PROBLEMS
CHAPTER CHALLENGE 3
A Click here for answers. 1. (a) Find the domain of the function f x S
Click here for solutions.
;
s1 s2 s 3 x . (b) Find f x . (c) Check your work in parts (a) and (b) by graphing f and f on the same screen.
CHAPTER 4
A Click here for answers. S
Click here for solutions.
1. Find the absolute maximum value of the function
fx
C D
1 2
1 1 x 1
1 x
2
2. (a) Let ABC be a triangle with right angle A and hypotenuse a
BC . (See the gure.) If the inscribed circle touches the hypotenuse at D, show that CD
1 2
(
BC
AC
AB
)
(b) If C, express the radius r of the inscribed circle in terms of a and . (c) If a is xed and varies, nd the maximum value of r.
A FIGURE FOR PROBLEM 2 B 3. A triangle with sides a, b, and c varies with time t, but its area never changes. Let
be the
angle opposite the side of length a and suppose always remains acute. (a) Express d dt in terms of b, c, , db dt, and dc dt. (b) Express da dt in terms of the quantities in part (a).
4. Let a and b be positive numbers. Show that not both of the numbers a 1
b and b 1
a
can be greater than 1. 4
5. Let ABC be a triangle with
BAC 120 and AB AC (a) Express the length of the angle bisector AD in terms of x (b) Find the largest possible value of AD .
1. AB .
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
CHAPTER 5
A Click here for answers. S
Click here for solutions.
1. Show that
1 17
y1
2
1 1 x4
dx
7 . 24
2. Suppose the curve y
f x passes through the origin and the point 1, 1 . Find the value of
the integral x01 f x dx.
3. In Sections 5.1 and 5.2 we used the formulas for the sums of the k th powers of the rst n
integers when k 1, 2, and 3. (These formulas are proved in Appendix E.) In this problem we derive formulas for any k. These formulas were rst published in 1713 by the Swiss mathematician James Bernoulli in his book Ars Conjectandi. (a) The Bernoulli polynomials Bn are dened by B0 x 1, Bn x Bn 1 x , and x01 Bn x dx 0 for n 1, 2, 3, . . . . Find Bn x for n 1, 2, 3, and 4. (b) Use the Fundamental Theorem of Calculus to show that Bn 0 Bn 1 for n 2.
1
2 CHALLENGE PROBLEMS
(c) If we introduce the Bernoulli numbers bn B0 x B2 x b0 x2 2! b1 x 1! 1! b2 2! B1 x B3 x
n! Bn 0 , then we can write x 1! x3 3! b1 1! b1 x 2 1! 2! b2 x 2! 1! b3 3!
and, in general, Bn x 1 n!
n k0
n bk x n k
k
where
n k
n! k! n k ! 2,
[The numbers ( n ) are the binomial coefcients.] Use part (b) to show that, for n k
n
bn
k0
n bk k
and therefore bn 1 n n b0 0 n b1 1 n b2 2 n n 2 bn
1
2
(d) (e)
;
(f) (g) (h)
This gives an efcient way of computing the Bernoulli numbers and therefore the Bernoulli polynomials. Show that Bn 1 x 1 nBn x and deduce that b2n 1 0 for n 0. Use parts (c) and (d) to calculate b6 and b8 . Then calculate the polynomials B5 , B6 , B7 , B8 , and B9 . Graph the Bernoulli polynomials B1, B2 , . . . , B9 for 0 x 1. What pattern do you notice in the graphs? Use mathematical induction to prove that Bk 1 x 1 Bk 1 x x k k! . By putting x 0, 1, 2, . . . , n in part (g), prove that 1k 2k 3k nk k! Bk
1
n
1
Bk
1
0
k! y
n1
0
Bk x dx
(i) Use part (h) with k 3 and the formula for B4 in part (a) to conrm the formula for the sum of the rst n cubes in Section 5.2. (j) Show that the formula in part (h) can be written symbolically as 1k 2k 3k nk 1 k 1 n 1 b
k1
bk
1
CHAPTER 6
A Click here for answers. S
Click here for solutions.
1. A solid is generated by rotating about the x-axis the region under the curve y
f is a positive function and x x 0 to x b is b 2 for all b
f x , where 0. The volume generated by the part of the curve from 0. Find the function f.
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
where the expression n 1 b k 1 is to be expanded formally using the Binomial Theorem and each power b i is to be replaced by the Bernoulli number bi. (k) Use part (j) to nd a formula for 15 2 5 3 5 n 5.equator that have exactly the same temperature.
CHALLENGE PROBLEMS 3
CHAPTER 8
A Click here for answers. S
Click here for solutions.
; 1. The Chebyshev polynomials Tn are dened by Tn x
cos n arccos x , n 0, 1, 2, 3, . . . . (a) What are the domain and range of these functions? (b) We know that T0 x 1 and T1 x x. Express T2 explicitly as a quadratic polynomial and T3 as a cubic polynomial. (c) Show that, for n 1, Tn 1 x 2x Tn x Tn 1 x . (d) Use part (c) to show that Tn is a polynomial of degree n. (e) Use parts (b) and (c) to express T4 , T5 , T6 , and T7 explicitly as polynomials. (f) What are the zeros of Tn ? At what numbers does Tn have local maximum and minimum values? (g) Graph T2 , T3 , T4 , and T5 on a common screen. (h) Graph T5 , T6 , and T7 on a common screen. (i) Based on your observations from parts (g) and (h), how are the zeros of Tn related to the zeros of Tn 1 ? What about the x-coordinates of the maximum and minimum values? (j) Based on your graphs in parts (g) and (h), what can you say about x1 1 Tn x dx when n is odd and when n is even? (k) Use the substitution u arccos x to evaluate the integral in part (j). (l) The family of functions f x cos c arccos x are dened even when c is not an integer (but then f is not a polynomial). Describe how the graph of f changes as c increases.
CHAPTER 11
A Click here for answers. S
Click here for solutions.
1. A circle C of radius 2r has its center at the origin. A circle of radius r rolls without slipping in
;
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
the counterclockwise direction around C. A point P is located on a xed radius of the rolling circle at a distance b from its center, 0 b r. [See parts (i) and (ii) of the gure.] Let L be the line from the center of C to the center of the rolling circle and let be the angle that L makes with the positive x-axis. (a) Using as a parameter, show that parametric equations of the path traced out by P are x b cos 3 3r cos , y b sin 3 3r sin . Note: If b 0, the path is a circle of radius 3r; if b r, the path is an epicycloid. The path traced out by P for 0 b r is called an epitrochoid. (b) Graph the curve for various values of b between 0 and r . (c) Show that an equilateral triangle can be inscribed in the epitrochoid and that its centroid is on the circle of radius b centered at the origin. Note: This is the principle of the Wankel rotary engine. When the equilateral triangle rotates with its vertices on the epitrochoid, its centroid sweeps out a circle whose center is at the center of the curve. (d) In most rotary engines the sides of the equilateral triangles are replaced by arcs of circles centered at the opposite vertices as in part (iii) of the gure. (Then the diameter of the rotor is constant.) Show that the rotor will t in the epitrochoid if b 3(2 s3 )r 2.
y
y
P P=P 2r r b
x
P
x
(i)
FIGURE FOR PROBLEM 1
(ii)
(iii)
4 CHALLENGE PROBLEMS
CHAPTER 12
S
Click here for solutions.
Show that, for n sin (b) Deduce that sin cos 2 cos 4 cos 8 1, 2, 3, . . . , 2 n sin cos cos cos cos
1. (a)
2n
2
4
8
2n
The meaning of this innite product is that we take the product of the rst n factors and then we take the limit of these partial products as n l . (c) Show that 2 s2 s2 s2 2 2
s2
s2
2
s2
This innite product is due to the French mathematician Francois Vite (15401603). Notice that it expresses in terms of just the number 2 and repeated square roots.
2. Suppose that a 1
cos , an
1
2
1 2
2, b1 an bn
1, and bn
1
sbn an
1
Use Problem 1 to show that
nl
lim an
nl
lim bn
sin
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
CHALLENGE PROBLEMS 5
ANSWERS
Chapter 3
S
Solutions
(b) 8 1 1 2 3x 2 3x 3x
1. (a) [1, 2]
Chapter 4
S
Solutions
4 3
1.
1 db 1 dc + 3. (a) tan c dt b dt
5. (a) y =
b (b) (b)
1 2
x ,x>0 x2 + 1
dc dc db db +c b +c sec dt dt dt dt b2 + c2 2bc cos
Chapter 5
S
Solutions
12 2x
3. (a) B1 (x) = x 1 , B2 (x) = 2
1x + 2
1 12 ,
B3 (x) = 1 x3 1 x2 + 6 4
1 12 x,
B4 (x) =
14 24 x
13 12 x
+
12 24 x
1 720
(e) b6 =
1 , 42
1 b8 = 30 ; 1 120 1 5040
B5 (x) = B7 (x) = B9 (x) =
x5 5 x4 + 5 x3 1 x , B6 (x) = 2 3 6
1 720
x6 3x5 + 5 x4 1 x2 + 2 2
1 40,320
1 42
,
1 30
x7 7 x6 + 7 x5 7 x3 + 1 x , B8 (x) = 2 2 6 6 x9 9 x8 + 6x7 2
21 5 5x
x8 4x7 +
14 6 x 3
7 x4 + 2 x2 3 3
,
1 362,880
+ 2x3
3 10 x
(f ) There are four basic shapes for the graphs of Bn (excluding B1 ), and as n increases, they repeat in a cycle of four. For n = 4m, the shape resembles that of the graph of cos 2x; for n = 4m + 1, that of sin 2x; for n = 4m + 2, that of cos 2x; and for n = 4m + 3, that of sin 2x. (k)
12 12 n (n
+ 1)2 (2n2 + 2n 1)
Chapter 6
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
S
Solutions
2x/
1. f (x) =
Chapter 7
S
Solutions
1. (a) [1, 1]; [1, 1] for n > 0
(b) T2 (x) = 2x2 1, T3 (x) = 4x3 3x (e) T4 (x) = 8x4 8x2 + 1, T5 (x) = 16x5 20x3 + 5x, T6 (x) = 32x6 48x4 + 18x2 1, T7 (x) = 64x7 112x5 + 56x3 7x (f ) x = cos k + n
2
, k an integer with 0 k < n; x = cos(k/n), k an integer with 0 < k < n
6 CHALLENGE PROBLEMS
(g)
(h)
(i) The zeros of Tn and Tn+1 alternate; the extrema also alternate ( j) When n is odd, and so as n gets larger.
1 1
Tn (x) dx = 0; when n is even, the integral is negative, but decreases in absolute value
(k)
0
(l ) As c increases through an integer, the graph of f gains a local extremum, which starts at x = 1 and moves rightward, compressing the graph of f as c continues to increase.
2 if n is even n2 1 cos(nu) sin u du = 0 if n is odd
Chapter 10
S
Solutions
1. (b)
b = 1r 5
b = 2r 5
b = 3r 5
b = 4r 5
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
CHALLENGE PROBLEMS 7
SOLUTIONS
E
Exercises
Chapter 3
1. (a) f (x) =
3x D = x | 3 x 0, 2 3 x 0, 1 1 2 = x | 3 x, 2 3 x, 1
2
3x0
2 3x = x | 3 x, 4 3 x, 1 2 3 x = x | x 3, x 1, 1 3 x
= {x | x 3, x 1, 1 3 x } = {x | x 3, x 1, x 2 } = {x | 1 x 2 } = [1, 2] (b) f (x) = f 0 (x) = 1 = 2 = (c) 1 8 1 1 2 1 3x 1 2 3x
d dx 2 3x 1 2 3x 2
2
1 d 2 3x 3 x dx
1 2 3x 2 3x 3x Note that f is always decreasing and f 0 is always negative.
E
Exercises
Chapter 4
1. f(x) =
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
We see that f 0 (x) > 0 for x < 0 and f 0 (x) < 0 for x > 2. For 0 < x < 2, we have f 0 (x) =
1 1 + 1 + |x| 1 + |x 2| 1 1 + 1 x 1 (x 2) 1 1 + = 1 (x 2) 1 + x 1 1 + 1+x 1 + (x 2)
if x < 0 if 0 x < 2 if x 2
1 1 + (1 x)2 (3 x)2 1 1 + f 0 (x) = (1 + x)2 (3 x)2 1 1 (1 + x)2 (x 1)2
if x < 0 if 0 < x < 2 if x > 2
x2 + 2x + 1 x2 6x + 9 1 1 8 (x 1) = , so f 0 (x) < 0 for 2 2= (3 x) (x + 1) (3 x)2 (x + 1)2 (3 x)2 (x + 1)2
0 < x < 1, f 0 (1) = 0 and f 0 (x) > 0 for 1 < x < 2. We have shown that f 0 (x) > 0 for x < 0; f 0 (x) < 0 for 0 < x < 1; f 0 (x) > 0 for 1 < x < 2; and f 0 (x) < 0 for x > 2. Therefore, by the First Derivative Test, the local maxima of f are at x = 0 and x = 2, where f takes the value 4 . Therefore, 3
4 3
is the absolute maximum value of f .
8 CHALLENGE PROBLEMS
3. (a) A =
1 bh 2
with sin = h/c, so A = 1 bc sin . But A is a 2
constant, so differentiating this equation with respect to t, we get dA 1 d dc db =0= bc cos +b sin + c sin dt 2 dt dt dt bc cos d dc db = sin b +c dt dt dt
d 1 dc 1 db = tan + . dt c dt b dt
(b) We use the Law of Cosines to get the length of side a in terms of those of b and c, and then we differentiate implicitly with respect to t: a2 = b2 + c2 2bc cos 2a
da db dc d dc db = 2b + 2c 2 bc( sin ) +b cos + c cos dt dt dt dt dt dt
1 db dc d dc db da = b +c + bc sin b cos c cos . Now we substitute our value of a from the Law dt a dt dt dt dt dt of Cosines and the value of d/dt from part (a), and simplify (primes signify differentiation by t): bb0 + cc0 + bc sin [ tan (c0/c + b0/b)] (bc0 + cb0 )(cos ) da = dt b2 + c2 2bc cos = bb0 + cc0 [sin2 (bc0 + cb0 ) + cos2 (bc0 + cb0 )]/ cos bb0 + cc0 (bc0 + cb0 )sec = b2 + c2 2bc cos b2 + c2 2bc cos
5. (a) Let y = |AD|, x = |AB|, and 1/x = |AC|, so that |AB| |AC| = 1.
We compute the area A of 4ABC in two ways. First, A=
1 2
|AB| |AC| sin 2 = 3
1 2
1
3 2
=
3 4.
Second,
A = (area of 4ABD) + (area of 4ACD) =
1 2
|AB| |AD| sin + 3
1 2
|AD| |AC| sin = 1 xy 3 2
3 y 4
3 2
+ 1 y(1/x) 2 1 x =
3 4
3 2
=
3 4 y(x
+ 1/x)
Equating the two expressions for the area, we get
x+
y=
x 1 =2 , x > 0. x + 1/x x +1
Another method: Use the Law of Sines on the triangles ABD and ABC. In 4ABD, we have A + B + D = 180 60 + + D = 180 D = 120 . Thus,
3 2
by a similar argument with 4ABC, y= x , x > 0. x2 + 1
3 2
cot = x2 + 1 . Eliminating cot gives 2
x = x2 + y
1 2
+
1 2
(b) We differentiate our expression for y with respect to x to nd the maximum: x2 + 1 x(2x) dy 1 x2 = = = 0 when x = 1. This indicates a maximum by the First Derivative Test, 2 + 1)2 dx (x (x2 + 1)2 since y 0 (x) > 0 for 0 < x < 1 and y 0 (x) < 0 for x > 1, so the maximum value of y is y(1) = 1 . 2
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
x sin(120 ) sin 120 cos cos 120 sin = = = y sin sin
cos + 1 sin 2 sin
x = y
3 2
cot + 1 , and 2
CHALLENGE PROBLEMS 9
E
Exercises
Chapter 5
1. For 1 x 2, we have x4 24 = 16, so 1 + x4 17 and
2 1 2 1
1 dx 1 + x4 1 dx < 1 + x4
2 1
2 1 2 1
1 1 . Thus, 1 + x4 17 1 1 1 1 < 4 and dx = . Also 1 + x4 > x4 for 1 x 2, so 17 17 1 + x4 x x3 3
2 1
x4 dx =
=
1 1 7 += . Thus, we have the estimate 24 3 24
1 17
7 1 dx . 1 + x4 24
0 3. (a) To nd B1 (x), we use the fact that B1 (x) = B0 (x)
B1 (x) =
1 (x 0
B0 (x) dx =
1 21 x0 2 1 2
1 dx = x + C. Now we
1
impose the condition that
1 0
B1 (x) dx = 0 0 =
+ C) dx = x
+ Cx
0
=
1 2
+C
C = 1 . So B1 (x) = x 1 . Similarly B2 (x) = 2 2
1 0
B1 (x) dx =
1 6
dx = 1 x2 1 x + D. But 2 2
1 12 ,
B2 (x) dx = 0 0 =
1 . 12
1 0
12 2x
1 x + D dx = 2 B2 (x) dx = 1 x2 + 4
1 x 12
1 4
+D
D=
1 12 1 12
so
1 x 12
B2 (x) = 1 x2 1 x + 2 2
1 0
B3 (x) =
1 0 13 x 6
12 x 2
+ 1x 2
1 24
dx = 1 x3 1 x2 + 6 4 +
1 24
+ E. But
B3 (x) dx = 0 0 =
1 x. 12
+ E dx =
13 x 6 12 x 24
+E
E = 0. So
14 x 24 1 72
B3 (x) = 1 x3 1 x2 + 6 4 But
1 0
B4 (x) =
1 0
B3 (x) dx =
13 x 12
1 x2 + 4
1 x 12 1 120
dx =
1 48
13 x 12
+
12 x 24
+ F.
B4 (x) dx = 0 0 =
14 24 x
14 x 24 1 720 .
+
+ F dx =
+
+F
1 F = 720 .
So B4 (x) =
13 12 x
+
12 24 x 1 0
(b) By FTC2, Bn (1) Bn (0) = Bn (0) = Bn (1) for n 2. (c) We know that Bn (x) = Bn (1) = Bn (0) = bn =
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved. n 0
0 Bn (x) dx =
1 0
Bn1 (x) dx = 0 for n 1 1, by denition. Thus,
1n n! k=0
n k
bk xnk . If we set x = 1 in this expression, and use the fact that
n k
n bn for n 2, we get bn = n! k=0 n n2
bk . Now if we expand the right-hand side, we get
n n
b0 +
n 1
b1 + +
n n1
bn2 +
n n1
bn1 +
n 0
bn . We cancel the bn terms, move the bn1 term to
n 1
the LHS and divide by
1 = n: bn1 = n
b0 +
b1 + +
n n2
bn2 for n 2, as required.
(d) We use mathematical induction. For n = 0: B0 (1 x) = 1 and (1)0 B0 (x) = 1, so the equation holds for n = 0 since b0 = 1. Now if Bk (1 x) = (1)k Bk (x), then since
d dx Bk+1 (1 0 d x) = Bk+1 (1 x) dx (1 x) = Bk (1 x), we have d dx Bk+1 (1
x) = (1)(1)k Bk (x) = (1)k+1 Bk (x). Integrating, we get
Bk+1 (1 x) = (1)k+1 Bk+1 (x) + C. But the constant of integration must be 0, since if we substitute x = 0 in the equation, we get Bk+1 (1) = (1)k+1 Bk+1 (0) + C, and if we substitute x = 1 we get Bk+1 (0) = (1)k+1 Bk+1 (1) + C, and these two equations together imply that Bk+1 (0) = (1)k+1 (1)k+1 Bk+1 (0) + C + C = Bk+1 (0) + 2C C = 0.
So the equation holds for all n, by induction. Now if the power of 1 is odd, then we have
10 CHALLENGE PROBLEMS
B2n+1 (1 x) = B2n+1 (x). In particular, B2n+1 (1) = B2n+1 (0). But from part (b), we know that Bk (1) = Bk (0) for k > 1. The only possibility is that B2n+1 (0) = B2n+1 (1) = 0 for all n > 0, and this implies that b2n+1 = (2n + 1)! B2n+1 (0) = 0 for n > 0.
1 (e) From part (a), we know that b0 = 0! B0 (0) = 1, and similarly b1 = 1 , b2 = 1 , b3 = 0 and b4 = 30 . 2 6
We use the formula to nd b6 = b71 = 1 7 7 b0 + 0 7 b1 + 1 7 b2 + 2 7 b3 + 3 7 b4 + 4 7 b5 5
The b3 and b5 terms are 0, so this is equal to Similarly, b8 = = = Now we can calculate B5 (x) = = = 15 5 bk x5k 5! k=0 k 1 1 x5 + 5 120 2
1 120
1 1 1+7 7 2
+
76 21
1 6
+
765 321 9 b2 + 2 1 6 +
1 30
=
1 7
1
7 7 7 + 2 2 6
=
1 42
1 9
9 b0 + 0
9 b1 + 1 + 98 21
9 b4 + 4
9 b6 6 1 30 + 987 321 1 42
1 1 1+9 9 2 1 9 1
9 21 +6 +2 2 5
=
1 30
9876 4321
x4 +
54 21
1 6
x3 + 5
1 30
x
x5 5 x4 + 5 x3 1 x 2 3 6 x5 + 65 21
1 42
B6 (x) = =
1 1 x6 + 6 720 2
1 720
1 6
x4 +
65 21
1 30
x2 +
1 42
x6 3x5 + 5 x4 1 x2 + 2 2 x6 +
=
1 5040
x7 7 x6 + 7 x5 7 x3 + 1 x 2 2 6 6 x7 + 87 21 1 6 x6 +
1 30
B8 (x) = =
1 1 x8 + 8 40,320 2
1 40,320
8765 4321
1 30
x4 +
87 21
1 42
x2 +
1 30
x8 4x7 +
14 6 3x
7 x4 + 2 x2 3 3 x8 + 98 21 1 6
B9 (x) =
1 1 x9 + 9 362,880 2
x7 +
9876 4321
1 30 +
x5 987 321 1 42 x3 + 9 1 30 x
=
1 362,880
x9 9 x8 + 6x7 2
21 5 5x
+ 2x3
3 10 x
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
B7 (x) =
1 1 x7 + 7 5040 2
76 21
1 6
x5 +
765 321
1 30
x3 + 7
1 42
x
CHALLENGE PROBLEMS 11
(f )
n=1
n=2
n=3
n=4
n=5
n=6
n=7
n=8
n=9
There are four basic shapes for the graphs of Bn (excluding B1 ), and as n increases, they repeat in a cycle of four. For n = 4m, the shape resembles that of the graph of cos 2x; For n = 4m + 1, that of sin 2x; for n = 4m + 2, that of cos 2x; and for n = 4m + 3, that of sin 2x. x0 = 1, so the equation holds for k = 0. We 0!
(g) For k = 0: B1 (x + 1) B1 (x) = x + 1 now assume that Bn (x + 1) Bn (x) =
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
1 2
x
1 2
= 1, and
xn1 . We integrate this equation with respect to x: (n 1)!
[Bn (x + 1) Bn (x)] dx = Bn+1 (x) =
xn1 dx. But we can evaluate the LHS using the denition (n 1)! 1n x n 1n x , since by part (b) Bn+1 (1) Bn+1 (0) = 0, and so the n!
Bn (x) dx, and the RHS is a simple integral. The equation becomes 1 (n 1)! =
Bn+1 (x + 1) Bn+1 (x) =
constant of integration must vanish. So the equation holds for all k, by induction.
(h) The result from part (g) implies that pk = k! [Bk+1 (p + 1) Bk+1 (p)]. If we sum both sides of this equation from p = 0 to p = n (note that k is xed in this process), we get
n p=0
pk = k!
n p=0
[Bk+1 (p + 1) Bk+1 (p)]. But the
12 CHALLENGE PROBLEMS
RHS is just a telescoping sum, so the equation becomes 1k + 2k + 3k + + nk = k! [Bk+1 (n + 1) Bk+1 (0)]. But from the denition of Bernoulli polynomials (and using the Fundamental Theorem of Calculus), the RHS is equal to k!
n+1 0
Bk (x) dx.
(i) If we let k = 3 and then substitute from part (a), the formula in part (h) becomes 13 + 23 + + n3 = 3! [B4 (n + 1) B4 (0)] =6 =
1 (n 24
+ 1)4
1 (n 12
+ 1)3 +
1 (n 24
+ 1)2
1 720
1 24
1 12
+
1 24
1 720 2
(n + 1)2 [1 + (n + 1)2 2(n + 1)] (n + 1)2 [1 (n + 1)]2 n(n + 1) = = 4 4 2
n+1
(j) 1k + 2k + 3k + + nk = k! = k!
Bk (x) dx [by part (h)]
0 n+1 0
1k k bj xkj dx = k! j=0 j
n+1 0
k j=0
k bj xkj dx j
Now view
k j=0
k bj xkj as (x + b)k , as explained in the problem. Then j
n+1 0
1k + 2k + 3k + + nk =
(x + b)k dx =
(x + b)k+1 k+1
n+1
=
0
(n + 1 + b)k+1 bk+1 k+1
(k) We expand the RHS of the formula in (j), turning the bi into bi , and remembering that b2i+1 = 0 for i > 0: 15 + 25 + + n5 = = = = = =
1 6 1 6 1 6
(n + 1)6 b6 (n + 1)6 + 6(n + 1)5 b1 +
65 (n 21
+ 1)4 b2 +
65 (n 21
+ 1)2 b4
(n + 1)6 3(n + 1)5 + 5 (n + 1)4 1 (n + 1)2 2 2 + 1)2 2(n + 1)4 6(n + 1)3 + 5(n + 1)2 1 + 1)2 [(n + 1) 1]2 2(n + 1)2 2(n + 1) 1 + 1)2 (2n2 + 2n 1)
1 12 (n 1 (n 12
12 n (n 12
E
Exercises
Chapter 6
b 0
1. The volume generated from x = 0 to x = b is
[f(x)]2 dx. Hence, we are given that b2 =
b 0
[f (x)]2 dx
for all b > 0. Differentiating both sides of this equation using the Fundamental Theorem of Calculus gives 2b = [f (b)]2 f (b) = 2b/, since f is positive. Therefore, f(x) = 2x/.
E
Exercises
Chapter 8
1. (a) Tn (x) = cos(n arccos x). The domain of arccos is [1, 1], and the domain of cos is R, so the domain of Tn (x)
is [1, 1]. As for the range, T0 (x) = cos 0 = 1, so the range of T0 (x) is {1}. But since the range of n arccos x is at least [0, ] for n > 0, and since cos y takes on all values in [1, 1] for y [0, ], the range of Tn (x) is [1, 1] for n > 0.
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
CHALLENGE PROBLEMS 13
(b) Using the usual trigonometric identities, T2 (x) = cos(2 arccos x) = 2 [cos(arccos x)]2 1 = 2x2 1, and T3 (x) = cos(3 arccos x) = cos(arccos x + 2 arccos x) = cos(arccos x) cos(2 arccos x) sin(arccos x) sin(2 arccos x) = x 2x2 1 sin(arccos x) [2 sin(arccos x) cos(arccos x)] = 2x3 x 2 sin2 (arccos x) x = 2x3 x 2x 1 cos2 (arccos x) = 2x3 x 2x 1 x2 = 4x3 3x (c) Let y = arccos x. Then Tn+1 (x) = cos[(n + 1)y] = cos(y + ny) = cos y cos ny sin y sin ny = 2 cos y cos ny (cos y cos ny + sin y sin ny) = 2xTn (x) cos(ny y) = 2xTn (x) Tn1 (x) (d) Here we use induction. T0 (x) = 1, a polynomial of degree 0. Now assume that Tk (x) is a polynomial of degree k. Then Tk+1 (x) = 2xTk (x) Tk1 (x). By assumption, the leading term of Tk is ak xk , say, so the leading term of Tk+1 is 2xak xk = 2ak xk+1 , and so Tk+1 has degree k + 1.
(e) T4 (x) = 2xT3 (x) T2 (x) = 2x 4x3 3x 2x2 1 = 8x4 8x2 + 1, T5 (x) = 2xT4 (x) T3 (x) = 2x 8x4 8x2 + 1 4x3 3x = 16x5 20x3 + 5x, T6 (x) = 2xT5 (x) T4 (x) = 2x 16x5 20x3 + 5x 8x4 8x2 + 1 = 32x6 48x4 + 18x2 1, T7 (x) = 2xT6 (x) T5 (x) = 2x 32x6 48x4 + 18x2 1 16x5 20x3 + 5x = 64x7 112x5 + 56x3 7x
(f ) The zeros of Tn (x) = cos(n arccos x) occur where n arccos x = k + cos(n arccos x) = cos k + continue: n arccos x = k +
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved. 2 2
2
for some integer k, since then
= 0. Note that there will be restrictions on k, since 0 arccos x . We arccos x = k + n
2
. This only has solutions for 0
k + n
2
0 < k +
2
< n
0 k < n. [This makes sense, because then Tn (x) has n zeros, and it is a polynomial of
degree n.] So, taking cosines of both sides of the last equation, we nd that the zeros of Tn (x) occur at x = cos k + n
2
, k an integer with 0 k < n. To nd the values of x at which Tn (x) has local extrema, we set sin(n arccos x) = 0
n n sin(n arccos x) 0 0 = Tn (x) = sin(n arccos x) = 2 1x 1 x2
n arccos x = k, k some integer arccos x = k/n. This has solutions for 0 k n, but we disallow the cases k = 0 and k = n, since these give x = 1 and x = 1 respectively. So the local extrema of Tn (x) occur at x = cos(k/n), k an integer with 0 < k < n. [Again, this seems reasonable, since a polynomial of degree n has at
14 CHALLENGE PROBLEMS
most (n 1) extrema.] By the First Derivative Test, the cases where k is even give maxima of Tn (x), since then n arccos [cos(k/n)] = k is an even multiple of , so sin (n arccos x) goes from negative to positive at x = cos(k/n). Similarly, the cases where k is odd represent minima of Tn (x). (g) (h)
(i) From the graphs, it seems that the zeros of Tn and Tn+1 alternate; that is, between two adjacent zeros of Tn , there is a zero of Tn+1 , and vice versa. The same is true of the x-coordinates of the extrema of Tn and Tn+1 : between the x-coordinates of any two adjacent extrema of one, there is the x-coordinate of an extremum of the other. ( j) When n is odd, the function Tn (x) is odd, since all of its terms have odd degree, and so
1 1
Tn (x) dx = 0. When
n is even, Tn (x) is even, and it appears that the integral is negative, but decreases in absolute value as n gets larger. (k)
1 1
Tn (x) dx =
1 1
cos(n arccos x) dx. We substitute u = arccos x x = cos u dx = sin u du,
x = 1 u = , and x = 1 u = 0. So the integral becomes
cos(nu) sin u du =
0 0
1 [sin(u 2
nu) + sin(u + nu)] du
0
=
(l ) From the graph, we see that as c increases through an
integer, the graph of f gains a local extremum, which starts at x = 1 and moves rightward, compressing the graph of f as c continues to increase.
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.
1 1 1 2 n1 n+1 = 1 1 1 2 n1 n+1 2 if n is even n2 1 = 0 if n is odd
1 cos[(1 n)u] cos[(1 + n)u] 2 n1 n+1
1 1 n1 n+1 1 1 n1 n+1
if n is even if n is odd
CHALLENGE PROBLEMS 15
E
Exercises
Chapter ` 1 1
1. (a) Since the smaller circle rolls without slipping around C, the amount of
arc traversed on C (2r in the gure) must equal the amount of arc of the smaller circle that has been in contact with C. Since the smaller circle has radius r, it must have turned through an angle of 2r/r = 2. In addition to turning through an angle 2, the little circle has rolled through an angle against C. Thus, P has turned through an angle of 3 as shown in the gure. (If the little circle had turned through an angle of 2 with its center pinned to the x-axis, then P would have turned only 2 instead of 3. The movement of the little circle around C adds to the angle.) From the gure, we see that the center of the small circle has coordinates (3r cos , 3r sin ). Thus, P has coordinates (x, y), where x = 3r cos + b cos 3 and y = 3r sin + b sin 3. (b)
b = 1r 5
b = 2r 5
b = 3r 5
b = 4r 5
(c) The diagram gives an alternate description of point P on the epitrochoid. Q moves around a circle of radius b, and P rotates one-third as fast with respect to Q at a distance of 3r. Place an equilateral triangle with sides of length 3 3r so that its centroid is at Q and one vertex is at P . (The distance from the centroid to a vertex is the equilateral triangle.) As increases by
2 3, 1 3
times the length of a side of
the point Q travels once around the circle of radius
b, returning to its original position. At the same time, P (and the rest of the triangle) rotate through an angle of
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved. 2 3
about Q, so P s position is
occupied by another vertex. In this way, we see that the epitrochoid traced out by P is simultaneously traced out by the other two vertices as well. The whole equilateral triangle sits inside the epitrochoid (touching it only with its vertices) and each vertex traces out the curve once while the centroid moves around the circle three times. (d) We view the epitrochoid as being traced out in the same way as in part (c), by a rotor for which the distance from its center to each vertex is 3r, so it has radius 6r. To show that the rotor ts inside the epitrochoid, it sufces to show that for any position of the tracing point P , there are no points on the opposite side of the rotor which are outside the epitrochoid. But the most likely case of intersection is when P is on the y-axis, so as long as the diameter of the rotor (which is 3 3r) is less than the distance between the y-intercepts, the rotor will t. The y-intercepts occur
16 CHALLENGE PROBLEMS
when = t if 3
2
or =
3 2
y = (3r b), so the distance between the intercepts is 6r 2b, and the rotor will
3 (2 3 ) 3r 6r 2b b r. 2
E
Exercises
Chapter 12
1. (a) sin = 2 sin
cos = 2 2 sin cos 2 2 4 4
cos
= 2 2 2 sin cos 2 8 8 cos 2n1 cos 8
cos
4 4
cos
2 2
= = 2 2 2 2 2 sin = 2n sin (b) sin = 2n sin
cos n 2n 2
cos
cos
cos cos cos cos n 2n 2 4 8 2 cos cos cos cos n 2n 2 4 8 2
x0
/2n sin = cos cos cos cos n . sin (/2n ) 2 4 8 2
Now we let n , using lim
n
sin x = 1 with x = n : x 2 sin = cos cos cos . 2 4 8
1 2 (1
lim
sin /2n = lim cos cos cos cos n n sin (/2n ) 2 4 8 2
2
(c) If we take =
in the result from part (b) and use the half-angle formula cos x =
+ cos 2x)
(see Formula 17a in Appendix D), we get cos + 1 4 +1 2 +1 2 2 2+ 2 2 2
cos sin /2 = cos 4 /2 cos + 1 4 2
2 2
4
+1 +1 2 2
2 2 = 2 = 2 2
2 2
+1 2 2 2+
+1 +1 2 2 = 2 2 2+ 2
2+ 2 2
+1
2+ 2
2
Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.