and g / / >0 on ( , ). / h (x)= f (g(x))g (x)
29 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph / / / / / / / / / h (x)= f (g(x))g (x)g (x)+ f (g(x))g
So h is CU if f is increasing.
/ 69. f (x)=tan x x
0, 2 f (x)=sec x 1>0 for 0<x< . Thus, f (x)> f (0)=0 for 0<x< 2 (x)= f x / / / 2 2 since sec x>1 for 0<x< tan x x>0 2 2 / (g(x)) g (x) + f (g(x))g 0 . 2 / / (x)>0 if f >0 . . So f is increasing on 2 tan x>x for 0<x< / / x 70. (a) Let f (x)=e 1 x . Now f (0)=e 1=0 , and for x 0 , we have f (x)=e 1 0 . Now, since
x x f (0)=0 and f is increasing on 0, ) , f (x) 0 for x 0 e 1 x 0 e 1+x .
1 2
x
/
x
x . Thus, f (x)=e 1 x , which is positive for x 0 by part (a). Thus, f (x) is
(b) Let f (x)=e 1 x
2
1 2
1 2
x
x
increasing on ( 0, ) , so on that interval, 0= f (0) f (x)=e 1 x
x e 1+x+ x .
2
2
(c) By part (a), the result holds for n=1 . Suppose that e
2 k x
f (x)=e 1 x
2!
x x
k! 2 k x
1+x+ +
2! x k+1 x
+
for x 0 . Let
k! k x
/
x
. Then f (x)=e 1 x
(k+1)! x
k! 0 by assumption. Hence, f (x)
k x
k! x is increasing on ( 0, ) . So 0 x implies that 0= f (0) f (x)=e 1 x k k+1 2 n x
x
x
x
e 1+x+
+ +
for x 0 . Therefore, for x 0 , e 1+x+ +
k! (k+1)!
2!
positive integer n , by mathematical induction.
x 3 2 / 2 k+1 x
, and hence
(k+1)!
x
+
for every
n! / / 71. Let the cubic function be f (x)=ax +bx +cx+d f (x)=3ax +2bx+c f (x)=6ax+2b . So f is
CU when 6ax+2b>0 x> b/(3a) , CD when x< b/(3a) , and so the only point of inflection occurs
when x= b/(3a) . If the graph has three x intercepts x , x and x , then the expression for f (x) must
1 2 3 ( ) ( x x ) ( x x ) . Multiplying these factors together gives us
f (x)=a x ( x +x +x ) x + ( x x +x x +x x ) x x x x . \ Equating the coefficients of the x terms
for the two forms of f gives us b= a ( x +x +x ) . Hence, the x coordinate of the point of inflection
a ( x +x +x ) x +x +x
b
.
is
=
=
factor as f (x)=a x x 1 2 3 3 2 1 2 3 2 1 2 1 3 2 3 1 1 3a 2 3 1 3a 4 3 2 72. P(x)=x +cx +x
parabola. \ If P / / 2 2 1 2 3 3 3 3 / 3 2 P (x)=4x +3cx +2x P / / 2 (x)=12x +6cx+2 . The graph of P / / (x) is a (x) has two roots, then it changes sign twice and so has two inflection points. This
30 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph / / 2 2 happens when the discriminant of P (x) is positive, that is, ( 6c ) 4 12 2>0 36c 96>0
2 6
2 6
2
/ /
, P (x) is 0 at one point, but there is still no inflection
c >
1.63. If 36c 96=0 c=
3
3
2 6
/ /
2
/ /
point since P (x) never changes sign, and if 36c 96<0 c <
, then P (x) never changes
3
sign, and so there is no inflection point. c=3 c=6 c=1.8 2 6
c=0
c= 2
3
For large positive c , the graph of f has two inflection points and a large dip to the left of the y axis.
As c decreases, the graph of f becomes flatter for x<0 , and eventually the dip rises above the x axis,
and then disappears entirely, along with the inflection points. As c continues to decrease, the dip and
the inflection points reappear, to the right of the origin.
c= 73. By hypothesis g= f / is differentiable on an open interval containing c . Since ( c,f(c) ) is a point of inflection, the concavity changes at x=c , so f
Derivative Test, f
74. f (x)=x 4 / / / / (x) changes signs at x=c . Hence, by the First has a local extremum at x=c . Thus, by Fermat’s Theorem f f (x)=4x 3 f / / / / 2 (x)=12x for x>0 , f (x)>0 , so f is also CU on ( 0,
an inflection point.
2 f / / (0)=0 . For x<0 , f / / / / (c)=0 . (x)>0 , so f is CU on ( ,0 ) ; ) . Since f does not change concavity at 0 , ( 0,0 ) is not
2 / 2 2 2 75. Using the fact that x = x , we have that g(x)=x x
g (x)= x + x =2 x =2 x
/ /
2 1/2 2x
/ /
g (x)=2x x
=
<0 for x<0 and g (x)>0 for x>0 , so ( 0,0 ) is an inflection point. But
x ( ) 31 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.3 How Derivatives Affect the Shape of a Graph g / / (0) does not exist. 76. There must exist some interval containing c on which f
and f / / / is continuous. On this interval, f / / / / / is positive, since f is increasing (since f / / / changes from negative to positive at c . So by the First Derivative Test, f / / / (c) is positive is positive), so f
/ / / / / =( f ) has a local minimum at
/ / x=c and thus cannot change sign there, so f has no maximum or minimum at c . But since f
changes from negative to positive at c , f has a point of inflection at c (it changes from concave down
to concave up). 32 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule f (x)
0
is an indeterminate form of type
.
0
x a g(x)
f (x)
(b) lim
=0 because the numerator approaches 0 while the denominator becomes large.
x a p(x)
h(x)
(c) lim
=0 because the numerator approaches a finite number while the denominator becomes
x a p(x)
large.
p(x)
= . If f (x) 0 through
(d) If lim p(x)= and f (x) 0 through positive values, then lim
x a
x a f (x)
p(x)
negative values, then lim
=
. If f (x) 0 through both positive and negative values, then the
x a f (x)
limit might not exist.
p(x)
is an indeterminate form of type
.
(e) lim
x a q(x)
1. (a) lim 2. (a) lim
x f (x) p(x) is an indeterminate form of type 0 . a (b) When x is near a , p(x) is large and h(x) is near 1 , so h(x) p(x) is large. Thus, lim h(x) p(x) =
x (c) When x is near a , p(x) and q(x) are both large, so p(x)q(x) is large. Thus, lim
x . a p(x)q(x) = . a 3. (a) When x is near a , f (x) is near 0 and p(x) is large, so f (x) p(x) is large negative. Thus,
lim f (x) p(x) =
.
x a (b) lim
x p(x) q(x) is an indeterminate form of type . a (c) When x is near a , p(x) and q(x) are both large, so p(x)+q(x) is large. Thus, lim
x 4. (a) lim
x f (x) g ( x) 0 is an indeterminate form of type 0 . p ( x) , then ln y= p(x)ln f (x) . When x is near a , p(x)
and ln f (x)
p ( x)
ln y
p
. Therefore, lim f (x)
=lim y=lim e =0 , provided f is defined.
x (c) lim h(x)
x p ( x) a x a x , so a is an indeterminate form of type 1 . a (d) lim
x . a (b) If y= f (x)
ln y p(x)+q(x) = a p(x) f ( x) is an indeterminate form of type 0 . a (e) If y= p(x) q ( x) , then ln y=q(x)ln p(x) . When x is near a , q(x) and ln p(x) , so
1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule . Therefore, lim ln y x
q(x) (f) lim
x p(x) =lim a x p(x) q ( x) =lim y=lim e a x p(x) 1/q ( x ) a x ln y = . a
0 is an indeterminate form of type . a 5. This limit has the form 0
. We can simply factor the numerator to evaluate this limit.
0 2 x 1
(x+1)(x 1)
lim
=lim
=lim (x 1)= 2
x+1
x
1 x+1 x
1
x
1
x+2 6. lim
x 2 =lim 2 x x +3x+2 x+2
1
=lim
= 1
2 (x+1)(x+2) x
2 x+1
9 8 x 1
9
9
0
4 9
9x
= lim x = (1)=
7. This limit has the form
. lim 5 =lim
4 5 x 1
5
5
0 x 1
x 1 x 1 5x
a x 1 8. lim
x b 1 =lim
x x 1 1 ax
bx a 1
b 1 9. This limit has the form = a
b 0
. lim
0
x 2 cos x
= lim
1 sin x + ( /2) x + ( /2) sin x
= lim
cos x
x tan x= . + ( /2) 2 x+tan x
1+sec x 1+1
10. lim
=lim
=
=2
1
x 0 sin x
x 0 cos x
t t 0
e 1
e
=lim
=
11. This limit has the form
. lim
3
0 t 0
t 0 3t 2
t 12. lim
t 0 e 3t t 1 t since e 1 and 3t 2 + 0 as t 0. 3t 3e
=lim
=3
t 0 1
2 2 tan px
psec px p(1)
p
0
13. This limit has the form
. lim
=lim
=
=
2
2
q
0 x 0 tan qx x 0
qsec qx q(1)
2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule 14. lim
/2 1 sin
csc = 0
=0 . L’Hospital’s Rule does not apply.
1 15. This limit has the form ln x
1/x
=lim
=0
x x
1 . lim
x x x e
e
x
=lim
=lim e =
16. lim
x x
1 x
x
17. lim (ln x)/x =
x + as x since ln x 0 and dividing by small values of x just increases the + 0 magnitude of the quotient (ln x)/x . L’Hospital’s Rule does not apply. 18. lim
x ln ln x
=lim
x
x 1 1
ln x x
1
=lim
=0
1
xln x
x
t t t t 5 3
5 ln 5 3 ln 3
5
0
=lim
=ln 5 ln 3=ln
19. This limit has the form
. lim
1
3
0 t 0 t
t 0
20. lim
x 1 ln x
=lim
sin x x 1 1/x
=
cos x 1 ( 1) = 1 x x x e 1
e 1
0
e 1 x
=lim
=lim
=
21. This limit has the form
. lim
2
2
0 x 0
x 0 2x
x 0 2
x
x 22. lim
x 0 2 e 1 x x /2
x 3 x =lim
x 0 23. This limit has the form 24. lim
x 0 e 1 x
2 3x x x e 1
e 1
=lim
=lim
=
6
x 0 6x
x 0 6
x x x x e
e
=lim
=lim
=lim
=
. lim
3 x
2 x
6x x
6
x
x
3x
e e sin x
cos x 1
=lim
= =1
sinh x x 0 cosh x 1
3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule 2 1 1/ 1 x
0
sin x
25. This limit has the form
. lim
=lim
=lim
0 x 0
x
1
x 0
x 0 26. lim
x 0 sin x x
x =lim 3 x cos x 1
2 0 =lim
x 3x 27. This limit has the form 0 1 =
2 1 x 1
=1
1 sin x
cos x
1
=lim
=
6x
6
6
x 0 0
sin x
cos x 1
1 cos x
. lim
=lim
=lim
=
2
0 x 0
2
2
x 0 2x
x 0
x 2 (ln x)
2(ln x)(1/x)
ln x
1/x
28. lim
=lim
=2lim
=2lim
=2(0)=0
x
1
x
1
x
x
x
x
29. lim
x 0 30. lim
x x+sin x 0+0 0
=
= =0 . L’Hospital’s Rule does not apply.
x+cos x 0+1 1
2 cos mx cos nx =lim 2 0 x x 0 2 msin mx+nsin nx
m cos mx+n cos nx 1 2 2
=lim
=
n m
2x
2
2
x 0
x . lim 31. This limit has the form x ( ln 1+2e x ) 1 =lim 1 x 1+2e 32. lim
x 0 x
1 tan (4x) =lim
x 0 x 2e x ( 1+2e =lim
x 2e x x =lim
x )
2e
2e x
x =1 2 1
1
2 4 1+16x
1
=lim
=
4
4
x 0 1+(4x) 1 x+ln x
0
=lim
33. This limit has the form
. lim
0 x 1 1+cos x x 1
2 34. lim
x x +2
2 2x +1 2 =lim
x x +2
2 2x +1 1+1/x
=lim
sin x x 1 2 = lim
x x +2
2 2x +1 2 1/x
2 cos x
2 = lim
x = 1+2/x 2 2+1/x = 1
2 ( 1) = 1
2 1
2 4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule a 1 a ax
a
a(a 1)x
0
x ax+a 1
=lim
=lim
35. This limit has the form
. lim
2
2
0 x 1
x 1 2(x 1)
x 1
(x 1) a 2 = a(a 1)
2 2x 1 e
1 1
=
=0 . L’Hospital’s Rule does not apply.
36. lim
1
x 0 sec x
37. This limit has the form 0 ( ) . We need to write this product as a quotient, but keep in mind
1
that we will have to differentiate both the numerator and the denominator. If we differentiate
,
ln x
we get a complicated expression that results in a more difficult limit. Instead we write the quotient as
ln x
.
1/2
x
lim
x ln x x ln x=lim
+ 0 + x x + 0 2 x 2 x 38. lim x e = lim
x =lim 1/2 x 0 x x e lim cot 2xsin 6x=lim
0 x 0 2x = lim
x e 39. This limit has the form x 1/x
1 3/2
x
2 x 2x
2x 3/2
3/2 x e 2 x ) =0 + x 2 = lim ( =lim
0 x x = lim 2e =0
x 0 . We’ll change it to the form 0
.
0 sin 6x
6cos 6x 6(1)
=lim
=
=3
2
2
tan 2x x 0
2sec 2x 2(1) 40.
lim sin xln x =lim
+ x 0 x = + 0 ln x
1/x
=lim
= lim
csc x
+ csc xcot x
+
x lim
x 0 x sin x
x
+ 0 lim tan x = 1 0=0 0 41. This limit has the form sin x
tan x
x x + 0 3 2 x 0 . lim x e =lim
x x x 3
2 e x 2 =lim
x 3x 2 2xe x =lim
x 3x
2 2e x =lim
x 3
2 4xe =0 x 42.
5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule lim (1 tan x)sec x= ( 1 1 ) 2 =0 . L’Hospital’s Rule does not apply.
x /4 43. This limit has the form 0 (
).
ln x
lim ln xtan ( x/2)=lim
=lim
+
+ cot ( x/2)
+
x 1 x 1 x 1/x ( 1 /2 ) csc ( x/2) ( 2 1 = 2 2 tan (1/x)
sec (1/x) 1/x
44. lim x tan (1/x)=lim
=lim
2
1/x
x
x
x
1/x 2 ( = 2 /2)(1) ) =lim sec 2(1/x)=12=1
x 45.
1
csc x
x lim
x 0 1
x =lim
x 0 =lim
x 0 0 x =lim
x
0 sin x x
xsin x cos x 1
sin x
0
=lim
= =0
xcos x+sin x x 0 2cos x xsin x 2
1
sin x 46. lim (csc x cot x)=lim
x 1
sin x 0 cos x
sin x =lim
x 0 1 cos x
sin x
=lim
=0
sin x
x 0 cos x 47. We will multiply and divide by the conjugate of the expression to change the form of the
expression.
lim ( 2 2 ) x +x x
1 x +x x =lim x x x =lim 2 x x +x +x
2 2 x +x +x
2 =lim ( x2+x) 2 x 2 x
x +x +x
x +x +x
1
1
1
=lim
=
= .
1+1/x +1
1 +1 2
x 2 As an alternate solution, write x +x x as x +x
( 1+1/x 1 ) /(1/x) , and apply l’Hospital’s Rule. 2 x , factor out 2 x , rewrite as 48.
lim
x 1 1
ln x 1
x 1 =lim
x 1 =lim
x 1 x 1 ln x
1 1/x
x
=lim
(x 1)ln x x 1 (x 1)(1/x)+ln x x
x 1
1
1
1
=lim
=
=
x 1+xln x x 1 1+1+ln x 2+0 2
6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule 49. The limit has the form
ln x
lim (x ln x)=lim x 1
x
x
x and we will change the form to a product by factoring out x .
ln x
1/x
= since lim
=lim
=0 .
x x
1
x
1/x 50. As x
, 1/x 0 , and e
a product by factoring out x . ( xe 1/x lim ) ( 1/x x =lim x e x 1/x 51. y=x x ) 1/x e =lim ( 2 + 2 lim x =lim e
x ln y + x and we will change the form to ) =lim ln x 0 x + 0 2 1/x 0 e =e =1 x 1/x 2 0 2 1/x
2 x ln y=x ln x , so lim ln y=lim x ln x=lim x + 1 1/x x x 0 e 1 =lim x
2 x 1 . So the limit has the form 1/x =lim 1/x + x 0 2/x 3 =lim
+ x 0 1 2
x =0
2 0 =e =1 . + 0 x ln y=x ln tan 2x , so
lim ln y lim x ln tan 2x=lim ln tan 2x
=
+
1/x
+
+
x 0 52. y=(tan 2x) x 0 x 0 2 2 (1/tan 2x)(2sec 2x) = lim
x 2 + x x lim (tan 2x) =lim e
x + 0 1/x ln y= 1/x ln y lim (1 2x) =lim e
0 a
54. y= 1+
x x ln y + 0 2 sin 2xcos 2x =lim
x 2x
x
lim
=1 0=0
+ sin 2x
+ cos 2x 0 x 0 0 =e =1 . + x 53. y=(1 2x) x =lim 1/x 0 2x cos 2x 0 1
ln (1 2x)
2/(1 2x)
ln (1 2x) , so lim ln y=lim
=lim
= 2
x
x
1
x 0
x 0
x 0 =e 2 . 0
bx ln y=bxln 1+
b bln (1+a/x)
lim ln y=lim
=lim
1/x
x
x
x a
x , so
1
1+a/x a
2 x
2 1/x =lim
x ab
=ab
1+a/x 7 . Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule bx a
1+
x lim
x =lim e ab . x ln y=xln x so lim 1+ 3 5
+
2
x
x 1+ lim ln y=lim
x 1+ 3 5
+
2
x
x
1/x ln x =e x 3 5
1+ +
2
x
x 55. y= ln y x 3 5
+
2
x
x =lim e =lim
x ln y 3 =e . 3
2 / 10
3 x x 1+ 3 5
+
2
x
x 10
x 3+
=lim 2 3 5
1+ +
2
x
x x 1/x =3 , x ln 2
ln x
1+ln x
(ln 2)(ln x)
(ln 2)(1/x)
lim ln y=lim
=lim
=lim ln 2=ln 2 ,
1+ln x
1/x
x
x
x
x
ln y ln 2
( ln 2 ) / ( 1+ln x )
so lim x
=lim e =e =2 .
56. y=x ( ln 2 ) / ( 1+ln x ) ln y= x x
1/x 57. y=x ln y=(1/x)ln x lim ln y=lim
x x x ) 1/x ln y= ) 1/x lim x =lim e
x x ( x 2 x 0 =e =1 x ) x x ln e +x
e +1
e
e
lim ln y=lim
=lim x =lim x =lim
=1 lim
x
x
x
x
x
x
x
x
e +x
e +1
e
x
x
x
59. y=
ln y=xln
x+1
x+1
x
ln x ln (x+1)
1/x 1/(x+1)
lim ln y =lim xln
=lim
=lim
2
x+1
1/x
x
x
x
x
1/x
=lim ln y 1
x
ln e +x , so
x ( ( 58. y= e +x ln x
1/x
=lim
=0
x x
1 x
x+
x+1 =lim
x ( ex+x) 1/x=lim e ln y 1 =e =e . x x
= 1
x+1 so
8 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule x x
x+1 lim
x =lim e x =e 1 x
x x
x+1 Or: lim ln y x 5/x 60. y=(cos 3x) ln y= 5/x x 1 x+1
x =lim = lim
x 5
ln (cos 3x)
x lim ln y=5lim
x 0 x 0 1 x 1
x 1+ =e 1 ln (cos 3x)
3tan 3x
=5lim
=0 ,
x
1
x 0 0 so lim (cos 3x) =e =1 .
x 0
2 1/x 61. y=(cos x) ln y= 1
2 ln cos x x
2 1/x lim (cos x)
x =lim e + 0 62. y= x 2x 3
2x+5 ln y =e ln cos x lim ln y=lim
x + 0 x 2 + 2 =lim x 0 x + 0 tan x
sec x
1
=lim
=
2x
2
2
+
x 0 1/2 =1/ e + 0 2x+1 2x 3
2x+5 ln y=(2x+1)ln 2 8(2x+1)
lim ln y =lim ln (2x 3) ln (2x+5) =lim 2/(2x 3) 2/(2x+5) =lim
x
2
1/(2x+1)
(2x 3)(2x+5)
x
x
x
2/(2x+1)
2 8(2+1/x)
=lim
= 8
(2 3/x)(2+5/x)
x lim
x 2x 3
2x+5 2x+1 =e 8 63. From the graph, it appears that lim x ln (x+5) ln x =5 .
x To prove this, we first note that
x+5
5
ln (x+5) ln x=ln
=ln 1+
x
x ln 1=0 as x . Thus, 9 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule lim x ln (x+5) ln x
x =lim
x 1
x+5 ln (x+5) ln x
=lim
1/x
x x 2 1/x 2 x (x+5)
x(x+5) =lim 1
x x
1 2 5x =lim
x 2 =5 x +5x 64. tan 2x From the graph, it appears that lim (tan x)
x 0.368 . /4
tan 2x The limit has the form 1 . Now y=(tan x)
ln y=tan 2xln (tan x) , so
2 ln (tan x)
sec x/tan x
2/1
lim ln y= lim
= lim
=
= 1
2
2(1)
x
/4
x
/4 cot 2x
x
/4 2csc 2x
tan 2x lim (tan x)
x /4 = lim e
x ln y 1 =e =1/e 0.3679 . /4 65. / f (x)
f (x)
From the graph, it appears that lim
=lim
=0.25. We calculate
/
x 0 g(x) x 0 g (x)
x x f (x)
e 1
e
1
lim
=lim 3
=lim
= .
2
x 0 g(x) x 0 x +4x x 0 3x +4 4
66. 10 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule / f (x)
f (x)
=lim
=4 . We calculate
From the graph, it appears that lim
/
x 0 g(x) x 0 g (x)
f (x)
2xsin x
2(xcos x+sin x)
lim
=lim
=lim
sec xtan x
x 0 g(x) x 0 sec x 1 x 0
2( xsin x+cos x+cos x)
4
=lim
= =4
2
x 0 sec x sec x +tan x(sec xtan x) 1 ( 67. lim
x e
x 68. lim
x ) x
n =lim
x ln x
x e p nx =lim
x x n 1 =lim
x 1/x
px p 1 i
n px 1+ n . Thus, as n p n 2 = e
=lim
=
n!
x =0 since p>0 . nt i
n , which is of the form 1 . y= 1+ i
n nt , so lim ln y=lim ntln
n 1 x n 1+ x x n(n 1)x =lim 69. First we will find lim
ln y=ntln e ln ( 1+i/n )
=t lim
=t lim
1/n
n
n i
1+
n , A=A 0 nt i
1+
n ( i/n2) =t lim
2
( 1+i/n ) ( 1/n ) n i
=ti
1+i/n it lim y=e
n it Ae .
0 70. (a)
lim v =lim mg 1 e
t
c
t ( = ct/m ) = mg lim ( 1 e ct/m)
c
t mg
mg
(1 0)=
,
c
c which is the speed the object approaches as time goes on, the so called limiting velocity.
(b) lim v
m
11 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule mg
= lim
1 e
c
m ( ct/m ) ct/m g
1 e
= lim
cm
1/m g
= lim
cm e ct/m ( ct/m2)
2 1/m g
ct/m
(ct) lim e
=gt(1)
c
m
The speed of a very heavy falling object is approximately proportional to the elapsed time t , provided
it can fall for time t in an environment where the given model continues to hold. .
= 71. We see that both numerator and denominator approach 0 , so we can use l’Hospital’s Rule:
1
1
3
4 1/2
3
3
2/3 2
4
3
3
2a x x
2a 4x a
( aax ) a
2a x x a aax
2
3
lim
= lim
4
1
3 3/4
2
3
x a
x a
ax
3ax
a
ax
4
1
1 3 2 2/3
3
4 1/2
3
3
2a a a
2a 4a
a aa
2
3
=
1
3 3/4
2
aa
3aa
4
1
1 3 3 2/3
4 1/2
3
a
a
a
a
a a
3
4
4
16
3
=
=
=
a =
a
3
3
3
9
3 3 4 3/4
a a
4
4 ( ) ( ) ( ) ( ( ) ) (
)
( )
( ) ( ) ( ) ( )
( )
( ) 72. Let the radius of the circle be r . We see that A( ) is the area of the whole figure (a sector of the
circle with
1 2
radius 1 ), minus the area of OPR . But the area of the sector of the circle is r (see Reference
2
1
1
1 2
Page 1), and the area of the triangle is r PQ = r(rsin )= r sin . So we have
2
2
2
1 2 1 2
1 2
A( )= r
r sin = r ( sin ) . Now by elementary trigonometry,
2
2
2
1
1
1
1 2
B( )= QR PQ = ( r OQ ) PQ = (r rcos )(rsin )= r (1 cos )sin .
2
2
2
2
So the limit we want is
1 2
r ( sin )
A( )
2
1 cos
lim
=lim
B( ) = lim + 1 2
+
+ (1 cos )cos +sin (sin )
0
r (1 cos )sin
0
0
2 12 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule 1 cos = lim + cos 0 cos 2 +sin sin
= lim + sin +4sin cos =lim 2 + 0 + 0 73. Since f (2)=0 , the given limit has the form
/ sin
1
1+4cos =lim 0 sin
2cos ( sin )+2sin (cos )
1
1
=
1+4cos 0 3 = 0
.
0 / f (2+3x)+ f (2+5x)
f (2+3x) 3+ f (2+5x) 5
/
/
/
lim
=lim
= f (2) 3+ f (2) 5=8 f (2)=8 7=56
x
1
x 0
x 0
sin 2x 74. L=lim
x 0 x 3 +a+ ( 2cos 2x+3ax +b)
2 x 2cos 2x+3ax 2
2 0 =lim 2 x x sin 2x+ax +bx 0 x 3 2 =lim
x 2cos 2x+3ax +b
2 0 . As x 2 0 , 3x 0 , and 3x b+2 , so the last limit exists only if b+2=0 , that is, b= 2 . Thus, 2 lim 3 b =lim 4sin 2x+6ax
8cos 2x+6a 6a 8
=lim
=
, which is equal to 0 if and
6x
6
6
x 0 x 0
3x
4
4
only if a= . Hence, L=0 if and only if b= 2 and a= .
3
3 75. Since lim [ f (x+h) f (x h)]= f (x) f (x)=0 ( f is differentiable and hence continuous) and lim 2h=0
h 0 h 0 , we use l’Hospital’s Rule:
/ / / / / f (x+h) f (x h)
f (x+h)(1) f (x h)( 1) f (x)+ f (x) 2 f (x)
/
lim
=lim
=
=
= f (x)
2h
2
2
2
h 0
h 0
f (x+h) f (x h)
is the slope of the secant line between ( x h,f(x h) ) and ( x+h,f(x+h) ) . As h
2h 0 , this / line gets closer to the tangent line and its slope approaches f (x) . 76. Since lim
h f (x+h) 2 f (x)+ f (x h) = f (x) 2 f (x)+ f (x)=0 ( f is differentiable and hence continuous) 0 and
13 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule 2 lim h =0 , we can apply l’Hospital’s Rule:
h 0 lim
h / 2 0 / f (x+h) f (x h)
=lim
=f
2h
h 0 f (x+h) 2 f (x)+ f (x h)
h / / (x) / At the last step, we have applied the result of Exercise to f (x) .
f (x) 77. (a) We show that lim
x 0 x n lim
x 0 lim
x 0 2n x
f (x)
x n e =lim
x n (x )
n y f (x) e y n 1 y e
f (x) n x x 0 ny =lim =lim x lim 2n 0 y =lim 2 n 0 =lim x
x 1/x 1
2 . Then x 2 f (x) =0 for every integer n 0 . Let y= x 2n 0 = y n! =lim
y e y =0
f (x) f (0)
f (x)
=lim
=0 .
x 0
x 0 x / =0 . Thus, f (0)=lim
x x 0 ( n)
(b) Using the Chain Rule and the Quotient Rule we see that f (x) exists for x 0 . In fact, we prove
by induction that for each n 0 , there is a polynomial p and a non negative integer k with
n f ( n) k n n (x)= p (x) f (x)/x for x 0 . This is true for n=0 ; suppose it is true for the n th derivative. Then
n / 3 f (x)= f (x)(2/x ) , so
f ( n+1 ) k (x) = x
k = / n n n k +3 x n 2k p (x) f (x) n n k x 2k f (x)x n k +2 / n n n p (x)
n x n p (x) n p (x)+2 p (x) k x
n n k 1 ( 3) / n k x n n x p (x)+ p (x) 2/x = k 1 / p (x) f (x)+ p (x) f (x) f (x)x n n ( 2kn+3) which has the desired form.
Now we show by induction that f
Then ( n) / ( 0 ) =0 for all n . By part (a), f (0)=0 . Suppose that f
k f ( n+1 ) f
(0) = lim
x 0 ( n) (x) f
x 0 ( n) (0) =lim
x 0 f ( n) x (x) p (x) f (x)/x
=lim
x 0 n x n ( n) (0)=0 . p (x) f (x)
=lim
x 0 n k +1 x n 14 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.4 Indeterminate Forms and L’Hospital’s Rule = f (x) lim p (x)lim
x 0 n x 0 k +1 x n = p (0) 0=0
n 78. (a) For f to be continuous, we need lim f (x)= f (0)=1 . We note that for x 0 ,
x 0 x ln f (x)=ln x =xln x . So lim ln f (x)=lim xln x =lim
x lim f (x)=lim e
x 0 x ln f ( x ) 0 x 0 x 0 ln x
1/x
=lim
=0 . Therefore,
1/x
x 0 1/x2 0 =e =1 . So f is continuous at 0 . 0 (b) From the graphs, it appears that f is differentiable at 0 . (c) To find f
/ / / , we use logarithmic differentiation: ln f (x)=xln x
x f (x)
=x
f (x) 1
x +ln x / as x 0 , so the curve has a vertical
f (x)= f (x) ( 1+ln x ) = x ( 1+ln x ) , x 0 . Now f (x)
tangent at ( 0,1 ) and is therefore not differentiable there. The fact cannot be seen in the graphs in part
(b) because ln x
very slowly as x 0 . 15 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching ( 3 2 ) 1. y= f (x)=x +x=x x +1 A. f is a polynomial, so D=R . B. x intercept =0 , y intercept = f (0)=0
C. f ( x)= f (x) , so f is odd; the curve is symmetric about the origin. D. f is a polynomial, so there is
/ 2 no asymptote. E. f (x)=3x +1>0 , so f is increasing on (
/ / ) . F. There is no critical number and
/ /
,0 ) ,
(x)=6x>0 on ( 0, ) and f (x)<0 on (
, hence, no local maximum or minimum value. G. f
so f is CU on ( 0, ) and CD on (
,0 ) . Since the concavity changes at x=0 , there is an inflection
point at ( 0,0 ) .
H. 3 2 2 2. y= f (x)=x +6x +9x=x(x+3) A. D=R B. x intercepts are 3 and 0 , y intercept =0 C. No
/ 2 symmetry D. No asymptote E. f (x)=3x +12x+9=3(x+1)(x+3)<0 3<x< 1 , so f is decreasing on
( 3, 1 ) and increasing on ( , 3) and ( 1, ) . F. Local maximum value f ( 3)=0 , local minimum
value f ( 1)= 4 G. f
IP at ( 2, 2 )
H. / / 2 (x)=6x+12=6(x+2)>0 3 ( x> 2 , so f is CU on ( 2, ) and CD on ( , 2) . ) 2 3. y= f (x)=2 15x+9x x = (x 2) x 7x+1 A. D=R B. y intercept: f (0)=2 ; x intercepts: f (x)=0
7
45
x=2 or (by the quadratic formula) x=
0.15 , 6.85
2
/ 2 ( 2 ) C. No symmetry D. No asymptote E. f (x)= 15+18x 3x = 3 x 6x+5 = 3(x 1)(x 5)>0 1<x<5
so f is increasing on ( 1,5) and decreasing on (
,1 ) and ( 5, ) . F. Local maximum value f (5)=27
, local minimum value f (1)= 5 G. f
on ( 3, ) . IP at ( 3,11 )
H. / / (x)=18 6x= 6(x 3)>0 x<3 , so f is CU on ( ,3) and CD 1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 2 4 2 ( 2 ) 4. y= f (x)=8x x =x 8 x A. D=R B. y intercept: f (0)=0 ; x intercepts: f (x)=0 x=0 , 2 2 (
2.83 ) C. f ( x)= f (x) , so f is even and symmetric about the y axis. D. No asymptote E.
/ ( 3 2 ) f (x)=16x 4x =4x 4 x =4x(2+x)(2 x)>0 x< 2 or 0<x<2 , so f is increasing on (
, 2 ) and
( 0,2 ) and decreasing on ( 2,0 ) and ( 2, ) . F. Local maximum value f ( 2)=16 , local minimum
2
2
2
/ /
2
2
/ /
value f (0)=0 G. f (x)=16 12x =4 4 3x =0 x=
. f (x)>0
<x<
, so f is CU
3
3
3
2
2
2
2
2 80
on
,
and CD on
,
and
,
. IP at
,
H.
3
3
3
3
3 9 ( 4 3 ) 3 5. y= f (x)=x +4x =x ( x+4 ) A. D=R B. y intercept: f (0)=0 ; x intercepts: f (x)=0
/ 3 2 2 x= 4,0 C. No symmetry D. No asymptote E. f (x)=4x +12x =4x ( x+3) >0 x> 3 , so f is increasing on ( 3,
and decreasing on (
, 3) . F. Local minimum value f ( 3)= 27 , no local maximum G.
/ / ) 2 f (x)=12x +24x=12x(x+2)<0 2<x<0 ,
so f is CD on ( 2,0 ) and CU on (
, 2 ) and ( 0,
IP at ( 0,0 ) and ( 2, 16 )
H. ). 2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 3 6. y= f (x)=x(x+2) A. D=R B. y intercept: f (0)=0 ; x intercepts: f (x)=0 x= 2,0 C. No symmetry
1
/
2
3
2
2
/
D. No asymptote E. f (x)=3x(x+2) + ( x+2 ) = ( x+2 ) 3x+ ( x+2 ) =(x+2) (4x+2) . f (x)>0 x>
,
2
1
1
/
and f (x)<0 x< 2 or 2<x<
, so f is increasing on
,
and decreasing on (
, 2 ) F.
2
2
1
27
Local minimum value f
=
, no local maximum
2
16
G.
/ / f f 2 (x) =(x+2) (4)+(4x+2)(2)(x+2)
=2(x+2) (x+2)(2)+4x+2
=2(x+2)(6x+6)=12(x+1)(x+2) / / 2<x< 1 , so f is CD on ( 2, 1 ) and CU on ( (x)<0
( 1, 1 )
H. 5 , 2 ) and ( 1, ) . IP at ( 2,0 ) and 2 7. y= f (x)=2x 5x +1 A. D=R B. y intercept: f (0)=1 C. No symmetry D. No asymptote E.
/ 4 3 2 / / f (x)=10x 10x=10x(x 1)=10x(x 1)(x +x+1) , so f (x)<0 0<x<1 and f (x)>0
Thus, f is increasing on (
,0 ) and ( 1, ) and decreasing on ( 0,1 ) .
F. Local maximum value f (0)=1 , local minimum value f (1)= 2 G. f
f / / 3 x=1/ 4 . f (x)=0 ( 3 and CU on 1/ 4, / / (x)>0 ) . IP at 3 x>1/ 4 and f
1
9
,1
3
2
3
4
2
4 ( ) / / (x)<0 / / 3 x<0 or x>1 . 3 3 (x)=40x 10=10(4x 1) so x<1/ 4 , < so f is CD on ( 3 ,1/ 4 ) ( 0.630, 0.786 ) H. 8.
3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 3 5 y= f (x)=20x 3x A. D=R B. y intercept: f (0)=0 ; x intercepts: f (x)=0 3x 3 2 x 20
3 =0 x=0 20/3
2.582 C. f ( x)= f (x), so f is odd;
or
the curve is symmetric about the origin. D. No asymptote E.
/ 2 4 2 2 2 / / f (x)=60x 15x = 15x (x 4)= 15x (x+2)(x 2) , so f (x)>0 2<x<0 or 0<x<2 and f (x)<0
x< 2 or x>2 . Thus, f is increasing on ( 2,0 ) and ( 0,2 ) and f is decreasing on (
, 2 ) and ( 2,
F. Local minimum value f ( 2)= 64 , local maximum value f (2)=64 G.
/ / 3 2 / / f (x)=120x 60x = 60x(x 2) . f (x)>0 x<
Thus, f is CU on (
, 2)
and ( 0, 2 ) , and f is CD on ( 2 ,0 ) and ( 2 ,
and ( 2 ,28 2 )
H. 2 or 0<x< 2 ; f ) . IP at ( / / 2 <x<0 or x> 2 . (x)<0 2 , 28 2 ) ). ( 1.414, 39.598 ) , ( 0,0 ) , 9. y= f (x)=x/(x 1) A. D= { x| x 1} = (
,1 ) ( 1, ) B. x intercept =0 , y intercept = f (0)=0 C. No
x
x
x
symmetry D. lim
=1 , so y=1 is a HA. lim
=
, lim
= , so x=1 is a VA. E.
x 1
x 1
+ x 1
x
x / f (x)= (x 1) x
2 1 = 2 (x 1)
G. f
/ / x <0 for x 1 , so f is decreasing on ( 1 ,1 ) and ( 1, ) . F. No extreme values (x 1)
2 (x)= 1 3 >0 x>1 , so f is CU on ( 1, ) and CD on ( ,1 ) . No IP (x 1)
H. 2 10. y=x/(x 1) A. D= { x| x 1} = (
,1 ) ( 1, ) B. x intercept =0 , y intercept = f (0)=0 C. No
x
x
symmetry D. lim
=0 , so y=0 is a HA. lim
= , so x=1 is a VA. E.
2
x
x 1 (x 1)2
(x 1) 4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 2 / f (x)= (x 1) (1) x(2)(x 1)
4 x 1 = . This is negative on ( 3 (x 1)
, so f (x) is decreasing on ( , 1 ) and ( 1, ) and positive on ( 1,1 ) (x 1)
, 1 ) and ( 1, ) and increasing on ( 1,1 ) .
1
F. Local minimum value f ( 1)=
, no local maximum. G.
4
f / / 3 (x)= 2 (x 1) ( 1)+(x+1)(3)(x 1) = 6 2(x+2) (x 1)
and ( 1, . This is negative on ( 4 , 2 ) , and positive on ( 2,1 ) (x 1) ) . So f is CD on ( , 2 ) and CU on ( 2,1 ) and ( 1, ) . IP at 2, 2
9 H. ( 2 ) 3} = ( 11. y= f (x)=1/ x 9 A. D= { x| x
intercept C. f ( x)= f (x) , 3) ( 3,3) ( 3, f is even; the curve is symmetric about the y axis. D. lim
x y=0 is a HA. lim
x 1 = x 9 + x 1 1 , lim
3 2 = , x 9
2x / >0 x<0 ( x 3 )
2
2
+
x 9
x 9
x
3
x
3
x 9
so f is increasing on (
, 3) and ( 3,0 ) and decreasing on ( 0,3) and ( 3, ) . F. Local maximum
lim 2 = 3 1
2 , lim 1
value f (0)=
. G. y
9
is CU on (
H. 2 / / = , 3) and ( 3, = 1
, no x
9
1
=0 , so
2
x 9 ) B. y intercept = f (0)= , so x=3 and x= 3 are VA. E. f (x)= ( 2 )2 ( 2 ) 2 x 9 +(2x)2 x 9 ( 2x ) ( x 9)
2 4 = ( 2 6 x +3 ( x 9)
2 ) >0
3 ( ) 2 x >9 x>3 or x< 3 , so f ) and CD on ( 3,3) . No IP 5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching ( ) 2 12. y= f (x)=x/ x 9 A. D= { x| x 3} = ( , 3) ( 3,3) ( 3, ) B. x intercept =0 , y intercept
x = f (0)=0. C. f ( x)= f (x) , so f is odd; the curve is symmetric about the origin. D. lim
x x y=0 is a HA. lim
+ x 3 = 2 x 9 , lim
x 3 x
2 = x 9 x , lim
x + 3 2 = x , lim x 9 x = 2 3 =0 , so 2 x 9 , so x=3 and x= 3 x 9 ( x 9) x(2x) = x +9 <0 ( x 3 ) so f is decreasing on ( , 3) , ( 3,3) ,
( x2 9) 2
( x2 9) 2
2
2
2
2
2
/ /
2x ( x 9 ) ( x +9 ) 2 ( x 9 ) ( 2x ) 2x ( x +27)
=
>0
) . F. No extreme values G. f (x)=
4
3
2
2
( x 9)
( x 9)
2 / 2 are VA. E. f (x)= and ( 3, when 3<x<0 or x>3 , so f is CU on ( 3,0 ) and ( 3,
H. ( 2 ) ; CD on ( , 3) and ( 0,3) > IP at ( 0,0 ) ) 13. y= f (x)=x/ x +9 A. D=R B. y intercept: f (0)=0 ; x intercept: f (x)=0 x=0 C. f ( x)= f (x) , / ( x +9) =0 , so y=0 is a HA; so f is odd and the curve is symmetric about the origin. D. lim x x no VA E. f / ( x2+9) ( 1 ) x(2x) = 9 x2 = (3+x)(3 x) >0
(x)=
( x2+9) 2
( x2+9) 2 ( x2+9) 2 2 3<x<3 , so f is increasing on ( 3,3) and decreasing on ( , 3) and ( 3, ) .
1
1
, local maximum value f (3)= G. f
F. Local minimum value f ( 3)=
6
6 ( x2+9) 2( 2x) ( 9 x2) 2 ( x2+9) (2x) = (2x) ( x2+9) ( x2+9) 2 ( 9 x2)
=
2
( x2+9) 4
( x2+9) 2
27= 3 3 f
on ( / / (x)>0 3 3 <x<0 or x>3 3 , so f is CU on ( / / = (x) ( 2 2x x 27 ( x +9)
2 3 ) =0 3 3,0 ) and ( 3 3, , 3 3 ) and ( 0,3 3 ) . There are three inflection points: ( 0,0 ) and 3 3, ) x=0 ,
, and CD 1
3
12 . H. 6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 2 14. y= f (x)=x / ( x +9)
2 A. D=R B. y intercept: f (0)=0 ; x intercept: f (x)=0
2 so f is even and symmetric about the y axis. D. lim x x f / ( ( x2+9) ( 2x ) x2(2x) = 18x >0
(x)=
( x2+9) 2
( x2+9) 2 / ( x +9)
2 ( x2+9) 2 (18) 18x 2 ( x2+9) 2x = 18 ( x2+9) ( x2+9)
=
2
( x2+9) 4
( x2+9) 2
54 ( x+ 3 ) ( x = ( x +9)
2 ( 3, ) 3) 3 2 4x 3 <x< 3 so f is CU on >0 =1 , so y=1 is a HA; no VA E. x>0 , so f is increasing on ( 0, ,0 ) . F. Local minimum value f (0)=0 ; no local maximum G. f . There are two inflection points: 3, 1
4 ( = x=0 C. f ( x)= f (x) , / / ) and decreasing on (x) ( 2 18 9 3x ) ( x2+9) 3 3, 3 ) and CD on ( , 3 ) and . H. 15. y= f (x)= x 1
2 A. D= { x| x 0} = ( ,0 ) ( 0, ) B. No y intercept; x intercept: f (x)=0 x=1 C. x x 1 No symmetry D. lim 2 x
/ f (x)= x 1 (x 1) 2x
2 2 (x ) x x 2 2 = x +2x
4 = (x 2)
3 / / 3 (x)= x ( 1) 0 / 3 2 2 2x 6x
x x<0 or x>2 . Thus, f ) . F. No local minimum, local maximum 3 = , so x=0 is a VA. E. 0<x<2 and f (x)<0 ,0 ) and ( 2, (x 2) 3x = x / 2 (x ) 2 , so f (x)>0 x
x
is increasing on ( 0,2 ) and decreasing on (
1
value f (2)= . G. f
4 x 1 =0 , so y=0 is a HA. lim 6 = 2(x 3)
x 4 .f / / (x) is negative on 7 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching ( ,0 ) and ( 0,3) and positive on ( 3,
2
3,
9
H. ) , so f is CD on ( ,0 ) and ( 0,3) and CU on ( 3, ) . IP at 2 16. y= f (x)= x 2
x 4 A. D= { x| x 0} = ( ,0 ) ( 0, ) B. No y intercept; x intercepts: f (x)=0
2 x= x 2 2 C. f ( x)= f (x) , so f is even; the curve is symmetric about the y axis. D. lim
x x 4 =0 , so 2 x 2 y=0 is a HA. lim
x f (x)= x 4 = , so x=0 is a VA. E. ( x 2) (4x3) =
( x4 ) 2 4 / 0
2 x 2x 5 2x +8x 3 8 2 2(x 4) = 5 2(x+2)(x 2) = 5 / . f (x) is negative on ( 2,0 ) and x
x
x
( 2, ) and positive on ( , 2 ) and ( 0,2 ) , so f is decreasing on ( 2,0 ) and ( 2, ) and increasing
1
, 2 ) and ( 0,2 ) . F. Local maximum value f ( 2)= , no local minimum. G.
on (
8
f / / (2 )
( x5) 2 5 (x)= x ( 4x)+2 x 4 5x , 20
3 and CU on , 20
3 20 21
,
3 200 4 = 2x 4 ( 2 2 2x +5 x 4
10 ) = x
20
,
3 and 2 x and negative on
20
,
3 ( 2 3x 20 and CD on ) 6 f / / 20
,0
3
20
,0
3 (x) is positive on and
and 20
3 0,
0, 20
3 , so f is
. IP at ( 2.5820,0.105) H. 8 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching ( x2+3)
= 2 17. y= f (x)= x
2 3 2 x +3 3 =1 x +3 A. D=R B. y intercept: f (0)=0 ; x intercepts: f (x)=0 2 x +3
2 x x=0 C. f ( x)= f (x) , so f is even; the graph is symmetric about the y axis. D. lim
2x / y=1 is a HA. No VA. E. Using the Reciprocal Rule, f (x)= 3
/ x<0 , so f is decreasing on ( and f (x)<0 value f (0)=0 , no local maximum. G. f = ( 2 6 x +3 f ( 1, / / ) ( x2+3)
( x2+3) 4 2 4x = / / 2 = ,0 ) and increasing on ( 0, ( x2+3)
(x)= 2 ( 2 ( x2+3) 2 x +3 6x ( x +3) ( x +3)
2 2 =1 , so 2 x / 2 . f (x)>0 x>0 ) . F. Local minimum ) 2 6 6x 2 x +3 2x ( 2) = 18(x+1)(x 1)
( x2+3) 3 ( x2+3) 3 6 3 3x (x) is negative on ( , 1 ) and ( 1,
1
1,
) and CU on ( 1,1 ) . IP at
4 ) and positive on ( 1,1 ) , so f is CD on ( , 1 ) and H. 3 18. y= f (x)= x 1 1} = ( A. D= { x| x 3 , 1 ) ( 1, ) B. x intercept =1 , y intercept = f (0)= 1 C. x +1
3 No symmetry D. lim
x x 1
3 x +1 = lim
x 1 1/x 3 1+1/x 3 =1 , so y=1 is a HA. 9 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 3 lim
x 3 x 1 = 3 1 x 1 and lim x +1 x + 1 3 = , so x= 1 is a VA. E. x +1 ( x +1) ( 3x ) ( x 1) ( 3x2) = 6x2 >0 ( x 1 ) so f is increasing on ( , 1 ) and
f (x)=
( x3+1) 2
( x3+1) 2
2
3
2
3
2
3
/ / 12x ( x +1 ) 6x 2 ( x +1 ) 3x
12x ( 1 2x )
=
>0 x< 1 or
( 1, ) . F. No extreme values G. y =
4
3
3
3
( x +1)
( x +1)
3 / 0<x< 1
3 ( 0, 1 ) , 2 3 , <so f is CU on ( 2
1
3 ,
2 , 1 ) and 0, 1
3 and CD on ( 1,0 ) and
2 1
3 , . IP at 2 1
3 H. 19. y= f (x)=x 5 x A. The domain is { x|5 x 0} = (
,5 B. y intercept: f (0)=0 ; x intercepts:
f (x)=0 x=0 , 5 C. No symmetry D. No asymptote E.
1
1
10 3x
10
/
1/2
1/2
1/2
f (x)=x (5 x) ( 1 ) +(5 x) 1= (5 x)
x+2(5 x) =
>0 x<
, so f is
2
2
3
2 5 x
10
10
increasing on
,
and decreasing on
,5 . F. Local maximum value
3
3
10
10
/ /
f
=
15 4.3 ; no local minimum G. f (x)
3
9
1
1/2
1/2
1/2
2(5 x) ( 3) (10 3x) 2
(5 x) ( 1)
(5 x)
6(5 x)+(10 3x)
/ /
2
3x 20
=
=
=
f (x)<0
2
3/2
4(5 x)
4(5 x)
(2 5 x )
for x<5 , so f is CD on (
,5) . No IP
H. 10 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 2 )] B. y intercept: f (0)=0 ; x intercepts: f (x)=0 2 x =x 4x=x
1
1
2
/
4x x =0 x(4 x)=0 x=0 , 4 C. No symmetry D. No asymptote E. f (x)=
1=
(1 x ) .
x
x
This is positive for x<1 and negative for x>1 , so f is increasing on ( 0,1 ) and decreasing on ( 1, ) .
/
1 3/2
/ /
1/2
1
1 =
x =
<0 for
F. Local maximum value f (1)=1 , no local minimum. G. f (x)= x
3/2
2
2x
x>0 , so f is CD on ( 0, ) . No IP
H.
20. y= f (x)=2 x x A. D=[0, ( ) 2 21. y= f (x)= x +1 x A. D=R B. No x intercept, y intercept =1 C. No symmetry
D. lim ( 2 ) and lim x +1 x = ( ) 2 x +1 x =lim x x
/ so y=0 is a HA. E. f (x)=
extreme values G. f / / (x)= x
2 ( x +1) x +1 x ) x 1= x +1
1
2 ( 2 x 2 x +1 +x
2 x +1 +x =lim
x 1
2 =0 , x +1 +x 2 x +1
2 / f (x)<0 , so f is decreasing on R. F. No x +1
3/2 >0 , so f is CU on R . No IP H. 22. y= f (x)= x/(x 5) A. D= { x| x/(x 5) 0} = (
,0 ( 5, ) . B. Intercepts are 0 . C. No symmetry
x
1
x
= lim
=1 , so y=1 is a HA. lim
= , so x=5 is a VA. E.
D. lim
x 5 x
1 5/x
x 5
+
x
x 1
/
f ( x) =
2 x
x 5 5 1/2 5
3 1/2
( 5)
=
x(x 5)
<0 , so f is decreasing on (
2
2
(x 5)
5
/ /
3 3/2
2
No extreme values G. f (x)=
x ( x 5)
(x 5) (4x 5)>0 for x>5 , and f
4 ,0 ) and ( 5,
/ / ) . F. (x)<0 for x<0 , so f
11 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching is CU on ( 5,
H. ) and CD on ( ,0 ) . No IP 2 23. y= f (x)=x/ x +1 A. D=R B. y intercept: f (0)=0 ; x intercepts: f (x)=0
so f is odd; the graph is symmetric about the origin. D. lim f (x) x=0 C. f ( x)= f (x) , x x =lim 2 x = lim
x x +1
x
2 x +1 x/x =lim =lim 2 x 2 x 2 x x +1 /x 2 x +1 / x
x/x = lim 1+1/x /( ) / / (x)= 2 2 x +1
2 x>0 . Thus, f is CU on ( 3 2
x +1
2 ( =
2 x x 1+1/x 1/2 2 2 = [(x +1) ]
increasing on R . f 1 = lim 1
= 1 so
1+0 2x 2 x +1 x
y= 1 are HA. No VA. E. f (x)= lim f (x)
2 x x 2 x +1 / 1 =lim 2 x x +1 /x
x/x = lim x/x ) 5/2 2x= 3x
2 5/2 2 x +1 x ( x +1)
2 , so f / / 3/2 = 1 ( x +1)
2 3/2 >0 for all x , so f is (x)>0 for x<0 and f / / (x)<0 for (x +1)
,0 ) and CD on ( 0, ) . IP at ( 0,0 ) H. 2 24. y= f (x)=x 2 x A. D=
2 , 2 B. y intercept: f (0)=0 ; x intercepts: f (x)=0
x=0 ,
2 . C. f ( x)= f (x) , so f is odd; the graph is symmetric about the origin. D. No asymptote
/ E. f (x)=x x 2 + 2 x =
2 2 2 x +2 x
2 = 2(1+x)(1 x) / . f (x) is negative for 2 <x< 1 and 2 2 x
2 x
2 x
1<x< 2 , and positive for 1<x<1 , so f is decreasing on ( 2 , 1 ) and ( 1, 2 ) and increasing on ( 1,1 ) . F. Local minimum value f ( 1)= 1 , local maximum value f (1)=1 . G f / / (x) 12 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 2 x 2 2 x ( 4x) (2 2x ) 2 2 x = 2 1/2 2 2 = 2 2 3/2 / / / / in
2 , 2 , f (x)>0 for 2 <x<0 and f
CD on ( 0, 2 ) . The only IP is ( 0,0 ) .
H. 2x 6x = (2 x ) (2 x ) 25. y= f (x)= 3 (2 x )( 4x)+(2 2x )x 2 3/2 2 = (2 x ) 2x(x 3)
2 3/2 2 . Since x 3<0 for x (2 x ) (x)<0 for 0<x< 2 . Thus, f is CU on 2 1 x /x A. D= { x |x| 1,x 0 } = 1,0 ) ( 0,1 B. x intercepts x
/ x=0 is a VA. E. f (x/
(x)=
2 2 1 x ) 2 1 x 2 ( 0,1 ) . F. No extreme values G. f 2 x
(x)= 2
3 1,
2
,
3 1
2 and 0, 2
3 3 ( 1 x2) 3/2 and CD on 2 1 x
=
x 1 x
=
x , lim
x 0 , so <0 , so f is decreasing on ( 1,0 ) and
2 1 x 2 2 3x
x CU on 0 1 = x / / + 2 ,0 ) and 1 , no y intercept C. 2 f ( x)= f (x) , so the curve is symmetric about ( 0,0 ) . D. lim ( >0 1<x<
2
,0
3 2
or 0<x<
3
and 2
,1
3 2
, so f is
3
. IP at H. 2 26. y= f (x)=x/ x 1 A. D= (
, 1 ) ( 1,
is symmetric about the origin. D. ) B. No intercepts C. f ( x)= f (x) , so f is odd; the graph
13 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching lim
x x x =1 and lim 2 x 1 = 1 , so y= 1 are HA. lim f (x)=+ 2 x x 1 x 2 x 1 / x= 1 are VA. E. f (x)= ( x 1)
( , 1 ) and ( 1,
/ / f (x)<0 on (
H. 1/3 27. y= f (x)=x 3x
3 x 27x=0 1/2 2 2 = / / (x)>0 on ( 1, 2 x 1 x ( x 1) ) . F. No extreme values G. f
, 1 ) and f 1 x 2 / / 3/2 = 1 ( x 1)
2 3
2 (x)= ( 1 ) ) , so f is CD on ( 3/2 2 x=0 , 1 <0 , so f is decreasing on ( x2 1) 5/2 2x= 3x ( x2 1) 5/2 , 1 ) and CU on ( 1, A. D=R B. y intercept: f (0)=0 ; x intercepts: f (x)=0 x(x 27)=0 , so x 2 x 1 x
2 and lim f (x)= + 1/3 x=3x ) . No IP 3 x =27x 3 3 C. f ( x)= f (x) , so f is odd; the graph is symmetric about the / origin. D. No asymptote E. f (x)=1 x 2/3 =1 1
2/3 x
, 1 ) and ( 1, 2/3 = x 1 2/3 / / . f (x)>0 when x >1 and f (x)<0 when x
0< x <1 , so f is increasing on (
) , and decreasing on ( 1,0 ) and ( 0,1 ) . F. Local
2 5/3
/ /
maximum value f ( 1)=2 , local minimum value f (1)= 2 G. f (x)= x <0 when x<0 and
3
/ / f (x)>0 when x>0 , so f is CD on (
H. 28. y= f (x)=x 5/3 2/3 ,0 ) and CU on ( 0, ) . IP at ( 0,0 ) 2/3 5x =x (x 5) A. D=R B. x intercepts 0 , 5 ; y intercept 0 C. No symmetry D.
5 2/3 10 1/3 5 1/3
/
2/3
, so there is no asymptote E. f (x)= x
x = x (x 2)>0 x<0 or
lim x (x 5)=
3
3
3
x
x>2 , so f is increasing on (
,0 ) , ( 2, ) and decreasing on ( 0,2 ) . F. Local maximum value
f (0)=0 , local minimum value f (2)= 3 3 4 G.
14 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching f / / ( 10 1/3 10 4/3 10
x +
x =
x
9
9
9
, 1 ) . IP at ( 1, 6 ) H. (x)= 4/3 x> 1 , so f is CU on ( 1,0 ) and ( 0, (x+1)>0 ) , CD on 29. y= f (x)=x+ x A. D=R B. x intercepts 0 , 1 ; y intercept 0 C. No symmetry D.
. No asymptote E. For x>0 , f (x)=x+ x
lim ( x+ x ) = , lim ( x+ x ) =
x x 1 / f (x)=1+ ) . For x<0,f(x)=x+ x 1 / f (x)=1 2 x >0 2 x >1 1
1
1
, so f increases on
,
and decreases on
,0 . F. Local maximum
4
4
4
1
1
1 3/2
/ /
/ /
value f
= , local minimum value f (0)=0 G. For x>0 , f (x)=
x
f (x)<0 , so f
4
4
4
1
/ /
3/2
/ /
is CD on ( 0, ) . For x<0 , f (x)=
( x)
f (x)<0 , so f is CD on (
,0 ) . No IP
4
H.
x> 1
4 2 x >0 , so f increases on ( 0, x< 3 30. y= f (x)= ( x2 1) 2 = ( x2 1) 2/3 A. D=R B. x the curve is symmetric about the y axis. D. lim 1 , y intercept 1 C. f ( x)= f (x) , so intercepts ( x2 1) 2/3 = , no asymptote E. x
1/3
4
2
/
/
x x 1
f (x)>0 x>1 or 1<x<0 , f (x)<0 x< 1 or 0<x<1 . So f is increasing on
3
( 1,0 ) , ( 1, ) and decreasing on ( , 1 ) , ( 0,1 ) . F. Local minimum values f ( 1)= f (1)=0 , local
1/3 4
4/3
4 2
1
/ /
2
maximum value f (0)=1 G. f (x)=
x 1
+ x
x 1
(2x)
3
3
3
4/3
4 2
2
, 3 ) , ( 3, ) and CD on ( 3, 1 ) ,
=
x 3 x 1
>0 x > 3 so f is CU on (
9
/ f (x)= ( ) ( ( )( ) ( ) ) ( 1,1 ) , ( 1, 3 ) . IPs at ( 3 3, 4 ) H.
15 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 3 2 31. y= f (x)=3sin x sin x A. D=R B. y intercept: f (0)=0 ; x intercepts: f (x)=0 sin x(3 sin x)=0 2 sin x=0 [ since sin x 1<3 ] x=n , n an integer.
C. f ( x)= f (x) , so f is odd; the graph (shown for 2 x 2 ) is symmetric about the origin and / ( 2 2 ) 3 periodic with period 2 . D. No asymptote E. f (x)=3cos x 3sin x cos x=3cos x 1 sin x =3cos x .
/ f (x)>0 cos x>0 cos x<0 x 2n 2 x 2n 2
3
2n + ,2n +
2
2 ,2n + 2 ,2n + / for each integer n , and f (x)<0 2 for each integer n . Thus, f is increasing on for each integer n , and f is decreasing on integer n . F. f has local maximum values f (2n +
f (2n +
G. f / / 2 2n + 2 ,2n + 3
2 for each )=2 and local minimum values 3
)= 2 .
2
2 2 2n + (x)= 9sin x cos x= 9sin x(1 sin x)= 9sin x(1 sin x)(1+sin x) . f sin x 1 x 2n ,2n + sin x 1 x (2n 1) ,(2n 1) + on the intervals 2 2n , 2n+ 1
2 2 ,2n + and 2n+ 1
and
2
inflection points at ( n ,0 ) for each integer n .
H.
CU on the intervals (2n 1) , 2n 32. y= f (x)=sin x tan x A. D= { x| x (2n+1) 2 } (x)<0 for some integer n . f (2n 1) + 2 / / 2 ,2n 1
2 1
2 (x)>0 sin x<0 and for some integer n . Thus, f is CD , ( 2n+1 )
2n / / sin x>0 and for each integer n , and f is
,2n B. y=0 sin x=tan x= for each integer n . f has sin x
cos x sin x=0 or cos x=1
16 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching x=n ( x intercepts), y intercept = f (0)=0 C. f ( x)= f (x) , so the curve is symmetric about ( 0,0 )
. Also periodic with period 2 D. lim (sin x tan x)=
and lim (sin x tan x)= , so
x x=n + / 2 (2n 1) + x ( /2) are VA. E. f (x)=cos x sec x 0 , so f decreases on each interval in its domain, that is, on
2 ,(2n+1) . F. No extreme values G. f 2 3 1
.f
2 3 Note that 1+2sec x 0 since sec x
1
2 n ( /2)
2 ,n and CD on / / / / ( 2 3 ) (x)= sin x 2sec x tan x=sin x 1+2sec x . (x)>0 for 2 3
<x<2 , so f is CU on
2 <x<0 and 1
2 . f has IPs at ( n ,0 ) . Note also that f (0)=0 , ,
2 2 n , n+ B. Intercepts are 0 C. f ( x)= f (x) , so the curve / / but f ( )= 2 .
H. 33. y= f (x)=x tan x , 2 <x< A. D= 2 is symmetric about the y axis. D. lim
x
/ 2 are VA. E. f (x)=tan x+x sec x>0
2 ,0 x tan x= ( /2) 0<x< lim
x 2 x tan x= , so x= + ( /2) , so f increases on 0, 2 2 and x= 2 and decreases on . F. Absolute and local minimum value f (0)=0 . G. y
f is CU on and ,
2 2 / / 2 2 =2sec x+2x tan x sec x>0 for 2 <x< 2 , so . No IP H. 34. y= f (x)=2x tan x , 2 <x< 2 A. D= ,
2 2 B. y intercept: f (0)=0 ; x intercepts: f (0)=0
17 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 2x=tan x
D.
x x=0 or x lim
( 2x tan x ) =
+
( /2 ) / 2 increasing on
4 = 2 and
x lim ( 2x tan x ) =
( /2 )
/ f (x)=2 sec x<0 f 1.17 C. f ( x)= f (x) , so f is odd; the graph is symmetric about the origin. sec x > 2 and f (x)>0
,
4 4 , so x= 2 = 4 ,
4 2
2 2 / / (x)<0 tan x>0 0<x< 2 . Thus, f is CU on 2 / / ,0 2 , 4 , F. Local maximum value +1 G. f = 2sec x sec x tan x= 2tan x sec x= 2tan x(tan x+1) so f
f are VA. No HA. E. sec x < 2 , so f is decreasing on , and decreasing again on 1 , local minimum value f 2 / / (x)>0 (x)
tan x<0 and CD on 0, 2
2 <x<0 , and
. f has an IP at ( 0,0 ) . H. 1
x sin x , 0<x<3 A. D= ( 0,3 ) B. No y intercept. The x intercept, approximately 1.9
2
1
/
, can be found using Newton’s Method. C. No symmetry D. No asymptote E. f (x)=
cos\:,x>0
2
1
5
7
5
7
cos x<
<x<
or
<x<3 , so f is increasing on
,
and
,3
and
2
3
3
3
3 3
3
3
5 7
decreasing on 0,
and
,
. F. Local minimum value f
=
, local
3
3 3
3
6
2
3
3
5
5
7
7
maximum value f
=
+
, local minimum value f
=
G.
3
6
2
3
6
2
35. y= f (x)= f / / (x)=sin x>0 IPs at , 2 0<x< or 2 <x<3 , so f is CU on ( 0, ) and ( 2 ,3 ) and CD on ( ,2 ) .
and ( 2 , ) . H. 18 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 2 36. y= f (x)=cos x 2sin x A. D=R B. y intercept: f (0)=1 C. No symmetry, but f has period 2 . D.
/ / No asymptote E. y =2cos x( sin x) 2cos x= 2cos x(sin x+1) . y =0
x=(2n+1)
(4n+1)
values f / 2
2 cos x=0 or sin x= 1 / . y >0 when cos x<0 since sin x+1 0 for all x . So y >0 and f is increasing on
,(4n+3) (4n+3) 2 / 2 ; y <0 and f is decreasing on
=2 , local minimum values f / y = 2cos x(sin x+1)= sin 2x 2cos x y / / (4n 1) (4n+1) 2 2 ,(4n+1) 2 . F. Local maximum = 2 G. ( 2 ) = 2cos 2x+2sin x= 2 1 2sin x +2sin x
1
5
2
/ /
+2n , or
=4sin x+2sin x 2=2(2sin x 1)(sin x+1)y =0 sin x= or 1 x= +2n ,
2
6
6
3
5
/ /
/ /
;y
0 and f is CD on
+2n . y >0 and f is CU on
+2n ,
+2n
2
6
6
5
1
5
1
+2n , +2(n+1)
.IPs at
+2n ,
and
+2n ,
6
6
6
4
6
4
. H. 37. y= f (x)=sin 2x 2sin x A. D=R B. y intercept = f (0)=0 . y=0 2sin x=sin 2x=2sin xcos x
sin x=0 or cos x=1 x=n ( x intercepts) C. f ( x)= f (x) , so the curve is symmetric about ( 0,0 ) .
Note: f is periodic with period 2 , so we determine E G for
x
. D. No asymptotes E.
1
2
/
2
<x<
or
f (x)=2cos 2x 2cos x=2 2cos x 1 cos x =2(2cos x+1)(cos x 1)>0 cos x<
2
3
2
2
2
2 2
<x< , so f is increasing on
,
,
,
and decreasing on
,
. F.
3
3
3
3 3
3 3
3 3
2
2
=
=
Local maximum value f
, local minimum value f
G.
3
2
3
2
1
/ /
11
f (x)= 4sin 2x+2sin x=2sin x(1 4cos x)=0 when x=0 ,
or cos x= . If =cos
, then f is
4
4
3 15
,
CU on ( ,0 ) and ( , ) and CD on ( , ) and ( 0, ) . IPs at ( 0,0 ) , (
,0 ) ,
,
8
3 15
,
.
8
H. ( ) 19 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 38. f (x)=sin x x A. D=R B. x intercept =0=y intercept C. f ( x)=sin ( x) ( x)= (sin x x)= f (x) ,
/ so f is odd. D. No asymptote E. f (x)=cos x 1 0 for all x , so f is decreasing on (
/ / , ) . F. No / / extreme values G. f (x)= sin x f (x)>0 sin x<0 (2n 1) <x<2n , so f is CU on
( (2n 1) ,2n ) and CD on ( 2n ,(2n+1) ) , n an integer. Points of inflection occur when x=n .
H. 39. y= f (x)= sin x
1+cos x when
cos x
1 = sin x
1 cos x sin x(1 cos x) 1 cos x
=
=
=csc x cot x
2
1+cos x 1 cos x
sin x
sin x A. The domain of f is the set of all real numbers except odd integer multiples of . B. y intercept:
f (0)=0 ; x intercepts: x=n , n an even integer. C. f ( x)= f (x) , so f is an odd function; the graph
is symmetric about the origin and has period 2 . D. When n is an odd integer, lim f (x)= and
x (n )
lim f (x)=
, so x=n is a VA for each odd integer n . No HA. E.
+
x (n )
1
/
/
(1+cos x) cos x sin x( sin x)
1+cos x
. f (x)>0 for all x except odd multiples
f (x)=
=
=
2
2 1+cos x
(1+cos x)
(1+cos x)
of , so f is increasing on ( (2k 1) ,(2k+1) ) for each integer k . F. No extreme values G.
/ /
/ /
sin x
f (x)=
>0 sin x>0 x ( 2k ,(2k+1) ) and f (x)<0 on ( (2k 1) ,2k ) for each
2
(1+cos x)
integer k . f is CU on ( 2k ,(2k+1) ) and CD on ( (2k 1) ,2k ) for each integer k . f has IPs at
( 2k ,0 ) for each integer k .
H 20 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 40. y= f (x)=cos x/(2+sin x) A. D=R Note: f is periodic with period 2 , so we determine B G on
3
1
[0,2 ]. B. x intercepts ,
, y intercept = f (0)= C. No symmetry other than periodicity D.
2 2
2
/
/
(2+sin x)( sin x) cos x(cos x)
2sin x+1
No asymptote E. f (x)=
=
. f (x)>0 2sin x+1<0
2
2
(2+sin x)
(2+sin x)
1
7
11
7 11
7
sin x<
<x<
,
, so f is increasing on
and decreasing on 0,
,
2
6
6
6
6
6
11
7
1
11
1
,2
. F. Local minimum value f
=
, local maximum value f
=
6
6
6
3
3
G. f / / 2 (2+sin x) (2cos x) (2sin x+1)2(2+sin x)cos x (x)= 4 (2+sin x)
3
,
and CD on
2 2 3
<x<
, so f is CU on
2
2
3
,0
2
H. ( 41. y=1/ 1+e ( lim 1/ 1+e x x ) A. D=R B. No x 1
) = 1+0 =1 and lim 1/ ( 1+e x) =0 (since lim
/ ( x f x
x 2 ) 2 ( x) =e x/ ( 1+e ) e
y=0 and y=1 . E. f (x)= 1+e
increasing on R . F. No extreme values
G.
/ / 2 2cos x(1 sin x)
3 (2+sin x)
3
and
,2
2 intercept; y intercept = f (0)= x x 0, = x e = >0 cos x<0 . IP at 2 ,0 , 1
. C. No symmetry D.
2
), so f has horizontal asymptotes . This is positive for all x , so f is ( 1+e x) 2 ( e x) e x ( 2 ) ( 1+e x) ( e x)
(x) =
( 1+e x) 4
21 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching = e ( e x 1)
( 1+e x) 3
x The second factor in the numerator is negative for x>0 and positive for x<0 , and the other factors are
1
always positive, so f is CU on (
,0 ) and CD on ( 0, ) . f has an inflection point at 0,
.
2
H. 2x 42. y= f (x)=e x 2x e A. D=R B. y intercept: f (0)=0 ; x intercepts: f (x)=0
2x x No symmetry D. lim e / 2x e =0 , so y=0 is a HA. No VA. E. f (x)=2e e =e
x x x ( x x x=0. C. e =1 ) / e =e 2e 1 , so f (x)>0 x 1
e>
2 1
1
x 1
/
x>ln = ln 2 and f (x)<0 e <
x<ln
, so f is decreasing on
2
2
2
1
1
2ln (1/2) ln (1/2)
increasing on ln ,
. F. Local minimum value f ln
=e
e
=
2
2
x / / 2x x x ( ) x ,ln
1
2 1
2
2 and
1
1
=
2
4 / / G. f (x)=4e e =e 4e 1 , so f (x)>0
1
1
/ /
x 1
e>
x>ln
and f (x)<0 x<ln
.
4
4
4
H. Thus, f is CD on
ln 1
,
4 1
4 2 ,ln
1
4 1
4 = ln 43. y= f ( x ) =xln x A. D= ( 0, and CU on
1
3
,
4 16 ln 1
,
4 . f has an IP at . ) B. x intercept when ln x=0 x=1 , no y intercept
22 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching C. No symmetry D. lim xln x= , x lim xln x=lim
x + 0 + x 1 0 ln x
/
1/x
=lim
=lim ( x ) =0 , no asymptote. E. f ( x ) =ln x+1=0 when ln x= 1
2
1/x
+
+
x 0 1/x
x 0 / x=e . f ( x ) >0 ln x> 1 1 x>e , so f is increasing on ( 1/e, f (1/e)= 1/e is an absolute and local minimum value. G. f
IP
H. / / ) and decreasing on ( 0,1/e ) . F.
( x ) =1/x>0 , so f is CU on ( 0, ) . No x x e
e
44. y= f (x)=e /x A. D= { x| x 0} B. No intercept C. No symmetry D. lim
=lim
=
x x
1
x
x x x lim x / x E. f (x)= xe e
2 x >0 x (x 1)e >0 x
so f is increasing on ( 1,
value.
G. f x e
e
=0 , so y=0 is a HA. \ lim
=
x
+ x x / / 2 (x)= , e
, lim
=
x 0 x , so x=0 is a VA. 0 x>1 , ) , and decreasing on ( ,0 ) and ( 0,1 ) . F. f (1)=e is a local minimum ( x) 2x ( xex ex) = ex ( x2 2x+2) >0 x xe x
2 4 x 3 x>0 since x 2x+2>0 for all x . So f is CU on ( 0,
H. 45. y= f (x)=xe x ) and CD on ( ,0 ) . No IP x A. D=R B. Intercepts are 0 C. No symmetry D. lim xe =lim
x x x
e x =lim
x 1
e x =0 , so y=0 is a HA.
23 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching x / lim xe = E. f (x)=e x x x xe =e (1 x)>0 x<1 , so f is increasing on ( ,1 ) and decreasing x on ( 1,
H. G. f ) . F. Absolute and local maximum value f (1)=1/e. / / x (x)=e ( x 2 ) >0 ( x>2 , so f is CU on ( 2, )
3x+2>0} = ( ( ) and CD on ( 2 ,2 ) . IP at 2,2/e ) 2 46. y= f (x)=ln x 3x+2 =ln (x 1)(x 2) { 2 A. D= xinR:x ,1 ) ( 2, ). B. y intercept: f (0)=ln 2 ; x intercepts: f (x)=0
2.62 C. No symmetry D. lim f (x)=lim f (x)=
x
/ f (x)= 2x 3 = 2 1 f 0 2 x 3x+1=0 x= 3 5 2
, so x=1 and x=2 are VAs. No HA. E. x 0.38 , + 2 2(x 3/2)
/
/
, so f (x)<0 for x<1 and f (x)>0
(x 1)(x 2) x 3x+2
for x>2 . Thus, f is decreasing on (
G.
/ / x 2 x 3x+2=e ,1 ) and increasing on ( 2, ) . F. No extreme values ( x2 3x+2) 2 (2x 3)2
(x) =
( x2 3x+2) 2
2 = 2 2x 6x+4 4x +12x 9 ( x2 3x+2) 2
2 = 2x +6x 5 ( x2 3x+2) 2 The numerator is negative for all x and the denominator is positive, so f
domain of f . Thus, f is CD on (
,1 ) and ( 2, ) . No IP
H. / / (x)<0 for all x in the 24 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 47. y= f (x)=ln (sin x)
A. ( 2n , ( 2n+1 ) ) D= { x in R|sin x>0} =
n= ( 4 , 3 ) ( 2 , ) ( 0, ) ( 2 ,3 ) = B. No y intercept; x intercepts: f (x)=0 0 C. f is periodic with period 2 . D. lim
x x=2n + ln (sin x)=0 sin x=e =1
f (x)=
+ (2n ) and lim
x f (x)= for each integer n .
2
, so the lines x=n [(2n+1) ] cos x
/
are VAs for all integers n . E. f (x)=
=cot x , so f (x)>0 when 2n <x<2n + for each
sin x
2
/ / integer n , and f (x)<0 when 2n +
decreasing on 2n + 2 , no local minimum. G. f
H. ,(2n+1)
/ / 2 <x<(2n+1) . Thus, f is increasing on 2n ,2n + for each integer n . F. Local maximum values f and 2 2n + 2 =0 2 (x)= csc x<0 , so f is CD on ( 2n ,(2n+1) ) for each integer n. No IP 2 48. y= f (x)=x(ln x) A. D= ( 0, ) B. x intercept =1 , no y intercept C. No symmetry
2 (ln x)
2ln x
2(ln x)(1/x)
2/x
=lim
=lim
=lim
=lim 2x=0 ,
, lim x(ln x) =lim
2
2
+
+ 1/x
+
+ 1/x
+
+
1/x
x 0
x 0
x 0
x 0
x 0 1/x
x 0 2 2 D. lim x(ln x) =
x
/ 2 no asymptote E. f (x)=(ln x) +2ln x=(ln x)(ln x+2)=0 when ln x=0
/ f (x)>0 when 0<x<e ( 2 x=1 and when ln x= 2 2 x=e . and when x>1 , so ) and ( 1, ) and decreasing on ( e 2,1) .
2
2
F. Local maximum value f ( e ) =4e ,
f is increasing on 0,e 2 25 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching local minimum value f (1)=0
/ / G. f (x)=2(ln x)(1/x)+2/x=(2/x)(ln x+1)=0 when ln x= 1
on ( 1/e, ) , CD on ( 0,1/e ) .
IP at ( 1/e,1/e )
H. 1 x=e . f / / x>1/e , so f is CU (x)>0 2 x 49. y= f (x)=xe A. D=R B. Intercepts are 0 C. f ( x)= f (x) , so the curve is symmetric
2
x
x
1
= lim
=0 , so y=0 is a HA. E.
about the origin. D. lim xe = lim
2
2
x 2 / f (x)=e x 2 2 2 x 2x e =e decreasing on , minimum value f
x> 3
or
2 x e ( 1 2x2) >0 1
2
1
2 x x 2 x< 1
2 1
,
2 and x 2xe
1
x <
, so f is increasing on
2 / / = 1/ 2e G. f and CD on 2 (x)= 2xe 3
,
2 IP are ( 0,0 ) and 3
e
2 3/2 1
2 x ( 1 2x2) 0, 3
2 2 2 x 4xe =2xe x and =1/ 2e , local ( 2x2 3) >0 3
,
2 3
2 , 1
1
,
2
2 . F. Local maximum value f 3
<x<0 , so f is CU on
2 3
,0
2 and x and . . H. x 50. y= f (x)=e 3e
D. lim (e x x 3e x x 4x A. D=R B. y intercept = 2 ; x intercept
x ) 4x =lim x
x e
e
3
x
x x x 4 = 2.22 C. No symmetry
x e
e
, since lim
=lim
=
x x
1
x . Similarly,
26 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching ( ex 3e x 4x) = lim . No HA; no VA x
/ x x x ( 2x ) x x ( )( x x ) x x E. f (x)=e +3e 4=e e 4e +3 =e e 3 e 1 >0 e >3 or e <1
x>ln 3 or x<0 . So f is increasing on (
,0 ) and ( ln 3, ) and decreasing on ( 0,ln 3) . F. Local
maximum value f (0)= 2 ,
1
/ /
x
x
x 2x
2x
local minimum value f (ln 3)=2 4ln 3 G. f (x)=e 3e =e e 3 >0 e >3 x> ln 3 , so f is
2
1
1
1
CU on
ln 3,
and CD on
, ln 3 . IP at
ln 3, 2ln 3 .
2
2
2
H. ( 3x 51. y= f (x)=e +e
asymptotes
/ E. f (x)=3e
x> 1
2
ln
5
3 3x 2e 2x 2x ) A. D=R B. y intercept = f (0)=2 ; no x intercept C. No symmetry D. No
/ 3x , so f (x)>0 3e >2e
/ 0.081 . Similarly, f (x)<0 2x x< 2x [ multiply by e 5x e > 2
3 5x>ln 2
3 1
2
ln
.
5
3 1
2
1
2
ln
and increasing on
ln ,
.
5
3
5
3
2/5
1
2
2 3/5
2
F. Local minimum value f
ln
=
+
1.96 ; no local maximum.
5
3
3
3
f is decreasing on G. f
H. / / 3x (x)=9e +4e , 2x , so f / / (x)>0 for all x , and f is CU on ( x 1
A. D= { x| x
x+1
B. x intercept =1 , y intercept
52. y= f (x)=tan 1 , ) . No IP 1}
27 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 1 = f (0)=tan ( 1)=
1 lim tan
x
1 lim tan
+ x 1 4 C. No symmetry D. x 1
1 1/x
1
1
= lim tan
=tan 1=
, so y=
is a HA. Also
x+1
1+1/x
4
4
x
x 1
x 1
1
=
and lim tan
= . E.
x+1
2
x+1
2
x / f (x) 1 = (x+1) (x 1)
2 / / IP at 2 1+ (x 1)/(x+1)
(x+1)
1
2
=
= 2 >0
2
2
(x+1) +(x 1) x +1 so f is increasing on (
H. G. f 1 , 1 ) and ( 1, 2 ( x ) = 2x/ ( x +1 ) >0
2 0, ) . F. No extreme values x<0 , so f is CU on ( , 1 ) and ( 1,0 ) , and CD on ( 0, ). 4
2 W 4 WL 3 WL 2
W 2 2
W 2
2
2
2
2
53. y=
x+
x
x=
x x 2Lx+L =
x (x L) =cx (x L) where
24EI
12EI
24EI
24EI
24EI
W
2
2
c=
is a negative constant and 0 x L . We sketch f (x)=cx (x L) for c= 1 . f (0)= f (L)=0
24EI
/ 2 ( ) 2 / f (x)=cx 2(x L) +(x L) (2cx)=2cx(x L) x+(x L) =2cx(x L)(2x L) . So for 0<x<L , f (x)>0
x(x L)(2x L)<0 (since c<0 ) / L/2<x<L and f (x)<0 0<x<L/2 . So f is increasing on ( L/2,L ) and ( )
/ /
2
2
f (x)=2c 1(x L)(2x L)+x(1)(2x L)+x(x L)(2) =2c ( 6x 6Lx+L ) =0
4 decreasing on ( 0,L/2 ) , and there is a local and absolute minimum at ( L/2,f(L/2) ) = L/2,cL /16 .
/ f (x)=2c x(x L)(2x L)
x= 6L 2 12L
1
= L
12
2 3
L , and these are the x coordinates of the two inflection points.
6 28 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 54. F(x)= k
2 k + / 2 , where k>0 and 0<x<2 . For 0<x<2 , x 2<0 , so F (x)= x
( x 2)
F is increasing. lim F(x)=
x and lim F(x)= + 0 x 2k
3 2k
3 >0 and x
( x 2)
, so x=0 and x=2 are vertical asymptotes. Notice that 2 when the middle particle is at x=1 , the net force acting on it is 0 . When x>1 , the net force is
positive, meaning that it acts to the right. And if the particle approaches x=2 , the force on it rapidly
becomes very large. When x<1 , the net force is negative, so it acts to the left. If the particle
approaches 0 , the force becomes very large to the left. 55.
2 x +1
y=
. Long division gives us:
x+1 x 1
2 x + 2x x 2 + 1 2 x +x
x+1
x+1
2
2
2
x +1
2
2
x
=x 1+
Thus, y= f (x)=
and f (x) (x 1)=
=
x+1
x+1
x+1
1
1+
x
a slant asymptote (SA). 0 as x . So the line y=x 1 is 56. 29 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 3 y= 2 2x +x +x+3
2 2x 3
. Long division gives us: 3 2 2 2x + x + x + 3 x + 2x x +2x 3 2 2x + 4x
2 3x + x
3x 2 6x
7x + 3
7 3
+
3 2
2
x
2x +x +x+3
7x+3
7x+3
x
=2x 3+ 2
=
Thus, y= f (x)=
and f (x) (2x 3)= 2
2
2
x +2x
x +2x
x +2x
1+
x
the line y=2x 3 is a SA. 0 as x . So 57.
3 y= 2 4x 2x +5
2 2x 2
. Long division gives us: 2 2x + x 3 2x +x 3 4x 3 3 2 2x 2 4x + 2x +3
6x 2 4x + 6x + 5
4x
3 Thus, y= f (x)= 2 4x 2x +5
2 =2x 2+ 2x +x 3 f (x) (2x 2)= 2 = and 1 x x 2 2+ 1 3 x 2x +x 3 2 2x + 6
8x 3 2x +x 3
8 8x 1 8x 1 2 0 as x . So the line y=2x 2 is a SA. x 2 58. 30 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 4 y= 2 5x +x +x
3 2 5x + 2
. Long division gives us: x x x +2 3 2 5x x +2 4 5x 2 +x
4 5x 3 +x + 10x 3 2 5x + x
5x 3 5x 9x 2 + 10
2 6x
4 Thus, y= f (x)= 2 5x +x +x
3 2 2 =5x+5+ x x +2 2 6x 9x 10
3 9x 10
6 9 10 and f (x) (5x+5)= 2 6x 9x 10
3 x x +2 2 = x x +2 x
1 2 x
1 + x
x x
2
x 3 0 as 3 . So the line y=5x+5 is a SA.
2 2x +5x 1
1
59. y= f (x)=
= x+2+
A. D=
2x 1
2x 1 { B. y intercept: f (0)=1 ; x intercepts: f (x)=0
No symmetry D. lim
x lim f (x)= (1/2) x x 1
, so f is decreasing on
2
/ f (x)= 1 2(2x 1) 2 f / / } 2 2x +5x 1=0
f (x)= , so x= + (1/2) = ,
x= 1
2 5 1
,
2
17 4 , 1
2 and 1
,
2 ( 2x 1 )
1
when x< . Thus, f is CU on
2
H. 1
,
2 and CD on 2 ( 2x 1 ) 2 <0 for . F. No extreme values G.
8 3 (x)= 2( 2)(2x 1) (2)= x 0.22 , 2.28 . C. 1
is a VA.
2 1
/
=0 , so the line y= x+2 is a SA. E. f (x)= 1
2x 1 f (x) ( x+2) = lim x x and lim 1
2 x R| x , so f 3 , 1
2 / / (x)>0 when x> 1
and f
2 / / (x)<0 . No IP 31 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 2 x +12
16
=x+2+
60. y= f (x)=
A. D= { x R| x 2} = (
,2 ) ( 2, ) B. y intercept: f (0)= 6 ; no x
x 2
x 2
intercepts. C. No symmetry D. lim f (x)=
and lim f (x)= , so x=2 is a VA.
x lim x
2 16 / f (x)=1 ( x 2) 2 = x 4x 12 ( x 2) x + 2 16
=0 , so the line y=x+2 is a slant asymptote. E.
x 2 f (x) (x+2) = lim x 2 2 = (x 6)(x+2) ( x 2) 2 / / , so f (x)>0 when x< 2 or x>6 and f (x)<0 when 2<x<2 or 2<x<6. Thus, f is increasing on (
, 2 ) and ( 6, ) and decreasing on ( 2,2 ) and ( 2,6 ) .
F. Local maximum value f ( 2)= 4 , local minimum value f (6)=12 G.
/ /
3
/ /
/ /
32
f (x)=16( 2)(x 2) =
, so f (x)>0 for x>2 and f (x)<0 for x<2 . f is CU on ( 2, ) and
3
(x 2)
CD on (
,2 ) . No IP
H. ( 2 ) 61. y= f (x)= x +4 /x=x+4/x A. D= { x| x 0} = (
,0 ) ( 0, ) B. No intercept C. f ( x)= f (x)
symmetry about the origin D. lim ( x+4/x ) = but f (x) x=4/x 0 as x
, so y=x is a slant
x asymptote. lim ( x+4/x ) =
x and lim ( x+4/x ) = + 0 or x< 2 , so f is increasing on ( x / 2 , so x=0 is a VA. E. f (x)=1 4/x >0 2 x >4 x>2 0 , 2 ) and ( 2, ) and decreasing on ( 2,0 ) and ( 0,2 ) . F. Local maximum value f ( 2)= 4 , local minimum value f (2)=4 G. f
( 0, ) and CD on ( ,0 ) . No IP
H. / / 3 (x)=8/x >0 x>0 so f is CU on 62.
32 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching ( ex x ) = x y= f (x)=e x A. D=R B. No x intercept; y intercept =1 C. No symmetry D. lim , x ( e x) =lim
x lim
x x e
1
x x x x x e
e
=lim
=
since lim
x x
1
x = ( ) x x y= x is a slant asymptote since e x ( x)=e 0 as x
so f is increasing on ( 0, ) and decreasing on (
,0 ) .
F. f (0)=1 is a local and absolute minimum value.
/ / G. f
H. x x e >1 x>0 , x 63. y= f (x)= 2 2x +x +1
2 2 2 A. D=R B. y intercept: f (0)=1 ; x intercept: f (x)=0 x +1
2 0=2x +x +1=(x+1)(2x x+1)
f (x) (2x+1) = lim x 2x =2x+1+ x +1
3 / / . E. f (x)=e 1>0 (x)=e >0 for all x , so f is CU on R . No IP 3 lim . x x= 1 C. No symmetry D. No VA
2x
2/x
= lim
=0 , so the line y=2x+1 is a slant asymptote. E.
2
2
x
x +1
1+1/x ( x2+1) ( 2) ( 2x)(2x) = 2 ( x4+2x2+1) 2x2 2+4x2 = 2x4+6x2 = 2x2(x2+3)
(x)=2+
2
2
2
2
(x +1)
(x +1)
( x2+1) 2
( x2+1) 2 x 0 . Thus, f is increasing on ( ,0 ) and ( 0, / so f (x)>0 if ) . Since f is continuous at 0 , f is increasing on R . 2 ( x2+1) ( 8x3+12x) ( 2x4+6x2) 2 ( x2+1) (2x)
F. No extreme values f (x)=
2
( x2+1) 2
2
2
2
4
2
2
/ /
4x ( x +1 ) ( x +1 ) ( 2x +3) 2x 6x ] 4x ( x +3)
=
=
so f (x)>0 for x< 3 and 0<x<
( x2+1) 4
( x2+1) 3
/ / f / / ( 3 <x<0 and x> 3 . f is CU on (
, 3 ) and ( 0, 3 ) , and CD on
3
. There are three IPs: ( 0,1 ) ,
3,
3 +1
( 1.73, 1.60 ) , and
2 (x)<0 for 3,
3, ) 3
3 +1
2 ( 3 , and
3,0 ) and ( 1.73,3.60 ) . H.
33 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 3 3 64. y= f (x)= (x+1) 2 = 2 x +3x +3x+1
2 =x+5+ x 2x+1
(x 1)
intercept: f (0)=1 ; x intercept: f (x)=0 12x 4 A. D= { x R| x 1} = ( 2 (x 1)
x= 1 C. No symmetry D. lim f (x)=
x 12
lim f (x) (x+5) = lim x 2 x 2 = 12x 4 2 x 2x+1
3 (x 1) 3(x+1) (x+1) 2(x 1)
2 2 = lim
x /
x x
=0 , so the line y=x+5 is a SA. f (x)
22 1
1
+
2
x
x (x 1)(x+1) [3(x 1) 2(x+1)]
4 (x 1) 2 = (x+1) (x 5)
3 / so f (x)>0 when x< 1 , (x 1) / 1<x<1 , or x>5 , and f (x)<0 when 1<x<5 . f is increasing on (
,1 ) and ( 5,
on ( 1,5) .
216 27
/ /
=
, no local maximum G. f (x)
F. Local minimum value f (5)=
16
2
= 2 , so x=1 is a VA. 1 2 (x 1) 3 ) B. y 4 2 = ,1 ) ( 1, 2 ) and decreasing 2 (x 1) (x 1) +(x 5) 2(x+1) (x+1) (x 5) 3(x 1)
3 2 (x 1)
2 = (x 1) (x+1){ (x 1)[(x+1)+2(x 5)] 3(x+1)(x 5)}
6 (x 1)
/ / </ so f (x)>0 if 1<x<1 or x>1 , and f
CD on (
, 1 ) . IP at ( 1,0 )
H. = ( { 4 (x 1)
/ / 2 (x+1) (x 1)[3x 9] 3 x 4x 5 )} = (x+1)(24)
4 (x 1) (x)<0 if x< 1 . Thus, f is CU on ( 1,1 ) and ( 1, ) and 34 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 1 2 1 / 65. y= f (x)=x tan x , f (x)=1 2 1+x 1 = 2 1+x
/ / f lim f (x) x x lim f (x) 2 , 2 1+x x+ 2 1+x 2 x 2 2 =0 , so y=x 2 is a SA. Also, tan x 2 x = tan x = . 1 = lim 2 2 1 =lim 2 x so y=x+ = ( 1+x ) (2x) x (2x) = 2x ( 1+x x ) = 2x
( 1+x2) 2
( 1+x2) 2 ( 1+x2) 2
2 (x)= 2 x 2 =0 2 / is also a SA. f (x) 0 for all x , with equality / / (x)
2
has the same sign as x , so f is CD on (
,0 ) and CU on ( 0, ) . f ( x)= f (x) , so f is an odd
function; its graph is symmetric about the origin. f has no local extreme values. Its only IP is at ( 0,0 )
. 2 4 or x 0 , so D= ( 66. y= f (x)= x +4x = x(x+4) . x(x+4) 0 x f (0)=0 ; x intercepts: f (x)=0 x +4x 2 2 x +4x ( x+2 )
1 =
lim x +4x 2 x= 4 , 0 . ( x+2 ) 2 2 = x=0 , so f is increasing on R . f , 4 0, ) . y intercept: ( x+2 ) 2 (x +4x) (x +4x+4)
2 = 4 so 2 x +4x ( x+2 )
x +4x ( x+2 )
x +4x ( x+2 )
f (x) (x+2) =0 . Thus, the graph of f approaches the slant asymptote y=x+2 as x and it x approaches the slant asymptote y= (x+2) as x / . f (x)= x+2
2 / , so f (x)<0 for x< 4 and x +4x
/ f (x)>0 for x>0 ; that is, f is decreasing on (
/ 2 extreme values. f (x)=(x+2)(x +4x) , 4 ) and increasing on ( 0, ) . There are no local 1/2
35 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 1
2
on D so f is CD on ( f / / 2 67. 2 (x)= ( x+2 ) y 2 2 a
lim b
a y= b
b
a x =1
2 x a which shows that y=
b
a lim
x 2 2 2 =(x +4x) 3/2 2 ( 2 2 (x+2) +(x +4x) = 4 x +4x ) 3/2<0 ) . No IP 2 ( 2 2 x a x 2
2 ) 2
2 x a +x
x a +x 2 b
= lim
a x a
2 =0 , 2 x a +x b
x is a slant asymptote. Similarly,
a
b
x
a 2 x a 3 3 2 b
=
lim
a x a
2 2 b
x is a slant asymptote.
a =0 , so y= x a +x 3 x +1 2 x +1 x
1
68. f (x) x =
x=
= , and lim
x
x
x
x
2 1/2 x a . Now b
b
x = lim
a
a x 2 2 (2x+4)+(x +4x) , 4 ) and ( 0, 2 x 3/2 (x +4x) 1
=0. Therefore, lim
x
x 2 f (x) x =0 , and so the 2 2 graph of f is asymptotic to that of y=x . For purposes of differentiation, we will use f (x)=x +1/x.
A. D= { x| x 0} B. No y intercept; to find the x intercept, we set y=0 x= 1.
3 x +1
C. No symmetry D. lim
=
x
+
x 0 3 x +1
and lim
=
x
x
2 , so x=0 is a vertical asymptote. Also, the 0 / 2 graph is asymptotic to the parabola y=x , as shown above. E. f (x)=2x 1/x >0
1 increasing on
f ( 1
3 3
3 = , and decreasing on ( 0, 2 3 3
, no local maximum G. f
2 2
, 1 ) and ( 0, ,0 ) and / / 3 (x)=2+2/x >0 1
3 x> 1
3 , so f is 2 . F. Local minimum value 2 x< 1 or x>0 , so f is CU on ) , and CD on ( 1,0 ) . IP at ( 1,0 ) H.
36 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 4 4 x +1
x 1
3
=0 , so the graph of f is asymptotic to that of y=x .
x
x
x
3 1
A. D= { x| x 0} B. No intercept C. f is symmetric about the origin. D. lim x +
=
and
x
3 69. lim f (x) x = lim x
= lim
x x x 1
x+
x
3 lim
x + 0 = , so x=0 is a vertical asymptote, and as shown above, the graph of f is asymptotic 3 / 2 2 4 to that of y=x . E. f (x)=3x 1/x >0
1
4 , 4 G. f
H. / / 4 5/4 1
3 ,0 x > 3 (x)=6x+2/x >0 1
4 and 0, 3 , local minimum value f 3 70. lim
x = 4 3 x>
1 and decreasing on 3
1 f 0 x>0 , so f is CU on ( 0, 1
4 , so f is increasing on , 3
1
4 1
4 and 3 . F. Local maximum value 3
=4 3 5/4 3 ) and CD on ( ,0 ) . No IP 2 f (x) cos x = lim 1/x =0 , so the graph of f is asymptotic to that of cos x . The
x intercepts can only be found approximately. f (x)= f ( x) , so f is even. lim
x 0 cos x+ 1
2 = , so x
x=0 is a vertical asymptote. We don’t need to calculate the derivatives, since we know the asymptotic
behavior of the curve.
37 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.5 Summary of Curve Sketching 38 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators 4 3 2 1. f (x)=4x 32x +89x 95x+29
f
f / /
/ / 2 (x)=48x 192x+178 . f (x)=0 / 3 2 f (x)=16x 96x +178x 95
/ x 0.5 , 1.60 ; f (x)=0 x 0.92 , 2.5 , 2.58 and x 1.46 , 2.54 . (x)=0 From the graphs of f / / , we estimate that f <0 and that f is decreasing on ( / , and that f >0 and f is increasing on ( 0.92,2.5) and ( 2.58, ,0.92 ) and ( 2.5,2.58 ) ) with local minimum values
/ f (0.92) 5.12 and f (2.58) 3.998 and local maximum value f (2.5)=4 . The graphs of f make it
clear that f has a maximum and a minimum near x=2.5 , shown more clearly in the fourth graph.
From the graph of f / / , we estimate that f / / >0 and that f is CU on ( ,1.46 ) and ( 2.54, ) , and / / that f <0 and f is CD on ( 1.46,2.54 ) . There are inflection points at about ( 1.46, 1.40 ) and
( 2.54,3.999 ) . 6 5 4 3 2. f (x)=x 15x +75x 125x x
f
f / /
/ / 4 3 / 2 (x)=30x 300x +900x 750x . f (x)=0
(x)=0 5 4 3 2 f (x)=6x 75x +300x 375x 1
/ x=0 or x 5.33 ; f (x)=0 x 2.50 , 4.95 , or 5.05; x=0 , 5 or x 1.38 , 3.62 . 1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators / From the graphs of f , we estimate that f is decreasing on (
,2.50 ) , increasing on ( 2.50,4.95) ,
decreasing on ( 4.95,5.05) , and increasing on ( 5.05, ) , with local minimum values
f (2.50) 246.6 and f (5.05) 5.03 , and local maximum value f (4.95) 4.965 (notice the second
/ / graph of f ). From the graph of f
, we estimate that f is CU on (
,0 ) , CD on ( 0,1.38 ) , CU on
( 1.38,3.62 ) , CD on ( 3.62,5) , and CU on ( 5, ) . There are inflection points at ( 0,0 ) and ( 5, 5) ,
and at about ( 1.38, 126.38 ) and ( 3.62, 128.62 ) .
3 3. f (x)= 1
f (x)=
3 2 2x 3 / x 3x 5 (x 2 3x 5 ) 2/3 f / / 2
(x)=
9 2 x 3x+24 ( x2 3x 5) 5/3 2 Note: With some CAS’s, including Maple, it is necessary to define f (x)= x 3x 5
2 2 x 3x 5 1/3 , x 3x 5
since the CAS does not compute real cube roots of negative numbers. We estimate from the graph of
/ f that f is increasing on ( 1.5,
value: f (1.5) 1.9 . ) , and decreasing on ( / / ,1.5) . f has no maximum. Minimum From the graph of f
, we estimate that f is CU on ( 1.2,4.2 ) and CD on (
IP at ( 1.2,0 ) and ( 4.2,0 ) .
4 4. f (x)= 3 2 x +x 2x +2
2 x +x 2 / f (x)=2 5 4 3 , 1.2 ) and ( 4.2, ). 2 x +2x 3x 4x +2x 1 ( x2+x 2) 2
2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators f / / 6 5 4 3 2 x +3x 3x 11x +12x +18x 2 (x)=2 ( x2+x 2) 3 / We estimate from the graph of f that f is increasing on ( 2.4, 2 ) , ( 2, 1.5) and ( 1.5,
, 2.4 ) , ( 1.5,1 ) and ( 1,1.5) . Local maximum value: f ( 1.5) 0.7 .
decreasing on ( ) and / / Local minimum values: f ( 2.4) 7.2 , f (1.5) 3.4 . From the graph of f
, we estimate that f is
, 2) , ( 1.1,0.1 ) and ( 1, ) and CD on ( 2, 1.1 ) and ( 0.1,1 ) .
CU on (
f has IP at ( 1.1,0.2 ) and ( 0.1, 1.1 ) . 5. f (x)= x
3 2 x x 4x+1 / f (x)= 3 2 2x +x +1 ( x3 x2 4x+1) 2 f / / (x)= ( 5 4 3 2 2 3x 3x +5x 6x +3x+4 ) ( x3 x2 4x+1) 3 We estimate from the graph of f that y=0 is a horizontal asymptote, and that there are vertical
/ asymptotes at x= 1.7 , x=0.24 , and x=2.46 . From the graph of f , we estimate that f is increasing
, 1.7) , ( 1.7,0.24 ) , and ( 0.24,1 ) , and that f is decreasing on ( 1,2.46 ) and ( 2.46, ) .
on (
1
/ /
There is a local maximum value at f (1)=
. From the graph of f
, we estimate that f is CU on
3
( , 1.7) , ( 0.506,0.24 ) , and ( 2.46, ) , and that f is CD on ( 1.7, 0.506 ) and ( 0.24,2.46 ) .
3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators There is an inflection point at ( 0.506, 0.192 ) .
/ 6. f (x)=tan x+5cos x 2 f (x)=sec x 5sin x f / / 2 (x)=2sec x tan x 5cos x . Since f is periodic with period 2 , and defined for all x except odd multiples of
3
2 2
, 2 , we graph f and its derivatives on . We estimate from the graph of f / that f is increasing on 2 ,0.21 , 1.07, 2 , 2 ,2.07 3
, and decreasing on ( 0.21,1.07) and ( 2.07,2.93) . Local minimum values:
2
f (1.07) 4.23 , f ( 2.93) 5.10 . Local maximum values: f (0.21) 5.10 , f (2.07) 4.23 .
3
/ /
From the graph of f
, we estimate that f is CU on 0.76,
and 2.38,
, and CD on
2
2
, and 2 2.93, ,0.76 and 2 ,2.38 . f has IP at ( 0.76,4.57) and ( 2.38, 4.57) . 2 / 7. f (x)=x 4x+7cos x , 4 x 4 . f (x)=2x 4 7sin x
/ f (x)=0 x 1.49 , 1.07 , or 2.89 ; f / / (x)=0 f / / (x)=2 7cos x . f (x)=0
2
1
x= cos
1.28
7 x 1.10 ; . / / From the graphs of f , we estimate that f is decreasing ( f <0) on ( 4, 1.49 ) , increasing on
( 1.49, 1.07) , decreasing on ( 1.07,2.89 ) , and increasing on ( 2.89,4 ) , with local minimum values
f ( 1.49) 8.75 and f (2.89) 9.99 and local maximum value f ( 1.07) 8.79 (notice the second
/ / / / , we estimate that f is CU ( f >0 ) on ( 4, 1.28 ) , CD on
graph of f ). From the graph of f
( 1.28,1.28 ) , and CU on ( 1.28,4 ) . There are inflection points at about ( 1.28,8.77) and
4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators ( 1.28, 1.48 ) .
8. f (x)= e x / x 2 f (x)= 2 e (x 2x 9) ( ) 2 f 2 / / x (x)= 4 3 2 ( )3 e (x 4x 12x +36x+99) 2
x 9
x 9
x 9
There are vertical asymptotes at x= 3 . It is difficult to show all the pertinent features in one viewing rectangle, so we’ll show f , f / , and f / / for x<3 and also for x>3 . For x<3 For x>3 / We estimate from the graphs of f and f that f is increasing on (
, 3) , ( 3, 2.16 ) , and
( 4.16, ) and decreasing on ( 2.16,3) and ( 3,4.16 ) . There is a local maximum value of
f ( 2.16) 0.03 and a local minimum value of f (4.16) 7.71 . From the graphs of f
f is CU on (
, 3) and ( 3, ) and CD on ( 3,3) . There is no inflection point.
3 2 9. f (x)=8x 3x 10 / 2 f (x)=24x 6x f 3 / / / / , we see that (x)=48x 6 2 From the graphs, it appears that f (x)=8x 3x 10 increases on (
,0 ) and ( 0.25, ) and decreases
on ( 0,0.25) ; that f has a local maximum value of f (0)= 10.0 and a local minimum value of
f (0.25) 10.1 ; that f is CU on ( 0.1, ) and CD on (
,0.1 ) ; and that f has an IP at ( 0.1, 10 ) .
/ 2 To find the exact values, note that f (x)=24x 6x=6x(4x 1) , which is positive ( f is increasing) for
5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators ( 1
,
4 ,0 ) and , and negative ( f is decreasing) on maximum at x=0 : f (0)= 10 ; and f has a local minimum at
f / / (x)=48x 6=6(8x 1) , which is positive ( f is CU) on
, 1
8 1
,f
8 . f has an IP at 1
8 1 321
,
8
32 = 0, 1
4 . By the FDT, f has a local 1
1
1 3
161
:f
=
10=
.
4
4
8 16
16
1
,
and negative ( f is CD) on
8
. 10. From the graphs, it appears that f increases on ( 0,3.6 ) and decreases on (
,0 ) and ( 3.6, ) ; that f
has a local maximum of f (3.6) 2.5 and no local minima; that f is CU on ( 5.5, ) and CD on
2 ( ,0 ) and ( 0,5.5) ; and that f has an IP at ( 5.5,2.3) . f (x)= x +11x 20
2 =1+ x
/ 2 3 3 f (x)= 11x +40x = x (11x 40) , which is positive ( f is increasing) on
f is decreasing) on ( f 40
11 40
11 = 2 +11
40
11 / 40
,
11 ,0 ) and on 2 40
11 20
= 2
3 / / 11
x 0, 40
11 20
2 x , and negative ( . By the FDT, f has a local maximum at x= 40
:
11 1600+11 11 40 20 121 201
=
; and f has no local
1600
80
3 4 4 f (x)=22x 120x =2x (11x 60) , which is positive ( f is CU)
minimum. f (x)= 11x +40x
60
60
on
,
, and negative ( f is CD) on (
,0 ) and 0,
. f has an IP at
11
11
60
60
60 211
,f
=
,
.
11
11
11 90
11. From the graph, it appears that f increases on ( 2.1,2.1 ) and decreases on ( 3, 2.1 ) and ( 2.1,3) ;
that f has a local maximum of f (2.1) 4.5 and a local minimum of f ( 2.1) 4.5 ; that f is CU on
2 ( 3.0,0 ) and CD on ( 0,3.0 ) , and that f has an IP at ( 0,0 ) . f (x)=x 9 x 6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators 2 x / f (x)= 2 2 + 9 x = 9 2x 2 2 9 x 9 x 3 2
2 f 3 2
2 3 2
=
2 3 2
2 3, negative ( f is decreasing) on
value of f 2 3 2
2 9 2 1
2 2 9 x ( 2x)+x
= = ( 9 x2) 1/2 ( 2x) ( x ( 2 ) . By the FDT, f has a local maximum 2 2 ) x 2 + 9 x f / / (x) 2 9 x 1/2 = 2x x 3 ( 9 x2) 1 x
2 9 x 3 2 3/2 and 9
; and f has a local minimum value of
2 x 9 x 2 9 x
= 3 2
,3
2 and 9
/
=
(since f is an odd function). f (x)=
2 3x 3 2 3 2
,
2
2 , which is positive ( f is increasing) on = ( 2 )
( 9 x2) 3/2 x 2x 27 which is positive ( f is CU) on ( 3,0 ) and negative ( f is CD) 9 x
9 x
on ( 0,3) . f has an IP at ( 0,0 ) . 12. From the graph, it appears that f increases on ( 5.2, 1.0 ) and ( 1.0,5.2 ) and decreases on
( 2 , 5.2 ) , ( 1.0,1.0 ) , and ( 5.2,2 ) ; that f has local maximum values of f ( 1.0) 0.7 and
f (5.2) 7.0 and local minimum values of f ( 5.2) 7.0 and f (1.0) 0.7 ; that f is CU on
( 2 , 3.1 ) and ( 0,3.1 ) and CD on ( 3.1,0 ) and ( 3.1,2 ) , and that f has IP at ( 0,0 ) , ( 3.1, 3.1 )
and ( 3.1,3.1 ) . f (x)=x 2sin x / f (x)=1 2cos x , which is positive ( f is increasing) when cos x< 1
,
2 5
5
5
,
and
,
, and negative ( f is decreasing) on
2 ,
,
3
3
3 3
3
5
,
, and
,2
. By the FDT, f has local maximum values of f
= + 3
3 3
3
3
3
5
5
5
5
and f
=
+ 3 , and local minimum values of f
=
3 and
3
3
3
3
that is, on f 3 = / 3 3 . f (x)=1 2cos x f / / (x)=2sin x , which is positive ( f is CU) on ( 2 , )
7 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators and ( 0, ) and negative ( f is CD) on ( ,0 ) and ( ,2 ) . f has IP at ( 0,0 ) , ( 2 13. (a) f (x)=x ln x . The domain of f is ( 0, ln x 2 (b) lim x ln x=lim
x + 0 + x 0 2 1/x + x 2/x 0 ). 2 1/x =lim ) and ( , ) . , x
2 =lim 3 x + 0 =0 . There is a hole at ( 0,0 ) .
2 (c) It appears that there is an IP at about ( 0.2, 0.06 ) and a local minimum at ( 0.6, 0.18 ) . f (x)=x ln x
1
/
2
1/2
, so f is increasing on ( 1/ e , ) ,
f (x)=x (1/x)+(ln x)(2x)=x(2ln x+1)>0 ln x>
x>e
2
decreasing on ( 0,1/ e ) . By the FDT, f ( 1/ e ) = 1/(2e) is a local minimum value. This point is
approximately ( 0.6065, 0.1839 ) , which agrees with our estimate.
3
/ /
3/2
3/2
, so f is CU on e ,
and CD on
f (x)=x(2/x)+(2ln x+1)=2ln x+3>0 ln x>
x>e
2 ( ( 0,e 3/2) . IP is ( e 3/2, 3/ ( 2e3) )
1/x 14. (a) f (x)=xe ( 0.2231, 0.0747) . . The domain of f is ( 1/x 1/x e
e
(b) lim xe =lim
=lim
+
+ 1/x
+
1/x x 0 x 0 x 0 ) ( ,0 ) ( 0, 2 1/x
2 1/x ) =lim e1/x=
x ). , so x=0 is a VA. + 0 Also
8 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators 1/x 1/x lim xe =0 since 1/x
x e 0. 0 ,0 ) and
(c) It appears that there is a local minimum at ( 1,2.7) . There are no IP and f is CD on (
CU on ( 0, ) .
1
1
/
1/x
1/x 1/x
1/x
1
f (x)=xe
+e =e
1
>0
<1 x<0 or x>1 , so f is increasing on
f (x)=xe
2
x
x
x
( ,0 ) and ( 1, ) , and decreasing on ( 0,1 ) . By the FDT, f (1)=e is a local minimum value, which
agrees with our estimate.
/ / 1/x ( 2) 1/x f (x)=e
1/x +(1 1/x)e
and CD on (
,0 ) . No IP. ( 2 ) ( 1/x 2 1/x = e /x ) (1 1+1/x)=e1/x/x3>0 x>0 , so f is CU on ( 0, ) 15.
2 f (x)= (x+4)(x 3)
4 has VA at x=0 and at x=1 since lim f (x)=
x x (x 1) 0 , lim f (x)=
x 1 and lim f (x)=
x . + 1 2 f (x) = x+4 (x 3)
2
x
x
x
x [dividing numerator and denominator by x 4
3 3 (x 1)
2 (1+4/x)(1 3/x)
=
x(x 1) 0 as x , so f is asymptotic to the x axis. Since f is undefined at x=0 , it
2 has no y intercept. f (x)=0 (x+4)(x 3) =0 x= 4 or x=3 , so f has x intercepts 4 and 3 . Note,
however, that the graph of f is only tangent to the x axis and does not cross it at x=3 , since f is
positive as x 3 and as x + 3 . 9 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators From these graphs, it appears that f has three maximum values and one minimum value. The
maximum values are approximately f ( 5.6)=0.0182 , f (0.82)= 281.5 and f (5.2)=0.0145 and we
know (since the graph is tangent to the x axis at x=3 ) that the minimum value is f (3)=0 . 16.
4 f (x)= 10x(x 1)
3 has VA at x= 1 and at x=2 since lim f (x)= 2 x (x 2) (x+1) 1 , lim f (x)=
x 2 and lim f (x)=
x . + 2 4 f (x)= 10(1 1/x)
3 2 10 as x , so f is asymptotic to the line y=10 . f (0)=0 , so f has a y (1 2/x) (1+1/x) 4 intercept at 0 . f (x)=0 10x(x 1) =0 x=0 or x=1 . So f has x intercepts 0 and 1 . Note, however,
that f does not change sign at x=1 , so the graph is tangent to the x axis and does not cross it. We
know (since the graph is tangent to the x axis at x=1 ) that the maximum value is f (1)=0 . From the
graphs it appears that the minimum value is about f (0.2)= 0.1 . 2 17. f (x)= 3 x (x+1)
2 / 4 (x 2) (x 4) From the graphs of f / f (x)= 2 x(x+1) ( x3+18x2 44x 16)
3 5 (from CAS). (x 2) (x 4) , it seems that the critical points which indicate extrema occur at x 20 , 0.3
10 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators , and 2.5 , as estimated in Example 3. (There is another critical point at x= 1 , but the sign of f
not change there.) We differentiate again, obtaining
f / / (x)=2 ( 6 5 4 3 2 (x+1) x +36x +6x 628x +684x +672x+64
4 6 ) / does . (x 2) (x 4) / / From the graphs of f
, it appears that f is CU on ( 35.3, 5.0 ) , ( 1, 0.5) , ( 0.1,2 ) , ( 2,4 ) and
( 4, ) and CD on ( , 35.3) , ( 5.0, 1 ) and ( 0.5, 0.1 ) . We check back on the graphs of f to find
the y coordinates of the inflection points, and find that these points are approximately
( 35.3, 0.015) , ( 5.0, 0.005) , ( 1,0 ) , ( 0.5,0.00001 ) , and ( 0.1,0.0000066 ) .
4 18. f (x)= 10x(x 1)
3 / 2 (x 2) (x+1) From the graphs of f / f (x)= 20 3 (x 1) (5x 1)
4 3 (from CAS). (x 2) (x+1) , we estimate that f is increasing on (
/ ( 1,0.2 ) , ( 1,2 ) and ( 2, ) . Differentiating f (x) , we get f , 1 ) and ( 0.2,1 ) and decreasing on
/ / 2 (x)=60 (x 1) ( 5x3 8x2+17x 6)
5 4 . (x 2) (x+1) 11 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators / / From the graphs of f
, it seems that f is CU on (
, 1.0 ) , ( 1.0,0.4 ) and ( 2.0,
( 0.4,2 ) . There is an inflection point at about ( 0.4, 0.06 ) .
2 sin x 19. y= f (x)= / with 0 x 3 . From a CAS, y = 2 ( ( x +1)
2 x +1
y / / ( 4x4+6x2+5) cos 2x 4x ( x2+1) sin xcos x 2x4 2x2 3
=
( x2+1) 5/2 / / ) 2 sin x 2 x +1 cos x xsin x
3/2 ) , and CD on and . / From the graph of f and the formula for y , we determine that y =0 when x= , 2 , 3 , or
x 1.3 , 4.6 , or 7.8 . So f is increasing on ( 0,1.3) , ( ,4.6 ) , and ( 2 ,7.8 ) . f is decreasing on
( 1.3, ) , ( 4.6,2 ) , and ( 7.8,3 ) . Local maximum values: f (1.3) 0.6 , f (4.6) 0.21 , and
/ / / / f (7.8) 0.13 . Local minimum values: f ( )= f (2 )=0 . From the graph of f
, we see that y =0
x 0.6 , 2.1 , 3.8 , 5.4 , 7.0 , or 8.6 . So f is CU on ( 0,0.6 ) , ( 2.1,3.8 ) , ( 5.4,7.0 ) , and ( 8.6,3 ) .
f is CD on ( 0.6,2.1 ) , ( 3.8,5.4 ) , and ( 7.0,8.6 ) . There are IP at ( 0.6,0.25) , ( 2.1,0.31 ) , ( 3.8,0.10 ) ,
( 5.4,0.11 ) , ( 7.0,0.061 ) , and ( 8.6,0.065) .
20. f (x)= 2x 1
4 4 x +x+1
/ f (x)= 3 4x +6x+9 ( 4 4 x +x+1 ) 5/4
12 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators f / / 6 (x)= 4 3 2 32x +96x +152x 48x +6x+21 ( 4 16 x +x+1 From the graph of f / ) 9/4 , f appears to be decreasing on ( , 0.94 ) and increasing on ( 0.94, ). / / There is a local minimum value of f ( 0.94) 3.01 . From the graph of f
, f appears to be CU on
( 1.25, 0.44 ) and CD on ( , 1.25) and ( 0.44, ) . There are inflection points at ( 1.25, 2.87)
and ( 0.44, 2.14 ) .
1/x 21. y= f (x)= 1 e 1/x 1/x 2e / . From a CAS, y = 1+e 2 ( 1/x 2 x 1+e f is an odd function defined on ( ) and y / / 1/x = 2e ( 1 e1/x+2x+2xe1/x)
4
1/x 3
x ( 1+e ) . ,0 ) ( 0, ) . Its graph has no x or y intercepts. Since
lim f (x)=0 , the x axis is a HA. f (x)>0 for x 0 , so f is increasing on (
,0 ) and ( 0, ) . It
/ x
/ / has no local extreme values. f (x)=0 for x
0.417 , so f is CU on (
, 0.417) , CD on
( 0.417,0 ) , CU on ( 0,0.417) , and CD on ( 0.417, ) . f has IPs at ( 0.417,0.834 ) and
( 0.417, 0.834 ) .
22. y= f (x)= 1
tan x . From a CAS, 1+e 13 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators tan x e / y = 2 tan x 2 ( cos x 1+e
function with period ) and y / / tan x tan x e = e (2sin xcos x 1)+2sin xcos x+1
4 tan x 3 ( ) . f is a periodic cos x 1+e
that has positive values throughout its domain, which consists of all real
3
5
numbers except odd multiples of
(that is,
,
,
, and so on). f has y intercept
2
2
2
2
1
/
/ /
, but no x intercepts. We graph f , f , and f
on one period,
,
.
2
2 2 / Since f (x)<0 for all x in the domain of f , f is decreasing on the intervals between odd multiples of
2 .f / / (x)=0 for x=0+n and for x 1.124+n , so f is CD on ( 1.124,0 ) , CD on ( 0,1.124 ) , and CU on
on every interval of length 1.124, 2 2 , 1.124 , CU on . Since f is periodic, this behavior repeats . f has IPs at ( 1.124+n ,0.890 ) , n , 1
2 , and ( 1.124+n ,0.110 ) . 1/x 23. (a) f (x)=x b (b) Recall that a =e bln a 1/x . lim x =lim e
x + 0 x ( 1/x ) ln x + 0 indicates that there is a hole at ( 0,0 ) . As x . As x + 0 , ln x
x 1/x (1/x)ln x , so x =e , we have the indeterminate form 0 0 . This .
14 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators 1/x lim x =lim e
x ( 1/x ) ln x , but lim
x x ln x
1/x
1/x 0
=lim
=0 , so lim x =e =1 . This indicates that y=1 is a
x x
1
x HA.
(c) Estimated maximum: ( 2.72,1.45) . No estimated minimum. We use logarithmic differentiation to
1/x find any critical numbers. y=x
/ 1 ln x 1/x y =x 2 =0 ln x=1 1
ln y= ln x
x / y
1 1
=
+ ( ln x )
y x x 1
2 x / / 1/e x=e . For 0<x<e , y >0 and for x>e , y <0 , so f (e)=e is a x
local maximum value. This point is approximately ( 2.7183,1.4447) , which agrees with our estimate. (d) / / From the graph, we see that f (x)=0 at x 0.58 and x 4.37 . Since f
values, they are x coordinates of inflection points. / / changes sign at these sin x 24. (a) f (x)=(sin x)
is continuous where sin x>0 , that is, on intervals of the form ( 2n ,(2n+1) )
, so we have graphed f on ( 0, ) . sin x (b) y=(sin x) ln y=sin xln sin x , so
ln sin x
cot x
lim ln y=lim sin xln sin x=lim
=lim
=lim ( sin x)=0
+
+
+ csc x
+ csc xcot x
+ x 0 x 0 x 0 x 0 x 0 0 lim y=e =1 .
x + 0 (c) It appears that we have a local maximum at ( 1.57,1 ) and local minima at ( 0.38,0.69 ) and
sin x ( 2.76,0.69 ) . y=(sin x) ln y=sin xln sin x 15 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators / y
=(sin x)
y cos x
sin x cos x=0 or ln sin x= 1
x=
3 sin 1 ( x ,f ( x ) )
1 1 +(ln sin x)cos x=cos x(1+ln sin x)
x=
2 / sin x y =(sin x) 1 2 or sin x=e . On ( 0, ) , sin x=e 1 / (cos x)(1+ln sin x) . y =0 x =sin
1 1 ( e 1) and ( e 1) . Approximating these points gives us ( ( )) ( 0.3767,0.6922 ) , x2,f x2 ( ( )) ( 1.5708,1 ) , and x3,f x3 ( 2.7649,0.6922 ) . The approximations confirm our estimates. (d)
/ / From the graph, we see that f (x)=0 at x 0.94 and x 2.20 . Since f
values, they are x coordinates of inflection points. / / changes sign at these 25. From the graph of f (x)=sin (x+sin 3x) in the viewing rectangle 0, by 1.2,1.2 , it looks like f / has two maxima and two minima. If we calculate and graph f (x)= cos (x+sin 3x) (1+3cos 3x) on
0,2 ,
we see that the graph of f
/ / / appears to be almost tangent to the x axis at about x=0.7 . The graph of
2 f = sin (x+sin 3x) (1+3cos 3x) +cos (x+sin 3x)( 9sin 3x) is even more interesting near this x
value: it seems to just touch the x axis. 16 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators If we zoom in on this place on the graph of f / / , we see that f / / actually does cross the axis twice
/ near x=0.65 , indicating a change in concavity for a very short interval. If we look at the graph of f
on the same interval, we see that it changes sign three times near x=0.65 , indicating that what we had
thought was a broad extremum at about x=0.7 actually consists of three extrema (two maxima and a
minimum). These maximum values are roughly f (0.59)=1 and f (0.68)=1 , and the minimum value is
roughly f (0.64)=0.99996 . There are also a maximum value of about f (1.96)=1 and minimum values
of about f (1.46)=0.49 and f (2.73)= 0.51 . The points of inflection on ( 0, ) are about ( 0.61,0.99998 )
, ( 0.66,0.99998 ) , ( 1.17,0.72 ) , ( 1.75,0.77) , and ( 2.28,0.34 ) . On ( ,2 ) , they are about
( 4.01, 0.34 ) , ( 4.54, 0.77) , ( 5.11, 0.72 ) , ( 5.62, 0.99998 ) , and ( 5.67, 0.99998 ) . There are also
IP at ( 0,0 ) and ( ,0 ) . Note that the function is odd and periodic with period 2 , and it is also
rotationally symmetric about all points of the form ( (2n+1) ,0 ) , n an integer. ( 3 2 26. f (x)=x +cx=x x +c ) / 2 f (x)=3x +c c= 6 f / / (x)=6x c=0 c=6 x intercepts: When c 0 , 0 is the only x intercept. When c<0 , the x intercepts are 0 and
y intercept = f (0)=0 . f is odd, so the graph is symmetric with respect to the origin. f
/ / x<0 and f (x)>0 for x>0 , so f is CD on (
inflection point. ,0 ) and CU on ( 0, / / c . (x)<0 for ) . The origin is the only / If c>0 , then f (x)>0 for all x , so f is increasing and has no local maximum or minimum.
/ If c=0 , then f (x) 0 with equality at x=0 , so again f is increasing and has no local maximum or
minimum.
/ 2 If c<0 , then f (x)=3 x ( c / 3) =3 ( x+ ( c / 3, ) / ; f ( x ) <0 on ( c / 3, c/3 ) / c / 3 ) , so f ( x ) >0 on (
2
. It follows that f (
c / 3 )=
c
3 c / 3 ) (x , c / 3 ) and c / 3 is a local 2
c c / 3 is a local minimum value. As c decreases (toward more
3
negative values), the local maximum and minimum move further apart. There is no absolute
maximum or minimum value. The only transitional value of c corresponding to a change in character
of the graph is c=0 .
maximum value and f ( c / 3 )= 17 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators 4 2 2 27. f (x)=x +cx =x ( x2+c ) . Note that f is an even function. For c 0 , the only x intercept is the 1
/ /
2
c
f (x)=12x +2c . If c 0 , x=0 is the
2
only critical point and there is no inflection point. As we can see from the examples, there is no
change in the basic shape of the graph for c 0 ; it merely becomes steeper as c increases. For c=0 ,
the graph is the simple curve
/ 3 2 point ( 0,0 ) . We calculate f (x)=4x +2cx=4x x+ 4 c . Also, there is a maximum at ( 0,0 ) , and
y=x . For c<0 , there are x intercepts at 0 and at
1
1 2
there are minima at
c,
c
. As c
, the x coordinates of these minima get
2
4
larger in absolute value, and the minimum points move downward. There are inflection points at
1
5 2
c,
c
, which also move away from the origin as c
.
6
36 2 2 2 28. We need only consider the function f (x)=x c x for c 0 , because if c is replaced by c , the
function is unchanged. For c=0 , the graph consists of the single point ( 0,0 ) . The domain of f is
c,c , and the graph of f is symmetric about the y axis.
2 2 2
3
2 2
3
3x x
c
/
2
2 2
2 2
2x
x
2x c x x
3
f (x)=2x c x +x
=2x c x
=
=
. So we
2 2
2 2
2 2
2 2
c x
c x
c x
2 c x ( ) / see that all members of the family of curves have horizontal tangents at x=0 , since f (0)=0 for all
c>0 .
c , since
Also, the tangents to all the curves become very steep as x
18 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators / / x c + x / / f (
(x)= 2 9x +2c 2 ) . We set f (x)=0 x=0 or x c 2
c
3 values are f 2 2
c =0 , so the absolute maximum
3 2 / and lim f (x)= lim f (x)= =
2 2 c x 2 3 3 3 ( c .
3 2 3x +2c x
2 )( x/ 2 2 c x 2 ) = 6x 2 2 9c x +2c
2 3/2 (c x ) c x Using the quadratic formula, we find that f 4 / / (x)=0 2 x= 9c 2 2 4 . 2 c 33
. Since c<x<c , we take
12 33 2
9 33
( 9 33 ) ( 33 3) 3
c,
c
.
c , so the inflection points are
12
12
144
From these calculations we can see that the maxima and the points of inflection get both horizontally
and vertically further from the origin as c increases. Since all of the functions have two maxima and
two inflection points, we see that the basic shape of the curve does not change as c changes.
2 x= 9 29. c=0 is a transitional value
we get the graph of y=1 . For c>0 , we see that there is a HA at y=1 ,
and that the graph spreads out as c increases. At first glance there appears to be a minimum at ( 0,0 ) ,
but f ( 0 ) is undefined, so there is no minimum or maximum. For c<0 , we still have the HA at y=1 ,
19 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators 2 but the range is ( 1,
2 / f (x)=e c/x ( ) rather than ( 0,1 ) . We also have a VA at x=0 . f (x)=e 2c/x 3 ) f / / ( 2 2c 2c 3x (x)= ) / c/x / . f (x) 0 and f (x) exists for all x 0 (and 0 is not 2 6 c/x xe in the domain of f ), so there are no maxima or minima. f
inflection points spread out as c increases,
and if c<0 , there are no IP. For c>0, there are IP at
the IP is constant. ( 2 ( / / (x)=0 2c/3,e ) 3/2 x= 2c/3 , so if c>0 , the ) . Note that the y coordinate of 2 x > c , and
30. We see that if c 0 , f (x)=ln x +c is only defined for x > c
lim f (x)= lim f (x)=
, since ln y
as y 0 . Thus, for c<0 , there are vertical
x c + x c asymptotes at x= c , and as c decreases (that is, c increases), the asymptotes get further apart.
, so there is a vertical asymptote at x=0 . If c>0 , there are no asymptotes. To
For c=0 , lim f (x)=
x 0 find the extrema ( 2 and inflection points, we differentiate: f (x)=ln x +c ) / f (x)= 1
2 (2x) , so by the First Derivative x +c
Test there is a local and absolute minimum at x=0 . Differentiating again, we get
f / / (x)= 1
2 x +c (2)+2x ( x +c )
2 2 (2x) = ( 2 2 c x ( x +c )
2 )
2 . Now if c 0 , f / / is always negative, so f is concave down on both of the intervals on which it is defined. If c>0 , then f
2 c=x x= c . So for c>0 there are inflection points at x=
points get further apart. / / changes sign when c , and as c increases, the inflection 31. Note that c=0 is a transitional value at which the graph consists of the x axis. Also, we can see
cx
that if we substitute c for c , the function f (x)=
will be reflected in the x axis, so we
2 2
1+c x
investigate only positive
values of c (except c= 1 , as a demonstration of this reflective property). Also, f is an odd function.
20 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators lim f (x)=0 , so y=0 is a horizontal asymptote for all c . We calculate x f / ( 1+c2x2) c cx ( 2c2x) = c ( c2x2 1)
(x)=
( 1+c2x2) 2
( 1+c2x2) 2 / . f (x)=0 2 2 c x 1=0 x= 1/c . So there 1
1
and an absolute minimum value of f ( 1/c)=
. These
2
2
extrema have the same value regardless of c , but the maximum points move closer to the y axis as
c increases.
is an absolute maximum value of f (1/c)= f f / / / / ) ( 1+c2x2) 2 ( c3x2+c )
(x)
( 1+c2x2) 4
( 2c3x) ( 1+c2x2) + ( c3x2 c ) ( 4c2x) = 2c3x ( c2x2 3)
=
( 1+c2x2) 3
( 1+c2x2) 3
(
= (x)=0 3 2c x 3 /c , so there are inflection points at ( 0,0 ) and at x=0 or ( 3 /c, 3 /4 ) . Again, the y coordinate of the inflection points does not depend on c , but as c increases, both
inflection points approach the y axis.
1
is an even function, and also that lim f (x)=0 for any value of c ,
32. Note that f (x)=
2 2
2
x
1 x +cx
so y=0 is a horizontal asymptote. We calculate the derivatives:
1
2
2
4x x +
c 1
/
4 1 x x+2cx
2
f (x)=
=
, and ( ( ) ) ( 1 x2) 2+cx2 2 2 ( 1 x2) 2+cx2
6
4
2
2
/ /
10x +(9c 18)x + ( 3c 12c+6 ) x +2 c
f (x)=2
4 2 x +(c 2)x +1 3 . 21 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators ( / 2 2 ) 2 We first consider the case c>0 . Then the denominator of f is positive, that is, 1 x +cx >0 for all
1
x , so f has domain R and also f >0 . If c 1 0 ; that is, c 2 , then the only critical point is f (0)=1
2
, a maximum. Graphing a few examples for c 2 shows that there are two IP which approach the y
axis as c
. c=2 and c=0 are transitional values of c at which the shape of the curve changes. For
1
1
0<c<2 , there are three critical points: f (0)=1 , a minimum value, and f
,
1
c =
2
c(1 c/4)
both maximum values. As c decreases from 2 to 0 , the maximum values get larger and larger, and the
x values at which they occur go from 0 to 1 . Graphs show that there are four inflection points for
+ 0<c<2 , and that they get farther away from the origin, both vertically and horizontally, as c 0 . For
c=0 , the function is simply asymptotic to the x axis and to the lines x= 1 , approaching + from
both sides of each. The y intercept is 1 , and ( 0,1 ) is a local minimum. There are no inflection
points. Now if c<0 , we can write
1
1
1
f (x)=
=
= 2
. So f has vertical
2
2
2 2
2
2 2
x
cx 1 x + cx 1
1 x +cx
1 x
( c x) ( ) ( . .
( ) 2 )( / ) / asymptotes where x
c x 1=0 x= (
c
4 c ) 2 or x= ( c
4 c ) 2 . As c decreases, the
two exterior asymptotes move away from the origin, while the two interior ones move toward it. We
graph a few examples to see the behavior of the graph near the asymptotes, and the nature of the
1
critical points x=0 and x=
1
c :
2 c= 1 c= 1 c= 1 We see that there is one local minimum value, f (0)=1 , and there are two local maximum values,
22 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators 1
1
c =
as before. As c decreases, the x values at which these maxima occur
2
c(1 c/4)
get larger, and the maximum values themselves approach 0 , though they are always negative.
f 1 / / / 33. f (x)=cx+sin x f (x)=c+cos x f (x)= sin x
f ( x)= f (x) , so f is an odd function and its graph is symmetric with respect to the origin.
f (x)=0 sin x= cx , so 0 is always an x intercept.
/ f (x)=0 cos x= c , so there is no critical number when c >1 . If c 1 , then there are infinitely
many critical numbers. If x is the unique solution of cos x= c in the interval 0, , then the critical
1 numbers are 2n
, x= 2 x , where n ranges over the integers. (Special cases: When c=1 , x =0 ; when c=0
1 1 ; and when c= 1 , x = .)
1 / / f (x)<0 sin x>0 , so f is CD on intervals of the form ( 2n ,(2n+1) ) . f is CU on intervals of the
form ( (2n 1) ,2n ) . The inflection points of f are the points ( 2n ,2n c ) , where n is an integer.
/ If c 1 , then f (x) 0 for all x , so f is increasing and has no extremum. If c / 1 , then f (x) 0 / for all x , so f is decreasing and has no extremum. If c <1 , then f (x)>0 cos x> c x is in an
interval of the form 2n x ,2n +x for some integer n . These are the intervals on which f is ( 1 1 ) increasing. Similarly, we
find that f is decreasing on the intervals of the form 2n +x ,2(n+1) ( 1 x 1 ) . Thus, f has local
2 maxima at the points 2n +x , where f has the values c(2n +x )+sin x =c(2n +x )+ 1 c , and f
1 has local minima at the points 2n
f (2n x )=c(2n
1 x ) sin x =c(2n
1 1 1 1 1 x , where we have
1 x)
1 2 1 c .The transitional values of c are 1 and 1 . The inflection points move vertically, but not horizontally, when c changes. When c 1 , there is no
extremum. For c <1 , the maxima are spaced 2 apart horizontally, as are the minima. The
horizontal spacing between maxima and adjacent minima is regular (and equals ) when c=0 , but
the horizontal space between a local maximum and the nearest local minimum shrinks as c
approaches 1 . 23 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators ( at bt ) 34. For f (t)=C e e
, C affects only vertical stretching, so we let C=1 . From the first figure,
we notice that the graphs all pass through the origin, approach the t axis as t increases, and approach
as t
. Next we let a=2 and produce the second figure. Here, as b increases, the slope of the tangent at the origin increases and the local maximum value
increases. f (t)=e
/ bt f (t)=0 2t be =2e e bt 2t / f (t)=be
b (b
=e
2 2)t bt 2t / 2e . f (0)=b 2 , which increases as b increases.
b
ln b ln 2
ln =(b 2)t t=t =
, which decreases as b increases
1
2
b 2
2/(b 2) (the maximum is getting closer to the y axis). f (t )= (b 2)2 1 1+2/(b 2) . We can show that this value b
increases as b increases by considering it to be a function of b and graphing its derivative with respect
to b , which is always positive.
x 35. If c<0 , then lim f (x)= lim
x x If c>0 , then lim f (x)= cx e 1 = lim cx x x If c=0 , then f (x)=x , so lim f (x)= x . x ce , and lim f (x)=lim x =0 , and lim f (x)=
1
cx =0 . ce
respectively. x So we see that c=0 is a transitional value. We now exclude the case c=0 , since we know how the
function behaves in that case. To find the maxima and minima of f , we differentiate: f (x)=xe
/ f (x)=x ( ce cx ) +e cx= ( 1 cx ) e cx . This is 0 when 1 cx=0 cx x=1/c . If c<0 then this / represents a minimum value of f ( 1/c ) =1/ ( ce ) , since f (x) changes from negative to positive at
x=1/c ;
24 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators and if c>0 , it represents a maximum value. As c increases, the maximum or minimum point gets
/ cx closer to the origin. To find the inflection points, we differentiate again: f (x)=e (1 cx)
/ / cx ( cx ) cx f (x)=e ( c)+(1 cx) ce
=(cx 2)ce . This changes sign when cx 2=0
increases, the points of inflection get closer to the origin. x=2/c . So as c 36. For c=0 , there is no inflection point; the curve is CU everywhere. If c increases, the curve simply
becomes steeper, and there are still no inflection points. If c starts at 0 and decreases, a slight upward
bulge appears near x=0 , so that there are two inflection points for any c<0 . This can be seen
4 2 algebraically by calculating the second derivative: f (x)=x +cx +x
f / / 2 / 3 f (x)=4x +2cx+1 / / (x)=12x +2c . Thus, f (x)>0 when c>0 . For c<0 , there are inflection points when
1
x=
c . For c=0 , the graph has one critical number, at the absolute minimum somewhere
6
around x= 0.6 . As c increases, the number of critical points does not change. If c instead decreases
from 0 , we see that the graph eventually sprouts another local minimum, to the right of the origin,
somewhere between x=1 and x=2 . Consequently, there is also a maximum near x=0 .
After a bit of experimentation, we find that at c= 1.5 , there appear to be two critical numbers: the
absolute minimum at about x= 1 , and a horizontal tangent with no extremum at about x=0.5 . For
any c smaller than this there will be 3 critical points, as shown in the graphs with c= 3 and with c= 5
/ 3 . To prove this algebraically, we calculate f (x)=4x +2cx+1 . Now if we substitute our value of
1
/
3
2
c= 1.5 , the formula for f (x) becomes 4x 3x+1=(x+1)(2x 1) . This has a double root at x= ,
2
1
indicating that the function has two critical points: x= 1 and x= , just as we had guessed from the
2
graph. 25 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators 4 2 2 37. (a) f (x)=cx 2x +1 . For c=0 , f (x)= 2x +1 , a parabola whose vertex, ( 0,1 ) , is the absolute
4 2 maximum. For c>0 , f (x)=cx 2x +1 opens upward with two minimum points. As c 0 , the
minimum points spread apart and move downward; they are below the x axis for 0<c<1 and above
for c>1 . For c<0 , the graph opens downward, and has an absolute maximum at x=0 and no local
minimum.
/ ( 3 ) 2 (b) f (x)=4cx 4x=4cx x 1/c ( c 0 ). If c 0 , 0 is the only critical number. f / / / / 2 (x)=12cx 4 , 2 so f (0)= 4 and there is a local maximum at ( 0,f(0) ) = ( 0,1 ) , which lies on y=1 x . If c>0 , the
critical numbers are 0 and 1/ c . As before, there is a local maximum at ( 0,f(0) ) = ( 0,1 ) , which lies
2 / / on y=1 x . f ( 1/ c ) =12 4=8>0 , so there is a local minimum at x= 1/ c . Here ( 2) f ( 1/ c ) =c 1/c 3 2/c+1= 1/c+1 . But 2 38. (a) f (x)=2x +cx +2x
critical points 2 c 12 0 / 2 ( 2 1/ c , 1/c+1 ) lies on y=1 x since 1 ( ) 2 / f (x)=6x +2cx+2=2 3x +cx+1 . f (x)=0
c / 2 3 . For c= 2 3 , f (x) 0 on (
2 change signs at c/6 , and there is no extremum. If c 12>0 , then f
negative at x= c 2 / x=
, c 2 ( 1/ c ) =1 1/c . 2 c 12
. So f has
6
/ ) , so f does not changes from positive to
2 c 12
c+ c 12
and from negative to positive at x=
. So f has a local
6
6 maximum at
26 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.6 Graphing with Calculus and Calculators x= c 2 2 c 12
c+ c 12
and a local minimum at x=
.
6
6
2 / 2 (b) Let x be a critical number for f (x) . Then f (x )=0 3x +cx +1=0 0 0 1 3x
0 0 c= x 0 . Now 0 2 1 3x f (x ) = 2x3+cx2+2x =2x3+x2
0
0 0 ( 0 ) 0 0 So the point is x ,y = x ,x x
0 0
3 0 0 x
3
0 0 3 3 3 +2x = 2x0 x0 3x0+2x0=x0 x0
0 0 ; that is, the point lies on the curve y=x x . 27 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 1. (a)
First Number
1
2
3
4
5
6
7
8
9
10
11 Second Number
22
21
20
19
18
17
16
15
14
13
12 Product
22
42
60
76
90
102
112
120
126
130
132 We needn’t consider pairs where the first number is larger than the second, since we can just
interchange the numbers in such cases. The answer appears to be 11 and 12 , but we have considered
only integers in the table.
(b) Call the two numbers x and y . Then x+y=23 , so y=23 x . Call the product P . Then
2 2 / P=xy=x(23 x)=23x x , so we wish to maximize the function P(x)=23x x . Since P (x)=23 2x , we
23
/
2
see that P (x)=0 x=
=11.5 . Thus, the maximum value of P is P(11.5)= ( 11.5) =132.25 and it
2
occurs when x=y=11.5 .
/ / Or: Note that P (x)= 2<0 for all x , so P is everywhere concave downward and the local maximum
at x=11.5 must be an absolute maximum.
2 / 2. The two numbers are x+100 and x . Minimize f (x)=(x+100)x=x +100x . f (x)=2x+100=0
. Since f / / x= 50 (x)=2>0 , there is an absolute minimum at x= 50 . The two numbers are 50 and 50 .
2 100
100
/
100 x 100
, where x>0 . Minimize f (x)=x+
. f (x)=1
=
.
3. The two numbers are x and
2
2
x
x
x
x
/ / The critical number is x=10 . Since f (x)<0 for 0<x<10 and f (x)>0 for x>10 , there is an absolute
minimum at x=10 . The numbers are 10 and 10 .
4. Let x>0 and let f (x)=x+1/x . We wish to minimize f (x) . Now
/
2
1 1
1
x 1 = (x+1)(x 1) , so the only critical number in ( 0,
f (x)=1
=
2
2
2
x
x
x ( ) ) is 1 .
1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems / / f (x)<0 for 0<x<1 and f (x)>0 for x>1 , so f has an absolute minimum at x=1 , and f (1)=2 .
/ / 3 Or: f (x)=2/x >0 for all x>0 , so f is concave upward everywhere and the critical point ( 1,2 ) must
correspond to a local minimum for f .
5. If the rectangle has dimensions x and y , then its perimeter is 2x+2y=100 m, so y=50 x . Thus, the
2 area is A=xy=x(50 x) . We wish to maximize the function A(x)=x(50 x)=50x x , where 0<x<50 .
/ / / Since A (x)=50 2x= 2(x 25) , A (x)>0 for 0<x<25 and A (x)<0 for 25<x<50 . Thus, A has an
2 2 absolute maximum at x=25 , and A(25)=25 =625 m . The dimensions of the rectangle that maximize
its area are x=y=25 m. (The rectangle is a square.)
2 6. If the rectangle has dimensions x and y , then its area is xy=1000 m , so y=1000/x . The perimeter
P=2x+2y=2x+2000/x . We wish to minimize the function P(x)=2x+2000/x for x>0 .
/ 2 ( 2) ( x2 1000) , so the only critical number in the domain of P is x= P (x)=2 2000/x = 2/x
/ / 1000 . 3 P (x)=4000/x >0 , so P is concave upward throughout its domain and P ( 1000 ) =4 1000 is an
absolute minimum value. The dimensions of the rectangle with minimal perimeter are
x=y= 1000 =10 10 m.
(The rectangle is a square.) 7. (a) 2 The areas of the three figures are 12 , 500 , 12 , 500 , and 9000 ft . There appears to be a maximum
2 area of at least 12 , 500 ft .
(b) Let x denote the length of each of two sides and three dividers.
Let y denote the length of the other two sides. 2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems (c) Area A= length width =y x
(d) Length of fencing =750 5x+2y=750
5
5
(e) 5x+2y=750 y=375
x A(x)= 375
x
2
2
/ 5 2
x
2 x=375x / / (f) A (x)=375 5x=0 x=75 . Since A (x)= 5<0 there is an absolute maximum when x=75 . Then
375
375
2
y=
=187.5 . The largest area is 75
=14 , 062.5 ft . These values of x and y are between
2
2
the values in the first and second figures in part (a). Our original estimate was low. 8. (a) 3 The volumes of the resulting boxes are 1 , 1.6875 , and 2 ft . There appears to be a maximum
3 volume of at least 2 ft .
(b) Let x denote the length of the side of the square being cut out. Let y denote the length of the
base. (c) Volume V = length width height 2 V =y y x=xy 3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems (d) Length of cardboard =3 x+y+x=3
2
(e)
y+2x=3 y=3 2x V ( x ) =x(3 2x)
2
(f)
V (x)=x(3 2x)
/ y+2x=3 2 V (x)=x 2(3 2x)( 2)+(3 2x) 1=(3 2x)[ 4x+(3 2x)]=(3 2x)( 6x+3) ,
3
1
3
3
and V (0)=V
=0 , so the
so the critical numbers are x= and x= . Now 0 x
2
2
2
2
1
1
3
2
maximum is V
=
( 2 ) =2 ft , which is the value found from our third figure in
2
2
part (a).
9. 6 6 xy=1.5 10 , so y=1.5 10 /x . Minimize the amount of fencing, which is ( ) 6 6 / 6 2 ( 2 6 ) 2 3x+2y=3x+2 1.5 10 /x =3x+3 10 /x=F(x) . F (x)=3 3 10 /x =3 x 10 /x . The critical
3 / 3 / 3 number is x=10 and F (x)<0 for 0<x<10 and F (x)>0 if x>10 , so the absolute minimum occurs
3 3 when x=10 and y=1.5 10 . The field should be 1000 feet by 1500 feet with the middle fence
parallel to the short side of the field.
2 10. Let b be the length of the base of the box and h the height. The volume is 32 , 000=b h
2 2 2 2 h=32 , 2 000/b . The surface area of the open box is S=b +4hb=b +4(32 , 000/b )b=b +4 ( 32,000 ) /b . So
/ ( 2 ) 3 2 3 S (b)=2b 4(32 , 000)/b =2 b 64,000 /b =0
/ b= 64,000 =40 . This gives an absolute minimum / since S (b)<0 if 0<b<40 and S (b)>0 if b>40 . The box should be 40 40 20 .
2 11. Let b be the length of the base of the box and h the height. The surface area is 1200=b +4hb
3 2
2
2
2
2
3
/
h= 1200 b /(4b) . The volume is V =b h=b 1200 b /4b=300b b /4 V (b)=300
b .
4
3 2
/
2
/
/
V (b)=0 300= b
b =400 b= 400 =20 . Since V (b)>0 for 0<b<20 and V (b)<0 for b>20 ,
4
there is an absolute maximum when b=20 by the First Derivative Test for Absolute Extreme Values ( ) ( ( 2 (see page 280 ). If b=20 , then h= 1200 20
2 2 ) ) / ( 4 20 ) =10 , so the largest possible volume is 3 b h=(20) (10)=4000 cm .
4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 12. 2 ( 2) 2 . 2 10=(2w)(w)h=2w h , so h=5/w . The cost is 10 2w +6 2(2wh)+2(hw) =20w +36wh , so
9
/
2
3 9
2
2
2
2
3
w
w
w=
is
C(w)=20w +36w 5/w =20w +180/w . C (w)=40w 180/w =40
2
2 V =lwh ( ) the critical number. There is an absolute minimum for C when w=
0<w< 3 9
/
3
and C (w)>0 for w>
2 9
.C
2 . 13. 2 9
2 3 2 3 =20 / 9
/
since C (w)<0 for
2 3 2 9
2 + 180
3 $163.54 . 9/2 ( 2)
(
)/ ( 2) 10=(2w)(w)h=2w h , so h=5/w . The cost is C(w)=10 2w +6 2(2wh)+2hw +6 2w =
45
2
2
/
2
3
2
3
32w +36wh=32w +180/wC (w)=64w 180/w =4 16w 45 w w=
is the critical
16
/ number. C (w)<0 for 0<w<
C 3 45
16 3 2/3 45
/
3
and C (w)>0 for w>
16 =32(2.8125) +180 / 2.8125 45
. The minimum cost is
16 $191.28 . 14. (a) Let the rectangle have sides x and y and area A , so A=xy or y=A/x . The problem is to
/ 2 ( 2 ) 2 minimize the perimeter =2x+2y=2x+2A/x=P(x) . Now P (x)=2 2A/x =2 x A /x . So the critical
/ / number is x= A . Since P (x)<0 for 0<x< A and P (x)>0 for x> A , there is an absolute minimum
at x= A . The sides of the rectangle are A and A/ A= A , so the rectangle is a square.
1
(b) Let p be the perimeter and x and y the lengths of the sides, so p=2x+2y 2y= p 2x y= p x .
2
1
1
1
1
1
2
/
The area is A(x)=x
p x = px x . Now A (x)=0
p 2x=0 2x= p x= p . Since
2
2
2
2
4
1
/ /
A (x)= 2<0 , there is an absolute maximum for A when x= p by the Second Derivative Test. The
4
sides of the rectangle are
5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 1
1
1
1
p and p
p= p , so the rectangle is a square.
4
2
4
4
15. The distance from a point ( x,y ) on the line y=4x+7 to the origin is
However, it is easier to work with the square of the distance; that is,
D(x)= ( 2 x +y ) =x +y =x + ( 4x+7)
2 2 2 2 2 2 2 2 2 (x 0) +(y 0) = x +y . 2 . Because the distance is positive, its minimum value will
occur at the same point as the minimum value of D .
28
/
/
.
D (x)=2x+2(4x+7)(4)=34x+56 , so D (x)=0 x=
17
28
/ /
. The
D (x)=34>0 , so D is concave upward for all x . Thus, D has an absolute minimum at x=
17
28
28
28 7
.
point closest to the origin is ( x,y ) =
,4
+7 =
,
17
17
17 17
16. The square of the distance from a point ( x,y ) on the line y= 6x+9 to the point ( 3,1 ) is
45
2
2
2
2
2
/
/
.
D(x)=(x+3) +(y 1) =(x+3) +( 6x+8) =37x 90x+73 . D (x)=74x 90 , so D (x)=0 x=
37
45
/ /
. The
D (x)=74>0 , so D is concave upward for all x . Thus, D has an absolute minimum at x=
37
45 63
point on the line closest to ( 3,1 ) is
,
.
37 37
17. From the figure, we see that there are two points that are farthest away from A(1,0) . The distance d
2 2 from A to an arbitrary point P(x,y) on the ellipse is d= (x 1) +(y 0) and the square of the distance
1
2 2
2 2
2
2
/
/
is S=d =x 2x+1+y =x 2x+1+ 4 4x = 3x 2x+5 . S = 6x 2 and S =0 x=
. Now
3
1
1
16
/ /
S = 6<0 , so we know that S has a maximum at x=
. Since 1 x 1 , S( 1)=4 , S
=
3
3
3
16
, and S(1)=0 , we see that the maximum distance is
. The corresponding y values are
3 ( ) 6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems y= 1
3 4 4 2 32
=
9 = 4
2
3 1
,
3 1.89 . The points are 4
2
3 . 18. 2 2 The distance d from ( 1,1 ) to an arbitrary point P(x,y) on the curve y=tan x is d= (x 1) +(y 1) and
2 2 2 / 2 / the square of the distance is S=d =(x 1) +(tan x 1) . S =2(x 1)+2(tan x 1)sec x . Graphing S on
,
gives us a zero at x 0.82 , and so tan x 1.08 . The point on y=tan x that is closest to
2 2
( 1,1 ) is approximately ( 0.82,1.08 ) .
19. 2 2 2 2 2 The area of the rectangle is (2x)(2y)=4xy . Also r =x +y so y= r x , so the area is
2 2 / A(x)=4x r x . Now A (x)=4 2 2 2 x 2 r x 2 =4
2 2 r x
x= 1
r . Clearly this gives a maximum. y=
2 r 2 2 r 2x . The critical number is
2 r x
1
r
2 2 = 1 2 1
r =
r=x , which tells us
2
2 that the rectangle is a square. The dimensions are 2x= 2 r and 2y= 2 r .
20. The area of the rectangle is (2x)(2y)=4xy . Now
7 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 2 x 2 2 + y 2 =1 gives y= a
b
4b
1 2 2
x
a x
a
2 ( b
a 2 2 a x , so we maximize A(x)=4 ) 1/2 ( 2x)+ ( a2 x2) 1/2 1 = b
/
2 2
x a x A (x)=
a 4b 2 2
a x
a ( ) 1/2 2 2 2 x +a x = 4b
2 2 2 a 2x So 2 a a x
1
1
the critical number is x=
a , and this clearly gives a maximum. Then y=
b , so the maximum
2
2
1
1
area is 4
a
b =2ab .
2
2
21. 3
2
2 2
L , since h +(L/2) =L
2
3
3
L y
L
3
2
2
3 x=
=
= 3
L y
x
L/2
2 The height h of the equilateral triangle with sides of length L is 2 2 h =L
y= 1 2 3 2
L= L
4
4 3
L
2 3x y= h= 3
L . Using similar triangles,
2 3
(L 2x) . The area of the inscribed rectangle is
2
2 / A(x)=(2x)y= 3 x(L 2x)= 3 Lx 2 3 x , where 0 x L/2 . Now 0=A (x)= 3 L 4 3 x
x= 3 L ( 4 3 ) =L/4 . Since A(0)=A(L/2)=0 , the maximum occurs when x=L/4 , and
3
3
3
3
y=
L
L=
L , so the dimensions are L/2 and
L.
2
4
4
4 / 22. ( 2 ) 3 / 2 The rectangle has area A(x)=2xy=2x 8 x =16x 2x , where 0 x 2 2 . Now A (x)=16 6x =0
8 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems x=2 2
. Since A ( 0 ) =A ( 2 2 ) =0 , there is a maximum when x=2
3 2
16
. Then y=
, so the
3
3 2
16
and
.
3
3 rectangle has dimensions 4
23. The area of the triangle is A(x)=
/ A (x)=r 2x
2 2 2 + r x +x
2 2 r x
2 2 0=2x +rx r =(2x r)(x+r)
x= 1
2 2
(2t)(r+x)=t(r+x)= r x (r+x) . 0=
2
2 2x
2 =
2 x +rx
2 2 2 x +rx 2 + r x 2 2 r x
r x
1
x= r or x= r . Now A(r)=0=A( r)
2 1
1
3
r , so the triangle has height r+ r= r and base 2
2
2
2 r 2 1
r
2 2 2 2 = r x 2 2 2 x +rx=r x 2 r x the maximum occurs where
2 =2 3 2
r = 3r .
4 24. The rectangle has area xy . By similar triangles 3 y 3
=
x
4 3
3 2
3
/
x+3 =
x +3x where 0 x 4 . Now 0=A (x)=
x+3
4
4
2
3
2
A(0)=A(4)=0 , the maximum area is A(2)=2
=3 cm .
2
A(x)=x 3
x+3 . So the area is
4
3
x=2 and y= . Since
2 4y+12=3x or y= 25. 9 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 2 2 2 The cylinder has volume V = y (2x) . Also x +y =r ( 2 2) (2 3 2 ) 2 2 2 y =r x , so (r 2 3x2) =0 / V (x)= r x (2x)=2 r x x , where 0 x r . V (x)=2
V (0)=V (r)=0 , so there is a maximum when x=r / 3 and (r 2 r 2/3) ( 2r / V (r / 3 ) = 3 ) =4 r 3 / (3 3 ) x=r / 3 . Now . 26. By similar triangles, y/x=h/r , so y=hx/r . The volume of the cylinder is
2 2 3 / 2 / x ( h y ) = hx ( h/r ) x =V (x) . Now V (x)=2 hx ( 3 h/r ) x = hx ( 2 3x/r ) . So V (x)=0
2
2
x= r . The maximum clearly occurs when x= r and then the volume is
3
3
2
2
2
4
2
3
2
2
hx ( h/r ) x = hx ( 1 x/r ) =
r h 1
=
r h.
3
3
27 x=0 or 27. 2 The cylinder has surface area 2( area of the base )+( lateral surface area )=2 ( radius ) +2 ( radius )(
2 2 2 height )=2 y +2 y(2x) . Now x +y =r
2 (r 2 x2) +4 =0 4 x+4 2 2 2 2 2 2 y =r x 2 2 ( ( 2 2 2 ) ) Thus, S (x) ) x r x , 0 x r =2 r 2 x +4 x r x
1 2 2 1/2
2 2 1/2
x
r x
1
( 2x ) + r x
2 ( 2 y= r x , so the surface area is S(x)=
/ 10 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 2 =4 x x 2 2 2 + r x 2 2 =4 2 / 2 2 2 4 2 2 r x 2 2 (x 2 x r x =r 2x ( * ) 4 2 x +r x 2 r x
S (x)=0 2 x r x 2 2 4 4 2 2 2 2 ) = (r 2 r x 2 2 2 2x ) 2 ( 2 2) 4 2 2 x r x =r 4r x +4x 4 4 2 r x x =r 4r x +4x
5x 5r x +r =0 . This is a quadratic equation in x . By the quadratic
5 2
2 5
r , but we reject the root with the + sign since it doesn’t satisfy ( * ). So
formula, x =
10
5 x=
2 x= 5
10 5
10
5 2
r r . Since S(0)=S(r)=0 , the maximum surface area occurs at the critical number and
2 y =r 2 5 5 5+ 5
2
r +4
10
5+ 5 2 20
2
= r
+
5
5 5 2 5+ 5 2
r
10 5 10 2 ( 5 5 ) ( 5+ 5 )
5+ 5 2
5+ 5
2
r = r
2
+4
10
10
10
5+ 5+2 2 5
5+5 5
2
2
= r
= r ( 1+ 5 ) .
5
5 r = 10
= r 2 the surface area is 28. Perimeter=30 x
2 2y+x+ =30 y= 1
2 30 x rectangle plus the area of the semicircle, or xy+
A(x)=x
A / / 15 (x)= 30
y=15
4+ x
2
1+ x
4 + 1
2 x
2
x
2 =15 x
2 x
. The area is the area of the
4 2 , so 1 2
1 2
2
/
x =15x
x
x . A (x)=15
8
2
8 1+ 4 x=0 <0 , so this gives a maximum. The dimensions are x= x= 15
60
=
.
1+ /4 4+ 60
ft and
4+ 4
15
60+15 30 15
30
=
=
ft, so the height of the rectangle is half the base.
4+
4+
4+ 29. 11 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems xy=384
/ y=384/x . Total area is A(x)=(8+x)(12+384/x)=12(40+x+256/x) , so ( 2 ) A (x)=12 1 256/x =0 / x=16 . There is an absolute minimum when x=16 since A (x)<0 for / 0<x<16 and A (x)>0 for x>16 . When x=16 , y=384/16=24 , so the dimensions are 24 cm and 36 cm.
30. xy=180 , so y=180/x . The printed area is (x 2)(y 3)=(x 2)(180/x 3)=186 3x 360/x=A(x) .
/ 2 2 A (x)= 3+360/x =0 when x =120 / x=2 30 . This gives an absolute maximum since A (x)>0 for / 0<x<2 30 and A (x)<0 for x>2 30 . When x=2 30 , y=180/(2 30 ) , so the dimensions are
2 30 in. and 90/ 30 in.
31. Let x be the length of the wire used for the square. The total area is
3
x 2 1
10 x
10 x
1 2 3
2
A(x)=
+
=
x+
(10 x) , 0 x 10
4
2
3
2
3
16
36
3
4 3
40 3
40 3
3
1
9
/
A (x)= x
(10 x)=0
x+
x
=0 x=
. Now A(0)=
8
18
72
72
72
36
9+4 3
40 3
100
, A(10)=
=6.25 and A
2.72 , so
16
9+4 3
(a) The maximum area occurs when x=10 m, and all the wire is used for the square.
40 3
(b) The minimum area occurs when x=
4.35 m.
9+4 3 100 4.81 32.
12 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 2 2 x 2
10 x 2 x
(10 x)
+
=
+
Total area is A(x)=
, 0 x 10 .
4
2
16
4
x 10 x
1 1
5
/
A (x)=
=
+
x
=0 x=40/(4+ ) . A(0)=25/
7.96 , A(10)=6.25 , and
8 2
2
8
A(40/(4+ )) 3.5 , so the maximum occurs when x=0 m and the minimum occurs when x=40/(4+ )
m.
33. 2 2 2 The volume is V = r h and the surface area is S(r)= r +2 rh= r +2 r V
r / S (r)=2 r 2V
r 2 3 =0 r= 2 r =2V V 3 / 3 V , h= V
r 2 = V
2/3 = 3 V 2 2V
.
r cm. This gives an absolute minimum since S (r)<0 for 0<r<
r= 2 = r + 3 V / and S (r)>0 for r> 3 V . When cm. (V / ) 34. 13 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems L=8 +4sec , 0< < 3 tan = 2 2 13 =tan when =tan tan =0 when sec tan =2 cot tan 3 =2 2 . dL/d <0 when 0< <tan
13 dL
= 8 cot +4sec
d , 13 2 , dL/d >0 when tan 13 2< < , so L has an absolute minimum 2 2/3 2 , and the shortest ladder has length L=8 1+2 1/3 +4 2/3 1+2 16.65 ft. 2
x
8
Another method: Minimize L =x + ( 4+y ) , where
= .
4+y y
2 2 2 35. 2 2 2 h +r =R V= ( R2 h2) h= 3 ( R2h h3) . V / ( h) = 3 ( R2 3h2) =0 when h= 2 3 r h= 3 / 1
R . This
3 1
1
/
R and V (h)<0 for h>
R . The
3
3
1
2
3
3
R =
R .
3 3
9 3 gives an absolute maximum, since V (h)>0 for 0<h<
1
R =
3
3 maximum volume is V 1 3
R
3 36. The volume and surface area of a cone with radius r and height h are given by V =
2 2 2 S= r r +h . We’ll minimize A=S subject to V =27 . V =27
2 2 2 2 A= r (r +h )=
3 h= r= 162 h= 3 3
6 6
r and h . 3 2 2 81
h 162 2.632 . A =3
/ / 3 1 2
r h=27
3 2 r = 81
(1) .
h 2 81 2
/
81
2 81
+h =
+81 h , so A =0
+81 =0
2
3
h
h
h
6
81
2 81
27
3.722 . From (1) , r =
=
=
3
3
h
2
3 6/
6
2 1 2
r h and
3 2 81 = 2 81
h 3 4 =6 81 /h >0 , so A and hence S has an absolute minimum at these values of 2 37.
14 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems H H h
1 2
=
(1) . The volume of the inner cone is V =
r h , so we’ll solve (1)
R
r
3
Hr
Hr HR Hr H
for h .
=H h h=H
=
= (R r) (2) .
R
R
R
R
H
H
H
2 H
2 3
/
2
Thus, V (r)= r
(R r)=
(Rr r ) V (r)=
(2Rr 3r )=
r(2R 3r) .
3
R
3R
3R
3R
2
H
2
H
1
1
/
V (r)=0 r=0 or 2R=3r r= R and from (2) , h=
R
R =
R = H.
3
R
3
R
3
3
2
/
V (r) changes from positive to negative at r= R , so the inner cone has a maximum volume of
3
2
1 2 1
2
1
4 1
2
V=
r h=
R
H =
R H , which is approximately 15% of the volume of
3
3
3
3
27 3
the larger cone.
By similar triangles, 3 aLv
38. (a) E(v)=
v u / E (v)=aL 2 (v u)3v v
2 3 3 2 =0 when 2v =3uv 2v=3u v= 3
u.
2 (v u)
The First Derivative Test shows that this value of v gives the minimum value of E . (b)
3 2
2 3
s cot +3s
2
2
dS 3 2 2
2 3
(a)
= s
3s
cot
d
2
2
dS
(b)
=0 when
3 cot =0
d
39. S=6sh 3 2
s
3 cot ) .
(
2
1
cos
1
3
=0 cos =
. The First Derivative Test
sin
sin
3
1
1
shows that the minimum surface area occurs when =cos
55 .
3
(c)
or 15 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 3
1
1
, then cot =
and =
, so the surface area is S < =
2
3
2
3
3 2 1
3 2 9 2
6 2
1
2 3
6sh
s
+3s
=6sh
s+
s =6sh+
s =6s h+
s
2
2
2
2
2 2
2 2
2 2
2 2
If cos = 40. Let t be the time, in hours, after 2:00 P.M. The position of the boat heading south at time t is ( 0, 20t )
. The position of the boat heading east at time t is ( 15+15t,0 ) . If D(t) is the distance between the
2 22 2 2 / boats at time t , we minimize f (t)=[D(t)] =20 t +15 (t 1) . f (t)=800t+450(t 1)=1250t 450=0 when
450
60min
/ /
t=
=0.36 h. 0.36 h
=21.6 min =21 min 36 s. Since f (t)>0 , this gives a minimum, so
1250
h
the boats are closest together at 2:21:36 P.M.
2 41. Here T (x)= x +25 5 x
+
,0 x 5
6
8 / T (x)= x
2 6 x +25 1
=0
8 8x=6 2 x +25 15
15
. But
>5 , so T has no critical number. Since T (0) 1.46 and
7
7
T (5) 1.18 , he should row directly to B .
2 ( 2 16x =9 x +25 ) x= 42. In isosceles triangle AOB , O=180
, so BOC=2 . The distance rowed is 4cos
distance walked is the length of arc BC=2(2 )=4 . The time taken is given by while the 16 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems T ( )= 4cos
2 + 4
=2cos + , 0
4 Check the value of T at =
T (0)=2 , T
when = 2
0 < = 3+ 6 2 2 6 1
2 = 6 . and at the endpoints of the domain of T ; that is, =0 and 6 2.26 , and T 6 sin = 2 = 2 = 2 . 1.57 . Therefore, the minimum value of T is ; that is, the woman should walk all the way. Note that T , so = 2 / . T ( )= 2sin +1=0 / / ( )= 2cos <0 for gives a maximum time. 43. The total illumination is I(x)= 3k
2 + x
3 6k(10 x) =2kx
3 ( 3 3 3 3(10 x) =x ) 10 3 = 1+ 3 x 3 k
2 (10 x)
3 6k / , 0<x<10 . Then I (x)= 3(10 x)=x x
3 10 3 3 3 x=x 3 x= 3 2k
3 =0 (10 x)
3 10 3 =x+ 3 x 3 10 3 3 + 5.9 ft. This gives a minimum since I / / (x)>0 for 0<x<10 . 1+ 3
44. The line with slope m (where m<0 ) through ( 3,5) has equation y 5=m(x 3) or y=mx+ ( 5 3m ) . The
y intercept is 5 3m and the x intercept is 5/m+3 . So the triangle has area
1
9
5
/
2 25
25 9
A(m)= (5 3m)( 5/m+3)=15 25/(2m)
m . Now A (m)=
=0 m =
m=
(since
2 2
2
2
9
3
2m
5
/ /
25
m<0 ). A (m)=
>0 , so there is an absolute minimum when m=
. Thus, an equation of the
3
3
m
5
5
line is y 5=
(x 3) or y=
x+10 .
3
3
45.
17 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems Every line segment in the first quadrant passing through ( a,b) with endpoints on the x and y axes
satisfies an equation of the form y b=m(x a) , where m<0 . By setting x=0 and then y=0 , we find its
b
,0 . The distance d from A to B is given by
endpoints, A(0,b am) and B a
m
2
b
2
d=
a
0 + 0 (b am) .
m
It follows that the square of the length of the line segment, as a function of m , is given by
2 b
S(m)= a
m
/ S (m) =
= 2 2b 2 m m 3 m 3 / b
a = a+b
2 3 3 ( (am b) b+am 3 ) b
. Since b/a>0 and m<0 , m must equal
a 3 m=b/a or m= <0 , we see that S (m)<0 for m< 3 2
m minimum value when m=
S ( abm b2+a2m4 abm3)
3 2 3 / m 3 2 +2a m 2ab= b(am b)+am (am b) = Thus, S (m)=0
2 2 2ab b
2 2
2
+
+a m 2abm+b . Thus,
2
m
m 2 +(am b) =a 2ab
m
2 2 b
/
and S (m)>0 for m>
a 3 b
. Since
a 3 3 b
. Thus, S has its absolute
a 2 3 b
. That value is
a 3 2 a
b + 4/3 2/3 a 2/3 4/3 2 b
a 3 = ( a+ b 4/3 2/3 3 2/3 4/3 2 3 3 ab ) + (
2 2 4/3 2/3 a b +b )
2 2 2/3 4/3 2 =a +2a b +a b +a b +2a b +b =a +3a b +3a b +b
3 2 2 The last expression is of the form x +3x y+3xy +y =(x+y)
2/3 2/3 with x=a 2/3 and y=b 2/3 3 , 2/3 2/3 3/2 so we can write it as (a +b ) and the shortest such line segment has length S =(a +b )
3 46. y=1+40x 3x
2 5 4 / 2 . 4 y =120x 15x , so the tangent line to the curve at x=a has slope
/ 3 2 / m(a)=120a 15a . Now m (a)=240a 60a = 60a(a 4)= 60a(a+2)(a 2) , so m (a)>0 for a< 2 ,
18 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems / and 0<a<2 , and m (a)<0 for 2<a<0 and a>2 . Thus, m is increasing on (
, 2 ) , decreasing on
2,0 ) , increasing on ( 0,2 ) , and decreasing on ( 2, ) . Clearly, m(a)
as a
, so the
(
maximum value of m(a) must be one of the two local maxima, m( 2) or m(2) . But both m( 2) and
2 4 m(2) equal 120 2 15 2 =480 240=240 . So 240 is the largest slope, and it occurs at the points
( 2, 223) and ( 2,225) . Note: a=0 corresponds to a local minimum of m .
47. 2 2 2 2 2 2 Here s =h +b /4 , so h =s b /4 . The area is A=
or s=( p b)/2 A(b)= 1
b
2 2 p 2 pb
4 / A (b)= 2 2 2 s b /4 . Let the perimeter be p , so 2s+b= p 2 ( p b ) /4 b /4 =b p 2 pb /4 . Now bp/4
2 2 3 pb+ p = p 2 pb / 2 1
b
2 / . Therefore, A (b)=0 2 4 2 3 pb+ p =0 b= p/3 . Since p 2 pb / A (b)>0 for b< p/3 and A (b)<0 for b> p/3 , there is an absolute maximum when b= p/3 . But then
2s+ p/3= p , so s= p/3 s=b the triangle is equilateral. 48.
See the figure. The area is given by
1
1
2 2
2 2
2 2 2
2 2
2 2 2
x +b a = a x x+ x +b a
A(x)=
2 a x x+
2 a x
2
2
/
2 2
2 2 2
x
x
A (x)= a x
1+
+ x+ x +b a
=0
2 2 2
2 2
x +b a
a x ( x
2 2 ) ( x+ 2 ( 2 2 x +b a )= )(
( 2 2 a x ) ( ) for 0 x a . Now ) 2 2 2 x+ x +b a
2 2 . 2 a x
x +b a
Except for the trivial case where x=0 , a=b and A(x)=0 , we have
19 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 2 2 2 x 2 x+ x +b a >0 . Hence, cancelling this factor gives = 2 2 2 a x
2 x ( x +b a ) =a
2 2 2 4 2 2 2a x +x 4 2 x ( b a ) =a
2 2 4 2 2 2 2a x x 2 2 2 2 2 x x +b a =a x 2 2 x +b a ( b +a ) =a
2 2 a x 2 2 4 x= a . 2 2 a +b
Now we must check the value of A at this point as well as at the endpoints of the domain to see which
2 2 gives the maximum value. A(0)=a b a , A(a)=0 and
2 a A 2 a =
2 2 a +b 2 2 2 + 2 a +b b 2 2 a +b
2 2 b a ,A a / 2 2 2 2 2 2 2 a 2 2 2 +b a = 2 a +b ) =ab=0 a +b a +b ( 2 = + a +b ab a +b 2 2 2 ( 2 a Since b 2 a a +b ab
2 2 2 a 2 2 a +b ) 2 a A ( 0 ) . So there is an absolute maximum when x= 2 .
2 a +b
2ab In this case the horizontal piece should be 2 and the vertical piece should be
2 a +b
2 2 a +b
2 2 2 = a +b .
2 a +b 49. Note that AD = AP + PD
PDB and PDC gives us
L(x) = 5=x+ PD
2 2 PD =5 x . Using the Pythagorean Theorem for
2 2 AP + BP + CP =x+ (5 x) +2 + (5 x) +3 = x+ x2 10x+29 + x2 10x+34
x 5
x 5
/
/
L (x)=1+
+
From the graphs of L and L , it seems that the
2
2
x 10x+29
x 10x+34
minimum value of L is about L(3.59)=9.35 m.
50. 20 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems We note that since c is the consumption in gallons per hour, and v is the velocity in miles per hour,
c gallons/hour gallons
then =
=
gives us the consumption in gallons per mile, that is, the quantity G
v miles/hour
mile
dc
dv
dc
v
c
v
c
dG d
c
dv
dv
dv
. To find the minimum, we calculate
=
=
=
. This is 0 when
2
2
dv dv
v
v
v
dc
dc c
v
c=0
= . This implies that the tangent line of c(v) passes through the origin, and this
dv
dv v
occurs when v 53 mi / h. Note that the slope of the secant line through the origin and a point
( v,c(v) ) on the graph is equal to G(v) , and it is intuitively clear that G is minimized in the case where
the secant is in fact a tangent.
51. The total time is
2 T (x) </ = ( time from A to C ) + ( time from C to B ) = v 1 x / T (x)=
v 1 2 sin d x
2 a +x v 2 = 2 2 b +(d x)
/ The minimum occurs when T (x)=0 sin 1 v 2 a +x 2 + 2 b +(d x)
v , 0<x<d 2 2 v 1 2 sin 1 v 1 sin
= 2 v . 2 52. 21 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems If d= QT , we minimize f
df
setting
d ( ) = PR + RS =a 1 2 cot 1 1 b +bcot 1 d 2 d a
=
b 1 cot 1 1 have 1
2 = 1 1 1 , we get a df
d . We substitute this into the expression for 2 b 1 csc
1 . 2 b 1 d 2
2 d 2 =0 1 2 cot 2 2 b =csc 2
2 1 cot 2 to get
1 2 2 csc 1 =0 2 a csc 1 cot +a 1 1 cot csc 2 cot
cot d 2 , and . 2 a
a 2 1 . We can do this by observing that QT = constant Differentiating this equation implicitly with respect to 2 cot 2 1 1 d d . Differentiating with respect to
d df
=0= a
equal to 0 , we get
d So we need to find an expression for
=acot +b 1 csc 1
1 cot
= csc 2 cos =cos 1 2 . Since 2 =0 2 1 and 2 are both acute, we 2 . 53. 2 2 2 y =x +z , but triangles CDE and BCA are similar, so z/8=x
2 2 2 3 / (4 x 4) z=2x / x 4 . Thus, we minimize f ( x ) =y =x +4x /(x 4)=x /(x 4) , 4<x 8 .
22 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems / f ( x) = ( 2) (x 4) 3x x 3 2 = 2 x 3(x 4) x
2 2 = 2x ( x 6 ) (x 4)
(x 4)
( x 4)
when x>6 , so the minimum occurs when x=6 in. 2 / / =0 when x=6 . f (x)<0 when x<6 , f (x)>0 54. Paradoxically, we solve this maximum problem by solving a minimum problem. Let L be the length
of the line ACB going from wall to wall touching the inner corner C . As , we have
2
L
and there will be an angle that makes L a minimum. A pipe of this length will just fit around
the corner.
From the diagram, L=L +L =9csc +6sec
dL/d = 9csc cot +6sec tan =0 when
1 6sec tan =9csc 2 csc =1+
L=9 1+ 2 cot tan 3 = 9
=1.5
6 2 tan = [3]1.5 . Then sec =1+ 3
2 2/3 and 2/3 3
2
3
2 Or, use =tan 0 or , so the longest pipe has length
1/2 2/3 +6 ( 1 3 1.5 ) 3
2 1+ 0.852 2/3 L=9 1/2 21.07 ft.
+6sec 21.07 ft. 55. 3t
tan +tan
t+tan
=tan ( + ) =
=
. So
1
1 tan tan
1 ttan
2
2t
2t
2t= 1+3t tan
tan =
. Let f (t)=tan =
2
2
1+3t
1+3t It suffices to maximize tan
3t ( 1 ttan ) =t+tan
/ f (t)= ( 2 1+3t 2 ) 2t(6t)
2 2 ( 1+3t ) . Now ( = ( ) 2 1 3t 2 ) =0 2 2 ( 1+3t ) 2 1 3t =0 t= 1
since t
3 0. Now
23 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems 1
1
1
/
and f (t)<0 for t>
, so f has an absolute maximum when t=
3
3
3
1
2 ( 1/ 3 )
=
and tan =
=
. Substituting for t and in 3t=tan ( + ) gives us
2
6
3
1+3 ( 1/ 3 )
/ f (t)>0 for 0 t< 3 =tan + = 6 6 . 56. We maximize the cross sectional area A ( ) =
1
10h+2
dh =10h+dh=10(10sin )+(10cos )(10sin ) > =100(sin +sin
2 ( / 2 A ( ) =100 cos +cos
1
cos =
= . (cos
2
3
Now A(0)=0 , A sin 2 ) =100 ( cos 1 since 0 =100 and A 2 +2cos
2 3 2 ) 1 =100(2cos cos ), 0 2 1)(cos +1)=0 when .) =75 3 129.9 , so the maximum occurs when = 3 . 57. 5
2
and tan =
. Since + + =180 = ,
x
3 x
2
1
tan
3 x
5
1
2
2
2
2
2
2
x
( 3 x)
1+
3 x From the figure, tan =
5
x
1
d =
5
dx
1+
x
= 1 tan 2 = x
2 5 ( 3 x) 2 2 x +25 x 2 2 ( 3 x ) +4 ( 3 x ) 2 . 24 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems Now d
=0
dx 5
2 x +25 2 2 = 2 2 2x +50=5x 30x+65 2 x 6x+13 2 3x 30x+15=0 x 10x+5=0 x=5 2 5 . We reject the root with the + sign,
since it is larger than 3 . d /dx>0 for x<5 2 5 and d /dx<0 for x>5 2 5 , so is maximized when
AP =x=5 2 5 0.53 .
58. Let x be the distance from the observer to the wall. Then, from the given figure, <
h+d
d
1
1
=tan
tan
, x>0
x
x
1
h+d
1
d
h+d
d
d
=
+
=
2
2
2
2
2
2
2 2
dx
1+ (h+d)/x
x
1+ ( d/x )
x
x + ( h+d )
x +d
2 = d x + ( h+d )
2 2 x + ( h+d )
2 2 2 2 ( h+d ) ( x +d
2 ( x +d ) 2 2 2 2 )= 2 2 2 h d+hd hx
2 x + ( h+d ) 2 ( x +d )
2 2 =0 2 hx =h d+hd x =hd+d x= d ( h+d ) . Since d /dx>0 for all x< d ( h+d ) and d /dx<0 for all
x> d ( h+d ) , the absolute maximum occurs when x= d ( h+d ) .
59. In the small triangle with sides a and c and hypotenuse W , sin = a
c
and cos =
. In the triangle
W
W d
b
and cos = . Thus, a=W sin
L
L
d=Lsin , and b=Lcos , so the area of the circumscribed rectangle is
A( ) = (a+b)(c+d)=(W sin +Lcos )(W cos +Lsin )=1 12pt
with sides b and d and hypotenuse L , sin = , c=W cos , = W 2sin cos +WLsin 2 +LW cos 2 +L2sin cos =1 12pt ( ) = LW sin 2 +LW cos 2 + L2+W 2 sin cos =1 12pt
2
2
2
2 1
= LW sin +cos
+ L +W
2sin cos =1 12pt
2
1
2
2
= LW +
L +W sin 2 , 0
2
2 ( ) ( ( ) ) 25 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems This expression shows, without calculus, that the maximum value of A( ) occurs when sin 2 =1
2 = 2
= 4 . So the maximum area is A 4 =LW + 1
1
1
2
2
2
2
2
L +W =
L +2LW +W = ( L+W ) .
2
2
2 ( ) ( ) b
BC 60. (a) Let D be the point such that a= AD . From the figure, sin = BC =b and BD
a AB
=
BC = ( a AB ) sec . Eliminating BC gives ( a AB ) sec =b
BC
BC
bcot =a AB
AB =a bcot . The total resistance is
AB
BC
a bcot
b csc
R( )=C
+C
=C
+
.
4
4
4
4
r
r
r
r
cos = 1 2 b csc / (b) R ( )=C r
/ R ( )=0 csc
r / R ( )>0 4 4 > r 1 r cot
r 4 r 2 r 4 4 4 r 1 . 2 4
2
4 = cot
csc =cos . 1
4
2
4 r / and R ( )<0 when cos >
r 1 4
2
4 , so there is an absolute 1 4 2 cot r r 2 csc 2 cos < 4 =bC csc 4 r cot 1 2 b csc cot 4 1 csc
r = 1 2 1 minimum when cos =r / r .
(c) When r =
2 2
r , we have cos =
3 1 2
3 4 , so =cos 1 2
3 4 79 . 61. (a)
If k= energy / km over land, then energy / km over water =1.4k . So the total energy is
dE
2
1.4kx
E=1.4k 25+x +k(13 x) , 0 x 13 , and so
=
k.
dx
2 1/2
25+x
1/2
dE
5
2
2 2
2
Set
=0 : 1.4kx=k 25+x
1.96x =x +25 0.96x =25 x=
5.1 . Testing against the
dx
0.96 ( ( ) ) 26 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems value of E at the endpoints: E(0)=1.4k(5)+13k=20k , E(5.1) 17.9k , E(13) 19.5k . Thus, to
minimize energy, the bird should fly to a point about 5.1 km from B .
(b) If W / L is large, the bird would fly to a point C that is closer to B than to D to minimize the
energy used flying over water. If W / L is small, the bird would fly to a point C that is closer to D than
dE
2
Wx
=
L=0 when
to B to minimize the distance of the flight. E=W 25+x +L(13 x)
dx
2
25+x
2 25+x
W
=
. By the same sort of argument as in part (a), this ratio will give the minimal
L
x
expenditure of energy if the bird heads for the point x km from B .
2 25+13
(c) For flight direct to D , x=13 , so from part (b), W / L=
1.07 . There is no value of
13
W / L for which the bird should fly directly to B . But note that lim (W / L)= , so if the point at
x + 0 which E is a minimum is close to B , then W / L is large.
(d) Assuming that the birds instinctively choose the path that minimizes the energy expenditure, we ( 2 1/2 can use the equation for dE/dx=0 from part (a) with 1.4k=c , x=4 , and k=1 : (c)(4)=1 25+4
c= 41 /4 1.6 .
strength of source 62. (a) I(x) ( distance from source )
I(x)= k
2 2 + k
2 2 = k
2 2 + 2 ) . Adding the intensities from the left and right lightbulbs,
k 2 2 . x +d
( 10 x ) +d x +d x 20x+100+d
(b) The magnitude of the constant k won’t affect the location of the point of maximum intensity, so
/
2x
2 ( x 10 )
for convenience we take k=1 . I (x)=
.
2
2 2
2
2 2
x +d
x 20x+100+d ( ) ( ) / Substituting d=5 into the equations for I(x) and I (x) , we get
1
1
/
2x
2(x 10)
I (x)= 2
+ 2
and I (x)=
5
5
2
2
2
x +25 x 20x+125
x +25
x 20x+125 ( ) ( )2 27 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems From the graphs, it appears that I (x) has a minimum at x=5 m.
5 1 / (c) Substituting d=10 into the equations for I(x) and I (x) gives I (x)=
10 I (x)=
10 2 ( x 10 ) 2x / 2 ( x +100) ( x
2 2 20x+200 ) 2 2 x +100 + 1
2 and x 20x+200 . From the graphs, it seems that for d=10 , the intensity is minimized at the endpoints, that is, x=0 and
x=10 . The midpoint is now the most brightly lit point!
(d) From the first figures in parts (b) and (c), we see that the minimal illumination changes from the
midpoint ( x=5 with d=5 ) to the endpoints ( x=0 and x=10 with d=10 ). 28 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.7 Optimization Problems So we try d=6 (see the first figure) and we see that the minimum value still occurs at x=5 . Next, we
let d=8 (see the second figure) and we see that the minimum value occurs at the endpoints. It appears
that for some value of d between 6 and 8 , we must have minima at both the midpoint and the
endpoints, that is, I(5) must equal I(0) . To find this value of d , we solve I(0)=I(5) (with k=1 ):
2
2
2
2
2
2
1
1
1
1
2
25+d
100+d +d 25+d =2d 100+d
+
=
+
=
2
2
2
2
2
d
100+d
25+d
25+d
25+d ( 2 4 2 4 2 4 )( 2 ) ( ) ( ) 2 2500+125d +d +25d +d =200d +2d
2500=50d
d =50 d=5 2 7.071 (for 0 d 10 ). The
third figure, a graph of I(0) I(5) with d independent, confirms that I(0) I(5)=0 , that is, I(0)=I(5) ,
when d=5 2 . Thus, the point of minimal illumination changes abruptly from the midpoint to the
endpoints when d=5 2 . 29 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.8 Applications to Business and Economics 1. (a) C(0) represents the fixed costs of production, such as rent, utilities, machinery etc., which are
incurred even when nothing is produced.
(b) The inflection point is the point at which C / / (x) changes from negative to positive; that is, the / marginal cost C (x) changes from decreasing to increasing. Thus, the marginal cost is minimized at
the inflection point.
/ (c) The marginal cost function is C (x) . We graph it as in Example 1 in Section. 2. (a) We graph C / as in Example 1 in Section. (b) By reading values of C(x) from its graph, we can plot c(x)=C(x)/x . (c) Since the graph in part (b) is decreasing, we estimate that the minimum value of c(x) occurs at x=7
. The average cost and the marginal cost are equal at that value. See the box preceding Example 1.
3. c(x)=21.4 0.002x and c(x)=C(x)/x 2 / C(x)=21.4x 0.002x . C (x)=21.4 0.004x and / C (1000)=17.4 . This means that the cost of producing the 1001 st unit is about $17.40 .
4. (a) Profit is maximized when the marginal revenue is equal to the marginal cost; that is, when R
and C have equal slopes. See the box preceding Example 2.
1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.8 Applications to Business and Economics (b) P(x)=R(x) C(x) is sketched. / (c) The marginal profit function is defined as P (x) . 2 5. (a) The cost function is C(x)=40 , 000+300x+x , so the cost at a production level of 1000 is
C(x) 40,000
C(1000)=$1 , 340 , 000 . The average cost function is c(x)=
=
+300+x and
x
x
/ / c(1000)=$1340/ unit. The marginal cost function is C (x)=300+2x and C (1000)=$2300/ unit.
40,000
/
(b) See the box preceding Example 1. We must have C ( x ) =c ( x ) 300+2x=
+300+x
x
40,000
2
x =40 , 000 x= 40,000 =200 . This gives a minimum value of the average cost
x=
x
/ /
80,000
function c ( x ) since c ( x ) =
>0 .
3
x
(c) The minimum average cost is c(200)=$700/ unit.
2 6. (a) C(x)=25 , 000+120x+0.1x , C(1000)=$245 , 000 . c(x)=
/ C(x) 25,000
=
+120+0.1x ,
x
x / c(1000)=$245/ unit. C (x)=120+0.2x , C (1000)=$320/ unit.
25,000
/
(b) We must have C (x)=c(x) 120+0.2x=
+120+0.1x
x 0.1x= 25,000
x 2 0.1x =25 , 000
2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.8 Applications to Business and Economics x= 250,000 =500 . This gives a minimum since c / / (x)= 50,000
x 3 >0 . (c) The minimum average cost is c(500)=$220.00/ unit.
7. (a) C(x)=16 , 000+200x+4x 3/2 , C(1000)=16 , 000+200 , 000+40 , 000 10 216 , 000+126 , 491
16,000
1/2
, so C(1000) $342 , 491 . c(x)=C(x)/x=
+200+4x , c(1000) $342.49/unit .
x
/ 1/2 / , C (1000)=200+60 10 $389.74/unit .
1/2 16,000
1/2
/
3/2
2/3
+200+4x
2x =16 , 000 x=(8 , 000) =400
(b) We must have C (x)=c(x) 200+6x =
x
2
/
16,000
2 3/2
units. To check that this is a minimum, we calculate c (x)=
+
= (x 8000) . This is
2
x x2
x C (x)=200+6x 2/3 negative for x<(8000) =400 , zero at x=400 , and positive for x>400 , so c is decreasing on ( 0,400 )
and increasing on ( 400, ) . Thus, c has an absolute minimum at x=400 .
(c) The minimum average cost is c(400)=40+200+80=$320/ unit.
2 3 8. (a) C(x)=10 , 000+340x 0.3x +0.0001x , C(1000)=$150 , 000 .
10,000
/
2
2
c(x)=C(x)/x=
+340 0.3x+0.0001x , c(1000)=$150/unit . C (x)=340 0.6x+0.0003x ,
x
/ C (1000)=$40/unit .
/ (b) We must have C (x)=c(x)
2 0.0002x = 10,000
+0.3x
x 2 340 0.6x+0.0003x =
3 10,000
2
+340 0.3x+0.0001x
x 2 0.0002x 0.3x 10 , 000=0 x 1521.60 1522 units. This gives a minimum since c 3 2 x 1500x 50 , 000 , 000=0
/ / (x)= 20,000 x
(c) The minimum average cost is about c(1521.60) $121.62/ unit.
2 3 / 3 +0.0002>0. 2 9. (a) C(x)=3700+5x 0.04x +0.0003x C (x)=5 0.08x+0.0009x (marginal cost).
C(x) 3700
2
c(x)=
=
+5 0.04x+0.0003x (average cost).
x
x (b)
The graphs intersect at ( 208.51,27.45) , so the production level that minimizes average cost is about
3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.8 Applications to Business and Economics 209 units.
/ (c) c (x)= 3700 2 2 3 3700+0.04x 0.0006x =0 0.04+0.0006x=0 ( ) x 208.51 . c x 1 x 1 $27.45/ unit. / (d) The marginal cost is given by C (x) , so to find its minimum value we’ll find the derivative of
800
/
/ /
/ /
/
. C (x)= 0.08+0.0018x=0 x =
=44.44 . C x =$3.22/ unit.
C ; that is, C
1
1 18 ( ) C / / / of C (x)=0.0018>0 for all x , so this is the minimum marginal cost. C / / / / is the second derivative / . cost is given by C (x) .
2 3 / 2 10. (a) C(x)=339+25x 0.09x +0.0004x C (x)=25 0.18x+0.0012x (marginal cost).
C(x) 339
2
c(x)=
=
+25 0.09x+0.0004x (average cost).
x
x (b)
The graphs intersect at ( 135.56,22.65) , so the production level that minimizes average cost is about
136 units.
/ (c) c (x)= 339
2 0.09+0.0008x=0 x 1 x
(d) C
C / / / / / (x)= 0.18+0.0024x=0 x= 2 12=4+0.02x R / /
/ / 1 $22.65/ unit. 1800
/
=75 . C (75)=$18.25/ unit.
24 (x)=0.0024>0 for all x , so this is the minimum marginal cost. 11. C(x)=680+4x+0.01x , p(x)=12 P ( ) 135.56 . c x (x)=R / / 0.02x=8 (x) C (x)=0<0.02=C / /
/ / / / R(x)=xp(x)=12x . If the profit is maximum, then R (x)=C (x) x=400 . The profit is maximized if P (x) , we can just check the condition R / / / / (x)<0 , but since (x)<C / / (x) . Now (x) , so x=400 gives a maximum.
2 2 12. C(x)=680+4x+0.01x , p(x)=12 x/500 . Then R(x)=xp(x)=12x x /500 . If the profit is maximum,
1000
/
/
then R (x)=C (x) 12 x/250=4+0.02x 8=0.024x x=8/0.024=
. The profit is maximized if
3
P / / (x)<0 , but since P / / (x)=R / / (x) C / / (x) , we can just check the condition R / / (x)<C / / (x) .
4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.8 Applications to Business and Economics / / 1
<0.02=C
250 (x)= / / 2 Now R 3 (x) , so x= 1000
gives a maximum.
3
2 13. C(x)=1450+36x x +0.001x , p(x)=60 0.01x . Then R(x)=xp(x)=60x 0.01x . If the profit is
/ / 2 maximum, then R (x)=C (x)
formula, x= 2 ( 1.98) +4(0.003)(24) 1.98 4.2084
. Since x>0 ,
=
2(0.003)
0.006 1.98 x (1.98+2.05)/0.006 672 . Now R
R / / (672)<C / / 2 0.003x 1.98x 24=0 . By the quadratic 60 0.02x=36 2x+0.003x / / (x)= 0.02 and C / / (x)= 2+0.006x C / / (672)=2.032 there is a maximum at x=672 . (672) 2 3 2 14. C(x)=16 , 000+500x 1.6x +0.004x , p(x)=1700 7x . Then R(x)=xp(x)=1700x 7x . If the profit
/ / is maximum, then R (x)=C (x)
2 x +900x 100 , 000=0
P / / (x+1000)(x 100)=0 (x)<0 , but since P Now R / / 2 1700 14x=500 3.2x+0.012x / / (x)=R / / (x)= 14< 3.2+0.024x=C
3 (x) C
/ / / / 2 0.012x +10.8x 1200=0 x=100 (since x>0 ). The profit is maximized if (x) , we can just check the condition R / / (x)<C / / (x) . (x) for x>0 , so there is a maximum at x=100 . 2 / 2 15. C(x)=0.001x 0.3x +6x+900 . The marginal cost is C (x)=0.003x 0.6x+6 .
/ C (x) is increasing when C
when x=100 .
3 / / (x)>0 0.006x 0.6>0 2 / x>0.6/0.006=100 . So C (x) starts to increase / 2 16. C(x)=0.0002x 0.25x +4x+1500 . The marginal cost is C (x)=0.0006x 0.50x+4 .
/ C (x) is increasing when C
increase when x=417 . / / 2 (x)>0 0.0012x 0.5>0 3 / x>0.5/0.0012 417 . So C (x) starts to 2 17. (a) C(x)=1200+12x 0.1x +0.0005x . R(x)=xp(x)=29x 0.00021x . Since the profit is maximized
/ / when R (x)=C (x) , we examine the curves R and C in the figure, looking for x values at which the
slopes of the tangent lines are equal. It appears that x=200 is a good estimate. 5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.8 Applications to Business and Economics / / (b) R (x)=C (x) 2 29 0.00042x=12 0.2x+0.0015x
/ / 2 0.0015x 0.19958x 17=0 x 192.06 (for / / x>0 ). As in Exercise 11, R (x)<C (x)
0.00042< 0.2+0.003x 0.003x>0.19958 x>66.5 . Our
value of 192 is in this range, so we have a maximum profit when we produce 192 yards of fabric.
3/4 2 18. (a) Cost=setupcost+manufacturingcost C(x)=500+m(x)=500+20x 5x +0.01x . We can solve
x( p)=320 7.7 p for p in terms of x to find the demand (or price) function. x=320 7.7 p 7.7 p=320 x
2 320 x
320x x
p(x)=
. R(x)=xp(x)=
.
7.7
7.7
15 1/4
320 2x
/
/
(b) C (x)=R (x) 20
x +0.02x=
x 81.53 planes, and p(x)=$30.97 million. The
4
7.7
maximum profit associated with these values is about $463.59 million.
19. (a) We are given that the demand function p is linear and p(27 , 000)=10 , p(33 , 000)=8 , so the
10 8
1
1
slope is
=
and an equation of the line is y 10=
( x 27,000 )
27,000 33,000
3000
3000
1
y= p(x)=
x+19=19 (x/3000) .
3000
2 (b) The revenue is R(x)=xp(x)=19x (x /3000)
R
. / / / R (x)=19 (x/1500)=0 when x=28 , 500 . Since (x)= 1/1500<0 , the maximum revenue occurs when x=28 , 500 the price is p(28 , 500)=$9.50 20. (a) Let p(x) be the demand function. Then p(x) is linear and y= p(x) passes through ( 20,10 ) and
1
1
1
and an equation of the line is y 10=
(x 20) y=
x+20 . Thus, the
( 18,11 ) , so the slope is
2
2
2
1
1 2
demand is p(x)=
x+20 and the revenue is R(x)=xp(x)=
x +20x .
2
2
1 2
/
(b) The cost is C(x)=6x , so the profit is P(x)=R(x) C(x)=
x +14x . Then 0=P (x)= x+14 x=14 .
2
1
/ /
Since P (x)= 1<0 , the selling price for maximum profit is p(14)=
(14)+20=$13 .
2
21. (a) As in Example 3, we see that the demand function p is linear. We are given that p(1000)=450
and deduce that p(1100)=440 , since a $10 reduction in price increases sales by 100 per week. The
440 450
1
1
1
slope for p is
=
, so an equation is p 450=
(x 1000) or p(x)=
x+550 .
1100 1000
10
10
10
1 2
1
/
(b) R(x)=xp(x)=
x +550x . R (x)=
x+550=0 when x=5(550)=2750 .
10
5
p(2750)=275 , so the rebate should be 450 275=$175 .
(c) C(x)=68 , 000+150x
6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.8 Applications to Business and Economics 1 2
1 2
1
/
x +550x 68 , 000 150x=
x +400x 68 , 000 , P (x)=
x+400=0 when
10
10
5
x=2000 . p(2000)=350 . Therefore, the rebate to maximize profits should be 450 350=$100 . P(x)=R(x) C(x)= 22. Let x denote the number of $10 increases in rent. Then the price is p(x)=800+10x , and the
number of units occupied is 100 x . Now the revenue is
R(x) = ( rental price per unit ) ( number of unit srented )
= (800+10x)(100 x)= 10x2+200x+80,000for0 x 100
/ / / R (x)= 20x+200=0 x=10 . This is a maximum since R (x)= 20<0 for all x . Now we must check
the value of R(x)= ( 800+10x ) ( 100 x ) at x=10 and at the endpoints of the domain to see which value
of x gives the maximum value of R . R(0)=80 , 000 , R(10)=(900)(90)=81 , 000 , and
R(100)=(1800)(0)=0 . Thus, the maximum revenue of $81 , 000/ week occurs when 90 units are
occupied at a rent of $900/ week.
800
orders per year. Storage costs for the
x
1
800
80,000
year are x 4=2x dollars. Handling costs are $100 per delivery, for a total of
100=
2
x
x
80,000
dollars. The total costs for the year are C(x)=2x+
. To minimize C(x) , we calculate
x
/
80,000 2 2
C (x)=2
= (x 40 , 000) . This is negative when x<200 , zero when x=200 , and positive
2
2
x
x
when x>200 , so C is decreasing on ( 0,200 ) and increasing on ( 200, ) , reaching its absolute
minimum when x=200 . Thus, the optimal reorder quantity is 200 cases. The manager will place 4
orders per year for a total cost of C(200)=$800 .
23. If the reorder quantity is x , then the manager places 24. She will have A/n dollars after each withdrawal and 0 dollars just before the next withdrawal, so
1
her average cash balance at any given time is (A/n+0)=A/(2n) . The transaction costs for n
2
withdrawals are nT . The lost interest cost on the average cash balance is A/(2n) R . Thus, the total
AR
/
/
2 AR
AR
AR
cost for n transactions is C(n)=nT +
. Now C (n)=T
and C (n)=0
=T n =
2
2
2n
2T
2n
2n
AR
/ /
AR
n=
, the value of n that minimizes total costs since C (n)=
<0 . Thus, the optimal
3
2T
n
AT
A A 2T
AT
.
average cash balance is
=
=
=
2n 2 AR
2R
2R 7 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 1. (a)
The tangent line at x=1 intersects the x axis at x 2.3 , so x 2 intersects the x axis at x 3 , so x 2.3 . The tangent line at x=2.3 3.0 . 3 (b) x =5 would not be a better first approximation than x =1 since the tangent line is nearly
1 1 horizontal. In fact, the second approximation for x =5 appears to be to the left of x=1 .
1 2. The tangent line at x=9 intersects the x axis at x 6.0 , so x 2 intersects the x axis at x 8.0 , so x 3 6.0 . The tangent line at x=6.0 8.0 . 3. Since x =3 and y=5x 4 is tangent to y= f (x) at x=3 , we simply need to find where the tangent line
1 intersects the x axis. y=0 5x 4=0
2 x=
2 4
.
5 4. (a)
If x =0 , then x is negative, and x is even more negative. The sequence of approximations does not
1 2 3 converge, that is, Newton’s method fails. (b)
1 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method If x =1 , the tangent line is horizontal and Newton’s method fails.
1 (c)
If x =3 , then x =1 and we have the same situation as in part (b). Newton’s method fails again.
1 2 (d)
If x =4 , the tangent line is horizontal and Newton’s method fails.
1 3 3 / 5. f (x)=x +2x 4 x +2x 4 2 f (x)=3x +2 , so x =x n+1 n n . Now x =1 2 n 1 3x +2 x =1
2 n 1+2 4
2 3 1 +2 =1 1
=1.2
5 3 (1.2) +2(1.2) 4 x =1.2 1.1797 . 2 3 3(1.2) +2
3 2 / 6. f (x)=x x 1 Now x =1
1 f (x)=3x 2x , so x
1 1 1
=2
3 2 x =1
2 4 / 7. f (x)=x 20 =x n+1
3 1 5 8. f (x)=x +2 2 20 x =2
2 3 / 2 n n . 2 3x 2x n n n =1.625 . 3 2 2 2 ( ) =x
(x ) f x
=x n+1 n f 4 x 20 n / n n 4x n 3 . n 4 =2.125 4(2)
/ 3 n x x 1 n 2 2 3 3 f 2 2 1 x =2 f (x)=4x , so x n 4 Now x =2 ( ) =x
(x ) f x 2 x =2.125
3 (2.125) 20
3 2.1148 . 4(2.125) 4 f (x)=5x , so 2 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 5 5 x +2
x =x n+1 n n 5x 1
x= 1
= 1
= 1.2
2
4
5
5 ( 1)
( 1) +2 . Now x = 1
1 4
n 5 x = 1.2
3 ( 1.2) +2 1.1529 . 4 5( 1.2) 3 3 9. f (x)=x +x+3 / x +x +3 2 f (x)=3x +1 , so x =x n+1 n n n 2 3x +1 . Now x = 1
1 n 3 x= 1 ( 1) +( 1)+3
2 2 = 1 1 1+3
1
= 1
= 1.25 . Newton’s method follows the tangent line at
3+1
4 3( 1) +1
( 1,1 ) down to its intersection with the x axis at ( 1.25,0 ) , giving the second approximation
x = 1.25 .
2 4 4 10. f (x)=x x 1 / 4 x x 1 3 f (x)=4x 1 , so x =x n+1 n n n 3 4x 1 . Now x =1
1 n x =1
2 1 1 1
3 =1 4 1 1 1 4
= .
3 3 Newton’s method follows the tangent line at ( 1, 1 ) up to its intersection with the x axis at
, giving the second approximation x =
2 3 4
,0
3 4
.
3 3 3 / 2 11. To approximate x= 30 (so that x =30 ), we can take f (x)=x 30 . So f (x)=3x , and thus, 3 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 3 x 30
x =x n+1 n n 2 3x . Since 3 27=3 and 27 is close to 30 , we’ll use x =3 . We need to find
1 n approximations until they agree to eight decimal places. x =3
1 x 4 3.10723251 x . So 3 5 x 2 3.11111111 , x 3.10723734 , 3 30 3.10723251 , to eight decimal places. Here is a quick and easy method for finding the iterations for Newton’s method on a programmable calculator. (The screens
3 shown are from the TI 83 Plus, but the method is similar on other calculators.) Assign f (x)=x 30 to
/ 2 Y , and f (x)=3x to Y . Now store x =3 in X and then enter X Y /Y
1 2 1 1 2 X to get x =3.1 . By
2 successively pressing the ENTER key, you get the approximations x , x , ... .
3 4 In Derive, load the utility file NEWTON ( x^3 30, x,3) and then APPROXIMATE to get . You can
request a specific iteration by adding a fourth argument. For example, NEWTON ( x^3 30, x,3, 2 )
gives [3,3.11111111,3.10723733] .
In Maple, make the assignments f:=x x^3 30; , g:=x x f (x)/D( f )(x); , and x:=3.; . Repeatedly
execute the command x:=g(x); to generate successive approximations.
/ In Mathematica, make the assignments f [x]:=x^3 30 , g[x]:=x f [x]/ f [x] , and x=3. Repeatedly
execute the command x=g[x] to generate successive approximations.
7 7 / 12. f (x)=x 1000 x 1000 6 f (x)=7x , so x n+1 agree to eight decimal places. x =3 x 1 x 5 2.68269580 x . So
6 3 2 13. f (x)=2x 6x +3x+1 7 =x 2 n n 7x 6 . We need to find approximations until they n 2.76739173 , x 3 2.69008741 , x 4 2.68275645 , 1000 2.68269580 , to eight decimal places.
/ 2 f (x)=6x 12x+3
4 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 3 2 n n 2x 6x +3x +1
x =x n+1 1 . We need to find approximations until they agree to six decimal places. 2 n 6x 12x +3
n n x =2.5 n x 2 2.285714 , x 3 2.228824 , x 2.224765 , x 4 2.224745 x . So the root is 2.224745 , 5 6 to six decimal places.
4 4 / 14. f (x)=x +x 4 x +x 4 3 f (x)=4x +1 x =x n+1 n n . x =1.5 3 n x 1 4x +1 1.323276 , x 2 3 1.285346 , n x 4 1.283784 , x 5 1.283782 x . So the root is 1.283782 , to six decimal places.
6 2 2 2 15. sin x=x , so f (x)=sin x x sin x x / f (x)=cos x 2x x =x n+1 n n cos x 2x
n 2 positive root of sin x=x is near 1 . x =1 x 1 2 0.891396 , x n . From the figure, the n 0.876985 , x 3 4 0.876726 x . So the
5 positive root is 0.876726 , to six decimal places. 4 16. 2cos x=x , so f (x)=2cos x x 4 / f (x)= 2sin x 4x 3 2cos x x
x =x n+1 n n the positive root of 2cos x=x is near 1.x =1
1 x 2 1.014184 , x 3 n 2sin x 4x
n 4 4
3 . From the figure, n 1.013958 x . So the positive root
4 is 1.013958 , to six decimal places. 17.
5 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method From the graph, we see that there appear to be points of intersection near x= 0.7 and x=1.2 . Solving
4 4 4 4 x =1+x is the same as solving f (x)=x x 1=0 . f (x)=x x 1 / 3 f (x)=4x 1 , so x x x 1
=x n+1 n n n 3 . 4x 1
n x = 0.7 x =1.2 x 0.725253 x 1.221380 x 0.724493 x 1.220745 x 0.724492 x x 1.220744 x 1
2 3 4 1
2 3 5 4 5 To six decimal places, the roots of the equation are 0.724492 and 1.220744 .
18. 5 From the graph, we see that reasonable first approximations are x=0.5 and x= 1.5 . f (x)=x 5x+2
5 / 4 f (x)=5x 5 , so x x 5x +2
=x n+1 n n n 4 . 5x 5
n 6 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method x = 1.5
1.593846 x 1.582241 3 x 4 1 1 x 2 x =1.5 x =0.5 1 x 0.402102 x 3 1.582036 x 1.373078 x 1.371885 2 2 x 1.396923 x x =0.4
4 3 4 5 x 5 1.371882 x 6 To six decimal places, the roots are 1.582036 , 0.402102 , and 1.371882 .
19. From the graph, we see that there appear to be points of intersection near x= 0.5 and x=1.5 . Solving
1 2/3
/
2
2
2
3
3
3
2x , so
x =x 1 is the same as solving f (x)= x x +1=0 . f (x)= x x +1 f (x)= x
3
3 x =x n+1 n x 2 n x +1
n 1 2/3
x
2x
n
3 n . x = 0.5 x =1.5 x 0.471421 x 1.461653 x 0.471074 x x 1.461070 x 1
2 3 1
2 4 3 4 To six decimal places, the roots are 0.471074 and 1.461070 .
20. From the graph, we see that there appear to be points of intersection near x= 1.2 and x=1.5 . Solving
7 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 2 2 2 x+3 =x is the same as solving f (x)=x
2 x
x =x n+1 n 2x 1
n x+3 =0 . f (x)=x / f (x)=2x x+3 1
, so
2 x+3 x +3
n n / . 2 x +3
n x = 1.2 x =1.5 x 1.164526 x 1.453449 x 1.164035 x x 1.452627 x 1
2 3 1
2 4 3 4 To six decimal places, the roots of the equation are 1.164035 and 1.452627 .
21. From the graph, there appears to be a point of intersection near x=0.6 . Solving cos x= x is the
same as solving f (x)=cos x
cos x
x =x n+1 n x n n sin x 1/ ( 2 x x 0.641928 , x 2 3 x / f (x)= sin x 1/ ( 2 x ) , so . Now x =0.6 ) n x =0 . f (x)=cos x 1 0.641714 x . To six decimal places, the root of the equation is 0.641714 .
4 2 22. From the graph, there appears to be a point of intersection near x=0.7 . Solving tan x= 1 x is
2 1 x =0 . f (x)=tan x the same as solving f (x)=tan x
tan x
x =x n+1 n / 2 2 f (x)=sec x+x/ 1 x , so 2 1 x n 2 sec x +x /
n 2 1 x n n . x =0.7
2 1 x 1 x 2 0.652356 , x 3 0.649895 , x 4 0.649889 x . To six
5 n decimal places, the root of the equation is 0.649889 . 8 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 23. 5 4 3 2 n n n n n x x 5x x +4x +3 4 3 2 / 4 3 2 x f (x)=5x 4x 15x 2x+4 =x n+1 n 4 3 n
2 n 5 f (x)=x x 5x x +4x+3 n n . From the 5x 4x 15x 2x +4 graph of f , there appear to be roots near 1.4 , 1.1 , and 2.7 .
x = 1.4 x =1.1 x =2.7 x 1.39210970 x 1.07780402 x 2.72046250 x 1.39194698 x 1.07739442 x 2.71987870 x 1.39194691 x x 1.07739428 x x 2.71987822 x 1
2 3 4 1
2 3 5 4 1
2 3 5 4 5 To eight decimal places, the roots of the equation are 1.39194691 , 1.07739428 , and 2.71987822 .
24. 2 Solving x ( 4 x2) = 4
2 2 is the same as solving f (x)=4x x x +1
2 4 =x n n n+1 4
2 x +1 / 3 =0 . f (x)=8x 4x + 8x ( x2+1) 2 2 4x x 4 /
x 4 x +1 3 2 n 8x 4x +8x /
n n n n x +1 2 . From the graph of f (x) , there appear to be roots near x= 1.9 and n 9 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method x= 0.8 . Since f is even, we only need to find the positive roots.
x =0.8 x =1.9 x 0.84287645 x 1.94689103 x 0.84310820 x 1.94383891 x 0.84310821 x x 1.94382538 x 1
2 3 4 1
2 3 5 To eight decimal places, the roots of the equation are 4 0.84310821 and 5 1.94382538 . 25. 2 2 From the graph, y=x 2 x x and y=1 intersect twice, at x 2 and at x 2 1 . f (x)=x 2 2 x x 1 2 1
2 1/2
2 1/2
=x
2 x x
( 1 2x)+ 2 x x
2x
f (x)
2
1
2 1/2
2
= x 2 x x
x( 1 2x)+4 2 x x
2 ( / ) )
2
x ( 8 5x 6x )
= ( ( ( 2 so x =x n+1 n x n 2
8 ) , 2 (2+x)(1 x) x ) x x n 5x n 2 2(2 + x )(1
n 2 1 n 6x 2 / . Trying x = 2 won’t work because f ( 2) is undefined, so
1 n x)
n we’ll try x = 1.95 .
1 10 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method x = 1.95 x = 0.8 x 1.98580357 x 0.82674444 x 1.97899778 x 0.82646236 x 1.97807848 x 0.82646233 x 1
2 3 4 x 1
2 3 4 5 1.97806682 5 x 1.97806681 x 6 7 To eight decimal places, the roots of the equation are 1.97806681 and 0.82646233 .
26. 2 From the equations y=3sin (x ) and y=2x and the graph, we deduce that one root of the equation
2 3sin (x )=2x is x=0 . We also see that the graphs intersect at approximately x=0.7 and x=1.4 .
2 2 f (x)=3sin (x ) 2x / 3sin (x ) 2x 2 f (x)=3cos (x ) 2x 2 , so x =x n+1 n n 2 n . 6x cos (x ) 2
n n x =0.7 x =1.4 x 0.69303689 x 1.39530295 x 0.69299996 x x 1.39525078 x 1.39525077 x 1 1
2 3 2 4 3 4 5 To eight decimal places, the roots of the equation are 0.69299996 and 1.39525077 .
27. 11 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 2 From the graph, we see that y=e x 2 and y=x x intersect twice. Good first approximations are x= 0.5
2 x
2 and x=1.1 . f (x)=e x 2 x +x
e 2 / x f (x)= 2xe 2x+1 , so x =x n+1 n 2 x +x
n n 2 n x 2x e
n n . 2x +1 x = 0.5 x =1.1 x 0.51036446 x 1.20139754 x 0.51031156 x x 1.19844118 x 1.19843871 x n 1 1
2 3 2
3 4 4 5 To eight decimal places, the roots of the equation are 0.51031156 and 1.19843871 .
28. ( 2 From the graph, y=ln 4 x ) and y=x intersect twice, at x
2 / f (x)= 2x
2 ln 4 x
1 , so x 4 x ( n+1 2 domain of y=ln 4 x =x n 2x n / x n n 2 ) x . Trying x = 2 won’t work because it’s not in the 2 4 x ( 2 and at x 1 . f (x)=ln 4 x 1 1 n ) . Trying x1= 1.9 also fails after one iteration because the approximation x2 is less than 2 . We try x = 1.99 .
1 x = 1.99 x =1.1 x 1.97753026 x 1.05864851 x 1.96741777 x 1.05800655 x 1.96475281 x 1.05800640 x 1
2 3 4 x
x 5 6 1
2 3 4 5 1.96463580
1.96463560 x 7
12 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method To eight decimal places, the roots of the equation are 1.96463560 and 1.05800640 .
2 / 29. (a) f (x)=x a f (x)=2x , so Newton’s method gives 2 x a
x =x n+1 n n 2x =x 1
1
1
a
a
x+
= x+
=
2 n 2x
2 n 2x
2 n n n a
x x+
n n . n (b) Using (a) with a=1000 and x = 900 =30 , we get x
1 x 31.666667 , x 2 31.622807 , and 3 31.622777 x . So 1000 31.622777 . 4 5 1
a
30. (a) f (x)=
x / f (x)= 1/x a 1 , so x 2 x =x n+1 n n n 2 1/x 2 2 =x +x ax =2x ax .
n n n n n 1
=0.5 , we get x =0.5754 , x 0.588485 , and
1 2
2
3
0.588789 x . So 1/1.6984 0.588789 . (b) Using (a) with a=1.6894 and x =
x 4 5 3 / 31. f (x)=x 3x+6 2 / f (x)=3x 3 . If x =1 , then f ( x ) =0 and the tangent line used for
1 1 approximating x is horizontal. Attempting to find x results in trying to divide by zero.
2 2 3 3 3 32. x x=1 3 / x x 1=0 . f (x)=x x 1 x x 1 2 f (x)=3x 1 , so x =x n+1 n n 2 n . 3x 1
n (a) x =1 , x =1.5 , x
1 2 3 1.347826 , x (b) x =0.6 , x =17.9 , x
1 x 2 1.820129 , x 8 ,x 7 11.946802 , x 1.461044 , x 9 10 (c) x =0.57 , x
1 3 54.165455 , x 2 7.165534 , x 8 1.325200 , x 4 7.985520 , x 1.339323 , x 36.114293 , x
9 5 6 5.356909 , x 1.324913 , x 11 3 4.801704 , x 4 1.324718 x 5 12 24.082094 , x 4 3.233425 , x 10 3.624996 , x 6 13 16.063387 , x 5 2.193674 , x 11 0.997546 , x 0.496305 , x 2.894162 , x 1.967962 , x x 0.870187 , x 0.249949 , x 1.192219 , x 0.731952 , x x 1.753322 , x 1.189420 , x 0.729123 , x x 1.320350 , x 0.851919 , x 0.200959 , x 17
22
27 13 18
23
28 14 19
24
29 15 20
25
30 2.505589 , 1.324718 x x 12 7 16 21 0.377844 , x 26 1.119386 , x 31 6 10.721483 1.496867 ,
1.341355 ,
0.355213 ,
1.937872 ,
0.654291 , 13 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method x 1.547010 , x 32 1.360051 , x 33 34 1.325828 , x 1.324719 , x 35 36 1.324718 x 37 . (d)
From the figure, we see that the tangent line corresponding to x =1 results in a sequence of
1 approximations that converges quite quickly ( x 5 x ). The tangent line corresponding to x =0.6 is
6 1 close to being horizontal, so x is quite far from the root. But the sequence still converges just a 2 little more slowly ( x 12 x 13 ). Lastly, the tangent line corresponding to x =0.57 is very nearly
1 horizontal, x is farther away from the root, and the sequence takes more iterations to converge (
2 x x 36 37 ). 1/3 33. For f (x)=x 1
, f (x)= x
3
/ 2/3 ( ) =x
(x ) f x
and x 1/3 x n n =x 3x = 2x . Therefore, each
n
1 2/3 n n
f
x
n
3 n
successive approximation becomes twice as large as the previous one in absolute value, so the
sequence of approximations fails to converge to the root, which is 0 . In the figure, we have x =0.5 ,
=x n+1 n / n 1 x = 2(0.5)= 1 , and x = 2( 1)=2 .
2 3 x
34. According to Newton’s Method, for x >0 , x
n x
x =x n+1 n 1/ 2 n x =x n 2 ( x)
n =x n+1 n n 1/ 2 x =x 2x = x and for x <0 ,
n n n n n = x . So we can see that after choosing any value x the
n 1 n
14 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method subsequent values will alternate between x and x and never approach the root.
1 4 3 2 / 35. (a) f (x)=3x 28x +6x +24x
1
solve f (x)=0 , try x =
1 2
/ try x =6
1 x
x 2 x =x 1 f / (x ) = 2
3
(x )
1 / / x 3 f 0.6455 / / x 2 (x)=36x 168x+12 . Now to 4 0.6452 x 5 0.6452 . Now 1 x =7.12
2 6.8353 3 3 f (x)=12x 84x +12x+24
f 2 1 x 4 6.8102 x 5 6.8100 . Finally try x = 0.5
1 x 2 0.4571 x 0.4552 3 0.4552 . Therefore, x= 0.455 , 6.810 and 0.645 are all critical numbers correct to three decimal 4 places.
(b) f ( 1)=13 , f (7)= 1939 , f (6.810) 1949.07 , f ( 0.455) 6.912 , f (0.645) 10.982 .
Therefore, f (6.810) 1949.07 is the absolute minimum correct to two decimal places.
2 36. f (x)=x +sin x / / f (x)=2x+cos x . f (x) exists for all x , so to find the minimum of f , we can examine the zeros of f / . From the graph of f / g(x)=2x+cos x and g (x)=2 sin x to obtain x 2 f / / (x)=2 sin x>0 for all x , f ( 0.450184) / , we see that a good choice for x is x = 0.5 . Use
1 0.450627 , x 3 1 0.450184 x . Since
4 0.232466 is the absolute minimum. 37. 15 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method From the figure, we see that y= f (x)=e cos x is periodic with period 2 . To find the x coordinates of the IP, we only need to approximate the zeros of y
/ / f (x)=e cos x ( sin 2x cos x) . Since ecos x / g (x)=2sin xcos x+sin x , and x =1 . x
1 / / / . f (x)= e cos x sin x
2 0 , we will use Newton’s method with g(x)=sin x cos x , 0.904173 , x 2 on 0, 3 0.904557 x . Thus, ( 0.904557,1.855277)
4 is the IP.
38. f (x)= sin x / / f (x)= cos x . At x=a , the slope of the tangent line is f (a)= cos a . The line through
sin a 0
x . If this line is to be tangent to f at x=a , then its slope must
the origin and ( a,f(a) ) is y=
a 0
sin a
/
equal f (a) . Thus,
= cos a tan a=a . To solve this equation using Newton’s method, let
a
g(x)=tan x x ,
/ tan x x 2 g (x)=sec x 1 , and x =x n+1 n n 2 n sec x 1 with x =4.5 (estimated from the figure). x
1 2 4.493614 , n x
f 4.493410 , x 3
/ (x )
5 4 4.493409 x . Thus, the slope of the line that has the largest slope is
5 0.217234 . 39. 1
4 3
2 2
3
r =30 r +
r .
2
3
3
From a graph of V , we see that V (r)=15 , 000 at r 11 ft. Now we use Newton’s method to solve the
dV
2
=60 r+2 r , so
equation V (r) 15 , 000=0 .
dr
2 The volume of the silo, in terms of its radius, is V (r)= r (30)+ 16 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 2 30 r +
r =r n+1 n n 2 3
r 15,000
3 n 60 r +2 r
n . Taking r =11 , we get r
1 2 2 11.2853 , r 11.2807 r . So in order 3 4 n
3 for the silo to hold 15 , 000 ft of grain, its radius must be about 11.2807 ft.
40. Let the radius of the circle be r . Using s=r , we have 5=r and so r=5/ . From the Law of
2 2 2 Cosines we get 4 =r +r 2 r r cos
Multiplying by 2 2 2 16=2r (1 cos )=2 ( 5/ ) (1 cos ) . 2 2 gives 16 =50(1 cos ) , so we take f ( )=16 +50cos 50 and 2 16 +50cos / f ( )=32 50sin . The formula for Newton’s method is graph of f , we can use =2.2 , giving us 1 places, the angle is 2.2622 radians 2.2662 , 2 3 = n+1 n n 32 2.2622 4 375
1 (1+x)
x 60 48x(1+x) 60 50sin . From the
n . So correct to four decimal 130 . 41. In this case, A=18 , 000 , R=375 , and n=5(12)=60 . So the formula A=
18 , 000= n 50 n 48x=1 (1+x) 60 R
1 (1+i)
i n becomes 60 [ multiply each term by (1+x) ] 60 (1+x) +1=0 . Let the LHS be called f (x) , so that /
59
60
59
f (x) = 48x(60)(1+x) +48(1+x) 60(1+x) = 12(1+x)59 4x(60)+4(1+x) 5 =12(1+x)59(244x 1)
x =x n+1 n 60 60 ( 1+x ) ( 1+x ) +1
12 ( 1+x ) ( 244x 1 ) 48x n n n 59 n . An interest rate of 1% per month seems like a reasonable n estimate for x=i . So let x =1%=0.01 , and we get x
1 x 5 2 0.0082202 , x 3 0.0076802 , x 4 0.0076291 , 0.0076286 x . Thus, the dealer is charging a monthly interest rate of 0.76286% (or 9.55% per
6 year, compounded monthly).
17 Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.9 Newton’s Method 5 4 3 2 42. (a) p(x)=x (2+r)x +(1+2r)x (1 r)x +2(1 r)x+r 1
/ 4 3 2 p (x)=5x 4(2+r)x +3(1+2r)x 2(1 r)x+2(1 r) . So we use
5 4 3 2 n n n n x (2+r)x +(1+2r)x (1 r)x +2(1 r)x +r 1
x =x 4 n 3 2 n n+1 n n n . We substitute in the value r 3.04042 10 6 5x 4(2+r)x +3(1+2r)x 2(1 r)x +2(1 r)
n in order to evaluate the approximations numerically. The libration point L is slightly less than 1 AU
1 from the Sun, so we take x =0.95 as our first approximation, and get x
1 x 0.98451 , x 4 5 2 0.98830 , x 0.98976 , x 6 0.98998 , x 7 0.96682 , x 3 0.97770 , 0.98999 x . 8 9 So, to five decimal places, L is located 0.98999 AU from the Sun (or 0.01001 AU from Earth).
1 (b) In this case we use Newton’s method with the function
2 5 4 3 2 p(x) 2rx =x (2+r)x +(1+2r)x (1+r)x +2(1 r)x+r 1
2 p(x) 2rx / 4 3 2 =5x 4(2+r)x +3(1+2r)x 2(1+r)x+2(1 r) . So 5 4 3 2 n n n n x (2+r)x +(1+2r)x (1+r)x +2(1 r)x +r 1
x =x n 4 3 2 n n+1 n n n . Again, we substitute r 3.04042 10 5x 4(2+r)x +3(1+2r)x 2(1+r)x +2(1 r) 6 . L is
2 n slightly more than 1 AU from the Sun and, judging from the result of part (a), probably less than 0.02
AU from Earth. So we take x =1.02 and get x 1.01422 , x 1.01118 , x 1.01018 ,
1 x 5 2 3 4 1.01008 x . So, to five decimal places, L is located 1.01008 AU from the Sun (or 0.01008 AU
6 2 from Earth). 18 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances 1. (a) Since f is increasing , we can obtain a lower estimate by using left endpoints. We are instructed
to use five rectangles, so n=5 .
5 ( ) x
= f ( x ) 2+ f ( x ) 2+ f ( x ) 2+ f ( x ) 2+ f ( x ) 2 L =
5 i=1 f x i 1 0 1 2 3 4 =2 f (0)+ f (2)+ f (4)+ f (6)+ f (8)
2(1+3+4.3+5.4+6.3)=2(20)=40 Since f is increasing , we can obtain an upper estimate by using right endpoints.
R =
5
=2 5 ( ) x
f (x )+ f (x )+ f (x )+ f (x )+ f (x ) i=1 f x i 1 2 3 4 5 =2 f (2)+ f (4)+ f (6)+ f (8)+ f (10)
2(3+4.3+5.4+6.3+7)=2(26)=52 Comparing R to L , we see that we have added the area of the rightmost upper rectangle, f (10) 2 ,
5 5 to the sum and subtracted the area of the leftmost lower rectangle, f (0) 2 , from the sum. aaaaa
(b)
L 10 = 10
i=1 ( ) f x i 1 x 1 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances ( ) ( ) ( ) =1 f x + f x +
0 +f x 1 9 = f (0)+ f (1)+
+ f ( 9)
1+2.1+3+3.7+4.3+4.9+5.4+5.8+6.3+6.7
=43.2 R 10 = 10
i=1 ( ) f x x= f (1)+ f (2)+ i =L +1 f (10) 1 f (0)
10 + f (10) [add rightmost upper rectangle, subtract leftmost lower rectangle ] =43.2+7 1=49.2 2. (a)
(i)
L =
6
=2 6 ( ) x
f (x )+ f (x )+ f (x )+ f (x )+ f (x )+ f (x ) i=1 f x i 1 0 1 2 3 4 5 =2[ f (0)+ f (2)+ f (4)+ f (6)+ f (8)+ f (10)]
2(9+8.8+8.2+7.3+5.9+4.1)
=2(43.3)=86.6 2 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances (ii) R =L +2 f (12) 2 f (0)
6 6 86.6+2(1) 2(9)=70.6 (iii) M =
6 6
i=1 f x * i x =2[ f (1)+ f (3)+ f (5)+ f (7)+ f (9)+ f (11)]
2(8.9+8.5+7.8+6.6+5.1+2.8)
=2(39.7)=79.4 (b) Since f is decreasing , we obtain an overestimate by using left endpoints; that is, L .
6 (c) Since f is decreasing , we obtain an underestimate by using right endpoints; that is, R .
6 (d) M gives the best estimate, since the area of each rectangle appears to be closer to the true area
6 than the overestimates and underestimates in L and R .
6 6 3. (a)
3 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances R =
4 4 ( ) f x x i i=1 ( ) 1+ f ( x ) 1+ f ( x ) 1+ f ( x ) 1 =f x 1 2 3 4 = f (2)+ f (3)+ f (4)+ f (5)
1 1 1 1 77
= + + + =
=1.283
2 3 4 5 60
Since f is decreasing on 1,5 , an underestimate is obtained by using the right endpoint
approximation, R .
4 (b)
L =
4 4
i=1 ( ) f x x i 1 = f (1)+ f (2)+ f (3)+ f (4)
1 1 1 25
=1+ + + =
=2.083
2 3 4 12
L is an overestimate. Alternatively, we could just add the area of the leftmost upper rectangle and
4 subtract the area of the rightmost lower rectangle; that is, L =R + f (1) 1 f (5) 1 .
4 4 4. (a)
R =
5 5
i=1 ( ) f x i x
4 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances ( ) 1+ f ( x ) 1+ f ( x ) 1+ f ( x ) 1+ f ( x ) 1 =f x 1 2 3 4 5 = f (1)+ f (2)+ f (3)+ f (4)+ f (5)
=24+21+16+9+0=70
Since f is decreasing on 0,5 , R is an underestimate.
5 (b)
L =
5 5
i=1 ( ) f x i 1 x = f (0)+ f (1)+ f (2)+ f (3)+ f (4)
=25+24+21+16+9=95
L is an overestimate.
5 2 5. (a) f (x)=1+x and
x= x= 2 ( 1)
=1
3 R =1 f (0)+1 f (1)+1 f (2)=1 1+1 2+1 5=8 .
3 2 ( 1)
=0.5
6 R =0.5[ f ( 0.5)+ f (0)+ f (0.5)+ f (1)+ f (1.5)+ f (2)]
6
=0.5(1.25+1+1.25+2+3.25+5)
=0.5(13.75)=6.875 5 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances (b)
L =1 f ( 1)+1 f (0)+1 f (1)=1 2+1 1+1 2=5
3
L =0.5[ f ( 1)+ f ( 0.5)+ f (0)+ f (0.5)+ f (1)+ f (1.5)]
6
=0.5(2+1.25+1+1.25+2+3.25)
=0.5(10.75)=5.375 6 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances (c)
M =1 f ( 0.5)+1 f (0.5)+1 f (1.5)
3
=1 1.25+1 1.25+1 3.25=5.75
M =0.5[ f ( 0.75)+ f ( 0.25)+ f (0.25)]
6
+ f (0.75)+ f (1.25)+ f (1.75)]
=0.5(1.5625+1.0625+1.0625+1.5625+2.5625+4.0625)
=0.5(11.875)=5.9375 (d) M appears to be the best estimate.
6 6. (a)
2 (b) f (x)=e
(i) x and x= 2 ( 2)
=1
4
1 1 R =1 f ( 1)+1 f (0)+1 f (1)+1 f (2)=e +1+e +e
4 4 1.754
7 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances (ii) M =1 f ( 1.5)+1 f ( 0.5)+1 f (0.5)+1 f (1.5)=e 2.25 +e 4 (c)
(i) . 0.25 +e 0.25 +e 2.25 R =0.5 f ( 1.5)+ f ( 1)+ f ( 0.5)+ f (0) + f (0.5)+ f (1)+ f (1.5)+ f (2) =e
8 +e 0.25 1 +e +e 2.25 +e 4 1.761 1.768 2.25 1 +e +e 0.25 +1 (ii) Due to the symmetry of the figure, we see that M =(0.5)(2)[ f (0.25)+ f (0.75)+ f (1.25)+ f (1.75)]
8 =e 0.0625 +e 0.5625 +e 1.5625 +e 3.0625 1.766 7. Here is one possible algorithm (ordered sequence of operations) for calculating the sums:
(a) Let SUM =0 , X_MIN =0 , X_MAX = , N =10 (or 30 or 50 , depending on which sum we are
calculating), DELTA_X = ( XMAX XMIN ) / N, and RIGHT_ENDPOINT = X_MIN + DELTA_X.
(b) Repeat steps 2a, 2b in sequence until RIGHT_ENDPOINT > X_MAX.
(c) Add sin ( RIGHTENDPOINT ) to SUM.
(d) Add DELTA_X to RIGHT_ENDPOINT.
At the end of this procedure, ( DELTAX ) ( SUM ) is equal to the answer we are looking for. We find
10
30
i
i
that R =
sin
1.9835 , R =
sin
1.9982 , and
10 10 i=1
30 30 i=1
10
30
50
i
R =
sin
1.9993 . It appears that the exact area is 2 .
50 50 i=1
50
Shown below is program SUMRIGHT and its output from a TI 83 Plus calculator. To generalize the
8 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances program, we have input (rather than assigned) values for Xmin, Xmax, and N. Also, the function,
sin x , is assigned to Y , enabling us to evaluate any right sum merely by changing Y and running
1 1 the program. 8. We can use the algorithm from Exercise 7 with X_MIN =1 , X_MAX =2 , and / 2 1 ( RIGHTENDPOINT ) instead of sin ( RIGHTENDPOINT ) in step 2a. We find that
1 10
1 30
1
1
R =
0.4640 , R =
0.4877 , and
10 10 i=1
30 30 i=1
2
2
1+i/10 )
1+i/30 )
(
(
1 50
1
1
R =
0.4926 . It appears that the exact area is
.
50 50 i=1
2
2
( 1+i/50 )
9. In Maple, we have to perform a number of steps before getting a numerical answer. After loading
the student package we use the command sum: =leftsum(x^(1 / 2) , x=1..4,10 [ or 30, or 50]) ;
which gives us the expression in summation notation. To get a numerical approximation to the sum,
we use evalf(left_sum);. Mathematica does not have a special command for these sums, so we must
type them in manually. For example, the first left sum is given by
(3 / 10)*Sum[Sqrt[1 + 3 (i 1) / 10] , {i, 1, 10}] , ], and we use the N command on the resulting
output to get a numerical approximation.
In Derive, we use the LEFT_RIEMANN command to get the left sums, but must define the right
sums ourselves. (We can define a new function using LEFT_RIEMANN with k ranging from 1 to n
instead of from 0 to n 1 .)
3 n
3(i 1)
(a) With f (x)= x , 1 x 4 , the left sums are of the form L =
1+
. Specifically,
n n i=1
n
L 10 4.5148 , L Specifically, R 30 10 4.6165 , and L
4.8148 , R 30 50 4.6366 . The right sums are of the form R = 4.7165 , and R n 50 3
n n
i=1 1+ 3i
.
n 4.6966 . (b) In Maple, we use the leftbox and rightbox commands (with the same arguments as leftsum and
rightsum above) to generate the graphs.
9 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances left endpoints, n=10 left endpoints, n=30 left endpoints, n=50 right endpoints, n=10 10 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances right endpoints, n=30 right endpoints, n=50 (c) We know that since x is an increasing function on ( 1,4 ) , all of the left sums are smaller than the
actual area, and all of the right sums are larger than the actual area. Since the left sum with n=50 is
about 4.637>4.6 and the right sum with n=50 is about 4.697<4.7 , we conclude that 4.6<L < exact
50 area <R <4.7 , so the exact area is between 4.6 and 4.7 .
50 10.
(a) With f (x)=sin (sin x) , 0 x
In particular, L 10 0.8251 , L 30 , the left sums are of the form L = 2
0.8710 , and L sin sin (i 1) 2n i=1
2n
0.8799 . The right sums are of the form
n 50 n . i
R=
sin sin
. In particular, R
0.9573 , R
0.9150 , and R
0.9064 .
n 2n i=1
10
30
50
2n
(b) In Maple, we use the leftbox and rightbox commands (with the same arguments as leftsum and
rightsum above) to generate the graphs.
n left endpoints, n=10 left endpoints, n=30 11 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances left endpoints, n=50 right endpoints, n=10 right endpoints, n=30 righnt endpoints, n=10
(c) We know that since sin (sin x) is an increasing function on , all of the left sums are
2
smaller than the actual area, and all of the right sums are larger than the actual area. Since the left sum
with n=50 is about 0.8799>0.87 and the right sum with n=50 is about 0.9064<0.91 , we conclude that
0.87<L < exact area <R <0.91 , so the exact area is between 0.87 and 0.91 .
50 0, 50 11. Since v is an increasing function, L will give us a lower estimate and R will give us an upper
6 6 estimate.
L = ( 0ft/s ) ( 0.5s ) + ( 6.2 ) ( 0.5) + ( 10.8 ) ( 0.5) + ( 14.9 ) ( 0.5) + ( 18.1 ) ( 0.5) + ( 19.4 ) ( 0.5)
6
=0.5 ( 69.4 ) =34.7 ft
R =0.5 ( 6.2+10.8+14.9+18.1+19.4+20.2 ) =0.5 ( 89.6 ) =44.8 ft
6 12 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances 12.
(a)
d L =(30 ft / s )(12 s )+28 12+25 12+22 12+24 12
5
=(30+28+25+22+24) 12=129 12=1548 ft
(b) d R = ( 28+25+22+24+27) 12=126 12=1512 ft
5 (c) The estimates are neither lower nor upper estimates since v is neither an increasing nor a
decreasing function of t .
13. Lower estimate for oil leakage: R =(7.6+6.8+6.2+5.7+5.3)(2)=(31.6)(2)=63.2 L.
5 Upper estimate for oil leakage: L =(8.7+7.6+6.8+6.2+5.7)(2)=(35)(2)=70 L.
5 14. We can find an upper estimate by using the final velocity for each time interval. Thus, the
distance d traveled after 62 seconds can be approximated by
d= 6
i=1 () v t i t = ( 185ft/s ) ( 10s ) +319 5+447 5+742 12+1325 27+1445 3=54 , 694 ft
i 15. For a decreasing function, using left endpoints gives us an overestimate and using right endpoints
results in an underestimate. We will use M to get an estimate. t=1 , so
6 M = 1[v(0.5)+v(1.5)+v(2.5)+v(3.5)+v(4.5)+v(5.5)]
6
55+40+28+18+10+4=155ft
For a very rough check on the above calculation, we can draw a line from ( 0,70 ) to ( 6,0 ) and
1
calculate the area of the triangle: (70)(6)=210 . This is clearly an overestimate, so our midpoint
2
estimate of 155 is reasonable.
16. For an increasing function, using left endpoints gives us an underestimate and using right
30 0
5
=5 s =
h
endpoints results in an overestimate. We will use M to get an estimate. t=
6
6
3600
1
=
h.
720
M = 1 [v(2.5)+v(7.5)+v(12.5)+v(17.5)+v(22.5)+v(27.5)]
6
720
1
1
=
(31.25+66+88+103.5+113.75+119.25)=
(521.75) 0.725km
720
720
For a very rough check on the above calculation, we can draw a line from ( 0,0 ) to ( 30,120 ) and
calculate the area
of the triangle:
13 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances 1
(30)(120)=1800 . Divide by 3600 to get 0.5 , which is clearly an underestimate, making our
2
midpoint estimate of 0.725 seem reasonable. Of course, answers will vary due to different readings of
the graph.
4 17. f (x)= x , 1 x 16 .
A=lim R =lim
n n n i=1 A=lim R =lim
n ( ) f x n n
i=1 A=lim R =lim
n n
n 20. lim i=1 n
10 y=(5+x) 2
n n
i=1 5+ n x=lim i i=1 n . ( ) f x i x=( 2 x=lim
n i=1 x=3+7i/n . ln (3+7i/n) 7
.
3+7i/n
n 0)/n=
n x=1+15i/n . 15
.
n i 2 /n and x =0+i
i i
cos
2n i
2n 2n x= 2 i/n . . 10 can be interpreted as the area of the region lying under the graph of
10 on the interval 0,2 , since for y=(5+x) * and x =x , the expression for the area is A=lim
i 15i
n x=(10 3)/n=7/n and x =3+i ( ) 2i
n 1+ i=1 n f x 2 n 4 x=lim i 19. f (x)=xcos x , 0 x n i ln x
, 3 x 10 .
x 18. f (x)= n n x=(16 1)/n=15/n and x =1+i i n
10 the answer is not unique. We could use y=x on 0,2 with n
i=1 * f (x ) x=lim
i n 2 0 2
2i
= , x =0+i x=
,
i
n
n
n
n
2i 10 2
5+
. Note that
i=1
n
n x= 10 on 5,7 or, in general, y=((5 n)+x) on n,n+2 . i
can be interpreted as the area of the region lying under the graph of y=tan x
i=1 4n
4n
n
/4 0
i
=
on the interval 0,
, since for y=tan x on 0,
with x=
, x =0+i x=
,
i
4
4
n
4n
4n
n
n
i
*
*
and x =x , the expression for the area is A=lim
f x
x=lim
tan
. Note
i
i
i
i=1
i=1
4n
4n
n
n
that this answer is not unique, since the expression for the area is the same for the function
21. lim n tan y=tan (x k ) on the interval k ,k + 4 , where k is any integer. 22. (a)
14 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances 1 0 1
x=
= and x =0+i
i
n
n
i n (b) lim i=1 n 3 n 1
=lim
n n 3 5 23. (a) y= f (x)=x . n (b) 2 n ( ) f x 2 ( n 5 2 2 i =lim i=1 4 n 2 ( i=1 n
i=1 ( ) f x 4 2 =lim ( n+1 )
4n ) 2 ( e2 1) =e 2 ( e2 1) 2/n e 1
lim
4n 1+ n 2 1
n 5 = 1
4 5 2
=lim
n n n
i=1 32i
n 5 5 2
64
=lim
6
n n
n i . i=1 )
( 2
24. From Example 3(a), we have A=lim
n
n
2 = 2 n )( 2 2 64
64 n (n+1) 2n +2n 1
n +2n+1 2n +2n 1
=
lim
(c) lim
6
2 2
12
12 n
n
n
n n
16
2 1
2 1
16
32
=
lim
1+ +
2+
=
1 2=
2
2
3 n
n
n
3
3
n
n 2 e
lim
n
n 2 1
.
n 2i
.
n x= 2i
n i=1 n 3 i
n n x=lim i n(n+1)
2 1
n n n n (n+1) 2n +2n 1
i=
i=1
12
n 3 x=lim i i=1 n 1 2 0 2
= and x =0+i
i
n
n x= n A=lim R =lim
n i
x= . A=lim R =lim
n
n
n
n n
i=1 e 2i/n ) n . Using a CAS, i=1 e 2i/n = e ( e2 1) 2 2/n e 0.8647 , whereas the estimate from Example 3(b) using M 1 and 1
10 was 0.8632 .
25. y= f (x)=cos x . A=lim R =lim
n n If b= 2 n n
i=1 b 0 b
= and x =0+i
i
n
n x= ( ) f x , then A=sin i 2 x=lim
n n cos i=1 x= bi
n bi
.
n b
=lim
n n bsin b 2nsin 1
+1
2n
b
2n b
2n =sin b =1 . 26. (a)
15 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.1 Areas and Distances The diagram shows one of the n congruent triangles, AOB , with central angle 2 /n . O is the
center of the circle and AB is one of the sides of the polygon. Radius OC is drawn so as to bisect
AOB . It follows that OC intersects AB at right angles and bisects AB . Thus, AOB is divided into
1
two right triangles with legs of length (AB)=rsin ( /n ) and rcos ( /n ) .
2
AOB has area 2
1
2 1
1 2
2
[rsin ( /n ) ][rcos ( /n ) ]=r sin ( /n)cos ( /n)= r sin (2 /n) , so A =n area
n
2
2 2 ( AOB) = nr sin (2 /n) .
sin (b) To use Equation 3.4.2, lim =1 , we need to have the same expression in the denominator as 0 we have in the argument of the sine function
in this case, 2 /n .
1 2
1 2 sin (2 /n) 2
sin (2 /n) 2
2
lim A =lim
nr sin ( 2 /n ) =lim
nr
=lim
r . Let =
.
n
2
2
2 /n
n n
2 /n
n
n
n
n
sin (2 /n) 2
sin
2
2
2
Then as n
,
0 , so lim
r =lim
r = ( 1) r = r .
2 /n
n
0 16 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 1.
R =
4 4
i=1 ( ) f x i x =0.5 f (0.5)+ f (1)+ f (1.5)+ f (2)
=0.5 1.75+1+( 0.25)+( 2)
=0.5(0.5)=0.25
The Riemann sum represents the sum of the areas of the two rectangles above the x axis minus the
sum of the areas of the two rectangles below the x axis; that is, the net area of the rectangles with
respect to the x axis. 2. L =
6 6
i=1 ( ) x=0.5 f (1)+ f (1.5)+ f (2)+ f (2.5)+ f (3)+ f (3.5) f x i 1 0.5( 1 0.5945349 0.3068528 0.0837093+0.0986123+0.2527630)=0.5( 1.6337217) 0.816861
The Riemann sum represents the sum of the areas of the two rectangles above the x axis minus the
sum of the areas of the four rectangles below the x axis; that is, the net area of the rectangles with
respect to the x axis. 3. . M =
5 5
i=1 ( ) f x i x =1 f (1.5)+ f (2.5)+ f (3.5)
+ f (4.5)+ f (5.5)
0.856759
The Riemann sum represents the sum of the areas of the two rectangles above the x axis minus the
sum of the areas of the three rectangles below the x axis.
1 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral . 4. (a)
R =
6 6
i=1 ( ) f x i x =0.5 f (0.5)+ f (1)+ f (1.5)+ f (2)
+ f (2.5)+ f (3)
5.353254
The Riemann sum represents the sum of the areas of the four rectangles above the x axis minus the
sum of the areas of the two rectangles below the x axis. (b) . M =
6 6
i=1 ( ) f x i x =0.5 f (0.25)+ f (0.75)+ f (1.25)+ f (1.75)
+ f (2.25)+ f (2.75)
4.458461
The Riemann sum represents the sum of the areas of the four rectangles above the x axis minus the
sum of the areas of the two rectangles below the x axis. 2 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 5. x=(b a)/n=(8 0)/4=8/4=2 .
8 (a) Using the right endpoints to approximate
4
i=1 ( ) f x i 4 ( ) f x i 1 f (x)dx , we have x=2[ f (2)+ f (4)+ f (6)+ f (8)] 2[1+2+( 2)+1]=4 . (b) Using the left endpoints to approximate
i=1 0 8
0 f (x)dx , we have x=2[ f (0)+ f (2)+ f (4)+ f (6)] 2[2+1+2+( 2)]=6 . (c) Using the midpoint of each subinterval to approximate
4
i=1 ( ) f x i i=1 ( ) g x i 0 f (x)dx , we have x=2[ f (1)+ f (3)+ f (5)+ f (7)] 2[3+2+1+( 1)]=10 .
3 6. (a) Using the right endpoints to approximate
6 8 g(x)dx , we have 3 x = 1[g( 2)+g( 1)+g(0)+g(1)+g(2)+g(3)]
1 0.5 1.5 1.5 0.5+2.5= 0.5 (b) Using the left endpoints to approximate
6
i=1 ( ) g x i 1 3 g(x)dx , we have 3 x = 1[g( 3)+g( 2)+g( 1)+g(0)+g(1)+g(2)]
2+1 0.5 1.5 1.5 0.5= 1 (c) Using the midpoint of each subinterval to approximate
6
i=1 ( ) g x i 3 g(x)dx , we have 3 x = 1[g( 2.5)+g( 1.5)+g( 0.5)+g(0.5)+g(1.5)+g(2.5)]
1.5+0 1 1.75 1+0.5= 1.75 7. Since f is increasing, L 25
5 0 f (x)dx R .
5 3 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 5 Lower estimate = L =
5 i=1 ( ) f x x=5[ f (0)+ f (5)+ f (10)+ f (15)+ f (20)] i 1 = 5( 42 37 25 6+15)=5( 95)= 475
5 Upper estimate = R =
5 i=1 ( ) f x x=5[ f (5)+ f (10)+ f (15)+ f (20)+ f (25)] i = 5( 37 25 6+15+36)=5( 17)= 85
6 8. (a) Using the right endpoints to approximate
3
i=1 ( ) f x i=1 f (x)dx , we have x=2[ f (2)+ f (4)+ f (6)]=2(8.3+2.3 10.5)=0.2 i (b) Using the left endpoints to approximate
3 0 ( ) f x 6
0 f (x)dx , we have x=2[ f (0)+ f (2)+ f (4)]=2(9.3+8.3+2.3)=39.8 i 1 6 (c) Using the midpoint of each interval to approximate
3
i=1 ( ) 0 x=2 f (1)+ f (3)+ f (5) =2(9.0+6.5 7.6)=15.8 . f x i The estimate using the right endpoints must be less than ( ) endpoint x of each interval, then f x
i ( ) i x i x 3 f (x)dx , and so the sum i=1 ( ) f x i * Similarly, if we take x to be the left endpoint x
i i 1 i 3 , and so i=1 i 1 ( ) f x i 1 6 x 0 * f (x)dx , since if we take x to be the right
i i 1 i i 1 x ,x 6 f (x) for all x on x ,x i x f x f (x)dx , we have 0 x , which implies that 3 x i=1 x i f (x)dx = i 1 ( ) of each interval, then f x i 1 6
0 f (x)dx . f (x) for all x on f (x)dx . We cannot say anything about the midpoint estimate.
9. x=(10 2)/4=2 , so the endpoints are 2 , 4 , 6 , 8 , and 10 , and the midpoints are 3 , 5 , 7 , and 9 .
The Midpoint Rule gives
10
2 3 x +1 dx 4
i=1 ( ) f x i x=2 ( 3 3 3 3 3 +1 + 5 +1 + 7 +1 + 9 +1 ) 124.1644 . 2
3
4
5
6
,
,
,
, and
, and the
6
6
6
6
6
6
6
3
5
7
9
11
midpoints are
,
,
,
,
, and
. The Midpoint Rule gives
12 12
12
12
12
12
6
3
5
7
9
11
sec ( x/3) dx
f x
x=
sec
+sec
+sec
+sec
+sec
+sec
i
0
i=1
6
36
36
36
36
36
36
10. x=( 0)/6= , so the endpoints are 0 , ( ) , 3.9379
4 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral .
11. x=(1 0)/5=0.2 , so the endpoints are 0, 0.2, 0.4, 0.6, 0.8, and 1, and the midpoints are 0.1, 0.3,
0.5, 0.7, and 0.9. The Midpoint Rule gives
1 5 2 sin (x )dx 0 i=1 2 ( ) f x 2 2 2 2 x=0.2 sin (0.1) +sin (0.3) +sin (0.5) +sin (0.7) +sin (0.9) i 0.3084 . 12. x=(5 1)/4=1 , so the endpoints are 1, 2, 3, 4, and 5, and the midpoints are 1.5, 2.5, 3.5, and 4.5.
The Midpoint Rule gives
5 2
1 4 x x e dx n=1 ( ) f x i 2 x=1 (1.5) e 1.5 2 +(2.5) e 2.5 2 +(3.5) e 3.5 2 +(4.5) e 4.5 1.6099 . 13. In Maple, we use the command with(student) to load the sum and box commands, then
m:=middlesum(sin(x^),x=0..1,5); which gives us the sum in summation notation, then M:=evalf(m);
which gives M 0.30843908 , confirming the result of Exercise 11. The command
5 middlebox(sin(x^),x=0..1,5) generates the graph. Repeating for n=10 and n=20 gives
M
0.30981629 and M
0.31015563 .
10 20 14. See the solution to Exercise 5.1.7 for a possible algorithm to calculate the sums. With
x=(1 0)/100=0.01 and subinterval endpoints 1, 1.01, 1.02, ... , 1.99, 2, we calculate that the left
Riemann sum is L
R = 100 100
i=1 sin = 100
2 x i 100
i=1 sin (x ) x 0.30607 , and the right Riemann sum is
i 1 x 0.31448 . ( 2) is an increasing function, we must have L100 1sin ( x2) dx R100 , so
0
1
2
sin ( x ) dx R <0.315 . Therefore, the approximate value 0.3084 0.31 in Exercise
100
0 Since f (x)=sin x
0.306<L 100 11 must be accurate to two decimal places.
15. We’ll create the table of values to approximate 0 sin xdx by using the program in the solution to Exercise 5.1.7 with Y =sin x , Xmin=0 , X = , and n=5 , 10 , 50 , and 100 .
1 The values of R appear to be approaching 2 .
n n R n
5 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 5
10
50
100 1.933766
1.983524
1.999342
1.999836
2 16. 0 2 e L n x dx with n=5 , 10 , 50 , and 100 .
R n n 5 1.077467
10 0.980007
50 0.901705
100 0.891896 0.684794
0.783670
0.862438
0.872262
2 The value of the integral lies between 0.872 and 0.892 . Note that f (x)=e 17. On 0, n , lim x sin x i=1 i n i x= 0 x 18. On 1,5 , lim i=1 n 1 n x= n 6
= lim n
n dx since f is increasing on ( 1,0 ) . xsin xdx . x i
2 8 x= 1 2 2x+x dx . 5 0 5 ( 1) 6
= and x = 1+i
i
n
n (1+3x)dx = lim x e
dx .
1+x i 2 e x= (4 3x +6x )dx . i=1 n 1 * 2 2x + x i=1 20. On [0,2] , lim 1 x= 5 * n n 5 e
1+x i 19. On 1,8 , lim 21. Note that i n is decreasing on ( 0,2 ) . 2 2 We cannot make a similar statement for x n
i=1 ( ) f x i n
i=1 x=lim
n 2+ 18i
n x= 1+
n
i=1 =lim
n 6i
.
n 1+3
6
n 1+
n 6i
n ( 2)+ i=1 6
n
n
i=1 18i
n 6 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 6
= lim n
n 2n+ 18
n n i =lim i=1 n 6
n 2n+ 18 n(n+1)
n
2 n+1
108 n(n+1)
=lim
12+54
2
2
n
n
n
1
12+54 1+
= 12+54 1=42
n = lim 12+ n = lim
n 22.
n 4 2 (x +2x 5)dx =lim 1 i=1 n 3
=lim
n
n
3
n =lim
n 3
n =lim
n =lim
n =lim
n n x= n 2 i=1 n =lim
n +2 1+ 3i
n 3
n 5 2 6i 9i
6i
1+ +
+2+
5
i=1
2
n
n
n
n
9 2 12
i+
i 2
i=1
2
n
n
n
9 n 2 12 n
i+
i
2
i=1
2 i=1
n i=1
n
27 n(n+1)(2n+1) 36 n(n+1)
+
3
2
6
2
n
n
9 n+1 2n+1
n+1
+18
6
2 n
n
n
1+ 1
n 1
n 2+ +18 1+ 1
n 6
n
n 6 = 9
1 2+18 1 6=21
2 2 0 2
2i
= and x =0+i x=
.
i
n
n
n ( 2 x ) dx =lim
0 2 2 n 9
2 =lim i 3i
n 1+ i=1 n x=3/n and x =1+3i/n ] i n =lim 23. Note that f (x ) x [ 2
n ( ) f x i n 2 i=1 n x=lim i=1 n 4
2 n n i i=1 2 2 4i 2 2
n 2 n
2
=lim
n
n 2n 4
2 n n i 2 i=1 7 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral =lim
n n
4
4
3 =lim
n n(n+1)(2n+1)
6 8 4 3 1
n 1+ 2+ =lim 4 n 1
n 4 n+1 2n+1
3 n
n 4
4
1 2=
3
3 =4 24.
n ( 1+2x3) dx =lim
0
n
5 i=1 ( ) f x n =lim 125i 1+2 i=1 n x i n 5
=lim
n
n 1 n+ =lim n i n 5
n n 1+ i=1 n =lim 2 =lim 5+312.5 1+ n n n = lim
n = lim 4 n
1
n n = lim
n n 3 4 1+ n = lim ( n3+3n2i+3ni2+i 3) =lim
i=1
n 2 n n +3n
3
2 n n i+3n i=1 n 2 i+ i=1 1
n
n n 3
2 3 i=1 4 i 3 i=1 i=1 n+ n 2 n 3 n+i
n n 3n i+ i=1 2 3ni + i=1 n i 3 i=1 3 i=1 2 2 n(n+1) 3 n(n+1)(2n+1) 1 n (n+1)
+
+
3
4
2
6
4
n
n 3 n+1 1 n+1 2n+1 1 ( n+1 )
1+
+
+
2
2 n
2 n
n
4
n
1+ i 2 1
n ( ) 1 3 n 2 =5+312.5=317.5
2 1 1
25. Note that x=
= and x =1+i x=1+i ( 1/n ) =1+i/n .
i
n
n
n
n
i 3 1
1
2 3
x=lim
1+
=lim
x dx = lim i=1 f xi
i=1
1
n
n
n
n
n
n
= lim 250 1250 n (n+1)
5+
4
4
n n 2 n =lim 2 3 i=1 3 ( n+1 ) 5+312.5 5
n 3 250
n 3 1+ 1
n + 1
2 1+ 1
n 2+ 1
n 2 + 1
4 1+ 1
n 2 =1+ 3 1
1
+ 2+ =3.75
2 2
4 8 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral . * x=(4 0)/8=0.5 and x =x =0.5i . 26. (a) i i ( x 3x) dx i=1 f xi x
2
2
=0.5 { 0.5 3(0.5) + 1.0 3(1.0) +
2
2
+ 3.5 3(3.5) + 4.0 3(4.0) }
4 8 2 * 0 = 1
2 5
9
5
7
2
2
+0+ +4 = 1.5
4
4
4
4 (b)
(c) ( x2 3x) dx=lim
0
4 4
n =lim
n 16
2 n n
n 2 i i=1 4i
n n
i=1 12
n n 2 3 4i
n 4
n i i=1 64 n(n+1)(2n+1) 48 n(n+1)
3
2
6
2
n
n
n
32
1
1
1
=lim
1+
2+
24 1+
3
n
n
n
n
32
8
=
2 24=
3
3
=lim (d) ( x2 3x) dx=A1
0
4 A , where A is the area marked + and A is the area marked .
2 1 2 27. 9 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral b a
xdx =lim
a
n
n b a
a+
i =lim
n
n n b i=1 a(b a)
n 2 n 1+ i=1 (b a) n i i=1 2 n 2 2 a(b a)
1
( b a)
(b a) n(n+1)
=lim
n+
=a ( b a ) +lim
1+
2
n
2
2
n
n
n
n
1
1
1
1
1 2 2
2
=a(b a)+ (b a) =(b a) a+ b
a =(b a) (b+a)=
b a
2
2
2
2
2 ( ) 28.
b a
x dx =lim
a
n
n n b 2 i=1 ( b a) =lim
n n b a
a+
i
n 3 n 2 i+ 2 b a
=lim
n
n 2a ( b a ) i=1 3 2 n 3 =lim
n n ( b a) =lim ( b a)
3 x 2 1+x 30. 5 x= 1+ 2 n 1
n 2+ 1
n 3 2 3 5 2a ( b a ) 2 2 2 2 1 i=1 n ( n+1 ) a ( b a )
+
n
2
n +a ( b a ) 1
2 n 1+ 1
n 2 +a ( b a ) 3 b 3ab +3a b a
2
2
3 2
3
+a ( b a ) +a ( b a ) =
+ab 2a b+a +a b a
3
2 3 , a=2 , b=6 , and dx=lim R =lim
n 1 + a
b a
2 2
2 2
ab +a b+ab a b=
3
3 x
1+x 6 3 3 b
=
3
29. f (x)= 3 6 n = 6 3 2 a ( b a)
i+
i=1
n 2 ( b a ) n ( n+1 ) ( 2n+1 ) 2 i=1 n 2 b a
( b a) 2
a +2a
i+
i
2
n
n n n n n
i=1 10 1 9
= and x =1+i
i
n
n x= 3 6 2 4
*
= . Using Equation 3, we get x =x =2+i
i
i
n
n 4i
n
4i
1+ 2+
n
2+ x=1+ 5 x=2+ 4i
, so
n 4
.
n 9i
, so
n
10 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 10
1 n n 31.
0 n (x 4ln x)dx=lim R =lim 4ln 1+ 9i
n 9
.
n * i i n
i=1 n ( sin 5x i ) n =lim n i=1 n sin 5 i
n n 1
cot
n = lim
n 5
2n = 2
5 = * x=(10 2)/n=8/n and x =x =2+8i/n .
i n 10 6
2 i=1 n 9i
n 0)/n= /n and x =x = i/n . x=( sin 5xdx=lim 32. 1+ x dx =lim i=1 n i 8i
2+
n 6 ( 8
n 1
=8lim
n
n 6 n
i=1 5 8i
2+
n 6 4 2 1 64 58,593n +164,052n +131,208n 27,776n +2048
=8 lim
5
n
n
21n
1,249,984
9,999,872
=8
=
1,428,553.1
7
7
33. (a) Think of
trapezoid is A= 2
0 ) f (x)dx as the area of a trapezoid with bases 1 and 3 and height 2 . The area of a 1
(b+B)h , so
2 2
0 f (x)dx= 1
(1+3)2=4 .
2 (b)
5
0 f (x)dx = 2
0 f (x)dx + 3
2 f (x)dx + 5
3 f (x)dx trapezoid
rectangle triangle
1
1
=
+
(1+3)2 + 3 1
2 3 =4+3+3=10
2
2
(c)
7
5 7
5 f (x)dx is the negative of the area of the triangle with base 2 and height 3 . f (x)dx= (d) 9
7 1
2 3= 3 .
2 f (x)dx is the negative of the area of a trapezoid with bases 3 and 2 and height 2 , so it equals 1
1
(B+b)h=
(3+2)2= 5 . Thus,
2
2 9
0 f (x)dx= 5
0 f (x)dx+ 7
5 f (x)dx+ 9
7 f (x)dx=10+( 3)+( 5)=2 . 34. (a)
11 2
5 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 1
4 2=4 (area of a triangle)
0
2
6
1
2
(b) g(x)dx=
(2) = 2 (negative of the area of a semicircle)
2
2
7
1
1
(c) g(x)dx= 1 1= (area of a triangle)
6
2
2
7
2
6
7
1
g(x)dx= g(x)dx+ g(x)dx+ g(x)dx=4 2 + =4.5 2
0
0
2
6
2
2 g(x)dx= 1
x 1 dx can be interpreted as the area of the triangle above the x axis minus the area of
0
2
1
1
1
1
3
the triangle below the x axis; that is, (1)
(2)(1)=
1=
.
2
2
2
4
4
35. 36. 3 2
2 2 2 4 x dx can be interpreted as the area under the graph of f (x)= 4 x between x= 2 and x=2 . This is equal to half the area of the circle with radius 2 , so 37. 0
3 ( 1+ 2 9 x 2
2 2 4 x dx= ) dx can be interpreted as the area under the graph of f (x)=1+ 1
2 2 2 =2 . 2 9 x between x= 3 and x=0 . This is equal to one quarter the area of the circle with radius 3 , plus the area of the
0
1
9
2
2
.
rectangle, so
1+ 9 x dx=
3 +1 3=3+
3
4
4 ( ) 12 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 38. 3 (3 2x)dx can be interpreted as the area of the triangle above the x axis minus the area of the 1 triangle below the x axis; that is, 39. 2
1 1
2 5
2 (5) 1
2 3
2 (3)= 25
4 9
=4 .
4 x dx can be interpreted as the sum of the areas of the two shaded triangles; that is, 1
1
1 4 5
(1)(1)+ (2)(2)= + = .
2
2
2 2 2 40. 10
0 1
2 2 x 5 dx can be interpreted as the sum of the areas of the two shaded triangles; that is,
(5)(5)=25 . 41.
4
9 t dt =
=
= 9
4
9
4 t dt
x dx 38
3
13 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 1 2 42. x cos xdx=0 since the limits of integration are equal. 1 ( 5 6x2) dx=
0 1 43. 1 0 0 ( 2ex 1) dx=2
1 3 x 3 1 1 3 x+2 2 3 44.
45. 1 e 1
3 1 2 5dx 6 x dx=5(1 0) 6 3 x e dx ( 3 1dx=2 e e 2 3 x dx= e e dx=e
1 1 2 e dx=e ) =5 2=3
3 1( 3 1 ) =2e 2e 2 ( e3 e) =e5 e3 46.
/2
0 (2cos x 5x)dx = /2 2cos xdx 0 2 /2
0 5xdx=2 2 ( /2) 0
5
=2(1) 5
=2
2
8 /2
0 cos xdx 5 /2
0 xdx 2 47.
2
2 f (x)dx+ 5
2 f (x)dx 1
2 f (x)dx =
= 48.
49. 4
1 f (x)dx= 5
1 f (x)dx 5
4 2
5
1 f (x)dx+ 2
1 f (x)dx f (x)dx f (x)dx=12 3.6=8.4 9 9 0 0 [2 f (x)+3g(x)]dx=2 5 9 f (x)dx+3 g(x)dx=2(37)+3(16)=122
0 { 5
3 for x<3
, then f (x)dx can be interpreted as the area of the shaded region,
0
x for x 3
which consists of a 5 by 3 rectangle surmounted by an isosceles right triangle whose legs have
5
1
length 2. Thus, f (x)dx=5(3)+ (2)(2)=17 .
0
2 50. If f (x)= 14 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 51. 0 sin x < 1 on 0, 3 2 , so sin x 4 sin x on 0, . Hence, 4 /4
0 /4 3 sin xdx 0 2 sin xdx (Property 7).
52. 5 x 3 x+1 on 1,2 , so 5 x
2 53. If 1 x 1 , then 0 x
1 1 1 ( 1) 54. 2 1
2 sin x 1 for /2 6 sin xdx 3 /6 55. If 1 x 2 , then 2 57. If 1
2 1
x
3 3 , then 1 /3 12 2 , so 1 , so tan xdx /4 1
2 x+1 dx . 1
2 2 , so 1 2 and 1+x 1
1 2 1+x dx 2 2 .
/2 2 1
(2 1)
2 sin xdx 1 6 2 /6 12 2
1 1
1
dx 1(2 1) or
x
2 3 3 8 , so 1 x +1 9 and 1
2 3 x 4 6 x x +1 dx 3(2 0) that is, 2 0 1 ; that is, 6 . 56. If 0 x 2 , then 0 x
1(2 0) 2 1 and 1 1+x 2 5 x dx 2 1 ( 1) ; that is, 2 1+x dx 1 2 x+1 and tan x x +1 2
1 1
dx 1 .
x 3 . Thus, 3 x +1 dx 6 . 0 3 , so 1 /3 3 tan xdx 4 /4 3 3 4 or 3 . 3 / 2 58. Let f (x)=x 3x+3 for 0 x 2 . Then f (x)=3x 3=3(x+1)(x 1) , so f is decreasing on ( 0,1 ) and
increasing on ( 1,2 ) . f has the absolute minimum value f (1)=1 . Since f (0)=3 and f (2)=5 , the
3 absolute maximum value of f is f (2)=5 . Thus, 1 x 3x+3 5 for x in 0,2 . It follows from
Property 8 that 1 (2 0) ( x3 3x+3) dx
0 2 5 (2 0) ; that is, 2 ( x3 3x+3) dx
0 2 x 59. The only critical number of f (x)=xe on 0,2 is x=1 . Since f (0)=0 , f (1)=e
f (2)=2e 2 10 .
1 0.368 , and 0.271 , we know that the absolute minimum value of f on 0,2 is 0 , and the absolute maximum is e 1 . By Property 8 ,0 xe x e 1 for 0 x 2 0(2 0) 2
0 x 1 xe dx e (2 0)
15 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral 2 0 0 60. If
1
2 x xe dx 2/e .
1
4
3
4 3
4 x
1
4 3 /4
/4 4 61. 4 x +1 62. 0 2
2 , then 3
4 2 sin xdx 1
3 2 x =x , so sin x 1 for 0 x ; that is, 3 2 4 1 x dx= 0 1
4 3 /4 2 sin xdx /4 1
2 . 1 3 3 26
3 1 =
.
3
3 ( /2 , so xsin x x 2 2 sin x 1 , so 1
4 x +1 dx 1 1
2 sin x 1 and ) /2 xsin xdx 0 2 1
xdx=
2 2 0 2 2 = 8 . 63. Using a regular partition and right endpoints as in the proof of Property 2, we calculate
b n cf (x)dx=lim a ( ) cf x i i=1 n n x =lim c
i i=1 n 64. As in the proof of Property 2, we write ( ) , so f x n x 0 and therefore i nonnegative by Theorem 2.3.2, so
65. Since f (x) f (x)
b
a i=1
b
a ( ) f x i b
a ( ) f x i n x =clim
i f (x)dx= n i=1 n
n
i=1 ( ) f x i ( ) x =c
i b
a f (x)dx . ( ) f x x . Now f x i i 0 and x 0 x 0 . But the limit of nonnegative quantities is f (x)dx 0 . f (x) , it follows from Property 7 that f (x) dx b
a f (x)dx b
a f (x) dx b
a b f (x)dx a f (x) dx Note that the definite integral is a real number, and so the following property applies: a b a
b a for all real numbers b and nonnegative numbers a .
2 66. 0 sin 2x 67. lim
n 2 f (x)sin 2xdx
1
n
i=1 f (x)sin 2x dx= 0 f (x) sin 2x
i 4 n 5 =lim
n 1
n 2
0 f (x) sin 2x dx 2
0 f (x) dx by Property 7 , since f (x) .
n
i=1 i
n 4 1 4 = x dx
0 68.
16 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.2 The Definite Integral lim
n 1
n 1 n
i=1 2 dx 1 = 0 1+(i/n) 2 1+x i
i 1
*
and x = x x =
1+
i
i 1 i
i
n
n
1 n
1
2 2
lim
n i=1
i 1
i
x dx = n
1
1+
1+
n
n
n
1
= lim n i=1 (n+i 1)(n+i)
n 69. Choose x =1+ = lim n
n i=1
n 1 = lim n i=0 n n n 1
n+i 1
n+i n
i=1 1
1
+
+
n n+1 = lim n
= lim n 1
n+i 1 n 1
n 1
2n =lim
n 1+ i
n . Then [by the hint] 1
n+i
+ 1
2n 1 1 1
2 1
+
n+1
= + 1
1
+
2n 1 2n 1
2 17 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus 1. The precise version of this statement is given by the Fundamental Theorem of Calculus. See the
statement of this theorem and the paragraph that follows it at the end of Section 5.3.
2. (a)
g(x) =
g(1) =
g(2) =
= x
0
1
0
2
0 1
2 f (t)dt , so g(0)=
f (t)dt= 0
0 f (t)dt=0. 1
1
1 1 [area of triangle] = .
2
2
1 f (t)dt= 0 f (t)dt+ 2
1 f (t)dt [below the x axis] 1
1 1=0.
2 1
1
1 1= .
2
2
2
4
1 1
g(4) =g(3)+ f (t)dt=
+ 1 1=0.
3
2 2
g(3) =g(2)+ g(5) =g(4)+
g(6) =g(5)+ 3 5
4
6
5 f (t)dt=0 f (t)dt=0+1.5=1.5.
f (t)dt=1.5+2.5=4. (b) g(7)=g(6)+ 7
6 f (t)dt 4+2.2 [estimate from the graph] =6.2 . (c) The answers from part (a) and part (b) indicate that g has a minimum at x=3 and a maximum at
x=7 . This makes sense from the graph of f since we are subtracting area on 1<x<3 and adding area
on 3<x<7 . (d)
3. (a)
g(x) =
g(0) = x
0
0
0 f (t)dt .
f (t)dt=0 1 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus g(1) =
g(2) = 1
0
2
0 f (t)dt=1 2=2 ,
f (t)dt= =2+1 2+
g(3) = 3
0 1
0 f (t)dt+ 6
3 3 2
1 f (t)dt 2 1
1 4=7 ,
2 f (t)dt=5+ f (t)dt
1
2 2+1 2
2 =7+ 1 f (t)dt=g(1)+ 1
1 2=5 ,
2 f (t)dt=g(2)+ g(6) =g(3)+ 2 =7 4=3 (b)
(c) g is increasing on ( 0,3) because as x increases from 0 to 3 , we keep adding more area.
(d) g has a maximum value when we start subtracting area; that is, at x=3 .
4. (a) g( 3)= 3
3 f (t)dt=0 , g(3)= 3
3 f (t)dt= 0
3 f (t)dt+ 3
0 f (t)dt=0 by symmetry, since the area above the x axis is the same as the area below the axis.
(b) From the graph, it appears that to the nearest 1
, g( 2)=
2 2
3 f (t)dt 1 , g( 1)= 1
3 f (t)dt 3 1
,
2 1
.
3
2
(c) g is increasing on ( 3,0 ) because as x increases from 3 to 0 , we keep adding more area.
(d) g has a maximum value when we start subtracting area; that is, at x=0 . and g(0)= 0 f (t)dt 5 (e)
/ (f) The graph of g (x) is the same as that of f (x) , as indicated by FTC1.
2 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus 5. x 2 2 / (a) By FTC1 with f (t)=t and a=1 , g(x)= t dt
1 1 3
t
3 x 2 (b) Using FTC2, g(x)= t dt=
1 x = 1 2 g (x)= f (x)=x . 1 3 1
x
3
3 / 2 g (x)=x . 6. (a) By FTC1 with f (t)=1+ t and a=0 , g(x)=
x ( 1+ t ) dt=
0 (b) Using FTC2, g(x)= x 7. f (t)= 1+2t and g(x)= t+ x / ( 1+ t ) dt
0 2 3/2
t
3 x g (x)= f (x)=1+ x . 2 3/2
x
3 =x+ 0 / 1/2 g (x)=1+x =1+ x . / 1+2t dt , so by FTC1, g (x)= f (x)= 1+2x . 0 x / 8. f (t)=ln t and g(x)= ln t dt , so by FTC1, g (x)= f (x)=ln x .
1 y2 2 / 2 9. f (t)=t sin t and g(y)= t sin t dt , so by FTC1, g (y)= f (y)=y sin y .
2 10. f (x)= 1
2 3 x+x
2 ( 2) 11. F(x)= cos t dt=
x 12. f ( )=tan and F(x)= 1 u and g(u)= 2 x+x x ( 2) cos t dt 2
10
x 1 / dx , so g (u)= f (u)= tan d = u+u
/ 2 . ( 2) F (x)= cos x
x / tan d , so by FTC1, F (x)= f (x)= tan x . 10 3 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus 1
du
dh dh du
1
. Then
=
. Also,
=
, so
2
x
dx
dx du dx
x
d 1/x
d u
du
du
/
arctan ( 1/x )
h (x)=
arctant dt=
arctant dt
=arctanu
=
.
2
dx 2
du 2
dx
dx
x 13. Let u= du
dh dh du
=2x . Also,
=
, so
dx
dx du dx
d u
du
3
3
3
1+r dr=
1+r dr
= 1+u (2x)=2x
du 0
dx 2 14. Let u=x . Then
2 d
h (x)=
dx x / 0 15. Let u= x . Then
/ y = d
dx 3 6 1+x . du
1
dy dy du
=
. Also,
=
, so
dx 2 x
dx du dx cos t
d
dt=
t
du x 23 1+(x ) =2x cos x
cos t
du cos u
1
dt
=
=
3 t
dx
u
2 x
x u 1
2 x = cos x
.
2x du
dy dy du
= sin x . Also,
=
, so
dx
dx du dx
d cos x
d u
du
/
y = dx 1 (t+sin t)dt= du 1(t+sin t)dt dx
= (u+sin u) ( sin x)= sin x cos x + sin ( cos x ) 16. Let u=cos x . Then dw
dy dy dw
= 3 . Also,
=
, so
dx
dx dw dx 17. Let w=1 3x . Then
d
y = dx 1 3x d
dw = u 1 / w
1 3 1+u
u 2 3 1+u 2 d
du=
dw u 1
w dw
du
=
dx 3 1+u
w 2 du dw
dx 3 3 2 1+w ( 3)= 3(1 3x) 2 1+(1 3x) du x
dy dy du
=e . Also,
=
, so
dx
dx du dx
d 0 3
du
d u 3
du
/ d 0
3
3
x
x
3 x
y =
sin t dt=
sin t dt
=
sin t dt
= sin u e = e sin e .
dx ex
du u
dx
du 0
dx
x 18. Let u=e . Then ( ) 19. 3
1 5 x dx= 6 x
6 3 6 3
=
1 6 6 ( 1) 729 1 364
=
=
6
6
3
4 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus 20. 21. 22. 23. 24. 25. 5 =6[5 ( 2)]=6(7)=42 2 (4x+3)dx= 2 ( 1+3y y2) dy=
0
4 1 4/5
0 x 83 8 1/3 x dx= x 1 1 3 2 3 1 = 0 2 2 t dt=3 t dt=3 0 = 1 1 2 5 3 t 30. 31. 1 5 4
1
dx= x
1
x
/4 0 1 5 2 =sin 2 2 1/2 6 2
x 1/2 dx= 2 sec t dt= tan t /4
0 =tan 4 3 has an infinite discontinuity at x=0 ; that 5,5 . sin =0 0=0
2 x
1/2 has an infinite discontinuity at x=0 ; that is, 2,3 . 1 7
29. x(2+x )dx= (2x+x )dx= x + x
0
0
7
2 4 3 dx does not exist because the function f (x)= cos d = sin 20
3 1
7
1 =
8
8 = 1 x dx does not exist because the function f (x)=x 5 2 2 1 5 2 ( 0) = 3 4/3 4/3 3 4
3
3
45
(8 1 )= (2 1)= (16 1)= (15)=
4
4
4
4
4
3
3 = x
is, f is discontinuous on the interval
28. 3
1
16
64
2
3 = 4+ 8 f is discontinuous on the interval
27. 4 2 3 3 1 ) ( 2 22+3 2) =152 14=138 5
5
0=
9
9 3 4/3
x
4 dx= 2 4 1 4 ( = 2 8 +3 8 3 2 1 3
y
y
2
3 y+ 5 9/5
x
9 dx= 8 4 2
x +3x
2 8 t 26. 5 6dx= 6x 2 4 = 4+ 0 128
7 ( 0+0 ) = 156
7 1/2 4 = 2x 1 2 =2 4 2 1 =4 2=2 1 tan 0=1 0=1 32.
5 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus 1 ( 3+x
0
2 33. x ) dx= csc 2 1 2 5/2
3x+ x
5 ( 3+x3/2) dx=
0 1 = 3+ 0 2
5 0 = d does not exist because the function f ( )=csc and =2 ; that is, f is discontinuous on the interval ,2 2 17
5 has infinite discontinuities at = . /6 34. csc cot d does not exist because the function f ( )=csc cot 0 has an infinite discontinuity at =0 ;
that is, f is discontinuous on the interval
9 35. 1 1
1
dx=
2x
2 1 36. 9
1 1
1
dx=
ln x
x
2
1 x x 10 dx= 0 10
ln 10 0 1 = 1 10
ln 10 3 /2 9 = 0, . 6 1
1
1/2
( ln 9 ln 1 ) = ln 9 0=ln 9 =ln 3
2
2 1
9
=
ln 10 ln 10 37.
6 3 /2 dt =6 1/2 1 t 2 1/2 1 t
=6 38. 4 1 0 2 1 dt=4 0 t +1 39. 1 u+1 1 e 3
1 1+t u+1 1 du= e 2 6 =6 6
1 2 3 /2 1 dt=6 sin t 1 =6 1/2 sin 2 0 sin 1 1
2 =
( ) 1 1 dt=4 tan t =4 tan 1 tan 0 =4
0 3
2 1 2 u+1 =e e =e 1 [ or start with e 0 = 4 u 1 =e e ] 1 40.
2
1 4+u
u 3 2 du =
= ( 4u 3+u 1) du=
1
2 1
+ln 2
2 4 2
u +ln u
2 ( 2+ln 1)= 2 = 1 2
u 2 2 +ln u
1 3
+ln 2
2 6 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus 41. 2
0 1 4
0 1 5
x
5 2 5
1 f (x)dx= x dx+ x dx= 1 + 0 1 6
x
6 2 1
0 +
5 = 1 64
6 1
6 =10.7 42.
f (x)dx = 0 xdx+ sin xdx=
0 2 = 2 1 2
x
2 2 0 cos x =
0 0 2 (cos cos 0) 2 ( 1 1)=2 2 43. From the graph, it appears that the area is about 60 . The actual area is
27 1/3
3 4/3 27 3
243
3
of the area of the viewing rectangle.
x dx=
x
= 81 0=
=60.75 . This is
0
4
0
4
4
4 44. From the graph, it appears that the area is about
6
1 4 x dx= x 3 3 6 = 1 1
3x 3 6 =
1 1
1 215
+ =
3 216 3 648 45. It appears that the area under the graph is about
2
3 2.1 . The actual area is 0 1
. The actual area is
3
0.3318 . 2
of the area of the viewing rectangle, or about
3 sin xdx= cos x = ( cos ) ( cos 0 ) = ( 1 ) +1=2 .
0 7 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus 46. Splitting up the region as shown, we estimate that the area under the graph is
/3
1
2
/3
+
3
1.8 . The actual area is
sec xdx= tan x = 3 0= 3 1.73 .
0
0
3 4
3 47. 48. 2
1 2 1 4
x
4 3 x dx= 1 5 /2 5 /2 /4 /4 sin xdx= cos x 49. g(x)=
/ g (x)= 3x
2x 2 u 1 du= 2 u +1 2x 2
2
=
2
2 2 u 1 du+ 2 u +1 3x
0 2 u 1
2 2x du= 0 u +1 2 (2x) 1 2 u 1
2 du+ u +1 2 3x
0 2 u 1
2 du u +1 2 d
(3x) 1 d
4x 1
9x 1
(2x)+
(3x)= 2
+3
2
2
2
2
dx
dx
(2x) +1
(3x) +1
4x +1
9x +1 50. g(x)= 1 x dt= tan x 2+t
g (x)= 0 =0+ 2 2 / 1 15
=
=3.75
4 4 =4 1
4 2+tan x 4 1
tan x d
(tan x)+
dx 2 dt
2+t +
4 1
2+x 8 x dt 1 2+t
d
2
x =
dx ( ) =
4 2 dt tan x
1 2+t
2 sec x
4 2+tan x + +
4 x dt 1 2+t 4 2x
2+x 8
8 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus 51. y=
/ y = x 3 t sin t dt= x
4 1 x ( sin x ) 3 1 x ) +x sin x =3x sin x 2 4 5x 53. F(x)=
/ / F 1 2 / F (x)= f (x)= f (t)dt 24 (x)= f (x)= x 1 2 ( 2) ( = ( 1+2x ) 1+x
2 since f (t)= 8 2x= 2 1+x
x x for all x . ( 1+2x ) <0
1 Section 5.2,
b
a t 2 e dt= b
0 b
0
t / f (x)dx= f (4) f (1) , so 17= f (4) 12
x
0
t 4 2 1+u
du
u 1 8 . So F >0 . y= x
0 / / 8 (2)= 1+2 = 257 . 1
1+t+t 1
. Thus, the curve is concave upward on
2 x< 2 56. (a) erf(x)= / / t ) 2 dt 1 / y = 2 1+x+x ( ( 1+x+x ) 4 ( 2 2 . For this expression to be positive, we must have ( 1+2x ) <0 , since 1+x+x 2 2 55. By FTC2, ) 4 1+u
du
u 54. For the curve to be concave upward, we must have y
/ / ( 3) ( 3x2) 3/2 +x sin x d
2
2
(cos x)=cos 25x 5 cos cos x ( sin x )
dx ( ) x y t sin t dt 1 x sin x
2 x 3 cos u du 0 d
2
x =
dx 1+(x ) / 4 ( ) 3 cos x cos u du= x t sin t dt+ x x 5x
( 2) 0 cos ( u2) du
cos x
d
2
2
/
y =cos ( 25x ) dx (5x) cos ( cos x )
2
2
=5cos ( 25x ) +sin xcos ( cos x ) 52. y= 1 d
3
x =
dx ( ) 3/2 ( t sin t dt= x sin x ( 3) 7/2 t sin t dt+ x d
dx x 2 e dt= 2 e dt a
0 t 2 x e dt
a
0
t t 0 2 e dt+ b
a t 2 2 . f (4)=17+12=29 . erf(x). By Property 5 of definite integrals in 2 e dt , so 2 e dt= 1
2 2 e dt=
t , ) >0 erf(b) 2 erf(a)= 1
2 erf(b) erf(a) . 2 x (b) y=e erf(x)
9 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus 2 / 2 x x 2 / y =2xe erf(x)+e erf (x)=2xy+e 2 x 2 e x 57. (a) The Fresnel function S(x)= sin (
0 / 0=S (x)=sin ( 2 2 2 2 x =(2n 1) x [ by FTC1] =2xy+2/
2 2 t )dt has local maximum values where / x ) and S changes from positive to negative. For x>0 , this happens when
2 x =2(2n 1) to negative where / x= 4n 2 , n any positive integer. For x<0 , S changes from positive 2 2 2 x =2n x =4n / x= 2 n . S does not change sign at x=0 . (b) S is concave upward on those intervals where S
, we get S
0< / / 2 2 x< 2 2 (x)=cos (
or . 2
1
2n
2 x) 2 2 x = xcos ( 1
< x < 2n+
2
2 any positive integer. For x<0 , S 2 / / / / / (x)>0 . Differentiating our expression for S (x) 2 2 x ) . For x>0 , S
, n any integer (x)>0 where cos ( (x)>0 where cos ( 2 2 x )>0 0<x<1 or 4n 1 <x< 4n+1 , n 2 2 / / x )<0 2n 3
2 < 2 2 x < 2n 1
2 ,n 2 any integer 4n 3<x <4n 1
4n 3 < x < 4n 1
4n 3 < x< 4n 1
4n 3 >x> 4n 1 , so
the intervals of upward concavity for x<0 are ( 4n 1 , 4n 3 ) , n any positive integer. To
summarize: S is concave upward on the intervals ( 0,1 ) , ( 3, 1 ) , ( 3, 5 ) , ( 7, 5 ) , ( 7,3)
, ... .
(c) In Maple, we use plot({int(sin(Pi*t^2/2),t=0..x),0.2},x=0..2); . Note that Maple recognizes the
Fresnel function, calling it FresnelS(x) . In Mathematica, we use
Plot[{Integrate[Sin[Pi*t^2/2],{t,0,x}],0.2},{x,0,2}] . In Derive, we load the utility file FRESNEL and
plot FRESNEL_SIN(x) . From the graphs, we see that x sin ( 0 2 2 t )dt=0.2 at x 0.74 . 58. (a) In Maple, we should start by setting si:=int(sin(t)/t,t=0..x); . In Mathematica, the command is
10 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus si=Integrate[Sin[t]/t,{t,0,x}] . Note that both systems recognize this function; Maple calls it Si(x) and
Mathematica calls it SinIntegral[x] . In Maple, the command to generate the graph is
plot(si,x= 4*Pi..4*Pi); . In Mathematica, it is plot[si,{x, 4*Pi.,4*Pi}] . In Derive, we load the utility
file EXP_INT and plot SI(x) . / (b) SI(x) has local maximum values where Si (x) changes from positive to negative, passing through
d x sin t
sin x
/
dt=
0 . From the Fundamental Theorem we know that Si (x)=
, so we must have
dx 0 t
x
/ sin x=0 for a maximum, and for x>0 we must have x=(2n 1) , n any positive integer, for Si to be
changing from positive to negative at x . For x<0 , we must have x=2n , n any positive integer, for a
/ maximum, since the denominator of Si (x) is negative for x<0 . Thus, the local maxima occur at
x= , 2 ,3 , 4 ,5 , 6 ,... .
cos x sin x
/ /
(c) To find the first inflection point, we solve Si (x)=
=0 . We can see from the graph
2
x
x
that the first inflection point lies somewhere between x=3 and x=5 . Using a root finder gives the
value x 4.4934 . To find the y coordinate of the inflection point, we evaluate Si(4.4934) 1.6556 .
So the coordinates of the first inflection point to the right of the origin are about ( 4.4934,1.6556 ) .
/ / Alternatively, we could graph S (x) and estimate the first positive x value at which it changes sign.
(d) It seems from the graph that the function has horizontal asymptotes at y 1.5 , with
lim Si(x) 1.5 respectively. Using the limit command, we get lim Si(x)= x x 2 . Since Si(x) is an odd function, lim Si(x)= . So Si(x) has the horizontal asymptotes y=
.
2
2
(e) We use the fsolve command in Maple (or FindRoot in Mathematica) to find that the solution is
x 1.1 . Or, as in Exercise (c), we graph y=Si(x) and y=1 on the same screen to see where they
intersect.
x / / 59. (a) By FTC1, g (x)= f (x) . So g (x)= f (x)=0 at x=1,3,5,7 , and 9 . g has local maxima at x=1 and
/ 5 (since f =g changes from positive to negative there) and local minima at x=3 and 7 . There is no
local maximum or minimum at x=9 , since f is not defined for x>9 .
(b) We can see from the graph that 1
0 f dt < 3
1 f dt < 5
3 f dt < 7
5 f dt < 9
7 f dt . So
11 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus g(1)= 1
0 f dt , g(5)= 5
0 3 f dt=g(1) 1 f dt + 5
3 f dt , and g(9)= 9
0 7 f dt=g(5) 5 9 f dt + 7 f dt . Thus, g(1)<g(5)<g(9) , and so the absolute maximum of g(x) occurs at x=9 .
/ / / / / / (c) g is concave downward on those intervals where g <0 . But g (x)= f (x) , so g (x)= f (x) ,
1
which is negative on (approximately)
,2 , ( 4,6 ) and ( 8,9 ) . So g is concave downward on
2
these intervals. (d)
/ / 60. (a) By FTC1, g (x)= f (x) . So g (x)= f (x)=0 at x=2 , 4 , 6 , 8 , and 10 . g has local maxima at x=2
/ and 6 (since f =g changes from positive to negative there) and local minima at x=4 and 8 . There is
no local maximum or minimum at x=10 , since f is not defined for x>10 .
2 (b) We can see from the graph that
g(2)=
g(10)= 2 0
10
0 , g(6)= f dt f dt=g(6) 6
0
8
6 0
4 f dt=g(2)
f dt + 2
10
8 f dt >
f dt + 4
2
6
4 6 f dt > 4 8 f dt > 6 f dt > 10
8 f dt . So , and f dt . Thus, g(2)>g(6)>g(10) , and so the absolute maximum of f dt g(x) occurs at x=2 .
/ / / / / / (c) g is concave downward on those intervals where g <0 . But g (x)= f (x) , so g (x)= f (x) ,
which is negative on ( 1,3) , ( 5,7) and ( 9,10 ) . So g is concave downward on these intervals. (d) 61. lim
n 1
62. lim
n
n n
i=1 i 3 n 1 0
=lim
4 n
n
1
+
n 2
+
n 3 i
n n
i=1 + 1 3 = x dx=
0 n
n 1 4 x
4 1 0
=lim
n
n = 0 n
i=1 1
4 i
=
n 1
0 3/2 x dx= 2x
3 1 = 0 2
2
0=
3
3 63. Suppose h<0 . Since f is continuous on x+h,x , the Extreme Value Theorem says that there are
12 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus numbers u and v in x+h,x such that f (u)=m and f (v)=M , where m and M are the absolute
minimum and maximum values of f on x+h,x . By Property 8 of integrals,
x m( h) x+h x+h M( h) ; that is, f (u)( h) f (t)dt x f (t)dt f (v)( h) . Since h>0 , we can divide 1 x+h
g(x+h) g(x) 1 x+h
f ( t ) dt f (v) . By Equation 2,
=
f (t)dt for
h x
h
h x
g(x+h) g(x)
f (v) , which is Equation 3 in the case where h<0 .
h this inequality by h : f (u)
h 0 , and hence f (u)
64.
d
dx h(x)
g(x) d
dx
d
=
dx f (t)dt = a
g(x) h(x) f (t)dt+ g(x)
a a f (t)dt + f (t)dt d
dx / (where a is in the domain of f ) h(x)
a / / f (t)dt = f ( g(x) ) g (x)+ f ( h(x) ) h (x) / = f ( h(x) ) h ( x ) f ( g(x) ) g (x)
65. (a) Let f (x)= x
so 1+x / ( 3 1 and since f is increasing, this means that f 1+x 2 g(t)=t t / g (t)=2t 1 ( 1+x3) 1+x / g (t)>0 when t
3 3 0 for x 0 . Therefore, 1 1 1 0 0 3 1+x dx x+ 1 4
x
4
x 66. (a) If x<0 , then g(x)=
If 0 x 1 , then g(x)= x
0 0 1+x 1 1 0 0 1dx (b) From part (a) and Property 7:
1 1 1 0 f (t)dt= x
0 3 ) f (1) 1 . Thus, g is increasing on ( 1, 1 . Now let t= 1+x , where x 0 . g(t) 0 when t x f is increasing on ( 0, f (x)=1/(2 x )>0 for x>0 3 1+x 1+x 3 3 0, 1 for x 0 . Next let ) . And since g(1)=0 , 1 (from above) t 1 g(t) 0 3 1+x for x 0 . 3 1+x dx
3 1+x dx 1+ 0 3 ) . If x 0 , then x ( 1+x3) dx
0 1 1
=1.25 .
4 0dt=0 . x f (t)dt= t dt=
0 1 2
t
2 x = 0 1 2
x .
2 If 1<x 2 , then
g(x) = x
0 f (t)dt=
x 1
0 f (t)dt+ =g(1)+ (2 t)dt=
1 x
1 f (t)dt 1 2
1 2
(1) + 2t
t
2
2 x
1 13 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus = 1
1 2
+ 2x
x
2
2 If x>2 , then g(x)= g(x)= { x
0 2 1
2 =2x 1 2
x 1.
2 x f (t)dt=g(2)+ 0dt=1+0=1 . So
2 if
0
1 2
x
if
2
1 2
2x
x 1 if
2
1
if x<0
0 x 1
1<x 2
x>2 (b) (c) f is not differentiable at its corners at x=0 , 1 , and 2 . f is differentiable on (
( 1,2 ) and ( 2, ) . g is differentiable on ( , ) .
67. Using FTC1, we differentiate both sides of 6+ x f (t) a 2 t dt=2 x to get 2 =2 x To find a , we substitute x=a in the original equation to obtain 6+ a f (t) a 2 t dt=2 a 1
2 x f (x)=x 3/2 . 6+0=2 a a=9 . 3= a 68. B=3A
b f (x) ,0 ) , ( 0,1 ) , a e =3e 2 b x a x 0 0 e dx=3 e dx ( a b=ln 3e 2 69. (a) Let F(t)= t
0 e x b x a 0 0 =3 e b ( a e 1=3 e 1 ) )
/ f (s)ds . Then, by FTC1, F (t)= f (t)= rate of depreciation, so F(t) represents the loss in value over the interval 0,t .
14 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus t
1
A+F(t)
A+ f (s)ds =
represents the average expenditure per unit of t during the
0
t
t
interval 0,t , assuming that there has been only one overhaul during that time period. The company
wants to minimize average expenditure.
t
t
1
1
/
1
(c) C(t)=
A+ f (s)ds . Using FTC1, we have C (t)=
A+ f (s)ds + f (t) .
0
0
2
t
t
t
t
t
1
/
A+ f (s)ds =C(t) .
C (t)=0 t f (t)=A+ f (s)ds f (t)=
0
0
t (b) C(t)= 70. (a) C(t)=
/ C (t)= 1
t t f (s)+g(s) ds . Using FTC1 and the Product Rule, we have 0 1
f (t)+g(t)
t
1
t f (t)+g(t) t
0 1 t 2 0 t / f (s)+g(s) ds . Set C (t)=0 : f (s)+g(s) ds=0 (b) For 0 t 30 , we have D(t)= So D(t)=V V
V 2
t
t =V
15 900 f (t)+g(t) C(t)=0
V
15 t
0
2 60t t =900 V
s
450 ds= 1
f (t)+g(t)
t 1 t 2 0 t f (s)+g(s) ds=0 C(t)= f (t)+g(t) .
V
V 2
s
s
15 900 t V
V 2
t
t .
15 900 = 0 2 t 60t+900=0 2 ( t 30 ) =0 t=30 . So the length of time T is 30 months.
(c)
1
t
1
=
t C(t) = t
0 V
15 V
V
1
V
V 2
V
2
3
s+
s ds=
s
s+
s
450
12,900
t
15 900
38,700
V
V 2
V
V
V
V
3
2
t
t +
t =
t+
t
15 900
38,700
15 900 38,700 t
0 V
V
1
1
t=21.5 .
+
t=0 when
t=
900 19,350
19,350 900
V
V
V
V
2
C(21.5)=
(21.5)+
(21.5) 0.05472V , C(0)=
0.06667V , and
15 900
38,700
15
V
V
V
2
C(30)=
(30)+
(30) 0.05659V , so the absolute minimum is C(21.5) 0.05472V .
15 900
38,700
V
V
V
2
(d) As in part (c), we have C(t)=
t+
t , so C(t)= f (t)+g(t)
15 900 38,700
V
V
V
V
V
1
1
1
1
2 V
2
2
t+
t =
t+
t t
=t
15 900 38,700
15 450 12,900
12,900 38,700
450 900
1/900
43
t=
=
=21.5 . This is the value of t that we obtained as the critical number of C in part (c),
2/38,700 2
/ C (t)= 15 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.3 The Fundamental Theorem of Calculus so we have verified the result of (a) in this case. 16 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem 1. 2. d
dx d
dx 2 x +1 +C = ( x2+1) 1/2+C = 1 2
x +1
2 ( x ) 1/2 2x= 2 x +1 d
xsin x+cos x+C =xcos x+ ( sin x ) 1 sin x=xcos x
dx d
3.
dx x
2 +C 2 a = 2 2 2 a x 1 x( x) 2 2 a a x / 2 2 a x = 2 a x 1
2 a ( a2 x2) +x2 = 1
( a2 x2) 3/2 ( a2 x2) 3 4.
2 2 x +a d
dx = +C 2 2 a x a ( x2+a2) 2 x = 2 2 2 a x 5. x 6. 7. 3 ( 3/4 3/4+1 2 2 2 x x/ x +a = ) 2 2 x +a 1 2 2 a x
1 =
2 x +a ( 2 2 x 2 x +a 1/4 x
x
1/4
dx=
+C=
+C=4x +C
3/4+1
1/4
1/3 x dx= x 4/3 x
3 4/3
dx=
+C= x +C
4/3
4
4 2 x
x
1 4 2
x +6x+1 dx= +6 +x+C= x +3x +x+C
4
2
4 ) 3 ( 4 ) 8. x 1+2x dx= ( 2 ) 9. (1 t) 2+t dt= 10. 2 x +a
x d
dx 1 2 x +1+ 1
2 x +1 ( 2 6 x
x
1 2 1 6
x+2x dx= +2 +C= x + x +C
2
6
2
3
5 ( ) 2 3 t
t
2 2t+t t dt=2t 2 +
2 3
2 3 ) 4 t
2 1 3 1 4
+C=2t t + t
t +C
4
3
4 3 x
1
dx= +x+tan x+C
3
1 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem ) 2 3/2 2 x
x
8 3/2 1 2
dx= ( 4 4 x +x ) dx=4x 4
+ +C=4x
x + x +C
3/2 2
3
2 11. (2 12. ( 3e u + sec2 u ) du = 3e u + tan u + C x sin x 13. 2 sin x dx= 2 1 sin x
14. dx= cos x 1
sin x
dx= sec x tan xdx=sec x+C
cos x cos x sin 2x
2sin xcos x
dx=
dx= 2cos xdx=2sin x+C
sin x
sin x 2 5/2
x +C .
5
The members of the family in the figure correspond to C=5 , 3 , 0 , 2 , and 4 .
15. x x dx= x 3/2 dx= 16. (cos x 2sin x)dx=sin x+2cos x+C .
The members of the family in the figure correspond to C=5 , 3 , 0 , 2 , and 4 . 1 3
1 2
17. (6x 4x+5)dx= 6
x 4
x +5x
0
3
2
2 2 2 3 2 2 = 2x 2x +5x =(16 8+10) 0=18
0 0 18.
3 3 1 2
1 4 3
2 4
x 4
x
= x+x x
2
4
1
=(3+9 81) (1+1 1)= 69 1= 70 (1+2x 4x )dx = x+2
1 19. ( 2x ex) dx=
1 0 2 x e x 0 = ( 0 1) 1 3
1 ( 1 e 1) = 2+1/e
2 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem 20. 0 5 3 0 1 6 1 4 1 3
u
u+ u
6
4
3 2 (u u +u )du= 2 32
8
4
3
3 =0 2 = 4 21.
2 2 1 3
1 2
u +6 u +u
3
2
=(24+12+2) ( 24+12 2)=38 ( 14)=52 2 (3u+1) du =
2 ( 9u2+6u+1) du=
2 2 9 3 2 = 3u +3u +u 2
2 2 22.
4 4 4 1 3
1 2
v +13 v 5v
3
2 2 (2v+5)(3v 1)dv = (6v +13v 5)dv= 6
0
0 3 = 2v + 0 4 13 2
v 5v
2 0 =(128+104 20) 0=212
23. 24. 25. 26. 27. 28.
29. 4 4 1/2 t (1+t)dt= (t 1 1 9 2t dt= 0
1
2 2 0 3 4y + y
x 9 3 2 t 0 2
y y+5y 1 1 +t )dt= dt= 2 2 4 1 1 0 4/3 1 y +5 2 16 64
+
3
5 = 1 1 y 2 1 5
y
5 0 1 1 4 = =(1 1) 2 y
2
1 16 2
2 1 5
+y
y = 1 3 7/3 4 9/4
x + x
7
9 5/4 (x +x )dx= 2 2
+
3 5 = = 0 1
4 1
+32
2 = 1 3 4
+
7 9 0= = 1 5/x dx= 5 4
1 x 1/2 dx= 5 2 x 63
4 ( 1+1 ) = 63
2 55
63 5
( 2ex+4cos x) dx= 2ex+4sin x 0= ( 2e5+4sin 5) ( 2e0+4sin 0) =2e5+4sin 5 2
0
5 4 14 62 256
+
=
3
5
15 2
27 0=18 2
3 = 2 1 4
1
y +2
y
4
2 dy= (y +5y )dy= ( 3 x + 4 x ) dx= 9 2 3/2
t
3 2 dy= 4 3 7 1/2 4 2 3/2 2 5/2
t + t
3
5 3/2 290.99 4 = 5 (2 2 2 1)=2 5 1 30.
9
1 3x 2
9
dx = (3x1/2 2x
1
x 1/2 )dx= 3 2 3/2
1/2
x 2 2x
3 9 = 2x 1 3/2 1/2 9 4x 1 3 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem =(54 12) (2 4)=44
31. 0 (4sin 3cos )d = 4cos /3 32. sec 0 /3 tan d = sec /4 3sin =sec sec 3 /4 = ( 4 0 ) ( 4 0 ) =8
=2 4 2 33.
1+cos /4
0 2 d 2 cos 1 /4 = 0 cos = tan + + 2 cos
cos 2
2 /4 = tan 0 d = 4 + /4
0 ( sec 2 +1) d ( 0+0 ) =1+ 4 4 34.
2 sin +sin /3
0 sec tan d = 2 sin (1+tan /3
0 sec
0 ) 2 d = /3 sin 0 2 sec sec /3 d = 2 sin d 0 1
1
( 1)=
2
2 /3 = cos 2 = 35.
64 3 1+ x 1 x 1 dx = 1
1/2 x 1 1 23 + x dx= 1/2 x 1 6 4 64
1 64 6 5/6
x
5
2 1/2 = 2x + 36. 1/3 64 = 16+ (x 1/2 3 0 )dx= 192
5 (1+x ) dx= (1+3x +3x +x )dx= x+x + 0 (1/3) (1/2) +x 2+ 6
5 3 5 1 7
x+ x
5
7 64
1 =14+ (x 1/2 1/6 +x )dx 186 256
=
5
5 1 = 1+1+ 0 3 1
+
5 7 0= 96
35 37.
e
1 2 x +x+1
dx=
x e
1 x+1+ 1
x dx= 1 2
x +x+ln x
2 e = 1 1 2
e +e+ln e
2 1
1 2
1
+1+ln 1 = e +e
2
2
2 38.
9
4 x+ 1
x 2 dx= 9
4 x+2+ 1
x dx= 1 2
x +2x+ln x
2 9 = 4 81
85
9
+18+ln 9 (8+8+ln 4)=
+ln
2
2
4
4 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem 39.
2 0 ( x 2 x ) dx =
1 2 [x 2( x)]dx+ [x 2(x)]dx= 1 =3 0 0 1
2 (2 0)= 0 2 3xdx+ ( x)dx=3 1 0 1 2
x
2 0
1 1 2
x
2 2
0 7
= 3.5
2 40.
3 /2
0 sin x dx = sin xdx+ 3 /2 0 ( sin x ) dx= cos x 0 + cos x 3 /2 = 1 ( 1 ) + 0 ( 1 ) =2+1=3
2 4 41. The graph shows that y=x+x x has x intercepts at x=0 and at x=a 1.32 . So the area of the
region that lies under the curve and above the x axis is ( x+x2 x4) dx =
0
a 1 2 1 3
x+ x
2
3
1 2 1 3
=
a+ a
2
3
0.84 a 1 5
x
5
1 5
a
5 4 0 0 6 42. The graph shows that y=2x+3x 2x has x intercepts at x=0 and at x=a 1.37 . So the area of the
region that lies under the curve and above the x axis is ( 2x+3x4 2x6) dx =
0
a 3 5
x
5
2 3 5
= a+ a
5
2.18
2 x+ 2 7
x
7
2 7
a
7 a
0 0 5 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem 43. A= ( 2y y2) dy=
0 2 2 y 44. y= x 2 8
3 = 4 0 x=y , so A= y dy=
0 0=
1 1 5
y
5 1 4 4 4 1 3
y
3 = 0 4
3 1
.
5 / 45. If w (t) is the rate of change of weight in pounds per year, then w(t) represents the weight in
pounds of the child at age t . We know from the Net Change Theorem that 10
5 / w (t)dt=w(10) w(5) , so the integral represents the increase in the child’s weight (in pounds) between the ages of 5 and 10 .
b 46. a b / I(t)dt= Q (t)dt=Q(b) Q(a) by the Net Change Theorem, so it represents the change in the
a charge Q from time t=a to t=b .
/ 47. Since r(t) is the rate at which oil leaks, we can write r(t)= V (t) , where V (t) is the volume of oil
at time t . Thus, by the Net Change Theorem,
120 120 0 0 r(t)dt= / V (t)dt= V (120) V (0) =V (0) V (120) , which is the number of gallons of oil that leaked from the tank in the first two hours (120 minutes).
48. By the Net Change Theorem,
population in 15 weeks. So 100+
49. By the Net Change Theorem, 15
0
15
0 / n (t)dt=n(15) n(0)=n(15) 100 represents the increase in the bee
/ n (t)dt=n(15) represents the total bee population after 15 weeks. 5000
1000 / R (x)dx=R(5000) R(1000) , so it represents the increase in revenue when production is increased from 1000 units to 5000 units.
/ 50. The slope of the trail is the rate of change of the elevation E , so f (x)=E (x) . By the Net Change
Theorem, 5
3 5 / f (x)dx= E (x)dx=E(5) E(3) is the change in the elevation E between x=3 miles and
3 x=5 miles from the start of the trail.
6 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem b 51. In general, the unit of measurement for a f (x)dx is the product of the unit for f (x) and the unit for x . Since
100 f (x) is measured in newtons and x is measured in meters, the units for 0 f (x)dx are newton meters. (A newton meter is abbreviated N m and is called a joule.)
52. The units for a(x) are pounds per foot and the units for x are feet, so the units for da/dx are
pounds per foot per foot, denoted (lb / ft) / ft. The unit of measurement for 8
2 a(x)dx is the product of pounds per foot and feet; that is, pounds.
3 3 2
t 5t
2 3 53. (a) displacement = (3t 5)dt=
0 = 0 27
3
15=
m
2
2 (b)
3 5/3 3 0 0 5/3 distance traveled = |3t 5|dt=
= 5t (5 3t)dt+ 5/3 3 2
t
2 0 3 2
t 5t
2 + 3 = 5/3 25
3 1 3 2
t t 8t
3 ( t2 2t 8) dt=
1
6 54. (a) displacement = (3t 5)dt
3 25 27
+
15
2 9
2 6 3 25
2 9 25
3 = 41
m
6 1
10
1 8 =
m
3
3 = ( 72 36 48 ) 1 (b)
6 2 6 1 1 distance traveled = |t 2t 8|dt= |(t 4)(t+2)|dt
= 1 3 2
t +t +8t
3 ( t2+2t+8) dt+ 6 ( t2 2t 8) dt=
1
4
4 64
+16+32
3 =
/ 55. (a) v (t)=a(t)=t+4 v(t)= 1
+1+8 + ( 72 36 48 )
3 1 2
t +4t+C
2 v(0)=C=5 v(t)= 4 6
1 3 2
+
t t 8t
1
3
4
64
98
16 32 =
m
3
3 1 2
t +4t+5 m / s
2 (b)
distance traveled =
=
/ 10
0 1 2
t +4t+5 dt=
2 10
0 |v(t)|dt= 1 3 2
t +2t +5t
6 56. (a) v (t)=a(t)=2t+3 2 10
0 = v(t)=t +3t+C 10
0 1 2
t +4t+5 dt
2 500
2
+200+50=416 m
3
3
v(0)=C= 4 2 v(t)=t +3t 4
7 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem (b)
distance traveled =
= 3
0 2 t +3t 4 dt= 3 (t+4)(t 1) dt 0 3
( t2 3t+4) dt+ 1 ( t2+3t 4) dt
0 1 1
3
1 3 3 2
1 3 3 2
t
t +4t +
t + t 4t
3
2
0
3
2
1
1 3
27
1 3
89
+4 + 9+
12
+
4 =
m
3 2
2
3 2
6 =
= / 57. Since m (x)= (x) , m= 4
0 4 3/2
(x)dx= ( 9+2 x ) dx= 9x+ x
0
3
4 4 =36+ 0 32
140
2
0=
=46 kg.
3
3
3 58. By the Net Change Theorem, the amount of water that flows from the tank is
10 10 0 0 r(t)dt= (200 4t)dt= 200t 2t 2 10
0 = ( 2000 200 ) 0=1800 liters. 59. Let s be the position of the car. We know from Equation 2 that s(100) s(0)= 100
0 v(t)dt . We use the Midpoint Rule for 0 t 100 with n=5 . Note that the length of each of the five time intervals is
20
1
20 seconds =
hour=
hour. So the distance traveled is
3600
180
100
1
v(t)dt
[v(10)+v(30)+v(50)+v(70)+v(90)]
0
180
1
=
(38+58+51+53+47)
180
247
=
1.4miles
180
60. (a) By the Net Change Theorem, the total amount spewed into the atmosphere is
6 Q(6) Q(0)= r(t)dt=Q(6) since Q(0)=0 . The rate r(t) is positive, so Q is an increasing function.
0 Thus, an upper estimate for Q(6) is R and a lower estimate for Q(6) is L .
6 R=
6 L=
6 (b) 6 6 t= b a 6 0
=
=1 .
n
6 ( ) t=10+24+36+46+54+60=230 tonnes.
r ( t ) t=R +r(0) r(6)=230+2 60=172 tonnes.
r t i=1
6 i i=1 i 1 6 b a 6 0
t=
=
=2 . Q(6) M =2[r(1)+r(3)+r(5)]=2(10+36+54)=2(100)=200 tonnes.
3
n
3
8 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.4 Indefinite Integrals and the Net Change Theorem 61. From the Net Change Theorem, the increase in cost if the production level is raised from 2000
yards to 4000 yards is C(4000) C(2000)=
4000 / C (x)dx = 2000 = 4000 / C (x)dx . 2000 ( 3 0.01x+0.000006x2) dx
2000
4000 2 3x 0.005x +0.000002x 3 4000 =60,000 2,000=$58,000 2000 62. By the Net Change Theorem, the amount of water after four days is
4 25 , 000+M 25 , 000+ r(t)dt
0 4 4 0
r(0.5)+r(1.5)+r(2.5)+r(3.5)
4
25 , 000+ 1500+1770+740+ ( 690 ) =28 , 320 liters =25 , 000+ 63. (a) We can find the area between the Lorenz curve and the line y=x by subtracting the area under
y=L(x) from the area under y=x . Thus,
1 area between Lorenz curve and line y=x
coefficient of inequality =
=
area under line y=x x L(x) dx 0 1
0 1 = 0 1 x L(x) dx
2 x /2 1 0 = x L(x) dx
=2 1/2 1 xdx x L(x) dx 0 0 5 2 7
1
5
7 19
x+
x L(50%)=L
=
+
=
=0.39583 , so the bottom 50% of the
12
12
2
48 24 48
households receive at most about 40% of the income. Using the result in part (a),
1
1
5 2 7
coefficientofinequality = 2 x L(x) dx=2
x
x
x dx
0
0
12
12
1
1 5
5
5 2
2
= 2
x
x dx=2
x x dx
0
0 12
12
12
(b) L(x)= ( = 5
6 1 1 2 1 3
x
x
2
3 = 0 5
6 1
2 3 1
3 ) = 5
6 1
6 = 5
36 2 64. (a) From Exercise 4.1. (a), v(t)=0.00146t 0.11553t +24.98169t 21.26872 .
(b) h(125) h(0)= 125
0 4 3 2 v(t)dt= 0.000365t 0.03851t +12.490845t 21.26872t 125
0 206 , 407 ft
9 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule 1
du . Thus,
3
1
1
1
1
cos 3xdx= cos u
du =
cos udu= sin u+C= sin 3x+C . Don’t forget that it is often very
3
3
3
3
easy to check an indefinite integration by differentiating your answer. In this case,
d
1
1
sin 3x+C = (cos 3x) 3=cos 3x , the desired result.
dx
3
3
1. Let u=3x . Then du=3dx , so dx= 1
du , so
2
1
1 1 11
1
2
du =
u +C=
4+x
2
2 11
22 2 2. Let u=4+x . Then du=2xdx and xdx=
2 10 ( ) x 4+x dx= u ( 10 3 2 2 3. Let u=x +1 . Then du=3x dx and x dx=
2 x ) 11+C . 1
du , so
3 3/2 3 x +1 dx= u 1
1 u
1 2 3/2
2 3 3/2
du =
+C=
u +C= (x +1) +C .
3
3 3/2
3 3
9 4. Let u= x . Then du=
sin x 1
2 x 1
dx=2du , so
x dx and dx= sin u (2du)=2( cos u)+C= 2cos x +C . x 5. Let u=1+2x . Then du=2dx and dx= 1
du , so
2 2 4
3 dx=4 u (1+2x) 6. Let u=sin 3 1
u
1
1
+C=
+C .
du =2
+C=
2
2
2
2
u
(1+2x)
sin . Then du=cos d , so e u u sin cos d = e du=e +C=e +C . 1 5
1 2 5
u +C= (x +3) +C .
5
5 2 2 3 1
du , so
3
1
1 1 10
1 3 10
du =
u +C=
(x +5) +C .
3
3 10
30 4 4 7. Let u=x +3 . Then du=2xdx , so 2x(x +3) dx= u du=
2 2 8. Let u=x +5 . Then du=3x dx and x dx=
2 3 9 x (x +5) dx= u 9 9. Let u=3x 2 . Then du=3dx and
1 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule 1
20
20
du , so (3x 2) dx= u
3 dx= 1
1 1 21
1
21
du =
u +C=
(3x 2) +C .
3
3 21
63
6 6 10. Let u=2 x . Then du= dx and dx= du , so (2 x) dx= u ( du)= 1 7
1
7
u +C=
(2 x) +C .
7
7 2 11. Let u=1+x+2x . Then du=(1+4x)dx , so
1+4x du
= u
u dx=
2 1+x+2x 1/2 1/2 u
2
du=
+C=2 1+x+2x +C .
1/2 2 1
du , so
2
1
1
1
+C=
+C=
+C .
2
u
2u
2(x +1) 12. Let u=x +1 . Then du=2xdx and xdx=
x
2 2 dx= u 1
1
du =
2
2 2 (x +1) 1
du , so
3
1
1
1
du =
ln u +C=
ln 5 3x +C .
3
3
3 13. Let u=5 3x . Then du= 3dx and dx=
dx
1
=
5 3x
u
2 14. Let u=x +1 . Then du=2xdx and xdx=
x dx= 2 x +1 1
du
2
1
1
1
2
2
2
= ln u +C= ln x +1 +C= ln x +1 +C [ since x +1>0 ]
u
2
2
2 ( 1/2 2 or ln (x +1) +C=ln 5 dy= 3u 5 (2y+1) ) 2 x +1 +C . 15. Let u=2y+1 . Then du=2dy and dy=
3 1
du , so
2 1
3
du =
2
2 1
du , so
2 1 4
u +C=
4 3
4 +C . 8(2y+1) 1
du , so
5
1
1
1.7
u +C=
u
1.7
8.5 16. Let u=5t+4 . Then du=5dt and dt=
1
2.7 (5t+4) dt= u 2.7 1
1
du =
5
5 1.7 +C= 2
1.7 +C . 17(5t+4) 2 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule 17. Let u=4 t . Then du= dt and dt= du , so
4 4 t dt= u 1/2 ( du)= 2 3/2
2
3/2
u +C=
(4 t) +C .
3
3 1
du , so
8
1
1 2 3/2
1
4
3/2
du =
u +C=
(2y 1) +C .
8
8 3
12
3 3 18. Let u=2y 1 . Then du=8y dy and y dy=
y 3 4 2y 1 dy= u 1/2 19. Let u= t . Then du= dt and dt=
sin t dt= sin u 1 du = 1 1 1 ( cos u)+C= 20. Let u=2 . Then du=2d and d =
tan u du , so
cos t+C . 1
du , so sec 2 tan 2 d = sec u
2 1
1
1
du = sec u+C= sec 2 +C .
2
2
2 dx
21. Let u=ln x . Then du=
, so
x
1 22. Let u=tan x . Then du= x 2
2 dx= u du=
1 dx
2 1+x
23. Let u= t . Then du= ( ln x ) , so tan x
2 1+x 1 3
1
3
u +C= ( ln x ) +C .
3
3
2 ( 1 u
tan x
dx= udu= +C=
2
2 dt
1
and
dt=2du , so
2 t
t cos t ) 2 +C . dt= cos u (2du)=2sin u+C=2sin t +C t .
24. Let u=1+x 3/2 3/2 . Then du= x sin (1+x )dx= sin u 25. Let u=sin 26. Let u=1+tan 3 1/2
2
x dx and x dx= du , so
2
3
2
2
2
3/2
du = ( cos u)+C=
cos (1+x )+C .
3
3
3 . Then du=cos d , so cos sin . Then du=sec 2 6 6 d = u du=
5 d , so (1+tan ) sec 2 1 7
1
7
u +C= sin +C .
7
7
5 d = u du= 1 6
1
6
u +C= (1+tan ) +C .
6
6
3 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule x x 27. Let u=1+e . Then du=e dx , so e
2 x x x 1+e dx= u du= 2 3/2
2
x 3/2
u +C= (1+e ) +C .
3
3 x x Or: Let u= 1+e . Then u =1+e and 2udu=e dx , so
2 3
2
x
x
x 3/2
e 1+e dx= u 2udu= u +C= (1+e ) +C .
3
3
28. Let u=cos t . Then du= sin t dt and sin t dt= du , so e
3 2 2 29. Let u=1+z . Then du=3z dz and z dz=
2 z dz= u 3 1+z cos t 2 2 ax +2bx+c cos t +C . 1
du , so
3 30. Let u=ax +2bx+c . Then du=2(ax+b)dx and (ax+b)dx= = u 1
1 3 2/3
1
3 2/3
du =
u +C= (1+z ) +C .
3
3 2
2 1/3 3 (ax+b)dx u sin t dt= e ( du)= e +C= e 1
du
2
1
=
u
2
u 31. Let u=ln x . Then du= x 1/2 1/2 1
du , so
2 2 du=u +C= ax +2bx+c +C . dx
, so
x dx
du
=
=ln u +C=ln ln x +C .
xln x
u x 32. Let u=e +1 . Then du=e dx , so e
x x dx= e +1
2 du
x
=ln u +C=ln e +1 +C .
u ( ) 2 33. Let u=cot x . Then du= csc xdx and csc xdx= du , so
2 3/2 cot x csc xdx= 34. Let u= x cos ( /x )
2 x u
2
3/2
+C=
(cot x) +C .
u ( du)=
3/2
3 . Then du=
dx= cos u 2 x
1 dx and
du = 1
2 x
1 dx= 1 du , so sin u+C= 1 sin x +C . 35.
4 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule cos x
1
dx . Let u=sin x . Then du=cos xdx , so cot xdx=
du=ln u +C=ln sin x +C .
sin x
u cot xdx= 36. Let u=cos x . Then du= sin xdx and sin xdx= du , so
sin x
1
1
du
dx=
= tan u+C= tan (cos x)+C .
2
2
1+cos x
1+u
37. Let u=sec x . Then du=sec x tan xdx , so
3 2 1 3
1
3
u +C= sec x+C .
3
3 2 sec x tan xdx= sec x(sec x tan x)dx= u du=
3 3 2 38. Let u=x +1 . Then x =u 1 and du=3x dx , so
3 3 3 5 x +1 x dx = 1
3 = 39. Let u=b+cx
x 3 3 1
1
4/3 1/3
du =
(u u )du
3
3
1 3 7/3 1 3 4/3
+C= (x +1)
(x +1) +C
7
4 2 1/3 x +1 x x dx= u (u 1)
3 7/3 3 4/3
u
u
7
4 a+1 a . Then du=(a+1)cx dx , so
1
1
2 3/2
2
a+1
1/2
a+1
b+cx dx= u
du=
u
+C=
b+cx
(a+1)c
(a+1)c
3
3c(a+1) ( a ) 3/2+C . 40. Let u=cos t. Then du= sin t dt and sin t dt= du , so
2 2 sin t sec (cos t)dt= sec u ( du ) = tan u+C= tan (cos t)+C .
2 41. Let u=1+x . Then du=2xdx , so
1+x
2 dx= 1+x 1 =tan x+ 1
2 1+x dx+ x 1 2 dx=tan x+ 1+x 1
du
2
1
1
=tan x+ ln u +C
u
2 1
1
2
2
1
2
ln 1+x +C=tan x+ ln 1+x +C [ since 1+x >0 ].
2
2 ( 2 42. Let u=x . Then du=2xdx , so x
1+x 4 ) dx= 1
du
2
1+u 2 = 1
1
1
tan u+C= tan
2
2 1 ( x2) +C . 43. Let u=x+2 . Then du=dx , so
5 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule x
4 x+2 u 2 dx = 4 du= u ( u3/4 2u 1/4) du= 4 u7/4 2
7 4 3/4
u +C
3 4
7/4 8
3/4
(x+2)
(x+2) +C
7
3 = 44. Let u=1 x . Then x=1 u and dx= du , so
2 2 x
dx =
1 x
= 45. f (x)= 2 (1 u)
1 2u+u
1/2
1/2 3/2
du=
u
2u +u
du
( du ) =
u
u
2 3/2 2 5/2
4
1/2
3/2 2
5/2
2u 2 u + u
+C= 2 1 x + (1 x)
(1 x) +C
3
5
3
5
3x 1 ( 3x 2 2x+1 ) 4 ( ) . 2 u=3x 2x+1
3x 1 ( du=(6x 2)dx=2(3x 1)dx , so
1
1
4
1
dx =
du =
u du
4
4
2
2
2
u
3x 2x+1
1 3
1
+C
= 6 u +C=
3
2
6 3x 2x+1 ) ( Notice that at x= 46. f (x)= x
2 ) 1
, f changes from negative to positive, and F has a local minimum.
3
2 . u=x +1 du=2xdx , so x +1
x
2 x +1 dx = 1
u 1
1
du =
u
2
2 1/2 du 6 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule = u1/2+C= x2+1 +C.
Note that at x=0 , f changes from negative to positive and F has a local minimum. 3 47. f (x)=sin x cos x . u=sin x 3 du=cos xdx,so 3 sin x cos xdx= u du= 1 4
1
4
u +C= sin x+C
4
4 Note that at x= , f changes from positive to negative and F has a local maximum. Also, both f
2
and F are periodic with period , so at x=0 and at x= , f changes from negative to positive and F
has local minima. 48. f ( )=tan 2 sec 2 . u=tan
tan du=sec
2 sec 2 2 d , so
2 d = u du= 1 3
1
3
u +C= tan +C
3
3 Note that f is positive and F is increasing. At x=0 , f =0 and F has a horizontal tangent. 49. Let u=x 1 , so du=dx . When x=0 , u= 1 ; when x=2 , u=1 . Thus,
Theorem 7(b), since f (u)=u 25 2 25 (x 1) dx= 0 1 25 u du=0 by 1 is an odd function.
7 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule 50. Let u=4+3x , so du=3dx . When x=0 , u=4 ; when x=7 , u=25 . Thus,
7 4+3x dx= 0 25
4 u 3 25 3/2 1
1
du =
3
3 u
3/2 4 = 2
234
3/2 3/2 2
(25 4 )= (125 8)=
=26
9
9
9 2 51. Let u=1+2x , so du=6x dx . When x=0 , u=1 ; when x=1 , u=3 . Thus, ( 1+2x3) 5 dx= 1 2
0 x 1
1
du =
6
6 3 5 u 1 3 1 6
u
6 = 1 2 52. Let u=x , so du=2xdx . When x=0 , u=0 ; when x=
1
1
1
2
xcos x dx= cos u
du =
sin u = (sin
0 2
0
0
2
2 ( ) 53. Let u=t/4 , so du=
/4 2 0 sec (t/4)dt= 0 1 6 6 1
728 182
(3 1 )=
(729 1)=
=
36
36
36
9
, u= . Thus,
1
sin 0)= (0 0)=0 .
2 1
dt . When t=0 , u=0 ; when t= , u= /4 . Thus,
4
2 sec u (4du)=4 tan u /4
0 =4 tan 4 tan 0 =4(1 0)=4 . 1
1
, u= ; when t= , u= . Thus,
6
6
2
2
/2
1
1
1
1
/2
csc u cot u
du =
csc u
=
(1 2)=
. 54. Let u= t , so du= dt . When t=
1/2 csc t cot t dt= 1/6 /6 55. tan 3 /6
2 56. 0 dx
2 /6 /6 d =0 by Theorem (b), since f ( )=tan
1 does not exist since f (x)= (2x 3) 2 3 is an odd function. has an infinite discontinuity at x= (2x 3)
2 57. Let u=1/x , so du= 1/x dx . When x=1 , u=1 ; when x=2 , u=
2
1 1/x e 2 x dx= 1/2 u
1 e ( du)= u 1/2 e 1 1/2 = (e e)=e 3
.
2 1
. Thus,
2 e . 2 58. Let u= x , so du= 2xdx . When x=0 , u=0 ; when x=1 , u= 1 . Thus,
2
1
1 u
1
1 u 1 1
1
x
1 0
e
e e = (1 1/e) .
xe dx= e
du =
=
0
0
0
2
2
2
2 ( ) 8 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule 59. Let u=cos
/3
0 sin
cos
/2 60. /2 , so du= sin d . When =0 , u=1 ; when = d = 2 1/2
1 du
u 2 = 1 2 u du= 1/2 1
u 1+x 6 1
. Thus,
2 1 = 1 ( 2)=1 . 1/2 2 x sin x 3 , u= 2 dx=0 by Theorem (b), since f (x)= x sin x
1+x 6 is an odd function. 61. Let u=1+2x , so du=2dx . When x=0 , u=1 ; when x=13 , u=27 . Thus,
13
27 2/3
1
1
1/3 27 3
dx
= u
du =
3u
= (3 1)=3 .
0 3
1
2
2
1
2
2
( 1+2x )
62. Let u=sin x , so du=cos xdx . When x=0 , u=0 ; when x=
/2
0 1 , u=1 . Thus, 1 0 2 0 cos x sin (sin x)dx= sin udu= cos u = (cos 1 1)=1 cos 1 . 63. Let u=x 1 , so u+1=x and du=dx . When x=1 , u=0 ; when x=2 , u=1 . Thus,
2
1
1 3/2 1/2
2 5/2 2 3/2 1 2 2 16
x x 1 dx= (u+1) u du= (u +u )du=
u + u
= + =
.
1
0
0
5
3
0 5 3 15
64. Let u=1+2x , so x=
4
0 1
(u 1)
9 2
1
2 3/2 1/2
du 1 9 1/2 1/2
=
(u u )du=
u 2u
1
2 4 1
4
3
u
1 2 3/2 1/2 9 1
20 10
=
u 3u
=
(27 9) (1 3) =
=
1 6
4 3
6
3 xdx
=
1+2x 65. Let u=ln x , so du=
e
e 4 1
(u 1) and du=2dx . When x=0 , u=1 ; when x=4 , u=9 . Thus,
2 4
dx
= u
1
x ln x
1 1/2 9
1 dx
4
. When x=e , u=1 ; when x=e ; u=4 . Thus,
x du=2 u 66. Let u=sin x , so du= 1/2 4 =2(2 1)=2 . 1 dx . When x=0 , u=0 ; when x=
2 1 x 1
, u= . Thus,
2
6 9 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule 1/2 1 sin x 0 dx= 2 udu= 0 1 x
67. dx 4
0 3 /6 2 /6 u
2 0 = 2 72 does not exist since f (x)= (x 2) . 1
3 has an infinite discontinuity at x=2 . (x 2)
2 2 2 68. Assume a>0 . Let u=a x , so du= 2xdx . When x=0 , u=a ; when x=a , u=0 . Thus,
2
2
a
1
1 a 1/2
1
2 3/2 a 1 3
2 2
0 1/2
du =
u du=
u
= a .
x a x dx= 2u
0
2
2 0
2
3
0 3
a
2 2 69. Let u=x +a , so du=2xdx and xdx=
2 a
0 2 2 x +a dx = x 2 a = 70. 2a a
a x 2 2 u 1/2 1
1
du =
2
2 1
2
2
du . When x=0 , u=a ; when x=a , u=2a . Thus,
2
2 3/2
u
3 2 2a
2 = a 1 3/2
u
3 2 2a
2 a 1
1
2 3/2
2 3/2
3
(2a ) (a ) = ( 2 2 1 ) a
3
3 x +a dx=0 by Theorem 7(b), since f (x)=x 2 2 x +a is an odd function. 71. From the graph, it appears that the area under the curve is about
1
1+ a little more than
1 0.7 , or about 1.4 . The exact area is given by A=
2
u=2x+1 , so du=2dx . The limits change to 2 0+1=1 and 2 1+1=3 , and
3
1
1
2 3/2 3 1
1
A=
u
du =
u
= ( 3 3 1) = 3
1.399 .
1
2
2
3
1 3
3 72. From the graph, it appears that the area under the curve is almost 1
2 1
0 2x+1 dx . Let 2.6 , or about 4 . The
10 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule exact area is given by
A= 0 (2sin x sin 2x)dx= 2 cos x 0 0 sin 2xdx = 2( 1 1) 0=4
Note: 0 that sin 2xdx=0 since it is clear from the graph of y=sin 2x
/2 sin 2xdx= /2 0 sin 2xdx . 73. First write the integral as a sum of two integrals: I=
x+ 2 2 3 4 x d x . I =0 by Theorem 7(b), since f (x)=x
1 2 2 2 2 (x+3) 4 x d x=I +I =
1 2 2
2 x 2 4 x d 2 4 x is an odd function and we are integrating from x= 2 to x=2 . We interpret I as three times the area of a semicircle with radius 2 ,
2 1
2 so I=0+3 ( 2 ) 2 =6 . 2 2 2 74. Let u=x . Then du=2xdx and the limits are unchanged ( 0 =0 and 1 =1 ), so
1
1 1
4
2
I= x 1 x dx=
1 u du . But this integral can be interpreted as the area of a quarter circle
0
2 0
1 1
1
2
with radius 1 . So I=
1 =
.
2 4
8 ( ) 75. First Figure
2 Let u= x , so x=u and dx=2udu . When x=0 , u=0 ; when x=1 , u=1 . Thus,
1 A= e
1 0 x 1 u 1 0 0 u dx= e (2udu)=2 ue du . Second Figure
1 x 1 u A = 2xe dx=2 ue du .
2 0 0 Third Figure
Let u=sin x , so du=cos xdx . When x=0 , u=0 ; when
11 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule x= 2 , u=1 . Thus, A =
3 /2 sin x e 0 sin 2xdx= /2 sin x
0 1 u 1 0 0 u (2sin xcos x)dx= e (2udu)=2 ue du . e Since A =A =A , all three areas are equal.
1 2 3 bt / 76. Let r(t)=ae with a=450.268 and b=1.12567 , and n(t)= population after t hours. Since r(t)=n (t)
3 r(t)dt=n(3) n(0) is the total change in the population after three hours. Since we start with 400 , 0 bacteria, the population will be
3 3 0 0 bt n(3)=400+ r(t)dt=400+ ae dt=400+ a bt 3
a 3b
e
=400+
e 1
0
b
b ( ) 400+11,313=11,713 bacteria 77. The volume of inhaled air in the lungs at time t is
t
t 1
2 t/5 1
2
5
2
2
V (t) =
f (u)du=
sin
u du=
sin v
dv v=
u,dv=
du]
0
02
0
5
2
2
5
5
5
2
5
2
2 t/5 5
=
cos v
=
cos
t +1 =
1 cos
t
liters
0
4
4
5
4
5
78.
4 Number of calculators =x(4) x(2)= 5000 1 100(t+10)
2 =5000 t+100(t+10) 79. Let u=2x . Then du=2dx , so
2 80. Let u=x . Then du=2xdx , so 2
0 f (2x)dx= 3
0 1 4 =5000 2
4
0 f (u) ( 2) xf x dx= 9
0 f (u) 2 dt
4+ 100
14 1
1
du =
2
2 2+
4
0 1
1
du =
2
2 100
12 f (u)du=
9
0 4048 1
(10)=5 .
2 f (u)du= 1
(4)=2 .
2 81. Let u= x . Then du= dx , so
b
a f ( x)dx= b
a f (u)( du)= a
b f (u)du= a
b f (x)dx From the diagram, we see that the equality follows from the fact that we are reflecting the graph of f ,
and the limits of integration, about the y axis. 12 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule 82. Let u=x+c . Then du=dx , so
b
a b+c f (x+c)dx= a+c f (u)du= b+c
a+c f (x)dx From the diagram, we see that the equality follows from the fact that
we are translating the graph of f , and the limits of integration, by a distance c . 83. Let u=1 x . Then x=1 u and dx= du , so
1 a
0 b x (1 x) dx= 84. Let u=
0 0 = 85. 0 1 b a a x . Then du= dx . When x= , u=0 and when x=0 , u= . So xf (sin x)dx = 2 1 b a b ( 1 u ) u ( du)= 0u (1 u) du= 0x (1 x) dx .
1 xf (sin x)dx= xsin x ( 0 u) f (sin (
f (sin u)du 0 f (sin x)dx sin x u))du=
0 0 ( u) f (sin u)du uf (sin u)du=
0 xf (sin x)dx= 2 =xf (sin x) , where f (t)= 0 2 f (sin x)dx
0 0 xf (sin x)dx f (sin x)dx . t . By Exercise 84,
2
1+cos x
2 sin x
2 t
xsin x
sin x
dx= xf (sin x)dx=
f (sin x)dx=
dx
2
2
0
0
2 0
2 0
1+cos x
1+cos x
Let u=cos x . Then du= sin xdx . When x= , u= 1 and when x=0 , u=1 . So
1 du
1 du
sin x
1 1
dx=
=
=
tan u =
2
1
2
2
2 0
2 1
2 1
2
1+cos x
1+u
1+u
2 =x 0 13 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.5 The Substitution Rule 1 2 2
1 tan 1 tan ( 1) = 2 4 4 = 4 14 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.6 The Logarithm Defined as an Integral 1. (a)
We interpret ln 1.5 as the area under the curve y=1/x from x=1 to x=1.5 . The area of the rectangle
1 2 1
1 1
2
5
BCDE is
= . The area of the trapezoid ABCD is
1+
=
. Thus, by comparing
2 3 3
2 2
3
12
1
5
areas, we observe that <ln 1.5<
.
3
12
(b) With f (t)=1/t , n=10 , and
ln 1.5 = x=0.05 , we have 1.5
1 (1/t)dt (0.05)[ f (1.025)+ f (1.075)+ = (0.05) 1
1
+
+
1.025 1.075 + 1
1.475 + f (1.475)]
0.4054 1
1/2 1
1
/
1
,y =
. The slope of AD is
=
. Let c be the t coordinate of the point on
2
t
2 1
2
t
1
1
1
2
1
y= with slope
. Then
=
c =2 c= 2 since c>0 . Therefore, the tangent line is given
2
t
2
2
c
1
1
1
by y
=
( t 2 ) y= 2 t+ 2 .
2
2
2. (a) y= (b)
Since the graph of y=1/t is concave upward, the graph lies above the tangent line, that is, above the
1
+ 2 and CD = 1+ 2 . So the area of the trapezoid ABCD is
line segment BC . Now AB =
2
1
1
3
+ 2 + ( 1+ 2 ) 1 =
+ 2 0.6642 . So ln 2> area of trapezoid ABCD>0.66 .
2
2
4
3. 1 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.6 The Logarithm Defined as an Integral The area of R is
i The area of S is
i 1 1
Thus, + +
2 3 1
1 1
and so + +
i+1
2 3 + 1
<
n n
1 1
dt=ln n .
t n1
1
1
1
and so 1+ +
+
>
dt=ln n .
i
2
n 1 1t
1
1
1
+ <ln n<1+ +
+
.
n
2
n 1 4. (a) From the diagram, we see that the area under the graph of y=1/x between x=1 and x=2 is less
2 than the area of the square, which is 1 . So ln 2= (1/x)dx<1 . To show the other side of the
1 inequality, we must find an area larger than 1 which lies under the graph of y=1/x between x=1 and
x=3 . One way to do this is to partition the interval 1,3 into 8 intervals of equal length and calculate
the resulting Riemann sum, using the right endpoints: 1
4 1
1
1
1
1
1
1
1
+
+
+ +
+
+
+
5/4 3/2 7/4 2 9/4 5/2 11/4 3 = 28,271
>1
27,720 3 and therefore 1< (1/x)dx=ln 3 .
1 A slightly easier method uses the fact that since y=1/x is concave upward, it lies above all its tangent
3 2
5 2
lines. Drawing two such tangent lines at the points
and
, we see that the area
,
,
2 3
2 5
under the curve from x=1 to x=3 is more than the sum of the areas of the two trapezoids, that is,
2 2 16
16 3
+ =
. Thus, 1<
< (1/x)dx=ln 3 .
3 5 15
15 1
(b) By part (a), ln 2<1<ln 3 . But e is defined such that ln e=1 , and because the natural logarithm
function is increasing, we have ln 2<ln e<ln 3 2<e<3 . ( r ) , then f /(x)= ( 1/xr ) (rxr 1) =r/x . But if g(x)=rln x , then g /(x)=r/x . So f and g
r
r
r
must differ by a constant: ln ( x ) =rln x+C . Put x=1 : ln ( 1 ) =rln 1+C C=0 , so ln ( x ) =rln x .
5. If f (x)=ln x 2 Stewart Calculus ET 5e 0534393217;5. Integrals; 5.6 The Logarithm Defined as an Integral 6. Using the second law of logarithms and Equation 10, we have ( x y ) x ( x y) . Since ln y x y is a one to one function, it follows that e /e =e ln e /e =ln e ln e =x y=ln e rx x x y ( x) r . Since ln 7. Using the third law of logarithms and Equation 10, we have ln e =rx=rln e =ln e
rx x r ( ) a one to one function, it follows that e = e .
is .
x 8. Using Definition 13 and the second law of exponents for e , we have
a ( x y ) ln a xln a
=e
=e x y yln a = e
e xln a
yln a = a x a y . x 9. Using Definition 13, the first law of logarithms, and the first law of exponents for e , we have
x (ab) =e xln (ab) =e x(ln a+ln b) =e xln a+xln b =e xln a xln b e x x =a b . r s 10. Let log x=r and log y=s . Then a =x and a =y .
a
r s r+s (a) xy=a a =a
r x a
r
(b)
= =a
s
y
a
y log (xy)=r+s=log x+log y
a s ( r ) y=ary (c) x = a a log
log a a a x
=r s=log x log y
a
a
y ( xy) =ry=ylog ax
a 3 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 1.
x=4 A=
x=0 4 ( y y ) dx= ( 5x x )
T B = 2x 4 4 0 64
3 = 32 0 ( 4x x2) dx x dx= 0 1 3
x
3 2 2 (0)= 32
3 2.
2 A= 0 2
2
3/2
(x+2) ln (x+1)
3
0
2 3/2
16
4
(2) ln 1 =
ln 3
2
3
3
3 1
x+1 x+2 dx= 2 3/2
(4) ln 3
3 =
3. y=1 A= y= 1 1 ( x x ) dy=
R L 1 ( ey y2+2) dy=
1 1 = e e y y ( y2 2) dy
1 1 3
y +2y
3 1 = e 1 1
+2
3 1 e + 1
1 10
2 =e
+
3
e 3 4.
3 A= ( 2y y ) ( y
2 2 4y 3 ) dy= 0 2 ) 2y +6y dy 0 2 3 2
y +3y
3 = ( 3 =( 18+27) 0=9 0 5.
2 A= ( 9 x2) ( x+1 ) dx 1
2 = ( 8 x x2) dx 1
2 = 3 x
8x
2 = 16 2 x
3
8
3 2
1 8 1 1
+
2 3
1 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves =22 3+ 1 39
=
2 2 6.
/2 A= 0 ( ex sin x) dx
/2 x = e +cos x ( = e
=e /2 /2 +0 ) 0 ( 1+1 ) 2 2 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 2 7. The curves intersect when x=x
1 2 x x=0 x(x 1)=0 x=0 , 1 . ( x x2) dx A=
0 1 2 1 3
x
x
2
3
1 1
=
2 3
1
=
6
= 1
0 8.
1 ( x2 x4) dx A=
1 1 =2 ( x2 x4) dx 0 =2
=2 1 3 1 5 1
x
x
3
5
0
1 1
4
=
3 5
15 3 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 9.
A= 2
1 1
x = ln 2+
=ln 2 1 dx= ln x+ 2 x
1
2 1
2 0
9 x
0 1 0.19 9 ( 1+ 2 ( ln 1+1 ) 3+x
x
10. 1+ x =
=1+
3
3
A= 1
x x ) 3+x
3 1
x dx=
3 x
x=
3 2 x
x=
9 2 9x x =0 9 ( 1+ dx= x ) 1+ 0 2 3/2 1 2
x
x
3
6 9 =18 0 x(9 x)=0
x
3 x=0 or 9 , so dx= 27 9
=
2 2 4 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 11.
1 A= ( 2 ) x x dx 0 2 3/2 1 3
=
x
x
3
3
2 1
=
3 3
1
=
3 3 3 1
0 3 2 12. x= x x =x x x=0
x= 1 , 0 , or 1 , so
1 0
3 A= x x dx= 1 1 (x 3 x(x 1)=0 ) 1 x dx+
0 ( 3 x(x+1)(x 1)=0 ) 1 x x dx=2 ( x1/3 x) dx 0 [by symmetry]
=2 3 4/3 1 2
x
x
4
2 1 =2 0 3
4 1
2 = 1
2 5 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 2 2 2 13. 12 x =x 6
3 2 2x =18 3 ( 12 x ) ( x 6)
2 A= 2 x =9 2 dx=2 3 x = 3 , so ( 18 2x2) dx [ by symmetry] 0 =2 18x 2 3
x
3 3 3 =2 ( 54 18 ) 0 =2(36)=72 0 3 2 14. x x=3x x 4x=0
By symmetry,
2 A= 3x ( x x)
3 x(x 4)=0
2 dx=2 2 3x x(x+2)(x 2)=0 ( x x)
3 2 dx=2 0 x=0 , 2 , or 2 . ( 4x x3) dx=2 0 2 2x 1 4
x
4 2
0 = 2(8 4)=8 ]
15. 1
x= x
2 1 2
x =x
4 2 x 4x=0 x(x 4)=0 x=0 or 4 , so 6 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 4 9 1
x
x dx+
2
4 A=
0 16
4
3 = 1
x
2 x 81
18
4 0 + 2 3/2 1 2
x
x
3
4 dx=
4 16
3 = 4 + 0 1 2 2 3/2
x
x
4
3 9
4 81 32
59
+
26=
4
3
12 16.
3 (8 x )
2 A= 3 2 x dx=2 3 2 2 8 2x dx=2
0 ( 8 2x ) dx+2 ( 2x2 8) dx 0 2 3
2 3
x 8x =2
0
3
2
32
32
64 92
=32
+20
=52
=
3
3
3
3 =2 2 2 3
x
3 8x 2 17. 2y =1 y
1/2 +2 2 ( 1 y ) 2y dy= A=
1 = 1/2 ( 1 y 2y2) dy= 1 1
2 1
8 1
12 1 1 2
+
2 3 = 7
24 16
3 16 2y +y 1=0 (2y 1)(y+1)=0
2 3 2 y y= 0 +2 ( 18 24 ) 16
16
3 1
1
or 1 , so x= or 2 and
2
2 1 2 2 3
y
y
2
3
5
6 = 1/2
1 7 20 27 9
+
=
=
24 24 24 8
7 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 2 18. 4x+x =12
2 (x+6)(x 2)=0 1 2
y +3
4 A=
6 x= 6 or x=2 , so y= 6 or y=2 and 2 2 19. The curves intersect when 1 y =y 1
1 A= ( 1 y2) ( y2 1) 2 1 3 1 2
y
y +3y
12
2 y dy= = 6 2 2=2y 2 y =1 2
2+6
3 (18 18 18)=22 2 64
=
.
3 3 y= 1 . dy 1
1 ( 2 ) = 2 1 y dy
1
1 =2 2 ( 1 y2) dy 0 =4 1 3
y
y
3 1 =4 0 1 1
3 = 8
3 8 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 1 20. A=2 sin
0 x
2 2 x dx=2 cos 1 2 x
2 x
2 =2 0 0 1
2 2 0 = 4 1 21. Notice that cos x=sin 2x=2sin xcos x
2sin xcos x cos x=0 cos x(2sin x 1)=0
2sin x=1 or cos x=0 x= or 6 /6
0 ( sin 2x cos x ) dx
/6 /6
1
cos 2x
+
2
0
1 1 1
1
= +
0+ 1 +
2 2 2
2 = . /2 ( cos x sin 2x ) dx+ A= 2 sin x+ 1
cos 2x sin x
2
1
1 1
1
2
2 2 /2
/6 1
2 = 1
2 22. sin x=sin 2x=2sin xcos x when sin x=0 and when cos x=
that is, when x=0 or 3 . /3 /2 ( sin 2x sin x ) dx+ A=
0 = 1
;
2 ( sin x sin 2x ) dx
/3 1
cos 2x+cos x
2 /3
0 + 1
cos 2x cos x
2 /2
/3
9 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 1
2 =
+ 1
2 + 1
0
2 1
2
1
2 1
2 1
+1
2
1
1
=
2
2 23. From the graph, we see that the curves intersect at x=0 , x=
A= cos x 1 /2 2x dx=2 0 =2 sin x x+ 1 2 x 1 dx=2 x ( x 2)
2 0 =2 1 2 0=x x 2
2 dx = 2 2 x 2 + dx=2
0 2 2x dx 2 4 2 0 2 1 0=(x 2)(x+1) ( x 2)
2 cos x 1+
0 /2 2 , and x= . By symmetry, /2 2x 0 24. For x>0 , x=x 2
2 cos x 2 0 =2 1 2 + 4 =2 2 x=2 . By symmetry, ( x x2+2) dx=2 1 2 1 3
x
x +2x
2
3 2
0 = 8
20
+4 =
3
3
10 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 2 25. The curves intersect when x =
A= 1
1 2
2 2 x dx=2 x +1 1
0 2
2 4 2 x +x =2 x +1
2
2
x
2
x +1 dx=2 4 2 x +x 2=0
1 2tan x 1 3
x
3 ( x2+2) ( x2 1) =0
1 =2 0 2 4 1
3 2 x =1
= 2
3 x= 1 .
2.47 26.
1 A= sin x
0 ( x x)
2 2 dx+ ( x2 x) sin x dx 1 11 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 1 = 1 =
= 4 cos x
1 1
+
3 2 1 3 1 2
x+ x
3
2
1 1 + 0 + 1 3 1 2 1
x
x + cos x
3
2
8
1
1 1 1
2+
3
3 2 2
1 +1 27. From the graph, we see that the curves intersect at x= a 1.02 , with 2 2cos x>x on ( a,a ) . So the area of the region bounded by the
curves is
a A= a ( 2cos x x ) dx=2 ( 2cos x x2) dx
2 a =2 0 2sin x 1 3
x
3 a
0 2.70 28. 12 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 2
0 4
x+5
5 2 5 A= 27
=
xdx+
0 10
2 27 2 2
x
+
20
0
27
0 +
5 =
= 5 7
x+5
2 dx+
2 4
x+5
5 ( x 4 ) dx 9
x+9 dx
5
9 2
x +9x
10
45
+45
2 5
2 18
27
+18 =
5
2 29.
1 3 A= x x dx
1
1 =2 ( x x3) dx [ by symmetry] 0 =2
=2 1 2 1 4
x
x
2
4
1 1
1
=
2 4
2 1
0 13 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 30. The curves intersect when x+2 =x
(x 2)(x+1)=0 x= 1 or 2 . 2 2 x+2=x x x 2=0 4 A= x+2 x dx
0
2 4 ( = x+2 x ) dx+ 0 (x x+2 ) dx 2 2
3/2
(x+2)
3
16
=
2
3
32 4
=4+
2
3 3
= 1 2 2
x
+
2
0
2
(2 2 )
3
44
4 6=
3 1 2 2
3/2 4
x
(x+2)
2
3
2
2
0 + 8
(6 6 )
3
4
4 6
2
3 x
2
sin
4
The shaded area is given by
31. Let f (x)=cos
1 A= f(x)dx M =
0 4 x
4 2 1
4 f 1
8 +f 3 32. The curves intersect when 3
8 and +f x=
5
8 2 16
3 1 0
.
4
+f 7
8 0.6407 3 16 x =x
14 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 3 16 x =x 3 3 3 2x =16
3 Let f (x)= 16 x 3 x =8
x and x=2.
2 0
x=
.
4 The shaded area is given by
2 M A= f (x)dx 4 0 = 2
1
f
4
4
2.8144 +f 3
4 +f 5
4 +f 7
4 33. From the graph, we see that the curves intersect at x= a 1.02 , with 2 2cos x>x on ( a,a ) . So the area of the region bounded by the
curves is
a A= a ( 2cos x x ) dx=2 ( 2cos x x2) dx
2 a =2 0 2sin x 1 3
x
3 a
0 2.70 34. 15 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 3 4 From the graph, we see that the curves intersect at x=0 and at x=a 1.17 , with 3x x >x on ( 0,a ) . So
the area of the region bounded by the curves is
a A= ( 3x x3) 4 x dx= 0 3 2 1 4 1 5
x
x
x
2
4
5 a
0 1.15
35. From the graph, we see that the curves intersect at x= a
So the area of the region bounded by the curves is
A=2 a
0 2 3 xcos (x ) x dx=2 0.86 . 1
2 1 4
sin (x )
x
2
4 a
0 0.40 36. From the graph, we see that the curves intersect at x=a
. So the area of the region bounded by the curves is
A= b
a ( 2 x2) x e dx= 2x 2 x 1.32 and x=b 0.54 , with 2 x >e on ( a,b ) 1 3 x
x e
3 b
a 1.45 37. As the figure illustrates, the curves y=x and
16 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 5 3 5 3 y=x 6x +4x enclose a four part region symmetric about the origin (since x 6x +4x and x are odd
5 3 4 2 functions of x ). The curves intersect at values of x where x 6x +4x=x ; that is, where x(x 6x +3)=0
36 12
2 6
. That happens at x=0 and where x =
=3 6 ; that is, at x= 3+ 6 , 3 6 , 0 ,
2
3 6 , and 3+ 6 .
The exact area is
3+ 6 2 (x 5 3 6x +4x ) 3+ 6
5 x dx =2 3 5 3 x 6x +3x dx 0 0
3 6 =2 (x 3+ 6 ) 6x +3x dx+2 0 3 ( 5 3 ) x +6x 3x dx 6 =12 6 9 2 2 38. The inequality x 2y describes the region that lies on, or to the right of, the parabola x=2y . The
inequality x 1 y describes the region
1 y if y 0
.
that lies on, or to the left of, the curve x=1 y =
1+y if y<0
So the given region is the shaded region that lies between the curves. { 2 2 The graphs of x=1 y and x=2y intersect when 1 y=2y
1
2
2y +y 1=0 (2y 1)(y+1)=0 y= (for y 0 ).
2 17 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves By symmetry,
1/2 A=2 2 (1 y) 2y dy=2
0 1/2 2 3 1 2
y
y +y
3
2 1
12 =2 0 1 1
+
8 2 7
24 0 =2 = 7
.
12 1
1
hour, so 10 s =
h. With the given data, we can take n=5 to use the
3600
360
1/360 0
1
Midpoint Rule. t=
=
, so
5
1800
39. 1 second = . 1/360 distance Kelly distance Chris = 1/360 v dt
0 K 1/360 v dt=
C 0 1
1800 (v K ) (1)+ ( v
+ ( v v ) (7)+ ( v v ) (9) M=
5 K C (v 0 K v C K K v C ) dt v C ) (3)+ ( v K v C ) (5) C 1
( 22 20 ) + ( 52 46 ) + ( 71 62 ) + ( 86 75) + ( 98 86 )
1800
1
1
1
1
=
(40)=
mile,or117 feet
( 2+6+9+11+12 ) =
1800
1800
45
3
= 40. If x= distance from left end of pool and w=w(x)= width at x , then the Midpoint Rule with n=4
16 and b a 8 2 0
2
x=
=
=4 gives Area = wdx 4(6.2+6.8+5.0+4.8)=4(22.8)=91.2 m .
n
4
0
x 41. We know that the area under curve A between t=0 and t=x is v (t)dt=s (x) , where v (t) is the
0 A A A velocity of car A and s is its displacement. Similarly, the area under curve B between t=0 and t=x is
A x v (t)dt=s (x) .
0 B B (a) After one minute, the area under curve A is greater than the area under curve B . So car A is ahead
after one minute.
(b) The area of the shaded region has numerical value s (1) s (1) , which is the distance by which A
A B is ahead of B after 1 minute.
(c) After two minutes, car B is traveling faster than car A and has gained some ground, but the area
under curve A from t=0 to t=2 is still greater than the corresponding area for curve B , so car A is still
ahead.
(d) From the graph, it appears that the area between curves A and B for 0 t 1 (when car A is going
18 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves faster), which corresponds to the distance by which car A is ahead, seems to be about 3 squares.
Therefore, the cars will be side by side at the time x where the area between the curves for 1 t x
(when car B is going faster) is the same as the area for 0 t 1 . From the graph, it appears that this
time is x 2.2 . So the cars are side by side when t 2.2 minutes.
/ 42. The area under R (x) from x=50 to x=100 represents the change in revenue, and the area under
/ C (x) from x=50 to x=100 represents the change in cost. The shaded region represents the difference
between these two values; that is, the increase in profit as the production level increases from 50 units
to 100 units. We use the
Midpoint Rule with n=5 and x=10 :
/ M =
5 / / / x{+[R (65) C (65)]+[R (75) C (75)]
/ / / / +[R (85) C (85)]+[R (95) C (95)]}
10(2.40 0.85+2.20 0.90+2.00 1.00+1.80 1.10+1.70 1.20)
= 10(5.05)=50.5 thousand dollars
Using M would give us 50(2 1)=50 thousand dollars.
1 43. To graph this function, we must first express it as a combination of explicit functions of y ; namely,
y= x x+3 . We can see from the graph that the loop extends from x= 3 to x=0 , and that by
symmetry, the area we seek is just twice the area under the top half of the curve on this interval, the
0 ( equation of the top half being y= x x+3 . So the area is A=2 x x+3 ) dx . We substitute u=x+3 , 3 so du=dx and the limits change to 0 and 3 , and we get
3 A= 2 3 (u 3) u du= 2
0 = 2 ( u3/2 3u1/2) du 0 2 5/2 3/2
u 2u
5 3 = 2 0 2 2
3 3
5 ( ) 2 (3 3 ) = 24
5 3 19 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 44. 2 / We start by finding the equation of the tangent line to y=x at the point ( 1,1 ) : y =2x , so the slope of
the tangent is 2(1)=2 , and its equation is y 1=2(x 1) , or y=2x 1 . We would need two integrals to
integrate with respect to x , but only one to integrate with respect to y .
1 A=
0 = 1
(y+1)
2 1 1
+
4 2 y dy= 1 2 1
2 3/2
y+ y
y
4
2
3 1
0 2 1
=
3 12 45. 2 By the symmetry of the problem, we consider only the first quadrant, where y=x
b 4 looking for a number b such that y dy=
0 3/2 b =4 2/3 b=4 b b 3/2 3/2 b =4 b 3/2 3/2 2b =8 4 1 46. (a) We want to choose a so that
1 a= 4 2.52 .
a 5 2
=
4 a 2 3/2 b 2 3/2
y
=
y
0 3
3 y dy x= y . We are 2 x 1 dx=
a 2 dx x 1
x a 1
x = 1 4
a 1
1 1
+1=
+
a
4 a 8
.
5
2 (b) The area under the curve y=1/x from x=1 to x=4 is 3
. Now the line y=b must intersect the
4
2 curve x=1/ y and not the line x=4 , since the area under the line y=1/4 from x=1 to x=4 is only 3
16 , which is less than half of
20 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 3
. We want to choose b so that the upper area in the diagram is half of the total area under the curve
4
1
y=
from x=1 to x=4 . This implies that
2
x
1 ( 1/ y 1 ) dy= b 1 3
2 4 1 2 y y =
b 3
8 1 2 b +b= 16
5
2
=0 8c 16c+5=0 . Thus, c=
8
6 1
3
2
b=c =1+
= ( 11 4 6 ) 0.1503 .
8 2
8
2 c 2c+ 3
8 b 2 b+ 256 160
=1
16 5
=0 . Letting c= b , we get
8 6
. But c= b <1
4 c=1 6
4 47. We first assume that c>0 , since c can be replaced by c in both equations without changing the
graphs, and if c=0 the curves do not enclose a region. We see from the graph that the enclosed area A
lies between x= c and x=c , and by symmetry, it is equal to four times the area in the first quadrant.
The enclosed area is
c A=4 ( c2 x2) dx=4 2 c x 0 =4 c 3 So A=576 1 3
c =4
3
8 3
c =576
3 1 3
x
3 c
0 2 3
8 3
c = c
3
3
3 c =216 3 c= 216 =6 . Note that c= 6 is another solution, since the graphs are the same. 48. 21 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves It appears from the diagram that the curves y=cos x and y=cos (x c) intersect halfway between 0 and
c , namely, when x=c/2 . We can verify that this is indeed true by noting that
cos (c/2 c)=cos ( c/2)=cos (c/2) .
The point where cos (x c) crosses the x axis is x= 2 +c . So we require that c/2 cos x cos (x c) dx= cos (x c)dx (the negative sign on the RHS is needed since the second 0 /2+c area is beneath the x axis)
sin x sin ( x c ) c/2
0 = sin ( x c ) sin (c/2) sin ( c/2)
2sin (c/2)=1 /2+c sin ( c) = sin ( sin (c/2)= 1
2 c/2= 6 c)+sin
c= 2 +c c 2sin (c/2) sin c= sin c+1 . . So . 3 49. The curve and the line will determine a region when they intersect at two or more points. So we
2
(2 )
x=x ( mx +m )
2
x ( mx +m 1 ) =0 solve the equation x/ x +1 =mx ( 2 x mx +m )
2 x=0 x=0 or mx +m 1=0 2 x=0 or x = 1 m
m 1
1 . Note that if m=1 , this has only the solution x=0 , and no region is determined.
m
But if 1/m 1>0 1/m>1 0<m<1 , then there are two solutions. [Another way of seeing this is to
x=0 or x= 2 / observe that the slope of the tangent to y=x/(x +1) at the origin is y =1 and therefore we must have
0<m<1 .] Note that we cannot just integrate between the positive and negative roots, since the curve
22 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.1 Areas between Curves 2 and the line cross at the origin. Since mx and x/(x +1) are both odd functions, the total area is twice
the area between the curves on the interval 0, 1/m 1 . So the total area enclosed is
2 1/m 1
0 x
2 mx x +1
ln (1/m) 1+m=m ln m 1 dx=2 1
2
ln x +1
2 ( ) 1 2
mx
2 1/m 1
0 = ln (1/m 1+1) m(1/m 1) (ln 1 0) = 23 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes ( x2) 2 . 2 1. A cross section is circular with radius x , so its area is A(x)=
1 1 2 2 ( x ) dx= V = A(x)dx=
0 0 1 4 x dx=
0 x 2. A cross section is a disk with radius e , so its area is A(x)=
V= 1
0 A(x)dx= 1
0 ( ex) 2 dx= 1 2x
0 e dx= 1
2 1 1 5
x
5 = 0 5 ( ex ) 2 .
2x 1 e = 0 2 ( e2 1) 2 3. A cross section is a disk with radius 1/x , so its area is A(x)= (1/x) . 1 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 2
2 2 1
x V = A(x)dx=
1 1 2 1 dx=
1 2 1
x dx= x 2
1 4. A cross section is circular with radius x 1 , so its area is A(x)=
5 5 V = A(x)dx=
2 (x 1)dx=
2 1 2
x x
2 5 = 2 5. A cross section is a disk with radius y , so its area is A(y)= 1
( 1)
2 = ( x 1 ) = 2 2 = (x 1) . 25
4
15
5
+2 =
2
2
2 ( 2 y) . 2 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 4 4 ( V = A(y)dy=
0 2 4 y ) dy= 0 ydy=
0 2 6. A cross section is a disk with radius y y , so its area is A(y)=
1 V = 1 A(y)dy=
0 2 2 (y y ) 1 dy= 0 1
5 ( y4 2y3+y2) dy= 0 1 1
+
2 3 = 1 2
y
2 4 =8 0 ( y y2) 2 . 1 5 1 4 1 3
y
y+ y
5
2
3 1
0 30 2 7. A cross section is a washer (annulus) with inner radius x and outer radius x , so its area is
A(x)= ( x ) 2 ( x2) 2= ( x x4) .
3 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 1 1 V = A(x)dx=
0 ( x x4) dx= 0 1 2 1 5
x
x
2
5 1 1
2 = 0 1
5 = 3
10 8. A cross section is a washer with inner radius 1 and outer radius sec x , so its area is
2 1 V = (1) =
1 A(x)dx=
1 ( sec 2x 1) .
1
2
( sec x 1) dx=2 ( sec 2x 1) dx=2
2 A(x)= (sec x) 1 0 1 tan x x =2 (tan 1 1)
0 3.5023 2 9. A cross section is a washer with inner radius y and outer radius 2y , so its area is
2 A(y)= (2y) ( y2) 2= ( 4y2 y4) .
4 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 2 2 V = A(y)dy=
0 2/3 10. y=x x=y 3/2
2 area is A(y)= (1) ( 4y2 y4) dy= 0 3/2 2 1 ( 1 y3) dy= V = A(x)dx=
0 ( 1 2x+x )
1
( 3x+x2+2 x ) dx=
x ) 2 = y 0 11. A cross section is a washer with inner radius 1
1 3/2 = 64
15 and outer radius 1 , and its ( y ) = (1 y ) .
0 (1 0 32
5 3 V = A(y)dy= A(x)= (1 x) 32
3 = , so a cross section is a washer with inner radius y 1 2 2 4 3 1 5
y
y
3
5 2 (1 1 1 4
y
4 = 0 3
4 x and outer radius 1 x , so its area is 2 x +x ) = ( 2 3x+x +2 x ). 0
5 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 1 3 2 1 3 4 3/2
x+ x+ x
2
3
3 3 5
+
2 3 = 0 = 6 2 12. A cross section is circular with radius 4 x , so its area is A(x)=
2 V = 2 2 A(x)dx=2 A(x)dx=2
2 2 0 32 64 32
+
3
5 ( 16 8x2+x4) dx=2 0 =64 1 2 1
+
3 5 =64 16x 8 3 1 5
x+ x
3
5 ( 4 x2) 2= ( 16 8x2+x4) .
2
0 = 8 512
=
15
15 4 13. A cross section is an annulus with inner radius 2 1 and outer radius 2 x , so its area is
6 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes ( 2 x4 ) 2 ( 3 4x4+x8) .
1
1
1
4 8
V = A(x)dx=2 A(x)dx=2 ( 3 4x +x ) dx=2 A(x)= 1 2 ( 2 1) = 0 2 0 4 1
+
5 9 3 3x = 4 5 1 9
x+ x
5
9 1
0 = 208
45 14.
V =
=
= { 3
1
3
1 1
( 1)
x
1
2 + 2
x x
1
+2ln 3
3 2 0 ( 1) dx= } 2 dx= 1
+2ln x
x ( 1+0 ) = 3
1 1
+1
x 2 2 1 dx 3
1 2ln 3+ 2
3 =2 ln 3+ 1
3 7 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 15.
1 2 2 (1 y ) V = 1 2 2 (1 y ) dy=2 1 1 dy=2 0 =2 y 16. y= x
1 0
1 2 3 1 5
y+ y
3
5 8 16
=
15 15 =2 0 2 2 x=y , so the outer radius is 2 y .
2 2 (2 y ) V = ( 1 2y2+y4) dy 2 1 ( 2 y ) dy= 0 ( 4 4y2+y4) ( 4 4y+y2) dy 0
1 =
0 ( y4 5y2+4y) dy= 1 5 5 3 2
y
y +2y
5
3 1 = 0 1
5 5
8
+2 =
3
15 8 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 2 17. y=x x= y for x 0 . The outer radius is the distance from x= 1 to x= y and the inner radius
2 is the distance from x= 1 to x=y .
1 { V = y ( 1) 2 2 y ( 1) } dy= 2 0 ( y +1 ) ( y2+1) 2 2 dy 0
1 = ( y+2 4 2 1 ) y +1 y 2y 1 dy= 0 = 1 ( y+2 4 2 ) y y 2y dy 0 1 2 4 3/2 1 5 2 3
y+ y
y
y
2
3
5
3 1 = 0 1 4
+
2 3 1
5 2
3 = 29
30 18. For 0 y<2 , a cross section is an annulus with inner radius 2 1 and outer radius 4 1 , the area of
which is A (y)= ( 4 1 )
1 2 2 ( 2 1 ) . For 2 y 4 , a cross section is an annulus with inner radius y 1 and outer radius 4 1 , the area of which is A (y)= ( 4 1 )
2 2 2 ( y 1) . 9 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 4 V = 2 4 2 ( 4 1 ) ( 2 1 ) dy+ A(y)dy=
0
4 8y +
0 2 2 ( 8+2y y2) dy=16 1 3
y
3 2 + 8y+y 2 = 16 + 2 ( 4 1 ) ( y 1 ) dy 0
2 = 2 64
3 32+16 8
3 16+4 = 1 1 0 2 76
3 1 32 19. R about OA (the line y=0 ): V = A(x)dx=
1 4 (x ) dx=
0 1 1 7
x
7 6 x dx=
0 = 0 7 20. R about OC (the line x=0 ):
1 1 1 V = A(y)dy=
0 ( 2 (1) 3 y ) 2 1 dy= 2/3 (1 y )dy= 0 y 0 3 5/3
y
5 1 = 0 1 3
5 = 2
5 21. R about AB (the line x=1 ):
1 1 1 V = A(y)dy= 0 0 = y (1 3 y ) 2 1 dy= 1/3 2/3 (1 2y +y )dy
0 3 4/3 3 5/3
y + y
2
5 1 = 1 0 3 3
+
2 5 = 10 22. R about BC (the line y=1 ):
1 1 1 V = A(x)dx=
0 2 (1)
0 32 1 (1 x ) dx= 3 6 1 (1 2x +x ) dx
0
10 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 1 = 3 1 1 4 1 7
x
x
2
7 6 (2x x )dx=
0 1
2 = 0 1
7 = 5
14 x 1 2
x
2 23. R about OA (the line y=0 ):
2 1 1 2 V = A(x)dx=
0 ( (1) x ) 2 1 dx= (1 x)dx= 0 0
1 1 24. R about OC (the line x=0 ): V = A(y)dy=
2 0 1 22 (y ) dy=
0 1 = 1 0 1
2 = 1 5
y
5 4 y dy=
0 2
1 = 0 5 25. R about AB (the line x=1 ):
2 1 1 2 V = A(y)dy=
0 1 22 (1) = 2 4 1 (1 2y +y ) dy 0
1 2 (1 y ) dy=
0
1 2 3 1 5
y
y
3
5 4 (2y y )dy=
0 2
3 = 0 1
5 = 7
15 26. R about BC (the line y=1 ):
2 1 1 (1 V = A(x)dx=
0 x dx= 0 1/2 (1 2x +x)dx
0 1 4 3/2 1 2
x
x + x
3
2 = ) 1 2 = 1 0 4 1
+
3 2 = 6 27. R about OA (the line y=0 ):
3 1 1 ( V = A(x)dx=
0 x ) 2 32 1 (x ) dx= 0 6 (x x )dx=
0 1 2 1 7
x
x
2
7 1 1
2 = 0 1
7 = 5
.
14 Note: Let = + mathpi5 +R . If we rotate about any of the segments OA , OC , AB , or BC , we
1 2 3 2 2 obtain a right circular cylinder of height 1 and radius 1 . Its volume is r h= (1) 1= . As a check
for Exercises 19, 23, and 27, we can add the answers, and that sum must equal . Thus,
5
2+7+5
+ +
=
= .
7 2 14
14
28.
11 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes R about OC (the line x=0 ):
3 1 1 ( V = A(y)dy=
0 3 ) y 2 1 22 (y ) dy= 0 4 y )dy 0 3 5/3 1 5
y
y
5
5 = 2/3 (y 1 3
5 = 0 1
5 = 2
5 Note: See the note in Exercise 27. For Exercises 20, 24, and 28, we have 2
2
+ +
= .
5
5
5 29. R about AB (the line x=1 ):
3 1 1 V = A(y)dy=
0 (1 22 (1 y ) 3 y ) 1 2 dy= 0
1 = 2 4 1/3 2/3 0
4 1/3 ( 2y +y +2y 2 3 1 5 3 4/3 3 5/3
y+ y+ y
y
3
5
2
5 2/3 y )dy= 0 2 1 3
+ +
3 5 2 = 2 (1 2y +y ) (1 2y +y ) dy 3
5 = 1
0 13
30 Note: See the note in Exercise 27. For Exercises 21, 25, and 29, we have
7
13
3+14+13
+
+
=
= .
10 15
30
30
30. R about BC (the line y=1 ):
3 1 1 V = A(x)dx=
0 32 (1 x ) (1 x ) 2 0
1 = 3 6 1 1/2 (1 2x +x ) (1 2x +x) dx= 3 6 1/2 ( 2x +x +2x 0 = dx x)dx 0 1 4 1 7 4 3/2 1 2
x+ x+ x
x
2
7
3
2 1 = 0 1 1 4
+ +
2 7 3 1
2 = 10
21 Note: See the note in Exercise 27. For Exercises 22, 26, and 30, we have
5
10
15+7+20
+ +
=
= .
14 6
21
42
31.
12 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes /4 V= 3 2 (1 tan x) dx
0 4 4 4 32. y=(x 2) and 8x y=16 intersect when (x 2) =8x 16=8(x 2) (x 2) 8(x 2)=0
3 x 2=0 or x 2=2
1
8x=y+16 x= y+2 .
8 (x 2) (x 2) 8 =0
8x y=16
16 V=
0 { 2 1
y+2
8 10 ( 4 x=2 or 4 . y=(x 2) 4 10 2+ y ) } 2 x 2= 4 y 4 x=2+ y [ since x 2 ]. dy 33.
2 2 ( 1 0 ) ( 1 sin x ) dx V =
0 2 1 ( 1 sin x ) = 2 dx 0 34.
13 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 2 2 ( sin x+2 ) 2 dx V=
0 35.
8 V = { 3 ( 2) 2 } dy 2 2 y +1 ( 2) 8
2 2 = 2 5 ( 1+y +2 )
2 2 dy 2 2 2 36. Solve the equations for x : ( y 1 ) =4 x 2 x=4 ( y 1 ) and 2x+3y=6 The points of intersection of the two curves are ( 3,0 ) and
7/2 V =
0
7/2 =
0 {
{ 2 4 ( y 1 ) ( 5)
9 ( y 1) 2 2 2 8 3
3
y
2 3
y ( 5)
2 } } 9 7
,
4 2 x=3 3
y.
2 . Therefore, 2 dy 2 dy 14 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 37.
2 y=x and y=ln (x+1) intersect at x=0 and at x=a 0.747 .
a V= 0 { ln (x+1) ( x ) } dx
2 2 2 0.132 38. 2 x/2 y=3sin (x ) and y=e +e 2x intersect at x=a 0.772 and at x=b 1.524 . V= b
a { 2 3sin (x ) 2 ( ex/2+e 2x) 2} dx 7.519 39. { V = 2 sin x ( 1 ) 2 0 ( 1) } dx 2 0 = 11
8 2 40. V
15 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 2 = 2 1 x/2 2 /2 41. 2 ) dx = (3 x) (3 xe 0 CAS 2e +24e 142
3 2 cos xdx describes the volume of the solid obtained by rotating the region
0 R= {( x,y ) |0 x 5 42. 2 5 2 of the xy plane about the x axis. 2 ( ydy= } ,0 y cos x y ) dy describes the volume of the solid obtained by rotating the region 2 R={ ( x,y ) |2 y 5,0 x
1 y } of the xy plane about the y axis. 1 ( y y ) dy= ( y2) 2 ( y4) 2 dy describes the volume of the solid obtained by rotating the
0
0
4
2
region R= { ( x,y ) |0 y 1,y x y } of the xy plane about the y axis.
4 43. 8 /2 2 2 ( 1+cos x ) 1 dx describes the volume of the solid obtained by rotating the region 44.
0 = {( x,y ) |0 x 2 } ,1 y 1+cos x of the xy plane about the x axis.
/ Or: The solid could be obtained by rotating the region = {( x,y ) |0 x 2 ,0 y } x about the line y= 1 .
45. There are 10 subintervals over the 15 cm length, so we’ll use n=10/2=5 for the Midpoint Rule.
15 V = A(x)dx M =
0 5 15 0
[A(1.5)+A(4.5)+A(7.5)+A(10.5)+A(13.5)]
5 =3(18+79+106+128+39)=3 370=1110 cm 3 16 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 46.
10 V = A(x)dx M =
5 0 10 0
[A(1)+A(3)+A(5)+A(7)+A(9)]
5 =2(0.65+0.61+0.59+0.55+0.50)=2(2.90)=5.80m 3 47. We’ll form a right circular cone with height h and base radius r by revolving the line y= r
x about
h the x axis.
h
h r
x
h V =
0 = r 2 2 r dx=
0 2
2 r 2 x dx= 2 1 3
x
3 2 h h h
0 1 3
1 2
h =
r h
3
3 2 h r
y+r about the y axis.
h Another solution: Revolve x=
h
h r
y+r
h V =
0 = r 2
2 3h
* 2 y 3 r dy=
2 u
r 2 h
r du= h
r y h
h 0 1 3
u
3 2 2r
2
y+r
h dy 1 2 2 2
1 2
r h r h+r h =
r h
3
3 = Or use substitution with u=r
0 2 2 0 r 2 2
y +r y
h 2 r
r
y and du=
dy to get
h
h
0
r = h
r 1 3
1 2
r =
r h.
3
3 17 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 48.
h V = Rr
y
h R
0
h = 2 dy 2R(R r)
y+
h 2 R
0 Rr
h 2 2 y dy R(R r) 2 1
Rr 2 3 h
=
R y
y+
y
h
3
h
0
1
2
2
=
R h R(R r)h+ (R r) h
3
1
1
2
2
2
2
=
h 3Rr+(R 2Rr+r ) =
h(R +Rr+r )
3
3
2 H H h
=
by similar triangles. Therefore,
R
r
hR
hR hR=H(R r) H=
. Now
Rr
1 2
2
RH
r (H h) [by Exercise 47]
3
1 2 rh
rH
rhR
2 hR
R
r
H h=
=
Rr 3
Rr
R R(R r) Another solution:
Hr=HR
1
3
1
=
3 V = 3 3 1
R r
1
2
2
=
h
=
h R +Rr+r
3
Rr
3 ( )
18 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes = 1
3 2 ( 2 R+ r + 2 R ) ( r 2) h= 1
3 A +A +
1 2 A A 1 2 h where A and A are the areas of the bases of the frustum. (See Exercise 50 for a related result.)
1 2 2 2 49. x +y =r
r V = (r 2 2 2 2 2 x =r y
2 ) y dy= r h = { r = { 2 3
r
3 1
3
1
=
3
1
=
3
1
=
3
= 3 r h 3 r
2
(r h)
r (r h)
3
3
1
2
2
(r h) 3r ( r h )
3 3 } } { 2r 3 (r h)
{ 2r 3 r 3 y
r y
3
2 3r 2 (r 2 2rh+h2) } 2 2 ( r h ) 2r +2rh h } ( 2r 3 2r 3 2r 2h+rh2+2r 2h+2rh2 h3)
( 3rh2 h3) = 1
3 2 2 h (3r h) , or, equivalently, h r h
3 50. An equation of the line is
19 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes x
a/2 b/2
b a b
b
y+(x intercept )=
y+ =
y+ .
y
h 0
2 2h
2 x= h h 2 V = A(y)dy= (2x) dy
0 0 h = a b
b
y+
2h
2 2
0 2 h a b
y+b
h dy=
0 2 dy h ( a b) = 2 y+ 2 h 0 = 2 ( a b)
2 3h 2 2b(a b)
2
y+b
h b(a b) 2 2
y+
y +b y
h dy
h 3 0 1
1 2
2
2
2
(a b) h+b(a b)h+b h=
a 2ab+b +3ab h
3
3
1 2
2
=
a +ab+b h
3 ( = ( ) ) Note that this can be written as 1
3 A +A +
1 2 A A 1 2
2 h , as in Exercise 48. If a=b , we get a rectangular solid with volume b h . If a=0 , we get a square pyramid with volume
1 2
b h.
3
51. For a cross section at height y , we see from similar triangles that /2 h y
=
, so =b
b/2
h 1 y
h .
Similarly, for cross sections having 2b as their base and replacing , =2b 1 y
h . So 20 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes h h V = A(y)dy=
0 b 1 0 y
h 2b 1 y
h dy h
h 2 y
1
h 2 = 2b
0 2 2y y
1
+
2
h
h 2 dy=2b 0 2 dy h 3 y
1
2
y
=2b
y
+
=2b h h+ h
2
h
3
0
3h
2 2
1
= b h [= Bh where B is the area of the base, as with any pyramid.]
3
3
2 52. Consider the triangle consisting of two vertices of the base and the center of the base. This
=a /b . Also by similar
triangle is similar to the corresponding triangle at a height y , so a/b= /
triangles, b/h= /(h y)
=b(h y)/h . These two equations imply that =a ( 1 y/h ) , and since the cross section is an
1
equilateral triangle, it has area A(y)=
2
2 h V = A(y)dy= a 4 0
2 = a 4 3 h
3 3 h
1
0 y
1
h 2 a ( 1 y/h )
=
4 2 3 , so 2 y
h
3 3
2 dy
h = 0 3 2
3 2
a h( 1)=
ah
12
12 21 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 53. A cross section at height z is a triangle similar to the base, so we’ll multiply the legs of the base
triangle, 3 and 4, by a proportionality factor of (5 z)/5 . Thus, the triangle at height z has area
1
5 z
5 z
z 2
A(z)= 3
4
=6 1
, so
2
5
5
5
5 5 2 A(z)dz=6 ( 1 z/5) dz V =
0 0
0 2 = 6 u ( 5du) u=1 z/5,du= 1
0 1 3
u
3 = 30 1
3 = 30 1 1
dz
5
=10cm 3 54. A cross section is shaded in the diagram.
A(x)=(2y) = ( 2
2 r V = 2 2 r x
r ) 2 , so ( 2 2) A(x)dx=2 4 r x dx
r 0 1 3 r
x
3
0
2 3
16 3
r =
r
3
3
2 =8 r x
=8 22 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 55. If l is a leg of the isosceles right triangle and 2y is the hypotenuse,
2 2 then l +l = ( 2y )
2 2 2 2 2 2 2 l =2y . 2l =4y 2 2 1
2
V = A(x)dx=2 A(x)dx=2
(l)(l)dx=2 y dx
2
0
0 2
0
2 1
9
2
=2
36 9x dx=
2
0 4
= 9
2 ( ) 1 3
x
3 4x 2 = 0 9
2 2 ( 4 x2) dx 0 8 8
3 =24 56. The cross section of the base corresponding to the coordinate y has length 2x=2 y . The
corresponding equilateral triangle with side s has area
3
3
3
2
2
2
A(y)=s
= ( 2x )
=(2 y )
=y 3 . Therefore,
4
4
4
1 1 V = A(y)dy= y 3 dy= 3
0 0 1 2
y
2 1 = 0 3
.
2 57. The cross section of the base corresponding to the coordinate y has length 2x=2 y . The square
2 1 1 2 1 has area A(y)= ( 2 y ) =4y , so V = A(y) dy= 4ydy= 2y
0 0 =2 . 0 23 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 58. A typical cross section perpendicular to the y axis in the base has length (y)=3 3
y. This
2 length is the diameter of a cross sectional semicircle in S , so
2 V = 2 A(y)dy=
0 0
0 = 2 3 12 1 3
u
3 8 = 2 dy= 2 2
3 u du 2 2 ( y) 3 12 0 3
y
2 2 dy 3
3
y,du=
dy
2
2 u=3 0 = 8 3 ( 9)= 3
4 59. A typical cross section perpendicular to the y axis in the base has length (y)=3 3
y. This
2 length is the leg of an isosceles right triangle, so
1
1
2
A(y) =
(y)
bh with base = height
2
2
= 1
2 3 1 2 1
y
2 = 9
2 1 1
y
2 2 Thus,
2 0 9 2
V = A(y)dy=
u ( 2du)
2 1
0
= 9 1 3
u
3 0 = 9 1 1
3 u=1 1
1
y,du=
dy
2
2 =3 24 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes r r r ( r ) 1
2 2
2 2
60. (a) V = A(x)dx=2 A(x)dx=2
h 2 r x dx=2h r x dx
r
0
0 2
0
(b) Observe that the integral represents one quarter of the area of a circle of radius r , so
1 2 1
2
V =2h
r =
hr .
4
2
2 2 2 61. (a) The torus is obtained by rotating the circle ( x R ) +y =r
about the y axis. Solving for x , we see that the right half of the
2 2 2 circle is given by x=R+ r y = f (y) and the left half by x=R
r { V = f (y) 2 g(y) 2 r y =g(y) . So } dy 2 r
r ( R +2R
2 =2 2 2 2 2 r y +r y ) (R 2 2 2 2 2 2R r y +r y ) dy 0
r 2 =2 2 r 4R r y dy=8 R
0 2 2 r y dy
0 (b) Observe that the integral represents a quarter of the area of a circle with radius r , so
r 2 2 r y dy=8 R 8 R
0 1 2
2 2
r =2 r R .
4 62. The cross sections perpendicular to the y axis in Figure 17 are rectangles. The rectangle
2 corresponding to the coordinate y has a base of length 2 16 y in the xy plane and a height of
1
1
2
2
y , since BAC=30 and BC =
AB . Thus, A(y)=
y 16 y and
3
3
3
4 2
V = A(y)dy=
3
0 4 2 16 y ydy
0
25 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 2
3 = 1
3 = 0 u 1
2 1/2 16
16 u 1/2 2 du[Putu=16 y ,sodu= 2ydy] 1 2 3/2
u
3 3 du= 0 16 2 0 3 3 = ( 64 ) = 128
3 3 h ( ) ( ) since the cross sectional area A(z) at height z is the 63. (a) Volume S = A(z)dz= Volume S
1 0 2 same for both solids.
(b) By Cavalieri’s Principle, the volume of the cylinder in the figure is the same as that of a right
2 circular cylinder with radius r and height h , that is, r h .
64. Each cross section of the solid S in a plane perpendicular to the x axis is a square (since the
edges of the cut lie on the cylinders, which are perpendicular). One quarter of this square and one
2 2 2 ( 2 2) eighth of S are shown. The area of this quarter square is PQ =r x . Therefore, A(x)=4 r x
and the volume of S is
r r V = A(x)dx=4
r
r =8 (r 2 x2) dx r (r 2 x2) dx=8 1 3
x
3 2 r x 0 r = 0 16 3
r
3 65. The volume is obtained by rotating the area common to two circles of radius r , as shown. The
volume of the right half is
r/2 V right = 2 r/2 y dx=
0 r
0 2 1
r+x
2 2 dx 26 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 1
r x
3 r/2 3 1 3 1 3
r
r
0
2
3
5
3
So by symmetry, the total volume is twice this, or
r .
12
= 1
r+x
2 2 = 1 3
r
24 0 = 5
3
r
24 Another solution: We observe that the volume is the twice the volume of a cap of a sphere, so we can
1
use the formula from Exercise 49 with h= r :
2
2
1 2
2
1
1
5
3
V =2
h (3r h)=
r
3r
r =
r .
3
3
2
2
12
66. We consider two cases: one in which the ball is not completely submerged and the other in which
it is.
Case 1: 0 h 10 The ball will not be completely submerged, and so a cross section of the water
parallel to the surface will be the shaded area shown in the first diagram. We can find the area of the
cross section at height x above the bottom of the bowl by using the Pythagorean Theorem:
2 2 2 2 2 2 R =15 ( 15 x ) and r =5 ( x 5) , so A(x)=
h h ( R2 r 2) =20
2 h depth h is then V (h)= A(x)dx= 20 xdx= 10 x
0 0 2 x . The volume of water when it has
3 =10 h cm , 0 h 10 . 0 Case 2: 10<h 15 In this case we can find the volume by simply subtracting the volume displaced by
the ball from the total volume inside the bowl underneath the surface of the water. The total volume
underneath the
surface is just the volume of a cap of the bowl, so we use the formula from Exercise 49:
1 2
V (h)=
h (45 h) . The volume of
cap
3
4
3 500
the small sphere is V =
(5) =
,
ball 3
3
1
3
2 3
so the total volume is V
V =
45h h 500 cm .
cap ball 3 ( ) 27 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 67. Take the x axis to be the axis of the cylindrical hole of radius r . A quarter of the cross section
through y, perpendicular to the y axis, is the rectangle shown. Using the Pythagorean Theorem
2 2 2 2 twice, we see that the dimensions of this rectangle are x= R y and z= r y , so
1
2 2
2 2
A(y)=xz= r y
R y , and
4
r r V = 2 2 2 A(y)dy= 4 r y
r 2 R y dy r r 2 =8 2 2 r y 2 R y dy 0 2 2 2 2 68. The line y=r intersects the semicircle y= R x when r= R x
2 2 x =R r 2 2 2 2 2 2 ( V =
2 2 R r . Rotating the shaded region about the x axis gives us x= R r 2 r =R x R r 2 2 R x ) 2 r 2 dx 2 28 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes 2 R r 2 ( R2 x2 r 2) dx [by symmetry] =2
0
2 R r =2 2 (R r )
2 2 2 x dx=2 (R r )
2 2 0 =2
=2 1 3
x
x
3 2 R r 2 0 ( R2 r 2) 3/2 1 ( R2 r 2) 3/2
3
2
2 2
R r
3 ( ) 3/2= 43 ( R2 r 2) 3/2
4
3
R , which is the volume of the full
3
0 , which makes sense because the hole’s radius is approaching that of the Our answer makes sense in limiting cases. As r
sphere. As r
sphere. R ,V 0 ,V 2 69. (a) The radius of the barrel is the same at each end by symmetry, since the function y=R cx is
even. Since the barrel is obtained by rotating the graph of the function y about the x axis, this radius
2
1
1
is equal to the value of y at x= h , which is R c
h =R d=r .
2
2 29 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes (b) The barrel is symmetric about the y axis, so its volume is twice the volume of that part of the
barrel for x>0 . Also, the barrel is a volume of rotation, so
h/2 V =2 h/2 2 y dx=2
0 ( R cx2) 2dx=2 2 Rx 0 2
3 1 2 5
Rcx + c x
3
5 h/2
0 1 2
1
1 2 5
3
Rh
Rch +
ch
2
12
160
Trying to make this look more like the expression we want, we rewrite it as
1
2
2 1
2 3 2 4
V=
h 2R + R
Rch +
ch
. But
3
2
80
1 2 2 1 2 4
1 2 2 2 2 2
2 1
2 3 2 4
2 2
R
Rch +
ch= R
ch
c h =( R d)
ch
=r
d .
2
80
4
40
5
4
5
1
2 2 2 2
Substituting this back into V , we see that V =
h 2R +r
d
, as required.
3
5
=2 70. It suffices to consider the case where is bounded by the curves y= f (x) and y=g(x) for a x b ,
where g(x) f (x) for all x in a,b , since other regions can be decomposed into subregions of this
type. We are concerned with the volume obtained when is rotated about the line y= k , which is equal
to
b V =
2 ( f (x)+k a 2 g(x)+k 2 b ) dx= (
a f (x) 2 g(x) 2 ) dx+2 b k f (x) g(x) dx
a = V 1+2 kA 30 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 1. If we were to use the "washer’’ method, we would first have to locate the local maximum point (a,b)
2 of y=x ( x 1 ) using the methods of Chapter 4. Then we would have to solve the equation y=x ( x 1 )
for x in terms of y to obtain the functions x=g (y) and x=g (y) shown in the first figure. This step
1 2 2 would be difficult because it involves the cubic formula. Finally we would find the volume using
b V=
0 { g (y) 2 g (y) 1 } dy . 2 2 Using shells, we find that a typical approximating shell has radius x , so its circumference is 2 x . Its
2 height is y , that is, x(x 1) . So the total volume is
1 2 1 V = 2 x x(x 1) dx=2
0 0 (x 4 3 2 ) 2x +x dx=2 5 4 3 x
x
x
2 +
5
4 3 1 = 0 15 2. ( 2) ( 2) . V = 2 xsin x dx . Let A typical cylindrical shell has circumference 2 x and height sin x 0
1 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 2 u=x . Then du=2xdx , so V = sin udu=
0 cos u = [1 ( 1 ) ]=2 .
0 ( 2) using the For slicing, we would first have to locate the local maximum point ( a,b) of y=sin x ( 2) methods of Chapter 4. Then we would have to solve the equation y=sin x for x in terms of y to
obtain the functions x=g (y) and x=g (y) shown in the second figure. Finally we would find the
1 volume using 2 . Using shells is definitely preferable to slicing. 3.
2 1
V = 2 x
dx=2
x
1
=2 2 1dx
1 2 x =2 (2 1)=2
1 4.
1 1 2 V = 2 x x dx=2
0 =2 3 x dx
0 1 4
x
4 1 =2 0 1
=
4 2 2 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 2 1 5. V = 2 xe x 0 2 dx . Let u=x . Thus, du=2xdx , so
1 V= 0 u e du= e u 1
0 = (1 1/e) 6.
3 V =2 3 { x[(3+2x x ) (3 x)]} dx=2
2 0
3 =2 2 x(3x x ) dx
0 2 3 (3x x )dx=2 x 0 3 3 1 4
x
4 =2 0 2 27 2 7. The curves intersect when 4(x 2) =x 4x+7
2 3(x 4x+3)=0
3 V =2
1 81
4 27
4 =2 2 2 = 27
2 4x 16x+16=x 4x+7 2 3x 12x+9=0 3(x 1)(x 3)=0 , so x=1 or 3 . {x (x 2 ) 2 4x+7 4(x 2) } dx=2 3 2 2 x(x 4x+7 4x +16x 16) dx
1 3 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 3 =2 3 2 3 1 4 4 3 3 2
x
x+ x
4
3
2 2 x( 3x +12x 9) dx=2 ( 3) (x 4x +3x)dx= 6
1 1 81
27
36+
4
2 = 6 1
4 4 3
+
3 2 = 6 20 36+12+ 4
3 = 6 3
1 8
3 =16 8. By slicing:
1 ( V= y) 2 2 2 (y ) 1 dy= 0 ( y y4) dy= 0 1 2 1 5
y
y
2
5 1 = 0 1
2 1
5 = 3
10 By cylindrical shells:
1 V = 2 x ( 2 1 ) x x dx=2 (x 0 =2 3/2 3 x )dx 0 2 5/2 1 4
x
x
5
4 1 =2 0 2
5 1
4 4 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells =2 3
20 = 3
10 9.
2 ( 2 2 ) 1 =2 1 2 1 4
y+ y
2
4 ( y+y3) dy=2 V = 2 y 1+y dy=2
1 1 1
+
2 4 ( 2+4 ) =2 21
4 = 2
1 21
2 10.
1 1 V = 2 y y dy=2 y 0 =2 3/2 dy 0 2 5/2
y
5 1 = 0 4
5 5 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 11.
8
3 V =2 y( y 0) dy
0
8 =2 4/3 y 3 7/3
y
7 dy=2 0 = 8
0 6
7/3 6
7 768
(8 )=
(2 )=
7
7
7 12.
4 V =2 2 3 y(4y y ) dy
0
4 =2 3 4 (4y y )dy
0
6 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 1 5 4
y
=2
5
0
256
512
=
5
5
4 =2 y =2 256 1024
5 2 13. The curves intersect when 4x =6 2x
2 equations for x gives us y=4x =2
V =2 4
0
4 { (y
0 y 1
2
y ) dy+2 2
=2
32 +2
5
128
433
=
+2
5
15 1
2 x=
1
2 y y 2 2x +x 3=0 (2x+3)(x 1)=0
y and 2x+y=6 } dy+2 x=
9
4 { y x= 1
y+3 .
2
1
y+3
2 1 2
1 3/2
2 5/2 4
y +3y+ y
dy=2
y
+2
4
2
2
5
0
243 243 243
32
32
+
+
+24+
2
2
5
3
5
1250
250
=
=
15
3 9 3
or 1 . Solving the
2 1
2 y } dy 1 3 3 2 1 5/2
y+ y+ y
6
2
5 9
4 14. 7 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 3 V = 2 y 4 ( y 1) 2 ( 3 y ) dy 0
3 =2 y ( ) 2 y +3y dy 0
3 =2 ( 3 2 0 =2 81
+27 =2
4 3 1 4 3
y +y
4 ) y +3y dy=2 27
4 = 0 27
2 15.
2 1 4 1 3
x
x
4
3 2 V = 2 (x 1)x dx=2
1 =2 4 8
3 1
4 1
3 = 2
1 17
6 16.
8 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 1
2 1
4 =2 1 1 4
x
4 2 V = 2 ( x) x dx=2 ( 4) = 2 15
2 17.
2
1 32
4
3 =2 2 4 3 1 4
x
x
3
4 2 V = 2 (4 x)x dx=2 4
3 1
4 = 1 67
6 18.
4 2 2 V = 2 [x ( 2)][(8x 2x ) (4x x )]dx
0
4 2 = 2 (2+x)(4x x )dx
0
4 =2 2 3 (8x+2x x )dx
0 =2
=2 2 3 1 4 4
x
x
3
4
0
128
256
64+
64 =
3
3
2 4x + 9 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 19.
2 V = 2 (3 y)(5 x)dy
0
2 ( ) 2 = 2 ( 3 y ) 5 y 1 dy
0
2 = 2 ( 12 4y 3y2+y3) dy 0 =2 2 3 12y 2y y + 1 4
y
4 2
0 =2 (24 8 8+4)=24 20.
1 V = 2 ( y+1 )
0 =2 ( 2 ) 1 y y dy=2 3/2 1/2 3 2 (y +y y y )dy 1 2 2
+
5 3 0 2 5/2 2 3/2 1 4 1 3
y + y
y
y
5
3
4
3 =2 0 1
4 1
3 =2 29
60 = 29
30 10 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 2 21. V = 2 xln xdx
1 3 22. V = 2 ( 7 x ) ( 4x x2) x dx 0 1 23. V = 2 x ( 1) sin 0 2 x x 4 dx 2 24. V = 2 (2 x)
0 1
2 dx 1+x 11 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 25. V = 2 (4 y) sin y dy
0 3 ( 26. V = 2 (5 y) 4 2 ) y +7 dy 3 27. x= /4 0
=
.
4
16 /4 V= 2 xtan xdx 2
0 16 32 tan 32 + 3
3
5
5
7
7
tan
+
tan
+
tan
32
32 32
32 32
32 1.142 12 2
*
=2 , n=5 and x =2+(2i+1) , where i=0 , 1 , 2 , 3 , 4 . The values of f (x) are taken
i
5
directly from the diagram.
28. x=
12 V = 2 xf (x)dx 2 3 f (3)+5 f (5)+7 f (7)+9 f (9)+11 f (11) 2 2 2 3(2)+5(4)+7(4)+9(2)+11(1) 2=332
12 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 3 3 5 29. 2 x dx=2
0 4 4 x(x )dx . The solid is obtained by rotating the region 0 y x , 0 x 3 about the
0 y axis using cylindrical shells.
2 2 y 30. 2 2 dy=2 1 y 2 dy . The solid is obtained by rotating the region 0 x 1+y
1+y
0
0 y 2 about the x axis using cylindrical shells.
0 1 1
2 , 1+y 2 2 31. 2 (3 y)(1 y )dy . The solid is obtained by rotating the region bounded by (i) x=1 y , x=0 , and
0
2 y=0 or (ii) x=y , x=1 , and y=0 about the line y=3 using cylindrical shells.
/4 32. 2 ( x)(cos x sin x)dx . The solid is obtained by rotating the region bounded by 0 (i) 0 y cos x sin x , 0 x 4 or (ii) sin x y cos x , 0 x 4 about the line x= using cylindrical shells.
33. 2 4 From the graph, the curves intersect at x=0 and at x=a 1.32 , with x+x x >0 on the interval ( 0,a ) .
So the volume of the solid obtained by rotating the region about the y axis is
a V =2 2 4 x(x+x x ) dx=2
0 =2 a 2 3 5 (x +x x )dx
0 1 3 1 4 1 6
x+ x
x
3
4
6 a
0 4.05 34.
13 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 3 4 From the graph, the curves intersect at x=0 and at x=a 1.17 , with 3x x >x on the interval ( 0,a ) .
So the volume of the solid obtained by rotating the region about the y axis is
a V =2 { x[(3x x ) x ]} dx=2
3 4 0 =2 a 2 4 5 (3x x x )dx
0 x 3 a 1 5 1 6
x
x
5
6 0 4.62 35.
/2 V =2 2 0 = 1
32 x ( sin 2x sin 4x) dx 3 36. { [x ( 1)](x3sin x)} dx=2 V =2 ( 4 + 3 12 2 6 +48) 0 =2 5 +2 4 24 3 12 2 +96 14 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 37. Use disks:
1 V = ( x +x 2)
2 2 1 ( x4+2x3 3x2 4x+4) dx dx= 2 2 1 5 1 4 3 2
x + x x 2x +4x
5
2
33 3
81
+
=
5 2
10 =
= 1 1 1
+
1 2+4
5 2 = 2 32
+8+8 8 8
5 38. Use shells:
2 V = 2 x ( 2 ) 2 x +3x 2 dx=2 1 ( 3 2 ) x +3x 2x dx 1 1 4 3 2
x +x x
4 =2 2 =2 1 ( 4+8 4 ) 1
+1 1
4 = 2 39. Use shells:
4 V= 2
1 2
2 (
1
4 x ( 1) 5 (x+4/x) dx =2
2 ) x +4x+1 4/x dx=2
64
+32+4 4ln 4
3 4 (x+1)(5 x 4/x)dx =2 1 ( 5x x2 4+5 x 4/x) dx =
1
4 4
1 3 2
=
x +2x +x 4ln x
3
1
1
+2+1 0
=2 (12 4ln 4)=8 (3 ln 4)
3 15 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells 40. Use washers:
1 {20 V = ( 2 4 2 1 y ) 2} dy 1
1 =2 ( 4 2 ) dy [by symmetry] ( 4 8 4 1+y
0
1 =2 4 1+2y +y 1 ) dy=2 0 0
1 2 5 1 9
3y
y
y
5
9 =2 ( 3 2y4 y8) dy =2 0 2 1 ( y 1) 41. Use disks: V = 2
5 3 2 1
9 2 2 dy= 0 =2 112
45 ( 2y y2) dy= = 2 y 0 224
45 2 1 3
y
3 = 0 4 8
3 = 4
3 42. Using shells, we have
2 V = 2 y ( 2 1 (y 1) 2 1 (y 1) ) dy 0
2 =2 y 2 1 2 1 ( y 1 ) dy=4 0 1 1 =4 2 (u+1) 1 u du [let u=y 1] u
1 2 1 1 u du+4 2 1 u du
1 The first definite integral equals zero because its integrand is an odd function. The second is the area
of a semicircle of radius 1 , that is, 2 . Thus, V =4 0+4 2 =2 2 . 43.
16 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells r 2 r 2 V =2 2 x r x dx= 2
0 2x ) dx= 2 0 4
3 = (r 2 x2) 1/2 ( ( 0 r 3) = 4
3 r 2 2 2
r x
3 ( r ) 3/2 0 3 44.
R+r V = 2 x 2 r 2 2 ( x R) dx Rr
r 2 2 = 4 (u+R) r u du [u=x R ]
r
r 2 =4 R 2 r r u du+4
r 2 2 u r u du
r The first integral is the area of a semicircle of radius r , that is, 1 2
r ,
2 and the second is zero since the integrand is an odd function. Thus, V =4 R 1 2
r +4
2 0=2 Rr 2 . r r 45. V =2 x
0 h
x+h dx=2 h
r
0 2 x
+x
r 3 dx=2 h 2 x
x
+
3r 2 r 2 2 r
r h
=2 h =
0
6
3 46. By symmetry, the volume of a napkin ring obtained by drilling a hole of radius r through a sphere
with radius R is twice the volume obtained by rotating the area above the x axis and below the curve
2 2 y= R x (the equation of the top half of the cross section of the sphere), between x=r and x=R ,
about the y axis.
This volume is equal to 17 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.3 Volumes by Cylindrical Shells outerradius 2 R 2 rhdx=2 2 innerradius 2 x 1 2 2
R x
3 ( 2 R x dx=4 r
2 2 But by the Pythagorean Theorem, R r = 1
h
2 ) 3/2 R
r = 4
3 ( R2 r 2) 3/2 2 , so the volume of the napkin ring is 3 1
4
1
3
h =
h , which is independent of both R and r ; that is, the amount of wood in a
3
2
6
napkin ring of height h is the same regardless of the size of the sphere used. Note that most of this
calculation has been done already, but with more difficulty, in Exercise 6.2.68.
Another solution: The height of the missing cap is the radius of the sphere minus half the height of the
1
cut out cylinder, that is, R
h . Using Exercise 6.2.49,
2 V =V napkinring sphere V cylinder 2V = cap 4
3
R
3 2 r h 2 3 R 1
h
2 2 3R R 1
h
2 = 1 3
h
6 18 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.4 Work 1. By Equation 2, W =Fd=(900)(8)=7200 J.
2. F=mg=(60)(9.8)=588 N; W =Fd=588 2=1176 J
3.
9 10 b 10 W = f (x)dx=
a ( 1+x ) 0 =10
2 4. W = cos
1 2 10 1
u 1 1 dx=10
1 u 2 du [u=1+x,du=dx] 1
+1 =9ft lb
10 =10 1
3
x dx=
3 1
x
3 sin 2 = 3 1 3
2 3
2 3
Interpretation: From x=1 to x= , the force does work equal to
2 =0 N m =0 J. 3/2 cos
1 1
x
3 dx= 3 1 3
2 3
to x=2 , the force opposes
2
the motion of the particle, decreasing its kinetic energy. This is negative work, equal in magnitude but
3
opposite in sign to the work done from x=1 to x= .
2
J in accelerating the particle and increasing its kinetic energy. From x= 5. The force function is given by F(x) (in newtons) and the work (in joules) is the area under the
8 4 8 curve, given by F(x)dx= F(x)dx+ F(x)dx=
0 0 4 1
(4)(30)+(4)(30)=180 J.
2 20 6. W = f (x)dx M = x[ f (6)+ f (10)+ f (14)+ f (18)]=
4 4 7. 10= f (x)=kx=
1/2 1
1
1
k [ 4 inches = foot], so k=30 lb / ft and f (x)=30x . Now 6 inches = foot, so
3
3
2
2 1/2 W = 30xdx= 15x
0 20 4
[5.8+8.8+8.2+5.2]=4(28)=112 J
4 0 = 15
ft lb.
4 8. 25= f (x)=kx=k(0.1) [10 cm =0.1 m], so k=250 N / m and f (x)=250x . Now 5 cm =0.05 m, so
0.05 W= 2 0.05 250xdx= 125x
0 0 =125(0.0025)=0.3125 0.31 J. 1 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.4 Work 0.12 1 2 0.12 1
2
2500
9. If
kxdx=2 J, then 2=
kx
= k(0.0144)=0.0072k and k=
=
2
0
2
0.0072
9
0
m. Thus, the work needed to stretch the spring from 35 cm to 40 cm is
0.10 2500
xdx=
0.05 9
1
0 0 0 1/20 1 =12 11. f (x)=kx , so 30= = 0 1250
9 1
100 1
400 = 25
24 1.04 J. 1
k , then k=24 lb / ft and the work required is
2 9 27
=
=6.75 ft lb.
16 4 2 3/4 24xdx= 12x = 1 2
kx
2 10. If 12= kxdx=
3/4 1/10 1250 2
x
9 277.78 N / 2500
270
x and x=
m =10.8 cm
9
2500 12. Let L be the natural length of the spring in meters. Then
0.12 L 6= kxdx= 0.10 L
0.14 L 0.12 L 1 2
kx
2 = 0.10 L 1
2
2
k ( 0.12 L ) ( 0.10 L ) and
2 1 2 0.14 L 1
2
2
kx
= k ( 0.14 L ) ( 0.12 L ) . Simplifying gives us
2
0.12 L 2
0.12 L
12=k(0.0044 0.04L) and 20=k(0.0052 0.04L) . Subtracting the first equation from the second gives
32
8=0.0008k , so k=10 , 000 . Now the second equation becomes 20=52 400L , so L=
m =8 cm.
400
10= kxdx= 13. (a) The portion of the rope from x ft to ( x+ x ) ft below the top of the building weighs
* and must be lifted x ft, so its contribution to the total work is
i 1
2 x lb 1 *
x x ft lb. The total work is
2 i 50 1 *
1
1 2 50 2500
W =lim
x x=
xdx=
x
=
=625 ft lb Notice that the exact height of the
i=1 2 i
4
0
4
n
0 2
building does not matter (as long as it is more than 50 ft).
(b) When half the rope is pulled to the top of the building, the work to lift the top half of the rope is
n 25 W =
1 0 1
xdx=
2 1 2
x
4 25
0 = 625
ft lb. The bottom half of the rope is lifted 25 ft and the work needed
4 50 to accomplish that is W =
2 25 1
25
25dx=
x
2
2 50 = 25 625
ft lb. The total work done in pulling half the
2
2 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.4 Work rope to the top of the building is W =W +W =
1 2 625 625 3
1875
+
= 625=
ft lb.
2
4
4
4 14. Assumptions : 1. After lifting, the chain is L shaped, with 4 m of the chain lying along the ground.
2. The chain slides effortlessly and without friction along the ground while its end is lifted.
3. The weight density of the chain is constant throughout its length and therefore equals
2 (8kg/m)(9.8m/s )=78.4N/m .
The part of the chain xm from the lifted end is raised 6 xm if 0 x 6 m, and it is lifted 0 m if x>6m
.
Thus, the work needed is
W =lim
n n * 6 (6 x ) 78.4 x= (6 x)78.4dx=78.4 i=1 i 6x 0 15. The work needed to lift the cable is lim
n n 2x i=1 * i 6 1 2
x
2 =(78.4)(18)=1411.2 J. 0 500 x= 2 500 2xdx= x 0 0 =250 , 000 ft lb. The work needed to lift the coal is 800 lb 500 ft =400 , 000 ft lb. Thus, the total work required is 250 ,
000+400 , 000=650 , 000 ft lb.
16. The work needed to lift the bucket itself is 4 lb 80 ft =320 ft lb. At time t (in seconds) the bucket
* is x =2t ft above its original 80 ft depth, but it now holds only ( 40 0.2t ) lb of water. In terms of
i
distance, the bucket holds 40 0.2 1 *
x
2 i depth. Moving this amount of water a distance * lb of water when it is x ft above its original 80 ft
i x requires 40 1 *
x
10 i x ft lb of work. Thus, the work needed to lift the water is
80 1 *
1
1 2
W =lim
40
x
x=
40
x dx= 40x
x
i=1
10 i
10
20
n
0
work of lifting the bucket gives a total of 3200 ft lb of work.
n 80
0 = ( 3200 320 ) ft lb Adding the 17. At a height of x meters ( 0 x 12 ), the mass of the rope is (0.8 kg / m )(12 x m )=(9.6 0.8x) kg
36
and the mass of the water is
kg/m (12 xm)=(36 3x) kg. The mass of the bucket is 10 kg, so
12
the total mass is (9.6 0.8x)+(36 3x)+10=(55.6 3.8x) kg, and hence, the total force is 9.8(55.6 3.8x)
N.
The work needed to lift the bucket * x m through the i th subinterval of 0,12 is 9.8(55.6 3.8x ) x ,
i so the total work is 3 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.4 Work W =lim
n n 12 * 2 12 9.8(55.6 3.8x ) x= (9.8)(55.6 3.8x)dx=9.8 55.6x 1.9x
i i=1 0 0 =9.8(393.6) 3857 J
25lb
=2.5 lb / ft. The part of the chain x ft below the ceiling (for
10ft
5 x 10 ) has to be lifted 2(x 5) ft, so the work needed to lift the i th subinterval of the chain is 2 ( 18. The chain’s weight density is
* x 5)(2.5 x) . The total work needed is
i W =lim
n =5 n 10 * 2(x 5)(2.5) x= [2(x 5)(2.5)]dx=5 (x 5)dx
i i=1 5 5 10 1 2
x 5x
2 =5 5 25
25
2 (50 50) =5 25
2 =62.5 ft lb
* * i 19. A ‘‘slice’’ of water (2 1 10 i x m thick and lying at a depth of x m (where 0 x 1
) has volume
2 3 x ) m , a mass of 2000 x kg, weighs about (9.8)(2000 x)=19 , 600 x N, and thus requires about 19 , 600x * i x J of work for its removal. So W =lim
n n 19 , 600x * x= i i=1 1/2
0 19 , 2 1/2 600xdx= 9800x 0 =2450 J. 20. A horizontal cylindrical slice of water ( weighs about 62.5lb/ft
pool (where 1 x * i ) ( 144 ) xft =9000 9000 x * i 5 * i * x . Thus, 2 5 x= 9000 xdx= 4500 x
1 21. A rectangular ‘‘slice’’ of water 3 x ft and x lb. If the slice lies x ft below the edge of the
i i=1 n 3 2 12 5 ), then the work needed to pump it out is about 9000 x
n W =lim 3 2 x ft thick has a volume of r h= =4500 (25 1)=108,000 ft lb 1 x m thick and lying x ft above the bottom has width x ft and 3 volume 8x x m . It weighs about ( 9.8 1000 ) ( 8x x ) N, and must be lifted ( 5 x ) m by the pump, ( 3 so the work needed is about 9.8 10 ) (5 x)(8x x) J. The total work required is 4 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.4 Work 3 W ( 9.8 3 10 ) (5 x)8xdx= ( 9.8 3 10 ) ( 40x 8x2) dx= ( 9.8 0 = ( 9.8 3 3 10 ) 8 3
x
3 2 20x 0
3 10 ) (180 72)= ( 9.8 3 10 ) (108)=1058.4 3 3
0 6 10 1.06 10 J 22. For convenience, measure depth x from the middle of the tank, so that 1.5 x 1.5 m. Lifting a
slice of water of thickness x at depth x requires a work contribution of
W
W ( 9.8 103) ( 2 ( 1.5) 2 x2 ) (6 x)(2.5+x) , so
1.5
( 9.8 103) 12 2.25 x2 (2.5+x)dx
1.5 = ( 9.8 3 10 3/2 ) 9 2
x dx+12
4 60
0 3/2 9 2
x dx
4 x
3/2 The second integral is 0 because its integrand is an odd function, and the first integral represents the
area of a quarter circle of radius
3
.Therefore,[W
2 ( 9.8 3 10 3/2 ) 60 0 9 2
3
x dx= 9.8 10 ( 60 )
4 ( 1
4 ) 3
2 2 =330,750 6 1.04 10 J] 23. Measure depth x downward from the flat top of the tank, so that 0 x 2 ft. Then ( 2 W = ( 62.5) 2 4 x ) (8 2 W (62.5)(16) (x+1) x)(x+1) ft lb, so
2 2 4 x dx=1000 0
4 = 1000 u 1/2 0 x 2 2 4 x dx+ 0 1
2 du+ 1
4 ( 22) 2 4 x dx
0 2 [Put u=4 x ,so du= 2xdx] 1 2 3/2 4
8
3
u
+
=1000
+
5.8 10 ft lb
2 3
0
3
Note: The second integral represents the area of a quarter circle of radius 2 .
= 1000 24. Let x be depth in feet, so that 0 x 5 . Then
5 W 62.5 2
x ( 25 x ) dx=62.5 0 25 2 1 4
x
x
2
4 W =(62.5)
5 =62.5 0 (
625
2 2 2 5 x 625
4 ) 2 x x ft lb and
=62.5 625
4 4 3.07 10 ft lb
5 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.4 Work 5 25. If only 4.7 10 J of work is done, then only the water above a certain level (call it h ) will be
pumped out. So we use the same formula as in Exercise 21 , except that the work is fixed, and we are
trying to find the lower limit of integration:
3 5 4.7 10 ( 9.8 10 ) (5 x)8xdx= ( 9.8 10 ) 3 3 20x h 4.7
2
2 8 3
10 48= 20 3
3
9.8
3
3 8 3
h
3 2 20h 3 8 3
x
3 h 3 2 2 2h 15h +45=0 . To find the solution of this 2 equation, we plot 2h 15h +45 between h=0 and h=3 . We see that the equation is satisfied for h 2.0
. So the depth of water remaining in the tank is about 2.0 m. 3/2 ( 9.8 920 ) 12 26. W 0
3/2 9 2
1
x dx=
4
4 Here
0 9 2
x
4
3
2 2 9
=
and
16 1 3/2 9/4 1
27
u
=
3
0
3
8
9
9
30
+12
=9016
16
8 du= 2xdx ] =
W 9016 3/2 5
+x dx=9016
2
3/2 30
0 3/2 9 2
x dx+12 x
4
0
9 2
x dx
4 . 9/4 9 2
1 1/2
9 2
x dx=
u du [ where u=
x , so
4
4
0 2 x
0 9
, so
8
135
27
+
8
2 = 5 6.00 10 J. 2 2 27. V = r x , so V is a function of x and P can also be regarded as a function of x . If V = r x and
1 1 2 V = r x , then
2 2 x 2 W = x 2 F(x)dx=
x 1 2 r P(V (x))dx x 1 x 2 = 2 2 P(V (x))dV (x)[LetV (x)= r x,sodV (x)= r dx.]
x 1 6 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.4 Work V 2 = P(V )dV by the Substitution Rule.
V 1 2 100
3 800
3
3
ft , and 800 in =
ft .
1728
1728
1.4
1.4
25
=23 , 040
426.5 . Therefore, P 426.5V
432
3 2 28. 160 lb / in =160 144 lb / ft , 100 in =
k=PV 1.4 = ( 160 144 ) 800/1728 W = 426.5V 100
1728
1.4 1
V
0.4 dV =426.5 100/1728 = (426.5)(2.5) 0.4 432
25 54
25 1.4 and 25/54 0.4 25/432 0.4 3 1.88 10 ft lb
b
b mm 29. W = F(r)dr= G
a a 1 2 r 2 dr=Gm m 1 2 1
r b =Gm m a 1 2 1
a 1
b 1
1
where M= mass of Earth in kg, R= radius of
R R+1,000,000
Earth in m, and m= mass of satellite in kg. (Note that 1000 km =1 , 000 , 000 m.) Thus,
30. By Exercise 29 , W =GMm ( W = 6.67 10 11 ) ( 5.98 24 10 ) (1000) 1 1
6 6.37 10 9 6 8.50 10 J 7.37 10 7 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.5 Average Value of a Function b 1
1. f =
ave b a
2. f = ave 1 4 4 1 1 (1/x)dx= ave cos xdx= 2 sin x 0 0 2 1 1
2
x dx= 2 x dx=
2 0
1
2 1 3
x
3 1 = 0 1
3 1
4 1
ln x = ln 4 0.46
1 3
3 /2 1 3. g = 1 1
f (x)dx=
1 ( 1)
a 2 /2
0 = (1 0)= 2 4.
g ave = 2 1
2 0 = 0 9 2 3/2
u
3
1 5 1 t 2 te dt= 1
5 ( 6. f 1 = ave 0 4 1 1
du
2 25 u e 0 25 ) 1 = 1
1 e
10 ( /4 sec tan d = 0 1
3
2
du [ u=1+x , du=3x dx ]
3 u 1
26
( 27 1 ) =
9
9 = 5 0 0
1
1
u 25
=
e
=
e
0
10
10
ave 9 3 x 1
=
6
5. f 1
1+x dx=
2 2 4 25 2 [ u= t , du= 2t dt , t dt= 1
du ]
2 )
/4 sec 0 = 4 ( 2 1) 7.
h ave 1 = = 4 0
1 1 cos xsin xdx=
0
4 1 1 4 u ( du) [ u=cos x , du= sin xdx ]
1 u du= 1 1 4 2 u du= 1 0 2 1 5
u
5 1 = 0 2
5 8.
6 h ave 1
=
6 1 7 3
1 ( 1+r ) 2 1
2
dr=
3u du [ u=1+r , du=dr ]
52 1 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.5 Average Value of a Function = 3
u
5 1 7 = 2 3
5 1
7 1
2 = 9.
(a)
f ave =
(b) = 3
5 1
5 2 1
2 1
7 5 2 = (x 3) dx=
2 3 5
3
=
5 14 14 1
3 5 1
3
(x 3)
3 2 1 3
1
3
2 ( 1) = (8+1)=1
9
9 f (c)= f 2 ave c 3= 1 (c 3) =1 c=2 or 4
(c) 10.
(a)
f ave = 1
4 0 4 x dx=
0 1
4 2 3/2
x
3 1 3/2 4 1
4
=
x
= [8 0]=
0 6
6
3 4
0 (b) f (c)= f ave c= 4
3 c= 16
9 (c) 11. 2 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.5 Average Value of a Function (a)
f ave 1 =
=
= 0
1 (2sin x sin 2x)dx
(b) 0 2cos x+ 1 2+ 1
cos 2x
2 1
2 2+ c 0 1
2 f (c)= f = ave 2sin c sin 2c= 1.238 or c 1 4 2.808 2 4 (c) 12.
(a) 2 f ave = 1 2x 2 0 0 22 dx (1+x ) (b) 5 1
=
2
= 1
1 1
2 u 2 1
u 2 du [ u=1+x , du=2xdx ]
5 = 1 1
2 f (c)= f
c 1 2c
ave 22 = 2
5 22 5c=(1+c ) (1+c )
0.220 or c 1.207
2 1
2
1 =
5
5 (c) 13. f is continuous on 1,3 , so by the Mean Value Theorem for Integrals there exists a number c in
3 1,3 such that f (x)dx= f (c)(3 1)
1 8=2 f (c) ; that is, there is a number c such that f (c)= 8
=4 .
2 14. The requirement is that 3 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.5 Average Value of a Function 1
b 0
b 1
b 0 b= b f (x)dx=3 . The LHS of this equation is equal to
0 ( 2+6x 3x2) dx= 1
b 2 2x+3x x 3 b 2 2 =2+3b b , so we solve the equation 2+3b b =3 0 2 b 3b+1=0 2 ( 3) 4 1 1 3 5
. Both roots are valid since they are positive.
=
2 1
2 3 15.
f ave 50
1
1
1 50 20
f (x)dx
M=
[ f (25)+ f (35)+ f (45)]
50 20 20
30 3 30
3
1
115
1
= (38+29+48)=
=38
3
3
3 = 1
16. (a) v =
ave 12 0 12 1
12 0
I . Use the Midpoint Rule with n=3 and t=
=4 to estimate I .
12
3
0
1
2
I M =4[v(2)+v(6)+v(10)]=4[21+50+66]=4(137)=548 . Thus, v
(548)=45 km / h.
3
ave 12
3
2
(b) Estimating from the graph, v(t)=45 when t 5.2 s.
3
v(t)dt= 17. Let t=0 and t=12 correspond to 9 A.M. and 9 P.M., respectively.
T ave 1
=
12 0
= 1
12 12 1
1
t dt=
12
12 50+14sin
0 50 12+14 12 +14 12 50t 14 = 50+ 28 12 cos 1
t
12 12
0 59 F F 18.
T ave 30
1
1
t/50
t/50
20+75e
dt=
20t 50 75e
30 0 0
30
1
3/5
3/5
=
4350 3750e
=145 125e
76.4 C
30 ( = ( 1
19.
=
ave 8 8
0 ) 30
0 = 1
30 ( 600 3750e 3/5) ( 3750 ) ) 12
3
dx=
2
x+1 8 ( x+1 )
0 1/2 dx= 3 x+1 8 =9 3=6 kg / m 0 4 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.5 Average Value of a Function 2 v
1 2
ds
2
20. s= gt
. We see that
t= 2s/g [ since t 0 ]. Now v= =gt=g 2s/g = 2gs v =2gs s=
2g
2
dt
v can be regarded as a function of t or of s : v=F(t)=gt and v=G(s)= 2gs . Note that v =F(T )=gT .
T 2 Displacement can be viewed as a function of t : s=s(t)= 2 1 2
v
[F(t)]
gt ; also s(t)=
.
=
2
2g
2g When t=T , these two formulas for s(t) imply that
1
2
2gs(T ) =F(T )=v =gT =2
gT
/T =2s(T )/T (*)
T
2
The average of the velocities with respect to time t during the interval 0,T is
v =F t ave ave 1 = T 0 T F(t)dt=
0 1
s(T ) s(0) [ by FTC]
T s(T )
1
[ since s(0)=0 ] = v [ by (*) ]
T
2 T = But the average of the velocities with respect to displacement s during the corresponding
displacement interval s(0),s(T ) = 0,s(T ) is
v =G s ave ave 1
=
s(T ) 0 s(T )
0 1
G(s)ds=
s(T ) 2g 2 3/2
s
s(T ) 3 = s(T )
0 = s(T )
0 2g
2
3 s(T ) s(T ) 2g
2gs ds=
s(T )
s(T ) 3/2 = 1/2 s ds 0 2
2
2gs(T ) = v [by(*)]
3
3 T 21.
5 V ave 1
1
=
V (t)dt=
50
5
= 1
4 t
R 5
0 5
4 5
sin
2 1
1
v(r)dr=
22. v =
ave R 0
R
0 2
t
5 1 cos
5 2
t
5
R
0 P
4 l = 0 ( 1
4 ( 5 0) 0 = P
R r dr=
4 lR
2 2 5 1
dt=
4 ) 1 cos
0 5
4 dt 0.4 L 1 3
Rr
r
3
2 2
t
5 R P
=
0 4 lR 2
3 2 PR
R=
.
6 l
3 2 PR
2
Since v(r) is decreasing on ( 0,R , v =v(0)=
. Thus, v = v
.
max
ave 3 max
4 l
23. Let
5 Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.5 Average Value of a Function x F(x)= f (t)dt for x in a,b . Then F is continuous on a,b and differentiable on ( a,b) , so by the
a
/ / Mean Value Theorem there is a number c in ( a,b) such that F(b) F(a)=F (c)(b a) . But F (x)= f (x)
b by the Fundamental Theorem of Calculus. Therefore, f (t)dt 0= f (c)(b a) .
a 24.
f ave a,b = 1
b a
c a
=
b a b 1
f (x)dx=
b a
a
1
c a c 1
f (x)dx+
b a
a c f (x)dx
a b c
+
b a b f (x)dx
c 1
b c b f (x)dx
c = c a
b c
f
a,c +
f
c,b
b a ave
b a ave 6 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 1 2
x . Then by Equation 2, udv=uv vdu ,
2
1 2
1 2
1
1 2
1 1 2
x ( dx/x ) = x ln x
xdx= x ln x
x +C
2
2
2
2
2 2 1. Let u=ln x , dv=xdx du=dx/x , v= 1 2
x ln x
2
1 2
1 2
=
x ln x
x +C
2
4 xln xdx = 2. Let u= , dv=sec
sec 2 2 d d = tan du=d , v=tan tan d = tan 3. Let u=x , dv=cos 5xdx
xcos 5xdx= 1
xsin 5x
5
x 4. Let u=x , dv=e dx 1
sin 5x . Then by Equation 2,
5
1
1
1
sin 5xdx= xsin 5x+
cos 5x+C .
5
5
25
du=dx , v= x x du=dr , v=2e 6. Let u=t , dv=sin 2t dt du=dt , v= x r/2 r/2 x . Then xe dx= xe + e dx= xe r/2 5. Let u=r , dv=e dr dt= +C . ln sec du=dx , v= e r/2 . Then . Then re dr=2re r/2 x x e +C .
r/2 2e dr=2re r/2 4e +C . 1
1
1
cos 2t . Then tsin 2t dt=
tcos 2t+
cos 2t
2
2
2 1
1
tcos 2t+ sin 2t+C .
2
4
2 7. Let u=x , dv=sin xdx
2 I= x sin xdx= 1 du=2xdx and v= 2 x cos x+ Next let U =x , dV =cos xdx
xcos xdx= 1 xsin x * ), we get
1 2
2
I=
x cos x+
where C= 2 1 1 2 cos x . Then xcos xdx ( * ). dU =dx , V =
sin xdx= xsin x+ 1 1
2 1 1 sin x , so xsin x+ cos x+C 1 1
2 = cos x+C . Substituting for xcos xdx in (
1 1 2 x cos x+ 2
2 xsin x+ 2
3 cos x+C , C .
1 1 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 2 8. Let u=x , dv=cos mxdx du=2xdx , v= 1
sin mx . Then
m 1 2
2
x sin mx
xsin mxdx ( * ). Next let U =x , dV =sin mxdx dU =dx ,
m
m
1
1
1
1
1
V=
cos mx , so xsin mxdx=
xcos mx+
cos mxdx=
xcos mx+
sin mx+C .
1
2
m
m
m
m
m
Substituting for xsin mxdx in ( * ), we get
1 2
2
1
1 2
1
2
2
I= x sin mx
xcos mx+
sin mx+C
= x sin mx+
xcos mx
sin mx+C ,
1
2
2
3
m
m
m
m
m
m
m
2
where C=
C .
m 1
2 I= x cos mxdx= 9. Let u=ln (2x+1) , dv=dx 2
dx , v=x . Then
2x+1
2x
(2x+1) 1
dx=xln (2x+1)
dx
2x+1
2x+1
1
1
1
dx=xln (2x+1) x+ ln (2x+1)+C
2x+1
2 du= ln (2x+1)dx =xln (2x+1)
=xln (2x+1)
= 1
(2x+1)ln (2x+1) x+C
2 1 10. Let u=sin x , dv=dx du= dx 1 1 , v=x . Then sin xdx=xsin x
2 2 1 x
= t 1
1
1/2
1/2
2
dt =
2t
+C=t +C= 1 x +C .
2
2 ( 1/2 2 1 x
1 1 dx . Setting
2 1 x
xdx t=1 x , we get dt= 2xdx , so x ) 2 Hence, sin xdx=xsin x+ 1 x +C .
11. Let u=arctan 4t , dv=dt 4 du= 2 dt= 1+(4t)
arctan 4t dt =t arctan 4t
=t arctan 4t 4t
1+16t 2 4 1+16t
1
dt=t arctan 4t
8 2 dt , v=t . Then
32t
1+16t 2 dt 1
2
ln (1+16t )+C
8 5 12. Let u=ln p , dv= p dp
2 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts du= 1
1 6
1 6
1 5
1 6
1 6
5
dp , v= p . Then p ln pdp= p ln p
p dp= p ln p
p +C .
p
6
6
6
6
36
2 13. First let u= ( ln x ) , dv=dx
2 2 1
dx , v=x . Then by Equation 2,
x du=2ln x 1
2
dx=x(ln x) 2 ln xdx . Next let U =ln x , dV =dx
x
x ( 1/x ) dx=xln x dx=xln x x+C . Thus, I= (ln x) dx=x(ln x) 2 xln x
to get ln xdx=xln x
2 1 ( I=x(ln x) 2 xln x x+C
3 dU =1/xdx , V =x t 14. Let u=t , dv=e dt 2 1 ) =x(ln x)
2 2xln x+2x+C , where C= 2C .
1 t 3 t 3 t du=3t dt , v=e . Then I= t e dt=t e 2 t 3t e dt . Integrate by parts twice t more with dv=e dt .
I = t 3et ( 3t2et t ) 3 t 2 t t t 6te dt =t e 3t e +6te ( 6e dt ) = t 3et 3t 2et+6tet 6et+C= t 3 3t 2+6t 6 et+C
More generally, if p(t) is a polynomial of degree n in t , then repeated integration by parts shows that
t
/
/ /
/ / /
n ( n)
t
p ( t ) e dt= p(t) p (t)+ p (t) p
(t)+
+ ( 1 ) p (t) e +C .
2 15. First let u=sin 3 , dv=e d du=3cos 3 d , v= 1 2
e . Then
2 1 2
3 2
2
e sin 3
e cos 3 d . Next let U =cos 3 , dV =e d
2
2
1 2
dU = 3sin 3 d , V = e to get
2
1 2
3 2
2
e cos 3 d = e cos 3 +
e sin 3 d . Substituting in the previous formula gives
2
2
1 2
3 2
9 2
1 2
3 2
9
I= e sin 3
e cos 3
e sin 3 d = e sin 3
e cos 3
I
2
4
4
2
4
4
13
1 2
3 2
1 2
4
I= e sin 3
e cos 3 +C . Hence, I =
e ( 2sin 3 3cos 3 ) +C , where C= C .
1
4
2
4
13
13 1
2 I= e sin 3 d = 16. First let u=e , dv=cos 2 d I= e cos 2 d = 1
e sin 2
2 Next let U =e , dV =sin 2 d 1
sin 2 . Then
2
1
1
1
sin 2
e d = e sin 2 +
e sin 2 d .
2
2
2
1
cos 2 , so
dU = e d , V =
2
du= e d , v= ( ) 3 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 1
1
1
1
e cos 2
cos 2
e d =
e cos 2
e cos 2 d .
2
2
2
2
1
1
1
1
1
1
sin 2 +
e cos 2
I = e sin 2
e cos 2
I
2
2
2
2
4
4
1
sin 2
e cos 2 +C
1
4
1
2
1
e sin 2
e cos 2 +C = e sin 2
e cos 2 +C .
1
4
5
5 ( e sin 2 d =
1
e
2
5
1
I= e
4
2
4
1
I=
5
2 So I= ) 17. Let u=y , dv=sinh ydy du=dy , v=cosh y . Then
ysinh ydy=ycosh y cosh ydy=ycosh y sinh y+C .
sinh ay
. Then
a
1
ysinh ay cosh ay
sinh aydy=
+C .
2
a
a
a 18. Let u=y , dv=cosh aydy
ycosh aydy= ysinh ay
a 1
cos 3t . Then
3
1
1
1
1
t cos 3t +
cos 3t dt=
0 +
sin 3t = .
0 3
3
0 3 0
3
9 19. Let u=t , dv=sin 3t dt
0 du=dy , v= t sin 3t dt= du=dt , v= 2 x dx= ( x +1) e 20. First let u=x +1 , dv=e dx ( x +1) e
0 1 2 ,V= e
1
0 x x ( x +1) e
2 x x 1 1
0 1 2 x . By (6), 0 x 1 1 x x + 2xe dx= 2e +1+2 xe dx . Next let U =x , dV =e dx x xe dx= dx= 2e +1+2 21. Let u=ln x , dv=x dx 1 0 2 . By (6) again, du=2xdx , v= e ( 1 xe x 1 1 0
1 x + e dx= e + e 0 ) 0 1 1 x 1 = e 0 1 1 dU =dx 1 e +1= 2e +1 . So 1 2e +1 = 2e +1 4e +2= 6e +3 . du= 1
dx , v= x
x 1 . By (6), 2 ln x
1 2 x dx= ln x
x 2 2 2 + x dx= 1 1 22. Let u=ln t , dv= t dt 1
ln 2+ln 1+
2 du=dt/t , v= 1
x 2 = 1 1
1
1
ln 2+0
+1=
2
2
2 1
ln 2 .
2 2 3/2
t . By Formula 6,
3 4 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 4
1 4 2 3/2
t ln t
3 t ln t dt= 1 dy 2
3 2y 2
t dt= 8 ln 4 0
3 4
1 4 2 2 3/2
t
3 3 = 1 16
4
16
28
ln 4
( 8 1 ) = ln 4
3
9
3
9 .
23. Let u=y , dv= 2y =e dy 1
e
2 du=dy , v= e
y 1 0 2y 1
ye
2 dy= e 1 2y + 0 1
2 2 24. Let u=x , dv=csc xdx
/2
/4 2 xcsc xdx =
4 0 1 2
e +0
2 2y e . Then dy= 1
e
4 2y 1 = 0 1
e
2 2 1 2 1 1
e + =
4
4 4 3 2
e .
4 du=dx , v= cot x . Then
/2 /2 /4 xcot x = 1 2y /4 + +ln 1 ln 1 25. Let u=cos x , dv=dx cot xdx= 2 1
= +0 ln 2
2 4
dx du= 0+ 4 1/2 = + 4 /2 1+ ln sin x [see Exercise 5.5.] /4 1
ln 2
2 , v=x . Then
2 1 x
1/2
1/2 1 1 I= cos xdx= xcos x
0 6 + 1
2 1 1/2 t dt= 3/4 + 2 1 x x x x5 dx = x5
ln 5 5
= ln 5 6 + t 1 = 6 3/4 ( x 26. Let u=x , dv=5 dx 0 0 3/4 xdx
0 Thus, I= 1 1/2 1
=
+ t
2 3 1
+1 3 1
=
2
6 ( 1
2 1/2 2 dt , where t=1 x dt= 2xdx . +6 3 3 ) . ) x du=dx , v= 5 /ln 5 . Then
1 1
0 0 x 5
5
1
dx=
0
ln 5
ln 5
ln 5 x 5
ln 5 1 = 0 5
ln 5 5 ( ln 5) 2 + 1 ( ln 5) 2 4 ( ln 5) 2 27. Let u= ln ( sin x ) , dv=cos xdx cos x
dx , v=sin x . Then
sin x
cos xdx=sin x ln ( sin x ) sin x+C . du= I= cos x ln ( sin x ) dx=sin x ln ( sin x ) 5 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts Another method: Substitute t=sin x , so dt=cos xdx . Then I= ln t dt=t ln t t+C (see Example 2) and
so I=sin x ( ln sin x 1 ) +C .
28. Let u=arctan(1/x) , dv=dx 1 du= 1
2 2 1+(1/x)
3 arctan(1/x)dx = xarctan
1 3 = 6 + 4 3 3 1
x 1 +
1 , v=x . Then 2 x +1 x x 2
1
2
x +1= 3
ln (x +1)
1
+
dx
6
4 2
3 1
(ln 4 ln 2)=
2 6 w 29. Let w=ln x dx dx= 2 + 3 1
4
ln =
2
2 6 2 + 3
1 1
ln 2
2 w dw=dx/x . Then x=e and dx=e dw , so
1 w
w
cos ( ln x ) dx = e cos wdw= e ( sin w+cos w ) +C [by the method of Example 4]
2
1
=
x sin ( ln x ) +cos ( ln x ) +C
2
r 2 30. Let u=r , dv= 2 dr 4+r du=2r dr , v= 4+r . By (6), 2 1
3 2 dr 4+r 0 r = r = 5 4+r 2 x (ln x) dx=
1 5 x
2
( ln x )
5
4 Then
1 4 x
ln xdx=
5 2
2 r 4+r dr= 5
3
0
2 5 1 10
3 + ( 4+r 2) 3/2
16 16
=
3
3 1
0 7
5
3 5 4 x
dx
Let U =ln x , dV =
5
2 0 1 2 3/2 2
(5) + (8)= 5
3
3 31. Let u=(ln x) , dv=x dx
4 1 2 2 2 2 x
ln x
du=2
dx , v=
. By (6),
5
x
2
1 2 2 4 4 x
32
x
2
2
ln xdx=
ln xdx .
( ln 2 ) 0 2
5
1 5
1 5
5 x
1
.
dU = dx , V =
25
x x
ln x
25 2 2
1 1 4 x
32
dx=
ln 2 0
25
25 5 x
125 2 = 1 32
ln 2
25 32
125 1
125 .
6 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 2 4 2 So x (ln x) dx=
1 32
2
(ln 2) 2
5 32
31
ln 2
25
125 s s 32. Let u=sin (t s) , dv=e ds du= cos (t s)ds , v=e . Then
t s t s 0 t s 32
62
2 64
(ln 2)
ln 2+
.
5
25
125 = 0 t 0 s I= e sin (t s)ds= e sin (t s) + e cos (t s)ds=e sin 0 e sin t+I . For I , let U =cos (t s) , dV =e ds
0 s t dU =sin (t s)ds , V =e . So I = e cos (t s)
1 t t I= sin t+e cos t I I= 2I=e cos t sin t 0 1
0 1
t t s 0 s e sin (t s)ds=e cos 0 e cos t I . Thus, 1 t
e cos t sin t .
2 ( ) 2 33. Let w= x , so that x=w and dx=2wdw . Thus, sin x dx= 2wsin wdw . Now use parts with
u=2w , dv=sin wdw , du=2dw , v= cos w to get
2wsin wdw = 2wcos w+ 2cos wdw= 2wcos w+2sin w+C
2 x cos x +2sin x +C=2 ( sin x = 4 2 34. Let w= x , so that x=w and dx=2wdw . Thus,
w w dv=e dw , du=2dw , v=e to get
2 35. Let x=
3 cos 2 w
1 e 2wdw= 2we x cos x ) +C
e 1
w 2
1 x 2 w dx= e 2wdw . Now use parts with u=2w ,
1 2 w ( 2 2 ) 2 2 e dw=4e 2e 2 e e =2e .
1 , so that dx=2 d . Thus, ( 2) d 2 = /2 /2 cos 1
1
(2 d )=
2
2 ( 2) /2 xcos x}{dx} . Now use parts with u=x , dv=cos xdx , du=dx , v=sin x to get
1
2 xcos xdx =
/2 = 1
2 xsin x /2 sin xdx = /2 1
1
( sin +cos )
2
2 2 sin 1
xsin x+cos x
2
2 +cos = 2 /2 1
(
2 0 1) 1
2 2 1+0 = 1
2 4 36.
2 5 x x e dx = 2 2 2 x (x ) e 2 t xdx= t e 1
dt=xdx ]
2 1
2
dt [ where t=x
2 2 1 2
1 4 2
t
x
=
t 2t+2 e +C [ by Example 3] =
x 2x +2 e +C
2
2 ( ) ( ) 7 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 37. Let u=x , dv=cos xdx du=dx , v= ( sin x ) / . Then
sin x
sin x
xsin x cos x
xcos xdx=x
dx=
+
+C . We see from the graph that this is
2 reasonable, since F has extreme values where f is 0 . 1
2 5/2
dx , v= x . Then
x
5
2 5/2
2 3/2
2 5/2
2 2 5/2
3/2
=
x ln x
x dx= x ln x
x +C
x ln xdx
5
5
5
5
2 5/2
4 5/2
=
x ln x
x +C.
5
25
We see from the graph that this is reasonable, since F has a minimum where f changes from negative
to positive.
38. Let u=ln x , dv=x 3/2 x 39. Let u=2x+3 , dv=e dx
x x du= dx x du=2dx , v=e . Then
x x x x (2x+3)e dx=(2x+3)e 2 e dx=(2x+3)e 2e +C= ( 2x+1 ) e +C . We see from the graph that this is
reasonable, since F has a minimum where f changes from negative to positive. 8 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 2 3 x 2 2 x 40. x e dx= x xe dx=I .
2 2 1 x
Let u=x , dv=xe dx du=2xdx , v= e . Then
2
2
2
2
2
2
1 2 x
1 2 x 1 x
1 x 2
x
I= x e
xe dx= x e
e +C= e x 1 +C . We see from the graph that this is
2
2
2
2
reasonable, since F has a minimum where f changes from negative to positive.
2 x ( ) 1
1
x sin 2x
cos xsin x+
1dx=
+C .
2
2
2
4
1
3
1
3
3
4
3
2
3
sin xdx=
cos xsin x+
sin xdx=
cos xsin x+ x
sin 2x+C .
4
4
4
8
16
2 41. (a) Take n=2 in Example 6 to get sin xdx=
(b) 42. (a) Let u=cos n 1 x , dv=cos xdx du= (n 1)cos n 2 xsin xdx , v=sin x in (2): n
n 1
n 2
2
cos xdx = cos xsin x+(n 1) cos xsin xdx ( ) = cos n 1 xsin x+(n 1) cos n 2 x 1 cos 2 x dx
= cos n 1 xsin x+(n 1) cos n 2 xdx (n 1) cos nxdx
n Rearranging terms gives n cos xdx=cos
n cos xdx= n 1 xsin x+(n 1) cos n 2 xdx or 1
n 1
n 1
n 2
cos xsin x+
cos xdx
n
n 1
1
x sin 2x
cos xsin x+
1dx= +
+C .
2
2
2
4
1
3
1
3
3
4
3
2
3
cos xdx= cos xsin x+
cos xdx= cos xsin x+ x+
sin 2x+C
4
4
4
8
16
2 (b) Take n=2 in part (a) to get cos xdx=
(c) n 43. (a) From Example 6, sin xdx= 1
n 1
n 1
n 2
cos xsin x+
sin xdx . Using (6),
n
n 9 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts /2
0 /2 n 1 n cos xsin x
n 1 /2 n 2
sin xdx =
+
sin xdx
n
0
n 0
n 1 /2 n 2
n 1 /2 n 2
= (0 0)+
sin xdx=
sin xdx
n 0
n 0 /2 2
2 /2
2
(b) Using n=3 in part (a), we have
sin xdx=
sin xdx=
cos x
= .
0
3 0
3
0
3
/2
4 /2 3
4 2 8
5
Using n=5 in part (a), we have
sin xdx=
sin xdx=
=
.
0
5 0
5 3 15
(c) The formula holds for n=1 (that is, 2n+1=3 ) by (b). Assume it holds for some k 1 . Then
/2
2 4 6
2k+1
( 2k )
sin
xdx=
. By Example 6,
0
3 5 7
( 2k+1 )
/2
2k+3
2k+2 /2 2k+1
2k+2 2 4 6
( 2k )
sin
xdx =
sin
xdx=
0
2k+3 0
2k+3 3 5 7
( 2k+1 )
2 4 6
(2k)[2 ( k+1 ) ]
=
,
3 5 7
(2k+1)[2 ( k+1 ) +1]
so the formula holds for n=k+1 . By induction, the formula holds for all n 1 .
/2 3 44. Using Exercise 43 (a), we see that the formula holds for n=1 , because
/2
1 /2
1
2
/2 1
sin xdx=
1dx=
x
=
.
0
0
2 0
2
2 2
/2
1 3 5
(2k 1)
2k
Now assume it holds for some k 1 . Then
sin xdx=
. By Exercise 43
0
2 4 6
( 2k ) 2
(a),
/2
2 ( k+1 )
2k+1 /2 2k
2k+1 1 3 5
(2k 1)
sin
xdx =
sin xdx=
0
2k+2 0
2k+2 2 4 6
( 2k ) 2
1 3 5
(2k 1)(2k+1)
=
,
2 4 6
(2k)(2k+2) 2
so the formula holds for n=k+1 . By induction, the formula holds for all n 1 .
n 45. Let u=(ln x) , dv=dx
n n n 1 (ln x) dx=x(ln x)
n nx(ln x)
x 46. Let u=x , dv=e dx ( 2 2 n 47. Let u= x +a ) n 1 du=n(ln x) du=nx , dv=dx ( dx/x ) , v=x . By Equation 2,
n n 1 ( dx/x ) =x(ln x) n (ln x) dx . n 1 x n x n x dx , v=e . By Equation 2, x e dx=x e n x ( 2 n 1 x e dx . 2 n 1 du=n x +a ) 2xdx , v=x . Then
10 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts ( x2+a2) ndx = x ( x2+a2) n 2n x2 ( x2+a2) n 1 dx
2 2 n
2
2 2 n
2
2 2 n 1
2 2
since x = ( x +a )
= x ( x +a ) 2n ( x +a ) dx a ( x +a ) dx
2 2 n
2 2 n
2
2 2 n 1
( 2n+1 ) ( x +a ) dx=x ( x +a ) +2na ( x +a ) dx , and
2 2 n
2
x ( x +a )
2na
2 2 n
( x +a ) dx= 2n+1 + 2n+1 ( x2+a2) n 1 dx [ provided 2n+1 0 ].
48. Let u=sec n 2 2 du= ( n 2 ) sec x , dv=sec xdx n 3 2 a xsec xtan xdx , v=tan x . Then by Equation 2, n
n 2
n 2
2
sec xdx = tan xsec x ( n 2 ) sec xtan xdx ( ) = tan xsec n 2 x ( n 2 ) sec n 2 x sec 2 x 1 dx
= tan xsec n 2 x ( n 2 ) sec nxdx+ ( n 2 ) sec n 2 xdx
n so ( n 1 ) sec xdx=tan xsec
n sec xdx= tan xsec
n 1 n 2 x + n 2 n 2 x+ ( n 2 ) sec xdx . If n 1 0 , then n 2
n 2
sec xdx .
n 1
3 3 2 3 2 49. Take n=3 in Exercise 45 to get (ln x) dx=x ( ln x ) 3 (ln x) dx=x(ln x) 3x(ln x) +6xln x 6x+C [
by Exercise 13 ].
Or : Instead of using Exercise 13 , apply Exercise 45 again with n=2 .
50. Take n=4 in Exercise 46 to get ( ) 4 x
3 x
4 x
3
2
x
4 x
x e dx = x e 4 x e dx=x e 4 x 3x +6x 6 e +C ( [by Exercise 14 ] ) = ex x4 4x3+12x2 24x+24 +C
Or: Instead of using Exercise 14 , apply Exercise 46 with n=3 , then n=2 , then n=1 .
5 0.4x 51. Area = xe dx . Let u=x , dv=e 0 du=dx , v= 2.5e
area = 2.5xe 0.4x 0.4x 5 5 0 0 +2.5 e 2
2 dx . Then = 12.5e +0+2.5
= 12.5e 0.4x 6.25(e 2.5e
2 0.4x dx 0.4x 5
0 1)=6.25 18.75e 2 or 25
4 75
e
4 2 11 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts ( x 5) ln x=0 ; that is, 52. The curves y=xln x and y=5ln x intersect when xln x=5ln x xln x 5ln x=0
when x=1 or x=5 . For 1<x<5 , we have 5ln x>xln x since ln x>0 . Thus, area
= 5 ( 5ln x xln x ) dx=
1 area =
= 5 (5 x)ln x dx . Let u=ln x , dv= ( 5 x ) dx 1 5 1 2
(ln x) 5x
x
2
25
ln 5
2 5x 1 2
x
4 1
5 = 1 5 1 2
x
2 1 25
ln 5
2 1
x 25 5x 25
4 dx=(ln 5)
5 1
4 du=dx/x , v=5x
25
2
= 0 5
1 1 2
x . Then
2
1
5
x dx
2 25
ln 5 14
2 2 53. The curves y=xsin x and y=(x 2) intersect at a 1.04748 and b 2.87307 , so
area =
= b
a 2 xsin x (x 2) dx 1
3 b
(x 2)
[ by Example 1]
3
a
2.81358 0.63075=2.18283
xcos x+sin x 54. The curves y=arctan3x and y=x/2 intersect at x= a
area = a arctan3x 1 x dx=2 a arctan3x 1 x dx
a
0
2
2
1
2 1 2
=2 xarctan3x
ln (1+9x )
x
6
4
2(1.39768)=2.79536 . 2.91379 , so a
0 [ see Example 5] 12 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 1 55. V = 2 xcos ( x/2)dx . Let u=x , dv=cos ( x/2)dx
0 2 V =2
=4+ 8 1 x
2 xsin
8 (0 1)=4 2 2 0 x
2 1 sin 0 2 du=dx , v= dx=2 2 sin ( x/2) . 0 2 4 cos x
2 1
0 . 56. ( 1 x Volume = 2 x e e
0 1 =2 0 57. Volume = =2 1 x xe dx 0 0 x xe dx [ both integrals by parts]
xe x e x ) 1 =2 [2/e 0]=4 /e 0 x x 2 (1 x)e dx . Let u=1 x , dv=e dx 1 x (1 x)( e )
xe ) dx=2 1 ( xex xe x) dx
0 ( xex ex) ( =2 V =2 x x 0 0
1 2 0
1 x e dx=2 x (x 1)(e )+e du= dx , v= e x . x 0
1 =2 (0+e)=2 e 1 58.
Volume = 1 2 y ln ydy=2 2 = 59. The average value of f (x)=x ln x on the interval 1,3 is f = 1 =2 (2ln
4 1 (0 1)
4 =2 1 2
y (2ln y 1)
4 1 2
1 2
y ln y
y
2
4 2 2 Let u=ln x , dv=x dx du=(1/x)dx , v= 1) ave 3
3 ln 1
3 1 2
3 2
1 + 2 x ln xdx= 1
I.
2 1 3
x . So
3
13 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 3 31 2
1 3
1 3 3
I=
x ln x
x dx= ( 9ln 3 0 )
x
=9ln 3
3
1 1 3
9
1
1
1
26
9
13
Thus, f = I=
9ln 3
= ln 3
.
ave 2
2
9
2
9 60. The rocket will have height H=
H = 60 m rt
m gt v ln
e 0 e 60
0 ln (m rt)dt = 0 26
.
9 =9ln 3 v(t)dt after 60 seconds.
60 1 2
t
2 v 0 60
0 e 60
0 ln (m rt)dt ln mdt 60 ln (m rt)dt e 0 1
( r)dt , v=t . Then
m rt
60 60 rt
tln (m rt) +
dt=60ln (m 60r)+
0
0 m rt Let u=ln (m rt) , dv=dt 1
9 60 dt= g = g(1800)+v (ln m)(60) v 3 du= = 60ln (m 60r)+ t m
ln (m rt)
r 60 1+ 0 m
m rt dt 60
0 m
m
ln (m 60r)+ ln m
r
r
m
m
So H= 1800g+60v ln m 60v ln (m 60r)+60v + v ln (m 60r)
v ln m . Substituting g=9.8 ,
e
e
e r e
r e
m=30 , 000 , r=160 , and v =3000 gives us H 14 , 844 m.
= 60ln (m 60r) 60 e t t 2 w 61. Since v(t)>0 for all t , the desired distance is s(t)= v(w)dw= w e dw .
0 2 w First let u=w , dv=e dw du=2wdw , v= e w Next let U =w , dV =e dw
2 s(t) =
=
= t t e +2 we w dU =dw , V = e w t t w 2 w 0 . Then s(t)= w e t + e dw = t e +2 0 0
w t t w +2 we dw . 0 0 . Then 0 t te +0+ e w t
0 ( te t e t+1) = t2e t 2te t 2e t+2
t 2
2 e ( t +2t+2 ) meters
2 t t e +2 62. Suppose f (0)=g(0)=0 and let u= f (x) , dv=g
a 2 f (x)g / / / (x)dx= f (x)g (x) a a 0 0 / / / / / (x)dx / du= f (x)dx , v=g (x) . Then
/ f (x)g (x)dx= f (a)g (a) a
0 / / / f (x)g (x)dx . Now let U = f (x) ,
14 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts / dV =g (x)dx
a
0 / / / dU = f / (x)dx and V =g(x) , so
a f (x)g (x)dx= f (x)g(x) a 0 / 0 Combining the two results, we get
4 / / 63. For I= xf
/ I= xf (x) 1
4 4 1 1 / / f a
0 / (x)g(x)dx= f (a)g(a) f (x)g
/ / (x)dx , let u=x , dv= f
/ / / / a
0 f / / / (x)g(x)dx . / (x)dx= f (a)g (a) f (a)g(a)+ a
0 f / / (x)g(x)dx . / (x)dx du=dx , v= f (x) . Then / f (x)dx=4 f (4) 1 f (1) [ f (4) f (1)]=4 3 1 5 (7 2)=12 5 5=2 . We used the fact that f / / is continuous to guarantee that I exists.
/ 64. (a) Take g(x)=x and g (x)=1 in Equation 1.
(b) By part (a), b
a b f (x)dx=bf (b) a f (a) / dy= f (x)dx . Then b
a f ( b) / x f (x)dx= f ( a) a / x f (x)dx . Now let y= f (x) , so that x=g(y) and g(y)dy . The result follows. (c) Part (b) says that the area of region ABFC is
f ( b) = bf (b) af (a) = ( area of rectangle OBFE ) ( area of rectangle OACD) 1 f ( a) x (d) We have f (x)=ln x , so f (x)=e , and since g= f
e ln e y 1 1 y ln 1 ln xdx=eln e 1ln 1 0 e dy=e e dy=e e 1 g(y)dy ( area of region DCFE ) y , we have g(y)=e . By part (b), y 1 =e (e 1)=1 . 0 65. Using the formula for volumes of rotation and the figure, we see that Volume
= d
0 2 b dy c
0 2 a dy d
c 2 2 and g(y)=x , so that V = b d
/ dv= f (x)dx 2 g(y) dy= b d
2 ac b 2
a 2 ac
/ d
c / 2 g(y) dy . Let y= f (x) , which gives dy= f (x)dx
2 x f (x)dx . Now integrate by parts with u=x , and du=2xdx , v= f (x) , and
15 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts b 2
a / 2 x f (x)dx= x f (x)
2 2 V= b d b b a a 2 ac 2 b 2 2x f (x)dx=b f (b) a f (a)
b b a 2 bd ac 2x f (x)dx , but f (a)=c and f (b)=d a a 2xf (x)dx = 2 xf (x)dx . 66. (a) We note that for 0 x , 0 sin x 1 , so sin 2
Comparison Property of the Integral, I 2n+2 I 2n+1 I 2n 2n+2 x sin 2n+1 x sin 2n x . So by the second . (b) Substituting directly into the result from Exercise 44 , we get
I 1 3 5
2 4 6
2n+2
=
1 3 5
I
2n
2 4 6 (c) We divide the result from part (a) by I
I 2n+2 I 2n 2n+1
2n+2 I 2n+1 I
I 2n 2n+1 I 2n I
I 2n 2(n+1) 1
2(n+1)
2
2(n+1) 1 2n+1
=
=
(2n 1)
2(n+1)
2n+2
( 2n ) 2 2n . The inequalities are preserved since I . Now from part (b), the left term is equal to 2n 2n is positive: 2n+1
, so the expression becomes
2n+2 I
2n+1
2n+1
=lim 1=1 , so by the Squeeze Theorem, lim
1 . Now lim
=1 .
2n+2 n
I
n
n
2n (d) We substitute the results from Exercises 43 and 44 into the result from part (c):
2 4 6
( 2n )
I
3 5 7
( 2n+1 )
2n+1
1 = lim
=lim
1 3 5
( 2n 1 )
I
n
n
2n
2 4 6
( 2n ) 2
2 4 6
2 4 6
2
( 2n )
( 2n )
= lim
3 5 7
1 3 5
( 2n+1 )
( 2n 1 )
n
2 2 4 4 6 6
= lim 1 3 3 5 5 7
n
Multiplying both sides by 2 2n
2n 2
2n 1 2n+1 [rearrange terms] gives us the Wallis product : 2 2 4 4 6 6
2 1 3 3 5 5 7
(e) The area of the k th rectangle is k . At the 2n th step, the area is increased from 2n 1 to 2n by
= 16 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts 2n
, and at the ( 2n+1 ) th step, the area is increased from 2n to 2n+1 by
2n 1
2n+1
2n
multiplying the height by
. These two steps multiply the ratio of width to height by
and
2n
2n 1
1
2n
2 2 4 4 6 6
=
respectively. So, by part (d), the limiting ratio is
=
.
(2n+1)/(2n) 2n+1
1 3 3 5 5 7
2
multiplying the width by 17 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 1.
3 2 2 2 ( 1 cos 2x) cos 2xsin xdx= ( 1 u2) u2( du)
( u4 u2) du= 1 u5 1 u3+C= 1 cos 5x 1 cos 3x+C
5
3
5
3 sin xcos xdx = sin xcos xsin xdx= ( u2 1) u2 du= =
2.
6 3 6 2 ( 6 2 ) 6 ( 2 ) sin xcos xdx = sin xcos xcos xdx= sin x 1 sin x cos xdx= u 1 u du
1 7 1 9
1
1
6 8
7
9
= u u du= u
u +C= sin x
sin x+C
7
9
7
9 ( ) 3.
3 /4 5 3 /4 3 sin xcos xdx =
/2 5 3 /4 2 sin xcos xcos xdx=
/2 s
= ( 2 ) /2 2 /2 u 5 ( 1 u ) du=
2 1 2 /2 ( u5 u7) du= 1 1/8
6 = 5 sin x 1 sin x cos xdx 1/16
8 1
6 1
8 1 6 1 8
u
u
6
8 2 /2
1 11
384 = 4.
/2 /2 5 cos xdx =
0 ( cos x)
2 2 /2 cos xdx= 0
1 = 1 ( 1 sin x) cos xdx= ( 1 u2) 2 du
2 2 0 ( 1 2u2+u4) du= 0 u 0 2 3 1 5
u+ u
3
5 1 = 1 0 2 1
+
3 5 0= 8
15 5.
5 4 4 4 ( 1 sin 2x) 2sin 4xcos xdx= ( 1 u2) 2u4 du
( u4 2u6+u8) du= 1 u5 2 u7+ 1 u9+C
5
7
9 cos xsin xdx = cos xsin xcos xdx=
=
= ( 1 2u2+u4) u4 du= 1
2
1
5
7
9
sin x
sin x+ sin x+C
5
7
9 6.
1 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 1
m ( 1 cos 2mx) sin mxdx= 3
sin (mx)dx = ( 1 u2) du [ u=cos mx , du= msin mxdx ] 1
1 3
1
1
3
u
u +C=
cos mx
cos mx +C
m
3
m
3
1
1
3
=
cos mx
cos mx+C
3m
m
= 7.
/2
0 cos 2 =
/2 8. 0 /2 d = 0 1
(1+cos 2 )d [ half angle identity]
2 1
2 +
/2 2 sin (2 )d = 0 /2 1
sin 2
2 0 = 1
2 1
1
(1 cos 4 )d =
2
2 ( 0+0 ) = +0 2 /2 1
sin 4
4 = 0 4 1
2 2 ( 0 0) = 0 4 9.
4 0 = 1
4 = sin (3t) dt= 1
4 1 2cos 6t+ 0 2 1
(1 cos 6t)
2 2 2 sin (3t)dt =
0 0 dt= 1
1
(1+cos 12t) dt=
2
4 3 1
1
t
sin 6t+
sin 12t
2 3
24 0 = 1
4 1
4 2 0 (1 2cos 6t+cos 6t)dt 3
1
2cos 6t+ cos 12t dt
2
2 0 3
2 ( 0 0+0 ) = 0+0 3
8 10.
0 cos 6 d =
= 0 1
8 (cos 2 + 1
(1+cos 2 )
2 3 ) d =
3
sin 2
2 0 0 + 1
3
1
+
+ sin 4
8
16
4
3
5
= +
+0=
8 16
16
= 1
8 d = 1
8 2 0 + 1
8 0 2 (1 u ) 0 3 (1+3cos 2 +3cos 2 +cos 2 )d 3
1
(1+cos 4 ) d +
2
8 0 0 3 1
du
2 2 0 (1 sin 2 )cos 2 d [ u=sin 2 , du=2cos 2 d ] 11. 2 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 1
(1+cos 2 )d
2
1
1
3
1
= +2sin +
+ sin 2 +C=
+2sin + sin 2 +C
2
4
2
4 2 = (1+2cos +cos (1+cos ) d 2 )d = +2sin + 1
1
1
(1+cos 2x)dx= x+ sin 2x , so
2
2
4
1
1
1 2 1
1 2 1
x+ sin 2x dx= x + xsin 2x
x + cos 2x+C
2
4
2
4
4
8
2 2 12. Let u=x , dv=cos xdx du=dx , v= cos xdx= 1
1
x+ sin 2x
2
4
1 2 1
1
= x + xsin 2x+ cos 2x+C
4
4
8 2
xcos xdx =x 13.
/4
0 4 2 2
1
sin 2x dx
0
0
2
1 /4
1 /4 2
1 /4 2
2
=
(1 cos 2x)sin 2xdx=
sin 2xdx
sin 2xcos 2xdx
8 0
8 0
8 0
/4 sin xcos xdx = 2 2 sin x(sin xcos x) dx= 1
16
1
=
16 /4 = 1
(1 cos 2x)
2 /4 1
1
3
sin 2x
16
3
1
=
(3 4)
192 (1 cos 4x)dx 0 4 0 1
3 /4
0 = 1
16 1
1
3
sin 4x
sin 2x
4
3 x /4
0 14.
/2
0 2 2 sin xcos xdx = /2
0 1
4
1
=
8
= 1
2
2
4sin xcos x dx=
4 ( /2
0 ) 1
1
(1 cos 4x)dx=
2
8 2 = /2
0 1
1
2
(2sin xcos x) dx=
4
4 /2
0 (1 cos 4x)dx= 1
8 x /2
0 2 sin 2xdx 1
sin 4x
4 /2
0 16 15.
3 ( 1 cos 2x) ( 2 ) 1/2 cos x sin xdx= 1 u u ( du ) =
2 7/2 2 3/2
2
7/2 2
3/2
= u
u +C= ( cos x )
( cos x ) +C
7
3
7
3
2
2
3
=
cos x
cos x
cos x +C
7
3 sin x cos x dx = ( u5/2 u1/2) du 3 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 16. Let u=sin . Then du=cos d and
5 5 cos cos (sin )d 2 2 2 2 = cos u du= (cos u) cos udu= (1 sin u) cos u du
2 4 = (1 2sin u+sin u)cos u du=I
Now let x=sin u . Then dx=cos udu and
2 3 1 5
2
1
2 4
3
5
I = (1 2x +x )dx=x 3 x + 5 x +C=sin u 3 sin u+ 5 sin u+C
2
1
3
5
=sin (sin )
sin (sin )+ sin (sin )+C
3
5
17. ( 3 2 ) sin x
1 u ( du )
1
cos xtan xdx =
dx=
=
+u du
cos x
u
u
1 2
1
2
= ln u + u +C= cos x ln cos x +C
2
2
2 3 5 4 18.
cot sin d = cos
sin = 5
5 sin 4 ( 1 u2) 2 du=
u 2 =ln u u + cos
d =
sin
2 5 4 cos
d =
sin 4 cos d = ( 1 sin 2 ) 2 cos
sin d 1 2u +u
du=
u 1 4
u +C=ln sin
4 1
3
2u+u du
u
1
2
4
sin + sin +C
4 19.
1 sin x
dx = ( sec x tan x ) dx=ln sec x+tan x ln sec x +C
cos x
=ln ( sec x+tan x ) cos x +C=ln 1+sin x +C
=ln ( 1+sin x ) +C since 1+sin x 0 by (1) and the boxed
formula above it Or: ( ) 2
1 sin x
1 sin x 1+sin x
1 sin x dx
cos xdx
dx =
dx=
=
cos x
cos x 1+sin x
cos x ( 1+sin x )
1+sin x
4 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals dw
[ where w=1+sin x , dw=cos xdx ]
w
=ln w +C=ln 1+sin x +C=ln ( 1+sin x ) +C
= 2 3 1 4
1
4
u +C=
cos x+C
2
2 3 20. cos xsin 2xdx=2 cos xsin x dx= 2 u du= 1 2
1
2
u +C= tan x+C .
2
2
1 2
1
2
2
Or: Let v=sec x , dv=sec xtan xdx . Then sec xtan xdx= vdv= v +C= sec x+C .
2
2
2 2 21. Let u=tan x , du=sec xdx . Then sec xtan xdx= udu= 22.
/2
0 4 sec (t/2)dt = /4 4 sec x(2dx) [ x=t/2 , dx= 0
1 1
dt ] =2
2 2 /4
0 2 =2 (1+u )du [ u=tan x , du=sec xdx ] =2
0 2 23. tan xdx= 2 2 sec x(1+tan x)dx
u+ 1 3
u
3 1 =2 0 1+ 1
3 = 8
3 ( sec 2x 1) dx=tan x x+C 1
3
tan x tan x+x+C
3
(Set u=tan x in the first integral and use Exercise 23 for the second.)
4 2 ( ) 2 2 2 2 24. tan xdx= tan x sec x 1 dx= tan xsec xdx tan xdx= 25.
6 2 4 2 2 2 2 2 sec t dt = sec t sec t dt= (tan t+1) sec t dt= (u +1) du
1 5 2 3
1
4
2
5 2
3
= (u +2u +1)du= u + u +u+C= tan t+ tan t+tan t+C
5
3
5
3
26.
/4
0 sec 4 tan 4 d /4 = (tan 0
1 6 2 +1)tan 4 = (u +u )du=
0 4 sec 2 1 2 4 d = (u +1)u du 1 7 1 5
u+ u
7
5 0 1 = 0 1 1 12
+ =
7 5 35 27.
/3
0 5 /3 4 tan xsec xdx = 0 5 2 2 tan x(tan x+1)sec xdx
5 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals =
= 3 5 2 u (u +1)du 0
3 7 0 3 1 8 1 6
u+ u
8
6 5 (u +u )du= = 0 81 27 81 9 81 36 117
+
=
+ =
+
=
8
6
8 2 8
8
8 Alternate solution:
/3
0 5 /3 4 tan xsec xdx = 4 /3 3 tan xsec xsec xtan xdx= 0
2 2 2 2 3 0 1
8 1 1
+
3 4 2 3 4 2 (sec x 1) sec xsec xtan xdx = (u 1) u du
1
2 2 3 7 5 3 = (u 2u +1)u du= (u 2u +u )du
1 1 1 8 1 6 1 4
u
u+ u
8
3
4 = 2 64
+4
3 = 32 1 = 117
8 28.
3 5 2 4 tan (2x)sec (2x)dx = tan (2x)sec (2x) sec (2x)tan (2x)dx
2
4 1
= (u 1)u ( du) [ u=sec (2x),du=2sec (2x)tan (2x)dx ]
2
1
1 7 1 5
1
1
6 4
7
5
=
(u u )du=
u
u +C=
sec (2x)
sec (2x)+C
2
14
10
14
10
29.
3 2 tan xsec xdx = tan xsec xtan xdx= ( sec 2x 1) sec xtan xdx 2 = (u 1)du
1 3
1
3
= u u+C= sec x sec x+C
3
3
30.
/3
0 5 6 tan xsec xdx =
=
= /3 5 0 3 5 ( u 1+u 0
3
0 4 2 2 ) /3 2 tan xsec xsec xdx= 0 5 ( 2 )2 3 5 2 du [ u=tan x , du=sec xdx ] = ( u5+2u7+u9) du= 2 tan x 1+tan x sec xdx 1 6 1 8 1 10
u+ u+
u
6
4
10 0
3
0 ( 2 4 ) u 1+2u +u du = 27 81 243 981
+
+
=
6
4
10
20
6 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals Alternate solution:
/3
0 5 6 tan xsec xdx =
=
= /3
0 4 /3 5 tan xsec xsec xtan xdx= 0 ( sec 2x 1) 2sec 5xsec xtan xdx ( u2 1) 2u5 du [ u=sec x , du=sec xtan xdx ]
1
2 4
( u 2u2+1) u5 du= 2 ( u9 2u7+u5) du
1
1
2 2 1 10 1 8 1 6
u
u+ u
10
4
6 = = 1 512
32
64+
5
3 1
10 1 1
+
4 6 = 981
20 31.
5 tan xdx = ( sec 2x 1) 2tan xdx= 4 2 sec xtan xdx 2 sec xtan xdx+ tan xdx 3 2 = sec xsec xtan xdx 2 tan xsec xdx+ tan xdx
1
1
4
2
4
2
= sec x tan x+ln sec x +C [ or sec x sec x+ln sec x +C ]
4
4
32.
6 4 ( 2 ) 4 2 4 tan aydy = tan ay sec ay 1 dy= tan aysec aydy tan aydy
1
5
2
2
=
tan ay tan ay sec ay 1 dy
5a
1
5
2
2
2
=
tan ay tan aysec aydy+ sec ay 1 dy
5a
1
1
1
5
3
=
tan ay
tan ay+ tan ay y+C
5a
3a
a ( ) ( ) 33.
tan 3 cos 4 d = tan 3 sec 4 d = tan 3 (tan 3 2 2 +1) sec 2 d 2 = u (u +1)duu=tan , du=sec d ]
1 6 1 4
1
1
5 3
6
4
= (u +u )du= u + u +C= tan + tan +C
6
4
6
4
34.
2 tan xsec xdx = ( sec 2x 1) sec xdx= 3 sec xdx sec xdx
7 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 1
( sec xtan x+ln sec x+tan x ) ln sec x+tan x +C [ by Example 8 and (1)]
2
1
= ( sec xtan x ln sec x+tan x ) +C
2
= /2 35. ( csc 2x 1) dx=
/6
/2 2 cot xdx= /6 cot x x /2 = 0 3 2 /6 = 3 6 3 36.
/2 3 ( /2 cot xdx = ) 2 /2 cot x csc x 1 dx= /4 /4 /4
/2 1
2
cot x ln sin x
2 = /2 2 cot xcsc xdx /4 cos x
dx
sin x 1
1
ln
2
2 = ( 0 ln 1 ) /4 = 1
1
1
+ln
= (1 ln 2)
2
2 2 37.
cot 3 csc 3 d = cot
2 2 csc 2 csc cot d = (csc 2 = (u 1)u ( du) [ u=csc , du= csc
1 3 1 5
1
2 4
3
= (u u )du= u
u +C= csc
3
5
3 2 1)csc 2 csc cot d cot d ]
1
5
csc +C
5 38.
4 6 6 2 2 csc xcot xdx = cot x(cot x+1)csc xdx
6 2 2 = u (u +1) ( du) [ u=cot x , du= csc xdx ]
1 9 1 7
1
1
8 6
9
7
= ( u u )du=
u
u +C=
cot x
cot x+C
9
7
9
7
csc x ( csc x cot x )
39. I= csc xdx=
dx=
csc x cot x ( 2 2 csc xcot x+csc x
dx . Let u=csc x cot x
csc x cot x ) du= csc xcot x+csc x dx . Then I= du/u=ln u =ln csc x cot x +C .
40.
4 6 6 2 2 csc xcot xdx = cot x(cot x+1)csc xdx
6 2 2 = u (u +1) ( du) [ u=cot x , du= csc xdx ]
8 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 8 6 = ( u u )du= 1 9 1 7
1
1
9
7
u
u +C=
cot x
cot x+C
9
7
9
7 41. Use Equation 2(b):
1
1
cos (5x 2x) cos (5x+2x) dx=
( cos 3x cos 7x ) dx
sin 5xsin 2xdx =
2
2
1
1
= sin 3x
sin 7x+C
6
14
42. Use Equation 2(a):
1
1
sin (3x+x)+sin (3x x) dx=
( sin 4x+sin 2x ) dx
sin 3xcos xdx =
2
2
1
1
=
cos 4x
cos 2x+C
8
4
43. Use Equation 2(c):
1
1
cos (7 5 )+cos (7 +5 ) d =
( cos 2 +cos 12 ) d
2
2
1
1
1
1
1
=
sin 2 +
sin 12
+C= sin 2 +
sin 12 +C
2
2
12
4
24 cos 7 cos 5 d = 44.
cos x+sin x
1 cos x+sin x
1
dx =
dx=
(csc x+sec x)dx
sin 2x
2 sin xcos x
2
1
= ( ln csc x cot x +ln sec x+tan x ) +C [ by Exercise 39 and (1)]
2
2 45. 1 tan x
2 dx= ( cos 2x sin 2x) dx= cos 2xdx= sec x 1
sin 2x+C
2 46.
1
cos x+1
cos x+1
cos x+1
dx
=
dx=
dx=
dx
2
2
cos x 1 cos x+1
cos x 1
cos x 1
sin x
= (
2 2 ) cot xcsc x csc x dx=csc x+cot x+C 47. Let u=tan (t ) 2 2 du=2tsec (t )dt . Then
9 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals tsec 2 ( t2) tan 4 ( t2) dt= u 1
1 5
1
5 2
du =
u +C=
tan (t )+C .
2
10
10 4 7 6 48. Let u=tan x , dv=sec xtan xdx
7 8 2 du=7tan xsec xdx , v=sec x . Then
7 tan xsec xdx = tan x sec xtan xdx=tan xsec x
7 6 7 ( 6 2 7tan xsec xsec xdx ) 2 8 =tan xsec x 7 tan x tan x+1 sec xdx
6 =tan xsec x 7 tan xsec xdx 7 tan xsec xdx .
8 7 dx= 1
7
tan xsec x
8 49. Let u=cos x
5
sin xdx = = /4
0 7
8 /4 6 Thus, 8 tan xsec xdx=tan xsec x 7 tan xsec xdx and
/4
0 6 tan xsec xdx= 2
8 0 8 tan xsec x 7
I.
8 du= sin xdx . Then ( 1 cos 2x) 2sin xdx= ( 1 u2) 2 ( du )
( 1+2u2 u4) du= 1 u5+ 2 u3 u+C
5
3 1
2
5
3
cos x+ cos x cos x+C
5
3
Notice that F is increasing when f ( x ) >0 , so the graphs serve as a check on our work.
= 50.
4 4 sin xcos xdx =
1
64
1
=
64
= 1
sin 2x
2 4 1
1
4
dx=
sin 2xdx=
16
16 1
( 1 cos 4x )
2 2 dx ( 1 2cos 4x+cos 24x) dx
x 1
1
sin 4x +
( 1+cos 8x ) dx
2
128 10 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 1
1
1
1
x
sin 4x +
x+ sin 8x +C
64
2
128
8
3
1
1
=
x
sin 4x+
sin 8x+C
128
128
1024
= Notice that f (x)=0 whenever F has a horizontal tangent. 51.
sin 3xsin 6xdx = 1
cos (3x 6x) cos (3x+6x) dx
2 1
(cos 3x cos 9x)dx
2
1
1
= sin 3x
sin 9x+C
6
18
= Notice that f (x)=0 whenever F has a horizontal tangent. 52.
sec 4 x
dx =
2 x
2 x
+1 sec
dx
2
2
x
1
2
2 x
= u +1 2du [ u=tan
, du= sec
dx ]
2
2
2
2 3
2
x
3 x
= u +2u+C= tan
+2tan +C
3
3
2
2
tan ( 2 ) Notice that F is increasing and f is positive on the intervals on which they are defined. Also, F has
11 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals no horizontal tangent and f is never zero. 53.
f ave 1
2
1
=
2
=0 2 = 3 sin xcos xdx=
0 2 ( 2 1
2 ( 2 2 ) sin x 1 sin x cos xdx ) u 1 u du [ where u=sin x ] 0 1 2
1
2
u +C=
cos x+C .
1
2
2
1 2
1
2
sin xcos xdx= udu= u +C= sin x+C .
2
2
2 54. (a) Let u=cos x . Then du= sin xdx
(b) Let u=sin x . Then du=cos xdx sin xcos xdx= u( du)= 1
1
sin 2xdx=
cos 2x+C
3
2
4
(d) Let u=sin x , dv=cos xdx . Then du=cos xdx , v=sin x ,
(c) sin xcos xdx= 2 so sin xcos xdx=sin x sin xcos xdx , by Equation .1.2, so sin xcos xdx= 1
2
sin x+C .
4
2 The answers differ from one another by constants. Since
1
1
1
1
1
2
2
2
2
cos 2x=1 2sin x=2cos x 1 , we find that
cos 2x= sin x
=
cos x+ .
4
2
4
2
4
55. For 0<x< 3 , we have 0<sin x<1 , so sin x<sin x . Hence the area is 2 ( 2 Then area = u ( du ) = u du=
1
0 1 3
u
3 /2
0 ( sin x sin 3x) dx=
0 2 56. sin x>0 for 0<x<
as that of sin x /2
0 ) sin x 1 sin x dx= 1 2 1 = 0 /2
0 2 cos xsin xdx . Now let u=cos x 1
.
3
2 2 du= sin xdx . , so the sign of 2sin x sin x [ which equals 2sin x sin x 1
2
. Thus 2sin x sin x is positive on
2 ,
6 2 and negative on 1
2
0, is the same
6 . The
desired area is
12 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals /6
0 = /6
0 ( sin x 2sin 2x) dx+ /2 ( 2sin 2x sin x) dx
/6
/2 ( sin x 1+cos 2x ) dx+ = cos x x+ 1
sin 2x
2 = 3
2 6 3
4 6 /6 ( 1 cos 2x sin x ) dx /6 /2 3
2 = 1+ + 0 ( 1) + 1
sin 2x+cos x
2 + x 2 /6 3
3
+
4
2 6 57. It seems from the graph that 2
0 3 cos xdx=0 , since the area below the x axis and above the graph looks about equal to the area above the axis and below the graph. By Example 1, the integral is
2
1
3
sin x
sin x
=0 . Note that due to symmetry, the integral of any odd power of sin x or cos x
3
0
between limits which differ by 2n ( n any integer) is 0 .
58. It seems from the graph that 2 sin 2 xcos 5 xdx=0 , since each bulge above the x axis seems to 0 have a corresponding depression below the x axis. To evaluate the integral, we use a trigonometric
identity:
1
1 2
sin 2 xcos 5 xdx =
sin (2 x 5 x)+sin (2 x+5 x) dx
0
2 0
1 2
=
sin ( 3 x)+sin 7 x dx
2 0
13 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 1
2
1
=
2
2 59. V = /2 sin xdx= 2 1
1
cos ( 3 x)
cos 7 x
3
7
1
1
(1 1)
(1 1) =0
3
7 = /2 1
(1 cos 2x)dx=
2 0 2 1
1
x
sin 2x
2
4 = 2 /2 0 4 +0 = 4 60.
Volume =
= ( tan 2x) 2dx= 0 /4tan 2x ( sec 2x 1) dx=
0
/4 2
/4
( sec 2x 1) dx
u du
0
0
/4 1 3
u
3 = /4 tan x x x=0 0 2 = /4 2 tan xsec xdx 1
3
tan x tan x+x
3 /4
0 /4 /4
0 0 1
1+
3
4 = 2 tan xdx = 4 2
3 61.
/2 Volume = 0 2 1
1
2sin x+ x+ sin 2x
2
4 = /2 2 ( 1+cos x ) 1 dx= 0 ( 2cos x+cos 2x) dx
2 /2 = 0 2+ =2 + 4 4 62.
/2 Volume = 0 t 63. s= f (t)= sin
0 1 cos
1 2 t 2 ucos y dy= 2 /2 dx= 1
1
2sin x
x
sin 2x
2
4 = s= 2 1 ( 1 cos x ) 0 ( 2cos x cos 2x) dx
2 /2 = 0 udu . Let y=cos 1 1 3
y
3 cos t = 1 1
3 2 u 4 dy= 0 0 =2 sin 4 udu . Then ( 1 cos 3 t ) . 64. (a) We want to calculate the square root of the average value of
2 2 2 2 E(t) = 155sin (120 t) =155 sin (120 t) . First, we calculate the average value itself, by
1
2
integrating E(t) over one cycle (between t=0 and t=
, since there are 60 cycles per second) and
60
14 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 1
0 :
60
1 1/60
2
2
2
=
155 sin (120 t) dt=60 155
1/60 0 dividing by
E(t) 2
ave = 60 1552 1
2 1/60
0 1
1 cos (240 t) dt
2 1/60 1
t
sin (240 t)
240 0 2 =60 155 The RMS value is just the square root of this quantity, which is
(b) 220= E(t) 2
220 = E(t)
2 = 30A
2 Thus, 220 = 2 t 1 2
A
2 1
0
60 2 ( 0 0) 155
=
2 110 V. 2
ave = ave 155
2 1
2 1
1/60 1/60 2
0 2 1/60 2 A sin (120 t)dt=60A 1
sin (240 t)
240 1/60
0 2 =30A 0 1
[1 cos (240 t)]dt
2 1
0
60 ( 0 0) = 1 2
A
2 A=220 2 311 V. 65. Just note that the integrand is odd .
Or: If m n , calculate
1
sin mxcos nxdx =
sin (m n)x+sin (m+n)x dx
2
1
cos (m n)x cos (m+n)x
2
m n
m+n
If m=n , then the first term in each set of brackets is zero.
= =0 1
[cos (m n)x cos (m+n)x]dx . If m n , this is equal to
2
1
sin (m n)x sin (m+n)x
=0 . If m=n , we get
2
m n
m+n
1
1
sin (m+n)x
[1 cos (m+n)x]dx=
x
= 0= .
2
2
2(m+n) 66. sin mxsin nxdx= 1
cos (m n)x+cos (m+n)x dx . If m n , this is equal to
2
1
sin (m n)x sin (m+n)x
+
=0 . If m=n , we get
2
m n
m+n
1
1
sin (m+n)x
[1+cos (m+n)x]dx=
x
+
= +0= .
2
2
2(m+n) 67. cos mxcos nxdx= 15 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals 68. 1 f (x)sin mxdx= 1 m a sin nx sin mx dx= n=1 n m a n sin mxsin nxdx . By n=1 a
Exercise 66 , every term is zero except the m th one, and that term is m =a .
m 16 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 1. Let x=3sec , where 0 < 2 or 3
. Then dx=3sec
2 < ( tan d and ) 2
2
2
2
x 9 =
9sec
9 = 9 sec
1 = 9tan
= 3 tan =3tan for the relevant values of 1
2 x 2 1 dx=
9sec x 9 Note that sec ( + )=sec 2. Let x=3sin , where 3sec 2 1
d =
cos
9 tan 3tan 1
d = sin
9 , so the figure is sufficient for the case 2 2 ( . Then dx=3cos d < 1
+C=
9 2 x 9
+C
x 3
.
2 and ) 2
2
2
2
9 x =
9 9sin
= 9 1 sin
= 9cos
= 3 cos =3cos for the relevant values of \italic{\theta}\,\,. x 3 3 2
9 x dx = 3 sin 3 3cos 3cos = 35 sin 2 cos 2 sin
= 35
5 = 3 5 = 3 5 d =3 sin
5 d =3 3 5 3 3 2 ( 1 cos 2 ) cos 2 ( 1 u2) u2 ( du)[u=cos ,du= sin d
( u4 u2) du=35 1 u5 1 u3 +C=35
5
3
2 5/2
2 3/2
1 (9 x )
1 (9 x )
+C
5 cos d
sin 1
5
cos
5 d 1
3
cos
3 +C 3 3 1 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 1
2
9 x
5 ( = 3. Let x=3tan , where 1 2
3/2
x +6 ( 9 x ) +C
5 ) 5/2 3 ( 9 x2) 3/2+Cor 2 < < ( . Then dx=3sec 2 ( 2 ) d ) 2
2
2
2
x +9 =
9tan +9 = 9 tan +1 = 9sec
= 3 sec =3sec for the relevant values of x 3 2 3 3 tan
3sec dx = x +9 2 3 d =3 tan ) 1 tan = 3 1 2
x +9
3 ( , where 2 16 x = 16 16sin
x=2 3 , 4sin =2 3 2 ) 3/2 9 = 16cos
3
sin =
2 sec d =3
1
3
sec
3 2 x +9 +C or /2 3 3 sec 1 3
3
u u +C=3
3 3 4. Let x=4sin 3sec ( sec 2 = 33 = 3 and . 3 d =3 tan 2 ( u2 1) du[u=sec
3 sec 1 2
x 18
3 ( +C=3 ) tan sec ,du=sec
1
3 d
tan ( x2+9) 3/2
3 3 d ]
2 x +9
3 +C 2 x +9 +C /2 . Then dx=4cos d and
2 =4 cos
= 3 =4cos . When x=0 , 4sin =0 =0 , and when . Thus, substitution gives 2 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 2 3 /3 x = 3 0 dx
2 /3 3 16 x 0 3 4 sin
4cos =4 3
3 4cos d =4 = 4 d 1 3 1/2
u
u
3
1
1
1
= 64
3 1 2 3 d ( 1 u2) du= 64
1
2 = 64 sin
0 ( 1 cos 2 ) sin 0
1/2
3 /3 1
24 5
24 = 40
3 2 Or: Let u=16 x , x =16 u , du= 2xdx .
5. Let t=sec , so dt=sec tan d , t= 2 2 = /3 1
2 t 3 2 1
3 sec /4 tan =
= 2 3
0 x 1
1
(1+cos 2 )d =
2
/4 2
1
2 3 , so dx=2sec
/4 3 2 x +4 dx = 0 2 tan
/4 5 =2 2 5 =2 1
2 5 =2 =32 (sec 0 1 2 , x=0 1)sec
2 2 sec 2 5 (u u )du=2 2
2
2+
15
15 2 1
1
2 = /4 5 d =2
tan d 12
= tan 0 1
2 2 4 sec + 3
4 1
2 = tan + 3
8 1
5 1
3 d 24 1
4 . Then
2 sec d ,du=sec 1 5 1 3
u
u
5
3
64
=
( 2 +1 )
15 2 /4 =0 , and x=2 2sec cos /4 4 + . Then /3 3
2 (u 1)u du [u=sec
4 d = 1
sin 2
2 1
2 2sec 2 + 2 d + 2 3 sec /4 3 /3 1 sec tan d = /3 6. Let x=2tan = /3 dt = t 1 , and t=2 4 tan
2
1 d ]
5 =2 1
1
4 2
2 2
5
3 3 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 2 2 Or: Let u=x +4 , x =u 4 , du=2xdx .
7. Let x=5sin
1
2 2 x , so dx=5cos . Then 1 dx = 2 5 sin 25 x 2 5cos d 5cos 1
1
2
csc d =
cot
25
25 = +C 2 25 x
+C
x 1
25 = 8. Let x=asec d , where 0 < 2 or < 3
. Then dx=asec
2 tan d and 2 2 x a =atan , so
2 2 x a
x 4 atan
dx = 4 a sec
1 = 4 sin 2 2 asec cos tan d d a
= 1
2 sin 3 ( x2 a2) 3/2 +C
+C=
2 3 3a 9. Let x=4tan , where 3a x 2 < < 2 . Then dx=4sec 2 d and
4 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 2 2 x +16 = 16tan 2 +16 = 16(tan +1) 2 = 16sec
=4 sec
=4sec for the relevant values of dx . 2 4sec d
=
2
4sec
x +16 d =ln sec = sec +tan +C 1 2 x +16 x
+
4
4 =ln ( =ln ) 2 ln 4 +C 1 x +16 +x +C , where C=C ln 4 .
1 2 10. Let t= 2 tan
2 t +2 = 2tan , where
2 = 2 sec dt = = 2 sec 4 2 tan t +2
=4 2 2 < < +2 = 2(tan 5 2 1 2 x +16 +x x +16 +x>0 , we don’t need the absolute value.) (Since t +C =ln 2 sec 2 2 . Then dt= 2 sec 2 +1) = 2sec d and 2 for the relevant values of . 5 2 sec 2 d =4 2 tan ( u2 1) 2du [ u=sec , du=sec 5 sec tan d d =4 2 =4 2 ( sec 2 )2 1 sec tan d ( u4 2u2+1) du
5 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 4 2
1 5 2 3
4
2
u
u +u +C=
u(3u 10u +15)+C
5
3
15 =4 2 2 4 2
=
15 3 2 4
15
1
=
15 2 = 11. Let 2x=sin 2 t +2 t +2 2 (t +2)
2 2 2 t +2
10
+15
2 +C 1
4
2
2
3(t +4t +4) 20(t +2)+60 +C
4 2 4 2 t +2 (3t 8t +32)+C , where 2 2 2 2 . Then x= 1
sin
2 , dx= 1
cos
2 d , and 1 4x = 1 ( 2x ) =cos
2
1 4x dx = .
1
cos
2
1
+ sin 2
2 cos 1
( 1+cos 2 ) d
4
1
+C= ( +sin cos )+C
4
d = 1
4
1
1
=
sin (2x)+2x
4
= 12. 1
0 x 2 x +4 dx= 13. Let x=3sec 5
4 u , where 0 2 1 4x 1
du
2
< +C 2 [ u=x +4 , du=2xdx ] = 2 or < 1
2 2 3/2 5 1
u
= ( 5 5 8)
4 3
3 3
. Then dx=3sec
2 tan d and 2 x 9 =3tan , so
2 x 9
x 3 dx = 3tan
27sec 3 3sec tan 1
d =
3 tan
sec 2
2 d 6 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution = 1
1
2
sin d =
3
3 1
= sec
6 x
3 1 14. Let u= 5 sin
du
2 =
u 5 u 1
1
(1 cos 2 )d =
2
6
2 x 9 3
1
+C= sec
x
x
6 1
6 , so du= 5 cos
5 sin
1
ln csc
5
1
ln
5 5
u = 1
ln
5 5 2 2 3/2 (a x )
2 2 = cot a sin 2
3 5 u +C 2 x
3 x 9
2 +C 2x d = 1
csc
5 d acos d
3 =tan 2 +C 2 2 ( sec 2 1) d cos +C [byExercise2.39
5 u
u 2 a cos
= 5 cos u , where 1 1
sin
6 . Then 5 cos = x dx d 1 = 15. Let x=asin 1
1
sin 2 +C=
12
6 +C . Then dx=acos d and = tan 2 d +C 7 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution x
= 2 sin 1 2 a x 16. Let 4x=3sec , where 0 x
+C
a < 2 or < 3
3
. Then dx= sec
2
4 tan d and 2 16x 9 =3tan 2 x , so
3
sec tan d
dx
4
=
2
3 2 2
16x 9
sec
3tan
4 4
=
cos
9 4
d = sin
9 2 2 16x 9
+C=
4x 4
+C=
9 x 2 17. Let u=x 7 , so du=2xdx . Then 2 16x 9
+C
9x dx= x 7 ( ax2) 2 2 2 , so ( ax ) =b sec
b
2
b =btan , dx= sec tan
a 18. Let ax=bsec 2 2 1
2 1
1
2
du= 2 u +C= x 7 +C .
2
u
2 ( ax ) b =b sec
d 2 2 2 b =b ( sec 2 ) 2 1 =b tan 2 . So , and 8 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution b
sec
a dx
2 = 2 3/2 ( ax ) b tan 3 3 b tan
1
= 19. Let x=tan 2 csc 2
sec
1+x
=
dx
tan
x 2 ab
1 +C= 2 2 sec 2 sec
tan (1+tan 2 sin
x +C=
2 2 2 cos 1
ab
2 ( ax ) b . Then dx=sec d = d = 2 2 ab < < tan
ax 2 ab , where sec 1 d = b d 2 2 csc cot d ab
+C 2 2 ( ax ) b
2 1+x =sec and 1 d = , so )d = (csc +sec tan )d
=ln csc cot +sec +C [ by Exercise 8.2.39]
2 =ln 20. Let t=5sin
t
25 t 2 1+x
x , where dt = 5sin
5cos 2
5cos 1
x 2 + 2 2 1+x
+C=ln
1 1+x
x . Then dt=5cos d =5 sin d and 1 2 + 1+x +C 2 25 t =5cos , so d 2 = 5cos +C= 5 25 t
+C=
5 2 25 t +C 2 Or: Let u=25 t , so du= 2t dt . 9 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 2/3 3
0 du= 18xdx . Then x =
1
1/2
(4 u)u
49 0 2 4 9x dx = x 1
162 = Or: Let 3x=2sin 22. Let x=tan
=
1
0 4 1
( 4 u ) and
9
1
1
du=
18
162 2 2 21. Let u=4 9x , where , where 4 8 3/2 2 5/2
u
u
3
5 2 2 0 1
162 . Then dx=sec 2 64
3 64
5 = 64
1215 . 2 < < = ( 4u1/2 u3/2) du
0
4 2 d , 2 x +1 =sec and x=0 =0 , x=1 , so
/4 2 x +1 dx = 0 sec sec 2 d = /4
0 sec 3 d 1
/4
sec tan +ln sec +tan
[by Example 8.2.8]
0
2
1
1
=
2 1+ln ( 1+ 2 ) 0 ln (1+0) =
2 +ln ( 1+ 2 )
2
2
= 2 2 2 23. 5+4x x = (x 4x+4)+9= (x 2) +9 . Let x 2=3sin
2 5+4x x dx =
= 2 9 (x 2) dx=
9cos 2 3cos 9 9sin 2 d = 9cos 3cos
2 , 2 2 , so dx=3cos d . Then d d 10 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 9
9
1
(1+cos 2 )d =
+ sin 2
+C
2
2
2
9
9
9
9
=
+ sin 2 +C=
+ (2sin cos )+C
2
4
2
4
= 9
= sin
2
9
= sin
2 (2 2 2 5+4x x
9 x 2
+
+C
2 3
3
1
2
+ (x 2) 5+4x x +C
2 x 2
3
x 2
3 1 1 ) 2 2 24. t 6t+13= t 6t+9 +4= ( t 3) +2 . Let t 3=2tan
dt
2 1 = 2sec
2 t 6t+13 2 2 ( 2tan ) +2
= sec d =ln sec +tan , so dt=2sec 2sec
d =
2sec
+C 1 2 d . Then 2 d [by Formula 8.2.1] 2 t 6t+13 t 3
+
2
2 = ln 1 2 = ln 2 +C t 6t+13 +t 3 +CwhereC=C ln 2
1 2 25. 9x +6x 8= ( 3x+1 ) 9 , so let u=3x+1 , du=3dx . Then dx
2 9x +6x 8
u=3sec = 1
du
3
2 . Now let u 9 , where
11 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 0 < or 2 1
du
3 < sec = 2 u 9 2 tan
3tan d ( x2 4x+4) +4=4 2 1
=
sec
3 d and 1
d = ln sec
3 2 u 9 =3tan , so 1
+C = ln
1 3 +tan 2 u+ u 9
3 +C 1 2 ( x 2 ) , so let u=x 2 . Then x=u+2 and dx=du , so 2 x dx tan 1
1
2
2
ln u+ u 9 +C= ln 3x+1+ 9x +6x 8 +C
3
3 = 26. 4x x = 3
. Then du=3sec
2 = ( u+2 ) du 2 4x x 4 u ( ( 2sin +2 ) = 2 2cos 2 2cos d ) 2 =4 sin +2sin +1 d
=2 ( 1 cos 2 ) d +8 sin d +4 d
=2 sin 2 8cos +4 +C
=6 8cos 2sin cos +C
1
1
2 1
2
=6sin
u 4 4 u
u 4 u +C
2
2
x 2
x 2
1
2
2
=6sin
4 4x x
4x x +C
2
2 2 2 27. x +2x+2=(x+1) +1 . Let u=x+1 , du=dx . Then
dx ( x +2x+2)
2 2 =
= du ( u +1)
2 cos 2 2 = d = sec 2 sec d
4 where u=tan ,du=sec 2 and u +1=sec 1
1
( 1+cos 2 ) d = ( +sin
2
2 cos 2 2 d , x )+C 12 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 1
= 2 1+u ( x2+4x+4) +9=9 2 28. 5 4x x =
dx ( 5 4x x2) 5/2 = +C= 2 x+1 1 tan (x+1)+ +C 2 x +2x+2 2 where u=3sin , du=3cos d , 3cos d = ( 9 u2) 5/2 5 (3cos ) 1
1
4
sec d =
81
81 1
= 243 1
2 ( x+2 ) . Let u=x+2 du=dx . Then du
= u 1 tan u+ u ( tan 2 +1) sec 2 3 (9 u ) d = 2 cos 2 9 u =3cos
1
81 1
3
tan +tan
3
3 1
+C=
243 d 3u + 2 3/2 and 9 u (x+2) 2 3/2 ( 5 4x x ) + +C
3(x+2) +C
2 5 4x x 2 29. Let u=x , du=2xdx . Then
x 4 1 x dx = 1 u whereu=sin 1
1
du =
cos
2
2 2 ,du=cos d , 2 and 1 u =cos 1 1
1
1
1
1
(1+cos 2 )d =
+ sin 2 +C=
+ sin
2 2
4
8
4
4
1
1
1
1
2
1 2 1 2
4
= sin u+ u 1 u +C= sin (x )+ x 1 x +C
4
4
4
4
= cos +C 30. Let u=sin t , du=cos tdt . Then
1 /2 cos t
2 dt = 1+sin t 0 dx
2 2 0 1
sec sec 2 ( 2 +1 ) ln (1+0)=ln , where 2 < < 2 2 x +a asec d
asec ( /4
0 [ by (1) in Section 8.2] 2 +1 ) . Then 2 2 x +a =asec and
2 = sec d , 2 2 = 2 and 1+u =sec sec d = ln sec +tan 0 where u=tan , du=sec d /4 =ln
31. (a) Let x=atan dt= 1+u 0 = /4 1 d =ln sec +tan +C =ln
1 2 x +a
x
+
a
a +C 1
13 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution ( 2 2 = ln x+ x +a ) +C whereC=C 1
2 ln a
2 (b) Let x=asinh t , so that dx=acosh t dt and x +a =acosh t . Then
acosh t dt
1 x
dx
=
=t+C=sinh
+C .
acosh t
a
2 2
x +a
32. (a) Let x=atan , 2 2 2 x I =
= < < 2 2 3/2 ( sec dx= cos ( x +a ) 2 a tan
3 a sec 2 asec 3 2 tan
d =
sec ) d =ln sec +tan
2 x +a
x
+
a
a = ln . Then 2 x sin ( 2 2 sec
d =
sec 2 d +C
2 2 +C=ln x+ x +a 2 1 ) x
2 x +a +C
2 1 x +a (b) Let x=asinh t . Then
2 I = 2 a sinh t
3 3 a cosh t
1 x
= sinh a 2 acosh t dt= tanh t dt=
x
2 ( 1 sech2t ) dt=t tanh t+C +C
2 a +x 2 33. The average value of f (x)= x 1 /x on the interval 1,7 is
7 1
7 1 1 2 1
x 1
dx = 6
x 0 tan
sec where x=sec
sec tan d 2 x 1 =tan ,dx=sec
,and tan d ,
1 =sec 7 1
1
2
2
tan d =
(sec
1)d
6 0
6 0
1
1
=
tan
= (tan
)
0 6
6
1
1
=
48 sec 7
6
= ( ) 14 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 2 2 34. 9x 4y =36
area = 2 =3 3
2 0 = 12
= 12 3
2 y= 35. Area of
x=rcos u
2 2 x 4 2 x 4 dx=3 3
2 2 x 4 dx
where x=2sec ,
dx=2sec tan d ,
1 3
=sec
2 2tan 2sec tan d ( sec 2 1) sec
0 d =12 ( sec 3
0 1
( sec tan +ln sec +tan
2 = 6 sec tan
=6 3
2 3 5
ln
4 POQ= ln sec +tan
5
3
+
2
2 r x dx =
= )d ) ln sec +tan = u 0 0 9 5
6ln
2 3+ 5
2 1
1 2
(rcos )(rsin )= r sin cos
2
2 dx= rsin udu for 2 sec r
rcos 2 2 r x dx . Let . Then we obtain 2 rsin u ( rsin u)du= r . Area of region PQR= 2 2 sin udu= 1 2
r (u sin ucos u)+C
2 1 2
1
1
2 2
r cos (x/r)+ x r x +C
2
2 so
area of region = 1 r 2cos 1(x/r)+x r 2 x2
2
1
2
=
0 r +rcos rsin
2 ( r
rcos ) 15 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution = 1 2
r
2 1 2
r sin cos
2 and thus, ( area of sector POR ) = ( area of 36. Let x= 2 sec
dx , where 0
2 sec = tan < 2 or POQ ) + ( area of region PQR ) = < 3
, so dx= 2 sec
2 d . Then d 4 4
2
2 tan
4sec
x x 2
1
1
3
2
=
cos d =
1 sin
cos d
4
4
1
1
3
=
sin
sin
+C [substitute u=sin
4
3 ( ) ( x2 2) 3/2 2 x 2
x 1
=
4 tan 1 2
r .
2 3x +C 3 From the graph, it appears that our answer is reasonable.
2 37. From the graph, it appears that the curve y=x
2 2 4 x >2 x on ( 0.81,2 ) . So the area bounded by the curve and the line is
2
1 2 2
2
2
2
4 x (2 x) dx=
x 4 x dx 2x
x
. To evaluate the integral, we put
0.81
2
0.81 and x=2 , with x
A 2
0.81 x=2sin 2 x 2 4 x and the line y=2 x intersect at about x=0.81 , where 2 2 . Then 16 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution dx=2cos
2
0.81 d 2 1 , x=2 =sin 1=
/2 2 x 4 x dx 0.417 4sin (2cos )(2cos d )=4 =2 2 0.417 /2 2 0.417 2 0 sin 2 d =4 0.417 1
2
0.81
2 2 0.81 , so that dx=bsec =sin 0.405 0.417 . So /2 1 2
2
2 2 2 38. Let x=btan 2 1
sin 4
4 = 2
Thus, A 2.81 2 1 , and x=0.81 2 d and 1
(0.995)
4 1
(1 cos 4 )d
0.417 2
/2 2.81 2.10 . 2 x +b =bsec . L a E ( P) = b
4 a 0
2 = 4 0 b 2 3/2 ( x +b )
2 0 d = 4 0 1 b 4 0 1 ( bsec ) 1 bsec 3 2 d 2 1
sec 2 cos d = b
1 L a x
= 4 dx= 2 b =
2 x +b a 4 0 b 1 L a
4 0 b 2 2 +
2 (L a) +b 2 2 sin a
2 2 a +b 2 39. Let the equation of the large circle be x +y =R . Then the equation of the small circle is
2 2 2 2 2 x +(y b) =r , where b= R r is the distance between the centers of the circles. The desired area is
17 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution A= ( b+ r
r 2 2 r x r r 0 0 = 2 bdx+2 ) 2 2 2 R x
2 r x dx 2 r 2 2 = a cos 2 1 2
a
2
2 2 2 0 2 2 2 2 2 R x ) dx 2 2 x
a a
+
2 2 R r +r 2 2 2 2 2 2 r 2 2 R arcsin(x/R)+x 0 2 2 R arcsin(r/R)+r 2 2 a x
a
+C=
arcsin
a
2 x
a R r + r arcsin(x/r)+x r x = 2r 2 r x R r . To evaluate the other two integrals, note that
1 2
d [x=asin ,dx=acos d ] = a ( 1+cos 2 ) d
2
1
1 2
+ sin 2
+C= a ( +sin cos )+C
2
2 = a arcsin
2
so the desired area is
A = 2r 0 ( b+ R x dx The first integral is just 2br=2r
2 2
a x dx = r dx=2 R r 2 =r 2 2 x
a 2 2 2 a x +C 0
2 2 x
2 r 2 R x R r + + 2 r R arcsin(r/R) 40. Note that the circular cross sections of the tank are the same everywhere, so the percentage of the
total capacity that is being used is equal to the percentage of any cross section that is under water.
The underwater area is
A=2 2
5 2 25 y dy = 25 arcsin(y/5)+y
= 25 arcsin 2 25 y 2
25
+2 21 +
5
2 2
5 y=5sin 2 58.72ft so the fraction of the total capacity in use is A
2 (5) 58.72
25 0.748 or 74.8% . 18 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution 2 2 2 41. We use cylindrical shells and assume that R>r . x =r (y R)
2 x= 2 2 r (y R) , so 2 g(y)=2 r (y R) and
V =
=4 R+r
Rr 2 2 2 y 2 r (y R) dy= r 2 r 1 2 2
r u
3 ( =4
= 2 Rr 2 u r u du+4 R 2 /2 ) 3/2 r r 2 r 2 r 2 /2 2 +4 R r cos /2 (1+cos 2 )d =2 Rr /2 2 [where u=y R] where u=rsin ,du=rcos
in the second tegral r u du r
r 2 4 (u+R) r u du + 2 d = 4
2
(0 0)+4 Rr
3 1
sin 2
2 Another method: Use washers instead of shells, so V =8 R
evaluate the integral using y=rsin /2 2 =2 /2
r
0 2 Rr d
/2 cos /2 2 d 2 2 r y dy as in Exercise 6.2.61(a) , but . 19 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 2x
A
B
=
+
(x+3)(3x+1) x+3 3x+1
1
1
A
B
1
C
(b) 3 2 =
=
= +
+
2
2
2
x x+1
x +2x +x x(x +2x+1) x(x+1)
(x+1)
1. (a) A B
C
+ +
2
3 2
2 x+1
x
x (x+1)
x +x
x
x 1
x 1
A Bx+C
(b) 3 =
= + 2
2
x
x +x x x +1
x +1
x 1 2. (a) x 1 = ( ) 2 3. (a) = = 2 x +3x 4 2
A
B
=
+
(x+4)(x 1) x+4 x 1
2 x 2 (b) x +x+1 is irreducible, so ( 2 (x 1) x +x+1 4. (a) x 3 13x+12 =x 4+ 2 2 x +4x+3
2x+1 =x 4+ = ) A
Bx+C
+ 2
.
x 1
x +x+1 13x+12
A
B
=x 4+
+
(x+1)(x+3)
x+1 x+3 x +4x+3
A
Dx+E Fx+G
B
C
(b)
=
+
+
+ 2
+
3 2
2 x+1
2
3
2
2
x +4 (x +4)
(x+1) (x +4)
(x+1) (x+1)
5. (a)
x 4 4 = 4 (x 1)+1
4 x 1 1 =1+ x 1 [ or use long division] =1+ 4 =1+ 2 (x 1)(x+1)(x +1)
4 (b) 2 t +t +1 ( t2+1) ( t2+4) 2 6. (a) x
3 = At+B
2 + t +1 2 (x +x)(x x+3) A
B Cx+D
+
+ 2
x 1 x+1
x +1 Ct+D
2 t +4 4 = 2 (x 1)(x +1) x 1
1 =1+ 1
2 x
2 + Et+F ( t2+4) 2 4
2 x(x +1)(x x+3) = x
2 3
2 (x +1)(x x+3) = Ax+B
2 x +1 + Cx+D
2 x x+3 (b)
1 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 1
6 x x 3 1 =
x 3 ( x3 1 ) = 1
3 x
(x 6)+6
dx=
dx=
x 6
x 6 7. ( 2 x (x 1) x +x+1
1+ ) 6
x 6 = A B C
D
Ex+F
+ + +
+ 2
2
3 x 1
x
x +x+1
x
x
dx=x+6ln x 6 +C 8.
2 2 r
dr =
r+4
= r 16 16
+
r+4 r+4 dr= r 4+ 16
r+4 dr [ or use long division] 1 2
r 4r+16ln r+4 +C
2 x 9
A
B
=
+
. Multiply both sides by (x+5)(x 2) to get x 9=A(x 2)+B(x+5) .
(x+5)(x 2) x+5 x 2
Substituting 2 for x gives 7=7B B= 1 . Substituting 5 for x gives 14= 7A A=2 . Thus,
x 9
2
1
dx=
+
dx=2ln x+5 ln x 2 +C
(x+5)(x 2)
x+5 x 2
9. 1
A
B
=
+
1=A(t 1)+B(t+4) .
(t+4)(t 1) t+4 t 1
1
1
t=1 1=5B B= . t= 4 1= 5A A=
. Thus,
5
5
1
1/5 1/5
1
1
1
dt=
+
dt=
ln t+4 + ln t 1 +C or ln
(t+4)(t 1)
t+4 t 1
5
5
5
10. 1 11. 2 = x 1 3 2 2 3 dx =
2 x 1 x 1
2 1/2 1/2
+
x+1 x 1 B=
dx= 1
1
ln 4+ ln 2
2
2 = 12. +C 1
A
B
=
+
. Multiply both sides by (x+1)(x 1) to get 1=A(x 1)+B(x+1) .
(x+1)(x 1) x+1 x 1 Substituting 1 for x gives 1=2B
1 t 1
t+4 = 1
. Substituting 1 for x gives 1= 2A
2
1
1
ln x+1 + ln x 1
2
2 A= 1
. Thus,
2 3
2 1
1
1
1
3
ln 3+ ln 1 = (ln 2+ln 3 ln 4) or ln
2
2
2
2
2 A
B
+
. Multiply both sides by (x+1)(x+2) to get x 1=A(x+2)+B(x+1) .
x+1 x+2 x +3x+2
Substituting 2 for x gives 3= B B=3 . Substituting 1 for x gives 2=A . Thus,
2 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 1 1 x 1
0 dx = 2 x +3x+2 2
3
+
x+1 x+2 0 =( 2ln 2+3ln 3) ( 2ln 1+3ln 2)=3ln 3 5ln 2
ax 13. x bx
14. If a b , 0 orln 27
32 ax
a
dx=
dx=aln x b +C
x(x b)
x b dx= 2 1 dx= 2ln x+1 +3ln x+2 1
1
=
(x+a)(x+b) b a 1
x+a 1
x+b , so if a b , then dx
1
1
=
ln
( ln x+a ln x+b ) +C=
(x+a)(x+b) b a
b a
dx If a=b , then (x+a)
15. 2x+3
2 = A
+
x+1 (x+1)
to get A=2 . Now
1 B ( x+1 ) 2 2
+
x+1 dx = (x+1) 0 = 2ln 2
3 16. 2x+3=A(x+1)+B . Take x= 1 to get B=1 , and equate coefficients of x 2 1 2x+3
0 x 4x 10
2 =x+1+ 1 ( x+1 ) 2 0 2 dx = x+1+
0 x x 6
= 1
0 3x 4
3x 4
A
B
. Write
=
+
. Then 3x 4=A(x+2)+B(x 3) .
(x 3)(x+2)
(x 3)(x+2) x 3 x+2 1 3 x 4x 10 1
dx= 2ln (x+1)
x+1 1
1
( 2ln 1 1 ) =2ln 2+
2
2 x x 6
Taking x=3 and x= 2 , we get 5=5A
1 +C 1
+C .
x+a = 2 x+a
x+b 1
2
+
x 3 x+2 1
+1+ln 2+2ln 3
2 A=1 and 10= 5B
dx= B=2 , so 1 2
x +x+ln x 3 +2ln (x+2)
2
3
2 3
2 1
0 ( 0+0+ln 3+2ln 2 ) = +ln 3 ln 2= +ln 3
2 3 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 2 4y 7y 12 A
B
C
= +
+
17.
y(y+2)(y 3) y y+2 y 3 2 4y 7y 12=A(y+2)(y 3)+By(y 3)+Cy(y+2) . Setting y=0 gives 12= 6A , so A=2 . Setting y= 2 gives 18=10B , so B= 9
1
. Setting y=3 gives 3=15C , so C= .
5
5 Now
2
1 2 4y 7y 12
dy =
y(y+2)(y 3) 2 2 9/5 1/5
+
+
y y+2 y 3 1 9
1
dy= 2ln y + ln y+2 + ln y 3
5
5 2
1 9
1
9
1
ln 4+ ln 1 2ln 1
ln 3
ln 2
5
5
5
5
18
1
9
27
9
9
9
8
= 2ln 2+
ln 2
ln 2
ln 3=
ln 2
ln 3= (3ln 2 ln 3)= ln
5
5
5
5
5
5
5
3
= 2ln 2+ 2 x +2x 1 18. 3 x x 2 x +2x 1
A
B
C
=
= +
+
. Multiply both sides by x(x+1)(x 1) to get
x(x+1)(x 1) x x+1 x 1 2 x +2x 1=A(x+1)(x 1)+Bx(x 1)+Cx(x+1) . Substituting 0 for x gives 1= A
for x gives 2=2B B= 1 . Substituting 1 for x gives 2=2C C=1 . Thus,
2 x +2x 1 1
x dx= 3 x x
19. 1 = 2 ( x+5) (x 1) 1
1
+
x+1 x 1 A
+
x+5 B ( x+5) 2 + x(x 1)
x+1 dx=ln x ln x+1 +ln x 1 +C=ln
C
x 1 A=1 . Substituting 1 +C . 2 1=A(x+5)(x 1)+B(x 1)+C(x+5) . Setting x= 5 gives 1
1
. Setting x=1 gives 1=36C , so C=
. Setting x= 2 gives
6
36
1 1
3
1
1
2
1=A(3)( 3)+B( 3)+C 3 = 9A 3B+9C= 9A+ + = 9A+ , so 9A=
and A=
. Now
2 4
4
4
36
1/36
1/36
1
1/6
+
dx
dx =
2
2
x+5
x 1
( x+5) (x 1)
( x+5)
1
1
1
=
ln x+5 +
+
ln x 1 +C
36
6(x+5) 36
1= 6B , so B= ( ) 2 20. x 2 (x 3)(x+2) = A
B
+
+
x 3 x+2 C 2 2 2 x =A(x+2) +B(x 3)(x+2)+C(x 3) . (x+2)
9
4
2
Setting x=3 gives A=
. Take x= 2 to get C=
, and equate the coefficients of x to get 1=A+B
25
5
4 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions B= 16
. Then
25
2 9/25 16/25
+
x 3
x+2 x dx =
2
(x 3)(x+2) 2 2 21. 3 2 = 5x +3x 2 (x+2) A B
C
2
+ +
. Multiply by x (x+2) to get
2 x+2
x
x = 2 x ( x+2 ) x +2x dx 2 9
16
4
ln x 3 +
ln x+2 +
+C
25
25
5(x+2) = 5x +3x 2 4/5 2 2 5x +3x 2=Ax(x+2)+B(x+2)+Cx . Set x= 2 to get C=3 , and take x=0 to get B= 1 . Equating the
2 coefficients of x gives 5=A+C
2 5x +3x 2
3 2 2
x dx= x +2x
1 22. 2 2 = s (s 1) 1
2 + x A=2 . So 3
x+2 dx=2ln x + A B C
D
+ +
+
2 s 1
2
s
s
(s 1) 1
+3ln x+2 +C .
x
2 2 2 2 1=As(s 1) +B(s 1) +Cs (s 1)+Ds . Set s=0 , giving B=1 .
3 Then set s=1 to get D=1 . Equate the coefficients of s to get 0=A+C or A= C , and finally set s=2 to
get 1=2A+1 4A+4 or A=2 . Now
2 1
2
1
1
ds
1
=
+
+
ds=2ln s
2ln s 1
+C .
2
2
2 s 1
2
s
s
s 1
s (s 1)
s
(s 1)
2 x 23. 3 = (x+1) A
+
x+1 B
2 C + (x+1) 3 2 2 . Multiply by ( x+1 ) to get x =A(x+1) +B(x+1)+C . Setting 3 (x+1) 2 x= 1 gives C=1 . Equating the coefficients of x gives A=1 , and setting x=0 gives B= 2 . Now
2 x dx
3 = (x+1) 1
x+1 2
2 1 + (x+1) 3 (x+1) x
(x+1) 1
1
24.
=
=1
, so
x+1
x+1
x+1
x 3
3 (x+1) dx= 1 3
+
x+1 dx=ln x+1 + 3 3 1
= 1
3
x+1
(x+1)
x 1
2 (x+1) 3 (x+1) 2
x+1
3 =1 dx=x 3ln x+1 1
2 +C . 2(x+1)
3
+
x+1 3 1
2 (x+1) 3
+
x+1 3 . Thus, (x+1) 1
2 +C . 2(x+1) 5 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 10 25. ( 2 (x 1) x +9 ( ) = A Bx+C
2
+ 2
. Multiply both sides by (x 1) x +9 to get
x 1
x +9 ( ) ) 2 10=A x +9 +(Bx+C)(x 1) ( * ). Substituting 1 for x gives 10=10A A=1 . Substituting 0 for x gives 2 10=9A C C=9(1) 10= 1 . The coefficients of the x terms in ( * ) must be equal, so 0=A+B
B= 1. Thus,
10
1
x 1
1
x
1
dx =
+ 2
dx=
dx
2
2
2
x 1
x 1
(x 1)(x +9)
x +9
x +9 x +9
1
1
x
2
2
1
=ln x 1
ln x +9 [let u=x +9]
tan
[Formula10] +C
2
3
3 ( 2 26. x x+6
3 x +3x 2 = x x+6 = x ( x +3)
2 ) A Bx+C
2
2
2
+ 2
. Multiply by x x +3 to get x x+6=A x +3 +(Bx+C)x .
x
x +3 ( ) ( ) 2 Substituting 0 for x gives 6=3A A=2 . The coefficients of the x terms must be equal, so 1=A+B
B=1 2= 1 . The coefficients of the x terms must be equal, so 1=C . Thus,
2 x x+6
3 2
x 1
+ 2
x
x +3 dx = x +3x 1
2
ln x +3
2 ( =2ln x
3 27. 2 x +x +2x+1 ( x2+1) ( x2+2) 3 2 3 2 Ax+B
2 x +1 ( 2 3 = 2 2
x dx= + 2 3 x +3
1 x +3 x
+C
3 ( 2 ) ( x2+2) to get x +2 ) ( 2 ) ( 2 3 )
2 x +x +2x+1= Ax +Bx +2Ax+2B + Cx +Dx +Cx+D
3 dx 2 . Multiply both sides by x +1 x +x +2x+1=(Ax+B) x +2 +(Cx+D) x +1 ( 1 2 1
tan
3 ) Cx+D x ) 2 x +x +2x+1=(A+C)x +(B+D)x +(2A+C)x+(2B+D) . Comparing coefficients gives us the following
system of equations:
A+C=1
2A+C=2 (1) B+D=1
(3) 2B+D=1 (2)
(4) Subtracting equation (1) from equation (3) gives us A=1 , so C=0 . Subtracting equation (2) from
3 equation (4) gives us B=0 , so D=1 . Thus, I= 2 x +x +2x+1 ( x +1) ( x +2)
2 2 dx= x
2 x +1 + 1
2 dx . For x +2
6 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions x x 2 dx , let u=x +1 so du=2xdx and then 2 x +1
1 2 x +1
1 dx , use Formula 10 with a= 2 . So 2 x +2
1
1
2
I= ln x +1 +
tan
2
2 ( ) 2 x 2x 1 28. 2 ( x 1) ( x2+1) = dx= 2 A
+
x 1 1
1
1
2
du= ln u +C= ln x +1 +C . For
u
2
2 ( 1 dx= 2 x +( 2 x +2
1 1
2 ) 2 dx= 1
tan
2 1 ) x
+C . Thus,
2 x
+C .
2
B
2 + Cx+D ( 2 ) ( 2 ) 2 2 x 2x 1=A(x 1) x +1 +B x +1 +(Cx+D) ( x 1 ) . 2 x +1 (x 1) 3 Setting x=1 gives B= 1 . Equating the coefficients of x gives A= C . Equating the constant terms
gives 1= A 1+D , so D=A ,
and setting x=2 gives 1=5A 5 2A+A or A=1 . We have
2 x 2x 1 ( x 1) 2 ( x +1)
2 1
x 1 dx = x 1 1 x +1 (x 1) = ln x 1 + 1
x 1 dx 2 2 1
2
1
ln x +1 +tan x+C
2 ( ) 29.
x+4
2 x +2x+5 3 2 2 dx= 1
2 (2x+2)dx + 2 3dx 2 3 = 2 x 2x +x+1 ( 2 )( ) 2 = Ax+B
2 + Cx+D 3 2 2 x 2x +x+1
4 2 dx dx= 2 x +5x +4
31. 1
3 x 1 = x 3 + 2 x +1 x +4 1 ( 2 (x 1) x +x+1 1 dx=tan x+ ) = A
Bx+C
+ 2
x 1
x +x+1 ( 2 +C ) ( 2 x 2x +x+1= ( Ax+B ) x +4 + (Cx+D ) x +1 2 x +5x +4
x +1 x +4
x +1
x +4
. Equating coefficients gives A+C=1 , B+D= 2 , 4A+C=1 , 4B+D=1
3 2 x +2x+5
x +2x+5
(x+1) +4
1
2du
2
where x+1=2u,
= ln x +2x+5 +3
2
and dx=2du
2
4(u +1)
1
3
1
3
x+1
2
1
2
1
= ln (x +2x+5)+ tan u+C= ln (x +2x+5)+ tan
2
2
2
2
2 2 4 3 dx+ x +2x+5 x 2x +x+1 30. x+1 dx = 1
2
ln x +4
2 ( ( 2 ) A=0 , C=1 , B=1 , D= 3 . Now 3
1
tan (x/2)+C .
2 )
) 1=A x +x+1 +(Bx+C)(x 1) .
7 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 1
2
. Equating coefficients of x and then comparing the constant terms, we get
3
1
1
1
2
0= +B , 1= C , so B=
, C=
3
3
3
3
1
1
2
1
x
3
3
3
1
1
x+2
dx =
3
dx+
dx= ln x 1
dx
2
2
x 1
x 1
3
3
x +x+1
x +x+1
1
1 x+1/2
1
( 3/2 ) dx
dx
= 3 ln x 1 3
2
2
3
x +x+1
( x+1/2 ) +3/4
1
x+
1
1
2
2
2
1
= 1
ln x 1
ln x +x+1
tan
+K
3
6
2
3
3 2
1
1
1
1
2
1
ln x 1
ln x +x+1
tan
(2x+1) +K
=
3
6
3
3
Take x=1 to get A= (
( . )
) / 32.
1 1 x dx = 2 x +4x+13 0 1
(2x+4)
2
2 18 dx dx 2 x +4x+13 0 1
=
2 1 0 2 (x+2) +9 1 1
18 2
1
ln y
tan u
13 3
2
1
18
2
1
= ln
+ tan
2
13 6 3
= 3 2 dy
3
2
2
9u +9
13 y
2/3 du ( 2 204 1 = 2/3 1
18
ln
2
13 2
3 4 tan 2
3 1 2
3 ) 1
ln u
3 where y=x +4x+13,dy=(2x+4)dx,
x+2=3u, and dx=3du 2 33. Let u=x +3x +4 . Then du=3 x +2x dx
5 2 x +2x
2 3 2 x +3x +4
3 ( x3+1)
= 204
24 1 du u= =1 1 24 = 1
1
204 1
17
(ln 204 ln 24)= ln
= ln
.
3
3
24 3
2 A
Bx+C
2
+ 2
1=A x x+1 +(Bx+C)(x+1) . Equate the
3
3
3
x+1
x +1
x +1
x +1
x x+1
terms of degree 2 , 1 and 0 to get 0=A+B , 0= A+B+C , 1=A+C . Solve the three equations to get
34. x 1
dx=
3 =1 ( ) 8 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 1
1
2
, B=
, and C= . So
3
3
3
1
1
2
3
x
x
3
3
3
dx =
1
+ 2
dx
3
x+1
x +1
x x+1
1
1
2x 1
1
x
ln x+1 +
dx
2
3
6
2
=
x x+1 A= 1
1
2
ln x+1 + ln x x+1
3
6 ( = x
1 35. 4 2 1 = x x = 2 x (x 1)(x+1) A B + 2 x C + ) + x 1 x dx
1 2 3
x
+
2
4
1
1
1
tan
( 2x 1 )
3
3 D +K 2 . Multiply by x (x 1)(x+1) to get x+1 2 2 1=Ax(x 1)(x+1)+B(x 1)(x+1)+Cx (x+1)+Dx (x 1) . Setting x=1 gives C= 1
, taking x= 1 gives
2 1
3
. Equating the coefficients of x gives 0=A+C+D=A . Finally, setting x=0 yields B= 1 . Now
2
x 1
dx
1 1/2 1/2
1 1
=
+
dx= + ln
+C .
4 2
2
2
x 1 x+1
x+1
x x
x
x D= 4 ( 2 36. Let u=x +5x +4
1 3 2x +5x
0 4 x 3 37. =
=
= 2 x +5x +4 1
dx=
2 ( x +2x+4)
2 udu ( u2+3) 2
1
( 2v ) 4 ( 10
4 du 1
=
ln u
u 2 2 x +2x+4 ) ( 2 ( x+1 ) +3
du ( u2+3)
2 3
9 10
4 x 3 dx= 1
=
2 2 4 3
2
cos d =
9
1 2 2 ) 3 ) 3 du= 4x +10x dx=2 2x +5x dx , so ( dv 2 4 2 v
1
2 ) 2 u +3
x+1
1
tan
3 = 1
1
5
(ln 10 ln 4)= ln
.
2
2
2 dx= u 4 ( u +3)
2 3sec 2 2 d 4 du [ with u=x+1 ]
2 v=u +3 in the first integral;
u= 3 tan in the second 9sec
2 3
( +sin cos ) +C
9
+ 3 (x+1)
2 +C x +2x+4 9 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 2 3
tan
9 1 = 2 ( x +2x+4)
2 4 x +1 38. ( 2 x x +1 ) 2 x+1
3 1 2(x+1)
2
3 ( x +2x+4) A Bx+C
Dx+E
+ 2
+
2
x
2
x +1
x +1 = ( ( 4 +C )2 2 ( ) 2 x +1=A x +1 +(Bx+C)x x +1 +(Dx+E)x . Setting x=0 ) 4 gives A=1 , and equating the coefficients of x gives 1=A+B , so B=0 . Now
C
2 + x +1 4 Dx+E ( x +1)
2 x +1 = 2 ( 2 x x +1 1 ) 2 = x ( x4+2x2+1)
( x2+1) 2 4 1 x +1 x 4 x +1 D= 2 , and E=0 . Hence, ( 2 x x +1 2 ) 1 2x x dx= ( x +1)
2 2 = 2x ( x +1)
2 2 dx=ln x + , so we can take C=0 ,
1
2 +C . x +1 2 39. Let u= x+1 . Then x=u 1 , dx=2udu
dx
2udu
du
u 1
=
=2 2 =ln
2
u+1
x x+1
u 1 u
u 1 ( ) u
u u 2 = A
B
+
u 2 u+1 x+1 1
x+1 +1 +C . dx
2udu
udu
= 2
=2 2
and
x x+2
u 2 u
u u 2
1
u=A(u+1)+B(u 2) . Substituting 1 for u gives 1= 3B B= and
3
2 40. Let u= x+2 . Then x=u 2 , dx=2udu 2 +C=ln substituting 2 for u gives 2=3A A= I= 2
. Thus,
3 2
2
1
2
+
du= ( 2ln u 2 +ln u+1 ) +C
3
u 2 u+1
3
2
=
2ln x+2 2 +ln ( x+2 +1 ) +C
3 I = 2 41. Let u= x , so u =x and dx=2udu . Thus,
16
9 4 x
dx =
x 4 4 u
3 2 u 2udu=2 u 4 3 2 4 2 du=2 u 4 1+
3 4
2 du [by long division] u 4 4 =2+8
3 du (u+2)(u 2). ( * )
10 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 1
A
B
=
+
by (u+2)(u 2) to get 1=A(u 2)+B(u+2) . Equating coefficients we
(u+2)(u 2) u+2 u 2
1
1
1
1/4 1/4
get A+B=0 and 2A+2B=1 . Solving gives us B= and A=
, so
=
+
and ( *
4
4
(u+2)(u 2) u+2 u 2
) is
Multiply 4 1/4 1/4
+
u+2 u 2 2+8
3 = 2+2 ln = 2+2ln
3 2
1
ln
6
5 =2+2 3 =2+2ln 5
or2+ln
3 3 4 = 2+ 2ln u 2 2ln u+2 3 4 1
1
ln u+2 + ln u 2
4
4 du = 2+8 4 u 2
u+2 3 2/6
1/5 2 5
3 ln =2+ln 25
9 2 42. Let u= x . Then x=u , dx=3u du
1 1 1
0 dx = 3 0 1+ x 2 = 3
3 43. Let u= 1 3u
1+u=
du
0 3u 3+ 3
1+u 3 2
u 3u+3ln (1+u)
2 du= 1
0 1
2 ln 2 2 2 3 2 x +1 . Then x =u 1 , 2xdx=3u du
3
3 2
3
x dx
u 1
u du
=
2
3
3 5 3 2
4
3 2
u u du=
=
u
u +C
x +1
u
2
10
4
5/3 3
2/3
3
2
2
=
x +1
x +1 +C
10
4 ( ) ( ( ) ( ) ) 3 x 2 44. Let u= x . Then x=u , dx=2udu
1/3
1 =2 tan u 3
1/ 3 =2 3 6 = 2 x +x 3 dx=
1/ 3 u 2udu
4 u +u 2 3 =2
1/ 3 du
2 u +1 3 45. If we were to substitute u= x , then the square root would disappear but a cube root would
remain. On the other hand, the substitution
11 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions u= 3 x would eliminate the cube root but leave a square root. We can eliminate both roots by means
6 of the substitution u= x . (Note that 6 is the least common multiple of 2 and 3 .)
6 6 5 3 3 Let u= x . Then x=u , so dx=6u du and x =u ,
5 dx
x 3 6u du = x 3 = 6 46. Let u= 12 3 x . Then x=u
11 dx =6 12u du 3 u
du=6
du
u 1 2 12 6 x 1 +C 11 , dx=12u du
8 u du
=12
=12
u+1 7 6 5 4 3 2 x+ x x 47. Let u=e . Then x=ln u , dx=
2x 2 e dx = x e +3e +2 u ( du/u )
2 u +3u+2 = u u +u u +u u +u 1+ 1
u+1 du
4 3
u +u
3 8 12 7
6 12 5
4
3
2
=
u
u +2u
u +3u 4u +6u 12u+12ln u+1 +C
2
7
5
3 2/3 12 7/12
12 5/12 3
4
6
12
=
x
x +2 x
x +3 x 4 x +6 x 12 x +12ln
2
7
5
= 4 2x 5 u (u 1)
1
2
u +u+1+
du [by long division]
u 1
1 3 1 2
3
6
u + u +u+ln u 1 +C=2 x +3 x +6 x +6ln
3
2 u u
= 6 2 u 2 x =u . Thus, ( 12 x +1) +C du
u
udu
=
(u+1)(u+2) 1
2
+
u+1 u+2 du = 2ln u+2 ln u+1 +C=ln (ex+2)2 /(ex+1) +C
48. Let u=sin x . Then du=cos xdx
cos xdx
du
du
=
=
=
2
2
u(u+1)
sin x+sin x
u +u
u
= ln
+C=ln
u+1 ( 2 ) 1
u 1
u+1 sin x
1+sin x 49. Let u=ln x x+2 , dv=dx . Then du= 2x 1
2 du
+C dx , v=x , and (by integration by parts) x x+2
12 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions ( 2 ( ) 2 ln x x+2 dx =xln x x+2 ( 2 =xln x x+2 ( 2 =xln x x+2 2 2x x ) 2 dx=xln x x+2 2 x x+2
1
(2x 1)
2
7
2x
dx+
2
2
x x+2 )
) ( x 4
2 dx x x+2
dx
1 2 7
x
+
2
4 1
7
2
ln x x+2 +
2
2 2x 2+ ) 7
du
2
7 2
(u +1)
4 7
1
=
u,
2
2
7
dx=
du,
2
1 2 7 7 2
x
+ = (u +1)
2
4 4
where x 1
2
1
= x
2
= x 1 50. Let u=tan x , dv=xdx ( 2 ) 2x+ 7 tan u+C ( 2 ) 2x+ 7 tan ln x x+2
ln x x+2 ( 1 2
1
1
Then xtan xdx= x tan x
2
2
observe that ( 1+x )
2 x 2 dx= 2 du=dx/ 1+x 2 1 2x 1
+C
7 ) , v= 1 x2 .
2 2 x 1 2 1 2 dx . To evaluate the last integral, use long division or 1+x
1 1 dx= 1dx 1 2 dx=x tan x+C . So
1 1+x
1+x
1+x
1 2
1
1 2
1
1
1
1
1
xtan xdx= x tan x
x tan x+C =
x tan x+tan x x +C .
1
2
2
2 ( ) 51. 13 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions From the graph, we see that the integral will be negative, and we guess that the area is about the same
as that of a rectangle with width 2 and height 0.3 , so we estimate the integral to be ( 2 0.3) = 0.6 .
1
1
A
B
1
Now 2
=
=
+
1=(A+B)x+A 3B , so A= B and A 3B=1 A= and
(x 3)(x+1) x 3 x+1
4
x 2x 3
1
B=
, so the integral becomes
4
2
0 1
x 2x 3 =
dx
4
2 1
4 = 0 1
x 3
dx
4
ln x 3
x+1 2
0 dx x+1=
2 = 0 1
4 1
ln x 3 ln x+1
4
ln 2
0 1
1
ln 3 =
ln 3
3
2 0.55 A B
C
2
+ +
1=(A+C)x +(B 2A)x 2B , so A+C=B 2A=0 and
2
3
2
2 x 2
x
x (x 2)
x 2x
x
1
1
1
1
2B=1 B=
, A=
, and C= . So the general antiderivative of
is
3
2
2
4
4
x 2x
1 dx 1 dx 1 dx
dx
=
+
3
2
2
4 x 2
4 x 2
x 2x
x
1
1
1
=
ln x
( 1/x ) + ln x 2 +C
4
2
4
1
x 2
1
= ln
+
+C
4
x
2x 52. 1 = 1 2 = We plot this function with C=0 on the same screen as y= 1
3 2 . x 2x 14 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 53.
dx dx = 2 x 2x du = 2 2 (x 1) 1
1
u 1
= ln
2
u+1 u 1
+C [ by Equation 6] = 1
ln
2 x 2
x +C 54.
(2x+1)dx
2 = 4x +12x 7
= 1
4 (8x+12)dx 2dx 2 2 4x +12x 7 (2x+3) 16
du 1
2
ln 4x +12x 7
4 2 u 16 1
2
ln 4x +12x 7
4
1
2
= ln 4x +12x 7
4 1
ln (u 4)/(u+4) +C
8
1
ln (2x 1)/(2x+7) +C
8 = x
2 55. (a) If t=tan
sin x
2 = t (b) cos x=cos 2 1+t
sin x=sin x
2
1= 2 2 =2cos
2
1+t x
2 x
2 = 1
1+t and
2 2 2 1 x
1
=tan t . The figure gives cos
2 . 1+t =2 , then =2sin 2 x
2 2 1= 1 t 2 1+t
x
2 1 2 cos x
2 =2 t
1+t 1
2 1+t =
2 2t
1+t 2 (c)
15 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions x
=arctant
2 x=2arctant 2 dx= 1+t 2 dt 2 56. Let t=tan (x/2) . Then, using Exercise 55 , dx= 1+t
dx
=
3 5sin x ( 2dt/ 1+t ( 2 )
2 ) 2dt = 3 10t/ 1+t
1
1
3
=
4
t 3 3t 1 ( 2 ) =2 2 dt , sin x= 2t
1+t 2 dt
2 3 1+t 10t
3t 10t+3
1
1
dt= ( ln t 3 ln 3t 1 ) +C= ln
4
4 tan (x/2) 3
3tan (x/2) 1 +C 57. Let t=tan (x/2) . Then, the expressions in Exercise 55 , we have
dt
1
2dt
dt
=2
=
2
1
2
2
2
2t +3t 2
2t
1 t
1+t
3(2t) 4 1 t
dx =
3
4
3sin x 4cos x
2
2
1+t
1+t
dt
2 1
1 1
=
=
dt [using partial fractions]
(2t 1)(t+2)
5 2t 1 5 t+2
1
1
2t 1
1
2tan ( x/2 ) 1
=
ln 2t 1 ln t+2 +C= ln
+C= ln
5
5
t+2
5
tan ( x/2 ) +2 ( ) +C 58. Let t=tan (x/2) . Then, by Exercise 55 ,
1 /2 dx
=
/3 1+sin x cos x 1
( 2)
2dt
=
2
2
2
2
2
3 1+2t/ ( 1+t ) ( 1 t ) / ( 1+t ) 1/ 3 1+t +2t 1+t 2dt/ 1+t 1/
1 =
1/ 3 1
t 1
t+1 dt= ln t ln ( t+1 ) 1
1/ 3 =ln 1
ln
2 1
=ln
3 +1 3 +1
2 59. Let t=tan (x/2) . Then, by Exercise 55 , 16 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 1
dx
= 2
2sin x+sin 2x dx
1
=
sin x+sin xcos x 2 ( 1+t2) dt = 1
2
2
4
t ( 1+t ) +t ( 1 t ) 1
=
2
=
2 1
1 2
1
ln t + t +C= ln
4
8
4 ( 2)
2
2
2 2
2t/ ( 1+t ) +2t ( 1 t ) / ( 1+t )
( 1+t2) dt = 1 1 +t dt
2dt/ 1+t t 1
x
2 tan 4 + t 1
2
tan
8 1
x +C
2 2 60. x 6x+8=(x 3) 1 is positive for 5 x 10 , so
10 area =
5 = 61. 1+
2 dx du 2 1
3
ln
2
4 1
ln
2 2 (x 3) 1= u 1 [putu=x 3] = 7 u 1
u+1 2 1
1 1
3
ln = (ln 3 2ln 2+ln 3)=ln 3 ln 2=ln
2
3 2
2 x+1
2
=1+
>0 for 2 x 3 , so area
x 1
x 1 3 = 7 2 2
x 1 dx= x+2ln x 1 3 = ( 3+2ln 2 ) ( 2+2ln 1 ) =1+2ln 2 . 2 62. (a) We use disks, so the volume is V = 2 1 1
0 1 dx 0 dx= 2 2 2 . To evaluate the x +3x+2
(x+1) (x+2)
A
C
1
B
D
integral, we use partial fractions:
=
+
+
+
2
2 x+1
2 x+2
2
(x+1) (x+2)
(x+1)
(x+2)
2 2 2 2 1=A(x+1)(x+2) +B(x+2) +C(x+1) (x+2)+D(x+1) . We set x= 1 , giving B=1 , then set x= 2 , giving
3 D=1 . Now equating coefficients of x gives A= C , and then equating constants gives
1=4A+4+2( A)+1 A= 2 C=2 . So the expression becomes
1 V =
= 0 2
+
x+1
2ln 3
2 1 ( x+1 )
1
2 2
+
+
2
( x+2 ) 1
3 1 ( x+2 ) 2ln 2 1 1
2 2 dx=
= 2ln
2ln x+2
x+1 3/2 2
+
2
3 (b) In this case, we use cylindrical shells, so the volume is V =2 1
x+1 0 xdx
2 0 2
9
+ln
3
16 =
1 1 1
x+2 x +3x+2 =2 xdx
. We
0 (x+1)(x+2) 1 17 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions x
A
B
=
+
(x+1)(x+2) x+1 x+2
A= 1 and B=2 . So the volume is use partial fractions to simplify the integrand:
A+B=1 and 2A+B=0
1
1
2
2
+
dx = 2
0
x+1 x+2 ln x+1 +2ln x+2 x=(A+B)x+2A+B . So 1
0 = 2 ( ln 2+2ln 3+ln 1 2ln 2)=2 (2ln 3 3ln 2)=2 ln
P+S
A
B
= +
P (r 1)P S
P (r 1)P S
A= 1 , B=r . Now 63. t= so t= ln P+ P+S=A (r 1)P S +BP= (r 1)A+B P AS P+S
dP=
P (r 1)P S 1
r
+
P (r 1)P S dP= dP
r
+
P r 1 9
8
(r 1)A+B=1 , A=1 r 1
dP
(r 1)P S r ln (r 1)P S +C . Here r=0.10 and S=900 , so
r 1
0.1
1
t = ln P+
ln 0.9P 900 +C= ln P
ln ( 1 0.9P+900 )
0.9
9
1
= ln P
ln (0.9P+900)+C
9 When t=0 , P=10 , 000 , so 0= ln 10 , 000 1
1
ln (9900)+C . Thus, C=ln 10 , 000+ ln 9900 , so our
9
9 equation becomes
1
1
10,000 1
9900
ln 9900
ln (0.9P+900)=ln
+ ln
9
9
P
9
0.9P+900
10,000 1
1100
10,000 1
11,000
= ln
+ ln
=ln
+ ln
P
9
0.1P+100
P
9
P+1000 t = ln 10,000 ln P+ 2 64. If we subtract and add 2x , we get ( ) 2 2x2= ( x2+1) 2
( x2+1) + 2 x = ( x2 4
4
2
2
2
x +1 = x +2x +1 2x = x +1 = ( x2+1) 2x
1 So we can decompose 4 x +1 ( 2 1= ( Ax+B ) x ) = Ax+B
2 x + 2 x+1 ( 2 + ( 2 x) 2 x+1
Cx+D 2 x 2 ) ( x2+ 2 x+1 ) 2 x+1 ) 2 x+1 + (Cx+D ) x + 2 x+1 . Setting the constant terms equal gives B+D=1 , then
18 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions 3 from the coefficients of x we get A+C=0 . Now from the coefficients of x we get A+C+ ( B D ) 2 =0
1
1
2
(1 D) D 2 =0 D=
B= , and finally, from the coefficients of x we get
2
2
2
2
1
and A=
C=
.
2 (C A) +B+D=0 C A=
4
4
2
So we rewrite the integrand, splitting the terms into forms which we know how to integrate:
2
1
x+
4
2 1
4 = x +1 2 + x + 2 x+1
2
8 = 2
1
x+
4
2
2 x 2 x+1 2 x + 2 x+1 Now we integrate: 4 x +1 2 x 1 2x+2 2 4 2 x + 2 x+1 2 2 2x 2x+ 2 dx = + 2 x+1 1
4 x + 2 x+1
2 x 2 2 x+1 x 1
1
x+
2 2 2
=
ln
8 2x 2 2 + 2 x+1 2
4 tan 2 +
1 ( 1
2 1 +
x 2 1
2 2 x+1 ) +tan 1 + ( 1
2 2 x 1 ) +C . 65. (a) In Maple, we define f (x) , and then use convert(f,parfrac,x) ; to obtain
24,110/4879 668/323 9438/80,155 (22,098x+48,935)/260,015
f (x)=
+
.
2
5x+2
2x+1
3x 7
x +x+5
In Mathematica, we use the command Apart, and in Derive, we use Expand.
(b)
24,110 1
668 1
9438 1
ln 5x+2
ln 2x+1
ln 3x 7
f (x)dx =
4879 5
323 2
80,155 3
1
22,098 x+
+37,886
1
2
+
dx+C
260,015
1 2 19
x+
+
2
4
24,110 1
668 1
9438 1
=
ln 5x+2
ln 2x+1
ln 3x 7
4879 5
323 2
80,155 3
+ 1
260,015 22,098 1
2
ln x +x+5 +37,886
2 ( ) 4
tan
19 1 1
19/4 x+ 1
2 4822
334
3146
11,049
2
ln 5x+2
ln 2x+1
ln 3x 7 +
ln x +x+5
4879
323
80,155
260,015
75,772
1
1
+
tan
( 2x+1 ) +C
260,015 19
19
= ( +C ) 19 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions Using a CAS, we get
4822ln (5x+2)
4879 ( 2 334ln (2x+1)
323 3146ln (3x 7)
80,155 ) 3988 19
11,049ln x +x+5
+
+
tan
260,015
260,015 19
(2x+1)
19 1 The main difference in this answer is that the absolute value signs and the constant of integration have
been omitted. Also, the fractions have been reduced and the denominators rationalized.
66. (a) In Maple, we define f (x) , and then use convert(f,parfrac ,x) ; to get
5828/1815 59,096/19,965 2(2843x+816)/3993 (313x 251)/363
f (x)=
+
+
.
2
2
2
5x 2
2
2x +1
( 5x 2 )
2x +1
In Mathematica, we use the command Apart, and in Derive, we use Expand.
(b) As we saw in Exercise 65, computer algebra systems omit the absolute value signs in
( 1/y ) dy=ln y . So we use the CAS to integrate the expression in part (a) and add the necessary
absolute value signs and constant of integration to get ( f (x)dx = ( 2 5828
59,096ln 5x 2
2843ln 2x +1
+
9075(5x 2)
99,825
7986
503
1 1004x+626
1
+
2 tan ( 2 x )
+C
2
15,972
2904
2x +1 ) ) (c) From the graph, we see that f goes from negative to positive at x 0.78 , then back to negative at
x 0.8 , and finally back to positive at x=1 . Also,
f (x)= . So we see (by the First Derivative
x 0.4 Test) that f (x)dx has minima at x 0.78 and x=1 , and a maximum at x 0.80 , and that f (x)dx is
unbounded as x 0.4 . Note also that just to the right of x=0.4 , f has large values, so f (x)dx
increases rapidly, but slows down as f drops toward 0 . f (x)dx decreases from about 0.8 to 1 , then
increases slowly since f stays small and positive.
67. There are only finitely many values of x where Q(x)=0 (assuming that Q is not the zero
20 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions polynomial). At all other values of x , F(x) / Q(x) =G(x) / Q(x) , so F(x)=G(x) . In other words, the
values of F and G agree at all except perhaps finitely many values of x . By continuity of F and G ,
the polynomials F and G must agree at those values of x too.
More explicitly: if a is a value of x such that Q(a)=0 , then Q(x) 0 for all x sufficiently close to a .
Thus,
F(a)=lim F(x)[by continuity of F] = lim G(x) [whenever Q(x) 0]
x a x a = G(a) [by continuity of G]
f (x) 2 68. Let f (x)=ax +bx+c . We calculate the partial fraction decomposition of 2 3 . Since f (0)=1 , x (x+1)
2 A B
C
D
E
. Now in order for the
+ +
+
+
2
3
2
3
2 x+1
2
3
x
x (x+1)
x (x+1)
x
(x+1) (x+1)
integral not to contain any logarithms (that is, in order for it to be a rational function), we must have
we must have c=1 , so f (x) 2 = 3 ax +bx+1 2 = 2 A=C=0 , so ax +bx+1=B(x+1) +Dx (x+1)+Ex . Equating constant terms gives B=1 , then equating
coefficients of x gives 3B=b / b=3 . This is the quantity we are looking for, since f (0)=b. 21 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration sin x+sec x
dx=
tan x 1. sin x sec x
+
tan x tan x dx= (cos x+csc x)dx=sin x+ln csc x cot x +C 2.
tan 3 d = (sec 2 = udu+
= 2 1)tan d = tan sec
u=tan , dv
v du=sec 2 sin
cos d d v=cos ,
dv= sin d d 1 2
1
2
u +ln v +C= tan +ln cos
2
2 +C 3.
2 1 2t ( t 3) 0 2 1 2 ( u+3) 2 6
+
du= 2ln u
2
2
u
u
u
3
3
= ( 2ln 1+6 ) ( 2ln 3+2 ) =4 2ln 3 or 4 ln 9 dt = du= xdx 2 4. Let u=x . Then du=2xdx 3 x 4 1
=
2 3 1
= sin
2 du
3 u 1 6
u 2 1 u
1
+C= sin
2
3 1 2 x
+C .
3 1 5. Let u=arctany . Then du= /4 arctany dy e
2 1+y 1 2 dy= 1+y u u e du= e /4 =e /4 /4 /4 e /4 . 6.
x csc x cot xdx u=x, dv=csc x cot xdx,
du=dx v= csc x = x csc x ( csc x)dx = xcsc x+ln csc x cot x +C
7.
3 4 r ln r dr
1 u=ln r,
dr
du=
r 4 dv=r dr
1 5
v= r
5 = = 1 5
r ln r
5
243
ln 3
5 3 3
1 1 243
25 1 4
243
r dr=
ln 3 0
5
5
1
25 = 1 5
r
25 3
1 243
242
ln 3
5
25
1 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration x 1 8. 2 = x 4x 5 x 1
A
B
=
+
(x 5)(x+1) x 5 x+1 4 x 1
2 0 x 4x 5 0 2/3 1/3
+
x 5 x+1 dx = 1
3 2
. Now
3 . Setting x=5 gives 4=6A , so A=
4 x 1=A(x+1)+B(x 5) . Setting x= 1 gives 2= 6B , so B= 4 2
1
ln x 5 + ln x+1
3
3 dx= 0 2
1
2
1
1
ln 1+ ln 5
ln 5
ln 1=
ln 5
3
3
3
3
3 =
9.
x 1 ( x 2 ) +1 dx = 2 u dx= 2 2 ( x 2 ) +1 x 4x+5
= + u +1 1 du [ u=x 2 , du=dx ] 2 u +1 1
1
2
1
2
ln u +1 +tan u+C= ln x 4x+5 +tan
2
2 ( ) ( ) 1 (x 2)+C 10.
x
4 2 1
du
2 dx = 2 [ u=x , du=2xdx ] = 2 x +x +1 u +u+1 = 3
dv
2 1
2 = 3 2
v +1
4 ( ) [ u+ 1
2 du
1 2 3
u+
+
2
4 3
3
3
1
=
v , du=
dv ] =
2
2
2
4 1
1
1
tan v+C=
tan
3
3 1 2
3 2 x+ 1
2 4
3 dv
2 v +1 +C 11.
sin 3 cos 5 d = cos
= 5 sin 5 2 2 sin u (1 u )du
7 5 = (u u )du= d = cos 5 (1 cos 2 )( sin )d u=cos ,
du= sin d 1 8 1 6
1
8
u
u +C= cos
8
6
8 1
6
cos +C
6 Another solution:
2 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration sin 3 cos 5 3 = sin d 2 (cos 3 2 ) cos u=sin ,
du=cos d 22 = u (1 u ) du
3 5 d = sin 7 = (u 2u +u )du= 3 (1 sin
3 13. Let x=sin . Then dx=cos dx
2 3/2 = cos (1 x ) d
3 (cos
=tan 2 = sec 2 2 d 4 1 4 1 6 1 8
1
4
u
u + u +C= sin
4
3
8
4 sin xcos (cos x)dx= 2 2 ) cos = u (1 2u +u )du 12. Let u=cos x . Then du= sin xdx
, where 2 1
1
6
8
sin + sin +C
3
8 cos udu= sin u+C= sin (cos x)+C .
2 1/2 d and (1 x ) =cos , so d ) x +C= +C
2 1 x 14. Let u=ln x . Then du=dx/x
1+u
du=
u 1+ln x
dx =
xln x
= 2 1+ v 2vdv 2 v 1 1
2 dv=2v+ln v 1
2 15. Let u=1 x 2 [put v= 1+u ,u=v 1,du=2vdv]
v 1
v+1 1+ln x 1 +C=2 1+ln x +ln 1+ln x +1 +C du= 2xdx . Then 1/2 x
2 0 1 x 1
dx=
2 3/4
1 1
1
du=
2
u 1 u
3/4 1/2 du= 1
1/2
2u
2 1 = 3/4 u 1 =1 3/4 3
2 3 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration 16.
2 /2
/4
2 x dx =
0 2 2 sin
cos cos d [ x=sin , }{dx}=cos d ] 1 x 0 = /4
0 1
1
(1 cos 2 )d =
2
2 /4 1
sin 2
2 0 = 1
2 4 1
2 (0 0) = 8 1
4 17.
2 dv=sin xdx u=x, 2 xsin xdx 2 du=dx v= sin xdx=
1 2
x
2
1 2
= x
2
= 1
1
1
(1 cos 2x)dx= x
sin xcos x
2
2
2 1
1
1
x sin x cos x
x
sin x cos x dx
2
2
2
1
1 2 1
1 2 1
1
2
2
x sin x cos x
x + sin x+C= x
x sin x cos x+ sin x+C
2
4
4
4
2
4 1 2
s +C where s=sin x,ds=cos xdx .
2
1
1
1
2
A slightly different method is to write x sin xdx= x (1 cos 2x)dx=
xdx
xcos 2xdx . If
2
2
2
we evaluate the second integral by parts, we arrive at the equivalent answer
1 2 1
1
x
xsin 2x
cos 2x+C .
4
4
8
Note: sin x cos xdx= sds= 2t 2t 18. Let u=e , du=2e dt . Then
1
1
2t
2t
(2e )dt
du
1
1
1
1 2t
e
2
2
dt =
=
= tan u+C= tan (e )+C .
4t
2t 2
2
2
2
1+e
1+(e )
1+u
x 19. Let u=e . Then e x+e 3 x 20. Let u= x . Then x=u
u v=e u 2 ( 2 u x e x u u e x dx= e e dx= e du=e +C=e +C .
3 e
u [3]x ) u 2 2 u dx= e 3u du . Now use parts: let w=u , dv=e du dw=2udu , u 3 e u du=3 u e 2 ue du . Now use parts again with W =u , dV =e du to get
4 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration u 2 u e 3u du=e 3 ( 3u2 6u+6) +C=3e x ( x2/3 2 3 x +2) +C .
3 2t 21. Integrate by parts three times, first with u=t , dv=e dt
1 3 2t 1
1 3 2t 3 2 2t 1
2 2t
3 2t
t e +
t e dt = 2 t e + 2 3t e dt= 2 t e
4
2
1 3 3 2
3 2t 3
2t
2t
2t
= e
t + t
te +
e dt= e
2
4
4
4
1 2t
3
2
=
e
4t +6t +6t+3 +C
8 ( 1 2 1 2
1
1
x sin x
2
2 x sin xdx = 3te 2t dt 1 3 3 2 3 3
t + t + t+
2
4
4 8 +C ) 22. Integrate by parts: u=sin x , dv=xdx 1 : 2 x dx
2 1 x ( du= 1 / 2 1 x 1 2
1
1
= x sin x
2
2 ) dx , v= 1 x
2 2 sin 2 cos
cos , so d where x=sin for 2 1 2
1
1 2
1
1
1
x sin x
( 1 cos 2 ) d = x sin x ( sin cos
2
4
2
4
1 2
1
1
1
1
2
2
1
=
x sin x
sin x x 1 x +C=
2x 1 sin x+x
2
4
4
= ( ) )+C
2 1 x +C 2 23. Let u=1+ x . Then x=(u 1) , dx=2(u 1)du
1 ( 1+
0 x ) 8 u 2(u 1)du=2 1 = ( 1024
5 2 1024
9 ) 24. Let u=ln x 1 , dv=dx 1 10
1 9
u 2 u
5
9 ( u9 u8) du=
1
2 2 8 dx = 2
1 1 2 4097
+ =
5 9
45
2x du= 2 , v=x . Then x 1 ( 2 ) ( 2 ln x 1 dx =x ln x 1 ) 2 2x
2 ( 2 dx=x ln x 1 ) x 1 ( 2 =x ln x 1 ) 2+ 1
x 1 1
x+1 2+ 2
(x 1)(x+1) dx dx
5 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration ( 2 =x ln x 1
2 3x 2 25. 2 =3+ ) 2x ln x 1 +ln x+1 +C 6x+22
A
B
=3+
+
(x 4)(x+2)
x 4 x+2 6x+22=A(x+2)+B(x 4) . Setting x=4 gives 46=6A , so x 2x 8
23
5
A=
. Setting x= 2 gives 10= 6B , so B=
. Now
3
3
2 3x 2 dx= 2 3+ x 2x 8
2 3x 2 26. 3 x 2x 8 23/3
x 4 5/3
x+2 dx=3x+ 23
ln x 4
3 5
ln x+2 +C .
3 3 du
dx=
u u=x 2x 8, ( 2 3 =ln u +C=ln x 2x 8 +C ) du= 3x 2 dx 27. Let u=ln (sin x) . Then du=cot xdx cot xln (sin x)dx= udu= 1 2
1
2
u +C=
ln (sin x) +C .
2
2 28.
2
2
2
udu [ u= at , u =at , 2udu=adt ] =
usin udu
a
a
2
2
ucos u+sin u +C [ integration by parts] =
at cos at + sin at +C
a
a sin at dt = sin u
= 2
a t
2
cos at + sin at +C
a
a = 2
29.
5 5 3w 1
dw =
w+2
0 0 7
w+2 3 dw= 3w 7ln w+2 5
0 =15 7ln 7+7ln 2=15+7(ln 2 ln 7)=15+7ln 2
7 2 30. x 4x<0 on 0,4 , so
2
2 2 x 4x dx =
=0 0 2 (x 4x)dx+ 2 ( 4x x2) dx=
0 2 8
8
8 + 8
3
3 1 3 2
x 2x
3 0 2 + 2x 2 1 3
x
3 2
0 0=16
6 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration 31. As in Example 5,
1+x 1+x 1 x 1+x
dx =
1 x 1+x = sin 1 x dx= 1+x dx dx=
2 xdx +
2 1 x 2 1 x 1 x 2 1 x +C Another method: Substitute u= (1+x)/(1 x) .
32.
u udu
4
2
2
2x 1
u= 2x 1 ,2x+3=u +4,u =2x 1,udu=dx =
1 2
dx =
2
2x+3
u +4
u +4
1
1
1
1
1
=u 4 tan
u +C= 2x 1 2tan
2x 1 +C
2
2
2
2 2 2 33. 3 2x x = (x +2x+1)+4=4 (x+1) . Let x+1=2sin , where 2 2 du . Then dx=2cos d and
2 2 3 2x x dx = 4 (x+1) dx=
2 =4 cos 4 4sin =2sin
=2sin 1 d 2cos d =2 (1+cos 2 )d =2 +sin 2 +C=2 +2sin
1 2 x+1
2
x+1
2 cos +C
2 3 2x x
x+1
+2
+C
2
2
x+1
2
+
3 2x x +C
2 34.
/2 1+4cot x
dx =
4 cot x
/4 /2
/4 (1+4cos x/sin x) sin x
(4 cos x/sin x) sin x /2 dx= sin x+4cos x
dx
4sin x cos x
/4
7 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration 4 =
3/ 2 1
du
u u=4sin x cos x,
du=(4cos x+sin x)dx 3
4
=ln
=ln
2
3/ 2 =ln 4 ln 4 = ln u 3/ 2 4
2
3 8 35. Because f (x)=x sin x is the product of an even function and an odd function, it is odd. Therefore,
1 8 1 x sin xdx=0 [ by (5.5. )(b)]. 1
(sin x+sin 7x) by Formula .2.2(a), so
2
1
1
1
1
1
sin 4xcos 3xdx=
(sin x+sin 7x)dx=
cos x
cos 7x +C=
cos x
cos 7x+C .
2
2
7
2
14 36. sin 4xcos 3x= 37.
/4
0 cos 2 tan 2 d = /4 sin 0 = 1
( 1 cos 2 ) d =
0 2
1
( 0 0) =
8 4 2 d = 1
4 8 1
2 /4 1
sin 2
4 /4
0 38.
/4
0 tan 5 sec 3 =
d /4
0 d = ( tan 2 ) 2sec 2
2
1 2 sec 2 2 (u 1) u du tan u=sec
du=sec tan d 2
1 7 2 5 1 3
u
u+ u
1
7
5
3
1
8
8
2
1 2 1
22
8
2
=
2
2+
2
+
=
2
=
(11 2 4)
7
5
3
7 5 3
105
105 105 = 2 6 4 2 (u 2u +u )du= 2 39. Let u=1 x . Then du= 2xdx
xdx
1
du
=
= 2
2
2
u+ u
1 x+ 1 x
= vdv
2 2 [v= u ,u=v ,du=2vdv] v +v dv
= ln v+1 +C= ln
v+1 ( 2 ) 1 x +1 +C
8 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration 2 2 2 40. 4y 4y 3=(2y 1) 2 , so let u=2y 1
1
dy
dy
=
=
2
2
2 2
4y 4y 3
(2y 1) 2 du=2dy . Thus,
du
2 2 u 2 1
2 2
ln u+ u 2
2
1
2
=
ln 2y 1+ 4y 4y 3 +C
2
= 41. Let u=
2 , dv=tan 2 = d 2 ) (tan (tan = tan ( d = sec tan 1
2 2 ) du=d 1 d ln sec 3 1 3
1
x tan xdx = 3 x tan x x
3 2 2 = 1 3
1
1
x tan x
3
3 x du=dx/ 1+x 1+x
1 3
1 2 1
1
2
=
x tan x
x + ln x +1 +C
3
6
6 ( x 2 +C ) , v= 1 x3 :
3 ( 2 dx ln sec . So
1
+
2 +C 1 1 2 )d = tan 42. Integrate by parts with u=tan x , dv=x dx
2 and v=tan x dx 2 x +1 ) x 43. Let u=1+e , so that du=e dx . Then
2 3/2
2
1/2
x 3/2
x
x
= u du= u +C= (1+e ) +C .
e 1+e dx
3
3
x 2 x x Or: Let u= 1+e , so that u =1+e and 2udu=e dx . Then
2 3
2
2
x 3/2
x
x
e 1+e dx = u 2udu= 2u du= 3 u +C= 3 (1+e ) +C .
x 2 x x 2 44. Let u= 1+e . Then u =1+e , 2udu=e dx=(u 1)dx , and dx= 2u
2 du , so u 1 9 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration x 1+e dx = u 2u 2u du= 2 2 du= 2 u 1 u 1 2 5 45. Let t=x . Then dt=3x dx
1 t 1
1
t
I=
te +
e dt=
te
3
3
3 I= x e 2 x ( x 1+e +ln 3 dx= 1 t
1
e +C=
e
3
3 t du= 2+ u 1 =2u+ln u 1 ln u+1 +C=2
3 2 2+ x 1+e x 1 1
u 1 1
u+1 ) ln ( du
x ) 1+e +1 +C 1
t
t
te dt . Now integrate by parts with u=t , dv=e dt :
3
3 ( x3+1) +C . x 46. Let u=e . Then x=ln u , dx=du/u
1+e
1 e x x (1+u)du
=
(1 u)u dx = (u+1)du
=
(u 1)u 2
u 1 1
u du = ln u 2ln u 1 +C=ln ex 2ln ex 1 +C=x 2ln ex 1 +C
47.
1
dx =
2 2
2
x +a
x+a =ln 2xdx
2 2 2 dx +a 2 x +a 2 2 x +a
1 x +a +tan = 1
1
2 2
ln x +a +a tan
2
a ( ) x
a 1 +C ( x/a ) +C 2 48. Let u=x . Then du=2xdx
1
du
xdx
2
1
=
=
ln
4 4
2
2
2 2
x a
4a
u a 2 u a 2 ( ) 49. Let u= 4x+1 2 u =4x+1 1
dx =
x 4x+1 = ln 2udu=4dx 1
udu
2
1 2
u 1 u
4 ( ) +C= u+a =2 4x+1 1
4x+1 +1 du
2 u 1 2 1
2 ln 4a
dx= =2 2 2 2 x a +C . x +a 1
udu . So
2
1
2 ln u 1
u+1 +C +C
10 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration dx 50. As in Exercise 49, let u= 4x+1 . Then 2 x
1 ( u2 1 ) 2 1 = 2 ( u+1 ) ( u 1 )
2 2 = A
+
u+1 B
2 (u+1) 2 +
= 1 2
u 1
4 ( 4x+1 C
+
u 1 2 1
udu
2 ) 2 =8 u du ( u 1) 2 2 D= 1
, u= 1
4 B= 1
. Equating
4 3 1
1
1
=A C . So A= and C=
. Therefore,
2
4
4
1/4
1/4
dx
1/4
1/4
+
+
+
du
= 8
2
2
2
u+1
u 1
x 4x+1
(u+1)
(u 1)
2
2
2
2
=
+2(u+1)
+2(u 1)
du
u+1
u 1
2
2
= 2ln u+1
2ln u 1
+C
u+1
u 1
2
2ln 4x+1 1
= 2ln ( 4x+1 +1 )
4x+1 +1
1
tan
2 , dx= 1
2
sec
d
2 1
2
sec
d
dx
2
=
1
2
tan sec
x 4x +1
2
= ln csc +cot +C 2 = sec
tan = ln orln 1
1
C+
4
4 , so d = csc 2 +C 1=A+ 2
+C
4x+1 1 4x +1 =sec , 2 4x +1
1
+
2x
2x . Now (u 1) coefficients of u gives A+C=0 , and equating coefficients of 1 gives 1=A+B C+D x= 2 D 1=A(u+1)(u 1) +B(u 1) +C(u 1)(u+1) +D(u+1) . u=1 51. Let 2x=tan 2 4x +1
2x d 1
2x +C 11 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration 2 52. Let u=x . Then du=2xdx
dx
xdx
1
=
=
4
2 4
2
x x +1
x x +1 ( ) ( du ) 1
2
=
ln x
2 ( ) ( u u +1 Or: Write I=
x 4 ) 1
2 1
u u
2 u +1 ( ) ( ) 1 2
2
53. x sinh (mx)dx= x cosh (mx)
xcosh (mx)dx
m
m 1 2
x cosh (mx)
m 2 ( 1
+C= ln
4 x ) 4 4 +C x +1 2 dv=sinh (mx)dx
1
du=2xdx v= m cosh (mx)
u=x , dV =cosh (mx)dx
1
V = sinh (mx)
dU =dx
m U =x, 1
1
xsinh (mx)
sinh (mx)dx
m
m 2 ) 1
2
ln u +1 +C
4 and let u=x . 2 = ( 1
ln u
2 4 ( x4+1) 1 2
2
= x cosh (mx)
m
m du= 1
1
4
4
4
ln x +1 +C=
ln x ln x +1
4
4 3 x dx 2 = xsinh (mx)+ m 2
m 3 cosh (mx)+C 54.
1 1 2
( x+sin x ) dx = ( x +2xsin x+sin x ) dx= 3 x +2(sin x xcos x)+ 2 (x sin xcos x)+C
2 = 2 3 1 3 1
1
x + x+2sin x
sin xcos x 2xcos x+C
3
2
2
2 55. Let u= x+1 . Then x=u 1
2udu
dx
=
=
2
x+4+4 x+1
u +3+4u 1
3
+
u+1 u+3 du = 3ln u+3 ln u+1 +C=3ln
2 ( / 2 ( ) 2 x+1 +3) ln
2 ( x+1 +1 ) +C 2 56. Let t= x 1 . Then dt= x
x 1 dx , x 1=t , x= t +1 , so
xln x
1
2
2
2
I=
dx= ln t +1 dt=
ln t +1 dt . Now use parts with u=ln t +1 , dv=dt :
2
2
x 1 ( ) ( ) 12 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration 1
2
I = tln t +1
2 ( t ) 2 dt= 2 t +1 1
2
tln t +1
2 ( ) ( dt 2 t +1 1
2
1
2
tln t +1 t+tan t+C= x 1 ln x
2 = 1 1 ) 2 1 x 1 +tan 2 x 1 +C ( Another method: First integrate by parts with u=ln x , dv= x ( x=sec 2 or u= x 1 ). / 2 x 1 ) dx and then use substitution 3 3 57. Let u= x+c . Then x=u c
x 3 ( u3 c ) u 3u2 du=3 ( u6 cu3) du= 3 u7
7 x+c dx = 3 4
cu +C
4 3
7/3 3
4/3
(x+c)
c(x+c) +C
7
4 = 2 58. Integrate by parts with u=ln (1+x) , dv=x dx du=dx/(1+x) , v= 1 3
x :
3 3 1 3
x dx
1 3
1
1
2
x ln (1+x)dx =
x ln (1+x)
= x ln (1+x)
x x+1
3
3(1+x) 3
3
x+1
1 3
1 3 1 2 1
1
=
x ln (1+x)
x+ x
x+ ln (1+x)+C
3
9
6
3
3
2 dx x 59. Let u=e . Then x=ln u , dx=du/u
du
du/u
dx
=
=
=
3
2
3x x
u u
(u 1)u (u+1)
e e
1 1
=
+ ln
u 2 u 1
u+1 3 3 1/2
u 1 1
+C=e + ln
2
x 1
u 2 1/2
u+1 du x e 1 +C x e +1 2 60. Let u= x . Then x=u , dx=3u du
dx
3 2 3u du = 3 x+ x u +u
5 = 3
2 2udu
2 u +1 = 3
3
2
2/3
ln u +1 +C= ln x +1 +C .
2
2 ( ) ( ) 4 61. Let u=x . Then du=5x dx
13 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration 4 x dx = 10 x +16 1
du
5
2 = u +16 1 1
tan
5 4 1
1
u +C=
tan
4
20 1 1 5
x +C .
4 1 62. Let u=x+1 . Then du=dx
x 3 3 10 dx = (u 1) (x+1) u ( u 7 3u 8+3u 9 u 10) du du= 10 1 6 3 7 3 8 1 9
u + u
u + u +C
6
7
8
9
1
1
9
3 3
2 3
= (x+1)
(x+1) + (x+1)
(x+1)+
6
7
8
9
= 63. Let y= x so that dy= 1 dx 2 x y +C dx=2 x dy=2ydy . Then
2 2 y y u=2y , dx = ye (2ydy)= 2y e dy
2 y y dv=e dy, du=4ydy xe x v=e
y 2 y =2y e U=4y, 4ye dy ( 4yey 2 y dV=e dy, dU=4dy =2y e y V=e ) 2 y y y y 4e dy =2y e 4ye +4e +C
y =2(y 2y+2)e +C=2 ( x 2 x +2 ) e x +C 64. Let u=tan x . Then
/3 ln (tan x)
sin xcos x =
dx
/4 /3 ln (tan x)
2
sec xdx=
tan x
/4 = 1
2
(ln u)
2 3
1 = 3
1 ln u
du
u 2 1
1
( ln 3 ) = 8 (ln 3)2
2 65.
dx
=
x+1 + x
= 1
x+1 + x x+1 x x+1 x dx= ( x+1 x ) dx 2
3/2 3/2
(x+1) x
+C
3
14 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration 3 2 u +1 66. 3 u u 2 du= u +1 1+ (u 1)u 2
u 1 du=u+ 2 1
u 1
u du=u+2ln u 1 ln u + 2 1
+C . Thus,
u 3
3 u +1
2 3 u u 2 du = u+2ln (u 1) ln u+ = 1+3ln 2 ln 3 3 arctan t dt = t 1 tan 1 = 3+2ln 2 ln 3+ 2 1 utan 1
2
ln (1+u )
2 u 1 3 tan =2 3 3 2+2ln 1 ln 2+ 1
2 ) u (2du)=2 =2 1
3 1 5
8
= +ln
6 6
3 67. Let u= t . Then du=dt/ ( 2 t
3 3 1
u 1 1
ln 4
2 3 ln 2 4 tan 1 1 1
ln 2
2 3
1 1
ln 2
2
2
=
3
3 [ Example 5 in Section 8.1] 1
2 ln 2 x 68. Let u=e . Then x=ln u , dx=du/u
dx
du/u
du
2/3
1/3
=
du
= 1+2u 1/u =
x
x
2
2u 1 u+1
1+2e e
2u +u 1
1
1
1
x
x
=
ln 2u 1
ln u+1 +C= ln 2e 1 / e +1
3
3
3 ( )( ) +C x 69. Let u=e . Then x=ln u , dx=du/u
2x e 1+e 2 dx =
x u du
u
=
du=
1+u u
1+u 1 ( 1
1+u du ) = u ln 1+u +C=ex ln 1+ex +C
2 70. Use parts with u=ln (x+1) , dv=dx/x :
ln (x+1)
dx
1
1
1
dx = 1 ln (x+1)+
=
ln (x+1)+
dx
2
x
x(x+1)
x
x x+1
x
1
1
=
ln (x+1)+ln x ln (x+1)+C= 1+
ln (x+1)+ln x +C
x
x
15 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration x 71. x = = Ax+B + Cx+D ( x2+3) ( x2+1) x2+3 x2+1
x = ( Ax+B ) ( x2+1 ) + (Cx+D ) ( x2+3) = ( Ax3+Bx2+Ax+B) + ( Cx3+Dx2+3Cx+3D)
4 2 x +4x +3 = ( A+C ) x3+ ( B+D ) x2+ ( A+3C ) x+ ( B+3D )
A+C=0 , B+D=0 , A+3C=1 , B+3D=0
x
4 1
x
2 dx = 2 x +4x +3 2 1
x
2 + x +3 A= 1
1
, C= , B=0 , D=0.Thus,
2
2 dx 2 x +1
2 1
1
1
2
2
ln x +3 + ln x +1 +C or
ln
4
4
4 ( = ) 6 6 ( x +1 ) 2 +C x +3 5 72. Let u= t . Then t=u , dt=6u du
3 t dt = 3 5 u 6u du
2 1+ t =6 u 8 2 6 du=6 4 2 u u +u 1+ 1
2 du u +1
u +1
1+u
1 7 1 5 1 3
1
= 6
u
u + u u+tan u +C
7
5
3
1 7/6 1 5/6 1 1/2 1/6
1 1/6
= 6
t
t + t t +tan t
+C
7
5
3
1
A Bx+C
2
2
73.
=
+ 2
1=A x +4 + ( Bx+C ) ( x 2 ) = ( A+B ) x + (C 2B ) x+ ( 4A 2C ) . So
2
( x 2 ) x +4 x 2 x +4
1
1
1
0=A+B=C 2B , 1=4A 2C . Setting x=2 gives A=
B=
and C=
. So
8
8
4
1
1
1
1
x
8
8
4
1 dx
1 2xdx 1
dx
dx =
2
+
dx=
2
2
2
( x 2 ) x +4
x 2
8 x 2 16
4
x +4
x +4
x +4
1
1
1
2
1
=
ln x 2
ln x +4
tan (x/2)+C
8
16
8 ( ( ) ( ) ) ( ) x 74. Let u=e . Then x=ln u , dx=du/u
x dx
x e e x = e dx
2x e 1 = u
2 u 1 du
=
u du
2 u 1 1
= ln
2 u 1
u+1 1
+C= ln
2 x e 1
x +C . e +1
16 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration 75.
1
1
cos (2x 3x) cos (2x+3x) dx=
( sin xcos x sin xcos 5x ) dx
2
2
1
1 1
=
sin 2xdx
sin (x+5x)+sin (x 5x) dx
4
2 2
1
1
1
1
1
=
cos 2x
(sin 6x sin 4x)dx=
cos 2x+
cos 6x
cos 4x+C
8
4
8
24
16 sin xsin 2xsin 3xdx = sin x 76. ( x2 bx) sin 2xdx = 1 ( x2 bx) cos 2x+ 1
2
2 (2x b)cos 2xdx = = 77. Let u=x
x
1+x 3 3/2 dx= 1 2
1
x bx cos 2x+
2
2 1 2
1
1
x bx cos 2x+ (2x b)sin 2x+ cos 2x+C
2
4
4 ( ) ( 2 1+u 2 du= sin 2xdx ) 3 so that u =x and du=
2
3 1
(2x b)sin 2x
2 3 1/2
x dx
2 x dx= 2
du . Then
3 2
2
1
1 3/2
tan u+C= tan (x )+C .
3
3 78.
sec xcos 2x
sec xcos 2x 2cos x
2cos 2x
dx =
dx=
dx
sin x+sec x
sin x+sec x 2cos x
2sin xcos x+2
2cos 2x
1
u=sin 2x+2,
=
dx=
du
du=2cos 2xdx
sin 2x+2
u
=ln u +C=ln sin 2x+2 +C=ln (sin 2x+2)+C
2 1
3
sin x . Then
3
1
1
1
3
3
3
2
xsin x
sin xdx= xsin x
(1 cos x)sin xdx
3
3
3
1
3
2
y=cos x,
xsin x+
(1 y )dy
dy= sin xdx
3 79. Let u=x , dv=sin xcos xdx
1
2
xsin xcos xdx = 3
1
=
3 du=dx , v= 17 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.5 Strategy for Integration = 1
1
1 3
1
1
1
3
3
3
xsin x+ y
y +C= xsin x+ cos x
cos x+C
3
3
9
3
3
9 80.
sin xcos x
4 4 sin xcos x dx = sin x+cos x 2 2 2 2 sin xcos x dx= 2 2 (sin x) +(cos x) = 2 2 u +(1 u)
1 dx (sin x) +(1 sin x)
2 1
du
2 1 = 2 2 u=sin x,
du=2sin xcos xdx
1 du= du
(4u 4u+1)+1
1
1
y=2u 1,
=
du=
dy
2
2
dy=2du
2
(2u 1) +1
y +1
1
1
1
1
1
1
2
= tan y+C= tan (2u 1)+C= tan (2sin x 1)+C
2
2
2
2 2 4u 4u+2
1 Another solution:
4 sin xcos x
4 dx = 4 sin x+cos x 2 (sin xcos x)/cos x
4 4 4 tan xsec x dx= 4 (sin x+cos x)/cos x
= 1
2 u +1
= dx tan x+1
2 1
du
2 u=tan x,
2 du=2tan x sec xdx 1
1
1
2
1
tan u+C= tan (tan x)+C
2
2 2 x 81. The function y=2xe does have an elementary antiderivative, so we’ll use this fact to help
evaluate the integral.
2 2 2 x 2 2 x 2 x x 2 x (2x +1)e dx = 2x e dx+ e dx= x(2xe )dx+ e dx
2 =xe x 2 x 2 x e dx+ e dx 2 u=x, x dv=2xe dx,
2 du=dx v=e 2 x =xe +C x 18 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems 1. We could make the substitution u = 2 x to obtain the radical 7 u 2 and then use Formula 33
7
.
with a= 7 . Alternatively, we will factor 2 out of the radical and use a=
2
7 2
x
2 2 7 2x dx = 2 2 1
x dx= 2 2 x x 1
x = 2 7 2x 2 sin 7 2
x sin
2 2
x
7 1 1 x +C
7
2 +C 2.
x
dx=3
3+( 2)x 3x
dx =3
3 2x
= 2
2 ( 2x 2 3) 3+( 2)x +C 3( 2) 1
( 2x 6) 3 2x +C= (x+3) 3 2x +C
2 3. Let u = x du= dx , so
1
1
3
3
sec u du=
sec ( x)dx =
= 1
1
sec u tan u + ln sec u +tan u
2
2 1
1
sec xtan x+
ln sec x+tan x +C
2
2
2 2 4. e sin 3 d = e 2 2 (2sin 3 3cos 3 )+C= 2 +3
1 5. 2xcos
0 2 1 +C xdx=2 x
2x 1
1
cos x
4 2 1 x
4 2 2
e sin 3
13 3 2
e cos 3 +C
13 1 =2 0 1
0 0
4 1
0
4 2 =2 8 = 4 6. 1 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems 6
3 1
2 2 x dx 2 1 = 4x 7 1
du
2 1 2 2
u u 7
2 4 6 2 6 =2 du 4 u 2 29
42 =2 u 7
7u 2 u 7 =2
3
28 29
21 = [u=2x, du=2dx] 4 3
14 7. By Formula 99 with a= 3 and b=4 ,
e 3x cos 4xdx= e 3x 3x
2 2 ( 3) +4 e
( 3cos 4x+4sin 4x ) +C=
( 3cos 4x+4sin 4x ) +C .
25 8. Let u =x/2 , so dx=2du , and we use Formula 72:
3
3
csc (x/2)dx = 2 csc u du= csc u cot u +ln csc u cot u +C
= csc (x/2)cot (x/2)+ln csc (x/2) cot (x/2) +C 9. Let u =2x and a=3 . Then du=2dx and
1
du
dx
du
2
=
=2
= 2
2
2
2
2
2
2
u
2 2
x 4x +9
a +u
u
u +a
4
2 = 2 4x +9
+C=
9 2x 2 a +u
2 2 +C au 2 4x +9
+C
9x 10. Let u = 2 y and a= 3 . Then du= 2 dy and 2 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems 2 2 2y 3
2 2 du
= 2
2 u a
1 2
u
2 dy = y 2 u 2 u 2 a du 2 2 u a
2 2
+ln u + u a
u = 2 +C 2 2y 3 = 2 2y 2 2 y+ 2y 3 +ln +C 2 2y 3
+ 2 ln
y = 2 2 y+ 2y 3 +C 11.
0 2 1 2
t e
1 t t e dt = 1 =e+2 0 t 2
1 1 0 t te dt=e+2 1 0 1 t te dt=e+2 1 ( 1) 2 ( t 1) e t 0
1 0 e +0 =e 2 12. Let u =3x . Then du=3dx , so
1
1
2
2
2
x cos 3xdx = 27 u cos u du= 27 u sin u 2 u sin u du
1 2
2
= x sin 3x
(sin 3x 3xcos 3x)+C
3
27
1
2
=
9x 2 sin 3x+6xcos 3x +C
27
1
1
6
2
2
2
Thus, x cos 3xdx=
9x 2 sin 3x+6xcos 3x =
=
.
( 0+6 ( 1) ) ( 0+0 ) =
0 27
0
27
27
9 ( ( ) ) ( ) 13.
3 tan (1/z)
2 z dz u=1/z,
2 du= dz/z =
= 3 tan u du=
1
2
tan
2 1
z 1
2
tan u ln cos u +C
2
ln cos 1
z +C 2 14. Let u = x . Then u =x and 2u du=dx , so
3 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems sin 1 1 x dx =2 usin udu=
= 2x 1
sin
2 x 2u 2 1 1 sin u+
2
x(1 x)
x+
+C
2 1 u 1 u
2 2 +C x 15. Let u =e . Then du=e dx , so
x ( x) e sech e dx= sechu du 107
= tan 1 sinh u +C=tan 1 ( x) sinh e +C 2 16. Let u =x , so that du=2xdx . Then
1
1 cos (1 3)u 1 cos (1+3)u
2
2
+C
xsin x cos 3x dx = 2 sin u cos 3u du= 2
2(1 3)
2
2(1+3)
1
1
1
1
2
2
= cos 2u
cos 4u +C= cos (2x )
cos (4x )+C
8
16
8
16 ( ) ( ) 2 2 2 2 2 17. Let z=6+4y 4y =6 (4y 4y+1)+1=7 (2y 1) , u =2y 1 , and a= 7 . Then z=a u , du=2dy , and y 1
2
2 1
(u+1) a u
du
2
2
6+4y 4y dy
1
1
2
2
2
2
=
u a u du+
a u du
4
4
1
1
2
2
2
2
=
a u du
( 2u) a u du
4
8
2 = y z dy= u
=
8
2y 1
8
2y 1
=
8
= 2 a
a u + sin
8
2 2 1 7
sin
8
2 7
6+4y 4y + sin
8
2 6+4y 4y + u
a
1 1 1
8
2y 1
7
2y 1
7 2 w dw 2 w=a u ,
dw= 2udu 1 2 3/2
w +C
8 3
1
2 3/2
(6+4y 4y ) +C.
12 This can be rewritten as 4 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems 1
1
7
2
1 2y 1
(2y 1)
(6+4y 4y ) + sin
+C
8
12
8
7
1 2 1
5
2y 1
2 7
1
=
y
y
6+4y 4y + sin
+C
3
12
8
8
7
1
2y 1
2
2 7
1
=
(8y 2y 15) 6+4y 4y + sin
+C
24
8
7
2 6+4y 4y 2 18. Let u =x . Then du=2xdx , so by Formula 48,
5 2 x dx 2
1
u
1 1
=
du=
( u + 2 ) 4 2 ( u + 2 ) +4ln u + 2
2
2 u+ 2
2 2
x+ 2
2
1
2
2
2
=
x+ 2
4 2 x + 2 +4ln x + 2 +C
4
1 4 1 2
2
x
x +ln x + 2 +K
=
4
2 ( ) ( ) ( ( +C ) ) 2 Or: Let u =x + 2 .
19. Let u =sin x . Then du=cos xdx , so
2 2 sin xcos xln (sin x)dx = u ln udu= u 2+1
2 (2+1)ln u 1 +C= (2+1)
= 1 3
u (3ln u 1)+C
9 1
3
sin x 3ln (sin x) 1 +C
9 x 20. Let u =e . Then x=ln u , dx=du/u , so
dx
du/u
du
1
=
+2ln
= u (1+2u ) =
x
x
2
u
e 1+2e
u (1+2u ) ( 1+2u
u ) = x ( x +C ) e +2ln e +2 +C x x 21. Let u =e and a= 3 . Then du=e dx and
e x
2x 3 e dx= du
2 a u 1
=
ln
2 2a u +a
u a +C= 1
2 3 x ln e+ 3
e x +C . 3 22. Let
5 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems 2 u =x and a=2 . Then du=2xdx and
2 3 x 0 2 4 4x x dx =
=
=
= 1
2 2 2 x 0 2u
2u
u 2 2 2 2 1
2 22 2 2 x (x ) 2xdx=
2 au 3a
12 2u 12
12 u 6
6
1 2 4u u + 0 3 2 8
cos
4 2 u 2au u du a
2au u + cos
4
2 4u u +2cos = (0+2cos 1)=2 4 1 2 u
2 1 0
4 2 u
2 1 4 a u
a
0
4
0 2 0=2 23.
1
3
1
3
1
1
3
3
3
5
tan xsec x+
sec xdx
sec xdx = 4 tan xsec x+ 4 sec xdx= 4 tan xsec x+ 4
2
2
1
3
3
3
= tan xsec x+ tan xsec x+ ln sec x+tan x +C
4
8
8
24. Let u =2x . Then du=2dx , so
1
1
1
5
6
5
4
6
sin u cos u +
sin u du
sin 2xdx = 2 sin u du= 2
6
6
1
5
1
3
5
3
2
=
sin u cos u +
sin u cos u +
sin u du
12
12
4
4
1
5
5
1
1
5
3
=
sin u cos u
sin u cos u +
u
sin 2u
12
48
16
2
4
1
5
5
5
5
3
=
sin 2xcos 2x
sin 2xcos 2x
sin 4x+
x+C
12
48
64
16
25. Let u =ln x and a=2 . Then du= +C dx
and
x
2 2
u
2
2
2
2 a
2
2
4+(ln x)
+C
dx = a +u du= 2 a +u + 2 ln u + a +u
x
1
2
2
= (ln x) 4+(ln x) +2ln ln x+ 4+(ln x) +C
2 ( ) 26.
6 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems 4 x 4 x 3 x 4 ( x x e dx = x e +4 x e dx= x e +4 =
=
1 4 x 2 x ) ( x4+4x3) e x+12 ( x2e x+2 xe x dx)
( x4+4x3+12x2) e x+24 ( x 1)e x +C
( x4+4x3+12x2+24x+24) e x+C = ( x4+4x3+12x2+24x+24) e x x x e dx =
0 So 3 x e +3 x e dx 1 1
0 0 = (1+4+12+24+24)e +24e =24 65e 1 . x 27. Let u =e . Then x=ln u , dx=du/u , so
2 2x e 1 dx= u 1
2
du= u 1 cos
u 1 2x ( 1/u ) +C= e 1 cos 1 ( e x) +C . 28. Let u= t 3 and assume that
0 . Then du= dt and
1
1 3/
(1/ )u
( u+3 ) /
t
e
sin udu= e
e
sin udu
e sin ( t 3)dt =
= 1 e e 3/ ( 1/ ) u
2 1
2 sin u cos u +C (1/ ) +1
= 1 e 3/ 2 (1/ )u e 2 1+
1 =
= 1+
1
1+ 2 e ( u+3 ) / 1 (sin u cos u)+C t 2 sin u cos u +C e sin ( t 3) cos ( t 3) +C 29.
4 4 x dx
10 x =
2 x dx
52 (x ) 2
= = 1
5 du
u 2 5 4 [ u =x , du=5x dx ]
2 1
1
2
5
10
ln u + u 2 +C= ln x + x 2 +C
5
5
7 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems and a=3 . Then du=sec 30. Let u =tan
sec 2 tan 9 tan 2 = d 2 u 2 2 a u
1
=
tan
2 2 2 d and u
du=
2
9 tan 2 a
a u + sin
2
2 2 + 2 9
sin
2 u
a 1 tan
3 1 +C +C 31. Using cylindrical shells, we get
2 V =2 2 2 x x 4 x dx=2 2 x 0 4 x dx=2 0
1 =2 x
2
2 16
(2x 4) 4 x +
sin
8
8 2 1 (0+2sin 1) (0+2sin 0 =2 2 x
2 2
0 2 =2 2 1 32. Using disks, we get
Volume = /4
0 1
1+
3
4 = d
33. (a)
du
= = 0 0 3 2 3 ( a+bu ) (b) Let t=a+bu
2 u du
2 = (a+bu) = b dt=bdu . Note that u =
1
b = 2 3 1
b 3 2 (t a)
t 2 dt= 1
b 3 = b
bu 3 2 b+ 3 (a+bu) 2 (a+bu)
u = ba /4
0 2ab
(a+bu) 2
2 (a+bu) t a
1
and du= dt .
b
b 2 2 t 2at+a
t 2 2a a
1
+
2
t
t +C 2 1 3 2 1 1
3
tan x tan x+x
3 2 tan xdx = 2 2 b /4 a
a+bu
2aln a+bu
a+bu 1 b(a+bu) +ba (a+bu)2ab 1
b /4 1
3
tan x
3
2
4 3 4 tan xdx= dt= 2 1
b 3 dt
2 t 2aln t a
t +C 8 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems = b
d
34. (a)
du
u
=
8
= ( 2u 2 u
8
2 a 3 ( 2u ) 2 2 8 a u 1 2 2
=
a u
2 ( 1 2 2
=
a u
2 ( ) 2 a u 2 2 ) 1/2 u
4 ) 1/2 2 2 ( 2u 2u a 2u 2 2 4 2 ) a +u
= u 2 2 (a 2 d 2 4 sin 4 a
=
8
= u
8 sin ( 2u 2 1 1 2 a + u 2 ) 2 2 a u 1 2 2
+
a u
4 ( 2 a u ) ( 2u 2 4 a
a +
4
2 ) 2 . Then
4 acos 2 ( 2 u
a a u
a u
a u
a a u
a 2 4 2 2 2u 1 ) ( 1 2sin 2 ) +C 2 +C 2 a
2 a
a u + sin
8
2 2 d =a sin cos
d
1 4
2
cos 2 )d = a
1 cos 2 d
4
1 4 1
1
) d = a
sin 4
+C
4
2
8
1 4 1
1
cos 2
+C= a
sin cos
4
2
2 u
a ) 2 1 u /a 4 2 8 1/a 2 2
2
2
2
2
a u du = a sin
a 1 sin
1
4 1
=a
(1+cos 2 ) (1
2
2
1 4
1
= a
1
(1+cos 4
4
2
1 4 1
1
= a
2sin 2
4
2
8 a
=
8 a ( a2 u 2) =u 2
a u du=acos 4 a
+
8 ) 2 u
2u a
+
2
8 2 u
+C
a 1 ( 2 2 +C u
2 2 1
(4u )+ 2u a
8
8 2 + a u 2 (b) Let u =asin
2 a 4 a
a u + sin
8
2 2 2 ( 2u 2 a2) +
2 2 u
a u u u 2 a
a+bu
2aln a+bu
a+bu 1 2 a 2u 2 2 +C a
1 u
+C
a 35. Maple, Mathematica and Derive all give
9 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems 1
1
2 3/2 5
2 25
1
x 5 x
+ x 5 x + sin
x . Using Formula 31, we get
4
8
8
5
1
1
2
2
2
2 1
2
1
x 5 x dx= x 2x 5
5 x +
5 sin
x +C . But
8
8
5
1
1
1
2 3/2 5
2 1
2
2
2
2
x 5 x
+ x 5 x = x 5 x 5 2 5 x = x 2x 5
5 x , and the sin terms are
4
8
8
8
the same in each expression, so the answers are equivalent.
2 x ( 2 5 x dx= ) ( ( ) ( ) ) ( 2 36. Maple and Mathematica both give x ) ( ) 1
( 1+x3) 4 dx= 15 x15+ 1 x12+ 2 x9+ 2 x6+ 1 x3 , while
3
3
3
3 1
( 1+x3) 4 dx= 15 ( x3+1) 5 . Using the substitution u =1+x3 du=3x2 dx , we get
1
1 5
1
2
3 4
4
( 1+x3) 5+C . We can use the Binomial Theorem or a
x ( 1+x ) dx= u
du =
u +C=
3
15
15
1
1
1
( 1+x3) 5+C= 15 + 1 x3+ 2 x6+ 2 x9+ 1 x12+ 15 x15+C
CAS to expand this expression, and we get
15
3
3
3
3
2 Derive gives x .
1
2
2
3
3
sin xcos x
cos x (although Derive factors
5
15
1
1
1
3
2
the expression), and Mathematica gives sin xcos xdx=
cos x
cos 3x+
cos 5x . We can
8
48
80
1
1
3
5
use a CAS to show that both of these expressions are equal to
cos x+ cos x . Using Formula
3
5
86, we write
1
2
1
2
1
2
3
2
2
3
3
3
2
cos x +C
sin xcos xdx = 5 sin xcos x+ 5 sin xcos xdx= 5 sin xcos x+ 5
3
1
2
2
3
3
=
sin xcos x
cos x+C
5
15
3 2 37. Maple and Derive both give sin xcos xdx= 3 3 1 sin x 2 sin x
38. Maple gives tan xsec dx=
+
, Mathematica gives
5
3
5
15
cos x
cos x
1
2
4
5
tan xsec dx=
sec x( 20sin x+5sin 3x+sin 5x) , and Derive gives
120
2
sin x
sin x
2
4
tan xsec dx=
tan x
+
. All of these expressions can be ‘‘simplified’’ to
3
5
15
15cos x 5cos x
2 ( 2 4 4 1 sin x cos x 2cos x 3
5
15
cos x ) 2 2 using Maple. Using the identity 1+tan x=sec x , we write 10 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems 2 4 2 ( 2 ) 2 ( tan 2x+tan 4x) sec 2xdx . Now we substitute u =tan x
( u 2+u 4) du= 1 u 3+ 1 u 5+C= 1 tan 3x+ 1 tan 5x+C . If
3
5
3
5 tan xsec xdx= tan x 1+tan x sec xdx=
2 du=sec xdx , and the integral becomes
5 ( 3 ) 2 5 we write sin x=sin x 1 cos x and substitute into the numerator of the tan x term, this becomes
3 3 ( 2 ) 3 1 sin x 1 sin x 1 cos x
1 sin x
+
+C=
+
3
5
5
3
5
5
cos x
cos x
cos x
which is the same as Maple’s expression. 1
3 1
5 3 sin x 3 3 1 sin x 2 sin x
+C=
+
+C ,
3
5
3
5
15
cos x
cos x
cos x 1
5/2 1
3/2
(1+2x)
(1+2x) , Mathematica gives
10
6
2 2 1
1
1
3/2
1+2x
x+
x
(1+2x) (3x 1) . The first two expressions can be
, and Derive gives
5
15
15
15
simplified to Derive’s result. If we use Formula 54, we get
1
3/2
3/2
2
(3 2x 2 1)(1+2x) +C=
(6x 2)(1+2x) +C
x 1+2x dx =
2
30
15 ( 2 )
1
3/2
=
(3x 1)(1+2x)
15
39. Maple gives x 1+2x dx= 4 40. Maple and Derive both give sin xdx=
gives 1
3
3
3
sin xcos x
cos xsin x+ x , while Mathematica
4
8
8 1
(12x 8sin 2x+sin 4x) , which can be expanded and simplified to give the other expression.
32 Now
1
3
1
3
1
1
3
2
3
4
x
sin 2x +C
sin xdx = 4 sin xcos x+ 4 sin xdx= 4 sin xcos x+ 4
2
4
1
3
3
3
=
sin xcos x
sin xcos x+ x+C since sin 2x=2sin xcos x
4
8
8
5 41. Maple gives tan xdx= 1
1
1
4
2
2
tan x
tan x+ ln 1+tan x , Mathematica gives
4
2
2 ( ) 1
5
4
tan xdx= [ 1 2cos (2x)]sec x ln (cos x) , and Derive gives
4
1
1
5
4
2
tan xdx= tan x
tan x ln (cos x) . These expressions are equivalent, and none includes
4
2
absolute value bars or a constant of integration. Note that Mathematica’s and Derive’s expressions
suggest that the integral is undefined where cos x<0 , which is not the case.
Using Formula 75,
11 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems 5 tan xdx= 1 tan 5 1 x tan 5 2 1
4
tan x
4 xdx= 3 tan xdx . Using Formula 69, 5 1
1
1
1
3
2
5
4
2
tan xdx= tan x+ln cos x +C , so tan xdx= tan x
tan x ln cos x +C .
2
4
2
1 4
4 2
8
5
2
2
2
2 1 6
2
42. Maple gives x x +1 dx=
x 1+x
x 1+x +
1+x + x 1+x . When we
35
105
105
7
1
2 3/2
4
2
1+x
15x 12x +8 . Mathematica
use the factor command on this expression, it becomes
105
8
4 2 1 4 1 6
2
gives 1+x
x+
x+ x
, which again factors to give the above expression, and
105 105
35
7 ( ) ( ) 4 2 ( ( ) ) ( ( ) ( x ) ( ( ) (
x 1 43. Derive gives I= 2 2 1 ) 2 , xdx=u du , ) ( x ( x= u Derive gives the factored form immediately. If we substitute u = x +1
then the integral
becomes
1 7 2 5 1 3
4
2
2
2
2
u 2u +1 u du= u
u + u +C
u 1 u (u du) =
7
5
3
3/2
2 2
1 2
1
2
2
= x +1
x +1
x +1 +
+C
7
5
3
3/2
2
1
2
2
2
=
x +1
15 x +1 42 x +1 +35 +C
105
3/2
1
2
4
2
=
x +1
15x 12x +8 +C
105
4 1 dx= 2 ) ) ( ) ) 2x 2 1
ln 2 ln ( 2x x 2 1 +2
2ln 2 ) immediately. Neither Maple nor
x x2 Mathematica is able to evaluate I in its given form. However, if we instead write I as 2 (2 ) 1 dx
, both systems give the same answer as Derive (after minor simplification). Our trick works because
x the CAS now recognizes 2 as a promising substitution.
2 44. None of Maple, Mathematica and Derive is able to evaluate (1+ln x) 1+(xln x) dx . However,
2 1+u du , which any CAS can
if we let u =xln x , then du=(1+ln x)dx and the integral is simply
1
1
2
2
evaluate. The antiderivative is ln xln x+ 1+ ( xln x ) + xln x 1+(xln x) +C .
2
2 ( ) 2 45. Maple gives the antiderivative F(x)= x 1
4 2 dx= 1
1
2
2
ln x +x+1 + ln x x+1 . We can
2
2 ( ) ( ) x +x +1
see that at 0 , this antiderivative is 0 . From the graphs, it appears that F has a maximum at x= 1 and
12 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems a minimum at x=1 , and that F has inflection points at x 1.7 , x=0 , and x 1.7 . 46. Maple gives the antiderivative which, after we use the simplify command, becomes
1 x
1
x
xe sin xdx=
e (cos x+xcos x+xsin x) . At x=0 , this antiderivative has the value
, so we use
2
2
1 x
1
F(x)=
e (cos x+xcos x+xsin x)+ to make F(0)=0 .
2
2 From the graphs, it appears that F has a minimum at x
has inflection points where f
4 / changes sign, at x 3.1 and a maximum at x 3.1 , and that F 2.5 , x=0 , x 1.3 and x 4.1 . 6 47. Since f (x)=sin xcos x is everywhere positive, we know that its antiderivative F is increasing.
Maple gives
1
3
1
1
3
3
3
7
7
5
3
f (x)dx=
sin xcos x
sin xcos x+
cos xsin x+
cos xsin x+
cos xsin x+
x
10
80
160
128
256
256
and this expression is 0 at x=0 . F has a minimum at x=0 and a maximum at x= . F has inflection points where f / changes sign, that
13 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.6 Integration Using Tables and Computer Algebra Systems is, at x 0.7 , x= /2 , and x 2.5 .
3 48. From the graph of f (x)= x x
6 , we can see that F has a maximum at x=0 , and minima at x 1. x +1
The antiderivative given by Maple is F(x)= 1
1
2
4 2
ln x +1 + ln x x +1 , and F(0)=0 . Note that f
3
6 ( ) ( ) is odd, and its antiderivative F is even. F has inflection points where f / changes sign, that is, at x 0.5 and x 1.4 . 14 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 1. (a)
L=
2 R=
2 M= x=(b a)/n=(4 0)/2=2
2
i=1
2
i=1
2 2 ( ) x= f ( x ) 2+ f ( x ) 2=2 f (0)+ f (2) =2(0.5+2.5)=6
f ( x ) x= f ( x ) 2+ f ( x ) 2=2 f (2)+ f (4) =2(2.5+3.5)=12 f x i=1 i 1 0 i 1 1 ( ) f x i 2 ( ) 2+ f ( x ) 2=2 f (1)+ f (3) x= f x 1 2 2(1.6+3.2)=9.6 (b)
L is an underestimate, since the area under the small rectangles is less than the area under the curve,
2 and R is an overestimate, since the area under the large rectangles is greater than the area under the
2 curve. It appears that M is an overestimate, though it is fairly close to I . See the solution to Exercise
2 45 for a proof of the fact that if f is concave down on a,b , then the Midpoint Rule is an
overestimate of b
a f (x)dx . 1
2
x
f x +2 f x + f x
=
f (0)+2 f (2)+ f (4) =0.5+2(2.5)+3.5=9 .
2
0
1
2
2
2
This approximation is an underestimate, since the graph is concave down. Thus, T =9<I . See the
(c) T = ( ) ( ) ( ) 2 solution to Exercise 45 for a general proof of this conclusion.
(d) For any n , we will have L <T <I<M <R .
n n n n 2. The diagram shows that L >T >
4 4 2
0 f (x)dx>R , and it appears that M is a bit less than
4 4 fact, for any function that is concave upward, it can be shown that L >T >
n n 2
0 2
0 f (x)dx . In f (x)dx>M >R .
n n (a) Since 0.9540>0.8675>0.8632>0.7811 , it follows that L =0.9540T =0.8675 , M =0.8632 , and
n n n 1 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration R =0.7811 .
n (b) Since M <
n 2 f (x)dx<T , we have 0.8632<
n 0 1 0 1
=
4
4
1
1
(a) T =
f (0)+2 f
+2 f
4 4 2
4
1
1
3
(b) M =
f
+f
+f
4 4
8
8 ( 2) , 3. f (x)=cos x 2
0 f (x)dx<0.8675 . x= 2
4
5
8 3
4
7
8 +2 f
+f + f (1) 0.895759 0.908907 The graph shows that f is concave down on 0,1 . So T is an underestimate and M is an
overestimate. We can conclude that 0.895759< 1 4
2 4 ( ) cos x dx<0.908907 . 0 4. (a) Since f is increasing on 0,1 , L will underestimate I (since the area of the darkest rectangle is
2 less than the area under the curve), and R will overestimate I . Since f is concave upward on 0,1 ,
2 M will underestimate I and T will overestimate I (the area under the straight line segments is
2 2 greater than the area under the curve).
(b) For any n , we will have L <M <I<T <R .
n (c) L =
5 R=
5 M=
5 5
i=1 5 n n n 1
x= [ f (0.0)+ f (0.2)+ f (0.4)+ f (0.6)+ f (0.8)] 0.1187
5 ( )
1
f ( x ) x= [ f (0.2)+ f (0.4)+ f (0.6)+ f (0.8)+ f (1)]
5 5
i=1 i=1 f x i 1 i ( ) f x i 0.2146 1
x= [ f (0.1)+ f (0.3)+ f (0.5)+ f (0.7)+ f (0.9)] 0.1622
5
2 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 1
x
0.1666
5
2
From the graph, it appears that the Midpoint Rule gives the best approximation. (This is in fact the
case, since I 0.16371405 .) T = b a
0
=
=
n
8
8
3
+f
+f
16
16 2 5. f (x)=x sin x ,
(a) M =
8 x= f 8 5
16 + +f 15
16 5.932957 (b)
S =
8 f (0)+4 f 8 3
5.869247 2
8 +2 f 8 3
8 +4 f +2 f 4
8 5
8 +4 f +2 f 6
8 7
8 +4 f Actual:
2 2 x sin xdx =
0 x cos x 0 0 xcos xdx= 2 +2[ ( 1+0 ) ( 1+0 ) ]= = 8 M E = actual S =
8 x sin xdx M 0 2 0 2 x sin xdx S 2 ( 1 ) 0 +2 cos x+xsin x 0 4 5.869604 2 Errors: E = actual M =
S +2 0.063353 8 0.000357 8 b a 1 0 1
=
=
n
6
6
1
1
3
5
7
9
11
(a) M =
f
+f
+f
+f
+f
+f
6 6
12
12
12
12
12
12
1
1
2
3
4
5
(b) S =
f (0)+4 f
+2 f
+4 f
+2 f
+4 f
+ f (1)
6 6 3
6
6
6
6
6
Actual:
x 6. f (x)=e 1
0 e x dx = , x= 1 u
0 0 Errors: E = actual M =
M 6 E = actual S =
S 6 1
0 e x x , u =x , 2udu=dx ] 1 u 0.533979 2 e 2uduu= =2 (u 1)e 0.525100 =2
1
0 dx S e 6 2e
x 1 ( dx M 0 1e 6 ) =2 4e 1 0.528482 0.003382 0.005497 3 + Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 4 2 0 1
=
8
4
1
1
(a) T =
f (0)+2 f
+2 f
8 4 2
4
1
1
3
(b) M =
f
+f
+
8 4
8
8
(c)
1
1
1
S=
f (0)+4 f
+2 f
8 4 3
4
2
7. f (x)= 2 1+x , x= 1
2
+f +4 f + +2 f
13
8
3
4 +f 3
7
+2 f
+ f (2)
2
4
15
2.411453
8 +2 f (1)+4 f 1
0
2
1
2
8. f (x)=sin x , x=
=
4
8
1
1
2
3
(a) T =
f (0)+2 f
+2 f
+2 f
+f
4 8 2
8
8
8
1
1
3
5
7
(b) M =
f
+f
+f
+f
4 8
16
16
16
16
1
1
2
3
(c) S =
f (0)+4 f
+2 f
+4 f
+f
4 8 3
8
8
8 5
4 +2 f 3
2 2.413790 +4 f 7
4 + f (2) 2.4122 ( ) 1
2 0.042743
0.040850 1
2 0.041478 ln x
2 1 1
, x=
=
1+x
10 10
1
(a) T =
f (1)+2 f (1.1)+2 f (1.2)+
+2 f (1.8)+2 f (1.9)+ f (2) 0.146879
10 10 2
1
(b) M =
f (1.05)+ f (1.15)+
+ f (1.85)+ f (1.95) 0.147391
10 10
(c)
S = 1 [ f (1)+4 f (1.1)+2 f (1.2)+4 f (1.3)+2 f (1.4)+4 f (1.5)+2 f (1.6)+4 f (1.7)]
10
10 3
+ 2 f (1.8)+4 f (1.9)+ f (2)]
0.147219
3 0 1
1
10. f (t)=
, t=
=
2 4
6
2
1+t +t
1
1
3
5
(a) T =
f (0)+2 f
+2 f (1)+2 f
+2 f (2)+2 f
+ f (3)
0.895122
6 2 2
2
2
2
1
1
3
5
7
9
11
(b) M =
f
+f
+f
+f
+f
+f
0.895478
6 2
4
4
4
4
4
4
1
1
3
5
(c) S =
f (0)+4 f
+2 f (1)+4 f
+2 f (2)+4 f
+ f (3)
0.898014
6 2 3
2
2
2
11.
9. f (x)= 4 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 1
0
2
1
t /2
=
f (t)=sin (e ) , t=
8
16
1
1
2
(a) T =
f (0)+2 f
+2 f
+
+2 f
8 16 2
16
16
1
1
3
5
(b) M =
f
+f
+f
+
+f
8 16
32
32
32
1
1
2
(c) S =
f (0)+4 f
+2 f
+
+4 f
8 16 3
16
16 7
+f
16
13
+f
32
7
+f
16 1
2
15
32
1
2 0.451948
0.451991
0.451976 4 0 1
=
8
2
1
1
7
(a) T =
f (0)+2 f
+2 f (1)+
+2 f (3)+2 f
+ f (4)
6.042985
8 2 2
2
2
1
1
3
13
15
(b) M =
f
+f
+
+f
+f
6.084778
8 2
4
4
4
4
(c)
1
1
3
5
7
S=
f (0)+4 f
+2 f (1)+4 f
+2 f (2)+4 f
+2 f (3)+4 f
+ f (4)
8 2 3
2
2
2
2
12. f (x)= 1+ x , 1/x 13. f (x)=e
(a) T = , x= 6.061678 2 1 1
=
4
4 x= 1 [ f (1)+2 f (1.25)+2 f (1.5)+2 f (1.75)+ f (2)] 2.031893
4 2
1
(b) M = [ f (1.125)+ f (1.375)+ f (1.625)+ f (1.875)] 2.014207
4 4
1
(c) S =
[ f (1)+4 f (1.25)+2 f (1.5)+4 f (1.75)+ f (2)] 2.020651
4 4 3
4 14. f (x)= x sin x , x= 4 0 1
=
8
2 (a)
T =
8 1
2 2 (b) M =
8 {
1
2 f (0)+2
f f 1
4 +f 1
2 3
2 + f (1)+ f
3
4 +f 5
4 + f (2)+ f +f 7
4 + 5
2
+f 13
4 } 7
2 + f (3)+ f
+f + f (4)
15
4 1.732865 1.787427 (c)
S=
8 1
2 3 15. f (x)= f (0)+4 f
cos x
,
x 1
+2 f (1)+4 f
2
5 1 1
x=
=
8
2 3
2 +2 f (2)+4 f 5
2 +2 f (3)+4 f 7
2 + f (4) 1.772142 5 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration (a) T = 1 8 2 2 1
2 f 3
2 f (1)+2 f +2 f (2)+ 9
2 +2 f (4)+2 f + f (5) 0.495333 (b)
M=
8 5
4 +f 7
4 +f 9
4 3
2 +2 f (2)+4 f 11
4 +f +f 13
4 15
4 +f +f 17
4 +f 9
2 19
4 + f (5) 0.5433 (c)
S = 1
f (1)+4 f
8
2 3
0.526123 ( ) 3 16. f (x)=ln x +2 ,
(a) T = x= 5
2 +2 f (3)+4 f 7
2 +2 f (4)+4 f 6 4 1
=
10 5 1 f (4)+2 f (4.2)+2 f (4.4)+
+2 f (5.6)+2 f (5.8)+ f (6)
5 2
1
(b) M =
f (4.1)+ f (4.3)+
+ f (5.7)+ f (5.9) 9.650912
10 5
(c)
S = 1 [ f (4)+4 f (4.2)+2 f (4.4)+4 f (4.6)+2 f (4.8)+4 f (5)]
10
5 3
+ 2 f (5.2)+4 f (5.4)+2 f (5.6)+4 f (5.8)+ f (6)]
9.650526
10 17. f (y)= 1
5 , y= 9.649753 3 0 1
=
6
2 1+y
1
(a) T =
f (0)+2 f
6 2 2
1
1
(b) M =
f
+f
6 2
4
1
(c) S =
f (0)+4 f
6 2 3 1
2 +2 f
3
4 1
2 +f
+2 f 2
+2 f
2
5
+f
4
2
+4 f
2 3
4
5
+2 f
+2 f
+ f (3)
1.064275
2
2
2
7
9
11
+f
+f
1.067416
4
4
4
3
4
5
+2 f
+4 f
+ f (3)
1.074915
2
2
2 x 18.
(a)
(b)
(c) e
4 2 1
f (x)=
, x=
=
x
10 5
1
T =
{ f (2)+2[ f (2.2)+ f (2.4)+ f (2.6)+ + f (3.8)]+ f (4)} 14.704592
10 5 2
1
M = [ f (2.1)+ f (2.3)+ f (2.5)+ f (2.7)+
+ f (3.7)+ f (3.9)] 14.662669
10 5
1
S =
[ f (2)+4 f (2.2)+2 f (2.4)+4 f (2.6)+
+2 f (3.6)+4 f (3.8)+ f (4)] 14.676696
10 5 3
6 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 2 19. f (x)=e
(a) T =
10 x , 1
5 2 2 0 1
=
10 5 x= { f (0)+2[ f (0.2)+ f (0.4)+ 1
M = [ f (0.1)+ f (0.3)+ f (0.5)+
10 5
2 x (b) f (x)=e
f / / / , f (x)= 2xe (x)=0 x x=0 or x= ( / / 2 3
2 / /
3 E 2 2 T 2 T n ( 2 ) 2 x / / (0) =2 , f / / (2) 0.2564 and 10 2 5 /2 E T 0.006 .
3 2(2 0)
2 10 3 2
5
n 10 n 365.1 ...
4 5 12n , again take K=2 in Theorem 3 to get E M M 10 5 3 2
5
n 10
2 n 259 . Take n=259 for M .
n 1 20. (a) T = 8 2
1
f
16 8 1
8 M
3 12n n 366 . Take n=366 for T . For E 8 . 3
. f
2 E K ( b a) (c) Take K=2 in Theorem 3. E M= ) / / / 0.8925 . Thus, taking K=2 , a=0 , b=2 , and n=10 in Theorem 3, we get 1
/ ( 12 10 ) = 75 =0.013 , and n 258.2 2 x (x)=4x 3 2x e .
, f (x)= 4x 2 e , f
3
/ /
. So to find the maximum value of f (x) on 0,2 , we need
2 only consider its values at x=0 , x=2 , and x=
f 0.881839 + f (1.7)+ f (1.9)] 0.882202 2 / + f (1.8)]+ f (2)} { f (0)+2 +f 3
16 1
8 f
+f 2
8 +f
5
16 + + +f +f 15
16 7
8 } + f (1) 0.902333 =0.905620 ( 2) , f /(x)= 2xsin ( x2) , f / /(x)= 2sin ( x2) 4x2cos ( x2) . For 0 x 1 , sin and
/ /
2
2
2
2
cos are positive, so f (x) =2sin ( x ) +4x cos ( x ) 2 1+4 1 1=6 since sin ( x ) 1 and
2
2
cos ( x ) 1 for all x , and x 1 for 0 x 1 . So for n=8 , we take K=6 , a=0 , and b=1 in Theorem
1
1
3
2
3, to get E
6 1 / ( 12 8 ) =
=0.0078125 and E
=0.00390625 .
T
M
128
256
3
2
2
5
2
5
(c) Using K=6 as in part (b), we have E
6 1 / ( 12n ) =1/ ( 2n ) 10
2n 10
T
1
3
2
5
5
n
10 or n 224 . To guarantee that E
0.00001 , we need 6 1 / ( 24n ) 10
M
2
(b) f (x)=cos x 2 4n 5 10 n 1
5
10 or n 159 .
4
7 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 1
{ f (0)+2[ f (0.1)+ f (0.2)+
10 2 21. (a) T =
10 S =
10 1
[ f (0)+4 f (0.1)+2 f (0.2)+4 f (0.3)+
10 3 Since I= 1 x e dx= e 0
x (b) f (x)=e 3 E T + f (0.9)]+ f (1)} e(1) have E / / f =e 1 1.71828183 , E =I T
T ( x ) =e x 10 0.00143166 and E =I S
S 0.00000095 . 10 e for 0 x 1 . Taking K=e , a=0 , b=1 , and n=10 in Theorem 3, we get
x
( 4)
0.002265>0.00143166 . f (x)=e <e for 0 x 1 . Using Theorem 4, we / ( 12 10 )
e(1) / ( 180 10 )
2 5 S +4 f (0.9)+ f (1)] 1.71828278 x 1 0 1.71971349 4 0.0000015>0.00000095 . We see that the actual errors are about two thirds the size of the error estimates.
3 K(b a) (c) From part (b), we take K=e to get E T 2 0.00001 ( 3) e 1
12(0.00001) 2 n 12n n 150.5 . 3 Take n=151 for T . Now E
n K(b a)
M 5 E K(b a)
S 4 0.00001 180n
for Simpson’s Rule). n 2 0.00001 4 76e/[180(0.00001)] 4 ( 5) e 1
180(0.00001) n 6.23 . Take n=8 for S (since n has to be even
n 5 S cos x (b) A CAS gives M 10 / ( 180n )
4 twice, and find that f From the graph, we see that the maximum value of f
. Since f 76e(1) 0.00001 n 18.4 . Take n=20 (since n must be even). 23. (a) Using a CAS, we differentiate f (x)=e interval 0,2 n 24n 22. From Example 7(b), we take K=76e to get E
n n 106.4 . Take n=107 for M . Finally, / / / / / / (x)=e cos x ( sin 2x cos x) . (x) occurs at the endpoints of the (0)= e , we can use K=e or K=2.8 . 7.954926518 . (In Maple, use student[middlesum]. ) (c) Using Theorem 3 for the Midpoint Rule, with K=e , we get
8 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration E 3 e(2 0) M 0.280945995 . With K=2.8 , we get E 2 24 10
(d) A CAS gives I 7.954926521 . 2.8(2
M 3 0)
2 =0.289391916 . 24 10 9 (e) The actual error is only about 3 10 , much less than the estimate in part (c).
(f) We use the CAS to differentiate twice more, and then graph
cos x
4
2
2
( 4)
f (x)=e
sin x 6sin xcos x+3 7sin x+cos x . ( ) (4) From the graph, we see that the maximum value of f (x) occurs
( 4)
at the endpoints of the interval 0,2 . Since f (0)=4e , we can use
K=4e or K=10.9 . (g) A CAS gives S 7.953789422 . (In Maple, use student[simpson ]. .) 10 (h) Using Theorem 4 with K=4e , we get E E 10.9(2
S 4e(2
S 5 0)
4 0.059153618 . With K=10.9 , we get 180 10 5 0) 0.059299814 . 4 180 10
(i) The actual error is about 7.954926521 7.953789422 0.00114 . This is quite a bit smaller than the
estimate in part (h), though the difference is not nearly as great as it was in the case of the Midpoint
Rule.
5 5 (j) To ensure that E
n 4 S 5 , 915 , 362 0.0001 , we use Theorem 4: E 4e(2 )
S 180 n 4 0.0001 n 49.3 . So we must take n 50 to ensure that I S n 4e(2 )
180 0.0001 n 4 0.0001 . ( K=10.9 leads to the same value of n .)
3 24. (a) Using the CAS, we differentiate f (x)= 4 x twice,
and find that f / / (x)= 9x ( 4 4 4 x 3x 3 3/2 ) 3 1/2 (4 x ) . 9 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration From the graph, we see that f (b) A CAS gives M 10 / / (x) <2.2 on 1,1 . 3.995804152 . (In Maple, use student[middlesum] .) (c) Using Theorem 3 for the Midpoint Rule, with K=2.2 , we get E 2.2 1 ( 1 )
M 2 3 0.00733 . 24 10
(d) A CAS gives I 3.995487677 .
(e) The actual error is about 0.0003165 , much less than the estimate in part (c).
(f) We use the CAS to differentiate twice more, and then graph f
(4) From the graph, we see that f (x) <18.1 on (g) A CAS gives S ( 4) 2 9 x
(x)=
16 ( x6 224x3 1280)
( 4 x3) 7/2 . 1,1 . 3.995449790 . (In Maple, use student[simpson] .) 10 (h) Using Theorem 4 with K=18.1 , we get E 18.1 1 ( 1 )
S 5 4 0.000322 . 180 10
(i) The actual error is about 3.995487677 3.995449790 0.0000379 . This is quite a bit smaller than
the estimate in part (h).
5 5 (j) To ensure that E
n 4 32 , 178
1 3 25. I= x dx=
0 0.0001 ,we use Theorem 4: E S 18.1(2)
S 180 n 4 n 13.4 . So we must take n 14 to ensure that I S
1 4
x
4 1 0.0001 n 18.1(2)
180 0.0001 n 4 0.0001 . 3 =0.25 . f (x)=x . 0 10 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration n=4 :
1
L=
4 4
1
R=
4 4
T = 3 1
4 3 0+ + 3 1
4 3 2
4 + 3 4 2
1
M=
4 4
4 =0.140625 3 3
4
3
1
2
3
0 +2
+2
4
4
3
3
1
3
5
+
+
8
8
8 1 2
4 + 3 3
4 + 3 +1 =0.390625 3 3
4
7
8 +2
3 + 3 3 +1 =0.265625 , 3 =0.2421875 , 1
1
0.140625=0.109375 , E =
0.390625= 0.140625 ,
L
4 4
R 4
1
1
E =
0.265625= 0.015625 , E =
0.2421875=0.0078125
T 4
M 4
E =I L = n=8 :
1
L=
8 8
1
R=
8 8
T = f (0)+ f 1
8 +f 1
8 +f 2
8 f
1 { f +
1
+f
8
3
+
16 + +f +f 7
8
2
8 + 7
8 0.191406 + f (1)
+f 0.316406
7
8 } + f (1)
0.253906
8 2
1
1
13
15
M=
f
+f
+f
+f
=0.248047
8 8
16
16
16
1
1
E
0.191406 0.058594 , E
0.316406 0.066406 ,
L 4
R 4
1
1
E
0.253906 0.003906 , E
0.248047 0.001953 .
T
M 4
4
8 f (0)+2 2
8 n=16 :
1
1
2
15
L =
f (0)+ f
+f
+
+f
0.219727
16 16
16
16
16
1
1
2
15
R =
f
+f
+
+f
+ f (1)
0.282227
16 16
16
16
16
1
1
2
15
T =
f (0)+2 f
+f
+
+f
+ f (1)
16 16 2
16
16
16
1
1
3
31
M =
f
+f
+
+f
0.249512
16 16
32
32
32
1
1
E
0.219727 0.030273 , E
0.282227 0.032227 ,
L 4
R 4 { } 0.250977 11 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration E T n 1
0.250977
4
L
R
n 0.000977 , E
T n n M 1
0.249512 0.000488 .
4
M
n 4 0.140625 0.390625 0.265625 0.242188
8 0.191406 0.316406 0.253906 0.248047
16 0.219727 0.282227 0.250977 0.249512 E n E L 4 0.109375
8 0.058594
16 0.030273 E R E T 0.140625
0.066406
0.032227 M 0.015625 0.007813
0.003906 0.001953
0.000977 0.000488 Observations:
(a) E and E are always opposite in sign, as are E and E
L R T M . (b) As n is doubled, E and E are decreased by about a factor of 2 , and E and E
L R T M are decreased by a factor of about 4 .
(c) The Midpoint approximation is about twice as accurate as the Trapezoidal approximation.
(d) All the approximations become more accurate as the value of n increases.
(e) The Midpoint and Trapezoidal approximations are much more accurate than the endpoint
approximations.
26. 2 x
0 e dx= e x 2 2 =e 1 6.389056 . f (x)=e x 0 n=4 :
x=(2 0)/4= 1
2 1 0 1/2 1 3/2
e +e +e +e
4.924346
4 2
1 1/2 1 3/2 2
R=
e +e +e +e
8.118874
4 2
1
0
1/2
1
3/2 2
T =
e +2e +2e +2e +e
6.521610
4 2 2
1 1/4 3/4 5/4 7/4
M=
e +e +e +e
6.322986 .
4 2
E 6.389056 4.924346 1.464710 , E 6.389056 8.118874= 1.729818 ,
L= L R 12 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration E 6.389056 6.521610 T 0.132554 , E M 6.389056 6.322986=0.0660706 . n=8 :
x=(2 0)/8= 1
4 1 0 1/4 1/2 3/4 1 5/4 3/2 7/4
e +e +e +e +e +e +e +e
5.623666
8 4
1 1/4 1/2 3/4 1 5/4 3/2 7/4 2
R=
e +e +e +e +e +e +e +e
7.220930
8 4
1
0
1/4
1/2
3/4
1
5/4
3/2
7/4 2
T =
e +2e +2e +2e +2e +2e +2e +2e +e
6.422298
8 4 2
1 1/8 3/8 5/8 7/8 9/8 11/8 13/8 15/8
M=
e +e +e +e +e +e +e +e
6.372448
8 4
E 6.389056 5.623666 0.765390 , E 6.389056 7.220930 0.831874 ,
L= L R E 6.389056 6.422298 T 0.033242 , E M 6.389056 6.372448 0.016608 . n=16 :
x=(2 0)/16=
1
16 8
1
R =
16 8 1
8 f (0)+ f L = 1
8 +f 2
8 + 1
8 +f 2
8 +f 3
8 f
1 +f
+ 14
8
+f +f 15
8 15
8 + f (2) 5.998057
6.796689 1
2
3
15
+f
+f
+
+f
+ f (2)
6.397373
16 8 2
8
8
8
8
1
1
3
5
29
31
M =
f
+f
+f
+
+f
+f
6.384899
16 8
16
16
16
16
16
E 6.389056 5.998057 0.390999 , E 6.389056 6.796689 0.407633 , T = f ( 0 ) +2 f L R E 6.389056 6.397373 n E T E L 4 1.464710
8 0.765390
16 0.390999 R 1.729818
0.831874
0.407633 0.008317 , E
E T E M 6.389056 6.384899 0.004158 . M 0.132554 0.066071
0.033242 0.016608
0.008317 0.004158 Observations:
(a) E and E are always opposite in sign, as are E and E
L R T M . (b) As n is doubled,
13 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration E and E are decreased by a factor of about 2 , and E and E
L R T M are decreased by a factor of about 4 . (c) The Midpoint approximation is about twice as accurate as the Trapezoidal approximation.
(d) All the approximations become more accurate as the value of n increases.
(e) The Midpoint and Trapezoidal approximations are much more accurate than the endpoint
approximations.
27. 4 2 3/2
x
3 x dx= 1 4 = 1 2
14
( 8 1) =
3
3 4.666667 n=6 :
x= ( 4 1 ) /6=
T = 1
2 1 1 +2 1.5+2 2 +2 2.5+2 3 +2 3.5+ 4
4.661488
2 2
1
M=
1.25+ 1.75+ 2.25+ 2.75+ 3.25+ 3.75
4.669245
6 2
1
S=
1 +4 1.5+2 2 +4 2.5+2 3 +4 3.5+ 4
4.666563
6 2 3
14
14
E
4.661488 0.005178 , E
4.669245 0.002578 ,
T
M
3
3
14
E
4.666563 0.000104 .
S
3
6 n=12 :
x= ( 4 1 ) /12= 1
4 1 ( f (1)+2 f (1.25)+ f (1.5)+ + f (3.5)+ f (3.75) + f (4) ) 4.665367
4 2
1
M =
f (1.125)+ f (1.375)+ f (1.625)+
+ f (3.875) 4.667316
12 4
1
S =
4.666659
12 4 3
14
14
E
4.665367 0.001300 , E
4.667316 0.000649 ,
T
M
3
3
14
E
4.666659 0.000007 .
S
3 T =
12 Note: These errors were computed more precisely and then rounded to six places. That is, they were
not computed by comparing the rounded values of T , M , and S with the rounded value of the
n n n actual integral.
14 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration n Tn M S n n 6 4.661488 4.669245 4.666563
12 4.665367 4.667316 4.666659
E n E T E M S 6 0.005178 0.002578 0.000104
12 0.001300 0.000649 0.000007
Observations:
(a) E and E are opposite in sign and decrease by a factor of about 4 as n is doubled.
T M (b) The Simpson’s approximation is much more accurate than the Midpoint and Trapezoidal
approximations, and seems to decrease by a factor of about 16 as n is doubled.
2 28. I= 1 x x xe dx= xe e x 2 2 x =e +2/e 8.124815 . f (x)=xe . 1 n=6 :
x=[2 ( 1)]/6= 1
2 1 { f ( 1)+2 f ( 0.5)+ f (0)+ + f (1.5) + f (2)} 8.583514
2 2
1
M=
f ( 0.75)+ f ( 0.25)+
+ f (1.75) 7.896632
6 2
1
S=
f ( 1)+4 f ( 0.5)+2 f (0)+4 f (0.5)+2 f (1)+4 f (1.5)+ f (2) 8.136885
6 2 3
E I 8.583514 0.458699 , E
I 7.896632 0.228183 , T =
6 T E S M 0.012070 . I 8.136885 n=12 :
x=[2 ( 1)]/12= 1
4 1 { f ( 1)+2 f ( 0.75)+ f ( 0.5)+
4 2
1
7
5
M =
f
+f
+
+f
12 4
8
8
1
S =
f ( 1)+4 f ( 0.75)+2 f ( 0.5)+
12 4 3 T =
12 E T I 8.240073 0.115258 , E M + f (1.75) + f (2)}
13
8 +f 15
8 8.240073
8.067259 +2 f (1.5)+4 f (1.75)+ f (2) 8.125593 I 8.067259 0.057556 ,
15 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration E S I 8.125593 n Tn 0.000778 M S n n 6 8.583514 7.896632 8.136885
12 8.240073 8.067259 8.125593
E n E T 6
12 E M 0.458699 0.228183
0.115258 0.057556 S 0.012070
0.000778 Observations:
(a) E and E are opposite in sign and decrease by a factor of about 4 as n is doubled.
T M (b) The Simpson’s approximation is much more accurate than the Midpoint and Trapezoidal
approximations, and seems to decrease by a factor of about 16 as n is doubled.
x= ( 4 0 ) /4=1
1
1
(a) T =
f (0)+2 f (1)+2 f (2)+2 f (3)+ f (4)
0+2(3)+2(5)+2(3)+1 =11.5
4 2
2
(b) M =1 f (0.5)+ f (1.5)+ f (2.5)+ f (3.5) 1+4.5+4.5+2=12
29. 4 (c) S =
4 1
f (0)+4 f (1)+2 f (2)+4 f (3)+ f (4)
3 1
0+4(3)+2(5)+4(3)+1 =11.6
3 30. If x= distance from left end of pool and w=w(x)= width at x , then Simpson’s Rule with n=8 and
2
16
2
x=2 gives Area = wdx
0+4(6.2)+2(7.2)+4(6.8)+2(5.6)+4(5.0)+2(4.8)+4(4.8)+0 84 m .
0
3
31. (a) We are given the function values at the endpoints of 8 intervals of length 0.4, so we’ll use the
Midpoint Rule with n=8/2=4 and x=(3.2 0)/4=0.8 .
3.2
0 f (x)dx M =0.8[ f (0.4)+ f (1.2)+ f (2.0)+ f (2.8)]
4 =0.8[6.5+6.4+7.6+8.8]
=0.8(29.3)=23.44
(b) 4 f / / (x) 1 f / / (x) 4 , so use K=4 , a=0 , b=3.2 , and n=4 in Theorem 3. So 16 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 3 E 4(3.2 0)
M 2 24(4) = 128
=0.3413 .
375 32. We use Simpson’s Rule with n=10 and distance= 5 v(t)dt S 10 0 = x= 1
:
2 1 f (0)+4 f (0.5)+2 f (1)+
2 3
1
=
6
1
= (268.41)=44.735m
6 +4 f (4.5)+ f (5) 33. By the Net Change Theorem, the increase in velocity is equal to 6
0 a(t)dt . We use Simpson’s Rule with n=6 and t=(6 0)/6=1 to estimate this integral:
6
1
a(t)dt S = [a(0)+4a(1)+2a(2)+4a(3)+2a(4)+4a(5)+a(6)]
6
0
3
1
1
[0+4(0.5)+2(4.1)+4(9.8)+2(12.9)+4(9.5)+0]= (113.2)=37.73 ft/s
3
3
34. By the Net Change Theorem, the total amount of water that leaked out during the first six hours is
6 0
6
equal to r(t)dt . We use Simpson’s Rule with n=6 and t=
=1 to estimate this integral:
0
6
6 r(t)dt S 0 6 1
r(0)+4r(1)+2r(2)+4r(3)+2r(4)+4r(5)+r(6)
3
1
4+4(3)+2(2.4)+4(1.9)+2(1.4)+4(1.1)+1
3
1
= (36.6)=12.2liters
3
= The function values were obtained from a high resolution graph.
35. By the Net Change Theorem, the energy used is equal to
n=12 and t=(6 0)/12= 6
0 P(t)dt . We use Simpson’s Rule with 1
to estimate this integral:
2
17 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 6
0 P(t)dt S 12 1/2
[P(0)+4P(0.5)+2P(1)+4P(1.5)+2P(2)+4P(2.5)]
3
+2P(3)+4P(3.5)+2P(4)+4P(4.5)+2P(5)+4P(5.5)+P(6)]
1
= [1814+4(1735)+2(1686)+4(1646)+2(1637)+4(1609)+2(1604)]
6
+4(1611)+2(1621)+4(1666)+2(1745)+4(1886)+2052]
1
= (61,064)=10,177.3 megawatt hours.
6
= 8 36. By the Net Change Theorem, the total amount of data transmitted is equal to 0 D(t)dt 3600 . We use Simpson’s Rule with n=8 and t=(8 0)/8=1 to estimate this integral:
8
1
D(t)dt S = [D(0)+4D(1)+2D(2)+4D(3)+2D(4)+4D(5)+2D(6)+4D(7)+D(8)]
8
0
3
1
[0.35+4(0.32)+2(0.41)+4(0.50)+2(0.51)+4(0.56)+2(0.56)+4(0.83)+0.88]
3
1
= (13.03)=4.343
3
Now multiply by 3600 to obtain 15 , 636 megabits.
37. Let y= f (x) denote the curve. Using cylindrical shells, V = . 10 10 2 2 2 xf (x)dx=2 xf (x)dx=2 I . Now use Simpson’s Rule to approximate I :
I S = 10 2 2 f (2)+4 3 f (3)+2 4 f (4)+4 5 f (5)+2 6 f (6)
8
3(8)
+4 7 f (7)+2 8 f (8)+4 9 f (9)+10 f (10)
1
[2(0)+12(1.5)+8(1.9)+20(2.2)+12(3.0)+28(3.8)+16(4.0)+36(3.1)+10(0)]
3
1
= (395.2)
3
1
Thus, V 2
(395.2) 827.7 or 828 cubic units.
3
38.
18 0
f (0)+4 f (3)+2 f (6)+4 f (9)+2 f (12)+4 f (15)+ f (18)
6 6 3
0
=1 9.8+4(9.1)+2(8.5)+4(8.0)+2(7.7)+4(7.5)+7.4 =148 joules Work = 18 f (x)dx S = 18 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 39. Volume = . ( 2 3
0 1+x 3 2 ) dx= ( 1+x ) 3 2/3 2 0 dx . V S 10 ( where f (x)= 1+x 3 2/3 ) and 1
. Therefore,
5
1
S =
f (0)+4 f (0.2)+2 f (0.4)+4 f (0.6)+2 f (0.8)+4 f (1)
10
5 3
+2 f (1.2)+4 f (1.4)+2 f (1.6)+4 f (1.8)+ f (2)
12.325078 x=(2 0)/10=
V 40. Using Simpson’s Rule with n=10 ,
2 k =sin 2 T =4 =4 1
2 2 41. I( )= dx /2
0 1
9.8 2 /2
10 3 2 f (0)+4 f 2 N sin k
k 2 , where k= 4 2 I( ( 10 ) sin 2k
)=
k = 0 2 2 , where k= Ndsin 20 +2 f ( 104) ( 10 4) sin
632.8 10 9 7 10 42. f (x)=cos ( x) , x= 2
20 +4 f 4 9
20 +f , and =632.8 10 . Now n=10 and = 10 6 2.07665 2
9 ( . So 10
10 6 ) =2 7 10 , +I(0.0000009)] 59.4 . 20 0
=2
10 2
{ f (0)+2[ f (2)+ f (4)+ + f (18)]+ f (20)}
2
= 1[cos 0+2(cos 2 +cos 4 +
+cos 18 )+cos 20 ]
= 1+2(1+1+1+1+1+1+1+1+1)+1=20
1
20
20 1
The actual value is
cos ( x)dx=
sin x = (sin 20
10 + , N=10 , 000 , d=10 so M =2 10 [I( 0.0000009)+I( 0.0000007)+ T 42
2
radians, g=9.8 m / s ,
180 L
S
g 10 4 1 k sin x 2 /2
, L=1 ,
10 , and f (x)=1/ 1 k sin x , we get 0 L
g x= = 0 0 sin 0)=0 . The discrepancy is due to the fact that the function is sampled only at points of the form 2n , where its value is
f (2n)=cos (2n )=1 .
43. Consider the function f whose graph is shown. The area
19 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration 2
0 f (x)dx is close to 2 . The Trapezoidal Rule gives T =
2 2 0
1
f (0)+2 f (1)+ f (2) =
1+2 1+1 =2 .
2 2
2 2 0
f (0.5)+ f (1.5) =1[0+0]=0 ,
2
2
so the Trapezoidal Rule is more accurate.
The Midpoint Rule gives M = 2 44. Consider the function f (x)= x 1 , 0 x 2 . The area 0 f (x)dx is exactly 1 . So is the right endpoint approximation: R = f (1) x+ f (2) x=0 1+1 1=1 . But Simpson’s Rule approximates f with
2 2 the parabola y=(x 1) , shown dashed, and S =
2 x
1
2
f (0)+4 f (1)+ f (2) =
1+4 0+1 = .
3
3
3 45. Since the Trapezoidal and Midpoint approximations on the interval a,b are the sums of the
Trapezoidal and Midpoint approximations on the subintervals x ,x , i=1,2,... ,n , we can focus
i 1 i our attention on one such interval. The condition f / / (x)<0 for a x b means that the graph of f is
b concave down as in Figure 5. In that figure, T is the area of the trapezoid AQRD ,
n area of the region AQPRD , and M is the area of the trapezoid ABCD , so T <
n n general, the condition f / / and since f
T <
n b
a a f (x)dx is the f (x)dx<M . In
n <0 implies that the graph of f on a,b lies above the chord joining the points ( a,f(a) ) and ( b,f(b) ) . Thus,
/ / b a b
a f (x)dx>T . Since M is the area under a tangent to the graph,
n n <0 implies that the tangent lies above the graph, we also have M >
n b
a f (x)dx . Thus, f (x)dx<M .
n 3 2 46. Let f be a polynomial of degree 3 ; say f (x)=Ax +Bx +Cx+D . It will suffice to show that
Simpson’s estimate is exact when there are two subintervals ( n=2 ), because for a larger even number
20 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration of subintervals the sum of exact estimates is exact. As in the derivation of Simpson’s Rule, we can
assume that x = h , x =0 , and x =h . Then Simpson’s approximation is
0 1 2 1
h[ f ( h)+4 f (0)+ f (h)]
h
3
1
3
2
3
2
=
h
Ah +Bh Ch+D +4D+ Ah +Bh +Ch+D
3
1
2 3
2
=
h 2Bh +6D = Bh +2Dh
3
3
The exact value of the integral is
h f (x)dx ( ) ( Ax3+Bx2+Cx+D) dx =
h h ( ) ( Bx2+D) dx [by Theorem 5.5. (a) and (b)]
0
h 2 h 1 3
Bx +Dx
3 = 2 = 0 2 3
Bh +2Dh
3 Thus, Simpson’s Rule is exact.
47. T =
n . 1
2 ( ) ( )
0 so n 1 and n ( ) + f ( x ) , where x = 1 ( x +x ) . Now
2
1
1
x
f ( x ) +2 f ( x ) +2 f ( x ) +2 f ( x ) +2 f ( x ) +
2
2
+2 f ( x ) +2 f ( x ) +2 f ( x ) + f ( x ) n 2n +2 f x 1 ( ) ( ) M = x f x +f x +
T ( )+ f (x ) x f x +2 f x + = 1 +f x 2 n 1 n 1 n 1 0 n 1 1 i 1 i i 2 2 n n 1
1
1
T +M =
T + M
n
n
2
2 n 2 n
1
=
x f x +2 f x +
+2 f x
+f x
0
1
n 1
n
4
1
+
x 2 f x +2 f x +
+2 f x
+2 f x
1
2
n 1
n
4 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = T 2n
48. T =
n x
2 ( ) f x +2
0 n 1
i=1 ( ) ( ) f x +f x
i 1
2
1
x
T + M=
T +2M =
n
n
3 n 3 n 3
3 2 ( ) n and M = x
n ( ) f x +2
0 n 1
i=1 n
i=1 f x
2 x i ( ) ( ) f x + f x +4
i n , so
n
i=1 f x i x
2 where
21 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.7 Approximate Integration . b a
b a
. Let x=
. Then x=2 x , so
n
2n
2
1
x
n 1
n
T + M =
f x +2
f x + f x +4
f x
0
i
n
i
i=1
i=1
3 n 3 n
3
1
=
x f x +4 f x x +2 f x +4 f x x
0
1
1
2
3
x= Since x ,x 0 1 ( )
( ) ( )
( x)
( ) ( ) ( ) ( )
+2 f ( x ) +
+2 f ( x ) +4 f ( x x ) + f ( x ) x,x ,x 1 2 2 n 1 x,x ,... ,x
2 ,x n 1 n is the width of the subintervals for S
S 2n . Therefore, 2n n n x,x are the subinterval endpoints for S
n , the last expression for 2n , and since x= b a
2n 1
2
T + M is the usual expression for
3 n 3 n 1
2
T + M =S .
3 n 3 n 2n 22 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 4 1. (a) Since xe x 4 dx has an infinite interval of integration, it is an improper integral of Type I. 1
/2 (b) Since y=sec x has an infinite discontinuity at x= 2 , sec xdx is a Type II improper integral.
0
2 (c) Since y= x
has an infinite discontinuity at x=2 ,
(x 2)(x 3) x
0 2 dx is a Type II improper x 5x+6 integral.
0 1 (d) Since dx has an infinite interval of integration, it is an improper integral of Type I. 2 x +5
2. (a) Since y=1/(2x 1) is defined and continuous on 1,2 , the integral is proper.
1 1
1
1
(b) Since y=
has an infinite discontinuity at x= ,
dx is a Type II improper integral.
2x 1
2 0 2x 1
sin x (c) Since 2 dx has an infinite interval of integration, it is an improper integral of Type I. 1+x 2 (d) Since y=ln (x 1) has an infinite discontinuity at x=1 , ln (x 1)dx is a Type II improper integral.
1
3 3. The area under the graph of y=1/x =x
t 3 between x=1 and x=t is 1 2 t 1 2
1
1
2
x
=
t
=
1 2t
. So the area for 1 x 10 is
2
1
2
2
2
1
A(10)=0.5 0.005=0.495 , the area for 1 x 100 is A(100)=0.5 0.00005=0.49995 , and the area for
1 x 1000 is A(1000)=0.5 0.0000005=0.4999995 . The total area under the curve for x 1 is
1
1
2
lim A(t)=lim
1 2t
= .
2
2
t
t /( ) 3 A(t)= x dx= /( ) 4. (a)
1 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals (b) The area under the graph of f from x=1 to x=t is
t t F(t) = f (x)dx= x
1 1
x
0.1 1.1 dx= 1 ( = ) 0.1 ( 0.1 t 0.1 1 ) 10 t
1 =10 1 t
and the area under the graph of g is
t t G(t) = g(x)dx= x
1 1 0.1
x
0.1 0.9 dx= 1 ( = 10 t 0.1 1
t
10
100 t
1 ) F(t)
2.06
3.69
6.02 G(t)
2.59
5.85
15.12 6 7.49 29.81 10 9 90 20 9.9 990 4 10
10
10
10 ( (c) The total area under the graph of f is lim F(t)=lim 10 1 t
t 0.1 ) =10 . t ( 0.1 1) = The total area under the graph of g does not exist, since lim G(t)=lim 10 t
t . t t 1 5. I=
1 2 t (3x+1) 1
2 (3x+1) dx = 1 dx=lim
1
3 1 1
u 2 2 dx . Now (3x+1) du [ u=3x+1 , du=3dx ] 2 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 1
1
+C=
+C ,
3u
3(3x+1) = t 1
3(3x+1) so I=lim
t
0 1 t 0 1
6.
dx= lim
2x 5
t
Divergent 1
1
+
3(3t+1) 12 =lim 1
dx= lim
2x 5
t t =0+ 1
1
=
. Convergent
12 12 0 1
ln 2x 5
2 1
1
ln 5
ln 2t 5
2
2 =lim t t = . 7.
1 1 1
dw = lim
2 w
t 1
dw= lim
2 w
t t = lim 2 2 w 2 3 +2 2 t = 1
t . Divergent t 8.
t x
2 2 dx t 0 (x +2) 0 = e 9. y/2 1
2 dy=lim
t 2 y/2 e 2 1
2 = dx=lim
t (x +2) 0+ t 4 x =lim t x +2 0 = 2 1
lim
2t 1
t +2 2 2 2 + 1
2 1
. Convergent
4 dy=lim y/2 t 2e =lim ( 2e 4 t 4 1 1
2 t/2 +2e )=0+2e =2e 2 . t Convergent
1 10. e 2t 1 dt= lim
x 11.
0 x
2
1+x =
dx
x
2
1+x = lim
dx
t e 2t x x
0 1
e
2 dt= lim 2t 1
x 1 2 1
e+ e
2
2 = lim
x 2x = . Divergent x
x
2
2
1+x +
1+x and
dx
0 dx
1
2
ln 1+x
2 ( ) 0 = lim t t 0 1
2
ln 1+t
2 ( ) = . Divergent
3 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 0 ( 2 v ) dv=I1+I2= ( 2 v4) dv+ ( 2 v4) dv , but
4 12. I= 0 I = lim
1 0 1 5
v
5 2v t = 1 5
t =
5 2t+ t t . Since I is divergent, I is divergent, and there is no
1 need to evaluate I . Divergent
2 0 2 13. xe x 2 dx= xe x 2 x dx+ xe dx . 0
0 2 x xe dx= lim
t 2 x xe t 2 14. xe xe x x e 2 dx= x xe x =lim
t t t 1
2 =lim 0 ( e 1) = 1 1= 1 , and
2
2
( e 1) = 1 ( 1)= 1 .
2
2 1
2 t t 2 2 1 1
+ =0 . Convergent
2 2 dx= 0 3 t
2 2 Therefore, x e 1
2 dx=lim 0 0 2 1
2 3 2 x dx+ x e 3 dx , and 0
0 2 xe x 3 1
e
3 dx= lim
t 15. 2 t sin d =lim 2 t x 3 0 1 1
=
+
t
3 3 sin d =lim t lim e 3 = . Divergent t
t cos 2 t integral is divergent. Divergent
t 1
2
cos d =lim
(1+cos 2 )d =lim
16.
0
02
t
t
1
1
1
sin 2t
for all t , but t
as t
4
4
2 =lim ( cos t+1) . This limit does not exist, so the
t 1
2 + 1
sin 2
4 t =lim 0 t 1 1
t+ sin 2t =
2 4 since . Divergent 17.
t x+1
1 2 x +2x dx =lim
t = 1
(2x+2)
2
2 x +2x
. Divergent
1 dx= 1
lim
2t 2 t ln (x +2x) =
1 1
lim
2t
2 ln (t +2t) ln 3 4 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 18.
t dz
2 t z +3z+2 0 1
z+1 =lim
0 =lim 1
z+2 dz=lim
t t+1
t+2 ln t 1
2 ln t z+1
z+2 ln 0 =ln 1+ln 2=ln 2 . Convergent 19.
se 5s t ds =lim se t 0 1
se
5 ds=lim
t 0 1
te
5 =lim
t = 5s 1
e
25 5s 1 5t 1
e +
25
25 5t t 5s by integration by
parts with u=s 0 =0 0+ 1
[ by l’Hospital’s Rule]
25 1
. Convergent
25 20.
6 6 r/3 re dr = lim r/3 r/3 re dr= lim t 3re r/3 6 9e t t t
2 2 t/3 t/3 by integration by
parts with u=r 2 = lim (18e 9e 3te +9e )=9e 0+0 [ by l’Hospital’s Rule]
t
2 =9e . Convergent ( ln x ) ln x
21.
dx=lim
x
t
1
Divergent
x e 22. 0 dx= 2 x x e dx+ e dx , 0 t e x t =lim 0 t
1 x 0 e dx=lim 2 (by substitution with u=ln x , du=dx/x ) =lim
t 0 x e dx= lim e t ( 1 e t ) =1 . Therefore, x 0
t = lim ( ln t )
2 2 = . ( 1 et ) =1 , and t e x dx=1+1=2 . Convergent t 23. 5 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 0
2 2 x 9+x 6 dx= 9+x
2 x dx Now 2 x 9+x 6 dx+ 9+x 0 u=x 6 3 = 2 6 2 x du=3x dx x dx=2 9+x 0 1
du
3
9+u 6 dx [ since the integrand is even]. u=3v
du=3dv 2 1
1
1
= tan v+C= tan
9
9 1
(3dv)
3 = 2 = 9+9v
u
3 1 1
9 1
+C= tan
9 dv
2 1+v
1 3 x
3 +C , t
2 2 x so 2
0 9+x 6 x dx =2lim
t 0 9+x 1
=2lim
tan
9
t 6
1 1
tan
9 dx=2lim
t
3 t
3 ( 2) . 3 du=dx/x , v= 1/ 2x t 1 x 3 t t ln x dx =lim 0 2
=
. Convergent
9 2 9 = 24. Integrate by parts with u=ln x , dv=dx/x
ln x t 3 x
3 1 3 t 1 dx=lim 2 t ln x + x
2x
1
1 ln t
1
1 1
=lim
+0
+
=
2 4
2 2
4
t
t
4t
1/t
ln t
1
since lim
=lim
=lim
=0 . Convergent
2 t
2
2t t
t
t
2t
1 2 25. Integrate by parts with u=ln x , dv=dx/x 1
2 1
1 x 3 dx du=dx/x , v= 1/x . t ln x
2 dx = ln x lim 2 ln x
x t
x
= 0 0+0+1=1
ln t
1/t
since lim
=lim
=0 . Convergent
t t
1
t
1 x t dx=lim 1 1
x t =lim 1 t ln t
t 1
+0+1
t 6 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals t x arctanx 26.
0 1
v=
2 22 (1+x )
2xdx
22 t = x arctanx dx=lim
1/2
2 22 xdx dx . Let u=arctanx , dv= 22 (1+x ) (1+x ) 0 . Then du= dx
2 , 1+x , and (1+x ) 1+x
x arctanx
1 arctanx 1
dx =
+
22
2
2
2
(1+x )
1+x dx
22 (1+x ) x=tan ,
dx=sec 2 d 2 1 arctanx 1 sec d
=
+
2
2 2
2
2
1+x
(sec )
1 arctanx 1
2
=
+
cos d
2
2
2
1+x
1 arctanx
sin cos
=
+ +
+C
2
2
4
4
1+x
1 arctanx 1
1 x
=
+ arctanx+
+C
2
2
2
4
4
1+x
1+x It follows that
x arctanx
0 22 dx =lim
t (1+x ) =lim
t 1 arctanx 1
1 x
+ arctanx+
2
2
2
4
4
1+x
1+x
1 arctant 1
1 t
+ arctant+
2
2
2
4
4
1+t
1+t t
0 =0+ 1
+0=
.
4 2
8 Convergent.
27. There is an infinite discontinuity at the left endpoint of 0,3 . 7 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 3 3 dx
dx
=lim
=lim 2 x
+t
+
x
x 0 t 0 t 3 = t + t 0 (2 3 2 t ) =2 3 . Convergent 0 28. There is an infinite discontinuity at the left endpoint of 0,3 .
3
3 dx
=lim
+
x x 0 t dx 0 x t 3/2 + t 3 2
x =lim
0 2
2
+lim
=
+
3
t =
t t 0 29. There is an infinite discontinuity at the right endpoint of
0 1,0 . t dx
1 . Divergent dx =lim 2 x t 0 2 x 1 t t 1
x =lim
0 1 1
+
t
1 =lim 1 t 0 = . Divergent 9
t dx 30. 3
1 3 =lim x 9 dx 9 1 t
0 t 9 3 t 3
2/3
(x 9)
2 x 9 =lim 1 t 9 3 t 0 3 dx 4 31.
2 dx x=
2 x 4 dx +
0 1 x 4 dx , but
2 x 4 x
3 =lim
t 0 dx t 1 x 0 dx =lim
2 1 =lim 2 t 3t 0 2 1 x 0 t 1 t 1 sin x =lim sin t=
0 1 t
33 33. There is an infinite discontinuity at x=1 . (x 1) 1/5 1
1 dx= (x 1) 0
1 0 1 =lim 1 3 = . Divergent 24 t 32. t (x 1) 3
2/3 3
(t 9)
(4) =0 6= 6 . Convergent
2
2 = 1/5 dx=lim
t 1 0 (x 1) 1/5 dx=lim
t 1 5
4/5
(x 1)
4 2 . Convergent 1/5 33 dx+ (x 1) 0
t =lim 0 t 1 1/5 dx . Here 1 5
4/5 5
(t 1)
4
4 = 5
and
4 8 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 33 (x 1) 1/5 33 dx=lim 1 (x 1) 1/5 (x 1) + t 33 5
4/5
(x 1)
4 dx=lim +
1 t t 33 1/5 1 t t 34. f (y)=1/(4y 1) has an infinite discontinuity at y= 1/4 t t 1/4 1 1
1
dy= lim
ln 4y 1
lim
4
+
+ t 4y 1
t ( 1/4 )
( 1/4 )
1
1
lim
ln 3
ln (4t 1) =
4
4
+
( 1/4 ) = so 1
.
4 1 1
dy =
4y 1 1 + 1 5
75
+20=
. Convergent
4
4 dx= 0 1 5
5
4/5
16
(t 1)
=20 . Thus,
4
4 =lim 1
dy diverges, and hence,
4y 1 1
0 t 1
dy diverges. Divergent
4y 1 35.
/2 sec xdx = /2 sec xdx+
0 0 4 dx 36.
0
2
0 2 x +x 6 sec xdx= lim /2 = lim
t sec xdx . t 0 ln sec x+tan x t t = lim ln sec t+tan t = . Divergent 0 /2 t 4 sec xdx /2 0 /2 2 4 dx
dx
dx
=
=
+
, and
0 (x+3)(x 2) 0 (x 2)(x+3) 2 (x 2)(x+3) dx
= lim
(x 2)(x+3)
t t 2 0 1
lim
=
5
t 2 1/5
x 2 1/5
x+3 dx [partial fractions =lim ln t 2
t+3 ln t 2
3 = 37. There is an infinite discontinuity at x=0 .
1 2 0
x x 2
x+3 t
0 .Divergent 1 e 1
ln
5 x e 1 1 e dx=
1 x x e 1 e dx+
0 x x dx . e 1
9 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals t 0 e
1 x e dx=lim x e 1 e 1 t t x ln e 1 dx=lim x 0 1 t x t 0 1 t ln e 1 =lim 1 ln e 1 1 = , 0 1 e so x e dx is divergent. The integral x e 1 1 dx also diverges since x e 1 0 1 x 1 e x dx=lim x e 1
Divergent
0 e e 1 0 t 2 ln e 1 t + t 3/2 1 x dx=lim x + t x 0 t =lim ln e 1 ln e 1 = . + t 0 2 x 3
x 3
x 3
dx=
dx+
dx and
38.
0 2x 3
0 2x 3
3/2 2x 3
x 3
1 2x 6
1
3
1
3
dx=
dx=
1
dx= x
ln 2x 3 +C , so
2x 3
2 2x 3
2
2x 3
2
4
3/2 x 3
dx= lim
2x 3 0 t 1
2x 3ln 2x 3
4 3/2 t = . Divergent 0 39.
2 2 2 I = z ln zdz=lim
0 0 t + 0 (3ln z 1) 3 0 3ln t 1 3 Now L=lim t (3ln t 1) =lim
t 2 + 2 3 t 8
1 3
8
8
(3ln 2 1)
t (3ln t 1) = ln 2
9
9
3
9 =lim
t z ln zdz=lim
+t t z 2 + 0 t + 0 t 3 t 3/t =lim
+ t 1
8
8
3
lim t (3ln t 1) = ln 2
9
3
9
+ 0 3/t 4 0 =lim
t 1
L.
9 + 0 ( t 3) =0 . Thus, L=0 and I= 8 ln 2
3 8
.
9 Convergent
40. Integrate by parts with u=ln x , dv=dx
1
0 / 1 ln x
ln x
dx=lim
dx = lim
+t
+
x
x
t 0
t 0
= ( lim
t du=dx/x , v=2 x . x 2 x ln x 1
t 1 2
t dx x =lim
t 2 t ln t 4
+ x 1
t 0 2 t ln t 4+4 t ) = 4 + 0 10 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals since lim
+ t ln t t ln t=lim 0 + t 0 1/2 t 1/t =lim
t + t 0 3/2 ( =lim
/2 t 2 t ) =0 . Convergent + 0 41. 1 x Area = e dx= lim e x 1
t t
t =e lim e =e
t 42. Area = e x/2 dx= 2lim x/2 t e 2 t 2 = 2lim e t/2 +2e=2e t 43. Area = 2
2 x +9 1 dx=2 2
0 2 dx x +9
11 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals t 1 =4lim
t = 2 t x +9 0 4
lim
3t 1
tan
3 dx=4lim tan 1 1 x
3 t
0 t
4
2
0 =
=
3
3 2
3 44. t x Area =
0 2 t = t x +9 =lim x dx=lim ( 1
lim
2t x +9 0 1
2
ln x +9
2 dx 2 t ) 0 (2 ) ln t +9 ln 9 = Infinite area
45. /2 Area = sec xdx= lim
0 ( /2) 2 sec xdx ( /2) 0 t = lim
t t 2 t tan x = lim (tan t 0)
0 t ( /2) =
Infinite area 12 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 46. 0 0 1
dx= lim
+t
x+2 Area =
2 t 2 x+2 = lim
t + 1
dx
x+2 2 0
t (2 = lim 2 t 2 2 t+2 ) + 2 =2 2 0=2 2
47. (a)
t t g(x)dx
1 2
5
10
100
1000
10 , 000 0.447453
0.577101
0.621306
0.668479
0.672957
0.673407
2 g(x)= sin x
2 . It appears that the integral is convergent. x (b) 1 sin x 1 2 2 0 sin x 1 0 sin x
2 x 1
2 x 1 . Since
1 2 dx is convergent x 2 sin x (Equation 2 with p=2>1 ),
1 2 dx is convergent by the Comparison Theorem. x (c)
13 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals Since f (x)dx is finite and the area under g(x) is less than the area under f (x) on any interval 1,t ,
1 g(x)dx must be finite; that is, the integral is convergent.
1 48. (a)
t t g(x)dx
2 5
10
100
1000
10 , 000 3.830327
6.801200
23.328769
69.023361
208.124560 1
. It appears that the integral is divergent.
x 1 g(x)= 1
1
<
. Since
x
x 1 (b) For x 2 , x > x 1 2 2 1
1
dx is divergent (Equation 2 with p=
2
x 1 ), 1
dx is divergent by the Comparison Theorem.
x 1 (c)
f (x)dx is infinite and the area under g(x) is greater than the area under f (x) on any interval Since
2 2,t , g(x)dx must be infinite; that is, the integral is divergent.
2 2 49. For x 1 , cos x
2 1+x 1
2 1+x < 1
2 x 1 .
1 2 dx is convergent by Equation 2 with p=2>1 , so x 14 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 2 cos x dx is convergent by the Comparison Theorem. 2 1+x 1 x 2+e
2
1
x
50. For x 1 ,
> [ since e >0 ] > .
x
x
x dx is divergent by the Comparison Theorem.
2x 1 2x 51. For x 1 , x+e >e >0 1
2x 2x x+e
e 2x 1
e
2 dx=lim
t 1 2x 1 t x 52. For x 1 , 0< 2x 1+x x
1+x 1
53.
x sin x
But ln t dx x <
6 x 2x 1 2t 1
e + e
2
2 =lim 1 =e =
6 on 1, ). e t by the Comparison Theorem, 1 1 x 2+e
x 1 1
dx is divergent by Equation 2 with p=1 1 , so
x x+e x
x 3 = 2 = 1
e
2 2 e . Therefore, 2x dx is convergent, and 1 is also convergent. 1
2 x 1 .
1 dx is convergent by Equation 2 with p=2>1 , so 2 x dx is convergent by the Comparison Theorem.
6 /2 1
on
x
as t 0, since 0 sin x 1 . 2 0
/2 + 0 , so
0 dx
=lim
x
+
t /2 0 t dx
=lim ln x
x
+
t /2
t . 0 dx
is divergent, and by the Comparison Theorem,
x /2
0 dx
is also
x sin x divergent. 54. For 0 x 1 , e x 1 e x x 1
.
x
15 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 1 1 1
1
dx=lim
dx=lim 2 x
+t
+
x
x 0 t 0 t 1 1
t (2 =lim
t 0 2 t ) =2 is convergent. Therefore, + e x 0 0 x dx is convergent by the Comparison Theorem.
1 dx
=
x (1+x) 0 55.
0 1 dx
+
x (1+x) 1 dx
=lim
+t
x (1+x)
t dx
+lim
x (1+x) t 0 t dx
. Now
x (1+x) 1 2
2udu
dx
=
[ u= x ,x=u ,dx=2udu ]
2
x (1+x)
u 1+u
1
1
du
=2
=2tan u+C=2tan
x +C ,
2
1+u ( ) so
dx
=lim 2tan
+
x (1+x) t 0 0 =lim 2 x 0 1
t +lim 1 2tan x t 2tan 4 + t 1 1 t +lim t
1
1 2tan t 2 t = 4 0+2 2 2 2 = .
3 dx 56. x 4 dx
= 2 x 4 x = dx
2 x dx = 2 x 2 2 3 2 x 2 x 4 2sec tan d
2sec 2tan
1
1
+C= sec
2
2 =lim
t + 2 1
sec
2 dx +
3 x 2 x 4 [ x=2sec
1 1 t dx =lim
t + 2 , where 0 x t 2 x 4 < /2 or dx +lim
t 3 x 2 . Now x 4 <3 /2 ] 1
x +C , so
2
1
x
2 3 +lim t t 1
sec
2 1 1
x
2 t
3 x 4 16 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals = 1
sec
2
1 57. If p=1 , then
0 3
2 1 1
2 1
sec
2 2 1 p dx 0+ 1 x =lim dx +t t x=lim ln x = 0 t = 4 . Divergent. t + 3
2 1 0 If p 1 , then
1 1
0 p p dx x =lim +
0 t t x =lim 1 p+1 p+1 + t dx x (note that the integral is not improper if p<0) 0 t 1 =lim
t + 0 1 p 1 If p>1 , then p 1>0 , so
t 1 If p<1 , then p 1<0 , so
t p 1 t p 1 + as t p 1 1 1 0 , and the integral diverges.
1 + 0 as t 0 and
0 p dx x = 1 ( 1 t 1 p) lim 1 p + t 0 Thus, the integral converges if and only if p<1 , and in that case its value is dx 58. Let u=ln x . Then du=dx/x x ( ln x ) e p du = u 1 p =
1 1
1 p . . 1 p . By Example 4, this converges to 1
if p>1
p 1 and diverges otherwise.
59. First suppose p= 1 . Then
1 1 1 ln x
ln x
x ln xdx=
dx=lim
dx=lim
x
+t
+
0
0 x
p t Now suppose p 1 p+1 x ln xdx=lim
0 t + 0 0 p p+1 x
x
dx=
ln x
p+1
p+1 p+1 p t 1
t = 1
2
lim (ln t) =
2
+
t , so the integral diverges. 0 1 . Then integration by parts gives x
x ln xdx=
ln x
p+1
p 0 1
2
(ln x)
2 x
ln x
p+1 x x p+1
2 ( p+1) 1 p+1
2 ( p+1) =
t +C . If p< 1 , then p+1<0 , so 1 ( p+1 ) 2 1
p+1 lim
t t
+ 0 p+1 ln t 1
p+1 = . If p> 1 , then p+1>0 and
17 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 1 1
p+1 1 p x ln xdx = 2 ( p+1) 0 1
= 1 + 2 ( p+1) 2 ( p+1) + t 0 = + t t + 0 ( p+1)t ( p+2 ) 2 ( p+1) 0 1 Thus, the integral converges to 1/t lim 1 p+1 lim t 1
p+1 ln t 1/( p+1)
1
=
2
( p+1 )
t
( p+1) lim 2 if p> 1 and diverges otherwise. ( p+1)
60. (a)
n=0 :
n t x x e dx =lim
t 0 x e dx=lim x t e 0 t 0
t =lim e +1 =0+1=1 t n=1 :
n t x x e dx=lim
t 0 x x xe dx . To evaluate xe dx , we’ll use integration by parts
0
x with u=x , dv=e dx
x So xe dx= xe
t lim
t x du=dx , v= e
x e dx= xe x x x . x x e +C=( x 1)e +C and
x t =lim ( x 1)e xe dx 0 t 0 t =lim ( t 1)e +1 =lim
t te t t e +1 t =0 0+1 [ use l’Hospital’s Rule] =1
n=2 :
n t x x e dx=lim
t 0 2 x 2 x x e dx . To evaluate x e dx , we could use integration by parts
0 again or Formula 97. Thus,
t lim
t 2 x x e dx
0 =lim
t 2 xe x t t +2lim 0 t x xe dx
0 =0+0+2(1) [ use l’Hospital’s Rule and the result for n=1 ] =2
18 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals n=3 :
n t x 3 x e dx =lim 3 x e dx=lim t 0 x xe +3lim 0 t 0 t x t 2 x x e dx t 0 =0+0+3(2) [ use l’Hospital’s Rule and the result for n=2 ] =6 n (b) For n=1 , 2 , and 3 , we have x x e dx=1 , 2 , and 6 . The values for the integral are equal to the
0 n factorials for n , so we guess x x e dx=n! .
0 k evaluate x
k+1 x x t lim k+1 x t 0
x e x k x k+1 x e dx=lim
t k+1 e dx =lim x x t e t =lim t k+1 k du=(k+1)x dx , v= e k+1 x
0
x x e dx . To . So x e +(k+1) x e dx and
t k +(k+1)lim 0 k , dv=e dx x (k+1)x e dx= x x 0 x t x 0
k+1 e dx , we use parts with u=x k+1 e dx= x k+1 x e dx=k! for some positive integer k . Then (c) Suppose that
k+1 x x x e dx t 0 t e +0 +(k+1)k!=0+0+(k+1)!=(k+1)! , t so the formula holds for k+1 . By induction, the formula holds for all positive integers. (Since 0!=1 ,
the formula holds for n=0 , too.)
0 61. (a) I= t xdx= xdx+ xdx , and
0 xdx=lim
t 0 xdx=lim
t 0 1 2
x
2 t =lim 0 t 1 2
t 0 =
2 , so I is divergent.
t xdx= (b)
t 1 2
x
2 t 1 2 1 2
= t
t =0 , so lim
t 2
2
t M
4 3/2 3
62. Let k=
so that v=
k
ve
2RT
0
2 Let =v , d =ve 2 kv dv t t xdx=0 . Therefore, xdx lim
t t xdx .
t 2 kv dv . Let I denote the integral and use parts to integrate I . 1
d =2vdv , =
e
2k 2 kv : 19 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 2 t 2 1 2 kv
1
I =lim
ve
+
2k
0 k
t
1
1
1
=
0
(0 1)=
2
2
2k
2k
2k Thus, v= 4 k 1 3/2 2k 1 r dx= lim
t dr=lim 2 t 24 M= mass of Earth =5.98 10 1 GMm
r R 2 Therefore, Work = 11 = dx x = lim
t 2 = lim 1 1 t t =GMm R
3 1
e
2k 2 kv t
0 8RT
.
M = t 1
r dr=lim GMm kt t M
1
x 2 2 2 2 RT t t 1
t = < . 1 1
+
t
R = GMm
, where
R
6 kg, m= mass of satellite =10 kg, R= radius of Earth =6.37 10 m, and
11 G= gravitational constant =6.67 10
6.67 10 1/2 M/ ( 2RT )
t (t e ) + 1 lim
k 1
dv=
lim
2k t 2 2 t GMm
R 1/2 ve 0 = (k ) 1
x 63. Volume = 64. Work = 2 2 = kv 2 N m / kg.
24 5.98 10 3 10 10 6.26 10 6 J. 6.37 10
t GmM 65. Work = F dr=lim
t R R r 2 1
R dr=lim GmM
t 1 2 GmM
mv =
2 0
R v= x(r)dr and x(r)= provides the work, so 1
t = GmM
. The initial kinetic energy
R 1
2
(R r)
2 0 2GM
.
R R 2r 66. y(s)= 2 s 2 r s 20 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals R R
2 y(s) =
lim 2 + t s t = lim
+ t s 2 2 r s 2 2 + s t R 3 r
dr 2 r 2Rr +R r dr=lim 2 r s t
R t 3 r(R r) 2 dr 2 r s R r
2R
t dr 2 2 2 r s +R t r
dr 2 2 r s 2 = lim + I 1 2RI 2+R I 3 =L
t s 2 2 2 2 2 2 2 2 u =r s , r =u +s , 2r dr=2udu , so, omitting limits and constant of For I : Let u= r s
1 integration, ( u2+s2) u du= ( u2+s2) du= 1 u3+s2u= 1 u ( u2+3s2)
u
3
3 I =
1
1
3 = 2 2 r s (r 2 s2+3s2) = 1
3 2 r
For I : Using Formula 44, I =
2
2 2
2 (r 2+2s2) 2 r s 2 s
2 2
.
r s + ln r+ r s
2
1 du 1
2 2
du=2r dr . Then I =
= 2 u= r s .
3 2
u 2 2 For I : Let u=r s
3 2 2 Thus, + 1
3 + 1
3 L = lim
t s = lim
t s =
= 1
3 + s 1
3 (r +2s )
2 2 2 ( R +2s ) 2 2 ( t +2s ) R s 2 t s
2 2 R s 2 2 2 2 ( R2+2s2) ( R +2s )
2 2 r
2 2R 2 R s
2 2 r s 1
3 lim
t 2 2 2R R
2 2R t
2 2 2 2 2 2 2 +R 2 2 s
2 2
t s + ln t+ t s
2
2 2 Rs ln 2 s
2 2
R s + ln R+ R s
2
2 r s 2 +R 2 +R 2 2 t 2 2 R s 2 t s 2 Rs ln R+ R s
2 R 2 s
2 2
r s + ln r+ r s
2
2 2 Rs ln s R+ R s
s 21 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 67. (a)
/ (b) r(t)=F (t) is the rate at which the fraction F(t) of burnt out bulbs increases as t increases. This
could be interpreted as a fractional burnout rate.
r(t)dt=lim F(x)=1 , since all of the bulbs will eventually burn out. (c) x 0 68.
1 kt I = te dt=lim
s 0 k 2 s kt ( kt 1 ) e [ Formula 96, or parts]
0 1 ks 1 ks
se
e
2
k
k =lim
s 1
k . 2 Since k<0 the first two terms approach 0 (you can verify that the first term does so with l’Hospital’s ( 2) 2 Rule), so the limit is equal to 1/k . Thus, M= kI= k 1/k = 1/k= 1/( 0.000121) 8264.5 years.
t 1 69. I=
a 2 t x +1
1 2 tan a<0.001
2 70. f (x)=e
4
0 f (x)dx 1 dx=lim x a 2 1 2 a 0.001 a>tan ( tan 1 1 0.001 ) t tan a = t 2 4 0 1
= .
8
2
1
S =
[ f (0)+4 f (0.5)+2 f (1)+
8
2 3
1
(5.31717808) 0.8862
6 and 1 2 tan a . I<0.001 1000 . x= 2 Now x>4 t tan x =lim t x +1 tan a> 1 dx=lim x x< x 4 x e <e +2 f (3)+4 f (3.5)+ f (4)] 2 4x
4 e x dx< 4 e 4x dx . 22 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 4 e 4x 1
e
4 dx=lim
t t 4x 1
0 e
4 ( = 4 st 16 ) =1 / ( 4e16)
e st 71. (a) F(s)= f (t)e dt= e dt=lim
0 if s>0 . Therefore F(s)= n st e =lim s n 0 0.0000000281<0.0000001 , as desired.
sn + s 0 n 1
s . This converges to 1
only
s 1
with domain { s|s>0} .
s (b)
st t n st F(s) = f (t)e dt= e e dt=lim
0 n 0 e =lim ( 1 s) n This converges only if 1 s<0
n st (c) F(s)= f (t)e dt=lim
n 0 t Then F(s)=lim
n F(s)= 1
2 n 0 n 1 t ( 1 s)
e
1 s dt=lim 0 1
1 s 1 s n t ( 1 s) e s>1 , in which case F(s)= 1
with domain { s|s>1} .
s 1 st st te dt . Use integration by parts: let u=t , dv=e dt du=dt , v= 0 e s st 1
2 e n st n =lim
0 n s 1 sn 2 sn se +0+ se 1
2 s = 1
2 e st s . only if s>0 . Therefore, s and the domain of F is { s|s>0} . s
72. 0 f (t) Me
at at 0
n st f (t)e
t ( a s) Me e dt=lim M e
n 0 0 st at Me e dt=M lim
n This is convergent only when a s<0 st for t 0 . Now use the Comparison Theorem: 1 t ( a s)
e
a s n =M lim 0 n 1
n ( a s)
e
1
a s s>a . Therefore, by the Comparison Theorem, st F(s)= f ( t ) e dt is also convergent for s>a .
0 / st 73. G(s)= f (t)e dt . Integrate by parts with u=e st / , dv= f (t)dt du= se st , v= f (t) : 0
23 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals G(s)=lim f (t)e st n
0 n But 0 f (t) Me st sn +s f (t)e dt=lim f (n)e at 0 f (0)+sF(s) n 0 f (t)e st at Me e t ( a s) st and lim Me =0 for s>a . So by the Squeeze Theorem, t
st lim f (t)e =0 for s>a G(s)=0 f (0)+sF(s)=sF(s) f (0) for s>a .f}(0) for s>a . t 74. Assume without loss of generality that a<b . Then
a a f (x)dx+ f (x)dx = lim f (x)dx+lim t a u f (x)dx u t a a = lim b f (x)dx+lim t t f (x)dx+ f (x)dx u a u b = lim a b
u f (x)dx+ f (x)dx+lim t t
a = lim b b f (x)dx+ f (x)dx t f (x)dx u a t + f (x)dx a b b = lim f (x)dx+ f (x)dx t t b b = f (x)dx+ f (x)dx
b
2 75. We use integration by parts: let u=x , dv=xe
2 xe
0 2 x dx = lim
t = lim
t 1
xe
2 2 x t / ( 2e )
t 2 t 1
+
0 2
1
+
2 x 1
du=dx , v=
e
2 dx 2 x . So 2 e x dx 0
2 e
0 x 1
dx=
2 2 e x dx 0 (The limit is 0 by l’Hospital’s Rule.)
76.
24 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 2 2 e x dx is the area under the curve y=e x 2 for 0 x< and 0<y 1 . Solving y=e x for x , we get 0
2 y=e x 2 2 ln y= x ln y=x ln y . Since x is positive, choose x= x= ln y , and the area is 1 represented by ln y dy . Therefore, each integral represents the same area, so the integrals are
0 equal.
77. For the first part of the integral, let x=2tan
1 2sec
2sec dx= 2 x +4 dx=2sec 2 d . 2 . From the figure, tan = d = sec d =ln sec +tan x
, and
2 2 x +4
. So
2 sec = t 2 1 I= C
x+2 2 x +4 0 dx = lim x +4 x
+
2
2 ln t Cln x+2 0 2 = lim t +4 +t
Cln (t+2) ( ln 1 Cln 2 )
2 ln t 2 = lim t +4 +t ln t 2 ( t+2 ) C +ln 2 C 2 = ln lim t+ t +4 t
2 Now L=lim
t t+ t +4 ( t+2 ) C ( t+2 ) 2 =lim
t 1+t/ t +4
C ( t+2 ) C 1 C C 1 +ln 2 2 = Clim ( t+2 ) C 1 . t If C<1 , L= and I diverges. If C=1 , L=2 and I converges to
25 Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals 0 ln 2+ln 2 =ln 2 . If C>1 , L=0 and I diverges to . 78.
x I= 2 x +1 0 C
3x+1 dx =lim
t =lim 1
2
ln x +1
2 ( ) 1
Cln (3x+1)
3 t
0 ( 2 ) 1/2 ln (3t+1)C/3 ln t +1 t =lim ln t For C ( t2+1) 1/2
( 3t+1 ) =ln C/3 lim
t ( 3t+1 ) C/3 0 , the integral diverges. For C>0 , we have
2 L=lim
t For C/3<1 C<3 , L=
to 2 t +1 t +1 =lim C/3 t
(3t+1) t / 2 t +1 C ( 3t+1 ) (C/3) 1 and I diverges. For C=3 , L= = 1
lim
Ct 1
( C/3 ) ( 3t+1 ) 1 . 1
1
and I=ln
. For C>3 , L=0 and I diverges
3
3 . 26 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth 1 1. y=2 3x L= 1 2 2 1+({dy /dx) italic{dx}= 1+( 3) dx= 10 1 ( 2) =3 10 . 2 2 The arc length can be calculated using the distance formula, since the curve is a line segment, so
2 2 L=[ distance from ( 2,8) to (1, 1) ]= [1 ( 2)] +[ ( 1 ) 8] = 90 =3 10
dy
=
dx 2 2. Using the arc length formula with y= 4 x
2 L= 2 dy
dx 1+
0 t 2 2 dx= 1+ sin (x/2) =2lim
0 2 t x 4 x 2 1 t 2dx dx=
0 4 x 1 sin (t/2) sin 0 =2 2 dx =2lim
2 4 x t 1 , we get
2 2 0 = 2lim x t 2 2 2 2 x 0 0 = 2 The curve is a quarter of a circle with radius 2 , so the length of the arc is 1
(2
4 2)= , as above. 3. From the figure, the length of the curve is slightly larger than the hypotenuse of the triangle formed
3/2
2 2
x 1
by the points ( 1,0 ) , ( 3,0 ) , and ( 3,f(3) ) ( 3,15) , where y= f (x)=
. This length is about
3
1/2
3/2
2 2
2 2
/
2
15 +2 15 , so we might estimate the length to be 15.5 . y=
y = x 1 (2x)
x 1
3 ( ( / 2 2 1+(y ) =1+4x
3 L= 2 = (18 3) ) ( ( x2 1) =4x4 4x2+1= ( 2x2 1) 2 , so, using the fact that 2x2 1>0 for 1
2 3 (2x 1) dx=
1 ) 2 3 2 2x 1 dx= (2x 1)dx=
1 1 2 3
x x
3 ) x 3, 3
1 2
46
1 =
=15.3
3
3
1 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth 4. From the figure, the length of the curve is slightly larger than the hypotenuse of the triangle formed
2
3
, where y= f (x)=x /6+1/(2x) . This length is
by the points ( 0.5,f(0.5) 1 ) , ( 1,f(0.5) 1 ) and 1,
3
1
2 about 2 2 1
3 + 0.6 , so we might estimate the length to be 0.65 . 3 x
1
y= +
6 2x 2 x
2 4 4 x
=1+
4 / 2 ( ) 1+ y 2 x
y =
2
/ 4 4 2 1 x
x
1 x
+
= + +
=
2 4
4 2 4 2 x
x
+
2
2 2 so, using the fact that the parenthetical expression is positive,
1 2 L=
1/2 1
6 = 5. y=1+6x 1
2
3/2 1 L= dx=
1/2 1/2 dy/dx=9x
82 0 1/2 1 1 2 3/2
u
81 3 2 3 1 2 2 x
x
+
2
2 3 dx= x
6 1
2x 1
1/2 1
31
1 =
=0.64583
48
48 1+81x dx= u = 2 2 x
x
+
2
2 82
1 6. y =4(x+4) , y>0 = 2 1+(dy/dx) =1+81x . So 1
81 du [ where u=1+81x and du=81dx ] 2
( 82 82 1 )
243
3/2 y=2(x+4) 1/2 dy/dx=3(x+4) 2 1+(dy/dx) =1+9(x+4)=9x+37 . So 2 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth 2 L= 9x+37 dx
0 1 2 3/2
u
9 3 = 55 u=9x+37,
du=9dx
= 37 5 1
9 1/2 37 du 2
( 55 55 37 37 )
27 55 x
1
7. y= +
3
6
10x = u dy 5 4 3
= x
x
dx 6
10 4 25 8 1
9
9
8 25 8 1
8
1+ ( dy/dx ) =1+
x
+
x =
x+ +
x =
36
2 100
36
2 100 5 4 3
x+
x
6
10 2 2 5 4 3
x+
x
6
10 L=
1 32
6 = 2 x
8. y=
2
4 1
80 2 1
10 = dy
1
=x
dx
4x dy
dx 4 2 4 dx= 1 5 1
x
x
6
10 9 L=
1 = 1
2 1 1/2 1
y + y
2
2
24 8
3 2 = 0 1/2 1
2 dy= = 8+ 2 1
2 64
3 dy/dx= tan x 2 2 =x + 1 1
1
+
. So
2
2
16x 2ln 2
4 2+ 9 2 3/2
1/2
y +2y
3
= = 1 ln 2
4 =6+ ln 2
.
4 2 . So 2
27+2 3
3 2
1+2 1
3 2 2 2 1+(dy/dx) =1+tan x=sec x . So sec xdx= ln sec x+tan x
0 1
2 1/2 32
3 /3 sec x dx= L= 2 y ( y 3) = 10. y=ln (cos x)
/3 3 . So 31 7 1261
+
=
6 80 240 1+ x
ln x
+
2
4 dx= 5 4 3
x+
x
6
10 dx= 1 3/2 1/2
1 1/2 1 1/2
y y
dx/dy= y
y
3
2
2
1
1 1 1 1
1 1 1
1 1/2 1
2
+ y = y+ + y =
y + y
1+ ( dx/dy ) =1+ y
4
2 4
4
2 4
2
2
9. x= 1
3 2 2 1 1
6 ln x
4 1
x+
4x L= 4 2 4 /3
0 =ln ( 2+ 3 ) ln (1+0)=ln ( 2+ 3 ) . 11. y=ln (sec x)
3 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth dy sec xtan x
=
=tan x
dx
sec x
/4 1+ sec x dx=
0 ( sec x dx= dy 1
=
dx x ( 0 2 +1 )
2 dy
dx 1+ 2 2 /4 sec xdx= ln (sec x+tan x)
0 2 +1 ) ln (1+0)=ln 12. y=ln x 2 /4 0 =ln 2 =1+tan x=sec x , so /4 2 L = 2 dy
dx = 1+ 1
x 3 2 2 1+x
x = . So L=
1 2 1+x
dx . Now let
x 2 v= 1+x , so v =1+x and vdv=xdx. Thus
2 L=
2 = 2 v v vdv= 2 v 1 2 1
ln
2 1/2
1+
v 1 1/2
v+1 2 v+1
v 1 2 1
dv= v+ ln v 1
2 1
ln 3
2 =2 2+ 1
ln v+1
2 2 +1 1
ln
2 2 1 =2 2 2 +ln 2 ( 2 +1 ) 1
ln 3
2 Or: Use Formula 23 in the table of integrals. ( / ) 2=1+sinh 2x=cosh 2x . / 13. y=cosh x y =sinh x 1 1+ y
1 So L= cosh xdx= sinh x =sinh 1=
0 0
2 14. y =4x , x= 1 2
y
4 dx 1
= y
dy 2
L=
= u 15. y=e
1 L= x / y =e
2x 1+e
0 x 1
(e 1/e) .
2 1+ 2
0 dx
dy
1+
2 2 =1+ 1 2
y dy=
4 1 2
y . So
4
1
0 2 1+u 2du 1+u +ln u+ 1+u 2 1 = 2 +ln ( 1+ 2 ) 0 ( / ) 2=1+e2x . So 1+ y
e 1+u dx=
1 2 du
x
u=e ,sox=ln u,dx=du/u
u 4 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth e 1+u = u 1 2 1+e 2 2 v udu= 2 2 2 vdv v= 1+u ,sov =1+u ,vdv=udu 2 v 1 2 2 1+e 1/2
1+
v 1 =
2 1/2
v+1
2 1+e 1
1+e + ln
2
2 = 2 = 1+e ( 2 1+e ( x 16. y=ln ( 1 2 2 1
2 +1 ) 1 ln ( 2 1) ) 2 1+u /u du , or substitute u=tan
x =ln e +1 x 1+e 1
ln
2 2 2 Or: Use Formula 23 for
e +1 1 1+e +1 2 +ln 2 1
v 1
dv= v+ ln
2
v+1 ) ln ( e 1)
x y = e 1 ( / ) 2=1+ 1+ y e / x e x x = x e +1
2x 2 ( e +1)
( e2x 1) 2 ( e2x 1) 2
2x 2x 4e e 1
x 2x e x 1 x 1+ y e 1 cosh x
b
dx= ln sinh x =ln sinh b ln sinh a=ln
So L=
a
a sinh x
2 dy/dx= sin x 2e ( / ) 2 = e2x+1 = e x+e x = cosh xx
sinh = b 17. y=cos x . e e sinh b
sinh a
2 2 . b b a a e e =ln e e . 2 1+(dy/dx) =1+sin x . So L= 1+sin x dx .
0 ( ) x x 18. y=2 dy/dx= 2 ln 2 19. x=y+y
2 20. x 2 a 3 2 dx/dy=1+3y 2 + y 2 b =1 , y= b 1 . 3 2 2x 1+ ( ln 2 ) 2 dx L=
0 2 22 4 2 4 4 2 9y +6y +2 dy . 1+(dx/dy) =1+(1+3y ) =9y +6y +2 . So L=
1
2 x /a =
2 b
a 2 2 a x . 5 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth b
y=
a 2 dy
=
dx 2 a x 2 2 a a x a 1/2 2 2 So L=2 1+ a a 21. y=xe b x
2 x dy
dx bx (a x )
2 dy/dx=e x 2 x x 2x 2 a . ( a2 x2) 1/2 ( b2 a2) x2+a4
2 dx . 2 a x 0 2 xe =e (1 x) 2 b x = a 4
dx=
a 2 2 2 1+(dy/dx) =1+e 2x 2 (1 x) . Let 5 2 (1 x) . Then L= f (x)dx . Since n=10 , f (x)= 1+(dy/dx) = 1+e x= 0 L S 10 1/2
1
3
[ f (0)+4 f
+2 f (1)+4 f
+2 f (2)+4 f
3
2
2
7
9
+4 f
+2 f (4)+4 f
+ f (5)] 5.115840
2
2 5
2 = 5 0 1
= . Now
10 2 +2 f (3)] The value of the integral produced by a calculator is 5.113568 (to six decimal places).
22. x=y+ y dx/dy=1+ 1
2 y 2 . Then L= g(y)dy . Since n=10 , y= 1 L S 10 1 2 1+ ( dx/dy ) =1+ 1+ 2 y 2 =2+ 1
1
2
+
. Let g(y)= 1+(dx/dy)
y 4y 2 1 1
=
. Now
10 10 1/10
[g(1)+4g(1.1)+2g(1.2)+4g(1.3)+2g(1.4)+4g(1.5)]
3
+2g(1.6)+4g(1.7)+2g(1.8)+4g(1.9)+g(2)] 1.732215 ,
= which is the same value of the integral produced by a calculator to six decimal places.
/3 23. y=sec x dy/dx=sec xtan x 2 2 f (x)dx , where f (x)= 1+sec xtan x . L=
0 Since n=10 ,
L S 10 = /30
3
+2 f /3 0
x=
=
. Now
10
30
f (0)+4 f
6
30 +4 f 30
7
30 +2 f
+2 f 2
+4 f
30
8
+4 f
30 3
+2 f
30
9
+f
30 4
30
3 +4 f 5
30 1.569619 . 6 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth The value of the integral produced by a calculator is 1.569259 (to six decimal places).
24. y=xln x 2 2 dy/dx=1+ln x . Let f (x)= 1+(dy/dx) = 1+(1+ln x) . 3 Then L= f (x)dx . Since n=10 , x= 1 3 1 1
= . Now
10 5 1/5
[ f (1)+4 f (1.2)+2 f (1.4)+4 f (1.6)+2 f (1.8)+4 f (2)]
3
+2 f (2.2)+4 f (2.4)+2 f (2.6)+4 f (2.8)+ f (3)] 3.869618 .
The value of the integral produced by a calculator is 3.869617 (to six decimal places).
L S 10 = 25. (a) (b)
3 Let f (x)=y=x 4 x . The polygon with one side is just the line segment joining the points
( 0,f(0) ) = ( 0,0 ) and ( 4,f(4) ) = ( 4,0 ) , and its length is 4 . The polygon with two sides joins the points ( 0,0 ) , ( 2,f(2) ) = ( 2,2 2 ) and ( 4,0 ) .
3 Its length is
2 ( (2 0) + 2 3 2 0 )2 + 2 ( (4 2) + 0 2 3 2 ) 2 =2 8/3 4+2 ( 3 6.43 ) ( 3 ) Similarly, the inscribed polygon with four sides joins the points ( 0,0 ) , 1, 3 , 2,2 2 , ( 3,3) ,
and ( 4,0 ) , so its length is
1+ ( 3 3 )2 + ( 1+ 2 3 2 3 3 )2 + ( 1+ 3 2 3 2 )2 + 1+9 7.50
7 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth (c) Using the arc length formula with dy
=x
dx 1
(4 x)
3 2/3 3 ( 1) + 4 x = 12 4x
2/3 , the length of the 3(4 x) 4
4 curve is L= 2 dy
dx 1+
0 dx= 2 12 4x 1+ dx . 2/3 3(4 x)
(d) According to a CAS, the length of the curve is L 7.7988 . The actual value is larger than any of
the approximations in part (b). This is always true, since any approximating straight line between two
points on the curve is shorter than the length of the curve between the two points.
0 26. (a) Let f (x)=y=x+sin x with 0 x 2 . (b) The polygon with one side is just the line segment joining the points ( 0,f(0) ) = ( 0,0 ) and
2 ( 2 ,f(2 ) ) = ( 2 ,2 ) , and its length is 2 (2 0) +(2 0) =2 2
8.9 .
The polygon with two sides joins the points ( 0,0 ) , ( ,f( ) ) = ( , ) , and ( 2 ,2 ) . Its length is
( 2 0) +( 2 0) + (2 2 2 ) +(2 ) = 2 + 2 =2 2 8.9 Note from the diagram that the two approximations are the same because the sides of the 2 sided
1
polygon are in fact on the same line, since f ( )= = f (2 ) .
2 The four sided polygon joins the points ( 0,0 ) , , 2 2 +1 ,( , ) , 3 3
,
2 2 1 , and ( 2 ,2 ) , so its length is
8 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth 2 2 + 2 +1 2 2 + 2 + 2 1 2 2 + 2 + 2 1 2 2 + 2 + 2 2 +1 9.4 (c) Using the arc length formula with dy/dx=1+cos x , the length of the curve is
2 2 2 L= 2 1+(1+cos x) dx= 2+2cos x+cos x dx 0 0 (d) The CAS approximates the integral as 9.5076 . The actual length is larger than the approximations
in part (b). ( 2 27. x=ln 1 y dx
2y
=
2
dy
1 y ) 1+ ( 1+y2) 2
=1+
=
2 2
( 1 y ) ( 1 y2) 2
2 2 dx
dy 4y . So 1/2
1/2 2 2 ( 1+y )
( 1 y2) 2 L=
0 28. y=x L= 1
0 81
=
64
81
=
64
205
=
128 4/3 1+ dy/dx= 4 1/3
x
3 16 2/3
x dx=
9 2 1+(dy/dx) =1+ 2 dy=
0 0 1+u 2 81 2
u du
64 u= ) 1
2
2 1
2
1+u
u 1+2u
ln u+ 1+u
8
8
1
32
25 1
4
25
1+
ln
+
6
9
9
8
3
9
81
ln 3 1.4277586
512 ( 2/3 ) 2/3 29. y =1 x ( 2/3 3/2 y= 1 x ) 1 y 4 1/3
4
x ,du= x
3
9 1
2 0.599 2/3 dx,dx= 9 2/3
9 9 2
81 2
x du=
u du=
u du
4
4 16
64 4/3
0 dy 3
2/3
=
1 x
dx 2 ( 2 dy=ln 3 16 2/3
x
9 ( 4/3 1+y = ) 1/2 81
64 2
x
3 1 41 5
6 9 3 1/3 = x 1/3 1
ln 3
8 ( 1 x2/3) 1/2
9 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth 2 dy
dx =x 1 2/3 ( L=4 ( 1 x2/3) =x 2/3 1 . Thus 2/3 1+ x 1 ) 1 dx=4 x 0 1/3 0 + t 1 3 2/3
x
2 dx=4lim
0 t =6 . 30. (a)
2/3 (b) y=x 1+ integral]. x=y 3/2 2 dy
dx 2
x
3 =1+
2 dx
dy 1+ 1/3 2 4
=1+ x
9 1 . So L= 1+
0 9
=1+ y . So L=
4
0 4 2
9
The second integral equals
1+ y
9 3
4
integral can be evaluated as follows: 4
x
9 1 2 3 1/2
y
2 =1+ 2/3 1 3/2 8
27 = 0 1+ 2/3 dx [ an improper 9
y dy .
4 13 13
13 13 8
1 =
. The first
8
27 1 1 4
1+ x
9 0 2/3 2/3 9x +4 dx = lim
t
9 =
0 + 0 t 1/3 3x u+4
1
du=
18
18 9 dx=lim
t + 0 9t 2/3 2
3/2
(u+4)
3 u+4
2/3
du u=9x ,du=6x
18
9 = 0 1/3 dx 1
3/2 3/2 13 13 8
(13 4 )=
27
27 (c)
L = length of the arc of this curve from ( 1,1 ) to ( 8,4 )
1 =
0 4 9
1+ y dy+
4
0 13 13 8 8
9
1+ y dy=
+
4
27
27 9
1+ y
4 3/2 4
0 10 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth = 13 13 8 8
13 13 +80 10 16
+
( 10 10 1 ) =
27
27
27 31. y=2x 3/2 / ( / ) 2=1+9x . The arc length function with starting point P0 ( 1,2 ) is 1/2 y =3x 1+ y
x s ( x) = 2
3/2
(1+9t)
27 1+9t dt=
1 x = 1 2
3/2
(1+9x) 10 10
27 32. (a)
dy
dx (b) 1+
x s ( x) = 2 2 4 =x + 1
1
+
,
4
2
16x ( 2) t +1/ 4t dt 1 1 3
t 1/(4t)
3 = x
1 1 3
1 1
x 1/(4x)
3
3 4
1 3
1
=
x 1/(4x)
for x 1
3
12
= (c)
33. The prey hits the ground when y=0
/ positive. y = 2
x
45 ( / ) 2=1+ 1+ y 4 180 1 2
x =0
45 2 x =45 180 x= 8100 =90 , since x must be 2 2 x , so the distance traveled by the prey is 45 11 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth L= 90
0 1
2
(x 50)
40 2 2 x dx= 45 4 1+u 0 45
du
2 2 4
1
2
1+u + ln u+ 1+u
0
2
1
45
2 17+ ln ( 4+ 17 ) =45 17+
ln ( 4+ 17 ) 209.1m
2
4 45
=
2
45
=
2
34. y=150 4 1+
1
u
2 / y = ( 2 1
(x 50)
20 ) ( / ) 2=1+ 1+ y 1 2 2 (x 50) , so the distance traveled by the 20 kite is
80 L= 1+ 1 3/2 2 2 1+u (20du) [u=
5/2 20 0 2 (x 50) dx= 1
1
(x 50),du=
dx]
20
20 3/2
1
2
=20
1+u + ln u+ 1+u
5/2
2
13
3
13
5
29
=10
+ln
+
+
ln
4
2
4
2
4
3+ 13
15
25
13 +
29 +10ln
=
122.8ft
2
2
5+ 29 1
u
2
3
2 ( 2 ) 5
+
2 29
4 35. The sine wave has amplitude 1 and period 14 , since it goes through two periods in a distance of
2
28 in., so its equation is y=1sin
x =sin
x . The width w of the flat metal sheet needed
14
7
to make the panel is the arc length of the sine curve from x=0 to x=28 . We set up the integral to
dy
evaluate w using the arc length formula with
= cos
x :
dx 7
7
28 L= 14 2 1+ cos 2 dx . This integral would be very
7
7
7
7
0
0
difficult to evaluate exactly, so we use a CAS, and find that L 29.36 inches.
36. (a) y=c+acosh x x
a dx=2 / y =sinh 1+ x
a cos x ( / ) 2=1+sinh 2 1+ y x
a =cosh 2 x
a . So 12 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth b L= cosh x
a 2 b b dx=2 cosh
0 x
a dx=2 asinh x
a b =2asinh 0 b
a (b) At x=0 , y=c+a , so c+a=20 . The poles are 50 ft apart, so b=25 , and L=51 51=2asinh (b/a) .
From the figure, we see that y=51 intersects y=2xsinh (25/x) at x 72.3843 for x>0 . So a 72.3843
and the wire should be attached at a distance of y=c+acosh (25/a)=20 a+acosh (25/a) 24.36 ft
above the ground. x 37. y= dy
3
= x 1 [ by FTC1]
dx 3 t 1 dt
1 4 3 L= 4 x dx= x
1 1 3/2 dx= 1+ dy
dx 2 =1+ ( 3 x 1 2 ) =x 3 2 5/2 4 2
62
x
= (32 1)=
=12.4
1 5
5
5 38. By symmetry, the length of the curve in each quadrant is the same, so we’ll find the length in the
2k 2k
2k
2k
2k 1/ ( 2k )
(in the first quadrant), so
y= 1 x
first quadrant and multiply by 4 . x +y =1 y =1 x
we use the arc length formula with
dy
1
2k 1/ ( 2k ) 1
2k 1
=
1 x
2kx
dx
2k
= x2k 1 1 x2k 1/ ( 2k ) 1
The total length is therefore ( ( ) ( ( ) ) ) 13 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.1 Arc Legth 1 L =4
2k 1+ 2k 1 x 0 2k 1/ ( 2k ) 1 2 (1 x ) 1 2 ( 2k 1 ) dx=4 1+x ( 1 x2k ) 1/k 2 dx Now from the graph, we 0 see that as k increases, the ‘‘corners’’ of these fat circles get closer to the points ( 1, 1 ) and
( 1, 1 ) , and the ‘‘edges’’ of the fat circles approach the lines joining these four points. It seems
, the total length of the fat circle with n=2k will approach the length of the
plausible that as k
perimeter of the square with sides of length 2 . This is supported by taking the limit as k
of the
2k 1/ ( 2k )
equation of the fat circle in the first quadrant: lim 1 x
=1 for 0 x<1 . So we guess that ( ) k lim L =4 2=8 .
k 2k 14 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution 2 3 2 ds= 1+(dy/dx) dx= 1+(1/x) dx 1. y=ln x 2 S= 2 (ln x) 1+(1/x) dx
1 2 2 2. y=sin x /2 2 ds= 1+(dy/dx) dx= 1+(2sin xcos x) dx S= 2 2 2 sin x 1+(2sin xcos x) dx [ by
0 (7)]
2 /4 2 ds= 1+(dy/dx) dx= 1+(sec xtan x) dx 3. y=sec x S= 2 2 x 1+(sec xtan x) dx [ by (8)] 0 4. y=e x 2 2x ds= 1+(dy/dx) dx= 1+e ln 2 S= dx 2 x 2x 1+e 2 dx [ by (8)] or 2 (ln y) 0 2 1+ ( 1/y ) dy 1 [ by (6)]
5. y=x 3 / 2 y =3x . So 2 / 2 ( ) S= 2 y 1+ y 2 dx=2 x 0 2
=
36 3 4 4 3 1+9x dx[u=1+9x ,du=36x dx] 0
145 u du= 2 3/2
u
3 18 1 145 = 1 27 ( 145 145 1 ) 2 6. The curve 9x=y +18 is symmetric about the x axis, so we only use its top half, given by y=3 x 2
3
9
2
. dy/dx=
, so 1+(dy/dx) =1+
. Thus,
4(x 2)
2 x 2
6 S= 2 3 x 2 2 =6 2
3 7. y= x
S= 2 y 3/2 1
x+
4 2 25
4 =4 2
2 1+ 6 6 1+(dy/dx) =1+ 1/ ( 2 x 9
4 9
1+
dx=6
4(x 2) dy
dx 2 ) 4 3/2 9
4 6 x+
2 3/2 =4 1/2 1
4 dx
125
8 27
8 =4 98
=49
8 2 =1+1/(4x) . So 9 dx= 2 9
x 2+ dx=6
4 x 1
1+
dx=2
4x 9 x+
4 1
dx
4 1 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution 2
3 =2 3/2 1
x+
4 4
3 = 4 2 /6 3 2 2 cos 2x 1+4sin 2x dx=2 1+u 2 ( 1
2
ln u+ 1+u
2 2 1+u + 2 2 1 1 ) du[u=2sin 2x,du=4cos 2xdx] 3
0 3
1
2+ ln
2
2 ( 3 +2 ) x
x x
1
+
+ +
3
12 12 4
4x = 3 dx 2 = 2 + 4 ln ( 2+ 3 ) 2 1
(1+cosh 2x)dx=
2 cosh xcosh xdx=2
0 0 1+ 1
sinh 2
2 3 x
1
10. y= +
6 2x or 1+ 2 dy x
=
dx 2 1 2
e e
4 ( 2 0 ) 2 2x 4 1+(dy/dx) = 1 1
x+ sinh 2x
2 1
2 2 x
1
1
+ +
=
4
4 2
4x 2 x
1
+
2
2
2x 1 2 x
1
= +
2
2
2x 1
3 2 x
1
+
6 2x S= 2
1/2
1 5 1/2 3 1 1
+
72 6 =2 1 2
y +2
3 ( ) 3/2 5 x
1
+
2
2
2x x
x x
+ +
12 3 4 =2 11. x= 37 17 17 ) 1+(dy/dx) =1+sinh x=cosh x . So 9. y=cosh x
S=2 ( 37 6 1
4 2 0 1
u
2 = = 4
2 0 = 9 1
3/2
(4x+1)
8 ds= 1+(dy/dx) dx= 1+( 2sin 2x) dx 8. y=cos 2x S= 9 dx=2
1/2
6 dx=2 1
8
dx/dy= 1
64 72
1 2
y +2
2 ( 2 x
x
+
72 6
+ 1
24 1
2 ) 1/2(2y)=y 1 2 x
8 1/2 =2
2 y +2 263
512 = 263
256
2 2 1+(dx/dy) =1+y ( y2+2) = ( y2+1) 2 . So
2 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution 2 S=2
1
2 2 12. x=1+2y 2 2 y 1+16y dy= 1 = 24 ( 65 3 13. y= x x=y x 3 = 21
2 16 1 2
2
16y +1
3 ( ) 3/2 2
1 2 4 1+(dx/dy) =1+9y . So
2 2 1+(dx/dy) dy=2
2
4
1+9y
3 ( 2 y 2
1+9y dy=
36 3 4 2 ) 3/2 = ( 145 27 1 2 14. y=1 x 2 4 3 1+9y 36y dy
1 ) 145 10 10 2 1+(dy/dx) =1+4x
1 S=2 1 2 x 1+4x dx= 2 15. x= a y 2 y 2 1+(dx/dy) =1+ 2 2 a y = 4x +1 dx= 1 2 2
(a y )
2 1/2 2 2 2 2 a y + a y 2 2
2
4x +1
3 ( ) 3/2 1 = 0 6 (5 5 1) 2 ( 2y)= y/ a y
2 y
2 4 0 2 dx/dy= 2 8x 4 0
2 1
2 2 1 18 1 ( 16y2+1) 1/232ydy= 16 1 = 1
4 4+2 65 17 17 ) 2 S=2 2 =2 1+(dx/dy) =1+(4y) =1+16y . So 2 S=2 2 1 4 1 2
y+ y
4
2 2
y ( y +1 ) dy=2 2 a y a = 2 2 a y a/2 S= 2 2 2 a y 2 0 a/2 a dy=2
2 ady=2 a y
0 a y a/2
0 =2 a a
2
0 = a . Note that this is
2 1
1
the surface area of a sphere of radius a , and the length of the interval y=0 to y=a/2 is
the
4
4
length of the interval y= a to y=a .
16. x=acosh (y/a) 2 2 2 1+(dx/dy) =1+sinh (y/a)=cosh (y/a) . So
3 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution a S=2 y
a acosh
a cosh a
y+ sinh
2 =2 a 17. y=ln x dy=4 a cosh =2 a 0 2 2y
a 1+cosh
0 3 2 1+1/x . Since n=10 , dy=2 a a
2
a+ sinh 2 =2 a
2 1+(dy/dx) =1+1/x 2 Let f (x)=ln x a y
a 2 0 a 2y
a dy/dx=1/x a y
a 1
1+ sinh 2
2 dy
2 or a ( e2+4 e 2)
2 2 S= 2 ln x 1+1/x dx . 1 3 1 1
= . Then
10 5 x= 1/5
f (1)+4 f (1.2)+2 f (1.4)+
+2 f (2.6)+4 f (2.8)+ f (3) 9.023754 .
10
3
The value of the integral produced by a calculator is 9.024262 (to six decimal places). S S =2 18. y=x+ x dy/dx=1+ 1
x
2 1/2 2 1+(dy/dx) =2+x 1/2 + 1
x
4 1 2 1
1
1
1
+
dx . Let f (x)=(x+ x ) 2+
+
.
x 4x
x 4x
1
2 1 1
Since n=10 , x=
=
. Then
10 10
1/10
S S =2
f (1)+4 f (1.1)+2 f (1.2)+
+2 f (1.8)+4 f (1.9)+ f (2) 29.506566 .
10
3
The value of the integral produced by a calculator is 29.506568 (to six decimal places).
S= 2 ( x+ x 19. y=sec x ) 2+ dy/dx=sec xtan x /3 S= 2 2 2 2 1+sec xtan x dx . Let f (x)=sec x 2 sec x 2 1+(dy/dx) =1+sec xtan x
2 2 1+sec xtan x . 0 Since n=10 , x=
/30
3 S S =2
10 /3 0
=
. Then
10
30
f (0)+4 f 2
30 +2 f 30 + +2 f 8
30 +4 f 9
30 +f 3 13.527296 .
The value of the integral produced by a calculator is 13.516987 (to six decimal places).
x 1/2 20. y=(1+e ) dy 1
x
= (1+e )
dx 2 1/2 x e= e x
x 1/2 2(1+e ) 4 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution 1+ 2x 2 dy
dx x e =1+ = x 2x 4+4e +e = x 4(1+e ) x 4(1+e ) 2 (e +2) x 4(1+e ) 1 S= 2 1+e 1 x e +2 x 2 0 dx= 1+e x x 1 x (e +2)dx= e +2x = [(e+2) (1+0)]= (e+1) .
0 0 1 x
1 0 1
(e +2) . Since n=10 , x=
=
. Then
2
10 10
1/10
S S =2
f (0)+4 f (0.1)+2 f (0.2)+
+2 f (0.8)+4 f (0.9)+ f (1) 11.681330 .
10
3
The value of the integral produced by a calculator is 11.681327 (to six decimal places).
Let f (x)= 2 2 ( 2 ) ds= 1+(dy/dx) dx= 1+ 1/x 21. y=1/x 4 dx= 1+1/x dx 2
2 S= 1
x 2 1+ 1
x 1 x +1 dx=2 4 1+u
u 1 x 1 4 = 4
4 ( 17
2
+ln ( 4+ 17 ) +
ln ( 1+ 2
4
1 = dy
=
dx 2 22. y= x +1 x ds= 2 u 1 1+u
2
+ln u+ 1+u
u du= 2 dx=2 1
2 2 2 du[u=x ,du=2xdx] 4 2 2 3 2 u +1 x +1 1 = dy
dx 1+ ) ) 17
+ln
4 2 2 4+ 17
1+ 2 2 dx= 1+ x
2 dx x +1 3 S = 2 2 2 x +1 1+ 2 1
x
2 2 0 x+ 2 2x +1 dx=2 2 dx=2 1 1
x + + ln
2 4
2 3 3 x +1 0 =2 2 x x+
0 1
x+
2
2 1
2 2 dx 3
0 5 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution =2 2 3
2 =2 2 3
2 9+
19 1 1
+ ln
2 4 3 / 23. y=x and 0 y 1
1 2 2 ( ) S= 2 x 1+ 3x 3 = 3 dx=2 1+u 0 3
1
10 + ln ( 3+ 10
2
2 24. y=ln (x+1) , 0 x 1 . ds=
1 = =3 19 + 2 3
2 =2 2 19 1
+ ln
2
4 19
2 3+ ln ( 3 2 + 19 ) 1
2
du[u=3x ,du=6xdx]
6 2 1
u
2 2 1+u du= [or use CAS ] 3 1
1
ln
4
2 2 0
3 1
2 y =3x and 0 x 1 . 0 = 9+ 1
ln ( 3 2 + 19 )
4 + 2 3+ ) 3
= 3 10 +ln ( 3+ 10 6 2 dy
dx 1+ ( 1
2
ln u+ 1+u
2 2 1+u + dx= 0 ) 1
x+1 1+ 3 ) 2 dx , so S 2 2 x 1 1+ ( x+1 ) 0
2 =2
1 2 (u 1) u
2 2 du[u=x+1,du=dx]
2 2 1+u
du=2
u 1 1 1+ 1
2 2 1+u
du 2
u u dx= 2 2 1+u
du
u 2 1+u du 2
1 1 2
1
2 1
2
= [or use CAS ] 2
u 1+u + ln u+ 1+u
2
1
2
2
1
1
1
5 ln
=2
5+ ln ( 2+ 5 )
2
ln ( 1+ 2 )
2
2
2
2
1+ 5
2 3
1
=2
ln ( 2+ 5 ) +ln
+
ln ( 1+ 2 )
2
2
2 2 ( 25. S=2 y
1 1+ dy
dx 2 1
x dx=2
1 ) 1+ 1
x 4 1+u
1+ 5
2 2 ln 1+ 1+u
u 2 2
1 2 +ln ( 1+ 2 ) 4 x +1 dx=2
1 x 3 dx . Rather than trying to 6 + 1
4 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution 4 4 2 x +1 > x =x for x>0 . Thus, if the area is finite, evaluate this integral, note that 4 2 x +1 S=2 x 1 x dx>2 3 x 1 dx=2 3 1 1
dx
x But we know that this integral diverges, so the area S is infinite.
2 26. S= 2 y 1+(dy/dx) dx=2 0 x2 1+( e ) dx . 0 Evaluate I= e
I= x e x x x2 x 1+( e ) dx by using the substitution u= e , du=e dx . ( ) 1
2 1
2
2
1+u du = 2 u 1+u + 2 ln u+ 1+u +C
1
x
2x 1
x
= ( e ) 1+e + ln e + 1+e
2
2 ( ) +C 2x Returning to the surface area integral, we have
t S =2 lim
t e 1
x
( e )
2 t =2 lim
=2
=2 {
{ x2 1+( e ) dx 0 =2 lim t x { 1+e 1
t
( e )
2 2x 1+e +
2t ( e + 1+e )
1
+ ln ( e + 1+e )
2 1
ln
2 1
1
(0) 1 + ln ( 0+ 1
2
2
1
[0]+
2 ln ( 2 1 )
2
2 x 2x t } =
2 2 ln 0 1
1
( 1) 1+1 + ln
2
2 2t 1
1
2 + ln
2
2 ) t ( ( 1+ 2 ) ( 1+ 1+1 ) } } 2 1)
2 2 27. Since a>0 , the curve 3ay =x(a x) only has points with x 0 . (3ay 0 x(a x) 0 x 0 . )
The curve is symmetric about the x axis (since the equation is unchanged when y is replaced by y ).
y=0 when x=0 or a , so the curve’s loop extends from x=0 to x=a . 7 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution d
d
2
(3ay )=
dx
dx 2 (a x) (a 3x) = 2 2 2 = dy (a x)[ 2x+a x]
=
dx
6ay 36a is 1/y x(a x)
2 (a 3x)
=
12ax 2 2 2 2 the last fraction 3a 2 2 2 2 (a x) (a 3x) 36a y 2 2 2 2 a 6ax+9x
12ax a 6ax+9x
a +6ax+9x (a+3x)
=1+
=
+
=
=
for x 0 .
12ax
12ax
12ax
12ax
12ax dy
dx 1+ dy
2
=x 2(a x)( 1)+(a x)
dx 2 2 dy
dx 6ay (a)
a S= a x (a x) 2 yds=2 x=0 3a 0
a a+3x
dx=2
12ax a
0 (a x)(a+3x)
dx
6a
2 a
=
(a +2ax 3x )dx=
a x+ax x =
(a +a a )=
a=
.
0 3a
3a 0
3a
3a
3
2 2 2 2 3 a 3 3 3 3 Note that we have rotated the top half of the loop about the x axis. This generates the full surface.
(b) We must rotate the full loop about the y axis, so we get double the area obtained by rotating the
top half of the loop:
a a a+3x
4
S =2 2
xds=4 x
dx=
12ax
2 3a
x=0
0
2
=
3a
= 2 a 3/2 0 3
3 1/2 (ax +3x )dx=
2 6
+
3 5 2 a= 2
3a 2 a 1/2 x (a+3x)dx
0 2 3/2 6 5/2
ax + x
3
5 3 28
15 3 2 a= 56 a = 2 0 3 3 a 2 5/2 6 5/2
a + a
3
5 2 3a
45 2 28. In general, if the parabola y=ax , c x c , is rotated about the y axis, the surface area it
generates is
c 2 x
0 2 2ac 1+(2ax) dx = 2
0 u
2a 1+u 2 1
du u=2ax,du=2adx
2a 8 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution 2ac = 2 4a
= 2 ( 1+u2) 1/22udu= 2
2
1+u
3 ( 2 0 4a ) 3/2 2ac
0 ( 1+4a2c2) 3/2 1 6a 2
. Thus, the surface area is
25 2 Here 2c=10 ft and ac =2 ft, so c=5 and a= 3/2
625
4
625
1+4
25
1 =
6 4
625
24
5
=
( 41 41 125) 90.01ft2
24 S= 2 29. x 2 2 + a
1+ y 2 y(dy/dx) =1 2 b dy
dx b 3/2 16
25 1 = 625
24 41 41
1
125 2 dy
b x
= 2
dx
a y x = 1+ 2 a ( 1 x2 /a2) = a4b2+b4x2 a2b2x2
1+
=
=
4 2
4 2
4 2
2 2
4 2 2 2 2
a y
a y
a b ( 1 x /a )
ab ab x
4 2 2 2 2
4
2 2 2
a +b x a x
a (a b ) x
=
4 2 2
2 2 2
a a x
a (a x )
4 2 2 b x = = 4 2 4 2 b x +a y 4 2 4 2 b x +a b The ellipsoid’s surface area is twice the area generated by rotating the first quadrant portion of the
ellipse about the x axis. Thus,
a S= a 2 2 y 1+ 0 dy
dx 2 b
a dx=4 a 2 a x 4 b
2 a a a 4 0 4 b =
2 a 2 2 a b 2 u
2 4 a
a u + sin
2
4 2 ( a2 b2) x2
2 dx 2 a x 2 a b ( a2 b2) x2 dx= 4 2b
a 4 a 0
a = 2 4 a u du 2 2 2 2 a b 0
1 2 [u= a b x] u
2 a a 2 2 a b 0 9 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution 2 2 2 a 2 a 4 b = 2 a b
2 2 a b 4 2 a a (a b )
2 2 2 4 a
+ sin
2 2 2 a b
a 1 2 =2 a bsin 1 b+
2 2 a b
a
2 a b
2 2 2 30. The upper half of the torus is generated by rotating the curve (x R) +y =r , y>0 , about the y
dy
axis. y = (x R)
dx 1+ 2 2 dy
dx (x R) =1+ 2 = 2 2 y +(x R) = 2 y r y r 2 2 ( x R) 2 . Thus, R+r S= R+r 2 2 x 2 dy
dx 1+ Rr rx dx=4
2 dx
2 r (x R) Rr
r = u+R 4 r 2 r u r du[u=x R]
2 r = r udu 4 r 2 r u r du +4 Rr
2 2 r u r 2 r = du 4 r 0+8 Rr 2 [since the first integrand is odd and the second is even ] r u 0 2 r 1 = 8 Rr sin (u/r) =8 Rr
0 31. The analogue of f x * i n S=lim 2 c f x i i=1 n 1/2 32. y=x * / y = 1
x
2 1/2 =4 2 2 Rr in the derivation of (4) is now c f x * i 1+ f / x * i 2 b x= 2 c f (x) , so
/ 1+ f (x) 2 dx . a ( / ) 2=1+1/4x , so by Exercise 31, 1+ y 10 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution 4 S= 2 (4 x ) 1+1/(4x) dx . Using a CAS, we get S=2 ln ( 17+4 ) + 0
2 2 / 2 ( 31 6 17+1 ) 80.6095 . 2 33. For the upper semicircle, f (x)= r x , f (x)= x/ r x . The surface area generated is
r S =
1 r 2 (r 2 2 r x 2 ) 1+ x
2 2 (r dx=4 r x r 2 2 r x ) r
2 dx
2 r x 0 r = r 4 2 2 r dx 2 r x 0 r For the lower semicircle, f (x)= 2 / 2 r x and f (x)= x r , so S =4 2 2 2 2 2 r x +r dx . 2 r x 0 Thus, the total area is
r r S=S +S =8
1 2 2 dx=8 2 2 r x 0 2 r sin 1 x
r r =8 r 0 2 2 2 2 =4 r . 1 2
d and let the intersecting planes be y=c and y=c+h , where
4
1
1
2 2 1 2
d c
d h . The sphere intersects the xy plane in the circle x +y = d . From this equation,
2
2
4
dx
dx
y
we get x
+y=0 , so
=
. The desired surface area is
dy
dy
x
2 2 2 34. Take the sphere x +y +z = 11 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.2 Area of a Surface of Revolution c+h S=2 xds=2 c+h 2 x 1+(dx/dy) dy=2 x c
c+h =2
c 2 c+h 2 1+y /x dy=2 c 1
d dy= d
2 2 2 x +y dy
c c+h dy= dh
c 35. In the derivation of (4), we computed a typical contribution to the surface area to be
y +y
i 1 i
2
P P , the area of a frustum of a cone. When f (x) is not necessarily positive, the
i 1 i
2 ( ) approximations y = f x
i ( ) y= f x
i i f x i f x * i 1 2 i i 1 i i 1 ( ) i 1 2 f x 1+ f i b as before, we obtain S= 2 / f (x) * 1+ f (x) x 2 * i 2 must be replaced by . Thus, i / * i f x i 1 * f x i 1 and y = f x i P P ( ) and y = f x i y +y
2 * x . Continuing with the rest of the derivation dx . a
/ / 36. Since g(x)= f (x)+c , we have g (x)= f (x) . Thus,
b S =
g 2 g(x) / 1+ g (x) 2 a 2 f (x)
a dx= 2 2 / f (x)+c 1+ f (x) dx a b = b / 1+ f (x) 2 b dx+2 c / 1+ f (x)
a 2 dx=S +2 cL
f 12 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 3 1. The weight density of water is =62.5 lb / ft . ( 62.5lb/ft3) ( 3ft ) =187.5 lb / ft 2
2
(b) F=P A ( 187.5lb/ft ) ( 5ft ) ( 2ft ) =1875 lb. ( A is the area of the bottom of the tank.) (a) P= d (c) As in Example 1, the area of the i th strip is 2( x) and the pressure is d= x . Thus,
i 3 3 F= 1 2
x
2 x 2dx (62.5)(2) xdx=125
0 0 3 9
2 =125 0 =562.5lb 4 2. (a) P= gd=1030(9.8)(2.5)=25 , 235 2.52 10 Pa =25.2 kPa ( 2.52 (b) F=P A 4 2 10 N/m ) ( 50m2) =1.26 2.5 (c) F= 6 10 N 2.5 gx 5dx=(1030)(9.8)(5)
0 4 xdx 2.52 10
0 2 2.5 x 0 5 1.58 10 N 3. Set up a vertical x axis as shown, with x=0 at the water’s surface and x increasing in the
downward direction. Then the area of the i th rectangular strip is 6 x and the pressure on the strip is
* 3 x (where i n hydrostatic force F = lim
n =6 * 62.5lb/ft ). Thus, the hydrostatic force on the strip is x 6 x and the total i n
i=1 * x 6 x . The total force
i i=1 6 * x 6 x= 1 2
x
2 i 6 x 6dx=6
2 xdx
2 6 =6 (18 2)=96 2 6000lb 4. Set up a vertical x axis as shown. Then the area of the i th rectangular strip is
* * i i pressure on the strip is x , so the hydrostatic force on the strip is x 4
*
(4 x ) x . The
i
3 4
*
(4 x ) x and the total
i
3 force on the plate
1 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering n
i=1 x 4
*
(4 x ) x . The total force
i
3 * i F =lim
= * i=1 n 4
3 4 4 4
4
4
*
x
(4 x ) x= x (4 x)dx=
i 3
i
3
3
1 n 2 2x 4 1 3
x
3 = 1 4
3 32 64
3
2 2
2 2 (4x x )dx
1 1
3 = 4
(9)=12
3 750 lb 2 2 5. Since an equation for the shape is x +y =10 ( x 0 ), we have y= 100 x . Thus, the area of the i
*2 th strip is 2 100 (x ) x i * and the pressure on the strip is gx , so the hydrostatic force on the
i * *2 strip is gx 2 100 (x ) i n plate gx 2 n
i=1 n
10 = * i=1 F = lim g i * gx 2
i x and the total force on the *2 100 (x )
i 2000
g
3 x . The total force
10 *2 100 (x ) x= 2 gx i ( 100 x2) 1/2( 2x)dx= 0 = i 2 100 x dx 0 g 2
2 3/2
(100 x )
3 2000
6
1000 9.8 6.5 10 N
3 10
0 = 2
g(0 1000)
3
3 2 1000kg/m andg 9.8m/s . 2 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering * 6. By similar triangles, w /4=x /5 , so w =
i i i 4 *
x and the area of the
5 i 4 *
*
x x . The pressure on the strip is gx , so the hydrostatic
i
5 i
n
* 4 *
* 4 *
force on the strip is gx
x x and the total force on the plate
gx
x x . The total
i 5 i
i 5 i
i=1
force i th strip is n F =lim i=1 n gx * i 4 *
x x=
5 i 5
0 gx 4
4
xdx=
g
5
5 1 3
x
3 5 = 0 4
125 100
g
=
g
5
3
3 100
5
1000 9.8 3.3 10 N.
3
4 ft wide a ft wide
1 *
= *
, so a= x and the width of the i th rectangular
8 ft high
2 i
x ft high 7. Using similar triangles, i * * i i strip is 12+2a=12+x . The area of the strip is 12+x
F = lim
n
8 = n
i=1 x * i 12+x 0 = (62.5) 1664
3 i 8 * x= i ( 12x+x ) dx=
2 * x . The pressure on the strip is x . x (12+x)dx
0
3 x
6x +
3
2 8 = 0 384+ 512
3 4 3.47 10 lb 8. In the figure, deleting a b h rectangle leaves a triangle with base a b and height h . By similar
triangles,
3 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering ( a b ) ft wide
h ft high d ft wide = h x * , so the width of the triangle is ft high i h x * x i d= (a b)= h x
x i (a b)=a b h * i h (a b) h (a b) . The area of the i th rectangular strip is * i a x * i and the width of the trapezoid is b+d=a 1 * * (a b) h n F = lim i=1 n x and the pressure on it is gx .
i gx x * i a i * h h h (a b) x= gx a
0 2 h x
(a b) dx
h
3 g(b a) 2
h
b a h
= ga xdx+
x dx= ga + g
h
2
h 3
0
0
a b a
+
2
3 2 = gh 2 = gh a+2b
6 500 2
gh (a+2b)N
3 9. From the figure, the area of the i th rectangular strip is 2
* g x +r
i F = lim
n
r = r 2 x * 2 i x and the pressure on it is .
n
i=1 * g x +r 2
i 2 r 2 x * 2 i x 2 g(x+r) 2 r x dx
r 4 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering r = g 2 2 r 2 r x 2xdx+2 gr
r 2 r x dx
r The first integral is 0 because the integrand is an odd function. The second integral can be interpreted
as the area of a semicircular disk with radius r , or we could make the trigonometric substitution
1 2
3
3
x=rsin . Continuing: F= g 0+2 gr
r = g r =1000g r N (SI units assumed).
2 10. The area of the i th rectangular strip is 2
8 F= 8 (8 y)2 2y dy=42 2 1/2 2 (8 y)y 0 * 2y i y and the pressure on it is d =
i * 8 y i . dy 0
8 = 84 2 ( 8y1/2 y3/2) dy=84 0 2 8 2 3/2 2 5/2
y
y
3
5 8
0 2
2
16 2
128 2
3
5
1 1
2
= 84 2 256 2
=43,008
=5734.4lb
3 5
15
= 84 2 8 5 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 11. By similar triangles, 2x w 8
4 3 i =
x w=
i * * i 3 2x
. The area of the i th rectangular strip is * i 3 x and i the pressure on it is g 4 3 x * . i 4 3 F=
0 4 3 2x
g ( 4 3 x)
dx=8 g
3
2 g 3
x
3 3 2 4 3 =4 g x 0 4 3
0 0 2 g
xdx
3 =192 g 4 3 2 x dx
0 2 g
64 3 3
3 3
5 =192 g 128 g=64 g 64(840)(9.8) 5.27 10 N 12.
2 F= 2 4 x dx g(10 x)2
0
2 2 4 x dx =20 g 2 0 1
=20 g
4 2 4 x 2xdx g
0 (2 )
2 4 g u 1/2 2 du [ u=4 x , du= 2xdx ] 0
6 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 2
3/2 4
=20
g u
0
3
16
=(1000)(9.8) 20
3
=20 g g 16
g= g
3 20 16
3 5 5.63 10 N 13. (a) The top of the cube has depth d=1 m 20 cm =80 cm =0.8 m.
2 F= gdA (1000)(9.8)(0.8)(0.2) =313.6 314N * (b) The area of a strip is 0.2 x and the pressure on it is gx .
i 1 F= 1 2
x
2 gx(0.2)dx=0.2 g 0.8 1 =(0.2 g)(0.18)=0.036 g=0.036(1000)(9.8) 0.8 =352.8 353N
2 2 14. The height of the dam is h= 70 25 cos 30 =15 19
8, the width of the trapezoid is 100
figure, cos 30 =
h F= x
0 200
=
3 x
z 50x
100
h
2 h
2 3
2 . From the solution for Exercise x
50x
(100 50)=100
. From the small triangle in the second
h
h z= xsec 30 =2 x/ 3 .
2
200
dx=
3
3
3 h 100
xdx
h 3
0 h 2 x dx
0 2 100 h 200 h 200(62.5) 12,825
=
=
4
h 3 3
3 3
3 3 6 7.71 10 lb 7 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 15. (a) The area of a strip is 20 x and the pressure on it is x .
i 3 F= 1 2
x
2 x20dx=20
0 3 =20 0 9
=90
2 = 90(62.5)=5625lb 5.63 103lb 9 1 2 9
81
4
x
=20
=810 =810(62.5)=50 , 625 lb 5.06 10 lb.
2
0
2
0
(c) For the first 3 ft, the length of the side is constant at 40 ft. For 3<x 9 , we can use similar
a 9 x
9 x
=
a=40
triangles to find the length a :
.
40
6
6
(b) F= x20dx=20 3 F= 9 x40dx+
0 = 180 + 3 20
3 9 x
x(40)
dx=40
6
9 2 1 3
x
x
2
3 1 2
x
2 9 =180 + 3 20
3 3 20
+
0 3 9 ( 9x x2) dx 3 729
243
2 81
9
2 = 180 +600 =780 =780(62.5)=48,750lb 4.88 104lb 8 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering (d) For any right triangle with hypotenuse on the bottom, csc =
2 9 F= x20
3 hypotenuse 2 40 +6
409
=
6
3 = xcsc = x x
hypotenuse x. 409
1
dx= ( 20 409 )
3
3 9 1 2
x
2 3 1
10 409 (81 9)
3 = 5 303,356lb 3.03 10 lb 16. Partition the interval a,b by points x as usual and choose x
i * x ,x i 1 i i for each i . The i th horizontal strip of the immersed plate is approximated by a rectangle of height
, so its area is A i P i w x * x . For small
i i x and width w x
i i i gx by Equation 1. The hydrostatic force F acting on the i th strip is F =P A
i i i Adding these forces and taking the limit as n
plate: F=lim
n F =lim i=1 i n i x , the pressure P on the i th strip is almost constant and * n i i * * i i gx w x x .
i , we obtain the hydrostatic force on the immersed n
i=1 * * i i gx w x b x=
i gxw(x)dx
a 5 17. F= * gx w(x)dx , where w(x) is the width of the plate at depth x . Since n=6 ,
2 x= 5 2 1
= ,
6
2 and
F S = g 1/2 [2 w(2)+4 2.5 w(2.5)+2 3 w(3)+4 3.5 w(3.5)]
6
3
+2 4 w(4)+4 4.5 w(4.5)+5 w(5)]
9 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 1
g(2 0+10 0.8+6 1.7+14 2.4+8 2.9+18 3.3+5 3.6)
6
1
5
= (1000)(9.8)(152.4) 2.5 10 N
6
= b b 1
18. (a) From Equation 8, x=
xw(x)dx
Aa b Ax= xw(x)dx gAx= g xw(x)dx a a b ( gx)A= gxw(x)dx=F by Exercise 16.
a (b) The centroid of a circle is its center. In this case, the center is at a depth of r meters, so x=r .
2 3 Thus, F=( gx)A=( gr)( r )= g r .
19. The moment M of the system about the origin is M=
The mass m of the system is m= 2 2 m x =m x +m x =40 2+30 5=230 . i=1 i i 1 1 2 2 m =m +m =40+30=70 . The center of mass of the system is i=1 i 1 2 230 23
M/m=
=
.
70
7
20. M=m x +m x +m x =25( 2)+20(3)+10(7)=80 ;
1 1 2 2 x=M/(m +m +m )=
1 2 3 21. m= 3 3 3 80 16
=
.
55 11 m =6+5+10=21 . M = i=1 i x 3 m y =6(5)+5( 2)+10( 1)=10 ; i=1 i i M M
1
x 10
M=
m x =6(1)+5(3)+10( 2)=1 . x=
=
and y=
=
, so the center of mass of the
y
i=1 i i
m 21
m 21
1 10
system is
,
.
21 21
3 22. M =
x m= 4 4 m y =6( 2)+5(4)+1( 7)+4( 1)= 3 , M = i=1 i i y M
m =16 , so x= i=1 i y y m = 4 m x =6(1)+5(3)+1( 3)+4(6)=42 , and i=1 i i M
42 21
x
3
and y=
; the center of mass is ( x,y ) =
=
=
16 8
16
m 21
3
,
8
16 . 23. Since the region in the figure is symmetric about the y axis, we know that x=0 . The region is
‘‘bottom heavy,’’ so we know that y<2 , and we might guess that y=1.5 . 10 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 2 2 2 2 A = (4 x )dx=2 (4 x )dx=2
2 4x 0 =2 8
3 8 = 1 3
x
3 2
0 32
3 2 1
2
2
x=
x(4 x )dx=0 since f (x)=x(4 x ) is an odd function (or since the region is symmetric about the
A 2
y axis). 2 1
y=
A
= 2 2 1
3 1
3
22
2 4
(4 x ) dx=
2 (16 8x +x )dx=
2
32 2 0
32 3
32 32 64 32
+
3
5 =3 2 1
+
3 5 1 Thus, the centroid is ( x,y ) = 0, 8
5 =3 8
15 16x
= 8 3 1 5
x+ x
3
5 2
0 8
5 . 24. The region in the figure is ‘‘left heavy’’ and ‘‘bottom heavy,’’ so we know x<1 and y<1.5 , and
we might guess that x=0.7 and y=1.2 .
3
3x+2y=6 2y=6 3x y=3
x.
2
2 A= 3
0
2 1
x=
x
A0
1
=
3 3
3 2
x dx= 3x
x
2
4
3
1
3
x dx=
2
3
3 2 1 3
x
x
2
2 2 = 0 2 =6 3=3 . 0 2 3x
0 3 2
x dx
2 1
2
(6 4)= ;
3
3
11 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 1
y=
A 2 1
2 3
3
x
2 2 2 1 1
9 2
1
dx=
9 9x+ x dx=
3 2 0
4
6
0
2
Thus, the centroid is ( x,y ) =
,1 .
3 9x 9 2 3 3
x+ x
2
4 2 = 0 1
( 18 18+6 ) =1 .
6 25. The region in the figure is "right heavy’’ and "bottom heavy,’’ so we know x>0.5 and y<1 , and
we might guess that x=0.6 and y=0.9 .
1 x A= e dx= e x 1 0 =e 1 , 0 1 1
1
x
x x
x=
xe dx=
xe e
A0
e 1
= 1
e 1 1
y=
A 1 0 ( 1) = 1
e 1 1
0 [ by parts] , 1 x 2
1 1 2x 1
1
e dx=
e
=
0 4 ( e 1)
e 1 4
0 2
1 e+1
( x,y ) = e 1 , 4
( 0.58,0.93) . ( ) ( e2 1) = e+1 . Thus, the centroid is
4 26. The region in the figure is "left heavy’’ and "bottom heavy,’’ so we know x<1.5 and y<0.5 , and
we might guess that
12 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering x=1.4 and y=0.4 .
2 2 1
1
1
1
1
2
2 1
A=
dx= ln x =ln 2 , x=
x
dx=
x = =
,
1
1 A ln 2
A1 x
A
1 x
1
y=
A
= 2 1
2 1 2 1
x 1
2ln 2 2 1
1
2
dx=
x dx=
2A 1
2A 1 ( 2
1 1
1
+1 =
.
2
4ln 2
1
1
,
ln 2 4ln 2 Thus, the centroid is ( x,y ) = 27. A= 1
x 2 3/2 1 2
x
x
3
2 x x ) dx= 0
1 1 = 0 2
3 ( 1.44,0.36 ) . 1 1
= .
2 6 1 1
3/2 2
x=
x( x x)dx=6 (x x )dx
A0
0
=6 1
y=
A
=3 1 2 5/2 1 3
x
x
5
3
1
0 1
2 ( x ) 1 2 1 3
x
x
2
3 2 2
5 =6 0 1
3 =6 1
15 = 2
;
5 1 1
2
x dx=6
(x x )dx
2 0
2 1 =3 0 Thus, the centroid is ( x,y ) = 1
2 1
3
2 1
,
5 2 = 1
.
2
. 13 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 28.
2 2 A = (x+2 x )dx=
1 = 2+4 1
x=
A 2 2
x(x+2 x )dx=
9
1 2
9
2
=
9 = 1
1
2+
2
3
2 1 3 2 1 4
x +x
x
3
4
8
+4 4
3 = 1
y=
A 8
3 2 2 = 1
9 1 9
.
2 ( x2+2x x3) dx 1
2
1 1
1
+1
3
4 1
2 1
2
22
(x+2) (x ) dx=
2
9 2 1 2 1 2
1 3
x +2x
x
2
3 8
32
+8+8
3
5 Thus, the centroid is ( x,y ) = 2
1 ( x2+4x+4 x4) dx= 1
9 1
1
+2 4+
3
5
1 8
,
2 5 (cos x sin x)dx= sin x+cos x
0 = 1
9 18+ 1 3 2
1 5
x +2x +4x
x
3
5
9
3 33
5 = 2
1 1 72 8
= .
9 5 5 . /4 29. A= 2 9 1
= ;
9 4 2 = /4
0 = 2 1, 14 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering x =A 1 /4 x(cos x sin x)dx
0 =A =A y=A 1 1 x(sin x+cos x)+cos x sin x 1 2 1 = 4 /4
0 1
4 ) 2 30. A= xdx+
0 1
x=
A
= 1
A 1
1 1
dx=
x 2 2 0 = 1 /4 cos 2xdx=
0 2 4 1 2 + ln x =
1 0 1
A 1 3
x
3 1
sin 2x
4A 1
2 1) , 4 ( 2 1) 4 ( 1 2
x
2 x dx+ 1dx [integration by parts] 2 1 Thus, the centroid is ( x,y ) =
1 0 2 1 1
1
2
2
cos x sin x dx=
2
2A ( /4 /4 = 0 1
1
=
4A 4 ( 2 1 ) (0.27,0.60) . 1
+ln 2 ,
2
1 + x 0 2
1 1
2
4
8
+1 =
=
,
3
1+2ln 2 3 3(1+2ln 2)
2 y=1
A
= 1
2A 1
0 1 2
x dx+
2
1 1
+
3 2 1
2 dx = 1
2A 2x
5
5
=
=
.
12A 6+12ln 2
1 1 3
x
3 1 + 0 1
x 2
1 15 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 8
5
,
(1.12,0.35) . The principle used in this
3(1+2ln 2) 6(1+2ln 2)
problem is stated after Example 3: the moment of the union of two nonoverlapping regions is the sum
of the moments of the individual regions.
Thus, the centroid is ( x,y ) = 31. From the figure we see that y=0 . Now
5 2
3/2
(5 x)
3 A = 2 5 x dx=2
0 =2 0+ 5
0 2 3/2
20
5
=
5
3
3 so 5 5 1
x=
x
A0
1
=
A
= 4
A 5 x 0 ( 2 5 u 2 ( 5 x ) 1
dx=
2x 5 x dx
A0 ) u( 2u)du[u= 2 2 5 x ,x=5 u ,u =5 x,dx= 2udu] 5
5 u
0 2 ( 5 u2) du= 4
A 5 3 1 5
u
u
3
5 5
0 = 3
5 5 25
5 5 5
3 =5 3=2 Thus, the centroid is (x,y)=(2,0) .
32. By symmetry, M =0 and x=0 ;
y 16 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering A= 1
2 2 1 +4 , so m= A=5( +4)= 2 1 1 5
( +8) ;
2 1
1
1 3
10
50
2 2
2
2
M= 2
[( 1 x ) ( 2) ]dx=5
x 3 dx= 5
x +3x = 5
=
;
x
2
3
0
3
3
0
0
1
2
50
20
20
y= M =
=
. Thus, the centroid is ( x,y ) = 0,
.
m x 5( +8) 3
3( +8)
3( +8) ( 33. By symmetry, M =0 and x=0 . A=
y 1 M =
x
1 1
2
(2 2x) dx=2
2
1 1 2
2 1
2
2 = 0 1
0 ) 1
1
bh= 2 2=2 .
2
2 1
2
(2 2x) dx
2
2 ( 1 x ) dx
0 2 = 4 u ( du)[u=1 x,du= dx]
1 =
y= 0 1 3
u
3 4 = 4 1 1
3 = 4
3 1
1
1 4 2
2
M=
M=
= . Thus, the centroid is ( x,y ) = 0,
m x A x 1 2 3 3
3 34. By symmetry about the line y=x , we expect that x=y . A=
r M=
x 0
r ; y= ( 2 2 r x
2 2 ) 2 r x r x dx= M=
y 1
2 0 1
2
M=
2
m x
r 1
dx=2
2 2 3
r
3 (r 2 x2) dx= 2 r x 0
r 1 3
x
3 r 1 2
1 2
r , so m= A=2A=
r .
4
2
2 3
r .
3 = 0 2
2 1
2 3
2
r . x= M =
y
2
m
0 3
0
r
4
4
4
=
r . Thus, the centroid is ( x,y ) =
r,
r .
3
3
3
2 1/2 (r2 x ) 0 r . 2xdx= u 1/2 du= 2 3/2
u
3 r = 2 3
4
r =
r
3
3 17 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 35.
2 A= x 2 0 4
ln 2 = 8
3 2 3 2
ln 2 ( 2 x ) dx=
x x
3 0 1
3
=
ln 2 ln 2 2 8
3 1.661418 . 2 1
1
x 2
x 3
x=
x(2 x )dx=
(x2 x )dx
A0
A0
x 1
=
A
1
A 8
ln 2 1
A 8
ln 2 =
= 1
y=
A
= 2
0 1
2 1 1
A 2 x
4 2 2 ( ln 2 )
4 ( ln 2 )
( ln 2 ) 2 0 1 ( ln 2 ) 2 2 2 32
5 2 1
(1.297453) 0.781
A 4 (2 ) ( x )
16
2ln 2 [use parts] 4+ 3 x 2 2 4 x x2
ln 2 1
dx=
A
1
2ln 2 2
0 = 1 2x 4
1 1
(2 x )dx=
2
A 2
1
A 3 15
4ln 2 16
5 36. The curves y=x+ln x and y=x x intersect at ( a,c )
( b,d ) ( 1.507397,1.917782 ) .
b A= 2x 2
2ln 2 5 x
5 2
0 1
(2.210106) 1.330
A ( 0.447141, 0.357742 ) and b ( x+ln x x +x) dx= ( 2x+ln x x3) dx
3 a a
2 = x +xln x x 1 4
x
4 b
a 0.709781
18 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering b 1
1
3
x=
x 2x+ln x x dx=
Aa
A
= 1
A 1
y=
A
= ( ) b ( 2x2+xln x x4) dx a 2 3 1 2
1 5
x + x ( 2ln x 1 )
x
3
4
5
b
a 1
2A 1
2 ( x+ln x )
2 x ln x 2 ( x x)
3 2 b
a 1
dx=
2A 1
(0.699489) 0.985501
A
b 2xln x+ ( ln x ) 6 4 x +2x dx a 1 2
1 7 2 5
2
x +x ( ln x ) 2xln x+2x
x+ x
2
7
5 Thus, the centroid is ( x,y ) 2 b
a 1
(0.765092) 0.538964
2A ( 0.986,0.539 ) . 37. Choose x and y axes so that the base (one side of the triangle) lies along the x axis with the
other vertex along the positive y axis as shown. From geometry, we know the medians intersect at a
2
point
of the way from each vertex (along the median) to the opposite side. The median from B
3
1
goes to the midpoint
(a+c),0 of side AC , so the point of intersection of the medians is
2
1
1
1
2 1
(a+c), b =
(a+c), b .
3 2
3
3
3 This can also be verified by finding the equations of two medians, and solving them simultaneously to
find their point of intersection. Now let us compute the location of the centroid of the triangle. The
1
area is A= (c a)b .
2 19 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 1
x=
A 0 c b
b
x
( a x ) dx+ x (c x)dx
a
c
a
0 = b
Aa = 2
a ( c a) 0 1 2 1 3
ax
x
2
3 + a b
Ac 3 1
=
A b
a
c 1 2 1 3
cx
x
2
3 = 0 0 ( a b
ax x dx+
c
2 b
Aa ) c ( cx x2) dx 0 1 3 1 3
b
a+ a +
2
3
Ac 1 3 1 3
c
c
2
3 3 a
2
+
6
c ( c a) c
1
a+c
2 2
=
(c a )=
6 3( c a)
3 and
1
y=
A
1
=
A
1
=
A 0
a 1
2 b
(a x)
a 2 0 b 2 2a 2 b 2 2a 2 (a 2 c 2 dx+
0
2 ) 2ax+x dx+ a 1
2 b
(c x)
c
2 c b 2c
1 3
a x ax + x
3
2 2 0 + a 2 2 dx ( c2 2cx+x2) dx 0
2 b 2c 2 1 3
c x cx + x
3
2 2 2 c
0
2 2 1 3
1
b
2
3 3 1 3
( c a) b b
b
a +a
a +
c c+ c
=
=
( a+c ) =
2
2
3
3
A
6
6
3
( c a) b
2a
2c
a+c b
Thus, the centroid is ( x,y ) =
,
, as claimed.
3 3
Remarks: Actually the computation of y is all that is needed. By considering each side of the triangle
1
in turn to be the base, we see that the centroid is
of the way from each side to the opposite vertex
3
and must therefore be the intersection of the medians.
The computation of y in this problem (and many others) can be simplified by using horizontal rather
than vertical approximating rectangles. If the length of a thin rectangle at coordinate y is (y) , then its
area is (y) y , its mass is (y) y , and its moment about the x axis is M = y (y) y . Thus,
1
=
A b 3 3 x In this problem, (y)= c a
( b y ) by similar triangles, so
b
20 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 1
y=
A b
0 c a
2
y(b y)dy=
2
b
b b
0 ( by y ) dy=
2 2
2 b 1 2 1 3
by
y
2
3 b = 0 2
2 b 3 b
b
=
6 3 Notice that only one integral is needed when this method is used.
38. Divide the lamina into three rectangles with masses 2 , 2 and 6 , with centroids 3
,1
2 , 1
3
and 2,
, respectively. The total mass of the lamina is 10 . So, using Formulas 5, 6,
2
2
and 7, we have
M
y
1 3
1
3
1
2
+2(0)+6(2) =
x=
=
mx=
(9) , and
2
10
m m i=1 i i 10
M
x
1 3
1
1
3
1
2(1)+2
+6
y=
=
m y=
=
(12) .
2
2
10
m m i=1 i i 10
9 6
,
.
Thus, the centroid is ( x,y ) =
10 5
0, 39. Divide the lamina into two triangles and one rectangle with respective masses of 2 , 2 and 4 , so
2
that the total mass is 8 . Using the result of Exercise 37, the triangles have centroids
1,
and
3
2
1
1,
. The centroid of the rectangle (its center) is 0,
. So, using Formulas 5 and 7, we
3
2
M
x
1 3
1
2
2
1
1
2
1
2
+2
+4
have y=
=
m y=
=
=
, and x=0 , since
3
3
2
8
3
12
m m i=1 i i 8
1
the lamina is symmetric about the line x=0 . Thus, the centroid is ( x,y ) = 0,
.
12
40. A sphere can be generated by rotating a semicircle about its diameter. By Example 4, the center of
4r
8r
=
, so by the Theorem of Pappus, the volume of the
mass travels a distance 2 y=2
3
3
r
sphere is V =Ad=
2 2 8r 4 3
=
r .
3 3 41. A cone of height h and radius r can be generated by rotating a right triangle about one of its legs
1
as shown. By Exercise 37, x= r , so by the Theorem of Pappus, the volume of the cone is
3
1
1
1
1 2
V =Ad=
base height (2 x)= rh 2
r =
r h.
2
2
3
3
21 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering 42. From the symmetry in the figure, y=4 . So the distance traveled by the centroid when rotating the
1
1
triangle about the x axis is d=2 4=8 . The area of the triangle is A= bh= (2)(3)=3 . By the
2
2
Theorem of Pappus, the volume of the resulting solid is Ad=3(8 )=24 . 43. Suppose the region lies between two curves y= f (x) and y=g(x) where f (x) g(x) , as illustrated in
* Figure 13. Choose points x with a=x <x <
i 0 <x =b and choose x to be the midpoint of the i th 1 n i 1
subinterval; that is, x =x =
x +x . Then the centroid of the i th approximating rectangle R is its
i 1 i
i
i
i 2
1
center C = x ,
f x +g x
. Its area is f x g x
x , so its mass is
i
i 2
i
i
i
i
f x g x
x . Thus, M R = f x g x
x x= x f x g x
x and
* ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
1
1
M ( R ) = f ( x ) g( x )
x
f ( x ) +g ( x ) =
f ( x ) g( x )
x . Summing over i and
2
2
taking the limit as n
, we get M =lim
x f ( x ) g( x )
x= x f (x) g(x) dx and
i y i i i i i i i
2 2 x i i i i i i i i b y M =lim
x n i 1
2
M 2 ( ) g( x ) f x i b 2 x= i a 1
2 i i 2 i y n a 2 f (x) g(x) dx . Thus, b
M
M
1
x
x
1
x=
=
=
x f (x) g(x) dx and y=
=
=
m
A Aa
m
A A
y M i n b
a 1
2 2 2 f (x) g(x) dx m 44. (a) Let 0 x 1 . If n<m , then x >x ; that is, raising x to a larger power produces a smaller
22 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.3 Applications to Physics and Engineering number. (b) Using Formulas 9 and the fact that the area of is
1 A=
0 1
( xn xm) dx= n+1 1
m n
=
, we get
m+1 (n+1)(m+1) 1 (n+1)(m+1)
(n+1)(m+1)
n m
x x x dx=
x=
m n
m n
0
(n+1)(m+1)
=
m n
= (n+1)(m+1)
2(m n) 1
n+2
1
2n+1 1
m+2 1 ( xn+1 xm+1) dx 0 (n+1)(m+1)
=
and
(n+2)(m+2) 1
2m+1 = (n+1)(m+1)
y=
m n 1
0 1
2 ( x n ) 2 ( xm ) 2 dx= (n+1)(m+1)
(2n+1)(2m+1) (c) If we take n=3 and m=4 , then ( x,y ) =
which lies outside R since 2
3 3 = 4 5 4 5
,
5 6 7 9 = 2 20
,
3 63 8 20
. This is the simplest of many possibilities.
<
27 63 23 (n+1
2( Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.4 Applications to Economics and Biology 2000 1. By the Net Change Theorem, C(2000) C(0)= / C (x)dx
0 2000 C(2000) =20 , 000+ 2 2 (5 0.008x+0.000009x )dx=20 , 000+ 5x 0.004x +0.000003x 3 2000 0 0 =20 , 000+10 , 000 0.004(4 , 000 , 000)+0.000003(8 , 000 , 000 , 000)=30 , 000 16 ,
000+24 , 000
=$38 , 000
5000 2. By the Net Change Theorem, R(5000) R(1000)= / R (x)dx 1000 R(5000) =12,400+ 5000 2 5000 1000 1000 (12 0.0004x)dx=12,400+ 12x 0.0002x =12,400+(60,000 5,000) (12,000 200)=$55,600
3. If the production level is raised from 1200 units to 1600 units, then the increase in cost is
1600 C(1600) C(1200) = 1600 / C (x)dx=
1200 ( 74+1.1x 0.002x2+0.00004x3) dx 1200 0.002 3
4 1600
x +0.00001x
3
1200
= 64,331,733.33 20,464,800=$43,866,933.33
2 = 74x+0.55x 4.
30 Consumer surplus = p(x) p(30) dx
0 30 = 5
0 = 3x 1
x
10
1 2
x
20 5 30
10 dx 30
0 =90 45=$45 1 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.4 Applications to Economics and Biology 5. p(x)=10 450
=10
x+8 x+8=45 x=37 . 37 Consumersurplus = 37 450
10
x+8 p(x) 10 dx=
0 0 = 450ln ( x+8 ) 10x dx 37
0 = (450ln 45 370) 450ln 8
45
= 450ln
370 $407.25
8 2 6. p (x)=3+0.01x . P= p (10)=3+1=4 .
S S 10 Producersurplus = [P p (x)]dx
0
10 = S 2 4 3 0.01x dx= x
0 0.01 3
x
3 10
0 10 3.33=$6.67 2 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.4 Applications to Economics and Biology 7. P= p (x)
S 400=200+0.2x 3/2 200=0.2x 100 100 Producer surplus = S 3/2 400 200+0.2x 3/2 2/3 x=1000 =100 . ) 0 2 5/2
x
25 = 200x 1000=x ( P p (x) dx=
0 8. p=50 3/2 100 dx= 200
0 1 3/2
x
dx
5 100
0 =20 , 000 8 , 000=$12 , 000 1
1
x and p=20+
x intersect at p=40 and x=200 .
20
10
200 Consumer surplus= 50
0 = 1
x 40 dx
20 1 2
10x
x
40 200
0 =$1000
200 Producer surplus=
= 20x 1 2
x
20 0 40 20 1
x dx
10 200
0 =$2000 x/5000 800,000e
9. p(x)=
x+20,000 =16 x=x 1 3727.04 .
x 1 Consumersurplus= [ p(x) 16]dx $37,753
0 3 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.4 Applications to Economics and Biology 0.5
1
=
and p(400)=7.5 , so an equation is
35
70
1
1
185
1
185
p 7.5=
(x 400) or p=
x+
. A selling price of $6 implies that 6=
x+
70
70
14
70
14
1
185 84 101
x=
=
x=505 .
70
14 14 14 10. The demand function is linear with slope 505 1
185
x+
6 dx=
70
14 Consumer surplus=
0
8 8 / 11. f (8) f (4)= f (t)dt=
4 2 3/2
t
3 t dt=
4 8 = 4 505 1 2 101
x+
x
140
14 $1821.61 0 2
( 16 2 8 ) $9.75 million
3 12.
9 n(9) n(5) = ( 2200+10e ) dt=
0.8t 5 ( =2200(9 5)+12.5 e
4 7.2 e 4 ) 4 4 PR
=
8 l
8 l
0 0 14. If the flux remains constant, then 0 R 0 3
R
4 0 5 3 4 P
=
P 5 5 9 1.19 10 cm / s
P R 3
R= R
4 0 25 0.8t
e
2 9 = 2200t + 24 , 860 4 PR
(4000)(0.008)
13. F=
=
8 l
8(0.027)(2) 9 0.8t 10e
2200t+
0.8 4 P=P 0 4
3 4 4 P R =PR
0 0 P
=
P
0 R 0 R 4 . 4 3.1605P >3P ; that is, the blood pressure is more
0 0 than tripled.
15.
4 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.4 Applications to Economics and Biology 12 12 12 1 2
3 2 1 3 12
c(t)dt=
3t
t dt=
t
t
= ( 216 144 ) =72 mg s / L.
4
2
12
0
0
0
A
8mg
1
60
Thus, the cardiac output is F= 12
=
= L/s=
L / min.
72mg s/L 9
9
c(t)dt
1
t(12 t)dt=
4
0 0 . 16. As in Example 2, we will estimate the cardiac output using Simpson’s Rule with
20 c(t)dt
0 t=2 . 2
1(0)+4(2.4)+2(5.1)+4(7.8)+2(7.6)
3 +4(5.4)+2(3.9)+4(2.3)+2(1.6)+4(0.7)+1(0)
2
=
(110.8) 73.87mg s/L
3
A
8
Therefore, F
=
0.1083 L / s or 6.498 L / min.
73.87 73.87 5 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability 40,000 1. (a) f (x)dx is the probability that a randomly chosen tire will have a lifetime between 30 ,
30,000 000 and 40 , 000 miles.
(b) f (x)dx is the probability that a randomly chosen tire will have a lifetime of at least 25 , 000
25,000 miles.
15 2. (a) The probability that you drive to school in less than 15 minutes is f (t)dt .
0 f (t)dt . (b) The probability that it takes you more than half an hour to get to school is
30 3. (a) In general, we must satisfy the two conditions that are mentioned before Example 1
namely, (1) f (x) 0 for all x , and (2) f (x)dx=1 . For 0 x 4 , we have f (x)= 3
x
64 2 16 x 0, so f (x) 0 for all x . Also,
4 3
f (x)dx =
x
0 64
= 1
64 4 3
3
2 1/2
16 x dx=
(16 x ) ( 2x)dx=
128 0
128
2 ( 16 x2) 3/2 4 = 0 2
2 3/2
(16 x )
3 4
0 1
(0 64)=1 .
64 Therefore, f is a probability density function.
(b)
2 P(X<2) = f (x)dx= 2
0 3
x
64 2 3
2 1/2
16 x dx=
(16 x ) ( 2x)dx
128 0
2 3
2
1
1
2 3/2 2
2 3/2 2
3/2
3/2
(16 x )
=
(16 x )
=
(12 16 )
0
128
3
0
64
64
1
1
3
=
( 64 12 12 ) = 64 ( 64 24 3 ) =1 8 3 0.350481
64
= 2 4. (a) For 0 x 1 , we have f (x)=kx (1 x) , which is nonnegative if and only if k 0 . Also,
1 2 1 2 3 f (x)dx= kx (1 x)dx=k (x x )dx=k
0 0 1 3 1 4
x
x
3
4 1 =k/12 . Now k/12=1 0 k=12 . Therefore, f is
1 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability a probability density function if and only if k=12 .
(b) Let k=12 .
P X 1
2 1 = 1 2 2 3 3 f (x)dx= 12x (1 x)dx= (12x 12x )dx= 4x 3x 1/2 1/2 1
2 =(4 3) 4 1
1/2 1/2 3
16 5 11
=
16 16 =1 (c) The mean
1 = 1 2 3 1 4 1 5
x
x
4
5 4 xf (x)dx= x 12x (1 x)dx=12 (x x )dx=12
0 =12 1
4 1
5 0 = 1
0 12 3
=
20 5 5. (a) In general, we must satisfy the two conditions that are mentioned before Example 1
namely, (1) f (x) 0 for all x , and (2) f (x)dx=1 . Since f (x)=0 or f (x)=0.1 , condition (1) is
10 satisfied. For condition (2), we see that f (x)dx= 0.1dx=
0 10 1
x
10 0 =1 . Thus, f (x) is a probability density function for the spinner’s values.
(b) Since all the numbers between 0 and 10 are equally likely to be selected, we expect the mean to
be halfway between the endpoints of the interval; that is, x=5 .
10 = xf (x)dx= x(0.1)dx=
0 1 2
x
20 10
0 = 100
=5 , as expected.
20
10 6. (a) As in the preceding exercise, (1) f (x) 0 and (2) f (x)dx= f (x)dx=
0 1
(10)(0.2)=1 . So f (x)
2 is a probability density function.
(b)
3
(a)
1
3
P(X<3)= f (x)dx= (3)(0.1)=
=0.15
2
20
0
(b) 3 P(X<3)= f (x)dx=
0 1
3
(3)(0.1)=
=0.15
2
20 (c) We first compute P(X>8) and then subtract that value and our answer in (i) from 1 (the total
2 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability 10 1
2
(2)(0.1)=
=0.10 . So P(3 X 8)=1 0.15 0.10=0.75 .
2
20 probability). P(X>8)= f (x)dx=
8 (d) We first compute P(X>8) and then subtract that value and our answer in (i) from 1 (the total
10 1
2
(2)(0.1)=
=0.10 . So P(3 X 8)=1 0.15 0.10=0.75 .
2
20 probability). P(X>8)= f (x)dx=
8 (c) We find equations of the lines from ( 0,0 ) to ( 6,0.2 )
1
x
30
f (x)=
1
1
x+
20
2
0 and from ( 6,0.2 ) to ( 10,0 ) , and find that { 6 = 10 1
x dx+ x
30
6 xf (x)dx= x
0 = 216
+
90 1000 100
+
60
4 1
1
x+
20
2 216 36
+
60
4 = if 0 x<6
if 6 x<10
otherwise dx= 1 3
x
90 6 + 0 x 200 P(0 X 200)=
0 (b) 200 P(0 X 200)=
0 (c)
P(X>800)=
800 (d)
P(X>800)=
800 1
e
1000 t/1000 1
e
1000 t/1000 1
e
1000 t/1000 1
e
1000 t/1000 dt= e dt= e dt=lim e x dt=lim
x e 6 16
=5.3
3 1
1 t/5
1
1
7. We need to find m so that f (t)dt=
lim
e dt=
lim
( 5)e
2 x
2 x
5
m
m 5
1
1
m/5 1
m/5 1
( 1)(0 e )=
e =
m/5=ln
m= 5ln =5ln 2 3.47 min.
2
2
2
2
8. (a)
(a) 10 1 3 1 2
x+ x
60
4 t/1000 200
0 = e t/1000 200
0
t/1000 x = e = m 1
2 +1 0.181 1/5 =0+e =0+e 800 x 1/5 +1 0.181 4/5 800
t/1000 x t/5 4/5 0.449 0.449 3 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability x 1
1
1
t/1000
(b) We need to find m so that f (t)dt=
lim
e
dt=
2 x
2
m
m 1000
1
1
m/1000 1
0+e
=
m/1000=ln
m= 1000ln =1000ln 2 693.1 h.
2
2
2 lim e t/1000 x = m x 1
2 9. We use an exponential density function with =2.5 min.
x (a) P(X>4)= f (t)dt=lim
x 4 4 1
e
2.5 2 (b) P(0 X 2)= f (t)dt= e t/2.5 e t/2.5 x =0+e = e 4/2.5 4 x t/2.5 2 0 dt=lim 0.202 2/2.5 +1 0.551 0 (c) We need to find a value a so that P(X a)=0.02 , or, equivalently, P(0 X a)=0.98
a f (t)dt=0.98 e t/2.5 a =0.98 0 0 e a/2.5 +1=0.98 e a/2.5 =0.02 a/2.5=ln 0.02 1
=2.5ln 50 9.78 min 10 min. The ad should say that if you aren’t served within 10
50
minutes, you get a free hamburger.
a= 2.5ln 73 1 10. (a) With =69 and =2.8 , we have P(65 X 73)= 2.8 2 exp ( x 69 ) 2 2 dx 0.847 2 2.8
65
(using a calculator or computer to estimate the integral).
(b) P(X>6 feet )=P(X>72 inches )=1 P(0 X 72) 1 0.858=0.142 , so 14.2% of the adult male
population is more than 6 feet tall. 1 11. P(X 10)=
10 4.2 2 exp ( x 9.4 ) 2 2 dx . To avoid the improper integral we approximate it 2 4.2 100 1 by the integral from 10 to 100 . Thus, P(X 10) 4.2 2 exp ( x 9.4 )
2 2 dx 0.443 (using a 2 4.2
calculator or computer to estimate the integral), so about 44 percent of the households throw out at
least 10 lb of paper a week.
Note : We can’t evaluate 1 P(0 X 10) for this problem since a significant amount of area lies to
the left of X=0 .
10 12. (a)
4 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability 480 1
12 2 P(0 X 480)= ( x 500 ) exp 2 2 dx 0.0478 (using a calculator or computer to 2 12
estimate the integral), so there is about a 4.78% chance that a particular box contains less than 480 g
of cereal.
(b) We need to find so that P(0 X<500)=0.05 . Using our calculator or computer to find
P(0 X 500) for various values of , we find that if =519.73 , P=0.05007 ; and if =519.74 ,
P=0.04998 . So a good target weight is at least 519.74 g.
0 +2 13. P( 2 X . . 1
2 +2 )=
2 exp 2 gives us 14. Let f (x)=
bx 2 bx 2 = 1
2 2 0
ce e if x<0
if x 0 cx t 2 / 2( 2 dx . Substituting t= 2 2 1
dt)=
2 e t 2 / 2 dt x and dt= 1 dx 0.9545 2 where c=1/ . By using parts, tables, or a CAS, we find that / b ) ( bx 1)
dx= ( e / b ) ( b x 2bx+2 ) (1): xe dx=
(2): x e
Now . { 2 (x ) (x ( e bx 2 bx 3 2 2 0 2 ) f (x)dx= (x 2 ) f (x)dx+ (x 2 ) f (x)dx 0
t = 0+lim c (x
t 2 )e cx t dx=c lim
t 0 ( x2e cx 2x cx e + 2 e cx ) dx 0 Next we use (2) and (1) with b= ctoget
2 =clim
t e
c cx
3 ( c x +2cx+2)
2 2 2 e
c cx
2 ( cx 1)+ 2 e cx c t 0 Using l’Hospital’s Rule several times, along with the fact that =1/c , we get 5 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability 2 =c 2 0 c
15. (a) First p(r)= 4
a 3 2 3 + 2
c 1
c + 2 1
c 2 1
c 1 =c c 1 = 3 c = 2 1
=
c 2r/a 0 r e 0 for r 0 . Next, 0 4 p(r)dr= a 3 2 2r/a 0 r e dr= 0 0 2 bx 4 t 2r/a 2 0 r e lim 3
a t
0 dr 0 bx 3 2 2 By using parts, tables, or a CAS , we find that x e dx=(e /b )(b x 2bx+2) . ( * )
a 4 Next, we use ( * ) (with b= 2/a ) and l’Hospital’s Rule to get
0 a 3 0 8 3 ( 2) =1 . This satisfies the 0 second condition for a function to be a probability density function.
(b) Using l’Hospital’s Rule, 4 r lim 3
a r
0 2 2r/a e = 0 4 2r lim 3
a r
0 2r/a 0 ( 2/a ) e
0 = 2 2 lim 2
a r
0 2r/a ( 2/a ) e =0 . 0 0 To find the maximum of p , we differentiate:
/ p (r)= 4
a 3 2r/a 2 2
a 0 r e / r=0 or 1= r
a 0 (2r) = 4
a 0 0 p (r)=0 2r/a +e 3 2r/a e 0 (2r) 0 r
+1
a
0 / r=a . p (r) changes from positive to negative at r=a , so p(r) has its
0 0 0 maximum value at r=a .
0 (c) It is fairly difficult to find a viewing rectangle, but knowing the maximum value from part (b)
helps. ( ) p a =
0 4
a 3 0 2 ae
0 2a /a 0 0 = 4
e
a 2 9,684,098,979 0 6 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability With a maximum of nearly 10 billion and a total area under the curve of 1 , we know that the
‘‘hump’’ in the graph must be extremely narrow. 4a r 0 4 (d) P(r)= a 3 2s/a 2 0 se ( ) ds P 4a =
0 0 0 4
0 a 2 3 2s/a 0 se ds . Using ( * ) from part (a) 0 (with b= 2/a ),
0 4a 2s/a 4 ( )= P 4a 0 a 0 e 3 8/a 0 (e) = 3 bx ( rp(r)dr= x e dx= e bx b
= 4
a 3 4 3 2 4 8 ) 8 2 =1 41e 3 r e lim 4
s+2
a
0 0 t 3
a t
0 2 s+ a 0 1
82e
2 = 4 0 = 4
a 0 3 a 3 0 8 8 e (64+16+2) 1(2) 0 0.986 2r/a 0 dr . Integrating by parts three times or using a CAS, we find that 0 ( b3x3 3b2x2+6bx 6) . So with b= 2
, we use l’Hospital’s Rule, and get
a
0 a 4 0 16 ( 6) = 3
a .
2 0 0 7 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.1 Modeling with Differential Equations 1 1. y=x x / 2 y =1+x . To show that y is a solution of the differential equation, we will substitute
/ the expressions for y and y in the left hand side of the equation and show that the left hand side is
equal to the right hand side. ( / LHS=xy +y=x 1+x 2 ) + ( x x 1) =x+x 1+x x 1=2x=RHS / 2 2 y =sin x( sin x)+cos x(cos x) ( sin x)=cos x sin x+sin x . 2. y=sin xcos x cos x LHS = y / +(tan x)y=cos 2 x sin 2 x+sin x+(tan x)(sin xcos x cos x)
= cos 2 x sin 2 x+sin x+sin 2 x sin x=cos 2 x=RHS,
so y is a solution of the differential equation. Also, y(0)=sin 0cos 0 cos 0=0 1 1= 1 , so the initial
condition is satisfied.
/ 3. (a) y=sin kt y =kcos kt y / / 2 = k sin kt . y / / ( 9 k2) sin kt=0 for all t 2 k sin kt+9sin kt=0 for all t
/ (b) y=Asin kt+Bcos kt y =Akcos kt Bksin kt / / +9y=0 2 y / / 2 k= 3 9 k =0
2 2 = Ak sin kt Bk cos kt . ( 9 k2) Asin kt+ ( 9 k2) Bcos kt=0 . 2 Thus, y +9y=0
Ak sin kt Bk cos kt+9(Asin kt+Bcos kt)=0
The last equation is true for all values of A and B if k= 3 .
rt / rt 4. y=e
y =re
r= 3 or 2
t y / / / t 5. (a) y=e y =e y
differential equation.
(b) y=e
(c) y=te t / y = e t LHS = y
=e / y =t
/ /
t t 2 rt / / +y t =e . LHS =y / / y / / =r e . y t / / / =e . LHS =y t / t y +2y +y=e (t 2)+2e (1 t)+te rt rt r e +re 6e =0 t t t t +2y +y=e +2e +e =4e / / ( e t ) +e t(1)=e t(1 t)
/ 2 rt 6y=0 / +2y +y=e
/ / t t t (r 2+r 6) ert=0 (r+3)(r 2)=0 t 0 , so y=e is not a solution of the
t 2e +e =0= RHS, so y=e is a solution. t =e (t 2) . t t ( t 2 ) +2(1 t)+t =e (0)=0=RHS, t so y=te is a solution.
2 (d) y=t e
LHS = y
=e t
/ /
t / t y =te ( 2 t )
/ +2y +y=e t y / / =e t ( t2 4t+2) . ( t2 4t+2) +2te t(2 t)+t2e t ( t2 4t+2) +2t(2 t)+t2 =e t(2) 0,
1 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.1 Modeling with Differential Equations 2 t so y=t e is not a solution.
2 6. (a) y=Ce x /2 / 2 2 y =Ce x /2 ( 2x/2 ) =xCe x /2 =xy . (b)
0 (c) y(0)=5 Ce =5
1/2 (d) y(1)=2 Ce =2 2 C=5 , so the solution is y=5e
C=2e 1/2 / x /2 .
2 , so the solution is y=2e ( x2 1 ) /2 .
=2e 1/2 x /2 e 2 7. (a) Since the derivative y = y is always negative (or 0 if y=0 ), the function y must be decreasing
(or equal to 0 ) on any interval on which it is defined.
2
1
1
/
/
2
1
1
y =
. LHS =y =
=
= y = RHS
(b) y=
2
2
x+C
x+C
( x+C )
( x+C )
/ 2 (c) y=0 is a solution of y = y that is not a member of the family in part (b).
1
1
1
1 1
1
(d) If y(x)=
, then y(0)=
=
. Since y(0)=0.5 ,
=
C=2 , so y=
.
x+C
0+C C
C 2
x+2
3 / 8. (a) If x is close to 0 , then xy is close to 0 , and hence, y is close to 0 . Thus, the graph of y must
3 have a tangent line that is nearly horizontal. If x is large, then xy is large, and the graph of y must
have a tangent line that is nearly vertical. (In both cases, we assume reasonable values for y .) ) 1/2 y /=x ( c x2) 3/2 .
3
2 1/2 3
2 3/2
/
RHS =xy =x ( c x )
=x ( c x ) =y = LHS
( 2 (b) y= c x (c)
/ When x is close to 0 , y is also close to 0 . As x gets larger, so does y / .
2 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.1 Modeling with Differential Equations (d) y(0)= ( c 0 ) 1/2 =1/ c and y(0)=2 1
c=
2 dP
P
dP
=1.2P 1
. Now
>0
dt
4200
dt
increasing for 0<P<4200 .
dP
(b)
<0 P>4200
dt
dP
(c)
=0 P=4200 or P=0
dt
9. (a) 10. (a) y=k / y =0 , so dy 4 3 2
=y 6y +5y
dt 1
c= , so y=
4
1 4 P
>0
4200 3 0=k 6k +5k 2 1/2 1 2
x
4 . P
<1
4200 k 2 P<4200 ( k2 6k+5) =0 the population is 2 k (k 1)(k 5)=0 k=0 , 1 , or 5
dy
>0
dt
dy
<0
dt (b) y is increasing
(c) y is decreasing 2 y (y 1)(y 5)>0 y ( ,0 ) ( 0,1 ) ( 5, ) y ( 1,5)
t 2 11. (a) This function is increasing and also decreasing. But dy/dt=e (y 1) 0 for all t , implying that
the graph of the solution of the differential equation cannot be decreasing on any interval.
(b) When y=1 , dy/dt=0 , but the graph does not have a horizontal tangent line.
12. The graph for this exercise is shown in the figure at the right.
/ / A. y =1+xy>1 for points in the first quadrant, but we can see that y <0 for some points in the first
quadrant. So equation A is incorrect.
/ / B. y = 2xy=0 when x=0 , but we can see that y >0 for x=0 . So equation B is incorrect.
/ C. y =1 2xy seems reasonable since: (1)
(2) / When x=0 , y could be 1 .
/ When x<0 , y could be greater than 1 . 3 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.1 Modeling with Differential Equations (3) 1 y
Solving y =1 2xy for y gives us y=
2x
/ ,y /
/ . If y takes on small negative values, then as x + 0 , as shown in the figure. Thus, the correct equation is C. 13. (a) P increases most rapidly at the beginning, since there are usually many simple, easily learned
sub skills associated with learning a skill. As t increases, we would expect dP/dt to remain positive,
but decrease. This is because as time progresses, the only points left to learn are the more difficult
ones.
dP
=k(M P) is always positive, so the level of performance P is increasing. As P gets close to M
(b)
dt
, dP/dt gets close to 0 ; that is, the performance levels off, as explained in part (a). (c)
14. (a) P increases most rapidly at the beginning, since there are usually many simple, easily learned
sub skills associated with learning a skill. As t increases, we would expect dP/dt to remain positive,
but decrease. This is because as time progresses, the only points left to learn are the more difficult
ones.
dP
=k(M P) is always positive, so the level of performance P is increasing. As P gets close to M
(b)
dt
, dP/dt gets close to 0 ; that is, the performance levels off, as explained in part (a). (c) 4 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method 1. (a) (b) It appears that the constant functions y=0 , y= 2 , and y=2 are equilibrium solutions. Note that
1 2
/
these three values of y satisfy the given differential equation y =y 1
y
.
4 2. (a)
/ (b) From the figure, it appears that y= is an equilibrium solution. From the equation y =xsin y , we
see that y=n ( n an integer) describes all the equilibrium solutions.
/ 3. y =y 1 . The slopes at each point are independent of x , so the slopes are the same along each line
/ parallel to the x axis. Thus, IV is the direction field for this equation. Note that for y=1 , y =0 .
/ 4. y =y x=0 on the line y=x , when x=0 the slope is y , and when y=0 the slope is x . Direction field
II satisfies these conditions.
/ 2 2 / 3 3 5. y =y x =0 y= x . There are horizontal tangents on these lines only in graph III, so this
equation corresponds to direction field III.
3 3 6. y =y x =0 on the line y=x , when x=0 the slope is y , and when y=0 the slope is x . The graph
is similar to the graph for Exercise 4, but the segments must get steeper very rapidly as they move
away from the origin, because x and y are raised to the third power. This is the case in direction field
I.
1 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method 7. (a) y(0)=1 (b) y(0)=0 (c) y(0)= 1 8. (a) y(0)=1 (b) y(0)=0 2 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method (c) y(0)= 1 9.
10.
x y
1
3
1
0.5 / 2 2 y =x y
3
8
1
8
0.5 0.75
1
0.75
/ / Note that y =0 for y= x . If x < y , then y <0 ; that is, the slopes are negative for all points in
quadrants I and II above both of the lines y=x and y= x , and all points in quadrants III and IV below
both of the lines y= x and y=x . A similar statement holds for positive slopes. 11.
x y y / =y 2x
2 2 2
2 2 6
2 2 2
2 2 6
Note that
3 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method / y =0 for any point on the line y=2x . The slopes are positive to the left of the line and negative to the
right of the line. The solution curve in the graph passes through (1,0) . 12.
x y
1
2
2 / y =1 xy
1 0
2 3
2 5
/ Note that y =0 for any point on the hyperbola xy=1 (or y=1/x ). The slopes are negative at points
‘‘inside’’ the branches and positive at points everywhere else. The solution curve in the graph passes
through (0,0) . 13.
x y
0
1
3 / y =y+xy
2 2
2 4
2 4
/ Note that y =y(x+1)=0 for any point on y=0 or on x= 1 . The slopes are positive when the factors y
and x+1 have the same sign and negative when they have opposite signs. The solution curve in the
graph passes through (0,1) . 4 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method 14.
x y y
2 0
2 3
2 1 / =x xy
2
4
4
/ Note that y =x(1 y)=0 for any point on x=0 or on y=1 . The slopes are positive when the factors x
and 1 y have the same sign and negative when they have opposite signs. The solution curve in the
graph passes through (1,0) . 15. In Maple, we can use either directionfield (in Maple’s share library) or plots [fieldplot] to plot the
direction field. To plot the solution, we can either use the initial value option in directionfield, or
actually solve the equation. In Mathematica, we use PlotVectorField for the direction field, and the
Plot [Evaluate[...]] construction to plot the solution, which . ( 1 cos 2x ) / 2
is y=e
. In Derive, use Direction_Field (in utility file ODE_APPR) to plot the direction
field. Then use DSOLVE1( y*SIN(2*x),1,x,y,0,1) (in utility file ODE1) to solve the equation.
Simplify each result.
5 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method 16. See Exercise 15 for specific CAS directions. The exact solution is 17. L=lim y(t) exists for 2 c 2 ; L= 2 for c= 2 and L=0 for 2<c<2 . For other values of c , L does
t not exist.
18. / Note that when f (y)=0 on the graph in the text, we have y = f (y)=0 ; so we get horizontal segments
at y= 1 , 2 . We get segments with negative slopes only for 1< y <2 . All other segments have
positive slope. For the limiting behavior of solutions:
If y(0)>2 , then lim y= and lim y=2 .
t t If 1<y(0)<2 , then lim y=1 and lim y=2 .
t t If 1<y(0)<1 , then lim y=1 and lim y= 1 .
t t If 2<y(0)< 1 , then lim y= 2 and lim y= 1 .
t t
6 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method If y< 2 , then lim y= 2 and lim y=
t 19. (a)
(a) h=0.4 and y =y +hF x ,y ( )
(b) h=0.4 and y =y +hF x ,y
( )
1 0
0 y =1+0.4 1=1.4 . x =x +h=0+0.4=0.4 , so y =y ( 0.4 ) =1.4 . 0 0 1 (c) h=0.2 . t 1 0 0 1 0 1 y =1+0.4 1=1.4 . x =x +h=0+0.4=0.4 , so y =y ( 0.4 ) =1.4 .
1
1 0
1 ( 0 0 ( 0 0 ) 0 ) 0 x =0.2 and x =0.4 , so we need to find y . y =y +hF x ,y =1+0.2y =1+0.2 1=1.2 ,
1 ( 2 ) 2 1 0 y =y +hF x ,y =1.2+0.2y =1.2+0.2 1.2=1.44 .
2 1 (d) h=0.2 1 1 1 x =0.2 and x =0.4 , so we need to find y . y =y +hF x ,y =1+0.2y =1+0.2 1=1.2 ,
1 ( 2 ) 2 1 0 y =y +hF x ,y =1.2+0.2y =1.2+0.2 1.2=1.44 .
2 1 (e) h=0.1 1 1 1 ( ) x =0.4 , so we need to find y . y =y +hF x ,y =1+0.1y =1+0.1 1=1.1 ,
4 4 1 0 0 0 0 ( )
y =y +hF ( x ,y ) =1.21+0.1y =1.21+0.1 1.21=1.331 ,
y =y +hF ( x ,y ) =1.331+0.1y =1.331+0.1 1.331=1.4641 .
h=0.1 x =0.4 , so we need to find y . y =y +hF ( x ,y ) =1+0.1y =1+0.1 1=1.1 ,
y =y +hF ( x ,y ) =1.1+0.1y =1.1+0.1 1.1=1.21 ,
y =y +hF ( x ,y ) =1.21+0.1y =1.21+0.1 1.21=1.331 ,
y =y +hF ( x ,y ) =1.331+0.1y =1.331+0.1 1.331=1.4641 .
y =y +hF x ,y =1.1+0.1y =1.1+0.1 1.1=1.21 ,
2 1 1 3 2 2 2 4 (f) 1 3 1 3 3 2 3 4 4 2 1 1 1 3 2 2 2 4 3 3 3 1 0 0 0 0 1 2 3 (b)
x We see that the estimates are underestimates since they are all below the graph of y=e .
(c)
7 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method (a) For h=0.4 : ( exactvalue ) ( approximatevalue ) =e (b) For h=0.2 : ( exactvalue ) ( approximatevalue ) =e 0.4
0.4 (c) 1.4 0.0918
1.44 0.0518 0.4 For h=0.1 : ( exactvalue ) ( approximatevalue ) =e 1.4641 0.0277
Each time the step size is halved, the error estimate also appears to be halved (approximately).
20. As x increases, the slopes decrease and all of the estimates are above the true values. Thus, all of the
estimates are overestimates.
21. h=0.5 , x =1 , y =0 , and F ( x,y ) =y 2x .
0 0 Note that x =x +h=1+0.5=1.5 , x =2 , and x =2.5 .
1 0 2 3 ( )
y =y +hF ( x ,y ) = 1+0.5F ( 1.5, 1 ) = 1+0.5[ 1 2(1.5)]= 3 .
y =y +hF ( x ,y ) = 3+0.5F ( 2, 3) = 3+0.5[ 3 2(2)]= 6.5 .
y =y +hF ( x ,y ) = 6.5+0.5F ( 2.5, 6.5) = 6.5+0.5[ 6.5 2(2.5)]= 12.25 .
y =y +hF x ,y =0+0.5F ( 1,0 ) =0.5[0 2(1)]= 1 .
1 0 0 0 2 1 1 1 3 2 2 2 4 3 3 3 22. h=0.2 , x =0 , y =0 , and F ( x,y ) =1 xy .
0 0 Note that x =x +h=0+0.2=0.2 , x =0.4 , x =0.6 , and x =0.8 .
1 0 2 3 4 ( )
y =y +hF ( x ,y ) =0.2+0.2F ( 0.2,0.2 ) =0.2+0.2[1 (0.2)(0.2)]=0.392 .
y =y +hF ( x ,y ) =0.392+0.2F ( 0.4,0.392 ) =0.392+0.2[1 (0.4)(0.392)]=0.56064 .
y =y +hF ( x ,y ) =0.56064+0.2[1 (0.6)(0.56064)]=0.6933632 .
y =y +hF ( x ,y ) =0.6933632+0.2[1 (0.8)(0.6933632)]=0.782425088 .
y =y +hF x ,y =0+0.2F ( 0,0 ) =0.2[1 (0)(0)]=0.2 .
1 0 0 0 2 1 1 1 3 2 2 2 4 3 3 3 5 4 4 4 Thus, y(1) 0.7824 .
8 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method 23. h=0.1 , x =0 , y =1 , and F ( x,y ) =y+xy .
0 0 Note that x =x +h=0+0.1=0.1 , x =0.2 , x =0.3 , and x =0.4 .
1 0 2 3 4 ( )
y =y +hF ( x ,y ) =1.1+0.1F ( 0.1,1.1 ) =1.1+0.1[1.1+(0.1)(1.1)]=1.221 .
y =y +hF ( x ,y ) =1.221+0.1F ( 0.2,1.221 ) =1.221+0.1[1.221+(0.2)(1.221)]=1.36752 .
y =y +hF ( x ,y ) =1.36752+0.1F ( 0.3,1.36752 ) =1.36752+0.1[1.36752+(0.3)(1.36752)]
y =y +hF x ,y =1+0.1F ( 0,1 ) =1+0.1[1+(0)(1)]=1.1 .
1 0 0 0 2 1 1 1 3 2 2 2 4 3 3 3 =1.5452976 . ( y =y +hF x ,y
5 4 4 4 ) =1.5452976+0.1F ( 0.4,1.5452976)
=1.5452976+0.1[1.5452976+(0.4)(1.5452976)]=1.761639264 . Thus, y(0.5) 1.7616 .
24. (a) h=0.2 , x =1 , y =0 , and F ( x,y ) =x xy .
0 0 We need to find y , because x =1.2 and x =1.4 .
2 1 2 ( )
y =y +hF ( x ,y ) =0.2+0.2F ( 1.2,0.2 ) =0.2+0.2[1.2 (1.2)(0.2)]=0.392
y =y +hF x ,y =0+0.2F ( 1,0 ) =0.2[1 (1)(0)]=0.2 .
1 0 0 0 2 1 1 1 y(1.4) . (b) Now h=0.1 , so we need to find y .
4 y =0+0.1[1 (1)(0)]=0.1 ,
1 y =0.1+0.1[1.1 (1.1)(0.1)]=0.199 ,
2 y =0.199+0.1[1.2 (1.2)(0.199)]=0.29512 , and
3 y =0.29512+0.1[1.3 (1.3)(0.29512)]=0.3867544 y(1.4) .
4 25. (a)
(a) H =1 , N =1 y(1)=3
(b) H =1 , N =1 y(1)=3
(c) H =0.1 , N =10 y(1) 2.3928
(d) H =0.1 , N =10 y(1) 2.3928
(e) H =0.01 , N =100 y(1) 2.3701
(f) H =0.01 , N =100 y(1) 2.3701
9 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method (g) H =0.001, N =1000
(h) H =0.001, N =1000 (b) y=2+e x 3 / 2 y = 3x e y(1) 2.3681
y(1) 2.3681
x / 3 2 2 x 3 2 LHS=y +3x y= 3x e +3x ( 2+e ) = 3x e
x 3 2 x 3 2 2 x 3 2 +6x +3x e =6x =RHS 0 y(0)=2+e =2+1=3
(c)
1
(a)
For h=1 : ( exactvalue ) ( approximatevalue ) =2+e 3 0.6321
1
(b)
For h=1 : ( exactvalue ) ( approximatevalue ) =2+e 3 0.6321
1
(c)
For h=0.1 : ( exactvalue ) ( approximatevalue ) =2+e 2.3928 0.0249
1
(d)
For h=0.1 : ( exactvalue ) ( approximatevalue ) =2+e 2.3928 0.0249
1
(e)
For h=0.01 : ( exactvalue ) ( approximatevalue ) =2+e 2.3701 0.0022
1
(f)
For h=0.01 : ( exactvalue ) ( approximatevalue ) =2+e 2.3701 0.0022
1
(g)
For h=0.001 : ( exactvalue ) ( approximatevalue ) =2+e 2.3681 0.0002
1
(h)
For h=0.001 : ( exactvalue ) ( approximatevalue ) =2+e 2.3681 0.0002
3 3 26. (a) We use the program from the solution to Exercise 25 with Y =x y , H =0.01 , and N
1 2 0
=
=200 . With x ,y = ( 0,1 ) , we get y(2) 1.9000 .
0 0
0.01 ( ) (b)
Notice from the graph that y(2) 1.9 , which serves as a check on our calculation in part (a).
27. (a) R dQ 1
1
/
/
+ Q=E(t) becomes 5Q +
Q=60 or Q +4Q=12 .
dt C
0.05 10 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method (b) From the graph, it appears that the limiting value of the charge Q is about 3 .
/ (c) If Q =0 , then 4Q=12 Q=3 is an equilibrium solution. (d)
/ / (e) Q +4Q=12 Q =12 4Q . Now Q(0)=0 , so t =0 and Q =0 .
0 0 ( )
= Q +hF ( t ,Q ) =1.2+0.1(12 4 1.2)=1.92
= Q +hF ( t ,Q ) =1.92+0.1(12 4 1.92)=2.352
= Q +hF ( t ,Q ) =2.352+0.1(12 4 2.352)=2.6112
= Q +hF ( t ,Q ) =2.6112+0.1(12 4 2.6112)=2.76672 Q = Q +hF t ,Q =0+0.1(12 4 0)=1.2
1
0
0 0
Q
Q
Q
Q 2
3
4
5 1 1 1 2 2 2 3 3 3 4 4 4 Thus, Q =Q(0.5) 2.77 C.
5 28. (a) From Exercise .1.14, we have dy/dt=k(y R) . We are given that R=20 C and dy/dt= 1 C /
1
min when y=70 C. Thus, 1=k(70 20) k=
and the differential equation becomes
50
1
dy/dt=
(y 20) .
50
(b) 11 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.2 Direction Fields and Euler’s Method The limiting value of the temperature is 20 C;
that is, the temperature of the room.
1
(y 20) . With t =0 , y =95 , and h=2 min, we get
0
0
50
1
=95+2
(95 20) =92
50
1
=92+2
(92 20) =89.12
50
1
=89.12+2
(89.12 20) =86.3552
50
1
=86.3552+2
(86.3552 20) =83.700992
50
1
=83.700992+2
(83.700992 20) =81.15295232
50 (c) From part (a), dy/dt= ( )
y =y +hF ( t ,y )
y =y +hF ( t ,y )
y =y +hF ( t ,y )
y =y +hF ( t ,y )
y =y +hF t ,y
1 0 0 0 2 1 1 1 3 2 2 2 4 3 3 3 5 4 4 4 Thus, y(10) 81.15 C. 12 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 1. dy y
=
dx x dy dx
=
y
x dy
dx
=
y
x ln y =ln x +C y =e ln x +C =e ln x C C e =e x y=Kx , C where K= e is a constant. (In our derivation, K was nonzero, but we can restore the excluded case
y=0 by allowing K to be zero.)
2x dy e
=
2.
3
dx
4y
2 3 2x 2 2 1 2x
e +C
2 dy xdx
= 2
[y 0]
y
x +1 dy
=
y 2x 4y dy= e dx dy
xy
= 2
dx
x +1 / 3. (x +1)y =xy 4 3 4y dy=e dx 1/2 ( C y= C 2 u=x +1 , du=2xdx ] =ln (x +1) +ln e =ln e x +1 ) 1 2x
e +C
2 4 y= xdx ln y = 2 x +1
C 2 y =e x +1 1
2
ln (x +1)+C [
2 y=K 2 x +1 , where C K= e is a constant. (In our derivation, K was nonzero, but we can restore the excluded case y=0 by
allowing K to be zero.)
/ dy 2
=y sin x
dx 2 4. y =y sin x
y= dy
2 2 y 1
= cos x+C
y = sin xdx y 1
=cos x C
y 1
, where K= C . y=0 is also a solution.
cos x+K
/ 2 5. (1+tan y) y =x +1 sin y
2
dy= (x +1)dx
cos y
y+ln sec y .
du 1+ r
=
dr 1+ u dy
7.
=
dt
2 ( 1+
y 1 3
x +x+C . Note: The left side is equivalent to
3 2 t 1+y dy=te dt sin y
cos y 2 dy=(x +1)dx 1/2 (1+u )du= (1+r y 2 t 2/3 xy
8. y =
2ln y y= 2ln y
dy=xdx
y t t 2/3 [3(te e +C)] )dr ( 2 t 1/2 u+ 1
2
1+y
3 t 1+y dy= te dt 1+y y 1+y =[3(te e +C)]
/ 1+ u ) du= ( 1+ r ) dr t te dy 2
=x +1
dx
y ln cos y = (1+tan y) 1 6. dy =sin xdx 2 3/2
2 3/2
u =r+ r +C
3
3 ) 3/2=tet et+C 1 2ln y
dy= xdx
y 2 x
( ln y ) = +C
2
2 ln y= 2 x /2+C 1 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 2 x /2+C y=e
9. du
=2+2u+t+tu
dt du
=(1+u)(2+t)
dt 2 t /2+2t +C 1+u =e 2 du
= (2+t)dt
1+u
2 C t /2+2t t /2+2t , where K=e =Ke ln 1+u = 1+u= Ke 1 2
t +2t+C
2
2 t /2+2t u= 1 Ke where K>0 . u= 1 is 2 t /2+2t also a solution, so u= 1+Ae
10. dz t + z
+e =0
dt dz
t z
= ee
dt 1 z=ln , where A is an arbitrary constant.
z e dz= t z e dt t e = e +C z t e =e C 1
e z t 1 z =e C e= t e C (t ) z= ln e C t e C
11. dy 2
=y +1 , y(1)=0 .
dx Thus, tan 1 dy
2 = dx tan 1 1 y=x+C . y=0 when x=1 , so 1+C=tan 0=0 C= 1 . y +1 y=x 1 and y=tan (x 1) .
2 dy ycos x
1
1+y
2
12.
=
, y(0)=1 . 1+y dy=ycos xdx
+y dy= cos xdx
dy=cos xdx
2
dx
y
y
1+y
1 2
1
1
1 2
1
ln y + y =sin x+C . y(0)=1 ln 1+ =sin 0+C C= , so ln y + y =sin x+ . We cannot
2
2
2
2
2
solve explicitly for y . ( 3y ) / 3y 3y 13. xcos x=(2y+e ) y
xcos xdx=(2y+e )dy
(2y+e )dy= xcos xdx
1 3y
2
y + e =xsin x+cos x+C [ where the second integral is evaluated using integration by parts]. Now
3
1
2
2
2 1 3y
y(0)=0 0+ =0+1+C C=
. Thus, a solution is y + e =xsin x+cos x
.
3
3
3
3
We cannot solve explicitly for y .
14. dP
= Pt
dt P 1/2 dP= t 1/2 dt 1/2 2P = 2
+C
3
1 3/2
1 2
t + 2
.
3
3 P(1)=2
P= 2 3/2
t +C .
3
2
2
1 3/2
1
1/2 2 3/2
C=2 2
, so 2P = t +2 2
P= t + 2
3
3
3
3
3 dP/ P = t dt 2 2= 2 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 2 du 2t+sec t
=
15.
, u(0)= 5 . 2udu=
dt
2u
2 2 ( 2t+sec 2t ) dt 2 C=( 5) =25 . Therefore, u =t +tan t+25 , so u=
u=
16. 2 2 2 2 u =t +tan t+C , where u(0) =0 +tan 0+C
2 t +tan t+25 . Since u(0)= 5 , we must have 2 t +tan t+25 .
dy
y
y
=te , y(1)=0 . e dy= t dt
dt y e = 1 2
0 1 2
t +C . Since y(1)=0 , e = 1 +C . Therefore,
2
2 1
3
y 1 2 3
y 3
C= 1
=
and e = t
. So e =
2
2
2
2
2 1 2 3 t
t =
2
2 2 y e= 2
3 t 2 ( y=ln 2 ln 3 t 2 ) for t < 3 .
/ 17. y tan x=a+y , 0<x< /2
dy
cos x
=
dx
a+y
sin x dy a+y
=
dx tan x ln a+y =ln sin x +C dy
=cot xdx [ a+y 0 ]
a+y
a+y =e ln sin x +C =e ln sin x C C e =e sin x C a+y=Ksin x , where K= e . (In our derivation, K was nonzero, but we can restore the excluded case
y= a by allowing K to be zero.) y( /3)=a
3
4a
4a
4a
2a=K
a+a=Ksin
K=
. Thus, a+y=
sin x and so y=
sin x a .
2
3
3
3
3
/ 2 18. xy +y=y x dy 2
=y y
dx 2 xdy=(y y)dx dy
2 = dx
x y y
dy
dx
1
1
dx
=
[y 0,1]
dy=
ln y 1 ln y =ln x +C
y(y 1)
x
y 1 y
x
y 1
y 1
y 1
1
1
C
C
C
ln
=ln e x
=e x
=Kx , where K= e
1
=Kx
=1 Kx
y
y
y
y
y
1
1
y=
. Now y(1)= 1
1=
1 K= 1 K=2 ,
1 Kx
1 K
1
so y=
.
1 2x
4
4
dy
dy
dy
4
ln y
x +C
x C
3
3
3
19.
=4x y , y(0)=7 .
=4x dx
= 4x dx ln y =x +C e
=e
y =e e
dx
y
y ( x ) 4 y=Ae ; y(0)=7 A=7 x 4 y=7e . 20.
3 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 2 dy y
=
, y(1)=1 .
3
dx
x dy
2 = y x / 2 (b) y(0)=0 2 +C . y(1)=1 1= 2x 1
+C
2 C= 1
. So
2 2x . 2 x +1
dy
=2x
dx 2 dy 2 1 y dy =2xdx
2 = 2xdx sin 1 2 y=x +C for 2 1 y 1 y 2 ( 2) for . 2 1 2 C=0 , so sin sin 0=0 +C 1 y=x and y=sin x /2 x /2 . 2 (c) For 1 y to be a real number, we must have 1 y 1 ; that is, 1 y(0) 1 . Thus, the initial
/ value problem y =2x
y y= 1 y 2 x +C 3 1 2 2 1
1 2+2x
1
=
+ =
2 2
2
y
2x
2 2x
21. (a) y =2x 1
=
y dx / 22. e y +cos x=0 2 1 y , y(0)=2 does not have a solution.
y e dy= cos xdx y e = sin x+C 1 y= ln (sin x+C) . The solution is periodic, with period 2 . Note that for C>1 , the domain of the solution is R , but for 1<C 1 it is only
defined on the intervals where sin x+C>0 , and it is meaningless for C 1 , since then sin x+C 0 ,
and the logarithm is undefined. 4 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations For 1<C<1 , the solution curve consists of concave up pieces separated by intervals on which the
solution is not defined (where sin x+C 0 ). For C=1 , the solution curve consists of concave up
pieces separated by vertical asymptotes at the points where sin x+C=0 sin x= 1 . For C>1 , the
curve is continuous, and as C increases, the graph moves downward, and the amplitude of the
oscillations decreases.
23. dy sin x
=
, y(0)=
. So sin ydy= sin xdx
dx sin y
2 initial condition, we need cos
that we cannot take cos 1 2 =cos 0 C 0=1 C cos y= cos x+C cos y=cos x C . From the C=1 , so the solution is cos y=cos x 1 . Note of both sides, since that would unnecessarily restrict the solution to the
1 case where 1 cos x 1 0 cos x , as cos is defined only on 1,1 . Instead we plot the graph
using Maple’s plots [implicitplot] or Mathematica’s Plot [Evaluate[..]]. 2 dy x
24.
=
dx x +1
ye y y ye dy= x 2 y x +1 dx . We use parts on the LHS with u=y , dv=e dy , and on
2 the RHS we use the substitution z=x +1 , so dz=2xdx . The equation becomes ye
y e (y 1)= 1 2
x +1
3 ( ) 3/2+C , so we see that the curves are symmetric about the y ( x,y ) in the plane lies on one of the curves, namely the one for which C=(y 1)e y y e dy= 1
2 z dz axis. Every point
y 1 2
x +1
3 ( ) 3/2 . For example,
1
4
, so the origin lies on the curve with C=
. We use Maple’s plots
3
3
command or Plot] in Mathematica to plot the solution curves for various values of C .
along the y axis, C=(y 1)e y 5 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations It seems that the transitional values of C are 4
and
3 1
4
. For C<
, the graph consists of left and
3
3 4
, the two branches become connected at the origin, and as C increases, the
3
1
graph splits into top and bottom branches. At C=
, the bottom half disappears. As C increases
3
further, the graph moves upward, but doesn’t change shape much.
right branches. At C= 25. (a)
x y / y =1/y
0 0.5 2
0 0.5 2
0 1
1
0 1
1
0 2
0.5
x y
0 2
0 4 / y =1/y
0.5
0.25
6 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 0 3
0.3
0 0.25 4
0 0.3 3 / (b) y =1/y
ydy=dx dy/dx=1/y
ydy= dx 1 2
y =x+c
2 2 y =2(x+c) or y= 2(x+c) . (c)
26. (a)
x y / 2 y =x /y
1
1
1
1
1
1
1
1
1
1 1
1
1 2
0.5
2
1 4
2
2
2
1
0.5 2
0.5 1
0.25
2
0.5 8 7 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations / 2 (b) y =x /y 2 ydy=x dx , so 1 2 1 3
y = x +c , or y=
1
2
3 2 3
x +c
3 1/2 . (c)
2 . / 27. The curves y=kx form a family of parabolas with axis the y axis. Differentiating gives y =2kx ,
2 / but k=y/x , so y =2y/x . Thus, the slope of the tangent line at any point ( x,y ) on one of the parabolas
/ / is y =2y/x , so the orthogonal trajectories must satisfy y = x / (2y)
2 2 y= 2ydy= xdx 2 x +2y =C . This is a family of ellipses. 2 2 2 x / 2+C 1 / 28. The curves x y =k form a family of hyperbolas. Differentiating gives 2x 2y ( dy/dx ) =0 or y =x/y
, the slope of the tangent line at ( x,y ) on one of the hyperbolas. Thus, the orthogonal trajectories must
/ satisfy y = y/x
C xy =e 1 dy/y= dx/x ln y = ln x +C 1 ln x +ln y =C 1 ln xy =C 1 xy=C . This is a family of hyperbolas. 8 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 29. Differentiating y= ( x+k ) 1 1 / gives y = ( x+k )
1 / orthogonal trajectories must satisfy y = 2 y 30. Differentiating y=ke x / = , but k= 2 1 1
/
x , so y =
y 2 3 y dy=dx y x ( 1/y ) 2 = y . Thus, the y
=x+C or y= 3(x+C)
3 2 2 1 x 1/3 / gives y = ke , but k=ye , so y = y . Thus, the orthogonal trajectories
1 2
/
1/2
must satisfy y = 1/( y)=1/y ydy=dx
y =x+C y= 2(C+x)
. This is a family of parabolas
2
with axis the x axis. 31. From Exercise .2.27,
12 4Q =e
Q(t)=3 3e 4t 4t 4C . As t dQ
=12 4Q
dt 12 4Q=Ke
, Q(t) 4t dQ
= dt
12 4Q
4Q=12 Ke 4t 1
ln 12 4Q =t+C
4 Q=3 Ae 4t . Q(0)=0 ln 12 4Q = 4t 4C 0=3 A A=3 3 0=3 (the limiting value).
9 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 32. From Exercise .2.28,
y 20=Ke t/50 y(t)=Ke dy
1
=
(y 20)
dt
50 t/50 +20 . y(0)=95 dy
=
y 20 1
50 95=K+20 K=75 dt ln y 20 = y(t)=75e 1
t+C
50 t/50 +20 . dP
dP
kt+C
kt
=k(M P)
= ( k)dt ln P M = kt+C
P M =e
P M=Ae
P=M+Ae
dt
P M
. If we assume that performance is at level 0 when t=0 , then P(0)=0 0=M+A A= M
33. P(t)=M Me kt kt . lim P(t)=M M 0=M .
t dx
1
1/(b a) 1/(b a)
=k ( a x ) ( b x ) , a b . Using partial fractions,
=
, so
dt
(a x)(b x)
a x
b x
dx
1
b x
= k dt
= ( b a ) ( kt+C ) . The
( ln a x +ln b x ) =kt+C ln
(a x)(b x)
b a
a x
b x
b x
b x
concentrations A =a x and B =b x cannot be negative, so
0 and
=
. We now
a x
a x
a x
b x
b
have ln
=(b a)(kt+C) . Since x(0)=0 , we get ln
=(b a)C . Hence,
a x
a
( b a ) kt
( b a ) kt
b x
b
b x b ( b a ) kt
b e
1
ab e
1
ln
=(b a)kt+ln
= e
moles /
x=
=
( b a ) kt
( b a ) kt
a x
a
a x a
be
a 1
be
a
L.
dx
1
1
2
dx
(b) If b=a , then
=k(a x) , so
= k dt and
=kt+C . Since x(0)=0 , we get C= .
2
dt
a x
a
( a x)
34. (a) . / 2 1
a
a kt moles
Thus, a x=
and x=a
=
.
kt+1/a
akt+1 akt+1
L
2 Suppose x= C =a/2 when t=20 . Then x(20)=a/2 a 20a k
=
2 20ak+1 2 2 40a k=20a k+a 2 20a k=a 2 1
a t/(20a)
at/20
at moles
k=
, so x=
=
=
.
20a
1+at/(20a) 1+t/20 t+20
L
35. (a) If a=b , then
(a x) 3/2 dx= k dt dx
dx
3/2
1/2
3/2
(a x) dx=k dt
=k(a x)(b x) becomes
=k(a x)
dt
dt
2
2
2
1/2
2(a x) =kt+C
= a x
=a x x(t)=a
kt+C
kt+C initial concentration of HBr is 0 , so x(0)=0 0=a 4
C 2 4
C 2 =a 2 4
C =
a 4
2 . The (kt+C) C=2/ a ( C is positive since
10 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations kt+C=2(a x) 4 1/2 >0 ). Thus, x(t)=a ( kt+2/
(b) dx
1/2
=k(a x)(b x)
dt a) 2 dx
=k dt
(a x) b x . dx
= k dt ( * ) . From the hint, u= b x
(a x) b x 2 u =b x 2udu= dx , so
dx
2udu
du
du
=
= 2
= 2
= 2
2
2
2 2
(a x) b x
a b u
u
a b+u
( a b ) +u
2
2
1 b x
* ) becomes
tan
=kt+C . Now x(0)=0 C=
tan
a b
a b
a b
b
2
b
2
2
1
1 b x
1
tan
tan
=kt
tan
a b
a b
a b
a b
a b
a b
2
b
b x
1
1
t(x)=
tan
tan
.
a b
a b
k a b ( ) 2 dT
dS d T
36. If S=
, then
=
. The differential equation
2
dr
dr
dr
dS 2
dS
2S
dS
2
1
+ S=0 . Thus,
=
=
dr
dS=
dr r
dr
r
S
r
S
S=dT /dr>0 and r>0 , we have S=e
dT 1
= k
2
dr
r
T (1)=15 1 dT = r 2 k dr 2ln r+C dT = 1
r 2 15= k+A (1) and T (2)=25 Now solve for k and A : 2 (2) + (1)
37. (a) dC
=r kC
dt
kt+M kC r =e
M =C r/k
4 0 dC
= (kC r)
dt 2 kC r=M e
3 ( 1
tan
a b ) kt =e ln r k dr 2 C 2 e =r k T (r)= 2 dT
2 + 1 b
a b
tan u
a b 1 . So ( and we have
b x
a b 1 =kt 2 dT
=0 can be written as
r dr dr
2
dr ln S = 2ln r +C . Assuming
r
1
S= k
2
r k
+A .
r 1
k+A (2) .
2
35= A , so A=35 and k=20 , and T (r)= 20/r+35 .
25= dC
=
kC r
kt kC=M e +r
3 dt (1/k)ln kC r = t+M ln kC r = kt+M 1 kt C(t)=M e +r/k . C(0)=C
4 0 2 C =M +r/k
0 4 kt C(t)= C r/k e +r/k .
0 (b) If C <r/k , then C r/k<0 and the formula for C(t) shows that C(t) increases and lim C(t)=r/k .
0 0 t As t increases, the formula for C(t) shows how the role of C steadily diminishes as that of r/k
0 increases.
11 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 38. (a) Use 1 billion dollars as the x unit and 1 day as the t unit. Initially, there is $10 billion of old
currency in circulation, so all of the $50 million returned to the banks is old. At time t , the amount of
new currency is x(t) billion dollars, so 10 x(t) billion dollars of currency is old. The fraction of
circulating money that is old is 10 x(t) /10 , and the amount of old currency being returned to the
10 x(t)
0.05 billion dollars. This amount of new currency per day is introduced
banks each day is
10
dx 10 x
into circulation, so
=
0.05=0.005(10 x) billion dollars per day.
dt
10
dx
dx
0.005t
c
(b)
=0.005dt
= 0.005dt ln (10 x)= 0.005t+c 10 x=Ce
, where C=e
10 x
10 x ( 0.005t 0.005t ) x(t)=10 Ce
. From x(0)=0 , we get C=10 , so x(t)=10 1 e
.
(c) The new bills make up 90% of the circulating currency when x(t)=0.9 10=9 billion dollars. ( 0.005t 9=10 1 e
1.26 years. ) 0.9=1 e 0.005t e 0.005t 0.005t= ln 10 =0.1 t=200ln 10 460.517 days 39. (a) Let y(t) be the amount of salt (in kg) after t minutes. Then y(0)=15 . The amount of liquid in
the tank is 1000 L at all times, so the concentration at time t (in minutes) is y(t)/1000 kg/L and
dy
y(t) kg
L
y(t) kg
dy
1
t
=
10
=
.
=
dt ln y=
+C , and y(0)=15
dt
1000 L
min
100 min
y
100
100
t
y
t
y
t/100
t/100
ln 15=C , so ln y=ln 15
. It follows that ln
=
and
=e
, so y=15e
kg.
100
15
100
15
(b) After 20 minutes, y=15e 20/100 =15e 0.2 12.3 kg. 40. (a) If y(t) is the amount of salt (in kg) after t minutes, then y(0)=0 and the total amount of liquid
in the tank remains constant at 1000 L.
dy
kg
L
kg
L
y(t) kg
L
=
0.05
5
+ 0.04
10
15
dt
L
min
L
min
1000 L
min
130 3y kg
= 0.25+0.40 0.015y=0.65 0.015y=
200 min
dy
dt
1
1
1
so
=
and
ln 130 3y =
t+C ; since y(0)=0 , we have
ln 130=C ,
130 3y
200
3
200
3
1
1
1
3
3t/200
so
ln 130 3y =
t
ln 130 ln 130 3y =
t+ln 130=ln 130e
, and
3
200 3
200 ( ) 3t/200 . Since y is continuous, y(0)=0 , and the right hand side is never zero, we
130
3t/200
3t/200
and y=
1 e
kg.
deduce that 130 3y is always positive. Thus, 130 3y=130e
3
130
130
3 60/200
0.9
(b) After one hour, y=
1 e
=
1 e
25.7 kg.
3
3
Note: As t
,
130 3y =130e ( ( ) ( ) ) 12 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations y(t) 130
1
=43 kg.
3
3
/ 41. Assume that the raindrop begins at rest, so that v(0)=0 . dm/dt=km and ( mv ) =gm
dv
/
/
/
/
= dt
mv +vm =gm mv +v(km)=gm v +vk=g dv/dt=g kv
( 1/k ) ln g kv =t+C
g kv
ln g kv = kt kC
t ,e kt kt g kv=Ae . v(0)=0 A=g . So kv=g ge ( kt v= ( g/k ) 1 e kt ) . Since k>0 , as 0 and therefore, lim v(t)=g/k .
t 42. (a) m
ln v
v 0 dv
= kv
dt
k
=
t
m dv
k
k
=
dt ln v =
t+C . Since v(0)=v , ln v =C . Therefore,
0
0
v
m
m
kt/m
kt/m
v
=e
. The sign is + when t=0 , and we assume v is
v(t)= v e
0
v
0 continuous, so that the sign is + for all t . Thus, v(t)=v e kt/m 0 mv
From s(0)=s , we get s =
0 k 0 0 +C / mv / , so C =s +
0 0 k . ds/dt=v e mv kt/m s(t)= 0 mv
and s(t)=s + e mv
traveled from time 0 to time t is s(t) s , so the total distance traveled is lim
0 s(t) s =
0 t Note: In finding the limit, we use the fact that k>0 to conclude that lim e kt/m +C / . ( 1 e kt/m) . The distance 0 k 0 k 0 k 0 . kt/m =0 . t (b) m dv
2
= kv
dt dv
2 k
dt
m = v 1
kt
=
+C
v
m 1 kt
1 kt 1
1
=
C . Since v(0)=v , C=
and = +
0
v m
v m v
v
0 mv
mv
0
ds
0
1
. Therefore, v(t)=
=
.
=
kt/m+1/v kv0t+m dt kv0t+m
0 Since s(0)=s , we get s =
0 0 m
ln m+C
k / / ( ) ( 0 ) kv dt
0 kv t+m
0 = m
ln kv t+m +C
0
k / . m
ln m
0 k
kv t+m C =s m
m
ln kv t+m ln m =s + ln
0 k
0
0 k
v
0
m
s(t) as v(t)=
and s(t)=s + ln
0 k
1+ kv /m t
s(t)=s + m
s(t)=
k 0 0 m
kv
0
1+
t
m . We can rewrite the formulas for v(t) and
. Remarks: This model of horizontal motion through a resistive medium was designed to handle the
case in which
13 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 2 v >0 . Then the term kv representing the resisting force causes the object to decelerate. The absolute
0 value in the expression for s(t) is unnecessary (since k , v , and m are all positive), and lim s(t)=
0 . t
2 In other words, the object travels infinitely far. However, lim v(t)=0 . When v <0 , the term kv
0 t increases the magnitude of the object’s negative velocity. According to the formula for s(t) , the
as t approaches m/k v : lim
s(t)=
. Again the
position of the object approaches ( )
0 t ( 0) m/ kv object travels infinitely far, but this time the feat is accomplished in a finite amount of time. Notice
v(t)=
when v <0 , showing that the speed of the object increases without limit.
also that lim
0 ( 0) t m/ kv 43. (a) The rate of growth of the area is jointly proportional to A(t) and M A(t) ; that is, the rate is
proportional to the product of those two quantities. So for some constant k , dA/dt=k A (M A) . We
are interested in the maximum of the function dA/dt (when the tissue grows the fastest), so we
differentiate, using the Chain Rule and then substituting for dA/dt from the differential equation:
d
dA
dA
1 1/2 dA
1
1/2 dA
= k
A ( 1)
+(M A)
A
= kA
2A+ ( M A)
dt
dt
dt
2
dt
2
dt
1
1 2
1/2
=
kA
k A(M A) M 3A = k (M A)(M 3A)
2
2
This is 0 when M A=0 and when M 3A=0 A(t)=M/3 . This represents a maximum by the First
d
dA
Derivative Test, since
goes from positive to negative when A(t)=M/3 .
dt
dt
(b) From the CAS, we get A(t)=M Ce
Ce M kt
M kt 2 1 . To get C in terms of the initial area A and the
0 +1 maximum area M , we substitute t=0 and A=A =A(0) : A =M
0 C A =C M A +
0 0 M M+ A =C M C
0 M+
M+ A =C
0 M A 0 0 A 2 A =(C 1) M (C+1) 0 A 0 A M C 1
C+1 C= 0 . (Notice that if A =0 , then C=1 .)
0 0 dv
dv dx
dv
=m
=mv
=
44. (a) According to the hint we use the Chain Rule: m
dt
dx dt
dx 2 mgR ( x+R) 2 14 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.3 Separable Equations 2 2 ( x+R) v 2 gR
0
v
gR
=
+C
=
+C . When x=0 , v=v , so
0
2 0+R
2 x+R gR dx vdv= 2 2 2 C= 1 2
v gR
2 0 2 1 2 1 2 gR
v
v=
gR . Now at the top of its flight, the rocket’s velocity will be 0 , and its height will
2
2 0 x+R
2 be x=h . Solving for v :
0 (b) v =lim v =lim
e h 0 h
2 1 2 gR
v=
gR
2 0 h+R
2gRh
=lim
R+h h (c) v = 2 32ft/s 3960mi 5280ft/mi
e 2 v 0 2 2 =g R
R ( R+h )
+
R+h
R+h = gRh
R+h v=
0 2gRh
.
R+h 2gR
= 2gR
( R/h ) +1
36 , 581 ft / s 6.93 mi / s 15 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.4 Exponential Growth and Decay 1 dP
dP
=0.7944 , so
=0.7944P and, by Theorem 2,
P dt
dt
0.7944 ( 6 )
234.99 or about 235 members.
. Thus, P(6)=2e 1. The relative growth rate is
0.7944t P(t)=P(0)e 0.7944t =2e kt 1
hour), there are 120 cells, so
3 kt 2. (a) By Theorem 2, P(t)=P(0)e =60e . In 20 minutes (
1
3 P k/3 k/3 =60e =120 (b) P(t)=60e ( ln 8 ) t e =2 k/3=ln 2 ( 3) k=3ln 2=ln 2 =ln 8 . t =60 8 8 24 (c) P(8)=60 8 =60 2 =1 , 006 , 632 , 960
/ (d) dP/dt=kP P (8)=kP(8)=(ln 8)P(8) 2.093 billion cells / h (e) P(t)=20 , 000 t t 60 8 =20 , 000 tln 8=ln ( 1000/3) 8 =1000/3 kt kt k(3) 3. (a) By Theorem 2, y(t)=y(0)e =500e . Now y(3)=500e
k= ( ln 16 ) /3 . So y(t)=500e =8000 t= ln ( 1000/3)
ln 8 3k e = 8000
500 2.79 h 3k=ln 16 t/3
( ln 16 ) t/3
=500 16 4/3 (b) y(4)=500 16 20 , 159
1
4/3
(c) dy/dt=ky y (4)=ky(4)= ln 16 500 16
18 , 631 cells / h
3
1
t/3
t/3
(d) y(t)=500 16 =30 , 000 16 =60
tln 16=ln 60 t=3(ln 60)/(ln 16) 4.4h
3 ( / kt ) 2k 4. (a) y(t)=y(0)e 8k y(2)=y(0)e =600 , y(8)=y(0)e =75 , 000 . Dividing these equations, we get
1
3
8k 2k
6k
3
e /e =75 , 000/600 e =125 6k=ln 125=ln 5 =3ln 5 k= ln 5= ln 5 . Thus,
6
2
2k
ln 5 600
y(0)=600/e =600/e =
=120 .
5
kt (ln 5)t/2 (b) y(t)=y(0)e =120e t/2 or y=120 5 5/2 (c) y(5)=120 5 =120 25 5=3000 5 6708 bacteria.
1
t/2
/
t/2
t/2
(d) y(t)=120 5
y (t)=120 5 ln 5 =60 ln 5 5 .
2
/ 5/2 y (5)=60 ln 5 5 =60 ln 5 25 5 5398 bacteria / hour.
(ln 5)t/2 5000
(ln 5)t/2
=
(e) y(t)=200 , 000 120e
=200 , 000 e
3
5000
t= 2ln
/ln 5 9.2 h.
3 (ln 5)t/2=ln 5000
3 1 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.4 Exponential Growth and Decay 5. (a) Let the population (in millions) in the year t be P(t) . Since the initial time is the year 1750, we
k ( t 1750 )
substitute t 1750 for t in Theorem 2, so the exponential model gives P(t)=P(1750)e
. Then
980 k(50)
980
1
980
k(1800 1750)
=e
ln
=50k k=
ln
0.0043104 . So with
P(1800)=980=790e
790
790
50
790
k(1900 1750) k(1950 1750) this model, we have P(1900)=790e
1508 million, and P(1950)=790e
million. Both of these estimates are much too low. 1871 k(t 1850) (b) In this case, the exponential model gives P(t)=P(1850)e
1650
1
1650
k(1900 1850)
P(1900)=1650=1260e
ln
=k(50) k=
ln
1260
50
1260
k(1950 1850) model, we estimate P(1950)=1260e
estimate of P(1950) in part (a). 0.005393 . So with this 2161 million. This is still too low, but closer than the
k(t 1900) k(1950 1900) P(1950)=2560=1650e
(c) The exponential model gives P(t)=P(1900)e
2560
1
2560
ln
=k(50) k=
ln
0.008785 . With this model, we estimate
1650
50
1650
k(2000 1900) P(2000)=1650e
3972 million. This is much too low. The discrepancy is explained by the
fact that the world birth rate (average yearly number of births per person) is about the same as always,
whereas the mortality rate (especially the infant mortality rate) is much lower, owing mostly to
advances in medical science and to the wars in the first part of the twentieth century. The exponential
model assumes, among other things, that the birth and mortality rates will remain constant.
6. (a) Let P(t) be the population (in millions) in the year t . Since the initial time is the year 1900, we
k(t 1900) substitute t 1900 for t in Theorem 2, and find that the exponential model gives P(t)=P(1900)e
1
92
k(1910 1900)
P(1910)=92=76e
k=
ln
0.0191 . With this model, we estimate
10
76
k(2000 1900) P(2000)=76e
514 million. This estimate is much too high. The discrepancy is explained
by the fact that, between the years 1900 and 1910, an enormous number of immigrants (compared to
the total population) came to the United States. Since that time, immigration (as a proportion of total
population) has been much lower. Also, the birth rate in the United States has declined since the turn
of the century. So our calculation of the constant k was based partly on factors which no longer exist.
(b) Substituting t 1980 for t in Theorem 2, we find that the exponential model gives
1
250
k(t 1980)
k(1990 1980)
P(t)=P(1980)e
P(1990)=250=227e
k=
ln
0.00965 . With this model,
10
227
k(2000 1980) we estimate P(2000)=227e
P(2010)=227e
(c) 30k 275.3 million. This is quite accurate. The further estimates are 303 million and P(2020)=227e 40k 334 million. 2 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.4 Exponential Growth and Decay The model in part (a) is quite inaccurate after 1910 (off by 5 million in 1920 and 12 million in 1930).
The model in part (b) is more accurate (which is not surprising, since it is based on more recent
information).
7. (a) If y= N O
(b) y(t)=Ce 2 5
0.0005t then by Theorem 2,
=0.9C e 0.0005t =0.9 dy
= 0.0005y
dt y(t)=y(0)e 0.0005t=ln 0.9
kt 0.0005t =Ce 0.0005t . t= 2000ln 0.9 211 s
kt 8. (a) The mass remaining after t days is y(t)=y(0)e =800e . Since the half life is 5.0 days,
5k
5k 1
y(5)=800e =400 e =
2
1
(ln 2)t/5
t/5
5k=ln
k= (ln 2)/5 , so y(t)=800e
=800 2 .
2
(b) y(30)=800 2
(c) 800e 30/5 =12.5 mg (ln 2)t/5 =1 (ln 2) t
1
ln 800
=ln
= ln 800 t=5
5
800
ln 2 48 days (d) kt kt 9. (a) If y(t) is the mass (in mg) remaining after t years, then y(t)=y(0)e =100e .
30k 1
30k 1
(ln 2)t/30
t/30
y(30)=100e = (100) e =
k= (ln 2)/30 y(t)=100e
=100 2
2
2
(b) y(100)=100 2
(c) 100e 100/30 9.92 mg (ln 2)t/30 =1
3 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.4 Exponential Growth and Decay (ln 2)t/30=ln 1
100 t= 30 ln 0.01
ln 2 199.3 years
kt 3k 10. (a) If y(t) is the mass after t days and y(0)=A , then y(t)=Ae . y(3)=Ae =0.58A
1
3k
( ln 0.58 ) t/3 1
e =0.58 3k=ln 0.58 k= ln 0.58 . Then Ae
= A
3
2
1
(ln 0.58)t
1
3ln 2
( ln 0.58 ) t/3
ln e
=ln
=ln
, so the half life is t=
3.82 days.
2
3
2
ln 0.58
(ln 0.58)t
1
3ln 10
( ln 0.58 ) t/3
=0.10A
(b) Ae
=ln
t=
12.68 days
3
10
ln 0.58
11. Let y(t) be the level of radioactivity. Thus, y(t)=y(0)e kt and k is determined by using the half
1
ln
1
1
2
ln 2
k(5730) 1
5730k 1
life: y(5730)= y(0) y(0)e
= y(0) e
=
5730k=ln
k=
=
. If
2
2
2
2
5730 5730
tln 2
14
t ( ln 2 ) /5730
74% of the C remains, then we know that y(t)=0.74y(0) 0.74=e
ln 0.74=
5730
5730(ln 0.74)
t=
2489 2500 years.
ln 2
12. From the information given, we know that
2(0) use the point ( 0,5) : 5=Ce dy
=2y
dx 2x y=Ce by Theorem 2. To calculate C we
2x C=5 . Thus, the equation of the curve is y=5e . dT
dT
=k(T T ) , we have
=k(T 75) .
s
dt
dt
Now let y=T 75 , so y(0)=T (0) 75=185 75=110 , so y is a solution of the initial value
13. (a) Using Newton’s Law of Cooling, kt kt problem dy/dt=ky with y(0)=110 and by Theorem 2 we have y(t)=y(0)e =110e .
15
1
15
30k
30k 75
y(30)=110e =150 75 e =
=
k=
ln
,
110 22
30
22
so y(t)=110e 1
tln
30 (b) T (t)=100
25
110
15
ln
22 15
22 and y(45)=110e y(t)=25 . y(t)=110e 1
tln
30 45
ln
30
15
22 15
22 =25 62 F. Thus, T (45) 62+75=137 F.
e 1
tln
30 15
22 = 25
110 1
15
25
tln
=ln
30
22
110 30ln
t= 116 min. 4 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.4 Exponential Growth and Decay 14. (a) Let T (t)= temperature after t minutes. Newton’s Law of Cooling implies that dT
=k(T 5) .
dt dy
kt
kt
kt
=ky , so y(t)=y(0)e =15e
T (t)=5+15e
dt
7
k 7
k
ln ( 7/15 ) t
2ln ( 7/15 )
k=ln
, so T (t)=5+15e
and T (2)=5+15e
T (1)=5+15e =12 e =
8.3 C.
15
15
1
ln
7
1
15
ln ( 7/15 ) t 1
ln ( 7/15 ) t
(b) 5+15e
=
ln
t=ln
t=
3.6 min.
=6 when e
15
15
15
7
ln
15
Let y(t)=T (t) 5 . Then dT
dy
kt
=k(T 20) . Letting y=T 20 , we get
=ky , so y(t)=y(0)e . y(0)=T (0) 20=5 20= 15 , so
dt
dt
25k
25k
25k
25k 2
y(25)=y(0)e = 15e , and y(25)=T (25) 20=10 20= 10 , so 15e = 10 e = . Thus,
3
2
1
2
kt
(1/25)ln (2/3)t
25k 2
25k=ln
and k=
ln
, so y(t)=y(0)e = 15e
. More simply, e =
3
25
3
3
1/25
t/25
t/25
2
2
2
k
kt
e=
e =
y(t)= 15
.
3
3
3
2 50/25
2 2
20
(a) T (50)=20+y(50)=20 15
=20 15
=20
=13.3 C
3
3
3
2 t/25
2 t/25
2 t/25 1
(b) 15=T (t)=20+y(t)=20 15
15
=5
=
3
3
3
3
2
1
1
2
(t/25)ln
=ln
t=25ln
ln
67.74 min.
3
3
3
3
15. / dT
dy
kt
=k(T 20) . Let y=T 20 . Then
=ky , so y(t)=y(0)e . y(0)=T (0) 20=95 20=75 , so
dt
dt
dT
dy
kt
y(t)=75e . When T (t)=70 ,
= 1 C / min. Equivalently,
= 1 when y(t)=50 . Thus,
dt
dt
dy
kt
1= =ky(t)=50k and 50=y(t)=75e . The first relation implies k= 1/50 , so the second relation says
dt
2
2
t/50
t/50 2
50=75e
. Thus, e
=
t/50=ln
t= 50ln
20.27 min.
3
3
3 16. kh kh 17. (a) Let P(h) be the pressure at altitude h . Then dP/dh=kP P(h)=P(0)e =101.3e .
87.14
1
87.14
1000k
k=
ln
P(h)=101.3
P(1000)=101.3e
=87.14 1000k=ln
101.3
1000
101.3
e 1
hln
1000 87.14
101.3 3ln , so P(3000)=101.3e 87.14
101.3 64.5 kPa.
5 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.4 Exponential Growth and Decay 6187
ln
1000 (b) P(6187)=101.3e r
1+
n 18. (a) Using A=A 0 87.14
101.3 39.9 kPa nt with A =500 , r=0.14 , and t=2 , we have:
0 1 2 A=500 0.14
1+
1 Quarterly: n=4 ; 1+ 0.14
4 4 2 A=500 (iii) Monthly: n=12 ; 1+ 0.14
12 12 2 A=500 (iv) Daily: n=365 ; 1+ 0.14
365 365 2 A=500 (i) Annually: n=1 ; (ii) (v) =$658.40
=$660.49
=$661.53
365 24 2 0.14
1+
365 24 Hourly: n=365 24 ; A=500 =$661.56 (0.14)2 (vi) Continuously: A=500e (b)
A (2)=$661.56 , A
0.14 =$649.80 =$661.56 (2)=$610.70 , and A 0.10 19. (a) Using A=A 0 1+ (2)=$563.75 . 0.06 r
n nt with A =3000 , r=0.05 , and t=5 , we have:
0 (i) Annually: n=1 ; (ii) Semiannually: n=2 ; A=3000 A=3000 0.05
1 1 5 0.05
1+
2 2 5 1+ =$3828.84
=$3840.25 (iii) Monthly: n=12 ; 1+ 0.05
12 12 5 A=3000 (iv) Weekly: n=52 ; 1+ 0.05
52 52 5 A=3000 =$3850.08
=$3851.61 6 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.4 Exponential Growth and Decay (v) Daily: n=365 ; A=3000 1+ 365 5 0.05
365 =$3852.01 (0.05)5 (vi) Continuously: A=3000e =$3852.08 (b) dA/dt=0.05A and A(0)=3000 .
0.06t 20. (a) A e
0 0.06t =2A e 0 =2 50
ln 2 11.55 , so the investment will double in
3 0.06t=ln 2 t= about 11.55 years.
t 0.06t (b) The annual interest rate in A=A (1+r) is r . From part (a), we have A=A e
0 t 0.06t 0.06 must be equal, so (1+r) =e
annual interest rate. 1+r=e m
k 0.06 r=e . These amounts 1 0.0618=6.18% , which is the equivalent m
dy dP
, so
=
and the differential equation becomes
k
dt dt
dy
m
m
m
m
kt
kt
kt
=ky . The solution is y=y e
P
= P
e
P(t)= + P
e .
0
0 k
0 k
dt
k
k
m
(b) Since k>0 , there will be an exponential expansion P
>0 m<kP .
0 k
0
m
m
(c) The population will be constant if P
=0 m=kP . It will decline if P
<0 m>kP .
0 k
0
0 k
0
(d) P =8 , 000 , 000 , k=
=0.016 , m=210 , 000 m>kP ( =128 , 000 ), so by part (c), the
21. (a) dP
=kP m=k
dt 0 P . Let y=P 0 0 population was declining. 22. (a) dy
1+c
=ky
dt y 1 c dy=k dt c y c y
=kt+C . Since y(0)=y , we have C= , or y =y c 0 1 c ckt . So y =
y c 0 (b) y(t) c as 1 cy kt
0 ckt y
= 0 c 0 c c c . Thus, y c c c y
=kt+ 0 c y 0
c and y(t)= 1 cy kt c 1 cy kt 0 1/c . 0 1 0 , that is, as t 0 c cy k
0 . Define T = 1
c cy k
0 . Then lim y(t)=
t . T (c) According to the data given, we have c=0.01 , y(0)=2 , and y(3)=16 , where the time t is given in
y
1
c
0
months. Thus, y =2 and 16=y(3)=
. Since T = c , we will solve for cy k .
0
0
1/c
c
cy k
1 cy k 3
0
0 7 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.4 Exponential Growth and Decay 16= 2 c c 1 3cy k 100 1 3cy k=
0 1
8 0.01 =8 0.01 c cy k=
0 1
0.01
1 8
. Thus, doomsday occurs
3 ( ) 0 when t=T = 1
c cy k
0 = 3
1 8 0.01 145.77 months or 12.15 years. 8 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.5 The Logistic Equation 2 1. (a) dP/dt=0.05P 0.0005P =0.05P(1 0.01P)=0.05P ( 1 P/100 ) . Comparing to Equation 1,
dP/dt=kP ( 1 P/K ) , we see that the carrying capacity is K=100 and the value of k is 0.05 .
(b) The slopes close to 0 occur where P is near 0 or 100 . The largest slopes appear to be on the line
P=50 . The solutions are increasing for 0<P <100 and decreasing for P >100 .
0 0 (c)
All of the solutions approach P=100 as t increases. As in part (b), the solutions differ since for
0<P <100 they are increasing, and for P >100 they are decreasing. Also, some have an IP and some
0 0 dont. It appears that the solutions which have P =20 and P =40 have inflection points at P=50 .
0 0 (d) The equilibrium solutions are P=0 (trivial solution) and P=100 . The increasing solutions move
.
away from P=0 and all nonzero solutions approach P=100 as t
2. (a) K=6000 and k=0.0015 dP/dt=0.0015P ( 1 P/6000 ) . (b)
All of the solution curves approach 6000 as t . (c)
The curves with P =1000 and P =2000 appear to be concave upward at first and then concave
0 0 downward. The curve with P =4000 appears to be concave downward everywhere. The curve with
0 P =8000 appears to be concave upward everywhere. The inflection points are where the population
0 grows the fastest.
1 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.5 The Logistic Equation (d) See the solution to Exercise .2.25 for a possible program to calculate P(50) . We find that
P(50) 1064 .
K
6000
=
(e) Using Equation 4 with K=6000 , k=0.0015 , and P =1000 , we have P(t)=
0
kt
0.0015t
1+Ae
1+Ae
K P 6000 1000
0
6000
=
=5 . Thus, P ( 50 ) =
, where A=
1064.1 , which is extremely
0.0015 ( 50 )
1000
P
1+5e
0
close to the estimate obtained in part (d). (f)
The curves are very similar.
3. (a) dy
=ky
dt 1 y
K K y(t)= 1+Ae kt with A= K y(0)
7
. With K=8 10 , k=0.71 , and
y(0) 7 8 10 7 y(0)=2 10 , we get the model y(t)= 1+3e
7 8 10 7 (b) y(t)=4 10 1+3e 0.71t 7 =4 10 0.71t 2=1+3e 7 , so y(1)= 8 10
1+3e 0.71t e 0.71t = 0.71 1
3 7 3.23 10 kg. 0.71t=ln 1
3 t= ln 3
0.71 1.55 years 4. (a)
From the graph, we estimate the carrying capacity K for the yeast population to be 680 .
1 dP 1 39 18 7
=
=
=0.583 .
(b) An estimate of the initial relative growth rate is
12
P dt 18 2 0
0 (c) An exponential model is P(t)=18e 7t/12 . A logistic model is P(t)= 680
1+Ae A= 7t/12 , where 680 18 331
=
.
18
9
2 Stewart Calculus ET 5e 0534393217;9. Differential Equations; 9.5 The Logistic Equation (d)
Time in
Hours
0
2
4
6
8
10
12
14
16
18 Observed
Values
18
39
80
171
336
509
597
640
664
672 Exponentia