CHALLENGE PROBLEMS

CHALLENGE PROBLEMS

CHAPTER CHAPTER 2 3 A Click here for answers.

Click here for solutions.

S

s1 ⫺ s2 ⫺ s 3 ⫺ x . (b) Find f ⬘共x兲. (c) Check your work in parts (a) and (b) by graphing f and f ⬘ on the same screen.

1. (a) Find the domain of the function f 共x兲苷

;

CHAPTER CHAPTER 34 A Click here for answers.


S

1. Find the absolute maximum value of the function

f 共x兲苷

1 1 ⫹ 1⫹ x 1⫹ x⫺2

ⱍ ⱍ

ⱍ

ⱍ ⱍ

ⱍ

2. (a) Let ABC be a triangle with right angle A and hypotenuse a 苷 BC . (See the

C

figure.) If the inscribed circle touches the hypotenuse at D, show that

ⱍ CD ⱍ 苷 (ⱍ BC ⱍ ⫹ ⱍ AC ⱍ ⫺ ⱍ AB ⱍ) 1 2

D

(b) If ␪ 苷 12 ⬔C, express the radius r of the inscribed circle in terms of a and ␪. (c) If a is fixed and ␪ varies, find the maximum value of r. A

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␪兾dt in terms of b, c, ␪, db兾dt, and dc兾dt. (b) Express da兾dt in terms of the quantities in part (a).

FIGURE FOR PROBLEM 2

4. Let a and b be positive numbers. Show that not both of the numbers a共1 ⫺ b兲 and b共1 ⫺ a兲

can be greater than 14.

ⱍ ⱍ ⱍ

ⱍ

5. Let ABC be a triangle with ⬔BAC 苷 120⬚ and AB ⴢ AC 苷 1.

ⱍ ⱍ

(a) Express the length of the angle bisector AD in terms of x 苷 AB . (b) Find the largest possible value of AD .

Stewart: Calculus, Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

ⱍ

ⱍ

CHAPTER 5 4 CHAPTER A Click here for answers.

1. Show that

S


1 2 1 7 艋y . 4 dx 艋 1 1 ⫹ x 17 24

2. Suppose the curve y 苷 f 共x兲 passes through the origin and the point 共1, 1兲. Find the value of

the integral x01 f ⬘共x兲 dx.

4.1 and 4.2 3. In Sections 5.1 5.2 we used the formulas for the sums of the k th powers of the first n

integers when k 苷 1, 2, and 3. (These formulas are proved in Appendix E.) In this problem we derive formulas for any k. These formulas were first published in 1713 by the Swiss mathematician James Bernoulli in his book Ars Conjectandi. (a) The Bernoulli polynomials Bn are defined by B0共x兲苷 1, Bn⬘共x兲苷 Bn⫺1共x兲, and x01 Bn共x兲 dx 苷 0 for n 苷 1, 2, 3, . . . . Find Bn共x兲 for n 苷 1, 2, 3, and 4. (b) Use the Fundamental Theorem of Calculus to show that Bn共0兲苷 Bn共1兲 for n 艌 2.

Stewart: Calculus, Seventh Edition. ISBN: 0538497815. (c) 2012. Brooks/Cole, Cengage Learning. All rights reserved.

1

2 ■ CHALLENGE PROBLEMS

(c) If we introduce the Bernoulli numbers bn 苷 n! Bn共0兲, then we can write B0共x兲苷 b0 B2共x兲苷

x2 b1 x b2 ⫹ ⫹ 2! 1! 1! 2!

B1共x兲苷

x b1 ⫹ 1! 1!

B3共x兲苷

x3 b1 x 2 b2 x b3 ⫹ ⫹ ⫹ 3! 1! 2! 2! 1! 3!

and, in general, Bn共x兲苷

1 n!

n

兺

k苷0

冉冊

n bk x n⫺k k

where

冉冊 n k

苷

n! k! 共n ⫺ k兲!

[The numbers ( nk ) are the binomial coefficients.] Use part (b) to show that, for n 艌 2, n

bn 苷

兺

k苷0

冉冊

n bk k

and therefore bn⫺1 苷 ⫺

(d) (e)

;

(f) (g) (h)

1 n

冋冉冊冉冊冉冊 n b0 ⫹ 0

n b1 ⫹ 1

n b2 ⫹ ⭈ ⭈ ⭈ ⫹ 2

冉冊册 n bn⫺2 n⫺2

This gives an efficient way of computing the Bernoulli numbers and therefore the Bernoulli polynomials. Show that Bn共1 ⫺ x兲苷共⫺1兲nBn共x兲 and deduce that b2n⫹1 苷 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 艋 x 艋 1. What pattern do you notice in the graphs? Use mathematical induction to prove that Bk⫹1共x ⫹ 1兲 ⫺ Bk⫹1共x兲苷 x k兾k! . By putting x 苷 0, 1, 2, . . . , n in part (g), prove that 1k ⫹ 2 k ⫹ 3 k ⫹ ⭈ ⭈ ⭈ ⫹ n k 苷 k! 关Bk⫹1共n ⫹ 1兲 ⫺ Bk⫹1共0兲兴苷 k! y

n⫹1

0

Bk共x兲 dx

(i) Use part (h) with k 苷 3 and the formula for B4 in part (a) to confirm the formula for the sum of the first n cubes in Section 4.2. 5.2. (j) Show that the formula in part (h) can be written symbolically as 1 关共n ⫹ 1 ⫹ b兲k⫹1 ⫺ b k⫹1 兴 k⫹1

where the expression 共n ⫹ 1 ⫹ b兲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 find a formula for 15 ⫹ 2 5 ⫹ 3 5 ⫹ ⭈ ⭈ ⭈ ⫹ n 5.equator that have exactly the same temperature.

CHAPTER CHAPTER 5 6 A Click here for answers.

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1. A solid is generated by rotating about the x-axis the region under the curve y 苷 f 共x兲, where

f is a positive function and x 艌 0. The volume generated by the part of the curve from x 苷 0 to x 苷 b is b 2 for all b ⬎ 0. Find the function f.



1k ⫹ 2 k ⫹ 3 k ⫹ ⭈ ⭈ ⭈ ⫹ n k 苷

CHALLENGE PROBLEMS ■ 3

CHAPTER 6 CHAPTER 8 A Click here for answers.

S


; 1. The Chebyshev polynomials Tn are defined by Tn共x兲苷 cos共n arccos x兲, n 苷 0, 1, 2, 3, . . . .

(a) What are the domain and range of these functions? (b) We know that T0共x兲苷 1 and T1共x兲苷 x. Express T2 explicitly as a quadratic polynomial and T3 as a cubic polynomial. (c) Show that, for n 艌 1, Tn⫹1共x兲苷 2x Tn共x兲 ⫺ Tn⫺1共x兲. (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? 1 (j) Based on your graphs in parts (g) and (h), what can you say about x⫺1 Tn共x兲 dx when n is odd and when n is even? (k) Use the substitution u 苷 arccos x to evaluate the integral in part (j). (l) The family of functions f 共x兲苷 cos共c arccos x兲 are defined 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 CHAPTER 9 A Click here for answers.

S


1. A circle C of radius 2r has its center at the origin. A circle of radius r rolls without slipping in

the counterclockwise direction around C. A point P is located on a fixed radius of the rolling circle at a distance b from its center, 0 ⬍ b ⬍ r. [See parts (i) and (ii) of the figure.] 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 苷 b cos 3␪ ⫹ 3r cos ␪, y 苷 b sin 3␪ ⫹ 3r sin ␪. Note: If b 苷 0, the path is a circle of radius 3r; if b 苷 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 figure. (Then the diameter of the rotor is constant.) Show that the rotor will fit in the epitrochoid if b 艋 3(2 ⫺ s3 )r兾2.


;

y

y

P P=P¸ 2r

r b

¨ x

P¸

x

(ii)

(i) FIGURE FOR PROBLEM 1


(iii)


CHAPTER 10 CHAPTER 12 S


1. (a)

Show that, for n 苷 1, 2, 3, . . . , sin ␪ 苷 2 n sin

␪ ␪ ␪ ␪ ␪ cos cos cos ⭈ ⭈ ⭈ cos n 2n 2 4 8 2

(b) Deduce that

␪ ␪ ␪ sin ␪ 苷 cos cos cos ⭈ ⭈ ⭈ ␪ 2 4 8 The meaning of this infinite product is that we take the product of the first n factors and then we take the limit of these partial products as n l ⬁. (c) Show that 2 s2 s2 ⫹ s2 苷 ␲ 2 2

s2 ⫹ s2 ⫹ s2 2

⭈⭈⭈

This infinite product is due to the French mathematician Franc᝺ ois Viète (1540–1603). Notice that it expresses ␲ in terms of just the number 2 and repeated square roots. 2. Suppose that a 1 苷 cos ␪, ⫺␲兾2 艋 ␪ 艋 ␲兾2, b1 苷 1, and

an⫹1 苷 2 共an ⫹ bn 兲 1

bn⫹1 苷 sbn an⫹1

Use Problem 1 to show that lim an 苷 lim bn 苷 nl⬁

sin ␪ ␪


nl⬁



ANSWERS CHAPTER Chapter 23

Solutions

S

1 (b) t s s 8 1 2 3{ 2 3{ 3{

1. (a) [1> 2]

CHAPTER Chapter 3 4

Solutions

S

1.

4 3

3. (a) tan

{ ,{A0 {2 + 1

5. (a) | =

CHAPTER 4 Chapter 5

S

1 gf 1 ge + f gw e gw

e (b) (b)

gf gf ge ge +f e +f sec gw gw gw gw e2 + f2 2ef cos

1 2

Solutions

3. (a) E1 ({) = { 12 , E2 ({) =

1 2 2{

12 { +

1 12 ,

E3 ({) = 16 {3 14 {2 +

1 12 {,

E4 ({) =

1 4 24 {

1 3 12 {

+

1 2 24 {

1 e8 = 30 ; 6 1 1 1 {5 52 {4 + 53 {3 16 { , E6 ({) = 720 { 3{5 + 52 {4 12 {2 + 42 , E5 ({) = 120 1 1 E7 ({) = 5040 {6 73 {4 + 23 {2 {7 72 {6 + 72 {5 76 {3 + 16 { , E8 ({) = 40,320 {8 4{7 + 14 3 9 9 8 5 3 1 3 { 2 { + 6{7 21 E9 ({) = 362,880 5 { + 2{ 10 {

(e) e6 =

1 720

1 , 42

1 30

,

(f ) There are four basic shapes for the graphs of Eq (excluding E1 ), and as q increases, they repeat in a cycle of four. For q = 4p, the shape resembles that of the graph of cos 2{; for q = 4p + 1, that of sin 2{; for q = 4p + 2, that of cos 2{; and for q = 4p + 3, that of sin 2{. (k)


CHAPTER6 5 Chapter

S

1 2 12 q (q

Solutions

1. i ({) =

CHAPTER7 6 Chapter

S

+ 1)2 (2q2 + 2q 1)

s 2{@

Solutions

1. (a) [1> 1]; [1> 1] for q A 0

(b) W2 ({) = 2{2 1, W3 ({) = 4{3 3{ (e) W4 ({) = 8{4 8{2 + 1, W5 ({) = 16{5 20{3 + 5{, W6 ({) = 32{6 48{4 + 18{2 1, W7 ({) = 64{7 112{5 + 56{3 7{ (f ) { = cos

n + q

2

, n an integer with 0 n ? q; { = cos(n@q), n an integer with 0 ? n ? q



(g)

(h)

(i) The zeros of Wq and Wq+1 alternate; the extrema also alternate ( j) When q is odd, and so

U1 1

Wq ({) g{ = 0; when q is even, the integral is negative, but decreases in absolute value

as q gets larger. ]

(k) 0

; ? 2 if q is even q2 1 cos(qx) sin x gx = = 0 if q is odd

(l ) As f increases through an integer, the graph of i gains a local extremum, which starts at { = 1 and moves rightward, compressing the graph of i as f continues to increase.

S

Solutions

1. (b)

e = 15 u

e = 25 u

e = 35 u

e = 45 u


CHAPTER10 9 Chapter



SOLUTIONS

E

Exercises

Chapter 32 CHAPTER

t s 1 2 3{ q r s G = { | 3 { 0, 2 3 { 0, 1 2 3 { 0 r q s = { | 3 {, 2 3 {, 1 2 3 { = { | 3 {, 4 3 {, 1 2 3 { = { | { 3, { 1, 1 3 {

1. (a) i ({) =

= {{ | { 3, { 1, 1 3 { } = {{ | { 3, { 1, { 2 } = {{ | 1 { 2 } = [1> 2] (b) i ({) =

t s 1 2 3{

s 1 g i 0 ({) = t 1 2 3{ s g{ 1 2 3{ =

g 1 1 t · s 2 3{ s g{ 2 2 3 { 2 1 2 3{

1 = t s s 8 1 2 3{ 2 3{ 3{ Note that i is always decreasing and i 0 is always negative.

(c)

E

Exercises

CHAPTER4 3 Chapter

1 1 + 1 + |{| 1 + |{ 2| ; 1 1 A A + A A 1 { 1 ({ 2) A A A ? 1 1 + = 1+{ 1 ({ 2) A A A A A 1 1 A A + = 1+{ 1 + ({ 2)


1. i({) =

if { ? 0 if 0 { ? 2 if { 2

; 1 1 A A 2 + A A (1 {) (3 {)2 A A A ? 1 1 i 0 ({) = 2 + A (1 + {) (3 {)2 A A A A 1 1 A A = (1 + {)2 ({ 1)2

if { ? 0 if 0 ? { ? 2 if { A 2

We see that i 0 ({) A 0 for { ? 0 and i 0 ({) ? 0 for { A 2. For 0 ? { ? 2, we have 2 { + 2{ + 1 {2 6{ + 9 1 1 8 ({ 1) i 0 ({) = = = , so i 0 ({) ? 0 for (3 {)2 ({ + 1)2 (3 {)2 ({ + 1)2 (3 {)2 ({ + 1)2 0 ? { ? 1, i 0 (1) = 0 and i 0 ({) A 0 for 1 ? { ? 2. We have shown that i 0 ({) A 0 for { ? 0; i 0 ({) ? 0 for 0 ? { ? 1; i 0 ({) A 0 for 1 ? { ? 2; and i 0 ({) ? 0 for { A 2. Therefore, by the First Derivative Test, the local maxima of i are at { = 0 and { = 2, where i takes the value 43 . Therefore,

4 3

is the absolute maximum value of i .



3. (a) D =

1 ek 2

with sin = k@f, so D = 12 ef sin . But D is a

constant, so differentiating this equation with respect to w, we get gD 1 g gf ge =0= ef cos +e sin + f sin gw 2 gw gw gw g gf ge g 1 ge 1 gf ef cos = sin e +f = tan + . gw gw gw gw f gw e gw (b) We use the Law of Cosines to get the length of side d in terms of those of e and f, and then we differentiate implicitly with respect to w: d2 = e2 + f2 2ef cos gd ge gf g gf ge 2d = 2e + 2f 2 ef( sin ) +e cos + f cos gw gw gw gw gw gw 1 ge gf g gf ge gd = e +f + ef sin e cos f cos . Now we substitute our value of d from the Law gw d gw gw gw gw gw of Cosines and the value of g@gw from part (a), and simplify (primes signify differentiation by w): ee0 + ff0 + ef sin [ tan (f0@f + e0@e)] (ef0 + fe0 )(cos ) gd = gw e2 + f2 2ef cos =

ee0 + ff0 [sin2 (ef0 + fe0 ) + cos2 (ef0 + fe0 )]@ cos ee0 + ff0 (ef0 + fe0 )sec = e2 + f2 2ef cos e2 + f2 2ef cos

5. (a) Let | = |DG|, { = |DE|, and 1@{ = |DF|, so that |DE| · |DF| = 1.

We compute the area A of 4DEF in two ways. First, A=

1 2

|DE| |DF| sin 2 3 =

1 2

·1·

3 2

=

3 4 .

Second,

A = (area of 4DEG) + (area of 4DFG) =

1 2

|DE| |DG| sin 3 +

1 2

|DG| |DF| sin 3 = 12 {|

Equating the two expressions for the area, we get

3 | 4

3 2

+ 12 |(1@{)

1 = {+ {

3 4

3 2

=

3 4 |({

|=

+ 1@{)

{ 1 = 2 , { A 0. { + 1@{ { +1

Another method: Use the Law of Sines on the triangles DEG and DEF. In 4DEG, we have 60 + + _G = 180

{ sin(120 ) sin 120 cos cos 120 sin = = = | sin sin by a similar argument with 4DEF, |=

3 2

_G = 120 . Thus,

3 2

cos + 12 sin sin

{ = 23 cot + 12 , and | { cot = {2 + 12 . Eliminating cot gives = {2 + 12 + 12 |

{ , { A 0. {2 + 1

(b) We differentiate our expression for | with respect to { to nd the maximum: 2 { + 1 {(2{) g| 1 {2 = = = 0 when { = 1. This indicates a maximum by the First Derivative Test, 2 2 g{ ({ + 1) ({2 + 1)2 since | 0 ({) A 0 for 0 ? { ? 1 and | 0 ({) ? 0 for { A 1, so the maximum value of | is |(1) = 12 .



_D + _E + _G = 180


E

Exercises

CHAPTER5 4 Chapter

1 1 . Thus, 1 + {4 17 ] 2 ] 2 1 1 1 1 1 g{ ? 4 and g{ = . Also 1 + {4 A {4 for 1 { 2, so 4 17 1 + {4 { 1 1+{ 1 17 3 2 ] 2 ] 2 { 1 1 1 7 4 g{ ? { g{ = = + = . Thus, we have the estimate 4 3 1 24 3 24 1 1+{ 1 ] 2 7 1 1 g{ . 4 17 1 + { 24 1

1. For 1 { 2, we have {4 24 = 16, so 1 + {4 17 and

3. (a) To nd E1 ({), we use the fact that E10 ({) = E0 ({)

U

U

1 . 2

E0 ({) g{ =

U

1 g{ = { + F. Now we 1 k l1 U1 U1 impose the condition that 0 E1 ({) g{ = 0 0 = 0 ({ + F) g{ = 12 {2 0 + F{ = 12 + F 12 .

E1 ({) =

0

U E1 ({) g{ = { 12 g{ = 12 {2 12 { + G. But

So E1 ({) = { Similarly E2 ({) = F= U1 U1 1 E2 ({) g{ = 0 0 = 0 12 {2 12 { + G g{ = 16 14 + G G = 12 , so 0 U U 1 1 2 1 1 g{ = 16 {3 14 {2 + 12 . E3 ({) = E2 ({) g{ = { 12 { + 12 { + H. But E2 ({) = 12 {2 12 { + 12 2 U 1 1 3 1 2 U1 1 1 1 1 E3 ({) g{ = 0 0 = 0 6 { 4 { + 12 { + H g{ = 24 12 + 24 + H H = 0. So 0 U U 1 1 3 1 1 4 1 3 1 2 {. E4 ({) = E3 ({) g{ = { 14 {2 + 12 { g{ = 24 { 12 { + 24 { + I. E3 ({) = 16 {3 14 {2 + 12 6 U1 1 4 U1 1 3 1 2 1 1 1 1 { 12 { + 24 { + I g{ = 120 48 + 72 + I I = 720 . But 0 E4 ({) g{ = 0 0 = 0 24 So E4 ({) =

1 4 24 {

1 3 12 {

+

1 2 24 {

(b) By FTC2, Eq (1) Eq (0) =

U1 0

1 720 .

Eq0 ({) g{ =

U1 0

Eq1 ({) g{ = 0 for q 1 1, by denition. Thus,

Eq (0) = Eq (1) for q 2. (c) We know that Eq ({) =

q 1 S q en {qn . If we set { = 1 in this expression, and use the fact that q! n=0 n

q S eq q e for q 2, we get eq = n n . Now if we expand the right-hand side, we get q! n=0 q q eq2 + q1 eq1 + q e . We cancel the eq terms, move the eq1 term to eq = q0 e0 + q1 e1 + · · · + q2 q q k l q q = q: eq1 = q1 q0 e0 + q1 e1 + · · · + q2 eq2 for q 2, as required. the LHS and divide by q1


Eq (1) = Eq (0) =

(d) We use mathematical induction. For q = 0: E0 (1 {) = 1 and (1)0 E0 ({) = 1, so the equation holds for q = 0 since e0 = 1. Now if En (1 {) = (1)n En ({), then since

g g{ En+1 (1

g g{ En+1 (1

0 g {) = En+1 (1 {) g{ (1 {) = En (1 {), we have

{) = (1)(1)n En ({) = (1)n+1 En ({). Integrating, we get

En+1 (1 {) = (1)n+1 En+1 ({) + F. But the constant of integration must be 0, since if we substitute { = 0 in the equation, we get En+1 (1) = (1)n+1 En+1 (0) + F, and if we substitute { = 1 we get En+1 (0) = (1)n+1 En+1 (1) + F, and these two equations together imply that En+1 (0) = (1)n+1 (1)n+1 En+1 (0) + F + F = En+1 (0) + 2F F = 0. So the equation holds for all q, by induction. Now if the power of 1 is odd, then we have



E2q+1 (1 {) = E2q+1 ({). In particular, E2q+1 (1) = E2q+1 (0). But from part (b), we know that En (1) = En (0) for n A 1. The only possibility is that E2q+1 (0) = E2q+1 (1) = 0 for all q A 0, and this implies that e2q+1 = (2q + 1)! E2q+1 (0) = 0 for q A 0. 1 (e) From part (a), we know that e0 = 0! E0 (0) = 1, and similarly e1 = 12 , e2 = 16 , e3 = 0 and e4 = 30 .

We use the formula to nd e6 = e71 =

1 7

%# $ # $ # $ # $ # $ # $ & 7 7 7 7 7 7 e0 + e1 + e2 + e3 + e4 + e5 0 1 2 3 4 5

The e3 and e5 terms are 0, so this is equal to

Similarly,

1 1 7·6 1 7·6·5 1 1 7 7 7 1 1+7 + + = 1 + = 7 2 2·1 6 3·2·1 30 7 2 2 6 42 %# $ # $ # $ # $ # $ & 9 9 9 9 9 e0 + e1 + e2 + e4 + e6 0 1 2 4 6 1 1 9·8 1 9·8·7·6 1 9·8·7 1 = 1+9 + + + 9 2 2·1 6 4·3·2·1 30 3 · 2 · 1 42 9 21 1 1 1 +6 +2 = = 9 2 5 30

1 e8 = 9

Now we can calculate # $ 5 1 S 5 E5 ({) = en {5n 5! n=0 n 1 1 5·4 1 1 5 4 3 = { +5 { + { +5 { 120 2 2·1 6 30 5 5 4 5 3 1 1 { 2{ + 3{ 6{ = 120 1 1 1 6·5 1 6·5 1 {6 + 6 {5 + {4 + {2 + 720 2 2·1 6 2·1 30 42 6 5 1 5 4 1 2 1 = 720 { 3{ + 2 { 2 { + 42

1 1 7·6 1 7·6·5 1 1 {7 + 7 {6 + {5 + {3 + 7 { 5040 2 2·1 6 3·2·1 30 42 7 7 6 7 5 7 3 1 1 { 2{ + 2{ 6{ + 6{ = 5040

E7 ({) =

1 1 1 8·7 1 8·7·6·5 8·7 1 1 {8 + 8 {7 + {6 + {4 + {2 + 40,320 2 2·1 6 4·3·2·1 30 2 · 1 42 30 8 7 1 14 6 7 4 2 2 1 = 40,320 { 4{ + 3 { 3 { + 3 { 30

E8 ({) =

E9 ({) =

=

1 1 9·8 1 9·8·7·6 1 {9 + 9 {8 + {7 + {5 362,880 2 2·1 6 4·3·2·1 30

1 362,880

9 9 8 { 2 { + 6{7

21 5 { 5

+ 2{3

9·8·7 + 3·2·1

1 42

1 { +9 { 30 3

3 { 10



E6 ({) =


(f )

q=1

q=2

q=3

q=4

q=5

q=6

q=7

q=8

q=9

There are four basic shapes for the graphs of Eq (excluding E1 ), and as q increases, they repeat in a cycle of four. For q = 4p, the shape resembles that of the graph of cos 2{; For q = 4p + 1, that of sin 2{; for q = 4p + 2, that of cos 2{; and for q = 4p + 3, that of sin 2{.

(g) For n = 0: E1 ({ + 1) E1 ({) = { + 1 now assume that Eq ({ + 1) Eq ({) = ]


] [Eq ({ + 1) Eq ({)] g{ = Eq+1 ({) =

U

1 2

{0 { 12 = 1, and = 1, so the equation holds for n = 0. We 0!

{q1 . We integrate this equation with respect to {: (q 1)!

{q1 g{. But we can evaluate the LHS using the denition (q 1)!

Eq ({) g{, and the RHS is a simple integral. The equation becomes 1 q 1 1 { = {q , since by part (b) Eq+1 (1) Eq+1 (0) = 0, and so the Eq+1 ({ + 1) Eq+1 ({) = (q 1)! q q! constant of integration must vanish. So the equation holds for all n, by induction.

(h) The result from part (g) implies that sn = n! [En+1 (s + 1) En+1 (s)]. If we sum both sides of this equation from s = 0 to s = q (note that n is xed in this process), we get

q S s=0

sn = n!

q S s=0

[En+1 (s + 1) En+1 (s)]. But the



RHS is just a telescoping sum, so the equation becomes 1n + 2n + 3n + · · · + qn = n! [En+1 (q + 1) En+1 (0)]. But from the denition of Bernoulli polynomials (and using the Fundamental Theorem of Calculus), the RHS is U q+1 equal to n! 0 En ({) g{. (i) If we let n = 3 and then substitute from part (a), the formula in part (h) becomes 13 + 23 + · · · + q3 = 3! [E4 (q + 1) E4 (0)] 1 1 = 6 24 (q + 1)4 12 (q + 1)3 +

1 (q 24

+ 1)2

1 720

1 24

1 12

+

1 24

1 720

2 (q + 1)2 [1 + (q + 1)2 2(q + 1)] (q + 1)2 [1 (q + 1)]2 q(q + 1) = = = 4 4 2

(j) 1n + 2n + 3n + · · · + qn = n!

]

q+1

En ({) g{ [by part (h)] # $ ] q+1 ] q+1 n # $ n S n 1 S n nm em { em {nm g{ = n! g{ = m n! m=0 m m=0 0 0 0

# $ n em {nm as ({ + e)n , as explained in the problem. Then Now view m m=0 n S

n

n

n

n

1 + 2 + 3 + · · · + q “=”

U q+1 0

({ + e)n+1 ({ + e) g{ = n+1 n

q+1 = 0

(q + 1 + e)n+1 en+1 n+1

(k) We expand the RHS of the formula in (j), turning the el into el , and remembering that e2l+1 = 0 for l A 0: 15 + 25 + · · · + q5 = 16 (q + 1)6 e6 = 16 (q + 1)6 + 6(q + 1)5 e1 + 62 ·· 51 (q + 1)4 e2 + 62 ·· 51 (q + 1)2 e4 = 16 (q + 1)6 3(q + 1)5 + 52 (q + 1)4 12 (q + 1)2 1 = 12 (q + 1)2 2(q + 1)4 6(q + 1)3 + 5(q + 1)2 1 1 (q + 1)2 [(q + 1) 1]2 2(q + 1)2 2(q + 1) 1 = 12 =

Exercises

+ 1)2 (2q2 + 2q 1)

CHAPTER Chapter 65

1. The volume generated from { = 0 to { = e is

Ue 0

[i({)]2 g{. Hence, we are given that e2 =

Ue 0

[i ({)]2 g{

for all e A 0. Differentiating both sides of this equation using the Fundamental Theorem of Calculus gives s s 2e = [i (e)]2 i (e) = 2e@, since i is positive. Therefore, i({) = 2{@.

E

Exercises

CHAPTER Chapter 86

1. (a) Wq ({) = cos(q arccos {). The domain of arccos is [1> 1], and the domain of cos is R, so the domain of Wq ({)

is [1> 1]. As for the range, W0 ({) = cos 0 = 1, so the range of W0 ({) is {1}. But since the range of q arccos { is at least [0> ] for q A 0, and since cos | takes on all values in [1> 1] for | [0> ], the range of Wq ({) is [1> 1] for q A 0.



E

1 2 q (q 12


(b) Using the usual trigonometric identities, W2 ({) = cos(2 arccos {) = 2 [cos(arccos {)]2 1 = 2{2 1, and W3 ({) = cos(3 arccos {) = cos(arccos { + 2 arccos {) = cos(arccos {) cos(2 arccos {) sin(arccos {) sin(2 arccos {) = { 2{2 1 sin(arccos {) [2 sin(arccos {) cos(arccos {)] = 2{3 { 2 sin2 (arccos {) { = 2{3 { 2{ 1 cos2 (arccos {) = 2{3 { 2{ 1 {2 = 4{3 3{ (c) Let | = arccos {. Then Wq+1 ({) = cos[(q + 1)|] = cos(| + q|) = cos | cos q| sin | sin q| = 2 cos | cos q| (cos | cos q| + sin | sin q|) = 2{Wq ({) cos(q| |) = 2{Wq ({) Wq1 ({) (d) Here we use induction. W0 ({) = 1, a polynomial of degree 0. Now assume that Wn ({) is a polynomial of degree n. Then Wn+1 ({) = 2{Wn ({) Wn1 ({). By assumption, the leading term of Wn is dn {n , say, so the leading term of Wn+1 is 2{dn {n = 2dn {n+1 , and so Wn+1 has degree n + 1. (e) W4 ({) = 2{W3 ({) W2 ({) = 2{ 4{3 3{ 2{2 1 = 8{4 8{2 + 1, W5 ({) = 2{W4 ({) W3 ({) = 2{ 8{4 8{2 + 1 4{3 3{ = 16{5 20{3 + 5{, W6 ({) = 2{W5 ({) W4 ({) = 2{ 16{5 20{3 + 5{ 8{4 8{2 + 1 = 32{6 48{4 + 18{2 1, W7 ({) = 2{W6 ({) W5 ({) = 2{ 32{6 48{4 + 18{2 1 16{5 20{3 + 5{ = 64{7 112{5 + 56{3 7{ (f ) The zeros of Wq ({) = cos(q arccos {) occur where q arccos { = n + 2 for some integer n, since then cos(q arccos {) = cos n + 2 = 0. Note that there will be restrictions on n, since 0 arccos { . We


continue: q arccos { = n + 0 ? n +

2

? q

2

arccos { =

n + q

2

. This only has solutions for 0

n + q

2

0 n ? q. [This makes sense, because then Wq ({) has q zeros, and it is a polynomial of

degree q.] So, taking cosines of both sides of the last equation, we nd that the zeros of Wq ({) occur at { = cos

n + q

2

, n an integer with 0 n ? q. To nd the values of { at which Wq ({) has local extrema, we set

q q sin(q arccos {) 0 = Wq0 ({) = sin(q arccos {) = 2 1{ 1 {2

sin(q arccos {) = 0

q arccos { = n, n some integer arccos { = n@q. This has solutions for 0 n q, but we disallow the cases n = 0 and n = q, since these give { = 1 and { = 1 respectively. So the local extrema of Wq ({) occur at { = cos(n@q), n an integer with 0 ? n ? q. [Again, this seems reasonable, since a polynomial of degree q has at



most (q 1) extrema.] By the First Derivative Test, the cases where n is even give maxima of Wq ({), since then q arccos [cos(n@q)] = n is an even multiple of , so sin (q arccos {) goes from negative to positive at { = cos(n@q). Similarly, the cases where n is odd represent minima of Wq ({). (g)

(h)

(i) From the graphs, it seems that the zeros of Wq and Wq+1 alternate; that is, between two adjacent zeros of Wq , there is a zero of Wq+1 , and vice versa. The same is true of the {-coordinates of the extrema of Wq and Wq+1 : between the {-coordinates of any two adjacent extrema of one, there is the {-coordinate of an extremum of the other. ( j) When q is odd, the function Wq ({) is odd, since all of its terms have odd degree, and so

U1 1

Wq ({) g{ = 0. When

q is even, Wq ({) is even, and it appears that the integral is negative, but decreases in absolute value as q gets larger. (k)

U1 1

Wq ({) g{ =

U1 1

cos(q arccos {) g{. We substitute x = arccos { { = cos x g{ = sin x gx,

{ = 1 x = , and { = 1 x = 0. So the integral becomes

0

] cos(qx) sin x gx =

0

1 [sin(x 2

qx) + sin(x + qx)] gx

1 cos[(1 q)x] cos[(1 + q)x] = 2 q1 q+1 0 ; 1 1 1 1 1 A A if q is even A ? 2 q1 q+1 q1 q+1 = A 1 1 1 1 1 A A if q is odd = 2 q1 q+1 q1 q+1 ; ? 2 if q is even q2 1 = = 0 if q is odd

(l ) From the graph, we see that as f increases through an integer, the graph of i gains a local extremum, which starts at { = 1 and moves rightward, compressing the graph of i as f continues to increase.



]


E

Exercises

CHAPTER 9 Chapter 11

1. (a) Since the smaller circle rolls without slipping around F, the amount of

arc traversed on F (2u in the gure) must equal the amount of arc of the smaller circle that has been in contact with F. Since the smaller circle has radius u, it must have turned through an angle of 2u@u = 2. In addition to turning through an angle 2, the little circle has rolled through an angle against F. Thus, S 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 {-axis, then S would have turned only 2 instead of 3. The movement of the little circle around F adds to the angle.) From the gure, we see that the center of the small circle has coordinates (3u cos > 3u sin ). Thus, S has coordinates ({> |), where { = 3u cos + e cos 3 and | = 3u sin + e sin 3. (b)

e = 15 u

e = 25 u

e = 35 u

e = 45 u

(c) The diagram gives an alternate description of point S on the epitrochoid. T moves around a circle of radius e, and S rotates one-third as fast with respect to T at a distance of 3u. Place an equilateral triangle with sides of length 3 3u so that its centroid is at T and one vertex is at S . (The distance from the centroid to a vertex is

1 3

times the length of a side of

the equilateral triangle.) As increases by

2 , 3

the point T travels once around the circle of radius

e, returning to its original position. At the same time, S (and the rest of the Stewart: Calculus, Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

triangle) rotate through an angle of

2 3

about T, so S ’s position is

occupied by another vertex. In this way, we see that the epitrochoid traced out by S 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 3u, so it has radius 6u. To show that the rotor ts inside the epitrochoid, it sufces to show that for any position of the tracing point S , 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 S is on the |-axis, so as long as the diameter of the rotor (which is 3 3u) is less than the distance between the |-intercepts, the rotor will t. The |-intercepts occur



when = t if 3

Exercises

or =

3 2

| = ±(3u e), so the distance between the intercepts is 6u 2e, and the rotor will

3 (2 3 ) 3u 6u 2e e u. 2

CHAPTER 10 Chapter 12

cos = 2 2 sin cos cos = 2 2 2 sin cos cos cos 2 2 4 4 2 8 8 4 2 cos cos = · · · = 2 2 2 · · · 2 2 sin q cos q cos q1 · · · cos 2 2 2 8 4 2

1. (a) sin = 2 sin

= 2q sin

cos cos cos · · · cos q 2q 2 4 8 2

(b) sin = 2q sin

cos cos cos · · · cos q 2q 2 4 8 2

@2q sin · = cos cos cos · · · cos q . sin (@2q ) 2 4 8 2

sin { = 1 with { = q : { 2 sin @2q cos · = lim cos cos · · · cos lim q q sin (@2q ) 2 4 8 2q Now we let q , using lim

{0

(c) If we take =

2

sin = cos cos cos · · · . 2 4 8

in the result from part (b) and use the half-angle formula cos { =

t

1 2 (1

+ cos 2{)

(see Formula 17a in Appendix D), we get yy xxu xx x x cos 4 + 1 yu xw x +1 x x cos 4 + 1 2 u x x +1 + 1 w cos 4 + 1 w 2 sin @2 2 = cos 4 ··· @2 2 2 2 yv x ys x 2 x v x x 2+ 2 2 +1 s x + 1 2 +1 w 2 2 2 2+ 2 w 2 2 +1 2 = ··· = ··· 2 2 2 2 2 2 t s s 2 2+ 2 2+ 2+ 2 ··· = 2 2 2 Stewart: Calculus, Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

E

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