Unformatted text preview: Section 1.1 Chapter 1
Prerequisites for Calculus Section 1.1 Exercises
1. 1. y 2 4(3 3) 2. 3
3
2x
x 3
3 2(x
2x 2 4(0) 1)
2 2 0 1
2 x
y 1
2 ( 3) 2
2
4 3. Quick Review 1.1 1
1 2. s Section 1.1 Lines (pp. 1–9) x
y x
y 2
4. 2
1 x
y 8
1 1 0 ( 3)
0 0
2 2
3 5 0
4 6 5. (a, c) y
5 3. m 2
5 3
4 1
1 4. m 2
3 ( 3)
( 1) 1
B 5. (a) 3(2) A 5 Yes 4( 1)
13 5
5 (b) m
No 9
(x2 7. d 2( 2)
No
x1)2 (0 1
2 ( 2)
1 3
1 6. (a, c) 2( 1) 5
2 5 Yes (b) 1
1 x 5 1 6 6. (a) 7
7 5 1
4 4 (b) 3(3) 5
4 3 y
5 5 (y2 1)2 y1)2 5 0)2 (1 x A
B 2 (x2 x1)2 (1 8. d = 2)2 (y2
1
3 y1)2
2 1 (b) m 2 ( 1)
1 ( 2) 7. (a, c) y B 16
9 1 1
3 5 42
3 ( 1)2 1
3 A
5 25
9 x 5
3 9. 4x 3y 7 3y 4x
4
x
3 y
10. 2x (b) m 5y 7
7
3 33
12 0
3 8. (a, c) 0 y
5 3 A 5y 2x 3 y 2
x
5 3
5 5
B x 1 2 Section 1.1 8. continued 26. The line contains (0, 0) and (5, 2).
32
11 (b) m 5
(undefined)
0 m This line has no slope.
9. (a) x
(b) y y 2
3 27. 3x 10. (a) x 1 (b) y 0
2
0 y 1) 1[x
1(x 3
3
4 1
( 1)] 1
1) 1 2(x 16. y
y y 12 (c) 1(x 14. y
y 17. m 3x (b) yintercept: 3 (b) y 15. y 12 (a) Slope: 12. (a) x
13. y 2
5 3
x
4 y (b) y 0
0 4y 4y 4
3 11. (a) x 2
5
2
x
5 0) 2[x
2(x 3
[ 10, 10] by [ 10, 10] ( 4)] 0
4) 0 28. x
y 30
3
20
2
3
(x 0) 0
2
3
x
2 2y 2y 2
x 2 (a) Slope: 1 (b) yintercept: 2
(c) 3x 3x y 18. m 0 1
2 1
1 y 0(x y [ 10, 10] by [ 10, 10] 0
1 0
29. 1) 1
2 19. m 1 2
(undefined)
0 0
( 2) 2 Vertical line: x
20. m
y 2
2 3
[x
4 4y 3(x 4y 3x 3x
21. y 2) 1
x
3
4
x
3 1
4
4
3 (b) yintercept: 4 3
4 (c) 1
4 2 4y
3x 3
4 ( 2)] y y
4 (a) Slope: 2 1
( 2) x
3
y
4 [ 10, 10] by [ 10, 10] 30. y 2 2x 4 (a) Slope: 2 2 22. y 1x 2 or y 23. y 1
x
2
1
x
3 2 (b) yintercept: 4 3 24. y x (c) 1
[ 10, 10] by [ 10, 10] 25. The line contains (0, 0) and (10, 25).
m
y 25
10
5
x
2 0
0 25
10 5
2 Section 1.1
31. (a) The desired line has slope
(0, 0):
y
1(x 0) 0 or y 1 and passes through
x.
1
1 (b) The desired line has slope
(0, 0):
y 1(x 0) 0 or y x. 1 and passes through 39. (a) y (c) 1
2 1
and passes through
2 1
(x
2 2) 1
x
2 2 or y (b) We seek a horizontal line through ( 2, 4): y line through 1, 1
:y
2 1
.
2
1
1, : x
2 (b) We seek a vertical line through
2
1 f (x) 1) 7
x
2 Since f (x)
4
4 3
,
2 ( 1)
2
3
(x
2 f (x) 2 7
(5)
2 Check: f (5) 36. m 7
3
x
2
2
3
16, as expected.
2
7
we have m
and b
2 3
2 2) 3
x
2 Since f (x) 4 2
(6)
3 y
y 3 1 y [1975, 1995] by [20,000, 35,000] (d) When x 2000,
y 1,060.4233(2000) 2,077,548.669 43,298.
In 2000, the construction workers’ average annual compensation will be about $43,298. 42. (a) When y (b) When y 2 When x 3
and b
2 2. 2(x
x 2
x 8) 4 8 2 x 1, so x c. 1, so y d. 2, so x 2c. 2, so y 2d. and y x 2
and
k so k ( 2)
( 8) 44. (a) m 3
k 1, respectively, so the slopes are 1. The lines are parallel when 2
k 1, 2. 68 69.5
0.4 0 (b) m 10 68
4 0.4 (c) m
6 2
x
k 43. (a) The given equations are equivalent to y (b) The lines are perpendicular when
k
2. 2 38. x
c
y
0, we have
d
x
0, we have
c
y
0, we have
d 0, we have The xintercept is 2c and the yintercept is 2d. 3 4 3
.
2 7, as expected. 2, we have m 3
( 2) 37. y 2 2,077,548.669 (c) When x
3
x
2 29.413. 41. y 1 (x 3) 4
yx34
yx1
This is the same as the equation obtained in Example 5. 3
2 ( 1) 3
(6)
2 Check: f (6) 2
3 1. 7
2 7
(x
2 1,060.4233x 9.013 4. 34. (a) The given line is horizontal, so we seek a horizontal 9
3 (d) When x 30, y 0.680(30)
She weighs about 29 pounds. (b) The slope is 1,060.4233. It represents the approximate
rate of increase in earnings in dollars per year. 3. 33. (a) The given line is vertical, so we seek a vertical line
through ( 2, 4): x
2. 35. m [15, 45] by [15, 45] 40. (a) y ( 2, 2):
y 9.013 (b) The slope is 0.68. It represents the approximate average
weight gain in pounds per month. 32. (a) The given equation is equivalent to y
2x 4.
The desired line has slope 2 and passes through
( 2, 2):
y
2(x 2) 2 or y
2x 2.
(b) The desired line has slope 0.680x 3 5 10
4.7 4 1.5
0.4
58
3.6
5
0.7 2
k 1
, so
1 3.75 degrees/inch
16.1 degrees/inch
7.1 degrees/inch (d) Best insulator: Fiberglass insulation
Poorest insulator: Gypsum wallboard
The best insulator will have the largest temperature
change per inch, because that will allow larger
temperature differences on opposite sides of thinner
layers. 4 Section 1.1
p
d 45. Slope: k 10.94 1
100 0 9.94
100 y
6 0.0994 atmospheres per meter
(–1, 4) At 50 meters, the pressure is
p 0.0994(50) 46. (a) d(t) 1 (2, 3) 5.97 atmospheres. 45t (–1, 1) (b) x 6 (2, 0) y
[0, 6] by [ 50, 300] 6 (c) The slope is 45, which is the speed in miles per hour.
(d) Suppose the car has been traveling 45 mph for several
hours when it is first observed at point P at time t 0.
(e) The car starts at time t
47. (a) y 5632x (2, 3) 0 at a point 30 miles past P. (–1, 1) 11,080,280 2732x 5,362,360 50. (d) The median price is increasing at a rate of about $5632
per year in the Northeast, and about $2732 per year in
the Midwest. It is increasing more rapidly in the
Northeast.
48. (a) Suppose x F is the same as x C.
9
x 32
5
9
x 32
5 x
1 4
x
5 y
(c , d ) W
(a , b ) x X Z
(g , h ) 32 x (–1, –2) (b) The rate at which the median price is increasing in
dollars per year
(c) y 6 (2, 0) (e , f ) Y Suppose that the vertices of the given quadrilateral are x 40 Yes, 40 F is the same as (a, b), (c, d), (e, f ), and (g, h). Then the midpoints of the
40 C. (b) cb d
c ed f
,
,X
,
,
2
2
2
2
e gf h
g ah b
Y
,
, and Z
,
. When these four
2
2
2
2 consecutive sides are W a points are connected, the slopes of the sides of the resulting
figure are:
[ 90, 90] by [ 60, 60] d It is related because all three lines pass through the
point ( 40, 40) where the Fahrenheit and Celsius
temperatures are the same.
49. The coordinates of the three missing vertices are (5, 2),
( 1, 4) and ( 1, 2), as shown below.
y
6
(2, 3)
(5, 2)
(–1, 1)
(2, 0) 6 x WX: c f d
2 e 2
h
2
XY: e g
2
fh
2
ZY: e g
2
hb
2
WZ: g a
2
f b 2
a c f
e b
a 2
d f
2 c e h
g d
c f
e b
a 2
h b
2 g a 2
bd
2
ac
2 h
g d
c Opposite sides have the same slope and are parallel. Section 1.1
4
3 51. The radius through (3, 4) has slope 0
0 4
.
3 The tangent line is tangent to this radius, so its slope is
3
(x
4
3
x
4
3
x
4 y
y
y 3) 1
4/3 3
. We seek the line of slope
4 3
that passes through (3, 4).
4 4 9
4
25
4 4 52. (a) The equation for line L can be written as
A
x
B y C
, so its slope is
B B
(x
A y a) A
. The perpendicular line has slope
B B
A B
(x
A a) 2 2 Ax B (x a) b for y in the equation for line L gives: b C a) ABb AC (A2 B B
and passes through (a, b), so its equation is
A b. (b) Substituting (x
Ax 1
A/B B2)x B2 a
B2 a x AC ABb AC ABb
A2 B 2 Substituting the expression for x in the equation for line L gives:
B2 a A
By
By
By
y AC ABb
By C
A2 B 2
A(B2a AC ABb)
C(A2 B2)
2
2
A
B
A2 B 2
2
2
2
2
AB a A C A Bb A C B2C
A2 B2
A2Bb A2 b B2C AB2a
A2 B2 BC ABa
A2 B2
B2a The coordinates of Q are
(c) Distance A a)2 (x
B2a (y B 2 A Bb A 2 a 2
B2
A AC BC 2 A(C Aa) 2 Bb
A2 (A2 B2)(C Aa
(A2 B2)2
Aa (C
A
C Bb)2 2 2 Aa
A2 Aa B Bb
B2 Bb
A 2 C
B 2 BC
ABa
A2 B 2 B2) 2 A2b ABa B2b 2
B2
A
2 B(C B2 A2(C Aa Bb)2
(A2 B2)2 A2 b a ABb a(A2
A2 B2 AC BC ABa
.
A2 B2 , 2 b)2 AC
ABb
A2 B 2 B2a ABb A2b AC
2 Bb) 2 Aa
A2 B2 B2(C Aa Bb)2
(A2 B2)2
Bb)2 BC 2 b
ABa b(A2
A2 B2 B2 ) 2 5 6 Section 1.2 s Section 1.2 Functions and Graphs 5. x2 16
Solutions to x2 16: x
4, x 4
Test x
6 ( 6)2 36 16
x2 16 is false when x
4
Test x 0: 02 0 16
x2 16 is true when 4 x
Test x 6: 62 36 16
x2 16 is false when x 4.
Solution set: ( 4, 4) (pp. 9–19)
Exploration 1
1. y3 g f, y4 Composing Functions
fg 2. Domain of y3: [ 2, 2] Range of y3: [0, 2]
y1: 6. 9 x2 0
Solutions to 9 x2 0: x
3, x 3
Test x
4: 9 ( 4)2 9 16
70
9 x2 0 is false when x
3.
Test x 0: 9 02 9 0
9 x2 0 is true when 3 x 3.
Test x 4: 9 42 9 16
70
9 x2 0 is false when x 3.
Solution set: [ 3, 3] [ 4.7, 4.7] by [ 2, 4.2] y2: 7. Translate the graph of f 2 units left and 3 units downward. [ 4.7, 4.7] by [ 2, 4.2] 8. Translate the graph of f 5 units right and 2 units upward. y3: 9. (a)
x2
x2
3)(x (x
[ 4.7, 4.7] by [ 2, 4.2] 3. Domain of y4: [0, ); Range of y4: (
y4: (b) , 4] f (x)
5
9
3)
x y1(x)
4 x2
2
4 (y2(x))
4 ( x)2 1
x 4 x, x 1
5 0 1
x 1. 3x 1 5x 3
2x 4
x
2
Solution: [ 2, ) 3. x 4
4x34
1x7
Solution set: [ 1, 7] 0 No solution
11. (a) 0.
x 2. (b) 2. 3 4. x 2
5
x2
5 or x 2 5
x
3 or x 7
Solution set: ( , 3] [7, ) 6 5 x 3 (b) f (x) = 0 Quick Review 1.2 2. x(x 2) 0
Solutions to x(x 2) 0: x 0, x 2
Test x
1: 1( 1 2) 3 0
x(x 2) 0 is true when x
Test x 1: 1(1 2)
10
x(x 2) 0 is false when 0
Test x 3: 3(3 2) 3 0
x(x 2) 0 is true when x
Solution set: ( , 0) (2, ) 3 or x 5 [ 2, 6] by [ 2, 6] y2 ( y1(x))
y1( y2(x)) 4
4
0
0 f (x)
6
x2
5
x2
1
No real solution 10. (a) f (x)
4. y3
y4 4 12. (a) (b) f (x) = 4
x7
4
x7
16
x
9
Check: 9 7
f (x)
x7
x7
x
Check:
3 3 f (x)
x1
x1
x
x
x f (x)
1
1
x 16 4; it checks. 1
1
1
6
6 7
2
2
8
7 3
3
27
28 1; it checks. Section 1.2
Section 1.2 Exercises
d2 r2 1. Since A 2 8. (a) Since we require
2 , the formula is A d
, where A
4 (b) ( x 0, the domain is ( 7 , 0]. , 0] (c) represents area and d represents diameter.
2. Let h represent height and let s represent side length.
s2
2 h2 s2 [ 10, 3] by [ 4, 2] (d) None 12
s
4 h2 s2 h2 32
s
4 x 0, the domain is ( 2 , 3]. 0, the domain is (b) [0, )
3 h 9. (a) Since we require 3 2 (c) s Since side length and height must be positive, the formula
3 is h 2 s.
[ 4.7, 4.7] by [ 6, 6] (d) None
s s
2 10. (a) Since we require x
( , 2) (2, ). s h (b) Since s
2 ( s 3. S 6e2, where S represents surface area and e represents
edge length. 1
x , 0) 2 can assume any value except 0, the range is (0, ). (c) 4 r3, where V represents volume and r represents
4. V
3
radius.
5. (a) (
(b) ( , ) or all real numbers
[ 4.7, 4.7] by [ 6, 6] , 4] (d) None (c) 11. (a) ( , ) or all real numbers (b) ( , ) or all real numbers (c)
[ 5, 5] by [ 10, 10] (d) Symmetric about yaxis (even)
6. (a) ( , ) or all real numbers
[ 6, 6] by [ 3, 3] (b) [ 9, ) (d) None (c) 12. (a) ( , ) or all real numbers (b) The maximum function value is attained at the point
(0, 1), so the range is ( , 1].
(c) [ 5, 5] by [ 10, 10] (d) Symmetric about the yaxis (even)
7. (a) Since we require x 1 0, the domain is [1, ). (b) [2, )
[ 6, 6] by [ 3, 3] (c) (d) Symmetric about the yaxis (even) [ 3, 10] by [ 3, 10] (d) None 8 Section 1.2 13. (a) Since we require x 0, the domain is ( , 0]. 18. (a) This function is equivalent to y
[0, ). (b) [0, ) x3, so its domain is (b) [0, ) (c) (c) [ 10, 3] by [ 1, 2]
[ 2, 5] by [ 2, 8] (d) None
14. (a) Since we require x
( , 0) (0, ).
(b) Note that (d) None 0, the domain is 19. Even, since the function is an even power of x. 1
can assume any value except 0, so 1
x 1
x 20. Neither, since the function is a sum of even and odd powers
of x.
21. Neither, since the function is a sum of even and odd powers
of x (x1 2x0). can assume any value except 1.
The range is ( , 1) (1, ). 22. Even, since the function is a sum of even powers of x
(x2 3x0). (c) 23. Even, since the function involves only even powers of x.
24. Odd, since the function is a sum of odd powers of x.
25. Odd, since the function is a quotient of an odd function (x3)
and an even function (x2 1). [ 4, 4] by [ 4, 4] (d) None
15. (a) Since we require 4 x2 26. Neither, since, (for example), y( 2)
0, the domain is [ 2, 2]. (b) Since 4 x2 will be between 0 and 4, inclusive (for x
in the domain), its square root is between 0 and
2, inclusive. The range is [0, 2].
(c) 41/3 and y(2) 0. 27. Neither, since, (for example), y( 1) is defined and y(1) is
undefined.
28. Even, since the function involves only even powers of x.
29. (a) [ 9.4, 9.4] by [ 6.2, 6.2] [ 4.7, 4.7] by [ 3.1, 3.1] Note that f (x)
x 3 2, so its graph is the graph
of the absolute value function reflected across the
xaxis and then shifted 3 units right and 2 units upward. (d) Symmetric about the yaxis (even)
16. (a) This function is equivalent to y
all real numbers. 3 x2, so its domain is (c) (b) ( ,) (c) ( (b) [0, ) , 2] 30. (a) The graph of f (x) is the graph of the absolute value
function stretched vertically by a factor of 2 and then
shifted 4 units to the left and 3 units downward. [ 2, 2] by [ 1, 2] (d) Symmetric about the yaxis (even)
17. (a) Since we require x2
( , 0) (0, )
(b) Since 1
x2 0, the domain is
[ 10, 5] by [ 5, 10] (b) (
0 for all x, the range is (1, ). , ) or all real numbers (c) [ 3, )
31. (a) (c) [ 4, 4] by [ 1, 5] (d) Symmetric about the yaxis (even) [ 4.7, 4.7] by [ 1, 6] (b) ( , ) or all real numbers (c) [2, ) Section 1.2
32. (a) 44. Line through ( 1, 0) and (0,
m 3
0 3
1 0
( 1) 3): 3, so y Line through (0, 3) and (2,
m [ 4, 4] by [ 2, 3] (b) ( 13
20 , ) or all real numbers 3 1): 4
2 2, so y 3,
3, 3x
2x f (x) (c) [0, ) 3x 1
0 2x x
x 3 0
2 33. (a)
45. Line through ( 1, 1) and (0, 0): y x Line through (0, 1) and (1, 1): y 1 Line through (1, 1) and (3, 0):
[ 3.7, 5.7] by [ 4, 9] (b) ( , ) or all real numbers m (c) ( , ) or all real numbers so y 34. (a) 0
3 1
1
1
(x
2 1
2 1) [ 2.35, 2.35] by [ 1, 3] 35. Because if the vertical line test holds, then for each
xcoordinate, there is at most one ycoordinate giving a
point on the curve. This ycoordinate would correspond to
the value assigned to the xcoordinate. Since there is only
one ycoordinate, the assignment would be unique.
36. If the curve is not y 0, there must be a point (x, y) on the
curve where y 0. That would mean that (x, y) and (x, y)
are two different points on the curve and it is not the graph
of a function, since it fails the vertical line test. Line through (1, f (x) x 3 1) and (3, m 40. No x,
x 2, 0
1 x
x 0
1
2
3 x
x
x
x 0 2
x
T T
2 1, x
1
x
3 2,
5
,
3 x x T x 3T
2 T
3T
2 2 2 x 5 5
3 2
1
2 1
3 1
,
3 1 T
2 T
2 (b) g( f (x)) 0
5 2
x
T 0 T x 49. (a) f (g(x)) x x T
2 T
2 0 A, 0 x A, 2 1
2
3
4 1 3 T
, 0 and (T, 1):
2
10
2
2
, so y
x
T (T/2)
T
T 48. f (x) Line through (2, 1) and (5, 0): m
2) x A, x 1
2 1
x
3 1 x 43. Line through (0, 2) and (2, 0): y 1
(x
3 x 1 0 1 A,
2,
0,
2,
0, x 0 2, 0, 41. Line through (0, 0) and (1, 1): y
Line through (1, 1) and (2, 0): y 1): y 2 47. Line through f (x) 39. Yes 2x 2 2x 1
x
2 1) and (0, 0): y 1, 38. Yes so y 1 1
x,
2 37. No 42. f (x) x Line through (0, 2) and (1, 0): y (c) [0, ) f (x) 0 1 3
,
2 x 0 1, 46. Line through ( 2, , ) or all real numbers 3
2 1 1
x
2 (b) ( 1
x
2 1 x,
f (x) f (x) 1
,
2 (x 2 x 2T 3) 2 2 3
(x 5)
(x 2 10x 25)
x 2 10x 22 (c) f (g(0)) 02 2 (d) g( f (0)) 02 10 0 (e) g(g( 2))
(f) f ( f (x)) x2 5 [( 2)
(x 5) 3 2
2 22
3] 5 2 x 22
3
10 12 3 2 9 10 Section 1.2 50. (a) f (g(x)) (x 1) 1 (b) g( f (x)) (x 1) 1 x (c) f (g(x)) 0 (d) g( f (0)) 0
1) 1 1 x g f: x (e) g(g( 2))
(f) f ( f (x)) (2
(x 1) 3 1 4
[ 4.7, 4.7] by [ 2, 4] 2 51. (a) Enter y1 f (x) x 7, y2 g(x)
y3 ( f g)(x) y1(y2(x)),
and y4 (g f )(x) y2(y1(x))
f g: Domain: ( ,
Range: [0, ) x, (b) ( f g)(x) g f:
(g f )(x) [ 3, 20] by [ 4, 4] Domain: [0, )
Range: [ 7, )
(b) ( f g)(x) x Domain: [7, )
Range: [0, )
7 (g f )(x) 52. (a) Enter y1 f (x) 1 x 2, y2 g(x)
y3 ( f g)(x) y1(y2(x)),
and y4 (g f )(x) y2(y1(x))
f g: x [1, ) ( x 2)2 3
(x 2) 3, x
x 1, x
2
(x2 3) 2 54. (a) Enter y1(x)
[ 10, 70] by [ 10, 3] 1] 2x
x f (x) 2
x2 1
,y
32 1 3x
2 1
,
x y3 ( f g)(x) y1(y2(x)),
and y4 (g f )(x) y2(y1(x)).
Use a “decimal window” such as the one shown.
f g: 7 x, [ 9.4, 9.4] by [ 6.2, 6.2] Domain: ( , 2) (2, )
Range: ( , 2) (2, )
g f:
[ 6, 6] by [ 4, 4] Domain: [0, )
Range: ( , 1]
g f: [ 9.4, 9.4] by [ 6.2, 6.2] Domain: ( , 3) ( 3, )
Range: ( , 3) ( 3, )
2
[ 2.35, 2.35] by [ 1, 2.1] Domain: [ 1, 1]
Range: [0, 1]
(b) ( f g)(x) 1 ( x)2
(g f )(x)
1 x2 (b) ( f g)(x) 1 x, x 53. (a) Enter y1 f (x) x 2 3, y2 g(x)
y3 ( f g)(x) y1(y2(x)),
and y4 (g f )(x) y2(y1(x)).
f g: 3x 1
2x
3x 1
2x 1
3 2(3x 1) (2
(3x 1) 3(2 0
x 7x
,x
7 2, x, x
3 (g f )(x) 2x
x 2 2
1
3
2x
x 1
1
3 7x
,x
7 Domain: [ 2, )
Range: [ 3, ) x, x 2 2 3(2x 1) (x
2(x 3) (2x
[ 10, 10] by [ 10, 10] x)
,x
x) 3
3 3)
,x
1) 3 Section 1.2
55. 11 58. [ 5, 5] by [ 2, 5] We require x 2
and x 2
is ( 4
, [ 2.35, 2.35] by [ 1.55, 1.55] 4 We require x 2 0 (so that the square root is defined) ( 0 (to avoid division by zero), so the domain
(2, ). For values of x in the domain, x2 2) x2 and hence 1 4 and 4 can attain any positive x2 , 1) x2 (1, ). For values of x in the 1 can attain any value in [ 1, 0) (0, ), so 1 can also attain any value in [ 1, 0) 4 value, so the range is (0, ). (Note that grapher failure may 0, so the domain is ( 1, 1) domain, x 2
3 1 Therefore, 1
3 x (0, ). can attain any value in
1 cause the range to appear as a finite interval on a ( grapher. (Note that grapher failure can cause the intervals in the 56. , 1] (0, ). The range is ( , 1] (0, ). range to appear as finite intervals on a grapher.)
59. (a) y
1.5 [ 5, 5] by [ 2, 5] x2 We require 9
x2 and 9 0 (so that the fourth root is defined) any value in (0, 9]. Therefore,
value in (0,
, 0 2 x 0 (to avoid division by zero), so the domain is ( 3, 3). For values of x in the domain, 9 2 –2 2 3], and 4 x2 9
2 . The range is 3 , 4 x 2 can attain any 9 –1.5 x 2 can attain
(b) y can attain any value in
1.5 or approximately [1.15, ). 3
–2 (Note that grapher failure may cause the range to appear as 0 a finite interval on a grapher.) 2 x –1.5 57.
60. (a) y
2 [ 4.7, 4.7] by [ 3.1, 3.1] x2 We require 9
( , 3) domain, 9 ( 3, 3) , 0) (0, 9], so ( , 0) (0, ( , 0) –2 (3, ). For values of x in the x 2 can attain any value in ( value in ( 0, so the domain is 3 , 0)
2
3 , 3 9 2
3 , 2
3 9 x2 can attain any . The range is 9 or approximately ( 1
–1
–2 x 2 can attain any value in 9]. Therefore, –1 , 0) [0.96, ). 9 (Note that grapher failure can cause the intervals in the
range to appear as finite intervals on a grapher.) 2 x 12 Section 1.2 60. continued
(b) (d) Since ( f g)(x) f ( x) x , f (x) x 2.
The completed table is shown. Note that the absolute
value sign in part (d) is optional. y
2 g(x) 1 –2 –1 1 2 x f (x) x2 x 1
x x2 5
1
x 1 1 ( f g)(x)
5 x, x 1 –2 61. (a) 1
x y
1.3 –2 1
x –1 1 2 0 x2 x x, x
x, x 0 64. (a) Note that the data in the table begins at x 20. (We do
not include the initial investment in the data.) The
power regression equation is y 27.1094x 2.651044. x (b)
–1.3 (b) y [0, 30] by [ 20,000, 180,000] 1.3 –2 –1 1 2 (c) When x 30, y 223,374. According to the power
regression equation, the investment will grow to
approximately $223,374. x (d) The linear regression equation is
y 12,577.97x 177,275.52. When
x 30, y 200,064. According to the linear regression
equation, the investment will grow to approximately
$200,064. –1.3 62. (a) y 65. (a) Because the circumference of the original circle was
8 and a piece of length x was removed. 1.3 (b) r
–2 0 2 8 x
2 x
2 4 x y
1.3 4
16 4x (b) r2 16 –1.3 16
16 (c) h x2
42 x2
2 0 2 16 x
42 x x2
42 x2
42 16 x
2 –2 4x x2 –1.3 63. (a) Since ( f g)(x) g(x) 5 x2 5, g(x) 1
x, we know that
g(x)
1
1
x 1, so g(x)
.
g(x)
x1
1
1
(c) Since ( f g)(x) f
x, f (x)
.
x
x (b) Since ( f g)(x) 1 x 2. (d) V 12
rh
3
18
x2
16 x
3
2
2
(8
x)2 16 x x 2
24 2 x2 Section 1.3
66. (a) Note that 2 mi 10,560 ft, so there are 8002
feet of river cable at $180 per foot and (10,560
feet of land cable at $100 per foot. The cost is
C(x) 180 8002 x2 100(10,560 x)
(b) x2
x) C(0) $1,200,000
C(500) $1,175,812
C(1000) $1,186,512
C(1500) $1,212,000
C(2000) $1,243,732
C(2500) $1,278,479
C(3000) $1,314,870
Values beyond this are all larger. It would appear that
the least expensive location is less than 2000 ft from
point P. 5. [ 5, 5] by [ 2, 5] [ 3, 3] by [ 1, 3] (c) The functions y1 y2, y2 y1, and y1 y2 all have
domain [0, 1], the same as the domain of y1 y2 found
in part (b).
Domain of
Domain of y2 y2
y1 3
3 x 2. Using a calculator, 3 1.5 3 5 x for x 0; 0.192. 17
17
2.5713
24
24
1.8882
5 1.4567
10
1.4567
1.0383 7. 500(1.0475)5 $630.58
$1201.16 (x 3y2)2
(x4y3)3 x 6y4
x12y9
6 12 4 9 x 18 y y 5 1
x18y5
a3b
c4 22 a4c
b3 1 a6b
c8
a6
b4c8 ( f g)(x), 4 a6 2 b 43 1 6 a 2b 4 b3
a4c 2
32 bc
a4 c c 82 a2
bc6 Section 1.3 Exercises
1. The graph of y 2x is increasing from left to right and has
the negative xaxis as an asymptote. (a)
2. The graph of y 1. x 3 6. x10
x
x 10. s Section 1.3 Exponential Functions 3 4.729. 3. Using a calculator, 3 x (b) The product will be even, since
( f g)( x) f ( x) g( x)
( f (x)) ( g(x))
f (x) g(x)
( f g)(x). x 0; 2
0. 2.924. The domain of a quotient of two functions is the
intersection of their domains with any zeros of the
denominator removed. Exponential Functions for x
for x 2 (d) The domain of a sum, difference, or product of two
functions is the intersection of their domains. Exploration 1 x 1. Using a calculator, 52/3 9. : (0, 1] (pp. 20–26) x 5
5 8. 1000(1.063)3 : [0, 1) 68. (a) Yes. Since
( f g)( x) f ( x) g( x) f (x) g(x)
the function ( f g)(x) will also be even. x
x Quick Review 1.3 5. x5
x
x (b) Domain of y1: [0, )
Domain of y2: (
, 1]
Domain of y3: [0, 1] y1 x 6. 2
2 4. x
x
x 67. (a) 13 3 x or, equivalently, y 1x
, is
3 decreasing from left to right and has the positive xaxis as
an asymptote. (d)
3. The graph of y
3 x is the reflection about the xaxis of
the graph in Exercise 2. (e) [ 5, 5] by [ 2, 5] 2. x 0 3. x 0 4. x 0 2x, is the
4. The graph of y
0.5 x or, equivalently, y
reflection about the xaxis of the graph in Exercise 1. (c)
5. The graph of y 2
and has the line y x 2 is decreasing from left to right
2 as an asymptote. (b) 6. The graph of y 1.5x 2 is increasing from left to right
and has the line y
2 as an asymptote. (f) 14 Section 1.3 7. 16. [ 4, 4] by [ 8, 6] [ 6, 6] by [ 2, 6] Domain: ( , )
Range: ( , 3)
xintercept: 1.585
yintercept: 2 x 1.3863 17. 8.
[ 6, 6] by [ 3, 5] x 0.6309 18. [ 4, 4] by [ 2, 10] Domain: ( , )
Range: (3, )
xintercept: None
yintercept: 4 [ 6, 6] by [ 3, 5] 9. x
19. 1.5850
x y y 1 1 2 1 3 3 4 5 2 [ 4, 4] by [ 4, 8] Domain: ( , )
Range: ( 2, )
xintercept: 0.405
yintercept: 1 2
2 10.
20. x y y 1 1 [ 4, 4] by [ 8, 4] 3 Domain: ( , )
Range: ( , 1)
xintercept: None
yintercept: 2
2x 2 2x 11. 9 (3 ) 12. 163x 3
3 1 2x
8 (2 3 2x 14. 1x
27 (3 3x ) 2 6x 21. x
1 ) 3 15. 8 4 212x 13. 5 3 4x 3 (24)3x 2 2 y y
1 3x 3
2 4
5 3 9
7 4
[ 6, 6] by [ 2, 6] x 2.3219 16 Section 1.3
22. x ratio y 1 8.155 31. Let A be the amount of the initial investment, and let t be
the number of years. We wish to solve 2.718
2 A1 22.167 3 60.257 2.718
2.718
4 163.794 23. Let t be the number of years. Solving
500,000(1.0375)t 1,000,000 graphically, we find that
t 18.828. The population will reach 1 million in about
19 years.
24. (a) The population is given by P(t) 6250(1.0275)t, where
t is the number of years after 1890.
Population in 1915: P(25) 12,315
Population in 1940: P(50) 24,265
(b) Solving P(t) 50,000 graphically, we find that
t 76.651. The population reached 50,000 about
77 years after 1890, in 1967.
25. (a) A(t) 6.6 1 t/14
2 (b) Solving A(t) 1 graphically, we find that t 38.1145.
There will be 1 gram remaining after about 38.1145
days.
26. Let t be the number of years. Solving 2300(1.06)t 4150
graphically, we find that t 10.129. It will take about
10.129 years. (If the interest is not credited to the account
until the end of each year, it will take 11 years.)
27. Let A be the amount of the initial investment, and let t be
the number of years. We wish to solve A(1.0625)t 2A,
which is equivalent to 1.0625t 2. Solving graphically, we
find that t 11.433. It will take about 11.433 years. (If the
interest is credited at the end of each year, it will take 12
years.)
28. Let A be the amount of the initial investment, and let t be
the number of years. We wish to solve
A1
1 15 0.0625 12t
2A, which is equivalent to
12
12t
0.0625
2. Solving graphically, we find that
12 t 11.119. It will take about 11.119 years. (If the interest
is credited at the end of each month, it will take 11 years 2
months.)
29. Let A be the amount of the initial investment, and let t be
the number of years. We wish to solve Ae 0.0625t 2A,
which is equivalent to e 0.0625t 2. Solving graphically, we
find that t 11.090. It will take about 11.090 years.
30. Let A be the amount of the initial investment, and let t be
the number of years. We wish to solve A(1.0575) t 3A,
which is equivalent to 1.0575t 3. Solving graphically, we
find that t 19.650. It will take about 19.650 years. (If the
interest is credited at the end of each year, it will take 20
years.) 1
t 0.0575 365t
3A, which is equivalent to
365
0.0575 365t
3. Solving graphically, we find that
365 19.108. It will take about 19.108 years. 32. Let A be the amount of the initial investment, and let t be
the number of years. We wish to solve Ae 0.0575t 3A,
which is equivalent to e 0.0575t 3. Solving graphically, we
find that t 19.106. It will take about 19.106 years.
33. After t hours, the population is P(t) 2 t/0.5 or, equivalently,
P(t ) 22t. After 24 hours, the population is
P(24) 2 48 2.815 1014 bacteria.
34. (a) Each year, the number of cases is 100% 20% 80%
of the previous year’s number of cases. After t years,
the number of cases will be C(t ) 10,000(0.8)t.
Solving C(t ) 1000 graphically, we find that
t 10.319. It will take about 10.319 years.
(b) Solving C(t ) 1 graphically, we find that t
It will take about 41.275 years. 41.275. 35. Since x 1, the corresponding value of y is equal to the
slope of the line. If the changes in x are constant for a linear function, then the corresponding changes in y are constant as well.
36. (a) When t 0, B 100e0 100. There were 100
bacteria present initially.
(b) When t 6, B 100e 0.693(6) 6394.351.
After 6 hours, there are about 6394 bacteria.
(c) Solving 100e 0.693t 200 graphically, we find that
t 1.000. The population will be 200 after about
1 hour. Since the population doubles (from 100 to 200)
in about 1 hour, the doubling time is about 1 hour.
37. (a) Let x 0 represent 1900, x 1 represent 1901, and so
on. The regression equation is P(x) 6.033(1.030)x. [0, 100] by [ 10, 90] (b) The regression equation gives an estimate of
P(0) 6.03 million, which is not very close to the
actual population.
(c) Since the equation is of the form P(x)
the annual rate of growth is about 3%.
38. (a) The regression equation is P(x) P(0) 1.030x, 4.831(1.019)x. [0, 100] by [ 5, 30] (b) P(90) 26.3 million (c) Since the equation is of the form P(x) P(0) 1.019x,
the annual rate of growth is approximately 1.9%. 16 Section 1.4 39. 5422(1.018)19 s Section 1.4 Parametric Equations
(pp. 26–31) 7609.7 million 40. (a) Exploration 1 Parametrizing Circles 1. Each is a circle with radius a . As a increases, the radius
of the circle increases.
[ 5, 5] by [ 2, 10] In this window, it appears they cross twice, although a
third crossing offscreen appears likely.
(b) x change in Y1 change in Y2
[ 4.7, 4.7] by [ 3.1, 3.1] 1
3
5 0 8 t : t : t 3
:
2 4 7 2. 0 2 2 2
3
[ 4.7, 4.7] by [ 3.1, 3.1] 4
It happens by the time x
(c) Solving graphically, x 4.
0.7667, x 2, x 4. (d) The solution set is approximately
( 0.7667, 2) (4, ).
41. Since f (1) 4.5 we have ka
we have ka 1 0.5.
Dividing, we have a 0.5 0 4.5
0.5 ka
ka 1 a2 4.5, and since f ( 1) [ 4.7, 4.7] by [ 3.1, 3.1] [ 4.7, 4.7] by [ 3.1, 3.1] 9 2 t 4: 3 Since f (x) k a x is an exponential function, we require
a 0, so a 3. Then ka 4.5 gives 3k 4.5, so k 1.5.
The values are a 3 and k 1.5.
42. Since f (1) 1.5 we have ka 1.5, and since f ( 1)
we have ka 1 6. Dividing, we have
ka
ka 1 6 [ 4.7, 4.7] by [ 3.1, 3.1] 0 t 4: 1.5
6 a2 0.25 a 0.5 Since f (x) k a x is an exponential function, we require
a 0, so a 0.5. Then ka 1.5 gives 0.5k 1.5, so
k 3. The values are a 0.5 and k 3. [ 4.7, 4.7] by [ 3.1, 3.1] Let d be the length of the parametric interval. If d
you get d
of a complete circle. If d
2 complete circle. If d 2, 2 , you get the 2 , you get the complete circle but portions of the circle will be traced out more than once. For
example, if d 4 the entire circle is traced twice. Section 1.4
Exploration 2 3. 1. a 2 2, b 3
2 t Parametrizing Ellipses 3: [ 12, 12] by [ 8, 8] initial point: (0, 3)
terminal point: (0, 3) a 2, b 4: [ 12, 12] by [ 8, 8] a 2, b 5: t2
initial point: ( 3, 0)
terminal point: (3, 0)
[ 12, 12] by [ 8, 8] a 3
2 t 2, b 6: 3 initial point: (0, 3)
terminal point: ( 3, 0) [ 12, 12] by [ 8, 8] 2. a 3, b 4: [ 9, 9] by [ 6, 6] t5
initial point: ( 3, 0)
terminal point: ( 3, 0)
4. For 0 t 2 the complete circle is traced once clockwise beginning and ending at (2, 0).
For
t 3 the complete circle is traced once clockwise beginning and ending at ( 2, 0).
For 2 t starting at (0, 3
the half circle below is traced clockwise
2 a 5, b 4: [ 9, 9] by [ 6, 6] a 6, b 4: 2) and ending at (0, 2).
[ 9, 9] by [ 6, 6] a 7, b 4: [ 9, 9] by [ 6, 6] 3. If a
b , then the major axis is on the xaxis and the
minor on the yaxis. If a
b , then the major axis is on
the yaxis and the minor on the xaxis. 17 18 Section 1.4 4. 0 t 2 Exploration 3 : [ 6, 6] by [ 4, 4] 0 t Graphing the Witch of Agnesi 1. We used the parameter interval [0, ] because our graphing
calculator ignored the fact that the curve is not defined
when t 0 or . The curve is traced from right to left
across the screen. x ranges from
to .
2. : t 2 2 : [ 5, 5] by [ 2, 4] [ 6, 6] by [ 4, 4] 0 t 0 3
:
2 t 2 : [ 5, 5] by [ 2, 4] [ 6, 6] by [ 4, 4] 0 t 4: 2 t : [ 6, 6] by [ 4, 4] [ 5, 5] by [ 2, 4] Let d be the length of the parametric interval. If d
you get d
of a complete ellipse. If d
2 2, 2 , you get the complete ellipse. If d 2 , you get the complete ellipse
but portions of the ellipse will be traced out more than
once. For example, if d 4 the entire ellipse is traced
twice.
5. 0 t For t 2 2 , the entire graph described in part 1 is drawn. The left branch is drawn from right to left across the
screen starting at the point (0, 2). Then the right branch is
drawn from right to left across the screen stopping at the 2:
point (0, 2). If you leave out 2 and 2 , then the point (0, 2) is not drawn.
[ 6, 6] by [ 4, 4] initial point: (5, 0)
terminal point: (5, 0)
t For 0 2 , the right branch is drawn from right to left across the screen stopping at the point (0, 2). If you leave
out 3: t For 2 , then the point (0, 2) is not drawn.
t 2 , the left branch is drawn from right to left across the screen starting at the point (0, 2). If you leave
[ 6, 6] by [ 4, 4] initial point: ( 5, 0)
terminal point: ( 5, 0)
2 t 3
:
2 [ 6, 6] by [ 4, 4] initial point: (0, 2)
terminal point: (0, 2)
Each curve is traced clockwise from the initial point to the
terminal point. out 2 , then the point (0, 2) is not drawn. 3. If you replace x 2 cot t by x
2 cot t, the same graph
is drawn except it is traced from left to right across the
screen. If you replace x 2 cot t by x 2 cot (
t), the
same graph is drawn except it is traced from left to right
across the screen. Quick Review 1.4
1. m
y
y 3
4 8
1
5
(x
3
5
x
3 5
3 1)
29
3 5
3 8 19 Section 1.4
2. y Section 1.4 Exercises 4 3. x 1. Graph (c). Window: [ 4, 4] by [ 3, 3], 0
2 4. When y 0, we have and 3. When x x
9 1, so the xintercepts are
y2
16 0, we have 3 0, we have and 4. When x x2
16 1, so the xintercepts are
y
9 4 1, which has no real solution, so there are no yintercepts.
6. When y 0, we have 0 1, so the xintercept is 0, we have 2y2 When x
and x 1, so the yintercepts are 2
t 2 4. Graph (b). Window: [ 15, 15] by [ 15, 15], 0 t 2 5. (a) The resulting graph appears to be the right half of a
hyperbola in the first and fourth quadrants. The
parameter a determines the xintercept. The parameter
b determines the shape of the hyperbola. If b is smaller,
the graph has less steep slopes and appears “sharper.” If
b is larger, the slopes are steeper and the graph appears
more “blunt.” The graphs for a 2 and b 1, 2, and 3
are shown. 2 0, we have 2 t 3. Graph (d). Window: [ 10, 10] by [ 10, 10], 0 1, so the yintercepts are 4 and 4.
5. When y t 2. Graph (a). Window: [ 2, 2] by [ 2, 2], 0 2 1.
1
2 1 .
2 [ 10, 10] by [ 10, 10] 7. (a) 2(1)2(1) + 12
3 3
3 Yes (b) 2( 1)2( 1) ( 2) 2 8. (a) 9(1)2 4 3
3 [ 10, 10] by [ 10, 10] This appears to be the left half of the same hyberbola.
(c)
Yes
27
27
27 Yes 18(1) 4( 3)
9 18 36
27 9( 1)2 18( 1) 4(3)2
9 18 36
63 [ 10, 10] by [ 10, 10] 27
27
27 Yes
27
27
27 No 5
2x 5 t 2x
3 5 2t
2t
2t
t tan2 t 1 by a standard trigonometric identity. Substituting
tan t gives 1
3y 1
3y 1
3y One must be careful because both sec t and tan t are
discontinuous at these points. This might cause the
grapher to include extraneous lines (the asymptotes of
the hyperbola) in its graph. The extraneous lines can be
avoided by using the grapher’s dot mode instead of
connected mode.
(d) Note that sec2 t 3t
3t (b) 3y No 2 (b) 9(1) 9. (a) 2x 3 18(1) 4(3)2
9 18 36
27 2 (c) ( 2)2 3 2 12 3
3
3 1 (c) ( 1)2
21
1 (b) x2
a y2
b x
y
for sec t and for
a
b 1. (e) This changes the orientation of the hyperbola. In this
case, b determines the yintercept of the hyperbola, and 1 a determines the shape. The parameter interval 2 10. (a) The equation is true for a 0. , 2 a or
(b) The equation is equivalent to “ a
a2
a.” Since a2 a is true for a 0 and
a2
a is true for a 0, at least one of the two
equations is true for all real values of a. Therefore, the
given equation a2
a is true for all real values
of a.
(c) The equation is true for all real values of a. 22 gives the upper half of the hyperbola. The parameter interval 2 , 3
2 gives the lower half. The same values of t cause discontinuities and may add
extraneous lines to the graph. Substituting
x
for tan t in the identity sec2 t
a
y2
x2
1.
b
a and y
for sec t
b tan2 t 1 gives 20 Section 1.4 6. (a) 10. (a) [ 6, 6] by [ 4, 4] [ 4.7, 4.7] by [ 3.1, 3.1] The graph is a circle of radius 2 centered at (h, 0). As h
changes, the graph shifts horizontally. Initial and terminal point: (4, 0)
(b) (b) x2
4 y2
2 cos2 t sin2 t 1 The parametrized curve traces all of the ellipse defined
by x2
4 y2
2 1. 11. (a) [ 6, 6] by [ 4, 4] The graph is a circle of radius 2 centered at (0, k). At k
changes, the graph shifts vertically.
(c) Since the circle is to be centered at (2, 3), we use
h 2 and k
3. Since a radius of 5 is desired, we
need to change the coefficients of cos t and sin t to 5.
x 5 cos t 2, y 5 sin t 3, 0 t 2
(d) x 5 cos t 3, y 2 sin t 4, 0 t [ 4.7, 4.7] by [ 3.1, 3.1] Initial point: (0, 2)
Terminal point: (0, 2 7. (a) (b) x2
4 y2
2 2) sin2 t cos2 t 1 The parametrized curve traces the right half of the
ellipse defined by
defined by x [ 3, 3] by [ 2, 2] Initial point: (1, 0)
Terminal point: ( 1, 0) x2
4 4 2 y2
2 1 (or all of the curve y2). 12. (a) (b) x 2 y 2 cos2 t sin 2 t 1
The parametrized curve traces the upper half of the
circle defined by x 2 y 2 1 (or all of the semicircle
defined by y
1 x 2). [ 9, 9] by [ 6, 6] 8. (a) Initial and terminal point: (0, 5)
(b) x2
4 y2
5 sin2 t cos2 t 1 The parametrized curve traces all of the ellipse defined
[ 3, 3] by [ 2, 2] by Initial and terminal point: (0, 1)
(b) x 2 y 2 sin2 (2 t) cos2 (2 t) 1
The parametrized curve traces all of the circle defined
by x 2 y 2 1. x2
4 y2
5 1. 13. (a) 9. (a)
[ 3, 3] by [ 1, 3] No initial or terminal point.
(b) y 9t 2 (3t)2 x 2
The parametrized curve traces all of the parabola
defined by y x 2. [ 3, 3] by [ 2, 2] Initial point: ( 1, 0)
Terminal point: (0, 1)
2 2 2 14. (a)
2 y
cos (
t) sin (
t) 1
(b) x
The parametrized curve traces the upper half of the
circle defined by x 2 y 2 1 (or all of the semicircle
defined by y
1 x2).
[ 3, 3] by [ 1, 3] Initial point: (0, 0)
Terminal point: None Section 1.4
(b) y t (
t)2 x 2
The parametrized curve traces the left half of the
parabola defined by y x 2 (or all of the curve defined
by x
y). 21 19. (a) 15. (a)
[ 9, 9] by [ 6, 6] No initial or terminal point.
(b) y 4t 7 2(2t 5) 3 2x 3
The parametrized curve traces all of the line defined by
y 2x 3. [ 1, 5] by [ 1, 3] Initial point: (0, 0)
Terminal point: None 20. (a) (b) y
t
x
The parametrized curve traces all of the curve defined
by y
x (or the upper half of the parabola defined
by x y 2). [ 6, 6] by [ 4, 4] 16. (a) No initial or terminal point.
(b) y 1 t 2 (1 t) 2 x
x2
The parametrized curve traces all of the line defined by
y
x 2.
21. (a)
[ 3, 9] by [ 4, 4] No initial or terminal point.
(b) x sec2 t 1 tan2 t y 2
The parametrized curve traces all of the parabola
defined by x y 2. [ 3, 3] by [ 2, 2] 17. (a) Initial point: (0, 1)
Terminal point: (1, 0)
(b) y 1 t 1 x
x1
The Cartesian equation is y
x 1. The portion
traced by the parametrized curve is the segment from
(0, 1) to (1, 0). [ 3, 3] by [ 2, 2] No initial or terminal point. Note that it may be
necessary to use a tinterval such as [ 1.57, 1.57] or
use dot mode in order to avoid “asymptotes” showing
on the calculator screen. 22. (a) (b) x 2 y 2 sec2 t tan2 t 1
The parametrized curve traces the left branch of the
hyperbola defined by x 2 y 2 1 (or all of the curve
defined by x
y2 1).
18. (a) [ 2, 4] by [ 1, 3] Initial point: (3, 0)
Terminal point: (0, 2)
(b) y 2t (2t 2) 2 The Cartesian equation is y 2
(3
3 3t)
2
x
3 2 2
x
3 2 2. The portion traced by the curve is the segment from (3, 0) to (0, 2).
[ 6, 6] by [ 5, 1] 23. (a) No initial or terminal point. Note that it may be
necessary to use a tinterval such as [ 1.57, 1.57] or
use dot mode in order to avoid “asymptotes” showing
on the calculator screen.
(b) y2
2 x2 sec2 t tan2 t [ 6, 6] by [ 2, 6] 1 The parametrized curve traces the lower branch of the
hyperbola defined by
defined by y y2
2 2 x2 x2
1). 1 (or all of the curve Initial point: (4, 0)
Terminal point: None
t 4 (4
t) 4 x
x4
(b) y
The parametrized curve traces the portion of the line
defined by y
x 4 to the left of (4, 0), that is, for
x 4. 22 Section 1.4 24. (a) 30. The vertex of the parabola is at ( 1, 1), so the left half of
the parabola is given by y x 2 2x for x
1.
Substituting t for x, we obtain one possible parametrization:
x t, y t 2 2t, t
1.
[ 1, 5] by [ 1, 3] Initial point: (0, 2)
Terminal point: (4, 0)
4 t2
4x
(b) y
The parametrized curve traces the right portion of the
curve defined by y
4 x, that is, for x 0.
25. (a) 31. For simplicity, we assume that x and y are linear functions
of t and that the point (x, y) starts at (2, 3) for t 0 and
passes through ( 1, 1) at t 1. Then x f (t), where
f (0) 2 and f (1)
1.
x
t Since slope
x f (t) g(0) 3t y
No initial or terminal point, since the tinterval has no
beginning or end. The curve is traced and retraced in
both directions.
(b) y cos 2t
cos2 t sin2 t
1 2 sin2 t
1 2x 2
2x 2 1
The parametrized curve traces the portion of the
parabola defined by y
2x 2 1 corresponding to
1 x 1. g(t) 4t x (b) x t 2 3 y 2 3
The parametrized curve traces the lower half of the
parabola defined by x y 2 3 (or all of the curve
defined by y
x 3).
27. Using ( 1, 3) we create the parametric equations
x
1 at and y
3 bt, representing a line which
goes through ( 1, 3) at t 0. We determine a and b so
that the line goes through (4, 1) when t 1.
Since 4
1 a, a 5.
Since 1
3 b, b 4.
Therefore, one possible parametrization is x
1 5t,
y
3 4t, 0 t 1.
28. Using ( 1, 3) we create the parametric equations
x
1 at and y 3 bt, representing a line which
goes through ( 1, 3) at t 0. We determine a and b so that
the line goes through (3, 2) at t 1.
Since 3
1 a, a 4.
Since 2 3 b, b
5.
Therefore, one possible parametrization is x
1 4t,
y 3 5t, 0 t 1.
29. The lower half of the parabola is given by x y 2 1
for y 0. Substituting t for y, we obtain one possible
parametrization: x t 2 1, y t, t 0. g(t), where 13
10 3 3 4, 4t. 2 3t, y 3 4t, t 0. 32. For simplicity, we assume that x and y are linear functions
of t and that the point (x, y) starts at ( 1, 2) for t 0 and
passes through (0, 0) at t 1. Then x f (t), where
f (0)
1 and f (1) 0.
Since slope x
t x ( 1) y Initial point: None
Terminal point: ( 3, 0) 3t. Also, y 1. y
t f (t) 1t g(t) 0
1 ( 1)
0 1 g(t), where g(0)
y
t Since slope [ 4, 5] by [ 4, 2] 2 3, One possible parametrization is: Also, y 26. (a) 2 3 and g(1) Since slope [ 3, 3] by [ 2, 2] 12
10 2t 0
1 2 t.
2 and g(1) 2
0 2 1, 0. 2,
2t. One possible parametrization is:
x 1 t, y 2 2t, t 0. 33. The graph is in Quadrant I when 0 y 2, which
corresponds to 1 t 3. To confirm, note that x(1)
and x(3) 0. 2 34. The graph is in Quadrant II when 2 y 4, which
corresponds to 3 t 5. To confirm, note that x(3)
and x(5)
2. 0 35. The graph is in Quadrant III when 6 y
4, which
corresponds to 5 t
3. To confirm, note that
x( 5)
2 and x( 3) 0.
36. The graph is in Quadrant IV when 4 y 0, which
corresponds to 3 t 1. To confirm, note that
x( 3) 0 and x(1) 2.
37. The graph of y x 2 2x 2 lies in Quadrant I for all
x 0. Substituting t for x, we obtain one possible
parametrization:
x t, y t 2 2t 2, t 0.
x 3 lies in Quadrant I for all x 0.
38. The graph of y
Substituting t for x, we obtain one possible parametrization:
x t, y
t 3, t 0. Section 1.5 s Section 1.5 Functions and Logarithms
(pp. 32–40) 39. Possible answers:
(a) x a cos t, y (b) x a cos t, y (c) x a cos t, y (d) x a sin t, 0 a cos t, y a sin t, 0 t
t a sin t, 0
a sin t, 0 2
2 t
t 23 Exploration 1 4 Testing for Inverses Graphically 1. It appears that ( f g)(x) (g f )(x)
and g may be inverses of each other. 4 40. Possible answers: x, suggesting that f (a) f and g: (a) x a cos t, y (b) x a cos t, y (c) x a cos t, y (d) x b sin t, 0 a cos t, y t b sin t, 0
b sin t, 0 t
t b sin t, 0 41. Note that m OAQ 2
2
4
t 4 [ 4.7, 4.7] by [ 3.1, 3.1] t, since alternate interior angles (b) f g: formed by a transversal of parallel lines are congruent.
OQ
AQ Therefore, tan t 2
, so x
x 2
tan t 2 cot t.
[ 4.7, 4.7] by [ 3.1, 3.1] Now, by equation (iii), we know that
2 (c) g f : (AQ)
AO
AQ
(AQ)
AO AB (cos t)(x)
[ 4.7, 4.7] by [ 3.1, 3.1] (cos t)(2 cot t) 2. It appears that f g
identity function. 2 cos2 t
.
sin t 2 AB sin t 2 g, suggesting that f may be the (a) f and g: Then equation (ii) gives
y gf 2 cos2 t
sin t sin t 2 2 cos2 t 2 sin2 t.
[ 4.7, 4.7] by [ 3.1, 3.1] The parametric equations are:
x 2 cot t, y 2 sin2 t, 0 (b) f g: t Note: Equation (iii) may not be immediately obvious, but it
may be justified as follows. Sketch segment QB. Then
[ 4.7, 4.7] by [ 3.1, 3.1] OBQ is a right angle, so
AB
AQ ABQ AQO, which gives (c) g f : AQ
.
AO 42. (a) If x2 x1 then the line is a vertical line and the first
parametric equation gives x x1, while the second will
give all real values for y since it cannot be the case that
y2 y1 as well.
Otherwise, solving the first equation for t gives
t (x x1)/(x2 x1).
Substituting that into the second equation gives
y y1 [(y2 y1)/(x2 x1)](x x1)
which is the pointslope form of the equation for the
line through (x1, y1) and (x2, y2).
Note that the first equation will cause x to take on all
real values, because (x2 x1) is not zero. Therefore, all
of the points on the line will be traced out. [ 4.7, 4.7] by [ 3.1, 3.1] 3. It appears that ( f g)(x) (g f )(x)
and g may be inverses of each other.
(a) f and g: [ 4.7, 4.7] by [ 3.1, 3.1] (b) f g: (b) Use the equations for x and y given in part (a), with
0 t 1.
[ 4.7, 4.7] by [ 3.1, 3.1] x, suggesting that f 24 Section 1.5
5. Substituting t for x, one possible answer is: 3. continued
(c) g f : x t, y 1
t 1 ,t 2. 6. Substituting t for x, one possible answer is:
x t, y t, t
3.
7. [ 4.7, 4.7] by [ 3.1, 3.1] 4. It appears that ( f g)(x) (g f )(x) x, suggesting that f
and g may be inverse of each other. (Notice that the domain
of f g is (0, ) and the domain of g f is ( , ).)
(a) f and g: [ 10, 10] by [ 10, 10] (4, 5)
8.
[ 4.7, 4.7] by [ 3.1, 3.1] (b) f g:
[ 10, 10] by [ 10, 10] 8
,
3 3 (2.67, 3) 9. (a) [ 4.7, 4.7] by [ 3.1, 3.1] (c) g f : [ 10, 10] by [ 10, 10] (1.58, 3) [ 4.7, 4.7] by [ 3.1, 3.1] Exploration 2 Supporting the Product Rule 1. They appear to be vertical translations of each other. (b) No points of intersection, since 2x
values of x. 0 for all 10. (a) [ 10, 10] by [ 10, 10]
[ 1, 8] by [ 2, 4] ( 1.39, 4) 2. This graph suggests that each difference (y3
constant. y1 y2) is a (b) No points of intersection, since e
of x. x 0 for all values Section 1.5 Exercises
1. No, since (for example) the horizontal line y
the graph twice. 2. Yes, since each horizontal line intersects the graph only
once. [ 1, 8] by [ 2, 4] 3. y3 y1 y2 ln (ax)
Thus, the difference y3 ln x ln a ln x ln x ln a
y1 y2 is the constant ln a. f (g(1)) 2. (g f )( 7)
3. ( f g)(x)
3 x 2 4. (g f )(x) f (2) g( f ( 7))
f (g(x)) x 1
g( 2) f (x 2 1) 2/3
3 g( f (x))
3 g( x (x 1) (x 2/3 1) 2 1
1 3. Yes, since each horizontal line intersects the graph at most
once.
4. No, since (for example) the horizontal line y
intersects the graph twice. Quick Review 1.5
1. ( f g)(1) 2 intersects 1) 5
3 (x 2 1) 1 0.5 5. Yes, since each horizontal line intersects the graph only
once.
6. No, since (for example) the horizontal line y
the graph at more than one point. 2 intersects Section 1.5
7. 13. y 2x 3 y 3 2x y 3 x 2 Interchange x and y.
[ 10, 10] by [ 10, 10] Yes, the function is onetoone since each horizontal line
intersects the graph at most once, so it has an inverse
function. x 3 y 2 f 1 x (x) . Verify.
1 )(x) (f f 8. 3
2 x f 3
2 x 2 3 3 2 (x 3) 3 (2x 3) x
[ 10, 10] by [ 10, 10] 1 (f f )(x) 1 f No, the function is not onetoone since (for example) the
horizontal line y 0 intersects the graph twice, so it does
not have an inverse function. (2x 3)
2 2x
2 3 x 9.
14. y 5 4x 4x 5 y 5 x y
4 [ 10, 10] by [ 10, 10] Interchange x and y. No, the function is not onetoone since (for example) the
horizontal line y 5 intersects the graph more than once,
so it does not have an inverse function. y 10. f 5
1 (x) x
45 x
4 Verify.
1 )(x) (f f 5 f x 5 4 5 (5 4 5 x
4 x) x
1 f )(x) x
3 15. y x y 1 1 (y 12. 4x) (5 5 11. No, the function is not onetoone since each horizontal line
intersects the graph twice, so it does not have an inverse
function. 4x) f Yes, the function is onetoone since each horizontal line
intersects the graph only once, so it has an inverse function. [ 10, 10] by [ 10, 10] 1 (5
4x 4
4 (f
[ 5, 5] by [ 20, 20] x3 1) 1/3 x Interchange x and y.
1)1/3 (x
f 1 (x) y
(x 1)1/3 or 1 )(x) 3 f( x
( (f 1 f )(x) 3 1 f 1)
1)3 x
1 3 (x 3 Yes, the function is onetoone since each horizontal line
intersects the graph at most once, so it has an inverse
function. x Verify.
(f f [ 9, 9] by [ 2, 10] 3 (x3 1 (x 1) 1 1)
1) 1 3 x3 x x 25 26 Section 1.5
x2 y 1, x 1 16. y 2 x ,x y 1 For x 0 (f 0 2 (the domain of f ), 1 f )(x) x
1 1 20. y x )(x) f( x 1) f )(x) 1 f (x (x
x
2 x ,x 17. y ), (x 2)2 1 1 2 f 1 1 1 or x 1/2 x For x 1 x 1 )(x) (f f x f( x 1)2
2 x Interchange x and y. For x
1 (f 1/2 x or x f )(x) f 1 )(x) (f f
For x f( f )(x) 1 f ),
x)2 ( (x 2) (x 2/3)3/2, x y 3/2 x2 x x 0 1 f x 3/2
1 0 (the domain of f
1 )(x) for x f (x 3/2 ) f )(x) 1 (x 3/2 2/3 f 2/3 (x ) 2)2, x 2)2 y, x ) x 2 (x ) x 1 x 0
y or 1
x1/2 )(x) f ), 1
(1/ x)2 x 0 (the domain of f ),
f )(x) 1
x3
1
y
31
y f 1 11 x2 x2 1/x 2 1
3 y 1 y 2 1 1 Interchange x and y. ( x)1/2 x or 2 f Verify. 3 1 x (x) 1
3 or
x 1
x1/3 Verify. For x
(f f (f x x (x) x x3 y 2
1 1 0 (the domain of f 1 22. y Interchange x and y.
f 1) 1 1 1 For x
2/3 3/2 y 2 y 1 x 2 x 1 0
1 (x) (f f 2 x 1 For x ), 0, (the domain of f ), (x (x 2x Verify. For x 19. y 1 1) 1)2
1 x Verify. (f 2 x (x) 1 x 1 y y (f f 2 Interchange x and y. Interchange x and y.
f 1
,x
x2
1
,x
y
1
y 21. y x x x 2x 2 x 0 y 3/2 3/2 (x 2 (x x x2 x 2/3, x 18. y x) 1 (x 0, (the domain of f ), 1 (f 1 0 (the domain of f x 1 (the domain of f ),
1 x Verify.
For x 1] ), 2( 2 x)2 x 1 1) x [( 0 (x) 1) 1 0 (the domain of f x) f x 1 y (x) ( 1 x x ( y 2) 1 y x (x 1 1 2 Verify. 1) 2 1) , x
x 2 1), x 2 (x x x 1) 2 x Interchange x and y. 0, (the domain of f ), 1 (f 1) 2x y 2 (x 2 y 1) x ( (x y
1 1 (the domain of f 1 For x 1)1/2 1 or (x Verify. For x
(f f 2)2) 2 y (x) ( (x 2 x Interchange x and y.
f 1 f 1 0 (the domain of f
1 )(x) f (2 (f f x)
x) [(2
( ) 2 x) 2]2
x (f
x 1 1 1 1 3 (1/ x)3
1 )(x) f f )(x) x
1
f13
x 3 3 1/x3 x
x x x Section 1.5
2x
x 23. y 1
3 x
x 24. y xy 3y 2x 2x 1 2)x 1 3y x 1
y 2y x 3 xy 3y (y xy 1 xy x 2y 3 x(y 1
x f 1 1
x 2x
x y
f 3x
2 1 )(x) f
2 1
x 1 3x
x2
1 3x
x2 5x
5
1 f )(x) f 3
1 3 2x
x
1
3 1 5x
5 2)
2) (f 1 f )(x) x 3
1
3
1 1
3 x
x
x
x Graph of f 3(x
2(x 1)
1) x
3
2 3
2
3
2 3
1 3)
3) 3(x 2)
(x 2) x
et
t : x2 Graph of y 3)
3) t, y1 25. Graph of f: x1
1 2 x 2(x
(x
5x
5 2 3 1x f
2 1)
3) 3
1 (2x
(2x
5x
5 3 (x 3) 3(2x
(2x 1) 2(x 2x
x f 1
3 x 2x
x )(x) 2x
x
2x
x x 1 2x 1 1 3x
2 2(1 3x) (x
(1 3x) 3(x (f 2x
x Verify.
(f f 1 3 3
1 (x) Verify.
(f f 2y
3
1 Interchange x and y. 3x
2 (x) 1)
2y
y x 3y
2 Interchange x and y.
y 3
2 e , y2 t t, y3 x: x3 t [ 6, 6] by [ 4, 4] 26. Graph of f: x1
Graph of f
Graph of y 1 t, y1 : x2
x: x3 [ 6, 6] by [ 4, 4] 3t 3t, y2
t, y3 t
t 27 28 Section 1.5 27. Graph of f: x1
Graph of f 1 t, y1 : x2 Graph of y t 2 33. t 2 , y2 t t, y3 t x: x3 [ 10, 5] by [ 7, 3] Domain: ( , 3)
Range: ( , )
34. [ 4.5, 4.5] by [ 3, 3] 28. Graph of f: x1
Graph of f 1 t, y1 3 t, y2 : x2 Graph of y t 3 t t, y3 t x: x3 [ 5, 10] by [ 5, 5] Domain: ( 2, )
Range: ( , )
35.
[ 4.5, 4.5] by [ 3, 3] 29. Graph of f: x1
Graph of f 1 t, y1 : x2 Graph of y ln t ln t, y2 t t, y3 t x: x3 [ 3, 6] by [ 2, 4] Domain: ( 1, )
Range: ( , )
36. [ 4.5, 4.5] by [ 3, 3] 30. Graph of f: x1
Graph of f 1 t, y1 : x2 Graph of y log t [ 2, 10] by [ 2, 4] log t, y2
t, y3 x: x3 t Domain: (4, )
Range: ( , ) t 37. (1.045)t 2
t ln(1.045) ln 2 t ln 1.045 ln 2 t
[ 4.5, 4.5] by [ 3, 3] 31. Graph of f: x1
Graph of f 1 Graph of y sin sin 1 1 t t, y2 t, y3 x: x3 15.75 Graphical support: t, y1 : x2 ln 2
ln 1.045 t t [ 2, 18] by [ 1, 3] 38. e 0.05t
ln e 0.05t
[ 3, 3] by [ 2, 2] 32. Graph of f: x1
Graph of f
Graph of y 1 0.05t t, y1 : x2
x: x3 tan 1 1 t, y2 t, y3 t t ln 3
0.05 3
ln 3
ln 3
20 ln 3 t tan t Graphical support: [ 5, 35] by [ 1, 4]
[ 6, 6] by [ 4, 4] 21.97 Section 1.5
39. e x
e x e x 3 3 x e e x(e x 3 (e x)2 3e x x ex 0
x e e x(0) ) 1 3 1 1
2 0 ( 3)2
2(1)
5 3 e 43. y 100
2 2 3 ln 100
y x 100
y x log2 x 5 x 0.96 or 0.96 2 1
100
1
y
100
log2
1
y
100
log2
1
y
100 y
log2
y
y
log2
100 y log2(2 x) 4(1)(1) 2 x x Graphical support: Interchange x and y.
y
[ 4, 4] by [ 4, 8] 40. 2x 2 2x x 5 f x 2x(2x 5 (2x)2 5(2x) 1 (x) x
100 x log2 x
100 x Verify. 5
2 log2 (f f 0 2 x) 1 )(x) f log2 x
100 x 2x(0)
100 2x
2x 5 x 0 2 5 1 4(1)(1) ( 5)
2(1)
21 2 1 2log2 100 2
5 log2 x
100 x log2 1 21 2.26 or 2.26 2 100 x
x Graphical support:
100
100 x
x 1 [ 4, 4] by [ 4, 8] 41. ln y 2t 4 e ln y
y e 2t (f 4 42. ln(y 1) ln 2 ln(y 1) x e ln(y 100x
100 x) x 4 e 2t 100x
(100 x 1) y 1 y 2xe x ex x f )(x) f 1 1 100
2 ln 2 x ln x ln x 1 ln x ln 2 log2 e x(x)(2) 1 100 100
2
1 x 100
2 x 1
log2 100(1 100
2 x) log2 1
2x log2(2 x) 100 x 29 30 Section 1.5 44. y 50
1.1 1 1 1.1 1.1 x 50
y (b) f ( f (x)) 50
log1.1
y log1.1(1.1 )
50
y
50
log1.1
y log1.1 f log1.1
1 (x) 46. (a) Amount 1
1 log1.1 50 y
y log1.1 y
50 y (b) 8 x
50 x log1.1 x
50 1 )(x) x f log1.1 t x
50 x 0 1
1/x x for all x 0 1 t/12
2 1
1
8
13
2 36 47. 500(1.0475)t log1.1 1.1 8 There will be 1 gram remaining after 36 hours. 50
1 1 t/12
2 1 t/12
2
1 t/12
2
t
3
12 Verify.
(f f x, since x 1
x f 1 Interchange x and y:
y x x 2) 1 x x (1 x2 50
y ( f (x))2 1 x x x 1 45. (a) f ( f (x)) 1000 x
50 x 1.0475t 2
t ln(1.0475 ) ln 2 t ln 1.0475 ln 2 50
1.1log1.1 1 50 x
x t 1 14.936 It will take about 14.936 years. (If the interest is paid at the
end of each year, it will take 15 years.) 50
50 ln 2
ln 1.0475 x
x 48. 375,000(1.0225)t
50x
(50 x 50x
50 x) 1.0225t x 8
3 ln(1.0225t)
t ln 1.0225
(f 1 f )(x) f 1 1 50
1.1 t x 1,000,000 8
3
8
ln
3 ln ln(8/3)
ln 1.0225 44.081 It will take about 44.081 years.
50
1 1.1 x
log1.1
50
50
1 1.1 log1.1 50
1.1 x) 50(1 49. (a) y 2539.852 636.896 ln x (b) When x 75, y 209.94. About 209.94 million metric
tons were produced. x (c) 636.896 ln x 400 636.896 ln x 50 2539.852 2939.852 ln x
log1.1 1
1.1 x log1.1(1.1x) x x 2939.852
636.896
2939.852 e 636.896 101.08 According to the regression equation, Saudi Arabian oil
production will reach 400 million metric tons when
x 101.08, in about 2001. 31 Section 1.6
50. (a) y 590.969 (b) The expression a logb(x c) d is defined when
x c 0, so the domain is (c, ).
Since a logb(x c) d attains every real value for
some value of x, the range is ( , ). 152.817 ln x (b) When x 85, y 87.94.
About 87.94 million metric tons were produced.
(c) 590.969 152.817 ln x 120 152.817 ln x
ln x 56. (a) Suppose f (x1) 710.969
710.969
152.817 ax1
cx1 104.84 51. (a) Suppose that f (x1) f (x2). Then mx1 b
so mx1 mx2. Since m 0, this gives x1
mx y b y b mx2
x2. b f 1 (x) x b 1
. The graphs of the
m d) bd adx2 bcx1 bd bcx1 (ad bc)x2 bc 0, this means that x1 ax x2. b b
d a)x dy m, respectively. Since each of 1
and
m dx b
cx a
dx b
f 1(x)
cx a slopes. a
(c
c ,f 1 (x) tal asymptote is y 52. (a) y2 is a vertical shift (upward) of y1, although it’s
difficult to see that near the vertical asymptote at x
One might use “trace” or “table” to verify this. 0. (b) Each graph of y3 is a horizontal line.
(c) The graphs of y4 and y
ln(e y2 y1) a,
ln a, y1 y2 a are the same.
ln a,
ln a 0). Since f (x) is undefined at dx b
→
cx a
d
(c
c 1
f (x1) d
.
c
d
, so the horizonc 0). Since f 1 (x) is a
, the vertical asymptote is x
c undefined at x a
.
c The horizontal asymptote of f becomes the vertical
asymptote of f 1 and vice versa due to the reflection of the graph about the line y
ln x x. ln a 53. If the graph of f (x) passes the horizontal line test, so will
the graph of g(x)
f (x) since it’s the same graph
reflected about the xaxis.
Alternate answer: If g(x1) g(x2) then
f (x1)
f (x2), f (x1) f (x2), and x1 x2
since f is onetoone.
g(x2). Then b
a
→ , so the horizontal
d
c d
, the vertical asymptote is x
c (d) As x → inverses will be perpendicular lines with nonzero ax
cx , f (x) asymptote is y
x m is the negative reciprocal of the other, the graphs of the 54. Suppose that g(x1) b dy b
cy a (c) As x → 1
,
m respectively, then the inverse functions will have slopes f (x1) bcx2 b)(cx1 y (d) If the original functions have slopes m and y1 (ax2 Interchange x and y: m inverses will be parallel lines with nonzero slope. y2 dy x inverse functions will have slope (d) ax
cx (cy (c) If the original functions both have slope m, each of the e y2 y1 d) adx2 bc)x1 cxy The slopes are reciprocals. 1
and
m bcx2 adx1 (b) y mx y m d adx1 Since ad Interchange x and y.
x cx2 b)(cx2 b x m b acx1x2 (ad b f (x2). Then: ax2 acx1x2 According to the regression equation, oil production
will reach 120 million metric tons when x 104.84, in
about 2005. (b) y d (ax1 710.969
e 152.817 x b 1
,
f (x2) f (x2), and x1 and x2 since f is onetoone. 55. (a) The expression a(b c x) d is defined for all values of
x, so the domain is ( , ). Since b c x attains all
positive values, the range is (d, ) if a 0 and the
range is ( , d) if a 0. s Section 1.6 Trigonometric Functions
(pp. 41–51)
Exploration 1
Functions Unwrapping Trigonometric 1. (x1, y1) is the circle of radius 1 centered at the origin (unit
circle). (x2, y2) is one period of the graph of the sine
function. 32 Section 1.6 2. The yvalues are the same in the interval 0 t 2. 3. The yvalues are the same in the interval 0 t 3. 4. 4. The x1values and the y2values are the same in each
interval.
5. y2 tan t: [ 3, 3] by [ 2, 2] Using trace, cos t and sin t are being computed for 0, 15,
30, … , 360 degrees. Quick Review 1.6
y2 csc t:
1. 180 2.
y2 180 40 4. 45 450 180 2.5 3. sec t: 60 3 2
9 143.24 4 180 5.
y2 cot t: [0, 2 ] by [ 1.5, 1.5] For each value of t, the value of y2
ratio y1
x1 tan t is equal to the x 0.6435, x 2.4981 6. . For each value of t, the value of y2 csc t is equal to the 1
ratio .
y1 [0, 2 ] by [ 1.5, 1.5] x For each value of t, the value of y2 sec t is equal to the 1.9823, x 4.3009 7. 1
ratio .
x1 For each value of t, the value of y2
ratio x1
y1 cot t is equal to the . Exploration 2 2 Finding Sines and Cosines 1. The decimal viewing window [ 4.7, 4.7] by [ 3.1, 3.1] is
square on the TI82/83 and many other calculators. There
are many other possibilities. x , 3
2 by [ 2, 2] 0.7854 or 4 ,x 3.9270 or 5
4 8. f ( x) 2( x)2 3 2x 2 3 f (x)
The graph is symmetric about the yaxis because if a point
(a, b) is on the graph, then so is the point ( a, b).
( x)3 3( x)
x 3 3x
(x 3 3x)
f (x)
The graph is symmetric about the origin because if a point
(a, b) is on the graph, then so is the point ( a, b). 9. f ( x) [ 4.7, 4.7] by [ 3.1, 3.1] 2. Using the Ask table setting for the independent variable on
the TI83 we obtain 10. x 0 Section 1.6 Exercises
1. Arc length 5
(2)
8
10 2. Radius 175 180 5
4
72
7 3.274 Section 1.6 3. Angle
4. Angle 16. (a) Period 1
radian or about 28.65
2 7
14
3 /2
6 4 2 33 2 (b) Amplitude 1 (c) [ 4, 4] by [ 2, 2] radian or 45 5. (a) The period of y sec x is 2 , so the window should
have length 4 .
One possible answer: [0, 4 ] by [ 3, 3] 17. (a) Period (b) Domain: Since csc (3x (b) The period of y csc x is 2 , so the window should
have length 4 .
One possible answer: [0, 4 ] by [ 3, 3]
(c) The period of y cot x is , so the window should
have length 2 .
One possible answer: [0, 2 ] by [ 3, 3]
6. (a) The period of y sin x is 2 , so the window should
have length 4 .
One possible answer: [0, 4 ] by [ 2, 2] 2
3 3x (k k , or x 1)
3 k
3 equivalent to x ) 1
sin (3x ) , we require . This requirement is for integers k. (c) Since csc (3x
)
1, the range excludes numbers
between 3 2
5 and 3 2 1. The range is
( , 5] [1, ).
(d) (b) The period of y cos x is 2 , so the window should
have length 4 .
One possible answer: [0, 4 ] by [ 2, 2]
(c) The period of y tan x is , so the window should
have length 2 .
One possible answer: [0, 2 ] by [ 3, 3]
7. Since
sin 6 8. Since
sin 6 is in the range , 22 0.5, sin 1(0.5) 4 is the range
2
2 4
180 , sin radian or 6 , 22
2 2
3 22
,
33 ,) (d)
1 x and radian or 1.3734 radians or
0.7954 radian or 45.5730 . 19. (a) Period 3 2 k
2 (b) Domain: We require 3x 1.5 (b) Amplitude (k 2)
6 for odd integers k. for odd integers k. This requirement is equivalent to x k
for odd integers k.
6 (c) Since the tangent function attains all real values, the
range is ( , ).
(d) by [ 4, 4]
2
2 (b) Amplitude 3 (c) [ 2 , 2 ] by [ 4, 4]
2
1/2 (b) Amplitude 4
5 2 (b) Amplitude ,
by [ 8, 8]
22
2
20. (a) Period
2 (b) Domain: ( ,) (c) Range: Since sin 2x (c) [ 4 , 4 ] by [ 10, 10]
15. (a) Period 30 . 2 (c) Since sin (4x
)
1, the range extends from
2 3 1 to 2 3 5. The range is [1, 5]. Therefore, x (c) [ 2 , 2 ] by [ 2, 2] 14. (a) Period 180 2
4 (b) Domain: ( 2
2 (b) Amplitude 13. (a) Period x and by [ 8, 8] ,
by [ 8, 8]
22 10. Using a calculator, cos 1(0.7) (c) 4 1 6 sin 2 1 9. Using a calculator, tan 1( 5)
78.6901 . 12. (a) Period of y sin 18. (a) Period 45 . 4 11. (a) Period of y 22
,
33 (d) 6 /3 4 (c) [ 3, 3] by [ 5, 5] [ , ] by [ 3, 3] 3 1, the range is [ 2, 2]. 34 Section 1.6
82 21. Note that 152 8
17 Since sin
cos and 1
tan 22. Note that 15
,
8 52 , 2 15
,
17 cos sin
cot
csc 1
cos sec 122 y
r
x
y sin
cot ( 3)
4
,
5 4
x
r 22 2 1 y
r 22 5
,
3 tan y
x 2
2 1, cot sec r
x 2 2 2
2 x 1.190 and x 7
6 11
6 30. The equation is equivalent to tan x 1
1 x 1 tan ( 1) 4
,
3
5
4 32. Let 2 1 22 , 17 22
2 2 2 x 2 3
4 6 2 1 cos2 9
13 . Then sin2 1
19 tan 13 2 1 1
,
2 2 are x 6 is y so the solutions in
5
.
6 and x 92
13 88
.
13 9/13 9 88/13 sin
cos so Therefore, 88 0.959. 33. (a) Using a graphing calculator with the sinusoidal
regression feature, the equation is
y 1.543 sin (2468.635x 0.494) 0.438. (b) The frequency is 2468.635 radians per second, which is
equivalent to 2468.635
2 392.9 cycles per second (Hz). The note is a “G.”
2
12 34. (a) b 6 cos 1 1
, so
3
1
3 are x 1.911 and x 80 (c) k 30
2 80 30
2 the solution
1.911. 2 (when the temperature is 30 F) and its maximum at t 8 (when the temperature is 80 F). The value of h is
2 1.911. 25. 55 (d) The function should have its minimum at t Since the cosine function is even, the solutions in the
x 9
,
13 and sin 2 [0, 0.01] by [ 2.5, 2.5] 28. This equation is equivalent to cos x interval 72
11 1 4.332. The solutions are 27. This equation is equivalent to sin x x so 0.771. 11
1 7
,
11 and cos (b) It’s half of the difference, so a in the interval 0 k , where k is any . Another solution in 4.332. x tan x is , k , where k is any 4 . Then 0
sin 11 cos
tan sin 1, 11 sin 2 2
2 17 cos 72
11 26. The angle cos 1( 0.7) 2.346 is the solution to this
equation in the interval 0 x
. Since the cosine
function is even, the value cos 1( 0.7)
2.346 is also
a solution, so any value of the form cos 1( 0.7) 2k
is a solution, where k is an integer. In 2
x 4 the
solutions are x cos 1( 0.7) 2
8.629 and
x
cos 1( 0.7) 4
10.220. the interval 0 is integer. r
y r
y 2 1, so the . Since the period of y integer. This is equivalent to x y
x x
y 4 x 2 all solutions are of the form x sin cos csc 2k , where k 1.190 is the solution to this 2 is tan 1(2.5) x 2k or x 31. Let x
r cos equation in the interval
0 , we 5
,
12
13
,
12 2, csc 25. The angle tan 1(2.5) 2 2. Then: 2 sin 2 In summary: tan r
x ( 2)2 24. Note that r 17
8 5. Then:
3
,
5 sec and 7
and
6 2 are x sin x has period 2 , the solutions are solution in the interval 2 cos
3
,
4 1
sin csc x is any integer. 8
,
15 13. 2 11
. Since y
6 all of the form x
sin
cos 17
,
15 sin
5/13
5
12/13
12
cos
12
5
sin
and cos
.
13
13
12
5
, cos
, tan
13
13
1
1
12
, sec
5
tan
cos
1
13
5
sin 23. Note that r 15
.
17 tan Since tan
have x 2 8
17 1
8
,
17 Therefore: sin 29. The solutions in the interval 0 2 sin2 1 cot 17. 8
2 5. Equation: y 25 sin 6 (x 5) 55 Section 1.6
(e) (b) Amplitude
Period 35. (a) Amplitude 2 (c) Horizontal shift
(d) Vertical shift (b) Average 365 (c) sin x (sin x) cos 4 101 37
37
62 ( 12)
2 (cos x) 4 1 2 1 62 F
12 F (cos x) sin 4 1 (sin x) 25 36. (a) Highest: 25
Lowest: 25 4 7
4 Vertical shift: 0 2
(2 /365) (b) Period 0.785 that is, or 5.498 that is, 37 2) 1.414 (that is, Horizontal shift
[ 1, 13] by [ 10, 100] 35 2 (sin x cos x) 2 Therefore, sin x
43. (a) 25 F This average is the same as the vertical shift because 2 sin ax cos x 2 sin x 4 . 4 (b) See part (a).
(c) It works. the average of the highest and lowest values of
y (d) sin ax sin x is 0. 37. (a) cot ( x) cos ( x)
sin ( x) (sin ax) cos (x)
sin (x) cot (x) f(x)
g(x) f ( x)
g( x) f (x)
g(x) so f
g csc (x) Then f ( x) so 39. Assume that f is even and g is odd.
Then f ( x)g( x) ( f (x))( g(x))
odd. cos (ax) 2 sin ax 4 tan . 1b a (c) It works.
(d) sin x 1
f (x) cos ax) (b) See part (a). 1
sin (x) 1
f (x) 4 2 (sin ax 44. (a) One possible answer:
y
a 2 b 2 sin x (b) Assume that f is odd.
1 (cos ax) sin 4
1 is odd. The situation g
f 38. (a) csc ( x) (cos ax)
2 1 So, sin (ax) is similar for .
1
sin ( x) 1 2 (b) Assume that f is even and g is odd.
Then (sin ax) cos 4 1
f is odd. tan 1b a
1b sin (x) cos tan
a sin (x) a2 1 f (x)g(x) so ( fg) is a2 b2 cos (x) sin tan a cos (x) b2 (a sin x 1b a b
a2 b2 b cos x) 40. If (a, b) is the point on the unit circle corresponding to the
angle , then ( a, b) is the point on the unit circle
corresponding to the angle (
) since it is exactly half
way around the circle. This means that both tan ( ) and
b
tan (
) have the same value, . and multiplying through by the square root gives the 41. (a) Using a graphing calculator with the sinusoidal
regression feature, the equation is
y 3.0014 sin (0.9996x 2.0012) 2.9999. sin tan a (b) y 3 sin ( x 2) 3 42. (a) desired result. Note that the substitutions
cos tan 1b a a
1b
a a2
b b2 a2 b2 and
depend on the requirement that a is positive. If a is negative, the formula does not
work.
45. Since sin x has period 2 , sin3 (x 2 ) sin3 (x). This
function has period 2 . A graph shows that no smaller
number works for the period. [ 2 , 2 ] by [ 2, 2] The graph is a sine/cosine type graph, but it is shifted
and has an amplitude greater than 1.
[ 2 , 2 ] by [ 1.5, 1.5] 36 Chapter 1 Review 46. Since tan x has period , tan (x
)
tan x . This
function has period . A graph shows that no smaller
number works for the period. 9. Since 4x 3y 4
.
3 4
(x
3
4
x
3 y
y
[ 2 , 2 ] by [ 1, 5] 4) 12 20
3 10. Since 3x
30 4, the slope of the given line (and hence the slope of the desired
line) is 2
47. The period is
60 4
x
3 12 is equivalent to y 5y 1 is equivalent to y . One possible graph:
the given line is 3
5 3
x
5 1
,
5 the slope of and the slope of the perpendicular line is 5
.
3
5
(x
3
5
x
3 19
3 1
2 1
y
3 y
y
, 60 60 by [ 2, 2] 2
48. The period is
60 1
.
30 2) 11. Since x
One possible graph: 3 1 is equivalent to y
3
2 of the given line is 3
x
2 and the slope of the perpendicular 2
3 line is .
2
(x
3
2
x
3 y
y
11
,
by [ 2, 2]
60 60 3(x
3x
1
(x
2
1
x
2 2. y
y
3. x
4. m 1)
9 m 2 2
1 6
( 3) y y
7. y 3) 8
4 2 2) 6 m
y 2 y
53
2
23
5
2
(x 3) 3
5
2
21
5
5 3x 2
5 4 3
1
(4)
2 3 0 ( 2)
34
2
(x 4)
7
2
6
x
7
7 2
7 2
7 2 15. [ 3, 3] by [ 2, 2] 3 8. Since 2x y
2 is equivalent to y 2x 2, the slope
of the given line (and hence the slope of the desired line)
is 2.
y 2(x 3) 1
y 2x 5 1
2 14. The line passes through (4, 2x y 2
4 4
( 2)
1
(x
2
1
x
2 f (x) 5) and (3, 0) 5
3 5
2 2 f (x) 3
2 ( 5)
30 Check: f (4) 2(x 6. m 5
x
3 13. m 0 y 5. y 0 y ( 6) 1) 2 8
3 12. The line passes through (0, s Chapter 1 Review Exercises
(pp. 52–53)
1. y
y 1) 3, the slope Symmetric about the origin.
16. [ 3, 3] by [ 2, 2] Symmetric about the yaxis. 1, as expected.
2) and ( 3, 0). Chapter 1 Review
17. 29. (a) Since the square root requires 16
is [ 4, 4]. x2 37 0, the domain (b) For values of x in the domain, 0 16 x 2
0
16 x 2 4. The range is [0, 4]. 16, so (c) [ 6, 6] by [ 4, 4] Neither
18.
[ 9.4, 9.4] by [ 6.2, 6.2] 30. (a) The function is defined for all values of x, so the
domain is ( , ).
[ 1.5, 1.5] by [ 0.5, 1.5] (b) Since 32
(1, ). Symmetric about the yaxis.
19. y( x)
Even ( x)2 1 x2 20. y( x)
Odd ( x)5 ( x)3 21. y( x)
Even 1 22. y( x) sec ( x) tan ( x) cos( x) 1 y(x) 1 cos x attains all positive values, the range is (c)
x5 ( x) x x3 x y(x) y(x)
[ 6, 6] by [ 4, 20] sin ( x)
cos2 ( x) 31. (a) The function is defined for all values of x, so the
domain is ( , ). sin x
cos2 x sec x tan x (b) Since 2e
( 3, ). y(x) x attains all positive values, the range is (c) Odd
23. y( x)
Odd x4
x3 ( x)4 1
( x)3 2( x) 24. y( x) 1 sin ( x)
Neither even nor odd 1 26. y( x)
Even ( x) 1 1
2x y(x)
[ 4, 4] by [ 5, 15] sin x 25. y( x)
x cos ( x)
Neither even nor odd
4 x4
x3 1
2x x 32. (a) The function is equivalent to y cos x 2x
x 4 1 k
4 x 27. (a) The function is defined for all values of x, so the
domain is ( , ).
(b) Since x attains all nonnegative values, the range is
[ 2, ). k
2 tan 2x, so we require for odd integers k. The domain is given by
for odd integers k. (b) Since the tangent function attains all values, the range
is ( , ).
(c) (c) , 22 [ 10, 10] by [ 10, 10] 28. (a) Since the square root requires 1
( , 1]. x 0, the domain is by [ 8, 8] 33. (a) The function is defined for all values of x, so the
domain is ( , ). (b) Since 1 x attains all nonnegative values, the range
is [ 2, ). (b) The sine function attains values from 1 to 1, so
2 2 sin (3x
) 2, and hence
3 2 sin (3x
) 1 1. The range is [ 3, 1]. (c) (c) [ 9.4, 9.4] by [ 3, 3] [ , ] by [ 5, 5] 38 Chapter 1 Review 34. (a) The function is defined for all values of x, so the
domain is ( , ).
5 x 2, which attains
(b) The function is equivalent to y
all nonnegative values. The range is [0, ).
(c) 39. First piece: Line through (0, 1) and (1, 0)
m 0
1 y 1
0 x 1
1 1 1 or 1 x Second piece:
Line through (1, 1) and (2, 0)
m 1
1 y [ 8, 8] by [ 3, 3] 35. (a) The logarithm requires x
(3, ). 0
2 3 0, so the domain is (b) The logarithm attains all real values, so the range is
( , ).
(c) (x y x 1
1 1) 1 2 or 2
1
2 f (x) 1 x,
x, x
0
1 x
x 1
2 40. First piece: Line through (0, 0) and (2, 5)
m
y 5
2
5
x
2 0
0 5
2 Second piece: Line through (2, 5) and (4, 0) [ 3, 10] by [ 4, 4] 36. (a) The function is defined for all values of x, so the
domain is ( , ).
(b) The cube root attains all real values, so the range is
( , ). m
y
y (c)
f (x) 0
4 5
2 5
2 5
(x
2
5
x
2
5x
,
2 2) 5
5x
2 10 or 10
0 x 2 2 5x
,
2 10
(Note: x 5
2 x 4 2 can be included on either piece.) [ 10, 10] by [ 4, 4] 37. (a) The function is defined for
is [ 4, 4]. 4 x 4, so the domain x , 4 x 4,
(b) The function is equivalent to y
which attains values from 0 to 2 for x in the domain.
The range is [0, 2]. 41. (a) ( f g)( 1) f (g( 1)) (b) (g f )(2) g( f (2))
f ( f (x)) f (d) (g g)(x) g(g(x)) [ 6, 6] by [ 3, 3] x 2, so the domain x 42. (a) ( f g)( 1) (b) See the graph in part (c). The range is [ 1, 1]. 3 f (0) 0
1
1/ x 2 2 2 2 1 1) 2 (b) (g f )(2) g( f (2)) (c) ( f f )(x)
[ 3, 3] by [ 2, 2] x, x f ( g( 1))
f( (c) 2
5 or 2 2 1 2.5 1
x x 1 2 1
1/x g 4 1 1/2 + 2 1
x (c) ( f f )(x) 2
1 1
2 g 1
1 f (1) 1 (c) 38. (a) The function is defined for
is [ 2, 2]. 1 f f ( f (x)) f (2 (d) (g g)(x) g(g(x)) g( 0 g(2 2
2) x)
3 x 2
1) 3 g(0) 3 3 0 1 x) (2 x x 1 1 1 Chapter 1 Review
43. (a) ( f g)(x) f (g(x))
f( x 2) (x 2 2) x, x
(g f )(x) cos2 t 2 x
4 2
x 2) y
4 2 y
4 in the identity 1 gives the Cartesian equation
1, or x 2 y2 16. The left half of the 47. (a) 2 (2 2 sin2 t and sin t circle is traced by the parametrized curve. g( f (x))
g(2 x
4 (b) Substituting cos t 39 x) 2 4 x 2 (b) Domain of f g: [ 2, )
Domain of g f: [ 2, 2]
[ 8, 8] by [ 10, 20] (c) Range of f g: (
, 2]
Range of g f: [0, 2]
44. (a) ( f g)(x) Initial point: (4, 15)
Terminal point: ( 2, 3)
The line segment is traced from right to left starting at
(4, 15) and ending at ( 2, 3). f (g(x))
f( 1 x) 1
4 (g f )(x) 1 (b) Substituting t 2 x into y 11 2t gives the
Cartesian equation y 11 2(2 x), or y 2x 7.
The part of the line from (4, 15) to ( 2, 3) is traced by
the parametrized curve. x
x g( f (x)) g( x) 1 x 48. (a) (b) Domain of f g: ( , 1]
Domain of g f: [0, 1]
(c) Range of f g: [0, )
Range of g f: [0, 1] [ 8, 8] by [ 4, 6] Initial point: None
Terminal point: (3, 0)
The curve is traced from left to right ending at the
point (3, 0). 45. (a) [ 6, 6] by [ 4, 4] Initial point: (5, 0)
Terminal point: (5, 0)
The ellipse is traced exactly once in a counterclockwise
direction starting and ending at the point (5, 0).
x
5 (b) Substituting cos t
cos2 t
x
5 2 sin2 t
y
2 and sin t y
2 in the identity 1 gives the Cartesian equation 2 1. (b) Substituting t x 1 into y
4 2t gives the
Cartesian equation y
4 2(x 1), or
y
6 2x. The entire curve is traced by the
parametrized curve.
49. (a) For simplicity, we assume that x and y are linear
functions of t, and that the point (x, y) starts at ( 2, 5)
for t 0 and ends at (4, 3) for t 1. Then x f (t),
where f (0)
2 and f (1) 4. Since
slope
x f (t) x
t 4
1 6t ( 2)
0 2 6,
2 6t. 46. (a) g(t), where g(0) slope The entire ellipse is traced by the curve. Also, y y
t y g(t) 3
1 5
0 2t 5 5 and g(1) 2,
5 2t. One possible parametrization is:
x
2 6t, y 5 2t, 0 t
[ 9, 9] by [ 6, 6] Initial point: (0, 4)
3
Terminal point: None (since the endpoint
is not
2
included in the tinterval)
The semicircle is traced in a counterclockwise direction
starting at (0, 4) and extending to, but not including,
(0, 4). 1 3. Since 40 Chapter 1 Review 50. For simplicity, we assume that x and y are linear functions
of t and that the point (x, y) passes through ( 3, 2) for
t 0 and (4, 1) for t 1. Then x f (t), where
f (0)
3 and f (1) 4. Since
x
t slope =
x f (t) 4 ( 3)
0 1 7t 7, 3 3 7t. Also, y
Since g(t), where g(0) slope y
t y g(t) 1
1 t 2 and g(1) ( 2)
0 2 1. 1 2 t. One possible parametrization is:
x
3 7t, y
2 t,
t . 51. For simplicity, we assume that x and y are linear functions
of t and that the point (x, y) starts at (2, 5) for t 0 and
passes through ( 1, 0) for t 1. Then x f (t), where
f (0) 2 and f (1)
1. Since
x
t slope 12
10 3, x Also, y g(t), where g(0) slope y
t 0
1 5
0 f (t)
g(t) 3x 3t. 5 5 [ 6, 12] by [ 4, 8] 5t. 55. Using a calculator, sin 1(0.6)
36.8699 . sin 2 40
,
7 sin x
2 (x) x csc
1 )(x) 1 f(f
f 2 x
3 3 2
1 f )(x) 2 x (2 f 1 3 x) 1 f ( f (x))
(2 2 (2
3 3x
3 3
40
7 3x)
3x) x Interchange x and y. 40
,
3
7
,
3 0.2 x y y = f (–x)
x
(3, 0) [ 6, 6] by [ 4, 4] x , sec sin
cos
1
cos 60. (a) The given graph is reflected about the yaxis. (–1, 0) 2)2, x
x2
y2 tan 59. e 0.2x 4
ln e 0.2x ln 4
0.2x ln 4
ln 4
x
5 ln 4 –3 y 40
.
7 (b) Since the period of sin x is 2 , the solutions are
x 3.3430 2k and x 6.0818 2k , k any
integer. (b) (x 40
49 40 3
(0, 2) 54. (a) y 32
7 1 58. (a) Note that sin 1( 0.2)
0.2014. In [0, 2 ), the
solutions are x
sin 1( 0.2) 3.3430 and
x sin 1( 0.2) 2
6.0818. (x)) 2 (f , 3
,
7 cos 1
tan
1
sin cot 3 Verify.
(f f 1.1607 radians or Therefore, 3
1 and 0 cos2 1 Interchange x and y. f 0.6435 radians or y
3 y 3
7 57. Since cos y 2 x 2 56. Using a calculator, tan 1( 2.3)
66.5014 . 3x 2 2 (b) 0. Since
5t 52. One possible parametrization is:
x t, y t(t 4), t 2.
2 3t 5 and g(1) 5, y One possible parametrization is:
x 2 3t, y 5 5t, t 0. 53. (a) y x2
y
f 1(x)
x2
Verify.
For x 0 (the domain of f 1)
( f f 1)(x) f ( f 1(x))
f ( x 2)
[( x 2) 2]2
( x)2 x
For x
2 (the domain of f ),
( f 1 f )(x) f 1( f (x))
f 1((x 2)2)
(x 2)2 2
x2
2
(x 2) 2 x 2
–3 Chapter 1 Review
(b) The given graph is reflected about the xaxis.
y 62. (a) V
(b) 100,000 10,000x, 0 x 41 10 V 55,000
10,000x 55,000
10,000x
45,000
x 4.5
The value is $55,000 after 4.5 years.
100,000 3 (1, 0) (–3, 0)
–3 3 x 63. (a) f (0)
(b) f (2)
(c) y = – f (x )
(0, –2)
–3 (c) The given graph is shifted left 1 unit, stretched
vertically by a factor of 2, reflected about the xaxis,
and then shifted upward 1 unit. 90 units
90 52 ln 3 32.8722 units [0, 4] by [ 20, 100] y
y = –2f (x + 1) + 1
(–4, 1) 64. 1500(1.08)t 3 1.08t
(0, 1) –4 5000
1500 10
3 ln (1.08)t ln t ln 1.08 x 2 ln t
(–1, –3) (d) The given graph is shifted right 2 units, stretched
vertically by a factor of 3, and then shifted downward
2 units.
4 y y = 3f (x – 2) – 2
(2, 4) 10
3
10
3 ln (10/3)
ln 1.08 t 15.6439
It will take about 15.6439 years. (If the bank only pays
interest at the end of the year, it will take 16 years.)
4 2t 65. (a) N(t) (b) 4 days: 4 24
1 week: 4 27
(c) N(t)
4 2t
2t
ln 2t
t ln 2 x –2 4 (–1, –2) (3, –2) 61. (a) 5000 y
(3, 2) –3 3 ln 500
ln 2 2000
2000
500
ln 500
ln 500
8.9658 There will be 2000 guppies after 8.9658 days, or after
nearly 9 days.
(d) Because it suggests the number of guppies will
continue to double indefinitely and become arbitrarily
large, which is impossible due to the finite size of the
tank and the oxygen supply in the water. 3
(–3, 2) t 64 guppies
512 guppies x 66. (a) y 20.627x 338.622 (–1, –1) (1, –1)
–3 (b) y [0, 30] by [ 100, 1000] 3
(3, 2)
(–1, 1)
–3 3
(1, –1) (–3, –2)
–3 x (b) When x 30, y 957.445. According to the
regression equation, about 957 degrees will be earned.
(c) The slope is 20.627. It represents the approximate
annual increase in the number of doctorates earned by
Hispanic Americans per year.
67. (a) y 14.60175 1.00232x (b) Solving y 25 graphically, we obtain x 232.
According to the regression equation, the population
will reach 25 million in the year 2132.
(c) 0.232% 42 Section 2.1 Chapter 2
Limits and Continuity 4. (a) lim p(s)
(b) lim p(s) (pp. 55–65) 3 5(2) 2 2. f (2) 5 4(2)
23 3. f (2) sin 4
2
2 sin 0 (b) lim F(x) 2 (c) lim F(x) does not exist, because the left and
righthand limits are not equal. 4 2
3
1 x
x 8. x
d2
d2 x
c
3x x 4 6. (a) lim G(x) 1 (b) lim G(x) 3
2
5 1 (c) lim G(x) 1 x→2 3 x→2 d2 c x2 (d) F(0) c2 x 7. x –3 x→0 c2
c 4 x→0 x→0 x 6. x d2
d2 c
x
18 (x 2x
x
2x2 x 1 c 3)(x 6)
x3 3 2 10. 5. (a) lim F(x) 1
3 1 3 0 4
4 9. 4 11
12 1
22 (d) p( 2)
2 2(2 ) 5. x 3 s→−2 s→ 2 Quick Review 2.1 4. f (2) 3 (c) lim p(s) s Section 2.1 Rates of Change and Limits 1. f (2) 3 s→ 2 x(2x 1)
(2x 1)(x 1) x→2 x 6, x
x x 1 ,x 3
1
2 Section 2.1 Exercises
1. (a) lim f (x) (d) G(2) 3 7. lim 3x 2(2x
x→ 1/2 1) 3 12
2 2 1
2 1 3 1
4 ( 2) 3
2 Graphical support: 3 x→3 (b) lim f (x) –2 x→3 (c) lim f (x) does not exist, because the left and [ 3, 3] by [ 2, 2] x→3 righthand limits are not equal.
(d) f (3) = 1 8. lim (x
x→ 4 3)1998 (4 3)1998 ( 1)1998 1 Graphical support: 2. (a) lim g(t) 5 (b) lim g(t) 2 t→ 4
t→ 4 (c) lim g(t) does not exist, because the left and [ 4.001, 3.999] by [0, 5] t→ 4 righthand limits are not equal.
(d) g( 4) 9. lim (x 3
x→1 3x 2 2x 3. (a) lim f (h) 4 (b) lim f (h) Graphical support: 4 h→0 h→0 –4 h→0 (d) f (0) 4 (1)3
1 2 (c) lim f (h) 17) [ 3, 3] by [ 25, 25] 3(1)2
3 2 2(1)
17 17
15 43 Section 2.1
10. lim y2 5y
y y→2 22 6
2 2 5(2)
2 6 20
4 5 16. lim ln (sin x) Graphical support: 11. lim y 4y [
2 ( 3)
4( 3)
( 3)2 3 3 y2 y→ 3 3 ln 1 2 0 Graphical support: [ 3, 3] by [ 5, 10]
2 ln sin x→ /2 3 0
6 , ] by [ 3, 1] 17. You cannot use substitution because the expression x
is not defined at x
2. Since the expression is not
defined at points near x
2, the limit does not exist. 0 Graphical support: 18. You cannot use substitution because the expression
not defined at x 0. Since 1
x2 2 1
is
x2 becomes arbitrarily large as x approaches 0 from either side, there is no (finite) limit. (As
[ 5, 5] by [ 5, 5] 12. lim int x int x→1/2 1
2 we shall see in Section 2.2, we may write lim 1 .) 2
x→0 x 0 Note that substitution cannot always be used to find limits
of the int function. Its use here can be justified by the
Sandwich Theorem, using g(x) h(x) 0 on the interval
(0, 1).
Graphical support: 19. You cannot use substitution because the expression
defined at x 0. Since lim
x→0 x
x x
x 1 and lim
x→0 x
x is not 1, the left and righthand limits are not equal and so the limit
does not exist.
20. You cannot use substitution because the expression
(4
(4 [ 4.7, 4.7] by [ 3.1, 3.1] 13. lim (x
x→ 2 6) 2/3 (2 6) 2/3 3 ( 8) 2 3 64 4 x)2
x
x)2
x 16 is not defined at x 16 x2 8x 0. Since and is equal to lim (8
x→0 8 x for all x x) x 8 0 0, the limit exists 8. Graphical support:
21. [ 4.7, 4.7] by [ 3.1, 3.1] [ 10, 10] by [ 10, 10] 14. lim x→2 x 3 2 3 5 lim 1
1 x 2
x→1 x 1
2 Algebraic confirmation: Graphical support: lim 1
1 x 2
x→1 x lim x→1 (x x1
1)(x 1) lim 1 x→1 x 1
2 1
1 1 1 22.
[ 4.7, 4.7] by [ 3.1, 3.1] 15. lim (e xcos x)
x→0 e 0 cos 0 Graphical support: 11 1
[ 4.7, 4.7] by [ 3.1, 3.1] lim t2 3t
t2 t→2 2
4 1
4 Algebraic confirmation:
lim
[ 4.7, 4.7] by [ 3.1, 3.1] t→2 t2 3t
t2 2
4 (t lim (t
t→2 1)(t
2)(t 2)
2) lim t t→2 t 1
2 2
2 1
2 1
4 44 Section 2.1
27. 23. [ 4.7, 4.7] by [ 3.1, 3.1]
8x2
16x2 5x3
lim 4
x→0 3x [ 4.7, 4.7] by [ 3.1, 3.1] 1
2 Algebraic confirmation:
8x2
16x2 5x3
3x4
x→0 lim sin x lim 2
x→0 2x 1 x Algebraic confirmation: x2(5x 8)
x2(3x2 16)
x→0
5x 8
lim 2
16
x→0 3x
5(0) 8
3(0)2 16
8
1
16
2 lim sin x lim 2
x→0 2x lim x x→0 sin x
lim
x→0 x sin x
x 1
2x 1 1 lim x→0 2x (1) 1 1
2(0) 1 28. 24. [ 4.7, 4.7] by [ 3.1, 3.1] 1
2 1 lim 2 x x→0 lim Algebraic confirmation:
1
2 1 2 lim x 2 x x→0 sin x
x lim 1 x→0 sin x
x lim 1 2 (2 x)
x→0 x(2 x)(2)
x
lim
x→0 x(2 x)(2)
1
lim
x→0 2(2 x)
1
1
2(2 0)
4 lim x x→0 sin x
x Algebraic confirmation: 1
4 x x→0 x lim [ 4.7, 4.7] by [ 3.1, 3.1] lim x→0 1 1 x→0 sin x
x 2 29. 25.
[ 4.7, 4.7] by [ 3.1, 3.1]
sin2 x
x→0 x lim [ 4.7, 4.7] by [ 5, 20] lim (2 x→0 x)3
x 8 Algebraic confirmation:
sin2 x
x→0 x 12 lim Algebraic confirmation:
lim (2 x→0 3 x)
x 8 lim lim sin x x→0 lim sin x 12x x→0 6x
x lim (12 2 x 6(0) lim x→0 3 (sin 0)(1) x→0 sin x
x sin x
x 0 2 6x x→0 12 0 x)
2 (0) 30.
12 26.
[ 2, 2] by [ 10, 10] lim 3 sin 4x x→0 sin 3x 4 [ 4.7, 4.7] by [ 3.1, 3.1] Algebraic confirmation: sin 2x
lim
x
x→0 x→0 sin 3x 2 lim Algebraic confirmation:
lim x→0 sin 2x
x 2 lim x→0 sin 2x
2x 3 sin 4x 4 lim x→0 2(1) 2 4(1) 3x
sin 4x
sin 3x
4x
sin 3x
lim
x→0 3x 4 lim x→0 sin 4x
4x (1) 4 1 45 Section 2.1
31. (a) True 42. Since (b) True
(c) False, since lim f (x)
x→0 x
x 1 for x 43. (a) lim (g(x) 0. x→4 (b) lim x f (x) (d) True, since both are equal to 0. x→4 (e) True, since (d) is true. (c) lim g2(x)
x→4 (f) True x
x 0, lim
x→0 3) 1. lim g(x) lim 3 x→4 lim x lim f (x) x→4 x→4
2 lim g(x) x→4 3 x→4 40 32 3 6 0 9 lim g(x) (g) False, since lim f (x)
x→0 0. (d) (h) False, lim f (x) 1, but lim f (x) is undefined. (i) False, lim f (x) x →4 g(x)
lim
x→4 f (x) 1 lim f (x) 0, but lim f (x) is undefined. x→1 x→1 x→1 44. (a) lim (f (x)
x→b g(x)) 0 x→4 lim f (x) 1 3 lim g(x) x→b 7 3 lim 1 x→4 x→b ( 3) 4 x→1 (j) False, since lim f (x)
x→2 (b) lim ( f (x) g(x))
x→b 0. lim f (x) lim g(x) x→b x→b (7)( 3) 21 32. (a) True
(b) False, since lim f (x)
(c) False, since lim f (x) (c) lim 4g(x) 1.
1. x→2
x→2 (d) True x→b f (x)
x→b g(x) (d) lim 4lim g(x) 4( 3) x→b lim f (x) x→b 7 7 lim g(x) x→b 12 3 3 (e) True
45. (a)
(f) True, since lim f (x)
x→1 lim f (x). x→1 (g) True, since both are equal to 0.
(h) True
(i) True, since lim f (x)
x→c 33. y1 x 2 x
x 2 (x 1 1 for all c in (1, 3). 1)(x 2)
x1 x 2, x [ 3, 6] by [ 1, 5] (b) lim f (x) 1 x→2 2; lim f (x)
x→2 1 (c) No, because the two onesided limits are different. (c) 46. (a)
34. y1 x2 x
x 2 (x 1 1)(x 2)
x1 (b)
35. y1 x2
x 2x 1
1 (x
x 1)2
1 x 1, x [ 3, 6] by [ 1, 5] 1 (b) lim f (x) (d)
36. y1 x→2 1; lim f (x)
x→2 1 (c) Yes. The limit is 1. x2 x
x 2 (x 1 1)(x 2)
x1 47. (a) (a)
37. Since int x
38. Since int x
39. Since int x
40. Since int x
41. Since x
x 0 for x in (0, 1), lim int x
x→0 0.
[ 5, 5] by [ 4, 8] 1 for x in ( 1, 0), lim int x
x→0 0 for x in (0, 1), lim int x
x→0.01 1 for x in (1, 2), lim int x
x→2 1 for x 0, lim
x0 x
x 1. (b) lim f (x)
x→1 0. 4; lim f (x) does not exist.
x→1 (c) No, because the lefthand limit does not exist.
48. (a) 1. 1.
[ 4.7, 4.7] by [ 3.1, 3.1] 46 Section 2.1 48. continued 54.
f (x) (b) lim x→ 1 0; lim f (x) x→ 1 0 (c) Yes. The limit is 0.
49. (a) [ 4.7, 4.7] by [ 5, 5] lim (x 2 sin x) 0 x→0 Confirm using the Sandwich Theorem, with g(x) =
h(x) x 2.
x 2 sin x
x 2 sin x
x 2 1 x 2.
2
2
2
x
x sin x x [ 2 , 2 ] by [ 2, 2] (b) ( 2 , 0)
(c) c (0, 2 ) 2 (d) c Because lim ( x 2) lim x 2 x→0
2 2 x→0 gives lim (x sin x) 50. (a) x 2 and 0, the Sandwich Theorem 0 x→0 55. [ (b) , ] by [ 3, 3] , 2 2 [ 0.5, 0.5] by [ 0.25, 0.25] , (c) c lim x 2 sin (d) c x→0 51. (a) 1
x2 0 Confirm using the Sandwich Theorem, with g(x)
and h(x) x 2.
1
1
x 2 sin 2
x 2 sin 2
x 2 1 x 2.
x x (c) c x 1
x sin 2
x
2 x 2 Because lim ( x 2) [ 2, 4] by [ 1, 3] (b) (0, 1) 2 x→0
2 (1, 2) give lim x sin 2 (d) c x→0 0 x2 lim x 2 x→0 1
x2 0, the Sandwich Theorem 0. 56. 52. (a) [ 0.5, 0.5] by [ 0.25, 0.25] [ 4.7, 4.7] by [ 3.1, 3.1] (b) ( , 1) ( 1, 1) (1, ) (c) None lim x 2 cos x→0 (d) None 1
x2 0 Confirm using the Sandwich Theorem, with g(x)
and h(x) x 2.
1
1
x2 cos 2
x2 cos 2
x2 1 x2. 53. x x
[ 4.7, 4.7] by [ 3.1, 3.1] lim (x sin x) x→0 x→0
2 give lim x cos Confirm using the Sandwich Theorem, with g(x)
and h(x) x .
x sin x
x sin x
x
x sin x
x
x→0 gives lim (x sin x)
x→0 x 1
x cos 2
x
2 x lim x x→0 0. x Because lim ( x 2) 0 Because lim ( x ) 2 1 x x 0, the Sandwich Theorem x→0 x2 1
x2 2 lim x 2 x →0 0, the Sandwich Theorem 0. 57. (a) In three seconds, the ball falls 4.9(3)2
44.1
average speed is
3 14.7 m/sec. 44.1 m, so its Section 2.1
(b) The average speed over the interval from time t
time 3 h is
4.9(3 h)2
(3 h) y
t 29.4 4.9(3)2
3 3 to h2) 4.9(6h
h 63. (a) Because the righthand limit at zero depends only on
the values of the function for positive xvalues near
zero.
1
1
sin
(base)(height)
(1)(sin )
2
2
2
(angle)(radius)2
(1)2
Area of sector OAP
2
2
2
1
1
tan
Area of OAT
(base)(height)
(1)(tan )
2
2
2 (b) Area of 4.9h Since lim (29.4 4.9h) h→0 47 29.4, the instantaneous speed is 29.4 m/sec. OAP (c) This is how the areas of the three regions compare.
58. (a) y gt (d) Multiply by 2 and divide by sin . 2 g(42) (e) Take reciprocals, remembering that all of the values
involved are positive. 20
16 20 (f) The limits for cos 5
or 1.25
4
20
(b) Average speed
5 m/sec
4 g sin (c) If the rock had not been stopped, its average speed over
the interval from time t 4 to time t 4 h is
1.25(4 h)2
(4 h) y
t 10 1.25(4)2
4 sin ( is between them, it must also have a limit of 1.
) sin sin ( ) (h) If the function is symmetric about the yaxis, and the
righthand limit at zero is 1, then the lefthand limit at
zero must also be 1. h2) 1.25(8h
h (g) and 1 are both equal to 1. Since 1.25h (i) The two onesided limits both exist and are equal to 1.
Since lim (10
h→0 1.25h) 10, the instantaneous speed
64. (a) The limit can be found by substitution. is 10 m/sec. lim f (x) x→2 59. (a)
x 0.1 0.01 0.001 0.0001 f (x) 0.054402 0.005064 0.000827 0.000031 f (2) (b) The graphs of y1
shown. 3(2)
f (x), y2 2 4 2 1.8, and y3 2.2 are (b)
0.1 x
f (x) 0.01 0.054402 0.001 0.005064 0.0001 0.000827 0.000031
[1.5, 2.5] by [1.5, 2.3] The limit appears to be 0.
60. (a) x 0.1 f (x)
(b) 0.8269 0.3056 0.01 0.1 f (x) 0.001 0.5064 0.5440 x 0.01 0.0001 0.001 0.0001 0.5064 0.8269 0.5440 0.3056 The intersections of y1 with y2 and y3 are at
x 1.7467 and x 2.28, respectively, so we may
choose any value of a in [1.7467, 2) (approximately)
and any value of b in (2, 2.28].
One possible answer: a 1.75, b 2.28.
(c) The graphs of y1
shown. f (x), y2 1.99, and y3 2.01 are There is no clear indication of a limit.
61. (a) x 0.1 0.01 0.001 f (x) 2.0567 2.2763 2.2999
(b) x 0.1 0.01 0.001 0.0001
2.3023
0.0001
[1.97, 2.03] by [1.98, 2.02]
f (x) 2.5893 2.3293 2.3052 2.3029
The limit appears to be approximately 2.3.
62. (a) x 0.1 0.01 f (x) 0.074398 0.001 0.0001 0.009943 0.000585 0.000021 (b)
x
f (x) 0.1
0.074398 0.01
0.009943 The limit appears to be 0. 0.001
0.000585 0.0001
0.000021 The intersections of y1 with y2 and y3 are at x 1.9867
and x 2.0134, respectively, so we may choose any
value of a in [1.9867, 2) and any value of b in
(2, 2.0134] (approximately).
One possible answer: a 1.99, b 2.01 48 Section 2.2 65. (a) f sin 6 s Section 2.2 Limits Involving Infinity 1
2 6 (b) The graphs of y1
shown. f (x), y2 0.3, and y3 0.7 are (pp. 65–73)
Exploration 1 Exploring Theorem 5 1. Neither lim f (x) or lim g(x) exist. In this case, we can
x→ x→ describe the behavior of f and g as x →
lim f (x) [0, 1] by [0, 1] The intersections of y1 with y2 and y3 are at x
and x 0.3047 0.7754, respectively, so we may choose any value of a in 0.3047,
6 and lim g(x) x→ One possible answer: a 0.305, b f (x), y2 sin x
x sin x
x lim 5 x→ 5 0 5, so the limit of the quotient exists.
2. Both f and g oscillate between 0 and 1 as x → , taking on
each value infinitely often. We cannot apply the sum rule
because neither limit exists. However, 0.775 0.49, and y3 5x lim approximate. (c) The graphs of y1
shown. Example 5, x→ , 0.7754 , where the interval endpoints are . We cannot apply the x→ quotient rule because both limits must exist. However, from and any value of b in 6 by writing 0.51 are lim (sin2 x cos2 x) x→ lim (1) x→ 1, so the limit of the sum exists.
3. The limt of f and g as x →
apply the difference rule to f
lim f (x) [0.49, 0.55] by [0.48, 0.52] The intersections of y1 with y2 and y3 are at x
and x 6 6 , and any value of b in , 0.5352 , where the interval endpoints are approximate.
One possible answer: a 0.513, b 0.535 x→ f (x) 0.5121 0.5352, respectively, so we may choose any value of a in 0.5121, lim g(x) x→ g(x) 66. Line segment OP has endpoints (0, 0) and (a, a ), so its
midpoint is
2 a
a
a
,
2 y
y 0 a2 a0 , 2 a a2
=,
and its slope is
22 2 0
a. The perpendicular bisector is the line through
0
2
a
1
with slope
, so its equation is
2
a
2
1
a
a
x
, which is equivalent to
a
2
2
2
1a
1 a2
1
x
. Thus the yintercept is b
. As
a
2
2 P→0 a→0 a2 1
2 02 1
2 1
.
2 1) 4. The fact that the limits of f and g as x → ln 2x
x 1 . do not exist does not necessarily mean that the limits of f g, f f
do not exist, just that Theorem 5 cannot be applied.
g 1. y 2x 3 y 3 2x y 3
2 x Interchange x and y.
x 3
2 f 1 (x) y
x 3
2 value of a approaches zero. Therefore,
lim ln (x We can use graphs or tables to convince ourselves that this
limit is equal to ln 2. the point P approaches the origin along the parabola, the lim b g. We can say that . We can write the difference as ln (2x) Quick Review 2.2
2 do not exist, so we cannot [ 12, 12] by [ 8, 8] g or Section 2.2
2. y e x
ln y x
Interchange x and y.
ln x y
f 1(x) ln x 7. (a) f ( x)
1
(b) f
x cos ( x) 8. (a) f ( x)
1
(b) f
x ex 1/x e 1
x ( x) e ln ( x)
x 9. (a) f ( x) (b) f ln 1/x
1/x ln ( x)
x x ln x [ 6, 6] by [ 4, 4] 10. (a) f ( x)
1 3. y tan x tan y x x, y 2 2 (b) f Interchange x and y.
tan x
1 f y, (x) x 2 tan x, 2 x 2 1
x 1
x 1
1/x 1. [ 5, 5] by [ 1.5, 1.5] 4. y cot 1x
cot y x, 0 x
Interchange x and y.
cot x y, 0 y
f 1(x) cot x, 0 x
(a) lim f (x)
x→ (b) lim f (x) x→ (c) y
[ 6, 6] by [ 4, 4] 1
1 1 2. 2
3 3x 3 0x2 8
x
3 10
3 5
x
3 7
3 1
[ 10, 10] by [ 1, 1] 2
3 q(x) 3x 2 r(x)
6. q(x)
r(x) x 3x2 x 3x 2 2x3 3 5 2x 3 4x x 2 5
x
3 2x2
1 2x 5
2x5 2x 2
x2 2x
x 7
3 2x 1
0x 4
x3
4
2x
0x3
2x4
x3
4
2x
2x3
x3
x3
1
2 (a) lim f (x)
0x 2
2x2
2x2
0x2
2x2
x2
x2 x 1 x→ (b) lim
x1
2x
x1
0x 1
x2 f (x) x→ (c) y 0 x ln x
x 1
( sin x)
x 1
sin x
x sin Section 2.2 Exercises
2 1 1
sin ( x)
x x [ 6, 6] by [ 4, 4] 5. cos x 1
cos
x 0
0 1
x 1
x x sin 1
x 49 50 Section 2.2 3. 6. [ 5, 5] by [ 10, 10] (a) lim f (x)
x→ [ 20, 20] by [ 4, 4] (a) lim f (x)
x→ 0 (b) lim f (x) 2 x→ (b) lim f (x) (c) y x→ (c) y 2 0 2, y 2 7. 4. [ 5, 5] by [ 2, 2]
[ 10, 10] by [ 100, 300] (a) lim f (x)
x→ 1 (b) lim f (x) (a) lim f (x)
x→ 1 x→ (c) y (b) lim f (x) 1, y 1 8. x→ (c) No horizontal asymptotes.
5. [ 5, 5] by [ 2, 2] [ 20, 20] by [ 4, 4] (a) lim f (x)
x→ (b) lim f (x)
(a) lim f (x)
x→ x→ 3 (c) y (b) lim f (x) 3 x→ (c) y 3, y 3 1 1
1 Section 2.2
9. 13. [ 2, 6] by [ 1, 5] lim x→2 [ 4, 4] by [ 3, 3] 1
x2 10. lim x→0 int x
x 0 14. [ 2, 6] by [ 3, 3] lim [ 4, 4] by [ 3, 3] x lim x→2 x 2 11. x→0 int x
x 15. [ 7, 1] by [ 3, 3] lim [ 3, 3] by [ 3, 3] 1 lim csc x x→ 3 x 3 12. x→0 16. [ 7, 1] by [ 3, 3] lim x x→ 3 x 3 [ lim x→( /2) , ] by [ 3, 3] sec x 51 52 Section 2.2 17. 22. [ 4, 4] by [ 3, 3] (a) x 2, x [ 2 , 2 ] by [ 3, 3] 2 (a) x (b) Lefthand limit at 2 is .
Righthand limit at 2 is
Lefthand limit at 2 is
.
Righthand limit at 2 is . . (b) If n is even:
Lefthand limit is .
Righthand limit is 2(x 1) x
x
x2
x2
x1
x 1 5 x2
5 x2
2
x2
x 3 2x 2
2
3
15 x
x
x 2 5x 5 2 23. y x
x [ 7, 5] by [ 5, 3] 2
An end behavior model for y is (b) Lefthand limit at 2 is
.
Righthand limit at 2 is . lim y 19. lim 1 x→ lim y
2
x 24. y
[ 6, 6] by [ 12, 6] 5x3 1 (b) Lefthand limit at 1 is
.
Righthand limit at 1 is . lim 1 lim y 5x2 1
x2
2
10x
x2
x3 2 1 lim 5 x→ x x 5x2 1
x2 lim y 5x3
x3 5. 5 x→ x→ 1. 1 x→ An end behavior model for y is 20. x3
x3 1 x→ x→ (a) x . If n is odd:
Lefthand limit is
.
Righthand limit is . 18. (a) x n , n any integer 2 lim 5 5 x→ 25. Use the method of Example 10 in the text.
[ 2, 4] by [ 2, 2]
1
,x 3
2 cos (a) x lim x→ 1
(b) Lefthand limit at
is .
2
1
Righthand limit at
is
2 . Righthand limit at 3 is cos lim x→ Lefthand limit at 3 is .
. 1 1
x
1
x 1 1
x
1
x x→ lim 2x lim 2 x→ Similarly, lim y
x→ [ 2 , 2 ] by [ 3, 3] (a) x k , k any integer (b) at each vertical asymptote:
Lefthand limit is
.
Righthand limit is . cos (0)
10 1
1 1 cos x
x cos (0)
10 1
1 1 0 2. x→0 1 26. Note that y
So, lim y 21. cos x
x lim x→0 1 sin x
x lim x→ 2. 2 sin x
.
x sin x
x 2 Section 2.2
sin x
2x 2 x
sin x
lim
0
x
x→
1
lim
0
2x 1
x→ sin x
x 27. Use y So, lim y lim 0. x is both a right end behavior lim ln x
x 1 x→ 1 lim 0. x→ x→ x2 sin x
x2 2x3
x lim x→ 1 sin x
x2 2x 2. (a) 30. An end behavior model is x5
2x2 0.5x 3. (c) 31. An end behavior model is 2x4
x 2x 3. (d) x
x2 32. An end behavior model is [ 4, 4] by [ 1, 3] x 2. (b) lim f (x) x→ 2 lim f (x) x→ (b) None f 1 1/x
e is shown.
x 0 44. 4x 3 34. (a) 1
x
1
lim f
x
x→0
1
lim f
x
x→0 The graph of y 4 (b) None
35. (a) x
2x2 (b) y
36. (a) 1
2x 3x2
x2 (b) y 1
x
1
lim f (x) lim f
x
x→
x→0
1
lim f (x) lim f
x
x→
x→0 The graph of y 3
3 3 37. (a) [ 4, 4] by [ 1, 3] 0 4x
x 4x 2 x4
x2 1 1/x
e
is shown.
x2 0 45. (b) None
38. (a) f x2 (b) None because lim ex 2x x→ e (b) The function y
ex 2x
2x e x 1 0 1 1. f x ln 1
is shown.
x 0
0 46. ex
x2 lim 1 x→ e
2 lim (x e x→ x2 (b) The function y x→ ex
2x lim 1
x
1
lim f (x) lim f
x
x→
x→0
1
lim f (x) lim f
x
x→
x→0 The graph of y 1. x is a right end behavior model x2 x→ x→
2x 0 2 40. (a) The function y because lim 1 2x is a left end behavior model x→ because lim 2x
ex lim 1 x x→ because lim [ 3, 3] by [ 2, 2] e x is a right end behavior model 39. (a) The function y x e
e 1) x 1 0 is a left end behavior model
2 x lim x x→ 0 1 x
e 1. x 1 1.
[ 5, 5] by [ 1.5, 1.5] 1
x
1
lim f (x) lim f
x
x→
x→0
1
lim f (x) lim f
x
x→
x→0 The graph of y f 1. x 2 is both a right end behavior 43. 29. An end behavior model is 33. (a) 3x 0 model and a left end behavior model because 0 and lim y x ln x
x 42. (a, b) The function y 1 sin x
xx So, lim y x x→
x→ sin x
x 2 41. (a, b) The function y
1 model and a left end behavior model because 0 and lim y x→ 28. y 1
2x sin x
is shown.
x 1
1 1. 53 54 Section 2.2 47. (a) lim f (x)
x→ (b) lim f (x)
x→ (c) lim f (x)
x→0 (d) lim f (x)
x→0 48. (a) lim f (x)
x→ (b) lim f (x)
x→ (c) lim f (x)
x→0 (d) lim f (x)
x→0 1
x lim x→ 53. (a) Using 1980 as x
y
2.2316x 3 0 lim ( 1) 0:
54.7134x 2 351.0933x 733.2224 1 x→ 1
x lim x→0 lim ( 1)
x
x lim x→ 2
1 1
lim 2
x→ x x
x lim x→ (b) Again using 1980 as x 0:
y 1.458561x 4 60.5740x 3 905.8877x 2
5706.0943x 12967.6288 1 0 x lim [0, 20] by [0, 800] 1 x→0 2
1 x→0 x 0
0 2
1 2 1 lim [0, 20] by [0, 800] 2
x→0 x (c) Cubic: approximately 2256 dollars
Quartic: approximately 9979 dollars 49. One possible answer:
y (d) Cubic: End behavior model is 2.2316x3.
This model predicts that the grants will
become negative by 1996. 5
y = f (x ) Quartic: End behavior model is 1.458561x4.
This model predicts that the size of the grants
will grow very rapidly after 1995. x 10 Neither of these seems reasonable. There is no reason
to expect the grants to disappear (become negative)
based on the data. Similarly, the data give no indication
that a period of rapid growth is about to occur. 50. One possible answer: 54. (a) Note that fg f (x)g(x) 1.
f→
as x → 0 , f → as x → 0 , g → 0, fg → 1 y
5 (b) Note that fg f (x)g(x)
f → as x → 0 , f → y = f (x ) 8 (c) Note that fg f (x)g(x) 3(x 2)2.
f→
as x → 2 , f → as x → 2 , g → 0, fg → 0 x 7 8.
as x → 0 , g → 0, fg → (d) Note that fg 5 f (x)g(x) 3)2 (x . f → , g → 0, fg →
(e) Nothing – you need more information to decide.
51. Note that f1(x)/f2(x) f1(x)g2(x) f1(x)/g1(x) g1(x)/g2(x) g1(x)f2(x)
f1
f2 f2(x)/g2(x) As x becomes large, g1 and using the above equation, g2
f1/f2 g1/g2 . both approach 1. Therefore, 55. (a) This follow from x 1 int x x, which is true for
all x. Dividing by x gives the result.
(b, c) Since lim x x→ must also approach 1. 52. Yes. The limit of ( f g) will be the same as the limit of g.
This is because adding numbers that are very close to a
given real number L will not have a significant effect on the
value of ( f g) since the values of g are becoming
arbitrarily large. 1 Theorem gives lim
x→ 56. For x 0, 0 e x lim 1 1, the Sandwich x→ x int x
x 1, so 0 int x
x lim x→ x e
x 1. 1
.
x 1
approach zero as x → , the Sandwich
x
ex
Theorem states that
must also approach zero.
x Since both 0 and 57. This is because as x approaches infinity, sin x continues to
oscillate between 1 and 1 and doesn’t approach any given
real number.
ln x2
x→ ln x 58. lim 2, because ln x2
ln x 2 ln x
x 2. 55 Section 2.3 ln x
x→ log x 59. lim ln x
log x ln (10), since (b) lim f (x) ln x
(ln x)/(ln 10) ln 10. (d) f (2)
1
x ln x 1 ln x ln 1 1
x ln x ln 1 1 ln x ln x (pp. 73–81) (g f )(x)
x
2x (x 3)(x 3). The domain of f is (
(3, ) or all x
3. , 3) [ 3, 6] by [ 2, 8] Thus, lim (x x→3 10
3 20
6 2), x 2 1. lim 3x x3 x→–1 2. (a) lim f (x) (b) lim f (x) x→ 1 x→ 1 1)(x 2)
for x
x3
10
.
3 2 3( 1)
2( 1)
( 1)3 4 lim int (x)
lim f (x) x x→2 lim (x 2 x→2 1 6
3 2 5 1 5) 1 9x
1)(x f (x) ( x
x 5 1 1) 2 x ,x 1. 0. Squaring both sides 1 1
x2 1, x
1 x 1 0.
1 1 0
5) 0 1
,x
2 5 0.453 The only solution to the original equation is x 22 0 4(2) 1. For x 3, f (x) 4 when x 2 6x 8 4, which gives
x 2 6x 12 0. The discriminant of this equation is
b 2 4ac ( 6)2 4(1)(12)
12. Since the
discriminant is negative, the quadratic equation has no
solution. 1
4x 1 1
5 9. For x 3, f (x) 4 when 5 x 4, which gives x
(Note that this value is, in fact, 3.) righthand limits are not equal. 3. (a) lim f (x) 1
2x
x 1
. Therefore, f (x)
x2
1 1
1 Solution: x x→ 1 int ( 1) 5 [ 5, 5] by [ 10, 10] 3. (c) lim f (x) does not exist, because the left and (d) f ( 1) 1 3. 2 x→ 1 x→ 1 1 8. g(3), so g is continuous at x 2x 1
4 1 9 Quick Review 2.3
2 f (x) Solutions: x 1)(x 1)(x 2)
x3 x→3 5. lim g(x) 10
.
3 (x 1
x
1
x g( f (x))
1
for x
x f (g(x)) (2x 3), so f (x) 1
5
4
,x
1 (g f )(x) ( f g)(x) 7. 2x 2 3)(x 1
x 1 x 3)(x 3x
2x 1
1 gives f (x) 1 (x g 2x
2x 5
1 Therefore, 2. It appears that the limit of f as x → 3 exists and is a little
more than 3. 6 2x
x g( f (x)) 6. Note that Removing a Discontinuity 3. f (3) should be defined as 2 5. Note that sin x 2 (g f )(x) g( f (x)) g(x 2).
Therefore: g(x) sin x, x 0
( f g)(x) f (g(x)) = f (sin x) (sin x)2 or sin2 x, x s Section 2.3 Continuity
Exploration 1 2 1
4. ( f g)(x) f (g(x)) f
1
x
2(1 x) x
x2
,x 0
(1 x) 5x
6x 1 1
x ln 1 4 2 1
,
x But as x → , 1 1/x approaches 1, so ln(1 1/x)
approaches ln (1) 0. Also, as x → , ln x approaches
infinity. This means the second term above approaches 0
and the limit is 1. (x 2 righthand limits are not equal. 1) ln (x 1)
ln x 7x 2 1 Since ln (x 4. x 3 4 x→2 ln (x 1)
ln x
x→ 1. x
9
( 3, 3) x) x→2 (c) lim f (x) does not exist, because the left and 60. lim 2 lim (4 x→2 5 1 1. 56 Section 2.3 10. x
is equivalent to
x 7. The function y
y 1, x
x 1, 0
0. It has a jump discontinuity at x
[ 2.7, 6.7] by [ 6, 6] A graph of f (x) is shown. The range of f (x) is
( , 1) [2, ). The values of c for which f (x) c has
no solution are the values that are excluded from the range.
Therefore, c can be any value in [1, 2). 1
2)2 (x is continuous because it is a cos x
, a quotient
sin x cot x is equivalent to y of continuous functions, so it is continuous. Its only points
of discontinuity occur where it is undefined. It has infinite
discontinuities at x Section 2.3 Exercises
1. The function y 8. The function y 0. k for all integers k. e 1/x is a composition ( f g)(x) of the 9. The function y 1
, so it is
x quotient of polynomials, which are continuous. Its only continuous functions f (x) point of discontinuity occurs where it is undefined. There is continuous. Its only point of discontinuity occurs at x an infinite discontinuity at x where it is undefined. Since lim e 1/x 2. e x and g(x) , this may be x→0 x 2. The function y x2 1
is continuous because it is a
4x 3 quotient of polynomials, which are continuous. Its only
points of discontinuity occur where it is undefined, that is,
where the denominator x 2 4x 3 (x 1)(x zero. There are infinite discontinuities at x 3) is considered an infinite discontinuity.
10. The function y ln (x 1) is a composition ( f g)(x) of
the continuous functions f (x) ln x and g(x) x 1, so it
is continuous. Its points of discontinuity are the points not
in the domain, i.e., x
1.
11. (a) Yes, f ( 1) 0. 1 and at
(b) Yes, lim x 0, x→ 1 3. 0. (c) Yes
1 3. The function y x2 1 is continuous because it is a (d) Yes, since quotient of polynomials, which are continuous. lim f (x) x→ 1 Furthermore, the domain is all real numbers because the
denominator, x 2 1, is never zero. Since the function is continuous and has domain ( 12. (a) Yes, f (1) 1 is a left endpoint of the domain of f and
f ( 1), f is continuous at x
1. (b) Yes, lim f (x)
x→1 , ), there are no points of 1. 2. (c) No
discontinuity. (d) No 4. The function y
x 1 is a composition ( f g)(x) of the
continuous functions f (x) x and g(x) x 1, so it is
continuous. Since the function is continuous and has
domain ( , ), there are no points of discontinuity.
5. The function y 2x 3 is a composition ( f g)(x) of the continuous functions f (x) x and g(x) 2x 3, so it is continuous. Its points of discontinuity are the points
not in the domain, i.e., all x
3 3
.
2 6. The function y
2x 1 is a composition ( f g)(x) of
3
the continuous functions f (x)
x and g(x) 2x 1, so
it is continuous. Since the function is continuous and has
domain ( , ), there are no points of discontinuity. 13. (a) No
(b) No, since x 2 is not in the domain. 14. Everywhere in [ 1, 3) except for x 0, 1, 2. 15. Since lim f (x) 0, we should assign f (2) 16. Since lim f (x) 2, we should reassign f (1) x→2 x→1 0.
2. 17. No, because the righthand and lefthand limits are not the
same at zero.
18. Yes. Assign the value 0 to f (3). Since 3 is a right endpoint
of the extended function and lim f (x)
x→3 function is continuous at x 3. 0, the extended Section 2.3
19. 27. Since lim x→0 sin x
x 1, the extended function is x 0 1, y sin x
,
x x 0. [ 3, 6] by [ 1, 5] (a) x 28. Since lim 2 x→0 (b) Not removable, the onesided limits are different. sin 4x
x 4 lim x→0 29. For x
f (x) sin 4x
,
x x 4, the extended 0 4, y (a) x 4(1) function is 20. [ 3, 6] by [ 1, 5] sin 4x
4x x 0. 4 (and x
x 4
x 2 0),
(x 2)( x
x 2 2) x 2. 2 (b) Removable, assign the value 1 to f (2).
The extended function is y 21. 30. For x
f (x)
[ 5, 5] by [ 4, 8] (a) x 1 (b) Not removable, it’s an infinite discontinuity.
22. 2 (and x 2. 2), x3 4x 2 11x 30
x2 4
(x 2)(x 5)(x 3)
(x 2)(x 2)
(x 5)(x 3)
x2
x 2 2x 15
.
x2 The extended function is y x2 2x
x 15
2 . 31. One possible answer:
Assume y x, constant functions, and the square root
function are continuous. [ 4.7, 4.7] by [ 3.1, 3.1] (a) x x 1 (b) Removable, assign the value 0 to f ( 1).
23. (a) All points not in the domain along with x 0, 1 (b) x 0 is a removable discontinuity, assign f (0) 0.
x 1 is not removable, the onesided limits are
different.
24. (a) All points not in the domain along with x 1, 2 (b) x 1 is not removable, the onesided limits are
different.
x 2 is a removable discontinuity, assign f (2) 1.
25. For x 3, f (x) x2
x 9
3 The extended function is y
x3
x2
(x (x x 3)(x 3)
x3 3. 1
1
1)(x 2 x 1)
(x 1)(x 1)
x2 x 1
.
x1
x2 x 1
The extended function is y
.
x1 26. For x 1, f (x) x 3. By the sum theorem, y x 2 is continuous. By the composite theorem, y
By the quotient theorem, y x 2 is continuous. 1
x is continuous.
2 Domain: ( 2, )
32. One possible answer:
Assume y x, constant functions, and the cube root
function are continuous.
By the difference theorem, y 4 x is continuous.
3
By the composite theorem, y
4 x is continuous.
By the product theorem, y x 2 x x is continuous.
3
By the sum theorem, y x 2
4 x is continuous.
Domain: ( , )
33. Possible answer:
Assume y x and y x are continuous. By the product theorem, y x 2 x x is continuous.
By the constant multiple theorem, y 4x is continuous.
By the difference theorem, y x 2 4x is continuous.
By the composite theorem, y
x 2 4x is continuous.
Domain: ( , ) 57 58 Section 2.3 34. One possible answer:
Assume y x and y 1 are continuous. 39. Use the product, difference, and quotient theorems. One
also needs to verify that the limit of this function as
x approaches 1 is 2.
Alternately, observe that the function is equivalent to
y x 1 (for all x), which is continuous by the sum
theorem.
Domain: ( , ) [ 3, 3] by [ 2, 2] Solving x x 4 1, we obtain the solutions
x
0.724 and x 1.221.
40. 35. One possible answer:
y
5
y = f (x ) [ 6, 6] by [ 4, 4] x3 Solving x
5 2, we obtain the solution x
lim (x 2 41. We require that lim 2ax
x→3 32 2a(3)
36. One possible answer: 6a 1 4
3 5 42. Consider f (x)
y = f (x )
x f (1) 1): x→3 8 a y 5 1.521. x 1 x 1
e e x . f is continuous, f (0) 1, and 0.5. By the Intermediate Value Theorem, for some c in (0, 1), f (c) 0 and e c c. 43. (a) Sarah’s salary is $36,500 $36,500(1.035)0 for the
first year (0 t 1), $36,500(1.035) for the second
year (1 t 2), $36,500(1.035)2 for the third year
(2 t 3), and so on. This corresponds to
y 36,500(1.035)int t. 37. One possible answer:
y
5 (b) y = f (x ) 5 x
[0, 4.98] by [35,000, 45,000] The function is continuous at all points in the domain
[0, 5) except at t 1, 2, 3, 4.
38. One possible answer: 44. (a) We require: y
5
y = f (x ) f (x)
5 x { 0
1.10,
2.20,
3.30,
4.40,
5.50,
6.60,
7.25, x
x
x
x
x
x
x 0
1
2
3
4
5
6 x 0
1
2
3
4
5
6
24. This may be written more compactly as
1.10 int( x), 0 x 6
f (x)
7.25,
6 x 24 Section 2.4 59 49. For any real number a, the limit of this function as x
approaches a cannot exist. This is because as x approaches
a, the values of the function will continually oscillate
between 0 and 1. (b) Section 2.4 Rates of Change and
Tangent Lines (pp. 82–90)
s [0, 24] by [0, 9] This is continuous for all values of x in the domain
[0, 24] except for x 0, 1, 2, 3, 4, 5, 6.
45. (a) The function is defined when 1 1
x Quick Review 2.4
0, that is, on , 1) 1. x
y 3
5 ( 5) 8
23 2. ( x
y a
b 1
3 (0, ). (It can be argued that the domain should also include certain values in the interval
( 1, 0), namely, those rational numbers that have odd 5 1 denominators when expressed in lowest terms.)
4. m
(b) y
[ 5, 5] by [ 3, 10] f ( 1)
f (0) 1
1 1
1 10
0 1 0 1 y 1
(undefined) and
0 (undefined). Since f is undefined at y y points of discontinuity.
(d) The discontinuity at x 0 is removable because the
righthand limit is 0. The discontinuity at x
1 is not
removable because it is an infinite discontinuity. 4
(x
3
4
x
3 y
y 9. Since 2x
2
.
3
2
[x
3
2
x
3 m
[0, 20] by [0, 3] y
y
b
4 3
2 b 10. 3 The limit is about 2.718, or e.
b
46. This is because lim f (a
h→0 h) lim f (x). x→a 47. Suppose not. Then f would be negative somewhere in the
interval and positive somewhere else in the interval. So, by
the Intermediate Value Theorem, it would have to be zero
somewhere in the interval, which contradicts the
hypothesis.
48. Since the absolute value function is continuous, this follows
from the theorem about continuity of composite functions. 2
3 3 6 1) 7
3 4 19
4
4
3 1
3/4 8. m (e) ( 2)] 3
(x
4
3
x
4 7. y these values due to division by zero, both values are 4
6 4
7 16
7
41
3
7
(x 1) 6
3
7
25
x
3
3 6. m (c) If we attempt to evaluate f (x) at these values, we obtain ( 1)
( 3) 3
[x
2
3
x
2 5. y 4
7 3
( 2) 3
3 3. m 1) 4 8
3 3y 5 is equivalent to y ( 1)] 2
x
3 3 7
3
5
3
10
3
19
3 Section 2.4 Exercises
1. (a) f
x f (3)
3 f (2)
2 (b) f
x f (1) f ( 1)
1 ( 1) 2. (a) f
x f (2)
2 (b) f
x f (12)
12 f (0)
0
f (10)
10 28 9 19 1
2 0 1 2
3 1 1 2
7 41
2 0.298 5
, we use
3 60 Section 2.4 3. (a) f
x f (0)
0 f ( 2)
( 2) 1 2 e
2 3 (b) f
x f (3)
3 f (1)
1 e 4. (a) f
x f (4)
4 f (1)
1 ln 4 0
3 (b) f
x f (103)
103 5. (a) e f
x f ( /2)
( /2) 6. (a) f
x f( ) (b) f
x f( ) f( )
() 1 h→0 h→0 1
103
ln
3
100 4 1.273 3 33 1 3 2 1.654 0.637 1 (b) The tangent line has slope
( 2, y( 2)) ( 2, 4).
y
4[x ( 2)] 4
y
4x 4
(c) The normal line has slope 0 2 (16.5, 475), y ( 2)] 4 h
4h h2
h h) 4
4 and passes through 1
4 1
and passes
4 ( 2, 4). 4 9
2 46
50
[ 8, 7] by [ 1, 9] 50 PQ1
PQ2 10. (a) lim 43
46 PQ3
PQ4 y (1 h)
h h→0 lim y (1)
h)2 [(1 lim h→0 50 lim [12 4h 3 4(1)] h
1 2h h2 4
h h2 h→0 2h
h lim (h 2) h→0 The appropriate units are meters per second. h)] 4(1 h→0 50 2 (b) Approximately 50 m/sec
8. We use Q1 (5, 20), Q2 (7, 38), Q3
Q4 (9.5, 72), and P (10, 80).
12
14 (8.5, 56), (b) The tangent line has slope
(1, y (1)) (1, 3).
y
2(x 1) 3
y
2x 1
(c) The normal line has slope 16 through (1, y (1)) 16 Secant
PQ1 12 PQ2 1) 3 7
2 16 PQ4 1
(x
2
1
x
2 14 PQ3 y Slope 16 The appropriate units are meters per second.
(b) Approximately 16 m/sec h2 4h ( 2)2 (d) 43 Slope Slope of PQ4: 1
[x
4
1
x
4 y (14, 375), Q3
(20, 650). 20
5
38
7
56
8.5
72
9.5 4 lim ( 4 h→0 through ( 2, y( 2))
1 h)2
h (2 h→0 lim /3 Secant Slope of PQ3: lim 0.462 ln 100 1 0 650 225
20 10
650 375
Slope of PQ2:
20 14
650 475
slope of PQ3:
20 16.5
650 550
Slope of PQ4:
20 18 Slope of PQ2: y( 2) h→0 /2 (a) Slope of PQ1: 80
10
80
10
80
10
80
10 h)
h = 7. We use Q1 (10, 225), Q2
Q4 (18, 550), and P (a) Slope of PQ1: y( 2 lim 3 f ( /6)
( /6)
f (0)
0 ln 4
3 ln 103 1
ln 1.03 0.0099
3
f (3 /4) f ( /4)
f
(3 /4) ( /4)
x (b) lim 8.684 2 f (100)
100 9. (a) 0.432 y
(d) [ 6, 6] by [ 6, 2] (1, 3). 2 and passes through 1
2 1
and passes
2 61 Section 2.4 11. (a) lim y(2 h)
h h→0 y(2) lim 1
h) (2 2 1 1
h (b) Near x h h→0 h)
h h→0 1 1 2, f (x) f (2 lim h h→0 lim 13. (a) Near x 1
1 h→0 1 lim (h 1)
h(h 1) 14. Near x h→0 h 1 1 and passes through h→0 1
1 (c) The normal line has slope
through (2, y(2))
y 1(x y x 2) f (0 x (3 lim 2 f (1) lim 1 h→0 h)
h 1 3 (x 2 x. (1 [2 lim 2)
h)]
h (2 1) h→0 1 lim 1 h→0 h)
h 2 x.
h→0 15. First, note that f (0)
lim x f ( 3) h→0 1
(b) The tangent line has slope
(2, y(2)) (2, 1).
y
(x 2) 1
y
x3 f (1 h)
h 1 1, f (x) h)
h
1h
lim
h
h→0 lim (2 lim h→0 1 h→0 1 lim f (2) h)
h h→0 lim x. 3, f (x) f( 3 lim x 1 2. f (0) (2 lim h→0
h→0 lim ( 2 (2, 1). h) h→0 1 2 h
h2 2h
h lim 1 and passes h 2) 2h 2 1 f(0 lim h→0 (d) h)
h f(0) (2h lim 2)
h h→0 2 lim 2 h→0 2
No, the slope from the left is 2 and the slope from the
right is 2. The twosided limit of the difference quotient
does not exist. [ 4.7, 4.7] by [ 3.1, 3.1] 12. (a) lim y(0 h)
h h→0 y(0) lim ( h2 3h h→0 lim 1) ( 1) h
h2 h→0 3h lim (h lim f (0 h→0 h h→0 16. First, note that f (0) 3) 3 lim f (0 h→0 lim (h h→0 (b) The tangent line has slope
(0, y(0))
y 3(x y (0, 3x 3 and passes through f (0)
f (0) lim 1 lim h→0 1) h)
h ( h2 1
2 f (2) 1 lim h→0 lim through (0, y(0))
y
y 1
(x
3
1
x
3 0) 1 1 (d) [ 6, 6] by [ 5, 3] (0, h)
h 0 1. h→0 (c) The normal line has slope 1 1 1
1
3 0
h h→0 Yes. The slope is f (2 h lim h→0 17. First, note that f (2) 1). 0) h)
h
h)
h 0. 1
and passes
3 lim h→0 lim 1). h→0 = 1
4 2 h 1
2 h
2 (2 h)
2h(2 h)
h
2h(2 h)
1
2(2 h) 62 Section 2.4
(b) The slope of the tangent steadily decreases as a
increases. 17. continued
4 lim f (2 h)
h h→0 f (2) (2
4 lim
[4 h→0 x→(3 /4) but f
19. (a) 3
4 f (x) cos lim f (a h→0 lim lim
lim 3
4 2
2
f (a) sin [(a 2ah 2ah h→0 19.6 2 (a 2 2]
h
h2
h 24. Let f (t) 2)
a2 2 2 lim f (10 h→0 lim h) 300 3t 2.
h)
h f (10) a lim
lim h→0 lim h→0 h→0 3h 2 60h lim (60 a h 25. Let f (r) h
2a 2(a h)
ah(a h)
2
a(a h) lim f (3 h→0 r 2, the area of a circle of radius r.
h)
h f (3) 1
h lim 1
1 a 1
1)(a h (a 43
r.
3
f (2 h) f (2)
lim
h
h→0 1 26. Let f (r) 1) (b) The slope of the tangent is always negative. The
tangents are very steep near x 1 and nearly
horizontal as a moves away from the origin.
[9 (a 9 a2 h→0 lim h→0 lim h→0 2ah
h lim ( 2a h→0 2a h)2] (9
h
2ah h 2
h
h2 h) lim (2 h→0 lim 8 h)3
h
h)3
h
12h h→0 lim (12 h→0 6h 4
(2)3
3 23
6h 2 + h 3
h 8 h 2) 12 16 a 2)
9 lim 4
(2
3 h→0 4
3
4
3
4
3
4
3 1) 2 (a h→0 h) The area is changing at a rate of 6 in2/in., that is,
6 square inches of area per inch of radius. 1 lim 9 h lim (6 h h→0 f (a) h2 6 h→0 h)
h (3)2 h→0 (a 1) (a h 1)
1)(a h 1)
h→0 h(a f (a 9 h)2
h
6h h→0 a lim (3 lim h→0 lim lim 22. (a) lim 300 The rocket’s speed is 60 ft/sec. (b) The slope of the tangent is always negative. The
tangents are very steep near x 0 and nearly
horizontal as a moves away from the origin. h→0 300 3h) 2
a2 21. (a) lim h) 2
h 3(10 lim 60 h→0 f (a) 19.6 h h→0 2a h)
h 100 4.9h) h→0 (b) The slope of the tangent steadily increases as a
increases.
2
2 f (a 4.9h 2
h 19.6 h lim (2a f (a) 4.9(2) 2] [100 2 h2 h→0 h)
h 100 lim ( 19.6 3
4 h) 2 a2 lim h→0 f (a h) 2]
h
19.6h 4.9(2 The object is falling at a speed of 19.6 m/sec. lim lim f (2) h→0 . h→0 h→0 h)
h
[100 h→0 sin x x→(3 /4) lim 20. (a) f (2 4.9t 2. 100 h→0 3
4 h)
h lim h→0 18. No. The function is discontinuous at x
because lim 23. Let f (t) 2 h
lim
h→0 4h
1
4 1
.
4 Yes. The slope is 1
2 h
(2 h)]
4h h→0 lim h) a2 The volume is changing at a rate of 16 in3/in., that is, 16
cubic inches of volume per inch of radius. 63 Section 2.4
s(1 27. lim h)
h h→0 s(1) h)2
h
3.72h 1.86(1 lim h→0 1.86 lim 1.86(1)2
1.86h2 31. (a) From Exercise 21, the slope of the curve at x
1
1.86 lim (3.72 or a 3.72
The speed of the rock is 3.72 m/sec.
s(2 h)
h h→0 s(2) lim 11.44(2 h→0 lim 45.76 h→0 lim (45.76 h)2 11.44(2)2
h
45.76h 11.44h 2
h lim 0: y
y 1(x 0)
x1 4(a h) 2: y
y 1(x 2)
x3 1 1
1 2ah h2 2ah 1] (a 2 4a lim (2a h 4) h→0 2a 4h
h 1 a2 4a 1 h) f (a)
h
[3 4(a h) 3 4a
4h h→0 lim ( 4 h→0 4 1(x 2)
x1 1 4 0, or (a 4h a 2 12
a2 3
a 2 2a 3
(a 1)(a 3)
a a. a2 2ah
h At a h2 2a h)2] (3 4a a 2)
h
2ah h 2 3 4a a 2
h h) 33. (a) 2a
2a(a 1)
2a 2 2a
0
0
1 or a 3 1 (or x
1), the slope is
y 2(x 1) 12
y 2x 10
3 (or x
y
y 2.1
1995 1.5
1993 2( 1) 3), the slope is 2(3)
6(x 1) 12
6x 18 2. 6. 0.3 The rate of change was 0.3 billion dollars per year. 2a The tangent at x a is horizontal when 4
a
2. The tangent line is horizontal at
( 2, f ( 2)) ( 2, 7). a2) 12
a1 9 At a h→0 lim 2: y
y (9 h→0 lim 1 There is only one such line. It is normal to the curve at
two points and its equation is y x 1. f (a lim 1(x 0)
x1 4 30. First, find the slope of the tangent at x
h→0 0: y
y 32. Consider a line that passes through (1, 12) and a point
(a, 9 a 2) on the curve. Using the result of Exercise 22,
this line will be tangent to the curve at a if its slope is 2a. The tangent at x a is horizontal when 2a
a
2. The tangent line is horizontal at
( 2, f ( 2)) ( 2, 5). lim 1) and (2, 1), this time using slope 1. At x 1) 4h h→0 h2
h 4a h→0 lim 1, so we again need to find lines through (0, a. h
a2 1) and (2, 1), 1 At x h→0 lim 1 passing through (0, At x f (a)
h)2 [(a 1, so we need to find the equations of (b) The normal has slope 1 when the tangent has slope 29. First, find the slope of the tangent at x
h)
h 0 1 and 1 At x 45.76 The speed of the rock is 45.76 m/sec. f (a 0 1, so a respectively. 45.76 lim 1 lines of slope 11.44h) h→0 h→0 1
2 1)2 1 2. Note that y(0) y(2) 1 when 1, which gives (a 1)2 (a 1.86h) h→0 . The tangent has slope 1 h h→0 28. lim 1)2 (a a, is (b)
2a 0, or 3.1
1997 2.1
1995 0.5 The rate of change was 0.5 billion dollars per year.
(c) y 0.0571x 2 0.1514x [0, 10] by [0, 4] 1.3943 64 Section 2.4 33. continued
(d) y(5)
5
y(7)
7 y(3)
3
y(5)
5 0.31
0.53 According to the regression equation, the rates were 0.31 billion dollars per year
and 0.53 billion dollars per year.
(e) lim y(7 h→0 h)
h y(7) lim h)2 [0.0571(7 0.1514(7 h→0 lim h 2)
h 0.0571(14h h→0 lim [0.0571(14) 1.3943]
h [0.0571(7)2 0.1514h 0.1514 h→0 h) 0.0571h ] 0.65
The funding was growing at a rate of about 0.65 billion dollars per year.
34. (a) [7, 18] by [0, 900] (b) Q from year Slope
440 225
17 8
440 289
17 9
440 270
17 10
440 493
17 11
440 684
17 12
440 763
17 13
440 651
17 14
440 600
17 15
440 296
17 16 1988
1989
1990
1991
1992
1993
1994
1995
1996 23.9
18.9
24.3
8.8
48.8
80.8
70.3
80.0
144.0 (c) As Q gets closer to 1997, the slopes do not seem to be approaching a limit value.
The years 1995–97 seem to be very unusual and unpredictable.
35. (a) f (1 h)
h f (1) e1 h e h (b) [ 4, 4] by [ 1, 5] Limit 2.718 (c) They’re about the same.
(d) Yes, it has a tangent whose slope is about e. 0.1514(7) 13943] Section 2.4
f (1 36. (a) h)
h f (1) 21 h 2 40. Let f (x) x2/3. The graph of y
is shown. h (b) [ 4, 4] by [ 1, 5] f (0 h)
h f (0) 65 f (h)
h [ 4, 4] by [ 3, 3] The left and righthand limits are
and , respectively.
Since they are not the same, the curve does not have a
vertical tangent at x 0. No.
Limit 41. This function has a tangent with slope zero at the origin. It
is sandwiched between two functions, y x 2 and y
x 2,
both of which have slope zero at the origin. 1.386 (c) They’re about the same.
Looking at the difference quotient, (d) Yes, it has a tangent whose slope is about ln 4.
37. Let f (x) x 2/5. The graph of y f (0 h)
h f (0) f (h)
h is shown. h f (0 h)
h f (0) h, so the Sandwich Theorem tells us the limit is 0.
42. This function does not have a tangent line at the origin. As
the function oscillates between y x and y
x infinitely
often near the origin, there are an infinite number of
difference quotients (secant line slopes) with a value of 1
and with a value of 1. Thus the limit of the difference
quotient doesn’t exist. [ 4, 4] by [ 3, 3]
The left and righthand limits are
and , respectively.
Since they are not the same, the curve does not have a
vertical tangent at x 0. No.
38. Let f (x) x 3/5. The graph of y
is shown. f (0 h)
h f (0) f (h)
h The difference quotient is
oscillates between 1 and
43. Let f (x)
f (1 h)
h f (0 h)
h f (0) sin 1
which
h 1 infinitely often near zero. sin x. The difference quotient is
f (1) sin (1 h)
h sin (1) . A graph and table for the difference quotient are shown. [ 4, 4] by [ 3, 3] Yes, the curve has a vertical tangent at x
lim f (0 h→0 h)
h f (0) 0 because [ 4, 4] by [ 1.5, 1.5] . 39. Let f (x) x1/3. The graph of y
is shown. f (0 h)
h f (0) f (h)
h Since the limit as h→0 is about 0.540, the slope of
y sin x at x 1 is about 0.540. [ 4, 4] by [ 3, 3] Yes, the curve has a vertical tangent at x
lim h→0 f (0 h)
h f (0) . 0 because 66 Chapter 2 Review s Chapter 2 Review Exercises 13. Since lim ( e (pp. 91–93) e 1. lim (x 3 2x 2 x→ 2 x2 2. lim x→−2 3x 2( 2)2 ( 2)2 1
3( 2)2 2( 2) 1
2x 5 2 ( 2)3 1) 3. No limit, because the expression
values of x near 4.
4. No limit, because the expression
values of x near 5.
1 5. lim 2 1
2 x lim x→0
2 2x
5x 2 6. lim
x→ 3
7 lim x→ 2x
5x 2 x→ 8. lim x→0 sin 2x
4x x 10. lim e x sin x
x→0 11. Let x sin x and x cos x, lim
x→ 2 9 x is undefined for 7
2 x→0 2x(2 21. Yes
23. No x3
x4
is
128
12x3 24. Yes 1
x.
12 25. (a) lim g(x) 1 1.5 (c) No, since lim g(x) x 1
(1)
2 sin x
x sin x
x→0 x e 0 sin 0 int 2 (d) g is discontinuous at x
domain). 1
2 26. (a) lim k(x)
x→1 sin x
x lim 1 1 (b) lim k(x)
x→1 2 (c) k(1)
10 0 1.5
0 0 (d) No, since lim k(x) because 6 int (6 2h) 1 (and at points not in the (f) No, the discontinuity at x 1 is not removable because
the onesided limits are different. 6,
27. lim 6 x→7/2 h, where h is in
7
2 k(1) (e) k is discontinuous at x
domain). 1
. Then
2 2h – 1 x→7/2 int 2 3 (and at points not in the (e) Yes, the discontinuity at x 3 can be removed by
assigning the value 1 to g(3). x→0 1 7
2 g(3). x→3 lim 1) 2h 6. 1
, 0 . Then
2 1 int (6 [ 4, 4] by [ 3, 3] (a) Vertical asymptote: x
2h) 5,
(b) Lefthand limit 2h is in (5, 6). lim 2
x x→–2 x Righthand limit: lim x x→–2 x Therefore, lim int (2x
x→7/2 1. 20. Limit exists. 1
4 (b) g(3) Therefore, lim int (2x – 1) int (2x x
x 19. Limit exists. x) 2h is in (6, 7). 7
2 lim x→ 22. No h, where h is in 0, because 6 sin x
cos x 18. Limit does not exist. x lim 1
x
x→ 12
1
lim
x
12
x→ lim 1) x
x 16. Limit exists. x→1 int (2x 12. Let x for all x, the Sandwich Theorem lim x→0 x→0 0, and 0. 2x is undefined for 9. Multiply the numerator and denominator by sin x. lim 1 x 17. Limit exists. x4
12x3 1
sin 2x
lim
2 x→0 2x x csc x 1
lim
x csc x
x→0 e cos x x→3 x4 x3
3
128
x→ 12x
x 4 x3
lim
12x3 128
x→ x lim e x→ 14. Since the expression x is an end behavior model for both 5
21 Therefore:
lim ) cos x
x gives lim e 15 2
5 7. An end behavior model for e x x 15. Limit exists.
4 lim 2 1 5 1 2 (2 x)
x
x)
x→0 2x(2
1
1
2(2 x)
2(2 0) x→0 x x→ 1) lim 5 x→7/2 5 3
2
3
2 Chapter 2 Review
28. 33. (a) End behavior model: 2x
2
, or
x2
x (b) Horizontal asymptote: y
34. (a) End behavior model:
[ 4, 4] by [ 3, 3] (a) Vertical asymptotes: x 0, x 0 (the xaxis) 2x2
, or 2
x2 (b) Horizontal asymptote: y 2 67 2 3 (b) At x 35. (a) End behavior model: 0: (b) Since the end behavior model is quadratic, there are no
horizontal asymptotes. x1
2
2)
x→0 x (x
x1
Righthand limit lim 2
2)
x→0 x (x Lefthand limit At x lim 36. (a) End behavior model: 2: 29. (a) At x Righthand limit
At x lim x→ x→
x x
ex 1 1, a right end behavior model is e .
lim f (x) x→ 1 lim f (x) x→ 1 lim f (x) x→0 Righthand limit lim (1) 1 x→ 1 lim ( x) x→ 1 x→ 1 ex x (b) Since lim lim ex
x 1 x→ x 1, a left end behavior model is x. lim f (x) x→0 lim ( x) x→0 lim ( x) x→ 1
ln x 0 x→0 lim f (x) x→1 lim f (x) x→1 lim ( x) 1 x→1 1
for all x
ln x 0 and 0. lim (1) x→1 1 (c) At x
1: Continuous because f ( 1) the limit.
At x 0: Discontinuous because f (0) the limit.
At x 1: Discontinuous because the limit does not
exist.
30. (a) Lefthand limit lim f (x) x→1 4(1) 3 Righthand limit
2(1) lim (x 2 x→1 2x 2) 3 39. lim f (x)
x→3 lim (x lim x2 2x – 15
x–3 x→3 3 5 x→0 5) lim x→3 1 31. Since f (x) is a quotient of polynomials, it is continuous and
its points of discontinuity are the points where it is
undefined, namely x
2 and x 2.
32. There are no points of discontinuity, since g(x) is
continuous and defined for all real numbers. 0 (x – 3)(x 5)
x–3 lim x→0 sin x
2x 8. 1
sin x
lim
2 x→0 x 1
.
2 Assign the value k
41. One possible answer: 1 1 8. x→3 (b) No, because the two onesided limits are different.
(c) Every place except for x sin x
ln x end behavior model. 40. lim f (x) lim f (x) 1 1, so ln x is both a right end behavior model and a left Assign the value k 4x 3 x→1 2 lim x 3 x→1 sin x
0. Hence
ln x
ln x
sin x
lim
= lim
ln x
x→
x→ lim x→ (b) At x
1: Yes, the limit is 1.
At x 0: Yes, the limit is 0.
At x 1: No, the limit doesn’t exist because the two
onesided limits are different. (1)2 sin x
ln x 1
ln x lim x→ Therefore, the Sandwich Theorem gives Righthand limit (1)3 1
ln x 38. (a, b) Note that lim 0 1: Lefthand limit (d) At x lim ex 0: Lefthand limit At x ex x 37. (a) Since lim 1: Lefthand limit x4
, or x
x3 (b) Since the end behavior model represents a
nonhorizontal line, there are no horizontal asymptotes. x1
2
2)
x→–2 x (x
x1
Righthand limit lim 2
2)
x→–2 x (x Lefthand limit x
, or x 2
x y
10
y = f (x ) 10 x 1
(1)
2 1
2 68 Chapter 2 Review 42. One possible answer: 48. At x y lim 5 a, the slope of the curve is
f (a h→0 h)
h f (a) lim [(a h)2 a2 2ah h2 lim (2a 3 The tangent is horizontal when 2a 3 lim 2ah h→0 2a V(a 44. lim h)
h h→0 1
3
1
3
1
3
2
3 45. lim 2 2 V(a) 1
(a
3 2 lim h h2 2ah 3
3
. Since f
2
2
3
9
is ,
.
2
4
200
49. (a) p(0)
1 7e 0.1(0) or x 12
aH
3 h)2H h→0 a H lim 1
/2 a2 h h→0 H lim (2a h) h→0 (b) lim p(t)
t→ aH
h)
h h→0 lim 6a 2 S(a) h)2
h 6(a lim h→0 6h 2 12ah lim (12a 6a2 6a 2 6h) h→0 y(a h)
h h→0 lim h) 2 [(a (a h) 0, at a 3
2 9
, the point where this occurs
4 200
8 25 lim t→ 1 200
7e 0.1t 200
1 200 (a 2 2] a 2) h
a2 2ah h2 a h a2 2 a 2 h h→0 lim 3 (b) y(a) h→0 lim 3a (Note that we cannot use the formula
f (x) 3.20 1.35 int x, because it gives incorrect
results when x is an integer.) 12a 46. lim a2 (c) Perhaps this is the maximum number of bears which
the reserve can support due to limitations of food,
space, or other resources. Or, perhaps the number is
capped at 200 and excess bears are moved to other
locations.
3.20 1.35 int ( x 1), 0 x 20
50. (a) f (x)
0,
x0 h h→0 3a – 3h
h Perhaps this is the number of bears placed in the
reserve when it was established. H(2a) S(a 3a) h) x y = f (x ) f ( /2) – f (0)
/2 – 0 (a 2 h2 h→0 3h
h h→0 5 h)]
h h→0 lim 43. 3(a h2
h h→0 lim (2a h→0 [0, 20] by [ 5, 32] h h 2ah 1) f is discontinuous at integer values of x: 0, 1, 2, ..., 19.
51. (a) Cubic: y
Quartic: y 2a 1 47. (a) lim f (1 h→0 lim h) f (1)
h
1 2h h 2 lim ( 1 h→0 3 3h 3(1
h h)] ( 2) 2 h h→0
h0 h)2 [(1 lim 52. Let A lim f (x) and B x→c 1 and passes through (c) The normal at P has slope 1 and passes through
(1, 2).
y 1(x 1) 2
yx3 B 3
2 1
(b) The tangent at P has slope
(1, 2).
y
1(x 1) 2
y
x1 (b) Cubic: 1.644x 3, predicts spending will go to 0
Quartic: 2.009x 4, predicts spending will go to A h) 1.644x 3 42.981x 2 254.369x
300.232
2.009x 4 102.081x 3 1884.997x 2
14918.180x 43004.464 B and lim x→c lim g(x). Then A x→c B 2 and 3
1. Adding, we have 2A 3, so A
, whence
2
1
3
2, which gives B
. Therefore, lim f (x)
2
2
x→c
1
g(x)
.
2 53. (a) [3, 12] by [ 2, 24] Section 3.1
(b) Year of Q 69 Slope of PQ
20.1
2000
20.1
2000
20.1
2000
20.1
2000
20.1
2000 1995
1996
1997
1998
1999 2.7
1995
4.8
1996
7.8
1997
11.2
1998
15.2
1999 3.48
3.825
4.1
4.45
4.9 (c) Approximately 5 billion dollars per year.
(d) y 0.3214x 2
lim y(10 h→0 1.3471x
h)
h y(10) 1.3857
[0.3214(10 h) 2 1.3471(10 0.3214(20h h 2)
h 1.3471h h) lim h→0 lim h→0 0.3214(20) 1.3857]
h [0.3214(10)2 1.3471(10) 1.3857] 1.3471 5.081
The predicted rate of change in 2000 is about
5.081 billion dollars per year. Chapter 3
Derivatives
s Section 3.1 Derivative of a Function (pp. 95–104)
Exploration 1 Reading the Graphs 1. The graph in Figure 3.3b represents the rate of change of the depth of the water in the puddle with respect to time. Since y is
measured in inches and x is measured in days, the derivative dy
would be measured in inches per day. Those are the units that
dx should be used along the yaxis in Figure 3.3b.
2. The water in the ditch is 1 inch deep at the start of the first day and rising rapidly. It continues to rise, at a gradually decreasing
rate, until the end of the second day, when it achieves a maximum depth of 5 inches. During days 3, 4, 5, and 6, the water level
goes down, until it reaches a depth of 1 inch at the end of day 6. During the seventh day it rises again, almost to a depth of
2 inches.
3. The weather appears to have been wettest at the beginning of day 1 (when the water level was rising fastest) and driest at the end
of day 4 (when the water level was declining the fastest).
4. The highest point on the graph of the derivative shows where the water is rising the fastest, while the lowest point (most
negative) on the graph of the derivative shows where the water is declining the fastest.
5. The ycoordinate of point C gives the maximum depth of the water level in the ditch over the 7day period, while the
xcoordinate of C gives the time during the 7day period that the maximum depth occurred. The derivative of the function
changes sign from positive to negative at C , indicating that this is when the water level stops rising and begins falling.
6. Water continues to run down sides of hills and through underground streams long after the rain has stopped falling. Depending
on how much high ground is located near the ditch, water from the first day’s rain could still be flowing into the ditch several
days later. Engineers responsible for flood control of major rivers must take this into consideration when they predict when
floodwaters will “crest,” and at what levels. 70 Section 3.1 Quick Review 3.1
1. lim (2
h→0 h)2
4 2. f (3) 4 lim (4 h 2) 4h lim f (3 4 h 4 4 0 3 lim lim 4 h→0 f (3)
1
3 1 h h→0 h)
h h→0 h h h→0 lim 3 (3 h)
h) h→0 3h(3 2. lim+ x 2 3
x→2 5
2 2( 2x 8 x 2 lim 2( h→4 h→0 3(3 y 0, lim
x 2)( x x→4 x 2) 3. f (3) lim x→3 2 f (x)
x 2) 2( 4 2) 8 1
x 1
3 3 lim 3 lim 2 through (0, 1) and another point (h, h
is 1) 1
h0 h. Since lim h x→3 (x 1) on the parabola 0, the slope of the line h→0 4. f (x) lim x→1 lim f (x) x→1 8. lim f (1
h→0 lim (x 1)2 lim (x 2) x→1
x→1 h) lim f (x) x→1 1)2 (1
1 2 0 3 f (x) [3(x h) 12]
h lim 3h 5. dy
dx lim h)
h
h)
h 7(x
7h 1 because lim f (x) does not
x→1 exist. 6. Let f (x)
d2
(x )
dx (b) The normal line has slope
(2, 3).
y
y 1
(x
5
1
x
5 2)
17
5 3 1
and passes through
5 7 h→0
f (x)
(x h→0 1. (a) The tangent line has slope 5 and passes through (2, 3).
y 5(x 2) 3
y 5x 7 7x x 2. lim Section 3.1 Exercises 12) y(x) lim 7 h→0 h (3x 3 h→0 y(x lim h→0 lim 0 lim 3 h→0 h h→0 9. No, the two onesided limits are different (see Exercise 7).
10. No, f is discontinuous at x h)
h lim h→0 1
9 f (x h→0 6. Use the graph of f in the window [ 6, 6] by [ 4, 4] to find
that (0, 2) is the coordinate of the high point and (2, 2) is
the coordinate of the low point. Therefore, f is increasing
on ( , 0] and [2, ). x
3)(x)(3) 1
3x lim x→3 tangent to the parabola at its vertex is 0. 7. lim f (x) f (3)
3 x→3 x x 5. The vertex of the parabola is at (0, 1). The slope of the line (h 2 h) 1. y→0 y lim 1
9 1 lim 1 for y y x→4 3
2 y 3. Since
4. lim 2 lim x2 h→0 lim (2x h→0 lim f (x h)
h h→0 h)2 x 2
h
2xh h 2
h h) f (x) x2 2x 7. The graph of y f1(x) is decreasing for x 0 and
increasing for x 0, so its derivative is negative for x
and positive for x 0. (b) 0 8. The graph of y f2(x) is always increasing, so its derivative
is always 0. (a)
9. The graph of y f3(x) oscillates between increasing and
decreasing, so its derivative oscillates between positive and
negative. (d)
10. The graph of y f4(x) is decreasing, then increasing , then
decreasing, and then increasing, so its derivative is
negative, then positive, then negative, and then positive. (c) Section 3.1 11. dy
dx lim h→0 lim y(x h) y(x)
h
[2(x h)2 13(x 15. (a) The amount of daylight is increasing at the fastest rate
5] (2x 2
h
13x 13h 5
h
h) h→0 2x 2 4xh 2h 2 4xh lim (4x 2h 13) 4(3) 13 13x h→0 lim h→0 dy
dx 4x slope 1 and passes through (3, y(3)) y x 12. Let f (x) 1, so the tangent line has
(3, 16). x 3.
lim h→0 lim h→0 lim f (1
(1
1 h→0 16 lim (3 h2 ) 3h (b) Yes, the rate of change is zero when the tangent to the
graph is horizontal. This occurs near the beginning of
the year and halfway through the year, around
January 1 and July 1.
(c) Positive: January 1 through July 1
Negative: July 1 through December 31
16. The slope of the given graph is zero at x
2 and at
x 1, so the derivative graph includes ( 2, 0) and (1, 0).
The slopes at x
3 and at x 2 are about 5 and the
slope at x
0.5 is about 2.5, so the derivative graph
includes ( 3, 5), (2, 5), and ( 0.5, 2.5). Connecting the
points smoothly, we obtain the graph shown. h) f (1)
h
h)3 13
h
3h 3h 2 h 3
h h→0 about onefourth of the way through the year, sometime
around April 1. The rate at this time is approximately 13 f (1) 3) 5 13 3, 1(x 13x 1
4 hours
or hour per day.
6
24 days At x y 2x 2 when the slope of the graph is largest. This occurs 5) 13h h→0 2h 2
h lim y
5 1 3
5 (a) The tangent line has slope 3 and passes through (1, 1).
Its equation is y 3(x 1) 1, or y 3x 2.
(b) The normal line has slope
(1, 1). Its equation is y
y 1
x
3 1
and passes through
3
1
(x 1) 1, or
3 4
.
3 13. Since the graph of y x ln x x is decreasing for
0 x 1 and increasing for x 1, its derivative is
negative for 0 x 1 and positive for x 1. The only
one of the given functions with this property is y ln x.
Note also that y ln x is undefined for x 0, which
further agrees with the given graph. (ii)
14. Each of the functions y
property that y(0) sin x, y x has the 0, so none of these functions can be its own derivative. The function y
derivative because y(1)
2 x 2 is not its own 1 but
2 (1 1 h→0 h)
h lim (2 y (1) x, y 0 but the graph has nonzero slope (or undefined slope) at x h) h2 2. lim h→0 71 lim h→0 2h 17. (a) Using Figure 3.10a, the number of rabbits is largest
after 40 days and smallest from about 130 to 200 days.
Using Figure 3.10b, the derivative is 0 at these times.
(b) Using Figure 3.10b, the derivative is largest after
20 days and smallest after about 63 days. Using
Figure 3.10a, there were 1700 and about 1300 rabbits,
respectively, at these times.
18. (a) The slope from x 4 to x The slope from x 0 to x The slope from x 1 to x The slope from x 4 to x 1
20
.
2
0 ( 4)
22
1 is
4.
10
2 ( 2)
4 is
0.
41
2 ( 2)
6 is
2.
64 0 is Note that the derivative is undefined at x 0, x 1,
and x 4. (The function is differentiable at x
4
and at x 6 because these are endpoints of the domain
and the onesided derivatives exist.) The graph of the
derivative is shown. h This leaves only e x, which can plausibly be its own
derivative because both the function value and the slope
increase from very small positive values to very large
values as we move from left to right along the graph. (iv) x y
5
4
3
2
–5–4–3–2 –1
–2
–3
–4
–5 1234567 x 72 Section 3.1 18. continued
(b) x
19. (b) 0, 1, 4
Midpoint of
Interval (x) 1.5
2.5
3.5
4.5
5.5
6.5
7.5
8.5
9.5 y
x 0.00 0
3.3
0
3.3
10.0
1
13.3
16.6
2
29.9
23.3
3
53.2
30.0
4
83.2
36.6
5
119.8
43.2
6
163.0
49.9
7
212.9
56.6
8
269.5
63.2
9 0.56 Slope
3.3
1
13.3
2
29.9
3
53.2
4
83.2
5
119.8
6
163.0
7
212.9
8
269.5
9
332.7
10 0.5 Midpoint of
Interval (x)
0.56
2
0.92
2
0.92 1.19
2 1.19 1.30
2 1.30 1.39
2 1.39 1.57
2 1.57 1.74
2 1.74 1.98
2 1.98 2.18
2 2.18 2.41
2 A graph of the derivative data is shown.
2.41 2.64
2 2.64 3.24
2 Slope y
x 0.28 1512
0.56 1577
0.00 116.07 0.74 1448
0.92 1512
0.56 177.78 1.055 1384
1.19 1448
0.92 237.04 1.245 1319
1.30 1384
1.19 590.91 1.345 1255
1.39 1319
1.30 711.11 1.48 1191
1.57 1255
1.39 355.56 1.655 1126
1.74 1191
1.57 382.35 1.86 1062
1.98 1126
1.74 266.67 2.08 998
2.18 1062
1.98 320.00 2.295 933
2.41 998
2.18 282.61 2.525 869
2.64 933
2.41 278.26 2.94 805
3.24 869
2.64 106.67 A graph of the derivative data is shown.
[0, 10] by[ 10, 80] (a) The derivative represents the speed of the skier.
(b) Since the distances are given in feet and the times are
given in seconds, the units are feet per second.
(c) The graph appears to be approximately linear and [0, 3.24] by [ 800, 100] (c) Since the elevation y is given in feet and the distance x
down river is given in miles, the units of the gradiant
are feet per mile. passes through (0, 0) and (9.5, 63.2), so the slope is
63.2
9.5 0
0 6.65. The equation of the derivative is approximately D (d) Since the elevation y is given in feet and the distance x 6.65t. downriver is given in miles, the units of the distance 20. (a) dy
dx are feet per mile.
(e) Look for the steepest part of the curve. This is where
the elevation is dropping most rapidly, and therefore the
most likely location for significant “rapids.”
[ 0.5, 4] by [700, 1700] (f) Look for the lowest point on the graph. This is where
the elevation is dropping most rapidly, and therefore the
most likely location for significant “rapids.”
21. [ , ] by [ 1.5, 1.5] The cosine function could be the derivative of the sine
function. The values of the cosine are positive where the
sine is increasing, zero where the sine has horizontal
tangents, and negative where sine is decreasing. Section 3.1
22. We show that the righthand derivative at 1 does not exist.
h) f (1)
3(1
lim
h
h→0
2 3h
2
lim
lim
3
h
h
h→0
h→0 lim f (1 23. lim f (0 h→0 lim h)
h f (0) lim 1 h→0 lim f (x) 2. x→1 (f) lim f (1 h→0 h h→0 (e) Yes, the onesided limits exist and are the same, so (1)3 h)
h h→0 0 h lim h→0 h h)
h 2h lim lim h)2
h (1 h→0 h2 h) f (1)
2(1
lim
h
h→0
h→0
1 2h
1
lim
lim
2
h
h
h→0
h→0 Thus, the righthand derivative at 0 does not exist. lim (2 12 2
h)
h (g) lim 24. Two parabolas are parallel if they have the same derivative
at every value of x. This means that their tangent lines are
parallel at each value of x.
Two such parabolas are given by y x 2 and y x 2 4.
They are graphed below. f (1) h) h→0 h h 73 h→0 h f (1 12 The righthand derivative does not exist.
(h) It does not exist because the righthand derivative does
not exist.
27. The yintercept of the derivative is b a. 28. Since the function must be continuous at x 1, we have
2.
lim (3x k) f (1) 1, so 3 k 1, or k
x→1 x 3,
3x This gives f (x) slope 1, the graph of y y 2( 1) 1 1): y 2x 1, so for x 1. Then 1, the graph of f (x) must lie on a line of slope 1 that passes through ( 1, 1): y
Thus f (x) 1(x 1) x 1 or y 2,
2x 1, f (1 h)
h f (1) x
x x 2. (1 3 h)
h
3h 2
h (1) lim (3 3h h 2) lim h→0 lim 3h h→0 h) f (1)
h
h→0
(1 3h) 1
lim
h
h→0 lim f (1 1. 3 h→0 lim h→0 f (x) must lie on a line of 2 that passes through (0, y( 1) 1
1. Now we confirm that f (x) is differentiable at x [ 4, 4] by [ 5, 20] The parabolas are “everywhere equidistant,” as long as the
distance between them is always measured along a vertical
line.
25. For x x
x 2, lim [3(1 h) lim 3 2] 3
(1)3] h h→0 h→0 h3 3 Since the righthand derivative equals the lefthand
derivative at x 1, the derivative exists (and is equal to 3)
when k
2. 1
1 29. (a) 1 y
5 364
365 363
365 0.992 Alternate method: P 365 3
3 365 0.992 (b) Using the answer to part (a), the probability is about
1 0.992 0.008.
x 5 (c) Let P represent the answer to part (b), P 0.008. Then the probability that three people all have different
birthdays is 1
26. (a) f (x) lim h→0 lim f (x
(x h→0 lim (2x h→0 (b) f (x) lim h→0 lim f (x
2(x h→0 (c) lim f (x)
x→1 (d) lim f (x)
x→1 h)
h
h)2
h x2 probability that all have different birthdays is
lim h 2x h)
h
h)
h h2 2xh h→0 h) f (x) lim 2x x→1 lim 2 x→1 f (x) 2x 2 (1
1 lim 2 h→0 2(1) P. Adding a fourth person, the 2 2 P)
(1 362
, so the probability of a shared birthday is
365
362
P)
0.016.
365 (d) No. Clearly February 29 is a much less likely birth
date. Furthermore, census data do not support the
assumption that the other 365 birth dates are equally
likely. However, this simplifying assumption may still
give us some insight into this problem even if the
calculated probabilities aren’t completely accurate. 74 Section 3.2 s Section 3.2 Differentiability (pp. 105–112) Quick Review 3.2
1. Yes Exploration 1 Zooming in to “See”
Differentiability 2. No (The f (h) term in the numerator is incorrect.) 1. Zooming in on the graph of f at the point (0, 1) always
produces a graph exactly like the one shown below,
provided that a square window is used. The corner shows
no sign of straightening out. 3. Yes 4. Yes 5. No (The denominator for this expression should be 2h).
6. All reals
7. [0, ) 8. [3, ) 9. The equation is equivalent to y
slope is 3.2.
10. f (3 0.001) f (3
0.002 (3.2 5), so the 5(3 0.001) 5(3
0.002
5(0.002)
5
0.002 0.001) [ 0.25, 0.25] by [0.836, 1.164] 2. Zooming in on the graph of g at the point (0, 1) begins to
reveal a smooth turning point. This graph shows the result
of three zooms, each by a factor of 4 horizontally and
vertically, starting with the window
[ 4, 4] by [ 1.624, 3.624]. 3.2x 0.001) Section 3.2 Exercises
1. Lefthand derivative:
lim f (0 h→0 h)
h f (0) lim h2 h→0 0 lim h 0 h→0 h Righthand derivative:
lim f (0 h→0 [ 0.0625, 0.0625] by [0.959, 1.041] 3. On our grapher, the graph became horizontal after 8 zooms.
Results can vary on different machines.
4. As we zoom in on the graphs of f and g together, the
differentiable function gradually straightens out to resemble
its tangent line, while the nondifferentiable function
stubbornly retains its same shape. Since 0
point P. h)
h f (0) lim h h→0 0 lim 1 1 h→0 h 1, the function is not differentiable at the 2. Lefthand derivative:
lim f (1 h→0 h)
h f (1) lim 2 h→0 2 lim 0 0 h→0 h Righthand derivative:
lim f (1 h→0 Since 0
point P. h)
h f (1) lim 2(1 h)
h h→0 2 lim 2 h→0 2 2, the function is not differentiable at the [ 0.03125, 0.03125] by [0.9795, 1.0205] 3. Lefthand derivative: Exploration 2 Looking at the Symmetric
Difference Quotient Analytically
1. f (10 h)
h f (10) f (10) 2 10 2 (10.01)
10
0.01 lim f (1 h→0 2 h)
h f (1) 20.01 (1 2. h) f (10 h) 2h lim (10.01)2 (9.99)2
0.02 f (10 h)
h f (10) 3 102 f (10) (10.01)3 103
0.01 h→0 h)
f (10 lim 20 300.3001. lim f (1 h→0 2h h)
h f (1) h 1 lim [2(1 1) 1)
1 lim 1
2 1
2 The symmetric difference quotient is 0.0001 away from Since f (10). point P. 1 2h
h lim 2 h→0 300.0001. h) 1]
h h→0
h→0 (10.01)3 (9.99)3
0.02 h h Righthand derivative: 300 h) 1
1 h→0 The difference quotient is 0.3001 away from f (10).
f (10 h( 1)( 1 h( 1 h
h) 1 (1 The symmetric difference quotient exactly equals f (10).
3. h h→0 20 1 h
h lim The difference quotient is 0.01 away from f (10).
f (10 1 lim h→0 2 2, the function is not differentiable at the 1) 75 Section 3.2
4. Lefthand derivative:
lim f (1 h)
h h→0 15. Note that y f (1) (1 lim h)
h h→0 Righthand derivative:
f (1 lim h)
h h→0 f (1) lim lim 1 1 h→0 1
1 lim lim 1 h lim h→0 lim h→0 16. lim lim h→0 lim (c) None
7. (a) All points in [ 3, 3] except x 0 0 8. (a) All points in [ 2, 3] except x 1, 0, 2 1
2
0 0 10. (a) All points in [ 3, 3] except x
2, x 2, 2 2 (c) None
11. Since lim tan 1 x→0 x tan 1 0 0 y(0), the problem is a discontinuity. lim y(0 h)
h
h)
h y(0) 4/5 h
h
h4/5
lim
h
h→0 1
h1/5
1
lim 1/5
h→0 h lim lim h→0 y(0) 13. Note that y
2,
2x 2,
lim y(0 lim y(0 h→0 h)
h
h)
h x
x
x
y(0) x2
0
0.
lim h→0 y(0) h→0 y(0 h→0 lim h→0 h) y(0)
h
1
h2/3 lim 2 x 1 3 lim h→0 3 h
h 0 h
h 0 3 lim h→0 lim h→0 h h 3 lim h→0 h h 1
h2/3 h(2 h(2) lim [ 3 3(2 k) 3 3 lim 6 x lim 2 h→0 g (0 2 h
(2h h→0 lim 0
2 lim 0 h→0 2)
h (3 3 h)
h h→0 lim 2 2 h→0 The problem is a vertical tangent. 3 5 1 2/3
k→0 k h)
h lim (h 2 5] k k→0 19. Note that the sine function is odd, so
sin x 1,
P(x) sin ( x ) 1
sin x 1,
The graph of P(x) has a corner at x
differentiable for all reals except x 3 lim h→0 h h g (0 g (0) h)
h lim (h h→0 2) h→0 h→0 lim lim 1 h→0 2. It is x0
x 0.
0. The function is
0. 21. The function is piecewisedefined in terms of polynomials,
so it is differentiable everywhere except possibly at x 0
and at x 3. Check x 0: The problem is a corner.
14. lim 5 20. Since the cosine function is even,
so Q(x) 3 cos ( x ) 3 cos x. The function is
differentiable for all reals. The problem is a cusp. h→0 h) y(0)
h
1
h2/3
h) y(0)
h The function has a vertical tangent at x
differentiable for all reals except x 2. (c) None h→0 (h lim h→0 k)
k
k→0
3
3k
lim
k→0 k lim 9. (a) All points in [ 1, 2] except x (b) x y(0) lim 5 h→0 18. The function is differentiable except possibly where
3x 6 0, that is, at x 2. We check for differentiability
at x 2, using k instead of the usual h, in order to avoid
confusion with the function h(x). (b) None 0, x 1) ( 1)
h
1) ( 1)
h h→0 0
0 17. Find the zeros of the denominator.
x2 4x 5 0
(x 1)(x 5) 0
x
1 or x 5
The function is a rational function, so it is differentiable for
all x in its domain: all reals except x
1, 5. (b) None (c) x (5h lim x
x The problem is a cusp. 6. (a) All points in [ 2, 3] y(0 h→0 lim (c) None h→0 y(0 h→0 (b) None 12. lim y(0 h→0 1, the function is not differentiable at the (b) x y(0) 5x 1,
x 1, 1 The problem is a corner. 5. (a) All points in [ 3, 2] (b) x y(0 h→0 h)
h
h)
h 2x 1 lim (c) x y(0 h→0 h
(1 h)
h(1 h)
h
h(1 h)
1
1
1h h→0
h→0 Since 1
point P. 1 3x 1)2
h 1 1)
h 1 lim h2 h→0 2h
h 2
g (0) lim h→0 (2h The function is differentiable at x lim 2 h→0 0. 2 76 Section 3.2 21. continued
Check x 27.
3: Since g(3)
lim g(x) x→3 3)2 (4 lim (2x x→3 1 and
1) 2(3) 1 7, the function is not continuous (and hence not differentiable) at x 3. The function is differentiable for all reals except x
2 22. Note that C(x) x,
x 2, xx x
x lim C(0 h→0 3. y NDER (x) actually has two asymptotes for each asymptote of y C(0) lim h→0 hh 0
h tan x. The asymptotes of y occur at x 0. 0:
h)
h tan x Note: Due to the way NDER is defined, the graph of 0
, so it is
0 differentiable for all x except possibly at x
Check x [ 2 , 2 ] by [ 4, 4] dy
dx k 2 NDER (x) 0.001, where k is an integer. A good window for viewing this behavior is [1.566, 1.576] by
lim h h→0 0 [ 1000, 1000]. The function is differentiable for all reals.
28.
23. (a) x 0 is not in their domains, or, they are both
discontinuous at x 0.
1
1
,0
x
x
1
1
For 2 : NDER 2 , 0
x
x (b) For : NDER 1,000,000
[ 2 , 2 ] by [ 20, 20] 0 (c) It returns an incorrect response because even though
these functions are not defined at x
defined at x 0, they are 0.001. The responses differ from each 1
other because 2 is even (which automatically makes
x
1
1
0) and is odd.
NDER 2 , 0
x
x 24. The graph of NDER (x) does not look like the graph of any
basic function.
29. (a) lim f (x) x→1 f (1) x) a(1)2 2 lim (3 x→1 a The relationship is a b(1) b
b 2. (b) Since the function needs to be continuous, we may
assume that a b 2 and f (1) 2.
lim f (1 h→0 [ 5, 5] by [ 10, 10]
dy
x3
dx f (1) f (1 h)
h f (1) lim h→0 lim h→0 lim h→0 dy
dx sin x 26. 3 (1 h) a(1 1 h)2 b(1 h) 2
h
a 2ah ah 2 b bh
h
2ah ah 2 bh (a b
h lim (2a h→0 2a 2 h lim ( 1) h→0 [ 2 , 2 ] by [ 1.5, 1.5] lim h→0
h→0 lim 25. h)
h ah 2
2) b) b Therefore, 2a b
1. Substituting 2 a for b gives
2a (2 a)
1, so a
3.
Then b 2 a 2 ( 3) 5. The values are
a
3 and b 5.
[ 6, 6] by [ 4, 4] dy
dx abs (x) or x 30. The function f (x) does not have the intermediate value
property. Choose some a in ( 1, 0) and b in (0, 1). Then
f (a) 0 and f (b) 1, but f does not take on any value
between 0 and 1. Therefore, by the Intermediate Value
Theorem for Derivatives, f cannot be the derivative of any
function on [ 1, 1]. 77 Section 3.3 31. (a) Note that x
1
x so lim x sin
x→0 1
x sin 0 by the Sandwich Theorem. Therefore, f is continuous at x
f (0 (b) h)
h 1
h sin
h f (0) x 10. (a) f (x) 0. 0 h)
h 1
h h 2 sin g(0) 0 1. 1
h h sin h As noted in part (a), the limit of this as x approaches
zero is 0, so g (0) lim h→0 x(x h lim 1 h→0 f (x) lim x h
h lim h→0 hx(x h)
2 x x2 h) x h h→0 Section 3.3 Exercises (d) No, because the limit in part (c) does not exist.
g(0 lim 1 x h h→0 f (x lim lim interval containing 0). (e) hx
h f (x) h→0 h→0 0 (that is, for h in any open h)
h h)
h
x
(x h)
hx(x h) (b) f (x) 1 and 1 an infinite number of times arbitrarily close to h lim x h→0 1
oscillates
h (c) The limit does not exist because sin f (x h→0 lim 1
h sin h between 9. These are all constant functions, so the graph of each
function is a horizontal line and the derivative of each
function is 0. x, for all x, 2. 0.
3. s Section 3.3 Rules for Differentiation
(pp. 112–121) dy
dx
d 2y
dx 2 d
( x 2)
dx
d
( 2x)
dx dy
dx
d 2y
dx 2 d 13
x
dx 3
d2
(x )
dx dy
dx
d 2y
dx 2 d
(2x)
dx
d
(2)
dx dy
dx
d 2y
dx 2 d2
(x )
dx
d
(2x)
dx dy
dx d 13
x
dx 3 d
(3)
dx 2x 0 2x 2
d
(x)
dx
d
(1)
dx x2
2x d
(1)
dx 1
0 2 2x 0 2 0 Quick Review 3.3
1. (x 2
x
2. 2)(x 1 1)
x 2 2x 1
1 2
x
2x 3
2x 2 3x 4
1 5. (x 6. 1 x x 2x 1 3x 4
2x 2 4
2 2)(x
x 3 x
x 5
x2 1)
2x 1 x2 1 x2
x 3x 2 1 3. 3x 2
4. x2 1 x x
x2 x 2x
2 1
x
1 x 1 x 1 4. 21 1 x 5.
2 5x x2 x 2 2x 3
2x 2
2 2x 32
x
2 4
2x 2 x 2 x 1 x 1 2x 2x
2 dy
dx 2 2 21 6. 2 dy
dx 3 x (x 3 1 x 2 ) x 2 x 1 7. dy
dx At x 1.173, 500x 6 1305.
At x 2.394, 500x 6 94,212
After rounding, we have:
At x 1, 500x 6 1305.
At x 2, 500x 6 94,212.
8. (a) f (10) 8. 7 (b) f (0) 7 (c) f (x h) f (x )
(d) lim
x
x→a 9.
7
f (a)
a 7
a lim 0 x→a 0 2x 3x 2 1 2 6x 2x d
(4x 3)
dx 4x 1 0 2x 1 42x 4 d3
(x )
dx 3x 2 2x 6x
d
(2x 2)
dx 0 4x 3 d
(21x 2)
dx d
(15)
dx 21x 2 d
(4x)
dx 4x
12x 2 dy
dx
d 2y
dx 2 d
d
(5x 3)
(3x 5) 15x 2 15x 4
dx
dx
d
d
(15x 2)
(15x 4) 30x 60x 3
dx
dx dy
dx d
(4x 2)
dx 2 dy
dx 2 1 d
(3x 2)
dx d
(7x 3)
dx 21x 2 2x d2
(x )
dx 2 8x
7
lim
x→a x d
(1)
dx d
(x)
dx d4
(x )
dx d 2y
dx 2 0 d
(x)
dx d
(2x)
dx 4x 3
[0, 5] by [ 6, 6] d
(x)
dx d
( 1)
dx 0 1 1 d
(1)
dx d 2y
dx 2 7. d 12
x
dx 2 d2
(x )
dx 0 2 d
d
(x)
(1) 2x
dx
dx
d
(1) 2 0 2
dx 3 8 d
(8x)
dx 0 d
( 8x 3)
dx 8x
d
(8)
dx d
(1)
dx
3 8
24x 4 0 24x 4 78 10. Section 3.3
dy
dx d1 4
x
dx 4
d
(3)
dx x 5 x 5 d 2y
dx 2 x 4 x 6 dy
dx d1 2
x
dx 2 d
(x 1)
dx (x 18.
x 4 3 x 3 d
( x 5)
dx 5x
11. (a) d1 3
x
dx 3 5 4x x 2 x 0 4 d
(x 3 )
dx 1)(2x) (x 2x 2 2x x2 3x 2 2x 1) (x 19. d (x
dx (x dy
dx d
[(x
dx 1)(x d3
(x
dx 3x (x 2 2 x2 d
1) (x
dx 1)
(2x 3 3x 2 dy
dx 2 1)] x 1) h lim f (x 3x
3x x( 1)2 x 2
2 2)(2x h)
h h→0 lim (1) 3) 5x 6) f (x) lim h→0 1 h→0 d
dx (x 2 3) d
( u)
dx u(x) is a function of x. x 2 u(x lim u(x 3
x2 d x2 3
x
dx
3
1
x2 d
(x
dx 3x 1 ) 1 2 3x f (x ) 0 dy
14.
dx d
dx 0 5x x2 5x
x2 2 2x d (x
dx
d
(1
dx (3x 5
2 x 1 ) 0 1) 3x d1x
dx 1 x 2
x 2 2x 1
(1 x 2)2 (1 d
x2
dx 1 x 3 (1 (2x
2)2 23. (a) At x 5)(3) 4 2)2 0, d
f (x)
dx 1 d
(uv)
dx u(0)v (0) ( 1)( 3) f (x)
[ f (x)]2 v(0)u (0) 13 x ) 0, (c) At x 0, (5)(2) x)(2x) (d) At x
7(2) x 3)(2x) x 2( 3x 2)
(1 x 3)2 d
(c)
dx v(0)u (0) u(0)v (0)
du
[v(0)]2
dx v
( 1)( 3) (5)(2)
7
( 1)2 (b) At x
2 d x3 1
dx
x3
3
x4 x 2)( 1) (1
(1 x 2)2 du
dx 19
(3x d
(1 5x 1
dx
5
2
x2
x3 3 1)(x 2 x
x3
3 2)(2)
(3x u(x) [ f (x)]2 (5)(2)
d 2x
dx 3x u(x) d
d
(c f (x)) c
f (x) f (x)
dx
dx
d
d
c
f (x) 0 c
f (x)
dx
dx d
1
22.
dx f (x) This is equivalent to the answer in part (a).
dy
13.
dx h)
h h→0 3) h) [ u(x)]
h h)
h h→0 lim (x 2 u(x lim h→0 3) (x) 21. dy
dx d x2
dx x 2 2)
2) f (x) h→0 h 1 x2 17. 1
1)2 x. d
(x)
dx 3
dx
x
dx x(2x) dy
dx x 12 6x 2
(x 2 3x 2)2 1 x2 16. 1) 2 d
dx dy
dx x( 5x 6) (2x 3 3x 2
(x 2 3x 2)2 1)(1) lim 2x x (x 2 15. x 2 2
2 2)(2x 3) (x 2 3x
(x 2 3x 2)2 3x (b) Note that u dy
dx x 1) 1)2 1)(x
1)(x 20. (a) Let f (x) (b) 2 1 dy
(b)
dx 12. (a) ( 1 1 1) d
(x 2)
dx 3 d
[(x 1)(x 2 1)]
dx
d
(x 1) (x 2 1) (x 2
dx (x (x x (x
x 2 1 2 x( x 2x 2 x d
dx 2 d
(x 4)
dx 3x dy
dx 1 1) x 4 2x
(1 x 3)2 0, u(0)v (0) v(0)u (0)
[u(0)]2 dv
dx u
( 1)( 3)
(5)2
d
(7v
dx 2u) 2( 3) 20 7
25 7v (0) 2u (0) (x h)
h x Section 3.3 24. (a) At x 2, d
(uv)
dx (3)(2) (1)( 4) u(2)v (2) v(2)u (2) 2 v(2)u (2) u(2)v (2)
du
[v(2)]2
dx v
(1)( 4) (3)(2)
10
(1)2 (b) At x 2, 29. y (x) 2, [ 3, 3] by [ 10, 30] d
(d) Use the result from part (a) for
(uv).
dx
d
At x 2,
(3u 2v 2uv)
dx
d
3u (2) 2v (2) 2 (uv)
dx 3( 4) 2(2) 30. y (x) 3x 2
y ( 2) 12
The tangent line has slope 12 and passes through ( 2,
so its equation is y 2(2)
xintercept is 3
x
2 3x
3
3(2)2 3 9 y 1
and passes through (2, 3).
9
1
(x
9
1
x
9 2) 12x 3 29
9 31. y (x) (x 2 1)(4) 4x(2x)
(x 2 1)2 At the origin: y (0)
The tangent is y
At (1, 2): y (1) 4x 2 4
(x 2 1)2 4
4x. 0 The tangent is y 2. Graphical support: Graphical support: [ 4.7, 4.7] by [ 3.1, 3.1]
[ 4.7, 4.7] by [ 2.1, 4.1] 28. y (x) 3x 2 1
The slope is 4 when 3x 2 1 4, at x
1. The tangent
at x
1 has slope 4 and passes through ( 1, 2), so its
equation is y 4(x 1) 2, or y 4x 2. The tangent at
x 1 has slope 4 and passes through (1, 2), so its equation
is y 4(x 1) 2, or y 4x 2. The smallest slope
occurs when 3x 2 1 is minimized, so the smallest slope is
1 and occurs at x 0. 32. y (x)
y (2) x 2)(0) 8(2x)
(4 x 2)2 (4 16x
x 2)2 1
2 The tangent has slope
equation is y 1
(x
2 Graphical support: Graphical support: [ 3, 5] by [ 2, 4]
[ 4.7, 4.7] by [ 3.1, 3.1] (4 8), 16. The [ 3, 3] by [ 20, 20] 2 The tangent line has slope 9, so the perpendicular line has y 8, or y 6, so the 3
slope is . (iii)
2 slope 2) Graphical support: 25. y (x) 2x 5
y (3) 2(3) 5 11
The slope is 11. (iii)
26. The given equation is equivalent to y 12(x 4
and the yintercept is 16.
3 12 27. y (x)
y (2) 1 Graphical support: u(2)v (2) v(2)u (2)
dv
[u(2)]2
dx u
(3)(2) (1)( 4)
10
(3)2
9 (c) At x 6x 2 6x 12
6(x 2 x 2)
6(x 1)(x 2)
The tangent is parallel to the xaxis when y
0, at x
and at x 2. Since y( 1) 27 and y(2) 0, the two
points where this occurs are ( 1, 27) and (2, 0). 79 1
and passes through (2, 1). Its
2
1
2) 1, or y
x 2.
2 80 33. Section 3.4
an 2
d nRT
V2
dV V nb
d
(V nb) (nRT) dP
dV (c)
(nRT) dV (V
0
(V nR T
nb)2 d
(V
dV nb) nb)2 2an 2V d
(an 2V 2)
dV 3 d
(uv)
dx (V 34. 35. ds
dt
d 2s
dt 2 d
(4.9t 2)
dt
d
(9.8t)
dt dR
dM 2an
V3 nRT
nb)2 CM Exploration 1 M2 37. If the radius of a sphere is changed by a very small amount
r, the change in the volume can be thought of as a very
thin layer with an area given by the surface area, 4 r 2, and
a thickness given by r. Therefore, the change in the
volume can be thought of as (4 r 2)( r), which means that
the change in the volume divided by the change in the
radius is just 4 r 2.
38. Let t(x) be the number of trees and y(x) be the yield per tree
x years from now. Then t(0) d
(ty)
dx 156, y(0) 12, t (0) 13, 1.5. The rate of increase of production is t(0)y (0) y(0)t (0) (156)(1.5) (12)(13) 39. Let m(x) be the number of members and c(x) be the
pavillion cost x years from now. Then m(0)
250, m (0) 6, and c (0) of each member’s share is
(65)(10) (250)(6)
(65)2 dc
dx m Growth Rings on a Tree 1. Figure 3.22 is a better model, as it shows rings of equal
area as opposed to rings of equal width. It is not likely that
a tree could sustain increased growth year after year,
although climate conditions do produce some years of
greater growth than others.
2. Rings of equal area suggest that the tree adds
approximately the same amount of wood to its girth each
year. With access to approximately the same raw materials
from which to make the wood each year, this is how most
trees actually grow.
3. Since change in area is constant, so also is
change in area
. If we denote this latter constant by k, we
2
k
have
r, which means that r varies
change in radius inversely as the change in the radius. In other words, the
change in radius must get smaller when r gets bigger, and
viceversa. Exploration 2 390 bushels of annual production per year. c(0) u du
u
v dv
v
(u du)(v) (u)(v dv)
(v dv)(v)
uv vdu uv udv
v 2 vdv
vdu udv
v2 s Section 3.4 Velocity and Other Rates
of Change (pp. 122–133) 36. If the radius of a circle is changed by a very small amount
r, the change in the area can be thought of as a very thin
strip with length given by the circumference, 2 r, and
width r. Therefore, the change in the area can be thought
of as (2 r)( r), which means that the change in the area
divided by the change in the radius is just 2 r. and y (0) u
v 9.8
M
3
13
M
3 dv
dx u . This is equivalent to the product (d) Because dx is ‘infinitely small,” and this could be
thought of as dividing by zero. 9.8t C
d
M2
2
dM
dC2
M
dM 2 du
dx rule given in the text. (e) d
2 v Modeling Horizontal Motion 1. The particle reverses direction at about t
t 2.06. 0.61 and 65, 10. The rate of change
m(0)c (0) c(0)m (0)
[m(0)]2 0.201 dollars per year. Each member’s share of the cost is decreasing by approximately
20 cents per year.
40. (a) It is insignificant in the limiting case and can be treated
as zero (and removed from the expression).
(b) It was “rejected” because it is incomparably smaller
than the other terms: v du and u dv. 2. When the trace cursor is moving to the right the particle is
moving to the right, and when the cursor is moving to the
left the particle is moving to the left. Again we find the
particle reverses direction at about t 0.61 and t 2.06. Section 3.4
3. When the trace cursor is moving upward the particle is
moving to the right, and when the cursor is moving
downward the particle is moving to the left. Again we find
the same values of t for when the particle reverses
direction. 81 Quick Review 3.4
1. The coefficient of x 2 is negative, so the parabola opens
downward.
Graphical support: [ 1, 9] by [ 300, 200] 4. We can represent the velocity by graphing the parametric
equations
x4(t) x1 (t) 12t 2 32t 15, y4(t) 2 (part 1),
x5(t) x1 (t) 12t 2 32t 15, y5(t) t (part 2),
x6 (t) t, y6(t) x1 (t) 12t 2 32t 15 (part 3) 2. The yintercept is f (0)
256.
See the solution to Exercise 1 for graphical support.
3. The xintercepts occur when f (x) 0.
16x 2 160x 256 0
16(x 2 10x 16) 0
16(x 2)(x 8) 0
x 2 or x 8
The xintercepts are 2 and 8. See the solution to Exercise 1
for graphical support.
4. Since f (x)
16(x 2 10x 16)
2
16(x
10x 25 9)
16(x 5)2 144,
the range is ( , 144].
See the solution to Exercise 1 for graphical support. [ 8, 20] by [ 3, 5] (x4, y4) 5. Since f (x)
16(x 2 10x 16)
2
16(x
10x 25 9)
16(x 5)2 144,
the vertex is at (5, 144). See the solution to Exercise 1 for
graphical support.
6. [ 8, 20] by [ 3, 5] (x5, y5) f (x) 80
16x 2 160x 256 80
16x 2 160x 336 0
16(x 2 10x 21) 0
16(x 3)(x 7) 0
x 3 or x 7
f (x) 80 at x 3 and at x 7.
See the solution to Exercise 1 for graphical support. [ 2, 5] by [ 10, 20] (x6, y6) dy 7. 100 dx For (x4, y4) and (x5, y5), the particle is moving to the right
when the xcoordinate of the graph (velocity) is positive,
moving to the left when the xcoordinate of the graph
(velocity) is negative, and is stopped when the xcoordinate
of the graph (velocity) is 0. For (x6, y6), the particle is
moving to the right when the ycoordinate of the graph
(velocity) is positive, moving to the left when the
ycoordinate of the graph (velocity) is negative, and is
stopped when the ycoordinate of the graph (velocity) is 0. 32x 160 100 60 32x x
dy 100 at x dx 15
8
15
8 Graphical support: the graph of NDER f (x) is shown. Exploration 3 Seeing Motion on a Graphing
Calculator
1. Let tMin 0 and tMax 10. 2. Since the rock achieves a maximum height of 400 feet, set
yMax to be slightly greater than 400, for example
yMax 420.
4. The grapher proceeds with constant increments of t (time),
so pixels appear on the screen at regular time intervals.
When the rock is moving more slowly, the pixels appear
closer together. When the rock is moving faster, the pixels
appear farther apart. We observe faster motion when the
pixels are farther apart. [ 1, 9] by [ 200, 200] dy 8. 0 dx 32x dy 160
32x
x 0 when x 0
160
5
5. dx See the solution to Exercise 7 for graphical support. 82 Section 3.4 9. Note that f (x)
lim f (3 h)
h h→0 32x
f (3) (e) 160. f (3) 32(3) 160 64 For graphical support, use the graph shown in the solution
to Exercise 7 and observe that NDER (f (x), 3)
10. f (x)
f (x)
At x 64. 30, so it was aloft 4. On Mars: 32x 160
32
7 (and, in fact, at any other of x), Velocity 2 dy
dx 2 s(t) 0
24t 0.8t 2 0
0.8t(30 t) 0
t 0 or t 30
The rock was aloft from t 0 to t
for 30 seconds. ds
dt Solving 3.72t 32. d
(1.86t 2)
dt 3.72t 16.6, the downward velocity reaches 16.6 m/sec after about 4.462 sec. Graphical support: the graph of NDER(NDER f (x)) is
shown. On Jupiter:
Velocity ds
dt Solving 22.88t d
(11.44t 2)
dt 22.88t 16.6, the downward velocity reaches 16.6 m/sec after about 0.726 sec.
[ 1, 9] by [ 40, 10] Section 3.4 Exercises
s 3, the instantaneous rate of change is 1. Since V 2. (a) Displacement = s(5) s(0)
10 m
5 sec (b) Average velocity 2 10 m 3 3s 2. 5 m/sec 2 m/sec (c) Velocity s (t) 2t 3
At t 4, velocity s (4)
(d) Acceleration 12 dV
ds 2(4) 2 m/sec 2 s (t) (e) The particle changes direction when
s (t) 2t 3 0, so t 3
sec.
2 (f) Since the acceleration is always positive, the position s
is at a minimum when the particle changes direction, at
t 3
3
sec. Its position at this time is s
2
2 3. (a) Velocity: v(t) ds
dt Acceleration: a(t) d
(24t 0.8t 2) 24
dt
dv
d
(24 1.6t)
dt
dt 1
m.
4 1.6t m/sec
1.6 m/sec2 (b) The rock reaches its highest point when
v(t) 24 1.6t 0, at t 15. It took 15 seconds.
(c) The maximum height was s(15)
(d) s(t)
24t 0.8t 2
0
t 180 meters. 1
(180)
2 90
0.8t 2
24 24t 90
2 ( 24)
4(0.8)(90)
2(0.8) 4.393, 25.607
It took about 4.393 seconds to reach half its maximum
height. 5. The rock reaches its maximum height when the velocity
s (t) 24 9.8t 0, at t 2.449. Its maximum height is
about s(2.449) 29.388 meters.
6. Moon:
s(t)
0
832t 2.6t 2 0
2.6t(320 t) 0
t 0 or t 320
It takes 320 seconds to return.
Earth:
s(t) 0
832t 16t 2 0
16t(52 t) 0
t 0 or t 52
It takes 52 seconds to return.
7. The following is one way to simulate the problem situation.
For the moon:
x1(t) 3(t 160) 3.1(t 160)
y1(t) 832t 2.6t 2
tvalues: 0 to 320
window: [0, 6] by [ 10,000, 70,000]
For the earth:
x1(t) 3(t 26) 3.1(t 26)
y1(t) 832t 16t 2
tvalues: 0 to 52
window: [0, 6] by [ 1000, 11,000]
8. The growth rate is given by
b (t) 104 2 103t 10,000 2000t.
At t 0: b (0) 10,000 bacteria/hour
At t 5: b (5) 0 bacteria/hour
At t 10: b (10)
10,000 bacteria/hour Section 3.4
9. Q(t) 200(30 t)2 200(900 60t t 2)
180,000 12,000t 200t 2
Q (t)
12,000 400t
The rate of change of the amount of water in the tank after
10 minutes is Q (10)
8000 gallons per minute.
Note that Q (10) 0, so the rate at which the water is
running out is positive. The water is running out at the rate
of 8000 gallons per minute. 13. a(t) v (t) 6t 2 18t 12
Find when acceleration is zero.
6t 2 18t 12 0
6(t 2 3t 2) 0
6(t 1)(t 2) 0
t 1 or t 2
At t 1, the speed is v(1)
0
0 m/sec.
At t 2, the speed is v(2)
1 1 m/sec.
14. (a) g (x) The average rate for the first 10 minutes is
Q(10)
10 Q(0)
0 80,000 180,000
10 h (x) 10,000 gallons per t (x) minute.
The water is flowing out at an average rate of 10,000 83 d3
(x ) 3x 2
dx
d3
(x
2) 3x 2
dx
d3
(x
3) 3x 2
dx (b) The graphs of NDER g(x), NDER h(x), and NDER t(x)
are all the same, as shown. gallons per minute over the first 10 min.
10. (a) Average cost c(100)
100 (b) c (x) 100 0.2x
Marginal cost c (100) 11,000
100 $110 per machine
[ 4, 4] by [ 10, 20] $80 per machine (c) Actual cost of 101st machine is
c(101) c(100) $79.90, which is very close to the
marginal cost calculated in part (b). (c) f (x) must be of the form f (x) = x 3
constant.
x3 (e) Yes. f (x) 11. (a) (d) Yes. f (x) x3 c, where c is a 3 15. (a) [0, 50] by [ 500, 2200] The values of x which make sense are the whole
numbers, x 0.
(b) Marginal revenue d
2000 1
dx r (x) d
2000
2000
dx
x1
(x 1)(0) (2000)(1)
0
(x 1)2 (c) r (5) 2000
(5 1)2 2000
36 1
x (c)
1 2000
(x 1)2 [0, 200] by [ 0.1, 0.2] 55.56 The increase in revenue is approximately $55.56.
(d) The limit is 0. This means that as x gets large, one
reaches a point where very little extra revenue can be
expected from selling more desks.
12. v(t) s (t) 3t 2 12t 9
a(t) v (t) 6t 12
Find when velocity is zero.
3t 2 12t 9 0
3(t 2 4t 3) 0
3(t 1)(t 3) 0
t 1 or t 3
At t 1, the acceleration is a(1)
At t 3, the acceleration is a(3) [0, 200] by [ 2, 12] (b) The values of x which make sense are the whole
numbers, x 0. 2 6 m/sec
6 m/sec2 P is most sensitive to changes in x when P (x) is
largest. It is relatively sensitive to changes in x between
approximately x 60 and x 160.
(d) The marginal profit, P (x), is greatest at x 106.44.
Since x must be an integer,
P(106) 4.924 thousand dollars or $4924.
(e) P
P
P
P
P
P (50) 0.013, or $13 per package sold
(100) 0.165, or $165 per package sold
(125) 0.118, or $118 per package sold
(150) 0.031, or $31 per package sold
(175) 0.006, or $6 per package sold
(300) 10 6, or $0.001 per package sold (f) The limit is 10. The maximum possible profit is
$10,000 monthly.
(g) Yes. In order to sell more and more packages, the
company might need to lower the price to a point
where they won’t make any additional profit. 84 Section 3.4 16. (a) 190 ft/sec
(b) 2 seconds
(c) After 8 seconds, and its velocity was 0 ft/sec then
(d) After about 11 seconds, and it was falling 90 ft/sec then
(e) About 3 seconds (from the rocket’s highest point)
(f) The acceleration was greatest just before the engine
stopped. The acceleration was constant from t 2 to
t 11, while the rocket was in free fall.
17. Note that “downward velocity” is positive when McCarthy
is falling downward. His downward velocity increases
steadily until the parachute opens, and then decreases to a
constant downward velocity. One possible sketch: (b) The particle speeds up when its speed is increasing,
which occurs during the approximate intervals
1.153 t 2.167 and t 3.180. It slows down during
the approximate intervals 0 t 1.153 and
2.167 t 3.180. One way to determine the
endpoints of these intervals is to use a grapher to find
the minimums and maximums for the speed,
NDER x(t) , using function mode in the window
[0, 5] by [0, 10].
(c) The particle changes direction at t
t 3.180 sec. 1.153 sec and at (d) The particle is at rest “instantaneously” at
t 1.153 sec and at t 3.180 sec. Downward velocity (e) The velocity starts out positive but decreasing, it
becomes negative, then starts to increase, and becomes
positive again and continues to increase.
The speed is decreasing, reaches 0 at t 1.15 sec, then
increases until t 2.17 sec, decreases until
t 3.18 sec when it is 0 again, and then increases after
that.
Time 18. (a) We estimate the slopes at several points as follows,
then connect the points to create a smooth curve.
t (days) 0 10 20 30 40 (f) The particle is at (5, 2) when
2t 3 13t 2 22t 5 5 at t 0.745 sec,
t 1.626 sec, and at t 4.129 sec.
21. (a) s (ft)
50 50 Slope (flies/day) 0.5 3.0 13.0 14.0 3.5 0.5
p′ (slope)
20 t (sec) 5 (b) s (1)
50 22. (a) t 18, s (2.5) 0, s (3.5) 12 s (ft)
30 Horizontal axis: Days
Vertical axis: Flies per day
(b) Fastest: Around the 25th day
Slowest: Day 50 or day 0
19. The particle is at (5, 2) when 4t 3
occurs at t 2.83. 5 16t 2 15t t (sec) 5, which 20. The motion can be simulated in parametric mode using
x1(t) 2t 3 13t 2 22t 5 and y1(t) 2 in [ 6, 8] by
[ 3, 5].
(a) It begins at the point ( 5, 2) moving in the positive
direction. After a little more than one second, it has
moved a bit past (6, 2) and it turns back in the negative
direction for approximately 2 seconds. At the end of
that time, it is near ( 2, 2) and it turns back again in
the positive direction. After that, it continues moving in
the positive direction indefinitely, speeding up as it
goes. (b) s (1) 6, s (2.5) 12, s (3.5) 24 23. (a) The body reverses direction when v changes sign, at
t 2 and at t 7.
(b) The body is moving at a constant speed, v
between t 3 and t 6. 3 m/sec, Section 3.4
(c) The speed graph is obtained by reflecting the negative
portion of the velocity graph, 2 t 7, about the
xaxis. 85 Velocity graph:
v (t) (cm/sec)
4 Speed(m/sec)
2 3
2 1 1
2 4 6 8 10 2 3 4 2 3 4 5 6 t (sec) –2 t (sec) –1
–4 –2 Speed graph: –3 (d) For 0 t 3
1 1: a For 1 t For 3 t 3
3
3
6
3(
8
0
10 3: a
6: a For 6 t 8: a For 8 t 10: a v (t) (cm/sec) 0
0 3 m/sec 2 Acceleration (m/sec2)
3
2
1 (1, 3) (6, 3) (8, 3) (3, 0)
2 (6, 0)
4 6
8
(8, –1.5) –1
–2
–3 10 t (sec) (10, –1.5) (3, –3)
(1, –3) 24. (a) The particle is moving left when the graph of s has
negative slope, for 2 t 3 and for 5 t 6.
The particle is moving right when the graph of s has
positive slope, for 0 t 1.
The particle is standing still when the graph of s is
horizontal, for 1 t 2 and for 3 t 5.
(b) For 0 t 1: v 2
1 0
0 Speed
For 1 t 2: v 2
2 Speed
For 2 t 3: v For 3 t 5: v
Speed For 5 t 2 cm/sec
v 2
1 2 cm/sec
0 cm/sec v 22
32 Speed 6: v
Speed 4 3
3 m/sec 2
1
( 3)
0 m/sec 2
3
3)
3 m/sec 2
6
3
1.5 m/sec 2
8 v
2 ( 2)
53 v
4 ( 2)
65 v 0 cm/sec
4 cm/sec
4 cm/sec
0 cm/sec
0 cm/sec
2 cm/sec
2 cm/sec 2 1 5 6 t (sec) –2
–4 25. (a) The particle moves forward when v 0, for
0 t 1 and for 5 t 7.
The particle moves backward when v 0, for
1 t 5.
The particle speeds up when v is negative and
decreasing, for 1 t 2, and when v is positive and
increasing, for 5 t 6.
The particle slows down when v is positive and
decreasing, for 0 t 1 and for 6 t 7, and when
v is negative and increasing, for 3 t 5.
(b) Note that the acceleration a
t 2, t 3, and t dv
is undefined at
dt 6. The acceleration is positive when v is increasing, for
3 t 6.
The acceleration is negative when v is decreasing, for
0 t 2 and for 6 t 7.
The acceleration is zero when v is constant, for
2 t 3 and for 7 t 9.
(c) The particle moves at its greatest speed when v is
maximized, at t 0 and for 2 t 3.
(d) The particle stands still for more than an instant when v
stays at zero, for 7 t 9. 86 Section 3.4 26. (a) To graph the velocity, we estimate the slopes at several
points as follows, then connect the points to create a
smooth curve.
t (hours) 0 2.5 5 7.5 10 12.5 v (km/hour) 0 56 75 56 0 94 (b) v s (t) 980t a s (t) 980 15 The velocity was s 225 4
7 560 cm/sec. The acceleration was s 4
7 980 cm/sec 2. v (km/hr) (c) Since there were about 16 flashes during 100
5 15 the light was flashing at a rate of about 28 flashes per t (hours) second. –100
–200
–300 To graph the acceleration, we estimate the slope of the
velocity graph at several points as follows, and then
connect the points to create a smooth curve.
t (hours) 0 v (km/hour ) 2.5 5 7.5 30 2 15 0 15 10
30 12.5
45 a (km/hr2)
50 15
60 28. Graph C is position, graph A is velocity, and graph B is
acceleration.
A is the derivative of C because it is positive, negative, and
zero where C is increasing, decreasing, and has horizontal
tangents, respectively. The relationship between B and A is
similar.
29. Graph C is position, graph B is velocity, and graph A is
acceleration.
B is the derivative of C because it is negative and zero
where C is decreasing and has horizontal tangents,
respectively.
A is the derivative of B because it is positive, negative, and
zero where B is increasing, decreasing, and has horizontal
tangents, respectively. 25 30. (a)
10 15 t (hours) dy
dt –25
–50 (b) ds
dt 4
of a second,
7 30t 3t 2 d
t2
61
dt
12
d
t
t2
61
dt
6
144
d
12
6t
t
dt
24
t
t
01
12
12 1 (b) The fluid level is falling fastest when
negative, at t 0, when falling slowest at t dy
dt 12, when dy
is the most
dt 1. The fluid level is
dy
dt 0. [0, 15] by [ 300, 100] d 2s
dt 2 (c)
30 6t [0, 12] by [ 2, 6] y is decreasing and
[0, 15] by [ 100, 50] interval. y decreases more rapidly early in the interval, The graphs are very similar.
27. (a) Solving 160
It took
160 cm
4
7 490t 2 gives t 4
.
7 4
of a second. The average velocity was
7 sec = 280 cm/sec. dy
is negative over the entire
dt and the magnitude of
t dy
dy
is larger then.
is 0 at
dt
dt 12, where the graph of y seems to have a horizontal tangent. Section 3.5 31. (a) dV
dr d4 3
r
dr 3
dV
When r 2,
dr 37. (a) Assume that f is even. Then, 4 r2
4 (2)2 f ( x) 16 cubic feet of 32. For t f( x h)
h f (x h) f (x)
, and substituting k
h
f (x k) f (x)
lim
k
k→0
f (x k) f (x)
lim
f (x)
k
k→0 11.092 cubic feet. So, f is an odd function. 20
3600 sec
t. Multiplying by
9
1h 1 km
,
1000 m (b) Assume that f is odd. Then, we find that this is equivalent to 8t kilometers per hour.
Solving 8t 200 gives t f ( x) 25 seconds. The aircraft takes lim f( x h→0 lim f (x h→0 25 seconds to become airborne, and the distance it travels
during this time is D(25) h)
h
h)
h and substituting k 694.444 meters.
lim 33. Let v0 be the exit velocity of a particle of lava. Then f (x 16t 2 feet, so the velocity is v0t ds
Solving
dt
v0 feet, is s 32 v0 0 gives t
v0 32 v0 1900 gives v0 64 v0 32t. v0 2 v0 2 32 64 38.
. Solving 348.712. The exit velocity was about 348.712 ft/sec. Multiplying by 3600 sec
1h 1 mi
, we
5280 ft find that this is equivalent to about 237.758 mi/h.
34. By estimating the slope of the velocity graph at that point. d
( fgh)
dx Rule gives revenue cost, the Sum and Difference d
(profit)
dx d
(revenue)
dx d
(cost), where x is
dx the number of units produced. This means that marginal
profit marginal revenue marginal cost. 36. (a) It takes 135 seconds.
(b) Average speed F
t 5
73 0
0 5
73 k)
k 2
26 f (x) Making a Conjecture with NDER 1. When the graph of sin x is increasing, the graph of
NDER (sin x) is positive (above the xaxis).
2. When the graph of sin x is decreasing, the graph of
NDER (sin x) is negative (below the xaxis).
3. When the graph of sin x stops increasing and starts
decreasing, the graph of NDER (sin x) crosses the xaxis
from above to below.
4. The slope of the graph of sin x matches the value of
NDER (sin x) at these points. speed is approximately
2
33 f (x) s Section 3.5 Derivatives of Trigonometric
Functions (pp. 134–141) (c) Using a symmetric difference quotient, the horse’s 4
59 k)
k f (x) 5. We conjecture that NDER (sin x) cos x. The graphs
coincide, supporting our conjecture. 0.068 furlongs/sec. F
t , d
d
d
[f (gh)] f
(gh) gh
( f)
dx
dx
dx
dh
dg
df
fg
h
gh
dx
dx
dx
dg
dh
df
gh f
h fg
dx
dx
dx Exploration 1
35. Since profit f (x f (x ) So, f is an even function. . Then the maximum height, in 16 32 ds
dt lim k→0 f ( x) h, k→0 v0 2 h, 0, the speed of the aircraft in meters per second after t seconds is s(t) f ( x) h→0 (b) The increase in the volume is
4
(2)3
3 lim h→0 lim volume per foot of radius. 4
(2.2)3
3 87 1
13 0.077 furlongs/sec. (d) The horse is running the fastest during the last furlong
(between 9th and 10th furlong markers). This furlong
takes only 11 seconds to run, which is the least amount
of time for a furlong.
(e) The horse accelerates the fastest during the first furlong
(between markers 0 and 1). [ 2 , 2 ] by [ 4, 4] 88 Section 3.5 6. When the graph of cos x is increasing, the graph of
NDER (cos x) is positive (above the xaxis).
When the graph of cos x is decreasing, the graph of
NDER (cos x) is negative (below the xaxis).
When the graph of cos x stops increasing and starts
decreasing, the graph of NDER (cos x) crosses the xaxis
from above to below.
The slope of the graph of cos x matches the value of
NDER (cos x) at these points.
We conjecture that NDER (cos x)
sin x. The graphs
coincide, supporting our conjecture. 5. d
(4
dx d
(4)
dx x 2 sin x) x 2 cos x
6. d
(3x
dx d
4
dx cos x 180° 2. 1.7
3. sin 2.356 (1 97.403° 3 cot x) (cot x)
(1 d
dx (1 d
cos x
10.
dx 1 sin x sin2 a 7. If tan a 1, then a
1 sin a ( 1)2 1
3
4 0 0 8. k for some integer k, so .
1 cos 2 h
h(1 cos h) cos h)(1 cos h)
h(1 cos h) 9. y (x) 6x
14x
y (3) 12
The tangent line has slope 12 and passes through (3, 1), so
its equation is y 12(x 3) 1, or y 12x 35.
v (t)
12 6t 2 14t Section 3.5 Exercises
1. d
(1
dx 2. d
(2 sin x
dx 3. d1
dx x 4. d
(x sec x)
dx x cos x)
tan x) 0 ( sin x) 2 cos x
1
x2 5 sin x 1 sec 2 x 5 cos x d
dx x (sec x)
x sec x tan x sin x) sin x) 2 1
sin x 11. y (x)
y( ) d
(sin x
dx 3) cos 1 cos x The tangent line has slope 2 10. a(t)
a(3) d
dx (cos x) (1 cos x)2 sin x)( sin x) (cos x)(cos x)
(1 sin x)2 1 (1
cos h
h
sin2 h
h(1 cos h) 1
(sin x (1 sin x)
(1 sin x)2 2
1 cot x) 2 sin x sin 2 x cos 2 x
(1 sin x)2 Range: All reals
1 cot x) sin x) (cos x)
(1 k
for odd integers k
2 6. cos a d
dx (cot x) (1 csc2 x sin2 x
(1 cot x)2 sin2 x (1 5. Domain: x cos x) 2 cot x)( csc x) (cot x)( csc 2 x)
(1 cot x)2 2 4. Domain: All reals
Range: [ 1, 1] d
(1
dx cos x) d
dx (1 csc2 x
(1 cot x)2 3 x 2 (1 306 ° tan x cos x x sin x
(1 cos x)2 [ 2 , 2 ] by [ 4, 4] 3
4 d
dx (tan x) (x) 4 sec x tan x cos x) (x) 1 180° d
dx x (tan x) d
dx (1 d
cot x
9.
dx 1 cot x (sin x)(2x)] x sec 2x d
(4 sec x)
dx x
cos x d
dx (sin x) (x 2) 2x sin x d
(3x)
dx x tan x) 3 d
8.
dx 1 1. 135° [x 2 cos x 0 7. Quick Review 3.5 d
dx x 2 (sin x) d
dx sec x (x)
sec x 1 sin x ( , sin 3) Its equation is y 1 and passes through ( , 3).
1(x ) 3, or y x 3. Section 3.5
d
dx 12. y (x) 8 2 16
2 4
2 The normal line has slope 8 tan and passes through , 4 16 16 4 4 , 4 d cos x
dx sin x d
cot x
dx 16. (a) 2 d
dx 8 2 16 2 y 16 x 8
3 8 x 64 4
4
4 32 d
dx (sin x) (cos x) (cos x) (sin x) (sin x)2 . (sin x)( sin x) (cos x)(cos x)
sin 2 x 4 Its equation is y d (cos x) (1) (1) (cos x)
d1
dx
dx
(cos x)2
dx cos x
(cos x)(0) (1)( sin x)
cos 2 x
sin x
sec x tan x
cos2 x d
sec x
(b)
dx 1 4 4 d tan x (x)
x2 x sec2 x tan x
x2
( 2)2 y d
dx x (tan x) d tan x
dx x 89 , or (sin2 x cos2 x)
sin2 x . 1
sin2 x csc 2 x Using decimals, this equation is approximately
d
csc x
dx (b)
y 1.081x 2.122. d1
dx sin x
d
dx (sin x) (1) 13. y (x) d2
(x sin x)
dx x 2 cos x
y (3) d
dx (sin x) (x 2) (sin x)(0) (1)(cos x)
sin 2 x 2x sin x 9 cos 3 6 sin 3 The tangent line has slope 9 cos 3 6 sin 3 and passes
through (3, 9 sin 3). Its equation is
y (9 cos 3 6 sin 3)(x 3) 9 sin 3, or
y (9 cos 3 6 sin 3)x 27 cos 3 9 sin 3. Using
decimals, this equation is approximately
y
8.063x 25.460.
14. lim h→0 cos (x (sin x)2 d
dx x 2 (sin x) h)
h lim cos x
sin2 x 17. sin x sin h)
h cos x(cos h 1)
h h→0 lim (cos x) h→0 cos h
h cos h
h
h→0 (cos x) lim cos x sin x sin h 1 (sin x) 1 d
tan x
dx
d
cot x
dx 19. y (x) sin h
h
sin h
h→0 h y 15. (a) (sin x)(1) d
d sin x
tan x
dx
dx cos x
(cos x)(cos x) (sin x)( sin x)
cos 2 x
1
cos2 x sin x which is 0 at x 0, 1
, which is never 0.
cos2 x
1
csc 2 x
, which is never 0.
sin2 x d
(
dx 2 cos x) 2 sin x 2 sin 2 4 1 1
2 The tangent line has slope 1 and passes through (sin x) lim
sin x d
(cos x) (sin x)
dx cos2 x sin2 x
cos2 x 0, so the slope of sec 2 x 4 4 (cos x)(0) sec x tan x which is 0 at x so the slope of the tangent line is 0.
18. (cos x cos h d
sec x
dx csc x cot x d
the tangent line is 0.
cos x
dx cos x h→0 lim d
dx (1) (sin x) sec 2 x y
d
(sin x) (cos x)
dx , 2 cos
1x 4
4 4 , 1 , so its equation is 1, or y x 4 1. The normal line has slope 1 and passes through (cos x)2 its equation is y 1x 4 1, or y x 1 4 , 1 , so
4 . 90 Section 3.5
d
tan x
dx 20. y (x) sec 2 x
sec x y (x) 0 2 csc x cot x d
(2x)
dx y (x) (b) 2 csc x 0 2 cos x
sin2 x sec 2 x 1
sin2 x 0 1) 0 2 2 1
(
sin2 x 2 2 cos x 1 cos x 2 On , , the solutions are x 22 curve are 4 , 1 and d
(4
dx cot x 0 2 21. y (x) 2 csc 2 x 2 csc 2( 2) cot 2 2 1 and passes through
1x Speed: 2 cos
2 0 2 csc x cot x 0 Jerk: 2 cos 2 cos x
sin2 x 0 1) 1
2 x
3 cot 3 1 4 2 csc
2 4 Greatest when cos t
at point Q 3 3 Zero when cos t 0 3 3.
3. 2( csc x cot x) (or t Zero when sin t ( csc 2 x) 2 csc 4 4 4 2( 2)(1)
2 2 y 4 , 4 . Its equation is y
4x 4. k) Greatest (in magnitude) when cos t
csc 2 4 4 and passes through
4x 0 or t 1 (or t k
, k odd
2 24. (a) Velocity: s (t) cos t sin t m/sec
Speed: s (t)
cos t sin t m/sec
Acceleration: s (t)
sin t cos t m/sec2
Jerk: s (t)
cos t sin t m/sec3 2)2 4 The tangent line has slope
P ( 0 (or t Jerk: csc 2 x
cot 1 k
, k odd)
2 Zero when cos t
(a) y k
, k odd), at the
2 Greatest (in magnitude) when sin t
4 cot x) 2 csc x cot x 0 (or t Acceleration:
,4 The equation of the horizontal line is y
2 csc x k ), at the center of endpoints of the interval of motion.
3 3 d
(1
dx 1 (or t the interval of motion. 3 4 The coordinates of Q are 22. y (x) 2 m/sec 3 Speed: 2 3 3 4 2 m/sec 2 4 (c) The body starts at 2, goes to 0 and then oscillates
between 0 and 4. 0 cos x 4 2 m/sec Acceleration: 2 sin 1
(2 cos x
sin2 x y 2 m/sec 4 y (x) 1
sin2 x 2. 23. (a) Velocity: s (t)
2 cos t m/sec
Speed: s (t)
2 cos t m/sec
Acceleration: s (t) 2 sin t m/sec 2
Jerk: s (t) 2 cos t m/sec 3
(b) Velocity: 2, or 2 2. csc 2 x 3
,2 .
4 The coordinates of Q are 1 (b) ( 1) The equation of the horizontal line is y 2(1)(0) 2 3
4 1 2
3
at point Q
4
3
cot
4 2 , 2 . Its equation is y y 3
4 1 ,1 . The tangent line has slope
2 y 2 csc x cot x 12 P . The points on the 2 csc x cot x csc 2 2 4 4 x 2 csc x) csc x
csc 2 x (a) y 1 cos x 4 4, or k) 91 Section 3.5 (b) Velocity: cos
Speed: 0 sin 4 0 m/sec 4 (d) 0 m/sec Acceleration: sin Jerk: sin cos 4 cos 4 2 m/sec2 4 0 m/sec3 4 (c) The body starts at 1, goes to
between
2. cos (x h) cos x
d
cos x lim
h
dx
h→0
cos x cos h sin x sin h cos x
lim
h
h→0
(cos x)(cos h 1) sin x sin h
lim
h
h→0
lim cos x 2 and then oscillates 3
4 Greatest when t
Zero when t 4 180 k
(e) k 1802
3
4 4 180 180 2 d
dx sin x 2 sin x 2 180 1802 180 cos x 3 1803 cos x 2 3
4 d
cos x
dx 2 k d
dx sin x 180 180 180 cos x 2 1802 k cos x
3 d
cos x
dx 3 25. (a) 2 d
dx 2 cos x 2 180 1802 180 sin x 3 1803 26. y
[ 360, 360] by [ 0.01, 0.02] The limit is 180 y because this is the conversion factor sin x d
csc x
csc x cot x
dx
d
( csc x cot x)
dx
d
d
(csc x) (cot x) (cot x) (csc x)
dx
dx for changing from degrees to radians. (csc x)( csc2 x) (b) csc3 x
27. y [ 360, 360] by [ 0.02, 0.02] sec2 h) sin x
h
sin x cos h cos x sin h sin x
lim
h
h→0
sin x(cos h 1) cos x sin h
lim
h
h→0
cos h 1
lim sin x lim
lim cos x
h
h→0
h→0
h→0
lim (sin x)(0)
180 sin (x y h→0 (cos x) 180 sin h
h→0 h
lim (cot x)( csc x cot x) csc x cot2 x d
( tan )
d
d
(tan )
d This limit is still 0.
d
sin x
dx sin h
h sin x d
sin x
dx 3 k k Greatest (in magnitude) when t (c) lim h→0 180 d
cos x
dx 180 3 Jerk: 4 lim sin x h→0 2 Greatest (in magnitude) when t Zero when t (sin x) 1 sin x d2
sin x
dx 2 Acceleration: Zero when t h→0 (cos x)(0) Speed: cos h
h lim h→0 (tan ) d
()
d tan d
( sec2
tan )
d
d
d
d
[(sec )(sec )] (sec2 ) ( )
(tan )
d
d
d
d
d
(sec ) (sec ) (sec ) (sec )
sec2
d
d sec2
2 sec2
(2 tan tan 2 sec2 2)(sec2 ) cos x
or, writing in terms of sines and cosines,
2 2 tan
cos2
2 cos
2 sin
cos3 92 Section 3.6 28. Continuous: 33. Note that g(0)
lim g(x) x→0 so b lim g(x) x→0 lim (x x→0 b) lim cos x cos (0) x→0 b. We require lim g(x)
x→0 1. The function is continuous if b d
sin 2x
dx d
(2 sin x cos x)
dx
d
2 (sin x cos x)
dx
d
2 (sin x) (cos x)
dx 1, and
g(0), 1. Differentiable:
For b 1, the lefthand derivative is 1 and the righthand
derivative is sin (0) 0, so the function
is not differentiable. For other values of b, the function is
discontinuous at x 0 and there is no
lefthand derivative. So, there is no value of b that will
make the function differentiable at x 0. 2[(sin x)( sin x)
2 d5
dx 5
d6
dx 6
d7
dx 7
d8
dx 8 sin x
cos x
sin x
cos x cos x 34. d
cos 2x
dx d
[(cos x)(cos x) (sin x)(sin x)]
dx
d
d
(cos x) (cos x) (cos x) (cos x)
dx
dx
d
d
(sin x) (sin x) (sin x) (sin x)
dx
dx cos x sin x when n 4k 2(2 sin x cos x) cos x 2 sin 2x 35. lim h→0 (cos h
h 1) lim h→0 4(249) 3, d 999
cos x
dx 999 1) cos2 h 1
1)
h→0 h(cos h
sin2 h
lim
1)
h→0 h(cos h sin x. lim h→0 sin h
h 0
(1)
2 30. Observe the pattern:
d
sin x
dx
2
d
sin x
dx 2
d3
sin x
dx 3
4
d
sin x
dx 4 (cos h 1)(cos h
h(cos h 1) lim 3 for any whole number k.
Since 999 2(sin x)(cos x) 4 sin x cos x sin x Continuing the pattern, we see that
dn
cos x
dx n 2(cos x)( sin x) cos x cos x sin x) 2 cos 2x sin x cos x (cos x)(cos x)] 2 2(cos x 29. Observe the pattern:
d
cos x
dx
d2
cos x
dx 2
3
d
cos x
dx 3
4
d
cos x
dx 4 d
dx (cos x) (sin x) sin h lim h→0 cos h 1 0 5 d
dx 5
d6
dx 6
d7
dx 7
d8
dx 8 cos x
sin x
cos x
sin x sin x cos x d
(A sin x
dx
d
(A cos x
dx 36. y sin x sin x y sin x cos x B cos x) A cos x B sin x Solve: sin x B sin x) B cos x
y ( A sin x sin x A sin x B cos x) 2A sin x
Continuing the pattern, we see that
dn
sin x
dx n At x
cos x when n 4k 2 1 for any whole number
At x , this gives 0, we have k.
Since 725 4(181) 1, d 725
sin x
dx 725 32. (a) Using y x, sin (0.12) 0.12. (b) sin (0.12) 0.1197122; The approximation is within
0.0003 of the actual value. 0, so B B cos x) sin x
sin x 1
.
2 1, so A 2B 1
and B
2 Thus, A
cos x. 31. The line is tangent to the graph of y sin x at (0, 0). Since
y (0) cos (0) 1,
the line has slope 1 and its equation is y x. 2A sin x 2B cos x (A sin x y 0. 0. s Section 3.6 Chain Rule (pp. 141–149)
Quick Review 3.6
1. f (g(x) f (x 2 1) sin (x 2 1) 2. f (g(h(x))) f (g(7x) f ((7x )2
sin [(7x)2 1] sin (49x 2
3. (g h)(x) g(h(x)) 4. (h g)(x)
5. f g(x)
h(x) h(g(x))
f x2 1
7x (7x)2 g(7x)
h(x 2 sin 1)
1) 1)
x2 1
7x 7(x 1
2 49x 2
1) 7x 2 1
7 Section 3.6
6. cos x 2 7. g(cos x) 3 cos2 x g(3 cos2 x) 2 9. g(h(cos x)) g(h( f (x))) dy
dx d
(x
dx f (27x 4) cos 3x
10. cos 2 3x
f (g(3x 2)) f (g(h(x))) dy
dx d
sin (3x
dx [cos (3x
2. dy
dx 1) 2 f ( 3x 2 f (h(h(x))) 10. 2) dy
dx cot x)
cot x) d
sin (7
dx [cos (7 1
d
dx 3 cos (3x 5x) [cos (7 5x)]( 5) 1) 5 cos (7 11.
5x) dy
dx dy
dx d
cos (
dx [ sin ( 3x)](
4. dy
dx d
tan (2x
dx [sec2 (2x
5. 6. dy
dx [ sin ( 3x)] (
3) ( 5 sin x 3)](2 3x 2) x 3)] (2x 2
2
d
5 cot
5 csc2
x
x
dx
10
22
2
5 csc
( 2x )
csc2
x
x2 dy
dx d
sin x
dx 1 cos x
sin x
2
1 cos x 2
2
2 1 sin x
cos x 2 2 (1 cos x) sin x
cos x cos x 1 sin x
cos x 1
(1 2 (1 1 d
dx sin x (1 dy
dx d
cos (sin x)
dx dy
dx cos x) cos x) d
sec (tan x)
dx 3 cos2 x sin x 5)3 (2x
5)3(2) 5)4 d
dx 5)4 (x 3) (2x
d
dx 5)3
5)3[8x 5)
3x 2(2x 3x 2(2x 5)4(3x 2) 5)4 5)4 3(2x (2x 5)] 3 5) (14x 15) d
(sin3 x tan 4x)
dx
d
d
(sin3 x) (tan 4x) (tan 4x) (sin3 x)
dx
dx
d
d
(sin3 x)(sec2 4x) (4x) (tan 4x)(3 sin2 x) (sin x)
dx
dx (sin3 x)(sec2 4x)(4) (tan 4x)(3 sin2 x)(cos x) 2 cos x sin x
cos x)2 4 sin3 x sec2 4x cos x
cos x)2 14. 1
cos x dy
dx d
(4
dx sec x
1 4 2 sec x
2
d
(sin x)
dx d
dx sec (tan x) tan (tan x) (tan x) sec (tan x) tan (tan x) sec2 x d
dx (3 cos2 x) (cos x) 5)4] sec x 2 sec x
2 sec x dy
dx d
dx d3
[x (2x
dx
d
(x 3) (2x
dx x (2x
13. csc2 x) csc x
csc x cot x x cos x 2 2 [ sin (sin x)] x) cot x) ( cot x csc x 6 sin (sin x) cos x 8. (csc x x) (sin x) x 2(2x 2 sin x
(1 cos x)2 7. 6 8x 3(2x d
(2x 1)
dx
2
x cos x)(cos x) (sin x)( sin x)
(1 cos x)2 1 sin x
2
1 cos x d
sin x
dx (1
(1 x 3) sin x
d
sin x
cos x dx 1 cos x 1 dy
dx x 3) 3x 2) sec2 (2x (2 dx (x 3)(4)(2x 12. d
dx [sec2 (2x 2d (x 3)(4)(2x 3x) 3 sin ( 3x) x 3) (x cos3 x) 5x)
d
dx 3x) cot x)2 d
(sin 5 x
dx 5 sin
3. dx 1 (csc x)(cot x csc x)
(csc x cot x)2 1)
d
5x)] (7
dx 3d 1 1 d
(csc x
dx (csc x 1)] (3x x) 2x f (h(3x 2)) 2 [cos (3x 1)](3) 3 x) Section 3.6 Exercises
1. 2(x (csc x f (3(3x 2)2) 2 2 x) 2(x 3(cos x 2) 3( cos x 2)2
2) h(g(cos x)) h(g( f (x))) 8. 3 cos x 6
h( cos x
9. cos 27x 4 g( f (x)) 93 3 sin2 x cos x tan 4x tan x)
d
(sec x
tan x dx (sec x tan x tan x
sec x tan x
sec x sec x tan x tan x tan x)
sec2 x) 94 15. Section 3.6
dy
dx 3 d
dx 19. 2x 1 dy
dx d
(1
dx cos2 7x)3 ( 2x
2x ( d
dx d
dx 1)(0) 2x d
(2x
1 dx 1 3 2 2x 2x
1 3 1) 20. dy
dx 3 16. dy
dx 1) 2x 1) 1 21. d
x ) (x)
dx d
x(
dx 1 21 21 x) x (1 x 2)( 1 (1 x2) 2 x sin x 2) 3 sin x2 (2x) 22. x2 2 (1 d
(1
dx x2
1 x
1 3t x) 2 ds
dt 2 3t 2 d
[t cos (
dt
d
(t) [cos (
dt 23. d
sin (3x
dx 2) 2 sin (3x 2) cos (3x 2) (3x 2 sin (3x 2) cos (3x 6 sin (3x 2) cos (3x 2) 3 sin (6x 4) ds
dt 2)(3) d
(1
dx cos 2x)2 2)
d
dx 2) 4t)]
d
dt d
dt cos ( 4t) (t) 4t)] ( 4t) 4t)]( 4) cos ( cos 2x) (1
d
dx d
dt
4
3
4
3
4 cos ( cos ( 4t)(1) 4t) 4t) 4
4
sin 3t
cos 5t
3
5
d
4
d
(cos 3t) (3t)
( sin 5t) (5t)
dt
5
dt
4
(cos 3t)(3)
( sin 5t)(5)
5
4 cos 3t ds
dt d
sin
dt
3
cos
2
3
cos
2 25. dr
d d
tan (2
d cos 2x) cos 2x)( sin 2x) (2x) 2(1 4t) 24. d
dx 2(1 2(1 cos 2x)( sin 2x)(2) 4(1 4t)] sin 2a. 2) 2 sin (3x dy
dx 3t t[ sin ( x 2) 17. The last step here uses the identity 2 sin a cos a 18. d
dt 2 t[ sin ( 3/2 d
sin2 (3x
dx sec2 5x 3t ( 3) 2 4t sin (
dy
dx 1/2 3t 2 1 x
1 x2 d
cos
2
dt x) x )(1) 1 tan 5x 22 2 1 ds
dt 5
2 or (tan 5x) sin x2 (1
( 1
d
tan 5x
2 tan 5x dx
1
d
(sec2 5x) (5x)
dx
2 tan 5x
1
2 tan 5x) 2 tan 5x
5 sec2 5x 2 (1 d
(
dx 2 1 cos2 7x)2 cos 7x sin 7x (sec 5x)(5) 3/2 x d
dx d
dx 42(1 1 3(2x cos2 7x)2(2 cos 7x)( sin 7x)(7) d
dx 1 2x (2x cos2 7x)2(2 cos 7x)( sin 7x) (7x) 1 cos2 7x) (2) 2x 2 1) cos2 7x)2(2 cos 7x) (cos 7x) 3(1 1)2 1) (3) cos2 7x)2 (1 3(1 3( 3(1
3(1 d
dx ( 2x sec2 (2 sin 5t 3
7
t
cos
t
2
4
7
d3
d7
t
t
sin
t
t
4 dt 4
dt 2
3
7
7
t
sin
t
2
4
4 ) )( 1) sec2 (2
sec2 (2 ) d
(2
d ) ) cos 2x)(sin 2x)
26. dr
d d
(sec 2 tan 2 )
d
d
d
(sec 2 ) (tan 2 ) (tan 2 ) (sec 2 )
d
d
d
d
(sec 2 )(sec2 2 ) (2 ) (tan 2 )(sec 2 tan 2 ) (2 )
d
d 2 sec3 2 2 sec 2 tan2 2 Section 3.6 27. dr
d d
d 1 sin sin
1 2
cos dr
d 2 (1 sec ) 2 ( sec ) d
(sec )
d sec
1 (2 )
( g (x) 1 (2 ) (sec y 2 (4) sec
2 sec 2) d
tan x sec2 x
dx
d
d
sec2 x (2 sec x) (sec x)
dx
dx 10 36. f (u) 3 csc (3x
y 3[2 csc (3x 18 csc (3x
32. y y d
9 tan
dx
x
3 sec2
3
d
3 sec2
dx
x
6 sec
3
x
2 sec2
3 (1) csc2 ud
u
10 du 10 5
x 2 f (5)g (1) 5
2 2 5
2 4
2 (cos u) d
2(cos u) 3 cos u
du
2 sin u
cos3 u 1)
1 1)
37. f (u)
d
dx 1)] (3x 9 sec2 xdx
3 dx 3 3 2 sec x
x
tan
3
3
x
tan
3 x
3 d
x
sec
dx
3
dx
dx 3 (u 2 1) 1) cot (3x 1) sec 1
g ( 1)
2 2
1 () 3 5 1) cot (3x x
3 f 2 1)][ csc (3x x
3 x) x)2 (1 f (g(1))g (1) d
[u
du 1)] 1) cot (3x 3[2 csc (3x (1 d
( x)
dx
1
1
1
( f g)
f (g( ))g
4
4
4
1
f
g
4
4 1)] [ csc (3x 2 1) (3x d
1)] csc (3x
dx 3[2 csc (3x dx 1 ( 1) 1) d
[ 3 csc2(3x
dx 2d 2 d
dx csc2 (3x 5
2 1
u2 x) 1
2 g (x) 2 csc2 x cot x
1) 1 2 u f (g( 1))g ( 1) csc2 1 ( 2 csc x)( csc x cot x) 2 (5) 1 1 d
cot x
csc2 x
dx
d
d
( csc2 x) ( 2 csc x) (csc x)
dx
dx d
cot (3x
dx 1
4 10 2 sec x tan x 31. y 2 x) ( f g) (1) 2 y x) ) u
d
cot
10
du
u
csc2
10
10
d
g (x)
(5 x)
dx (2 sec x)(sec x tan x) 30. y f (1)g (1) (1 1 35. f (u) sec sec 2 u ( f g) ( 1) d
(2 )
d tan ) sec )(tan )
sec ( tan 29. y d
(1
du
d
(1
dx 34. f (u) sec ) d
(
d f (g(1))g (1) sin ) sin d
(2
d (2 ) ( f g) (1) sin 2 28. g (x) sin ( cos d5
(u
1) 5u 4
du
1
d
( x)
dx
2x 33. f (u) sin
2
d
d
(sin ) (sin ) ( )
d
d 1
2 d
( sin )
d 1)](3) g (x) 2u
d
du u 2 1 (u 2 1) d
(2u)
du (u 2 (2u)
1) d2
(u
du 1) 2 1)(2) (2u)(2u)
2u 2 2
(u 2 1)2
(u 2 1)2
d
(10x 2 x 1) 20x 1
dx ( f g) (0) f (g(0))g (0) f (1)g (0) (0)(1) 0 95 96 Section 3.6 38. f (u)
u
2
u
u
2
u g (x) du
du u 12
1 (u
1
1
1 (u
1 1) ( f g) ( 1) d
(u
du 1du
1 du u 1) (u 1) 1
1
d
(u
du 1) 3 2x f (g( 1))g ( 1) dy
dt
dx
dt 2 cos t 2 cos t
2 sin t cot t and has slope cot 2, or y x dx
dt d
(sin 2 t)
dt (cos 2 t) (2 t) dy
dt 42.
2) 6 sin u d
(cos 2 t)
dt
dy
dt
dx
dt 2 sin 2 t
2 cos 2 t and has slope tan 1)
y 2 43. 6 sin 2u dy
40. (a)
dx dx
dt d
(sec 2 t
dt 2 , cos 31
,
2 2
6 2 3. Its equation is 6
1
, or y
2 3x 2. d
dt (2 sec t) (sec t)
(2 sec t)(sec t tan t) 2) dy
dt 1)(1) 2x 2x cos (x 2 d
tan t
dt dy
dx d
1) (x 2)
dx 2x cos (u dy
dt
dx
dt 1) sec 2 y 1) 1
cot t.
2 1, tan 4
1
cot
2
1
(x
2 1) 44.
2 sec2 t
2 sec2 t tan t slope (cos u)(2x)
2x cos u sec 2 t The line passes through 1) dy du
du dx
d
d
(sin u) (x 2
du
dx 2x cos (x 6 2 sec 2 t tan t dy du
du dx
d
sin (u
du cos (u dy
(b)
dx 1) 2 sin 2 t tan 2 t 2 3 3x ( sin 2u)(2) (3) 2 cos 2 t d
dt 2) dy du
du dx
d
d
(cos 2u) (3x
du
dx 6 sin (6x 2) 2. ( sin 2 t) (2 t) d
dt The line passes through sin
dy
dx 2 (x dy
dx dy du
du dx
d
d
(cos u) (6x
du
dx 6 sin (6x ( 2, 4 8 ( sin u)(6) (b) , 2 sin 1. Its equation is 4 2) 4 y ( 4)(2)
dy
dx d
(2 sin t)
dt 2 sin t The line passes through 2 cos f (0)g ( 1) 39. (a) d
(2 cos t)
dt dy
dx 1) dx
dt
dy
dt 41. (u 1)2
(u 1)
4(u 1)
1)2
(u 1)3 1)
(u d
(x 2
dx u
u 2 dx
dt
dy
dt d
sec t
dt
d
tan t
dt dy
dx dy
dt
dx
dt 1, or y 2x 1
.
2 sec 2 t
sec t
tan t The line passes through sec or y 1
x
2 sec t tan t sec 2 t
sec t tan t has slope csc 1) and has 1
. Its equation is
2 4 1) (1, 4 1
sin t 6 , tan csc t
2
6 2. Its equation is y 6 1 ,
3 2x 2 1
3 3. and
3 ,
3 Section 3.6
dx
dt d
t
dt dy
dt d
dt dy
dx 45. dy
dt
dx
dt 1 t 1/(2 t)
1 1
2 y 1
4 (b) 11
,
and has slope
42
1
4 1x dx
dt
dy
dt dy
dt
dx
dt 1
, or
2 (2t 4t 4t 3 Then dy
dx dy
dt
dx
dt 1 sin t) 1 cos t) 1 y 1)(sin t) 2 cos t
(2t 1)2 du dx
du
, so
dx dt
dx du
dt dx
. Therefore,
dt d dy
dt dx
(2t dx
dt
1)(sin t) 2 cos t
(2t 1)2
1)(sin t) 2 cos t
(2t 1)3 cos t sin t (2t 1) (d) The expression in part (c). sin t
cos t 50. Since the radius passes through (0, 0) and (2 cos t, 2 sin t),
it has slope given by tan t. But the slope of the tangent is sin 3 ,1 cos dy
dt
dx
dt dy
dx 3 31
,
and has slope
2 3 2 cos t
2 sin t cot t, which is the negative reciprocal of tan t. This means that the radius and the 2 tangent are perpendicular. (The preceding argument breaks 3 cos du
dt d dy
dx dx The line passes through sin 1) (2t d
(t
dt
d
(1
dt 1) 2 dy
.
dx (c) Let u
t2 d
dt (cos t) (2t 1)( sin t) (cos t)(2)
(2t 1)2 (2t 3) 4t 3
4t dx
dt
dy
dt 3 d cos t
dt 2t 1
d
(2t 1) (cos t)
dt The line passes through (2( 1)2 3, ( 1)4) (5, 1) and
has slope ( 1)2 = 1. Its equation is y 1(x 5) 1, or
y x 4.
47. d dy
dt dx (2t d
(2t 2
dt
d4
(t )
dt 1 cos t
2t 1 1
.
4 x dy
dx 46. dy
dt
dx
dt t 1. Its equation is y 1
4 d2
(t
t) 2t
dt
d
sin t cos t
dt t 2 1
The line passes through ,
4
1
2 dx
dt
dy
dt
dy
dx 49. (a) 1 3. Its equation is
down when t 3 1
, or
2 3 3x 3 2 k
, where k is an integer. At these values,
2 either the radius is horizontal and the tangent is vertical or
the radius is vertical and the tangent is horizontal, so the y 3x 2 . result still holds.) 3 48. 97 dx
dt
dy
dt d
cos t
sin t
dt
d
(1 sin t) cos t
dt dy
dx dy
dt
dx
dt cos t
sin t cot 2 ds
dt ds d
d dt ( sin )
When cot t The line passes through cos
has slope 51. 2 ,1 sin 0. Its equation is y (0, 2) and 2 2. 52. dy
dt d
d
(cos )
dt
d
d
dt
3
d
and
2
dt dy dx
dx dt (2x
When x 7) d2
(x
dx 5, ds
dt 7x 5) sin 3
(5)
2 5. dx
dt dx
dt 1 and dy
dt 1 dy
,
3 dt [2(1) 7] 1
3 3. 98 53. 54. Section 3.6
dy
dx 1
x
cos
2
2
1
x
Since the range of the function f(x)
cos is
2
2
dy
1
the largest possible value of
is .
dx
2
d
x
sin
dx
2 dy
dx xdx
2 dx 2 cos d
(sin mx)
dx d
dx (cos mx) (mx) The desired line has slope y (0) (e)
11
,,
22 through (0, 0), so its equation is y At x f (g(x))g (x) 2, the derivative is f (g(2))g (2) m cos mx m cos 0 d
f (g(x))
dx (f) 1
( 3)
3 f (2)g (2)
1 d
dx f (x) f (x) d
f (x)
f (x) dx 2 At x f (x ) 2 m and passes mx. 2, the derivative is f (2) 55. dy
dx x
d
2 tan
4
dx
x
sec 2
2
4 y (1) 2 sec 2 2 sec 2 (g) 4 2 2)2 ( The tangent line has slope
1, 2 tan
y . x (x 1) (h) 1 equation is y 12 2 d1
dx g 2(x) d
[g(x)] 2
dx
2(5)
( 4)3 1) 2, or y x f 2(x) g 2(x) 2. g 2(x) 2, the derivative is 2 f (2) At x
(c) g(x)] g (x) g(x)] f (x)g (x) g (3) 2 g(x) f (x) 3, the derivative is f (3)g (3) g(3) f (3) 5. 5 f (1)
(b) 15 ( 4)(2 )
8. 10
3 2 At x g(x) f (x) f (x)g (x)
[g(x)]2 2, the derivative is g(2) f (2) f (2)g (2)
[g(2)]2
74
3 4 37
.
6 (2) 1
3 (8)( 3)
(2)2 22 3 17 5 f (x) g (x) 1, the derivative is
g (1) d
f (x)g 3(x)
dx 1
3 5 8
3 d g 3(x) f (x) dx 3 f (x)g 2(x)g (x) 1.
d
dx f (x) g 3(x)
d
dx At x (2)( 3) 5 17 g(x)] g 3(x) f (x)
g 3(x) f (x) 0, the derivative is 3 f (0)g 2(0)g (0) 3(1)(1)2
d f (x)
(d)
dx g(x) d
dx 2g(x) g(x)] 82 f (x)[3g 2(x)] g(x) (3)(5) 1
3 g 2(2) d
[5 f (x)
dx At x 3, the derivative is f (3) d
[ f (x)
dx At x f (x) (8) g(2)g (2) 68 57. (a) d
f (x)
dx g 2(x)] g 2(x) 10
3 2 f (x)
2
.
3 d2
[ f (x)
g 2(x) dx 2, the derivative is f 2(2) 1
2
3 2g (x)
[g(x)]3 g(x) g(x)g (x) f 2(x) [ 4.7, 4.7] by [ 3.1, 3.1] d
(b) [ f (x)
dx [2f (x) 2 f (x) f (2) f (2) At x 2 f 2(x) 1 At x d
[2 f (x)]
dx dx 5
.
32 1 d
dx 2 Graphical support: 56. (a) 10
64 f (x) f (x) (x 1 3d 2[g(x)] 3, the derivative is and passes through (1, 2). Its
1 . 2, or 2.
1 1 6(2 2) 2g (3)
[g(3)]3 and passes through The normal line has slope 1 28 At x (1, 2). Its equation is y 4 1
3 2 f (2) xd
x
4 dx 4 1. 1
3 (1)3(5) 6. g 3(0) f (0) 99 Section 3.6 [g(x) d d 1] f (x) f (x) [g(x) dx d
f (x)
(c)
dx g(x) 1 1]2 [g(x)
[g(x) 58. For y 1] d
dx sin 2x, y (cos 2x) (2x) dx 2 cos 2x and the slope at the origin is 2. 1] f (x) f (x)g (x)
[g(x) 1]2 For y x
2 sin , y cos 1
x
cos and the
2
2 xdx
2 dx 2 1
. Since the slopes of the two
2
1
tangent lines are 2 and
, the lines are perpendicular and
2 slope at the origin is At x 1, the derivative is [g(1) 1] f (1) f (1)g (1)
[g(1) 1]2 (4 At x and y 1
3 1
3 8
(5)
3 g (1) f (0) f (x)] 2 2[g(x) f (x)] 59. Because the symbols 40
3
3d dx 2[g (1) f (1)]
[g(1) f (1)]3 8
3 (4 1
3 values.
[g(x) 6
1 3)3
d
dx 6. 60. Velocity: s (t)
2 bA sin (2 bt)
acceleration: s (t)
4 2b 2A cos (2 bt)
33
jerk: s (t) 8 b A sin (2 bt)
The velocity, amplitude, and jerk are proportional to b, b 2,
and b 3, respectively. If the frequency b is doubled, then the
amplitude of the velocity is doubled, the amplitude of the
acceleration is quadrupled, and the amplitude of the jerk is
multiplied by 8.
d
2
37 sin
(x
dt
365 61. (a) y (t) d
(25)
dt 101) f (x g(x))(1 g (x)) 37 cos 2
(x
365 101) d2
(x
dx 365 At x 0, the derivative is 37 cos 2
(x
365 101) 2
365 f (0 g(0))(1 g (0)) f (0 1) 1 4
f (1)
3
14
33 g(x)] f (x)] g(x))] f (x g(x)) [x dy dy
du
, , and
are not fractions. The
dx du
dx individual symbols dy, dx, and du do not have numerical 1, the derivative is
2 d
[ f (x
dx 1
x is shown.
2 [ 4.7, 4.7] by [ 3.1, 3.1] g ( f (x)) f (x) 2[g (x) f (x)]
[g(x) f (x)]3 (g) 2x 1
.
9 0, the derivative is d
[g(x)
dx At x the curves are orthogonal.
A graph of the two curves along with the tangents y 1. f (1)g (0) g ( f (0)) f (0) (f) 1)2 0, the derivative is d
g( f (x))
dx At x 8
3 f (g(x))g (x) f (g(0))g (0)
(e) (3) (4
9
9 d
(d)
f (g(x))
dx 1
3 1) 1
3 74
2
cos
(x
365
365 101) Since cos u is greatest when u
4
.
9 y (t) is greatest when
x 101) 2
(x
365 0,
101) 2 , and so on,
0, or 101. The temperature is increasing the fastest on day 101 (April 11).
(b) The rate of increase is
y (101) 74
365 0.637 degrees per day. 0 100 Section 3.6
d
dt 62. Velocity: s (t) 1 1 4t 4
2 At t d
(1
4t dt 21 66. dT
du dT dL
dL du 4t 1 1 2
m/sec
5 4(6) 1 2 4t 2 6, the velocity is 1 4t
d
dt (1 4t) (2)
(
21 1 d
dt 1 4t 4t)2 1 d
(1
4t dt 1 2 2 4t) 4t y
4t 1 4t 4
4t)3/2 (1 L
g dL
(kL)
dL g 1
(kL)
g k d
f (g(x))
dx 6, the acceleration is f (g(x)) has a horizontal tangent at x f (g(1))g (1) dv
dt 63. Acceleration
k
2 (k 4
4(6)]3/2 [1 dv ds
ds dt dv
(v)
ds 4
m/sec 2
125
d
(k
ds 0, so either g (1) s tangent at x 1, or the graph of y tangent at u 1, then 0 or f (g(1)) 0. This g(x) has a horizontal
f (u) has a horizontal g(1). s) (k s) k2
, a constant.
2 s) kT
2 f (g(x))g (x). If the graph of means that either the graph of y
At t L
g 67. No, this does not contradict the Chain Rule. The Chain
Rule states that if two functions are differentiable at the
appropriate points, then their composite must also be
differentiable. It does not say: If a composite is
differentiable, then the functions which make up the
composite must all be differentiable.
68. Yes. Note that 4 1 2 L
g 2 d
dt L
(kL)
g d
2
dL 2 1 Acceleration: s (t) 4t) 69. For h 1: 64. Note that this Exercise concerns itself with the slowing
down caused by the earth’s atmosphere, not the acceleration
[ 2, 3.5] by [ 3, 3] caused by gravity. For h
Given: v 0.5: k
s Acceleration dv
dt dv ds
ds dt
k dv
(v)
ds (v) dv
ds dk
s ds s
d [ 2, 3.5] by [ 3, 3] s (k) k ds k d
ds For h 0.2: s ( s)2 s
k
(2 s) k s s
k2
2s 2 [ 2, 3.5] by [ 3, 3]
As h → 0, the second curve (the difference quotient) ,s 0 Thus, the acceleration is inversely proportional to s 2.
65. Acceleration dv
dt df (x)
dt df (x) dx
dx dt f (x) f (x) approaches the first (y 2 cos 2x). This is because
2 cos 2x is the derivative of sin 2x, and the second curve is
the difference quotient used to define the derivative of
sin 2x. As h → 0, the difference quotient expression should
be approaching the derivative. 101 Section 3.7
70. For h 1: 72. dG
dx d
dx
1
2 2x
uv
2 uv uv d
dx x(x d
(x 2
dx
cx
A
G cx) d
dx c)
2x
2x x2 c
2 cx
x cx (x c) 2 x(x c) [ 2, 3] by [ 5, 5] For h 0.7: s Section 3.7 Implicit Differentiation
(pp. 149–157)
Exploration 1
1. 2x
dy
dx [ 2, 3] by [ 5, 5] For h 2y An Unexpected Derivative 2xy 2yy 1 (provided y 0. Solving for y , we find that x). 2. With a constant derivative of 1, the graph would seem to be
a line with slope 1. 0.3: 3. Letting x 0 in the original equation, we find that
y
2. This would seem to indicate that this equation
defines two lines implicitly, both with slope 1. The two
lines are y x 2 and y x 2.
[ 2, 3] by [ 5, 5] As h → 0, the second curve (the difference quotient)
approaches the first (y
2x sin (x 2)). This is because
2x sin (x 2) is the derivative of cos (x 2), and the second
curve is the difference quotient used to define the derivative
of cos (x 2). As h → 0, the difference quotient expression
should be approaching the derivative.
71. (a) Let f (x) 4. Factoring the original equation, we have
[(x y) 2][(x y) 2] 0
x y 2 0 or x y 2 0
y x 2 or y x 2.
The graph is shown below. x. Then
d
u
dx d
f (u)
dx f (u) du
dx d
u
du du
dx u
u.
u The derivative of the absolute value function is
positive values,
at 0. So f (u) 1 for negative values, and undefined
1,
1, u
u 0
0. g (x) u
evaluates.
u (2x)(x 2 9)
d
x2 9
(x 2 9)
x2 9
dx
x2 9
d
( x sin x)
dx
d
d
x (sin x) (sin x)
x
dx
dx
x sin x
x cos x
x Note: The expression for g (x) above is undefined at
x 0, but actually g (0) lim g(0 h→0 h)
h g(0) lim h→0 h sin h
h Therefore, we may express the derivative as
x cos x
g (x)
0, x sin x
,
x 5. At each point (x, y) on either line, x 0 x 0. 0. dy
dx 1. The condition y x is true because both lines are parallel to the line
y x. The derivative is surprising because it does not
depend on x or y, but there are no inconsistencies. Quick Review 3.7 But this is exactly how the expression (b) f (x) 1 for [ 4.7, 4.7] by [ 3.1, 3.1] 1. x y2
x
x
y1 0
y2
y
x, y2 [ 6, 6] by [ 4, 4] x 102 Section 3.7 2. 4x2 9y 2 36 9y 2 36 7. y sin x
4x 2 36 y2 4x 2
3
2
3 y1 4
(9
9 y xy y (sin x x)y y x cos x y x cos x x 2) x2 9 y
2
3 2 9 xy y sin x 2 9 y x cos x x , y2 9 x cos x sin x x y) 8. x(y 2 y (x 2 y) xy 2 y (x 2 y 9. [ 4.7, 4.7] by [ 3.1, 3.1] x2 4y 2
2y) 0 10. y x
,y
22 3 x2 x3
x x
2 y1 x+ x 1/2 x x 2/3 x
x x 2/3
x 3/2 x 3/2 x
2 y x 1/2(x x 1/3)
x 1/2x x 1/2x 1/3
x 3/2 x 5/6 x) 0 2y)(x (x x2
3 x(x x) xy2 y 3. x 2 3/2 5/6 x Section 3.7 Exercises
1.
[ 6, 6] by [ 4, 4] 4. x 2 2 y
y2
y
y1 9
9 2. x2
9 x2
9 x2, y2 3.
9 y2 4.
5. [ 4.7, 4.7] by [ 3.1, 3.1] 5. x 2 2 y
y2
y
y1 2x
2x 3
3 x2
2x 3 x2
2x 3 x2, y2 6. 2x 3 dy
dx
dy
dx
dy
dx
dy
dx
dy
dx dy
dx x2 [ 4.7, 4.7] by [ 3.1, 3.1] 6. x 2y 2xy
x 2y
y 4x y 4x y 4x y
x2 2xy dy
dx 9 5/4
x
4 3 ( 3/5)
x
5 d
x 3/5
dx 3 8/5
x
5 1 d3
x
dx d 1/3
x
dx 1 (1/3)
x
3 1 1 2/3
x
3 d4
x
dx d 1/4
x
dx 1 (1/4)
x
4 1 1 3/4
x
4 1
d
d
(2x 5) 1/2
(2x 5)( 1/2) 1 (2x
2
dx
dx
1
(2x 5) 3/2(2)
(2x 5) 3/2
2
d
(1
dx
2
(1
3
2
(1
3 4(1
7. 9 (9/4) 1
x
4 d 9/4
x
dx 6x)2/3
1d 6x)(2/3)
6x) 1/3 6x) dx (1 ( 6) 1/3 d
(x x 2 1)
dx
d
x
x2 1
x2
dx
d
x (x 2 1)1/2 (x 2
dx
12
x
(x
1) 1/2(2x)
2 x 2(x 2 1) 6x) 1/2 (x 2 1 d
(x)
dx 1)1/2
(x 2 1)1/2 1)1/2 2xy Note: This answer is equivalent to 2x 2 1 x2 1 . 5) Section 3.7 dy x d
dx dx x2 (x 2 8. 1)1/2 (x d
dx 1)1/2 x x (x 2 x2 1
12
(x
2 x
x2 x2 d
dx (x 2 1) 1/2 1)1/2 x x2 12. d2
(x )
dx 1 y xy
dx y
dx x y (2x) (x 1 x
1
(x 2 (x 2x
x 2y 9.
d2
(x y)
dx x 2 dy y(2x) dx x(2y)
x xy 2 6 d
(xy 2)
dx
dy y 2(1) dx
2 dy 2xy dx dy x
d3
(x )
dx
2 d3
(y )
dx
2 dy 2 dy dx
dy 3x
3y y 3 dx (3y 2 3y 18x
18x) dx
dy
dx
dy d2
y
dx
dy 2y
2y x
x
d
dx
(x dx
dy
dx
dy
dx y 2) (2xy
2xy 18x
18y x(x
x y)2 x y y) x y x3 x 2y x y x2 2 1) 3x 2 y2 13. dy
dx 6x 14.
(x
1)2 y (x x d
(x)
dx y)2 dy dx 3x 2 1 dx
dy dy
dx 1 3x d
3(2x 1/2
dx (2x x2 1/2 (2x 2 x
dy
dx 2xy
1 1/2 1)
1)
1) 1/3 4/3 d dx
4/3 (2x (x 3/2 (2x 1/2 1) 1/2 3/2 (x 15. 2xy 2 d
(1 x 1/2)1/2
dx
1
d
(1 x 1/2) 1/2 (1 x 1/2)
2
dx
1
1 1/2
(1 x 1/2) 1/2
x
2
2
1
(1 x 1/2) 1/2x 1/2
4 1)(1) 1)2
1
y(x 1)2 d
(y)
dx dy 1 dx 3x 2 18y y)2 y(2x) dx 2 18y dx d2
(x y)
dx dy (x
3x y)2 y x x2 3y 2 18x
6y x 2 1
1
1)(1)
(x y x 2(x 18y (1) dx x dx x(x dx
dy y2 2xy x dx y)2
dy y dy y 2) 1
1
x
x dy Alternate solution: 3x 2 dx y2 (2xy dx dy x dx 18xy dx
dy 11. y y d3
(x )
dx d
(18xy)
dx
dy 2x (x y)2
x 0 dx 10. y dx x2 x(x d
(6)
dx dx
dy
x 2)
dx
dy (2xy 3 dy (x
2y 3/2 1) x dy y) 1 y)2 2x 1)3/2 2 (x dx (x 1 x2
1)(x 2 1)1/2 2 dy y) 1 2x 4/3 d
3(csc x)3/2
dx
9
d
(csc x)1/2 (csc x)
2
dx
9
(csc x)1/2( csc x cot x)
2
9
(csc x)3/2 cot x
2 ) 1) dy
dx y y dy
dx 103 104 16. Section 3.7
dy
dx 17. d
[sin (x
dx
5
[sin (x
4
5
[sin (x
4 x (c) Differentiating both sides of the given equation 5)]5/4
5)]1/4 d
sin (x
dx 5)]1/4 cos (x f (x) 5) dy
dx x 22. (a) If g (t) sin y d
(x)
dx
dy
dx 4 4 t 19. d
(4t 1/4
dx 4) 3/4 t x 1
t 3/4 g (t) tan xy t 3/4 dy
dx
dy
dx
dy
dx (c) If g (t) 0 g (t) 0 4 1 (sec2 xy)(y) 1 d
(x)
dx 1 sin y y sec2 xy (cos y) (cos y x) x2 x 1/3 d2
(x )
dx 2x 1 y 1 x x y2 y
cos y d2
(y )
dx 2yy 0
2x
x
y y x
y d
dx 1 4/3
x
3 y x 1/3 x x
y y2 . y2
y3 7, then 3 2/3
x and f (x)
2 (x)(y )
y2 x2 f (x) 1 3/4
t
, which
16 d
(1)
dx (y)(1) which contradicts the given equation f (x)
(b) If f (x) 4t 1/4 and 1 3, then and f (x) 9 5/3
x
10 1 , which matches the given 1 1/4
t , then g (t)
4 2yy y cos y t 3/4 y 1 3 2/3
x
2 21. (a) If f (x)
f (x) dy
dx
dy
dx 3/4 16 5/4
t , then g (t)
5
1 Conclusion: (a) and (c) could be true. y
x d
(xy)
dx
dy
x
(y)(1)
dx dx 7 .
t contradicts the given equation. xy d
(sin y)
dx
dy t (d) If g (t) 23.
x 4 equation. 0 x sec2 xy
1
cos2 xy
x t 3 7/4
t
, which is not
4 gives g (t) 0 d
(tan xy)
dx
d
1 sec2 (xy) (xy)
dx
dy
(sec2 xy)[x
(y)(1)]
dx 20. 1
, which matches the
t 3/4 (b) Differentiating both sides of the given equation d
(x)
dx (sec2 xy)(x) , which matches 4, then consistent with g (t) 1 1/3 given equation. d
(sin y)
dx
dy
cos y
dx
1
sec y
cos y 1 x Conclusion: (b), (c), and (d) could be true. g (t)
18. 6, then f (x) the given equation. d
(tan y)
dx
dy
sec2 y
dx
1
cos2 y
sec2 y 1 1 4/3
x
.
3
3 2/3
x
2 (d) If f (x) 1 4/3
x
, so it must be true
3 gives f (x) that f (x) 5) tan y d
(x)
dx 1/3 x x 1/3 , which matches the given equation. Since our original equation was x 2
substitute 1 for x 2 y 2, giving y y2
1
.
y3 1, we may Section 3.7
x2/3 24.
d 2/3
(x )
dx y2/3 1 d 2/3
(y )
dx 2 1/3
x
3 0 1 /3 y 1/3 1 /3 x
y y x
y 1/3 d
dx y x 1y
3x 2/3 d 1y
3x 2/3 xy x2
y 1/3
x Slope at (3, 3x 4/3y 1/3 substitute 1 for x
y2 25. x2 d2
(y )
dy y 2/3 , giving y 3x4/3y1/3 . 2x y y (x 2)(2y) (x 1) x 1
y (y 2)(2x) 0 2 y2
d2
(y )
dx d
(2x)
dx y (2y
y
y 2y 2x
2y) 2)y 2 Slope at ( 2, 1):
(a) Tangent: y
(b) Normal: y (y 1) 2 (y 1) 2 1
1)3 y
1
y 1 1) 3 or y 1)
4y 1 d
(1)
dx
dy
24
0
dx
dy
(2y 4)
dx
dy
dx dy
dx 1
1 1(x 6
1
x
3 8
3 0
d
(0)
dx 0
2
1
y 2 1 2 (x 3x 3 or y d
(4y)
dx 1) 1
y1
d1
dx y 1 (y 2y y
x 3 2x 1
d
(2x
dx 2xy 2
2x 2y 1
(x
3 30.
2x, we may
2x 1)
1, 2xy 2 3(x (b) Normal: y 1)2 (x
y3 dy
dx 4
x
3 ( 4) or y d
(9)
dx (a) Tangent: y Since our original equation was y
x
write y 2 (x 1)2 (x 2 2x) (x 2
1
which gives y
3. d2
(y
dx 25
4 9 dy
dx
dy
dx
3
Slope at ( 1, 3):
1 2 26. y 3
x
4 3) 2x 2y y2 2 3) x 2y 2 2 y2 ( 4) or y 3
4 3
4 d 22
(x y )
dx 2x 2
x1
2y
y
dx 1
dx y
(y)(1) (x 1)y
y2 y 29
7 x
y 4
(x
3 29. 4
x
7 0 (b) Normal: y
d
(2x)
dy 1
2 3 or y 2) 3
(x
4 (a) Tangent: y 7
x
4 3 or y y
x d
(25)
dx 4): 1, we may
1 2x d2
(x )
dy 2yy y 2/3 2x
2y 25 d2
(y )
dx
dy
2x 2y
dx
dy
dx y 2/3 Since our original equation was x 2/3 y2 d2
(x )
dx 2/3 1/3
y
1x y
3 x4/3y2/3 2/3 x2 28. y xy x 2/3 1 2) 4
(x
7 (b) Normal: y (y)(1) 4/3 2/3 3 7
(x
4 (a) Tangent: y y dx x (x) 1 y2 xy d2
d
d2
d
(x )
(xy)
(y )
(1)
dx
dx
dx
dx
dy
dy
2x x
(y)(1) 2y
0
dx
dx
dy
(x 2y)
2x y
dx
dy
2x y
dx
x 2y
2(2) 3
7
Slope at (2, 3):
2(3) 2
4 d
(1)
dx 2 1/3
y
y
3 x2 27. 2)
2) 1 or y
1 or y x
x 1
3 105 106 Section 3.7
6x 2 31.
d
(6x 2)
dx d
(3xy)
dx
dy
12x 3x
dx
dy
3x
dx 2y 2 3xy 17y 6 0 34. x sin 2y d
d
d
(17y)
(6)
(0)
dx
dx
dx
dy
dy
(3y)(1) 4y
17
00
dx
dx
dy
dy
4y
17
12x 3y
dx
dx
dy
(3x 4y 17)
12x 3y
dx
dy
12x 3y
dx
3x 4y 17 y cos 2x d
(2y 2)
dx d
(x sin 2y)
dx (x)(cos 2y)(2) dy
dx (sin 2y)(1)
dy
dx
dy
(cos 2x)
dx
dy (y)( sin 2x)(2)
(2x cos 2y) d
(y cos 2x)
dx (cos 2x) dy
dx 2y sin 2x sin 2y 2y sin 2x
2x cos 2y sin 2y
cos 2x dx 6
(x
7 (a) Tangent: y 1) 7
(x
6 (b) Normal: y 2x 23
3 (a) Tangent: y d
(sin y)
dx
dy
cos y
dx
dy
cos y)
dx
dy
dx 2y(1)
(2x 2 d
(5)
dx (b) Normal: y y 3 : 2
2 (x (x 1
2 4 0
2 cos ( x Slope at (1, 0): d
(2 )
dx x 4 2 2 (x or y 2x
1
x
2 or y 2 2 sin ( x 2 cos ( x 5
8 y) d
2 sin ( x
dx y)
dy
dx y) 2 cos ( x y) 2 cos ( x y)
1 2 cos ( x y) 2 cos
2 cos (a) Tangent: y 1 cos 2 dy
dx
dy
dx
dy
y)]
dx
dy
dx 4y [1 sin ( ) 2 cos ( ) y 2 1 1) 2 ( 1)
2( 1) 0 or y 2
2x 2 0
1
(x
2 (b) Normal: y 2y 1) x
2 0 or y 1
2 2y
2x 1)
1) sin 2x 2x x3 4 35. 2x 3 y 2 2 0 0 2( /2)
2(1)
cos( /2) (a) Tangent: y
(b) Normal: y 0 ( 1) (a) Tangent: y 4(2) sin y d
2 (xy)
dx Slope at 1, 2 2 3 2xy dy
dx 3 42 ( )(1) 7
6 : 2 (b) Normal: x
33. 23 6
7 , 5 d
2 (y 2 )
dx
dy
3(y)(1) 4y
dx
dy
( x 3 4y)
dx
dy
dx Slope at ( 3, 2): 2x 2y 2 3xy 2 Slope at 7
x
6 0 or y d
3 (xy)
dx dy
3(x)
dx 6
7 6
x
7 0 or y 1) x2 32.
d2
(x )
dx 12
14 12( 1) 3(0)
3( 1) 4(0) 17 Slope at ( 1, 0): 2 2 2
2 x sin y 0 d2
(x cos2 y)
dx d
(sin y)
dx
dy
(x 2)(2 cos y)( sin y)
dx 2 or y or y 36. x 2 cos2 y cos y x (2x 2 cos y sin y 2
2 dy
dx cos y)
2 cos y d
(0)
dx (cos2 y)(2x)
dy
dx 2x cos y
2x 2 cos y sin y Slope at (0, ): 1 2(0) cos
2(0)2 sin (a) Tangent: y
(b) Normal: x 0 (cos y) 2x cos2 y
1 2x cos y
2x2 sin y 0 dy
dx 0 107 Section 3.7
y4 37. (a) y2 d4
(y )
dx
dy
4y3
dx
dy
3
(4y
2y)
dx
dy
dx
3
3
At
,
:
4
2 x2 39. (a) ( 1)3(1)2 d2
d2
(y )
x
dx
dx
dy
2y
2x
dx (x 3)(2y) 2x
2y x (2x 3y 2y3 y 2 2 2
4
3
4
3 4 At 33
4 2 3 y3 2y 1 3
4
13
2 1
2 1
4 4
4 1 5 t y (2 Tangent: y
Normal: y x) dy
dx
dy
x)
dx
dy
dx 1
(x
2 1)
1) xy . Hence,
5 2 , and 1 xy (y)(1) 0 x)y y (3y 2 y
3y 2 x
y
d
dx 3y 2 x
(3y 2 x)(y ) (y)(6y y
(3y 2 x)2
y xy
3y 2 y
2
(3y
x)2 y 3x 2
2 y y2
2 3x
2y(2 1 or y 5
2
1 3 3x 2 4
2 1). The d
( 1)
dx d3
(x )
dx x)] 3(1)2 (1)2
2(1)(2 1) 2(x 1 d
(xy)
dx 3y 2y
x x)(2y) 2y(2 Slope at (1, 1): 1 y 1 = 0 are d3
(y )
dx 2 d2
[y (2
dx y y3 [ 1.8, 1.8] by [ 1.2, 1.2] 1 1)(y 2 2y + 1 as (y . (b) Parameter interval: 3
2 1 is one solution, and there are three possible yvalues: 1,
2 (2 3
2 1, or (1)
4(1)( 1)
2(1) 3
1 (y2)( 1) 3x 2y 2
sin y 2y 2 1 y 3 (b) 38. (a) 2x 3y 0. Clearly, y solutions of y 2 2 3x 2y 2 1 we may factor y 3 3
4
1
2 dy
dx
dy
dx 2, we have y 3 40. (a) When x 1 31
,:
42 Slope sin y) dy
dx 3( 1)2(1)
2( 1)3(1)
sin
3
The slope of the tangent line is .
2 33 3
3
2 ( sin y)( ) Slope at ( 1, 1): 4
3 d
cos ( y)
dx (y 2)(3x 2) dy
dx 1. cos( y) d 32
(x y )
dx 3 Slope x 3y 2 (b) 2x
4y3 cos ( ) is true since both sides equal y
x) 2
2x 1 or y (b) One way is to graph the equations y Since we are working with numerical information,
there is no need to write a general expression for y in
terms of x and y. 1
1
x
2 3
2
x3
2 1) x . To evaluate f (2), evaluate the expression for y using
x 2 and y f (2) 1:
1 3(1)2 2 1
1 1 To evaluate f (2), evaluate the expression for y using
x 2, y f z (2) 1, and y
(1) 2(1)
[3(1)2 1:
3(1)2(1)
2]2 4
1 4 108 Section 3.7 41. Find the two points:
The curve crosses the xaxis when y 0, so the equation
becomes x 2 0x 0 7, or x 2 7. The solutions are
x
7, so the points are (
7, 0).
Show tangents are parallel:
x2 y2 xy (2x y) 0 2( 7) 2x 2
d
(2x 2)
dx 2
0 2 2(0) d2
(x )
dx 2x x d
(xy)
dx dy
dx y2 xy 2y d
(7)
dx dy
dx 0 d
(3y 2)
dx
dy
4x 6y
dx
dy
dx 5
d
(5)
dx 0
4x
6y (2x dy
dx y) 3x 2
3x 2
2y the slopes are 2x y
x 2y 2x
3y d3
x
dx At (1, 1), the slopes are dy
2y)
dx (x 3y 2 x3 d2
y
dx
dy
2y
dx
dy
dx 7 d2
(y )
dx (y)(1) 7
.
3 Second curve:
y2 x2 2 43. First curve: 2x y
x 2y The tangents at these points are parallel because they have
the same slope. The common slope is 2.
42. 7
7
and 2 ,
3
3 Note that these are the same points that would be
obtained by interchanging x and y in the solution to
part (a). 2(0)
7 7
3 The points are 0 27 7, 0): 7
,
3 7
7
7
7 d
(7)
dx 7 Slope at ( ( 2y) y2
y2
y2
3y 2
y 7 d2
d
d2
(x )
(xy)
(y )
dx
dx
dx
dy
dy
2x x
(y)(1) 2y
dx
dx
dy
(x 2y)
dx
dy
dx Slope at ( 7, 0): x 2 xy
( 2y)(y)
4y 2 2y 2 2 2
and
3 2
3
and respectively. At (1,
3
2 1), 3
respectively. In both cases, the
2 tangents are perpendicular. To graph the curves and normal
lines, we may use the following parametric equations for (a) The tangent is parallel to the xaxis when
dy
dx t 2x y
x 2y Substituting 0, or y : 2x.
First curve: x 2x for y in the original equation, we have
Second curve: x
2 x2 2 x
xy y
(x)( 2x) ( 2x)2
x 2 2x 2 4x 2
3x 2
x The points are 7
7
7
7 7
7
,2
and
3
3 5
cos t, y
2
32 t ,y 7
3
7
,
3 2 7
.
3 dx
can be obtained by interchanging x and y
dy
dy
dx
2y x
in the expression for . That is,
. The
dx
dy
x 2y
dx
tangent is parallel to the yaxis when
0, or
dy equation, 2y. Substituting equation, we have: t Tangents at (1, 1):
x 1 3t, y
x 1 2t, y 1 3t
Tangents at (1, 1):
x 1 3t, y
x 1 2t, y
1 3t (b) Since x and y are interchangeable in the original x 5
sin t
3 2y for x in the original [ 2.4, 2.4] by [ 1.6, 1.6] 1 2t
1 2t Section 3.7 s (t) a(t) d
(4
dt 9(4 44. v(t) 3
(4
2 6t)1/2 At t (6) 27(4 2, the velocity is v(2) 6t) 1/2 dv
dt t) 4(s t) 4(s t) 1/2 (v 4(s t) 1/2 1] [(8(s 1/2 x
3 0 x6
27 3x 3 0 54) 0 9x x 1)
t)1/2 1) 1] At x 1/2 (s 0 2 x
3 1 dt 9xy 23 x3 3
(x
27 x3 1/2 1/2 ds t) y3 x3 d
[8(s
dt 32(s 3y
y2 x3 36 m/sec and the 27
m/sec2.
4 acceleration is a(2)
45. Acceleration x2
x2
0, or y
.
3x
3
x2
Substituting for y in the original equation, we have:
3 6t)1/2] 1/2 6t) (b) The tangent is horizontal when 6t)1/2(6) dy
dx d
[9(4
dt v (t)
9
(4
2 6t)3/2 109 t) 3 0 or x
02
3 0, we have y 54 32 ft/sec2 y 13
(9 4)
3
3 2 0, which gives the point (0, 0), which is the origin. At x
132
(3 2)
3 3 3 3 3 3 3 2, we have 4, so the point other 3 than the origin is (3 2, 3 4) or approximately
46. y 4 4y 2 x d4
(y )
dx d
(4y 2)
dx
dy
dy
8y
4y 3
dx
dx 4 9x d4
(x )
dx 4x 3
4y 3 47. (a) 2): (c) The equation x 3 9(3)
4(2) 2(3)3
2( 2)3 x3
d3
d3
(x )
(y )
dx
dx
dy
dy
3x 2 3y 2
9x
dx
dx (3y 2 3(2)
Slope at (4, 2): 2
(2)
3(4)
Slope at (2, 4): 2
(4) 2x 3
2y 3 the line y
9x
4y 3 9(3)
4( 2) d
dx x2 48. 27
8 2) or 2x d
(0)
dx 9(y)(1) 0 dy
dx
dy
dx 9y 3y2 0 d
d
d
2 (xy)
(3y2)
(0)
dx
dx
dx
dy
dy
2x
2(y)(1) 6y
0
dx
dx
dy
(2x 6y)
2x 2y
dx
dy
2x 2y
xy
dx
2x 6y
3y x 0 9 (xy) 2xy d2
(x )
dx 27
8 9xy (4)2
3(4)
(2)2
3(2) 3 approximately (4.762, 3.780).
27
8 9( 3)
4( 2) 9x) x and we may find the desired point by to part (b). The desired point is (3 4, 3 9( 3)
4(2) y3 9xy is not affected by interchanging the xvalue and the yvalue in the answer 27
8 2( 3)3
2( 2)3 y3 interchanging x and y, so its graph is symmetric about 18x
8y 2( 3)3
Slope at ( 3, 2):
2(2)3 2): d
(9x 2)
dx 18x 2(3)3
Slope at (3, 2):
2(2)3 Slope at (3, (3.780, 4.762). 4x 3 dy
dx Slope at ( 3, 2 At (1, 1) the curve has slope
normal line is y 3x 2 9y 3x 2
3y 2 9x
10
5
8
4
4
8
5
10 3y
y2 x2
3x Substituting x 1(x 11
3(1) 1 1) + 1 or y 2
2 1, so the
x 2. 2 for y in the original equation, we have: x 2 2xy 3y 2 0
x
2x( x 2) 3( x 2)2 0
2
x
2x 2 4x 3(x 2 4x 4) 0
4x 2 16x 12 0
4(x 1)(x 3) 0
x 1 or x 3
Since the given point (1, 1) had x 1, we choose x
and so y
(3) 2
1. The desired point is (3,
2 3
1). 110 Section 3.7 49. xy
d
(xy)
dx
dy
x
dx 2x y 0 d
(2x)
dx d
(y)
dx
dy
(y)(1) 2
dx
dy
(x 1)
dx
dy
dx 0 b 2x 2 0
2 y 2
1 2, we wish to
2, that is, where 1 2y 1 0
2b 2x
2a 2y
b 2x1
a 2y1 2y Substituting 2y d
(x)
dx 1
dy
dx y b 2x1x But a 2y12 3 .
b 2x1 y1 a 2y1 (x x1). This gives: 2 a 2y12 b 2x12 b 2x12. a 2b 2 since (x1, y1) is on the dividing by a 2b 2 gives
x2
a2 (b) b 2x 2 b2 a 2b 2 d 22
(b x )
dx d 22
(a y )
dx
dy
2b 2x 2a 2y
dx
dy
dx d 22
(a b )
dx 0
2b 2x
2a 2y
b 2x 1
a 2y 1 y1 a 2y1y a2y12 b2x1x b2x12 a 2y12 b 2x1x a 2y1y hyperbola. Therefore, b 2x1x 2b3 b
2b b2 1
greater than if b
2 2b(x 1
or b
2
1
12
2
2 b2 ) b. This line intersects the xaxis at
1
, which is the value of a and must be
2 b) will be perpendicular when 1, which gives 2y 1 or 1
. The corresponding value of a is
2
1
3
b2
. Thus, the two nonhorizontal
2
4
3
normals are perpendicular when a
.
4 a 2y12 a 2b 2 since (x1, y1) is on the dividing by a 2b 2 gives y2
b2
y2
b2 y1y 2 b2 a y 2 y 1
x2
a2 a 2y1y x1x 52. (a) Solve for y:
x2
a2 0. The two normals at (b 2,
they have slopes x1). b2x12 b, or x (x This gives: line at (b 2, b) is y
2b b 2x1
a 2y1 But b 2x12 2bx b 2x
a 2y . 2y. Thus, the normal y 1. 1 The normal line at (x, y) has slope 3 a 2b 2, and y1y a2 a 2y 2 The slope at (x1, y1) is b 2x1x x1x y2
b2 The tangent line is y d2
(y )
dx
dy
2y
dx
1
2y b 2x12 b 2x1x ellipse. Therefore, a 2y1y 3 in the original equation, we have: 3)y x a 2y12 a 2y1y xy 2x y 0
2( 2y 3) y 0
2y 2 8y 6 0
2(y 1)(y 3) 0
y
1 or y
3
At y
1, x
2y 3 2 3
1.
At y
3: x
2y 3 6 3 3.
The desired points are ( 1, 1) and (3, 3).
Finally, we find the desired normals to the curve, which are
the lines of slope 2 passing through each of these points.
At ( 1, 1), the normal line is y
2(x 1) 1 or
y
2x 3. At (3, 3), the normal line is
y
2(x 3) 3 or y
2x 3.
( 2y y b 2x
a 2y x a 2y1y x 50. d 22
(a b )
dx The tangent line is y x 4 a 2b 2 The slope at (x1, y1) is 1
2 y) a 2y 2 dy
dx 1
the tangent has slope . Thus, we have
2 2(2 1 d 22
(a y )
dx
dy
2b 2x 2a 2y
dx y
x 0 is find points where the normal has slope y
x y2
b2 d 22
(b x )
dx y 2y
x1 Since the slope of the line 2x 2
1 x2
a2 51. (a) 1 b2 2
(x
a 2)
a2
b
x2 a2
a 1. a 2b 2, and Section 3.8
b f (x)
x→ g(x) (b) lim x2 lim a
lim x2 x→ lim x→ (c) lim
x→ f (x)
g(x) lim x→ x5 x dy
at
dx 1, which is 7. 2x 1 at the point (1, 2). The slope is x2
a2
x2 1
b
a x2 7. The slope of L is the reciprocal of the slope of its reflection 1 x2 8. 1
7 a2
x 1 y
gets reflected to become
x since a2 b
x
a x→ lim y
a2 x→ lim 6. The reflection of line L is the tangent line to the graph of a2 b
x
a x→ 111 Quick Review 3.8 2 a2
x2 1. Domain: [ 1, 1] 1 Range: s Section 3.8 Derivatives of Inverse
Trigonometric Functions (pp. 157–163)
Exploration 1 Finding a Derivative on an
Inverse Graph Geometrically
1. The graph is shown at the right. It appears to be a onetoone function At 1: , 22 2 2. Domain: [ 1, 1]
Range: [0, ]
At 1: 0
3. Domain: all reals
Range:
At 1: , 22 4 4. Domain: ( [ 4.7, 4.7] by [ 3.1, 3.1]
4 2. f (x) 5x
2. The fact that this function is always
positive enables us to conclude that f is everywhere
increasing, and hence onetoone. Range: 0, , 1] 2 2 At 1: 0 1 3. The graph of f is shown to the right, along with the
graph of f. The graph of f 1 is obtained from the graph of f
by reflecting it in the line y x. 5. Domain: all reals
Range: all reals
At 1: 1
6. f (x) y 3x y 8 3x x [ 4.7, 4.7] by [ 3.1, 3.1] 4. The line L is tangent to the graph of f 1 at the point (2, 1). y 8 8
3 Interchange x and y:
y
f 1 7. f (x) (x)
y [ 4.7, 4.7] by [ 3.1, 3.1] y 5. The reflection of line L is tangent to the graph of f at the
point (1, 2). 3 x x 8
3 x 8
3 3 x
y x 5
5 3 5 Interchange x and y:
y
f [ 4.7, 4.7] by [ 3.1, 3.1] 1 x3 5 (x) 3 5 x [1, )
, x
1
. It is .
y
7 112 Section 3.8 8. f (x) 8
x
8
y y
x 5. dy
ds 1 f
9. f (x) 2s 6.
2
x xy 3x 3)x 7. 2 x 3 1 4s 2s d
csc 1 (x 2
dx 3 1 y
(x 2 (x2
2x 1) x s
1 d
(5s)
1 ds (5s)2 5s s2 1 1 dy
dx x2 s 25s 2 1)
d
(x 2
1 dx 1)2 1)
2 4 2x 2 (x Note that the condition x x 2 x2 1) 2 0 is required in the last step. 2
3 x 8. x
arctan
3 y
x
,
3 tan y 4s d
sec 1 5s
ds 2 2 (x) 10. f (x) 1 1) (2) 2 1 2
y3 y 1) dy
ds Interchange x and y: 1 (2s 2 x f 1 d
(2s
ds
1 2 1 3x y (y 2s 8
x
8
x (x) 1) 1 Interchange x and y:
y d
sec 1 (2s
ds 3 tan y, 2 y 2 1 d
x
csc 1
dx
2 dx
1 dx 2 x2
2 x
2 2 y 2 dy
dx x2 x 4 2 Interchange x and y:
y
f 3 tan x,
1 (x) x 2 3 tan x, 2 9. 2 x dy
dt 1 1
d
sec 1
t
dt 1
t 2
1 Section 3.8 Exercises
1. dy
dx 1 d
cos 1 (x 2)
dx
1
x4 1 d2
(x )
dx
(x ) x4 1 1 1
t2 1 dy
dx 10. 1 1
d
cos 1
x
dx d
1 2 dx
x 1
1
1 1
x 2 1
x2 0 is required in the last step. dy
dt 1 d
3
sin 1 2
dt
t 1
x 1
1 1
x2 x t2 1 11. 1 dy
dt d
cot 1
dt 3
t2 6 6
t3 9
t4 d
3 2 dt
t2 1 2. 1 Note that the condition t 22 1
2x (2x) 12
t 1
t 1
t d1
dt t 2 t t4
1 t
1 ( 9
d
t)2 dt t 1 3. 4. dy
dt d
sin 1
dt dy
dt d
sin 1 (1
dt
1
2t t2 1 2t ( 2 t)2 1 1 t)
1 (1 d
(
dt 2 2 t) d
(1
2 dt
t) 1 t) 2 t (t 1) 2t 2 12. dy
dt d
cot 1
dt 1 1
t t 1 1
1 ( 1
12t 1 ) d
dt
1)
2 (t
1
2t t 1 t 1 1 113 Section 3.8 13. dy
ds d
(s
ds 1 (s)
2 d
(cos 1s)
ds s2) 1 ( 2s) 2 1 1 s (1 s) 1 14. dy
ds d
ds [sin 1 1 (2) 4x 2 1 (2x)]2 1 dy
dx 4x 2 sec 2 x, the slope at The tangent line is given by y
d
sec 1 s
ds 1 y 1 (2s) s2 (2x) 1 2 1 s2 1 1 sin 2 (2x)] s 19. (a) Since s2 2 s 2 1 dx 2 s2 2s 2d (2x)] 2 1
2 1 [sin [sin s 2 d
[sin 1(2x)] 1
dx 1 s2 1 2 dy
dx 1 s2)(1) 1 s s2
1 ( 18. 1 (b) Since s2 s 2x 1 2x 1
x2 1 , the slope at 1, s2 s
dy
15.
dx 1 d
(tan 1
dx x 1
1
1 1 x2 2 2 1) x2 x x 2 1 1 x2 x y 1 f x Note that the condition x dy
dx 1
d
cot 1
x
dx 5x 4 1 1) 4 1
.
2 , or 6x 2 1. Thus f (1) 3 and f (x) includes the point (1, 3) and (x) will include (3, 1) and the slope will be
1
. (We have
12 (x) is defined and differentiable at 5x 4 6x 2 1, which is never zero.) 21. (a) Note that f (x)
sin x 3, which is always between
2 and 4. Thus f is differentiable at every point on the
interval( , ) and f (x) is never zero on this interval,
so f has a differentiable inverse by Theorem 3. d
(tan 1 x)
dx
d1
dx x ) (3) 3. This is true by Theorem 3, because f (x) 1 is required in the last step. 1 1 and ( f 1 assumed that f 1 0 16. 1
1 + 12 . 1
. Thus, f 1(3)
12 x2 x is the slope of the graph is 12 at this point, the graph of 1 x x2 4 1 1 1 4 (b) Since the graph of y 1 1) (2x) x2 1
2 20. (a) Note that f (x)
f (1) 12. d
(csc 1 x)
dx d
(
1)2 dx x2 ( 1
x
2 y 1 2. 4 1, or 4 1
(x
2 The tangent line is given by y
ss , 1 is sec 2 1. 2 dy
dx 4 1 1
x2 1 ( 1
1
x2 1 ) 1
x2 1
x2 (b) f (0) cos 0 3(0)
f (0)
sin 0 3 x2 1 1;
3 (c) Since the graph of y 1 f (x) includes the point (0, 1) 1
x2 1 and the slope of the graph is 3 at this point, the graph
of y 1
1 0, x 1 x2 f 1 (x) will include (1, 0) and the slope will be 1
. Thus, f 1(1)
3 22. 0 The condition x 0 is required because the original
function was undefined when x 0.
17. dy
dx d
(x sin 1 x)
dx
1 (x) 1 sin 1 x x2 d
(
dx (sin 1
1 x 2) [ 2 , 2 ] by [ 4, 4] 1 x)(1)
2 1 x2 ( 2x) (a) All reals
(b) , 22 0 and (f 1 ) (1) 1
.
3 114 Section 3.8 22. continued (b) y (c) At the points x
(d) k , where k is an odd integer.
2 y 2 (c) None, since
30. (a) y 0. x2 x 1 0 (b) y 1 d
sec 1 x
dx 0 2 (c) None, since 1 d
csc 1 x
dx 0. x2 x 1 2π 1 x is
1. (b) None, since sin –2π 31. (a) None, since sin
undefined for x x 1 x is undefined for x (c) None, since –2 1 x is undefined for x (b) None, since cos 1 x is undefined for x 1 d
sin x
1 sin2 x dx
cos x which is (c) None, since 23. (a) v(t)
(b) a(t)
(c)
24. 1 which is always positive. 1
1 So cos t2
2t
which is always negative.
(1 t 2)2 x, x sin
1 sin (b) β 1 x x 2 . x α 2 d
cos 1 (x)
dx d
dx sin 2 1 x 1 tan
1 So tan x, x cot
cot (c) x2 1
d
cot 1 x
dx 1 d
sin 1 (x)
dx
1 0 25. 0. x cos
1 x2 1. α cos x is positive or negative.
dx
dt
dv
dt 1 1. β 1 1 depending on whether 1 d
cos 1 x
dx 33. (a) sin2 x 1 0. x2 1 32. (a) None, since cos d
sin 1 (sin x)
dx (e) f (x) 1 d
sin 1 x
dx 1. 1 1 x x 2 . β x
α d
tan 1 (x)
dx 2
d
0
tan 1 (x)
dx
1
1 x2 1
1 sec
So sec 1 x x, csc
csc 1 x 1 x
2 . 34.
d
26.
csc 1 (x)
dx d
dx 2 x (x) x 2 A
B 1 C 2 (b) y 2 (c) None, since
28. (a) y 1 d
sec 1 (x)
dx
1 0 27. (a) y sec d
tan 1 x
dx 1 0. x2 1 0 (b) y
(c) None, since
29. (a) y 2 d
cot 1 x
dx 1
1 x2 0. The “straight angle” with the arrows in it is the sum of the
three angles A, B, and C.
A is equal to tan 1 3 since the opposite side is 3 times as
long as the adjacent side.
B is equal to tan 1 2 since the side opposite it is 2 units
and the adjacent side is one unit.
C is equal to tan 1 1 since both the opposite and adjacent
sides are one unit long.
But the sum of these three angles is the “straight angle,”
which has measure radians. Section 3.9
35. s 6. Fold 1 log4 x 15 15 log4 x 12 12 log4 x log4 x ln Fold 2
s
2 Fold 3 s
s
s tan 1, so tan 1 tan 1 ln 19
ln 3 5t ln 5 1 1 tan 2 tan 1 t ln 5 s Section 3.9 Derivatives of Exponential
and Logarithmic Functions (pp. 163–171)
Exploration 1 Leaving Milk on the Counter 55 and solve;
72 30(0.98)t 4. ln (x 2
ln (x dy
dx d
(2e x)
dx 2e x dy
dx d 2x
(e )
dx e 2x dy
dx d
ex
dx dy
dx d
e 5x
dx e 5. dy
dx d 2x/3
e
dx e 2x/3 6. dy
dx d
e x/4
dx e dy
dx d
(xe 2)
dx dx
(e )
dx dy
dx d 2x
(x e )
dx d
(xe x)
dx (e x)(2x) tan x x 2e x
2 ln (x 2) ln x
x 4
2 ln (x 2)(x 2)
x2 2) 5. log2 (8x 5) log2 (23)x 5 log2 23x 15 2.71 Section 3.9 Exercises e x ln 7 4) ln 3
ln 2 ln 3 x d
(2x)
dx xd e dx 2e 2x ( x) 5x d dx x e ( 5x) d 2x
dx 3 5e 5x 2 2x/3
e
3 , ln 8
ln 5 ) ln 3 (x 2)(e x) ln (0.98) Quick Review 3.9 3. ln (e ln 2) 7. 17
30 0.343 degrees/minute. tan x x ln 2 4. 30 ln (0.98) (0.98)t. At t e ln 7 1) ln 3 x(ln 3 28.114 ln (0.98) ln 2. 7x ln 2x 3. 17
30 The milk reaches a temperature of 55 F after about
28 minutes. x 1 1.50 2x ln 3x
(x 1 8. ln 1. log5 8 ln (ln 5)
ln 5 2. 17
ln
30 t ln (0.98) dy
5.
dt
dy
dt ln 18 1. 55 17
30 (0.98)t t ln (ln 5) 3x 10. 2. The temperature of the room is 72 F, the limit to which y
tends as t increases.
3. The milk is warming up the fastest at t 0. The second
derivative y
30(ln(0.98))2(0.98)t is negative, so y (the
rate at which the milk is warming) is maximized at the
lowest value of t. ln 18 t 1. The temperature of the refrigerator is 42 F, the temperature
of the milk at time t 0. 4. We set y 18
ln 5
18
ln
ln 5 ln 5t 3. 2.68 18 5t
1 ln (12x 2) ln 19 From Exercise 34, we have
tan ln 3x ln 19 x
9. ln x 3 19 x ln 3 2. 0 ln (4x 4) 1 and 2, so s
2 (x 3)(12x 2)
3x ln 3x If s is the length of a side of the square, then
tan ln (12x 2) 3x 8. 5
,x
4 ln 3x 7. 3 ln x
s 15
12 3x 15 9.
10. x/4 d xe x 1 x/4
e
4 x
4 dx e2 ex [(x)(e x) (e x)(1)] ex dy
dx d
ex
dx e dy
dx d (x 2)
e
dx e (x ) xd dx
2 ( x) d2
(x )
dx e x 2x 2xe (x )
2 115 116 Section 3.9 11. dy
dx d
(x )
dx 12. dy
dx d1
(x
dx 13. dy
dx d
x
dx 14. dy
dx d 1e
x
dx 15. dy
dx dx
8
dx 16.
17. dy
dx 26. 2 2x e1 (1 dy
dx d
ln (2
dx
sin x
2 cos x dy
dx d
ln (ln x)
dx 30. 2 d
ln (2x 2)
dx
1
,x
1
x1 27. 2)x dy
dx 21 e)x 1 (1 21 dy
dx 29. 2)x 1 (1 dy
dx 28. 2) (1 e e)x 8x ln 8 d
9x
dx dy
dx 25. 1 x 9 d csc x
3
dx x d
dx (ln 9) ( x) 9 x ln 9 d
dx 3csc x (ln 3) (csc x)
3csc x (ln 3)( csc x cot x) 18. d cot x
3
dx 3 cot x 31. d
(ln 3) (cot x)
dx dy
dx 3cot x (ln 3)( csc 2 x)
dy
32.
dx 19. Use logarithmic differentiation.
x ln x ln y
ln y
d
(ln y)
dx
1 dy
y dx
dy
dx
dy
dx ln x ln x 33. ln x ln x
d
(ln x)2
dx
1
(2 ln x)
x
2y ln x
x
2x ln x ln x
x 34. 36. dy
dx ln x 1/ln x ln y 1
ln x
ln x ln y 1 y dy
dx dy
dx x 1/ln x ln y dy
dx 35. 20. Use logarithmic differentiation.
y e dy
dx d
(e)
dx 21. dy
dx d
ln (x 2)
dx 22. dy
dx d
(ln x)2
dx 23. dy
dx d
ln (x 1)
dx 24. dy
dx 10
d
ln
x
dx 37. dy
dx 38.
0, x 2 ln x dy
dx 39. dy
dx 0 1d 2
(x )
x 2 dx 1
(2x)
x2 d
(ln x)
dx d
( ln x)
dx
d
(ln 10
dx ln x) 2
x 1 d2
(x
1 dx x2 1d
ln x
ln x dx 1 x) (x) 1
x ,x 2 2
2x 2 cos x) 2x 1) 1
x 1
ln x 2 2) 1
d
(2
cos x dx 2 1) x d
(2x
2 dx 2x x2 1 1
x ln x (ln x)(1) 1 ln x d
d ln x 2
(log4 x 2)
dx
dx ln 4
1
2
2
1
ln 4 x
x ln 4
x ln 2 d
dx 2
(ln x)
ln 4 ln x
d2
dx ln 5
1
,x
2x ln 5 d
d ln x 1/2
(log5 x)
dx
dx ln 5
1
1
d
1
(ln x)
2 ln 5 dx
2 ln 5 x
d
log2 (3x
dx
3
,x
(3x 1) ln 2 1) (3x 1
,x
x 40. 0
1
x 1
,x
x 0 dy
dx 0 1
d
(3x
1) ln 2 dx 1) 1
3 1d
d
log10 (x 1)1/2
log10 (x 1)
2 dx
dx
1
1
d
1
(x 1)
,x
2 (x 1) ln 10 dx
2(x 1) ln 10
1
d
log2
x
dx d
( log2 x)
dx d
1
1
d
(log2 x)
dx log2 x
(log2 x)2 dx
1
1
1
or
(log2 x)2 x ln 2
x(ln 2)(log2 x)2
d
(ln 2 log2 x)
dx
1
1
(ln 2)
,x
x
x ln 2 1 1
,x
x ln 2 0 ln 2
x(ln x)2 d
dx (ln 2) (log2 x)
0 d
1
d
log3 (1 x ln 3)
(1
dx
(1 x ln 3) ln 3 dx
ln 3
1
1
,x
(1 x ln 3) ln 3
1 x ln 3
ln 3
d
(log10 e x)
dx d
(x log10 e)
dx log10 e 1
ln 10 2 ln x
x 0 1 1 2) 1 3cot x (ln 3)(csc 2x) y d
(x
2 dx x cos x) d
(x ln x
dx ln x 1 2) d
ln (x 2
dx 1 3csc x (ln 3)(csc x cot x)
dy
dx d
ln (x
dx d
ln 10x
dx d
(x ln 10)
dx ln 10 ln e
ln 10 x ln 3) Section 3.9
41. The line passes through (a, e a) for some value of a and has
e a. Since the line also passes through the origin, slope m y ea
, so a
a xe x, we have y (x)(e x) (e x)(1) (x has slope m (a 1
(x
1)e a (a includes the point (0, 0), so we have: need to use a
y
43. y 0) 46. 0e 0, or y ln y y ln (sin x)(1) 47. dA
dt ln (sin x)] ln y (tan x)(ln x) At t d
[(tan x)(ln x)]
dx
1
(tan x)
(ln x)(sec 2 x)
x
tan x
y
(ln x)(sec 2 x)
x
tan x
x tan x
(ln x)(sec 2 x)
x dA
dt 1
1
(2x)
2 x2 1
1
x 2d
ln (x
3 dx 21
(1)
3x 1 x
x2 1) 2
1 3(x xx
11
(x 1)2/3 x 1) x
x2 2
1 3(x 1) d 1 t/140
dt 2
d
20 2 t/140
dt (2 ln (x tan x) 1d
ln (x 2
2 dx 1) 20 20(2 x tan x 2
ln (x
3 1) 2 20 (2 ln y d
ln y
dx
1 dy
y dx
dy
dx
dy
dx y dy
dx ln (sin x)] (sin x)x[x cot x 1
x dy
dx x(x 2 1)1/2
(x 1)2/3 1
ln (x 2
2 d
ln x
dx 1 dy
y dx d
[x ln (sin x)]
dx
1
(x)
(cos x)
sin x y[x cot x ln x d
ln y
dx x ln (sin x) d
ln y
dx
1 dy
y dx
dy
dx
dy
dx ln ln y ln (sin x)x 6
5(2x 5) x(x 2 1)1/2
(x 1)2/3 ln y x. 6
5(2x 5) 3)4(x 2 1) 1/5
(2x 5)3 x x2 1
(x 1)2/3 y t/140 )(ln 2) d
dt t/140 )(ln 2) t
140
1
140 t/140 )(ln 2)
7 2 days, we have
(2 1/70 )(ln 2)
7 0.098 grams/day. This means that the rate of decay is the positive rate of
approximately 0.098 grams/day.
48. (a)
(b) d
ln (kx)
dx 5)] 3
1
(2)
5 2x 5 2x
5(x 2 1) 3) 2x
5(x 2 1) 3) 3 ln (2x 5) 1
1
(2x)
5 x2 1 4
5(x
(x 4
5(x (sin x)x ln y 44. dy
dx 0. The equation of the normal line is 1
(x
(0 1)e 0 41
5x 3 y 1) 3) 3d
ln (2x
5 dx 1) dy
dx ln (x 2 3) 4d
ln (x
5 dx 1 dy
y dx ae a. The desired normal line 1
0
(0 a) ae a
(a 1)e a
a
0
ae a
(a 1)e a
1
0a
ea
(a 1)e a
1
a 0 or
ea 0
(a 1)e a
1
The equation
e a 0 has no solution, so we
(a 1)e a 1
[4 ln (x
5 1d
ln (x 2
5 dx 1
and its equation is
1)e a a) 1
(x 3)4(x 2 1)
ln
5
(2x 5)3 d
(ln y)
dx 1)e x, so the normal line through the point (a, ae a) y 3)4(x 2 1) 1/5
(2x 5)3 (x ln ln y 1. Hence, the slope is e and the equation is 3)4(x 2 1) 1/5
(2x 5)3 (x ln y 0
and we have
0 ex. 42. For y 3)4(x 2 1)
(2x 5)3 (x 5 y ln y the slope is also given by m
ea ea
a 45. 117 1d
kx
kx dx d
d
ln (kx)
(ln k
dx
dx
1
d
0
ln x
x
dx k
kx ln x) 1
x 1) 118 Chapter 3 Review
2x ln 2, f (0) 49. (a) Since f (x)
(b) f (0) lim f (h) h→0 f (0) 20 ln 2 lim 2h 20 h→0 h s Chapter 3 Review Exercises ln 2.
lim h→0 h (pp. 172–175) 2h 1 (c) Since quantities in parts (a) and (b) are equal,
lim 2h h→0 1 dy
dx d5
x
dx dy
dx d
(3
dx 3. dy
dx d
(2 sin x cos x)
dx
d
2(sin x) (cos x)
dx ln 2. h (d) By following the same procedure as above using
7x, we may see that lim g(x) 1.
2. h 7h h→0 1 ln 7. h y3 3 dy
dx 4 dy
dx d 2x
dx 2x 1, or a
dx
a
dx (d) y2 1. y1
dx
ln a, then
a will equal a x if and only if
dx So if y3
ln a y2 5. ds
dt 1, or a 12
x
2 d
dx k x and ds
dt 2
d
cot
t
dt
2
22
csc
t
t2 7. dy
dx a x if and d
dx 8. dy
dx 1
.
x c) 9. dr
d Therefore, at any given value of x, these two curves will
10. dr
d 53. (a) Since the line passes through the origin and has slope (b) The graph of y
y positive x e. Therefore, ln x x
for all
e 11. to see that 2d2
t dt t 1 3/2
x
2 e 1)2 x. e . Therefore, e is bigger. 1 d
sec (1
d 2x 3) 1 2
t 2t) 2
t2 1/2 x ) ( 2x 1)(1) 1 1 3 ) tan (1 d
tan 2 (3
d (2) 2x sec (1 3 ) tan (1 3 )(3) 3) 2 ) 2d ) 2 ) sec 2 (3
2 2 tan (3 d 2 ) )( 2 ) 2 2 ) sec (3 ) d2
(x csc 5x)
dx 5x 2 csc 5x cot 5x
dy
dx d
ln
dx 13. dy
dx d
ln (1
dx 14. dy
dx d
(xe x)
dx 12. 2 sin (1 csc 2 1 d
(x 2x 1) (x)
dx
2
x (2x 1)
3x 1 (x 2)( csc 5x cot 5x)(5)
e (d) Exponentiating both sides of ln x e x, we have
e
e ln x
e x, or x e e x for all positive x e.
(e) Let x dy
dx 2t)( 2) d 1/2
(x
dx
1
1
2x3/2
2x 4 tan (3 e.
x or ln x 4
(2x x 2 tan (3 ln x lies below the graph of the line (c) Multiplying by e, e ln x 1)(2) 1 1 2 tan (3 x
.
e x
for all positive x
e (cos 2x)(2) (2x
1)2 sin (1
csc 2 3 sec (1 have perpendicular tangent lines. 1
, its equation is y
e 1)(2)
(2x 2t) x 1 1/2
x
2 e.
d
(ln x
dx d
sin 2x
dx (2x 1
1 2x 52. 2 cos 2 x d
cos (1
dt e.
a x ln a. This will equal y1 only if ln a d
dx 5 We conclude that the graph of y3 is a horizontal line at
y ln a.
a x if and only if y3 21x 6 2(cos x) (sin x) d
(2 sin x cos x)
dx 4. 0.693147 1.098613 1.386295 1.609439 dx
a
dx 21x 2 2 cos 2x ln a 0.693147 1.098612 1.386294 1.609438 (c) 3x 7) 1
4 Alternate solution: 6. 2 1
x
4 5x 4 a. (b) The graph of y3 is always a horizontal line.
a 1
x
4 7x 3 2 sin 2 x 50. Recall that a point (a, b) is on the graph of y e x if and
only if the point (b, a) is on the graph of y ln x.
Since there are points (x, e x) on the graph of y e x with
arbitrarily large xcoordinates, there will be points (x, ln x)
on the graph of y ln x with arbitrarily large ycoordinates.
51. (a) The graph y4 is a horizontal line at y 12
x
8 x 1 2x csc 5x d
x dx e x)
(x)(e x x 1 1
x 1
1 (csc 5x)(2x) d
(1
e x dx )( 1) (e 2 e x)
x 1
,x
2x x )(1) 0 ex
1 ex xe x e x Chapter 3 Review 15. dy
dx d 1 ln x)
(e
dx dy
16.
dx d 1 ln x
(e e )
dx d
ln (sin x)
dx d
(ex)
dx Alternate solution, using logarithmic differentiation: e (2x)2x y cos x
sin x 1d
(sin x)
sin x dx x2 cot x, for 1 ln (2x) dr
17.
dx 1) ), where k is even. d
1
d
ln (cos 1 x)
cos 1 x
dx
cos 1 x dx
1
1
1
cos 1 x
1 x2
cos 1x 1 x 2 18. dr
d d
log2 ( 2)
d 19. ds
dt d
log5 (t
dt t ln y 2 7) 1
d2
()
ln 2 d (t 2
ln 2 1
d
(t
7) ln 5 dt 7) (t 1
,
7) ln 5 7
23.
d
(8 t )
dt d
8 (ln 8) ( t)
dt
t 8 t dy
dx 24. ds
20.
dt x d tan 1x
e
dx e tan 1 dy
du 1 u2 d
sin 1
du
1
1 25. d
(ln x)2
dx
d
2 ln x
ln x
dx
2y ln x
x
2x ln x ln x
x dy
dt x2 26. x2 dy
dt 1 27. (2x)(2x) 1
2 x2 dy
dz t 2) cot
t 2)
2t 2
4t 2 x 2 1)(2 )(2x ln 2 2)
(x 2 1)3/2 (2 2x)[(x 2 x 3 1)(x ln 2
(x 2 1)3/2 2x (2 ) 1) 2 (2 2 )(x ln 2 x
x ln 2
(x 2 1)3/2 cos
28. x 2] 1 2 dy
dx d
(2
dx (2 t 1 1 2t)(2t) 2t
z 2) 1 1 (cos 1 z)(1)
2 1 1 z z z2 1 x 1 1 csc x)
1 1) x) x
1 x) 1
2 ( x) x
x)2 1
x
1 csc
x x
1 1
1 csc
1 1 1
2x 1) 1
x (cot z x ( 1
2t t)(1) 2t] cos z2 (2 csc
(2 2x)(x 3 ln 2 x ln 2
(x 2 1)3/2 1 z2 z
1 u2 1 1
2t 2t cot 1 1 x 1 d
(z cos 1 z
dz
1 1 1 1
(2)
(2t)2 1 1
(x (sec (z) (2x) u
u 1 d
[(1
dt u2 1 ( 2u) sec t2 2
1 1
(2x)(2)] x 1 t 1[(2x)(2x)(ln 2) etan x
1 x2 1 tan dx 1 t d
(2x)(2x)
dx 1 xd 1
ln t)
2 t2 t x2
2 u2 d
(t sec 1 t
dt 1 x2
x2 1 (1 d
1 dx [(2x)2x] x2 1 (t) (2x)2x d
dx x ln 2 d
u 2)2 du 1 u2 2 (ln x)(ln x) d
ln y
dx
1 dy
y dx
dy
dx
dy
dx ( 1 ln (x ln x) ln y 1 1
ln (x 2 1)
2
1
d
[ln 2 ln x x ln 2
ln (x 2 1)]
2
dx
1
1
1
0
ln 2
(2x)
x
2 x2 1
1
x
y
ln 2
x
x2 1
(2x)2x 1
x
ln 2
x2 1
x2 1 x ln x ln y dy
22.
dx ln x x2 ln ln 8 21. Use logarithmic differentiation.
y ln 2 d
ln y
dx
1 dy
y dx
dy
dx
dy
dx 2
ln 2 2 ln (2x) ln y values of x in the intervals (k , (k 119 x x
1 2 x z2 ( 2z) 120 29. Chapter 3 Review
dy
dx 32. Since y d
csc 1 (sec x)
dx
1 d
(sec x)
1 dx 2 sec x sec x
1
tan 2 x sec x dy
dx sin x
cos x
sin x
cos x sin x
sin x 2 cos 2 1 x ,, 2 ,x 2 3
, we may
2 ,x (sec x) 0 x , d1
d1 x 1, sin
cos
sin
c os (1 1
2
1 sin
c os cos 1
1 sin
c os cos
(1 2 (2x 7) dy
dx x xy 2
3
2 2,x x , x 2,x x 2
2 cos )(cos ) (1 sin )(sin )
(1 cos )2
cos2
sin
(1 cos )2 all x 0. x
2x 5) 7)(1) (x
7)2 2x
x2 3y 5x 4/5
d
(5x 4/5)
dx 4x 1/5 10y 6/5 5
is defined for all
7
5)(2)
17
, the
(2x 7)2
7 . 1
d
(1)
dx 0
(y
y
x 2)
2
3 15 d
(10y 6/5)
dx
dy
12y 1/5
dx
dy
dx d
(15)
dx 0
4x 1/5
12y 1/5 1
3(xy)1/5 37. Use implicit differentiation.
xy
d
dx
1 [x
xy dy
dx 0 and 1 xy d
(1)
dx (y)(1)] 0 dy
dx 0 y
dy
dx y
x Alternate method:
Since xy Therefore, 1, we have xy
dy
dx 1
.
x2 1, 1. 36. Use implicit differentiation. 2 2
, the function is differentiable for
x 2x d
d
d
(xy)
(2x)
(3y)
dx
dx
dx
dy
dy
x
(y)(1) 2 3
dx
dx
dy
(x 3)
dx
dy
dx sin2 sin
1
cos )2 ln x is defined for all x 1d 2
(x )
x 2 dx (x (2x x
2 31. Since y 1 (2x 7
dy
and
2
dx 2 1
1 2 34. Since y 2
3
2 Note that the derivative exists at 0 and 2 only because
these are the endpoints of the given domain; the twosided
derivative of y csc 1(sec x) does not exist at these
points.
dr
d , which is defined only for x 35. Use implicit differentiation. x 2,x
, 1, 0
dy
Therefore,
dx 30. x 2)3/2 x (1 2 x x, 2 1 function is differentiable for all x x x), x, 2 2x 1 and x)(2x) (cos x)
0 ( 2 x 2)( 1) (1
(1 x 2)2 (1 x
x2 21 3
2 x x, 2 1
2
1 x2 (sec x) sec 2 x sin x, the x
is defined for all x
x2 1 dy
dx rewrite the function as follows: 1 (cos x)(1) the function is differentiable for all x On the domain 0 1 1
1 33. Since y Alternate method: csc (x)( sin x) sec x tan x sign (sin x), x y cos x x cos x is defined for all real x and function is differentiable for all real x. sec x tan x
sec x tan x
1
cos x
1
cos x sin x 1 and y 1
.
x 121 Chapter 3 Review
38. Use implicit differentiation.
y y3 41. y 2 cos x x 2 x1
d
x
dx x 1
(x 1)(1) (x)(1)
(x 1)2
1
2y(x 1)2 d2
y
dx
dy
2y
dx
dy
dx 39. x
d3
(x )
dx 3x 2 3 y 3 d3
(y )
dx d
(y)
dx 3y 2y
(3y 2 y 2 sin x 1)y 2 sin x 1 d3
(y )
dx 3y 2y 0 y
x2
y2 2 sin x
3y 2 1 d
dx (3y 2 1)(2 cos x) (2 sin x)(6yy )
(3y 2 1)2 (3y 2 1)(2 cos x) 2 x
y2 d
dx y 2 sin x
3y 2 1 y d
(1)
dx y d
(2 cos x)
dx (y 2)(2x) (x 2)(2y)(y )
y4 (y 2)(2x) (x 2)(2y) (3y 2 2xy 2x x
y2 2
x 1/3 42. 4 5 y
2x(x 3 y 3)
y5
2x
y5 since x 3 y3 (3y 2 d 1/3
(x )
dx
1 2/3
x
3 y 1/3 d 1/3
(y )
dx
1 2/3
y
y
3 y 1
y 40. y2
d2
(y )
dx 2yy
y
y 1 2
x d
d2
(1)
dx
dx x
2
x2
2
1
x 2(2y)
x 2y
d1
dx x 2y
1
d2
(x y)
(x 2y)2 dx
1
[(x 2)(y )
(x 2y)2
1
1
(x 2) 2
(x 2y)2
xy
11
2xy
x 4y 2 y
1 2xy 2
x 4y 3 1)2 1)2 cos x 12y sin 2 x
(3y 2 1)3 4
d
(4)
dx 0
x
y y 2/3
x 2/3
2/3 d
dx y 2/3
x 2y
3x 1/3 2y
3x 1/3 xy
(x) (y)(1)
x2
y 2/3
x 43. y
2x 3 3x 1,
y
6x 2 3,
y
12x,
y (4) 12, and the rest are all zero. 2xy x4
,
24
3
x
,
6
2
x
,
2 44. y
y
y
y (4)
y (5) x,
1, and the rest are all zero. y x2 2 1/3 1/3
xy
( x 5/3y 2/3
3
2 4/3 1/3
2 5/3 2/3
x
y
x
y
3
3 (y)(2x)] 2 sin x
3y2 1 2 y4
3 (12y sin x) x 2 y) 122 Chapter 3 Review 45. dy d
dx dx x2 (2x x2 32 2(3) 3 1 2 32 2(3) x 2) 2x 3, we have y At x 2 dy
dx and 1 2x (x 3 3 or y
46. dy
dx 2 d
(4
dx 3) 2 3 or y csc 2 x
At x
y 2 4 3 3) dy
dx At t 3 y 2 csc x) 2 csc x cot x csc 2 2 csc 2 y 2 (a) Tangent: y
(b) Normal: y 4 2 2 csc cot 2 1x 0 2 and 1 2(1)(0) 2 or y
2 or y 2 50. 2 2 2 1x x
x 1.
2 2 dy
dx At t
y 2y 2 d
(2y 2)
dx
dy
2x 4y
dx
dy
dx Slope at (1, 2): y d
(9)
dx dy
51.
dx 0
2x
4y x
2y (a) Tangent: y At t 1
4 1
2(2)
1
(x
4 11
5 1) dy
dt
dx
dt 2 sin t
2 cos t tan t 3
, we have x
4 2 cos 3
4 3
4 2 sin 2, and 2, dy
dx tan 3
4 1. 1
x
4 2 or y 1(x 2) 2), or y 4 cos t
3 sin t dy
dt
dx
dt ( 4
cot t
3 3
, we have x
4 4 sin 3
4 2 3 cos 2, and x 2 3
4 2. 3 2
2 dy
dx , 4
3
cot
3
4 4
.
3 The equation of the tangent line is 9 d2
(x )
dx 4
x
5 1 or y 6 2 2 47. Use implicit differentiation.
x2 4) 5
x
4 1 or y The equation of the tangent line is , we have cot dy
dx x 49. 3 cot x 4) . 3
(x
2
53
2 x 4
(x
5 (b) Normal: y 3 (b) Normal: y 5
(x
4 (a) Tangent: y 2x 3 2 (a) Tangent: y 1 x2 9
4 y 4
x
3 32
2 dy
dt
dx
dt 6 2 5 sec 2 t
3 sec t tan t , we have x 5 tan 5 sec t
3 tan t 3 sec 4 2. 5
3 sin t 2 6 53
dy
, and
dx
3 6 4
x
3 2, or y 3,
10
.
3 5
3 sin 6 The equation of the tangent line is
(b) Normal: y 4(x 1) 2 or y 4x 2
y 48. Use implicit differentiation.
xy x
d
(x)
dx
1 1
2 (x)
xy d
(
dx dy
dx xy) (y)(1)
x
2 dy
xy dx
dy
dx 6
d
(6)
dx 0
y 1
2 xy
x 2 1
4 1
4 2
2 y 1
2 y
2
x Slope at (4, 1): xy 2 1
4 y
x
5
4 xy 10
(x
3 2 3) 53
, or y
3 10
x
3 5 3. Chapter 3 Review
(c) Nowhere in its domain dy
dt
dx
dt dy
52.
dx At t 1 cos t
sin t 59. y
2 4 , we have x 2 cos 4 2 y = f (x ) ,
–3 y dy
dx 2 sin 4
1 123 4 cos 4 4 sin 2 2 2
2 4 y 1.
y = f (x ) The equation of the tangent line is
y (2 y (1 2 1) x
2)x 2
4 2 2 1 This is approximately y 2
4 x –2 , and
60. 2
2 1 3 , or
x . 2.414x 3.200. 61. (a) iii (b) i (c) ii 53. (a) 62. The graph passes through (0, 5) and has slope
x 2 and slope 0.5 for x 2. 2 for y
7
[ 1, 3] by 5
3 1, (b) Yes, because both of the onesided limits as x → 1 are
equal to f (1) 1.
(c) No, because the lefthand derivative at x
the righthand derivative at x 1 is 1. 1 is 1 and 54. (a) The function is continuous for all values of m, because
the righthand limit as x → 0 is equal to f (0) 0 for
any value of m.
(b) The lefthand derivative at x 0 is 2 cos (2 0) 2,
and the righthand derivative at x 0 is m, so in order
for the function to be differentiable at x 0, m must
be 2.
55. (a) For all x 0 (b) At x y = f (x ) 7 x 63. The graph passes through ( 1, 2) and has slope 2 for
x 1, slope 1 for 1 x 4, and slope 1 for 4 x 6.
y
7 0 (c) Nowhere
56. (a) For all x y = f (x ) (b) Nowhere (c) Nowhere 7 57. Note that lim f (x) lim (2x x→0 lim f (x) x→0 x→0 lim (x 3) x→0 3) x 3 and 3. Since these values agree with f (0), the function is continuous at x 0. On the other 64. i. If f (x)
f (x) 9 7/3
x
28 9, then f (x) 3 4/3
x and
4 x 1/3, which matches the given equation. hand,
2, 1 x 0 f (x) , so the derivative is undefined at
1, 0 x x 4 (0, 4] (b) At x 0 58. Note that the function is undefined at x 0. (c) Nowhere in its domain
(a) [ 2, 0) (0, 2] 9 7/3
x
28 3 4/3
x , which
4 2, then f (x) contradicts the given equation f (x) 0. (a) [ 1, 0) ii. If f (x) (b) Nowhere iii. If f (x) 3 4/3
x
4 6, then f (x) the given equation. x 1/3. x 1/3, which matches 124 Chapter 3 Review 64. continued (d)
3 4/3
x
4 iv. If f (x) 4, then f (x) x 1/3 and f (x) 5 tan 0)( 5 sec 2 0) (e) however that i and iii could not simultaneously be true. (2 (f)
[ 1, 5] by [ 10, 80] (b) t interval avg. vel.
38
0.5
58
1
70
1.5
74
2
70
2.5
58
3
38
3.5
10
4 [0, 0.5]
[0.5, 1]
[1, 1.5]
[1.5, 2]
[2, 2.5]
[2.5, 3]
[3, 3.5]
[3.5, 4] 10
0
38
0.5
58
1
70
1.5
74
2
70
2.5
58
3
38
3.5 1
2 67. (a) 8 f (1) 1 At x 2 d
f(
dx (b) 56 f(x) At x d
dx
g (x) g ( 1) 3(2) 1 5. d2
[f (x)g 3(x)]
dx (c) d
g(f (x))
dx At x f (x) g (f (x))f (x) 1, the derivative is x
1
( 3)
2 13
.
10 (d) d
f (g(x))
dx g (0)f ( 1) (4)(2) 8. f (g(x))g (x) 1, the derivative is f (x) f (x) f ( x) 3f (x) 2 5 ( 3)2(0) f 2(x) 3g 2(x)g (x) g 3(x) 2f (x)f (x)
f (x)g 2(x)[3f (x)g (x) 2g(x)f (x)]
At x 0, the derivative is
f (0)g 2(0)[3f (0)g (0) 2g(0)f (0)]
( 1)( 3)2[3( 1)(4) 2( 3)( 2)]
9[ 12 12] 0. 1
5 x f (g( 1))g ( 1) 2 f (x)
f (0) 0, the derivative is x) 5 f 2(1) cos 1
(1)
5 g (f ( 1))f ( 1) 2 (c) 2 1, the derivative is 3f ( 1) 40 At x
1 f(x) 2 x
f (x) cos
2
2 g(x)] d
[3f (x)
dx At x 24 1 1 x
2 12. 1, the derivative is 1 f (1) 10f 2(x) cos 8 (d) Average velocity is a good approximation to velocity. 2 2
.
3 1, the derivative is 24 [ 1, 5] by [ 80, 80] x f (x) x2
f (x)]
2
x
10 sin
(2f (x)f (x))
2
x
20 f (x)f (x) sin
5
2 3f (0)
32 d
[10 sin
dx 20( 3) (c) x f (x)] 0, the derivative is 20 f (1) f (1) sin 40 cos x)( f (x)) ( f (x))( sin x)
(2 cos x)2 cos 0)(f (0)) (f (0))( sin 0)
(2 cos 0)2 At x 56 (2 d
f (x)
dx 2 cos x At x 65. (a) d
dx 1
( 5)
5 f (1)( 5) 1. Answer is D: i and iii only could be true. Note, (b) 5 tan x)( 5 sec 2 x) 0, the derivative is x 1/3. At x f (1 f (1 1 2/3
x
, which contradicts the given equation
3 d
[
dx 5 tan x) At x f (x) 66. (a) d
f (1
dx 2 f (0) 29 f( 1
.
3 d
f (x)
dx g(x) 2 (g(x) 1
5 2 2. 0, the derivative is (g(0)
1
.
10 (2)(1) 2)f (x) f (x) g (x)
(g(x) 2)2 At x x) 2x
f ( 1)
f (1)
1, the derivative is
2
21 (e) f ( 1)g ( 1) 2)f (0) f (0)g (0)
(g(0) 2)2 6. (3 2)( 2) ( 1)(4)
( 3 2)2 Chapter 3 Review
d
g(x
dx (f) f (x)) g (x f (x))(1 At x dw
ds f (x)) (x 1)[1 dw dr
dr ds ( 2)] d
[sin (
dr
1 cos ( r f (0))[1 (1)( 1) r 2) 0, we have r dw
ds cos ( 4
cos 0
4 2)] 1 4 t 0, the particle first reaches the origin at 2 . The velocity is given by v(t) 1 10 sin t
4 is 10 sin speed at t
6 t 4 4 , so the velocity at
10, and the 2 is 4 by a(t) 6 10 10. The acceleration is given 10 cos t is 10 cos , so the acceleration at 4 0. 2 4 and so 6 71. (a) 8 cos 0 6 24
183
4
2 6 f (0)] d
8 sin s
ds 8 cos s
8 sin 0 2) 8 cos (d) Since cos
t 2r At s f (x)) f (x)) 0, the derivative is g (0 g (0 68. d
dx g (x 3 ds
dt
d 2s
dt 2 d
(64t
dt
d
(64
dt 16t 2) 64 32t) 32t 32 (b) The maximum height is reached when
69. Solving 2 t 1 for t, we have 1 t 2 1 2 d2
(
d
12
(
3 7) 2(2
3( 2 dr
dt 7) 2 4( 2
3( ) At t
and so dr
dt 2 0, we may solve
2(1)4(12
3(1 2 t 7)
2) 2 (b) s(0) 10 cos t
10 cos 4 5 4 2/3 2(8) 2/3 3 1, (b) Since v(t) 1
.
6 (980) , y(t) 12.3 sec. 393.8 ft.
4
sec.
7 280 cm/sec. 980t, the velocity is 4
= 560 cm/sec. Since a(t)
7 dV
dx 73. 2 d
dx 10 (20x 1 dv
dt 980, the 1, we have s(t) 10. 1, we have s(t) 10. 3 x2
x
3 x2
x
40 d
dx 13
x
3 x 2)
9x 10x 2 32
x
20 13
x
1600 (b) The marginal revenue is
r (x) 4 ds
dt 5.2t, the
64
5.2 acceleration is 980 cm/sec 2. Farthest right:
When cos t 64
5.2 4
7 74. (a) r(x) 4 64 The average velocity is 160 (c) Farthest left:
When cos t 2.6t 2) 490t 2, it takes 72. (a) Solving 160
7)
2) 0, which 64, so the velocity is 64 ft/sec. d
(64t
dt ) 70. (a) One possible answer:
x(t) ds
dt ds
dt The maximum height is s 1 to obtain 2/3 0, ) 3 ds
dt 2 sec. maximum height is reached at t 1 dr
(2
dt 2/3
2 (d) Since dr d
(
dt d (2 ) 3 (c) When t dr dt
dt d 7)1/3
2/3 occurs at t , and we may write: dr
d 125 9 3
(x 2
1600
3
(x
1600 3
x
10 160x
40)(x 32
x
1600 4800)
120), which is zero when x 40 or x 120. Since the bus
holds only 60 people, we require 0 x 60. The
marginal revenue is 0 when there are 40 people, and the
40 2
corresponding fare is p(40)
3
$4.00.
40 126 Chapter 3 Review 74. continued 78. (a) P(0) (c) One possible answer:
If the current ridership is less than 40, then the
proposed plan may be good. If the current ridership is
greater than or equal to 40, then the plan is not a good
idea. Look at the graph of y r(x). 1 200
e5 (b) lim P(t) lim t→ t→ 1 student 1 200
e5 d
200(1
dt (c) P (t) 200(1 dx
dt tan , we have (sec 2 ) d
dt (b) 0.6 rad
sec 2 0.6 sec 0 1 revolution
2 rad e5 t)2(200e5 t)( 1) (200e5 t)(2)(1
(1 e5 t)4
e 5 t)( 200e 5 t) 400(e 5 t)2
(1 e 5 t)3 0.6 km/sec. f (x) Since P 0 when t cos (x
cos (x
cos x sin x) (x sin x)(1
sin x) and P (t) cos x). This derivative is zero when t 0 (which we need not solve) or when of these values, f (x)
0) P (t) 5. To confirm that this corresponds to the maximum value of P (t), note that P (t) sin x) 1,which occurs at x sin (2k 5, the critical point of y sin x). Then
d
dx cos (x e5 t)(e5 t)( 1) (200e 5 t)(e 5 t 1)
(1 e 5 t)3 60 sec 18
=
revolutions per
1 min occurs at t
sin (x e 5 t) 2(e 5 t)( 1) (1 minute or approximately 5.73 revolutions per minute.
76. Let f (x) 1 (1 P (t) 0.6 sec 2 . At point A, we have dx
0 and
dt e 5 t) 200 students 200e 5 t
(1 e 5 t)2 [0, 60] by [ 50, 200] 75. (a) Since x 200
1 t f (2k )
0. Thus, f (x) sin (2k sin 2k ) f (x) 0 for x 5 5. The maximum rate occurs at 5, and this rate is
200e 0
(1 e 0)2 P (5) 2k for integers k. For each 0 for t 0 for t 200
22 50 students per day. Note: This problem can also be solved graphically.
2k ,
79. which means that the graph has a horizontal tangent at each
of these values of x. 77. y (r)
y (l)
y (d)
1
2rl y (T)
1
2rl d1
dr 2rl
d1
dl 2rl
d1
dd 2rl
1
T
d
2
d1
dT 2rl
1
1
d2 T Since y (r) 1
T d1
1
2l
d dr r
2r 2l
1
T d1
1
2r
d dl l
2rl 2
1
Td
(d 1/2)
2rl
dd
T
1
d3
4rl
1
1d
( T)
2rl
d dT
1 T
d
T
d
T
d
3/2 T
d
4rl 0, y (l) T
d
T
d dT 0, and y (d) 0, increasing r, l, or d would decrease the frequency. Since y (T)
increasing T would increase the frequency. 0, , ] by [ 4, 4] [ (a) x
(b) k , where k is an odd integer
4 , 22 (c) Where it’s not defined, at x
(d) It has period 2 in this window. k , k an odd integer
4 and continues to repeat the pattern seen 127 Section 4.1
Quick Review 4.1 80. Use implicit differentiation.
x2 y2 1 d2
(y )
dx 2x 2yy
y
y 1. f (x) d
(1)
dx 2. f (x) d
2(9
dx 24 0
2x
x
2y
y
dx
dx y
(y)(1) (x)(y )
y2 y 12
(x
3 2 x 6. g (x) 1
y3 y2
1 ( 3) 1
3 8. h (x) x 2 → 0 . Therefore, lim f (x) . (9 12
(x
1) 4/3
3
2x
3(x 2 1)4/3 (2x) d2
(x
dx d ln x
e
dx 2x
x 2)3/2 1) d
ln x
dx 2x
x2 1
sin (ln x)
x 2e 2x 1)
. d
x
dx 1 33 Chapter 4
Applications of Derivatives
s Section 4.1 Extreme Values of Functions 9. As x → 3 ,
10. As x → 9 3, x→3 x 2 → 0 . Therefore, lim 9 Finding Extreme Values 1. From the graph we can see that there are three critical
points: x
1, 0, 1.
Critical point values: f ( 1) 0.5, f (0) 0, f (1) 0.5
Endpoint values: f ( 2) 0.4, f (2) 0.4
Thus f has absolute maximum value of 0.5 at x
1 and
x 1, absolute minimum value of 0 at x 0, and local
minimum value of 0.4 at x
2 and x 2. d3
(x
dx 2x) 3x 2 3(1)2 2 d
(x
dx 2) f (3) (b) f (x) x→ 3 f (1) 11. (a) (pp. 177– 185)
Exploration 1 1) d
2x
dx e 2x 3/2 d2
(x
dx sin (ln x) 7. h (x) (since the given equation is x 2
dy
dx 2 4/3 1 x 2) (9 1/3 1) 2 1
y3 3), x2 x x 2) 1/2 ( 2x) 1) 4 2 3/2 1 5. g (x) y3 At (2, 1 x) d
(9
dx x 2) d2
(x
dx 4. f (x) y2 x x 2) (9 x
x
y y 2 d
(4
dx 3 1/4
x
4 3. f (x) d2
(x )
dx 1 . 1 2
1 1 (c) Lefthand derivative:
lim f (2 h→0 lim h)
h h3 f (2) lim (h 2 h)3 [(2 h) 2(2
h h)] 4 10h 6h h→0 [(2 h→0 6h 2
h h→0 lim 10) 10 [ 2, 2] by [ 1, 1] 2. The graph of f has zeros at x
1 and x 1 where the
graph of f has local extreme values. The graph of f is not
defined at x 0, another extreme value of the graph of f. Righthand derivative:
lim f (2 h→0 lim h→0 h)
h f (2) lim h→0 2] 4 h h
h lim 1 h→0 [ 2, 2] by [ 1, 1] 1 d
3. Using the chain rule and
x
dx
x
df
1 x2
.
x (x 2 1)2
dx () x
, we find
x Since the left and righthand derivatives are not equal,
f (2) is undefined. 128 Section 4.1 12. (a) The domain is x 2. (See the solution for 11.(c)).
3x 2 2, x 2
(b) f (x)
x2
1, 14. The first derivative k (x) Section 4.1 Exercises
1. Maximum at x b, minimum at x c2;
The Extreme Value Theorem applies because f is
continuous on [a, b], so both the maximum and minimum
exist. 0. Since k(0) 15. The first derivative f (x)
x Endpoint values: 4. No maximum, no minimum;
The Extreme Value Theorem does not apply, because the
function is not continuous or defined on a closed interval. Maximum value is 1 at x 6. Maximum at x a, minimum at x c;
The Extreme Value Theorem does not apply since the
function is not continuous. 4 minimum value is 2 ln 0.5 1
4 f (4)
Maximum value is 1
4 x e ( 1) e 1
at x
e 13. The first derivative h (x) x ;
5
;
4 sec x tan x has zeros at x and is undefined at x 0 and x 2 2 . Since g(x) 0 sec x is , the critical points occur only at .
x
x 0 g(x)
g(x) 1
1 Since the range of g(x) is ( , 1] [1, ), these values
must be a local minimum and local maximum, respectively.
Local minimum at (0, 1); local maximum at ( , 1)
2 3/5
x
is never zero but is
5 17. The first derivative f (x)
undefined at x has no zeros, so we need g(1) e 1 1
e 0. 1 has no zeros, so we need only consider the endpoints.
h(0) ln 1 0
h(3) ln 4
Maximum value is ln 4 at x 3;
minimum value is 0 at x 0. f (x)
3 f (x) 0
( 3)2/5
32/5 1.552 Since f (x) 0 for x 0, the critical point at x 0 is a
local minimum, and since f (x) ( 3)2/5 for 3 x 1,
the endpoint value at x
3 is a global maximum.
Maximum value is 32/5 at x
3;
minimum value is 0 at x 0. undefined at x 1.
1 0 3 2/5
x
is never zero but is
5 18. The first derivative f (x) 1; x 4 4; ln 2 Maximum value is e at x
minimum value is 1.307 2 0 ; Critical point value: x
Endpoint value:
x 12. The first derivative g (x)
e
only consider the endpoints.
g( 1) 1 1 7
,0
4 1. 1; 1
,2
2 1 Critical point values: 1.636 ln 4 at x minimum value is 1 at x
local maximum at ln 4 f (x) 1 2 also undefined at x Endpoint values: f (0.5) f (x) 7
4 1 f (x) 1 at x local minimum at 0, and x ln 1 0 x 9. Maximum at (0, 5) Note that there is no minimum since the
endpoint (2, 0) is excluded from the graph. 1 f (x) x 8. Minima at ( 2, 0) and (2, 0), maximum at (0, 2) Critical point value: f (1) , has zeros at 4 4
5
4 16. The first derivative g (x) 10. Local maximum at ( 3, 0), local minimum at (2, 0),
maximum at (1, 2), minimum at (0, 1) 0, 0. x local maximum at 7. Local minimum at ( 1, 0), local maximum at (1, 0) x→ 5
.
4 ,x 3. Maximum at x c, no minimum;
The Extreme Value Theorem does not apply, because the
function is not defined on a closed interval. 11. The first derivative f (x) 0. 1 and lim k(x) cos x Critical point values: x 5. Maximum at x c, minimum at x a;
The Extreme Value Theorem does not apply, because the
function is not continuous. 02 e the maximum value is 1 at x 2. Maximum at x c, minimum at x b;
The Extreme Value Theorem applies because f is
continuous on [a, b], so both the maximum and minimum
exist. 1
has a zero at x
x has a zero at x Since the domain has no endpoints, any extreme value must
occur at x 1
x2 x2 2xe 0. Critical point value: x
Endpoint value:
x 0
3 f (x)
f (x) 0
33/5 1.933 Since f (x) 0 for x 0 and f (x) 0 for x 0, the
critical point is not a local minimum or maximum. The
maximum value is 33/5 at x 3. Section 4.1
19. 129 24. [ 2, 6] by [ 2, 4] [ 4.7, 4.7] by [ 3.1, 3.1] Minimum value is 1 at x 2. To confirm that there are no “hidden” extrema, note that 20. y
x (x 2 1) 2 (2x) 3x 2 To find the exact values, note that y 2
,4
3 2, which is 2
. Local maximum at
3 46
9
46
9 1). 25. [ 6, 6] by [ 2, 7] 2
,4
3 2x
which is zero only at
1)2 0 and is undefined only where y is undefined. There is a local maximum at (0, zero when x (x 2 [ 1.5, 1.5] by [ 0.5, 3] ( 0.816, 5.089); local minimum at The minimum value is 1 at x 0. 26. (0.816, 2.911) 21. [ 4.7, 4.7] by [ 3.1, 3.1] The actual graph of the function has asymptotes at
x
1, so there are no extrema near these values. (This is
an example of grapher failure.) There is a local
minimum at (0, 1). [ 6, 6] by [ 5, 20] To find the exact values, note that
y 3x 2 2x 8 (3x 4)(x 2), which is zero when 27. 4
. Local maximum at ( 2, 17);
3
4
41
local minimum at ,
3
27 x 2 or x 22. [ 4.7, 4.7] by [ 3.1, 3.1] Maximum value is 2 at x
minimum value is 0 at x 1;
1 and at x 28.
[ 6, 6] by [ 4, 4] Note that y
3x 2 6x 3 3(x 1)2, which is zero at
x 1. The graph shows that the function assumes lower
values to the left and higher values to the right of this point,
so the function has no local or global extreme values. [ 4, 4] by [ 80, 30] 23. Minimum value is 115
at x
2 local maximum at (0, 10);
local minimum at 1,
29. [ 4, 4] by [ 2, 4] Minimum value is 0 at x 13
2 1 and at x 1. [ 5, 5] by [ 0.7, 0.7] 1
at x 1;
2
1
minimum value is
at x
1.
2 Maximum value is 3; 3. 130 Section 4.1 30. x3 36. Note that f (x) x 9x,
9x, 3 x 3x 2 9,
3x 2 9, Therefore, f (x)
[ 5, 5] by [ 0.8, 0.6] 3 or 0 x
x 0 or x 3
x 3
3. 3 or 0 x
x 0 or x 3 3
3. (a) No, since the left and righthand derivatives at x
are 9 and 9, respectively. 31. 0 (b) No, since the left and righthand derivatives at x
are 18 and 18, respectively. 1
at x 0;
2
1
minimum value is
at x
2.
2 Maximum value is 3 (c) No, since the left and righthand derivatives at x
are 18 and 18, respectively.
(d) The critical points occur when f (x)
(at x
[ 6, 6] by [0, 12] x Maximum value is 11 at x 5;
minimum value is 5 on the interval [ 3, 2];
local maximum at ( 5, 9) 0 3) and when f (x) is undefined (at x
3). The minimum value is 0 at x and at x 3, at x ( 3, 6 3). [ 4, 4] by [ 3, 3]
[ 3, 8] by [ 5, 5] Maximum value is 4 on the interval [5, 7];
minimum value is 4 on the interval [ 2, 1]. x2/3(1) y 33. crit. pt. x
[ 6, 6] by [ 6, 6] Maximum value is 5 on the interval [3, );
minimum value is 5 on the interval ( , 0 5x 2) 3 4
3 x derivative extremum 4
5 x 2 1/3
x
(x
3 0 local max value
12 1/3
10
25 undefined local min 2]. 34.
[ 4, 4] by [ 3, 3] x2/3(2x) y
[ 6, 6] by [0, 9] 2 1/3 2
x
(x
3 4) 8x 2
3 8
3 x Minimum value is 4 on the interval [ 1, 3]
35. (a) No, since f (x)
x 2. 2
(x
3 2) 1/3 , which is undefined at (b) The derivative is defined and nonzero for all x
Also, f (2) 0 and f (x) 0 for all x 2. crit. pt. derivative extremum value
x 2. (c) No, f (x) need not not have a global maximum because
its domain is all real numbers. Any restriction of f to a
closed interval of the form [a, b] would have both a
maximum value and a minimum value on the interval.
(d) The answers are the same as (a) and (b) with 2 replaced
by a. 1 x 0 x 1 0 minimum undefined local max
0 minimum 1.034
0 38. 3
0
3 0 or
0, 3, 6 3) and 3; local maxima occur at ( 37. 32. 3 131 Section 4.1
39. 42. [ 2.35, 2.35] by [ 3.5, 3.5] y 1 x x2 24
x 2 ( 2x) 2 (4 [ 4, 4] by [ 1, 6] x) 1, y 2 x
x 2x, 0
0 crit. pt. derivative 2x 2 4 x2 x2 (1) 4 x2 4 extremum value x crit. pt. undefined local max undefined local min 3 1 0 local max 4 derivative extremum value x 0 x 4 2 0 minimum 2 0 maximum 2 undefined local min x 0 2 x 2
x 2 43. 0 [ 4, 6] by [ 2, 6] 40. 2x
2x y 2,
6, x
x 1
1 crit. pt. derivative extremum value
x 1 x2 y 23 3 crit. pt.
x 3 3 x 5x 2 23 0
0 0 5
1 3 0 maximum 12x
x value minimum
local max [ 4, 6] by [ 5, 5] 0
144 1/2
15
125 undefined minimum 0 4.462 We begin by determining whether f (x) is defined at x
where
12
x
4 f (x) x3 41. 1
x
2 6x 2 15
,
4 lim f (1 h)
h h→0 f (1) lim h→0 [ 4.7, 4.7] by [0, 6.2] 2,
1, x
x 1 h→0 1
1 crit. pt. derivative
x lim undefined lim h→0 2 1 1
(1
4 h)2 1
(1
2 h h2 h
4h
1
(h
4 h) 4h) Righthand derivative:
lim f (1 h→0 lim h→0 lim h) f (1)
h
(1 h)3
h3 h→0 lim (h 2 h→0 1
Thus f (x) 3h 2
h 3h
1
x
2 3x 2 6(1 h)2
h 8(1 h) h 1)
1
,
2 12x 1, 1 1 extremum value
minimum x
x 8x, Lefthand derivative: y 5 44. derivative extremum 12
5 x x)
x 0 x 2x x 4x(3 2 x 2 ( 1) 1 x [ 4.7, 4.7] by [ 1, 5] maximum undefined local min 1 x x
8, 1 x 1 3 15
4 3 132 Section 4.2 44. continued
1
x
2 Note that
3x 2 12x 12 8
48 2 3 crit. pt.
x
x 1, and
122 4(3)(8)
2(3) 12 23
.
3 0.845 3 occur at x 0 when x 0 when x
2 6 But 2 1
2 1 and x [ 3, 3] by [ 5, 5] 2 2 3
3 derivative extremum 3.155.
value 0 3.155 local max 4 0 1 The function f (x) x 3 x has no critical points.
(b) The function can have either two local extreme values
or no extreme values. (If there is only one critical point,
the cubic function has no extreme values.)
52. (a) By the definition of local maximum value, there is an
open interval containing c where f (x) f (c),
so f (x) f (c) 0. 1, so the only critical points local max 3.079 (b) Because x → c ,we have (x c) 0, and the sign of
the quotient must be negative (or zero). This means the
limit is nonpositive. 45. Graph (c), since this is the only graph that has positive
slope at c. (c) Because x → c , we have (x c) 0, and the sign of
the quotient must be positive (or zero). This means the
limit is nonnegative. 46. Graph (b), since this is the only graph that represents a
differentiable function at a and b and has negative slope
at c. (d) Assuming that f (c) exists, the onesided limits in (b)
and (c) above must exist and be equal. Since one is
nonpositive and one is nonnegative, the only possible
common value is 0.
(e) There will be an open interval containing c where
f (x) f (c) 0. The difference quotient for the
lefthand derivative will have to be negative (or zero),
and the difference quotient for the righthand derivative
will have to be positive (or zero). Taking the limit, the
lefthand derivative will be nonpositive, and the
righthand derivative will be nonnegative. Therefore,
the only possible value for f (c) is 0. 47. Graph (d), since this is the only graph representing a
function that is differentiable at b but not at a.
48. Graph (a), since this is the only graph that represents a
function that is not differentiable at a or b.
49. (a) V(x) 160x 52x 2 4x 3
V (x) 160 104x 12x 2 4(x 2)(3x 20)
The only critical point in the interval (0, 5) is at x
The maximum value of V(x) is 144 at x 2. 2. (b) The largest possible volume of the box is 144 cubic
units, and it occurs when x 2.
50. (a) P (x) 2 200x 2
The only critical point in the interval (0, ) is at
x 10. The minimum value of P(x) is 40 at x 10. 53. (a) (b) The smallest possible perimeter of the rectangle is
40 units and it occurs at x 10, which makes the
rectangle a 10 by 10 square. [ 0.1, 0.6] by [ 1.5, 1.5] 2 2bx c is a quadratic, so it can have
51. (a) f (x) 3ax
0, 1, or 2 zeros, which would be the critical points of f.
Examples: f (0) 0 is not a local extreme value because in any
open interval containing x 0, there are infinitely
many points where f (x) 1 and where f (x)
1.
(b) One possible answer, on the interval [0, 1]: f (x) The function f (x) x
x
1 and x 1. x) cos 1 , 0 x x 1 x 1 This function has no local extreme value at
x 1. Note that it is continuous on [0, 1]. [ 3, 3] by [ 5, 5]
3 1 0, (1 3x has two critical points at s Section 4.2 Mean Value Theorem
(pp. 186–194)
Quick Review 4.2
1. [ 3, 3] by [ 5, 5] The function f (x)
x 0. x3 1 has one critical point at 2x 2 6
2x 2
x2
3
Interval: ( 0
6
3
x
3, 3
3) Section 4.2
2. 3x 2 6
3x 2
x2
x
Intervals: ( 0
6
2 Since h (x) is never zero and is undefined only where
h(x) is undefined, there are no critical points. Also, the
domain ( , 0) (0, ) has no endpoints. Therefore,
h(x) has no local extrema. 2 or x
2
2) ( 2, ) 2x 2 0
8 2x 2
4 x2
2x2
The domain is [ 2, 2]. 3. Domain: 8 (b) Since h (x) is never positive, h(x) is not increasing on
any interval.
(c) Since h (x) 0 on ( , 0)
on ( , 0) and on (0, ). 4. f is continuous for all x in the domain, or, in the interval
[ 2, 2]. 6. We require x 2 1 0, so the domain is x Since k (x) is never zero and is undefined only where
k(x) is undefined, there are no critical points. Also, the
domain ( , 0) (0, ) has no endpoints. Therefore,
k(x) has no local extrema. 1. 7. f is continuous for all x in the domain, or, for all x 1. 10. 1
1
C 2( 2)
C 1. C 5 (c) Since f (x) is never negative, f (x) is not decreasing on
any interval. 2x 6. (a) f (x) 0 on f (x) 5
,
2 0 on , 5
, f (x)
2 0 at x 5
, and
2 , we know that f (x) has a local (b) Since f (x) is never positive, f (x) is not increasing on
any interval. 25
maximum at x
, the local
4
5 25
maximum occurs at the point ,
. (This is also a
24 (c) Since f (x) is always negative, f (x) is decreasing on
( , ). global maximum.)
(b) Since f (x) , 0 on 2x 5
,
2 0 on g (x) 1
,
2 0 on minimum at x
1
2 1
,
2 . , 1
, g (x)
2 0 at x (c) Since y is negative on ( 2, ), y is decreasing on
[ 2, ). 1
, and
2 8. (a) y , we know that g(x) has a local (c) Since g (x) 0 on 1
,
2 , g(x) is increasing on
, 1
,
2 1
, g(x) is decreasing on
2 2 In the domain [ 2, ), y is never zero and is
undefined only at the endpoint x
2. The function y
has a local maximum at ( 2, 4). (This is also a global
maximum.)
(b) Since y is never positive, y is not increasing on any
interval. 49
. (This is also a global minimum.)
4 0 on 1
.
2 5
,
2 1
.
2
49
, the local minimum occurs at the
4 (b) Since g (x) , , f (x) is decreasing on 1 Since g (x) point 2x 5
, f (x) is increasing on
2 5
.
2 (c) Since f (x) 1 7. (a) y 0 on 0.5x 0.5e Since f (x) is never zero or undefined, and the domain
of f (x) has no endpoints, f (x) has no extrema. 5
5
. Since f
2
2 Since g 0 on (0, ), k(x) is decreasing on (0, ). 2e 2x (b) Since f (x) is always positive, f (x) is increasing on
( , ). Since f (x) 2. (a) g (x) , 0), k(x) is increasing on Since f (x) is never zero or undefined, and the domain
of f (x) has no endpoints, f (x) has no extrema. C Section 4.2 Exercises , 0 on ( 5. (a) f (x) (1)2 2(1)
3C
4 1. (a) f (x) (b) Since k (x)
( , 0).
(c) Since k (x) 8. f is differentiable for all x in the domain, or, for all x
4
3 (0, ), h(x) is decreasing 2
x3 4. (a) k (x) 5. f is differentiable for all x in the interior of its domain, or,
in the interval ( 2, 2). 9. 7
7
C 2
x2 3. (a) h (x) , 133 . 4x 3 20x 4x(x 5)(x 5) 5, x 0,
The function has critical points at x
and x
5. Since y
0 on ( ,
5) and
(0, 5) and y
0 on (
5, 0) and ( 5, ), the
points at x
5 are local minima and the point at
x 0 is a local maximum. Thus, the function has a
local maximum at (0, 9) and local minima at
(
5, 16) and ( 5, 16). (These are also global
minima.)
(b) Since y
0 on (
5, 0) and ( 5, ), y is increasing
on [
5, 0] and [ 5, ).
(c) Since y
0 on (
,
decreasing on ( , 5) and (0, 5), y is
5] and [0, 5]. 134 Section 4.2 9. (b) Since h (x) 0 on ( , 2) and (2, ), h(x) is
increasing on ( , 2] and [2, ).
(c) Since h (x)
[ 2, 2].
12. [ 4.7, 4.7] by [ 3.1, 3.1] (a) f (x) 1 x
2
3x 4 x 8 4 ( 1) 4 x 2 x
[ 4.7, 4.7] by [ 3.1, 3.1] 8
and
3 The local extrema occur at the critical point x
at the endpoint x (a) k (x) 4. There is a local (and absolute) and a local minimum at (4, 0). , 0 on , 0 on 4)(1) x(2x)
(x 2 4)2 x2
(x 2 4
4)2 (b) Since k (x) is never positive, k(x) is not increasing on
any interval. 8
, f (x) is increasing on
3 (c) Since k (x) is negative wherever it is defined, k(x) is
decreasing on each interval of its domain: on
( , 2), ( 2, 2), and (2, ). 8
.
3 (c) Since f (x) (x 2 Since k (x) is never zero and is undefined only where
k(x) is undefined, there are no critical points. Since
there are no critical points and the domain includes no
endpoints, k(x) has no local extrema. 8 16
maximum at ,
or approximately (2.67, 3.08),
33 3 (b) Since f (x) 0 on ( 2, 2), h(x) is decreasing on 8
, 4 , f (x) is decreasing on
3 13. 8
,4 .
3 10.
[ 4, 4] by [ 6, 6] (a) f (x) 3x 2 2 2 sin x
Note that 3x 2 2 2 for x
1.2 and 2 sin x
2 for
all x, so f (x) 0 for x
1.2. Therefore, all critical
points occur in the interval ( 1.2, 1.2), as suggested by
the graph. Using grapher techniques, there is a local
maximum at approximately ( 1.126, 0.036), and a
local minimum at approximately (0.559, 2.639). [ 5, 5] by [ 15, 15] (a) g (x) 1 2/3
x
(x
3 x 1/3(1) 8) 4x 8
3x 2/3 The local extrema can occur at the critical points
x 2 and x occurs at x 0, but the graph shows that no extrema minimum at ( 2,
( 2, (b) f (x) is increasing on the intervals ( , 1.126] and
[0.559, ), where the interval endpoints are
approximate. 0. There is a local (and absolute)
6 3 2) or approximately 7.56). (b) Since g (x) 0 on the intervals ( 2, 0) and (0, ), and
g(x) is continuous at x 0, g(x) is increasing on
[ 2, ).
(c) Since g (x) 0 on the interval (
decreasing on ( , 2]. , (c) f (x) is decreasing on the interval [ 1.126, 0.559],
where the interval endpoints are approximate.
14. 2), g(x) is 11.
[ 6, 6] by [ 12, 12] (a) g (x) 2 sin x
Since 1 g (x) 3 for all x, there are no critical
points. Since there are no critical points and the domain
has no endpoints, there are no local extrema. [ 5, 5] by [ 0.4, 0.4] (a) h (x)
(x (x 2
2)(x
(x 2 4)2 4)( 1) ( x)(2x)
(x 2 4)2
2) x2
(x 2 (b) Since g (x) 4
4)2 and a local (and absolute) minimum at 2, , ). (c) Since g (x) is never negative, g(x) is not decreasing on
any interval. The local extrema occur at the critical points, x
There is a local (and absolute) maximum at 0 for all x, g(x) is increasing on ( 2.
1
2,
4 1
.
4 135 Section 4.2
15. (a) f is continuous on [0, 1] and differentiable on (0, 1).
(b) f (1) f (0)
10
2 ( 1)
1 2c
c The equation is y 1
2 2 (3, f (3)) 1 f (c)
2c 1 2
3 2 1/3
c
3
1/3 c c
c (b) f (c) 1 c2
c2 1
1 c 1
c 4
2 f (2)
2
ln 1 2
ln 3 2
2 c
f (c) 1
2 31
,
. Its equation is y
2
2
1
1 x , or y f (c)
1 2/3
c
3
1 2/3
c
3 2.820 c
(0.5, 2.5)
2.5. 0
1
1
f (1) 2 The tangent line has slope 0 and passes through (1, 2),
so its equation is y 2. 2/3 c 1 and passes through
2
1 x
2 0.707x 3
2 1 or
2 0.354. 2 2 1 2/3
x
, f is not differentiable at x
3 [ 1, 1] by [ 1, 1] (c) (b) The slope of the secant line is 0, so we need to find c
such that f (c) 0.
c
c 1
2 0.771 19. (a) The secant line passes through (0.5, f (0.5))
and (2, f (2)) (2, 2.5), so its equation is y 1 3
2 2 2
ln 3 1 f (b) 2 c . 1
2
3
2 4 1 f (4)
4
ln 3 1 2 18. (a) f is continuous on [2, 4] and differentiable on (2, 4).
f (c) 1 2 f (c) y 2 c 0.707. 2 1 21. (a) Since f (x) 1 0 0.707x The tangent line has slope 4 2 1) 1
1 c 2 1 , or y c 2 c (x 2 c 2 2 1 (b) x 1
2 1 1
2c 2 1 .
2 2 f ( 1)
( 1) 2 c 1 2
2 (b) We need to find c such that f (c) 17. (a) f is continuous on [ 1, 1] and differentiable on
( 1, 1).
f (1)
1 (1, 0) and 2), so its slope is 2 f (1) f (0)
10
10
1
3
2
33
2
8
27 f (c) (3, 0
1 or y 16. (a) f is continuous on [0, 1] and differentiable on (0, 1).
(b) 20. (a) The secant line passes through (1, f (1)) f (1) f ( 1)
1 ( 1)
1 ( 1)
2 1
3
3 3/2 0.192 0. 136 Section 4.2 22. (a) Since f (x) 1, 0
3 1, x
x (b) 1
1 , f is not differentiable at x 1. (If f were differentiable
at x 1, it would violate the Intermediate Value
Theorem for Derivatives.) [ (b) , by [ 1, 2] sin x,
cos x, (c) Note that f (x) f( ) We require f (c)
For c [0, 3] by [ 1, 3] f (3)
3 (c) We require f (c)
f (x) f (0)
0 2 1 c 1
, but
3 3 1, x 1 x 0
0 11 25. f (x) x2
2 26. f (x) 2x x2 x cos x
ex
ln (x f ( 1)
( 1) 0 0
2 1
2 1 for all x where f (x) is defined. Therefore, 1 C there is no such value of c. f (x)
24. (a) We test for differentiability at x 0, using the limits
32. given in Section 3.5.
Lefthand derivative:
f (0 h)
h f (0) lim h→0 cos h
h 1 0 Righthand derivative:
lim h→0 f (0 h)
h f (0) lim h→0 lim h→0 (1
sin h
h sin h)
h 1 f (x)
f (1)
11/4 C
1C
C
f (x) 33.
ln ( 1 1
Since the left and righthand derivatives are not equal,
f is not differentiable at x 0. 34. C 1 C 1) 1
x f (2) 0, but C C f (x) 31. [ 1, 1] by [ 1, 2] lim , so c C x 30. f (x) h→0 1
cos 1 C 3 29. f (x) f (x) sin c 2.818. For 0
1 C 28. f (x) f (1)
1 ( 1) f (x)
f (0)
0C
C
f (x) 1
2
1
x C, x 0 1
,x
2 0 x 1/4
2
2
2
3
1/4
x 1 2 0.324, and 1.247. 27. f (x)
(b) . 1 . , so 0.324 or sin 2.818, , f is not differentiable at x 0. (If f were differentiable
at x 1, it would violate the Intermediate Value
Theorem for Derivatives.) (c) We require f (c) 1 possible values of c are approximately 1
1, ) 0 ) 0, this occurs when occurs when cos c there is no such value of c. f(
( 11 c 1 for all x where f (x) is defined. Therefore, 23. (a) Since f (x) sin x
x 0 C 3 f (x)
f ( 1)
2) C
0C
C
f (x) ln (x
3
3
3
3
ln (x 2) C 2) 3 x2
3
3
3
x2 x sin x C x sin x 3 c , this
1.247. The 137 Section 4.2
35. Possible answers:
(a) (b) [ 2, 4] by [ 2, 4] [ 1, 4] by [0, 3.5] 42. The runner’s average speed for the marathon was
approximately 11.909 mph. Therefore, by the Mean Value
Theorem, the runner must have been going that speed at
least once during the marathon. Since the initial speed and
final speed are both 0 mph and the runner’s speed is
continuous, by the Intermediate Value Theorem, the
runner’s speed must have been 11 mph at least twice.
43. (a) Since v (t) 1.6, v(t) 1.6t C. But v(0) 0, so
C 0 and v(t) 1.6t. Therefore,
v(30) 1.6(30) 48. The rock will be going
48 m/sec. (c) (b) Let s(t) represent position.
Since s (t) v(t) 1.6t, s(t) 0.8t 2 D. But
s(0) 0, so D 0 and s(t) 0.8t 2. Therefore,
s(30) 0.8(30)2 720. The rock travels 720 meters in
the 30 seconds it takes to hit bottom, so the bottom of
the crevasse is 720 meters below the point of release. [ 1, 4] by [0, 3.5] 36. Possible answers:
(a) (b) [ 1, 5] by [ 2, 4] (c) [ 1, 5] by [ 1, 8] (d) [ 1, 5] by [ 1, 8] [ 1, 5] by [ 1, 8] (c) The velocity is now given by v(t) 1.6t C, where
v(0) 4. (Note that the sign of the initial velocity is
the same as the sign used for the acceleration, since
both act in a downward direction.) Therefore,
v(t) 1.6t 4, and s(t) 0.8t 2 4t D, where
s(0) 0 and so D 0. Using s(t) 0.8t 2 4t and the
known crevasse depth of 720 meters, we solve
s(t) 720 to obtain the positive solution t 27.604,
and so v(t) v(27.604) 1.6(27.604) 4 48.166.
The rock will hit bottom after about 27.604 seconds,
and it will be going about 48.166 m/sec. 37. One possible answer:
44. (a) We assume the diving board is located at s
water at s 0 and the 10, so that downward velocities are positive. The acceleration due to gravity is 9.8 m/sec2,
[ 3, 3] by [ 15, 15] 38. One possible answer: so v (t) 9.8 and v(t) have v(t) [ 3, 3] by [ 70, 70] 39. Because the trucker’s average speed was 79.5 mph, and by
the Mean Value Theorem, the trucker must have been going
that speed at least once during the trip.
40. Let f (t) denote the temperature indicated after t seconds.
We assume that f (t) is defined and continuous for
0 t 20. The average rate of change is 10.6 F/sec.
Therefore, by the Mean Value Theorem, f (c) 10.6 F/sec
for some value of c in [0, 20]. Since the temperature was
constant before t 0, we also know that f (0) 0 F/min.
But f is continuous, so by the Intermediate Value Theorem,
the rate of change f (t) must have been 10.1 F/sec at some
moment during the interval.
41. Because its average speed was approximately 7.667 knots,
and by the Mean Value Theorem, it must have been going
that speed at least once during the trip. C. Since v(0) 0, we 9.8t. Then the position is given by s(t) where s (t)
s(0) 9.8t v(t) 4.9t 2 9.8t, so s(t) 0, we have s(t) 4.9t 2. Solving s(t) D. Since
10 gives 100
10
t2
, so the positive solution is t
. The
49
7
10
10
velocity at this time is v
9.8
14 m/sec.
7
7
10
4.9 (b) Again v(t) 9.8t C, but this time v(0) v(t) 9.8t s(t) 4.9t 2 2t s(t) 4.9t 2 2t. Solving s(t) solution t 2. Then s (t) 2 9.8t 2, so D. Since s(0) 10 2
9.8 2 and so 0, we have 10 gives the positive 1.647 sec. The velocity at this time is
v 2 10 2
9.8 9.8 about 14.142 m/sec. 2 10 2
9.8 2 10 2 m/sec or 138 Section 4.3 45. Because the function is not continuous on [0, 1]. The
function does not satisfy the hypotheses of the Mean Value
Theorem, and so it need not satisfy the conclusion of the
Mean Value Theorem. 52. (a) Toward: 0 t 2 and 5
away: 2 t 5 (c) Regression equation:
y
0.0820x 3 0.9163x 2 47. f (x) must be zero at least once between a and b by the
Intermediate Value Theorem. Now suppose that f (x) is zero
twice between a and b. Then by the Mean Value Theorem,
f (x) would have to be zero at least once between the two
zeros of f (x), but this can’t be true since we are given that
f (x) 0 on this interval. Therefore, f (x) is zero once and
only once between a and b. ln (x 1 1
x 1 53. , which is never zero on [0, 3]. Now f (0) 0, so x 0 is
one solution of the equation. If there were a second
solution, f (x) would be zero twice in [0, 3], and by the
Mean Value Theorem, f (x) would have to be zero
somewhere between the two zeros of f (x). But this can’t
happen, since f (x) is never zero on [0, 3]. Therefore,
f (x) 0 has exactly one solution in the interval [0, 3].
50. Consider the function k(x) f (x) g(x). k(x) is continuous
and differentiable on [a, b], and since
k(a) f (a) g(a) 0 and k(b) f (b) g(b) 0, by the
Mean Value Theorem, there must be a point c in (a, b)
where k (c) 0. But since k (c) f (c) g (c), this
means that f (c) g (c), and c is a point where the graphs
of f and g have parallel or identical tangent lines. 3.3779 (d) Using the unrounded values from the regression
equation, we obtain
f (t)
0.2459t 2 1.8324t 2.5126. According to
the regression equation, Priya is moving toward the
motion detector when f (t) 0 (0 t 1.81 and
5.64 t 8), and away from the detector when
f (t) 0 (1.81 t 5.64). 1). Then f (x) is continuous and differentiable everywhere on [0, 3]. f (x) 2.5126x [ 0.5, 8.5] by [ 0.5, 5] 48. Let f (x) x 4 3x 1. Then f (x) is continuous and
differentiable everywhere. f (x) 4x 3 3, which is never
zero between x
2 and x
1. Since f ( 2) 11 and
f ( 1)
1, exercise 47 applies, and f (x) has exactly one
zero between x
2 and x
1.
x 8; (b) A local extremum in this problem is a time/place where
Priya changes the direction of her motion. 46. Because the Mean Value Theorem applies to the function
y sin x on any interval, and y cos x is the derivative of
sin x. So, between any two zeros of sin x, its derivative,
cos x, must be zero at least once. 49. Let f (x) t f(b)
b 1
b Thus, c
f (b)
b f (c) a 1
, so
c2 f (c) 54. 1
a b f(a)
a 1
ab 1
c2 1
and c 2
ab ab. ab. f (a)
a b2
b 2c, so 2c a2
a b b a a and c a b
2 . 55. By the Mean Value Theorem,
sin b sin a (cos c)(b a) for some c between a and b.
Taking the absolute value of both sides and using
cos c
1 gives the result.
56. Apply the Mean Value Theorem to f on [a, b].
Since f (b) f (a), f(b)
b f(a)
is negative, and hence f (x)
a must be negative at some point between a and b.
57. Let f (x) be a monotonic function defined on an interval D.
For any two values in D, we may let x1 be the smaller value
and let x2 be the larger value, so x1 x2. Then either
f (x1) f (x2) (if f is increasing), or f (x1) f (x2) (if f is
decreasing), which means f (x1) f (x2). Therefore, f is
onetoone. [ 1, 1] by [ 2, 2] 51. (a) Increasing: [ 2, 1.3] and [1.3, 2];
decreasing: [ 1.3, 1.3];
local max: x
1.3
local min: x 1.3
(b) Regression equation: y 3x 2 5 s Section 4.3 Connecting f and f with the
Graph of f (pp. 194–206)
Exploration 1
[ 2.5, 2.5] by [ 8, 10] (c) Since f (x) 3x 2 5, we have f (x) x 3 5x
But f (0) 0, so C 0. Then f (x) x 3 5x. C. Finding f from f 1. Any function f (x) x 4 4x 3 C where C is a real
number. For example, let C 0, 1, 2. Their graphs are all
vertical shifts of each other.
2. Their behavior is the same as the behavior of the function f
of Example 8. Section 4.3 Exploration 2 Finding f from f and f Section 4.3 Exercises 1. f has an absolute maximum at x 0 and an absolute
minimum of 1 at x 4. We are not given enough
information to determine f (0).
2. f has a point of inflection at x 1. (a) Zero: x
1;
positive: ( , 1) and (1, );
negative: ( 1, 1) 2. (b) Zero: x 0;
positive: (0, );
negative: (
, 0) 3. 2. (a) Zero: x 0, 1.25;
positive: ( 1.25, 0) and (1.25, );
negative: (
, 1.25) and (0, 1.25)
(b) Zero: x
0.7;
positive: ( , 0.7) and (0.7, );
negative: ( 0.7, 0.7) [ 3, 5] by [ 5, 20] Quick Review 4.3
1.
(x x2
3)(x Intervals
Sign of
(x 3)(x 9
3) 0
0 3. (a) ( x 3 3 x 33 x x(x x3
2)(x 3) (b) ( 4x
2) 2 2 x 0 0 x 2 xe x f :x 2 x 5. (a) [0, 1], [3, 4], and [5.5, 6]
(b) [1, 3] and [4, 5.5] (c) Local maximum: x 3;
local minimum: x 0
7. y 0, since f (x) 6;
5.5 (b) Nowhere
3 2/5
x
5
(x 2)(1) (x)(1)
(x 2)2 (x 2
2)2 2x 1
x Intervals 6. f: all reals
f :x 2;
2 (a) [0, 3] 2
2, since f (x) 2] and [2, ) 6. If f is continuous on the interval [0, 3]: ex 4. f: all reals 5. f: x 2; (c) Local maxima: x 1, x 4
(if f is continuous at x 4), and x
local minima: x 0, x 3, and x (2, ) 3. f: all reals
f : all reals, since f (x) 0, since f (x) , (c) Local maximum: x
local minimum: x 0
0 Sign of
x(x 2)(x 2)
Solution set: ( 2, 0) f :x 2 and x
0 4. (a) [ 2, 2] x Intervals 2] and [0, 2] (c) Local maxima: x
local minimum: x Solution set: ( 3, 3)
2. , (b) [ 2, 0] and [2, ) 1
2 x 1
2 Sign of y
2 3/5
x
5 Behavior of y Decreasing Increasing 7. Left end behavior model: 0
Right end behavior model: x 2e x
Horizontal asymptote: y 0
2 8. Left end behavior model: x e
Right end behavior model: 0
Horizontal asymptote: y 0 y 2 (always positive: concave up) Graphical support: x 9. Left end behavior model: 0
Right end behavior model: 200
Horizontal asymptotes: y 0, y
10. Left end behavior model: 0
Right end behavior model: 375
Horizontal asymptotes: y 0, y [ 4, 4] by [ 3, 3] 200
(a)
375 1
,
2
1
2 (b) , (c) ( ,) (d) Nowhere
(e) Local (and absolute) minimum at
(f) None 1
,
2 5
4 139 140 Section 4.3
6x 2 8. y 12x 6x(x Intervals x 2) 0 0 xe 1/x( x 10. y
x 2 2 2 e 1/x ) 1
x e 1/x 1 x
Intervals Sign of y x 0 0 x 1 1 x Sign of y Behavior of y Decreasing Increasing Decreasing Behavior of y Increasing Decreasing Increasing
y 12x 12 12(x Intervals x 1) 1 x e 1/x(x y 1 ) Behavior of y 0 x Graphical support: [ 8, 8] by [ 6, 6] (a) [0, 2]
(b) (
(c) ( (a) ( , 0] and [2, )
, 1) , 0) and [1, ) (b) (0, 1] (c) (0, ) (d) ( (f) None (e) Local maximum: (2, 5);
local minimum: (0, 3) 11. y 1 x 28 (f) At (1, 1)
8x 8x(x x 1)(x 1 Intervals 1) 1 x x2 00 x 1 1 x ( 2x) ( 8 8( 3x 1)( 3x 1 x Intervals Increasing x 2 2 x 1
3 x Decreasing ( x2)( 4x) 8 Increasing 3 1
3 x
3 Sign of y (
2x 3 24x
(8 x 2)3/2 Concave up Concave down Concave up Graphical support: x 8 8 x2 0 x Decreasing ( 2x) x2)2 8 2x(x 2 12)
(8 x 2)3/2 Intervals Behavior of y 1 2x2) (8 2 y
1 22 Sign of y Decreasing Increasing
1) x2 8 8 Behavior of y
Behavior of y Decreasing 2x2 8 x 2)(1) 8 Sign of y 24x 2 , 0) (e) Local minimum: (1, e) (d) (1, ) y 0 Behavior of y Concave down Concave up [ 4, 4] by [ 6, 6] Intervals x Sign of y Concave up Concave down Graphical support: 8x 3 e 1/x
x3 1 1/x
e ( x 2)
x 1 Intervals Sign of y 9. y 2 8 x 0 8 Sign of y
Behavior of y Concave up Concave down Graphical support: [ 4, 4] by [ 3, 3] (a) [ 1, 0] and [1, )
(b) ( , (c) , [ 3.02, 3.02] by [ 6.5, 6.5] 1] and [0, 1] (a) [ 2, 2]
1 and
3 (d) 1 ,
3 1 , 1 ,
3 1
9 1) and (1, 1) 8, 0) (d) (0, 3 8, (c) ( 1 (e) Local maximum: (0, 1);
local (and absolute) minima: ( 1,
(f) (b) [ 3 2] and [2, 8] 8) 8, 0) and (2, 4);
(e) Local maxima: (
local minima: ( 2, 4) and ( 8, 0)
Note that the local extrema at x
2 are also absolute
extrema.
(f) (0, 0) Section 4.3 2x,
2x, 12. y x
x 141 0
0 Intervals x 0 x 0 Sign of y
Behavior of y Increasing Increasing
2, y x
x 2, 0
0 Intervals x 0 x 0 Sign of y
Behavior of y Concave down Concave up
Graphical support: [ 4, 4] by [ 3, 6] (a) ( , 0) and [0, ) (b) None
(c) (0, )
(d) ( , 0) (e) Local minimum: (0, 1)
(f) Note that (0, 1) is not an inflection point because the graph has no tangent line at this point. There are no inflection points.
12x 2 13. y 42x 36 Intervals 2)(2x 2 x 6(x 2 3)
3
2 x 3
2 x Sign of y
Behavior of y
y 24x 42 Increasing
6(4x Increasing 7)
7
4 x Intervals Decreasing 7
4 x Sign of y
Behavior of y Concave down Concave up
Graphical support: [ 4, 4] by [ 80, 20] (a) ( , (b) 2, (c) , 3
,
2 3
2 7
,
4 (d) 2] and 7
4 (e) Local maximum: ( 2,
(f) 7
,
4 321
8 40); local minimum: 3
,
2 161
4 142 Section 4.3 14. y
4x 3 12x 2 4
Using grapher techniques, the zeros of y are x
Intervals x 0.53 0.53 x 0.53, x 0.65 0.65 x 0.65, and x
2.88 2.88. 2.88 x Sign of y
Behavior of y
y 12x 2 Increasing
24x 12x(x Intervals x Decreasing Increasing Decreasing 2) 0 0 x 2 2 x Sign of y
Behavior of y Concave down Concave up Concave down
Graphical support: [ 2, 4] by [ 20, 20] (a)
(b)
(c)
(d)
(e) ( , 0.53] and [0.65, 2.88]
[ 0.53, 0.65] and [2.88, )
(0, 2)
( , 0) and (2, )
Local maxima: ( 0.53, 2.45) and (2.88, 16.23);local minimum: (0.65, 0.68)
Note that the local maximum at x 2.88 is also an absolute maximum.
(f) (0,1) and (2, 9)
2 4/5
x
5 15. y Intervals x 0 0 x Sign of y
Behavior of y Increasing Increasing 8
x 9/5
25 y Intervals x 0 0 x Sign of y
Behavior of y Concave up Concave down
Graphical support: [ 6, 6] by [ 1.5, 7.5] (a) ( ,) (b) None
(c) ( , 0) (d) (0, )
(e) None
(f) (0, 3)
16. y 1 2/3
x
3 Intervals x 0 0 x Sign of y
Behavior of y Decreasing Decreasing Section 4.3 143 2 5/3
x
9 y Intervals x 0 0 x Sign of y
Behavior of y Concave down Concave up
Graphical support: [ 8, 8] by [0, 10] (a) None
(b) ( ,) (c) (0, )
(d) ( , 0) (e) None
(f) (0, 5)
17. This problem can be solved using either graphical or analytic methods. For a graphical solution, use NDER to obtain the graphs
shown.
y
y
y [ 10, 20] by [0, 5] [ 10, 20] by [0, 0.3] [ 10, 20] by [ 0.02, 0.02] An analytic solution follows.
y (e x 3e 0.8x)(5e x) 5e x(e x
(e x 3e 0.8x)2 5e 2x (e x 2.4e 0.8x) 15e 1.8x 5e 2x 12e 1.8x
(e x 3e 0.8x)2 3e 1.8x
3e 0.8x)2 Since y 0 for all x, y is increasing for all x. (e x 3e 0.8x)2(5.4e 1.8x) (e x y 3e 0.8x)(5.4e 1.8x) (6e 1.8x)(e x
(e x 3e 0.8x)3 ( 0.6e x
(e x
0.6(3
(e x (3e 1.8x)(2)(e x
(e x 3e 0.8x)4 e 0.2x)e 2.6x
3e 0.8x)3 e 0.2x 0: 3 0
ln 3 x
Intervals 2.4e 0.8x) 1.8e 0.8x)e 1.8x
3e 0.8x)3 0.2x Solve y 3e 0.8x)(e x x 5 ln 3 5 ln 3 5 ln 3 x Sign of y
Behavior of y Concave up Concave down 2.4e 0.8x) 144 Section 4.3 17. continued
(a) ( ,) (b) None
(c) ( , 5 ln 3) ( (d) (5 ln 3, ) , 5.49) (5.49, ) (e) None
(f) 5 ln 3, 5
2 (5.49, 2.50) 18. This problem can be solved using either graphical or analytic methods. For a graphical solution, use NDER to obtain the graphs
shown.
y
y
y [ 6, 10] by [0, 4] [ 6, 10] by [0, 0.6] [ 6, 10] by [ 0.15, 0.15] An analytic solution follows.
x 8e y 2e x (2 y 1.5x 5e 8
5e 2 0.5x 0.5x 5e 0.5x )(0) (8)( 2.5e
(2 5e 0.5x)2 ) 20e 0.5x
(2 5e 0.5x)2 Since y 0 for all x, y is increasing for all x. (2 y 0.5x 2 5e 0.5x ) ( 10e )
(2 0.5x (2 5e 10e 0.5x (5e
5e (2 (20e
5e )(10e 0.5x) (40e
(2 5e 0.5x)3
0.5x 5e )( 2.5e 0.5x 2) ) 5e 0.5x 2 0 0.5x 2
5 0.5x 0: 0.5x )(2)(2
) 0.5x 4 0.5x 3 e Solve y 0.5x ln 2
5 x
x Intervals 2 ln 2 ln 2
5 2 ln 5
2 2 ln 5
2 x Sign of y
Behavior of y
(a) ( Concave up ,) (b) None
(c) , 2 ln 5
2 ( , 1.83) 5
2 (1.83, ) 5
2 (1.83, 2) (d) 2 ln ,
(e) None
(f) 2 ln , 2 Concave down 5
2 ) 0.5x )( 2.5e 0.5x ) Section 4.3 2, 19. y 2x, x
x 1
1 Intervals x 1 1 x Sign of y
Behavior of y
0, y 2, Increasing
x
x Decreasing 1
1 Intervals x Sign of y 1 1 x 0 Behavior of y Linear Concave down Graphical support: [ 2, 3] by [–5, 3] (a) ( , 1) (b) [1, )
(c) None
(d) (1, )
(e) None
(f) None
20. y
ex
y
ex
Since y and y are both positive on the entire domain, y is increasing and concave up on the entire domain.
Graphical support: [0, 2 ] by [0, 20] (a) [0, 2 ]
(b) None
(c) (0, 2 )
(d) None
(e) Local (and absolute) maximum: (2 , e2 ); local (and absolute) minimum: (0, 1)
(f) None
21. y 2 xe 1/x ( 2x
2 e 1/x (1
Intervals 3 2x
x 2 (e 1/x )(1) )
2 ) e 1/x
2 2 x2 2
x2 2 x 00 x 2 2 x Sign of y
Behavior of y Increasing Decreasing Decreasing Increasing 145 146 Section 4.3 21. continued
2 (e 1/x )(4x y (e 1/x 2 2e 3 (1 3 )(2x 1/x 2 x ) 2 2 2x 2 )(e 1/x )( 2x 3 ) 5 4x ) 2
x5 Intervals x 0 0 x Sign of y
Behavior of y Concave down Concave up
Graphical support: [ 12, 12] by [ 9, 9] (a) ( , 2] and [ 2, ) 2, 0) and (0, (b) [ 2] (c) (0, )
(d) ( , 0)
2, (e) Local maximum: ( 2e) ( 1.41, 2.33); local minimum: ( 2, 2e) (1.41, 2.33) (f) None
22. This problem can be solved using either graphical or analytic methods. For a graphical solution, use NDER to obtain the graphs
shown.
y
y
y [ 4.7, 4.7] by [ 3, 11] [ 4.7, 4.7] by [ 10, 10] [ 4.7, 4.7] by [ 10, 10] An analytic solution follows.
1 x2 y 2 ( 2x) 9 x2
3x 3 18x
9 x 3x(x 2 2 Intervals x 2(2x) 9 9 3 6)
x 2 6 x 6 x 00 x 6 6 x 3 Sign of y
Behavior of y Increasing x 2)( 9x 2 (9 Decreasing
( 3x 3 18) 1 18x) 29 y (
(9 x 2)( 9x 2 6x 4 81x 2
(9 x 2)3/2 162 x 2)2 ( 3x 3 18)
(9 9 2 3/2 x) Increasing 18x)(x) x2 ( 2x) Decreasing Section 4.3
Find the zeros of y :
3(2x 4 27x 2 54)
(9 x 2)3/2 2x 4 27x 2 0 54 0 x2 27 27 2 4(2)(54)
2(2) 27 3
4
27 x
27 Note that we do not use x
Intervals 3 x 1.56 3
4 33 1.56 33 3 33
4 1.56 3.33, because these values are outside of the domain.
x 1.56 1.56 x 3 Sign of y
Behavior of y Concave down 6] and [0,
(a) [ 3,
and [0, 2.45]
6, 0] and [ (b) [ Concave up 6] or, 6, 3] or, [ 3, 2.45] [ 2.45, 0] and [2.45, 3] (c) Approximately ( 1.56, 1.56)
(d) Approximately ( 3, 1.56) and (1.56, 3) 6, 6 3) ( 2.45, 10.39);
(e) Local maxima: (
local minima: (0, 0) and ( 3, 0)
(f) ( 1.56, 6.25)
1 23. y x2 1 Since y 0 for all x, y is always increasing. d
(1
dx y x 2) 1 Intervals (1
x 0 x 2) 2(2x)
0 (1 x Sign of y
Behavior of y Concave up Concave down
Graphical support: [ 4, 4] by [ 2, 2] (a) ( ,) (b) None
(c) ( , 0) (d) (0, )
(e) None
(f) (0, 0) Concave down 2x
x 2 )2 147 148 Section 4.3
x 3/4(5 24. y 5x 3/4 x) 15 1/4
x
4 y 7 3/4
x
4 0 Intervals x 7/4 (a) [1, )
(b) ( 15
7 x 15
7 , 2) and (0, ) (d) ( 2, 0) x (e) Local minimum: (1,
3 Sign of y (f) ( 2, 6 2) Behavior of y Increasing 15 5/4
x
16
3(7x 5)
16x 5/4 y , 1] (c) ( 15 7x
4x 1/4 Decreasing x 1/4(x 26. y 21 1/4
x
16 0, y is always increasing on its 0. 5
x 3/4
16 y Graphical support: 5x 3
4x 3/4 0 for all x domain x 0, the graph of y is concave down 3x 1/4 3 3/4
x
4 Since y Since y
0 for all x
for x 0. x 5/4 3) 5 1/4
x
4 y 3) ( 2, 7.56) and (0, 0) 5x 9
16x 7/4 9
x 7/4
16 0 Intervals x 9
5 9
5 x Sign of y
Behavior of y
[0, 8] by [ 6, 6] (a) Concave up Graphical support: 15
0,
7 (b) Concave down 15
,
7 (c) None [0, 6] by [0, 12] (d) (0, ) (a) [0, ) (e) Local (and absolute) maximum: (b) None 15 15 3/4
,
7
7 20
7 15
, 5.06 ;
7 (c) local minimum: (0, 0) (d) 0, (f) None
25. y x 1/3(x y 4 1/3
x
3 9
,
5
9
5 (e) Local (and absolute) minimum: (0, 0)
x 4/3 4) 4 2/3
x
3 Intervals x 4x 1/3 (f) 4x 4
3x 2/3 0 0 x 1 1 9 24
,
55 4 9
5 9
, 5.56
5 27. We use a combination of analytic and grapher techniques to x Sign of y
Behavior of y Decreasing Decreasing Increasing
y solve this problem. Depending on the viewing window
chosen, graphs obtained using NDER may exhibit strange 4 2/3
x
9 8 5/3
x
9 Intervals x behavior near x 4x 8
9x 5/3 2 NDER (y, 2)
2 x 0 0 x x 2 because, for example,
1,000,000 while y is actually undefined at 2. The graph of y Sign of y
Behavior of y Concave up Concave down Concave up
Graphical support: [ 4,7, 4.7] by [ 5, 15] [ 4, 8] by [ 6, 8] x3 2x 2 x
x2 1 is shown below. Section 4.3
2)(3x 2 (x y 2x 3 4x 8x 2 8x
(x 2)2 (x 3
2)2 1)
(x 2x 2 x 1)(1) 1 The graph of y is shown below. [ 4,7, 4.7] by [ 10, 10] The zeros of y are x
Intervals x 0.15, x
0.15 1.40, and x 0.15 x 2.45. 1.40 1.40 x 22 x 2.45 2.45 x Sign of y
Behavior of y Decreasing
(x 2)2(6x 2 (x y 2)(6x 2 16x
16x 2x 3 12x 2 24x
(x 2)3 2(x 1)(x 2 5x
(x 2)3 8) Increasing
(2x 3 8x 2
(x 2)4 8) 2(2x 3
(x 2)3 Decreasing
8x
8x 8x 2 1)(2)(x Decreasing Increasing 2) 1) 14
7) The graph of y is shown below. [ 4,7, 4.7] by [ 10, 10] Note that the discriminant of x 2
Intervals x 1 7 is ( 5)2 5x
1 x 2 4(1)(7) 2 3, so the only solution of y x Sign of y
Behavior of y Concave up Concave down Concave up
(a) Approximately [0.15, 1.40] and [2.45, )
(b) Approximately (
(c) ( , 0.15], [1.40, 2), and (2, 2.45] , 1) and (2, ) (d) (1, 2)
(e) Local maximum:
(f) (1, 1) (1.40, 1.29);local minima: (0.15, 0.48) and (2.45, 9.22) 0 is x 1. 149 150 Section 4.3
(x 2 28. y x2 1)(1) x(2x)
(x 2 1)2 Intervals (x x 1
1)2 2 1 1 x 1 1 x Sign of y
Behavior of y Decreasing Increasing (x 2 1)2( 2x) (x 2 1)( 2x) 4x( x 2
(x 2 1)3 2x 3
(x 2 6x
1)3 y ( x 2 1)(2)(x 2
(x 2 1)4 Decreasing
1)(2x) 1) 2x(x 2 3)
(x 2 1)3 Intervals 3 x 3 x 0 0 x 3 3 x Sign of y
Behavior of y Concave down Concave up Concave down Concave up Graphical support: [ 4.7, 4.7] by [ 0.7, 0.7] (a) [ 1, 1]
(b) ( , 1] and [1, )
3, 0) and ( (c) (
(d) ( , 3, ) 3) and (0, 3) 1
;
2
1
1,
2 (e) Local maximum: 1,
local minimum:
(f) (0, 0),
29. y 3,
1)2(x (x
Intervals 3
4 , and 3 3, 4 2)
x 1 1 x 2 2 x Sign of y
Behavior of y Decreasing Decreasing Increasing
y (x
(x
(x Intervals 1)2(1) (x 2)(2)(x 1)
1)[(x 1) 2(x 2)]
1)(3x 5)
x 1 1 x 5
3 5
3 x Sign of y
Behavior of y Concave up Concave down Concave up
(a) There are no local maxima.
(b) There is a local (and absolute) minimum at x
(c) There are points of inflection at x 1 and at x 2.
5
.
3 Section 4.3
30. y (x 1)2(x 2)(x Intervals x 4)
1 1 x 2 2 x 4 4 x Sign of y
Behavior of y
d
[(x
dx y Increasing 1)2(x 2 Increasing 6x 8)] (x 2 1) (2x 6) (x 1)[(x 1)(2x 6) (x 1)(4x 2 20x 22) 1)(2x 2 2(x (x 10x 2 6x 10 2 10 8)(2)(x 2(x 2 1) 6x 8)] 11) Note that the zeros of y are x
x Decreasing Increasing 4(2)(11) 1 and 10 12 4 4 5 3
2 1.63 or 3.37. The zeros of y can also be found graphically, as shown. [ 3, 7] by [ 8, 4] Intervals x 1 1 x 1.63 1.63 x 3.37 3.37 x Sign of y
Behavior of y Concave down
(a) Local maximum at x 4 (c) Points of inflection at x
31. 1, at x y y = f ′(x ) y = f (x )
P
x y = f ′′(x) 32. Concave down 2 (b) Local minimum at x Concave up y
P y = f (x ) 0 y = f ′(x )
x
y = f ′′(x) 1.63, and at x 3.37. Concave up 151 152 Section 4.3 33. (a) Absolute maximum at (1, 2);
absolute minimum at (3, 2) 38. (a) v(t) (b) None s (t) 2 (b) a(t) v (t) 2 (c) It begins at position 6 and moves in the negative
direction thereafter. (c) One possible answer: 39. (a) v(t)
y = f (x ) 2 3 3t 2 v (t) 6t 3 (c) It begins at position 3 moving in a negative direction. It
moves to position 1 when t 1, and then changes
direction, moving in a positive direction thereafter. 1
1 s (t) (b) a(t) y
2 2t x 40. (a) v(t) –2 34. (a) Absolute maximum at (0, 2);
absolute minimum at (2, 1) and ( 2, 1) (b) At (1, 0) and ( 1, 0) 6t 6t 2 (b) a(t) –1 s (t)
v (t) 6 12t (c) It begins at position 0. It starts moving in the positive
direction until it reaches position 1 when t 1, and
then it changes direction. It moves in the negative
direction thereafter.
41. (a) The velocity is zero when the tangent line is horizontal,
at approximately t 2.2, t 6, and t 9.8. (c) One possible answer: (b) The acceleration is zero at the inflection points,
approximately t 4, t 8, and t 11. y
2 42. (a) The velocity is zero when the tangent line is horizontal,
at approximately t
0.2, t 4, and t 12. y = f (x )
–3 3 x –1
–2 (d) Since f is even, we know f (3) f ( 3). By the
continuity of f, since f (x) 0 when 2 x 3, we
know that f (3) 0, and since f (2)
1 and f (x) 0
when 2 x 3, we know that f (3)
1. In
summary, we know that f (3) f ( 3), 1 f (3) 0,
and 1 f ( 3) 0. (b) The acceleration is zero at the inflection points,
approximately t 1.5, t 5.2, t 8, t 11,
and t 13.
43. No. f must have a horizontal tangent at that point, but f
could be increasing (or decreasing), and there would be no
local extremum. For example, if f (x) x 3,
f (0) 0 but there is no local extremum at x 0.
44. No. f (x) could still be positive (or negative) on both sides
of x c, in which case the concavity of the function would
not change at x c. For example, if f (x) x 4, then
f (0) 0, but f has no inflection point at x 0.
45. One possible answer: 35. y y 4 5 3
y = f (x ) 2
1 –5
1 2 3 4 5 6 x x –1 36. 5 –5
y 46. One possible answer: 5
4
3
2
1 y
5 2 –1
–2
–3 3 x
–5 37. (a) v(t) s (t) 2t (b) a(t) v (t) 2 4 (c) It begins at position 3 moving in a negative direction. It
moves to position 1 when t 2, and then changes
direction, moving in a positive direction thereafter. 5 –5 x Section 4.4
47. One possible answer: (1 51. (a) f (x) y
(–2, 8) 10 ) bx 2 ) bx 5 (2, 0) abce
,
(e bx a)2 x so the sign of f (x) is the same as the sign of abc. –10 (e bx a)2(ab 2ce bx) (abce bx)2(e bx
(e bx a)4 (e bx (b) f (x) a)(ab 2ce bx) (abce bx)(2be bx)
(e bx a)3 48. One possible answer:
y
8 (6, 7) (4, 4) 4 Since a 2 (2, 1) –2 2 4 a)(be bx) ab 2ce bx(e bx a)
(e bx a)3 6 6 8 0, this changes sign when x the e bx x ln a
due to
b a factor in the numerator, and f (x) has a point of inflection at that location. –2 52. (a) f (x) 4ax 3 3bx 2 2cx d
f (x) 12ax 2 6bx 2c
Since f (x) is quadratic, it must have 0, 1, or 2 zeros. If
f (x) has 0 or 1 zeros, it will not change sign and the
concavity of f (x) will not change, so there is no point
of inflection. If f (x) has 2 zeros, it will change sign
twice, and f (x) will have 2 points of inflection. 49. (a) Regression equation:
y bx bx abce
(1 ae
(0, 4) –5 bx
)(0) (c)( abe
(1 ae bx)2 ae 153 1 2161.4541
28.1336e 0.8627x [0, 8] by [ 400, 2300] (b) At approximately x 3.868 (late in 1996), when the
sales are about 1081 million dollars/year (b) If f has no points of inflection, then f (x) has 0 or 1
zeros, so the discriminant of f (x) is 0. This gives
(6b)2 4(12a)(2c) 0, or 3b 2 8ac.
If f has 2 points of inflection, then f (x) has 2 zeros and
the inequality is reversed, so 3b 2 8ac. In summary, f
has 2 points of inflection if and only if 3b 2 8ac. (c) 2161.45 million dollars/year s Section 4.4 Modeling and Optimization b
7
,
3a
4
which is the xvalue where the point of inflection 50. (a) In exercise 13, a 4 and b 21, so occurs. The local extrema are at x
which are symmetric about x 2 and x 7
.
4 (pp. 206–220)
3
,
2 b
(b) In exercise 8, a
2 and b 6, so
1, which
3a
is the xvalue where the point of inflection occurs. The
local extrema are at x 0 and x Exploration 1 Constructing Cones 1. The circumference of the base of the cone is the
circumference of the circle of radius 4 minus x, or 8
Thus, r 8 x
2 x. . Use the Pythagorean Theorem to find h, and the formula for the volume of a cone to find V. 2, which are
2. The expression under the radical must be nonnegative, that symmetric about x 1.
is, 16 (c) f (x) 3ax 2 2bx c and
f (x) 6ax 2b.
The point of inflection will occur where b
.
3a
If there are local extrema, they will occur at the zeros
of f (x). Since f (x) is quadratic, its graph is a parabola
and any zeros will be symmetric about the vertex which
will also be where f (x) 0. f (x) x2 8
2 0. Solving this inequality for x gives: 0 0, which is at x [0, 16 ] by [ 10, 40] x 16 . 154 Section 4.4 3. The circumference of the original circle of radius 4 is 8 .
Thus, 0 x 8 . 9. x 2
x2 y 2 4 and y
3x
( 3x)2
4
x 2 3x 2
4
4x 2
4
x
1
Since y
3x, the solutions are:
3, or, x
1 and y
3.
x 1 and y
In ordered pair notation, the solutions are (1, 3) and
( 1,
3). [0, 8 ] by [ 10, 40] 4. The maximum occurs at about x
volume is about V 25.80.
5. Start with dV
dx 2
dr
rh
3 dx 4.61. The maximum dh
dh
dr
. Compute
and ,
dx
dx
dx
dV
set
0, and solve for x to
dx r2 3
dV
substitute these values in ,
dx
8(3
6)
obtain x
4.61.
3
128
3
Then V
25.80.
27 10. x2
4 y2
9 1 and y x x2
4 3)2 4(x 3)2 36 24x 36 36 13x 2 2. y
6x 2 6x 12 6(x 2)(x 1)
y
12x 6
The critical points occur at x
2 or x 1, since y
0
at these points. Since y ( 2)
18 0, the graph has a
local maximum at x
2. Since y (1) 18 0, the graph
has a local minimum at x 1. In summary, there is a local
maximum at ( 2, 17) and a local minimum at (1, 10).
1
(5)2(8)
3 r 2h 4. V
SA 200
3 24x 0 x(13x 24) 0 2 r2 600
1000
.
r2 Substituting into the surface area equation gives r 2 r2 600. Solving graphically, we have 11.14, r 4.01, or r value and using h x x 3, the solutions are: 0 and y 24
and y
13 3, or, x 15
.
13 In ordered pair notation, the solutions are
(0, 3) and 24 15
,
.
13 13 x, where 0 x 20. (a) The sum of the squares is given by
f (x) x 2 (20 x)2 2x 2 40x 400. Then
f (x) 4x 40. The critical point and endpoints occur
at x 0, x 10, and x 20. Then f (0) 400,
f (10) 200, and f (20) 400. The sum of the squares
is as large as possible for the numbers 0 and 20, and is
as small as possible for the numbers 10 and 10.
Graphical support: 7.13. Discarding the negative 1000
to find the corresponding values
r2 of h, the two possibilities for the dimensions of the cylinder
are: [0, 20] by [0, 450] (b) The sum of one number plus the square root of the r 4.01 cm and h 19.82 cm, or, r 7.13 cm and h 6.26 cm. sin x is an odd function, sin ( 6. Since y cos x is an even function, cos ( sin ( 8. cos ( ) ) other is given by g(x)
g (x) 5. Since y
7. Since y 24
13 0 or x 1. Represent the numbers by x and 20 Solving the volume equation for h gives h 2000
r x Section 4.4 Exercises
cm3 1000 2 rh 1 9 4x 2 9x 2 1. y
3x 2 12x 12 3(x 2)2
Since y
0 for all x (and y
0 for x 2), y is
increasing on ( , ) and there are no local extrema. 12
rh
3 3 9x 2 Quick Review 4.4 3. V (x ) sin .
) cos . 1
2 2 20 x 1 x 20 20 x. Then . The critical point occurs when
x 1, so 20 x 1
and x
4 sin cos
cos sin
0 cos
( 1) sin
sin endpoints and critical point, we find cos cos
( 1) cos
cos g(20) 79
. Testing the
4 sin sin
0 sin g(0) 20
20. 4.47, g 79
4 81
4 20.25, and Section 4.4
The sum is as large as possible when the numbers are
79
1
79
and
summing
+
4
4
4 155 Graphical support: 1
, and is as small as
4 possible when the numbers are 0 and 20
(summing 0 + 20).
[0, 20] by [0, 40] Graphical support: 4. Let x represent the length of the rectangle in meters
(0 x 4). Then the width is 4 x and the area is
A(x) x(4 x) 4x x 2. Since A (x) 4 2x, the
critical point occurs at x 2. Since A (x) 0 for
0 x 2 and A (x) 0 for 2 x 4, this critical point
corresponds to the maximum area. The rectangle with the
largest area measures 2 m by 4 2 2m, so it is a square. [0, 20] by [ 10, 25] 2. Let x and y represent the legs of the triangle, and note that
5. Then x 2 y2 0 x y 0). The area is A dA
dx 1
x
22 1
2x 2 25 x2 25 2 x2 25 25, so y 1
xy
2 1
x
2
1
2 ( 2x) 25 x 2 (since x 2, so 25
x2 25 Graphical support: [0, 4] by [ 1.5, 5] 5. (a) The equation of line AB is y
ycoordinate of P is x 1. . (b) A(x) The critical point occurs when 25 2x 2 0, which means 2x(1 x 1, so the x) d
(2x 2x 2) 2 4x, the critical point
dx
1
1
occurs at x
. Since A (x) 0 for 0 x
and
2
2
1
A (x) 0 for
x 1, this critical point corresponds
2 (c) Since A (x)
5 x , (since x 0). This value corresponds to the 2 largest possible area, since
dA
dx 5 0 for 2
5 25 y 1
xy
2 A x 0 for 0 5 x and
2 5. When x 5 to the maximum area. The largest possible area is , we have 1
square unit, and the dimensions of the
2
1
rectangle are unit by 1 unit.
2 2
2 2
52 1
2 dA
dx 2 5 A and 2
25
. Thus, the largest possible area is
4 1
2 Graphical support: 25
5
5
cm2, and the dimensions (legs) are
cm by
cm.
4
2
2 Graphical support:
[0, 1] by [ 0.5, 1] [0, 5] by [ 2, 7] 3. Let x represent the length of the rectangle in inches (x
16
Then the width is
and the perimeter is
x
16
32
P(x) 2 x
2x
.
x
x
2(x2 16)
Since P (x) 2 32x 2
the critical point
x2 occurs at x
P (x) 4. Since P (x) 0 for x 0 for 0 x 0). 6. If the upper right vertex of the rectangle is located at
(x, 12 x 2) for 0 x
12, then the rectangle’s
dimensions are 2x by 12 x 2 and the area is
A(x) 2x(12 x 2) 24x 2x 3. Then
A (x) 24 6x 2 6(4 x 2), so the critical point
(for 0 x
12) occurs at x 2. Since A (x) 0 for
0 x 2 and A (x) 0 for 2 x
12, this critical
point corresponds to the maximum area. The largest
possible area is A(2) 32, and the dimensions are 4 by 8.
Graphical support: 4 and 0, this critical point corresponds to the minimum perimeter. The smallest possible perimeter is
P(4) 16 in., and the rectangle’s dimensions are 4 in. by 4 in. [0, 12] by [ 10, 40] 156 Section 4.4 7. Let x be the side length of the cutout square (0
Then the base measures 8 2x in. by 15 x 4). 2x in., and the volume is
V(x) 2x)(15 x(8 2x) 4x 3 92x 120 4(3x 12x 2 V (x) 46x 2 120x. Then 5)(x 6). Then the 5
critical point (in 0 x 4) occurs at x
. Since
3
5
5
V (x) 0 for 0 x
and V (x) 0 for
x
3
3 4, the 9. Let x be the length in meters of each side that adjoins the
river. Then the side parallel to the river measures
800 2x meters and the area is
A(x) x(800 2x) 800x 2x 2 for 0 x 400.
Therefore, A (x) 800 4x and the critical point occurs at
x 200. Since A (x) 0 for 0 x 200 and A (x) 0
for 200 x 400, the critical point corresponds to the
maximum area. The largest possible area is
A(200) 80,000 m2 and the dimensions are 200 m
(perpendicular to the river) by 400 m (parallel to the river).
Graphical support: critical point corresponds to the maximum volume.
2450
27
5
14
35
dimensions are in. by
in. by
in.
3
3
3 The maximum volume is V 5
3 90.74 in3, and the
[0, 400] by [ 25,000, 90,000] Graphical support:
10. If the subdividing fence measures x meters, then the
216
m and the amount of fence
x pea patch measures x m by
needed is f (x)
f (x) [0, 4] by [ 25, 100] 8. Note that the values a and b must satisfy a 2
and so b
A
dA
da 400 a 2. Then the area is given by a 2) (400 2 400 when a 2 a 200 2 400 200. Since 0 for 200 a dA
da a2
a2 . The critical point occurs 0 for 0 a 400 a2 f (x) 2 432x occurs at x 2 216
x 12. Since f (x) 0 for x 432x 1 . Then and the critical point (for x
0 for 0 x 0) 12 and 12, the critical point corresponds to the minimum total length of fence. The pea patch will measure
12 m by 18 m (with a 12m divider), and the total amount
of fence needed is f (12) 72 m. Graphical support: 200 and 200 then 200, so the maximum area occurs [0, 40] by [0, 250] 11. (a) Let x be the length in feet of each side of the square
base. Then the height is when a 3x 20, this critical point corresponds to the maximum area. Furthermore, if a
b 202 1
1
ab
a 400 a 2 for 0 a 20, and
2
2
1
1
1
a
( 2a)
400 a 2
2 2 400 a 2
2
a2 dA
da b2 3 3x b. 500
ft and the surface area
x2 (not including the open top) is
Graphical support:
S(x)
S (x) x2
2x 4x 500
x2 2000x point occurs at x
[0, 20] by [ 30, 110] and S (x) x2 0 for x 2 2000x 1. Therefore, 2(x 3 1000)
and the critical
x2 10. Since S (x) 0 for 0 x 10 10, the critical point corresponds to the minimum amount of steel used. The dimensions
should be 10 ft by 10 ft by 5 ft, where the height is 5 ft.
(b) Assume that the weight is minimized when the total
area of the bottom and the four sides is minimized. Section 4.4
12. (a) Note that x 2y 5(x 2 5x 2
10x 30x 33,750x 33,750x 10xy (for 0 0 x dc
dx 0 for 15, the critical point 5 ft. (b) The material for the tank costs 5 dollars/sq ft and the
excavation charge is 10 dollars for each square foot of
the crosssectional area of one wall of the hole.
13. Let x be the height in inches of the printed area. Then the
50
in. and the overall
x
50
8 in. by
4 in. The amount of
x width of the printed area is paper used is
400 2
in . Then
x (x A (x) 4 (for x 0) occurs at x 10. Since A (x) 0 10 and A (x) 0 for x 400x 4
2 4x
4(x 2 82
100) x2 and the critical point
0 for 10, the critical point 8 and 50
x 4 for x 10, the overall dimensions are 18 in. high by 9 in. wide.
14. (a) s(t)
v(t)
At t , the critical point 2 2 (or 90 ). 16. Let the can have radius r cm and height h cm. Then
1000
. The area of material used is
r2
2000
A
r 2 2 rh
r2
, so
r
dA
2 r 3 2000
2
2 r 2000r
. The critical point
dr
r2
3 1000
dA
occurs at r
10 1/3 cm. Since
0 for
dr
dA
0 r 10 1/3 and
0 for r 10 1/3, the critical
dr r 2h 1000, so h point corresponds to the least amount of material used and
hence the lightest possible can. The dimensions are
r 1/3 10 6.83 cm and h 10 1/3 6.83 cm. In Example 2, because of the top of the can, the “best” design 17. Note that r 2h
A
dA
dr 8r 2
16r dA
dr 1000
. Then
r2 0 for r 8r 2 5, the critical point corresponds to the least amount of aluminum used or wasted and hence the
most economical can. The dimensions are r 2 16t
96t 112
s (t)
32t 96
0, the velocity is v(0) and 1000, so h 2000
, so
r
3
16(r
125)
2000r 2
. The critical point
r2
3
dA
125 5 cm. Since
0 for 0 r 5
dr 2 rh occurs at r
corresponds to the minimum amount of paper. Using
x 0 for is less big around and taller.
50
x A(x) x 8) 0 for maximizes the triangle’s area is 3375) 15. Since 0 for x 15 ft and y dimensions are x and A ( ) . Since A ( ) x2 corresponds to the minimum cost. The values of x and
y are x 2 2 corresponds to the maximum area. The angle that 1 The critical point occurs at x
dc
15 and
dx 1
ab cos . The critical point
2 and A ( )
) occurs at 0 1125
x2
10(x 3 2 1
ab sin
2 A( ) 30xy 5x 2 dc
dx 4xy) 5x 2 c 15. We assume that a and b are held constant. Then 1125
. Then
x2 1125, so y 157 h
96 ft/sec. (b) The maximum height occurs when v(t) 0, when
t 3. The maximum height is s(3) 256 ft and it
occurs at t 3 sec.
(c) Note that
s(t)
16t 2 96t 112
16(t 1)(t 7),
so s 0 at t
1 or t 7. Choosing the positive
value of t, the velocity when s 0 is
v(7)
128 ft/sec. 40 , so the ratio of h to r is 8 to 1. 5 cm and 158 Section 4.4 18. (a) The base measures 10 2x in. by 15 2x
2 (d) V (x) 24x 2 336x
The critical point is at in., so the volume formula is
x(10 V(x) 2x)(15
2 2x) 2x 3 25x 2 ( 14)2
2(1) 14 x
75x. 24(x 2 864 4(1)(36) 14x 14 52
2 36) 7 13, that is, x 3.39 or x 10.61. We discard the larger
value because it is not in the domain. Since
V (x) 24(2x 14), which is negative when x 3.39,
the critical point corresponds to the maximum volume.
The maximum value occurs at x 7
13 3.39,
which confirms the results in (c). (b) We require x 0, 2x 10, and 2x 15. Combining
these requirements, the domain is the interval (0, 5). (e) 8x 3 168x 2 864x 1120
8(x 3 21x 2 108x 140) 0
8(x 2)(x 5)(x 14) 0
Since 14 is not in the domain, the possible values of x
are x 2 in. or x 5 in. [0, 5] by [ 20, 80] (c) (f) The dimensions of the resulting box are 2x in.,
(24 2x) in., and (18 2x) in. Each of these
measurements must be positive, so that gives the
domain of (0, 9).
[0, 5] by [ 20, 80] 20. 6 mi The maximum volume is approximately 66.02 in3 when
x 1.96 in. x
2 mi (d) V (x) 6x 2 50x 6–x Village 4 + x2 miles 75
Jane The critical point occurs when V (x)
x 50
25 5
6 ( 50)2
2(6)
7 4(6)(75) , that is, x 0, at 50 Let x be the distance from the point on the shoreline nearest 700
12 1.96 or x Jane’s boat to the point where she lands her boat. Then she 6.37. We discard needs to row x 2 mi at 2 mph and walk 6 4 x mi at the larger value because it is not in the domain. Since 5 mph. The total amount of time to reach the village is V (x) f (x) 12x 50, which is negative when x 1.96, x2 4 6 x
5 2 the critical point corresponds to the maximum volume. f (x) The maximum volume occurs when f (x) hours (0 1
1
(2x)
2 2 4 x2 x 6). Then 0, we have: x 25 57
6 x 1
5 x2 24 19. (a) The “sides” of the suitcase will measure 24 2x in. by
18 2x in. and will be 2x in. apart, so the volume
formula is
V(x) 2x(24 2x)(18 2x) 8x 3 168x 2 864x. 2 24 x 2 1
. Solving
5 1
5 5x 1.96, which confirms the result in (c). x 25x 2 4(4 21x 2 x2 4 16 (b) We require x 0, 2x 18, and 2x 24. Combining
these requirements, the domain is the interval (0, 9). x 2) 4 x 21 We discard the negative value of x because it is not in the
domain. Checking the endpoints and critical point, we have
[0, 9] by [ 400, 1600] (c) f (0) 2.2, f land her boat 4
21
4 2.12, and f(6) 0.87 miles down the shoreline from 21 the point nearest her boat. [0, 9] by [ 400, 1600] The maximum volume is approximately 1309.95 in3
when x 3.39 in. 3.16. Jane should Section 4.4
21. If the upper right vertex of the rectangle is located at
(x, 4 cos 0.5x) for 0 x
, then the rectangle has width
2x and height 4 cos 0.5x, so the area is A(x) 8x cos 0.5x.
Then A (x) 8x( 0.5 sin 0.5x) 8(cos 0.5x)(1)
4x sin 0.5x 8 cos 0.5x.
Solving A (x) graphically for 0 x
, we find that
x 1.72. Evaluating 2x and 4 cos 0.5x for x 1.72, the
dimensions of the rectangle are approximately
3.44 (width) by 2.61 (height), and the maximum area is
approximately 8.98.
22. Let the radius of the cylinder be r cm, 0 2 r 2 100 1 2 r3
2 r(200 100 r
3r 2) The critical point for 0
200
3 0 10 r Cubic:
A 1.74 a 3 ( 2r) (2 100 20 h
4000
3 10 2
3 1.86a 0.19 r 2)(2r) r 2) [ 0.5, 1.5] by [ 0.2, 0.6] 10 occurs at Quartic:
A
1.92a 4 5.96a 3 6.87a 2 2.71a 0.12 0 for 0 for 10 2
< r < 10, the
3 critical point corresponds to the maximum volume. The
dimensions are r 3.78a 2 2 2
. Since V (r)
3 2
10
and V (r)
3 r 0.34 [ 0.5, 1.5] by [ 0.2, 0.6] r2 100 r 0.54a r and the volume is 100 r 2
4 r(100 2 [0, 1.1] by [ 0.2, 0.6] (d) Quadratic:
A
0.91a 2 10. Then r 2 cm3. Then 2 r2 V (r) (c) 2 the height is 2 100
V(r) r 159 8.16 cm and [ 0.5, 1.5] by [ 0.2, 0.6] (e) Quadratic:
’ 11.55 cm, and the volume is 3 2418.40 cm3. 3 23. (a) f (x) x( e x) e x(1) e x(1 x)
The critical point occurs at x 1. Since f (x) 0 for
0 x 1 and f (x) 0 for x 1, the critical point
corresponds to the maximum value of f. The absolute
maximum of f occurs at x 1.
(b) To find the values of b, use grapher techniques to solve
xe x 0.1e 0.1, xe x 0.2e 0.2, and so on. To find
the values of A, calculate (b a)ae a, using the
unrounded values of b . (Use the list features of the
grapher in order to keep track of the unrounded values
for part (d).)
a b A 0.1 3.71 0.33
0.2 2.86 0.44
0.3 2.36 0.46 [ 0.5, 1.5] by [ 0.2, 0.6] According to the quadratic regression equation, the
maximum area occurs at a 0.30 and is approximately
0.42.
Cubic: [ 0.5, 1.5] by [ 0.2, 0.6] According to the cubic regression equation, the
maximum area occurs at a 0.31 and is approximately
0.45.
Quartic: 0.4 2.02 0.43
0.5 1.76 0.38
0.6 1.55 0.31
0.7 1.38 0.23
0.8 1.23 0.15
0.9 1.11 0.08
1.0 1.00 0.00 [ 0.5, 1.5] by [ 0.2, 0.6] According to the quartic regression equation the
maximum area occurs at a 0.30 and is approximately
0.46. 160 Section 4.4 24. (a) f (x) is a quadratic polynomial, and as such it can have
0, 1, or 2 zeros. If it has 0 or 1 zeros, then its sign
never changes, so f (x) has no local extrema.
If f (x) has 2 zeros, then its sign changes twice, and
f (x) has 2 local extrema at those points. the radius is 36, we know that y 18 x
and the height is 18
2 x. In part (a), x, and so the volume is given by
x2
(18
2 r 2h radius is x and the height is 18
2 2 given by r h x (18 x). In part (b), the x, and so the volume is x). Thus, each problem requires
x 2(18 us to find the value of x that maximizes f (x)
in the interval 0 x x) 18, so the two problems have the same answer.
To solve either problem, note that f (x) 18x 2 x 3 and so
f (x) 36x 3x 2
3x(x 12). The critical point
occurs at x 12. Since f (x) 0 for 0 x 12 and
f (x) 0 for 12 x 18, the critical point corresponds to
the maximum value of f (x). To maximize the volume in
either part (a) or (b), let x 12 cm and y 6 cm.
27. Note that h 2 r2 3 and so r volume is given by V
for 0 h critical point (for h
0 h dV
1 and
dh h 2. Then the 3 r 2h 3
dV
3, and so
dh 3 (3 h 2)h
h2 h 3 h3 h 2). The (1 dV
dh 0) occurs at h 1. Since 0 for 1 3, the critical point h 0 for corresponds to the maximum volume. The cone of greatest
volume has radius ax , we have f (2) requirement is that 4 a
4 4 0. Therefore, a 16. To and so f (x) 2 32x 3 , so f (2) positive as expected. So, use a 2x 16x 2 6, which is 16. (b) We require f (1) 0. Since f (x) 2 2ax 3, we
have f (1) 2 2a, so our requirement is that
2 2a 0. Therefore, a
1. To verify that x 1 is
in fact an inflection point, note that we now have
f (x) 2 2x 3, which is negative for 0 x 1 and
positive for x 1. Therefore, the graph of f is concave
down in the interval (0, 1) and concave up in the
interval (1, ), So, use a
1.
2x 3 a
, so the only sign change in
x2
a 1/3
f (x) occurs at x
, where the sign changes from
2 29. f (x) 2x ax 2 negative to positive. This means there is a local 12
x (18
4 x) 2x 2. Since a
and so our
4 minimum, note that we now have f (x) 25. Let x be the length in inches of each edge of the square
end, and let y be the length of the box. Then we require
4x y 108. Since our goal is to maximize volume, we
assume 4x y 108 and so y 108 4x. The volume is
V(x) x 2(108 4x) 108x 2 4x 3, where 0 x 27.
Then V
216x 12x 2
12x(x 18), so the critical
point occurs at x 18 in. Since V (x) 0 for 0 x 18
and V (x) 0 for 18 x 27, the critical point
corresponds to the maximum volume. The dimensions of
the box with the largest possible volume are
18 in. by 18 in. by 36 in.
2y f (x) 2 verify that the critical point corresponds to a local (b) Possible answers:
No local extrema: y x 3;
2 local extrema: y x 3 3x 26. Since 2x 28. (a) We require f (x) to have a critical point at x 2 m, height 1 m, and volume 2
m3.
3 minimum at that point, and there are no local maxima.
30. (a) Note that f (x) 3x 2 2ax b. We require
f ( 1) 0 and f (3) 0, which give 3 2a b 0
and 27 6a b 0. Subtracting the first equation
from the second, we have 24 8a 0 and so a
3.
Substituting into the first equation, we have 9 b 0,
so b
9. Therefore, our equation for
f (x) is f (x) x 3 3x 2 9x. To verify that we have a
local maximum at x
1 and a local minimum at
x 3, note that
f (x) 3x2 6x 9 3(x 1)(x 3), which is
positive for x
1, negative for 1 x 3, and
positive for x 3. So, use a
3 and b
9.
(b) Note that f (x) 3x 2 2ax b and f (x) 6x 2a.
We require f (4) 0 and f (1) 0, which give
48 8a b 0 and 6 2a 0. By the second
equation, a
3, and so the first equation becomes
48 24 b 0. Thus b
24. To verify that we
have a local minimum at x 4, and an inflection
point at x 1, note that we now have f (x) 6x 6.
Since f changes sign at x 1 and is positive at x 4,
the desired conditions are satisfied. So, use a
3 and
b
24. 161 Section 4.4
31. Refer to the illustration in the problem statement. Since
x2 y2 9, we have x 33. (a) Note that w 2 y 2. Then the volume of 9 dS
dw 12
12
rh
x (y 3)
3
3
1
2
(9 y )(y 3)
3
3 for 3 y 3y 9y 27), 3 ( 3y 2 9) 3)(y (y y 2 2y 3) 6 1), so the critical point in dV
1. Since
dy the interval ( 3, 3) is y
0 for 1 6y w 2)3/2, so
k(144 144 w 2)(w 2 0 for 3 y 0 for 0 w w w 2) 12) occurs at w dS
6 and
dw 6. 0 for 12, the critical point corresponds to the maximum stiffness. The dimensions are 6 in. wide by 1 3, the critical point does w w2)3/2(1) 36) 144 The critical point (for 0 3. (y dV
and
dy 2 dS
Since
dw dV
dy Thus (y 3 kw(144 w 2)( 3w 2 (k 144
( 4k w 2. Then we 144 w2)1/2( 2w) 3
(144
2 kw 122, so d kwd 3 may write S the cone is given by
V d2 6 3 in. deep.
(b) 32
3 correspond to the maximum value, which is V(1)
cubic units.
32. (a) Note that w 2 d2
kwd 2 may write S
for 0
so w dS
dw dS
Since
dw 43 w 144
w 2) kw(144 w 2. Then we
144kw The graph of S w(144 w2)3/2 is shown. The
maximum stiffness shown in the graph occurs at w
which agrees with the answer to part (a). kw 3 3kw 2
w 3k(w 2 48). The critical 12) occurs at w 0 for 0 w 4 dS
3 and
dw 48 4 6, (c) 3. 0 for 12, the critical point corresponds to the maximum strength. The dimensions are 4 3 in. wide
by 4 [0, 12] by [ 2000, 8000] 12, 144k point (for 0 122, so d 6 in. deep. (b) [0, 12] by [ 2000, 8000] The graph of S d 3 144 d 2 is shown. The
maximum stiffness shown in the graph occurs at
d 6 3 10.4 agrees with the answer to part (a),
and its value is the same as the maximum value found
in part (b), as expected.
Changing the value of k changes the maximum
stiffness, but not the dimensions of the stiffest beam.
The graphs for different values of k look the same
except that the vertical scale is different. [0, 12] by [ 100, 800] The graph of S 144w w 3 is shown. The maximum
strength shown in the graph occurs at w 4 3 6.9,
which agrees with the answer to part (a).
(c) 34. (a) v(t) s (t) 10 sin t The speed at time t is 10 sin t . The maximum speed
is 10 cm/sec and it occurs at t
t 1
,t
2 7
sec. The position at these times is s
2 3
,t
2 5
, and
2 0 cm (rest position), and the acceleration
[0, 12] by [ 100, 800]
2 a(t) v (t) 10 2 cos t is 0 cm/sec2 at these 2 The graph of S d 144 d is shown. The
maximum strength shown in the graph occurs at
d 4 6 9.8, which agrees with the answer to part
(a), and its value is the same as the maximum value
found in part (b), as expected.
Changing the value of k changes the maximum
strength, but not the dimensions of the strongest beam.
The graphs for different values of k look the same
except that the vertical scale is different. times.
(b) Since a(t)
10 2 cos t, the greatest magnitude of
the acceleration occurs at t 0, t 1, t 2, t 3, and
t 4. At these times, the position of the cart is either
s
10 cm or s 10 cm, and the speed of the cart is
0 cm/sec. 162 Section 4.4 35. Since di
dt 2 sin t 2 cos t, the largest magnitude of the current occurs when 2 sin t 2 cos t x 2 is positive, it has an absolute minimum at the point 0, or cos t. Squaring both sides gives sin2 t sin t we know that sin2 t cos2 t 1, so sin2 t Thus the possible values of t are cos2 t, and
1
.
2 cos2 t 35
, , and so on.
444 , Eliminating extraneous solutions, the solutions of
sin t cos t are t times i k for integers k, and at these 4 2 cos t 2 sin t 2 2. The peak current is 2 2 amps.
36. The square of the distance is
D(x) 32
2 x so D (x) 2x Since D (x) ( 0)2 x x2 2 and the critical point occurs at x
0 for x 1 and D (x) where f (x) 0 for x 5 D(1) 2 1 13
,.
24 0, and that point is (b) No. Since f (x) is continuous on [0, 2 ], its absolute
minimum occurs at a critical point or endpoint.
Find the critical points in [0, 2 ]:
f (x)
4 sin x 2 sin 2x 0
4 sin x 4 sin x cos x 0
4(sin x)(1 cos x) 0
sin x 0 or cos x
1
x 0, , 2
The critical points (and endpoints) are (0, 8), ( , 0),
and (2 , 8). Thus, f (x) has an absolute minimum at
( , 0) and it is never negative. 1. 1, the critical point corresponds to the minimum distance. The
minimum distance is 2x 39. (a) Because f (x) is periodic with period 2 . 40. (a) 9
,
4 2x 38. No. Since f (x) is a quadratic function and the coefficient of 2 sin t sin 2t
2 sin t 2 sin t cos t
2(sin t)(1 cos t) 0
sin t 0 or cos t 1
t k , where k is an integer
The masses pass each other whenever t is an integer
multiple of seconds. (b) The vertical distance between the objects is the
absolute value of f (x) sin 2t 2 sin t.
Find the critical points in [0, 2 ]: . 37. Calculus method:
The square of the distance from the point (1,
(x, (x 1)2 ( x2 2x 1 2x
D (x) 2 48 2 48 3x 2 2(2 cos2 t 3)2 x2 16 2 2 x2 16 20 2 48 3x 2 2(2 cos t 3 0 cos t 1) 0 1)(cos t 1) 0 ( 6x) The distance from the origin to (1, 6x 2 48 3x 2 . or cos t 1 t 2 , 4 3 , 0, 2 3 The critical points (and endpoints) are (0, 0),
2
,
3 33
4 33
,
,
, and (2 , 0)
3
2
2
2
The distance is greatest when t
sec and when
3
4
33
t
sec. The distance at those times is
meters.
3
2 41. (a) sin t sin (t sin t sin t cos sin t
1
sin t
2 tan t 3 )
cos t sin 3
1
3
sin t
cos t
2
2
3
cos t
2 3 3 3) is 2. The shortest distance from the point to the semicircle is the distance along the radius containing
3). That distance is 4 1
2 The semicircle is centered at the origin and has radius 4. the point (1, 2 cos t 1) 0 cos t Geometry method: ( 3)2 2 cos t 3x 2. Then Solving D (x) 0, we have:
6x 2 48 3x 2
36x 2 4(48 3x 2)
9x 2 48 3x 2
12x 2 48
x
2
We discard x
2 as an extraneous solution, leaving
x 2. Since D (x) 0 for 4 x 2 and D (x) 0 for
2 x 4, the critical point corresponds to the minimum
distance. The minimum distance is D(2) 2. 12 2 cos 2t 2(2 cos2 t x 2) is given by 16 D(x) f (x) 3) to 2 2. Solving for t, the particles meet at t
t 4
sec.
3 3 sec and at 163 Section 4.4
(b) The distance between the particles is the absolute value
of f (t) sin t sin t 3 3 cos t 2 1
sin t. Find
2 3
2 sin t 1
cos t
2
3
sin t
2 tan t measure 1 ft and (1 2 sin ) ft, so the volume is given by 1
(cos )(1
2 V( ) the critical points in [0, 2 ]:
f (t) 42. The trapezoid has height (cos ) ft and the trapezoid bases 1 20(cos )(1 0 2 sin )(20) sin ). Find the critical points for 0 1
cos t
2
1 V( ) 20(cos )(cos ) 2 20(1 20 cos2 3
5
11
The solutions are t
and t
, so the critical
6
6
5
11
, 1 and
, 1 , and the interval
points are at
6
6
3
3
, and 2 ,
. The particles
endpoints are at 0,
2
2
5
11
sec and at t
sec, and
are farthest apart at t
6
6 sin )( sin )
20 sin2 0 20 sin 20 sin2 0 20(2 sin2 sin 0 1)(sin 20(2 sin 1)
1) 0 sin 1
or sin
2 The critical point is at
3 f (t) 2 1
sin t
2
1
sin t
2 cos t 0. 0 0 3 For the function y sec and t 6 6 and V ( ) , 15 0 for 3 . Since V ( )
6 2 0 for , the critical point corresponds to the maximum possible trough volume. The
3
2 cos t volume is maximized when This is the same equation we solved in part (a), so the
solutions are t 1
6 the maximum distance between the particles is 1 m.
(c) We need to maximize f (t), so we solve f (t) 0 20 sin sin2 ) 20(1 : 4
sec.
3 43. (a) D 6 . C
8.5 R S y f (t), the critical points occur at
Q 4
, 1 , and the interval endpoints are at
3
3
1
1
0,
and 2 ,
.
2
2
4
Thus, f (t) is maximized at t
and t
. But
3
3 , 1 and these are the instants when the particles pass each
other, so the graph of y
points and f (t ) has corners at these d
f (t) is undefined at these instants. We
dt t or 3
4
the distance is changing faster than at any other
3 time in the interval. L
x
x A P B Sketch segment RS as shown, and let y be the length of
segment QR. Note that PB 8.5 x, and so
QB
x 2 (8.5 x)2
8.5(2x 8.5).
Also note that triangles QRS and PQB are similar.
QR
RS cannot say that the distance is changing the fastest at
any particular instant, but we can say that near t y y
8.5
y2
8.52 PQ
QB
x
8.5(2x 8.5)
x2
8.5(2x 8.5) y2 8.5x 2
2x 8.5 L2 x2 y2 L2 x2 8.5x 2
2x 8.5 L2
L2 x 2(2x 8.5) 8.5x 2
2x 8.5 2x 3
2x 8.5 164 Section 4.4 43. continued 46. (b) Note that x
y 2x L2 4.25, and let f (x) 2x 8.5 11, the approximate domain of f is 5.20 x 2 3 8.5)(6x ) (2x )(2)
(2x 8.5)2 x (8x 51)
(2x 8.5)2 and f (x) 0 for x 0 for 5.20 x 6.375 6.375. Therefore, the value of x that minimizes L 2 is x 6.375 in. (c) The minimum value of L is
2(6.375)3
2(6.375) 8.5 M2 44. Since R
dR
dM f (M) CM
C C
2 11.04 in.
M
3 C2
M
2 2 M . Let f (M) 13
M , we have
3 CM x c–x
R Let P be the foot of the perpendicular from A to the mirror,
and Q be the foot of the perpendicular from B to the mirror.
Suppose the light strikes the mirror at point R on the way
from A to B. Let:
a distance from A to P
b distance from B to Q
c distance from P to Q
x distance from P to R
To minimize the time is to minimize the total distance the
light travels going from A to B. The total distance is
D(x)
x2 a2
(c x)2 b 2
Then 2 M . Then 1 D (x) 2 x2
x 2M, and the critical point for f occurs at C
.This value corresponds to a maximum because
2
C
C
f (M) 0 for M
and f (M) 0 for M
. The value
2
2
C
dR
of M that maximizes
is M
.
2
dM M 45. The profit is given by
P(x) (n)(x c) 2bx
b(100 a2 x2 a b(100
2 (100
c (100 x)(x c) x c
a2 50 A selling price of 50 x)2 b2 [ 2(c x)] b2 0 gives the equation
x
x)2 (c c)bx
(a b2 x2 50 x)2
x)2 b 2 b 2] (c x)2(x 2 x 2b 2 (c x)2x 2 (c x)2a 2 x 2b 2 [c 2 2xc 0 (a 2 b 2)x 2 0 x 2(c
100 c
2 (c [(a b)x x2 x 2[(c c)b a2 x 2b 2 100bc). x)2
x)2 c
, and this
2 value corresponds to the maximum profit because
0 for x (c x)2 (c
x we have: 2x). The critical point occurs at x P (x) 2
c a2 Solving D (x) x2 1 (2x) which we will refer to as Equation 1. Squaring both sides, bx
Then P (x) Q
c 6.375 in., and this corresponds to a minimum value of f (x) because f (x) b u1 u2
P 2 5.20, the critical point occurs at 51
8 x A 8.5.
a (2x For x B . Since Then
f (x) Normal 3 c
c
and P (x) 0 for x 50
.
2
2
c
will bring the maximum profit.
2 (c ac x
Note that the value x
because x and c a
ac a b b a 2)
x)2a 2 (c x 2]a 2
2a 2cx
ac][(a a 2c 2
b)x ac or x a b is an extraneous solution x have opposite signs for this value. The only critical point occurs at x ac
a b . ac] 165 Section 4.4
To verify that this critical point represents the minimum
distance, note that x2 ( a 2)(1) D (x) x 2 x)2 ( (c (x)
a ( x
x2 a2 ) (c (c x) 2 (c
(x 2 a 2) x2
(x 2 a 2)3/2 (x 2 a2
a 2)3/2 x) [(c b x)
x)2 (c b2 2 50. (a) Since A (q) x)2 b 2] (c
[(c x)2 b 2]3/2 x)2 when km
q2 kmq 2 h
, or q
2 h
, the critical point occurs
2
2km
. This corresponds to the
h minimum value of A(q) because A (q) b2
,
x)2 b 2]3/2 [(c [200 2(x 50)]x
2x 2 300x
6000 32x
r(x) c(x)
2x 2 268x 6000, 50 x 80 r(x)
c(x)
p(x) Since p (x)
4x 268
4(x 67), the critical point
occurs at x 67. This value represents the maximum
because p (x)
4, which is negative for all x in the
domain. The maximum profit occurs if 67 people go on the
tour. 2 b 2)( 1) 49. Revenue:
Cost:
Profit: which is positive for q
which is always positive. 3 , 0. (b) The new formula for average weekly cost is We now know that D(x) is minimized when Equation 1 is
true, or, equivalently, 2kmq PR
AR (k B(q) QR
. This means that the two
BR bq)m
q cm bm cm km
q right triangles APR and BQR are similar, which in turn A(q) hq
2
hq
2 bm implies that the two angles must be equal.
47. Since B(q) differs from A(q) by a constant, the dv
dx minimum value of B(q) will occur at the same qvalue ka 2kx
ka
2k The critical point occurs at x as the minimum value of A(q). The most economical a
, which represents a
2 quantity is again maximum value
because d 2v
dx 2 2k, which is negative for all x. The maximum value of v is
kx 2 kax
48. (a) v
dv
dr ka cr0r 2
2cr0r a
2 k a2
2 ka 2
.
4 cr 3
3cr 2 cr(2r0 3r) The critical point occurs at r 2r0
3 . (Note that r 0 is not in the domain of v.) The critical point represents a
maximum because d 2v
dr 2 2cr0 which is negative in the domain
(b) We graph v (0.5 6cr
r0
2 2c(r0
r 51. The profit is given by
p(x) r(x) c(x)
6x (x 3 6x 2 15x)
x 3 6x 2 9x, for x 0.
Then p (x)
3x 2 12x 9
3(x 1)(x 3), so the
critical points occur at x 1 and x 3. Since p (x) 0 for
0 x 1, p (x) 0 for 1 x 3, and p (x) 0 for
x 3, the relative maxima occur at the endpoint x 0 and
at the critical point x 3. Since p(0) p(3) 0, this
means that for x 0, the function p(x) has its absolute
maximum value at the points (0, 0) and (3, 0). This result
can also be obtained graphically, as shown. 3r), r0.
[0, 5] by [ 8, 2] r)r 2, and observe that the maximum indeed occurs at v 2
0.5
3 2km
.
h 1
.
3 52. The average cost is given by
a(x) c(x)
x a (x) 2x x2 20x 20,000. Therefore, 20 and the critical value is x represents the minimum because a (x) 10, which 2, which is positive for all x. The average cost is minimized at a
[0, 0.5] by [ 0.01, 0.03] production level of 10 items. 166 Section 4.4 53. (a) According to the graph, y (0) 0. (b) According to the graph, y ( L)
(c) y(0) 0, so d c 2bx r
h 3ax 2 bx 2 and 3aL 2 and y ( L) 2bL bL 2 H 0, so we have two linear 2. and let m y 3aL
. Substituting into the first
2
3aL 3
aL 3
equation, we have aL 3
H, or
H, so
2
2
H
H
a 2 3 . Therefore, b 3 2 and the equation for y is
L
L
H
H
x3
x2
y(x) 2 3 x 3 3 2 x 2, or y(x) H 2
3
.
L
L
L
L m(0 equation gives b 2a x
and so the
2
2a x2
.
2 a2 height is h r2 a2 r 2h 3 x2 2a
2 3 x2 2a
2 a2 m(x x0) x0) m(x x0) a a 3 d2
(r
3 dr f (r)
= a2 a2 2
r 3 3 a2 2a 2r
3 a2
r(2a 2 2 2 a2 h
r a 6
3 r 2)(2r) mx0 m x0 have A(x) y2 2 2a 2
, which gives
3 2 a6
. Then
3 a2 mx0 a m a m
a2 m
a2
m f(x) 2
.
f (x) f (x) x To graph, let y1 3r 2) r2 mx0 mx0) (b) The domain is the open interval (0, 10). 2 The critical point occurs when r
r ( a2 r2
r (a m m a. 3r 3 2 . 1
(xintercept of line RT)(yintercept of line RT)
2
1 mx0 a 2
2 r 2) r 2 3a 2
a
3 r a m Substituting x for x0, f (x) for m, and f (x) for a, we
( 2r) r2 2r(a
a r 2, where 0 mx0 0, or x 2(Area of triangle ORT) . r 2) 1 r2 3 r2 mx0, (Area of triangle RST) m
f(r) a Let O designate the origin. Then (b) To simplify the calculations, we shall consider the
volume as a function of r:
volume a. The yintercept of this line is and the xintercept is the solution of Therefore,
V(x) f (x0) be the slope of line RT. Then the equation of line RT is equations in the two unknowns a and b. The second 54. (a) The base radius of the cone is r a3
, the relationship is
3 and h P has coordinates (x0, a), aL 3 2bx. Then y( L) 6
3 55. (a) Let x0 represent the fixed value of x at point P, so that c, so y (0) implies that
ax 3 0. Therefore, y(x) y (x) 0. 0. 3ax 2 Now y (x) a (c) Since r f (x) y3 A(x) f (x) 5 51 x2
,
100 NDER(y1), and
y2 x y1 2
y2 . The graph of the area function y3 A(x) is shown below.
2a
3 2 2 a
3 a 3
3 . Using a3
,
3 and h we may now find the values of r and h for the given
values of a.
[0, 10] by [ 100, 1000] When a 4: r when a 5: r when a 6: r when a 8: r 46
,h
3
56
,h
3 2 4 3
3 5 3
3 ; ; 6, h
6 8
3 2 3;
8 ,h 3
3 The vertical asymptotes at x 0 and x 10
correspond to
horizontal or vertical tangent lines, which do not form
triangles. Section 4.4 167 (c) Using our expression for the yintercept of the tangent line, the height of the triangle is
a mx f (x) f (x) x
1
2
1
2 5
5 x x2 100 2 100
x2
2 100 x2 100 x2 x x2 We may use graphing methods or the analytic method in part (d) to find that the minimum value of A(x) occurs at x 8.66.
Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the ycoordinate of the
center of the ellipse.
(d) Part (a) remains unchanged. The domain is (0, C). To graph, note that
f (x)
f (x) x2
C2 B1 B
B
C2 1
C2 x2 B
C B C2 x 2 and Bx ( 2x) C C2 . x2 Therefore, we have A(x) f (x) 2
f (x) f (x) x C2 C B
C B Bx x x2 Bx
C C2 Bx
C C2 x2 1
BCx C2 BCx C2 BCx C2 x2 1
x2 1 C2 x2)
Bx [Bx 2 (BC B C2 x 2)( [Bx 2 BC C 2 x2 x2 x C2 C2 [BC(C C2 x2 2 B(C 2 C2 BC(C x2 C2 x 2)]2 x 2)]2 x 2)]2 x 2)2
x2 (x C 2 A (x) B (BC x x2 2 C2 x 2)(2)(C C2 x x 2) C2 BC x 2 (C 2 C 2 x 2)
BC(C
22
x (C
x 2) 2x 2 C 2 x 2)
BC(C
22
x (C
x 2) [ BC(C
C 2 x 2)
22
x (C
x 2) ( C2 (C Cx2 2x 2 C2 Cx 2
C2 BC(C
x 2(C 2 C 2 x 2)
[Cx 2
x 2)3/2 BC 2(C
x 2(C 2 C 2 x 2)
(2x 2
x 2)3/2 x2 x2 C C(C 2 C2 C2 x 2) C C2 x 2) C2 (C x2 ( x2
C 2 C2 x2 x2 ) C2 C2 C2 x 2) x
C2 x2 x 2) C C2 x2 x 2 )2 x x 2] x2 (C 2 ) x2 x 2) C2 x 2(1) 168 Section 4.5 55. continued
To find the critical points for 0
2x 2
4x 4 C2 C 4C 2x 2 C4 C4 4x 4 3C 2x 2
3C 2) x2 C 2x 2 0 The minimum value of A(x) for 0
corresponding triangle height is
a C, we solve: 0 x 2(4x 2 C2 x mx f (x) C occurs at the critical point x C3
, or x 2
2 3C 2
. The
4 f (x) x B
C B x C2 x2 Bx2
C C2 x2 2
B 3C B B
C B BC
C2 B B
2 C2 3C 2
4 4 CC 2 3C 2
4 3BC 2
4
C2
2 3B
2 3B
This shows that the triangle has minimum area when its height is 3B. s Section 4.5 Linearization and Newton’s Method (pp. 220–232)
Exploration 1
1. f (x) Approximating with Tangent Lines 2x, f (1) 2, so an equation of the tangent line is y 1 2(x 1) or y 2x 1. 3. Since (y1 y2)(1) y1(1) y2(1) 1 1 0, this view shifts the action from the point (1, 1) to the point (1, 0). Also
(y1 y2) (1) y1 (1) y2 (1) 2 2 0. Thus the tangent line to y1 y2 at x 1 is horizontal (the xaxis). The measure of
how well y2 fits y1 at (1, 1) is the same as the measure of how well the xaxis fits y1 y2 at (1, 0).
4. These tables show that the values of y1 y2 near x
x 1. Here are two tables with Table 0.0001. 1 are close to 0 so that y2 is a good approximation to y1 near Section 4.5
Exploration 2
Grapher Using Newton’s Method on the 1–3. Here are the first 11 computations. 7. (a) x
x 1 169 0
1 (b) 2ex
2ex
x e
e 1
(e 1 0
1)
0.684 2e 8. f (x) 3x 2 4
f (1) 3(1)2 4
1
Since f (1)
2 and f (1)
1, the graph of g(x) passes
through (1, 2) and has slope 1. Its equation is
g(x)
1(x 1) ( 2), or g(x)
x 1.
x 2. dy
dx (x 1)(1 1) sin x) (x
(x 1)2 x x sin x 1
(x sin x
1)2 1 x 2x cos (x 2 2 2.069 2.1 2.072 2.2 2.003 2.3 9. f (x) cos x
f (1.5) cos 1.5
Since f (1.5) sin 1.5 and f (1.5) cos 1.5, the tangent
line passes through (1.5, sin 1.5) and has slope cos 1.5. Its
equation is y (cos 1.5)(x 1.5) sin 1.5, or
approximately y 0.071x 0.891 Quick Review 4.5
d2
(x
dx 1.9 2 1.3 1) 1.871 1.2 cos (x 2 1.8 1.1 dy
dx 1.7 1.688 1.0 1. 1.457 0.8
0.9 2. g(x) 0.7
4. Answers will vary. Here is what happens for x1 f (x) 1) cos x)(1)
cos x cos x (x 1) sin x
(x 1)2 [0, ] by [ 0.2, 1.3] 10. For x , and so f (4) 2 1
. Since
2 x3
1
, the tangent line passes through
2
1
1
(4, 1) and has slope . Its equation is y
(x 4) 1, or
2
2
1
y
x 1.
2 f (4) 3. 1 3, f (x)
1 and f (4) [ 2, 6] by [ 3, 3] x 0.567 4. [ 1, 7] by [ 2, 2]
[ 4, 4] by [ 10, 10] x 0.322 5. f (x) (x)( e x) (e x)(1) e x xe x
f (0) 1
The lines passes through (0, 1) and has slope 1. Its equation
is y x 1.
6. f (x) (x)( e x) ( e x)(1) e x xe x
f ( 1) e 1 ( e 1) 2e
The lines passes through ( 1, e 1) and has slope 2e.
Its equation is y 2e(x 1) ( e 1), or
y 2ex e 1. Section 4.5 Exercises
1. (a) f (x) 3x 2 2
We have f (2) 7 and f (2) 10.
L(x) f (2) f (2)(x 2)
7 10(x 2)
10x 13
(b) Since f (2.1) 8.061 and L(2.1) 8, the
approximation differs from the true value in absolute
value by less than 10 1. 170 Section 4.5
1 2. (a) f (x)
2 x 9 We have f ( 4)
L(x) x 8. (a) f (x) 2 9 4
.
5 5 and f ( 4) f ( 4) 4
x
5 x)6 (1
2 (b) f (x) 1 2 9
5 (b) Since f ( 3.9) 4.9204 and L( 3.9) 4.92, the
approximation differs from the true value by less than
10 3. 1 2[1 ( x)] 1 1/2 x (1 ( 4)) 4) ( x)]6 [1 1 6( x) 1
x
2 2[1 1 6x ( 1)( x)] 2x (c) f (x) f ( 4)(x 4
(x
5 5 x (2x) 2 (d) f (x) x) x2 2
21 x 2 1/2
2
x2
4 21 1 x2
22 x
2 1 21 2 3. (a) f (x) 1 x
We have f (1) 2 and f (1)
L(x) f (1) f (1)(x 1)
2 0(x 1)
2 0. (e) f (x) 41/3 1 (b) Since f (1.1) 2.009 and L(1.1) 2, the
approximation differs from the true value by less than
10 2.
4. (a) f (x) (f) f (x) x 1 2
3 1 2x We have f (0)
L(x) (b) Since f (0.1) 0.0953 and L(0.1) 0.1 the
approximation differs from the true value by less than
10 2.
5. (a) f (x) sec2 x
We have f ( ) 0 and f ( )
L(x) f ( ) f ( )(x
)
0 1(x
)
x f (0) f (0)(x
( 1)(x 2 x x 3
3 and f (0) 2 x 2 1 x 1. 0)
0) (b) Since f (0.1) 1.47063 and L(0.1) 1.47080, the
approximation differs from the true value in absolute
value by less than 10 3. x 6 3x cos x
1 1 and f (0)
f (0)(x 3
2 0) 3
x
2
1
x
2 1.
0.002)100 1 0.021 10 0.009)1/3 1 1.009 (1 1.009 1.003 9 10 6 (100)(0.002) 1 1
(0.009)
3 10 f (x) 5 0(x 1) 5 1 2/3
x
3 We have f (8)
L(x) 1.003; 8 2 and f (8) k1 k. 2
2 11. Center
1
f (x) 4x 4
We have f ( 1)
5 and f ( 1) 0
L(x) f ( 1) f ( 1)(x ( 1))
5
12. Center 2 7. f (x) k(1 x)
We have f (0) 1 and f (0)
L(x) f (0) f (0)(x 0)
1 k(x 0)
1 kx 2/3 1 1 1 10. (a) (1.002)100 (1
1.2;
1.002100 1.2 x2
2 2/3 1 The linearization is the sum of the two individual for 1 We have f (0) f (0)
1 (b) 1 41/3 1 linearizations, which are x for sin x and 1 1. (b) Since f (
0.1) 0.10033 and L(
0.1) 0.1, the
approximation differs from the true value in absolute
value by less than 10 3.
6. (a) f (x) 1 3x
34 3x 1/3
4
x
4 41/3 1 1 9. f (x) 1. 3x)1/3 1 1 We have f (0) 0 and f (0)
L(x) f (0) f (0)(x 0)
0 1x
x L(x) (4 f (8) f (8)(x 8) 1
.
12 2 1
(x
12 8) x
12 4
3 Section 4.5
13. Center 1
(x 1)(1) (x)(1)
1
f (x)
(x 1)2
(x 1)2
1
1
We have f (1)
and f (1)
2
4
1
1
L(x) f (1) f (1)(x 1)
(x
2
4 Then f (x)
xn
1
x
4 1) 1
4 f 3
2 14. Center f 1 0 and f 2 f
x3 x 4
.
25
4
x
25 1x 1. Then f (x) x 3x 2 f (xn) xn 2 xn3 xn 3xn2 1
1 2 x Solution: x 1 f (xn) x4 x 2xn 2xn 2 1 sin xn cos xn f (x) shows that f (x) xn f (xn) xn xn4 xn 4xn 3 1 and 0 has two x1
x2
x3
x4
x5 1 0 has two x4 xn 1 f (xn) xn xn4 xn f (xn) 19. (a) Since dy Solutions: x 2 4xn3 . Note that f (x) 0 clearly has two solutions, namely
4
x
2. We use Newton’s method to find the decimal
equivalents.
x1 1.5
x2 1.2731481
x3 1.1971498
x4 1.1892858
x5 1.1892071
x6 1.1892071 20. (a) Since [ 3, 3] by [ 4, 4] 4x 3 and 2. Then f (x) 1.189207 dy
dx 3x 2 3, dy (b) At the given values,
dy (3 22 3)(0.05) solutions. 1.5
1.455
1.4526332
1.4526269
1.4526269 2
1.9624598
1.9615695
1.9615690
1.9615690 0.386237, 1.961569 Solutions: x 3 f (x) shows that f (x) 2 0.3
0.3825699
0.3862295
0.3862369
0.3862369 18. Let f (x) . 4x 3 3. Then f (x) f (xn) x1
x2
x3
x4
x5 Solutions: x 0.682328 The graph of y x1
x2
x3
x4
x5 xn2 xn 1 and Note that f is cubic and f is always positive, so there is
exactly one solution. We choose x1 0.
x1 0
x2 1
x3 0.75
x4 0.6860465
x5 0.6823396
x6 0.6823278
x7 0.6823278 xn xn cos x and 9
25 1. 2 0 2 f (xn) xn 16. Let f (x) 2 f (xn) sin x. [ 4, 4] by [ 3, 3] 2 15. Let f (x)
xn 3
2
3
2 sin x We have f
L(x) 3
and f
5
4
x
25 2 f (x) 2x 1 solutions 3
3
, we have f
2
2
3
3
3
f
x
2
2
5 Using center 1 2x The graph of y Alternate solution: L(x) x2 17. Let f (x) 171 (3x 2 3) dx. 9(0.05) 0.45. (1 x 2)(2) (2x)(2x)
dy
(1 x 2)2
dx
2
2 2x
dx.
(1 x2)2 2
(1 2x 2
,
x2)2 (b) At the given values,
x1
x2
x3
x4
x5 1.2
1.6541962
1.1640373
1.1640351
1.1640351 1.452627, 1.164035 2
[1 dy 2( 2)2
(0.1)
( 2)2]2 2 8
52 (0.1) 0.024.
21. (a) Since
dy dy
dx 1
x (x 2 ) (2x ln x (ln x)(2x) 2x ln x x, x) dx. (b) At the given values,
dy [2(1) ln (1) 1](0.01) 1(0.01) 0.01 172 Section 4.5
dy
dx
x2 22. (a) Since 1 dy 2
x 1 1 (x) (1 x2 x2 1 2 ( 2x) x2 1 x 2) (1
1 2x
1 x 2x 2 1 2 29. (a) 1 x2 , f f (0.55) 20
11 f (0.5) (b) Since f (x) 2 x 2
11 2 , f (0.5) 4. 2 x 2 Therefore, df dx. (0)2 1 e sin x cos x, dy f df 30. (a) ( 0.2) 4 dx f f (1.01) 0.2. 4(0.05) 1
5 2
11 (c) 2(0)2 1 (b) At the given values, dy
dy
23. (a) Since
dx x 2)(1) f (1) 1
55 1.04060401 (b) Since f (x) 4x 3, f (1) 4.
Therefore, df 4 dx 4(0.01) (cos x)e sin x dx. (c) f 1
5 0.2 df 0.04060401 dV
dr 4 r 2, dV 0.04 1 0.04060401 0.04.
0.00060401 (b) At the given values,
(cos )(e sin )( 0.1) dy ( 1)(1)( 0.1) 0.1. 31. Note that
a to a dy
24. (a) Since
dx x
3 csc 1
cot 1
3
x
x
csc 1
cot 1
,
3
3
x
x
dy csc 1
cot 1
dx.
3
3 1
3 x
3 25. (a) y(1 x Since
dy a to a x a to a x1
1)(1) (x)(1)
(x 1)2 dy
(x
dx
dx
.
(x 1)2 (x dV
dx dS
dx dy
dx 1) dV
dr 36. Note that
sec (x 2 1) tan (x 2 1) tan (x 2 2(1.5) sec (1.52 1) (2x), 1 ) dx. 1) tan (1.52 0.15 sec 1.25 tan 1.25
27. (a) f f (0.1) (c) f df f f (1.1) f (0) 0 0.21 0.2 f (1) 1) (0.05) 0.21 0.01 0.231 0 1, f (1)
(b) Since f (x) 3x
Therefore, df 2dx 2(0.1)
f df dr, the change in volume is approximately 0.231 0.2 0.231 2.
0.2. 0.031 a to a dS
dh 2 r, so dS 2 r dh. When h changes from dh, the change in lateral surface area is 37. (a) Note that f (0) cos 0 1.
L(x) f (0) f (0)(x 0) 1 1.431663. 0.21 2 (c) 2 rh dr. When r changes approximately 2 r dh. (b) Since f (x) 2x 2, f (0) 2.
Therefore, df 2 dx 2(0.1) 0.2.
28. (a) 2 rh, so dV 2 ah dr. (b) At the given values,
dy 12x dx. When x changes from 0.01. 2x sec (x 2 dy 12x, so dS 2, from a to a 26. (a) Since 3x 2 dx. When x changes from dx, the change in surface area is approximately 35. Note that 0.01
(0 1)2 3x 2, so dV 12a dx. 1 (b) At the given values,
dy 8 r dr. When r changes from dx, the change in volume is approximately 3a 2 dx. 34. Note that
x y 8 r, so dS dr, the change in surface area is approximately 33. Note that 0 x) dS
dr 8 a dr. 1
1
cot 1
(0.1)
3
3
2
2
0.1 csc cot
0.205525
3
3 xy 4 a 2 dr. a to a csc 1 y dr, the change in volume is approximately 32. Note that (b) At the given values,
dy 4 r 2 dr. When r changes from (b) f (0.1) L(0.1) 1x x 1 1.1 (c) The actual value is less than 1.1. This is because the
derivative is decreasing over the interval [0, 0.1], which
means that the graph of f (x) is concave down and lies
below its linearization in this interval. Section 4.5
dA
dr r 2 and 38. (a) Note that A 2 r, so dA When r changes from a to a 2 r dr. dr, the change in area is approximately 2 a dr. Substituting 2 for a and 0.02 for 43. Let angle of elevation and h
dh
30 tan , so
d h want dh 2 30 sec height of building. Then 0.04h, which gives: dA
(b)
A 0.08
4 39. Let A
D 0.08 dA D2
2 (2) C dD
1
1
and dD
dC.
dC
C2
dA
C
, so
and
4
dC
2 40. Let x 10
(2)
2 edge length and V
3x2 dx. With x have V 103 0.04 sin 5
radians, we have
12
5
5
0.04 sin
cos
0.01 radian. The angle should
12
12 75 be measured with an error of less than 0.01 radian (or
approximately 0.57 degrees), which is a percentage error of
approximately 0.76%.
44. Note that dV
dh 3 h 2, so dV 3 h 2 dh. We want 0.01V, which gives 3 h 2 dh dV 0.01( h 3), or 0.01h
. The height should be measured with an error
3
1
of no more than %.
3 dh 2 10 in .
volume. Then V x 3, and so
r 2h 45. (a) Note that V
10 cm and dx 1000 cm3 and dV cos in. to 0.6366 in. and the area increases by approximately dA dV , so C2
2 2 0.04 sin
cos d circumference, and 2 in. the diameter increases by
1 d For C
dC. When C increases from 10
2 10
dD 2% cross section area, C Also, A 0.04 (30 tan ) 0.2513 0.02 diameter. Then D d 1
cos2 d 2 (2)(0.02) 30 sec2 d . We and dh 30 sec2 dr, the change in area is approximately 173 0.01x 10 r 2 2.5 D2, where D is the 0.1 cm, we 3(10)2(0.1) 30 cm3, interior diameter of the tank. Then
dV 5 D dD. We want dV dV
dD 5 D, so 0.01V, which gives so the percentage error in the volume measurement is
approximately
41. Let x
dA
dx dV
V 30
1000 side length and A
2x, so dA gives 2x dx 5 D dD
0.03 0.01(2.5 D 2), or dD 0.005D. The 3%.
interior diameter should be measured with an error of area. Then A 2x dx. We want dA
0.02x 2, or dx x2 and no more than 0.5%. 0.02A, which
(b) Now we let D represent the exterior diameter of the 0.01x. The side length
tank, and we assume that the paint coverage rate should be measured with an error of no more than 1%.
(number of square feet covered per gallon of paint) is
42. The volume of a cylinder is V
fixed, we have
For h dV
dr 30 in., r 2 r h. When h is held 2 rh, and so dV
6 in., and dr 2 rh dr. paint within 5%, we need to calculate the lateral 0.5 in., the thickness of the shell is approximately
dV 2 rh dr 2 (6)(30)(0.5) known precisely. Then, to determine the amount of surface area S with an error of no more than 5%. Note
that S 180 3 565.5 in . dS 2 rh 10 D, so 10 dD. We want dS 10 dD dS
dD 10 and
0.05S, which gives 0.05(10 D), or dD 0.05D. The exterior diameter should be measured with an error of no more
than 5%. 174 Section 4.5
r 2h, we have 46. Since V
We want dV dV
dr 2 rh, and dV 52. (a) i. Q(a) f (a) implies that b0 f (a). ii. Since Q (x) b1 2b2(x
implies that b1 f (a). 0.0005r. The variation of a), Q (a) iii. Since Q (x) 0.001V, which gives 0.001 r 2h, or dr 2 rh dr 2 rh dr. f (a) implies that 1
the radius should not exceed
of the ideal radius, that
2000 2b2, Q (a) f (a) f (a)
2 b2 is, 0.05% of the ideal radius.
In summary, b0
47. We have
Then dW
dg 2 bg dWmoon , so dW b(5.2)
b(32) dWearth 2
2 bg 2 dg. 2 dg
dg 32
5.22 37.87. The ratio is Since f (0)
2 L 1/2g L 1/2g dT 3/2 1/2 , so dT
dg L 1/2g 3/2 L1/2g
(100)1/2(980) 49. If f (x) h x1 f (x1) 0
f (x1) x1 f (x1) 1
and x2
2h 1/2 h, then f (x1) f (x1) x1. h h h 1/2 x x 2. As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(d) g(x) x 1
g (x)
x
g (x) 2x 1
2h 1/2 and x2 h h 1
2h 1/2 1, g (1) 1, and g (1)
1, b1 1, and b2 2, the
2
2 1. The quadratic approximation is
Q(x) h 1/2 2
3 coefficients are bo h, then h. If x1
2 1 1. The [ 2.35, 2.35] by [ 1.25, 3.25] 979.0235. Since g(1) 1 2
2 1, and b2 3/2 x1, and all later approximations are also 2h 1, b1 2, the (c) equal to x1.
50. If x1 1, and f (0) 3/2 0, we have x2 Therefore x2 (1 x) 2
2(1 x) 3 quadratic approximation is Q(x) dg. dg dT
dg 0.001
dg
0.9765
0.9765, g 980 0.9765 Since dg coefficients are b0 and (b) Note that dT and dg have opposite signs. Thus, if g
increases, T decreases and the clock speeds up.
(c) 1, f (0) f (a)
.
2 f (a), and b2 (b) f (x) (1 x) 1
f (x)
1(1 x) 2( 1)
f (x)
2(1 x) 3( 1) about 37.87 to 1.
48. (a) Note that T f (a), b1 2h 1 (x 1) (x 1)2. h [ 1.35, 3.35] by [ 1.25, 3.25] As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical. [ 3, 3] by [ 0.5, 2] 51. Note that f (x)
xn xn1/3
xn 2/3 1 2/3
x
and so xn 1
3 xn 3xn xn 2xn. For x1 f (xn)
f (xn) 1, we have 3 x2 2, x3 4, x4 [ 10, 10] by [ 3, 3] 8, and x5 16; xn 2n 1. Section 4.6
(e) h(x)
h (x)
h (x) Since h(0) x)1/2 (1 55. g(a) 1
(1 x) 1/2
2
1
(1 x) 3/2
4
1
1, h (1)
, and h (1)
2 Then E(x) f (x) 0, then g(a)
g(x) f (x) f (a) and c
f (a) m(x f (a).
a). f (x) f (a)
E(x)
m.
xa
xa
f (x) f (a)
E(x)
lim
f (a), so lim
f (a) m.
xa
a
x→a
x→a x
E(x)
Therefore, if the limit of
is zero, then m f (a) and
xa Thus, 1
, the
4
1
4 1
, and b2
2 1, b1 coefficients are b0 c, so if E(a) 175 The quadratic approximation is Q(x) 1
.
8 2 g(x) L(x). x2
.
8 x
2 1 s Section 4.6 Related Rates (pp. 232–241)
Exploration 1 Sliding Ladder 1. Here the xaxis represents the ground and the yaxis
represents the wall. The curve (x1, y1) gives the position of [ 1.35, 3.35] by [ 1.25, 3.25] the bottom of the ladder (distance from the wall) at any As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(f) The linearization of any differentiable function u(x) at
x a is L(x) u(a) u (a)(x a) b0 b1(x a), where b0 and b1 are the coefficients of the constant and
linear terms of the quadratic approximation. Thus, the
linearization for f (x) at x
for g(x) at x 1 is 1 0 is 1 (x linearization for h(x) at x x; the linearization 1) or 2 13
. The curve (x2, y2) gives the position
3
13
of the top of the ladder at any time in 0 t
.
3
13
2. 0 t
3 time t in 0 3. This is a snapshot at t 3.1. The top of the ladder is
moving down the yaxis and the bottom of the ladder is
moving to the right on the xaxis. Both axes are hidden
from view. x; and the
x
.
2 0 is 1 53. Just multiply the corresponding derivative formulas by dx.
d
(a) Since (c)
dx 0, d(c) du
dx d
(c) Since (u + v)
dx
d
(d) Since (u
dx (e) Since du
dx v d
(f) Since u n
dx 54. tan x
x→0 x lim 0. v)
du
v
dx v2 nu n 1 du dx v) du v) , d(u ) n1 sin x/cos x
x
x→0
1 sin x
lim
x→0 cos x x
sin x
1
lim
lim
x→0 cos x x→0 x 0.348 ft/sec , y (1) 0.712 ft/sec2, 1.107 ft/sec2, y (2) 1.561 ft/sec2. u dv
v2 nu du. . v du. 1. , the speed of the top of the ladder t→(13/3) is infinite as it hits the ground. Quick Review 4.6
1. D
2. D (7 0)2 (b 2 0) (0 5)2 49 25 (0 2 2 2 a) a 3. Use implicit differentiation. lim (1)(1) 9t 2
2 dv u dv v du n 132 Since lim y (t) u
v ,d 9t y (t) y (1.5) du
v , d(u
dx dv
u
dx dy
dt y (0.5) c du. dv
, d(u
dx dv
u
dx [ 1, 15] by [ 1, 15] 4. du
c , d(cu)
dx d
(b) Since (cu)
dx t d
(2xy
dx
dy
2x
dx 2y(1)
(2x 2y y 2)
dy
2y
dx
dy
1)
dx
dy
dx d
(x
dx (1)
1 2y
1 2x y)
dy
dx
2y
2y 1 b 74 176 Section 4.6 4. Use implicit differentiation.
d
(x sin y)
dx (x)(cos y) dy
dx (x (sin y)(1)
x cos y) 3. (a) Since
d
(1
dx
dy
x
dx dy
dx
dy
dx
dy
dx y 2x
dy
dx
dy
dx dV dh
dV
, we have
dh dt
dt (b) Since dV
dt dV dr
dV
, we have
dr dt
dt xy)
y(1)
sin y (c) y sin y
x x cos y
y sin y
x x cos y 5. Use implicit differentiation.
d2
x
dx dV
dt 4. (a) d
tan y
dx
dy
sec2 y
dx
2x
sec2 y 2x cos2 y dV
dt
dV
dt
dV
dt d2
rh
dt
dh
r2
dt
dh
r2
dt dP
dt
dP
dt
dP
dt
dP
dt d
ln (x y)
dx
dy
1
1
dx
xy
dy
1
dx
dy
dx dh
.
dt
dr
dt 2 rh . d
(RI 2)
dt
dR
d
R I2 I2
dt
dt
dI
dR
R 2I
I2
dt
dt
dI
dR
2RI
I2
dt
dt d2
(r h)
dt
dr
h(2r)
dt
dr
2 rh
dt (b) If P is constant, we have 6. Use implicit differentiation. r2 2RI d
(2x)
dx dI
dt dR
dt 0, or y2 I2 dP
dt 0, which means dR
dt 2R dI
I dt 2P dI
.
I 3 dt z2 2
2(x
2x 5. y)
2y 8. Using A(0, 4), we create the parametric equations
x 0 at and y
4 bt, which determine a line
passing through A at t 0. We now determine a and b so
that the line passes through B(5, 0) at t 1. Since
5 0 a, we have a 5, and since 0
4 b, we have
b 4. Thus, one parametrization for the line segment is
x 5t, y
4 4t, 0 t 1. (Other answers are
possible.) 10. One possible answer: 2
3
2 d
dt x2 1
ds
dt 7. Using A( 2, 1) we create the parametric equations
x
2 at and y 1 bt, which determine a line
passing through A at t 0. We determine a and b so that
the line passes through B(4, 3) at t 1. Since
4
2 a, we have a 6, and since 3 1 b, we
have b
4. Thus, one parametrization for the line
segment is x
2 6t, y 1 4t, 0 t 1. (Other
answers are possible.) 9. One possible answer: ds
dt t dA
dt dA dr
dA
, we have
dr dt
dt 2. Since dS
dt dS dr
dS
, we have
dr dt
dt 6. 2x 2 y y 2 2 dx
x
dt dy
y
dt x2 dA
dt
dA
dt
dA
dt y2 (x 2 y2 dx
dt 2y z 2) z dt 1
z
z 2 2x dy
dt 2z dz
dt dz
dt z2 d1
ab sin
dt 2
1 da
db
b sin
a
2 dt
dt
1
da
db
b sin
a sin
2
dt
dt dV
dt 2 sin ab ab cos d
sin
dt d
dt 1 volt/sec. (b) Since I is decreasing at the rate of
dI
dt Section 4.6 Exercises
1. Since ds
dt 2x d
2 7. (a) Since V is increasing at the rate of 1 volt/sec, 3
2 t ds
dt 1
2 dr
dt 2r .
dr
dt 8r . 1
amp/sec.
3 (c) Differentiating both sides of V
dV
dt 1
amp/sec,
3 I dR
dt dI
dt R. IR, we have Section 4.6
(d) Note that V IR gives 12 2R, so R 6 ohms. Now
substitute the known values into the equation in (c).
1 2 dR
dt (b) P 1
3 6 dR
2
dt
3
ohms/sec
2 3
dR
dt 3
ohms/sec. Since this
2 At the instant in question, dr
dt 0.01 cm/sec, r dA
.
dt Step 5:
2r dr
dt 52 At the instant in question,
cm2/sec 2 (50)(0.01) At the instant in question, the area is increasing at the rate
cm2/sec. At the instant in question,
dw
dt 2 cm/sec, 2 cm/sec, l 12 cm, 5 cm. dy
dt y = 3 m, and z We want to find 2 m/sec, dz
dt 1 m/sec, x 4 m, 2 m. dA dP
dD
, , and
.
dt dt
dt dV dS
ds
, , and .
dt dt
dt Steps 4, 5, and 6:
(a) V xyz
xy dz
dt xz (4)(3)(1) dy
dt yz dx
dt (4)(2)( 2) dS
dt
dS
dt 2(xy xz dy
2x
dt yz)
dx
y
dt x dz
dt z dx
dt y lw 2[(4)( 2) (3)(1)
(3)(1) (2)( 2)] dz
dt z dy
dt (4)(1) (2)(1) Steps 4, 5, and 6: 2 m3/sec (3)(2)(1) The rate of change of the volume is 2 m3/sec.
(b) S Step 3: dw
l
dt 1 m/sec, dV
dt
dV
dt Step 2: We want to find dx
dt Step 3: 9. Step 1:
l length of rectangle
w width of rectangle
A area of rectangle
P perimeter of rectangle
D length of a diagonal of the rectangle dA
dt
dA
dt dl
2l
2
2
dt
2l
w
(12)( 2) (5)(2)
122 dw l
w
dt
dt
dw
2w
2
dt
l
w2
14
cm/sec
13 1 Step 2: Step 6: (a) A w2 10. Step 1:
x, y, z edge lengths of the box
V volume of the box
S surface area of the box
s diagonal length of the box r2 and w 0 cm/sec (d) The area is increasing, because its derivative is positive.
The perimeter is not changing, because its derivative is
zero. The diagonal length is decreasing, because its
derivative is negative. Step 4: dl
dt 2(2) 14
cm/sec.
13 50 cm. Step 3: of 2( 2) The rate of change of the length of the diameter is Step 2: dA
dt dw
dt dl dD
dt
dD
dt 8. Step 1:
r radius of plate
A area of plate dA
dt 2 l2 (c) D value is positive, R is increasing. A 2w dl
2
dt The rate of change of the perimeter is 0 cm/sec. R is changing at the rate of We want to find 2l dP
dt
dP
dt 177 0 m2/sec The rate of change of the surface area is 0 m2/sec.
dl
w
dt (12)(2) (5)( 2) 14 cm2/sec The rate of change of the area is 14 cm2/sec. 178 Section 4.6
13. Step 1:
x distance from wall to base of ladder
y height of top of ladder
A area of triangle formed by the ladder, wall, and ground
angle between the ladder and the ground 10. continued
x2 (c) s y2 z2 1 ds
dt 2 x 2 y2 2x z2 dx
dt 2y dy
dt 2z dz
dy Step 2:
dx
x
dt dy
y
dt x 2 y dz
z
dt 2 z (4)(1) ds
dt At the instant in question, x 2 (2)(1) 32 0 22 29 Step 4, 5, and 6: 11. Step 1:
s (diagonal) distance from antenna to airplane
x horizontal distance from antenna to airplane ds
dt Step 4:
x 2 49 y 300 mph. dx
.
dt s2 x2 169 169 122 5. rate of 12 ft/sec. (Note that the downward rate of 49 motion is positive.)
2s s2 49 ds
dt s
s2 ds
49 dt 10
102 (300) 3000 49 mph 1
xy
2
1 dy
x
2 dt (b) A Step 6:
dx
dt 0 The top of the ladder is sliding down the wall at the s 2 or x 1
2 169
dy
2y
dt dy
Then 2(12)(5) 2(5)
0
dt
dy
dy
12 ft/sec or
12 ft/sec
dt
dt Step 5:
dx
dt y2 To evaluate, note that, at the instant in question, Step 3:
We want to find (a) x 2
dx
2x
dt Step 2:
At the instant in question,
10 mi and 5 ft/sec. dy dA
d
, , and .
dt dt
dt We want to find 0 m/sec The rate of change of the diagonal length is 0 m/sec. s dx
dt Step 3: (3)( 2)
42 12 ft and dA
dt 420.08 mph y dx
dt Using the results from step 2 and from part (a), we 51 The speed of the airplane is about 420.08 mph. have 12. Step 1:
h height (or depth) of the water in the trough
V volume of water in the trough dA
dt 1
[(12)( 12)
2 (5)(5)] 119
ft/sec.
2 The area of the triangle is changing at the rate of
59.5 ft2/sec. Step 2:
At the instant in question, dV
dt 2.5 ft3/min and h 2 ft. y
x (c) tan
2 x d
dt Step 3: sec dh
We want to find .
dt Since tan Step 4:
4
3 cos The width of the top surface of the water is h, so we have
V 1
4
(h) h (15), or V
2
3 10h 2 20h y dx
dt x2
5
, we have for 0
12
1 12
and so sec2
13 dh
dt d
dt Step 6: of
dh
dt 2.5 20(2) dh
dt 0.0625 1
ft/min
16 The water level is increasing at the rate of 1
ft/min.
16 2 12 2
13 169
.
144 Combining this result with the results from step 2 and
from part (a), we have Step 5:
dV
dt dy
dt 169 d
144 dt (12)( 12) (5)(5)
, so
122 1 radian/sec. The angle is changing at the rate
1 radian/sec. Section 4.6
14. Step 1: 179 Step 5:
Kite
s Inge dV
dt 12 r dr
dt Step 6: 300 ft dV
dt x 1
3000 12 (1.900) 19
2500 0.0076 3 0.0239 in /min
s
x length of kite string
horizontal distance from Inge to kite The volume is increasing at the rate of approximately
0.0239 in3/min. Step 2:
At the instant in question, dx
dt 25 ft/sec and s 500 ft 16. Step 1: Step 3:
We want to find h ds
.
dt Step 4:
x 2 r 300 2 s 2 r
h
V Step 5:
2x dx
dt 2s ds
dx
or x
dt
dt s ds
dt base radius of cone
height of cone
volume of cone Step 2: Step 6:
At the instant in question, since x 2
x s2 3002 5002
ds
dt Thus (400)(25) (500) , so 3002 3002
ds
dt s 2, we have At the instant in question, h 4 m and dV
dt 10 m3/min. Step 3: 400.
20 ft/sec. Inge must let the string out at the rate of 20 ft/sec. We want to find dh
dr
and .
dt
dt Step 4: 15. Step 1: Since the height is 3
of the base diameter, we have
8 4
h.
3
12
142
16 h 3
We also have V
rh
hh
. We will
3
33
27
16 h 3
4
use the equations V
and r
h.
27
3 h
6 in. 3
(2r) or r
8 Steps 5 and 6:
r (a) The cylinder shown represents the shape of the hole.
r radius of cylinder
V volume of cylinder
Step 2:
At the instant in question, dr
dt 0.001 in.
3 min and (since the diameter is 3.800 in.), r
Step 3: Step 4:
r 2(6) 1.900 in. 10
dh
dt 16 h 2 dh
9
dt
16 (4)2 dh
9
dt
45
m/min
128 1125
cm/min
32 The height is changing at the rate of
1125
32 11.19 cm/min. (b) Using the results from Step 4 and part (a), we have dV
We want to find .
dt V 1
in./min
3000 dV
dt dr
dt 4 dh
3 dt 4 1125
3 32 375
cm/min.
8 The radius is changing at the rate of
6 r2 375
8 14.92 cm/min. 180 Section 4.6 17. Step 1: Step 6:
45 m
r 6
6 6m 144 dy
dt h or
r
h
V radius of top surface of water
depth of water in reservoir
volume of water in reservoir r dV
dt At the instant in question, 50 m3/min and h 5 m. Step 3:
We want to find 0.01326 m/min 25
6 1.326 cm/min
y)2 (13 169 132, y)2 (13 26y dh
dr
and .
dt
dt Step 2:
At the instant in question, Note that
Then V h
6
by similar cones, so r 7.5h.
r
45
12
1
rh
(7.5h)2h 18.75 h3
3
3 We want to find the value of
18.75 h 3, dV
dt 56.25 h 2 dh
.
dt From part (b), r 7.5h, dr
dt dr
dt dh
dt 80
cm/min. The rate
3 or of change of the radius of the water’s surface is
80
3 18. (a) Step 1:
y depth of water in bowl
V volume of water in bowl 1
y2 26y 2 (26 2y) dy
dt 13
26y y dy
y 2 dt 13 8 26(8) 5
288
125
72 1
24 82 5
12 1
24 0.00553 m/min
0.553 cm/min Step 2:
No numerical information is given. Step 2:
dV
dt y 2. 26y 19. Step 1:
r radius of spherical droplet
S surface area of spherical droplet
V volume of spherical droplet 8.49 cm/min. At the instant in question, dr
.
dt Step 6:
dr
dt 7.5 1
m/min.
24 Step 5: falling is positive.)
(b) Since r dy
dt 8 m, Step 4: 56.25 (52) 50 6 m3/min, y Step 3: dh
, and so
dt
dh
8
32
m/min
cm/min.
dt
225
9
32
The water level is falling by
1.13 cm/min.
9
dh
(Since
0, the rate at which the water level is
dt Thus dV
dt and therefore (from part (a)) Step 5 and 6:
(a) Since V y 2. (c) Step 1:
y depth of water
r radius of water surface
V volume of water in bowl Step 4: y dy
dt dy
dt 1
24 (b) Since r 2 Step 2: (82)] [26 (8) 6 m3/min and Step 3:
We want to show that 8 m. Step 3: Step 4: dy
We want to find the value of .
dt S Step 4: dr
is constant.
dt Steps 5 and 6: V 3 y 2(39 y) or V Step 5:
dV
dt (26 y y 2) dy
dt 13 y 2 3 y3 4 r 2, V 4 3 dV
r,
3
dt kS for some constant k 43
dV
dr
r , we have
4 r2 .
3
dt
dt
dV
Substituting kS for
and S for 4 r 2, we have kS
dt
dr
or
k.
dt Differentiating V dr
dt S, Section 4.6
20. Step 1:
r radius of spherical balloon
S surface area of spherical balloon
V volume of spherical balloon (b) 16 cos l 1 d
dt Step 2:
At the instant in question, dV
dt 100 ft3/min and r 1 d
dt dr
dS
We want to find the values of and .
dt
dt 6 dl
l 2 dt 62
l 5 ft. Step 3: 1
0.6 1 6
( 2)
102 2 The rate of change of angle 3
radian/sec
20
3
radian/sec.
20 is Steps 4, 5, and 6:
(a) V
dV
dt dr
dt 4 (5)2 100
dr
dt 22. Step 1:
x distance from origin to bicycle
y height of balloon (distance from origin to balloon)
s distance from balloon to bicycle 43
r
3 4 r2 dr
dt Step 2: 1 ft/min dS
dt
dS
dt
dS
dt 4 r2
8r dy
dx
is a constant 1 ft/sec and
is a constant
dt
dt We know that The radius is increasing at the rate of 1 ft/min.
(b) S 17 ft/sec. Three seconds before the instant in question, the
values of x and y are x 0 ft and y 65 ft. Therefore, at
the instant in question x 51 ft and y 68 ft.
Step 3: dr
dt 8 (5)(1) We want to find the value of 40 ft2/min ds
at the instant in question.
dt Step 4:
x2 The surface area is increasing at the rate of s 40 ft2/min. y2 Step 5:
ds
dt 21. Step 1:
l length of rope
x horizontal distance from boat to dock
angle between the rope and a vertical line dl
At the instant in question,
dt 2 ft/sec and l ds
dt 10 ft. 2 dx
dt
dx
dt 2 51 dc
dt l2
102 dx
dt x2 y dy
dt y2 11 ft/sec 682 6x 2
12x [3(2)2 15x)
15) 12(2) dx
dt 15](0.1) 0.3
dr
dt
dp
dt dl
36 dt 10 dy
2y
dt (68)(1) d3
(x
dt (3x 2 dx
d
and .
dt
dt 36
l x2 (51)(17) Step 4, 5, and 6:
l2 dx
2x
2
dt
y The distance between the balloon and the bicycle is
increasing at the rate of 11 ft/sec.
23. (a) Step 3:
We want to find the values of x
1 Step 6: Step 2: (a) x 181 ( 2) 2.5 ft/sec 36 The boat is approaching the dock at the rate of (b) dx
d
(9x) 9
9(0.1) 0.9
dt
dt
dr
dc
0.9 0.3 0.6
dt
dt dc
dt d3
x
dt 3x 2 2.5 ft/sec. 45
x
45 dx
x 2 dt 6x 2
12x 3(1.5)2 12(1.5) 45
(0.05)
1.52 1.5625
dr
dt
dp
dt d
(70x)
dt
dr
dc
dt
dt 70 dx
dt 3.5 70(0.05)
( 1.5625) 3.5
5.0625 182 Section 4.6 24. (a) Note that the level of the coffee in the cone is not
needed until part (b).
Step 1:
V1 volume of coffee in pot
y depth of coffee in pot
Step 2:
dV1
dt Step 2: 10 in3/min At the instant in question, Q Step 3:
We want to find the value of dy
dt dD
dt . dt y dy
dt dy
dt Q
D Step 5: Step 6:
9 0 mL/min . Step 4: 9 10 41 mL/L, 2 dy
.
dt We want to find the value of 9y Step 5:
dV1 233 mL/min, D dQ
2 (mL/L)/min, and
dt Step 3: Step 4:
V1 25. Step 1:
Q
rate of CO2 exhalation (mL/min)
D
difference between CO2 concentration in blood
pumped to the lungs and CO2 concentration in blood
returning from the lungs (mL/L)
y
cardiac output dy
dt 10
9 dy
dt 0.354 in./min D dQ
dt Q dD
dt D2 The level in the pot is increasing at the rate of Step 6: approximately 0.354 in./min. dy
dt (b) Step 1:
V2 volume of coffee in filter
r radius of surface of coffee in filter
h depth of coffee in filter (41)(0) 466
1681 0.277 L/min2 The cardiac output is increasing at the rate of
approximately 0.277 L/min2.
26. Step 1:
y Step 2:
At the instant in question,
h (233)( 2)
(41)2 dV2
dt 10 in3/min and
(x , y ) 5 in. Step 3:
We want to find dh
.
dt x Step 4:
Note that 3
, so r
6 r
h 12
rh
3 Then V2 h
.
2 h3
.
12 x
y Step 5:
dV2
dt h 2 dh
4 dt xcoordinate of particle’s location
ycoordinate of particle’s location
angle of inclination of line joining the particle to the
origin. Step 2: Step 6: At the instant in question, (5)2 dh
10
4 dt
dh
8
in./min
dt
5
dh
Note that
0, so the rate at which the level is
dt dx
dt Step 3: falling is positive. The level in the cone is falling at the Step 4: 8
rate of
5 Since y 0.509 in./min. 10 m/sec and x We want to find x 0, 3 m. d
.
dt x 2, we have tan
tan 1 x. y
x x2
x x and so, for Section 4.6
Step 5:
d
dt 1
1 28. Step 1:
x xcoordinate of particle
y ycoordinate of particle
D distance from origin to particle dx
x 2 dt Step 6:
d
dt 1
32 1 183 (10) Step 2: 1 radian/sec At the instant in question, x The angle of inclination is increasing at the rate of dx
dt 1 radian/sec. 1 m/sec, and dy
dt 5 m, y 12 m, 5 m/sec. Step 3: 27. Step 1: We want to find y dD
.
dt Step 4:
x2 D
(x , y ) y2 Step 5:
θ dD
dt x x
1 2 dx
2x
dt
y2 x2 dy
2y
dt dx
dt x2 y dy
dt y2 Step 6:
x
y (5)( 1) dD
dt xcoordinate of particle’s location
ycoordinate of particle’s location
angle of inclination of line joining the particle to the
origin (12)( 5) 52 5 m/sec 122 The particle’s distance from the origin is changing at the
rate of 5 m/sec. 29. Step 1: Step 2:
At the instant in question, dx
dt 8 m/sec and x Street
light 4 m. Step 3:
We want to find d
.
dt 16 ft Step 4: 6 ft Since y x, we have tan and so, for x y
x x 0,
1 tan [ ( x) 1/2 x tan 1 ( x) ( x) 1/2 . Shadow 1/2 , x x
s s distance from streetlight base to man
length of shadow Step 2:
Step 5:
d
dt 1 1
[( x)
1
1
x 1 1/2 2 x(x 1
dx
( x) 3/2( 1)
2
dt dx
1
2( x)3/2 dt 1
2 10 ft. Step 3:
We want to find ds
.
dt By similar triangles, dx
1) dt 16s 6s 6x, or s s s
6 x
16 . This is equivalent to 3
x.
5 Step 5:
1 2 5 ft/sec and x Step 4: Step 6:
d
dt dx
dt At the instant in question, 4( 4 ( 8)
1) 2
radian/sec
5 ds
dt 3 dx
5 dt The angle of inclination is increasing at the rate of Step 6: 2
radian/sec.
5 ds
dt 3
( 5)
5 3ft/sec The shadow length is changing at the rate of 3 ft/sec. 184 Section 4.6
Step 2: 30. Step 1:
s distance ball has fallen
x distance from bottom of pole to shadow At the instant in question,
dV
dt Step 2:
At the instant in question, s
ds
dt 1
32
2 16 12
2 4 ft and Step 4: dx
.
dt 43
r and S
3 We have V Step 4:
x 30
By similar triangles,
50 s x 1500 50x
1 1500s x
. This is equivalent to
50 sx, or sx 1500. We will use 2 ds dS
dt 1500(4) 2(16) 1500 ft/sec. 31. Step 1:
x
position of car (x 0 when car is right in front of
you)
camera angle. (We assume is negative until the car
passes in front of you, and then positive.) At the first instant in question, x
1
(264)
2 A half second later, x 0 ft and dx
dt 264 ft/sec. 2 1/3 dV 3V
4 dt 4
4000
(10)3
.
3
3
1/3
4000
( 8)
3 2 33. Step 1:
p xcoordinate of plane’s position
x xcoordinate of car’s position
s distance from plane to car (lineofsight) At the instant in question,
dp
dt 0, 120 mph, s d
at each of the two instants.
dt Step 4:
(x Step 4:
1 x
132 p)2 32 s2 Step 5:
2(x Step 5:
1 1 dx
132 dt x2
132 dx
dt p) dp
dt 2s ds
dt Step 6:
Note that, at the instant in question, Step 6:
1 d
0:
dt 132: 5 mi, and dx
.
dt We want to find We want to find When x 3 dV
4 dt Step 3: Step 3: When x 1/3 3V
4 area is decreasing at the rate of 1.6 cm2/min. p 132 ft and 264 ft/sec. 1 2
3 4 Step 2: Step 2: d
dt 3V 2/3
4 16
3
1.6 cm2/min
3
4
1000
dS
Since
0, the rate of decrease is positive. The surface
dt 1500 ft/sec The shadow is moving at a velocity of tan 4 Step 5: Note that V dt Step 6: dx
dt 3V 1/3
, so S
4 Step 6:
1500s dx
dt 4 r 2. These equations can be combined by noting that r dS
dt . Step 5:
dx
dt dS
.
dt We want to find Step 3: 50x 1
(20) = 10 cm.
2 Step 3: 16 ft/sec. We want to find 8 cm3/min and r 8 mL/min 02
132 1
d
dt 1
(264)
132 1
1 132
132 2 32. Step 1:
r radius of balls plus ice
S surface area of ball plus ice
V volume of ball plus ice 1
(264)
132 52 x
2 radians/sec 1 radian/sec 2(4 32 dx
dt
dx
8
dt
dx
dt 0) 4 mi.
120 2(5)( 160) 120
120
dx
dt The car’s speed is 80 mph. 1600
200
80 mph ds
dt 160 mph. 185 Section 4.6
Step 6: 34. Step 1:
s shadow length
sun’s angle of elevation d
dt (20)( 2) (10)(1)
102 202 0.1 radian/sec
5.73 degrees/sec Step 2: To the nearest degree, the angle is changing at the rate At the instant in question,
s 60 ft and d
dt 0.27 /min of 0.0015 radian/min. 6 degrees per second. 36. Step 1: Step 3: A ds
.
dt We want to find
Step 4: c
a 80
or s
s tan 80 cot 120° Step 5:
ds
dt O 80 csc d
dt 2 a
b
c Step 6:
80
and
60 Note that, at the moment in question, since tan
0 2 4
and so csc
5 , we have sin distance from O to A
distance from O to B
distance from A to B At the instant in question, a
b 5 nautical miles, da
dt 14 knots, and c2
c2 3 nautical miles, a2
a2 db
dt 21 knots. Step 3: 2.25 in./min We want to find 7.1 in./min
ds
Since
dt B Step 2: 5
.
4 52
80
(0.0015 )
4
12 in
ft
0.1875
1 ft
min ds
dt b dc
.
dt Step 4:
Law of Cosines: 0, the rate at which the shadow length is b2
b2 2ab cos 120
ab decreasing is positive. The shadow length is decreasing at
Step 5: the rate of approximately 7.1 in./min. 2c 35. Step 1:
a distance from origin to A
b distance from origin to B
angle shown in problem statement 10 m, and b da
dt 2 m/sec, a2 db
dt dc
dt
dc
14
dt
dc
dt 2(7) 1 m/sec, 20 m. Step 3:
We want to find d
.
dt tan 37.
a
tan 1
b Step 5:
d
dt b 1
1 a2
b da
dt a
b2 2b db
dt a db
dt b da
dt b2 (5)2 ab 2(5)(14) 2(3)(21) (3)2 (5)(3) (5)(21) 49 (3)(14) 413
29.5 knots The ships are moving apart at a rate of 29.5 knots. Step 4:
a
or
b da
dt Note that, at the instant in question,
c At the instant in question, 2a Step 6: Step 2: a dc
dt db
dt b da
dt a db
dt a2 b2 dy
dt dy dx
dx
10(1 x 2) 2(2x)
dx dt
dt
dx
Since
3 cm/sec, we have
dt
dy
60x
cm/sec.
dt
(1 x2)2 (a) dy
dt [1 60( 2)
( 2)2]2 (b) dy
dt (1 60(0)
0 2)2 (c) dy
dt 60(20)
(1 202)2 120
52 (1 24
cm/sec
5 0 cm/sec
0.00746 cm/sec dx
20x
x 2 ) 2 dt 7 186 38. Chapter 4 Review dy
dt dy dx
dx dt
dx
Since
dt (3x 2 4) 41. (a) 2 cm/sec, we have dy
dt 8 (b) dy
dt 8 6(1)2 (c) dy
dt 8 6(4)2 (a) dx
dt 6( 3)2 dy
dt dx
dt
dy
dt 2 cm/sec
(b) At
dx
dt
dy
dt 2 dx
dt
dy
dt cos 2) 32 cos d
(uv)
dt 1. y x2
1 x ( 1) 2
3x 4
22 32 2 2 100.531 ft/sec dx
dt
dy
dt x)(1) x 4
3 x :
32 sin 4
3 x Endpoint values: 0 ft/sec Critical point value: 0 ft/sec 32 cos y 2 The global maximum value is 32 100.531 ft/sec minimum value is 4 at x 46
9 y 2 x 4 y
4 6 0
4
, and the global
3 at x 9 1.09 2. 2. Since y is a cubic function with a positive leading
30 sin coefficient, we have lim y and lim y x→ . There are x→ no global extrema.
3. y 2 2 (x 2)(e 1/x )( 2x 3)
2e 1/x 1
x 2 2e 1/x (x In general: At t 2 x 2 dx
dt
dy
dt ( 22 x
x 2(2 x) 71.086 ft/sec (b) Since the ferris wheel makes one revolution every
10 sec, we may let
0.2 t and we may write
x 30 cos 0.2 t, y 40 30 sin 0.2 t.
(This assumes that the ferris wheel revolves
counterclockwise.) At t du
dt v The first derivative has a zero at . 40. (a) One possible answer:
x 30 cos , y 40 dx
dt
dy
dt dv
dt x 2 16 ( 2) u s Chapter 4 Review (pp. 242–245) 71.086 ft/sec : 32 sin At 16 ( dy
dt u(0.03v) v( 0.02u)
0.01uv
0.01y
The total production is increasing at the rate of 1% per
year. y 4 0.09y, the rate of growth of total production 32 sin : 32 cos v(0.04u) is 9% per year. 88 cm/sec 4 du
dt v 0.09y
dy
Since
dt d
2 sin
2(sin )(16 )
dt
d
2 cos
2(cos )(16 ) 32
dt 32 sin dv
dt u(0.05v) 46 cm/sec (c) In general, assuming counterclockwise motion: 4 u 0.09uv (b) One possible answer is
16 t, where t is in seconds.
(An arbitrary constant may be added to this expression,
and we have assumed counterclockwise motion.) At d
(uv)
dt 6x 2 cm/sec. 8 39. (a) The point being plotted would correspond to a point on
the edge of the wheel as the wheel turns. dx
dt
dy
dt dy
dt (e 1/x )(2x) x
1)(x 1) x 30(sin 0.2 t)(0.2 ) 6 sin 0.2 t
Intervals 30(cos 0.2 t)(0.2 ) 6 cos 0.2 t 5:
6 sin
6 cos 1 0 ft/sec
6 ( 1) y
18.850 ft/sec x 00 x 1 x 1 Sign of y
Increasing 2
d
[2e 1/x ( x 1
dx 2 1) (x 2 2 1 2x (2e 1/x )(x
17.927 ft/sec 1/x 2 (x 4 x 2 4 2) x4 5.825 ft/sec
2 2e 1/x [(x 2 Decreasing Increasing x)] 2 (2e 1/x )(x 2e 6 cos 1.6 1 Behavior of y Decreasing 8:
6 sin 1.6 x 0.5)2
x4 1.75] 1 2 x)(2e 1/x )( 2x 3)
2x 2 ) Chapter 4 Review
5. y
1 2x 4x 3
Using grapher techniques, the zero of y is x The second derivative is always positive (where defined), so
the function is concave up for all x 0.
Graphical support: Intervals x 0.385 0.385 187 0.385. x Sign of y
Behavior of y Increasing Decreasing (a) [ 1, 0) and [1, ) y
2 12x 2
2(1 6x 2)
The second derivative is always negative so the function is
concave down for all x. (b) ( , Graphical support: (c) ( , 0) and (0, ) [ 4, 4] by [ 1, 5] 1] and (0, 1] (d) None
(e) Local (and absolute) minima at (1, e) and ( 1, e)
(f) None
[ 4, 4] by [ 4, 2] 4. Note that the domain of the function is [ 2, 2].
y x 1 ( 2x) 2 4 x2
x 2 (4 x 2 ) ( (a) Approximately ( 4 x )(1) (c) None
(d) ( 4 x2
2x 2 4
4 x ,) (e) Local (and absolute) maximum at (0.385, 1.215) (f) None 2 ex 6. y
Intervals , 0.385] (b) Approximately [0.385, ) 2 2 2 x 2 x 2 2 x 2 1 1 Intervals Sign of y x 1 1 x Sign of y Behavior of y
( 4 Decreasing
x 2)( 4x) (4 Increasing
1 2x 2) 24 y 4 x x2 Decreasing ( 2x) 2 Behavior of y Decreasing Increasing
ex y 1 The second derivative is always positive, so the function is
concave up for all x. 2x(x 2 6)
(4 x 2)3/2 Graphical support: 6 are not zeros of y because
Note that the values x
they fall outside of the domain.
Intervals 2 x 0 0 x 2 Sign of y
Behavior of y Concave up Concave down [ 4, 4] by [ 2, 4] (a) [1, )
Graphical support: (b) ( , 1] (c) ( ,) (d) None
(e) Local (and absolute) minimum at (1, 0)
(f) None
[ 2.35, 2.35] by [ 3.5, 3.5] (a) [ 2, (b) [ 2, 2]
2] and [ 2, 2] (c) ( 2, 0)
(d) (0, 2)
(e) Local maxima: ( 2, 0), ( 2, 2)
Local minima: (2, 0), (
2, 2)
Note that the extrema at x
2 are also absolute
extrema.
(f) (0, 0) 188 Chapter 4 Review
Graphical support: 7. Note that the domain is ( 1, 1).
y x 2) (1 1/4 1
(1
4 y x 2) x 5/4 ( 2x) Intervals 1 x 2)5/4 2(1 x 00 x [ 4.7, 4.7] by [ 3.1, 3.1] 1 (a) (
Sign of y , 1/3 (b) [ 2 Behavior of y Decreasing 5
(x)(2) (1
4 2 5/4 2(1 x ) (1) (1 2 1/4 y Increasing (c) ( 2 , 1) (d) ( 2 , 1) x ) ( 2x) ( , 0.794] [ 0.794, 1) and (1, ) 21/3) , ( , 1.260) and (1, ) ( 1.260, 1) (e) Local maximum at 5x 2] x ) [2 2x
4(1 x 2)5/2 1/3 2 1/4 x 2)5/2 4(1 1/3 2 2 3x 2 2
4(1 x 2)9/4 , 3 1
3 21/3, (f) The second derivative is always positive, so the function is
concave up on its domain ( 1, 1).
Graphical support: 1/3 2 1/3 2
21/3 ( 0.794, 0.529)
( 1.260, 0.420) 9. Note that the domain is [ 1, 1].
1 y
1 x2 Since y is negative on ( 1, 1) and y is continuous, y is
decreasing on its domain [ 1, 1].
d
[ (1 x 2) 1/2]
dx
1
(1 x 2) 3/2( 2x)
2 y [ 1.3, 1.3] by [ 1, 3] (a) [0, 1)
(b) ( 1, 0]
(c) ( 1, 1) Intervals 1 x (1 0 x
x 2)3/2 0 x 1 (d) None
Sign of y (e) Local minimum at (0, 1) Behavior of y (f) None
(x 3 8. y 1)(1) (x)(3x 2)
(x 3 1)2 Intervals x 2 2x 3 1
(x 3 1)2 1/3 2 1/3 Concave up Concave down Graphical support: x 1 1 x Sign of y
Behavior of y Increasing Decreasing Decreasing
[ 1.175, 1.175] by y (x 3
(x 3 1)2(6x 2)
2 1)(6x )
(x 3 (2x 3 1)(2)(x 3
(x 3 1)4
3 (2x
1)3 5
4 (b) [ 1, 1]
(c) ( 1, 0) 2 1)(6x ) (d) (0, 1)
(e) Local (and absolute) maximum at ( 1, );
local (and absolute) minimum at (1, 0)
(f) x , (a) None 1)(3x 2) 6x 2(x 3 2)
(x 3 1)3 Intervals 4 21/3 21/3 x 0 0 x 1 1 x Sign of y
Behavior of y Concave up Concave down Concave down Concave up 0, 2 Chapter 4 Review
10. This problem can be solved graphically by using NDER to obtain the graphs shown below.
y
y
y [ 4, 4] by [ 1, 0.3] [ 4, 4] by [ 0.4, 0.6] [ 4, 4] by [ 0.7, 0.8] An alternative approach using a combination of algebraic and graphical techniques follows.
Note that the denominator of y is always positive because it is equivalent to (x
(x 2 y 2x 3)(1)
(x 2 2x
2
x
3
2x 3)2 (x 2 Intervals (x)(2x
3)2 2) 3 x 1)2 3 x 3 3 x Sign of y
Behavior of y Decreasing
(x 2 2x 3)2( 2x) (x 2 y 2x 3)( 2x)
(x 2x 3
(x 2 2 Increasing Decreasing ( x 2 3)(2)(x 2
(x 2 2x 3)4
2)( x 2 2(2x
2x 3) 2x 3)(2x 2) 3) 3 18x 12
2x 3)3 Using graphing techniques, the zeros of 2x 3
x
2.584, x
0.706, and x 3.290.
Intervals ( , 2.584) ( 2.584, 18x 12 (and hence of y ) are at 0.706) ( 0.706, 3.290) (3.290, ) Sign of y
Behavior of y Concave down
(a) [
(b) ( 3,
, Concave up 3]
3] and [ 3, ) (c) Approximately ( 2.584,
(d) Approximately ( , (e) Local maximum at 0.706) and (3.290, ) 2.584) and ( 0.706, 3.290)
3, 31
4 (1.732, 0.183);
local minimum at
( 1.732,
(f) Concave down 3, 3
4 1 0.683) ( 2.584, 0.573), ( 0.706,
(3.290, 0.161) 0.338), and Concave up 2. 189 190 Chapter 4 Review 11. For x 0, y For x 0: y d
1
ln x
dx
x
d
ln ( x)
dx 1
( 1)
x 1
x 1
for all x in the domain.
x Thus y Intervals (0, 2) ( 2, 0) Sign of y
Behavior of y Decreasing Increasing y
x2
The second derivative is always negative, so the function is concave down on
each open interval of its domain.
Graphical support: [ 2.35, 2.35] by [ 3, 1.5] (a) (0, 2]
(b) [ 2, 0)
(c) None
(d) ( 2, 0) and (0, 2)
(e) Local (and absolute) maxima at ( 2, ln 2) and (2, ln 2)
(f) None
12. y 3 cos 3x 4 sin 4x Using graphing techniques, the zeros of y in the domain
0 x 2 are x x 2.148, and x Intervals 0 0.176, x
2.965, x x 0.994, x
3.834, x 0.176 0.176 2
3
,x
2 x 1.57,
5.591 0.994 0.994 x 22 x 2.148 2.148 x 2.965 Sign of y
Behavior of y Intervals Increasing 2.965 x Decreasing 3.834 3.834 x Increasing 3
2 3
2 x Decreasing 5.591 5.591 Increasing x Sign of y
Behavior of y Decreasing Increasing Decreasing Increasing 2 Chapter 4 Review
y
9 sin 3x 16 cos 4x
Using graphing techniques, the zeros of y in the domain
0 x 2 are x 0.542, x 1.266, x 1.876,
x 2.600, x 3.425, x 4.281, x 5.144 and x 6.000.
Intervals 0 x 0.542 0.542 x 1.266 1.266 x 1.876 1.876 x 2.600 2.600 x 3.425 Sign of y
Behavior of y Concave down Intervals 3.425 x Concave up 4.281 4.281 x Concave up Concave down 5.144 5.144 x 6.000 6.000 x Concave down 2 Sign of y
Behavior of y Concave up Concave down Concave up Concave down Graphical support: 4 , 9
4 by [ 2.5, 2.5] (a) Approximately [0, 0.176], 0.994,
(b) Approximately [0.176, 0.994], 2 2 , [2.148, 2.965], 3.834, , 2.148 , [2.965, 3.834], and 3
, and 5.591, 2
2
3
, 5.591
2 (c) Approximately (0.542, 1.266), (1.876, 2.600), (3.425, 4.281), and (5.144, 6.000)
(d) Approximately (0, 0.542), (1.266, 1.876), (2.600, 3.425), (4.281, 5.144), and (6.000, 2 )
(e) Local maxima at (0.176, 1.266), ,0 2
3
and (2.965, 1.266),
, 2 , and (2 ,1);
2 local minima at
(2.148, (0, 1), (0.994, 0.513), (3.834, and (5.591, 1.806), 1.806) Note that the local extrema at x
and x
(f) 0.513), 3.834, x 3
,
2 5.591 are also absolute extrema. (0.542, 0.437), (1.266, 0.267), (1.876,
(5.144, 0.120), and (6.000, 0.329) 0.267), (2.600, 0.437), (3.425, 0.329), (4.281, 0.120), 191 192 Chapter 4 Review
x e 13. y ,
3x 2, 4 x
x Intervals 0
0
x 0 0 2 2 x x
3 3 Sign of y
Behavior of y Decreasing
e y x ,
6x , x
x Intervals Increasing Decreasing 0
0
x 0 0 x Sign of y
Behavior of y Concave up Concave down
Graphical support: [ 4, 4] by [ 2, 4] (a) 0, 2
3 (b) ( 2 , 0] and ,
3 (c) ( , 0) (d) (0, )
(e) Local maximum at 2 16 ,
33 (1.155, 3.079) 3 (f) None. Note that there is no point of inflection at x
at this point.
14. y
5x 4 7x 2 10x 4
Using graphing techniques, the zeros of y are x
Intervals x 0.578 0.578 x 0 because the derivative is undefined and no tangent line exists 0.578 and x 1.692 1.692 x Sign of y
Behavior of y Decreasing Increasing y
20x 3 14x 10
Using graphing techniques, the zero of y is x
Intervals x 1.079 1.079 x Sign of y
Behavior of y Concave up Concave down
Graphical support: [ 4, 4] by [ 10, 25] Decreasing 1.079. 1.692. Chapter 4 Review
(a) Approximately [ 0.578, 1.692]
(b) Approximately ( , 0.578] and [1.692, ) (c) Approximately ( , 1.079) (d) Approximately (1.079, )
(e) Local maximum at
(f) 2x 4/5 15. y (1.692, 20.517);local minimum at (1.079, 13.601)
x 9/5 8 1/5
x
5 y 9 4/5
x
5 8 x 0 Intervals 9x 5 5 x 0 8
9 x 8
9 x Sign of y
Behavior of y Decreasing
36 1/5
x
25 8
x 6/5
25 y 2
9 x Intervals Increasing Decreasing 4(2 9x)
25x 6/5
2
9 x 0 0 x Sign of y
Behavior of y Concave up Concave down Concave down
Graphical support: [ 4, 4] by [ 3, 3] (a) 0, 8
9 (b) ( , 0] and (c) , (d) 8
,
9 2
9 2
, 0 and (0,
9 ) (e) Local maximum at 8 10
,
99 8 4/5
9 local minimum at (0, 0)
(f) 2 20
,
99 2 4/5
9 2
, 0.667
9 (0.889, 1.011); ( 0.578, 0.972) 193 194 Chapter 4 Review 16. We use a combination of analytic and grapher techniques to solve this problem. Depending on the viewing windows chosen,
graphs obtained using NDER may exhibit strange behavior near x
is actually undefined at x 5 2. The graph of y 4x 4x
x2 2 x [ 5.875, 5.875] by [ 50, 30] (x y 2)( 4
2x 3 3x 2) (5
(x 2)2
16x 3 8x 10x2
(x 2)2 4x 2 4x x 3)(1) The graph of y is shown below. [ 5.875, 5.875] by [ 50, 30] The zero of y is x
Intervals 0.215. x 0.215 0.215 x 2 2 x Sign of y
Behavior of y Increasing Decreasing (x 2)2( 6x 2 20x 16) ( 2x 3
(x 2)4 (x y 2)( 6x 2 20x 16)
(x Decreasing 2( 2x 3
2)3 2(x 3 6x 2 12x
(x 2)3 10x 2
10x 2 16x
16x 3)(2)(x
3) 13) The graph of y is shown below. [ 5.875, 5.875] by [ 20, 20] The zero of x 3 6x 2
(and hence of y ) is x
Intervals x 12x 13
3.710.
2 2 x 3.710 3.710 x Sign of y
Behavior of y Concave down
(a) Approximately ( Concave up , 0.215] (b) Approximately [0.215, 2) and (2, )
(c) Approximately (2, 3.710)
(d) ( , 2) and approximately (3.710, ) (e) Local maximum at
(f) (3.710, 3.420) (0.215, 2.417) Concave down 2 because, for example, NDER (y, 2) 3 2) is shown below. 5,000,000 while y Chapter 4 Review
17. y 6(x 2)2 1)(x Intervals f (x) 3 4/3
x
4 f (1) 0 C 0 24. x 1 1 x 2 2 x Sign of y
Behavior of y Decreasing Increasing 3
4 Increasing 1
3 1
2 1 C
y Intervals x f (x) 6(x 2) (1)
(x 2)] 0 0 x 2 2 x Sign of y
Behavior of y Concave up Concave down Concave up
(a) There are no local maxima.
(b) There is a local (and absolute) minimum at x 1. (c) There are points of inflection at x 2. 18. y 6(x 0 and at x 2) Intervals x 1 1 x 2 2 x Sign of y
Behavior of y Increasing
d
6(x 2
dx y 25. v(t)
s(t)
s(0)
C
s(t) s (t)
4.9t 2
10
10
4.9t 2 26. a(t)
v(t)
v(0)
C1 x 2) 6(2x v(t)
s(t)
s(0)
C2
s(t) e 20. Since d
sec x
dx 21. Since d
2 ln x
dx L(x) f (x)
23.
1 x 5 x 5 f e , sec 4 2x 2x
28. f (x) 13
x
3
13
x
3 x 4 4 2 1 1
.
2
x f 4 4 x 4 4 1 2 sec x f (x) sec x tan x
f C. x sec x 2
x x x2 C. f 4 sec 1 , 29. f (x) x cos x sin x sin x 2 (1 tan 4 4 2 2 4 tan x) 2 (sec2 x) 1
cos2 x(1 tan x)2
1
(cos x sin x)2 C cos x 4 1
tan x 1 f (x) x 2(1) x 2x x 4 sec 4 2 1, C. d 2 3/2
x
2x 1/2
dx 3
2 3/2
x
2x 1/2 C.
3 f (x)
f( )
0C
C
f (x) 20t tan 2. sec x tan x, f (x) 2 ln x 22. Since 20
C2 L(x) 14
x
4 f (x) 32t
20t x (c) There is a point of inflection at x f (x) 10 sec2 x 1. (b) There is a local minimum at x e 5t tan x f (x) 1
2 (a) There is a local maximum at x x 31
12 Increasing Behavior of y Concave down Concave up 14
x
4 x 9.8t 5
5t C s (t)
16t 2
5
5
16t 2 Sign of y d
19. Since
dx 12
x
2 1) 1
2 x Intervals Decreasing C v (t) 32
32t C1
20
20 27. f (x) 1)(x x 13
x
3 3 4/3
x
4 12
x
2 31
12 2 6(x 1)(2)(x 2)
6(x 2)[(2x 2)
18x(x 2) 13
x
3 3
3
2
L(x) f (0)
1 f (0)(x
1(x x 1 0) 0) 4 x 4 195 196 Chapter 4 Review 30. f (x) e x sin x
f (x) e x cos x
L(x) f (0) f (0)(x
1 2(x 0)
2x 1 (b) f (x)
Using a 0) 1
x (x) 0.5 and b 1
occurs at x
2 2. 32. (a) The values of y and y are both negative where the
graph is decreasing and concave down, at T. 1 0.5 ln 0.5
2.5 ln c 0.4 ln(27 2) c 35. e c (c) The local extreme values occur only at the endpoints of
the domain. A local maximum value of 1 occurs at
x
3, and a local maximum value of 3 occurs at
x 2. 1 2.5 (b) The function is decreasing on the interval [ 3, 0). 34. The 24th day 33
0.50.5 ln c 33. (a) The function is increasing on the interval (0, 2]. 3, we solve as follows. 3 ln 3 ln c ln x f (0.5)
0.5 ln (b) The value of y is negative and the value of y is positive where the graph is decreasing and concave up, at
P. 1 f (3)
3 f (c) 31. The global minimum value of (ln x)(1) e 1 1 (27 2)0.4 15 1458 1.579 (c) The slope of the line is
y m 2 f (b)
b f (a)
a 0.4 ln (27 2) 0.2 ln 1458, and the line passes through (3, 3 ln 3). Its equation is
–3 3 x y
y y = f (x )
–3 36. (a) We know that f is decreasing on [0, 1] and increasing
on [1, 3], the absolute minimum value occurs at x 1
and the absolute maximum value occurs at an endpoint.
Since f (0) 0, f (1)
2, and f (3) 3, the absolute
minimum value is 2 at x 1 and the absolute maximum value is 3 at x 3.
(b) The concavity of the graph does not change. There are
no points of inflection.
(c) y
3 0.2(ln 1458)(x
1.457x 1 3 4 –3 37. (a) f (x) is continuous on [0.5, 3] and differentiable on
(0.5, 3). s (t) (b) a(t) –1 3 ln 3, or approximately 1.075. (d) The slope of the line is m 0.2 ln 1458, and the line
passes through
5
5
(c, f (c)) (e 1 1458, e 1 1458( 1 0.2 ln 1458))
(1.579, 0.722).
Its equation is
y 0.2(ln 1458)(x c) f (c),
5
y 0.2 ln 1458(x e 1 1458)
5
e 1 1458( 1 0.2 ln 1458),
5
y 0.2(ln 1458)x e 1 1458,
or approximately y 1.457x 1.579.
38. (a) v(t) x 3) 4 v (t) 3t 2 6t
6 6t (c) The particle starts at position 3 moving in the positive
direction, but decelerating. At approximately t 0.528,
it reaches position 4.128 and changes direction, beginning to move in the negative direction. After that, it
continues to accelerate while moving in the negative
direction.
39. (a) L(x) f (0)
1 (b) f (0.1) f (0)(x 0)
0(x 0)
1 L(0.1) 1 (c) Greater than the approximation in (b), since f (x) is
actually positive over the interval (0, 0.1) and the
estimate is based on the derivative being 0.
40. (a) Since
dy
(b) dy dy
dx (2x (x 2)( e x)
x 2)e x (e x)(2x) dx. [2(1) (1)2](e 1)(0.01)
0.01e 1
0.00368 (2x x 2)e x, 197 Chapter 4 Review 41. (a) Regression equation y 43. Let t represent time in seconds, where the rocket lifts off at
t 0. Since a(t) v (t) 20 m/sec2 and v(0) 0 m/sec,
we have v(t) 20t, and so v(60) 1200 m/sec. The speed
after 1 minute (60 seconds) will be 1200 m/sec. 2701.73
17.28e 0.36x 1 44. Let t represent time in seconds, where the rock is blasted
upward at t 0. Since a(t) v (t)
3.72 m/sec2 and
v(0) 93 m/sec, we have v(t)
3.72t 93. Since
s (t)
3.72t 93 and s(0) 0, we have
s(t)
1.86t 2 93t. Solving v(t) 0, we find that the
rock attains its maximum height at t 25 sec and its height
at that time is s(25) 1162.5 m. [0, 20] by [ 300, 2800] (b) Note that
y d
2701.73(1
dx 2701.73(1 0.36x 17.28e ) 0.36x 1 ) 2(17.28)( 0.36e 17.28e 0.36x ) 45. Note that s 0.36x 16,806.9e
(1 17.28e r2 A 0.36x 2 ) 100 2r and the sector area is given by 1
rs
2 s
2r the domain of A(r) The graph of y is shown below. 0 s 1
r(100
2 50r 2r) r 2, note that r
50 2 r, which gives 12.1 A (r) 50 r 2. To find 50r 1 0 and
r 50. Since 2r, the critical point occurs at r 25. This value is in the domain and corresponds to the maximum
[0, 20] by [ 75, 275] area because A (r) Using graphing techniques, y has its maximum at
x 7.92. This corresponds to the year 1998 and
represents the inflection point of the logistic curve. The
logistic regression equation predicts that the rate of
increase in debit card transactions will begin to
decrease in 1998, and since y(7.92) 1351, there are
approximately 1351 million transactions that year. 2, which is negative for all r. The greatest area is attained when r
46. y (x, 27 – x2) –4 f (x) 2 cos x 1 For 0 21 xn 1 xn x A(x) f (xn)
2 cos xn
2 sin xn The graph of y
near x 1. 1 A xn
1 2 1 0 has one solution, 1
0.8361848
0.8283814
0.8283608
0.8283608 Solution: x 0.828361 27, the triangle with vertices at (0, 0) and
x 2) has an area given by 1
(2x)(27
2 27 3x 2 x 2)
3(3 27x
x)(3 x 3. Since
x) and A 6x, the 27) occurs at x 3 and corresponds to the maximum area because A (x) is negative
in this interval. The largest possible area is
A(3) x1
x2
x3
x4
x5 x critical point in the interval (0,
xn f (x) shows that f (x) [ 2, 10] by [ 6, 2] x ( x, 27 f (xn) xn 4 x
1 2 sin x 50 ft. 27 (c) As x increases, the value of y will increase toward
2701.73. The logistic regression equation predicts a
ceiling of approximately 2702 million transactions per
year.
42. f (x) 25 ft and s 54 square units. 198 Chapter 4 Review 47. If the dimensions are x ft by x ft by h ft, then the total
amount of steel used is x 2
x 2 4xh 108 and so h
x 2h by V(x)
V (x) 108x x 3
4 0.75x 2 27 V (x) 4xh ft2. Therefore, 27x 0.75(6 V x) and 1.5x. The critical point occurs at x 108 62
0. The corresponding height is
4(6) 12 2r. 12
rh
3 0 0.25x 3. Then x)(6 12 h
, so h
12 r
6 The volume of the smaller cone is given by 108 x 2
. The volume is given
4x 12
23
r (12 2r) 4 r 2
r for
3
3
dV
6. Then
8 r 2 r 2 2 r(4 r), so the
dr r critical point occurs at r
6, and it corresponds to the maximum volume because V (x)
x 50. Note that, from similar cones, 4. This critical point corresponds to the maximum volume because 0 for 0 3 ft. The r 4 and dV
dr 0 for 4 r dV
dr 0 for 6. The smaller cone has the largest possible value when r 4 ft and h 4 ft. base measures 6 ft by 6 ft, and the height is 3 ft.
51.
48. If the dimensions are x ft by x ft by h ft, then we have
x 2h Lid 32
. Neglecting the quarterinch
x2 32 and so h x thickness of the steel, the area of the steel used is
A(x) x2 x2 4xh 10 in.
Base 128
. We can minimize the weight
x
x of the vat by minimizing this quantity. Now
A (x) 2x A (x) 2 128x 2 23
(x
x2 x 43) and x
15 in. (a) V(x) 256x 3. The critical point occurs at x 4 and x(15 2x)(5 (b, c) Domain: 0 x x)
5 corresponds to the minimum possible area because
A (x)
32
42 0 for x 0. The corresponding height is 2 ft. The base should measure 4 ft by 4 ft, and the
[0, 5] by [ 20, 70] height should be 2 ft.
h2 49. We have r 2 2 3, so r 2 The maximum volume is approximately 66.019 in3 and
it occurs when x 1.962 in. 2 3 h
. We wish to
4 minimize the cylinder’s volume
V r 2h Since dV
dh d 2V
dh 2 3
3 h2
h 3h
4
2
3h
3
(2
4
4 h3
for 0
4 h)(2 h 2 3. h) and 3h
, the critical point occurs at h
2 (d) Note that V(x) 2x 3 25x 2
so V (x) 6x 2 50x 75.
Solving V (x) 0, we have 50 2 and it d 2V
corresponds to the maximum value because 2
dh
22
h 0. The corresponding value of r is 3
4 0 for The largest possible cylinder has height 2 and radius ( 50)2
2(6) 50 x 2.
2. 10 7
12 4(6)(75) 75x, 50 700
12 25 5
6 7 . These solutions are approximately x 1.962 and
x 6.371, so the critical point in the appropriate
domain occurs at
x 25 5
6 7 . 199 Chapter 4 Review
52. 54. The length of the track is given by 2x y
10 2x 2r 400 and therefore x 2 r, so we have 200 r. Then the area of the rectangle is
(x, 8 cos 0.3x) A(r) 2rx
2r(200 –2π –π For 0
A(x) π 2π x 400r
Therefore, A (r) 5
, the area of the rectangle is given by
3 x (2x)(8 cos 0.3x) Then A (x) 16(cos 0.3x
Solving A (x) because A (r) 0.3x sin 0.3x) x 2.868 and the corresponding area is 30(20 Then C (x) y)
30 40
2 Solving C (x) x x2 600 (2x)
144 40 30 200 0 for all r. 100 100 m. 100 100 m and r x 144. 30x
x2 2k dollars per hundred grade A tires. .
144 Then the profit is given by
P(x) 2kx 144 9x 2
2304 4 16x 2 144 2k 2304 x2 2k 7x 2 P (x) 2k Choose the positive solution:
x 2k
18.142 mi The solutions of P (x) 7 y x 2 m. 55. Assume the profit is k dollars per hundred grade B tires and
2 40
3x 48 m 0, we have: 30x
x2 40x 100 4 , so the The rectangle will have the largest possible area when 53. The cost (in thousands of dollars) is given by
40x 4 r and A (r) . The corresponding value of x is approximately 29.925 square units. C(x) 400 200 r and this point corresponds to the maximum rectangle area 16(cos 0.3x)(1) 0 graphically, we find that the critical point occurs at x 2 r 2, for 0 critical point occurs at r 16x cos 0.3x. 16x( 0.3 sin 0.3x) r) 12 2 36
7 13.607 mi x 10 2 ( 10)
2(1) 40
5 k 10x
x 5x)
5 x(5
x (20
20
5 x2
x (5 x)( 2x)
(5 x) x2
(5 (20
x )2 x 2)( 1) 10x 20
x)2 0 are 4(1)(20) the appropriate domain is x 5
5 5, so the solution in
5 2.76. Check the profit for the critical point and endpoints:
Critical point:
x 2.76
P(x) 11.06k
Endpoints:
x0
P(x) 8k
x4
P(x) 8k
The highest profit is obtained when x
which corresponds to
276 grade A tires and 553 grade B tires. 2.76 and y 5.53, 200 Chapter 4 Review 56. (a) The distance between the particles is f (t) where
cos t f (t) cos t sin t f (t) sin t Solving f (t ) 4 . Then 4 0 graphically, we obtain t 1.178, t 4.230, and so on. [0, 2 ] by [ 2, 2] Alternatively, f (t )
f (t) t sin 8 sin t
2 sin 0 may be solved analytically as follows. cos 8
8 sin 8 cos t 8 cos t t 8 8 8 sin 8 sin t 8 8 cos 8 cos t 8 , so the critical points occur when
cos t
f (t) 0, or t 8 2 cos 3
8 3
8 k . At each of these values, 0.765 units, so the maximum distance between the particles is 0.765 units.
(b) Solving cos t cos t 4 graphically, we obtain t 2.749, t 5.890, and so on. [0, 2 ] by [ 2, 2] Alternatively, this problem may be solved analytically as follows.
cos t
cos
cos t 8 cos 8 t 8 sin t 8 2 sin t 8 sin t cos 8 sin
sin 8
8 7
8 4 t cos t 8
8 8 cos 8 sin t 8 cos 8 0
0 8 t
The particles collide when t cos t 7
8 k 2.749 (plus multiples of if they keep going.) sin 8 Chapter 4 Review
57. The dimensions will be x in. by 10 2x in. by 16 2x in.,
so V(x) x(10 2x)(16 2x) 4x 3 52x 2 160x
for 0 x 5.
Then V (x) 12x 2 104x 160 4(x 2)(3x 20), so
the critical point in the correct domain is x 2.
This critical point corresponds to the maximum possible
volume because V (x) 0 for 0 x 2
and V (x) 0 for 2 x 5. The box of largest volume
has a height of 2 in. and a base measuring
6 in. by 12 in., and its volume is 144 in3. 201 59. Step 1:
x xcoordinate of particle
y ycoordinate of particle
D distance from origin to particle
Step 2:
At the instant in question, x
dx
dt dy
1 m/sec, and
dt 5 m, y 12 m, 5 m/sec. Step 3: Graphical support: We want to find dD
.
dt Step 4:
x2 D y2 [0, 5] by [ 40, 160] Step 5: 58. Step 1:
r radius of circle
A area of circle dD
dt At the instant in question, dr
dt 2 m/sec and r Step 3:
dA
.
dt x2 2 dy
2y
dt dx
dt x2 y dy
dt y2 10 m. (5)( 1) dD
dt (12)( 5) 5 m/sec 52 122
dD
Since
is negative, the particle is approaching the origin
dt at the positive rate of 5 m/sec.
60. Step 1:
x edge of length of cube
V volume of cube Step 4:
A dx
2x
2
dt
y Step 6: Step 2: We want to find x
1 r2 Step 5: Step 2: dA
dt At the instant in question, dr
2r
dt dV
dt Step 6:
dA
dt 2 (10) 2 40 The area is changing at the rate of 1200 cm3/min and x 20 cm. Step 3:
2 40 m /sec. We want to find dx
.
dt Step 4:
V x3 Step 5:
dV
dt 3x 2 dx
dt Step 6:
1200
dx
dt 3(20)2 dx
dt 1 cm/min The edge length is increasing at the rate of 1 cm/min. 202 Chapter 4 Review
63. Step 1:
r radius of outer layer of cable on the spool
clockwise angle turned by spool
s length of cable that has been unwound 61. Step 1:
x xcoordinate of point
y ycoordinate of point
D distance from origin to point Step 2: Step 2:
At the instant in question, x 3 and dD
dt 11 units per sec. At the instant in question, ds
dt 6 ft/sec and r 1.2 ft Step 3: Step 3: We want to find dx
We want to find .
dt d
.
dt Step 4: Step 4:
s
Since D 2 x2 x2 D y 2 and y x 3 for x x 3/2, we have Step 5: 0. Since r is essentially constant, 1 (2x 2 x2 x3
2x 3x 2 dx
2x 1 x dt 3x 2)
3x
2 6 dx
dt d
dt 2 dx
1 x dt 3(3) 2 dx
2 4 dt dx
dt 4 units per sec 62. (a) Since h
r 10
, we may write h
4 5r
or r
2 2h
.
5 (b) Step 1:
h depth of water in tank
r radius of surface of water
V volume of water in tank 1.2 5 radians/sec 64. a(t) v (t)
g
32 ft/sec2
Since v(0) 32 ft/sec, v(t) s (t)
32t 32.
Since s(0)
17 ft, s(t)
16t 2 32t 17.
The shovelful of dirt reaches its maximum height when
v(t) 0, at t 1 sec. Since s(1)
1, the shovelful of dirt
is still below ground level at this time. There was not
enough speed to get the dirt out of the hole. Duck!
65. We have V 12
dV
r h, so
3
dr 2
rh and dV
3 When the radius changes from a to a dV
dt 5 ft3/min and h 6 ft. dS Step 3:
We want to find dh
.
dt 6x 2, which means 12x dx. We want dS 12x dx 2 0.02(6x ) or dx surface area of dS
dx 12x and 0.02S, which gives
0.01x. The edge should be measured with an error of no more than 1%. Step 4:
12
rh
3 dr, the volume edge of length of cube and S cube. Then S 2
rh dr.
3 2
ah dr.
3 change is approximately dV
66. (a) Let x At the instant in question, 4
h3
75 Step 5:
dV
dt d
.
dt d
dt Step 2: V r The spool is turning at the rate of 5 radians per second. Step 6:
11 ds
dt Step 6: Step 5:
dD
dt r dV
dx dh
4
h2
dt
25 volume of cube. Then V
3x 2 and dV dh
4
(6)2
dt
25
dh
125
0.276 ft/min
dt
144
dh
Since
is negative, the water level is dropping at the
dt 5 0.276 ft/min. so dV x 3, which means 3x 2 dx. We have dx which means 3x 2 dx Step 6: positive rate of (b) Let V 3x 2(0.01x) 0.01x, 0.03V, 0.03V. The volume calculation will be accurate to within approximately 3% of the correct
volume. 203 Chapter 4 Review
67. Let C circumference, r (a) Since C radius, S
dC
dr 2 r, we have Therefore, dC
C 2 dr
2r surface area, and V 2 and so dC
dr
r 0.4 cm
10 cm volume. 2 dr. 0.04 The calculated radius will be within approximately 4% of the correct radius.
dS
dr
dS
dS 8 r dr. Therefore,
S
2 dr
2(0.04) 0.08. The
r (b) Since S 4 r 2, we have 8 r and so
8 r dr
4 r2 calculated surface area will be within approximately 8% of the correct surface area. 43
dV
r , we have
3
dr
dV
2
4 r dr. Therefore
V 4 r 2 and so (c) Since V
dV 4 r 2 dr 3 dr
r 43
r
3 3(0.04) 0.12. The calculated volume will be within approximately 12% of the correct volume.
68. By similar triangles, we have
approximately 6
have dh
69. dy
dx 120(15) 120a 2 sin x cos x Therefore, dy
dx 2 2 da 1 a
6 a 20
h , which gives ah 6a 120, or h 120a 14 ft. The estimated error in measuring a was da 120(15) 21 12 2
ft, so the estimated possible error is
45 3. Since sin x and cos x are both between 1 and
3 6 1 for all values of x. dy
Since
is always negative, the function decreases on every interval.
dx 1 . The height of the lamp post is dh
1
ft. Since
da
12
2
8
ft or
in.
45
15 1 in. 120a 2 , we 1, the value of 2 sin x cos x is never greater than 2. 204 Section 5.1 Chapter 5
The Definite Integral
s Section 5.1 Estimating with Finite Sums
(pp. 247–257)
Exploration 1 Which RAM is the Biggest? y y y 5 5 5 4 4 4 3 3 3 2 2 2 1 1 1 1 2 1. LRAM x 3 MRAM 1 2 x 3 1 2 x 3 RRAM y y y 5 5 5 4 4 4 3 3 3 2 2 2 1 2 2. MRAM x 3 RRAM 1 2 x 3 1 2 3 x LRAM 3. RRAM MRAM
function. LRAM, because the heights of the rectangles increase as you move toward the right under an increasing 4. LRAM MRAM
function. RRAM, because the heights of the rectangles decrease as you move toward the right under a decreasing Quick Review 5.1
1. 80 mph 5 hr 400 mi 2. 48 mph 3 hr 144 mi 3. 10 ft/sec2 10 sec 100 ft/sec 3600 sec
1 mi
68.18 mph
1h
5280 ft
3600 sec 24 hr 365 days
4. 300,000 km/sec
1 hr
1 yr
1 day
12 100 ft/sec 9.46 10 5. (6 mph)(3 h) (5 mph)(2 h) 6. 20 gal/min 1 h 60 min
1h 7. ( 1 C/h)(12 h) 18 mi (1.5 C)(6 h) 8. 300 ft3/sec 3600 sec
1h
3600 sec
1h 10 mi 28 mi 1200 gal 24 h
1 day 9. 350 people/mi2 50 mi2
10. 70 times/sec 1 yr km 3C
1 day 25,920,000 ft3 17,500 people
1 h 0.7 176,400 times Section 5.1
Section 5.1 Exercises
1. (a) y
2 R
2 (b) x y
2 2 x 1
2 LRAM: [2(0)
2. (a) x (0)2] 1
2 12 1
2
2 1
2 2 [2(1) (1)2] 1
2 2 32 1
2
2 3
2 5
4 1.25 1
2 5
4 1.25 y
2 2 RRAM: 2
(b) 1
2 12 1
2
2 x [2(1) (1)2] 1
2 2 32 1
2
2 3
2 [2(2) (2)2] y
2 2 MRAM: 2
3. 12 1
4
2 1
4 x 2 3
4 n LRAMn MRAMn RRAMn 10 1.32 1.34 1.32 50 1.3328 1.3336 1.3328 100 1.3332 1.3334 1.3332 500 1.333328 1.333336 1.333328
4. The area is 1.333 4
.
3 32 1
4
2 2 5
4 52 1
4
2 2 7
4 72 1
4
2 11
8 1.375 205 206
5. Section 5.1
n LRAMn MRAMn RRAMn 10 12.645 13.4775 14.445 RRAM: 10 (44 50 13.3218 13.4991 13.6818 Average 100 13.41045 12. LRAM: 10 (0 13.499775 13.59045 Estimate the area to be 13.5.
n 1.16823 1.09714 1.03490 50 1.11206 1.09855 1.08540 100 1.10531 1.09860 1.09198 500 1.09995 1.09861 1.09728 1000 1.09928 1.09861 15 3490 ft 3840 ft
2 3490 ft 35) 3840 ft 3665 ft 1.09795
n MRAM 10 526.21677 LRAMn MRAMn RRAMn 20 524.25327 10 0.98001 0.88220 0.78367 40 523.76240 50 0.90171 0.88209 0.86244 80 523.63968 100 0.89190 0.88208 0.87226 160 523.60900 500 0.88404 0.88208 0.88012 1000 0.88306 0.88208 0.88110 n 4
(5)3
3 15. V 500
3 error n Estimate the area to be 0.8821. 523.59878 % error 10 2.61799 0.5 n LRAMn MRAMn RRAMn 20 0.65450 0.125 10 1.98352 2.00825 1.98352 40 0.16362 0.0312 50 1.99934 2.00033 1.99934 80 0.04091 0.0078 100 1.99984 2.00008 1.99984 500 1.99999 2.00000 8. … 30 ) 14. Use f (x)
25 x 2 and approximate the volume using
r 2h
( 25 ni2)2 x, so for the MRAM program, use
(25 x 2) on the interval [ 5, 5]. Estimate the area to be 1.0986.
7. 35 … (b) The halfway point is 0.4845 mi. The average of LRAM
and RRAM is 0.4460 at 0.006 h and 0.5665 at 0.007 h.
Estimate that it took 0.006 h 21.6 sec. The car was
going 116 mph. LRAMn MRAMn RRAMn 10 15 13. (a) LRAM: 0.001(0 40 62 … 137) 0.898 mi
RRAM: 0.001(40 62 82 … 142) 1.04 mi
Average 0.969 mi 500 13.482018 13.499991 13.518018 6. 44 1.99999 160 0.01023 0.0020
16. (a) Use LRAM with (16 x 2).
S8 146.08406
S8 is an overestimate because each rectangle is above
the curve. Estimate the area to be 2.
9. LRAM:
Area
f (2) 2 f (4) 2
2 (0 0.6 1.4
44.8 (mg/L) sec
RRAM:
Area
f (4) 2 f (6) 2
2(0.6 1.4 2.7
44.8 (mg/L) sec f (6) 2 …
… 0.5) f (22) 2 (b) … f (8) 2
… 0) f (24) 2 (b) 60 sec
1 min Note that estimates for the area may vary. (b) RRAM: 1 (12 22 10 5
6 0) 87 in. 7.25 ft
11. 5 min 5 13 11 6 (b) RRAM: 300 (1.2 1.2
1.7 1.7
2.0 9% V S8
V 0.10 10%
x 2) on the 6 2 (b) V S8
V 0.11 11% 11 …
… 1.2) 5220 m 0) 4920 m 19. (a) (5)(6.0 8.2 9.1 … 12.7)(30) 15,465 ft3 (b) (5)(8.2 13 300 sec (a) LRAM: 300 (1 0.09 18. (a) Use LRAM with (64
interval [4, 8], n 8.
S 372.27873 m3 6.7 L/min 10. (a) LRAM: 1 (0 12 22 10
2 6) 87 in. 7.25 ft S8
V 17. (a) Use RRAM with (16 x 2).
S8 120.95132
S8 is an underestimate because each rectangle is below
the curve. Patient’s cardiac output:
5 mg
44.8 (mg/L) sec V 9.1 9.9 … 13.0)(30) 16,515 ft3 20. Use LRAM with x on the interval [0, 5], n 5.
1(0
2
3
4 ) 10
31.41593 Section 5.1
21. Use MRAM with x on the interval [0, 5], n
1
1
2 3
2 5
2 7
2 22. (a) LRAM5:
32.00 19.41
(b) RRAM5:
19.41 11.77 9
2 25
2 5. triangles with a hypotenuse of length 1 and an acute 39.26991 7.14 7.14
4.33 4.33
2.63 74.65 ft/sec (5 sec)(32 ft/sec2) Area 16 45.28 ft/sec (c) The upper estimates for speed are 32.00 ft/sec for the
first sec, 32.00 19.41 51.41 ft/sec for the second
sec, and 32.00 19.41 11.77 63.18 ft/sec for the
third sec. Therefore, an upper estimate for the distance
fallen is 32.00 51.41 63.18 146.59 ft.
23. (a) 400 ft/sec (b) Think of the octagon as a collection of 16 right angle measuring
11.77 2 97
70 (b) Upper 70 Lower 50 97 136 190 265 369 516
720 2363 gal
70 97 136 190 265 369
516 1693 gal . 8 4 2 2.828 (c) Think of the 16gon as a collection of 32 right triangles
with a hypotenuse of length 1 and an acute angle
2
.
32
16
1
32
sin
cos
2
16
16 measuring (b) Use RRAM with 400 32x on [0, 5], n 5.
368 336 304 272 240 1520 ft
70
50 2
16 1
sin
cos
2
8
8 4 sin 240 ft/sec 24. (a) Upper
Lower 207 Area 136 190 265 758 gal
97 136 190 543 gal 8 sin 3.061 8 (d) Each area is less than the area of the circle, . As n
increases, the area approaches .
28. The statement is false. We disprove it by presenting a (c) 25,000 2363 22,637 gal 22,637
720 31.44 h (worst case) 25,000 1693 23,307
720 23,307 gal 32.37 h (best case) 25. (a) Since the release rate of pollutants is increasing, an
upper estimate is given by using the data for the end of
each month (right rectangles), assuming that new
scrubbers were installed before the beginning of
January. Upper estimate:
30(0.20 0.25 0.27 0.34 0.45 0.52)
60.9 tons of pollutants
A lower estimate is given by using the data for the end
of the previous month (left rectangles). We have no
data for the beginning of January, but we know that
pollutants were released at the newscrubber rate of
0.05 ton/day, so we may use this value.
Lower estimate:
30(0.05 0.20 0.25 0.27 0.34 0.45)
46.8 tons of pollutants
(b) Using left rectangles, the amount of pollutants released
by the end of October is
30(0.05 0.20 0.25 0.27 0.34 0.45
0.52 0.63 0.70 0.81) 126.6 tons.
Therefore, a total of 125 tons will have been released
into the atmosphere by the end of October.
26. The area of the region is the total number of units sold, in
millions, over the 10year period. The area units are
(millions of units per year)(years) (millions of units).
27. (a) The diagonal of the square has length 2, so the side
length is 2. Area ( 2)2 2 counterexample, the function f (x)
0 x 1, with n LRAM1 RRAM1 1. MRAM1
1f (0) 2 x 2 over the interval
1f (0.5) 0.25 1f (1)
2 0 1
2 0.5 MRAM1 ( x)[ f (x1) f (x2) … f (xn 1) f (xn)]
( x)[ f (x0) f (x1) f (x2) … f (xn 1)]
( x)[ f (xn) f (x0)]
LRAMn f ( x)[ f (xn) f (x0)]
But f (a) f (b) by symmetry, so f (xn) f (x0) 0.
Therefore, RRAMn f LRAMn f. 29. RRAMn f 30. (a) Each of the isosceles triangles is made up of two right
triangles having hypotenuse 1 and an acute angle
2
. The area of each isosceles triangle
2n
n
1
1
2
2
sin
cos
sin .
2
n
n
2
n measuring
is AT (b) The area of the polygon is
AP nAT lim AP n→ n
2
sin , so
2
n
n
2
lim sin
n
n→ 2 (c) Multiply each area by r 2:
AT
AP 12
r sin
2
n2
r sin
2 lim AP n→ 2
n
2
n r2 208 Section 5.2 s Section 5.2 Definite Integrals
(pp. 258–268)
Exploration 1
1. 7. 0. (The equal areas above and below the xaxis sum to
zero.) Finding Integrals by Signed Areas 2. (This is the same area as sin x dx, but below the
0 xaxis.) [ 2 , 2 ] by [ 3, 3] 8. 4. (Each rectangle in a typical Riemann sum is twice as
wide as in sin x dx.)
0 [ 2 , 2 ] by [ 3, 3] 2. 0. (The equal areas above and below the xaxis sum to
zero.)
[ 2 , 2 ] by [ 3, 3] 9. 0. (The equal areas above and below the xaxis sum to
zero.) [ 2 , 2 ] by [ 3, 3] 3. 1. (This is half the area of sin x dx.)
0 [ 2 , 2 ] by [ 3, 3] 10. 0. (The equal areas above and below the xaxis sum to zero,
since sin x is an odd function.) [ 2 , 2 ] by [ 3, 3] 4. 2 2. (The same area as sin x dx sits above a rectangle
0 of area 2.) [ 2 , 2 ] by [ 3, 3] Exploration 2 More Discontinuous Integrands 1. The function has a removable discontinuity at x 2. [ 2 , 2 ] by [ 3, 3] 5. 4. (Each rectangle in a typical Riemann sum is twice as tall
sin x dx.) as in [ 4.7, 4.7] by [ 1.1, 5.1] 0 2. The thin strip above x 2 has zero area, so the area under
3 the curve is the same as (x
0 [ 2 , 2 ] by [ 3, 3] 6. 2. (This is the same region as in
2 units to the right.) sin x dx, translated
0 [ 4.7, 4.7] by [ 1.1, 5.1] [ 2 , 2 ] by [ 3, 3] 2) dx, which is 10.5. Section 5.2
3. The graph has jump discontinuities at all integer values, but n the Riemann sums tend to the area of the shaded region ∑
P →0 shown. The area is the sum of the areas of 5 rectangles (one 1 tition of [0, 1]. 1 k1 4 x 2 dx where P is any par 0 n 6. lim 5 0 ck2 xk 4 ∑ (sin3 ck)
P →0 of them with height 0):
int(x) dx 5. lim 2 3 4 10. 0 partition of [ sin3 x dx where P is any xk k1 , ]. 1 5 dx 7. 5[1 ( 2)] 15 ( 20)(7 3) 2
7 [ 2.7, 6.7] by [ 1.1, 5.1] 3 (1)2 ( 160) dt 9. 5 ∑ n2 (2)2 (3)2 (4)2 (5)2 ( 160)(3 0) 55 1 4 ∑ (3k 2) [3(0) 2] [3(1) 2] 480 0 n1 2. 80 3 Quick Review 5.2
1. ( 20) dx 8. [3(2) 2] 10. 2 4 d 2 [1 3
2 ( 4)] k0 [3(3) 2] [3(4) 2] 20 3.4 11. 0.5 ds 3. 0.5[3.4 ( 2.1)] ∑ 100( j 1)2 100[(1)2 (2)2 (3)2 (4)2 (5)2] j0 18 2 dr 12. 5500 2( 18 2) ∑k x
2 13. Graph the region under y k1
25 5. 4 2 99 4. 2.75 2.1 4 ∑ 2k 3 for 2 x 4. y
5 k0
500 6. ∑ 3k 2 k1 50 50 7. 2∑ x 2 3∑ x x1 x1 50 ∑ (2x 2 x x1 8 20 20 k0 8. 5 3x) k9 k0 ∑ xk ∑ xk ∑ xk
n 9. ∑( 1)k 0 if n is odd. 4 k0 2 n 10. ∑( 1)k 1
(6)(2
2 5) 14. Graph the region under y Section 5.2 Exercises
n ∑ ck 2 2 xk P →0 k 1 x dx where P is any partition of [0, 2]. ∑ k1 1) 2 5 (x 2 3x) dx where P is any 7 3
2
1 1
x
ck k n ∑
P →0 4. lim 4 for 4 5 partition of [ 7, 5]. P →0 k 1 2x 0 3ck) xk k1 n 21 y 2 n ∑ (ck2
P →0 2. lim 3. lim 3 dx 1 if n is even. k0 1. lim x
2 1 1
dx where P is any partition of [1, 4].
x
3 1
1 4 ck xk 2 1
1 x 5 3/2 ( 2x
of [2, 3]. x x=1 x=3
2
2 dx where P is any partition
1/2 4) dx 1
(1)(3
2 209 1
2 x 3
.
2 210 Section 5.2 15. Graph the region under y x 2 for 9 3 x 3. 18. Graph the region under y 1 x 1. x for 1 x 1. 2 5 5 x 2 1 This region is half of a circle of radius 3.
9 x for y y 3 1 2 x dx 3 1
(3)2
2 1
(2)(1)
2 1 19. Graph the region under y 2 (1 x ) dx 1 9
2 16. Graph the region under y x 16 2 x for 4 x 0. y
2 y
5 2
5 x 1 (2 The region is one quarter of a circle of radius 4.
0 16 x 2 dx 4 1
(4)2
4 17. Graph the region under y x x ) dx 1 1
(1)(1
2 2)
1 20. Graph the region under y 4
x for 2 x 1
(1)(1
2 1 2)
x 2 for 3
1 y 1. 2 y
2
2 2 x x 1 (1 1 x 2) dx 1
(1)2
2 (2)(1) 1
1 x dx
2 1
(2)(2)
2 1
(1)(1)
2 5
2 21. Graph the region under y for 2 y
2 2
2 d 1
(2
2 )(2 ) 2 2 3
2 2 x 1. 211 Section 5.2
22. Graph the region under y r for 2 r 5 33. Observe from the graph below that the region under the
graph of f (x) 1 x 3 for 0 x 1 cuts out a region R
from the square identical to the region under the graph of
g(x) x 3 for 0 x 1. 2. y
10 y
1
y = f (x ) √2
5 r 5√2 2 r dr
2 10 2 2)( 1
(5
2 2 5 2) 1 24
1 b 23. 1
(b)(b)
2 x dx
0
b 2s ds 1
(b
2 a)(2b 2a) b2 3t dt 1
(b
2 a)(3b 3a) 32
(b
2 b 25.
a
b 26.
a 2a 27. 1
(2a
2 x dx
a
3a 28. x dx
a 1
(
2 a)(2a a) a)( 3a x x 3 dx 1 0 1
4 3
4 34. Observe from the graph of f (x) ( x 1)3 for
1 x 2 that there are two regions below the xaxis and
one region above the axis, each of whose area is equal to
the area of the region under the graph of g(x) x 3 for
0 x 1. 2b 2 3a 1 0 1
(b)(4b)
2 0 1 x 3) dx (1 12
b
2 4x dx 24. R a2 y a 2) 2 3a 2
2 a) 1
(3a 2
2 a 2) a2 y = f (x ) 2 x 29. Observe that the graph of f (x) x 3 is symmetric with
respect to the origin. Hence the area above and below the
xaxis is equal for 1 x 1.
1 x 3 dx (area below xaxis) 1 (area above xaxis) 2 0 3 is three units higher than the
30. The graph of f (x) x
graph of g(x) x 3. The extra area is (3)(1) 3.
1 (x 3 1
4 3) dx 0 13
4 3 31. Observe that the region under the graph of f (x) (x 2)3
for 2 x 3 is just the region under the graph of
g(x) x 3 for 0 x 1 translated two units to the right.
3 (x 1 2)3 dx 1
4 x 3 dx 2 0 32. Observe that the graph of f (x) x 3 is symmetric with
respect to the yaxis and the right half is the graph of
g(x) x 3.
1
1 x 3 dx 1 2 x 3 dx
0 1
2 1)3 dx (x 3 2 1 1
4 1
4 35. Observe that the graph of f (x) 1
4
x3
for 0
2 a horizontal stretch of the graph of g(x)
by a factor of 2. Thus the area under f (x)
0 x for 0
2
0 x 2 is just x 3 for 0 x 2 is twice the area under the graph of g(x)
x x3
dx
2 1 x3
for
2 x3 1.
1 2 x 3 dx
0 1
2 36. Observe that the graph of f (x) x 3 is symmetric with
respect to the origin. Hence the area above and below the
xaxis is equal for 8 x 8.
8
8 x 3 dx (area below xaxis) (area above xaxis) 0 212 Section 5.2 37. Observe from the graph below that the region between the
graph of f (x) x 3 1 and the xaxis for 0 x 1 cuts
out a region R from the square identical to the region under
the graph of g(x) x 3 for 0 x 1. 44. (a) The function has discontinuities at
x
5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5.
(b) y
1 x
[ 6, 5] by [ 18, 4]
5 2 int(x 3) dx ( 18) ( 12) ( 10) ( 16) ( 8) ( 14) 6 y = f (x ) R 0 –1
1 (x 3 1) dx 1
4 1 0 2 ( 6) ( 4) 88 45. (a) The function has a discontinuity at x 3
4 ( 2) 1. (b) 38. Observe from the graph below that the region between the
3 graph of f (x) x and the xaxis for 0 x 1 cuts out a
[ 3, 4] by [ 4, 3] region R from the square identical to the region under the
4 x 3 for 0 graph of g(x) x 1. 3 x2
x 1
(4)(4)
2 1
dx
1 1
(3)(3)
2 46. (a) The function has a discontinuity at x y 7
2 3. (b) 1
R y = f (x ) [ 5, 6] by [ 9, 2]
6 x 1
1 x dx 1 0 x 39. NINT
40. 3 x2 4 1
4 x 2 , x, 42. NINT(x 2e x, x, 2, 2)
1, 3) 1
(9)(9)
2 3 1 n 1
dx
x2 lim ∑ n→ ∑ n 10.6667 n→ 1.8719 k1 n
k2 n→ (b) … 1
22 lim n 1 0. 1
n 1
k2
n k1 lim 0 4.3863 43. (a) The function has a discontinuity at x 1
.
n2 1
1
…
22
n2
1
…
→.
n2 n and n → , so Note that n 1
1
22 n1 48. (a) [ 2, 3] by [ 2, 2]
3
2 x dx 2 3 77
2 (b) Using right endpoints we have
0.9905 2 NINT tan x, x, 0, 41. NINT(4 1
(2)(2)
2 47. (a) As x approaches 0 from the right, f (x) goes to . 3
4 , x, 0, 5 x2
dx
3 9
5x 1
,x
nk x RRAM 1 x n ∑ k1
n (b) k2
n ∑ k1 (c) n ∑ k2 1
n3 k 1 k
n 12
n
k2 1
n
n n ∑ 1
n k1 1
n3 22
n 1
n n(n k2
n3 … 1
n n2
n 1
n n ∑ k2 1
n3 k
1)(2n
6 1 1) n(n 1)(2n
6n 3 1) Section 5.3 (d) lim
n→ k2
n ∑ k1 1
n lim n lim n(n
2n 3 n→ 2
6 1)(2n 1)
6n3
3n 2 n
6n 3 1
3 Exploration 2
Integral 213 Finding the Derivative of an Pictures will vary according to the value of x chosen.
(Indeed, this is the point of the exploration.) We show a
typical solution here.
y 1 x 2 dx equals the limit of any Riemann sum (e) Since
0 over the interval [0, 1] as n approaches , part (d)
1 1
.
3 x 2 dx proves that
0 x
0a b x s Section 5.3 Definite Integrals and
Antiderivatives (pp. 268–276)
Exploration 1 How Long is the Average Chord
of a Circle?
1. The chord is twice as long as the leg of the right triangle in
the first quadrant, which has length r2 x2 by the
Pythagorean Theorem. 1. We have chosen an arbitrary x between a and b.
2. We have shaded the region using vertical line segments.
x 3. The shaded region can be written as f (t) dt using the
a definition of the definite integral in Section 5.2. We use t as y a dummy variable because x cannot vary between a and
r2 − x2 r itself.
4. The area of the shaded region is our value of F(x). x x y 2. Average value
3. Average value 1
( r) r
2
2r
1
r r x r r2 2 x2 dx. 0a r2 x2 dx (area of semicircle of radius r)
r2
2 5. We have drawn one more vertical shading segment to
represent F.
6. We have moved x a distance of x so that it rests above the
new shading segment. r
2 y 4. Although we only computed the average length of chords ∆x perpendicular to a particular diameter, the same
computation applies to any diameter. The average length of
a chord of a circle of radius r is
5. The function y 2 x ∆F r 1
r b r 2 r x
0a r
.
2 b x ∆F 2 x is continuous on [ r, r], so the Mean Value Theorem applies and there is a c in [a, b]
so that y(c) is the average value r
.
2 7. Now the (signed) height of the newlyadded vertical
segment is f (x).
8. The (signed) area of the segment is F
so F (x) lim x→0 F
x f (x) x f (x), 214 Section 5.3 Quick Review 5.3
dy
1.
dx
dy
2.
dx 9 (c) 9 [2 f (x) 3h(x)] dx 7 sin x 7 cos x
1 sec x tan x
sec x 4. dy
dx c os x
sin x 5. dy
dx sec x tan x sec2 x
sec x tan x (d) tan x (e) f (x) dx
7 f (x) dx
9
9 f (x) dx f (x) dx 1 7 1
ln x 5 6 7 (f) ( n 1) x n
n1 f (x) dx 1
9 sec x 1 1 9 f (x) dx
1 ln x 7 [h(x) f (x)] dx 7 h(x) dx 9 xn 2 1 7 1
x
x dy
dx 7 3(4) 9 f (x) dx
9 cot x 9 f (x) dx
9 9 dy
dx 9. dy
dx xe x 10. dy
dx 2 1
(2x 2x ln 2
(2x 1)2 (ln 2)2x 1)2 9 h(x) dx f (x) dx 7 8. 9 3 h(x) dx 7 2(5) dy
dx 7. 3h(x) dx
7 9 2 f (x) dx 3. dy
6.
dx 9 2f (x) dx 7 4 5 1 3 f (z) dz 5 2 3. (a) ex f (u) du 5 1
2 1
x 2 (b) 3f (z) dz
1 1 1 (c) Section 5.3 Exercises 3 1
2 f (t) dt f (t) dt 2 5 1 2 1. (a) g(x) dx 2 0 (d) 2 2 [ f (x)] dx f (x) dx 1
1 (b) g(x) dx g(x) dx 5 3 8 4. (a) 1 0 g(t) dt g(t) dt 0
2 (c) 5 1 5 2 3 2 3 f (x) dx 3 1 f (x) dx 3( 4) 0 12 (b) 1 g(u) du 2 3
5 (d) 1 f (x) dx 5 f (x) dx 2 2 3 5 [ f (x) g(x)] dx 1 1 8 5 [4f (x) g(x)] dx g(x) dx
g(x) dx [ f (x) 7 7 5 0 3 f (t) dt
0 3 4 2
6. (a) 1 h(r) dr
1 h(x) dx 3 h(r) dr
1 h(r) dr
1
3 1 7 4 f (t) dt
0 4 f (t) dt 9 f (x) dx 4
3 4 3 h(x)] dx f (z) dz f (t) dt 16
2( 1) 9 4
0 0 f (t) dt
4 1 9 f (z) dz
0 3 7 2 f (x) dx 1 7 3 9 2f (x) dx (b) 3 (b) 1 8 4 f (z) dz 3 1
5 5
1 9 3 1 f (z) dz 4 f (x) dx 2. (a) 0 f (z) dz 5 4(6) g(r) dr
3 2 0 4f (x) dx
1 2 2 5 1 5. (a) 1 6
(f) g(x) dx 2 0 1 dr 4 5 f (x) dx g(x) dx
3 g(r) (d) 10 5 (e) 0 1 6 [ g(x)] dx
3 f (x) dx 1 0 (c) 5 f (x) dx
4 0 f (x) dx
1 2 h(r) dr
1 9
1 (b) 1 h(u) du
3 h(r) dr
1 h(u) du
3 6 1 3 h(u) du
1 h(u) du
1 1 h(u) du
1 6 Section 5.3
7. An antiderivative of 7 is F(x) 17. Divide the shaded area as follows. 7x. 1 7 dx F(1) F(3) 7 215 21 y 14 3 3 52
x
2 8. An antiderivative of 5x is F(x)
2 5x dx F(2) F(0) 10 0 R2
R1 10 y = x2 – 1 0 x2
.
16
9
1
16 x
is F(x)
8
25
F(3)
16 9. An antiderivative of
5 x
dx
8
3 F(5) 10. An antiderivative of 2t t2 3 is F(t) x 2 R3 Note that an antiderivative of x 2 3t. 1 is F(x) 2 (2t 3) dt F(2) F(0) 2 0 Area of R1 3(1) Area of R2 2 0 (3)(1) 11. An antiderivative of t 2 is F(t) t 2. (x 2 (t 2) dt F( 2) F(0) 3 1 z
is F(z)
2
5
F(2)
4 1 z
dz
2 1
2 F(1) 13. An antiderivative of
1 1
x2 11 dx 1
x2 1 F(1) 1/2
1/2 dx
1 x2 is F(x) F( 1)
1 14. An antiderivative of x2 1 1
F
2 F e x dx F(2) e2 F(0) tan 4 1
2 1 sin 6 F(0)] 2
3 2
1 6 0 3 Total shaded area x. x 3. 1) dx 5
3 0 x. 2
3 5
3 2
3 16
3 18. Divide the shaded area as follows.
y 3
3 e x.
1 (x 2 Area of R3 4 is F(x) 15. An antiderivative of e x is F(x)
2 3 2
3 1 z2
.
4
7
4 z F(1)] 2
3 1 0 12. An antiderivative of 1 [F(2) [F(1) 0 3x 1) dx 1 3 2 x. 3
2 12
t
2 13
x
3 y = 3 – 3x 2 R1 x 2 6.389 0 16. An antiderivative of
3
0 3 dx
x1 3
x 1 F(3) R2 1. R3 0 3 ln 4 3 ln x F(0) 3 ln 4 is F(x) 4.159 –9 3x 2 is F(x) Note that an antiderivative of 3
1 Area of R1
Area of R2 3x 2) dx (3 F(1) 0 (9)(1) 9
2 Area of R3 (9)(1) 3x 2) dx (3
1 9 [F(2) 9 (2 Total shaded area 2 F(1)]
2)
9 5
5 16 F(0) 2 216 Section 5.3 19. Divide the shaded area as follows. 2 3 (x 2 (b) Area 6x 8) dx 0 y
2.25 [F(2) y = 3x – x 2 20
3 R1 (x 2 6x 8) dx 2 F(0)]
0 [F(3) 6 F(2)] 20
3 22
3 x R2 4 22. –4 Note that an antiderivative of 3x
3 Area of R1 x 2) dx (3x
0 4 Area of R2 32
13
x
x.
2
3
9
9
0
2
2 x 2 is F(x) F(3) F(0) [0, 2] by [ 5, 3] x 2) dx (3x 13
x
3 F(x) 3 [F(4)
8
3 F(3)]
9
2 2 (a) 11
6 9
2 Total shaded area x2 An antiderivative of
52
x
2 ( x2 5x 5x 4 is 4x.
4) dx F(2) 2
3 F(0) 11
6 0
1 19
3 (b) Area 2 ( x2 5x ( x2 4) dx 0 5x 4) dx 1 [F(1)
20. Divide the shaded area as follows. 2
3 0 F(0)] [F(2) F(1)] 0 2
3 11
6 3 x2 13
x.
3 11
6 y 23. 3
y= x2 – 2x R2 [0, 3] by [ 3, 2] x 3 R1 x 2 is F(x) An antiderivative of 2x Note that an antiderivative of x 2
2 Area of R1 (x 2 13
x
3 2x is F(x) 3 x 2. (a) 2x) dx [F(2) 2 (x 2 F(0) 0 3 x 2 ) dx (2x 0 0 (2x 4
3 2x) dx x 2) dx 2 [F(2) 4
3 0 F(3) 0 F(0)] 4
3 3 x 2 ) dx (b) Area 0 Area of R2 (2x
0 F(0)]
0 [F(3)
4
3 0 F(2)] 8
3 2 F(3)
0 24. F(2)
4
3
4
3 Total shaded area 4
3
4
3 8
3
[0, 5] by [ 5, 5] 21. An antiderivative of x 2
5 (a)
0 [0, 3] by [ 1, 8] An antiderivative of x 2
3 (a)
0 (x 2 6x 8) dx 6x
F(3) 13
x
3 8 is F(x)
F(0) 6 0 3x 2
6 8x. (x 2 4x) dx 4x is F(x) F(5) F(0) 13
x
3
25
3 2x 2.
0 25
3 Section 5.3
4 (b) Area 5 (x 2 4x) dx 0 30. The region between the graph and the xaxis is a rectangle 4x) dx 4 [F(4) F(0)] [F(5)
25
3 32
3 with a half circle of radius 1 cut out. The area of the region F(4)] 0 32
3 25. An antiderivative of x 2 x. (x 2
3 F( 3) 14
2
2 f (t) dt
1 0 2 1
2 av( f ) F(0) 0) . 2 4
4 . 31. There are equal areas above and below the xaxis. 1) dx 0 1 4 1 1
2 av( f ) 13
x
3 1 is F(x) 1
(1)2
2 is 2(1) 13 3 1 av (x 2 217 1
2 f (t) dt
0 0 0 3
1 (0 32. Since tan is an odd function, there are equal areas above 3 Find x c in [0, 3] such that c2
c2 1
c
1
Since 1 is in [0, 3], x 1. 1 and below the xaxis. 0 33. min f
2 3 x
x
26. An antiderivative of
is F(x)
.
2
6
3
1
x2
1
1
av
dx
[F(3) F(0)]
30
2
3
3
c2
3
Find x c in [0, 3] such that
.
2
2 c2 9
2 1
2 3
2 1
2
8
17
1
2
1
4
8
17
49
68 c 3 Since 3 is in [0, 3], x 27. An antiderivative of
1
1 av 3. 3x 2 1 is F(x) x 3 x. 1 ( 3x 2 1) dx F(1) F(0) 2 0 Find x 3c2 c in [0, 1] such that 3c2 1 2 1 c 1 0 0 1 16
17
0.5 16
17
0.5 1
x4 1 0 1
1 1
(1)
2 dx
1
2 dx x4
1
1
1
2
x4
0.5 1
1
1
dx
x4
0.5 1
1
1
1
4
x4
01
1
1
dx
x4
01
0 35. max sin (x 2 ) 1
3 2 2 f( ) d
/4 1
and max f
2
1
1
dx 1
x4
01 34. f (0.5) 3 /4 1
/2 av( f ) 1 16
2 17 dx
8
17 dx 1
2 8
17 33
34 sin (1) on [0, 1] sin (x 2 ) dx sin (1) 1 0 1 c 36. max 3
1 Since 1 is in [0, 1]. x
3 1
3 av 3 1)2 dx (x . 22 0 3 and min
x 1
(x 1)3.
3
18
1
F(0)]
33
3 37. (b
1 a) min f (x) Find x c in [0, 3] such that (c 1)
1.
c1
1
c 2 or c 0.
Since both are in [0, 3], x 0 or x 2.
29. The region between the graph and the xaxis is a triangle of
height 3 and base 6, so the area of the region
1
2 av( f ) 1
6 9.
2 f (x) dx
4 9
6 3
2 8 8 dx 3 0 on [a, b]
b 0 (b a) min f (x) f (x) dx
a 2 is (3)(6) x 0 1)2 is F(x)
1
[F(3)
3 8
1 3 28. An antiderivative of (x x 38. (b a) max f (x) 0 on [a, b] b f (x) dx
a (b a) max f (x) 0 2 2 on [0, 1] 218 Section 5.4
b 39. Yes, a 46. An antiderivative of F (x) is F(x) and an antiderivative of b av ( f ) dx f (x) dx. a G (x) is G(x). This is because av( f ) is a constant, so
b av( f ) dx a b a b a
b F (x) dx av( f ) b av( f ) a F(b) F(a) G (x) dx av ( f ) x G(b) G(a) a
b (b a)av( f ) (b Since F (x) b a a b F (x) dx G (x) dx, so a b 1 a) G (x), f (x) dx F(b) F(a) G(b) a G(a). b
a f (x) dx s Section 5.4 Fundamental Theorem of
Calculus (pp. 277–288) 40. (a) 300 mi
(b) 150 mi
30 mph 150 mi
50 mph 300 mi
(c)
8h 8h Exploration 1 37.5 mph 2. The function y (d) The average speed is the total distance divided by the
total time. Algebraically,
1 d1
2 t1 d2 d1 d2 t1 t2 1000 m3
100 min
10 m3/min
1000 m3
Time for second release
50 min
20 m3/min
total released
2000 m3
1
Average rate
13 m3/min
total time
150 min
3 41. Time for first release 1 42. 1 sin x dx x dx 0 0 1 43. 1 sec x dx 1 0 0 1 21
x
20 multiples of . There are six of these between 3. In attempting to find F( 10) 10 and 10. tan (t) dt 5, the 3 calculator must find a limit of Riemann sums for the
integral, using values of tan t for t between 10 and 3. The
large positive and negative values of tan t found near the
asymptotes cause the sums to fluctuate erratically so that no
limit is approached. (We will see in Section 8.3 that the
“areas” near the asymptotes are infinite, although NINT is
not designed to determine this.)
4. y tan x 1
2 x2
dx
2 x3 1
60 x 7
6
[1.6, 4.7] by [ 2, 2] 1
bh
2 44. (a) Area 2 tan x has vertical asymptotes at all odd 10 . The driver computed . The two expressions are not equal. t2 Graphing NINT f 5. The domain of this continuous function is the open interval
(b) h2
x
2b C
2 b (c) y (x) dx
0 45. av(x k)
Graph y1 1
k h 2b
x
2b 0 k x k dx 0 hb 2
2b 1
bh
2 11
k
xk 1
kk 1
0 xx 1
and y2
x(x 1) 3
.
2 6. The domain of F is the same as the domain of the
kk 1
k(k 1) x on a graphing calculator and find the point of intersection for x , 1. continuous function in step 4, namely 2 , 3
.
2 7. We need to choose a closed window narrower than
2 , 3
2 to avoid the asymptotes. [1.6, 4.7] by [0, 16]
[1, 3] by [0, 3] 8. The graph of F looks like the graph in step 7. It would be
Thus, k 2.39838 decreasing on 2 , and increasing on vertical asymptotes at x 2 and x 3
.
2 , 3
, with
2 219 Section 5.4
Exploration 2 The Effect of Changing a in 10. x f (t) dt
a dy
dx 1
dx/dy 1
3x Section 5.4 Exercises 1. 3 1. 3 1
dx
x 2
1/2 2x ln x
1/2 (6 [ 4.7, 4.7] by [ 3.1, 3.1] 2. ln 3) 1 ln 1
2 1
2 5 ln 3 ln 5 ln 3 ln 2 5 ln 6 3.208 [ 4.7, 4.7] by [ 3.1, 3.1] 3. Since NINT(x 2, x, 0, 0) 0, the xintercept is 0. 2 4. Since NINT(x , x, 5, 5) 1 0, the xintercept is 5. 5. Changing a has no effect on the graph of y
It will always be the same as the graph of y d
dx 1 1
3x
ln 3 3x dx 2.
2
x 2 11
ln 3 3 f (t) dt.
a 9 26
3 ln 3 f (x). 7.889 x 6. Changing a shifts the graph of y f (t) dt vertically in
a 1 3. a dy
dx dy
2.
dx 3. dy
dx f (t) dt. a 4. 2 5/2
x
5 x 3/2 dx 0 2
32 Quick Review 5.4
1. 5
1 13
x
3 x) dx 0 such a way that a is always the xintercept. If we change
from a1 to a2, the distance of the vertical shift is (x 2 6/5 x 5. dx 5 2
(25
5 0 1/5 5x 1 cos(x 2) 2x 1 6.
2 5 2 x 2 32 5 dx 1 dy
dx 3
3x 5. dy
dx 1 1/2
x
2 5
2
1
2 21 1 0 7. sin x dx cos x 1 0 ( 1) 2 0 8. (1 cos x) dx x sin x
0 2x ln 2 dy
6.
dx 1 1 22.361 2 0 4. 1 5 0) 2 2(tan x)(sec2 x) 2 tan x sec2 x (0 10 1
2 x 2 2
3 0) 2 sin x cos x 2(sec x)(sec x tan x)
2 sec2 x tan x 2
dx
x2 1 1
3 0 1 2x cos (x 2) 2(sin x)(cos x) 1 2 3/2
x
3 7. dy
dx 8. dy
dt
dy
dx 7
7x ( 0 (0 0) 3.142
/3 /3 dx
dt dy/dt
dx/dt 0 2( sin t x sin x cos x
x2 2
5 /6 10. 2y)
y
x (y
1
2y y
2y csc2 3
3 1
x 3.464
5 /6 d cot
/6 3 1) 0) /6 cot t dy
2y
dx 1 2 tan 0 cos t
sin t (1)y d x 2 ( sin x)(x) (cos x)(1)
x2 cos t, 2 sec2 9. 1 9. Implicitly differentiate:
dy
x
dx
dy
(x
dx
dy
dx 0) 2 (
3 3)
3.464
3 /4 3 /4 11. csc x cot x dx csc x ( /4 4 sec x tan x dx
0 ( 2) /3 /3 12. 2) /4 4 sec x 4(2
0 1) 4 0 220 Section 5.4 1 13. (r 1)2 dr 1
(r
3 u 4 1
4 14. 1 0 du 1)3 1 8
3 1 1/2 (u x3 17. Graph y 8
3 0 3x 2 2x. 1) du 0 u 4 2u1/2 u [0, 2] by [ 1, 1]
0 (4
15. Graph y 2 4) (0 0) Over [0, 1]: 0 1 x. (x 3 3x 2 14
x
4 0 x3 14
x
4
1
2 2x) dx x3 1 1
4 0 0 1
4 0 ( 4) 4 4 0 x2 0 1
4 Over [1, 2]:
2 (x 3 3x 2 2x) dx 1 2 Over [0, 2]: (2 2x
2x x) dx 12
x
2
12
x
2 0
3 Over [2, 3]: (2 x) dx 2 Total area
16. Graph y 1
2 2
3x 2 1
4 Total area [0, 3] by [ 2, 3] x3 18. Graph y 2 1
4 x2 2
1 1
4 4x. 2
0
3 3
2 2 2 1
2 5
2 [ 2, 2] by [ 4, 4] 3. Over [ 2, 0]:
0 (x 3 14
x
4 4x) dx 2 2x 2 0
2 Over [0, 2]:
2 [ 2, 2] by [ 4, 10] (x 3 14
x
4 4x) dx 0 Over [ 2,
1 (3x 2 1]:
3) dx Total area
x3 1 3x 2 2 ( 2) 4 2 (b)
(3x 2 3) dx x3 3x 1 1 2 2 (3x 2 Over [1, 2]: 3) dx 1
1 x x3 8 1
is discontinuous at x
1 1, x 1
2 2 ( 2) 4 9
. The area
2 between the graph of f and the xaxis over (1, 3] where 1 1
2 f is positive is (2)(2)
4 1. 1 2 3x 4 0 [ 2, 1) where f is negative is (3)(3) 1 Total area 2 The area between the graph of f and the xaxis over 4 1
2 x2
x 4
x2
x 19. (a) No, f (x) Over [ 1, 1]:
1 4 2x2 4 4 12 3
2 2 x
x 20. (a) No, f (x)
(b) 9 x2
3x 9 9
2 1
dx
1 2.
2 5
2 9 x2
is discontinuous at x
3x 9
3 x
3 ,x 3. 3 Note that f is negative for x in [0, 5]. f (0) 1 and 8
. The area between the graph of f and the
3
1
8
55
xaxis over [0, 5] is (5 0) 1
2
3
6
5
55
9 x2
dx
6
9
0 3x f (5) Section 5.4 21. (a) No, f (x) tan x is discontinuous at x 3
.
2 and x 2 26. First find the area under the graph of y
1 (b) The integral does not have a value. If 0 b 2 0 , then tan x dx ln cos x ln cos b since the 0 2 b ln cos b → cos b → 0 so or 2
3 0 2 2
3 x 2 dx 1 tan x dx → . 8
3 x 2. 7
3 Area of the shaded region , 13
x
3 2 1
3 0 Fundamental Theorem applies for [0, b]. As b → 1 x. Next find the area under the graph of y b b 2 3/2
x
3 x 1/2 dx 221 1 7
3 3 0 Hence the integral does not exist over a subinterval of 27. First find the area under the graph of y
(1 [0, 2 ], so it doesn’t exist over [0, 2 ]. 0 x
x2 b 1
dx
1 x
since 2
x 0 1
1 x 1 1
x 1 dx ln x 0 x
x2 1
dx →
1 b 0 xaxis for 5 /6 or The area of the rectangle is sin
Area of the shaded region 31.
1, 2 2.55. The integral exists since 2x 4
x4 1
, x,
1 0.8, 0.8 cos x, x, 1
NINT(
2 1, 1) 2x 2 8 32. sin x
is bounded.
x the area is finite because 1
, x, 0, 10
2 sin x 3 24. (a) No, f (x) 0. 2
3 6 3 3 3.802
1.427
0.914 2x 2, x, 8.886 2, 2) t2 , t, 0, x), y2 2 0.6 in a [0, 1] by [0,1] window, then use the intersect function to find
(b) NINT 1 cos x
, x,
x2 2, 3 2.08. The integral exists since the area is finite because 1 cos x
is bounded.
x2 1 x 2 dx 0 13
x
3 1
0 2 (2
1 x) dx 2x 12
x
2 Area of the shaded region 2 3
2 2
1 1
3 2 1
2 1
2
5
6 y 3 1
3 0, x
x 1
3 NINT( 1 1
3 Next find the area under the graph of y 0.699. 34. When y
y x 2. 25. First, find the area under the graph of y x x. 3 3 2 and x NINT(e 33. Plot y1 2 0 between x NINT( 8
cos x
is discontinuous at x
x2 3 2 /6 30. NINT 0. 3 cos x /6 29. NINT
1 5
.
6 sin x dx . Hence the integral does not exist sin x
, x,
x , 6 5 /6 1→ sin x and the 1, and the Fundamental Theorem sin x
is discontinuous at x
x . 28. First, find the area of the region between y
ln b [0, 2].
23. (a) No, f (x) 2 1, then over a subinterval of [0, 2], so it does not exist over (b) NINT 0 Area of the shaded region 1 applies for [0, b]. As b → 1 , ln b
b sin x 1. b 1 cos x. The area of the rectangle is 2 . (b) The integral does not have a value. If 0
b x 0 1
is discontinuous at x
1 x
x2 22. (a) No, f (x) cos x) dx 1 1. 3 x3
x 3, x, 0, 1) 0.883 222 Section 5.4
x 35. 43. Choose (d). x f (t) dt K f (t) dt a b
x f (t) dt K dy
dx x f (t) dt a b a t2 e dt b dy
dx 1 (t 2 K 3t 1) dt 13
t
3 3
2 that cos 2x
0 3 sec t dt
sec t dt t
8
3 ( 1) 6 dy
dx 1
2 1
cos 2x.
2 sec t dt y (0) 0 1
sin (2x)
4 sec t dt 1 x d
dx t2 e
1
1 t2 e x 2 48. f (x)
0
2 d
dx sin 2 cos 2
2 dy
dx cos (2x)
x d
dx x d
dx 1 d
(
dx 1
dt
t x) d
(2x)
dx 2 1.189 f (0) (sin x) x cos (2t) dt 1
x
1 1/2
x
2 dt 0
0 3x cos (2x ) 2 x
0 cos x
sin x
cos x 10
1 t 1) 2x 2 10 dt 10
1 t dt 1 x 2 x 50. f (x)
sin x
2x d
f (t) dt
dx 0
d
(x cos x)
dx x( sin x) f (4) 4 sin 4 1 cos x
cos x
cos 4 1 cos (2x ) 2x
2x cos (2x 2) 51. One arch of sin kx is from x
/k sin kx dx
0 t 2 dt 2
sin x t 2 dt 0 t 2 dt 0 cos2 x ( sin x)
sin x cos2 x 2x 10x Area
d
dx
d
dx 3 2 cos (2x ) 3x
3 3 d2
(x
dx f (t) dt x sin x 0 2 x2 e 3 1 2 2 cos (2t) dt 0 2 4 10 2 cos (2x) 3 3 3 x 0 sin ( x)2 dt d
2
dx 49. f (x) f (0) dy
dx 4 a 2 1
dt
t 0 f (t) dt a since L (x)
1 d
dx 4 a 47. x sin 2 cos 2
2 x sec x 46. Choose (a). 0 dy
37.
dx 4 1 0 1
sin x cos x
2 1 4 0 1
cos (2x) dx
2 1
x
2 4 0
0 y (1) 1
x
2 0 x d
dx dy
dx sin t dt 2 4 45. Choose (b). 3
2 2 2 sin2 x, so sin2 x 1 1
2 sec x 2 2 0 4 1 2 dy
dx 3 1 36. To find an antiderivative of sin2 x, recall from trigonometry 42. 0 x d
dx y ( 1) 1 32
t
2 1
3 dy
41.
dx 3 1 2 40. x2 e 44. Choose (c). f (t) dt 39. 3 b a dy
38.
dx dt f (t) dt x K t2 e y( ) x f (t) dt x d
dx sin2 x cos x sin2 x cos x 52. (a) (6 x x 2) dx 0 to x 1
cos kx
k 6x 3 22
3
125
6 /k
0 12
x
2
27
2 k .
1
k 13
x
3 1
k
2
3 2
k Section 5.4
( 1)
2( 1) (b) The vertex is at x ax 2 vertex of a parabola y
y 1
2 1
. Recall that the
2 bx b
.
2a c is at x 25
25
, so the height is .
4
4 55. (a) s (3)
(b) s (3) f (3) 0 units/sec f (3) 0 so acceleration is positive. 3 (c) s (3) 1
( 6)(3)
2 f (x) dx
0 (c) The base is 2 ( 3) 2
(base)(height)
3 5. 2
25
(5)
3
4 6 (d) s (6) 125
6 223 9 units 1
( 6)(3)
2 f (x) dx
0 1
(6)(3)
2 particle passes through the origin at t 0, so the
6 sec. 0 53. (a) H(0) f (t) dt (e) s (t) 0 0 (b) H (x)
H (x) x d
dx f (t) dt f (x) 0 0 when f (x) 0. H is increasing on [0, 6].
(c) H is concave up on the open interval where
H (x) f (x) 0.
f (x) 0 when 9 x 12.
H is concave up on (9, 12).
f (t) dt 0 at t 7 sec (g) The particle is on the positive side since
9 s(9) f (x) dx 0 (the area below the xaxis is 0 smaller than the area above the xaxis). 12 (d) H(12) f (t) (f) The particle is moving away from the origin in the
negative direction on (0, 3) since s (0) 0 and
s (t) 0 on (0, 3). The particle is moving toward the
origin on (3, 6) since s (t) 0 on (3, 6) and s (6) 0.
The particle moves away from the origin in the positive
direction for t 6 since s (t) 0. 0 because there is more area
0 0
x Si( x) H(12) is positive. 0 (e) H (x) f (x) 0 at x 6 and x 12. Since
H (x) f (x) 0 on [0, 6), the values of H are
increasing to the left of x 6, and since
H (x) f (x) 0 on (6, 12], the values of H are
decreasing to the right of x 6. H achieves its
maximum value at x 6.
(f) H(x) 0 on (0, 12]. Since H(0)
minimum value at x 0.
54. (a) s (t) f (t). The velocity at t 0, H achieves its 5 is f (5) 2 units/sec. 0 (b) Si(0)
0 (c) Si (x) sin (t)
dt
t
sin (t)
dt
t
x
x
sin (t)
dt
t
0 x sin (t)
dt
t
x 56. (a) f (t) is an even function so above the xaxis than below the xaxis. 0 0 sin t
dt
t f (t) Si(x) 0 0 when t k, k a nonzero integer. (d) (b) s (t) f (t) 0 at t 5 since the graph is decreasing,
so acceleration at t 5 is negative.
3 (c) s (3) f (x) dx
0 1
(3)(3)
2 4.5 units [ 20, 20] by [ 3, 3] (d) s has its largest value at t 6 sec since
s (6) f (6) 0 and s (6) f (6) 0.
(e) The acceleration is zero when s (t) f (t)
occurs when t 4 sec and t 7 sec. 100 57. (a) c(100) c(1)
1
100 0. This (f) Since s (0) 0 and s (t) f (t) 0 on (0, 6), the
particle moves away from the origin in the positive
direction on (0, 6). The particle then moves in the
negative direction, towards the origin, on (6, 9) since
s (t) f (t) 0 on (6, 9) and the area below the xaxis
is smaller than the area above the xaxis. dc
dx
dx
1 1 9 s (9) f (x) dx 0 (the area below the xaxis is 0 smaller than the area above the xaxis). x x 2 10 1
400 c(100)
100
400 1 9 or $9
dc
dx
dx
1 400 dx 100 (g) The particle is on the positive side since 100 dx (b) c(400) sin (t)
dt.
t 20 2 x 10 x
100 10 or $10 224 Section 5.5 3 58. 2 2 (x 0 1) dx 2 2x 2(x
1
2 6 1 1) 9
2 2 3 64. Solving NINT 0 x 30 30 0 x2
dx
2 450
0 x3
6 1
450x
30 (b) (300 drums)($0.02 per drum) local minima at x
f (x). (b) True, because h and h are both differentiable by part
(a).
0. f (1) 1) , (2k 2) ], where 0, Si(x) has its 2k , where k is a positive integer. Furthermore, each arch of y f (x) is smaller in height than (2k 1) (2k 2) the previous one, so f (x) dx f (x) dx. 2k (2k 1) This means that
(2k 2) Si((2k
(d) True, because h (1) 1) ] and k is a nonnegative integer. Thus, for x $6 f (1) sin x
, Si(x) is
x f (x) decreasing on each interval [(2k 0 f (x) and therefore h (x) (c) True, because h (1) d
Si(x)
dx increasing on each interval [2k , (2k 30 300 drums 60. (a) True, because h (x) 1.0648397. We now argue that there are no other Exercise 56. Since The company should expect $4500.
1 1 graphically, the solution is solutions, using the functions Si(x) and f (t) as defined in 4.5 thousand 59. (a) sin t
, t, 0, x
t 0 and h (1) f (1) 2) ) Si(2k ) f (x) dx 0, so each 2k 0. successive minimum value is greater than the previous one.
(e) False, because h (1) f (1) 0.
Since f (2 ) (f) False, because h (1) f (1) 0 0 1.42 and Si(x) is 0, this means Si(x) Si(x) 2 . Now, Si(x) 1) for x 1.42 (and hence 1 has exactly one solution in the interval [0, ] because Si(x) is increasing on x f (t) dt sin x
, x, 0, 2
x continuous for x (g) True, because h (x) f (x), and f is a decreasing
function that includes the point (1, 0).
61. Since f (t) is odd, NINT f (t) dt because the area x 0 this interval and x 1.065 is a solution. Furthermore, between the curve and the xaxis from 0 to x is the opposite
Si(x)
of the area between the curve and the xaxis from 1 has no solution on the interval [ , 2 ] because x to 0,
Si(x) is decreasing on this interval and Si(2 ) 1.42 1. but it is on the opposite side of the xaxis.
x 0 x f (t) dt
0 f (t) dt
x x Thus Thus, Si(x) x f (t) dt 1 has exactly one solution in the interval f (t) dt 0 0 [0, ). Also, there is no solution in the interval ( , 0] f (t) dt is even.
0 because Si(x) is odd by Exercise 56 (or 62), which means
0 62. Since f (t) is even, x f (t) dt f (t) dt because the area x that Si(x) 0 for x 0 (since Si(x) 0 for x 0). 0 between the curve and the xaxis from 0 to x is the same as
the area between the curve and the xaxis from
x 0 f (t) dt
0 Thus f (t) dt
x x x to 0. f (t) dt Exploration 1 0 f (t) dt is odd.
0
x 63. If f is an even continuous function, then
d
dx f (t) dt is odd, but
0 x f (t) dt s Section 5.5 Trapezoidal Rule (pp. 289–297) x f (x). Therefore, f is the derivative of the odd 0 Area Under a Parabolic Arc 1. Let y f (x) Ax 2 Bx C
Then y0 f ( h) Ah 2 Bh
y1 f (0) A(0)2 B(0) C
y2 f (h) Ah 2 Bh C.
4y1
2. y0 y2 x continuous function f (t) dt. h 0 Similarly, if f is an odd continuous function, then f is the
x derivative of the even continuous function f (t) dt.
0 3. Ap
3 (Ax 2 h 2 Ah 2 Bh C
2Ah 2 6C.
Bx x
x
B
Cx
3
2
3
2
h
h
A
B
Ch
3
2
3
h
2A
2Ch
3
h
(2Ah 2 6C)
3 A C,
C, and
4C Ah 2 h2
2 Ch C) dx
h
h A h3
3 B Bh C Section 5.5
4. Substitute the expression in step 2 for the parenthetically
enclosed expression in step 3: Section 5.5 Exercises
1. (a) f (x) h
(2Ah 2 6C)
3
h
(y
4y1 y2).
30 Ap 2 x, h x 0 Quick Review 5.5
1. y
sin x
y
cos x
y
0 on [ 1, 1], so the curve is concave down on
[ 1, 1].
2. y
y
y 3 4x
12
12x 2
0 on [8, 17], so the curve is concave up on [8, 17]. 3. y
12x 2 6x
y
24x 6
y
0 on [ 8, 0], so the curve is concave down on
[ 8, 0].
4. y
y
y 2 (c) 5. y
y
y 2e
4e 2x
0 on [ 5, 5], so the curve is concave up on [ 5, 5]. 6. y y y
y
8. y
y 0
0 1
2
1
4 1
0
4 T y 2 (c) 2 2 4 2.75 1
2
3
2
9
4 1
1 1
4 2(1) 2 2
4
9
4 x ,h 0 1
2
1
8 0 1
0
4 T 0 1
2 4 0 f (x) 8
3 2 3 3. (a) f (x) 2 13
x
3 x 2 dx 0 2 3
2
27
8 1
1 1
8 2(1) 2 2
8
27
8 8 4.25 (b) f (x) 3x 2, f (x) 6x 0 on [0, 2]
The approximation is an overestimate.
2 (c) 2 14
x
4 x 3 dx 0 0 on [3, 6], so the curve is concave up on [3, 6]. cos x sin x
sin x cos x
0 on [1, 2], so the curve is concave down. 2 (b) f (x) 2x, f (x) 2 0 on [0, 2]
The approximation is an overestimate. 4
0 1
,h
x 2 x 1 f (x) 1 5
4
4
5 4. (a) f (x) 9. y
100x 9
y
900x 8
y
0 on [10, 1010], so the curve is concave down on
[10, 1010].
10. y
y
y 0
4 0 on [100, 200], so the curve is concave down on csc x cot x
( csc x)( csc2 x) (csc x cot x)(cot x)
csc3 x csc x cot2 x
0 on [0, ], so the curve is concave up on [0, ]. 3
2 2 f (x) 1
x2 2
x3 2 0 2 x 1
x2 2 2 2 x [100, 200].
7. y 2(1) x 2, h 2. (a) f (x) 1
x y 1 1
2 12
x
2 x dx
0 [48 , 50 ].
2x 2 3
2
3
2 1 (b) f (x) 1, f (x) 0
The approximation is exact. 1
x
cos
2
2
1
x
sin
4
2 0 on [48 , 50 ], so the curve is concave down on 1
0
4 T 1
2 1
2
1
2 0 f (x) 0
4 1
1
8 T
(b) f (x) 1 1
4 4 2 4
5 3
2
2
3 2 1
, f (x)
x2 2
3
2
x3 7
4
4
7 2 2
1
2
4
7 1
2 0 on [1, 2] The approximation is an overestimate.
2 (c)
1 1
dx
x 2 ln x ln 2
1 0.693 0.697 225 226 Section 5.5 5. (a) f (x) 4 x, h 0 1 4 x 0 1 f (x) 0 1 1
(0
2 T 2 3 2( 4 x 3 2 f (x) 2( 2 2(1) 3) 2) 2) 1 3/2
x
4 1 4
0 6. (a) f (x) 0 f (x) 0 8 (b) f (x) 0 16
3 0 4(0) 4 (x 3 2x) dx 2 2 2 14
x
4 2 0 sin x dx cos x 7. 5
(6.0
2 1.896 1
(126
2 2 65
1
12 … 2 66 841 3.13791, S100 2(800) 1.08943, S100 15. S50 1.37066, S100
limit
S50 1.37076, S100
lower limit 26,360,000 ft3 2(860) 0.0001 as lower 1.37076 using a 0.000000001 as 0.82812 17. (a) T10 1.983523538
T100 1.999835504
T1000 1.999998355
(b) ET n
10 988.5, the town can sell at most 988 licenses. 2 0.016476462 1.6476462 100 1.64496 1000 1
to change seconds
3600 Tn 1.645 (3.2 2.2)(30 (4.5 3.2)(40 50) (5.9 (7.8 5.9)(60 70) (10.2 (12.7 10.2)(80 90) 16.0)(100
20.6)(110
26.2)(120 130)) (d) b 80) 12.7)(90 a 100) ET n ET 10n 1
3600 2 10 10 ET n , h2
2 120) (37.1 7.8)(70 60) 110) (26.2 10n 4.5)(50 (16.0 (20.6 (c) ET 40) 0.9785 miles 12 n 2 3 12(10n)2 2 ,M
n2
3 1 12n 2 10 2 4 10 to hours
30) 841 1.37066 using a 0.82812, S100 16. S50 19,770 fish to be caught. Since 9. Sum the trapezoids and multiply by 110) 1.08943 … 2(1000) 2 58 3.14029 14. S50 (b) You plan to start with 26,360 fish. You intend to have 1
(2.2(0
2 0. 70.08 13. S50 0)(20) 19,770
20 1 0 since f (4) 2x, Mf (4) 13.0)(30) 2(12.7) 2(520) (0.75)(26,360) 1
4 9 (b) We are approximating the area under the temperature
graph. By doubling the endpoints, the error in the first
and last trapezoids increases. 15,990 ft3
200
8. (a)
(0
2 12. (a) 0 on [0, ] 2
… 2(9.1) x3 av( f ) 0 2 0 2(8.2) 3 11. The average of the 13 discrete temperatures gives equal
weight to the low values at the end. 2 2 sin x 0 12 (d) Simpson’s Rule for cubic polynomials will always give
exact values since f (4) 0 for all cubic polynomials. The approximation is an underestimate.
(c) x2 21) 0 (c) For f (x) 2 cos x, f (x) 4(4) 12 4 2(1) 2 21 1 3
4 1 2 4 81
4 Es 2 3 2( 1) 5.333 4 2 1 1 0 sin x, h x T 4 2 3/2
x
3 x dx 1 0 The approximation is an underestimate.
(c) ( 1)
4 1 0 1
(1
3 (b) 0 on [0, 4] 3 2x, h
1 S 5.146 3 1 1/2
x
, f (x)
2 (b) f (x) x3 10. (a) f (x) ET n 6 10 2 Chapter 5 Review
22. Note that the tank crosssection is represented by the
shaded area, not the entire wing crosssection. Using
Simpson’s Rule, estimate the crosssection area to be 18. (a) S10 2.000109517
S100 2.000000011
S1000 2.000000000
(b) ES n
10 2 1.09517 100 10 1000
(c) ES 10 a ,h 4 ES 4 n4 ,M n ES 10n 1 h[ y0 yn 1 180(10n)4 y6]
2(2.0) 2yn 1 10.63 ft
yn] h[ y1 y2 … 4y3 … 2y2n RRAMn LRAMn
4 10 ES 2
n 2x cos (x 2)
2x
2x sin (x 2) 2 cos (x 2)
4x 2 sin (x 2) 2 cos (x 2) 24. S2n h
(y
30 4y1 2y2 4y2n
1
[h(y0
3 (b) y2n ) 2y2 2y1 1 … (2h)(y1
2T2n MRAMn
3 y3 2y2n
… y5
b , where h 1 a
2n y2n)
y2n 1)] . [ 1, 1] by [ 3, 3] (c) The graph shows that
for 1 x 1.
1 (d) ET
(e) For 0
1 (f) n 3 h2
2 0.1, ET ( 1)
h 2
0.1 f (x) 2 so f (x) 3 s Chapter 5 Review (pp. 298–301)
1. h2
2 ( 1) 2
(h )(3)
12 h f (4)(x) y
4 0.12
2 0.005 0.01 20
2 4x2 2x cos (x 2) 8x sin (x 2) 4x sin (x 2)
8x3 cos (x 2) 12x sin (x 2)
8x3
2x sin (x 2) 24x2 cos (x 2)
12x 2x cos (x 2) 12 sin (x 2)
(16x 4 12)sin (x 2) 48x 2 cos (x 2) 20. (a) f (x) 1 2. 2 1 (b) 2 x y
4 [ 1, 1] by [ 30, 10] (c) The graph shows that 30 f (4)(x)
so f (4)(x)
30 for 1 x 1.
1 (d) Es
(e) For 0
(f) n
21. h ( 1) 4
(h )(30)
180 h 1 24 in.
6 2
0.4 4 h
3 0.4, ES ( 1)
h h4
3 2 10 4 0.4
3 0.00853 0.01
LRAM4: 5 4 in. Estimate the area to be
4
[0
3 0] 4(18.75)
466.67 in2 yn] 2 180n 4 5 19. (a) f (x)
f (x) … 5 180 n 4 4(1.9) 1
1
42 lb/ft3 11.2 ft2 2y2 … 4y5 11.2 ft2 2.1] 2y1 y1 2y4 2(1.8) (5000 lb) h
[y
20 23. Tn n 4 4y3 4(2.1)
Length 4 ES 8 0 10n (d) b 4 10 1.1 1
[y
4y1 2y2
30
1
[1.5 4(1.6)
3 Sn 2(24) 4(26) 2(24) 4(18.75) 1
0
2 15
8 3 x 21
8 15
4 3.75 2 227 228 Chapter 5 Review 3. 2 y 5 f (x) dx 9. (a) f (x) dx 5 4 3 2 The statement is true.
5 (b) 2 [ f (x)
2 g(x)] dx 5 5 f (x) dx g(x) dx 2
2 1 MRAM4:
4. 1 63
2 64 165
64 f (x) dx x 2 2
5 195
64 4 105
64 5 f (x) dx 2 g(x) dx 2 3 2 2 9 4.125
The statement is true. y 5 (c) If f (x) 4 5 g(x) on [ 2, 5], then f (x) dx g(x) dx, 2 2 but this is not true since
5 2 5 f (x) dx 2 f (x) 2
5 2 g(x) dx f (x) 4 3 7 and 2 2. The statement is false. 2 1 1 15
RRAM4:
28 21
8 3 10. (a) Volume of one cylinder: r 2h x 2 15
4 0 Total volume: V lim n→ 3.75 y ∑ sin2(mi ) x i1 sin2 x on [0, ]. (b) Use
5. sin2 (mi ) x n NINT( sin2 x, x, 0, ) 4.9348 4 11. (a) Approximations may vary. Using Simpson’s Rule, the
area under the curve is approximately 2 1
[0
3 4(0.5) 2(4.75) 1
(LRAM4
2 T4
2 6. 2 (4x x 3) dx 2x 2 7. 14
x
4 1 15
24 15
4 3.75 0] 26.5 s 4 4 0 LRAMn MRAMn RRAMn
1.78204 1.60321 1.46204 20 1.69262 1.60785 1.53262 30 1.66419 1.60873 1.55752 50 10 1.62557 1.60937 1.59357 The curve is always increasing because the velocity is
always positive, and the graph is steepest when the
velocity is highest, at t 6. 1000 1.61104 1.60944 1.60784
1
dx
x 5 ln x ln 5
1 ln 1 ln 5 t Time (sec) 1.64195 1.60918 1.57795 100 1 4(2) 4(4.5) 30 8 10 5 (b) 2 n 8. 2(3.5) 2(3.5) The body traveled about 26.5 m. RRAM4) 0 4(4.5) 4(2) Position (m) 1 x 2(1) 1.60944 229 Chapter 5 Review
10 e x 3 dx 12. (a) 1 10 (b) 1 x sin x dx 25. 0
10 ln x 1 36(2x dx 1)3 1) 9(2x 1 2 10 (e) 1) ( 9) 0 1 13. The graph is above the xaxis for 0
xaxis for 4 x x 2 1
dx
x2 x 26. x
sin
dx
10
2 9 8 x 2) dx (x
1 12
x
2 4 and below the 1 x
1
2 2
0 0 Total area 6 (4 x) dx (4 0 27. x) dx 2
1 3
2 6 4 sec x tan x dx sec x 1 /3 4 12
x
2 [8 0] 0 [6 8] 1 4 28. 10 x 2) cos (1 1 1 x 2 and below the 29. 1 0 2 2
y 0 1
1 1 dy 2 ln y 1 2 ln 3 0 2 ln 3 0 30. Graph y x 2 x 2) dx 2x sin (1
2 14. The graph is above the xaxis for 0 1 6 12
x
2 4x 2 /3 4 4x xaxis for dx
1 dx x2 1 2 0 1 10 3 0 2)2 dx x(3x 0 1 1 36 0 (d) 0 1 (2x 0 (c) e 1
dx
x 24. 0 4 x 2 on [0, 2]. /2 Total area cos x dx cos x dx 0 /2 /2 sin x sin x
0 (1 /2 0) (0 1) 2 [ 1.35, 3.35] by [ 0.5, 2.6] The region under the curve is a quarter of a circle of radius
2. 2 2 15. 5 dx 5x 10 ( 10) 50 2 8 20 2
2 5 4x dx 16. 2x 2 2 42 0 2 31. Graph y
/4 /4 cos x dx 17. 2 sin x 0
1 (3x 2 4x x3 7) dx 1 2x 2 7x
1 6
1 (8s 3 12s 2 ( 10)
2s4 5) ds 16 20.
1 5s 4
x 4
dx
x2 y 4
1 4/3 8 x dx 2 ( 4) 2 32. Graph y 1 dy 3y 1/3 1
(4)(4)
2 64 1
(8)(8)
2 40 x 2 on [ 8, 8]. 27 1 ( 3) 2 1 ( 2) 1 dt
tt sec2 0 3 4 4 t 3/2 dt 2t 1 1/2 4
1 d tan 3
0 1 [ 9.4, 9.4] by [ 3.2, 9.2] The region under the curve y
of radius 8. /3 /3 23. 0 2 1 22. 3
0 27 21. [ 4, 8] by [0, 8] The region under the curve consists of two triangles. 1 4s3 0
2 x dx on [ 4, 8]. 2 1 19. 1
(2)2
4 2 0 2 0 18. x 2 dx 4 5 0 3 8 2
8 64 x 2 dx 8 2 64
8 64 x 2 dx x 2 is half a circle 2 1
(8)2
2 64 230 Chapter 5 Review 33. (a) Note that each interval is 1 day 24 hours
Upper estimate:
24(0.020 0.021 0.023 0.025 0.028
0.031 0.035) 4.392 L
Lower estimate:
24(0.019 0.020 0.021 0.023 0.025
0.028 0.031) 4.008 L
24
(b) [0.019
2 2(0.020)
0.035] 3
[5.30
2 x 2 25 2(0.031) dt 1 0 1
t2 1 dt 1 2 x2 1 1
x2 dt
t 4t1/2 1 50 x 50
25 4
1.11) … 0) 2(1.11) 20 50 4 x 30 4 2500 87.15 ft … x c(2500) … 2(5.04) 2(5.25) t 0 x 1
2 1
(2x)2 1
2
4x 2 1 4.2 L 34. (a) Upper estimate:
3(5.30 5.25 5.04
Lower estimate:
3(5.25 5.04 4.71
(b) 2x d
dx 43. c(x) … 2(0.021) dy
dx 42. 103.05 ft 30 230 0]
The total cost for printing 2500 newsletters is $230. 95.1 ft
14 44. av(I) 35. One possible answer:
The dx is important because it corresponds to the actual
physical quantity x in a Riemann sum. Without the x,
our integral approximations would be way off.
4 36. 0 f (x) dx 4 0 0 x 2 dx 2) dx 4 13
x
3 2x
4 [0 64
3 16] 2 1 since min sin x
0) 16
3 1 1 1 x4 sin x dx sin2 x dx 1 (max f )(1 0) 2 4 1
4 0 0 2t (24
0 14
t
4
14
x
4 3) dt 4 1 16
43 0 0 3x a 1
a 0 a 1 2 3/2
ax
a3 x dx 0 39. dy
dx 2 4
3 x2 3x 4 dy
dx 2 cos3 (7x 2) 4x 2 12x 16 dy
41.
dx d
dx 1 6
3 t4 x2 3x 0 0
3.09131 1.63052.
f (x).
f (1) (d) False, because g (1) f (1) (e) True, because g (1) f (1)
f (1) 0.
0.
0 and g (1) f (1) 0. (g) True, because g (x) f (x), and f is an increasing
function which includes the point (1, 0).
d
(7x 2)
dx 14x 2 cos3 (7x 2) 1 x 4 dx 1 47. F(1) 0 x 3t (b) True, because g is differentiable. 2 3/2
a
3 0 192
0 x t2 (f) False, because g (1) a 14 0 46. (a) True, because g (x) cos3 x 40. 12t 2 4 (c) True, because g (1)
(b) av( y) 1
24t
14 24t) dt x2 or x 1 2 3/2
x
43 x dx 24t Using a graphing calculator, x 0 38. (a) av( y) (t 3 14
x
4
14
x
4 2 24 14 0 0 1 0 45. 0 1 (min f )(1 x 0 0 2 since max sin2 x min f 4 sin2 x 1 1
14 Rich’s average daily holding cost is $192.
We could also say (0.04)4800 192. 0 0 12
x
2 0.04 I(t) av(c) 4 (x max f c(t) f (x) dx 4 37. Let f (x) Rich’s average daily inventory is 4800 cases. 4 f (x) dx 1
(600 600t) dt
14 0
14
1
[600t 300t 2
4800
14
0 dt 6
3 x4 x 48. y (x)
5 sin t
dt
t 3 F(0) 0. Chapter 5 Review 49. y 1
x
1
x2 2x y 2 b (b) Let h
S2n a
2n h
[y
30 1 y(1) 1
1 y (1) 1
dt
t 1 2 1 3 0 1 1
[h(y0
3 2 h( y2i 2t dt x t2 4 (x 2 4 1 1) x2 4 1
[2.5 2(2.4)
24
29
2.42 gal
12 4y2i 2 h(y2 4y2n 1
h.
12 54. (a) g(1)
2(2.4) 2 MRAMn 1 f (t) dt 2 1 (c) g( 1) 24.83 mi/gal (2)(1) 2704 ft. When her parachute opens, her
4296 ft. For t
SA(t) 6144 6208
16 6208 388 sec. For t 45 4296 16(t so B lands at t 5224
16 53. (a) Area of the trapezoid 58) 16t, so A lands at
13 58 sec, B’s 5224 0 and the absolute minimum is
3 1 f (t) dt f (t) dt
3 1
(2)2
2 The range of g is [ 2 , 0].
55. (a) NINT(e x 2/2 , x, 10, 10) , x, 2.506628275 20, 20) x 2/2 2.506628275 16t,
(b) The area is 2. 326.5 sec. B lands first.
56. First estimate the surface area of the swamp. 1
(2h)(y1
2 (2h)y2 h(y1 h(y1 2(2hy2) (f) g (x) f (x), f (x) 0 at x
1 and f (x) is not
defined at x 2. The inflection points are at x
1
and x 2. Note that g (x) f (x) is undefined at
x 1 as well, but since g (x) f (x) is negative on
both sides of x 1, x 1 is not an inflection point. NINT(e Area of the rectangle
y3) (e) g ( 1) f ( 1) 2
The equation of the tangent line is
y ( ) 2(x 1) or y 2x 2 1 position is given by
SB(t) 1
(2)2
4 (d) g (x) f (x); Since f (x) 0 for 3 x 1 and
f (x) 0 for 1 x 3, g(x) has a relative maximum at
x 1. g( 3) 4 sec, A’s position is given by t f (t) dt
1 (g) Note that the absolute maximum is g(1) (c) Let t represent the number of seconds after A jumps. 4) 1 1 f (t) dt
1 (b) The distance B falls in 13 seconds is 16(t . 0 3 (b) g(3) 2704 Tn 3 f (t) dt 2.3] 6144 ft. altitude is 7000 … 1 … When her parachute opens, her altitude is 1
(32)(132)
2 y4) 1 12
52. (a) Using the freefall equation s
gt from Section 3.4,
2
1
the distance A falls in 4 seconds is (32)(42) 256 ft.
2 256 4y3
y2n )] 1 1 6400 2 the ith of n rectangles plus the area of the ith of n
trapezoids, S2n 2(2.3) 12
(b) (60 mi/h)
h/gal
29 2y2n y2i ) is equal to twice the area of 1 3 1 51. (a) Each interval is 5 min … 2y4 Since each expression of the form 50. Graph (b).
x y2) 2 4y3 y2n] 1 4y1 h(y2n Thus, it satisfies condition ii. y 2y2 4y2n 1 1
1 2 .
4y1 Thus, it satisfies condition i. 231 y3)
2hy2 4y2 y3) h(y1 y3) 20
[146
2 13] 2(122) 2(76) 2(54) 8030 ft2 (5 ft)(8030 ft2) 1 yd3
27 ft3 1500 yd3 2(40) 2(30) 2. 232 Section 6.1 57. (a) V 2 (V max)2 sin2 (120 t) 5. Again h 1
1 av(V ) 7
2 2 (Vmax) 2 2 sin (120 t) dt 240 21 (Vmax)2 2 2 2
1
2 Vmax ,3
1
2 ,5
1 2 , ..., 19
1 2 2
1 2 339.41 volts Chapter 6
Differential Equations and
Mathematical Modeling 1
2 3
2 7
,0
2 x 1, x 3
2 7
,1
2 x 2, x 5
2 7
,2
2 x 3, 1
2 y3 1
2 1
3 y2 s Section 6.1 Antiderivatives and
Slope Fields (pp. 303–315) x 2 y1 Exploration 1 2 The 10 graphs are graphs of the functions. 2 2 19
. From left to right the
2 slopes of the line segments are (Vmax) 0 2 (b) Vmax 2 (Vmax) sin (120 t) dt (Vmax)2 Vrms 135
222 each line segment are , , , ..., 1
0
1 1. The ycoordinate of the midpoint of each line segment is . The xcoordinates of the midpoint of Using NINT:
2 k 5
2 1 Constructing a Slope Field 1. As i and j vary from 1 to 10, 100 ordered pairs are produced. Each ordered pair represents a distinct point in the
viewing window. 2 y10 19
2 x 19
2 1 7
,9
2 x 10. 2. The distance between the points with j fixed and i r and
i r 1 is the distance between their xcoordinates.
Xmin 2(r (Xmin 1) Xmin) 1 h
2 (2r Xmin
2 1 (2r
2r 1) 1) h
2 h
2 [0, 10] by [0, 10] h 3. The distance between the points with i fixed and j r and
j r 1 is the distance between their ycoordinates.
Ymin (2(r k
2 1) Ymin) (Ymin 1) (2r Ymin
2 1 (2r
2r 1) 1) k
2 k
2 6. For each line segment in part (5), make a column of parallel line segments as in part (4).
7. WL Quick Review 6.1
1. 100(1.06) k $106.00 2. 100 1
4. Here h k 0.06 4
4 3. 100 1 0.06 12
12 4. 100 1 0.06 365
365 1. Each line segment in the third column has 4
7 slope , because the xcoordinate of the midpoint of each
line segment is 2.5. The ycoordinates are , , , …,
135
222 19
.
2 The 10 graphs are graphs of the functions
y $106.14
$106.17
$106.18 n
,2
2 x 3, for n 1, 3, 5, …, 19. The length of the line segment can be increased or
decreased by adjusting the restriction 2 [0, 10] by [0, 10] x dy
dx d
sin 3x
dx (cos 3x)(3) 6. dy
dx 5
d
tan x
2
dx sec 2 x dy
dx d
Ce 2x
dx (Ce 2x)(2) 8. 2.5) 5. 7. 4
(x
7 dy
dx d
ln (x
dx 2) 5
2 3 cos 3x
5
2 5
5
sec 2 x
2
2 3.
2Ce 2x 1
x 2 Section 6.1
9. 6. 1
x dx 3 ln x 3 C Check:
d
[ln x
dx [0.01, 5] by [ 3, 3] By setting the left endpoint at x 0.01 instead of x 7. (x 5 0, we avoid an error that occurs when our calculator attempts
1
to calculate NINT , x, 1, 0 . The graph appears to be the
x same as the graph of y 3 ln x. 6x 8. ( x 3 9. x x 3 x6
6 3) dx 5
t2 e t/2 1 C] 3x 2
2 x 1) dx 2e t/2 10.
[ 5, 0.01] by [ 3, 3] 11. By setting the right endpoint at x 0.01 instead of x 43
t dt
3 1
dx
x3 x3 (x 3 1
, x,
x to calculate NINT 1, 0 . The graph appears to be the same as the graph of y 1. (x 2x 13. x3
3 1) dx x 2 x C 3 dx
dx 3 x2 x 4 2. ( 3x ) dx x2 C
3 x 2x 1 15. 17. d
(x 3
dx C) 3x x) dx (x 2 4x x3
3 1/2 ) dx 8 3/2
x
3 Check:
d x3
dx 3 4. (8 8 3/2
x
3 cos 1
x 1 C 1
x 1 4x
e
4 2 x 2 8x 4x 1/2 x csc x 2 C dx cos 3x
3 C C C 2 ln x
e cos 5x
5 tan 5r
cot 7t
7 x C
cos2 x dx 8 csc x cot x C
22. sin 2 x dx Check:
d 1 4x
e
dx 4 x 2 2 sec t sin 5x 2 ln x 21. 5. e 4x dx C 3 cos x sin 20. csc 2 7t dt C C) ) dx C x dx 2 1/3 3 2/3
x
2 sin 3x) dx Check:
csc x x 3 4/3
x
4 19. 5 sec 2 5r dr csc x cot x) dx d
(8x
dx (x 1/3 x 1/3 2
x C 1 ln x C 4 18.
4 2 C 2
1
2x 2 dx 1 2/3
x
dx
3 C 2 x x 16. 2 sec t tan t dt C Check: 3. (x 3 14. (3 sin x Check: 2 1 x ln ( x). Section 6.1 Exercises
2 3 12. C x 3) dx x
4
x4
4 we avoid an error that occurs when our calculator attempts C C t 4/3 4 0, 1 5
t 4 1/3
t dt
3 x ) dt 5t t/2 C 2 5t 10.
2e x2
2 2 (e t/2 dt 3x 23. tan 2 d dx
2x e 2 C C
C 1 cos 2x
dx
2
1
cos 2x
dx
2
2
sin 2x
x
C
4
2
1
1
2 e 4x 2x cos 2x
dx
2
cos 2x
dx
2 (sec 2 1) d x
2 tan sin 2x
4 C
C 233 234 Section 6.1 24. cot 2 t dt (csc 2 t 1) dt cot t t C dy
dx 29. dy
dx
dx 25. (a) Graph (b)
(b) The slope is always positive, so (a) and (c) can be ruled
out. 1 when x is positive since slope dy
dx and (c) don’t show this slope pattern. y 2x 1 2 x. Graphs (a) tan 1 negative, zero slope when x is zero and negative slope dy
dx
dx 1 4 C C 4 C Solution: y (2x tan x 2 x2 1) dx
x C
, 0 by [ 8, 8] dy
dx x 22 30. dy
dx
dx 2 y 1
x2 2/3 x 2/3 3x 1/3 dx
C Initial condition: y ( 1)
5 3( 1)1/3
5
3C
2C
y 3x 1/3 2 [ 4, 4] by [ 3, 3] dy
dx C 1 Initial condition: y (2)
0 22 2 C
02C
2C
Solution: y x 2 x 28. tan x Initial condition: y (b) The solution should have positive slope when x is dy
dx sec 2 x dx y 26. (a) Graph (b) 27. sec 2 x 5
C x dy
dx
dx (x 2 x) dx y x 1 x2
2 y x2
2 1
x 1
1
1
2 Solution: y 2
2
3
2 [ 4, 4] by [ 8, 4] C Initial condition: y (2)
2 C 1
2 1 dy
dx
dy
dx
dx C C y C
x2
2 31. 1
x [ 6, 6] by [ 4, 4] 1
2 9x 2
(9x 2
3x 3 4x
4x
2x 2 5
5) dx
5x C Initial condition: y ( 1) 0
0 3( 1)3 2( 1)2 5( 1) C
0
10 C
10 C
Solution: y 3x 3 2x 2 5x 10 Section 6.1
dy
dx 32. cos x dy
dx
dx (cos x y sin x cos x dy
dt 2e dy
dt
dt C y t ln 2 0 2
2 0 1 y 0 C dy
dx
dx y t 1 1
x cos
1) d sin sin 0 C Initial condition: y(e 3) 0
0 ln (e 3) C
03C
3C
Solution: y ln x 3 0 3 C2 C2 d 2y
dx
dx 2
2
dy
dx
dx 2
dy
dx 1
dx
x 1 C2 Solution: y
36. ln x dy
d ( cos 3 dy
dx C1 Initial condition: y (0) C
2e 0 C1 dy
d
d C 3 Solution: y C1 C1 C 1 34. cos 0 First derivative: Initial condition: y (ln 2)
2e 0 dt
t d cos 1 2e 0 sin Initial condition: y (0) t 2e sin d 2y
d
d2
dy
d sin x) dx Initial condition: y ( ) 1
1 sin
cos
C
11C
0C
Solution: y sin x cos x
33. d 2y
d2 35. sin x sin
2 3 6x
(2 6x) dx
3x 2 2x C1 Initial condition: y (0)
4
4 3(0)2 2(0) 4 C1 First derivative:
dy
dx
dx y x2 dy
dx 4x 02 1 03 4) dx C2 Initial condition: y (0)
1 3x 2 2x 3x 2 (2x
x3 C1 1 C2 4(0) C2 Solution: y x2 x3 4x x3 x2 4x 1 or y 1 4 235 236 37. Section 6.1
d 3y
1
dt 3
t3
d 3y
dt
t 3 dt
dt 3
d 2y
12
t
C1
dt 2
2 38.
4 1
(1) 2
2
1
C1
2 2
5
2 cos sin cos 0 dy
dt
dt 2
dy
11
t
dt
2 3 1 2 dy
dt 2 12
t
2 12
5
t
2
2
5
t C2
2 1
(1) 1
2 3 3 5
(1)
2 5
2 C1 C1 dy
d
d3
d 2y
d2 sin 0 1 1 0 C2 2 5
t
2 C2 cos 0 2(0) d 2y
d2 cos 0
1 2 cos cos 2 2 )d 2 sin 1 C3
1 2 sin 0 0 C3
dy
d cos dy
d
d y C3 C3 First derivative: 1
4 C2 sin ( sin 1 C3 52
t
4 1 Initial condition: y (0) 1 C2 C2 dy
d
d2
dy
cos
d C3
1
ln t
2 2 2) d cos 2 dy
11
First derivative:
t
dt
2
dy
11
5
dt
t
t dt
dt
2
2
1
52
y
ln t
t
C3
2
4 sin sin sin Second derivative: 1
52
ln 1
(1)
2
4
5
C3
4 cos ( cos 1 3 C2 Solution: y d 3y
d3 Initial condition: y (0) C2 Initial condition: y (1) C1 3 dt Initial condition: y (1)
3 C1
3 sin 0 Third derivative: 2 1
4 cos ) d 3
C1 Second derivative: 1 (sin Initial condition: y (3)(0) 2 1 cos 2 C1 0 sin dy
d
d4
d 3y
d3 Initial condition: y (1)
2 d 4y
d4 (cos 2 sin
2 sin 2 2) d 3 sin cos 2 3 Initial condition: y (0)
3 sin 0 3 1 4 cos 0 C4 3
03
3 C4 2(0) C4 C4
3 Solution: y
39. ds
v
dt
ds
dt
dt s 4.9t sin 9.8t
(9.8t
2 5t cos
5
5) dt
C Initial condition: s(0) 10
10 4.9(0)2 5(0) C
10 C
Solution: s 4.9t 2 5t 10 3 2 4 Section 6.1 40. ds
v sin t
dt
ds
dt
sin t dt
dt
1 s cos t 42. v C Initial condition: s (1)
1 0
1 0
1 cos t dt sin t C1 Initial condition: v(0)
1 C C
C sin 0 1 (1 cos t s cos t) C1 ds
dt v sin t (sin t ds
dt
dt 1 1) dt cos t t 1 32
1 cos 0 1 1 2 0 C2 C2 32 dt 32t C1 Initial condition: v(0) 20 C2 Solution: s
20 32(0) 20 C1 Velocity: C1 ds
dt ds
dt
dt 16t 1 C2 Initial condition: s (0) dv
a
dt
dv
dt
dt s 1 C1 Velocity: 1 or s v 0 cos t 1 cos Solution: s 41. dv
a
dt
dv
dt
dt t 2 43. v (32t
2 cos t 32t 20 20) dt 20t [ 2, 2] by [ 3, 3] C2 Initial condition: s (0)
0 16(0)2 + 20(0) 0 44.
0 0 C2
[ 2, 3] by [ 3, 3] Solution: s 16t 2 20t
45. d
(tan 1 x
dx C) 46. d
(sin 1 x
dx C) d
(sec 1 x
dx C) 47.
48. dy
x
dx
dy
dx
dx
x2
y
2 x2
1
x2 1
1
x d
( cos 1 x
dx 49. (a) 1
1 x2
1 C) 2
2
1
2 1
2
3
2 x2 1
1
x2 (x
x x 1 2 ) dx
x2
2 C Initial condition: y (1)
2 1 1
1 1
x C 2 C C C Solution: y x2
2 1
x 1
,x
2 0 237 238 Section 6.1 49. continued
(b) Again, y 52. (a) 1
x ( 1)2
1
2
( 1)
1
C
2 1
3
2 (b) (c) For x d
(x sin x) dx
dx [ f (x)] dx f (x) dx x 2e x [ g(x)] dx g(x) dx x sin x (e) C x2
2
dy
dx 0, g(x) dx (c) 1 C Solution: y d 2x
(x e ) dx
dx [ f (x) C. Initial condition: y ( 1)
1 f (x) dx (d) x2
2 1
x 3
,x
2
x2
2 d1
dx x
1
x2 0
C1 g(x)] dx (f) 1
.
x2
dy
x2
d1
For x 0,
C2
dx
2
dx x
1
x
x2
1
x
.
x2
dy
And for x 0,
is undefined.
dx [ f (x) g(x)] dx (g) [x (h) [g(x) 53. (a) (d) Let C1 be the value from part (b), and let C2 be the
value from part (a). Thus, C1 1
.
2 f (x)] dx d 2s
dt
dt 2
ds
kt
dt 1
1
7
2 1 y ( 2)
22
2 1
2
5
2 50.
r x 2
1
2 C2 (3x 3 1
and C2
2 3x 2 2 6x 03 12x 0 3(0)2 ( 2)2
1
2
( 2)
3
C1
2 ( k)(0) 88
C1 0
0 C dc
dx
dx c (3x 2 x3 6x 2 12x
15x Initial condition c(0)
400 03 400 6(0)2 0 when t kt 2
2 C2 kt 88 88t 0
88
k 15) dx
(c) C k 88 2
2k 400
15(0) C 88
k
88
88
k
3872
k s k
x3 0 0 12x C Solution: c(x) C2 88(0) ds
dt t
51. 88) dt Solution: s 3x 2 88 6x 2 15x 400 242
242
242
16 ft/sec 2 x 2e x x sin x 0 C2 (b)
x3 kt 88t k2
(0)
2 C Solution: r (x) 88 when t Initial condition: s 0 4 dx C1 ( kt
k2
t
2 s C 12(0) ds
dt ds
dt ds
dt
dt 7
.
2 x2
2 C1 Velocity: C1 C C1 2 12) dx Initial condition: r (0)
0 2 C2 Thus, C1
dr
dx
dx C2 g(x) dx k dt 88
(e) y (2) C x sin x g(x) dx Initial condition: C f (x) dx x dx 4] dx C x sin x x 2e x C g(x) dx f (x) dx x C x sin x f (x) dx
x 2e x x 3
and C2
2 x 2e x C
4x C Section 6.1 54. We first solve
s (0) d 2s
dt 2 k with the initial conditions 2 44 and s (0)
k ( k)(0) ds
dt kt ( kt
k2
t
2 44) dt 44t k2
(0)
2 0 kt C2 s0 44 44
, so it takes
k 0 when t 44
k
44
k V 45
45
968
45 k 21.5 It requires a constant deceleration of approximately 0 ds
dt 4 25
t
24
125t
48 0 5.2t 2.6t 2 2.6(0)2 s0 C2 43
h
75 10 when t h 5/2 0. 10 5/2 125t
48
125t
48 The height is given by h 105/2
105/2 2/5 125t
48 105/2 volume is given by
4 when t 0
V C2 C2 Position: s (t) v0t 2
(10)5/2 C
5
2
(10)5/2
5
2 5/2
2
h
(10)5/2
5
5 h 5.2 dt Initial condition: s
4 0 when t C1 ds
dt C2 122
hh
35
d4 3
h
dt 75
4 2 dh
h
25 dt
dh
h 3/2
dt
3/2 dh
h
dt
dt
2 5/2
h
C
5 h5/2 ds
dt
dt s C C1 C1 Velocity: 12
rh
3
dV
dt
1
h
6
25
24
25
dt
24
25
t
24 25
(0)
24 5.2 dt 5.2(0) a2
t
2 Initial condition: h 21.5 ft/sec 2. Initial condition: (v0 )(0) 57. We use the method of Example 7. 45 968
k d 2s
dt
dt 2
ds
5.2t
dt s0 44t 44 2k 0 C2 C2 s 55. v0t a2
(0)
2 44
seconds to stop, and we require:
k k 44 2 v0 v0) dt Position: s ds
dt at (at s0 C2 Now, ds
dt 0 k2
t
2 C1 Initial condition: s (0) C2 44(0) Position: s v0 C1 ds
dt
dt
a2
s
t
2 44 Initial condition: s (0)
0 (a)(0) Velocity: ds
dt
dt s C1 v0: C1 v0 44 C1 Velocity: s0, and v(0) a dt v0 Initial condition: s (0) 44 a, s(0) Initial condition: s (0) C1 44 d 2s
dt 2 ds
dt
dt 2
ds
at
dt 0. 2 ds
dt
dt 2
ds
kt
dt 56. Solving 2.6t 2 4
4
, so the positive
2.6 Solving s (t) 0, we have t 2 solution is t 1.240 sec. They took about 1.240 sec to fall. 43
h
75 4
75 125t
48 105/2 6/5 . 2/5 and the 239 240 Section 6.1 58. (a) y 500e 0.0475t (b) 1000
2 63. Use differential equation graphing mode.
For reference, the equations of the solution curves are as
follows.
(1, 1): y e (x 1)/2
( 1, 2): y 2e (x 1)/2
(0, 2): y
2e x /2
( 2, 1): y
e (x 4)/2 500e 0.0475t
e 2 0.0475t 2 2 ln 2
t 0.0475t 2 ln 2
0.0475 14.6 It will take approximately 14.6 years.
59. (a) y 1200e 0.0625t (b) 3600 [ 6, 6] by [ 4, 4] 1200e 0.0625t 3
ln 3 The concavity of each solution curve indicates the sign of
y. e 0.0625t
0.0625t t ln 3
0.0625 64. Use differential equation graphing mode.
For reference, the equations of the solution curves are as
follows.
(0, 1): y e x
(0, 2): y 2e x
(0, 1): y
ex 17.6 It will take approximately 17.6 years.
x x 2 cos x dx 60. (a) t 2 cos t dt C 0
0 (b) We require t 2 cos t dt C 0 1, so C
x The required antiderivative is 1. t 2 cos t dt 1. 0
x xe x dx 61. (a) te t dt C 0
0 (b) We require te t dt C 1, so C 0 x The required antiderivative is 65. Use differential equation graphing mode.
For reference, the equations of the solution curves are as
follows.
(0, 1): y
3e 2x 4
(0, 4): y 4
(0, 5): y e 2x 4 1. te t dt 1. 0 62. (a) d 2y
dx
dx 2
dy
3x 2
dx 6x dx
C1 Initial condition (horizontal tangent): y (0)
0 3(0)2 0 0 C1 C1 First derivative:
dy
dx
dx y x3 [ 3, 3] by [ 4, 10] dy
dx 3x 2 dx
C2 (0)3 1 The concavity of each solution curve indicates the sign of
y. 3x 2 Initial condition (contains (0, 1)): y (0)
1 1 C2 Solution: y C2 x3 1 (b) Only one function satisfies the differential equation on
( [ 4, 4] by [ 3, 3] The concavity of each solution curve indicates the sign of
y. , ) and the initial conditions. 241 Section 6.2
66. Use differential equation graphing mode.
For reference, the equations of the solution curves are as Section 4.2, y1 and y2 must differ by a constant. We find follows. that constant by evaluating the two functions at x 0. 1 (0, 1): y x 1
2 (0, 2): y
(0, 2
. By Corollary 3 to the Mean Value Theorem of
3 y2 3. y1 2x 1
1
x1 1): y (0, 0): y 0
[ 3, 3] by [ 2, 8] 4. [ 2.35, 2.35] by [ 1.55, 1.55] [ 10, 10] by [ 30, 30] The concavity of each solution curve indicates the sign of
y.
67. (a)
(b) d
(ln x
dx d
[ln ( x)
dx for x 1
for x
x C) 0 Exploration 2 Two Routes to the Integral 1d
( x)
x dx C] 5. The derivative with respect to x of each function graphed in
part (4) is equal to 1 x 2. 1
( 1)
x 1
x 0 1 3x 2 1. C ln x C, which is a solution to the differential equation, as we showed in part (a).
For x 0, ln x C ln ( x) C, which is a 3x 2 2 1. 0, we have y ln ( x) 1 dx x3 0, we have y 0 5 x 1 dx (x 1 part (b). Thus, dy
dx s Section 6.2 Integration by Substitution 1 1 15
(0)
5 1)1/2 dx 1 3. dy
dx 2
(x
3 dy
dx 3x 5. dy
dx 4(x 3 2x 2 3)3(3x 2 4x) (pp. 315–323)
6. 1 x 2 2x dx u du 2 sin (4x 5) cos (4x 5) 4 5) cos (4x 5) 7. 2 3/2
u
C
3
2
(1 x 2)3/2
3 2. Their derivatives are equal: dy
dx 8 sin (4x Exploration 1 Supporting Indefinite Integrals
Graphically
1. dy1 dy2 dx dx 1 x 2 2x. dy
dx 1
cos x 8. C 0 42
3 23
(x
1)3/2 so
3
42
.
3 32
5 3x 4. 1
for all x except 0.
x 1)3/2 2 3/2
2 3/2
(4)
(0)
3
3
2
16
(8)
3
3 C1, which is a solution to the differential equation, as we showed 15
(2)
5 0 5 C2, which is a ln x 23
(x
3 2 solution to the differential equation, as we showed in
part (a). For x 2 3/2
u
3 u du 1 dx 15
x
5 x 4 dx 2.
(d) For x x 3 2 Quick Review 6.2 1
for all x except 0.
x d
ln x
dx 2 1 solution to the differential equation, as we showed in
part (b). Thus, 2 3/2
u
3 u du
0 2. 3x 0, ln x 1 dx 1 1 (c) For x 2 x3 dy
dx 1
sin x sin x
cos x tan x
cot x 1)3/2 5
1 242 9. Section 6.2
dy
dx 1
(sec x tan x
sec x tan x
sec x tan x sec2 x
sec x tan x
sec x(tan x sec x)
sec x tan x 4. u sec2 x) du dy
dx dx d
Check: [(7x
dx 1
( csc x cot x
csc x cot x
csc x cot x csc2 x
csc x cot x
csc x(cot x csc x)
csc x cot x csc2 x) 3 du dx d
Check:
dx 1
sin u du
3
1
cos u C
3
1
cos 3x C
3
1
cos 3x
3 C 1
( sin 3x)(3)
3 6. u du 1 1 r 3 9 1
3 3u 1
cos u du
4
1
sin u C
4
1
sin (2x 2) C
4
1
d1
Check:
sin (2x 2) C
cos (2x 2)(4x)
4
dx 4 Check:
x cos (2x2) du d
(6
dx 1 C
r3 r3 C
C) 21
r3 1 du 2 dx 1
du
2 Check:
2x d1
sec 2x
dx 2 1 cos t
2 1
t
sin dt
2
2
t
2 du sin dt
2
t2
t
1 cos
sin dt
2
2 du dx sec 2x tan 2x dx 1 6
9r 2 7. u 1
sec u tan u du
2
1
sec u C
2
1
sec 2x C
2
1
C
sec 2x tan 2x
2 2 sec 2x tan x2 u
1/2 61 2x 1
9 du 3(2)u1/2 x cos (2x 2) dx 3. u 1
3 x2
3 r3 1
du r 2 dr
3
9r2 dr 1
du
4 x dx 1 1
31 3r 2 dr du 4x dx 2)3 3u
9 sin 3x 2x 2 2. u 28(7x C 9u 2 9
du
3
9 u2 1
1
du
3 u2 1
1
tan 1 u C
3
1
x
tan 1
C
3
3
d1
x
Check:
tan 1
C
dx 3
3 3 dx 2)3(7) 4(7x 2)4 (7x dx dx 3x sin 3x dx C] C 1
dx
3 du Section 6.2 Exercises 1
du
3 2)4 u4 x
3 5. u x2 du 1
28u3 du
7 2)3 dx 28(7x csc x 1. u 2 7 dx 1
du
7 sec x
10. 7x 2 u2 du
23
u
3
2
1
3 t3 d2
1 cos
dx 3
2
t2
t1
2 1 cos
sin
2
22
2
t
t
1 cos
sin
2
2 Check: C C
cos t3
2 C r3 ( 3r 2) Section 6.2
y4 8. u 4y 2 14. Let u du (4y 3 8y) dy du 3 tan x du 1 sec2 x dx 2y) dy 4( y /4 1
du
4 (y3 8( y 4 2y) dy
4y 2 /4 1)2(y 3 2y) dy 8 1
4 d24
(y
dx 3 4y 2 2( y 4 4y 2 1)2(4y 3 8( y 4 4y 2 1)2( y 3 1)3 13
u
3 C
4y 2 1)3 C 1
3 8y)
15. Let u
du 9. Let u 1 du x dx 1 1 x du 2 sec (x 2) dx C
2) 2 3/2
u
C
3
2
3/2
(cot x)
3 C
17. Let u tan x du 2 tan x sec x dx 1/2 u du 2 3/2
u
C
3
2
(tan x)3/2
3 du C 2 tan 2 2 d sec u tan u du
sec u ln x 1
du
dx
x
6
dx
e x ln x ln 6
1 du
u ln 6 ln u ln (ln 6)
1 2 C tan 2 u7 du
18
u
8
1
x
tan8
4
2 2 C sec
13. Let u u6 du x
2
1 2x
du
sec dx
2
2
x
7x
tan
sec2 dx
2
2 d sec 1
dx
x
17
u
C
7
17
(ln x)
7 18. Let u
12. Let u ln x ln6 x
dx
x sec x dx
2 u1/2 du sec u du tan (x C C cot x 2 tan u du 1
cos u du
3
1
sin u C
3
1
sin (3z 4)
3 cot x csc2 x dx 2 2
3 csc2 x dx du dx 11. Let u 4) dz C
16. Let u x 1
3 4 C 1 1
( 1)3
3 dz cos (3z u 10. Let u 3z du
u2 x)2 (1 1 3 dz 1
du
3 dx 1 1
(1)
3 C 2y) u2 du 1 u2 du 23
u
3
24
(y
3 Check: 1 tan2 x sec2 x dx C
C C 243 244 Section 6.2
s4/3 19. Let u
du
3
du
4 3 /4 cot x dx
/4 /4 Let u s1/3 ds dx
sin2 3x Let u 23. 4 1/3
s ds
3 s1/3 cos (s4/3 20. 3 /4 8 3
cos u du
4
3
sin u C
4
3
sin(s4/3 8)
4 8) ds du cos x
dx
sin x sin x
cos x dx 3 /4 C
/4 x 3 /4 cos x
dx
sin x csc2 3x dx 1 /4 u x x ln u 3 /4 x 3x du /4
3 /4 du
1
du
3 ln sin x 3 dx /4 dx 24. Let u
du du cos (2t
sin (2t 1
du sin (2t
2
sin (2t 1)
dt
cos2 (2t 1) 1) 2 2 1) dt
1
u 2 du
2 x2 25. Let u ln 2 1) 1) cos t dt 1 C
C du 2x dx
1
du
2 x dx x dx
2 1 1
2 10
2 1
du
u 1
ln u
2 6u
2 2 1.504 C 1x 6u 0 1
du
u9 ln 9 sin t 6 cos t
dt
(2 sin t)2 2
2 ln u 3 du 9 2 1)(2) dt 1
2 cos (2t
1
sec (2t
2 2 ln 2 dx
x 0 11
u
2 22. Let u x 2 dx 7 21. Let u 2 ln 1
csc2 u du
3
1
cot u C
3
1
cot (3x) C
3 csc2 3x dx du 1 6
sin t 10
2 1
(ln 10
2 C
C ln 2) 1
ln 5
2 0.805 Section 6.2 5 26.
0 40
25 5 40 dx
x 2 25 30. csc x dx x2
5 0 25 2
25 5 40
25 1 dx dx x2
5 Let u 5 du csc x cot x 0 cot x
csc2 x dx 1
du
u csc x dx ln u dx
5 40 dx
x 2 25 cot x
dx
cot x 1 1
dx
5 du csc x du 0 csc x
csc x csc2 x csc x cot x
dx
csc x cot x x
5 Let u csc x 8
(5)
5 1
0 1
u2 1 C ln csc x du 1 31. Let u 8 arctan u y cot x C 1 0 du 8(arctan 1) dy 3 8
27. dx
cot 3x 4 4 y 2 0 1 4 2 3/2
u
3
1
2 3/2
2 3/2
(4)
(1)
3
3
2
2
14
(8)
3
3
3 sin 3x
dx
cos 3x Let u cos 3x du 3 sin 3x dx 1
du
3
dx
cot 3x sin 3x dx
1
3
1
3
1
3 32. Let u 1
du
u ln u ln cos 3x du 5x r dr 1 1
ln sec 3x
3 0 1
u1/2 du
21
0
1 2 3/2
u
23
1
1
1
(0)
(1)
3
3 r 2 dr r1 C.) 0 8 5 dx 1
du
5 dx
dx 5x 8 29. sec x dx Let u
du 2r dr 1
du
2 C r2 1 du
C (An equivalent expression is
28. Let u u1/2 du 1 dy 33. Let u sec2 x dx 0 0 tan x sec2 x dx u du /4 1 12
u
2 sec x tan x
dx
sec x tan x
sec2 x sec x tan x
dx
sec x tan x sec x sec x 1
du
u 0
1 1
(0)
2 1
( 1)2
2 1
2 tan x sec x tan x sec x dx tan x du 1
u 1/2 du
5
1
2u1/2 C
5
2
5x 8 C
5 sec2 x dx
ln u 34. Let u
C ln sec x tan x C 1
3 du
1
du
21
1 (4 4 r2 2r dr
r dr
5r
dr
r 2)2 5
2 5 u
5 2 du 0 245 246 Section 6.2 35. Let u 39. 3 1/2
d
2 du
2
du
3
1 1/2 dy
( y 5)(x 2)
dx
dy
(x 2)dx
y5 Integrate both sides.
d 10
3/2 2 (1 0 3/2 1 ) dy
2
(10)
3 d y 2 2 u du 2 let u y 1 20 1
32 5 du
1 20
3 2) dx On the left, 1 20 1
u
3 (x 5 1
2 dy 1
du
u 10
3 12
x
2
12
x
2
12
x
2 ln u 2x C 2x C 3 sin x 2x C 5
5 e (1/2)x y du 4 ln y
y 36. Let u 5 e Ce(1/2)x 3 cos x dx 1
du
3 cos x dx
cos x dx 1
3 1/2 u du 0 2 2x eC or C We now let C 4 2 2x C e C, depending on whether (5t 4 du
1 t5 4 2t
2) dt 2t (5t 4 u1/2 du 2) dt 2 2x C e (1/2)x 2 2x value C 3 2 3/2
u
3 2
3 27 2 3 40. dy
dx dy cos 2 sin 2 d 5.
y x dx y cos2 y Integrate both sides.
dy x dx y cos2 sin 2 d
3 2 2x y cos2 x 2 sin 2 d /6 0 gives a solution to the original differential Ce (1/2)x y cos 2 1
du
2 5 equation), we may write the solution as 0 2 3/2
(3)
3 0 C e (1/2)x Since C represents an arbitrary constant (note that even the
0 du 5 3 0 38. Let u 5) is positive or negative. Then y t5 37. Let u (y
y 3 sin x 4 y On the left, let u
1
2
1
2 1/2 u 3 du 1 11
42
1
(3)
4 1
u2
2
2 1
3
4 y 1 1/2
y
dy
2 du
1/2 2 du 1/2 y dy 1 12
x
C
2
12
2 sec2 u du
x
C
2
12
2 tan u
x
C
2
12
2 tan y
x
C
2
12
tan y
x
C
4 2 du
cos2 u (Note: technically, C is now C
y
y tan
tan 2
1x 4 1x C
2 2 4 C C
. But C’s are generic.)
2 247 Section 6.2 41. dy
dx
dy
dx
dy
ey (cos x)(e y esin x) dy
dx
dy cos x esin x dx Integrate both sides. (cos x)e y sin x 44. dy On the right, let u ln x
dx
x 4 y cos x e sin x dx du y ln x
x
ln x
4
dx
x y Integrate both sides.
dy
ey 4 On the right, let u sin x du
2y1/2 4 u du 2y1/2 cos x dx ln x 1
dx
x 4 u2 1
2 e y e u du e y e u e y e sin x C y (ln x) C e sin x C y [(ln x)2 C]2 y e 1/2 C y 1/2 C]2 1 (1 2 1 C 0 e sin x) x y e xe y e C [(ln e)2 y (e) ) e ln (C dy
42.
dx
dy
dx
dy
ey C.) 2 sin x ln (C y 2(ln x) 2y (Note: technically C is now C C
2 C) y (ln x)4
Note: Absolute value signs are not needed because the
original problem involved ln x, so we know that x 0. e x dx 45. (a) Let u x 1 Integrate both sides.
dy
ey
y e e y e dx
x ln (e C) dy
43.
2xy 2
dx
dy
2x dx
y2
dy
2x dx
y2 y 1 y x2
x y (1) C C 3 y dy1 x x dx 1 and antiderivatives of dy2 x dx 1. x 1, so both are 1. x 0 1 2 y1 0.25 4 0 1.219 2.797 y2 4.667
y2 y1 1
2 x (c) Using NINT to find the values of y1 and y2, we have:
C
1 C C C 1 1 C (b) By Part 1 of the Fundamental Theorem of Calculus, 1
2 u1/2 du 1 dx 2 3/2
u
C
3
2
(x 1)3/2
3
d2
Alternatively,
(x 1)3/2
dx 3 C
x dx
x x e dy
y du e x dx 3 4.667 3.448
4.667 1.869
4.667 2
3 C 4 (d) C y1 y2 x x x 1 dx 0
x x 1 dx x 1 dx 0
3 x x
0 x 1 dx 3
3 1 dx 3 4 4.667 6.787
0 2.120 4.667 4.667 248 46. (a) Section 6.2
d
[F(x)
dx x4 47. Let u C] should equal f (x). 1 (b) The slope field should help you visualize the solution
curve y F(x). 10 x 3 dx (a) x4 0 9 x (c) The graphs of y1 F(x) and y2 4x 3 dx. 9, du f (t) dt should
0 9 1
2
1
2 1 1/2
1 1/2
u
du
u
4
2
1
10
9
2
3
10
0.081
2 10
9 differ only by a vertical shift C.
x3 (b) (d) A table of values for y1 y2 should show that
y1 y2 C for any value of x in the appropriate
domain. x 4 9 1 1/2
u
2 (e) The graph of f should be the same as the graph of
NDER of F(x). x 0 x3
x 4 9 dx u1/2
x2 1 48. Let u C 9 1
2 x4 9 10 1
2 1
2 2x dx. x4 1
2 1, du 1 1/2
u
du
2 dx
1 C 1
2
1 x2 (f) First, we need to find F(x). Let u
x2 1 1/2
u
du
4 dx 10 3
2 C
1
0 1 cos 3x, du (1 9
0.081 3 sin 3x dx. cos 3x) sin 3x dx /3 x2 Therefore, we may let F(x) (a) 1. 2 /6 a) d
(
dx x2 1 1 C) 12
(2)
6 (2x) 2 x2
x
x2 1 1
u du
3 1 f (x)
1 (b) (1 12
(1)
6 12
u
6 cos 3x)2 /3 (1 cos 3x) sin 3x dx /6 c) 1
(1
6
12
(2)
6 49. We show that f (x)
f (x) [ 4, 4] by [ 3, 3] f (x) d) x 1 2 y1 1.000 1.414 2.236 3.162 4.123 y2
y1 0 0.000 0.414 1.236 2.162 3.123 1 1 1 y2 3 1 4 1
f (3) e) [ 4, 4] by [ 3, 3] tan x and f (3) cos 3
5.
cos x
cos 3
d
ln
5
cos x
dx
d
(ln cos 3 ln cos x
5)
dx
d
ln cos x
dx
1
( sin x) tan x
cos x
cos 3
ln
5 (ln 1) 5 5
cos 3 ln 1 1
2 C 1
(1
6
[ 4, 4] by [ 3, 3] 2 1
u du
3 cos 3x) sin 3x dx b) 12
u
6 C
/3 cos 3x)2 /6 12
(1)
6 5, where 1
2 Section 6.3
50. (a) u 2 csc2 2 d cot 2 , du 1
u du
2
1 u2
C
2
2
u2
C
4
1
cot2 2
C
4 csc2 2 cot 2 d F1( ) csc 2 , du 2 csc 2 cot 2 d
1
u du
2
1 u2
C
2
2
u2
C
4
1
csc2 2
C
4 F2 ( ) d
(sin2 x C) 2 sin x cos x
dx
d
( cos2 x C) ( 2 cos x)( sin x)
dx
1
1
d
cos 2x C
sin 2x (2)
2
dx 2 s Section 6.3 Integration by Parts
(pp. 323–329)
Exploration 1
Integrals
1. u Evaluating and Checking
dx
and dv
x ln x ⇒ du 2 csc 2 cot 2
(d) F1( ) F2( ) b 51. (a) u sin x, du 1
(csc2 2
cot2 2 )
4
1 1 cos2 2
1 sin2 2
sin2 2
4
4 sin2 2 2. cos x dx cos x, du
2 sin x cos x dx dx x ln x
1
4 d
(x ln x
dx x) x
ln x 3. The slope field of
2 2u du u C 2 sin x C dy
dx ( 2u) du
C cos2 x C C
x 1
x 1 ln x ln x shows the direction of the curve as it is graphed from left to right across the window. sin x dx u2 x. Thus, v du x ln x b dx ⇒ v u dv
uv 1
csc2 2
4 2 sin x cos x dx
(b) u ln x dx b 1
cot2 2
4 2 sin x cos x 2 sin x cos x 1
cot 2 ( 2 csc2 2 ) csc2 2 cot 2
2
1
csc 2 ( 2 csc 2 cot 2 )
2 (c) F1 ( ) sin 2x dx sin 2x 1
csc2 2
4 F2( ) 2 dx 1
sin u du
2
1
cos u C
2
1
cos 2x C
2 (d) csc2 2 cot 2 du 2x, du 2 sin x cos x dx 1
cot2 2
4 (b) u (c) u 249 [0, 6] by [ 2, 5] 250 Section 6.3 4. The graph of y2 x ln x x appears to be a vertical shift of the graph of y1 1 ln t dt (down 1 unit). Thus, y2 appears to be an antiderivative of ln x which
supports x ln x x dy
dx x sin x dy 9. x (x sin x)dx Integrate both sides. C as the set of all antiderivatives of dy ln x. (x
12
x
2 y
y (0) sin x) dx
cos x 10. [0, 6] by [ 2, 5] dy
dx (x 3)(cos 2x)(2)
2x 3 cos 2x 2. dy
dx 1
1 (2x)2
2
1 4x 2 dy
dx 5. y tan tan y 1)(2e 2x) 2
x du dv 1 3) 2. Let u
du cos x dx
sin x
sin x C
C) (cos x)( 1) cos x x2 1 2x dx 1 cos e 2x dx Integrate both sides. C sin x
2 x sin x dx 1 2[ x cos x
(x 2 cos 0 (2x)( sin x) sin x] 2)sin x d
Check: [2x cos x
dx cos x e 2x x 2 sin x 2x cos x
1 ( 1) cos x dx v x 2 sin x (x 2 C C
2)sin x (2 cos x)(1)
(sin x)(2x) 2 x 2 cos x e 2x dx dv Using the result from Exercise 1, 0 1 2x
e
2 x cos x x 2 cos x dx 1 y cos x 2 3x 1 dy v dx sin x dx ( x)( sin x) (x sin x dx dy 1x
e sin x
2 Section 6.3 Exercises x cos x cos 1 (x 1)
x1
cos y 1 dy
dx 1x
e sin x
2 1
2
1x
e cos x
2 cos x) e x (sin x d
Check: ( x cos x
dx 0 8. sin x) x sin x 1 7. cos x) x sin x dx 1
tan y
3 y
cos y
x 3 1) 3x x
6. cos x 1
1 3 d1x
e (sin x
dx 2
1x
e (cos x
2
1x
e cos x
2 1. Let u ln (3x 3x 1
3e 2x
2e 2x ln (3x
3x 1 dy
3.
dx 4. (sin 2x)(3x 2) 3x 2 sin 2x 3 (e 2 x) 2 e x sin x Quick Review 6.3
1. 12
x
2 C
C y 1 C (x 2 C]
2)(cos x) Section 6.3
3. Let u ln y dv 1
dy
y du 12
y
2 6. Let u 12
y
2 v
12
y ln y
2
12
y ln y
2
12
y ln y
2 y ln y dy y dy 1 sin dv 1 du 1
dy
y sin d d v 2 1
1 sin 1 1 2 1 Let w 1
y dy
2
12
y
C
4 2d
1 1 1 d 1
1 sin 2 sin
sin sin 12
d12
Check:
y ln y
y
C
4
dy 2
12 1
1
y
(ln y)(y)
y
2
y
2 d 2 1 dw 1 u 1 dy y tan 1 dy v
y y
y 1 y2 du 2y dy
1 y Check: 5. Let u
du dy y2 tan 1 y x dv dx v x sec2 x dx 1 y 1 y 1 y 1
2
1
2
1
2 1
ln (1 y 2)
2
1
1
(2y)
2 1 y2 d
y tan 1 y
dy y
1 y2 1 y tan y tan y t2 7. Let u y tan y tan [ 1, 1] by [ 0.5, 2] dy y2 1 dw dv y2 1 y dy Let w y 1 du
tan 1 tan 1
dw
w ln (1 y 2) Let u C tan C x tan x t cos t
(2 cos t dt
sin t 2 t cos t dt
2t sin t 2 sin t dt 2t sin t 2 cos t t 2) cos t 2t sin t C dv
v C csc2 t dt
cot t tan x dx x tan x v 2 tan x 2 (cos t)(t) dt dv t 2 cos t sec2 x dx t 2 cos t dt du y cos t t t 2 cos t
1 sin t dt v t 2 sin t dt ln w C dv 2t dt sin x
dx
cos x x tan x ln cos x C [ 3, 3] by [0, 8] 8. Let u
du t
dt t csc2 t dt
[ 1.5, 1.5] by [ 1, 4] t cot t cot t dt t cot t cos t
dt
sin t t cot t [0, 3] by [ 4, 4] 1 1 2 y ln y
4. Let 251 ln sin t C dw w
1/2 C 1 2 w C 252 Section 6.3 9. Let u ln x
1
dx
x du x 3 dx dv 14
x
4 v
14
x ln x
4
14
x ln x
4
14
x ln x
4 x 3 ln x dx f (x) and its
derivatives f (x) and its
derivatives 1 41
x
dx
4
x
13
x dx
4
14
x
C
16 3x2 (–) 6x x and g (x) (+) 6 (–) e. x4 x 3e e–x (+) 4x 3 (–)
(+) e–x 24x (–) –e–x (+) e–x 24 x 4e x x xe
(x 4 4x e
4x 3 x2 11. Let u
du
(x 2 3 x 2 12x e 12x 2 24)e dv 24e
C 5x)e x x C du e y sin y dy u 2x du
(x 2
(x 2 5 5x)e x (x 2 5x)e x (x 2 7x e x(2x x e 5)e x
5)e x (2x
C e du Let u e y cos y
1y
e (sin y
2 y e C e du 2e x dx
2e x v
e y cos y sin y e y sin y sin ye y dy e y sin y
cos y) C dv
y e y e cos y dy
y 2e
y dy v sin y e y e cos y dy cos y dy cos y dy e ysin y e ycos y dy 5) dx (2x 7)ex e dx v 2 dx 5x)e x x dv cos y dy 5) dx
14. Let u Let e dy e y sin y dy
e x(2x cos y e y dy
dv y 2 e y sin y dy ex v (x 2 x x e x dx 5) dx 5x)e x dx 24xe 24x 5x (2x x cos y e y cos y ey C sin y dy v e y sin y dy
Let u 3 2x
e
8 dv e y dy du dx 4 ey 13. Let u –e–x 0 dx 1 3 2x
3 2 2x
3
xe
xe
xe 2x
2
4
4
3
2
x
3x
3x
3
e 2x C
2
4
4
8 –e–x 12x 2 2x . 1 –2x
e
2
1 –2x
e
4
1 –2x
e
8
1 –2x
e
16 0 g(x) and its
integrals 2x e–2x (+) x e g(x) and its
integrals x3 4 10. Use tabular integration with f (x) x 3 and g(x) 12. Use tabular integration with f (x) sin y
y dv
y y e
e sin y dy v dy cos y sin y
y dy sin y 1y
e (sin y
2 e ycos y
e y e cos y cos y) C y cos y dy Section 6.3
x 2 and g(x) 15. Use tabular integration with f (x)
f (x) and its
derivatives sin 2x. g(x) and its
integrals x2 (+) 2x (–) 2 (+) sin 2x 1
cos 2x
2
1
sin 2x
4
1
cos 2x
8 0 12
x cos 2x
2 x 2 sin 2x dx 1
x sin 2x
2 2x 2
cos 2x
4 1
/2 1 x 2 sin 2x dx 0 1
cos 2x
4 x
sin 2x
2 2x 2
cos 2x
4 C C x
/2
sin 2x
2
0 2 1 2 2 ( 1) 4
2 8 1
2 0 0.734 Check: NINT x 2 sin 2x, x, 0, 0.734 2 x 3 and g(x) 16. Use tabular integration with f (x)
f (x) and its
derivatives 1
(1)
4 0 cos 2x. g(x) and its
integrals x3 (+) 3x2 (–) 6x (+) 6 (–) cos 2x 1
sin 2x
2
1
cos 2x
4
1
sin 2x
8
1
cos 2x
16 0 x 3 cos 2x dx 13
x sin 2x
2
x3
2 /2 3x
sin 2x
4
x3
2 x 3cos 2x dx 0 0
3
4 32
x cos 2x
4
3x 2
4 3
( 1)
8 32
16 Check: NINT x 3cos 2x, x, 0, 3
cos 2x
8 3x 2
4 3x
sin 2x
4
32
16 3
x sin 2x
4 0 1.101 2 1.101 3
cos 2x
8 C 3
/2
cos 2x
8
0
3
(1)
8 253 254 Section 6.3
e 2x 17. Let u dv 2e 2x dx du e 2x cos 3x dx Let 1
sin 3x
3
1
1
(e 2x) sin 3x
sin 3x (2e2x dx)
3
3
1 2x
2 2x
e sin 3x
e sin 3x dx
3
3 v e2x u
du cos 3x dx dv 2e 2x dx sin 3x dx
1
cos 3x
3
2 2x
(e )
3 v 1 2x
1
1
e sin 3x
cos 3x
cos 3x (2e 2x dx)
3
3
3
1 2x
4 2x
e (3 sin 3x 2cos 3x)
e cos 3x dx
9
9
13 2x
1 2x
e cos 3x dx
e (3 sin 3x 2 cos 3x)
9
9
1 2x
e 2x cos 3x dx
e (3 sin 3x 2 cos 3x)
13
3
1 2x
3
e 2x cos 3x dx
e (3 sin 3x 2 cos 3x)
13
2
2 e 2x cos 3x dx 16
[e (3 sin 9
13
16
[e (2 cos 9
13 2 cos 9) e 4 3 sin 9) e 4 (3 sin ( 6) (2 cos 6 2 cos ( 6)] 3 sin 6)] 18.186
Check: NINT(e 2x cos 3x, x,
18. Let u e du
2x e 2x 2, 3) 18.186 dv
2x 2e sin 2x dx (e v dx
2x sin 2x dx
1
cos 2x
2 1
cos 2x
2 ) 1
cos 2x ( 2e 2x dx)
2 1 2x
e cos 2x
2 Let u e du 2x dv
2x 2e 2x e cos 2x dx cos 2x dx
1
sin 2x
2 v e sin 2x dx 2x 2e 2x 2 e 1 2x
e cos 2x
2 sin 2x) 1 2x
e (cos 2x
2 sin 2x dx
e sin 2x dx
2x (e 1 2x
e (cos 2x
2 2x e 1
2 (cos 2x e sin 2x)
sin 2x) 4 (cos 2x sin 2x) 4 e sin 2x dx C 4 3 sin 4) e6
[cos ( 6)
4 (cos 4 sin 4) e6
(cos 6
4 4 e 2 (cos 4 4 125.028
2x 2x C 2x e 3 Check: NINT(e 1
sin 2x ( 2e 2x dx)
2 ) sin 2x 2x 4 sin 2x dx 2x sin 2x, x, 3, 2) 125.028 sin ( 6)] sin 6) 255 Section 6.3
23. Let x 2e4x dx 19. y x2 Let u
du
y 1
4 1 2 4x
xe
4 y
y dx y v
1
1
(x) e4x
2
4
1 4x
1 4x
xe
e
8
32
1 4x
e
C
32 Let u 2 x cos x du 2 v 1 x dx 2 1 sec 2 2 Let w
2 y ) sec 2 2 y
22. y
Let
y
y 2 sec x2 (x 2 2 ln sec x 2x v x 1)e 2 dx 1)e
x e v 2 x 1)e x (2x 1)e x (x 2 1 dx dx (x 2 d dx x e x 1)e x (2x 1)e x (x 2 1 x dx x e (2x
dv x x e dx. dx 1 1)e x 1)e dv x x x 1 x 1)e du 1 3x 4)e x x 2e
2e x dx
C C [ 3, 3] by [ 3, 3] dv sec
v sec tan d The graph shows that the two curves intersect at x
where k 1.050. The area we seek is
k (x 2 x 1)e k x x 2 dx 0 d
tan 4 1) dx (2x Let u d sec x (x 2 12
2 x sin x dx 0 Let u C Note: In the last step, we used the result of Exercise 29 in
Section 6.2. 2 x sin x dx 24. We begin by evaluating (x 2 1
w 1/2 dw
4
1 1/2
w
C
2
1
2
1C
2 sec tan
u
du d sec 0 3 2d 1 sec ( 1) 0 d 12
2
2d 1 sec 2 0 x sin x dx (x 2 12
2
1
4 1 2 y 0 2 2 (c) 1, so no absolute value is needed 1, dw 1 0 0(1) 2 dv 1 (sec 0 x sin x dx sin x 2 (1) in the expression for du. y x sin x dx du Note that we are told y sin x 2 (b) C d
1 du x sin x dx
0 ( 1) 1 4x
e
4
1 4x
e dx
4 dv 1 sec C x cos x 1
13
du
dx
v
x
x
3
1
13 1
(ln x) x 3
x
dx
3
3
x
13
12
x ln x
x dx
3
3
13
13
x ln x
x
C
3
9 sec sin x 3 ln x 21. y cos x cos x dx e4x dx x ln x dx Let u y x sin x dx
0 2 20. y y (a)
dv 1 2 4x
xe
4
1 2 4x
xe
4
x2
x
4
8 y x cos x
x cos x 1
xe4xdx
2 x du v dx sin x dx 1 4x
e (2x dx)
4 (x 2) e4x Let u dv x sin x dx 1 4x
e
4 v x du e4xdx dv 2x dx u 0 (x 2
( 2.888
0.726 3x 4)e
4) xk 0 (0.386 13k
x
30 0) k, 256 Section 6.3 t 25. First, we evaluate e (c) Using the result from part (b): cos t dt. x3 Let u
Let u e t du
e t t e dt e sin t du dv
t e t dt t e cos t dt or [x n e t(sin t cos t) C e tcos t dt 1t
e (sin t
2 cos t) C 0 t ( 1)n n(n 1)x n 1 1)!x ( 1)n (n d nx
xe
dx n C 2 ( 1)n(n!)]e x C 2e tcos t for integration.
Alternately, show that the derivative of the answer to 1 part (d) is x ne x: 2 1
2 Average value 2e tcos t dt d
xn
dx 0
2 [x n 1
e t(sin t
2 2
cos t)
0 1
[e 2 ( 1)
2
1 dx …
[xn … ex
1)e x … C 2 ( 1)nn!]e x 1 1)x n n(n 2 ( 1)nn!] dv 2x dx v x2 ex 2x 1)x n n(n 2 ( 1)n 1(n!)x ( 1)nn!]e x
1)x n 1 n(n 2)x n 1)(n 2 3 ( 1)n 1n!]e x sin e x dx x dx Let u ex du w
dw w sin w dw 1)e x
2)e x dx x. Then dw , so dx 2 x dw 2x 2x e x dx
2(x 1 x ne x C x2 (x 2 nx n n(n (b) Using the result from part (a): x 2e x 1)x n n(n C [nx n 27. Let w x 2 e x dx 2 ( 1)nn! e x (n!)x ( 1)n 1(n!)x nx n e x dx (x du 1 ( 1)n 1(n!)x d
dx ex v xe x u 1 ex [x n e0( 1)] e x dx dv xe x nx n 1)x n n(n ( 1)n … 0.159 x xe x dx 1 2 e
2 du nx n … e tcos t dt 0 Let … C
n (e) Use mathematical induction or argue based on tabular 2. 26. (a) Let u 1 C 6)e x 6x dn
x
dx 2 nx n 2)e x 2x 3x 2
2 dn
x
dx … 2 e tcos t dt Now we find the average value of y 3(x 2 (x 3
(d) x n e cos t 3x 2 e x dx x 3e x cos t ex v x3 ex dt t e sin t x 3 e x dx sin t dt v
t e cos t dt t e x dx dv 3x 2 dx du sin t sin t e t e cos t dt v t cos t dt Let u dv dv
v sin x dx 2 w sin w dw sin w dw
cos w w cos w
w cos w C
C (sin w)(2w dw) cos w dw
sin w C 2 w sin w dw
2w cos w 2 sin w 2 x cos x 2 sin C
x C 2w dw. Section 6.3 28. Let w
2
3 dx
e 3x 9. Then dw 3x 3x 9 Let u 9 dw 2
(e ) w dw
3 dx dw v w ew dw xe dx nx dx x n cos x dx cos x dx v sin x x n sin x 2
w ew dw
3 (sin x)(nx n 1dx) x n sin x
32. Let ew
w ew n xn 1 dv sin x dx v cos x xn u
du ew dw nx x n sin x dx ew n1 x ncos x 2
w ew dw
3
2
(w 1)ew
3
2
( 3x 9
3 2 3 x2 (x ) e xn 33. Let u sin x dx 1)e 3x 9 C w 3 and g(w) e w. g(w) and its
integrals
(+) 34. Let (–)
(+) ew 6 (–) 1 ax
e (nx n 1 dx)
a dv 0 dx n1 dx v n x
n(ln x)n
x
x (ln x)n(x)
x(ln x) ew 6w (ln x)n dx cos x dx n n 1 ax
x e dx, a
a n(ln x)
x dx) 1 ax
e
a v (ln x)n u ew 3w 2 dx x ne ax
a du w3 1 1 1 e ax dx (x n) e ax 2x dx.
x dx n xn 1
a x ne ax dx 1 3w
w e dw.
2 ( cos x)(nx n dv nx n du Use tabular integration with f (x)
f (w) and its
derivatives dx (x n)( cos x) 1)ew x 2. Then dw 29. Let w
7 x2 dx du 9 dv
n1 e dw (w
3x 9 (3) dx, so w w ew e 2 3x w w dv du 2
w dw.
3 xn 31. Let u 1 n1 n (ln x) 1 dx
dx ew 35. (a) Let y f 1 (x). Then x 1 Hence, f
(b) Let u (x) dx f ( y) dy. (y)[ f ( y) dy] y dv y f ( y) dy f ( y)dy ew 0 du
w 3 e w dw w3 ew
(w 3 2 x 7e x dx 3w 2e w
3w 2 6w 6w e w
6)e w 6e w 1
dr, and so dr
r r dy Using the result of Exercise 13, we have:
sin (ln r) dr v y f ( y)
f C f ( y) dy (x)(x) (x) dx 1 f e y dy. f 1 (x) xf 37. (a) Using y 1 (x) dv d
f 1(x) dx
dx (x) dx f ( y) dy y f ( y) dy
xf 36. Let u f ( y) 1 1 Hence, f 1 3w
w e dw
2
13
(w
3w 2 6w 6)e w C
2
2
(x 6 3x 4 6x 2 6)ex
C
2 ln r. Then dy dy y f ( y) dy C du
30. Let y f (y), so dx 1 f 1 v dx
x d
x
f 1(x) dx
dx (x) (x) f ( y) dy. sin 1 x and f (y) sin y, y (sin y)e dy
1y
e (sin y cos y) C
2
1 ln r
e [sin (ln r) cos (ln r)] C
2
r
[sin (ln r) cos (ln r)] C
2 y 2 sin 1 x dx 2 , we have:
x sin 1 x x sin 1 x cos y x sin 1 x cos (sin sin y dy
C
1 x) C 257 258 Section 6.4 37. continued 40. (a) Using y 1 sin x sin x dx 1 x 1 x d
sin 1 x dx
dx
1 x x sin (b) x x2 1 u x 2, du 1 1 1 f x y 2 (x) 1/2 (b) C
x2 1 log2 x dx x log2 x
x log2 x C x log2 x
x log2 x 1 x and f (y) x tan 1 (c) 2log2 x x 1 x 2 y, we have 2 y dy
2y
C
ln 2
1 log2 x
2
ln 2
d
log2 x dx
dx
1
x
dx
x ln 2
dx
ln 2
1
xC
ln 2 x tan y, , we have: x dx log2 x and f (y) x log2 x x2 x tan tan u tan (x) x log2 x dx 1
u 1/2 du
2 x 1 1 x log2 x x x) 38. (a) Using y 1 1 x sin 2 1 x sin 1 log2 x dx 2x dx x sin (c) cos (sin f tan y dy
ln sec y x s Section 6.4 Exponential Growth and
Decay (pp. 330–341) C (Section 6.2, Example 7) Exploration 1 x tan u 1 x ln cos (tan x tan x dx ln cos y 1 x x tan 1 tan x x tan
(b) 1 1 x x 2, du 1 1
1 x 1 x) f ( y) 1 x
1 ln (x) cos x, 0
1 1 C x cos 1 x 1 x sin y 1 x x cos
x cos 1 0 2 x) v (u) du traveled by the object at time t. cos y dy
C
1 x) C
[0, 70] by [0, 1500] (b) cos 1 x dx x
x d
x
cos 1 x dx
dx
1 x 1 u 1 x 2, du 2x dx x cos 1 x x cos 1 x u1/2 C x cos 1 x 1 x2 (c) sin (cos 1 x) 1 x2 x2 dx Quick Review 6.4
1. a eb 2. c ln d 3. ln (x
x 1
u 1/2 du
2 2x C 0, the integral gives the distance 0 3. The total distance traveled is about 200 units for m 1,
about 400 units for m 2, about 800 units for m 4, and
about 1200 units for m 6. sin (cos 1 distance traveled by the object over the time interval [0, t].
Since s (0) , we have: x cos x dx v (u) du gives the
0 x and cos
x t 2. As we saw in Section 5.1, s (t) 1
ln (1
2 x2 x cos cos 1. As m increases the velocity of the object represented by the
graph slows down more slowly. That is, the ycoordinates
of the graphs decrease to 0 more slowly as m increases. [0, 20] by [0, 120] 1
u 1 du
2
1
ln u
C
2
1
ln (1 x 2)
2 x 1 f C 2x dx x tan 39. (a) Using y x) dx d
tan 1 x dx
dx
1
x
dx
1 x2 x tan (c) ln cos (tan 1 Slowing Down More Slowly x x tan 1 C 4. 100e
e2x
2x
x 3)
3
x 2
e2
e2 600
6
ln 6
1
ln 6
2 3 Section 6.4
0.85x 5. 5. Doubling time: 2.5 ln 0.85x
x ln 0.85 A0ert A(t) ln 2.5
ln 2.5 1000e0.086t 2000
2 ln 2.5
ln 0.85 x ln 2 5.638 e0.086t
0.086t
ln 2
0.086 t
2 k1 ln 2 k1 6. (k 3 k Amount in 30 years: ln 3 k A 1)ln 2 k ln 3 ln 2 k(ln 3 1.1t 7. 1.710 10 ln 1.1t ln 10 t ln 1.1 ln 10 t ln 10
ln 1.1 2t ln 24.159 1
4 2t 8. e 1
log 1.1 9. ln ( y
y
10. ln y
y
y 1)
1
y 1
ln 4
2 2x 3
e2x 3
1 e2x 2
2
2
y 3t 1
e3t 1
e3t 1
2 e3t A(t)
ln 2 2898.44
A0 1. y(t)
y(t) A(t)
1200
1 2 y0e kt
200e ln 2 y0e kt
100e1.5t 2. y(t)
y(t) A0e rt
A0e(0.0525)(30)
2898.44
e1.575 t A0e rt
600e0.0525t
e0.0525t
0.0525t
ln 2
0.0525 A(t)
10,405.37
104.0537
12
104.0537
ln
12 50 20.2t A0e rt
1200e(r)(30)
e30r
30r
104.0537
1
ln
12
30 r y(t)
y(10)
1
2
1
ln
2 k kt e A(t)
60e 10k 2400 10k Solution: y(t) A0e rt
1200e0.072t 2
1
2 0.1 ln 2
60e (0.1 ln 2)t e0.072t ln 2 10k
0.1 ln 0.072 Doubling time: 60e kt
30 13.2 years 8. Annual rate: 05t 3. y(t) y0e kt
y(t) 50e kt
y(5) 100 50e 5k
2 e 5k
ln 2 5k
k 0.2 ln 2
Solution: y(t) 50e(0.2 ln 2)t or y(t)
y0e $600.00 Doubling time: 3 Section 6.4 Exercises 4. y(t) $13,197.10 7. Initial deposit: 1
4
1
1
ln
2
4 t 1000e(0.086)(30) 6. Annual rate:
A(t) A0e rt
4000 2000e(r)(15)
2 e15r
ln 2 15r
ln 2
r
0.0462 4.62%
15
Amount in 30 years:
A(t) A0e rt
A 2000e[(ln 2)/15](30)
2000e2 ln 2
2000 22
$8000 ln 2) ln 2
ln 3 ln 2 k 8.06 yr 0.072 t t
or y(t) 60 2 t/10 ln 2
0.072 9.63 years 7.2% 259 260 Section 6.4 9. (a) Annually:
2 1.0475 t
ln 2 t ln 1.0475
ln 2
t
14.94 years 11. (a) Since there are 48 halfhour doubling times in 24
hours, there will be 248 2.8 1014 bacteria.
(b) The bacteria reproduce fast enough that even if many
are destroyed there are still enough left to make the
person sick. ln 1.0475 0.0475 12t
12
0.0475
12t ln 1
12
ln 2 2 t Hence
e2k
14.62 years 0.0475
12 12 ln 1 y0 ln 2 ln 2
4 ln 1.011875 13. ln 2
0.0475 0.9
ln 0.9
t 14.59 years (b) e 0.05 ln 2
ln 1.0825 0.005t
ln 0.05
0.005 8.74 years 15. Since y0 y (0) 5 0.0825
12 12t ln 1 k 12 ln 1 0.0825
12 8.43 years t t ln 2 0.5 ln 2.5 2 2e (0.5 ln 2.5)t y (0) 0.0825 4t
4 y 4t ln 1.020625 3 1.1e(k)( or y 2e0.4581t 1.1, we have: 1.1e kt 1 ln 2
4 ln 1.020625 8.49 years (d) Continuously: ln 2 2k ln 5 Function: y ln 3
k 2 ln 2 16. Since y0 (c) Quarterly: ln 2 2e(k)(2) ln 5 12 ln 2 2 2e 0.0825 12t 1 t 2, we have: kt y ln 2 599.15 days The sample will be useful for about 599 days. (b) Monthly:
2 0.005t e t t ln 1.0825 t 1250. 0.18t 1.0825t ln 2 23 0.18t ln 0.05 10. (a) Annually:
2 10,0000 ln 0.9
0.585 days
0.18
ln 2
ln 2
14. (a) Halflife
138.6 days
k
0.005 0.0475t t y0e3ln 2, we have 1250. There were 1250 bacteria initially. population was 14.68 years e0.0475t ln 2 ln 2. Solving 10,000 population would have doubled 3 times, so the initial (d) Continuously:
2 4, or k , which gives 2 hrs, so the doubling time is 1 hr. Thus in 3 hrs the 4t ln 1.011875 t y0e3k population increased by a factor of 4, i.e. doubled twice, in 0.0475 4t
4 1 y0e5k 40,000
10,000 y0e5k. We could solve this more quickly by noticing that the (c) Quarterly:
2 y0e3k and 40,000 10,000 1 ln 2 y0e kt, we have 12. Using y (b) Monthly: e 0.0825t
ln 2
0.0825 ln 1.1
1
3 8.40 years 3k (ln 1.1 Function: y 0.0825t 3) ln 3) 1.1e(ln 1.1 ln 3)t/3 or y 1.1e 0.3344t Section 6.4
3 17. At time t
kt y0e k y0e 20. First, we find the value of k. , the amount remaining is
k(3/k) 3 y0e 0.0499y0. This is less than 5% of the original amount, which means that over 95% has Taking “right now” as t (T0 Ts) e 35 65 (T0 65)e 65 (T0 65)e (T0 Ts)e 60e(
e20k k 7
1
ln
20 6 (k)(20) Dividing the first equation by the second, we have: k (a) T e10k kt k)( 20) (k)(10) 50 Ts 7
6 kt 0, 60 above room temperature 60. Thus, we have 70 Ts 2 Ts means T0
T decayed already.
18. T 261 Ts (T0 Ts)e kt 60e( (1/20)ln (7/6))(15) 53.45 1
ln 2
10 It will be about 53.45 C above room temperature. Substituting back into the first equation, we have:
30 (b) T [(ln 2)/10](10) (T0 65)e 30 (T0
T0 5 T0 Ts)e kt 60e( (1/20)ln (7/6))(120) 23.79 1
65)
2 60 (T0 Ts It will be about 23.79 above room temperature. 65 Ts (T0 10 60e( (c) T The beam’s initial temperature is 5 F. ln 19. (a) First, we find the value of k.
T Ts (T0 Ts)e 20 (90 20)e kt (1/20)ln (7/6))t 7
1
ln t
6
20
20 ln (1/6)
ln (7/6) t kt 60 1
6 Ts)e 232.47 min (k)(10) 4
7 e It will take about 232.47 min or 3.9 hr. 10k 4
1
ln
7
10 k 20
15
ln 70e (4/7)]t 3
14 10 ln
4
ln
7 0.445
ln 0.445
ln 0.445
k t [(1/10) ln (4/7)]t 5700 ln 0.445
ln2 6658 years Crater Lake is about 6658 years old. 4
1
ln t
7
10 3
14 t 20)e[(1/10) ln (90 kt kt e When the soup cools to 35 , we have:
35 ln 2
(see Example 3).
5700 21. Use k ln 2
(see Example 3).
5700 22. Use k
(a) e kt 0.17 27.53 min
kt It takes a total of about 27.53 minutes, which is an
additional 17.53 minutes after the first 10 minutes. t ln 0.17
ln 0.17
k 5700 ln 0.17
ln 2 14,571 years The animal died about 14,571 years before A.D. 2000,
(b) Using the same value of k as in part (a), we have:
Ts (T0 ( 15) [90 T
35 50
10
ln
21 t 105e Ts)e kt ( 15)]e[(1/10) ln (4/7)]t in 12,571 B.C.
(b) e kt 0.18 [(1/10) ln (4/7)]t kt 4
1
ln t
7
10
10
10 ln
21
4
ln
7 t ln 0.18
ln 0.18
k 5700 ln 0.18
ln 2 14,101 years The animal died about 14,101 years before A.D. 2000,
13.26 It takes about 13.26 minutes in 12,101 B.C. 262 Section 6.4 22. continued 26. kt 0.16 kt ln 0.16 (c) e ln 0.16
k t 5700 ln 0.16
ln 2 15,070 years 0.2 e 0.1t
ln 0.2
0.1t
t
10 ln 0.2 16.09 yr
It will take about 16.09 years. 27. (a) The animal died about 15,070 years before A.D. 2000,
in 13,070 B.C.
23. Note that the total mass is 66
v v0e (k/m)t
v 9e 3.9t/73 7 73 kg. 2190 3.9t/73
e
C
13
2190
Since s (0) 0 we have C
and
13
2190
2190
lim s(t) lim
(1 e 3.9t/73)
13
13
t→
t→ (a) s(t) 3.9t/73 9e ln p 1 p 168.5 Ae kh p0
p 1013 millibars, so:
1013e kh v 9e (59,000/51,000,000)t 90 v 9e 59t/51,000 90
1013 459,000 59t/51,000
e
59
459,000
Since s (0) 0, we have C
and
59
459,000
lim s (t) lim
(1 e 59t/51,000)
59
t→
t→
459,000
7780 m
59
59t/51,000 9e dt The ship will coast about 7780 m, or 7.78 km.
(b) 1 9e 59t
51,000 C 1013e(k)(20)
e20k 1
90
ln
20
1013 k Thus, we have p 1013e 1013e((1/20) ln (90/1013))(50)
(c) 1 0.121h 2.383 millibars. 1013e kh 900 e kh
1
900
ln
k
1013 h 51,000 ln 9
59 0.121 km (b) At 50 km, the pressure is 900
1013 59t/51,000 ln 9 t p0e kh Using the giving altitudepressure data, we have (k/m)t (a) s (t) 0 p0 Solution: p
41.13 sec p0 when h Ae0 A 73 ln 9
3.9 v0e C eCe kh p0 It will take about 41.13 seconds.
24. v C Initial condition: p ln 9 t e kh p p 3.9t/73 9e 3.9t
73 kh eln dt The cyclist will coast about 168.5 meters.
(b) dp
kp
dn
dp
k dh
p
dp
k dh
p 20 ln (900/1013)
ln (90/1013) 0.977 km The pressure is 900 millibars at an altitude of about
1899.3 sec 0.977 km. It will take about 31.65 minutes.
25. y
800
0.8
k
At t
y y0e 28. By the Law of Exponential Change, y 100e
t 1 hour, the amount remaining will be
100e 0.6(1) 54.88 grams. kt
(k)(10) 1000e ln 0.8
10 10 (b) 0.1 14 1000e
1000e . At 29. (a) By the Law of Exponential Change, the solution is
V V0e (1/40)t. 10k e 0.6t 24 h: ln 0.1 ( ln 0.8/10)24
2.4 ln 0.8 585.4 kg About 585.4 kg will remain. t e (1/40)t t
40 40 ln 0.1 92.1 sec It will take about 92.1 seconds. Section 6.4
30. (a) A(t) A0e t
It grows by a factor of e each year. 34. (a) T Ts 79.47(0.932)t (b) T 10 263 79.47(0.932)t t (b) 3 e
ln 3 t
It will take ln 3 1.1 yr. (c) In one year your account grows from A0 to A0e, so you
can earn A0e A0, or (e 1) times your initial
amount. This represents an increase of about 172%.
(k/m)t v0e 31. (a) s (t) v0m dt k Initial condition: s (0)
v0m 0 (k/m)t (c) Solving T 12 and using the exact values from the
regression equation, we obtain t 52.5 sec. C 0 (d) Substituting t 0 into the equation we found in part
(b), the temperature was approximately 89.47 C. C k v0m e [0, 35] by [0, 90] k(T dT
Ts k dt T dT
Ts k dt 35. (a)
v0m s (t) dT
dt
T C k k
v0m
k (b) lim s (t) (k/m)t e v0m
k 1 e (k/m)t v0 m lim 1 e v0m (k/m)t ln T e(r)(100) T Ts ln 90 100r T Ts r 32. (a) ln 90
100 T Ts 90 (b) t→ k k 0.045 or 4.5% e(r)(100) 131
ln 131 k
(0.80)(49.90)
k k
We know that v0m Ae Ts Ae T0 coasting distance Ts kt kt T0 when t 0 (k)(0) A Solution: T (T0 Ts Ts)e kt 1.32
(b) lim T 998
33
v0m lim [Ts t→ 1.32 and k (k/m)t k
m 998
33(49.9) e 1.32(1 e 36. (a) 2y0 20t/33 1.32(1 e 0.606 t ) (T0 kt Ts Ts y0e rt rt
ln 2
r t ) Ts)e e rt ln 2 ) A graph of the model is shown superimposed on a graph of
the data. t→ Horizontal asymptote: T 20
.
33 2 (1 k eCe 0.049 or 4.9% Using Equation 3, we have:
s (t) kt C e T0 ln 131
100 v0m C Initial condition: T 100r r
33. kt Ts t→ Ts) (b) [0, 0.1] by [0, 100] [0, 4.7] by [0, 1.4] (c) ln 2 0.69, so the doubling time is almost the same as the rules.
(d) 70
5 14 years or 72
= 14.4 years
5 0.69
which is
r 264 Section 6.4 36. continued r 0.5
x 0.5 x
x 1 (e) 3y0 y0e rt
3
ln 3 rt t 10
100
1000
10,000
100,000 ert ln 3
r Since ln 3 1.099, a suitable rule is 108
108
or
.
i
100r (We choose 108 instead of 110 because 108 has more
factors.)
37. (a) 1 10
100
1000
10,000
100,000 Graphical support: 2.5937
2.7048
2.7169
2.7181
2.7183 Graphical support:
1x
, y2
x 1 [0, 10] by [0, 3] e
38. (a) To simplify calculations somewhat, we may write: [0, 50] by [0, 4] 2
x 1 10
100
1000
10,000
100,000
e2 2x
x 6.1917
7.2446
7.3743
7.3876
7.3889 mg
k
mg
k
mg
k y1 1 [0, 500] by [0, 10] e 2 2
e2at 1 The left side of the differential equation is:
m dv
dt m mg
(2)(e 2at
k 4ma mg 2at
(e
k 4m
Graphical support: e at e at eat
eat e at eat
e2at 1
e2at 1
(e2at 1) 2
e2at 1 mg
1
k 7.389 2x
, y2
x e0.5 (c) As we compound more times, the increment of time
between compounding approaches 0. Continuous
compounding is based on an instantaneous rate of
change which is a limit of average rates as the
increment in time approaches 0. v(t) (b) r 0.5 x
, y2
x 1 1x
x 2.7183 y1 1.6487 y1
x e e 0.5 1.6289
1.6467
1.6485
1.6487
1.6487 gk
m 4mge 2at
(e 2at 1)2 2 1)
1) mg 2at
(e
k 2 (2ae 2at) (e 2at) 1) 2 (e 2at) Section 6.5
The right side of the differential equation is:
mg kv 2 mg k mg 1 1 mg 1 2
2
e 2at 1 mg
1
k 2
2
e 2at 1 4
e2at 1
1 4
(e 2at 1)2 2at mg 4(e
1) 4
(e 2at 1)2 4 mg e2at
(e2at 1)2 Since the left and right sides are equal, the differential equation is satisfied.
mg e 0
k e0 And v (0) e0
e0 0, so the initial condition is also satisfied. mg eat e at e
k ea t e a t e
t→
t→
mg 1 e 2 a t
lim
k 1 e 2at
t→
mg 1 0
mg
k1 0
k
mg
The limiting velocity is
.
k (b) lim v (t) (c) lim mg
k 160
0.005 at
at 179 ft/sec The limiting velocity is about 179 ft/sec,
or about 122 mi/hr. s Section 6.5 Population Growth
(pp. 342–349)
Quick Review 6.5
1. All real numbers
2. lim f (x)
x→ lim f (x) x→ 3. y 0, y 50
1 0 50 0
50 4. In both f and f , the denominator will be a power of
1 5e 0.1x, which is never 0. Thus, the domains of both are all real numbers.
5. [ 30, 70] by [ 10, 60] f (x) has no zeros. 265 266 Section 6.5 6. Use NDER f (x), or calculate the derivative as follows.
d
50
dx 1 5e 0.1x
(1 5e 0.1x)(0) (50)(5e
(1 5e 0.1x)2
0.1x
25e
(1 5e 0.1x)2 f (x) 0.1x )( 0.1) [ 30, 70] by [ 0.5, 2] (a) ( ,) (b) None
7. Use NDER(NDER f (x)), or calculate the second derivatives as follows.
d
25e 0.1x
dx (1 5e 0.1x)2
(1 5e 0.1x)2(25e f (x) 0.1x )( 0.1) (25e
(1 5e
2.5e 0.1x[(1 5e 0.1x) 2(5e 0.1x]
(1 5e 0.1x)3
12.5e 0.2x 2.5e 0.1x
(1 5e 0.1x)3 0.1x )(2)(1
) 5e 0.1x )(5e 0.1x )( 0.1) 0.1x 4 [ 30, 70] by [ 0.08, 0.08] Locate the inflection point using graphical methods, or analytically as follows.
f (x)
12.5e
(1 2.5e 0.2x 0.1x 2.5e
) 0 0.1x 3 5e (5e 0 0.1x 0.1x 1) e 0 0.1x 1
5 0.1x ln 5 x 10 ln 5 16.094 (a) Since f (x) 0 for x 10 ln 5, the graph of f is concave up on the interval ( (b) Since f (x) 0 for x 10 ln 5, the graph of f is concave down on the interval (10 ln 5, ), or approximately (16.094, ). 8. Using the result of the previous exercise, the inflection point occurs at x
Since f (10 ln 5) 1 50
5e ln 5 25, the point of inflection is (10 ln 5, 25), or approximately (16.094, 25).
9. x
x2 12
4x A
x x 12 A( x 4) Bx x 12 (A B)x 4A Since A
B 2. B B
x 4 1 and 4A 12, we have A 3 and 10 ln 5. , 10 ln 5), or approximately ( , 16.094). Section 6.5 10. 2x 16
x2 x 6 2x A
x 16 A( x When x B
3 x 2) 4. (a) 2 B(x 3) 3, the equation becomes 10 x 2, the equation becomes 20 B 5A, and when 5B. Thus, A dP
dt
dP
dt
dP
dt 4. Section 6.5 Exercises M
Ae kt
150
Ae 0.02t 1 P dP
1. (a)
dt k
P(M P)
m
0.02
P(150 P)
150
1
P(150 P)
7500 2 and
(b) P 1 Initial condition: P(0) 0.025P
1 A
A (c) 15 150
1 Ae0
150
10
15 15 (b) Using the Law of Exponential Change from Section
6.4, the formula is P 75,000e 0.025t 267 9 Formula: P 150
9e 0.02t 1 (c)
[0, 100] by [0, 1,000,000] 2. (a) dP
dt 0.019P
[0, 200] by [0, 200] (b) Using the Law of Exponential Change from Section
6.4, the formula is P 110,000e0.019t. 5. The growth rate is 0.3 or (c) 7. dP
dt 0.0004P2 0.04P 0.0004P(100 dP
dt
dP
dt
dP
dt (b) P
P k
P(M P)
M
0.05
P(200 P)
200 0.00025P(200 1
1 10
1 A
A Thus, k
P) M
Ae kt
200
Ae 0.05t Initial condition: P(0) 200
19e 50
P
1
P dP
dt
dP
dt Thus, k 0.04 and the carrying capacity is M
2 100. P
250 1
2
50
1
1
25 P
250
P
500
P
0.04 1
500
P
k1
M 0.04 and the carrying capacity is M 500. 9. Choose the slope field that shows slopes that increase as y
increases. (d) 19
1 8. 10 200
1 Ae0
200
20
10 Formula: P P) 0.04
P(100 P)
100
k
P(M P)
M [0, 100] by [0, 1,000,000] 3. (a) 30%. 6. The growth rate is 0.075 or 7.5%. 0.05t 10. Choose the slope field that matches a logistic differential
equation with M 100. (b)
11. Choose the only slope field whose slopes vary with x as
well as with y. (c) (c) 12. Choose the slope field that matches a logistic differential
equation with M 150. (a)
[0, 100] by [0, 250] 268 Section 6.5
1000
e4.8 0.7t
1000
1 e4.8e 0.7t
M
1 Ae kt 13. (a) P (t) 17. (a) 1 (b) P (0) 0.7 and Thus, k 1 6
1 1 25 A 24 Formula: P 1 and 1 dP
dT 1 person
14 sec 0.225t
0.225t e 365 24 3600 sec
1 yr 24e
24e Initially 1 student has the measles. 15. (a) Note that 0.225t 0.225t 2,252,571 people per year. t
125 The relative growth rate is
dP
dT 2,252,571
257,313,431 P 1 100
1 0.00875 or 0.875% (b) The population after 8 years will be approximately
P0e rt 257,313,431e 8r
275,980,017, where r is the unrounded rate
from part (a). 24e 0.225t 24e 0.225t e 1 0.225t 0.225t
t 16. (a) Let t be the number of years.
1000
0.1
ln 0.1
t 0.8 t
t ln 0.8
10.32 It will take about 10.32 years.
10,000(0.8)t. So that f (t) will round to less (b) Let f (t) than 1, we actually require f (t)
0.5
0.00005
ln 0.00005
t 10,000(0.8)t
0.8t
t ln 0.8
ln 0.00005
ln 0.8 150
24e 0.225t
150
1 24e 0.225t
3
2
1
2
1
48 ln 48
ln 48
17.21 weeks
0.225
150
1 24e 0.225t
6
5
1
5
1
120 ln 120
ln 120
0.225 21.28 It will take about 17.21 weeks to reach 100 guppies,
and about 21.28 weeks to reach 125 guppies. 10,000(0.8)t ln 0.1
ln 0.8 6 150
Ae0 A (b)
200
e 5.3 150. Initial condition: P(0) This is a logistic growth model with k
M 200. 1 P) M
Ae kt
150
Ae 0.225t 1 8 200
1 e 5.3 t
200
1 e 5.3e t
M
1 Ae kt (b) P (0) P) 0.225 and M
P Initially there are 8 rabbits.
14. (a) P (t) 0.0015P(150
0.225
P(150
150
k
P(M P)
M This is a logistic growth model with k
M 1000.
1000
1 e4.8 dP
dt 44.38 It will take about 44.4 years. 0.5. Section 6.5 18. (a) dP
dt 0.0004P(250 Now find the time to grow from 10,000 bees to 25,000
bees. P) 0.1
P(250 P)
250
k
P(M P)
M Thus, k
P 1
1 dP
dt M
Ae
250
Ae 10,000e t/12 2.5
ln 2.5 0.1t Initial condition: P(0)
year 1970. 28(1 25,000 250. 10,000e t/12 kt 28 1 28, where t 0 represents the 250
28 t
12 t dx
dt 21. (a) 250 A 12 ln 2.5 e t/12 The total time required is 4 ln 2 250
Ae0 A) 1
P
12 P(t) 0.1 and M Formula: P(t) 111
14 1 1000 7.9286 1 249.5 10 ln 1000 0.1x t ln 1000 0.1x 0.1(t 1000 0.1x e0.1(t 0.1x 111e 0.1t
14
(249.5)(111e 0.1t)
14 249.5 1 0.1t e 0.1t
t (b) y0e 0.01(t/1000) Solution: x ln 14) 82.8 0.00001t (c) y y0e (0.00001)(20,000) 0.819y0
The tooth size will be about 81.9% of our present tooth
size.
20. First find the time to grow from 5000 bees to 10,000 bees. P(t) 5000e0.25t 10,000 5000e0.25t 2 e0.25t ln 2 0.25t 4 ln 2 t 10,000 Ae0.1t 11,000 14
55,389
14
ln
55,389 0.9 e
ln 0.9
0.00001t
t
100,000 ln 0.9 10,536
It will take about 10,536 years. 1
P
4 e0.1C e0.1t
10e0.1C e0.1t 1000 1000 0.5 0.00001t dP
dt C) 10,000 Initial condition: x(0) 250 y0e C) x It will take about 83 years.
19. (a) y C x 250
111e 0.1t/14 10(ln 55,389 0.10x 1000 249.5.
1 13.8 years. dt 0.1x (b) The population P(t) will round to 250 when
P(t) 12 ln 2.5 1000 dx 250
, or approximately
111e 0.1t/14
250
7.9286e 0.1t 1 P(t) 269 10,000 Ae0 A 11,000e0.1t 10,000 (b) 100,000 11,000e0.1t 10,000
10 e0.1t
ln 10 0.1t
t 10 ln 10 23 yr
It will take about 23 years.
22. (a) Using the Law of Exponential Change in Section 6.4,
the solution is p(x) p0e x/100, where p0 p(0).
Initial condition: p(100) 20.09
20.09 p0e 1
20.09e p0
Solution: p(x) (20.09e)e x/100
or p(x) 20.09e1 0.01x
(b) p(10)
p(90) 20.09e0.9
20.09e0.1 $49.41
$22.20 (c) r(x) x p(x) 20.09x e1 0.01x
r (x) 20.09[(x)(e1 0.01x)( 0.01) (e1 0.01x)(1)]
20.09e1 0.01x(1 0.01x)
The derivative is zero at x 100, positive for x 100,
and negative for x 100, so r (x) has its maximum
value at x 100. 270 Section 6.5 23. (a) Note that the given years correspond to x
x
y 50, x
1 70, and x 0, x 20, 26. (a) y 16.90
5.132e 0.0666x 1 80.
(b) Carrying capacity 18.70
1.075e 0.0422x lim y x→ 16.90 representing 16.9 million people.
27. dy
dx (cos x)e sin x dy
dy [0, 100] by [0, 20] y 18.70, representing (c) Using NDER twice and solving graphically, we find
that y
0 when x 1.7, corresponding to the year
1912. The population at this time was about
y (1.7) 9.35 million.
1 eu du
eu y (b) Carrying capacity lim y
18.7 million people. x→ 24. (a) y (cos x)e sin x dx e sin x 0 esin 0 1 C dy
dx
dy
y3 ln y
[0, 50] by [0, 15] y
lim y 24.76, representing x→ ln P ln M
ln
ln P M
M P
P k
P(M
M M P
P M P
M
P 5 Ae0 3
A Solution: y
dy
dx
2 y
2 3 2e 2x x
y x dx
2 x
2 C Initial condition: kt C kt y (0) C Ae
APe C This gives kt e 02
2 2 C e 2 2
2 C
kt kt C y2
2 x2
2 2, or y 2 x2 4.
4. The solution of the initial value problem is the function that
satisfies the initial condition, namely y
Ae M
Ae x2 But this equation represents two functions, y kt P(1
1 2x Ae y dy k dt P
M 3 C 2x C e 2 29. k dt P 2x 5 P) k dt P 3) 2 dx 3 2 P 2( y 1 Initial condition: y (0) (c) Using NDER twice and solving graphically, we find
that y
0 when x 38.44, corresponding to the year
1988. The population at this time was about
y(38.44) 12.38 million.
dP
dt
M dP
P(M P)
(M P) P
dP
P(M P)
1
1
dP
P
MP e sin x 3 y 24.76 million people. 0 C Solution: y
28. 25. C Initial value: y(0) 24.76
7.195e 0.0513x (b) Carrying capacity C kt kt ) x2 4. Section 6.5 30. dy
dx
dy
y y x (c) x1/2 dx ln y 2 3/2
x
3 y A) 1200A 100 300 100 1200A 300A 200 900A e(2/3)x 3/2 C
3/2 eCe(2/3)x y Ae(2/3)x y 3/2 P(t)
1
P(t) Ae0 1 A P(t)
e(2/3)x Solution: y 31. (a) Note that k (d) 3/2 0 and M 0, so the sign of same as the sign of (M
both M P and P positive. For P
and P P)(P dP
is the
dt m). For m P M,
[0, 75] by [0, 1500] m are positive, so the product is m or P M, the expressions M Note that the slope field is given by P dP
dt m have opposite signs, so the product is negative. (b) 2
9
1200(2/9)e11kt/12 100
1 (2/9)e11kt/12
1200(2)e11kt/12 100(9)
9 2e11k/12
300(8e11kt/12 3)
9 2e11kt/12 A Initial condition: y(0)
1 1200Ae0 100
1 Ae0 300
300(1 C 0.1
(1200
1200 P)(P k
(M
M dP
dt k
(1200
1200 P)(P M 1200
P)(P (1200 1100
P)(P M 100) dP
100) dt 1 11
k
12 P dP
1
P 100 dt P P P ln P 11
k
12 1
dP
P 100 P ln M (P 100) (1200 P) dP
(1200 P)(P 100) dt P M 11
k
12 P m 11
k dt
12 (1200 1
1200
1
1200 ln 1200
ln P P 100
1200 P
P 100
1200 P ln P 100 11
kt
12 1 1
m ln 11
kt
12 C
P(1 Ae (M dP
dt M dP
dt M m
) P C 11kt/12 ee P(0)
P 100
1200 P P
P(1 Ae11kt/12
1200Ae11kt/12 100 Ae11kt/12)
P 1200Ae11kt/12 m
M M 1200Ae11kt/12 100
1 Ae11kt/12 C
C
m)kt/M Ae (M m)kt/M (M P)Ae (M AMe (M m)kt/M m)kt/M m AMe (M m)kt/M m
1 Ae (M m)kt/M
AM e 0 m
1 Ae 0 AM M) m A k dt e C e (M A(P(0) 100 m kt A) m P(0)
P(0) M k M
Mm
kt
M P(0)(1
APe11kt/12 k M
m M m m)kt/M m
M dP Pm
MP
Pm
MP
Pm
MP P C m) k 1 M dP
100) dt dP
dt P)(P k M
m
m (M P)(P m)
(P m) (M P)
(M P)(P m) k k
(M
M dP
m) dt M
P)(P (M m)
P)(P 100). dP
dt (e) dP
dt 271 AM m
1A m
P(0) P(0) m
M P(0) Therefore, the solution to the differential equation is
P AMe (M m)kt/M m
where A
1 Ae (M m)kt/M P(0) m
.
M P(0) 272 Section 6.6
dp
dt
dp
p 32. (a) s Section 6.6 Numerical Methods k(t)p (pp. 350–356)
k(t) dt Quick Review 6.6 t ln p k(u) du 1. f (x)
f (2) t k(u) du eCe 0
t p Ae 0 p0 . 0 Ae 0 p0 Ae
A f Solution: p(t)
9 (b) p0e
9 k(u) du
0 0 sec2 4 4. L(x) t f k(u) du
0 f 4 1 x 2 4 4
2 5. f (x)
f (4) 0 ln 1) 0.2x 5x 2
0.2(4) 5(4) 6. L(x) 9 u) 0.04(ln 10 4 2x 2x 0.04 ln (1 2) ( 2)2 4 1 0.04
du
1u 9 sec2 x 3. f (x) k(u) du 0 p0 f (2) f (2)(x
2 9(x 2)
9x 16 k(u) du Initial condition: p(0)
p0 3x 2 3
3(2)2 3 2. L(x) 0 p C f (4) f (4)(x 4)
2.85 0.4875(x 4)
0.4875x 0.9 0.04 ln 10 2 0.4875 9 p(9) 7. L(4.1) k(u) du p0 e 0 100e0.04 ln 10 f (4.1) 109.65 0.4875(4.1)
0.1(4.1)2 After 9 years during which the inflation rate is (a) L(4.1) 0.04
per year, the price of an item which originally
1t (b) f (4.1) L(4.1) f (4.1)
f (4.1) 0.9
5
4.1 2.89875
2.900512 0.001762
0.00061 0.061% cost $100 will be increased to $109.65.
8. L(4.2) 9 (c) p(9) p0e 0.04 du
0 0.04(9) 100e 143.33 The price will be $143.33.
9
0 (0.04 0.004u) du
0.002u2 k(u) du
0 p(9) p0 e dP
dt P 2 dP P 1 P 100e 0.522 168.54 9. L(4.5)
f (4.5) (b) k dt
kt C 0.4875(4.5)
2 0.1(4.5) f (4.5) L(4.5) f (4.5)
f (4.5) 10. L(3.5) 0.4875(3.5) 1
kt 1
C
1
P0 Solution: P L(4.2) f (4.2)
f (4.2) (a) L(4.5) f (3.5) C Initial condition: P(0) C 0.522 kP2 33. (a) f (4.2) P0 1
(1/P0) 0.1(3.5)2 (a) L(3.5)
(b) kt 0.9
5
4.2 2.9475
2.954476 0.006976
0.00236 0.236% 9
0 9 P0 (b) 0 0.04u 0.1(4.2)2 (a) L(4.2) 9 k(u) du (d) f (4.2) 0.4875(4.2) P0
1 (b) There is a vertical asymptote at t kP0t
1
kP0 f (3.5) L(3.5) f (3.5)
f (3.5) 0.9
5
4.5 3.09375
3.136111 0.042361
0.01351
0.9
5
3.5 1.351% 2.60625
2.653571 0.047321
0.01783 1.783% Section 6.6
Section 6.6 Exercises y
x y 2e x) 1 x (x Therefore, y 2e x( 1) 1
x 1 2e ) x 1 1 x 2e 2e 0 1 y. dy
dx
dy
1y 6. (0) 2e 1 2 1 1 Initial condition: y (0) 1
1 Ae0 1
2A
Solution: y 2 e x 1 x Check the initial condition:
y (0) Ae x y 1. Check the differential equation:
d
(x
dx eC ex y 1 ln 1 x(1 y) x dx
12
x
2 y 2. Check the differential equation: C (x2/2) y x 1 x (x Therefore, y e x( 1) e) x 1 e 1 e y e (x2/2) C y x 1 y 1 e x) 1 e C 1 d
(x
dx y e C x x e x2/2 y.
y Ae x2/2 1 1 Check the initial condition:
y (0) 0 1 e (0) 1 1 Initial condition: y ( 2) 0
2
0 Ae ( 2) /2 1
2
0 Ae
1
e2 A
2
Solution: y
e2 e x /2 1 or y 2 3. Check the differential equation:
2x y de
dx 2y 2 sin x
5 sin x e2x 2 2e2x
2 e2x cos x 2e 2x 2 sin x cos x
5
4 sin x 2 cos x
5
2 cos x sin x
5 2 cos x
5 sin x dy
dx
dy
y
dy
y 7. sin x
5 sin x 2y(x 1) 2(x 1) dx (2x
x2 ln y 2) dx
2x C y 2
e x 2x C sin x y eC e x Check the initial condition: y Therefore, y 2y e2(0) y (0) 2 sin 0
5 cos 0 1 1 4. Check the differential equation:
y dx
(e
dx y 2x e e2x 1 (e Therefore, y ex 1)
x y 2x e
e 2x 1) 2e2x
2x e e x y (0) e e 2(0) 1 2e y 1. 1 2x) 2 dy (1 2x) dx 1 x2 x 1 1
x x2 ( 1)2
1
C y
C x 1 y ex 1 y C
C ex C 1
C 1 Solution: y y C 1
( 1) dx ln 1 C Initial condition: y ( 1) 1
1 2 y 2(1 y 5. Note that we are finding an exact solution to the initial
value problem discussed in Examples 1–4.
dy
dx
dy
1y 2x y 2x Check the initial condition:
0 2 dy
dx 8.
1 A ex 2x Initial condition: y ( 2)
2
2 Ae( 2) 2( 2)
2A
2
Solution: y 2e x 2x 0 5 2 x2 1
x 1 e (x2/2) 2 1 273 274 Section 6.6 9. To find the approximate values, set y1 2y sin x and use
EULERT with initial values x 0 and y 0 and step size
0.1 for 10 points. The exact values are given by
1 2x
(e
5 y
x 2 sin x cos x). y (Euler) y (exact) Error 11. To find the approximate values, set y1 2y(x 1) and use
IMPEULT with initial values x
2 and y 2 and
step size 0.1 for 20 points. The exact values are given by
2
y 2e x 2x.
x y improved
Euler y (exact) Error 0 0 0 0 0.1 0 0.0053 0.0053 2 2 2 0 0.2 0.0100 0.0229 0.0129 1.9 1.6560 1.6539 0.0021 1.3983 1.3954 0.0030 0.3 0.0318 0.0551 0.0233 1.8 0.4 0.0678 0.1051 0.0374 1.7 1.2042 1.2010 0.0032 1.0578 1.0546 0.0032 0.5 0.1203 0.1764 0.0561 1.6 0.6 0.1923 0.2731 0.0808 1.5 0.9478 0.9447 0.0031 0.8663 0.8634 0.0029 0.7 0.2872 0.4004 0.1132 1.4 0.8 0.4090 0.5643 0.1553 1.3 0.8077 0.8050 0.0027 0.7683 0.7658 0.0025 0.9 0.5626 0.7723 0.2097 1.2 1.0 0.7534 1.0332 0.2797 1.1 0.7456 0.7432 0.0024 1.0 0.7381 0.7358 0.0023 0.9 0.7455 0.7432 0.0023 0.8 0.7682 0.7658 0.0024 0.7 0.8075 0.8050 0.0024 Error 0.6 0.8659 0.8634 0.0025 0 0.5 0.9473 0.9447 0.0026 1.8048 0.0048 0.4 1.0572 1.0546 0.0026 1.2036 1.2010 0.0026 10. To find the approximate values, set y1 x y and use
EULERT with initial values x 0 and y
2 and step
size 0.1 for 10 points. The exact values are given
by y x 1 e x.
x y (Euler) y (exact) 0 2 0.1 1.8000 2 0.2 1.6100 1.6187 0.0087 0.3 0.3 1.4290 1.4408 0.0118 0.2 1.3976 1.3954 0.0022 1.6553 1.6539 0.0014 1.9996 2 0.0004 0.4 1.2561 1.2703 0.0142 0.1 0.5 1.0905 1.1065 0.0160 0 0.6 0.9314 0.9488 0.0174 0.7 0.7783 0.7966 0.0183 0.8 0.6305 0.6493 0.0189 0.9 0.4874 0.5066 0.0191 1.0 0.3487 0.3679 0.0192 Section 6.6
12. To find the approximate values, set y1 x(1 y) and use IMPEULT with initial values x
2
step size 0.1 for 20 points. The exact values are given by y
e (x /2) 2 1.
x
2 y improved
Euler y (exact) Error 0 0 0 1.9 0.2140 0.2153 0.0013 1.8 0.4593 0.4623 0.0029 1.7 0.7371 0.7419 0.0049 1.6 1.0473 1.0544 0.0071 1.5 1.3892 1.3989 0.0097 1.4 1.7607 1.7732 0.0125 1.3 2.1585 2.1740 0.0155 1.2 2.5780 2.5966 0.0186 1.1 3.0131 3.0350 0.0219 1.0 3.4565 3.4817 0.0252 0.9 3.9000 3.9283 0.0284 0.8 4.3341 4.3656 0.0315 0.7 4.7491 4.7834 0.0344 0.6 5.1348 5.1719 0.0370 0.5 5.4815 5.5208 0.0394 0.4 5.7796 5.8210 0.0413 0.3 6.0210 6.0639 0.0430 0.2 6.1986 6.2427 0.0441 0.1 6.3073 6.3522 0.0449 0 6.3438 6.3891 0.0452 2 and y 0 and 275 276 Section 6.6 13. To find the approximate values, set y1 x y and use EULERT and IMPEULT with initial values x
y 1 and step size 0.1 for 20 points. The exact values are given by y x 1 2e x.
x y (Euler) y improved
Euler y (exact) Error (Euler) Error improved
Euler 0 1 1 1 0 0 0.1 0.9000 0.9100 0.9097 0.0097 0.0003 0.2 0.8200 0.8381 0.8375 0.0175 0.0006 0.3 0.7580 0.7824 0.7816 0.0236 0.0008 0.4 0.7122 0.7416 0.7406 0.0284 0.0010 0.5 0.6810 0.7142 0.7131 0.0321 0.0011 0.6 0.6629 0.6988 0.6976 0.0347 0.0012 0.7 0.6566 0.6944 0.6932 0.0366 0.0012 0.8 0.6609 0.7000 0.6987 0.0377 0.0013 0.9 0.6748 0.7145 0.7131 0.0383 0.0013 1.0 0.6974 0.7371 0.7358 0.0384 0.0013 1.1 0.7276 0.7671 0.7657 0.0381 0.0013 1.2 0.7649 0.8037 0.8024 0.0375 0.0013 1.3 0.8084 0.8463 0.8451 0.0367 0.0013 1.4 0.8575 0.8944 0.8932 0.0357 0.0012 1.5 0.9118 0.9475 0.9463 0.0345 0.0012 1.6 0.9706 1.0050 1.0038 0.0332 0.0012 1.7 1.0335 1.0665 1.0654 0.0318 0.0011 1.8 1.1002 1.1317 1.1306 0.0304 0.0011 1.9 1.1702 1.2002 1.1991 0.0290 0.0010 2.0 1.2432 1.2716 1.2707 0.0275 0.0010 0 and Section 6.6
14. To find the approximate values, set y1 y e2x 1 and use EULERT and IMPEULT with initial values x
y
1 and step size 0.1 for 20 points. The exact values are given by y e x e2x 1.
y (Euler) x improved
Euler y y (exact) Error (Euler) Error 277 0 and improved
Euler 0 1 1 1 0 0 0.1 1.1000 1.1161 1.1162 0.0162 0.0002 0.2 1.2321 1.2700 1.2704 0.0383 0.0004 0.3 1.4045 1.4715 1.4723 0.0677 0.0007 0.4 1.6272 1.7325 1.7337 0.1065 0.0012 0.5 1.9125 2.0678 2.0696 0.1571 0.0018 0.6 2.2756 2.4954 2.4980 0.2224 0.0026 0.7 2.7351 3.0378 3.0414 0.3063 0.0037 0.8 3.3142 3.7224 3.7275 0.4133 0.0050 0.9 4.0409 4.5832 4.5900 0.5492 0.0068 1.0 4.9499 5.6616 5.6708 0.7209 0.0092 1.1 6.0838 7.0087 7.0208 0.9370 0.0121 1.2 7.4947 8.6872 8.7031 1.2084 0.0159 1.3 9.2465 10.7738 10.7944 1.5480 0.0206 1.4 11.4175 13.3628 13.3894 1.9719 0.0267 1.5 14.1037 16.5696 16.6038 2.5001 0.0342 1.6 17.4227 20.5358 20.5795 3.1568 0.0437 1.7 21.5182 25.4345 25.4902 3.9720 0.0556 1.8 26.5664 31.4781 31.5486 4.9822 0.0705 1.9 32.7829 38.9262 39.0153 6.2324 0.0891 2.0 40.4313 48.0970 48.2091 7.7778 0.1121 dy
dx
dy
y2 2y 2(x 2 dy (2x y 1 15. (a) y 1) 2(x 1)dx x2 2) dx
2x C
1
2 Initial value: y (2)
2 22 2 C 2(2) Solution:
y (3) y
32 C 1 1
2(3) x2 2x
2 2 or y
1
5 1
2x 2 0.2 (b) To find the approximation, set y1
5 points. This gives y (3) x2 2y 2(x 0.1851; error (c) Use step size 0.1 for 10 points. This gives
y (3)
0.1929; error 0.0071.
(d) Use step size 0.05 for 20 points. This gives
y (3)
0.1965; error 0.0035. 1) and use EULERT with initial values x
0.0149. 2 and y 1
and step size 0.2 for
2 278 Section 6.6
dy
dx
dy
y1 16. (a) y
dx 5 ln y 1 x y 1 ex y 1
y 19. Set y1 2y sin x and use EULERG with initial values
x 0 and y 0 and step size 0.1. The exact solution is
1 2x
y
(e
2 sin x cos x). 1 C
C eCe x
Ae x 1 [ 0.1, 1.1] by [ 0.13, 0.88] Initial condition: y (0)
3 Ae0 2 20. Set y1 x y and use EULERG with initial values
x 0 and y
2 and step size 0.1. The exact solution is
y x 1 e x. 3 A 1 Solution: y
y (1) 2e 2e x
1 1
6.4366 (b) To find the approximation, set y1 y 1 and use
EULERT with initial values x 0 and y 3 and step
size 0.2 for 5 points. This gives y (1) 5.9766;
error 0.4599.
(c) Use step size 0.1 for 10 points.
This gives y (1) 6.1875; error [ 0.1, 1.1] by [ 2.3, 0.3] 21. Set y1 2y(x
x
2 and y
2
y 2e x 2x. 1) and use IMPEULG with initial values
2 and step size 0.1. The exact solution is 0.2491. (d) Use step size 0.05 for 20 points. This gives
y (1) 6.3066; error 0.1300.
17. The exact solution is y x2 1
2x (a) To find the approximation, set y1
IMPEULT with initial values x 2 , so y (3)
2y 2(x 2 and y step size 0.2 for 5 points. This gives y (3)
error 0.2.
1) and use
1
and
2 [ 2.2, 0.2] by [ 0.2, 2.2] 22. Set y1 x(1 y) and use IMPEULG with initial values
x
2 and y 0 and step size 0.1. The exact solution is
2
y
e (x /2) 2 1. 0.2024; 0.0024. (b) Use step size 0.1 for 10 points. This gives
y (3)
0.2005; error 0.0005.
(c) Use step size 0.05 for 20 points. This gives
y (3)
0.2001; error 0.0001.
(d) As the step size decreases, the accuracy of the method
increases and so the error decreases.
18. The exact solution is y 2e x
y (1) 2e 1 6.4366. 1, so [ 2.2, 0.2] by [ 7.3, 1.1] 23. To find the approximate values, set y1 x y and use
EULERT with initial values x 0 and y 1 and step size
0.1 for 10 points. The exact values are given by
y 2e x x 1.
x Error 1 1.0 0 0.1 0.9000 0.9097 0.0097 0.2 0.8200 0.8375 0.0175 0.3 0.7580 0.7816 0.0236 (b) Use step size 0.1 for 10 points. This gives
y (1) 6.4282; error 0.0084 0.4 0.7122 0.7406 0.0284 (c) Use step size 0.05 for 20 points. This gives
y1 6.4344; error 0.0022. 0.5 0.6810 0.7131 0.0321 0.6 0.6629 0.6976 0.0347 0.7 0.6566 0.6932 0.0366 0.8 0.6609 0.6987 0.0377 0.9 0.6748 0.7131 0.0383 1.0 0.6974 0.7358 0.0384 (a) To find the approximation, set y1 y 1 and use
IMPEULT with initial values x 0 and y 3 and step
size 0.2 for 5 points. This gives y (1) 6.4054;
error 0.0311. (d) As the step size decreases, the accuracy of the method
increases and so the error decreases. 0 y (Euler) y (exact) Section 6.6
24. To find the approximate values, set y1 x y and use
IMPEULT with initial values x 0 and y 1 and step size
0.1 for 10 points. The exact values are given by
y 2e x x 1.
x y improved
Euler 0 y (exact) (b) [0, 10] by [0, 3] Error 1 1.0 0 0.1 0.9100 0.9097 0.0003 0.2 0.8381 0.8375 0.7824 0.7816 0.0008 0.4 0.7416 0.7406 0.0010 0.5 0.7142 0.7131 0.0011 0.6 0.6988 0.6976 0.0012 0.7 0.6944 0.6932 0.0012 0.8 0.7000 0.6987 0.0013 0.9 0.7145 0.7131 0.0013 1.0 0.7371 0.7358 28. Set y sin (2x y) and use IMPEULG with initial values
x 0 and y 1 and step sizes 0.1 and 0.05. 0.0006 0.3 279 0.0013 x 25. Set y1 y e
2 and EULERG, with initial values
x 0 and y 2 and step sizes 0.1 and 0.05.
(a) (a) [0, 10] by [0, 5] (b) [0, 10] by [0, 5] 29. To find the approximate values, let y1 y and use
EULERT with initial values x 0 and y 1 and step size
0.05 for 20 points. This gives y (1) 2.6533.
Since the exact solution to the initial value problem is
y e x, the exact value of y (1) is e. [0, 4.7] by [0, 100] 30. To find the approximate values, let y1 3y and use
IMPEULT with initial values x 0 and y 1 and step size
0.05 for 20 points. This gives y (1) 19.8845.
Since the exact solution to the initial value problem is
y e3x, the exact value of y (1) is e3. (b) [0, 4.7] by [0, 100] 26. Set y1 cos (2x y) and use EULERG with initial values
x 0 and y 2 and step sizes 0.1 and 0.05. 31. To find the approximate values, let y1 1 y and use
RUNKUTT with initial values x 0 and y 1 and step
size 0.1 for 10 points. The exact values are given by
y 2e x 1.
x y (RungeKutta) y (exact) Error 0 and y 1
and step size 0.1 and 0.05.
3 (a) [0, 10] by [0, 3] 0.0000006 1.9836 1.9836 0.0000009 0.5 2.2974 2.2974 0.0000013 2.6442 2.6442 0.0000017 3.0275 3.0275 0.0000022 3.4511 3.4511 0.0000027 0.9 x 0.0000004 1.6997 0.8 ln y and use IMPEULG with initial values 1.4428 1.6997 0.7 27. Set y1 1.4428 0.6 1
y
2 0
0.0000002 0.4 [0, 10] by [0, 6] 1
1.2103 0.3 (b) 1
1.2103 0.2 [0, 10] by [0, 6] 0
0.1 (a) 3.9192 3.9192 0.0000034 1.0 4.4366 4.4366 0.0000042 280 Chapter 6 Review 32. (a) Set y1 x y and use RUNKUTT with initial values
x 0 and y 1 and step size 0.1. 5. Let u
du sin x
cos x dx /2 1 5 sin 3/2 x cos x dx 5u 3/2 du 0 0 2 5/2
u
5 5
2(1
[0, 10] by [ 3, 10] (b) Use RUNKUTT with initial values x
and step size 0.1. 0 and y 2 2
4 x2 6. 3x
x 1/2 4 dx (x 3) dx (x
4 12
x
2 11
24 3(4) s Chapter 6 Review Exercises 20 d tan tan 0 0 2 2. 12
x
2 1
dx
x2 x
1 x 1 3 tan 0 13
8 7. Let u
du 1
2 1
2 1 tan x
sec 2 x dx /4 1 e tan x sec 2 x dx e u du 0 3
2 3. Let u
du
1
du
2
1 eu 2 e 1
8. Let u dx du
3 3 1) 1
du
u3
1
u2
18
2
1
9
1
9
8
9
9 dx 18 e 1 3
1 du
du 1 ln r
dr
r u 1/2 du 0 2 3/2
u
3
2
(1
3 9. Let u
du
du 2x dx 2x sin (1
1 x 2) dx 0 sin u du
0 1
0 0) 2
3 2x dx 1 1 1
dr
r 1 x2 1 e0 ln r 8
4. Let u 1
0 e1 2 dx 36
(2x 0 2x 0 1
2 4
2 12
8 147
8 2
1 1
(4)
2 3 1
8 20 /3 sec 2 1/2 3
2 (pp. 358 – 361)
/3 3x 1
(16)
2 [0, 10] by [ 3, 10] 1. 0) 1/2 0
2 2 sin x
cos x dx cos x dx cos x
dx
sin x 1
du
u ln u C ln 2 sin x C 0) 1
0 Chapter 6 Review
10. Let u 3x du 1
dx
x
1
du
u dx
x ln x dx ln u dx
3x ln x du 3 dx 1
du
3
3 15. Let u 4 1
u 1/3 du
3
1 3 2/3
u
C
32
1
(3x 4)2/3 C
2 4 dt du
1
du
2 t2 3/2 12. Let u 1
2 sec 2 1
du
u 1
ln u
2 ln t 2 5 ln (t 2 5) f (x) and its
derivatives tan (+) cos x 3x2 (–) sin x 6x (+) –cos x 6 1 2 (–) –sin x d
1 d sec u tan u du 0 C 1 C x 3 sin x 1
dy
y 18. Let u
du sin u du
1
dw
w ln w C ln cos u C ln cos (ln y)
14. Let u
du C x e
e x dx e x sec (e x) dx 6x sin x ln x dv 1
dx
x v x 4 ln x dx cos u dw 3x 2 cos x 6 cos x x 4 dx
15
x
5 tan u du
sin u
du
cos u Let w cos x x 3 cos x dx ln y tan (ln y)
dy
y cos x. g(x) and its
integrals x3 1 x 3 and g (x) C sec du C 17. Use tabular integration with f (x) C C sec u 13. Let u C t 1
2
1
2
1
2 1 1/2 2t t dt du dt 5 2t dt t dt
t2 5 C dt
t 3/2 tt t
11. Let u C ln ln x 16. 281 sec u du
ln sec u tan u C
ln sec (e x) tan (e x) C 15
x ln x
5
15
x ln x
5
15
x ln x
5 151
x
dx
5
x
14
x dx
5
15
x
C
25 C 282 Chapter 6 Review
e 3x 19. Let u e sin x dx e 3x sin x dx v 3e 3x dx du
3x dv cos x 12
x
12
x
dx
x
2
1
x
dx
x
1
x2 2
dx
x2
13
x
2x x 1 C
3
1
21C1
3 dy
dx 22. x dy 3 cos x e 3x dx cos x dy
Integrate by parts again.
3e 3x Let u dv 9e 3x dx du v y cos x dx y sin x y (1)
e 3x 10 e 3x e
e sin x dx
sin x dx 3x x2 20. Let u
du cos x
cos x dv 2x dx
dx 1 2 3x
xe
3 dv Let u
du e sin x 3e 3x sin x 9e 3x 3x sin x dx 4
3 1
1
3 C
C x3
3 y 2x 1
x Graphical support: dx 2
e 3xx dx
3 e 3x dx [ 2, 2] by [ 10, 10] 23. 1 3x
e
3
1 2 3x
2
1
1
xe
xe 3x
e 3x dx
3
3
3
3
1 2 3x
2
2
xe
xe 3x
e 3x dx
3
9
9
1 2 3x
2
2
xe
xe 3x
e 3x C
3
9
27
x2
2x
2
e 3x C
3
9
27 dx C C 1 3x
e
3 v x 3x 3e 3x 1
[ e 3x cos x 3e 3x sin x]
10
3 sin x
cos x 3x
e
C
10
10 e 3x sin x dx x 2e 3x dy
dt 1
t 4
1 dy v t dy
y dt 4
1
t ln t 4 dt 4 y ( 3) C ln (1) C ln (t 2 2 y C 4) 2 Graphical Support:
x2
1x
2
x2
dy
1x
dx
2
2
x
dy
1x
dx
2
12
13
yx
x
x
C
2
6 dy
21.
dx y (0)
y C
x3
6 x 24. dy
d csc 2 cot 2 dy 1
x2
2 [ 4.5, 5] by [ 2, 5] csc 2 cot 2 d 1
dy csc 2 cot 2 d Graphical support: 1
csc 2
C
2
1
C1
2 y
y
C
[ 4, 4] by [ 3, 3] y 4
3
2 1
csc 2
2 3
2 [0, 1.57] by [ 5, 3] 1
3 283 Chapter 6 Review
d(y )
dx 25. d (y )
d (y ) x2 y
y (1) x 2 1 C 26. d(r )
dt cos t d (r ) 1
x2
1
2x
dx
x2
1
2x
dx
x2 2x cos t dt d (r ) cos t dt r C r (0) 1 sin t
C r
C sin t d(r ) y (1)
2
3 C
C
y x (x 2 dy
y 1 1
1 x r
1) dx 13
x
ln x
3
1
01
3 r (0) x3
3 t C C C 0 r cos t r
ln x 1) dt 1 2 dr (cos t
sin t 2
t t2
2 2) dt
C 2t 1 We first show the graph of y r r C
sin t 1
t2
2 Graphical support: 2
.
3 x t 2t r (0) 2
3 x ln x 1 C 0
2
3
x3
3 cos t C We first show the graph of y
x 1 ( sin t x Graphical support:
Let f (x) 1 1
x2 y C f (x) 0, along with the slope field for y x2 x
f (x) 1 1,
2x with the slope field for y sin t r 1 along cos t. 1
.
x2 [ 6, 4] by [ 3, 3]
[ 0.5, 4.21] by [ 9, 21] Next, we show the graph of y We now show the graph of y
for y f (x) x 2 x f (x) along with the slope field
1 with the slope field for y r cos t r sin t t 2 along 1. 1. [ 6, 4] by [ 3, 3]
[ 0.5, 4.21] by [ 9, 21] Finally we show the graph of y r along with the slope field for y r [ 6, 4] by [ 8, 2] sin t
cos t t2
2 2t
t 2. 1 284 27. Chapter 6 Review dy
y2
dx
dy
dx
y2
dy
dx
y2 ln y 2
2 y Since 1 C C x x2
2 C g (x)] dx x C. 2 f (x) dx
2(1 2 g (x) dx x) (x x C 2 4e x C C is an arbitrary constant, we may write the 31. [2f (x) 2 x) 1 4 y f (x) dx
(1 indefinite integral as 2 C x dx
x
2
x2
2 x Ce x f (x)] dx 2 Ce x y y (0) 30. [x 2 32. [g (x) 4] dx x g (x) dx Graphical support: C 4 dx (x 2) 4x 2
Since 2 2) 3x C C C is an arbitrary constant, we may write the indefinite integral as 3x C. [ 5, 5] by [ 5, 20] 28. dy
dx
dy
y1
dy
y1 33. We seek the graph of a function whose derivative is
(2x 1)(y 1)
Graph (b) is increasing on [ (2x 1) dx (2x ln y 1 x2 y 1 Ce x 2 sin x
is positive,
x and oscillates slightly outside of this interval. This is the 1) dx
x , ], where sin x
.
x correct choice, and this can be verified by graphing C NINT x sin x
, x, 0, x .
x 34. We seek the graph of a function whose derivative is e x .
Since e x
0 for all x, the desired graph is increasing for
all x. Thus, the only possibility is graph (d), and we may
verify that this is correct by graphing NINT(e x , x, 0, x).
2 2 y Ce x x 1 1 2 1 2 y ( 1) C C 2 y 2e x 35. (iv) The given graph looks like the graph of y
2 x satisfies
1
36. Yes, y Graphical support: 37. (a) dy
dx 2 dv [ 3, 3] by [ 10, 40] 29. 2t 1
Since 1 4 x)
x C x C. 0 4 C C is an arbitrary constant, we may write the indefinite integral as 6t) dt 3t 2 C Initial condition: v f (x) dx
(1 6t
(2 2t 4 when t C v f (x) dx 1. x is a solution. dv
dt v 2x and y (1) C
3t 2 4 1 1 v (t) dt (b)
0 3t 2 (2t
0 t2 t3 4) dt
1 4t
0 6 0 6
The particle moves 6 m. 0 x 2, which Chapter 6 Review
38. 285 40. Set y1 (2 y)(2x 3) and use IMPEULT with intial
values x
3 and y 1 and step size 0.1 for 20 points.
x y 3
2.9 x 0.6680 2.8 [ 10, 10] by [ 10, 10] 39. Set y1 y cos x and use EULERT with initial values
x 0 and y 0 and step size 0.1 for 20 points. 1
0.2599 2.7 0.2294 y 2.6 0.8011 0 0 2.5 1.4509 0.1 0.1000 2.4 2.1687 0.2 0.2095 2.3 2.9374 0.3 0.3285 2.2 3.7333 0.4 0.4568 2.1 4.5268 0.5 0.5946 2.0 5.2840 0.6 0.7418 1.9 5.9686 0.7 0.8986 1.8 6.5456 0.8 1.0649 1.7 6.9831 0.9 1.2411 1.6 7.2562 1.0 1.4273 1.5 7.3488 1.1 1.6241 1.4 7.2553 1.2 1.8319 1.3 6.9813 1.3 2.0513 1.2 6.5430 1.4 2.2832 1.1 5.9655 1.5 2.5285 1.0 5.2805 1.6 2.7884 1.7 3.0643 1.8 3.3579 1.9 3.6709 2.0 4.0057 x
x 41. To estimate y (3), set y1
initial values x 0 and y points. This gives y (3) 1 and step size 0.05 for 60
0.9063. 42. To estimate y (4), set y1
with initial values x x2 2y
x 1 and y 60 points. This gives y (4)
43. Set y1 e (x
x 0 and y 2y
and use IMPEULT with
1 y 2) 1 and use EULERT 1 and step size 0.05 for 4.4974. and use EULERG with initial values
2 and step sizes 0.1 and 0.1. (a) [ 0.2, 4.5] by [ 2.5, 0.5] 286 Chapter 6 Review 43. continued
(b) Note that we choose a small interval of xvalues
because the yvalues decrease very rapidly and our
calculator cannot handle the calculations for x
1.
(This occurs because the analytic solution is
y
2 ln (2 e x ), which has an asymptote at
x
ln 2
0.69. Obviously, the Euler
approximations are misleading for x
0.7.) kt 47. T Ts (T0 Ts)e
We have the system:
39 Ts (46 Ts )e 10k 33 Ts (46 Ts )e 20k Thus, 39 Ts 46 Ts Since (e e 10k 2 ) 10k
20k e and 33 Ts 46 Ts (39 Ts)2 (33 78Ts Ts2 1518 Ts 3 39 x2
ey 44. Set y1
values x Ts 2 33 Ts 46 1521 Ts 46 Ts y
and use IMPEULG with initial
x 0 and y 0 and step sizes 0.1 and 0.1. (a) . , this means: ()
[ 1, 0.2] by [ 10, 2] 20k e Ts)(46 Ts) 79Ts Ts2 The refrigerator temperature was 3 C. 48. Use the method of Example 3 in Section 6.4.
kt 0.995 kt ln 0.995 e [ 0.2, 4.5] by [ 5, 1] 1
ln 0.995
k t (b) 5700
ln 0.995
ln 2 41.2 The painting is about 41.2 years old.
49. Use the method of Example 3 in Section 6.4.
Since 90% of the carbon14 has decayed, 10% remains. [ 4.5, 0.2] by [ 1, 5] 2.645
k
(b) Mean life
46. T
T Ts
40 (T0 1
k 180 50. Use t kt 250 e rt kt e rt 180 and t (220 15k 180
140
1
ln
15 k (k)(15) 9
7 (220 40)e ((1/15) ln (9/7))t 70 40 (220 40)e ((1/15) ln (9/7))t ln 6 t 15 ln 6
ln (9/7) 6 107 min It took a total of about 107 minutes to cool from 220 F to
70 F. Therefore, the time to cool from 180 F to 70 F was
about 92 minutes. 18,935 1924 64 years. 7500
30 rt ln 30
ln 30
t ln 30
64 0.053 The rate of appreciation is about 0.053, or 5.3%. 9
7 180
30 1988 r 40)e 40 9
1
ln t
7
15 5700
ln 0.1
ln 2 15 to find k. T e((1/15) ln (9/7))t 1
ln 0.1
k 3.81593 years 40
e ln 0.1 The charcoal sample is about 18,935 years old. 40)e Use the fact that T 0.1 t 0.262059 Ts)e (220 kt kt e ln 2
k
ln 2
k
ln 2
2.645 45. (a) Halflife Chapter 6 Review
51. Using the Law of Exponential Change in Section 6.4 with (c) 1 appropriate changes of variables, the solution to the
differential equation is L(x) kx L0e the surface intensity. We know 0.5 , where L0
e 150
e 4.3 125 t 6
5
1
5 L(0) is 18k , so 1 e 4.3 e 4.3 t t ln 5 4.3 t t ln 0.5
k
and our equation becomes
18
1 x/18
L(x) L0e (ln 0.5)(x/18) L0
. We now find the depth
2 4.3 ln 5 5.9 days It took about 6 days. where the intensity is onetenth of the surface value.
54. Use the Fundamental Theorem of Calculus. 1 x/18
2
1
x
ln
2
18
18 ln 0.1
ln 0.5 0.1
ln 0.1
x ln c kA
(c y)
V
kA
dt
V
kA
tC
V
kA
tC
V y ln c y c y c y
c e y c De Initial condition y c Solution: y k y(0) c 150
1 e 4.3
1 3(0)2 sin(t 2) dt kt P), or 1 150
e 4.3 0.002P P
800 800 P
800 0.002 dt (800 P) P
dP
P(800 P) 0 0.002 dt 1
dP
800 P 0.002 dt ln 800 0.002t C c)e 0.002t C P (kA/V)t P
800 P
800 P
ln
P ln (kA/V)t c)e
150
e 4.3e c 800 P
P t where M 150, A 4.3 e , and dP
dt 1
P(150
150 P). The 1
P Initial condition: P(0)
50 2 Initially there were 2 infected students. 0.002t 1 50 800
1 Ae 0 A 16 A 15 Solution: P 1 800
15e 0.002t e
Ae
1 C 0.002t C e 800 P
P
800
P carrying capacity is 150.
(b) P(0) 0.002P 1 1
P (y0 1 2 800
dP
P(800 P) differential equation.
k
P(M
M 2 dP
dt 1. Therefore, it is a solution of the logistic dP
dt 0 03 dP
dt (kA/V)t (y0 M
Ae 1 55. ln P t 1 0 D t→ This is P 0 y0 when t lim [c t→ (sin 02) y (0) D y0 (b) lim y(t) Verify the initial conditions: (kA/V)t C y c 6x (kA/V)t C e y0 1) Thus, the differential equation is satisfied. (kA/V)t C e 2) 6x 2x cos (x 2)
dy
dt
dy
cy x 1) 3x 2 (cos x 2)(2x) 59.8 feet. 53. (a) P(t) (3x 2 d
(sin x 2
dx y d3
(x
dx sin t 2 dt 0 (sin x 2)
59.8 ft You can work without artificial light to a depth of about 52. (a) x d
dx y C e 0.002t 0.002t 800
Ae 0.002t 287 288 Section 7.1 56. Method 1–Compare graph of y1
y2 x 3 ln x
3 NDER x 3 ln x
3 79.961(0.9273)t 60. (a) T x3
. The graphs should be the same.
9 Method 2–Compare graph of y1
y2 x 2 ln x with NINT(x 2 ln x) with x3
. The graphs should be the same or differ
9
[ 1, 33] by [ 5, 90] only by a vertical translation. (b) Solving T(t) 40 graphically, we obtain t 9.2 sec.
The temperature will reach 40 after about 9.2 seconds. 10,000(1.063)t 57. (a) 20,000 (c) When the probe was removed, the temperature was
about T(0) 79.96 C. 1.063t 2
ln 2 t ln 1.063
ln 2
ln 1.063 t 61.
11.345 v0m It will take about 11.3 years.
(b) 20,000 k ln 2
0.063 11.002 It will take about 11.0 years.
d
dx
d
dx 58. (a) f (x)
g (x) v0m (1 e 0.97(1 e (27.343/30.84)t s(t) 0.063t t 27.343 s(t) 0.063t e 0.063t ln 2 0.97 s(t) 10,000e 2 coasting distance k
(0.86)(30.84)
k 0.97(1 e 0.8866t k (k/m)t )
) ) A graph of the model is shown superimposed on a graph of
the data. x u(t) dt u(x) u(t) dt u(x) 0
x
3 [0, 3] by [0,1] (b) C f (x) g(x) x x u(t) dt
0
x u(t) dt
3
3 u(t) dt
0
3 u(t) dt
x u(t) dt
0 59. (a) y 1 Chapter 7
Applications of Definite Integrals 56.0716
5.894e 0.0205x s Section 7.1 Integral as Net Change
(pp. 363–374)
Exploration 1
1. s(t)
s(0) [ 20, 200] by [ 10, 60] (b) The carrying capacity is about 56.0716 million people.
(c) Use NDER twice to solve y
0. The solution is
x 86.52, representing (approximately) the year 1887.
The population at this time was approximately
P(86.52) 28.0 million people. 8 t2
03
3 Thus, s(t)
2. s(1) Revisiting Example 2 13
3 (t 1) 8
0 C 1 t3
3 8
t
8 1 1 t3
3 dt 2 1 1 9⇒C 8
t 1 C 1 1.
16
. This is the same as the
3 answer we found in Example 2a.
3. s(5) 53
3 8
5 1 found in Example 2b. 1 44. This is the same answer we Section 7.1
Quick Review 7.1 7. 1. On the interval, sin 2x 0 when x 2 4 , sin 2x 3
, sin 2x
4 x
and 2 1; for x 4 –
– 3x + – 0 (x 1)(x 2 f (x ) 32
,x
2 for x
3x 3x 2 2 x2
x2 0 when x x2
x2
x2
x2
x2
x2 1 or 2. Test 0, x 2 3x 2 2; 3, –2 1 17
x2 2
; for x
1.9, 2
4.13; for x 0,
9
x
4
2
1
x
2
5
; for x 1.9, 2
4.13; and for x
,
2
x
4
2
17
. The function changes sign at 2,
2, 2
9 2
4
2
4
2
4 – + – f (x ) 1)2(2x 1) 3x 2 1 0, 2x 3 4; for x 1
or 1.
2 0 when x 3x 2 1 0 when x function changes sign at
– 0
4
x 2 ; and for x + 6. xe
0 when x
always positive. – + +
3
4 , x cos 2x 4, x cos 2x
4 , 3
5
0, , , or .
44
4 k or 2.1783 + – 32.7984. The 8
2 k , where k is an f (x )
x –2.1783 –0.9633 0.9633 2.1783 1
x sin 1
x 1
x 0.15, sin 1
1
or . Test one
3
2
1
0.1, sin
0.54; for
x 0 when x 0.37; and for x ; for x , The graph is
+
3 f (x )
x 54
4 0. On the rest of the interval, xe –
0.1 1 0.58. The x is 0.2, 0.96. The graph changes sign at , 3
5
, and . The graph is
4
4
– sin2 1) 1 – x Test one point on each subinterval: for x
; for x 2.4030. point on each subinterval: for x x
2 5. On the interval, x cos 2x x cos 2x 1: sec (1 is undefined when k for any integer k. Test for sin2 0) 1 changes sign at 0.9633 1; and f (x ) +
1 16 0: sec (1 10. On the interval, sin –1
2 2 x 1. The function changes sign at + x cos 2x k or 2.1783 cos x ) integer. The graph is 1, 1
. The graph is
2
– 0.9633 cos (1 sign alternates over successive subintervals. The function Test one point on each subinterval: for x
1 x Test for x x (x x
3 1 sin2 x) 1 9. sec (1 4 2 1 f (x ) + √2 2 x 2 –4 –2 5
,
2 f (x ) + + 3
, 2x 3
2 2 and is undefined when and 2. The graph is 3. x 2 2x 3 0 has no real solutions, since
b 2 4ac ( 2)2 4(1)(3)
8 0. The function is
always positive. The graph is x x 0 when x + – 3x 2 1
. The
2 2. The function changes sign at 1 and 2. + 2x 3 1 30 –3 –2 –√2 3x 2 x
x2 2. Test one point on each subinterval: for x The graph is 4. 2x 3 1, f (x ) + 2
4 x 1
; and for x
4 1
; for x
2 1 –5 0 x 2) x2 – , 0, 2 2 1, 1; and for 2 2 0. Test one point on each subinterval: x function changes sign at 0. The graph is 1. The function changes sign at one point on each subinterval: for x x 1; , sin 2x 0 when x 1 for x 8.
+ 2 . Test one . The graph is –3 2. x 2 2 3
, sin 2x
4 point on each subinterval: for x
for x , 0, or x
x2 289 –
1
2 f (x )
x
0.2 1
1
, and .
3
2 290 Section 7.1 Section 7.1 Exercises
1. (a) Right when v(t)
when 0 0, which is when cos t
or 2 2 i.e., when t 6t 2 v(t) 0, i.e., 3
t 2 . Left when cos t
2
3
t
. Stopped when cos t 0,
2
3
or .
2
2 t i.e., when 4. (a) Right when
18t 12 6(t 1)(t 2) i.e., when 0 t 1. Left when 6(t i.e., when 1 0, t 0, 2. Stopped when 6(t 1)(t 2) 1)(t 0, i.e., when x 2) 0, 1, or 2. 2
0 2 2 5 cos t dt 5 sin t 5[sin 2 0 sin 0] 2t 3 0 9t 2 12t
2 3 /2 5 cos t dt 0 10 5 t 3 2 t 3 36 . Left when sin 3t . Stopped when sin 3t 18t 0, i.e., when 0, i.e., when t /2 6 sin 3t dt 1
cos 3t
3 6 0 3
2 cos 0 0 ( 6t2 12) dt 1 /2 cos t 0, which is when sin t 0, i.e., when 0 when sin t 0 and cos t 2 0 (b) Displacement 2 3
or
2
2 t 0] 0 and 0, i.e., when 2 . Left
t 2 or 0 or cos t 5 sin2 t cos t dt 5 0, 1
sin3 t
3 2 6 sin 3t dt 4 2 3. (a) Right when v(t)
Left when 49 /2 49 9.8t 9.8t Stopped when 49 0, i.e., when 0 0, i.e., when 5
9.8t t (49 0, i.e., when t 5. 5 sin2 t cos t dt 0 2 49[(10 10) 0] 0 0 5 sin2 t cos t dt 5 sin2 t cos t dt 3 /2 5
3 10
3 when 0 10 3 /2
/2 5
3 20
3 6. (a) Right when v(t) 9.8t) dt 0 4.9t 2 5. 0 10. 10 (b) Displacement t 5 sin2 t cos t dt (c) Distance 6 /3 t 0, which is when 4 4. Left: never, since negative. Stopped when 4 t 4 t 0, i.e.,
t cannot be 0, i.e., when t 4. 10 (c) Distance
(49 49
0 9.8t dt
( 49
5 122.5 4 (b) Displacement 10 9.8t) dt 0 122.5 0 0 /2 0 5 2 6 sin 3t dt 6 sin 3t dt 49t 12) dt t 0 5[0 0 /3 18t 6 3
. Stopped when sin t
2
3
i.e., when t 0, , , , or 2 .
2
2 /2 (c) Distance 4 12 dt t 2 cos 0] 2 18t 5. (a) Right when v(t) 0, . (b) Displacement 24) 1 5 0, which is when sin 3t 6t 2 0 3 /2 20 2. (a) Right when v(t)
i.e., when 0 (6t 2 5 cos t dt /2 5 [(16 0 1 2 5 cos t dt 3 (c) Distance 5 cos t dt /2 12) dt 0 2
0 18t 2 0 (c) Distance or (6t 2 (b) Displacement (b) Displacement 245 9.8t) dt 4 t dt 0 2
[0
3 8] 16
3 4 (c) Distance 4
0 t dt 16
3 2
(4
3 t)3/2 4
0 291 Section 7.1
7. (a) Right when v(t)
i.e., when 0 0, which is when cos t
3
or
2
2 t cos t 0, i.e., when cos t e sin t cos t dt e 0 [e 0 0 e] 16t
/2 0 e sin t cos t dt e sin t cos t dt C1, where 6 ft/sec.
90t 0. Solve s(t) 90t when t 90 16t2 2t( 8t
45
8 0 or t 0 2 /2 esin t cos t dt 32(3) s(0)
2 32t 90. v(t) dt C2 0 esin t cos t dt (c) Distance
3 /2 (b) s(t) 0
2 v(0) 32 dt Then v(3) 2 sin t a(t) dt C1 3
. Stopped when
2
3
or .
2
2 0, i.e., when t (b) Displacement 11. (a) v(t) 2 . Left when t 2 2 t 0, C2, where 0:
45) 0 5.625 sec. The projectile hits the ground at 5.625 sec. 3 /2 (e 1) 1
e e 1
e 1 2
e 2e 4.7 (c) Since starting height 8. (a) Right when v(t ) 0, which is when 0 t 3. Left:
never, since v(t ) is never negative. Stopped when t 0.
3 (b) Displacement 1 0 1
[ln (10)
2 t
t 3 t 0 t 2) 0. 3 (d) Max. Height
0 1.15 ln 10
2 dt t2 1 1
ln (1
2 dt ln 10
2 ln (1)] (c) Distance 2 Displacement 16 s 5.625 2
2 Distance 1.15 ending height, 5.625
2
5.625
90
2 126.5625, and 2(Max. Height) 253.125 ft. c 9. (a) v(t) a(t) dt v(t) t 2t 3/2 C, and since v(0) 2t 3/2. Then v(9) t 9 2(27) 0, 63 mph. 12. Displacement v(t) dt 4 5 24 23 cm 0
c 13. Total distance v(t) dt 4 5 24 33 cm v(t) dt 15 4 11. v(t) dt 15 4 5 16. v(t) dt 15 4 5 24 0
a (b) First convert units: 14. At t t
t 3/2
t 2t 3/2 mph
mi/sec. Then
3600
1800
9
3/2
t
t
Distance
dt
1800
0 3600
2
5/2
t
t
9
27
9
0
0.06525 mi
7200
4500 0
800
500 344.52 ft. a, s s(0)
0
b At t b, s s(0)
0
c At t c, s s(0) 8. 0 15. At t a, where dv
is at a maximum (the graph is steepest
dt upward).
4 10. (a) Displacement (t 2) sin t dt 0 sin t 16. At t 4 t cos t 2 cos t 4 cos 4 2 cos 4) 2] 1.44952 m
17. Distance (b) Because the velocity is negative for 0
for 2 t , and negative for t t 2, positive 4, 2 Distance (t
0 2) sin t dt [(2 sin 2) (
( 2 2 cos 4 (t 2) sin t dt 2 4 (t 2) sin t dt
sin 2 2 cos 4
2 sin 2 dv
is at a maximum (the graph is steepest
dt upward). 0 [(sin 4 c, where sin 4 2)]
2 1.91411 m. 4 1
2 (a) Final position Initial position
2 4 6; ends at x 6. 12 4 Distance (b) 4 meters
18. (a) Positive and negative velocities cancel: the sum of
signed areas is zero. Starts and ends at x 2.
(b) Distance
meters 2) sin 4 Area under curve Sum of positive areas 4(1 1) 4 292 Section 7.1
25. (Answers may vary.)
Plot the speeds vs. time. Connect the points and find the
area under the line graph. The definite integral also gives
the area under the curve. 7 19. (a) Final position 2 v(t) dt
0 1
(1)(2)
2
1
(2)(2)
2 2 1
(1)(2)
2
1
(2)(1)
2 1(2) 26. (a) Sum of numbers in Sales column (b) Enter the table in a graphing calculator and use
QuadReg: B(x) 1.6x 2 2.3x 5.0. 5;
ends at x 5. 11 (1.6x 2 (b) 1
1
(1)(2)
(1)(2)
2
2
1
(2)(1)
2 v(t) dt
0 v(t) dt
0 1
1
(2)(3)
(1)(3)
2
2
1
(3)(3)
2 2 1
(1)(3)
2 (3)(3) (1.6x 2 1.6 3
x
3 10 v(t) dt
1
(1)(3)
2 3) 1
(1)(3)
2 3(3) 1
(3)(3)
2 21. 27.08 e t/25 b dt 27.08 25e 10 t/25 27.08[25e 0 0.4 25] 332.965 billion barrels
3.9 22.
0 93.6 t
2.4 sin
dt
12
28.8
28.8 23. (a) Solve 10,000(2
(b) Width 3.9t r) 0: r 20,000 29. F(x) r2 20,000
80,000
3 13
r
3 20,000 4 r ) dr 8
3 30. F(x) 2 r: Area 3 (c) 8(10 (2 r) r in2 flow in 396 in3
sec x2 2x dx
0 kx; 10,000 2x. 12
kx
2 kx dx 9 81 N cm
0 k(1), so k (b) For total distance: W 16 (10r 3 16
0 1244.07 in3
sec 45 81
4 W in3
sec r 3) dr 0 14
r
4 2 and F(x) 9 0 2rr 3 r2)(2 r) dr
5r 2 2(110) 18N F(x) dx (a) W Cross section area 0 16 2(112) d 10,000.
12
kd
2 0 1
(10,000)(0.5)2
2 1
(10,000)(1)2
2 5000 For second half of distance: Inches per second
in.
sec 2(119) 1250 inchpounds r, Length r 2) 2(115) 2(115) k(3), so k d 0 (b) Volume per second 8(10 2(120) 0 2 83,776 24. (a) Width 2(115) 2(9) (b) W (2r 0 kx; 6 (a) F(9) 0 2 2(110) 9 2 r)(2 r) dr f (xn) 121] Area 0 ∑ 2 f (xi) 1156.5 2rr Population density 10,000(2 798.97 thousand
0.5 i1 2(120) 0 2 miles. 2 r: Area 2 (d) 5.0x n1 f (x0) 24 t
cos
12 93.6 kilowatthours r, Length (c) Population 28.8 a 2n
18 8
[120
2(10) 0 24 10.5 2.3 2
x
2 28. Treat 6 P.M. as 18 o’clock: 19.5 meters
10 5.0) dx (b) The answer in (a) corresponds to the area of midpoint
rectangles. The curve now gives a better approximation
since part of each rectangle is above the curve and part
is below. 0 1
(2
2 2.3x 0.5 2.5. (b) Distance 0 10.5 27. (a) 2.5;
ends at x 11 (d) The answer in (a) corresponds to the area of left hand
rectangles. These rectangles lie under the curve B(x).
The answer in (c) corresponds to the area under the
curve. This area is greater than the area of rectangles. 10 2 5.0x 904.02 thousand 7 meters
20. (a) Final position 5.0) dx 2.3 2
x
2 0 1
(2)(2)
2 1(2) 2.3x 1.6 3
x
3 (c)
7 797.5 thousand 31. 5000 (12 0)
[0.04
2(12) 1250 3750 inchpounds 2(0.04) 2(0.05) 2(0.06) 2(0.05) 2(0.04) 2(0.04) 2(0.05) 2(0.04) 2(0.06) 2(0.06) 2(0.05) 0.05] 0
0.585
The overall rate, then, is
12 0.04875. 0.585 Section 7.2 32. (12 0)
[3.6
2(12) 2(4.0) 2(3.1) 2(2.8) 2(2.8) 2(3.2) s Section 7.2 Areas in the Plane
(pp. 374–382) 2(3.3) 2(3.1) 2(3.9) 4.0] 2(3.2) 2(3.4) 2(3.4) 40 thousandths or 0.040 Exploration 1
1. For k ∑mkxk
. Taking dm
∑mk My 33. (a) x 293 M A Family of Butterflies 1: [(2 sin x) sin x] dx (2 0 dA as mk and letting 2 sin x) dx 0 2x 2 cos x 2 4 0 dA → 0, k → x dm yields . For k dm 2: /2 ∑mky
. Taking dm
∑mk My (b) y M [(4
0 (4 dA as mk and letting (2 sin 2x)] dx 4 sin 2x) dx 0 /2 4x
dA → 0, k → yields 2 sin 2x) /2 2 cos 2x y dm 2 4 0 .
dm 2. It appears that the areas for k
2
4. 3 will continue to be /k 34. By symmetry, x 0. For y, use horizontal strips: 3. Ak [(2k k sin kx) k sin kx] dx 0
/k y dm y dA (2k y dA 2k sin kx) dx 0 y
dm dA dA If we make the substitution u kx, then du k dx and the 4 y(2 y) dy ulimits become 0 to . Thus, 0
4 2 /k y dy 2
5 2k sin kx) dx 0
/k 4
0 0 2 3/2 4
y
3
0 0 2 y 5/2
2 (2k Ak 0 (2
(2
4. 2 12
5 2 sin kx)k dx
2 sin u) du. 4 5. Because the amplitudes of the sine curves are k, the kth
butterfly stands 2k units tall. The vertical edges alone have
lengths (2k) that increase without bound, so the perimeters
are tending to infinity. 35. By symmetry, y 0. For x, use vertical strips: Quick Review 7.2
x dm x dA x dA 1. x
dm dA sin x dx 0 x(2x) dx [1 1] 2 0 1
2 cos x 0 dA 2. 1 2x
e
2 e2x dx 0 1
0 12
(e
2 1) 3.195 2
0 2x dx /4 3. sec2 x dx tan x /4 1 ( 1) 2 /4 2 32
x
30 x2 2
0 /4
2 4. x 3) dx (4x
0
3 4
3 5. 9
3 radius 3.) x 2 dx 2x 2 14
x
4 2 (8 4) 0 4 0 9
(This is half the area of a circle of
2 294 Section 7.2 6. Solve x 2 4x x 6.
x 2 5x 6 0
(x 6)(x 1) 0
x 6 or x
1
y 6 6 12 or y
1
(6, 12) and ( 1, 5)
7. Solve e x 1 3.
1 6 [(12y 2 (x 1)] 0. Test: e0 e
0 x ( 12y 3 (x 10
3 1) is at a minimum. x2 2y)] dy 2y) dy
1 y2 0 4
3 1 1, i.e., when
5. Use the region’s symmetry: 1. So the solution is (0, 1). 2 2x 10y 2
10 3
y
3 3y 4
3 8. Inspection of the graphs shows two intersection points:
(0, 0), and ( , 0). Check: 02
0 sin 0 0 and
2
2
sin
0.
9. Solve (2y 2 2 [2x 2 (x 4 2 2x 2)] dx 2 ( x4 0 2 15
x
5 43
x
3 2 32
5 2 2 x4 x4
x 1 2 (x 2 x2 x2 2 0 0 25
x
5 2 x3 1 2
0 1
3 2
5 2 or 1
1 Throw out the negative solution. 0 (2 y 4 3/2
y
3 y) dy 12
y
2 1
0 1
x3 y 4
3 1 (0, 0), ( 1, 1
2 5
6 0 8. Integrate with respect to y: 1) and (1, 1) 1 10. Use the intersect function on a graphing calculator: 0 [(2 y) y] dy 12
y
2 2y 2 3/2
y
3 1 1
2 2
0 2
3 0 9. Integrate in two parts:
0 [(2x 3 x2 [( x 2 3x) [ 2, 2] by [ 2, 2] ( 0.9286, 0.8008), (0, 0), and (0.9286, 0.8008) 0 (1 2 cos x) dx 0 1
x
2 1
sin 2x
4 /3
0 1
sec2 t
2 /3 (sec2 t 8 sin2 t) dt 0 /3 tan t 4t 2 sin 2t
0 3
4
3 4
3 3 0 x2
2 8x) dx 3x)] dx
5x)] dx ( 2x 3 8x) dx 0 14
x
2 2 0 [0
4 sin2 t dt (2x 3 (2x 3 2 2. Use symmetry:
2 0 ( x2 5x) 2
2 Section 7.2 Exercises
1. 128
15 22
15 7. Integrate with respect to y:
8 1
2 x 1
3 2x 4) dx 0 1 2 0 6. Use the region’s symmetry: x2 1 2 32
3 (0, 0) is a solution. Now divide by x.
x2 4x 2) dx 0 x 3. 1 1
12 0 0 1, so that if they are 1 is zero when e x
1 12y 3) 1 14
y
4 0
1 always greater than or equal to x x 4. 5 1. From the graphs, it appears that e is ever equal, this is when e x 13
y
3 y 3) dy 0 x x dx
[e
dx ( y2 4x 2
(8 0
2 16)] 14
x
2 [( 8 4x 2
16) 2
0 0] 16 5
6 Section 7.2
10. Integrate in three parts: x
1 [( x 2) 2x 2 x2 4: x 2 2x 2) (x 2 4)] dx 13. Solve 7 1, so the curves intersect at 1. 2 (4 x )] dx 1 2 [(7 1 1 2 x 2) [(4 295 (x ( 3x 2 3) dx 1 2)] dx 1 1 3 x 2) dx (1
1 3 [( x 2) x 2)] dx (4 1 (x 2 2 x 2) dx (x 2 2 x 2) dx (x 2 x 2
3 3 1 3 13
x
3 3x 2 1
1 2
3 4 14. 2) dx 2 13
x
3 1 12
x
2 1 x3
3 2 1
2 2 1 x2
2 2x
1 [ 3, 3] by [ 1, 5] 3 1 x2
2 1
3 13
x
3 2x 2x The curves intersect at x 2 8
3 2 2 1 2 1
3 4 4x 2 9
2 49
6 8 2 1 2 (x 4 2 (x 4 4x 2 4)] dx 5x 2 2 2 ( x4 4) dx 1
5 1
2
5 5x 2 4) dx 1 53
x
3 2 x5 4 1
6 5
3 4 1 4x 15
x
5 2
0 32
5 2 40
3 2 53
x
3 4x
1 1
5 8 5
3 4 8 15. 11. Solve x 2
x 8
3 6 2 [x 2
1 0 9 2 x 2] dx 4) 0 1
2 2. Use the region’s symmetry: 4 2 [(x 4
8
3 1 and x 2 2: x 2 4, so the curves intersect at 2.
2 [2 2 (x 2 2)] dx x 2) dx (4 2 33
a, a by [ a 2, a 2]
22 2 4x 1 32
x
3
2 8 x2 3: x 2 12. Solve 2x 8
3 8
3 8 2x 3 (x 32
3 3)(x 10 1) 2
3 The curves intersect at x
0, 3 (2x x 2 3) dx x 1 and x
13
x
3 2 1 (9 9 32
3 10 9)
2
3 2 x a2 3. 3x
1 x 2 dx 2 0 3
1 20
1
3 3 a. Use the region’s symmetry:
a so the curves intersect at x 0 and x 23
a
3 12
(a
3 x 2)3/2 a
0 13
a
3 296 Section 7.2
18. Solve y 2 16. 2: y2
y y curves intersect at y
2 (y [ 2, 12] by [0, 3.5] 12
y
2 1 The curves intersect at three points:
x
1, x 4 and x 9.
Because of the absolute value sign, break the integral up at
x 0 also:
0 x 4 6
5 1 x x dx 9 x x 6 12
x
2 2
( x)3/2
3 5 6x
5 1 y 2 3/2
x
3 20 5 6x y
4 4
5 18 189
10 16
15 13
30 32
5
16
3 1
6 5
3 1 16
3 (y 2 1
3 4.
y
4 4) 5)( y 0 y2
4 y3
12 32
5 y2
4 4 4 2
3 y
4 y2
4 0. 4: The curves intersect at y 5 11
10 y
4 1 1
2 1
2 1 and x 1 2 13
y
3 5 0, 4 and y 5. 0 4 0 y2
4 0, so the 4 9 12
x
2 2 3/2
x
3 y2 12
x
2 y2
4 4 1) 2. 8
3 4 9
2 19. Solve for x: x 2)(y 2y 2 Now solve 0 6x x dx dx 5 4 6
5 0 (y 1 and y y 2) dy 2 2 y
4 2
3 5 5y
4 25
8 20. Solve for x: x dy 5 dy y2
8 125
12 1 16
3 25 y 2 and x 2 243
8 20 30 3
8 2y 2. Now solve 3 17.
y2 2y 2: y 2 3 1, so the curves intersect at y 1. Use the region’s symmetry:
1 [ 5, 5] by [ 1, 14] 2 0 2y 2 (3 1 y 2) dy 2 The curves intersect at x 0 and x
4. Because of the
absolute value sign, break the integral up
at x
2 also (where x 2 4 turns the corner). Use the
graph’s symmetry:
2 2
0 x2
2
2 2
0 2 4 (4 4 2
2 4 2 3x
dx
2 x3 2
20 2[4] x 2) dx 2
2 2
2 x3
6
32
3 x2
2 4 (x 2 3y 2) dy 0 (1 y 2) dy
13
y
3 6y 4) dx 1
0 1
3 61 0 4 8 dx
4 y 2 and x 21. Solve for x: x 8x
2 32 (3 1 6 2 x
2 0 4
3 16 64
3 21 1
3 y2 Now solve
at y
0 3y 2 (2
1 4 3y 2: y 2 2 1, so the curves intersect 1. Use the region’s symmetry: 1 2 3y 2. 2 0 (1 y 2) dy y 2) dy 4y 1 2 0 2y 2) dy (2 13
y
3 1 41
0 1
3 0 8
3 Section 7.2
22. Solve for y: y 4x 2 Now solve 4
x4 4x 2 4x 2 and y 4 x4 (x 2 5 x4 297 27. 1. 1: 1)(x 2 5) 0.
[ 1.5, 1.5] by [ 1.5, 1.5] The curves intersect at x = 1.
The curves intersect at x 0 and x 1. Use the area’s Use the region’s symmetry:
symmetry: 1 2 0 4x 2) [(4 1)] dx 1 0 ( x4 2 4
3 6 4 2
y and x
y2 2
:y
4 2 0 2 2. 0 2 3 0 y 2 /4 2 y2
dy
4 2 y3
4 2(6
(2 sin x 0.273 [sec2 x (sec2 x 1)] dx /4 dx 2x 2 0 0 (tan2 y tan2 y) dy /4 4 /4
0 0 8 41 1
cos 2x
2 2 cos x
1
2 2 4 /2 3 sin y 30. cos y dy 2 (8 cos x 0 sec x) dx 2 8 sin x tan x 0.858
2
(cos y)3/2
3 3 2
3 30 /3 y 3 and x 31. Solve for x: x 0 4 4 0 1
2 0 2 y 0 25. Use the region’s symmetry:
/3 tan2 y dy 0 4 tan y 0 2 tan2 x) dx 2 2) sin 2x) dx 2 3y 2
dy
4 3 0 2 3y 24. 0 29. Use the region’s symmetry: Use the region’s symmetry:
2 (sec2 x /4 4, so the curves intersect at y 0 /4 y2
.
4 2 y2 Now solve 3 1 28. Use the region’s symmetry, and simplify before integrating: 14
15 3 12
x
2 2 0 0 x
2 cos 1
2 2 5x 5 23. Solve for x: x 2 2 1 /4 104
15 x dx 5) dx 43
x
3 1
5 2 0 4x 2 15
x
5 x
2 sin 2
1 2 (x 4 /2
0 2 y. 0 2[(4 3 3) 0] 6 3 26.
[ 1.5, 1.5] by [ 1.5, 1.5] The curves intersect at x [ 1.1, 1.1] by [ 0.1, 1.1] 1 The curves intersect at x
cross at x
1 2 cos 2x 13
x
3 2 21 1
3 1 1, but they do not 0.
x2 0 0 and x 2 x
2 sin
0 dx
x
2 1
0 4
3 4 0.0601 symmetry: 2 ( y
0 y 3) dy 0 and x
1
2 2 y2 1. Use the area’s
14
y
4 1
0 1
2 298 Section 7.2 32. (b) The two areas in Quadrant I, where x
c y, are equal: 4 y dy y dy 0 c 4 2 3/2 c
2 3/2
y
y
3
3
0
c
2 3/2
2 3/2
2 3/2
c
4
c
3
3
3 [ 0.5, 2.5] by [ 0.5, 1.5] y 1
intersect at x
x2 x and y
1 2 x dx
0 1 12
x
2 1
dx
x2 1. Integrate in two parts: 1 1
x 0 1
2 1
2 8 c 3/2 1 ( 1) 33. The curves intersect when sin x 2c 3/2 2 1 4
42/3 c cos x, i.e., at x 4 . 24/3 (c) Divide the upper right section into a (4 c)by c /4 /4 (cos x sin x) dx sin x rectangle and a leftover portion: cos x 0 0 2 1 0.414 c 2 x 2) dx (c
34. (4 c) c
c 13
x
3 cx c c
[ 3, 3] by [ 2, 4] c c3/2 4c 2 2 0 1) dx 2 2 3/2
c
3 0 13
x
3 2 4x 8
3 (b) Solve y x 2 for x: x 3 yintercepts are
3 y dy 16
3 4 c3/2 c 4 2 1 20
35. (a) 0
3 32
3 y. The c 42/3 5
y=4 2
(3
3 36. 3 y)3/2
1 16
3 32
3 y = x2
(2, 4) (–2, 4) [ 1, 5] by [ 1, 3] The key intersection points are at x 0, x 1 and x Integrate in two parts:
1 1 (–√c, c) 0 (√ c , c ) 3 ( x2 c, then x c, c) and ( c, c). x
dx
4 x x If y 1 3/2
c
3 24/3 y y=c c 1 and 3. 3 2 1 3/2
c
3 4c 0 8
3 28 c 16
3 c3/2 2 4 4 3/2
c
3 x 2) dx (4 2 c 2. 2 x2 (3 4x 0 8 Use the region’s symmetry: 13
x
3 3/2 4 1 3/2
c
3 3/2 (a) The curves intersect at x x 2) dx (4 0 1 2
3 x c. So the points are x2
8 2 3/2
x
3
1
8 4
1 2 x
dx
4 x 1 4 x2
8 x 0 (8 2) 4 4
1 1
8 11
3 4. 299 Section 7.2
37. First find the two areas. 43. First graph y 1
2 For the triangle, (2a)(a 2)
a For the parabola, 2 (a 2
The ratio, then, is 2 a 2x 13
x
3 a
0 43
a
3 3
, which remains
4 a 43
a
3 a3
x 2) dx 0
3 x 2. cos x and y [ 1.5, 1.5] by [ 0.5, 1.5] constant as a The curves intersect at x
0.8241 approaches zero. 2 0.8241. Use NINT to find x 2) dx (cos x 1.0948. Multiplying both 0
b 38. functions by k will not change the xvalue of any b [2f (x) f (x)] dx a f (x) dx, which we already know
a intersection point, so the area condition to be met is equals 4. 0.8241 2 2 kx 2) dx (k cos x
0 39. Neither; both integrals come out as zero because the
1to0 and 0to1 portions of the integrals cancel each
other. ⇒2
⇒2 k(1.0948) 40. Sometimes true, namely when dA [ f (x) g(x)] dx is
always nonnegative. This happens when f (x) g(x) over
the entire interval. ⇒k 1.8269. 0.8241 k2 x 2) dx (cos x
0 44. (a) Solve for y:
41. x2
a2 y2 The curves intersect at x
symmetry:
2
0 2x
x2 1 x 3 b2 1 y [ 1.5, 1.5] by [ 1.5, 1.5] 1 y2
b2 dx 0 and x ln 4 14
x
4 1
1
2 2 ln 2
1
2 x2
a2 1. Use the area’s
a 2 ln x x2
a2 b1 b1 1 x2
dx or 4
a2 a
a 2 0 0.886 x2
a2 b1 (b)
2 1 a dx or a
0 x2
dx
a2 b1 (c) Answers may vary.
a (d, e) 2 42. x2
a2 b1 b1
a x2
dx
a2 2b x
2 1 x2
a2 a
2b sin 1 (1)
2 a a
x
sin 1
2
a a a
sin 1 ( 1)
2 ab
[ 1.5, 1.5] by [ 1.5, 1.5] The curves intersect at x
0.9286 to find 2 (sin x 0 and x x 3) dx 0.9286. Use NINT 0.4303. 45. By hypothesis, f (x) g(x) is the same for each region, where f (x) and g(x) represent the upper and lower edges. 0 b But then Area [f (x)
a each. g(x)] dx will be the same for 300 Section 7.3
Quick Review 7.3 1
:
2 46. The curves are shown for m 1. x 2
x 2. s x2
.
2 s2 , so Area
2 3. 12
x2
r or
2
2 4. 1
2 [ 1.5, 1.5] by [ 1, 1] In general, the intersection points are where
which is where x x
x2 mx, 1 (1/m) 1 x
x2 0 1) 1
m 1
m
m 1 1 0 2x m ln (m) 2x 1. s Section 7.3 Volumes x (pp. 383–394)
b Exploration 1 x and h
1
bh
2 12
x
2 (2x)2 xk2. 3xk Area 2. Unrolling the cylinder, the circumference becomes one
dimension of a rectangle, and the height becomes the other.
The thickness x is the third dimension of a slab with
dimensions 2 (xk 1) by 3xk xk2 by x. The volume is
obtained by multiplying the dimensions together.
3 3. The limit is the definite integral 2 (x 1)(3x x 2) dx. 15
x, so
2 15 2
x.
4 Volume by Cylindrical Shells 1. Its height is f (xk) x2
.
4 8. 1 ln 1
bh
2 , so Area 32
x. 4 x2
.
2 1
bh
2 x, so Area
x 1
bh
2 x, so Area 2 (1/m) 1 12
mx
2 h 2 1. Then, mx dx 1 1
2 ln (x 2
2 h 3 x and h 7. b 0 or else x 5. b
6. b 1
m because of symmetry, the area is
2 d2
x2
or
2
8 9. This is a 345 right triangle. b
Area 1
bh
2 4x, h 3x, and 2 6x . 10. The hexagon contains six equilateral triangles with sides of 0 4. length x, so from Exercise 5, Area 45
2 Exploration 2 Surface Area b 1. 2y 1 (a) A r 2, where r [a, b].
(b) A
sin x, so dy
dx b
a 2 sin x x, so (d) A
cos2 x dx 1 dy
dx 1
2x
12 4 2
0 s 2, where s w2
2 w
, so A(x)
2 w2 w, so A(x)
w , so A(x) 2 14.424. 0 3. y s 2, where s (c) A cos x and dy 2
dx
dx 2y 1 x1
2 x 3 32
x.
2 1. In each case, the width of the cross section is
w 2 1 x 2. The limit will exist if f and f are continuous on the interval 2. y 32
x 4 Section 7.3 Exercises dy 2
dx
dx a 6 and
dx 36.177. A(x) 4(1
w2 (1
x 2).
2(1 2 32
w (see Quick Review Exercise 5), so 4 3
4 (2 1 x 2)2 3(1 x 2). x 2). x 2). Section 7.3
2. In each case, the width of the cross section is w
w
, so A(x)
2 r 2, where r (a) A
(b) A s 2, where s (c) A s 2, where s w, so A(x)
w 2 x. w2 x. 2 w 2 8. A cross section has width w 4x.
/3 w2 , so A(x) 2 V 2x. /3 4 2 /3 32
w (see Quick Review Exercise 5), so (d) A 3
4 x)2 (2 4 4 x2 2x dx
0 4 x and area 2 2 2 w s2 (sec2 x tan x 2 sec x tan x r x 2) (2 1 x ) dx (x 2x 2 2 2 2x x ) . The volume is 2 23
x
3 w2 21 1 4 x 2) dx (1 1 1 2 x 2 and area 4x 2 x ) dx 1 2 2 (1 1 1
1 16
.
3 x 2 and area x ) dx 2x 1 7. A cross section has width w
32
w (a) A(x) 13
x
3 13
x
3 1
1 8
.
3 2 sin x. 54
y dy
4 y5 4 8.
0 10. A cross section has width w
12
s
21 12
w
2 1 y 2 and area y 2). The volume is 2(1 y 2) dy 2 2y 13
y
3 1
1 8
.
3 11. (a) The volume is the same as if the square had moved
without twisting: V Ah s 2h.
(b) Still s 2h: the lateral distribution of the square cross
sections doesn’t affect the volume. That’s Cavalieri’s
Volume Theorem. from y 3 sin x dx 12
2 6 to y 12, for a diameter of 6 and a radius of 3, the solid has the same cross sections as the right 0 3 54
y . The volume is
42 2 12. Since the diameter of the circular base of the solid extends 3 sin x, and 4 V w2 2(1 x 2). The volume is 2(1 2 5y 2 and area 9. A cross section has width w 1 2 w 21 2
.
3 x 1 6. A cross section has width w tan x)2 dx, which by same method as (sec x x 2). The volume is 4(1 x 2) dx 2(1 6 tan x)2, and (sec x in part (a) equals 4 3 1 0 1 2 /3 r2 5. A cross section has width w w2 /3 V 1) dx 16
.
15 s2 3 6 . 6 s2 22 (1 4 15
x
5 x2 (b) A(x) w2
2 2 1 A(x) /3 2 1 4(1 /3 1
x
2 sec x 3 2 16. 22 s2 x
/3 2 and area A(x) 1 tan x 2x. The volume is 4. A cross section has width w A(x) tan2 x) dx 2 sec x tan x /3 3 (1 tan x)2, and (sec x tan x)2 dx (sec x 0 1 4 tan x. /3 3x. 3. A cross section has width w
A(x) 4 4 A(x) w2
2 r2 (a) A(x) sec x 301 sin x dx circular cone. The volumes are equal by Cavalieri’s 0 3 Theorem. cos x
0 2 s2 (b) A(x)
V 13. The solid is a right circular cone of radius 1 and height 2. 3.
w2 4 sin x dx
0 V 4 sin x, and
4 sin x dx
0 4 cos x 8. 1
Bh
3 1
( r 2)h
3 1
( 12)2
3 2
3 14. The solid is a right circular cone of radius 3 and height 2. 0 V 1
Bh
3 1
( r 2)h
3 1
( 32)2
3 6 302 Section 7.3 15. A cross section has radius r
r2 A( y )
1 tan2 4 0 tan2
y dy 4
4 tan 4 y and area 19. y . The volume is
1 tan 4 4 y y
[ 6, 6] by [ 4, 4] 0 1
The solid is a sphere of radius r 4 . 43
r
3 16. A cross section has radius r sin x cos x and area A(x) r2 from x 3. The volume is 36 . 20. sin2 x cos2 x. The shaded region extends 0 to where sin x cos x drops back to 0, i.e., where
[ 0.5, 1.5] by [ 0.5, 0.5] x 2 2 cos2 x . Now, since cos 2x 1, we know The parabola crosses the line y 1 cos 2x
cos x
and since cos 2x 1 2 sin2 x, we
2
/2
1 cos 2x
know sin2 x
.
sin2 x cos2 x dx
0
2
/2
1 cos 2x 1 cos 2x
dx
0
2
2
2 /2 4 (1 0 /2 4
8 1 0 x /2 cos2 2x) dx
cos 4x
dx
2 1
sin 4x
4 4 0 0 x(1 x) r2 (x 2 2x 3 x 4). 14
x
2 15
x
5 The volume is
1 cos 4x) dx 1. A cross x and area (x 2 2x 3 13
x
3 x 4) dx 0 (1 0 or x 2 x 2)2 (x 0 when 0, i.e., when x
x A(x) /2 8 x section has radius r sin2 2x dx /2
0 x 2 1
0 30 . 21. 2 8 2 0 0 16 . 17. [ 1, 2] by [ 1, 2] Use cylindrical shells: A shell has radius y and height y.
The volume is
1 2 ( y)( y) dy [ 2, 4] by [ 1, 5] 2 0 A cross section has radius r
A(x)
2 r2 0 1 2
.
3 0 22. x4. The volume is
15
x
5 x 4 dx x 2 and area 13
y
3 2
0 32
.
5
[ 1, 3] by [ 1, 3] 18. Use washer cross sections: A washer has inner radius r
outer radius R
1 3 x 2 dx The volume is [ 4, 6] by [ 1, 9] 0 A cross section has radius r
A(x)
2
0 x 6 dx r2 x 3 and area x 6. The volume is
17
x
7 2
0 128
.
7 (R 2 2x, and area A(x)
3 13
x
3 r 2) 1 .
0 3 x 2. x, Section 7.3
23. 303 26. [ 2, 3] by [ 1, 6] [ 1, 5] by [ 3, 1] The curves intersect when x 2 1 x 3, which is when
x 2 x 2 (x 2)(x 1) 0, i.e., when
x
1 or x 2. Use washer cross sections: a washer has
inner radius r x 2 1, outer radius R x 3, and area
A(x)
(R 2 r 2)
[(x 3)2 (x2 1)2]
( x 4 x 2 6x 8). The volume is
2 ( x4 x2 6x The curves intersect where
x
2, which is where
x 4. Use washer cross sections: a washer has inner radius
r
x, outer radius R 2, and area
A(x)
(R 2 r 2)
(4 x).
4 The volume is (4 x) dx 4x 0 12
x
2 4 8
0 27. 8) dx 1 15
x
5 2 13
x
3 32
5 8
3 3x 2 8x
1 12 1
5 16 1
3 3 8 [ 0.5, 1.5] by [ 0.5, 2] 117
.
5 The curves intersect at x 24. radius r 2
r2 A(x) The curves intersect when 4 x
2 x, which is when
x2 x 2 (x 2)(x 1) = 0, i.e., when x
1 or
x 2. Use washer cross sections: a washer has inner radius
r 2 x, outer radius R 4 x 2, and area
A(x)
(R 2 r 2)
[(4 x 2)2 (2 x)2]
(12 4x 9x 2 x 4). 4x [ 1, 3] by [ 1, 3] The curve and horizontal line intersect at x
9x 2 section has radius 2 x 4) dx 1 r2 A(x)
12x
24 2x 2
8 2.301. 28. The volume is
(12 sec x tan x)2. Use NINT to find 0 2 2 2 sec x tan x)2 dx (2 [ 2, 3] by [ 1, 5] sec x tan x and area
( 0.7854 0.7854. A cross section has 15
x
5 3x 3
24 sin x)2 4 (1 4 (1 2 sin x The volume is 32
5 /2 12 2 3 1
5 2 sin x sin2 x) dx 2 cos x 4 (1 108
.
5 1
sin 2x
4 0 4 25. 4 3
x
2
3
4 2 (3 /2
0 8) 29.
3 , 3 by [ 0.5, 2] Use washer cross sections: a washer has inner radius
r sec x, outer radius R
2, and area
A(x)
(R 2 r 2)
(2 sec2 x).
The volume is
/4 (2 sec2 x) dx [ 1, 3] by [ 1.5, 1.5] A cross section has radius r
/4 2x A(y) tan x /4 2
2 2. 2 The volume is
1 5y 2 and area 5 y 4.
1 /4 1 r2 5 y 4 dy 1 . A cross 2 sin x and area 2
1 2 y5 1 2.
1 sin2 x). 304 Section 7.3 30. 35. [ 1, 4] by [ 1, 3] [ 1, 5] by [ 1, 3] A cross section has radius r
r2 A(y)
2 The curved and horizontal line intersect at (4, 2). y 3/2 and area (a) Use washer cross sections: a washer has inner radius y 3. The volume is
14
y
4 y 3 dy 0 2 r 4.
0 x, outer radius R A(x) 31. (R 2 2, and area 2 r) (4 4 (4 0 x) dx x). The volume is
12
x
2 4x 4 8
0 y 2 and area (b) A cross section has radius r
[ 1.2, 3.5] by [ 1, 2.1] r2 A(y) y 4.
2 Use washer cross sections. A washer has inner radius r
outer radius R y (R 2 A(y) 1 ( y2 0 2 32
.
5 0 1, and area r 2) volume is 15
y
5 y 4 dy The volume is 1, 1)2 [(y 13
y
3 2y) dy 0 (c) A cross section has radius r (y2 1]
y2 1
0 2 x and area 2y). The
r2 A(x) 4
.
3 x)2 (2 (4 4 x x). The volume is 32. 4 (4 4 x x) dx 8 3/2
x
3 4x 0 12
x
2 4
0 8
.
3 (d) Use washer cross sections: a washer has inner radius
[ 1.7, 3] by [ 1, 2.1] r
Use cylindrical shells: a shell has radius x and height x. The
1 volume is 2 (x)(x) dx 2 0 13
x
3 1
0 y 2, outer radius R 4 (R 2 A( y ) 2
.
3 (8y 2 r 2) 4, and area [16 83
y
3 15
y
5 y 2)2] (4 y4). 33.
The volume is
2
0 (8y 2 y 4) dy 2 224
15 0 36. [ 2, 4] by [ 1, 5] Use cylindrical shells: A shell has radius x and height x 2.
2 2 (x)(x 2) dx The volume is 2 0 14
x
4 2 8.
0 [ 1, 3] by [ 1, 3] 34. The slanted and vertical lines intersect at (1, 2)
(a) The solid is a right circular cone of radius 1 and
height 2. The volume is
[ 0.5, 1.5] by [ 0.5, 1.5] 1
Bh
3 The curves intersect at x 0 and x shells: a shell has radius x and height
1 is 2 (x)(
0 x x) dx 2 2 5/2
x
5 1. Use cylindrical
x
13
x
3 x. The volume
1
0 2
.
15 1
( r 2)h
3 1
( 12)2
3 2
.
3 (b) Use cylindrical shells: a shell has radius 2
height 2x. The volume is
1 1 2 (2 x)(2x) dx 4 0 (2x x 2) dx 0 4 x2 13
x
3 1
0 8
.
3 x and 305 Section 7.3
40. 37. [ 2, 2] by [ 1, 2] [ 2, 2] by [ 1, 3] The curves intersect at ( 1, 1). x2 (a) A cross section has radius r
r2 A(x) 1 x and area (1 x 2)2 (1 2x 2 2 x 4). 2x 2 (1 1. A shell has radius x and height x 2. The volume is x 2 (x)(2 x 2) dx x 13
x
3 x2 2 0 x 4) dx 23
x
3 x 1 1 5
.
6 0 16
.
15 1 y and
[ 1, 5] by [ 1, 3] height 2 y. The volume is
1 1 y)(2 y) dy 14
x
4 41. 1 15
x
5 (b) Use cylindrical shells: a shell has radius 2 2 (2 x at x 1 The volume is
1 2 2 4 (2 0 y y 3/2) dy 0 4 3/2
y
3 4 2 5/2
y
5 1 A shell has radius x and height
4 56
.
15 0 2 (x)( x) dx 2 5/2
x
5 2 0 x. The volume is 4 128
.
5 0 42.
(c) Use cylindrical shells: a shell has radius y 1 and height 2 y. The volume is
1 1 2 (y 1)(2 y) dy ( y3/2 4 0 2 5/2
y
5 4
38. (a) A cross section has radius r
r2 A(x)
b y) dy 0 2 h1
0 h2 1
x2
dx
b 2 3/2
y
3 1
0 [ 2, 2] by [ 2, 2] 64
.
15 x
and area
b h1 x2
. The volume is
b
b
b
x3
h2
1
bh 2.
3
3
b0 The functions intersect where 2x x
. The volume is
b
b
x
2 (x)h 1
dx 2 h
b
0 x (2x 1) x 2x x2
dx
b
b
x3
b 2h.
3
3b 0 x
0 1
2 2 h x2 1. The volume is 1 1 2 (x)( x 2x 1) dx (x 3/2 2 0 2x 2 x) dx 23
x
3 12
x
2 0 2 5/2
x
5 2
7
.
15 43. A shell has height 12( y 2
b x, i.e., at x A shell has radius x and height (b) Use cylindrical shells: a shell has radius x and height
h1 1 (a) A shell has radius y. The volume is
1 2 ( y)12( y 2 y 3) dy 1 ( y3 24 0 y 4) dy 0 24
(b) A shell has radius 1
1 y)12( y 2 2 (1
0 1 ( y4 24 14
y
4 15
y
5 24 2 3
2 2 (x) x dx The volume is
0 2y 3 y. The volume is
y 3) dy
y 2) dy 0 [ 2, 3] by [ 2, 3] x
2 x3 3
x.
2 2 8.
0 0 y 3). 39. A shell has radius x and height x 1 15
y
5 14
y
2 13
y
3 1
0 4
.
5 1
0 6
.
5 1. 306 Section 7.3 43. continued 45. (c) A shell has radius
1 2
0 8
5 8
5 y. The volume is y 12( y 2
1 24 y4 0 15
y
5 24 y 3) dy 2 y 0
1 24
0 24 The functions intersect at (2, 8).
(a) Use washer cross sections: a washer has inner radius A(x) 2
12( y 2 y 3) dy
5
22
33
y4
y
y dx
5
5
1
15
34
23
y
y
y
2.
5
20
15 0 y2
44. A shell has height
2
y2
y4
y2
.
2
4 2 2 ( y) y 2 0 (b) A shell has radius 2
2 2 (2
2
0 2 16
y
24 16
y
24 y4
y 3 2y 2 dy
2
14
23
15
y
y
y
4
3
10 2
0 2
0 2 y5
4 16
y
24 2 y
2 2
0 2 5
8
y5
4
16
y
24 512
.
21 0 y and 8
.
3 46. The functions intersect at (0, 0) and (1, 1).
2 8
.
5 0 y. The volume is 5y 4
4
15
y
4 y3 (a) Use cylindrical shells: a shell has radius x and height
x2 1 2 (x)(x x x 2. The volume is x
x 2) dx 2 0 53
y
3 13
x
3 y2 height 2x 2 8. x 2 1 2 (1 0 5
. The volume is
8 5y 4
32 x)(x x x 15
y
32 14
y
4 5y 2
dy
8
53
y
24 0 6 . x 2) dx 2 4.
0 x and x . The volume is
1 (x 3 2 2x 2 x) dx 23
x
3 12
x
2 0 6 dy
y3 1 2 14
x
4 2 4 y
4 14
x
4 (b) Use cylindrical shells: a shell has radius 1 5y 2 dy 14
y
4 (d) A shell has radius y 0 2 y
. The volume is
4
8
y
2 (8 y) y 1/3
dy
4
0
8
y2
2
8y 1/3 2y y 4/3
dy
4
0
8
3 7/3
832
13
2 6y 4/3 y 2
y
y
.
7
21
12 0 0 2 17
x
7 [ 0.5, 1.5] by [ 0.5, 1.5] y4
dy
4 y) y 2 0 2 16 3
x
3 height y 1/3 2x 2 (5 x 6). The volume is dy (c) A shell has radius 5
2 (16x 2 y. The volume is y
4 y) y 2
y5
4 r) 4 0 2 14
y
4 2 (16x 2 2 (b) Use cylindrical shells: a shell has a radius 8 y4
4 y4
dy
4 4x, and area x 6) dx (R 2 0 (a) A shell has radius y. The volume is
2 x 3, outer radius R r 2
. The volume is
5 (d) A shell has radius y
1 [ 1, 3] by [ 1.4, 9.1] 13 3
82
y
y dy
5
5
1
13 4
83
y
y
2.
20
15 0 . 1
0 307 Section 7.3
47. (a) Solve
d
dc 2 22 c 4c
2 2 0 2 c 4 0 [ 0.5, 2.5] by [ 0.5, 2.5] 1
1
, 1 , , 2 , and (1, 1).
4
4 The intersection points are and area (R
2 1
y4 1 1
y4
1
3y 3 2 r) 1
dy
16 1 (b) A shell has radius x and height 1. The volume is 2 0
2 c 1
1
, outer radius R
,
4
y2
1
. The volume is
16
2
11
1
y
.
48
16 1 (a) A washer has inner radius r
2 c This value of c gives a minimum for V because
a 2V 2 dc 2 2 0.
2
22 Then the volume is 1 2 (x)
1/4 1 dx 2 3/2
x
3 2 x 48. (a) For 0 x x(sin x)
x , x f (x) For x 0, x f (x) 01 x f (x) sin x for 0 1
1/4 2 anywhere else besides c 11
.
48 2 4 , the maximum must occur
2 at c sin x. sin 0 x 12
x
2 2 2 (b) Since the derivative with respect to c is not zero x
1 2 2 4 0 or c and for c sin x. So 1. The volume for c
(3 1 it is 8) 0 is 2.238. c 2 2 4.935, 0 maximizes the volume. . (c) (b) Use cylindrical shells: a shell has radius x and height y.
The volume is 2 xy dx, which from part (a) is
0 2 sin x dx 2 cos x 0 49. (a) A cross section has radius r
r2 A(x)
6 144 0 (b) [0, 2] by [0, 6] 4. The volume gets large without limit. This makes sense,
since the curve is sweeping out space in an
everincreasing radius. 0 144 (36x 2 (36x 2 x 4) dx x
12 36 x 2 and area x 4). The volume is
144 36
cm3 (8.5 g/cm3)
5 15
x
5 12x 3 6
0 51. (a) Using d
36
cm3.
5 C d2
2 , and A C2
yields the
4 following areas (in square inches, rounded to the
nearest tenth): 2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7, 192.3 g 10.7, 9.3, 6.4, 3.2. 50. A cross section has radius r c sin x and area
(b) If C( y) is the circumference as a function of y, then the r2 A(x) sin x)2 (c (c 2 2c sin x sin2 x).
area of a cross section is The volume is C(y)/
2 A(y)
(c2 sin2 x) dx 2c sin x and the volume is 0 1
x
2 c2x 2c cos x 2 1
2 c 22 c 2c 1
sin 2x
4 2c 0 (c) 1
4 6 2 . 0 5.12
11.62 C 2( y)
,
4
6 1
4 C 2( y) dy. 0
6 A( y) dy 2 4c 2 1
C 2( y) dy
40
16 0
[5.42
24
4 6.32
10.82 7.82 9.42 9.02) 2(4.52
10.82 6.32] 4.42
11.62 34.7 in.3 308 Section 7.3 52. (a) A cross section has radius r
5 r2 2 y. The volume is 2 y dy
0 dV
dt
dh
so
dt For h
dh
so
dt 1
8 0 and x a. For revolution about the xaxis, a cross section has radius
x 2 and area ax A(h). r2 A(x) dh
,
dt x 2)2 (ax (a 2x 2 2ax 3 x 4). The volume is
a 8, units3
3
3
=
sec
8 53. (a) 25 .
0 4, the area is 2 (4) 0: This is true at x 5 r dV
A(h) dh, so
dh
dV dh
A(h)
dh dt
dV
1
A(h) dt (b) V(h) y2 x2 55. Solve ax 2y and area (a 2x 2 2ax 3 123
ax
3 x 4) dx 0 units3
.
sec 14
ax
2 15
x
5 a
0 1
a 5.
30 y For revolution about the yaxis, a cylindrical shell has
2
r x 2. The volume is radius x and height ax
x a x 2) dx 2 (x)(ax 13
ax
3 2 0 The remaining solid is that swept out by the shaded
region in revolution. Use cylindrical shells: a shell has
r2 radius x and height 2 . x 2. The volume is 1
a5
30 14
1
a yields a
6
30 2 (x)(2 r 2 2 x 2) dx 22 1
, so a
6 5. 56. The slant height s of a tiny horizontal slice can be
x2 written as s
r2 0 14
a.
6 Setting the two volumes equal, –2 .r a 14
x
4 y2 ( g ( y))2 y. So the 1 surface area is approximated by the Riemann sum
n 22
(r
3
4
( 8)
3 Σ2
k1 r 2 3/2 x) r2 4 32
.
3 integral.
dx
dy 57. g ( y) (b) The answer is independent of r. (g ( y))2 y. The limit of that is the g( yk) 1 1 2 2 y1 0 , and 2y
1 2 2 dy 4y and measure the radius r (x) of the shadow region at these 6 points. Then use an approximation such as the trapezoidal (4y 13
3 b rule to estimate the integral
a r 2(x) dx. dx
dy 58. g ( y)
1 2
0 y3
3 3 2 1 dx
dy y1/2 1
3 2
1 1)3/2 2
0 13.614 y 2, and
2
3 ( y 2)2 dy 9 59. g ( y) 1 dy 0 2y 54. Partition the appropriate interval in the axis of revolution y1/2 1
(1
6 (2 2 y 4)3/2 0 1) 1 1/2
y
, and
2
1 3/2
3
1 3/2
3 1 1/2 2
y
dy
2 1
1 1
dy.
4y Using NINT, this evaluates to 1 16.110 0.638. Section 7.3 309 66. Use washer cross sections: a washer has inner radius
dx
dy 2y 2y 11 60. g ( y) 1 , and 1 2 r 1 5/8 area (R2 2 1 dy 2y 1 2y dy 2
4 2 3/2
y
3 2
2 dy
dx 61. f (x)
2 2 x2 a 1 a2 y 2)2 5/8 y 2)2] 4 b a2 a2 a y 2 dy 4b y 2 dy a a2
2 2.997. 22 2 ab 2x, and
67. (a) Put the bottom of the bowl at (0, 2 (2x)2 dx 1 2 x2 0 4x2 dx evaluates, 1 using NINT, to (a2 y 2) dy 13
y
3 a2y a 2 2 (3x x) 1 2 (3 2x) dx evaluates, using NINT, (52 0 to dh
dt 44.877.
dy
dx 63. f (x) 1 2 x
x2 2x 1.5 1 0.5 dx x2 2 1 dx
0.5 0.5 1
2 2 5 3/2
4 2
x
3 2
4
3 a 6.283 25 3/2 x Since A1 (a2 x 2) dx 49
3 x 2 and area ( a2 13
x
3 a2x x2 r2 1 1 x 2). The a
a 13
a
3 a3 13
a
3 h 2 volume of cone y
and area
h r1 y2
h r2 1 2y
h y2
. The
h2 volume is 51.313
22 h) A1. A2 23
R.
3 a2 x 2)2 a A( y ) theorem. Thus, volume of hemisphere R3 a2 5 volume of cylinder m/sec. 43
a.
3 A2, the two volumes are equal by Cavalieri’s 13
R
3 ( dx Right circular cylinder with cone removed cross sectional
h2 r2 1 65. Hemisphere cross sectional area: ( R area: R2 1
120 (b) A cross section has radius x 9 3/2
4 4 1
(0.2)
24 2 5
dx
4 x
1 1 dV
A dt a3 1 11 5 . 24 . The rate of rise is 1 1 2 a h) 1 and the area of a cross section is , and x 5 x h2(3a
3 x 2 2 ha volume is 1.5 2 dy
dx 12) A(x) 1.5 2 x 2x 64. f (x) 4, y 68. (a) A cross section has radius r , and x2 1 2x y 2). 2x, and
(b) For h 3 ( a2 The volume for height h is 53.226. 3 y 2)2 horizontal cross section is ( a ha dy
dx a). The area of a
2 0 62. f (x) y 2, and y 2. The volume is 4b 5
2 a2 b a2 (b a 5
16 1 3 a2 4b 5/8 y 2, outer radius R r 2) [(b 1 2 a2 b 0 r2 1 2y
h y2
dy
h2 r2 y
12
r h.
3 y2
h y3
3h2 h
0 310 Section 7.4 s Section 7.4 Lengths of Curves 3. (a) x (pp. 395–401) cos2 y dy. 1 cos y, so Length
0 (b) Quick Review 7.4
1. 1 1 x 1
2. x)2, which, since x (1 x2
4 1, equals x2
, which, since x
2 1 1 (tan x)2
equals sec x. [ 1, 2] by [ 1, 4] 2, equals (c) Length 2x
x
or
.
2
2 1
3. x2
1. 2x
x or x 4. (a) x (sec x)2, which, since 0 x 2 3.820
y 2) y(1 , 1/2 , so 1/2 Length y2 1
1/2 4. 12
x x
4 1
since x 5. 1 1
2 12
x
16 x2 0, equals 4
4x (x 2 1
4 1
x2 which, x 2 2 10
3 3 [ 1, 2] by [ 1, 1] (c) Length 6. f (x) has a corner at x
d
7. (5x 2/3)
dx (b) , 2 cos x. equals 8. d
(
dx is undefined at x 2y 2x 2 2y 1 y 0. f (x) has a cusp x2 1
is undefined for x
5(x 3)4/5 10. d
1
dx 4x
3 1)2 (y 4 x 2 Length 3. 2x 2, and
1 1. Then y 2x 1 1 2x 1 2 , and
2 dx. (b) 2 has a corner at x 2. cos x
is undefined for x
3(sin x)2/3 sin x 2x 7 f (x) has a vertical tangent there.
9. 1, so x y
x 3) 1.047 5. (a) y 2 4. there.
5 dy. . 2 cos2 x, which, since cos 2x 4)2
x2 y2 1 k,
[ 1, 7] by [ 2, 4] where k is any integer. f (x) has vertical tangents at these (c) Length 9.294 values of x.
6. (a) y Section 7.4 Exercises cos x x sin x Length cos x x sin x, so x 2 sin2 x dx. 1
0 1. (a) y 2x, so
2 Length 1 (b) 2 (2x)2 dx 4x 2 dx. 1 1 1 (b)
[0, by [ 1, 4] (c) Length
7. (a) y [ 1, 2] by [ 1, 5] (c) Length
2. (a) y 6.126
2 4.698 /6
0 (b) y
0 sec x, so Length 1 tan x dx ln ( sec x ) 4 sec x dx. /3 (b) 0,
3 , 0 by [ 3, 1] (c) Length 2.057 1 tan x, so Length 6 (c) Length by [ 0.1, 0.2] 0.549 tan2 x dx. Section 7.4
/4 sec2 y 8. (a) x 1, so Length sec y dy.
/3 (e x 10. y e x)
2 3 , so Length ex 1 x2 e dx. 2 3 (b)
sec2 y (b) x 1 tan y , ln (cos y)
ln (cos y) y 3 so x
0 y 0
4 [ 3, 3] by [ 2, 12] (c) Length
12
(x
2 11. y
3 1
0 [ 2.4, 2.4] by (c) Length 2 , (x x 2 sec x tan x, so
/3 1 2 sec x tan x dx. 3
2 3
2 1
0 , (c) Length 3 x 12. 4 dx 9x
dx
4
9x 3/2
4
8 1
0 8
1
27
80 10
27 y2 by [ 1, 3] 3
1 3.139 2 1 4
0 . 3 1
, so the length is
4y 2
12
13
y2
dy
y
3
4 y2 12
dy
4 y2 y2 1
1 1
4y 3 53
.
6 1 1
, so the length is
4y 3 y3 14. x 1 dx 0 2 x (b) 13. x 2x 2 x, so the length is /3 3 x4 3 13
x
3 1) dx 4 2 2, so the length is 3
0 (x 2 2 x x2 2)2 dx 0 2.198 Length 2)1/2(2x) 3 12. y
9. (a) y 20.036 2 12
dy
4 y3 y3 1 12
dy
4 y3 y3 1 14
y
4 1
8y 2 2
1 123
.
32
y2
2 15. x 1
, so the length is
2y 2 2
1 2 12
dy
2 y2 y2
2 1 1 13
y
6 x2 16. y 2x 1 1
2y 4 (x 4)2 (4x so the length is
2 1 (x 2 4(x 0 1
(x
3 1)3 2 4(x 1)2
2 1
1) dx 1)2 4(x
1 1)2 (x 2 1 1)2 0 0 12
dy
2y 2 y2
2 dx
53
.
6 2
1 1)2 17
.
12
1
4(x 1)2 311 312 Section 7.4
sec4 y 17. x
/4 /4 (sec4 y 1 24. The area is 300 times the length of the arch. 1, so the length is
1) dy /4 y sec2 y dy /4 x
, so the length is
50
2
x
sin2
dx, which evaluates, using NINT,
2
50 sin 2
25 /4 tan y 1 2. 25 /4 to
3x 4 18. y
1 the cost (rounded to the nearest dollar): $38,422. 1 (3x 4 1 73.185. Multiply that by 300, then by $1.75 to obtain 1, so the length is
3x 2 dx 1) dx 2 2 1 1
3 3 x3 73
.
3 2 dy 2
dy
1
1
corresponds to
here, so take
as
. Then
dx
dx
4x
2x 19. (a)
y x C, and, since (1, 1) lies on the curve, C 0.2x y1
10 1 0. x. dx 2
dx
1
1
corresponds to 4 here, so take
as 2 .
dy
dy
y
y
1
Then x
C and, since (0, 1) lies on the curve,
y
1
C 1. So y
.
1x (b) Only one. We know the derivative of the function and
the value of the function at one value of x.
21. y cos 2x, so the length is
/4 5 22.3607, and . NINT fails to evaluate (y1 )2 dx because of the undefined slope at 22.36 (y1 )2 dx 1 2
0 52.548. Then, pretending the last little stretch at each end is a straight line, add
0.2(22.36)2 2 100
track 1 as 0.156 to get the total length of 52.704. Using a similar strategy, find the length of the right half of track 2 to be 2 cos x dx
0 2 sin x 32.274 + NINT Y2 /4 1. 32.274. Now 52.704 and 2 cos 2x dx 0 from the upper limit a little, and find enter Y1 /4 1 5 10 the limits, so use the track’s symmetry, and “back away” (b) Only one. We know the derivative of the function and
the value of the function at one value of x.
20. (a) 0.2x 2 100 5 10 So y 0 at x 25. For track 1: y1 2 0.2t 1
150 0.2t 2 , t, x, 0 and graph in a [ 30, 0] by [0, 60] window to see the effect of 0 the xcoordinate of the lane2 starting position on the length
22. y x 2/3)1/2x (1 1/3 , so the length is of lane 2. (Be patient!) Solve graphically to find the 1 8 1 x 2/3)x (1 2/3 intersection at x dx 2/4 coordinates ( 19.909, 8.410). 1 8 2/3 x dx 24 1 2/3
x
3 26. f (x) 1 8 x 1/3 dx 2 1 2/4 1 3
2 31
22 x 0. So, instead solve for x in terms of y using the quadratic formula. (x1/3)2 6. 23. Find the length of the curve y 20
0 to (f (x))2 dx because of the undefined slope at 2/4 1 x1/3 y 2 1/3
x
, but NINT fails on
3 0 3
2 8 x 2/3
8 19.909, which leads to starting point 3
sin x for 0
20 x 20. 3
3
cos x, so the length is
20
20
2
3
3
1
cos x dx, which evaluates, using NINT,
20
20 21.07 inches. 1
(1
8
3
(1
8 x
x
2 1
2 1/3 4y
4y 2/3
2 y 0, and . Using the positive values, 1)3. Then,
2 1)2
1 1
0 4y x1/3 (x )2 dy , and
4y 3.6142. 313 Section 7.5
(4x 2 27. f (x) 1) (8x 2
(4x 2 1)2 1
1/2 using NINT, to partition goes to zero will always be the length (b 4x 2 8x 1 2
dx which evaluates,
(4x 2 1)2 1 length is 31. Because the limit of the sum Σ xk as the norm of the 4x 2 8x 1
, so the
(4x 1)2 8x) a) of the interval (a, b). 2.1089.
32. No; the curve can be indefinitely long. Consider, for 28. There is a corner at x 0: 1
3 example, the curve sin 1
+ 0.5 for 0
x x 1. 33. (a) The fin is the hypotenuse of a right triangle with leg
lengths xk and
[ 2, 2] by [ 1, 5] Σ
k1 (b) lim
n→ x3
x3 ( xk)2 2
0 x
x n 0
and
1 Σ
k1 lim n→ xk 1 ( f (x))2 dx b 2 3x
3x 2 y 5
5 x
x 1 The length is 0
.
1 (y )2 dy 2 0 (3x 2 1 34. Yes. Any curve of the form y 5)2 dx a 1 (3x 2 5)2 dx, x)2, 0 (1 x (y )2 dx 1 c, where c is a 1, so that a 2 dx 0 0 which evaluates, using NINT for each part, to
29. y x constant, has constant slope 1 2 ( f (xk 1))2 a 2
0
1 f (xk 1) xk. ( f (xk 1) xk)2 1 5x
5x xk
1 n Break the function into two smooth segments:
y df
dx x xk a 2. 0 13.132. s Section 7.5 Applications from Science
and Statistics 1 (pp. 401–411)
Quick Review 7.5
1 1. (a) [ 0.5, 2.5] by [ 0.5, 1.5] x y 1 e into two equal segments by solving
1
. The total length is 2
4 which evaluates, using NINT, to x y 1 1 2. (a) 12 x 1/4 ex 1 e 1 1.718
/2 /2 dx, 1
e
1 0 (b)
3. (a) sin x dx x (b) 2 cos x /4 1.623. 2 /4 0.707
3 30. e x dx 0 1 with 1 0.632
1 0. So, split the curve x 0 (b) , but NINT may fail using y over the entire interval because y is undefined at x x: x dx 0 x y x e 4. (a) (x 2 13
x
3 2) dx 0 3 2x 15
0 (b) 15
2 5. (a)
1 x2
x 3 1 [0, 16] by [0, 2] 1 3/4
x
, but NINT may fail using y over the entire
4 y interval, because y is undefined at x
4 x y ,0
2 1 y 2: x 0. So, use (b) 3 4y and the length is 6. 0.501 2 (x 2) sin x dx 0 0 7 16.647. 1
ln (x 3 1)
3
1
[ln 9 ln 2]
3
1
9
ln
3
2 7 (4y 3)2 dy, which evaluates, using NINT, to dx 7. (1
0 x 2)(2 x) dx 2
1 314 Section 7.5
7 cos2 x dx 8.
7 y2
(10
2 9.
0
7 3 10. 4 0 ks, so 150 k 1
and k
16 1
,F
8 1
8 300 lb. 10. (a) F 0 s y) dy 2400 1/8 sin2 x dx 1/8 1200x 2 2400x dx (b) 2400 lb/in. Then for 18.75 in.lb 0 0 Section 7.5 Exercises
5 1. xe x/3 dx 5 x/3 3e (3 0 0 3 2. x
dx
4 x sin
0 44 x
4 sin 4 22 11. When the end of the rope is x m from its starting point, the 24
e 5/3 x) 9 4.4670 J
(50 x
4 x cos F(x) 0 3. 3
2 (0.624)(50 x 2 dx x9 (esin x cos x 4. 9J
(b) p1V11.4 10 esin x 19 19.5804 J 109,350
F(x) 490 1 490 20 W (490 24.5x) dx 12.25x 2 490x 0 0.4 V 4900 J. 20 4x/5
20 x
25 490 1 490 W (490 19.6x) dx 490x 9.8x 0 k1 F(x) ∑ 124.8yk 144 x/2 x
36 144 1 18 W (144 4x) dx 144x 2x (b) ks, so 800 k(14 10) and k 2 (b) F(x) 200x, and 100x 2 200x dx (c) F
9. (a) F ks, so 21,714 400 in.lb. n Σ (62.4yk) 6
k1 7238 lb/in. (b) 374.4y 1
0 1/2 (b) F(x) 7238x. W 3619x 2 7238x dx
0 1/2
0 1 904.75 905 in.lb, and W 7238x dx
1/2 3619x 2 1 2714.25
1/2 1, it follows that a thin
y2
y, where y is
64 force is approximately 8 5) and k y2
82 distance from the top, and pressure 62.4y. The total 8 in. k(8 3 2
0 1600
200 200s, so s x2
32 horizontal rectangle has area 6 1 200 lb/in. 0 y 2)3/2 0 14. (a) From the equation 0 8. (a) F 41.6(9 1123.2 lb 1944 ftlb 0 9 yk2) y yk2 y. 0 4x lb. 18 2 y 2 dy 3 Then
18 124.8y 5880 J. 144 9 k1 0 18 y 2 y, where y is n 20 7. When the bag is x ft off the ground, the sand weighs 32, it follows that a thin force is approximately ∑ (62.4yk)(2 9
n 2 y2 distance from the top, and pressure 62.4y. The total 19.6x N. Then
20 243 horizontal rectangle has area 2 9 0 490 109,350
and
V1.4 32 V 6. When the bucket is x m off the ground, the water weighs
F(x) p dV (p , V )
11 dV 2.5V 13. (a) From the equation x 2 20 (p , V )
22 pA dx 37,968.75 in.lb 24.5x N. Then 780 J.
0 109,350, so p 5. When the bucket is x m off the ground, the water weighs
x
20 50 (p , V )
11 (p , V )
2 2 109,350
V1.4
(p , V )
11 Work 0 esin 10 (p , V )
22 F(x) dx (50)(243)1.4 2x 0 20 x
490
20 12
x
2 0.624 50x (p , V )
11 0 2) dx x) dx
(p , V )
22 3 x 2)3/2 10 x) N. The total work is 0 12. (a) Work 1
(9
3 0 (0.624)(50 50 3.8473 J
3 x) m of rope still to go weigh 3 2714 in.lb. 1 yk2
64 y2
dy
64 n y Σ 374.4yk
k1
7987.2 1 7987.2 lb yk2 1 y 2 3/2
64 64
8
0 y. 315 Section 7.5 3
y, it follows that a thin
8
3
horizontal rectangle has area y y, where y is the
4 18. The work needed to raise a thin disk is (10)2(51.2)y y, 15. (a) From the equation x distance from the top of the triangle, the pressure is
62.4(y ∑ 62.4(yk ∑ 46.8(yk y 2 3yk) y. k1 46.8(y 22 3y) dy 3494.4 3 ( 210.6) 15.6y 3 70.2y 2 8 5120 0 12
y
2 30 7,238,229 ftlb
0 4 (62.4)( y 12
y
2 6 15y
0 84,687.3 ftlb x2
, it follows that a thin
2 for the whole tank. Work to pump over the rim is
y). The total force is approximately (2)2(62.4)(6 15) y for a thin disk and 6
0 n 4 (62.4)(21) dy 4 (62.4)(21)(6) 98,801.8 ftlb for the whole tank. Through a hose attached to a valve in the yk)(2 2yk) y n 2(4 249.6 0 distance from the bottom, and pressure 62.4(4 Σ 124.8
k1 15) dy 3705 lb horizontal rectangle has area 2 2y y, where y is Σ 62.4(4
k1 15) y for a thin disk and
6 3 16. (a) From the equation y bottom is faster, because it takes more time for a pump with yk3/2) y. yk a given power output to do more work. 4 (b) 100 (51.2)y dy 19. Work to pump through the valve is (2)2(62.4)( y n 3
3) yk
4 k1 (b) 30 3). The total force is approximately n 8 where y is height up from the bottom. The total work is 3/2 124.8 2(4 y y ) dy 124.8 2 y3/2 0 8
3 1064.96 2 2 5/2
y
5 4 20. The work is the same as if the straw were initially an inch 0 long and just touched the surface, and lengthened as the 1506.1 lb 17. (a) Work to raise a thin slice 62.4(10 20 Total work 12)( y)y. 62.4 60y 2 62.4(120)y dy
0 liquid level dropped. For a thin disk, the volume is 20
0 1,497,600 ftlb
(b) (1,497,600 ftlb)
100 min (250 ftlb/sec) 5990.4 sec 10 62.4 60y 2 7
0 (c) Work to empty half the tank 17.5 2
14
y 17.5 2
14
y 17.5
14
y to y and the work to raise it is
4
(8 y) y. The total work is
9
24
(8 y) dy, which using NINT evaluates
9 91.3244 in.oz. 62.4(120)y dy
0 10 374,400 ftlb, and 21. The work is that already calculated (to pump the oil to the 0 374,400 250 1497.6 sec 25 min rim) plus the work needed to raise the entire amount 3 ft
higher. The latter comes to (d) The weight per ft3 of water is a simple multiplicative
factor in the answers. So divide by 62.4 and multiply 12
r h (57)(3)
3 total is 22,921.06 57 (4)2(8)
30,561.41 22,921.06 ftlb, and the
53,482.5 ftlb. by the appropriate weightdensity
For 62.26:
62.26
1,494,240 ftlb and
62.4
62.26
5990.4
5976.96 sec 100 min.
62.4 1,497,600 For 62.5:
62.5
1,500,000 ftlb and
62.4
62.5
5990.4
6000 sec 100 min.
62.4 1,497,600 22. The weight density is a simple multiplicative factor: Divide
by 57 and multiply by 64.5.
30,561.41 64.5
57 34,582.65 ftlb. 316 Section 7.5
(d) 0 if we assume a continuous distribution. Between 23. The work to raise a thin disk is
r 2(56)h ( 102 56 (12 y 2)2(56)(10 2 59.5 in. and 60.5 in., the proportion is y) y 60.5 y 2) y. The total work is y)(100 f (x) dx 56 (12 y 2) dy, which evaluates using NINT y)(100 0 to 0.071 (7.1%) 59.5 10 1 28. Use f (x) (x 498)2/(2 1002) e 100 2 967,611 ftlb. This will come to
500 (967,611)($0.005) (a) $4838, so yes, there’s enough money f (x) dx 0.34 (34%) 400 to hire the firm. (b) Take 1000 as a conveniently high upper limit:
1000 24. Pipe radius 1
ft;
6 700
360 12
(62.4)y dy
6 Work to fill pipe
0
385 0.217(300) 58,110,000 ftlb. 30. The proportion of lightbulbs that last between 100 and 800
hours. 58,222,320 ftlb, which will take 35,780,000 31.
58,222,320 1650 110,855 sec 6.5 people 29. Integration is a good approximation to the area (which
represents the probability), since the area is a kind of
Riemann sum. 360 Total work 0.217, which means about 112,320 ftlb. (10)2(62.4)y dy Work to fill tank f (x) dx 6,370,000 31 hr. M
25. (a) The pressure at depth y is 62.4y, and the area of a thin 1000MG
dr
r2 1024, G 5.975 1
r 1000 MG
6.6726 10 35,780,000 , which for
6,370,000 11 evaluates to 1010 J. 5.1446 11
ft, so
horizontal strip is 2 y. The depth of water is
6 32. (a) The distance goes from 2 m to 1 m. The work by an the total force on an end is
11/6 (62.4y)(2 dy) external force equals the work done by repulsion in 209.73 lb. 0 moving the electrons from a 1m distance to a 2m
(b) On the sides, which are twice as long as the ends, the
initial total force is doubled to 419.47 lb. When the tank is upended, the depth is doubled to
11/3 force on a side becomes 11
ft, and the
3 (62.4y)(2 dy) distance:
2 Work 23 dr 2 1 r 23 838.93 lb, 29 10 10 0 1.15 which means that the fluid force doubles.
26. 3.75 in. 5
ft, and 7.75 in.
16 31
ft.
48 31/48 Force on a side p dA (64.5y)
0 5
dy
16 4.2 lb. 27. (a) 0.5 (50%), since half of a normal distribution lies
below the mean.
65 (b) Use NINT to find f (x) dx, where
63 f (x) 1 e 2 2 (x 63.4) /(2 3.2 ) . The result is 3.2 2 0.24 (24%).
(c) 6 ft 72 in. Pick 82 in. as a conveniently high upper
82 limit and with NINT, find f (x) dx. The result is
72 0.0036 (0.36%). 28 10 2 1
r 29 1 J (b) Again, find the work done by the fixed electrons in
pushing the third one away. The total work is the sum
of the work by each fixed electron. The changes in
distance are 4 m to 6 m and 2 m to 4 m, respectively.
6 Work 23 10
r2 4 23 10 7.6667 4 29 dr 29 10 1
r
29 6
4 J. 2 23 10
r2
4
1
r2 29 dr 317 Chapter 7 Review 33. F dv
dt m x
2 W x
1
x
2
x
1
v
2 mv s Chapter 7 Review Exercises dv
, so
dx (pp. 413–415) F(x) dx 5 1. 0.2t 3) dt 0 13
t
3 mv dv 7 v
1 2. 1
mv 2
22 (t 2 0 dv
dx
dx mv 5 v (t) dt 0.001t 4) dt (4 0 1
mv 2
21 10.417 ft
0 7 c(t) dt 5 0.05t 4 0 0.0002t 5 4t 7 31.361 gal
0 34. Work Change in kinetic energy 12
mv .
2 100 3. 0.3125 lb
0.009765625 slug, and
32 ft/sec2
5280 ft
1 hr
90 mph 90
132 ft/sec, so
1 mi
3600 sec
1
Work change in kinetic energy
(0.009765625)(132)2
2 0 1.6 oz 1
(0.003125)(280)2
2 Work
37. 2 oz 2 2 4. (x) dx (11 0 so Work 24 Work
39. 6.5 oz
Work cos t
12 sin t
12 0 12 dt
24 14,400
0 6. 122.5 ftlb.
[ 1, 3] by [ 1, 2] The curves intersect at x
2 x dx 0.02832 slug, so 0 1 12
x
2 1
dx
x2 1
2 109.7 ftlb. 1 lb
/(32 ft/sec2)
16 oz 1
(0.01270)(132)2
2 14 g 0.003125 slug, so 1 lb
14.5 oz
/(32 ft/sec2)
16 oz 6.5 oz 2
0 300 2 0 1 lb
1
/(32 ft/sec2)
slug, and
16 oz
256
5280 ft
1 hr
124 mph
181.867 ft/sec,
1 mi
3600 sec
11
(181.867)2 64.6 ftlb.
2 256 1
(0.02832)(88)2
2 2x 2 11x 24 E(t) dt 5. 1 38. 14.5 oz 4x) dx 0 2 oz 124 mph 1464
0 300 2t
1 lb
/(32 ft/sec2)
16 oz 100 33.333e0.03x 21x 85.1 ftlb.
36. 1.6 oz e0.03x) dx (21 0 2 oz
1/8 lb
1
m
slug, so
32 ft/sec2
32 ft/sec2
256
11
Work
(160)2 50 ftlb.
2 256 35. 0.3125 lb 100 B(x) dx 0.01270 slug, so 1. The area is
1 1
x 0 1
2 2
1 1 1. 7. 110.6 ftlb. 1
1
lb
slug. Compression energy of spring
8
256
12
1
12
ks
(18)
0.5625 ftlb, and final height is
2
2
4
0.5625
given by mgh 0.5625 ftlb, so h
4.5 ft.
(1/256)(32) 40. 2 oz [ 4, 4] by [ 4, 4] The curves intersect at x
1 [3 x2 2 and x
1 (x 1)] dx 2 ( x2 1. The area is
x 2) dx 12
x
2 2x 2 13
x
3
1
3
9
.
2 1
2 2 1
2 8
3 2 4 318 Chapter 7 Review 8. x y 1 implies y x)2 (1 1 2 x x. 12. 2 [ 0.5, 2] by [ 0.5, 1]
1 The area is (1 2 x x) dx 4 3/2
x
3 x 0 12
x
2 1
.
6 by [ 3, 3] The area is 1 (2 sin x 0 sin 2x) dx 2 cos x 0 1
cos 2x
2 4.
0 13. x
.
2 2y 2 implies y 9. x 3
2 , [ 5, 5] by [ 5, 5] The curves intersect at x [ 1, 19] by [ 1, 4] 2.1281 (4 x 2.1281 The curves intersect at x
18 x
dx
2 3
0
3 or 0 y2 10. 4x 1
y
4 x 18
4 x 3/2 3x
23
y
3 2y 2 dy 18. The area is
32 2 2.1281. The area is cos x) dx, which using NINT evaluates to 18, 8.9023. 14. 0 3 18.
0 12
y
4 4 implies x 1, and 4x y 16 implies
[ 4, 4] by [ 4, 4] 4. The curves intersect at x
0.8256 (3 0.8256. The area is sec2 x) dx, x 0.8256 which using NINT evaluates to 2.1043. [ 6, 6] by [ 6, 6] 15. Solve 1
The curves intersect at (3, cos x 2 cos x for the xvalues at the two 4) and (5.25, 5). The area is
ends of the region: x 5 1
y
4 4
5 12
y
4 4
12
y
4 4 12
y
8 425
24 38
3 3 , i.e., symmetry of the area: dy 7 /3 2 1
y
4 13
y
12 1 2 [(1
2 5 dy cos x) 2 cos x)] dx (2 cos x 1) dx 2 5 5y 2 2 sin x x 4 243
8 (2 7 /3 2 30.375. 2
3 3 7 /3
2 1.370. 5 /3 11. 16. [(2
/3 cos x) (1 5 /3 (1 2 cos x) dx /3 x
[ 0.1, 1] by [ 0.1, 1]
/4 The area is (x
0 sin x) dx 2
/4 12
x
2 cos x 2 2 32 0 2 0.0155. 1 2 sin x
3 4
3 5 /3
/3 7.653 cos x)] dx 5
7
or . Use the
3
3 Chapter 7 Review
17. Solve x 3
x x x x2 to find the intersection points at 1 (c) Use cylindrical shells. A shell has radius 4
height 2 x 21/4. Then use the area’s symmetry: 0 and x 2 (4
0 2 0 (x3 1 1
ln (x 2
2 2 x)(2 x x) dx (8 4x 2x 3/2 x 2) dx 2x 2 x 4 5/2
x
5 13
x
3 0 x
x2 x. The total volume is 4 1/4 2 ln ( 2 x) dx
14
x
4 1)
1) 2 1 16 3/2
x
3 2 12
x
2 x and 4 the area is
2 21/4 4 64
.
5 0 (d) Use cylindrical shells. A shell has radius 4 0 319 1.2956. y and y2
. The total volume is
4
4
y2
2 (4 y) y
dy
4
0
4
y3
2
4y 2y 2
dy
4
0
4
23
32
14
2 2y 2
y
y
.
3
3
16
0 height y
18. Use the intersect function on a graphing calculator to
determine that the curves intersect at x 1.8933. The area is
1.8933 31 x2 3
dx,
10 x2 1.8933 which using NINT evaluates to 5.7312. 22. (a) Use disks. The volume is
2 19. Use the x and yaxis symmetries of the area:
4 x sin x dx 4 sin x x cos x 0 4. 0 k (b) 0 0 A(x) r
1 V 1 k, 9 x. 9 x dx x 9 When k 1 y2 2y dy 2 4.
0 k2 0 8 8 k 2 dV (c) Since V
2 0 y2 2y dy 3x 4 and area 20. A cross section has radius r 2 2y)2 dy ( 1, 2. dt dk
dt dk
dt
1 dV
2 k dt 2k .
1
(2)
2 1
1 so the depth is increasing at the rate of 1 21. , unit per second. 23. The football is a solid of revolution about the xaxis. A
[ 5, 5] by [ 5, 5] cross section has radius The graphs intersect at (0, 0) and (4, 4).
r2
(a) Use cylindrical shells. A shell has radius y and height
2 y
. The total volume is
4
4
4
y2
y3
2 ( y) y
dy 2
y2
dy
4
4
0
0
4
13
14
2
y
y
3
16
0
32
.
3 y 12 1 11/2 2 12
0 1 4x 2
and area
121 12 1 4x 2
. The volume is, given the symmetry,
121
11/2
4x 2
4x 2
dx 24
1
dx
121
121
0
11/2
21
2
24 x
x3
3
11
0
11
11
24
2
6 88 276 in3. 24. The width of a cross section is 2 sin x, and the area is
(b) Use cylindrical shells. A shell has radius x and height
2x x. The total volume is 4 4 2 (x)(2 x x) dx 2 0 (2x 3/2 2 x ) dx 0 2 4 5/2
x
5 128
.
15 13
x
3 4
0 12
1
r
sin2 x. The volume is
2
2
1
1
1
sin2 x dx
x
sin 2x
22
4
02 2 0 4 . 320 Chapter 7 Review
29. 25. [ 1, 2] by [ 1, 2] [ 1, 3] by [ 3, 3] Use washer cross sections. A washer has inner radius r
e x/2, and area (R2 outer radius R r 2) (e x 1, 1). 3x 2 y is at x 6x equals zero when x 0, the minimum at x = 2. The distance between
2 them along the curve is The volume is
ln 3 (e x ex 1) dx using NINT evaluates to x
0 (3 ln 3 (2 4.5920
2 ln 3). 1) 26. Use cylindrical shells. Taking the hole to be vertical, a shell
has radius x and height 2 22 3 2x
0 4 2
(4
3 2 x 2 dx x4 27. The curve crosses the xaxis at x
3 ( 2x)2 dx 1
3 1 2 x 4 dx 2
5 x 2 dx 2 2x, so the 13
x
3 5 39.
2 32. (a) (100 N)(40 m) = 4000 J
(b) When the end has traveled a distance y, the weight of
the remaining portion is (40 4x 2 dx, which 3 using NINT evaluates to 5 (F (x))2 dx 1 0 3. y
3 1, so 3 x2)3/2 4
(1 8)
3
28
29.3215 ft3.
3 length is 30. If (b) were true, then the curve y k sin x would have to
get vanishingly short as k approached zero. Since in fact the
curve’s length approaches 2 instead, (b) is false and (a) is
true. 5 3 2 0 4.5920. The time taken is about 2.296 sec. 31. F (x)
x 2) dx y)(0.8) 40 28. 0.8y) dy 0.4y 2 32y 0 The curves intersect at x 0 and x 1. Use the graphs’ 1 4 1
0 (3x 2 640 is (800)(8)
4750 x and yaxis symmetry:
3x 40 640 J. 4640 J 33. The weight of the water at elevation x (starting from x [ 2, 2] by [ 2, 2] x) 0.8y. 0 (c) 4000 2 32 The total work to lift the rope is 19.4942. (32 d3
(x
dx 6x)2 dx, which x 2. The volume of the piece cut out is
2 (x)(2 22 (3x 2 1
0 ln 3 0 0 or 2. The maximum is
0 4750 x/2
4750 128
4750
95 128
4750
95 1
x dx
2 1, and the total perimeter is 0) 1
x . The total work
2 128
4750x
95 12
x
4 4750
0 22,800,000 ftlb. 1)2 dx, which using NINT evaluates to
34. F ks, so k 5.2454. 0.3 Work
0 F
800
80
N/m. Then
s
3
0.3
0.3
800
800 2
x dx
x
12 J.
3
6
0 To stretch the additional meter,
1.3 Work
0.3 800
x dx
3 800 2
x
6 1.3 213.3 J.
0.3 35. The work is positive going uphill, since the force pushes in
the direction of travel. The work is negative going downhill. Chapter 7 Review
82 36. The radius of a horizontal cross section is y 2, where 1 42. Use f (x) 321 x 2/2 e . 2 y 2), the y is distance below the rim. The area is (64 1 (a) y )( y) y. The total work is y ) dy 0.04 3 (64y 2 y ) dy 2 36 320y(2 y) dy 320 ( y 0 2 13
y
3 1280
3
23
ft, 3.5 in.
48 38. 5.75 in. f (x) dx 1 A shell has radius x and height 2x x
2 4y, and the
[ 1, 3] by [ 1, 3] 2y) dy
2 y2 0 and 44. 0 320 0.9973 (99.73%) 43. Because f (x) 113.097 in.lb. 2 f (x) dx
3 y)4y y. The total force is 2 0.9545 (95.45%) 3 2 37. The width of a thin horizontal strip is 2(2y)
force against it is 80(2 f (x) dx (c) 8 14
y
4 32y 2 0.04 (b)
2 8 2 0.04 y(64 0.6827 (68.27%).
2 2 the rim is 0.04 (64
8 f (x) dx evaluates, using NINT, to
1 y 2) y, and the work to lift it over weight is 0.04 (64 1 volume is 2 (x)
0 0 3
x dx
2 1 x3 3
x. The total
2 .
0 45. 426.67 lb.
5
ft.
6 7
ft, and 10 in.
24 For the base,
Force 57 23
48 [ 3, 3] by [ 3, 3] 5
6 7
24 6.6385 lb. 5/6 Force
0 2 5/6 399 1 2
y
24 2 7
57
y dy
24 5/6 57
0 23
y dy
48 39. A square’s height is y
y2 ( 0 1/2 2 1
dx
x 2 dx 2 2x 1/2 3.
1/2 46.
5/6 1311 1 2
y
48 2 9.4835 lb.
0 x)2, and its area is (6 x)4. The volume is 6 2 (x) 5.7726 lb. For the sides,
Force 1
x A shell has radius x and height . The total volume is For the front and back, 6 ( 6 x)4 dx, 2 , 3
2 by [ 2, 2] 0 A shell has radius x and height sin x. The total volume is which using NINT evaluates to exactly 14.4. 2 (x)(sin x) dx
40. Choose 50 cm as a conveniently large upper limit.
50
20 1 e (x 17.2)2/(2 3.42) x cos x 2 2 . 0 47. dx, evaluates, 0.2051 (20.5%).
[ 1, 4] by [ 4, 1] 41. Answers will vary. Find , then use the fact that 68% of the
class is within of to find , and then choose a conveniently large number b and calculate
10 sin x 3.4 2 using NINT to b 2 0 1
2 e (x )2/(2 2 ) dx. The curves intersect at x
x and height x 3 (x 2 1 and x 3. A shell has radius 3x) 2 x 4x 3. The total volume is
3 2 (x)( x 2 3 4x 3) dx ( x3 2 1 4x 2 3x) dx 14
x
4 43
x
3 32
x
2 1 2
16
.
3 3
1 322 Section 8.1 48. Use the intersect function on a graphing calculator to
determine that the curves intersect at x
has radius x and height 3 1x x 2 2 3 10 1.8933. A shell that l’Hôpital’s Rule does not say that lim . The volume, which lim 0 evaluates to 9.7717. 5
(x
4 49. (a) y . x2 3
dx, which using NINT
10 x2 2 (x) 31 x→0 y2 y1 x→0 y2 is calculated using the right half of the area, is
1.8933 x cos x sin x .
The graphs of y3 and y5 clearly show
x2
y1 3. y5 2)(x 2) 52
x
4 5 [ 3, 3] by [ 2, 2] Quick Review 8.1
(b) Revolve about the line x 4, using cylindrical shells. 1.
x A shell has radius 4 52
x . The total
4 x and height 5 1 2 2 (4
2 10
2 320
3 1 100 x
12
x
2 4x
2 dL
dx 1
x f (x) must equal 1
x
2 2
f (x)
x 1
x2 2. 12
x
4 51. y 12
x
4 1
ln x
2 3
4 sec x, so the area is 2 (tan x) 2 2.7183 1 2.7183
2.7183
2.7183 0.00001 2.7183
As x→0 , x1/(ln x) approaches 2.7183. 2 (sec x) dx, 0 which using NINT evaluates to x1/(ln x) 0.0001 3
. The
4
2 ln x 3
.
4 x2 x 0.001 C, and the /4 2 0.1 x
approaches 1.1052.
x 0.01 1
ln x
2 requirement to pass through (1, 1) means that C
function is f (x) 1.1052 0.1 ( f (x))2, and 1
. Then f (x)
2x 1.1052 As x→ , 1
( f (x))2, 1 1.1052 10,000 2 1.1051 1,000,000 4 dx 335.1032 in3. ( f (x))2 f (x) x 2 13
x
3 14
x
16 10 50. Since 13
x
4 1.1046 1000 52
x dx
4 x) 5 1.1000 10 volume is
2 0.1 x
x 1 3.
x 3.84. 1 1x
x 1
1
52. x
and x
y
2
1
2
1
y
1 0.5 0.1 1
, so the area is
y2
12
dy,
y2 0.78679 Chapter 8
L’Hôpital’s Rule, Improper Integrals,
and Partial Fractions 0.99312 0.0001 5.02. 0.95490
0.99908 0.00001 which using NINT evaluates to 0.01
0.001 0.99988 0.000001 0.99999
1x
approaches 1.
x As x→0 , 1
4. 1x
x x s Section 8.1 L’Hôpital’s Rule (pp. 417–425)
Exploration 1 Exploring L’Hôpital’s Rule
Graphically
1. lim x→0 sin x
x lim x→0 cos x
1 1 1.1 13.981 2. The two graphs suggest that lim y1 x→0 y2 lim y1 x→0 y2 . 105.77
1007.9 1.0001 1 1.01
1.001 10010 As x→ 1 , 1 1x
goes to
x . is equal to Section 8.1
5. 323 3. [0, 2] by [0, 3] t As t→1, 1
t [0, 3] by [0, 3] From the graph, the limit appears to be 1. The limit leads to
the indeterminate form 0. approaches 2.
1 6.
ln 1 ln 1
[0, 500] by [0, 3] lim As x→ , 1
x x ln 1 1 lim 1
x2 1 lim x→0 lim 1
x2 1/x x→0 7. 1
x 1
x x→0 x [ 5, 5] by [ 1, 4] 1
x 1 1
x 1
x x→0 4x 2 1
approaches 2.
x1 1
x ln 1 1x
x 0 1 Therefore, sin 3x
As x→0,
approaches 3.
x 1x
x lim 1 x→0 8. lim f (x) x→0 lim e ln f (x) e0 x→0 1. 4. [0, ] by [ 1, 2] tan
→,
2 2 tan As [0, 1000] by [0, 1] approaches 1. From the graph, the limit appears to be about 0.714.
5x 2
2
x→ 7x lim 1
sin h
h 9. y
10. y lim x→ x3 1
4x 3 x 3
x→1 h)1/h (1 3x
1 10x 3
14x lim x→ 3x 2
12x 2 1
x→1 5. lim lim 10
14 5
7 3
11 Section 8.1 Exercises
1. [0, 2] by [0, 1] [0, 2] by [0, 1] 1
4 The graph supports the answer. From the graph, the limit appears to be .
x
2
x→2 x lim 2
4 1
x→2 2x lim 1
4 6. lim x→0 1 cos x
x2 lim x→0 sin x
2x lim x→0 2. [ 2, 2] by [ 2, 6] [ 5, 5] by [ 1, 1] From the graph, the limit appears to be 5.
lim x→0 sin 5x
x lim x→0 5 cos 5x
1 5 The graph supports the answer. cos x
2 1
2 0.71429 324 Section 8.1 7. The limit leads to the indeterminate form 1 .
ln(e x Let ln f (x)
ln(e x
x
x→0 lim lim (e x x) ex
ex lim 1
x x) . t ex
x
x→0 e 1
x lim 1 x→0 x)1/x x→0 ln(e x
x x)1/x 2
1 lim e ln f (x) lim f (x) x→0 1
t 1
sin t 12. Let f (t) 2 e2 x→0 100 f (t) 0.1884 t sin t
.
t sin t
1 10 10 0.0167 2 10 0.0017 3 0.00017 Estimate the limit to be 0.
1
t 1 lim t→0 sin t lim t t→0 lim
t→0 lim
t→0 [0, 5] by [0, 10] 13. Let f (x) sin t
t sin t
1 cos t
t cos t sin t
sin t
t sin t cos t 0 cos t x)1/x. (1 The graph supports the answer.
8. lim
x→ 3x
2x
x3 x 1 3
1 4x
3x 2 lim x→ lim x→ 4
6x 0 102 10 x
2 103 104 105 f (x) 1.2710 1.0472 1.0069 1.0009 1.0001
Estimate the limit to be 1.
ln (1 x)
1
x
1x
ln (1 x)
lim
lim
1
x
x→
x→ ln f (x) [ 5, 25] by [ 1, 2] The graph supports the answer.
9. (a) 102 10 x x)1/x lim (1 x→ 103 104 105 0
1 0 lim e ln f (x) e0 102 103 104 105 0.6525 0.6652 0.6665 0.6667 lim f (x) x→ x→ 1 x 2x2
.
3x2 5x 14. Let f (x) f (x) 1.1513 0.2303 0.0345 0.00461 0.00058
10 x
Estimate the limit to be 0.
ln x 5
x
x→ 5 ln x
x
x→ 10. (a) 100 x 5/x
1
x→ lim (b) lim 1 10 10 2 f (x) 0
1 lim 10 3 10 4 f (x) 0.1585 0.1666 0.1667 0.1667 0.1667
1
6 Estimate the limit to be .
(b) lim
x→0 x sin x
x3 0.5429 0 x 2x 2
2
5x
x→ 3x lim 15. lim sin 2 2 cos
1
→0 x→0 lim x→0 1 16. lim → /2 1 sin
cos 2 1
6 sin 3
.
sin 4 17. lim
10
f( ) 0 10 1 10 3
4 sin 3 →0 sin 4 2 10 3 10 4 0.1865 0.7589 0.7501 0.7500 0.7500 Estimate the limit to be .
lim lim 3 cos 3 →0 4 cos 4 4
6 (2)(0) cos (0)2 2
.
3 0 1 lim x→0 lim 11. Let f ( ) 2 lim →0 cos x
3x 2
sin x
6x
cos x
6 2
.
3
1 4x
lim
lim
5 x→
x→ 6x Estimate the limit to be 3
4 cos t
t t
t→0 e 1
1 cos
2 sin 2
sin
lim
→ /2 4 cos 2
sin /2
4 cos
1
4 lim → /2 lim t
t→0 e sin t
1 lim
t→0 c os t
et 1 Section 8.1 18. lim t t→1 ln t 1
sin t 1 lim t→1 1 cos t t y→ /2 1
1 22. lim y tan y 2 ( 1) lim x→ 2 x ( 1)(1)
(1) lim ln 2 x→0 x→0 ln 2 log3 (x 3) lim x→ lim (x lim lim x→0 tan
1
24. lim x tan
x
x→ lim x→ ln 3
ln 2
ln (y 2 2y)
ln y
y→0 21. lim x x→0 x ln 3 3 ln 3
x ln 2 lim x→ y→0 lim y→0 2
2y 1
1
sec2
x2
x
1
x2 lim sec2 1
x sec2 0 1 y(2y 2)
y 2 2y 4y y→0 2y 4(0)
2(0) 1
x x→ 1
y 2y 2
lim 2
y→0 y lim 0 1
x lim 2y
y2 1 x2
x lim 3) ln 3
x ln 2 ln 3
x→ ln 2 2 1
x lim lim ( 1) sin 2 1
x
1
x2 x→0 1
x ln 2
1
(x 3) ln 3 x→ x→ cos lim ln x 23. lim x ln x x→ x→ 2 sin 1
x ln 2 ( 1) sin y sin y 1 x ln 2
1
x→ x 20. lim y cos y y→ /2 lim log2 x 2 lim 1 y sin y
cos y y→ /2 1
+1 ln (x 1)
19. lim
log2 x
x→ 2 lim 2
2 2y
2y
2
2 25. lim (csc x cot x x→0 2
2 1 lim x→0 lim x→0 lim x→0 1
sin x
1 cos x
sin x cos x) cos x cos x cos x sin x
sin x sin x cos x cos x
cos x sin x sin x 1 2 325 326 Section 8.1 26. lim (ln 2x ln (x x→ 2x Let f (x) x 1 x→ x lim 1 x→ x 1 ln (x x→ 27. lim (ln x 1)) ln sin x) x→0 x
lim
x→0 sin x lim ln f (x) ln (ln x)
lim
x
x→ ln 2 x→ x→0 Let f (x) 1 ln sin x) 1
x x→0 1 lim ln f (x) x→0 1 lim ln 1 ln (e
lim x x x)1/x. x) ln (e x→0 x x) lim x)1/x x→0 ln lim x→0 x→0 1x
x2
x→0 31. lim
x→ 32. lim 2x 2
sin 7x x→0 tan 11x 5
x 2 lim (x 2 ln 2/x 3
1/x 2 2x x lim x→ 2
x 1 lim eln f (x) lim x→ 2x e1/2 x→ lim 1)x e 1 1 1) ln (x 2 (x ln (x 2 2x
x e2
0 . 2x 1) 1) 2x 1) lim x2 x→1 1 4x 0 7 cos 7x 2
x→0 11 sec 11x 1)2 2(x x→1 lim (x 2 x→1 2x 1) x1 lim e 1) ln f (x) x→1 0
e 0 1 36. The limit leads to the indeterminate form 00. 1 Let f (x)
1 1 2(x 1)
(x 1)2
lim
1
x→1
(x 1)2 x→0 3 2
2x
1 (x 1 lim e0 1
2x 2
lim ln (x 1
x2
1
x lim 2x 1/x 2 x→ 2x
1)x 2x x x→0 3x lim 1 x→ 2x)1/(2 ln x) x→1 lim e ln f (x) lim ln (1 2x)
2 ln x 1 x→0 lim . 2 1 lim e ln f (x) 1
x ln 2
x 1
x2
1
x lim (1 x→ ln (x 2
1
x 1x
.
x2 1x
ln 2
x 2x) 2 Let f (x) 30. The limit leads to the indeterminate form
Let f (x) . x) ex
ex x→0 x lim (e x 0 1 35. The limit leads to the indeterminate form 00. ln (e x
x 1/x (1 ln (1 2x)
lim
2 ln x
x→ 29. The limit leads to the indeterminate form 1 .
(e x e0 0 1/(2 ln x) 0 x x→0 x Let f (x) x→ x→ 2x)1/(2 ln x) ln (1 28. lim 1
x ln x lim 1 lim eln f (x) x→ Therefore,
lim (ln x lim 34. The limit leads to the indeterminate form 1
lim
x→0 cos x x→0 1/x
ln x x→ lim (ln x)1/x x
sin x lim ln x
.
sin x Let f (x) ln (ln x)
x ln (ln x)1/x 2 . (ln x)1/x. Let f (x) Therefore,
lim (ln 2x 0 33. The limit leads to the indeterminate form 2x lim ln . 2
x→ 1 2x lim 1)) 0
7
11 (cos x)cos x. ln (cos x)cos x ln (cos x)
sec x (cos x) ln (cos x)
sin x ln (cos x)
cos x
lim
sec x
x→ /2
x→ /2 sec x tan x lim lim tan x x→ /2 sec x tan x lim x→ /2 lim (cos x)cos x x→ /2 cos x lim e ln f (x) x→ /2 0
e0 1 1
2 1
2 Section 8.1
37. The limit leads to the indeterminate form 1 .
Let f (x) 1/x (1 x x) . x) lim x 1 1 lim eln f (x) e1 x→0 1/(x 1) ln x1/(x
lim x e ln x
1 x→1 x lim x1/(x lim . x→ 1) e1 x→1 lim x→0 (c) lim ln (sin x) x→ 1
x x 2 cos x
sin x x→0 x 2 sin x 2x cos x
lim
cos x
x→0 lim e ln f (x) x→0 x→ 1 1/2 1) (1/2)(x 1) lim 1/2 x→ e0 x→0 9x 1 x 1 1 lim x→ 1
x
1
x 9 lim sec x x→ /2 tan x 0 ln (sin x) lim x→ /2 sec x tan x
sec2 x tan x lim x→ /2 sec x (b) lim (sin x) [0, ] by [ 1, 5] ln (sin x)
cot x tan x ln (sin x)
cos x
sin x ln (sin x)
lim
cot x
x→0 lim csc x 1 x→0 ln f (x) 0 lim e x→0 The limit appears to be 1. lim ( sin x cos x) 2 x→0 tan x e x→0 0 sec x
x→ /2 tan x (c) lim ln x1/(1 ln x
lim
x
x→1 1
x→1 lim x→1
x) lim (b) f (x)
1
x 1 lim e ln f (x) x→1 . 7(x
f (x) x→3 g(x) . ln x
1x x) lim x1/(1 x) (x f (x)
lim
x→3 g(x) 1
e 1 1 lim x→ /2 sin x 1 x→ /2 sin x 45. Possible answers: 1 (a) f (x)
x1/(1 lim cos x
cos x 41. The limit leads to the indeterminate form 1
Let f (x) 1
1 3 1 (sin x)tan x
tan x x
9x 1 40. The limit leads to the indeterminate form 00.
Let f (x) 9 44. (a) L’Hôpital’s Rule does not help because applying
L’Hôpital’s Rule to this quotient essentially “inverts”
the problem by interchanging the numerator and
denominator (see below). It is still essentially the same
problem and one is no closer to a solution. Applying
L’Hôpital’s Rule a second time returns to the original
problem. 1
x2 lim lim (sin x)x (9/2)(9x lim 9 cos x
sin x 1
x 1 [0, 100] by [0, 4] x ln (sin x) lim ln (sin x) 9x
x (sin x)x. x→0 ln 2 The limit appears to be 3. 39. The limit leads to the indeterminate form 0 . ln (sin x)x lim ln 2 x→ e
0 Let f (x) 2x
x lim ln x→ 1 lim e ln f (x) x→1 dt
t x 2x
x ln (b) ln x
x1
1/x
lim
x→1 1 1) 2x ln x 43. (a) L’Hôpital’s Rule does not help because applying
L’Hôpital’s Rule to this quotient essentially “inverts”
the problem by interchanging the numerator and
denominator (see below). It is still essentially the same
problem and one is no closer to a solution. Applying
L’Hôpital’s Rule a second time returns to the original
problem. 38. The limit leads to the indeterminate form 1 .
Let f (x) ln 2x
x x→ 1 x→0 x)1/x lim (1 ln t
lim 1 ln (1 x)
lim
x
x→0
x→0 2x dt
t ln (1 x)
x 1/x ln (1 2x 42. 327 1
e (c) f (x)
lim x
f (x) x→3 g(x) 3); g(x)
lim x→3 7(x
x x 3)
3 3)2; g(x) 3
lim 7 7 x→3 1 x 3 (x
lim
x→3 x 3)2
3 x→3 3; g(x) (x 3)3 x 3
3)3 lim x→3 (x lim lim 2(x x→3 3(x 3)
1 1
3)2 0 328 Section 8.1 46. Answers may vary.
(a) f (x) 3x 1; g(x) f (x)
lim
x→ g(x) (b) f (x) lim x lim x→ 1 lim x→ x 1; g(x) f (x)
lim
x→ g(x) x x lim x
x 0 lim x→ 1 1
1 f (x)
g (x) 0, f (x ) 1. 1 x→0 g (x) 2
1 f (x)
lim
x→0 g (x) 2 (b) This does not contradict L’Hôpital’s Rule since
lim f (x) 2 and lim g (x) 1. 1 x lim x→ 50. (a) For x
lim 2 f (x)
g(x) 3 1
lim
x→ 2x x2 x→ 3
1 2 1 x 2; g(x) (c) f (x) x 3x x→ (b) Part (a) shows that as the number of compoundings per
year increases toward infinity, the limit of interest compounded k times per year is interest compounded continuously. x→0 2x
1 x→0 t 51. (a) A (t) e x dx 0 47. Find c such that lim f (x)
lim f (x) 3 sin 3x
5x3
9 9 cos 3x
15x2
27 sin 3x
30x
81 cos 3x
81
30
30 x→0 lim x→0 lim x→0 lim x→0 1) t 27
10 e
c f (0), so f is lim x→0 x→0 ln x lim x x x→0 lim e x→0 lim x→0 x ln x e x→0 0 x 0 V(t)
t→0 A (t) (c) lim lim (e
lim 2 r
k lim 1 lim A0 1 k→ lim k→ rt
1 rt lim k→ t 1
k
r
k r kt
k 0.1 1 0.00495
0.00050 0.0001 0.00005 The limit appears to be 0.
(b) lim sin x
2x x→0 1 rt ) t 0.04542 0.001 r
r
1
k2
k
1
k2 1)
1 f (x) 0.01 1
k 2 (2) x r
k t ln 1 1 2t
t (1) 2t (2e
e 1 2 1 52. (a) r kt
.
k
r
kt ln 1
k t ln 1
k→ e t→0 1) t e t→0 49. (a) The limit leads to the indeterminate form 1 .
1 2t (e t→ 2 ln f (k) 2 lim 2 0. Extend the 1
2 1) 1 Thus, f has a removable discontinuity at x
definition of f by letting f (0) 1. Let f (k) dx 2t (e V(t)
lim
t→ A (t) 1
x lim 1
x 2x 1 2t
e
2 0. lim f (x) leads to the indeterminate 1
x
1
x2 1 1 2x t
e
2
0 2 ln x t 0 form 0 .
x ln x 1 lim ( e t→ (e x)2 dx t 0 ln x x 1 0 continuous.
48. f (x) is defined at x t→ 1
et lim t→ (b) V (t) 27
. This works since lim f (x)
10
x→0 Thus, c t lim A (t) 9x lim x→0 e
0 c. x→0 t x e 0
1 0 L’Hôpital’s Rule is not applied here because the limit is
A0 lim 1
k→ r kt
k A0 lim e ln f (k)
k→ A0 e rt not of the form
limit 1. 0
or
0 , since the denominator has Section 8.2
e x ln (1 53. (a) f (x)
1
x 1 1/x) (b) f (x) 0 when x Domain: ( , 1 or x 1) f 0 sin x, g (x)
f (0) 2 tan c 1
x (c) lim x ln 1
x→− lim 56. (a) ln f (x)g(x) 1
x lim e ln f (x) x→c 1 lim f (x) x→− lim e x ln (1 ( 1/x) x→− e x→0 cos x 6
x12 Exploration 1
x→
ax
2
x→ x 1. lim x→0 lim x→0 lim x→0 lim x→0 2. lim ax
x
x→ b 3. lim ln x
1. lim x
e
x→
ex
3
x→ x 2. lim 3. lim
x→ 55. (a) f (x) 3x , g (x) f (1) f ( 1) 2x 2, g(1) 3c 2
2c 1 (3c
c 2c
1)(c 1
or c
3 2c 1 x2
2x
x→ e 2
5. 1 (ln a)2 a x
2
x→ , so a x grows lim lim 1.5x x→ lim x→ ax
b a
b because 0 f (x)
x→ g(x) 8. lim 1 The value of c that satisfies the property is c 1
.
3 lim x→ 1
x 1. 0 ex ex
2
x→ 3x ex
x→ 6x lim ex
x→ 6 lim lim lim x→ 2x
2e2x lim x→ 2
4e2x 3x 4 f (x)
7. lim
x→ g(x) 0 1) (ln a)(a x)
2x
x→ lim x2
e2x 4. lim 2
2 3c 2
3c 2 1
g( 1) Comparing Rates of Growth as Quick Review 8.2 (d) The graph and/or table on a grapher show the value of
the function to be 0 for xvalues moderately close to 0,
but the limit is 1/2. The calculator is giving unreliable
information because there is significant roundoff error
in computing values of this function on a limited
precision device.
2 e faster than x 2 as x→ .
3x
2x
x→ 6x 5 sin x 6
12x 11
sin x 6
2x 6
6x 5 cos x 6
12x 5
cos x 6
1
2
2 lim lim eln f (x) x→c (pp. 425–433) (b) Same reason as in part (a) applies.
1 ) s Section 8.2 Relative Rates of Growth 54. (a) Because the difference in the numerator is so small
compared to the values being subtracted, any calculator
or computer with limited precision will give the
incorrect result that 1 cos x 6 is 0 for even moderately
small values of x. For example, at x 0.1,
cos x 6 0.9999999999995 (13 places), so on a
10place calculator, cos x 6 1 and 1 cos x 6 0. (c) lim x→c )(
g(x) lim f (x) g(x) x→c 0 lim g(x) lim ln f (x) x→c 1 1
x 1 e x→c x→c x→− x→c )
g(x) lim f (x) g(x) (b) lim (g(x) ln f (x))
lim 2 lim g(x) lim ln f (x) x→c 1 1
x on 0, 4 (
1
x2 x→− 1 g(x) ln f (x) x→c 1
1
x2 lim 1 tan lim (g(x) ln f (x)) 1
x x→− 1 1 c ln 1 g(0) 2 1
1 (b) The form is 0 1, so lim f (x)
x→ 1 cos x 1, g sin c
cos c (0, ) 329 0
6. lim x→ lim x→ x ln x
x
4x 2 5x
2x 1
x 1 lim x→ lim x→ 2x 3
x 1 1 2x 2
1
5
4x 1 330 Section 8.2
ex 9. (a) f (x)
f (x)
2x x x x2
ex
2x 1
2x xe
e2 x 2 x x ln x
6. lim
ex
x→ 2 ex x) f (x) 0 or x 2 1; f (2) 4
e2 1 ecos x
ex
x→ ecos x grows slower than e x as x→ . (2, 1.541) and has a local xe x
x
x→ e 8. lim x1000
ex
x→ 9. lim , 0] and [2, ). sin x
sin x
1
,x 0
x
x
sin x
1 since sin x
x for x
Observe that
x
sin x
112
lim f (x) 1 lim
x→0
x→0 x gets lim x x→ 0. 10. lim
ex doesn’t have an absolute maximum value. f is not defined at
0. 11. lim x 3x 3x 2 3
ex
x→ lim x→
x 6x
ex lim x→ 6
ex 0 1x
x2 lim x→ x 4 grows at the same rate as x 2. 1
1 lim x 4 lim x4 x→
2 x→ 1
x4 1 x2
x
x→ e lim 2x
x
x→ e lim 2
x
x→ e lim than e as x→ .
lim x→ 4x
e since 4
e (5/2)x
5x
5
lim
0 since
x
e
2e
2e
x→
x→
5x
grows slower than e x as x→ .
2 4. lim ex 1
ex
x→
1 lim e x→ 3
x2 lim x x→ 15x 3
x2 15. lim
x→ 15
x lim x→ 3
x2 0 3 grows slower than x 2 as x→ . 15x x 4 5x
x2 lim x4 5x lim x4 x→ x→ 5
x3 1 1 5x grows at the same rate as x 2 as x→ . ln x
x2 2x
2
x→ x 1. 16. lim 4x grows faster than e x as x→ . 5. lim 1 lim x→ 1/x
2x lim x→ 1
2x 2 0 ln x grows slower than x 2 as x→ . x 4x
x
x→ e x→ x4
x 4 grows slower 4
x lim 1 x→ 3 0
1 1
2 1
2e 2x 3 grows faster than x 2 as x→ . x x→ x 2 grows slower than e x as x→ , so 3. lim 4x x2 14. lim Next compare x with e . 1
2 lim x→ grows at the same rate as e x as x→ . x3 1 x 0. x1000 grows slower than e x as x→ . e x)/2
ex x2 x→
3 13. lim
4 1000!
ex 4x grows at the same rate as x 2 as x→ . 1 grows slower than e as x→ . 2. First observe that 0 Repeated application of L’Hôpital’s Rule x2 x→
2 12. lim lim x e
2 x Section 8.2 Exercises ex (e x x→ Thus the values of f get close to 2 as x gets close to 0, so f x→
3 lim x x→ xe xgrows faster than e x as x→ . (c) f is decreasing on ( 1 e for all x. 1.541 (b) f is increasing on [0, 2] 3x
ex 0 0 since ecos x 7. lim minimum at (0, 1). x3 1 x x grows slower than e x as x→ . x ln x f has a local maximum at 1. lim e x→ 0 for x ln x ln x
ex
1/x
ex lim 0
2 1
x x→ lim The graph decreases, increases, and then decreases. 10. f (x) lim x→ 0 or x f (0) x x 0 ex x(2 x2 ex
2xe x (ln 2)2 x
2x
x→ lim (ln 2)2 2 x
2
x→ lim . 2 x grows faster than x 2 as x→ .
1. e grows at the same rate as e x as x→ . 17. lim
x→ log2 x 2
ln x lim x→ 2 log2 x
ln x lim x→ 2(ln x)/(ln 2)
ln x 2
ln 2 log2 x 2 grows at the same rate as ln x as x→ .
18. lim
x→ log log x
ln x lim x→ log x
2 ln x lim x→ (ln x)/(ln 10)
2 ln x x grows at the same rate as ln x as x→ . 1
2 ln 10 Section 8.2
1/ x
ln x
x→
1 19. lim 1 lim x→ 25. Compare f1 to f2. 0 x ln x lim grows slower than ln x as x→ . x→ x
ex
x→ ln x
x e lim 0 x 2 ln x
ln x x→ lim (x lim x
ln x lim x→ 2 lim x→ 1
1/x 22. lim
x→ 5 ln x
ln x lim x→ e
xx ex lim x→ lim x→ Compare e x to (ln x)x.
x lim e
(ln x)x 10 f3(x)
f1(x) lim x x→ 1 lim 1
x 1 x→ x 1 lim x→ f1(x) lim x→ x4 x
x2 lim x→ 1 1
x3 1 e
ln x f3(x)
f1(x) lim x→ x4 x3
x2 lim x→ 1 1
x 1 Thus f1 and f3 grow at the same rate.
x 0 By transitivity, f2 and f3 grow at the same rate, so all three e grows slower than (ln x)x.
x f2(x) Compare f1 to f3. 0 x e grows slower than x x. x→
x 1
x 10 Thus f1 and f2 grow at the same rate. 23. Compare e x to x x.
lim lim x→ 26. Compare f1 to f2. 5 ln x grows at the same rate as ln x as x→ . x→
x 1 x functions grow at the same rate as x→ . 5 x 10x By transitivity, f2 and f3 grow at the same rate, so all three 2 ln x grows faster than ln x as x→ . x lim x→ Thus f1 and f3 grow at the same rate. 2 2) x→ f1(x) Compare f1 to f3. grows slower than ln x as x→ . x→ 21. lim f2(x) Thus f1 and f2 grow at the same rate. 1
x
x→ e ln x 20. lim 331 functions grow at the same rate as x→ . x/2 Compare e to e .
ex
x/2
x→ e lim e x/2 lim 27. Compare f1 to f2. x→ x x/2 e grows faster than e .
lim x→ Compare x x to (ln x)x.
x x
x
x→ (ln x) lim lim x→ x
ln x x x
x→ ln x since lim 1
x→ 1/x lim f2(x)
f1(x) lim 24. Compare 2x to x 2.
(ln 2)2 x
2x
x→ x→ lim x→ Compare 2 x to (ln 2)x.
2x
(ln 2)x
x→ lim x→ 2x
ln 2 since 2
ln 2 lim x→
x 2
ex lim x→ e 0 since since lim x 2
x→ lim x→ lim x→ 1 2
e 9x 4x
3x
9x 4x
9 1 x 4x
9 1 Thus f1 and f3 grow at the same rate.
By transitivity, f2 and f3 grow at the same rate, so all three
functions grow at the same rate as x→ . 1. Compare x 2 to (ln 2)x.
x2
(ln 2)x
x→ 2x
9 1 x→ 2 grows slower than e x. lim 9 1. Compare 2 x to e x.
2x f3(x)
f1(x) lim 2 x grows faster than (ln 2)x.
x 2x
x Compare f1 to f3. (ln 2)2 2 x
2
x→ lim 2 x grows faster than x 2. lim 9x Thus f1 and f2 grow at the same rate. get e x/2, e x, (ln x)x, x x. lim 9x 2x
3x x→ Thus, in order from slowestgrowing to fastestgrowing, we 2x
2
x→ x x→ lim . x x grows faster than (ln x)x. lim lim and lim (ln 2)x
x→ 0. x 2 grows faster than (ln 2)x.
Thus, in order from slowestgrowing to fastestgrowing, we
get (ln 2)x, x 2, 2 x, e x. 332 Section 8.2 28. Compare f1 to f2.
x4 lim x→ f2(x)
f1(x) 2x 2
x lim
x 2x x→ x→ 1 1
x lim f1(x) lim x→ x→ lim x→ 2x 5 1 5 lim
1 x→ 1
x5
1 (c) True, since lim
x→ x2 x 1 x
5 1. ex
e2x x→ lim x 1
x 1 f (x)
g(x) , so lim
x→ g(x)
f (x) 1
2 0. 0. lim 5 1 1 x→ lim x2
x
x f (x)
g(x) 1 and not equal to zero. Thus, f 1 lim lim x→ 1
x 1. 1
x2 5
x2 1 3 1 1
x2 1
x
1
x lim 1
1
x 1 lim x→ ex
.
xn
x→ … ex
x→ n! lim 1.
(b) The nth derivative of a x, a 1 1, is (ln a)n a x. We can
ax
n.
x→ x apply L’Hôpital’s Rule n times to find lim 1. ax
n
x→ x
1
x 1
x→ 1
x 3 and not equal to zero. Thus, f Thus e x grows faster than x n as x→ for any positive 0. 1 x
x ex
n
x→ x lim lim 1
x f (x)
g(x) integer n. x→ 3 o(g), so i is true. L’Hôpital’s Rule n times to find lim 1 x→ 2x
x2 35. (a) The nth derivative of x n is n!, a constant. We can apply 0. 1/x
x→ 1/x ln x
x→ ln 2x (g) False, since lim (c) False, since lim 1/x lim x→ 1. and g grow at the same rate, so iii is true. 1 ln x
x x→ x→ 0 0. Thus f 34. From the graph, lim 1
ex x→ 1
ln x f (x)
g(x) x→ (e) True, since lim (b) True, since lim 1
x and g grow at the same rate, so iii is true. 1. 30. (a) True, since lim 1
x ln x o( f ), so ii is true. x→ 0. 1 5 x x→ Thus g 0. x (h) True, since lim x→ 0. 33. From the graph, lim 1 1
2 (f) True, since lim 31. From the graph, lim x→ x
2x x→ 1/x
1 lim x→ 32. From the graph, lim (d) True, since lim
x→ 1. x2 1
lim
2x 2
x→ 2 x
29. (a) False, since lim
x→ x
x→ ln x
ln (x 2 1) (h) False, since lim x3 Thus f1 and f3 grow at the same rate.
By transitivity, f2 and f3 grow at the same rate, so all three
functions grow at the same rate. (b) False, since lim ln x
x lim 1 1 1 x→ x3 x 1
ex 1 x2 2
x→ lim x→ ln (ln x)
ln x x→ x→ lim x ln x
x2 (f) True, since lim (g) True, since lim
5 1
ex x→ x→ 1 Compare f1 and f3. x→ e lim 1
x4 1 2x 3
.
2 ex lim 1 Thus f1 and f2 grow at the same rate. f3(x) x
x x3 2
x2 1 ex (e) True, since lim 2 x4 lim 3
2 lim x→ 1 4 x→ cos x
2 x→ x3 x→ lim 2 (d) True, since lim 1 … (ln a)n ax
n!
x→ lim Thus a x grows faster than x n as x→ for any positive
1 1. 1 0. integer n. Section 8.2
39. Compare n log2 n to n3/2 as n→ . 36. (a) Apply L’Hôpital’s Rule n times to find
ex lim anx x→ n an 1x … n1 ex lim anx n x→ an 1x n a1 x … 1 a0 a1 x lim . n log2 n
n3/2 n→ ex
x→ ann! lim a0 log2 n lim n1/2 n→ (ln n)
(ln 2) lim n→ an 1x … 2
n1/2(ln 2) n→ a0 as x→ . a1x 1
n ln 2
1
2n1/2 lim anx n1 n1/2 lim Thus e x grows faster than
n n→ x→ lim x→ ax
anx n an 1x … n1 ax
anx n an 1x a1 x … n1 a0 a1 x Compare n log2 n to n(log2 n)2 . lim … a0 n→ Thus ax grows faster than
an 1x n ln x
37. (a) lim 1/n
x→ x … 1 1
x lim x→ lim x→ 1 (1/n) 1
x
n n 1
x lim ax a x→ 38. lim
x→ 1
a
x→ ax lim 1 0 ln x
an 1x n 1 x→ nanx n 1 x→ f (x)
g(x) 1 M for some integer M. lim … a1 x a0 1)an 1x n (n (n 1
L 1 N for x5
2
x→ x 43. (a) lim 1)an 1 … 2 a1 xn 1 … a1 x 0 Thus ln x grows slower than any nonconstant polynomial as x→ . lim x 3 x→ x 5 grows faster than x 2. 1
nanx n g(x)
f (x) 42. (a) The limit will be the ratio of the leading coefficients of
the polynomials since the polynomials must have the
same degree. 5x 3
3
x→ 2x (b) lim
x→ L L. Then for sufficiently (b) By the same reason as in (a), the limit will be the ratio
of the leading coefficients of the polynomial. 1
x lim number L such that lim some integer N. 0. anx n 19.9; it might take 20 binary Similarly, for sufficiently large x, Thus ln x grows slower than x a as x→ for any number
a 0 f (x)
large x,
g(x) positive integer n.
ln x
xa
x→ 1
log2 n 41. Since f and g grow at the same rate, there exists a nonzero Thus ln x grows slower than x1/n as x→ for any (b) lim n→ (b) log2 1,000,000
searches. 0 x1/n lim n(log2 n)2 40. (a) It might take 1,000,000 searches if it is the last item in
the search. a0 as x→ . a1x n log2 n Thus n log2 n grows slower than n(log2 n)2 as n→ .
The algorithm of order of n log2 n is likely the most
efficient because of the three functions, it grows the most
slowly as n→ . (ln a)na x
lim
ann!
x→ anx n 0 Thus n log2 n grows slower than n3/2 as n→ . (b) Apply L’Hôpital’s Rule n times to find
lim 333 5
2 lim x→
3 5
2 5x 3 and 2x have the same rate of growth.
(c) m n since lim xm
xn x→ (d) m n since lim xm
xn x→ x→ x→ lim x m n lim x m n .
is nonzero and finite. (e) Degree of g degree of f (m n) since lim g(x)
f (x) (f) Degree of g degree of f (m n) since lim g(x)
is
f (x) nonzero and finite. x→ x→ . 334 Section 8.3 44. (a) f f (x)
x→a g(x) o(g) as x→a if lim 3. If p 0 1 Suppose f and g are both positive in some open interval
containing a. Then f 0 1, then
1 dx
xp lim c→0 O(g) as x→a if there is a positive integer M for which f (x)
g(x) lim c→0 where M is a bound for the absolute value of f
h (b 4 as h→0, so ES M
a)
180 4. If 0 b a4
hM
180 (b) From Section 5.5, we know that ES [a, b]. Thus, int (b (4) 1 on M
a)
180 0 b (c) From Section 5.6, we know that ET 12 f(x) 45. (a) lim
x→ (b h2 h→0, so ET c→0 a) M
12 int (b a) x→ (b) lim
x→ lim x→ g(x) f (x)
g(x) 1.
1 as f (x)
x→ g(x) ln x 3 1 x 1 2 1 x 1
4 4 L
4. f ( x)
g( x) x→ lim x→ Thus f ( x) grows faster than g( x) by definition.
f ( x)
(b) lim
x→ g( x) f (x)
lim
g(x)
x→ L 1 cos x 1 Investigating
0 1
0 dx
x f (x)
x→ g(x) 1 lim
c 0. 10. lim
x→ dx
x lim ln x c→0 C 1
c x2 1. 1
x2 x2 1 x for x 1 4
3 4
3 1
x
4e x
x
x→ 3e 4e x
x
x→ 3e 5
7 lim f (x)
g(x) lim x→ lim ( ln c) c→0 C
3 lim lim x→ Thus f and g grow at the same rate as x→ . dx
xp 1
has an infinite discontinuity at x
xp c→0 1 9. lim (pp. 433–444) 2. 1 x 2 so 1 1. Because dx
x2
2 x2 1
x2 s Section 8.3 Improper Integrals 1
ln 2
2 1 1, so cos x cos x cos x
x2 8. x 2 definition. ln 2 6. x 1 0 for x 1
The domain is (1, ).
7. Thus f ( x) grows at the same rate as g(x) by Exploration 1 1 1
x
tan 1
C
2
2
13
x 4 dx
x
3 dx
x4 ln 3 1 5. 9 x 2 0 for 3 x
The domain is ( 3, 3). f (x)
g(x) p ln 6 1
ln x 2
2 dx
2 3 1
x
2 tan 1
4
2 Thus f grows at the same rate as g as x→ by 46. (a) lim 0 1
1 0 x dx 2. 3. definition. 1
ln 2
2 p1 3 dx
x 0 lim f (x)
lim
x→ g(x) 0. dx
xp 1
p 1c
1 cp1
p1 c→0 Thus f grows faster than g as x→ by definition.
f(x) 1) Quick Review 8.3 O(h ). Thus as h→0, ET→0.
f (x)
g(x) x lim 3 M
12 c lim 2 lim g(x) lim c→0 a2
hM because ( p 1, then
1 dx
xp 1 where M is a bound for the absolute value of f on
[a, b]. Thus p O(h 4). Thus as h→0, ES→0. ET p1
1
p 1c
1 cp1
p1 c→0 M for x sufficiently dx
xp x lim close to a. ES c lim x→ 2x 1 x 3 2x
x 1
3
2 lim x→ 1 1
x
3
x 2 Section 8.3
Section 8.3 Exercises
1. (a) The integral is improper because of an infinite limit of
integration.
x2 0 lim b→ 1 0 4. (a) The integral is improper because of two infinite limits
of integration.
0 2x dx
2x dx
2x dx
2
1)2
(x2 1)2
(x 2 1)2
0 (x
0
2x dx
2x dx
lim
2
1)2
(x 2 1)2 b→
b (x (b) b dx (b) dx 0 x2 1
1 lim tan b→ b x 1 lim (tan b (x 2 lim 0 b→ b→ 0) b b→ 2x dx
(x 2 1)2 0 The integral converges. b 1) 2x dx
(x 2 1)2 lim b→ 0 lim ( x 2 1) 1 lim [ (b2 1) 2x dx
(x 2 1)2 1 0 b→ 2. (a) The integral is improper because the integrand has an
infinite discontinuity at x 0. 1 b 1 b→ 2 0 1 1)
(b 2 lim [ 1 2 (c) 335 1 1 1] 1 0 The integral converges.
1 dx (b)
0 x 1 dx lim b→0 b (c) 0 x 5. (a) The integral is improper because the integrand has an
infinite discontinuity at 0. 1 lim b→0 2 x
b lim (2 2 b→0 b) ln 2 2 (b) x 2 1/x e ln 2 dx 0 x b→0 b e1/x lim The integral converges. b→0 2 1/x e b 1 (b)
8
0
8 0 dx
x1/3
dx
x1/3 8 b b→0
b→0 lim b→0 0 dx
x1/3 0 lim
lim 1 1 dx
x1/3
8 lim b→0 lim b→0 lim b→0 b (c) No value
dx
x1/3 (b) b 8 dx
x1/3 d b→0 6 6 3
2 /2 b lim b→0 dx
x1/3 3 2/3
x
2 cot
b lim b→0 b→0
1 cos d
sin /2 b ln sin b ) The integral diverges. b (c) No value 3 2/3
b
2 dx 7. 1.001 x b lim b→ lim 3
2 d ln sin lim (0 1 6 lim 8 The integral converges.
9
2 /2 cot
0 3
2
1 6. (a) The integral is improper because the integrand has an
infinite discontinuity at x 0.
/2 3 2/3
b
2
1 The integral diverges. dx
x1/3 3 2/3
x
2 e1/b] b→0 3. (a) The integral involves improper integrals because the
integrand has an infinite discontinuity at x 0. dx ln 2 lim [ e1/ln 2 (c) 2 (c) lim 9
2 b→ 1 dx
x1.001 1000 x lim ( 1000b b→ 0.001 b
1 0.001 1000) 1000 336 Section 8.3
1 0 dx
x 2/3 8.
1
0 1 0 b dx
x 2/3 1 1 dx
x 2/3
1 2 b→ 1 b→0 b lim 1
4
0 4 b 2 b→4 0 dr lim b→0 r b lim b→0 1 11.
0 1 x 2 b 4) r 0.001 1000b0.001) x 1 sin 1 b→1 lim b→ lim b→ lim b→ 4 b tan 4
2 2 tan
3
4 2
b s
4 s 2 ds b b→2
1b b b 1 2b) 3 0 b→2 0 lim 4 b→2 4 2 s2 1 sin 2 lim 4 b→2 sin 1 1 2 1
s 1
s2 4 s lim s lim b 3 2 2
0 2 2
1 2 b→0 16. (1/2) dx
(x/2)2 1
1x 2) d
2 2 lim ( 0) 2
2 1 (2 b 1
2 3 ln 2 b→0 x
b 3 ln b lim b
0 1 1 3 ln 2 b→0 2 2 b lim 2 3 ln t
b t 1 1
2 0 dx 0 1
1 2 3 ln 1 1000
15. lim (sin 12. 1 b→ 1 lim 2 dx
x2 4 t lim 3 ln 1
b b→1 2 1 b→ 4 dr lim b
b b 1000r b→1 b 3
dt
t 1 lim 3 ln t 0.999 b dx b t 2 b→ b→0 2 ln 3 3 b→ r 4 lim (1000 1 (t 1)] dt
t(t 1) 2 lim 3 ln 1 0.999 3[t lim 0 b→4 ln 1 lim r 4 lim ( 2 10. ln b→ dr
4 lim 1 3 dt
t2 t 14. 0 ln x 3 6 b→4 ln 3 ln 3 3b1/3) lim r dx 1 b b→ 2 dr 9. 3 1
1 lim 1
b 3 x
ln
x b→ 3x b→0 dx
x 2/3 x 1 3 dx
x 2/3 lim (3 1)] dx
1) 1
1 ln x b→ 3) 1/3 b→0 1 1
x b lim 1 lim (x
1)(x 2 b→0 b 1)
(x lim lim (3b1/3 dx
x 2/3 [(x b lim b→0 1 2 2 dx
lim
x 2 1 b→ b 3x1/3 lim 13. dx
x 2/3 lim b→0 2 dx
x 2/3 b2 2 sin
2 (s/2)2
1s 2
1b 2 2 4 ds b
0 2 s2 ds Section 8.3
dx 17. First integrate
(1 x) 20.
1 x b d
5 2 b→ 6 dx. b→ (1 x) 1 2 tan 1
2 2 ln lim ln 2
3
2
3 ln C
x C dx 0 x) x
1 b
b b (1 lim 2 tan lim (2 tan 1 b→0
b→0 1 lim (2 tan c→ 2
1 x x2 1 x 1) b 16 tan 1 x
dx
1 x2 0 1 x2 1 b→ lim 1 b→1 b x x2 1 8 4 x 1 22.
1 2 1 sec b) 0 2 sec b
2 1 lim (sec dx x 3 b→0 3 1 2 x
1 lim ( 2 x
4 b sec x x b b→0 b 1 2) 0 4 dx lim b→0 x 2) b 4 dx b x 2 b→0 6 6 4 b ds
s s2 sec 1 lim (sec 1 lim b→1 1 1
2 s
b b→1 1 2 2 sec sec
1 1 1 b) 3 dx
x 2 x
b 4 b→0 2 2 lim sec 1 lim lim 3 b→1 dx lim x dx 1 2 4 2 0 dx 0 b lim ds b)2 b 1 2 1 4 b→0 x x2 b→ 1 1 2 2 dx 1 x dx 1 0 dx 1 1 b 1 b b→ x)2 0 2 1 2 1 sec lim 1 2 x2 x lim sec s s2 0 b→ dx x b→ 1 16 tan 1 x
dx
1 x2 lim dx 2 1 dx 2 C lim 8(tan dx sec 19. x)2 b→ lim (sec x x2 1 8(tan C b) b→1 1 8u2 16u du 2
2 tan . 1 1 b→1 x x2 c 1 x2 1 16 tan 1 x
dx
1 x2 x c→ b lim 2 x) lim 2 tan 2 tan c (1 dx du x) x lim 8 (tan dx
x x2 1 2 tan 1 dx
1 1 x 0 2 18. c→ x 1 lim (1
dx c x) 16 tan 1 x
dx by letting u
1 x2 1
2 dx
1 x) x dx lim b→0 dx
(1 21. Integrate 3
1 lim ln Now evaluate the improper integral. Note that the integrand
is infinite at x 0. (1 d b
1 b→ u b→ 0 3 b 1 1 1 1 b→ 2 tan 2)
d
2) lim ln x
2 du
1 u2 x 3) (
3)( ( 1 lim 2
dx ( lim
b 1 x, so du u by letting 4 2 6 2 ln 2 1 x, so 337 338 Section 8.3 23. Integrate e d by parts. x e 25. 0 e x dx dx x e dx 0 u
du d 0 ed dv
v x e ed 0 e e b→− lim b→ 0 ed lim b→− dx b b lim (1 e b→ 1) b→ 0
b lim 0 lim ( e ed b→ x e ex lim b
b dx 0 C e x dx lim e
e ed 0 e x dx b→−
x eb) b
0 1 0 e lim b→ e
be b lim ( 1 b→ Note that lim be b lim b→ lim c→ 1
ec u eb) ce c→ 0 and lim e b dx 1 1 sin 2 sin c 26. Integrate x ln x dx by parts. c
ec lim c→
c u 0. c→ e dv 1
dx
x v x ln x dx d 2 cos d v e sin d 2e Integrate 2e cos sin 2e cos d lim 2 cos du 2 sin d e v 12
x ln x
2 b→0 1
4 lim d cos d cos 2e sin b→0 d
lim b→0 Thus,
2e sin b2
2 27.
2e cos 2e sin b tan
0 sin d lim sin d 2e 2e sin d e sin
sin 2e sin d lim b→ 2e
e lim b→ 2e cos lim cos C1 The integral diverges. d
b lim ( e b→ sin
sin e sin b e cos
0 b b cos b sin 1) lim b→0 d ln cos lim [ ln cos b b→ /2 C 0 e 12
b
4 b d b
0 b ln b
1/b 2 lim b→0 b→ /2 0 cos
b→ /2 2 2e 1 0. /2 d
2e 12
x
4 1
4 e
2e 12
x
4 12
b ln b
2 Note that lim b 2 ln b
2e 12
x ln x
2 b lim d by parts.
dv 1
x dx
2 x ln x dx b→0 b→0 u 12
x
2 1 x ln x
0 2e x dx 12
x ln x
2 du d by parts.
dv ln x 1 du 2 1 lim e b→ 24. Integrate 2e x e
b 1 0 0] 1/b
2/b 3 C 1 Section 8.3
sin 28. On [0, ], 0 1 , so 1 32. 0 1
x sin 1 d 0 4 0
b 1 d lim b→ 0 1 b lim b→ x 4 b lim 2 x b→ x lim [2 b b→ 4 4] Since this integral diverges, the given integral diverges. d 0 on [4, )
1
dx x dx d 339 b lim 2 b→ 1 33. 0 x3 0 lim ( 2 b b→ 2 )
1 1
on [1,
x3 1 b dx
x3 lim b→ 1 lim 2 12
x
2 2 b→ Since this integral converges, the given integral converges. 2xe x2 dx x2 2xe 0 dx 2xe x2 dx 0 x2 2xe b dx lim b→ 0 x2 2xe
e lim [ e dx 2xe x2 lim e x 0 x lim b→1 b→1 b→0 b2 1 dx 2 lim [ 2e 2 35. 1
ln 1
2 x 1
ln 1
2 1
1
ln
2
1 x
x 1
1
ln
2
1 b
b 0 1 0 b] x dt
0 x t lim b→0 b 2 lim [2 b→0 x lim b→1
b→1 2 0 1 dx
b x
0 b
0 0 dx
1 x dx
1 x ln 1 lim ( ln 1 x
0 b 0) Since this integral diverges, the given integral diverges. on (0, ] since sin t 0 on [0, ]. dt lim b→0 x) b t sin t 1 lim The integral converges. t 2 dx
b dx
1 0 b→1 1 1
2(1 1 dx 1 2e b→0 1 1
x) b→1 b 2 x) (1 x)]
dx
x)(1 x) 2(1 b→1 x 2e 2e x2 lim dx x b→0 1/2[(1
(1 0 0 e x b lim dx
1 1 Since this integral diverges, the given integral diverges. e lim 2 lim 1 4 31. 0 0 b→1 4 dx x lim The integral converges.
e 1 1 0 2 dx
b b b→ 1
2 lim x2 x2 lim [ 1 4 x 2 dx b lim 1
2 b 1] 2xe b→
b→ 30. x 1 2 0 0 dx 0 1 dx 1 b b→
0 2 34.
0 b2 lim b→ 1 Since this integral converges, the given integral converges. 0 x2 b 12
b
2 b→ 29. dx
x3 lim 20 ) t t
b 2 b] 2
Since this integral converges, the given integral converges. 340 Section 8.3
1 1 36. ln x dx 2 ln x dx by symmetry of ln x about the 1 0 40. 1 dx
0 yaxis. Integrate ln x dx by parts.
1 dx 0 x
dx dx
1 x
b lim b→ x x dx 1 x
b u ln x dv 1
dx
x du ln x dx v lim 2
lim (2 x ln x dx x ln x x 2 lim
b→0 0 b→0 1 x ln x x
b 2 lim [ 1 b ln b b→0 Note that lim b ln b b] b→0 1/b
1/b 2 b→0 1 42. 0 0. 2 1 on [2, ) x2
b lim b→ 2 2x 1 2b 2 1
dx
x lim (ln b b→ dx
b 2 2 b lim ln x b→ lim (ln b b→ ln 2) 1
ex e
1 x. ex
e x(e2x 1)
1 (e x)2
e x dx
Integrate
by letting u e x so du
1 (e x)2
dx
e x dx
x
x
e
e
1 (e x)2
du
1 u2
e x tan 1 u tan 2 1 ex C Since this integral diverges, the given integral diverges. 39. Let f (x) x1
and g(x)
x2 1
. Both are continuous on
x3/2 [1, ).
x1
f (x)
lim
lim
x→ g(x)
x→
x→
x
b
1
dx lim
x 3/2 dx
x3/2
1
b→ 1
b→ lim ( 2b b→ 1/2 x dx
e ex x 0 x 1 lim b→ tan 1 1 b→ 4 2) 2 Since this integral converges, the given integral converges. 0 dx
ex e 0
b x lim b→ x 0 ex 0 4
dx
ex e
1 ex lim [tan 1 b b→ tan 1 lim tan 1 e b] tan 1 1] x b
0 b→ 2 dx
e ex b b→ b 0 dx
e ex b lim 1 1/2 dx
e
dx
ex e C lim [tan 1 lim 2x 1
x 0 ex
0 lim ln ) ) 1 1 ex Since this integral converges, the given integral converges. dx
x 2x b→ 43. First rewrite 1
e 1
x b Since this integral converges, the given integral converges. e 1] b→ 38. 0 b lim lim 1 b 2
on [ ,
x2 b→ b lim [ e sin x
x2 b→ e b→ cos x
on [ , )
x
b
dx
lim
lim ln x
x
b→
b→ lim 1
1
on [1, )
1e
e
b
1
d
lim
ed
e
1
b→ lim 2 2 dx
x2 The integral converges. 1 2) Since this integral diverges, the given integral diverges. 2 lim ln b
1/b lim b→0 b 1
x
dx
x 41. 0 ln x dx
b 2 lim 37. 0 1 Since this integral diverges, the given integral diverges. C 1 lim b b→ ln x dx b→0 x b→ x 1 2 dx 4 e 4 Thus, the given integral converges. x e x dx. 341 Section 8.3
dx 44. 1 x4 1 dx 1 0 x4 1 1
on [1,
x2 1 u lim 2 x b→ dx
b 1 x
1
b lim b→ dv ln x
x2 Area 1 1 ln x
dx
x2 dx
x2 v 1
dx
x 1 lim 1 1
x ). 1 b→ b→ ln x du
b 1
dx
x2 1 b lim ln x
dx by parts.
x2 exists on [0, 1]. x4 1 1 Integrate 1 0 on [1, ). ln x
dx
x2 1 x4 1 1 exists because x4 0 0, y Area dx x4 47. For x 1 dx 0 1 by symmetry about the yaxis x4 0 1 dx
0 dx 2 x4 ln x
x dx
x2 ln x
x b→ Note that lim
b→ C b ln x
x lim 1
x 1
x ln b
b b→ ln b
b lim b→ 1 lim 1/b
1 1
b 1 1 0. Since this integral converges, the given integral converges.
48. For x
45. Integrate (1 dy
y )(1 tan
2 1 y) by letting u 1 y2
dy
y 2)(1 tan 1 (1 du
1 y) 1 u ln 1 (1 C ln 1
dy
y 2)(1 tan u
tan 1 b 1 lim y) b→ 0 (1 y C ln x
ln x
dx lim
dx
x
1
b→ 1 x
ln x
dx
Integrate
dx by letting u ln x so du
.
x
x
ln x
12
1
dx
u du
u
C
(ln x)2 C
x
2
2
b
1
1
Area lim (ln x)2
lim (ln b)2
2
b→
b→ 2
1 Area 49. (a) The integral in Example 1 gives the area of region R. dy
y 2)(1 tan 1 y) Area
1 lim ln 1 b→ tan 1 tan 1 lim (ln 1 b→ b 1
x2 y
0) The surface area of the solid is given by the following
integral. 2 The integral converges. 2
1 e y dy
y2 1 46.
0 lim y→ 0 e y dy
y2 1 0 y 1 lim 2
y→ y ey
1 1
x 1 12
dx
x2 1 ey
2y
y→ lim ey
y→ 2 lim 1
x 1 x4 1
on [1,
x3 1
x4 dx x4 1
dx
x3 2
Since 0 x4 1
x 2 e y dy
y2 1 e y dy
diverges since
y2 1
e
y2 dx
x (b) Refer to Exploration 2 of Section 7.3. b y
0 ln 1 0 on [1, ).
b y so dy du 0 tan 0, y ), the direct comparison test shows that the integral for the surface
area diverges. The surface area is . Thus the given integral diverges.
(c) Volume
1 12
dx
x 1 1
dx
x2
b lim b→ lim b→ lim b→ 1 1
dx
x2
1
x
1
b b
1 1 (d) Gabriel’s horn has finite volume so it could only hold a
finite amount of paint, but it has infinite surface area so
it would require an infinite amount of paint to cover
itself. 342 Section 8.3
1 50. (a) f (x) x 2/2 e 6, x 2 51. (a) For x x2 6x, so e 6x e 2 e x2 dx 6x e 6 dx 6
b lim , 0]. f is decreasing on [0, ). 1 (b) NINT e 2
1 NINT x 2/2 1, 1 x 2/2 2, 2 1 36
e
6 0.954 3, 3 b→ 0.683 e 2
1 NINT 0, 0.997 , x,
, x, x 2/2 e , x, (b) x2 6 dx 4 x2 e 1 1
6 1
e 36
6
17 10 x2 dx 10 dx 17 e
6 x2 e dx 4 1 2 Thus, from part (a) we have shown that the error is (c) Part (b) suggests that as b increases, the integral
approaches 1. We can make 1 large enough. Also, we can x2 NINT(e (c) . e dx x2 , x, 1, 6) 0.1394027926 1 (This agrees with Figure 8.16.) b f (x) dx and 10 f (x) dx as close to 1 as
b we want by choosing b 17 bounded by 4 b make e 6 1
e 6b
6 lim 1
2 f has a local maximum at (0, f (0)) b 1 6x
e
6 b→ f is increasing on ( dx 6 lim [ 3, 3] by [0, 0.5] 6x e b→ f (x) dx as small as we want b by choosing b large enough. This is because
0 f (x) 0 e f (x) e x/2
x/2 x/2 dx b for x 3 dx x2 e e x2 c lim c→ e x/2 dx b lim 2e x/2 lim [ 2e c/2 e e 3x 3 c dx 3.
b dx 3x 3 3x for x
lim e b→ 3x dx 3 b lim 2e 1
e 3x
3 lim 1 3b
e
3 19
e
3 b c→ 2e since x 2 dx dx 0 dx x2 3 3 1) Thus, x/2 e dx 0 1. (Likewise, for x c→ As b→ , 2e x2 0 b e e e f (x) dx
b (d) 0.000041 b→ b/2 b/2 b→ →0, so for large enough b, b/2 f (x) dx
b 3 19
e
3 0.000042 is as small as we want. Likewise, for large enough b,
52. (a) Since f is even, f ( x) b f (x) dx is as small as we want. f (x). Let u x, du dx. 0 f (x) dx f (x) dx f (x) dx
0 0 f ( u)( 1) du f (x) dx
0 f (u) du f (x) dx 0 0 2 f (x) dx
0 (b) Since f is odd, f ( x) f (x). Let u 0 x, du 0 f (x) dx f (x) dx f (x) dx 0 f ( u)( 1) du f (x) dx
0 f (u) du
0 f (x) dx
0 0 dx Section 8.3
b 2x dx
x2 1 53. (a)
0 lim b→ 0 55. Suppose 0 2x dx
x2 1 1) b→ b Thus the integral diverges.
0 2x dx
2x dx
and
must converge in order
x2 1
x2 1
2x dx
to converge.
x2 1 (b) Both
0 for
b (c) lim
b→ b b 2x dx
x2 1 lim ln (x 2 1) b→ lim [ln (b 1) b→ lim 0 2x Note that x2 ln (b If the infinite integral of g converges, then taking the limit
in the above inequality as b→ shows that the infinite integral of f is bounded above by the infinite integral of g.
Therefore, the infinite integral of f must be finite and it converges. If the infinite integral of f diverges, it must grow to
infinity. So taking the limit in the above inequality as b→
shows that the infinite integral of g must also diverge to
infinity. 1)] 0:
b x dx e x e x lim [ e e b lim b→ 0
b
b b→ 2x dx
= 0.
x2 1 (d) Because the determination of convergence is not made
using the method in part (c). In order for the integral to
converge, there must be finite areas in both directions
(toward and toward
). In this case, there are
infinite areas in both directions, but when one
computes the integral over an interval [ b, b], there is
cancellation which gives 0 as the result. For n 1:
ux
du dx
xe x b dx 0 lim b→ xe y 2/3 xe y 1 x dy
dx 3
(1
2 x dx x 1/3 (1 x 2/3)1/2 x2 e 2/3 x 2/3) (1 2/3 (x x b dx lim b→ 1)
b→ dy 2 1/3 2/3 x dx 1/3 lim b→0 lim b→0 lim b→0 3 2/3
x
2 Thus, the perimeter is 4 1 x2 e x dx 0 x2 e x b b b b 2
eb 2 3 2/3
b
2
3
2 3
2 6. b→ x 2x e dx 0 0 lim 1 dx e x dx
ex 2(1) b→ x 1 2b
eb dx dx 0 lim 1/3 b 3
2 e b→ 2 lim b→ x b lim b2
eb lim 1 dx x x 0 dv
v lim 1 0 x 0 dy 2 1 For n 2:
u x2
du 2x dx
2 1/3
x
3 b e 1
eb lim x 2/3)1/2 b x b
eb b→ x 2/3)3/2 (1 dx 0 b→ 2/3 x 0 b→ 0. 1 e x dx
ex dv
v lim 1, y 1] b→ 54. By symmetry, find the perimeter of one side, say for
x b
0 = lim 0 dx 0 lim is an odd function so 1 2 0. b→ a 56. (a) For n b 2 a, g(x) dx. a 1) b→ a. b f (x) 0 lim ln (b 2 g(x) for all x From the properties of integrals, for any b b lim ln (x 2 f (x) 343 b→ xe
0 2 x dx 344 Section 8.4 56. continued c n x (b) Evaluate x e
u xn
du nx n 1 (b)
dx using integration by parts
dv e x dx
v
ex 0 f (x) dx c f (x) dx f (x) dx
0 0 f (x) dx f (x) dx c c f (x) dx
0 Thus,
n x xe n dx x xe nx n1 x e dx c f (x) dx f (x) dx
c f (n x ne 1) x 0 dx b→ x ne b x nx n 1e 0 lim b→ x n 1e (c) Since f (n
f (n x n xe 0 0 nf (n),
1) … f (1) n!; thus dx converges for all integers n 0. 0 sin x
, x, 0, x or create a table
x 57. (a) On a grapher, plot NINT of values. For large values of x, f (x) appears to
approach approximately 1.57.
(b) Yes, the integral appears to converge.
1 58. (a) 1 dx
1 x 2 dx lim b→ x2 1 b tan lim (tan 1 b→ 4
1 1 1 lim b→ 1 lim [tan 1 b x b→ 1 b→ 2
1 x2 b) x2
1 3
4 1 dx
1 lim tan dx tan 3
4 2
b x x
b b→ 2 1 1 lim dx 0 c f (x) dx
c c f (x) dx
0 f (x) dx 0. 0 s Section 8.4 Partial Fractions and Integral
Tables (pp. 444–453)
Quick Review 8.4 n(n x f (x) dx, because f (x) dx dx 0.
1) 1) f (x) dx
c bn
eb f (x) dx
0 0 Note: apply L’Hôpital’s Rule n times to show that b→ f (x) dx
c 0 dx 0 n x nf (n) bn
eb 0 f (x) dx
0 lim lim c f (x) dx 0 4 b 4
4 tan 1 1] 1. Solving the first equation for B yields B
3A
Substitute into the second equation.
2A 3( 3A 5) 7
2A 9A 15 7
11A 22
A
2
Substituting A
2 into B
3A 5 gives B
solution is A
2, B 1. 5. 1. The 2. Solve by Gaussian elimination. Multiply first equation by
3 and add to second equation. Multiply first equation by
1 and add to third equation.
A 2B C 0
7B 5C 1
B 2C 4
Multiply third equation by 7 and add to second equation.
A 2B C 0
9C
27
B 2C 4
Solve the second equation for C to get C 3. Solve for B
by substituting C 3 into the third equation.
B 2(3) 4
B
2
B2
Solve for A by substituting B 2 and C 3 into the first
equation.
A 2(2) 3 0
A10
A
1
The solution is A
1, B 2, C 3.
2x 1
3. x 2 3x 4 2x 3 5x 2 10x 7
2x 3 6x 2
8x
x2
2x 7
x2
3x 4
x3
2x 1 x
x2 3
3x 4 Section 8.4 4. x 2 3x 2
5. x x2 3 11x
8x
3x x 6. y 7. 8. 5y 2 3 4 2
x x (x (y
(y x 2 3t
t2 2 3) 2 3)(x 2 (x 3) 1) 4)(y
1)
2)(y 2)(y A
1)(y x2 3. 2 (x 2)(x A A x 1) B(x 4. s 3 B)x (A x 7. 2
x B 0, B 2, B s2 C(t (B C 1) C)t 1, and 2, C C C 1. 1.
2
t 2
t 1 s(s 2 6s A
s 4
s2 A( s 3)(s
s 6s (A 1 Solving the system simultaneously yields A
2 1) B)t 2 A( s 2 Equating coefficients of like terms gives 3 C
t2 Bt(t At 2 4 2B) x 1)2 (x
B
t 1 s3 2) 2 1 A t1
t 2(t 1) 1) A( x 7 4 x t 2, B Solving the system simultaneously yields s3 7 3x B) 2. 2
1 1 1 B B C 2
,B
3 A 2 5x B 1
t2 s 6) s(s B
s 2) C)s 2 2) C
3 s 2 B(s)(s
B(s 2 6) 3)(s (A 2) 2s)
2B C(s)(s
C(s 2 3) 3s) 3C)s 0, A 2B 3C 0, 6A Solving the system simultaneously yields (A x2 B 2
2x t1
t 2(t 1) A 3x 2B 1)2 (x Equating coefficients of like terms gives B 5 and A B
1 6A Equating coefficients of like terms gives A
2 A (A 3 7 B (A t 1 5x A 1) 2x Section 8.4 Exercises
3x A( x 2 and x2 1
(x 1)
(x 1)
2(x 1)2
3(x 1)
1
(x 1)3
(x 1)3
(x 1)3
2x 2 4x 2 3x 3 1
(x 1)3
2
2x
7x 6
(x 1)3 5x 2 A Solving the system simultaneously yields A 1) (3t 4)(t 2 2)
(t 1)(t 2 1)
(t 2 2)(t 2 1)
(t 2 2)(t 2 1)
3
2
t
t
t 1 3t 3 4t 2 6t 8
(t 2 2)(t 2 1)
2t 3 5t 2 5t 9
(t 2 2)(t 2 1)
2t 3 5t 2 5t 9
(t 2 2)(t 2 1) 4
1 3 1. x 2 x 2 Equating coefficients of like terms gives 2 2 x 1 2x 1
2
4x 5
x5
(x 1)(x 5)
2(x 2 4x 5)
(x 5)(x 2 4x 5)
(x 5)(x 2 4x 5)
x 2 4x 5 2x 2 8x 10
(x 5)(x 2 4x 5)
x2 12x 15
(x 5)(x 2 4x 5)
1
2 2x 2x x x2 (x x2 3(x 3)
2(x 2)
(x 2)(x 3)
(x 2)(x 3)
2x 4 3x 9
(x 2)(x 3)
x 13
(x 2)(x 3)
x 13
x2 x 6 3
3 t
9. 2
t 10. 2 1)2 1 5 2 (x
4 2x 4
4x 3x 2. x 2 2
06
10
04 Ax 5 2x 2
2x 2 4x 345 3, B 2. 4
s2 6s 4
,C
15
2
3s 2
.
5
4
15(s 3) 2
5(s 2) 4 4. 346 Section 8.4 5. x 2 6 x2
x2 5x x 2 8 x2 x2 5x 6 (x 5x 6 x 5x 2 A( x
(A 3)(x
x A 2 B(x B 5 and 2A 3B 17, B
x2 x 2 x2 0 and 2A 5x 6. y 2 y3
y2 6 4 y3
y3 x 0y 2
0y 2 9. y 2 (1 x2 1 1
1 x)(1 1 B y B( y 1 3) B)y (A 3B) 3B 0.
3
,B
4 Solving the system simultaneously yields A
y dy
2y 3 y 3/4
dy
3 3
ln y
4 1 1/4
dy
y1
1
ln y 1
4 3 x) B(1 B)x y2 (A A
y 4 A( y 1) 1
.
4 C By (A B) y(y 4
y y x) y y
y2 10. x 0 and A B 1. B)y A 1
.
2
1/2
1/2
dx
dx
1x
1x
1
1
ln 1 x
ln 1
2
2
1
1x
ln
C
2
1x 1)
B
y 1 Equating coefficients of like terms gives Solving the system simultaneously yields
1
A
,B
2
dx
1 x2 1) 1 and A Equating coefficients of like terms gives
A 3 B
x A(1
(A y A( y x) A 1)
B 3 B y2 x2 3)(y Equating coefficients of like terms gives The rational function cannot be decomposed any further.
7. 1 1
.
2 C y 4y 1
y2 4 y (y
A 2 y
1
0
1 0y
4y
4y 3 y
2y y2 A
1
4 2y 2 (A x 1
,B
2 1/2
dx
2 x 1
1
ln x
ln x
2
2
1
x
ln
C
2
x2 12
3 1 1/2
dx
x 2 17 1 2A 2x 12. 8 B)x dx 3B) Solving the system simultaneously yields
A Bx Solving the system simultaneously yields A 3) ( 2A B Equating coefficients of like terms gives
A 2 2) Equating coefficients of like terms gives B
3 B
x (A 6 2) A 2) A( x 2 B)x 2 A
x 1 2) 5x
x2 x(x 2x x2 5x 6 2x
1 5x 1 5x x2 8. x 2 1
8
6
2 0x
5x
5x A B 1 and A 4. Solving the system simultaneously yields A x C y
y2 4
dy
y 4
dy
y 4 ln y y 3
dy
1 3 ln y 1 C 4, B 3. Section 8.4
11. t 3 t2
1
t2 t3 1 t(t 2 2t A
t 2t A( t 2)(t A( t 2 t (A t 2) t(t B 1) 2 t 1 5x 2
5x 2 x 1 B(t)(t 1) C(t)(t B( t 2 t) C(t 2 2t) C)t 2 (A B 2C)t 2A x 5x 2
x 2 5x 5
x2 x 1 5 1 5
5
5
5 5x
5x
5x 2) 2) B 15. x 2 C t 1) 2)(t 5x 2 dx
x
x1 5 dx 2 x
5 5x
B C 0, A B 2C 0, and 2A 1
1
1
,B
,C
.
2
6
3
1/2
dt
dt
t
t 2 2t
1
ln t
2 t 3 12. 2t 3 2t(t 2 8t t3
2t 3 8t t A
t 3 4) 1/6
dt
t2
1
ln t
6 t t 2A(t 2)(t 2A(t 2 4) (2A x2 1/3
dt
t1
1
ln t 1
3 2 x C 2)(t 2) 2B 2C 2B(t 2 2) 0, 4B 3
,B
8
3
dt
8t t
2t 3 13. s 2 4 s 3s 3
s3 2C(t 2 2t)
(4B 4C)t 2t)
8A 4C 1, 8A s2
s2 5
1
,C
.
16
16
3/8
5/16
dt
dt
t
t2
3
5
ln t
ln t 2
8
16 s 4
s3 (x 12
2 1
2 (x x
x x2 4 ds 1/16
dt
t2
1
ln t
16 2 1)(x (s 2
13
s
3 2 x 2 ln (s ds (A 4) 2s
s2 1
1 s 2 1) ds
s ln (s 2 1/2
3/2 2x 1 tan 1
3 5 1) tan 2x 1 1 3 2x 1) (x 3 C 1) 1)2
2 x 2x x C
(x 1 1)2 1)(x 1) B)x 2 B
1 B(x 1)2 1)(x A
2x A( x
A( x 4 ( 3/2)2
x 1 tan 3 1)(x 2 (x 2 1/2)2 3 1 1) x2 s2 2 dx 3/4
dx C
16. (x s ds 1 1
2 1) 5
ln (x 2
2 5x 1) C(x 1) 2 1) C(x 1) A B B(x (2A C)x C Equating coefficients of like terms gives s2 1
4
3
2
1 s + 0s + 0s
2s + 0
s 4 + s3 s 2+ s + 0
s 2 2s + 0
s2 + s 1
2s 1 s 4 2s
s2
s2 1
s 4 2s
ds
s2 1 3
4 1/2
1/2)2 5x 2 dx
x1 4s
s2 4
12
s
2 14. s 2 x 1/2
dx
1/2)2 3/4 The second integral was evaluated by using Formula 16
from the Brief Table of Integrals. 3 s
4s
4s
4s 4s 3
4 dx 1
ln x
2 2) 12
2 x x1
dx
1/2)2 3/4 x2 s3 1 3
4 1
ln (x 2
2 2C(t)(t Solving the system simultaneously yields
A x 1
4 x 1
ln (x 2
2 2B(t)(t dx 1 (x 2) 2 2C)t 2 2B x2 1
1 x2 Equating coefficients of like terms gives
2A x C
2 x (x 2t(t B 1 x2 dx 1 To evaluate the second integral, complete the square in the
denominator. Solving the system simultaneously yields
A x x 5 1. 1 x2 Equating coefficients of like terms gives
A A B 1, 2A C 0, and A B C 0. Solving the system simultaneously yields
1
,B
4 A 1 2s
s2 2s ds
s2 1 1) 347 s2 1
ds
s tan 1 2 1 C 1
.
2 x 2 dx
1)(x 2 2x (x
1
x 1 s 3
,C
4 1/4
dx
1 1
ln x
4 1 1)
x 3/4
dx
1 3
ln x
4 (x 1 1/2
dx
1)2
1 2(x 1) C C 348 Section 8.4 17. (x 2 1)2 1)2(x (x 1 1)2 A (x 2 1)2 21. x 3 B C
1)2 1)(x 1)2 B(x 1)2 1)(x 2 x2 2x 2 A 1)2 x A(x A(x 3 1 x2 C)x x2 x 2B B C 2x 1)2 x3 1)2(x x2 1) C 2D)x 2x C
A 0,
2B A B C C 2D D) x
(A B D 18. x 2 5x
x x2 x 6 4
5x 6 4 1
,D
4 C 1, B
2 6)(x A
x B C D 1 A( x 1) 2
1 dx B)x 1 (x A 4x 4 x3 x2 1 4x r 2 2 dr
2r A 1, r2 4r 5 3 dr
4r 5 1) C x C
x 1 1) B)x 2 A 3, B
x2 B (Bx (A C 2, C 4x
x3 C)(x
B 1) C)x 4, and A 4 (A C) C 4. 1 1.
3 dx x 1 dx 3 ln x
3x 2 2x 12
(x 2 4)2 3x 2 4. 2x Ax
x2 x2 1 2x 1
dx
x1 ln (x 2 x 1) B)(x 2 4) (Cx D) Bx 2 (4A C)x 4B C D
4)2 (Ax 12 Cx
(x 2 Ax 3 23. B
4 D Equating coefficients of like terms gives
5/7
dx
x1
5
ln x 1
7 6 A
C 0, B 2r 1 2 dr
(r 1)2 1 2 tan 1 1)2 (r
1 (r A 0, B r2 4r
(r 3 dr
2)2 4 1
1 2)2 (r
3 tan 1 (r 2 (x 2 1
1) 3, 4A C 2, and 4B D 12 Solving the system simultaneously yields C 3, C 2x 12
4)2 dx 2, D 0. 3
x2 4 dx 3
x
tan 1
2
2 2x
dx
(x 2 4)2
1
C
x2 4 The first integral was evaluated by using Formula 16 from
the Brief Table of Integrals. 20. Complete the square in the denominator.
r2 x 1) x2 A(x 2 4 B A 2/7
dx
x6 r2 2 2. Equating coefficients of like terms gives 3x 2 C) Solving the system simultaneously yields 19. Complete the square in the denominator.
2r x 1 6B) 2
ln x
7 r2 (A C ln (x 2 Bx x Solving the system simultaneously yields
2
5
A
,B
.
7
7
x4
dx
x 2 5x 6 C)x 1) 2x 1
dx
x2 x 1 1 A x 6) 6B B 1
dx
1 1)(x 2 1
.
4 (A (A C)(x 2, and A x 2 1 B(x (Bx 1. 22. x 3 1) x 1 1) C ln x B
6 x 2, C 2x
x3 1/4
1/4
1/4
dx
dx
dx
(x 1)2
x1
(x 1)2
1
1
1
ln x 1
C
4
4(x 1)
4(x 1) (x 1 and 1, A A D) Equating coefficients of like terms gives
B B x B C
x B)x 2 (A (A A A( x 2 2 0, 0, and A x2 1 Solving the system simultaneously yields Solving the system simultaneously yields
1
1
A
,B
,C
4
4
dx
(x 2 1)2
1/4
dx
x1
1
ln x 1
4 x 2x A 1) 2 Equating coefficients of like terms gives
A 1 1) Equating coefficients of like terms gives 1) D(x 2 1) (A (A (x C(x B(x 2 1) 3 1 x
Bx (A x C(x 3
(A (x D x D(x 1 (x 1 1
2) C Section 8.4 24. x 3 2x 2 2
(x 2 1)2 x3 2x 2 Ax
(x 2 Cx
(x 2 B
1) (Ax B)(x 2 Ax 3 2 Bx 2 D
1)2 1) (Cx (A 1
dy
y2 y
1
dy
y(y 1)
1
y(y 1) 27. C)x D)
(B D) Equating coefficients of like terms gives
A 1, B 2, A C 0, B D e x dx
e x dx
A
y 1 ex
B y 1 A( y 1) (A 2 C B(y) B)y A Solving the system simultaneously yields Equating coefficients of like terms gives A A 1, B 2, C 1, D 0. x 3 2x 2 2
dx
(x 2 1)2 x
x2 1
ln (x 2
2 25. x
(x 2 1)2 2
x2 1) 1 y(y dx dx
1 2 tan (x 2 1
1 1
2(x 2 1) x 1 1 d
0 0 1 1 2
0 1 1 1 y ex 1 1, B 1. dy 1 C2 C 2. C or C ln y 1 1 ln 2 dy ex 1 sin 1 ln 2. d 1)2
1
dy
sin d
( y 1)2
1
cos
C
y1 Substitute x 0 1 0 2 ,y 0. C or C 1 The solution to the initial value problem is
1
y
2 1 d 2 d
0 2 0 2 tan
0 2 1
1 1 1 tan 2 1 1
cos 1
1 y d cos 1 1 2 x2 dx
3x 2 3x 2 (x 29. dy
0 1 cos 1 y 1
2 1 1 ln y 1
0, y 0 (y ln 2 2
2 ln 2 1 1
+1
1
1 2
2 ln y Substitute x C d 1 0 1 2 ln y 1 ln 1 dx 28. d 2 1
dy
y dy 1) ln y 1 0 1 The solution to the initial value problem is 1 2 1)2 1 1 A ln y
x 1
+1
1
1 1 0 and 1 dx 1 B Solving the system simultaneously yields A 2
dx
1 x
x2 26. 349 x 2 2)(x x2 1
3x 1 A( x 1) B(x 1 (A B)x A 1) A
2 B x 2 x 1 2)
2B Equating coefficients of like terms gives
A B 0, A 2B 1 Solving the system simultaneously yields A
dy
y dx
3x x2 ln x 2 Substitute x
0 0 dx
2 x 2 ln x 1 C 3, y ln 2 dx
x 1 0. C or C ln 2 The solution to the initial value problem is
y ln x 2 ln x 1 ln 2 1, B 1. 350 Section 8.4
ds 2 dt
t 2 2t
1 ds
2s 1 2t t(t 2) 2t A
t t 1 A( t 2) Bt Let u 1 30. (A B)t 2A with x 2s 2
ds 2s t2
1
t2 32. (a) Complete the square in the denominator.
x2 C1 B 0 and 2A 2t 1
,B
2 1/2
dt
2 t ln s 1 ln t ln t Substitute t 1, s 0 ln 3 1 ln s 2 C or C s ln t 1 ln 2 1 t ln t t 6t
2 ln 3 2 ln 6 . x with x (x 2 1 (x 2 4x (x 2)2 2 so du B 0.5 5 (b) x2 x
d1
ln
x
dx 2a a
a 1
1
2a x a
1
2a (x a 1
x2 C
C a a
(x 1
a2 3
3
1
5 ln x xa
a)(x a)
a2 x2 dx
x2 3x 0.5 x x) Bx B)x 0 and 3A 3A 1 2.5 dx
3x x 2 3
0.5 3 1
x x 9 2.5 dx
x 0.5 dx
3 x 2.5 C 1d
ln x
2a dx 2.5 dx 2 Solving the system simultaneously yields A
9 du
9 u2
1
u
ln
6
u
1
x
ln
6
x xa
a)(x a) 1
.
3 x 2)2 B
3 A(3 2.5 3. 1
,B
3 1 Equating coefficients of like terms gives
A dx, and then use Formula 18 x2 x) (A
4) C 1
1/a
2a 3 1 (x/a)2 (a 2 3x A
x x2 C 1
a2 9 x(3 4x) u and a dx
4x 2 1
3x 9
x x2
x 2)2 33. Volume 1) a2
1
2a 4 a 2 x 2 2a 2
1
2a 2 (a 2 x 2)2 3x 5 1
x
tan 1
2a 3
a x 2) 0.5 9 Let u (1 2.5 ln 6 2 x2 4x dx and then use Formula 17 du
u 2)2
1
u
tan 1 u C
2
2(1 u 2)
x1
1
tan 1 (x
2
2(x 2 2x 2) 2)2 1
a2
2a 2 (a 2 31. (a) Complete the square in the denominator.
5 1)2 1. a2 x2
2a 2(a 2 x 2)2 C 6t ln 1 1 1 (a 2 x 2) x(2x)]
(a 2 x 2)2
2a 2 The solution to the initial value problem is
ln s u and a x
d
dx 2a 2(a 2 1. ln 2 (x 1 so du 1
.
2 (b) 1
1
ln t
ln t 2 C2
2
2
1
1
1
ln t
ln t 2 C3
2
2 1
ln s
2 dx
2x (x 2 x 1 1/2
dt
t dt 1)2 1) 2 Solving the system simultaneously yields A
t2 2 B Equating coefficients of like terms gives
A 2x 2x 1 1 (x 2
(x 1
ln s
2 ln x 2.5 ln 3 x 0.5 3 (ln 2.5 ln 0.5 3 ln 25 6 ln 5 0.5 ln 0.5 ln 2.5) 351 Section 8.4
1 34. Volume 2x 2
1)(2 (x 0
1 4
0 x
1)(2 (x x)
B
2x A
x) x x 1 A(2 x) (A 5 sin , dt 25 x dx
1)(2 (x 36. t dx x) 25 25 sin t 2 dt 25 B(x B)x t 2 1) (2A B 1 and 2A B B) 1 x dx
1)(2 (x 0 4
3 1
0 1
0 ln x 35. y y2 9 25
sin
2 C
cos C 25 t
25
t 25
2 25
5
t2 Use Figure 8.18(b) from the text with a 0 get 3 sec2 d, 9 tan2 t2 C C 5 and x t to x 4 ln 2
3 2 ln 2) 2 t2 25 7
sec , dx
2 37. x
2 9 sec2 3 sec
y 2 4x 2 5 cos . 7
sec
2 49 sec 2 49 d dx 3 sec sec ln 4x 2 d ln sec tan
49 d ,0 2 49 tan 2 tan 1
2x
ln
2
7 4x 2
7 1
ln 2x
2 4x 2 49 1
ln 7
2 1
ln 2x
2 4x 2 49 C C1 9 y2 y ln 3 ln 9 y2 y C Use Formula 88 for sec d with x Figure 8.18(a) from the text with a
y2 3 sec . d 1
ln sec
2 y
3 3 7/2 sec tan
7 tan
1
sec
2 y2 9 49 C1 tan ln 9 25 sin 2
4 2 dy
9 cos 2
d
2 1 2 ln 2
0 3 tan , dy 9 2 dx
2x 1 4
( ln 2
3 1 d 25
2 2
.
3 2 (5 cos ) d 25
t
sin 1
2
5
25
t
sin 1
2
5 1 4
3 1
,B
3 2 25 cos 25
2 0 x) dx
x1 2 2 5 cos 25 Solving the system simultaneously yields A
4 d, 25 cos 2 Equating coefficients of like terms gives
A 5 cos C1 and a 1. Use 3 to get d
C1
49 Use Figure 8.18(c) from the text with a
x2 49
4 7
tan .
2 C1 7
to get
2 C1 352 Section 8.4 38. x sec 2 tan , dx x2 tan 2 1 x2 2 40. x 2 x sec 2 1 sec , dx
2 1 1 sec sec 2 x3 x2 tan 2 cos 2
1
2
1 csc cos 2 d 1
sin 2
2 C x2 1 sec Use Figure 8.18(a) from the text with a 39. x
1 sin , dx
x2 1 1 . 41. z cos d, sin 2 2 2 3 cos sin 3
(1 d d
cos 2 ) sin d 1 x x2 1
3 C 1
x2 1
(1
3
2 C
x) 1 x2
3 Use Figure 18.8(b) from the text with a
1 x2 cos . C
C
1 4 ln 4 sin ) d
cot 4 cos 4
z 16
z z2 4 16
z z2 from the text with a 4 2 sin , dw
w2 4
4 z2 16
4 4 z2 16 4 sin 2 z2 C
. Use Figure 8.18(b) 16
4 and cos d, 2 z2 . 2 4 cos 2 8(2 cos ) d
4 sin 2 2 cos w2 2 csc 2 d 2 cot
2 4
w C
w 2 C Use Figure 8.18(b) from the text with a
get cot C 4 to get 2 cos 8 dw
w2 C 1 and x 16
z 4
, cot
z 42. w 16 cos 2 d Use Formula 89 with a csc 2 d 4 sin2
sin 4 ln d ,0 C (4 cos ) d
4 sin 4 ln csc 2 x2 1 1
x
x
x2 1
C
x2 16 sin 2 (4 csc 1
cos 2
3 1
2 4 4 cos C dz 4 cos 2
sin 1
cos 3
3 cos z2 16
z cos x 16 4 cos cos to get z2 16 cos x2 1 C x 4 sin , dz cos 2 sin x 3 dx 1 to get 1 sec C x x2 d C cos sin x d d (sin ) 1 2 2 2d
sec 2 (sin ) csc d ,0 2 sec tan d
sec 3 tan 1 sec d
tan 2
cos
sin 2 tan 1 2 dx sec 2 d
tan 2 sec dx
x2 d, 4 w
w 2 . 2 and x w to Section 8.4 x x
x 45. For x dx 43. dy 2 9 9 9 sec 3 sec
2 tan 9 d ,0 9 tan 3 sin , dx 3 cos d ,0 9 x 2 9 3 d 9 0,
x 2 9 cos /2 dx
0 /2 0 3 cos2 x2
3 x
3 ln Substitute x 5, y 5
ln
3 ln 3 9 C 1 cos 2
d
2
/2
sin 2
4
0 2 C or C 0 1 0 x x2 1 2 x sec 2 tan , dx 1 2.356 2 46. Volume 4
3 d 1 3
4 ln 3. dx 2 4
0 d ,0 4 The solution to the initial value problem is
y x2
3 x
3 ln 9 1 44. (x dy
x
x2 1) 2 dy x dx
dx
(x 2 1)3/2 tan , dx 1 1
0 0, cos d, 2 2 x C
0, y 1 0 47. (a) 4 1. dx
x(1000
1
x(1000 1 The solution to the initial value problem is
x 1. 1) 1, 4 . d 8 x) x)
B)x 1 2 1
dt
250
A
B
x
1000 x) A(1000
(A cos 2
d
2
/4
sin 2
4
0 1
4
1
4 8 C x2 cos 2 1
2 1 Substitute x x d /4 sec 2 1 sec 2
sec 4 0 C x2 (since 0 0 Volume y /4 dx
(1 x 2)2 d sin 1 0 and when x dx
(x 2 1)3/2
2
sec d
sec 3 y dx
x 2)2 sec 2 /4 sec 2 tan 2 1 2 tan 2 (1 .
When x 2 . d /2 3 2 3 cos 0 C 3, 3 cos
3 3
tan 2 0 and when x 2 3 0 9 sin d ln sec 2 2 When x x2 9
3 sec tan
3 tan sec x dx 3 0 2 x2 9 Area dx y 0 on [0, 3] 3 3 sec , dx
2 0, y 353 8.076 x Bx
1000A Equating the coefficients and solving for A and B gives 1 A 1
,B
1000
dx
x(1000 x) 1
1000
(1/1000) dx
(1/1000) dx
x
1000 x
1
1
ln x
ln (1000 x)
1000
1000
1
x
ln
C1
1000
1000 x
t
C2
x
250 1
x
ln
1000
1000
x
ln
4t C
1000 x
x
e 4t C Ae 4t
1000 x C1 354 Section 8.4 47. continued
When t 49. (a) From the figure, tan
0, x 2
998 (b) 500 4t 499e cos x) sin x cos x)2 sin 2 x z 2(1 cos x)2 1 z 2(1 cos x)2 1 cos x 1 ds 1 sin 2 x 1 cos 2 x 1
x(1000 (1
(1 500 which occurs when z 2)2
z 2)2 z 2)2 (1 z 2)2
(1 z 2)2 (1 x) is greatest. 1 2z 2 (1 Only sin x (d) z 1
dx for 0
1 4z 2
z 2 )2 A
1 x where A
x
x2 1
dx
1 dz 1
1
2 tan 2 1
(1
2 z 2) dx 1 dx
2
x2 1
B
x1 1 and B
2 dz 1
1
1
2 1
dx
1 1
1/2 1 1 x 0 x 1
1 ln x 1 1
2 ln 1
2 ln ln 3 1
2 x ln x
3
2 1 dx
1/2 1
0 1 1
x
sec 2
dx
2
2 1 2 dz
z2 z2 makes sense in this case. x
2 1
2 x x2
x2 0 1 z2 dz 1/2 Arc length 1
1 tan 2z
1 z 4 1 2z 2
(1 z 2)2 2z sin x 2x 2
dx
1 x2 x2 1
dx
1 x2 x2
x2
2 z2 z2
z2 1
1 x2 1 x2
x2 1 1 does not make sense in this case. 2x x2
x2 0 z2
z2 1
1 cos x 0 1 1.553 x4 2x2 1
(1 x 2)2 0 1) cos x 1.553 as shown in part (b). dy
48.
dx 1/2 (z 2 or cos x) (c) From part (b), cos x This occurs when x x2 1 cos x) cos x 4t dx
1
(c)
x(1000 x)
dt
250
dx
will be greatest when y
dt x2 cos x)(1 z 2cos x 0 Half the population will have heard the rumor in about
1.553 days. t (1 cos x)(z 2 (1 cos 2 x 2 1
1
ln
4
499 t sin x
cos x z 2(1 1
499 4t e z(1 1 cos x 1000
1 499e 1 (b) From part (a), z 1
499 1 4t
e
x
499
1000 4t
x 4t
x
e
e
499
499
4t
4t
1000e
e
499
499
1000e 4t
x
499 e 4t
1000
or x
1 499e 4t x1 sin x
.
cos x 1 2. A or A x
1000 x
2 x
dx
2 z4 355 Chapter 8 Review 50. s Chapter 8 Review Exercises 2 dz
1 z2 dx
1 sin x (pp. 454–455) 2z 1 z2 1 z2 (z 1. lim 2 dz
2z
2 dz
1)2 1
2
z 2 3. lim x→0 1 1 C x cos x sin x
sin x
x sin x cos x
lim
cos x
x→0 x sin x
cos x lim x→0 x1/(1 cos x 2 2 dz
2z z lim x 1
e 1 e 5. The limit leads to the indeterminate form
f (x) . ln x
x ln x
x lim x→ 1/x 1/x
1 0 ln f (x) lim x C 0 x1/x lim e0 lim e x→ 2
tan lim e x→1 ln f (x) 1 1 ln f (x) x→1 1 2 x) ln x
1x
1/x
lim
x→1 1 ln x lim 2 dz
1 z2
2z
1
1 z2 x→ 1 C
1 2 2 6. The limit leads to the indeterminate form 1 .
C tan 2
1 cos t for t→0 and 3
5 2
t→0 5 sec 5t x
x→1 1
1/(1 x) x→ 1 3 sec2 3t lim ln f (x) C x
tan
2 2 dz
(z 1)2 53. 2t 2t t→0 f (x) z2 dt
sin t lim 4. The limit leads to the indeterminate form 1 . 1
z 1 tan 3t t→0 tan 5t dz
z2 52. 1 The limit does not exist. C 1 C x
1
2
2 dz
1 z2
1 z2
1
1 z2 d
1 sin 2 1 2t) for t→0 tan dx
1 cos x ln (1
t2 t→0 2. lim 51. t 2 dz
z2 1 z z2
z2 1
1 2z 1 f (x) 2 3x
x 1 ln f (x) x ln 1 1
x dz
z 1 lim ln z 3
x ln 1 1 t
ln tan
2 C 1 C lim 1 x→ 7. lim
r→ 8. lim → /2 lim 1
x x→ x→ 3x
x cos r
ln r 1 3/x 2
3/x
1
x2 lim eln f (x) x→ 0 since cos r 2 sec 3
x ln 1 3
x lim lim x→ 3x
x 3 3 e3
1 and ln r→ as r→ . → /2 cos 2 lim → /2 1
sin 1 356 Chapter 8 Review
1 9. lim x→1 x 1
ln x 1 ln x x 1
(x 1) ln x lim x→1 1
x
lim x 1
x→1
x x→ ln x 1
2 ln x 19. lim
x→ 0 10. The limit leads to the indeterminate form ln f (x) lim x→0 . f (x)
x→ g(x) 20. lim ln (1 1/x)
1/x
1x
x lim x→
1 x x x→ csc 1 x
1/x
x→ lim 1 /x 2
1 1/x lim x→ lim 1/x 2 x→0 x→0 lim eln f (x) e0 x→0 lim x2
1 lim x
x 0 1 x lim x2 1
x2
x2 1 lim 1 1 x2 x→
x→
x→ 11. The limit leads to the indeterminate form 0 . x→0 since x and lim x 2 lim 1
1/x 2 1 1 f grows at the same rate as g. (tan ) ln f ( ) 1 tan x ln (1 1/x)
1/x 0 f( ) lim 1 1
x x ln 1 lim 1 1 f grows faster than g. 1x
x x→0 1
1/x 2 1 ex
x→ 100 lim f (x)
g(x) lim tan 1 lim x→ 1/x f grows faster than g. x→ f (x) x
x x/100
x
x→ xe f (x)
x→ g(x) x
x ln x x/x lim x→ 18. lim 1 lim x→1 1 f (x)
g(x) f grows at the same rate as g. 1 1
1 lim x→1 x 17. lim ln (tan )
1/ f (x)
x→ g(x) ln (tan )
1/ ln (tan ) 21. lim lim x ln x lim x→ lim x (ln x)(1 2 x→0 (ln x)/ln 2 x→ 2 lim log2x lim xln x sec2
tan
1 x→0 xln x
log x
x→ x 2 lim sin x→ cos lim x→ 2 lim sin2 x→0 cos2 Note that 1 0 1/ln 2) 1 (ln x)(1/ln 2 x
1
ln 2 1) 0 0 since ln 2 1. f grows slower than g.
ln f ( ) lim (tan ) lim e x→0 12. lim
→ 2 x 3 3x 2
2
x
x→ 2x 1
3 3x 2 x
4
x3
x→ x 1
2 14. lim x→ f (x)
g(x) lim x→ 1 f (x)
x→ g(x) 22. lim sin t
lim 2
t→0 t 13. lim 15. lim e x→0 1 sin 0 cos t
lim
2t
t→0 3x 2
4x
x→ 6x
1 6x
3
x→ 4x 1
3x 2 lim lim x
5x lim x→ lim x→ 3
x→ 2 lim 6x 6 23. lim 4 x→ 6
12x 2 6x 0 lim x→ lim x→ log2 x
log3 x f (x)
x→ g(x) lim x→ lim lim x→ x→ lim f grows at the same rate as g. (ln x)/(ln 2)
(ln x)/(ln 3) ln 3
ln 2 ln 2x
ln x 2 lim x→ lim x→ 2x
3 ln x ln 2
2 ln x f grows at the same rate as g.
24. lim 1
5 f (x)
g(x) x→ f (x)
x→ g(x) 2x
x
x→ 3 lim x 0 since 2
3 1/x
2/x 1
2 f grows slower than g. f grows at the same rate as g.
16. lim x lim x→ 10x 3 2x 2
ex 30x2 4x
ex
60x 4
ex
60
0
ex f grows slower than g. lim x→ 1. Chapter 8 Review
f (x)
x→ g(x) 25. lim 1
(1/x)
1/x tan lim x→ 1
1 lim 34. lim 2( 1 (1/x) ) 1 2 x→ 2 (x 2 x lim
2 sec 1 x
1
x→ 28. (a) lim f (x) sin x 1 ex x→0 (b) Define f (0) lim x→0 1 (ln 2)(cos x)2
ex ln 2. ln 2 39. x 9 ln x
x→0 1/x
1/x
lim
1/x 2
x→0 lim ( x) 2 2 3 cos x 2 0 sin
1
x2 lim 1 d 1 3 1 0 x
3
b dx
9 1 lim x 2 b→3 0 C
dx
x2 9 True
1
x2 30. lim
x→ 1
x4
1
x4 lim lim (x 2 b→3 1) x→ lim False
31. lim
x→ x b→3 x
ln x 1 lim x→ 1
x 1 1 0 lim 1
ln x False
32. lim
x→ ln (ln x)
ln x lim 1
x ln x x→ 1
x True
tan 1 x
1
x→ 33. lim True 2 2 2 C 0. x→ d, 3 cos 1 x2 1 d x→0 2
4
29. lim x 1 x
x→ 1 3 cos dx lim 1 1
x
1
x 3 sin , dx x→0 (b) Define f (0) 0 1 True lim x ln x x→0 lim x→ 38. lim 2 lim lim True sin x x→0 0 1 (1/x)2 1 f grows faster than g. 27. (a) lim f (x) x2 1
x x→ ln 2x
37. lim
x→ ln x 3 2x x→ 1 lim ) (1/x ) x→ 1 True 1
1 1
x2 ln x
36. lim
1
x→ x lim lim x→ 0 1 True sin 1 (1/x)
(1/x 2)
x→ f (x)
x→ g(x) 1
x4 x2 x→ 1
x4 35. lim f grows at the same rate as g. 26. lim 1 lim 1
x4 1
x2 True 1 x→ x→ 2 x lim 2 x (1/x) x→ 1
x4 x→ 0 sin sin 1x 3 b
0 1b 3 2 2 357 358 Chapter 8 Review 40. u ln x dv ln x dx 43.
3 1
dx
x du dx v
x ln x x
dx 1 x lim ln x dx b→0 1 x ln x b→0 b ln b 1 1 dy
y 2/3 0 b→0 1 lim 3y 1/3 3 3x dy 3 b→ 0 where A
3
1 6
1 d 2
1
2 1)3/5 ( 2 1)3/5 1 B
x 2 5
(
2 lim 5
(b
2 b→ 1
0
1 0 d
1)3/5 ( lim b→ 1 lim b→ 1 lim b→ 1
0
2 d
( 1)3/5 5
2 1) 4B)x 2 1
dx
x2 3x
4x 3 ln 3 C(4x (B 1, C 1) 4C)x C 1)3/5 ( b b→ b 1)2/5 ln
2 1) 5
2 2/5 5
2 du 5
(
2 1)2/5 0 u x 1 dv dx 2x x e v
x 2e b 5
2 5
(b
2 5
2 5
2 du 2 dx 0 1)2/5 x 2e x x x e v e x 2e
0 x dx 1 dx x x 2e x 2x e x x 2x e x b x 2e x lim x 2e x lim b2
eb lim b→
b→
b→ 1 dx x 2e dx 1 dx 2x e
dv 3
4 b x e x ln 3 ln 1
dx
x2 1
x ln x
1
b 1 ln 3 2x dx x 2e 1)3/5 4b x2 45. u d
( 1
4 1 1 b lim ln 1
x 4
4x 1 ln 4x b→ b 1 lim b→ d 1 1)3/5 ( lim b→ 1 1
3 C
x2 Bx(4x 4, B d lim 1)3/5 b→ 1 ( 0 d
(
b d ln 3x 1
dx
x 2(4x 1) 1 lim 42. 2
b A (A 3 b 1
b b→0 2 lim 4x Ax 2 1 2 b x b 1
dx
x2 3x
4x 3 3x 1
x 2(4x 1) 3b1/3] lim [3 3 1 3] 2/3 y lim 1 44. dy ln x x b→ b b→0 dy
y 2/3 1 1 b→0 1 lim ln b lim [3b1/3 dx 2 3 b→ 1 3y1/3 1
x ln x b→ dy
y 2/3 2/3 y 1
x 3 lim ln 1 lim b→0 b→ b lim ( b) b dy
y 2/3 2A 1. lim b→0 b→0 0 B)x lim ln b
1/b
1/b
lim
1/b 2
b→0 lim 1 (A b 2 dx
x(x 2) lim 1 dy
y 2/3 2 1, B b→0 1 1
0 b)
3 1 dy
y 2/3 Bx B b b→0 41. 2) where A x lim ( 1 0 x b lim 1 3 A( x 2 1 ln x dx
0 b→ A
x x(x C 2 dx
x(x 2) lim 2) 2 x ln x b 2 dx
2x
x
2 x 2e
2e dx x C dx 0 2x e
2b
eb 2
eb x 2e x b
0 2 2 359 Chapter 8 Review
46. u x du e 3x dx dv
dx 0 xe 1 3x
e
3 v
1
x
3
1
x
3 x e3x dx dx has infinite discontinuities at t 1 3x
e dx
3 e3x 1 3x
e
9 3x e 0 3x Note that the integral must be broken up since the integrand lim xe b→ 0 b 3x lim b→0 dx lim b→ B
t2 b→ C ln (t 1)2
t2 1 tan lim 1
b (b 1)2
ln 2
b
1 1 3b
e
9 1
9 tan b→0 b
4t 3 t 1
47. 2
t (t 1)(t 2 1) t 1 A
t t 1)(t 2 At(t 1) 22 Ct (t B E)t (t
B E)t 3 D
(A B)t 0 4 dx
x 2 16 A D B 0, C A E B 0, D A E B 1x tan 2 0 lim 1.
0 b 4 dx
x 2 16 Solving the system simultaneously yields
0, B 1, C 2, D 2, E 0 4 dt
t2 2 dt
t1 2t
t dt
t2 2 dt
t1 4 dx
x 2 16 2t dt
t2 1 2 1
dt
1 1 ln t (t 1)
t2 1 ln 1 tan 1 t C
lim 2 1
t tan 1 t → C t1
4t
dt
t 2(t 1)(t 2 1)
1 t1
4t
dt
t 2(t 1)(t 2 1) 2
1 4t 3 t 1
dt
t (t 1)(t 2 1)
2 1
1/2 3 t1
4t
dt
t 2(t 1)(t 2 1) 1 2 1 lim lim → g( ) d → 1
1 b→ 3 1 t1
4t
dt
t 2(t 1)(t 2 1) and
1 1 1 1 2 d
1 1 d
b lim ln b→ 4t 3 t 1
dt
t (t 1)(t 2 1) 1 lim ln b 2 b→ 3 2 1
2 . Both are positive continuous functions on [1, ). b
0 2 2 lim 3 1/2 f( )
g( ) Since 3 0 1 g( )
2 ln t 4 49. Use the limit comparison test with f( ) 1 dt
t2 1
2 2 0 1b lim tan t1
4t
dt
t 2(t 1)(t 2 1) 1
t 4 b→ 3 2 b 1x lim tan 1. b 1b b→ A 4 4 dx
x 2 16 lim b→ 4 tan b→ B 4 dx
x 2 16 0 1x tan b→ 4, 0 4 dx
x 2 16 b lim 1 and 1 C using Formula 16 with a 4 4 dx
lim
16 b→
x B Equating coefficients of like terms gives
C 4 dx
x 2 16 2 0 A 1 4 4 dx
16
x 48. 1) E) t 2 C ln 2 t 1) 2 (A 1 1 Since this limit diverges, the given integral diverges. 1)(t 2 (Dt D)t 4 (A E
1 B( t 1) C (A Dt
t2 1 b 1
t b→0 b 1 3b
be
3 4t 3 t 1
dt
t (t 1)(t 2 1)
2 lim 0 1 3x
e
9 1
9 lim 4t 3 1 b 1 3x
xe
3 1. 4t 3 t 1
dt
t 2(t 1)(t 2 1) 1 C 0 and t t1
4t
dt
t 2(t 1)(t 2 1) ,
we know that 1 g( ) d diverges and so 1 f( ) d diverges. This means that the given integral diverges. 360 Chapter 8 Review x 50. Evaluate e
u cos x dx using integration by parts. cos x du
e dv sin x dx
x v cos x dx x Evaluate sin x e
u cos x x sin x e ex dx dv cos x dx v x x x e x dx
e 0 ex e dx x x lim b→ sin x e dx e sin x e cos x dx lim e x cos x dx x e cos x x e sin x e x b→ cos x dx
0 x 2e cos x dx x e sin x x e cos x dx
ex e b→ C1 0 x u e x e cos x dx sin x e x cos x b cos u du lim e b→ 0 x 1 eb cos x dx lim x 2 e x cos x b
0 e sin x b cos b 2 b→ e lim b sin b
2 b→ 1
2 1
2 4 x eb 4 4 4 Since these two integrals converge, the given integral
converges. 54. The integral has an infinite discontinuity at x Note that we cannot use a comparison test since
e x cos x 0 for some values on [0, ). 1 dx
x (1 e x ) ln z
on [e, )
z
b
dz
b
lim
lim ln z
lim (ln b 1)
e b→
b→ e z
b→
1
dz diverges, so the given
Since this integral diverges,
ez 2 1
4x 2
1 0 dx
4x 2 1
x 2(1 ex )
1 dx
4x 2 lim b→0 b 1 11
4x b lim b→0 1
0 0. dx
x 2(1 e x )
dx
x 2(1 e x ) ex on (0, 1] since 1 Since this integral diverges, integral diverges. 1 dx
x 2(1 e x ) 0 0 0 dx
x (1 e x ) 2 1 1
51. 0
z
dz
z
e C b
0 ex b→ ex x 1 0 e 1 lim tan C 2 dx
e ex 1 0
b lim tan b→ e ex dx
ex e lim x tan x tan 4
b C 0 dx
e
1 tan b→ u x ex b lim x dx
e ex 0 dx
e 1 tan 0 dx
e
e dx 1) du
u2 1 x e x dx
(e x)2 1 x e , du x x e dx
e (e 2 x x x Let u dx using integration by parts. sin x du x dx x e x e x e dx
e
e 53. 4 on (0, 1].
1
4 lim b→0 1
4b dx
diverges, so the
x 2(1 e x ) given integral diverges.
t e 52. 0 e t on [1, ) t e
1 t 55. x 2 b dt lim b→ e
e lim e b→ dt 1 lim b→ t x2 tb 2x 1
b 1
e 7x 2x 1
7x 12 1 (x 4)(x A
x 3)
B 4 x 3 A( x 3) B(x 4) (A 1
e Since this integral converges, the given integral converges. 12 B)x 3A 4B Equating coefficients of like terms gives
A B 2 and 3A 4B 1. Solving the system simultaneously yields A
2x 1
dx
x 2 7x 12 9 ln x 9, B 7 dx
x3 9 dx
x4 4 7 ln x 3 C 7. 361 Chapter 8 Review 56. 2 B
x C
x2 Bx 2(x 2) Cx(x 8 A x 3(x 2) x Ax 3 8 B)x 3 (A D
x3 2) D(x
D)x B 0, 2B C 0, 2C x3
x3 1
x1
1
x
x3 x
A
x1
x
x(x 1)(x 1) D 8 1, B 1, C 8 dx
x3(x 2) 2, D
dx
x dx
x2 ln x
57. t 3
3t 2 4t
t 3t 3 A
t 4 4t 4 Bt
t2 A( t 2 (A 4 2 dx
x2
2
ln x
x 4 dx
x3
2
x2 A C B A Bx(x C)x 2 B C 0, B (Bt 3, C (B C C)t 1, C 1)
C)x Cx(x 1
1)(x 1) 1) A 1, and A 1. 0 dx dx
x x B)t 2 4 and A 1, B
1
dx
x x3
x3 Ct ln x 3 x3
x3 4x 2
4x 2 dx
x 1 ln x 1 C A Equating coefficients of like terms gives
B 1)(x x(x Equating coefficients of like terms gives C
1 1) (A A A( x x 1 1) t 2 1) 1 1 Solving the system simultaneously yields t(t 2 t 2 x1
x(x 2 1)
B
C
x1
x1 x 0, and 2D Solving the system simultaneously yields
A 1
x1
x+0
x1 0x2
0x2 2D Equating coefficients of like terms gives
A x x3
x3 2) (2C C)x 2 (2B 59. x 3 60. x 2 4x 4. x
0x
3x
3x Solving the system simultaneously yields
x3 A 4, B
3t 2 1, C 4t
t3 4
t 4 x 4 dt
t
4 dt
t dt t4
dt
t2 1
t dt
4 dt
t2 1
t2 1
1
ln (t 2 1) 4 tan 1 t
2 4 ln t
58. t 4
(t 2 4t 2 (t 2 3 1
1)(t 2 3) (At C)t 3 At
t2 B)(t 2 (A 1 3)(t 2
B
1 (B C 0, B D D
3 D)t 2 (3A C 0, B
(t 2 dt
3)(t 2 1
,C
2
1) 0, D 1
tan 1 t
2 C)t 3B 0, and 3B D D 1/2
dt
t2 3
1
1t tan 2 3 1 C
3 Evaluate the integrals using Formula 16, with x
in the first integral and a (A B)x 3A (x 3x
1)(x 3 in the second. t, a 1 3) 1)
B Equating coefficients of like terms gives
B x 1
2 1/2
dt
t2 1 B(x x3 1) Solving the system simultaneously yields
A 3) x 3, 3A B 0
3
,B
2 Solving the system simultaneously yields A D)(t 2 0, 3A A( x 3x
C Equating coefficients of like terms gives
A 4x 2
3x
x
4x 3
x 2 4x 3
3x
A
B
1)(x 3)
x1
x3 3x A (Ct 3) (x 1)
Ct
t2 2 2 4x 2
dx
4x 3 x dx
x2
2 x 3/2
dx
1 3
ln x
2 1 x 9
.
2 9/2
dx
3 9
ln x
2 3 C 362 61. Chapter 8 Review dy
y(500 y)
1
y(500 y) 1 0.002 dx
A
y A(500 63. y B
500 y y) 1 1
tan , dy
3 9y 2 A)y By 9y 2 1/500
dy
500 y
1
ln 500
500 1
y
ln
500
500 y C1 y
500 y C1 y
500 y 20
480 0.002x 0, y y
500 y 20. 1x
e
24 24y 500e x (e x 24)y C2 64. t yex
500e x 500 e x
e x 24 y
y 1
dy y2 65. x 500
24 e x y2 x 1
dx 1 x tan 1 y C1 tan 1 y ln x Substitute x
tan 1 tan 1 y tan ln x 4 y 9t 2 9t 2 dt 1
C 0, y C2 1
1 4 1 and x . 1
.
3 2 2 1
cos
3 9 sec2 9 2 3
sec
5 d tan , 0 1 4 2 9 tan2 9 3 sec tan d 3 tan 9 sec ln tan
tan C cos2 d ln sec C
ln x 3y d, cos 3
sec , dx
5 25x 1 1 1
cos
3 sin2 1 5 dx ln x 9y 2 1 1
sin , dt
3 25x 2 dx 1
dy C 1
cos2 d
3
1 cos 2
d
6
sin 2
C
6
12
sin cos
C
6
6
sin 1 3t
3t 1 9t 2
C
6
6
1
1
sin 1 3t
t 1 9t 2 C
6
2
1
Use Figure 8.18(b) with a
and x t.
3 1
24 ke0 or k tan C1 1 ke x Substitute x d Use Figure 8.18(a) from the text with a 1 x sec Integrate by using Formula 88 with a C ln 62. ln
y 2 d ln sec 1
ln y
500 2 sec 1
.
500 1
y
ln
500
500 y d,
2 sec 1/500
dy
y dy
y(500 y) 2 sec2 500A 1
,B
500 where A tan 3 dy
1 (B 1 1
sec2
3 5x
3 ln (5x tan
25x 2
3 25x 2 C1
9 C1
9) C 1 4 Integrate by using Formula 88 with a
Use Figure 8.18(c) with a 3
.
5 1 and x . Chapter 8 Review
68. For x
66. x sin , dx cos d, 2 0, y Area 2 0 on [0, ).
b x xe dx lim x2 1
(1 cos2 4x 2 dx
x 2)3/2 4 sin2 cos u d 3 cos
4(1 cos2 )
d
cos2
2 (4 sec du dv
dx x xe 4 tan 4 x C 0 1 1. 69. (a) 1 b→0 b 2 u (ln x) dv 2 ln x
dx
x du (ln x)2 dx 2 v dx x (ln x) 1
a 2 ln x dx 2 ln x du dv 2
dx
x v dx
x 2x ln x 2 dx (ln x)2 dx x (ln x)2 2x ln x lim b→0 x(ln x)2 b→0 2
2
2
2 2x ln x
2x 2x C C
1 2x ln x 2x
b lim [2
lim b→0 b(ln b)2
(ln b)2
1/b 2b ln b
b→0 2 (ln b)(1/b)
1/b2 lim 2 (ln b)
1/b lim 2 /b
1/b2 b→0
b→0 /b
1/b2 2 lim
b→0 2 lim (
b→0 2 2b] ln b
1/b 2 lim lim b→0 a x
1
kt 1/a x 2 ln x dx Area kt C1 1 b) lim 2 b b→0 2 1 a 1 C
0, t C
1 Evaluate 2 ln x dx by using integration by parts.
u C1 b k dt x)2
dx
(a x)2
1
C2
ax
1
kt
ax Substitute x x 2 (ln x) dx kt xe x)2 k(a
dx (ln x)2 dx by using integration by parts. Evaluate dx
dt
(a (ln x) dx
0 k dt b
eb
1
lim b
b→ e lim 2 lim 1] dx 0 b→ ( ln x)2 dx Volume e b b b→ 0 on (0, 1].
1 x lim [ be 1 4 sin x2 Use Figure 8.18(b) with a
0, y e x b→ C x e
x xe lim 1
a kt
a x 1
kt dx dx e xe 4) d 4x
1 x x e v dx x 0 dx by using integration by parts. x Area 67. For x x Evaluate xe xe b→ 0 1/a 0 x e x C 363 364 Section 9.1 Chapter 9
Infinite Series 69. continued
dx
(b)
k dt
(a x)(b x)
dx
k dt
(a x)(b x)
1
A
(a x)(b x)
ax 1 A( b x) (A B(a B)x kt C1 s Section 9.1 Power Series (pp. 457–468) B
b x Exploration 1 x) bA 1. 1 aB 2. x
3. 1 Equating coefficients of like terms gives
A B 0 and bA aB 4. 1 5. A a b dx
x)(b (a ,B a 1/(a b)
dx
ax x) a
a a
b x
x b C2 a
b
a
b ab (x 3 x 4x 8x 1/(a b)
dx
bx ln b x
ab
a
b x
x C2 1n
(x
3 C2 abe (a b)kt ab(e (a
ae (a b)kt 1)
1
(x
3 1)2 … 3 … 1)3 …. 1)n x 1 x 1 1, 2. The interval of conver Exploration 2
1. 1
2. tan x 2 1 x A Power Series for tan 4 x
x x 1 01
x
0 … 6 t2 t3
3
x3
3 t
0. x …. n 2n ( 1) x dt t2 (1 1 t4
t5
5
x5
5 … t6
t7
7
x7
7 …
… ( 1)nt 2n …) dt t 2n 1
…
2n 1
2n 1
x
….
( 1)n
2n 1 ( 1)n x
0 3. The graphs of the first four partial sums appear to be converging on the interval ( 1, 1).
b)kt axe (a ab(e (a b)
b)kt (2x) gence is (0, 2). C a (a b)kt
e
b bx 1
(x
3 1) …. n 2 which is equivalent to 0 C1 D
x
x ( 1) x
… 3 …. nn1 (x
1) (x 1)
( 1)n(x 1)n …. 1
(x
3 …. ( x)n … 4 2 x x(ae (a
x 0, x ln
kt b
ax
ln
(a b)kt
bx
ax
De(a b)kt
bx Substitute t 2x … x3 This geometric series converges for b 1 ln 1
3 x 1 ln a x
ab 1 2 1 Solving the system simultaneously yields
1 x Power Series for Other Functions x2 x b)kt b)kt 1) 1)
b [ 5, 5] by [ 3, 3] 4. When x
e bkt
Multiply the rational expression by bkt .
e
ab(e akt e bkt)
x
akt
bkt
ae
be 1 1
3 1, the series becomes
1
5 1
7 … ( 1)n
2n 1 …. This series does appear to converge. The terms are getting
smaller, and because they alternate in sign they cause the
partial sums to oscillate above and below a limit. The two
calculator statements shown below will cause the successive partial sums to appear on the calculator each time the
ENTER button is pushed. The partial sums will appear to
be approaching a limit of /4 (which is
tan 1(1)), although very slowly. Section 9.1
Exploration 3 A Series with a Curious Property
2 1. f (x) 1 2. f (0) x
2! x 1 3 0 x
3! … 0 … n x
n! …. u2 1. u3 3. Since this function is its own derivative and takes on the
value 1 at x 0, we suspect that it must be e x.
4. If y dy
f (x), then
dx y and y ( 1)1
1
( 1)2
2
( 1)3
3
( 1)4
4
( 1)30
30 2. u1 1 when x 0. u4
u30 3. (a) Since
5. The differential equation is separable.
dy
dx
y
dy
dx
y ln y x (c) an 1 Ke0 ⇒ K
y e x. 1
30
18
6
1 54
18 3, the common ratio is 3. ) 4
8 (c) an 1 1
4 8 1
2 2
4 19
2 k Ke 1
3 2(3n (b) 8
y 1
2 39,366 4. (a) Since
C 1 6
2 (b) 2(39) 1
, the common ratio is
2 1
.
2 1
64
1n1
2 8( 0.5)n 5. (a) We graph the points n, 6. The first three partial sums are shown in the graph below. It
is risky to draw any conclusions about the interval of
convergence from just three partial sums, but so far the
convergence to the graph of y e x only looks good on
( 1, 1). Your answer might differ. 365 1 1 n
n2 for n 1, 2, 3, … . (Note that there is a point at (1, 0) that does not show
in the graph.) [0, 25] by [ 0.5, 0.5]
[ 5, 5] by [ 3, 3] 7. The next three partial sums show that the convergence
extends outside the interval ( 1, 1) in both directions, so
( 1, 1) was apparently an underestimate. Your answer in
#6 might have been better, but unless you guessed “all real
numbers,” you still underestimated! (See Example 3 in
Section 9.3.) (b) lim an
n→ lim n→ 1 n
n2 0 6. (a) We graph the points n, 1 1n
for n
n 1, 2, 3, … . [0, 23.5] by [ 1, 4] (b) lim an [ 5, 5] by [ 3, 3] n→ Quick Review 9.1
1. u1 1 2
4 u2 2 u3 3 u4
u30 4
3
4
4
4
5
4
6 4 2
4
2
4 4
30 2
4
2 lim 1 n→ 1n
n e 7. (a) We graph the points (n, ( 1)n) for n 1, 2, 3, … . 1 2
3
4
32 [0, 23.5] by [ 2, 2] 1
8 (b) lim an does not exist because the values of an
n→ oscillate between 1 and 1. 366 Section 9.1 8. (a) We graph the points n, 1
1 2n
for n
2n 1, 2, 3, … . 2. (a) Note that a0 1
,a
32 1, a1 1n
.
3 an 1
, and so on. Thus
9 1, a2 1
,a
23 (c) Note that a0 5, a1 0.5, a2 5(0.1)n 5
.
10n (b) Note that a1
( 1)n
n an 1 1
, and so on. Thus
3 . [0, 23.5] by [ 2, 2] 1
n→ 1 (b) lim an 2n
2n lim n→ 2
2 1 an
1
for n
n 9. (a) We graph the points n, 2 1, 2, 3, … . 0.05, and so on. Thus 1n1
alternate
2 3. Different, since the terms of ∑ n1 between positive and negative, while the terms of
1n1
are all negative.
2 ∑ n1 4. The same, since both series can be represented as [0, 23.5] by [ 1, 3] (b) lim an 1
n lim 2 n→ n→ 10. (a) We graph the points n, 1
2 1 2
ln (n 1)
for n
n 1, 2, 3, … . 1
4 …. 1
8 5. The same, since both series can be represented as
1
2 1 1
4 …. 1
8 6. Different, since ∑ ∑
n [0, 23.5] by [ 1, 1]
1 ln (n 1)
lim
n
n→ (b) lim an
n→ lim n→ (n 1) 1
u4 and 1
u1 1, 1
, so
16 u1 1, u2 un n2, or * ( 1)
2n 1 1 2n
3 1 1
8 4, u3 11
,
4 u3 1
u2 9, and u4 1
,
9 9. Converges; ∑ n0 5
4 2n
3 5
4 1 5
4 3 2
3 15
4 10. Diverges, because the common ratio is 1 and the terms do not approach zero.
11. Diverges, because the terms alternate between 1 and
and do not approach zero. 16. We may write n2. 12. Converges; ∑ 3( 0.1)n 1 n0 3
( 0.1) 30
11 3 … (b) Let un represent the value of * in the nth term, starting
with n 0. Then 1
u3 u3 1
u0 1, 1
, so u0
16 1, u1 16. We may write un (c) If *
( 1)3
( 1)6 11
,
4 u2 1
u1 4, u2
(n 1
1 1
4 2 ( 1)4 n0 1)2, or * (n 1)2. 1
2 ( 1)5 1
3 3. 1 1 1 2 2(
(2 2 1 2 1 1 2 n0 1 2
n 1 2 1 2 ∑ 2 n 4 9, and … , which is the same as the desired series. Thus let * 13. Converges; ∑ sin n 1
, and
9 3, the series is
2 … but 3 2
3 1 1
4 8. Diverges, because the terms do not approach zero. 1. (a) Let un represent the value of * in the n th term, starting
1. Then 1 n0 Section 9.1 Exercises with n n 7. Converges; ∑ 0 1 n1 1n1
1
1
2
2
1
1
1
….
2
4
8 2 2 2
1)( 2 1) 2
1) 2
2 1 2 2 1 14. Diverges, because the terms do not approach zero. 1 Section 9.1 e 15. Converges; since 1, ∑ 0.865 16. Converges; ∑ n0 6n 17. Since ∑ 2 x n0 n0 1 1 function f (x) 1 2x 18. Since ∑ ( 1)n(x 1 2x ∑[ 1)n n0 (c) The partial sums alternate between positive and
negative while their magnitude increases toward
infinity. 1
.
2 x 1)]n, the series (x 24. Since ∑ n0 converges when (x 1) 1 1
, the series represents the
1 tan x
1
,
k
x
k.
4
4
tan x (b) The partial sums are alternately 1 and 0. , the series represents the 1
2 , k , where k is any integer. Since the 4 23. (a) Since the terms are all positive and do not approach
zero, the partial sums tend toward infinity. 11
, . Since
22 1 and the interval of convergence is the sum of the series is x function f (x) n n0 k sum of the series is 1 5
6 1 n0 1. Thus, the series converges for 4 ∑ (2x) , the series converges when nn 2x 1 tan x 1
6 15n
66 ∑ 5n ∑ (tan x)n, the series converges when n0 e 1 n1 e 22. Since ∑ tan n x 1 en ∑ en
ne n0 1 and the interval of n e
e , this is a geometric series with n0 e common ratio r 1.03, which is greater than one. e convergence is ( 2, 0). Since the sum of the series is
1
[ (x 1 1
1)]
1 f (x) x 2 x , 2 2 25. , the series represents the function x ∑ xn 20 n0 1 0.
1 20, x x 1
1n
(x
2 19. Since ∑ n0 converges when ∑ 3)n 2 n0 3 x
2 xn 3 20 20x 19
20 1 and the interval of convergence 1 x 1 . 2 2 the series represents the function f (x) x) x 1 ,1 x 5. For any real number a
use a
2 a
4
1
2 use 1
20. For ∑ 3
n0 x 1n x
2 a
16
1
8 ∑ 2r n 1…
.
32 1 1 3 x 5r , the series 3 x , 1 x Series: ∑ 2 3. n1 21. Since ∑ sin x
n n0 sin x ∑ (sin x) , the series converges when
n n0 1. Thus, the series converges for all values of x except odd integer multiples of 2 , that is, x for integers k. Since the sum of the series is
series represents the function f (x)
(2k 1) .
2 3
5 1 1
1
,
sin x (2k 1) (b) ∑ n1 3n1
5
13
2 13 n 1
r
2 1 r
13
2 1 5 5r 5r
r
Series:
n1 13
2 5r 5, r 2 1
, the
sin x 1 3 2 6 represents the function f (x) 1) 5 r 6 3
(x 1: 5, r r 2 Since the sum of the series is x … . To get 0, a
32
1
16 2 1 n1 1 and the interval of convergence is ( 1, 3). 2 a
8
1
4 0, 27. Assuming the series begins at n , the series converges when (a) 1 20x 26. One possible answer: 2 1
(3 1 19 x , the series is (1, 5). Since the sum of the series is 367 3n1
10 3
2
3
10 368 Section 9.1 28. Let a
0.21 21
and r
100 0.21 1
, giving
100 33. 1.414 0.21(0.01)2 0.21(0.01) 1 0.414 0.414(0.001)2 0.414(0.001) ∑ 0.414(0.001)n 1 n0 … 0.21(0.01)3 ∑ 0.21(0.01)n 1 n0 157
111 0.21
1 0.01
0.21
0.99
7
33 29. Let a
0.234 34. 1.24123 1
, giving
1000 0.234(0.001) 0.234(0.001)3 0.234(0.001)2 … 124
100
41,333
33,300 0.234
1 0.001
0.234
0.999
26
111 0.7(0.1) 35. 3.142857 0.7(0.1)2 0.7(0.1)3 3
3
3 36. Total distance …) 0.142857 ∑ 0.000001n
(0.142857) 1 1
0.000001 0.142857
0.999999
1
7 0.06(0.1) ∑ 0.06(.1)n 4(0.6)2 2[4(0.6) 4 2 4(0.6)3 …] 2 ∑ 2.4(0.6)n 4 …] 4 4 0.13 26 n0 ∑ 0.06
1 0.1
0.06
0.9
1
15 0.000001 22
7 d
[1 0.1 0.12
10
d
(0.1)n
10 n 0
d
1
10 1 0.1
d1
10 0.9
d
9 n0 0.142857(1 n0 … 0.7
1 0.1
0.7
0.9
7
9 0.06 3 3 ∑ 0.7(0.1)n 32. 0.06 … 0.00123
1 0.001
n1
0.00123
0.999
41
33,300 0.0000012 n0 31. 0.d 0.00123(0.001) ∑ 1.24 ∑ 0.234(0.001)n 0.7 0.00123 1.24 n0 30. 0.7 1.24 0.00123(0.001)2 234
and r
1000 0.234 0.414
1 0.001
46
111 1 1 2.4
0.6 16m 37. Total time
0.06(0.1)2 0.06(0.1)3 4
4.9 … 4(0.6)2
4.9 4(0.6)
4. 9 2
4(0.6)3
4.9 … 4
4.9 2 4(0.6)
[1
4. 9 4
4.9 2 4(0.6)
4.9 7.113 sec 0.6
1
1 0.6 ( 0.6)2 …] 369 Section 9.1
38. The area of each square is half of the area of the preceding
square, so the total of all the areas is ∑ 4
1 8 m 2. 1
2 1n
2 n1 n1 ∑ 2 ∑ 45. Comparing a
a S a ar n
1r 41. Using the notation Sn ar a 1
4 1, then lim r n
n→ ar 2 ar ar 3 1 r 4) , the first term is a (x 4).
… 4)2 (x a ar n
1r
n→ a
1 r ∑ ar n 4 1 3x a
1 r the common ratio is r
3x 9x 1 … 4)n 1, so the 1) with a
1 r , the leading term is a ( 3x) .
1 and (x 1
(x
4 1) … 1)2 x
2x with the common ratio is r
2x 2 4x 3 3x 1
2 x as (x 1
(x 1 1) a
r , the first term is a 11
,.
22 1, so the 2n 1 x n a
1 1 and the common ratio is r
… r x , … 1) (x 1) Rewriting 1
2 2 (x 1) 1 n 1. 1, so the 1
1
x
21 and comparing with x
2
1
and the common ratio is r
2 the first is a
1
2 as 1
x
4 12
x
8 … a
1 r , x
.
2 … 1n
x
2n 1 x and 48. 1 eb e 2b e 3b … x
2 1, so the interval ∑ (e b)n n0 1 Interval: The series converges when 2x
of convergence is 1 … of convergence is ( 2, 2). 2x.
… x. 1)n and comparing with 1, so the 11
,.
33 1 1 1
( 1)n(x
4 Interval: The series converges when 1 1) interval of convergence is (0, 2). Series: … n Interval: The series converges when
interval of convergence is , the first term is Alternate solution: 3x.
… 2 1
(x Interval: The series converges when x 1 1, then the nth partial sum is na, which goes to
with 1
(x
4 Series: 1 . a ar n
has no finite limit and
1r
n 1 1
41 The first term is a 1 n1 diverges. Series: x ( 1)n(x 1
and the common ratio is r
4 47. Rewriting 1, then r n has no finite limit as n → , so the expression 43. Comparing 1 ar n 1, ... a ar
.
1r 0 and so ∑ ar n lim 1 or r Series: 1 a with interval of convergence is (0, 2). the formula from Exercise 40 is Sn 42. Comparing 4) Interval: The series converges when x n If r 1, so the interval Series: n rS S (1 n→ (x 46. Comparing r: ar n r) S If r … 3x 3n interval of convergence is (3, 5). (a ar ar 2 ar 3 ... ar n 2 ar n 1)
(ar ar 2 ar 3 ar 4 ... ar n 1 ar n)
a ar n a lim Sn … Interval: The series converges when x (b) Just factor and divide by 1 If r 3 and x 3. 3x 6 1
(x 1 Series: 1
1
2 2 rS , the first term is a and the common ratio is r 4 40. (a) S r of convergence is ( 1, 1). 1n
42 1 a
1 Interval: The series converges when x 3 12
2n 1
2 n0 3x 3 Series: 3 ∑ 2n 39. Total area with x3 1 the common ratio is r 2 n0 4 3 44. Comparing 1n …
1, so the interval 1
eb 9 1 9 9e b 8 eb 8
9 b ln 9e b 8
9 ln 8 ln 9 370 Section 9.1 49. (a) When t n0 1 1 t. x (1
1 t t 1 t, which is true when t For t 1
.
2 (1 ∑ n0 2(1 n t
1 10 when t 1 1 t t 1 1 (1 t
t) 1 t t, so 4
1 t2 with a
1 r , the first term First four terms: 4 4t
n 4t (1 4t 2n 0, so the constant term of the power series for G (x) will be 0. Integrate the terms for f (x) to
obtain the terms for G (x).
43
45
47
x
x
x
3
5
7
4
General term: ( 1)n
x 2n 1
2n 1 First four terms: 4x (c) The series in part (a) converges when 2 x) C. 4x 3 … nx n x) d
(1
dx 2 2x 3x 2 4x 3 … 3 x) 0
2 2 2 …) 12x 2 6x n (n 1)x n ... … 12x 2 6x 1)x n n (n (n …. 1)x n 2)(n 1, so the interval (b) No, because if you integrate it again, you would have
the original series for f, but by Theorem 2, that would
have to converge for 2 x 2, which contradicts
the assumption that the original series converges only
for 1 x 1. t2 1, so the lim an. Then by definition of convergence, for n→ there corresponds an N such that for all m and n,
n, m N ⇒ am L 2 and an L 2 Now,
1, which result in the am an 4 4
3 4
5 4
7 … ( 1)n 4
2n 1 … and
G ( 1)
4 4
3 respectively. 4
5 4
7 … ( 1)n 4 1 2n 1 …, whenever m am L am convergent series
G (1) …. 53. (a) No, because if you differentiate it again, you would
have the original series for f, but by Theorem 1, that
would have to converge for 2 x 2, which
contradicts the assumption that the original series
converges only for 1 x 1. 54. Let L interval of convergence is ( 1, 1).
(d) The two numbers are x 2 … 2 2, this may be written as
12x 2 6x 1 … 1 of convergence is ( 1, 1). 6 General term: ( 1) (4t )
(b) Note that G (0) dx Interval: The series converges when x t 2. 4 and the common ratio is r
4 3 3x 2 Replacing n by n t 2 x) 2x 1 Thus, f (x) f (x) is a 2 x) 9. 50. (a) Comparing f (t) … 4 nx n 1
, we have
2 (c) For t 2(1 d
(1
dx 1
.
2 Thus, S converges for all t 1)4 (x 1
dt
t 1 Using the result from Example 4, we have: t, which is always true. S 1)2
(x 1)3
2
3
( 1)n(x 1)n
n
(x 1 f (x) dx
1 (x 52. To determine our starting point, we note that 0, the inequality is equivalent to 0, the inequality is equivalent to t x 1) 1)2 … 1)3 (x 1)n t), which is always false. For (x … , we may write ln x 1 1, this inequality is equivalent to t S 1
x ( 1)n(x 1, or t t 51. Since 2. 1
2 1 t (b) S converges when
For t 1 1n
2 ∑ 1, S L L
N and n an an L
N. 2 2 = . 2 Section 9.2
55. Given an
0, by definition of convergence there
and
corresponds an N such that for all n N, L1 an
L 2 an
. (There is one such number for each series,
and we may let N be the larger of the two numbers.) Now
L2 L1
L2 an an L1
an L1
L 2 an 1 L1, lim ai(n) L2, and L1 n→ ak(n) L1 , and an N2 such that for i(n) ai(n) L2 0 N1, max {N1, N2}. Then for n an L1 and an implies that lim an L2 N2, N, we have that for infinitely many n. This L1 and lim an n→ n→ L2 where L1 (b) The line y L2. 1. f (x)
f (x)
f (x)
f (x)
f (n)(x) 1 (x f (x) 3x
x 1
, which means lim f (x)
1
x→ 3. an for all positive integers n,
n→ s Section 9.2 Taylor Series (pp. 469–479)
Exploration 1 Designing a Polynomial to
Specifications
1, we know that the constant coefficient is 1. Since P (0) 1) 4 ( 1)nn!(x 1) 1)! x f (x) nx n f (x) n(n 1)x n (The 2 in the denominator is needed to cancel the factor of f (x) n(n 1)(n 2 that results from differentiating x 2.) Similarly, we find the f (k)(x) 4
5
coefficients of x 3 and x 4 to be and .
6
24
32
43
54
Thus, P(x) 1 2x
x
x
x.
2
6
24 n for n 1 xn 5. f (x) 3
3, we know that the coefficient of x is .
2 (n 1) 3x
3x ln 3
3x(ln 3)2
3x(ln 3)3
3x(ln 3)n 4. f (x) ln x
f (x) x 1
f (x)
x2
f (x) 2x 3
f (4)(x)
6x 4
f (n)(x) ( 1)n 1(n 2, we know that the coefficient of x is 2. Since P (0) 3. f (x)
f (x)
f (x)
f (x)
f (n)(x) 3 1) 6(x f (n)(x) 2 1) 2(x f (x) it follows that lim an must also be 3. 1. Since P(0) x f (x) 3 is a horizontal asymptote of the graph of Because f (n) e2x
2e2x
4e2x
8e2x
2ne2x
1 2. f (x) 3 the function f (x) Approximating sin 13 4. 20 terms. an does not converge and hence diverges.
1
1 x6
6! x6
, and the Taylor series is
6!
2n
… ( 1)n x
….
(2n)! 1. 0.4201670368… Since the limit of a sequence is unique (by Exercise 55), 3n
57. (a) lim
n→ n x4
4! x4
4! Quick Review 9.2 . Assume an converges. Let N x2
2 x2
2 Exploration 3 L2. Given an there corresponds an N1 such that for k(n) 1 2. A clever shortcut is simply to differentiate the previouslydiscovered series for sin x termbyterm! 56. Consider the two subsequences ak(n) and ai(n), where
lim ak(n) A Power Series for the Cosine 1. cos (0) 1
cos (0)
sin (0) 0
cos (0)
cos (0)
1
cos(3) (0) sin (0) 0
etc.
The pattern 1, 0, 1, 0 will repeat forever. Therefore,
P6(x) 2.
L2 L1
2 says that the difference between two fixed
values is smaller than any positive number 2 . The only
nonnegative number smaller than every positive number
is 0, so L2 L1 0 or L1 L2. n→ Exploration 2 371 2 f (n)(x) 1
2 2)x n 3 n!
xn k
(n k)!
n! 0
x
n!
0! 6. dy
dx d xn
dx n! nx n
n! 7. dy
dx d 2n(x a)n
n!
dx 1 xn 1
(n 1)!
2nn(x a)n
n! 1 2n(x
(n a)n
1) ! 1 372 8.
9.
10. Section 9.2
d
x 2n 1
( 1)n
dx
(2n 1)! dy
dx 2n dy
dx d (x a)
dx (2n)! dy
dx 2n(x d (1 x)n
n!
dx n(1 ( 1)n (2n 1)x 2n
(2n 1)! 2n 1 ( 1)nx 2n
(2n)! 5. cos (x (x a)
(2n 1)! x)n 1( 1)
n! (1 x)n 1
( n 1) ! (cos 2) at the end of Section 9.2.
(2x)
3!
4x 3
3 2x
2x 5 2n 1 (2x)
… ( 1)n (2x)
5!
(2n 1)!
4x 5
( 1)n(2x)2n 1
…
…
15
(2n 1)! … x for x in the Maclaurin series for ln (1 x) x) x5
…
5!
(cos 2)x 2
2! (sin 2)x ( x)2
( x)3
…
2
3
n
( x)
…
( 1)n 1
n
2
3
n
x
x
…x
…
x
2
3
n 1 x 1, so the interval … ( 1)n(cos 2)
where 2n
(2n)! 2n 1 k. Thus the coefficient is B sin 2 if n is even and cos 2 if n is odd. 1 cos 2 sin 2
x , 2 cos 2 cos (2 and so on, so the general term is
2 2n 1
(x 2)3
(x 2)5
… ( 1)n (x )
3
5
2n 1
n 4n 2
x6
x 10
… ( 1) x
…
3
5
2n 1 x2
x2 This series converges when x 2 7x(1
7x x
7x 2 … n
x2
…x
…)
2!
n!
3
n1
7x
… 7x
2!
n! This series converges for all real x. ),
n
2 1
cos 2
n! x n. The series converges for all real x.
6. x 2 cos x x2
2! x2 1
x4
2 x2 1, so the interval of convergence is [ 1, 1].
4. 7xe x k. Thus the coefficient is shown at the end of Section 9.2.
x2 x 2n 1
…
(2n 1)!
(sin 2)x 3
(cos 2)x 4
3!
4! ( 1)n Another way to handle the general term is to observe that 3. Substitute x 2 for x in the Maclaurin series for tan 1 … We need to write an expression for the coefficient of x k. and B of convergence is [ 1, 1). tan x 2n
(2n)! ( 1)int[(k 1)/2](cos 2)
( 1)(k 1)/2(sin 2)
, which is the same as
.
k!
(2n 1)!
An
( 1) Bx
n1
, where A int
,
Hence the general term is
n!
2 ( x) This series converges when ( 1)n ( 1)int[(k 1)/2](cos 2)
( 1)k/2(cos 2)
, which is the same as
.
k!
k!
n1
( 1) (sin 2)
where
If k is odd, the coefficient is
(2n 1)! shown at the end of Section 9.2.
ln (1 (sin 2)(sin x) … If k is even, the coefficient is This series converges for all real x.
2. Substitute x4
4!
x3
3! (sin 2)x 5
5! 1. Substitute 2x for x in the Maclaurin series for sin x shown sin 2x x2
2! (sin 2) x Section 9.2 Exercises 3 (cos 2)(cos x) (cos 2) 1 2n a)
(2n)! 2) x6
24 … x4
…
4!
( 1)nx 2n
(2n)! ( 1)n
1 x 2n
(2n)! … … The series converges for all real x.
7. Factor out x and substitute x 3 for x in the Maclaurin series
for 1
1 x shown at the end of Section 9.2.
x 1
1 x3 x[1 1 x3 x x3 x x4 (x 3)2
x7 … The series converges for x 3
convergence is ( 1, 1). …
x 3n (x 3)n
1 …] … 1, so the interval of Section 9.2
2x for x in the Maclaurin series for e x shown at 8. Substitute x1/2 12. f(4) the end of Section 9.2.
2x 1
2x ( 2x)2
…
2!
nnn
… ( 1) 2 x
n! ( 2x) 1 e 2x 2 ( 2x)n
n! … f (4) 9. f (2) 1
xx 2 f (2)
f (2) 2x f (2)
P0(x)
P1(x)
P2(x)
P3(x)
10. f
f
f 4 1 3 x2
4
4
6x
x2 1
2
1
2
1
2
1
2 x f (2)
1
2!
8
3
f (2)
, so
8
3! 4
2 8 x 2 (x 4 2)3
16 (x (b) f (1) 2 /4 2
x cos x 2 /4 f 2
x 2 /4 4 , so 2 2!
f 4 , so 2
12 3! 2 P1(x) 2 P2(x)
P3(x) 2 2 x 2
2 2 2
2 2 4 f
f f cos x 4 4 4 P0(x) 4 x3 2x 6x f (1) 6 P3(x) 3 4 4 2 1 x1 f (1)
2! 3 6, so f (1)
3! 1 1) 3(x 1)2 6, so x1 x1 4 (x 4 2x 3 x2 3x 6x 2 2x 12x 2 f (1) 12 P3(x) 3 2 8 14, so f (1)
2! f (1)
12, so
3! 2 x1 x1 11(x 1) 2 /4 2
x 2 /4 2 /4 2 P2(x) 2 P3(x) 2 7 1)2 7(x 2(x f , so 4 2 2! 4 (b) f (1) x4 1 x1 f , so f (1) 3! 12x 2
24x
1 4(x 1) 16. (a) P3(x) 2
12 4x 3 P3(x) 4 f (1) 4 5x 82
x
2! 4 5x x1 x1 4
12, so f (1)
2! 2
2 2 2
2 2 2
2
2
x
12 x 4
2 x 4 4 4 4 2 x 3 4 x1 6 24, so f (1)
3! 4 2 x 4
2 x 6(x 4x 2 4 f (0.2) P3(0.2) 4.848 1)2 63
x
3! x3 1)3 0, the Taylor 2
x 2
2 11 x1 2 P1(x) 2 x1 15. (a) Since f (0) f (0) f (0) f (0)
polynomial of order 3 is P3(0) 0. 2 1)3 (x 14. (a) Since f is a cubic polynomial, it is its own Taylor
polynomial of order 3.
P3(x) 2x 3 x 2 3x 8 or 8 3x x 2 2x 3 f (1) 2 x 4 2
x 4)3
512 3 f (1) sin x (x x1 4 3 /4 cos x 4
4 f (1) 2
x sin x 4 x 2 x 2 2 2 2
x
12 11. f 4
x 3x 2 (b) f (1) 4 x 2 2 (x 4 2 P0(x) 4)2
64
(x 4)2
64 4 f (1) 2 sin x x 2 /4 x 4
4 f (1) 2 cos x 4 2)2 8 x 2 x 1
512 13. (a) Since f is a cubic polynomial, it is its own Taylor
polynomial of order 3.
P3(x) x 3 2x 4 or 4 2x x 3 2)2 (x 4 4 1
16 2 P3(x) , so 2 x sin x 4 f x2 2 P2(x) 1
4 2 x f (4)
3
, so
3!
256 1
64 2 P1(x) 1
2 f (4)
1
, so
2!
32 3 5/2
x
8
x4 P0(x) The series converges for all real x. 1
4 1 3/2
x
4
x4 f (4) … 2 x4 1 1/2
x
2
x4 f (4) 373 4(x 1)3 374 Section 9.2 16. continued (x)(e x)
d ex 1
x
dx
xe x e x 1
x2
ee1
1
1 (c) g (x) (b) Since the Taylor series of f (x) can be obtained by
differentiating the terms of the Taylor series of f (x), the
g (1)
second order Taylor polynomial of f (x) is given by
5 3x 2. Evaluating at x 8x f (0.2) g (x)
3
(x
2! 4 ( 1)(x 1) 4 (x 3
(x
2 f (1.2) 1) P3(1.2) 2
(x
3! 1)2
1
(x
3 1)2 1)3 3.863 ∑ second order Taylor polynomial of f (x) is given by
1) f (1.2) 1)2. Evaluating at x (x 1.2, n1 n1 n
(n x
ex g(x)
(c) g(x) x2
2! x3
3! x4
4! f (t) t2 … … x
for x in the Maclaurin series for e x shown
2 x
2
x
2 1 ex (b) g(x) x2
8 3 n! … 2 1 xn
2 … 2 G(x) … x
2! … (t 2)3 2t 4 2x 3
3 x)1/2 (1 f (0) f (0) xn
2n n! 1
(1
2 … 2t 6 P4(x) 2x 5
5 (t 2)n 2t 2n …] … x) 5/2
x0 x2
8 x
2 2x 2n 1
2n 1 1
f (0)
, so
4
2! 3/2
x0 x) … … 1
2 1/2
x0 3
(1
8 1 2x 7
7 1 x0 x) 1
(1
4 3
f (0)
, so
8
3! 1
8
1
16 x3
16 x
x2
2! x
2!
x3
3!
x2
3! x2
3! f (x 2), the first four terms are (b) Since g(x)
n x
3! ... x
n! … xn
n! … xn 1
n! … … … xn
(n 1)! … 1 1 x2
2 x4
8 (c) Since h (0) x6
.
16 5, the constant term is 5. The next three terms are obtained by integrating the first three terms of This can also be written as
1 , which means 1 x
2! 1 1)! 0, the constant term is zero and we may 2x f (0) x 1
1
x
1
x
x … find G(x) by integrating the terms of the series for f (x).
xn 1
(n 1)! at the end of Section 9.2
e x/2 n
(n (t 2)2 2t 2 (b) Since G(0) 21. (a) f (0) x2
2 nx n 1
(n 1)! … … 1)! 1
at the end of Section 9.2.
1x
2
1 t2
1
2
1 t2 1 19. (a) Substitute (n series for 2 (b) Multiply each term of f (x) by x. xn … 1. 1)! 2[1
1
.
11 x3
4! 20. (a) Factor out 2 and substitute t 2 for x in the Maclaurin 0.36 1
x
1
18. (a) Since f (0)x
, f (0)
.
2
2!
2!
f (10)(0) 10
10!
x 10 (10)
Since
x
, f (0)
11!
10 !
11! ∑ Therefore, g (1) differentiating the terms of the Taylor series of f (x), the 3(x x2
d
x
1
3!
dx
2!
2x
3x 2
1
3!
4!
2!
nxn 1
1)!
n 1 (n ∑ 1)3 (b) Since the Taylor series of f (x) can be obtained by 1 1)(1) From the series, 0.2, 3.52 17. (a) P3(x) (e x
x2 the answer to part (b). The first four terms of the series
…. for h(x) are 5 x x3
6 x5
.
40 375 Section 9.2
22. (a) a0 1
3
a
10
3
a
21
3
a
32 a1
a2
a3 31 3 3
2 9
2 3 (x 2 a2 we have f (0) 3
n 3
.
n! 1 92
x
2 3x … 93
x
2 (b) Since the series can be written as ∑ n0 3 and 3e 3x 3n
x
n! …
L2(x) 1
(x
2 3 n (3x)
, it represents
n! 1
x
2 3) 3
2 4x and 3, respectively. e.
3e 3 x1 23. First, note that cos 18 ∑( n 3x the function f (x) Using cos x 3) 4, f ( 1
, so the linearizations are L1(x)
2 f ( 3) n0 (c) f (1) 0, f (0) n term by , an
n 1)(4) (4x)(2x)
(x 2 1)2 4 4x 2
,
(x 2 1)2 9
2 Since each term is obtained by multiplying the previous ∑ anx 4x
d
dx x 2 1 29. (a) Since f (x) 0.6603. 1)n n0 x 2n
, enter the following twostep
(2n)! commands on your home screen and continue to hit [ 2, 4] by [ 3, 3] (b) f (a) must be 0 because of the inflection point, so the
second degree term in the Taylor series of f at x a is
zero. ENTER.
30. The series represents tan
tan 1 1 4 . When x tan 1( 1)
The sum corresponding to N 25 is about 0.6582 (not
withing 0.001 of exact value), and the sum corresponding
to N 26 is about 0.6606, which is within 0.001 of the
exact value. Since we began with N 0, it takes a total of
27 terms (or, up to and including the 52nd degree term).
24. One possible answer: Because the end behavior of a
polynomial must be unbounded and sin x is not unbounded.
Another: Because sin x has an infinite number of local
extrema, but a polynomial can only have a finite number.
25. (1) sin x is odd and cos x is even
(2) sin 0 0 and cos 0 1 (3x)5
35 81
so = .
5!
5! 40 x. When x 1, it converges to 1, it converges to . 1
(sin x)
x
2n 1
1
x3
x5
… ( 1)n x
…)
(x
x
3!
5!
(2n 1)!
2
4
2n
x
x
… ( 1)n x
…
1
3!
5!
(2n 1)! 31. (a) f (x) (b) Because f is undefined at x
(c) k 0. 1 32. Note that the Maclaurin series for
1 26. Replace x by 3x in series for sin x. Therefore, we have 4 1 … x2 x 1
1 x is …. If we differentiate this series xn and multiply by x, we obtain the desired Maclaurin series
x 2x 2 … 3x 3 …. Therefore, the desired nx n function is
3 27. Since
is 1
4 3! d
ln x
dx 3 2x 3 , which is 1
at x
4 2, the coefficient 1
.
24 28. The linearization of f at a is the first order Taylor
polynomial generated by f at x a. f (x) x d
1
dx 1 x 33. (a) f (x) (1 x 1
(x 1)2 . x)m f (x) m(1 x)m f (x) m( m 1)(1 f (x) x
x)2 (1 m(m 1 1)(m x)m 2 2)(1 x)m (b) Differentiating f (x) k times gives
f (k)(x) m(m 1)(m 2) … (m
Substituting 0 for x, we have
f (k)(0) m(m 1)(m 2) … (m 3 k 1)(1 k 1). x)m k. 376 Section 9.3 33. continued 3. Since f (x) is increasing and positive on [ 3, 0],
M f (0) 1. (c) The coefficient is
f (k)(0)
k! m( m (d) f (0) 2) … (m
k! 1)(m 1, f (0) k 1) m, and we’re done by part (c). 1
,M
2 maximum value of f (x) is f (1) m 34. Because f (x) (1 x) is a polynomial of degree m.
Alternately, observe that f (k)(0) 0 for k m 1. s Section 9.3 Taylor’s Theorem (pp. 480–487)
Exploration 1 1. We need to consider what happens to Rn(x) as n → .
By Taylor’s Theorem, Rn(x)
f (n 1) (c) is the (n 1) (c)
(x
1)! 1) 0) , where 1 and 1 inclusive. Therefore, no Rn(x) 1) (c)
(x
1)! 0) 1 n (n 1)! n x f (x) xn
→ 0 for all x. This means
(n 1)! that Rn(x) → 0 for all x, which completes the proof. 1. e ix 1
1 ix
ix … f (0) (i) n n! … 2. If we isolate the terms in the series that have i as a factor,
we get:
e ix 1. f (0) f (0) nx f (4)(0)
P4(x)
f (0.2) 1 ix 1 x2
2! ix
cos x x2
2!
x4
4!
x3
3! x3
i
3!
x6
6!
x5
5! x4
4! x5
i
5! … … ( 1)n x 2n
(2n)! … x7
7! … ( 1)n x 2n 1
(2n 1)! (i) nx n n! 2x e 1 x0 2e
4e 2x 2. f (0)
… f (0)
f (0) i sin x. 2x 1 0 2 cos (3x) 2 on [ 2 , 2 ] and f (0) 2. Since f (x) is increasing and positive on [1, 2],
M f (2) 7. f (0)
2! 2 f (0)
3!
f (4)(0)
16e 2x
16, so
x0
4!
43
24
2
1 2x 2x
x
x
3
3 8e 2x P4(0.2) 4
3 8, so x0 2
3 0.6704 f (0)
f (4)(0) x
1
2x0
x
sin
0
2
2x0
2
2
2
x
f (0)
cos
, so
x0
4
2
4
2!
8
3
x
f (0)
sin
0, so
0
8
2x0
3!
4
4
4
x
f (4)(0)
cos
, so
16
2x0
16
4!
384 cos 2 x2 P4(x) 1 Quick Review 9.3
1. Since f (x)
M 2. 4, so x0 2, 1 f (0.2) i sin
10 2 x0 … (We are assuming here that we can rearrange the terms of a
convergent series without affecting the sum. It happens to
be true in this case, but we will see in Section 9.5 that it is
not always true.)
3. ei
cos
Thus, e i 3 1/2
x
, so f (0) is undefined.
4 f (0) Euler’s Formula
n
(ix)2
(ix)3
… (ix)
2!
3!
n!
x2
x3
x4
x5
…
i
i
2!
3!
4!
5! 3 1/2
x and
2 Section 9.3 Exercises Example 3, eventually outstrips the power growth in the Exploration 2 x 3/2, we have f (x) 10. No, since f (x) xn
.
(n 1)! The factorial growth in the denominator, as noted in numerator, and we have 4 has a corner at x 9. Yes, since the function f (x) e x has derivatives of the
form f (n)(x)
e x for odd values of n and f (n)(x) e
for even values of n, and both of these expressions are
defined for all values of x. matter what x is, we have
f (n
(n x2 2. 8. Yes, since the derivatives of all orders for sin x and cos x
are defined for all values of x. 1)st derivative of cos x evaluated at (c) lies between 5. On [ 3, 1], the minimum value of f (x) is f ( 3)
7 and
the maximum value of f (x) is f (0) 2. On (1, 3], f is
increasing and positive, so the maximum value of f is
f (3) 5. Thus f (x)
7 on [ 3, 3] and M 7. 7. No, since the function f (x) n some c between x and 0. As with sin x, we can say that
f (n 1
.
2 6. Yes, since each expression for an nth derivative given by
the Quotient Rule will be a rational function whose
denominator is a power of x 1. Your Turn f (n
(n 1
and the
2 4. Since the minimum value of f (x) is f ( 1) P4(0.2) 8 4 384 x4 0.9511 x 377 Section 9.3 5 sin ( x) x 3. f(0) 0 5 sin x x 5 cos x x 0 0 5 sin x x f (0)
f (4)(0)
P4(x)
f (0.2) 4x 2
2 2! 1 x2 (2x)2
2! 1
1
2 1 0, so 0 1
cos (2x)
2 1
2 f (0)
0
2!
f (0)
5
5 cos x x 0 5, so
3!
6
(4)
f (0)
5 sin x x 0 0, so
0
4!
53
5x
x
6
149
P4(0.2)
0.9933
150 f (0) 1
2 8. cos2 x 5 f (0) 0 16x 4
2 4!
x4
3 … 4x 2
2 2! x 2) ( 1)n 2n
(x 2)2
(x 2)3
… ( 1)n 1 (x )
…
2
3
n
2n
x4
x6
… ( 1)n 1 x
…
2
3
n
x4
x2
and f (0.2) P(0.2) 0.0392.
2 Therefore, P4(x)
5. f (0) (1 2 x) f (0) 2(1 f (0) x) 6(1 3 f (0)
f (4)(0) 4 3x 2 P4(x) 1 f (0.2) P4(0.2) 6. xe x 2x x1
x 7. sin x
x
x5
5! x
x2 x3
x
3!
x3
x5
3!
5!
x7
x9
7!
9! x3
2! x2
1 2x 4x 3 … (2x)6
6! (2x)2n
(2n)! … …
…
… 2 (2n)! … ( 1)n 2n 1 2n
x
12 1, the general term can be 1
2x 2x
2x 3 … (2x)2
… 4x 4 sin x. Then P4(x) …] (2x)n
2n x n P3(x) 2 … x x3
, so we use
6 the Remainder Estimation Theorem with n
f (5)(x) 5x 4 … (2n)! 22n 1 x 2n 2
.
(2n 2)! 1 11. Let f (x)
5 … cos x 4. Since 1 for all x, we may use M r 1, x5
giving R4(x)
, so we may assure that
5!
x5
4
R4(x)
5 10 by requiring
5 10 4, or
5!
xn
n! …
… x2 x2 1.56
x2
2! written as ( 1)n 6, so x0 (2x)4
4! 2n 2n
12 x 2x 6
45 x 2[1 f (0)
3
2!
f (0)
4
24(1 x) 5 x 0 24, so
3!
(4)
f (0)
120(1 x) 6 x 0 120, so
4! x) 22n 1x 2n
(2n)! Note: By replacing n with n 2 x0 x4
3 x2 10. 1 x0 … 22nx 2n
(2n)! 64x 6
2 6! 16x 4
2 4! x) at the end of Section 9.2, we have x2 (2x)2
2! ( 1)n ln (1 2 (2x)2n
(2n)! 1
cos (2x)
2
1
1
2 ( 1)n ( 1)n ( 1)n 1
2 ln (1 … … 1
2 9. sin2 x 4. Substituting x 2 for x in the Maclaurin series given for x2 (2x)4
4! … xn 1
n! 5 x 0.06 0.5697. Thus, the absolute error is no greater than 5 … 0.56 x 10 4 when 0.56 (approximately). Alternate method: Using graphing techniques,
…
… x 2n 1
(2n 1)!
2n 1
nx
( 1)
(2n 1)! ( 1)n Note: By replacing n with n … x x3
3! … 2, the general term can be sin x x3
6 x 12. Let f (x) 5 10 cos x. Then P3(x) 4 when 0.57 P2(x) 1 x 0.57. x2
, so we may
2 use the Remainder Estimation Theorem with n 3. Since 2n 5 x
written as ( 1)n
(2n 5)! f (4)(x) cos x 1 for all x, we may use M r 1, 4 giving R3(x)
than (0.5)4
4! x
. For x
4! 0.5, the absolute error is less 0.0026 (approximately). 378 Section 9.3
Alternate method: 12. continued Using graphing techniques, we find that when x Alternate method: Using graphing techniques, we find that
when x
error 1 cos x x2
2 1 cos 0.5 0.52
2 1 x 1 error 0.5, 0.01 1.26
15. Note that 1 0.002583.
x2
tends to be too small, as shown by the
2
x2
cos x and y 1
.
2 x
2 1 0.01
2 1
5 10 . x2
is the second order Taylor polynomial
2 x e x at x for f (x) 0, so we may use the Remainder The quantity 1
graphs of y Estimation Theorem with n
is less than e0.1 when x
0.1 e 0.01, 2. Since f (x) 0.1 and r e x, which 1, giving 3 x
. Thus, for x
0.1, the maximum possible
3!
e0.1(0.1)3
error is about
1.842 10 4.
3! R2(x) [ , 16. Note that e x by [ 1.5, 1.5] 13. Let f (x) sin x. Then P2(x) P1(x) the Remainder Estimation Theorem with n
f (x) cos x e x, so we may use r Alternate method: error sin x x The inequality x
by graphing y sin 10 10 sin x is true for x
sin x 3 1.67 … … x
2! 1x
(e
e
2
1x
for cosh x
(e
2
x3
sinh x x
3!
x2
cosh x 1
2! x3
. Thus, for x
10 3, the maximum
3!
(10 3)3
possible error is about
1.67 10 10.
3! 3 x sinh x 1, giving R2(x) Using graphing techniques, we find that when x 2 1 x2
2! x ( 1)n n xn
n! … and
… x
n! Thus the terms with n even will cancel for 2. Since 1 for all x, we may use M x 1 10 3,
10 10 0, as we may see . x ), and the terms with n odd will cancel
e x).
5 x
5!
x4
4! …
… x 2n 1
…
(2n 1)!
2n
x
…
(2n)! 17. All of the derivatives of cosh x are either cosh x or sinh x.
For any real x, cosh x and sinh x are both bounded by e x .
e x and r So for any real x, let M 1 in the Remainder Estimation Theorem. This gives Rn(x) x. ex x n 1
1)!
n → (n But for any fixed value of x, lim ex x n 1
(n 1)! 0. It follows that the series converges to cosh x for all real values of x.
18. For n
3 [ 10 , 10 3 by [ 2 14. Let f (x) 1 10 10 ,2 10 x. Then P1(x) 10 has derivatives of all orders in an open interval I containing 1 x
, so we may use
2 the Remainder Estimation Theorem with n
f (x)
x 1
(1
4 x) 3/2 a, then for each x in I,
f (x) f (a) R(x), where R(x) f (x) f (a) f (c)(x Letting b possible absolute error is about 1, giving 0.01 the maximum x this equation is f (b) 10 5 . a), so f (b)
b f (a) f (c)(b a), f (a)
for some c
a between a and b. Thus, for the class of functions that have
derivatives of all orders in an open interval containing a
and b, the Mean Value Theorem can be considered a special
case of Taylor’s Theorem. 1.27 f (c)(x a) for some c between a and x. which is equivalent to f (c)
0.2538 and r 0.2538 x 2
. Thus, for x
2! 0.2538(0.01)2
2! 1. Since , which is less than 0.2538 for 0.01, we may use M R1(x) 0, Taylor’s Theorem with Remainder says that if f Section 9.3 19. f (0) ln (cos x) ln 1 x0 1
( sin x) x 0
cos x f (0) sec2 x x f (0)
(a) L(x) tan x x 1 so 0 22. f (0) 0 f (0)
2! sec x tan x x f (0) 1
2 2 so 0 (c) The graphs of the linear and quadratic approximations
fit the graph of the function near x 0. 0 0 (sec x)(sec x) f (0)
2! (tan x)(sec x tan x) x 1, 0 1
2 (a) L(x)
12
x
2 (b) P2(x) 1 0 f (0) 0 0 sec x x 379 1 (b) P2(x) x2
2 1 (c) The graphs of the linear and quadratic approximations
fit the graph of the function near x 0. [ 3, 3] by [ 3, 1]
[ 3, 3] by [ 1, 3] 20. f (0) sin x e 0 e x0 1
23. f (0) sin x f (0) e cos x x 1 0 (esin x)( sin x) f (0) 1 (b) P2(x) sec2 x x f (0) (2 sec x)(sec x tan x) x 1, 1
2 (a) L(x) (a) L(x)
x 1 x 0 0 f (0) (cos x)(esin x cos x)
x0 f (0)
so
2! tan x x (c) The graphs of the linear and quadratic approximations
fit the graph of the function near x 0. 0 0, so f (0)
2! 0 x (b) P2(x) x2
2 1 0 x (c) The graphs of the linear and quadratic approximations
fit the graph of the function near x 0. [ 3, 3] by [ 2, 2]
[ 3, 3] by [ 1, 3] 24. f (0)
21. f (0) (1 x)
1
(1
2 f (0)
f (0)
so 2 (a) L(x)
(b) P2(x) 1 x0
2 x)
3
(1
2 (x) f (0)
2! 1/2 x2) 5/2 0 x(1 ( 2x) (1 2 x)
x 2) 3/2
x0
3/2 0
1, k(1 x) f (0) k(k 1)(1 so x0 1
2 1 0 k1 f (0) 3/2 ( 2x) x x)k x (1 f (0)
2! k(k x)k 1 (c) The graphs of the linear and quadratic approximations
fit the graph of the function near x 0. k(k x0 1), 1)
k(k 1 For k
x2
2 2 2 P2(x) 1 k x0 kx 1) 2
x 3, we have f (x) 2 x)3 and f (x) (1 6. We may use the Remainder Estimation Theorem with n
1, giving R2(x) and r 6x3
3! 2, M x . (In this particular case it is actually true that R2(x) x 3, since f (x) is a cubic polynomial.) Thus the absolute error is less than
whenever x 3
when 0 x 1
100 0.01. In the interval [0, 1], this occurs
3 0.01 0.215. Alternate method:
[ 3, 3] by [ 1, 3] Note that P2(x) 1 3x 3x 2. Using graphing techniques,
(1 x)3 (1 3x 3x 2) 6, 3 1
when x
100 0.215. 380 Section 9.3
e x. Then P3(x) 25. Let f (x) 1 x2
2 x x3
, so we may
6 use the Remainder Estimation Theorem with n
f (4)(x) 3. Since e x, which is no more than e0.1 when x
0.1 may use M
for x e and r 1, giving R3(x) 27. (a) No
(b) Yes, since
dy
dx 0.1, we 0.1 10 0.1, y x
6 0.01
2 x2 … P3(x) 1 x x2 x 3. Since f (4) 24(0.9) 5 24(1 when x …, xn x x3
3 … 0.1, we may use M
24(0.9)
4! 5 x4 0.14
0.95 1.694 (2n x 2n 1
1)(n Using graphing techniques, when x
x
1
1 0.1 (1 x 1.111 1)! . x for x in the Maclaurin series for ln (1 24(0.9) 5 and x) (b) ln 1
1 x
x x2
2 x ln (1 x)
2 x x
2 x
2x 10 4. Rounding up x3
3 2x 3
3 ln (1
3 x 2 29. (a) 0.1,
2 3 x) 1.11 , 2 by [ 2, 2] The series approximates tan x.
10 4. (b) 2 , 2 … xn
n … x) n
x
… ( 1)n 1 x
…
3
n
n
x2
x3
…x
2
3
n
2n 1
2x 5
… 2x
…
5
2n 1 Alternate method: 1 … x) given at the end of Section 9.2. x4
. Thus, for
0.95 10 4. to be safe, an upper bound is 1.70 error x 2n 1
(2n 1)n! 1 0.1, an upper bound for the magnitude of the 1 ( 1)n ln (1 x) , which is no more than approximation error is x5
10 1 for n, the general term may also 28. (a) Substitute 5 1, giving R3(x) x 2, and we may obtain (c) The power series equals the function y for all real
2
values of x. This is because the series for e x
converges for all real values of x, so Theorem 2 of
Section 9.1 implies that the new series also converges
for all x. . x r …. x
n! be written as ( 1)n 0.001
6 26. Since the Maclaurin series is (x) … By substituting n 6 1 2 3 x
2 x
1 4.251 2n series. 1 e0.1 ( x 2 )n
n! the remaining terms of y by integrating the above 2 ex x ( 1)n x
2! The constant term of y is y (0)
10 6. 4.605 Using graphing techniques, when x 1 … … 4 x2 1 Alternate method: error ( x 2)2
2! ( x 2) 1 e0.1 x 4
. Thus,
4! 0.1, the maximum possible absolute error is about e0.1(0.1)4
24 1 x2 e by [ 1, 4] The series approximates sec x. 381 Section 9.4
1
(1 cos 2x)
2
1
1
(2x)2
1
2
2
2! 30. (a) sin2 x (b) e ax cos bx dx (2x)4
(2x)6
…
4!
6!
(2x)2n
…
( 1)n
(2n)!
256x 8
4x 2
16x 4
64x 6
2 8!
2 2!
2 4!
2 6!
1024x10
…
2 10!
x4
2x 6
x8
2x10
…
x2
3
45
315
14,175 (b) derivative
(c) part (b) 4x
3
(2x)3
3! 2x
2x 3 5 4x
15
(2x)5
5! e(a a2 sin P e b2 [(a cos bx b sin bx) i(a sin bx b cos bx)] Separating the real and imaginary parts gives sin 2x e ax
(a cos bx
a2 b2
ax
e
(a sin bx
a2 b2 e ax cos bx dx
e ax sin bx dx b sin bx) and
b cos bx) (pp. 487–496)
x) sin ( x) x
sin x sin x Exploration 1
Test x3
.
6 ei e x3
is less than .
6 sin P and
i (cos i sin ) cos i sin )) 2
2 cos
2 (cos i sin ) (b)
(cos (
2i
(cos ) i sin ( )) d
33. [eax(cos bx
dx (eax)[(bi2 sin bx bi cos bx) a(cos bx i sin bx) bi)(e )(cos bx (a n
n 1. 1 1
dx
x 1 n→ 1
dx
x2 1)2 (n lim ln x k→ lim n→ k
1 x k→ n2 lim 1
n2 (n lim ln k . k→
1k 1
k lim k→ 1 1. 1)2 1 1 1.
1
dx. Since
x 1
is less than 1
n2 1 1
dx.
x2 Since the integral converges, so must the series.
(aeax)(cos bx (a 1 lim Figure 9.14b shows that ∑ bi cos bx) ax lim n→ the integral diverges, so must the series. sin (e )( b sin bx (eax)[bi(cos bx 1
1
n 1
n i sin bx)] ax lim 3. Figure 9.14a shows that ∑ is greater than i sin ) 2i
2i sin
2i n n→ 1
n 2. (a) i sin ) 1 For ∑ 2 : L cos (cos i e
2i 1. For ∑ 1
:L
n Finishing the Proof of the Ratio 1 (cos ( ) i sin (
2
cos
i sin 2 a(cos bx i sin bx) i sin bx)]
i sin bx)] (a bi)x i sin bx) bi)e 34. (a) The derivative of the righthand side is
a
a2 bi cos bx s Section 9.4 Radius of Convergence ( P (b) b sin bx ax a2 Therefore, the difference between the new estimate ei (a cos bx x where x is the error in the original But by the Remainder Theorem, x 32. (a) b2 i sin bx) ai sin bx) estimate. Then
P ea x 7 8x
…
315
(2x)7
…
7! dx bi (a bi)x
e
b2
bi ax
e (cos bx
b2 a
a2
a
a2 31. (a) It works. For example, let n 2. Then P 3.14 and
P sin P 3.141592653, which is accurate to more
than 6 decimal places.
(b) Let P bi)x i e ax sin bx dx bi
(a bi)e(a bi)x
b2
a 2 (bi)2 (a bi)x
e
a2 b2
a 2 b 2 (a bi)x
e
e(a bi)x,
a2 b2 which confirms the antiderivative formula. 4. These two examples prove that L 1 can be true for either
a divergent series or a convergent series. The Ratio Test
itself is therefore inconclusive when L 1. Exploration 2
1. L lim n→ x
n Revisiting a Maclaurin Series n1 n
xn 1 lim n→ n
n converges absolutely when x 1 x x . The series 1, so the radius of convergence is 1.
2. When x
1 1
2 1
3 1, the series becomes
… 1 ….
n Each term in this series is the negative of the corresponding
term in the divergent series of Figure 9.14a. Just as ∑
diverges to , this series diverges to . 1
n 382 Section 9.4 3. Geometrically, we chart the progress of the partial sums as
in the figure below: Section 9.4 Exercises
1. Diverges by the nthTerm Test, since lim 1 –2 n→ n
n 1 +3 n→ n 0.
. (The 1 Ratio Test can also be used.)
1 L 1 1–2 2n 2. Diverges by the nthTerm Test, since lim 1
–4
1
+5 1 1 3. Converges by the Ratio Test, since etc. lim 1
1
1– 2 + 3 1
1
1
1– 2 + 3 – 4 n→ 4. The series converges at the righthand endpoint. As shown
in the picture above, the partial sums are closing in on
some limit L as they oscillate left and right by constantly
decreasing amounts. an 1 lim n→ an 1) 2 (n
2 2n 1 n1 n 2 1
2 1 1. 4. Converges, because it is a geometric series with r
so r 1
,
8 1. 5. Converges by the Ratio Test, since
5. We know that the series does not converge absolutely at the
righthand endpoint, because ∑ 1
diverges (Exploration 1
n of this section). n→ nx
n n2 x
n→ n(n 3. lim
n→ n→ 3
1) 2. lim xn
n! 1 2n lim n
n→ (3 an 1 1 1)
2n n x ∑ Comparison Test. n x lim 1 an ∑ Quick Review 9.4
1. lim 3n 1
2
1.
3
2n
n→
n
2
Alternately, note that n
for all n.
3
3
1
n
n
2
2
Since
converges,
converges by the Direct
n
1
n 13
n 13 lim x 1 6. Diverges by the nthTerm Test, since n2 3 lim
n→ n 2 x n 3 lim n sin n→ 0 1
n 1 0 7. Converges by the Ratio Test, since (Note: This limit is similar to the limit which is discussed lim n→ at the end of Example 3 in Section 9.3.) an 1 lim (n n→ an 1)2e
n 2e n n1 1 1. 1
10 1. e 8. Converges by the Ratio Test, since
4. lim
n→ (n 1) 4 x 2
(2n)4 2 x lim
21 6. Since n 2 5n for n 2
1n lim 2x 2 6, an ln n for n n 2, bn
n 2, an n→ 1 lim n→ 5n, and N
5 5 , bn 1, an n and N
n, bn an 1 an (n 1)10
10 n 1
n→ 10 n
n10 lim an 1 (n 4)!
3!(n 1)!3n
n4
lim
1)
n→ 3(n
1
1.
3 lim n→ an 6.
6. ln n, and 1 3!n! 3n
(n 3)! 10. Diverges by the nthTerm Test, since 1
n! and hence n
9. Since 10
10
1
1
an
,b
, and N 25.
10 n n
n! n lim 2 1
for n
n! n 10. Since n 2 1 9. Converges by the Ratio Test, since 1 6, an n for n 8. Since n
N 1. 2x n→ 5 4n x
16 16 2x 1
5. lim n 1
2x
n→ 2 6n 2
16n 4 2 n1n 7. Since 5 4n 3 n→ x n n4 n 3 and hence
1
,b
n2 n n 3 1
n2 25, 1n
n lim 1 n→ e 0.
2
,
3 11. Converges, because it is a geometric series with r
n , and N 3 for so r 1. 2.
12. Diverges by the Ratio Test, since
lim n→ an
an 1 lim n→ (n 1)!e
n!e n n1 lim (n n→ (The nthTerm Test can also be used.) 1)e 1 . 383 Section 9.4
13. Diverges by the Ratio Test, since
lim an n→ 3n 1
lim
1)3 2n
n→ (n
3n 3
lim
(n 1)3(2)
n→
3
1.
2 1 an (The nth 22. lim n32 n
3n 1 an n→ 1 diverges for x lim 1 (n lim n→ an 1
lim
n→ 2
1
1.
2 x an n→ lim
lim n→ lim n→ 2 1)ln (n 1)
2
2n 1
n ln n
n 1 ln (n 1)
n
ln n 24. lim
n→ lim n→ 1, or x n→ lim n→ lim n→ lim n→ 2 10, so the radius of convergence is an 1 lim (n
n n→ an 1) x n
3 1 n2
n xn lim x x n→ 1 and diverges for x 1, so 25. lim
n→ (2n 1)!
1)!
n!
3)!
n1
(2n 3)(2n 2)
1
0 1.
2(2n 3) an 1 xn lim n→ an (n
x
lim
n→ 3 1 n
n1 1) n
x
3 13 The series converges for x n 3n
xn 3 and diverges for x 3, so the radius of convergence is 3. lim an 2
3 1, or x the radius of convergence is 1. 26. lim n 1 1
, and
3
1
1
, so the radius of convergence is .
3
3 2 The series converges for x 16. Converges by the Ratio Test, since
an 2
2
3 10. (n
n→ (2n 1 an 3x 2n n 15. Converges by the Ratio Test, since
lim n
3x 23. This is a geometric series which converges only for Term Test can also be used.) 14. Converges by the Ratio Test, since
n→ 1 The series converges for 3x 10 an 3x 2 n
n1 lim n→ an (n 1)!
n
(n 1)n 1 n!
(n 1)n n
(n 1)(n 1)n
nn
n1
1
(1 1/n)n 1
e n→ an 1 an x 2n 3
1)!
n→ (n n! lim x lim 2n 1 n→ x2
n 0 1 The series converges for all values of x, so the radius of
convergence is . 27. lim
n→ an 1 lim (n 1) x
5n n→ an lim 1 n→ x 3 3n 1 5n 1 nx x 5 3n 3
5 The series converges for x 3 5 and diverges for 17. One possible answer:
x ∑ 1
diverges (see Exploration 1 in this section) even
n 1n
1
though lim
0.
n→ n 28. lim
n→ 3
an 5, so the radius of convergence is 5. 1 an 18. One possible answer:
Let an 2 n and bn n 3 a n 20. This is a geometric series which converges only for
x5
1, so the radius of convergence is 1. 1] 4 and diverges for x 1, 29. lim
n→ an
an 1 lim n→ 1xn n 1 3n n1 3 The series converges for x nx (4x 1) 1, or x
1
4 convergence is . 1
4 1
, so the radius of
4 n lim n→ x
3 3 and diverges for x the radius of convergence is 3.
21. This is a geometric series which converges only for 4n(n2 1)
nxn 4, so the radius of convergence is 4. 3n
is a divergent geometric series.
2 19. This is a geometric series which converges only for x
so the radius of convergence is 1. 1 n1 The series converges for x Then ∑an and ∑bn are convergent geometric series, but ∑ bn ∑ (n 1) x n
[(n 1)2
n→ 4
x
x
lim
4
n→ 4 lim x
3 3, so 384 Section 9.4 30. lim an n→ 1 lim 1)! x 4 n
n! x 4 n (n n→ an lim (n 1) x n→ 35. This is a geometric series with first term a 1 common ratio r 4 (x 4) 1) 1 The series converges only for x 4 2 x 1 a
1 r 31. lim
n→ 4 n1 lim ( 2) (n 2) x 1
2n (n 1) x 1 n lim 2 x
1 1
and diverges for
2
1
1
, so the radius of convergence is .
2
2 The series converges for x
x 1
an n→ 1 4x
(n 5
1)3/2 n→ n 3 4
2x 3 1)2 (x 4 1 1)2 (x common ratio r 9 1 and . It converges only when 1, so the interval of convergence is 9 5 2n 4x 4
2x 36. This is a geometric series with first term a 3/2 5)2 lim n→ an 1 2n 3 lim (4x 32. lim 1)2 x2 1 n→ 2x 4
(x 4 n1 x2 n→ an 1)2 (x 1 convergence is 0. 1 . It converges only when 3. 4, so the radius of
Sum an 1 and 1, so the interval of convergence is 4 (x 1)2 (x x 2.
a Sum 1 r 5)2 (4x 1 The series converges for (4x
to 4x
x 33. lim
n→ 5 5
4
an 1 1, which is equivalent 1
and diverges for
4
1
1
. The radius of convergence is .
4
4 lim n1 x
n x 0 n→ lim n→ 2 2n x
2n 3 1 Sum 2n 1
(x
2)2
n→ 2
1
(x
2)2
2
1
The series converges for (x
2 x x equivalent to x
x 2 2 1 and 1. It converges only when 2 1, so the interval of convergence is
16.
1 a
1 r 2 x 1 1 2 4 x 1 and 1 common ratio r ln x. It converges only when ln x lim so the interval of convergence is
Sum
2)2 8 38. This is a geometric series with first term a 2 2n x 9
2x x2 8 1 and diverges for 1, so the radius of convergence is 1. 1 1 2 The series converges for x an 9
2x common ratio r x an 1)2 37. This is a geometric series with first term a n n→ 34. lim 9
(x x2 lim x x 9 9 n
x 1 1)2 (x 1 5
4 1, or x n→ an 5) 2 1, which is 2, and diverges for 2. The radius of convergence is a
1 r x e. 1
ln x 39. This is a geometric series with first term a
common ratio 2. 1 1
e x2 1
3 a
1 1
r 1 1 and
x2 . It converges only when so the interval of convergence is
Sum 1, x2 1
3 3 2 x 3
(x 2 1) 1
3 2.
3
4 x2 1, Section 9.4
40. This is a geometric series with first term a
sin x
sin x
common ratio
. Since
2
2 1 and 46. 1 for all x, the interval ∑ n 1 (2n 6
1)(2n
3
3 1) s1 3 s2 (3 1) 1 s3 (3 1) 1 41. Almost, but the Ratio Test won’t determine whether there is
convergence or divergence at the endpoints of the interval. sn 3 42. (a) For k S of convergence is x a Sum 1 1
r 2 sin x
2 1 2
sin x N, it’s obvious that
… a1 . ak … a1 For all k ak ∑ nN1 cn. 47. N, … aN a1 … a1 a1 … aN a1 … aN aN
… cN 1 ∑ aN … ak cn (b) Since all of the an are nonnegative, the partial sums of
the series form a nondecreasing sequence of real
numbers. Part (a) shows that the sequence is bounded
above, so it must converge to a limit.
43. (a) For k N, it’s obvious that
… dk For all k
… dk d1 d1 d1 an. dN d1 … dN 1 … dN d1 … dN … ∑ ak dN sn
S 48. N, d1 … nN1 ∑
s1 … 1
1 an sn (b) If ∑an converged, that would imply that ∑dn was also S nN1 convergent.
49. s1 44. Answers will vary. S lim sn n→ 1 1) ∑ n1 4n 1
3 4n s2 1 s3
1
5
1
5 1
9
1
9 1
1
9 1
9 sn
1
13 1 s2 ∑ ∑ 2n
(n 2 s3 dk 1 1 lim sn 1 1
13 S
50. s1
s2
s3 sn
S 1 ∑ 1
1)2
1
4
1
4
1
4 n1 ∑ n1 3
7 3
7 3 5 5
1)2 (2n 5
25
5
25 1
n2 1
4
1
4 1)2 (2n 5
25 5
5
25 5
49 5
49 5 1
9
1
9 1
1)2 (n 1
9 1
1
9 1
16 1
16 1 1
1)2 (n lim sn 1 n→ 1
1 1
2
1 1 1 2 1
1 1 2
1 lim sn 3
1 2 1 1 2 1 n
n→ 3
5
3
5 1 5 n→ n 1n aN 4
45.
3)(4n
n 1 (4n
1
s1 1
5
1
s2
1
5
1
s3
1
5
1
sn 1
4n 1 40n
1)2(2n 1)2
5
5
9
5
5
5
9
9
5
5
5
9
9
5
5
(2n 1)2 s3 nN1 3 3
5
3
5 2n 3 n 1 (2n s2 3
1 1 lim sn s1 ck 1 3
2n 3
2n n→ ∑ ∑ n1 385 3
1 1 3 3 4 1
ln 3
1
ln 3 1
ln 4
1
ln 5 1
ln 2
1
ln 4 1 1 1
1
ln 3
ln 2
1
1
1
ln 3
ln 2
ln 4
1
1
1
ln 3
ln 2
ln 4
1
1
ln 5
ln 2
1
1
ln(n 2)
ln 2
1
lim sn
ln 2
n→ 1 1
4 386 Section 9.5
1 51. s1 tan s2 (tan s3 tan 1 tan 1 (tan 1 sn 4 S
52. x 1 (tan 2)
1 tan 1 1 2 2
tan 1 3) tan 1 (tan 1 2 tan 1 3) 1. We first note that the Integral Test applies to any series of lim tan n→ 1 n 4 2 4 If p 1 1
dx
xp n0 k→
k→ n1 ∑ nx x)2 1 is negative for all x 1
dx
xp 1 k→ kp 1p
1
p1 1
p1 n lim 1 1 0 Multiply by x:
(1 k lim n0 x is continuous and positive for all x xp p = lim ∑ nx x)2 p x 0, 0. 1: Differentiate:
(1 1
where p is positive. This is because the
np and f (x) 1) ∑ xn 1 The pSeries Test function f (x) (n 4 s Section 9.5 Testing Convergence at
Endpoints (pp. 496–508) the form ∑ 4) 4 lim sn 1 2) 1 tan 4 Exploration 1 3 tan n→ 1 1 tan
1 2 3 1 (tan
4 1 tan 4 1 1 1 x p1k p1 1 1 (since p 1 0) . The series converges by the Intergral Test. n0 Differentiate:
(1 d
x
dx (1 x)2 (1 x)2(1)
x) (1 x )
x1
(1 x)3 x
(1 ∑ n2x n 1
x)3 (1
2x (x)(2)(1
x)4 x)( 1) 2. If 0 p
1
dx
xp 3
1 13
2 lim ∑ n 2x n 1
:
2 ∑ n2 n0 ∑ nn n 02 The sum is 6. 1n
2 1
dx
xp 1 p1k x k→ 1 k→ n0 2 6 k→ p1 n0 x(x 1)
(1 x)3 13
22 k lim
lim Multiply by x: Let x 1: 1 1 (k 1 1p p 1) (since 1 p 0).
The series diverges by the Integral Test.
If p 0, the series diverges by the nth Term Test. This
completes the proof for p 1.
3. If p 1 1: 1
dx
xp k lim k→ 1 1
dx
x
k lim ln x k→ 1 lim ln k k→ . The series diverges by the Integral Test. Exploration 2 The Maclaurin Series of a
Strange Function
1. Since f (n)(0) 0 for all n, the Maclaurin Series for f has all zero coefficients! The series is simply ∑ 0 x n 0. n0 2. The series converges (to 0) for all values of x.
3. Since f (x) 0 only at x 0, the only place that this series
actually converges to its fvalue is at x 0. 387 Section 9.5
Quick Review 9.5 10. Converges by the Limit Comparison Test, since 1
p dx with p
1x
1
2. Diverges, limit comparison test with integral of
x 1. Converges, since it is of the form 1 lim n→ 1, and ∑ n1 1
n2 2. 11. Diverges by the nthTerm Test, since 2
4. Converges, comparison test with integral of 2
x 3n lim 1 1 1
3 3n n→ 0. 1 5. Diverges, limit comparison test with integral of
0
22 {un } is a decreasing sequence of positive terms with 8. No, neither positive nor decreasing for x lim un 3 0. n→ 9. No, oscillates
10. No, not positive for x 13. Diverges by the nthTerm Test, since lim 1 n→ Section 9.5 Exercises
5
x 1 3∑ 3
n n1 n1 1 dx diverges. 1
, which diverges by
n1/2 3. Diverges by the Direct Comparison Test, since
2 and ∑ n2 ln n
n 1
for
n 1
diverges.
n lim un n→ 1
2x 1 x1
and observe that
x1
1
(x 1)
( x 1)(1) f (x) 1 dx diverges. (x 1)2 6. Converges, since it is a geometric series with 1
n n→ Then an 1
n 0 and bn e 0, and ∑ e for n n converges as a n0 geometric series with r e 1 an lim 1. 8. Converges by the Direct Comparison Test, since
n ln n
2 ln n 1
,
2 1
.
2 0.37. n→ lim n→ bn 1
.
n 1
and bn
n2 0.91. 7. Diverges by the nthTerm Test, since lim n sin en
1 e 2n ln n
ln n2 16. Diverges by the Limit Comparison Test. 1.44. Let an
1
ln 3 , 1.) 15. Diverges by the nthTerm Test since 5. Diverges, since it is a geometric series with r 1
1 1 x 2x
,
2(x 1)2 x 2x f (x) which means each term is
1
ln 2 n
n 0. (To show that un is decreasing, let which is negative, at least for x 4. Diverges by the Integral Test, since r . then {un} is a decreasing sequence of positive terms with the pseries Test. n 10 n
n10 14. Converges by the Alternating Series Test. If un 1. Diverges by the Integral Test, since
2. Diverges because ∑ 1
, then
ln n 12. Converges by the Alternating Series Test. If un x 7. Yes, for N 5n 3 3n
2)(n 2 5)
1
n2 converges as a pseries with p 1
3. Diverges, comparison test with integral of
x 6. Yes, for N n 2(n 1
n2 1
n
1
n 0 for n lim n→ 2 and n 1
n lim 1 n→ 1
n 1. Since ∑ bn diverges, ∑ an also diverges.
n1 n1 17. Converges absolutely, because, absolutely, it is a geometric
series with
r 0.1.
18. Converges conditionally: 9. Converges by the Direct Comparison Test, since
n
n2 1 with p 1
for n
n3/2
3
.
2 1, and ∑ n0 1
converges as a pseries
n3/2 1 n 1
n 1
, then {un} is a decreasing sequence
n2
1n
of positive terms with lim un 0, so
( 1)n 1 2
n
n→
n1 If un n2 ∑ converges by the Alternating Series Test.
But ∑ n1 1 n n2 1 n
diverges by the Direct Comparison Test, since
n2
1
1
for n 1 and
diverges.
n
n 1n ∑ 388 Section 9.5
2n
converges by the
3 19. Converges absolutely, since ∑ n2
n1 Ratio Test:
an lim n→ (n 1) n1
22 1 3n
n2 2 3 2
3 (c) None 1. 28. This is a geometric series which converges only for
x5
1, or 6 x
4. 20. Converges conditionally. (a) ( 6, ∑ 29. This is a geometric series which converges only for
4x 1
diverges by the integral test, since
n ln n lim ln ln x b→ n→ n!
2n and so the terms do not approach 0. n1 n1 sin n
converges by direct
n2 1
, which converges as a pseries with
n2 23. Converges conditionally: n 1 n1 n ( 1)n
1 converges 24. Converges absolutely, since ∑ n1 n1 1
.
2 1
,
n1/2 ∑ nn n1 n1 ∑( cos n
n n1 1)n+1
.
n x ∑( 1
:
3n 1. Check x 1
n 1 3x 2n 1 1: ∑ 2
2 1, or n 1)
converges
n
1n
1
diverges.
n
n 1n ∑ 1
,1
3
1
,1
3
1
3 (c) At x x 2 1, or 10 8 x 12. (b) ( 8, 12)
(c) None
n→ an 1 lim (n n→ an n 1) x n
3 1 n2
nxn The series converges absolutely when x , then {un } is a decreasing sequence of positive terms with lim un
n→ 0, so ∑ ( 1) n1 n n 1 n lim
n x 1. For x 1/2 n Since ∑ vn diverges as a pseries with p nthTerm Test.
(a) ( 1, 1) But ∑ un diverges by the Limit Comparison Test:
n→ 1 n converges by the Alternating Series Test. u
1
, then lim n
n1/2
n → vn (a) 32. lim 26. Converges conditionally: diverges. n
3x The series converges absolutely when 3x 1
, which
n3/2 (See Examples 2 and 4.) n1 1 (a) ( 8, 12) 25. Converges conditionally, since ∑ If vn n→ an 3x 2 n
n1 31. This is a geometric series which converges only for cos n converges as a pseries. n lim n diverges by direct comparison to ∑ which diverges as a pseries with p If un 1 (b) 0, so ∑ n→ n1 n→ , then {un} is a decreasing sequence of by the Alternating Series Test. 1 an conditionally. Check x positive terms with lim un n1 30. lim n1 1 But ∑ 1
,0
2 1
3 2. 1 0. 1
,0
2 (b) x (c) None 22. Converges absolutely, since ∑
comparison to ∑ 1
2 1, or (a) .
2 21. Diverges by the nthTerm Test, since lim If un 1 b 1
dx
x ln x p 4) (c) None converges by the Alternating Series Test. 2 4) (b) ( 6, 1
, then {un} is a decreasing sequence of
n ln n
1
positive terms with lim un 0, so
( 1)n 1
n ln n
n→
n2 If un n2 (a) ( 1, 1)
(b) ( 1, 1) 1 an But ∑ 27. This is a geometric series which converges only for x 1
.
2
1
1
,
u also
2n 1 n ∑ (b) ( 1, 1)
(c) None x
1, or 1, the series diverges by the 1. 389 Section 9.5 33. lim an n→ 1 xn lim n→ an (n 1 1) n 3n 1 3, Σ when x n1 x n3
3
.
2 n x n 38. lim
n→ 3 (a) Only at x
(b) At x 1) x 1 4 4
4 4 4 (c) None
an n→ 1 n→ an 0, x
,x 39. lim
lim n→ an 1)! x 4 n
n! x 4 n (n lim (n (c) None
an lim n→ (b) [ 3, 3] n→ 1 3. Furthermore, (a) [ 3, 3] 34. lim an Σ1 n1 , which also converges
3/2
n n as a pseries with p x 1 The series converges absolutely for x
n n 3n n x
(n 2n 3 n!
x 2n 1)! 1 lim n→ x 2 n 2 n 1(n
2 n (n lim n→ 1n
1n 2) x
1) x 1 2(x The series converges absolutely for 2(x 0 1 1 an The series converges absolutely for all real numbers. 1
2 (a) All real numbers 1) 1, or nthTerm Test. x (b) All real numbers
(a) x3
5n
5
nx 3n
n→
n→
x3
The series converges absolutely for
1,
5
x3
or 8 x 2. For
1, the series diverges by
5
an 1 lim an (n 1) x
5n 3n (c) None
40. lim n→ (n 1) x n
4n 1[(n 1)2
n→ lim 1 Check x 4n(n2 1)
nxn 1]
x
4 1, or 4 x 4. 4: n
∑ (n2 1) 1 converges by the Alternating Series Test. (a) 1, ∑ n2 4: an n→ diverges by the Limit Comparison Test 1 lim n→ an 1 (a) [ 4, 4)
(b) ( 4, 4) ∑( 37. lim
n→ 4 1 lim n→ an 1 xn
3n 1 1 n xn x
3 ∑ 3 3, the series diverges by the nthTerm Test. (a) ( 3, 3)
(b) ( 3, 3)
(c) None 1 n
x n x x 3. 1, or 1.
1: converges by Alternating Series Test. n
1 n1 3, or 1 n Check x
3n The series converges absolutely for x
For x 1) n1 n n x Check x an ∑ The series converges absolutely for x with ∑ (c) At x n1 x 1: diverges as a pseries with p
n (a) [ 1, 1) (b) ( 1, 1) (c) At x 1 1
.
2 5)2 1, or
1
n3/2
3
:
2 3
2 41. lim 1
1
.
1n n 5)2 2n 1 (c) None n0 n0 ∑ (4x 1 3
2 n n n 3/2
5 2n 4x ∑ x
4 (b) 1, The series converges absolutely for Check x n→ 5 2n 3
1)3/2 4x
(n ( 1)
1x
1:
n3/2
n1
n
3
converges as a pseries with p
. Check x
2
12n 1
3
.
3/2 converges as a pseries with p
2
n1n (c) None
1 lim 3
. Check x
2 (b) ( 8, 2) an 1 an The series converges absolutely for (4x (a) ( 8, 2) an an n→ the nthTerm Test. 36. lim 1, the series diverges by the 13
,
22 1 1 1) 13
,
22 (b) (c) None
35. lim 3
. For 2(x
2 1) 390 Section 9.5
46. (a) Diverges by the Limit Comparison Test. 42. This is a geometric series which converges only for
ln x 1, or (a) x 1
. Then ak 0 and
k1/2
ak
k1/2
1
bk 0 for k 1 and lim
lim
.
bk k→
k→
2
2k 7
1
Since
bk diverges as a pseries with p
,
2
k
1
1 Let ak e. 1
,e
e (b) 1
e 1
,e
e 2k and bk 7 ∑ ∑ ak also diverges. k1 (c) None
43. n 13 109 365 24 3600
ln (n 1) sum 1 ln n
ln (4.09968 1017 1) sum
1017)
40.5548... sum 41.5548...
40.554 sum 41.555 1 f (x) dx
1 a1 k→ e 0. (c) Converges absolutely by the Comparison Test, since
cos k
k2 1, 1 and ∑ 1
for k
k2 k pseries with p k1 1
converges as a
k2 2. n … an a1 f (x) dx.
1 If the integral diverges, it must go to infinity, and the first
inequality forces the partial sums of the series to go to
infinity as well, so the series is divergent. If the integral
converges, then the second inequality puts an upper bound
on the partial sums of the series, and since they are a
nondecreasing sequence, they must converge to a finite sum
for the series. (See the explanation preceding Exercise 42 in
Section 9.4.) (d) Diverges by the integral test, since
3 b 18
dx
x ln x lim 18 ln ln x b→ 3 47. One possible answer: ∑ n3 1
n ln n This series diverges by the integral test, since
3 45. 1k
k lim 1 ln (4.09968 44. Comparing areas in the figures, we have for all n
n1 (b) Diverges by the nthTerm Test, since 1017 4.09968 b 1
dx
x ln x lim ln ln x . Its partial sums are b→ 3 roughly ln (ln n), so they are much smaller than the partial y sums for the harmonic series, which are about ln n.
y = f (x )
an – 1
0 ( 1)k 48. (a) ak a N aN + 1
N N+1 N+2 n an
n+1 x 1/k 1 6(kx)2 dx 0 ( 1)k 1 2k 2x 3 1/k
0 y ( 1)
y = f (x ) (c) The first few partial sums are: an
N–1 N N+1 N+2 n–1 x n Comparing areas in the figures, we have for all n
n1 f (x) dx
N aN … S1
N, n an k (b) The series converges by the Alternating Series Test. a N a N + 1 aN + 2
0 k12 aN f (x) dx.
N If the integral diverges, it must go to infinity, and the first
inequality forces the partial sums of the series to go to
infinity as well, so the series is divergent. If the integral
converges, then the second inequality puts an upper bound
on the partial sums of the series, and since they are a
nondecreasing sequence, they must converge to a finite sum
for the series. (See the explanation preceding Exercise 42 in
Section 9.4.) 2, S2 S7 319
,S
210 8 5
,S
34 1, S3 533
,S
420 9 7
47
37
,S
,S
,
65
30 6
30
1879
. For an alternating
1260 series, the sum is between any two adjacent partial
sums, so 1 S8 sum S9 3
.
2 49. (a) Diverges by the Limit Comparison Test. Let
an
n n
and bn
3n2 1
an 1, and lim n→ bn 1
. Then an
n
n2
3n2 1
n→ lim ∑ bn diverges, ∑ an diverges. n1 n1 0 and bn
1
. Since
3 0 for 391 Section 9.5 ∑ 3n2n (b) S 1 n1 ∑ 3n23 3
n n1 53. ln (1 . 1 Σ This series converges by the Direct Comparison Test,
since n1 ∑ n12 is convergent as a 3
3n2 1 1
and
n2
n pseries with p Σ(
n1 x) ( 1)
n n1 (b) 1 x 1, the series is Test. 2. ∑( 54. arctan x x2 x3 2 x , so at x . This series converges by the Alternating Series 1)n n0 x) n 1 At x 50. (a) From the list of Maclaurin series in Section 9.2,
ln (1 n
1x 1)n 3 … ( 1)n 1x n ∑( …. n n0 x 2n 1
2n 1 1, the sequence is
1)n( 1)2n
2n 1 ∑ 2(n 1) 1 , which converges by the 1 n n0 Alternating Series Test. At x 1 1, the sequence is ∑ 2(n 1) 1 , which converges by the Alternating Series Test.
n 3
2 1
2 (c) To estimate ln , we would let x n0 55. (a) It fails to satisfy un The truncation error is less than the magnitude of the x6
6 1
26 6 ∑ (b) The sum is sixth nonzero term, or
1
384 un 1 for all n ∑ 21n 1
n n 13 N. 1
1
2 n1 0.002605 n1 51. lim
k→ ak ∑ ( 1)n 1x 2n
2n 1
2n k1 1 ( 1)n 1(x 2)n
n k1 2x
lim
3)
k→ ln (k 1 ak ln (k 2)
2k x k 1
ln (1
2 1
2 Check x
k (b) lim 1
,
2 (c) pseries when p (b) For p 1, the series is ∑ Integral Test, since
2 ∑ n2 p 1
converges as a
np 1
, which diverges by the
n ln n 1
dx
x ln x 1
1, we have p
n ln n b lim ln (ln x) b→ .
2 1
, so
n ln n 1
diverges by the Direct Comparison Test with
n p ln n 1
from part (b).
n ln n n ( lim n→ nn
2n 1 n lim n→ n) 2
2 1
2 n lim n→ 2n n an n lim n→ , n odd 1 1
2 n
2n lim n→ , n odd
n lim n→ , n even Thus, lim
n→ n an n an n 1
2 n 2 lim n→ , n even n x Check x 1
2 1
2n 1
, so the series converges.
2 an
lim 1n nx n→ n 4 x 1
4 The series converges absolutely if
3 1. n2 (c) For 0 n3 n→ n→ , n odd n→ 52. (a) The series converges by the Direct Comparison Test,
3, and ∑ n2
2n n lim an lim 58. (a) lim
1
for n
np an The series converges. ∑ 1
n ln n n n→ ∑ p 1 The series converges. ( 1)
converges by the Alternating Series Test.
2)
k 0 ln (k
1
Check x
:
2
1
diverges by the Direct Comparison Test, since
2)
k 0 ln (k
1
1
1
for k 2 and
diverges. The original
k
ln (k 2)
k 2k
1
1
series converges for
x
.
2
2 since n n→ x) 1
.
2
1
:
2 x ∑ 57. (a) lim 2 2x The series converges absolutely for x
or 1/2
1/2 56. Answers will vary. 0.002605. ∑ 1 1
.
2 Thus, a bound for the (absolute) truncation error is (d) 1/3
1/3 x 1
4 5.
3: ∑ ( 1)n diverges.
n0 Check x 5: ∑ 1n diverges.
n0 The interval of convergence is ( 3, 5). 1, or 392 Chapter 9 Review 58. continued
n (b) lim
n→ an 2n nx lim x lim n 3n n→ 2 x x n n→ n3
x2
The series converges absolutely if
3 1 x4
n3 n
an
3
x 4n
n→
x4
The series converges absolutely for
1,
3
an 2. lim
2
3 1, or or 7 lim x 1. Check x 7: ∑ Check x 5. 1: ∑ n1 1: ∑ Check x n1 5: ∑ Check x ( 1)n
converges.
n n1 1
diverges.
n n1 n n→ an n lim (b) [ 7, n→ 1) (c) ( 7, 1) (d) At x
2n x n ( 1)n
converges
n
1
diverges.
n (a) 3 The interval of convergence is [ 1, 5).
(c) lim 4n 1
1)3 n 1 x
n → (n 1 7 2x
3. This is a geometric series, so it converges absolutely when The series converges absolutely if
2x r ∑ 1 diverges. (a) ∑ Check x 1
:
2n Check x 1 n n→ an x n→ ln x n 1) 1, or 1, or ∑ ln 0 1n
e e: ∑ (ln e) n n0 ∑( 1)n diverges. (d) None
an n→ 1 an x 1 2n
(2n 1)!
1)! x 1 2n 2
n→ (2n
x 12
lim
0
1)(2n)
n→ (2n lim The series converges absolutely for all x. n0 The interval of convergence is 5
.
2 15
,
22 n0 ∑ 1n diverges. x 15
,
22 4. lim
1
:
en 1
2 3
2 ln x e. Check x 1), the series converges absolutely when (c) n lim Check: x 2
(x
3 (b) The series converges absolutely if ln x
1
e 1 and diverges for all other values of x. Since 2
(x
3 11
,.
22 The interval of convergence is
(d) lim r 1
1
x
.
2
2
1
:
( 1)n diverges.
2n 1 1, or (a) 1
,e .
e (b) All real numbers
(c) All real numbers s Chapter 9 Review Exercises (d) None (pp. 509–511)
1. lim
n→ an 1 an lim n→ x
(n n1 1)! 5. lim
n!
xn lim n→ x
n 1 0 an n→ 1 an 3x (d) None n2
3x 1n 3x 1 3x 2
. Furthermore, when
3
1
1, we have an
and
n2 1 1, or 0 1 (a)
(c) All real numbers n→ 1n 1
1)2 3x
(n The series converges absolutely for The series converges absolutely for all x. (b) All real numbers lim ∑ n1 x 1
converges as a pseries with p
n2 2, so ∑ an also
n1 converges absolutely at the interval endpoints.
(a) 1
3 (b) 0, 2
3 (c) 0, 2
3 (d) None 393 Chapter 9 Review
an 6. lim
n→ (n 2) x 3n 3
(n 1) x 3n
n→ 1 x3 lim an n→ The series converges absolutely for x 3
1 x 1. When x an 9. lim 1 xn lim n→ an n 1 n
xn 1 x The series converges absolutely for x 1, or 1, the series diverges by the nth Check x ∑ Term Test. n1 ∑ (b) ( 1, 1)
(c) ( 1, 1) 1 x 1. 1: ( 1)n
converges by the Alternating Series Test.
n Check x (a) 1 1, or 1 1:
1
.
2 diverges as a pseries with p
n n1 (a) 1 (d) None (b) [ 1, 1)
an 7. lim
n→ 1 lim (n n→ an 2x 2) 2x 1 n
(2n 3)2n 1 1 (2n 1)2n
(n 1) 2x 1 n (c) ( 1, 1)
(d) At x 1
2 The series converges absolutely for
3
2 1
2x 1
. When
2
2 x 2x 1 10. lim
n→ 1, or 2 1, the series diverges by the lim 1
e 1 an x lim n→ (n 1, we have an ∑ 1
and
ne
n 1 1
ne e, so ∑ an also converges
n1 (a) (d) None
an 1, absolutely at the interval endpoints. 31
,
22 (c) ex 1
.
e x converges as a pseries with p 31
,
22 (b) ne
e xn
n The series converges absolutely for e x Furthermore, when e x (a) 1 n→ en 1 x n 1
(n 1)e
n→ 1 an or nthTerm Test. 8. lim an 1 xn 1
1)n
n→ (n
1 lim 1) 1 1n
n 1 nn
xn x lim
n→ (n x
1
lim
e n→ n 1 nn
1)(n 1
e (b) 0 1) 11
,
ee (c) 11
,
ee n (d) None The series converges absolutely for all x.
Another way to see that the series must converge is to
observe that for n 2x, we have xn
nn 1n
, so the terms
2 3n
an
(n 1) x 2n 1
n→
n→
x2
The series converges absolutely when
1,
3 11. lim an 1 lim are (eventually) bounded by the terms of a convergent or geometric series. When x A third way to solve this exercise is to use the nth Root (a) Test (see Exercises 57–58 in Section 9.5). (b) ( 3, 3) (c) ( 3, 3) (a)
(b) All real numbers
(c) All real numbers
(d) None 3 x (n 2) x 2 n
3n 1 1 x2
3 3.
3, the series diverges by the nth Term Test. 3 (d) None 394 Chapter 9 Review 12. lim an n→ 1 lim x n→ an 1 2n
2n 3 3 1 2n
x 1 12 The series converges absolutely when x
or 0 x 12 x 2n 1 1 1, or 2 n0 n ∑ 2n 1 2: ∑ n0 ( 1) ( 1)
2n 1 n0 ( 1)
converges
2n 1 (a) ( 1)n
converges conditionally by the
2n 1 (b) ( 3,
3, 3) (d) None
1
1 (a) 1 1 x n→ 1 lim 1)! x 2n
2n 1 (n n→ (n lim n→ 18. f (x) 2 1)x 2
2 0, x
,x 2 2n
n! x 2n ln (1 x) 19. f (x) sin x 20. f (x)
0 only (c) x 10x n 1
ln (n 1)
n→ 1 lim an 1
10 ln n
10x n x ∑ 1
:
10 n 2 ∑ 1
:
10 n
1
Test, since
ln n 2 1x n n , 5
.
3 x 2n 1
(2n 1)! …, 0. tan 1 x cos x2
2! ( 1)n n ( 1)
converges by the Alternating
ln n xn
n! x5
5 … . Sum …, evaluated at tan 1 ( 1)n ∑ x 2n 1
2n 1 1 …, . (Note that 6 3 when n is replaced by n 1
diverges by the Direct Comparison
ln n
1
1
for n 2 and
diverges.
n
n 2n …, 1
.
2 3 … x3
3 n 1x x 2n
(2n)! 2. x
1 … 3 1, the general term of tan 1 x 2n 1 2n 1 , which matches the general term given in the exercise.)
23. Replace x by 6x in the Maclaurin series for 1
1 x given at the end of Section 9.2. 1
10 1 (b)
(c) 11
,
10 10 1
n→ The series converges only at x 0. 1 an lim (a) 0
(b) x 0 only (c) x 0 (d) None 6x (6x)2 …
… 36x 2 … (6x)n
(6x)n …
1
1 x given at the end of Section 9.2. (n 2)! x n 1
(n 1)! x n
n→ an (6x) 24. Replace x by x 3 in the Maclaurin series for
1
10 (d) At x 6x 1
1 1 11
,
10 10 n ln ( 1)n sin
x4
4! e ln 2 ln 2. Sum becomes ( 1) Check n 2
3 … . Sum x evaluated at x Series Test. 15. lim 3 1 22. f (x) 1, x2
2! 1 x 10x 1
.
10 Check n (a) ex 21. f (x) The series converges absolutely for 10x
or cos x 0 an n→ . Sum evaluated at x (d) None
14. lim x5
5! ( 1)n 0. (a) 0
(b) x … ln 1 x3
3! x evaluated at x
The series converges only at x x3
3 2
. Sum
3 evaluated at x 0
0 x2
2 x 4
.
5 1
4 1 …, ( 1)nx n 1 (c) (0, 2)
0 and x … x2 x
1
. Sum
4 evaluated at x (b) [0, 2] an 3. It 3) (c) ( 17. f (x) an x 3 Alternating Series Test. 13. lim 3 , so it converges n conditionally by the Alternating Series Test. (d) At x 1
2 diverges for all other values of x. 0: ∑ Check x x2 absolutely when 1, 2. Check x x2 16. This is a geometric series with r lim (n 2) x (x 0) x3 1 (x 3) 1 1 x3 (x 3 ) 2
x6 … … ( x 3)n ( 1)nx 3n …
… 25. The Maclaurin series for a polynomial is the polynomial
itself: 1 2x 2 x 9. 395 Chapter 9 Review 26. 4x
1 4x x 33. Use the Maclaurin series for e x given at the end of 1
1 x 4x(1 x
4x 2 4x … x2
4x 3 …) xn … 4x n Section 9.2.
… 1 x2 xe 5 27. Replace x by x in the Maclaurin series for sin x given at ( x)3
3! ( x)5
5! … ( 1)n ( x)2n 1
(2n 1)! … 2x
in the Maclaurin series for sin x given at
3 the end of Section 9.2 ( 2x
sin
3 30. x e x 2 5! 4x 5
3645 … ( ( 1)n … 2x 2n+1
1)n 3 (2n 1 (2n x2
2! x
1
1
2
x2
2! 1 3x ) ln (1 x
x4
4! ( 5x)2
2! … xn
n! 1 32. Replace x by end of Section 9.2.
e x/2 1
1 (3x)5
5 … ( 1)n (3x)2n 1
2n 1 ( 2x)2
( 2x)3
…
2
3
n
( 2x)
…
( 1)n 1
n
n
8x 3
… (2x)
2x 2x 2
3
n 36. Use the Maclaurin series for ln (1 …
x) …. x) given at the end of Section 9.2.
x ln (1
… x x2
2! … … x 2n
(2n)! x) ( 1)n xn
n! … 37. f (2) x2
2 x3
2 (3 2!
22 x
8 … n!
xn 1
n! 2 ( x)] ( x)
3 x4
3
1 x) (3
2(3 x) 6(3 x) 4 n!(3
1 x 38. f ( 1) x2 2)n f ( 1) (6x f ( 1) 6x (x 2x 2 (3x 2 1 5) x
1
1 … 2) 2 1 1
1 (x 2) 3 1
… 0 for n 10, so 5 7(x f ( 1)
2! 5 1 4. x3 2 7 f ( 1)
6, so
3! f (n)( 1)
… … … 4x) x
4) x n f (n)(2)
n!, so
n! n1 2) f ( 1) n
1 ( x) … f (2)
2!
f (2)
6, so
3! x2 (x (x 3 1 2, so x2 x) ( 1)n 1 x2 (x 3 … xn
n 3 f (2) … 1 x2
2 x) 3 … f (2) ( … 2 f (2) … 1 xn
2 x ln [1 ( x)
2 x
x2 x
in the Maclaurin series for e x given at the
2 x
2
x
2 x given at 2x 1)! … 5x)4
…
4!
( 5x)2n
…
( 1)n
(2n)!
n
5x
(5x)2
… ( 1)n (5x)
2!
4!
(2n)! 1 1 2x in the Maclaurin series for ln (1 2x) f (n)(2) 5x … n! given at the end of Section 9.2. the end of Section 9.2.
cos (3x)3
3 3x 35. Replace x by 5x in the Maclaurin series for cos x given at 31. Replace x by x … x x 1
1
2 1)! 1 2x 2n 1
3 x3
x5
x7
…
3!
5!
7!
x 2n 1
…
( 1)n
(2n 1)!
2n 1
x3
x5 x7
… ( 1)n x
3!
5! 7!
(2n 1)! sin x ex 2x 5
3 4x 3
81 2x
3 29. 2x 3
3 3! 2x
3 ( 1)n 2n 1 the end of Section 9.2.
tan 28. Replace x by ( x 2 )n
n! … 34. Replace x by 3x in the Maclaurin series for tan the end of Section 9.2.
x … x
2! x3 x sin x ( x 2)2
2! ( x 2) x1 2x 2 2 1) 5(x 1)2 (x This is a finite series and the general term for n 1)3
4 is 0. 396 Chapter 9 Review
1
xx 3 39. f (3)
f (3) 2x f (n)(3)
n!
1
1
x
3 40. f ( ) x3 f (3)
2
1
, so
2!
27
27
f (3)
2
1
6x 4 x 3
, so
3!
27
81
( 1)n
3n+1
1
1
1
(x 3)
(x 3)2
(x 3)3
9
27
81
n
(x 3)
( 1)n n+1
3 f (3) ∑ 3 x3 sin x x f( ) 1
, then {un} is a decreasing sequence of
ln (n 1)
( 1)n
positive terms with lim un 0, so
converges
ln(n 1)
n→
n1 If un 1
9 2 x f (3) 45. Converges conditionally: 1
3 by the Alternating Series Test. The convergence is
conditional because
… 1 cos x x 0, so (x 1
(x
3! ) lim 1
(x
7! ( 1)n (2n 3n 1
1)!
n → (n 1 n!
3n lim an lim n→ 3
n 1 0. 1n
for n
2 2n3n
nn )5 12 and ∑ n1 1n
is a convergent
2 geometric series. Alternately, we may use the Ratio Test or
the nthRoot Test (see Exercise 57 and 58 in Section 9.5). 1 1 1
.
ln 2 48. Converges absolutely by the Direct Comparison Test, since … )7 an n→ 1
(x
5! )3 1b
ln x 2 lim b→ 47. Converges absolutely the the Ratio Test, because 0, if k is even
1, if k 2n 1, n even
1, if k 2n 1, n odd sin x 1
dx
x(ln x)2 2 f( )
0
2!
f()
1
1, so
3!
6 sin x x f (k)( ) n 46. Converges absolutely by the Integral Test, because cos x x f() ∑ 1 diverges, so ∑ n 1n Comparison Test. 0 f( ) 1
1
for n 1 and
n
ln (n 1)
1
diverges by the Direct
1)
1 ln (n 1)! )2n (x 1 …
49. Diverges by the nthTerm Test, since 41. Diverges, because it is ∑ n1 5∑ 5
n 5 times the harmonic series: 1 ( 1)n(n2
2n2 n
n→ 1)
does not exist.
1 lim n 1n 50. Converges absolutely by the Direct Comparison Test, since
42. Converges conditionally. ∑ n1 converges as a pseries
3/2 1 1 If un n , then {un} is a decreasing sequence of positive terms with lim un
n→ 0, so ∑ ( 1)n converges by the n n1 1)(n
3
.
with p
2
n(n 1
and
n3/2
n 2) 1 Alternating Series Test. The convergence is conditional
because ∑ n1 1 is a divergent pseries p
n 1
.
2 51. Converges absolutely by the Limit Comparison Test. 43. Converges absolutely by the Direct Comparison Test, since
0 ln n
n3 with p 1
for n
n2 1 and ∑
n 1
converges as a pseries
n2
1 2. 1 Let an
Then lim n n2
an n→ bn 1 lim n→ 1
.
n2 and bn 1 and ∑ n2
n n2 n2 lim n→ an
an 1 n
lim
n→ (n 2
1)! n!
n 1 n
lim
n→ (n 52. Diverges by the nthTerm Test, since
2
1)2 0. lim n→ n n
n 1 2 2). Therefore ∑ an converges. pseries ( p 44. Converges absolutely by the Ratio Test, since 1 n 2n 1 lim 1 n→ 1
n n 1
e 0. converges as a 397 Chapter 9 Review
53. This is a telescoping series. ∑ (2n n3 s1
s2
s3
sn
S (c) The fourth order Taylor polynomial for g(x) at x
x ∑ 1
3)(2n 1) n
1
2(2 3 3)
2(2
1
1
1
6
10
10
1
1
1
6
10
10
1
1
6
2(2n 1)
1
lim sn
6
n→ 1
1
3)
2(2n 1)
3 2(2n
1
1
1
3 1)
6
10
1
1
1
6
14
14
1
1
1
1
6
14
14
18 [7 3(t 4) 4)2 5(t 4 is 4)3] dx 2(t 4 7t 4)2 5
(t
3 4)3 1
(t
2 4)4 7(x
1
18 3
(t
2 4) 3
(x
2 4)2 5
(x
3 4)3 1
(x
2 x
4 4)4 (d) No. One would need the entire Taylor series for f (x),
and it would have to converge to f (x) at x 3. 57. (a) Use the Maclaurin series for sin x given at the end of
Section 9.2. 54. This is a telescoping series. ∑ n(n 1) n2 2
2 s1
s2
s3 1 sn 2
n n2 2
n 2
2
1
3
3
2
2
3
3
2
2
3
3
2
n2 1 S ∑ 2 1
lim sn (x/2)3
(x/2)5
3!
5!
5x 3
x5
…
48
768 x
2 5 1 5x
2
2
4
2
4 1 2
4 2
4 2
5 1 2
5 f (3) an lim n→ 1 5 lim f (3)
(x
3! 1
f (3.2) 4(x 3) P3(3.2) f (3)
(x
2! 3)
3)
3(x 3)2 (2n 5 1)! 2 2n+1
x 0 3)3 5
2n 1, 2, 3, …. for all x and all n
2(x (2n (c) Note that the absolute value of f (n)(x) is bounded by 3 3)2 x 2n 3
3)! 2
x/2 2
3)(2n 2) n→ (2n an lim f (3)(x ( 1)n the Ratio Test: n→ 55. (a) P3(x) (x/2)2n 1
…
(2n 1)!
5
x 2n+1 …
( 1)n
(2n 1)! 2 … (b) The series converges for all real numbers, according to 1 n→ x
2 5 sin We may use the Remainder Estimation Theorem with 1.936 M
So if (b) Since the Taylor series for f can be obtained by term 1
.
2 5 and r
2 x 2, the truncation error using byterm differentiation of the Taylor Series for f, the Pn is bounded by second order Taylor polynomial for f at x 5
2n 4 6(x f (2.7) 3) 6(x 3)2. Evaluated at x 3 is 1 2n 1
(n 1)! .
4. So, two terms (up through degree 4) are needed. means the graph of f is concave up near x 6, which 58. (a) Substitute 2x for x in the Maclaurin series for 1
1 x given at the end of Section 9.2. 3. 1
1 7. Since 1 2x 2x (2x)2 1 56. (a) Since the constant term is f (4), f (4)
f (4)
, f (4)
3! 1)! To make this less than 0.1 requires n 2.7, 2.74. (c) It underestimates the values, since f (3) 2 5
(n 2x 4x 2 … (2x)3
… 8x 3 … (2x)n
(2x)n … 12. (b) Note that
P4 (x)
3 10(x 4) 6(x 4)2 24(x 4)3.
The second degree polynomial for f at x 4 is given
by the first three terms of this expression, namely
3 10(x 4) 6(x 4)2. Evaluating at x 4.3,
f (4.3)
0.54. (b) 11
1
, . The series for
is known to converge for
22
1t 1 t 1, so by substituting t resulting series converges for 1 2x, we find the
2x 1. 398 Chapter 9 Review 58. continued
(c) f (b) Integrate term by term. 1
4 2
, so one percent is approximately 0.0067. It
3 x ln x 1
(t
2 t 1
, the series is the
4 n0 18
2 through degree 7) are sufficient since dt
1
(t
3 1
(t
4 3)3 0.0067. … 3)4 x 3)n+1
…
n1
3
(x 3)2
(x 3)3
3)
2
3
3)n 1
n (x
…
( 1)
n1 (x Series Estimation Theorem, it shows that 8 terms (up 2 3)2 ( 1)n 1n
. If you use the Alternating
2 alternating series ∑ 1
t 3 takes 7 terms (up through degree 6). This can be found
by trial and error. Also, for x 2 (t 3)4 (x … 4 It is also a geometric series, and you could use the
(c) Evaluate at x
remainder formula for a geometric series to determine
the number of terms needed. (See Example 2 in 1
2 3.5. This is the alternating series 1
22 2 1
23 3 … ( 1)n 2n 1 1
(n 1) … By the Alternating Series Estimation Theorem, since
Section 9.3.)
the size of the third term is
an 59. (a) lim
n→ 1
(n 1)n 1
n!
(n 1)!
x nnn
x (n 1)n 1
lim
(n 1)nn
n→
n 1n
x lim
xe
n
n→ 1 xn lim terms will suffice. The estimate for ln given at the end of Section 9.2.
1
, so the
e 1, or x e 2x 2 ( 2x 2)2
( 2x 2)3
2!
3!
2n
… ( 2x )
…
n!
4x 6
…
2x 2 2x 4
3
n 2n
2x
…
( 1)n
n! ( 2x 2) 1 1
e 1
11
12
31
3
1
2
1
3
9
6
5
0.278
18 1
3 13
3 22
2! 33
3! (b) Use the Ratio Test:
lim n→ (c) By the Alternating Series Estimation Theorem the error
is no more than the magnitude of the next term, which
14
3 44
4! (x 2) 60. (a) f (3)
f (3)
f (3) (x
2(x f (3) f (x) 1 2 2) 1 (x 3) ( 1)n(x f (3)
2! x3 6, so 4 ( 1)nn!, so The series converges for all real numbers, so the f (n) (3)
n! (x
3)n , ). bounded by the magnitude of the fifth term, which is = 2, so
3 x lim n→ (c) This is an alternating series. The difference will be
1 x3
3 2n 1x 2n 2
n!
(n 1)!
2nx 2n
2x 2
lim
0
n1
n→ 1 an interval of convergence is ( x3 2) an 0.132. 1 2) 6(x f (n)(3) 32
243 3
is 0.375.
2 2x 2 for x in the Maclaurin series for e x 61. (a) Substitute radius of convergence is . is 0.05, the first two n→ an The series converges for x e (b) f 1
24 1
f (3)
3! 1 ( 1)n
3)2
… (x 3)3 … (2x 2)4
2x 8
. Since 0.6 x 0.6, this term is less
4!
3
2(0.6)8
than
which is less than 0.02.
3 62. (a) f (x) x2 1
1 x x 2(1 x x2
(b) No. At x x3 x2
x4 …
… ( x)n
( 1)nx n+2 …)
… 1, the series is ∑ ( 1)n and the partial
n0 sums form the sequence 1, 0, 1, 0, 1, 0, …, which has
no limit. Chapter 9 Review
63. (a) Substituting x 2 for x in the Maclaurin series for sin x
given at the end of Section 9.2,
sin x 2 6 10 x
3! x2 … x
5! du
( 1)n 4n 2 x
(2n 1)! . Integrating termbyterm and observing that the
constant term is 0,
x x3
3 sin t 2 dt 0 x7
7(3!) x1 1
11(5!) ( 1)n
1 (4n … x 4n 3
3)(2n 1)! x2 Let u
du e x dx dv 2x dx x 2e x dx x 2e x ex v
x 2e x 2x 2x e x dx
e x dx dv 2 dx ex v 2xe x dx x 2e x 2xe x x 2e x … 1
7(3!) suffices to use the first two nonzero terms (through
degree 7). 1 x 2e x dx (x 2 2)e x 0 2e x C 2)e x 2x 2x 2e x dx 2xe x (x 2 1
…
(b) sin x dx
11(5!)
0
1
….
( 1)n
(4n 3)(2n 1)!
1
1
Since the third term is
0.001, it
11(5!)
1320
1
3 (e) Let u 399 C e 2 1
0 65. (a) Because [$1000(1.08) n](1.08)n
available after n years. 0.71828 $1000 will be (b) Assume that the first payment goes to the charity at the
end of the first year.
1000(1.08) 1 1000(1.08) 2 1000(1.08) 3 …
(c) This is a geometric series with sum equal to (c) NINT(sin x 2, x, 0, 1)
(d) 1
3 1
7(3!) 0.31026830 1
11(5!) 258,019
831,600 1
15(7!) This is within 1.5 10 7 1000
0.08 1000/1.08
1 (1/1.08) 0.31026816 of the answer in (c). 12,500. This means that $12,500 should be invested today in order to completely fund
the perpetuity forever.
66. We again assume that the first payment occurs at the end of 64. (a) Let f (x)
1 x 2e x dx. x 2e x dx 0 the year. 1 f (x) dx Present value 0 h
f (0)
2 2f (0.5) 1
0
4 2 e0.5
8 e0.5
4 f (1) e
4 x2 1 x x2
2! 1 x3 1000(1.06) … xn
n! …
… 0.68333 (c) Since f is concave up, the trapezoids used to estimate
the area lie above the curve, and the estimate is too
large.
(d) Since all the derivatives are positive (and x 0), the
remainder, Rn(x), must be positive. This means that
Pn(x) is smaller than f (x). 3 1000
1.06 1 The present value is $16,666.67. n2
x4
…x
2!
n!
x4
2
3
x
P4(x) x
2
1
3
x
x4
41
x5 1
P4(x)
3
4
60
10 0
0 x2 1 1000(1.06)
1000/1.06
(1/1.06) e 0.88566
(b) x 2e x 1000(1.06) 2 …
16,666.67 400 Section 10.1 67. (a) Sequence of
Tosses Payoff
($) Probability Term of
Series T 0 1
2 0 70. (a) Note that ∑ x n 1 12
2 2 13
2 2 13
2 14
2 3 identity ∑ x n 12
1
2 HHT 14
2 HHHT 3 is a geometric series with first term x 2 and common ratio r a 1
2 HT 1 n1 x 1 1 n1 x, which explains the 2 x (for x 1). Differentiate. ∑ (n x)(2x) (x 2)( 1)
(1 x)2 (1 1)x n n1 2x
(1 x2
x)2 Differentiate again. ∑ n(n 1)x n 1 n1 Expected payoff
1
2 0
(b)
(c) 1 1 13 3 2 … 2 … 3x 2 2x 14 (2x x 2)(2)(1
(1 x)4
x)(2 2x) 2(2x x 2)
(1 x)3 nx n (1 … 1 x 2(1 x)2 x2 2x 3 … 3x 2 2x … 3x4 nx n
nx n …) 1 x)3 ∑ n(n 2x 1)x n (1 n1 1
, the formula in part (c) matches the
2 1
x (d)
If x
nonzero terms of the series in part (a). Since
2
(1/2)
1, the expected payoff is $1.
2
[1 ∑ n(nxn 1) 68. (a) The area of an equilateral triangle whose sides have
1
2 s2 3s
2 3
4 x)3 Replace x by . (1/2)] length s is (s) x)( 1) Multiply by x. … 1 2x) 2
(1 x2
(1 2 2 1 x)2 (1 x)2(2 (1 12 1 n1 . The sequence of areas (b) Solve x 2
x (x 13
x (x 2x 2
to get x
1)3 2x 2
,x
1)3 1 2.769 for x 1. removed from the original triangle is
b2 3
4 3
3n b2 3
4 b2 3
2
4 9 b2 3
2n
4 3b2 3
42 b2 3
4
4 Chapter 10
Vectors … … or 32b2 3
43 … 3nb2
4n …. 3
1 (b) This is a geometric series with initial term a
and common ratio r
b2 3
b
, so the sum is
4
1 2 b2 3
4 3/4
(3/4) s Section 10.1 Parametric Functions
(pp. 513–520)
Exploration 1 Investigating Cycloids
1. 3, which is the same as the area of the original triangle.
[0, 20] by [ 1, 8] (c) No, not every point is removed. For example, the
vertices of the original triangle are not removed. But
the remaining points are “isolated” enough that there
are no regions and hence no area remaining.
69. 1
1 x 1 x x2 x3 … 1
x)2 1 Substitute x 2x 3x 2 4x 3 2na for any integer n. 3. a 0 and 1 cos t 0 so y 0. 4. An arch is produced by one complete turn of the wheel.
Thus, they are congruent.
5. The maximum value of y is 2a and occurs when
x (2n 1)a for any integer n. Differentiate both sides.
(1 2. x 5x 4 1
to get the desired result.
2 … 6. The function represented by the cycloid is periodic with
period 2a , and each arch represents one period of the
graph. In each arch, the graph is concave down, has an
absolute maximum of 2a at the midpoint, and an absolute
minimum of 0 at the two endpoints. Section 10.1
Quick Review 10.1
1. (cos (0), sin (0)) 2. (a) (1, 0) 3
3
, sin
2
2 2. cos
3. x 2 y2 (0, 1 (since cos2 t (b) 1)
sin2 t 1) 4. The portion in the first three quadrants, moving counterclockwise as t increases.
5. x t2 t, y 1, 1 t 3. (a) dy
dx
d 2y
dx 2 dy
dx dy/dt
dx/dt dy
dx which at t 3t
3
t 3 (b) 3
.
2 2/2) 0 3/(2 3t)
1/(2 t 1)
3t 1 3 cos t
,
2 sin t
2/2) 0
sin t 3 dy/dt
dx/dt y 3 3
3(
equals
4
2( 3 sin t
sin t dy /dt
dx/dt 6. The graph is a circle with radius 2 centered at (2, 3).
Modify the x cos t, y sin t parameterization
correspondingly:
x 2 cos t 2, y 2 sin t 3, 0 t 2 .
7. dy/dt
dx/dt y 401 d 2y
dx 2 3/(2t 2 3
1/(2 t dy /dt
dx/dt
3 1/t
1/t 2 3t 3/t)
1) 1 t t2 3 3/t t 3/2
3
8. y
x C. For t
2
32
3
(
2)
2
2
3
Thus, y
x3
2
2
x
3 9. y
32
2 10. y 3 3 2
2 , so
4. (a) 2.
(b) 2.
3
,x
4 2)
2
x
3 Thus, y 2 and y C and C C. For t
2
(
3 3
2 3
,x
4 C and C 2 and y
5 2
6 32
, so
2 5. (a) . (b) 52
.
6 dy
dx
d 2y
dx 2
dy
dx 3 d 2y
dx 2 3
2 1 9
x dx
4 0 8
1
27 x 9 3/2
x
4 dx
6. (a)
3
0 1
1/t 2 dy/dt
dx/dt dy /dt
dx/dt
[(2t 3)(6t)
12t 2 18t
(2t 3)3
6t 2 18t
(2t 3)3 2 1
0
3 dy /dt
dx/dt y x, so Length dy/dt
dx/dt y 313/2 8
.
27 (b) dy
dx dy/dt
dx/dt y d 2y
dx 2 3t 2
2t 3 (3t 2)(2)]/(2t
2t 3
6t 2 2t
2t dy /dt
dx/dt
[(2t 1)(2) (b) dy
dx
d 2y
dx 2 y dy/dt
dx/dt
dy /dt
dx/dt 2 sin t
4 cos t
1
sec2 t
2 4 cos t 1
1 1)2 4 1
tan t
2
1
sec3 t
8 3)2 (2t 1)(2)]/(2t
2t 1 Section 10.1 Exercises
1. (a) t2 1)3 (2t 7. dy
dx dy/dt
dx/dt cos t
sin t (a) cot 0 when t (x, y)
(2,
(b) 2
1 cos cot t
k (k any integer). Then 2 k 2 , (2 sin k 2 1). The points are (2, 0) and (2, cot is undefined when t
(x, y) 1 cos (k ), The points are (1, 2). k (k any integer). Then
1 sin (k )) 1) and (3, 1). (2 1, 1). 402 8. Section 10.1 dy
dx sec2 t
sec t tan t dy/dt
dx/dt 13. x csc t t 2, y t, so
1 (a) Nowhere, since csc t never equals zero. 9. dy
dx dy/dt
dx/dt 3t 3t 2 (a) 4 4 2 2 2 2 , 12
(t
3 3 , 8 33 (0.845, 1 1) 0 1 0.609 2 4 8 1)3/2 22
3 .
3 3 3 which evaluates to
(3.155, 3.079). 3 2 3
2 1 dt 1 3/2
(2
3 4
3 0 when t Then (x, y) t t2 0 3t 2 4 1 t 2 dt 0
1 (b) csc t is undefined when t k (k any integer). Then
(x, y) (sec (k ), tan (k )) ( 1, 0).
The points are (1, 0) and ( 1, 0).
2 (t 2)2 Length 14. x 8t cos t, y , 8t sin t, so /2 3 (8t cos t)2 Length
0 3.079) and (8t sin t)2 dt /2 8t dt
0 2 (b) Nowhere, since 4 /2 4t 2 3t is never undefined. 2 0 dy
dx dy/dt
dx/dt 3 cos t
3 sin t (a) 10. cot t 0 when t cot t 15. x
y k (k any integer). 2 /3 ( 2, 1 2 k ,1 3 sin 2 k 3). The points are ( 2, 4) and ( 2, sec2 t /3 0 cos t)2 cot t is undefined when t
(x, y) 1 dt 0 /3 ln 2
0 k (k any integer). Then 3 cos (k ), 1 3 sin (k )) (2 3, 1). The points are (1, 1) and ( 5, 1). 16. x et 2t, y e t, so 1 2 (e t Length 2t )2 (1 e sin t, y 1 cos t)2 dt (1 17. x sin t, y 2(1
4 cos2 Area cos t) dt
t
dt
2 0 3 Length ( 2t 2 0
3 (t
0 2) dt 12
t
2 sin t) dt
2 2t cos t 8 2 0 18. x 1 t, y , so
t 2 2 3/2
t
3 2
0 t, so
3)2 (2
0 Area
1 cos2 t dt 2 2 t
2 cos
dt
2
0
t
2 2 sin
4
20 3, y sin t) ( sin t)2 2 (2
0 0 2t 4.497. cos t, so 2 0 12. x ) dt , cos t, so ( sin t)2 length t2 1 which using NINT evaluates to
11. x ( sin t)2 dt tan t dt 2). (2 cos t, /3 ln sec t
(b) sec t (sec t
0 3 cos cos t sin t, so Length Then (x, y)
2 sec t tan t sec2 t
sec t tan t 4
3 t)2 dt (1 3 2t
0 2 t t2 1 dt
1)3/2
1) 1, y dt 0 4 (5 19. x 12
t 4 12
(t
33 21
2 t)2 ( 2
0 5
9 14.214 2t, so
3 Area 2 (t 1) 1 (2t)2 dt, 0 which using NINT evaluates to 178.561. Section 10.1
20. x sec t cos t (see Ex. 15), y sin t, so /3 Length cos t)2 2 (cos t) (sec t
0 /3 26. Parameterize the curve as x sin t dt
0 /3 2 cos t (b) x 2t, y (t)
2, y t 1, 0 Area 2 (t formula. Then 1 27. x
22 1) 0 dy
dy
becomes
dt
dy 2t t, y 5 (t 12 dt 2
3 (1 2) ( (t 1)2 dt 2t 4 12
t
2 0 12 5 1. 1) dt t 22 3 12
m
2 5
r
x,
h 4 12.
0 t
0 6, or m2 m
2 m t
m 12
t
2 Now solve 5, so 22. (a) Because these values for x(t) and y (t) satisfy y 48 2 12
for m:
2 2m 12 solution. The midpoint is at t (h, r), and this range of tvalues gives the correct initial (x, y) and terminal points. h, y 13 1)2 1
, (2
3
2 ( 13 13 1, which gives 1)3/2 (3.394, 5.160). 28. r, so
1 2 (rt) h2 Area r 2 dt 0 r h2 [ 4, 4] by [0, 10] 1 r2 t 2 Use the right half of the curve, 0 0 r2 r h2.
r2 (c) Slant height x
h2, so Area r r2 h2 3 cos t, y Area 2 sin 2t, y 2 sin 2t)2 (
0 . 2 (3 sin t) (3 cos t)2 (6 cos 2t)2 dt, 0 159.485. 2 cos 2t, so /2 Length t 6 cos 2t, so which using NINT evaluates to
23. (a) x 0, and 13. Take the positive 1 which is the equation of the line through the origin and (b) x d. The 0 5 (c) Slant height y 0
4 1 1
5 t2
2 t2 Total length 1) dt 0 y, c 1, so 4 1 2 Area t 1, so
1 g( y), y parameter is y itself, so replace t with y in the general .
0 21. (a) x(t) 1. tan t dt /3 2 dx
dx becomes 2 2 cos t
0 dx
dt 25. In the first integral, replace t with x. Then ( sin t)2 dx 403 (2 cos 2t)2 dt 29. /2 2 dt . 0 (b) x cos t, y
1/2 Length [0, 9] by [ 3, 3] sin t, so
( cos t)2 ( sin t)2 dt 1/2
1/2 dt Use the top half of the curve, and make use of the shape’s
symmetry. . 1/2 x 3 cos t, y
/2 24. x 3 sin t, y 4 cos t, so Area 2 6 cos 2t, so 2 (3 sin 2t) (3 cos t)2 (6 cos 2t)2 dt 0
2 Length ( 3 sin t)2 (4 cos t)2 dt 0 which using NINT evaluates to 22.103. which using NINT, evaluates to 144.513. 404
30. y Section 10.1
0 for t 0 and t 2 .x a a cos t, y (b) x a sin t, t cos t, y t sin t, so 2 2 2 [a(1 t dt 0
2 2 a2 0
2 2 8 a2
8a cos t) 2 t
2 sin2
2
t
sin3
dt
2 cos2 1 0 t
2 cos
2 2 64 a
3
dx
dt 0
2 2 2 (a sin t)2 dt 0 34. All distances are a times as big as before.
(1 2a 8a a cos t)2 cos t)] (a 0
2 2 2 2 cos t dt (a) x a(cos t (b) Length t
4 sin2
dt
2 t sin t), y t
2 y t
dt
2
2
2
t
cos3
3
20 sin 0 for t v sin /16
0
0 v0 sin . To find the path length, evaluate 32 (v0 cos )2 32t, and . The maximum height is attained 16
v0 sin in midflight at t t cos t) and y v0 cos
v0 sin 0 or t a(sin t 2 2a For exercises 35–38, x 32t)2 dt using NINT. To (v0 sin find the
maximum height, calculate a(1 cos t) (Note: integrate with respect to x from 0 to 2a ; integrate ymax (v0 sin ) 2a Area y y dx 0 0 a 2 cos t)a(1 0 or t x
(1 cos t) dt a2 t 1
sin 2t
4 t
2 2 sin t 2 .
0. 0
3.206 150 sin 20 (150 cos 20 )2 Length 0 16t) 150 cos 20 , y
(75 sin 20 )/8 2 2 cos t 32 75
sin 20
8 cos t) dt 2 v0 sin 16 32 t(150 sin 20 t 2a a(1 v0 sin 35. (a) The projectile hits the ground when y with respect to t from 0 to 2 .) dx
32.
dt 2 (t sin t)2 dt 0 Area 31. (t cos t)2 Length so 32t
32t)2 (150 sin 20 0 2 3 a2 dt which, using NINT, evaluates to 461.749 ft 0 (b) The maximum height of the projectile occurs when
a(1 cos t), so y 2 Volume [a(1
0 2 cos t)] a(1 cos t) dt 2 3 a (1 3 cos t 2 3 cos t so t
3 cos t) dt 0 3 at 3
t
2
1
sin3
3 3 sin t
sin t 5 3
sin 2t
4 36. (a)
(b) 2 t
0 a (b) 33. (a) QP has length t, so P can be obtained by starting at Q
and moving t sin t units right and t cos t units
downward.
(If either quantity is negative, the corresponding
direction is reversed.) Since Q (cos t, sin t), the
coordinates of P are
x cos t t sin t and y sin t t cos t. 75
75
sin 20 , y
sin 20
16
16 5625
64 87.891 ft 840.421 ft
16,875
64 263.672 ft 38. (a) It is not necessary to use NINT.
75/8 Length 5625
16 (150 0 5625
8 (b) t
t
t t cos t
P (x , y ) (1, 0) x 16t 2 75/8
0 2y
a dx 2
dt and y with f (x).
Then 150t 351.5625 39. In the integral t sin t 32t) dt 703.125 ft b Q 41.125 ft 641.236 ft 37. (a) 23 y 0, dx
dx
becomes
dt
dx 1. dy 2
dt, replace t with x
dt 405 Section 10.2 40. dy
dx 3 e x, so Area 2 ex 1 (e x)2 dx which using 0 NINT evaluates to 1273.371. 5. (a) 2u
3v
2u
(b) dy
41.
dx 4 1
, so Area
x2 1
x 2
1 using NINT, evaluates to 12
dx, which,
x2 1 6. (a) 9.417.
(b) dy
42.
dx (ln 2)2
2 x (ln 2)2 2 (2 x Area x 2 x) (ln 2)(2 x 2 ), so
7. (a) 116.687. (pp. 520–529) 2. 1)2 3
5 2
1 3. Solve 2)2 (3 8. (a) 17 1
4
3
5 b
3 4: b 0
0
1
1 4
4
( 2), (5)
5
5
4
v
5 9. 2
2 4 32 14. cos 52 2(3)(5) cos (30 ) 34 10. 242 15 272 c 3 192 34 15 1, 2(27)(19) cos , so 272 192
257
and
2(27)(19)
513
257
cos 1
1.046 radians or 59.935 .
513 242 Section 10.2 Exercises
(b)
2. (a)
(b) 92 2( 2),
42 3. (a) 3
(b) ( 6)2 ( 2),
12 4. (a) 3 ( 2),
52 2 3 4, 2(5) 32 6
117 ( 10) 2 10 116
5 13
2 1, 3 10
2 ( 7)2 5 5,
74 12
v
13 15 10
,
13 13 5
( 2)
13 ( 4)
2 15
13 3
0, 1 3 0 2 3
1, 1
3 1, ( 1), 1 2
2
, sin
3
3
3
, sin
4 1, 1 ( 1) ( 1), 2
1 60
13 4 2, 1, 0 CD
E 24 10
,
13 13
6421
13 1, 3 2 24 60
,
13 13 70 2
13 1 1 14
,
55 4 ( 1) 1
0, 0 1
3
,
22
3
4 1 ,
2 1
2 15. This is the unit vector which makes an angle of 2.832 1. (a) 3(3), 3( 2) = 9, 3, so 6
5 197
5 5
(3),
13 ( 3)2 2 13. cos 8
,
5 12
12
( 2), (5)
13
13 12. E
AB 45 6
5
8
,4
5 14 2
5 5
u
13 2, 0 AB
E 9
,
5 9
5 12
5 11. 0 6 (b) 16, 29 1097 3
3
(3), ( 2)
5
5 CD
E 8. (a) (b) (b) 10. 30 (b) cos 0
and a 4.
a
b
and b 6.
8 2
3 7. (a) 9. c2 3
5
2
6 120 (b) 29 3
u
5 5
u
13 11. 3
1
5
5. Slope of AB = Slope of CD, so
3 19 505 2 ( 16) 12
v
13 4. Slope of AB = Slope of CD, so 6. (a) ( 19) 2 (b) Quick Review 10.2
(5 12 12, 2u
2(3), 2( 2)
6, 4
5v
5( 2), 5(5)
10, 25
2u 5v
6 ( 10), 4 25 3
u
5 s Section 10.2 Vectors in the Plane 1. 2 4
v
5 2 x)2 dx which 2 using NINT evaluates to 2 x (ln 2)2(2 x 1 2(3), 2( 2)
6, 4
3( 2), 3(5)
6, 15
3v
6 ( 6), 4 15 7 29 120 90 210 with the positive xaxis; cos 210 , sin 210
16. cos 135 , sin 135 3
2
1 ,
2 1
2 ,
1 2 3, 70
13 406 Section 10.2 17. The vector v is horizontal and 1 in. long. The vectors u and
11
w are
in. long. w is vertical and u makes a 45 angle
16
with the horizontal. All vectors must be drawn to scale.
(a) (d)
w v v
u u+v u u+v+w=0 (b) 1
3, 4
5 19. 42 42 ( 3)2 21. ( 15)2 82 22. ( 5)2 u –v 34
,
55 ( 2)2 5; w v (c) 32 20. u+v+w u 1 u–v 5, 1
4,
5 5; 1
17 17; 1 5 2 –w ,y (x )2 (a) 1
,2
17 2
2
1
2,
2
17 u 1
t 1 ,y (b) 5 (y )2 Tangent: u Normal 2 –v 25. x
y 1, 4 sin t, y
5
, and
2 2, and , 17
1 , 17 . 17
1
,y
2 3, x 1, and . 1
,1
52
2 4 , 17
4 2 5 u–v+w 1
,y
2 1, x 1 1; for t u–v (x )2 ; for t 2 Normal: 24. x 29 17
.
2 (y )2 Tangent: –v 2 , t u–w 18. The angles between the vectors is 120 and vector u is
horizontal. They are all 1 in. long. Draw to scale. 1 1 t 2 15 8
,
17 17 15, 8 29 23. x u 3
5 29; 29 (d) 4
,
5 3 1 2 ,
5 1
2 ,
5 2 1 ,
5 5 cos t; for t (x )2 .
5 3 ,x 2 3, 73
.
2 (y )2 w Tangent 2 2 3, 73 (c) 12 , 21 9 0.811, 0.585 , 2u Normal
–v
2u – v 5
2 2 5
,2
73 2 3 0.585, 0.811 . 5
73 , 12
219 5
73 Section 10.2 26. x 3 sin t, y
3 y 3 cos t; for t
(x )2 , and (y )2 4 3 ,x ,
2 3. 2 32. Since u and v are nonzero, we know that u ≠ 0 and v ≠ 0.
Therefore, the dot product u v
u v cos is 0 if and
only if cos
0, which occurs if and only if u and v are
.
orthogonal
2 1
3 3 1
3 3 Tangent:
Normal: 3 ,
2 1 , 2 , 2 3 2 1 1
2 33. ,
2 (r1, r2) 1 ,
2 v .
2
u 27. AB
E 3, 1 , BC
E BA
E
AB
E BA
E AC
E CA
E 3 , and AC
E 1, 1 , CB
E 3, 2, CB
E 2. 1 cos 1 cos 1 v 10, and r1
r2 AB AC
EE
AB AC
EE
3(2) 1( 2)
( u 2, 2 . 2 2.
cos (u1, u2)
u+v (v1, v2) 1, 3 , and CA
E 10, BC
E Angle at A v1
v2 u1 so r1
u2 so r2 u1
u2 v1
v2 34. (a) To find u v, place both vectors with their initial
points at the origin. The vector drawn from the terminal
point of v to the terminal point of u is u v. Or, add u
and v according to the parallelogram law. 10)(2 2) (b) 1 63.435 ,
5 BC E
E BA
cos 1
BC BA
EE Angle at B 1 cos (u1, u2) ( 1)( 3) 1 3
5 10) 1 cos 53.130 , and v 1 cos u –v
(–v1, –v2) 1( 2)
( u–v u CB CA
EE
cos 1
CB CA
EE Angle at C –v u–v ( 3)( 1) ( 10)( cos 3(2) 10)(2 2)
1 r1
r2 63.435 . ( v1)
( v2) u1 → r1
u2 → r2 u1
u2 v1
v2 5 28. AC
E 2, 4 and BD
E AC BD
EE 2(4) 4, 35. (a) Let P 2. 4( 2) 0, so the angle measures 90 . 29. (a) u (v w) u1(v1 w1) u2(v2
(u1v1 u1w1) (u2v2 u2w2)
(u1v1 u2v2) (u1w1 u2w2)
uv uw
(b) (u u2 2 v) (u 31. (u v) ( u1
u1 2 (u12
2
u w2) v) w (u1 v1)w1 (u2 v2)w2
(u1w1 v1w1) (u2w2 v2w2)
(u1w1 u2w2) (v1w1 v2w2)
uw vw
u12 30. u u 407 v1)(u1
v1 2 u22)
v2 ( u12
v1) u22
(v12 u22)2 u2 (u2 v2)(u2 v2) v22
v22) 1
OP
E
2
(a (a, b) and Q
1
OQ
E
2
c) (b d)
,
2
2 (c, d ). Then
1
a, b
2 OM
E (b) OM
E 2
OP
E
3 1
OQ
E
3 (c) OM
E 1
OP
E
3 2
OQ
E
3 1
c, d
2 ( r 1, r2) 408 Section 10.2 35. continued 40. The indicated diagonal is (u (d) M is a fraction of the way from P to Q. Let d be this d OQ
E Proof: PM
E
so PQ
E d)OP.
E (1 dPQ and MQ
E
E 1
PM and PQ
E
E
d
1
d 1 Therefore, PM
E
But PM
E
1
OM
E
d OM
E 1 d 1
1 d d)PQ,
E (1
1 1 d MQ.
E MQ.
E OQ
E OQ
E
1
1 d OM, so
E OM.
E Therefore,
1
OM
E
d ⇒ OM
E 1
1 d
1 d(1 ⇒ OM
E 1
OP
E
d OM
E
d) 1
OP
E
d d) OP
E (1 (u v) v
uv
v2
.
u vv
u vv
If u and v are the same length then these two quantities are
equal, and the diagonal makes the same angle with both
sides.
41. The slopes are the same. OP and MQ
E
E 1
OP
E
d between the diagonal and u is
(u v) u
u2 v u
.
u vu
u vu
But the cosine of the angle between the diagonal and v is fraction. Then
OM
E 1
OQ.
E
1d
1
OQ
E
1d 42. v 0 since any other vector has positive magnitude. 43. 25 west of north is 90
800 cos 115 , sin 115 25 44. 10 east of south is 270
600 cos 280 , sin 280 10 280 “north” of east.
104.189, 590.885 v
(u u v and CB
E 325 cos 70 , sin 70 111.157, 305.400 . Wind velocity is 130 north of east: dOQ.
E
u v. Since u , these vectors are orthogonal, as
v) (u v) v2 u2 115 north of east.
338.095, 725.046 45. Initial velocity is 70 north of east: 40 cos 130 , sin 130
36. CA
E v). The cosine of the angle 0. 37. Two adjacent sides of the rhombus can be given by two
vectors of the same length, u and v.
Then the diagonals of the rhombus are (u v) and (u v).
These two vectors are orthogonal since u
v so
(u v) (u v)
u2
v 2 0.
38. Two adjacent sides of a rectangle can be given by two vectors u and v. The diagonals are then
(u v) and (u v). These two vectors will be orthogonal
if and only if u and v are the same length, since
(u v) (u v)
u2
v 2.
39. Let two adjacent sides of the parallelogram be given by two
vectors u and v. The diagonals are then (u v) and
(u v). So the lengths of the diagonals satisfy
u v 2 (u v) (u v)
u 2 2u v
v2
2
and u v
(u v) (u v)
u 2 2u v
v 2.
The two lengths will be the same if and only if
u v 0, which means that u and v are perpendicular and
the parallelogram is a rectangle. 25.712, 30.642 . Add the two vectors to get 85.445, 336.042 . The speed is the magnitude, 346.735 mph. 336.042
The direction is tan 1
85.445 75.734 north of east, or 14.266 east of north.
46. w cos (33 15 ) 2.5 lb, so w 2.5 lb
.
cos 18 2.5 lb
cos 33 , sin 33
cos 18 Then w 2.205, 1.432 . 47. Juana’s pull 23 cos 18 , sin 18
21.874, 7.107 ;
Diego’s pull 18 cos ( 15 ), sin ( 15 )
17.387, 4.659 . Add to get the combined pull of the
children:
39.261, 2.449 . The puppy pulls with an
opposite force of the same magnitude:
39.2612 2.4492 39.337 lb.
48. (a) 7 cos 45 , sin 45 has its terminal point at
(4.950, 4.950).
(b) 7 cos 45 , sin 45
terminal point at
49. AB
E 3 0, 4 50. AB
E 2 ( 4), 2,
3 5
1, 4 8 cos 210 , sin 210 has its
( 1.978, 0.950). 0 3, 4
2 1 3
CD
E 1 4, 5 1 CD
E Section 10.3
51. u u1, u2 , v (iii)
(iv)
(v)
(vi)
(vii)
(viii) (ix) 3. dy/dt
dx/dt 5 cos t
, which for t
4 sin t dy
dx dy/dt
dx/dt 5 cos t
, which for t
4 sin t v 52. Write the two vectors as a 1, 1 and b 1,
sum is a b, a 7
,b
2 a dy
dx 4. w1, w2 u1 v1, u2 v2
v1 u1, v2 u2
vu
(u v) w
u1 v1, u2 v2
w1, w2
(u1 v1) w1, (u2 v2) w2
u1 (v1 w1), u2 (v2 w2)
u (v w)
u0
u1, u2
0, 0
u1 0, u2 0
u1, u2
u
u ( u)
u1 , u 2
u1, u2
u1 u1, u2 u2
0, 0
0
0u 0 u1, u2
0u1, 0u2
0, 0
0
1u 1 u1, u2
1u1, 1u2
u1, u2
u
a(bu) a(b u1, u2 ) a bu1, bu2
abu1, abu2
ab u1, u2
(ab)u
a(u v) a u1 v1, u2 v2
au1 av1, au2 av2
au1, au2
av1, av2
a u1, u2
a v1, v2
au av
(a b)u (a b) u1, u2
(a b)u1, (a b)u2
au1 bu1, au2 bu2
au1, au2
bu1, bu2
au bu (i) u
(ii) v1, v2 , w b , so solve a 1
. So 3, 4
2 b 77
,
22 b 5. dy
dx dy/dt
dx/dt 1
1 y 1 1, so y Also, at t 6 ,x 2 53
x
4 y
6. dy
dx dy/dt
dx/dt
6 7. lim 3 ,x 2 2
4 x 2
x→2 x (x (b) Slope
y1 m(x x 1, so y y1 m(x
x 2 or y x 3. 54. The slopes of the lines are 4 3 x1) becomes cos cos 7 2
10 52 5
. The equation for the
2 9
.
10 (x 2 3), or x2
2)(x 2) lim 2x, and Length 1 using NINT evaluates to t cos t sin t, and y Length t2 ( t sin t 10. y
2
y xe x e x C (use integration by parts), so
0 e0 C and C 3:
xe x e x 3 Section 10.3 Exercises
( 1)]i
3)i 3. E
AB
CD
E [0
(0 (1 [0 4)j 6i ( 4)] j 3j
3i 4j ( 3)]i (2 0)j
4)i ( 3 0)j 3i
4i (a) [3 ( 4)]i [2 ( 3)] j ( 4)]i [2 ( 3)] j 2j and
3j.
i 7i 1 f (x) . Then y 1, f (x)
3 3 y 1 (x 4 x
4 1 x2 1) or . , so for x . Then y
3 1, f (x)
3 3(x 3 and
1) or y 4] j 8i 4]j 2i 3x. j
5j 2j [( 2) 6j (c) 3(5)i 3 2. f (x) [( 2) 3)i (d) [2(5)i 3 x
3 f (x) 1 3 and 3)i (b) (5
, so for x cos t)2 dt 2.958. 4. (a) (5 Quick Review 10.3
x2 cos t, and 1 dt, (b) [3 x 2x)2 dx, which (3 t sin t sin t)2 (t cos t
0
2 2. (0 8.130 . (pp. 529–539) 4 2 3.400. which using NINT evaluates to s Section 10.3 Vectorvalued Functions 1. f (x) 1
4 1 lim x→2 x 0 x1) becomes 1 53
.
4 equals 0 9. x 3
and 1, which means that
4 31 6 53 x→2 (x 1. [5
41 3), or 4 to get vectors 4, 3 and 1, 1 are parallel to the respective lines.
1 2 2 8. y 1. 2) or y (x 3 and y 5
2 x 15 53
.
4 equals 10. normal line is y
y 3
4 5 cos t
, which for t
4 sin t Also, at t 4 5 6 5
. The equation for the
2 3 and y 5
2 tangent line is y 11
,.
22 y1 is undefined: 5 cos t
, which for t
4 sin t 2 53. (a) Slope equals zero. the tangent line is vertical. 1 . Then their
3, a 2 409 3( 2)j
2( 2)j] 15i 6j [3(3)i 3(4)j] i 16j 410 Section 10.3 5. (a) 8. (a) [ 6, 6] by [ 4, 4] [ 6, 6] by [ 3, 5] a(t) d
(2 cos t)i
dt d
(3 sin t)j
dt ( 2 sin t)i (b) v(t) (3 cos t)j d
( 2 sin t)i
dt d
(3 cos t)j
dt ( 2 cos t)i
(c) v 2 direction
(d) Velocity 1
2 2 a(t) (3 sin t)j
( 2)2 2, 0 ; speed
2, 0 d
d2
(2 ln (t 1))i
(t )j
dt
dt
2
i (2t)j
t1
d
2
d
2
i
(2t)j
i
dt t 1
dt
(t 1)2 (b) v(t) 02 1, 2 ; speed (c) v(1) 2, 12 22 1 1 2 direction 1, 0 1, 2
5 1, 0 (d) Velocity 1 5 , (2t 5, 5 2 5 (cos t)i ,
5 6. (a)
9. v(t) 2j 5 sin t)j, r(0) j and v(0) i. So the slope is zero (the velocity vector is horizontal).
(a) The horizontal line through (0, [ 4.5, 4.5] by [ 3, 3] a(t) d
(cos 2t)i
dt (2 cos t)j d
( 2 sin 2t)i
dt ( 4 cos t)i
(c) v(0) (b) The vertical line through (0, d
(2 sin t)j
dt ( 2 sin 2t)i (b) v(t) d
(2 cos t)j
dt (d) Velocity r (2 sin t)j v 02 0, 2 ; speed direction 10. v(t) 1
0, 2
2 22 2, ( 2 sin t)i
(2 4 ( 3 3)i
3 3
3
x
2 y 2
[x
3 1 2
x
3 y (sec t tan2 t sec3 t)i 22
42
3
3
1
2 24
, ; speed
33
3 24
,
direction
2 533 (c) v 6 (d) Velocity 25
3 1 ,
5 , 5
2
5 5 25
,
3 6t) dt i 2 3)] or 3 t dt j 1 2 2 3t 2 i 6t 2t 3/2 j 1 3i 1 2 (4 2)j /4 12. ( 2 (6
1 (2 sec2 t tan t)j 3)] or 5 2 18
6 2 (sec2 t)j (2 7 2 d
d
(sec t)i
(tan t)i (sec t tan t)i
dt
dt
d
d
(sec t tan t)i
(sec2 t)j
dt
dt 3/ 2
2 3
[x
2 62
2 3 (b) y a(t) j. So the slope is 1 0, 1 7. (a) (b) v(t) 1 j and 2 2 11. 0. (3 cos t)j, 2)i (a) y [ 6, 6] by [ 4, 4] 1): x 1. 2 4 2 0, 1 1): y /4 sin t dt i (1 /4 cos t) dt j /4 /4 /4 cos t i
/4 2 2 j t sin t j
/4 3
.
2 411 Section 10.3 sec t tan t dt i (sec t)i 1
5 (ln t )i t
4
t4
4 r(t) ( 32t)j
C2 r(t) 18. C t
j
2 100)i ( 2 cos t)i ( sin t)j. So v C, and r(0)
t2
2 ti tj i a j r(0) C2 dr
dt t 0 C i j, so ( 16t 2)j
C1 8j)t C1t C2 . 100i 2 8i 8j. So 2)i j.
2 5
,
2 a 3
, and
2
va
cos 1
va 1 cos 3
5 53.130 . 24. v(t) 3i (2t)j, and a(t) 2j. So v(0) 3i, and
a(0) 2j. These are perpendicular, i.e., the angle between
them measures 90 .
25. (a) Both components are continuous at t 8t)j. is 3i t2
j
2 3
32 2 9
j
3(3) 10i C1 C1t C2 10j, and 2i sin t cos t ( sin t)i. ( 3 sin t)i 0 or 0, i.e., t 3
3 ( 3 cos t)i 2i. (c) Discontinuous when t
(
1] {0}. dx
dt (sin t)i (1 sin t, and Distance 1 2 0 or cos t 0, i.e., for k
, k any nonnegative integer.
2 1 0, i.e.,
0, i.e., cos t)2 dt (1 2 cos t dt
t
dt
2 2 sin
0, is 1 cos t (sin t)2 0
2 /3 (16 sin t cos t) 0 or t 0
2 /3 ( 4 sin t)j. 0: (9 sin t cos t) 0 and t cos t)j, i.e., dy
dt 2 /3 (4 cos t)j, and a(t) 0j (b) Continuous so long as t
( 1, 0) (0, ). 27. v(t) 0, which is true for 21. v(t) true when sin t 0, sin 2t
i lim(ln (t 1))j
t
t→0
2 cos 2t
i lim(ln (t 1))j
lim
1
t→0
t→0 k
, k any nonnegative integer.
2 Solve v a 3t 0, i.e., t t→0 j, and a(t)
0: 3t lim t
t2
i
j (10i 10j)
2
2
t2
t2
10 i
10 j
2
2 Solve v a 2 26. (a) Use L’Hôpital’s Rule for the icomponent: 0, so (cos t)i 3, so the limit 3i. (c) Discontinuous when t 19. v(t) (1 cos t)i (sin t)j and a(t) (sin t)i (cos t)j.
Solve v a 0: (sin t sin t cos t) (sin t cos t) 0
implies sin t 0, which is true for t 0, , or 2 . t j, and
2
1 va 2 t 1 2)i ( 4 Then v C1, and t2
i
2 r(t) 20. v(t) ( 4 1 j. dr
dt t 0 ( 16t 0, (cos t)j, and 2 (8i ( 25sin t cos t) ( 2 sin t)i (b) Continuous so long as t 2 dr
dt r(t) 0: (25 sin t cos t) a(t) C2)j e j C, and
0, so C
(i j)
1)i (e t 1)j 100i, and (8t t 1i ( 16t2)j ( 5 sin t)j. which is true for all values of t. C1 and r(t) r(0) ( 5 cos t)i t 2t 2 i
2t 2 (5 cos t)j, and 23. v(t) t )j (t 1) i
ijC
((t 1)3/2 16. r(t) ( 5 sin t)i Solve v a C ( ln 5 (ln 5 4 C2)j dt j 3/2 15. r(t)
r(0)
r(t) dr
dt t C1)i (ln t 17. (ln sec t (ln sec t )j 1
dt i
t 14. tan t dt j C1)i (sec t 22. v(t)
a(t) 13. 0 4 cos t
2 2 /3 2
0 412 Section 10.3
10
e
4 28. (a) r(0) 1
i
4 4t (b) v(t)
dy
dt (e 18
e
4 (4 v(0)
e 2 2t 2 4t (e Length
0
2 1) 4t 4t 6 0 i C1 and C2. r(0) C2 i 2j, 1) (1
2 2)j
2)2 (1 j 10 3 10
i
5
32
t
So r(t)
2 (2e ) dt 1) dt (e 4 t 1 4t
e
4
e8 7
4 32. (a) 1) dt (cos t)i (b) v(t)
at t So v(t) 0 (b) 746.989 d 2y
dx 2 3
; sin 2t 0 at t
2
3
0 at t
,.
22 (2 0, 2 ,, 0. cos t 12
6t
2 t2
2t 2 1 2 j. 2, 6). 12, t 2. dy
dx 6t
, which for t
2/t 2 2 equals 3
(40)(40
64 (b) 0 3
, and 2 .
2 4
2t (b) Horizontal tangents: t
4 0 for t
2.
Vertical tangents: 2t 2 2t 0 for t 0, 1.
Plugging the tvalues into x 2t 3 3t 2 and
y t 3 12t produces the x and ycoordinates of the
critical points.
t
2: horizontal tangent at ( 28, 16)
t 0: vertical tangent at (0, 0)
t 1: vertical tangent at ( 1, 11)
t 2: horizontal tangent at (4, 16) 160) 0 and t 160: 225 m
15
, which for t
2 3
3
t(t 160)
t
64
32
15
equals
m per second.
4
3
t
32 6. 24. d
dt (d) v(t) 1 equals 2/t 2)6 (4/t 3)6t
,
(1 2/t 2)3 (1 dy /dt
dx/dt 33. (a) The jcomponent is zero at t
160 seconds. (c) x sin t, y cos 2t. Relate the two using the identity
cos 2u 1 2 sin2 u: y 1 2x 2, where as t ranges
over all possible values, 1 x 1. When t increases
from 0 to 2 , the particle starts at (0, 1), goes to
(1, 1), then goes to ( 1, 1), and then goes to (0, 1),
tracing the curve twice.
3t 2
6t 2 10
t
5 2. That corresponds to , 3( 2)2 dy/dt
dx/dt y which for t (c)
0 and sin 2t dy
dx (c) When y (2 sin 2t)j and dy/dt
dx/dt 0 when t
2 2 12
t
2 1i 2 0 when both cos t
2 2
t2 1 point 2 1 4t
e
t (e4t 1)2 (2e2t)2 dt
4
2
1 4t
e
t (e4t 1) dt
2
04
2
1 8t
1 4t
e
e
te4t t dt
2
4
04
122
1 8t
1 4t
1
e
e
(4t 1)e4t
t
2
20
32
16
16
e16 12e8 69
1,737,746.456
16 29. (a) v(t) dx
dt 10
j
5
3 10
t
5 t 2 dy
30. (a)
dx C1 t 1)i (4
2 10 0
2 2 2 C1 1, and 2 (e 0 12
tj
2 tj 2e2t.
2 (c) 32
ti
2 (3t)i with magnitude 2, 2, e4 dx
(2e )j;
dt
2t 1)i j, so v(t) and since v(0) must point directly from (1, 2) toward (4, 1) (e4)j 2i 1
, 1 , terminal
4 Initial 3i r(t) j, 18
e
4 r(2) 31. a(t) (e0)j 0i 15
equals 0 at t
2 40 80 seconds (and is negative after that time).
34. (a) Solve t
(t 3)2 3
3t
2 3t
2 4: t 2 for t (b) First particle: v1(t) i 2. Then check that
2: it does.
2(t 3)j, so v1(2) and the direction unit vector v1 is
Second particle: v2(t) 3
i
2 the direction unit vector is 1 , 2 i 2j . 5
5
3
j, which is constant, and
2
1
1 , 2 . 2 Section 10.3
35. (a) Referring to the figure, look at the circular arc from the
point where t 0 to the point “m”. On one hand, this
arc has length given by r0 , but it also has length given
by vt. Setting those two quantities equal gives the
result.
(b) v(t) v sin vt
i
r0 v cos v2
vt
cos
i
r0
r0 a(t) v2
r0 vt
j, and
r0 c (b) vt
sin
j
r0
r0 So, by Newton’s second law, F fu
41. u
(a) 2 r0 2
T
r02 2 d
(u
dt T (b) d
(u
dt , c dt du
dt fu
v1, v2
v1, u2 423
r
GM 0 u1 v2 ) d
(u
dt 2 v1),
v 1 , u2 v1, u2 v 1 , u2 u1 , u 2 du
dt dv
dt v2 ) d
(u
dt 2 v1), v2) v2 v1 , v2 d
(u
dt 1
d
(u
dt 1 v) GM
4 2r03 T dt f u1, f u2 u1 , u 2 GM
r0 2 f u2 u1 GM
r0 2 du1 du2 dt d
(u
dt 1
d
(u
dt 1 v) GM
and solve for T 2:
r0 for v in v 2 1
T2 T u 1 , u2 , v 2 and solve for vT.
2 r0 du2 fu1 , fu2 v2
m
r.
r0 result. (e) Substitute ,c c f u1, fu2 dt d
fu , fu
dt 1 2 r(t). Substituting for F in the law of gravitation gives the 4 d
( fu)
dt du1 fu1 v2 (c) From part (b) above, a(t) vT
(d) Set
r0 d
cu (t), cu2(t)
dt 1
d
d
(cu (t)), (cu2(t))
dt 1
dt v2
vt
sin
j
r0
r0 vt
cos
i
r0 u1(t), u2(t) . 40. (a) Suppose u
d
(cu)
dt 413 v2) v2 v1 , v2 x 36. Solve both equations for t: t e
1 and t
y 1.
Now eliminate the t and solve for y:
ex 1
y 1, y (e x 1)2 1, or y e 2x 2e x,
where t 0 so x 0.
37. (a) Apply Corollary 3 to each component separately. If the
components all differ by scalar constants, the
difference vector is a constant vector.
(b) Follows immediately from (a) since any two
antiderivatives of r(t) must have identical derivatives,
namely r(t).
d
38. v 2
dt d
(v v)
dt v v+v v 2v v Therefore, v is constant.
39. Let C dC
C1, C2 .
dt dC1 dC2
dt , dt 0, 0 . 0. du
dt 42. dr
dt
dr
dt df
i
dt
dt
ds dv
dt dg
j
dt
df
i
dt dg
j
dt df
dt dt
i
ds df
i
ds dt
ds
dg
dt dg
j
ds dt
j
ds dr
ds 43. f (t) and g(t) differentiable at c ⇒ f (t) and g(t) continuous at
c ⇒ r (t) f (t)i g(t)j is continuous at c. 414 Section 10.4 44. (a) Let r(t) s Section 10.4 Modeling Projectile Motion x(t), y(t) . b b b kr(t) dt kx(t), ky(t) dt kx(t) dt, a a b a
b a b k x(t) dt, k y(t) dt
a
b k a
b x(t) dt, b y(t) dt a
b k x(t), y(t) dt a a k r(t) dt
a (b) Let r1(t) x1(t), y1(t) and r2(t) b
a (pp. 539–552) ky(t) dt Exploration 1 Hitting a Home Run
1. The graphs of the parametric equations
x (152 cos 20
8.8)t, y 3 (152 sin 20 )t 16t 2
and the fence are shown in the window [0, 450] by
[ 20, 60]. The fence was obtained using the line command
“Line(“. You can zoom in as shown in the second figure to
see that the ball does just clear the fence. x2(t), y2(t) . b (r1(t) r2(t)) dt a ( x1(t), y1(t) x2(t), y2(t) ) dt b x1(t) a x2(t), y1(t) (x1(t) x2(t)) dt, y2(t) dt b
a
b
a
b
a
b b
a (y1(t) b x1(t) dt a
b x1(t) dt, a y2(t)) dt b x2(t) dt, a b y1(t) dt y2(t) dt a b y1(t) dt b x2(t) dt, a a y2(t) dt 2. angle (degrees) b r1(t) dt You can also use algebraic methods to show that t 2.984
when x 400, and that y 15.647 for this value of t. r2(t) dt (c) Let C a 25 30 45 C1, C2 , r(t) b range (ft)
x(t), y(t) . C r(t) dt
a (C1x(t) C2y(t)) dt b C1 588.279 665.629 4.061 4.789 6.745 3. Using the same window of part (1) we can see that the ball
clears the fence. b a 523.707 flight time (sec) a b x(t) dt C2 a y(t) dt
a b C1, C2 b x(t) dt, y(t) dt a a 4. angle (degrees) b C r(t) dt
a 25 30 45 45. (a) Let r(t)
d
dt f (t)i t 559.444 630.424 724.988 4.061 4.789 6.745 g(t)j. Then
d
dt r(q) dq
a range (ft)
flight time (sec) Exploration 2 Hitting a Baseball t [ f (q)i g(q)j] dq a
t d
dt f (q) dq i g(q) dq j a a t d
dt d
dt f (q) dq i
a f (t)i 1. x t g(t)j t g(q) dq j
a y 152
(1 e 0.05t ) cos 20
0.05
152
3
(1 e 0.05t ) sin 20
0.05
32
(1 0.05t e 0.05t )
0.052 r(t). t r(q) dq. Then part (a) shows that S(t) is an (b) Let S(t)
a antiderivative of r(t). Let R(t) be any antiderivative of
r(t). Then according to 37(b), S(t)
Letting t a, we have 0 Therefore, C S(a) C. R(a) b. C. R(t) R(a) and S(t) The result follows by letting t R(t) R(a). [0, 450] by [–20, 60] 2. The ball reaches a maximum height of about 43.07 ft when
t is about 1.56 sec.
3. The range is about 425.47 ft and the flight time is about
3.23 sec. Section 10.4
Quick Review 10.4 Section 10.4 Exercises 1. 50 cos 25 , 50 sin 25 45.315, 21.131 1. Solve vxt (840 cos 60°) t 40, 40 3 2. 80 cos 120 , 80 sin 120 3. To find the xintercepts, solve 2x 2 11x 40 0 using 112 4(2)( 40)
2(2) 11 the quadratic formula: x for v0: v0 5
8. The xintercepts are , 0 and ( 8, 0). For the
2 2(0)2 yintercept is (0, 2v0 sin 4x 11 2 11 11
4 11
. Then the
4 6. At the vertex, g (x) 20 2x 0 and x
vertex is (10, 20(10) 102) (10, 100).
7. y cos x C. y y cos x cos 10. Then the C C 2 8. y t C1 and y
2 y ( 1) ( 1) y ( 1) 1
( 1)3
3 t3
3 16 y y y(0) 16
16 4 (so the ground is t 32 y 2 y(0)
2 k 1
(32)t 2
2 2 seconds
(v0 cos ) t 5, so
5. Use y
2 (v0 sin ) t
22 2t 11 2 6.5
346 16 (32 cos 30°)2 12
gt
2 32 3 55.426 6.5. 0
2.135 seconds (by the quadratic (44 sin 45°) t to obtain x (v0 cos ) t 66.4206. 66.421 feet from the stopboard.
4 6. With the origin at the launch point, use
y
t 12
x
2 k x2
e
2
k
2
1 so k
2 (v0 sin )t
(44 sin 40 ) At t (44 cos 40 )t 66.5193 ft
0.0987 feet farther.
2 1.974, x 12
gt . 6.5 (44 sin 40 )t 16t 2
2
(44 sin 40 )2 416
1.974 sec
32 Thus the shot would have gone C x2 2 32), use t2 t ke 6377.551 m feet away (horizontally). C 20 so k 2y x (32 sin 30°) t
t t x dx e C2 3. 2 when y
12
gt .
2 (v0 sin ) t y 16t t 4e 2y y C2 2g Then x 4, so C1 (500 sin 45°)2
2(9.8) 4020 m 4. With the origin at the launch point 2 10
3 (v0 sin )2 (c) ymax 2, so C2 (tan ) x formula). Substitute that into x
t ke dy
4 2y
1
ln 4
2 10. 3( 1) C1 dt ln 16 y 1 25
3 3t dy
16 y 9. C1 C1t 25
3 C2
y 13
t
3 x2 9.8
50002 + (tan 45°)5000
2(500)2 cos2 45° t
2 g
2v02 cos2 (b) y 2. 2 25,510 m 25.510 km downrange 40 5. To find the xintercepts, solve 20x x 2 0: x 0 or 20.
The xintercepts are (0, 0) and (20, 0). For the yintercept,
find y (0): it is already known to be 0. So the yintercept is
(0, 0). sin 90° 72.154 seconds; 5002
sin 90°
9.8 v0
g sin 2 40. The 0 and x 11 2
4 11
,2
4
441
.
8 11
,
4 40 2(500)sin 45°
9.8 g R 9.8 50 seconds. 490 m/sec. 40). 4. At the vertex, f (x)
vertex is 11(0) v02 v0
sin 2 ; solve 24,500
g 3. (a) t yintercept, find f (0) 21,000 for t: t 2 2. Use R 5
or
2 415 1.18 inches 416 Section 10.4
v02
sin 2 ; solve 10
g 7. (a) Use R
v0 7 2 1 2 sin 2 9.8 v02
sin 2 : sin 2
g
1
sin 0.98
39.261 or
2 15. Use R sin 90° for v0: 9.899 m/sec.
(7 (b) Solve 6 v0 2 2)2
sin 2 for : sin 2
9.8 180° 0.6 36.870 and
1 sin 40 10 2 m
5 106 m/sec
1
positive) is gt 2
2 8. t 0.6 8 10 8 1
(9.8)(8
2 meters or 3.136 12 10 71.565 . sec. Then y (taking down as
10 ) 3.136 14 10 cm. sin 18° 32 ft/sec 2 is 189.556 mph.
80 10 2
3 v02
sin 2 ⇒ 200
g 32 1
sin 1 0.9
2 10 2 3 ymax and c2 31.339, which is well below y x 90 ft/sec, and x 135 ft, (tan ) x evaluates to d
r(t)
dt c1i ( gt c2)j. The initial
v0 cos v0 sin . Integrating again,
c3)i 12
gt
2 (v0 sin )t c4 j. The initial condition on the position gives
x0 and c4 y0. 20. With the origin at the launch point, ymax
29.942 . Substituting g , which is threefourths of the maximum height. ((v0 cos )t r(t)
c3 2 and it occurs when t condition on the velocity gives c1 sin 2 32.079 30°, v0
32
2v02 cos 2 8g 46.597 ft/sec. into the equation for y(t ) gives a height of 19. Integrating, the ceiling height.
11. No. For 2g
v0 sin 0.9. 32.079 . Then 2(32) (v0 sin )2 5
6 73 and 12
gt , and we know the maximum height
2
v0 sin v0(sin )t 2g
3(v0 sin )2 sin 2 ⇒ sin 2 (tan 40°) x. Setting x 6.5 and solving for v0 yields v0 t Taking the smaller of the two possible angles, 80 32
x2
2v02 cos 2 40° 18. y(t) 278.016 ft/sec or 10. R 17. With the origin at the launch point, y v02 (248.8 yd)(3 ft/yd)
v0 (b) To increase the range (and height) by a factor of 2,
increase v0 by a factor of 2 1.41. That is an
increase of 41%. y v02
sin 2
g 9. R 0.98 2 16. (a) Substitute 2v0 for v0 in the formula for range. 18.435 or 82 90 0.98;
1 50.739 . 0.6, so 143.130 and (9.8)(16,000)
4002
sin 2ymaxg v0 2(68)(32)
sin 56.505° v0 sin 68 ft. Then 79.107 ft/sec. feet above the ground, which is not quite high enough.
21. The horizontal distance is 30 yd
32
x2
2(116)2 cos 2 45° 12. Use y
22
x
841 x. Set y (tan 45°) x 84 45, then solve for x using the quadratic formula and taking the larger of the two values:
481
841 1 x 4
841 369.255 ft, which is 0.255 ft y 32
x2
2v02 cos 2 20° y 37 3 (tan 20°) x. Set x 315 and 34, then solve to find cos 20° 315 tan 20° (b) Solve v0(cos 20°) t
t 34 149.307 ft/sec. 149.307(cos 20°) t 315 to find 2.245 seconds. 14. In the formula for range, sin 2 sin 2(90 ). 68
45 56.5° and 1 17 ) = 84
v0 cos 1.924 seconds. Then 15 17 12
gt
2
119 15 1260 v0 4 17 (v0 sin ) t
(16 13. (a) With the origin at the launch point, use tan 84 ft. Then 16 17
(from Exercise 20). So t
sin
68
21
45
84 tan
119
16 17 3.059 inches beyond the pin. y v0 (v0 cos ) t, where 6 ft 17 1
119 2
(32)
2
15 17 67.698 ft. The height above the ground is 6 ft more than that,
73.698, and the height above the rim is about
73.698 70 3.698 feet.
22. The projectile rises straight up and then falls straight down,
returning to the firing point. Section 10.4
23. Angle is 62° (measurements may vary slightly). For
2v0 sin
g flight time t (v0 sin ) 1 sec, ymax 2 (b) v0 12
gt
8 2g
1
(32)(1)2 4 ft (independent of the measured angle).
8
gt
gt
v0
, so speed of engine v0 cos
2 sin
2 tan
32(1)
8.507 ft/sec (changes with the angle).
2 tan (62°) 24. The height of A is given by yA
height of B is given by yB (Rcos , Rsin ) 12
gt and the
2 (v sin )t 12
gt . The second
2 R tan 417 12
gt are equal for any value of t. But
2
R
A moves R units horizontally to B’s line of fall in
v cos R terms in yA and yB time units, and the first terms in yA and yB are also equal at
R
that time: (v sin )
v cos R tan . Therefore, A and B will always be at the same height when A reaches B’s line 2v02 R cos g cos2 tan sin ( cot ( so ) is maximum when ), 90 . The initial velocity vector bisects the angle between the
hill and the vertical for max range. of fall.
26. (a) r(t)
x(t)
y(t) 25. (a) (x(t))i (y(t))j, where
(145 cos 23
14)t and
2.5 (145 sin 23 )t 16t 2.
(v0 sin )2 (b) ymax (145 sin 23°)2
64 2.5 2g 52.655 feet , which is reached at t
v0 145 sin 23°
32 2.5
v0 sin
g 1.771 seconds. (c) For the time, solve
R (Rcos , –Rsin ) y 2.5 (145 sin 23°) t 16t 2 0 for t, using the quadratic formula:
x (v0 cos )t y (v0 sin )t x R cos ⇒t
⇒
⇒R t 1
2 x (v0 cos )t R cos
. Then y
R sin
v0 cos
(v0 sin ) R cos
g R 2 cos2
R sin
v0 cos
2 v02 cos2
2v02 Let f ( ) cos cos sin ( sin ( cos f() 0 ⇒ tan tan ( cot ( ⇒ 90 ) ). velocity bisects angle AOR. 14)(3.585)
2.5 428.262 feet.
16t 2 (145 sin 23°) t t 145 sin 23° (145 sin 23°)2
32 1120 0.342 and 3.199 seconds. At those times the ball is about
sin
) sin ( ) 1 x (0.342) (145 cos 23° and x (3.199) 14)(0.342) (145 cos 23° 40.847 feet 14)(3.199) 382.208 feet from home plate. . Note that f ( ) maximum when (145 cos 23° 20 for t, using the quadratic formula: ) ⇒ tan cos ( 3.585 sec. 3.585 is about (d) For the time, solve y ). f() 160 32 Then the range at t gt 2 ⇒ R cos g cos2 (145 sin 23°)2 145 sin 23° 90 0, so R is
. Thus the initial (e) Yes. According to part (d), the ball is still 20 feet above
the ground when it is 382 feet from home plate. 418 Section 10.4 27. (a) (Assuming that “x” is zero at the point of impact.)
r(t) (x(t))i (y(t))j, where
x(t) (35 cos 27 )t and
y(t) 4 (35 sin 27 )t 16t 2.
(b) ymax (v0 sin )2 4 2g which is reached at t (d) Graph Y2 30 and use the intersect function to find
that y 30 for t 0.689 and 2.305 seconds, at which
times the ball is about x (0.689) 94.513 feet and
x (2.305) 287.628 feet from home plate. (35 sin 27°)2
4 7.945 feet,
64
v0 sin
35 sin 27°
=
g
32 0.497 seconds. 29. (a) r(t) (c) For the time, solve y 4 16t 2 (35 sin 27°) t 0 ( 35 sin 27°)2
32 35 sin 27° (x(t))i 256 1.201 seconds.
Then the range is about x (1.201) (35 cos 27°)(1.201) 37.406 feet. (y(t))j, where 1
(1 e 0.08t)(152 cos 20
17.6) and
0.08
152
3
(1 e 0.08t)(sin 20 )
0.08
32
(1 0.08t e 0.08t)
0.082 x(t)
y(t) for t, using the quadratic formula:
t (e) Yes, the batter has hit a home run since a graph in
parametric mode shows that the ball is more than
14 feet above the ground when it passes over the fence. (b) Solve graphically: enter y (t) for Y1 (where X stands in
for t), then use the maximum function to find that at
t 1.527 seconds the ball reaches a maximum height
of about 41.893 feet.
(c) Use the zero function to find that y (d) For the time, solve y 4 16t 2 (35 sin 27°) t 7 for t, using the quadratic formula:
( 35 sin 27°)2
32 35 sin 27° t 192 has traveled for 0 when the ball 3.181 seconds. The range is about x (3.181)
0.254 and 0.740 seconds. At those times the ball is about
x (0.254) (35 cos 27°)(0.254)
(35 cos 27°)(0.740) 23.064 feet from the 0.08(3.181) e )(152 cos 20° 17.6) 351.734 feet 7.906 feet and x (0.740) 1
(1
0.08 impact point, or about 37.460
and 37.460 23.064 7.906 29.554 feet 14.396 feet from the landing (d) Graph Y2
that y 35 and use the interesect function to find 35 for t times the ball is about
x (2.190) spot.
(e) Yes. It changes things because the ball won’t clear the
net (ymax 7.945 ft). 0.877 and 2.190 seconds, at which
x(0.877) 106.028 feet and 251.530 feet from home plate. (e) No; the range is less than 380 feet. To find the wind
needed for a home run, first use the method of part (d) 28. (a) r(t)
x(t)
y(t) (x(t))i (y(t))j, where to find that y 152
(1 e 0.12t)(cos 20 ) and
0.12
152
(1 e 0.12t)(sin 20 )
3
0.12
32
(1 0.12t e 0.12t)
0.122 (b) Solve graphically: enter y (t) for Y1 (where X stands in
for t), then use the maximum function to find that at
t 1.484 seconds the ball reaches a maximum height
of about 40.435 feet.
(c) Use the zero function to find that y
has traveled for
x (3.126) 3.126 seconds. The range is about 152
(1
0.12 372.323 feet. 0 when the ball e 0.12(3.126) )(cos 20°) 20 at t 0.376 and 2.716 seconds. Then define
x (w) 1
(1
0.08 and solve x (w) e 0.08(2.716) )(152 cos 20° 380 to find w w), 12.846 ft/sec. This is the speed of a wind gust needed in the direction of the
hit for the ball to clear the fence for a home run. Section 10.5 419
30. (a) To save time, enter one expression for x (t) in the main
window, then use the ENTRY function to repeat it six
times in the parametric Y menu. Do the same for
y (t), then make appropriate modifications. 32. dr
dt
d 2r
dt (v0e kt cos )i
kt ( kv0e
gj k kt (v0e cos )i g kt
e
k sin ( kv0e kt sin g
)j
k
kt ge )j dr
dt The initial conditions are also satisfied, since
r(0) v0
k e0)(cos )i (1 v0 [0, 500] by [0, 50] k Now replace T with X in all the y (t) expressions in the
parametric Y menu, then change to function mode
and enter Y1T for Y1, Y2T for Y2, and so on. Use the
functions in the CALC menu to fill in the tables for
parts (b) and (c).
(b) drag coeff time at max ht 0.01
0.02
0.10
0.15
0.20
0.25 t
t
t
t
t
t 3.289
3.273
3.153
3.088
3.028
2.974 462.152
452.478
386.274
352.983
324.410
299.661 lim 1 − e t, and kt e
k2 kt R
,y
. So,
2 max 31. The points in question are (x, y)
x v02 sin cos , and y g
v0 v02 sin cos g2
v0 4
g2
v0 4
g2 2g 2 4 (v0 sin )2 sin4
4 sin2
4 cos sin2 cos2 4 sin2 cos2 (sin2 )(1 v04
4g2 4g
12
4 sin 2 v02 2 2g sin2
4
2 2 v0 4 1
g2 4 . Then 4g
g v0 4 (v0 sin )2 22 4y x2 (v0e0 cos )i
(v0 cos )i v0e0 sin g0
e
k g
j
k (v0 sin )j cos2 ) , so the point (x, y) lies on the ellipse. 1
16 sin2 s Section 10.5 Polar Coordinates and
Polar Graphs (pp. 552–559)
Exploration 1 Investigating Polar Graphs 1. The graph is drawn in the decimal window [ 4.7, 4.7] by
[ 3.1, 3.1] with 2 ≤ ≤ 2 2 . . )
4. If (r, ) is a solution of r 2 4 cos , then so is (r,
because cos ( ) cos . Thus, the graph is symmetric
about the xaxis.
If (r, ) is a solution of r 2 4 cos , then so is ( r,
)
because ( r)2 r 2 and cos ( ) cos . Thus, the graph
is symmetric about the yaxis.
The graph is symmetric about the origin because it is
symmetric about both the x and yaxes. You can also give a
direct proof by showing that ( r, ) lies on the graph if
(r, ) does. t2
.
2
k→0
As k → 0, the air resistance approaches 0. lim 1 e0) j 3. kt k dr
dt t 0 0 0, 2. r1 and r2 are 0 for (d) This follows from the following two limits
(as k → 0):
k→0 and 0j g
(1
k2 max ht k 0.01
t 1.612
44.777
k 0.02
t 1.599
44.336
k 0.10
t 1.505
41.149
k 0.15
t 1.454
39.419
k 0.20
t 1.407
37.854
k 0.25
t 1.363
36.431
(c) After flight times using the zero function, plug the
xintercepts into T and read the ranges out as X1T, X2T,
etc.
drag coeff
flight time
range
k
k
k
k
k
k 0i e0)(sin ) (1 1
4 420 Section 10.5 Exploration 2 Graphing Rose Curves 4. (a) No; y is a function of x and is not the zero function. All graphs are drawn in the window [ 4.7, 4.7] by [ 3.1, 3.1]. (b) No;
y( x) 1. The graphs are rose curves with 4 petals when n
2,
8 petals when n
4, and 12 petals when n
6. ( x)3 x3 ( x) (c) Yes; y( x) (x 3 x x) y(x) y(x) (see part (b)) 5. (a) No; y is a function of x and is not the zero function.
( x)2 (b) No; y( x)
(c) No; y( x) ( x) x2 x y(x) y(x) (see part (b)) 6. (a) No; y is a function of x and is not the zero function.
n 2 (b) Yes; y( x) cos ( x) (c) No; y( x) cos x y(x) y(x) (see part (b)) 7. (a) Yes; Substitute y for y in the equation to get the
original equation.
(b) Yes; Substitute x for x in the equation to get the
original equation.
n 4 n (c) Yes; since the curve is symmetric with respect to both
the xaxis and yaxis, it is symmetric with respect to the
origin. (Also, substitute x for x and y for y in the
equation to get the original equation.)
8. Solve for y : y (x 2)1/2 or (x 2)1/2.
Enter the first expression for Y1, the second for Y2. 6 3. The graph is a rose curve with 2 n petals.
4. The graphs are rose curves with 3 petals when n
5 petals when n
5, and 7 petals when n
7. x 2 1/2 4 9. Solve for y : y 2. 2 3 or x 2 1/2 4
3 . Enter the first expression for Y1, the second for Y2. 3, 10. (x 2 4x) ( y 2 6y 9) 0
(x 2 4x 4) ( y 2 6y 9)
(x 2)2 (x 3)2 22.
Center (2, 3), Radius 2. 4 Section 10.5 Exercises
n For Exercises 1 and 2, two pairs of polar coordinates label
the same point if the rcoordinates are the same and the
coordinates differ by an even multiple of , or if the
rcoordinates are opposites and the coordinates differ by
an odd multiple of . 3 n 1. (a) and (e) are the same.
(b) and (g) are the same.
(c) and (h) are the same.
(d) and (f) are the same. 5 2. (a) and (f) are the same.
(b) and (h) are the same.
(c) and (g) are the same.
(d) and (e) are the same.
n 7 3. 5.
6. The graph is a rose cuve with n petals. Quick Review 10.5
1 1. Slope
so y 3 4
( 2) 4 1[x 2. (x 0)2 3. [x ( 2)]2 (y 0)2
(y [ 3, 3] by [ 2, 2] 1,
( 2)] or y
32, or x 2
4)2 x
y2 (a) 2.
9. 22, or (x 2)2 2 cos 4 , 2 sin (b) (1 cos 0, 1 sin 0)
(y 4)2 4. (c)
(d) 0 cos 2 , 0 sin 2 cos 4 , 2 4 = (1, 1) (1, 0)
= (0, 0)
2 sin 4 ( 1, 1) 421 Section 10.5
6. 4. [ 9, 9] by [ 6, 6] (a) 3 cos 5
,
6 3 sin
1 (b) 5 cos tan
(c) ( 1 cos 7 ,
(d) 2 3 cos [ 9, 9] by [ 6, 6] 4
3 5
6 33
,
2 , 5 sin tan 1 sin 7 ) 2
,2
3 3 sin 1 3
2 (a) r
with (3, 4) 2, ( ( 1)2 2, tan 1 1
3 4
3 (1, 0)
2
3 3)2 ( 3, 3) in quadrant III. The coordinates are 2,
6 3
7
.
6 also works, since r has the opposite sign and differs by . 5. 4
with in quadrant I.
3
4
4
The coordinates are 5, tan 1 .
5,
tan 1
3
3 32 (b) r 42 5, tan also works, since r has the opposite sign and differs [ 6, 6] by [ 4, 4] ( 1)2 (a) r 12 quadrant II. The coordinates are
5
4 2, by . 1
1 2, tan 2, 1 with in
2
is undefined with
0
3
on the negative yaxis. The coordinates are 2,
.
2 3
.
4 (c) r also works, since r is the same and 2, differs by 2 .
12 (b) r ( 3)2 2
3 2 ( 2)2 also works, since r is the same and 3 with 1 (d) r
3 22 02 . 3, tan 3
is undefined with
0 the positive yaxis. The coordinates are 3,
also works, since r is the same and
(d) r 0
2 works, since r is the same and 32 ( 1)2 02 1, tan 0
1 2 on . 3, 5
2 differs by 2 . [ 6, 6] by [ 4, 4] 8.
0 with on the negative xaxis. The coordinates are (1, ). ( 1, 0) also
works, since r has the opposite sign and differs 0 with on the positive xaxis. The coordinates are (2, 0). (2, 2 ) also 7.
02 2, tan also works, since r has the opposite sign and differs by .
(c) r 2, tan by 2 . 3 2, tan in quadrant IV. The coordinates are 2,
2, 0 differs by .
[ 6, 6] by [ 4, 4] 9. [ 3, 3] by [ 2, 2] 10. [ 3, 3] by [ 2, 2] differs by 2 . 422 Section 10.5
25. x 2 y 2
x2 y2
(center [ 9, 9] by [ 6, 6] r 2 and y r sin , so the equation is
4y ⇒ x 2 ( y 2)2 4, a circle
(0, 2), radius 2). 26. r 11. 5
2 cos sin r sin 12. x 2r cos r cos and y a line (slope
27. r 2 sin 2
[ 6, 6] by [ 4, 4] r sin , so the equation is y 2, yintercept 2r sin cos 2
1 x r sin , so the equation is xy r cos or, y
[ 3, 3] by [ 2, 2] 28. r y and y 1 1
, a hyperbola.
x cot r sin [ 3, 3] by [ 2, 2] 5, 5). (r sin )(r cos ) 14. 2x 2 2 13. 5 csc
cot r sin and x
y cot , so the equation is y 2 x, a parabola. 15. 29. r csc ercos
r sin
ercos
x r cos and y r sin , so the equation is y
exponential curve. e x, the sin2
30. cos2
2
(r cos )
(r sin )2
x r cos and y r sin , so the equation is x 2
y
x, the union of two lines. [ 3, 3] by [ 2, 2] 16. y 2 or 31. r sin
ln r ln cos
r sin
ln (r cos )
y ln x, the logarithmic curve. [ 3, 3] by [ 2, 2] 17. 1
32. r 2 2r 2 cos sin
r 2 2(r cos )(r sin ) 1
x 2 y 2 2xy 1
(x y)2 1
xy
1, the union of two lines.
[ 1.8, 1.8] by [ 1.2, 1.2] 33. r 2
x2
(x 18. [ 3, 3] by [ 2, 2] 19. y r sin , so the equation is y 0, which is the xaxis. 20. x r cos , so the equation is x 0, which is the yaxis. 21. r 4 csc
r sin
4
y r sin , so the equation is y
22. r
3 sec
r cos
3
x r cos , so the equation is x 4, a horizontal line. 3, a vertical line. 23. x r cos and y r sin , so the equation is x
line (slope
1, yintercept 1).
24. x 2 y 2
(center r 2, so the equation is x 2
(0, 0), radius 1). y2 y 1, a circle 1, a 4r cos
y2
4x
2)2 y 2 4, a circle (center 34. r 8 sin
r 2 8r sin
x 2 y 2 8y
x2 ( y 4)2 16, a circle (center 35. r 2 cos
2 sin
2r sin
r 2 2r cos
x 2 y 2 2x 2y
(x 1)2 ( y 1)2 2,
a circle (center (1, 1), radius ( 2, 0), radius 2). (0, 4), radius 4). 2). Section 10.5 36. r sin 6 r sin
3
2
3
2 cos 1
x
2 y 3y x cos sin 1
r cos
2 r sin 42. x 2 y 2
r 2 cos2
r 2(cos2 2
6 1
1 2
4, a line slope 1 , yintercept
3 4 [ 4.7, 4.7] by [ 3.1, 3.1] .
3 43. 37. x 7
r cos r 2 sin2
sin2 ) 2 6 2 1 7. The graph is a vertical line. x2
y2
9
4
r 2 cos2
9
2 1
r 2 sin2
4
2 r (4 cos 9 sin 1
2 ) 36 [ 9.4, 9.4] by [ 6.2, 6.2] 38. y 1
r sin
1
The graph is a horiztonal line. [ 4.7, 4.7] by [ 3.1, 3.1] 44. xy 2
(r cos )(r sin ) 2
r 2 cos sin
2
r 2 2 cos sin
4
r 2 sin 2
4
[ 4.7, 4.7] by [ 3.1, 3.1] 39. x y ⇒ r cos r sin More generally, 4 ⇒ tan 1⇒ 4 2k for any integer k. The graph is a slanted line. [ 4.7, 4.7] by [ 3.1, 3.1] 45. y 2 4x
r 2 sin2
r sin2 4r cos
4 cos [ 4.7, 4.7] by [ 3.1, 3.1] 40. x y
r cos 3
r sin [ 4.7, 4.7] by [ 3.1, 3.1] 3 46. x 2 xy y 2 1
(r cos )2 (r cos )(r sin )
r 2(1 cos sin ) 1 [ 9.4, 9.4] by [ 6.2, 6.2] 41. x 2
r2 y2 4
4 or r 2 (or r 2)
[ 3, 3] by [ 2, 2] [ 4.7, 4.7] by [ 3.1, 3.1] (r sin )2 1 423 424 Section 10.5 47. x 2 ( y 2)2 4
r 2 cos2
(r sin
2)2 4
r 2 cos2
r 2 sin2
4r sin
44
r 2 4r sin
0
r 4 sin .
The graph is a circle centered at (0, 2) with radius 2. 52. (a) [ 3, 3] by [ 2, 2] (b) Length of interval 2 53. (a) [ 4.7, 4.7] by [ 1.1, 5.1]
[ 3.75, 3.75] by [ 2, 3] 48. (x 3)2 1)2 3)2 (r cos (y (r sin r 2 cos2
r 2 r
r 6r cos 6r cos
6 cos 3 cos 4 (b) Length of interval
1)2 2r sin
2 sin sin 6
(6 cos
2 (3 cos 54. (a) 4 r 2 sin2 9 2 2r sin 1 4 0
2 sin )2 24 [ 1.5, 1.5] by [ 1, 1] (b) Length of interval
sin )2 6 4 55. (a) [ 15, 15] by [ 10, 10] (b) Required interval [ 6, 6] by [ 4, 4] In Exercises 49–58, find the minimum interval by trying
different intervals on a graphing calculator. ( ,) 56. (a) 49. (a)
[ 3, 3] by [ 2, 2] (b) Length of interval
[ 3, 3] by [ 2, 2] (b) Length of interval 2 57. (a)
2 50. (a)
[ 3, 3] by [ 2, 2] (b) Length of interval
58. (a) [ 6, 6] by [ 4, 4] (b) Length of interval 2 51. (a)
[ 3, 3] by [ 1, 3] (b) Length of interval
[ 1.5, 1.5] by [ 1, 1] (b) Length of interval 2 2 59. If (r, ) is a solution, so is ( r, ). Therefore, the curve is
symmetric about the origin. And if (r, ) is a solution, so is
(r,
). Therefore, the curve is symmetric about the xaxis.
And since any curve with xaxis and origin symmetry also
has yaxis symmetry, the curve is symmetric about the
yaxis.
60. If (r, ) is a solution, so is ( r, ). Therefore, the curve is
symmetric about the origin. The curve does not have xaxis
or yaxis symmetry. 425 Section 10.5
61. If (r, ) is a solution, so is (r,
). Therefore, the curve
is symmetric about the yaxis. The curve does not have
xaxis or origin symmetry.
62. If (r, ) is a solution, so is ( r, ). Therefore, the curve is
symmetric about the origin. And if (r, ) is a solution, so is
(r,
). Therefore, the curve is symmetric about the xaxis.
And since any curve with xaxis and origin symmetry also
has yaxis symmetry, the curve is symmetric about the
yaxis. 66. (a) [ 9, 9] by [ 6, 6] The graph of r2 is the graph of r1 rotated by angle
counterclockwise about the origin.
(b) 63. (a) Because r a sec is equivalent to r cos
a, which
is equivalent to the Cartesian equation x a.
(b) r a csc is equivalent to y a. 64. (a) The graph is the same for n 2 and n
2, and in
general, it’s the same for n 2k and
n
2k. The graphs for n 2, 4, and 6 are roses with
4, 8, and 12 “petals” respectively.
The graphs for n
2 and n
6 are shown below.
n
2 [ 9, 9] by [ 6, 6] The graph of r2 is the graph of r1 rotated by angle
clockwise about the origin.
(c) The graph of r2 is the graph of r1 rotated
counterclockwise about the origin by the
angle .
67. d [(r2 cos [ 3, 3] by [ 2, 2] n 6 [r22 cos2
r22
r12 [ 3, 3] by [ 2, 2] x1)2 (x2 r1 cos 2 y2)2 ( y1
1 )2 2r2 r1 cos 2
2 sin 2 r22 2r1r2 cos ( (r2 sin
2 cos 2r2r1 sin
1 2 2 2 1 r12 cos2 1 sin r1 sin 1 r12 )2]1/2 1
2 sin
1/2 1 ) 68. (a) (b) 2
(c) The graph is a rose with 2 n “petals”.
(d) The graphs are roses with 3, 5, and 7 “petals”
respectively. The “center petal” points upward if
n
3, 5, or 7.
The graphs for n 3 and n
3 are shown below. [ 9, 9] by [ 6, 6] The graphs are ellipses.
(b) Graphs for 0 k 1 are ellipses. As k → 0 , the
graph approaches the circle of radius 2 centered at the
origin.
69. (a) [ 3, 3] by [ 2, 2] [ 5, 25] by [ 10, 10] The graphs are hyperbolas.
[ 3, 3] by [ 2, 2] (e)
(f) The graph is a rose with n “petals”.
65. (a) We have x
we have r r cos and y r sin . By taking t
,
f (t), so x f (t) cos t and y f (t) sin t. (b) x 3 cos t, y (c) x (1 3 sin t (d) x (3 sin 2t) cos t, y cos t) cos t, y (1 cos t) sin t (3 sin 2t) sin t (b) Graphs for k 1 are hyperbolas. As k → 1 , the right
branch of the hyperbola goes to infinity and
“disappears”. The left branch approaches the parabola
y 2 4 4x. 426 Section 10.6
Section 10.6 Exercises 70. (a) dy
dx [ 9, 9] by [ 6, 6] The graphs are parabolas. f ( ) sin
f ( ) cos f ( ) cos
f ( ) sin cos
cos 1. (1
(1 (b) As k → 0 , the limit of the graph is the negative xaxis. sin
cos 2 sin cos cos2 s Section 10.6 Calculus of Polar Curves dy
dx (pp. 559–568)
Quick Review 10.6 2. dy
dy/dt
5 cos t
dx
dx/dt
3 sin t
5
2.
cot 2 0.763
3 5
cot t
3 1. 3. Solve cot t 0: t 2 or dy
dx 3
;
2 and the corresponding points are 3 cos
3
3
and 3 cos , 5 sin
2
2 (0, 5
cot t is undefined when t
3 4. dy
dx 2 , 5 sin 2 3. 5. Length
0 9 sin2 t 1, 1
1 1 2 sin 2 sin
2 sin 2 cos f ( ) cos
f ( ) sin dy
dx /2 1
, which is undefined.
0 dy
dx f ( ) sin
f ( ) cos f ( ) cos
f ( ) sin 3 cos
3 cos sin
cos 2 cos
2 sin (2
(2 6 sin
3(cos2 3 sin ) cos
3 sin ) sin
cos
sin2 ) (3, 0) and
dy
dx (2, 0) dy
dx dy
dx (2, ) 25 cos2 t dt, 0 dy
dx 2
,
3
0
1 /2 dy
dx dy
dx (5, 3 /2) 12.763. 2
3 0 dy
dx ( 1, /2) dy 2
dt
dt which using NINT evaluates to dy
dx 1
dy
, which is undefined;
0
dx 0 0, , or 2 ; ( 3, 0). dx 2
dt sin (0, 5) 5) (3 cos 2 , 5 sin 2 ) (3 cos , 5 sin ) 1
1 f ( ) sin
f ( ) cos the corrresponding points are
(3 cos 0, 5 sin 0) cos sin2 0 sin )cos
sin )sin 0, 2
, and
3 dy
dx 0
5 3 /2 0. For questions 6–8, the graph is:
dy
dx f ( ) sin
f ( ) cos f ( ) cos
f ( ) sin 3 sin2
3 sin cos 4. 3 cos (1 cos )
3 sin (1 cos ) 3(cos2
sin2 )
cos
3 sin 3 cos
6 sin [ 2, 4] by [ 2, 2] 6. The upper half of the outer loop
7. The inner loop dy
dx (1.5, /3) 8. The lower half of the outer loop 1
2 1
2
3 3
2 2 9. y 0 for x
6 (6x Area 0 or 6.
x 2) dx 3x 2 0 136
x
3
0 1
2 dy
dx (4.5, 2 /3) 36 3
2 , which is undefined;
1
2
3
2 0; 10. Use a graphing calculator’s intersect function to find that
the curves cross at x 0.270 and x 2.248, then use NINT to find
2.248 Area [2 sin x
0.270 (x 2 2x 1)] dx 2.403. dy
dx (6, )
dy
dx (3, 3 /2) 11
, which is undefined; and
00
0
0 ( 1)
( 1) 1. cos 2 cos
.
cos 2 sin
0
1 0; Section 10.6 427 7. 5. [ 3.8, 3.8] by [ 2.5, 2.5] [ 1.5, 1.5] by [ 1, 1] The graph passes through the pole when r
which occurs when 2 3 cos 0, The polar solutions are 0, 3
. Since the
2 and when only consider x k
. This can be confirmed analytically by
5
dy
noting that the slope of the curve, , equals the slope of
dx
k
the line, tan . So the tangent lines are
0 [ y 0],
5
2
2
y
tan x ,
y
tan
x,
5
5
5
5
3
3
4
4
y
tan
x , and
y
tan
x.
5
5
5
5 (or x 2 curve at 0, 0). y 3 cos2 (3 cos ) cos
(3 cos ) sin
( 3 sin )sin 3(cos2 (3 cos )cos sin2 ) 6 cos ( sin ). 8. dx
,
0, and
At 0,
/2
2
d
dy
dx
3(02 12)
3. So at 0, ,
0
/2
2d
d
dy
dy
0, so
is undefined and the tangent line is
and
dx
d [ 3, 3] by [ 2, 2] The polar solutions are 0, vertical. a given k, the line 6. [ 3, 3] by [ 2, 2] A trace of the graph suggests three tangent lines, one with
positive slope for 6 , a vertical one for 2 5
.
6 with negative slope for , and one Confirm analytically:
dy
d 6 sin 3 sin 2 cos 3 cos dx
d 6 sin 3 cos 2 cos 3 sin . dy
dx
dy
dx 6 , 0, 2 , and 0, are all solutions. dy/d
, and so
dx/d
6(1)(1/2) 2(0)( 3/2)
6(1)( 3/2) 2(0)(1/2)
/6 dy
dx
dy
dx 6( 1)(1)
6( 1)(0) /2 5 /6 1 6(1)(1/2) 2(0)(
3/2)
6(1)(
3/2) 2(0)(1/2) 5
6 y ;
3 2(0)(0)
, which is undefined; and
2(0)(1) lines have equations
and 5
6 and 6
1 x.
3 y 1 1 x,
3 . The tangent
3
2 [x 0], k
2 for k 0, 1, 2, 3, 4, and for k
appears to be tangent to the
2 k
. This can be confirmed analytically by
2
dy
noting that the slope of the curve, , equals the slope of
dx
k
the line, tan . So the tangent lines are
0 [ y 0] and
2
3
[x 0].
,
and
2 are duplicate
2
2 curve at 0, 0, k
appears to be tangent to the
5 . At this point, there appears to be a 2 Confirm analytically: dx
d 0, 1, 2, 3, 4, and for a given k, the line vertical tangent line with equation and for k 1 produces the entire graph, we need interval 0 dy
d k
5 solutions. 428 9. Section 10.6 dy
d cos sin (1 cos (2 sin (1 dy
d 3
(cos
22 , 5
(2 sin
66
1 1 9 0). the line y
the line y 1
sin
2
6 2 , for 7 11
,
, or
6
6
3
,r
2 6 1
2 dy
,
2d dx
d d dy
dd
d dx
dd 4 sin 7
,r
6 sin 1 for cos 2 ,r or cos 3 3
4 2 0 for 0 [the line x 1
,
26 3 y 3
2 y 2,
Vertical at: 0, 2 [x . and
2 0]. 1
,
4 2
0], 37
,
26 x 3 11
,
26 x 5
dx
.
3d 5
,r
3 33
,
4
33
4 1
2 1 or , i.e., when 0 when 0, , 2 24
,
33 0). So there is a horizontal tangent line
3
the line y
2 ,r 3
the line y
2 2 cos 0 2 cos 2 3
sin
2
3 33
and for
4 3
5
sin
2
3 0, r 3 , so by . There is a 2 2], for 2 ,r
2
,r
3 2], for 3
4 2 [again, the
1
2 1
2
1
4
1
cos and for
,r
2
3
4
3
2
1
2
1
again, the line x
cos
.
2
3
4
dy
dx
For
,
0, but
d
d
d dy
4 cos sin
sin
0 for
, and
dd
d dx
2 cos 2
cos
1 for
,
dd
dy
so by L’Hôpital’s rule
0 and the tangent line is horidx zontal at ,r 0 [the line y 0]. This information can be summarized as follows.
Horizontal at: 3
,
23 3 y 35
,
23 Vertical at: 2
3
4
3 , 0],
33
4 y (2, 0) [x
1
,
2
1
,
2 3
4 (0, ) [y 1
,
4 y 15
,
26 9
4 0) or when This information can be summarized as follows.
Horizontal at: 1 [the line x 3
2 dy
is undefined and the tangent line is
L’Hôpital’s rule
dx vertical at 2 cos ) 2 cos line x 0, but 2 cos 2 3 [the line x 1
.
4 33
and for
4
3
3
11
the line x
cos
2
2
6 11
,r
6 cos )sin vertical tangent line for 5
,r
6 3
7
cos
2
6 the line x for
2 1
2 ,r 1
and for
4 There is a vertical tangent line for For , (then 1 1
5
sin
2
6 again, the line y (1 0 when . So there is a horizontal tangent line for
3
2 sin
2 1 sin (then sin dx
d sin2 0 when cos 0) or when 1
or 1, i.e., when
2 4 cos sin 2 1 cos )cos cos sin (1 sin
sin 0 when (1
cos sin 2 , 2 dx
d sin )sin sin 2 sin2 sin sin2 2 cos2 cos2
2 dy
d cos2 cos cos dy
d 10. 1) sin 2
dx
d sin )cos 2],
1
,
4
1
,
4 x
x (2, 2 ) [x 2] 429 Section 10.6
11. y
dy
d 2 sin2 12. 4 sin dy
d cos 4 sin2
4(sin 2 sin 2
x 2 sin dx
d cos 0, dx
, and
d , 2 , at (0, 0) [ y
(0, ) [ y 0 sin 0
0 sin 0], 2, 2, [x 4 2 sin 2 , and 2
3
,
the curve has
44 0]. For 3
2 cos
4 x y 2 2 cos 4 1. This information can be summarized as follows.
Horizontal at: (0, 0) [y
2, 2 (0, ) [y
Vertical at: 3) 2, 4
3
2,
4 1] and (then sin
2 [x 3 sin
3 137
16 dx
d 0 when 0) or when
3
8 1 cos [the line y cos 5.097 (then 8 cos r 5.176 [the line y 5.176 sin 2.146 0, r 2 2.146, 4.343], for 0.267]. There is a vertical 1 [the line x 7 [the line x 1 cos 0 7 cos 1 [again, the line x
cos 0.676 0.676 tangent for 1 0). So there 4.343], and for 0.676 sin 5.878 r 3 0.267], for 5.176 sin 4.137 ,r 1.186 or 0.405, r [the line y for or 2 5.176 5.878, r 1] 0,
3
8 1 0.676 sin 0.405 [the line y 1], , i.e., when is a horizontal tangent line for 0]
[x 4 cos )sin 0 when cos 4.137, r 2], 4 (3 graphing calculator). 0], [y 3 cos 0.405, 2.146, 4.137, or 5.878 (values solved for with a the curve has horizontal asymptotes vertical asymptotes at
3
2,
4 cos dy
d 0 when 3
, . They are never both zero.
44
2 4 sin ) 3 cos 4 sin 2 0 when 0, cos sin (8 cos 2 cos 2 For 4 cos )cos
2 8 cos2 sin 2
dy
d
dy
d (3 2 7], for 1 cos 2 1], for 3
,r
8 3
9
the line x
, and for
2
16
3
3
9
cos 1
,r
again, the line x
.
8
2
16 This information can be summarized as follows.
Horizontal at: ( 0.676, 0.405) [y
(5.176, 2.146) [y 0.267],
4.343], (5.176, 4.137) [y
( 0.676, 5.878) [y
Vertical at: 4.343],
0.267] ( 1, 0) [x 1], (1.5, 1.186) x (7, ) [x 7], (1.5, 5.097) x 9
,
16 ( 1, 2 ) [x 9
,
16 1] 1],
2, 430 Section 10.6 13. The curve is complete for 0 2 (as can be verified 19. by graphing). The area is
2 1
(4
2 0 2 cos )2 d 2 2 (4 [ 2.5, 5.2] by [ 2, 3.1] cos2 ) d 4 cos 0 24 1
2 4 sin 2 1
sin 2
4 The circles intersect at (x, y) coordinates (0, 0) and 18
0 2 (as can be verified 14. The curve is complete for 0 (1, 1). The area shared is twice the area inside the
circle r 2 sin between by graphing). The area is
2
0 12
a (1 cos )2 d
2
2
12
a
(1 2 cos
20
12
1
a
2 sin
2
2 15. Use r
4 2 2
0
/4 cos2 ) d 4 2 1
sin 2
4 32
a
2 0 4
4 . The total area is
2 2 2a cos 2 ) d 2a 1
(2 sin )2 d
2 4 sin2 d 1
2 1
4 8 /4 [ 3, 3] by [ 2, 2] The circles intersect at 1,
1
cos2 2 d
/4 2 17. Use r . Its area is 4 4
1
sin 4
16 1
4 /6 is 2 /4 8 /4 4 sin 2 . One loop is complete for 0
/2 Its area is
0 1
(
2 /2 4 sin 2 )2 d . 0 0 2 . 4 0 2 /2 2.
0 d 1
sin 2
4
3
2 2
3 and 1,
/6 5
. The shared area
6 12
(1) d
2 /2 sin2 1
2 6 /2 1
(2 sin )2 d
2 /6 4 2 sin 2 d cos 2
18. Use r 1. 2 /4 1
sin 2
2 2a2
(Integrating from 0 to 2 will not work, because r is not
defined over the entire interval.) /4 0 cos 2 d
/4 16. One leaf covers /4 1
sin 2
4 20. /4 2 2a2 . 4 0 2a2 cos 2 . One lobe is complete for /4 1
(
/4 2 /4 Shared area 0 and /6 d /6 /2 0 3
3 /6 2 3 21. 2 sin 3 . One leaf is complete for 0 3 . The total area is
/3 6
0 1
(
2 2 sin 3 )2 d /3 6 sin 3 d
0 2 [ 4.7, 4.7] by [ 3.1, 3.1] /3 cos 3 4.
0 The shared area is half the circle plus two lobelike regions:
/2 1
(2)2
2 2
0
/2 2 (4 1
[2(1
2 cos )]2 d
4 cos2 ) d 8 cos 0 2 4 5 8 2 sin 1
2 1
sin 2
4 /2
0 Section 10.6
22. 431 25. [ 4.7, 4.7] by [ 3.1, 3.1] [ 6, 6] by [ 4, 4] The area in question is half the circle minus two lobelike
regions: Use the symmetries of the graphs: the shared area is
/2 4
0 1
[2(1
2/2 8 cos )]2 d (1 2 2 cos cos )d 0 8 1
2 2 sin 23. For a 1
sin 2
4 /2 1
(2)2
2 2
0
/2 2 /2 6 1
[2(1
2 (4 sin )]2 d
4 sin2 ) d 8 sin 0 16 0 2 1: 4 2 cos 1
2 1
sin 2
4 /2 8
0 26. [ 3, 3] by [ 2, 2] [ 3, 3] by [ 2, 2] (a) To find the integration limits, solve
The curves intersect at the origin and when
3a cos a(1 2 cos 2 cos cos ) 0 Because of the curve’s symmetry, the area inside the
outer loop is 1
3 1
2
.
3 . 2 /3 2 Use the symmetries of the curves: the area in question is 0 1
(2 cos
2 1)2 d 2 /3
/3 2
0 1
[(3a cos )2
2 a (1 cos ) ] d 0 /3 (8 cos 1 2 2 cos 1) d 1) d 2 sin sin 2 4 sin
0 3 3
2 /3 2 sin 2 2 cos2 ) d 2 cos 0 a2 4 4 cos 0 2 /3 (9 cos2 2 (4 cos2 2 /3 a2
a 2 2 (b) Again, use the curve’s symmetry. The inner loop’s area
is a2 0 24.
2 1
2 /3 2 2 (2 cos
sin 2 1)2 d
4 sin
2 /3 Subtract this from the answer in (a) to get 3 3 [ 3, 3] by [ 2, 2] The curves intersect when
6 cos 2
6 3
or 5
.
6 Use the symmetries of the curves. The area in question is
/6 4
0 1
(6 cos 2
2 /6 3) d 33
.
2 6 sin 2 3
0 3 . . 432 Section 10.6 27. 30. The integral given is incorrect because r
out the circle twice as cos sweeps goes from 0 to 2 . Or, you can’t use equation (2) from the text on the interval [0, 2 ]
because r [ 9, 9] by [ 6, 6] To find the integration limits, solve
3 csc cos is negative for 3
5
. The correct area is , which can be found
2
2
4
3
by computing the areas of the cardioid
and the circle
2
4 6 separately and subtracting. 5
. The area in question is
66
5 /6
12
(6
32 csc2 ) d
/6 2
5 /6
1
36
9 cot
2
/6 , 12 9 31. dr
d 2 , so
5 ( 2)2 Length (2 )2 d 0 3 5 2 28. 4d 0 12
(
3 5 4)3/2
0 1
(27
3 19
3 8) [ 3, 3] by [2, 2] 32. To find the intersection points, solve
9
sec2
4 6 cos 2
48 cos4 , so
2 9 6 2 e 2 0 d
2 ed
0 e e . 1 0 By the symmetry of the curves, the area in question is
/6 2 e
0 3
4 cos 0 e Length 24 cos2 2 2 dr
d 1
6 cos 2
2 9
sec2
4
/6 9
tan
4 3 sin 2 33. dr
d sin , so d 2 3 3
4 0 0 1
tan2
2 /2 d
/4 /4 1
tan
2
1
1
2 1
(0
4 4 11
csc2
22 2
2 cos y sin lim → lim x → /2 x 1, /2 1, 3
2 lim lim y /2 8
4 sin sin2
cos → → /2 2 cos d,
2d 2 d 2 0 cos 4 3
4 1) ⇒y tan d . 8 sin .
34. dr
d a sin 2 2 cos 8
0 2 a2 sin4 Length sin2
0 2a cos , so
a2 sin2 2 0 a
y d 2 0 /4 ⇒x tan 4 cos2 0
2 cot Total area is twice that, or
(b) Yes. x 2 cos 0
2 /2 1
4 0 ( sin )2 d 0
2 . 29. (a) Find the area of the right half in two parts, then double
the result: Right half area
/4 cos )2 (1 Length 2
2 d
2a
0 2 cos2 2 d 433 Section 10.6 35. dr
d 6 sin
, so
cos )2 (1 41. /2 (1 62
cos )2 NINT evaluates to 6.887. Length
0 62 sin2
d , which using
(1 cos )4
[ 1.5, 1.5] by [ 1, 1]
3 (Note: the integrand can simplify to 3 sec
dr
36.
d 2 .) sin 2 22 sin2
d , which using
(1 cos )4 0
2 2.296. (Note: the integrand can simplify to csc 37. 2 dr
d cos2 sin 3
/4 3 cos6 2 sin d 0 /4 4 /4 cos4 cos (4 2 2) 3.681 3 sin2 d 3 3 /4
8 0 [ 3, 3] by [ 2, 2] .
dr
d , so
2 cos2 2
1 sin 2 0
2 2 16a2
43. 2 2d 2. 2 sin 2 ( sin 2 )(2) 2
0 cos
0 ( f ( ))2
2 2 e 44. (a)
, so surface area 2 2e /2 sin ( 2e (b) 0
/2 2 e sin 5d 0 25
5 (e 1
e (sin
2
/2 1) /2 cos )
0 40.818 22 /2 2 ) 2 (e ( f ( ) sin )2 sin2 ) 2 0 0 /2 2 )d
(c) /2 dr 2
d r2 a(1
0 1
( a
2 ad
0 /2) sin a
0 2 a
0 /2 a cos 2 a
2 cos ) d 2 1
2 sin2 ) 2 1 /2 /2 ( f ( ) sin )2 ( f ( ))2 4.443. 0 2 f ( ) cos )2 ( f ( ))2(cos2 /4 2 f ( ) sin )2 ( f ( ))2(cos2 sin2 2 d d sin dr
d d 4a2 0 ( f ( ) cos )2 2 /4 2
2 cos2 2 cos
0 sin 2
c os 2 /4 2 /2 1
sin 2
4 ( f ( ) cos )2 ( cos 2 )2 cos 2 cos d d ( f ( ) sin c os 2 cos 2 4a2 sin2 dy 2
d , so Surface area
/4 dx 2
d 1
2 ( f ( ) cos 0 1 cos2 0 (1 sin 2 )
d
sin 2 1 /2 16a2 sin 2 d
2 cos 2 0 1 4a2 cos2 2 (2a cos ) cos
0 Length dr
d 2a sin , so surface area
/2 2 sin 2 1 1: d 3
2
sin
4
3 cos 2 dr
38.
d 42. For a
cos4 3 0 1
2 , so 3 0 40. sin2 2
d
cos 2 cos 2 /4 2 ). cos 2 sin 0 Length 39. 2
0 3 : Surface area 4 /4 NINT evalutes to . cos 2 Use the curve’s symmetry and note that r is defined for 22
cos )2 (1 /2 2 sin 2 dr
d 2 sin
, so
(1 cos )2 Length dr
d 2r d /2 45. If g( ) 2f ( ), then
(g( )2 g ( )2) 2 ( f ( )2
g is 2 times the length of f. a /2 2a sin
/2 f ( )2), so the length of 434 Chapter 10 Review 46. If g( ) 2f ( ), then
(g( )2 g ( )2)
2 g( ) sin
4[2 f ( ) sin
( f ( )2 f ( )2)],
so the area generated by g is 4 times that of f.
47. (a) Let r 0.06
2 1.75
0 evaluates to 741.420 cm 2 b 2 1/2
2 0.06 2
d
2 7.414 m. (pp. 569–572) b since 2. (a)
(b) 2 741.420 cm (from part (c)),
80 La 1.75 0 0.03
2 1.75 0.06
2 d
236 741.416 cm 0 bn 2 the integral, La
bL
r02 b 1 2 r0n. Solving for n gives (c) Since L is proportional to time, the formula in part (a)
shows that n will grow roughly as the square root of
time.
3 3 a (1 5. 6 2 8
2 23
a
3 cos ) cos 25 a2 15
sin
8 2 cos 1 7. 2 42 8. 9. (a) 4, 12 (1 cos4 ) d 15
8 cos2 sin (b) 2 So x 3
2 53
a
2 3 a2 By symmetry, y 1
2 , dy
dx dy/dt
dx/dt
3 d 2y
dx 2 3
2 2 , dy
dx 3, (1/2)sec t tan t
(1/2)sec2 t
3 ;x x 17 34
,
55 (4/5)2 which for t
10. (a) cos )2 d 1
sin 2
4 2 sin 3
2 2 ,y 1 sin t
dy
dx 1, and
3
2 4 3
2
3 x 2 . So the or 1
.
4
dy /dt
dx/dt cos t
(1/2)sec2 t
1
equals .
3
4 2 cos3 t, 0 0 a2 29 17 (3/5)2 y cos2 ) d 2 cos 8 1 1 5 0 2 5 2 sin a2(1 25 725 tangent line is y
3 cos3 1
cos3
4
53
a , and
2 10, 31
,
2 6. 0 23
a 3 sin
3 10
2 10 8 [assuming counterclockwise]. d 3 cos2 (cos 6, radians below the negative xaxis: 0
2 1 2
2(4) For t
2
49.
3 1, 1 6 1. (b) The take up reel slows down as time progresses. 2 20 2 80 made n complete turns, then the angle is 2 n. So from n 2 2 (b) 8, 12 1313 2( 3), (b) 9 5 4. (a) 5(2), 5( 5) 48. (a) Use the approximation, La, from 47(e). If the reel has r0 2, 4 1 3. (a)
(e) L 5 322 3 is a very small quantity squared. 2r 3, 4
4 2,
17, 32
172 (b) 2 0, so the centroid is s Chapter 10 Review Exercises r 2r 0 4a
. By symmetry, x
3 a2
4a
0,
.
3 2 1/2 b r1 43
a , and
3 cos a2. So 1. (a) 3
(d) r 2 23
a
3 d 43
a
3 y length of the curve.
80 a3 sin
0 a2 d b
, this is just Equation 4 for the
2 (c) Using NINT, 2
3
0 0.06
.
2 1.75 dr
(b) Since
d 50. 2 3 a2. 0 (b) 5a
.
6 0, so the centroid is 5a
,0 .
6 dy
dx 3/t 2
2/t 3
5
,y
4 3
t
2
1
dy
For t 2: x
, and
2
dx
1
So the tangent line is y
3x
2
13
y
3x
.
4
d 2y
dx 2 dy/dt
dx/dt dy
dx equals 6. dy /dt
dx/dt 3/2
2/t 3 3.
5
or
4 3t 3
, which for t
4 2 Chapter 10 Review 11. dy
dt 1
tan t sec t equals zero for t k , where k is any
2
dx
1
integer.
sec2 t never equals zero.
dt
2
1
1
tan 0, sec 0
2
2 (a) Horizontal tangents at
1
tan
2 , 1
sec
2 0, 17. (a) 1
and
2 [ 1.5, 1.5] by [ 1, 1] (b) 2 1
.
2 0, 435 18. (a) (b) There are no vertical tangents, since dx
never equals
dt zero.
12. dy
dt integer. dx
dt [ 3, 3] by [ 2, 2] k
, where k is any odd
2 2 cos t equals zero for 2 sin t equals zero for t k , where k is any (b)
19. (a) integer.
2 cos (a) Horizontal tangents at
2 cos and 3
3
, 2 sin
2
2 2 , 2 sin (0, [ 1.5, 1.5] by [ 1, 1] 2).
(b) (b) Vertical tangents at ( 2 cos 0, 2 sin 0)
( 2 cos , 2 sin ) (2, 0).
13. dy
dt 2 sin t cos t where k is any integer. (0, 2) 2 ( 2, 0) and
k
,
2 sin 2t equals zero for t
dx
dt 2 20. (a) sin t equals zero for t k,
[ 1.5, 1.5] by [ 1, 1] where k is any integer. Where they are both zero, use (b) L’Hôpital’s rule:
dy/dt
t→k dx/dt lim lim t→k sin 2t
sin t (a) Horizontal tangent at 2 cos 2t
cos t lim t→k cos 2 , cos 21. 2. 2 (0, 0). 2 dy
dt 9 cos t equals zero for t dx
integer.
dt 0, 3
3
, 9 sin
2
2 2 , 9 sin 2 , 0, cos 2 cos
cos 2 sin 3
5
, 0,
4
4 /4 dy
dx k , where k is any integer.
(a) Horizontal tangents at 4 cos 4 dy
dx k
, where k is any odd
2 4 sin t equals zero for t f ( ) cos
f ( ) sin 2 sin 2 sin
2 sin 2 cos (b) There are no vertical tangents.
14. f ( ) sin
f ( ) cos dy
dx 2/ 2
2/ 2
2/ 2 5 /4 2/ 2 and 0, 1,
1, dy
dx dy
dx 7
4 2/ 2
3 /4 7 /4 The Cartesian equations are y (0, 9) and are polar solutions. 2/ 2
22
2 (0, 9). (b) Vertical tangents at (4 cos 0, 9 sin 0)
(4 cos , 9 sin ) ( 4, 0). 22. x. f ( ) sin
f ( ) cos f ( ) cos
f ( ) sin 2 sin 2 sin
2 sin 2 cos (1
(1 cos
2 cos3
cos
sin
2 cos2 sin
sin dy
dx (4, 0) and 15. 4 sin2
6 cos cos 2 )cos
cos 2 ) sin 2 cos2
sin 4 sin2
2 cos2
.
3 sin 2 [ 7.5, 7.5] by [ 5, 5] 16.
0,
dy
dx
[ 7.5, 7.5] by [ 5, 5] 1. 2 4 sin2 cos
4 cos2 sin 4 cos 1, 2 and 0, /2 dy
dx 3
2 are polar solutions. 3 /2 4
is undefined, so the tangent lines
0 are vertical with equation x 0. 436 23. Chapter 10 Review dy
d d
1 cos
sin
2
d
1
sin
sin
cos
cos
cos
2
2
2
dx
d
1 cos
cos
2
d
d
1
sin
cos
sin
cos
sin
2
2
2
dy
Solve
0 for with a graphing calculator: the solutions
d are 0, 2.243, 4.892, 7.675, 25. [ 1.5, 1.5] by [ 1, 1] 1 , 1 2 r sin horizontal tangent lines at y 0.443 and y Solve dx
d are 0, 0 for 1.070, reveals 9.035, 1.739. 1 Where r cos reveals 0.067, and x dy
dx
and
both equal zero (
dt
dt 1 y 0, 4 ), close y
y 0. (This can be confirmed 2 cos y 4 sin2 2 sin y , 62 , 5
3
, and .
6
2 dx
d reveals horizontal tangent lines at y 0 for 1
and
2 (by first using the quadratic formula to find sin ): the solutions are
two solutions to find x
3 lines at x
zero x 1 x or
2 2 [ 3, 3] by [ 2, 2] 4. Solve 2, and 2 Using the first, third, and fourth solutions to find
r sin x
1 or
2 0 for : the solutions are y 1 26. 2 sin2 2 sin or or x 2 sin )cos ] 2 cos2 dy
Solve
d cos 1, and 1, respectively. 2, 1 y 4 sin 1, 1, 2 2 dy
d
dx
d sin )sin ] 1 x x y d
[2(1
d
d
[2(1
d . From the 2 2, 2 using L’Hôpital’s rule.)
24. ,
2 2 x 1.104. inspection of the plot shows that the tangent lines are
horizontal, with equation y 1 , and
2 1 x
2 y
2, x 1 , 2
1 So the equations of the tangent lines are
y vertical tangent lines at x ,
2 those points have slopes of 11.497, and 4 . Using the middle five solutions to find x 1 ,
2 1 curve’s symmetries, it is evident that the tangent lines at with a graphing calculator: the solutions
3.531, 2 , ,
2 10.323, and 4 . Using the middle four solutions to y 1 The tips have Cartesian coordinates 3
2 2 , r cos 7
11
, and
. Using the last
6
6 reveals vertical tangent 2.598. Where dy
dx
and
both equal
dt
dt , inspection of the plot shows that the tangent As the plot shows, the curve crosses the xaxis at
(x, y)coordinates ( 1, 0) and (1, 0), with slope 1 and 1,
respectively. (This can be confirmed analytically.) So the
equations of the tangent lines are
y0
(x 1)
y
x 1 and
y0x1
y x 1.
27. r cos
r sin
x y, a line
28. r 3 cos line is vertical, with equation x r2 3r cos x2 2 y2 3x x2 3x 9
4 0. (This can be confirmed using L’Hôpital’s rule.) x 32
2 y2 a circle center y2 9
4
32
2
3
, 0 , radius
2 3
2 437 Chapter 10 Review
29. r 4 tan sec 37. r sin
4
r cos r cos y
4 or x 2
x x 30. r cos dr
d sin , so
2 2
2 0
2 3 2 cos cos 1
r cos
2
1
3
x
y
2
2 x r sin 3
3 r sin 2 2 2 3 2 3
38. 3 dr
d 31. x
r2
r 3 4 0 x 3 or y 2 y
5y
5r sin
5 sin 4, a line 32. x 2 y 2 2y
r 2 2r sin
r 2 sin
4y 39. 8 sin2 cos 3 /4 /4 2 4r 2 sin2 dx
= 2e2t
dt 16
16, or r 2 ln 2 8 sin2 1 2t
e
2 4t 4e 2e2t 8 1 2 cos 2
1 cos2 2
cos 2 sin2 2 2d
41. 2 dx
dt 2 sin t,
/2 dy
dt 0 24 3.087. 2t, so ( 2 sin t)2 Length
8 t /2 2 t2 (2t)2 dt sin2 t dt, 0 2 1, so
42. 3 Length d /2 ln 2 dy
2t,
dt d cos 2 sin2 2
d
1 cos 2 which using NINT evaluates to
dx
36.
dt 2 sin 2 /2 1 ln 2
8 3
0 cos 2 )2 cos 2 )2
cos 2 /2 dt d 3 , so Length (1
1 /2 e dt (2t) 2 (t 2 2 1) dt 3 dx
dt 3 cos t,
3 dy
dt Length 3 3.183. t, so (3 cos t)2 (3 4t 2 t4 2t 2 1 dt 3 t)2 dt 0 3 3 t 0 which using NINT evaluates to 12.363. 3
2 3 (t 2 43. Area 2 1) dt 0 3 3 3 t
3 d /4 1 /2 2t cos2 3 3 c os 2 1 /2 (et )2 dt 0 ln 2
8 sin 2 dr
d ln 2 t
8 e 40. cos 3 d 3 3
2
sin
4
3 (1 1
64 12 0 2t 16 sin2 /2 12
8 0 0
ln 2 cos2 2 8 sin2 3 3 8 sin2 1
2 8 16
4 sin2 e t, so
2e2t Length d 2 8 sin3 /2 1 dy
,
8 dt ln 2 0 5)2 16
(r sin
5)2 (y
2)2 2 sin )2 d (2 cos , so 3 0 r 2 cos2 3 dr
d /4 4(r sin ) 4 2. Length (r cos ) 35. 2d 16 2 34. (x 2)2
(r cos 8 cos2 0 2 0 /2 2 0
0 33. x 2 cos )2 8 sin2 /2 8. 2 0 0
0 2 4 cos 2 sin , so Length /2
0 3
2 2 cos d (2 sin 2 3y sin d
2 4 sin2 0 r cos ( sin )2 d 0
2 4y, a parabola 3 cos )2 (1 Length t 4
3 3. 1
(2
2
2 cos )2 d 1
(4 4 cos
20
1
4
4 sin
2 cos2 ) d
1
2 1
sin 2
4 2
0 9
2 cos2 t dt, 438 Chapter 10 Review /3 1
sin2 3 d
2 44. Area
0 11
22 1
sin (6 )
12 50. /3 12 0 45.
[ 1.5, 1.5] by [ 1, 1] r [ 3, 3] by [ 2, 2] cos 2 dr
d sin 2 and
/2 The curves cross where cos 2 0, such as 4 . Using Area 1
[(1
2
/4 4
0 cos 2 )2 2 2 (cos 2 4
0 /4 1
2 cos2 2
d
sin 2 sin 2 d sin 4 d
[(4 cos t)i
dt 51. (a) v(t) sin 2
0 ( 2 sin t)j] ( 4 sin t)i 2 4 cos 0 2 cos 2 ) d 1
sin 4
8 , so /2 1] d 0 2 2 /2 4 /4 sin 2 cos 0 the curves’ symmetry,
Length 2 2 , where 0 sin 2 d
[( 4 sin t)i
dt a(t) 46. ( 2 cos t)j
( 2 cos t)j] ( 4 cos t)i ( 2 sin t)j
2 (b) v 4 sin 4 [ 4.5, 4.5] by [ 2, 4] 8 2 2 cos 4 1 3 2 2i 4 Since the two curves are covered over different intervals,
(c) At t find the two areas separately. Then
2 Area
0 1
[2(1
2
2 2 sin )]2 d (1 r2
2 2 sin sin 47. dx
dt t, 1
2 2 cos dy
dt cos 1
sin 2
4 cos
2 5
0 5 2 (2t) t 2 Area ( 22 dt 0 4
(t 2
3 48. dx
dt 2t
1 4) 2 t 1
2t 2 2 1/ 2t sin 2 cos 2 sin cos 2 0 (c) At t sin2 2
d
cos 2 /4
0 2 sin d
/4 2 cos (2
0 2) ( 3 sec2 t)j] 3 sec t tan t)i 0, v sec3 t)i 3j, a
( , so 2 ( 3 sec2 t)j 3 sec t tan t)i 3 sec2 t tan t)j (2 3 sec4 0 0 3 3 4 dt, 10.110. /4 3 tan t)j] 2 cos 2 Area ( va
cos 1
va which using NINT evaluates to
dr
49.
d 3 sec t)i 3 sec2 0 tan2 0 (b) v(0) 12
2t 2 j, and va 3(sec t tan2 t 4, so 2 2i 18 d
[(
dt a(t) 76
3 0 1 dy
,
2t 2 dt Area 5 3/2 j, a 1 va d
[(
dt 52. (a) v(t) 2, so ,v 1
(3)(3)
7
cos 1
38.94 .
9 )d 0 2 4 1.840 53. v(t)
dr
dt (1 v(t) t
i
t 2)3/2 3i 0
3)( 0 1 cos 0 90 3) 1
j
(1 t 2)3/2
2
t
(1 t 2)3/2 which is at a maximum of 1 when t (1 2
1
t 2)3/2 0. 1
1 t2 , 439 Chapter 10 Review
dr
= (e t cos t
dt
dx
(e t cos t
dt 54. v(t)
a (t) t e t sin t)i
e t sin t
t (e sin t (e t sin t ( 2e t sin t)i e t sin t
t e cos t dx dy
,
dt dt 61. (a) v(t) e t cos t)i v(3)
(e t sin t)(2e t cos t) 0 for all t. The angle between r and a is always 90 .
1 6t) dt i 6 cos t dt j 0 0 3t 3t 1 2 6 sin t j 0
2
e 56.
e 2
e
e 1
dt j
t ln t 2 e e i r(0) j 3i x
3
x2
9 cos dx
dt 4 2 5 j, so C (sin t)j Length i (cos t 1)i r(0) j r(t) (tan C
1 1 (tan j, and 10 i
t j, so C 100 3
t
3 10 i dr
dt d 2r
dt
dt 2
dr
dt
dt r(t)
dr
dt t 1 t 2j C1 t C2 C2. And r(0) C2 2i r(t) ( t2 r(1) 5i r(t) ( t2 t 2i
2j
6t )i j C2
6t ( 2t)j
t 2j C1 3i
2)i C1t 6i 2j and C2 3j, so C2
( t2 (b) 2t e t sin t
e t cos t dy/dt
dx/dt dy
dx t C2 2t)j dy
dx 1
1 12
2 1250
3 t)2 dt, which (5 2)j 2j, and 1040.728. e t cos t
e t sin t cos t
cos t sin t
sin t 1 dy
dx
e t(sin t cos t),
e t(cos t sin t)
dt
dt
dy 2
e 2t(sin2 t 2 sin t cos t cos2 t)
dt e 2t(1
2i dx 2
dt 2 1 or t 4) dt 1 5 10
t
50 5t 4 using NINT evaluates to C1 , 4i, so C1 ( t2 20t 3 t(10 t)
2 2
0 63. (a)
( 2t)i 5 i, so t 2j r(t)
60. (c) Area dr
dt
dt 3 2 t)2 dt, which using NINT (5 (100t 2 8 1j dr
d 2r
59.
dt 2tj C1, r(t)
dt
dt 2
dr
C1 0, so r(t) t 2j
dt t 0 y2 0 C i and t2 1)i 0
10 8 1j 16 x2 dx
dt
dt
t(10 t) 2
dt
2 10 t2 3 4 2
5 y2 0 1)j t)i 2 25.874. 2 dr
dt
dt 58. r(t) (sin t sin 3
16 2 t, so 4 3 3.238 t so 2 evaluates to C 5
16 12 (b) Volume
r(t) 32
cos
16
4 sin 1
dy
and
2
dt
10 (ln 2)j 17
4 2 1. 0 (cos t)i C y
5 t and y
25 e dr
dt
dt i 62. (a) 2 ln (ln t) e 57. r(t) 6i 0 2 ln t
dt i
t ln2 t (c) 1 i 34
4 d 2y
ycomponent: 2
dt t 3 1 (3 , and 42
25 2
32 d 2x
dt 2 t 3 (b) xcomponent: 55. 5 , 42
92
32 e sin t)j (2e t cos t)j 3
5
sin t,
cos t ,
4
44
4 3 v(3) t e cos t (e t cos t)( 2e t sin t) r(t) a(t) e t cos t )j 2 sin t cos t) e 2t(cos2 t
e 2(1 2 cos t sin t 2 cos t sin t) v(t ) et 2 v(3) e3 2
3 (c) Distance v(t) dt
0
3 et 2 0 2 et
(e3 3
0 1) 2 sin2 t) 440 Chapter 10 Review dx dy
,
dt dt 64. (a) v(t) 6
5 2t, t 2 , v(4)
96 2
5 (2t)2 4 (b) Distance 69. (a) 96
, and
5 104
5
622
t
dt
5 82 v(4) 8, 0
4 [ 2, 10] by [ 2, 6] dx dy
,
dt dt
d 2x d 2y
,
dt 2 dt 2 (b) v(t) 2
t 25 9t 2 dt
5
2
4
(25 9t 2)3/2
135
0 0 (c) t x 2, so dy
dx 6t 2/5
2t dy/dt
dx/dt 65. x degrees east of north is (90 a(t) 4144
135
3
t
5 3
5 x 540 cos 10 , 540 sin 10
595 cos 10 , 485 sin 10
585.961, 84.219 .
585.9612 Speed
Direction 84.2192 585.961
tan 1
84.219 v(1)
a(1) 2 ,0
0, 2 0, 0
0, 2 v(3)
a(3) 2 ,0
0, 2 (c) Topmost point: 2 ft/sec
center of wheel: ft/sec
Reasons: Since the wheel rolls half a circumference, or
feet every second, the center of the wheel will move
feet every second. Since the rim of the wheel is
turning at a rate of ft/sec about the center, the
velocity of the topmost point relative to the center
is ft/sec, giving it a total velocity of 2 ft/sec.
Rg
, where
sin 2 70. v0
R 66. Add the vectors: 45 , g 32, and range for 4325 yds 120 cos 20 , 120 sin 20 300 cos ( 5 ), 300 sin ( 5 ) 411.622, 14.896 .
14.896
411.622 1 Direction tan Length 411.6222 12,975 ft: v0 644.360 ft/sec for 4752 yds 14,256 ft: v0 675.420 ft/sec Rg
sin 2 71. (a) v0 14.8962 411.891 lbs (b) The cork lands at y
Solve y
45 : v0 7 4, x 59.195 ft/sec g
2v02 cos2 Solve y 177.75. g
x 2 (tan )x for v0, with
2v02 cos2
gx 2
74.584 ft/sec
yx 72. (a) The javelin lands at y 2 (80 sin 45 )
2(32) (109.5)(32) 2.073 67. Taking the launch point as the origin,
y (44 sin 45 )t 16t 2 equals 6.5 when t 2.135 sec
(as can be determined graphically or using the quadratic
formula). Then x (44 cos 45 )(2.135) 66.421
horizontal feet from where it left the thrower’s hand.
Assuming it doesn’t bounce or roll, it will still be there 3
seconds after it was thrown.
68. ymax cos t 0, 0
0, 2 591.982 mph.
81.821 east of north 2 sin t, v(2)
a(2) x) degrees north of east. 55 cos ( 10 ), 55 sin ( 10 ) 2 sin t v(0)
a(0) 2 Add the vectors: cos t, 6.5, x
x2 262 5
.
12 (tan )x for v0, with 40 : 57 feet gx 2
(2 cos 40 )( y x tan 40 ) v0 2 (b) ymax 73. We have x
gt 2
2 y
x 2 y (v0 sin )2 6.5 2g
(91.008 sin 40 )2
64 (v0t) cos 91.008 ft/sec 6.5 59.970 ft and (v0t) sin . Squaring and adding gives
gt 2 2
2 (v0t)2(cos2 sin2 ) v02t 2. Cumulative Review
74. (a) r(t)
x(t)
y(t)
(b) ymax
tmax (155 cos 18
11.7)ti
(4 155 sin 18 t
(155 cos 18
11.7)t
4 155 sin 18 t 16t 2 Cumulative Review Exercises
(pp. 573–576) 16t 2)j 1. Since the function has no discontinuity at x (155 sin 18 )2
4 39.847 feet, reached at
2(32)
155 sin 18
1.497 sec
32 (c) y (t) 0 when t 3.075 sec (found using the quadratic
formula), and then
x (155 cos 18
11.7)(3.075) 417.307 ft.
(d) Solve y (t) 25 using the quadratic formula:
1552 sin2 18
32 155 sin 18 t 72.406 and 11.7)t equals 333.867 feet from home plate. (155 cos 18 11.7) 1
(1
0.09 155 sin 18
(1
0.09 4
32
(1
0.092 x(t) (155 cos 18 y(t) 4 0.09t e
e )i e 0.09t ) (c) AC
E AB
E 1
ex
ex 1
x→ 1 lim 1.
t (1
t 5. By l’Hôpital’s Rule, lim
t→0 t sin t (1 cos t)
1 cos t
t sin t 3 cos t
lim
3
cos t
t→0 lim
t→0 cos t)
sin t
t cos t 2 sin t
lim
sin t
t→0 ex ln (e x 1)
6. By l’Hôpital’s Rule, lim
ln x
x→0
xe x lim x lim xe x x→0 1 (e x 0.09t ln (e x
x
x→0 ) 8. lim 3x x→0 1
AD
E
2 AD, so by part (b), AP
E
E x) lim ex
e lim (e x 1) x→0 1
x ln (e x
x x) 1 x x)1/x. Then ln f (x) 1
AC.
E
2 77. The widths between the successive turns are constant and
are given by 2 a. lim x)1/x
1 x
(3x x→0 lim x→0 lim (e x 1)/(e x
1 x→0 x→0 ) AB
E
1
1
BD = AB
E
E
2
2 ex
ex So lim (e x 0.09t (e) No, the batter has not hit a home run. If the drag
coefficient k is less than 0.011, the hit will be a home
run.
(This result can be found by trying different kvalues
until the parametrically plotted curve has y 10 for
x 380.) AB
E ex
ex 1)2 (x x→0 )j (d) Plot y (t) and find that y (t) 30 at t 0.753 and
2.068 seconds. At those times, x 98.799 and
256.138 feet (from home plate). (b) AP
E x
x→ x lim x 4. By l’Hôpital’s Rule, lim 7. Use f (x) (c) Plot y (t) and find that y (t) 0 at t 2.959, then plug
this into the expression for x(t) to find
x(2.959) 352.520 ft. AD
E x→0 3
.
4
1 1 1 ) (b) Plot y (t) and use the maximum function to find
y 36.921 feet at t 1.404 seconds. 76. (a) BD
E 1
x 3. By l’Hôpital’s Rule, lim 3 cos 3x
4 lim x→0 0.09t 1
(1
11.7)
0.09 155 sin 18
(1
0.09
32
(1 0.09t
0.092 0.09t sin 3x
4x x→0 x→0 e e
0.09t e e 0. 2. By l’Hôpital’s Rule, lim x→ (e) Yes, the batter has hit a home run. When the ball is 380
feet from home plate (at t 2.800 seconds), it is
approximately 12.673 feet off the ground and therefore
clears the fence by at least two feet.
75. (a) r(t) 11
2(1)
12 1 12 lim (155 cos 18 1, the limit is 2 4(16)(21) 0.534 and 2.460 seconds.
At those times, x 441 (3x 1
sin x
1) cos x
x cos x lim e ln f(x)
(3x x→0 3 sin x
sin x 1) sin x 6 cos x
x sin x 2 cos x 2. e 2. x→0 lim x) 1) sin x
x sin x 1 3 x , and 1. 442 Cumulative Review
12 9. (a) 2(1)
(b) 2 1 1 19. y
20. y (d) Yes, since lim f (x) f (1) x→1 f (1 h)
h h→0 21. y f (1) lim h) h→0 2 h)2 (1
h 2h 1 2h 3 lim h2 lim h h→0 2 lim (1 h→0 h) h
h 1 1 cos x 2 and x
, 2x x 0 to find x x) 8
,
x 3 1
.
2 ,x
x [ 10, 10] by [ 4, 4] 0. 23. 14. y (x 15. y (x
2)2 sin ( 1
3 sin
2 16. y 4 1
5 5
2)(1)
(x 1
1 3x) y tan x cos x 18. y 1
x 2 (e x 2 1
x )(2x x 1) 2 (2x y2 2 x 1 x y) x sin (xy) 1
x 2yy y sin (xy) 0 1 xy sin (xy)
2xy x 2 sin (xy) 2y 1
d
x 1/2 x
2
dx
dy/dt
dx/dt d
(0)
dx ln x] 1
x 2 cos t
sin t x x
x cot t 2x
x
1 x 1
y ln[(cos x)x]
x ln (cos x) 1 dy
y dx
dy
dx x 1
( sin x)
cos x ln cos x x sin x
cos x
x sin x
x
(cos x) ln (cos x)
cos x y ln (cos x) (cos x)x 1 [cos x ln (cos x) x sin x] sin x
27. By the Fundamental theorem of Calculus,
y
1 x 3.
cos t cos (x 2) 2x 1)e x tan x2 ln y 28. y 1 cos x)(1 2 cos x)
sin 2 x 1 x2 2x (2x) 2 cot 2 x) cos x 2 cos2 x
sin2 x d
dx 2 x ln y 3x (sin x)(1 cos 2 x)
cos 2 x 17. y dy
dx ( 3) sin x
cos 2 x 1 1 2 x 2 2)2
1/2 sin 1
x 3
(x cot x csc x cot x csc x (1 sin (xy)(xy 3x sin x sec x csc 2 x 1 d
[cos (xy)
dx 26. 1)(1) 1
3x) (1
2 2 csc 2 x 2 25. 9 3 csc2 x
(1
cos x)4 d
dx 2 22. y 24. y f (0)
0 cot 2 x) while 0, x
(2 12. One possible function is y f (5)
5 cot x csc x 1 1 3 13. 3 csc2 x
(1
cos x)4 1 x→ 2 csc x sin x cos x csc x cot x) 2. 1, the end behavior at both ends is y Vertical: solve 2x 2 )e x 3(1 2 cos x)
(sin 4 x)(1 cos x)3 Since the left and righthand derivatives are not equal,
f is not differentiable at x 1. 11. Horizontal: since as x → 4 csc x cot x 3
(sin 2 x)(1 cos x)4 1 h lim 0: all x 3x cos x)( csc x cot x)
(1 cos x)2 3
(sin x)(1 cos x)4 h→0 x2 2 (1
csc x
1 cos x 2 f (1) 3 (x 3 csc2 x
(csc 2 x
(1 cos x)4 0 Righthand derivative: 10. Solve 4 3 (1 1 h2
h h→0 h)
h e xx e 3 csc 2 x
(1
(1 cos x)4 1 h h→0 f (1 4x (1 2(1 lim h→0 x2 1 1 (e) No.
Lefthand derivative: lim x2 x 1 (c) 1 [from (a) and (b)] lim 3x 1 2x tan y 2x sin (x 2) 2 sin (2x) cos (2x); cot 2 x) Cumulative Review 29. d2
(y
dx 2yy 2y y (1 3)( 1) (2)(3)
(1 3)2 a 3t 2 12t 6t (b) Solve v
t 3. 3 12 36. At t
1)(t 3) 0; t
t 1 or 4 r :r dy
dx 5 32. For x
dy
dx 1, y
6(1) (a) y 1 (b) y 1 12(1) 33. For x 3 dy
dx ,y 3 sin 3 3 x 1
34. (2x)
4 1, y (a) y
y
(b) y
or y 3
2 33
2
3
x
2
33
2
2 3 2
2 3 3
6 (x (x x
2 j and j 2i 2j, so that 1 (x 0.707x 1 1.414x 1 2) or 2 2x 1 2(3) C2 C1
x
2x 4, x
5, x 3
, choose C1, C2 so that
3
1. 3
.
3 . 0, 2 6
x
2
2y
9 (b) x
2.458x 9x
.
4y 2 (d) Absolute maximum of 2 at x
absolute minimum of 0 at x 0:
2, 2, 3 39. According to the Mean Value Theorem the driver’s speed at
3
2 some time was
. 1) or
0.866x 2.050 0 (c) x 1.155x
3 2 [ 3, 6] by [ 1, 5] 3.464 1) 5 x 0.950 or 3 3 3
3 1
2 0.407x , the slope is y 2 2i or 6 0; y 3 4 j 2) or y C1, x
C2, x f (x) 3 3 3 1
(2yy )
9 x 3 x 3 4 sec 2 i 4 3 and
6 x 6 tan x
2x 38. (a) x 3 y 6 3 6
6 i 4 37. With f (x) 1 3
2 3 3 6 (b) y 2x 3 6 3 1.443 2 1
x
2 3 cos 3 6 y At x cos 1.155x 2(x 2 1) or y 3 (a) y 1 y 1) or y 1
(x
2 1) or 2 4 2(x 3.464 2.
1 1 and 2 0.866x 2 (a) y 2, and at that instant
12(2) 9
3 m/sec . 3 tan 2 3
2 1) or (x sec 4 (x 3
5
2 sec , and 3 2 x (b) y 0 at t
3(2)2 2 3 (c) Right: v 0 for 0 t 1, 3
left: v 0 for 1 t 3
(d) a
v x 33
2
2 9 0 for t: 3(t 3
2 2 (b) y 3
2 3 cos ( /3)
2 sin ( /3) y 1 3 1, y 33
2 y
dx
dt
dv
dt 31. (a) v :x dy/dt
dx/dt (a) y sec x tan 2 x) 2y sec x tan x
(2y 2)2
2
3
2) (sec x sec x tan 2 x) 2 sec 2 x tan 2 x
(2y 2)3 v)u
uv
x0
(1 v)2 3 dy
dx sec x tan x, (2y (1 35. At t sec x tan x
,
2y 2
(2y 2)(sec 3 x y 30. d
(sec x)
dx 2y) 443 1.443 111
1.5 74 mph. 444 Cumulative Review 40. (a) Increasing in [ 0.7, 2] (where f
0), decreasing in
[ 2, 0.7] (where f
0), and has a local minimum
at x
0.7.
2x 2 (b) y 3x (e) Local (and absolute) maximum of approximately
3.079 at 3 2 x 6
3 2 and x 6
3 ; local (and absolute) minimum of 0 at x
at x
(f) 0 and 2 ( 1.042, 1.853) [ 3, 3] by [ 15, 10] 23
x
3 (c) f (x)
f (0)
41. f (x)
f (x) x2
x2 1: f (x)
3x
3x 43. (a) f has an absolute maximum at x
minimum at x 3. 32
x
3x C; choose C so that
2
23
32
x
x
3x 1.
3
2 cos x
cos x C; choose C so that f (0)
1. (b) f has a point of inflection at x 1 and an absolute
2. 1
(x 1)2
2
7
x2
2 2:
(c) The function f (x) 3, 1 x 2 3
,
2 2 x 3 42.
is one example of a function with the given properties. [ 2.35, 2.35] by [ 0.5, 3.5] f (x) is defined on [ 2, 2].
f (x) 2x 4 x x2 4 f (x) 0 for x to find x The graph of y 0, x [ 3.7, 5.7] by [ 3, 5] 3x 3 8x x3 2 4 x2 ; solve
44. y = 2 1 26
.
3 A(x) 4x f (x) is shown. x2
, and the area of the rectangle for x
16
x2
1
x 16 x 2.
16
x2
2(8 x 2)
2 16 A (x) x x2 16 A (x) (b) 2) 2 by 4 [ 2.35, 2.35] by [ 10, 10] 2,
2 possible area is A(2 (a) 0 when x = 26
26
, 0,
3
3 45. f 26
26
,0 ,
,2
3
3 equation is y Use NDER to plot f (x) and find that f (x)
x
1.042. 0 for y 2x 2 and y , and so 2. The maximum 8, with dimensions 2.
2 and f 4 x2 16 0 is sec 4 2
4 2x
2 4
4 1.414x tan 4 2. The or
0.303 46. V s 3
dV 3s 2 ds
Since ds 0.01s, the error of the volume calculation is
approximately dV 3s 2(0.01s) 0.03s 3 0.03V, or 3%.
[ 2.35, 2.35] by [ 15, 5] (c) Approximately ( 1.042, 1.042)
(d) Approximately ( 2, 1.042), (1.042, 2) Cumulative Review
47. Let s be the rope length remaining and x be the horizontal
distance from the dock.
s2 (a) x 52, ds
dt 1.5, and which means that for s
dx
dt speed s 8 s2 s 1.9 ft/sec ds
, which for
25 dt ( 1.5) 64 1 dx
x(ln x)2
2e Therefore, 0.15 rad/sec. e 1 e (3 0.179 in./min. t 2)i 2 V dV
h , so
dt
12
4
( 9)
25 1
1 1
ln 2 h 4
36
25 ex 57. Let u x 2 e cot (e (b) (1)(1.8
1 … 6.4 6.4
0 50. x dx x dx 2 1
2 165 in.
12
x
2 x dx 2 2 0) 1
0 0 12
x
2 x x
2 x dx
( 2 1
0 ) 4 x 2 2 sin 1 x
2 2, 2
2 2. Alternately, observe that the region under the curve and
above the xaxis is a semicircle of radius 2, so the area is
1
(2)2
2
3 52. 1 x2
ln 3
/4 53. 0 1. 2 1) dx cot u du
(csc2 u Since 1) du
u 1 C
(e x 1) 1) 2.
3 1
13
dx
x
x
3
26
9.765
3 sec2 x dx ln x 9
1 /4 tan x 1
0 ln 3 C C is an arbitrary constant, we may redefine C
cot (e x ex 1) ds
s
, so du
.
2
2
ds
1
ds
ds
2 2[(s/2)2
s2 4
4(s/2)2 4
1
1
s
tan 1 u C
tan 1
C
2
2
2 59. Let u 2 4 2 (ln 3)j C. 1
2 1] 2.5 51. Using Number 29 in the Table of Integrals, with a
2 2i 58. Let u 165 in. … 12.6 0.409 2i csc2 u and write the solution as 16.2) ln 2
ln 2 1 cot (e x the rate of about 0.458 in./min.
1.8 C. e x dx. 1, so du cot u dh
is negative, the level in the cone is falling at
dt 49. (a) (1)(0 1
ln x C (ln 3)j 0.458 in./min.
Since 1
3 5 3 (ln t)j dh
dh
and
dt
dt 2 1
u du Use the identity cot2 u
V be the volume of the coffee in the cone.
3 12
1 1 (b) Now let h be the level of the coffee in the cone, and let 1 h2
h
32
dV
4
h 2 dt x 1
j dt
t 2t)i (3t 56.
9
16 x dx
x(ln x)2 2e 1
ln x 25 dV/dt
16 4 1
dx.
x u the volume of the coffee in the pot.
dh
V
, so
dt
16 1 dx
x ln x, so du 48. (a) Let h be the level of the coffee in the pot, and let V be h 4 2 dx x Then
5 4 x 7 25 5
8 ds
,
25 dt 2 1 55. Let u ( 1.5) 64 8 ft becomes s
s2 8 ft, d
s
sec 1
, so
dt
5 (b) dx
dt 4 54. 445 sin (x
cos3 (x cos (x
3)
dx
3) 1
2 cos2 (x 3) 3) , so du
(u
C 3 ) du sin (x
12
u
2 3) dx.
C du
u2 1 446 Cumulative Review 60. Use integration by parts.
u e x du dv e
x e 65. Use integration by parts. x dx 1
sin 2x
2 v du 1x
e sin 2x
2 cos 2x dx 1x
e sin 2x dx
2 e 1
sin 2x dx
2
1
cos 2x
4 dv
x u cos x x 2 cos x x dx v du dv x 2x cos x dx 2 cos x dx dx v x 2 sin x dx e Then 2 sin x x 2 cos x 2x sin x 2 x cos x cos 2x dx 1x
e sin 2x
2 1x
e cos 2x
4 1
e x cos 2x dx
4 2x sin x x 2) cos x (2 so 2 sin x dx
2 cos x 2x sin x C C The graph of the slope field of the differential equation
x e
61. v 2x dx sin x dx Now let x du e dv x 2 sin x dx Now let
u x2 u cos 2x dx x
x2 x cos 2x dx 2
5x 6 2 Solving A x2
1)(x 6) 6)
B (2 sin 2x 5 (x A (x x e B (x
1, B cos 2x)
A
x 1)
6A dy
dx C 1 x x 2) cos x (2 2x sin x is shown below. B (A x 2 sin x and the antiderivative y 6 B) x (B 6A) 1
,B
7 2 yields A 2
8
1
. Then
5x 6
7(x 6)
7(x 1)
x2
8
1
dx
dx
x 2 5x 6
7(x 6)
7(x 1)
8
1
1
(x 6)8
ln x 6
ln x 1
C
ln
7
7
7
x1 8
so
7 x x2 5
[3
2 62. Area
Volume 2(8.3) 25 359 2(9.9) … 2(8.3) [ 5, 5] by [ 10, 10] 66. Use integration by parts.
C 3] u 359; x du dv
v ex xe x ex dx x e x dx 8975 ft 3 e x dx xe x e x dx C e x(x 1) C Confirm by differentiation:
63. y
C (t 1) 7
and y
2 1 1 2t
e
C; y (0)
2
1 2t
7
1
e
.
2
2
t1 1
2 1 C 2, so dx
[e (x
dx 67. (a) y
64. y
y
y
⇒y 1
cos 2
sin
2
1
cos 2
sin
2
1
sin 2
cos
4
1
sin 2
cos
4 C1, and y
1
.
2
1
2 1) ex C] (x Ce kt, with 6,000 1)e x xe x Ce k(2) and 10,000 Ce k(5). 0⇒ 2 Then 10,000
6,000 e k(5 2) , so 5
3 e 3k and therefore 5 C2, and y
1
2 4 2 0 k ln 3
3 Furthermore, C 0.170.
6,000
e 2k of bacteria is given by y
(b) About 4268 4268. The approximate number
4268e 0.170t. Cumulative Review
68. Let t be the time in minutes where t 0 represents right 72. The region has four congruent portions, so
/2 Area now, and let T(t) be the number of degrees above room 4 sin 2x dx 1
cos 2x
2 4 0 temperature. Then we may write T(t)
T(0) 50 and T( 15)
13
1
ln
10
15 k kt where 50 and kt gives t ln 0.1
k 2 74. Solve y dy
y
0.08y 1
dx
500
500 dy
0.08 dx
y(500 y)
(500 y) y
dy 0.08 dx
y(500 y)
1
1
dy 0.08 dx
y
500 y ln y 9
2 0 y
500 y 0.08x y (1
y
70. C2e dy
dx
dy
y4
dy
y4 ) 500C2e 2 cos 1 2 cos 3
2 cos 4)(x 3) (x 3) dx 77. Solve 4x
x x3 2
dx
2 x2 0 or x ln y 4 x2
2 y 4 e C1e(x y Ce(x 3x 2 2 /2) 3x /2) 3x 0 17
x
47 1 14 1 0.224 4 14
x
4 4 2 0 (4x 2 x 3) dx 128
3 0 134.041. 4
4 2 x 2) dx 2 x(4x 43
x
3 2 d 0 to find the limit of integration: 0 C1 d 4. By the cylindrical shell method, Volume 3) dx cos 2
2 42.412 4 (x 16.039. cos )2 d 9(1 1
sin 2
4 0) 1 0 1 27
2 1 2 1
cos 2
2 2 sin 76. Volume (y 3) dy cos2 ) d (1 9
(3
2 0.08x 500
Ce 0.08x 1 0 93
22 C2e 0.08x 1
2 r2 d 2 2 . Then (y 2 2) 2 Integrate both sides.
C1 2 21)/2
2 1
2 0 21 21)/2 [(y 9
2 0.08x 1 0⇒y (1 75. Area 2x 2) dx (8 2 to find the integration limits: Area 0 y y 5 y 9
2 ln 500 3
(1 y2 3)] dx 64
3 2 131.6 minutes, or about 2 hours and 12 minutes from now. 2 (x 2 2 2 23
x
3
2 4
0 3 to find the integration limits: x 2) [(5 8x /2 2. Then 2 6.13°C above room temperature.
50e x2 8⇒x 2x 2
Area (b) Solving 5 69. x2 73. Solve 5 0.0175. k(120) (a) 50e 65, giving T0 T0e 78. The average value is the integral divided by the interval
length. Using NINT, 71. Use EULERT.
x y 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0 0
0.1
0.2095
0.3285
0.4568
0.5946
0.7418
0.8986
1.0649
1.2411
1.4273 447 1 average value sin x dx 0.763 0 79. y sec 2 x, so we may use NINT to obtain
/4 Length (sec 2 x)2 dx 1 2.556. /4 80. dx
dt cos t and dy
dt 1 sin t, so we may use NINT to obtain
/2 Length cos 2 t (1 sin t)2 dt /2
/2 2
/2 2 sin t dt 4. 448 Cumulative Review 81. Using NINT, 88. (a) The work required to raise a thin disk at height y from
dr 2 2 Length d 0 dy
82.
dx 2 83. dx
dt 2 1 d 6.110. the bottom is 0 e x/2 4 cos t and
/2 Area x e
dy
dt 15 15 y 2(12 1 (y 22,500 sin t, so 3 y) dy 2 12y ) dy 14
y
4 15 2 cos t 2 sin t 2 2 (1 4y 3 10
0 70,686 ftlb. 22,500
275 (b) y). 0 10 8.423. 257 sec 4 min, 17 sec sin t) dt /2 0 Total work
0 dx 2 sin t
0 10 y2
dy (12
2 60 (weight)(distance) 1 x/2
e
, so we may use NINT to obtain
2
2
e x/2 2
2 e x/2 1
dx
2
0 Area
0 d 2 sin t dt 3.470. 89. The sideways force exerted by a thin disk at depth y is its
edge area times the pressure, or dr
84.
d 1, so we may use NINT to obtain Area 2 2 sin +1d (2 dy)(849y) 1698 y dy.
H 32.683. Total force /2 849 H 2, where H is depth. 1698 y dy
0 1 x2 2 x 85. Volume 2 0
1
0 4 12
x
42
9
280 dx 2x 5/2 (x 849 H 2 Solve: 40,000 40,000
and V
849 ⇒H 12.166 ft 3. H x 4) dx 4 7/2
x
7 15
x
5 ln x 90. Use l’Hôpital’s Rule: lim 1 x→ 0 f (x) 0.101. 1
x lim x 1
t2 (2 tan x) dx 2
0 lim /4 8 tan x dx 0 2 (sec x 0 tan x
1 b x lim b→ 8 4 2 2 93. 2 3 dt
t2 3 5.394. dx
x
x e
Since lim (ln b
0 e dx b→ F kx ⇒ 200
300 N, x k (0.8) ⇒ k
300
250 1.2 (b) Work 250x dx
0 125x 2 x dx 2 e x dx 0 x e dx 0 lim e b→ x b 1, the
0 original integral converges. 250 N/m, so for 1.2 m. . Both integrals diverge. e lim 1
, and
x 1
ln x 0 b x 2, ln 2) e x dx 0 87. (a) F converges. 4 b→ dx ), g(t) dt converges, we conclude that 92. Use the comparison test: for x 0 8 1, and 1) dx
/4 8 f (t)
g(t) f (t) dt 3 /4 8 t→ 2 and 4 1
. Since f and g are both continuous on [3,
t2 g(t)
2 0.
x x. 91. Use the limit comparison test with f (t) Volume x→ 2 ln x grows slower than g (x) 86. Use the region’s symmetry:
/4 2 lim 1/2 x→ x 1
0 1.2 180 J b 4r dr 94. 1 r b→1 0 1 b2 lim ( 4 b→1 0 10 95.
0 1 dx
1 x 0 1 Since 0 1 4) 1 x r lim b→1 2 4 1 x dx
1x
b
dx
x
01 1 lim b→1 b lim b→1 diverges. ln (1 x) , the original integral
0 r2 b
0 4. The integral converges. 10 dx dx
1 4r dr lim 2 449 Cumulative Review
2 1 dx 96. 3 0 x 0 1
b 3 0 b→1 1 1 lim x a→1 1 1) x x b 2/3 1 1 3
(x
2 lim a→1 0 x6
. By the Alternating
6!
x7
1
Series Estimation Theorem, error
0.001.
7!
7! 1 dx
3 a 101. The first six terms of the Maclaurin series are dx
3 2 3
(x
2 lim x dx lim b→1 2 dx
3 2 1)2/3 102. f (0) 1 x for 1 … ( 1)nx n 2x 1 1
2 1
,
3 1)2/3 3(0 2
,
9
2 5 … (3n
( 1)n 1
3n
2 1)2/3 9(0 … f (n)(0) 2x 4x 2 … 8x 3 ( 1)n2nx n 4) , so the Taylor series is … 1
x
3 2584
253
x
x
4! 34
3! 33
2 5 … (3n 4) n …
( 1)n 1
x
.
n! 3n 1 1, so the interval of convergence is 1
.
2 x 98. (a) 2
x2
2! 32 … Since by the Ratio Test cos t 2 (t 2)2
(t 2)4
(t 2)6 …
(t 2)2n
( 1)n
2!
4!
6!
(2n)!
4n
t 12
t4
t8
nt
… ( 1)
6!
2!
4!
(2n)! 1
1 …
lim n→ lim 9 x
5(2!) 13 x
9(4!) … x
13(6!) ( 1)n x 4n 1 (4n 2 5 … (3n 4)(3n
(n 1)!3n 1
(3n 1)x
(n 1)(3) n→ yields
5 an 1
an lim Integrating each term with respect to t from 0 to x x x5
5! 1. Substituting 2x for x yields 1 2x for … x3 x 1
1 x2 x x4
4! 1 1, f (0) f (0) 97. We know that
1 x3
3! 0.
a The whole integral converges. 1 x2
2! x …. 1)(2n)! n→ x 103. Using the Ratio Test, lim ; Since the cosine series converges for all 4)x n an 1
an 2 lim n1 n→ 3 3n
2 1
, so
3 the series converges. real numbers, so does the integrated series, by, the
104. Note that an termbyterm integration theorem (Section 9.1, n!3n
2 5 … (3n 1 x , the radius of convergence is 1. n→ (b) 1)x n Test, since
n1 Theorem 2). 1
for every n. By the Direct Comparison
n
1
2
diverges, so does
.
n
n1
n 105. Use the alternating series test.
99. ln (2 2x) ln [2(x
2 ln 2 x
2 x 1)]
3 x
3 ln 2 4 … x
4 an 1
Since by the Ratio Test lim
an
n→ ln (x
( 1)n
lim n→ 1)
1x n n
n n … 1 x x , the Note that ∑ n0 ( 1)n
n1 ∑( n0 1)nun, where un Since each un is positive, un
lim un n→ un 1 1
n 1 . for all n, and 0, the original series converges. series
converges for n1 1 x ( 1)n
converges, but
n
n 100. Let f (x) sin x. Then f (x)
cos x, f (4)(x) f (x) 106. Using the Ratio Test, 1.
1
does not.
n 1 cos x, f (x) sin x, and so on. At x lim n→ sin x,
2 the sine terms are zero and the cosine terms alternate between
1 and
(x 1, so the Taylor series is
2 )3
(x 2 )5
3!
5!
(x 2 )2n 1
….
( 1)n
(2n 1)! 2) (x … an 1
an lim n→ series converges. 3n 1
(n 1)! n!
3n lim n→ 3
n 1 0, and the 450 Cumulative Review 107. (a) Using the Ratio Test,
a lim n→ n1 n1 (x lim n→ an n 2)
1 n 3 x x 2)n (x means that the series converges for
or dr
dt
dv
dt 112. (a) v(t) 1 x 1. Furthermore, at x a(t) 2 , which
2 1, 3 x 3 x (c) At x 2 x n→ lim n→ x (n
(n nx ln 2 (n)
1) ln 2 (n n→ n 2t. 2t and 50 16t 2 100(sin 45 )t
13 130, we have t
2 13 and so 52
13 y n ln2 n
xn to easily clear the 35ft tree. 1) 1 50 2 2 75.92 ft, high enough 52 114. Since r cos
x, r sin
y, the Cartesian equation is
x y 2. The graph is a line with slope 1 and
yintercept 2. 1) that the series converges for 16
5 lim 1 50 4. an 1
an 2
ln n
1)
n→ ln (n n lim 16t 2 When x xn 1
1)ln2 (n 2 sin t dt 100(cos 45 )t y 1 n→ sin t)2 dt 113. Yes. The path of the ball is given by 1 and the radius of 108. (a) Using the Ratio Test, lim (1 /2 1 lim ( cos t)2 /2
3 /2 convergence is 1.
(b) 3 /2 /2 convergence is sin t) j (cos t)j v(t) dt n1 ∑ (sin t)i 3 /2 3, the series 1
, which diverges, and at x
1, the series is
n
( 1)n
, which converges. The interval of
n (1 (b) Using NINT, the distance traveled is is ∑
n1 ( cos t)i x , which means
x 1. At x 115. 1, the series converges by the Integral Test: 2 1
dx
x(ln x)2 b lim b→ 1
ln b lim b→ 2 1
ln 2 1
dx
x(ln x)2
1
ln 2 convergence interval is 1
ln x [ 3, 3] by [ 0.5, 3.5] b The shortest possible interval has length 2 . 2 116. x . So the 1 x 1 and the radius of convergence is 1.
(b) 1 x r cos y r sin dy
dx 1 cos 2 , cos sin
cos
sin dy/d
dx/d sin cos sin 2
cos 2
2 sin cos (c) Nowhere
109. 1
22 110. 1 cos 2, ( 3)2
3 2 3 13 , 1 sin 3 Zeros of
cos 3
13 1
3
,
22 dy
:
d sin 2 cos , 1 (2 cos
111. dy
dx t 3 /4 dy/dt
dx/dt t 3 /4 tangent vectors are 4
,
5 3 sin t
4 cos t t 3 /4 1
32 42 4, 3 3
3
. The normal vectors are ,
5
5 3
. The
4 4
and
5 sin
34
,.
55 sin 0
1) 2
,
3 0, 0 4
, or
3 2 dx
:
d 2 sin cos , 0
1
2 0 or cos
0, 0 2 cos2
1)(cos Zeros of 43
, and
55 cos 2 3 or 5
,
3 2 451 Appendix A2
dx
0 at
d
2
4
dx
dy
and at
, and vertical tangents
0,
0
3
3
d
d
5
at
, at
, and at
.
3
3
dy
0
For
0 (or 2 ),
becomes and l’Hôpital’s Rule
dx
0 There are horizontal tangents dy
d 0, leads to
dy
dx s Appendix A2
(pp. 581–584)
1. Step 1: The formula holds for n
…x
x
Step 2: Suppose x
1 2 x2
x1 x1 k …
x2 Then, x1 1, because x1 xk
… x2 … x1 .
xk . 1 xk xk 1 by the triangle inequality.
sin (0) 4 sin (0) cos (0)
cos (0) 2 cos 2 (0) 2 sin 2 (0) 0 0, so this is So, by the transitivity of ,
x1
x1 x2 … xk 1 … x2 xk 1 . another horizontal tangent line. The mathematical induction principle now guarantees the Horizontal tangents: original formula for all n. At 0 or 2 , we have r coordinates are (0, 0), so the tangent is y
2
, we have r
3 At
are 33 3
,
, so the tangent is
44 3
4 3
,
4 3 3 , so 4 1 rk 1 rk 1 rk rk 1
. Then
1r rk 1
rk 1
1r
r)
1 rk 2
.
1r
1 1 r k 1(1
r 3. Step 1: The formula holds for n x kx k 1 1, because d
(x)
dx 1. kx k 1. Then
x k) x dk
(x )
dx xk kx k xk xk d
(x)
dx 1)x k. (k The mathematical induction principle now guarantees the
original formula for any positive integer n. 1
1
cos , sin
2
32
3 At 1
3
,
, so the tangent is x
44 , we have r (2 cos , 2 sin ) ( 2, 0), so the tangent is x 1
51
5
cos , sin
2
32
3 x 1
. In summary, the horizontal tangents are y
4
3
4 , and y 2 and x 1
.
4 2. 1
and the Cartesian coordinates
2 are 3 1
.
4 2 and the Cartesian coordinates are 5
, we have r
3 At x … r2 r dk
(x )
dx
d k1
d
(x )
(x
dx
dx are y 1 r2
.
r 1
1 … r2 r Step 2: Suppose . 1
and the Cartesian coordinates
2 , we have r 3 Step 2: Suppose 1 r) original formula for every positive integer n. Vertical tangents:
At r)(1
1r 1, because The mathematical induction principle now guarantees the
3
and the Cartesian
2 3
43
4
cos , sin
2
32
3
3 (1 r 1 33
.
y
4
4
, we again have r
At
3 the tangent is y 1 0. 3
and the Cartesian coordinates
2 3
23
2
cos , sin
2
32
3 coordinates are 2. Step 1: The formula holds for n 0 and the Cartesian 1
,
4 3 3
4 3
4 , so the tangent is
0, , and the vertical tangents are 4. Step 1: The formula holds for n 1, because f (x1) f (x1).
Step 2: Suppose
f (x1x2 … xk) f (x1) f (x2) … f (xk).
Then by the given property,
f (x1x2 … xk 1) f (x1x2 … xk) f (xk 1)
f (x1) f (x2) … f (xk 1).
The mathematical induction principle now guarantees the
original formula for every positive integer n.
5. Step 1: The formula holds for n
Step 2: Suppose 2
31 2
32 2
32 … 1 2
31 32
3k 1 1, because …
2
3k 1 1 2
3k
1
3k 1
1
3k 1 2
3 1 1
.
3 1
. Then
3k
2
3k 1 1. The mathematical induction principle now guarantees the
original formula for all positive integers n. 452 Appendix A2 6. Experiment: 9. Step 1: The formula holds for n
5 6 1(1 12 n 1 2 3 4 7 n! 1 2 6 24 120 720 5040 n3 1 8 27 64 125 216 1/2)(1
3 … k(k k 3. Then (k Step 2: Suppose k!
4, k 1), and so k (k 1)k of 3 , (k 2
k 1
3 1)! 2
1
(since
k
k2 (9/2)k 2 k 2 2k 1 2 and 1)2 (13/2)k 1)(k 3/2)(k
3 (k The mathematical induction principle now guarantees the
original inequality for all n . 1)2 3(k (k 1)3. (k (k 1)
3 1) . So 1) , and thus by the transitivity 1)! 1) 3 3 3 (k 1) 2 (k 1/2)(k
3 1)2 (k 1)k 3. (k k(k k2 1/2)(k
3
1/2)(k k3 6, because 216. 1
k2 … k(k Step 1: The inequality holds for n For k 22 1. 22 Then 12
720 1) Step 2: Suppose 12 343 1, because 1)[(k 1) 2) 1/2][(k
3 1) 1] . The mathematical induction principle now 6. guarantees the original formula for all positive 7. Experiment: integers n. n 1 2 3 4 5 6 2n 2 4 8 16 32 64 n2 1 4 9 16 25 36 10. Step 1: The formula holds for n Step 1: The inequality holds for n
Step 2: Suppose 2k
since 1
k k 2. For k 1 3, k 1 , and so k2 2k 2 2k 25. Step 2: Suppose 13 1
k ,2 2k k Then 13 2k 1 (k by the transitivity of 1
. Then 2k 1
8 3, because 2
2 2k and the fact that 1
4 3 2
8 3. 1
.
8 1
, and
4
1 k1
1
,2
.
8
8 The mathematical induction principle now guarantees the
original inequality for n 1)2 k2
4
(k k 2 1)3
1)(k 1 (k
2) 2 1)2 1)2 (k 1)(k
2 . 1)3 (k 4 1) 2 k (k k3 (k 1) 2 k 2 (k 1)2. 5. 8. Step 1: The inequality holds for n
Step 2: Suppose 2k … 23
2 The mathematical induction principle now guarantees the
original inequality for all n … 23 1. Then by the 2k + 1. And thus k2 1. 2 k(k transitivity of
2k 5, because 32 1) 2 1(1 13 1, because (k 2)2
4 (k 1)2 . The mathematical induction principle now guarantees the
original formula for all positive integers n. 11. (a) Step 1: The formula holds for n
1 ∑ ( ak k1 1 ak) 1 k1 k1
i ∑ ak ∑ bk i Step 2: Suppose ∑ (ak bk ) k1 i1 ∑ ( ak k1 1, because
a1 b1 . i ∑ ak ∑ bk. Then k1 k1 i bk) i ∑ ak ∑ ( ak k1
i ∑ bk k1
i1 k1 ai 1 (ai
bi 1 bi 1) 1 i1 k1 bk) k1 ∑ ak ∑ bk. The mathematical induction principle now guarantees
the original formula for every positive integer n. Appendix A3
(b) Step 1: The formula holds for n
1 1 ∑ ( ak k1 k1 (pp. 584–592) k1 b1 . 1. k1 a c b 4
9 i 1
2 4
7 x ∑ ak ∑ bk. Then bk ) i1 a1 k1
i i Step 2: Suppose ∑ (ak s Appendix A3 1, because
1 ∑ ak ∑ bk bk) 453 k1 i ∑ ( ak ∑ ( ak bk) k1 bk) k1
i i ∑ ak ∑ bk k1
i1 ai k1 1 bi 1) 1 bi 1
2 Step 1: x ⇒ k1 ∑ ak ∑ bk. 1
2 Step 2: The value of The mathematical induction principle now guarantees ⇒ 4
9 1
2 x 1
x
2
4
⇒
9 ⇒ 1 i1 k1 (ai 1
2
1
2 1
, or
18 4
⇒
7
1
2 which assures x
4
is the smaller value,
7 x 1
.
14 1
.
18 the original formula for every positive integer n.
2. a c b 2.7591 3 3.2391 3 ⇒ x (c) Step 1: The formula holds for n
1 1, because 1 ∑ cak ∑ ak c k1 Step 1: x ca1. k1
i i Step 2: Suppose ∑ cak
k1 k1 k1 i 1 i ∑ cak cak 1 1 ∑ ak c ak k1 c ∑ ak k1 cak k1
i1 i c ⇒ i1 ∑ ak. Then ∑ cak c ∑ ak k1 1 3 x 3 2.7591 ⇒ 0.2409, 3 3.2391 ⇒ 0.2391. or The value of . 3 3 Step 2: 1 x ⇒ 2.7591 which assures x
x 3 3.2391 is the smaller value, The mathematical induction principle how guarantees
0.2391.
the original formula for every positive integer n.
3. Step 1: x n (d) Step 1: The formula ∑ c n c holds for n 1, ⇒ 3 ⇒ 3 x 3 Step 2: From the graph, 3 k1 3 x because
1 ∑c 1c k1 c.
or i Step 2: Suppose ∑ c ∑c ic c (i 4. Step 1: x 1) c. k1 the original formula for every positive integer n.
12. Step 1: The formula holds for n ⇒ ( 1) ⇒ The mathematical induction principle now guarantees 1 x 0.39. 0.77, or 1 (and every real 9
25 x 1 1 Step 2: From the graph, number x), because
x1 0.41; thus 0.39, i c. Then k1 i1 x1 3.41 ⇒ 3 2.61 ⇒ 16
7
⇒
9
9
16
9
⇒
0.36; thus
25
25 1 1
0.36. The mathematical induction principle now guarantees the 5. Step 1: (2x 2) ( 6)
0.02 ⇒ 2x 4
0.02
⇒ 0.02 2x 4 0.02
⇒ 4.02 2x
3.98 ⇒ 2.01 x
1.99
Step 2: x ( 2)
⇒
x2
⇒
2x
2⇒
0.01. original formula for every positive integer n (and every real 6. Step 1: Step 2: Suppose x
xk number x). 1 x.
k xk x x k k x . Then
xk x xk x x k 1. x11
0.1 ⇒ 0.1
⇒ 0.9
x 1 1.1 ⇒ 0.81
⇒ 0.19 x 0.21
Step 2: x 0
⇒
x
⇒ x 1 1 0.1
x 1 1.21
0.19. 454 Appendix A3 19 x 3
1⇒ 1
19 x 3 1
19 x 4 ⇒ 4 19 x 16
⇒2
⇒ 4 x 19
16 ⇒ 15 x 3
or 3 x 15
Step 2: x 10
⇒
x 10
⇒
10 x
10.
Then
10 3 ⇒
7, or
10 15
⇒
5; thus
5. 7. Step 1: 8. Step 1: x 2 4
0.5 ⇒ 0.5 x 2 4 0.5
2
⇒ 3.5 x
4.5 ⇒ 3.5
x
4.5
⇒
4.5 x
3.5, for x near 2.
Step 2: x ( 2)
⇒
x2
⇒
2x
2.
Then
2
4.5
⇒
4.5 2 0.1213,
or
2
3.5
2
3.5 0.1292;
thus
4.5 2 0.121.
9. (a) lim x 2 x→ 5 6x 5
x5 4, x lim x→ 5 lim (x x→ 5 4.05 x 1 5.05 Step 2: x x 5 lim f (x) lim (6x 4) x→1 1: (4 ⇒ sin 1 ⇒1 1 ⇒ sin
1 sin ⇒ sin 1 Choose 0.5 ⇒ 2 2
0.5 4) 0.5 ⇒ 6x 0.5 ⇒ 3
⇒
4
1
. Choose
12 0.05. x
1 2.
2x 0.5 6 0.5 13
.
12 6 1 .
1
, or 1
4
1
.
12 1 1 (sin 1 (sin 1 (sin 1 (sin 1 0.01)
0.01) 0.0182, or 0.01)
0.01) 1 0.0188. 0.018.
sin x, y2 sin 1 1, and 1. The curve intersects the lines at 0.98175 1 0.01825 and at 1.01878 1 0.01878. We may choose [0.9, 1.1] by [0.78, 0.88] 3
;
4 2 6 x 0.05, or x→1 2 ⇒x sin 1 0.01) .
1 sin
1 1 (sin 1 1 0.01 x x x Then 1 sin 1 0.01) ⇒ 1 0.01 sin x (sin 1
sin Step 2: x sin 1 0.01 5 2, so lim f (x) 2 1: (6x ⇒1 ⇒ sin 1 0.018. 2 and 2x) ⇒x 1 sin x 5. 0.05; thus 2x) Step 2: x 0.01 0.01 x 5 5.05 ⇒ lim (4 x sin 1 5. 4.95 ⇒ x→1 (b) Step 1: x Then 1 x x
5 5 x→1 0.841 x
0.05 3.95, x ⇒ Then x→1 ⇒ y3 4 4.95, x ( 5) ⇒ (b) Step 1: sin x 1) 0.05 5)(x 1)
x5 ⇒ 10. (a) lim f (x) ( 4)
(x 0.05 ⇒ sin 1 x→1 Alternately, graph y1 6x 5
x5 ⇒ ⇒ 5)(x 1)
x5 5.
x2 (b) Step 1: (x 11. (a) lim (sin x) 13
12 455 Appendix A3
1
( 1)2 4 x 12. (a) lim 2
x→ 1 x (b) Step 1: 4 x2 4 ⇒ 0.1
x2
x 4 13
7
⇒
30
30
x2 4
28
13 2
72
⇒x
x
x
30
30
30
28
7
Then x 2 x
0
30
30 15 421 15 x 421 15 1.1551 13 3.4628. Thus
Step 2: x 1 or
⇒ 1 x
x 1
3 421
7 ⇒ . ⇒ 1 901 4 , y2 1
3 0.78833 1
1 3 1 1 , 1. , 1 1}, 3 0.15513 and at 1 3
3 1 1 for x near 3 1 ⇒ 3 3. 3 3 1 3 ⇒ 3 3 . 3
1
3 1 3 3 3 3 x2 3 , x x 3 3 ⇒
3 ⇒ 3 Then 3 x 3 1
3 3 x 3 ⇒ 0.1, and 1
3
1 3 1 Step 2: x 1 3 , 3. 3 Choose
min
15. (a) (5
⇒I 3 3 1 3 , 3
1 3. 3 ⇒ 2 ⇒ 2 2 x ⇒ ⇒ )5
(5, 5 0 ) 0.21167. We may choose
(b) lim
x→5 0.155.
16. (a) 4
⇒I 5 (4
(4 (b) lim )
2 4 x→4 [ 1.5, 0] by [ 0.15, 0.55] 1 1 1
x2 1
x2
1
x2 3 0.1. The curve intersects the lines at 1.15513 1 ⇒ min {1 3 or x ⇒ 1 ⇒ 1
3
1 or 13 x2 1. 1 1
3 ⇒ 0.155. Alternately, graph y1 1. 1 x 1 1
x2 14. Step 1: 1.
15 near x x 1 or 901
13
15 901 28
0.1551,
13
15
421
1
7
22
421
0.2117.
7 Choose y3 x x x Then
⇒ 901 1 that is, the smaller of the two distances. 15 13 1 1 1 ⇒ Choose x ⇒ ( 1) ⇒ 15 x Then 5.0740, and also 7
901 1 ⇒ 0.7883 or 7 x 1 x2 1 1 Step 2: x
52
for x near 1.
30
13 2
52
x
x
,
30
30 which using the quadratic formula implies
x x2 ⇒ 0.1 ⇒ 1 ⇒
1
3 x 0.1 1, x 2 13. Step 1: For x
⇒1 1
3 x 1
3 , 4) x 0 17. If L, c, and k are real numbers and lim f (x)
x→c for any 0, there is a ⇒ k f (x) kL Proof: For any
is a . Therefore, 0 x c .
0, let 0 such that 0
k 0 such that 0 L, show that k x
x c
c . Since lim f (x)
x→c ⇒ f (x)
⇒ k f (x) L, there L
kL . 456 Appendix A5.1 18. If L, M, and c are real numbers and lim f (x) L and
x→c lim g(x) M, show that for any x→c x Proof: f (x) g(x) (L f (x) M g(x) L 0, there is a ⇒ f (x) such that 0 c g(x) M) f (x) f (x) there exist 1 , x c 1 ⇒ f (x) 0 x c 2 ⇒ g(x) min { 1, 2 2
2 M, for f (x) f (x) . g(x) (L 2 M) 2 g( f (c)) x x c f (c) f2(x) x→c L1 L2 f3(x)] lim [ f1(x) x→c ⇒ g( f (x)) c L3 L3, by two applications of the Sum Rule. To generalize:
Step 1 (n 1): lim f1(x) that g( f (x)) g f is continuous at c. Step 2: Suppose lim [ f1(x)
L1 … L2 lim [ f1(x) … h)2 (y
0)2 (y
(y 2)2
y (0, 4) fk(x)] (0, 2) … … fk(x)] Lk 1 –3 (0, 0) 1): lim f1(x) 2. (x
[x
(x L1, as given. x→c x→c y
(0, 8) lim (f1(x)) f2(x) … fk(x)) Lk
1 n factors
lim (anxn an 1xn lim anx x→c
n lim an 1x x→c an lim x n aa 1 cn 1 an c an 1 lim x n x→c cc…c
1 n factors
… ax
1 … n1 x→c x→c
n , by the Product Rule. lim (x x … x) x→c 1 (–1, 5) …L
k x→c 22. lim f (x) 1 …a c
1 …
a0 lim a1x x→c a1 lim x
x→c (0, 2) cn
–5 a0)
lim a0 x→c lim a0 x→c f (c), where in addition to the items given in the problem, the Constant
Multiple Rule was used (to move the coefficients out of the
scope of the limit signs). 10)2 10 lim ( f1(x) f2(x) … fk 1(x)) L1 L2 x h)2 (y k)2 a 2
( 1)]2 (y 5)2 (
1)2 (y 5)2 10 x→c x→c 3 Lk 1, by the Sum Rule. L1 L2 … Lk. Then x→c a2
22 5 Step 2: Suppose lim ( f1(x) f2(x) … fk(x)) 21. lim x n k)2
2)2
4 fk 1(x)] f2(x) L2 20. Step 1 (n 1. (x
(x
x2 Lk. Then lim[ f1(x)
L1 … f2(x) x→c f2(x) x→c (pp. 593–606) L1 as given. x→c g( f (c)) . f (c), so
. But from the . So for any g( f (c)) . f2(x)] 0 0 0 such that s Appendix A5.1
19. lim [ f1(x) 0 there is a 0 such that ⇒ f (x) there is a M x→c ⇒ g( f (x)) f (c) ⇒ g( f (x)) f (c) 0 g(x) lim g(x) The second inequality also holds when f (x) 0 L ⇒ f (x) such that 0 and }. Then ⇒ f (x) c M by the x→c M x g(x) g(x) f (c)
.
g(c) x→ c 24. From the continuity of g, for any continuity of f, there is a L 0 M . such that 2 0 Choose L 0
M) L, lim g(x) x→c any (L L triangle inequality. Since lim f (x) lim f (x) f (x)
x→c g(x) 23. By the Quotient Rule, lim 5 x , which means Appendix A5.1
3. Complete the squares.
x 2 y 2 4x 4y 4 0
x 2 4x 4 y 2 4y 4 4
(x 2)2 (y 2)2 22
Center ( 2, 2); radius 2 13. x2
2 1⇒c y2 vertices are (
14. y y2
4 2 1 ⇒ foci are ( 1, 0); 4 1 5 ⇒ foci are (0, 2, 0) 1⇒c x2 1 457 vertices are (0, 2); asymptotes are y 5); 2x 4 12x ⇒ 4p 15. y 2
2 –4 x –2 2 12 ⇒ p 3; focus is (3, 0), directrix is 3
y x
4 4. Complete the squares.
x 2 y 2 4x 4y 0
x 2 4x 4 y 2 4y 4 8
(x 2)2 (y 2)2 (2 2)2
Center (2, 2); radius 2 2 x = –3 y 2 = 12x 2
F –4 –2 2 x 4 –2
–4 y
5 5 1
y ⇒ 4p
4 4x 2 ⇒ x 2 16. y
x focus is 0, 1
⇒p
4 1
, directrix is y
16 1
16 y 5. The circle with center at (1, 0) and radius 2 plus its interior.
6. The region exterior to the unit circle and interior to the
circle with center at (0, 0) and radius 2. 1
4
1
8 –1
4 7. y 2 8x ⇒ 4p 8⇒p directrix is x
8. y 2 directrix is x
9. x 2 directrix is y
10. x 2 17. 16x 2 y2
9 y2
9 2⇒p 1 25 16 3 foci are ( 3, 0) 1
1
; focus is 0, ,
2
2 1
2 1⇒c 2
F1 4 9 –4 13 F2 –2 2
–2 13, 0); vertices are ( 2, 0);
3
x
2 1⇒c vertices are (0, y2
16 b2 a2 x2
25 y asymptotes are y
x2
4 400 ⇒ 25y 2 ⇒c 3
,
2 3
2 ⇒ foci are ( 12. 3
; focus is 0,
2 6⇒p 2y ⇒ 4p x –1
8
–1
4 1; focus is ( 1, 0), 1 directrix is y
x2
11.
4 y=– 1
16 4⇒p 6y ⇒ 4p 1
4 F 2; focus is (2, 0), 2 4x ⇒ 4p y = 4 x2 9
3) 4 5 ⇒ foci are (0, 5); 4 x 1
;
16 458 Appendix A5.1 18. 3x 2 6⇒ 2y 2 ⇒c a2 x2
2 b2 foci are (0, y2
3 3 2 b
and
a 1 4
⇒b
3 2 4
(3)
3 x
4⇒
9 y y F2 5 –1 1 x 2 (y + 2)2 = 8(x – 1) x = –1 F1
–5
V 2, 0), Vertices: ( 2, 0) ⇒ a ⇒ b2 a2
20. Foci: (0,
⇒ b2 c2 ( 2)2 4 5) ⇒ a 4), Vertices: (0,
25 9⇒ 16 21. x 2 y 2 1 ⇒ c
asymptotes are y
foci are (
2, 0) 2 2, c x2
2⇒
4 5, c y
25 a2
x, 1 26. (a) 1 x2
16 y2
9 1 ⇒ center is (0, 0), vertices are ( 4, 0) and (
1 a2 and (4, 0), c 4 1 b2 F 2 y2
2 2 x
9 x 5 –2 19. Foci: ( 3 25. (a) y 2 8x ⇒ 4p 8 ⇒ p 2 ⇒ directrix is x
2,
focus is (2, 0), and vertex is (0, 0); therefore the new
directrix is x
1, the new focus is (3, 2), and the
new vertex is (1, 2). 1) 2 –2 4
x⇒a
3
y2
1
16 24. Vertices: ( 3, 0), Asymptotes: y 1 7 ⇒ foci are ( 7, 0) b2 7, 0); therefore the new center is (4, 3), the new vertices are (0, 3) and (8, 3), and the new foci are 2; (4 7, 3). (b) y y
8 –x x
y= y= 4
2 F1
–4 F2 –2 2 V1 x 4 (4, 3) F1 F2 V2 –2
8 –4 22. 8y 2 16 ⇒ 2x 2 ⇒c a2 b2 asymptotes are y y2
2 x2
8 1
27. (a) 2 8 10;
10) c a2 b2 x 5 x (b) y ⇒a x⇒c 2), Asymptotes: y
1⇒a 1⇒b 4 (x – 2)2
y2
– =1
16
9 F1 b
a y = 3(x – 2) 10 F1 and 5 ⇒ foci are ( 5, 0) and 25 and (7, 0), and the new asymptotes are y y= 2 23. Foci: (0, 3x
,
4 vertices are ( 2, 0) and (6, 0), the new foci are ( 3, 0) F2
2 1 ⇒ center is (0, 0), vertices are ( 4, 0) (5, 0); therefore the new center is (2, 0), the new 5 –x y2
9 and (4, 0), and the asymptotes are y x
, foci are (0,
2 y y= x2
16 x b ⇒ c2
1 ⇒ y2 a2
x2 b2
1 –10 2 2a 2 ⇒ 2 F2 C
V1 10 V2 2a 2
–10 y = – 3(x – 2)
4 x 3(x 2)
4 . 459 Appendix A5.1
28. Original parabola: y 2 4x; vertex is (0, 0); y 2 4x
⇒ 4p 4 ⇒ p 1, so focus is (1, 0) and directrix is
x
1.
New parabola: (y 3)2 4(x 2); vertex is ( 2, 3),
focus is ( 1, 3), directrix is x
3.
2 x
6 29. Original ellipse:
(0, 3); c 2 2 y
9 3 ⇒ foci are (0, 3); center is (0, 0).
New ellipse: 2)2 (x ( 2, 2h x2
2 b/2 x4
b2 2h 4x 3
dx
b2 x 0 hb 2
;
8 0 b2
h
2 1
3 Volume of the Cone: V2 b2
h
4 1
3 hb 2
;
12 3
V
22 1; vertices are 9 4); foci are ( 2, 1 3); center is 38. (a) y 2 y2
; the volume of the solid formed by
k kx ⇒ x revolving A about the yaxis is 1). kx x2
30. Original hyperbola:
4 ( 2, 0); c 2 b/2 4h 2
x dx
b2 2 xh 0 therefore V1 1)2 (y 6 ( 2, 2) and ( 2, b/2 V1 1; vertices are (0, 3) and 6⇒c 9 37. Volume of the Parabolic Solid: y2
5 5⇒c 4 1; vertices are (2, 0) and
3 ⇒ foci are (3, 0) and
5 ( 3, 0); center is (0, 0); asymptotes are y
New hyperbola: 2)2 (x 2)2 (y 4 2 kx y2 2
dy
k 0 k2 0 x 2 kx
; the
5 y 4 dy volume of the right circular cylinder formed by x. revolving the rectangle about the yaxis is 1; vertices are (4, 2) 5 V1 V2 x2 kx ⇒ the volume of the solid formed by and (0, 2); foci are (5, 2) and ( 1, 2); center is (2, 2);
5 asymptotes are y 2 (x 2) revolving B about the yaxis is V3 2. 31. Original hyperbola: y 2 x 2 1; vertices are (0, 1) and
(0, 1); c 2 1 1 ⇒ c
2 ⇒ foci are (0,
2);
center is (0, 0); asymptotes are y
x.
New hyperbola: (y 1)2 (x 1)2 1; vertices are
( 1, 2) and ( 1, 0); foci are ( 1, 1
2); center is
( 1, 1); asymptotes are y
(x 1) 1.
32. x 2 4x y 2 12 ⇒ x 2 4x 4 y 2 12 4
⇒ (x 2)2 y 2 16; this is a circle: center at C( 2, 0),
a4
33. 2x 2 2y 2 28x 12y 114 0
⇒ x 2 14x 49 y 2 6y 9
57 49 9
⇒ (x 7)2 (y 3)2 1; this is a circle: center at
C(7, 3), a 1
4y
34. x 2 2x 4y 3 0 ⇒ x 2 2x 1
⇒ (x 1)2
4(y 1); this is a parabola:
V( 1, 1), F( 1, 0) 3 1 ⇒ x2 1 35. x 2
⇒ (x 5y 2 4x 2)2 5y 2 5⇒ 4x 2)2 (x 5y 2 4 5 y2 4 ellipse: the center is ( 2, 0), the vertices are
(2 5, 0); c a2 b2 5 1 4:1.
(b) The volume of the solid formed by revolving B about
x the xaxis is V1 2 ⇒ the foci are ( 4, 0) and (0, 0)
36. x 2 y 2 2x 4y 4
⇒ x 2 2x 1 (y 2 4y 4) 1
⇒ (x 1)2 (y 2)2 1; this is a hyperbola: the center
is (1, 2), the vertices are (2, 2) and (0, 2);
c
a2 b2
11
2 ⇒ the foci are
(1
2, 2); the asymptotes are y 2
(x 1) ( kt)2 dt 0 x k 0 t dt kx 2
.
2 The volume of the right circular cylinder formed by
revolving the rectangle about the xaxis is
( kx)2x kx 2 ⇒ the volume of the solid formed by revolving A about the xaxis is
V3 1; this is an V1 4 x 2 kx
. Therefore we can see the ratio of V3 to V1 is
5 V2
1 V2 V2 V1 kx 2 ratio of V3 to V1 is 1:1. kx 2
2 kx 2
. Therefore the
2 460 Appendix A5.2 39. Let P1( p, y1) be any point on x p, and let P(x, y) be a point where a tangent intersects y 2
4px ⇒ 2y y2 dy
dx 4p ⇒ tangent line from P1 is
⇒ y2
y2 yy1
yy1 1
2 ⇒ y2 y2
4p dy
dx ( p) 2 y2
, we have
4p
12
y
2 yy1 42. 9x 2 2p 2 y12 y1 2 2p
y12 y1 4p 2
4p
(y12 y1 2 4
2 9
4 4p 2. Therefore y12 y1 4p 2 43. x 2 24) 1 ⇒ the lines are 4p 2) x2
on the interval 0
4 x2
4 4x 1 and the height is 2y) ⇒ A (x) x 2. The area of x2
(since the length is 2x
4
x2
x2
41
x2 .
4
1 2 x
4 ⇒4 1 x2 x x2
4 0⇒4 1 0 ⇒ x2 2 0⇒0 2⇒x 3 2 2 0
3 ( y3
30 y C⇒C 0 we have that A( 2) C; y drA drB dt dt ⇒ rA 0; therefore y ⇒ rB d
(r
dt A rB) 0 a constant s Appendix A5.2 4 is the maximum ⇒c ⇒V
2 9
2
9x 0 36 ⇒ y 2 92
x
4 92
x and we use the positive root
4
2
9 22
92
9
x dx 2
9
x dx
0
4
4 3 32
x
40 x2
25 y2
16 1 b2 a2
c
a 2. 9 400 ⇒ 25 16 3 25y 2 x 2 wx 2
is the
2H 46. PF will always equal PB because the string has constant
length AB FP PA AP PB. ⇒e ⇒y 0 when equation of the cable’s curve.
45. 0 3
; F( 3, 0); directrices are
5
5
a
25
3=
e
3
5 2. 2x 2
⇒c 1 b2 2 1 ; F(0, a2 1); directrices are 1 c
a ⇒e y2
2 2 ⇒ x2 y2 2 24
y 0 a
e 2 2 1
2 3 y 2)2 dy 0 area, when the length is 2 2 and the height is
4y 2 y 24 wx 2
2H C 1 3 2 (only the x2
4 1 w x2
H2
w(0)2
2H w
x dx
H 1. 16x 2 41. (a) Around the xaxis: 9x 2 dx 8 (pp. 606–611) A(2) 4 y 2 on the interval y 2)2 dy y 2) dy (1 positive square root lies in the interval). Since
A(0) 2 x2 9 56
43 1 (1 3 0 44. y
x 4 Thus A (x) 4 on 4
2 4x 8 1⇒x ⇒V the inscribed rectangle is given by
2x 2 1 3
2 2 x2 24 3 3 A(x) 8
3 16 y2 2 1 4 3
2 ⇒y 9 x3
43 4) dx 64
3 36
4 4⇒V x (x 2 3
(56
4 perpendicular.
40. Let y 9x 2 36 ⇒ y 2 4y 2 2p and m2 2 ⇒ m1m2 0 the interval 2 the slopes of the two tangents from P1 are
m1 4y 9
4 16p 2 3 2 2p
y 0 4y12 2y1 ⇒y x y1 2p 2 ⇒ y 2 2p 2 yy1 y 42
y
9 4 4 ⇒V 2p
; then the slope of a
y 36 ⇒ x 2 4y 2 42
y and we use the positive root
9
3
4 22
42
4
y dy 2
4
y dy
0
9
9
4 33
y
16
27 0 ⇒x 4px. Now 2p 2. Since x 2px
2p dy
dx (b) Around the yaxis: 9x 2 1 Appendix A5.2
3. 3x 2
⇒c a2 F(0, x2
2 6⇒ 2y 2 y2
3 b2 3 ⇒ 16 ; 2 therefore
3 1 4. 6x 9y x2
54 ⇒
9 2 ⇒c a 1 2 b ; F( y
9 x
25 1 9 c
a 3⇒e 3
–4 –2 36 ⇒b 1600 64 ⇒ ae
x2
100 y2
94.24 and c
⇒ 100 0.1 ⇒ a
7 ⇒ b2 y
4900 9 ⇒
5 Then PF 5)2 ⇒ (x 5)2
2 4
9 ⇒ x2 y2 4⇒ ae
e2 ae x
16 y
15 1 is a model of Pluto’s orbit. 5 94.24 x 4851 2 4)2 y2 ⇒ x2 8x 16 3 y2 Therefore, a 5 2 52
x
9 1)2 (x 3, b 2 and the major axis is vertical. The ⇒c
18 x 5 81
5 y 4
e2 16 ⇒ e 2
(y 0)2 ae 1; c2 9 a2 4 c
a 5
3 4 4 5
3 y
8 16)2
32x 256) 1
–5 5
–2 32 22 x 95
.
5 5 5), the , and the directrices are 3 a
e 4 and 1
1
⇒e
. Then
4
2
1
x 16
2 b2 5; therefore the foci are F(1, 4 eccentricity is e 1
16 ⇒ c 4)2 (y 4 5 12
(x
4
2
x
y2
48 ⇒
64
48 y2 3 other axis is from (3, 4) to ( 1, 4) and is 4 units long.
. center is the point C(1, 4) and the ellipse is given by y
4 1
(x
4 5 9 x 4)2 (x 5 and
5
⇒e
9 ⇒ e2 2 x
9 13. One axis is from (1, 1) to (1, 7) and is 6 units long; the 5 9 y2 16 ⇒ ⇒ (x 3
4 15; therefore, 5 ⇒c 5 5
x
9 5 1
PD ⇒
2 ⇒ x2 5.76 49 5
9 0)2 (y
y2 5x 16 ⇒ PF 15 ⇒ b b2 16
2 y 4900 9 10. Focus: ( 4, 0), Directrix: x
a
e b2 ⇒ 1
2 70 ae
9
5
⇒2
e2
e
5
5
PD
3 (x ⇒ x2 a2 1 9. Focus: ( 5, 0), Directrix: x ⇒ ⇒ b2 4; c2 1 and a 1
4 2 x
4851 a
e take c
40 0.25 1 70(0.1) ae
2 c
a 12. The eccentricity e for Pluto is 0.25, ⇒ e 10 and 2.4 ⇒ b 2 70), e 8. Vertices: (0, 6 c
8
e
0.2
y2
1
1536 8 and a 0.24 ⇒ a 10(0.24) 3
0.5 1 x2
1536 ⇒
1600 7. Vertices: ( 10, 0), e
c y
36 0.2 ⇒ c 6. Foci: ( 8, 0), e
2 2 x
27 ⇒
27 9 c
e 3 and a 2 ⇒ b2 x 4 –4 0.5 ⇒ c 3), e 2
–2 3 5. Foci: (0, 1 2 3 33 1 3; 4 6 3 a
e 0 2 b2 y 3, 0); directrices are 3 x y2
6 9⇒b a2 2 3
2 5; c 2 4 and a b2 ⇒ b2 25 3
3 a
e 0 1 c
a 1⇒e 2 1); directrices are y 4
⇒ take c
5 11. e 1 461 462 Appendix A5.2 14. Using PF e PD, we have ⇒ (x 4) ⇒ x2 8x 5
9 ⇒ x2
15. 9x 2 2 y 20 ⇒ 5x 2 a 2 b 2 a
e 0 16. y 2 a2 x2
8 b2 a
e 8 16 ⇒ ⇒c x2
2 a2 2 2 x2
8 2⇒ 2 ⇒ (x 4) ⇒ x2 8x x2 2 y2 5(y 2 12y x
a2 y
b2 1)
45 4 4 180 c2
e2a 2 1⇒ c
⇒c
a a 2; e
a2 2 x
a2 y
a 2(e2 a 2(e2 ea
1); 2 1; the asymptotes 1) e2 1x. As e increases, the slopes of the asymptotes increase and the hyperbola 5 10 approaches a single straight line.
25. The ellipse must pass through (0, 0) ⇒ c 8 10 ( 1, 2) lies on the ellipse ⇒
x
;
2
2 0 c2 c
a 1 and e a2 9 1 2 5 a
e 10 3⇒c c2 a2 9 1 ⇒ x2 8 2 ⇒ e2 4)2 2) y2 2(x 2
x
8 y
8 ae (y 4x
2 y
8 1 3a 2 1
a
e 2 2x 2 4 and 2. 2 2(x 8⇒ 8 ⇒ x2 2⇒e (x 2 3⇒c 2⇒c 4) 0)2 2b 4 ⇒ the equation is 4x 2 and b
4x 2 3 c
a 3 and e a 0; the point
8. The ellipse is tangent to the xaxis ⇒ its center is on the yaxis, so a ⇒c 16 36) 1 y2 4y (y 2) 4 ⇒ 4x 2 4
2 4 1⇒a 3⇒e c
a y2 a2 b2 3
2 3
2 . 4y 2)2 (y 2 and b standard symbols) ⇒ c2 2 0 2 1 4
e2 4) 2 a
e 0 3⇒a 1) and e y 4 4y 2 e2a 2 ⇒ b2
2 92
(y
4 9 2x; 5; asymptotes are y 2PD ⇒ Then PF (x 1 60y b2 ⇒ b2 a2 10 1 21. Focus (4, 0) and directrix x ⇒ 6)
36 thus,
8 x
8 b2 1 ⇒ b2 ae
e2 1) 1
2 2x 2 y
2 20. Foci ( 3, 0) and e ⇒ 2x 0)2 2 6y 8x 2 y
3 of this hyperbola are y 3 ⇒ b2 3a ⇒a 5y 2 2. 1
4 x
2 3
y
2 3)2
y2 2 and 12
2 3 ⇒ x2 1 ae 4⇒e
(y 4 x2 (y 2 2 19. Vertices (0, ⇒ y2 ⇒ y2 y2 ⇒ 4(x 2 2)2 4x 2x (y 1
⇒ e2
2 (x 1)2 (x ⇒ c2 1 10); directrices are y ⇒c 8 5; asymptotes are y 10 c
a F(0, 3x 2 24. c 2 y2
8 b2 16 ⇒ 2x 2 ⇒e 2; 4); directrices are 2 ⇒c 4 c
a 4⇒e 10, 0); directrices are x 18. 8y 2 4 23. 2 10 c
a F( 4x 2 a2 ⇒e y2 ⇒ 5
;
4 c
a 2)2 ⇒ 4x 2 x; F(0, 2PD ⇒ ⇒ x2 5⇒e 8 1
2
⇒2
2
e ⇒ (x 1. 1
8 2y 2 17. 8x 2 y2
20 ⇒ x2 y2
8 8⇒ 0 x2
36 1 9 1
ae
⇒2
2
e Then PF 3
x; F( 5, 0); directrices are
4 asymptotes are y
y 81) 180 or 1
⇒c
2 22. Focus ( 2, 0) and directrix x 16
.
5 x2 ⇒c 18x y2
9 16 asymptotes are y
x 9 a
e 9y 2 x2
16 2
x
3 y2 2 42
(x
9 144 ⇒ 16y 2 9) y2 16 y2 ⇒c 4
(x
9 2 4)2 (x 0. Next,
4 1 (now using the
4 1 3 0 Appendix A5.2
26. We first prove a result which we will use: let m1 and m2 be b2
. Since tan
cy0 Similarly, tan the slopes of two nonparallel, nonperpendicular lines. Let are both less than 90 , we have tan , and 463 and . be the acute angle between the lines. Then tan
y m2 m1 m1m2 1 let 1 and . (To see this result for positiveslope lines,
α P(x0, y0)
β be the angle of inclination of the line with slope m1,
2 F1(–c, 0) be the angle of inclination of the line with slope m2. Assume m1 m2. Then 1 Then tan tan ( 2 tan since m1 1 1 ) and m2 2 and we have tan
1 tan tan tan 1 1 2 tan 2 1 2 m1 . m2 1 m1 m2 , 27. To prove the reflective property for hyperbolas:
b 2x 2 .) 2 Now we prove the reflective property of ellipses (see the x F2(c, 0) a 2y 2 2b 2x a 2b 2 2a 2yy 0 2 y
accompanying figure):
2 2b x 2 2a yy Let P(x0, y0) be a point of tangency (see the accompanying b 2x
.
a2y ⇒y 0 figure). The slope from P to F( c, 0) is
and from P to F2(c, 0) it is Let P(x0, y0) be any point on the ellipse a a b x0 2 x02 a 2y0 F2( c, 0) be the foci. Then mPF
mPF 2 y0
x0 c . Let and . Let F1(c, 0) and
y0 1 x0 c and F1PA will show that tan tan . be the angles between the y0 a 2y0 tan x0 y F1(–c, 0) c 2 b x0y0 1 a 2y0(x0 b 2x02
2 c) b 2x0c a 2y02 a y0x0 a y0c b 2x0y0 b 2x0c (b 2x02 a 2y02) 2 2 2 a y0 c (a b 2x0c a 2b 2 2 a y0c 2 c x0y0 b 2)x0y0
b2
.
cy0 x0 and define the angles tangent line and PF1 and PF2 respectively. Then
b 2x0 y0
c y0
x0 c . Let the tangent through P meet the xaxis in point A, 2 bx0 ⇒ y (x0) bx
a 2y P(x0, y0)
αβ
A
x
F2(c, 0) and F2 PA . We 464 Appendix A5.3
10. 3x 2 6xy 3y 2 4x 5y 12
⇒ B 2 4AC 62 4(3)(3) 0 ⇒ Parabola 27. continued
From the preliminary result in Exercise 26,
x0b 2
x0 c x0b 2 1 y0 y0a 2 x0 x0 b 2 c y02a 2 x0y0a 2 y0 a 2 c x0y0b 2 2 2 21xy
14. 25x
⇒ B 2 4AC 4y
212 2 x0b 2c x0y0c 2 x0 y0 1 and y0a 2 c tan
x0 x0b
c y0a 2 are acute angles, we have tangent to the hyperbola at P bisects A 17. cot 2 C therefore x APB. Since APC 2
2 ⇒ APC, and the m APB 2 12
2 A m LAPB
2 2 ⇒x 2
2 ⇒ 1. x
3xy y
⇒ B 2 4AC 2 2 2 x 2
2 2
2 2 2
2
2 y 2
2
2 y 4 ⇒ Hyperbola 2 y sin, y
y ,y y cos 2 0⇒2 2 x y 2 y 1 2
2 ; ⇒ 4 x sin
2 x 2
2 x 2 y cos y
2 y ; 2 x 2
2 y 2 y 1 12
12
12
12
12
xy
y
x
y
x
xy
2
2
2
2
2
12
3
12
y
1⇒ x2
y
1 ⇒ 3x 2 y 2 2
2
2
2 ⇒x
x0
( 3)2 4(1)(1) 5 0 ⇒ Hyperbola 2 18xy 27y
5x 7y
4
2. 3x
⇒ B 2 4AC ( 18)2 4(3)(27) 0 ⇒ Parabola
3. 3x 2 7xy
⇒ B 2 4AC
⇒ Ellipse 17y 2 1
( 7)2 4(3) 17 4. 2x 2
⇒ B2 2y 2 x y 0
(
15)2 4(2)(2) 15xy
4AC 1 2 s Appendix A5.3
(pp. 612–618) 2⇒x 2 4 x sin x
2 x x cos x 2 2 1 therefore x hyperbola are perpendicular. 2 B 90 , the tangents to the ellipse and ⇒ 2 2 y ,y
2 C 0 ⇒ Ellipse y sin , y y 12
y
2 ⇒x 180 and m APC
2 2
2 18. cot 2 m APC 2 x x 2 0⇒2 x cos 2 ⇒x APB are a linear pair, so that 2 0 ⇒ Hyperbola 41 435x 9y 72 0
4(3)(12) 0 ⇒ Parabola 0
1 B tan , and . 0 0 ⇒ Ellipse 3 350x 0
4(25)(4) 16. 3x 2 12xy 12y 2
⇒ B 2 4AC 122 b2
. Since tan
y0c 2 28. The tangent to the ellipse of P bisects and 0.01 2 x0b 2 y0 0 ⇒ Hyperbola 3xy 2y
17y 2 0
15. 6x
⇒ B 2 4AC 32 4(6)(2)
39 b2
. In a similar manner,
y0c y0a 2c 7
4(2)(3) 13. x 2 3xy 3y 2 6y 7
⇒ B 2 4AC ( 3)2 4(1)(3) c x02b 2 a 2b 2 14y
1
4(3)(2) 1 12. 2x 2 4.9xy 3y 2 4x
⇒ B 2 4AC ( 4.9)2
⇒ Hyperbola y0 y0a 2 tan 11. 3x 2 5xy 2y 2 7x
⇒ B 2 4AC ( 5)2 0.477 ⇒ Ellipse
A 19. cot 2 0 C 5. x 2 2xy y 2 2x y 2
⇒ B 2 4AC 22 4(1)(1) 0 ⇒ Ellipse 0
0 ⇒ Parabola ⇒3 6. 2x 2 y 2 4xy 2x 3y 6 ⇒ B 2 4AC
42 4(2)( 1) 24 0 ⇒ Hyperbola
7. x 2 4xy 4y 2 3x 6
⇒ B 2 4AC 42 4(1)(4)
8. x 2 y 2 3x
⇒ B 2 4AC 2y
02 9. xy y 2 3x
⇒ Hyperbola 5 ⇒ B2 10
4(1)(1)
4AC 3 ⇒x 2
3
2 4
12 3 3 2 1 0 3
1
2 ⇒ 4x 3x 2 16y ⇒ 3 x sin 1
x
2 1
y
2 x 2 ⇒2
3 3
2 y 2 1
y
2 8 0 ⇒ Ellipse (circle)
4(0)(1) 1
y ,y
2 x 1
x
2 0 ⇒ Parabola 1 y sin , y x cos x 2 1 23 therefore x
1 3 B 1
x
2 3 2 2 y 8
3
2 y 3
2 x 0 0 ⇒ Parabola y 1
y
2 6 ; y cos 465 Appendix A5.3
A 20. cot 2 C therefore x 3 3 12
2 52
y
2
A 21. cot 2 2
2
2 x 2 2 A 22. cot 2 y 3
3 x 2 2 x y A 23. cot 2 2 1 y 3 2 2 2 2 x sin 3 3 y 2 2 2
2 2x 2
2
2 2
2 2 x 1
y
2 2 8 2
2
2 ⇒ 4x 2 + 2y 2
2 y
2 y 2 2 x 2 1 2
2
2
2
2
2 y 2 y 2 0 ⇒ hyperbola
0⇒2 ⇒ 2 4 y sin ,
2 ⇒x 2 2 x 2 y, 2 2
2
2 x 2 2 y 2 2 x 2 y 2 y 19 19 ⇒ Ellipse 4 C 3 B 3
2 12
y
2 2 3
2 3y ⇒2 3 3 y sin , y 1
y ,y
2 x x 1 43 1
x
2 ⇒ 5x ( 1) x cos 3
2 1
x
2 4 3 3
2
3
2 x sin 6 y cos y
1
y
2 x ⇒ 1
x
2 3
2 2 y 7 7 ⇒ Hyperbola 2 y cos
27. cot 2 A C 14 2
16 B 3
⇒ cos 2
4 3
(if we
5 choose 2 in Quadrant I); 2 2 y 8 2
2 x y
2
2 y cos
or sin 2 4y 1 thus sin
0
0⇒x ; y x 2 A ; y 2 x y 0 y x 2y ⇒ x sin x 2
2 2
2 y cos y 2 x therefore x 2 x x 2 2
2 ; ⇒3 0⇒2 2 y ,y 2 2
8 2 y sin , y 2
2 2
2 ⇒x 22 2 3
2 x x sin ; y cos 1 2 x 2
2 26. cot 2
3 3x 12
y
2 x 2 3 y cos 3 1
y
2 x 1
2 2 x cos 2 2 2x C 4 1 x cos 2 2
⇒
3 3 3 y ,y B 2 y 2x B x sin y ⇒2 y sin , y C therefore x 2 2 ⇒ 2 x
2 x 2 therefore x 1 ⇒ Parallel horizontal lines 2 2
2 2 12
y
2 A 25. cot 2 2 y ,y 2 2 y ⇒3 1 y x 2 2 y cos 2 1 ⇒ Parallel horizontal lines 2 3 2 x sin 2 3 2 1
2 ⇒2 2 x cos ⇒3 x 2 2 y 2 1
x
2 ⇒x 2 4 ⇒x ; 0⇒2
y sin , y y x 12
2 2 2 B ⇒x ⇒ 4y 2
2
2 C therefore x 2 ⇒ 2 y sin , y x 2⇒y 2 2
2 0⇒ x 2 ⇒ ellipse 2 0⇒2 y ,y 2 x 2 y x 2 2 2 x 2 y 2
2
2 2
2 ⇒ y 0
1 2 2 x 3
2 0 x cos 2 5y 1 x cos 2 ⇒ 2y 2 2 2 therefore x
⇒x 1
x
2 C
B 1 1 B 2 y cos y 1
y
2 x A 24. cot 2 ; 2 y 1⇒x C therefore x ⇒ 2 3 2 6 x sin 2
3 3 1
2 ⇒x 1
x
2 ⇒ 3 y sin a, y 12
y
2 2x ⇒2
3 1
y ,y
2 x ⇒x 1 3 x 2
2 ⇒ 2 x cos 3 ⇒x
⇒ 1 B 0 ⇒ Parabola 1 cos 2
2 cos 2
2
2 and cos
5 1
1 (3/5)
2 (3/5)
2
1
5 1 and
5 2
5 y ; 466 Appendix A5.3
A 28. cot 2 C 4 1 B 3
⇒ cos 2
4 4 3
(if we
5 1 1 cos 2
2
1 (3/5)
2 cos 2
2
1 or sin 1 ⇒ sin
C 1
⇒2
2 0.23, cos ⇒ 0.88x E 3.12y 13.28
0.88, B 1.20, and F
0.74x 1.20y 0.00, C ⇒ sin
B 1
⇒2
5 0.10, cos 0.00, C 12
2 0, an F 11.31 ⇒ 1
,C
2 2.05, 2 3.05y 2 2.99x 0.30, and 0.30y 7 0, a hyperbola
4 31. tan 2 1 ⇒ sin
C 4
⇒2
3 0.45, cos
5.00, D ⇒ 5.00y 2 32. tan 2 0, and F 0 or y 0.32, cos
20.00, D ⇒ 20.00y 26.57
0.00, B 0.00, 5 1.00, parallel lines 3
⇒2
4 12
18 2 53.13 ⇒
0.89; then A 0, E
5 ⇒ sin
C 4 36.87 ⇒ 0.95; then A 0, E
49 18.48, 18.48x 7.65y 86 0, 0, and F 18.43
0.00, B 2 B 2 y 2 1⇒x 1
,B
2 2 2 2
1
,F
2 2 y 0.00, 0, 1 2 2 1
(see part (a) above),
2
1
0, F
a⇒ x 2
2 12
y
2 a 2a
0 we have 37. The one curve that meets all three of the stated criteria is
the ellipse x 2 4xy 5y 2 1 0. The reasoning: The
symmetry about the origin means that ( x, y) lies on the
graph whenever (x, y) does. Adding
Ax 2 Bxy Cy 2 Dx Ey F 0 and
A( x)2 B( x)( y) C( y)2 D( x) E( y) F 0
and dividing by the result by 2produces the equivalent
equation Ax 2 Bxy Cy 2 F 0. Substituting x 1,
y 0 (because the point (1, 0) lies on the curve) shows
further that A
F. Then Fx 2 Bxy Cy 2 F 0.
By implicit differentiation,
2Fx By Bxy
2Cyy
0, so substituting x
2,
y 1, and y
0 (from Property 3) gives 4F B 0
⇒B
4F ⇒ the conic is Fx 2 4Fxy Cy 2 F 0.
Now substituting x
2 and y 1 again gives
4F 8F C F 0 ⇒ C
5F ⇒ the equation is
now Fx 2 4Fxy 5Fy2 F 0. Finally, dividing
through by F gives the equation x 2 4xy 5y 2 1 0. 49
38. If A 2 2 36. Yes, the graph is a hyperbola: with AC
4AC 0 and B 2 4AC 0. 7 ⇒ 2.05x E ⇒x 5.65 2.99, E 12
y
2 ⇒x 3
3 0.995; then A 3.05, D 10.45y sin 45 cos 45 D
1
( 3) 10.45, D –86 cos 45 sin 45 (b) A 2 2 0.92; 0.00, C 7.65, and F 35. (a) A ellipse
30. tan 2 0.55, B 22.5 an ellipse 0.97; then A 2 45 ⇒ 0.38, cos ⇒ 0.55x 26.57 ⇒ 0.74, E 2 1⇒2 9 then A and cos
5 5 3 3.12, D 2 5
2 and cos 1
1 (3/5)
2 1 5 29. tan 2 2 ⇒ sin choose 2 in Quadrant II);
thus sin 7 34. tan 2 C, then B B cos 2 (C A) sin 2 0, parallel lines
B cos 2 . 5 33. tan 2 3 ⇒ sin
C 2 5⇒2 0.63, cos
0.05, D ⇒ 5.05x 2 a hyperbola 0.05y 78.69 ⇒
0.77; then A 5.07, E
2 5.07x 39.35
5.05, B 6.19, and F
6.19y 1 Then
0.00,
1 0, 4 ⇒2 2 ⇒B B cos 2 0 so the xyterm is eliminated.
39. 90 ⇒ x x cos 90
y sin 90
and y x sin 90
y cos 90
x
x2
b2
y2
(b) 2
a (a) (c) x 2 y2
a2
x2
b2 y y 1
1 2 a2 (d) y mx ⇒ x (e) y mx m( y ) ⇒ y b⇒x m( y ) 1
x
m b⇒y 1
x
m b
m 467 Appendix A6
180 ⇒ x 40. and y x cos 180 x sin 180
2 y sin 180 y cos 180 43. (a) B 2 4AC 42 4(1)(4) 0, so the discriminant
indicates that this conic is a parabola. x y (b) The lefthand side of
x 2 4xy 4y 2 6x 12y 9 0 factors as a
perfect square:
(x 2y 3)2 0 ⇒ x 2y 3 0
⇒ 2y
x 3; thus the curve is a degenerate
parabola (i.e., a straight line). 2 (a) x
a2 y
b2 1 (b) x2
a2 y2
b2 1 2 (c) x 2 y (d) y mx ⇒ (e) y mx 41. (a) B 2 b⇒ 2x x y 1 1 ⇒ b⇒y mx b 1 ⇒ hyperbola 4(0)(0) dy
dx mx m( x ) 0 ⇒ y(x y
2x (c) y m( x ) ⇒ y y 4AC (b) xy 44. (a) B 2 4AC 62 4(9)(1) 0, so the discriminant
indicates that this conic is a parabola. a2 2
2 (x 1) 2x 2x ⇒ y 1) x 1 2, the slope of y ⇒ 1)2 (x 2 2 ⇒ (x 1
A 1)2 Since B2 4 3 or x 1; x 3⇒y 3 ⇒ (3, 2 AC 4AC B 2 Cy 2 1 1
⇒ the area is
C
2 1
C (because B
⇒x 1
and
A ⇒ the semiaxes are
2x 0 45. Assume the ellipse has been rotated to eliminate the
xyterm ⇒ the new equation is A x and we want 1
dy
dx (b) The lefthand side of
9x 2 6xy y 2 12x 4y 4 0 factors as a
perfect square: (3x y 2)2 0 ⇒ 3x y 2
⇒y
3x 2; thus the curve is a degenerate
parabola (i.e., a straight line). . 4A C 4A C 4A C
2 0) we find that the area is 4AC 3) is a B2 as claimed.
point on the hyperbola where the line with
2 is normal ⇒ the line is slope m
y
x 3 46. (a) A
C
(A cos2
B cos sin
C sin2 )
2
2
(A sin
B cos sin
C cos )
A(cos2
sin2 ) C(sin2
cos2 ) A C 2(x 3) or y 1⇒y 2x 1 ⇒ ( 1, 3; 1) is a point on the hyperbola where the line with slope m 2x E 2 is normal ⇒ the line is y E sin )2 ( D sin
(b) D 2 E 2 (D cos
2
cos )
D 2 cos2
2DE cos sin
E2 sin2
D 2 sin2
2DE sin cos
E 2 cos2
2
2
2
2
2
D (cos
sin ) E (sin
cos2 )
2
2
D
E s Appendix A6
1 2(x 1) or y 3 (pp. 618–627)
1. sinh x
1 [ 9.4, 9.4] by [ 6.1, 6.1] 42. (a) False: let A
⇒ parabola C 1, B 2 ⇒ B2 4AC 0 (b) False: see part (a) above
(c) True: AC 0 ⇒
⇒ hyperbola 4AC 0 ⇒ B2 4AC 0 3
4
5
4 3
⇒ cosh x
1 sinh2 x
4
9
25
5
sinh x
, tanh x
16
4
16
cosh x
3
, coth x
5 4
, and csch x
5 1
sin x 1
tanh x
4
3 1 5
, sech x
3 32
4 1
cosh x 468 Appendix A6 4
⇒ cosh x
3 2. sinh x 16
9 1 25
9 tanh x 17
,x
15 3. cosh x 17 2
15 289
225 17
, sech x
8 169
25 13
, sech x
12 2
dy
dt 16. y x 1
(4e 0)
4 1
(4)
4 dy
dx tanh t
1
t t 2 tanh t 2 tanh t 1 2 )]( t 1
t 1 e 2t tanh ln x x2 1
sinh x 5
12 e ln x 2 ln x e 2 ln x 1
e ln x ln x 2 lnx e e 7. cosh 5x sinh 5x 1
t
cosh z
sinh z coth z ln (cosh z) ⇒ dy
dz sinh z
cosh z tanh z (sech )(1 ln sech ) sech tanh
sech ( sech
sech (sech ) tanh )(1 tanh (sech ln sech )
tanh )(1 (sech 1
x x 20. y 8. cosh 3x sinh 3x 9. (sinh x ⇒ cosh x)4 e3x sinh x) ln (cosh2 x e 5x e 5x e 3x 2
e 3x e 3x 2
ex e x ex ln (cosh x
ln 1
x) (b) cosh 2x cosh (x x)
cosh2 x sinh2 x e x4 2 (csch )(1
dy
d ln sech )] csch coth
csch csch ln csch )( csch
coth (1 coth ) ln csch )(csch coth ) 3x (e x)4 (csch 21. y
⇒
cosh x sinh x coth )(1 (csch e 4x 0 coth )(ln csch ) sinh x sinh x 1 ln csch ) 1
tanh2 x
2
1
sinh x
(2 tanh x)(sech2 x)
2
cosh x ln cosh x
dy
dx tanh x (tanh x)(sech2 x) (tanh x)(tanh2 x)
cosh x cosh x (1 ln csch ) (csch )
(1 sinh x) sinh x cosh x tanh )(ln sech ) e 5x
e 2 2 sinh2 x) 11. (a) sinh 2x sinh (x
2 sinh x cosh x 5x e 2 tanh )[1 ln sech ) 2 2 e 5x ) ) dy
dz dy
⇒
d x4 1
2x 2 2 10. ln (cosh x e 1/2 1 (2t)(tanh t 19. y 2 1)](2) dy
dt 18. y e ln x x x
3 (tanh t 1/2)(t ln (sinh z) ⇒ 1
tanh x 2 1
x2 ⇒ )(t 2) (sech
6. sinh (2 ln x) 2 cosh 1
[cosh (2x
2 1 1/2
t
(2t 1/2)
2
t [sech2(t 1/2)] sech2 In Exercises 5–10, graphical support may consist of showing
that the graph of the original expression minus the simplified
one is the line y 0.
2 e 2t 1/2 tanh t 1/2 t 1 5
, and csch x
13 (e x ) 1 1
3 x
3 6 cosh x e 17. y 12
, coth x
13 1
cosh x 5. 2 cosh (ln x) ) (e x 1) [sech2 (t 15
8 12
,
5 12
5
13
5 x2 e t 1
sinh x cosh2 x ex
2 e 1) ⇒ t tanh 2 1
tanh x 15
, and csch x
17 144
25 1 ⇒ 8
,
15 8
, coth x
17 0 ⇒ sinh x sinh x
cosh x tanh x 64
225 15. y 1 sech 1
cosh x 13
,x
5 4. cosh x cosh2 x 1 8
15
17
15 (e x ) 1
sinh (2x
2 cosh (2x x2 e
2 dy
x
⇒
dx
3 6 sinh 14. y 3
,
5 1
cosh x 0 ⇒ sinh x sinh x
cosh x tanh x x e 13. y 3
4 1 ex sinh2 x 1
(2e x )(2e x )
4 4
,
5 1
sinh x and csch x 12. cosh2 x
1x
(e
4 5
, sech x
4 1
tanh x coth x 42
3 1 5
,
3 4
3
5
3 sinh x
cosh x sinh2 x 1 tanh3 x (tanh x)(1 sech2 x) ) 469 Appendix A6 22. y
⇒ 1
coth2 x
2
cosh x
1
(2 coth x)( csch2 x)
sinh x
2 dy
dv 2 coth x (coth x)(csch x)
2 23. y (x
(x 2 (4x 2 24. y (4x 1) dy
⇒
dx (x 2
(2x) 2x csch x) 1) (4x 2
(4x 1 ln x e e ln 2x ⇒ e ln 2x 4x 1) 2 4x 2 4x 1 dy
dx 1
x 33. y 4 1
1 25. y
⇒ sinh x 1 sinh (x 1/2 1
x 1/2
2 dy
dx ⇒ 1 (x 1/2 )2 1 ) 2 x(1 x 1 cosh (2 (2)
dy
⇒
dx x 1) 1
(x
2 1) 1 x 1 4x 4x 27. y 7x (1 dy
⇒
d
1
1 (1 ) 1 sinh 36. y (sec x)(tan x) ⇒ 2 (
dy
d 2 ( 2
2 2
2 ( 1) 37. (a) If y 1 2) 1 (2 1)2 ( 1 t) coth 1 1) 1 2) tanh 1 ( 1 ( 1)
1 2) tanh (2 2) tanh (2 1 2 ) tanh ( 1) tan dy
dx (sec x)(tan x) (sinh x) (1 t (1 t) coth x 2 dy
dx sech x, which verifies the formula.
1 sin (tanh x) sech2 x 29. y 1 sec x, 0 C, then cosh x
cosh2 x cosh x
sinh2 x (b) If y
1 sec2 x (sec x)(tan x)
tan x
1 22 1 1 (tan x)2
sec x sec x
sec x
sec x (sec x) ⇒ 1 cosh (2 )
sec2 x 1 tanh x ln 2
2 dy
dx 1 ( 1) tanh 2 1 2 sec2 x
sec x tan2 x 28. y (ln 2)2 dy
d (tan x) ⇒ sec2 x 1 1 sech 12
2 1 2⇒ 1 sec2 x
1 x2 ln 2 12
2 3 1 x 2 1
2 1 3 ) tanh x x 1
2 ln (2) csch 35. y 1 1 1
2 1
2 1 1
2 1 ( 2x) sech 2 dy
d 34. y 1/2 sech x) 1 x 11 csch 1)1/2 ) (2(x x)
x2 1/2 1)1/2]2 [2(x 1 cosh 2 1 ln (1) 26. y 1 x2 1 x 1
(1
2
x 1 x1 2 sech 1 x 2)1/2 ln
1 x x x (1 1
x 1 1 x 1
(1 1
x x 2 sech x 2)1/2 sech ln x (1) sech x2 sech x2 1 ln x dy
2x ⇒
dx 1
x1 1
x2 2 1) 2 32. y x x x2 1 2 e ln x
2x
1) 2
1
x 2 (x 1 1 1 2 1 x sech 1 2 (coth x)(1 1) csch (ln 2x) 2 dy
dx coth x 1) sech (ln x)
2
1)
xx ⇒ 3 (coth x)(coth x)
2 cos 1x 31. y ln sinh x (t 1/2 sech2 x
sech x 2 tanh x 1 ) C, then dy
dx sech x, which verifies the formula.
⇒ 1/2 dy
dt (1 1 (1/2)t
1 (t 1/2 )2 t) coth 1 ( 1) coth 1 38. If y
dy
dx t t 2 (t 1/2 ) x2
sech 1 x
2 x sech
x sech 30. y
⇒ (1
dy
dt 1 t 2) coth
(1
2t coth 1 t 1 x x2 1 C, then x
1
2 x 1 x2 2x
41 x2 x, which verifies the formula. t
1 t 2)
1 1 1 2 1
2 t2 ( 2t) coth 1 t 39. If y
dy
dx x2 1
2 x coth coth
1 x 1 x
2
1 x
x 2 2 which verifies the formula. C, then
1
1 x2 1
2 x coth 1 x, 470 Appendix A6 40. If y 1 x tanh dy
then
dx 1
ln (1
2 x
1 tanh x x2) 1
2x
2 1 x2 1 x x2 1 1 tanh sech x, 2x and du 2 dx. x
dx
5 sinh x
2
x
6 cosh
2 ln 3 dx C x
5 12 sinh u 2 sech 51. Let u sinh x, du cosh x dx, the lower limit is e ln 2 e ln 4 4 cosh (3x ln 2) dx 4
sinh (3x
3 ln 2) 4
sinh u
3 C 52. Let u 1
4 4 ln 4 15/8 cosh x
dx
sinh x
15
3
ln
ln
8
4 cosh 2x, du 15
.
8 2 2 1
du
u
15 4
ln
8
3 ln 2 15/8
3/4 ln u C 1
x
and du
dx.
7
7
x
sinh u
tanh dx 7
du
7
cosh u
x
7 ln cosh
C1
7
x/7
x/7
e
e
7 ln
C1
2 e ln 4 coth x dx 3
and the upper
4 2 2 ln 4
ln 2 1
2 2 ln 2 e C 3 dx.
4
cosh u du
3 C C C ln 2 and du C csch u coth u du csch (ln t) limit is sinh (ln 4) 3x t dt
.
t C sinh (ln 2) 12 cosh u du ln 3 5 cosh C csch (ln t) coth (ln t)
dt
t C 1
dx.
2 ln 3 and du x
12 sinh
2 44. Let u 5 cosh u 2 sech u tanh u du ln t and du csch u 5 sinh u du 43. Let u cosh 2x
2 C 1
dx.
5 x
and du
5 42. Let u 50. Let u
cosh u
2 dt .
t 2 t 2( sech u) 1
sinh u du
2 sinh 2x dx t tanh
t which verifies the formula.
41. Let u dt t 1/2 and du t 49. Let u C, 3/4 ln 5
2 0.916 2 sinh (2x) dx, the lower limit is 45. Let u 7 ln e x/7 e 7 ln e x/7 e x/7
x/7 3 coth d 3 ln sinh C1 3 3 ln e / 3 e 3 ln e / 3 e /3 2 sech x tanh x
48. Let u 1
and du
2 x 1
dx
2
1
C
2 (5 csch2 (5
coth u
coth (5 C
C e /3 e ln 4 ln 2 C1 /3 ln 4 (e 2 e ln 2
ln 2 1
32 3
32 ln 2 54.
tanh u 0 sinh 2 0 (1 2 ln 2 d C 2 ln 2 2 e
e ln 4
3
32 ln 4 1
8 0 )d 1
2 2 ln 2 e e 2 2 2 ln 2 d 2 2
0 ln 2 e 4e 2 ln 2 2 dx.
( coth u) 2 ln 4 ln2 4e
ln 2 C d 0.787 ln 2 dx. 0.377 ln 2
ln 4 2 1
8 C1 e
2 e
2 1) d 1
du
u
1
17
ln
2
8 1 e 2e 2 2 C 2 d 2 ln 2 e C1 2 3 ln 2 csch2 u du x) dx ln 2 2e cosh ln 4 sech u du x) and du x) 3 ln sinh u 3 ln /3 47. Let u 53. 17/8 1
sinh 2x
dx
0
2
cosh 2x
1
17
ln
ln 1
2
8 ln 2 . 2 ln 2 tanh 2x dx
1
17/8
ln u
2
1 3
cosh u
3
du
sinh u 3 ln 2 C ln 4 e 17
.
8 2 C1 e ln 4 cosh (ln 4) 1
4 4 0 d 1 and the upper limit is C1
cosh (2 ln 2) 7 ln 2 and du 46. Let u cosh 0
7 ln cosh u ln 2
0 0 e
2 1
4 1 ln 4 3
4 0.636 Appendix A6
/4 55. 1 sinh u
e1 1
1 e e e d sinh (1)
1 e ln 10 e 1 e 1 e 4 sinh2
ln 10 2 (sinh 0 0)] sec d , the 9.9 e e 1 d 2 2(cosh 1
2 2
1 1.086, where u 2 2 ln 10 2 ln 10 ln e
2 1 62. 1
2 1 sin , du cos ln d, 2 1, so (cosh2 x 0 sinh2 x) dx 2 ln 1 8 cosh ln 2 ln 2
sinh u
0 cosh u du 0 e ln 2 sinh (0) x dx 16 16(sinh 2
e2 1
dt, the lower limit is
t 2 e cosh u du e
2 2 e e 3 e e ln ln 3 e ln 3 ln 3 ) x
1 3 3 ln 63. 199 16 sinh u 199 (1 ln 1 2 tanh x tanh2 x) dx (2 0 2 tanh x)2 dx (1 0 2 tanh x sech2 x) dx 199 ln 199 2 ln (cosh x) tanh x
0 39.227, where
dx 2 ln 2 cosh x
dx
2 0 1
(cosh x
2 ln 2
1
[(sinh 0 0)
2
1
e
(0 0)
2
1
2
1
1
2 (1/2)
2
1
4 2 ln 2 , the lower limit is 2 ln 199 2 ln [cosh (ln 199 2 ln 199)] tanh (ln 199) x 2 1 and the upper limit is 0 3
1 2 2 ln u 3 1 3 ln 3 e 3 e e ln 1
1 2x 1 1/2
x
dx
2 x 1/2, du ln 3 1 1 e ln 3 e 0 2 8(e 2 3
3 ln sinh 1)
e e 0 ln t, du 1 3 e ln 3 2 tanh x
ln 3 e ln ln 2 e
2 2 x 16 3 e ln 3 0 and the upper limit is ln 2 1 ln 3 sech2 x dx e ln 0 and the upper limit is 3
, where u
4 2 59. 2 ln 10 5.295 sinh2 x
1 dx 10 3 2 cosh (ln t)
dt
t
2 58. 2 e 1 sinh (ln 2) 4 sinh u du 0 cosh 0) the lower limit is sin 0 57. ln 10 1
10 1 1
2 cosh u
0 sin e 2 2 sinh (sin ) cos 0 ln 10
0 ln 10) 0 56. dx
x 61. cosh2 x /2 1 2 sinh x 1 and the upper limit is 4 cosh x
2 1) dx 1. 4 4 0 (cosh x 0 e ln 10
2 tan , du ln 10 x
dx
2 2[(sinh (ln 10) e 2 2.350, where u lower limit is tan 0 sinh ( 1) 2
1 60. cosh u du 1 e1 2 tan 1 cosh (tan ) sec2 /4 471 4 2. e ln 199 e ln 199 2 0 cosh x 1
dx
ln 2
2
1
0
1) dx
sinh x x
2
ln 2 e ln 2
2 e ln 199 e ln 199 ( 199 ln 199 ln 2 1
ln 2
2 3
8 ln 2 0.722 ln ln 199 ln 2)] ln 2
3
8 199 e ln 199 ln ln 199 (sinh ( ln 2)
ln 2 e ln ln 1.214 199 10,000
199 1/ 199)2
4 2 1/199
4
99
100 199
199
199
199 1
1 2 ln 199
100 1/ 199
1/ 199 99
100 472 Appendix A6
1
cosh 2x ⇒ y
2 64. (a) y ln 5 ln (sinh 2x)2 dx 1 0 sinh ax b
a
0 ln 5
0 ak sin kt bk cos kt ak 2 cos kt k 2(a cos kt 1 sinh2 ax bk 2 sin kt
k 2 s(t) ⇒ acceleration is sin kt) k 2 implies cosh2 ax
that the acceleration is directed toward the origin. b
0 cosh ax dx sinh ab
a f (x) (b) s(t) f ( x) ds
dt
d 2s
⇒2
dt f (x) and O(x) a cosh kt ⇒ f ( x)
. Then
2
2
f (x) f ( x)
f (x) f ( x)
2 f (x)
E(x) O(x)
2
2
2
f ( x) f ( ( x))
f (x) f ( x)
f (x). Also, E( x)
2
2
f ( x) f ( ( x))
E(x) ⇒ E(x) is even, and O( x)
2
f (x) f ( x)
O(x) ⇒ O(x) is odd.
2 65. (a) Let E(x) b sin kt proportional to s. The negative constant
( y )2 cosh2 ax dx 0 ds
dt
d 2s
⇒2
dt cosh 2x dx
2x a cos kt ⇒ 5 0 1 e 2x e
2
2 1
ln 5
sinh 2x
2
0
1
1
6
5
4
5
5
1
(b) y
cosh ax ⇒ 1
a
b ⇒L 68. (a) s(t) sinh 2x ⇒ L b sinh kt ak sinh kt bk cosh kt ak 2 cosh kt k 2 (a cosh kt bk 2 sinh kt
k 2 s(t) ⇒ acceleration is sinh kt) proportional to s. The positive constant k 2 implies that
the acceleration is directed away from the origin. Consequently, f (x) can be written as a sum of an even
69. and an odd function.
ex (b) Even part:
odd part: e x e
2
e x 2 f (x)
2 x
1 ⇒y f (x) f (x) f (x) 2 0⇒C 70. y f ( x)
2 4 cosh f (x) f (x) mg
sech2
k g sech2
Then m dv
dt mg kv 2 mg tanh2 mg sech2 4
4 sech C; x 1 dy 2
dx 1 and (x) sinh2 1 x 2 sinh ln 81 4
4 gk
t.
m kv 2 are equal to the same quantity, so the differential equation is satisfied. Furthermore, the initial
condition is satisfied because v(0) mg
tanh 0
k 0. 2 ln 36
4 ln 6 4 4 ln 6 cosh2 ln 81 4 x
;
4 dy 2
dx
dx ln 16 ln 16 1 cosh x
dx
2 ln 81
2 2 sinh ln 16
2 2 sinh 4 [ln (9 4)2 2
gk
t
m x
4 x ln 81
2 ln 16 4 [ln (81 16) 2
gk
t
m x2 1 2y 1 x
dx
4 cosh2 ln 16 gk
t and
m
mg
tanh
k k dv
and mg
dt gk
t
m f (x)
gk
m 0 ln 16 gk
t.
m mg sech2 mg 1 Thus, m f (x)
2 dv
dt 8 f ( x)
2 f (x) 2 67. Note that 1 dx ln 81 2
f (x) x2 (x) the surface area is S
ln 81 f ( x) x2 1 x
⇒1
4 0 (b) If f is odd, then
f (x) x dx 0⇒y sech y f (x) 1 x2 1
x2 1 x sinh x 2 x
2 1 x x 2 f ( x) 1 ⇒y cosh x 66. (a) If f is even, then
f (x) dy
dx 2 sinh (ln 9)
(e ln9
1
9 9
80
9 e 2 sinh (ln 4)] ln 9 ) 4 (e ln 4 e ln 4 )] 1
4 15
4 320 135
36 16 ln 6 455
9 248.889 Appendix A6
a cosh (x/a) ⇒ y 71. y
⇒y sinh (x/a) (1/a) cosh (x/a) (1/a) sinh2 (x/a) 1 Also, y (0) sinh (0) cosh2 (x/a) (1/a) (y )2. (1/a) 1
0 and y(0) a cosh (0) a. 72. (a) Let the point located at (cosh x, 0) be called T. Then
A(u) area of the triangle the curve y 1 from A to T
cosh u 1
cosh u sinh u
2 ⇒ A(u)
(b) A(u) x2 OTP minus the area under 1
cosh u sinh u
2 x2 1
cosh u
1 x2 1 dx.
1 dx ⇒ A (u)
1
(cosh2 u
2
1
cosh2 u
2
1
(cosh2 u
2 (c) A (u)
A(0) sinh2 u) cosh2 u 1
sinh2 u
2 sinh2 u) 1
(1)
2 1)(sinh u) sinh u 1
⇒ A(u)
2 0⇒C ( u
2 2 1
2 C, and from part (a) we have 0 ⇒ A(u) u
⇒u
2 2A 473 ...
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This note was uploaded on 02/08/2012 for the course HJKJK k101 taught by Professor Bjhj during the Spring '11 term at Westminster Theological Seminary.
 Spring '11
 bjhj
 The Land

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