Howard Anton Student Solutions Manual Set For Calculus Early Transcendentals Single Variable 8th E

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Table of Contents
EXERCISE SET FOR INTRODUCTION
CHAPTER 1
- EXERCISE SET
  - 1.1
  - 1.2
  - 1.3
  - 1.4
  - 1.5
  - 1.6
  - 1.7
- SUPPLEMENTARY EXERCISES
- HORIZON MODULE
CHAPTER 2
- EXERCISE SET
  - 2.1
  - 2.2
  - 2.3
  - 2.4
  - 2.5
- SUPPLEMENTARY EXERCISES
CHAPTER 3
- EXERCISE SET
  - 3.1
  - 3.2
  - 3.3
  - 3.4
  - 3.5
  - 3.6
- SUPPLEMENTARY EXERCISES
- HORIZON MODULE
CHAPTER 4
- EXERCISE SET
  - 4.1
  - 4.2
  - 4.3
  - 4.4
  - 4.5
  - 4.6
  - 4.7
- SUPPLEMENTARY EXERCISES
CHAPTER 5
- EXERCISE SET
  - 5.1
  - 5.2
  - 5.3
- SUPPLEMENTARY EXERCISES
- HORIZON MODULE
CHAPTER 6
- EXERCISE SET
  - 6.1
  - 6.2
  - 6.3
  - 6.4
  - 6.5
- SUPPLEMENTARY EXERCISES
CHAPTER 7
- EXERCISE SET
  - 7.1
  - 7.2
  - 7.3
  - 7.4
  - 7.5
  - 7.6
  - 7.7
  - 7.8
  - 7.9
- SUPPLEMENTARY EXERCISES
- HORIZON MODULE
CHAPTER 8
- EXERCISE SET
  - 8.1
  - 8.2
  - 8.3
  - 8.4
  - 8.5
  - 8.6
  - 8.7
  - 8.8
- SUPPLEMENTARY EXERCISES
CHAPTER 9
- EXERCISE SET
  - 9.1
  - 9.2
  - 9.3
  - 9.4
  - 9.5
  - 9.6
  - 9.7
  - 9.8
- SUPPLEMENTARY EXERCISES
- HORIZON MODULE
CHAPTER 10
- EXERCISE SET
- SUPPLEMENTARY EXERCISES
CHAPTER 11
- EXERCISE SET
  - 11.1
  - 11.2
  - 11.3
  - 11.4
  - 11.5
  - 11.6
  - 11.7
  - 11.8
  - 11.9
  - 11.10
- SUPPLEMENTARY EXERCISES
CHAPTER 12
- EXERCISE SET
  - 12.1
  - 12.2
  - 12.3
  - 12.4
  - 12.5
- SUPPLEMENTARY EXERCISES
- HORIZON MODULE
CHAPTER 13
- EXERCISE SET
  - 13.1
  - 13.2
  - 13.3
  - 13.4
  - 13.5
  - 13.6
  - 13.7
  - 13.8
- SUPPLEMENTARY EXERCISES
CHAPTER 14
- EXERCISE SET
  - 14.1
  - 14.2
  - 14.3
  - 14.4
  - 14.5
  - 14.6
  - 14.7
- SUPPLEMENTARY EXERCISES
CHAPTER 15
- EXERCISE SET
  - 15.1
  - 15.2
  - 15.3
  - 15.4
  - 15.5
  - 15.6
  - 15.7
  - 15.8
  - 15.9
- SUPPLEMENTARY EXERCISES
CHAPTER 16
- EXERCISE SET
  - 16.1
  - 16.2
  - 16.3
  - 16.4
  - 16.5
  - 16.6
  - 16.7
  - 16.8
- SUPPLEMENTARY EXERCISES
CHAPTER 17
- EXERCISE SET
  - 17.1
  - 17.2
  - 17.3
  - 17.4
  - 17.5
  - 17.6
  - 17.7
  - 17.8
- SUPPLEMENTARY EXERCISES
- HORIZON MODULE
APPENDICES
- APPENDIX A
- APPENDIX B
- APPENDIX C
- APPENDIX D
- APPENDIX E
- APPENDIX F
Wiley Help

CONTENTS

Introduction .............................................................. 1

Chapter 1. Functions ..................................................... 2

Chapter 2. Limits and Continuity ......................................... 43

Chapter 3. The Derivative ................................................ 65

Chapter 4. Logarithmic and Exponential Functions ......................... 99

Chapter 5. Analysis of Functions and Their Graphs ........................139

Chapter 6. Applications of the Derivative .................................177

Chapter 7. Integration ...................................................209

Chapter 8. Applications of the Deﬁnite Integral

in Geometry, Science, and Engineering .........................256

Chapter 9. Principles of Integral Evaluation ...............................292

Chapter 10. Mathematical Modeling with Differential Equations ..............343

Chapter 11. Inﬁnite Series ................................................361

Chapter 12. Analytic Geometry in Calculus .................................408

Chapter 13. Three-Dimensional Space; Vectors .............................448

Chapter 14. Vector-Valued Functions ......................................490

Chapter 15. Partial Derivatives ............................................524

Chapter 16. Multiple Integrals .............................................573

Chapter 17. Topics in Vector Calculus .....................................608

Appendix A. Real Numbers, Intervals, and Inequalities .......................640

Appendix B. Absolute Value ...............................................647

Appendix C. Coordinate Planes and Lines ...................................650

Appendix D. Distance, Circles, and Quadratic Equations ......................658

Appendix E. Trigonometry Review .........................................668

Appendix F. Solving Polynomial Equations .................................674

CALCULUS:

A New Horizon from Ancient Roots

EXERCISE SET FOR INTRODUCTION

1. (a) x=0.123123123 ...; 1000x= 123.123123123 ...= 123 + x; 999x= 123; x=123

999 =41

333

(b) x=12.7777 ...;10x= 127.7777 ...,so9x=10x−x= 115; x=115

x=3769.74

99 =376974

9900 =41886

1100 =20943

550

(d) 0.4296000 ...=0.4296 = 4296

10000 =537

1250

2. (a) πis irrational, and thus has a nonrepeating decimal expansion, whereas 22

7=3.

repeats

z }| {

142857 ...

(b) 22

7>π

3. (a) 223

71 <333

106 <63

25 Ã17+15

√

7+15

√

5!<355

113 <22

7(b) 63

25 Ã17+15

√

7+15

√

106 (d) 63

25 Ã17+15

√

7+15

√

4. (a) If ris the radius, then D=2rso µ8

9D¶2

=µ16

9r¶2

=256

81 r2. The area of a circle of radius ris

πr2so 256/81 was the approximation used for π.

(b) 256/81 ≈3.16049, 22/7≈3.14268, and π≈3.14159 so 256/81 is worse than 22/7.

5. The ﬁrst series, taken to ten terms, adds to 3.0418; the second, as printed, adds to 3.1416.

6. (a) 1

9=0.111111 ...=1

10 +1

100 +1

1000 +1

10000 +1

100000 +1

1000000 +...

(b) 2

27 =0.185185 ...=1

10 +8

100 +5

1000 +1

10000 +8

100000 +5

1000000 +...

45 =0.311111 ...=3

10 +1

100 +1

1000 +1

10000 +1

100000 +1

1000000 +...

7. (a) 7

11 =0.636363 ...=6

10 +3

100 +6

1000 +3

10000 +6

100000 +3

1000000 +...

(b) 8

33 =0.242424 ...=2

10 +4

100 +2

1000 +4

10000 +2

100000 +4

1000000 +...

12 =0.416666 ...=4

10 +1

100 +6

1000 +6

10000 +6

100000 +6

1000000 +...

8. (a) 1, 2, 1.75, 1.7321 (b) 1, 3, 2.33, 2.238, 2.2361

9. (a) 1, 4, 2.875, 2.6549, 2.6458 (b) 1, 25.5, 13.7, 8.69, 7.22, 7.0726, 7.0711

10. (a) Let x1=1

2(a+b), x2=1

2(a+x1), x3=1

2(a+x2), etc. Then b>x

1>x

2>···>x

n−1>x

n>a

so all the xi’s are distinct, there are inﬁnitely many of them and they all lie between aand b.

(b) x=0.99999 ...,10x=9.99999 ...,9x=9,x=1

(d) 10x=9+x,sox=9/9 = 1. They are equal.

CHAPTER 1

Functions

EXERCISE SET 1.1

1. (a) around 1943 (b) 1960; 4200

(e) news of health risk; social pressure, antismoking campaigns, increased taxation

2. (a) 1989; $35,600 (b) 1983; $32,000 (c) the ﬁrst two years; the curve is steeper (downhill)

3. (a) −2.9,−2.0,2.35,2.9(b) none (c) y=0

(d) −1.75 ≤x≤2.15 (e) ymax =2.8atx=−2.6; ymin =−2.2atx=1.2

4. (a) x=−1,4(b) none (c) y=−1

(d) x=0,3,5(e) ymax =9atx=6;y

min =−2atx=0

5. (a) x=2,4(b) none (c) x≤2; 4 ≤x(d) ymin =−1; no maximum value

6. (a) x=9 (b) none (c) x≥25 (d) ymin = 1; no maximum value

7. (a) Breaks could be caused by war, pestilence, ﬂood, earthquakes, for example.

(b) Cdecreases for eight hours, takes a jump upwards, and then repeats.

8. (a) Yes, if the thermometer is not near a window or door or other source of sudden temperature

change.

(b) The number is always an integer, so the changes are in movements (jumps) of at least one unit.

9. (a) If the side adjacent to the building has length xthen L=x+2y. Since A=xy = 1000,

L=x+ 2000/x.

(b) x>0 and xmust be smaller than the width of the building, which was not given.

20 80

(d) Lmin ≈89.44

10. (a) V=lwh =(6−2x)(6 −2x)x(b) From the ﬁgure it is clear that 0 <x<3.

(d) Vmax ≈16

3 Chapter 1

11. (a) V= 500 = πr2hso h=500

πr2. Then

C=(0.02)(2)πr2+(0.01)2πrh =0.04πr2+0.02πr 500

πr2

=0.04πr2+10

r;Cmin ≈4.39 at r≈3.4,h≈13.8.

1.5 6

(b) C=(0.02)(2)(2r)2+(0.01)2πrh =0.16r2+10

r. Since

0.04π<0.16, the top and bottom now get more weight.

Since they cost more, we diminish their sizes in the solution,

and the cans become taller.

1.5 5.5

12. (a) The length of a track with straightaways of length Land semicircles of radius ris

P= (2)L+ (2)(πr) ft. Let L= 360 and r=80togetP= 720 + 160π= 1222.65 ft.

Since this is less than 1320 ft (a quarter-mile), a solution is possible.

(b) P=2L+2πr = 1320 and 2r=2x+ 160, so

L=1

2(1320 −2πr)=1

2(1320 −2π(80 + x)) = 660 −80π−πx.

450

0 100

(d) The longest straightaway occurs when x=0,soL= 660 −80π= 408.67 ft.

EXERCISE SET 1.2

1. (a) f(0) = 3(0)2−2=−2; f(2) = 3(2)2−2 = 10; f(−2)=3(−2)2−2 = 10; f(3) = 3(3)2−2 = 25;

f(√2)=3(

√

2)2−2=4;f(3t) = 3(3t)2−2=27t

2−2

(b) f(0) = 2(0) = 0; f(2) = 2(2) = 4; f(−2)=2(−2)=−4; f(3) = 2(3) = 6; f(√2)=2

√

f(3t)=1/3tfor t>1 and f(3t)=6tfor t≤1.

2. (a) g(3) = 3+1

3−1=2; g(−1) = −1+1

−1−1=0;g(π)= π+1

π−1;g(−1.1) = −1.1+1

−1.1−1=−0.1

−2.1=1

21;

g(t2−1) = t2−1+1

2−1−1=t

2−2

(b) g(3) = √3+1=2;g(−1)=3;g(π)=√

π+1;g(−1.1)=3;g(t

2−1)=3ift

2<2 and

g(t2−1) = √t2−1+1=|t|if t2≥2.

3. (a) x6=3 (b) x≤−

√

3orx≥√

Exercise Set 1.2 4

x= 0, then x2−2x+5=5>0; domain: (−∞,+∞).

(d) x6=0 (e) sin x6=1,sox6=(2n+1

2)π,

n=0,±1,±2,...

4. (a) x6=−7

(b) x−3x2must be nonnegative; y=x−3x2is a parabola that crosses the x-axis at x=0,1

3and

opens downward, thus 0 ≤x≤1

x−4>0, so x2−4>0 and x−4>0, thus x>4; or x2−4<0 and x−4<0, thus

−2<x<2

(d) x6=−1

(e) cos x≤1<2, 2 −cos x>0, all x

5. (a) x≤3(b) −2≤x≤2(c) x≥0(d) all x(e) all x

6. (a) x≥2

3(b) −3

2≤x≤3

2(c) x≥0(d) x6=0 (e) x≥0

7. (a) yes (b) yes

8. The sine of θ/2is(L/2)/10 (side opposite over hypotenuse), so that L= 20 sin(θ/2).

9. The cosine of θis (L−h)/L (side adjacent over hypotenuse), so h=L(1 −cos θ).

10. T

11.

h12.

51015

13. (a) If x<0, then |x|=−xso f(x)=−x+3x+1=2x+1. If x≥0, then |x|=xso

f(x)=x+3x+1=4x+1;

f(x)=½2x+1,x<0

4x+1,x≥0

(b) If x<0, then |x|=−xand |x−1|=1−xso g(x)=−x+1−x=1−2x.If0≤x<1, then

|x|=xand |x−1|=1−xso g(x)=x+1−x=1. Ifx≥1, then |x|=xand |x−1|=x−1

so g(x)=x+x−1=2x−1;

g(x)=





1−2x, x < 0

1,0≤x<1

2x−1,x≥1

14. (a) If x<5/2, then |2x−5|=5−2xso f(x) = 3+(5−2x)=8−2x.Ifx≥5/2, then |2x−5|=2x−5

so f(x)=3+(2x−5)=2x−2;

f(x)=½8−2x, x < 5/2

2x−2,x≥5/2

5 Chapter 1

(b) If x<−1, then |x−2|=2−xand |x+1|=−x−1sog(x) = 3(2 −x)−(−x−1)=7−2x.If

−1≤x<2, then |x−2|=2−xand |x+1|=x+1 sog(x) = 3(2 −x)−(x+1)=5−4x.If

x≥2, then |x−2|=x−2 and |x+1|=x+1so g(x)=3(x−2) −(x+1)=2x−7;

g(x)=





7−2x, x < −1

5−4x, −1≤x<2

2x−7,x≥2

15. (a) V=(8−2x)(15 −2x)x(b) −∞ <x<+∞,−∞ <V <+∞(c) 0<x<4

(d) minimum value at x= 0 or at x= 4; maximum value somewhere in between (can be approxi-

mated by zooming with graphing calculator)

16. (a) x= 3000 tan θ(b) θ6=nπ +π/2 for nan integer, −∞ <n<∞

6000

17. (i) x=1,−2 causes division by zero (ii) g(x)=x+ 1, all x

18. (i) x= 0 causes division by zero (ii) g(x)=√

x+ 1 for x≥0

19. (a) 25◦F(b) 2◦F(c) −15◦F

20. If v= 48 then −60=WCI=1.6T−55; thus T=(−60 + 55)/1.6≈−3

◦F.

21. If v= 8 then −10=WCI=91.4 + (91.4−T)(0.0203(8) −0.304√8−0.474); thus

T=91.4+(10+91.4)/(0.0203(8) −0.304√8−0.474) and T=5

◦F

22. The WCI is given by three formulae, but the ﬁrst and third don’t work with the data. Hence

−15=WCI=91.4 + (91.4−20)(0.0203v−0.304√v−0.474); set x=√vso that v=x2and obtain

0.0203x2−0.304x−0.474+(15+91.4)/(91.4−20) = 0. Use the quadratic formula to ﬁnd the two

roots. Square them to get vand discard the spurious solution, leaving v≈25.

23. Let tdenote time in minutes after 9:23 AM. Then D(t) = 1000 −20tft.

EXERCISE SET 1.3

1. (e) seems best, though only (a) is bad.

-0.5

0.5

-1 1 x

2. (e) seems best, though only (a) is bad and (b) is

not good.

-0.5

0.5

-1 1 x

Exercise Set 1.3 6

3. (b) and (c) are good; (a) is very bad.

-2 -1 0 1 2 x

4. (b) and (c) are good; (a) is very bad.

-14

-13

-12

-11

-2 -1 0 1 2 x

5. [−3,3] ×[0,5]

-3 -2 -1 1 2 3 x

6. [−4,2] ×[0,3]

-3 -2 -1 1 x

7. (a) window too narrow, too short (b) window wide enough, but too short

-500

-400

-300

-200

-100

-5 5 101520

(d) window too narrow, too short

(e) window too narrow, too short

8. (a) window too narrow (b) window too short

-250

-200

-150

-100

-50

-16 -12 -8 -4 4

(d) window too narrow, too short

(e) shows one local minimum only, window too narrow, too short

7 Chapter 1

9. [−5,14] ×[−60,40]

-60

-40

-20

-5 5 10 x

10. [6,12] ×[−100,100]

-100

-50

100

81012

11. [−0.1,0.1] ×[−3,3]

-3

-2

-1

-0.1 x

0.1

12. [−1000,1000] ×[−13,13]

-10

-5

-1000 x

1000

13. [−250,1050] ×[−1500000,600000]

-500000

-1000 1000 x

14. [−3,20] ×[−3500,3000]

-2000

-1000

1000

51015

15. [−2,2] ×[−20,20]

-20

-10

-2 -1 1 2

16. [1.6,2] ×[0,2]

0.5

1.5

1.6 1.7 1.8 1.9 2 x

17. depends on graphing utility

-6

-4

-2

-4 -2 2 4 x

18. depends on graphing utility

-6

-4

-2

-4 -2 2 4 x

Exercise Set 1.3 8

19. (a) f(x)=√

16 −x2

-4 -2 2 4 x

(b) f(x)=−

√

16 −x2

-4

-3

-2

-1

-4 -2 2 4

(c)

-4

-2

-4 -2 2 4

(d)

1234

(e) No; the vertical line test fails.

20. (a) y=±3p1−x2/4(b) y=±√x2+1

-4

-2

-4 -2 2 4 x

21. (a)

-1 1 x

(b)

y(c)

-1 1x

-1

(d)

2π

(e)

−π

—1

(f)

-1

22.

-1 1

-1

23. The portions of the graph of y=f(x) which lie below the x-axis are reﬂected over the x-axis to give the

graph of y=|f(x)|.

9 Chapter 1

24. Erase the portion of the graph of y=f(x) which lies in the left-half plane and replace it with the reﬂection

over the y-axis of the portion in the right-half plane (symmetry over the y-axis) and you obtain the graph

of y=f(|x|).

25. (a) for example, let a=1.1

(b)

123

26. They are identical.

27.

-1 123

28. This graph is very complex. We show three views, small (near the origin), medium and large:

(a) y

-1

-1 2

(b)

-50

(c)

1000

40 x

29. (a)

0.5

1.5

-3 -2 -1 123x

(b)

-1

-0.5

0.5

-3 -2 -1 1 23

(c)

0.5

1.5

-1 1 2 3 x

(d)

0.5

1.5

-2 -1 1 x

Exercise Set 1.3 10

30. y

31. (a) stretches or shrinks the graph in the

y-direction; ﬂips it if cchanges sign

-2

4c = 2

c = 1

c = -1.5

-1 1 x

(b) As cincreases, the parabola moves down

and to the left. If cincreases, up and right.

-2 -1 1 2 x

c = 2

c = 1

c = -1.5

-2 -1 1 2 x

c = +2

c = .5

c = -1

32. (a) y

—2

12 x

(b) x-intercepts at x=0,a,b. Assume a<band let aapproach b. The two branches of the curve

come together. If amoves past bthen aand bswitch roles.

-3

-2

-1

123

a = 0.5

b = 1.5

-3

-2

-1

123

a = 1

b = 1.5

-3

-2

-1

123

a = 1.5

b = 1.6

11 Chapter 1

33. The curve oscillates between the lines

y=xand y=−xwith increasing ra-

pidity as |x|increases.

-30

-20

-10

-30 -20 -10 10 20 30 x

34. The curve oscillates between the lines

y= +1 and y=−1, inﬁnitely many times

in any neighborhood of x=0.

—1

4—2

—1

35. Plot f(x)on[−10,10]; then on [−1,0], [−0.7,−0.6], [−0.65,−0.64], [−0.646,−0.645]; for the other

root use [4,5], [4.6,4.7], [4.64,4.65], [4.645,4.646]; roots −0.6455,4.6455.

36. Plot f(x)on[−10,10]; then on [−4,−3], [−3.7,−3.6], [−3.61,−3.60], [−3.606,−3.605]; for the other

root use [3,4], [3.6,3.7], [3.60,3.61], [3.605,3.606]; roots 3.6055,−3.6055.

EXERCISE SET 1.4

1. (a)

-1

-1 1 2 x

(b)

123x

(c)

-1 1 2 x

(d)

-4 -2 2 x

2. (a)

-2

(b)

(c)

-1

(d)

-1

Exercise Set 1.4 12

3. (a)

-1

-2 -1 1 2 x

(b)

-1

-0.5 0.5 1 1.5 x

(c)

-1

-1 1 2 3 x

(d)

-1

-1 1 2 3 x

-2 2 x

5. Translate right 2 units, and up one unit.

-2 2 4 6x

6. Translate left 1 unit, reﬂect over x-axis, and

translate up 2 units.

–2

7. Translate left 1 unit, stretch vertically by

a factor of 2, reﬂect over x-axis, translate

down 3 units.

-100

-80

-60

-40

-20

-8 -6 -4 -2 2 4 6 x

8. Translate right 3 units, compress vertically

by a factor of 1

2, and translate up 2 units.

9. y=(x+3)

2−9; translate left 3 units and

down 9 units.

-5

-8 -6 -4 -2 2 x

13 Chapter 1

10. y=(x+3)

2−19; translate left 3 units

and down 19 units.

–5

–4

11. y=−(x−1)2+ 2; translate right 1 unit,

reﬂect over x-axis, translate up 2 units.

-6

-4

-2

-2 -1 1 2 3 4 x

12. y=1

2[(x−1)2+ 2]; translate left 1 unit

and up 2 units, compress vertically by

a factor of 1

13. Translate left 1 unit, reﬂect over x-axis,

translate up 3 units.

2 4 6 8 10 12 x

14. Translate right 4 units and up 1 unit.

410

15. Compress vertically by a factor of 1

translate up 1 unit.

123

16. Stretch vertically by a factor of √3 and

reﬂect over x-axis.

-1

17. Translate right 3 units.

-10

246

Exercise Set 1.4 14

18. Translate right 1 unit and reﬂect

over x-axis.

-4

-2 2

19. Translate left 1 unit, reﬂect over x-axis,

translate up 2 units.

-8

-6

-4

-2

-4 -3 -2 -1 1 2 x

20. y=1−1/x; reﬂect over x-axis, translate

up 1 unit.

-5

21. Translate left 2 units and down 2 units.

-2

-4 -2 x

22. Translate right 3 units, reﬂect over x-axis,

translate up 1 unit.

-1

23. Stretch vertically by a factor of 2, translate

right 1 unit and up 1 unit.

24. y=|x−2|; translate right 2 units.

25. Stretch vertically by a factor of 2, reﬂect

over x-axis, translate up 2 units.

-1

-2 2 x

15 Chapter 1

26. Translate right 2 units and down 3 units.

–2

27. Translate left 1 unit and up 2 units.

-3 -2 -1 1 x

28. Translate right 2 units, reﬂect over x-axis.

–1

29. (a)

-1 1 x

(b) y=½0ifx≤0

2xif 0 <x

30. y

–5

31. x2+2x+ 1, all x;2x−x

2−1, all x;2x

3+2x, all x;2x/(x2+ 1), all x

32. 3x−2+|x|, all x;3x−2−|x|, all x;3x|x|−2|x|, all x;(3x−2)/|x|, all x6=0

33. 3√x−1, x≥1; √x−1, x≥1; 2x−2, x≥1; 2, x>1

34. (2x2+1)/x(x2+ 1), all x6=0;−1/x(x2+ 1), all x6=0;1/(x

2+ 1), all x6=0;x

/(x

2+ 1), all x6=0

35. (a) 3(b) 9(c) 2(d) 2

36. (a) π−1(b) 0(c) −π2+3π−1(d) 1

Exercise Set 1.4 16

37. (a) t4+1 (b) t2+4t+5 (c) x2+4x+5 (d) 1

x2+1

(e) x2+2xh +h2+1 (f) x2+1 (g) x+1 (h) 9x2+1

38. (a) √5s+2 (b) p√x+2 (c) 3√5x(d) 1/√x

(e) 4

√x(f) 0(g) 1/4

√x(h) |x−1|

39. 2x2−2x+1, all x;4x

2+2x, all x40. 2−x6, all x;−x6+6x

4−12x2+ 8, all x

41. 1−x,x≤1; √1−x2,|x|≤142. p√x2+3−3, |x|≥√

6; √x,x≥3

43. 1

1−2x,x6=1

2,1; −1

2x−1

2,x6=0,144. x

x2+1,x6=0; 1

x+x,x6=0

45. x−6+1 46. x

x+1

47. (a) g(x)=√

x,h(x)=x+2 (b) g(x)=|x|,h(x)=x

2−3x+5

48. (a) g(x)=x+1,h(x)=x

2(b) g(x)=1/x,h(x)=x−3

49. (a) g(x)=x

,h(x) = sin x(b) g(x)=3/x,h(x) = 5 + cos x

50. (a) g(x)=3sinx,h(x)=x

2(b) g(x)=3x

2+4x,h(x) = sin x

51. (a) f(x)=x

,g(x) = 1 + sin x,h(x)=x

2(b) f(x)=√

x,g(x)=1−x,h(x)= 3

√

52. (a) f(x)=1/x,g(x)=1−x,h(x)=x

2(b) f(x)=|x|,g(x)=5+x,h(x)=2x

53. y

-4

-3

-2

-1

-3 -2 -1 1 2 3

54. {−2,−1,0,1,2,3}

55. Note that f(g(−x)) = f(−g(x)) = f(g(x)), so f(g(x)) is even.

f(g(x))

–3 –1 –1

–3

17 Chapter 1

56. Note that g(f(−x)) = g(f(x)), so g(f(x)) is even.

–1

–2

–1

–3

g(f(x))

57. f(g(x)) = 0 when g(x)=±2, so x=±1.4; g(f(x)) = 0 when f(x)=0,sox=±2.

58. f(g(x))=0atx=−1 and g(f(x))=0atx=−1

59. 3(x+h)2−5−(3x2−5)

h=6xh +3h

h=6x+3h

60. (x+h)2+6(x+h)−(x

2+6x)

h=2xh +h2+6h

h=2x+h+6

61. 1/(x+h)−1/x

h=x−(x+h)

xh(x+h)=−1

x(x+h)

62. 1/(x+h)2−1/x2

h=x2−(x+h)2

x2h(x+h)2=−2x+h

x2(x+h)2

63. (a) the origin (b) the x-axis (c) the y-axis (d) none

64. (a)

y(b)

y(c)

65. (a) x−3−2−1 0 1 2 3

f(x) 1 −5−1 0 −1−5 1 (b) x−3−2−1 0 1 2 3

f(x) 1 5 −1 0 1 −5 -1

66. (a)

y(b)

67. (a) even (b) odd (c) odd (d) neither

68. neither; odd; even

69. (a) f(−x)=(−x)

2=x

2=f(x), even (b) f(−x)=(−x)

3=−x

3=−f(x), odd

(e) f(−x)=(−x)

3−(−x)

1+(−x)

2=−x

3+x

1+x

2=−f(x), odd

(f) f(−x)=2=f(x), even

Exercise Set 1.4 18

70. (a) x-axis, because x=5(−y)

2+ 9 gives x=5y

2+9

(b) x-axis, y-axis, and origin, because x2−2(−y)2=3,(−x)

2−2y

2= 3, and

(−x)2−2(−y)2= 3 all give x2−2y2=3

71. (a) y-axis, because (−x)4=2y

3+ygives x4=2y

3+y

(b) origin, because (−y)= (−x)

3+(−x)

2gives y=x

3+x

y2=|x|−5

72. 3

–3

-4 4

73. 2

-2

-3 3

74. (a) Whether we replace xwith −x,ywith −y, or both, we obtain the same equation, so by Theorem

1.4.3 the graph is symmetric about the x-axis, the y-axis and the origin.

(b) y=(1−x

2/3

)

3/2

to use the tricks that precede Exercise 29 in Section 1.3.)

75.

-1 1 2 3 4 x

76. y

77. (a)

Oco

(b)

Cco

19 Chapter 1

78. (a)

-1

(b)

-1

(c)

-1

(d)

-1

π/2

−π

79. Yes, e.g. f(x)=x

kand g(x)=x

nwhere kand nare integers.

80. If x≥0 then |x|=xand f(x)=g(x). If x<0 then f(x)=|x|

p/q if pis even and f(x)=−|x|p/q if p

is odd; in both cases f(x) agrees with g(x).

EXERCISE SET 1.5

1. (a) 3−0

0−2=−3

2,3−(8/3)

0−6=−1

18,0−(8/3)

2−6=2

(b) Yes; the ﬁrst and third slopes above are negative reciprocals of each other.

2. (a) −1−(−1)

−3−5=0, −1−3

5−7=2, 3−3

7−(−1) =0, −1−3

−3−(−1) =2

(b) Yes; there are two pairs of equal slopes, so two pairs of parallel lines.

3. III <II <IV <I

4. III <IV <I<II

5. (a) 1−(−5)

1−(−2) =2, −5−(−1)

−2−0=2, 1−(−1)

1−0= 2. Since the slopes connecting all pairs of points are

equal, they lie on a line.

(b) 4−2

−2−0=−1, 2−5

0−1=3, 4−5

−2−1=1

. Since the slopes connecting the pairs of points are not

equal, the points do not lie on a line.

6. The slope, m=−2, is obtained from y−5

x−7, and thus y−5=−2(x−7).

(a) If x= 9 then y=1. (b) If y= 12 then x=7/2.

7. The slope, m= 3, is equal to y−2

x−1, and thus y−2=3(x−1).

(a) If x= 5 then y= 14. (b) If y=−2 then x=−1/3.

Exercise Set 1.5 20

8. (a) Compute the slopes: y−0

x−0=1

2or y=x/2. Also y−5

x−7=2ory=2x−9. Solve simultaneously

to obtain x=6,y =3.

9. (a) The ﬁrst slope is 2−0

1−xand the second is 5−0

4−x. Since they are negatives of each other we get

2(4 −x)=−5(1 −x)or7x= 13, x=13/7.

10. (a) 27◦(b) 135◦(c) 63◦(d) 91◦

11. (a) 153◦(b) 45◦(c) 117◦(d) 89◦

12. (a) m= tan φ=−√3/3, so φ= 150◦(b) m= tan φ=4,soφ=76

◦

13. (a) m= tan φ=√3, so φ=60

◦(b) m= tan φ=−2, so φ= 117◦

14. y= 0 and x= 0 respectively 15. y=−2x+4

-1 1

16. y=5x−3

–8

–1 2

17. Parallel means the lines have equal slopes, so

y=4x+7.

-1 1

18. The slope of both lines is −3/2, so y−2=(−3/2)(x−(−1)),

or 3x+2y=1

–4

-2 1

19. The negative reciprocal of 5 is −1/5, so y=−1

5x+6. 12

-9 9

21 Chapter 1

20. The slope of x−4y= 7 is 1/4 whose negative reciprocal

is −4, so y−(−4) = −4(x−3) or y+4x=8.

–3

018

21. y−(−7)

x−1=4−(−7)

2−1,ory=11x−18 7

-9

22. y−1

x−(−2) =6−1

−3−(−2),ory=−5x−915

–5

–4 0

23. (a) m1=m2= 4, parallel (b) m1=2=−1/m2, perpendicular

(d) If A6= 0 and B6= 0 then m1=−A/B =−1/m2, perpendicular; if A=0orB= 0 (not both)

then one line is horizontal, the other vertical, so perpendicular.

(e) neither

24. (a) m1=m2=−5, parallel (b) m1=2=−1/m2, perpendicular

1=m

2=−A/B;ifB= 0 both

are vertical, so parallel

(e) neither

25. (a) m=(0−(−3))/(2 −0))=3/2soy=3x/2−3

(b) m=(−3−0)/(4 −0) = −3/4soy=−3x/4

26. (a) m=(0−2)/(2 −0)) = −1soy=−x+2

(b) m=(2−0)/(3 −0)=2/3soy=2x/3

27. (a) The velocity is the slope, which is 5−(−4)

10 −0=9/10 ft/s.

(b) x=−4

x=−2.2.

(d) t=80/9

Exercise Set 1.5 22

28. (a) v=5−1

4−2= 2 m/s (b) x−1=2(t−2) or x=2t−3(c) x=−3

29. (a) The acceleration is the slope of the velocity, so a=3−(−1)

1−4=−4

3ft/s2.

(b) v−3=−4

(t−1), or v=−4

3t+13

3(c) v=13

3ft/s

30. (a) The acceleration is the slope of the velocity, so a=0−5

10 −0=−5

10 =−1

2ft/s2.

(b) v= 5 ft/s (c) v= 4 ft/s (d) t=4s

31. (a) It moves (to the left) 6 units with velocity v=−3 cm/s, then remains motionless for 5 s, then

moves 3 units to the left with velocity v=−1 cm/s.

(b) vave =0−9

10 −0=−9

10 cm/s

save =9

10 cm/s.

32. It moves right with constant velocity v= 5 km/h; then accelerates; then moves with constant, though

increased, velocity again; then slows down.

33. (a) If x1denotes the ﬁnal position and x0the initial position, then v=(x

1−x

)/(t

1−t

) = 0 mi/h,

since x1=x0.

(b) If the distance traveled in one direction is d, then the outward journey took t=d/40 h. Thus

save =total dist

total time =2d

t+(2/3)t=80t

t+(2/3)t= 48 mi/h.

34. (a) down, since v<0(b) v=0att=2 (c) It’s constant at 32 ft/s2.

35. (a) v

100

20 40 60 80 100 120

(b) v=





10tif 0 ≤t≤10

100 if 10 ≤t≤100

600 −5tif 100 ≤t≤120

36. x

23 Chapter 1

37. (a) y=20−15 = 5 when x= 45, so 5 = 45k,

k=1/9, y=x/9

(b) y

0.2

0.4

0.6

246

x= 135 lb.

38. (a) Since y=0.2 = (1)k,k=1/5 and y=x/5(b) y

246

x=7.5 tons

39. Each increment of 1 in the value of xyields the increment of 1.2 for y, so the relationship is linear. If

y=mx +bthen m=1.2; from x=0,y= 2, follows b=2,soy=1.2x+2

40. Each increment of 1 in the value of xyields the increment of −2.1 for y, so the relationship is linear.

If y=mx +bthen m=−2.1; from x=0,y=10.5 follows b=10.5, so y=−2.1x+10.5

41. (a) With TFas independent variable, we have TC−100

TF−212 =0−100

32 −212,soT

C=5

F−32).

(b) 5/9 (c) Set TF=TC=5

9(TF−32) and solve for TF:TF=TC=−40◦(F or C).

(d) 37◦C

42. (a) One degree Celsius is one degree Kelvin, so the slope is the ratio 1/1=1. ThusT

C=T

K−273.15.

(b) TC=0−273.15 = −273.15◦C

43. (a) p−1

h−0=5.9−1

50 −0,orp=0.098h+1 (b) when p=2,orh=1/0.098 ≈10.20 m

44. (a) R−123.4

T−20 =133.9−123.4

45 −20 ,soR=0.42T+ 115.

(b) T=32.38◦C

45. (a) r−0.80

t−0=0.75 −0.80

4−0,sor=−0.0125t+0.8

(b) 64 days

46. (a) Let the position at rest be y0. Then y0+y=y0+kx; with x=11wegety

0+kx =y0+11k= 40,

and with x=24wegety

0+kx =y0+24k= 60. Solve to get k=20/13 and y0= 300/13.

(b) 300/13 + (20/13)W= 30, so W= (390 −300)/20=9/2g.

Exercise Set 1.6 24

47. (a) For xtrips we have C1=2xand

C2=25+x/4

510 15 20 25 30

(b) 2x=25+x/4, or x= 100/7, so the com-

muter pass becomes worthwhile at x= 15.

48. If the student drives xmiles, then the total costs would be CA= 4000 + (1.25/20)xand

CB= 5500 + (1.25/30)x. Set 4000 + 5x/80 = 5500 + 5x/120 and solve for x=72,000 mi.

49. (a) H≈20000/110 ≈181

(b) One light year is 9.408 ×1012 km and t=d

v=1

H=1

20km/s/Mly =9.408 ×1018km

20km/s

=4.704 ×1017 s=1.492 ×1010 years.

EXERCISE SET 1.6

1. (a) y=3x+b(b) y=3x+6

(c)

-10

-5

-2 -1 1 2 x

y = 3x + 6

y = 3x + 2

y = 3x - 4

2. Since the slopes are negative reciprocals, y=−1

3x+b.

3. (a) y=mx +2 (b) m= tan φ= tan 135◦=−1, so y=−x+2

(c)

-1

-2 -1 1 2 x

m = –1

m = 1

m = 1.5

4. (a) y=mx (b) y=m(x−1)

25 Chapter 1

5. (a) The slope is −1.

-3

-2

-1

-2 -1 1 2 x

(b) The y-intercept is y=−1.

-6

-4

-2

-2 -1 1 2 x

-2

-6 -4 -2

(d) The x-intercept is x=1.

-3

-2

-1

0.5 11.5 2x

6. (a) horizontal lines

(b) The y-intercept is y=−1/2.

-1

(d) They pass through (−1,1).

-2

(-1,1)

7. Let the line be tangent to the circle at the point (x0,y

0) where x2

0+y2

0= 9. The slope of the tangent

line is the negative reciprocal of y0/x0(why?), so m=−x0/y0and y=−(x0/y0)x+b. Substituting

the point (x0,y

0) as well as y0=±p9−x2

0we get y=±9−x0x

p9−x2

8. Solve the simultaneous equations to get the point (−2,1/3) of intersection. Then y=1

3+m(x+ 2).

Exercise Set 1.6 26

9. The x-intercept is x= 10 so that with depreciation at 10% per year the ﬁnal value is always zero, and

hence y=m(x−10). The y-intercept is the original value.

246810

10. A line through (6,−1) has the form y+1 = m(x−6). The intercepts are x= 6+1/m and y=−6m−1.

Set −(6+1/m)(6m+1)=3,or36m

2+15m+1 = (12m+1)(3m+1) = 0 with roots m=−1/12,−1/3;

thus y+1=−(1/3)(x−6) and y+1=−(1/12)(x−6).

11. (a) VI (b) IV (c) III (d) V(e) I(f) II

12. In all cases kmust be positive, or negative values would appear in the chart. Only kx−3decreases, so

that must be f(x). Next, kx2grows faster than kx3/2, so that would be g(x), which grows faster than

h(x) (to see this, consider ratios of successive values of the functions). Finally, experimentation (a

spreadsheet is handy) for values of kyields (approximately) f(x)=10x

−3

,g(x)=x

/2, h(x)=2x

1.5

13. (a)

-30

-20

-10

-2 -1 1 2 x

-60

-50

-40

-30

-20

-10

-2 -1 1 2 x

(b)

-4

-2

-4 -2 2 4 x

(c)

-2

-1

-1 1 2 3 x

0.5

1.5

123

27 Chapter 1

14. (a)

-2

-40

-2

(b)

-2

-4

-2

-4

(c)

-1 -2

-3

-1

-2

15. (a)

-10

-5

-2 -1 1 2 x

(b)

-2

-2 -1 1 2 x

(c)

-10

-5

-2 -1 1 2 x

16. (a)

(b)

(c)

17. (a)

-3 -2 -1 1 x

(b)

-80

-60

-40

-20

-1 1 2 3 4 5x

Exercise Set 1.6 28

(c)

-50

-40

-30

-20

-10

-5 -4 -3 -2 -1 1 2 3 x

(d)

-40

-20

12 3 4 5

18. (a)

(b)

-2

(c)

-2

-4

(d)

-2

-6

19. (a)

-1.5

-1

-0.5

0.5

1.5

-3 -2 -1 1 x

(b)

-1

-0.5

0.5

123456x

(c)

-30

-20

-10

-1 1 2 3 x

(d)

-30

-20

-10

-2 -1 1 2 x

20. (a)

-10

134

(b)

-30

-20

-10

-1 1 2 3 x

29 Chapter 1

(c)

-106

106

-1 1 x

(d)

-4 -2 2 x

21.

-4 -2 2 x

22. (a)

0.5

1.5

-2 2 x

(b)

0.5

1.5

-2 2 x

23. t=0.445√d

2.3

025

24. (a) t=0.373r1.5(b) 238,000 km (c) 1.89 days

25. (a) N·m(b) 20 N·m

P(N/m2) 80 ×10340 ×10320 ×10313.3×10310 ×103

26. If the side of the square base is xand the height of the container is ythen V=x2y= 100; minimize

A=2x

2+4xy =2x

2+ 400/x. A graphing utility with a zoom feature suggests that the solution is a

cube of side 1001

3cm.

27. (a) F=k/x2so 0.0005 = k/(0.3)2and

k=0.000045 N·m2.

(b) 0.000005 N

Exercise Set 1.6 30

(c)

510

10-5

(d) When they approach one another, the force

becomes inﬁnite; when they get far apart it

tends to zero.

28. (a) 2000 = C/(4000)2,soC=3.2×1010 lb·mi2(b) W=C/50002=(3.2×1010)/(25 ×106)=

1280 lb.

(c)

5000

10000

15000

20000

2000 4000 6000 8000 10000

(d) No, but Wis very small when xis large.

29. (a) II; y=1,x=−1,2(b) I; y=0,x=−2,3

30. The denominator has roots x=±1, so x2−1 is the denominator. To determine kuse the point (0,−1)

to get k=1,y=1/(x

2−1).

31. Order the six trigonometric functions as sin, cos, tan, cot, sec, csc:

(a) pos, pos, pos, pos, pos, pos (b) neg, zero, undef, zero, undef, neg

(e) neg, neg, pos, pos, neg, neg (f) neg, pos, neg, neg, pos, neg

32. (a) neg, zero, undef, zero, undef, neg (b) pos, neg, neg, neg, neg, pos

(e) neg, neg, pos, pos, neg, neg (f) neg, neg, pos, pos, neg, neg

33. (a) sin(π−x) = sin x; 0.588 (b) cos(−x) = cos x; 0.924

(e) sin 2x=±2 sin xp1−sin2x; use the +

sign for xsmall and positive; 0.951

(f) cos2x=1−sin2x; 0.654

34. (a) sin(3π+x)=−sin x;−0.588 (b) cos(−x−2π) = cos x; 0.924

tive sign for xsmall and negative; −0.195

(e) cos(3π+3x)=−4 cos3x+ 3 cos x;−0.384 (f) tan2x=1−cos2x

cos2x; 0.172

35. (a) −a(b) b

(e) −b(f) −a

(g) ±2b√1−b2(h) 2b2−1

(i) 1/b (j) −1/a

(k) 1/c (l) (1 −b)/2

31 Chapter 1

36. (a) The distance is 36/360=1/10th of a great

circle, so it is (1/10)2πr =2,513.27 mi.

(b) 36/360=1/10

37. If the arc length is 1, then solve the ratio x

1=2πr

29.5to get x≈80,936 km.

38. The distance travelled is equal to the length of that portion of the circumference of the wheel which

touches the road, and that is the fraction 225/360 of a circumference, so a distance of (225/360)(2π)3 =

11.78 ft

39. The second quarter revolves twice (720◦) about its own center.

40. Add rto itself until you exceed 2πr; since 6r<2πr < 7r, you can cut oﬀ 6 pieces of pie, but there’s

not enough for a full seventh piece. We conclude that there is no exact solution of the equation

‘One pie = 2πr’.

41. (a) y= 3 sin(x/2) (b) y= 4 cos 2x(c) y=−5 sin 4x

42. (a) y= 1 + cos πx (b) y=1+2sinx(c) y=−5 cos 4x

43. (a) y= sin(x+π/2)

(b) y= 3 + 3 sin(2x/9)

44. V= 120√2 sin(120πt)

45. (a) 3,π/2,0

-2

π/2

(b) 2,2,0

-2

2π4π6π

46. (a) 4,π,0

-4

-2

π/4 3π/4 5π/4 7π/4

(b) 1/2,2π/3,π/3

-0.2

0.4

π/3 2π/3 π

-4

-2

3π/2 9π/2 15π/2 21π/2

47. (a) Asin(ωt +θ)=Asin(ωt) cos θ+Acos(ωt) sin θ=A1sin(ωt)+A

2cos(ωt)

(b) A1=Acos θ, A2=Asin θ,soA=pA

1+A

2and θ= tan−1(A2/A1).

√

13/2, θ= tan−11

2√3;

x=5√13

2sin µ2πt + tan−11

2√3¶

-10

Exercise Set 1.7 32

48. three; x=0,x=±1.8955

–3

-3 3

EXERCISE SET 1.7

1. (a) x+1=t=y−1, y=x+2

t=0

t=1

t=2

t=3

t=4

t=5

x−101234

y123456

2. (a) x2+y2=1

-1 1 x

t=1

t=.5

t=.75 t=.25

t=0

x1 0.7071 0.00 −0.7071 −1

y0 0.7071 1.00 0.7071 0

3. t=(x+4)/3; y=2x+10

-8 6

4. t=x+3;y=3x+2,−3≤x≤0

(0, 2)

(-3, -7)

5. cos t=x/2,sin t=y/5; x2/4+y

/25=1

-5 5

-5

6. t=x2;y=2x

2+4,x≥0

33 Chapter 1

7. cos t=(x−3)/2,sin t=(y−2)/4;

(x−3)2/4+(y−2)2/16=1

-2

8. sec2t−tan2t=1;x

−y

2=1,

x≤−1 and y≥0

-1 x

9. cos 2t=1−2 sin2t;x=1−2y

,−1≤y≤1

-1 1

-1

10. t=(x−3)/4; y=(x−3)2−9

(3, -9)

11. x/2+y/3=1,0≤x≤2,0≤y≤3

12. y=x−1,x≥1,y ≥0

13. x= 5 cos t,y=−5 sin t,0≤t≤2π

-5

-7.5 7.5

14. x= cos t,y= sin t,π≤t≤3π/2

–1

-1 1

15. x=2,y=t

-1

16. x= 2 cos t,y= 3 sin t,0≤t≤2π

–3

-2 2

Exercise Set 1.7 34

17. x=t2,y=t,−1≤t≤1

-1

18. x=1+4cost,y=−3+4sint,0≤t≤2π

-8

-7 8

19. (a) IV, because xalways increases whereas yoscillates.

(b) II, because (x/2)2+(y/3)2= 1, an ellipse.

(d) VI; examine the cases t<−1 and t>−1 and you see the curve lies in the ﬁrst, second and

fourth quadrants only.

(e) III because y>0.

(f) I; since xand yare bounded, the answer must be I or II; but as truns, say, from 0 to π,xgoes

directly from 2 to −2, but ygoesfrom0to1to0to−1 and back to 0, which describes I but

not II.

20. (a) from left to right

(b) counterclockwise

(d) As ttravels from −∞ to −1, the curve goes from (near) the origin in the third quadrant and

travels up and left. As ttravels from −1to+∞the curve comes from way down in the second

quadrant, hits the origin at t= 0, and then makes the loop clockwise and ﬁnally approaches the

origin again as t→+∞.

(e) from left to right

(f) Starting, say, at (1,0), the curve goes up into the ﬁrst quadrant, loops back through the origin

and into the third quadrant, and then continues the ﬁgure-eight.

21. (a) 14

-35 8

(b) t0 1 2 3 4 5

x0 5.5 8 4.5−8−32.5

y1 1.5 3 5.5 9 13.5

√

3. (d) for 0 <t<2

√

2(e) at t=2

22. (a) 5

-2 14

(b) yis always ≥1 since cos t≤1

35 Chapter 1

23. (a) 3

-5

020

(b) o

-1 1

24. (a) 1.7

-1.7

-2.3 2.3

(b)

-10 10

25. (a) x−x0

x1−x0

=y−y0

y1−y0

(b) Set t= 0 to get (x0,y

0); t= 1 for (x1,y

1).

26. (a) x=−3−2t,y=−4+5t(b) x=at,y=b(1 −t)

27. (a) |R−P|2=(x−x

)

2+(y−y

2=t

2[(x1−x0)2+(y

1−y

2] and |Q−P|2=(x

1−x

)

2+(y

1−y

so r=|R−P|=|Q−P|t=qt.

(b) t=1/2(c) t=3/4

28. x=2+t,y=−1+2t

(a) (5/2,0) (b) (9/4,−1/2) (c) (11/4,1/2)

29. The two branches corresponding to −1≤t≤0 and 0 ≤t≤1 coincide.

30. (a) Eliminate t−t0

t1−t0

to obtain y−y0

x−x0

=y1−y0

x1−x0

(b) from (x0,y

0)to(x

1)(c) x=3−2(t−1), y=−1+5(t−1)

-2

31. (a) x−b

a=y−d

c(b)

123

Exercise Set 1.7 36

32. (a) If a= 0 the line segment is vertical; if c= 0 it is horizontal.

(b) The curve degenerates to the point (b, d).

33.

0.5

1.5

0.5 1 x

34. x=1/2−4t, y =1/2 for 0 ≤t≤1/4

x=−1/2,y=1/2−4(t−1/4) for 1/4≤t≤1/2

x=−1/2+4(t−1/2),y=−1/2 for 1/2≤t≤3/4

x=1/2,y=−1/2+4(t−3/4) for 3/4≤t≤1

35. (a) x= 4 cos t,y= 3 sin t(b) x=−1+4cost,y=2+3sint

-3

-4 4

-1

-5 3

36. (a) t=x/(v0cos α), so y=xtan α−gx2/(2v2

0cos2α).

(b)

2000

4000

6000

8000

10000

12000

40000 80000 x

37. (a) From Exercise 36, x= 400√2t,y= 400√2t−4.9t2.(b) 16,326.53 m (c) 65,306.12 m

38. (a) 15

–15

-25 25

(b) 15

–15

-25 25

–15

-25 25

a = 3, b = 2

–15

-25 25

a = 2, b = 3

–15

-25 25

a = 2, b = 7

37 Chapter 1

39. Assume that a6= 0 and b6= 0; eliminate the parameter to get (x−h)2/a2+(y−k)

2/b2=1. If|a|=|b|

the curve is a circle with center (h, k) and radius |a|;if|a|6=|b|the curve is an ellipse with center

(h, k) and major axis parallel to the x-axis when |a|>|b|, or major axis parallel to the y-axis when

|a|<|b|.

(a) ellipses with a ﬁxed center and varying axes of symmetry

(b) (assume a6= 0 and b6= 0) ellipses with varying center and ﬁxed axes of symmetry

40. Refer to the diagram to get bθ =aφ,θ=aφ/b but θ−α=φ+π/2

so α=θ−φ−π/2=(a/b −1)φ−π/2

x=(a−b) cos φ−bsin α

=(a−b) cos φ+bcos µa−b

b¶φ,

y=(a−b) sin φ−bcos α

=(a−b) sin φ−bsin µa−b

b¶φ.

φφ

a − b

41. (a)

-a

-a a

(b) Use b=a/4 in the equations of Exercise 40 to get

x=3

4acos φ+1

4acos 3φ,y=3

4asin φ−1

4asin 3φ;

but trigonometric identities yield cos 3φ= 4 cos3φ−3 cos φ, sin 3φ= 3 sin φ−4 sin3φ,

so x=acos3φ,y=asin3φ.

2/3

42.

-50

-3 -2 -1 x

a=−2

-50

-2 -1 x

a=−1

-1

-1 1 x

a=0

-50

a=1

-50

12 3

a=2

Supplementary Exercises 1 38

CHAPTER 1 SUPPLEMENTARY EXERCISES

1. 1940-45; the greatest ﬁve-year slope

2. (a) f(−1)=3.3,g(3)=2 (b) x=−3,3

−5≤y≤4

(e) the domain is −4≤x≤4.1, the range is

−3≤y≤5

(f) f(x)=0atx=−3,5; g(x)=0at

x=−3,2

0246

T4. x

58 13 t

5. If the side has length xand height h, then V=8=x

h,soh=8/x2. Then the cost C=5x

2(4)(xh)=5x

2+64/x.

6. Assume that the paint is applied in a thin veneer of uniform thickness, so that the quantity of paint

to be used is proportional to the area covered. If P is the amount of paint to be used, P=kπr2. The

constant kdepends on physical factors, such as the thickness of the paint, absorption of the wood,

etc.

7. y

-1

-5 5

8. Suppose the radius of the uncoated ball is rand that of the coated ball is r+h. Then the plastic has

volume equal to the diﬀerence of the volumes, i.e. V=4

3π(r+h)3−4

3πr3=4

3πh[3r2+3rh +h2]in

9. (a) The base has sides (10 −2x)/2 and 6 −2x, and the height is x,soV=(6−2x)(5 −x)xft3.

(b) From the picture we see that x<5 and 2x<6, so 0 <x<3.

10. {x6=0}and ∅(the empty set)

11. impossible; we would have to solve 2(3x−2) −5 = 3(2x−5) −2, or −9=−17

12. (a) (3 −x)/x (b) no; f(g(x)) can be deﬁned at x=1,

whereas g, and therefore f◦g, requires

x6=1

13. 1/(2 −x2)14. g(x)=x

2+2x

15. x−4−3−2−1 0 1 2 3 4

f(x) 0 −1 2 1 3 −2−3 4 −4

g(x) 3 2 1 −3−1−4 4 −2 0

(f◦g)(x) 4 −3−2−1 1 0 −4 2 3

(g◦f)(x)−1−3 4 −4−2 1 2 0 3

39 Chapter 1

16. (a) y=|x−1|,y=|(−x)−1|=|x+1|,

y=2|x+1|,y=2|x+1|−3,

y=−2|x+1|+3

(b) y

-1

-3 -1 2

17. (a) even ×odd=odd (b) a square is even

18. (a) y= cos x−2 sin xcos x=(1−2 sin x) cos x,sox=±π

,±3π

2,π

,5π

6,−7π

6,−11π

(b) (±π

2,0),(±3π

2,0),(π

6,√3/2),(5π

6,−√3/2),(−7π

6,−√3/2),(−11π

6,√3/2)

19. (a) If xdenotes the distance from Ato the base of the tower, and ythe distance from Bto the

base, then x2+d2=y2. Moreover h=xtan α=ytan β,sod

2=y

2−x

2=h

(cot2β−cot2α),

h2=d2

cot2β−cot2α=d2sin2αsin2β

sin2αcos2β−cos2αsin2β. The trigonometric identity

sin(α+β) sin(α−β) = sin2αcos2β−cos2αsin2βyields h=dsin αsin β

psin(α+β) sin(α−β).

(b) 295.72 ft.

20. (a)

-20

100 200 300 t

(b) when 2π

365(t−101) = 3π

2,ort= 374.75,

which is the same date as t=9.75, so dur-

ing the night of January 10th-11th

about 122 days

21. Cis the highest nearby point on the graph; zoom to ﬁnd that the coordinates of Care (2.0944,1.9132).

Similarly, Dis the lowest nearby point, and its coordinates are (4.1888,1.2284). Since f(x)=1

2x−sin x

is an odd function, the coordinates of Bare (−2.0944,−1.9132) and those of Aare (−4.1888,−1.2284).

22. Let y=A+Bsin(at +b). Since the maximum and minimum values of yare 35 and 5, A+B=35

and A−B=5,soA= 20, B= 15. The period is 12 hours, so 12a=2πand a=π/6. The

maximum occurs at t= 2, so 1 = sin(2a+b) = sin(π/3+b), π/3+b=π/2, b=π/2−π/3=π/6 and

y= 20 + 15 sin(πt/6+π/6).

23. (a) The circle of radius 1 centered at (a, a2); therefore, the family of all circles of radius 1 with

centers on the parabola y=x2.

(b) All parabolas which open up, have latus rectum equal to 1 and vertex on the line y=x/2.

24. (a) x=f(1 −t),y =g(1 −t)25.

-2

-1

-2 -1 1 2 x

Supplementary Exercises 1 40

26. Let y=ax2+bx +c. Then 4a+2b+c=0,64a+8b+c=18,64a−8b+c= 18, from which b=0

and 60a= 18, or ﬁnally y=3

10x2−6

27.

-2

-1

-1 1 2 x

28. (a) R=R0is the R-intercept, R0kis the slope,

and T=−1/k is the T-intercept

(b) −1/k =−273, or k=1/273

(1+20/273), or R0=1.025 (d) T= 126.55◦C

29. d=p(x−1)2+(

√

x−2)2;

d=9.1atx=1.358094

30. d=p(x−1)2+1/x2;

d=0.82 at x=1.380278

0.8

1.2

1.4

1.6

1.8

0.5 1 1.5 2 2.5 3

31. w=63.9V,w=63.9πh2(5/2−h/3); h=0.48 ft when w= 108 lb

32. (a)

1000

2000

3000

4000

12345

(b) w=63.9πh2(5/2−h/3); at h=5/2,

w= 2091.12 lb

33. (a)

100

150

200

10 20 30 40 50 t

(b) N= 80 when t=9.35 yrs

41 Chapter 1

34. (a) T

(b) T=17

◦F, 27◦F, 32◦F

35. (a)

-20

WCI

10 20 30 40 50 v

(b) T=3

◦F, −11◦F, −18◦F, −22◦F

36. The domain is the set of all x, the range is −0.1746 ≤y≤0.1227.

37. The domain is the set −0.7245 ≤x≤1.2207, the range is −1.0551 ≤y≤1.4902.

38. (a) The potato is done in the interval

27.65 <t<32.71.

(b) 91.54 min.

39. (a)

12345

(b) As t→∞,(0.273)t→0, and thus

v→24.61 ft/s.

close).

(e) 3.013 s

CHAPTER 1 HORIZON MODULE

1. (a) 0.25,6.25 ×10−2,3.91 ×10−3,1.53 ×10−5,2.32 ×10−10,5.42 ×10−20,2.94 ×10−39,8.64 ×10−78,

7.46 ×10−155,5.56 ×10−309;

1,1,1,1,1,1,1,1,1,1;

4,16,256,65536,4.29 ×109,1.84 ×1019,3.40 ×1038,1.16 ×1077,1.34 ×10154,1.80 ×10308

2. 2,2.25,2.2361111,2.23606798,2.23606798,...

3. (a) 1

2,1

4,1

8,1

16,1

32,1

64 (b) yn=1

4. (a) yn+1 =1.05yn

(b) y0=$1000, y1=$1050, y2=$1102.50, y3=$1157.62, y4=$1215.51, y5=$1276.28

Horizon Module 1 42

5. (a) x1/2,x

1/4,x

1/8,x

1/16,x

1/32 (b) They tend to the horizontal line y= 1, with

a hole at x=0.

1.8

6. (a) 1

144

(b) 1,2,3,5,8,13,21,34,55,89 ;

each new numerator is the sum of the previous two numerators.

233,233

377,377

610,610

987,987

1597,1597

2584,2584

4181,4181

6765,6765

10946,10946

17711

(d) F0=1,F

1=1,F

n=F

n−1+F

n−2for n≥2.

(e) the positive solution

7. (a) y1=cr,y2=cy1=cr2,y3=cr3,y4=cr4

(b) yn=crn

(tends to +∞).

8. The ﬁrst point on the curve is (c, kc(1 −c)), so y1=kc(1 −c) and hence y1is the ﬁrst iterate. The

point on the line to the right of this point has equal coordinates (y1,y

1), and so the point above it on

the curve has coordinates (y1,ky

1(1 −y1)); thus y2=ky1(1 −y1), and y2is the second iterate, etc.

9. (a) 0.261,0.559,0.715,0.591,0.701

(b) It appears to approach a point somewhere near 0.65.

CHAPTER 2

Limits and Continuity

EXERCISE SET 2.1

1. (a) −1(b) 3(c) does not exist

(d) 1(e) −1(f) 3

2. (a) 2(b) 0(c) does not exist

(d) 2(e) 0(f) 2

3. (a) 1(b) 1(c) 1(d) 1(e) −∞ (f ) +∞

4. (a) 3(b) 3(c) 3(d) 3(e) +∞(f ) +∞

5. (a) 0(b) 0(c) 0(d) 3(e) +∞(f ) +∞

6. (a) 2(b) 2(c) 2(d) 3(e) −∞ (f ) +∞

7. (a) −∞ (b) +∞(c) does not exist

(d) undef (e) 2(f) 0

8. (a) +∞(b) +∞(c) +∞(d) undef (e) 0(f ) −1

9. (a) −∞ (b) −∞ (c) −∞ (d) 1(e) 1(f ) 2

10. (a) 1(b) −∞ (c) does not exist

(d) −2(e) +∞(f) +∞

11. (a) 0(b) 0(c) 0

(d) 0(e) does not exist (f) does not exist

12. (a) 3(b) 3(c) 3

(d) 3(e) does not exist (f) 0

13. for all x06=−414. for all x06=−6,3

15. (a) At x= 3 the one-sided limits fail to exist.

(b) At x=−2 the two-sided limit exists but is not equal to F(−2).

16. (a) At x= 2 the two-sided limit fails to exist.

(b) At x= 3 the two-sided limit exists but is not equal to F(3).

Exercise Set 2.1 44

17. (a) 2 1.5 1.1 1.01 1.001 0 0.5 0.9 0.99 0.999

0.1429 0.2105 0.3021 0.3300 0.3330 1.0000 0.5714 0.3690 0.3367 0.3337

The limit is 1/3.

(b) 2 1.5 1.1 1.01 1.001 1.0001

0.4286 1.0526 6.344 66.33 666.3 6666.3

The limit is +∞.

−1−1.7143 −7.0111 −67.001 −667.0−6667.0

-50

The limit is −∞.

18. (a) −0.25 −0.1−0.001 −0.0001 0.0001 0.001 0.1 0.25

0.5359 0.5132 0.5001 0.5000 0.5000 0.4999 0.4881 0.4721

0.6

-0.25 0.25

The limit is 1/2.

45 Chapter 2

(b) 0.25 0.1 0.001 0.0001

8.4721 20.488 2000.5 20001

100

0 0.25

The limit is +∞.

−7.4641 −19.487 −1999.5−20000

-100

-0.25 0 The limit is −∞.

19. (a) −0.25 −0.1−0.001 −0.0001 0.0001 0.001 0.1 0.25

2.7266 2.9552 3.0000 3.0000 3.0000 3.0000 2.9552 2.7266

-0.25 0.25

The limit is 3.

(b) 0−0.5−0.9−0.99 −0.999 −1.5−1.1−1.01 −1.001

1 1.7552 6.2161 54.87 541.1−0.1415 −4.536 −53.19 −539.5

-60

-1.5 0

The limit does not exist.

Exercise Set 2.1 46

20. (a) 0−0.5−0.9−0.99 −0.999 −1.5−1.1−1.01 −1.001

1.5574 1.0926 1.0033 1.0000 1.0000 1.0926 1.0033 1.0000 1.0000

1.5

-1.5 0

The limit is 1.

(b) −0.25 −0.1−0.001 −0.0001 0.0001 0.001 0.1 0.25

1.9794 2.4132 2.5000 2.5000 2.5000 2.5000 2.4132 1.9794

2.5

-0.25 0.25

The limit is 5/2.

21. (a) −100,000,000 −100,000 −1000 −100 −10 10 100 1000

2.0000 2.0001 2.0050 2.0521 2.8333 1.6429 1.9519 1.9950

100,000 100,000,000

2.0000 2.0000

-40

-14 6

asymptote y=2asx→±∞

(b) −100,000,000 −100,000 −1000 −100 −10 10 100 1000

20.0855 20.0864 20.1763 21.0294 35.4013 13.7858 19.2186 19.9955

100,000 100,000,000

20.0846 20.0855

-160 160

asymptote y=20.086.

47 Chapter 2

−100,000,001 −100,000 −1001 −101.0−11.2 9.2 99.0 999.0 99,999 99,999,999

–50

-20 20

no horizontal asymptote

22. (a) −100,000,000 −100,000 −1000 −100 −10 10 100 1000 100,000 100,000,000

0.2000 0.2000 0.2000 0.2000 0.1976 0.1976 0.2000 0.2000 0.2000 0.2000

0.2

-1.2

-10 10 asymptote y=1/5asx→±∞

(b) −100,000,000 −100,000 −1000 −100 −10 10 100

0.0000 0.0000 0.0000 0.0000 0.0016 1668.0 2.09 ×1018

1000 100,000 100,000,000

1.77 ×10301 ? ?

-6 6

asymptote y=0asx→−∞, none as

x→+∞

0.0000 0.0000 0.0008 −0.0051 −0.0544 −0.0544 −0.0051

1000 100,000 100,000,000

0.0008 0.0000 0.0000

1.1

-0.3

-30 30

asymptote y=0asx→±∞

23. (a) lim

x→0+

sin x

x(b) lim

x→0+

x−1

x+1 (c) lim

x→0−(1+2x)

1/x

Exercise Set 2.2 48

24. (a) lim

x→0+

cos x

x(b) lim

x→0+

x+1 (c) lim

x→0−µ1+ 2

x¶x

25. (a) y

(b) yes; for example f(x) = (sin x)/x

26. (a) no

(b) yes; tan xand sec xat x=nπ +π/2, and cot xand csc xat x=nπ,n=0,±1,±2,...

29. (a) The plot over the interval [−a, a] becomes subject to catastrophic subtraction if ais small enough

(the size depending on the machine).

EXERCISE SET 2.2

1. (a) −6(b) 13 (c) −8(d) 16 (e) 2(f ) −1/2

(g) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.

(h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.

2. (a) 0

(b) The limit doesn’t exist because lim fdoesn’t exist and lim gdoes.

(f) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.

(g) The limit doesn’t exist because pf(x) is not deﬁned for 0 ≤x<2.

(h) 1

3. (a) 7(b) −3(c) π(d) −6(e) 36 (f ) −∞

4. (a) 1(b) −1(c) 1(d) −1(e) 1(f ) −1

5. 06. 3/4 7. 88. −39. 4

10. 12 11. −4/512. 013. 3/2 14. 4/3

15. 016. 017. 018. 5/3 19. −√5

20. 3

p3/221. 1/√622. √523. √324. −1/√6

25. +∞26. √327. does not exist 28. −∞ 29. −∞

30. +∞31. +∞32. does not exist 33. does not exist 34. −∞

35. +∞36. −∞ 37. −∞ 38. does not exist 39. −1/7

40. +∞41. 642. +∞43. +∞44. 4

49 Chapter 2

45. +∞46. +∞47. −∞ 48. +∞

49. (a) 2(b) 2(c) 2

50. (a) −2(b) 0(c) does not exist

51. (a) 3(b) y

52. (a) −6(b) F(x)=x−3

53. (a) Theorem 2.2.2(a) doesn’t apply; moreover one cannot add/subtract inﬁnities.

(b) lim

x→0+µ1

x−1

x2¶= lim

x→0+µx−1

x2¶=−∞

54. lim

x→0−µ1

x+1

x2¶= lim

x→0−

x+1

2=+∞55. lim

x→0

x¡√x+4+2

¢=1

56. lim

x→0

x¡√x+4+2

¢=0

57. lim

x→+∞(px2+3−x)√

2+3+x

√

2+3+x= lim

x→+∞

√x2+3+x=0

58. lim

x→+∞(px2−3x−x)√x2−3x+x

√x2−3x+x= lim

x→+∞−3x

√x2−3x+x=−3/2

59. lim

x→+∞³px2+ax −x´√x2+ax +x

√x2+ax +x= lim

x→+∞

√x2+ax +x=a/2

60. lim

x→+∞³px2+ax −px2+bx´√x2+ax +√x2+bx

√x2+ax +√x2+bx = lim

x→+∞

(a−b)x

√x2+ax +√x2+bx =a−b

61. lim

x→+∞p(x)=(−1)n∞and lim

x→−∞ p(x)=+∞

62. If m>nthe limits are both zero. If m=nthe limits are both 1. If n>mthe limits are (−1)n+m∞

and +∞, respectively.

63. If m>nthe limits are both zero. If m=nthe limits are both equal to am, the leading coeﬃcient of

p.Ifn>mthe limits are ±∞ where the sign depends on the sign of amand whether nis even or

odd.

64. (a) p(x)=q(x)=x(b) p(x)=x,q(x)=x

,q(x)=x(d) p(x)=x+3,q(x)=x

65. The left and/or right limits could be plus or minus inﬁnity; or the limit could exist, or equal any

preassigned real number. For example, let q(x)=x−x

0and let p(x)=a(x−x

)

nwhere ntakes on

the values 0,1,2.

Exercise Set 2.3 50

66. If m>nthe limit is zero. If m=nthe limit is cm/dm.Ifn>mthe limit is ±∞, where the sign

depends on the signs of cnand dm.

EXERCISE SET 2.3

1. (a) |f(x)−f(0)|=|x+2−2|=|x|<0.1 if and only if |x|<0.1

(b) |f(x)−f(3)|=|(4x−5) −7|=4|x−3|<0.1 if and only if |x−3|<(0.1)/4=0.0025

which corresponds to δ=0.000124998; and f(x)=16−²=15.999 at x=3.999874998, for

which δ=0.000125002. Use the smaller δ:thus|f(x)−16|<²provided |x−4|<0.000125 (to

six decimals).

2. (a) |f(x)−f(0)|=|2x+3−3|=2|x|<0.1 if and only if |x|<0.05

(b) |f(x)−f(0)|=|2x+3−3|=2|x|<0.01 if and only if |x|<0.005

3. (a) x1=(1.95)2=3.8025,x

2=(2.05)2=4.2025

(b) δ= min ( |4−3.8025|,|4−4.2025|)=0.1975

4. (a) x1=1/(1.1)=0.909090 ...,x

2=1/(0.9)=1.111111 ...

(b) δ= min( |1−0.909090|,|1−1.111111|)=0.0909090 ...

5. |2x−8|=2|x−4|<0.1if|x−4|<0.05, δ=0.05

6. |x/2+1|=(1/2)|x−(−2)|<0.1if|x+2|<0.2, δ=0.2

7. |7x+5−(−2)|=7|x−(−1)|<0.01 if |x+1|<1

700,δ=1

700

8. |5x−2−13|=5|x−3|<0.01 if |x−3|<1

500,δ=1

500

9. ¯¯¯¯

x2−4

x−2−4¯¯¯¯

=¯¯¯¯

x2−4−4x+8

x−2¯

=|x−2|<0.05 if |x−2|<0.05, δ=0.05

10. ¯¯¯¯

x2−1

x+1 −(−2)¯¯¯¯

=¯¯¯¯

x2−1+2x+2

x+1 ¯

=|x+1|<0.05 if |x+1|<0.05, δ=0.05

11. if δ<1 then ¯¯x2−16¯¯=|x−4||x+4|<9|x−4|<0.001 if |x−4|<1

9000,δ=1

9000

12. if δ<1 then |√x−3|¯¯¯¯

√x+3

√

x+3¯

=|x−9|

√

x+3|<|x−9|

√

8+3 <1

4|x−9|<0.001 if |x−9|<0.004, δ=0.004

13. if δ≤1 then ¯¯¯¯

x−1

5¯¯¯¯

=|x−5|

5|x|≤|x−5|

20 <0.05 if |x−5|<1, δ=1

14. |x−0|=|x|<0.05 if |x|<0.05, δ=0.05 15. |3x−15|=3|x−5|<²if |x−5|<1

3²,δ=1

3²

16. |(4x−5) −7|=|4x−12|=4|x−3|<²if |x−3|<1

4²,δ=1

4²

17. |2x−7−(−3)|=2|x−2|<²if |x−2|<1

2²,δ=1

2²

51 Chapter 2

18. |2−3x−5|=3|x+1|<²if |x+1|<1

3²,δ=1

3²

19. ¯¯¯¯

x2+x

x−1¯¯¯¯

=|x|<²if |x|<²,δ=²20. ¯¯¯¯

x2−9

x+3 −(−6)¯¯¯¯

=|x+3|<²if |x+3|<²,δ=²

21. if δ<1 then |2x2−2|=2|x−1||x+1|<6|x−1|<²if |x−1|<1

6²,δ= min(1,1

6²)

22. if δ<1 then |x2−5−4|=|x−3||x+3|<7|x−3|<²if |x−3|<1

7²,δ= min(1,1

7²)

23. if δ<1

6then ¯¯¯¯

x−3¯¯¯¯

3|x−1

|x|<18|x−1

3|<²if |x−1

3|<1

18²,δ= min( 1

6,1

18 ²)

24. If δ<1

2and |x−(−2)|<δthen −5

2<x<−3

,x+1<−1

2,|x+1|>1

2; then

¯¯¯¯

x+1−(−1)¯¯¯¯

=|x+2|

|x+1|<2|x+2|<²if |x+2|<1

2²,δ= min( 1

2,1

2²)

25. |√x−2|=¯¯¯¯

(√x−2)√x+2

√

x+2¯

=¯

x−4

√

x+2¯

2|x−4|<²if |x−4|<2²,δ=2²

26. |√x+3−3|¯

√

x+3+3

√

x+3+3¯

=|x−6|

√

x+3+3 ≤1

3|x−6|<²if |x−6|<3²,δ=3²

27. |f(x)−3|=|x+2−3|=|x−1|<²if 0 <|x−1|<²,δ=²

28. If δ<1 then |(x2+3x−1) −9|=|(x−2)(x+5)|<8|x−2|<²if |x−2|<1

8²,δ= min (1,1

8²)

29. (a) |f(x)−L|=1

x2<0.1ifx>√

10, N=√10

(b) |f(x)−L|=|x

x+1−1|=|1

x+1|<0.01 if x+1>100, N=99

x3¯¯¯¯

1000 if |x|>10, x<−10, N=−10

(d) |f(x)−L|=¯¯¯¯

x+1−1

=¯

x+1¯

<0.01 if |x+1|>100, −x−1>100, x<−101, N=−101

30. (a) ¯¯¯¯

x3¯¯¯¯

<0.1, x>101/3,N=10

1/3(b) ¯¯¯¯

x3¯¯¯¯

<0.01, x>1001/3,N= 1001/3

x3¯¯¯¯

<0.001, x>10, N=10

31. (a) x2

1+x

=1−²,x

1=−

r1−²

²;x

1+x

=1−²,x

2=r1−²

(b) N=r1−²

²(c) N=−r1−²

32. (a) x1=−1/²3;x2=1/²3(b) N=1/²3(c) N=−1/²3

33. 1

x2<0.01 if |x|>10, N=10

Exercise Set 2.3 52

34. 1

x+2 <0.005 if |x+2|>200, x>198, N= 198

35. ¯¯¯¯

x+1−1

=¯

x+1¯

<0.001 if |x+1|>1000, x>999, N= 999

36. ¯¯¯¯

4x−1

2x+5−2

=¯

2x+5¯

<0.1if|2x+5|>110, 2x>105, N=52.5

37. ¯¯¯¯

x+2−0

<0.005 if |x+2|>200, −x−2>200, x<−202, N=−202

38. ¯¯¯¯

x2¯¯¯¯

<0.01 if |x|>10, −x>10, x<−10, N=−10

39. ¯¯¯¯

4x−1

2x+5−2

=¯

2x+5¯

<0.1if|2x+5|>110, −2x−5>110, 2x<−115, x<−57.5, N=−57.5

40. ¯¯¯¯

x+1−1

=¯

x+1¯

<0.001 if |x+1|>1000, −x−1>1000, x<−1001, N=−1001

41. ¯¯¯¯

x2¯¯¯¯

<²if |x|>1

√²,N=1

√²42. ¯¯¯¯

x¯¯¯¯

<²if |x|>1

²,−x>1

²,x<−1

²,N=−1

43. ¯¯¯¯

x+2¯

<²if |x+2|>1

²,−x−2<1

²,x>−2−1

²,N=−2−1

44. ¯¯¯¯

x+2¯

<²if |x+2|>1

²,x+2>1

²,x>1

²−2, N=1

²−2

45. ¯¯¯¯

x+1−1

=¯

x+1¯

<²if |x+1|>1

²,x>1

²−1, N=1

²−1

46. ¯¯¯¯

x+1−1

=¯

x+1¯

<²if |x+1|>1

²,−x−1>1

²,x<−1−1

²,N=−1−1

47. ¯¯¯¯

4x−1

2x+5−2

=¯

2x+5¯

<²if |2x+5|>11

²,−2x−5>11

²,2x<−11

²−5, x<−11

2²−5

2,N=−5

2−11

2²

48. ¯¯¯¯

4x−1

2x+5−2

=¯

2x+5¯

<²if |2x+5|>11

²,2x> 11

²−5, x>11

2²−5

2,N=11

2²−5

49. (a) 1

x2>100 if |x|<1

10 (b) 1

|x−1|>1000 if |x−1|<1

1000

(x−3)2<−1000 if |x−3|<1

10√10 (d) −1

x4<−10000 if x4<1

10000,|x|<1

50. (a) 1

(x−1)2>10 if and only if |x−1|<1

√10 (b) 1

(x−1)2>1000 if and only if

|x−1|<1

10√10

(x−1)2>100000 if and only if |x−1|<1

100√10

51. if M>0 then 1

(x−3)2>M,0<(x−3)2<1

M,0<|x−3|<1

√

M,δ=1

√

53 Chapter 2

52. if M<0 then −1

(x−3)2<M,0<(x−3)2<−1

M,0<|x−3|<1

√

−M,δ=1

√

−M

53. if M>0 then 1

|x|>M,0<|x|<1

M,δ=1

54. if M>0 then 1

|x−1|>M,0<|x−1|<1

M,δ=1

55. if M<0 then −1

x4<M,0<x

4<−1

M,|x|<1

(−M)

1/4,δ=1

(−M)

1/4

56. if M>0 then 1

x4>M,0<x

4<1

M,x< 1

1/4,δ=1

1/4

57. if x>2 then |x+1−3|=|x−2|=x−2<²if 2 <x<2+²,δ=²

58. if x<1 then |3x+2−5|=|3x−3|=3|x−1|= 3(1 −x)<²if 1 −x<1

3²,1−1

²<x<1, δ=1

3²

59. if x>4 then √x−4<²if x−4<²

,4<x<4+²

,δ=²

60. if x<0 then √−x<²if −x<²

,−²

2<x<0, δ=²2

61. if x>2 then |f(x)−2|=|x−2|=x−2<²if 2 <x<2+²,δ=²

62. if x<2 then |f(x)−6|=|3x−6|=3|x−2|= 3(2 −x)<²if 2 −x< 1

3²,2−1

²<x<2, δ=1

3²

63. (a) if M<0 and x>1 then 1

1−x<M,x−1<−1

M,1<x<1−1

M,δ=−1

(b) if M>0 and x<1 then 1

1−x>M,1−x< 1

M,1−1

M<x<1, δ=1

64. (a) if M>0 and x>0 then 1

x>M,x< 1

M,0<x< 1

M,δ=1

(b) if M<0 and x<0 then 1

x<M,−x<−1

M,1

M<x<0, δ=−1

65. (a) Given any M>0 there corresponds N>0 such that if x>N then f(x)>M,x+1>M,

x>M−1, N=M−1.

(b) Given any M<0 there corresponds N<0 such that if x<N then f(x)<M,x+1 <M,

x<M−1, N=M−1.

66. (a) Given any M>0 there corresponds N>0 such that if x>N then f(x)>M,x

2−3>M,

x>√

M+3,N=√

M+3.

(b) Given any M<0 there corresponds N<0 such that if x<N then f(x)<M,x

3+5<M,

x<(M−5)1/3,N=(M−5)1/3.

67. if δ≤2 then |x−3|<2, −2<x−3<2, 1 <x<5, and |x2−9|=|x+3||x−3|<8|x−3|<²if

|x−3|<1

8²,δ= min ¡2,1

8²¢

68. (a) We don’t care about the value of fat x=a, because the limit is only concerned with values of

xnear a. The condition that fbe deﬁned for all x(except possibly x=a) is necessary, because

if some points were excluded then the limit may not exist; for example, let f(x)=xif 1/x is

not an integer and f(1/n) = 6. Then lim

x→0f(x) does not exist but it would if the points 1/n were

excluded.

(b) when x<0 then √xis not deﬁned (c) yes; if δ≤0.01 then x>0, so √xis deﬁned

Exercise Set 2.4 54

69. |(x3−4x+5)−2|<0.05, −0.05 <(x3−4x+5)−2<0.05, 1.95 <x

3−4x+5 <2.05; x3−4x+5=1.95

at x=1.0616, x3−4x+5=2.05 at x=0.9558; δ= min (1.0616 −1,1−0.9558) = 0.0442

2.2

1.9

0.9 1.1

70. √5x+1=3.5atx=2.25, √5x+1=4.5atx=3.85, so δ= min(3 −2.25,3.85 −3)=0.75

EXERCISE SET 2.4

1. (a) no, x=2 (b) no, x=2 (c) no, x=2 (d) yes

(e) yes (f) yes

2. (a) no, x=2 (b) no, x=2 (c) no, x=2 (d) yes

(e) no, x=2 (f) yes

3. (a) no, x=1,3(b) yes (c) no, x=1 (d) yes

(e) no, x=3 (f) yes

4. (a) no, x=3 (b) yes (c) yes (d) yes

(e) no, x=3 (f) yes

5. (a) 3(b) 3

6. −2/5

7. (a) y

(b) y

55 Chapter 2

-1

(d) y

8. f(x)=1/x,g(x)=(0ifx=0

sin 1

xif x6=0

9. (a) C

(b) One second could cost you one dollar.

10. (a) no; disasters (war, ﬂood, famine, pestilence, for example) can cause discontinuities

(b) continuous

(d) continuous

11. none 12. none 13. none

14. fis not deﬁned at x=±115. fis not deﬁned at x=±4

16. fis not deﬁned at x=−7±√57

17. fis not deﬁned at x=±3

18. fis not deﬁned at x=0,−419. none

20. fis not deﬁned at x=0,−3

21. none; f(x)=2x+ 3 is continuous on x<4 and f(x)=7+16

xis continuous on 4 <x;

lim

x→4−f(x) = lim

x→4+f(x)=f(4)=11sofis continuous at x=4

22. lim

x→1f(x) does not exist so fis discontinuous at x=1

23. (a) fis continuous for x<1, and for x>1; lim

x→1−f(x)=5, lim

x→1

+f(x)=k,soifk= 5 then fis

continuous for all x

(b) fis continuous for x<2, and for x>2; lim

x→2−f(x)=4k, lim

x→2+f(x)=4+k,soif4k=4+k,

k=4/3 then fis continuous for all x

24. (a) no, fis not deﬁned at x=2 (b) no, fis not deﬁned for x≤2

Exercise Set 2.4 56

25. (a) y

(b) y

26. (a) f(c) = lim

x→cf(x)

(b) lim

x→1f(x)=2

-1

lim

x→1g(x)=1

27. (a) x= 0, lim

x→0−f(x)=−16=+1= lim

x→0

+f(x) so the discontinuity is not removable

(b) x=−3; deﬁne f(−3) = −3 = lim

x→−3f(x), then the discontinuity is removable

x→2f(x) = 1, so deﬁne f(2) = 1 and fbecomes continuous

there; at x=−2, lim

x→−2does not exist, so the discontinuity is not removable

28. (a) fis not deﬁned at x= 2; lim

x→2f(x) = lim

x→2

x+2

2+2x+4 =1

3, so deﬁne f(2) = 1

3and fbecomes

continuous there

(b) lim

x→2−f(x)=16=4= lim

x→2

+f(x), so fhas a nonremovable discontinuity at x=2

x→1f(x)=86=f(1), so fhas a removable discontinuity at x=1

29. (a) discontinuity at x=1/2, not removable; at

x=−3, removable

-5

(b) 2x2+5x−3=(2x−1)(x+3)

57 Chapter 2

30. (a) there appears to be one discontinuity near

x=−1.52

–4

-3 3

(b) one discontinuity at x=−1.52

31. For x>0, f(x)=x

3/5=(x

)

1/5is the composition (Theorem 2.4.6) of the two continuous functions

g(x)=x

3and h(x)=x

1/5and is thus continuous. For x<0, f(x)=f(−x) which is the composition

of the continuous functions f(x) (for positive x) and the continuous function y=−x. Hence f(−x)is

continuous for all x>0. At x=0,f(0) = lim

x→0f(x)=0.

32. x4+7x

2+1≥1>0, thus f(x) is the composition of the polynomial x4+7x

2+ 1, the square root

√x, and the function 1/x and is therefore continuous by Theorem 2.4.6.

33. (a) Let f(x)=kfor x6=cand f(c)=0;g(x)=lfor x6=cand g(c)=0. Ifk=−lthen f+gis

continuous; otherwise it’s not.

(b) f(x)=kfor x6=c,f(c)=1;g(x)=l6= 0 for x6=c,g(c)=1. Ifkl = 1, then fg is continuous;

otherwise it is not.

34. A rational function is the quotient f(x)/g(x) of two polynomials f(x) and g(x). By Theorem 2.4.2 f

and gare continuous everywhere; by Theorem 2.4.3 f/g is continuous except when g(x)=0.

35. Since fand gare continuous at x=cwe know that lim

x→cf(x)=f(c) and lim

x→cg(x)=g(c). In the

following we use Theorem 2.2.2.

(a) f(c)+g(c) = lim

x→cf(x) + lim

x→cg(x) = lim

x→c(f(x)+g(x)) so f+gis continuous at x=c.

(b) same as (a) except the + sign becomes a −sign

g(c)=

lim

x→cf(x)

lim

x→cg(x)= lim

x→c

f(x)

g(x)so f

gis continuous at x=c

36. h(x)=f(x)−g(x) satisﬁes h(a)>0, h(b)<0. Use the Intermediate Value Theorem or Theorem

2.4.9.

37. Of course such a function must be discontinuous. Let f(x)=1on0≤x<1, and f(x)=−1on

1≤x≤2.

38. A square whose diagonal has length rhas area f(r)=r

/2. Note that

f(r)=r

/2<πr

/2<2r

2=f(2r). By the Intermediate Value Theorem there must be a value

cbetween rand 2rsuch that f(c)=πr2/2, i.e. a square of diagonal cwhose area is πr2/2.

39. The cone has volume πr2h/3. The function V(r)=πr2h(for variable rand ﬁxed h) gives the volume

of a right circular cylinder of height hand radius r, and satisﬁes V(0) <πr

h/3<V(r). By the

Intermediate Value Theorem there is a value cbetween 0 and rsuch that V(c)=πr2h/3, so the

cylinder of radius c(and height h) has volume equal to that of the cone.

40. If f(x)=x

3−4x+ 1 then f(0)=1,f(1) = −2. Use Theorem 2.4.9.

41. If f(x)=x

3+x

2−2xthen f(−1)=2,f(1) = 0. Use the Intermediate Value Theorem.

Exercise Set 2.4 58

42. Since lim

x→−∞ p(x)=−∞ and lim

x→+∞p(x)=+∞(or vice versa, if the leading coeﬃcient of pis negative),

it follows that for M=−1 there corresponds N1<0, and for M= 1 there is N2>0, such that

p(x)<−1 for x<N

1and p(x)>1 for x>N

. Choose x1<N

1and x2>N

2and use Theorem 2.4.9

on the interval [x1,x

2] to ﬁnd a solution of p(x)=0.

43. For the negative root, use intervals on the x-axis as follows: [−2,−1]; since f(−1.3) <0 and

f(−1.2) >0, the midpoint x=−1.25 of [−1.3,−1.2] is the required approximation of the root.

For the positive root use the interval [0,1]; since f(0.7) <0 and f(0.8) >0, the midpoint x=0.75 of

[0.7,0.8] is the required approximation.

44. x=−1.25 and x=0.75.

-5

-2 -1

-1

0.7 0.8

45. For the negative root, use intervals on the x-axis as follows: [−2,−1]; since f(−1.7) <0 and

f(−1.6) >0, use the interval [−1.7,−1.6]. Since f(−1.61) <0 and f(−1.60) >0 the midpoint

x=−1.605 of [−1.61,−1.60] is the required approximation of the root. For the positive root use

the interval [1,2]; since f(1.3) >0 and f(1.4) <0, use the interval [1.3,1.4]. Since f(1.37) >0 and

f(1.38) <0, the midpoint x=1.375 of [1.37,1.38] is the required approximation.

46. x=−1.605 and x=1.375.

-2

-1.7 -1.6

-0.5

1.3 1.4

47. x=2.24.

48. Set f(x)= a

x−1+b

x−3

. Since lim

x→1+f(x)=+∞and lim

x→3−f(x)=−∞ there exist x1>1 and x2<3

(with x2>x

) such that f(x)>1 for 1 <x<x

1and f(x)<−1 for x2<x<3. Choose x3in (1,x

and x4in (x2,3) and apply Theorem 2.4.9 on [x3,x

4].

49. The uncoated sphere has volume 4π(x−1)3/3 and the coated sphere has volume 4πx3/3. If the volume

of the uncoated sphere and of the coating itself are the same, then the coated sphere has twice the

volume of the uncoated sphere. Thus 2(4π(x−1)3/3)=4πx3/3, or x3−6x2+6x−2 = 0, with the

solution x=4.847 cm.

50. Let g(t) denote the altitude of the monk at time tmeasured in hours from noon of day one, and let f(t)

denote the altitude of the monk at time tmeasured in hours from noon of day two. Then g(0) <f(0)

and g(12) >f(12). Use Exercise 36.

51. We must show lim

x→cf(x)=f(c). Let ²>0; then there exists δ>0 such that if |x−c|<δthen

|f(x)−f(c)|<². But this certainly satisﬁes Deﬁnition 2.3.3.

59 Chapter 2

EXERCISE SET 2.5

1. none 2. x=π3. x=nπ,n=0,±1,±2,...

4. x=nπ +π/2, n=0,±1,±2,... 5. x=nπ,n=0,±1,±2,...

6. none 7. none 8. x=nπ +π/2, n=0,±1,±2,...

9. 2nπ +π/6,2nπ +5π/6, n=0,±1,±2,... 10. none

11. (a) sin x, x3+7x+1 (b) |x|,sin x(c) x3,cos x, x +1

(d) √x, 3+x, sin x, 2x(e) sin x, sin x(f) x5−2x3+1,cos x

12. (a) Use Theorem 2.4.6. (b) g(x) = cos x,g(x)= 1

2+1,g(x)=x

2+1

13. cos µlim

x→+∞

x¶=cos0=1 14. sin µlim

x→+∞

x¶=sin0=0

15. sin µlim

x→+∞

πx

2−3x¶= sin ³−π

3´=−√3

216. 1

2lim

h→0

sin h

h=1

17. 3 lim

θ→0

sin 3θ

3θ=3 18. µlim

θ→0+

θ¶lim

θ→0+

sin θ

θ=+∞

19. −lim

x→0−

sin x

x=−120. 1

3µlim

x→0

sin x

x¶2

21. 1

5lim

x→0+√xlim

x→0+

sin x

x=0 22. sin 6x

sin 8x=6

sin 6x

sin 8x, so lim

x→0

sin 6x

sin 8x=6

8=3

23. tan 7x

sin 3x=7

3 cos 7x

sin 7x

sin 3xso lim

x→0

tan 7x

sin 3x=7

3(1)(1)(1) = 7

24. µlim

θ→0sin θ¶lim

θ→0

sin θ

θ=0 25. µlim

h→0cos h¶lim

h→0

sin h=1

26. sin h

1−cos h=sin h

1−cos h

1 + cos h

1 + cos h=sin h(1 + cos h)

1−cos2h=1 + cos h

sin h; no limit

27. θ2

1−cos θ

1 + cos θ

1 + cos θ=θ2(1 + cos θ)

1−cos2θ=µθ

sin θ¶2

(1 + cos θ) so lim

θ→0

θ2

1−cos θ= (1)22=2

28. cos(1

2π−x) = sin(1

2π) sin x= sin x, so lim

x→0

cos ¡1

2π−x¢=1

29. 030. t2

1−cos2t=µt

sin t¶2

,solim

t→0

1−cos2t=1

31. 1−cos 5h

cos 7h−1=(1 −cos 5h)(1 + cos 5h)(1 + cos 7h)

(cos 7h−1)(1 + cos 5h)(1 + cos 7h)=−25

49 µsin 5h

5h¶2µ7h

sin 7h¶21 + cos 7h

1 + cos 5hso

lim

h→0

1−cos 5h

cos 7h−1=−25

Exercise Set 2.5 60

32. lim

x→0+sin µ1

x¶= lim

t→+∞sin t;

limit does not exist

33. lim

x→0+cos µ1

x¶= lim

t→+∞cos t;

limit does not exist

34. lim

x→0x−3 lim

x→0

sin x

x=−335. 2 + lim

x→0

sin x

x=3

36. k=f(0) = lim

x→0

sin 3x

x= 3 lim

x→0

sin 3x

3x=3,sok=3

37. lim

x→0−f(x)=klim

x→0

sin kx

kx cos kx =k, lim

x→0+f(x)=2k

,sok=2k

,k=1

38. No; sin x/|x|has unequal one-sided limits.

39. (a) lim

t→0+

sin t

t=1 (b) lim

t→0−

1−cos t

t= 0 (Theorem 2.5.3)

x→π

π−x

sin x= lim

t→0

sin t=1

40. cos ³π

2−t´= sin t, so lim

x→2

cos(π/x)

x−2= lim

t→0

(π−2t) sin t

4t= lim

t→0

π−2t

4lim

t→0

sin t

t=π

41. t=x−1; sin(πx) = sin(πt +π)=−sin πt; and lim

x→1

sin(πx)

x−1=−lim

t→0

sin πt

t=−π

42. t=x−π/4; tan x−1= 2 sin t

cos t−sin t; lim

x→π/4

tan x−1

x−π/4= lim

t→0

2 sin t

t(cos t−sin t)=2

43. −x≤xcos µ50π

x¶≤x44. −x2≤x2sin µ50π

√x¶≤x2

45. lim

x→0f(x) = 1 by the Squeezing Theorem

-1

-1 1 x

y = cos x

y = 1 – x2

y = f(x)

46. lim

x→+∞f(x) = 0 by the Squeezing Theorem

-1

47. Let g(x)=−1

xand h(x)= 1

; thus lim

x→+∞

sin x

x= 0 by the Squeezing Theorem.

48. y

61 Chapter 2

49. (a) sin x= sin twhere xis measured in degrees, tis measured in radians and t=πx

180.Thus

lim

x→0

sin x

x= lim

t→0

sin t

(180t/π)=π

180.

50. cos x= cos twhere xis measured in degrees, tin radians, and t=πx

180.Thus

lim

x→0

1−cos x

x= lim

t→0

1−cos t

(180t/π)=0.

51. (a) sin 10◦=0.17365 (b) sin 10◦= sin π

18 ≈π

18 =0.17453

52. (a) cos θ= cos 2α=1−2 sin2(θ/2)

≈1−2(θ/2)2=1−1

2θ

(b) cos 10◦=0.98481

2³π

18´2≈0.98477

53. (a) 0.08749 (b) tan 5◦≈π

36 =0.08727

54. (a) h=52.55 ft (b) Since αis small, tan α◦≈πα

180 isagood

approximation.

55. (a) Let f(x)=x−cos x;f(0) = −1, f(π/2)=π/2. By the IVT there must be a solution of

f(x)=0.

(b) y

0.5

1.5

y = cos x

π/2

y = x

56. (a) f(x)=x+ sin x−1; f(0) = −1, f(π/6) = π/6−1/2>0. By the IVT there must be a solution

of f(x)=0.

(b) y

0.5

π/6

y = x

y = 1– sin x

57. (a) There is symmetry about the equatorial plane.

(b) Let g(φ) be the given function. Then g(38) <9.8 and g(39) >9.8, so by the Intermediate Value

Theorem there is a value cbetween 38 and 39 for which g(c)=9.8 exactly.

58. (a) does not exist

(b) the limit is zero

Supplementary Exercises 2 62

satisfying |x|<δand there are inﬁnitely many irrational numbers satisfying the same condition.

Thus if the limit were to exist, it could not be zero because of the rational numbers, and it could

not be 1 because of the irrational numbers, and it could not be anything else because of all the

numbers. Hence the limit cannot exist. For part (b) use the Squeezing Theorem with +xand

−xas the ‘squeezers’.

CHAPTER 2 SUPPLEMENTARY EXERCISES

1. (a) 1(b) no limit (c) no limit (d) 1(e) 3

(f) 0(g) 0(h) 2(i) 1/2

2. (a) f(x)=2x/(x−1) (b) y

4. f(x)=−1 for a≤x< a+b

2and f(x)=1for a+b

2≤x≤b.

5. (a) 0.222 ...,0.24390,0.24938,0.24994,0.24999,0.25000; for x6=2,f(x)= 1

x+2, so the limit is 1/4;

the limit is 4.

(b) 1.15782,4.22793,4.00213,4.00002,4.00000,4.00000; to prove, use tan 4x

x=sin 4x

xcos 4x=4

cos 4x

sin 4x

4x.

6. (a) y=0 (b) none (c) y=2

7. (a) x 1 0.1 0.01 0.001 0.0001 0.00001 0.000001

f(x) 1.000 0.443 0.409 0.406 0.406 0.405 0.405

(b) y

0.5

-1 1

8. (a) 0.4 amperes (b) [0.3947,0.4054] (c) ·3

7.5+δ,3

7.5−δ¸

(d) 0.0187 (e) It becomes inﬁnite.

9. (a) y

0.4

0.2 0.8

(b) Let g(x)=x−f(x). Then g(1) ≥0 and

g(0) ≤0; by the Intermediate Value Theo-

rem there is a solution cin [0,1] of g(c)=0.

63 Chapter 2

10. (a) lim

θ→0tan µ1−cos θ

θ¶= tan µlim

θ→0

1−cos θ

θ¶= tan µlim

θ→0

1−cos2θ

θ(1 + cos θ)¶=tan0=0

(b) t−1

√t−1=t−1

√t−1

√t+1

√

t+1 =(t−1)(√t+1)

t−1=√

t+ 1; lim

t→1

t−1

√t−1= lim

t→1(√t+1)=2

(3x2+2x−7)(x3−9x)=(2 −1/x)5

(3+2/x −7/x2)(1 −9/x2)→25/3=32/3asx→+∞

(d) sin(θ+π) = sin θcos π−cos θsin π=−sin θ, so lim

θ→0cos µsin(θ+π)

2θ¶= lim

θ→0cos µ−sin θ

2θ¶

= cos µlim

θ→0−sin θ

2θ¶= cos ¡−1

2¢

11. If, on the contrary, f(x0)<0 for some x0in [0,1], then by the Intermediate Value Theorem we would

have a solution of f(x) = 0 in [0,x

0], contrary to the hypothesis.

12. For x<2fis a polynomial and is continuous; for x>2fis a polynomial and is continuous. At

x=2,f(2) = −13 6=13= lim

x→2

+f(x)sofis not continuous there.

13. f(−6) = 185, f(0) = −1, f(2) = 65; apply Theorem 2.4.9 twice, once on [−6,0] and once on [0,2]

14. 3.317

15. Let ²=f(x0)/2>0; then there corresponds δ>0 such that if |x−x0|<δthen |f(x)−f(x0)|<²,

−²<f(x)−f(x

)<²,f(x)>f(x

)−²=f(x

)/2>0 for x0−δ<x<x

0+δ.

16. y

17. (a) 1.449 (xmust be ≥−1) (b) x=0,±1.896

18. Since lim

x→0sin(1/x) does not exist, no conclusions can be drawn.

19. (a) √5, no limit, √10, √10, no limit, +∞,

no limit

(b) 5,10,0,0,10,−∞,+∞

20. (a) −1/5,+∞,−1/10,−1/10, no limit, 0, 0 (b) −1,+1,−1,−1, no limit, −1,+1

21. a/b 22. 123. does not exist

24. 225. 026. k227. 3−k

28. The numerator satisﬁes: |2x+xsin 3x|≤|2x|+|x|=3|x|. Since the denominator grows like x2, the

limit is 0.

30. (a) √x2+4−2

√

2+4+2

√

2+4+2=x

(

√

2+4+2) =1

√

2+4+2,so

lim

x→0

√x2+4−2

2= lim

x→0

√x2+4+2 =1

Supplementary Exercises 2 64

(b) x1 0.1 0.01 0.001 0.0001 0.00001

f(x) 0.236 0.2498 0.2500 0.2500 0.25000 0.00000

The division may entail division by zero (e.g. on an HP 42S), or the numerator may be inaccurate

(catastrophic subtraction, e.g.).

31. x0.1 0.01 0.001 0.0001 0.00001 0.000001

f(x) 2.59 2.70 2.717 2.718 2.7183 2.71828

32. x3.1 3.01 3.001 3.0001 3.00001 3.000001

f(x) 5.74 5.56 5.547 5.545 5.5452 5.54518

33. x1.1 1.01 1.001 1.0001 1.00001 1.000001

f(x) 0.49 0.54 0.540 0.5403 0.54030 0.54030

34. x0.1 0.01 0.001 0.0001 0.00001 0.000001

f(x) 99.0 9048.8 368063.3 4562.7 3.9×10−34 0

35. x100 1000 104105106107

f(x) 0.48809 0.49611 0.49876 0.49961 0.49988 0.49996

36. For large values of x(not much more than 100) the computer can’t handle 5xor 3x, yet the limit is 5.

37. δ=0.07747 38. $2,001.60, $2,009.66, $2,013.62, $2013.75

39. (a) x3−x−1=0,x

3=x+1,x=3

√

x+1. (b) y

-1

-1 1

x1x2x3

(d) 1,1.26,1.31,1.322,1.324,1.3246,1.3247

40. (a) y

-1 1 32

x1x2

(b) 0,−1,−2,−9,−730

41. x=5

√x+ 2; 1.267168 42. x= cos x; 0.739085 (after 33 iterations!).

CHAPTER 3

The Derivative

EXERCISE SET 3.1

1. (a) msec =f(4) −f(3)

4−3=(4)2/2−(3)2/2

1=7

(b) mtan = lim

x1→3

f(x1)−f(3)

x1−3= lim

x1→3

1/2−9/2

x1−3

= lim

x1→3

1−9

2(x1−3) = lim

x1→3

(x1+ 3)(x1−3)

2(x1−3) = lim

x1→3

x1+3

2=3

x1→x0

f(x1)−f(x0)

x1−x0

= lim

x1→x0

1/2−x2

0/2

x1−x0

= lim

x1→x0

1−x2

2(x1−x0)

= lim

x1→x0

x1+x0

2=x0

(d)

Tangent

Secant

2. (a) msec =f(2) −f(1)

2−1=23−13

1=7

(b) mtan = lim

x1→1

f(x1)−f(1)

x1−1= lim

x1→1

1−1

x1−1= lim

x1→1

(x1−1)(x2

1+x1+1)

1−1

= lim

x1→1(x2

1+x1+1)=3

x1→x0

f(x1)−f(x0)

x1−x0

= lim

x1→x0

1−x3

x1−x0

= lim

x1→x0

(x2

1+x1x0+x2

=3x

(d)

Secant

Tangent

3. (a) msec =f(3) −f(2)

3−2=1/3−1/2

1=−1

(b) mtan = lim

x1→2

f(x1)−f(2)

x1−2= lim

x1→2

1/x1−1/2

x1−2

= lim

x1→2

2−x1

2x1(x1−2) = lim

x1→2−1

2x1

=−1

x1→x0

f(x1)−f(x0)

x1−x0

= lim

x1→x0

1/x1−1/x0

x1−x0

= lim

x1→x0

x0−x1

x0x1(x1−x0)

= lim

x1→x0

−1

x0x1

=−1

(d)

Secant

Tangent

Exercise Set 3.1 66

4. (a) msec =f(2) −f(1)

2−1=1/4−1

1=−3

(b) mtan = lim

x1→1

f(x1)−f(1)

x1−1= lim

x1→1

1/x2

1−1

x1−1

= lim

x1→1

1−x2

1(x1−1) = lim

x1→1−(x1+1)

=−2

x1→x0

f(x1)−f(x0)

x1−x0

= lim

x1→x0

1/x2

1−1/x2

x1−x0

= lim

x1→x0

0−x2

0x2

1(x1−x0)

= lim

x1→x0

−(x1+x0)

0x2

=−2

(d)

Tangent Secant

5. (a) mtan = lim

x1→x0

f(x1)−f(x0)

x1−x0

= lim

x1→x0

(x2

1+1)−(x

0+1)

1−x

= lim

x1→x0

1−x2

x1−x0

= lim

x1→x0

(x1+x0)=2x

(b) mtan =2(2)=4

6. (a) mtan = lim

x1→x0

f(x1)−f(x0)

x1−x0

= lim

x1→x0

(x2

1+3x

1+2)−(x

0+3x

0+2)

1−x

= lim

x1→x0

(x2

1−x2

0)+3(x

1−x

)

1−x

= lim

x1→x0

(x1+x0+3)=2x

0+3

(b) mtan =2(2)+3=7

7. (a) mtan = lim

x1→x0

f(x1)−f(x0)

x1−x0

= lim

x1→x0

√x1−√x0

x1−x0

= lim

x1→x0

√x1+√x0

2√x0

(b) mtan =1

2√1=1

8. (a) mtan = lim

x1→x0

f(x1)−f(x0)

x1−x0

= lim

x1→x0

1/√x1−1/√x0

x1−x0

= lim

x1→x0

√x0−√x1

√x0√x1(x1−x0)= lim

x1→x0

−1

√x0√x1(√x1+√x0)=−1

2x3/2

(b) mtan =−1

2(4)3/2=−1

9. (a) mtan = (50 −10)/(15 −5)

=40/10

= 4 m/s

(b)

t (s)

10 20

Velocity

(m/s)

67 Chapter 3

10. (a) (10 −10)/(3 −0) = 0 cm/s

(b) t=0,t= 2, and t=4.2 (horizontal tangent line)

(d) (3 −18)/(4 −2) = −7.5 cm/s (slope of estimated tangent line to curve at t=3)

11. From the ﬁgure:

t0t1t2

(a) The particle is moving faster at time t0because the slope of the tangent to the curve at t0is

greater than that at t2.

(b) The initial velocity is 0 because the slope of a horizontal line is 0.

(d) The particle is slowing down because the slope decreases as tincreases from t1to t2.

12.

t0t1

13. It is a straight line with slope equal to the velocity.

14. (a) decreasing (slope of tangent line decreases with increasing time)

(b) increasing (slope of tangent line increases with increasing time)

(d) decreasing (slope of tangent line decreases with increasing time)

15. (a) 72◦F at about 4:30 P.M. (b) about (67 −43)/6=4

◦F/h

of estimated tangent line to curve at 9 P.M.)

16. For V= 10 the slope of the tangent line is about −0.25 atm/L, for V= 25 the slope is

about −0.04 atm/L.

17. (a) during the ﬁrst year after birth

(b) about 6 cm/year (slope of estimated tangent line at age 5)

(d)

t (yrs)

Growth rate

(cm/year)

5101520

Exercise Set 3.2 68

18. (a) The rock will hit the ground when

16t2= 576, t2= 36, t= 6 s (only t≥0

is meaningful)

(b) vave =16(6)2−16(0)2

6−0= 96 ft/s

3−0= 48 ft/s (d) vinst = lim

t1→6

16t2

1−16(6)2

t1−6= lim

t1→6

16(t2

1−36)

t1−6

= lim

t1→616(t1+ 6) = 192 ft/s

19. (a) 5(40)3= 320,000 ft (b) vave = 320,000/40=8,000 ft/s

so t3= 27, t=3s;

ave = 135/3 = 45 ft/s.

(d) vinst = lim

t1→40

5t3

1−5(40)3

t1−40 = lim

t1→40

5(t3

1−403)

t1−40

= lim

t1→40 5(t2

1+40t

1+ 1600) = 24,000 ft/s

20. (a) vave =[3(3)2+3]−[3(1)2+1]

3−1= 13 mi/h

(b) vinst = lim

t1→1

(3t2

1+t1)−4

t1−1= lim

t1→1

(3t1+ 4)(t1−1)

t1−1= lim

t1→1(3t1+ 4) = 7 mi/h

21. (a) vave =6(4)4−6(2)4

4−2= 720 ft/min

(b) vinst = lim

t1→2

6t4

1−6(2)4

t1−2= lim

t1→2

6(t4

1−16)

t1−2

= lim

t1→2

6(t2

1+ 4)(t2

1−4)

t1−2= lim

t1→26(t2

1+ 4)(t1+ 2) = 192 ft/min

EXERCISE SET 3.2

1. f0(1)=2,f

0(3)=0,f

0(5) = −2, f0(6) = −1/2

2. f0(4) <f

0(0) <f

0(2) <0<f

0(−3)

3. (b) f0(2) = m=3 (c) the same, f0(2)=3

4. f0(−1) = m=4−3

0−(−1) =1

5. y

6. y

7. y−(−1)=5(x−3),y=5x−16 8. y−3=−4(x+2),y=−4x−5

9. f0(x) = lim

h→0

f(x+h)−f(x)

h= lim

h→0

3(x2+2xh +h2)−3x2

h= lim

h→03(2x+h)=6x;f(3) = 3(3)2= 27,

f0(3)=18soy−27 = 18(x−3), y=18x−27

69 Chapter 3

10. f0(x) = lim

h→0

f(x+h)−f(x)

h= lim

h→0

(x+h)2−(x+h)−(x2−x)

h= lim

h→0(2x+h−1)=2x−1;

f(2) = 22−2=2,f

(2) = 3 so y−2=3(x−2),y=3x−4

11. f0(x) = lim

h→0

f(x+h)−f(x)

h= lim

h→0

(x+h)3−x3

h=3x

;f(0)=0

3=0,

0(0) = 0 so y−0 = (0)(x−0), y=0

12. f0(x) = lim

h→0

f(x+h)−f(x)

h= lim

h→0

2(x+h)3+1−(2x3+1)

h= lim

h→0(6x2+6xh +2h

)=6x

;

f(−1)=2(−1)3+1=−1, f0(−1)=6soy+1=6(x+ 1), y=6x+5

13. f0(x) = lim

h→0

f(x+h)−f(x)

h= lim

h→0

√x+1+h−√

x+1

= lim

h→0

√x+1+h−√

x+1

√

x+1+h+√

x+1

√

x+1+h+√

x+1 = lim

h→0

√x+1+h+√

x+1 =1

√

x+1;

f(8) = √8+1=3,f

0(8) = 1

6so y−3=1

(x−8), y=1

6x+5

14. f0(x) = lim

h→0

f(x+h)−f(x)

h= lim

h→0

(x+h)4−x4

h= lim

h→0(4x3+6x

h+4xh2+h3)=4x

;

f(−2)=(−2)4= 16, f0(−2) = −32 so y−16 = −32(x+ 2), y=−32x−48

15. f0(x) = lim

∆x→0

x+∆x−1

∆x= lim

∆x→0

x−(x+∆x)

x(x+∆x)

∆x

= lim

∆x→0−∆x

x∆x(x+∆x)= lim

∆x→0−1

x(x+∆x)=−1

16. f0(x) = lim

∆x→0

(x+∆x)

2−1

∆x= lim

∆x→0

x2−(x+∆x)

(x+∆x)

∆x

= lim

∆x→0

x2−x2−2x∆x−∆x2

x2∆x(x+∆x)

2= lim

∆x→0−2x∆x−∆x2

x2∆x(x+∆x)

2= lim

∆x→0−2x−∆x

x2(x+∆x)

2=−2

17. f0(x) = lim

∆x→0

[a(x+∆x)

2+b]−[ax2+b]

∆x= lim

∆x→0

ax2+2ax∆x+a(∆x)2+b−ax2−b

∆x

= lim

∆x→0

2ax∆x+a(∆x)2

∆x= lim

∆x→0(2ax +a∆x)=2ax

18. f0(x) = lim

∆x→0

(x+∆x)+1 −1

x+1

∆x= lim

∆x→0

(x+1)−(x+∆x+1)

(x+ 1)(x+∆x+1)

∆x

= lim

∆x→0

x+1−x−∆x−1

∆x(x+ 1)(x+∆x+1) = lim

∆x→0−∆x

∆x(x+ 1)(x+∆x+1)

= lim

∆x→0−1

(x+ 1)(x+∆x+1) =−1

(x+1)

19. f0(x) = lim

∆x→0

√x+∆x−1

√

∆x= lim

∆x→0

√x−√x+∆x

∆x

√

x+∆x

= lim

∆x→0

x−(x+∆x)

∆x

√

x+∆x(

√

x+√

x+∆x)= lim

∆x→0−1

√x√x+∆x(

√

x+√

x+∆x)=−1

3/2

Exercise Set 3.2 70

20. f0(x) = lim

∆x→0

(x+∆x)

1/3−x

1/3

∆x, but a3−b3=(a−b)(a2+ab +b2) so with a=(x+∆x)

1/3

and b=x1/3

f0(x) = lim

∆x→0

(x+∆x)−x

∆x[(x+∆x)

2/3+(x+∆x)

1/3

1/3+x

2/3

]

= lim

∆x→0

(x+∆x)

2/3+(x+∆x)

1/3

1/3+x

2/3=1

2/3

21. f0(t) = lim

h→0

f(t+h)−f(t)

h= lim

h→0

[4(t+h)2+(t+h)] −[4t2+t]

= lim

h→0

4t2+8th +4h

2+t+h−4t

2−t

= lim

h→0

8th +4h

2+h

h= lim

h→0(8t+4h+1)=8t+1

22. dV

dr = lim

h→0

3π(r+h)3−4

3πr3

h= lim

h→0

3π(r3+3r

h+3rh2+h3−r3)

= lim

h→0

3π(3r2+3rh +h2)=4πr2

23. (a) D(b) F(c) B(d) C(e) A(f ) E

24. Any function of the form f(x)=x+khas slope 1, and thus the derivative must be equal to 1

everywhere.

-2

25. (a)

y(b)

-1

(c)

26. (a)

y(b)

y(c)

27. (a) f(x)=x

2and a=3 (b) f(x)=√

xand a=1

71 Chapter 3

28. (a) f(x) = cos xand a=π(b) f(x)=x

7and a=1

29. dy

dx = lim

h→0

[4(x+h)2+1]−[4x2+1]

h= lim

h→0

4x2+8xh +4h

2+1−4x

2−1

h= lim

h→0(8x+4h)=8x

dx¯¯¯¯x=1

=8(1)=8

30. dy

dx = lim

h→0µ5

x+h+1

¶−µ5

x+1

h= lim

h→0

x+h−5

h= lim

h→0

5x−5(x+h)

x(x+h)

= limh→0

5x−5x−5h

hx(x+h)= lim

h→0−5

x(x+h)=−5

dx¯¯¯¯x=−2

=−5

(−2)2=−5

31. y=−2x+1

-2 2

-3

32. 1.5

0 2.5

33. (b) h0.5 0.1 0.01 0.001 0.0001 0.00001

(f(1+h)−f(1))/h 1.6569 1.4355 1.3911 1.3868 1.3863 1.3863

34. (b) h0.5 0.1 0.01 0.001 0.0001 0.00001

(f(1+h)−f(1))/h 0.50489 0.67060 0.70356 0.70675 0.70707 0.70710

35. (a) dollars/ft

(b) As you go deeper the price per foot may increase dramatically, so f0(x) is roughly the price per

additional foot.

(d) From the approximation 1000 = f0(300) ≈f(301) −f(300)

301 −300 we see that f(301) ≈f(300) + 1000,

so the extra foot will cost around $1000.

36. (a) gallons/dollar

(b) The increase in the amount of paint that would be sold for one extra dollar.

sold.

(d) From −100 = f0(10) ≈f(11) −f(10)

11 −10 we see that f(11) ≈f(10) −100, so an increase of one

dollar would decrease the amount of paint sold by around 100 gallons.

37. (a) F≈200 lb, dF/dθ ≈60 lb/rad (b) µ=(dF/dθ)/F ≈60/200=0.3

38. (a) dN/dt ≈34 million/year; in 1950 the world population was increasing at the rate of about 34

million per year.

(b) dN/dt

N≈34

2490 ≈0.014=1.4 %/year

Exercise Set 3.2 72

39. (a) T≈120◦F, dT/dt ≈−4.5

◦F/min

(b) k=(dT /dt)/(T−T0)≈(−4.5)/(120 −75) = −0.1

41. lim

x→0f(x) = lim

x→0

√x=0=f(0), so fis continuous at x=0.

lim

h→0

f(0+h)−f(0)

h= lim

h→0

√h−0

h= lim

h→0

h2/3=+∞,so

(0) does not exist. x

2-2

42. lim

x→2f(x) = lim

x→2(x−2)2/3=0=f(2) so fis continuous at x=2.

lim

h→0

f(2+h)−f(2)

h= lim

h→0

h2/3−0

h= lim

h→0

h1/3which does not exist

so f0(2) does not exist.

43. lim

x→1−f(x) = lim

x→1+f(x)=f(1), so fis continuous at x=1.

lim

h→0−

f(1+h)−f(1)

h= lim

h→0−

[(1 + h)2+1]−2

h= lim

h→0−(2+h)=2;

lim

h→0+

f(1+h)−f(1)

h= lim

h→0+

2(1 + h)−2

h= lim

h→0+2=2,sof

0(1)=2.

3-3

44. lim

x→1−f(x) = lim

x→1+f(x)=f(1) so fis continuous at x=1.

lim

h→0−

f(1+h)−f(1)

h= lim

h→0−

[(1 + h)2+2]−3

h= lim

h→0−(2+h)=2;

lim

h→0+

f(1+h)−f(1)

h= lim

h→0+

[(1 + h)+2]−3

h= lim

h→0+1=1,sof

0(1)

does not exist. x

3-3

45. fis continuous at x= 1 because it is diﬀerentiable there, thus lim

h→0f(1+h)=f(1) and so f(1)=0

because lim

h→0

f(1+h)

hexists; f0(1) = lim

h→0

f(1+h)−f(1)

h= lim

h→0

f(1+h)

h=5.

46. Let x=y= 0 to get f(0) = f(0) + f(0) + 0 so f(0)=0. f

(x) = lim

h→0

f(x+h)−f(x)

h, but (with

y=h)f(x+h)=f(x)+f(h)+5xh so f(x+h)−f(x)=f(h)+5xh and f0(x) = lim

h→0

f(h)+5xh

lim

h→0(f(h)

h+5x)=3+5x.

47. f0(x) = limh→0

f(x+h)−f(x)

h= lim

h→0

f(x)f(h)−f(x)

h= lim

h→0

f(x)[f(h)−1]

h=f(x) lim

h→0

f(h)−f(0)

=f(x)f0(0) = f(x)

73 Chapter 3

EXERCISE SET 3.3

1. 28x62. −36x11 3. 24x7+2 4. 2x3

5. 06. √27. −1

3(7x6+2) 8. 2

9. 3ax2+2bx +c10. 1

aµ2x+1

b¶11. 24x−9+1/

√

x12. −42x−7−5

2√x

13. −3x−4−7x−814. 1

2√x−1

15. f0(x)=(3x

2+6) d

dx µ2x−1

4¶+µ2x−1

4¶d

dx(3x2+6)=(3x

2+ 6)(2) + µ2x−1

4¶(6x)

=18x

2−3

2x+12

16. f0(x)=(2−x−3x

3)d

dx(7+x5)+(7+x

dx(2 −x−3x3)=(2−x−3x

)(5x4)+(7+x

)(−1−9x2)

=−24x7−6x5+10x

4−63x2−7

17. f0(x)=(x

3+7x

2−8) d

dx(2x−3+x−4)+(2x

−3+x

−4

dx(x3+7x

2−8)

=(x

3+7x

2−8)(−6x−4−4x−5)+(2x

−3+x

−4

)(3x2+14x)=−15x−2−14x−3+48x

−4+32x

−5

18. f0(x)=(x

−1+x

−2)d

dx(3x3+27)+(3x

3+ 27) d

dx(x−1+x−2)

=(x

−1+x

−2

)(9x2)+(3x

3+ 27)(−x−2−2x−3)=3+6x−27x−2−54x−3

19. 12x(3x2+1) 20. f(x)=x

10 +4x

6+4x

,f0(x)=10x

9+24x

5+8x

21. dy

dx =

(5x−3) d

dx(1) −(1) d

dx(5x−3)

(5x−3)2=−5

(5x−3)2;y0(1) = −5/4

22. dy

dx =

(√x+2) d

dx(3) −3d

dx(√x+2)

(

√

x+2)

2=−3/(2√x(√x+2)

0(1) = −3/18 = −1/6

23. dx

dt =

(2t+1)d

dt(3t)−(3t)d

dt(2t+1)

(2t+1)

2=(2t+ 1)(3) −(3t)(2)

(2t+1)

2=3

(2t+1)

24. dx

dt =

(3t)d

dt(t2+1)−(t

2+1)d

dt(3t)

(3t)2=(3t)(2t)−(t2+ 1)(3)

9t2=t2−1

3t2

25. dy

dx =

(x+3) d

dx(2x−1) −(2x−1) d

dx(x+3)

(x+3)

2=(x+ 3)(2) −(2x−1)(1)

(x+3)

2=7

(x+3)

2;dy

dx¯¯¯¯x=1

26. dy

dx =

(x2−5) d

dx(4x+1)−(4x+1) d

dx(x2−5)

(x2−5)2=(x2−5)(4) −(4x+ 1)(2x)

(x2−5)2=−4x2+2x+20

2−5)2;

dx¯¯¯¯x=1

=13

Exercise Set 3.3 74

27. dy

dx =µ3x+2

x¶d

dx ¡x−5+1

¢+¡x

−5+1

¢d

dx µ3x+2

x¶

=µ3x+2

x¶¡−5x

−6

¢+¡x

−5+1

¢·x(3) −(3x+ 2)(1)

x2¸=µ3x+2

x¶¡−5x

−6

¢+¡x

−5+1

¢µ−2

2¶;

dx¯¯¯¯x=1

=5(−5)+2(−2) = −29

28. dy

dx =(2x

7−x

dx µx−1

x+1¶+µx−1

x+1¶d

dx(2x7−x2)

=(2x

7−x

)·(x+ 1)(1) −(x−1)(1)

(x+1)

2¸+µx−1

x+1¶(14x6−2x)

=(2x

7−x

)·2

(x+1)

2+µx−1

x+1¶(14x6−2x);

dx¯¯¯¯x=1

=(2−1)2

4+ 0(14 −2) = 1

29. 32t30. 2π31. 3πr232. −2α−2+1

33. (a) dV

dr =4πr2(b) dV

dr ¯¯¯¯r=5

=4π(5)2= 100π

35. (a) g0(x)=√

xf0(x)+ 1

√

xf(x), g0(4) = (2)(−5) + 1

4(3) = −37/4

(b) g0(x)=xf0(x)−f(x)

x2,g0(4) = (4)(−5) −3

16 =−23/16

36. (a) g0(x)=6x−5f

0(x), g0(3) = 6(3) −5(4) = −2

(b) g0(x)=2f(x)−(2x+1)f0(x)

(x),g

0(3) = 2(−2) −7(4)

(−2)2=−8

37. (a) F0(x)=5f

0(x)+2g

0(x),F0(2) = 5(4) + 2(−5)=10

(b) F0(x)=f

0(x)−3g

0(x),F0(2)=4−3(−5)=19

0(x)+g(x)f0(x),F0(2)=(−1)(−5) + (1)(4) = 9

(d) F0(x)=[g(x)f

0(x)−f(x)g

0(x)]/g2(x),F0(2) = [(1)(4) −(−1)(−5)]/(1)2=−1

38. dy

dx =(1+x)(−1) −(1 −x)(1)

(1+x)2=−2

(1+x)2,dy

dx¯¯¯¯x=2

=−2

9and y=−1

3for x= 2 so an equation of

the tangent line is y−µ−1

3¶=−2

9(x−2), or y=−2

9x+1

39. y−2=5(x+ 3), y=5x+17

40. d

dλ ·λλ0+λ6

2−λ0¸=1

2−λ0

dλ(λλ0+λ6)= 1

2−λ

(λ

0+6λ

)=λ

0+6λ

2−λ

41. (a) dy/dx =21x

2−10x+1,d

y/dx2=42x−10

(b) dy/dx =24x−2, d2y/dx2=24

(d) y=35x

5−16x3−3x,dy/dx = 175x4−48x2−3, d2y/dx2= 700x3−96x

75 Chapter 3

42. (a) y0=28x

6−15x2+2,y

00 = 168x5−30x

(b) y0=3,y

00 =0

5x2,y00 =−4

5x3

(d) y=2x

4+3x

3−10x−15, y0=8x

3+9x

2−10, y00 =24x

2+18x

43. (a) y0=−5x−6+5x

00 =30x

−7+20x

000 =−210x−8+60x

(b) y=x−1,y0=−x−2,y00 =2x

−3

000 =−6x−4

44. (a) dy/dx =10x−4, d2y/dx2= 10, d3y/dx3=0

(b) dy/dx =−6x−3−4x−2+1,d

y/dx2=18x

−4+8x

−3

y/dx3=−72x−5−24x−4

y/dx3=24ax

45. (a) f0(x)=6x,f

00(x)=6,f

000(x)=0,f

000(2)=0

(b) dy

dx =30x

4−8x,d

dx2= 120x3−8, d2y

dx2¯¯¯¯x=1

= 112

dx £x−3¤=−3x−4,d2

dx2£x−3¤=12x

−5

dx3£x−3¤=−60x−6,d4

dx4£x−3¤= 360x−7,

dx4£x−3¤¯¯¯¯x=1

= 360

46. (a) y0=16x

3+6x

00 =48x

2+12x,y

000 =96x+ 12, y000(0)=12

(b) y=6x

−4

,dy

dx =−24x−5,d2y

dx2= 120x−6,d3y

dx3=−720x−7,d4y

dx4= 5040x−8,

d4y

dx4¯¯¯¯x=1

= 5040

47. y0=3x

2+3,y

00 =6x, and y000 =6so

000 +xy00 −2y0=6+x(6x)−2(3x2+3)=6+6x

2−6x

2−6=0

48. y=x−1,y0=−x−2,y00 =2x

−3so

x3y00 +x2y0−xy =x3(2x−3)+x

(−x

−2

)−x(x

−1

)=2−1−1=0

49. F0(x)=xf0(x)+f(x), F00(x)=xf 00(x)+f0(x)+f0(x)=xf00(x)+2f0(x)

51. The graph has a horizontal tangent at points where dy

dx = 0, but

dx =x2−3x+2=(x−1)(x−2)=0ifx=1,2. The

corresponding values of yare 5/6 and 2/3 so the tangent line is

horizontal at (1,5/6) and (2,2/3).

1.5

Exercise Set 3.3 76

52. dy

dx =9−x2

(x2+9)

2;dy

dx = 0 when x2=9sox=±3. The points are

(3,1/6) and (−3,−1/6).

0.2

-0.2

-5.5 5.5

53. f0(1) ≈f(1.01) −f(1)

0.01 =0.999699 −(−1)

0.01 =0.0301, and by diﬀerentiation, f0(1) = 3(1)2−3=0

54. f0(1) ≈f(1.01) −f(1)

0.01 =1.01504 −1

0.01 =1.504, and by diﬀerentiation,

f0(1) = µ√x+x

2√x¶¯¯¯x=1 =1.5

55. f0(1)=0 56. f0(1)=1

57. The y-intercept is −2 so the point (0,−2) is on the graph; −2=a(0)2+b(0) + c,c=−2. The

x-intercept is 1 so the point (1,0) is on the graph; 0 = a+b−2. The slope is dy/dx =2ax +b;at

x= 0 the slope is bso b=−1, thus a= 3. The function is y=3x

2−x−2.

58. Let P(x0,y

0) be the point where y=x2+kis tangent to y=2x. The slope of the curve is dy

dx =2x

and the slope of the line is 2 thus at P,2x

0=2sox

0= 1. But Pis on the line, so y0=2x

0=2.

Because Pis also on the curve we get y0=x2

0+kso k=y0−x2

0=2−(1)2=1.

59. The points (−1,1) and (2,4) are on the secant line so its slope is (4 −1)/(2 + 1) = 1. The slope of

the tangent line to y=x2is y0=2xso 2x=1,x=1/2.

60. The points (1,1) and (4,2) are on the secant line so its slope is 1/3. The slope of the tangent line to

y=√xis y0=1/(2√x)so1/(2√x)=1/3, 2√x=3,x=9/4.

61. y0=−2x,soatanypoint(x

0)ony=1−x

2the tangent line is y−y0=−2x0(x−x0), or

y=−2x0x+x2

0+ 1. The point (2,0) is to be on the line, so 0 = −4x0+x2

0+1, x

0−4x

0+1=0. Use

the quadratic formula to get x0=4±√16 −4

2=2±√

62. Let P1(x1,ax

1) and P2(x2,ax

2) be the points of tangency. y0=2ax so the tangent lines at P1and P2

are y−ax2

1=2ax1(x−x1) and y−ax2

2=2ax2(x−x2). Solve for xto get x=1

2(x1+x2) which is the

x-coordinate of a point on the vertical line halfway between P1and P2.

63. y0=3ax2+b; the tangent line at x=x0is y−y0=(3ax2

0+b)(x−x0) where y0=ax3

0+bx0. Solve

with y=ax3+bx to get

(ax3+bx)−(ax3

0+bx0)=(3ax2

0+b)(x−x0)

ax3+bx −ax3

0−bx0=3ax2

0x−3ax3

0+bx −bx0

x3−3x2

0x+2x

0=0

(x−x

)(x2+xx0−2x2

0)=0

(x−x

2(x+2x

)=0,so x=−2x0.

64. Let (x0,y

0) be the point of tangency. Refer to the solution to Exercise 65 to see that the endpoints of

the line segment are at (2x0,0) and (0,2y0), so (x0,y

0) is the midpoint of the segment.

77 Chapter 3

65. y0=−1

x2; the tangent line at x=x0is y−y0=−1

(x−x0), or y=−x

. The tangent line

crosses the x-axis at 2x0, the y-axis at 2/x0, so that the area of the triangle is 1

2(2/x0)(2x0)=2.

66. f0(x)=3ax2+2bx +c; there is a horizontal tangent where f0(x) = 0. Use the quadratic formula on

3ax2+2bx +c= 0 to get x=(−b±√

2−3ac)/(3a) which gives two real solutions, one real solution,

or none if

(a) b2−3ac > 0(b) b2−3ac =0 (c) b2−3ac < 0

67. F=GmMr−2,dF

dr =−2GmMr−3=−2GmM

68. dR/dT =0.04124 −3.558 ×10−5Twhich decreases as Tincreases from 0 to 700. When T=0,

dR/dT =0.04124 Ω/◦C; when T= 700, dR/dT =0.01633 Ω/◦C. The resistance is most sensitive to

temperature changes at T=0

◦C, least sensitive at T= 700◦C.

69. f0(x)=1+1/x2>0 for all x6

-6 6

-6

70. f0(x)=−5x

2−4

2+4)

f0(x)>0 when x2<4, i.e. on −2<x<2

1.5

-1.5

-5 5

71. (f·g·h)0=[(f·g)·h]

0=(f·g)h

0+h(f·g)

0=(f·g)h

0+h[fg0+f0g]=fgh0+fg0h+f0gh

72. (f1f2···f

0=(f

2···f

n)+(f

2···f

n)+···+(f

2···f0

73. (a) 2(1 + x−1)(x−3+7)+(2x+ 1)(−x−2)(x−3+7)+(2x+ 1)(1 + x−1)(−3x−4)

(b) (x7+2x−3)3=(x

7+2x−3)(x7+2x−3)(x7+2x−3) so

dx(x7+2x−3)3=(7x

6+ 2)(x7+2x−3)(x7+2x−3)

+(x7+2x−3)(7x6+ 2)(x7+2x−3)

+(x7+2x−3)(x7+2x−3)(7x6+2)

= 3(7x6+ 2)(x7+2x−3)2

74. (a) −5x−6(x2+2x)(4 −3x)(2x9+1)+x

−5

(2x+ 2)(4 −3x)(2x9+1)

−5

2+2x)(−3)(2x9+1)+x

−5

2+2x)(4 −3x)(18x8)

Exercise Set 3.3 78

(b) (x2+1)

50 =(x

2+ 1)(x2+1)···(x

2+ 1), where (x2+ 1) occurs 50 times so

dx(x2+1)

50 = [(2x)(x2+1)···(x

2+1)]+[(x

2+ 1)(2x)···(x

2+ 1)]

+···+[(x

2+ 1)(x2+1)···(2x)]

=2x(x

2+1)

49 +2x(x

2+1)

49 +···+2x(x

2+1)

= 100x(x2+1)

49 because 2x(x2+1)

49 occurs 50 times.

75. fis continuous at 1 because lim

x→1−f(x) = lim

x→1+f(x)=f(1), also lim

x→1−f0(x) = lim

x→1−2x= 2 and

lim

x→1+f0(x) = lim

x→1+

2√x=1

2so fis not diﬀerentiable at 1.

76. fis continuous at 1/2 because lim

x→1/2−f(x) = lim

x→1/2+f(x)=f(1/2), also

lim

x→1/2−f0(x) = lim

x→1/2−3x2=3/4 and lim

x→1/2+f0(x) = lim

x→1/2+3x/2=3/4sof

(1/2)=3/4.

77. If fis diﬀerentiable at x= 1, then fis continuous there;

lim

x→1+f(x) = lim

x→1−f(x)=f(1)=3,a+b= 3; lim

x→1+f0(x)=aand

lim

x→1−f0(x)=6soa= 6 and b=3−6=−3.

78. (a) lim

x→0−f0(x) = lim

x→0−2x= 0 and lim

x→0+f0(x) = lim

x→0+2x=0;f

(0) does not exist because fis not

continuous at x=0.

(b) lim

x→0−f0(x) = lim

x→0+f0(x)=0andfis continuous at x=0,sof

0(0)=0;

lim

x→0−f00(x) = lim

x→0−(2) = 2 and lim

x→0+f00(x) = lim

x→0+6x=0,sof

00(0) does not exist.

79. (a) f(x)=3x−2ifx≥2/3, f(x)=−3x+2 if x<2/3sofis diﬀerentiable everywhere except

perhaps at 2/3. fis continuous at 2/3, also lim

x→2/3−f0(x) = lim

x→2/3−(−3)=−3 and lim

x→2/3+f0(x)=

lim

x→2/3+(3) = 3 so fis not diﬀerentiable at x=2/3.

(b) f(x)=x

2−4if|x|≥2, f(x)=−x

2+4 if |x|<2sofis diﬀerentiable everywhere except

perhaps at ±2. fis continuous at −2 and 2, also lim

x→2−f0(x) = lim

x→2−(−2x)=−4 and lim

x→2+f0(x)=

lim

x→2+(2x)=4sofis not diﬀerentiable at x= 2. Similarly, fis not diﬀerentiable at x=−2.

80. (a) f0(x)=−(1)x−2,f00(x)=(2·1)x−3,f000(x)=−(3 ·2·1)x−4

f(n)(x)=(−1)nn(n−1)(n−2) ···1

n+1

(b) f0(x)=−2x

−3

00(x)=(3·2)x−4,f000(x)=−(4 ·3·2)x−5

f(n)(x)=(−1)n(n+ 1)(n)(n−1) ···2

n+2

81. (a) d2

dx2[cf(x)] = d

dx ·d

dx[cf (x)]¸=d

dx ·cd

dx[f(x)]¸=cd

dx ·d

dx[f(x)]¸=cd2

dx2[f(x)]

dx2[f(x)+g(x)] = d

dx ·d

dx[f(x)+g(x)]¸=d

dx ·d

dx[f(x)] + d

dx[g(x)]¸=d2

dx2[f(x)] + d2

dx2[g(x)]

(b) yes, by repeated application of the procedure illustrated in part (a)

82. (f·g)0=fg0+gf0,(f·g)00 =fg00 +g0f0+gf00 +f0g0=f00g+2f0g

0+fg00

83. (a) f0(x)=nxn−1,f00(x)=n(n−1)xn−2,f000(x)=n(n−1)(n−2)xn−3,...,

f(n)(x)=n(n−1)(n−2) ···1

(b) from part (a), f(k)(x)=k(k−1)(k−2) ···1sof

(k+1)(x) = 0 thus f(n)(x)=0ifn>k

79 Chapter 3

n(n−1)(n−2) ···1

84. lim

h→0

f0(2+h)−f0(2)

h=f00(2); f0(x)=8x

7−2, f00(x)=56x

,sof

00(2) = 56(26) = 3584.

85. (a) If a function is diﬀerentiable at a point then it is continuous at that point, thus f0is continuous

on (a, b) and consequently so is f.

(b) fand all its derivatives up to f(n−1)(x) are continuous on (a, b)

EXERCISE SET 3.4

1. f0(x)=−2 sin x−3 cos x

2. f0(x) = sin x(−sin x) + cos x(cos x) = cos2x−sin2x= cos 2x

3. f0(x)=x(cos x)−sin x(1)

x2=xcos x−sin x

4. f0(x)=x

(−sin x) + (cos x)(2x)=−x

2sin x+2xcos x

5. f0(x)=x

(cos x) + (sin x)(3x2)−5(−sin x)=x

3cos x+(3x

2+ 5) sin x

6. f(x)=cot x

x(because cos x

sin x= cot x), f0(x)= x(−csc2x)−(cot x)(1)

x2=−xcsc2x+ cot x

7. f0(x) = sec xtan x−√2 sec2x

8. f0(x)=(x

2+ 1) sec xtan x+ (sec x)(2x)=(x

2+ 1) sec xtan x+2xsec x

9. f0(x) = sec x(sec2x) + (tan x)(sec xtan x) = sec3x+ sec xtan2x

10. f0(x)= (1 + tan x)(sec xtan x)−(sec x)(sec2x)

(1 + tan x)2=sec xtan x+ sec xtan2x−sec3x

(1 + tan x)2

=sec x(tan x+ tan2x−sec2x)

(1 + tan x)2=sec x(tan x−1)

(1 + tan x)2

11. f0(x) = (csc x)(−csc2x) + (cot x)(−csc xcot x)=−csc3x−csc xcot2x

12. f0(x)=1+4cscxcot x−2 csc2x

13. f0(x)=(1 + csc x)(−csc2x)−cot x(0 −csc xcot x)

(1 + csc x)2=csc x(−csc x−csc2x+ cot2x)

(1 + csc x)2but

1 + cot2x= csc2x(identity) thus cot2x−csc2x=−1so

(x)= csc x(−csc x−1)

(1 + csc x)2=−csc x

1 + csc x

14. f0(x)=tan x(−csc xcot x)−csc x(sec2x)

tan2x=−csc x(1 + sec2x)

tan2x

15. f(x) = sin2x+ cos2x= 1 (identity) so f0(x)=0

16. f(x)= 1

cot x= tan x,sof

(x) = sec2x

Exercise Set 3.4 80

17. f(x)= tan x

1+xtan x(because sin xsec x= (sin x)(1/cos x) = tan x),

f0(x)= (1+xtan x)(sec2x)−tan x[x(sec2x) + (tan x)(1)]

(1+xtan x)2

=sec2x−tan2x

(1+xtan x)2=1

(1+xtan x)2(because sec2x−tan2x=1)

18. f(x)= (x

2+ 1) cot x

3−cot x(because cos xcsc x= (cos x)(1/sin x) = cot x),

f0(x)= (3 −cot x)[2xcot x−(x2+ 1) csc2x]−(x2+ 1) cot xcsc2x

(3 −cot x)2

=6xcot x−2xcot2x−3(x2+ 1) csc2x

(3 −cot x)2

19. dy/dx =−xsin x+ cos x,d2y/dx2=−xcos x−sin x−sin x=−xcos x−2 sin x

20. dy/dx =−csc xcot x,d2y/dx2=−[(csc x)(−csc2x) + (cot x)(−csc xcot x)] = csc3x+ csc xcot2x

21. dy/dx =x(cos x) + (sin x)(1) −3(−sin x)=xcos x+ 4 sin x,

d2y/dx2=x(−sin x) + (cos x)(1) + 4 cos x=−xsin x+ 5 cos x

22. dy/dx =x2(−sin x) + (cos x)(2x)+4cosx=−x

2sin x+2xcos x+ 4 cos x,

d2y/dx2=−[x2(cos x) + (sin x)(2x)]+2[x(−sin x) + cos x]−4 sin x=(2−x

) cos x−4(x+ 1) sin x

23. dy/dx = (sin x)(−sin x) + (cos x)(cos x) = cos2x−sin2x,

d2y/dx2= (cos x)(−sin x) + (cos x)(−sin x)−[(sin x)(cos x) + (sin x)(cos x)] = −4 sin xcos x

24. dy/dx = sec2x;d2y/dx2= 2 sec2xtan x

26. Let f(x) = sin x, then f0(x) = cos x.

(a) f(0) = 0 and f0(0) = 1 so y−0 = (1)(x−0), y=x

(b) f(π)=0andf

0(π)=−1soy−0=(−1)(x−π), y=−x+π

4´=1

√2and f0³π

4´=1

√2so y−1

√2=1

√2³x−π

4´,y=1

√2x−π

4√2+1

√2

27. Let f(x) = tan x, then f0(x) = sec2x.

(a) f(0) = 0 and f0(0) = 1 so y−0 = (1)(x−0), y=x.

(b) f³π

4´= 1 and f0³π

4´=2soy−1=2³x−π

4´,y=2x−π

2+1.

4´=−1 and f0³−π

4´=2soy+1=2³x+π

4´,y=2x+π

2−1.

28. (a) If y= cos xthen y0=−sin xand y00 =−cos xso y00 +y=(−cos x) + (cos x)=0;

if y= sin xthen y0= cos xand y00 =−sin xso y00 +y=(−sin x) + (sin x)=0.

(b) y0=Acos x−Bsin x,y00 =−Asin x−Bcos xso

y00 +y=(−Asin x−Bcos x)+(Asin x+Bcos x)=0.

29. (a) f0(x) = cos x=0atx=±π/2,±3π/2.

(b) f0(x)=1−sin x=0atx=−3π/2,π/2.

(d) f0(x) = sec xtan x= 0 when sin x=0,x=±2π, ±π, 0

81 Chapter 3

30. (a) 0.5

-0.5

02p

(b) y= sin xcos x=(1/2) sin 2xand y0= cos 2x.Soy

0= 0 when 2x=(2n+1)π/2 for n=0,1,2,3

or x=π/4,3π/4,5π/4,7π/4

31. x= 10 sin θ,dx/dθ = 10 cos θ;ifθ=60

◦, then

dx/dθ = 10(1/2) = 5 ft/rad = π/36 ft/deg ≈0.087 ft/deg

32. s= 3800 csc θ, ds/dθ =−3800 csc θcot θ;ifθ=30

◦, then

ds/dθ =−3800(2)(√3) = −7600√3 ft/rad = −380√3π/9 ft/deg ≈−230 ft/deg

33. D= 50 tan θ, dD/dθ = 50 sec2θ;ifθ=45

◦, then

dD/dθ = 50(√2)2= 100 m/rad = 5π/9m/deg ≈1.75 m/deg

34. (a) From the right triangle shown, sin θ=r/(r+h)sor+h=rcsc θ,h=r(csc θ−1).

(b) dh/dθ =−rcsc θcot θ;ifθ=30

◦, then

dh/dθ =−6378(2)(√3) ≈−22,094 km/rad ≈−386 km/deg

35. (a) d4

dx4sin x= sin x,so d

dx4ksin x= sin x;d87

dx87 sin x=d3

dx3

d4·21

dx4·21 sin x=d3

dx3sin x=−cos x

(b) d100

dx100 cos x=d4k

dx4kcos x= cos x

36. d

dx[xsin x]=xcos x+ sin xd2

dx2[xsin x]=−xsin x+ 2 cos x

dx3[xsin x]=−xcos x−3 sin xd4

dx4[xsin x]=xsin x−4 cos x

By mathematical induction one can show

d4k

dx4k[xsin x]=xsin x−(4k) cos x;d4k+1

dx4k+1 [xsin x]=xcos x+(4k+ 1) sin x;

d4k+2

dx4k+2 [xsin x]=−xsin x+(4k+ 2) cos x;d4k+3

dx4k+3 [xsin x]=−xcos x−(4k+ 3) sin x;

Since 17 = 4 ·4+1, d

dx17 [xsin x]=xcos x+ 17 sin x

37. (a) all x(b) all x

(e) x6=π/2+nπ,n=0,±1,±2,... (f) x6=nπ,n=0,±1,±2,...

(g) x6=(2n+1)π,n=0,±1,±2,... (h) x6=nπ/2, n=0,±1,±2,...

(i) all x

Exercise Set 3.5 82

38. (a) d

dx[cos x] = lim

h→0

cos(x+h)−cos x

h= lim

h→0

cos xcos h−sin xsin h−cos x

= lim

h→0·cos xµcos h−1

h¶−sin xµsin h

h¶¸= (cos x)(0) −(sin x)(1) = −sin x

(b) d

dx[sec x]= d

dx ·1

cos x¸=cos x(0) −(1)(−sin x)

cos2x=sin x

cos2x= sec xtan x

dx[cot x]= d

dx hcos x

sin xi=sin x(−sin x)−cos x(cos x)

sin2x

=−sin2x−cos2x

sin2x=−1

sin2x=−csc2x

(d) d

dx[csc x]= d

dx ·1

sin x¸=sin x(0) −(1)(cos x)

sin2x=−cos x

sin2x=−csc xcot x

39. f0(x)=−sin x,f00(x)=−cos x,f000(x) = sin x, and f(4)(x) = cos xwith higher order derivatives

repeating this pattern, so f(n)(x) = sin xfor n=3,7,11,...

40. (a) lim

h→0

tan h

h= lim

h→0µsin h

cos h¶

h= lim

h→0µsin h

h¶

cos h=1

1=1

(b) d

dx[tan x] = lim

h→0

tan(x+h)−tan x

h= lim

h→0

tan x+ tan h

1−tan xtan h−tan x

= lim

h→0

tan x+ tan h−tan x+ tan2xtan h

h(1 −tan xtan h)= lim

h→0

tan h(1 + tan2x)

h(1 −tan xtan h)

= lim

h→0

tan hsec2x

h(1 −tan xtan h)= sec2xlim

h→0

tan h

1−tan xtan h

= sec2x

lim

h→0

tan h

lim

h→0(1 −tan xtan h)= sec2x

41. lim

x→0

tan(x+y)−tan y

x= lim

h→0

tan(y+h)−tan y

h=d

dy (tan y) = sec2y

43. Let tbe the radian measure, then h=180

πtand cos h= cos t, sin h= sin t.

(a) lim

h→0

cos h−1

h= lim

t→0

cos t−1

180t/π =π

180 lim

t→0

cos t−1

t=0

(b) lim

h→0

sin h

h= lim

t→0

sin t

180t/π =π

180 lim

t→0

sin t

t=π

180

dx[sin x] = sin xlim

h→0

cos h−1

h+ cos xlim

h→0

sin h

h= sin x(0) + cos x(π/180) = π

180 cos x

EXERCISE SET 3.5

1. f0(x) = 37(x3+2x)

36 d

dx(x3+2x) = 37(x3+2x)

36(3x2+2)

2. f0(x) = 6(3x2+2x−1)5d

dx(3x2+2x−1) = 6(3x2+2x−1)5(6x+ 2) = 12(3x2+2x−1)5(3x+1)

83 Chapter 3

3. f0(x)=−2µx

3−7

−3d

dx µx3−7

x¶=−2µx3−7

x¶−3µ3x2+7

x2¶

4. f(x)=(x

5−x+1)

−9

f0(x)=−9(x5−x+1)

−10 d

dx(x5−x+1)=−9(x5−x+1)

−10(5x4−1) = −9(5x4−1)

(x5−x+1)

5. f(x) = 4(3x2−2x+1)

−3

f0(x)=−12(3x2−2x+1)

−4d

dx(3x2−2x+1)=−12(3x2−2x+1)

−4

(6x−2) = 24(1 −3x)

(3x2−2x+1)

6. f0(x)= 1

√

3−2x+5

dx(x3−2x+5)= 3x

2−2

√

3−2x+5

7. f0(x)= 1

p4+3

√

dx(4+3

√

x)= 3

√

p4+3

√

8. f0(x)=3sin

2xd

dx(sin x)=3sin

2xcos x

9. f0(x) = cos(x3)d

dx(x3)=3x

2cos(x3)

10. f0(x) = 2 cos(3√x)d

dx[cos(3√x)] = −2 cos(3√x) sin(3√x)d

dx(3√x)=−3 cos(3√x) sin(3√x)

√x

11. f0(x) = sec2(4x2)d

dx(4x2)=8xsec2(4x2)

12. f0(x)=12cot

3xd

dx(cot x)=12cot

3x(−csc2x)=−12 cot3xcsc2x

13. f0(x)=20cos

4xd

dx(cos x)=20cos

4x(−sin x)=−20 cos4xsin x

14. f0(x)=−csc(x3) cot(x3)d

dx(x3)=−3x

2csc(x3) cot(x3)

15. f0(x) = cos(1/x2)d

dx(1/x2)=−2

3cos(1/x2)

16. f0(x)=4tan

dx[tan(x3)] = 4 tan3(x3) sec2(x3)d

dx(x3)=12x

2tan3(x3) sec2(x3)

17. f0(x) = 4 sec(x7)d

dx[sec(x7)] = 4 sec(x7) sec(x7) tan(x7)d

dx(x7)=28x

6sec2(x7) tan(x7)

18. f0(x)=3cos

2µx

x+1¶d

dx cos µx

x+1¶= 3 cos2µx

x+1¶·−sin µx

x+1¶¸(x+ 1)(1) −x(1)

(x+1)

=−3

(x+1)

2cos2µx

x+1¶sin µx

x+1¶

19. f0(x)= 1

pcos(5x)

dx[cos(5x)] = −5 sin(5x)

2pcos(5x)

20. f0(x)= 1

p3x−sin2(4x)

dx[3x−sin2(4x)] = 3−8 sin(4x) cos(4x)

2p3x−sin2(4x)

Exercise Set 3.5 84

21. f0(x)=−3£x+ csc(x3+3)

¤−4d

dx £x+ csc(x3+3)

=−3£x+ csc(x3+3)

¤−4·1−csc(x3+ 3) cot(x3+3) d

dx(x3+3)

=−3£x+ csc(x3+3)

¤−4£1−3x

2csc(x3+ 3) cot(x3+3)

22. f0(x)=−4£x

4−sec(4x2−2)¤−5d

dx £x4−sec(4x2−2)¤

=−4£x4−sec(4x2−2)¤−5·4x3−sec(4x2−2) tan(4x2−2) d

dx(4x2−2)¸

=−16x£x4−sec(4x2−2)¤−5£x2−2 sec(4x2−2) tan(4x2−2)¤

23. f0(x)=x

2·−2x

√

5−x

2+2x

p5−x

2=x(10 −3x2)

√5−x2

24. f0(x)=√

1−x

(1) −x(−x/√1−x2)

1−x2=1

(1 −x2)3/2

25. dy

dx =x3(2 sin 5x)d

dx(sin 5x)+3x

2sin25x=10x

3sin 5xcos 5x+3x

2sin25x

26. dy

dx =√x·3 tan2(√x) sec2(√x)1

2√x¸+1

2√xtan3(√x)=3

2tan2(√x) sec2(√x)+ 1

√

xtan3(√x)

27. dy

dx =x5sec µ1

x¶tan µ1

x¶d

dx µ1

x¶+ sec µ1

x¶(5x4)=x

5sec µ1

x¶tan µ1

x¶µ−1

2¶+5x

4sec µ1

x¶

=−x3sec µ1

x¶tan µ1

x¶+5x

4sec µ1

x¶

28. dy

dx =sec(3x+ 1) cos x−3 sin xsec(3x+ 1) tan(3x+1)

sec2(3x+1) =cos x−3 sin xtan(3x+1)

sec(3x+1)

29. dy

dx =−sin(cos x)d

dx(cos x)=−sin(cos x)(−sin x) = sin(cos x) sin x

30. dy

dx = cos(tan 3x)d

dx(tan 3x)=3sec

23xcos(tan 3x)

31. dy

dx = 3 cos2(sin 2x)d

dx[cos(sin 2x)] = 3 cos2(sin 2x)[−sin(sin 2x)] d

dx(sin 2x)

=−6 cos2(sin 2x) sin(sin 2x) cos 2x

32. dy

dx =(1 −cot x2)(−2xcsc x2cot x2)−(1 + csc x2)(2xcsc2x2)

(1 −cot x2)2=−2xcsc x21 + cot x2csc x2

(1 −cot x2)2

33. dy

dx =(5x+8)

1312(x3+7x)

11 d

dx(x3+7x)+(x

3+7x)

1213(5x+8)

12 d

dx(5x+8)

= 12(5x+8)

13(x3+7x)

11(3x2+7)+65(x

3+7x)

12(5x+8)

34. dy

dx =(2x−5)23(x2+4)

(2x)+(x

2+4)

2(2x−5)(2) = 6x(2x−5)2(x2+4)

2+ 4(2x−5)(x2+4)

= 2(2x−5)(x2+4)

(8x2−15x+8)

85 Chapter 3

35. dy

dx =3·x−5

2x+1¸2d

dx ·x−5

2x+1¸=3·x−5

2x+1¸2

·11

(2x+1)

2=33(x−5)2

(2x+1)

36. dy

dx =17µ1+x

1−x

2¶

16 d

dx µ1+x

1−x

2¶=17µ1+x

1−x

2¶

16 (1 −x2)(2x)−(1+x2)(−2x)

(1 −x2)2

=17µ1+x

1−x

2¶

16 4x

(1 −x2)2=68x(1+x2)16

(1 −x2)18

37. dy

dx =(4x2−1)8(3)(2x+3)

(2) −(2x+3)

(8)(4x2−1)7(8x)

(4x2−1)16

=2(2x+3)

(4x2−1)7[3(4x2−1) −32x(2x+ 3)]

(4x2−1)16 =−2(2x+3)

(52x2+96x+3)

(4x2−1)9

38. dy

dx = 12[1 + sin3(x5)]11 d

dx[1 + sin3(x5)]

= 12[1 + sin3(x5)]113 sin2(x5)d

dx sin(x5) = 180x4[1 + sin3(x5)]11 sin2(x5) cos(x5)

39. dy

dx =5£xsin 2x+ tan4(x7)¤4d

dx £xsin 2xtan4(x7)¤

=5£xsin 2x+ tan4(x7)¤4·xcos 2xd

dx(2x) + sin 2x+ 4 tan3(x7)d

dx tan(x7)¸

=5£xsin 2x+ tan4(x7)¤4£2xcos 2x+ sin 2x+28x

6tan3(x7) sec2(x7)¤

40. dy

dx = cos(3x2)d

dx(3x2)=6xcos(3x2),

d2y

dx2=6x(−sin(3x2)) d

dx(3x2) + 6 cos(3x2)=−36x2sin(3x2) + 6 cos(3x2)

41. dy

dx =x(−sin(5x)) d

dx(5x) + cos(5x)−2 sin xd

dx(sin x)

=−5xsin(5x) + cos(5x)−2 sin xcos x=−5xsin(5x) + cos(5x)−sin(2x),

d2y

dx2=−5xcos(5x)d

dx(5x)−5 sin(5x)−sin(5x)d

dx(5x)−cos(2x)d

dx(2x)

=−25xcos(5x)−10 sin(5x)−2 cos(2x)

42. dy

dx =xsec2µ1

x¶d

dx µ1

x¶+ tan µ1

x¶=−1

xsec2µ1

x¶+ tan µ1

x¶,

d2y

dx2=−2

xsec µ1

x¶d

dx sec µ1

x¶+1

x2sec2µ1

x¶+ sec2µ1

x¶d

dx µ1

x¶=2

x3sec2µ1

x¶tan µ1

x¶

43. dy

dx =(1 −x)+(1+x)

(1 −x)2=2

(1 −x)2= 2(1 −x)−2and d2y

dx2=−2(2)(−1)(1 −x)−3= 4(1 −x)−3

45. dy

dx =−3xsin 3x+ cos 3x;ifx=πthen y=−πand dy

dx =−1soy+π=−(x−π), y=−x

46. dy

dx =3x

2cos(1 + x3); if x=−3 then y= sin(−26) = −sin 26 and dy

dx = 27 cos 26

so y+ sin 26 = 27(cos 26)(x+3)

Exercise Set 3.5 86

47. dy

dx =−3 sec3(π/2−x) tan(π/2−x); if x=−π/2 then y=−1 and dy

dx =0

so y+ 1 = (0)(x+π/2), y=−1

48. dy

dx =3(x−1/x)2(1+1/x2); if x= 2 then y=27/8 and dy

dx = 135/16

so y−27

8=135

16 (x−2), y=135

16 x−27

49. y= cot3(π−θ)=−cot3θso dy/dx = 3 cot2θcsc2θ

50. 6µau +b

cu +d¶5ad −bc

(cu +d)2

51. d

dω [acos2πω +bsin2πω]=−2πa cos πω sin πω +2πb sin πω cos πω

=π(b−a)(2 sin πω cos πω)=π(b−a) sin 2πω

52. 2 csc2(π/3−y) cot(π/3−y)

53. (a) 2

-2 2

-2

√

4−x

2+p4−x

2=4−2x

√

4−x

-2 2

-6

(d) f(1) = √3 and f0(1) = 2

√3so the tangent line has the equation y−√3= 2

√

(x−1).

54. (a) 0.5

1.2

-1.2

87 Chapter 3

(d) f(1) = sin 1 cos 1 and f0(1) = 2 cos21−sin21, so the tangent line has the equation

y−sin 1 cos 1 = (2 cos21−sin21)(x−1).

0.8

55. (a) dy/dt =−Aω sin ωt, d2y/dt2=−Aω2cos ωt =−ω2y

(b) one complete oscillation occurs when ωt increases over an interval of length 2π,oriftincreases

over an interval of length 2π/ω

(d) amplitude = 0.6 cm, T=2π/15 s/oscillation, f=15/(2π) oscillations/s

56. dy/dt =3Acos 3t, d2y/dt2=−9Asin 3t,so−9Asin 3t+2Asin 3t= 4 sin 3t,

−7Asin 3t= 4 sin 3t, −7A=4,A=−4/7

57. (a) p≈10 lb/in2,dp/dh ≈−2 lb/in2/mi

(b) dp

dt =dp

dt ≈(−2)(0.3) = −0.6 lb/in2/s

58. (a) F=45

cos θ+0.3 sin θ,dF

dθ =−45(−sin θ+0.3 cos θ)

(cos θ+0.3 sin θ)2;

if θ=30

◦, then dF/dθ ≈10.5 lb/rad ≈0.18 lb/deg

(b) dF

dt =dF

dθ

dt ≈(0.18)(−0.5) = −0.09 lb/s

59. With u= sin x,d

dx(|sin x|)= d

dx(|u|)= d

du(|u|)du

dx =d

du(|u|) cos x=½cos x, u > 0

−cos x, u < 0

=½cos x, sin x>0

−cos x, sin x<0=½cos x, 0<x<π

−cos x, −π<x<0

60. d

dx(cos x)= d

dx[sin(π/2−x)] = −cos(π/2−x)=−sin x

61. (a) for x6=0,f

0(x)=xµcos 1

x¶µ−1

2¶+ sin 1

x=−1

xcos 1

x+ sin 1

(b) lim

x→0xsin 1

x=0=f(0)

h→0

f(0+h)−f(0)

h= lim

h→0

hsin 1

h= lim

h→0sin 1

h, which does not exist

62. (a) f0(x)=x

2µcos 1

x¶µ−1

2¶+2xsin 1

x=−cos 1

x+2xsin 1

x,x6=0

(b) lim

x→0x2sin 1

x=0=f(0)

Exercise Set 3.5 88

h→0

f(0+h)−f(0)

h= lim

h→0

h2sin 1

h= lim

h→0hsin 1

h=0

(d) lim

x→0f0(x) does not exist because cos 1

xoscillates between 1 and −1

63. (a) g0(x)=3[f(x)]2f0(x), g0(2)=3[f(2)]2f0(2) = 3(1)2(7)=21

(b) h0(x)=f

0(x

)(3x2), h0(2) = f0(8)(12) = (−3)(12) = −36

64. (a) F0(x)=f

0(g(x))g0(x), F0(−1) = f0(g(−1))g0(−1) = f0(2)(−3) = (4)(−3) = −12

(b) G0(x)=g

0(f(x))f0(x), G0(−1) = g0(f(−1))f0(−1) = g0(2)(3) = (−5)(3) = −15

65. (f◦g)0(x)=f

0(g(x))g0(x)so(f◦g)

(0) = f0(g(0))g0(0) = f0(0)(3) = (2)(3) = 6

66. F0(x)=f

0(g(x))g0(x)=p3(x2−1) + 4(2x)=2x

√

2+1

67. F0(x)=f

0(g(x))g0(x)=f

√

3x−1) 3

2√3x−1=√3x−1

(3x−1)+1

√

3x−1=1

68. d

dx[f(x2)] = f0(x2)(2x), thus f0(x2)(2x)=x

2so f0(x2)=x/2ifx6=0

69. d

dx[f(3x)] = f0(3x)d

dx(3x)=3f

0(3x)=6x,sof

(3x)=2x. Let u=3xto get f0(u)=2

dx[f(x)] = f0(x)= 2

70. (a) If f(−x)=f(x), then d

dx[f(−x)] = d

dx[f(x)], f0(−x)(−1) = f0(x), f0(−x)=−f

0(x)so

0is odd.

(b) If f(−x)=−f(x), then d

dx[f(−x)] = −d

dx[f(x)], f0(−x)(−1) = −f0(x), f0(−x)=f

0(x)so

0is even.

71. For an even function, the graph is symmetric about the y-axis; the slope of the tangent line at

(a, f(a)) is the negative of the slope of the tangent line at (−a, f(−a). For an odd function, the graph

is symmetric about the origin; the slope of the tangent line at (a, f(a)) is the same as the slope of the

tangent line at (−a, f(−a).

f(x)

f'(x)

f(x)

f'(x)

72. dy

dx =dy

73. d

dx[f(g(h(x)))] = d

dx[f(g(u))],u=h(x)

du[f(g(u))] du

dx =f0(g(u))g0(u)du

dx =f0(g(h(x)))g0(h(x))h0(x)

89 Chapter 3

EXERCISE SET 3.6

1. (a) dy =f0(x)dx =2xdx = 4(1) = 4 and

∆y=(x+∆x)

2−x

2=(2+1)

2−2

2=5

(b)

dy ∆y

2. (a) dy =3x

dx = 3(1)2(1) = 3 and

∆y=(x+∆x)

3−x

3=(1+1)

3−1

3=7

(b)

1 2

(1,1) dx

∆y

3. (a) dy =(−1/x2)dx =(−1)(−0.5)=0.5 and

∆y=1/(x+∆x)−1/x =1/(1 −0.5) −1/1=2−1=1

(b) y

0.5 1

dy=0.5

y∆=1

4. (a) dy =(1/2

√

x)dx =(1/(2 ·3))(−1) = −1/6≈−0.167 and

∆y=√x+∆x−√

x=p9+(−1) −√9=√

8−3≈−0.172

(b) y

dy = –0.167

∆y = –0.171

5. dy =3x

dx;

∆y=(x+∆x)

3−x

3=x

3+3x

∆x+3x(∆x)2+(∆x)

3−x

3=3x

∆x+3x(∆x)2+(∆x)

6. dy =8dx;∆y= [8(x+∆x)−4] −[8x−4]=8∆x

7. dy =(2x−2)dx;

∆y=[(x+∆x)

2−2(x+∆x)+1]−[x

2−2x+1]

2+2x∆x+(∆x)

2−2x−2∆x+1−x

2+2x−1=2x∆x+(∆x)

2−2∆x

8. dy = cos xdx;∆y= sin(x+∆x)−sin x

Exercise Set 3.6 90

9. (a) dy = (12x2−14x)dx

(b) dy =xd(cos x) + cos xdx=x(−sin x)dx + cos xdx =(−xsin x+ cos x)dx

10. (a) dy =(−1/x2)dx (b) dy = 5 sec2xdx

11. (a) dy =µ√1−x−x

2√1−x¶dx =2−3x

2√1−xdx

(b) dy =−17(1 + x)−18dx

12. (a) dy =(x3−1)d(0) −(1)d(x3−1)

(x3−1)2=(x3−1)(0) −(1)3x2dx

(x3−1)2=−3x2

(x3−1)2dx

(b) dy =(2 −x)(−3x2)dx −(1 −x3)(−1)dx

(2 −x)2=2x3−6x2+1

(2 −x)2dx

13. (a) f(x)≈f(1) + f0(1)(x−1)=1+3(x−1) (b) f(1+∆x)≈f(1) + f0(1)∆x=1+3∆x

14. (a) f(x)≈f(2) + f0(2)(x−2)=1/2+(−1/2

)(x−2)=(1/2) −(1/4)(x−2)

(b) f(2+∆x)≈f(2) + f0(2)∆x=1/2−(1/4)∆x

1/2.05 ≈0.5−0.25(0.05)=0.4875.

15. (a) f(x)≈f(x0)+f

)(x−x0) = 1+(1/(2√1)(x−0) = 1+(1/2)x, so with x0= 0 and

x=−0.1, we have √0.9=f(−0.1) ≈1+(1/2)(−0.1)=1−0.05=0.95. With x=0.1 we have

√1.1=f(0.1) ≈1+(1/2)(0.1)=1.05.

(b) y

0.1

–0.1

∆y

∆ydy

16. (a) f(x)≈f(x0)+f

0(x

)(x−x0)=1/2−£1/(2 ·43/2)¤(x−4)=1/2−(x−4)/16, so with x0=4

and x=3.9wehave1/

√

3.9=f(3.9) ≈0.5−(−0.1)/16=0.50625. If x0= 4 and x=4.1 then

1/√4.1=f(4.1) ≈0.5−(0.1)/16=0.49375

(b)

43.9 4.1

∆y

17. f(x)=(1+x)

15 and x0=0. Thus(1+x)

15 ≈f(x0)+f0(x

)(x−x0) = 1 + 15(1)14(x−0)=1+15x.

18. f(x)= 1

√

1−xand x0=0,so 1

√

1−x≈f(x

)+f0(x

)(x−x0)=1+ 1

2(1 −0)3/2(x−0)=1+x/2

91 Chapter 3

19. tan x≈tan(0) + sec2(0)(x−0) = x20. 1

1+x≈1+ −1

(1+0)

2(x−0)=1−x

21. x4≈(1)4+ 4(1)3(x−1). Set ∆x=x−1; then x=∆+1 and(1+∆x)

4=1+4∆x.

22. √x≈√1+ 1

√

1(x−1), and x=1+∆x,so√

1+∆x≈1+∆x/2

23. 1

2+x=1

2+1−1

(2+1)

2(x−1), and 2 + x=3+∆x,so 1

3+∆x=1

3−1

9∆x

24. (4+x)3=(4+1)

3+3(4+1)

(x−1) so, with 4 + x=5+∆xwe get (5 + ∆x)3= 125 + 75∆x

25. (a) The local linear approximation sin x≈xgives sin 1◦= sin(π/180) ≈π/180=0.0174533 and

a calculator gives sin 1◦=0.0174524. The relative error |sin(π/180) −(π/180)|/(sin π/180) =

0.000051 is very small, so for such a small value of xthe approximation is very good.

(b) Use x0=45

◦(this assumes you know, or can approximate, √2/2).

180 radians, and 45◦=45π

180 =π

4radians. With x=44π

180 and x0=π

4we obtain

sin 44◦= sin 44π

180 ≈sin π

4+³cos π

4´µ44π

180 −π

4¶=√2

2+√2

2µ−π

180¶=0.694765. With a

calculator, sin 44◦=0.694658.

26. (a) tan x≈tan 0 + sec20(x−0)=x, so tan 2◦= tan(2π/180) ≈2π/180=0.034907, and with a

calculator tan 2◦=0.034921

(b) use x0=π/3 because we know tan 60◦= tan(π/3) = √3

3=60π

180 and x=61π

180 we have tan 61◦= tan 61π

180 ≈tan π

3+³sec2π

3´µ61π

180 −π

3¶=

√3+4 π

180 =1.8019, and with a calculator tan 61◦=1.8040

27. f(x)=x

0(x)=4x

0=3,∆x=0.02; (3.02)4≈34+ (108)(0.02)=81+2.16=83.16

28. f(x)=x

0(x)=3x

0=2,∆x=−0.03; (1.97)3≈23+ (12)(−0.03)=8−0.36=7.64

29. f(x)=√

x,f

0(x)= 1

√

0= 64, ∆x=1;√

65 ≈√64 + 1

16(1)=8+ 1

16 =8.0625

30. f(x)=√

x,f

0(x)= 1

√

0= 25, ∆x=−1; √24 ≈√25 + 1

10(−1)=5−0.1=4.9

31. f(x)=√

x,f

0(x)= 1

√

0= 81, ∆x=−0.1; √80.9≈√81 + 1

18(−0.1) ≈8.9944

32. f(x)=√

x,f

0(x)= 1

√

0= 36, ∆x=0.03; √36.03 ≈√36 + 1

12(0.03)=6+0.0025 = 6.0025

33. f(x) = sin x,f0(x) = cos x,x0=0,∆x=0.1; sin 0.1≈sin 0 + (cos 0)(0.1)=0.1

34. f(x) = tan x,f0(x) = sec2x,x0=0,∆x=0.2; tan 0.2≈tan 0 + (sec20)(0.2)=0.2

35. f(x) = cos x,f0(x)=−sin x,x0=π/6, ∆x=π/180;

cos 31◦≈cos 30◦+µ−1

2¶³π

180´=√3

2−π

360 ≈0.8573

Exercise Set 3.6 92

36. (a) Let f(x)=(1+x)

kand x0= 0. Then (1 + x)k≈1k+k(1)k−1(x−0)=1+kx. Set k= 37 and

x=0.001 to obtain (1.001)37 ≈1.037.

(b) With a calculator (1.001)37 =1.03767.

is due to the relative largeness of f0(1)∆x= 37(0.1)=3.7.

37. f(x)=√

x+ 3 and x0=0,so

√

x+3≈√

3+ 1

√

3(x−0) = √3+ 1

√

3x, and

¯¯¯¯

f(x)−µ√3+ 1

√

¶¯

<0.1if|x|<1.692.

-0.1

-2 2

f(x) – ( 3 + x)

2 3

38. f(x)= 1

√

9−xso 1

√9−x≈1

√9+1

2(9 −0)3/2(x−0) = 1

3+1

54x,

and ¯¯¯¯

f(x)−µ1

3+1

54x¶¯¯¯¯

<0.1if|x|<5.5114

0.06

-6 6

f(x) – ( + x)

39. tan x≈tan 0 + (sec20)(x−0) = x, and |tan x−x|<0.1if

|x|<0.6316

0.06

-0.8 0.8

|f(x) – x|

40. 1

(1+2x)

5≈1

(1+2·0)5+−5(2)

(1+2·0)6(x−0)=1−10x, and

|f(x)−(1 −10x)|<0.1if|x|<0.0372

0.06

-0.8 0.8

|f(x) – x|

41. dy =3

2√3x−2dx,x=2,dx =0.03; ∆y≈dy =3

4(0.03)=0.0225

42. dy =x

√x2+8dx,x=1,dx =−0.03; ∆y≈dy =(1/3)(−0.03) = −0.01

93 Chapter 3

43. dy =1−x2

(x2+1)

2dx,x=2,dx =−0.04; ∆y≈dy =µ−3

25¶(−0.04)=0.0048

44. dy =µ4x

√8x+1+√

8x+1

¶dx,x=3,dx =0.05; ∆y≈dy = (37/5)(0.05)=0.37

45. (a) A=x2where xis the length of a side; dA =2xdx= 2(10)(±0.1) = ±2ft

(b) relative error in xis ≈dx

x=±0.1

10 =±0.01 so percentage error in xis ≈±1%; relative error in

Ais ≈dA

A=2xdx

2=2

x=2(±0.01) = ±0.02 so percentage error in Ais ≈±2%

46. (a) V=x3where xis the length of a side; dV =3x

dx = 3(25)2(±1) = ±1875 cm3.

(b) relative error in xis ≈dx

x=±1

25 =±0.04 so percentage error in xis ≈±4%; relative error in V

is ≈dV

V=3x2dx

x3=3

x=3(±0.04) = ±0.12 so percentage error in Vis ≈±12%

47. (a) x= 10 sin θ,y= 10 cos θ(see ﬁgure),

dx = 10 cos θdθ =10¡cos π

6¢¡±π

180 ¢=10³√

2´¡±π

180 ¢

≈±0.151 in,

dy =−10(sin θ)dθ =−10 ¡sin π

6¢¡±π

180 ¢=−10 ¡1

2¢¡±π

180 ¢

≈±0.087 in

10″x

(b) relative error in xis ≈dx

x= (cot θ)dθ =³cot π

6´³±π

180´=√3³±π

180´≈±0.030

so percentage error in xis ≈±3.0%;

relative error in yis ≈dy

y=−tan θdθ =−³tan π

6´³±π

180´=−1

√3³±π

180´≈±0.010

so percentage error in yis ≈±1.0%

48. (a) x= 25 cot θ,y= 25 csc θ(see ﬁgure);

dx =−25 csc2θdθ =−25 ³csc2π

3´³±π

360´

=−25 µ4

3¶³±π

360´≈±0.291 cm,

dy =−25 csc θcot θdθ =−25 ³csc π

3´³cot π

3´³±π

360´

=−25 µ2

√3¶µ1

√3¶³±π

360´≈±0.145 cm

25 cm

(b) relative error in xis ≈dx

x=−csc2θ

cot θdθ =−4/3

1/√3³±π

360´≈±0.020 so percentage

error in xis ≈±2.0%; relative error in yis ≈dy

y=−cot θdθ =−1

√3³±π

360´≈±0.005

so percentage error in yis ≈±0.5%

49. dR

R=(−2k/r3)dr

(k/r2)=−2dr

r, but dr

r≈±0.05 so dR

R≈−2(±0.05) = ±0.10; percentage error in Ris

≈±10%

50. h= 12 sin θthus dh = 12 cos θdθ so, with θ=60

◦=π/3 radians and dθ =−1◦=−π/180 radians,

dh = 12 cos(π/3)(−π/180) = −π/30 ≈−0.105 ft

Supplementary Exercises 3 94

51. A=1

4(4)2sin 2θ= 4 sin 2θthus dA = 8 cos 2θdθ so, with θ=30

◦=π/6 radians and

dθ =±150=±1/4◦=±π/720 radians, dA = 8 cos(π/3)(±π/720) = ±π/180 ≈±0.017 cm2

52. A=x2where xis the length of a side; dA

A=2xdx

2=2

x, but dx

x≈±0.01 so

A≈2(±0.01) = ±0.02; percentage error in Ais ≈±2%

53. V=x3where xis the length of a side; dV

V=3x2dx

x3=3

x, but dx

x≈±0.02

so dV

V≈3(±0.02) = ±0.06; percentage error in Vis ≈±6%.

54. dV

V=4πr2dr

4πr3/3=3

r, but dV

V≈±0.03 so 3dr

r≈±0.03, dr

r≈±0.01; maximum permissible

percentage error in ris ≈±1%.

55. A=1

4πD2where Dis the diameter of the circle; dA

A=(πD/2)dD

πD2/4=2

D, but dA

A≈±0.01 so

2dD

D≈±0.01, dD

D≈±0.005; maximum permissible percentage error in Dis ≈±0.5%.

56. V=x3where xis the length of a side; approximate ∆Vby dV if x= 1 and dx =∆x=0.02,

dV =3x

dx = 3(1)2(0.02)=0.06 in3.

57. V= volume of cylindrical rod = πr2h=πr2(15) = 15πr2; approximate ∆Vby dV if r=2.5 and

dr =∆r=0.001. dV =30πr dr =30π(2.5)(0.001) ≈0.236 cm3.

58. P=2π

√g√L,dP =2π

√g

2√LdL =π

√g√LdL,dP

P=1

Lso the relative error in P≈1

2the relative

error in L. Thus the percentage error in Pis ≈1

2the percentage error in L.

59. (a) α=∆L/(L∆T)=0.006/(40 ×10)=1.5×10−5

/◦C

(b) ∆L=2.3×10−5(180)(25) ≈0.1 cm, so the pole is about 180.1 cm long.

60. ∆V=7.5×10−4(4000)(−20) = −60 gallons; the truck delivers 4000 −60 = 3940 gallons.

CHAPTER 3 SUPPLEMENTARY EXERCISES

4. (a) dy

dx = lim

h→0p9−4(x+h)−√9−4x

h= lim

h→0

(9 −4(x+h)−(9 −4x)

h(p9−4(x+h)+√

9−4x)

= lim

h→0−4h

h(p9−4(x+h)+√

9−4x)=−4

√

9−4x=−2

√

9−4x

(b) dy

dx = lim

h→0

x+h

x+h+1−x

x+1

h= lim

h→0

(x+h)(x+1)−x(x+h+1)

h(x+h+ 1)(x+1)

= lim

h→0

h(x+h+ 1)(x+1) =1

(x+1)

95 Chapter 3

5. Set f0(x)=0: f

(x) = 6(2)(2x+7)

(x−2)5+ 5(2x+7)

(x−2)4=0,so2x+7=0orx−2=0

or, factoring out (2x+7)

(x−2)4, 12(x−2) + 5(2x+ 7) = 0. This reduces to x=−7/2, x=2,or

22x+ 11 = 0, so the tangent line is horizontal at x=−7/2,2,−1/2.

6. Set f0(x)=0: f

0(x)=4(x2+2x)(x−3)3−(2x+ 2)(x−3)4

(x2+2x)

2, and a fraction can equal zero only if its

numerator equals zero. So either x−3 = 0 or, after factoring out (x−3)3,4(x

+2x)−(2x+2)(x−3)=0,

2+12x+ 6 = 0, whose roots are (by the quadratic formula) x=−6±√36 −4·3

2=−3±√6. So

the tangent line is horizontal at x=3,−3±√

7. Set f0(x)= 3

√

3x+1(x−1)2+2

√

3x+1(x−1)=0. Ifx= 1 then y0=0. Ifx6= 1 then divide out

x−1 and multiply through by 2√3x+ 1 (at points where fis diﬀerentiable we must have √3x+16=0)

to obtain 3(x−1) + 4(3x+1)=0,or15x+ 1 = 0. So the tangent line is horizontal at x=1,−1/15.

8. f0(x)=3

µ3x+1

2¶2d

3x+1

2=3

µ3x+1

2¶2x

(3) −(3x+ 1)(2x)

x4=−3µ3x+1

2¶23x

2+2x

4=0.

If f0(x) = 0 then (3x+1)

(3x2+2x) = 0. The tangent line is horizontal at x=−1/3,−2/3(x=0is

ruled out from the deﬁnition of f).

9. (a) x=−2,−1,1,3(b) (−∞,−2), (−1,1), (3,+∞)(c) (−2,−1), (1,3)

(d) g00(x)=f

00(x) sin x+2f0(x) cos x−f(x) sin x;g00(0)=2f

0(0) cos 0 = 2(2)(1) = 4

10. (a) f0(1)g(1) + f(1)g0(1)=3(−2)+1(−1) = −7

(b) g(1)f0(1) −f(1)g0(1)

g(1)2=−2(3) −1(−1)

(−2)2=−5

2pf(1)f0(1) = 1

2√13=3

2(d) 0 (because f(1)g0(1) is constant)

11. The equations of such a line has the form y=mx. The points (x0,y

0) which lie on both the line and

the parabola and for which the slopes of both curves are equal satisfy y0=mx0=x3

0−9x2

0−16x0,

so that m=x2

0−9x0−16. By diﬀerentiating, the slope is also given by m=3x

0−18x0−16.

Equating, we have x2

0−9x0−16=3x

0−18x0−16, or 2x2

0−9x0= 0. The root x0= 0 corresponds

to m=−16,y

0= 0 and the root x0=9/2 corresponds to m=−145/4,y

0=−1305/8. So the line

y=−16xis tangent to the curve at the point (0,0), and the line y=−145x/4 is tangent to the curve

at the point (9/2,−1305/8).

12. The slope of the line x+4y=10ism

1=−1/4, so we set the negative reciprocal

4=m

2=d

dx(2x3−x2)=6x

2−2xand obtain 6x2−2x−4 = 0 with roots x=1±√1+24

6=1,−2/3.

13. The line y−x= 2 has slope m1= 1 so we set m2=d

dx(3x−tan x)=3−sec2x= 1, or sec2x=2,

sec x=±√2sox=nπ ±π/4 where n=0,±1,±2,....

14. f(x) is continuous and diﬀerentiable at any x6= 1, so we consider x=1.

(a) lim

x→1−(x2−1) = lim

x→1+k(x−1)=0=f(1), so any value of kgives continuity at x=1.

(b) lim

x→1−f0(x) = lim

x→1−2x= 2, and lim

x→1+f0(x) = lim

x→1+k=k, so only if k=2isf(x) diﬀerentiable at

x=1.

Supplementary Exercises 3 96

15. The slope of the tangent line is the derivative

y0=2x

¯x=1

2(a+b)=a+b. The slope of the secant is

a2−b2

a−b=a+b, so they are equal.

a+b

(a,a)

(b,b)

16. To average 60 mi/h one would have to complete the trip in two hours. At 50 mi/h, 100 miles are

completed after two hours. Thus time is up, and the speed for the remaining 20 miles would have to

be inﬁnite.

17. (a) ∆x=1.5−2=−0.5; dy =−1

(x−1)2∆x=−1

(2 −1)2(−0.5)=0.5; and

∆y=1

(1.5−1) −1

(2 −1) =2−1=1.

(b) ∆x=0−(−π/4) = π/4; dy =¡sec2(−π/4)¢(π/4) = π/2; and ∆y= tan 0 −tan(−π/4)=1.

√25 −x2=−0

p25 −(0)2(3) = 0; and

∆y=√25 −32−√25 −02=4−5=−1.

18. (a) 43−23

4−2=56

2=28 (b) (dV /d`)|`=5 =3`

¯`=5 = 3(5)2=75

19. (a) dW

dt = 200(t−15); at t=5, dW

dt =−2000; the water is running out at the rate of 2000 gal/min.

(b) W(5) −W(0)

5−0=10000 −22500

5=−2500; the average rate of ﬂow out is 2500 gal/min.

20. cot 46◦= cot 46π

180 ; let x0=π

4and x=46π

180 . Then

cot 46◦= cot x= cot π

4−³csc2π

4´³x−π

4´=1−2µ46π

180 −π

4¶=0.9651;

with a calculator, cot 46◦=0.9657.

21. (a) h= 115 tan φ,dh = 115 sec2φdφ; with φ=51

◦=51

180πradians and dφ =±0.5◦=±0.5³π

180´

radians, h±dh = 115(1.2349) ±2.5340 = 142.0135 ±2.5340, so the height lies between

139.48 m and 144.55 m.

(b) If |dh|≤5 then |dφ|≤ 5

115 cos251

180 π≈0.017 radians, or |dφ|≤0.98◦.

22. (a) dT

dL =2

√g

2√L=1

√gL (b) s/m

dL >0 an increase in Lgives an increase in T, which is the period. To speed up a clock,

decrease the period; to decrease T, decrease L.

(d) dT

dg =−√L

g3/2<0; a decrease in gwill increase Tand the clock runs slower

(e) dT

dg =2

√

Lµ−1

2¶g

−3/2=−√

3/2(f) s3/m

97 Chapter 3

23. (a) f0(x)=2x, f0(1.8)=3.6

(b) f0(x)=(x

2−4x)/(x−2)2,f0(3.5) ≈−0.777778

24. (a) f0(x)=3x

2−2x, f 0(2.3)=11.27 (b) f0(x)=(1−x

)/(x

2+1)

,f0(−0.5)=0.48

25. f0(x)=2

xln 2; f0(2) ≈2.772589

26. f0(x)=x

sin x(cos xln x+ sin x/x); f0(2)=0.312141

27. vinst = lim

h→0

3(h+1)

2.5+ 580h−3

10h=58+ 1

dx3x2.5¯¯¯¯x=1

=58+ 1

10(2.5)(3)(1)1.5=58.75 ft/s

28. 164 ft/s

2500

120

29. Solve 3x2−cos x= 0 to get x=±0.535428.

30. When x4−x−1>0, f(x)=x

4−2x−1; when x4−x−1<0,

f(x)=−x

4+ 1, and fis diﬀerentiable in both cases. The roots of

x4−x−1=0arex

1=−0.724492, x2=1.220744. So x4−x−1>0

on (−∞,x

1) and (x2,+∞), and x4−x−1<0on(x

2). Then

lim

x→x−

f0(x) = lim

x→x−

(4x3−2)=4x

1−2 and

lim

x→x+

f0(x) = lim

x→x+

1−4x3=−4x3

1which is not equal to 4x3

1−2, so

fis not diﬀerentiable at x=x1; similarly fis not diﬀerentiable at

x=x2.

1.5

-1.5

-1.5 2

31. (a) f0(x)=5x

4(b) f0(x)=−1/x2(c) f0(x)=−1/2x

3/2

(d) f0(x)=−3/(x−1)2(e) f0(x)=3x/√3x2+5 (f ) f0(x) = 3 cos 3x

32. f0(x)=2xsin x+x2cos x33. f0(x)= 1−2

√

xsin 2x

2√x

34. f0(x)=6x

2+8x−17

(3x+2)

235. f0(x)=(1+x2) sec2x−2xtan x

(1+x2)2

36. f0(x)=x

2cos √x−2x3/2sin √x

2x7/2

37. f0(x)=−2x

5sin x−2x4cos x+4x

4+6x

2sin x+6x−3xcos x−4xsin x+ 4 cos x−8

2x2√x4−3+2(2−cos x)2

Horizon Module 3 98

CHAPTER 3 HORIZON MODULE

1. x1=l1cos θ1,x

2=l

2cos(θ1+θ2), so x=x1+x2=l1cos θ1+l2cos(θ1+θ2) (see Figure 3 in text);

similarly y1=l1sin θ1+l2sin(θ1+θ2).

2. Fix θ1for the moment and let θ2vary; then the distance rfrom (x, y) to the origin (see Figure 3 in

text) is at most l1+l2and at least l1−l2if l1≥l2and l2−l1otherwise. For any ﬁxed θ2let θ1vary

and the point traces out a circle of radius r.

(a) {(x, y):0≤x

2+y

2≤2l

}

(b) {(x, y):l

1−l

2≤x

2+y

2≤l

1+l

}

2−l

1≤x

2+y

2≤l

1+l

}

3. (x, y)=(l

1cos θ+l2cos(θ1+θ2),l

1sin θ1+l2sin(θ1+θ2))

= (cos(π/4) + 3 cos(5π/12),sin(π/4) + 3 sin(5π/12)) = Ã√2+3

√

4,7

√

2+3

√

4. x= (1) cos 2t+ (1) cos(2t+3t) = cos 2t+ cos 5t,

y= (1) sin 2t+ (1) sin(2t+3t) = sin 2t+ sin 5t

5. y

-2

-2 2

v1 = 3, v2 = 5

-2

-2 2

v1 = 1, v2 = 4

-2

= 4, v2 = 1

6. x= 2 cos t,y= 2 sin t, a circle of radius 2

7. (a) 9 = [3 sin(θ1+θ2)]2+ [3 cos(θ1+θ2)]2=[5−3 sin θ1]2+[3−3 cos θ1]2

=25−30 sin θ1+ 9 sin2θ1+9−18 cos θ1+ 9 cos2θ1=43−30 sin θ1−18 cos θ1,

so 15 sin θ1+ 9 cos θ1=17

(b) 1 = sin2θ1+ cos2θ2=µ17 −9 cos θ1

15 ¶2

+ cos θ1, or 306 cos2θ1−306 cos θ1=−64

2±5√17

102

(e) If θ1=0.792436 rad, then θ2=0.475882 rad ≈27.2660◦;

if θ1=1.26832 rad, then θ2=−0.475882 rad ≈−27.2660◦.

8. dx

dt =−3 sin θ1

dθ1

dt −(3 sin(θ1+θ2)) µdθ1

dt +dθ2

dt ¶=−3dθ1

dt (sin θ1+ sin(θ1+θ2)) −3 (sin(θ1+θ2)) dθ2

=−ydθ1

dt −3(sin(θ1+θ2))dθ2

dt ; similarly dy

dt =xdθ1

dt + 3(cos(θ1+θ2))dθ2

dt . Now set dx

dt =0, dy

dt =1.

9. (a) x= 3 cos(π/3) + 3 cos(−π/3)=6

2= 3 and y= 3 sin(π/3) −3 sin(π/3) = 0; equations (4)

become 3 sin(π/3)dθ2

dt =0,3

dθ1

dt + 3 cos(π/3)dθ2

dt = 1 with solution dθ2/dt =0,dθ1/dt =1/3.

(b) x=−3, y=3,so−3

dθ1

dt = 0 and −3dθ1

dt −3dθ2

dt = 1, with solution dθ1/dt =0,dθ2/dt =−1/3.

CHAPTER 4

Logarithmic and Exponential Functions

EXERCISE SET 4.1

1. (a) f(g(x))=4(x/4) = x,g(f(x))=(4x)/4=x,fand gare inverse functions

(b) f(g(x)) = 3(3x−1)+1=9x−26=xso fand gare not inverse functions

p(x3+2)−2=x,g(f(x))=(x−2)+2=x,fand gare inverse functions

(d) f(g(x))=(x

1/4

)

4=x,g(f(x))=(x

)

1/4=|x|6=x,fand gare not inverse functions

2. (a) They are inverse functions. 2

-2

-2 2

(b) The graphs are not reﬂections of each other

about the line y=x.

-2

-2 2

gis restricted to [0,+∞)

(d) They are inverse functions provided the domain of f(x)

is restricted to [0,+∞)

3. (a) yes; all outputs (the elements of row two) are distinct

(b) no; f(1) = f(6)

4. (a) no; it is easy to conceive of, say, 8 people in line at two diﬀerent times

(b) no; perhaps your weight remains constant for more than a year

weight

Exercise Set 4.1 100

5. (a) yes (b) yes (c) no (d) yes (e) no (f ) no

6. (a) no, the horizontal line test fails

-2

-3 3

(b) yes, horizontal line test

-10

-1 3

7. (a) no, the horizontal line test fails

(b) no, the horizontal line test fails

8. (a) no, the horizontal line test fails

(b) no, the horizontal line test fails

9. (a) fhas an inverse because the graph passes the horizontal line test. To compute f−1(2) start at

2onthey-axis and go to the curve and then down, so f−1(2) = 8; similarly, f−1(−1) = −1 and

f−1(0)=0.

(b) domain of f−1is [−2,2], range is [−8,8] (c)

-2 -1 1 2

-8

-4

10. (a) the horizontal line test fails

(b) −∞ <x≤−1; −1≤x≤2; and 2 ≤x<+∞.

11. (a) f0(x)=2x+8;f0<0on(−∞,−4) and f0>0on(−4,+∞); not one-to-one

(b) f0(x)=10x

4+3x

2+3≥3>0; f0(x) is positive for all x,sofis one-to-one

12. (a) f0(x)=3x

2+6x=x(3x+ 6) changes sign at x=−2,0, so fis not one-to-one

(b) f0(x)=5x

4+24x

2+2≥2>0; f0is positive for all x,sofis one-to-one

(x+1)

2;fis one-to-one because:

if x1<x

2<−1 then f0>0on[x

2], so f(x1)6=f(x2)

if −1<x

1<x

2then f0>0on[x

2], so f(x1)6=f(x2)

if x1<−1<x

2then f(x1)>1>f(x

) since f(x)>1on(−∞,−1) and f(x)<1on(−1,+∞)

13. y=f−1(x), x=f(y)=y

,y=x

1/5=f

−1

(x)

14. y=f−1(x), x=f(y)=6y,y=1

x=f

−1

(x)

15. y=f−1(x), x=f(y)=7y−6, y=1

7(x+6)=f−1

(x)

16. y=f−1(x), x=f(y)=y+1

y−1,xy −x=y+1,(x−1)y=x+1,y=x+1

x−1=f−1

(x)

101 Chapter 4

17. y=f−1(x), x=f(y)=3y

3−5, y=3

p(x+5)/3=f

−1

(x)

18. y=f−1(x), x=f(y)= 5

√

4y+2,y=1

4(x

5−2) = f−1(x)

19. y=f−1(x), x=f(y)= 3

√

2y−1, y=(x

3+1)/2=f

−1

(x)

20. y=f−1(x), x=f(y)= 5

2+1,y=r5−x

x=f−1

(x)

21. y=f−1(x), x=f(y)=3/y2,y=−p3/x =f−1(x)

22. y=f−1(x), x=f(y)=½2y, y ≤0

y2,y>0

,y=f

−1

(x)=½x/2,x≤0

√

x, x > 0

23. y=f−1(x),x=f(y)=½5/2−y, y < 2

1/y, y ≥2,y=f

−1

(x)=½5/2−x, x > 1/2

1/x, 0<x≤1/2

24. y=p−1(x), x=p(y)=y

3−3y

2+3y−1=(y−1)3,y=x1/3+1=p

−1

(x)

25. y=f−1(x), x=f(y)=(y+2)

4for y≥0, y=f−1(x)=x

1/4−2 for x≥16

26. y=f−1(x), x=f(y)=√

y+ 3 for y≥−3, y=f−1(x)=x

2−3 for x≥0

27. y=f−1(x), x=f(y)=−

√

3−2yfor y≤3/2, y=f−1(x)=(3−x

)/2 for x≤0

28. y=f−1(x), x=f(y)=3y

2+5y−2 for y≥0, 3y2+5y−2−x= 0 for y≥0,

y=f−1(x)=(−5+√

12x+ 49)/6 for x≥−2

29. y=f−1(x), x=f(y)=y−5y

2for y≥1, 5y2−y+x= 0 for y≥1,

y=f−1(x)=(1+√

1−20x)/10 for x≤−4

30. (a) C=5

9(F−32)

(b) how many degrees Celsius given the Fahrenheit temperature

C≥−273.15

31. (a) y=f(x)=(6.214 ×10−4)x(b) x=f−1(y)= 104

6.214y

32. fand f−1are continuous so f(3) = lim

x→3f(x) = 7; then f−1(7) = 3, and

lim

x→7f−1(x)=f

−1³lim

x→7x´=f−1(7)=3

33. (a) f(g(x)) = f(√x)

√

2=x, x > 1;

g(f(x)) = g(x2)

=√x2=x, x > 1

(b)

y = f(x)

y = g(x)

Exercise Set 4.1 102

34. y=f−1(x), x=f(y)=ay2+by +c,ay2+by +c−x= 0, use the quadratic formula to get

y=−b±pb2−4a(c−x)

2a;

(a) f−1(x)=−b+pb

2−4a(c−x)

2a(b) f−1(x)= −b−pb

2−4a(c−x)

35. (a) f(f(x)) =

3−3−x

1−x

1−3−x

1−x

=3−3x−3+x

1−x−3+x=xso f=f−1

(b) symmetric about the line y=x

36. y=m(x−x0) is an equation of the line. The graph of the inverse of f(x)=m(x−x

) will be the

reﬂection of this line about y=x. Solve y=m(x−x0) for xto get x=y/m +x0=f−1(y)so

y=f

−1

(x)=x/m +x0.

37. (a) f(x)=x

3−3x

2+2x=x(x−1)(x−2) so f(0) = f(1) = f(2) = 0 thus fis not one-to-one.

(b) f0(x)=3x

2−6x+2, f0(x) = 0 when x=6±√36 −24

6=1±√

3/3. f0(x)>0(fis increasing)

if x<1−√

3/3, f0(x)<0(fis decreasing) if 1 −√3/3<x<1+√3/3, so f(x) takes on values

less than f(1 −√3/3) on both sides of 1 −√3/3thus1−√

3/3 is the largest value of k.

38. (a) f(x)=x

(x−2) so f(0) = f(2) = 0 thus fis not one to one.

(b) f0(x)=4x

3−6x

2=4x

(x−3/2),f0(x) = 0 when x= 0 or 3/2; fis decreasing on (−∞,3/2]

and increasing on [3/2,+∞)so3/2 is the smallest value of k.

39. if f−1(x) = 1, then x=f(1) = 2(1)3+ 5(1) + 3 = 10

40. if f−1(x) = 2, then x=f(2) = (2)3/[(2)2+1]=8/5

41. 6

-2

-2 6

42. 10

-5

-5 10

43. 3

44. 6

45. f(f(x)) = xthus f=f−1so the graph is symmetric about y=x.

46. (a) Suppose x16=x2where x1and x2are in the domain of gand g(x1), g(x2) are in the domain of f

then g(x1)6=g(x2) because gis one-to-one so f(g(x1)) 6=f(g(x2)) because fis one-to-one thus

f◦gis one-to-one because (f◦g)(x1)6=(f◦g)(x2)ifx

16=x

103 Chapter 4

(b) f,g, and f◦gall have inverses because they are all one-to-one. Let h=(f◦g)

−1then

(f◦g)(h(x)) = f[g(h(x))] = x, apply f−1to both sides to get g(h(x)) = f−1(x), then apply g−1

to get h(x)=g

−1

(x))=(g

−1◦f

−1

)(x), so h=g−1◦f−1

47.

48. Suppose that gand hare both inverses of fthen f(g(x)) = x,h[f(g(x))] = h(x), but h[f(g(x))] = g(x)

because his an inverse of fso g(x)=h(x).

49. F0(x)=2f

(2g(x))g0(x)soF

(3)=2f

(2g(3))g0(3). By inspection f(1)=3,sog(3) = f−1(3)=1

and g0(3)=(f

−1

)

(3)=1/f 0(f−1(3)) = 1/f 0(1)=1/7 because f0(x)=4x

3+3x

.Thus

(3)=2f

0(2)(1/7) = 2(44)(1/7)=88/7.

EXERCISE SET 4.2

1. (a) −4(b) 4(c) 1/4

2. (a) 1/16 (b) 8(c) 1/3

3. (a) 2.9690 (b) 0.0341

4. (a) 1.8882 (b) 0.9381

5. (a) log216 = log2(24)=4 (b) log2µ1

32¶= log2(2−5)=−5

6. (a) log10(0.001) = log10(10−3)=−3(b) log10(104)=4

7. (a) 1.3655 (b) −0.3011

8. (a) −0.5229 (b) 1.1447

9. (a) 2lna+1

2ln b+1

2ln c=2r+s/2+t/2(b) ln b−3lna−ln c=s−3r−t

10. (a) 1

3ln c−ln a−ln b=t/3−r−s(b) 1

2(ln a+3lnb−2lnc)=r/2+3s/2−t

11. (a) 1 + log x+1

2log(x−3) (b) 2ln|x|+ 3 ln sin x−1

2ln(x2+1)

12. (a) 1

3log(x+2)−log cos 5x(b) 1

2ln(x2+1)−1

2ln(x3+5)

13. log 24(16)

3= log(256/3) 14. log √x−log(sin32x) + log 100 = log 100√x

sin32x

Exercise Set 4.2 104

15. ln

√x(x+1)

cos x16. 1+x=10

3= 1000, x= 999

17. √x=10

−1=0.1, x=0.01 18. x2=e4,x=±e2

19. 1/x =e−2,x=e220. x=7

21. 2x=8,x=4 22. log10 x3= 30, x3=10

30,x=10

23. log10 x=5,x=10

24. ln 4x−ln x6=ln2,ln 4

5=ln2, 4

5=2,x

5=2,x=5

√

25. ln 2x2=ln3,2x

2=3,x

2=3/2, x=p3/2 (we discard −p3/2 because it does not satisfy the original

equation)

26. ln 3x=ln2,xln3=ln2,x=ln 2

ln 3

27. ln 5−2x=ln3,−2xln5=ln3,x=−ln 3

2ln5

28. e−2x=5/3,−2x= ln(5/3),x=−

ln(5/3)

29. e3x=7/2,3x= ln(7/2),x=1

ln(7/2)

30. ex(1 −2x)=0soe

x= 0 (impossible) or 1 −2x=0,x=1/2

31. e−x(x+2)=0soe

−x= 0 (impossible) or x+2=0,x=−2

32. e2x−ex−6=(e

x−3)(ex+2)=0soe

x=−2 (impossible) or ex=3,x=ln3

33. e−2x−3e−x+2=(e

−x−2)(e−x−1)=0soe

−x=2,x=−ln 2 or e−x=1,x=0

34. (a) y

-2

(b) y

-2 2

35. (a) y

-2 2 4

(b) y

-4

-4 2

105 Chapter 4

36. (a) y

-10

-1

(b) y

-1

37. log27.35 = (log 7.35)/(log 2) = (ln 7.35)/(ln 2) ≈2.8777;

log50.6 = (log 0.6)/(log 5) = (ln 0.6)/(ln 5) ≈−0.3174

38. 10

-5

39. 2

-3

40. (a) Let X= logbxand Y= logax. Then bX=xand aY=xso aY=bX,ora

Y/X =b, which means

logab=Y/X. Substituting for Yand Xyields logax

logbx= logab, logbx=logax

logab.

(b) Let x=ato get logba= (logaa)/(logab)=1/(logab) so (logab)(logba)=1.

(log281)(log332) = (log2[34])(log3[25]) = (4 log23)(5 log32) = 20(log23)(log32)=20

41. x=3.6541,y =1.2958

0.6

42. Since the units are billions, one trillion is 1,000 units. Solve 1000 = 0.051517(1.1306727)xfor xby

taking common logarithms, resulting in 3 = log 0.051517 + xlog 1.1306727, which yields x≈77.4, so

the debt ﬁrst reached one trillion dollars around 1977.

43. (a) no, the curve passes through the origin (b) y=2

x/4

−x(d) y=(

√

5)x

-1 2

Exercise Set 4.2 106

44. (a) As x→+∞the function grows very slowly,

but it is always increasing and tends to +∞.

As x→1+the function tends to −∞.

(b) y

-2

150

45. log(1/2) <0 so 3 log(1/2) <2 log(1/2)

46. Let x= logbaand y= logbc,soa=b

xand c=by.

First, ac =bxby=bx+yor equivalently, logb(ac)=x+y= logba+logbc.

Secondly, a/c =bx/by=bx−yor equivalently, logb(a/c)=x−y= logba−logbc.

Next, ar=(b

)

r=b

rx or equivalently, logbar=rx =rlogba.

Finally, 1/c =1/by=b−yor equivalently, logb(1/c)=−y=−logbc.

47. 75e−t/125 =15,t=−125 ln(1/5) = 125 ln 5 ≈201 days.

48. (a) If t= 0, then Q= 12 grams (b) Q=12e

−0.055(4) =12e

−0.22 ≈9.63 grams

−0.055t=0.5,t=−(ln 0.5)/(0.055) ≈12.6 hours

49. (a) 7.4; basic (b) 4.2; acidic (c) 6.4; acidic (d) 5.9; acidic

50. (a) log[H+]=−2.44,[H+]=10

−2.44 ≈3.6×10−3mol/L

(b) log[H+]=−8.06,[H+]=10

−8.06 ≈8.7×10−9mol/L

51. (a) 140 dB; damage (b) 120 dB; damage

52. Suppose that I1=3I

2and β1= 10 log10 I1/I0,β2= 10 log10 I2/I0. Then

I1/I0=3I

/I0, log10 I1/I0= log10 3I2/I0= log10 3 + log10 I2/I0,β1= 10 log10 3+β

1−β

2= 10 log10 3≈4.8 decibels.

53. Let IAand IBbe the intensities of the automobile and blender, respectively. Then

log10 IA/I0= 7 and log10 IB/I0=9.3, IA=10

0and IB=10

9.3

,soI

/IA=10

2.3≈200.

54. The decibel level of the nth echo is 120(2/3)n;

120(2/3)n<10 if (2/3)n<1/12,n> log(1/12)

log(2/3) =log 12

log 1.5≈6.13 so 6 echoes can be heard.

55. (a) log E=4.4+1.5(8.2)=16.7,E =10

16.7≈5×1016 J

(b) Let M1and M2be the magnitudes of earthquakes with energies of Eand 10E,

respectively. Then 1.5(M2−M1) = log(10E)−log E= log 10 = 1,

M2−M1=1/1.5=2/3≈0.67.

56. Let E1and E2be the energies of earthquakes with magnitudes Mand M+ 1, respectively. Then

log E2−log E1= log(E2/E1)=1.5,E

2/E1=10

1.5≈31.6.

57. If t=−2x, then x=−t/2 and lim

x→0(1 −2x)1/x = lim

t→0(1+t)−2/t = lim

t→0[(1 + t)1/t]−2=e−2.

58. If t=3/x, then x=3/t and lim

x→+∞(1+3/x)x= lim

t→0+(1+t)3/t = lim

t→0+[(1 + t)1/t]3=e3.

107 Chapter 4

EXERCISE SET 4.3

1. y=(2x−5)1/3;dy/dx =2

3(2x−5)−2/3

2. dy/dx =1

3£2 + tan(x2)¤−2/3sec2(x2)(2x)=2

xsec2(x2)£2 + tan(x2)¤−2/3

3. dy/dx =3

2·x−1

x+2¸1/2d

dx ·x−1

x+2¸=9

2(x+2)

2·x−1

x+2¸1/2

4. dy/dx =1

2·x2+1

2−5¸−1/2d

dx ·x2+1

2−5¸=1

2·x

2+1

2−5¸−1/2−12x

(x2−5)2=−6x

(x2−5)2·x2+1

2−5¸−1/2

5. dy/dx =x3µ−2

3¶(5x2+1)

−5/3

(10x)+3x

(5x2+1)

−2/3=1

(5x2+1)

−5/3

(25x2+9)

6. dy/dx =

x24

3(3 −2x)1/3(−2) −(3 −2x)4/3(2x)

x4=2(3 −2x)1/3(2x−9)

3x3

7. dy/dx =5

2[sin(3/x)]3/2[cos(3/x)](−3/x2)=−15[sin(3/x)]3/2cos(3/x)

2x2

8. dy/dx =−1

2£cos(x3)¤−3/2£−sin(x3)¤(3x2)=3

2sin(x3)£cos(x3)¤−3/2

9. (a) 3x2+xdy

dx +y−2=0, dy

dx =2−3x2−y

(b) y=1+2x−x

x=1

x+2−x

2so dy

dx =−1

x2−2x

dx =2−3x2−y

x=2−3x2−(1/x +2−x

)

x=−2x−1

10. (a) 1

2y−1/2dy

dx −ex=0or dy

dx =2e

√

(b) y=(2+e

)

2=2+4e

x+e

2xso dy

dx =4e

x+2e

dx =2e

√

y=2e

(2+ex)=4e

x+2e

11. 2x+2ydy

dx =0so dy

dx =−x

12. 3x2−3y2dy

dx =6(x

dx +y), −(3y2+6x)dy

dx =6y−3x

2so dy

dx =x2−2y

y2+2x

13. x2dy

dx +2xy +3x(3y2)dy

dx +3y

3−1=0

2+9xy2)dy

dx =1−2xy −3y3so dy

dx =1−2xy −3y3

x2+9xy2

14. x3(2y)dy

dx +3x

2−5x

2dy

dx −10xy +1=0

(2x3y−5x2)dy

dx =10xy −3x2y2−1so dy

dx =10xy −3x2y2−1

2x3y−5x2

Exercise Set 4.3 108

15. −1

dx −1

x2=0so dy

dx =−y2

16. 2x=(x−y)(1 + dy/dx)−(x+y)(1 −dy/dx)

(x−y)2,

2x(x−y)2=−2y+2xdy

dx so dy

dx =x(x−y)2+y

17. cos(x2y2)·x2(2y)dy

dx +2xy2¸=1, dy

dx =1−2xy2cos(x2y2)

2x2ycos(x2y2)

18. 2x=(1 + csc y)(−csc2y)(dy/dx)−(cot y)(−csc ycot y)(dy/dx)

(1 + csc y)2,

2x(1 + csc y)2=−csc y(csc y+ csc2y−cot2y)dy

dx,

but csc2y−cot2y=1,so dy

dx =−2x(1 + csc y)

csc y

19. 3 tan2(xy2+y) sec2(xy2+y)µ2xy dy

dx +y2+dy

dx¶=1

so dy

dx =1−3y2tan2(xy2+y) sec2(xy2+y)

3(2xy + 1) tan2(xy2+y) sec2(xy2+y)

20. (1 + sec y)[3xy2(dy/dx)+y

]−xy3(sec ytan y)(dy/dx)

(1 + sec y)2=4y

3dy

dx,

multiply through by (1 + sec y)2and solve for dy

to get dy

dx =y(1 + sec y)

4y(1 + sec y)2−3x(1 + sec y)+xy sec ytan y

21. dy

dx =3x

4y,d2y

dx2=(4y)(3) −(3x)(4dy/dx)

16y2=12y−12x(3x/(4y))

16y2=12y2−9x2

16y3=−3(3x2−4y2)

16y3,

but 3x2−4y2=7so d

dx2=−3(7)

16y3=−21

16y3

22. dy

dx =−x2

y2,d2y

dx2=−y2(2x)−x2(2ydy/dx)

y4=−2xy2−2x2y(−x2/y2)

y4=−2x(y3+x3)

y5,

but x3+y3=1so d

dx2=−2x

23. dy

dx =−y

x,d2y

dx2=−x(dy/dx)−y(1)

x2=−x(−y/x)−y

x2=2y

24. dy

dx =y

y−x,

d2y

dx2=(y−x)(dy/dx)−y(dy/dx −1)

(y−x)2=

(y−x)µy

y−x¶−yµy

y−x−1¶

(y−x)2

=y2−2xy

(y−x)3but y2−2xy =−3,so d2y

dx2=−3

(y−x)3

25. dy

dx = (1 + cos y)−1,d2y

dx2=−(1 + cos y)−2(−sin y)dy

dx =sin y

(1 + cos y)3

109 Chapter 4

26. dy

dx =cos y

1+xsin y,

d2y

dx2=(1+xsin y)(−sin y)(dy/dx)−(cos y)[(xcos y)(dy/dx) + sin y]

(1+xsin y)2

=−2 sin ycos y+(xcos y)(2 sin2y+ cos2y)

(1+xsin y)3,

but xcos y=y, 2 sin ycos y= sin 2y, and sin2y+ cos2y=1so

dx2=−sin 2y+y(sin2y+1)

(1+xsin y)3

27. By implicit diﬀerentiation, 2x+2y(dy/dx)=0,dy

dx =−x

y;at(1/

√

2,1/

√

2), dy

dx =−1; at (1/√2,−1/√2),

dx = +1. Directly, at the upper point y=√1−x2,dy

dx =−x

√1−x2=−1 and at the lower point

y=−√1−x2,dy

dx =x

√1−x2= +1.

28. If y2−x+ 1 = 0, then y=√x−1 goes through the point (10,3) so dy/dx =1/(2√x−1). By

implicit diﬀerentiation dy/dx =1/(2y). In both cases, dy/dx|(10,3) =1/6. Similarly y=−√x−1

goes through (10,−3) so dy/dx =−1/(2√x−1) = −1/6 which yields dy/dx =1/(2y)=−1/6.

29. 4x3+4y

3dy

dx =0,so dy

dx =−x3

y3=−1

153/4≈−0.1312.

30. 3y2dy

dx +x2dy

dx +2xy +2x−6ydy

dx =0,so dy

dx =−2xy+1

2+x

2−6y=0atx=0

31. 4(x2+y2)µ2x+2ydy

dx¶=25µ2x−2ydy

dx¶,

dx =x[25 −4(x2+y2)]

y[25+4(x

2+y

)] ;at(3,1) dy

dx =−9/13

32. 2

3µx−1/3+y−1/3dy

dx¶=0, dy

dx =−y1/3

x1/3=√3at(−1,3

√

34. (a) y

–2

12 x

(b) 2ydy

dx =(x−a)(x−b)+x(x−b)+x(x−a)=3x

2−2(a+b)x+ab.If

dx = 0 then

3x2−2(a+b)x+ab = 0. By the quadratic formula

x=2(a+b)±p4(a+b)2−4·3ab

6=1

3ha+b±(a2+b2−ab)1/2i.

necessary that all three of the factors x,x−a,x−bbe nonnegative, or that two of them be

nonpositive. If, for example, 0 <a<bthen the function is deﬁned on the disjoint intervals

0<x<aand b<x<+∞, so there are two parts.

Exercise Set 4.3 110

35. (a) y

-2 2

(b) ±1.1547

dx −y+2ydy

dx = 0. Solve for dy

dx =y−2x

2y−x.Ifdy

dx = 0 then

y−2x=0ory=2x.Thus4=x

2−xy +y2=x2−2x2+4x

2=3x

,x=±2

√

36. 1

2u−1/2du

dv +1

2v−1/2=0so du

dv =−√u

√v

37. 4a3da

dt −4t3=6µa

2+2at da

dt ¶, solve for da

dt to get da

dt =2t3+3a

3−6at

38. 1 = (cos x)dx

dy so dx

dy =1

cos x= sec x39. 2a2ωdω

dλ +2b

λ=0so dω

dλ =−b2λ

a2ω

40. Let P(x0,y

0) be the required point. The slope of the line 4x−3y+1=0is4/3 so the slope of

the tangent to y2=2x

3at Pmust be −3/4. By implicit diﬀerentiation dy/dx =3x

/y,soatP,

/y0=−3/4, or y0=−4x2

0. But y2

0=2x

0because Pis on the curve y2=2x

. Elimination of y0

gives 16x4

0=2x

(8x0−1) = 0, so x0= 0 or 1/8. From y0=−4x2

0it follows that y0= 0 when

x0= 0, and y0=−1/16 when x0=1/8. It does not follow, however, that (0,0) is a solution because

dy/dx =3x

/y (the slope of the curve as determined by implicit diﬀerentiation) is valid only if y6=0.

Further analysis shows that the curve is tangent to the x-axis at (0,0), so the point (1/8,−1/16) is

the only solution.

41. The point (1,1) is on the graph, so 1 + a=b. The slope of the tangent line at (1,1) is −4/3; use

implicit diﬀerentiation to get dy

dx =−2xy

x2+2ay so at (1,1), −2

1+2a=−4

3, 1+2a=3/2, a=1/4 and

hence b=1+1/4=5/4.

42. Use implicit diﬀerentiation to get dy/dx =(y−3x

)/(3y2−x), so dy/dx =0ify=3x

. Substitute

this into x3−xy +y3= 0 to obtain 27x6−2x3=0,x

3=2/27, x=3

√2/3 and hence y=3

√4/3.

43. Let P(x0,y

0) be a point where a line through the origin is tangent to the curve

x2−4x+y2+ 3 = 0. Implicit diﬀerentiation applied to the equation of the curve gives

dy/dx =(2−x)/y.AtPthe slope of the curve must equal the slope of the line so

(2 −x0)/y0=y0/x0,ory

0=2x

0−x

. But x2

0−4x0+y2

0+ 3 = 0 because (x0,y

0) is on the curve,

and elimination of y2

0in the latter two equations gives x2

0−4x0+(2x

0−x

)+3=0,x

0=3/2 which

when substituted into y2

0=2x

0−x

0yields y2

0=3/4, so y0=±√3/2. The slopes of the lines are

(±√3/2)/(3/2) = ±√3/3 and their equations are y=(

√

3/3)xand y=−(√3/3)x.

44. By implicit diﬀerentiation, dy/dx =k/(2y) so the slope of the tangent to y2=kx at (x0,y

0)is

k/(2y0)ify

06= 0. The tangent line in this case is y−y0=k

2y0

(x−x0), or 2y0y−2y2

0=kx −kx0.

But y2

0=kx0because (x0,y

0) is on the curve y2=kx, so the equation of the tangent line becomes

2y0y−2kx0=kx −kx0which gives y0y=k(x+x0)/2. If y0= 0, then x0= 0; the graph of y2=kx

has a vertical tangent at (0,0) so its equation is x= 0, but y0y=k(x+x0)/2 gives the same result

when x0=y0=0.

111 Chapter 4

45. By the chain rule, dy

dx =dy

dx. Use implicit diﬀerentiation on 2y3t+t3y= 1 to get

dt =−2y3+3t

6ty2+t3, but dt

dx =1

cos tso dy

dx =−2y3+3t

(6ty2+t3) cos t.

46. 2x3ydy

dt +3x

2dx

dt +dy

dt =0,dy

dt =−3x2y2

2x3y+1

47. 2xy dy

dt =y2dx

dt = 3(cos 3x)dx

dt ,dy

dt =3 cos 3x−y2

2xy

48. (a) f0(x)= 4

1/3

00(x)=4

−2/3

(b) f0(x)=7

4/3

00(x)=28

9x1/3,f000(x)=28

27x−2/3

49. y0=rxr−1,y

00 =r(r−1)xr−2so 3x2£r(r−1)xr−2¤+4x¡rxr−1¢−2xr=0,

3r(r−1)xr+4rxr−2xr=0,(3r2+r−2)xr=0,

2+r−2=0,(3r−2)(r+1)=0;r=−1,2/3

50. y0=rxr−1,y

00 =r(r−1)xr−2so 16x2£r(r−1)xr−2¤+24x¡rxr−1¢+xr=0,

16r(r−1)xr+24rxr+xr=0,(16r2+8r+1)x

r=0,

16r2+8r+1=0,(4r+1)

2=0;r=−1/4

51. We shall ﬁnd when the curves intersect and check that the slopes are negative reciprocals. For the

intersection solve the simultaneous equations x2+(y−c)

2=c

2and (x−k)2+y2=k2to obtain

cy =kx =1

2(x2+y2). Thus x2+y2=cy +kx,ory

2−cy =−x2+kx, and y−c

x=−x−k

Diﬀerentiating the two families yields (black) dy

dx =−x

y−c, and (gray) dy

dx =−x−k

y. But it was

proven that these quantities are negative reciprocals of each other.

52. Diﬀerentiating, we get the equations (black) xdy

dx +y= 0 and (gray) 2x−2ydy

dx = 0. The ﬁrst says

the (black) slope is = −y

xand the second says the (gray) slope is x

y, and these are negative reciprocals

of each other.

53. y=f−1(x), x=f(y)=5y

3+y−7, dx

dy =15y

2+1, dy

dx =1

15y2+1;

check: 1 = 15y2dy

dx +dy

dx,dy

dx =1

15y2+1

54. y=f−1(x), x=f(y)=1/y2,dx

dy =−2y−3,dy

dx =−y3/2;

check: 1 = −2y−3dy

dx,dy

dx =−y3/2

55. y=f−1(x), x=f(y)=2y

5+y

3+1, dx

dy =10y

4+3y

,dy

dx =1

10y4+3y

check: 1 = 10y4dy

dx +3y

2dy

dx,dy

dx =1

10y4+3y

Exercise Set 4.4 112

56. y=f−1(x), x=f(y)=5y−sin 2y,dx

dy =5−2 cos 2y,dy

dx =1

5−2 cos 2y;

check: 1 = (5 −2 cos 2y)dy

dx,dy

dx =1

5−2 cos 2y

EXERCISE SET 4.4

1. 1

2x(2)=1/x 2. 1

x3(3x2)=3/x

3. 2(ln x)µ1

x¶=2lnx

x4. 1

sin x(cos x) = cot x

5. 1

tan x(sec2x)=sec2x

tan x6. 1

2+√

xµ1

√

x¶=1

√

x(2+√x)

7. 1

x/(1+x2)·(1+x2)(1) −x(2x)

(1+x2)2¸=1−x2

x(1+x2)8. 1

ln xµ1

x¶=1

xln x

9. 3x2−14x

x3−7x2−310. x3µ1

x¶+(3x

)lnx=x

2(1+3lnx)

11. 1

2(ln x)−1/2µ1

x¶=1

2x√ln x12.

22(ln x)(1/x)

p1+ln

2x=ln x

xp1+ln

13. −1

xsin(ln x)14. 2 sin(ln x) cos(ln x)1

x=sin(ln x2)

15. 3x2log2(3 −2x)+ −2x

(ln 2)(3 −2x)16. £log2(x2−2x)¤3+3x£log2(x2−2x)¤22x−2

(x2−2x)ln2

17. 2x(1 + log x)−x/(ln 10)

(1 + log x)218. 1/[x(ln 10)(1 + log x)2]

19. 7e7x20. −10xe−5x2

21. x3ex+3x

x=x

(x+3) 22. −1

x2e1/x

23. dy

dx =(ex+e−x)(ex+e−x)−(ex−e−x)(ex−e−x)

(ex+e−x)2

=(e2x+2+e

−2x

)−(e

2x−2+e

−2x

)

x+e

−x

)

2=4/(e

x+e

−x

)

24. excos(ex)

25. (xsec2x+ tan x)extan x26. dy

dx =(ln x)ex−ex(1/x)

(ln x)2=ex(xln x−1)

x(ln x)2

27. (1 −3e3x)e(x−e3x)28. 15

2x2(1+5x

)

−1/2exp(p1+5x

)

29. (x−1)e−x

1−xe−x=x−1

ex−x30. 1

cos(ex)[−sin(ex)]ex=−extan(ex)

113 Chapter 4

31. dy

dx +1

xy µxdy

dx +y¶=0, dy

dx =−y

x(y+1)

32. dy

dx =1

xtan yµxsec2ydy

dx + tan y¶,dy

dx =tan y

x(tan y−sec2y)

33. d

dx ·ln cos x−1

2ln(4 −3x2)¸=−tan x+3x

4−3x2

34. d

dx µ1

2[ln(x−1) −ln(x+ 1)]¶=1

2µ1

x−1−1

x+1¶

35. ln |y|=ln|x|+1

3ln |1+x

|,dy

dx =x3

p1+x

2·1

x+2x

3(1 + x2)¸

36. ln |y|=1

5[ln |x−1|−ln |x+1|], dy

dx =1

rx−1

x+1·1

x−1−1

x+1¸

37. ln |y|=1

3ln |x2−8|+1

2ln |x3+1|−ln |x6−7x+5|

dx =(x2−8)1/3√x3+1

6−7x+5 ·2x

3(x2−8) +3x2

2(x3+1) −6x

5−7

6−7x+5¸

38. ln |y|=ln|sin x|+ln|cos x|+3ln|tan x|−1

2ln |x|

dx =sin xcos xtan3x

√x·cot x−tan x+3 sec2x

tan x−1

2x¸

39. f0(x)=2

xln 2; y=2

,lny=xln 2, 1

yy0=ln2,y

0=yln2=2

xln 2

40. f0(x)=−3

−xln 3; y=3

−x

,lny=−xln 3, 1

yy0=−ln 3, y0=−yln 3 = −3−xln 3

41. f0(x)=π

sin x(ln π) cos x;

y=πsin x,lny= (sin x)lnπ,1

0= (ln π) cos x,y0=πsin x(ln π) cos x

42. f0(x)=π

xtan x(ln π)(xsec2x+ tan x);

y=πxtan x,lny=(xtan x)lnπ,1

0= (ln π)(xsec2x+ tan x)

y0=πxtan x(ln π)(xsec2x+ tan x)

43. ln y= (ln x) ln(x3−2x), 1

dx =3x2−2

x3−2xln x+1

xln(x3−2x),

dx =(x

3−2x)

ln x·3x2−2

x3−2xln x+1

xln(x3−2x)¸

44. ln y= (sin x)lnx,1

dx =sin x

x+ (cos x)lnx,dy

dx =xsin x·sin x

x+ (cos x)lnx

45. ln y= (tan x) ln(ln x), 1

dx =1

xln xtan x+ (sec2x) ln(ln x),

dx = (ln x)tan x·tan x

xln x+ (sec2x) ln(ln x)¸

Exercise Set 4.4 114

46. ln y= (ln x) ln(x2+ 3), 1

dx =2x

x2+3ln x+1

xln(x2+ 3),

dx =(x

2+3)

ln x·2x

x2+3ln x+1

xln(x2+3)

47. y=Ae2x+Be−4x,y0=2Ae2x−4Be−4x,y00 =4Ae2x+16Be−4xso

y00 +2y

0−8y=(4Ae2x+16Be−4x) + 2(2Ae2x−4Be−4x)−8(Ae2x+Be−4x)=0

48. y=Aekt,dy/dt =kAekt =k(Aekt)=ky

49. (a) f0(x)=kekx,f00(x)=k

kx,f000(x)=k

kx,...,f(n)(x)=k

(b) f0(x)=−ke−kx,f00(x)=k

−kx,f000(x)=−k

−kx,...,f(n)(x)=(−1)nkne−kx

50. dy

dt =e−λt(ωA cos ωt −ωB sin ωt)+(−λ)e

−λt(Asin ωt +Bcos ωt)

=e−λt[(ωA −λB) cos ωt −(ωB +λA) sin ωt]

51. f0(x)= 1

√2πσ exp "−1

2µx−µ

σ¶2#d

dx "−1

2µx−µ

σ¶2#

√2πσ exp "−1

2µx−µ

σ¶2#·−µx−µ

σ¶µ1

σ¶¸

=−1

√2πσ3(x−µ) exp "−1

2µx−µ

σ¶2#

52. (a) y0=−xe−x+e−x=e−x(1 −x), xy0=xe−x(1 −x)=y(1 −x)

(b) y0=−x2e−x2/2+e−x2/2=e−x2/2(1 −x2), xy0=xe−x2/2(1 −x2)=y(1 −x2)

53. (a) logxe=ln e

ln x=1

ln x,d

dx[logxe]=−1

x(ln x)2

(b) logx2= ln 2

ln x,d

dx[logx2] = −ln 2

x(ln x)2

54. β= 10 log I−10 log I0,dβ

dI =10

Iln 10

(a) dβ

dI ¸I=10I0

I0ln 10 db/W/m2(b) dβ

dI ¸I=100I0

10I0ln 10 db/W/m2

dI ¸I=100I0

100I0ln 10 db/W/m2

55. dk

dT =k0exp ·−q(T−T0)

2T0T¸³−q

2T2´=−qk0

2T2exp ·−q(T−T0)

2T0T¸

56. (a) because xxis not of the form axwhere ais constant

(b) y=xx,lny=xln x,1

yy0=1+lnx,y

0=x

(1+lnx)

57. f0(x)=exe−1

115 Chapter 4

58. Let P(x0,y

0)beapointony=e

3xthen y0=e3x0.dy/dx =3e

3xso mtan =3e

0at Pand an equation

of the tangent line at Pis y−y0=3e

0(x−x

), y−e3x0=3e

0(x−x

). If the line passes through

the origin then (0,0) must satisfy the equation so −e3x0=−3x0e3x0which gives x0=1/3 and thus

y0=e. The point is (1/3,e).

59. (a) f(x)=lnx;f

0(1) = lim

h→0

ln(1 + h)−ln 1

h= lim

h→0

ln(1 + h)

h=1

x¯¯¯¯x=1

(b) f(x)=10

0(0) = lim

h→0

10h−1

h=d

dx(10x)¯¯¯¯x=0

=10

xln 10¯¯¯¯x=0

=ln10

60. (a) f(x)=lnx;f

0(e

) = lim

h→0

ln(e2+h)−2

h=d

dx(ln x)¯¯¯¯x=e2

x¯¯¯¯x=e2

=e−2

(b) f(x)=2

0(1) = lim

x→1

2x−2

x−1=d

dx(2x)¯¯¯¯x=1

xln 2¯¯¯¯x=1

= 2 ln 2

61. (b) y

1234

dx =1

2−1

xso dy

dx <0at

x= 1 and dy

dx >0atx=e

(d) The slope is a continuous function which goes from a negative value to a positive value; therefore

it must take the value zero in between, by the Intermediate Value Theorem.

(e) dy

dx = 0 when x=2

62. (a) 100

-100

-5 5

-25

-5 5

(d) x=−2,5/3

63. (a) excos πx oscillates between +exand −ex

as cos πx oscillates between −1 and +1.

(b) y

-20

123

Exercise Set 4.5 116

64. (a) 12

(b) Ptends to 12 as tgets large; lim

t→+∞P(t) = lim

t→+∞

5+7e

−t=60

5+7 lim

t→+∞e

−t=60

5=12

3.2

65. (a) 100

(b) as ttends to +∞, the population tends to 19

lim

t→+∞P(t) = lim

t→+∞

5−4e−t/4=95

5−4 lim

t→+∞e−t/4=95

5=19

-80

EXERCISE SET 4.5

1. (a) −π/2(b) π(c) −π/4(d) 0

2. (a) π/3(b) π/3(c) π/4(d) 4π/3

3. θ=−π/3; cos θ=1/2, tan θ=−√3, cot θ=−1/√3, sec θ= 2, csc θ=−2/√3

117 Chapter 4

4. θ=π/3; sin θ=√3/2, tan θ=√3, cot θ=1/

√

3, sec θ= 2, csc θ=2/

√

5. tan θ=4/3, 0 <θ<π/2; use the triangle shown to get sin θ=4/5,

cos θ=3/5, cot θ=3/4, sec θ=5/3, csc θ=5/4

6. domain range

sin−1[−1,1] [−π/2,π/2]

cos−1[−1,1] [0,π]

tan−1(−∞,+∞) (−π/2,π/2)

cot−1(−∞,+∞) (0,π)

sec−1(−∞,−1] ∪[1,+∞) [0,π/2) ∪[π, 3π/2)

csc−1(−∞,−1] ∪[1,+∞) (0,π/2] ∪(−π, −π/2]

7. (a) π/7(b) sin−1(sin π) = sin−1(sin 0) = 0

(d) Note that π/2<630 −200π<πso

sin(630) = sin(630 −200π) = sin(π−(630 −200π)) = sin(201π−630) where

0<201π−630 <π/2; sin−1(sin 630) = sin−1(sin(201π−630)) = 201π−630.

8. (a) π/7(b) π

(d) Note that −π/2<200 −64π<0 so cos(200) = cos(200 −64π) = cos(64π−200) where

0<64π−200 <π/2; cos−1(cos 200) = cos−1(cos(64π−200)) = 64π−200.

9. (a) 0≤x≤π(b) −1≤x≤1

10. Let θ= sin−1(−3/4) then sin θ=−3/4, −π/2<θ<0 and

(see ﬁgure) sec θ=4/

√

−3

÷7

11. Let θ= cos−1(3/5), sin 2θ= 2 sin θcos θ= 2(4/5)(3/5)=24/25

Exercise Set 4.5 118

12. (a) sin(cos−1x)=√

1−x

1 − x

cos−1x

(b) tan(cos−1x)=√

1−x

cos−1x

1 − x

1+x

tan−1x

1 + x

(d) sin(tan−1x)= x

√

1+x

tan−1x

1 + x

13. (a) cos(tan−1x)= 1

√

1+x

tan−1x

1 + x

(b) tan(cot−1x)= 1

cot−1x

1 + x

2−1

sec−1x

− 1

(d) cot(csc−1x)=px

2−1

csc−1x

− 1

14. (a) x−1.00 −0.80 −0.6−0.40 −0.20 0.00 0.20 0.40 0.60 0.80 1.00

sin−1x−1.57 −0.93 −0.64 −0.41 −0.20 0.00 0.20 0.41 0.64 0.93 1.57

cos−1x3.14 2.50 2.21 1.98 1.77 1.57 1.37 1.16 0.93 0.64 0.00

(b) y

-1

0.5 1

119 Chapter 4

15. (a) y

-10 10

-π/2

π/2

-5 51

(b) The domain of cot−1xis (−∞,+∞), the range is (0,π); the domain of csc−1xis

(−∞,−1] ∪[1,+∞), the range is [−π/2,0) ∪(0,π/2].

16. (a) y= cot−1x,x= cot y, tan y=1/x,y= tan−1(1/x)

(b) y= sec−1x,x= sec y, cos y=1/x,y= cos−1(1/x)

17. (a) 55.0◦(b) 33.6◦(c) 25.8◦

18. (a) x=π−sin−1(0.37) ≈2.7626 rad (b) θ= 180◦+ sin−1(0.61) ≈217.6◦

19. (a) x=π+ cos−1(0.85) ≈3.6964 rad (b) θ=−cos−1(0.23) ≈−76.7◦

20. (a) x= tan−1(3.16) −π≈−1.8773 (b) θ= 180◦−tan−1(0.45) ≈155.8◦

21. (a) 1

p1−x2/9(1/3)=1/

p9−x

2(b) −2/p1−(2x+1)

22. (a) 2x/(1+x4)(b) −1

1+xµ1

−1/2

¶=−1

2(1 + x)√x

23. (a) 1

x7√x14 −1(7x6)= 7

√

14 −1(b) −1/√e2x−1

24. (a) y=1/tan x= cot x,dy/dx =−csc2x

(b) y= (tan−1x)−1,dy/dx =−(tan−1x)−2µ1

1+x

2¶

25. (a) 1

p1−1/x2(−1/x2)=−1

|x|

√

2−1(b) sin x

√1−cos2x=sin x

|sin x|=½1,sin x>0

−1,sin x<0

26. (a) −1

(cos−1x)√1−x2(b) −1

2√cot−1x(1+x2)

27. (a) ex

x√x2−1+exsec−1x(b) 3x2(sin−1x)2

√1−x2+2x(sin−1x)3

28. (a) 0(b) 0

29. x3+xtan−1y=ey,3x

2+x

1+y

0+ tan−1y=eyy0,y0=(3x2+ tan−1y)(1 + y2)

(1+y2)ey−x

Exercise Set 4.5 120

30. sin−1(xy) = cos−1(x−y), 1

p1−x2y2(xy0+y)=−1

p1−(x−y)

2(1 −y0),

y0=yp1−(x−y)2+p1−x2y2

p1−x2y2−xp1−(x−y)2

31. (a) y

-1 1

32. (a) y

33. (a) y

−2π4π

34. (a) sin−10.9>1, so it is not in the domain of sin−1x

(b) −1≤sin−1x≤1 is necessary, or −0.841471 ≤x≤0.841471

35. (a)

-0.5 0.5

-6

y(b)

-6

36. (a) x=2π−cos−1k(b) x=π+ tan−1k

2sin−1kor x=π/2−1

2sin−1k

37. (b) θ= sin−1R

R+h= sin−16378

16,378 ≈23◦

38. (a) If γ=90

◦, then sin γ=1,p1−sin2φsin2γ=p1−sin2φ= cos φ,

D= tan φtan λ= (tan 23.55◦)(tan 65◦)≈0.934684245 so h≈21.2 hours.

(b) If γ= 270◦, then sin γ=−1, D=−tan φtan λ≈−0.934684245 so h≈2.8 hours.

39. sin 2θ=gR/v2=(9.8)(18)/(14)2=0.9, 2θ= sin−1(0.9) or 2θ= 180◦−sin−1(0.9) so

θ=1

2sin−1(0.9) ≈32◦or θ=90

◦−1

2sin−1(0.9) ≈58◦. The ball will have a lower

parabolic trajectory for θ=32

◦and hence will result in the shorter time of ﬂight.

40. 42=2

2+3

2−2(2)(3) cos θ, cos θ=−1/4, θ= cos−1(−1/4) ≈104◦

121 Chapter 4

41. y= 0 when x2= 6000v2/g,x=10v

p60/g = 1000√30 for v= 400 and g= 32;

tan θ= 3000/x =3/

√

30, θ= tan−1(3/√30) ≈29◦.

42. θ=α−β, cot α=x

a+band cot β=x

bso

θ= cot−1x

a+b−cot−1x

αβ

43. (a) Let θ= sin−1(−x) then sin θ=−x,−π/2≤θ≤π/2. But sin(−θ)=−sin θand

−π/2≤−θ≤π/2 so sin(−θ)=−(−x)=x,−θ= sin−1x,θ=−sin−1x.

(b) proof is similar to that in part (a)

44. (a) Let θ= cos−1(−x) then cos θ=−x,0≤θ≤π. But cos(π−θ)=−cos θand

0≤π−θ≤πso cos(π−θ)=x,π−θ= cos−1x,θ=π−cos−1x

(b) Let θ= sec−1(−x) for x≥1; then sec θ=−xand π/2<θ≤π.So0≤π−θ<π/2 and

π−θ= sec−1sec(π−θ) = sec−1(−sec θ) = sec−1x,orsec

−1

(−x)=π−sec−1x.

45. (a) sin−1x= tan−1x

√1−x2

÷1

− x

sin−1x

(b) sin−1x+ cos−1x=π/2; cos−1x=π/2−sin−1x=π/2−tan−1x

√1−x2

46. tan(α+β)= tan α+ tan β

1−tan αtan β,

tan(tan−1x+ tan−1y)= tan(tan−1x) + tan(tan−1y)

1−tan(tan−1x) tan(tan−1y)=x+y

1−xy

so tan−1x+ tan−1y= tan−1x+y

1−xy

47. (a) tan−11

2+ tan−11

3= tan−11/2+1/3

1−(1/2) (1/3) = tan−11=π/4

(b) 2 tan−11

3= tan−11

3+ tan−11

3= tan−11/3+1/3

1−(1/3) (1/3) = tan−13

2 tan−11

3+ tan−11

7= tan−13

4+ tan−11

7= tan−13/4+1/7

1−(3/4) (1/7) = tan−11=π/4

48. sin(sec−1x) = sin(cos−1(1/x)) = s1−µ1

x¶2

=√x2−1

|x|

Exercise Set 4.6 122

EXERCISE SET 4.6

1. (b) A=x2(c) dA

dt =2x

(d) Find dA

dt ¯¯¯¯x=3

given that dx

dt ¯¯¯¯x=3

= 2. From part (c), dA

dt ¯¯¯¯x=3

= 2(3)(2) = 12 ft2/min.

2. (b) A=πr2(c) dA

dt =2πr dr

(d) Find dA

dt ¯¯¯¯r=5

given that dr

dt ¯¯¯¯r=5

= 2. From part (c), dA

dt ¯¯¯¯r=5

=2π(5)(2) = 20πcm2/s.

3. (a) V=πr2h,so dV

dt =πµr2dh

dt +2rhdr

dt ¶.

(b) Find dV

dt ¯¯¯¯h=6,

r=10

given that dh

dt ¯¯¯¯h=6,

r=10

= 1 and dr

dt ¯¯¯¯h=6,

r=10

=−1. From part (a),

dt ¯¯¯¯h=6,

r=10

=π[102(1) + 2(10)(6)(−1)] = −20πin3/s; the volume is decreasing.

4. (a) `2=x2+y2,sod`

dt =1

`µxdx

dt +ydy

dt ¶.

(b) Find d`

dt ¯¯¯¯x=3,

y=4

given that dx

dt =1

2and dy

dt =−1

From part (a) and the fact that `= 5 when x= 3 and y=4,

dt ¯¯¯¯x=3,

y=4

5·3µ1

2¶+4µ−1

4¶¸=1

10 ft/s; the diagonal is increasing.

5. (a) tan θ=y

x,sosec

2θ

dθ

dt =

xdy

dt −ydx

x2,dθ

dt =cos2θ

x2µxdy

dt −ydx

dt ¶

(b) Find dθ

dt ¯¯¯¯x=2,

y=2

given that dx

dt ¯¯¯¯x=2,

y=2

= 1 and dy

dt ¯¯¯¯x=2,

y=2

=−1

When x= 2 and y= 2, tan θ=2/2=1soθ=π

4and cos θ= cos π

4=1

√2.Thus

from part (a), dθ

dt ¯¯¯¯x=2,

y=2

=(1/√2)2

22·2µ−1

4¶−2(1)¸=−5

16 rad/s; θis decreasing.

6. Find dz

dt ¯¯¯¯x=1,

y=2

given that dx

dt ¯¯¯¯x=1,

y=2

=−2 and dy

dt ¯¯¯¯x=1,

y=2

=3.

dt =2x

ydy

dt +3x

2dx

dt ,dz

dt ¯¯¯¯x=1,

y=2

= (4)(3) + (12)(−2) = −12 units/s; zis decreasing

7. Let Abe the area swept out, and θthe angle through which the minute hand has rotated.

Find dA

dt given that dθ

dt =π

30 rad/min; A=1

2r2θ=8θ,so dA

dt =8

dθ

dt =4π

15 in2/min.

123 Chapter 4

8. Let rbe the radius and Athe area enclosed by the ripple. We want dA

dt ¯¯¯¯t=10

given that dr

dt =3. We

know that A=πr2,so dA

dt =2πrdr

dt . Because ris increasing at the constant rate of 3 ft/s, it follows

that r= 30 ft after 10 seconds so dA

dt ¯¯¯¯t=10

=2π(30)(3) = 180πft2/s.

9. Find dr

dt ¯¯¯¯A=9

given that dA

dt = 6. From A=πr2we get dA

dt =2πr dr

dt so dr

dt =1

2πr

dt .IfA= 9 then

πr2=9,r=3/

√

πso dr

dt ¯¯¯¯A=9

2π(3/√π)(6)=1/

√

πmi/h.

10. The volume Vof a sphere of radius ris given by V=4

3πr3or, because r=D

2where D

is the diameter, V=4

3πµD

2¶3

6πD3. We want dD

dt ¯¯¯¯r=1

given that dV

dt = 3. From

V=1

6πD3we get dV

dt =1

2πD2dD

dt ,dD

dt =2

πD2

dt ,so dD

dt ¯¯¯¯r=1

π(2)2(3) = 3

2πft/min.

11. Find dV

dt ¯¯¯¯r=9

given that dr

dt =−15. From V=4

3πr3we get dV

dt =4πr2dr

dt so

dt ¯¯¯¯r=9

=4π(9)2(−15) = −4860π. Air must be removed at the rate of 4860πcm3/min.

12. Let xand ybe the distances shown in the diagram. We want to ﬁnd

dt ¯¯¯¯y=8

given that dx

dt = 5. From x2+y2=17

2we get

2xdx

dt +2ydy

dt =0,so dy

dt =−x

dt . When y=8,x

2+8

2=17

2= 289 −64 = 225, x=15so dy

dt ¯¯¯¯y=8

=−15

8(5) = −75

8ft/s; the

top of the ladder is moving down the wall at a rate of 75/8 ft/s.

13. Find dx

dt ¯¯¯¯y=5

given that dy

dt =−2. From x2+y2=13

2we get

2xdx

dt +2ydy

dt =0so dx

dt =−y

dt . Use x2+y2= 169 to ﬁnd that

x= 12 when y=5so dx

dt ¯¯¯¯y=5

=−5

12(−2) = 5

6ft/s.

14. Let θbe the acute angle, and xthe distance of the bottom of the plank from the wall. Find dθ

dt ¯¯¯¯x=2

given that dx

dt ¯¯¯¯x=2

=−1

2ft/s. The variables θand xare related by the equation cos θ=x

10 so

−sin θdθ

dt =1

dt ,dθ

dt =−1

10 sin θ

dt . When x= 2, the top of the plank is √102−22=√96 ft above

the ground so sin θ=√96/10 and dθ

dt ¯¯¯¯x=2

=−1

√96 µ−1

2¶=1

2√96 ≈0.051 rad/s.

Exercise Set 4.6 124

15. Let xdenote the distance from ﬁrst base and ythe distance from

home plate. Then x2+60

2=y

2and 2xdx

dt =2ydy

dt . When x=50

then y=10

√

61 so dy

dt =x

dt =50

10√61(25) = 125

√61 ft/s.

60 ft

Home

First

16. Find dx

dt ¯¯¯¯x=4

given that dy

dt ¯¯¯¯x=4

= 2000. From x2+5

2=y

2we get

2xdx

dt =2ydy

dt so dx

dt =y

dt . Use x2+25=y

2to ﬁnd that y=√41

when x=4so dx

dt ¯¯¯¯x=4

=√41

4(2000) = 500√41 mi/h.

Rocket

5 mi

Radar

station

17. Find dy

dt ¯¯¯¯x=4000

given that dx

dt ¯¯¯¯x=4000

= 880. From y2=x2+ 30002

we get 2ydy

dt =2x

dt so dy

dt =x

dt .Ifx= 4000, then y= 5000 so

dt ¯¯¯¯x=4000

=4000

5000(880) = 704 ft/s.

Rocket

Camera

3000 ft

18. Find dx

dt ¯¯¯¯φ=π/4

given that dφ

dt ¯¯¯¯φ=π/4

=0.2. But x= 3000 tan φso

dt = 3000(sec2φ)dφ

dt ,dx

dt ¯¯¯¯φ=π/4

= 3000 ³sec2π

4´(0.2) = 1200 ft/s.

19. (a) If xdenotes the altitude, then r−x= 3960, the radius of the Earth. θ= 0 at perigee, so

r= 4995/1.12 ≈4460; the altitude is x= 4460 −3960 = 500 miles. θ=πat apogee, so

r= 4995/0.88 ≈5676; the altitude is x= 5676 −3960 = 1716 miles.

(b) If θ= 120◦, then r= 4995/0.94 ≈5314; the altitude is 5314 −3960 = 1354 miles. The rate of

change of the altitude is given by

dt =dr

dθ

dt =4995(0.12 sin θ)

(1+0.12 cos θ)2

dθ

dt .

Use θ= 120◦and dθ/dt =2.7

◦/min = (2.7)(π/180) rad/min to get dr/dt ≈27.7 mi/min.

20. (a) Let xbe the horizontal distance shown in the ﬁgure. Then x= 4000 cot θand

dt =−4000 csc2θdθ

dt ,so dθ

dt =−sin2θ

4000

dt . Use θ=30

◦and

dx/dt = 300 mi/h = 300(5280/3600) ft/s = 440 ft/s to get

dθ/dt =−0.0275 rad/s ≈−1.6

◦/s; θis decreasing at the rate of 1.6◦/s.

(b) Let ybe the distance between the observation point and the aircraft. Then y= 4000 csc θso

dy/dt =−4000(csc θcot θ)(dθ/dt). Use θ=30

◦and dθ/dt =−0.0275 rad/s to get dy/dt ≈381

ft/s.

125 Chapter 4

21. Find dh

dt ¯¯¯¯h=16

given that dV

dt = 20. The volume of water in the tank

at a depth his V=1

3πr2h. Use similar triangles (see ﬁgure) to get

h=10

24 so r=5

12hthus V=1

3πµ5

12h¶2

h=25

432πh3,

dt =25

144πh2dh

dt ;dh

dt =144

25πh2

dt ,dh

dt ¯¯¯¯h=16

=144

25π(16)2(20) = 9

20π

ft/min.

22. Find dh

dt ¯¯¯¯h=6

given that dV

dt =8. V=1

πr2h, but r=1

2hso

V=1

3πµh

2¶2

h=1

12πh3,dV

dt =1

4πh2dh

dt ,dh

dt =4

πh2

dt ,

dt ¯¯¯¯h=6

π(6)2(8) = 8

9πft/min.

23. Find dV

dt ¯¯¯¯h=10

given that dh

dt =5. V=1

πr2h, but r=1

2hso

V=1

3πµh

2¶2

h=1

12πh3,

dt =1

4πh2dh

dt ,dV

dt ¯¯¯¯h=10

4π(10)2(5) = 125πft3/min.

24. Let rand hbe as shown in the ﬁgure. If Cis the circumference of

the base, then we want to ﬁnd dC

dt ¯¯¯¯h=8

given that dV

dt = 10. It is

given that r=1

2h,thusC=2πr =πh so

dt =πdh

dt (1)

Use V=1

3πr2h=1

12πh3to get dV

dt =1

4πh2dh

dt ,so

dt =4

πh2

dt (2)

Substitution of (2) into (1) gives dC

dt =4

dt so

dt ¯¯¯¯h=8

64(10) = 5

8ft/min.

25. With sand has shown in the ﬁgure, we want to ﬁnd dh

dt given that

dt = 500. From the ﬁgure, h=ssin 30◦=1

2sso

dt =1

2(500) = 250 mi/h.

Ground

30°

Exercise Set 4.6 126

26. Find dx

dt ¯¯¯¯y=125

given that dy

dt =−20. From x2+10

2=y

2we get

2xdx

dt =2ydy

dt so dx

dt =y

dt . Use x2+ 100 = y2to ﬁnd that

x=√15,525=15

√

69 when y= 125 so

dt ¯¯¯¯y=125

=125

15√69(−20) = −500

3√69. The boat is approaching the

dock at the rate of 500

3√69 ft/min.

xBoat

Pulley

27. Find dy

dt given that dx

dt ¯¯¯¯y=125

=−12. From x2+10

2=y

2we get

2xdx

dt =2ydy

dt so dy

dt =x

dt . Use x2+ 100 = y2to ﬁnd that

x=p15,525=15

√

69 when y= 125 so

dt =15√69

125 (−12) = −36√69

25 . The rope must be pulled at the rate

of 36√69

25 ft/min.

xBoat

Pulley

28. (a) Let xand ybe as shown in the ﬁgure. It is required to ﬁnd

dt , given that dy

dt =−3. By similar triangles, x

6=x+y

18 ,

18x=6x+6y,12x=6y,x=1

y,so

dt =1

2(−3) = −3

2ft/s. 6

Man

Shadow

Light

(b) The tip of the shadow is z=x+yfeet from the street light, thus the rate at which it is

moving is given by dz

dt =dx

dt +dy

dt . In part (a) we found that dx

dt =−3

2when dy

dt =−3so

dt =(−3/2)+(−3) = −9/2 ft/s; the tip of the shadow is moving at the rate of 9/2 ft/s toward

the street light.

29. Find dx

dt ¯¯¯¯θ=π/4

given that dθ

dt =2π

10 =π

5rad/s. Then

x= 4 tan θ(see ﬁgure) so dx

dt = 4 sec2θdθ

dt ,

dt ¯¯¯¯θ=π/4

=4³sec2π

4´³π

5´=8π/5 km/s.

Ship

127 Chapter 4

30. If x,y, and zare as shown in the ﬁgure, then we want dz

dt ¯¯¯¯x=2,

y=4

given

that dx

dt =−600 and dy

dt ¯¯¯¯x=2,

y=4

=−1200. But z2=x2+y2so

2zdz

dt =2x

dt +2ydy

dt ,dz

dt =1

zµxdx

dt +ydy

dt ¶. When x= 2 and

y=4,z

2=2

2+4

2= 20, z=√20=2

√

5so

dt ¯¯¯¯x=2,

y=4

2√5[2(−600) + 4(−1200)] = −3000

√5=−600√5 mi/h; the

distance between missile and aircraft is decreasing at the rate of

600√5 mi/h.

Aircraft

Missile

31. We wish to ﬁnd dz

dt ¯¯¯¯x=2,

y=4

given dx

dt =−600 and dy

dt ¯¯¯¯x=2,

y=4

=−1200 (see

ﬁgure). From the law of cosines,

z2=x2+y2−2xy cos 120◦=x2+y2−2xy(−1/2) = x2+y2+xy,so

dt =2x

dt +2ydy

dt +xdy

dt +ydx

dt ,

dt =1

2z·(2x+y)dx

dt +(2y+x)dy

dt ¸. When x= 2 and y=4,

2=2

2+4

2+ (2)(4) = 28, so z=√28=2

√

7, thus

dt ¯¯¯¯x=2,

y=4

2(2√7)[(2(2) + 4)(−600) + (2(4) + 2)(−1200)] = −4200

√7=

−600√7 mi/h; the distance between missile and aircraft is

decreasing at the rate of 600√7 mi/h.

Aircraft

Missile

120°

32. (a) Let x,y, and zbe the distances shown in the ﬁrst ﬁgure. Find dz

dt ¯¯¯¯x=2,

y=0

given that dx

dt =−75 and

dt =−100. In order to ﬁnd an equation relating x,y, and z, ﬁrst draw the line segment that

joins the point Pto the car, as shown in the second ﬁgure. Because triangle OPC is a right

triangle, it follows that PC has length px2+(1/2)2; but triangle HPC is also a right triangle

so z2=³px2+(1/2)2´2+y2=x2+y2+1/4 and 2zdz

dt =2x

dt +2ydy

dt +0,

dt =1

zµxdx

dt +ydy

dt ¶. Now, when x= 2 and y=0,z

2= (2)2+ (0)2+1/4=17/4, z=√17/2

so dz

dt ¯¯¯¯x=2,

y=0

(√17/2)[2(−75)+0(−100)] = −300/√17 mi/h

West East

North

Car

Helicopter

(b) decreasing, because dz

dt <0.

Exercise Set 4.6 128

33. (a) We want dy

dt ¯¯¯¯x=1,

y=2

given that dx

dt ¯¯¯¯x=1,

y=2

= 6. For convenience, ﬁrst rewrite the equation as

xy3=8

5+8

5y2then 3xy2dy

dt +y3dx

dt =16

5ydy

dt ,dy

dt =y3

5y−3xy2

dt so

dt ¯¯¯¯x=1,

y=2

=23

5(2) −3(1)22

(6) = −60/7 units/s.

(b) falling, because dy

dt <0

34. Find dx

dt ¯¯¯¯(2,5)

given that dy

dt ¯¯¯¯(2,5)

= 2. Square and rearrange to get x3=y2−17

so 3x2dx

dt =2ydy

dt ,dx

dt =2y

3x2

dt ,dx

dt ¯¯¯¯(2,5)

=µ5

6¶(2) = 5

3units/s.

35. The coordinates of Pare (x, 2x), so the distance between Pand the point (3,0) is

D=p(x−3)2+(2x−0)2=√5x2−6x+ 9. Find dD

dt ¯¯¯¯x=3

given that dx

dt ¯¯¯¯x=3

=−2.

dt =5x−3

√5x2−6x+9

dt ,so dD

dt ¯¯¯¯x=3

=12

√36(−2) = −4 units/s.

36. (a) Let Dbe the distance between Pand (2,0). Find dD

dt ¯¯¯¯x=3

given that dx

dt ¯¯¯¯x=3

=4.

D=p(x−2)2+y2=p(x−2)2+x=√x2−3x+4 so dD

dt =2x−3

2√x2−3x+4;

dt ¯¯¯¯x=3

2√4=3

4units/s.

(b) Let θbe the angle of inclination. Find dθ

dt ¯¯¯¯x=3

given that dx

dt ¯¯¯¯x=3

=4.

tan θ=y

x−2=√x

x−2so sec2θdθ

dt =−x+2

√

x(x−2)2

dt ,dθ

dt =−cos2θx+2

√

x(x−2)2

dt .

When x=3,D= 2 so cos θ=1

2and dθ

dt ¯¯¯¯x=3

=−1

2√3(4) = −5

2√3rad/s.

37. Solve dy

dt =3

dt given y=xln x. Then dy

dt =dy

dt =(1+lnx)dx

dt ,so1+lnx=3,lnx=2,x=e

38. 32xdx

dt +18ydy

dt =0;ifdy

dt =dx

dt 6= 0, then (32x+18y)

dt =0,32x+18y=0,y=−

9xso

16x2+9256

81 x2= 144, 400

9x2= 144, x2=81

25,x=±9

5.Ifx=9

, then y=−16

5=−16

5. Similarly,

if x=−9

5, then y=16

5. The points are (9

5,−16

5) and (−9

5,16

5).

39. Find dS

dt ¯¯¯¯s=10

given that ds

dt ¯¯¯¯s=10

=−2. From 1

s+1

S=1

6we get −1

dt −1

dt =0,so

dt =−S2

dt .Ifs= 10, then 1

10 +1

S=1

6which gives S= 15. So dS

dt ¯¯¯¯s=10

=−225

100(−2)=4.5 cm/s.

The image is moving away from the lens.

129 Chapter 4

40. Suppose that the reservoir has height Hand that the radius at the top is R. At any instant of time

let hand rbe the corresponding dimensions of the cone of water (see ﬁgure). We want to show that

dt is constant and independent of Hand R, given that dV

dt =−kA where Vis the volume of water,

Ais the area of a circle of radius r, and kis a positive constant. The volume of a cone of radius rand

height his V=1

3πr2h. By similar triangles r

h=R

H,r=R

Hhthus V=1

3πµR

H¶2

h3so

dt =πµR

H¶2

h2dh

dt (1)

But it is given that dV

dt =−kA or, because A=πr2=πµR

H¶

h2,

dt =−kπ µR

H¶

h2, which when substituted into equation (1) gives

−kπ µR

H¶

h2=πµR

H¶

h2dh

dt ,dh

dt =−k.

41. Let rbe the radius, Vthe volume, and Athe surface area of a sphere. Show that dr

dt is a constant

given that dV

dt =−kA, where kis a positive constant. Because V=4

3πr3,

dt =4πr2dr

dt (1)

But it is given that dV

dt =−kA or, because A=4πr2,dV

dt =−4πr2kwhich when substituted into

equation (1) gives −4πr2k=4πr2dr

dt ,dr

dt =−k.

42. Let xbe the distance between the tips of the minute and hour hands, and αand βthe angles shown

in the ﬁgure. Because the minute hand makes one revolution in 60 minutes,

dα

dt =2π

60 =π/30 rad/min; the hour hand makes one revolution in 12 hours (720 minutes), thus

dβ

dt =2π

720 =π/360 rad/min. We want to ﬁnd dx

dt ¯¯¯¯α=2π,

β=3π/2

given that dα

dt =π/30 and dβ

dt =π/360.

Using the law of cosines on the triangle shown in the ﬁgure,

x2=3

2+4

2−2(3)(4) cos(α−β)=25−24 cos(α−β), so

2xdx

dt = 0 + 24 sin(α−β)µdα

dt −dβ

dt ¶,

dt =12

xµdα

dt −dβ

dt ¶sin(α−β). When α=2πand β=3π/2,

x2=25−24 cos(2π−3π/2) = 25, x=5;so

dt ¯¯¯¯α=2π,

β=3π/2

=12

5(π/30 −π/360) sin(2π−3π/2) = 11π

150 in/min. 3

43. Extend sides of cup to complete the cone and let V0be the volume

of the portion added, then (see ﬁgure) V=1

3πr2h−V0where

h=4

12 =1

3so r=1

3hand V=1

3πµh

3¶2

h−V0=1

27πh3−V0,

dt =1

9πh2dh

dt ,dh

dt =9

πh2

dt ,dh

dt ¯¯¯¯h=9

π(9)2(20) = 20

9πcm/s.

Exercise Set 4.7 130

EXERCISE SET 4.7

1. (a) lim

x→2

x2−4

x2+2x−8= lim

x→2

(x−2)(x+2)

(x+ 4)(x−2) = lim

x→2

x+2

x+4 =2

(b) lim

x→+∞

2x−5

3x+7 =

2−lim

x→+∞

3 + lim

x→+∞

2. (a) sin x

tan x= sin xcos x

sin x= cos xso lim

x→0

sin x

tan x=0

(b) x2−1

x3−1=(x−1)(x+1)

(x−1)(x2+x+1) =x+1

2+x+1 so lim

x→1

x2−1

x3−1=2

3. lim

x→1

1/x

1=1 4. lim

x→0

2 cos 2x

5 cos 5x=2/5

5. lim

x→0

cos x=1 6. lim

x→3

6x−13 =1/5

7. lim

θ→0

sec2θ

1=1 8. lim

t→0

tet+et

−et=−1

9. lim

x→π+

cos x

1=−110. lim

x→0+

cos x

2x=+∞

11. lim

x→+∞

1/x

1=0 12. lim

x→+∞

3e3x

2x= lim

x→+∞

9e3x

2=+∞

13. lim

x→0+−csc2x

1/x = lim

x→0+−x

sin2x= lim

x→0+−1

2 sin xcos x=−∞

14. lim

x→0+−1/x

(−1/x2)e1/x = lim

x→0+

e1/x =0

15. lim

x→+∞

100x99

ex= lim

x→+∞

(100)(99)x98

ex=···= lim

x→+∞

(100)(99)(98) ···(1)

ex=0

16. lim

x→0+

cos x/ sin x

sec2x/ tan x= lim

x→0+cos2x=1 17. lim

x→0

2/√1−4x2

1=2

18. lim

x→0

1−1

1+x

2= lim

x→0

3(1 + x2)=1

319. lim

x→+∞xe−x= lim

x→+∞

ex= lim

x→+∞

ex=0

20. lim

x→π(x−π) tan(x/2) = lim

x→π

x−π

cot(x/2) = lim

x→π

−(1/2) csc2(x/2) =−2

21. lim

x→+∞xsin(π/x) = lim

x→+∞

sin(π/x)

1/x = lim

x→+∞

(−π/x2) cos(π/x)

−1/x2= lim

x→+∞πcos(π/x)=π

22. lim

x→0+tan xln x= lim

x→0+

ln x

cot x= lim

x→0+

1/x

−csc2x= lim

x→0+−sin2x

x= lim

x→0+−2 sin xcos x

1=0

23. lim

x→(π/2)−sec 3xcos 5x= lim

x→(π/2)−

cos 5x

cos 3x= lim

x→(π/2)−−5 sin 5x

−3 sin 3x=−5(+1)

(−3)(−1) =−5

131 Chapter 4

24. lim

x→π(x−π) cot x= lim

x→π

x−π

tan x= lim

x→π

sec2x=1

25. y=(1−3/x)x, lim

x→+∞ln y= lim

x→+∞

ln(1 −3/x)

1/x = lim

x→+∞−3

1−3/x =−3, lim

x→+∞y=e−3

26. y=(1+2x)

−3/x, lim

x→0ln y= lim

x→0−3 ln(1 + 2x)

x= lim

x→0−6

1+2x=−6, lim

x→0y=e−6

27. y=(e

x+x)

1/x, lim

x→0ln y= lim

x→0

ln(ex+x)

x= lim

x→0

ex+1

x+x= 2, lim

x→0y=e2

28. y=(1+a/x)bx, lim

x→+∞ln y= lim

x→+∞

bln(1 + a/x)

1/x = lim

x→+∞

1+a/x =ab, lim

x→+∞y=eab

29. y=(2−x)

tan(πx/2), lim

x→1ln y= lim

x→1

ln(2 −x)

cot(πx/2) = lim

x→1

2 sin2(πx/2)

π(2 −x)=2/π, lim

x→1y=e2/π

30. y= [cos(2/x)]x2,lim

x→+∞ln y= lim

x→+∞

ln cos(2/x)

1/x2= lim

x→+∞

(−2/x2)(−tan(2/x))

−2/x3= lim

x→+∞−tan(2/x)

1/x

= limx→+∞

(2/x2) sec2(2/x)

−1/x2=−2,lim

x→+∞y=e−2

31. lim

x→0µ1

sin x−1

x¶= lim

x→0

x−sin x

xsin x= lim

x→0

1−cos x

xcos x+ sin x= lim

x→0

sin x

2 cos x−xsin x=0

32. lim

x→0

1−cos 3x

x2= lim

x→0

3 sin 3x

2x= lim

x→0

2cos 3x=9

33. lim

x→+∞

(x2+x)−x2

√x2+x+x= lim

x→+∞

√x2+x+x= lim

x→+∞

p1+1/x +1 =1/2

34. lim

x→0

ex−1−x

xex−x= lim

x→0

ex−1

xex+ex−1= lim

x→0

xex+2e

x=1/2

35. lim

x→+∞[x−ln(x2+ 1)] = lim

x→+∞[ln ex−ln(x2+ 1)] = lim

x→+∞ln ex

x2+1,

lim

x→+∞

x2+1 = lim

x→+∞

2x= lim

x→+∞

2=+∞so lim

x→+∞[x−ln(x2+1)]=+∞

36. lim

x→+∞ln x

1+x= lim

x→+∞ln 1

1/x +1 = ln(1) = 0

38. (a) lim

x→+∞

ln x

xn= lim

x→+∞

1/x

nxn−1= lim

x→+∞

nxn=0

(b) lim

x→+∞

ln x= lim

x→+∞

nxn−1

1/x = lim

x→+∞nxn=+∞

39. (a) L’Hˆopital’s Rule does not apply to the problem lim

x→1

3x2−2x+1

2−2xbecause it is not a 0

0form

(b) lim

x→1

3x2−2x+1

2−2x=2

40. lim

x→1

4x3−12x2+12x−4

3−9x

2+6x−1= lim

x→1

12x2−24x+12

12x2−18x+6 = lim

x→1

24x−24

24x−18 =0

Exercise Set 4.7 132

41. lim

x→+∞

1/(xln x)

1/(2√x)= lim

x→+∞

√xln x=0 0.15

100 10000

42. y=xx, lim

x→0+ln y= lim

x→0+

ln x

1/x = lim

x→0+−x= 0, lim

x→0+y=1 1

0 0.5

43. y= (sin x)3/ln x, lim

x→0+ln y= lim

x→0+

3lnsinx

ln x= lim

x→0+(3 cos x)x

sin x=3,

lim

x→0+y=e3

0 0.5

44. lim

x→π/2−

4 sec2x

sec xtan x= lim

x→π/2−

sin x=4 4

0.5 p/2

45. ln x−ex=lnx−1

−x=e

−xln x−1

e−x;

lim

x→+∞e−xln x= lim

x→+∞

ln x

ex= lim

x→+∞

1/x

ex= 0 by L’Hˆopital’s Rule, so

lim

x→+∞[ln x−ex] = lim

x→+∞

e−xln x−1

e−x=−∞

-16

46. lim

x→+∞[ln ex−ln(1 + 2ex)] = lim

x→+∞ln ex

1+2e

= lim

x→+∞ln 1

e−x+2 =ln1

;

horizontal asymptote y=−ln 2

-0.6

-1.2

012

133 Chapter 4

47. y= (ln x)1/x, lim

x→+∞ln y= lim

x→+∞

ln(ln x)

x= lim

x→+∞

xln x=0;

lim

x→+∞y=1,y= 1 is the horizontal asymptote

1.02

100 10000

48. y=µx+1

x+2¶

lim

x→+∞ln y= lim

x→+∞

ln x+1

x+2

1/x = lim

x→+∞−x2

(x+ 1)(x+2) =−1;

lim

x→+∞y=e−1is the horizontal asymptote

050

49. (a) 0(b) +∞(c) 0(d) −∞ (e) +∞(f ) −∞

50. (a) y=xln a

1+ln x; lim

x→0+ln y= lim

x→0+

(ln a)lnx

1+lnx= lim

x→0+

(ln a)/x

1/x = lim

x→0+ln a=lna, lim

x→0+y=eln a=a

(b) same as part (a) with x→+∞

x, lim

x→0ln y= lim

x→0

(ln a) ln(x+1)

x= lim

x→0

ln a

x+1 =lna, lim

x→0y=eln a=a

51. lim

x→+∞

1 + 2 cos 2x

1does not exist, nor is it ±∞; lim

x→+∞

x+ sin 2x

x= lim

x→+∞µ1+sin 2x

x¶=1

52. lim

x→+∞

2−cos x

3 + cos xdoes not exist, nor is it ±∞; lim

x→+∞

2x−sin x

3x+ sin x= lim

x→+∞

2−(sin x)/x

3 + (sin x)/x =2

53. lim

x→+∞(2 + xcos 2x+ sin 2x) does not exist, nor is it ±∞; lim

x→+∞

x(2 + sin 2x)

x+1 = lim

x→+∞

2 + sin 2x

1+1/x , which

does not exist because sin 2xoscillates between −1and1asx→+∞

54. lim

x→+∞µ1

x+1

2cos x+sin x

2x¶does not exist, nor is it ±∞;

lim

x→+∞

x(2 + sin x)

x2+1 = lim

x→+∞

2 + sin x

x+1/x =0

55. lim

R→0+

−Rt/L

1=Vt

56. (a) lim

x→π/2(π/2−x) tan x= lim

x→π/2

π/2−x

cot x= lim

x→π/2−1

−csc2x= lim

x→π/2sin2x=1

(b) lim

x→π/2µ1

π/2−x−tan x¶= lim

x→π/2µ1

π/2−x−sin x

cos x¶= lim

x→π/2

cos x−(π/2−x) sin x

(π/2−x) cos x

= lim

x→π/2−(π/2−x) cos x

−(π/2−x) sin x−cos x

= lim

x→π/2

(π/2−x) sin x+ cos x

−(π/2−x) cos x+ 2 sin x=0

Supplementary Exercises 4 134

1/(π/2−1.57) −tan 1.57 ≈0.000265

57. (b) lim

x→+∞x(k1/x −1) = lim

t→0+

kt−1

t= lim

t→0+

(ln k)kt

1=lnk

√0.3−1¢=−1.20327; ln 2 = 0.69315, 1024 ¡1024

√2−1¢=0.69338

58. (a) No; sin(1/x) oscillates as x→0. (b) 0.05

-0.05

-0.35 0.35

x2sin(1/x)≤x2.Forx→0

−do the same; thus lim

x→0f(x)=0.

59. If k6=−1 then lim

x→0(k+ cos `x)=k+16= 0, so lim

x→0

k+ cos `x

x2=±∞. Hence k=−1, and by the rule

lim

x→0−1 + cos `x

x2= lim

x→0−`sin `x

2x= lim

x→0−`2cos `x

2=−`2

2=4if`=±2

√

60. (a) Apply the rule to get lim

x→0−cos(1/x)+2xsin(1/x)

cos xwhich does not exist (nor is it ±∞).

(b) Rewrite as lim

x→0hx

sin xi[xsin(1/x)], but lim

x→0

sin x= lim

x→0

cos x= 1 and lim

x→0xsin(1/x)=0,thus

lim

x→0hx

sin xi[xsin(1/x)] = (1)(0) = 0

61. lim

x→0+

sin(1/x)

(sin x)/x, lim

x→0+

sin x

x= 1 but lim

x→0+sin(1/x) does not exist because sin(1/x) oscillates between −1

and1asx→+∞, so lim

x→0+

xsin(1/x)

sin xdoes not exist.

CHAPTER 4 SUPPLEMENTARY EXERCISES

1. (a) f(g(x)) = xfor all xin the domain of g, and g(f(x)) = xfor all xin the domain of f.

(b) They are reﬂections of each other through the line y=x.

(d) The equation y=f(x) can always be solved for xas a function of y. Functions with no inverses

include y=x2,y= sin x.

(e) Yes, gis continuous; this is evident from the statement about the graphs in part (b) above.

(f) Yes, gmust be diﬀerentiable (where f06= 0); this can be inferred from the graphs. Note that if

f0= 0 at a point then g0cannot exist (inﬁnite slope).

2. (a) For sin x,−π/2≤x≤π/2; for cos x,0≤x≤π; for tan x,−π/2<x<π/2; for sec x,

0≤x<π/2orπ/2<x≤π.

135 Chapter 4

(b) y

-1

π/2

y = sin-1x

y = sinx

-1

y = cos-1x

y = cosx

-2

π/2

-π/2

y = tan-1x

y = tanx

-1

π/2

y = sec-1xy = sec x

y = sec-1x

y = secx

3. (a) when the limit takes the form 0/0or∞/∞

(b) Not necessarily; only if lim f(x) = 0. Consider g(x)=x; lim

x→0g(x)=0. Forf(x) choose

cos x,x2, and |x|1/2. There are three possibilities; lim

x→0

cos x

xdoes not exist; lim

x→0

x= 0, and

lim

x→0|x|1/2

x2=+∞.

4. In the case +∞−(−∞) the limit is +∞; in the case −∞−(+∞) the limit is −∞, because large positive

(negative) quantities are being added to large positive (negative) quantities. The cases +∞−(+∞)

and −∞−(−∞) are indeterminate; large numbers of opposite sign are being subtracted, and more

information about the sizes is needed.

5. (a) x=f(y)=8y

3−1; y=f−1(x)=µx+1

8¶1/3

2(x+1)

1/3

(b) f(x)=(x−1)2;fdoes not have an inverse because fis not one-to-one, for example

f(0) = f(2)=1.

)

2+1;y=f−1

(x)=ln√

x−1=1

2ln(x−1)

(d) x=f(y)=y+2

y−1;y=f−1

(x)= x+2

x−1

6. f0(x)= ad −bc

(cx +d)2;ifad−bc = 0 then the function represents a horizontal line, no inverse. If ad−bc 6=0

then f0(x)>0orf

(x)<0sofis invertible. If x=f(y)= ay +b

cy +dthen y=f−1(x)= b−xd

xc −a.

7. (a) Diﬀerentiating, 2

3x−1/3−2

3y−1/3y0−y0=0. Atx= 1 and y=−1, y0= 2. The tangent line is

y+1=2(x−1).

(b) (xy0+y) cos xy =y0. With x=π/2 and y= 1 this becomes y0= 0, so the equation of the

tangent line is y−1=0(x−π/2) or y=1.

8. Draw equilateral triangles of sides 5, 12, 13, and 3, 4, 5. Then sin[cos−1(4/5)]=3/5,

sin[cos−1(5/13)] = 12/13, cos[sin−1(4/5)]=3/5, cos[sin−1(5/13)] = 12/13

(a) cos[cos−1(4/5) + sin−1(5/13)] = cos(cos−1(4/5)) cos(sin−1(5/13)) −sin(cos−1(4/5)) sin(sin−1(5/13))

13 −3

13 =33

65.

Supplementary Exercises 4 136

(b) sin[sin−1(4/5) + cos−1(5/13)] = sin(sin−1(4/5)) cos(cos−1(5/13)) + cos(sin−1(4/5)) sin(cos−1(5/13))

13 +3

13 =56

65.

9. 3ln¡e

2x(e

3¢+ 2 exp(ln 1) = 3 ln e2x+ 3 ln(ex)3+2·1 = 3(2x)+(3·3)x+2=15x+2

10. Y= ln(Cekt)=lnC+lne

kt =lnC+kt, a line with slope kand Y-intercept ln C

11. (a) lim

x→+∞(ex−x2) = lim

x→+∞x2(ex/x2−1), but lim

x→+∞

x2= lim

x→+∞

2x= lim

x→+∞

2=+∞

so lim

x→+∞(ex/x2−1)=+∞and thus lim

x→+∞x2(ex/x2−1)=+∞

(b) lim

x→1

ln x

x4−1= lim

x→1

1/x

4x3=1

4; lim

x→1rln x

x4−1=rlim

x→1

ln x

x4−1=1

x→0axln a=lna

12. y0=aeax sin bx +beax cos bx and y00 =(a

2−b

ax sin bx +2abeax cos bx,soy

00 −2ay0+(a

2+b

=(a

2−b

ax sin bx +2abeax cos bx −2a(aeax sin bx +beax cos bx)+(a

2+b

ax sin bx =0.

13. sin(tan−1x)=x/√1+x

2and cos(tan−1x)=1/

√

1+x

, and y0=1

1+x

2,y

00 =−2x

(1+x2)2, hence

y00 + 2 sin ycos3y=−2x

(1+x2)2+2 x

√

1+x

(1+x2)3/2=0.

14. ln y=2xln3+7xln 5; dy

dx/y = 2 ln 3 + 7 ln 5, or dy

dx = (2 ln 3 + 7 ln 5)y

15. Find dθ

dt ¯¯¯¯x=1

y=1

given dz

dt =aand dy

dt =−b. From the ﬁgure

sin θ=y/z; when x=y=1,z=√

2. So θ= sin−1(y/z) and

dθ

dt =1

p1−y2/z2µ1

dt −y

dt ¶=−b−a

√2when x=y=1.

16. (a) f0(x)=−3/(x+1)

.Ifx=f(y)=3/(y+ 1) then y=f−1(x)=(3/x)−1, so d

dxf−1(x)= −3

and 1

f0(f−1(x)) =−(f−1(x)+1)

3=−(3/x)2

3=−3

x2.

(b) f(x)=e

x/2,f0(x)=1

x/2.Ifx=f(y)=e

y/2then y=f−1(x)=2lnx,so d

dxf−1(x)= 2

; and

f0(f−1(x)) =2e

−f

−1

(x)/2=2e

−ln x=2x

−1=2

17. (a) y

-2

(b) The curve y=e−x/2sin 2xhas x-intercepts

at x=0,π/2. It intersects the curve y=

e−x/2at x=π/4, and it intersects the curve

y=−e−x/2at x=−π/4,3π/4.

137 Chapter 4

18. (a) y

π/2

(b) y

π/2

−π/2

(d) y

π/2

19. (a) ln y=ln(1 + x)

x,y0

y=x/(1+x)−ln(1 + x)

x2=1

x(1+x)−ln(1 + x)

x2,

dx =1

x(1+x)(1/x)−1−(1+x)(1/x)

x2ln(1 + x)

(b) ln y=exln x,y0

y=exµ1

x+lnx

¶,dy

dx =xexexµ1

x+lnx

¶=e

x£x

x−1+x

xln x¤

0=3x

(d) y0=abe−x

(1+be−x)2

(e) 2

3xy−1/3dy

dx +y2/3+2

3yx−1/3+x2/3dy

dx =2x. Multiply by 3x1/3y1/3:

2x4/3dy

dx +3x

1/3

y+2y

4/3+3xy1/3dy

dx =6x

4/3

1/3

. Regroup:

dx ³2x4/3+3xy1/3´=6x

4/3

1/3−3x

1/3

y−2y

4/3

,dy

dx =6x4/3y1/3−3x1/3y−2y4/3

2x4/3+3xy1/3.

(f) y=1

2ln x+1

3ln(x+1)−ln sin x+ ln cos x,so

0=1

2x+1

3(x+1) −cos x

sin x−sin x

cos x=5x+3

6x(x+1) −cot x−tan x.

20. (a) Find xwhen y=5·12 = 60 in. Since y= log x,x=10

y=10

60 in. This is approximately

2.68 ×1042 light-years, so even in astronomical terms it is a fabulously long distance.

(b) Find xwhen y= 100(5280)(12) in. Since y=10

,x= log y=6.80 in or 0.57 ft, approximately.

21. (a) The function ln x−x0.2is negative at x= 1 and positive at x= 4, so it must be zero in between

(IVT).

(b) x=3.654

Supplementary Exercises 4 138

22. (a) If xk=exthen kln x=x,or ln x

x=1

k. The steps are reversible.

(b) By zooming it is seen that the maximum value of yis

approximately 0.368 (actually, 1/e), so there are two distinct

solutions of xk=exwhenever k>1/0.368 ≈2.717.

-2

23. Set y= logbxand solve y0=1: y

0=1

xln b=1sox=1

ln b. The

curves intersect when (x, x) lies on the graph of y= logbx,so

x= logbx. From Formula (9), Section 4.2, logbx=ln x

ln bfrom which

ln x=1,x=e,lnb=1/e,b=e1/e ≈1.4447.

24. (a) Find the point of intersection: f(x)=√

x+k=lnx. The

slopes are equal, so m1=1

x=m2=1

2√x,√x=2,x=4.

Then ln 4 = √4+k,k=ln4−2.

(b) Since the slopes are equal m1=k

2√x=m2=1

x,sok

√

x=2.

At the point of intersection k√x=lnx,2=lnx,x=e

k=2/e.

CHAPTER 5

Analysis of Functions and Their Graphs

EXERCISE SET 5.1

1. (a) f0>0 and f00 >0

(b) f0>0 and f00 <0

(d) f0<0 and f00 <0

2. (a) y

(b) y

(d) y

3. A:dy/dx < 0,d

y/dx2>0

B:dy/dx > 0,d

y/dx2<0

C:dy/dx < 0,d

y/dx2<0

4. A:dy/dx < 0,d

y/dx2<0

B:dy/dx < 0,d

y/dx2>0

C:dy/dx > 0,d

y/dx2<0

139

Exercise Set 5.1 140

5. An inﬂection point occurs when f00 changes sign: at x=−1,0,1 and 2.

6. (a) f(0) <f(1) since f0>0on(0,1). (b) f(1) >f(2) since f0<0on(1,2).

(e) f00(0) <0 since f0is decreasing there. (f) f00(2) = 0 since f0has a minimum there.

7. (a) [4,6] (b) [1,4] and [6,7] (c) (1,2) and (3,5)

(d) (2,3) and (5,7) (e) x=2,3,5

8. (1,2) (2,3) (3,4) (4,5) (5,6) (6,7)

f0− − − ++−

f00 +−+ + − −

9. f0(x)=2x−5

00(x)=2

(a) [5/2,+∞)(b) (−∞,5/2]

(e) none

10. f0(x)=−2(x+3/2)

f00(x)=−2

(a) (−∞,−3/2] (b) [−3/2,+∞)

(e) none

11. f0(x)=3(x+2)

f00(x)=6(x+2)

(a) (−∞,+∞)(b) none

(e) −2

12. f0(x) = 3(4 −x2)

f00(x)=−6x

(a) [−2,2] (b) (−∞,−2], [2,+∞)

(e) 0

13. f0(x)=12x

(x−1)

f00(x)=36x(x−2/3)

(a) [1,+∞)(b) (−∞,1]

(e) 0,2/3

14. f0(x)=4x(x

2−4)

f00(x) = 12(x2−4/3)

(a) [−2,0], [2,+∞)(b) (−∞,−2], [0,2]

(e) −2/√3, 2/√3

15. f0(x)= 4x

2+2)

2f00(x)=−43x

2−2

2+2)

(a) [0,+∞)(b) (−∞,0] (c) (−p2/3,+p2/3)

(d) (−∞,−p2/3), (+p2/3,+∞)(e) −p2/3,p2/3

16. f0(x)= 2−x

2+2)

2f00(x)=2x(x

2−6)

(x2+2)

(a) [−√2,√2] (b) (−∞,−√2], [√2,+∞)(c) (−√6,0), (√6,+∞)

(d) (−∞,−√6), (0,√6) (e) −√6,0,√6

17. f0(x)=1

3(x+2)

−2/3

f00(x)=−2

9(x+2)

−5/3

(a) (−∞,+∞)(b) none

(e) −2

141 Chapter 5

18. f0(x)=2

−1/3

00(x)=−2

−4/3

(a) [0,+∞)(b) (−∞,0]

(e) none

19. f0(x)=4(x+1)

2/3

f00(x)=4(x−2)

9x5/3

(a) [−1,+∞)(b) (−∞,−1]

(e) 0,2

20. f0(x)=4(x−1/4)

3x2/3

f00(x)=4(x+1/2)

9x5/3

(a) [1/4,+∞)(b) (−∞,1/4]

(e) −1/2,0

21. f0(x)=−xe−x2/2

f00(x)=(−1+x

−x

2/2

(a) (−∞,0] (b) [0,+∞)

(e) −1,1

22. f0(x)=(2x

2+1)e

f00(x)=2x(2x2+3)e

(a) (−∞,+∞)(b) none

(e) 0

23. f0(x)= 2x

1+x

f00(x)=2 1−x

(1+x2)2

(a) [0,+∞)(b) (−∞,0]

(e) −1,1

24. f0(x)=x(2 ln x+1)

f00(x)=2lnx+3

(a) [e−1/2,+∞)(b) (0,e

−1/2]

−3/2)

(e) e−3/2

25. f0(x)=−sin x

f00(x)=−cos x

(a) [π, 2π](b) [0,π]

(e) π/2, 3π/2

-1

02p

26. f0(x) = 2 sin 4x

f00(x) = 8 cos 4x

(a) (0,π/4], [π/2,3π/4] (b) [π/4,π/2], [3π/4,π]

(e) π/8, 3π/8, 5π/8, 7π/8

Exercise Set 5.1 142

27. f0(x) = sec2x

f00(x)=2sec

2xtan x

(a) (−π/2,π/2) (b) none

(e) 0

-10

28. f0(x)=2−csc2x

f00(x)=2csc

2xcot x=2cos x

sin3x

(a) [π/4,3π/4] (b) (0,π/4],[3π/4,π)

(e) π/2

-2

29. f0(x) = cos 2x

f00(x)=−2 sin 2x

(a) [0,π/4], [3π/4,π](b) [π/4,3π/4]

(e) π/2

0.5

-0.5

30. f0(x)=−2 cos xsin x−2 cos x=−2 cos x(1 + sin x)

f00(x)=2sinx(sin x+1)−2 cos2x= 2 sin x(sin x+1)−2+2sin

2x= 4(1 + sin x)(sin x−1/2)

Note: 1 + sin x≥0

(a) [π/2,3π/2] (b) [0,π/2], [3π/2,2π]

(e) π/6, 5π/6

-2

31. (a)

y(b)

y(c)

143 Chapter 5

32. (a)

y(b)

y(c)

33. (a) f0(x)=3(x−a)

00(x)=6(x−a); inﬂection point is (a, 0)

(b) f0(x)=4(x−a)

00(x) = 12(x−a)2; no inﬂection points

34. For n≥2, f00(x)=n(n−1)(x−a)n−2; there is a sign change of f00 (point of inﬂection) at (a, 0) if and

only if nis odd. For n=1,y=x−a, so there is no point of inﬂection.

35. f0(x)=1/3−1/[3(1 + x)2/3]sofis increasing on [0,+∞)thusif

x>0, then f(x)>f(0)=0,1+x/3−3

√1+x>0,

√1+x<1+x/3.

2.5

010

36. f0(x) = sec2x−1sofis increasing on [0,π/2) thus if 0 <x<π/2,

then f(x)>f(0) = 0, tan x−x>0, x<tan x.

37. x≥sin xon [0,+∞): let f(x)=x−sin x. Then f(0) = 0 and

f0(x)=1−cos x≥0, so f(x) is increasing on [0,+∞).

-1

38. (a) Let h(x)=e

x−1−xfor x≥0. Then h(0) = 0 and h0(x)=e

x−1≥0 for x≥0, so h(x)is

increasing.

(b) Let h(x)=e

x−1−x−1

. Then h(0) = 0 and h0(x)=e

x−1−x. By part (a), ex−1−x≥0

for x≥0, so h(x) is increasing.

Exercise Set 5.1 144

39. Points of inﬂection at x=−2,+2. Concave up on (−5,−2) and

(2,5); concave down on (−2,2). Increasing on [−3.5829,0.2513] and

[3.3316,5] , and decreasing on [−5,−3.5829] and [0.2513,3.3316].

250

-250

-5 5

40. Points of inﬂection at x=±1/√3. Concave up on (−5,−1/√3) and

(1/√3,5), and concave down on (−1/√3,1/√3). Increasing on

[−5,0] and decreasing on [0,5].

-2

-5 5

41. Break the interval [−5,5] into ten subintervals and check f00(x) at each endpoint. We ﬁnd f00(−1) >0

and f00(0) <0. Reﬁne [−1,0] into ten subintervals; f00(−0.2) >0, f00(−0.1) <0; repeat, f00(−0.18) >0,

f00(−0.17) <0, so x=−0.175 is correct to two decimal places. Note also that f00(1) = 0 so there are

two inﬂection points.

42. Break the interval [−5,5] into ten subintervals and check f00(x) at each endpoint. We discover

f00(−1) >0, f00(0) <0 and f00(1) >0. Reﬁne [−1,0] into ten subintervals and we see that f00(−0.6) >0,

f00(−0.5) <0. Subdivide [−0.6,−0.5] into 10 subintervals and we see that f00(−0.58) >0 and

f00(−0.57) <0. Thus x=−0.575 is within 0.005 of the true root and is thus correct to two dec-

imal places. For the other root we could proceed in a similar manner, but it easier to note that f00(x)

is an even function and thus the other root is x=0.575 to two decimal places.

43. f00(x)=2

90x3−81x2−585x+ 397

(3x2−5x+8)

3.The denominator has complex roots, so is always positive; hence

the x-coordinates of the points of inﬂection of f(x) are the roots of the numerator (if it changes sign).

A plot of the numerator over [−5,5] shows roots lying in [−3,−2], [0,1], and [2,3]. Breaking each of

these intervals into ten subintervals locates the roots in [−2.5,−2.4], [0.6,0.7] and [2.7,2.8]. Thus to

one decimal place the roots are x=−2.45,0.65,2.75.

44. f00(x)=2x

5+5x

3+14x

2+30x−7

2+1)

5/2. Points of inﬂection will occur when the numerator changes sign,

since the denominator is always positive. A plot of y=2x

5+5x

3+14x

2+30x−7 suggests that there

is only one root and it lies in [0,1]. Subdivide into ten subintervals and determine that the root lies

between x=0.2 and x=0.3. Thus to one decimal place the point of inﬂection is located at x=0.25.

45. f(x1)−f(x2)=x

1−x

2=(x

1+x

)(x1−x2)<0ifx

1<x

2for x1,x2in [0,+∞), so f(x1)<f(x

)

and fis thus increasing.

46. f(x1)−f(x2)= 1

1−1

2−x

>0ifx

1<x

2for x1,x2in (0,+∞), so f(x1)>f(x

) and thus f

is decreasing.

145 Chapter 5

47. (a) If x1<x

2where x1and x2are in I, then f(x1)<f(x

) and g(x1)<g(x

), so f(x1)+g(x

f(x

)+g(x

), (f+g)(x1)<(f+g)(x2). Thus f+gis increasing on I.

(b) Case I: If fand gare ≥0onI, and if x1<x

2where x1and x2are in I, then 0 <f(x

)<f(x

)

and 0 <g(x

)<g(x

), so f(x1)g(x1)<f(x

)g(x

), (f·g)(x1)<(f·g)(x2). Thus f·gis

increasing on I.

Case II: If fand gare not necessarily positive on Ithen no conclusion can be drawn: for example,

f(x)=g(x)=xare both increasing on (−∞,0), but (f·g)(x)=x

2is decreasing there.

48. (a) f(x)=x,g(x)=2x(b) f(x)=x,g(x)=x+6 (c) f(x)=2x,g(x)=x

49. (a) f00(x)=6ax +2b=6a(x+b

), f00(x) = 0 when x=−b

3a.fchanges its direction of concavity

at x=−b

3aso −b

3ais an inﬂection point.

(b) If f(x)=ax3+bx2+cx +dhas three x-intercepts, then it has three roots, say x1,x2and x3,so

we can write f(x)=a(x−x

)(x−x2)(x−x3)=ax3+bx2+cx +d, from which it follows that

b=−a(x1+x2+x3). Thus −b

3a=1

3(x1+x2+x3), which is the average.

2−3x

2+2)=x(x−1)(x−2) so the intercepts are 0, 1, and 2 and the average is 1.

f00(x)=6x−6=6(x−1) changes sign at x=1.

50. f00(x)=6x+2b, so the point of inﬂection is at x=−b

3. Thus an increase in bmoves the point of

inﬂection to the left.

51. (a) Let x1<x

2belong to (a, b). If both belong to (a, c] or both belong to [c, b) then we have

f(x1)<f(x

) by hypothesis. So assume x1<c<x

. We know by hypothesis that f(x1)<f(c),

and f(c)<f(x

). We conclude that f(x1)<f(x

(b) Use the same argument as in part (a), but with inequalities reversed.

52. By Theorem 5.1.2, fis increasing on any interval ((2n−1)π, 2(n+1)π)(n=0,±1,±2,···), because

f0(x) = 1+ cos x>0on[(2n−1)π, (2n+1)π]. By Exercise 51 (a) we can piece these intervals together

to show that f(x) is increasing on (−∞,+∞).

53. t=7.67 1000

015

54. By zooming on the graph of y0(t), maximum increase is at x=−0.577 and maximum decrease is at

x=0.577.

55.

infl pt

56. y

infl pts

Exercise Set 5.2 146

EXERCISE SET 5.2

1. (a)

f(x)

y(b)

f(x)

(c)

f(x)

y(d)

f(x)

2. (a) y

(b) y

(d) y

3. (a) f0(x)=6x−6 and f00(x) = 6, with f0(1) = 0. For the ﬁrst derivative test, f0<0 for x<1 and

f0>0 for x>1. For the second derivative test, f00(1) >0.

(b) f0(x)=3x

2−3 and f00(x)=6x.f

(x)=0atx=±1. First derivative test: f0>0 for x<−1

and x>1, and f0<0 for −1<x<1, so there is a relative maximum at x=−1, and a relative

minimum at x= 1. Second derivative test: f00 <0atx=−1, a relative maximum; and f00 >0

at x= 1, a relative minimum.

4. (a) f0(x)=2sinxcos x= sin 2x(so f0(0) = 0) and f00(x) = 2 cos 2x. First derivative test: if xis

near 0 then f0<0 for x<0 and f0>0 for x>0, so a relative minimum at x= 0. Second

derivative test: f00(0)=2>0, so relative minimum at x=0.

(b) g0(x)=2tanxsec2x(so g0(0) = 0) and g00(x)=2sec

2x(sec2x+ 2 tan2x). First derivative test:

g0<0 for x<0 and g0>0 for x>0, so a relative minimum at x= 0. Second derivative test:

g00(0)=2>0, relative minimum at x=0.

at x= 0, so that must be a relative minimum.

5. (a) f0(x)=4(x−1)3,g0(x)=3x

2−6x+3 sof0(1) = g0(1)=0.

(b) f00(x) = 12(x−1)2,g00(x)=6x−6, so f00(1) = g00(1) = 0, which yields no information.

(x)=3(x−1)2>0 on both sides of x= 1, so there is no relative extremum at x=1.

147 Chapter 5

6. (a) f0(x)=−5x

0(x)=12x

3−24x2so f0(0) = g0(0)=0.

(b) f00(x)=−20x3,g00(x)=36x

2−48x,sof

00(0) = g00(0) = 0, which yields no information.

(x−2) <0

on both sides of x= 0 (for xnear 0), so again there is no relative extremum there.

7. (a) f0(x)=3x

2+6x−9=3(x+ 3)(x−1), f0(x) = 0 when x=−3,1 (stationary points).

(b) f0(x)=4x(x

2−3), f0(x) = 0 when x=0,±

√

3 (stationary points).

8. (a) f0(x)=6(x

2−1), f0(x) = 0 when x=±1 (stationary points).

(b) f0(x)=12x

3−12x2=12x

(x−1), f0(x) = 0 when x=0,1 (stationary points).

9. (a) f0(x)=(2−x

)/(x

2+2)

,f0(x) = 0 when x=±√2 (stationary points).

(b) f0(x)=2

−1/3=2/(3x1/3), f0(x) does not exist when x=0.

10. (a) f0(x)=8x/(x2+1)

,f0(x) = 0 when x= 0 (stationary point).

(b) f0(x)=1

3(x+2)

−2/3

,f0(x) does not exist when x=−2.

11. (a) f0(x)= 4(x+1)

2/3,f0(x) = 0 when x=−1 (stationary point), f0(x) does not exist when x=0.

(b) f0(x)=−3 sin 3x, f0(x) = 0 when sin 3x=0,3x=nπ, n =0,±1,±2,···

x=nπ/3,n=0,±1,±2,··· (stationary points)

12. (a) f0(x)=4(x−3/2)

3x2/3,f0(x) = 0 when x=3/2 (stationary point), f0(x) does not exist when x=0.

(b) f(x)=|sin x|=½sin x, sin x≥0

−sin x, sin x<0so f0(x)=½cos x, sin x>0

−cos x, sin x<0and f0(x)doesnot

exist when x=nπ,n=0,±1,±2,···(sin x= 0) because limx→nπ−f0(x)6= limx→nπ+f0(x) (see

Theorem preceding Exercise 75, Section 3.3). Now f0(x) = 0 when ±cos x= 0 provided sin x6=0

so x=π/2+nπ,n=0,±1,±2,···are stationary points.

13. (a) x= 2 because f0(x) changes sign from −to + there.

(b) x= 0 because f0(x) changes sign from + to −there.

14. (a) x=1 (b) x=5 (c) x=−1,0,3

15. (a) critical points x=0,±

√

5; f0:00

−−−− ++

−

x= 0: relative maximum; x=±√5: relative minimum

(b) critical point x=0;f

0:0

−+−+−+

x= 0: relative minimum

16. (a) critical points x=0,−1/2,1; f0:00

−1

−−−+++ +

−

x= 0: neither; x=−1/2: relative maximum; x= 1: relative minimum

Exercise Set 5.2 148

(b) critical points: x=±3/2,−1; f0:0?

−1

++++ −−

−−

−

x=±3/2: relative maximum; x=−1: relative minimum

17. f0(x)=−2(x+ 2); critical point x=−2; f0(x): 0

-2

−−+++−

00(x)=−2; f00(−2) <0, f(−2) = 5; relative max of 5 at x=−2

18. f0(x)=6(x−2)(x−1); critical points x=1,2; f0(x): 00

−+++ ++

+−−

00(x)=12x−18; f00(1) <0, f00(2) >0, f(1)=5,f(2) = 4; relative min of 4 at x= 2, relative max

of5atx=1

19. f0(x)=2sinxcos x= sin 2x; critical points x=π/2, π,3π/2; f0(x): 00

++++ −−

−−

ππ

00(x) = 2 cos 2x;f00(π/2) <0, f00(π)>0, f00(3π/2) <0, f(π/2) = f(3π/2)=1,

f(π) = 0; relative min of 0 at x=π, relative max of 1 at x=π/2, 3π/2

20. f0(x)=1/2−cos x; critical points x=π/3, 5π/3; f0(x): 00

++−− −−

+++

ππ

3 3

00(x)=−sin x;f00(π/3) <0, f00(5π/3) >0

f(π/3) = π/6−√3/2, f(5π/3)=5π/6+√

3/2;

relative min of π/6−√3/2atx=π/3, relative max of 5π/6+√

3/2atx=5π/3

21. f0(x)=3x

2+ 5; no relative extrema because there are no critical points.

22. f0(x)=4x(x

2−1); critical points x=0,1,−1

00(x)=12x

2−4; f00(0) <0, f00(1) >0, f00(−1) >0

relative min of 6 at x=1,−1, relative max of 7 at x=0

23. f0(x)=(x−1)(3x−1); critical points x=1,1/3

00(x)=6x−4; f00(1) >0, f00(1/3) <0

relative min of 0 at x= 1, relative max of 4/27 at x=1/3

24. f0(x)=2x

(2x+ 3); critical points x=0,−3/2

relative min of −27/16 at x=−3/2 (ﬁrst derivative test)

25. f0(x)=4x(1 −x2); critical points x=0,1,−1

00(x)=4−12x2;f00(0) >0, f00(1) <0, f00(−1) <0

relative min of 0 at x= 0, relative max of 1 at x=1,−1

26. f0(x) = 10(2x−1)4; critical point x=1/2; no relative extrema (ﬁrst derivative test)

27. f0(x)=4

−1/5

; critical point x= 0; relative min of 0 at x= 0 (ﬁrst derivative test)

28. f0(x)=2+2

−1/3

; critical points x=0,−1/27

relative min of 0 at x= 0, relative max of 1/27 at x=−1/27

29. f0(x)=2x/(x2+1)

; critical point x= 0; relative min of 0 at x=0

149 Chapter 5

30. f0(x)=2/(x+2)

; no critical points (x=−2 is not in the domain of f) no relative extrema

31. f0(x)=2x/(1+x2); critical point at x= 0; relative min of 0 at x= 0 (ﬁrst derivative test)

32. f0(x)=x(2 + x)ex; critical points at x=0,−2; relative min of 0 at x= 0 and relative max of 4/e2at

x=−2 (ﬁrst derivative test)

33. f0(x)=2xif |x|>2, f0(x)=−2xif |x|<2,

f0(x) does not exist when x=±2; critical points x=0,2,−2

relative min of 0 at x=2,−2, relative max of 4 at x=0

34. f0(x)=−1ifx<3, f0(x)=2xif x>3, f0(3) does not exist;

critical point x= 3, relative min of 6 at x=3

35. f0(x) = 2 cos 2xif sin 2x>0, f0(x)=−2 cos 2xif sin 2x<0,

f0(x) does not exist when x=π/2,π,3π/2;

critical points x=π/4,3π/4,5π/4,7π/4,π/2,π,3π/2

relative min of 0 at x=π/2,π,3π/2; relative max of 1 at

x=π/4,3π/4,5π/4,7π/4

36. f0(x)=√

3+2cosx; critical points x=5π/6,7π/6

relative min of 7√3π/6−1atx=7π/6; relative max of

5√3π/6+1atx=5π/6

37. f0(x)=−sin 2x; critical points x=π/2,π,3π/2

relative min of 0 at x=π/2,3π/2; relative max of 1 at x=π

38. f0(x)=(2cosx−1)/(2 −cos x)2; critical points x=π/3,5π/3

relative max of √3/3atx=π/3, relative min of

−√3/3atx=5π/3

0.8

-0.8

Exercise Set 5.2 150

39. f0(x)=lnx+1, f00(x)=1/x;f0(1/e)=0,f

00(1/e)>0;

relative min of −1/e at x=1/e

2.5

-0.5

0 2.5

40. f0(x)=−2e

x−e

−x

x+e

−x

)

2= 0 when x= 0. By the ﬁrst derivative test

f0(x)>0 for x<0 and f0(x)<0 for x>0; relative max of 1 at

x=0

-2 2

41. f0(x)=2x(1 −x)e−2x=0atx=0,1. f00(x)=(4x

2−8x+2)e

−2x

;

f00(0) >0 and f00(1) <0, so a relative min of 0 at x= 0 and a

relative max of 1/e2at x=1.

0.14

-0.3 4

42. f0(x)=10/x −1=0atx= 10; f00(x)=−10/x2<0;

relative max of 10(ln(10) −1) ≈13.03 at x=10

-4

020

43. Relative minima at x=−3.58,3.33; relative max at x=0.25 250

-250

-5 5

44. Relative min at x=−0.84; relative max at x=0.84 1.2

-1.2

-66

151 Chapter 5

45. relative max at x=0.255 46. relative max at x=0.845

47. Relative min at x=−1.20 and a relative max at x=1.80 y

-2

-1

-4 -2 2 4

f'(x)

f''(x)

48. Relative max at x=−0.78 and a relative min at x=1.55 y

-5

-4 -2 2 4

f'(x)

f''(x)

49. (a) Let f(x)=x

2+k

, then f0(x)=2x−k

2=2x

3−k

2.fhas a relative extremum when 2x3−k=0,

so k=2x

3= 2(3)3= 54.

(b) Let f(x)= x

2+k

, then f0(x)= k−x

2+k)

.fhas a relative extremum when k−x2=0,so

k=x

2=3

2=9.

50. (a) one relative maximum, located at x=n0.3

014

(b) f0(x)=cxn−1(−x+n)e−x=0atx=n. Since f0(x)>0 for x<nand f0(x)<0 for

x>nit’s a maximum.

51. (a) f0(x)=−xf(x). Since f(x)isalways

positive, f0(x)=0atx=0,f

0(x)>0

for x<0 and f0(x)<0 for x>0, so

x= 0 is a maximum.

(b) y

µ,

()

2π

Exercise Set 5.3 152

52. (a) relative minima at x=±0.6436,

relative max at x=0

1.2

1.4

1.6

1.8

-1.5 -1 -0.5 0.5 1 1.5

(b) x=±0.6436,0

53. f0(x)=3ax2+2bx +cand f0(x) has roots at x=0,1, so f0(x) must be of the form f0(x)=3ax(x−1);

thus c= 0 and 2b=−3a,b=−3a/2. f00(x)=6ax +2b=6ax −3a,sof

00(0) >0 and f00(1) <0

provided a<0. Finally f(0) = d,sod= 0; and f(1) = a+b+c+d=a+b=−a/2soa=−2. Thus

f(x)=−2x

3+3x

54. (a) Because hand ghave relative maxima at x0,h(x)≤h(x0) for all xin I1and g(x)≤g(x0) for all

xin I2, where I1and I2are open intervals containing x0.Ifxis in I1∩I2then both inequalities

are true and by addition so is h(x)+g(x)≤h(x

0)+g(x

0) which shows that h+ghas a relative

maximum at x0.

(b) By counterexample; both h(x)=−x

2and g(x)=−2x

2have relative maxima at x= 0 but

h(x)−g(x)=x

2has a relative minimum at x= 0 so in general h−gdoes not necessarily have

a relative maximum at x0.

55. (a)

)( x0

f(x0) is not an extreme value.

(b)

)( x0

f(x0) is a relative maximum.

(c)

)( x0

f(x0) is a relative minimum.

EXERCISE SET 5.3

1. y=x2−2x−3;

y0=2(x−1);

y00 =2

-4

(1, -4)

153 Chapter 5

2. y=1+x−x

;

0=−2(x−1/2);

y00 =−2

()

3. y=x3−3x+1;

0=3(x

2−1);

y00 =6x

(1, -1)

(0, 1)

(-1, 3)

4. y=2x

3−3x

2+12x+9;

0=6(x

2−x+ 2);

y00 = 12(x−1/2)

229

()

5. y=x4+2x

3−1;

y0=4x

(x+3/2);

y00 =12x(x+1) x

243

, -

(-1, -2)

(0, -1)

()

6. y=x4−2x2−12;

y0=4x(x

2−1);

y00 = 12(x2−1/3)

113

-1 113

(1, -13)

(-1, -13)

(0, -12)

, - , -

()()

Exercise Set 5.3 154

7. y=x3(3x2−5);

y0=15x

2−1);

y00 =30x(2x2−1)

(-1, 2)

(0, 0)

(1, -2)

178

, -

()

8. y=3x

(x+4/3);

y0=12x

(x+ 1);

y00 =36x(x+2/3)

(-1, -1)

(0, 0)

316

-, -

()

9. y=x(x−1)3;

y0=(4x−1)(x−1)2;

y00 = 6(2x−1)(x−1) 0.1

-0.1

427

256

(1, 0)

, -

()

, -

()

10. y=x4(x+ 5);

y0=5x

(x+ 4);

y00 =20x

(x+3)

300

(0, 0)

(-4, 256)

(-3, 162)

11. y=2x/(x−3);

y0=−6/(x−3)2;

y00 =12/(x−3)3

y = 2

x = 3

155 Chapter 5

12. y=x

x2−1;

y0=−x2+1

2−1)2;

y00 =2x(x2+3)

2−1)3x

x = 1

x = -1

(0, 0)

13. y=x2

x2−1;

y0=−2x

(x2−1)2;

y00 =2(3x2+1)

2−1)3x

x = 1

y = 1

x = -1

(0, 0)

14. y=x2−1

x2+1;

0=4x

2+1)

00 =4(1 −3x2)

(x2+1)

211

(0, -1)

y = 1

, -

(

()

, -

)

3 3

15. y=x2−1

x=x3−1

y0=2x3+1

0= 0 when x=−3

2≈−0.8;

y00 =2(x3−1)

≈(-0.8, 1.9)

(1, 0)

16. y=2x2−1

x2;

y0=2

x3;

y00 =−6

x4x

y = 2

Exercise Set 5.3 156

17. y=x3−1

x3+1;

0=6x

3+1)

00 =12x(1 −2x3)

(x3+1)

(0, -1)

(1/ 2 , -1/3)

y = 1

x = -1

18. y=8

4−x2;

y0=16x

(4 −x2)2;

y00 =16(3x2+4)

(4 −x2)3x

x = 2x = -2

19. y=x−1

x2−4;

y0=−x2−2x+4

2−4)2

x = 2

x = -2 4

20. y=3−4

x−4

0=4(x+2)

00 =−8(x+3)

(-2, 4)

y = 3

-3,

()

21. y=(x−1)2

x2;

y0=2(x−1)

x3;

y00 =2(3 −2x)

(1, 0)

y = 1

()

157 Chapter 5

22. y=2+3

x−1

0=3(1 −x2)

x4;

y00 =6(x2−2)

x5x

(–1, 0)

(1, 4)

y = 2

2, 2 + 2

)(

−2, 2 – 2

)(

23. (a) VI (b) I(c) III (d) V(e) IV (f ) II

24. (a) When nis even the function is deﬁned only for x≥0; as nincreases the graph approaches the

line y= 1 for x>0.

(b) When nis odd the graph is symmetric with respect to the origin; as nincreases the graph

approaches the line y= 1 for x>0 and the line y=−1 for x<0.

25. y=√x2−1;

y0=x

√x2−1;

y00 =−1

(x2−1)3/2-1 1 x

26. y=3

px2−4;

y0=2x

3(x2−4)2/3;

y00 =−2(3x2+4)

9(x2−4)5/3-2

(-2, 0) (2, 0)

(0, -2)

Exercise Set 5.3 158

27. y=2x+3x

2/3

;

0=2+2x

−1/3

;

00 =−2

3x−4/3

(0, 0)

28. y=4x−3x

4/3

;

0=4−4x

1/3

;

00 =−4

3x−2/3

(1, 1)

29. y=x(3 −x)1/2;

y0=3(2 −x)

2√3−x;

y00 =3(x−4)

4(3 −x)3/2x

(2, 2)

30. y=x1/3(4 −x);

y0=4(1 −x)

3x2/3;

y00 =−4(x+2)

5/3

-10

(1, 3)

(-2, -6 )2

31. y=8(√x−1)

y0=4(2 −√x)

x2;

y00 =2(3√x−8)

x315 x

(4, 2) 64

915

()

159 Chapter 5

32. y=1+√

1−√

0=1

√

x(1−√x);

y00 =3√x−1

2x3/2(1 −√x)3

-2

x = 1

y = -1

9, 2

()

33. y=x+ sin x;

y0= 1 + cos x,y0= 0 when x=π+2nπ;

y00 =−sin x;y00 = 0 when x=nπ

n=0,±1,±2,... c

34. y=x−cos x;

y0= 1 + sin x;

y0= 0 when x=−π/2+2nπ;

y00 = cos x;

y00 = 0 when x=π/2+nπ

n=0,±1,±2,...

35. y= sin x+ cos x;

y0= cos x−sin x;

y0= 0 when x=π/4+nπ;

y00 =−sin x−cos x;

y00 = 0 when x=3π/4+nπ

-oo

-2

36. y=√3 cos x+ sin x;

y0=−√3 sin x+ cos x;

y0= 0 when x=π/6+nπ;

y00 =−√3 cos x−sin x;

y00 = 0 when x=2π/3+nπ

-2

Exercise Set 5.3 160

37. y= sin2x, 0≤x≤2π;

y0= 2 sin xcos x= sin 2x;

y00 = 2 cos 2x

38. y=xtan x, −π/2<x<π/2;

y0=xsec2x+ tan x;

y0= 0 when x=0;

00 = 2 sec2x(xtan x+ 1), which is always

positive for −π/2<x<π/2-6 6

39. (a) lim

x→+∞xex=+∞, lim

x→−∞ xex=0

(b) y=xex;

y0=(x+1)e

;

00 =(x+2)e

-1

-1-3-5

(-2, -0.27)

(-1, -0.37)

40. (a) lim

x→+∞xe−2x= 0, lim

x→−∞ xe−2x=−∞

(b) y=xe−2x;y0=−2µx−1

2¶e−2x;y00 =4(x−1)e−2x

-3 -1 1 3

-0.3

-0.1

0.1

0.3

(0.5, 0.18)

(1, 0.14)

41. (a) lim

x→+∞

e2x= 0, lim

x→−∞

e2x=+∞

(b) y=x2/e2x=x2e−2x;

y0=2x(1 −x)e−2x;

y00 = 2(2x2−4x+1)e

−2x

;

00 = 0 if 2x2−4x+ 1 = 0, when

x=4±√16 −8

4=1±√

2/2≈0.29,1.71

123

0.3

(0, 0)

(0.29, 0.05)

(1, 0.14)

(1.71, 0.10)

42. (a) lim

x→+∞x2e2x=+∞, lim

x→−∞ x2e2x=0.

161 Chapter 5

(b) y=x2e2x;

y0=2x(x+1)e

;

00 = 2(2x2+4x+1)e

;

00 = 0 if 2x2+4x+1=0,when

x=−4±√16 −8

4=−1±√2/2≈−0.29,−1.71

-3 -2 -1

0.2

0.3

(-1.71, 0.10)

(-1, 0.14)

(0, 0)

(-0.29, 0.05)

43. (a) lim

x→+∞f(x)=+∞, lim

x→−∞ f(x)=−∞

(b) y=xex2;

y0=(1+2x

00 =2x(3+2x

no relative extrema, inﬂection point at (0,0)

-100

100

-2 2

(0,0)

44. (a) lim

x→±∞ f(x)=1

(b) f0(x)=2x

−3

−1/x2so f0(x)<0 for x<0 and f0(x)>0 for

x>0. By L’Hˆopital’s Rule limx→0f0(x) = 0, so (by the ﬁrst

derivative test) f(x) has a minimum at x=0.

00(x)=(−6x

−4+4x

−6

−1/x2,sof(x) has points of inﬂection

at x=±p2/3

0.4

0.8

-10 -5 5 10

(0, 0)

2/3, e−3/2

−

()

2/3, e−3/2

()

45. (a) lim

x→0+y= lim

x→0+xln x= lim

x→0+

ln x

1/x = lim

x→0+

1/x

−1/x2=0;

lim

x→+∞y=+∞

(b) y=xln x,

y0=1+lnx,y

00 =1/x,

y0= 0 when x=e−1

(e-1, -e-1)

46. (a) lim

x→0+y= lim

x→0+

ln x

1/x2= lim

x→0+

1/x

−2/x3=0,

lim

x→+∞y=+∞

(b) y=x2ln x, y0=x(1+2lnx),

00 =3+2lnx,

y0=0ifx=e

−1/2

00 =0ifx=e

−3/2

lim

x→0+y0=0

-0.2

-0.1

0.1

0.2

(

e-1/2, - e-1

)

(

e-3/2, - e-3

)

Exercise Set 5.3 162

47. (a) lim

x→0+y= lim

x→0+

ln x

x2=−∞;

lim

x→+∞y= lim

x→+∞

ln x

x2= lim

x→+∞

1/x

2x=0

(b) y=ln x

x2,y

0=1−2lnx

00 =6lnx−5

0=0ifx=e

1/2

00 =0ifx=e

5/6

123

-0.4

-0.3

-0.2

-0.1

0.1

0.2

0.3

0.4

(

e1/2, e-1

)

(

e5/6, e-5/3

)

48. (a) lim

x→0+(ln x)/√x=−∞ by inspection, lim

x→+∞(ln x)/√x= lim

x→+∞

1/x

1/2√x= lim

x→+∞

√x= 0, L’Hˆopital’s

Rule.

(b) y=ln x

√x,y0=2−ln x

2x3/2

y00 =−8+3lnx

5/2

0=0ifx=e

00 =0ifx=e

8/3

-1

0.5

2 6 10 14

(e2, 2/e)

(

e8/3, e-4/3

)

49. (a) lim

x→−∞ y=−∞, lim

x→+∞y=+∞;

curve crosses x-axis at x=0,1,−1

-6

-4

-2

-2 -1 1

(b) lim

x→±∞ y=+∞;

curve never crosses x-axis

0.1

0.2

-1 1

x→−∞ y=−∞, lim

x→+∞y=+∞;

curve crosses x-axis at x=−1

-0.2

0.2

0.4

-1 1

163 Chapter 5

(d) lim

x→±∞ y=+∞;

curve crosses x-axis at x=0,1

0.2

0.4

-1 1

50. (a) y

(b) y

51. (a) horizontal asymptote y=3asx→±∞, vertical asymptotes of

x=±2

-5

-5 5

Exercise Set 5.3 164

(b) horizontal asymptote of y=1asx→±∞, vertical asymptotes

at x=±1

-10

-5 5

asymptotes at x=−2,1

-10

-5 5

(d) horizontal asymptote of y=1asx→±∞,

vertical asymptote at x=−1,2

-10

-5 5

52. y

53. (a) 0.4

-0.2

-0.5 3

(b) y0=(1−bx)e−bx,y00 =b2(x−2/b)e−bx;

relative max at x=1/b,y=1/be;pointof

inﬂection at x=2/b,y=2/be2. Increasing

bmoves the relative max and the point of

inﬂection to the left and down, i.e. towards

the origin.

54. (a) 1

-2 2

(b) y0=−2bxe−bx2,y00 =2b(−1+2bx2)e−bx2;

relative max at x=0,y= 1; points

of inﬂection at x=±p1/2b,y=1/

√

Increasing bmoves the points of inﬂection

towards the y-axis; the relative max doesn’t

move.

165 Chapter 5

55. (a) The oscillations of excos xabout zero

increase as x→±∞so the limit does not

exist.

(b) y

-2

-2 -1 1 2

(0,1)

(1.52, 0.22)

increases when bincreases.

-5

-1 1 2

a = 1

b = 1

b = 2

b = 3

-1 0.5 1

b = 1

a = 1

a = 2

a = 3

56. lim

x→±∞ ·P(x)

Q(x)−(ax +b)¸= lim

x→±∞

R(x)

Q(x)= 0 because the degree of R(x) is less than

the degree of Q(x).

57. y=x2−2

x=x−2

xso

y=xis an oblique asymptote;

y0=x2+2

00 =−4

y = x

58. y=x2−2x−3

x+2 =x−4+ 5

x+2 so

y=x−4 is an oblique asymptote;

y0=x2+4x−1

(x+2)

2,y

00 =10

(x+2)

≈(-4.24, -10.48)

≈(0.24, -1.52)

x = -2

y = x - 4

-10 10 x

59. y=(x−2)3

x2=x−6+12x−8

x2so

y=x−6 is an oblique asymptote;

y0=(x−2)2(x+4)

00 =24(x−2)

-10

(-4, -13.5)

y = x - 6

(2, 0)

Exercise Set 5.3 166

60. y=4−x3

x2,

y0=−x3+8

00 =24

(-2, 3)

y = -x

61. y=x+1−1

x−1

2=(x−1)(x+1)

y=x+ 1 is an oblique asymptote;

y0=(x+ 1)(x2−x+2)

00 =−2(x+3)

y = x + 1

(-1, 0)

-3, -

)(

62. The oblique asymptote is y=2xso (2x3−3x+4)/x2=2x,−3x+4=0,x=4/3.

63. lim

x→±∞[f(x)−x2] = lim

x→±∞(1/x)=0

y=x

2+1

x=x

3+1

x,y

0=2x−1

2=2x

3−1

00 =2+ 2

3=2(x3+1)

3,y

0= 0 when x=1/3

√

2≈0.8,

y=33

√

2/2≈1.9; y00 = 0 when x=−1,y =0

y = x2

64. lim

x→±∞[f(x)−(3 −x2)] = lim

x→±∞(2/x)=0

y=3−x

2+2

x=2+3x−x

x,y

0=−2x−2

2=−2(x3+1)

00 =−2+ 4

3=−2(x3−2)

x3,y0= 0 when x=−1,y =0;

00 = 0 when x=3

√2≈1.3, y=3

y = 3 - x2

65. Let ybe the length of the other side of the rectangle, then

L=2x+2yand xy = 400 so y= 400/x and hence L=2x+ 800/x.

L=2xis an oblique asymptote (see Exercise 48)

L=2x+800

x=2(x2+ 400)

x,L0=2−800

x2=2(x2−400)

x2,

L00 =1600

x3,L0= 0 when x=20,L =80 20

100

167 Chapter 5

66. Let ybe the height of the box, then S=x2+4xy and x2y= 500 so

y= 500/x2and hence S=x2+ 2000/x.

The graph approaches the curve S=x2asymptotically

(see Exercise 63)

S=x2+2000

x=x3+ 2000

x,S0=2x−2000

x2=2(x3−1000)

x2,

S00 =2+4000

x3=2(x3+ 2000)

x3,S00 = 0 when x=10,S = 300

1000

67. y0=0.1x

(6x−5);

critical points: x=0,x=5/6;

relative minimum at x=5/6,

y≈−6.7×10−3

-1 1

0.01

68. y0=0.1x

(x+ 1)(7x+ 5);

critical points: x=0,x=−1, x=−5/7,

relative maximum at x=−1, y=0;

relative minimum at x=−5/7, y≈−1.5×10−3

0.001

69. (a) P0(t)= kL2Ae−kLt

(1+Ae−kLt)2S,soP

(0) = kL2A

(1+A)2

(b) The rate of growth increases to its maximum, which occurs when Pis halfway between 0 and

L, or when t=1

Lk ln A; it then decreases back towards zero.

dt is maximized when Plies half way between 0 and L, i.e. P=L/2.

This follows since the right side of (6) is a parabola (with Pas independent variable) with

P-intercepts P=0,L. The value P=L/2 corresponds to t=1

Lk ln A, from (8).

70. Since 0 <P <Lthe right-hand side of (7) can change sign only if the factor L−2Pchanges sign,

which it does when P=L/2. From (5) we have L

2=L

1+Ae−kLt ,1=Ae−kLt,t=1

Lk ln A.

SUPPLEMENTARY EXERCISES FOR CHAPTER 5

4. (a) False; an example is y=x3

3−x2

2on [−2,2]; x= 0 is a relative maximum and x= 1 is a relative

minimum, buty=0isnotthelargest value of yon the interval, nor is y=−1

6the smallest.

Supplementary Exercises 168

(b) true

6. (a)

y(b)

y(c)

7. (a) f0(x)= 7(x−7)(x−1)

3x2/3; critical points at x=0,1,7;

neither at x= 0, relative max at x= 1, relative min at x= 7 (ﬁrst derivative test)

(b) f0(x)=2cosx(1 + 2 sin x); critical points at x=π/2,3π/2,7π/6,11π/6;

relative max at x=π/2,3π/2, relative min at x=7π/6,11π/6

√

x−1

2; critical points at x= 5; relative max at x=5

8. (a) f0(x)= x−9

18x3/2,f00(x)=27 −x

36x5/2; critical point at x=9;f

00(9) >0, relative min at x=9

(b) f0(x)=2

3−4

2,f

00(x)=2

3+8

critical point at x=4

1/3

00(41/3)>0, relative min at x=4

1/3

2x−2 sin2x+ cos x; critical points at x=2π/3,π,4π/3;

f00(2π/3) <0, relative max at x=2π/3; f00(π)>0, relative min at x=π;f00(4π/3) <0, relative

max at x=4π/3

9. lim

x→−∞ f(x)=+∞, lim

x→+∞f(x)=+∞

0(x)=x(4x2−9x+ 6), f00(x) = 6(2x−1)(x−1)

relative min at x=0,

points of inﬂection when x=1/2,1,

no asymptotes

(0,1)

)(

(1,2)

2,23

10. lim

x→−∞ f(x)=−∞, lim

x→+∞f(x)=+∞

f(x)=x

(x−2)2,f0(x)=x

(5x−6)(x−2),

f00(x)=4x(5x2−12x+6)

critical points at x=0,8±2

√

relative max at x=8−2√31

5=−0.63

relative min at x=8+2

√

5=3.83

points of inﬂection at x=0,6±√

5=0,−0.42,2.82

no asymptotes

-200

-100

-1 1234

(0,0)

(-0.42,0.16)

(2.82,-165.00) (3.83,-261.31)

(-0.63,0.27)

169 Chapter 5

11. lim

x→±∞ f(x) doesn’t exist

f0(x)=2xsec2(x2+ 1),

f00(x)=2sec

2+1)£1+4x

2tan(x2+1)

critical point at x= 0; relative min at x=0

point of inﬂection when 1 + 4x2tan(x2+1)=0

vertical asymptotes at x=±qπ(n+1

2)−1, n=0,1,2,...

-4

-2

-2 -1 1 2

12. lim

x→−∞ f(x)=−∞, lim

x→+∞f(x)=+∞

0(x) = 1 + sin x,f00(x) = cos x

critical points at x=2nπ +π/2, n=0,±1,±2,...,

no extrema because f0≥0 and by Exercise 51 of Section 5.1,

fis increasing on (−∞,+∞)

inﬂections points at x=nπ +π/2, n=0,±1,±2,...

no asymptotes

-6

-4

-2

−π π

13. f0(x)=2 x(x+5)

2+2x+5)

2,f00(x)=−2

3+15x

2−25

(x2+2x+5)

critical points at x=−5,0;

relative max at x=−5,

relative min at x=0

points of inﬂection at x=−7.26,−1.44,1.20

horizontal asymptote y=1asx→±∞

0.2

0.4

0.6

0.8

-20 -10 10 20

14. f0(x)=3

2−25

x4,f00(x)=−6

2−50

critical points at x=±5√3/3;

relative max at x=−5√3/3,

relative min at x=+5

√

3/3

inﬂection points at x=±5p2/3

horizontal asymptote of y=0asx→±∞,

vertical asymptote x=0

-5

-4 6

15. lim

x→−∞ f(x)=+∞, lim

x→+∞f(x)=−∞

f0(x)=½x

−2xif ½x≤0

x>0

critical point at x= 0, no extrema

inﬂection point at x=0(fchanges concavity)

no asymptotes

-2

-2 1

Supplementary Exercises 170

16. f0(x)= 5−3x

3(1 + x)1/3(3 −x)2/3,

f00(x)= −32

9(1 + x)4/3(3 −x)5/3

critical point at x=5/3;

relative max at x=5/3

cusp at x=−1;

point of inﬂection at x=3

oblique asymptote y=−xas x→±∞

-3

-1

-4 -2 2

17. lim

x→+∞f(x)=+∞

0(x)=1+lnx,f00(x)=1/x

lim

x→0+f(x)=0, lim

x→0

0(x)=−∞

critical point at x=1/e;

relative min at x=1/e

no points of inﬂection, no asymptotes

18. lim

x→+∞f(x)=+∞

0(x)=x(2 ln x+ 1), f00(x)=2lnx+3

lim

x→0+f(x)=0, lim

x→0

0(x)=0

critical point at x=e−1/2;

relative min at x=e−1/2

point of inﬂection at x=e−3/2

-0.1

0.2 0.4 0.6

19. f0(x)=1−2lnx

3,f00(x)=6lnx−5

critical point at x=e1/2,

relative max at x=e1/2

point of inﬂection at x=e5/6

horizontal asymptote y=0asx→+∞

-2

-1

20. lim

x→±∞ f(x)=+∞

0(x)= 2x

2+1,f00(x)=2 1−x

2+1)

critical point at x=0;

relative min at x=0

points of inﬂection at x=±1

no asymptotes

-2 2

171 Chapter 5

21. lim

x→+∞f(x)=+∞

0(x)=e

xx−1

2,f

00(x)=e

2−2x+2

critical point at x=1;

relative min at x=1

no points of inﬂection

vertical asymptote x=0,

horizontal asymptote y= 0 for x→−∞

-2

-2 1 3

22. f0(x)=(1−x)e

−x

00(x)=(x−2)e−x

critical point at x= 1; relative max at x=1

point of inﬂection at x=2

horizontal asymptote y=0asx→+∞, lim

x→−∞ f(x)=−∞

-0.8

0.2

23. f0(x)=x(2 −x)e1−x,f00(x)=(x

2−4x+2)e

1−x

critical points at x=0,2;

relative min at x=0,

relative max at x=2

points of inﬂection at x=2±√

horizontal asymptote y=0asx→+∞,

lim

x→−∞ f(x)=+∞

0.6

1.8

1234

24. f0(x)=x

(3+x)ex−1,f00(x)=x(x

2+6x+6)e

x−1

critical points at x=−3,0;

relative min at x=−3

points of inﬂection at x=0,−3±√

horizontal asymptote y=0asx→−∞

lim

x→+∞f(x)=+∞

-0.4

0.4

-4 -2 1

25. (a) 40

-40

-5 5

(b) f0(x)=x

2−1

400,f00(x)=2x

critical points at x=±1

20;

relative max at x=−1

20,

relative min at x=1

Supplementary Exercises 172

a much smaller x-window.

0.0001

-0.0001

-0.1 0.1

26. (a) 200

-200

-5 5

(b) critical points at x=±√2,3

2,2;

relative max at x=−√2,

relative min at x=√2,

relative max at x=3

relative min at x=2

-4

-2.2 3.5

-2.909

-2.912

1.3 1.6

27. (a) 6

-6

-5 5

(b) Divide y=x2+ 1 into y=x3−8 to get the

asymptote ax +b=x

28. (a) p(x)=x

3−x(b) p(x)=x

4−x

5−x

4−x

3+x

2(d) p(x)=x

5−x

29. f0(x)=4x

3−18x2+24x−8, f00(x) = 12(x−1)(x−2)

f00(1)=0,f

0(1)=2,f(1)=2;f

00(2)=0,f

0(2)=0,f(2)=3,

so the tangent lines at the inﬂection points are y=2xand y=3.

30. cos x−(sin y)dy

dx =2

dx;dy

dx = 0 when cos x= 0. Use the ﬁrst derivative test: dy

dx =cos x

2 + sin yand

2 + sin y>0, so critical points when cos x= 0, relative maxima when x=2nπ +π/2, relative minima

when x=2nπ −π/2, n=0,±1,±2,...

173 Chapter 5

31. f(x)= (2x−1)(x2+x−7)

(2x−1)(3x2+x−1) =x2+x−7

3x2+x−1,x6=1/2

horizontal asymptote: y=1/3,

vertical asymptotes: x=(−1±√

13)/6

-5

-4 -2 2 4

32. (a) f(x)= (x−2)(x2+x+ 1)(x2−2)

(x−2)(x2−2)2(x2+1)

2+x+1

2−2)(x2+1)

(b)

-2

-2 1 3

33. (a) sin x=−1 yields the smallest values, and

sin x= +1 yields the largest

(b) f0(x)=e

sin xcos x; relative maxima at x=2nπ +π/2, y=e; relative minima at x=2nπ −π/2,

y=1/e;n=0,±1,±2,... (ﬁrst derivative test)

2+t−1=0,

t=−1±√

2; sin x=−1−√5

2is impossible. So the points of inﬂection on 0 <x<2πoccur

when sin x=−1+√

2,orx=0.66624,2.47535

34. f0(x)=3ax2+2bx +c;f0(x)>0orf

(x)<0on(−∞,+∞)iff

(x) = 0 has no real solutions so

from the quadratic formula (2b)2−4(3a)c<0, 4b2−12ac < 0, b2−3ac < 0. If b2−3ac = 0, then

f0(x) = 0 has only one real solution at, say, x=cso fis always increasing or always decreasing on

both (−∞,c] and [c, +∞), and hence on (−∞,+∞) because fis continuous everywhere. Thus fis

always increasing or decreasing if b2−3ac ≤0.

35. (a) relative minimum −0.232466 at x=0.450184 2

-0.5

-1 1.5

Supplementary Exercises 174

(b) relative maximum 0 at x=0;

relative minimum −0.107587 at x=±0.674841

0.2

-0.15

-1.2 1.2

relative minimum −0.355977 at x=−1.244155

-0.4

-1.5 1.5

36. (a) f0(x) = 2+3x2−4x3has one real root at x=1.14, a relative max; so fis one-to-one for x≤1.14

(b) f(1.14)=3.07 so the domain of f−1is (−∞,3.07) and the range is (−∞,1.14); f−1(−1) = −0.70

37. (a) 0.5

(b) y=0atx= 0; lim

x→+∞y=0

(d) As aincreases, the x-coordinate of the maximum and the inﬂection point move towards the

origin.

38. (a) 4

-2 2

(b) y=1atx=a; lim

x→−∞ y=+∞, lim

x→+∞y=0

no relative extrema and no inﬂection points.

(d) An increase in bmakes the graph ﬂatter.

(e) An increase in ashifts the graph to the right.

175 Chapter 5

39. f0(x) = ln(1 + 1/x)−1

xand f00(x)= 1

(x+1);sof

00 >0ifx>1 and therefore f0is increasing on

[1,+∞). Next, f0(1)=ln2−1<0. Then by L’Hˆopital’s Rule,

lim

x→+∞xln(1 + 1/x) = lim

x→+∞

ln(1 + 1/x)

1/x = lim

x→+∞−1/x2

(1+1/x)(−1/x2)=1

and thus lim

x→+∞f0(x) = lim

x→+∞

xln(1 + 1/x)−1

xis indeterminate.

By L’Hˆopital’s Rule lim

x→+∞f0(x) = lim

x→+∞·ln µ1+ 1

x¶−1

x+1¸=0.

Thus on [1,+∞) the function f0starts negative and increases towards zero, so it is negative

on the whole interval. So f(x) is decreasing, and f(x)>f(x+ 1). Set x=nand obtain

ln(1 + 1/n)n+1 >ln(1 + 1/(n+ 1))n+2. Since ln xand its inverse function exare both increasing,

it follows that (1 + 1/n)n+1 >(1+1/(n+ 1))n+2.

CALCULUS HORIZON MODULE CHAPTER 5

1. The sum of the squares for the residuals for line I is approximately 12+12+12+02+22+12+12+12= 10,

and the same for line II is approximately 02+(0.4)2+(1.2)2+0

2+(2.2)2+(0.6)2+(0.2)2+0

2=6.84;

line II is the regression line.

2. (a) y=5.035714286x−4.232142857 (b) y

1234

4. r=0.9907002406

5. (a) S=2.155239850t+ 190.3600714; r=0.9569426456

(b) yes, because ris close to 1

(d) It is assumed that the line still gives a good estimate in the year 2000.

6. (a) Y=lny=bx +lnahas slope band Y-intercept ln a.

(b) y=a+bX has slope band y-intercept a.

(d) The same algebraic rules hold.

7. (a) y=3.923208367 + e0.2934589528x(b) y

1234567

Calculus Horizon Module 176

8. (a) It appears that log T=a+blog d,soT=10

, an exponential model.

(b) log T=1.719666407 ×10−4+1.499661719 log d

(d) “The squares of the periods of revolution of the planets are proportional to the cubes of their

mean distances”

9. (a) T=27+57.8e

−0.046t(b) T0=84.9

◦C(c) 53.19 min

CHAPTER 6

Applications of the Derivative

EXERCISE SET 6.1

1. relative maxima at x=2,6; absolute maximum at x= 6; relative and absolute minimum at x=4

2. relative maximum at x= 3; absolute maximum at x= 7; relative minima at x=1,5; absolute minima

at x=1,5

3. (a) y

(b) y

(c)

53 7

4. (a) y

(b) y

-5 5

5. f0(x)=8x−4, f0(x) = 0 when x=1/2; f(0)=1,f(1/2)=0,f(1) = 1 so the maximum value is 1

at x=0,1 and the minimum value is 0 at x=1/2.

6. f0(x)=8−2x,f

(x) = 0 when x=4;f(0)=0,f(4) = 16, f(6) = 12 so the maximum value is 16 at

x= 4 and the minimum value is 0 at x=0.

7. f0(x)=3(x−1)2,f0(x) = 0 when x=1;f(0) = −1, f(1)=0,f(4) = 27 so the maximum value is 27

at x= 4 and the minimum value is −1atx=0.

8. f0(x)=6x

2−6x−12=6(x+ 1)(x−2), f0(x) = 0 when x=−1,2; f(−2) = −4, f(−1)=7,

f(2) = −20, f(3) = −9 so the maximum value is 7 at x=−1 and the minimum value is −20 at x=2.

9. f0(x)=3/(4x2+1)

3/2

, no critical points; f(−1)=−3/√5, f(1)=3/

√

5 so the maximum value is

3/√5atx= 1 and the minimum value is −3/√5atx=−1.

10. f0(x)= 2(2x+1)

3(x2+x)1/3,f0(x) = 0 when x=−1/2 and f0(x) does not exist when x=−1,0;

f(−2)=2

2/3

,f(−1)=0,f(−1/2)=4

−2/3

,f(0)=0,f(3)=12

2/3so the maximum value is 122/3at

x= 3 and the minimum value is 0 at x=−1,0.

11. f0(x)=1−sec2x,f0(x)=0forxin (−π/4,π/4) when x=0;f(−π/4)=1−π/4, f(0)=0,

f(π/4) = π/4−1 so the maximum value is 1 −π/4atx=−π/4 and the minimum value is π/4−1

at x=π/4.

177

Exercise Set 6.1 178

12. f0(x) = cos x+ sin x,f0(x)=0forxin (0,π) when x=3π/4; f(0) = −1, f(3π/4)=√2, f(π)=1so

the maximum value is √2atx=3π/4 and the minimum value is −1atx=0.

13. f(x)=1+|9−x

|=½10 −x2,|x|≤3

−8+x

,|x|>3,f0(x)=½−2x, |x|<3

2x, |x|>3thus f0(x) = 0 when x=0,

(x) does not exist for xin (−5,1) when x=−3 because lim

x→−3−f0(x)6= lim

x→−3+f0(x) (see Theorem

preceding Exercise 75, Section 3.3); f(−5) = 17, f(−3)=1,f(0) = 10, f(1) = 9 so the maximum

value is 17 at x=−5 and the minimum value is 1 at x=−3.

14. f(x)=|6−4x|=½6−4x, x ≤3/2

−6+4x, x > 3/2,f0(x)=½−4,x<3/2

4,x>3/2

(x) does not exist when

x=3/2thus3/2 is the only critical point in (−3,3); f(−3) = 18, f(3/2)=0,f(3) = 6 so the

maximum value is 18 at x=−3 and the minimum value is 0 at x=3/2.

15. f0(x)=2x−3; critical point x=3/2. Minimum value f(3/2) = −13/4, no maximum.

16. f0(x)=−4(x+ 1); critical point x=−1. Maximum value f(−1) = 5, no minimum.

17. f0(x)=12x

(1 −x); critical points x= 0, 1. Maximum value f(1) = 1, no minimum because

lim

x→+∞f(x)=−∞.

18. f0(x)=4(x

3+ 1); critical point x=−1. Minimum value f(−1) = −3, no maximum.

19. No maximum or minimum because lim

x→+∞f(x)=+∞and lim

x→−∞ f(x)=−∞.

20. No maximum or minimum because lim

x→+∞f(x)=+∞and lim

x→−∞ f(x)=−∞.

21. f0(x)=x(x+2)/(x+1)

; critical point x=−2in(−5,−1). Maximum value f(−2) = −4, no

minimum.

22. f0(x)=−6/(x−3)2; no critical points in [−5,5] (x= 3 is not in the domain of f). No maximum or

minimum because lim

x→3+f(x)=+∞and lim

x→3−f(x)=−∞.

23. (x2−1)2can never be less than zero because it is the square of

x2−1; the minimum value is 0 for x=±1, no maximum because

lim

x→+∞f(x)=+∞.

-2 2

24. (x−1)2(x+2)

2can never be less than zero because it is the product

of two squares; the minimum value is 0 for x=1or−2, no

maximum because lim

x→+∞f(x)=+∞.

-3 2

179 Chapter 6

25. f0(x)=5(8 −x)

3x1/3,f0(x) = 0 when x= 8 and f0(x) does not exist

when x=0;f(−1) = 21, f(0)=0,f(8) = 48, f(20) = 0 so the

maximum value is 48 at x= 8 and the minimum value is 0 at

x=0,20.

-1 20

26. f0(x)=(2−x

)/(x

2+2)

,f0(x)=0forxin the interval (−1,4)

when x=√2; f(−1) = −1/3, f(√2) = √2/4, f(4)=2/9sothe

maximum value is √2/4atx=√

2 and the minimum value is

−1/3atx=−1.

0.4

-0.4

-1 4

27. f0(x)=−1/x2; no maximum or minimum because there are no

critical points in (0,+∞).

010

28. f0(x)=(1−x

)/(x

2+1)

; critical point x= 1. Maximum value

f(1)=1/2, minimum value 0 because f(x) is never less than zero

on [0,+∞) and f(0)=0.

0.5

020

29. f0(x)=2secxtan x−sec2x= (2 sin x−1)/cos2x,f0(x)=0forxin

(0,π/4) when x=π/6; f(0)=2,f(π/6) = √3, f(π/4)=2

√

2−1so

the maximum value is 2 at x= 0 and the minimum value is √3at

x=π/6.

1.5

30. f0(x)=2sinxcos x−sin x= sin x(2 cos x−1), f0(x)=0forxin

(−π, π) when x=0,±π/3; f(−π)=−1, f(−π/3)=5/4, f(0)=1,

f(π/3)=5/4, f(π)=−1 so the maximum value is 5/4at

x=±π/3 and the minimum value is −1atx=±π.

1.5

-1.5

Exercise Set 6.1 180

31. f0(x)=x

(2x−3)e−2x,f0(x)=0forxin [1,4] when x=3/2;

if x=1,3/2,4, then f(x)=e

−2

,27

8e−3,64e−8;

critical point at x=3/2; absolute maximum of 27

8e−3at x=3/2,

absolute minimum of 64e−8at x=4

0.2

32. f0(x)=1−ln x

x2,f0(x) = 0 when x=e;

absolute minimum of 0 at x=1;

absolute maximum of 1/e at x=e

0.4

33. f0(x)=−[cos(cos x)] sin x;f0(x) = 0 if sin x= 0 or if cos(cos x)=0.

If sin x= 0, then x=πis the critical point in (0,2π); cos(cos x)=0

has no solutions because −1≤cos x≤1. Thus f(0) = sin(1),

f(π) = sin(−1) = −sin(1), and f(2π) = sin(1) so the maximum

value is sin(1) ≈0.84147 and the minimum value is

−sin(1) ≈−0.84147.

-1

34. f0(x)=−[sin(sin x)] cos x;f0(x) = 0 if cos x= 0 or if sin(sin x)=0.

If cos x= 0, then x=π/2 is the critical point in (0,π);

sin(sin x) = 0 if sin x= 0, which gives no critical points in (0,π).

Thus f(0)=1,f(π/2) = cos(1), and f(π) = 1 so the maximum

value is 1 and the minimum value is cos(1) ≈0.54030.

1.5

35. f0(x)=½4,x<1

2x−5,x>1so f0(x) = 0 when x=5/2, and f0(x) does not exist when x= 1 because

lim

x→1−f0(x)6= lim

x→1+f0(x) (see Theorem preceding Exercise 75, Section 3.3); f(1/2)=0,f(1)=2,

f(5/2) = −1/4, f(7/2)=3/4 so the maximum value is 2 and the minimum value is −1/4.

36. f0(x)=2x+pwhich exists throughout the interval (0,2) for all values of pso f0(1) = 0 because

f(1) is an extreme value, thus 2 + p=0,p=−2. f(1) = 3 so 12+(−2)(1) + q=3,q= 4 thus

f(x)=x

2−2x+ 4 and f(0)=4,f(2) = 4 so f(1) is the minimum value.

37. sin 2xhas a period of π, and sin 4xa period of π/2sof(x) is periodic with period π. Consider the

interval [0,π]. f0(x) = 4 cos 2x+ 4 cos 4x,f0(x) = 0 when cos 2x+ cos 4x= 0, but cos 4x= 2 cos22x−1

(trig identity) so

2 cos22x+ cos 2x−1=0

(2 cos 2x−1)(cos 2x+1)=0

cos 2x=1/2 or cos 2x=−1.

From cos 2x=1/2, 2x=π/3or5π/3sox=π/6or5π/6. From cos 2x=−1, 2x=πso x=π/2.

f(0)=0,f(π/6)=3

√

3/2, f(π/2)=0,f(5π/6) = −3√3/2, f(π) = 0. The maximum value is 3√3/2

at x=π/6+nπ and the minimum value is −3√3/2atx=5π/6+nπ,n=0,±1,±2,···.

181 Chapter 6

38. cos x

3has a period of 6π, and cos x

2a period of 4π,sof(x) has a period of 12π. Consider the interval

[0,12π]. f0(x)=−sin x

3−sin x

2,f0(x) = 0 when sin x

3+ sin x

2= 0 thus, by use of the trig identity

sin a+ sin b= 2 sin a+b

2cos a−b

2, 2 sin µ5x

12 ¶cos ³−x

12´=0sosin

12 =0orcosx

12 = 0. Solve

sin 5x

12 = 0 to get x=12π/5, 24π/5, 36π/5, 48π/5 and then solve cos x

12 = 0 to get x=6π. The

corresponding values of f(x) are −4.0450,1.5450,1.5450,−4.0450,1,5,5 so the maximum value is 5

and the minimum value is −4.0450 (approximately).

39. Let f(x)=x−sin x, then f0(x)=1−cos xand so f0(x) = 0 when cos x= 1 which has no solution

for 0 <x<2πthus the minimum value of fmust occur at 0 or 2π.f(0)=0,f(2π)=2πso0isthe

minimum value on [0,2π]thusx−sin x≥0, sin x≤xfor all xin [0,2π].

40. Let f(x)=lnx−x+ 1, then f0(x)=1/x −1 and so f0(x)=0atx= 1. Since lim

x→0+f(x)=−∞ and

lim

x→+∞f(x)=−∞,f(x) has a maximum of f(1) = 0 at x= 1 and so f(x)≤0 for 0 <x<+∞,so

ln x≤x−1on(0,+∞).

41. Let m= slope at x, then m=f0(x)=3x

2−6x+5, dm/dx =6x−6; critical point for mis x=1,

minimum value of mis f0(1)=2

42. (a) f0(x)=−64 cos x

sin2x+27 sin x

cos2x=−64 cos3x+ 27 sin3x

sin2xcos2x,f0(x) = 0 when

27 sin3x= 64 cos3x, tan3x=64/27, tan x=4/3 so the critical point is x=x0where

tan x0=4/3 and 0 <x

0<π/2. To test x0ﬁrst rewrite f0(x)as

(x)= 27 cos3x(tan3x−64/27)

sin2xcos2x=27 cos x(tan3x−64/27)

sin2x;

if x<x

0then tan x<4/3 and f0(x)<0, if x>x

0then tan x>4/3 and f0(x)>0sof(x

)is

the minimum value. fhas no maximum because lim

x→0+f(x)=+∞.

(b) If tan x0=4/3 then (see ﬁgure)

sin x0=4/5 and cos x0=3/5

so f(x0)=64/sin x0+27/cos x0

=64/(4/5)+27/(3/5)

=80+45=125

43. f0(x)=2x(x

3−24x2+ 192x−640)

(x−8)3; real root of x3−24x2+ 192x−640 at x= 4(2 + 3

√2). Since

lim

x→8+f(x) = lim

x→+∞f(x)=+∞and there is only one relative extremum, it must be a minimum.

44. (a) dC

dt =K

a−b¡ae−at −be−bt¢so dC

dt =0att=ln(a/b)

a−b. This is the only stationary point and

C(0) = 0, lim

x→+∞C(t)=0,C(t)>0 for 0 <t<+∞, so it is an absolute maximum.

(b) 0.7

010

Exercise Set 6.2 182

45. The slope of the line is −1, and the slope of the tangent to y=−x2is −2xso −2x=−1, x=1/2.

The line lies above the curve so the vertical distance is given by F(x)=2−x+x

;F(−1)=4,

F(1/2)=7/4, F(3/2)=11/4. The point (1/2,−1/4) is closest, the point (−1,−1) farthest.

46. The slope of the line is 4/3; and the slope of the tangent to y=x3is 3x2so 3x2=4/3, x2=4/9,

x=±2/3. The line lies below the curve so the vertical distance is given by F(x)=x

3−4x/3+1;

F(−1)=4/3, F(−2/3)=43/27, F(2/3)=11/27, F(1)=2/3. The closest point is (2/3,8/27), the

farthest is (−2/3,−8/27).

47. The absolute extrema of y(t) can occur at the endpoints t=0,12 or when dy/dt = 2 sin t= 0, i.e.

t=0,12,kπ,k=1,2,3; the absolute maximum is y=4att=π, 3π; the absolute minimum is y=0

at t=0,2π.

48. (a) The absolute extrema of y(t) can occur at the endpoints t=0,2πor when

dy/dt = 2 cos 2t−4 sin tcos t= 2 cos 2t−2 sin 2t=0,t=0,2π, π/8,5π/8,9π/8,13π/8;

the absolute maximum is y=3.4142 at t=π/8,9π/8; the absolute minimum is y=0.5859

at t=5π/8,13π/8.

(b) The absolute extrema of x(t) occur at the endpoints t=0,2πor when dx

dt =−2 sin t+1

(2 + sin t)2=0,

t=7π/6,11π/6. The absolute maximum is x=0.5774 at t=11π/6 and the absolute minimum

is x=−0.5774 at t=7π/6.

49. f0(x)=2ax +b; critical point is x=−b

f00(x)=2a>0sofµ−b

¶is the minimum value of f, but

fµ−b

2a¶=aµ−b

2a¶2

+bµ−b

2a¶+c=−b2+4ac

4athus f(x)≥0 if and only if

fµ−b

2a¶≥0, −b2+4ac

4a≥0, −b2+4ac ≥0, b2−4ac ≤0

50. Use the proof given in the text, replacing “maximum” by “minimum” and “largest” by “smallest” and

reversing the order of all inequality symbols.

EXERCISE SET 6.2

1. Let x= one number, y= the other number, and P=xy where x+y= 10. Thus y=10−xso

P=x(10 −x)=10x−x

2for xin [0,10]. dP/dx =10−2x,dP/dx = 0 when x=5. Ifx=0,5,10

then P=0,25,0soPis maximum when x= 5 and, from y=10−x, when y=5.

2. Let xand ybe nonnegative numbers and zthe sum of their squares, then z=x2+y2. But x+y=1,

y=1−xso z=x2+(1−x)

2=2x

2−2x+ 1 for 0 ≤x≤1. dz/dx =4x−2, dz/dx = 0 when x=1/2.

If x=0,1/2,1 then z=1,1/2,1so

(a) zis as large as possible when one number is 0 and the other is 1.

(b) zis as small as possible when both numbers are 1/2.

3. If y=x+1/x for 1/2≤x≤3/2 then dy/dx =1−1/x2=(x

2−1)/x2,dy/dx = 0 when x=1. If

x=1/2,1,3/2 then y=5/2,2,13/6so

(a) yis as small as possible when x=1. (b) yis as large as possible when x=1/2.

183 Chapter 6

4. A=xy where x+2y= 1000 so y= 500 −x/2 and A= 500x−x2/2

for xin [0,1000]; dA/dx = 500 −x,dA/dx = 0 when x= 500. If

x= 0 or 1000 then A=0,ifx= 500 then A= 125,000 so the area

is maximum when x= 500 ft and y= 500 −500/2 = 250 ft.

Stream

5. Let xand ybe the dimensions shown in the ﬁgure and Athe area,

then A=xy subject to the cost condition 3(2x) + 2(2y) = 6000, or

y= 1500 −3x/2. Thus A=x(1500 −3x/2) = 1500x−3x2/2 for xin

[0,1000]. dA/dx = 1500 −3x,dA/dx = 0 when x= 500. If x=0or

1000 then A=0,ifx= 500 then A= 375,000 so the area is greatest

when x= 500 ft and (from y= 1500 −3x/2) when y= 750 ft.

Heavy-duty

Standard

6. Let xand ybe the dimensions shown in the ﬁgure and Athe area of

the rectangle, then A=xy and, by similar triangles,

x/6=(8−y)/8, y=8−4x/3soA=x(8 −4x/3)=8x−4x

/3 for

xin [0,6]. dA/dx =8−8x/3, dA/dx = 0 when x=3. Ifx=0,3,6

then A=0,12,0 so the area is greatest when x= 3 in and (from

y=8−4x/3) y= 4 in.

7. Let x,y, and zbe as shown in the ﬁgure and Athe area of the

rectangle, then A=xy and, by similar triangles, z/10 = y/6,

z=5y/3; also x/10=(8−z)/8=(8−5y/3)/8thus

y=24/5−12x/25 so A=x(24/5−12x/25)=24x/5−12x2/25 for

xin [0,10]. dA/dx =24/5−24x/25, dA/dx = 0 when x=5. If

x=0,5,10 then A=0,12,0 so the area is greatest when x= 5 in.

and y=12/5 in.

8. A=(2x)y=2xy where y=16−x

2so A=32x−2x

3for 0 ≤x≤4;

dA/dx =32−6x

,dA/dx = 0 when x=4/

√

3. If x=0,4/

√

3,4

then A=0,256/(3√3),0 so the area is largest when x=4/

√

3 and

y=32/3. The dimensions of the rectangle with largest area are

8/√3by32/3.

-4 4

9. A=xy where x2+y2=20

2= 400 so y=√400 −x2and

A=x√400 −x2for 0 ≤x≤20; dA/dx = 2(200 −x2)/√400 −x2,

dA/dx = 0 when x=√200=10

√

2. If x=0,10√2,20 then

A=0,200,0 so the area is maximum when x=10

√

2 and

y=√400 −200=10

√

2. x

10. Let xand ybe the dimensions shown in the ﬁgure, then the area of

the rectangle is A=xy.

But ³x

2´2+y2=R2,thusy=pR

2−x

/4= 1

p4R

2−x

2so

A=1

2xp4R2−x2for 0 ≤x≤2R.dA/dx =(2R

2−x

√

2−x

dA/dx = 0 when x=√2R.Ifx=0,√

2R, 2Rthen A=0,R

2,0so

the greatest area occurs when x=√2Rand y=√2R/2.

Exercise Set 6.2 184

11. Let x= length of each side that uses the $1 per foot fencing,

y= length of each side that uses the $2 per foot fencing.

The cost is C= (1)(2x) + (2)(2y)=2x+4y, but A=xy = 3200 thus y= 3200/x so

C=2x+ 12800/x for x>0,

dC/dx =2−12800/x2, dC/dx = 0 when x=80,d

C/dx2>0so

Cis least when x= 80, y= 40.

12. A=xy where 2x+2y=pso y=p/2−xand A=px/2−x2for x

in [0,p/2]; dA/dx =p/2−2x,dA/dx = 0 when x=p/4. If x=0or

p/2 then A=0,ifx=p/4 then A=p2/16 so the area is maximum

when x=p/4 and y=p/2−p/4=p/4, which is a square. x

13. Let xand ybe the dimensions of a rectangle; the perimeter is p=2x+2y. But A=xy thus

y=A/x so p=2x+2A/x for x>0, dp/dx =2−2A/x2=2(x

2−A)/x2,dp/dx = 0 when x=√A,

d2p/dx2=4A/x3>0ifx>0sopis a minimum when x=√Aand y=√Aand thus the rectangle

is a square.

14. With x,y,r, and sas shown in the ﬁgure, the sum of the enclosed

areas is A=πr2+s2where r=x

2πand s=y

4because xis the

circumference of the circle and yis the perimeter of the square, thus

A=x2

4π+y2

16. But x+y= 12, so y=12−xand

A=x2

4π+(12 −x)2

16 =π+4

16πx2−3

2x+ 9 for 0 ≤x≤12.

dx =π+4

8πx−3

2,dA

dx = 0 when x=12π

π+4.Ifx=0,12π

π+4,12

then A=9,36

π+4,36

πso the sum of the enclosed areas is

x y

cut

(a) a maximum when x= 12 in. (when all of the wire is used for the circle)

(b) a minimum when x=12π/(π+ 4) in.

15. (a) dN

dt = 250(20 −t)e−t/20 =0att= 20, N(0) = 125000, N(20) ≈161788, and N(100) ≈128,369;

the absolute maximum is N= 161788 at t= 20, the absolute minimum is N= 125000 at t=0.

(b) The absolute minimum of dN

dt occurs when d2N

dt2=12.5(t−40)e−t/20 =0,t= 40.

16. The area of the window is A=2rh +πr2/2, the perimeter is

p=2r+2h+πr thus h=1

2[p−(2+π)r]so

A=r[p−(2+π)r]+πr2/2

=pr −(2+π/2)r2for 0 ≤r≤p/(2+π),

dA/dr =p−(4+π)r,dA/dr = 0 when r=p/(4+π) and

d2A/dr2<0, so Ais maximum when r=p/(4+π). 2r

17. V=x(12 −2x)2for 0 ≤x≤6; dV/dx = 12(x−2)(x−6),

dV/dx = 0 when x= 2 for 0 <x<6. If x=0,2,6 then

V=0,128,0 so the volume is largest when x= 2 in.

12 − 2x

185 Chapter 6

18. The dimensions of the box will be (k−2x)by(k−2x)byxso V=(k−2x)

x=4x

3−4kx2+k2xfor

xin [0,k/2]. dV/dx =12x

2−8kx +k2=(6x−k)(2x−k), dV /dx = 0 for xin (0,k/2) when x=k/6.

If x=0,k/6,k/2 then V=0,2k

/27,0soVis maximum when x=k/6. The squares should have

dimensions k/6byk/6.

19. Let xbe the length of each side of a square, then V=x(3 −2x)(8 −2x)=4x

3−22x2+24xfor

0≤x≤3/2; dV/dx =12x

2−44x+ 24 = 4(3x−2)(x−3), dV/dx = 0 when x=2/3 for 0 <x<3/2.

If x=0,2/3,3/2 then V=0,200/27,0 so the maximum volume is 200/27 ft3.

20. Let x= length of each edge of base, y= height. The cost is

C= (cost of top and bottom) + (cost of sides) = (2)(2x2) + (3)(4xy)=4x

2+12xy, but

V=x2y= 2250 thus y= 2250/x2so C=4x

2+ 27000/x for x>0, dC/dx =8x−27000/x2,

dC/dx = 0 when x=3

√3375 = 15, d2C/dx2>0soCis least when x= 15, y= 10.

21. Let x= length of each edge of base, y= height, k=$/cm2for the sides. The cost is

C=(2k)(2x2)+(k)(4xy)=4k(x

2+xy), but V=x2y= 2000 thus y= 2000/x2so

C=4k(x

2+ 2000/x) for x>0dC/dx =4k(2x−2000/x2), dC/dx = 0 when

x=3

√1000 = 10, d2C/dx2>0soCis least when x= 10, y= 20.

22. Let xand ybe the dimensions shown in the ﬁgure and Vthe

volume, then V=x2y. The amount of material is to be 1000 ft2,

thus (area of base) + (area of sides) = 1000, x2+4xy = 1000,

y=1000 −x2

4xso V=x21000 −x2

4x=1

4(1000x−x3) for

0<x≤10√10.

dx =1

4(1000 −3x2), dV

dx = 0 when x=p1000/3=10

p10/3.

If x=0,10p10/3,10√10 then V=0,5000

3p10/3,0;

the volume is greatest for x=10

p10/3ftandy=5

p10/3 ft.

23. Let x= height and width, y= length. The surface area is S=2x

2+3xy where x2y=V,soy=V/x2

and S=2x

2+3V/x for x>0; dS/dx =4x−3V/x2,dS/dx = 0 when x=3

p3V/4, d2S/dx2>0soS

is minimum when x=3

r3V

4,y=4

r3V

24. Let rand hbe the dimensions shown in the ﬁgure, then the volume

of the inscribed cylinder is V=πr2h. But

r2+µh

2¶2

=R2thus r2=R2−h2

so V=πµR2−h2

4¶h=πµR2h−h3

4¶

for 0 ≤h≤2R.dV

dh =πµR2−3

4h2¶,dV

dh =0

when h=2R/√3. If h=0,2R/√3,2Rthen V=0,4π

√

,0 so the volume is largest when

h=2R/√3 and r=p2/3R.

Exercise Set 6.2 186

25. Let rand hbe the dimensions shown in the ﬁgure, then the surface

area is S=2πrh +2πr2.

But r2+µh

2¶2

=R2thus h=2

√

2−r

2so

S=4πr√R2−r2+2πr2for 0 ≤r≤R,

dr =4π(R2−2r2)

√R2−r2+4πr;dS

dr = 0 when

R2−2r2

√R2−r2=−r(i)

R2−2r2=−r√R2−r2

R4−4R2r2+4r

4=r

2−r

)

4−5R

2+R

4=0

and using the quadratic formula r2=5R2±√25R4−20R4

10 =5±√5

10 R2,r=s5±√5

10 R, of which

only r=s5+√

10 Rsatisﬁes (i). If r=0,s5+√

10 R, 0 then S=0,(5 + √5)πR2,2πR2so the surface

area is greatest when r=s5+√

10 Rand, from h=2

2−r

,h=2

s5−√

10 R.

26. Let Rand Hbe the radius and height of the cone, and rand hthe

radius and height of the cylinder (see ﬁgure), then the volume of the

cylinder is V=πr2h. By similar triangles (see ﬁgure) H−h

H=r

thus

h=H

R(R−r)soV=π

(R−r)r

2=π

(Rr2−r3) for 0 ≤r≤R.

dr =πH

R(2Rr −3r2)=πH

Rr(2R−3r), dV

dr = 0 for 0 <r<R

when r=2R/3. If r=0,2R/3,R then V=0,4πR2H/27,0sothe

maximum volume is 4πR2H

27 =4

3πR2H=4

9·(volume of cone).

27. From (13), S=2πr2+2πrh. But V=πr2hthus h=V/(πr2) and so S=2πr2+2V/r for r>0.

dS/dr =4πr −2V/r2, dS/dr =0ifr=3

pV/(2π). Since d2S/dr2=4π+4V/r3>0, the minimum

surface area is achieved when r=3

pV/2πand so h=V/(πr2)=[V/(πr3)]r=2r.

28. V=πr2hwhere S=2πr2+2πrh so h=S−2πr2

2πr ,V=1

2(Sr −2πr3) for r>0.

dr =1

2(S−6πr2)=0ifr=pS/(6π), d2V

dr2=−6πr < 0soVis maximum when

r=pS/(6π) and h=S−2πr2

2πr =S−2πr2

2πr2r=S−S/3

S/3r=2r, thus the height is equal to the

diameter of the base.

29. The surface area is S=πr2+2πrh where V=πr2h= 500 so

h= 500/(πr2) and S=πr2+ 1000/r for r>0;

dS/dr =2πr −1000/r2=(2πr3−1000)/r2,dS/dr = 0 when

r=3

p500/π,d2S/dr2>0 for r>0soSis minimum when

r=3

p500/π and h=500

πr2=500

πr3r=500

π(500/π)

p500/π

p500/π.

187 Chapter 6

30. The total area of material used is

A=Atop +Abottom +Aside =(2r)

2+(2r)

2+2πrh =8r

2+2πrh.

The volume is V=πr2hthus h=V/(πr2)soA=8r

2+2V/r for r>0,

dA/dr =16r−2V/r2= 2(8r3−V)/r2,dA/dr = 0 when r=3

√V/2. This is the only critical point,

d2A/dr2>0 there so the least material is used when r=3

√V/2, r

h=r

V/(πr2)=π

Vr3and, for

r=3

√V/2, r

h=π

8=π

31. Let xbe the length of each side of the squares and ythe height of the frame, then the volume is V=x2y.

The total length of the wire is Lthus 8x+4y=L,y=(L−8x)/4soV=x

(L−8x)/4=(Lx2−8x3)/4

for 0 ≤x≤L/8. dV /dx =(2Lx −24x2)/4, dV/dx = 0 for 0 <x<L/8 when x=L/12. If

x=0,L/12,L/8 then V=0,L

3/1728,0 so the volume is greatest when x=L/12 and y=L/12.

32. (a) Let x= diameter of the sphere, y= length of an edge of the cube. The combined volume is

V=1

6πx3+y3and the surface area is S=πx2+6y

2= constant. Thus y=(S−πx2)1/2

61/2and

V=π

6x3+(S−πx2)3/2

63/2for 0 ≤x≤rS

π;

dx =π

2x2−3π

63/2x(S−πx2)1/2=π

2√6x(√6x−pS−πx2). dV

dx = 0 when x= 0, or when

√6x=√S−πx2,6x

2=S−πx2,x2=S

6+π,x=rS

6+π.Ifx=0,rS

6+π,rS

π, then

V=S3/2

63/2,S3/2

6√6+π,S

3/2

√

πso that Vis smallest when x=rS

6+π, and hence when y=rS

6+π,

thus x=y.

(b) From part (a), the sum of the volumes is greatest when there is no cube.

33. Let hand rbe the dimensions shown in the ﬁgure, then the volume

is V=1

3πr2h. But r2+h2=L2thus r2=L2−h2so

V=1

3π(L2−h2)h=1

3π(L2h−h3) for 0 ≤h≤L.

dh =1

3π(L2−3h2). dV

dh = 0 when h=L/√3. If h=0,L/

√3,0

then V=0,2π

√

,0 so the volume is as large as possible when

h=L/√3 and r=p2/3L.

34. Let rand hbe the radius and height of the cone (see ﬁgure). The

slant height of any such cone will be R, the radius of the circular

sheet. Refer to the solution of Exercise 33 to ﬁnd that the largest

volume is 2π

9√3R3.hR

Exercise Set 6.2 188

35. The area of the paper is A=πrL =πr√r2+h2, but

V=1

3πr2h=10thush=30/(πr2)soA=πrpr2+ 900/(π2r4).

To simplify the computations let S=A2,

S=π2r2µr2+900

π2r4¶=π2r4+900

r2for r>0,

dr =4π

3−1800

r3=4(π2r6−450)

r3,dS/dr = 0 when

r=6

p450/π2,d2S/dr2>0, so Sand hence Ais least when

r=6

p450/π2,h=30

pπ2/450.

36. The area of the triangle is A=1

2hb. By similar triangles (see ﬁgure)

b/2

h=R

√h2−2Rh,b=2Rh

√h2−2Rh so A=Rh2

√h2−2Rh for h>2R,

dh =Rh2(h−3R)

(h2−2Rh)3/2,dA

dh = 0 for h>2Rwhen h=3R, by the ﬁrst

derivative test Ais minimum when h=3R.Ifh=3Rthen

b=2

√

3R(the triangle is equilateral).

h − R

bb/2

h2 − 2Rh

37. The volume of the cone is V=1

3πr2h. By similar triangles (see

ﬁgure) r

h=R

√h2−2Rh,r=Rh

√h2−2Rh so

V=1

3πR2h3

h2−2Rh =1

3πR2h2

h−2Rfor h>2R,

dh =1

3πR2h(h−4R)

(h−2R)2,dV

dh = 0 for h>2Rwhen h=4R,bythe

ﬁrst derivative test Vis minimum when h=4R.Ifh=4Rthen

r=√2R.

h − R

h2 − 2Rh

38. The area is (see ﬁgure)

A=1

2(2 sin θ)(4 + 4 cos θ)

= 4(sin θ+ sin θcos θ)

for 0 ≤θ≤π/2;

dA/dθ = 4(cos θ−sin2θ+ cos2θ)

= 4(cos θ−[1 −cos2θ] + cos2θ)

= 4(2 cos2θ+ cos θ−1)

= 4(2 cos θ−1)(cos θ+1)

dA/dθ = 0 when θ=π/3 for 0 <θ<π/2. If θ=0,π/3,π/2 then

A=0,3

√

3,4 so the maximum area is 3√3.

42 cos

4 cos

2 sin

39. Let band hbe the dimensions shown in the ﬁgure, then the

cross-sectional area is A=1

2h(5+b). But h= 5 sin θand

b= 5 + 2(5 cos θ)=5+10cosθso A=5

2sin θ(10 + 10 cos θ)

= 25 sin θ(1 + cos θ) for 0 ≤θ≤π/2.

dA/dθ =−25 sin2θ+ 25 cos θ(1 + cos θ)

= 25(−sin2θ+ cos θ+ cos2θ)

= 25(−1 + cos2θ+ cos θ+ cos2θ)

= 25(2 cos2θ+ cos θ−1) = 25(2 cos θ−1)(cos θ+1).

dA/dθ = 0 for 0 <θ<π/2 when cos θ=1/2, θ=π/3. If

θ=0,π/3,π/2 then A=0,75√3/4,25 so the cross-sectional area is

greatest when θ=π/3.

5 cos

h = 5 sin

189 Chapter 6

40. I=kcos φ

`2,kthe constant of proportionality. If his the height of the lamp above the table then

cos φ=h/` and `=√h2+r2so I=kh

`3=kh

(h2+r2)3/2for h>0, dI

dh =kr2−2h2

(h2+r2)5/2,dI

dh =0

when h=r/√2, by the ﬁrst derivative test Iis maximum when h=r/√2.

41. Let L,L1, and L2be as shown in the ﬁgure, then

L=L1+L2= 8 csc θ+ sec θ,

dθ =−8 csc θcot θ+ sec θtan θ, 0<θ<π/2

=−8 cos θ

sin2θ+sin θ

cos2θ=−8 cos3θ+ sin3θ

sin2θcos2θ;

dθ = 0 if sin3θ= 8 cos3θ, tan3θ= 8, tan θ= 2 which gives the

absolute minimum for Lbecause lim

θ→0+L= lim

θ→π/2−L=+∞.If

tan θ= 2, then csc θ=√5/2 and sec θ=√5so

L=8(

√

5/2)+√5=5

√

5 ft.

42. Let x= number of steers per acre

w= average market weight per steer

T= total market weight per acre

then T=xw where w= 2000 −50(x−20) = 3000 −50x

so T=x(3000 −50x) = 3000x−50x2for 0 ≤x≤60,

dT/dx = 3000 −100xand dT/dx = 0 when x= 30. If x=0,30,60 then T=0,45000,0 so the total

market weight per acre is largest when 30 steers per acre are allowed.

43. (a) The daily proﬁt is

P= (revenue) −(production cost) = 100x−(100,000+50x+0.0025x2)

=−100,000+50x−0.0025x2

for 0 ≤x≤7000, so dP/dx =50−0.005xand dP/dx = 0 when x=10,000. Because 10,000

is not in the interval [0,7000], the maximum proﬁt must occur at an endpoint. When x=0,

P=−100,000; when x= 7000, P= 127,500 so 7000 units should be manufactured and sold

daily.

(b) Yes, because dP/dx > 0 when x= 7000 so proﬁt is increasing at this production level.

44. (a) R(x)=px but p= 1000 −xso R(x) = (1000 −x)x

(b) P(x)=R(x)−C(x) = (1000 −x)x−(3000 + 20x)=−3000 + 980x−x2

that the proﬁt is a maximum when x= 490.

(d) P(490) = 237,100

(e) p= 1000 −x= 1000 −490 = 510.

45. The proﬁt is

P= (proﬁt on nondefective) −(loss on defective) = 100(x−y)−20y= 100x−120y

but y=0.01x+0.00003x2so P= 100x−120(0.01x+0.00003x2)=98.8x−0.0036x2for x>0,

dP/dx =98.8−0.0072x,dP/dx = 0 when x=98.8/0.0072 ≈13,722, d2P/dx2<0 so the proﬁt is

maximum at a production level of about 13,722 pounds.

Exercise Set 6.2 190

46. The total cost Cis

C=c·(hours to travel 3000 mi at a speed of vmi/h)

=c·3000

v=(a+bvn)3000

v= 3000(av−1+bvn−1) for v>0,

dC/dv = 3000[−av−2+b(n−1)vn−2] = 3000[−a+b(n−1)vn]/v2,

dC/dv = 0 when v=·a

b(n−1)¸1/n

. This is the only critical point and dC/dv changes sign from −to

+ at this point so the total cost is least when v=·a

b(n−1)¸1/n

mi/h.

47. The distance between the particles is D=p(1 −t−t)2+(t−2t)

2=√

2−4t+ 1 for t≥0. For

convenience, we minimize D2instead, so D2=5t

2−4t+1, dD2/dt =10t−4, which is 0 when t=2/5.

d2D2/dt2>0soD

2and hence Dis minimum when t=2/5. The minimum distance is D=1/

√

48. The distance between the particles is D=p(2t−t)2+(2−t

)

2=√

4−3t

2+ 4 for t≥0. For

convenience we minimize D2instead so D2=t4−3t2+4,dD2/dt =4t

3−6t=4t(t

2−3/2), which is

0 for t>0 when t=p3/2. d2D2/dt2=12t

2−6>0 when t=p3/2soD

2and hence Dis minimum

there. The minimum distance is D=√7/2.

49. Let P(x, y) be a point on the curve x2+y2= 1. The distance between P(x, y) and P0(2,0) is

D=p(x−2)2+y2, but y2=1−x

2so D=p(x−2)2+1−x

2=√

5−4xfor −1≤x≤1,

dx =−2

√5−4xwhich has no critical points for −1<x<1. If x=−1,1 then D=3,1sothe

closest point occurs when x= 1 and y=0.

50. Let P(x, y)beapointony=√

x, then the distance Dbetween Pand (2,0) is

D=p(x−2)2+y2=p(x−2)2+x=√x2−3x+ 4, for 0 ≤x≤3. For convenience we ﬁnd the

extrema for D2instead, so D2=x2−3x+4, dD2/dx =2x−3 = 0 when x=3/2. If x=0,3/2,3

then D2=4,7/4,4soD=2,√

7/2,2. The points (0,0) and (3,√3) are at the greatest distance, and

(3/2,p3/2) the shortest distance from (2,0).

51. Let (x, y) be a point on the curve, then the square of the distance between (x, y) and (0,2) is

S=x2+(y−2)2where x2−y2=1,x

2=y

2+1so

S=(y

2+1)+(y−2)2=2y

2−4y+ 5 for any y,dS/dy =4y−4, dS/dy = 0 when y=1,

S/dy2>0soSis least when y= 1 and x=±√2.

52. The square of the distance between a point (x, y) on the curve and the point (0,9) is

S=x2+(y−9)2where x=2y

2so S=4y

4+(y−9)2for any y,

dS/dy =16y

3+2(y−9) = 2(8y3+y−9), dS/dy = 0 when y= 1 (which is the only real solution),

d2S/dy2>0soSis least when y=1,x=2.

53. If P(x0,y

0) is on the curve y=1/x2, then y0=1/x2

0.AtPthe slope of the tangent line is −2/x3

0so its

equation is y−1

=−2

(x−x0), or y=−2

x+3

. The tangent line crosses the y-axis at 3

, and

the x-axis at 3

2x0. The length of the segment then is L=s9

4x2

0for x0>0. For convenience,

we minimize L2instead, so L2=9

4x2

0,dL2

dx0

=−36

2x0=9(x6

0−8)

2x5

, which is 0 when x6

0=8,

0=√

2. d2L2

dx2

>0soL

2and hence Lis minimum when x0=√2, y0=1/2.

191 Chapter 6

54. If P(x0,y

0) is on the curve y=1−x

, then y0=1−x

.AtPthe slope of the tangent line is −2x0so

its equation is y−(1 −x2

0)=−2x

(x−x

), or y=−2x0x+x2

0+ 1. The y-intercept is x2

0+ 1 and the

x-intercept is 1

2(x0+1/x0) so the area Aof the triangle is A=1

4(x2

0+1)(x0+1/x0)=1

+2x0+1/x0)

for 0 ≤x0≤1.

dA/dx0=1

4(3x2

0+2−1/x2

0)=1

(3x4

0+2x

0−1)/x2

0which is 0 when x2

0=−1 (reject), or

when x2

0=1/3sox

0=1/

√

3. d2A/dx2

0=1

4(6x0+2/x3

0)>0atx

0=1/

√

3 so a relative minimum and

hence the absolute minimum occurs there.

55. At each point (x, y) on the curve the slope of the tangent line is m=dy

dx =−2x

(1+x2)2for any x,

dx =2(3x2−1)

(1+x2)3,dm

dx = 0 when x=±1/√3, by the ﬁrst derivative test the only relative maximum

occurs at x=−1/√3, which is the absolute maximum because lim

x→±∞ m= 0. The tangent line has

greatest slope at the point (−1/√3,3/4).

56. Let xbe how far Pis upstream from where the man starts

(see ﬁgure), then the total time to reach Tis

t= (time from Mto P) + (time from Pto T)

=√x2+1

+1−x

for 0 ≤x≤1,

where rRand rWare the rates at which he can row and walk,

respectively. M

(a) t=√x2+1

3+1−x

5,dt

dx =x

3√x2+1−1

5so dt

dx = 0 when 5x=3

2+1,

25x2=9(x

2+1),x

2=9/16, x=3/4. If x=0,3/4,1 then t=8/15,7/15,√2/3 so the time is a

minimum when x=3/4 mile.

(b) t=√x2+1

4+1−x

5,dt

dx =x

4√x2+1−1

5so dt

dx = 0 when x=4/3 which is not in the interval

[0,1]. Check the endpoints to ﬁnd that the time is a minimum when x= 1 (he should row

directly to the town).

57. With xand yas shown in the ﬁgure, the maximum length of pipe

will be the smallest value of L=x+y. By similar triangles

8=x

√x2−16,y=8x

√x2−16 so

L=x+8x

√x2−16 for x>4, dL

dx =1−128

(x2−16)3/2,dL

dx = 0 when

(x2−16)3/2= 128

x2−16 = 1282/3= 16(22/3)

x2= 16(1 + 22/3)

x=4(1+2

2/3

)

1/2

− 16

d2L/dx2= 384x/(x2−16)5/2>0ifx>4soLis smallest when x=4(1+2

2/3

)

1/2

For this value of x,L=4(1+2

2/3

)

3/2ft.

Exercise Set 6.2 192

58. s=(x

1−¯x)

2+(x

2−¯x)

2+···+(x

n−¯x)

ds/d¯x=−2(x1−¯x)−2(x2−¯x)−···−2(xn−¯x),

ds/d¯x= 0 when

(x1−¯x)+(x

2−¯x)+···+(x

n−¯x)=0

1+x

2+···x

n)−(¯x+¯x+···+¯x)=0

1+x

2+···+x

n)−n¯x=0

¯x=1

1+x

2+···+x

n),

2s/d¯x2=2+2+···+2=2n>0, so sis minimum when ¯x=1

n(x1+x2+···+x

n).

59. Let x= distance from the weaker light source, I= the intensity at that point, and kthe constant of

proportionality. Then

I=kS

x2+8kS

(90 −x)2if 0 <x<90;

dx =−2kS

x3+16kS

(90 −x)3=2kS[8x3−(90 −x)3]

x3(90 −x)3=18

kS(x−30)(x2+ 2700)

x3(x−90)3,

which is 0 when x= 30; dI

dx <0ifx<30, and dI

dx >0ifx>30, so the intensity is minimum at a

distance of 30 cm from the weaker source.

60. If f(x0) is a maximum then f(x)≤f(x0) for all xin some open interval containing x0thus

pf(x)≤pf(x0) because √xis an increasing function, so pf(x0) is a maximum of pf(x)at

. The proof is similar for a minimum value, simply replace ≤by ≥.

61. Let v= speed of light in the medium. The total time required for the light to travel from Ato Pto

Bis

t= (total distance from Ato Pto B)/v =1

v(p(c−x)2+a2+px2+b2),

dx =1

v"−c−x

p(c−x)2+a2+x

√x2+b2#

and dt

dx = 0 when x

√x2+b2=c−x

p(c−x)2+a2. But x/√x2+b2= sin θ2and

(c−x)/p(c−x)2+a2= sin θ1thus dt/dx = 0 when sin θ2= sin θ1so θ2=θ1.

62. The total time required for the light to travel from Ato Pto Bis

t= (time from Ato P) + (time from Pto B)= √

2+a

+p(c−x)

2+b

dx =x

v1√x2+a2−c−x

v2p(c−x)2+b2but x/√x2+a2= sin θ1and

(c−x)/p(c−x)2+b2= sin θ2thus dt

dx =sin θ1

v1−sin θ2

so dt

dx = 0 when sin θ1

=sin θ2

63. (a) The rate at which the farmer walks is analogous to the speed of light in Fermat’s principle.

(b) the best path occurs when θ1=θ2

(see ﬁgure).

x1 − x

House

Barn

x/(1/4) = (1 −x)/(3/4)

3x=1−x

4x=1

x=1/4mi.

193 Chapter 6

EXERCISE SET 6.3

1. (a) positive, negative, slowing down (b) positive, positive, speeding up

2. (a) positive, slowing down (b) negative, slowing down

3. (a) left because v=ds/dt < 0att

(b) negative because a=d2s/dt2and the curve is concave down at t0(d2s/dt2<0)

(d) v<0 and a>0att

1so the particle is slowing down because vand ahave opposite signs.

4. (a) C(b) A(c) B

t(s)

s(m)

6. (a) when s≥0, so 0 <t<2 and 4 <t≤8(b) when the slope is zero, at t=3

–15

8. (a) v≈(30 −10)/(15 −10)=20/5=4m/s

(b)

2525

(1) (2)

9. (a) At 60 mi/h the slope of the estimated tangent line is about 4.6 mi/h/s. Use 1 mi = 5,280 ft

and 1 h = 3600 s to get a=dv/dt ≈4.6(5,280)/(3600) ≈6.7 ft/s2.

(b) The slope of the tangent to the curve is maximum at t=0s.

Exercise Set 6.3 194

10. (a) t1 2 3 4 5

s0.71 1.00 0.71 0.00 −0.71

v0.56 0.00 −0.56 −0.79 −0.56

a−0.44 −0.62 −0.44 0.00 0.44

(b) to the right at t= 1, stopped at t= 2, otherwise to the left

11. (a) v(t)=3t

2−12t,a(t)=6t−12

(b) s(1) = −5 ft, v(1) = −9 ft/s, speed = 9 ft/s, a(1) = −6 ft/s2

(d) for t≥0, v(t) changes sign at t= 4, and a(t) changes sign at t= 2; so the particle is speeding

up for 0 <t<2 and 4 <tand is slowing down for 2 <t<4

(e) total distance = |s(4) −s(0)|+|s(5) −s(4)|=|−32 −0|+|−25 −(−32)|=39ft

12. (a) v(t)=4t

3−4, a(t)=12t

(b) s(1) = −1 ft, v(1) = 0 ft/s, speed = 0 ft/s, a(1) = 12 ft/s2

(d) speeding up for t>1, slowing down for 0 <t<1

(e) total distance = |s(1) −s(0)|+|s(5) −s(1)|=|−1−2|+|607 −(−1)|= 611 ft

13. (a) v(t)=−(3π/2) sin(πt/2), a(t)=−(3π2/4) cos(πt/2)

(b) s(1) = 0 ft, v(1) = −3π/2 ft/s, speed = 3π/2 ft/s, a(1) = 0 ft/s2

(d) vchanges sign at t=0,2,4 and achanges sign at t=1,3,5, so the particle is speeding up for

0<t<1, 2 <t<3 and 4 <t<5, and it is slowing down for 1 <t<2 and 3 <t<4

(e) total distance = |s(2) −s(0)|+|s(4) −s(2)|+|s(5) −s(4)|

=|−3−3|+|3−(−3)|+|0−3|=15ft

14. (a) v(t)= 4−t

2+4)

2,a(t)= 2t(t

2−12)

(t2+4)

(b) s(1)=1/5 ft, v(1)=3/25 ft/s, speed = 3/25 ft/s, a(1) = −22/125 ft/s2

(d) achanges sign at t=2

√

3, so the particle is speeding up for 2 <t<2

√

3 and it is slowing down

for 0 <t<2 and for 2√3<t

(e) total distance = |s(2) −s(0)|+|s(5) −s(2)|=¯¯¯¯

4−0¯¯¯¯

+¯¯¯¯

29 −1

4¯¯¯¯

=19

58 ft

15. v(t)= 5−t

2+5)

2,a(t)= 2t(t

2−15)

(t2+5)

0.25

020

s(t)

0.2

-0.05

020

v(t)

0.01

-0.15

010

a(t)

195 Chapter 6

(a) v=0att=√

5(b) s=√5/10 at t=√5

15 and slowing down

for 0 <t<√

5 and √15 <t

16. v(t)=(1−t)e

−t

,a(t)=(t−2)e−t

0.4

010

s(t)

-0.2

v(t)

0.1

-2

a(t)

(a) v=0att=1 (b) s=1/e at t=1

0<t<1 and 2 <t

17. s=−3t+2

v=−3

a=0

Constant speed t = 0 s

18. s=t3−6t2+9t+1

v=3(t−1)(t−3)

a=6(t−2) 531

t = 1 (Stopped)

t = 0

(Stopped) t = 3t = 2

Speeding up

Slowing down

19. s=t3−9t2+24t

v=3(t−2)(t−4)

a=6(t−3)

2018160

t = 2 (Stopped)

t = 0

(Stopped) t = 4

t = 3

Speeding up

Slowing down

20. s=t+9

t+1

v=(t+ 4)(t−2)

(t+1)

a=18

(t+1)

(Stopped) t = 2

Speeding up

Slowing down

21. s=½cos t, 0≤t≤2π

1,t>2π

−sin t, 0≤t≤2π

0, t>2π

−cos t, 0≤t<2π

0,t>2π

10-1 s

Speeding up

t = π

t = π/2

t = 3π/2 t = 2π

t = 0

(Stopped

permanently)

Slowing down

22. v(t)=5t

2−6t+2

√

tis always positive, a(t)=15t2−6t−2

2t3/2has a positive root at t=3+√

Slowing down Speeding up

3 +

15 39

Exercise Set 6.3 196

23. (a) v=10t−22, speed = |v|=|10t−22|.d|v|/dt does not exist at t=2.2 which is the only critical

point. If t=1,2.2,3 then |v|=12,0,8. The maximum speed is 12 ft/s.

(b) the distance from the origin is |s|=|5t2−22t|=|t(5t−22)|, but t(5t−22) <0 for

1≤t≤3so|s|=−(5t2−22t)=22t−5t

,d|s|/dt =22−10t, thus the only critical point is

t=2.2. d2|s|/dt2<0 so the particle is farthest from the origin when t=2.2. Its position is

s= 5(2.2)2−22(2.2) = −24.2.

24. v=−200t

(t2+ 12)2, speed = |v|=200t

(t2+ 12)2for t≥0. d|v|

dt =600(4 −t2)

(t2+ 12)3= 0 when t= 2, which is the

only critical point in (0,+∞). By the ﬁrst derivative test there is a relative maximum, and hence an

absolute maximum, at t= 2. The maximum speed is 25/16 ft/s to the left.

25. s(t)=s

0−1

2gt2=s0−4.9t2m, v=−9.8tm/s, a=−9.8 m/s2

(a) |s(1.5) −s(0)|=11.025 m

(b) v(1.5) = −14.7 m/s

(d) s(t)−s0=−100 when 4.9t2= 100, t=4.5175 s

26. (a) s(t)=s

0−1

2gt2= 800 −16t2ft, s(t) = 0 when t=r800

16 =5

√

(b) v(t)=−32tand v(5√2) = −160√2≈226.27 ft/s = 154.28 mi/h

27. s(t)=s

0+v

t−1

2gt2=60t−4.9t

2m and v(t)=v

0−gt =60−9.8tm/s

(a) v(t) = 0 when t=60/9.8≈6.12 s

(b) s(60/9.8) ≈183.67 m

s=60t−4.9t

2about the line t=6.12 in the t-splane

(d) also 60 m/s, as seen from the symmetry of the parabola (or compute v(6.12))

28. (a) they are the same

(b) s(t)=v

t−1

2gt2and v(t)=v

0−gt;s(t) = 0 when t=0,2v

/g;

v(0) = v0and v(2v0/g)=v

0−g(2v0/g)=−v

0so the speed is the same

at launch (t= 0) and at return (t=2v

/g).

29. Ifg=32ft/s2,s0= 7 and v0is unknown, then s(t)=7+v

t−16t2and v(t)=v

0−32t;s=smax when

v=0,ort=v

/32; and smax = 208 yields 208 = s(v0/32)=7+v

/32) −16(v0/32)2=7+v

/64,

so v0=8

√

201 ≈113.42 ft/s.

30. (a) Use (6) and then (5) to get v2=v2

0−2v0gt +g2t2=v2

0−2g(v0t−1

2gt2)=v

0−2g(s−s

(b) Add v0to both sides of (6): 2v0−gt =v0+v,v0−1

2gt =1

2(v0+v);

from (5) s=s0+t(v0−1

2gt)=s

0+1

2(v

0+v)t

2gt =1

2(v0+v); from part (b), s=s0+1

2(v0+v)t=

s0+vt +1

2gt2

31. v0= 0 and g=9.8, so v2=−19.6(s−s0); since v= 24 when s= 0 it follows that 19.6s0=24

2or

s0=29.39 m.

32. s= 1000 + vt +1

2(32)t2= 1000 + vt +16t

2;s= 0 when t=5,sov=−(1000 + 16 ·52)/5=−280 ft/s.

197 Chapter 6

33. (a) s=smax when v=0,so0=v

0−2g(s

max −s0), smax =v2

0/2g+s0.

(b) s0=7,s

max = 208, g= 32 and v0is unknown, so from part (a) v2

0=2g(208 −7)=64·201,

v0=8

√

201 ≈113.42 ft/s.

34. s=t3−6t2+1,v=3t

2−12t,a=6t−12.

(a) a= 0 when t=2;s=−15, v=−12.

(b) v= 0 when 3t2−12t=3t(t−4)=0,t=0ort=4. Ift= 0, then s= 1 and a=−12; if t=4,

then s=−31 and a= 12.

35. (a) 1.5

(b) v=2t

√2t2+1, lim

t→+∞v=2

√2=√2

36. (a) a=dv

dt =dv

dt =vdv

ds because v=ds

(b) v=3

2√3t+7 =3

2s;dv

ds =−3

2s2;a=−9

4s3=−9/500

37. (a) s1=s2if they collide, so 1

2t2−t+3= −1

2+t+1, 3

2−2t+ 2 = 0 which has no real solution.

(b) Find the minimum value of D=|s1−s2|=¯¯3

4t2−2t+2

¯. From part (a), 3

4t2−2t+2

is never zero, and for t= 0 it is positive, hence it is always positive, so D=3

4t2−2t+2.

dt =3

2t−2 = 0 when t=4

3.d2D

dt2>0soDis minimum when t=4

3,D=2

2t+1. v

1<0if0≤t<1, v1>0ift>1; v2<0ift>2, v2>0if0≤t<2.

They are moving in opposite directions during the intervals 0 ≤t<1 and t>2.

38. (a) sA−sB=20−0=20ft

(b) sA=sB,15t

2+10t+20=5t

2+40t,10t

2−30t+20=0,(t−2)(t−1)=0,t=1ort=2s.

so car Bis ahead of car A.

39. (a) From the estimated tangent to the graph at the point where v= 2000, dv/ds ≈−1.25 ft/s/ft.

(b) a=v dv/ds ≈(2000)(−1.25) = −2500 ft/s2

40. r0(t)=2v(t)v

0(t)/[2pv2(t)] = v(t)a(t)/|v(t)|so r0(t)>0 (speed is increasing) if vand ahave the same

sign, and r0(t)<0 (speed is decreasing) if vand ahave opposite signs.

EXERCISE SET 6.4

1. f(x)=x

2−2, f0(x)=2x,x

n+1 =xn−x2

n−2

2xn

x1=1,x

2=1.5, x3=1.416666667,···,x

5=x

6=1.414213562

Exercise Set 6.4 198

2. f(x)=x

2−7, f0(x)=2x,x

n+1 =xn−x2

n−7

2xn

x1=3,x

2=2.666666667, x3=2.645833333,···,x

5=x

6=2.645751311

3. f(x)=x

3−6, f0(x)=3x

n+1 =xn−x3

n−6

3x2

x1=2,x

2=1.833333333, x3=1.817263545,···,x

5=x

6=1.817120593

4. xn−a=0

5. f(x)=x

3−x+3,f0(x)=3x

2−1, xn+1 =xn−x3

n−xn+3

n−1

1=−2, x2=−1.727272727, x3=−1.673691174,···,x

5=x

6=−1.671699882

6. f(x)=x

3+x−1, f0(x)=3x

2+1,x

n+1 =xn−x3

n+xn−1

3x2

n+1

1=1,x

2=0.75, x3=0.686046512,···,x

5=x

6=0.682327804

7. f(x)=x

5+x

4−5, f0(x)=5x

4+4x

n+1 =xn−x5

n+x4

n−5

5x4

n+4x

1=1,x

2=1.333333333, x3=1.239420573,···,x

6=x

7=1.224439550

8. f(x)=x

5−x+1,f0(x)=5x

4−1, xn+1 =xn−x5

n−xn+1

n−1

1=−1, x2=−1.25, x3=−1.178459394,···,x

6=x

7=−1.167303978

9. f(x)=x

4+x−3, f0(x)=4x

3+1,x

n+1 =xn−x4

n+xn−3

4x3

n+1

1=−2, x2=−1.645161290,

x3=−1.485723955,···,x

6=x

7=−1.452626879

-6

-2 2

10. f(x)=x

5−5x

3−2, f0(x)=5x

4−15x2,xn+1 =xn−x5

n−5x3

n−2

5x4

n−15x2

x1=2,x

2=2.5, x3=2.327384615,···,x

7=x

8=2.273791732

-20

-2.5 2.5

11. f(x)=2sinx−x,f

0(x)=2cosx−1, xn+1 =xn−2 sin xn−xn

2 cos xn−1

x1=2,x

2=1.900995594, x3=1.895511645, x4=x5=1.895494267

–7

199 Chapter 6

12. f(x) = sin x−x2,f0(x) = cos x−2x,xn+1 =xn−sin xn−x2

cos xn−2xn

x1=1,x

2=0.891395995,

x3=0.876984845,···,x

5=x

6=0.876726215

0.3

–1.3

0 1.5

13. f(x)=x−tan x,f0(x)=1−sec2x=−tan2x,

xn+1 =xn+xn−tan xn

tan2xn

x1=4.5, x2=4.493613903, x3=4.493409655,

x4=x5=4.493409458

100

–100

14. f(x)=1−e

xcos x,f0(x)=e

(sin x−cos x),

xn+1 =xn+1−excos x

ex(sin x−cos x)

x1=1,x

2=1.572512605, x3=1.363631415, x7=x8=1.292695719

-5

15. At the point of intersection, x3=0.5x−1, x3−0.5x+1=0. Let

f(x)=x

3−0.5x+ 1. By graphing y=x3and y=0.5x−1itis

evident that there is only one point of intersection and it occurs in

the interval [−2,−1]; note that f(−2) <0 and f(−1) >0.

f0(x)=3x

2−0.5so

n+1 =xn−x3

n−0.5x+1

n−0.5;x

1=−1, x2=−1.2,

x3=−1.166492147,···,

5=x

6=−1.165373043

-2

-2 2

16. The graphs of y=e−xand y=lnxintersect near x=1.3; let

f(x)=e

−x−ln x,f0(x)=−e

−x−1/x,x1=1.3,

xn+1 =xn+e−xn−ln xn

e−xn+1/xn

,x2=1.309759929, x4=x5=1.309799586

-4

17. The graphs of y=x2and y=√2x+ 1 intersect at points near

x=−0.5 and x=1;x

2=√

2x+1,x

4−2x−1=0. Let

f(x)=x

4−2x−1, then f0(x)=4x

3−2so

n+1 =xn−x4

n−2xn−1

4x3

n−2.

If x1=−0.5, then x2=−0.475, x3=−0.474626695,

x4=x5=−0.474626618; if x1= 1, then x2=2,

3=1.633333333,···,x

8=x

9=1.395336994.

-0.5 2

Exercise Set 6.4 200

18. The graphs of y=x3/8+1and y= cos 2xintersect at x= 0 and at

a point near x=−2;

x3/8 + 1 = cos 2x,x3−8 cos 2x+8=0. Letf(x)=x

3−8 cos 2x+8,

then f0(x)=3x

2+ 16 sin 2xso xn+1 =xn−x3

n−8 cos 2xn+8

n+ 16 sin 2xn

x1=−2, x2=−2.216897577,

x3=−2.193821581,···,x

5=x

6=−2.193618950.

-2

-3 2

19. (a) f(x)=x

2−a,f

0(x)=2x,x

n+1 =xn−x2

n−a

2xn

2µxn+a

xn¶

(b) a= 10; x1=3,x

2=3.166666667, x3=3.162280702, x4=x5=3.162277660

20. (a) f(x)= 1

x−a,f

0(x)=−1

2,x

n+1 =xn(2 −axn)

(b) a= 17; x1=0.05, x2=0.0575, x3=0.058793750, x5=x6=0.058823529

21. f0(x)=x

3+2x+ 5; solve f0(x) = 0 to ﬁnd the critical points. Graph y=x3and y=−2x−5tosee

that they intersect at a point near x=−1; f00(x)=3x

2+2so x

n+1 =xn−x3

n+2x

n+5

n+2 .

1=−1, x2=−1.4, x3=−1.330964467,···,x

5=x

6=−1.328268856 so the minimum value of f(x)

occurs at x≈−1.328268856 because f00(x)>0; its value is approximately −4.098859132.

22. From a rough sketch of y=xsin xwe see that the maximum occurs at a point near x= 2, which will

be a point where f0(x)=xcos x+ sin x=0. f

00(x)=2cosx−xsin xso

xn+1 =xn−xncos xn+ sin xn

2 cos xn−xnsin xn

=xn−xn+ tan xn

2−xntan xn

x1=2,x

2=2.029048281, x3=2.028757866, x4=x5=2.028757838; the maximum value is approxi-

mately 1.819705741.

23. Let f(x) be the square of the distance between (1,0) and any point (x, x2) on the parabola, then

f(x)=(x−1)2+(x

2−0)2=x4+x2−2x+ 1 and f0(x)=4x

3+2x−2. Solve f0(x) = 0 to ﬁnd

the critical points; f00(x)=12x

2+2 so x

n+1 =xn−4x3

n+2x

n−2

12x2

n+2 =x

n−2x

n+x

n−1

n+1 .x

1=1,

2=0.714285714, x3=0.605168701,···,x

6=x

7=0.589754512; the coordinates are approximately

(0.589754512,0.347810385).

24. The area is A=xy =xcos xso dA/dx = cos x−xsin x. Find xso that dA/dx =0;

A/dx2=−2 sin x−xcos xso xn+1 =xn+cos xn−xnsin xn

2 sin xn+xncos xn

=xn+1−xntan xn

2 tan xn+xn

x1=1,x

2=0.864536397, x3=0.860339078, x4=x5=0.860333589; y≈0.652184624.

25. (a) Let sbe the arc length, and Lthe length of the chord, then s=1.5L. But s=rθ and

L=2rsin(θ/2) so rθ =3rsin(θ/2), θ−3 sin(θ/2)=0.

(b) Let f(θ)=θ−3 sin(θ/2), then f0(θ)=1−1.5 cos(θ/2) so θn+1 =θn−θn−3 sin(θn/2)

1−1.5 cos(θn/2).

θ1=3,θ

2=2.991592920, θ3=2.991563137, θ4=θ5=2.991563136 rad so θ≈171◦.

26. r2(θ−sin θ)/2=πr2/4soθ−sin θ−π/2=0. Letf(θ)=θ−sin θ−π/2, then f0(θ)=1−cos θso

θn+1 =θn−sin θn−π/2

1−cos θn

θ1=2,θ

2=2.339014106, θ3=2.310063197,···,θ

5=θ

6=2.309881460 rad; θ≈132◦.

201 Chapter 6

27. If x= 1, then y4+y=1,y

4+y−1 = 0. Graph z=y4and z=1−yto see that they intersect near

y=−1 and y= 1. Let f(y)=y

4+y−1, then f0(y)=4y

3+1so y

n+1 =yn−y4

n+yn−1

4y3

n+1 .

If y1=−1, then y2=−1.333333333, y3=−1.235807860,···,y

6=y

7=−1.220744085;

if y1= 1, then y2=0.8, y3=0.731233596,···,y

6=y

7=0.724491959.

28. If x= 1, then 2y−cos y= 0. Graph z=2yand z= cos yto see that they intersect near y=0.5. Let

f(y)=2y−cos y, then f0(y) = 2 + sin yso yn+1 =yn−2yn−cos yn

2 + sin yn

y1=0.5, y2=0.450626693, y3=0.450183648, y4=y5=0.450183611.

29. S(25) = 250000 = 5000

i£(1+i)25 −1¤; set f(i)=50i−(1 + i)25 +1, f

0(i)=50−25(1 + i)24; solve

f(i)=0. Seti

0=.06 and ik+1 =ik−£50i−(1+i)25 +1

¤/£50 −25(1 + i)24¤. Then i1=0.05430,

i2=0.05338, i3=0.05336, ···,i=0.053362.

30. (a) x1=2,x

2=5.3333,

x3=11.055, x4=22.293,

x5=44.676

0.5

015

(b) x1=0.5, x2=−0.3333, x3=0.0833, x4=−0.0012, x5=0.0000 (and xn= 0 for n≥6)

31. (a) x1x2x3x4x5x6x7x8x9x10

0.5000 −0.7500 0.2917 −1.5685 −0.4654 0.8415 −0.1734 2.7970 1.2197 0.1999

(b) The sequence xnmust diverge, since if it did converge then f(x)=x

2+ 1 = 0 would have a

solution. It seems the xnare oscillating back and forth in a quasi-cyclical fashion.

EXERCISE SET 6.5

1. f(0) = f(4)=0;f

0(3)=0;[0,4], c=3 2. f(−3) = f(3)=0;f

0(0)=0

3. f(2) = f(4)=0,f

0(x)=2x−6, 2c−6=0,c=3

4. f(0) = f(2)=0,f

0(x)=3x

2−6x+2,3c

2−6c+2=0;c=6±√

36 −24

6=1±√

3/3

5. f(π/2) = f(3π/2)=0,f

0(x)=−sin x,−sin c=0,c=π

6. f(−1) = f(1)=0,f

0(x)=x

2−4x+1

(x−2)2,c2−4c+1

(c−2)2=0,c

2−4c+1=0

c=4±√

16 −4

2=2±√

3, of which only c=2−√

3isin(−1,1)

7. f(0) = f(4)=0,f

0(x)=1

2−1

√

2−1

√

c=0,c=1

8. f(1) = f(3)=0,f

0(x)=−2

3+4

2,−2

3+4

2=0,−6+4c=0,c=3/2

Exercise Set 6.5 202

9. f(8) −f(0)

8−0=6

8=3

4=f0(1.54); c=1.54 10. f(4) −f(0)

4−0=1.19 = f0(0.77)

11. f(−4) = 12, f(6) = 42, f0(x)=2x+1, 2c+1= 42 −12

6−(−4) =3,c=1

12. f(−1) = −6, f(2)=6,f

0(x)=3x

2+1, 3c

2+1= 6−(−6)

2−(−1) =4,c

2=1,c=±1 of which only c=1is

in (−1,2)

13. f(0)=1,f(3)=2,f

0(x)= 1

√

x+1,1

√

c+1 =2−1

3−0=1

3,√

c+1=3/2, c+1=9/4, c=5/4

14. f(3)=10/3, f(4)=17/4, f0(x)=1−1/x2,1−1/c2=17/4−10/3

4−3=11/12, c2= 12, c=±2√3of

which only c=2

√

3isin(3,4)

15. f(−5)=0,f(3)=4,f

0(x)=−x

√

25 −x2,−c

√25 −c2=4−0

3−(−5) =1

2,−2c=√25 −c2,

4c2=25−c

2=5,c=−

√

(we reject c=√5 because it does not satisfy the equation −2c=√25 −c2)

16. f(2)=1,f(5)=1/4, f0(x)=−1/(x−1)2,−1

(c−1)2=1/4−1

5−2=−1

4,(c−1)2=4,c−1=±2,

c=−1 (reject), or c=3

17. (a) f(−2) = f(1)=0 (b) c=−1.29

-2

-2 1

18. (a) m=f(−2) −f(1)

−2−1=−16 −5

−3=7soy−5=7(x−1),

y=7x−2

(b) f0(x)=3x

2+ 4 = 7 has solutions x=±1;

discard x=1,soc=−1

(d) 5

-20

-2 1

19. (a) f0(x) = sec2x, sec2c= 0 has no solution (b) tan xis not continuous on [0,π]

20. (a) f(−1)=1,f(8)=4,f

0(x)=2

−1/3

−1/3=4−1

8−(−1) =1

3,c1/3=2,c= 8 which is not in (−1,8).

(b) x2/3is not diﬀerentiable at x= 0, which is in (−1,8).

203 Chapter 6

21. (a) Two x-intercepts of fdetermine two solutions aand bof f(x) = 0; by Rolle’s Theorem there

exists a point cbetween aand bsuch that f0(c) = 0, i.e. cis an x-intercept for f0.

(b) f(x) = sin x=0atx=nπ, and f0(x) = cos x=0atx=nπ +π/2, which lies between nπ and

(n+1)π,(n=0,±1,±2,...)

22. f(x1)−f(x0)

x1−x0

is the average rate of change of ywith respect to xon the interval [x0,x

1]. By the

Mean-Value Theorem there is a value cin (x0,x

1) such that the instantaneous rate of change f0(c)=

f(x

)−f(x

)

1−x

23. Let s(t) be the position function of the automobile for 0 ≤t≤5, then by the Mean-Value Theorem

there is at least one point cin (0,5) where

s0(c)=v(c)=[s(5) −s(0)]/(5 −0)=4/5=0.8 mi/min = 48 mi/h.

24. Let T(t) denote the temperature at time with t= 0 denoting 11 AM, then T(0) = 76 and T(12) = 52.

(a) By the Mean-Value Theorem there is a value cbetween 0 and 12 such that

T0(c)=[T(12) −T(0)]/(12 −0) = (52 −76)/(12) = −2◦F/h.

(b) Assume that T(t1)=88

◦F where 0 <t

1<12, then there is at least one point cin (t1,12) where

T0(c)=[T(12) −T(t1)]/(12 −t1) = (52 −88)/(12 −t1)=−36/(12 −t1). But 12 −t1<12 so

T0(c)<−36/12 = −3◦F/h.

25. Let f(t) and g(t) denote the distances from the ﬁrst and second runners to the starting point, and let

h(t)=f(t)−g(t). Since they start (at t= 0) and ﬁnish (at t=t1) at the same time, h(0) = h(t1)=0,

so by Rolle’s Theorem there is a time t2for which h0(t2) = 0, i.e. f0(t2)=g

); so they have the

same velocity at time t2.

26. f(x)=x

6−2x

2+xsatisﬁes f(0) = f(1) = 0, so by Rolle’s Theorem f0(c) = 0 for some cin (0,1).

27. (a) By the Constant Diﬀerence Theorem f(x)−g(x)=kfor some k; since f(x0)=g(x

), k=0,so

f(x)=g(x) for all x.

(b) Set f(x) = sin2x+ cos2x,g(x) = 1; then f0(x)=2sinxcos x−2 cos xsin x=0=g

(x). Since

f(0)=1=g(0), f(x)=g(x) for all x.

28. (a) By the Constant Diﬀerence Theorem f(x)−g(x)=kfor some k; since f(x0)−g(x0)=c,k=c,

so f(x)−g(x)=cfor all x.

(b) Set f(x)=(x−1)3,g(x)=(x

2+ 3)(x−3). Then

f0(x)=3(x−1)2,g0(x)=(x

2+3)+2x(x−3)=3x

2−6x+3=3(x

2−2x+1)=3(x−1)2,

so f0(x)=g

0(x) and hence f(x)−g(x)=k. Expand f(x) and g(x)toget

h(x)=f(x)−g(x)=(x

3−3x

2+3x−1) −(x3−3x2+3x−9)=8.

3−3x

2+3x−1−(x

3−3x

2+3x−9)=8

29. (a) If x, y belong to Iand x<ythen for some cin I,f(y)−f(x)

y−x=f0(c),

so |f(x)−f(y)|=|f0(c)||x−y|≤M|x−y|;ifx>yexchange xand y;ifx=ythe inequality

also holds.

(b) f(x) = sin x,f0(x) = cos x,|f0(x)|≤1=M,so|f(x)−f(y)|≤|x−y|or |sin x−sin y|≤|x−y|.

30. (a) If x, y belong to Iand x<ythen for some cin I,f(y)−f(x)

y−x=f0(c),

so |f(x)−f(y)|=|f0(c)||x−y|≥M|x−y|;ifx>yexchange xand y;if

x=ythe inequality also holds.

(b) If xand ybelong to (−π/2,π/2) and f(x) = tan x, then |f0(x)|= sec2x≥1 and

|tan x−tan y|≥|x−y|

Exercise Set 6.5 204

31. (a) Let f(x)=√

x. By the Mean-Value Theorem there is a number cbetween xand ysuch that

√y−√x

y−x=1

2√c<1

2√xfor cin (x, y), thus √y−√x<y−x

√

(b) multiply through and rearrange to get √xy < 1

2(x+y).

32. Suppose that f(x) has at least two distinct real solutions r1and r2in I. Then

f(r1)=f(r

) = 0 so by Rolle’s Theorem there is at least one number between r1and r2where

f0(x) = 0, but this contradicts the assumption that f0(x)6=0,sof(x) = 0 must have fewer than two

distinct solutions in I.

33. (a) If f(x)=x

3+4x−1 then f0(x)=3x

2+ 4 is never zero, so by Exercise 32 fhas at most one real

root; since fis a cubic polynomial it has at least one real root, so it has exactly one real root.

(b) Let f(x)=ax3+bx2+cx +d.Iff(x) = 0 has at least two distinct real solutions r1and r2, then

f(r1)=f(r

) = 0 and by Rolle’s Theorem there is at least one number between r1and r2where

f0(x)=0. Butf

0(x)=3ax2+2bx +c= 0 for

x=(−2b±√

2−12ac)/(6a)=(−b±√

2−3ac)/(3a), which are not real if b2−3ac < 0

so f(x) = 0 must have fewer than two distinct real solutions.

34. f0(x)= 1

√

c=√

4−√

4−3=2−√

3. But 1

4<1

2√c<1

2√3for cin (3,4) so

4<2−√3<1

2√3,0.25 <2−√3<0.29, −1.75 <−√3<−1.71, 1.71 <√3<1.75.

35. (a) d

dx[f2(x)+g

(x)]=2f(x)f

0(x)+2g(x)g

0(x)=2f(x)g(x)+2g(x)[−f(x)]=0,

so f2(x)+g

(x) is constant.

(b) f(x) = sin xand g(x) = cos x

36. (a) d

dx[f2(x)−g2(x)]=2f(x)f

0(x)−2g(x)g

0(x)=2f(x)g(x)−2g(x)f(x)=0sof

(x)−g

(x)

is constant.

(b) f0(x)=1

2(e

x−e

−x

)=g(x) and g0(x)=1

2(e

x+e

−x

)=f(x)

37. If f0(x)=g

0(x), then f(x)=g(x)+k. Let x=1,

f(1) = g(1) + k= (1)3−4(1)+6+k=3+k=2,sok=−1. f(x)=x

3−4x+5.

38. Let h=f−g, then his continuous on [a, b], diﬀerentiable on (a, b), and h(a)=f(a)−g(a)=0,

h(b)=f(b)−g(b) = 0. By Rolle’s Theorem there is some cin (a, b) where h0(c)=0. But

0(c)=f

0(c)−g

0(c)sof

(c)−g

(c)=0,f

0(c)=g

0(c).

39. y

205 Chapter 6

40. (a) Suppose f0(x) = 0 more than once in (a, b), say at c1and c2. Then f0(c1)=f

)=0and

by using Rolle’s Theorem on f0, there is some cbetween c1and c2where f00(c) = 0, which

contradicts the fact that f00(x)>0sof

(x) = 0 at most once in (a, b).

(b) If f00(x)>0 for all xin (a, b), then fis concave up on (a, b) and has at most one relative

extremum, which would be a relative minimum, on (a, b).

41. similar to the proof of part (a) with f0(c)<0

42. similar to the proof of part (a) with f0(c)=0

CHAPTER 6 SUPPLEMENTARY EXERCISES

3. (a) If fhas an absolute extremum at a point of (a, b) then it must, by Theorem 6.1.4, be at a critical

point of f; since fis diﬀerentiable on (a, b) the critical point is a stationary point.

(b) It could occur at a critical point which is not a stationary point: for example, f(x)=|x|on

[−1,1] has an absolute minimum at x= 0 but is not diﬀerentiable there.

4. No; speeding up means the velocity and acceleration have the same sign, i.e. av > 0; the velocity is

increasing when the acceleration is positive, i.e. a>0. These are not the same thing. An example is

s=t−t2at t= 1, where v=−1 and a=−2, so av > 0 but a<0.

5. Yes; by the Mean Value Theorem there is a point cin (a, b) such that f0(c)=f(b)−f(a)

b−a=0.

7. (a) f0(x)=−1/x26= 0, no critical points; by inspection M=−1/2atx=−2; m=−1atx=−1

(b) f0(x)=3x

2−4x

3=0atx=0,3/4; f(−1) = −2,f(0)=0,f(3/4)=27/256,f(3/2) = −27/16,

so m=−2atx=−1, M=27/256 at x=3/4

3(x−2)2/3, critical points at x=12/7,2; m=f(12/7) = 144

49 µ−2

7¶1/3

≈−1.9356 at

x=12/7, M=9atx=3

(d) lim

x→0+f(x) = lim

x→+∞f(x)=+∞and f0(x)= e

(x−2)

x3, stationary point at x= 2; by Theorem

6.1.5 f(x) has an absolute minimum at x= 2, and m=e2/4.

8. (a) f0(x)=2 3−x

2+3)

2, critical point at x=√3. Since lim

x→0+f(x)=0,f(x) has no minimum, and

M=√3/3atx=√

(b) f0(x)=10x

(x−2), critical points at x=0,2; lim

x→3−f(x) = 88, so f(x) has no maximum; m=−9

at x=2

(d) f0(x)=(1+lnx)x

, critical point at x=1/e; lim

x→0+f(x) = lim

x→0+exln x= 1, lim

x→+∞f(x)=+∞;no

absolute maximum, absolute minimum m=e−1/e at x=1/e

9. x=2.3561945 10. x=−2.11491,0.25410,1.86081

11. (a) yes; f0(0)=0 (b) no, fis not diﬀerentiable on (−1,1)

12. (a) no, fis not diﬀerentiable on (−2,2) (b) yes, f(3) −f(2)

3−2=−1=f

0(1+√2)

Supplementary Exercises 6 206

x→1−f(x) = 2, lim

x→1+f(x)=2sofis continuous on [0,2]; lim

x→1−f0(x) = lim

x→1−−2x=−2 and

lim

x→1+f0(x) = lim

x→1+(−2/x2)=−2, so fis diﬀerentiable on (0,2); and f(2) −f(0)

2−0=−1=f

√

13. Let kbe the amount of light admitted per unit area of clear glass. The total amount of light admitted

by the entire window is

T=k·(area of clear glass) + 1

2k·(area of blue glass) = 2krh +1

4πkr2.

But P=2h+2r+πr which gives 2h=P−2r−πr so

T=kr(P−2r−πr)+1

4πkr2=khPr−³2+π−π

4´r

=k·Pr−8+3π

¸for 0 <r< P

2+π,

dr =kµP−8+3π

¶,dT

dr = 0 when r=2P

8+3π.

This is the only critical point and d2T/dr2<0 there so the most light is admitted when

r=2P/(8+3π) ft.

14. If one corner of the rectangle is at (x, y) with x>0, y>0, then A=4xy,y=3

1−(x/4)2,

A=12x

p1−(x/4)2=3x

√

16 −x2,dA

dx =6 8−x

√

16 −x2, critical point at x=2

√

2. Since A= 0 when

x=0,4 and A>0 otherwise, there is an absolute maximum A=24atx=2

√

15. (a) If a=k, a constant, then v=kt +bwhere bis constant; so

the velocity changes sign at t=−b/k.

b-b/k

(b) Consider the equation s=5−t

/6, v=−t2/2, a=−t. Then

for t>0, ais decreasing and av > 0, so the particle is

speeding up.

16. s(t)=t/(t2+ 5), v(t)=(5−t

)/(t

2+5)

,a(t)=2t(t

2−15)/(t2+5)

(a) 0.25

020

s(t)

0.2

-0.05

020

v(t)

0.01

-0.15

010

a(t)

(b) vchanges sign at t=√5(c) s=√5/10, v=0,a=−

√

5/50

207 Chapter 6

(d) achanges sign at t=√15, so the particle is speeding up for √5<t<√

15, and it is slowing

down for 0 <t<√

5 and √15 <t

(e) v(0)=1/5, lim

t→+∞v(t)=0,v(t) has one t-intercept at t=√5 and v(t) has one critical point at

t=√15. Consequently the maximum velocity occurs when t= 0 and the minimum velocity

occurs when t=√15.

17. (a) s(t)=s

0+v

t−1

gt2=v0t−4.9t2,v(t)=v

0−9.8t;s

max occurs when v= 0, i.e. t=v0/9.8,

and then 0.76 = smax =v0(v0/9.8) −4.9(v0/9.8)2=v2

0/19.6, so v0=√0.76 ·19.6=3.86 m/s and

s(t)=3.86t−4.9t2. Then s(t) = 0 when t=0,0.7878, s(t)=0.15 when t=0.0410,0.7468, and

s(t)=0.76−0.15=0.61 when t=0.2188,0.5689, so the player spends 0.5689−0.2188 = 0.3501 s

in the top 15.0 cm of the jump and 0.0410 + (0.7878 −0.7468) = 0.0820 s in the bottom

15.0 cm.

(b) The height vs time plot is a parabola that opens down, and the slope is smallest near the top

of the parabola, so a given change ∆hin height corresponds to a large time change ∆tnear the

top of the parabola and a narrower time change at points farther away from the top.

18. (a) s(t)=s

0+v

0−4.9t

; assume s0=v0=0,sos(t)=−4.9t

,v(t)=−9.8t

t0 1 2 3 4

s0−4.9−19.6−44.1−78.4

v0−9.8−19.6−29.4−39.2

(b) The formula for vis linear (with no constant term).

19. (a) 2.1

-0.5

-10 10

(b) minimum: (−2.111985,−0.355116)

maximum: (0.372591,2.012931)

20. (a) y

-2

-1.5

-1

-0.5

0.2 0.6 1

(b) The distance between the boat and the origin is px2+y2, where y=(x

10/3−1)/(2x2/3). The

minimum distance is 0.8247 mi when x=0.6598 mi. The boat gets swept downstream.

dy/dt =−4 and ﬁnd the value of dy/dx for the value of xobtained in part (b) to get

dx/dt =−3 mi/h.

21. (a) v=−2t(t4+2t

2−1)

(t4+1)

2,a=2

8+10t

6−12t4−6t2+1

4+1)

Supplementary Exercises 6 208

(b) s

0.2

0.4

0.6

0.8

1.2

123456

-1

-0.8

-0.6

-0.4

-0.2

0.2

0.4

123456

-3

-2

-1

123456

(d) Find tso that the velocity v=ds/dt > 0. The particle is moving in the positive direction for

0≤t≤0.64 s.

(e) It is speeding up when a, v > 0ora, v < 0, so for 0 ≤t<0.36 and 0.64 <t<1.1, otherwise it

is slowing down.

(f) Find the maximum value of |v|to obtain: maximum speed = 1.05 m/s when t=1.10 s.

22. Find tso that N0(t) is maximum. The size of the population is increasing most rapidly when t=8.4

years.

23. Solve φ−0.0167 sin φ=2π(90)/365 to get φ=1.565978 so

r= 150 ×106(1 −0.0167 cos φ) = 149.988 ×106km.

24. Solve φ−0.0934 sin φ=2π(1)/1.88 to get φ=3.325078 so

r= 228 ×106(1 −0.0934 cos φ) = 248.938 ×106km.

CHAPTER 7

Integration

EXERCISE SET 7.1

1. A= 1(1)/2=1/2; ∆x=(b−a)/n =1/n,x∗

k=k/n,f(x∗

k)=k/n,An=·1

n+2

n+···+n

n¸1

n12345678910

An1.0000 0.7500 0.6666 0.6250 0.6000 0.5833 0.5714 0.5625 0.5556 0.5500

2. A= 4(2)/2=4;∆x=(b−a)/n =2/n,x∗

k=2k/n,f(x∗

k)=4−2(2k/n)=4−4k/n = 4(1 −k/n),

An=4·µ1−1

n¶+µ1−2

n¶+···+³1−n

n´¸2

n1 2 3 4 5 6 7 8 9 10

An0.0000 2.0000 2.6667 3.0000 3.2000 3.3333 3.4286 3.50000 3.5556 3.6000

3. A= 2(2 + 14)/2 = 16; ∆x=(b−a)/n =2/n,x∗

k=2k/n,f(x∗

k)=2+12k/n = 2(1+6k/n),

An=2·µ1+ 6

n¶+µ1+ 12

n¶+···+µ1+6n

n¶¸2

n12345678910

An28.0000 22.0000 20.0000 19.0000 18.4000 18.0000 17.7143 17.5000 17.3333 17.2000

4. A=π(1)2/4=π/4; ∆x=(b−a)/n =1/n,x∗

k=k/n,f(x∗

k)=p1−(k/n)2,

An=hp1+(1/n)2+p1+(2/n)2+···+p1+(n/n)2i1

n1 2 3 4 5 6 7 8 9 10

An0.0000 0.4330 0.5627 0.6239 0.6593 0.6822 0.6982 0.7100 0.7190 0.7261

5. A(1) −A(0)=1/26. A(2) −A(0)=4

7. A(2) −A(0)=16 8. A(1) −A(0) = 1

2=π/4

9. A(x)=e

, area = A(1) −A(0) = e−110. A(x)=−cos x, area = A(π)−A(0)=2

EXERCISE SET 7.2

1. (a) Zx

√1+x

2dx =p1+x

2+C(b) Z(x+1)e

dx =xex+C

2. (a) d

dx(sin x−xcos x+C) = cos x−cos x+xsin x=xsin x

(b) d

dx µx

√1−x2+C¶=√1−x2+x2/√1−x2

1−x2=1

(1 −x2)3/2

3. d

dx hpx3+5

i=3x

√

3+5 so Z3x2

2√x3+5dx =px3+5+C

4. d

dx ·x

x2+3¸=3−x

2+3)

2so Z3−x2

(x2+3)

2dx =x

x2+3+C

209

Exercise Set 7.2 210

5. d

dx £sin ¡2√x¢¤=cos (2√x)

√xso Zcos (2√x)

√xdx = sin ¡2√x¢+C

6. d

dx[sin x−xcos x]=xsin xso Zxsin xdx= sin x−xcos x+C

7. (a) x9/9+C(b) 7

12x12/7+C(c) 2

9x9/2+C

8. (a) 3

5x5/3+C(b) −1

5x−5+C=−1

5x5+C(c) 8x1/8+C

9. (a) 1

2Zx−3dx =−1

4x−2+C(b) u4/4−u2+7u+C

10. 3

5x5/3−5x4/5+4x+C

11. Z(x−3+x1/2−3x1/4+x2)dx =−1

2x−2+2

3x3/2−12

5x5/4+1

3x3+C

12. Z(7y−3/4−y1/3+4y

1/2

)dy =28y

1/4−3

4/3+8

3/2+C

13. Z(x+x4)dx =x2/2+x

/5+C

14. Z(4+4y

2+y

)dy =4y+4

3+1

5+C

15. Zx1/3(4 −4x+x2)dx =Z(4x1/3−4x4/3+x7/3)dx =3x

4/3−12

7x7/3+3

10x10/3+C

16. Z(2 −x+2x

2−x

)dx =2x−1

2+2

3−1

4+C

17. Z(x+2x

−2−x

−4

)dx =x2/2−2/x +1/(3x3)+C

18. Z(t−3−2)dt =−1

2t−2−2t+C19. 2lnx+3e

x+C

20. Z·1

2t−1−√2et¸dt =1

2ln t−√2et+C21. −4 cos x+ 2 sin x+C

22. 4 tan x−csc x+C23. Z(sec2x+ sec xtan x)dx = tan x+ sec x+C

24. Z(sec xtan x+1)dx = sec x+x+C25. ln θ−2eθ+ cot θ+C

26. Zsin ydy =−cos y+C27. Zsec xtan xdx= sec x+C

28. Z(φ+ 2 csc2φ)dφ =φ2/2−2 cot φ+C29. Z(1 + sin θ)dθ =θ−cos θ+C

211 Chapter 7

30. Z2 sin xcos x

cos xdx =2Zsin xdx=−2 cos x+C

31. Z1−sin x

1−sin2xdx =Z1−sin x

cos2x=Z¡sec2x−sec xtan x¢dx = tan x−sec x+C

33. (a) y

-1 1 2

(b) f(x)=x

/2+5

34. (a) y

-1 1 2

(b) f(x)=e

/2+1/2

35. y

-5

π/4π/2

36. y

12345

37. f0(x)=m=−sin xso f(x)=Z(−sin x)dx = cos x+C;f(0)=2=1+Cso C=1,f(x) = cos x+1

38. f0(x)=m=(x+1)2,sof(x)=Z(x+1)2dx =1

3(x+1)3+C;f(−2)=8= 1

(−2+1)3+C=−1

3+C,

C=8+1

3=25

3,f(x)=1

(x+1)

3+25

39. (a) y(x)=Zx

1/3

dx =3

4x4/3+C,y(1) = 3

4+C=2,C=5/4; y(x)=3

4/3+5

(b) y(t)=Zt

−1

dt =ln|t|+C,y(−1) = C=5,C=5;y(t)=ln|t|+5

1/2+x

−1/2

)dx =2

3x3/2+2x

1/2+C,y(1)=0=8

3+C,C=−8

y(x)=2

3/2+2x

1/2−8

40. (a) y(x)=Zµ1

−3

¶dx =−1

16x−2+C,y(1)=0=−1

16 +C,C=1

16;y(x)=−1

16x−2+1

Exercise Set 7.2 212

(b) y(t)=Z(sec2t−sin t)dt = tan t+ cos t+C,y(π

4)=1=1+

√

2+C,C=−

√

y(t) = tan t+ cos t−√2

7/2

dx =2

9x9/2+C,y(0)=0=C,C=0;y(x)=2

9/2

41. f0(x)=2

3/2+C

;f(x)= 4

15x5/2+C1x+C2

42. f0(x)=x

/2 + sin x+C1, use f0(0) = 2 to get C1=2sof

0(x)=x

/2 + sin x+2,

f(x)=x

/6−cos x+2x+C

, use f(0) = 1 to get C2=2sof(x)=x

/6−cos x+2x+2

43. dy/dx =2x+1,y =Z(2x+1)dx =x2+x+C;y= 0 when x=−3

so (−3)2+(−3)+C=0,C =−6thusy=x

2+x−6

44. dy/dx =x2,y =Zx

2dx =x3/3+C;y= 2 when x=−1so(−1)3/3+C=2,C =7/3

thus y=x3/3+7/3

45. dy/dx =Z6xdx =3x

2+C

. The slope of the tangent line is −3sody/dx =−3 when x=1. Thus

3(1)2+C1=−3, C1=−6sody/dx =3x

2−6, y=Z(3x2−6)dx =x3−6x+C2;Ifx= 1, then

y=5−3(1) = 2 so (1)2−6(1) + C2=2,C

2= 7 thus y=x3−6x+7.

46. dT/dx =C1,T=C1x+C2;T= 25 when x=0soC

2= 25, T=C1x+ 25. T= 85 when x=50so

50C1+ 25 = 85, C1=1.2, T=1.2x+25

47. (a) F0(x)=G

0(x)=3x+4

(b) F(0)=16/6=8/3, G(0)=0,soF(0) −G(0)=8/3

2+24x+ 16)/6=3x

/2+4x+8/3=G(x)+8/3

48. (a) F0(x)=G

0(x)=10x/(x2+5)

(b) F(0)=0,G(0) = −1, so F(0) −G(0)=1

2+5 =(x

2+5)−5

2+5 =1−5

2+5 =G(x)+1

49. Z(sec2x−1)dx = tan x−x+C50. Z(csc2x−1)dx =−cot x−x+C

51. (a) 1

2Z(1 −cos x)dx =1

2(x−sin x)+C(b) 1

2Z(1 + cos x)dx =1

2(x+ sin x)+C

52. (a) F0(x)=G

0(x)=f(x), where f(x)=½1,x>0

−1,x<0

(b) G(x)−F(x)=½2,x>0

3,x<0so G(x)6=F(x) plus a constant

53. v=1087

2√273 ZT−1/2dT =1087

√273T1/2+C,v(273) = 1087 = 1087 + Cso C=0,v=1087

√273T1/2ft/s

213 Chapter 7

EXERCISE SET 7.3

1. (a) Zu23du =u24/24+C=(x

2+1)

24/24+C

(b) −Zu3du =−u4/4+C=−(cos4x)/4+C

(d) 3

8Zu−1/2du =3

4u1/2+C=3

4p4x2+5+C

(e) 1

3Zu−1du =1

3ln u+C=1

3ln(x3−4)+C

2. (a) 1

4Zsec2udu=1

4tan u+C=1

4tan(4x+1)+C

(b) 1

4Zu1/2du =1

6u3/2+C=1

6(1+2y

)

3/2+C

πZu1/2du =2

3πu3/2+C=2

3πsin3/2πθ +C

(d) Zu4/5du =5

9u9/5+C=5

9(x2+7x+3)

9/5+C

(e) Zdu

u=lnu+C= ln(1 + ex)+C

3. (a) −Zudu=−1

2+C=−1

2cot2x+C

(b) Zu9du =1

10u10 +C=1

10(1 + sin t)10 +C

udu =ln|u|+C=ln|ln x|+C

(d) −1

5Zeudu =−1

5eu+C=−1

5e−5x+C

(e) −1

3Z1

udu =−1

3ln |u|+C=−1

3ln |(1 + cos 3θ)|+C

4. (a) Z(u−1)2u1/2du =Z(u5/2−2u3/2+u1/2)du =2

7u7/2−4

5u5/2+2

3u3/2+C

7(1+x)7/2−4

5(1+x)5/2+2

3(1+x)3/2+C

(b) Zcsc2udu=−cot u+C=−cot(sin x)+C

(d) 1

2Zu1/2du =1

3u3/2+C=1

3(1+e2t)3/2+C

(e) Z1

udu =ln|u|+C=ln¯

¯x

5+1

¯+C

5. u=2x,du =2dx;1

2Zeudu =1

2eu+C=1

2e2x+C

Exercise Set 7.3 214

6. u=2x,du =2dx;1

2Z1

udu =1

2ln |u|+C=1

2ln |2x|+C

7. u=2−x

,du =−2xdx;−1

2Zu

3du =−u4/8+C=−(2 −x2)4/8+C

8. u=3x−1, du =3dx;1

3Zu5du =1

18u6+C=1

18(3x−1)6+C

9. u=8x,du =8dx;1

8Zcos udu=1

8sin u+C=1

8sin 8x+C

10. u=3x,du =3dx;1

3Zsin udu=−1

3cos u+C=−1

3cos 3x+C

11. u=4x,du =4dx;1

4Zsec utan udu=1

4sec u+C=1

4sec 4x+C

12. u=5x,du =5dx;1

5Zsec2udu=1

5tan u+C=1

5tan 5x+C

13. u=7t

2+ 12, du =14tdt;1

14 Zu1/2du =1

21u3/2+C=1

21(7t2+ 12)3/2+C

14. u=4−5x

,du =−10xdx;−1

10 Zu−1/2du =−1

5u1/2+C=−1

5p4−5x2+C

15. u=x3+1,du =3x

dx;1

3Zu−1/2du =2

3u1/2+C=2

3px3+1+C

16. u=1−3x,du =−3dx;−1

3Zu−2du =1

3u−1+C=1

3(1 −3x)−1+C

17. u=4x

2+1,du =8xdx;1

8Zu

−3du =−1

16u−2+C=−1

16(4x2+1)

−2+C

18. u=3x

,du =6xdx;1

6Zcos udu=1

6sin u+C=1

6sin(3x2)+C

19. u= sin x,du = cos xdx;Ze

udu =eu+C=esin x+C

20. u=x4,du =4x

dx;1

4Zeudu =1

4eu+C=1

4ex4+C

21. u=−2x3,du =−6x2,−1

6Zeudu =−1

6eu+C=−1

6e−2x3+C

22. u=ex−e−x,du =(e

x+e

−x

)dx,Z1

udu =ln|u|+C=ln¯

¯e

x−e

−x

¯+C

23. u=5/x,du =−(5/x2)dx;−1

5Zsin udu=1

5cos u+C=1

5cos(5/x)+C

24. u=√x,du =1

2√xdx;2

Zsec2udu= 2 tan u+C= 2 tan √x+C

25. u=x3,du =3x

dx;1

3Zsec2udu=1

3tan u+C=1

3tan(x3)+C

215 Chapter 7

26. u= cos 2t,du =−2 sin 2tdt;−1

2Zu

3du =−1

8u4+C=−1

8cos42t+C

27. Ze−xdx;u=−x,du =−dx;−Zeudu =−eu+C=−e−x+C

28. Zex/2dx;u=x/2, du =dx/2; 2 Zeudu =2e

u+C=2e

x/2+C=2

√

x+C

29. u= sin 3t,du = 3 cos 3tdt;1

3Zu

5du =1

18u6+C=1

18 sin63t+C

30. u= 5 + cos 2θ,du =−2 sin 2θdθ;−1

2Zu

−3du =1

4u−2+C=1

4(5 + cos 2θ)−2+C

31. u=2−sin 4θ,du =−4 cos 4θdθ;−1

4Zu

1/2du =−1

6u3/2+C=−1

6(2 −sin 4θ)3/2+C

32. u= tan 5x,du = 5 sec25xdx;1

5Zu

3du =1

20u4+C=1

20 tan45x+C

33. u= sec 2x,du = 2 sec 2xtan 2xdx;1

2Zu

2du =1

6u3+C=1

6sec32x+C

34. u= sin θ,du = cos θdθ;Zsin udu =−cos u+C=−cos(sin θ)+C

35. u=√y,du =1

2√ydy,2

du =2e

u+C=2e

√

y+C

36. u=√y,du =1

2√ydy,2

udu =2Ze

−u

du =−2e−u+C=−2e−√y+C

38. u=a+bx,du =bdx,dx =1

bdu

bZu1/ndu =n

b(n+1)u

(n+1)/n +C=n

b(n+1)(a+bx)(n+1)/n +C

39. u= sin(a+bx), du =bcos(a+bx)dx

bZundu =1

b(n+1)u

n+1 +C=1

b(n+1)sinn+1(a+bx)+C

41. u=x−3, x=u+3,dx =du

Z(u+3)u

1/2

du =Z(u3/2+3u

1/2

)du =2

5u5/2+2u

3/2+C=2

5(x−3)5/2+2(x−3)3/2+C

42. u=y+1,y=u−1, dy =du

Zu−1

u1/2du =Z(u1/2−u−1/2)du =2

3u3/2−2u1/2+C=2

3(y+1)

3/2−2(y+1)

1/2+C

43. u=3θ,du =3dθ

3Ztan2udu=1

3Z(sec2u−1)du =1

3(tan u−u)+C=1

3(tan 3θ−3θ)+C

44. Zsin22θsin 2θdθ=Z(1 −cos22θ) sin 2θdθ;u= cos 2θ,du =−2 sin 2θdθ,

−1

2Z(1 −u2)du =−1

2u+1

6u3+C=−1

2cos 2θ+1

6cos32θ+C

Exercise Set 7.3 216

45. Zµ1+1

t¶dt =t+ln|t|+C

46. e2lnx=e

ln x2=x2,x>0, so Ze2lnxdx =Zx2dx =1

3x3+C

47. ln(ex) + ln(e−x) = ln(exe−x)=ln1=0soZ[ln(ex) + ln(e−x)]dx =C

48. Zcos x

sin xdx;u= sin x,du = cos xdx;Z1

udu =ln|u|+C=ln|sin x|+C

49. (a) with u= sin x,du = cos xdx;Zudu =1

2+C

1=1

2sin2x+C1;

with u= cos x,du =−sin xdx;−Zudu =−1

2+C

2=−1

2cos2x+C2

(b) because they diﬀer by a constant:

µ1

2sin2x+C1¶−µ−1

2cos2x+C2¶=1

2(sin2x+ cos2x)+C

1−C

2=1/2+C

1−C

50. (a) First method: Z(25x2−10x+1)dx =25

3x3−5x2+x+C1;

second method: 1

5Zu2du =1

15u3+C2=1

15(5x−1)3+C2

(b) 1

15(5x−1)3+C2=1

15(125x3−75x2+15x−1)+C2=25

3x3−5x2+x−1

15 +C2;

the answers diﬀer by a constant.

51. y(x)=Z√

3x+1dx =2

9(3x+1)

3/2+C,

y(1) = 16

9+C=5,C=29

9so y(x)=2

(3x+1)

3/2+29

52. y(x)=Z(6 −5 sin 2x)dx =6x+5

2cos 2x+C,

y(0) = 5

2+C=3,C=1

2so y(x)=6x+5

2cos 2x+1

53. f0(x)=m=√

3x+1, f(x)=Z(3x+1)

1/2

dx =2

9(3x+1)

3/2+C;f(0)=1=2

9+C,C=7

,so

f(x)=2

(3x+1)

3/2+7

54. y

-4 4

217 Chapter 7

55. p(t)=Z(4+0.15t)3/2dt =8

3(4+0.15t)5/2+C;p(0) = 100,000 = 8

345/2+C=256

3+C,

C= 100,000 −256

3≈99,915, p(t)=8

(4+0.15t)5/2+99,915, p(5) = 8

3(4.75)5/2+99,915 ≈100,416

56. dr

dt =−k√t,r=−kZt1/2dt =−2

3kt3/2+C;r(0)=10,000 = Cso C=10,000 and

r=−2

3kt3/2+10,000; r(25) = 9,000 = −2

3k(25)3/2+10,000=10,000 −250

3k,so

k= 12, r=−8t3/2+10,000, and r(60) = 6281.94 m

EXERCISE SET 7.4

1. (a) 1+8+27=36 (b) 5+8+11+14+17=55

(e) 1−2+4−8+16=11 (f) 0+0+0+0+0+0=0

2. (a) 1+0−3+0=−2(b) 1−1+1−1+1−1=0

2=14e

(14 terms)

(d) 24+2

5+2

6= 112

(e) ln 1 + ln 2 + ln 3 + ln 4 + ln 5 + ln 6 = ln(1 ·2·3·4·5·6) = ln 720

(f) 1−1+1−1+1−1+1−1+1−1+1=1

k=1

k4.

k=1

3k5.

k=1

k(k+1) 6.

k=0

k=1

2k8.

k=1

(2k−1) 9.

k=1

(−1)k+1(2k−1)

10.

k=1

(−1)k+1 1

k11.

k=1

(−1)k1

k12.

k=0

cos kπ

13. (a)

2k(b)

(2k−1)

14. (a)

k=1

(−1)k+1ak(b)

k=0

(−1)k+1bk(c)

k=0

akxk(d)

k=0

a5−kbk

15. 1

2(100)(100 + 1) = 5050

16.

100

k=1

k−

k=1

2(100)(100 + 1) −(1 + 2) = 5050 −3 = 5047

17. 1

6(20)(21)(41) = 2,870 18. 7

100

k=1

100

k=1

1=7

(100)(101) + 100 = 35,450

19. 4

k=1

k3−2

k=1

1=4·1

(6)2(7)2¸−2·1

2(6)(7)¸+ 6 = 1728

Exercise Set 7.4 218

20.

k=1

k2−

k=1

k2=2,870 −14=2,856

21.

k=1

k(k2−4) =

k=1

(k3−4k)=

k=1

k3−4

k=1

4(30)2(31)2−4·1

2(30)(31) = 214,365

22.

k=1

k−

k=1

k3=1

2(6)(7) −1

4(6)2(7)2=−420

23.

k=1

(4k−3)=4

k=1

k−

k=1

3=4·1

n(n+1)−3n=2n

2−n

24.

n−1

k=1

k2=1

6(n−1)[(n−1) + 1][2(n−1)+1]= 1

6n(n−1)(2n−1)

25.

k=1

n=3

k=1

k=3

n·1

2n(n+1)= 3

2(n+1)

26.

n−1

k=1

n=1

n−1

k=1

k2=1

n·1

6(n−1)(n)(2n−1) = 1

6(n−1)(2n−1)

27.

n−1

k=1

n2=1

n−1

k=1

k3=1

n2·1

4(n−1)2n2=1

4(n−1)2

28.

k=1 µ5

n−2k

n¶=5

k=1

1−2

k=1

k=5

n(n)−2

n·1

2n(n+1)=4−n

30. S−rS =

k=0

ark−

k=0

ark+1

=(a+ar +ar2+···+arn)−(ar +ar2+ar3+···+arn+1)

=a−arn+1 =a(1 −rn+1)

so (1 −r)S=a(1 −rn+1), hence S=a(1 −rn+1)/(1 −r)

31. (a)

k=0

3k+1 =

k=0

3(3k)=3(1 −320)

1−3=3

2(320 −1)

(b)

k=0

2k+5 =

k=0

252k=25(1 −226)

1−2=2

31 −25

(c)

100

k=0

(−1) µ−1

2¶k

=(−1)(1 −(−1/2)101)

1−(−1/2) =−2

3(1+1/2

101)

32. (a) 1.999023438,1.999999046,2.000000000; 2 (b) 2.831059456,2.990486364,2.999998301; 3

33. 1+2+3+···+n

k=1

n2=1

k=1

n2·1

2n(n+1)= n+1

2n; lim

n→+∞

n+1

2n=1

219 Chapter 7

34. 12+2

2+3

2+···+n

k=1

n3=1

k=1

k2=1

n3·1

6n(n+ 1)(2n+1)= (n+ 1)(2n+1)

lim

n→+∞

(n+ 1)(2n+1)

2= lim

n→+∞

6(1+1/n)(2+1/n)=1

35.

k=1

n2=5

k=1

k=5

n2·1

2n(n+1)= 5(n+1)

2n; lim

n→+∞

5(n+1)

2n=5

36.

n−1

k=1

2k2

n3=2

n−1

k=1

k2=2

n3·1

6(n−1)(n)(2n−1) = (n−1)(2n−1)

3n2;

lim

n→+∞

(n−1)(2n−1)

3n2= lim

n→+∞

3(1 −1/n)(2 −1/n)=2

37. (a)

j=0

2j(b)

j=1

2j−1(c)

j=2

2j−2

38. (a)

k=1

(k+ 4)2k+8 (b)

k=9

(k−4)2k

39. (a)

k=1

sin ³π

k´(b)

k=0

ek=e7−1

e−1

40. 1+3+5+···+(2n−1) =

k=1

(2k−1)=2

k=1

k−

k=1

1=2·1

n(n+1)−n=n

41. For 1 ≤k≤nthe k-th L-shaped strip consists of the corner square, a strip above and a strip to the

right for a combined area of 1 + (k−1)+(k−1)=2k−1, so the total area is

k=1

(2k−1) = n2.

42. n(n+1)

2= 465, n2+n−930=0,(n+ 31)(n−30)=0,n= 30.

43. (35−34)+(3

6−3

)+···+(3

17 −316)=3

17 −34

44. µ1−1

2¶+µ1

2−1

3¶+···+µ1

50 −1

51¶=50

45. µ1

22−1

12¶+µ1

32−1

22¶+···+µ1

202−1

192¶=1

202−1=−399

400

46. (22−2)+(2

3−2

)+···+(2

101 −2100)=2

101 −2

47. (a)

k=1

(2k−1)(2k+1)=1

k=1 µ1

2k−1−1

2k+1¶

2·µ1−1

3¶+µ1

3−1

5¶+µ1

5−1

7¶+···+µ1

2n−1−1

2n+1¶¸

2·1−1

2n+1¸=n

2n+1

(b) lim

n→+∞

2n+1 =1

Exercise Set 7.4 220

48. (a)

k=1

k(k+1)=

k=1 µ1

k−1

k+1¶

=µ1−1

2¶+µ1

2−1

3¶+µ1

3−1

4¶+···+µ1

n−1

n+1¶

=1−1

n+1 =n

n+1

(b) lim

n→+∞

n+1 =1

49. both are valid 50. none is valid

51.

i=1

(xi−¯x)=

i=1

xi−

i=1

¯x=

i=1

xi−n¯xbut ¯x=1

i=1

xithus

i=1

xi=n¯xso

i=1

(xi−¯x)=n¯x−n¯x=0

52.

k=1

(ak−bk)=(a

1−b

1)+(a

2−b

)+···+(a

n−b

)

=(a

1+a

2+···+a

n)−(b

1+b

2+···+b

n)=

k=1

ak−

k=1

53.

k=1 £(k+1)

4−k

¤=(n+1)

4−1 (telescoping sum), expand the

quantity in brackets to get

k=1

(4k3+6k

2+4k+1)=(n+1)

4−1,

k=1

k3+6

k=1

k2+4

k=1

1=(n+1)

4−1

k=1

k3=1

4"(n+1)

4−1−6

k=1

k2−4

k=1

k−

k=1

4[(n+1)

4−1−n(n+ 1)(2n+1)−2n(n+1)−n]

4(n+ 1)[(n+1)

3−n(2n+1)−2n−1]

4(n+ 1)(n3+n2)=1

(n+1)

54. If n=2mthen 2m+2(m−1)+···+2·2+2=2

k=1

k=2·m(m+1)

2=m(m+1)= n

2+2n

if n=2m+ 1 then (2m+1)+(2m−1)+···+5+3+1=

m+1

k=1

(2k−1)

m+1

k=1

k−

m+1

k=1

1=2·(m+ 1)(m+2)

2−(m+1)=(m+1)

2=n

2+2n+1

55. 50·30+49·29+···+22·2+21·1=

k=1

k(k+20) =

k=1

k2+20

k=1

k=30 ·31 ·61

6+2030 ·31

2=18,755

221 Chapter 7

EXERCISE SET 7.5

1. (a) 1

2bh =1

2·4·4=8 y

(b) The approximation is greater than A,

as the rectangles extend beyond the area.

k=1

f(x∗

k)∆x=(4+3+2+1)(1) = 10

as the rectangles lie inside the area.

k=1

f(x∗

k)∆x=(3+2+1+0)(1) = 6

(d) The approximation is equal to A,

as can be seen by measuring congruent triangles.

k=1

f(x∗

k)∆x=(3.5+2.5+1.5+0.5)(1) = 8

2. (a) 1

2bh =1

24·12=24 y

Exercise Set 7.5 222

(b) The approximation is less than A,

as the rectangles lie inside the area.

k=1

f(x∗

k)∆x=(6+9+12+15)(1) = 42

as the rectangles extend beyond the area.

k=1

f(x∗

k)∆x= (9 + 12 + 15 + 18)(1) = 54

(d) The approximation is equal to A,

as can be seen by measuring congruent triangles.

k=1

f(x∗

k)∆x=(7.5+10.5+13.5+16.5)(1) = 48

3. (a) x∗

k=0,1,2,3,4

k=1

f(x∗

k)∆x=(1+2+5+10+17)(1) = 35

(b) x∗

k=1,2,3,4,5

k=1

f(x∗

k)∆x=(2+5+10+17+26)(1) = 60

k=1/2,3/2,5/2,7/2,9/2

k=1

f(x∗

k)∆x=(5/4+13/4+29/4+53/4+85/4)(1) = 185/4=46.25

4. (a) x∗

k=1,2,3,4,5

k=1

f(x∗

k)∆x=(1+8+27+64+125)(1) = 225

(b) x∗

k=2,3,4,5,6

k=1

f(x∗

k)∆x= (8 + 27 + 64 + 125 + 216)(1) = 440

k=3/2,5/2,7/2,9/2,11/2

k=1

f(x∗

k)∆x= (27/8 + 125/8 + 343/8 + 729/8 + 1331/8)(1) = 2555/8 = 319.375

223 Chapter 7

5. (a) x∗

k=−π/2,−π/4,0,π/4

k=1

f(x∗

k)∆x=(0+1/

√

2+1+1/

√

2)(π/4)=(1+√

2)π/4≈1.896

(b) x∗

k=−π/4,0,π/4,π/2

k=1

f(x∗

k)∆x=(1/

√

2+1+1/

√

2 + 0)(π/4)=(1+√

2)π/4≈1.896

k=−3π/8,−π/8,π/8,3π/8

k=1

f(x∗

k)∆x=·cos 3π

8+ cos π

8+ cos 3π

8¸π

4=πcos π

4cos π

8=³π√2 cos π

8´/2≈2.052

6. (a) x∗

k=0,1,2,3,4

k=1

f(x∗

k)∆x=¡e0+e1+e2+e3+e4¢(1)=(1−e

)/(1 −e)=85.791

(b) x∗

k=1,2,3,4,5

k=1

f(x∗

k)∆x=¡e1+e2+e3+e4+e5¢(1) = e(1 −e5)/(1 −e) = 233.204

k=1/2,3/2,5/2,7/2,9/2

k=1

f(x∗

k)∆x=³e1/2+e3/2+e5/2+e7/2+e9/2´(1) = e1/2(1 −e5)/(1 −e) = 141.446

7. left endpoints: x∗

k=1,2,3,4;

k=1

f(x∗

k)∆x=(2+3+2+1)(1) = 8

right endpoints: x∗

k=2,3,4,5;

k=1

f(x∗

k)∆x=(3+2+1+2)(1) = 8

8. (a) A=1

2(h1+h2)w=1

2(4 + 16)(4) = 40 y

(b) (left) x∗

k=1,2,3,4;

k=1

f(x∗

k)∆x=(4+7+10+13)(1) = 34

(right) x∗

k=2,3,4,5;

k=1

f(x∗

k)∆x= (7 + 10 + 13 + 16)(1) = 46; the average is 1

2(34 + 46) = 40

the left endpoint approximation by four diﬀerent, but congruent, triangles.

9. 0.718771403, 0.668771403, 0.692835360 10. 0.761923639, 0.584145862, 0.663501867

11. 0.919403170, 1.07648280, 1.001028825 12. 4.884074732, 5.684074732, 5.347070728

13. 0.351220577, 0.420535296, 0.386502483 14. 1.63379940, 1.805627583, 1.717566087

Exercise Set 7.5 224

15. n1/x 1/x2sin x√xln x ex

(a) 25 0.693097198 0.666154270 1.000164512 5.336963538 0.386327689 1.718167282

(b) 50 0.693134682 0.666538346 1.000041125 5.334644416 0.386302694 1.718253191

17. (a) A=1

2(3)(3) = 9/2

(b) −A=−1

2(1)(1 + 2) = −3/2

-2 -1 x

2+8=15/2

-1

(d) −A1+A2=0

-5

18. (a) A=1

2(1)(2) = 1

(b) A=1

2(2)(3/2+1/2)=2

-1 1

2(1/2)(1) = −1/4

(d) A1−A2=1−1/4=3/4

225 Chapter 7

19. (a) A=2(5)=10

12345

(b) 0; A1=A2by symmetry

2(5)(5/2) + 1

2(1)(1/2)

=13/2

-1

A1A2

(d) 1

2[π(1)2]=π/2

-1 1

20. (a) A= (6)(5) = 30

-10 -5

(b) −A1+A2= 0 because

A1=A2by symmetry

2(2)(2) + 1

2(1)(1) = 5/2

A1A2

(d) 1

4π(2)2=π

21. (a) 0.8(b) −2.6(c) −1.8(d) −0.3

22. (a) Z1

f(x)dx =Z1

2xdx =x2¸1

(b) Z1

−1

f(x)dx =Z1

−1

2xdx =x2¸1

−1

2−(−1)2=0

Exercise Set 7.5 226

f(x)dx =Z10

2dx =2x

¸10

=18

(d) Z5

1/2

f(x)dx =Z1

1/2

2xdx +Z5

2dx =x2¸1

1/2

+2x

2−(1/2)2+2·5−2·1=3/4+8=35/4

23. Z2

−1

f(x)dx +2Z2

−1

g(x)dx =5+2(−3) = −1

24. 3Z4

f(x)dx −Z4

g(x)dx = 3(2) −10 = −4

25. Z5

f(x)dx =Z5

f(x)dx −Z1

f(x)dx =1−(−2)=3

26. Z−2

f(x)dx =−Z3

−2

f(x)dx =−·Z1

−2

f(x)dx +Z3

f(x)dx¸=−(2 −6)=4

27. (a) Z1

xdx +2Z1

0p1−x

dx =1/2+2(π/4)=(1+π)/2

(b) 4Z3

−1

dx −5Z3

−1

xdx =4·4−5(−1/2+(3·3)/2) = −4

28. (a) Z0

−3

2dx +Z0

−3p9−x2dx =2·3+(π(3)2)/4=6+9π/4

(b) Z2

−2

dx −3Z2

−2|x|dx =4·1−3(2)(2 ·2)/2=−8

29. (a) √x>0, 1 −x<0on[2,3] so the integral is negative

(b) x2>0, 3 −cos x>0 for all xso the integral is positive

30. (a) x4>0, √3−x>0on[−3,−1] so the integral is positive

(b) x3−9<0, |x|+1>0on[−2,2] so the integral is negative

31. Z10

0p25 −(x−5)2dx =π(5)2/2=25π/2

32. Z3

0p9−(x−3)2dx =π(3)2/4=9π/4

33. (a) Z3

−3

4x(1 −3x)dx (b) Z1

exdx

34. (a) Z2

x3dx (b) Zπ/2

sin2xdx

35. Z1

(3x+1)dx =5/236. Z2

−2p4−x2dx =π(2)2/2=2π

37. (a) lim

max ∆xk→0

k=1

2x∗

k∆xk;a=1,b=2 (b) lim

max ∆xk→0

k=1

x∗

k+1∆x

k;a=0,b=1

227 Chapter 7

38. (a) lim

max ∆xk→0

k=1

ln x∗

k∆xk,a=1,b=2

(b) lim

max ∆xk→0

k=1

(1 + cos x∗

k)∆x

k,a=−π/2, b=π/2

39. (a) x∗

k+1 =x∗

k+1

nand x∗

1=1+1

,sox

∗

2=x

∗

1+1

n=1+2

∗

k+1 =x∗

k+1

n=1+k+1

nfor

k=2,3,···,n−1

(b) f(x∗

k)=1+k

nand ∆x=1

k=1

1+ 1

k=1

nn+1

n(n+1)

2=3

2+1

(d) lim

n→+∞µ3

2+1

2n¶=3

2which is the area of the trapezoid with base 1 and sides 1 and 2

40. f(x∗

k)=1+k−1

k=1

f(x∗

k)∆x=

k=1 µ1+k−1

n¶1

n=1

nn+1

(n−1)n

2=3

2−1

2n;

lim

n→+∞µ3

2−1

2n¶=3

41. (a) (right) ∆x=b−a

n=1

n,x∗

k=k

k=1

f(x∗

k)∆x=

k=1 µk

n¶21

n=1

k=1

k2=1

n(n+ 1)(2n+1)

6; lim

n→+∞

k=1

f(x∗

k)∆x=2

6=1

(b) (left) ∆x=b−a

n=1

n,x∗

k=k−1

k=1

f(x∗

k)∆x=

k=1 µk−1

n¶21

n=1

k=1

(k−1)2=1

(n−1)n(2n−1)

lim

n→+∞

k=1

f(x∗

k)∆x=2

6=1

42. (a) (right) ∆x=b−a

n=3

n,x∗

k=3k

k=1

f(x∗

k)∆x=

k=1 Ã4−1

4µ3k

n¶2!3

n=12

k=1

1−27

4n3

k=1

k2=12−27

4n3

n(n+ 1)(2n+1)

lim

n→+∞

k=1

f(x∗

k)∆x=12−27(2)

4(6) =39

(b) (left) ∆x=b−a

n=3

n,x∗

k=3(k−1)

k=1

f(x∗

k)∆x=

k=1 Ã4−1

4µ3(k−1)

n¶2!3

n=12

k=1

1−27

4n3

k=1

(k−1)2=12−27

4n3

(n−1)n(2n−1)

lim

n→+∞

k=1

f(x∗

k)∆x=12−27(2)

4(6) =39

Exercise Set 7.5 228

43. (a) (right) ∆x=b−a

n=4

n,x∗

k=2+4k

k=1

f(x∗

k)∆x=

k=1 µ2+4k

n¶34

n= 32+192

n(n+1)

2+384

n(n+ 1)(2n+1)

6+256

n4µn(n+1)

2¶2

;

lim

n→+∞

k=1

f(x∗

k)∆x=32+192

2+384(2)

6+256

4= 320

(b) (left) ∆x=b−a

n=4

n,x∗

k=2+4(k−1)

k=1

f(x∗

k)∆x=

k=1 µ2+4(k−1)

n¶34

n= 32+192

(n−1)n

2+384

(n−1)n(2n−1)

6+256

n4µ(n−1)n

2¶2

;

lim

n→+∞

k=1

f(x∗

k)∆x=32+192

2+384(2)

6+256

4= 320

44. (a) (right) ∆x = b−a

n=2

n,x∗

k=−3+2k

k=1

f(x∗

k)∆x=

k=1 Ã1−µ2k

n−3¶3!2

n=56

k=1

1−108

k=1

k+72

k=1

k2−16

k=1

=56−108

n(n+1)

2+72

n(n+ 1)(2n+1)

6−16

n4µn(n+1)

2¶2

;

lim

n→+∞

k=1

f(x∗

k)∆x=56−54 + 72(2)

6−16

4=22

(b) (left) ∆x = b−a

n=2

n,x∗

k=−3+2(k−1)

k=1

f(x∗k)∆x=

k=1 Ã1−µ2(k−1)

n−3¶3!2

=56

k=1

1−108

k=1

(k−1) + 72

k=1

(k−1)2−16

k=1

(k−1)3

=56−108

(n−1)n

2+72

(n−1)n(2n−1)

6−16

n4µ(n−1)n

2¶2

;

lim

n→+∞

k=1

f(x∗

k)∆x=56−108

2+72(2)

6−16

4=22

45. (a) fis continuous on [−1,1] so fis integrable there by part (a) of Theorem 7.5.8

(b) |f(x)|≤1sofis bounded on [−1,1], and fhas one point of discontinuity, so by part (b) of

Theorem 7.5.8 fis integrable on [−1,1]

x→0f(x)=+∞,sofis not integrable on [0,1]

(d) f(x) is discontinuous at the point x= 0 because lim

x→0sin 1

xdoes not exist. fis continuous

elsewhere. −1≤f(x)≤1 for xin [−1,1] so fis bounded there. By part (b), Theorem 7.5.8, f

is integrable on [−1,1].

46. Each subinterval of a partition of [a, b] contains both rational and irrational numbers. If all x∗

kare

chosen to be rational then

k=1

f(x∗

k)∆xk=

k=1

(1)∆xk=

k=1

∆xk=b−aso lim

max ∆xk→0

k=1

f(x∗

k)∆xk=b−a.

If all x∗

kare irrational then lim

max ∆xk→0

k=1

f(x∗

k)∆xk=0. fis not integrable on [a, b] because the

preceding limits are not equal.

229 Chapter 7

47. (a) Let Sn=

k=1

f(x∗

k)∆xkand S=Zb

f(x)dx then

k=1

cf(x∗

k)∆xk=cSnand we want to prove

that lim

max ∆xk→0cSn=cS.Ifc= 0 the result follows immediately, so suppose that c6= 0 then for

any ²>0, |cSn−cS|=|c||Sn−S|<²if |Sn−S|<²/|c|. But because fis integrable on [a, b],

there is a number δ>0 such that |Sn−S|<²/|c|whenever max ∆xk<δso |cSn−cS|<²and

hence lim

max ∆xk→0cSn=cS.

(b) Let Rn=

k=1

f(x∗

k)∆xk,Sn=

k=1

g(x∗

k)∆xk,Tn=

k=1

[f(x∗

k)+g(x

∗

)]∆xk,R=Zb

f(x)dx, and

S=Zb

g(x)dx then Tn=Rn+Snand we want to prove that lim

max ∆xk→0Tn=R+S.

|Tn−(R+S)|=|(Rn−R)+(S

n−S)|≤|R

n−R|+|S

n−S|

so for any ²>0|T

n−(R+S)|<²if |Rn−R|+|Sn−S|<².

Because fand gare integrable on [a, b], there are numbers δ1and δ2such that

|Rn−R|<²/2 for max ∆xk<δ

1and |Sn−S|<²/2 for max ∆xk<δ

If δ= min(δ1,δ

2) then |Rn−R|<²/2 and |Sn−S|<²/2 for max ∆xk<δthus

|Rn−R|+|Sn−S|<²and so |Tn−(R+S)|<²for max ∆xk<δwhich shows that

lim

max ∆xk→0Tn=R+S.

48. For the smallest, ﬁnd x∗

kso that f(x∗

k) is minimum on each subinterval: x∗

1=1,x

∗

2=3/2, x∗

3=3so

(2)(1)+(7/4)(2) +(4)(1) = 9.5. For the largest, ﬁnd x∗

kso that f(x∗

k) is maximum on each subinterval:

x∗

1=0,x

∗

2=3,x

∗

3= 4 so (4)(1) + (4)(2) + (8)(1) = 20.

EXERCISE SET 7.6

1. (a) Z2

(2 −x)dx =(2x−x

/2)¤2

0=4−4/2=2

(b) Z1

−1

2dx =2x

¸1

−1

= 2(1) −2(−1)=4

(x+1)dx =(x

/2+x)

1=9/2+3−(1/2+1)=6

2. (a) Z5

xdx =x2/2¸5

=25/2(b) Z9

5dx =5x

¸9

= 5(9) −5(3) = 30

−1

(x+3)dx =(x

/2+3x)

¸2

−1

=4/2+6−(1/2−3)=21/2

3. Z3

x3dx =x4/4¸3

=81/4−16/4=65/44. Z1

−1

x4dx =x5/5¸1

−1

=1/5−(−1)/5=2/5

5. Z9

√xdx =2

3x3/2¸9

3(27 −1)=52/36. Z4

x−3/5dx =5

2x2/5¸4

2(42/5−1)

7. Z3

exdx =ex¸3

=e3−e8. Z5

xdx =lnx

¸5

=ln5−ln1=ln5

9. µ1

3x3−2x2+7x

¶¸0

−3

=48 10. µ1

2x2+1

5x5¶¸2

−1

=81/10

Exercise Set 7.6 230

11. Z3

x−2dx =−1

x¸3

=2/312. Z2

x−6dx =−1

5x5¸2

=31/160

13. 4

5x5/2¸9

= 844/514. µ3x5/3+4

x¶¸8

= 179/2

15. −cos θiπ/2

−π/2=0 16. tan θiπ/4

0=1

17. sin xiπ/4

−π/4=√218. µ1

2x2−sec x¶¸1

=3/2−sec(1)

19. 5exi3

ln 2 =5e

3−5(2) = 5e3−10 20. (ln x)/2i1

1/2= (ln 2)/2

21. µ6√t−10

3t3/2+2

√t¶¸4

=−55/322. µ8√y+4

3y3/2−2

3y3/2¶¸9

= 10819/324

23. µ1

2x2−2 cot x¶¸π/2

π/6

=π2/9+2

√

24. ³ln x+√2ex+ csc x´i2

1=ln2+√

2(e2−e) + csc 2 −csc 1

26. µa1/2x−2

3x3/2¶#4a

=−5

3a3/2

27. (a) Z3/2

(3 −2x)dx +Z2

3/2

(2x−3)dx =(3x−x

)

¸3/2

+(x

2−3x)

¸2

3/2

=9/4+1/4=5/2

(b) Zπ/2

cos xdx+Z3π/4

π/2

(−cos x)dx = sin x¸π/2

0−sin x¸3π/4

π/2

=2−√

2/2

28. (a) Z0

−1

√2−xdx+Z2

√2+xdx=−2

3(2 −x)3/2¸0

−1

3(2+x)3/2¸2

=−2

3(2√2−3√3) + 2

3(8 −2√2) = 2

3(8 −4√2+3

√

(b) Z0

−1

(1−ex)dx+Z1

(ex−1)dx =(x−e

)

¸0

−1

+(ex−x)¸1

=−1−(−1−e−1)+e−1−1=e+1/e−2

29. Z0

−2

x2dx +Z3

(−x)dx =1

3x3¸0

−2−1

2x2¸3

=−11/6

30. Z1

√xdx+Z4

2dx =2

3x3/2¸1

0−1

x¸4

=17/12

31. 0.665867079; Z3

x2dx =−1

x¸3

=2/3

32. 1.000257067; Zπ/2

sin xdx =−cos x¸π/2

231 Chapter 7

33. 1.098242635; Z3

xdx =lnx

¸3

=ln3≈1.098612289

35. A=Z3

(x2+1)dx =µ1

3x3+x¶¸3

=12

36. A=Z2

(−x2+3x−2)dx =µ−1

3x3+3

2x2−2x¶¸2

=1/6

37. A=Z2π/3

3 sin xdx=−3 cos x#2π/3

=9/238. A=−Z−1

−2

x3dx =−1

4x4¸−1

−2

=15/4

39. A1=Z−2

−3

(x2−3x−10)dx =µ1

3x3−3

2x2−10x¶¸−2

−3

=23/6,

2=−Z5

−2

2−3x−10)dx = 343/6,

A3=Z8

(x2−3x−10)dx = 243/6, A=A1+A2+A3= 203/2

-3

-2

40. (a) the area is positive

(b) Z5

−2µ1

100x3−1

20x2−1

25x+1

5¶dx =µ1

400x4−1

60x3−1

50x2+1

5x¶¸5

−2

=343

1200

41. (a) the area between the curve and the x-axis breaks into equal parts, one above and one below the

x-axis, so the integral is zero

(b) Z1

−1

x3dx =1

4x4¸1

−1

4(14−(−1)4)=0;

Zπ/2

−π/2

sin xdx =−cos x#π/2

−π/2

=−cos(π/2) + cos(−π/2)=0+0=0

−a

f(x)dx =2Za

f(x)dx

(d) Z1

−1

x2dx =1

3x3¸1

−1

3(13−(−1)3)=2

3=2Z1

dx;

Zπ/2

−π/2

cos xdx = sin x#π/2

−π/2

= sin(π/2) −sin(−π/2)=1+1=2=2Zπ/2

cos xdx

42. The numerator is an odd function and the denominator is an even function, so the integrand is an

odd function and the integral is zero.

43. (a) x3+1 (b) F(x)=µ1

4+t

¶¸x

4x4+x−5

4;F0(x)=x

3+1

44. (a) cos 2x(b) F(x)= 1

2sin 2t¸x

π/4

2sin 2x−1

2,F0(x) = cos 2x

45. (a) sin √x(b) ex2

Exercise Set 7.6 232

46. (a) 1

1+√

x(b) ln x

47. −x

cos x48. |u|

49. F0(x)=√

2+1,F00(x)= 3x

√

2+1

(a) 0(b) √13 (c) 6/√13

50. F0(x)= cos x

x2+3,F00(x)=−(x

2+ 3) sin x−2xcos x

(x2+3)

(a) 0(b) 1/3(c) 0

51. (a) F0(x)= x−3

2+7 = 0 when x= 3, which is a relative minimum, and hence the absolute minimum,

by the ﬁrst derivative test.

(b) increasing on [3,+∞), decreasing on (−∞,3]

2+7)

2=(7 −x)(1 + x)

(x2+7)

2; concave up on (−1,7), concave down on (−∞,−1) and

on (7,+∞)

52. F

-20 -10 10 20

53. (a) (0,+∞) because fis continuous there and 1 is in (0,+∞)

(b) at x= 1 because F(1)=0

54. (a) (−3,3) because fis continuous there and 1 is in (−3,3)

(b) at x= 1 because F(1)=0

55. (a) fave =1

9Z9

x1/2dx =2;√

∗=2,x

∗=4

(b) fave =1

e−1Ze

xdx =1

e−1ln x¸e

e−1;1

x∗=1

e−1,x∗=e−1

56. (a) fave =1

2πZπ

−π

sin xdx= 0; sin x∗=0,x

∗=−π, 0,π

(b) fave =1

2Z3

x2dx =1

3;1

(x∗)2=1

3,x∗=√3

57. √2≤√x3+2≤√

29, so 3√2≤Z3

0px3+2dx ≤3√29

58. Let f(x)=xsin x,f(0) = f(1)=0,f

(x) = sin x+xcos x= 0 when x=−tan x,x≈2.0288,

so fhas an absolute maximum at x≈2.0288; f(2.0288) ≈1.8197, so 0 ≤xsin x≤1.82 and

0≤Zπ

xsin xdx ≤1.82π=5.72

233 Chapter 7

59. 0≤ln x≤ln 5 for xin [1,5], so 0 ≤Z5

ln xdx ≤4ln5

60. (a) £cF (x)¤b

a=cF (b)−cF (a)=c[F(b)−F(a)] = c£F(x)¤b

(b) £F(x)+G(x)

¤b

a=[F(b)+G(b)] −[F(a)+G(a)]

=[F(b)−F(a)]+[G(b)−G(a)] = F(x)¤b

a+G(x)¤b

a=[F(b)−G(b)] −[F(a)−G(a)]

=[F(b)−F(a)] −[G(b)−G(a)] = F(x)¤b

a−G(x)¤b

EXERCISE SET 7.7

1. (a) the increase in height in inches, during the ﬁrst ten years

(b) the change in the radius in centimeters, during the time interval t=1tot= 2 seconds

◦

Fto

t= 100◦F

(d) the displacement of the particle in cm, during the time interval t=t1to t=t2seconds

2. (a) Z1

V(t)dt gal

(b) the change f(x1)−f(x2) in the values of fover the interval

3. (a) displ = s(3) −s(0)

=Z3

v(t)dt =Z2

(1 −t)dt +Z3

(t−3)dt =(t−t

/2)¸2

+(t

/2−3t)

¸3

=−1/2;

dist = Z3

0|v(t)|dt =(t−t

/2)¸1

+(t

/2−t)

¸2

+(t

/2−3t)

¸3

=1/2

(b) displ = s(3) −s(0)

=Z3

v(t)dt =Z1

tdt +Z2

dt +Z3

(5 −2t)dt =t2/2¸2

+t¸3

+(5t−t

)

¸3

=5;

dist = Z1

tdt +Z2

dt +Z5/2

(5 −2t)dt +Z3

5/2

(2t−5)dt

=t2/2¸1

+t¸2

+(5t−t

)

¸5/2

+(t

2−5t)

¸3

5/2

=5/2

4. v

-1

246810

5. (a) v(t)=20+Zt

a(u)du; add areas of the small blocks to get

v(5) ≈20 + 1

2(1.5+2.7+4.6+6.2+7.6)=31.3

Exercise Set 7.7 234

(b) v(10) = v(4) + Z10

a(u)du ≈31.3+1

2(8.6+9.3+9.7+10+10.1)=55.15

6. a>0 and therefore (Theorem 7.5.6(a)) v>0, so the particle is always speeding up for 0 <t<10

7. (a) s(t)=Z(t

3−2t

2+1)dt =1

4t4−2

3t3+t+C,

s(0) = 1

4(0)4−2

3(0)3+0+C=1,C=1,s(t)= 1

4−2

3+t+1

(b) v(t)=Z4 cos 2tdt= 2 sin 2t+C1,v(0) = 2 sin 0 + C1=−1, C1=−1,

v(t) = 2 sin 2t−1, s(t)=Z(2 sin 2t−1)dt =−cos 2t−t+C2,

s(0) = −cos 0 −0+C

2=−3, C2=−2, s(t)=−cos 2t−t−2

8. (a) s(t)=Z(1 + sin t)dt =t−cos t+C,s(0)=0−cos 0 + C=−3, C=−2, s(t)=t−cos t−2

(b) v(t)=Z(t

2−3t+1)dt =1

3t3−3

2t2+t+C1,

v(0) = 1

3(0)3−3

2(0)2+0+C

1=0,C

1=0,v(t)=1

3−3

2+t,

s(t)=Zµ1

3−3

2+t

¶dt =1

12t4−1

2t3+1

2t2+C2,

s(0) = 1

12(0)4−1

2(0)3+1

2(0)2+C2=0,C

2=0,s(t)= 1

12t4−1

2t3+1

2t2

9. (a) s(t)=Z(2t−3)dt =t2−3t+C,s(1) = (1)2−3(1) + C=5,C=7,s(t)=t

2−3t+7

(b) v(t)=Zcos tdt = sin t+C1,v(π/2)=2=1+C

1=1,v(t) = sin t+1,

s(t)=Z(sin t+1)dt =−cos t+t+C2,s(π/2)=0=π/2+C2,C2=−π/2, s(t)=−cos t+t−π/2

10. (a) s(t)=Zt

2/3

dt =3

5t5/3+C,s(8)=0=3

32+C,C=−96

5,s(t)=3

5/3−96

(b) v(t)=Z√

tdt =2

3t3/2+C1,v(4)=1=2

8+C

1=−13

3,v(t)=2

3/2−13

s(t)=Zµ2

3/2−13

3¶dt =4

15t5/2−13

3t+C2,s(4) = −5= 4

1532 −13

34+C

2=−44

5+C2,

C2=19

5,s(t)= 4

15t5/2−13

3t+19

11. (a) displacement = s(π/2) −s(0) = Zπ/2

sin tdt =−cos t¸π/2

distance = Zπ/2

0|sin t|dt =1

(b) displacement = s(2π)−s(π/2) = Z2π

π/2

cos tdt = sin t¸2π

π/2

=−1

distance = Z2π

π/2|cos t|dt =−Z3π/2

π/2

cos tdt +Z2π

3π/2

cos tdt =3

235 Chapter 7

12. (a) displacement = s(6) −s(0) = Z6

(2t−4)dt =(t

2−4t)

¸6

=12

distance Z6

0|2t−4|dt =Z2

(4 −2t)dt +Z6

(2t−4)dt =(4t−t

)

¸2

+(t

2−4t)

¸6

=20

(b) displacement = Z5

0|t−3|dt =Z3

0−(t−3)dt +Z5

(t−3)dt =13/2

distance = Z5

0|t−3|dt =13/2

13. (a) v(t)=t

3−3t

2+2t=t(t−1)(t−2)

displacement = Z3

(t3−3t2+2t)dt =9/4

distance = Z3

0|v(t)|dt =Z1

v(t)dt +Z2

1−v(t)dt +Z3

v(t)dt =11/4

(b) displacement = Z3

(et−2)dt =e3−7

distance = Z3

0|v(t)|dt =−Zln 2

v(t)dt +Z3

ln 2

v(t)dt =e3−9+4ln2

14. (a) displacement = Z3

2−1

t)dt =1−ln 3

distance = Z3

1|v(t)|dt =−Z2

v(t)dt +Z3

v(t)dt = 2 ln 2 −ln 3

(b) displacement = Z9

3t−1/2dt =6

distance = Z9

4|v(t)|dt =Z9

v(t)dt =6

15. v(t)=−2t+3

displacement = Z4

(−2t+3)dt =−6

distance = Z4

1|−2t+3|dt =Z3/2

(−2t+3)dt +Z4

3/2

(2t−3)dt =13/2

16. v(t)= 1

2−2t

displacement = Z5

1µ1

2t2−2t¶dt =−10/3

distance = Z5

1¯¯¯¯

2t2−2t¯¯¯¯

dt =Z4

1−µ1

2t2−2t¶dt +Z5

4µ1

2t2−2t¶dt =17/3

17. v(t)= 2

5√5t+1+8

displacement = Z3

0µ2

5√5t+1+8

5¶dt =4

75(5t+1)

3/2+8

¸3

= 204/25

distance = Z3

0|v(t)|dt =Z3

v(t)dt = 204/25

Exercise Set 7.7 236

18. v(t)=−cos t+2

displacement = Zπ/2

π/4

(−cos t+2)dt =(π+√

2−2)/2

distance = Zπ/2

π/4|−cos t+2|dt =Zπ/2

π/4

(−cos t+2)dt =(π+√

2−2)/2

19. (a) s=Zsin 1

2πt dt =−2

πcos 1

2πt +C

s= 0 when t= 0 which gives C=2

πso s=−2

πcos 1

2πt +2

π.

a=dv

dt =π

2cos 1

2πt. When t=1:s=2/π,v=1,|v|=1,a=0.

(b) v=−3Ztdt =−3

2+C

1,v= 0 when t= 0 which gives C1=0sov=−3

s=−3

2Zt

dt =−1

2t3+C2,s= 1 when t= 0 which gives C2=1sos=−1

3+1.

When t=1:s=1/2, v=−3/2, |v|=3/2, a=−3.

20. (a) negative, because vis decreasing

(b) speeding up when av > 0, so 2 <t<5; slowing down when 1 <t<2

v<0 compared to where v>0

21. A=A1+A2=Z1

(1 −x2)dx +Z3

(x2−1)dx =2/3+20/3=22/3

22. A=A1+A2=Zπ

sin xdx −Z3π/2

sin xdx =2+1=3

23. A=A1+A2=Z0

−1

(1 −ex)dx +Z1

(ex−1)dx =1/e +e−2

24. A=A1+A2=Z1

1/2

1−x

xdx +Z2

x−1

xdx =−µ1

2−ln 2¶+(1−ln 2) = 1/2

25. s(t)=20

3t3−50t2+50t+s

,s(0) = 0 gives s0=0,sos(t)= 20

3t3−50t2+50t,a(t)=40t−100

150

-100

150

-100

237 Chapter 7

26. v(t)=2t

2−30t+v0,v(0)=3=v

,sov(t)=2t

2−30t+3, s(t)=2

3−15t2+3t+s

0,s(0) = −5=s

so s(t)=2

3−15t2+3t−5

1200

-1200

025

s(t)

500

-200

025

v(t)

-30

025

a(t)

27. (a) from the graph the velocity is positive, so the displacement is

always increasing and is therefore positive

0.2

0.3

0.4

0.5

12345

(b) s(t)=t/2+(t+1)e

−t

28. (a) If t0<1 then the area between the velocity curve and the

t-axis, between t= 0 and t=t0, will always be negative, so

the displacement will be negative.

-0.2

-0.1

0.1

0.2 0.4 0.6 0.8 1

(b) s(t)=µt

2−1

200¶ln(t+0.1) −t2

4+t

20 −1

200 ln 10

29. (a) a(t)=½0,t<4

−10,t>4

-10

-5

24 12

(b) v(t)=½25,t<4

65 −10t, t > 4

-40

-20

2 4 6 8 10 12

65t−5t2−80,t>4

,sox(8) = 120, x(12) = −20

(d) x(6.5) = 131.25

Exercise Set 7.7 238

30. (a) From exercise 30 part (a), in Section 3 of Chapter 6, v2=v2

0−2g(s−s0), so a=−g=v2−v2

2(s−s0)

(b) From exercise 30 part (b), in Section 3 of Chapter 6, s=s0+1

2(v0+v)t,sot=2(s−s0)

v0+v

2gt2=s0+vt −1

2at2

31. (a) a=−1 mi/h/s = −22/15 ft/s2(b) a= 30 km/h/min = 1/7200 km/s2

32. Take t= 0 when deceleration begins, then a=−10 so v=−10t+C1, but v= 88 when t= 0 which

gives C1=88thusv=−10t+ 88, t≥0

(a) v= 45 mi/h = 66 ft/s, 66 = −10t+ 88, t=2.2s

(b) v= 0 (the car is stopped) when t=8.8s

s=Zvdt =Z(−10t+ 88)dt =−5t2+88t+C

, and taking s= 0 when t=0,C

2=0so

s=−5t

2+88t.Att=8.8, s= 387.2. The car travels 387.2 ft before coming to a stop.

33. a=a0ft/s2,v=a0t+v0=a0t+ 132 ft/s, s=a0t2/2 + 132t+s0=a0t2/2 + 132tft; s= 200 ft

when v= 88 ft/s. Solve 88 = a0t+ 132 and 200 = a0t2/2 + 132tto get a0=−121

5when t=20

11,so

s=−12.1t2+ 132t,v=−121

5t+ 132.

(a) a0=−121

5ft/s2(b) v= 55 mi/h = 242

3ft/s when t=70

33 s

11 s

34. dv/dt =3,v=3t+C

, but v=v0when t=0soC

1=v

,v=3t+v

. From ds/dt =v=3t+v

0we

get s=3t

/2+v

t+C

2and, with s= 0 when t=0,C

2=0sos=3t

/2+v

t.s= 40 when t=4

thus 40 = 3(4)2/2+v

(4), v0= 4 m/s

35. Suppose s=s0=0,v=v

0=0att=t

0=0;s=s

1= 120, v=v1at t=t1; and s=s2,v=v2=12

at t=t2. From Exercise 30(a),

2.6=a=v

1−v

2(s1−s0),v

1=2as1=5.2(120) = 624. Applying the formula again,

−1.5=a=v

2−v

2(s2−s1),v

2=v

1−3(s2−s1), so

s2=s1−(v2

2−v2

1)/3 = 120 −(144 −624)/3 = 280 m.

36. a(t)=½4,t<2

0,t>2

, so, with v0=0,v(t)=½4t, t < 2

8,t>2

and, since s0=0,s(t)=½2t

,t<2

8t−8,t>2

s= 100 when 8t−8 = 100, t= 108/8=13.5s

37. The truck’s velocity is vT= 50 and its position is sT=50t+ 5000. The car’s acceleration is aC=2,

so vC=2t,s

C=t

2(initial position and initial velocity of the car are both zero). sT=sCwhen

50t+ 5000 = t2,t2−50t−5000 = (t+ 50)(t−100) = 0, t= 100 s and sC=sT=t2=10,000 ft

38. Let t= 0 correspond to the time when the leader is 100 m from the ﬁnish line; let s= 0 correspond to

the ﬁnish line. Then vC= 12, sC=12t−115; aL=0.5 for t>0, vL=0.5t+8, s

L=0.25t2+8t−100.

sC=0att= 115/12 ≈9.58 s, and sL=0att=−16+4

√

41 ≈9.61, so the challenger wins.

39. s= 0 and v= 112 when t=0sov(t)=−32t+ 112, s(t)=−16t2+ 112t

(a) v(3) = 16 ft/s, v(5) = −48 ft/s

(b) v= 0 when the projectile is at its maximum height so −32t+112=0,t=7/2s,

s(7/2) = −16(7/2)2+ 112(7/2) = 196 ft.

239 Chapter 7

is when it is at ground level on its way down. v(7) = −112, |v|= 112 ft/s.

40. s= 112 when t=0sos(t)=−16t2+v0t+ 112. But s= 0 when t= 2 thus

−16(2)2+v0(2) + 112 = 0, v0=−24 ft/s.

41. (a) s(t) = 0 when it hits the ground, s(t)=−16t2+16t=−16t(t−1) = 0 when t=1s.

(b) The projectile moves upward until it gets to its highest point where v(t)=0,

v(t)=−32t+ 16 = 0 when t=1/2s.

42. (a) s(t) = 0 when the rock hits the ground, s(t)=−16t2+ 555 = 0 when t=√555/4s

(b) v(t)=−32t,v(√555/4) = −8√555, the speed at impact is 8√555 ft/s

43. (a) s(t) = 0 when the package hits the ground,

s(t)=−16t2+20t+ 200 = 0 when t=(5+5

√

33)/8s

(b) v(t)=−32t+ 20, v[(5+5

√

33)/8] = −20√33, the speed at impact is 20√33 ft/s

44. (a) s(t) = 0 when the stone hits the ground,

s(t)=−16t2−96t+ 112 = −16(t2+6t−7) = −16(t+ 7)(t−1) = 0 when t=1s

(b) v(t)=−32t−96, v(1) = −128, the speed at impact is 128 ft/s

45. s(t)=−4.9t

2+49t+ 150 and v(t)=−9.8t+49

(a) the projectile reaches its maximum height when v(t)=0,−9.8t+49=0,t=5s

(b) s(5) = −4.9(5)2+ 49(5) + 150 = 272.5m

−4.9t(t−10)=0,t=10s

(d) v(10) = −9.8(10) + 49 = −49 m/s

(e) s(t) = 0 when the projectile hits the ground, −4.9t2+49t+ 150 = 0 when (use the quadratic

formula) t≈12.46 s

(f) v(12.46) = −9.8(12.46)+49≈−73.1, the speed at impact is about 73.1 m/s

46. take s= 0 at the water level and let hbe the height of the bridge, then s=hand v= 0 when t=0

so s(t)=−16t2+h

(a) s= 0 when t= 4 thus −16(4)2+h=0,h= 256 ft

(b) First, ﬁnd how long it takes for the stone to hit the water (ﬁnd tfor s=0):−16t2+h=0,

t=√

h/4. Next, ﬁnd how long it takes the sound to travel to the bridge: this time is h/1080

because the speed is constant at 1080 ft/s. Finally, use the fact that the total of these two

times must be 4 s: h

1080 +√h

4=4,h+ 270√h= 4320, h+ 270√h−4320 = 0, and by the

quadratic formula √h=−270 ±p(270)2+ 4(4320)

2, reject the negative value to get √h≈15.15,

h≈229.5 ft.

47. g=9.8/6=4.9/3 m/s2,sov=−(4.9/3)t,s=−(4.9/6)t2+5,s= 0 when t=p30/4.9 and

v=−(4.9/3)p30/4.9≈−4.04, so the speed of the module upon landing is 4.04 m/s

48. s(t)=−1

2gt2+v0t;s= 1000 when v=0,so0=v=−gt +v0,t=v0/g,

1000 = s(v0/g)=−1

2g(v

/g)2+v0(v0/g)=1

/g,sov

0= 2000g,v0=√2000g.

The initial velocity on the Earth would have to be √6 times faster than that on the Moon.

Exercise Set 7.7 240

49. fave =1

3−1Z3

3xdx=3

2¸3

=6 50. fave =1

2−(−1) Z2

−1

x2dx =1

9x3¸2

−1

51. fave =1

π−0Zπ

sin xdx=−1

πcos x¸π

=2/π

52. fave =1

π−0Zπ

cos xdx=1

πsin x¸π

53. fave =1

e−1=Ze

xdx =1

1−e(ln e−ln 1) = 1

e−1

54. fave =1

ln 5 −(−1) Zln 5

−1

exdx =1

ln5+1(5 −e−1)= 5−e

−1

1+ln5

55. (a) fave =1

2−0Z2

x2dx =4/3(b) (x∗)2=4/3,x

∗=±2/

√3,

but only 2/√3isin[0,2]

(c)

56. (a) fave =1

4−0Z4

2xdx=4 (b) 2x∗=4,x

∗=2

(c)

57. (a) vave =1

4−1Z4

(3t3+2)dt =1

789

4=263

(b) vave =s(4) −s(1)

4−1=100 −7

3=31

58. (a) aave =1

5−0Z5

(t+1)dt =7/2

(b) aave =v(π/4) −v(0)

π/4−0=√2/2−1

π/4=(2

√

2−4)/π

59. time to ﬁll tank = (volume of tank)/(rate of ﬁlling) = [π(3)25]/(1)=45π, weight of water in tank at

time t= (62.4) (rate of ﬁlling)(time) = 62.4t,

weightave =1

45πZ45π

62.4tdt= 1404πlb

241 Chapter 7

60. (a) If xis the distance from the cooler end, then the temperature is T(x)= (15+1.5x)

◦C, and

Tave =1

10 −0Z10

(15+1.5x)dx =22.5

◦C

(b) By the Mean-Value Theorem for Integrals there exists x∗in [0,10] such that

f(x∗)= 1

10 −0Z10

(15+1.5x)dx =22.5, 15 + 1.5x∗=22.5, x∗=5

61. (a) amount of water = (rate of ﬂow)(time) = 4tgal, total amount = 4(30) = 120 gal

(b) amount of water = Z60

(4+t/10)dt = 420 gal

(10 + √t)dt = 1200 + 160√30 ≈2076.36 gal

62. (a) The maximum value of Roccurs at 4:30 P.M. when t=0.

(b) Z60

100(1 −0.0001t2)dt = 5280 cars

63. (a) Zb

[f(x)−fave]dx =Zb

f(x)dx −Zb

favedx =Zb

f(x)dx −fave(b−a)=0

because fave(b−a)=Zb

f(x)dx

(b) no, because if Zb

[f(x)−c]dx = 0 then Zb

f(x)dx −c(b−a)=0so

c=1

b−aZb

f(x)dx =fave is the only value

EXERCISE SET 7.8

1. (a) Z3

u7du (b) −1

2Z4

u1/2du (c) 1

πZπ

−π

sin udu (d) Z0

−3

(u+5)u

20du

2. (a) 1

2Z1

−1

eudu (b) Z2

udu

u2du (d) 1

2Z4

(u−3)u1/2du

3. u=2x+1, 1

2Z3

du =1

10u5¸3

= 121/5, or 1

10(2x+1)

¸1

= 121/5

4. u=4x−2, 1

4Z6

u3du =1

16u4¸6

= 80, or 1

16(4x−2)4¸2

=80

5. u=1−2x,−1

2Z1

du =−1

8u4¸1

= 10, or −1

8(1 −2x)4¸0

−1

=10

6. u=4−3x,−1

3Z−2

du =−1

27u9¸−2

= 19, or −1

27(4 −3x)9¸2

=19

Exercise Set 7.8 242

7. u=1+x,Z9

(u−1)u1/2du =Z9

(u3/2−u1/2)du =2

5u5/2−2

3u3/2¸9

= 1192/15,

or 2

5(1+x)5/2−2

3(1+x)3/2¸8

= 1192/15

8. u=4−x,Z4

(u−4)u1/2du =Z4

(u3/2−4u1/2)du =2

5u5/2−8

3u3/2¸4

=−506/15

or 2

5(4 −x)5/2−8

3(4 −x)3/2¸0

−5

=−506/15

9. u=x/2, 8 Zπ/4

sin udu=−8 cos u¸π/4

=8−4

√

2, or −8 cos(x/2)¸π/2

=8−4

√

10. u=3x,2

3Zπ/2

cos udu=2

3sin u¸π/2

=2/3, or 2

3sin 3x¸π/6

=2/3

11. u=ex+4,du =exdx,u=e−ln 3 +4= 1

3+4= 13

3when x=−ln 3,

u=eln 3 + 4 = 3 + 4 = 7 when x=ln3,Z7

13/3

udu =lnu

¸7

13/3

= ln(7) −ln(13/3) = ln(21/13)

12. u=3−4e

,du =−4exdx,u=−1 when x=0,u=−17 when x=ln5

−1

4Z−17

−1

udu=−1

2¸−17

−1

=−36

13. 1

3Z5

0p25 −u2du =1

3·1

4π(5)2¸=25

12π

14. 1

2Z4

0p16 −u2du =1

2·1

4π(4)2¸=2π

15. −1

2Z0

1p1−u2du =1

2Z1

0p1−u2du =1

2·1

4[π(1)2]=π/8

16. Z6

−6p36 −u2du =π(6)2/2=18π

17. Z1

sin πxdx =−1

πcos πx¸1

=−1

π(−1−1)=2/π

18. A=Zπ/8

3 cos 2xdx=3

2sin 2x¸π/8

√

2/4

19. Z7

(x+5)

−2=−(x+5)

−1

¸7

=−1

12 +1

8=1

20. A=Z1

(3x+1)

2=−1

3(3x+1)¸1

21. fave =1

4−0Z4

e−2xdx =−1

8e−2x¸4

=1−e−8

243 Chapter 7

22. fave =1

1/4−(−1/4) Z1/4

−1/4

sec2πxdx =2

πtan πx#1/4

−1/4

23. 2

3(3x+1)

1/2

¸1

=2/324. 2

15(5x−1)3/2¸2

=38/15

25. 2

3(x3+9)

1/2

¸1

−1

√

10 −2√2) 26. 1

10(t3+1)

20¸0

−1

=1/10

27. u=x2+4x+7, 1

2Z28

u−1/2du =u1/2¸28

=√28 −√12=2(

√

7−√

28. Z2

(x−3)2dx =−1

x−3¸2

=1/2

29. 1

2sin2x¸π/4

−3π/4

=0 30. 2

3(tan x)3/2¸π/4

=2/331. 5

2sin(x2)¸√π

32. u=√x,2

Z2π

sin udu=−2 cos u¸2π

=−4

33. u=3θ,1

3Zπ/3

π/4

sec2udu=1

3tan u¸π/3

π/4

√

3−1)/3

34. u= sin 3θ,1

3Z−1

u2du =1

9u3¸−1

=−1/9

35. u=4−3y,y=1

(4 −u), dy =−1

3du

−1

27 Z1

16 −8u+u2

u1/2du =1

27 Z4

(16u−1/2−8u1/2+u3/2)du

27 ·32u1/2−16

3u3/2+2

5u5/2¸4

= 106/405

36. u=5+x,Z9

u−5

√

udu =Z9

(u1/2−5u−1/2)du =2

3u3/2−10u1/2¸9

=8/3

37. ln(x+e)ie

0= ln(2e)−ln e=ln2 38. −1

2e−x2¸√2

=(e

−1−e

−2

)/2

40. Z2

−2p4−u2du =1

2[π(2)2]=2π

41. (a) u=3x+1,1

3Z4

f(u)du =5/3(b) u=3x,1

3Z9

f(u)du =5/3

f(u)du =−1/2Z4

f(u)du =−1/2

42. u=1−x,Z1

(1 −x)ndx =−Z0

(1 −u)mundu =Z1

un(1 −u)mdu =Z1

xn(1 −x)mdx

Exercise Set 7.8 244

43. sin x= cos(π/2−x),

Zπ/2

sinnxdx=Zπ/2

cosn(π/2−x)dx =−Z0

π/2

cosnudu (u=π/2−x)

=Zπ/2

cosnudu=Zπ/2

cosnxdx (by replacing uby x)

44. u=1−x,−Z0

(1 −u)undu =Z1

(un−un+1)du =1

n+1−1

n+2 =1

(n+ 1)(n+2)

45. y(t) = (802.137) Ze1.528tdt = 524.959e1.528t+C;y(0) = 750 = 524.959 + C,C= 225.041,

y(t) = 524.959e1.528t+ 225.041, y(12) = 48,233,525,650

46. Vave =275000

10 −0Z10

e−0.17tdt =−161764.7059e−0.17t¸10

= $132,212.96

47. s(t)=Z(25+10e

−0.05t)dt =25t−200e−0.05t+C

(a) s(10) −s(0) = 250 −200(e−0.5−1) = 450 −200/√e≈328.69 ft

(b) yes; without it the distance would have been 250 ft

48. Zk

e2xdx =3,1

¸k

=3,1

2k−1)=3,e

2k=7,k =1

2ln 7

49. (a) V2

rms =1

1/f −0Z1/f

psin2(2πft)dt =1

2fV2

pZ1/f

[1 −cos(4πft)]dt

2fV2

p[t−1

4πf sin(4πft)]¸1/f

2V2

p,so Vrms =Vp/√2

(b) Vp/√2 = 120,V

p= 120√2≈169.7V

50. Let u=t−x, then du =−dx and

f(t−x)g(x)dx =−Z0

f(u)g(t−u)du =Zt

f(u)g(t−u)du;

the result follows by replacing uby xin the last integral.

51. (a) I=−Z0

f(a−u)

f(a−u)+f(u)du =Za

f(a−u)+f(u)−f(u)

f(a−u)+f(u)du

=Za

du −Za

f(u)

f(a−u)+f(u)du, I =a−Iso 2I=a, I =a/2

(b) 3/2(c) π/4

52. x=1

u,dx =−1

u2du,I=Z1

−1

1+1/u2(−1/u2)du =−Z1

−1

u2+1du =−Iso I= 0 which is impossible

because 1

1+x

2is positive on [−1,1]. The substitution u=1/x is not valid because uis not continuous

for all xin [−1,1].

53. Z1

sin πxdx =2/π

245 Chapter 7

55. (a) Let u=−xthen

−a

f(x)dx =−Z−a

f(−u)du =Za

−a

f(−u)du =−Za

−a

f(u)du

so, replacing uby xin the latter integral,

−a

f(x)dx =−Za

−a

f(x)dx,2

−a

f(x)dx =0,Za

−a

f(x)dx =0

The graph of fis symmetric about the origin so Z0

−a

f(x)dx is the negative of Za

f(x)dx thus

−a

f(x)dx =Z0

−a

f(x)+Za

f(x)dx =0

(b) Za

−a

f(x)dx =Z0

−a

f(x)dx +Za

f(x)dx, let u=−xin Z0

−a

f(x)dx to get

−a

f(x)dx =−Z0

f(−u)du =Za

f(u)du =Za

f(x)dx

so Za

−a

f(x)dx =Za

f(x)dx +Za

f(x)dx =2Za

f(x)dx

The graph of f(x) is symmetric about the y-axis so there is as much signed area to the left of

the y-axis as there is to the right.

EXERCISE SET 7.9

1. (a) y

123

(b) y

0.5 1

1e2

2. y

3. (a) ln tiac

1= ln(ac)=lna+lnc=7 (b) ln ti1/c

1= ln(1/c)=−5

1= ln(a/c)=2−5=−3(d) ln tia3

1=lna

3=3lna=6

Exercise Set 7.9 246

4. (a) ln ti√a

1=lna

1/2=2 (b) ln ti2a

1=ln2+4

1=ln2−4(d) ln tia

2=4−ln 2

5. ln 5 ≈1.603210678; ln 5 = 1.609437912; magnitude of error is <0.0063

6. ln 3 ≈1.098242635; ln 3 = 1.098612289; magnitude of error is <0.0004

7. (a) x−1,x>0(b) x2,x6=0

(e) x3,x>0(f) ln x+x,x>0

(g) x−3

√x,−∞ <x<+∞(h) ex

x,x>0

8. (a) f(ln 3) = e−2ln3 =e

ln(1/9) =1/9

(b) f(ln 2) = eln 2 +3e

−ln 2 =2+3e

ln(1/2) =2+3/2=7/2

9. (a) 3π=eπln 3 (b) 2√2=e√2ln2

10. (a) π−x=e−xln π(b) x2x=e2xln x

11. (a) lim

x→+∞·µ1+ 1

x¶x

¸2

=·lim

x→+∞µ1+ 1

x¶x

¸2

(b) y=2x, lim

y→0(1+y)2/y = lim

y→0h(1+y)1/yi2=e2

12. (a) y=3x, lim

y→+∞µ1+ 1

y¶y/3

= lim

y→+∞·µ1+ 1

y¶y¸1/3

=·lim

y→+∞µ1+ 1

y¶y¸1/3

1/3

(b) lim

x→0(1+x)1/3x= lim

x→0h(1+x)1/xi1/3=e1/3

13. g0(x)=x

2−x14. g0(x)=1−cos x

15. (a) 1

x3(3x2)= 3

x(b) eln x1

x=1

16. (a) 2x√x2+1 (b) −µ1

x2¶sin µ1

x¶

17. F0(x)= cos x

x2+3,F00(x)=−(x

2+ 3) sin x−2xcos x

(x2+3)

(a) 0(b) 1/3(c) 0

18. F0(x)=√

2+1,F00(x)= 3x

√

2+1

(a) 0(b) √13 (c) 6/√13

19. (a) d

dx Zx2

t√1+tdt =x2p1+x

(2x)=2x

p1+x

(b) Zx2

t√1+tdt =−2

3(x2+1)

3/2+2

5(x

2+1)

5/2−4

√

247 Chapter 7

20. (a) d

dx Za

f(t)dt =−d

dx Zx

f(t)dt =−f(x)

(b) d

dx Za

g(x)

f(t)dt =−d

dx Zg(x)

f(t)dt =−f(g(x))g0(x)

21. (a) −sin x2(b) −tan2x

1 + tan2xsec2x=−tan2x

22. (a) −(x2+1)

40 (b) −cos3µ1

x¶µ−1

2¶=cos3(1/x)

23. −33x−1

9x2+1+2xx

2−1

4+1

24. If fis continuous on an open interval Iand g(x), h(x), and aare in Ithen

Zg(x)

h(x)

f(t)dt =Za

h(x)

f(t)dt +Zg(x)

f(t)dt =−Zh(x)

f(t)dt +Zg(x)

f(t)dt

so d

dx Zg(x)

h(x)

f(t)dt =−f(h(x))h0(x)+f(g(x))g0(x)

25. (a) sin2(x3)(3x2)−sin2(x2)(2x)=3x

2sin2(x3)−2xsin2(x2)

(b) 1

1+x(1) −1

1−x(−1) = 2

1−x2

26. F0(x)= 1

(3) −1

x(1) = 0 so F(x) is constant on (0,+∞). F(1) = ln 3 so F(x) = ln 3 for all x>0.

27. from geometry, Z3

f(t)dt =0,Z5

f(t)dt =6,Z7

f(t)dt =0;andZ10

f(t)dt =Z10

(4t−37)/3dt =−3

(a) F(0)=0,F(3)=0,F(5)=6,F(7)=6,F(10) = 3

(b) Fis increasing where F0=fis positive, so on [3/2,6] and [37/4,10], decreasing on [0,3/2] and

[6,37/4]

−9/4atx=3/2

(d) F(x)

-2

246810

28. fave =1

10 −0Z10

f(t)dt =1

10F(10) = 0.3

29. x<0:F(x)=Zx

−1

(−t)dt =−1

2t2¸x

−1

2(1 −x2),

x≥0:F(x)=Z0

−1

(−t)dt +Zx

tdt=1

2+1

2;F(x)=((1 −x2)/2,x<0

(1+x2)/2,x≥0

Exercise Set 7.9 248

30. 0≤x≤2:F(x)=Zx

tdt=1

x>2:F(x)=Z2

tdt+Zx

2dt =2+2(x−2)=2x−2; F(x)=(x

/2,0≤x≤2

2x−2,x>2

31. y(x)=2+Zx

1/3

dt =2+3

4/3

¸x

4+3

4/3

32. y(x)=Zx

1/2+t

−1/2

)dt =2

3x3/2−2

3+2x

1/2−2

33. y(x)=1+Zx

π/4

(sec2t−sin t)dt = tan x+ cos x−√2/2

34. y(x)=Zx

tet2dt =1

2e−x2−1

235. P(x)=P

0+Zx

r(t)dt individuals

36. s(T)=s

1+ZT

v(t)dt

37. II has a minimum at x= 1, and I has a zero there, so I could be the derivative of II; on the other

hand I has a minimum near x=1/3, but II is not zero there, so II could not be the derivative of I

38. (b) lim

k→0

k(xk−1)=lnxlim

k→0xk=lnxby L’Hˆopital’s rule (with respect to k)

39. (a) where f(t) = 0; by the ﬁrst derivative test, at t=3

(b) where f(t) = 0; by the ﬁrst derivative test, at t=1

(d) at t=0,3 or 5; from the graph it is evident that it is at t=3

(e) Fis concave up when F00 =f0is positive, i.e. where fis increasing, so on (0,1/2) and (2,4);

it is concave down on (1/2,2) and (4,5)

(f) F(x)

-1

-0.5

0.5

12345

40. (a)

-1

-4 -2 2 4

erf(x)

(e) erf00(x)=−4xe−x2/√πchanges sign only at x= 0 so that is the only point of inﬂection

(g) lim

x→+∞erf(x) = +1, lim

x→−∞ erf(x)=−1

249 Chapter 7

41. C0(x) = cos(πx2/2), C00(x)=−πx sin(πx2/2)

(a) cos tgoes from negative to positive at 2kπ −π/2, and from positive to negative at t=2kπ +π/2,

so C(x) has relative minima when πx2/2=2kπ −π/2, x=±√4k−1, k=1,2,..., and C(x)

has relative maxima when πx2/2=(4k+1)π/2, x=±√4k+1,k=0,1,....

(b) sin tchanges sign at t=kπ,soC(x) has inﬂection points at πx2/2=kπ,x=±√2k,k=1,2,...;

the case k= 0 is distinct due to the factor of xin C00(x), but xchanges sign at x= 0 and

sin(πx2/2) does not, so there is also a point of inﬂection at x=0

42. Let F(x)=Zx

ln tdt,F0(x) = lim

h→0

F(x+h)−F(x)

h= lim

h→0

hZx+h

ln tdt; but F0(x)=lnxso

lim

h→0

hZx+h

ln tdt =lnx

43. Diﬀerentiate: f(x)=3e

,so2+Zx

f(t)dt =2+Zx

dt =2+e

¸x

= 2+e

3x−e

3a=e

3xprovided

e3a=2,a= (ln 2)/3.

44. (a) The area under 1/t for x≤t≤x+ 1 is less than the area of the rectangle with altitude 1/x and

base 1, but greater than the area of the rectangle with altitude 1/(x+ 1) and base 1.

(b) Zx+1

tdt =lnt

¸x+1

= ln(x+1)−ln x= ln(1 + 1/x), so

1/(x+1)<ln(1 + 1/x)<1/x for x>0.

ln(1+1/x)<e

1/x,e1/(x+1) <1+1/x<e

1/x,

ex/(x+1) <(1+1/x)x<e; by the Squeezing Theorem, lim

x→+∞(1+1/x)x=e.

(d) Use the inequality ex/(x+1) <(1+1/x)xto get e<(1+1/x)x+1 so

(1+1/x)x<e<(1+1/x)x+1.

45. From Exercise 44(d) ¯¯¯¯¯

e−µ1+ 1

50¶50¯¯¯¯¯

<y(50), and from the graph y(50) <0.06

0.2

0 100

46. F0(x)=f(x), thus F0(x) has a value at each xin Ibecause fis continuous on Iso Fis continuous

on Ibecause a function that is diﬀerentiable at a point is also continuous at that point

CHAPTER 7 SUPPLEMENTARY EXERCISES

5. If the acceleration a= const, then v(t)=at +v0,s(t)=1

2at2+v0t+s0

6. (a) Divide the base into nequal subintervals. Above each subinterval choose the lowest and highest

points on the curved top. Draw a rectangle above the subinterval going through the lowest point,

and another through the highest point. Add the rectangles that go through the lowest points to

obtain a lower estimate of the area; add the rectangles through the highest points to obtain an

upper estimate of the area.

Supplementary Exercises 7 250

(b) n= 10: 25.0 cm, 22.4cm

7. (a) 1

2+1

4=3

4(b) −1−1

2=−3

4¶=−35

4(d) −2

(e) not enough information (f) not enough information

8. (a) 1

2+2= 5

2(b) not enough information

2=13

9. (a) Z1

−1

dx +Z1

−1p1−x2dx = 2(1) + π(1)2/2=2+π/2

(b) 1

3(x2+1)

3/2

¸3

0−π(3)2/4=1

(103/2−1) −9π/4

2Z1

0p1−u2du =1

2π(1)2/4=π/8

10. 1

0.2

0.4

0.6

0.8

0.2 0.6 1

11. The rectangle with vertices (0,0), (π, 0), (π, 1) and (0,1) has area πand is much too large; so is the

triangle with vertices (0,0), (π, 0) and (π, 1) which has area π/2; 1 −πis negative; so the answer is

35π/128.

12. Divide ex+ 3 into e2xto get e2x

ex+3 =e

x−3e

x+3 so

Ze2x

ex+3dx =Zexdx −3Zex

ex+3dx =ex−3 ln(ex+3)+C

13. Since y=exand y=lnxare inverse functions, their graphs are

symmetric with respect to the line y=x; consequently the areas A1

and A3are equal (see ﬁgure). But A1+A2=e,so

ln xdx +Z1

exdx =A2+A3=A2+A1=e

14. (a) 1

k=1 rk

k=1

f(x∗

k)∆xwhere f(x)=√

x,x

∗

k=k/n, and ∆x=1/n for 0 ≤x≤1. Thus

lim

n→+∞

k=1 rk

n=Z1

x1/2dx =2

251 Chapter 7

(b) 1

k=1 µk

n¶4

k=1

f(x∗

k)∆xwhere f(x)=x

∗

k=k/n, and ∆x=1/n for 0 ≤x≤1. Thus

lim

n→+∞

k=1 µk

n¶4

=Z1

x4dx =1

(c)

k=1

ek/n

k=1

f(x∗

k)∆xwhere f(x)=e

∗

k=k/n, and ∆x=1/n for 0 ≤x≤1. Thus

lim

n→+∞

k=1

ek/n

n= lim

n→+∞

k=1

f(x∗

k)∆x=Z1

exdx =e−1.

15. Since f(x)= 1

xis positive and increasing on the interval [1,2], the left endpoint approximation

overestimates the integral of 1

xand the right endpoint approximation underestimates it.

(a) For n= 5 this becomes

0.2·1

1.2+1

1.4+1

1.6+1

1.8+1

2.0¸<Z2

xdx < 0.2·1

1.0+1

1.2+1

1.4+1

1.6+1

1.8¸

(b) For general nthe left endpoint approximation to Z2

xdx =ln2is

k=1

1+(k−1)/n =

k=1

n+k−1=

n−1

k=0

n+kand the right endpoint approximation is

k=1

n+k. This yields

k=1

n+k<Z2

xdx <

n−1

k=0

n+kwhich is the desired inequality.

n−1

2n=1

2nso 1

2n≤0.1, n≥5

(d) n≥1,000

16. The direction ﬁeld is clearly an even function, which means that the solution is even, its derivative

is odd. Since sin xis periodic and the direction ﬁeld is not, that eliminates all but x, the solution of

which is the family y=x2/2+C.

17. (a) 1·2+2·3+···+n(n+1)=

k=1

k(k+1)=

k=1

k2+

k=1

6n(n+ 1)(2n+1)+ 1

2n(n+1)= 1

3n(n+ 1)(n+2)

(b)

n−1

k=1 µ9

n−k

n2¶=9

n−1

k=1

1−1

n−1

k=1

k=9

n(n−1) −1

n2·1

2(n−1)(n)=17

2µn−1

n¶;

lim

n→+∞

2µn−1

n¶=17

(c)

i=1 "2

j=1

j#=

i=1 ·2i+1

2(2)(3)¸=2

i=1

3=2·1

(3)(4) + (3)(3) = 21

18. (a)

k=0

(k+ 4)(k+1) (b)

k=5

(k−1)(k−4)

Supplementary Exercises 7 252

19. (a) If u= sec x,du = sec xtan xdx,Zsec2xtan xdx =Zudu =u2/2+C

1= (sec2x)/2+C

;

if u= tan x,du = sec2xdx,Zsec2xtan xdx =Zudu =u2/2+C

2= (tan2x)/2+C

(b) They are equal only if sec2xand tan2xdiﬀer by a constant, which is true.

20. 1

2sec2xiπ/4

0=1

2(2 −1)=1/2 and 1

2tan2xiπ/4

0=1

2(1 −0)=1/2

21. Zp1+x

−2/3

dx =Zx−1/3px2/3+1dx;u=x2/3+1,du =2

3x−1/3dx

2Zu1/2du =u3/2+C=(x

2/3+1)

3/2+C

22. (a) Zb

k=1

fk(x)dx =

k=1 Zb

fk(x)dx

(b) yes; substitute ckfk(x) for fk(x) in part (a), and then use Zb

ckfk(x)dx =ckZb

fk(x)dx from

Theorem 7.5.4

23. (a) Zx

1+t

2dt

(b) Zx

tan(π/4−2)

1+t

2dt

24. (a) F0(x)= x−3

2+7; increasing on [3,+∞), decreasing on (−∞,3]

(b) F00(x)=7+6x−x

2+7)

2=(7 −x)(1 + x)

(x2+7)

2; concave up on (−1,7), concave down on (−∞,−1) and

(7,+∞)

2+7 = 0 when x= 3, which is a relative minimum, and hence the absolute minimum,

by the ﬁrst derivative test.

(d) F(x)

-10 10 20

25. F0(x)= 1

1+x

2+1

1+(1/x)2(−1/x2)=0soFis constant on (0,+∞).

26. (−3,3) because fis continuous there and 1 is in (−3,3)

27. (a) The domain is (−∞,+∞); F(x)is0ifx= 1, positive if x>1, and negative if x<1, because

the integrand is positive, so the sign of the integral depends on the orientation (forwards or

backwards).

(b) The domain is [−2,2]; F(x)is0ifx=−1, positive if −1<x≤2, and negative if −2≤x<−1;

same reasons as in part (a).

253 Chapter 7

28. The left endpoint of the top boundary is ((b−a)/2,h) and the right endpoint of the top boundary is

((b+a)/2,h)so

f(x)=





2hx/(b−a),x<(b−a)/2

h, (b−a)/2<x<(b+a)/2

2h(x−b)/(a−b),x>(a+b)/2

The area of the trapezoid is given by

Z(b−a)/2

2hx

b−adx +Z(b+a)/2

(b−a)/2

hdx +Zb

(b+a)/2

2h(x−b)

a−bdx =(b−a)h/4+ah +(b−a)h/4=h(a+b)/2.

29. (a) Z24

0³2000e−t/48 + 500 sin(πt/12)´dt = 96000(1 −1/√e)≈37,773.06

(b) 1

8−0Z8

0³2000e−t/48 + 500 sin(πt/12)dt´= 1125/π + 12000(1 −e−1/6)≈2,200.32

2000

(d) maximum rate is 2285.32 kW/h at t=4.8861

30. wave =1

52 −26 Z52

(t/7)dt =39/7; t∗/7=39/7, t∗=39

31. (a) no, since the velocity curve is not a straight line

(b) 25 <t<40

(d) 141.5ft

(e) no since the velocity is positive and the acceleration is never negative

(f) need the position at any one given time (e.g. s0)

32. (a) x=aekt +be−kt,dx/dt =akekt −bke−kt,

d2x/dt2=ak2ekt +bk2e−kt =k2(aekt +be−kt)=k

(b) At t=0,v=ak −bk =(a−b)k=v

0so k=v0/(a−b) and a=k2x=v2

0x/(a−b)2.

33. u= 5 + 2 sin 3x,du = 6 cos 3xdx;Z1

6√udu =1

3u1/2+C=1

3√5 + 2 sin 3x+C

34. u=3+√

x,du =1

2√xdx;Z2√udu =4

3u3/2+C=4

3(3+√x)3/2+C

35. u=ax3+b,du =3ax2dx;Z1

3au2du =−1

3au +C=−1

3a2x3+3ab +C

36. u=ax2,du =2axdx;1

2aZsec2udu =1

2atan u+C=1

2atan(ax2)+C

37. ln(ex) + ln(e−x) = ln(exe−x)=ln1=0soZ[ln(ex) + ln(e−x)]dx =C

Supplementary Exercises 7 254

38. µ−1

3u3−3

u+1

4u4¶¸−1

−2

= 389/192

39. u=lnx,du =(1/x)dx;Z2

udu =lnu

¸2

=ln2

40. Z1

e−x/2dx = 2(1 −1/√e)

41. u=e−2x,du =−2e−2xdx;−1

2Z1/4

(1 + cos u)du =3

8+1

2µsin 1 −sin 1

4¶

42. 1

3πsin3πx¸1

43. With b=1.618034, area = Zb

(x+x2−x3)dx =1.007514.

44. (a) f(x)= 1

2sin 3x−2

27 sin 3x+2

9xcos 3x−0.251607

(b) f(x)=p4+x

2+4

√

4+x

2−6

45. (a) Solve 1

4k4−k−k2+7

4= 0 to get k=2.073948.

(b) Solve −1

2cos 2k+1

3k3+1

2= 3 to get k=1.837992.

46. F(x)=Zx

−1

√

2+t

3dt,F0(x)= x

√

2+x

3,soFis increasing on [1,3]; Fmax =F(3) ≈1.152082854

and Fmin =F(1) ≈−0.07649493141

47. (a)

12345678

-0.5

0.5

y = J0(x)

y(b) 0.7651976866 (c) J0(x)=0ifx=2.404826

48. (a) A=Z1

x2dx =1

3x3¸1

=1/3

(b)

k=1 µk−1

n¶21

n=1

n3"n

k=1

k2−2

k=1

1#=1

n3µn(n+ 1)(2n+1)

6−2n(n+1)

2+n

¶,

lim

n→+∞

k=1 µk−1

n¶21

n=2

6=1

(c)

k=1 µk

n¶21

n=1

k=1

k2=1

n(n+ 1)(2n+1)

6and lim

n→+∞

k=1 µk

n¶21

n=2

6=1

49. 100,000/(ln 100,000) ≈8,686; Z100,000

ln tdt ≈9,629, so the integral is better

255 Chapter 7

CHAPTER 7 HORIZON MODULE

1. vx(0) = 35 cos α, so from Equation (1), x(t) = (35 cos α)t;vy(0) = 35 sin α, so from Equation (2),

y(t) = (35 sin α)t−4.9t2.

2. (a) vx(t)= dx(t)

dt = 35 cos α,vy(t)= dy(t)

dt = 35 sin α−9.8t

(b) vy(t)=35sinα−9.8t,v

(t) = 0 when t= 35 sin α/9.8;

y=vy(0)t−4.9t2= (35 sin α)(35 sin α)/9.8−4.9((35 sin α)/9.8)2=62.5 sin2α,so

max =62.5 sin2α.

3. t=x/(35 cos α)soy= (35 sin α)(x/(35 cos α)) −4.9(x/(35 cos α))2= (tan α)x−0.004

cos2αx2;

the trajectory is a parabola because yis a quadratic function of x.

4. 15◦25◦35◦45◦55◦65◦75◦85◦

no yes no no no yes no no

0 120

5. y(t) = (35 sin αs)t−4.9t2= 0 when t= 35 sin α/4.9, at which time

x= (35 cos α)(35 sin α/4.9) = 125 sin 2α; this is the maximum value of x,soR= 125 sin 2αm.

6. (a) R= 95 when sin 2α=95/125=0.76, α=0.4316565575,1.139139769 rad ≈24.73◦,65.27◦.

(b) y(t)<50 is required; but y(1.139) ≈51.56 m, so his height would be 56.56 m.

7. 0.4019 <α<0.4636 (radians), or 23.03◦<α<26.57◦

256

CHAPTER 8

Applications of the Deﬁnite Integral

in Geometry, Science, and Engineering

EXERCISE SET 8.1

1. A=Z2

−1(x2+1−x)dx =(x

/3+x−x

2/2)2

−1

=9/2

2. A=Z4

0(√x+x/4)dx =(2x

3/2

/3+x

2/8)4

=22/3

3. A=Z2

1(y−1/y2)dy =(y

/2+1/y)2

4. A=Z2

0(2 −y2+y)dy =(2y−y

/3+y

2/2)2

=10/3

5. (a) A=Z4

0(4x−x2)dx =32/3(b) A=Z16

0(√y−y/4)dy =32/3

(4, 16)

y = 4x

y = x2

(4, 4)

(1, -2)

y2 = 4x

y = 2x – 4

6. Eliminate xto get y2=4(y+4)/2,y

2−2y−8=0,

(y−4)(y+2)=0;y=−2,4 with corresponding

values of x=1,4.

(a) A=Z1

0[2√x−(−2√x)]dx +Z4

1[2√x−(2x−4)]dx

=Z1

04√xdx +Z4

1(2√x−2x+4)dx =8/3+19/3=9

(b) A=Z4

−2[(y/2+2)−y

2/4]dy =9

(1, 1)

y = x2

y = √x

7. A=Z1

1/4(√x−x2)dx =49/192

257 Chapter 8

8. A=Z2

0[0 −(x3−4x)]dx

=Z2

0(4x−x3)dx =4

y = 2x3 – 4x

9. A=Zπ/2

π/4(0 −cos 2x)dx

=−Zπ/2

π/4cos 2xdx=1/2

-1

y = cos 2x

10. Equate sec2xand 2 to get sec2x=2,

y = sec2 x

(#, 2)(3

, 2)

sec x=±√2, x=±π/4

A=Zπ/4

−π/4(2 −sec2x)dx =π−2

11. A=Z3π/4

π/4sin ydy=√2

x = sin y

12. A=Z2

−1[(x+2)−x

2]dx =9/2

(2, 4)

(–1, 1)

y = x2

x = y – 2

13. A=Zln 2

0e2x−exdx =1

2e2x−ex#ln 2

=1/2

ln 2

y = e2x

y = ex

1/e1

14. A=Ze

y=lny

1=1

Exercise Set 8.1 258

(–5, 8)

(5, 6)

y = 3 – xy = 1 + x

y = – x + 7

15. y=2+|x−1|=(3−x, x ≤1

1+x, x ≥1,

A=Z1

−5−1

5x+7

−(3 −x)dx

+Z5

1−1

5x+7

−(1+x)dx

=Z1

−54

5x+4

dx +Z5

16−6

5xdx

=72/5+48/5=24

16. A=Z2/5

0(4x−x)dx

+Z1

2/5(−x+2−x)dx

=Z2/5

03xdx+Z1

2/5(2 −2x)dx =3/5

(1, 1)

(, )

y = 4x

y = x

y = −x + 2

17. A=Z1

0(x3−4x2+3x)dx

+Z3

1[−(x3−4x2+3x)]dx

=5/12+32/12=37/12

-8

-1 4

-2

-1 3

18. Equate y=x3−2x2and y=2x

2−3x

to get x3−4x2+3x=0,

x(x−1)(x−3)=0;x=0,1,3

with corresponding values of y=0,−1.9.

A=Z1

0[(x3−2x2)−(2x2−3x)]dx

+Z3

1[(2x3−3x)−(x3−2x2)]dx

=Z1

0(x3−4x2+3x)dx +Z3

1(−x3+4x

2−3x)dx

12 +8

3=37

-1

19. From the symmetry of the region

A=2Z5π/4

π/4(sin x−cos x)dx =4

√

259 Chapter 8

20. The region is symmetric about the

origin so

A=2Z0

−2

3−4x)dx =8

3.1

-3.1

-3 3

21. A=Z0

−1(y3−y)dy +Z1

0−(y3−y)dy

=1/2

-1

-1 1

4.1

-2.2 12.1

22. A=Z1

0y3−4y2+3y−(y

2−y)

dy

+Z4

1y2−y−(y3−4y2+3y)

dy

=7/12+45/4=71/6

23. Solve 3−2x=x6+2x5−3x4+x2to ﬁnd the real roots x=−3,1; from a plot it is seen that the line

is above the polynomial when −3<x<1, so A=Z1

−3(3−2x−(x6+2x5−3x4+x2)) dx = 9152/105

24. Solve x5−2x3−3x=x3to ﬁnd the roots x=0,±1

q6+2

√

21. Thus, by symmetry,

A=2Z√

(6+2√21)/2

0(x3−(x5−2x3−3x)) dx =27

4+7

4√21

25. Zk

02√ydy =Z9

2√ydy

0y1/2dy =Z9

y1/2dy

3k3/2=2

3(27 −k3/2)

k3/2=27/2

k= (27/2)2/3=9/3

√

y = 9

y = k

26. Zk

0x2dx =Z2

x2dx

3k3=1

3(8 −k3)

k3=4

k=3

√

x = √y

x = k

Exercise Set 8.1 260

27. (a) A=Z2

0(2x−x2)dx =4/3

(b) y=mx intersects y=2x−x

2where mx =2x−x

2+(m−2)x=0,x(x+m−2)=0so

x=0orx=2−m. The area below the curve and above the line is

Z2−m

0(2x−x2−mx)dx =Z2−m

0[(2 −m)x−x2]dx =1

2(2 −m)x2−1

3x32−m

6(2 −m)3

so (2 −m)3/6=(1/2)(4/3)=2/3,(2 −m)3=4,m=2−3

√

( )

y = sin x

28. The line through (0,0) and (5π/6,1/2) is y=3

5πx;

A=Z5π/6

0sin x−3

5πxdx =√3

2−5

24π+1

29. (a) It gives the area of the region that is between fand gwhen f(x)>g(x) minus the area of

the region between fand gwhen f(x)<g(x), for a≤x≤b.

(b) It gives the area of the region that is between fand gfor a≤x≤b.

30. (b) lim

n→+∞Z1

0(x1/n −x)dx = lim

n→+∞n

n+1x

(n+1)/n −x2

21

= lim

n→+∞n

n+1−1

2=1/2

31. The curves intersect at x= 0 and, by Newton’s Method, at x≈2.595739080 = b,so

A≈Zb

0(sin x−0.2x)dx =−cos x+0.1x

2b

0≈1.180898334

32. By Newton’s Method, the points of intersection are at x≈±0.824132312, so with

b=0.824132312 we have A≈2Zb

0(cos x−x2)dx = 2(sin x−x3/3)b

0≈1.094753609

33. distance = R|v|dt,so

(a) distance = Z60

0(3t−t2/20) dt = 1800 ft.

(b) If T≤60 then distance = ZT

0(3t−t2/20) dt =3

2T2−1

60T3ft.

34. Since a1(0) = a2(0)=0,A=

0(a

(t)−a

(t)) dt =v2(T)−v1(T) is the diﬀerence in the velocities

of the two cars at time T.

35. Solve x1/2+y1/2=a1/2for yto get

y=(a

1/2−x

1/2

)

2=a−2a

1/2

1/2+x

A=Za

0(a−2a

1/2

1/2+x)dx =a2/6

261 Chapter 8

36. Solve for yto get y=(b/a)√a2−x2for the upper half of the ellipse; make use of symmetry to

get A=4Za

2−x

dx =4b

aZa

0pa2−x2dx =4b

a×1

4πa2=πab.

37. Let Abe the area between the curve and the x-axis and ARthe area of the rectangle, then

A=Zb

0kxmdx =k

m+1x

m+1b

=kbm+1

m+1,A

R=b(kbm)=kbm+1,soA/AR=1/(m+ 1).

EXERCISE SET 8.2

1. V=πZ3

−1(3 −x)dx =8π2. V=πZ1

0[(2 −x2)2−x2]dx

=πZ1

0(4 −5x2+x4)dx

=38π/15

3. V=πZ2

4(3 −y)2dy =13π/64. V=πZ2

1/2(4 −1/y2)dy =9π/2

5. V=πZ2

0x4dx =32π/5

y = x2

6. V=πZπ/3

π/4sec2xdx=π(

√3−1)

-2

-1

y = sec x

-1

y = √cos x

7. V=πZπ/2

π/4cos xdx=(1−√

2/2)π

1(1, 1)

y = x2

y = x3

8. V=πZ1

0[(x2)2−(x3)2]dx

=πZ1

0(x4−x6)dx =2π/35

Exercise Set 8.2 262

9. V=πZ4

−4[(25 −x2)−9]dx

=2πZ4

0(16 −x2)dx = 256π/3

yy = √25 – x2

y = 3

10. V=πZ3

−3(9 −x2)2dx

=πZ3

−3(81 −18x2+x4)dx = 1296π/5

-3 3

y = 9 – x2

11. V=πZln 3

0e2xdx =π

2e2xln 3

=4π

12. V=πZ1

0e−4xdx =π

4(1 −e−4)

-1

13. V=πZ4

0[(4x)2−(x2)2]dx

=πZ4

0(16x2−x4)dx = 2048π/15

16 (4, 16)

y = x2

y = 4x

14. V=πZπ/4

0(cos2x−sin2x)dx

=πZπ/4

0cos 2xdx=π/2

-1

y = cos x

y = sin x

15. V=πZ1

0y2/3dy =3π/5

-1 1

y = x3

263 Chapter 8

16. V=πZ1

−1(1 −y2)2dy

=πZ1

−1(1 −2y2+y4)dy =16π/15

-1 1

-1

x = 1 – y2

17. V=πZ3

−1(1+y)dy =8π

x = √1 + y

18. V=πZ3

0[22−(y+ 1)]dy

=πZ3

0(3 −y)dy =9π/2

3(2, 3)

yy = x2 – 1

19. V=πZ3π/4

π/4csc2ydy=2π

-2 -1 1 2

x = csc y

20. V=πZ1

0(y−y4)dy =3π/10

-1 1

-1

1(1, 1)

y = x2

x = y2

21. V=πZ2

−1[(y+2)

2−y

4]dy =72π/5

(4, 2)

x = y2

x = y + 2

(1, –1)

-1

yx = 2 + y2

x = 1 – y2

22. V=πZ1

−1(2+y2)2−(1 −y2)2dy

=πZ1

−1(3+6y

2)dy =10π

Exercise Set 8.2 264

23. V=πZa

−a

a2(a2−x2)dx =4πab2/3

–aa

y = √a2 – x2

24. V=πZ2

x2dx =π(1/b −1/2); π(1/b −1/2)=3,b=2π/(π+6)

25. V=πZ0

−1(x+1)dx

+πZ1

0[(x+1)−2x]dx

=π/2+π/2=π

-1 1

y = √2x

(1, √2)

y = √x + 1

26. V=πZ4

0xdx+πZ6

4(6 −x)2dx

=8π+8π/3=32π/3

y = √xy = 6 – x

27. V=πZ3

0(9 −y2)2dy

=πZ3

0(81 −18y2+y4)dy

= 648π/5

x = y2

28. V=πZ9

0[32−(3 −√x)2]dx

=πZ9

0(6√x−x)dx

= 135π/2

y = √x

y = 3

x = y2

x = y

y = −1

29. V=πZ1

0[(√x+1)

2−(x+1)

2]dx

=πZ1

0(2√x−x−x2)dx =π/2

265 Chapter 8

30. V=πZ1

0[(y+1)

2−(y

2+1)

2]dy

=πZ1

0(2y−y2−y4)dy =7π/15

x = y2

x = y

x = –1

31. A(x)=π(x

/4)2=πx4/16,

V=Z20

0(πx4/16)dx =40,000πft3

32. V=πZ1

0(x−x4)dx =3π/10

33. V=Z1

0(x−x2)2dx

=Z1

0(x2−2x3+x4)dx =1/30

Square

(1, 1)

y = x

y = x2

34. A(x)=1

π1

√



πx,

V=Z4

8πx dx =π

y = √x

35. On the upper half of the circle, y=√1−x2, so:

(a) A(x) is the area of a semicircle of radius y,so

A(x)=πy2/2=π(1 −x2)/2; V=π

2Z1

−1(1 −x2)dx =πZ1

0(1 −x2)dx =2π/3

-1

y = √1 – x2

(b) A(x) is the area of a square of side 2y,so

A(x)=4y

2= 4(1 −x2); V=4Z1

−1

(1 −x2)dx =8Z1

0(1 −x2)dx =16/3

-1

y = √1 – x2x

Exercise Set 8.2 266

A(x)=√

4(2y)2=√3y2=√3(1 −x2);

V=Z1

−1

√3(1 −x2)dx =2

√

3Z1

0(1 −x2)dx =4

√

3/3

-1

y = √1 – x2

2y2y

36. By similar triangles, R/r =y/h so

R=ry/h and A(y)=πr2y2/h2.

V=(πr2/h2)Zh

0y2dy =πr2h/3

37. The two curves cross at x=b≈1.403288534, so

V=πZb

0((2x/π)2−sin16 x)dx +πZπ/2

(sin16 x−(2x/π)2)dx =0.710172176.

38. V=πZe

1(1 −(ln y)2)dy =π

39. (a) V=πZr

r−h

(r2−y2)dy =π(rh2−h3/3) = 1

3πh2(3r−h)

x2 + y2 = r2

(b) By the Pythagorean Theorem,

r2=(r−h)

2+ρ

,2hr =h2+ρ2; from part (a),

V=πh

3(3hr −h2)=πh

33

2(h2+ρ2)−h2)

6πh(h2+3ρ

2).

h – 10

−10

x = √100 – y2

40. Find the volume generated by revolving

the shaded region about the y-axis.

V=πZ−10+h

−10 (100 −y2)dy =π

3h2(30 −h)

Find dh/dt when h= 5 given that dV/dt =1/2.

V=π

3(30h2−h3), dV

dt =π

3(60h−3h2)dh

dt ,

2=π

3(300 −75)dh

dt ,dh

dt =1/(150π) ft/min

267 Chapter 8

41. (b) ∆x=5

10 =0.5; {y0,y

1,···,y

10}={0,2.00,2.45,2.45,2.00,1.46,1.26,1.25,1.25,1.25,1.25};

left = π

i=0 yi

22∆x≈11.157;

right = π

i=1 yi

22∆x≈11.771; V≈average = 11.464 cm3

x = √r2 – y2

√3r

–

42. If x=r/2 then from y2=r2−x2we get y=±√3r/2

as limits of integration; for −√3≤y≤√3,

A(y)=π[(r2−y2)−r2/4] = π(3r2/4−y2), thus

V=πZ√3r/2

−√3r/2(3r2/4−y2)dy

=2πZ√

3r/2

0(3r2/4−y2)dy =√3πr3/2.

43. (a)

-4

0 ≤ h < 2

h – 4

(b)

-4

-2 h

2 ≤ h ≤ 4

h – 4 x

If the cherry is partially submerged then 0 ≤h<2 as shown in Figure (a); if it is totally sub-

merged then 2 ≤h≤4 as shown in Figure (b). The radius of the glass is 4 cm and that of the

cherry is 1 cm so points on the sections shown in the ﬁgures satisfy the equations x2+y2=16

and x2+(y+3)

2= 1. We will ﬁnd the volumes of the solids that are generated when the shaded

regions are revolved about the y-axis.

For 0 ≤h<2,

V=πZh−4

−4[(16 −y2)−(1 −(y+3)

2)]dy =6πZh−4

−4(y+4)dy =3πh2;

for 2 ≤h≤4,

V=πZ−2

−4[(16 −y2)−(1 −(y+3)

2)]dy +πZh−4

−2(16 −y2)dy

=6πZ−2

−4(y+4)dy +πZh−4

−2(16 −y2)dy =12π+1

π(12h2−h3−40)

3π(12h2−h3−4)

V=





3πh2if 0 ≤h<2

π(12h2−h3−4) if 2 ≤h≤4

Exercise Set 8.3 268

44. x=h±pr2−y2,

V=πZr

−rh(h+pr2−y2)2−(h−pr2−y2)2idy

=4πh Zr

−rpr2−y2dy

=4πh 1

2πr2=2π

(x – h2) + y2 = r2

45. tan θ=h/x so h=xtan θ,

A(y)=1

hx =1

2x2tan θ=1

2(r2−y2) tan θ

because x2=r2−y2,

V=1

2tan θZr

−r

(r2−y2)dy

= tan θZr

0(r2−y2)dy =2

3r3tan θ

46. A(x)=(xtan θ)(2pr2−x2)

= 2(tan θ)xpr2−x2,

V= 2 tan θZr

0xpr2−x2dx

3r3tan θ

√r2 – x2

x tan u

47. Each cross section perpendicular to the

y-axis is a square so

A(y)=x

2=r

2−y

V=Zr

0(r

2−y

)dy

V= 8(2r3/3)=16r

x = √r2 – y2

48. The regular cylinder of radius rand height hhas the same circular cross sections as do those of

the oblique clinder, so by Cavalieri’s Principle, they have the same volume: πr2h.

EXERCISE SET 8.3

1. V=Z2

12πx(x2)dx =2πZ2

dx =15π/2

2. V=Z√2

02πx(p4−x2−x)dx =2πZ√

0(x

p4−x

2−x

)dx =8π

3(2 −√2)

3. V=Z1

02πy(2y−2y2)dy =4πZ1

0(y

2−y

)dy =π/3

269 Chapter 8

4. V=Z2

02πy[y−(y2−2)]dy =2πZ2

0(y

2−y

3+2y)dy =16π/3

5. V=Z1

02π(x)(x3)dx

=2πZ1

dx =2π/5

-1 1

-1

y = x3

6. V=Z9

42πx(√x)dx

=2πZ9

3/2

dx = 844π/5

-9 -4 4 9

y = √x

7. V=Z3

12πx(1/x)dx =2πZ3

1dx =4π

-3 -1 1 3

y = x

8. V=Z√π/2

02πx cos(x2)dx =π/√2

√p

y = cos (x2)

9. V=Z2

12πx[(2x−1) −(−2x+ 3)]dx

=8πZ2

1(x

2−x)dx =20π/3

(2, 3)

(2, –1)

(1, 1)

10. V=2πZ1

2+1dx

=πln(x2+1)

0=πln 2

-1 1

y = 1

x2 + 1

-√3-1 1 √

y = ex2

11. V=Z√3

12πxex2dx =πex2i√3

1=π(e3−e)

Exercise Set 8.3 270

12. V=Z2

02πx(2x−x2)dx

=2πZ2

0(2x2−x3)dx =8

3π

y = 2x – x2

13. V=Z1

02πy3dy =π/2

x = y2

14. V=Z3

22πy(2y)dy =4πZ3

dy =76π/3

x = 2y

15. V=Z1

02πy(1 −√y)dy

=2πZ1

0(y−y

3/2

)dy =π/5

y = √x

16. V=Z4

12πy(5 −y−4/y)dy

=2πZ4

1(5y−y2−4)dy =9π

(1, 4)

(4, 1)

x = 5 – y

x = 4/y

17. V=2πZπ

0xsin xdx =2π

218. V=2πZπ/2

0xcos xdx =π2−2π

19. (a) V=Z1

02πx(x3−3x2+2x)dx =7π/30

(b) much easier; the method of slicing would require that xbe expressed in terms of y.

-1 1

y = x3 – 3x2 + 2x

271 Chapter 8

20. V=Z2

12π(x+ 1)(1/x3)dx

=2πZ2

1(x

−2+x

−3

)dx =7π/4

-1 x21

y = 1/x3

x + 1

1 – yx = y1/3

21. V=Z1

02π(1 −y)y1/3dy

=2πZ1

0(y

1/3−y

4/3

)dy =9π/14

22. (a) Zb

2πx[f(x)−g(x)]dx (b) Zd

2πy[f(y)−g(y)]dy

(0, r)

(h, 0)

23. x=h

r(r−y) is an equation of the line

through (0,r) and (h, 0) so

V=Zr

02πy h

r(r−y)dy

=2πh

rZr

0(ry −y2)dy =πr2h/3

x = k/2

k/2 – x

x = k/4

y = √kx

y = –√kx

24. V=Zk/4

02π(k/2−x)2√kxdx

=2π

√

kZk/4

0(kx1/2−2x3/2)dx =7πk3/60

y = √r2 – x2

y = –√r2 – x2

25. V=Za

02πx(2pr2−x2)dx =4πZa

0x(r

2−x

)

1/2

=−4π

3(r2−x2)3/2a

=4π

3hr3−(r2−a2)3/2i

Exercise Set 8.4 272

26. V=Za

−a

2π(b−x)(2pa2−x2)dx

=4πb Za

−apa2−x2dx −4πZa

−a

xpa2−x2dx

=4πb ·(area of a semicircle of radius a)−4π(0)

=2π

a–a

√

– x2

–√a2 – x2

b − x

x = b

27. Vx=πZb

1/2

x2dx =π(2 −1/b), Vy=2πZb

1/2dx =π(2b−1);

Vx=Vyif 2 −1/b =2b−1, 2b2−3b+ 1 = 0, solve to get b=1/2 (reject) or b=1.

EXERCISE SET 8.4

1. (a) dy

dx =2,L=Z2

√

1+4dx =√5

(b) dx

dy =1

2,L=Z4

2p1+1/4dy =2

√

5/2=√

2. dx

dt =1, dy

dt =5,L=Z1

0p1

2+5

2dt =√26

3. f0(x)=9

1/2

,1+[f0(x)]2=1+81

4x,

L=Z1

0p1+81x/4dx =8

243 1+81

4x3/2#1

= (85√85 −8)/243

4. g0(y)=y(y

2+2)

1/2,1+[g

0(y)]2=1+y

2+2)=y

4+2y

2+1=(y

2+1)

L=Z1

0p(y

2+1)

2dy =Z1

0(y2+1)dy =4/3

5. dy

dx =2

3x−1/3,1+dy

dx2

=1+4

−2/3=9x

2/3+4

2/3,

L=Z8

√

2/3+4

1/3dx =1

18 Z40

13 u1/2du, u =9x

2/3+4

27u3/240

27(40√40 −13√13) = 1

27(80√10 −13√13)

or (alternate solution)

x=y3/2,dx

dy =3

2y1/2,1+dx

dy 2

=1+9

y=4+9y

L=1

2Z4

1p4+9ydy=1

18 Z40

13 u1/2du =1

27(80√10 −13√13)

273 Chapter 8

6. f0(x)=1

3−x

−3

, 1+[f0(x)]2=1+1

16x6−1

2+x−6=1

16x6+1

2+x−6=1

4x3+x−32

L=Z3

2s1

4x3+x−32

dx =Z3

21

4x3+x−3dx = 595/144

7. f0(x)=1

2e

x−e

−x

,1+[f0(x)]2=1+1

4e

2x−2+e

−2x

=1

4e

2x+2+e

−2x

,so

L=Z3

0p1+(f0(x))2dx =1

2Z3

0ex+e−xdx =(e

3−e

−3

)/2

8. g0(y)=1

3−1

−3

,1+[g

0(y)]2=1+1

6−1

2+1

−6

=1

3+1

−3



L=Z4

11

3+1

−3

dy = 2055/64

9. (dx/dt)2+(dy/dt)2=(t

)

2+(t)

2=t

2(t

2+ 1), L=Z1

0t(t2+1)

1/2dt =(2

√

2−1)/3

10. (dx/dt)2+(dy/dt)2= [2(1 + t)]2+ [3(1 + t)2]2=(1+t)

[4+9(1+t)2],

L=Z1

0(1+t)[4 + 9(1 + t)2]1/2dt = (80√10 −13√13)/27

11. (dx/dt)2+(dy/dt)2=(−2 sin 2t)2+ (2 cos 2t)2=4,L=Zπ/2

02dt =π

12. (dx/dt)2+(dy/dt)2=(−sin t+ sin t+tcos t)2+ (cos t−cos t+tsin t)2=t2,

L=Zπ

0tdt=π

2/2

13. (dx/dt)2+(dy/dt)2=[e

(cos t−sin t)]2+[e

(cos t+ sin t)]2=2e

L=Zπ/2

√2etdt =√2(eπ/2−1)

14. (dx/dt)2+(dy/dt)2=(2e

tcos t)2+(−2e

tsin t)2=4e

,L=Z4

12e

dt =2(e

4−e)

15. dy/dx =sec xtan x

sec x= tan x,p1+(y

2=√

1 + tan2x= sec xwhen 0 <x<π/4, so

L=Zπ/4

0sec xdx= ln(1 + √2)

16. dy/dx =cos x

sin x= cot x,p1+(y

2=√

1 + cot2x= csc xwhen π/4<x<π/2, so

L=Zπ/2

π/4csc xdx=−ln(√2−1) = −ln √2−1

√2+1(

√

2+1)

!= ln(1 + √2)

17. (a) (dx/dθ)2+(dy/dθ)2=(a(1 −cos θ))2+(asin θ)2=a2(2 −2 cos θ), so

L=Z2π

0p(dx/dθ)2+(dy/dθ)2dθ =aZ2π

0p2(1 −cos θ)dθ

Exercise Set 8.4 274

18. (a) Use the interval 0 ≤φ<2π.

(b) (dx/dφ)2+(dy/dφ)2=(−3acos2φsin φ)2+(3asin2φcos φ)2

=9a

2cos2φsin2φ(cos2φ+ sin2φ)=(9a

/4) sin22φ, so

L=(3a/2) Z2π

0|sin 2φ|dφ =6aZπ/2

0sin 2φdφ=−3acos 2φπ/2

=6a

19. (a)

(-1, 1)

(8, 4)

y(b) dy/dx does not exist at x=0.

3/2

0(y)=3

1/2

L=Z1

0p1+9y/4dy (portion for −1≤x≤0)

+Z4

0p1+9y/4dy (portion for 0 ≤x≤8)

27 13

8√13 −1+8

27(10√10 −1) = (13√13+80

√

10 −16)/27

20. For (4), express the curve y=f(x) in the parametric form x=t, y =f(t)sodx/dt = 1 and

dy/dt =f0(t)=f

0(x)=dy/dx. For (5), express x=g(y)asx=g(t),y =tso

dx/dt =g0(t)=g

0(y)=dx/dy and dy/dt =1.

21. L=Z2

0p1+4x

2dx ≈4.645975301 22. L=Zπ

0p1 + cos2ydy≈3.820197789

23. Numerical integration yields: in Exercise 21, L≈4.646783762; in Exercise 22, L≈3.820197788.

24. 0≤m≤f0(x)≤M,so m

2≤[f

(x)]2≤M2, and 1 + m2≤1+[f0(x)]2≤1+M2;thus

√

1+m

2≤p1+[f0(x)]2≤√1+M2,

ap1+m

2dx ≤Zb

ap1+[f0(x)]2dx ≤Zb

ap1+M2dx, and

(b−a)p1+m

2≤L≤(b−a)

p1+M2

25. f0(x) = cos x,√2/2≤cos x≤1 for 0 ≤x≤π/4so

(π/4)p1+1/2≤L≤(π/4)√1+1, π

4p3/2≤L≤π

4√

26. (dx/dt)2+(dy/dt)2=(−asin t)2+(bcos t)2=a2sin2t+b2cos2t

=a2(1 −cos2t)+b

2cos2t=a2−(a2−b2) cos2t

=a21−a2−b2

a2cos2t=a2[1 −k2cos2t],

L=Z2π

0ap1−k2cos2tdt=4aZπ/2

0p1−k2cos2tdt

275 Chapter 8

27. (a) (dx/dt)2+(dy/dt)2= 4 sin2t+ cos2t= 4 sin2t+(1−sin2t)=1+3sin

2t,

L=Z2π

0p1+3sin

2tdt=4Zπ/2

0p1+3sin

2tdt

(b) 9.69

1.5p1+3sin

2tdt≈5.16 cm

28. The distance is Z4.6

0p1+(2.09 −0.82x)2dx ≈6.65 m

1 2 1.84 1.83 1.832

3.8202 5.2704 5.0135 4.9977 5.0008

29. L=Zπ

0p1+(kcos x)2dx

Experimentation yields the values in the table, which by the Intermediate-Value Theorem show

that the true solution kto L= 5 lies between k=1.83 and k=1.832, so k=1.83 to two decimal

places.

EXERCISE SET 8.5

1. S=Z1

02π(7x)√1+49dx =70π

√

2Z1

0xdx=35π

√

2. f0(x)= 1

√

,1+[f0(x)]2=1+ 1

S=Z4

12π

√

r1+ 1

4xdx =2πZ4

1px+1/4dx =π(17√17 −5√5)/6

3. f0(x)=−x/√4−x2,1+[f0(x)]2=1+ x

4−x

2=4

4−x

S=Z1

−12π

p4−x

(2/p4−x2)dx =4πZ1

−1dx =8π

4. y=f(x)=x

3for 1 ≤x≤2, f0(x)=3x

S=Z2

12πx3p1+9x

4dx =π

27(1+9x

3/22

=5π(29√145 −2√10)/27

5. S=Z2

02π(9y+1)

√

82dy =2π

√

82 Z2

0(9y+1)dy =40π

√

6. g0(y)=3y

,S=Z1

02πy3p1+9y

4dy =π(10√10 −1)/27

7. g0(y)=−y/p9−y2,1+[g

0(y)]2=9

9−y2,S=Z2

−22πp9−y2·3

p9−y2dy =6πZ2

−2dy =24π

8. g0(y)=−(1 −y)−1/2,1+[g

0(y)]2=2−y

1−y,

S=Z0

−12π(2p1−y)√2−y

√1−ydy =4πZ0

−1p2−ydy=8π(3√3−2√2)/3

Exercise Set 8.5 276

9. f0(x)=1

−1/2−1

1/2

,1+[f0(x)]2=1+1

−1−1

2+1

x=1

−1/2+1

1/2



S=Z3

12πx

1/2−1

3/2

1

−1/2+1

1/2dx =π

3Z3

1(3+2x−x

2)dx =16π/9

10. f0(x)=x

2−1

−2

,1+[f0(x)]2=1+x

4−1

2+1

16x−4=x2+1

4x−22

S=Z2

12π1

3x3+1

4x−1x

2+1

−2dx =2πZ2

11

5+1

x+1

16x−3dx = 515π/64

11. x=g(y)=1

4+1

−2

0(y)=y

3−1

−3

1+[g

0(y)]2=1+y

6−1

2+1

16y−6=y3+1

4y−32

S=Z2

12π1

4y4+1

8y−2y

3+1

−3dy =π

16 Z2

1(8y7+6y+y

−5)dy =16,911π/1024

12. x=g(y)=√

16 −y;g0(y)=−1

√

16 −y,1+[g

0(y)]2=65 −4y

4(16 −y),

S=Z15

02πp16 −ys65 −4y

4(16 −y)dy =πZ15

0p65 −4ydy = (65√65 −5√5)π

13. f0(x)=e

, 1+[f0(x)]2=1+e

,S=Z1

02πexp1+e

2xdx ≈22.94

14. f0(x) = cos x, 1+[f0(x)]2= 1 + cos2x,S=Zπ

02πsin xp1 + cos2xdx≈14.42

15. x=g(y)=lny, g0(y)=1/y, 1+[g

0(y)]2=1+1/y2;S=Ze

12πp1+1/y2ln ydy ≈7.05

16. x=g(y) = tan y, g0(y) = sec2y, 1+[g

0(y)]2= 1 + sec4y;

S=Zπ/4

02πtan yp1 + sec4ydy ≈3.84

17. Revolve the line segment joining the points (0,0) and (h, r) about the x-axis. An equation of the

line segment is y=(r/h)xfor 0 ≤x≤hso

S=Zh

02π(r/h)xp1+r

2/h2dx =2πr

h2pr2+h2Zh

0xdx=πrpr2+h2

18. f(x)=√

2−x

0(x)=−x/√r2−x2,1+[f0(x)]2=r2/(r2−x2),

S=Zr

−r

2πpr2−x2(r/pr2−x2)dx =2πr Zr

−r

dx =4πr2

19. g(y)=pr

2−y

0(y)=−y/pr2−y2,1+[g

0(y)]2=r2/(r2−y2),

(a) S=Zr

r−h

2πpr2−y2pr2/(r2−y2)dy =2πr Zr

r−h

dy =2πrh

(b) From part (a), the surface area common to two polar caps of height h1>h

2is

2πrh1−2πrh2=2πr(h1−h2).

277 Chapter 8

20. For (4), express the curve y=f(x) in the parametric form x=t, y =f(t)sodx/dt = 1 and

dy/dt =f0(t)=f

0(x)=dy/dx. For (5), express x=g(y)asx=g(t),y =tso

dx/dt =g0(t)=g

0(y)=dx/dy and dy/dt =1.

21. x0=2t, y0=2,(x

2+(y

2=4t

2+4

S=2πZ4

0(2t)p4t2+4dt =8πZ4

2+1dt =8π

3(17√17 −1)

22. x0=et(cos t−sin t),y

0=e

t(cos t+ sin t),(x0)2+(y

2=2e

S=2πZπ/2

0(etsin t)√2e2tdt =2

√

2πZπ/2

0e2tsin tdt

√

2π1

(2 sin t−cos t)π/2

=2√2

5π(2eπ+1)

23. x0=1,y

0=4t,(x

)

2+(y

2=1+16t

2,S=2πZ1

p1+16t

2dt =π

24(17√17 −1)

24. x0=−2 sin tcos t, y0= 2 sin tcos t, (x0)2+(y

2= 8 sin2tcos2t

S=2πZπ/2

0cos2tp8 sin2tcos2tdt=4

√

2πZπ/2

0cos3tsin tdt=√2π

25. x0=−rsin t,y0=rcos t,(x

)

2+(y

2=r

S=2πZπ

0rsin t√r2dt =2πr2Zπ

0sin tdt=4πr2

26. dx

dφ =a(1 −cos φ), dy

dφ =asin φ,dx

dφ2

+dy

dφ2

=2a

(1 −cos φ)

S=2πZ2π

0a(1 −cos φ)p2a2(1 −cos φ)dφ =2

√

2πa2Z2π

0(1 −cos φ)3/2dφ,

but 1 −cos φ= 2 sin2φ

2so (1 −cos φ)3/2=2

√

2 sin3φ

2for 0 ≤φ≤πand, taking advantage of the

symmetry of the cycloid, S=16πa2Zπ

0sin3φ

2dφ =64πa2/3.

27. (a) length of arc of sector = circumference of base of cone,

`θ =2πr, θ =2πr/`;S= area of sector = 1

2`2(2πr/`)=πr`

(b) S=πr2`2−πr1`1=πr2(`1+`)−πr1`1=π[(r2−r1)`1+r2`];

Using similar triangles `2/r2=`1/r1,r

2=r

1,r

1(`

1+`)=r

,(r

2−r

1=r

so S=π(r1`+r2`)=π(r

1+r

)`.

Exercise Set 8.6 278

28. 2πkp1+[f0(x)]2≤2πf(x)p1+[f0(x)]2≤2πKp1+[f0(x)]2,so

2πkp1+[f0(x)]2dx ≤Zb

2πf(x)p1+[f0(x)]2dx ≤Zb

2πKp1+[f0(x)]2dx,

2πk Zb

ap1+[f0(x)]2dx ≤S≤2πK Zb

ap1+[f0(x)]2dx,2πkL ≤S≤2πKL

29. (a) 1≤p1+[f0(x)]2so 2πf(x)≤2πf(x)p1+[f0(x)]2,

2πf(x)dx ≤Zb

2πf(x)p1+[f0(x)]2dx,2π

f(x)dx ≤S, 2πA ≤S

(b) 2πA =Sif f0(x) = 0 for all xin [a, b]sof(x) is constant on [a, b].

EXERCISE SET 8.6

1. (a) W=F·d= 30(7) = 210 ft·lb

(b) W=Z6

1F(x)dx =Z6

1x−2dx =−1

x6

=5/6ft·lb

2. W=Z5

0F(x)dx =Z2

040 dx −Z5

3(x−5) dx =80+60=140J

3. distance traveled = Z5

0v(t)dt =Z5

5dt =2

5t2i5

0= 10 ft. The force is a constant 10 lb, so the

work done is 10 ·10 = 100 ft·lb.

4. (a) F(x)=kx, F(0.05)=0.05k=45,k = 900 N/m

(b) W=Z0.03

0900xdx=0.405 J (c) W=Z0.10

0.05 900xdx=3.375 J

5. F(x)=kx,F(0.2)=0.2k= 100, k= 500 N/m, W=Z0.8

0500xdx = 160 J

6. F(x)=kx,F(1/2) = k/2=6,k= 12 N/m, W=Z2

012xdx=24J

7. W=Z1

0kx dx =k/2 = 10, k= 20 lb/ft

9 - x

8. W=Z6

0(9 −x)62.4(25π)dx

= 1560πZ6

0(9 −x)dx =56,160πft·lb

9. W=Z6

0(9 −x)ρ(25π)dx = 900πρ ft·lb

279 Chapter 8

15 - x

1510

10. r/10 = x/15, r=2x/3,

W=Z10

0(15 −x)62.4(4πx2/9)dx

=83.2

3πZ10

0(15x2−x3)dx

= 208,000π/3ft·lb

3 - x

w(x)

11. w/4=x/3,w =4x/3,

W=Z2

0(3 −x)(9810)(4x/3)(6)dx

= 78480 Z2

0(3x−x2)dx

= 261,600 J

-2

3 - x

w(x)

12. w=2

p4−x

W=Z2

−2

(3 −x)(50)(2p4−x2)(10)dx

= 3000 Z2

−2p4−x2dx −1000 Z2

−2xp4−x2dx

= 3000[π(2)2/2] −0 = 6000πft·lb

910 - x

20 15

13. (a) W=Z9

0(10 −x)62.4(300)dx

=18,720 Z9

0(10 −x)dx

= 926,640 ft·lb

(b) to empty the pool in one hour would require

926,640/3600 = 257.4ft·lb of work per second

so hp of motor = 257.4/550=0.468

3000

Rocket

14. When the rocket is xft above the ground

total weight = weight of rocket

+ weight of fuel

= 3 + [40 −2(x/1000)]

=43−x/500 tons,

W=Z3000

0(43 −x/500)dx = 120,000 ft·tons

Exercise Set 8.7 280

15. W=Z100

015(100 −x)dx

=75,000 ft·lb

100

100 - x

Pulley

Chain

16. Let F(x) be the force needed to hold

charge Aat position x, then

F(x)= c

(a−x)

2,F(−a)= c

2=k,

so c=4a

W=Z0

−a

k(a−x)

−2

dx =2ak J

0−aax

17. (a) 150 = k/(4000)2,k=2.4×109,w(x)=k/x2=2,400,000,000/x2lb

(b) 6000 = k/(4000)2,k=9.6×1010,w(x)=9.6×1010/(x+ 4000)2lb

4000 9.6(1010)x−2dx =4,800,000 mi·lb = 2.5344 ×1010 ft·lb

18. (a) 20 = k/(1080)2,k=2.3328 ×107, weight = w(x+ 1080) = 2.3328 ·107/(x+ 1080)2lb

(b) W=Z10.8

0[2.3328 ·107/(x+ 1080)2]dx = 213.86 mi·lb = 1,129,188 ft·lb

19. W=F·d=(6.40 ×105)(3.00 ×103)=1.92 ×109J; from Theorem 8.6.4,

f=2W/m +v2

i= 2(1.92 ·109)/(4 ·105)+20

2=10,000, vf= 100 m/s

20. W=F·d=(2.00 ×105)(2.00 ×105)=4×1010 J; from Theorem 8.6.4,

f=2W/m +v2

i=8·1010/(2 ·103)+10

8≈11.832 m/s.

21. (a) The kinetic energy would have decreased by 1

2mv2=1

24·106(15000)2=4.5×1014 J

(b) (4.5×1014)/(4.2×1015)≈0.107 (c) 1000

13 (0.107) ≈8.24 bombs

EXERCISE SET 8.7

1. (a) F=ρhA =62.4(5)(100) = 31,200 lb

P=ρh =62.4(5) = 312 lb/ft2

(b) F=ρhA = 9810(10)(25) = 2,452,500 N

P=ρh = 9810(10) = 98.1kPa

2. (a) F=PA =6·105(160) = 9.6×107N(b) F=PA = 100(60) = 6000 lb

3. F=Z2

062.4x(4)dx

= 249.6Z2

0xdx= 499.2lb

281 Chapter 8

4. F=Z3

19810x(4)dx

= 39240 Z3

1xdx

= 156,960 N 3

xy = √25 – x2

2√25 – x2

5. F=Z5

09810x(2p25 −x2)dx

=19,620 Z5

0x(25 −x2)1/2dx

=8.175 ×105N

2√3

xw(x)

6. By similar triangles

w(x)

4=2√3−x

2√3,w(x)= 2

√

(2√3−x),

F=Z2√3

062.4x2

√3(2√3−x)dx

=124.8

√3Z2√3

0(2√3x−x2)dx = 499.2lb

xw(x)

7. By similar triangles

w(x)

6=10 −x

w(x)=3

(10 −x),

F=Z10

29810x3

4(10 −x)dx

= 7357.5Z10

2(10x−x2)dx =1,098,720 N

w(x)

u(x)

8. w(x)=16+2u(x),but

u(x)

4=12 −x

8so u(x)=1

(12 −x),

w(x)=16+(12−x)=28−x,

F=Z12

462.4x(28 −x)dx

=62.4Z12

4(28x−x2)dx =77,209.6 lb.

9. Yes: if ρ2=2ρ

1then F2=Zb

ρ2h(x)w(x)dx =Zb

2ρ1h(x)w(x)dx =2Zb

h(x)w(x)dx =2F

Exercise Set 8.7 282

10. F=Z2

050x(2p4−x2)dx

= 100 Z2

0x(4 −x2)1/2dx

= 800/3lb

√4 – x2

y = √4 – x2

√2a

√2a/2

w1(x)

w2(x)

11. Find the forces on the upper and lower halves and add them:

w1(x)

√2a=x

√2a/2,w1(x)=2x

1=Z√

2a/2

0ρx(2x)dx =2ρZ√

2a/2

0x2dx =√2ρa3/6,

w2(x)

√2a=√2a−x

√2a/2,w2(x)=2(

√

2a−x)

2=Z√

√

2a/2ρx[2(√2a−x)]dx =2ρZ√

√

2a/2(√2ax −x2)dx =√2ρa3/3,

F=F1+F2=√2ρa3/6+√

2ρa3/3=ρa3/√2lb

xh(x)

200

100

60°

12. h(x)=xsin 60◦=√3x/2,

F=Z100

09810(√3x/2)(200)dx

= 981000√3Z100

0xdx

=4,905,000√3N

4√17

h(x)

13. √162+4

2=√

272=4

√

17 is the

other dimension of the bottom.

(h(x)−4)/4=x/(4√17)

h(x)=x/√17+4,

F=Z4

√

062.4(x/√17 + 4)10dx

= 624 Z4√17

0(x/√17+4)dx

=14,976√17 lb

h + 2

14. F=Zh+2

ρ0x(2)dx

=2ρ

0Zh+2

xdx

=4ρ

(h+1)

283 Chapter 8

15. (a) From Exercise 14, F=4ρ

(h+ 1) so (assuming that ρ0is constant) dF/dt =4ρ

(dh/dt)

which is a positive constant if dh/dt is a positive constant.

(b) If dh/dt = 20 then dF/dt =80ρ

0lb/min from part (a).

EXERCISE SET 8.8

1. (a) sinh 3 ≈10.0179

(b) cosh(−2) ≈3.7622

(d) sinh−1(−2) ≈−1.4436

(e) cosh−13≈1.7627

(f) tanh−13

4≈0.9730

2. (a) csch(−1) ≈−0.8509

(b) sech(ln 2) = 0.8

(d) sech−11

2≈1.3170

(e) coth−13≈0.3466

(f) csch−1(−√3) ≈−0.5493

3. (a) sinh(ln 3) = 1

2(eln 3 −e−ln 3)=1

23−1

=4

(b) cosh(−ln 2) = 1

2(e−ln 2 +eln 2)=1

21

2+2

=5

−2ln5

2ln5 +e

−2ln5 =25 −1/25

25+1/25 =312

313

(d) sinh(−3ln2)= 1

2(e

−3ln2 −e

3ln2)=1

21

8−8

=−63

4. (a) 1

2(eln x+e−ln x)=1

2x+1

=x

2+1

2x,x>0

(b) 1

2(eln x−e−ln x)=1

2x−1

=x

2−1

2x,x>0

−2lnx

2lnx+e

−2lnx=x

2−1/x2

x2+1/x2=x4−1

x4+1,x>0

(d) 1

2(e−ln x+eln x)=1

21

x+x

=1+x

2x,x>0

5. sinh x0 cosh x0 tanh x0 coth x0 sech x0 csch x0

2√52

/√

√

/√

(a)

(b)

(a) cosh2x0= 1 + sinh2x0= 1 + (2)2= 5, cosh x0=√5

(b) sinh2x0= cosh2x0−1=25

16 −1= 9

16, sinh x0=3

4(because x0>0)

2

=1−16

25 =9

25, sech x0=3

cosh x0=1

sech x0=5

3, from sinh x0

cosh x0= tanh x0we get sinh x0=5

34

5=4

Exercise Set 8.8 284

6. d

dx cschx=d

sinh x=−cosh x

sinh2x=−coth xcsch xfor x6=0

dx sech x=d

cosh x=−sinh x

cosh2x=−tanh xsech xfor all x

dx coth x=d

cosh x

sinh x=sinh2x−cosh2x

sinh2x=−csch2xfor x6=0

7. (a) y= sinh−1xif and only if x= sinh y;1=dy

dy =dy

dx cosh y;so

dx[sinh−1x]= dy

dx =1

cosh y=1

p1 + sinh2y=1

√1+x

2for all x.

(b) Let x≥1. Then y= cosh−1xif and only if x= cosh y;1=dy

dy =dy

dx sinh y,so

dx[cosh−1x]= dy

dx =1

sinh y=1

pcosh2y−1=1

x2−1for x≥1.

1=dy

dy =dy

dx sech2y=dy

dx(1 −tanh2y)=1−x

,so d

dx[tanh−1x]= dy

dx =1

1−x2.

9. 4 cosh(4x−8) 10. 4x3sinh(x4)11. −1

xcsch2(ln x)

12. 2sech22x

tanh 2x13. 1

x2csch(1/x) coth(1/x)14. −2e2xsech(e2x) tanh(e2x)

15. 2 + 5 cosh(5x) sinh(5x)

q4x+ cosh2(5x)16. 6 sinh2(2x) cosh(2x)

17. x5/2tanh(√x) sech2(√x)+3x

2tanh2(√x)

18. −3 cosh(cos 3x) sin 3x19. 1

p1+x

2/91

3=1/

p9+x

20. 1

p1+1/x2(−1/x2)=−1

|x|

√

2+1 21. 1/(cosh−1x)√x2−1

22. 1/q(sinh−1x)2−1√1+x

223. −(tanh−1x)−2/(1 −x2)

24. 2(coth−1x)/(1 −x2)25. sinh x

pcosh2x−1=sinh x

|sinh x|=1,x>0

−1,x<0

26. (sech2x)/p1 + tanh2x27. −ex

2x√1−x+exsech−1x

28. 10(1 + xcsch−1x)9−x

|x|√1+x

2+ csch−1x

31. 1

7sinh7x+C32. 1

2sinh(2x−3)+C33. 2

3(tanh x)3/2+C

285 Chapter 8

34. −1

3coth(3x)+C35. ln(cosh x)+C36. −1

3coth3x+C

37. −1

3sech3xln 3

ln 2

=37/375 38. ln(cosh x)iln 3

0=ln5−ln 3

39. u=3x,1

3Z1

√

1+u

2du =1

3sinh−13x+C

40. x=√2u,Z√2

√2u2−2du =Z1

√u2−1du = cosh−1(x/√2)+C

41. u=ex,Z1

u√1−u2du =−sech−1(ex)+C

42. u= cos θ,−Z1

√1+u

2du =−sinh−1(cos θ)+C

43. u=2x,Zdu

u√1+u

2=−csch−1|u|+C=−csch−1|2x|+C

44. x=5u/3, Z5/3

√25u2−25du =1

3Z1

√u2−1du =1

3cosh−1(3x/5)+C

45. tanh−1xi1/2

0= tanh−1(1/2) −tanh−1(0) = 1

2ln 1+1/2

1−1/2=1

2ln 3

46. sinh−1ti√3

0= sinh−1√3−sinh−10 = ln(√3+2)

49. A=Zln 3

0sinh 2xdx=1

2cosh 2xln 3

2[cosh(2 ln 3) −1],

but cosh(2 ln 3) = cosh(ln 9) = 1

2(eln 9 +e−ln 9)=1

(9+1/9)=41/9soA=1

[41/9−1]=16/9.

50. V=πZln 2

0sech2xdx=πtanh xln 2

=πtanh(ln 2) = 3π/5

51. V=πZ5

0(cosh22x−sinh22x)dx =πZ5

0dx =5π

52. Z1

0cosh ax dx =2,1

asinh ax1

=2,1

asinh a=2,sinh a=2a;

let f(a) = sinh a−2a, then an+1 =an−sinh an−2an

cosh an−2,a

1=2.2,...,a

4=a

5=2.177318985.

53. y0= sinh x,1+(y

2= 1 + sinh2x= cosh2x

L=Zln 2

0cosh xdx= sinh xln 2

= sinh(ln 2) = 1

2(eln 2 −e−ln 2)=1

22−1

=3

Exercise Set 8.8 286

54. y0= sinh(x/a),1+(y

2= 1 + sinh2(x/a) = cosh2(x/a)

L=Zx1

0cosh(x/a)dx =asinh(x/a)x1

=asinh(x1/a)

55. sinh(−x)=1

−x−e

)=−1

x−e

−x

)=−sinh x

cosh(−x)=1

−x+e

)=1

x+e

−x

) = cosh x

56. (a) cosh x+ sinh x=1

2(ex+e−x)+1

2(e

x−e

−x

)=e

(b) cosh x−sinh x=1

2(ex+e−x)−1

2(ex−e−x)=e

−x

4(ex−e−x)(ey+e−y)+1

4(e

x+e

−x

)(ey−e−y)

4[(ex+y−e−x+y+ex−y−e−x−y)+(e

x+y+e

−x+y−e

x−y−e

−x−y)]

2[e(x+y)−e−(x+y)] = sinh(x+y)

(d) Let y=xin part (c).

(e) The proof is similar to part (c), or: treat xas variable and yas constant, and diﬀerentiate

the result in part (c) with respect to x.

(f) Let y=xin part (e).

(g) Use cosh2x= 1 + sinh2xtogether with part (f).

(h) Use sinh2x= cosh2x−1 together with part (f).

57. (a) Divide cosh2x−sinh2x= 1 by cosh2x.

(b) tanh(x+y)=sinh xcosh y+ cosh xsinh y

cosh xcosh y+ sinh xsinh y=

sinh x

cosh x+sinh y

cosh y

1+ sinh xsinh y

cosh xcosh y

=tanh x+ tanh y

1 + tanh xtanh y

58. (a) Let y= cosh−1x; then x= cosh y=1

2(ey+e−y), ey−2x+e−y=0,e

2y−2xey+1=0,

y=2x±√

2−4

2=x±px

2−1. To determine which sign to take, note that y≥0

so e−y≤ey,x=(e

y+e

−y

)/2≤(e

y+e

)/2=e

, hence ey≥xthus ey=x+√x2−1,

y= cosh−1x= ln(x+√x2−1).

(b) Let y= tanh−1x; then x= tanh y=ey−e−y

ey+e−y=e2y−1

e2y+1,xe2y+x=e2y−1,

1+x=e

2y(1 −x), e2y=(1+x)/(1 −x), 2y=ln1+x

1−x,y=1

2ln 1+x

1−x.

59. (a) d

dx(cosh−1x)=1+x/√x2−1

x+√x2−1=1/

2−1

(b) d

dx(tanh−1x)= d

dx 1

2(ln(1 + x)−ln(1 −x))=1

21

1+x+1

1−x=1/(1 −x2)

287 Chapter 8

60. Let y= sech−1xthen x= sech y=1/cosh y, cosh y=1/x,y= cosh−1(1/x); the proofs for the

remaining two are similar.

61. If |u|<1 then, by Theorem 8.8.6, Zdu

1−u2= tanh−1u+C.

For |u|>1,Zdu

1−u2= coth−1u+C= tanh−1(1/u)+C.

62. (a) d

dx(sech−1|x|)= d

dx(sech−1√x2)=−1

√

1−x

√

2=−1

√

1−x

(b) Similar to solution of part (a)

63. (a) lim

x→+∞sinh x= lim

x→+∞

2(ex−e−x)=+∞−0=+∞

(b) lim

x→−∞ sinh x= lim

x→−∞

2(ex−e−x)=0−∞=−∞

x→+∞tanh x= lim

x→+∞

ex−e−x

ex+e−x=1

(d) lim

x→−∞ tanh x= lim

x→−∞

ex−e−x

ex+e−x=−1

(e) lim

x→+∞sinh−1x= lim

x→+∞ln(x+px2+1)=+∞

(f) lim

x→1−tanh−1x= lim

x→1−

2[ln(1 + x)−ln(1 −x)]=+∞

64. (a) lim

x→+∞(cosh−1x−ln x) = lim

x→+∞[ln(x+px2−1) −ln x]

= lim

x→+∞ln x+√x2−1

x= lim

x→+∞ln(1 + p1−1/x2)=ln2

(b) lim

x→+∞

cosh x

ex= lim

x→+∞

ex+e−x

2ex= lim

x→+∞

2(1+e−2x)=1/2

65. For |x|<1, y= tanh−1xis deﬁned and dy/dx =1/(1 −x2)>0; y00 =2x/(1 −x2)2changes sign

at x= 0, so there is a point of inﬂection there.

66. Let x=−u,Z1

√u2−1du =−Z1

√x2−1dx =−cosh−1x+C=−cosh−1(−u)+C.

−cosh−1(−u)=−ln(−u+pu2−1)=ln 1

−u+√

2−1

= ln(−u−pu2−1)=ln|u+pu

2−1|

67. Using sinh x+ cosh x=ex(Exercise 56a), (sinh x+ cosh x)n=(e

)

n=e

nx = sinh nx + cosh nx.

68. Za

−a

etxdx =1

tetxa

−a

t(eat −e−at)=2 sinh at

tfor t6=0.

69. (a) y0= sinh(x/a),1+(y

2= 1 + sinh2(x/a) = cosh2(x/a)

L=2Zb

0cosh(x/a)dx =2asinh(x/a)b

=2asinh(b/a)

(b) The highest point is at x=b, the lowest at x=0,

so S=acosh(b/a)−acosh(0) = acosh(b/a)−a.

Chapter 8 Supplementary Exercises 288

70. From part (a) of Exercise 69, L=2asinh(b/a)so120=2asinh(50/a),asinh(50/a) = 60. Let

u=50/a, then a=50/u so (50/u) sinh u=60,sinh u=1.2u.Iff(u) = sinh u−1.2u, then

un+1 =un−sinh un−1.2un

cosh un−1.2;u1=1,...,u

5=u

6=1.064868548 ≈50/a so a≈46.95415231.

From part (b), S=acosh(b/a)−a≈46.95415231[cosh(1.064868548) −1] ≈29.2 ft.

71. From part (b) of Exercise 69, S=acosh(b/a)−aso 30 = acosh(200/a)−a. Let u= 200/a,

then a= 200/u so 30 = (200/u)[cosh u−1],cosh u−1=0.15u.Iff(u) = cosh u−0.15u−1,

then un+1 =un−cosh un−0.15un−1

sinh un−0.15 ;u1=0.3,...,u

4=u

5=0.297792782 ≈200/a so

a≈671.6079505. From part (a), L=2asinh(b/a)≈2(671.6079505) sinh(0.297792782) ≈405.9 ft.

72. (a) When the bow of the boat is at the point (x, y) and the person has walked a distance D,

then the person is located at the point (0,D), the line segment connecting (0,D) and (x, y)

has length a;thusa

2=x

2+(D−y)

2,D=y+√

2−x

2=asech−1(x/a).

(b) Find Dwhen a= 15, x= 10: D= 15 sech−1(10/15)=15ln 1+p5/9

2/3!≈14.44 m.

x√a2−x2+x

√a2−x2=1

√a2−x2−a2

x+x=−1

xpa2−x2,

1+[y

2=1+a

2−x

2=a

2; with a= 15 and x=5,L=Z15

225

x2dx =−225

x15

=30m.

CHAPTER 8 SUPPLEMENTARY EXERCISES

6. (a) A=Z2

0(2+x−x2)dx (b) A=Z2

0√ydy+Z4

2[(√y−(y−2)] dy

0[(2 + x)2−x4]dx

(d) V=2πZ2

√

ydy+2πZ4

2y[

√

y−(y−2)] dy

(e) V=2πZ2

0x(2+x−x2)dx (f) V=πZ2

0ydy+Z4

2π(y−(y−2)2)dy

7. (a) A=Zb

(f(x)−g(x)) dx +Zc

(g(x)−f(x)) dx +Zd

(f(x)−g(x)) dx

(b) A=Z0

−1(x3−x)dx +Z1

0(x−x3)dx +Z2

1(x3−x)dx =1

4+1

4+9

4=11

8. (a) S=Z8/27

02πxp1+x

−4/3dx (b) S=Z2

02πy3

27p1+y

4/81 dy

02π(y+2)

p1+y

4/81 dy

9. By implicit diﬀerentiation dy

dx =−y

x1/3,so1+dy

dx2

=1+y



2/3=x

2/3+y

2/3

2/3=a

2/3

2/3,

L=Z−a/8

−a

a1/3

(−x1/3)dx =−a1/3Z−a/8

−a

x−1/3dx =9a/8.

289 Chapter 8

10. The base of the dome is a hexagon of side r. An equation of the circle of radius rthat lies in a

vertical x-yplane and passes through two opposite vertices of the base hexagon is x2+y2=r2.

A horizontal, hexagonal cross section at height yabove the base has area

A(y)=3

√

2=3

√

2(r

2−y

), hence the volume is V=Zr

3√3

2(r2−y2)dy =√3r3.

11. Let the sphere have radius R, the hole radius r. By the Pythagorean Theorem, r2+(L/2)2=R2.

Use cylindrical shells to calculate the volume of the solid obtained by rotating about the y-axis

the region r<x<R, −

√

2−x

2<y<√

2−x

V=ZR

(2πx)2pR2−x2dx =−4

3π(R2−x2)3/2R

3π(L/2)3,

so the volume is independent of R.

12. V=2ZL/2

0π16R2

L4(x2−L2/4)2=4π

15 LR2

13. Set a=68.7672, b=0.0100333, c= 693.8597, d= 299.2239.

(a) 650

-300 300

(b) L=2Zd

0p1+a

2sinh2bx dx

= 1480.2798 ft

14. y=0atx=b=30.585; distance = Zb

0p1 + (12.54 −0.82x)2dx = 196.306 yd

15. Let u=ax then du =adx,Z1

√a

2+u

2du =Za

√a2+a2x2dx =Z1

√1+x

2dx = sinh−1(u/a)+C,

√u2−a2du =Z1

√x2−1dx = cosh−1(u/a)+C,u>a,

2−u

2du =1

aZ1

1−x2dx =









atanh−1(u/a)+C, |u|<a

acoth−1(u/a)+C, |u|>a

2aln 

a+u

a−u+C

(a) sinh−1(x/2)+C

(b) cosh−1(x/3)+C









√2tanh−1(x/√2)+C, |x|<√2

√2coth−1(x/√2)+C, |x|>√2

2√2ln 

√2+x

√

2−x



+C

(d) Zdx

√16+5x

2=1

√

5Zdx

p16/5+x

2dx =1

√5sinh−1 √5x

4!+C

Chapter 8 Supplementary Exercises 290

16. (a) cosh 3x= cosh(2x+x) = cosh 2xcosh x+ sinh 2xsinh x

= (2 cosh2x−1) cosh x+ (2 sinh xcosh x) sinh x

= 2 cosh3x−cosh x+ 2 sinh2xcosh x

= 2 cosh3x−cosh x+ 2(cosh2x−1) cosh x= 4 cosh3x−3 cosh x

(b) from Theorem 8.8.2 with xreplaced by x

2: cosh x= 2 cosh2x

2−1,

2 cosh2x

2= cosh x+ 1, cosh2x

2=1

2(cosh x+ 1),

cosh x

2=r1

2(cosh x+ 1) (because cosh x

2>0)

2: cosh x= 2 sinh2x

2+1,

2 sinh2x

2= cosh x−1, sinh2x

2=1

2(cosh x−1), sinh x

2=±r1

2(cosh x−1)

17. (a) F=kx,1

2=k1

4,k=2,W=Z1/4

0kx dx =1/16 J

(b) 25 = ZL

0kx dx =kL2/2, L=5m

18. F=30x+ 2000, W=Z150

0(30x+ 2000) dx =15·1502+ 2000 ·150 = 637,500 lb·ft

19. (a) F=Z1

0ρx3dx N

h(x) = 1 + x

w(x)

(b) By similar triangles w(x)

4=x

2,w(x)=2x,so

F=Z4

1ρ(1+x)2xdx lb/ft2.

125x2−10, so F=Z0

−10 9810|y|2r125

8(y+ 10) dy N.

20. (a)

r(b) r= 1 when t≈0.673080 s.

291 Chapter 8

21. (a)

100 200

-1.6

-1.2

-0.8

-0.4

y(b) The maximum deﬂection occurs at

x= 96 inches (the midpoint

of the beam) and is about 1.42 in.

Z192

0p1+(dy/dx)2dx = 192.026 in.

22. The x-coordinates of the points of intersection are a≈−0.423028 and b≈1.725171; the area is

(2 sin x−x2+1)dx ≈2.542696.

23. Let (a, k), where π/2<a<π, be the coordinates of the point of intersection of y=kwith

y= sin x.Thusk= sin aand if the shaded areas are equal,

0(k−sin x)dx =Za

0(sin a−sin x)dx =asin a+ cos a−1=0

Solve for ato get a≈2.331122, so k= sin a≈0.724611.

24. The volume is given by 2πZk

0xsin xdx=2π(sin k−kcos k) = 8; solve for kto get

k=1.736796.

25. (a) Za+2

√1+x

3dx

(b) Use the result in Exercise 24, Section 7.9, to obtain

da Za+2

√1+x

3dx=a+2

p1+(a+2)

3−a

√

1+a

3= 0; solve for ato get a≈0.683772.

The maximum work is Za+2

√1+x

3dx ≈1.347655 J.

292

CHAPTER 9

Principles of Integral Evaluation

EXERCISE SET 9.1

1. u=3−2x, du =−2dx, −1

2Zu3du =−1

8u4+C=−1

8(3 −2x)4+C

2. u=4+9x, du =9dx, 1

9Zu1/2du =2

3·9u3/2+C=2

27(4+9x)

3/2+C

3. u=x2,du=2xdx, 1

2Zsec2udu=1

2tan u+C=1

2tan(x2)+C

4. u=x2,du=2xdx, 2Ztan udu=−2ln|cos u|+C=−2ln|cos(x2)|+C

5. u= 2 + cos 3x, du =−3 sin 3xdx, −1

3Zdu

u=−1

3ln |u|+C=−1

3ln(2 + cos 3x)+C

6. u=3x

2,du=3

2dx, 2

3Zdu

4+4u

2=1

6Zdu

1+u

2=1

6tan−1u+C=1

6tan−1(3x/2)+C

7. u=ex,du=e

xdx, Zsinh udu= cosh u+C= cosh ex+C

8. u=lnx, du =1

xdx, Zsec utan udu= sec u+C= sec(ln x)+C

9. u= cot x, du =−csc2xdx, −Zeudu =−eu+C=−ecot x+C

10. u=x2,du=2xdx, 1

2Zdu

√1−u2=1

2sin−1u+C=1

2sin−1(x2)+C

11. u= cos 7x, du =−7 sin 7xdx, −1

7Zu5du =−1

42u6+C=−1

42 cos67x+C

12. u= sin x, du = cos x dx, Zdu

u√u2+1 =−ln 

1+√

1+u

u



+C=−ln 

1+p1 + sin2x

sin x+C

13. u=ex,du=e

xdx, Zdu

√4+u

2=lnu+pu

2+4

+C=lne

x+pe

2x+4

+C

14. u= tan−1x, du =1

1+x

2dx, Zeudu =eu+C=etan−1x+C

15. u=√x−2,du=1

√x−2dx, 2Zeudu =2e

u+C=2e

√

x−2+C

16. u=3x

2+2x, du =(6x+2)dx, 1

2Zcot udu=1

2ln |sin u|+C=1

2ln sin |3x2+2x|+C

293 Chapter 9

17. u=√x, du =1

2√xdx, Z2 cosh udu= 2 sinh u+C= 2 sinh √x+C

18. u=lnx, du =dx

x,Zdu

u=ln|u|+C=ln|ln x|+C

19. u=√x, du =1

2√xdx, Z2du

3u=2Ze

−uln 3 du =−2

ln 3e−uln 3 +C=−2

ln 33−√x+C

20. u= sin θ, du = cos θdθ, Zsec utan udu= sec u+C= sec(sin θ)+C

21. u=2

x,du=−2

2dx, −1

2Zcsch2udu=1

2coth u+C=1

2coth 2

x+C

22. Zdx

√x2−3=ln



x+px

2−3



+C

23. u=e−x,du=−e

−xdx, −Zdu

4−u2=−1

4ln 

2+u

2−u



+C=−1

4ln 

2+e

−x

2−e

−x



+C

24. u=lnx, du =1

xdx, Zcos udu= sin u+C= sin(ln x)+C

25. u=ex,du=e

xdx, Zexdx

√1−e2x=Zdu

√1−u2= sin−1u+C= sin−1ex+C

26. u=x−1/2,du=−1

3/2dx, −Z2 sinh udu=−2 cosh u+C=−2 cosh(x−1/2)+C

27. u=x2,du=2xdx, 1

2Zdu

sec u=1

2Zcos udu=1

2sin u+C=1

2sin(x2)+C

28. 2u=ex,2du =exdx, Z2du

√4−4u2= sin−1u+C= sin−1(ex/2)+C

29. 4−x2=e−x2ln 4,u=−x

2ln 4,du=−2xln 4 dx =−xln 16 dx,

−1

ln 16 Zeudu =−1

ln 16eu+C=−1

ln 16e−x2ln 4 +C=−1

ln 164−x2+C

30. 2πx =eπx ln 2,Z2πx dx =1

πln 2eπx ln 2 +C=1

πln 22πx +C

31. (a) With u=xwe get

from (15), Zdx

√1−x2= sin−1x+C(|x|<1)

from (17), Zdx

1+x

2= tan−1x+C

from (19), Zdx

x√x2−1= sec−1x+C(x>1)

Exercise Set 9.2 294

(b) With u=ax, du =adx, we get (for a>0)

Zdu

√a2−u2=Zadx

a√1−x2= sin−1x+C= sin−1(u/a)+C

Zdu

a2+u2=1

a2Zadx

1+x

2=1

atan−1x+C=1

atan−1(u/a)+C

Zdu

u√u2−a2=Zadx

a2x√x2−1=1

asec−1x+C=1

asec−1(u/a)+C

EXERCISE SET 9.2

1. u=x,dv =e−xdx,du =dx,v=−e−x;Zxe−xdx =−xe−x+Ze−xdx =−xe−x−e−x+C

2. u=x,dv =e3xdx,du =dx,v=1

3e3x;Zxe3xdx =1

3xe3x−1

3Ze3xdx =1

3xe3x−1

9e3x+C

3. u=x2,dv =exdx,du =2xdx,v=e

x;Zx

xdx =x2ex−2Zxexdx.

For Zxexdx use u=x,dv =exdx,du =dx,v=exto get

Zxexdx =xex−ex+C1so Zx2exdx =x2ex−2xex+2e

x+C

4. u=x2,dv =e−2xdx,du =2xdx,v=−1

−2x;Zx

−2xdx =−1

2x2e−2x+Zxe−2xdx

For Zxe−2xdx use u=x,dv =e−2xdx to get

Zxe−2xdx =−1

2xe−2x+1

2Ze−2xdx =−1

2xe−2x−1

4e−2x+C

so Zx2e−2xdx =−1

2x2e−2x−1

2xe−2x−1

4e−2x+C

5. u=x,dv = sin 2xdx,du =dx,v=−1

2cos 2x;

Zxsin 2xdx=−1

2xcos 2x+1

2Zcos 2xdx=−1

2xcos 2x+1

4sin 2x+C

6. u=x,dv = cos 3xdx,du =dx,v=1

3sin 3x;

Zxcos 3xdx=1

3xsin 3x−1

3Zsin 3xdx=1

3xsin 3x+1

9cos 3x+C

7. u=x2,dv = cos xdx,du =2xdx,v= sin x;Zx2cos xdx=x

2sin x−2Zxsin xdx

For Zxsin xdx use u=x,dv = sin xdx to get

Zxsin xdx=−xcos x+ sin x+C1so Zx2cos xdx=x

2sin x+2xcos x−2 sin x+C

295 Chapter 9

8. u=x2,dv = sin xdx,du =2xdx,v=−cos x;

Zx2sin xdx=−x

2cos x+2Zxcos xdx; for Zxcos xdx use u=x,dv = cos xdx to get

Zxcos xdx=xsin x+ cos x+C1so Zx2sin xdx=−x

2cos x+2xsin x+ 2 cos x+C

9. u=lnx,dv =√xdx,du =1

xdx,v=2

3x3/2;

Z√xln xdx=2

3/2ln x−2

3Zx1/2dx =2

3x3/2ln x−4

9x3/2+C

10. u=lnx,dv =xdx,du =1

xdx,v=1

2x2;Zxln xdx=1

2ln x−1

2Zxdx=1

2ln x−1

4x2+C

11. u= (ln x)2,dv =dx,du =2

ln x

xdx,v=x;Z(ln x)2dx =x(ln x)2−2Zln xdx.

Use u=lnx,dv =dx to get Zln xdx=xln x−Zdx =xln x−x+C1so

Z(ln x)2dx =x(ln x)2−2xln x+2x+C

12. u=lnx,dv =1

√xdx,du =1

xdx,v=2

√

x;Zln x

√xdx =2

√

xln x−2Z1

√xdx =2

√

xln x−4√x+C

13. u= ln(2x+ 3), dv =dx,du =2

2x+3dx,v=x;Zln(2x+3)dx =xln(2x+3)−Z2x

2x+3dx

but Z2x

2x+3dx =Z1−3

2x+3dx =x−3

2ln(2x+3)+C

1so

Zln(2x+3)dx =xln(2x+3)−x+3

2ln(2x+3)+C

14. u= ln(x2+ 4), dv =dx,du =2x

x2+4dx,v=x;Zln(x2+4)dx =xln(x2+4)−2Zx

2+4dx

but Zx2

x2+4dx =Z1−4

x2+4dx =x−2 tan−1x

2+C1so

Zln(x2+4)dx =xln(x2+4)−2x+ 4 tan−1x

2+C

15. u= sin−1x,dv =dx,du =1/

√

1−x

dx,v=x;

Zsin−1xdx=xsin−1x−Zx/p1−x2dx =xsin−1x+p1−x2+C

16. u= cos−1(2x), dv =dx,du =−2

√1−4x2dx,v=x;

Zcos−1(2x)dx =xcos−1(2x)+Z2x

√

1−4x

2dx =xcos−1(2x)−1

2p1−4x2+C

Exercise Set 9.2 296

17. u= tan−1(2x), dv =dx,du =2

1+4x

2dx,v=x;

Ztan−1(2x)dx =xtan−1(2x)−Z2x

1+4x

2dx =xtan−1(2x)−1

4ln(1 + 4x2)+C

18. u= tan−1x,dv =xdx,du =1

1+x

2dx,v=1

2x2;Zxtan−1xdx=1

2tan−1x−1

2Zx2

1+x

2dx

but Zx2

1+x

2dx =Z1−1

1+x

2dx =x−tan−1x+C1so

Zxtan−1xdx=1

2tan−1x−1

2x+1

2tan−1x+C

19. u=ex,dv = sin xdx,du =exdx,v=−cos x;Zexsin xdx=−e

xcos x+Zexcos xdx.

For Zexcos xdx use u=ex,dv = cos xdx to get Zexcos x=exsin x−Zexsin xdx so

Zexsin xdx=−e

xcos x+exsin x−Zexsin xdx,

2Ze

xsin xdx=e

x(sin x−cos x)+C

1,Ze

xsin xdx=1

x(sin x−cos x)+C

20. u=e2x,dv = cos 3xdx,du =2e

dx,v=1

3sin 3x;

Ze2xcos 3xdx=1

2xsin 3x−2

3Ze2xsin 3xdx. Use u=e2x,dv = sin 3xdx to get

Ze2xsin 3xdx=−1

2xcos 3x+2

3Ze2xcos 3xdx so

Ze2xcos 3xdx=1

2xsin 3x+2

9e2xcos 3x−4

9Ze2xcos 3xdx,

9Ze2xcos 3xdx=1

2x(3 sin 3x+2 cos 3x)+C1,Ze2xcos 3xdx=1

13e2x(3 sin 3x+2 cos 3x)+C

21. u=eax,dv = sin bx dx,du =aeaxdx,v=−1

bcos bx (b6= 0);

Zeax sin bx dx =−1

beax cos bx +a

bZeax cos bx dx. Use u=eax,dv = cos bx dx to get

Zeax cos bx dx =1

beax sin bx −a

bZeax sin bx dx so

Zeax sin bx dx =−1

beax cos bx +a

b2eax sin bx −a2

b2Zeax sin bx dx,

Zeax sin bx dx =eax

a2+b2(asin bx −bcos bx)+C

22. From Exercise 21 with a=−3,b=5,x =θ, answer = e−3θ

√34 (−3 sin 5θ−5 cos 5θ)+C

297 Chapter 9

23. u= sin(ln x), dv =dx,du =cos(ln x)

xdx,v=x;

Zsin(ln x)dx =xsin(ln x)−Zcos(ln x)dx. Use u= cos(ln x), dv =dx to get

Zcos(ln x)dx =xcos(ln x)+Zsin(ln x)dx so

Zsin(ln x)dx =xsin(ln x)−xcos(ln x)−Zsin(ln x)dx,

Zsin(ln x)dx =(x/2)[sin(ln x)−cos(ln x)]+C

24. u= cos(ln x), dv =dx,du =−1

xsin(ln x)dx,v=x;

Zcos(ln x)dx =xcos(ln x)+Zsin(ln x)dx. Use u= sin(ln x), dv =dx to get

Zsin(ln x)dx =xsin(ln x)−Zcos(ln x)dx so

Zcos(ln x)dx =xcos(ln x)+xsin(ln x)−Zcos(ln x)dx,

Zcos(ln x)dx =1

2x[cos(ln x) + sin(ln x)]+C

25. u=x,dv = sec2xdx,du =dx,v= tan x;

Zxsec2xdx=xtan x−Ztan xdx=xtan x−Zsin x

cos xdx =xtan x+ln|cos x|+C

26. u=x,dv = tan2xdx= (sec2x−1)dx,du =dx,v= tan x−x;

Zxtan2xdx=xtan x−x2−Z(tan x−x)dx

=xtan x−x2+ln|cos x|+1

2x2+C=xtan x−1

2x2+ln|cos x|+C

27. u=x2,dv =xex2dx,du =2xdx,v=1

2dx =1

2x2ex2−Zxex2dx =1

2x2ex2−1

2ex2+C

28. u=xex,dv =1

(x+1)

2dx,du =(x+1)e

xdx,v=−1

x+1;

Zxex

(x+1)

2dx =−xex

x+1+Ze

dx =−xex

x+1+e

x+C=e

x+1+C

Exercise Set 9.2 298

29. u=x,dv =e−5xdx,du =dx,v=−1

5e−5x;

0xe−5xdx =−1

5xe−5x1

5Z1

0e−5xdx

=−1

5e−5−1

25e−5x1

=−1

5e−5−1

25(e−5−1)=(1−6e

−5

)/25

30. u=x,dv =e2xdx,du =dx,v=1

2e2x;

0xe2xdx =1

2xe2x2

0−1

2Z2

0e2xdx =e4−1

4e2x2

=e4−1

4(e4−1)=(3e

4+1)/4

31. u=lnx,dv =x2dx,du =1

xdx,v=1

3x3;

1x2ln xdx=1

3ln xe

1−1

3Ze

1x2dx =1

3e3−1

9x3e

3e3−1

9(e3−1)=(2e

3+1)/9

32. u=lnx,dv =1

x2dx,du =1

xdx,v=−1

√e

ln x

x2dx =−1

xln xe

√e

+Ze

√e

x2dx

=−1

e+1

√eln √e−1

xe

√e

=−1

e+1

2√e−1

e+1

√e=3√e−4

33. u= ln(x+ 3), dv =dx,du =1

x+3dx,v=x;

−2ln(x+3)dx =xln(x+3)

2

−2−Z2

−2

x+3dx = 2 ln 5 + 2 ln 1 −Z2

−21−3

x+3dx

= 2 ln 5 −[x−3 ln(x+ 3)]2

−2= 2 ln 5 −(2 −3ln5)+(−2−3ln1)=5ln5−4

34. u= sin−1x,dv =dx,du =1

√1−x2dx,v=x;

Z1/2

0sin−1xdx=xsin−1x1/2

0−Z1/2

√1−x2dx =1

2sin−11

2+p1−x21/2

2π

6+r3

4−1= π

12 +√3

2−1

35. u= sec−1√θ,dv =dθ,du =1

2θ√θ−1dθ,v=θ;

2sec−1√θdθ =θsec−1√θ4

2−1

2Z4

√θ−1dθ = 4 sec−12−2 sec−1√2−√θ−14

=4π

3−2π

4−√

3+1= 5π

6−√

3+1

299 Chapter 9

36. u= sec−1x,dv =xdx,du =1

x√x2−1dx,v=1

2x2;

1xsec−1xdx=1

2sec−1x2

1−1

2Z2

√x2−1dx

2[(4)(π/3) −(1)(0)] −1

2px2−12

=2π/3−√3/2

37. u=x,dv = sin 4xdx,du =dx,v=−1

4cos 4x;

Zπ/2

0xsin 4xdx=−1

4xcos 4xπ/2

4Zπ/2

0cos 4xdx=−π/8+ 1

16 sin 4xπ/2

=−π/8

38. Zπ

0(x+xcos x)dx =1

2x2π

+Zπ

0xcos xdx=π

2+Zπ

0xcos xdx;

u=x,dv = cos xdx,du =dx,v= sin x

Zπ

0xcos xdx=xsin xπ

0−Zπ

0sin xdx= cos xπ

=−2soZπ

0(x+xcos x)dx =π2/2−2

39. u= tan−1√x,dv =√xdx,du =1

2√x(1+x)dx,v=2

3x3/2;

√xtan−1√xdx =2

3x3/2tan−1√x3

1−1

3Z3

1+xdx

3x3/2tan−1√x3

1−1

3Z3

11−1

1+xdx

=2

3x3/2tan−1√x−1

3x+1

3ln |1+x|

3

=(2

√

3π−π/2−2+ln2)/3

40. u= ln(x2+ 1), dv =dx,du =2x

x2+1dx,v=x;

0ln(x2+1)dx =xln(x2+1)

2

0−Z2

2+1dx = 2 ln 5 −2Z2

01−1

x2+1dx

= 2 ln 5 −2(x−tan−1x)i2

0= 2 ln 5 −4+2tan

−12

41. t=√x, t2=x, dx =2tdt

(a) Ze√xdx =2Ztetdt;u=t, dv =etdt, du =dt, v =et,

Ze√xdx =2tet−2Zetdt =2(t−1)et+C=2(

√

x−1)e√x+C

(b) Zcos √xdx=2Ztcos tdt;u=t, dv = cos tdt, du =dt, v = sin t,

Zcos √xdx=2tsin t−2Zsin tdt =2tsin t+ 2 cos t+C=2

√

xsin √x+ 2 cos √x+C

Exercise Set 9.2 300

43. (a) A=Ze

1ln xdx=(xln x−x)e

(b) V=πZe

1(ln x)2dx =πh(x(ln x)2−2xln x+2x)

1=π(e−2)

44. A=Zπ/2

0(x−xsin x)dx =1

2x2π/2

0−Zπ/2

0xsin xdx=π

8−(−xcos x+ sin x)π/2

=π2/8−1

45. V=2πZπ

0xsin xdx=2π(−xcos x+ sin x)π

=2π

46. V=2πZπ/2

0xcos xdx=2π(cos x+xsin x)π/2

=π(π−2)

47. distance = Z5

0t2e−tdt;u=t2,dv =e

−tdt, du =2tdt, v =−e−t,

distance = −t2e−ti5

0+2Z5

0te−tdt;u=2t, dv =e−tdt, du =2dt, v =−e−t,

distance = −25e−5−2te−ti5

0+2Z5

−t

dt =−25e−5−10e−5−2e−ti5

=−25e−5−10e−5−2e−5+2=−37e−5+2

48. u=2t, dv = sin(kωt)dt, du =2dt, v =−1

kω cos(kωt); the integrand is an even function of tso

Zπ/ω

−π/ω

tsin(kωt)dt =2Zπ/ω

0tsin(kωt)dt =−2

kω tcos(kωt)π/ω

+2Zπ/ω

kω cos(kωt)dt

=2π(−1)k+1

kω2+2

k2ω2sin(kωt)π/ω

=2π(−1)k+1

kω2

49. (a) Zsin3xdx =−1

3sin2xcos x+2

3Zsin xdx=−1

3sin2xcos x−2

3cos x+C

(b) Zsin4xdx=−1

4sin3xcos x+3

4Zsin2xdx,Zsin2xdx=−1

2sin xcos x+1

2x+C1so

Zπ/4

0sin4xdx=−1

4sin3xcos x−3

8sin xcos x+3

8xπ/4

=−1

4(1/√2)3(1/√2) −3

8(1/√2)(1/√2)+3π/32=3π/32 −1/4

50. (a) Zcos5xdx=1

5cos4xsin x+4

5Zcos3xdx=1

5cos4xsin x+4

51

3cos2xsin x+2

3sin x+C

5cos4xsin x+4

15 cos2xsin x+8

15 sin x+C

301 Chapter 9

(b) Zcos6xdx=1

6cos5xsin x+5

6Zcos4xdx

6cos5xsin x+5

61

4cos3xsin x+3

4Zcos2xdx



6cos5xsin x+5

24 cos3xsin x+5

81

2cos xsin x+1

2x+C,

1

6cos5xsin x+5

24 cos3xsin x+5

16 cos xsin x+5

16xπ/2

=5π/32

51. u= sinn−1x,dv = sin xdx,du =(n−1) sinn−2xcos xdx,v=−cos x;

Zsinnxdx=−sinn−1xcos x+(n−1) Zsinn−2xcos2xdx

=−sinn−1xcos x+(n−1) Zsinn−2x(1 −sin2x)dx

=−sinn−1xcos x+(n−1) Zsinn−2xdx−(n−1) Zsinnx dx,

nZsinnxdx=−sinn−1xcos x+(n−1) Zsinn−2x dx,

Zsinnxdx=−1

nsinn−1xcos x+n−1

nZsinn−2xdx

52. (a) u= secn−2x,dv = sec2xdx,du =(n−2) secn−2xtan xdx,v= tan x;

Zsecnxdx= secn−2xtan x−(n−2) Zsecn−2xtan2xdx

= secn−2xtan x−(n−2) Zsecn−2x(sec2x−1)dx

= secn−2xtan x−(n−2) Zsecnxdx+(n−2) Zsecn−2x dx,

(n−1) Zsecnxdx= secn−2xtan x+(n−2) Zsecn−2xdx,

Zsecnxdx=1

n−1secn−2xtan x+n−2

n−1Zsecn−2xdx

(b) Ztannxdx=Ztann−2x(sec2x−1) dx =Ztann−1xsec2xdx−Ztann−2xdx

n−1tann−1x−Ztann−2xdx

53. (a) Ztan4xdx =1

3tan3x−Ztan2xdx=1

3tan3x−tan x+Zdx =1

3tan3x−tan x+x+C

(b) Zsec4xdx=1

3sec2xtan x+2

3Zsec2xdx=1

3sec2xtan x+2

3tan x+C

Exercise Set 9.2 302

=x3ex−3x2ex+6xex−Zexdx=x3ex−3x2ex+6xex−6ex+C

54. (a) u=3x,

dx =1

27 Zu2eudu =1

27 u2eu−2Zueudu=1

27u2eu−2

27 ueu−Zeudu

27u2eu−2

27ueu+2

27eu+C=1

3x2e3x−2

9xe3x+2

27e3x+C

(b) u=−√x,

0xe−√xdx =2Z−1

du,

Zu3eudu =u3eu−3Zu2eudu =u3eu−3u2eu−2Zueudu

=u3eu−3u2eu+6ueu−Zeudu=u3eu−3u2eu+6ueu−6eu+C,

2Z−1

0u3eudu =2(u

3−3u

2+6u−6)eu−1

=12−32e−1

55. u=x,dv =f00(x)dx,du =dx,v=f0(x);

−1xf00(x)dx =xf 0(x)1

−1−Z1

−1f0(x)dx

=f0(1) + f0(−1) −f(x)1

−1

=f0(1) + f0(−1) −f(1) + f(−1)

56. (a) u=f(x),dv =dx, du =f0(x),v =x;

f(x)dx =xf(x)b

a−Zb

xf0(x)dx =bf(b)−af(a)−Zb

xf0(x)dx

(b) Substitute y=f(x),dy =f0(x)dx, x =awhen y=f(a), x=bwhen y=f(b),

xf0(x)dx =Zf(b)

f(a)xdy =Zf(b)

f(a)f−1(y)dy

bf(b)−af(a)=βf−1(β)−αf−1(α); then

Zβ

f−1(x)dx =Zβ

f−1(y)dy =Zf(b)

f(a)f−1(y)dy,

which, by part (b), yields

a = f–1(a) b = f–1(b)

Zβ

f−1(x)dx =bf(b)−af(a)−Zb

f(x)dx

=βf−1(β)−αf−1(α)−Zf−1(β)

f−1(α)f(x)dx

303 Chapter 9

Note from the ﬁgure that A1=Zβ

f−1(x)dx, A2=Zf−1(β)

f−1(α)f(x)dx, and

A1+A2=βf−1(β)−αf−1(α), a ”picture proof”.

57. (a) Use Exercise 56(c);

Z1/2

0sin−1xdx=1

2sin−11

2−0·sin−10−Zsin−1(1/2)

sin−1(0) sin xdx=1

2sin−11

2−Zπ/6

0sin xdx

(b) Use Exercise 56(b);

Ze2

ln xdx=e

2ln e2−eln e−Zln e2

ln e

f−1(y)dy =2e

2−e−Z2

ydy =2e

2−e−Z2

xdx

58. (a) Zxexdx =x(ex+C1)−Z(ex+C1)dx =xex+C1x−ex−C1x+C=xex−ex+C

(b) u(v+C1)−Z(v+C1)du =uv +C1u−Zvdu−C

1u=uv −Zvdu

EXERCISE SET 9.3

1. u= cos x,−Zu5du =−1

6cos6x+C2. u= sin 3x,1

3Zu4du =1

15 sin53x+C

3. u= sin ax,1

aZudu=1

2asin2ax +C,a6=0

4. Zcos23xdx=1

2Z(1 + cos 6x)dx =1

2x+1

12 sin 6x+C

5. Zsin25θdθ=1

2Z(1 −cos 10θ)dθ =1

2θ−1

20 sin 10θ+C

6. Zcos3at dt =Z(1 −sin2at) cos at dt

=Zcos at dt −Zsin2at cos at dt =1

asin at −1

3asin3at +C(a6=0)

7. Zcos5θdθ =Z(1 −sin2θ)2cos θdθ =Z(1 −2 sin2θ+ sin4θ) cos θdθ

= sin θ−2

3sin3θ+1

5sin5θ+C

8. Zsin3xcos3xdx=Zsin3x(1 −sin2x) cos xdx

=Z(sin3x−sin5x) cos xdx=1

4sin4x−1

6sin6x+C

9. Zsin22tcos32tdt=Zsin22t(1 −sin22t) cos 2tdt=Z(sin22t−sin42t) cos 2tdt

6sin32t−1

10 sin52t+C

Exercise Set 9.3 304

10. Zsin32xcos22xdx=Z(1 −cos22x) cos22xsin 2xdx

=Z(cos22x−cos42x) sin 2xdx=−1

6cos32x+1

10 cos52x+C

11. Zsin2xcos2xdx=1

4Zsin22xdx=1

8Z(1 −cos 4x)dx =1

8x−1

32 sin 4x+C

12. Zsin2xcos4xdx=1

8Z(1 −cos 2x)(1 + cos 2x)2dx =1

8Z(1 −cos22x)(1 + cos 2x)dx

8Zsin22xdx+1

8Zsin22xcos 2xdx=1

16 Z(1 −cos 4x)dx +1

48 sin32x

16x−1

64 sin 4x+1

48 sin32x+C

13. Zsin xcos 2xdx=1

2Z(sin 3x−sin x)dx =−1

6cos 3x+1

2cos x+C

14. Zsin 3θcos 2θdθ =1

2Z(sin 5θ+ sin θ)dθ =−1

10 cos 5θ−1

2cos θ+C

15. Zsin xcos(x/2)dx =1

2Z[sin(3x/2) + sin(x/2)]dx =−1

3cos(3x/2) −cos(x/2)+C

16. u= cos x,−Zu1/5du =−5

6cos6/5x+C

17. Zπ/4

0cos3xdx=Zπ/4

0(1 −sin2x) cos xdx

=sin x−1

3sin3xπ/4

√

2/2) −1

3(√2/2)3=5

√

2/12

18. Zπ/2

0sin2(x/2) cos2(x/2)dx =1

4Zπ/2

0sin2xdx=1

8Zπ/2

0(1 −cos 2x)dx

8x−1

2sin 2xπ/2

=π/16

19. Zπ/3

0sin43xcos33xdx=Zπ/3

0sin43x(1 −sin23x) cos 3xdx=1

15 sin53x−1

21 sin73xπ/3

20. Zπ

−π

cos25θdθ=1

2Zπ

−π

(1 + cos 10θ)dθ =1

2θ+1

10 sin 10θπ

−π

=π

21. Zπ/6

0sin 2xcos 4xdx=1

2Zπ/6

0(sin 6x−sin 2x)dx =−1

12 cos 6x+1

4cos 2xπ/6

=[(−1/12)(−1)+(1/4)(1/2)] −[−1/12+1/4]=1/24

22. Z2π

0sin2kx dx =1

2Z2π

0(1 −cos 2kx)dx =1

2x−1

2ksin 2kx2π

=π−1

4ksin 4πk (k6=0)

305 Chapter 9

23. 1

3tan(3x+1)+C24. −1

5ln |cos 5x|+C

25. u=e−2x,du=−2e

−2xdx;−1

2Ztan udu=1

2ln |cos u|+C=1

2ln |cos(e−2x)|+C

26. 1

3ln |sin 3x|+C27. 1

2ln |sec 2x+ tan 2x|+C

28. u=√x, du =1

2√xdx;Z2 sec udu=2ln|sec u+ tan u|+C=2ln|sec √x+ tan √x|+C

29. u= tan x,Zu2du =1

3tan3x+C

30. Ztan5x(1 + tan2x) sec2xdx=Z(tan5x+ tan7x) sec2xdx=1

6tan6x+1

8tan8x+C

31. Ztan34x(1 + tan24x) sec24xdx=Z(tan34x+ tan54x) sec24xdx=1

16 tan44x+1

24 tan64x+C

32. Ztan4θ(1 + tan2θ) sec2θdθ=1

5tan5θ+1

7tan7θ+C

33. Zsec4x(sec2x−1) sec xtan xdx=Z(sec6x−sec4x) sec xtan xdx=1

7sec7x−1

5sec5x+C

34. Z(sec2θ−1)2sec θtan θdθ =Z(sec4θ−2 sec2θ+ 1) sec θtan θdθ =1

5sec5θ−2

3sec3θ+ sec θ+C

35. Z(sec2x−1)2sec xdx=Z(sec5x−2 sec3x+ sec x)dx =Zsec5xdx−2Zsec3xdx+Zsec xdx

4sec3xtan x+3

4Zsec3xdx−2Zsec3xdx+ln|sec x+ tan x|

4sec3xtan x−5

41

2sec xtan x+1

2ln |sec x+ tan x|+ln|sec x+ tan x|+C

4sec3xtan x−5

8sec xtan x+3

8ln |sec x+ tan x|+C

36. Z[sec2(x/2) −1] sec3(x/2)dx =Z[sec5(x/2) −sec3(x/2)]dx

=2Zsec5udu−Zsec3udu

(u=x/2)

=21

4sec3utan u+3

4Zsec3udu

−Zsec3udu

(equation (20))

2sec3utan u−1

2Zsec3udu

2sec3utan u−1

4sec utan u−1

4ln |sec u+ tan u|+C(equation (20), (22))

2sec3x

2tan x

2−1

4sec x

2tan x

2−1

4ln sec x

2+ tan x

2+C

Exercise Set 9.3 306

37. Zsec22t(sec 2ttan 2t)dt =1

6sec32t+C38. Zsec4x(sec xtan x)dx =1

5sec5x+C

39. Zsec4xdx=Z(1 + tan2x) sec2xdx =Z(sec2x+ tan2xsec2x)dx = tan x+1

3tan3x+C

40. Using equation (20),

Zsec5xdx=1

4sec3xtan x+3

4Zsec3xdx

4sec3xtan x+3

8sec xtan x+3

8ln |sec x+ tan x|+C

41. Use equation (19) to get Ztan4xdx=1

3tan3x−tan x+x+C

42. u=4x, use equation (19) to get

4Ztan3udu=1

41

2tan2u+ln|cos u|+C=1

8tan24x+1

4ln |cos 4x|+C

43. Z√tan x(1 + tan2x) sec2xdx=2

3tan3/2x+2

7tan7/2x+C

44. Zsec1/2x(sec xtan x)dx =2

3sec3/2x+C

45. Zπ/6

0(sec22x−1)dx =1

2tan 2x−xπ/6

=√3/2−π/6

46. Zπ/6

0sec2θ(sec θtan θ)dθ =1

3sec3θπ/6

=(1/3)(2/√3)3−1/3=8

√

3/27 −1/3

47. u=x/2,

2Zπ/4

0tan5udu=1

2tan4u−tan2u−2ln|cos u|π/4

=1/2−1−2 ln(1/√2) = −1/2+ln2

48. u=πx, 1

πZπ/4

0sec utan udu=1

πsec uπ/4

√

2−1)/π

49. Z(csc2x−1) csc2x(csc xcot x)dx =Z(csc4x−csc2x)(csc xcot x)dx =−1

5csc5x+1

3csc3x+C

50. Zcos23t

sin23t·1

cos 3tdt =Zcsc 3tcot 3tdt=−1

3csc 3t+C

51. Z(csc2x−1) cot xdx=Zcsc x(csc xcot x)dx −Zcos x

sin xdx =−1

2csc2x−ln |sin x|+C

52. Z(cot2x+ 1) csc2xdx=−1

3cot3x−cot x+C

307 Chapter 9

53. (a) Z2π

0sin mx cos nx dx =1

2Z2π

0[sin(m+n)x+sin(m−n)x]dx =−cos(m+n)x

2(m+n)−cos(m−n)x

2(m−n)2π

but cos(m+n)x2π

=0,cos(m−n)x2π

=0.

(b) Z2π

0cos mx cos nx dx =1

2Z2π

0[cos(m+n)x+ cos(m−n)x]dx;

since m6=n, evaluate sin at integer multiples of 2πto get 0.

0sin mx sin nx dx =1

2Z2π

0[cos(m−n)x−cos(m+n)x]dx;

since m6=n, evaluate sin at integer multiples of 2πto get 0.

55. y0= tan x,1+(y

2= 1 + tan2x= sec2x,

L=Zπ/4

√sec2xdx=Zπ/4

0sec xdx=ln|sec x+ tan x|π/4

= ln(√2+1)

56. V=πZπ/4

0(1 −tan2x)dx =πZπ/4

0(2 −sec2x)dx =π(2x−tan x)π/4

2π(π−2)

57. V=πZπ/4

0(cos2x−sin2x)dx =πZπ/4

0cos 2xdx=1

2πsin 2x#π/4

=π/2

58. V=πZπ

0sin2xdx=π

2Zπ

0(1 −cos 2x)dx =π

2x−1

2sin 2xπ

=π2/2

59. With 0 <α<β,D=D

−D

α=L

2πZβ

sec xdx=L

2πln |sec x+ tan x|#β

2πln 

sec β+ tan β

sec α+ tan α

60. (a) D=100

2πln(sec 25◦+ tan 25◦)=7.18 cm (b) D=100

2πln 

sec 50◦+ tan 50◦

sec 30◦+ tan 30◦=7.34 cm

61. (a) Zcsc xdx =Zsec(π/2−x)dx =−ln |sec(π/2−x) + tan(π/2−x)|+C

=−ln |csc x+ cot x|+C

(b) −ln |csc x+ cot x|=ln 1

|csc x+ cot x|=ln |csc x−cot x|

|csc2x−cot2x|=ln|csc x−cot x|,

−ln |csc x+ cot x|=−ln 

sin x+cos x

sin x=ln



sin x

1 + cos x

=ln



2 sin(x/2) cos(x/2)

2 cos2(x/2) =ln|tan(x/2)|

62. sin x+ cos x=√2[(1/√2) sin x+(1/

√

2) cos x]

=√2[sin xcos(π/4) + cos xsin(π/4)] = √2 sin(x+π/4),

Zdx

sin x+ cos x=1

√2Zcsc(x+π/4)dx =−1

√2ln |csc(x+π/4) + cot(x+π/4)|+C

=−1

√2ln 

√2 + cos x−sin x

sin x+ cos x+C

Exercise Set 9.4 308

63. asin x+bcos x=pa2+b2a

√a2+b2sin x+b

√a2+b2cos x=pa2+b2(sin xcos θ+ cos xsin θ)

where cos θ=a/√a2+b2and sin θ=b/√a2+b2so asin x+bcos x=√a2+b2sin(x+θ)

and Zdx

asin x+bcos x=1

√a2+b2Zcsc(x+θ)dx =−1

√a2+b2ln |csc(x+θ) + cot(x+θ)|+C

=−1

√a2+b2ln 

√a2+b2+acos x−bsin x

asin x+bcos x+C

64. (a) Zπ/2

0sinnxdx=−1

nsinn−1xcos xπ/2

+n−1

nZπ/2

0sinn−2xdx=n−1

nZπ/2

0sinn−2xdx

(b) By repeated application of the formula in part (a)

Zπ/2

0sinnxdx=n−1

nn−3

n−2Zπ/2

0sinn−4xdx

=









n−1

nn−3

n−2n−5

n−4···1

2Zπ/2

0dx, n even

n−1

nn−3

n−2n−5

n−4···2

3Zπ/2

0sin x dx, n odd

=









1·3·5···(n−1)

2·4·6···n·π

2,n even

2·4·6···(n−1)

3·5·7···n,n odd

65. (a) Zπ/2

0sin3xdx=2

3(b) Zπ/2

0sin4xdx=1·3

2·4·π

2=3π/16

0sin5xdx=2·4

3·5=8/15 (d) Zπ/2

0sin6xdx=1·3·5

2·4·6·π

2=5π/32

66. Similar to proof in Exercise 64.

EXERCISE SET 9.4

1. x= 2 sin θ,dx = 2 cos θdθ,

4Zcos2θdθ=2Z(1 + cos 2θ)dθ =2θ+ sin 2θ+C

=2θ+ 2 sin θcos θ+C= 2 sin−1(x/2) + 1

2xp4−x2+C

2. x=1

2sin θ,dx =1

2cos θdθ,

2Zcos2θdθ=1

4Z(1 + cos 2θ)dθ =1

4θ+1

8sin 2θ+C

4θ+1

4sin θcos θ+C=1

4sin−12x+1

2xp1−4x2+C

309 Chapter 9

3. x= 3 sin θ,dx = 3 cos θdθ,

9Zsin2θdθ=9

2Z(1 −cos 2θ)dθ =9

2θ−9

4sin 2θ+C=9

2θ−9

2sin θcos θ+C

2sin−1(x/3) −1

2xp9−x2+C

4. x= 4 sin θ,dx = 4 cos θdθ,

16 Z1

sin2θdθ =1

16 Zcsc2θdθ=−1

16 cot θ+C=−√16 −x2

16x+C

5. x= 2 tan θ,dx = 2 sec2θdθ,

8Z1

sec2θdθ =1

8Zcos2θdθ=1

16 Z(1 + cos 2θ)dθ =1

16θ+1

32 sin 2θ+C

16θ+1

16 sin θcos θ+C=1

16 tan−1x

2+x

8(4 + x2)+C

6. x=√5 tan θ,dx =√5 sec2θdθ,

5Ztan2θsec θdθ=5Z(sec3θ−sec θ)dθ =51

2sec θtan θ−1

2ln |sec θ+ tan θ|+C1

2xp5+x

2−5

2ln √5+x

2+x

√

5+C

1=1

p5+x

2−5

2ln(p5+x

2+x)+C

7. x= 3 sec θ,dx = 3 sec θtan θdθ,

3Ztan2θdθ=3Z(sec2θ−1)dθ = 3 tan θ−3θ+C=px2−9−3 sec−1x

3+C

8. x= 4 sec θ,dx = 4 sec θtan θdθ,

16 Z1

sec θdθ =1

16 Zcos θdθ=1

16 sin θ+C=√x2−16

16x+C

9. x=√2 sin θ,dx =√2 cos θdθ,

√2Zsin3θdθ=2

√

2Z1−cos2θsin θdθ

√

2−cos θ+1

3cos3θ+C=−2p2−x2+1

3(2 −x2)3/2+C

10. x=√5 sin θ,dx =√5 cos θdθ,

25√5Zsin3θcos2θdθ=25

√

5−1

3cos3θ+1

5cos5θ+C=−5

3(5 −x2)3/2+1

5(5 −x2)5/2+C

11. x=3

2sec θ,dx =3

2sec θtan θdθ,2

9Z1

sec θdθ =2

9Zcos θdθ=2

9sin θ+C=√4x2−9

9x+C

12. t= tan θ,dt = sec2θdθ,

Zsec3θ

tan θdθ =Ztan2θ+1

tan θsec θdθ=Z(sec θtan θ+ csc θ)dθ

= sec θ+ln|csc θ−cot θ|+C=p1+t

2+ln√

1+t

2−1

|t|+C

Exercise Set 9.4 310

13. x= sin θ,dx = cos θdθ,Z1

cos2θdθ =Zsec2θdθ= tan θ+C=x/p1−x2+C

14. x= 5 tan θ,dx = 5 sec2θdθ,1

25Zsec θ

tan2θdθ =1

25Zcsc θcot θdθ =−1

25 csc θ+C=−√x2+25

25x+C

15. x= sec θ,dx = sec θtan θdθ,Zsec θdθ=ln|sec θ+ tan θ|+C=ln



x+px

2−1



+C

16. 1+2x

2+x

4=(1+x

)

,x= tan θ,dx = sec2θdθ,

sec2θdθ =Zcos2θdθ=1

2Z(1 + cos 2θ)dθ =1

2θ+1

4sin 2θ+C

2θ+1

2sin θcos θ+C=1

2tan−1x+x

2(1 + x2)+C

17. x=1

3sec θ,dx =1

3sec θtan θdθ,

3Zsec θ

tan2θdθ =1

3Zcsc θcot θdθ=−1

3csc θ+C=−x/p9x2−1+C

18. x= 5 sec θ,dx = 5 sec θtan θdθ,

25 Zsec3θdθ=25

2sec θtan θ+25

2ln |sec θ+ tan θ|+C1

2xpx2−25 + 25

2ln |x+px2−25|+C

19. ex= sin θ,exdx = cos θdθ,

Zcos2θdθ=1

2Z(1 + cos 2θ)dθ =1

2θ+1

4sin 2θ+C=1

2sin−1(ex)+1

p1−e

2x+C

20. u= sin θ,Z1

√2−u2du = sin−1sin θ

√2+C

21. x= 4 sin θ,dx = 4 cos θdθ,

1024 Zπ/2

0sin3θcos2θdθ= 1024 −1

3cos3θ+1

5cos5θπ/2

= 1024(1/3−1/5) = 2048/15

22. x=2

3sin θ,dx =2

3cos θdθ,

24 Zπ/6

cos3θdθ =1

24 Zπ/6

0sec3θdθ=1

48 sec θtan θ+1

48 ln |sec θ+ tan θ|π/6

48[(2/√3)(1/√3)+ln|2/

√

3+1/

√

3|]= 1

48 2

3+1

2ln 3

23. x= sec θ,dx = sec θtan θdθ,Zπ/3

π/4

sec θdθ =Zπ/3

π/4cos θdθ= sin θπ/3

π/4

√

3−√

2)/2

24. x=√2 sec θ,dx =√2 sec θtan θdθ,2

Zπ/4

0tan2θdθ=2 tan θ−2θπ/4

=2−π/2

311 Chapter 9

25. x=√3 tan θ,dx =√3 sec2θdθ,

9Zπ/3

π/6

sec θ

tan4θdθ =1

9Zπ/3

π/6

cos3θ

sin4θdθ =1

9Zπ/3

π/6

1−sin2θ

sin4θcos θdθ=1

9Z√3/2

1/2

1−u

4du (u= sin θ)

9Z√3/2

1/2(u−4−u−2)du =1

9−1

3u3+1

u√3/2

1/2

=10√3+18

243

26. x=√3 tan θ,dx =√3 sec2θdθ,

√3

3Zπ/3

tan3θ

sec3θdθ =√3

3Zπ/3

0sin3θdθ=√3

3Zπ/3

01−cos2θsin θdθ

=√3

3−cos θ+1

3cos3θπ/3

=√3

3−1

2+1

24−−1+1

3=5

√

3/72

27. u=x2+4,du =2xdx,

2Z1

udu =1

2ln |u|+C=1

2ln(x2+4)+C;orx= 2 tan θ,dx = 2 sec2θdθ,

Ztan θdθ=ln|sec θ|+C1=ln√

2+4

2+C

1= ln(x2+4)

1/2−ln 2 + C1

2ln(x2+4)+Cwith C=C1−ln 2

29. y0=1

x,1+(y

2=1+ 1

2=x

2+1

L=Z2

1rx

2+1

2dx =Z2

√x2+1

xdx;x= tan θ,dx = sec2θdθ,

L=Ztan−1(2)

π/4

sec3θ

tan θdθ =Ztan−1(2)

π/4

tan2θ+1

tan θsec θdθ=Ztan−1(2)

π/4(sec θtan θ+ csc θ)dθ

=sec θ+ln|csc θ−cot θ|tan−1(2)

π/4

=√5+ln √

2−1

2!−h√

2+ln|

√

2−1|

=√

5−√

2+ln2+2

√

1+√

30. y0=2x,1+(y

2=1+4x

L=Z1

0p1+4x

2dx;x=1

2tan θ,dx =1

2sec2θdθ,

L=1

2Ztan−12

0sec3θdθ=1

21

2sec θtan θ+1

2ln |sec θ+ tan θ|tan−12

4(√5)(2) + 1

4ln |√5+2|=1

2√

5+1

4ln(2 + √5)

31. y0=2x,1+(y

2=1+4x

S=2πZ1

p1+4x

2dx;x=1

2tan θ,dx =1

2sec2θdθ,

Exercise Set 9.4 312

S=π

4Ztan−12

0tan2θsec3θdθ=π

4Ztan−12

0(sec2θ−1) sec3θdθ=π

4Ztan−12

0(sec5θ−sec3θ)dθ

=π

41

4sec3θtan θ−1

8sec θtan θ−1

8ln |sec θ+ tan θ|tan−12

=π

32[18√5−ln(2 + √5)]

32. V=πZ1

0y2p1−y2dy;y= sin θ,dy = cos θdθ,

V=πZπ/2

0sin2θcos2θdθ=π

4Zπ/2

0sin22θdθ=π

8Zπ/2

0(1 −cos 4θ)dθ =π

8θ−1

4sin 4θπ/2

=π2

33. (a) x= 3 sinh u,dx = 3 cosh udu,Zdu =u+C= sinh−1(x/3)+C

(b) x= 3 tan θ,dx = 3 sec2θdθ,

Zsec θdθ=ln|sec θ+ tan θ|+C= ln(px2+9/3+x/3)+C

but sinh−1(x/3) = ln(x/3+px

2/9+1)=ln(x/3+√

2+9/3) so the results agree.

Zsinh2udu=1

2Z(cosh 2u−1)du =1

4sinh 2u−1

2u+C

2sinh ucosh u−1

2u+C=1

2xpx2−1−1

2cosh−1x+C

because cosh u=x, and sinhu=pcosh2u−1=√

2−1

34. A=4b

aZa

0pa2−x2dx;x=acos θ, dx =−asin θdθ,

A=−4b

aZ0

π/2a2sin2θdθ=4ab Zπ/2

0sin2θdθ=2ab Zπ/2

0(1 −cos 2θ)dθ =πab

35. Z1

(x−2)2+9dx =1

3tan−1x−2

3+C36. Z1

p1−(x−1)2dx = sin−1(x−1)+C

37. Z1

p9−(x−1)2dx = sin−1x−1

3+C

38. Z1

16(x+1/2)2+1dx =1

16 Z1

(x+1/2)2+1/16dx =1

4tan−1(4x+2)+C

39. Z1

p(x−3)2+1dx =lnx−3+p(x−3)2+1

+C

40. Zx

(x+3)

2+1dx, let u=x+3,

Zu−3

2+1du =Zu

u2+1−3

2+1du =1

2ln(u2+1)−3 tan−1u+C

2ln(x2+6x+ 10) −3 tan−1(x+3)+C

313 Chapter 9

41. Zp4−(x+1)

2dx, let x+1=2sinθ,

4Zcos2θdθ=2θ+ sin 2θ+C=2θ+ 2 sin θcos θ+C

= 2 sin−1x+1

2+1

2(x+1)

p3−2x−x

2+C

42. Zex

p(ex+1/2)2+3/4dx, let u=ex+1/2,

pu2+3/4du = sinh−1(2u/√3)+C= sinh−12ex+1

√

3+C

Alternate solution: let ex+1/2=√

2tan θ,

Zsec θdθ=ln|sec θ+ tan θ|+C=ln 2

√

2x+e

x+1

√

3+2e

x+1

√

3!+C

= ln(2pe2x+ex+1+2e

x+1)+C

43. Z1

2(x+1)

2+5dx =1

2Z1

(x+1)

2+5/2dx =1

√10 tan−1p2/5(x+1)+C

44. Z2x+3

4(x+1/2)2+4dx, let u=x+1/2,

Z2u+2

2+4du =1

2Zu

u2+1+1

2+1du =1

4ln(u2+1)+1

2tan−1u+C

4ln(x2+x+5/4) + 1

2tan−1(x+1/2)+C

45. Z2

√4x−x2dx =Z2

p4−(x−2)2dx = sin−1x−2

22

=π/6

46. Z1

0p4x−x2dx =Z1

0p4−(x−2)2dx, let x−2=2sinθ,

4Z−π/6

−π/2cos2θdθ =2θ+ sin 2θ−π/6

−π/2

=2π

3−√3

48. u=xsin x, du =(xcos x+ sin x)dx;

Zp1+u

2du =1

2up1+u

2+1

2sinh−1u+C=1

2xsin xp1+x

2sin2x+1

2sinh−1(xsin x)+C

49. u= sin2x, du = 2 sin xcos xdx;

2Zp1−u

2du =1

4hup1−u2+ sin−1ui+C=1

4hsin2xp1−sin4x+ sin−1(sin2x)i+C

Exercise Set 9.5 314

50. u=3

x=e

xln 3,du= (ln 3)3xdx;

ln 3 Z3

1pu2−1du =1

2ln3u

2−1−ln u+pu2−13

=6√2−ln(3 + 2√2)

2ln3

EXERCISE SET 9.5

1. A

(x−2) +B

(x+5) 2. 5

x(x−3)(x+3) =A

x+B

x−3+C

x+3

3. 2x−3

x2(x−1) =A

x+B

x2+C

x−14. A

x+2+B

(x+2)

2+C

(x+2)

5. A

x+B

x2+C

x3+Dx +E

x2+1 6. A

x−1+Bx +C

x2+5

7. Ax +B

x2+5 +Cx +D

(x2+5)

28. A

x−2+Bx +C

x2+1 +Dx +E

(x2+1)

9. 1

(x+ 4)(x−1) =A

x+4+B

x−1;A=−1

5,B=1

5so

−1

5Z1

x+4dx +1

5Z1

x−1dx =−1

5ln |x+4|+1

5ln |x−1|+C=1

5ln 

x−1

x+4



+C

10. 1

(x+ 1)(x+7) =A

x+1+B

x+7;A=1

6,B=−1

6so

6Z1

x+1dx −1

6Z1

x+7dx =1

6ln |x+1|−1

6ln |x+7|+C=1

6ln 

x+1

x+7



+C

11. 11x+17

(2x−1)(x+4) =A

2x−1+B

x+4;A=5,B=3so

5Z1

2x−1

dx +3Z1

x+4dx =5

2ln |2x−1|+3ln|x+4|+C

12. 5x−5

(x−3)(3x+1) =A

x−3+B

3x+1;A=1,B=2so

x−3

dx +2Z1

3x+1dx =ln|x−3|+2

3ln |3x+1|+C

13. 2x2−9x−9

x(x+ 3)(x−3) =A

x+B

x+3+C

x−3;A=1,B=2,C=−1so

dx +2Z1

x+3dx −Z1

x−3dx =ln|x|+2ln|x+3|−ln |x−3|+C=ln



x(x+3)

x−3



+C

Note that the symbol Chas been recycled; to save space this recycling is usually not mentioned.

315 Chapter 9

14. 1

x(x+ 1)(x−1) =A

x+B

x+1+C

x−1;A=−1, B=1

2,C=1

2so

−Z1

xdx +1

2Z1

x+1dx +1

2Z1

x−1dx =−ln |x|+1

2ln |x+1|+1

2ln |x−1|+C

2ln 

(x+ 1)(x−1)

x2+C=1

2ln |x2−1|

x2+C

15. x2+2

x+2 =x−2+ 6

x+2,Zx−2+ 6

x+2dx =1

2x2−2x+6 ln|x+2|+C

16. x2−4

x−1=x+1−3

x−1,Zx+1−3

x−1dx =1

2x2+x−3ln|x−1|+C

17. 3x2−10

x2−4x+4 =3+ 12x−22

x2−4x+4,12x−22

(x−2)2=A

x−2+B

(x−2)2;A= 12, B=2so

Z3dx +12Z1

x−2dx +2Z1

(x−2)2dx =3x+12ln|x−2|−2/(x−2)+C

18. x2

x2−3x+2 =1+ 3x−2

2−3x+2,3x−2

(x−1)(x−2) =A

x−1+B

x−2;A=−1, B=4so

Zdx −Z1

x−1dx +4Z1

x−2dx =x−ln |x−1|+4ln|x−2|+C

19. x5+2x

2+1

3−x=x

2+1+2x

2+x+1

3−x,

2+x+1

x(x+ 1)(x−1) =A

x+B

x+1+C

x−1;A=−1, B=1,C=2so

Z(x

2+1)dx −Z1

xdx +Z1

x+1dx +2Z1

x−1dx

3x3+x−ln |x|+ln|x+1|+2ln|x−1|+C=1

3+x+ln



(x+ 1)(x−1)2

x+C

20. 2x5−x3−1

x3−4x=2x

2+7+28x−1

x3−4x,

28x−1

x(x+ 2)(x−2) =A

x+B

x+2+C

x−2;A=1

4,B=−57

8,C=55

8so

Z(2x2+7)dx +1

4Z1

xdx −57

8Z1

x+2dx +55

8Z1

x−2dx

3x3+7x+1

4ln |x|−57

8ln |x+2|+55

8ln |x−2|+C

21. 2x2+3

x(x−1)2=A

x+B

x−1+C

(x−1)2;A=3,B=−1, C=5so

3Z1

dx −Z1

x−1dx +5Z1

(x−1)2dx =3ln|x|−ln |x−1|−5/(x−1)+C

Exercise Set 9.5 316

22. 3x2−x+1

2(x−1) =A

x+B

x2+C

x−1;A=0,B=−1, C=3so

−Z1

2dx +3Z1

x−1dx =1/x +3ln|x−1|+C

23. x2+x−16

(x+ 1)(x−3)2=A

x+1+B

x−3+C

(x−3)2;A=−1, B=2,C=−1so

−Z1

x+1dx +2Z1

x−3dx −Z1

(x−3)2dx

=−ln |x+1|+2ln|x−3|+1

x−3+C=ln(x−3)2

|x+1|+1

x−3+C

24. 2x2−2x−1

x2(x−1) =A

x+B

x2+C

x−1;A=3,B=1,C=−1so

dx +Z1

x2dx −Z1

x−1dx =3ln|x|−1

x−ln |x−1|+C

25. x2

(x+2)

3=A

x+2+B

(x+2)

2+C

(x+2)

3;A=1,B=−4, C=4so

x+2dx −4Z1

(x+2)

2dx +4Z1

(x+2)

3dx =ln|x+2|+4

x+2−2

(x+2)

2+C

26. 2x2+3x+3

(x+1)

3=A

x+1+B

(x+1)

2+C

(x+1)

3;A=2,B=−1, C=2so

2Z1

x+1dx −Z1

(x+1)

2dx +2Z1

(x+1)

3dx =2ln|x+1|+1

x+1−1

(x+1)

2+C

27. 2x2−1

(4x−1)(x2+1) =A

4x−1+Bx +C

x2+1 ;A=−14/17, B=12/17, C=3/17 so

Z2x2−1

(4x−1)(x2+1)dx =−7

34 ln |4x−1|+6

17 ln(x2+1)+ 3

17 tan−1x+C

28. 1

x(x2+1) =A

x+Bx +C

x2+1 ;A=1,B=−1, C=0so

3+x

dx =ln|x|−1

2ln(x2+1)+C=1

2ln x2

x2+1+C

29. x3+3x

2+x+9

2+ 1)(x2+3) =Ax +B

x2+1 +Cx +D

x2+3 ;A=0,B=3,C=1,D=0so

3+3x

2+x+9

2+ 1)(x2+3)dx = 3 tan−1x+1

2ln(x2+3)+C

30. x3+x2+x+2

2+ 1)(x2+2) =Ax +B

x2+1 +Cx +D

x2+2 ;A=D=0,B=C=1so

3+x

2+x+2

2+ 1)(x2+2)dx = tan−1x+1

2ln(x2+2)+C

317 Chapter 9

31. x3−3x2+2x−3

2+1 =x−3+ x

2+1,

3−3x

2+2x−3

2+1 dx =1

2x2−3x+1

2ln(x2+1)+C

32. x4+6x

3+10x

2+x

2+6x+10 =x

2+x

2+6x+10,

2+6x+10dx =Zx

(x+3)

2+1dx =Zu−3

u2+1du, u =x+3

2ln(u2+1)−3 tan−1u+C1

so Zx4+6x

3+10x

2+x

2+6x+10 dx =1

3x3+1

2ln(x2+6x+ 10) −3 tan−1(x+3)+C

33. Let x= sin θto get Z1

x2+4x−5dx, and 1

(x+ 5)(x−1) =A

x+5+B

x−1;A=−1/6,

B=1/6soweget−1

6Z1

x+5dx +1

6Z1

x−1dx =1

6ln 

x−1

x+5



+C=1

6ln 1−sin θ

5 + sin θ+C.

34. Let x=et; then Zet

e2t−4dt =Z1

x2−4dx,

(x+ 2)(x−2) =A

x+2+B

x−2;A=−1/4, B=1/4so

−1

4Z1

x+2dx +1

4Z1

x−2dx =1

4ln 

x−2

x+2



+C=1

4ln 

et−2

et+2



+C.

35. V=πZ2

(9 −x2)2dx,x4

x4−18x2+81 =1+ 18x2−81

x4−18x2+81,

18x2−81

(9 −x2)2=18x2−81

(x+3)

2(x−3)2=A

x+3+B

(x+3)

2+C

x−3+D

(x−3)2;

A=−9

4,B=9

4,C=9

4,D=9

4so

V=πx−9

4ln |x+3|− 9/4

x+3+9

4ln |x−3|− 9/4

x−32

=π19

5−9

4ln 5

36. Let u=exto get Zln 5

−ln 5

1+e

x=Zln 5

−ln 5

exdx

ex(1+ex)=Z5

1/5

u(1+u),

u(1+u)=A

u+B

1+u;A=1,B=−1; Z5

1/5

u(1+u)= (ln u−ln(1 + u))5

1/5

=ln5

37. x2+1

2+2x+3)

2=Ax +B

x2+2x+3+Cx +D

(x2+2x+3)

2;A=0,B=1,C=D=−2so

2+1

2+2x+3)

2dx =Z1

(x+1)

2+2dx −Z2x+2

2+2x+3)

2dx

√2tan−1x+1

√

2+1/(x

2+2x+3)+C

Exercise Set 9.5 318

38. x5+x4+4x

3+4x

2+4x+4

2+2)

3=Ax +B

x2+2 +Cx +D

(x2+2)

2+Ex +F

(x2+2)

A=B=1,C=D=E=F=0so

Zx+1

2+2dx =1

2ln(x2+2)+ 1

√

2tan−1(x/√2)+C

39. x4−3x3−7x2+27x−18=(x−1)(x−2)(x−3)(x+ 3),

(x−1)(x−2)(x−3)(x+3) =A

x−1+B

x−2+C

x−3+D

x+3;

A=1/8, B=−1/5, C=1/12, D=−1/120 so

Zdx

x4−3x3−7x2+27x−18 =1

8ln |x−1|−1

5ln |x−2|+1

12 ln |x−3|− 1

120 ln |x+3|+C

40. 16x3−4x2+4x−1=(4x−1)(4x2+ 1),

(4x−1)(4x2+1) =A

4x−1+Bx +C

4x2+1;A=4/5, B=−4/5, C=−1/5so

Zdx

16x3−4x2+4x−1=1

5ln |4x−1|− 1

10 ln(4x2+1)−1

10 tan−1(2x)+C

41. (a) x4+1=(x

4+2x

2+1)−2x

2=(x

2+1)

2−2x

=[(x

2+1)+√

2x][(x2+1)−√

2x]

=(x

2+√

2x+ 1)(x2−√2x+ 1); a=√2,b =−√2

(b) x

(x2+√2x+ 1)(x2−√2x+1) =Ax +B

x2+√2x+1+Cx +D

x2−√2x+1;

A=0,B=−√

4,C=0,D=√

4so

x4+1dx =−√2

4Z1

x2+√2x+1dx +√2

4Z1

x2−√2x+1dx

=−√2

4Z1

(x+√2/2)2+1/2dx +√2

4Z1

(x−√2/2)2+1/2dx

=−√2

4Z1+√2/2

√2/2

u2+1/2du +√2

4Z1−√2/2

−√2/2

u2+1/2du

=−1

2tan−1√2u1+√2/2

√2/2

2tan−1√2u1−√2/2

−√2/2

=−1

2tan−1(√2+1)+1

2π

4+1

2tan−1(√2−1) −1

2−π

4

=π

4−1

2[tan−1(√2+1)−tan−1(√2−1)]

=π

4−1

2[tan−1(1+√2) + tan−1(1 −√2)]

=π

4−1

2tan−1"(1+√2)+(1−√

1−(1+√2)(1 −√2)#(Exercise 46, Section 4.5)

=π

4−1

2tan−11=π

4−1

2π

4=π

319 Chapter 9

42. 1

a2−x2=A

a−x+B

a+x;A=1

2a,B =1

2aso

2aZ1

a−x+1

a+xdx =1

2a(−ln |a−x|+ln|a+x|)+C=1

2aln 

a+x

a−x+C

EXERCISE SET 9.6

1. Formula (60): 3

16h4x+ln|−1+4x|i+C2. Formula (62): 1

92

2−3x+ln|2−3x|

+C

3. Formula (65): 1

5ln 

5+2x



+C4. Formula (66): −1

x−5ln



1−5x

x



+C

5. Formula (102): 1

5(x+ 1)(−3+2x)

3/2+C6. Formula (105): 2

3(−x−4)√2−x+C

7. Formula (108): 1

2ln 

√4−3x−2

√4−3x+2



+C8. Formula (108): tan−1√3x−4

2+C

9. Formula (69): 1

2√5ln 

x+√5

x−√5+C10. Formula (70): 1

6ln 

x−3

x+3



+C

11. Formula (73): x

2px2−3−3

2ln x+px2−3+C

12. Formula (93): −√x2+5

x+ ln(x+px2+5)+C

13. Formula (95): x

2px2+4−2 ln(x+px2+4)+C

14. Formula (90): −√x2−2

2x+C15. Formula (74): x

2p9−x2+9

2sin−1x

3+C

16. Formula (80): −√4−x2

x−sin−1x

2+C

17. Formula (79): √3−x2−√3ln



√3+√

9−x

x



+C

18. Formula (117): −√6x−x2

3x+C19. Formula (38): −1

10 sin(5x)+1

2sin x+C

20. Formula (40): −1

14 cos(7x)+1

6cos(3x)+C

21. Formula (50): x4

16 [4 ln x−1]+C22. Formula (50): 4√x1

2ln x−1+C

23. Formula (42): e−2x

13 (−2 sin(3x)−3 cos(3x))+C

Exercise Set 9.6 320

24. Formula (43): ex

5(cos(2x) + 2 sin(2x))+C

25. u=e2x,du=2e

dx, Formula (62): 1

2Zudu

(4 −3u)2=1

18 4

4−3e2x+ln

4−3e



+C

26. u= sin 2x, du = 2 cos 2xdx, Formula (116): Zdu

2u(3 −u)=1

6ln 

sin 2x

3−sin 2x+C

27. u=3

√

x, du =3

2√xdx, Formula (68): 2

3Zdu

u2+4 =1

3tan−13√x

2+C

28. u= sin 4x, du = 4 cos 4xdx, Formula (68): 1

4Zdu

9+u

2=1

12 tan−1sin 4x

3+C

29. u=3x, du =3dx, Formula (76): 1

3Zdu

√u2−4=1

3ln 3x+p9x2−4+C

30. u=√2x2,du=2

√

2xdx, Formula (72):

2√2Zpu2+3du =x2

4p2x4+3+ 3

√

2ln √2x2+p2x4+3

+C

31. u=3x

,du=6xdx, u2du =54x

dx, Formula (81):

54 Zu2du

√5−u2=−x2

36p5−9x4+5

108 sin−13x2

√5+C

32. u=2x, du =2dx, Formula (83): 2 Zdu

u2√3−u2=−1

3xp3−4x2+C

33. u=lnx, du =dx/x, Formula (26): Zsin2udu=1

2ln x+1

4sin(2 ln x)+C

34. u=e−2x,du=−2e

−2x, Formula (27): −1

2Zcos2udu=−1

−2x−1

8sin 2e−2x+C

35. u=−2x, du =−2dx, Formula (51): 1

4Zueudu =1

4(−2x−1)e−2x+C

36. u=5x−1,du=5dx, Formula (50): 1

5Zln udu=1

5(uln u−u)+C=1

5(5x−1)[ln(5x−1)−1]+C

37. u= cos 3x, du =−3 sin 3x, Formula (67): −Zdu

u(u+1)

2=−1

31

1 + cos 3x+ln



cos 3x

1 + cos 3x+C

38. u=lnx, du =1

xdx, Formula (105): Zudu

√4u−1=1

12(2 ln x+1)

√

4lnx−1+C

39. u=4x

,du=8xdx, Formula (70): 1

8Zdu

u2−1=1

16 ln 

4x2−1

4x2+1



+C

40. u=2e

,du=2e

dx, Formula (69): 1

2Zdu

3−u2=1

4√3ln 

2ex+√3

2ex−√3+C

321 Chapter 9

41. u=2e

,du=2e

dx, Formula (74):

2Zp3−u2du =1

4up3−u2+3

4sin−1(u/√3)+C=1

2exp3−4e2x+3

4sin−1(2ex/√3)+C

42. u=3x, du =3dx, Formula (80):

3Z√4−u2du

u2=−3√4−u2

u−3 sin−1(u/2)+C=−√4−9x2

x−3 sin−1(3x/2)+C

43. u=3x, du =3dx, Formula (112):

3Zr5

3u−u2du =1

6u−5

6r5

3u−u2+25

216 sin−1u−5

5+C

=18x−5

36 p5x−9x2+25

216 sin−118x−5

5+C

44. u=√5x, du =√5dx, Formula (117):

Zdu

uq(u/√5) −u2=−q(u/√5) −u2

u/(2√5) +C=−2√x−5x2

x+C

45. u=3x, du =3dx, Formula (44):

9Zusin udu=1

9(sin u−ucos u)+C=1

9(sin 3x−3xcos 3x)+C

46. u=√x, u2=x, 2udu =dx, Formula (45): 2 Zucos udu= 2 cos √x+2

√

xsin √x+C

47. u=−√x, u2=x, 2udu =dx, Formula (51): 2 Zueudu =−2(√x+1)e

−

√

x+C

48. u=2−3x

,du=−6xdx, Formula (50):

−1

6Zln udu=−1

6(uln u−u)+C=−1

6((2 −3x2) ln(2 −3x2)+1

6(2 −3x2)+C

49. x2+4x−5=(x+2)

2−9; u=x+2,du=dx, Formula (70):

Zdu

u2−9=1

6ln 

u−3

u+3



+C=1

6ln 

x−1

x+5



+C

50. x2+2x−3=(x+1)

2−4,u=x+1,du=dx, Formula (77):

Zp4−u2du =1

2up4−u2+ 2 sin−1(u/2)+C

2(x+1)

p3−2x−x

2+ 2 sin−1((x+1)/2)+C

51. x2−4x−5=(x−2)2−9,u =x−2,du=dx, Formula (77):

Zu+2

√

9−u

2du =Zudu

√9−u

2+2Zdu

√9−u2=−p9−u2+ 2 sin−1u

3+C

=−p5+4x−x

2+ 2 sin−1x−2

3+C

Exercise Set 9.6 322

52. x2+6x+13=(x+3)

2+4,u=x+3,du=dx, Formula (71):

Z(u−3) du

u2+4 =1

2ln(u2+4)−3

2tan−1(u/2)+C=1

2ln(x2+6x+ 13) −3

2tan−1((x+3)/2)+C

53. u=√x−2, x=u2+2,dx =2udu;

Z2u

2(u

2+2)du =2Z(u

4+2u

2)du =2

5u5+4

3u3+C=2

5(x−2)5/2+4

3(x−2)3/2+C

54. u=√x+1,x=u

2−1, dx =2udu;

2Z(u

2−1)du =2

3u3−2u+C=2

3(x+1)

3/2−2

√

x+1+C

55. u=√x3+1,x

3=u

2−1, 3x2dx =2udu;

3Zu

2(u

2−1)du =2

3Z(u4−u2)du =2

15u5−2

9u3+C=2

15(x3+1)

5/2−2

9(x

3+1)

3/2+C

56. u=√x3−1, x3=u2+1,3x

2dx =2udu;

3Z1

2+1du =2

3tan−1u+C=2

3tan−1px3−1+C

57. u=x1/6,x=u6,dx =6u

du;

Z6u5

u3+u2du =6Zu

u+1du =6Zu

2−u+1−1

u+1du

=2x

1/2−3x

1/3+6x

1/6−6 ln(x1/6+1)+C

58. u=x1/5,x=u5,dx =5u

du;Z5u4

u5−u3du =5Zu

2−1

du =5

2ln |x2/5−1|+C

59. u=x1/4,x=u4,dx =4u

du;4

u(1 −u)du =4Z1

u+1

1−u

du =4ln x

1/4

|1−x

1/4

|+C

60. u=x1/3,x=u3,dx =3u

du;3

3+1du =3Z(u−u

3+1)du,

u3+1 =u

(u+ 1)(u2−u+1) =−1/3

u+1+(1/3)u+1/3

2−u+1 so

3Zu−u

u3+1du =Z3u+1

u+1−u+1

2−u+1du

2u2+ln|u+1|−1

2ln(u2−u+1)−√

3 tan−12u−1

√3+C

2x2/3+ln|x

1/3+1|−1

2ln(x2/3−x1/3+1)−√

3 tan−12x1/3−1

√3+C

61. u=x1/6,x=u6,dx =6u

du;

6Zu3

u−1du =6Zu

2+u+1+ 1

u−1du =2x

1/2+3x

1/3+6x

1/6+6 ln|x

1/6−1|+C

323 Chapter 9

62. u=√x,x=u2,dx =2udu;

−2Zu

2+u

u−1du =−2Zu+2+ 2

u−1du =−x−4√x−4ln|√x−1|+C

63. u=√1+x

2,x

2=u

2−1, 2xdx=2udu,xdx=udu;

Z(u

2−1)du =1

3(1+x2)3/2−(1+x2)1/2+C

64. u=(x+3)

1/5,x=u

5−3, dx =5u

du;

5Z(u8−3u3)du =5

9(x+3)

9/5−15

4(x+3)

4/5+C

65. u=√x,x=u2,dx =2udu, Formula (44): 2 Zusin udu= 2 sin √x−2√xcos √x+C

66. u=√x,x=u2,dx =2udu, Formula (51): 2 Zueudu =2

√

xe√x−2e√x+C

67. Z1

1+ 2u

1+u

2+1−u

1+u

2du =Z1

u+1du =ln|tan(x/2)+1|+C

68. Z1

2+ 2u

1+u

2du =Z1

u2+u+1du

=Z1

(u+1/2)2+3/4du =2

√3tan−12 tan(x/2)+1

√

3+C

69. u= tan(θ/2), Zdθ

1−cos θ=Z1

u2du =−1

u+C=−cot(θ/2)+C

70. u= tan(x/2),

3u2+8u−3du =2

3Z1

(u+4/3)2−25/9du =2

3Z1

z2−25/9dz (z=u+4/3)

5ln 

z−5/3

z+5/3



+C=1

5ln 

tan(x/2) −1/3

tan(x/2)+3 



+C

71. u= tan(x/2), 2 Z1−u2

(3u2+ 1)(u2+1)du;

1−u2

(3u2+ 1)(u2+1) =(0)u+2

2+1 +(0)u−1

u2+1 =2

2+1−1

2+1 so

Zcos x

2−cos xdx =4

√3tan−1[√3 tan(x/2)] −x+C

72. u= tan(x/2), 1

2Z1−u2

udu =1

2Z(1/u −u)du =1

2ln |tan(x/2)|−1

4tan2(x/2)+C

Exercise Set 9.6 324

73. Zx

t(4 −t)dt =1

4ln t

4−tx

(Formula (65), a=4,b =−1)

4ln x

4−x−ln 1=1

4ln x

4−x,1

4ln x

4−x=0.5,ln x

4−x=2,

4−x=e

,x=4e

2−e

x,x(1+e2)=4e

,x=4e

/(1+e2)≈3.523188312

74. Zx

t√2t−1dt = 2 tan−1√2t−1x

(Formula (108), a=−1,b=2)

=2tan−1√2x−1−tan−11=2tan−1√2x−1−π/4,

2(tan−1√2x−1−π/4) = 1, tan−1√2x−1=1/2+π/4, √2x−1 = tan(1/2+π/4),

x= [1 + tan2(1/2+π/4)]/2≈6.307993516

75. A=Z4

0p25 −x2dx =1

2xp25 −x2+25

2sin−1x

54

(Formula (74), a=5)

=6+25

2sin−14

5≈17.59119023

76. A=Z2

2/3p9x2−4dx;u=3x,

A=1

3Z6

2pu

2−4du =1

31

2upu2−4−2ln



u+pu

2−4



6

(Formula (73), a2=4)

33

√

32 −2 ln(6 + √32) + 2 ln 2=4

√

2−2

3ln(3 + 2√2) ≈4.481689467

77. A=Z1

25 −16x2dx;u=4x,

A=1

4Z4

25 −u2du =1

40 ln 

u+5

u−5



4

40 ln 9 ≈0.054930614 (Formula (69), a=5)

78. A=Z4

√xln xdx=4

3/23

2ln x−14

(Formula (50), n=1/2)

9(12 ln 4 −7) ≈4.282458815

79. V=2πZπ/2

0xcos xdx=2π(cos x+xsin x)π/2

=π(π−2) ≈3.586419094 (Formula (45))

80. V=2πZ8

√

x−4dx =4π

15 (3x+ 8)(x−4)3/28

(Formula (102), a=−4,b=1)

=1024

15 π≈214.4660585

81. V=2πZ3

0xe−xdx;u=−x,

V=2πZ−3

0ueudu =2πeu(u−1)−3

=2π(1 −4e−3)≈5.031899801 (Formula (51))

325 Chapter 9

82. V=2πZ5

1xln xdx=π

2(2 ln x−1)5

=π(25 ln 5 −12) ≈88.70584621 (Formula (50), n=1)

83. L=Z2

0p1+16x

2dx;u=4x,

L=1

4Z8

0p1+u

2du =1

4u

2p1+u

2+1

2ln u+p1+u

28

(Formula (72), a2=1)

=√

65 + 1

8ln(8 + √65) ≈8.409316783

84. L=Z3

1p1+9/x2dx =Z3

√x2+9

xdx = px2+9−3ln



3+√

2+9

x



!#3

√

2−√

10+3ln3+√

1+√

2≈3.891581644 (Formula (89), a=3)

85. S=2πZπ

0(sin x)p1 + cos2xdx;u= cos x,

S=−2πZ−1

1p1+u

2du =4πZ1

0p1+u

2du =4πu

2p1+u

2+1

2ln u+p1+u

21

a2=1)

=2π[

√

2+ln(1+√2)] ≈14.42359945 (Formula (72)

86. S=2πZ4

p1+1/x4dx =2πZ4

√

4+1

3dx;u=x2,

S=πZ16

√u2+1

2du =π −√u2+1

u+lnu+pu

2+1!#16

=π √2−√257

16 +ln16+√257

1+√

2!≈9.417237485 (Formula (93), a2=1)

87. (a) s(t)=2+Zt

020 cos6usin3udu

=−20

9sin2tcos7t−40

63 cos7t+166

(b)

3 6 9 12 15

s(t)

88. (a) v(t)=Zt

0a(u)du =−1

10e−tcos 2t+1

5e−tsin 2t+1

74e−tcos 6t−3

37e−tsin 6t+1

10 −1

s(t)=10+Zt

0v(u)du

=−3

50e−tcos 2t−2

25e−tsin 2t+35

2738e−tcos 6t+6

1369e−tsin 6t+16

185t+343866

34225

Exercise Set 9.7 326

(b)

2 6 10 14 18

s(t)

89. (a) Zsec xdx =Z1

cos xdx =Z2

1−u2du =ln



1+u

1−u



+C=ln



1 + tan(x/2)

1−tan(x/2)+C

=ln



cos(x/2) + sin(x/2)

cos(x/2) −sin(x/2)

cos(x/2) + sin(x/2)

cos(x/2) + sin(x/2)+C=ln



1 + sin x

cos x+C

=ln|sec x+ tan x|+C

(b) tan π

4+x

2=tan π

4+ tan x

1−tan π

4tan x

=1 + tan x

1−tan x

90. Zcsc xdx=Z1

sin xdx =Z1/u du =ln|tan(x/2)|+Cbut

ln |tan(x/2)|=1

2ln sin2(x/2)

cos2(x/2) =1

2ln (1 −cos x)/2

(1 + cos x)/2=1

2ln 1−cos x

1 + cos x; also,

1−cos x

1 + cos x=1−cos2x

(1 + cos x)2=1

(csc x+ cot x)2so 1

2ln 1−cos x

1 + cos x=−ln |csc x+ cot x|

91. Let u= tanh(x/2) then cosh(x/2)=1/sech(x/2)=1/

q1−tanh2(x/2)=1/

√

1−u

sinh(x/2) = tanh(x/2) cosh(x/2) = u/√1−u2, so sinh x= 2 sinh(x/2) cosh(x/2)=2u/(1 −u2),

cosh x= cosh2(x/2) + sinh2(x/2)=(1+u

)/(1 −u2), x= 2 tanh−1u, dx =[2/(1 −u2)]du;

Zdx

2 cosh x+ sinh x=Z1

u2+u+1du =2

√3tan−12u+1

√

3+C=2

√

3tan−12 tanh(x/2)+1

√

3+C.

EXERCISE SET 9.7

1. exact value = 14/3≈4.666666667

(a) 4.667600663, |EM|≈0.000933996

(b) 4.664795679, |ET|≈0.001870988

2. exact value = 2

(a) 1.998377048, |EM|≈0.001622952

(b) 2.003260982, |ET|≈0.003260982

3. exact value = 2

(a) 2.008248408, |EM|≈0.008248408

(b) 1.983523538, |ET|≈0.016476462

4. exact value = sin(1) ≈0.841470985

(a) 0.841821700, |EM|≈0.000350715

(b) 0.840769642, |ET|≈0.000701343

5. exact value = e−1−e−3≈0.318092373

(a) 0.317562837, |EM|≈0.000529536

(b) 0.319151975, |ET|≈0.001059602

6. exact value = 1

2ln 5 ≈0.804718956

(a) 0.801605339, |EM|≈0.003113617

(b) 0.811019505, |ET|≈0.006300549

327 Chapter 9

7. f(x)=√

x+1,f00(x)=−1

(x+1)

−3/2,f(4)(x)=−15

16(x+1)

−7/2;K

2=1/4, K4=15/16

(a) |EM|≤ 27

2400(1/4)=0.002812500 (b) |ET|≤ 27

1200(1/4)=0.005625000

180 ×104(15/16) ≈0.000126563

8. f(x)=1/

√

x,f

00(x)=3

−5/2

(4)(x)=105

16 x−9/2;K2=3/4, K4= 105/16

(a) |EM|≤ 27

2400(3/4)=0.008437500 (b) |ET|≤ 27

1200(3/4)=0.016875000

180 ×104(105/16) ≈0.000885938

9. f(x) = sin x,f00(x)=−sin x,f(4)(x) = sin x;K2=K4=1

(a) |EM|≤ π

2400(1) ≈0.012919282 (b) |ET|≤ π

1200(1) ≈0.025838564

180 ×104(1) ≈0.000170011

10. f(x) = cos x,f00(x)=−cos x,f(4)(x) = cos x;K2=K4=1

(a) |EM|≤ 1

2400(1) ≈0.000416667 (b) |ET|≤ 1

1200(1) ≈0.000833333

180 ×104(1) ≈0.000000556

11. f(x)=e

−x

00(x)=f

(4)(x)=e

−x

2=K

4=e

−1

(a) |EM|≤ 8

2400(e−1)≈0.001226265 (b) |ET|≤ 8

1200(e−1)≈0.002452530

180 ×104(e−1)≈0.000006540

12. f(x)=1/(2x+ 3), f00(x) = 8(2x+3)

−3,f(4)(x) = 384(2x+3)

−5;K

2=8,K

4= 384

(a) |EM|≤ 8

2400(8) ≈0.026666667 (b) |ET|≤ 8

1200(8) ≈0.053333333

180 ×104(384) ≈0.006826667

13. (a) n>(27)(1/4)

(24)(5 ×10−4)1/2

≈23.7; n=24 (b) n>

(27)(1/4)

(12)(5 ×10−4)1/2

≈33.5; n=34

(180)(5 ×10−4)1/4

≈7.1; n=8

14. (a) n>(27)(3/4)

(24)(5 ×10−4)1/2

≈41.1; n=42 (b) n>

(27)(3/4)

(12)(5 ×10−4)1/2

≈58.1; n=59

(180)(5 ×10−4)1/4

≈11.5; n=12

Exercise Set 9.7 328

15. (a) n>(π

)(1)

(24)(10−3)1/2

≈35.9; n=36 (b) n>(π

)(1)

(12)(10−3)1/2

≈50.8; n=51

)(1)

(180)(10−3)1/4

≈6.4; n=8

16. (a) n>(1)(1)

(24)(10−3)1/2

≈6.5; n=7 (b) n>(1)(1)

(12)(10−3)1/2

≈9.1; n=10

(180)(10−3)1/4

≈1.5; n=2

17. (a) n>(8)(e−1)

(24)(10−6)1/2

≈350.2; n= 351 (b) n>(8)(e−1)

(12)(10−6)1/2

≈495.2; n= 496

(180)(10−6)1/4

≈15.99; n=16

18. (a) n>(8)(8)

(24)(10−6)1/2

≈1632.99; n= 1633 (b) n>(8)(8)

(12)(10−6)1/2

≈2309.4; n= 2310

(180)(10−6)1/4

≈90.9; n=92

19. 0.746824948,

0.746824133 20. 1.137631378,

1.137630147 21. 2.129861595,

2.129861293

22. 2.418388347,

2.418399152 23. 0.805376152,

0.804776489 24. 1.536963087,

1.544294774

25. (a) 3.142425985, |EM|≈0.000833331

(b) 3.139925989, |ET|≈0.001666665

26. (a) 3.152411433, |EM|≈0.010818779

(b) 3.104518326, |ET|≈0.037074328

27. S14 =0.693147984, |ES|≈0.000000803 = 8.03 ×10−7; the method used in Example 5 results

in a value of nwhich ensures that the magnitude of the error will be less than 10−6, this is not

necessarily the smallest value of n.

28. (a) greater, because the graph of e−x2is concave up on the interval (1,2)

(b) less, because the graph of e−x2is concave down on the interval (0,0.5)

29. f(x)=xsin x,f00(x)=2cosx−xsin x,|f00(x)|≤2|cos x|+|x||sin x|≤2+2=4soK

2≤4,

n>(8)(4)

(24)(10−4)1/2

≈115.5; n= 116 (a smaller nmight suﬃce)

30. f(x)=e

cos x,f00(x) = (sin2x)ecos x−(cos x)ecos x,|f00(x)|≤e

cos x(sin2x+|cos x|)≤2eso

K2≤2e,n>(1)(2e)

(24)(10−4)1/2

≈47.6; n= 48 (a smaller nmight suﬃce)

31. f(x)=√

x,f

00(x)=−1

3/2, lim

x→0+|f00(x)|=+∞

329 Chapter 9

32. f(x) = sin √x,f00(x)=−√

xsin √x+ cos √x

4x3/2, lim

x→0+|f00(x)|=+∞

33. L=Zπ

0p1 + cos2xdx≈3.820187623 34. L=Z3

1p1+1/x4dx ≈2.146822803

35. 0

58.67

107.07

123.2

t (s)

v (mi/hr)

v (ft/s)

Z20

0vdt≈20

(3)(4)[0 + 4(58.67) + 2(88) + 4(107.07) + 123.2] ≈1604 ft

36. 0

0.02

0.08

0.20

0.40

0.60

0.70

0.60

0adt≈8

(3)(8)[0 + 4(0.02) + 2(0.08) + 4(0.20) + 2(0.40) + 4(0.60) + 2(0.70) + 4(0.60)+0]

≈2.7 cm/s

37. Z180

0vdt≈180

(3)(6)[0.00 + 4(0.03) + 2(0.08) + 4(0.16) + 2(0.27) + 4(0.42)+0.65]=37.9mi

38. Z1800

0(1/v)dx ≈1800

(3)(6) 1

3100 +4

2908 +2

2725 +4

2549 +2

2379 +4

2216 +1

2059≈0.71 s

39. V=Z16

0πr2dy =πZ16

0r2dy ≈π16

(3)(4)[(8.5)2+ 4(11.5)2+ 2(13.8)2+ 4(15.4)2+ (16.8)2]

≈9270 cm3≈9.3L

40. A=Z600

0hdx≈600

(3)(6)[0 + 4(7) + 2(16) + 4(24) + 2(25) + 4(16) + 0] = 9000 ft2,

V=75A≈75(9000) = 675,000 ft3

41. Zb

f(x)dx ≈A1+A2+···+A

n=b−a

n1

2(y

0+y

1)+1

2(y

1+y

2)+···+1

2(y

n−1+y



=b−a

2n[y

0+2y

1+2y

2+···+2y

n−1+y

]

42. right endpoint, trapezoidal, midpoint, left endpoint

43. (a) The maximum value of |f00(x)|is approximately 3.844880.

(b) n=18

44. (a) The maximum value of |f00(x)|is approximately 1.467890.

(b) n=12

Exercise Set 9.8 330

45. (a) The maximum value of |f(4)(x)|is approximately 42.551816.

(b) n=8

46. (a) The maximum value of |f(4)(x)|is approximately 7.022710.

(b) n=8

EXERCISE SET 9.8

1. (a) improper; inﬁnite discontinuity at x=3

(b) continuous integrand, not improper

(d) improper; inﬁnite interval of integration

(e) improper; inﬁnite interval of integration and inﬁnite discontinuity at x=1

(f) continuous integrand, not improper

2. (a) improper if p>0(b) improper if 1 <p<2

3. lim

`→+∞(−e−x)`

= lim

`→+∞(−e−`+1)=1

4. lim

`→+∞

2ln(1 + x2)`

−1

= lim

`→+∞

2[ln(1 + `2)−ln 2] = +∞, divergent

5. lim

`→+∞ln x−1

x+1`

= lim

`→+∞ln `−1

`+1−ln 3

5=−ln 3

5=ln5

6. lim

`→+∞−1

2e−x2`

= lim

`→+∞

2−e−`2+1

=1/2

7. lim

`→+∞−1

2ln

2x`

= lim

`→+∞−1

2ln

2`+1

2=1

8. lim

`→+∞2√ln x`

= lim

`→+∞(2√ln `−2√ln 2) = +∞, divergent

9. lim

`→−∞ −1

4(2x−1)20

= lim

`→−∞

4[−1+1/(2`−1)2]=−1/4

10. lim

`→−∞

2tan−1x

22

= lim

`→−∞

2π

4−tan−1`

2=1

2[π/4−(−π/2)]=3π/8

11. lim

`→−∞

3e3x0

= lim

`→−∞ 1

3−1

3e3`=1

12. lim

`→−∞ −1

2ln(3 −2ex)0

= lim

`→−∞

2ln(3 −2e`)=1

2ln 3

331 Chapter 9

13. Z+∞

−∞

x3dx converges if Z0

−∞

x3dx and Z+∞

0x3dx both converge; it diverges if either (or both)

diverge. Z+∞

0x3dx = lim

`→+∞

4x4`

= lim

`→+∞

4`4=+∞so Z+∞

−∞

x3dx is divergent.

14. Z+∞

√x2+2dx = lim

`→+∞px2+2

`

= lim

`→+∞(p`2+2−√

2)=+∞

so Z∞

−∞

√x2+2dx is divergent.

15. Z+∞

(x2+3)

2dx = lim

`→+∞−1

2(x2+3)`

= lim

`→+∞

2[−1/(`2+3)+1/3] = 1

similarly Z0

−∞

(x2+3)

2dx =−1/6soZ∞

−∞

(x2+3)

2dx =1/6+(−1/6)=0

16. Z+∞

e−t

1+e

−2tdt = lim

`→+∞−tan−1(e−t)`

= lim

`→+∞h−tan−1(e−`)+π

4i=π

−∞

e−t

1+e

−2tdt = lim

`→−∞ −tan−1(e−t)0

= lim

`→−∞ h−π

4+ tan−1(e−`)i=π

Z+∞

−∞

e−t

1+e

−2tdt =π

4+π

4=π

17. lim

`→3+−1

x−34

= lim

`→3+−1+ 1

`−3=+∞, divergent

18. lim

`→0+

2x2/38

= lim

`→0+

2(4 −`2/3)=6

19. lim

`→π/2−−ln(cos x)`

= lim

`→π/2−−ln(cos `)=+∞, divergent

20. lim

`→9−−2√9−x`

= lim

`→9−2(−√9−`+3)=6

21. lim

`→1−sin−1x`

= lim

`→1−sin−1`=π/2

22. lim

`→−3+−p9−x21

= lim

`→−3+(−√8+p9−`

2)=−

√

23. lim

`→π/6−−√1−2 sin x`

= lim

`→π/6−(−√1−2 sin `+1)=1

24. lim

`→π/4−−ln(1 −tan x)`

= lim

`→π/4−−ln(1 −tan `)=+∞, divergent

25. Z2

x−2= lim

`→2−ln |x−2|`

= lim

`→2−(ln |`−2|−ln 2) = −∞, divergent

Exercise Set 9.8 332

26. Z2

x2= lim

`→0+−1/x2

= lim

`→0+(−1/2+1/`)=+∞so Z2

−2

x2is divergent

27. Z8

0x−1/3dx = lim

`→0+

2x2/38

= lim

`→0+

2(4 −`2/3)=6,

−1x

−1/3

dx = lim

`→0−

2x2/3`

−1

= lim

`→0−

2(`2/3−1) = −3/2

so Z8

−1x−1/3dx =6+(−3/2)=9/2

28. Z2

(x−2)2/3= lim

`→2−3(x−2)1/3`

= lim

`→2−3[(`−2)1/3−(−2)1/3]=33

√

similarly Z4

(x−2)2/3= lim

`→2+3(x−2)1/34

=33

√

2soZ4

(x−2)2/3=63

√

29. Deﬁne Z+∞

x2dx =Za

x2dx +Z+∞

x2dx where a>0; take a= 1 for convenience,

x2dx = lim

`→0+(−1/x)1

= lim

`→0+(1/` −1)=+∞so Z+∞

x2dx is divergent.

30. Deﬁne Z+∞

x√x2−1=Za

x√x2−1+Z+∞

x√x2−1where a>1,

take a= 2 for convenience to get

x√x2−1= lim

`→1+sec−1x2

= lim

`→1+(π/3−sec−1`)=π/3,

Z+∞

x√x2−1= lim

`→+∞sec−1x`

=π/2−π/3soZ+∞

x√x2−1=π/2.

31. Z+∞

e−√x

√xdx =2Z+∞

−u

du = 2 lim

`→+∞−e−u`

= 2 lim

`→+∞1−e−`=2

32. Z+∞

√x(x+4) =2Z+∞

u2+4 = 2 lim

`→+∞

2tan−1u

2`

= lim

`→+∞tan−1`

2=π

33. Z+∞

e−x

√1−e−xdx =Z1

√u= lim

`→0+2√u1

= lim

`→0+2(1 −√`)=2

34. Z+∞

e−x

√1−e−2xdx =−Z0

√1−u2=Z1

√1−u2= lim

`→1sin−1u`

= lim

`→1sin−1`=π

36. A=Z+∞

0xe−3xdx = lim

`→+∞−1

9(3x+1)e

−3x

`

=1/3

37. lim

`→+∞Z`

0e−xcos xdx= lim

`→+∞

2e−x(sin x−cos x)`

=1/2

39. (a) 2.726585 (b) 2.804364 (c) 0.219384 (d) 0.504067

333 Chapter 9

40. 1+dy

dx2

=1+ 16x2

9−4x2=9+12x

9−4x

2; the arc length is Z3/2

0s9+12x

9−4x

2dx ≈3.633168

41. Zln xdx=xln x−x+C,

0ln xdx= lim

`→0+Z1

ln xdx= lim

`→0+(xln x−x)1

= lim

`→0+(−1−`ln `+`),

but lim

`→0+`ln `= lim

`→0+

ln `

1/` = lim

`→0+(−`)=0soZ1

0ln xdx=−1

42. Zln x

x2dx =−ln x

x−1

x+C,

Z+∞

ln x

x2dx = lim

`→+∞Z`

ln x

x2dx = lim

`→+∞−ln x

x−1

x`

= lim

`→+∞−ln `

`−1

`+1

,

but lim

`→+∞

ln `

`= lim

`→+∞

`=0soZ+∞

ln x

x2=1

43. Zxe−3xdx =−1

3xe−3x−1

9e−3x+C,

Z+∞

0xe−3xdx = lim

`→+∞Z`

0xe−3xdx = lim

`→+∞−1

3xe−3x−1

9e−3x`

= lim

`→+∞−1

3`e−3`−1

9e−3`+1

9

but lim

`→+∞`e−3`= lim

`→+∞

e3`= lim

`→+∞

3e3`=0soZ+∞

0xe−3xdx =1/9

44. A=Z+∞

x2−4dx = lim

`→+∞2lnx−2

x+2`

= lim

`→+∞2ln `−2

`+2−ln 1

5= 2 ln 5

45. (a) V=πZ+∞

0e−2xdx =−π

2lim

`→+∞e−2x`

=π/2

(b) S=2πZ+∞

−x

p1+e

−2x

dx, let u=e−xto get

S=−2πZ0

1p1+u

2du =2πu

2p1+u

2+1

2ln |u+p1+u

1

=πh√

2+ln(1+√2)i

47. (a) For x≥1,x

2≥x, e−x2≤e−x

(b) Z+∞

1e−xdx = lim

`→+∞Z`

1e−xdx = lim

`→+∞−e−xi`

1= lim

`→+∞(e−1−e−`)=1/e

1e−x2dx is convergent and is ≤1/e.

48. (a) If x≥0 then ex≥1,1

2x+1 ≤e

2x+1

(b) lim

`→+∞Z`

2x+1 = lim

`→+∞

2ln(2x+1)

`

=+∞

2x+1dx is divergent.

Exercise Set 9.8 334

49. V= lim

`→+∞Z`

1(π/x2)dx = lim

`→+∞−(π/x)i`

1= lim

`→+∞(π−π/`)=π

A= lim

`→+∞Z`

12π(1/x)p1+1/x4dx; use Exercise 46(a) with f(x)=2π/x,g(x)=(2π/x)p1+1/x4

and a= 1 to see that the area is inﬁnite.

50. (a) 1≤√x3+1

xfor x≥2, Z+∞

21dx =+∞

(b) Z+∞

x5+1dx ≤Z+∞

x4= lim

`→+∞−1

3x3`

=1/24

2x+1 ≤e

2x+1 for x≥0, Z+∞

2x+1dx = lim

`→+∞

2ln(2x+1)

`

=+∞

51. Z2x

0p1+t

3dt ≥Z2x

0t3/2dt =2

5t5/22x

5(2x)5/2,

lim

x→+∞Z2x

0t3/2dt = lim

x→+∞

5(2x)5/2=+∞so Z+∞

0p1+t

3dt =+∞; by L’Hˆopital’s Rule

lim

x→+∞Z2x

0p1+t

3dt

x5/2= lim

x→+∞

2p1+(2x)

(5/2)x3/2= lim

x→+∞

2p1/x3+8

5/2=8

√

2/5

52. (b) u=√x, Z+∞

cos √x

√xdx =2Z+∞

0cos udu;Z+∞

0cos udu diverges by part (a).

53. Let x=rtan θto get Zdx

(r2+x2)3/2=1

r2Zcos θdθ=1

2sin θ+C=x

r2√r2+x2+C

so u=2πNIr

klim

`→+∞

r2√r2+x2`

=2πNI

kr lim

`→+∞(`/pr2+`2−a/pr2+a2)

=2πNI

kr (1 −a/pr2+a2).

54. Let a2=M

2RT to get

(a) ¯v=4

√πM

2RT 3/21

2M

2RT −2

√πr2RT

M=r8RT

πM

(b) v2

rms =4

√πM

2RT 3/23√π

8M

2RT −5/2

=3RT

Mso vrms =r3RT

55. (a) Satellite’s weight = w(x)=k/x2lb when x= distance from center of Earth; w(4000) = 6000

so k=9.6×1010 and W=Z4000+`

4000 9.6×1010x−2dx mi·lb.

(b) Z+∞

4000 9.6×1010x−2dx = lim

`→+∞−9.6×1010/x`

4000

=2.4×107mi·lb

56. (a) L{1}=Z+∞

0e−stdt = lim

`→+∞−1

se−st`

(b) L{e2t}=Z+∞

0e−ste2tdt =Z+∞

0e−(s−2)tdt = lim

`→+∞−1

s−2e−(s−2)t`

s−2

335 Chapter 9

0e−st sin tdt= lim

`→+∞

e−st

s2+1(−ssin t−cos t)`

s2+1

(d) L{cos t}=Z+∞

0e−st cos tdt= lim

`→+∞

e−st

s2+1(−scos t+ sin t)`

s2+1

57. (a) L{f(t)}=Z+∞

0te−st dt = lim

`→+∞−(t/s +1/s2)e−st`

(b) L{f(t)}=Z+∞

0t2e−st dt = lim

`→+∞−(t2/s +2t/s2+2/s3)e−st`

3e−stdt = lim

`→+∞−1

se−st`

=e−3s

58. 10 100 1000 10,000

0.8862269 0.8862269 0.8862269 0.8862269

59. (a) u=√ax, du =√a dx, 2Z+∞

0e−ax2dx =2

√aZ+∞

0e−u2du =pπ/a

(b) x=√2σu, dx =√2σ du, 2

√2πσ Z+∞

0e−x2/2σ2dx =2

√πZ+∞

0e−u2du =1

60. (a) Z3

0e−x2dx ≈0.8862; √π/2≈0.8862

(b) Z+∞

0e−x2dx =Z3

0e−x2dx +Z+∞

3e−x2dx so E=Z+∞

3e−x2dx < Z+∞

3xe−x2dx =1

2e−9<7×10−5

61. (a) Z4

x6+1dx ≈1.047; π/3≈1.047

(b) Z+∞

x6+1dx =Z4

x6+1dx +Z+∞

x6+1dx so

E=Z+∞

x6+1dx < Z+∞

x6dx =1

5(4)5<2×10−4

62. If p= 0, then Z+∞

0(1)dx = lim

`→+∞x`

=+∞,

if p6= 0, then Z+∞

0epxdx = lim

`→+∞

pepx`

= lim

`→+∞

p(ep` −1) = −1/p, p < 0

+∞,p>0

63. If p= 1, then Z1

x= lim

`→0+ln x1

=+∞;

if p6= 1, then Z1

xp= lim

`→0+

x1−p

1−p1

= lim

`→0+[(1 −`1−p)/(1 −p)] = 1/(1 −p),p<1

+∞, p>1

64. u=√1−x, u2=1−x, 2udu=−dx;

−2Z0

1p2−u2du =2Z1

0p2−u

2du =hup2−u2+ 2 sin−1(u/√2)i1

0=√2+π/2

65. 2Z1

0cos(u2)du ≈1.809 66. −2Z0

1sin(1 −u2)du =2Z1

0sin(1 −u2)du ≈1.187

Chapter 9 Supplementary Exercises 336

CHAPTER 9 SUPPLEMENTARY EXERCISES

1. (a) integration by parts, u=x,dv = sin xdx (b) u-substitution: u= sin x

(e) u-substitution: u=x3+1 (f) u-substitution: u=x+1

(g) integration by parts: dv =dx, u = tan−1x(h) trigonometric substitution: x= 2 sin θ

(i) u-substitution: u=4−x

2. (a) x= 3 tan θ(b) x= 3 sin θ(c) x=1

3sin θ

(d) x= 3 sec θ(e) x=√3 tan θ(f) x=1

9tan θ

5. (a) #40 (b) #57 (c) #113

(d) #108 (e) #52 (f) #71

6. (a) u=x2,dv =x

√x2+1dx,du =2xdx,v=√x

2+1;

√

2+1dx =x2px2+1

0−2Z1

0x(x

2+1)

1/2dx

=√2−2

3(x2+1)

3/21

=√

2−2

3[2√2−1]=(2−√

2)/3

(b) u2=x2+1,x

2=u

2−1, 2xdx=2udu,xdx=udu;

√x

2+1dx =Z1

√x2+1xdx=Z√2

2−1

uudu

=Z√2

1(u

2−1)du =1

3u3−u√2

=(2−√

2)/3

7. (a) u=2x,

Zsin42xdx=1

2Zsin4udu=1

2−1

4sin3ucos u+3

4Zsin2udu



=−1

8sin3ucos u+3

8−1

2sin ucos u+1

2Zdu

=−1

8sin3ucos u−3

16 sin ucos u+3

16u+C

=−1

8sin32xcos 2x−3

16 sin 2xcos 2x+3

8x+C

(b) u=x2,

Zxcos4(x2)dx =1

2Zcos4udu=1

21

4cos3usin u+3

4Zcos2udu



8cos3usin u+3

81

2cos usin u+1

2Zdu

8cos3usin u+3

16 cos usin u+3

16u+C

8cos3(x2) sin(x2)+ 3

16 cos(x2) sin(x2)+ 3

16x2+C

8. (a) With x= sec θ:

x3−xdx =Zcot θdθ=ln|sin θ|+C=ln√

2−1

|x|+C; valid for |x|>1.

337 Chapter 9

(b) With x= sin θ:

x3−xdx =−Z1

sin θcos θdθ =−Z2 csc 2θdθ

=−ln |csc 2θ−cot 2θ|+C=ln|cot θ|+C=ln√

1−x

|x|+C, 0<|x|<1.

Z−1

x+1/2

x+1+1/2

x−1dx =−ln |x|+1

2ln |x+1|+1

2ln |x−1|+C

=lnp|x

2−1|

|x|+C; valid for all xexcept x=0,±1.

9. (a) With u=√x:

√x√2−xdx =2Z1

√

2−u

2du = 2 sin−1(u/√2)+C= 2 sin−1(px/2)+C;

with u=√2−x:

√x√2−xdx =−2

√2−u2du =−2 sin−1(u/√2)+C=−2 sin−1(√2−x/√2)+C;

completing the square:

p1−(x−1)2dx = sin−1(x−1)+C.

(b) In the three results in part (a) the antiderivatives diﬀer by a constant, in particular

2 sin−1(px/2) = π−2 sin−1(√2−x/√2) = π/2 + sin−1(x−1).

10. A=Z2

3−x

x3+x2dx,3−x

x2(x+1) =A

x+B

2+C

x+1;A=−4, B=3,C=4

A=−4ln|x|−3

x+4ln|x+1|

2

=(−4ln2−3

2+ 4 ln 3) −(−4ln1−3 + 4 ln 2) = 3

2−8ln2+4ln3= 3

2+4ln3

11. Solve y=1/(1+x2) for xgetting

x=r1−y

yand integrate with respect to

to yto get A=Z1

0r1−y

ydy (see ﬁgure)

1y = 1/(1 + x2)

12. A=Z+∞

ln x−1

x2dx = lim

`→+∞−ln x

x`

=1/e

13. V=2πZ+∞

0xe−xdx =2πlim

`→+∞−e−x(x+1)

`

=2πlim

`→+∞1−e−`(`+1)



but lim

`→+∞e−`(`+ 1) = lim

`→+∞

`+1

`= lim

`→+∞

e`=0soV=2π

14. Z+∞

x2+a2= lim

`→+∞

atan−1(x/a)`

= lim

`→+∞

atan−1(`/a)= π

2a=1,a=π/2

Chapter 9 Supplementary Exercises 338

15. u= cos θ,−Zu1/2du =−2

3cos3/2θ+C

16. Use Endpaper Formula (31) to get Ztan7θdθ =1

6tan6θ−1

4tan4θ+1

2tan2θ+ln|cos θ|+C.

17. u= tan(x2), 1

2Zu2du =1

6tan3(x2)+C

18. x=(1/

√

2) sin θ,dx =(1/

√

2) cos θdθ,

√2Zπ/2

−π/2cos4θdθ =1

√2(1

4cos3θsin θπ/2

−π/2

4Zπ/2

−π/2cos2θdθ

)

√2(1

2cos θsin θπ/2

−π/2

2Zπ/2

−π/2dθ)=3

4√2

2π=3π

8√2

19. x=√3 tan θ,dx =√3 sec2θdθ,

3Z1

sec θdθ =1

3Zcos θdθ=1

3sin θ+C=x

3√3+x

2+C

20. Zcos θ

(sin θ−3)2+3dθ, let u= sin θ−3, Z1

u2+3du =1

√3tan−1[(sin θ−3)/√3]+C

21. Zx+3

p(x+1)

2+1dx, let u=x+1,

Zu+2

√

2+1du =Zu(u2+1)

−1/2+2

√

2+1du =pu2+1+2sinh−1u+C

=px2+2x+2+2sinh−1(x+1)+C

Alternate solution: let x+ 1 = tan θ,

Z(tan θ+ 2) sec θdθ=Zsec θtan θdθ+2Zsec θdθ= sec θ+2ln|sec θ+ tan θ|+C

=px2+2x+2+2ln(

2+2x+2+x+1)+C.

22. Let x= tan θto get Z1

x3−x2dx.

x2(x−1) =A

x+B

x2+C

x−1;A=−1, B=−1, C=1so

−Z1

dx −Z1

x2dx +Z1

x−1dx =−ln |x|+1

x+ln|x−1|+C

x+ln



x−1

x



+C= cot θ+ln



tan θ−1

tan θ+C= cot θ+ln|1−cot θ|+C

23. 1

(x−1)(x+ 2)(x−3) =A

x−1+B

x+2+C

x−3;A=−1

6,B=1

15,C=1

10 so

−1

6Z1

x−1dx +1

15 Z1

x+2dx +1

10 Z1

x−3dx

=−1

6ln |x−1|+1

15 ln |x+2|+1

10 ln |x−3|+C

339 Chapter 9

24. 1

x(x2+x+1) =A

x+Bx +C

x2+x+1;A=1,B=C=−1so

Z−x−1

2+x+1dx =−Zx+1

(x+1/2)2+3/4dx =−Zu+1/2

2+3/4du, u =x+1/2

=−1

2ln(u2+3/4) −1

√3tan−12u/√3+C

so Zdx

x(x2+x+1) =ln|x|−1

2ln(x2+x+1)−1

√

3tan−12x+1

√

3+C

25. u=√x−4, x=u2+4,dx =2udu,

2+4du =2Z2

01−4

2+4du =2u−4 tan−1(u/2)2

=4−π

26. u=√x,x=u2,dx =2udu,

2Z3

2+9du =2Z3

01−9

2+9du =2u−6 tan−1u

33

=6−3

27. u=√ex+1,e

x=u

2−1, x= ln(u2−1), dx =2u

u2−1du,

u2−1du =Z1

u−1−1

u+1du =ln|u−1|−ln |u+1|+C=ln√

x+1−1

√

x+1+1+C

28. u=√ex−1, ex=u2+1,x= ln(u2+ 1), dx =2u

u2+1du,

2u2

u2+1du =2Z1

01−1

2+1du =(2u−2 tan−1u)1

=2−π

29. lim

`→+∞−1

2(x2+1)`

= lim

`→+∞−1

2(`2+1)+1

2(a2+1)=1

2(a2+1)

30. lim

`→+∞

ab tan−1bx

a`

= lim

`→+∞

ab tan−1b`

a=π

2ab

31. Let u=x4to get 1

4Z1

√1−u2du =1

4sin−1u+C=1

4sin−1(x4)+C.

32. Z(cos32 xsin30 x−cos30 xsin32 x)dx =Zcos30 xsin30 x(cos2x−sin2x)dx

230 Zsin30 2xcos 2xdx=sin31 2x

31(231)+C

33. Zqx−px2−4dx =1

√2Z(√x+2−√

x−2)dx =√2

3[(x+2)

3/2−(x−2)3/2]+C

34. Z1

x10(1+x−9)dx =−1

9Z1

udu =−1

9ln |u|+C=−1

9ln |1+x

−9|+C

Chapter 9 Supplementary Exercises 340

35. (a) (x+ 4)(x−5)(x2+1)

2;A

x+4+B

x−5+Cx +D

x2+1 +Ex +F

(x2+1)

(b) −3

x+4+2

x−5−x−2

2+1−3

2+1)

2ln(x2+1)−3

2x

2+1+ tan−1x

36. (a) Γ(1) = Z+∞

0e−tdt = lim

`→+∞−e−t`

= lim

`→+∞(−e−`+1)=1

(b) Γ(x+1)=Z+∞

−t

dt; let u=tx,dv =e−tdt to get

Γ(x+1)= −t

−t

+∞

0+xZ+∞

x−1e

−t

dt =−txe−t+∞

0+xΓ(x)

lim

t→+∞txe−t= lim

t→+∞

et= 0 (by multiple applications of L’Hˆopital’s rule)

so Γ(x+1)=xΓ(x)

It appears that Γ(n)=(n−1)! if nis a positive integer.

(d) Γ1

2=Z+∞

0t−1/2e−tdt =2Z+∞

−u

2du (with u=√t)=2(

√

π/2) = √π

(e) Γ3

2=1

2Γ1

2=1

2√π,Γ

5

=3

3

=3

√

37. (a) t=−ln x, x =e−t,dx=−e

−tdt,

0(ln x)ndx =−Z0

+∞

(−t)ne−tdt =(−1)nZ+∞

0tne−tdt =(−1)nΓ(n+1)

(b) t=xn,x=t

1/n,dx=(1/n)t1/n−1dt,

Z+∞

0e−xndx =(1/n)Z+∞

0t1/n−1e−tdt =(1/n)Γ(1/n) = Γ(1/n +1)

38. (a) pcos θ−cos θ0=q2sin2(θ0/2) −sin2(θ/2)=q2(k2−k2sin2φ)=p2k

2cos2φ

=√2kcos φ;ksin φ= sin(θ/2) so kcos φdφ=1

2cos(θ/2) dθ =1

2q1−sin2(θ/2) dθ

2q1−k2sin2φdθ,thusdθ =2kcos φ

p1−k2sin2φdφ and hence

T=s8L

gZπ/2

√2kcos φ·2kcos φ

p1−k2sin2φdφ =4

gZπ/2

p1−k2sin2φdφ

(b) If L=1.5ftandθ

0=(π/180)(20) = π/9, then

T=√3

2Zπ/2

dφ

q1−sin2(π/18) sin2φ≈1.37 s.

341 Chapter 9

CHAPTER 9 HORIZON MODULE

1. The depth of the cut equals the terrain elevation minus the track elevation. From Figure 2, the

cross sectional area of a cut of depth Dmeters is 10D+2·1

2=D

2+10Dsquare meters.

Distance from

town A (m) Terrain elevation

(m)

2000

4000

6000

8000

10,000

12,000

14,000

16,000

18,000

20,000

100

105

108

110

104

106

120

122

124

128

130

Track elevation

(m)

100

101

102

103

104

105

106

107

108

109

110

Depth of cut

(m)

Cross-sectional area

f(x) of cut (m2)

119

336

375

416

551

600

The total volume of dirt to be excavated, in cubic meters, is Z2000

0f(x)dx.

By Simpson’s Rule, this is approximately

20,000 −0

3·10 [0+4·56+2·96+4·119+2·0+4·11+2·336+4·375+2·416+4·551 + 600]

=4,496,000 m3.

Excavation costs $4 per m3, so the ttal cost of the railroad from kAto Mis about

4·4,496,000=17,984,000 dollars.

2. (a) Distance from

town A (m) Terrain elevation

(m)

20,000

20,100

20,200

20,300

20,400

20,500

20,600

20,700

20,800

20,900

21,000

130

135

139

142

145

147

148

146

143

139

133

Track elevation

(m)

110

109.8

109.6

109.4

109.2

109

108.8

108.6

108.4

108.2

108

Depth of cut

(m)

25.2

29.4

32.6

35.8

39.2

37.4

34.6

30.8

Cross-sectional area

f(x) of cut (m2)

300

887.04

1158.36

1388.76

1639.64

1824

1928.64

1772.76

1543.16

1256.64

875

The total volume of dirt to be excavated, in cubic meters, is Z21,000

20,000 f(x)dx.

By Simpson’s Rule this is approximately

21,000 −20,000

3·10 [600 + 4 ·887.04+2·1158.36+...+4·1256.64 + 875] = 1,417,713.33 m3.

The total cost of a trench from Mto Nis about 4 ·1,417,713.33 ≈5,670,853 dollars.

Chapter 9 Horizon Module 342

(b) Distance from

town A (m) Terrain elevation

(m)

21,000

22,000

23,000

24,000

25,000

26,000

27,000

28,000

29,000

30,000

31,000

133

120

106

108

106

100

102

Track elevation

(m)

108

106

104

102

100

Depth of cut

(m)

Cross-sectional area

f(x) of cut (m2)

875

336

144

The total volume of dirt to be excavated, in cubic meters, is Z31,000

21,000 f(x)dx. By Simpson’s

Rule this is approximately

31,000 −21,000

3·10 [875 + 4 ·336+2·24+...+4·11+0]=1,229,000 m3.

The total cost of the railroad from Nto Bis about 4 ·1,229,000 ≈4,916,000 dollars.

3. The total cost if trenches are used everywhere is about

17,984,000+5,670,853+4,916,000=28,570,853 dollars.

4. (a) The cross-sectional area of a tunnel is AT=80+1

π5

2≈119.27 m2. The length of the

tunnel is 1000 m, so the volume of dirt to be removed is about 1000AT≈1,119,269.91 m3,

and the drilling and dirt-piling costs are 8 ·1000AT≈954,159 dollars.

(b) To extend the tunnel from a length of xmeters to a length of x+dx meters, we must move

a volume of ATdx cubic meters of dirt a distance of about xmeters. So the cost of this

extension is about 0.06 ×ATdx dollars. The cost of moving all of the dirt in the tunnel is

therefore

Z1000

00.06 ×ATdx =0.06AT

21000

=30,000AT≈3,578,097 dollars.

5. The total cost of the railroad, using a tunnel, is 17,894,000+4,532,257+4,916,000+27,432,257

dollars, which is smaller than the cost found in Exercise 3. It will be cheaper to build the railroad

if a tunnel is used.

343

CHAPTER 10

Mathematical Modeling with Differential

Equations

EXERCISE SET 10.1

1. y0=2x

/3=x

yand y(0) = 2 by inspection.

2. y0=x3−2 sin x, y(0) = 3 by inspection.

3. (a) ﬁrst order; dy

dx =c;(1+x)dy

dx =(1+x)c=y

(b) second order; y0=c1cos t−c2sin t, y00 +y=−c1sin t−c2cos t+(c

1sin t+c2cos t)=0

4. (a) ﬁrst order; 2dy

dx +y=2

−c

−x/2+1

+ce−x/2+x−3=x−1

(b) second order; y0=c1et−c2e−t,y

00 −y=c1et+c2e−t−c1et+c2e−t=0

5. 1

dx =xdy

dx +y, dy

dx(1 −xy)=y

,dy

dx =y2

1−xy

6. 2x+y2+2xy dy

dx = 0, by inspection.

7. (a) IF: µ=e3Rdx =e3x,d

dxye3x=0,ye

3x=C, y =Ce−3x

separation of variables: dy

y=−3dx, ln |y|=−3x+C1,y =±e

−3xe

1=Ce−3x

(b) IF: µ=e−2Rdt =e−2t,d

dt[ye−2t]=0,ye

−2t=C, y =Ce2t

separation of variables: dy

y=2dt, ln |y|=2t+C

,y=±e

2t=Ce2t

8. (a) IF: µ=e−4Rxdx =e

−2x

2,d

dx hye−2x2i=0,y=Ce2x2

separation of variables: dy

y=4x dx, ln |y|=2x

2+C

,y=±e

2=Ce2x2

(b) IF: µ=eRdt =et,d

dt yet=0,y=Ce−t

separation of variables: dy

y=−dt, ln |y|=−t+C1,y =±e

−t=Ce−t

9. 1

ydy =1

xdx,ln|y|=ln|x|+C

,ln



y



=C

x=±e

1=C,y=Cx

10. dy

1+y

2=x

2dx, tan−1y=1

3x3+C, y = tan 1

3x3+C

11. dy

1+y=−x

√

1+x

2dx, ln |1+y|=−

p1+x

2+C

1,1+y=±e

−

√

1+x2eC1=Ce−√1+x2,

y=Ce−√1+x2−1

Exercise Set 10.1 344

12. ydy =x

3dx

1+x

4,y

2=1

4ln(1 + x4)+C

1,2y

2= ln(1 + x4)+C, y =±p[ln(1 + x4)+C]/2

13. 1

y+ydy =exdx, ln |y|+y2/2=e

x+C; by inspection, y=0

14. dy

y=−x dx, ln |y|=−x2/2+C

1,y=±e

−x

/2=Ce−x2/2

15. eydy =sin x

cos2xdx = sec xtan xdx,e

y= sec x+C,y= ln(sec x+C)

16. dy

1+y

2=(1+x)dx, tan−1y=x+x2

2+C, y = tan(x+x2/2+C)

17. dy

y2−y=dx

sin x,Z−1

y+1

y−1dy =Zcsc x dx, ln 

y−1

y=ln|csc x−cot x|+C1,

y−1

y=±eC1(csc x−cot x)=C(csc x−cot x),y=1

1−C(csc x−cot x);

by inspection, y= 0 is also a solution.

18. 1

tan ydy =3

sec xdx,cos y

sin ydy = 3 cos xdx,ln|sin y|= 3 sin x+C1,

sin y=±e3 sin x+C1=±eC1e3 sin x=Ce3 sin x

19. µ=eR3dx =e3x,e3xy=Zexdx =ex+C,y=e−2x+Ce−3x

20. µ=e2Rxdx =e

2,d

dxhyex2i=xex2,ye

2=1

2+C, y =1

2+Ce−x2

21. µ=eRdx =ex,exy=Zexcos(ex)dx = sin(ex)+C,y=e

−xsin(ex)+Ce−x

22. dy

dx +2y=1

2,µ=e

R2dx =e2x,e2xy=Z1

2e2xdx =1

4e2x+C,y=1

4+Ce−2x

23. dy

dx +x

x2+1y=0,µ=e

R(x/(x2+1))dx =e1

2ln(x2+1) =px2+1,

dx hypx2+1

i=0,y

2+1=C, y =C

√x2+1

24. dy

dx +y=1

1+e

x,µ=e

Rdx =ex,exy=Zex

1+e

xdx = ln(1 + ex)+C,y=e

−xln(1 + ex)+Ce−x

25. dy

dx +1

xy=1,µ=e

R(1/x)dx =eln x=x, d

dx[xy]=x, xy =1

2x2+C, y =x/2+C/x

(a) 2=y(1) = 1

2+C, C =3

2,y =x/2+3/(2x)

(b) 2=y(−1) = −1/2−C, C =−5/2,y =x/2−5/(2x)

345 Chapter 10

26. dy

y=x dx, ln |y|=x2

2+C1,y=±e

/2=Cex2/2

(a) 1=y(0) = Cso C=1,y=e

(b) 1

2=y(0) = C,soy=1

27. µ=e−Rxdx =e

−x

2/2,e

−x

2/2y=Zxe−x2/2dx =−e−x2/2+C,

y=−1+Cex2/2,3=−1+C,C=4,y=−1+4e

2/2

28. µ=eRdt =et,ety=Z2etdt =2e

t+C,y=2+Ce−t,1=2+C,C=−1, y=2−e

−t

29. (y+ cos y)dy =4x

2dx, y2

2+ sin y=4

3x3+C, π2

2+ sin π=4

3(1)3+C, π2

2=4

3+C,

C=π2

2−4

3,3y2+ 6 sin y=8x

3+3π

2−8

30. dy

dx =(x+2)e

y,e

−ydy =(x+2)dx,−e−y=1

2x2+2x+C,−1=C,

−e

−y=1

2+2x−1, e−y=−1

2x2−2x+1,y=−ln 1−2x−1

2x2

31. 2(y−1) dy =(2t+1)dt, y2−2y=t2+t+C, 1+2=C, C =3,y

2−2y=t

2+t+3

32. y0+sinh x

cosh xy= cosh x,µ=eR(sinh x/ cosh x)dx =eln cosh x= cosh x,

(cosh x)y=Zcosh2xdx=Z1

2(cosh 2x+1)dx =1

4sinh 2x+1

2x+C=1

2sinh xcosh x+1

2x+C,

y=1

2sinh x+1

2xsech x+Csech x,1

4=C,y=1

2sinh x+1

2xsech x+1

4sech x

-2 2

-2

y = –0.5x2

y = –1.5x2

y = x2

y = 2x2

y = 2.5x2

y = 0

y = –3x2

33. (a) dy

y=dx

2x,ln |y|=1

2ln |x|+C1,|y|=C|x|1/2;

by inspection y= 0 is also a solution.

(b) 2=C(1)2,C =2,x =2y

-3 3

-3

yy = √9 – x2

y = –√6.25 – x2

y = –√4 – x2

y = –√1 – x2

y = √2.25 – x2

y = √0.25 – x2

34. (a) ydy =−x dx, y2

2=−x2

2+C1,y =±pC2−x

(b) y=√25 −x2

Exercise Set 10.1 346

35. dy

y=−xdx

2+4,

ln |y|=−1

2ln(x2+4)+C

y=C

√

2+4

1.5

-1

-2 2

C = –1

C = 1

C = 0 C = 2

C = –2

36. y0+2y=3e

,µ=e

2Rdt =e2t,

dtye2t=3e

,ye

2t=e

3t+C,

y=et+Ce−2t

C = 2

C = 1

C = 0

C = –1

C = –2

100

-100

-2 2

37. (1 −y2)dy =x2dx,

y−y3

3=x3

3+C1,x

3+y

3−3y=C

-2 2

-2

38. 1

y+ydy =dx, ln |y|+y2

2=x+C1,

yey2/2=±eC1ex=Cex

-5 5

-3

41. dy

dx =xey,e

−ydy =x dx, −e−y=x2

2+C, x = 2 when y=0so−1=2+C, C =−3,x

2+2e

−y=6

0 1.6

42. dy

dx =3x2

2y,2ydy =3x

2dx, y2=x3+C, 1=1+C, C =0,

2=x

,y=x

3/2passes through (1,1).

43. dy

dt = rate in −rate out, where yis the amount of salt at time t,

dt = (4)(2) −y

50(2)=8−1

25y, so dy

dt +1

25y= 8 and y(0) = 25.

µ=eR(1/25)dt =et/25,et/25y=Z8et/25dt = 200et/25 +C,

y= 200 + Ce−t/25, 25 = 200 + C,C=−175,

(a) y= 200 −175e−t/25 oz (b) when t= 25, y= 200 −175e−1≈136 oz

347 Chapter 10

44. dy

dt = (5)(10) −y

200(10) = 50 −1

20y, so dy

dt +1

20y= 50 and y(0)=0.

µ=e

20 dt =et/20,et/20y=Z50et/20dt = 1000et/20 +C,

y= 1000 + Ce−t/20, 0 = 1000 + C,C=−1000;

(a) y= 1000 −1000e−t/20 lb (b) when t= 30, y= 1000 −1000e−1.5≈777 lb

45. The volume Vof the (polluted) water is V(t) = 500 + (20 −10)t=500+10t;

if y(t) is the number of pounds of particulate matter in the water,

then y(0) = 50, and dy

dt =0−10 y

V=−1

50+ty, dy

dt +1

50+ty=0; µ=e

Rdt

50+t=50+t;

dt[(50 + t)y]=0,(50 + t)y=C, 2500 = 50y(0) = C, y(t) = 2500/(50 + t).

The tank reaches the point of overﬂowing when V=500+10t= 1000, t= 50 min, so

y= 2500/(50 + 50) = 25 lb.

46. The volume of the lake (in gallons) is V= 246πr2h= 246π(15)23 = 166,050π. Let y(t) denote

the number of pounds of mercury salts at time t, then dy

dt =0−103y

V=−y

166.05πlb/h and

y0=10

−5

V=1.6605πlb; dy

y=−dt

166.05π,ln y=−t

166.05π+C1,y=Ce−t/(166.05π), and

C=y(0) = y0=1.6605π,y=1.6605πe−t/(166.05π).

y(t)

5.2066

5.1967

5.1867

5.1768

5.1669

5.1570

5.1471

5.1372

5.1274

5.1176

5.1078

5.0980

47. (a) dv

dt +c

mv=−g, µ =e(c/m)Rdt =ect/m,d

dt hvect/mi=−gect/m,ve

ct/m =−gm

cect/m +C,

v=−gm

c+Ce−ct/m, but v0=v(0) = −gm

c+C, C =v0+gm

c,v =−gm

c+v0+gm

ce−ct/m

(b) Replace mg

cwith vτand −ct/m with −gt/vτin (23).

τt−(v

0+v

τ)

−gt/vτ;

s0=s(0) = C−(v0+vτ)vτ

g,C=s

+(v0+vτ)vτ

g,s(t)=s

0−v

τt+v

g(v

0+v

τ)1−e

−gt/vτ

48. Given m= 240,g =32,v

τ=mg/c: with a closed parachute vτ= 120 so c= 64, and with an open

parachute vτ= 24, c= 320.

(a) Let tdenote time elapsed in seconds after the moment of the drop. From Exercise 47(b),

while the parachute is closed

v(t)=e

−gt/vτ(v0+vτ)−vτ=e−32t/120 (0 + 120) −120 = 120 e−4t/15 −1and thus

v(25) = 120 e−20/3−1≈−119.85, so the parachutist is falling at a speed of 119.85 ft/s

when the parachute opens. From Exercise 47(c), s(t)=s

0−120t+120

32 120 1−e−4t/15,

s(25) = 10000 −120 ·25 + 450 1−e−20/3≈7449.43 ft.

(b) If tdenotes time elapsed after the parachute opens, then, by Exercise 47(c),

s(t) = 7449.43 −24t+24

32 (−119.85 + 24) 1−e−32t/24= 0, with the solution (Newton’s

Method) t= 310.42 s, so the sky diver is in the air for about 25 + 310 = 335 s.

Exercise Set 10.1 348

49. dI

dt +R

LI=V(t)

L,µ=e

(R/L)Rdt =eRt/L,d

dt(eRt/LI)=V(t)

Rt/L,

IeRt/L =I(0) + 1

LZt

0V(u)eRu/Ldu, I(t)=I(0)e−Rt/L +1

Le−Rt/L Zt

0V(u)eRu/Ldu.

(a) I(t)=1

−5t/2Zt

012e5u/2du =6

5e−5t/2e5u/2t

51−e−5t/2A.

(b) lim

t→+∞I(t)=6

50. From Exercise 49 and Endpaper Table #42,

I(t)=15e

−2t+1

−2tZt

03e

2usin udu=15e

−2t+e

−2te

5(2 sin u−cos u)t

=15e

−2t+1

(2 sin t−cos t)+1

−2t

51. (a) dv

dt =ck

m0−kt −g, v =−cln(m0−kt)−gt +C;v= 0 when t=0so0=−cln m0+C,

C=cln m0,v =cln m0−cln(m0−kt)−gt =cln m0

m0−kt −gt.

(b) m0−kt =0.2m

0when t= 100 so

v= 2500 ln m0

0.2m0−9.8(100) = 2500 ln 5 −980 ≈3044 m/s.

52. (a) By the chain rule, dv

dt =dv

dxvso mdv

dt =mv dv

dx.

(b) mv

kv2+mg dv =−dx, m

2kln(kv2+mg)=−x+C;v=v

0when x=0so

C=m

2kln(kv2

0+mg),m

2kln(kv2+mg)=−x+m

2kln(kv2

0+mg),x=m

2kln kv2

0+mg

kv2+mg .

max =m

2klnkv2

0+mg

mg =3.56 ×10−3

2(7.3×10−6)ln(7.3×10−6)(988)2+(3.56 ×10−3)(9.8)

(3.56 ×10−3)(9.8) ≈1298 m

53. (a) A(h)=π(1)2=π, π dh

dt =−0.025√h, π

√hdh =−0.025dt, 2π√h=−0.025t+C;h= 4 when

t=0,so4π=C, 2π√h=−0.025t+4π, √h=2−0.025

2πt, h ≈(2 −0.003979t)2.

(b) h= 0 when t≈2/0.003979 ≈502.6s≈8.4 min.

h − 22

2√4 − (h − 2)2

54. (a) A(h)=6h2

p4−(h−2)2i=12

p4h−h

12p4h−h2dh

dt =−0.025√h,12

√

4−hdh=−0.025dt,

−8(4 −h)3/2=−0.025t+C;h= 4 when t=0soC=0,

(4 −h)3/2=(0.025/8)t,4−h=(0.025/8)2/3t2/3,

h≈4−0.021375t2/3ft

(b) h= 0 when t=8

0.025(4 −0)3/2= 2560 s ≈42.7 min

349 Chapter 10

55. dv

dt =−0.04v2,1

v2dv =−0.04dt, −1

v=−0.04t+C;v= 50 when t=0so−1

50 =C,

−1

v=−0.04t−1

50,v =50

2t+1 cm/s. But v=dx

dt so dx

dt =50

2t+1,x = 25 ln(2t+1)+C

1;x=0

when t=0soC

1=0,x = 25 ln(2t+ 1) cm.

56. dv

dt =−0.02√v, 1

√vdv =−0.02dt, 2√v=−0.02t+C;v= 9 when t=0so6=C,

√

v=−0.02t+6,v =(3−0.01t)2cm/s. But v=dx

dt so dx

dt =(3−0.01t)2,

x=−100

3(3 −0.01t)3+C1;x= 0 when t=0soC

1= 900,x= 900 −100

3(3 −0.01t)3cm.

57. Diﬀerentiate to get dy

dx =−sin x+e−x2,y(0)=1.

58. Zh(y)dy =Zh(y(x))y0(x)dx; since h(y)dy

dx =g(x) it follows that Zh(y)dy =Zg(x)dx.

EXERCISE SET 10.2

1234

y2.

1234

y3.

-1

y(0) = 1

y(0) = 2

y(0) = –1

4. dy

dx +y=1,µ =e

Rdx =ex,

dx[yex]=e

yex=ex+C, y =1+Ce−x

(a) −1=1+C, C =−2,y =1−2e

−x

(b) 1=1+C, C =0,y =1

−x

-2 2

-10

y(0) = –1

y(–1) = 0 y(1) = 1

6. dy

dx −2y=−x, µ =e−2Rdx =e−2x,d

dx ye−2x=−xe−2x,ye

−2x=1

(2x+1)e

−2x+C,

y=1

4(2x+1)+Ce2x

(a) 1=3/4+Ce2,C=1/4e

,y=1

(2x+1)+1

2x−2

(b) −1=1/4+C, C =−5/4,y =1

4(2x+1)−5

/4,y=1

(2x+1)+1

2x+2

Exercise Set 10.2 350

7. lim

x→+∞y=1 8. lim

x→+∞y=(+∞if y0≥1/4

−∞,if y0<1/4

9. (a) IV, since the slope is positive for x>0 and negative for x<0.

(b) VI, since the slope is positive for y>0 and negative for y<0.

(d) II, since the slope changes sign when crossing the lines y=±1.

(e) I, since the slope can be positive or negative in each quadrant but is not periodic.

(f) III, since the slope is periodic in both xand y.

11. (a) y0=1,

n+1 =yn+(x

n+y

)(0.2)=(x

n+6y

)/5 n

0.2

1.20

2345

0.4

1.48

0.6

1.86

0.8

2.35

1.0

2.98

(b) y0−y=x, µ =e−x,d

dx ye−x=xe−x,

ye−x=−(x+1)e

−x+C, 1=−1+C,

C=2,y =−(x+1)+2e

y(x

)

abs. error

perc. error

0.2

1.24

0.04

7 9 11 13

0.4

1.58

0.10

0.6

2.04

0.19

0.8

2.65

0.30

1.0

3.44

0.46

(c)

0.2 0.4 0.6 0.8 1

12. h=0.1,y

n+1 =(x

n+11y

)/10

1.00

0.1

1.10

2345

0.2

1.22

0.3

1.36

0.4

1.53

0.5

1.72

0.6

1.94

0.7

2.20

0.8

2.49

0.9

2.82

1.0

3.19

In Exercise 11, y(1) ≈2.98; in Exercise 12, y(1) ≈3.19; the true solution is y(1) ≈3.44; so the

absolute errors are approximately 0.46 and 0.25 respectively.

2413

13. y0=1,y

n+1 =yn+√yn/2

0.5

1.50

2.11

1.5

2.84 3.68

2.5

4.64

5.72

3.5

6.91

8.23

351 Chapter 10

14. y0=1,y

n+1 =yn+(x

n−y

)/4

0.25

0.75

2345

0.50

0.67

0.75

0.68

1.00

0.75

1.25

0.86

1.50

0.99

1.75

1.12

2.00

1.24

0.5 1 1.5 2

0.5

1.5

15. y0=1,y

n+1 =yn+1

2sin yn

0.5

1.42

1.92

1.5

2.39 2.73

16. y0=0,y

n+1 =yn+e−yn/10

0.1

0.10

2345

0.2

0.19

0.3

0.27

0.4

0.35

0.5

0.42

0.6

0.49

0.7

0.55

0.8

0.60

0.9

0.66

1.0

0.71

17. h=1/5,y

0=1,y

n+1 =yn+1

5cos(2πn/5)

1.00

0.2

1.06

0.4

0.90

234

0.6

0.74

0.8

0.80

1.0

1.00

18. (a) By inspection, dy

dx =e−x2and y(0)=0.

(b) yn+1 =yn+e−x2

n/20 = yn+e−(n/20)2/20 and y20 =0.7625. From a CAS, y(1)=0.7468.

19. (b) ydy =−x dx, y2/2=−x

/2+C

1,x

2+y

2=C;ify(0) = 1 then C=1soy(1/2) = √3/2.

20. (a) y0=1,y

n+1 =yn+(

√

/2)h

h=0.2: y

n+1 =yn+√yn/10; y5≈1.5489

h=0.1:y

n+1 =yn+√yn/20; y10 ≈1.5556

h=0.05 : yn+1 =yn+√yn/40; y20 ≈1.5590

√y=1

2dx, 2√y=x/2+C, 2=C,

√y=x/4+1,y =(x/4+1)

y(1)=25/16=1.5625

Exercise Set 10.3 352

EXERCISE SET 10.3

1. (a) dy

dt =ky2,y(0) = y0,k > 0(b) dy

dt =−ky2,y(0) = y0,k > 0

3. (a) ds

dt =1

2s(b) d2s

dt2=2

4. (a) dv

dt =−2v2(b) d2s

dt2=−2ds

dt 2

5. (a) dy

dt =0.01y, y0=10,000 (b) y=10,000et/100

kln 2 = 1

0.01 ln 2 ≈69.31 hr (d) 45,000=10,000et/100,

t= 100 ln 45,000

10,000 ≈150.41 hr

6. k=1

Tln 2 = 1

20 ln 2

(a) dy

dt = ((ln 2)/20)y, y(0)=1 (b) y(t)=e

t(ln 2)/20 =2

t/20

t/20,

t=20

ln 106

ln 2 ≈398.63 min

7. (a) dy

dt =−ky, y(0)=5.0×107;3.83 = T=1

kln 2, so k=ln 2

3.83 ≈0.1810

(b) y=5.0×107e−0.181t

(d) y(t)=0.1y

0=y

−kt,−kt =ln0.1,t=−ln 0.1

0.1810 =12.72 days

8. (a) k=1

Tln 2 = 1

140 ln 2 ≈0.0050, so dy

dt =−0.0050y, y0= 10.

(b) y=10e

−0.0050t

−0.35 ≈7 mg.

(d) 0.3y0=y0e−kt,t=−ln 0.3

0.0050 ≈243.2days

9. 100e0.02t= 5000, e0.02t= 50, t=1

0.02 ln 50 ≈196 days

10. y=10,000ekt, but y=12,000 when t=10so10,000e10k=12,000, k=1

10 ln 1.2. y=20,000 when

2=e

kt,t=ln 2

k=10 ln 2

ln 1.2≈38, in the year 2025.

11. y(t)=y

−kt =10.0e

−kt,3.5=10.0e

−k(5),k=−

ln 3.5

10.0≈0.2100,T=1

ln 2 ≈3.30 days

353 Chapter 10

12. dy

dt =y0e−kt,0.6y0=y0e−5k,k=−

ln 0.6≈0.10

(a) T=ln 2

k≈6.8yr

(b) y(t)≈y0e−0.10t,y

y0≈e−0.10t,soe

−0.10t×100 percent will remain.

13. (a) k=ln 2

5≈0.1386; y≈2e0.1386t(b) y(t)=5e

0.015t

,100 = y0e10k. Divide: 100 = e9k,k=1

ln 100 ≈0.5117,

y≈y0e0.5117t; also y(1)=1,so y0=e−0.5117 ≈0.5995,y ≈0.5995e0.5117t.

(d) k=ln 2

T≈0.1386,1=y(1) ≈y0e0.1386,y

0≈e

−0.1386 ≈0.8706,y≈0.8706e0.1386t

14. (a) k=ln 2

T≈0.1386,y≈10e−0.1386t(b) y=10e

−0.015t

−10k. Divide: e9k= 100,k=1

ln 100 ≈0.5117;

y0=e10k≈e5.117 ≈166.81,y= 166.81e−0.5117t.

(d) k=ln 2

T≈0.1386,10 = y(1) ≈y0e−0.1386,y

0≈10e0.1386 ≈11.4866,y≈11.4866e−0.1386t

16. (a) None; the half-life is independent of the initial amount.

(b) kT =ln2,soTis inversely proportional to k.

17. (a) T=ln 2

k; and ln 2 ≈0.6931. If kis measured in percent, k0= 100k,

then T=ln 2

k≈69.31

k0≈70

k0.

(b) 70 yr (c) 20 yr (d) 7%

18. Let y=y0ekt with y=y1when t=t1and y=3y

1when t=t1+T; then y0ekt1=y1(i) and

y0ek(t1+T)=3y

1(ii). Divide (ii) by (i) to get ekT =3,T=1

kln 3.

19. From (12), y(t)=y

−0.000121t.If0.27 = y(t)

y0=e−0.000121tthen t=−ln 0.27

0.000121 ≈10,820 yrs,

and if 0.30 = y(t)

y0then t=−ln 0.30

0.000121 ≈9950, or roughly between 9000 B.C. and 8000 B.C.

20. (a) 1

0 50000

(b) t= 1988 yields

y/y0=e−0.000121(1988) ≈79%.

21. y0≈2,L≈8; since the curve y=2·8

2+6e

−kt passes through the point (2,4), 4 = 16

2+6e

−2k,

−2k=2,k=1

ln 3 ≈0.5493.

Exercise Set 10.3 354

22. y0≈400,L≈1000; since the curve y=400,000

400 + 600e−kt passes through the point (200,600),

600 = 400,000

400 + 600e−200k,600e−200k=800

3,k=1

200 ln 2.25 ≈0.00405.

23. (a) y0=5 (b) L=12 (c) k=1

(d) L/2=6= 60

5+7e

−t,5+7e

−t=10,t=−ln(5/7) ≈0.3365

(e) dy

dt =1

12y(12 −y),y(0)=5

24. (a) y0=1 (b) L= 1000 (c) k=0.9

(d) 750 = 1000

1 + 999e−0.9t,3(1 + 999e−0.9t)=4,t=1

0.9

ln(3 ·999) ≈8.8949

(e) dy

dt =0.9

1000y(1000 −y),y(0)=1

25. See (13):

(a) L=10 (b) k=10

dt = 10(1 −0.1y)y=25−(y−5)2is maximized when y=5.

26. dy

dt =50y1−1

50,000y; from (13), k=50,L=50,000.

(a) L=50,000 (b) k=50

dt is maximized when 0 = d

dy dy

dt =50−y/500,y =25,000

27. Assume y(t) students have had the ﬂu tdays after semester break. Then y(0)=20,y(5) = 35.

(a) dy

dt =ky(L−y)=ky(1000 −y),y

0=20

(b) Part (a) has solution y=20000

20 + 980e−1000kt =1000

1+49e

−1000kt ;

35 = 1000

1+49e

−5000k,k=0.000115,y≈1000

1+49e

−0.115t.

y(t)

(d)

36912

100

355 Chapter 10

28. (a) dp

dh =−kh, p(0) = p0

(b) p0=1,sop=e

−kh, but p=0.83 when h= 5000 thus e−5000k=0.83,

k=−ln 0.83

5000 ≈0.0000373,p≈e

−0.0000373hatm.

29. (a) dT

dt =−k(T−21),T(0)=95,dT

T−21 =−k dt, ln(T−21) = −kt +C1,

T=21+e

−kt =21+Ce−kt,95 = T(0)=21+C, C =74,T=21+74e

−kt

(b) 85 = T(1)=21+74e

−k

,k=−ln 64

74 =−ln 32

37,T=21+74e

tln(32/37) =21+7432

37t

T= 51 when 30

74 =32

37t

,t=ln(30/74)

ln(32/37) ≈6.22 min

30. dT

dt =k(70 −T),T(0) = 40; −ln(70 −T)=kt +C, 70 −T=e−kte−C,T= 40 when t=0,so

30 = e−C,T=70−30e−kt;52=T(1)=70−30e−k,k=−ln 70 −52

30 =ln5

3≈0.5,

T≈70 −30e−0.5t

31. Let Tdenote the body temperature of McHam’s body at time t, the number of hours elapsed after

10:06 P.M.; then dT

dt =−k(T−72),dT

T−72 =−kdt, ln(T−72) = −kt +C, T =72+e

−kt,

77.9=72+e

C=5.9,T=72+5.9e

−kt,75.6=72+5.9e

−k

,k=−ln 3.6

5.9≈0.4940,

T= 72+5.9e−0.4940t. McHam’s body temperature was last 98.6◦when t=−ln(26.6/5.9)

0.4940 ≈−3.05,

so around 3 hours and 3 minutes before 10:06; the death took place at approximately 7:03 P.M.,

while Moore was on stage.

32. If T0<T

athen dT

dt =k(Ta−T) where k>0. If T0>T

athen dT

dt =−k(T−Ta) where k>0;

both cases yield T(t)=T

a+(T

0−T

−kt with k>0.

33. k/m =0.25/1=0.25

(a) From (21), y=0.3 cos(t/2) (b) T=2π·2=4πs, f=1/T =1/(4π)Hz

(c)

6cio

-0.3

0.3

y(d) y= 0 at the equilibrium position,

so t/2=π/2,t=πs.

(e) t/2=πat the maximum position below

the equlibrium position, so t=2πs.

34. 64 = w=−mg, m =2,k/m=0.25/2=1/8,pk/m =1/(2√2)

(a) From (21), y= cos(t/(2√2)) (b) T=2π

k=2π(2√2)=4π

√

2s,

f=1/T =1/(4π√2) Hz

Exercise Set 10.3 356

(c)

-1

2p6p10p

(d) y= 0 at the equilibrium position,

so t/(2√2) = π/2,t=π

√2s

(e) t/(2√2) = π, t =2π

√

35. l=0.05,k/m=g/l =9.8/0.05 = 196 s−2

(a) From (21), y=−0.12 cos 14t.(b) T=2π

pm/k =2π/14 = π/7s,

f=7/π Hz

(c)

-0.15

0.15

(d) 14t=π/2,t=π/28 s

(e) 14t=π, t =π/14 s

36. l=0.5,k/m=g/l =32/0.5=64,pk/m =8

(a) From (21), y=−1.5 cos 8t.(b) T=2π

pm/k =2π/8=π/4s;

f=1/T =4/π Hz

(c)

-2

y(d) 8t=π/2,t=π/16 s

(e) 8t=π, t =π/8s

37. Assume y=y0cos rk

mt, so v=dy

dt =−y0rk

msin rk

(a) The maximum speed occurs when sin rk

mt=±1,rk

mt=nπ +π/2,

so cos rk

mt=0,y=0.

(b) The minimum speed occurs when sin rk

mt=0,rk

mt=nπ,socos

mt=±1,y=±y

38. (a) T=2π

k,k=4π

2m=4π

,sok=4π

9=4π

w+4

25 ,25w=9(w+4),

25w=9w+36,w =9

4,k =4π

9=4π

4=π2

(b) From part (a), w=1

357 Chapter 10

39. By Hooke’s Law, F(t)=−kx(t), since the only force is the restoring force of the spring. Newton’s

Second Law gives F(t)=mx00(t), so mx00(t)+kx(t)=0,x(0) = x0,x

0(0)=0.

40. 0=v(0) = y0(0) = c2rk

m,soc

2=0; y

0=y(0) = c1,soy=y

0cos rk

mt.

41. (a) y=y0bt=y0etln b=y0ekt with k=lnb>0 since b>1.

(b) y=y0bt=y0etln b=y0e−kt with k=−ln b>0 since 0 <b<1.

tln 2 (d) y= 4(0.5t)=4e

tln 0.5=4e

−tln 2

42. If y=y0ekt and y=y1=y0ekt1then y1/y0=ekt1,k=ln(y1/y0)

t1;ify=y

−kt and

y=y1=e−kt1then y1/y0=e−kt1,k=−

ln(y1/y0)

t1.

CHAPTER 10 SUPPLEMENTARY EXERCISES

4. The diﬀerential equation in part (c) is not separable; the others are.

5. (a) linear (b) linear and separable (c) separable (d) neither

6. IF: µ=e−2x2,d

dx hye−2x2i=xe−2x2,ye

−2x

2=−

−2x

2+C, y =−1

4+Ce2x2

Sep of var: dy

4y+1 =x dx, 1

4ln |4y+1|=x

2+C

1,4y+1 = ±e

2=C

2;y=−1

4+Ce2x2

7. The parabola ky(L−y) opens down and has its maximum midway between the y-intercepts, that

is, at the point y=1

2(0+L)=L/2, where dy

dt =k(L/2)2=kL2/4.

8. (a) If y=y0ekt,then y1=y0ekt1,y

2=y

kt2, divide: y2/y1=ek(t2−t1),k=1

2−t

1ln(y2/y1),

T=ln 2

k=(t2−t1)ln2

ln(y2/y1).Ify=y

−kt, then y1=y0e−kt1,y

2=y

−kt2,

y2/y1=e−k(t2−t1),k =−1

2−t

1ln(y2/y1),T=ln 2

k=−(t2−t1)ln2

ln(y2/y1).

In either case, Tis positive, so T=

(t2−t1)ln2

ln(y2/y1).

(b) In part (a) assume t2=t1+ 1 and y2=1.25y1. Then T=ln 2

ln 1.25 ≈3.1h.

9. dV

dt =−kS; but V=4π

3r3,dV

dt =4πr2dr

dt ,S=4πr2,sodr/dt =−k, r =−kt +C, 4=C,

r=−kt +4,3=−k+4,k =1,r =4−tm.

Chapter 10 Supplementary Exercises 358

10. Assume the tank contains y(t) oz of salt at time t. Then y0= 0 and for 0 <t<15,

dt =5·10 −y

100010 = (50 −y/100) oz/min, with solution y= 5000 + Ce−t/100. But y(0) = 0 so

C=−5000, y= 5000(1 −e−t/100) for 0 ≤t≤15, and y(15) = 5000(1 −e−0.15). For 15 <t<30,

dt =0−y

10005,y=C

−t/200,C

−0.075 =y(15) = 5000(1−e−0.15),C

1= 5000(e0.075 −e−0.075),

y= 5000(e0.075 −e−0.075)e−t/100,y(30) = 5000(e0.075 −e−0.075)e−0.3≈556.13 oz.

11. (a) Assume the air contains y(t)ft

3of carbon monoxide at time t. Then y0= 0 and for

t>0, dy

dt =0.04(0.1) −y

1200(0.1)=1/250 −y/12000,d

dt hyet/12000i=1

250et/12000,

yet/12000 =48e

t/12000 +C, y(0)=0,C=−48; y= 48(1 −e−t/12000). Thus the percentage

of carbon monoxide is P=y

1200100 = 4(1 −e−t/12000) percent.

(b) 0.012 = 4(1 −e−t/12000),t=35.95 min

12. dy

y2+1 =dx, tan−1y=x+C, π/4=C;y= tan(x+π/4)

13. 1

y5+1

ydy =dx

x,−1

4y−4+ln|y|=ln|x|+C;−1

4=C, y−4+ 4 ln(x/y)=1

14. dy

dx +2

xy=4x, µ =eR(2/x)dx =x2,d

dx yx2=4x

,yx

2=x

4+C, y =x2+Cx−2,

2=y(1)=1+C, C =1,y=x

2+1/x2

15. dy

y2= 4 sec22x dx, −1

y= 2 tan 2x+C, −1 = 2 tan 2π

8+C= 2 tan π

4+C=2+C, C =−3,

y=1

3−2 tan 2x

16. dy

y2−5y+6 =dx, dy

(y−3)(y−2) =dx, 1

y−3−1

y−2dy =dx, ln 

y−3

y−2=x+C1,

y−3

y−2=Cex;y=ln2ifx=0,so C=ln 2 −3

ln 2 −2;y=3−2Cex

1−cex=3ln2−6−(2 ln 2 −6)ex

ln 2 −2−(ln 2 −3)ex

17. (a) µ=e−Rdx =e−x,d

dx ye−x=xe−xsin 3x,

ye−x=Zxe−xsin 3xdx =−3

10x−3

50e−xcos 3x+−1

10x+2

25e−xsin 3x+C;

1=y(0) = −3

50 +C, C =53

50,y=

−3

10x−3

50cos 3x+−1

10x+2

25sin 3x+53

50ex

(c)

-10 -2

-2

359 Chapter 10

19. (a) Let T1= 5730 −40 = 5690,k

1=ln 2

T1≈0.00012182; T2= 5730 + 40 = 5770,k

2≈0.00012013.

With y/y0=0.92,0.93,t

1=−1

1ln y

y0= 684.5,595.7; t2=−1

k2ln(y/y0) = 694.1,604.1; in

1988 the shroud was at most 695 years old, which places its creation in or after the year 1293.

(b) If the true half-life is Twith decay rate kand solution y(t)=y

−kt, and if the half-life is

taken to be T1=T(1+r/100) with decay rate k1and solution y1(t)=y

−k

, then

k1=ln 2

T1=ln 2

T(1+r/100) =k

1+r/100 =100k

100 + r;k−k1=k−100k

100 + r=kr

100 + rand the

percentage error is given by 100 

y1−y

y= 100 e(k−k1)t−1= 100 ektr/(100+r)−1percent.

20. (a) yn+1 =yn+0.1(1+5t

n−y

),y

0=5

5.00

1.1

5.10

2345

1.2

5.24

1.3

5.42

1.4

5.62

1.5

5.86

1.6

6.13

1.7

6.41

1.8

6.72

1.9

7.05

7.39

(b) The true solution is y(t)=5t−4+4e

1−t

, so the percentage errors are given by

y(tn)

abs. error

rel. error (%)

5.00

0.00

1.1

5.10

5.12

0.02

0.38

1.2

5.24

5.27

0.03

0.66

1.3

5.42

5.46

0.05

0.87

1.4

5.62

5.68

0.06

1.00

1.5

5.86

5.93

0.06

1.08

1.6

6.13

6.20

0.07

1.12

1.7

6.41

6.49

0.07

1.13

1.8

6.72

6.80

0.08

1.11

1.9

7.05

7.13

0.08

1.07

7.39

7.47

0.08

1.03

21. (b) y=C1ex+C2e−x

0(0) = C1−C2;C2=0,C

1=1,y =e

22. (a) 2ydy =dx, y2=x+C;ify(0) = 1 then C=1,y

2=x+1,y =√x+1; ify(0) = −1 then

C=1,y

2=x+1,y =−√x+1.

-1 1

-1

-1 1

-1

-1 1

(b) dy

y2=−2x dx, −1

y=−x2+C, −1=C, y =1/(x

2+1)

23. (a) Use (15) in Section 10.3 with y0=19,L= 95: y(t)= 1805

19+76e

−kt ,25 = y(1) = 1805

19+76e

−k,

k≈0.3567; when 0.8L=y(t)= y

19+76e

−kt ,19+76e

−kt =5

4y0=95

4,t≈7.77 yr.

Chapter 10 Supplementary Exercises 360

(b) From (13), dy

dt =k1−y

95y, y(0) = 19.

24. (a) y0=y(0) = c1,v

0=y

(0) = c2rk

m,c

,y=y

0cos rk

mt+v0rm

ksin rk

1.1

-1.1

0 3.1

(b) l=0.5,k/m=g/l =9.8/0.5=19.6,

y=−cos(√19.6t)+0.25 1

√19.6sin(√19.6t)

19.6 sin √19.6t+0.25 cos √19.6t= 0 when tan √19.6t=−0.25

√19.6,so

sin √19.6t=±0.25

√19.6625,cos √19.6t=∓√19.6

√19.6625,

|y(t)|=√19.6

√19.6625 +0.25

√19.6

0.25

√19.6625 =√19.6625

√19.6

≈1.0016 m is the maximum displacement.

25. y=y0cos rk

mt, T =2π

k,y=y

0cos 2πt

(a) v=y0(t)=−2π

0sin 2πt

Thas maximum magnitude 2π|y0|/T and occurs when

2πt/T =nπ +π/2,y=y

0cos(nπ +π/2)=0.

(b) a=h00(t)=−4π

0cos 2πt

Thas maximum magnitude 4π2|y0|/T 2and occurs when

2πt/T =jπ, y =y0cos jπ =±y0.

26. (a) In tyears the interest will be compounded nt times at an interest rate of r/n each time. The

value at the end of 1 interval is P+(r/n)P=P(1 + r/n), at the end of 2 intervals it is

P(1+r/n)+(r/n)P(1+r/n)=P(1+r/n)2, and continuing in this fashion the value at the

end of nt intervals is P(1+r/n)nt.

(b) Let x=r/n, then n=r/x and

lim

n→+∞P(1+r/n)nt = lim

x→0+P(1+x)rt/x = lim

x→0+P[(1 + x)1/x]rt =Pe

rt.

27. (a) A= 1000e(0.08)(5) = 1000e0.4≈$1,491.82

(b) Pe

(0.08)(10) =10,000, Pe

0.8=10,000, P=10,000e−0.8≈$4,493.29

361

CHAPTER 11

Inﬁnite Series

EXERCISE SET 11.1

1. (a) 1

3n−1(b) (−1)n−1

3n−1(c) 2n−1

2n(d) n2

π1/(n+1)

2. (a) (−r)n−1;(−r)

n(b) (−1)n+1rn;(−1)nrn+1

3. (a) 2,0,2,0(b) 1,−1,1,−1(c) 2(1+(−1)n); 2 + 2 cos nπ

4. (a) (2n)! (b) (2n−1)!

5. 1/3, 2/4, 3/5, 4/6, 5/7,...; lim

n→+∞

n+2 = 1, converges

6. 1/3, 4/5, 9/7, 16/9, 25/11,...; lim

n→+∞

2n+1 =+∞, diverges

7. 2,2,2,2,2,...; lim

n→+∞2 = 2, converges

8. ln 1, ln 1

2,ln1

,ln1

,...; lim

n→+∞ln(1/n)=−∞, diverges

9. ln 1

1,ln 2

2,ln 3

3,ln 4

4,ln 5

5,...;

lim

n→+∞

ln n

n= lim

n→+∞

n=0apply L’Hˆopital’s Rule to ln x

x, converges

10. sin π, 2 sin(π/2), 3 sin(π/3), 4 sin(π/4), 5 sin(π/5),...;

lim

n→+∞nsin(π/n) = lim

n→+∞

sin(π/n)

1/n = lim

n→+∞

(−π/n2) cos(π/n)

−1/n2=π, converges

11. 0,2,0,2,0,...; diverges

12. 1, −1/4, 1/9, −1/16, 1/25,...; lim

n→+∞

(−1)n+1

n2= 0, converges

13. −1, 16/9, −54/28, 128/65, −250/126,...; diverges because odd-numbered terms approach

−2, even-numbered terms approach 2.

14. 1/2, 2/4, 3/8, 4/16, 5/32,...; lim

n→+∞

2n= lim

n→+∞

2nln 2 = 0, converges

15. 6/2, 12/8, 20/18, 30/32, 42/50,...; lim

n→+∞

2(1+1/n)(1+2/n)=1/2, converges

16. π/4, π2/42,π3/43,π4/44,π5/45,...; lim

n→+∞(π/4)n= 0, converges

17. cos(3), cos(3/2), cos(1), cos(3/4), cos(3/5),...; lim

n→+∞cos(3/n) = 1, converges

18. 0, −1, 0, 1, 0,...; diverges

Exercise Set 11.1 362

19. e−1,4e

−2

,9e

−3

,16e

−4

,25e

−5

,...; lim

x→+∞x2e−x= lim

x→+∞

ex= 0, so lim

n→+∞n2e−n= 0, converges

20. 1, √10 −2, √18 −3, √28 −4, √40 −5,...;

lim

n→+∞(pn2+3n−n) = lim

n→+∞

√n2+3n+n= lim

n→+∞

p1+3/n +1 =3

2, converges

21. 2, (5/3)2,(6/4)3,(7/5)4,(8/6)5,...; let y=x+3

x+1x

, converges because

lim

x→+∞ln y= lim

x→+∞

ln x+3

x+1

1/x = lim

x→+∞

2x2

(x+ 1)(x+3) = 2, so lim

n→+∞n+3

n+1n

22. −1, 0, (1/3)3,(2/4)4,(3/5)5,...; let y=(1−2/x)x, converges because

lim

x→+∞ln y= lim

x→+∞

ln(1 −2/x)

1/x = lim

x→+∞−2

1−2/x =−2, lim

n→+∞(1 −2/n)n= lim

x→+∞y=e−2

23. 2n−1

2n+∞

n=1

; lim

n→+∞

2n−1

2n= 1, converges

24. n−1

n2+∞

n=1

; lim

n→+∞

n−1

n2= 0, converges 25. 1

3n+∞

n=1

; lim

n→+∞

3n= 0, converges

26. {(−1)nn}+∞

n=1; diverges because odd-numbered terms tend toward −∞,

even-numbered terms tend toward +∞.

27. 1

n−1

n+1+∞

n=1

; lim

n→+∞1

n−1

n+1= 0, converges

28. 3/2n−1+∞

n=1; lim

n→+∞3/2n−1= 0, converges

29. √n+1−√

n+2

+∞

n=1; converges because

lim

n→+∞(√n+1−√

n+ 2) = lim

n→+∞

(n+1)−(n+2)

√

n+1+√

n+2 = lim

n→+∞−1

√n+1+√

n+2 =0

30. (−1)n+1/3n+4+∞

n=1; lim

n→+∞(−1)n+1/3n+4 = 0, converges

32. lim

n→+∞

√n= 1, so lim

n→+∞

√n3=1

3=1

33. (a) 1,2,1,4,1,6(b) an=(n, n odd

1/2n,neven (c) an=(1/n, n odd

1/(n+1),neven

(d) In part (a) the sequence diverges, since the even terms diverge to +∞and the odd terms

equal 1; in part (b) the sequence diverges, since the odd terms diverge to +∞and the even

terms tend to zero; in part (c) lim

n→+∞an=0.

34. The even terms are zero, so the odd terms must converge to zero, and this is true if and only if

lim

n→+∞bn=0,or−1<b<1.

363 Chapter 11

35. lim

n→+∞yn+1 = lim

n→+∞

2(yn+p/yn), L=1

2(L+p/L), L2=p,L=±√p;

L=−√p(reject, because the terms in the sequence are positive) or L=√p; lim

n→+∞yn=√p.

36. (a) an+1 =√6+a

(b) lim

n→+∞an+1 = lim

n→+∞√6+a

,L=√

6+L,L

2−L−6=0,(L−3)(L+2)=0,

L=−2 (reject, because the terms in the sequence are positive) or L= 3; lim

n→+∞an=3.

37. (a) 1, 1

4+2

4,1

9+2

9+3

9,1

16 +2

16 +3

16 +4

16 =1,3

n2(1+2+···+n)= 1

n(n+1)= 1

n+1

n,lim

n→+∞an=1/2

38. (a) 1, 1

8+4

8,1

27 +4

27 +9

27,1

64 +4

64 +9

64 +16

64 =1,5

,14

27,15

n3(12+2

2+···+n

2)= 1

n(n+ 1)(2n+1)= 1

(n+ 1)(2n+1)

lim

n→+∞an= lim

n→+∞

6(1+1/n)(2+1/n)=1/3

39. Let an=0,b

n=sin2n

n,c

n=1

n; then an≤bn≤cn, lim

n→+∞an= lim

n→+∞cn= 0, so lim

n→+∞bn=0.

40. Let an=0,b

n=1+n

2nn

n=3

4n

; then (for n≥2), an≤bn≤n/2+n

2nn

lim

n→+∞an= lim

n→+∞cn= 0, so lim

n→+∞bn=0.

41. (a) a1=(0.5)2,a

2=a

1=(0.5)4,...,a

n=(0.5)2n

n→+∞an= lim

n→+∞e2nln(0.5) = 0, since ln(0.5) <0.

(d) Replace 0.5 in part (a) with a0; then the sequence converges for −1≤a0≤1, because if

a0=±1, then an= 1 for n≥1; if a0= 0 then an= 0 for n≥1; and if 0 <|a0|<1 then

a1=a2

0>0 and lim

n→+∞an= lim

n→+∞e2n−1ln a1= 0 since 0 <a

1<1.

42. f(0.2)=0.4,f(0.4)=0.8,f(0.8)=0.6,f(0.6)=0.2 and then the cycle repeats, so the sequence

does not converge.

43. (a) 30

(b) Let y=(2

x+3

)

1/x, lim

x→+∞ln y= lim

x→+∞

ln(2x+3

)

x= lim

x→+∞

2xln2+3

xln 3

2x+3

= lim

x→+∞

(2/3)xln2+ln3

(2/3)x+1 = ln 3, so lim

n→+∞(2n+3

)

1/n =eln 3 =3

Exercise Set 11.2 364

44. Let f(x)=1/(1+x), 0 ≤x≤1. Take ∆xk=1/n and x∗

k=k/n then

an=

k=1

1+(k/n)(1/n)=

k=1

1+x

∗

∆x

kso lim

n→+∞an=Z1

1+xdx = ln(1 + x)1

=ln2

45. an=1

n−1Zn

xdx =ln n

n−1, lim

n→+∞an= lim

n→+∞

ln n

n−1= lim

n→+∞

n=0,

apply L’Hˆopital’s Rule to ln n

n−1, converges

46. (a) If n≥1, then an+2 =an+1 +an,soa

n+2

an+1 =1+ a

n+1 .

n→+∞(an+2/an+1) = lim

n→+∞(an+1/an), L=1+1/L,L2−L−1=0,

L=(1±√

5)/2, so L=(1+√

5)/2 because the limit cannot be negative.

47. 

n−0=1

n<if n>1/

(a) 1/ =1/0.5=2,N=3 (b) 1/ =1/0.1 = 10, N=11

48. 

n+1−1



=1

n+1 <if n+1>1/,n>1/ −1

(a) 1/ −1=1/0.25 −1=3,N=4 (b) 1/ −1=1/0.1−1=9,N=10

49. (a) 

n−0=1

n<if n>1/, choose any N>1/.

(b) 

n+1−1



=1

n+1 <if n>1/ −1, choose any N>1/ −1.

50. If |r|<1 then lim

n→+∞rn=0;ifr>1 then lim

n→+∞rn=+∞,ifr<−1 then rnoscillates between

positive and negative values that grow in magnitude so lim

n→+∞rndoes not exist for |r|>1; if r=1

then lim

n→+∞1n=1;ifr=−1 then (−1)noscillates between −1 and 1 so lim

n→+∞(−1)ndoes not exist.

EXERCISE SET 11.2

1. an+1 −an=1

n+1−1

n=−1

n(n+1) <0 for n≥1, so strictly decreasing.

2. an+1 −an=(1−1

n+1)−(1 −1

n)= 1

n(n+1) >0 for n≥1, so strictly increasing.

3. an+1 −an=n+1

2n+3−n

2n+1 =1

(2n+ 1)(2n+3) >0 for n≥1, so strictly increasing.

365 Chapter 11

4. an+1 −an=n+1

4n+3−n

4n−1=−1

(4n−1)(4n+3) <0 for n≥1, so strictly decreasing.

5. an+1 −an=(n+1−2

n+1)−(n−2n)=1−2

n<0 for n≥1, so strictly decreasing.

6. an+1 −an=[(n+1)−(n+1)

2]−(n−n

2)=−2n<0 for n≥1, so strictly decreasing.

7. an+1

=(n+1)/(2n+3)

n/(2n+1) =(n+ 1)(2n+1)

n(2n+3) =2n

2+3n+1

2+3n>1 for n≥1, so strictly increasing.

8. an+1

=2n+1

1+2

n+1 ·1+2

n=2+2

n+1

1+2

n+1 =1+ 1

1+2

n+1 >1 for n≥1, so strictly increasing.

9. an+1

=(n+1)e

−(n+1)

ne−n=(1+1/n)e−1<1 for n≥1, so strictly decreasing.

10. an+1

=10n+1

(2n+ 2)! ·(2n)!

10n=10

(2n+ 2)(2n+1) <1 for n≥1, so strictly decreasing.

11. an+1

=(n+1)

n+1

(n+ 1)! ·n!

nn=(n+1)

n=(1+1/n)n>1 for n≥1, so strictly increasing.

12. an+1

=5n+1

2(n+1)2·2n2

5n=5

22n+1 <1 for n≥1, so strictly decreasing.

13. f(x)=x/(2x+ 1), f0(x)=1/(2x+1)

2>0 for x≥1, so strictly increasing.

14. f(x)=3−1/x,f0(x)=1/x2>0 for x≥1, so strictly increasing.

15. f(x)=1/(x+lnx), f0(x)=−1+1/x

(x+lnx)

2<0 for x≥1, so strictly decreasing.

16. f(x)=xe−2x,f0(x)=(1−2x)e

−2x<0 for x≥1, so strictly decreasing.

17. f(x)=ln(x+2)

x+2 ,f0(x)=1−ln(x+2)

(x+2)

2<0 for x≥1, so strictly decreasing.

18. f(x) = tan−1x,f0(x)=1/(1+x2)>0 for x≥1, so strictly increasing.

19. f(x)=2x

2−7x,f

0(x)=4x−7>0 for x≥2, so eventually strictly increasing.

20. f(x)=x

3−4x

0(x)=3x

2−8x=x(3x−8) >0 for x≥3, so eventually strictly increasing.

21. f(x)= x

2+10,f0(x)= 10 −x2

(x2+ 10)2<0 for x≥4, so eventually strictly decreasing.

22. f(x)=x+17

x,f0(x)=x

2−17

x2>0 for x≥5, so eventually strictly increasing.

23. an+1

=(n+ 1)!

3n+1 ·3n

n!=n+1

3>1 for n≥3, so eventually strictly increasing.

24. f(x)=x

−x

0(x)=x

(5 −x)e−x<0 for x≥6, so eventually strictly decreasing.

Exercise Set 11.2 366

25. (a) Yes: a monotone sequence is increasing or decreasing; if it is increasing, then it is increas-

ing and bounded above, so by Theorem 11.2.3 it converges; if decreasing, then use Theo-

rem 11.2.4. The limit lies in the interval [1,2].

(b) Such a sequence may converge, in which case, by the argument in Part (a), its limit is ≤2.

But convergence may not happen: for example, the sequence {−n}+∞

n=1 diverges.

26. (a) an+1 =|x|n+1

(n+ 1)! =|x|

n+1|x|

n!=|x|

n+1a

(b) an+1/an=|x|/(n+1)<1ifn>|x|−1.

Theorem 11.2.4 it converges.

(d) If lim

n→+∞an=Lthen from Part (a), L=|x|

lim

n→+∞(n+1)L=0.

(e) lim

n→+∞|x|n

n!= lim

n→+∞an=0

27. (a) √2, q2+√

2, r2+q2+√

(b) a1=√2<2soa

2=√

2+a

1<√

2+2=2,a

3=√

2+a

2<√

2 + 2 = 2, and so on

indeﬁnitely.

n+1 −a2

n=(2+a

)−a

n=2+a

n−a

n=(2−a

)(1 + an)

(d) an>0 and, from Part (b), an<2so2−a

n>0 and 1 + an>0 thus, from Part (c),

n+1 −a2

n>0, an+1 −an>0, an+1 >a

;{a

}is a strictly increasing sequence.

(e) The sequence is increasing and has 2 as an upper bound so it must converge to a limit L,

lim

n→+∞an+1 = lim

n→+∞√2+a

,L=√

2+L,L

2−L−2=0,(L−2)(L+1)=0

thus lim

n→+∞an=2.

28. (a) If f(x)=1

2(x+3/x), then f0(x)=(x

2−3)/(2x2) and f0(x)=0forx=√

3; the minimum

value of f(x) for x>0isf(

√

3) = √3. Thus f(x)≥√3 for x>0 and hence an≥√3 for

n≥2.

(b) an+1 −an=(3−a

)/(2an)≤0 for n≥2 since an≥√3 for n≥2; {an}is eventually

decreasing.

n→+∞an+1 = lim

n→+∞

2(an+3/an),

L=1

2(L+3/L), L2−3=0,L=√

29. (a) The altitudes of the rectangles are ln kfor k=2ton, and their bases all have length 1 so

the sum of their areas is ln 2 + ln 3 + ···+lnn= ln(2 ·3···n)=lnn!. The area under the

curve y=lnxfor xin the interval [1,n]isZn

1ln xdx, and Zn+1

1ln xdx is the area for xin

the interval [1,n+ 1] so, from the ﬁgure, Zn

1ln xdx< ln n!<Zn+1

1ln xdx.

(b) Zn

1ln xdx=(xln x−x)n

=nln n−n+ 1 and Zn+1

1ln xdx=(n+ 1) ln(n+1)−nso from

Part (a), nln n−n+1<ln n!<(n+ 1) ln(n+1)−n,e

nln n−n+1 <n!<e

(n+1) ln(n+1)−n,

enln ne1−n<n!<e

(n+1) ln(n+1)e−n,nn

en−1<n!<(n+1)

n+1

367 Chapter 11

en−11/n

√n!<(n+1)

n+1

en1/n

e1−1/n <n

√n!<(n+1)

1+1/n

e,1

e1−1/n <

√n!

n<(1+1/n)(n+1)

1/n

but 1

e1−1/n →1

eand (1+1/n)(n+1)

1/n

e→1

eas n→+∞(why?), so lim

n→+∞

√n!

n=1

30. n!>nn

en−1,n

√n!>n

e1−1/n , lim

n→+∞

e1−1/n =+∞so lim

n→+∞

√n!=+∞.

EXERCISE SET 11.3

1. (a) s1=2,s

2=12/5, s3=62

25,s4=312

125 sn=2−2(1/5)n

1−1/5=5

2−5

2(1/5)n,

lim

n→+∞sn=5

2, converges

(b) s1=1

4,s2=3

4,s3=7

4,s4=15

4sn=(1/4) −(1/4)2n

1−2=−1

4+1

4(2n),

lim

n→+∞sn=+∞, diverges

(k+ 1)(k+2) =1

k+1−1

k+2,s

1=1

6,s

2=1

4,s

3=3

10,s4=1

sn=1

2−1

n+2, lim

n→+∞sn=1

2, converges

2. (a) s1=1/4,s

2=5/16,s

3=21/64,s

4=85/256

sn=1

4 1+1

4+···+1

4n−1!=1

1−(1/4)n

1−1/4=1

31−1

4n

; lim

n→+∞sn=1

(b) s1=1,s

2=5,s

3=21,s

4= 85; sn=4n−1

3, diverges

2=1/12,s

3=3/28,s

4=1/8;

sn=

k=1 1

k+3−1

k+4=1

4−1

n+4, lim

n→+∞sn=1/4

3. geometric, a=1,r=−3/4, sum = 1

1−(−3/4) =4/7

4. geometric, a=(2/3)3,r=2/3, sum = (2/3)3

1−2/3=8/9

5. geometric, a=7,r=−1/6, sum = 7

1+1/6=6

6. geometric, r=−3/2, diverges

7. sn=

k=1 1

k+2−1

k+3=1

3−1

n+3, lim

n→+∞sn=1/3

Exercise Set 11.3 368

8. sn=

k=1 1

2k−1

2k+1 =1

2−1

2n+1 , lim

n→+∞sn=1/2

9. sn=

k=1 1/3

3k−1−1/3

3k+2=1

6−1/3

3n+2, lim

n→+∞sn=1/6

10. sn=

n+1

k=2 1/2

k−1−1/2

k+1=1

2"n+1

k=2

k−1−

n+1

k=2

k+1#

2"n+1

k=2

k−1−

n+3

k=4

k−1#=1

21+1

2−1

n+1−1

n+2; lim

n→+∞sn=3

11. ∞

k=3

k−2=∞

k=1

1/k, the harmonic series, so the series diverges.

12. geometric, a=(e/π)4,r=e/π < 1, sum = (e/π)4

1−e/π =e4

π3(π−e)

13. ∞

k=1

4k+2

7k−1=∞

k=1

64 4

7k−1

; geometric, a= 64, r=4/7, sum = 64

1−4/7= 448/3

14. geometric, a= 125,r = 125/7, diverges

15. 0.4444 ···=0.4+0.04+0.004 + ···=0.4

1−0.1=4/9

16. 0.9999 ···=0.9+0.09+0.009 + ···=0.9

1−0.1=1

17. 5.373737 ···=5+0.37+0.0037 + 0.000037 + ···=5+ 0.37

1−0.01 =5+37/99 = 532/99

18. 0.159159159 ···=0.159+0.000159 + 0.000000159 + ···=0.159

1−0.001 = 159/999=53/333

19. 0.782178217821 ···=0.7821 + 0.00007821 + 0.000000007821 + ···=0.7821

1−0.0001 =7821

9999 =869

1111

20. 0.451141414 ···=0.451+0.00014 + 0.0000014 + 0.000000014 + ···=0.451 + 0.00014

1−0.01 =44663

99000

22. (a) geometric; 18/5 (b) geometric; diverges (c) ∞

k=1

21

2k−1−1

2k+1=1/2

23. d=10+2·3

4·10+2·3

4·3

4·10+2·3

4·3

4·10+···

=10+203

4+203

42

+203

43

+···=10+20(3/4)

1−3/4= 10 + 60 = 70 meters

369 Chapter 11

24. volume = 13+1

23

+1

43

+···+1

n3

+···=1+1

8+1

2

+···+1

8n

+···

1−(1/8) =8/7

25. (a) sn=ln1

2+ln2

3+ln3

4+···+ln n

n+1 =ln1

2·2

3·3

4··· n

n+1=ln 1

n+1 =−ln(n+1),

lim

n→+∞sn=−∞, series diverges.

(b) ln(1 −1/k2)=lnk

2−1

2=ln(k−1)(k+1)

2=lnk−1

k+lnk+1

k=lnk−1

k−ln k

k+1,

n+1

k=2 ln k−1

k−ln k

k+1

=ln 1

2−ln 2

3+ln 2

3−ln 3

4+ln 3

4−ln 4

5+···+ln n

n+1−ln n+1

n+2

=ln1

2−ln n+1

n+2,lim

n→+∞sn=ln1

2=−ln 2

26. (a) ∞

k=0

(−1)kxk=1−x+x

2−x

3+···=1

1−(−x)=1

1+xif |−x|<1, |x|<1, −1<x<1.

(b) ∞

k=0

(x−3)k=1+(x−3)+(x−3)2+···=1

1−(x−3) =1

4−xif |x−3|<1, 2 <x<4.

k=0

(−1)kx2k=1−x

2+x

4−x

6+···=1

1−(−x

2)=1

1+x

2if |−x2|<1, |x|<1, −1<x<1.

27. (a) Geometric series, a=x,r=−x2. Converges for |−x

2|<1, |x|<1;

S=x

1−(−x2)=x

1+x

(b) Geometric series, a=1/x2,r=2/x. Converges for |2/x|<1, |x|>2;

S=1/x2

1−2/x =1

x2−2x.

S=e−x

1−e−x=1

ex−1.

28. √k+1−√

√

2+k=√

k+1−√

√

k+1 =1

√

k−1

√

k+1,

k=1 1

√k−1

√k+1=1

√

1−1

√

2+1

√

2−1

√

3+1

√

3−1

√

4

+···+1

√n−1

√n+1=1−1

√

n+1; lim

n→+∞sn=1

29. sn=(1−1/3)+(1/2−1/4)+(1/3−1/5)+(1/4−1/6)+···+[1/n −1/(n+ 2)]

=(1+1/2+1/3+···+1/n)−(1/3+1/4+1/5+···+1/(n+ 2))

=3/2−1/(n+1)−1/(n+2),lim

n→+∞sn=3/2

Exercise Set 11.3 370

30. sn=

k=1

k(k+2) =

k=1 1/2

k−1/2

k+2=1

2"n

k=1

k−

k=1

k+2#

2"n

k=1

k−

n+2

k=3

k#=1

21+1

2−1

n+1−1

n+2; lim

n→+∞sn=3

31. sn=

k=1

(2k−1)(2k+1) =

k=1 1/2

2k−1−1/2

2k+1=1

2"n

k=1

2k−1−

k=1

2k+1#

2"n

k=1

2k−1−

n+1

k=2

2k−1#=1

21−1

2n+1; lim

n→+∞sn=1

32. Geometric series, a= sin x,r=−1

2sin x. Converges for |−1

2sin x|<1, |sin x|<2,

so converges for all values of x.S=sin x

1+1

2sin x

=2 sin x

2 + sin x.

33. a2=1

2a1+1

2,a3=1

2a2+1

2=1

22a1+1

22+1

2,a4=1

2a3+1

2=1

23a1+1

23+1

22+1

a5=1

2a4+1

2=1

24a1+1

24+1

23+1

22+1

2,...,a

n=1

n−1a

1+1

n−1+1

n−2+···+1

lim

n→+∞an= lim

n→+∞

2n−1+∞

n=1 1

2n

=0+ 1/2

1−1/2=1

34. 0.a1a2···a

n9999 ···=0.a1a2···a

n+0.9 (10−n)+0.09 (10−n)+···

=0.a1a2···a

n+0.9 (10−n)

1−0.1=0.a1a2···a

n+10

−n

=0.a1a2···(a

n+1)=0.a1a2···(a

n+ 1) 0000 ···

35. The series converges to 1/(1 −x) only if −1<x<1.

36. P0P1=asin θ,

P1P2=asin θcos θ,

P2P3=asin θcos2θ,

P3P4=asin θcos3θ,...

(see ﬁgure)

Each sum is a geometric series.

a sin u cos3 u

a sin u cos2 u

a sin u cos u

a sin u

P0P2P4

(a) P0P1+P1P2+P2P3+···=asin θ+asin θcos θ+asin θcos2θ+···=asin θ

1−cos θ

(b) P0P1+P2P3+P4P5+···=asin θ+asin θcos2θ+asin θcos4θ+···

=asin θ

1−cos2θ=asin θ

sin2θ=acsc θ

=asin θcos θ

1−cos2θ=asin θcos θ

sin2θ=acot θ

371 Chapter 11

37. By inspection, θ

2−θ

4+θ

8−θ

16 +···=θ/2

1−(−1/2) =θ/3

38. A1+A2+A3+···=1+1/2+1/4+···=1

1−(1/2) =2

39. (b) 2kA

3k−2k+2kB

3k+1 −2k+1 =2k3k+1 −2k+1A+2

k3

k−2

B

(3k−2k)(3

k+1 −2k+1)

=3·6k−2·22kA+6k−22kB

(3k−2k)(3

k+1 −2k+1)=(3A+B)6k−(2A+B)22k

(3k−2k)(3

k+1 −2k+1)

so 3A+B= 1 and 2A+B=0,A= 1 and B=−2.

k=1 2k

3k−2k−2k+1

3k+1 −2k+1 =

k=1

(ak−ak+1) where ak=2k

3k−2k.

But sn=(a

1−a

)+(a

2−a

3)+(a

3−a

4)+···+(a

n−a

n+1) which is a telescoping sum,

sn=a1−an+1 =2−2

n+1

3n+1 −2n+1 , lim

n→+∞sn= lim

n→+∞2−(2/3)n+1

1−(2/3)n+1 =2.

EXERCISE SET 11.4

1. (a) ∞

k=1

2k=1/2

1−1/2=1; ∞

k=1

4k=1/4

1−1/4=1/3; ∞

k=1 1

2k+1

4k=1+1/3=4/3

(b) ∞

k=1

5k=1/5

1−1/5=1/4; ∞

k=1

k(k+1) = 1 (Example 5, Section 11.3);

∞

k=1 1

5k−1

k(k+1)=1/4−1=−3/4

2. (a) ∞

k=2

k2−1=3/4 (Exercise 10, Section 11.3); ∞

k=2

10k−1=7/10

1−1/10 =7/9;

so ∞

k=2 1

k2−1−7

10k−1=3/4−7/9=−1/36

(b) with a=9/7,r =3/7, geometric, ∞

k=1

7−k3k+1 =9/7

1−(3/7) =9/4;

with a=4/5,r =2/5, geometric, ∞

k=1

2k+1

5k=4/5

1−(2/5) =4/3;

∞

k=1 7−k3k+1 −2k+1

5k=9/4−4/3=11/12

3. (a) p=3, converges (b) p=1/2, diverges (c) p=1, diverges (d) p=2/3, diverges

4. (a) p=4/3, converges (b) p=1/4, diverges (c) p=5/3, converges (d) p=π, converges

Exercise Set 11.4 372

5. (a) lim

k→+∞

k2+k+3

2+1 =1

2; the series diverges. (b) lim

k→+∞1+1

kk

=e; the series diverges.

k→+∞cos kπ does not exist;

the series diverges.

(d) lim

k→+∞

k!= 0; no information

6. (a) lim

k→+∞

ek= 0; no information (b) lim

k→+∞ln k=+∞; the series diverges.

k→+∞

√k= 0; no information (d) lim

k→+∞

√k

√k+3 = 1; the series diverges.

7. (a) Z+∞

5x+2 = lim

`→+∞

5ln(5x+2)

`

=+∞, the series diverges by the Integral Test.

(b) Z+∞

1+9x

2dx = lim

`→+∞

3tan−13x`

3π/2−tan−13,

the series converges by the Integral Test.

8. (a) Z+∞

1+x

2dx = lim

`→+∞

2ln(1 + x2)`

=+∞, the series diverges by the Integral Test.

(b) Z+∞

1(4+2x)

−3/2dx = lim

`→+∞−1/√4+2x

`

=1/

√

the series converges by the Integral Test.

9. ∞

k=1

k+6 =∞

k=7

k, diverges because the harmonic series diverges.

10. ∞

k=1

5k=∞

k=1

51

k, diverges because the harmonic series diverges.

11. ∞

k=1

√k+5 =∞

k=6

√k, diverges because the p-series with p=1/2≤1 diverges.

12. lim

k→+∞

e1/k = 1, the series diverges because lim

k→+∞uk=16=0.

13. Z+∞

1(2x−1)−1/3dx = lim

`→+∞

4(2x−1)2/3`

=+∞, the series diverges by the Integral Test.

14. ln x

xis decreasing for x≥e, and Z+∞

ln x

x= lim

`→+∞

2(ln x)2`

=+∞,

so the series diverges by the Integral Test.

15. lim

k→+∞

ln(k+1) = lim

k→+∞

1/(k+1) =+∞, the series diverges because lim

k→+∞uk6=0.

16. Z+∞

1xe−x2dx = lim

`→+∞−1

2e−x2`

=e−1/2, the series converges by the Integral Test.

373 Chapter 11

17. lim

k→+∞(1+1/k)−k=1/e 6= 0, the series diverges.

18. lim

k→+∞

k2+1

2+3 =16= 0, the series diverges.

19. Z+∞

tan−1x

1+x

2dx = lim

`→+∞

2tan−1x2`

=3π

/32, the series converges by the Integral Test, since

tan−1x

1+x

2=1−2xtan−1x

1+x

2<0 for x≥1.

20. Z+∞

√x2+1dx = lim

`→+∞sinh−1x`

=+∞, the series diverges by the Integral Test.

21. lim

k→+∞k2sin2(1/k)=16= 0, the series diverges.

22. Z+∞

1x2e−x3dx = lim

`→+∞−1

3e−x3`

=e−1/3,

the series converges by the Integral Test (x2e−x3is decreasing for x≥1).

23. 7∞

k=5

k−1.01,p-series with p>1, converges

24. Z+∞

1sech2xdx= lim

`→+∞tanh x`

=1−tanh(1), the series converges by the Integral Test.

25. 1

x(ln x)pis decreasing for x≥e−p, so use the Integral Test with Z+∞

e−p

x(ln x)pto get

lim

`→+∞ln(ln x)`

e−p

=+∞if p= 1, lim

`→+∞

(ln x)1−p

1−p`

e−p

=









+∞if p<1

−1

(−1)ppp(p−1) if p>1

Thus the series converges for p>1.

26. Set g(x)=x(ln x)[ln(ln x)]p,g

0(x)=(1+lnx) ln(ln x)+p, so for ﬁxed pthere exists A>0 such

that g0(x)>0,1/g(x) is decreasing for x>A; use the Integral Test with Z+∞

x(ln x)[ln(ln x)]p

to get

lim

`→+∞ln[ln(ln x)]`

=+∞if p= 1, lim

`→+∞

[ln(ln x)]1−p

1−p`

=









+∞if p<1,

(p−1)[ln(ln A)]p−1if p>1

Thus the series converges for p>1.

27. (a) 3∞

k=1

k2−∞

k=1

k4=π2/2−π4/90 (b) ∞

k=1

k2−1−1

22=π2/6−5/4

k=2

(k−1)4=∞

k=1

k4=π4/90

Exercise Set 11.4 374

28. (a) Suppose Σ(uk+vk) converges; then so does Σ[(uk+vk)−uk], but Σ[(uk+vk)−uk]=Σv

so Σvkconverges which contradicts the assumption that Σvkdiverges. Suppose Σ(uk−vk)

converges; then so does Σ[uk−(uk−vk)]=Σv

kwhich leads to the same contradiction as

before.

(b) Let uk=2/k and vk=1/k; then both Σ(uk+vk) and Σ(uk−vk) diverge; let uk=1/k and

vk=−1/k then Σ(uk+vk) converges; let uk=vk=1/k then Σ(uk−vk) converges.

29. (a) diverges because ∞

k=1

(2/3)k−1converges and ∞

k=1

1/k diverges.

(b) diverges because ∞

k=1

1/(3k+ 2) diverges and ∞

k=1

1/k3/2converges.

k=2

k(ln k)2(Exercise 25) and ∞

k=2

1/k2converge.

30. (a) If S=∞

k=1

ukand sn=

k=1

uk, then S−sn=∞

k=n+1

uk. Interpret uk,k=n+1,n+2,...,as

the areas of inscribed or circumscribed rectangles with height ukand base of length one for

the curve y=f(x) to obtain the result.

(b) Add sn=

k=1

ukto each term in the conclusion of part (a) to get the desired result:

sn+Z+∞

n+1 f(x)dx <

+∞

k=1

uk<s

n+Z+∞

f(x)dx

31. (a) In Exercise 30 above let f(x)= 1

2. Then Z+∞

f(x)dx =−1

x+∞

use this result and the same result with n+ 1 replacing nto obtain the desired result.

(b) s3=1+1/4+1/9=49/36; 58/36 = s3+1

4<1

6π2<s

3+1

3=61/36

(d) 1/11 <1

6π2−s10 <1/10

32. Apply Exercise 30 with f(x)= 1

(2x+1)

2,Z+∞

f(x)dx =1

2(2n+1):

(a) 1

46 <∞

k=1

(2k+1)

2−s

10 <1

(b) π/2−tan−1(11) <∞

k=1

k2+1−s

10 <π/2−tan−1(10)

k=1

ek<11e−10

33. (a) Z+∞

x3dx =1

2n2; use Exercise 30(b)

(b) 1

2n2−1

2(n+1)

2<0.01 for n=5.

375 Chapter 11

34. (a) Z+∞

x4dx =1

3n3; choose nso that 1

3n3−1

3(n+1)

3<0.005,n=4; S≈1.084

35. (a) Let F(x)= 1

, then Zn

xdx =lnnand Zn+1

xdx = ln(n+ 1), u1=1so

ln(n+1)<s

n<1+lnn.

(b) ln(1,000,001) <s

1,000,000 <1 + ln(1,000,000), 13 <s

1,000,000 <15

9= 1 + 9 ln 10 <22

(d) sn>ln(n+1)≥100, n≥e100 −1≈2.688 ×1043;n=2.69 ×1043

36. p-series with p=lna; convergence for p>1,a>e

37. x2e−xis decreasing and positive for x>2 so the Integral Test applies:

Z∞

1x2e−xdx =(x

2+2x+2)e

−x

∞

=5e

−1so the series converges.

38. (a) f(x)=1/(x

3+ 1) is decreasing and continuous on the interval [1,+∞], so the Integral Test

applies.

0.681980

0.685314

0.685966

0.686199

0.686307

0.686367

0.686403

0.686426

0.686442

100

0.686454

(e) Set g(n)=Z+∞

3+1dx =√3

6π+1

n3+1

(n+1)

3−√

3tan−12n−1

√3; for n≥10,

g(n)−g(n+1) ≤0.001; s10 +(g(10) + g(11))/2≈0.6865, so the sum ≈0.6865 to three

decimal places.

EXERCISE SET 11.5

1. (a) f(k)(x)=(−1)ke−x,f

(k)

(0)=(−1)k;e−x≈1−x+x2/2 (quadratic), e−x≈1−x(linear)

(b) f0(x)=−sin x, f00(x)=−cos x, f(0)=1,f0(0)=0,f00(0) = −1,

cos x≈1−x2/2 (quadratic), cos x≈1 (linear)

sin x≈1−(x−π/2)2/2 (quadratic), sin x≈1 (linear)

(d) f(1)=1,f0(1)=1/2,f00(1) = −1/4;

√x=1+1

(x−1) −1

8(x−1)2(quadratic), √x≈1+1

2(x−1) (linear)

2. (a) p2(x)=1+x+x

/2, p1(x)=1+x

(b) p2(x)=3+1

(x−9) −1

216(x−9)2,p1(x)=3+1

(x−9)

3+√

6(x−2) −7

72√3(x−2)2,p1(x)=π

3+√

6(x−2)

(d) p2(x)=x,p

(x)=x

Exercise Set 11.5 376

3. (a) f0(x)=1

−1/2

00(x)=−1

−3/2

;f(1)=1,f0(1) = 1

2,f00(1) = −1

√x≈1+1

2(x−1) −1

8(x−1)2

(b) x=1.1,x

0=1,√

1.1≈1+1

2(0.1) −1

8(0.1)2=1.04875, calculator value ≈1.0488088

4. (a) cos x≈1−x2/2

(b) 2◦=π/90 rad, cos 2◦= cos(π/90) ≈1−π2

2·902≈0.99939077, calculator value ≈0.99939083

5. f(x) = tan x, 61◦=π/3+π/180 rad; x0=π/3,f

(x) = sec2x, f00(x)=2sec

2xtan x;

f(π/3) = √3,f0(π/3)=4,f00(x)=8

√

3; tan x≈√3+4(x−π/3)+4

√

3(x−π/3)2,

tan 61◦= tan(π/3+π/180) ≈√3+4π/180+4

√

3(π/180)2≈1.80397443,

calculator value ≈1.80404776

6. f(x)=√

x, x0=36,f0(x)=1

−1/2

00(x)=−1

−3/2

;

f(36) = 6,f0(36) = 1

12,f00(36) = −1

864;√x≈6+ 1

12(x−36) −1

1728(x−36)2;

√36.03 ≈6+0.03

12 −(0.03)2

1728 ≈6.00249947917, calculator value ≈6.00249947938

7. f(k)(x)=(−1)ke−x,f(k)(0)=(−1)k;p0(x)=1,p

(x)=1−x, p2(x)=1−x+1

(x)=1−x+1

2−1

3!x3,p

(x)=1−x+1

2−1

3!x3+1

4!x4;∞

k=0

(−1)k

k!xk

8. f(k)(x)=a

ax,f(k)(0) = ak;p0(x)=1,p

(x)=1+ax, p2(x)=1+ax +a2

2x2,

p3(x)=1+ax +a2

2x2+a3

3! x3,p

(x)=1+ax +a2

2x2+a3

3! x3+a4

4! x4;∞

k=0

k!xk

9. f(k)(0) = 0 if kis odd, f(k)(0) is alternately πkand −πkif kis even; p0(x)=1,p

(x)=1,

(x)=1−π

2! x2;p3(x)=1−π

2! x2,p

(x)=1−π

2! x2+π4

4! x4;∞

k=0

(−1)kπ2k

(2k)! x2k

10. f(k)(0) = 0 if kis even, f(k)(0) is alternately πkand −πkif kis odd; p0(x)=0,p

(x)=πx,

p2(x)=πx;p3(x)=πx −π3

3! x3,p

(x)=πx −π3

3! x3;∞

k=0

(−1)kπ2k+1

(2k+ 1)! x2k+1

11. f(0)(0) = 0; for k≥1, f(k)(x)=(−1)k+1(k−1)!

(1+x)k,f(k)(0)=(−1)k+1(k−1)!; p0(x)=0,

(x)=x, p2(x)=x−1

(x)=x−1

2+1

(x)=x−1

2+1

3−1

;∞

k=1

(−1)k+1

kxk

12. f(k)(x)=(−1)kk!

(1+x)k+1 ;f(k)(0)=(−1)kk!; p0(x)=1,p

(x)=1−x,

p2(x)=1−x+x

(x)=1−x+x

2−x

(x)=1−x+x

2−x

3+x

;∞

k=0

(−1)kxk

377 Chapter 11

13. f(k)(0) = 0 if kis odd, f(k)(0) = 1 if kis even; p0(x)=1,p

1(x)=1,

(x)=1+x

/2,p

(x)=1+x

/2,p

(x)=1+x

/2+x

4/4!; ∞

k=0

(2k)!x2k

14. f(k)(0) = 0 if kis even, f(k)(0) = 1 if kis odd; p0(x)=0,p

(x)=x, p2(x)=x,

p3(x)=x+x

/3!,p

(x)=x+x

/3!; ∞

k=0

(2k+ 1)!x2k+1

15. f(k)(x)=((−1)k/2(xsin x−kcos x)keven

(−1)(k−1)/2(xcos x+ksin x)kodd ,f

(k)

(0) = ((−1)1+k/2kkeven

0kodd

p0(x)=0,p

(x)=0,p

(x)=x

3(x)=x

(x)=x

2−1

;∞

k=0

(−1)k

(2k+ 1)!x2k+2

16. f(k)(x)=(k+x)e

(k)

(0) = k;p0(x)=0,p

(x)=x, p2(x)=x+x

(x)=x+x

2+1

(x)=x+x

2+1

3+1

3!x4;∞

k=1

(k−1)!xk

17. (a) f(0)=1,f0(0)=2,f00(0) = −2,f000(0)=6,f(k)(0) = 0 for k≥4;

the Maclaurin series for f(x)isf(x).

(b) f(k)(0) = k!ckfor k≤n, and f(k)(0) = 0 for k>n; the Maclaurin series for f(x)isf(x).

19. f(k)(x0)=e;p

(x)=e, p1(x)=e+e(x−1),

p2(x)=e+e(x−1) + e

2(x−1)2,p

(x)=e+e(x−1) + e

2(x−1)2+e

3!(x−1)3,

p4(x)=e+e(x−1) + e

2(x−1)2+e

3!(x−1)3+e

4!(x−1)4;∞

k=0

k!(x−1)k

20. f(k)(x)=(−1)ke−x,f

(k)

(ln 2) = (−1)k1

2;p0(x)=1

(x)=1

2−1

(x−ln 2),

p2(x)=1

2−1

(x−ln 2)+ 1

2·2(x−ln 2)2,p

(x)=1

2−1

(x−ln 2)+ 1

2·2(x−ln 2)2−1

2·3!(x−ln 2)3,

p4(x)=1

2−1

(x−ln 2) + 1

2·2(x−ln 2)2−1

2·3!(x−ln 2)3+1

2·4!(x−ln 2)4;

∞

k=0

(−1)k

2·k!(x−ln 2)k

21. f(k)(x)=(−1)kk!

xk+1 ,f(k)(−1) = −k!; p0(x)=−1; p1(x)=−1−(x+ 1);

p2(x)=−1−(x+1)−(x+1)

2;p

3(x)=−1−(x+1)−(x+1)

2−(x+1)

4(x)=−1−(x+1)−(x+1)

2−(x+1)

3−(x+1)

4;∞

k=0

(−1)(x+1)

Exercise Set 11.5 378

22. f(k)(x)= (−1)kk!

(x+2)

k+1 ,f(k)(3) = (−1)kk!

5k+1 ;p0(x)=1

(x)=1

5−1

25(x−3);

p2(x)=1

5−1

25(x−3) + 1

125(x−3)2;p3(x)=1

5−1

25(x−3) + 1

125(x−3)2−1

625(x−3)3;

p4(x)=1

5−1

25(x−3) + 1

125(x−3)2−1

625(x−3)3+1

3025(x−3)4;∞

k=0

(−1)k

5k+1 (x−3)k

23. f(k)(1/2)=0ifkis odd, f(k)(1/2) is alternately πkand −πkif kis even;

p0(x)=p

(x)=1,p

2(x)=p

(x)=1−π

2(x−1/2)2,

p4(x)=1−π

2(x−1/2)2+π4

4! (x−1/2)4;∞

k=0

(−1)kπ2k

(2k)! (x−1/2)2k

24. f(k)(π/2)=0ifkis even, f(k)(π/2) is alternately −1and1ifkis odd; p0(x)=0,

(x)=−(x−π/2),p

(x)=−(x−π/2) + 1

3!(x−π/2)3,

p4(x)=−(x−π/2) + 1

3!(x−π/2)3;∞

k=0

(−1)k+1

(2k+ 1)!(x−π/2)2k+1

25. f(1) = 0, for k≥1,f(k)(x)=(−1)k−1(k−1)!

xk;f(k)(1)=(−1)k−1(k−1)!;

p0(x)=0,p

(x)=(x−1); p2(x)=(x−1) −1

2(x−1)2;p3(x)=(x−1) −1

2(x−1)2+1

3(x−1)3,

p4(x)=(x−1) −1

2(x−1)2+1

3(x−1)3−1

4(x−1)4;∞

k=1

(−1)k−1

k(x−1)k

26. f(e)=1,fork≥1,f(k)(x)=(−1)k−1(k−1)!

xk;f(k)(e)=(−1)k−1(k−1)!

ek;

p0(x)=1,p

(x)=1+1

(x−e); p2(x)=1+1

(x−e)−1

2(x−e)

;

(x)=1+1

(x−e)−1

2(x−e)

2+1

3(x−e)

(x)=1+1

(x−e)−1

2(x−e)

2+1

3(x−e)

3−1

4(x−e)

;1+

∞

k=1

(−1)k−e

kek(x−e)k

27. (a) f(1)=1,f0(1)=2,f00(1) = −2,f000(1)=6,f(k)(1) = 0 for k≥4;

the Taylor series for f(x)isf(x).

(b) f(k)(x0)=k!c

kfor k≤n; for k > n, f(k)(x0) = 0; the Taylor series for f(x)isf(x).

379 Chapter 11

29. f(k)(0)=(−2)k;p0(x)=1,p

(x)=1−2x,

p2(x)=1−2x+2x

2,p

(x)=1−2x+2x

2−4

-1

-0.6 0.6

30. f(k)(π/2)=0ifkis odd, f(k)(π/2) is alternately 1

and −1ifkis even; p0(x)=1,p

(x)=1−1

(x−π/2)2,

p4(x)=1−1

(x−π/2)2+1

24(x−π/2)4,

p6(x)=1−1

(x−π/2)2+1

24(x−π/2)4−1

720(x−π/2)6

1.25

-1.25

31. f(k)(π)=0ifkis odd, f(k)(π/2) is alternately −1

and1ifkis even; p0(x)=−1,p

(x)=−1+1

2(x−π)

4(x)=−1+1

2(x−π)

2−1

24(x−π)4,

p6(x)=−1+1

2(x−π)

2−1

24(x−π)4+1

720(x−π)6

1.25

-1.25

32. f(0) = 0; for k≥1,f(k)(x)=(−1)k−1(k−1)!

(x+1)

f(k)(0)=(−1)k−1(k−1)!; p0(x)=0,p

(x)=x,

(x)=x−1

2+1

1.5

-1.5

-1 1

33. p(0)=1,p(x) has slope −1atx= 0, and p(x) is concave up at x= 0, eliminating I, II and III

respectively and leaving IV.

34. Let p0(x)=2,p

1(x)=p

(x)=1−3x, p3(x)=1−3x+x

, and, for any arbitrary integer n≥4

and constants c4,c

5,...,c

n, let pn(x)=2−3(x−1)+(x−1)3+

k=4

ck(x−1)k; then any of the

polynomials p0,p

1,...,p

nis a possible Taylor polynomial for fabout x=1.

35. f(k)(ln 4) = 15/8 for keven, f(k)(ln4) = 17/8 for kodd, which can be written as

f(k)(ln 4) = 16 −(−1)k

8;∞

k=0

16 −(−1)k

8k!(x−ln 4)k

Exercise Set 11.6 380

36. (a) cos α≈1−α2/2; x=r−rcos α=r(1 −cos α)≈rα2/2

(b) In Figure Ex-36 let r= 4000 mi and α=1/80 so that the arc has length 2rα = 100 mi.

Then x≈rα2/2= 4000

2·802=5/16 mi.

37. From Exercise 2(a), p1(x)=1+x,p

(x)=1+x+x

(a) 3

-1

-1 1

(b) x

f(x)

p1(x)

p2(x)

–1.000

0.431

0.000

0.500

–0.750

0.506

0.250

0.531

–0.500

0.619

0.500

0.625

–0.250

0.781

0.750

0.781

0.000

1.000

0.250

1.281

1.250

1.281

0.500

1.615

1.500

1.625

0.750

1.977

1.750

2.031

1.000

2.320

2.000

2.500

for −0.14 <x<0.14

0.01

-0.15 0.15

(d) |esin x−(1+x+x2/2)|<0.01

for −0.50 <x<0.50

0.015

-0.6 0.6

EXERCISE SET 11.6

1. (a) 1

5k2−k≤1

5k2−k2=1

4k2,∞

k=1

4k2converges

(b) 3

k−1/4>3

k,∞

k=1

3/k diverges

2. (a) k+1

2−k>k

2=1

k,∞

k=2

1/k diverges (b) 2

k4+k<2

k4,∞

k=1

k4converges

3. (a) 1

3k+5 <1

k,∞

k=1

3kconverges (b) 5 sin2k

k!<5

k!,∞

k=1

k!converges

4. (a) ln k

k>1

kfor k≥3, ∞

k=1

kdiverges

(b) k

k3/2−1/2>k

k3/2=1

√k;∞

k=1

√kdiverges

381 Chapter 11

5. compare with the convergent series ∞

k=1

1/k5,ρ= lim

k→+∞

4k7−2k6+6k

7+k−8=1/2, converges

6. compare with the divergent series ∞

k=1

1/k,ρ= lim

k→+∞

9k+6 =1/9, diverges

7. compare with the convergent series ∞

k=1

5/3k,ρ= lim

k→+∞

3k+1 = 1, converges

8. compare with the divergent series ∞

k=1

1/k,ρ= lim

k→+∞

k2(k+3)

(k+ 1)(k+ 2)(k+5) = 1, diverges

9. compare with the divergent series ∞

k=1

k2/3,

ρ= lim

k→+∞

k2/3

(8k2−3k)1/3= lim

k→+∞

(8 −3/k)1/3=1/2, diverges

10. compare with the convergent series ∞

k=1

1/k17,

ρ= lim

k→+∞

k17

(2k+3)

17 = lim

k→+∞

(2+3/k)17 =1/2

17, converges

11. ρ= lim

k→+∞

3k+1/(k+ 1)!

3k/k!= lim

k→+∞

k+1 = 0, the series converges

12. ρ= lim

k→+∞

4k+1/(k+1)

/k2= lim

k→+∞

4k2

(k+1)

2= 4, the series diverges

13. ρ= lim

k→+∞

k+1 = 1, the result is inconclusive

14. ρ= lim

k→+∞

(k+ 1)(1/2)k+1

k(1/2)k= lim

k→+∞

k+1

2k=1/2, the series converges

15. ρ= lim

k→+∞

(k+ 1)!/(k+1)

k!/k3= lim

k→+∞

(k+1)

2=+∞, the series diverges

16. ρ= lim

k→+∞

(k+1)/[(k+1)

2+1]

k/(k2+1) = lim

k→+∞

(k+ 1)(k2+1)

k(k

2+2k+2) = 1, the result is inconclusive.

17. ρ= lim

k→+∞

3k+2

2k−1=3/2, the series diverges

18. ρ= lim

k→+∞k/100=+∞, the series diverges

19. ρ= lim

k→+∞

k1/k

5=1/5, the series converges

20. ρ= lim

k→+∞(1 −e−k) = 1, the result is inconclusive

Exercise Set 11.6 382

21. Ratio Test, ρ= lim

k→+∞7/(k+ 1) = 0, converges

22. Limit Comparison Test, compare with the divergent series ∞

k=1

1/k

23. Ratio Test, ρ= lim

k→+∞

(k+1)

2=1/5, converges

24. Ratio Test, ρ= lim

k→+∞(10/3)(k+1)=+∞, diverges

25. Ratio Test, ρ= lim

k→+∞e−1(k+1)

50/k50 =e−1<1, converges

26. Limit Comparison Test, compare with the divergent series ∞

k=1

1/k

27. Limit Comparison Test, compare with the convergent series ∞

k=1

1/k5/2,ρ= lim

k→+∞

k3+1 =1,

converges

28. 4

2+3

k<4

k,∞

k=1

3kkconverges (Ratio Test) so ∞

k=1

2+k3

kconverges by the Comparison Test

29. Limit Comparison Test, compare with the divergent series ∞

k=1

1/k,ρ= lim

k→+∞

√k2+k=1,

diverges

30. 2+(−1)k

5k≤3

5k,∞

k=1

3/5kconverges so ∞

k=1

2+(−1)k

5kconverges

31. Limit Comparison Test, compare with the convergent series ∞

k=1

1/k5/2,

ρ= lim

k→+∞

k3+2k

5/2

3+3k

2+3k= 1, converges

32. 4+|cos k|

k3<5

k3,∞

k=1

5/k3converges so ∞

k=1

4+|cos k|

k3converges

33. Limit Comparison Test, compare with ∞

k=1

1/√k

34. Ratio Test, ρ= lim

k→+∞(1+1/k)−k=1/e < 1, converges

35. Ratio Test, ρ= lim

k→+∞

ln(k+1)

eln k= lim

k→+∞

e(k+1) =1/e < 1, converges

36. Ratio Test, ρ= lim

k→+∞

k+1

2k+1 = lim

k→+∞

2e2k+1 = 0, converges

37. Ratio Test, ρ= lim

k→+∞

k+5

4(k+1) =1/4, converges

383 Chapter 11

38. Root Test, ρ= lim

k→+∞(k

k+1)

k= lim

k→+∞

(1+1/k)k=1/e, converges

39. diverges because lim

k→+∞

4+2

−k=1/46=0

40. ∞

k=1

√kln k

k3+1 =∞

k=2

√kln k

k3+1 because ln1 = 0, √kln k

k3+1 <kln k

k3=ln k

k2,

Z+∞

ln x

x2dx = lim

`→+∞−ln x

x−1

x`

2(ln 2+1) so ∞

k=2

ln k

k2converges and so does ∞

k=1

√kln k

k3+1.

41. tan−1k

k2<π/2

k2,∞

k=1

π/2

k2converges so ∞

k=1

tan−1k

k2converges

42. 5k+k

k!+3 <5

k+5

k!=25



k!,∞

k=1

25k

k!converges (Ratio Test) so ∞

k=1

5k+k

k!+3 converges

43. Ratio Test, ρ= lim

k→+∞

(k+1)

(2k+ 2)(2k+1) =1/4, converges

44. Ratio Test, ρ= lim

k→+∞

2(k+1)

(2k+ 4)(2k+3) =1/2, converges

45. uk=k!

1·3·5···(2k−1), by the Ratio Test ρ= lim

k→+∞

k+1

2k+1 =1/2; converges

46. uk=1·3·5···(2k−1)

(2k−1)! , by the Ratio Test ρ= lim

k→+∞

2k= 0; converges

47. Root Test: ρ= lim

k→+∞

3(ln k)1/k =1/3, converges

48. Root Test: ρ= lim

k→+∞

π(k+1)

1+1/k = lim

k→+∞πk+1

k=π, diverges

49. (b) ρ= lim

k→+∞

sin(π/k)

π/k = 1 and ∞

k=1

π/k diverges

50. (a) cos x≈1−x2/2,1−cos 1

k≈1

2k2(b) ρ= lim

k→+∞

1−cos(1/k)

1/k2= 2, converges

51. Set g(x)=√

x−ln x;d

dxg(x)= 1

√

x−1

x= 0 when x= 4. Since lim

x→0+ g(x) = lim

x→+∞g(x)=+∞

it follows that g(x) has its minimum at x=4,g(4) = √4−ln 4 >0, and thus √x−ln x>0 for

x>0.

(a) ln k

k2<√k

k2=1

k3/2,∞

k=1

k3/2converges so ∞

k=1

ln k

k2converges.

(b) 1

(ln k)2>1

k,∞

k=2

kdiverges so ∞

k=2

(ln k)2diverges.

Exercise Set 11.7 384

52. By the Root Test, ρ= lim

k→+∞

(k1/k)α=α

1α=α, the series converges if α<1 and diverges

if α>1. If α= 1 then the series is ∞

k=1

1/k which diverges.

53. (a) If Pbkconverges, then set M=Pbk. Then a1+a2+···+a

n≤b

1+b

2+···+b

n≤M;

apply Theorem 11.4.6 to get convergence of Pak.

(b) Assume the contrary, that Pbkconverges; then use part (a) of the Theorem to show that

Pakconverges, a contradiction.

54. (a) If lim

k→+∞(ak/bk) = 0 then for k≥K,ak/bk<1, ak<b

kso Pakconverges by the Comparison

Test.

(b) If lim

k→+∞(ak/bk)=+∞then for k≥K,ak/bk>1, ak>b

kso Pakdiverges by the

Comparison Test.

EXERCISE SET 11.7

1. ak+1 <a

,lim

k→+∞ak=0,a

k>0

2. ak+1

3(k+1) <1

3for k>0, so {ak}is decreasing and tends to zero.

3. diverges because lim

k→+∞ak= lim

k→+∞

k+1

3k+1 =1/36=0

4. diverges because lim

k→+∞ak= lim

k→+∞

k+1

√

k+1 =+∞6=0

5. {e−k}is decreasing and lim

k→+∞e−k= 0, converges

6. ln k

kis decreasing and lim

k→+∞

ln k

k= 0, converges

7. ρ= lim

k→+∞

(3/5)k+1

(3/5)k=3/5,converges absolutely

8. ρ= lim

k→+∞

k+1 = 0, converges absolutely

9. ρ= lim

k→+∞

3k2

(k+1)

2= 3, diverges

10. ρ= lim

k→+∞

k+1

5k=1/5, converges absolutely

11. ρ= lim

k→+∞

(k+1)

ek3=1/e, converges absolutely

12. ρ= lim

k→+∞

(k+1)

k+1k!

(k+ 1)!kk= lim

k→+∞(1+1/k)k=e, diverges

385 Chapter 11

13. conditionally convergent, ∞

k=1

(−1)k+1

3kconverges by the Alternating Series Test but ∞

k=1

3kdiverges

14. absolutely convergent, ∞

k=1

k4/3converges

15. divergent, lim

k→+∞ak6=0

16. absolutely convergent, Ratio Test for absolute convergence

17. ∞

k=1

cos kπ

k=∞

k=1

(−1)k

kis conditionally convergent, ∞

k=1

(−1)k

kconverges by the Alternating Series

Test but ∞

k=1

1/k diverges.

18. conditionally convergent, ∞

k=3

(−1)kln k

kconverges by the Alternating Series Test but ∞

k=3

ln k

diverges (Limit Comparison Test with P1/k).

19. conditionally convergent, ∞

k=1

(−1)k+1 k+2

k(k+3) converges by the

Alternating Series Test but ∞

k=1

k+2

k(k+3) diverges (Limit Comparison Test with P1/k)

20. conditionally convergent, ∞

k=1

(−1)k+1k2

k3+1 converges by the Alternating Series Test but

∞

k=1

k3+1 diverges (Limit Comparison Test with P(1/k))

21. ∞

k=1

sin(kπ/2)=1+0−1+0+1+0−1+0+···, divergent ( lim

k→+∞sin(kπ/2) does not exist)

22. absolutely convergent, ∞

k=1

|sin k|

k3converges (compare with P1/k3)

23. conditionally convergent, ∞

k=2

(−1)k

kln kconverges by the Alternating Series Test but ∞

k=2

kln kdiverges

(Integral Test)

24. conditionally convergent, ∞

k=1

(−1)k

pk(k+1) converges by the Alternating Series Test but

∞

k=1

pk(k+1) diverges (Limit Comparison Test with P1/k)

25. absolutely convergent, ∞

k=2

(1/ln k)kconverges by the Root Test

Exercise Set 11.7 386

26. conditionally convergent, ∞

k=1

(−1)k+1

√k+1+√

kconverges by the Alternating Series Test but

∞

k=1

√k+1+√

kdiverges (Limit Comparison Test with P1/√k)

27. conditionally convergent, let f(x)=x

2+1

3+2 then f0(x)=x(4 −3x−x3)

(x3+2)

2≤0 for x≥2so

}

+∞

k=2 =k2+1

3+2+∞

k=2

is nonincreasing, lim

k→+∞ak= 0; the series converges by the

Alternating Series Test but ∞

k=2

k2+1

3+2 diverges (Limit Comparison Test with P1/k)

28. ∞

k=1

kcos kπ

k2+1 =∞

k=1

(−1)kk

k2+1 is conditionally convergent, ∞

k=1

(−1)kk

k2+1 converges by the

Alternating Series Test but ∞

k=1

k2+1 diverges

29. absolutely convergent by the Ratio Test, ρ= lim

k→+∞

k+1

(2k+ 1)(2k)=0

30. divergent, lim

k→+∞ak=+∞31. |error|<a

8=1/8=0.125

32. |error|<a

6=1/6! <0.0014 33. |error|<a

100 =1/

√

100=0.1

34. |error|<a

4=1/(5 ln 5) <0.125

35. |error|<0.0001 if an+1 ≤0.0001, 1/(n+1)≤0.0001, n+1≥10,000, n≥9,999, n=9,999

36. |error|<0.00001 if an+1 ≤0.00001, 1/(n+ 1)! ≤0.00001, (n+ 1)! ≥100,000. But 8! = 40,320,

9! = 362,880 so (n+ 1)! ≥100,000 if n+1≥9, n≥8, n=8

37. |error|<0.005 if an+1 ≤0.005, 1/√n+1≤0.005, √n+1 ≥200, n+1≥40,000, n≥39,999,

n=39,999

38. |error|<0.05 if an+1 ≤0.05, 1/[(n+ 2) ln(n+ 2)] ≤0.05, (n+ 2) ln(n+2)≥20. But 9 ln 9 ≈19.8

and 10 ln 10 ≈23.0so(n+ 2) ln(n+2)≥20 if n+2≥10, n≥8, n=8

39. ak=3

2k+1 ,|error|<a

11 =3

212 <0.00074; s10 ≈0.4995; S=3/4

1−(−1/2) =0.5

40. ak=2

3k−1

,|error|<a

11 =2

310

<0.01735; s10 ≈0.5896; S=1

1−(−2/3) =0.6

41. ak=1

(2k−1)!,an+1 =1

(2n+ 1)! ≤0.005, (2n+ 1)! ≥200, 2n+1≥6, n≥2.5; n=3,

3=1−1/6+1/120 ≈0.84

42. ak=1

(2k−2)!,an+1 =1

(2n)! ≤0.005, (2n)! ≥200, 2n≥6, n≥3; n=3,s

3≈0.54

387 Chapter 11

43. ak=1

k2k,an+1 =1

(n+ 1)2n+1 ≤0.005, (n+ 1)2n+1 ≥200, n+1≥6, n≥5; n=5,s

5≈0.41

44. ak=1

(2k−1)5+ 4(2k−1),an+1 =1

(2n+1)

5+ 4(2n+1) ≤0.005,

(2n+1)

5+ 4(2n+1)≥200, 2n+1≥3, n≥1; n=1,s

1=0.20

45. (c) ak=1

2k−1,an+1 =1

2n+1 ≤10−2,2n+1≥100, n≥49.5; n=50

46. P(1/kp) converges if p>1 and diverges if p≤1, so ∞

k=1

(−1)k1

kpconverges absolutely if p>1,

and converges conditionally if 0 <p≤1 since it satisﬁes the Alternating Series Test; it diverges

for p≤0 since lim

k→+∞ak6=0.

47. 1+ 1

2+1

2+···=1+ 1

2+1

2+···−1

2+1

2+···

=π

6−1

21+ 1

2+1

2+···=π

6−1

6=π

48. 1+ 1

4+1

4+···=1+ 1

4+1

4+···−1

4+1

4+···

=π

90 −1

241+ 1

4+1

4+···=π

90 −1

π4

90 =π4

49. Every positive integer can be written in exactly one of the three forms 2k−1or4k−2or4k,

so a rearrangement is

1−1

2−1

4+1

3−1

6−1

8+1

5−1

10 −1

12+···+1

2k−1−1

4k−2−1

4k+···

=1

2−1

4+1

6−1

8+1

10 −1

12+···+1

4k−2−1

4k+···=1

2ln 2

50. (a) 1.5

010

(b) Yes; since f(x) is decreasing for x≥1

and lim

x→+∞f(x) = 0, so the series

satisﬁes the Alternating Series Test.

51. (a) The distance dfrom the starting point is

d= 180 −180

2+180

3−···− 180

1000 = 180 1−1

2+1

3−···− 1

1000.

From Theorem 11.7.2, 1 −1

2+1

3−···− 1

1000 diﬀers from ln 2 by less than 1/1001 so

180(ln 2 −1/1001) <d<180 ln 2, 124.58 <d<124.77.

Exercise Set 11.8 388

(b) The total distance traveled is s= 180 + 180

2+180

3+···+180

1000, and from inequality (2) in

Section 11.4,

Z1001

180

xdx<s<180 + Z1000

180

xdx

180 ln 1001 <s<180(1 + ln 1000)

1243 <s<1424

52. (a) Suppose Σ|ak|converges, then lim

k→+∞|ak|=0so|a

|<1 for k≥Kand thus |ak|2<|ak|,

k<|ak|hence Σa2

kconverges by the Comparison Test.

(b) Let ak=1

k, then Pa2

kconverges but Pakdiverges.

EXERCISE SET 11.8

1. geometric series, ρ= lim

k→+∞

uk+1

uk=|x|, so the interval of convergence is −1<x<1, converges

there to 1

1+x(the series diverges for x=±1)

2. geometric series, ρ= lim

k→+∞

uk+1

uk=|x|2, so the interval of convergence is −1<x<1, converges

there to 1

1−x2(the series diverges for x=±1)

3. geometric series, ρ= lim

k→+∞

uk+1

uk=|x−2|, so the interval of convergence is 1 <x<3, converges

there to 1

1−(x−2) =1

3−x(the series diverges for x=1,3)

4. geometric series, ρ= lim

k→+∞

uk+1

uk=|x+3|, so the interval of convergence is −4<x<−2,

converges there to 1

1+(x+3) =1

4+x(the series diverges for x=−4,−2)

5. (a) geometric series, ρ= lim

k→+∞

uk+1

uk=|x/2|, so the interval of convergence is −2<x<2,

converges there to 1

1+x/2=2

2+x; (the series diverges for x=−2,2)

(b) f(0)=1; f(1)=2/3

6. (a) geometric series, ρ= lim

k→+∞

uk+1

uk=

x−5

3, so the interval of convergence is 2 <x<8,

converges to 1

1+(x−5)/3=3

x−2(the series diverges for x=2,8)

(b) f(3)=3,f(6)=3/4

389 Chapter 11

7. ρ= lim

k→+∞

k+1

k+2|x|=|x|, the series converges if |x|<1 and diverges if |x|>1. If x=−1,

∞

k=0

(−1)k

k+1 converges by the Alternating Series Test; if x=1, ∞

k=0

k+1 diverges. The radius of

convergence is 1, the interval of convergence is [−1,1).

8. ρ= lim

k→+∞3|x|=3|x|, the series converges if 3|x|<1or|x|<1/3 and diverges if |x|>1/3. If

x=−1/3, ∞

k=0

(−1)kdiverges, if x=1/3, ∞

k=0

(1) diverges. The radius of convergence is 1/3, the

interval of convergence is (−1/3, 1/3).

9. ρ= lim

k→+∞|x|

k+1 = 0, the radius of convergence is +∞, the interval is (−∞,+∞).

10. ρ= lim

k→+∞

k+1

2|x|=+∞, the radius of convergence is 0, the series converges only if x=0.

11. ρ= lim

k→+∞

5k2|x|

(k+1)

2=5|x|, converges if |x|<1/5 and diverges if |x|>1/5. If x=−1/5, ∞

k=1

(−1)k

converges; if x=1/5, ∞

k=1

1/k2converges. Radius of convergence is 1/5, interval of convergence is

[−1/5,1/5].

12. ρ= lim

k→+∞

ln k

ln(k+1)|x|=|x|, the series converges if |x|<1 and diverges if |x|>1. If x=−1,

∞

k=2

(−1)k

ln kconverges; if x=1, ∞

k=2

1/(ln k) diverges (compare to P(1/k)). Radius of convergence

is 1, interval of convergence is [−1,1).

13. ρ= lim

k→+∞

k|x|

k+2 =|x|, converges if |x|<1, diverges if |x|>1. If x=−1, ∞

k=1

(−1)k

k(k+1) converges;

if x=1, ∞

k=1

k(k+1) converges. Radius of convergence is 1, interval of convergence is [−1,1].

14. ρ= lim

k→+∞2k+1

k+2|x|=2|x|, converges if |x|<1/2, diverges if |x|>1/2. If x=−1/2, ∞

k=0

−1

2(k+1)

diverges; if x=1/2, ∞

k=0

(−1)k

2(k+1) converges. Radius of convergence is 1/2, interval of convergence

is (−1/2,1/2].

15. ρ= lim

k→+∞

√k

√k+1|x|=|x|, converges if |x|<1, diverges if |x|>1. If x=−1, ∞

k=1

−1

√kdiverges; if

x=1, ∞

k=1

(−1)k−1

√kconverges. Radius of convergence is 1, interval of convergence is (−1,1].

Exercise Set 11.8 390

16. ρ= lim

k→+∞|x|2

(2k+ 2)(2k+1) = 0, radius of convergence is +∞, interval of convergence is (−∞,+∞).

17. ρ= lim

k→+∞|x|2

(2k+ 3)(2k+2) = 0, radius of convergence is +∞, interval of convergence is (−∞,+∞).

18. ρ= lim

k→+∞

k3/2|x|3

(k+1)

3/2=|x|

3, converges if |x|<1, diverges if |x|>1. If x=−1, ∞

k=0

k3/2converges;

if x=1, ∞

k=0

(−1)k

k3/2converges. Radius of convergence is 1, interval of convergence is [−1,1].

19. ρ= lim

k→+∞

3|x|

k+1 = 0, radius of convergence is +∞, interval of convergence is (−∞,+∞).

20. ρ= lim

k→+∞

k(ln k)2|x|

(k+ 1)[ln(k+ 1)]2=|x|, converges if |x|<1, diverges if |x|>1. If x=−1, then, by

Exercise 11.4.25, ∞

k=2

−1

k(ln k)2converges; if x=1, ∞

k=2

(−1)k+1

k(ln k)2converges. Radius of convergence

is 1, interval of convergence is [−1,1].

21. ρ= lim

k→+∞

1+k

1+(k+1)

2|x|=|x|, converges if |x|<1, diverges if |x|>1. If x=−1, ∞

k=0

(−1)k

1+k

converges; if x=1, ∞

k=0

1+k

2converges. Radius of convergence is 1, interval of convergence is

[−1,1].

22. ρ= lim

k→+∞

2|x−3|=1

2|x−3|, converges if |x−3|<2, diverges if |x−3|>2. If x=1, ∞

k=0

(−1)k

diverges; if x=5, ∞

k=0

1 diverges. Radius of convergence is 2, interval of convergence is (1,5).

23. ρ= lim

k→+∞

k|x+1|

k+1 =|x+1|, converges if |x+1|<1, diverges if |x+1|>1. If x=−2, ∞

k=1

−1

diverges; if x=0, ∞

k=1

(−1)k+1

kconverges. Radius of convergence is 1, interval of convergence is

(−2,0].

24. ρ= lim

k→+∞

(k+1)

(k+2)

2|x−4|=|x−4|, converges if |x−4|<1, diverges if |x−4|>1. If x=3,

∞

k=0

1/(k+1)2converges; if x=5, ∞

k=0

(−1)k/(k+1)2converges. Radius of convergence is 1, interval

of convergence is [3,5].

391 Chapter 11

25. ρ= lim

k→+∞(3/4)|x+5|=3

|x+5|, converges if |x+5|<4/3, diverges if |x+5|>4/3. If

x=−19/3, ∞

k=0

(−1)kdiverges; if x=−11/3, ∞

k=0

1 diverges. Radius of convergence is 4/3, interval

of convergence is (−19/3,−11/3).

26. ρ= lim

k→+∞

(2k+ 3)(2k+2)k

(k+1)

3|x−2|=+∞, radius of convergence is 0,

series converges only at x=2.

27. ρ= lim

k→+∞

k2+4

(k+1)

2+4|x+1|2=|x+1|2, converges if |x+1|<1, diverges if |x+1|>1. If x=−2,

∞

k=1

(−1)3k+1

k2+4 converges; if x=0, ∞

k=1

(−1)k

k2+4 converges. Radius of convergence is 1, interval of

convergence is [−2,0].

28. ρ= lim

k→+∞

kln(k+1)

(k+1)lnk|x−3|=|x−3|, converges if |x−3|<1, diverges if |x−3|>1. If

x=2, ∞

k=1

(−1)kln k

kconverges; if x=4, ∞

k=1

ln k

kdiverges. Radius of convergence is 1, interval of

convergence is [2,4).

29. ρ= lim

k→+∞

π|x−1|2

(2k+ 3)(2k+2) = 0, radius of convergence +∞, interval of convergence (−∞,+∞).

30. ρ= lim

k→+∞

16|2x−3|=1

16|2x−3|, converges if 1

16|2x−3|<1or|x−3/2|<8, diverges if

|x−3/2|>8. If x=−13/2, ∞

k=0

(−1)kdiverges; if x=19/2, ∞

k=0

1 diverges. Radius of convergence

is 8, interval of convergence is (−13/2,19/2).

31. ρ= lim

k→+∞

p|uk|= lim

k→+∞|x|

ln k= 0, the series converges absolutely for all xso the interval of

convergence is (−∞,+∞).

32. ρ= lim

k→+∞

2k+1

(2k)(2k−1)|x|=0

so R=+∞.

33. (a) 10

-1

-1 1

Exercise Set 11.8 392

34. Ratio Test: ρ= lim

k→+∞|x|2

4(k+ 1)(k+2) =0,R=+∞

35. By the Ratio Test for absolute convergence,

ρ= lim

k→+∞

(pk +p)!(k!)p

(pk)![(k+ 1)!]p|x|= lim

k→+∞

(pk +p)(pk +p−1)(pk +p−2) ···(pk +p−[p−1])

(k+1)

p|x|

= lim

k→+∞pp−1

k+1p−2

k+1···p−p−1

k+1|x|=p

|x|,

converges if |x|<1/pp, diverges if |x|>1/pp. Radius of convergence is 1/pp.

36. By the Ratio Test for absolute convergence,

ρ= lim

k→+∞

(k+1+p)!k!(k+q)!

(k+p)!(k+ 1)!(k+1+q)!|x|= lim

k→+∞

k+1+p

(k+ 1)(k+1+q)|x|=0,

radius of convergence is +∞.

37. (a) By Theorem 11.4.3(b) both series converge or diverge together, so they have the same radius

of convergence.

(b) By Theorem 11.4.3(a) the series P(ck+dk)(x−x0)kconverges if |x−x0|<R;if|x−x

|>R

then P(ck+dk)(x−x0)kcannot converge, as otherwise Pck(x−x0)kwould converge by

the same Theorem. Hence the radius of convergence of P(ck+dk)(x−x0)kis R.

|<min(R1,R

then Pck(x−x0)kand Pdk(x−x0)kconverge, so P(ck+dk)(x−x0)kconverges. Hence

r≥min(R1,R

2) (to see that r>min(R1,R

2) is possible consider the case ck=−dk= 1).

If in addition R16=R2, and R1<|x−x0|<R

2(or R2<|x−x0|<R

) then

P(ck+dk)(x−x0)kcannot converge, as otherwise all three series would converge. Thus

in this case r= min(R1,R

2).

38. By the Root Test for absolute convergence,

ρ= lim

k→+∞|ck|1/k|x|=L|x|,L|x|<1if|x|<1/L so the radius of convergence is 1/L.

39. By assumption ∞

k=0

ckxkconverges if |x|<Rso ∞

k=0

ckx2k=∞

k=0

ck(x2)kconverges if |x2|<R,

|x|<√

R. Moreover, ∞

k=0

ckx2k=∞

k=0

ck(x2)kdiverges if |x2|>R,|x|>√

R.Thus∞

k=0

ckx2k

has radius of convergence √R.

40. The assumption is that ∞

k=0

ckRkis convergent and ∞

k=0

ck(−R)kis divergent. Suppose that ∞

k=0

ckRk

is absolutely convergent then ∞

k=0

ck(−R)kis also absolutely convergent and hence convergent

because |ckRk|=|ck(−R)k|, which contradicts the assumption that ∞

k=0

ck(−R)kis divergent so

∞

k=0

ckRkmust be conditionally convergent.

393 Chapter 11

EXERCISE SET 11.9

1. sin 4◦= sin π

45=π

45 −(π/45)3

3! +(π/45)5

5! −···

(a) Method 1: |Rn(π/45)|≤(π/45)n+1

(n+ 1)! <0.000005 for n+1=4,n =3;

sin 4◦≈π

45 −(π/45)3

3! ≈0.069756

(b) Method 2: The ﬁrst term in the alternating series that is less than 0.000005 is (π/45)5

5! ,so

the result is the same as in part (a).

2. cos 3◦= cos π

60=1−(π/60)2

2+(π/60)4

4! −···

(a) Method 1: |Rn(π/60)|≤(π/60)n+1

(n+ 1)! <0.0005 for n= 2; cos 3◦≈1−(π/60)2

2≈0.9986.

(b) Method 2: The ﬁrst term in the alternating series that is less than 0.0005 is (π/60)4

4! , so the

result is the same as in part (a).

3. f(k)(x)=e

,|f

(k)

(x)|≤e

1/2<2on[0,1/2],let M=2,e

1/2=1+1

2+1

8+1

48 +1

24 ·16 +···;

n(1/2)|≤ M

(n+ 1)!(1/2)n+1 ≤2

(n+ 1)!(1/2)n+1 ≤0.00005 for n=5;

1/2≈1+1

2+1

8+1

48 +1

24 ·16 ≈1.64844, calculator value 1.64872

4. f(x)=e

,f(k)(x)=e

,|f

(k)

(x)|≤1on[−1,0], |Rn(x)|≤ 1

(n+ 1)!(1)n+1 =1

(n+ 1)! <0.5×10−3

if n=6,soe

−1≈1−1+ 1

2! −1

3! +1

4! −1

5! +1

6! ≈0.3681, calculator value 0.3679

5. |Rn(0.1)|≤(0.1)n+1

(n+ 1)! ≤0.000005 for n= 3; cos 0.1≈1−(0.1)2/2=0.99500, calculator value

0.995004 ...

6. (0.1)3/3<0.5×10−3so tan−1(0.1) ≈0.100, calculator value ≈0.0997

7. Expand about π/2 to get sinx=1−1

2!(x−π/2)2+1

4!(x−π/2)4−···,85

◦=17π/36 radians,

|Rn(x)|≤|x−π/2|n+1

(n+ 1)! ,|Rn(17π/36)|≤|17π/36 −π/2|n+1

(n+ 1)! =(π/36)n+1

(n+ 1)! <0.5×10−4

if n= 3, sin 85◦≈1−1

2(−π/36)2≈0.99619, calculator value 0.99619 ...

8. −175◦=−π+π/36 rad; x0=−π, x =−π+π/36,cos x=−1+(x+π)

2−(x+π)

4! −···;

n|≤(π/36)n+1

(n+ 1)! ≤0.00005 for n= 3; cos(−π+π/36) = −1+(π/36)2

2≈−0.99619,

calculator value −0.99619 ...

Exercise Set 11.9 394

9. f(k)(x) = cosh xor sinh x, |f(k)(x)|≤cosh x≤cosh 0.5=1



0.5+e

−0.5

<1

(2+1) = 1.5

so |Rn(x)|<1.5(0.5)n+1

(n+ 1)! ≤0.5×10−3if n= 4, sinh 0.5≈0.5+(0.5)3

3! ≈0.5208, calculator

value 0.52109 ...

10. f(k)(x) = cosh xor sinh x, |f(k)(x)|≤cosh x≤cosh 0.1=1

2e

0.1+e

−0.1

<1.06 so |Rn(x)|<

1.06(0.1)n+1

(n+ 1)! ≤0.5×10−3for n= 2, cosh 0.1≈1+(0.1)2

2! =1.005, calculator value 1.0050 ...

11. f(x) = sin x,f(n+1)(x)=±sin xor ±cos x,|f(n+1)(x)|≤1, |Rn(x)|≤|x−π/4|n+1

(n+ 1)! ,

lim

n→+∞|x−π/4|n+1

(n+ 1)! = 0; by the Squeezing Theorem, lim

n→+∞|Rn(x)|=0

so lim

n→+∞Rn(x) = 0 for all x.

12. f(x)=e

(n+1)(x)=e

;ifx>1 then |Rn(x)|≤ e

(n+ 1)!|x−1|n+1;ifx<1 then

|Rn(x)|≤ e

(n+ 1)!|x−1|n+1. But lim

n→+∞|x−1|n+1

(n+ 1)! = 0 so lim

n→+∞Rn(x)=0.

13. (a) Let x=1/9 in series (17).

(b) ln 1.25 ≈21/9+(1/9)3

3= 2(1/9+1/3

7)≈0.223, which agrees with the calculator value

0.22314 ... to three decimal places.

14. (a) Let x=1/2 in series (17).

(b) ln 3 ≈21/2+(1/2)3

3= 2(1/2+1/24)=13/12 ≈1.083; the calculator value is 1.099 to

three decimal places.

15. (a) (1/2)9/9<0.5×10−3and (1/3)7/7<0.5×10−3so

tan−11/2≈1/2−(1/2)3

3+(1/2)5

5−(1/2)7

7≈0.4635

tan−11/3≈1/3−(1/3)3

3+(1/3)5

5≈0.3218

(b) From Formula (21), π≈4(0.4635 + 0.3218) = 3.1412

2,b= tan−11

3; then |a−0.4635|<0.0005 and |b−0.3218|<0.0005, so

|4(a+b)−3.140|≤4|a−0.4635|+4|b−0.3218|<0.004, so two decimal-place accuracy is

guaranteed, but not three.

16. (27+x)1/3= 3(1+x/33)1/3=31+ 1

4x−1·2

82x

2+1·2·5

123! x3+...

, alternates after ﬁrst term,

3·2

382<0.0005, √28 ≈31+ 1

4≈3.0370

395 Chapter 11

17. (a) sin x=x−x3

3! +0·x

4+R

4(x),

4(x)|≤|x|

5! <0.5×10−3if |x|5<0.06,

|x|<(0.06)1/5≈0.569,(−0.569,0.569)

(b) 0.0001

-0.7 0.7

18. (a) f(k)(x)=e

x≤e

(x)|≤e

3! <0.0005,

ebb3<0.003 if b<0.137 (by trial and

error with a hand calculator), so [0,0.136].

(b) 0.0004

-0.2 0.2

19. (a) cos x=1−x

2! +x4

4! + (0)x5+R5(x),

|R5(x)|≤|x|

6! ≤(0.2)6

6! <9×10−8

(b) 0.000000005

-0.2 0.2

20. (a) f00(x)=−1/(1+x)2,

f00(x)<1/(0.99)2≤1.03,

|R1(x)|≤1.03|x|2

2≤1.03(0.01)2

≤5.11 ×10−5for −0.01 ≤x≤0.01

(b) 0.00005

-0.01 0.01

21. (a) (1+x)−1=1−x+−1(−2)

2! x2+−1(−2)(−3)

3! x3+···+−1(−2)(−3) ···(−k)

k!x

k+···

=∞

k=0

(−1)kxk

(b) (1+x)1/3=1+(1/3)x+(1/3)(−2/3)

2! x2+(1/3)(−2/3)(−5/3)

3! x3+···

+(1/3)(−2/3) ···(4 −3k)/3

k!xk+···=1+x/3+ ∞

k=2

(−1)k−12·5···(3k−4)

3kk!xk

2! x2+(−3)(−4)(−5)

3! x3+···+(−3)(−4) ···(−2−k)

k!x

k+···

=∞

k=0

(−1)k(k+ 2)!

2·k!xk=∞

k=0

(−1)k(k+ 2)(k+1)

Exercise Set 11.9 396

22. (1+x)m=m

0+∞

k=1 m

kxk=∞

k=0 m

kxk

23. (a) d

dx ln(1 + x)= 1

1+x,d

dxkln(1 + x)=(−1)k−1(k−1)!

(1+x)k; similarly d

dx ln(1 −x)=−(k−1)!

(1 −x)k,

so f(n+1)(x)=n!(−1)n

(1+x)n+1 +1

(1 −x)n+1 .

(b) f(n+1)(x)≤n!

(−1)n

(1+x)n+1 +n!

(1 −x)n+1 =n!1

(1+x)n+1 +1

(1 −x)n+1 

(n+ 1)! 1

3n+1

(d) If 0 ≤x≤1/3 then 1 + x≥1,1−x≥2/3,f(n+1)(x)≤M=n!1+ 1

(2/3)n+1 .

(e) 0.000005 ≥M

(n+ 1)! 1

3n+1

n+1"1

3n+1

+(1/3)n+1

(2/3)n+1 #=1

n+1"1

3n+1

+1

2n+1#

24. Set x=1/4 in Formula (17). Follow the argument of Exercise 23: parts (a) and (b) remain

unchanged; in part (c) replace (1/3) with (1/4): M

(n+ 1)! 1

4n+1

≤0.000005 for xin the interval

[0,1/4]. From part (b), together with 0 ≤x≤1/4,1+x≥1,1−x≥3/4, follows

part (d): M=n!1+ 1

(3/4)n+1 . Part (e) now becomes 0.000005 ≥M

(n+ 1)! 1

4n+1

n+1"1

4n+1

+1

3n+1#, which is true for n=9.

25. f(x) = cos x, f(n+1)(x)=±sin xor ±cos x, |f(n+1)(x)|≤1,set M=1,

(x)|≤ 1

(n+ 1)!|x−a|n+1,lim

n→+∞|x−a|n+1

(n+ 1)! = 0 so lim

n→+∞Rn(x) = 0 for all x.

26. f(x) = sin x, f(n+1)(x)=±sin xor ±cos x, |f(n+1)(x)|≤1, follow Exercise 25.

27. (a) From Machin’s formula and a CAS, π

4≈0.7853981633974483096156609,accurate to the 25th

decimal place.

(b) n

1/p

0.3183098 78 . . .

0.3183098 861837906 067 . . .

0.3183098 861837906 7153776 695 . . .

0.3183098 861837906 7153776 752674502 34 . . .

0.3183098 861837906 7153776 752674502 87 . . .

28. (a) f0(0) = lim

h→0

f(h)−f(0)

h= lim

h→0

e−1/h2

h, let t=1/h then h=1/t and

lim

h→0+

e−1/h2

h= lim

t→+∞te−t2= lim

t→+∞

et2= lim

t→+∞

2tet2= 0, similarly lim

h→0−

e−1/h2

h=0so

(0)=0.

397 Chapter 11

(b) The Maclaurin series is 0 + 0 ·x+0·x

2+···= 0, but f(0) = 0 and f(x)>0ifx6= 0 so the

series converges to f(x) only at the point x=0.

EXERCISE SET 11.10

1. (a) Replace xwith −x:1

1+x=1−x+x

2−···+(−1)kxk+···;R=1.

(b) Replace xwith x2:1

1−x2=1+x

2+x

4+···+x

2k+···;R=1.

1−2x=1+2x+4x

2+···+2

k+···;R=1/2.

(d) 1

2−x=1/2

1−x/2; replace xwith x/2: 1

2−x=1

2+1

2x+1

2+···+1

k+1 xk+···;R=2.

2. (a) Replace xwith −x: ln(1 −x)=−x−x

/2−x

/3−···−x

k/k −···;R=1.

(b) Replace xwith x2: ln(1 + x2)=x

2−x

/2+x

6/3−···+(−1)k−1x2k/k +···;R=1.

(d) ln(2 + x) = ln 2 + ln(1 + x/2); replace xwith x/2:

ln(2 + x)=ln2+x/2−(x/2)2/2+(x/2)3/3+···+(−1)k−1(x/2)k/k +···;R=2.

3. (a) From Section 11.9, Example 5(b), 1

√1+x=1−1

x+1·3

2·2!x2−1·3·5

23·3! x3+···,so

(2+x)−1/2=1

√2p1+x/2=1

21/2−1

25/2x+1·3

29/2·2!x2−1·3·5

213/2·3!x3+···

(b) Example 5(a): 1

(1+x)2=1−2x+3x

2−4x

3+···,so 1

(1 −x2)2=1+2x

2+3x

4+4x

6+···

4. (a) 1

a−x=1/a

1−x/a =1/a +x/a2+x2/a3+···+x

k/ak+1 +···;R=|a|.

(b) 1/(a+x)2=1

(1+x/a)2=1

a21−2(x/a)+3(x/a)2−4(x/a)3+···

2−2

3x+3

2−4

3+···

5. (a) 2x−23

3! x3+25

5! x5−27

7! x7+···;R=+∞(b) 1−2x+2x

2−4

3+···;R=+∞

2+1

2!x4+1

3!x6+···;R=+∞(d) x2−π2

2x4+π4

4! x6−π6

6! x8+···;R=+∞

6. (a) 1−22

2! x2+24

4! x4−26

6! x6+···;R=+∞

(b) x21+x+1

2!x2+1

3!x3+···=x

2+x

3+1

2!x4+1

3!x5+···;R=+∞

2!x2−1

3!x3+···=x−x

2+1

2!x3−1

3!x4+···;R=+∞

(d) x2−1

3!x6+1

5!x10 −1

7!x14 +···;R=+∞

Exercise Set 11.10 398

7. (a) x21−3x+9x

2−27x3+···=x

2−3x

3+9x

4−27x5+···;R=1/3

(b) x2x+23

3! x3+25

5! x5+27

7! x7+···=2x

2+2

3! x4+25

5! x6+27

7! x8+···;R=+∞

x−3

2x3+3

8x5+1

16x7+···;R=1

8. (a) x

x−1=−x

1−x=−x1+x+x

2+x

3+···=−x−x

2−x

3−x

4−···;R=1.

(b) 3+ 3

2!x4+3

4!x8+3

6!x12 +···;R=+∞

2−10x3+···,so

x(1+2x)

−3=x−6x

2+24x

3−80x4+···;R=1/2

9. (a) sin2x=1

2(1 −cos 2x)=1

21−1−2

2! x2+24

4! x4−26

6! x6+···

=x2−23

4! x4+25

6! x6−27

8! x8+···

(b) ln (1+x3)12= 12 ln(1 + x3)=12x

3−6x

6+4x

9−3x

12 +···

10. (a) cos2x=1

2(1 + cos 2x)=1

21+1−2

2! x2+24

4! x4−26

6! x6+···

=1−x

2+2

4! x4−25

6! x6+···

(b) In Equation (17) of Section 11.9 replace xwith −x:ln



1−x

1+x=−2x+1

3+1

5+···

11. (a) 1

x=1

1−(1 −x)=1+(1−x)+(1−x)

2+···+(1−x)

k+···

=1−(x−1)+(x−1)2−···+(−1)k(x−1)k+···

(b) (0,2)

12. (a) 1

x=1/x0

1+(x−x

0)/x0=1/x0−(x−x0)/x2

0+(x−x

2/x3

0−···+(−1)k(x−x0)k/xk+1

0+···

(b) (0,2x0)

13. (a) (1+x+x2/2+x

3/3!+x4/4!+···)(x−x3/3!+x5/5! −···)=x+x

2+x

/3−x

/30+···

(b) (1+x/2−x2/8+x

3/16 −(5/128)x4+···)(x−x2/2+x

3/3−x

4/4+x

5/5−···)

=x−x

3/24+x4/24 −(71/1920)x5+···

14. (a) (1 −x2+x4/2−x6/6+···)1−1

2+1

24x4−1

720x6···=1−3

2+25

24x4−331

720x6+···

(b) 1+4

2+···1+1

3x−1

2+5

81x3−···=1+1

x+11

9x2+41

81x3+···

399 Chapter 11

15. (a) 1

cos x=1

1−1

2!x2+1

4!x4−1

6!x6+···=1+1

2+5

24x4+61

720x6+···

(b) sin x

ex=x−x3

3! +x5

5! −···1+x+x

2! +x3

3! +x4

4! +···=x−x

2+1

3−1

30x5+···

16. (a) tan−1x

1+x=x−x

3/3+x

5/5−···/(1+x)=x−x

2+2

3−2

4···

(b) ln(1 + x)

1−x=x−x2/2+x

3/3−x

4/4+···/(1 −x)=x+1

2+5

3+7

12x4+···

17. ex=1+x+x

/2+x

3/3! + ···+x

k/k!+···,e

−x=1−x+x

/2−x

/3! + ···+(−1)kxk/k!+···;

sinh x=1

2ex−e−x=x+x3/3!+x5/5!+···+x

2k+1/(2k+1)!+···,R=+∞

cosh x=1

2ex+e−x=1+x

/2+x

4/4!+···+x

2k/(2k)!+···,R=+∞

18. tanh x=x+x3/3!+x5/5!+···

1+x

2/2+x

4/4!+··· =x−1

3+2

15x5−17

315x7+···

19. 4x−2

x2−1=−1

1−x+3

1+x=−1+x+x

2+x

3+x

4+···+31−x+x

2−x

3+x

4+···

=2−4x+2x

2−4x

3+2x

4+···

20. x3+x2+2x−2

2−1=x+1−1

1−x+2

1+x

=x+1−1+x+x

2+x

3+x

4+···+21−x+x

2−x

3+x

4+···

=2−2x+x

2−3x

3+x

4−···

21. (a) d

dx 1−x2/2!+x4/4! −x6/6!+···=−x+x

3/3! −x5/5!+···=−sin x

(b) d

dx x−x2/2+x

3/3−···=1−x+x

2−···=1/(1+x)

22. (a) d

dx x+x3/3!+x5/5!+···=1+x

/2!+x4/4!+···= cosh x

(b) d

dx x−x3/3+x

5/5−x

7/7+···=1−x

2+x

4−x

6+···=1

1+x

23. (a) Z1+x+x

2/2!+···dx =(x+x

/2!+x3/3!+···)+C

=1+x+x

2/2!+x3/3!+···+C

1−1=e

x+C

(b) Zx+x3/3!+x5/5!+···=x

2/2!+x4/4!+···+C

=1+x

/2!+x4/4!+···+C

1−1 = cosh x+C

Exercise Set 11.10 400

24. (a) Zx−x3/3!+x5/5! −···dx =x2/2! −x4/4!+x6/6! −···+C

=−1−x

2/2!+x4/4! −x6/6!+···+C

1+1

=−cos x+C

(b) Z1−x+x2−···dx =x−x2/2+x

3/3−···+C= ln(1 + x)+C

(Note: −1<x<1, so |1+x|=1+x)

25. (a) Substitute x2for xin the Maclaurin Series for 1/(1 −x) (Table 11.9.1)

and then multiply by x:x

1−x2=x∞

k=0

(x2)k=∞

k=0

x2k+1

(b) f(5)(0)=5!c

5=5,f

(6)(0)=6!c

6=0 (c) f(n)(0) = n!cn=(n!ifnodd

0ifneven

26. x2cos 2x=∞

k=0

(−1)k22k

(2k)! x2k+2;f(99)(0) = 0 because c99 =0.

27. (a) lim

x→0

sin x

x= lim

x→01−x2/3!+x4/5! −···=1

(b) lim

x→0

tan−1x−x

x3= lim

x→0x−x3/3+x

5/5−x

7/7+···−x

3=−1/3

28. (a) 1−cos x

sin x=1−1−x2/2!+x4/4! −x6/6!+···

x−x

3/3!+x5/5! −··· =x

2/2! −x4/4!+x6/6! −···

x−x

3/3!+x5/5! −···

=x/2! −x3/4!+x5/6! −···

1−x

2/3!+x4/5! −··· ,x6= 0; lim

x→0

1−cos x

sin x=0

1=0

(b) lim

x→0

xln √1+x−sin 2x= lim

x→0

x1

2ln(1 + x)−sin 2x

= lim

x→0

x1

2x−1

2x2+1

3x3−···−2x−4

3+4

15x5−···

= lim

x→0−3

2−1

4x+3

2x2+···=−3/2

29. Z1

0sin x2dx =Z1

0x2−1

3!x6+1

5!x10 −1

7!x14 +···dx

3x3−1

7·3!x7+1

11 ·5!x11 −1

15 ·7!x15 +···1

3−1

7·3! +1

11 ·5! −1

15 ·7! +···,

but 1

15 ·7! <0.5×10−3so Z1

0sin(x2)dx ≈1

3−1

7·3! +1

11 ·5! ≈0.3103

401 Chapter 11

30. Z1/2

0tan−12x2dx =Z1/2

02x2−8

3x6+32

5x10 −128

7x14 +···dx

3x3−8

21x7+32

55x11 −128

105x15 +···1/2

3−8

27+32

211 −128

105

215 −···,

but 128

105 ·215 <0.5×10−4so Z1/2

0tan−1(2x2)dx ≈2

3·23−8

21 ·27+32

55 ·211 ≈0.0806

31. Z0.2

01+x

41/3dx =Z0.2

01+1

4−1

8+···dx

=x+1

15x5−1

81x9+···0.2

=0.2+ 1

15(0.2)5−1

81(0.2)9+···,

but 1

15(0.2)5<0.5×10−3so Z0.2

0(1+x4)1/3dx ≈0.200

32. Z1/2

0(1+x2)−1/4dx =Z1/2

01−1

4x2+5

32x4−15

128x6+···dx

=x−1

12x3+1

32x5−15

896x7+···1/2

=1/2−1

12(1/2)3+1

32(1/2)5−15

896(1/2)7+···,

but 15

896(1/2)7<0.5×10−3so Z1/2

0(1+x2)−1/4dx ≈1/2−1

12(1/2)3+1

32(1/2)5≈0.4906

33. (a) x

(1 −x)2=xd

dx 1

1−x=xd

dx "∞

k=0

xk#=x"∞

k=1

kxk−1#=∞

k=1

kxk

(b) −ln(1 −x)=Z1

1−xdx −C=Z"∞

k=0

xk#dx −C

=∞

k=0

xk+1

k+1−C=∞

k=1

k−C, −ln(1 −0)=0soC=0.

+∞

k=1

(−1)k

kxk=

+∞

k=1

(−1)k+1

kxk

(d)

+∞

k=1

(−1)k+1

kconverges by the Alternating Series Test.

(e) By parts (c) and (d) and the remark,

+∞

k=1

(−1)k+1

kxkconverges to ln(1 + x) for −1<x≤1.

Exercise Set 11.10 402

34. (a) In Exercise 33(a), set x=1

3,S=1/3

(1 −1/3)2=3

(b) In part (b) set x=1/4,S = ln(4/3) (c) In part (e) set x=1,S =ln2

35. (a) sinh−1x=Z1+x

2−1/2dx −C=Z1−1

2x2+3

8x4−5

16x6+···dx −C

=x−1

6x3+3

40x5−5

112x7+···−C; sinh−10=0soC=0.

(b) 1+x

2−1/2=1+ ∞

k=1

(−1/2)(−3/2)(−5/2) ···(−1/2−k+1)

k!(x

=1+ ∞

k=1

(−1)k1·3·5···(2k−1)

2kk!x2k,

sinh−1x=x+∞

k=1

(−1)k1·3·5···(2k−1)

2kk!(2k+1) x

2k+1

36. (a) sin−1x=Z(1 −x2)−1/2dx −C=Z1+1

2+3

4+5

16x6+···dx −C

=x+1

6x3+3

40x5+5

112x7+···−C, sin−10=0soC=0

(b) 1−x2−1/2=1+ ∞

k=1

(−1/2)(−3/2)(−5/2) ···(−1/2−k+1)

k!−x

2k

=1+ ∞

k=1

(−1)k(1/2)k(1)(3)(5) ···(2k−1)

k!(−1)kx2k

=1+ ∞

k=1

1·3·5···(2k−1)

2kk!x2k

sin−1x=x+∞

k=1

1·3·5···(2k−1)

2kk!(2k+1) x

2k+1

37. (a) y(t)=y

∞

k=0

(−1)k(0.000121)ktk

(b) y(1) ≈y0(1 −0.000121t)it=1 =0.999879y0

38. (a) If ct

m≈0 then e−ct/m ≈1−ct

m, and v(t)≈1−ct

mv0+mg

c−mg

c=v0−cv0

m+gt.

(b) The quadratic approximation is

v0≈1−ct

m+(ct)2

2m2v0+mg

c−mg

c=v0−cv0

m+gt+c2

2m2v0+mg

ct2.

403 Chapter 11

39. θ0=5

◦=π/36 rad, k= sin(π/72)

(a) T≈2πrL

g=2π

p1/9.8≈2.00709 (b) T≈2πrL

g1+k

4≈2.008044621

40. The third order model gives the same result as the second, because there is no term of degree three

in (5). By the Wallis sine formula, Zπ/2

0sin4φdφ=1·3

2·4

2, and

T≈4rL

gZπ/2

01+1

2sin2φ+1·3

222!k4sin4φdφ =4

gπ

2+k

4+3k

3π

16 

=2π

g1+k

4+9k

64 

41. (a) F=mgR2

(R+h)2=mg

(1+h/R)2=mg 1−2h/R +3h

2/R2−4h3/R3+···

(b) If h= 0, then the binomial series converges to 1 and F=mg.

(d) mg −2mgh/R

mg =1−2h

R=1−2·29,028

4000 ·5280 ≈0.9973, so about 0.27% less.

42. (a) We can diﬀerentiate term-by-term:

y0=∞

k=1

(−1)kx2k−1

22k−1k!(k−1)! =∞

k=0

(−1)k+1x2k+1

22k+1(k+ 1)!k!,y

00 =∞

k=0

(−1)k+1(2k+1)x

2k+1(k+ 1)!k!, and

xy00 +y0+xy =∞

k=0

(−1)k+1(2k+1)x

2k+1

22k+1(k+ 1)!k!+∞

k=0

(−1)k+1x2k+1

22k+1(k+ 1)!k!+∞

k=0

(−1)kx2k+1

22k(k!)2,

xy00 +y0+xy =∞

k=0

(−1)k+1x2k+1

22k(k!)22k+1

2(k+1)+1

2(k+1)−1

=0

(b) y0=∞

k=0

(−1)k(2k+1)x

2k+1k!(k+ 1)! ,y

00 =∞

k=1

(−1)k(2k+1)x

2k−1

(k−1)!(k+ 1)! .

Since J1(x)= ∞

k=0

(−1)kx2k+1

22k+1k!(k+ 1)! and x2J1(x)= ∞

k=1

(−1)k−1x2k+1

22k−1(k−1)!k!, it follows that

x2y00 +xy0+(x

2−1)y

=∞

k=1

(−1)k(2k+1)x

2k+1

22k(k−1)!(k+ 1)! +∞

k=0

(−1)k(2k+1)x

2k+1

22k+1(k!)(k+ 1)! +∞

k=1

(−1)k−1x2k+1

22k−1(k−1)!k!

−∞

k=0

(−1)kx2k+1

22k+1k!(k+ 1)!

2−x

2+∞

k=1

(−1)kx2k+1

22k−1(k−1)!k!2k+1

2(k+1)+2k+1

4k(k+1)−1−1

4k(k+1)=0.

Chapter 11 Supplementary Exercises 404

0(x)= ∞

k=0

(−1)k+1x2k+1

22k+1(k+ 1)!k!=−J1(x).

43. If ∞

k=0

akxk=∞

k=0

bkxkfor xin (−r,r) then ∞

k=0

(ak−bk)xk= 0 so by Theorem 11.10.6 ∞

k=0

(ak−bk)xk

is the Taylor series for f(x) = 0 about 0 and hence ak−bk=0,a

k=b

kfor all k.

CHAPTER 11 SUPPLEMENTARY EXERCISES

4. (a) ∞

k=0

f(k)(0)

k!xk(b) ∞

k=0

f(k)(x0)

k!(x−x0)k

9. (a) always true by Theorem 11.4.2

(b) sometimes false, for example the harmonic series diverges but P(1/k2) converges

(d) always true by Example 3(d) of Section 11.1

(e) sometimes false, for example an=1

2+(−1)n1

(f) sometimes false, for example uk=1/2

(g) always false by Theorem 11.4.3

(h) sometimes false, for example uk=vk=(2/k)

(i) always true by the Comparison Test

(j) always true by the Comparison Test

(k) sometimes false, for example P(−1)k/k

(l) sometimes false, for example P(−1)k/k

10. (a) false, f(x) is not diﬀerentiable at x= 0, Deﬁnition 11.5.4

(b) true: sn=1ifnis odd and s2n=1+1/(n+ 1); lim

n→+∞sn=1

11. (a) geometric, r=1/5, converges (b) 1/(5k+1)<1/5

, converges

√k+1 ≥9

√

k+√

k=9

√

k,∞

k=1

2√kdiverges

12. (a) converges by Alternating Series Test

(b) absolutely convergent, ∞

k=1 k+2

3k−1k

converges by the Root Test.

2 + sin2k>k−1

2+1 =1

3k,∞

k=1

3kdiverges

13. (a) 1

k3+2k+1 <1

3,∞

k=1

1/k3converges, so ∞

k=1

k3+2k+1 converges by the Comparison Test

(b) Limit Comparison Test, compare with the divergent series ∞

k=1

k2/5, diverges

405 Chapter 11

cos(1/k)

k2<1

k2,∞

k=1

k2converges, so ∞

k=1

cos(1/k)

k2converges absolutely

14. (a) ∞

k=1

ln k

k√k=∞

k=2

ln k

k√kbecause ln1 = 0,

Z+∞

ln x

x3/2dx = lim

`→+∞−2lnx

1/2−4

1/2`

=√2(ln2 + 2) so ∞

k=2

ln k

k3/2converges

(b) k4/3

8k2+5k+1 ≥k

4/3

2+5k

2+k

2=1

14k2/3,∞

k=1

14k2/3diverges

k=1

k2+1 converges (compare with P1/k2)

15. ∞

k=0

5k−

k=0

5k=∞

k=100

5k=1

5100

∞

k=0

5k=1

4·599

16. no, lim

k→+∞ak=1

26= 0 (Divergence Test)

17. (a) p0(x)=1,p

1(x)=1−7x, p2(x)=1−7x+5x

3(x)=1−7x+5x

2−4x

(x)=1−7x+5x

2−4x

(b) If f(x) is a polynomial of degree nand k≥nthen the Maclaurin polynomial of degree kis

the polynomial itself; if k<nthen it is the truncated polynomial.

18. ln(1 + x)=x−x

/2+···;so|ln(1 + x)−x|≤x

/2 by Theorem 11.7.2.

19. sin x=x−x3/3!+x5/5! −x7/7!+···is an alternating series, so

|sin x−x+x3/3! −x5/5!|≤x

/7! ≤π7/(477!) ≤0.00005

20. Z1

1−cos x

xdx =x2

2·2! −x4

4·4! +x6

6·6! −···1

2·2! −1

4·4! +1

6·6! −···, and 1

6·6! <0.0005,

so Z1

1−cos x

xdx =1

2·2! −1

4·4! =0.2396 to three decimal-place accuracy.

21. (a) ρ= lim

k→+∞2k

k!1/k

= lim

k→+∞

√k!= 0, converges

(b) ρ= lim

k→+∞u1/k

k= lim

k→+∞

√k!=e, diverges

22. (a) 11≥1!; if kk≥k!,then (k+1)k+1 ≥(k+1)kk≥(k+1)k!=(k+1)!; mathematical induction

(b) X1

kk≤X1

k!, converges

Chapter 11 Supplementary Exercises 406

k→+∞1

kk1/k

= lim

k→+∞

k= 0, converges

23. (a) u100 =

100

k=1

uk−

k=1

uk=2−1

100−2−1

99=1

9900

(b) u1= 1; for k≥2,u

k=2−1

k−2−1

k−1=1

k(k−1),lim

k→+∞uk=0

k=1

uk= lim

n→+∞

k=1

uk= lim

n→+∞2−1

n=2

24. (a) ∞

k=1 3

2k−2

3k=∞

k=1

2k−∞

k=1

3k=3

21

1−(1/2) −2

31

1−(1/3) = 2 (geometric series)

(b)

k=1

[ln(k+1)−ln k] = ln(n+ 1), so ∞

k=1

[ln(k+1)−ln k] = lim

n→+∞ln(n+1)=+∞, diverges

n→+∞

k=1

21

k−1

k+2= lim

n→+∞

21+1

2−1

n+1−1

n+2=3

(d) lim

n→+∞

k=1 tan−1(k+1)−tan−1k= lim

n→+∞tan−1(n+1)−tan−1(1)=π

2−π

4=π

25. (a) e2−1(b) sin π=0 (c) cos e(d) e−ln 3 =1/3

26. ak+1 =√ak=a1/2

k=a1/4

k−1=···=a

1/2

1=c

1/2

k+1

(a) If c=1/2 then lim

k→+∞ak=1. (b) if c=3/2 then lim

k→+∞ak=1.

27. Expand in a Maclaurin Series about x= 100 : e−[(x−100)/16]2=1−(x−100)2

162+(x−100)4

2·164,so

p≈1

16√2πZ110

100 1−(x−100)2

162+(x−100)4

2·164dx =1

16√2π10 −103

3·162+105

2·5·164

≈0.220678, or about 22.07%.

28. f(x)=xex=x+x2+x3

2! +x4

3! +···=∞

k=0

xk+1

k!,

f0(x)=(x+1)e

x=1+2x+3x

2! +4x3

3! +···=∞

k=0

k+1

k!x

;∞

k=0

k+1

k!=f0(1)=2e.

29. Let A=1−1

2+1

2−1

2+···; since the series all converge absolutely,

π2

6−A=21

2+21

2+···=1

21+ 1

2+1

2+···=1

6,soA=1

6=π

12.

30. Compare with 1/kp: converges if p>1, diverges otherwise.

407 Chapter 11

31. (a) x+1

2x2+3

14x3+3

35x4+···;ρ= lim

k→+∞

k+1

3k+1|x|=1

3|x|,

converges if 1

3|x|<1, |x|<3soR=3.

(b) −x3+2

3x5−2

5x7+8

35x9−···;ρ= lim

k→+∞

k+1

2k+1|x|

2=1

2|x|

converges if 1

2|x|2<1, |x|2<2, |x|<√2soR=√

32. By the Ratio Test for absolute convergence, ρ= lim

k→+∞|x−x0|

b=|x−x0|

b; converges if

|x−x0|<b, diverges if |x−x0|>b.Ifx=x

−b,∞

k=0

(−1)kdiverges; if x=x0+b,

∞

k=0

1 diverges. The interval of convergence is (x0−b, x0+b).

33. If x≥0, then cos √x=1−(

√

2! +(√x)4

4! −(√x)6

6! +···=1−x

2! +x2

4! −x3

6! +···;ifx≤0, then

cosh(√−x)=1+(

√

−x)

2! +(√−x)4

4! +(√−x)6

6! +···=1−x

2! +x2

4! −x3

6! +···.

34. By Exercise 70 of Section 3.5, the derivative of an odd (even) function is even (odd); hence all

odd-numbered derivatives of an odd function are even, all even-numbered derivatives of an odd

function are odd; a similar statement holds for an even function.

(a) If f(x) is an even function, then

2f0(0) = lim

h→0+

f(0+h)−f(0)

h+ lim

h→0+

f(0) −f(0 −h)

h= lim

h→0+

f(0+h)−f(0 −h)

h=0,so

0(0)=0. Iff(x) is even then so is f(2k),thusf

(2k+1)(0)=0,k=0,1,2,.... Hence

c2k+1 =f(2k+1)(0)/(2k+1)!=0.

(b) If f(x) is an odd function, then f(2k−1) is even (k=1,2,...), and thus by Part (a),

f(2k)(0)=0,c

2k=f

(2k)(0)/(2k)!=0.

35. 1−v2

c2−1/2

≈1+ v

2,soK=m

2"1

p1−v

/c2−1#≈m0c2(v2/(2c2)=m

36. (a) Z+∞

x3.7dx < (0.5)10−5if n>63.7; let n= 64.

(b) sn≈1.10628342; CAS: 1.10628824

408

CHAPTER 12

Analytic Geometry in Calculus

EXERCISE SET 12.1

(1, 6)

(3, 3)

(4, e)(–1, r)

p/2

(5, 8)

(–6, –p)

( , L)

p/2

(–3, i)

(–5, @)

(2, $)

(0, c)

(2, g)

3. (a) (3√3,3) (b) (−7/2,7√3/2) (c) (3√3,3)

(d) (0,0) (e) (−7√3/2,7/2) (f) (−5,0)

4. (a) (−4√2,−4√2) (b) (7√2/2,−7√2/2) (c) (4√2,4√2)

(d) (5,0) (e) (0,−2) (f) (0,0)

5. (a) both (5,π)(b) (4,11π/6),(4,−π/6) (c) (2,3π/2),(2,−π/2)

(d) (8√2,5π/4),(8√2,−3π/4) (e) both (6,2π/3) (f) both (√2,π/4)

6. (a) (2,5π/6) (b) (−2,11π/6) (c) (2,−7π/6) (d) (−2,−π/6)

7. (a) (5,0.6435) (b) (√29,5.0929) (c) (1.2716,0.6658)

8. (a) (5,2.2143) (b) (3.4482,2.6260) (c) (p4+π

2/36,0.2561)

9. (a) r2=x2+y2= 4; circle (b) y= 4; horizontal line

(d) 3rcos θ+2rsin θ=6,3x+2y= 6; line

10. (a) rcos θ=5,x= 5; vertical line

(b) r2=2rsin θ,x2+y2=2y,x

2+(y−1)2= 1; circle

(d) r=1

cos θ

sin θ

cos θ,rcos2θ= sin θ,r2cos2θ=rsin θ,x2=y; parabola

11. (a) rcos θ=7 (b) r=3

(d) 4(rcos θ)(rsin θ)=9,4r

2sin θcos θ=9,r

2sin 2θ=9/2

12. (a) rsin θ=−3(b) r=√5

(d) r4cos2θ=r2sin2θ,r2= tan2θ,r= tan θ

409 Chapter 12

13.

p/2

-3

r= 3 sin 2θ

14. 2

-2

-2 2

r= 2 cos 3θ

15.

p/2

-4 4

-1

r=3−4 sin 3θ

16.

p/2

r=2+2sinθ

17. (a) r=5

(b) (x−3)2+y2=9,r= 6 cos θ

18. (a) From (8-9), r=a±bsin θor r=a±bcos θ. The curve is not symmetric about the y-axis,

so Theorem 12.2.1(a) eliminates the sine function, thus r=a±bcos θ. The cartesian point

(−3,0) is either the polar point (3,π)or(−3,0), and the cartesian point (−1,0) is either

the polar point (1,π)or(−1,0). A solution is a=1,b =−2; we may take the equation as

r=1−2 cos θ.

(b) x2+(y+3/2)2=9/4,r =−3 sin θ

19. (a) Figure 12.1.18, a=3,n =2,r = 3 sin 2θ

(b) From (8-9), symmetry about the y-axis and Theorem 12.1.1(b), the equation is of the form

r=a±bsin θ. The cartesian points (3,0) and (0,5) give a=3and5=a+b,sob= 2 and

r=3+2sinθ.

20. (a) Example 6 rotated through π/2 radian: a=3,r =3−3 sin θ

(b) Figure 12.1.18, a=1,r = cos 5θ

Exercise Set 12.1 410

21.

Line

22.

Line

(

23.

Circle

24. 4

Circle

25.

Circle

26.

Cardioid

27.

Circle

28.

Cardioid

29.

Cardioid

30.

Cardioid

31. 4

Cardioid

32.

Limaçon

33.

Cardioid

34.

Limaçon

35.

Limaçon

36.

Limaçon

37.

Limaçon

38.

Limaçon

411 Chapter 12

39.

Limaçon

40.

Limaçon

41.

Lemniscate

42. 1

Lemniscate

43.

Lemniscate

444.

Spiral

45.

Spiral

2p4p

46. 2p

Spiral

47.

Four-petal rose

48.

Four-petal rose

49.

Eight-petal rose

50.

Three-petal rose

52. 1

-1

-1 1

53. 3

-3

-3 3

54. 2

-2

-2 2

55. 1

-1

-1 1

Exercise Set 12.1 412

56. 0≤θ≤8π57. (a) −4π<θ<4π

58. In I, along the x-axis, x=rgrows ever slower with θ.InIIx=rgrows linearly with θ.

Hence I: r=√θ;II:r=θ.

59. (a) r=a/ cos θ, x =rcos θ=a, a family of vertical lines

(b) r=b/ sin θ, y =rsin θ=b, a family of horizontal lines

60. The image of (r0,θ

0) under a rotation through an angle αis (r0,θ

0+α). Hence (f(θ),θ) lies on

the original curve if and only if (f(θ),θ+α) lies on the rotated curve, i.e. (r, θ) lies on the rotated

curve if and only if r=f(θ−α).

61. (a) r= 1 + cos(θ−π/4)=1+√

2(cos θ+ sin θ)

(b) r= 1 + cos(θ−π/2) = 1 + sin θ

(d) r= 1 + cos(θ−5π/4)=1−√

2(cos θ+ sin θ)

62. r2= 4 cos 2(θ−π/2) = −4 cos 2θ

63. Either r−1=0orθ−1=0,

so the graph consists of the

circle r= 1 and the line θ=1.

r = 1

u = 1

64. (a) r2=Ar sin θ+Br cos θ,x2+y2=Ay +Bx, (x−B/2)2+(y−A/2)2=(A

2+B

)/4, which

is a circle of radius 1

4pA2+B2.

(b) Formula (4) follows by setting A=0,B =2a, (x−a)2+y2=a2, the circle of radius aabout

(a, 0). Formula (5) is derived in a similar fashion.

65. y=rsin θ= (1 + cos θ) sin θ= sin θ+ sin θcos θ,

dy/dθ = cos θ−sin2θ+ cos2θ= 2 cos2θ+ cos θ−1=(2cosθ−1)(cos θ+ 1);

dy/dθ = 0 if cos θ=1/2orifcosθ=−1; θ=π/3orπ.

If θ=π/3,π, then y=3

√

3/4,0 so the maximum value of yis 3√3/4 and the polar coordinates

of the highest point are (3/2,π/3).

66. x=rcos θ= (1 + cos θ) cos θ= cos θ+ cos2θ,dx/dθ =−sin θ−2 sin θcos θ=−sin θ(1 + 2 cos θ),

dx/dθ = 0 if sin θ= 0 or if cos θ=−1/2; θ=0,2π/3, or π.Ifθ=0,2π/3, π, then x=2,−1/4,0

so the minimum value of xis −1/4. The leftmost point has polar coordinates (1/2,2π/3).

67. (a) Let (x1,y

1) and (x2,y

2) be the rectangular coordinates of the points (r1,θ

1) and (r2,θ

2) then

d=p(x2−x1)2+(y

2−y

2=p(r

2cos θ2−r1cos θ1)2+(r

2sin θ2−r1sin θ1)2

=pr2

1+r2

2−2r1r2(cos θ1cos θ2+ sin θ1sin θ2)=pr

1+r

2−2r

2cos(θ1−θ2).

(b) Let Pand Qhave polar coordinates (r1,θ

1),(r

2,θ

2), respectively, then the perpendicular

from OQ to OP has length h=r2sin(θ2−θ1) and A=1

2hr1=1

2r1r2sin(θ2−θ1).

413 Chapter 12

(d) A=1

22 sin(5π/6−π/3)=1

68. (a) 0=(x

2+y

2+a

)

2−a

4−4a

2=x

4+y

4+a

4+2x

2+2x

2+2y

2−a

4−4a

4+y

4+2x

2−2x

2+2y

2=(x

2+y

)

2+2a

2(y

2−x

4+2a

2(sin2θ−cos2θ)=r

4−2a

2cos 2θ, so r2=2a

2cos 2θ

(b) (x2+a2+y2)2−4x2a2=a4;(x−a)2+y2(x+a)2+y2=a4;

p(x+a)2+y2p(x−a)+y

2=a

69. lim

θ→0+y= lim

θ→0+rsin θ= lim

θ→0+

sin θ

θ=1

70. lim

θ→0±y= lim

θ→0±rsin θ= lim

θ→0±

sin θ

θ2= lim

θ→0±

sin θ

θlim

θ→0±

θ=1·lim

θ→0±

θ, so lim

θ→0±ydoes not exist.

71. Note that r→±∞as θapproaches odd multiples of π/2;

x=rcos θ= 4 tan θcos θ= 4 sin θ,

y=rsin θ= 4 tan θsin θ

so x→±4 and y→±∞as θapproaches

odd multiples of π/2.4

-4

72. lim

θ→(π/2)−x= lim

θ→(π/2)−rcos θ= lim

θ→(π/2)−2 sin2θ=2,x= 2 is a vertical asymptote.

73. Let r=asin nθ (the proof for r=acos nθ is similar). If θstarts at 0, then θwould have to increase

by some positive integer multiple of πradians in order to reach the starting point and begin to

retrace the curve. Let (r, θ) be the coordinates of a point Pon the curve for 0 ≤θ<2π.Now

asin n(θ+2π)=asin(nθ +2πn)=asin nθ =rso Pis reached again with coordinates (r, θ +2π)

thus the curve is traced out either exactly once or exactly twice for 0 ≤θ<2π. If for 0 ≤θ<π,

P(r, θ) is reached again with coordinates (−r, θ +π) then the curve is traced out exactly once for

0≤θ<π, otherwise exactly once for 0 ≤θ<2π. But

asin n(θ+π)=asin(nθ +nπ)=asin nθ, n even

−asin nθ, n odd

so the curve is traced out exactly once for 0 ≤θ<2πif nis even, and exactly once for 0 ≤θ<π

if nis odd.

EXERCISE SET 12.2

1. (a) dy/dx =1/2

2t=1/(4t); dy/dxt=−1=−1/4; dy/dxt=1 =1/4

(b) x=(2y)

2+1, dx/dy =8y, dy/dxy=±(1/2) =±1/4

2. (a) dy/dx = (4 cos t)/(−3 sin t)=−(4/3) cot t;dy/dxt=π/4=−4/3, dy/dxt=7π/4=4/3

(b) (x/3)2+(y/4)2=1,2x/9+(2y/16)(dy/dx)=0, dy/dx =−16x/9y,

dy/dxx=3/√2

y=4/√2

=−4/3; dy/dxx=3/√2

y=−4/√2

=4/3

Exercise Set 12.2 414

3. d2y

dx2=d

dx =d

dt dy

dxdt

dx =−1

4t2(1/2t)=−1/(8t3); positive when t=−1,

negative when t=1

4. d2y

dx2=d

dt dy

dxdt

dx =−(4/3)(−csc2t)

−3 sin t=−4

9csc3t; negative at t=π/4, positive at t=7π/4.

5. dy/dx =2

1/(2√t)=4

√

t,d

y/dx2=2/√t

1/(2√t)=4,dy/dxt=1 =4, d

y/dx2t=1 =4

6. dy/dx =t2

t=t,d2y/dx2=1

t,dy/dxt=2 =2, d

y/dx2t=2 =1/2

7. dy/dx =sec2t

sec ttan t= csc t,d2y/dx2=−csc tcot t

sec ttan t=−cot3t,

dy/dxt=π/3=2/

√

3, d2y/dx2t=π/3=−1/(3√3)

8. dy/dx =sinh t

cosh t= tanh t,d2y

dx2= sech2t/ cosh t= sech3t,dy/dxt=0 =0,d

2y/dx2t=0 =1

9. dy

dx =dy/dθ

dx/dθ =−cos θ

2−sin θ;d2y

dx2=d

dθ dy

dx/dx

dθ =2

(2 −sin θ)2

2−sin θ=1

(2 −sin θ)3;

dxθ=π/3=−1/2

2−√3/2=−1

4−√3;d2y

dx2θ=π/3=1

(2 −√3/2)3=8

(4 −√3)3

10. dy

dx =3 cos φ

−sin φ=−3 cot φ;d2y

dx2=d

dφ(−3 cot φ)dφ

dx =−3(−csc2φ)(−csc φ)=−3 csc3φ;

dxφ=5π/6

√

3; d2y

dx2φ=5π/6

=−24

11. (a) dy/dx =−e−t

et=−e−2t; for t=1,dy/dx =−e−2,(x, y)=(e, e−1); y−e−1=−e−2(x−e),

y=−e−2x+2e

−1

(b) y=1/x, dy/dx =−1/x2,m=−1/e2,y−e

−1=−1

2(x−e),y =−1

2x+2

12. dy/dx =16t−2

2=8t−1; for t=1,dy/dx =7,(x, y)=(6,10); y−10=7(x−6), y=7x−32

13. dy/dx =4 cos t

−2 sin t=−2 cot t

(a) dy/dx = 0 if cot t=0,t=π/2+nπ for n=0,±1,···

(b) dx/dy =−1

2tan t= 0 if tan t=0,t=nπ for n=0,±1,···

14. dy/dx =2t+1

2−30t+24 =2t+1

6(t−1)(t−4)

(a) dy/dx =0ift=−1/2

(b) dx/dy =6(t−1)(t−4)

2t+1 =0ift=1,4

415 Chapter 12

15. x=y= 0 when t=0,π;dy

dx =2 cos 2t

cos t;dy

dxt=0

=2,dy

dxt=π

=−2,the equations of the tangent

lines are y=−2x, y =2x.

16. y(t) = 0 has three solutions, t=0,±π/2; the last two correspond to the crossing point.

For t=±π/2, m=dy

dx =2

±π; the tangent lines are given by y=±2

π(x−2).

17. If y= 4 then t2=4,t=±2, x= 0 for t=±2so(0,4) is reached when t=±2.

dy/dx =2t/(3t2−4). For t=2,dy/dx =1/2 and for t=−2, dy/dx =−1/2. The tangent lines

are y=±x/2+4.

18. If x= 3 then t2−3t+5=3,t

2−3t+2=0,(t−1)(t−2)=0,t= 1 or 2. If t= 1 or 2 then

y= 1 so (3,1) is reached when t= 1 or 2. dy/dx =(3t

2+2t−10)/(2t−3). For t=1,dy/dx =5,

the tangent line is y−1=5(x−3), y=5x−14. For t=2,dy/dx = 6, the tangent line is

y−1=6(x−3), y=6x−17.

19. (a) 1

-1

-1 1

(b) dx

dt =−3 cos2tsin tand dy

dt = 3 sin2tcos tare both zero when t=0,π/2,π,3π/2,2π,

so singular points occur at these values of t.

20. (a) when y=0

(b) dy

dx =asin θ

a−acos θ= 0 when θ=2nπ +π/2,n=0,1,... (which is when y= 0).

21. Substitute θ=π/3, r= 1, and dr/dθ =−√3 in equation (7) gives slope m=1/

√

22. As in Exercise 21, θ=π/4, dr/dθ =√2/2, r=1+√

2/2, m=−1−√2

23. As in Exercise 21, θ=2,dr/dθ =−1/4, r=1/2, m=tan 2 −2

2 tan 2 + 1

24. As in Exercise 21, θ=π/6, dr/dθ =4

√

3a,r=2a,m=3

√

3/5

25. As in Exercise 21, θ=3π/4, dr/dθ =−3√2/2, r=√2/2, m=−2

26. As in Exercise 21, θ=π,dr/dθ =3,r=4,m=4/3

27. m=dy

dx =rcos θ+ (sin θ)(dr/dθ)

−rsin θ+ (cos θ)(dr/dθ)=cos θ+ 2 sin θcos θ

−sin θ+ cos2θ−sin2θ;ifθ=0,π/2,π,

then m=1,0,−1.

28. m=dy

dx =cos θ(4 sin θ−1)

4 cos2θ+ sin θ−2;ifθ=0,π/2,π then m=−1/2,0,1/2.

Exercise Set 12.2 416

29. dx/dθ =−asin θ(1 + 2 cos θ), dy/dθ =a(2 cos θ−1)(cos θ+1)

(a) horizontal if dy/dθ = 0 and dx/dθ 6=0.dy/dθ = 0 when cos θ=1/2orcosθ=−1soθ=π/3,

5π/3, or π;dx/dθ 6= 0 for θ=π/3 and 5π/3. For the singular point θ=πwe ﬁnd that

lim

θ→πdy/dx = 0. There is a horizontal tangent line at (3a/2,π/3),(0,π), and (3a/2,5π/3).

(b) vertical if dy/dθ 6= 0 and dx/dθ =0. dx/dθ = 0 when sin θ= 0 or cos θ=−1/2soθ=0,π,

2π/3, or 4π/3; dy/dθ 6= 0 for θ=0,2π/3, and 4π/3. The singular point θ=πwas discussed

in part (a). There is a vertical tangent line at (2a, 0),(a/2,2π/3), and (a/2,4π/3).

30. dx/dθ =a(cos2θ−sin2θ)=acos 2θ, dy/dθ =2asin θcos θ=asin 2θ

(a) horizontal if dy/dθ = 0 and dx/dθ 6=0. dy/dθ = 0 when θ=0,π/2,π,3π/2,2π;dx/dθ 6=0

for (0,0),(a, π/2),(0,π),(−a, 3π/2),(0,2π); in reality only two distinct points

(b) vertical if dy/dθ 6= 0 and dx/dθ =0. dx/dθ = 0 when θ=π/4,3π/4,5π/4,7π/4; dy/dθ 6=0

there, so vertical tangent line at (a/√2,π/4),(a/√2,3π/4),(−a/√2,5π/4),(−a/√2,7π/4),

only two distinct points

31. dy/dθ =(d/dθ)(sin2θcos2θ) = (sin 4θ)/2=0atθ=0,π/4,π/2,3π/4,π; at the same points,

dx/dθ =(d/dθ)(sin θcos3θ) = cos2θ(4 cos2θ−3). Next, dx

dθ =0atθ=π/2, a singular point; and

θ=0,π both give the same point, so there are just three points with a horizontal tangent.

32. dx/dθ = 4 sin2θ−sin θ−2, dy/dθ = cos θ(1 −4 sin θ). dy/dθ = 0 when cos θ= 0 or sin θ=1/4so

θ=π/2, 3π/2, sin−1(1/4), or π−sin−1(1/4); dx/dθ 6= 0 at these points, so there is a horizontal

tangent at each one.

33.

p/2

θ0=π/6,π/2,5π/6

34.

p/2

θ0=π/2

35.

p/2

θ0=±π/4

36.

p/2

θ0=0,π/2

37.

p/2

θ0=2π/3,4π/3

38.

p/2

θ0=0

39. r2+(dr/dθ)2=a2+0

2=a

2,L=Z2π

0adθ =2πa

417 Chapter 12

40. r2+(dr/dθ)2=(2acos θ)2+(−2asin θ)2=4a

,L=Zπ/2

−π/22adθ =2πa

41. r2+(dr/dθ)2=[a(1 −cos θ)]2+[asin θ]2=4a

2sin2(θ/2), L=2Zπ

02asin(θ/2)dθ =8a

42. r2+(dr/dθ)2= [sin2(θ/2)]2+ [sin(θ/2) cos(θ/2)]2= sin2(θ/2), L=Zπ

0sin(θ/2)dθ =2

43. r2+(dr/dθ)2=(e

3θ

)

2+(3e

3θ)

2=10e

6θ

,L=Z2

√

10e3θdθ =√10(e6−1)/3

44. r2+(dr/dθ)2= [sin3(θ/3)]2+ [sin2(θ/3) cos(θ/3)]2= sin4(θ/3),

L=Zπ/2

0sin2(θ/3)dθ =(2π−3

√

3)/8

45. (a) From (3), dy

dx =3 sin t

1−3 cos t

(b) At t=10,dy

dx =3 sin 10

1−3 cos 10 ≈−0.4640,θ ≈tan−1(−0.4640) = −0.4344

46. (a) dy

dx = 0 when dy

dt = 2 sin t=0,t=0,π,2π, 3π

(b) dx

dt = 0 when 1 −2 cos t=0,cos t=1/2,t =π/3,5π/3,7π/3

47. (a) r2+(dr/dθ)2= (cos nθ)2+(−nsin nθ)2= cos2nθ +n2sin2nθ

=(1−sin2nθ)+n

2sin2nθ =1+(n

2−1) sin2nθ,

L=2Zπ/(2n)

0q1+(n

2−1) sin2nθdθ

(b) L=2Zπ/4

0p1+3sin

22θdθ ≈2.42

2.42211

2.22748

2.14461

2.10100

2.07501

2.05816

2.04656

2.03821

2.03199

2.02721

2.02346

2.02046

2.01802

2.01600

2.01431

2.01288

2.01167

2.01062

2.00971

48. (a)

p/2

(b) r2+(dr/dθ)2=(e

−θ

)

2+(−e

−θ)

2=2e

−2θ

,L=

Z+∞

√

−θ

dθ

θ0→+∞Zθ0

√2e−θdθ = lim

θ0→+∞

√2(1 −e−θ0)=√

Exercise Set 12.2 418

49. x0=2t,y

0=2,(x

2+(y

2=4t

2+4

S=2πZ4

0(2t)p4t2+4dt =8πZ4

2+1dt =8π

3(t2+1)

7/24

=8π

3(17√17 −1)

50. x0=et(cos t−sin t), y0=et(cos t+ sin t), (x0)2+(y

2=2e

S=2πZπ/2

0(etsin t)√2e2tdt =2

√

2πZπ/2

0e2tsin tdt

√

2π1

(2 sin t−cos t)π/2

=2√2

5π(2eπ+1)

51. x0=−2 sin tcos t,y0= 2 sin tcos t,(x

)

2+(y

2= 8 sin2tcos2t

S=2πZπ/2

0cos2tp8 sin2tcos2tdt=4

√

2πZπ/2

0cos3tsin tdt=−√2πcos4tπ/2

=√2π

52. x0=1,y

0=4t,(x

)

2+(y

2=1+16t

2,S=2πZ1

p1+16t

2dt =π

24(17√17 −1)

53. x0=−rsin t,y0=rcos t,(x

)

2+(y

2=r

2,S=2πZπ

0rsin t√r2dt =2πr2Zπ

0sin tdt=4πr2

54. dx

dφ =a(1 −cos φ), dy

dφ =asin φ,dx

dφ2

+dy

dφ2

=2a

(1 −cos φ)

S=2πZ2π

0a(1 −cos φ)p2a2(1 −cos φ)dφ =2

√

2πa2Z2π

0(1 −cos φ)3/2dφ,

but 1 −cos φ= 2 sin2φ

2so (1 −cos φ)3/2=2

√

2 sin3φ

2for 0 ≤φ≤πand, taking advantage of the

symmetry of the cycloid, S=16πa2Zπ

0sin3φ

2dφ =64πa2/3

55. (a) dr

dt = 2 and dθ

dt =1sodr

dθ =dr/dt

dθ/dt =2

1=2,r=2θ+C,r= 10 when θ=0so

10 = C, r =2θ+ 10.

(b) r2+(dr/dθ)2=(2θ+ 10)2+ 4, during the ﬁrst 5 seconds the rod rotates through an angle

of (1)(5) = 5 radians so L=Z5

0p(2θ+ 10)2+4dθ, let u=2θ+10toget

L=1

2Z20

10 pu2+4du =1

2hu

2pu2+4+2ln|u+pu

2+4|

i20

2"10√404 −5√104 + 2 ln 20+√404

10+√104#≈75.7mm

56. x=rcos θ, y =rsin θ, dx

dθ =dr

dθ cos θ−rsin θ, dy

dθ =rcos θ+dr

dθ sin θ,

dx

dθ 2

+dy

dθ 2

=r2+dr

dθ 2

, and Formula (6) of Section 8.4 becomes

L=Zβ

αsr2+dr

dθ 2

dθ

419 Chapter 12

EXERCISE SET 12.3

1. (a) Zπ

π/2

2(1 −cos θ)2dθ (b) Zπ/2

24 cos2θdθ (c) Zπ/2

2sin22θdθ

(d) Z2π

2θ2dθ (e) Zπ/2

−π/2

2(1 −sin θ)2dθ (f) 2Zπ/4

2cos22θdθ

2. (a) 3π/8+1 (b) π/2(c) π/8

(d) 4π3/3(e) 3π/4(f) π/8

3. (a) A=Z2π

2a2dθ =πa2(b) A=Zπ

24a2sin2θdθ=πa2

−π/2

24a2cos2θdθ =πa2

4. (a) r2=rsin θ+rcos θ, x2+y2−y−x=0,x−1

2

+y−1

2

(b) A=Z3π/4

−π/4

2(sin θ+ cos θ)2dθ =π/2

5. A=2Zπ

(2 + 2 cos θ)2dθ =6π6. A=Zπ/2

2(1 + sin θ)2dθ =3π/8+1

7. A=6Zπ/6

2(16 cos23θ)dθ =4π

8. The petal in the ﬁrst quadrant has area Zπ/2

24 sin22θdθ=π/2, so total area = 2π.

9. A=2Zπ

2π/3

2(1 + 2 cos θ)2dθ =π−3√3/210. A=Z3

θ2dθ =4/3

11. area = A1−A2=Zπ/2

24 cos2θdθ−Zπ/4

2cos 2θdθ=π/2−1

12. area = A1−A2=Zπ

2(1 + cos θ)2dθ −Zπ/2

2cos2θdθ=5π/8

13. The circles intersect when cos t=√3 sin t, tan t=1/

√

3,t =π/6, so

A=A1+A2=Zπ/6

2(4√3 sin t)2dt+Zπ/2

π/6

2(4 cos t)2dt =2π−3

√

3+4π/3−√3=10π/3−4√3.

14. The curves intersect when 1 + cos t= 3 cos t, cos t=1/2,t=±π/3, and hence

total area = 2 Zπ/3

2(1+cos t)2dt+2 Zπ/2

π/3

29 cos2tdt=2(π/4+9√3/16+3π/8−9√3/16)=5π/4.

Exercise Set 12.3 420

15. A=2Zπ/2

π/6

2[25 sin2θ−(2 + sin θ)2]dθ =8π/3+√

16. A=2Zπ

[16 −(2 −2 cos θ)2]dθ =10π

17. A=2Zπ/3

2[(2 + 2 cos θ)2−9]dθ =9

√

3/2−π

18. A=2Zπ/4

2(16 sin2θ)dθ =2π−4

19. A=2"Z2π/3

2(1/2 + cos θ)2dθ −Zπ

2π/3

2(1/2 + cos θ)2dθ#=(π+3

√

3)/4

20. A=2Zπ/3

2(2 + 2 cos θ)2−9

4sec2θdθ =2π+9

√

21. A=2Zcos−1(3/5)

2(100 −36 sec2θ)dθ = 100 cos−1(3/5) −48

22. A=8Zπ/8

2(4a2cos22θ−2a2)dθ =2a

23. (a) ris not real for π/4<θ<3π/4 and 5π/4<θ<7π/4

(b) A=4Zπ/4

2a2cos 2θdθ=a

2h4 cos 2θ−2idθ =2

√

3−2π

24. A=2Zπ/2

2sin 2θdθ=1 25. A=Z4π

2π

2a2θ2dθ −Z2π

2a2θ2dθ =8π

26. (a) x=rcos θ, y =rsin θ,

(dx/dθ)2+(dy/dθ)2=(f

0(θ) cos θ−f(θ) sin θ)2+(f0(θ) sin θ+f(θ) cos θ)2=f0(θ)2+f(θ)2;

S=Zβ

2πf(θ) sin θpf0(θ)2+f(θ)2dθ if about θ= 0; similarly for θ=π/2

(b) f0,g

0are continuous and no segment of the curve is traced more than once.

27. r2+dr

dθ 2

= cos2θ+ sin2θ=1,

so S=Zπ/2

−π/22πcos2θdθ =π

421 Chapter 12

28. S=Zπ/2

02πeθcos θ√2e2θdθ

√

2πZπ/2

0e2θcos θdθ=2

√2π

5(e

π−2)

29. S=Zπ

02π(1 −cos θ) sin θp1−2 cos θ+ cos2θ+ sin2θdθ

√

2πZπ

0sin θ(1 −cos θ)3/2dθ =2

52√2π(1 −cos θ)5/2

0=32π/5

30. S=Zπ

02πa sin(θ)adθ =4πa2

31. (a) r3cos3θ−3r2cos θsin θ+r3sin3θ=0,r=3 cos θsin θ

cos3θ+ sin3θ

32. (a) A=2Zπ/(2n)

2a2cos2nθ dθ =πa2

4n(b) A=2Zπ/(2n)

2a2cos2nθ dθ =πa2

2n×total area = πa2

4n(d) 1

n×total area = πa2

33. If the upper right corner of the square is the point (a, a) then the large circle has equation r=√2a

and the small circle has equation (x−a)2+y2=a2,r=2acos θ,so

area of crescent = 2 Zπ/4

2h(2acos θ)2−(√2a)2idθ =a2= area of square.

Exercise Set 12.4 422

34. A=Z2π

2(cos 3θ+2)

2dθ =9π/2

-3

-3 3

35. A=Zπ/2

24 cos2θsin4θdθ=π/16

–1

EXERCISE SET 12.4

1. (a) 4px =y2,point(1,1),4p=1,x=y

2(b) −4py =x2,point(3,−3),12p=9,−3y=x

9+y

4=1 (d) a=3,b =2,x

4+y

9=1

(e) asymptotes: y=±x,soa=b;point(0,1), so y2−x2=1

(f) asymptotes: y=±x,sob=a;point(2,0), so x2

4−y2

4=1

2. (a) part (a), vertex (0,0),p=1/4; focus (1/4,0), directrix: x=−1/4

part (b), vertex (0,0),p=3/4; focus (0,−3/4), directrix: y=3/4

(b) part (c), c=√a2−b2=√5, foci (±√5,0)

part (d), c=√a2−b2=√5, foci (0,±√5)

part (f), c=√a2+b2=√8=2

√

2, foci at (±2√2,0); asymptotes: x2

4−y2

4=0,y =±x

3. (a)

-3 3

-3

F( , 0)x

x = –

(b)

F(0, –

)

y =

-5 5

-5

4. (a)

x =

F(– , 0)x

y(b)

F(0, 1)

y = –1

423 Chapter 12

5. (a)

x =

V(2, 3)

F( , 3)

(b)

-4

F(–2, –

)

y = –

V(–2, –2)

6. (a) 23

x =

F( , –1)V(4, –1)

(b)

y =

F( , )

V( , 1)

7. (a)

V(2,

)

F(2, 2)

y = 3

(b)

V(–2, 2) 7

F(– , 2)

x = –

8. (a)

x = –

F(– , 3)

V(–4, 3)

(b)

y =

F(–1, )

V (–1, 1)

9. (a) c2=16−9=7,c=√

(4, 0)

(0, 3)

(0, –3)

(–4, 0) x

(–√7, 0)

(√7, 0)

(b) x2

1+y2

9=1

2=9−1=8,c=√8

(0, 3)

(0, –3)

(–1, 0) (1, 0)

(0, √8)

(0, –√8)

Exercise Set 12.4 424

10. (a) c2=25−4 = 21, c=√21

(0, 5)

(0, –5)

(–2, 0) (2, 0)

(0, –√21)

(0, √21)

(b) x2

9+y2

4=1

2=9−4=5,c=√5

(0, 2)

(0, –2)

(–3, 0)

(3, 0)

(–√5, 0)

(√5, 0)

11. (a) (x−1)2

16 +(y−3)2

9=1

2=16−9=7,c=√7

(1, 6)

(1, 0)

(1 – √7, 3) (1 + √7, 3)

(5, 3)(–3, 3)

(b) (x+2)

4+(y+1)

3=1

2=4−3=1,c=1

(–4, –1) (–3, –1)

(–1, –1) (0, –1)

(–2, –1 + √3)

(–2, –1 – √3)

12. (a) (x+3)

16 +(y−5)2

4=1

2=16−4=12,c=2

√

(1, 5)

(–3, 7)

(–3, 3)

(–7, 5)

(–3 –2√3, 5)

(–3 +2√3, 5)

(b) x2

4+(y+2)

9=1

2=9−4=5,c=√5

(0, –5)

(0, 1) (0, –2 + √5)

(0, –2 – √5)

(–2, –2) (2, –2)

425 Chapter 12

13. (a) (x+1)

9+(y−1)2

1=1

2=9−1=8,c=√8

(–1, 2)

(–1, 0)

(–4, 1)

(2, 1)

(–1 – √8, 1)

(–1 + √8, 1)

(b) (x+1)

4+(y−5)2

16 =1

2=16−4=12,c=2

√

(1, 5)(–3, 5)

(–1, 1)

(–1, 9)

(–1, 5 + 2√3)

(–1, 5 – 2√3)

14. (a) (x+1)

4+(y−3)2

9=1

2=9−4=5,c=√5

(–1, 3 + √5)

(–1, 3 – √5)

(1, 3)

(–1, 6)

(–1, 0)

(–3, 3)

(b) (x−2)2

9+(y+3)

5=1

2=9−5=4,c=2

(2, –3 + √5)

(2, –3 – √5)

(4, –3)

(–1, –3) (5, –3)

(0, –3)

15. (a) c2=a2+b2=16+4=20,c=2

√

(0, –2)

(0, 2)

y = – x2

y = x

(0, √13)

(0, –√13)

(b) y2/4−x2/9=1

2=4+9=13,c=√13

(–4, 0) (4, 0)

y = – x1

y = x

(–2√5, 0) (2√5, 0)

Exercise Set 12.4 426

16. (a) c2=a2+b2=9+25=34,c=√34

(0, –3)

(0, 3)

(0, √34)

(0, –√34)

y = – x3

y = x

(b) x2/25 −y2/16=1

2=25+16=41,c=√41

(–5, 0) (5, 0)

y = – x4

y = x

(–√41, 0) (√41, 0)

17. (a) c2=9+4=13,c=√13

(5, 4)

(2 – √13, 4) (2 + √13, 4)

(–1, 4)

y – 4 = – (x – 2)

y – 4 = (x – 2)

(b) (y+3)

2/36 −(x+2)

2/4=1

2=36+4=40,c =2

√

(–2, 3)

(–2, –9)

y + 3 = –3(x + 2) y + 3 = 3(x + 2)

(–2, –3 + 2√10)

(–2, –3 – 2√10)

18. (a) c2=3+5=8,c=2

√

(2, –4 + √3)

(2, –4 – √3)

(2, –4 – 2√2)

(2, –4 + 2√2)

√

y + 4 = (x – 2)

√

y + 4 = – (x – 2)

(b) (x+1)

2/1−(y−3)2/2=1

2=1+2=3,c =√3

(–2, 3)

(-1 + √3, 3)(-1 − √3, 3)

(0, 3)

y − 3 = √2(x + 1)

y − 3 = −√2(x + 1)

427 Chapter 12

19. (a) (x+1)

2/4−(y−1)2/1=1

2=4+1=5,c =√5

(–3, 1)

(1, 1)

y – 1 = – (x + 1)

y – 1 = (x + 1)

(–1 – √5, 1)

(–1 + √5, 1)

(b) (x−1)2/4−(y+3)

2/64=1

2=4+64=68,c =2

√

(–1, –3) (3, –3) x

y + 3 = –4(x –1)

y + 3 = 4(x –1)

(1 + 2√17, –3)(–1 – 2√17, –3)

20. (a) (y−3)2/4−(x+2)

2/9=1

2=4+9=13,c=√13

(-2, 5)

(-2, 1)

y – 3 = – (x + 2)

y – 3 = (x + 2)

(–2, 3 – √13)

(–2, 3 + √13)

(b) (y+5)

2/9−(x+2)

2/36=1

2=9+36=45,c =3

√

(–2, –8)

(–2, –2) x

y + 5 = – (x + 2)

y + 5 = (x + 2)

(–2, –5 + 3√5)

(–2, –5 – 3√5)

21. (a) y2=4px,p=3,y

2=12x(b) y2=−4px,p=7,y

2=−28x

22. (a) x2=−4py,p=4,x

2=−16y(b) x2=−4py,p=1/2, x2=−2y

23. (a) x2=−4py,p=3,x

2=−12y

(b) The vertex is 3 units above the directrix so p=3,(x−1)2= 12(y−1).

24. (a) y2=4px,p=6,y

2=24x

(b) The vertex is half way between the focus and directrix so the vertex is at (2,4), the focus is 3

units to the left of the vertex so p=3,(y−4)2=−12(x−2)

25. y2=a(x−h), 4 = a(3 −h) and 9 = a(2 −h), solve simultaneously to get h=19/5, a=−5so

2=−5(x−19/5)

26. (x−5)2=a(y+ 3), (9 −5)2=a(5+3)soa=2,(x−5)2=2(y+3)

27. (a) x2/9+y

2/4=1

(b) a=26/2 = 13, c=5,b

2=a

2−c

2= 169 −25 = 144; x2/169 + y2/144=1

Exercise Set 12.4 428

28. (a) x2+y2/5=1

(b) b=8,c=6,a

2=b

2+c

2= 64 + 36 = 100; x2/64+y2/100=1

29. (a) c=1,a

2=b

2+c

2=2+1=3;x

/3+y

2/2=1

(b) b2=16−12=4;x

/16+y2/4=1andx

/4+y

2/16=1

30. (a) c=3,b

2=a

2−c

2=16−9=7;x

/16+y2/7=1

(b) a2=9+16=25;x

2/25+y2/9=1andx

/9+y

2/25=1

31. (a) a=6,(2,3) satisﬁes x2/36+y2/b2= 1 so 4/36+9/b2=1,b

2=81/8; x2/36+y2/(81/8)=1

(b) The center is midway between the foci so it is at (1,3), thus c=1,b=1,a

2=1+1=2;

(x−1)2+(y−3)2/2=1

32. (a) Substitute (3,2) and (1,6) into x2/A +y2/B = 1 to get 9/A +4/B = 1 and 1/A +36/B =1

which yields A= 10, B= 40; x2/10+y2/40=1

(b) The center is at (2,−1) thus c=2,a=3,b

2=9−4=5;(x−2)2/5+(y+1)

2/9=1

33. (a) a=2,c=3,b

2=9−4=5;x

/4−y

/5=1

(b) a=1,b/a =2,b=2;x

2−y

/4=1

34. (a) a=3,c=5,b

2=25−9 = 16; y2/9−x2/16=1

(b) a=3,a/b =1,b=3;y

/9−x

/9=1

35. (a) vertices along x-axis: b/a =3/2soa=8/3; x2/(64/9) −y2/16=1

vertices along y-axis: a/b =3/2soa=6;y

/36 −x2/16=1

(b) c=5,a/b = 2 and a2+b2= 25, solve to get a2= 20, b2=5;y

/20 −x2/5=1

36. (a) foci along the x-axis: b/a =3/4 and a2+b2= 25, solve to get a2= 16, b2=9;

/16 −y2/9=1

foci along the y-axis: a/b =3/4 and a2+b2= 25 which results in y2/9−x2/16=1

(b) c=3,b/a = 2 and a2+b2=9soa

2=9/5, b2=36/5; x2/(9/5) −y2/(36/5)=1

37. (a) the center is at (6,4), a=4,c=5,b

2=25−16=9;(x−6)2/16 −(y−4)2/9=1

(b) The asymptotes intersect at (1/2,2) which is the center, (y−2)2/a2−(x−1/2)2/b2=1is

the form of the equation because (0,0) is below both asymptotes, 4/a2−(1/4)/b2= 1 and

a/b = 2 which yields a2=3,b

2=3/4; (y−2)2/3−(x−1/2)2/(3/4)=1.

38. (a) the center is at (1,−2); a=2,c= 10, b2= 100 −4 = 96; (y+2)

2/4−(x−1)2/96=1

(b) the center is at (1,−1); 2a=5−(−3)=8,a =4,(x−1)2

16 −(y+1)

16 =1

39. (a) y=ax2+b, (20,0) and (10,12) are on the curve so

400a+b= 0 and 100a+b= 12. Solve for bto get

b= 16 ft = height of arch.

(b) x2

a2+y2

b2=1,400 = a2,a = 20; 100

400 +144

b2=1,

b=8

√

3 ft = height of arch.

-20 -10 10 20

(10, 12)

429 Chapter 12

40. (a) (x−b/2)2=a(y−h), but (0,0) is on the parabola so b2/4=−ah,a=−b2

4h,

(x−b/2)2=−b2

4h(y−h)

(b) As in part (a), y=−4h

b2(x−b/2)2+h,A=Zb

0−4h

b2(x−b/2)2+hdx =2

3bh

41. We may assume that the vertex is (0,0) and the parabola opens to the right. Let P(x0,y

0)bea

point on the parabola y2=4px, then by the deﬁnition of a parabola, PF = distance from Pto

directrix x=−p,soPF =x

0+pwhere x0≥0 and PF is a minimum when x0= 0 (the vertex).

42. Let p= distance (in millions of miles) between

the vertex (closest point) and the focus F,

then PD =PF,2p+ 20 = 40, p= 10 million miles.

40 60°

40 cos 60° = 20

Directrix

43. Use an xy-coordinate system so that y2=4px is an equation of the parabola, then (1,1/2) is a

point on the curve so (1/2)2=4p(1), p=1/16. The light source should be placed at the focus

which is 1/16 ft. from the vertex.

44. (a) Substitute x2=y/2intoy

2−8x

2= 5 to get y2−4y−5=0;

y=−1,5. Use x2=y/2 to ﬁnd that there is no solution if

y=−1 and that x=±p5/2ify= 5. The curves intersect

at (p5/2,5) and (−p5/2,5), and thus the area is

A=2Z√

5/2

p5+8x

2−√

2x)dx

=5√2

2(√5−1) + 5

4√2 ln(2 + √5)

(

–

√

, 5

)

(

√

, 5

)

(b) Eliminate xto get y2=1,y=±1. Use either equation

to ﬁnd that x=±2ify= 1 or if y=−1. The curves

intersect at (2,1), (2,−1), (−2,1), and (−2,−1),

and thus the area is

A=4Z√

5/3

p1+2x

2dx

+4Z2

√

5/31

p1+2x

2−1

√

7p3x

2−5

dx

3√2 ln(2√2+3)+10

21√21 ln(2√3+√

7) −5

21 ln 5

(–2, 1)

(–2, –1) (2, –1)

(2, 1)

Exercise Set 12.4 430

Use either equation to ﬁnd that y=±√3ifx=2

or if x=−2. The curves intersect at

(2,√3), (2,−√3), (−2,√3), (−2,−√3) and thus

A=4Z1

0p7−x

2dx +4Z2

1hp7−x

2−px

2−1

idx

√

3+14sin

−12

7√

−4

√

3 + 2 ln(2 + √3)

(–2, √3) (2, √3)

(–2, –√3) (2, –√3)

45. (a) P:(bcos t, b sin t); Q:(acos t, a sin t); R:(acos t, b sin t)

(b) For a circle, tmeasures the angle between the positive x-axis and the line segment joining

the origin to the point. For an ellipse, tmeasures the angle between the x-axis and OPQ,

not OR.

46. (a) Foranypoint(x, y),the equation

y=bsinh thas a unique solution t,

−∞ <t<+∞. On the hyperbola,

a2=1+y

2= 1 + sinh2t

= cosh2t,sox=±acosh t.

(b) 3

-3

-3 3

47. (a) Foranypoint(x, y), the equation y=btan thas a unique solution twhere −π/2<t<π/2.

On the hyperbola, x2

a2=1+y

2= 1 + tan2t= sec2t,sox=±asec t.

(b) 3

-3

-3 3

48. By Deﬁnition 12.4.1, (x+1)

2+(y−4)2=(y−1)2,(x+1)

2=6y−15,(x+1)

2=6(y−5/2)

49. (4,1) and (4,5) are the foci so the center is at (4,3) thus c=2,a=12/2=6,b

2=36−4 = 32;

(x−4)2/32+(y−3)2/36=1

50. From the deﬁnition of a hyperbola, p(x−1)2+(y−1)2−px2+y2=1,

p(x−1)2+(y−1)2−px2+y2=±1, transpose the second radical to the right hand side of the

equation and square and simplify to get ±2px2+y2=−2x−2y+ 1, square and simplify again

to get 8xy −4x−4y+1=0.

51. Let the ellipse have equation 4

81x2+y2

4= 1, then A(x)=(2y)

2=161−4x

81 ,

V=2Z9/2

016 1−4x2

81 dx =96

431 Chapter 12

52. See Exercise 51, A(y)=√

2=√

41−y2

4,V =√381

2Z2

01−y2

4dy =54

√

53. Assume x2

a2+y2

b2=1,A =4Za

p1−x

/a2dx =πab

54. (a) Assume x2

a2+y2

b2=1,V =2Za

0πb21−x2/a2dx =4

3πab2

(b) In Part (a) interchange aand bto obtain the result.

55. Assume x2

a2+y2

b2=1,dy

dx =−bx

a√a2−x2,1+dy

dx2

=a4−(a2−b2)x2

a2(a2−x2),

S=2Za

2πb

ap1−x2/a2sa4−(a2−b2)x2

a2−x2dx =2πab b

a+a

csin−1c

a,c=pa

2−b

56. As in Exercise 55, 1 + dx

dy 2

=b4+(a

2−b

2)y

2(b

2−y

2),

S=2Zb

02πap1−y2/b2sb4+(a

2−b

2)y

2(b

2−y

2)dy =2πab a

b+b

cln a+c

b,c=pa

2−b

57. Open the compass to the length of half the major axis, place the point of the compass at an end

of the minor axis and draw arcs that cross the major axis to both sides of the center of the ellipse.

Place the tacks where the arcs intersect the major axis.

58. Let Pdenote the pencil tip, and let R(x, 0) be the point below Qand Pwhich lies on the line L.

Then QP +PF is the length of the string and QR =QP +PR is the length of the side of the

triangle. These two are equal, so PF =PR. But this is the deﬁnition of a parabola according to

Deﬁnition 12.4.1.

59. Let Pdenote the pencil tip, and let kbe the diﬀerence between the length of the ruler and that

of the string. Then QP +PF

2+k=QF1, and hence PF

2+k=PF

1,PF

1−PF

2=k. But this

is the deﬁnition of a hyperbola according to Deﬁnition 12.4.3.

60. In the x0y0-plane an equation of the circle is x02+y02=r2where ris the radius of the cylinder. Let

P(x, y) be a point on the curve in the xy-plane, then x0=xcos θand y0=yso x2cos2θ+y2=r2

which is an equation of an ellipse in the xy-plane.

61. L=2a=pD

2+p

2=D

p1+p

2(see ﬁgure), so a=1

2Dp1+p

2, but b=1

2D,

T=c=pa2−b2=r1

4D2(1+p2)−1

4D2=1

2pD.

62. y=1

4px2,dy/dx =1

2px,dy/dxx=x0

2px0, the tangent line at (x0,y

0) has the formula

y−y0=x0

2p(x−x0)=x

x−x

, but x2

2p=2y

0because (x0,y

0) is on the parabola y=1

4px2.

Thus the tangent line is y−y0=x0

2px−2y0,y=x0

2px−y0.

Exercise Set 12.4 432

63. By implicit diﬀerentiation, dy

dx(x0,y0)

=−b2

y0if y06= 0, the tangent line is

y−y0=−b2

y0(x−x0), a2y0y−a2y2

0=−b2x0x+b2x2

0,b2x0x+a2y0y=b2x2

0+a2y2

but (x0,y

0) is on the ellipse so b2x2

0+a2y2

0=a2b2; thus the tangent line is b2x0x+a2y0y=a2b2,

x0x/a2+y0y/b2=1. Ify

0= 0 then x0=±aand the tangent lines are x=±awhich also follows

from x0x/a2+y0y/b2=1.

64. By implicit diﬀerentiation, dy

dx(x0,y0)

=b2

y0if y06= 0, the tangent line is y−y0=b2

y0(x−x0),

b2x0x−a2y0y=b2x2

0−a2y2

0=a2b2,x0x/a2−y0y/b2=1. Ify

0= 0 then x0=±aand the tangent

lines are x=±awhich also follow from x0x/a2−y0y/b2=1.

65. Use x2

a2+y2

b2= 1 and x2

A2−y2

B2= 1 as the equations of the ellipse and hyperbola. If (x0,y

0)is

a point of intersection then x2

a2+y2

b2=1= x

2−y

2,sox

01

2−1

=y

01

2+1

and

a2A2y2

0(b2+B2)=b

2−A

)

. Since the conics have the same foci, a2−b2=c2=A2+B2,

so a2−A2=b2+B2. Hence a2A2y2

0=b2B2x2

0. From Exercises 63 and 64, the slopes of the

tangent lines are −b2x0

a2y0and B2x0

A2y0, whose product is −b2B2x2

a2A2y2

=−1. Hence the tangent lines are

perpendicular.

66. Use implicit diﬀerentiation on x2+4y

2= 8 to get dy

dx(x0,y0)

=−x0

4y0where (x0,y

0) is the point

of tangency, but −x0/(4y0)=−1/2 because the slope of the line is −1/2sox

0=2y

.(x

0)is

on the ellipse so x2

0+4y2

0= 8 which when solved with x0=2y

0yields the points of tangency (2,1)

and (−2,−1). Substitute these into the equation of the line to get k=±4.

67. Let (x0,y

0) be such a point. The foci are at (−√5,0) and (√5,0),the lines are perpendicular if

the product of their slopes is −1so y

0+√

5·y

0−√

5=−1,y

0=5−x

0and 4x2

0−y2

0=4.Solve

to get x0=±3/√5,y

0=±4/

√5. The coordinates are (±3/√5,4/√5),(±3/√5,−4/√5).

68. Let (x0,y

0) be one of the points; then dy/dx(x0,y0)=4x

/y0, the tangent line is y=(4x

/y0)x+4,

but (x0,y

0) is on both the line and the curve which leads to 4x2

0−y2

0+4y

0= 0 and 4x2

0−y2

0= 36,

solve to get x0=±3√13/2, y0=−9.

69. Let d1and d2be the distances of the ﬁrst and second observers, respectively, from the point

where the gun was ﬁred. Then t= (time for sound to reach the second observer) −(time for

sound to reach the ﬁrst observer) = d2/v −d1/v so d2−d1=vt. For constant vand tthe

diﬀerence of distances, d2and d1is constant so the gun was ﬁred somewhere on a branch of a

hyperbola whose foci are where the observers are. Since d2−d1=2a, a =vt

2,b

2=c

2−v

4, and

v2t2/4−y2

c2−(v2t2/4) =1.

70. As in Exercise 69, d2−d1=2a=vt = 299,792,458 ×10−7,a

2=(vt/2)2≈224.6888; c2= (50)2

= 2500,b

2=c

2−a

2≈2275.3112,x2

224.6888 −y2

2275.3112 = 1. But y= 200 km, so x≈64.6 km.

433 Chapter 12

71. (a) Use x2

9+y2

4=1,x=3

p4−y

V=Z−2+h

−2(2)(3/2)p4−y2(18)dy =54Z−2+h

−2p4−y2dy

=54y

2p4−y

2+ 2 sin−1y

2−2+h

−2

=274 sin−1h−2

2+(h−2)p4h−h2+2π

ft3

(b) When h= 4 ft, Vfull = 108 sin−11+54π= 108πft3, so solve for hwhen V=(k/4)Vfull,

k=1,2,3, to get h=1.19205,2,2.80795 ft or 14.30465,24,33.69535 in.

72. We may assume A>0, since if A<0 then one can multiply the equation by −1, and if A=0

then one can exchange Awith C(Ccannot be zero simultaneously with A). Then

Ax2+Cy2+Dx +Ey +F=Ax+D

2A2

+Cy+E

2C2

+F−D2

4A−E2

4C=0.

(a) Let AC > 0. If F<D

4Cthe equation represents an ellipse (a circle if A=C);

if F=D2

4A+E2

4C, the point x=−D/(2A),y =−E/(2C); and if F>D

4A+E

4Cthen there is

no graph.

(b) If AC < 0 and F=D2

4A+E2

4C, then

√Ax+D

2A+√−Cy+E

2C√Ax+D

2A−√−Cy+E

2C= 0, a pair of lines;

otherwise a hyperbola

If this polynomial has roots x=x1,x

2with x16=x2then a pair of parallel lines; if x1=x2

then one line; if no roots, then no graph. If A=0,C 6= 0 then a similar argument applies.

73. (a) (x−1)2−5(y+1)

2= 5, hyperbola

(b) x2−3(y+1)

2=0,x =±√3(y+ 1), two lines

2+8(y+1)

2= 4, ellipse

(d) 3(x+2)

2+(y+1)

2= 0, the point (−2,−1) (degenerate case)

(e) (x+4)

2+2y= 2, parabola

(f) 5(x+4)

2+2y=−14, parabola

74. distance from the point (x, y) to the focus (0,p) = distance to the directrix y=−p,sox

2+(y−p)2

=(y+p)

2=4py

75. distance from the point (x, y) to the focus (0,−c) plus distance to the focus (0,c) = const = 2a,

px2+(y+c)

2+px

2+(y−c)

2=2a, x2+(y+c)

2=4a

2+x

2+(y−c)

2−4a

2+(y−c)

2=a−c

ay, and since a2−c2=b2,x2

b2+y2

a2=1

76. distance from the point (x, y) to the focus (−c, 0) less distance to the focus (c, 0) is equal to 2a,

p(x+c)2+y2−p(x−c)2+y2=±2a, (x+c)2+y2=(x−c)

2+y

2+4a

2±4a

p(x−c)

2+y

p(x−c)

2+y

2=±cx

a−a, and, since c2−a2=b2,x2

a2−y2

b2=1

Exercise Set 12.5 434

EXERCISE SET 12.5

1. (a) r=3/2

1−cos θ,e=1,d=3/2

-2

(b) r=3/2

1+1

2sin θ,e=1/2,d=3

–2

–2 2

1+3

2cos θ,e=3/2,d=4/3

-5 10

-7

(d) r=5/3

1 + sin θ,e=1,d=5/3

-11

-7 7

2. (a) r=4/3

1−2

3cos θ,e=2/3,d=2

(b) r=1

1−4

3sin θ,e=4/3,d=3/4

1 + sin θ,e=1,d=1/3

(d) r=1/2

1+3sinθ,e =3,d =1/6

435 Chapter 12

3. (a) e=1,d=8,parabola, opens up

-10

-15 15

(b) r=4

1+3

4sin θ,e=3/4,d=16/3,

ellipse, directrix 16/3 units

above the pole

-20

-8 8

1−3

2sin θ,e=3/2,d=4/3,

hyperbola, directrix 4/3 units

below the pole

-8

-6 6

(d) r=3

1+1

4cos θ,e=1/4,d=12,

ellipse, directrix 12 units

to the right of the pole

-4

-5 3

4. (a) e=1,d=15,parabola, opens left

-20

-20 20

(b) r=2/3

1 + cos θ,e=1,

d=2/3,parabola, opens left

-10

-15 5

1−12

7sin θ,e=12/7,d=16/3,

hyperbola, directrix 16/3 units below pole

-40

-30 30

(d) r=4

1−2

3cos θ,e=2/3,d=6,

ellipse, directrix 6 units left of the pole

-6

-3 13

Exercise Set 12.5 436

5. (a) d=1,r =ed

1+ecos θ=2/3

1+2

3cos θ=2

3+2cosθ

(b) e=1,d =1,r =ed

1−ecos θ=1

1−cos θ

1+esin θ=3/2

1+3

2sin θ=3

2+3sinθ

6. (a) e=2/3,d =1,r =ed

1−esin θ=2/3

1−2

3sin θ=2

3−2 sin θ

(b) e=1,d =1,r =ed

1+esin θ=1

1 + sin θ

1−ecos θ=4/3

1−4

3cos θ=4

3−4 cos θ

7. (a) r=ed

1±ecos θ,θ =0:6= ed

1±e,θ =π:4= ed

1∓e,6±6e=4∓4e, 2=∓10e, use bottom

sign to get e=1/5,d =24,r=24/5

1−cos θ=24

5−5 cos θ

(b) e=1,r =d

1−sin θ,1=d

2,d=2,r =2

1−sin θ

1±esin θ,θ =π/2:3= ed

1±e,θ =3π/2:−7= ed

1∓e,ed=3±3e=−7±7e, 10 = ±4e,

e=5/2,d =21/5,r =21/2

1+(5/2) sin θ=21

2+5sinθ

8. (a) r=ed

1±esin θ,1= ed

1±e,4= ed

1∓e,1±e=4∓4e, upper sign yields e=3/5,d=8/3,

r=8/5

1+3

5sin θ=8

5+3sinθ

(b) e=1,r =d

1−cos θ,3=d

2,d=6,r=6

1−cos θ

2,r =√2d

1+√

2 cos θ;r= 5 when θ=0,sod=5+ 5

√

r=5

√

2+5

1+√

2 cos θ.

9. (a) r=3

1+1

2sin θ,e=1/2,d= 6, directrix 6 units above pole; if θ=π/2:r

0=2;

if θ=3π/2:r

1=6,a =(r

0+r

)/2=4,b =√r

1=2

√

3, center (0,−2) (rectangular

coordinates), x2

12 +(y+2)

16 =1

(b) r=1/2

1−1

2cos θ,e=1/2,d= 1, directrix 1/2 unit left of pole; if θ=π:r0=1/2

3/2=1/3;

if θ=0:r

1=1,a =2/3,b =1/

√

3, center = (1/3,0) (rectangular coordinates),

4(x−1/3)2+3y

2=1

437 Chapter 12

10. (a) r=6/5

1+2

5cos θ,e=2/5,d= 3, directrix 3 units right of pole, if θ=0:r

0=6/7,

if θ=π:r1=2,a =10/7,b=2

√

7, center (−4/7,0) (rectangular coordinates),

100(x+4/7)2+7

12y2=1

(b) r=2

1−3

4sin θ,e=3/4,d=8/3, directrix 8/3 units below pole, if θ=3π/2:r

0=8/7,

if θ=π/2; r1=8,a=32/7,b =8/

√

7, center: (0,24/7) (rectangular coordinates),

64x2+49

1024 y−24

72

11. (a) r=2

1+3sinθ,e =3,d =2/3, hyperbola, directrix 2/3 units above pole, if θ=π/2:

0=1/2; θ=3π/2:r

1= 1, center (0,3/4),a=1/4,b =1/

√

2,−2x

2+16y−3

42

(b) r=5/3

1−3

2cos θ,e=3/2,d=10/9, hyperbola, directrix 3/2 units left of pole, if θ=π:

r0=2/3; θ=0:r

1=5/3

1/2=10/3, center (−2,0),a=4/3,b =p20/9,9

16(x+2)2−9

20y2=1

12. (a) r=4

1−2 sin θ,e=2,d= 2, hyperbola, directrix 2 units below pole, if θ=3π/2:r

0=4/3;

θ=π/2:r

1=



1−2



= 4, center (0,−8/3),a=4/3,b =4/

√

3,9

16 y+8

32

−3

16x2=1

(b) r=15/2

1+4cosθ,e =4,d =15/8, hyperbola, directrix 15/8 units right of pole, if θ=0:

0=3/2; θ=π:r1=−5

2=5/2,a=1/2,b =√15

2, center (2,0),4(x−2)2−4

15y2=1

13. (a) r=1

1+1

2cos θ=d

2 + cos θ,ifθ=0:r

0=d/3; θ=π, r1=d,

8=a=1

1+r

)=2

d, d =12,r=12

2 + cos θ

(b) r=3

1−3

5sin θ=3d

5−3 sin θ,ifθ=3π/2:r

0=3

d;θ=π/2,r

1=3

2d,

4=a=1

1+r

)=15

16d, d =64

15,r =3(64/15)

5−3 sin θ=64

25 −15 sin θ

1−3

5cos θ=3d

5−3 cos θ,ifθ=π:r

0=3

d;θ=0,r

1=3

2d, 4=b=3

d=16/3,r=16

5−3 cos θ

(d)

1+1

5sin θ=d

5 + sin θ,ifθ=π/2:r

0=d/6; θ=3π/2,r

1=d/4,

5=c=1

d1

4−1

=1

24d, d = 120,r=120

5 + sin θ

Exercise Set 12.5 438

14. (a) r=1

1+1

2sin θ=d

2 + sin θ,ifθ=π/2:r

0=d/3; θ=3π/2:r

1=d,

10 = a=1

2(r0+r1)=2

d, d =15,r=15

2 + sin θ

(b) r=1

1−1

5cos θ=d

5−cos θ,ifθ=π:r

0=d/6,θ =0:r

1=d/4,

6=a=1

1+r

)=1

d1

4+1

=5

24d, d = 144/5,r=144/5

5−cos θ=144

25 −5 cos θ

1−3

4sin θ=3d

4−3 sin θ,ifθ=3π/2:r

0=3

d, θ =π/2:r

1=3d, 4=b=3d/√7,

d=4

3√7,r=4

√

4−3 sin θ

(d) r=4

1+4

5cos θ=4d

5+4cosθ,ifθ=0:r

0=4

d;θ=π:r

1=4d,

c=10=1

1−r

)=1

d4−4

=16

9d, d =45

8,r=45/2

5+4cosθ=45

10 + 8 cos θ

15. (a) e=c/a =1

2(r1−r0)

2(r1+r0)=r1−r0

r1+r0

(b) e=r1/r0−1

r1/r0+1,e(r

1/r0+1)=r

1/r0−1,r1

r0=1+e

1−e

16. (a) e=c/a =1

2(r1+r0)

2(r1−r0)=r1+r0

r1−r0

(b) e=r1/r0+1

1/r0−1,e(r

1/r0−1) = r1/r0+1,r

0=e+1

e−1

17. (a) T=a3/2=39.5

1.5≈248 yr

(b) r0=a(−e)=39.5(1 −0.249) = 29.6645 AU ≈4,449,675,000 km

r1=a(1+e)=39.5(1+0.249) = 49.3355 AU ≈7,400,325,000 km

1+ecos θ≈39.5(1 −(0.249)2)

1+0.249 cos θ≈37.05

1+0.249 cos θAU

(d)

p/2

-50

-30 20

18. (a) In yr and AU, T=a3/2; in days and km, T

365 =a

150 ×1063/2

so T= 365 ×10−9a

1503/2days.

439 Chapter 12

(b) T= 365 ×10−957.95 ×106

150 3/2

≈87.6days

1+ecos θ=37.82

1+0.205 cos θ

(d)

p/2

20-50

-40

19. (a) a=T2/3= 23802/3≈178.26 AU

(b) r0=a(1 −e)≈0.8735 AU,r

1=a(1+e)≈355.64 AU

1+ecos θ≈1.74

1+0.9951 cos θAU

(d)

p/2

-400

-20

20. (a) By Exercise 15(a), e=r1−r0

r1+r0≈0.093

(b) r=1

2(a0+a1) = 225,400,000 km ≈1.503 AU, so T=a3/2≈1.84 yr

1+ecos θ≈1.49

1+0.093 cos θAU

(d)

p/2

1.3632

1.49

1.6428

21. r0=a(1 −e)≈7003 km, hmin ≈7003 −6440 = 563 km,

r1=a(1+e)≈10,726 km, hmax ≈10,726 −6440 = 4286 km

Chapter 12 Supplementary Exercises 440

22. r0=a(1 −e)≈651,736 km, hmin ≈581,736 km; r1=a(1+e)≈6,378,102 km,

hmax ≈6,308,102 km

23. Since the foci are ﬁxed, ais constant; since e→0, the distance a

e→+∞.

Let dbe the distance between the directrix and the focus, then d=a1

e−eand lim

e→0d=+∞;

but the center and the focus come together, so distance between the directrix and the center also

tends to +∞.

24. (a) From Figure 12.4.22, x2

a2−y2

b2=1,x

2−y

2−a

2=1,(x−c)

2+y

2=c

x−a

2,

p(x−c)

2+y

2=c

x−afor x>0.

(b) From Part (a), PF =c

aPD, PF

PD =c/a.

CHAPTER 12 SUPPLEMENTARY EXERCISES

2. (a) (√2,3π/4) (b) (−√2,7π/4) (c) (√2,3π/4) (d) (−√2,−π/4)

3. (a) circle (b) rose (c) line (d) lima¸con

(e) lima¸con (f) none (g) none (h) spiral

4. (a) r=1/3

1+1

3cos θ, ellipse, right of pole, distance = 1

(b) hyperbola, left of pole, distance = 1/3

1 + sin θ, parabola, above pole, distance = 1/3

(d) parabola, below pole, distance = 3

5. (a) p/2

(b) p/2

(1, 9)

(1, #)

(d) p/2

(–1, 3)

(e) p/2

(1, 3)

(2, 3)

441 Chapter 12

6. Family I: x2+(y−b)

2=b

,b < 0, or r=2bsin θ; Family II: (x−a)2+y2=a2,a < 0, or

r=2acos θ

7. (a) r=2a/(1 + cos θ),r+x=2a, x2+y2=(2a−x)

2=−4ax +4a

2, parabola

(b) r2(cos2θ−sin2θ)=x

2−y

2=a

, hyperbola

√

2/2)r(sin θ−cos θ)=4,y−x=4

√

2, line

(d) r2=4rcos θ+8rsin θ, x2+y2=4x+8y, (x−2)2+(y−4)2= 20, circle

9. (a) c

a=e=2

7and 2b=6,b =3,a

2=b

2+c

2=9+ 4

49a2,45

49a2=9,a=7

√5,5

49x2+1

9y2=1

(b) x2=−4py, directrix y= 4, focus (−4,0),2p=8,x

2=−16y

2=a

2−b

2=16−3 = 13, foci (±√13,0);

for the hyperbola, c=√13, b/a =2/3,b =2a/3,13 = c2=a2+b2=a2+4

9a2=13

9a2,

a=3,b =2,x

9−y

4=1

10. (a) e=4/5=c/a, c =4a/5, but a=5soc=4,b=3,(x+3)

25 +(y−2)2

9=1

(b) directrix y=2,p =2,(x+2)

2=−8y

4−16(x+1)2=1

11. (a)

-4 8

-10

y(b)

210

-3

y(c)

-8 8

-12

13. (a) The equation of the parabola is y=ax2and it passes through (2100,470), thus a=470

21002,

y=470

21002x2.

(b) L=2Z2100

0s1+2470

21002x2

dx ≈4336.3ft

14. (a) As truns from 0 to π, the upper portion of the curve is traced out from right to left; as t

runs from πto 2πthe bottom portion of the curve is traced out from right to left. The loop

occurs for π+ sin−11

4<t<2π−sin−11

(b) lim

t→0+x=+∞,lim

t→0+y= 1; lim

t→π−x=−∞,lim

t→π−y= 1; lim

t→π+x=+∞,lim

t→π+y=1;

lim

t→2π−x=−∞,lim

t→2π−y= 1; the horizontal asymptote is y=1.

vertical tangent line when dx/dt =0,so−csc2t−4 sin t=0,t=π+sin−11

√4,2π−sin−11

√4,

t=3.823,5.602

Chapter 12 Supplementary Exercises 442

(d) r2=x2+y2= (cot t+ 4 cos t)2+(1+4sint)

2= (4 + csc t)2,r = 4 + csc t; with t=θ,

f(θ) = 4 + csc θ;m=dy/dx =(f(θ) cos θ+f0(θ) sin θ)/(−f(θ) sin θ+f0(θ) cos θ); when

θ=π+ sin−1(1/4),m =√15/15, when θ=2π−sin−1(1/4),m =−√15/15, so the tangent

lines to the conchoid at the pole have polar equations θ=±√15/15.

15. A

2=Zπ/6

2(2 sin θ)2dθ +Zπ/4

π/6

212,dθ =2θ−1

4sin 2θπ/6

+π

24,A =5π

12 −√3

16. The circle has radius a/2 and lies entirely inside the cardioid, so

A=Z2π

2a2(1 + sin θ)2dθ −πa2/4=3a

2π−a

4π=5a

4π

17. (a) r=1/θ, dr/dθ =−1/θ2,r

2+(dr/dθ)2=1/θ2+1/θ4,L=

Zπ/2

π/4

θ2p1+θ

2dθ ≈0.9457 by

Endpaper Table Formula 93.

(b) The integral Z+∞

θ2p1+θ

2dθ diverges by the comparison test (with 1/θ), and thus the

arc length is inﬁnite.

18. (a) When the point of departure of the thread from the circle has traversed an angle θ, the

amount of thread that has been unwound is equal to the arc length traversed by the point of

departure, namely aθ. The point of departure is then located at (acos θ, a sin θ), and the tip of

the string, located at (x, y), satisﬁes the equations x−acos θ=aθ sin θ, y−asin θ=−aθ cos θ;

hence x=a(cos θ+θsin θ),y =a(sin θ−θcos θ).

(b) Assume for simplicity that a= 1. Then dx/dθ =θcos θ, dy/dθ =θsin θ;dx/dθ = 0 has

solutions θ=0,π/2,3π/2; and dy/dθ = 0 has solutions θ=0,π,2π.Atθ=π/2, dy/dθ > 0,

so the direction is North; at θ=π, dx/dθ < 0, so West; at θ=3π/2, dy/dθ < 0, so South; at

θ=2π, dx/dθ > 0, so East. Finally, lim

θ→0+dy/dx = lim

θ→0+tan θ= 0, so East.

p/2

p

–1

3p/2

–3p/2

–1

–2p

Note that the parameter θin these equations does not satisfy equations (1) and (2) of Section

12.1, since it measures the angle of the point of departure and not the angle of the tip of the

thread.

-8

-5 2

443 Chapter 12

19. (a) V=Z√a2+b2

πb2x2/a2−b2dx

=πb2

3a2(b2−2a2)pa2+b2+2

3ab2π

(b) V=2πZ√

/a2−b2dx =(2b

/3a)π

20. (a)

-5 5

-5

p/2(b) θ=π/2,3π/2,r =1

−rsin θ+(dr/dθ) cos θ;atθ=π/2,m

1=(−1)/(−1)=1,m

2=1/(−1) = −1,

m1m2=−1; and at θ=3π/2,m

1=−1,m

2=1,m

2=−1

22. The tips are located at r=1,θ =π/6,5π/6,3π/2 and, for example,

d=p1+1−2 cos(5π/6−π/6) = p2(1 −cos(2π/3)) = √3

23. (a) x=rcos θ= cos θ+cos2θ, dx/dθ =−sin θ−2 sin θcos θ=−sin θ(1 +2 cos θ) = 0 if sin θ=0

or cos θ=−1/2, so θ=0,π,2π/3,4π/3; maximum x=2atθ= 0, minimum x=−1/4at

θ=π

(b) y=rsin θ= sin θ+ sin θcos θ, dy/dθ = 2 cos2θ+ cos θ−1 = 0 at cos θ=1/2,−1, so

θ=π/3,5π/3,π; maximum y=3

√

3/4atθ=π/3, minimum y=−3√3/4atθ=5π/3

24. (a) y=rsin θ= (sin θ)/√θ, dy/dθ =2θcos θ−sin θ

2θ3/2=0if2θcos θ= sin θ, tan θ=2θwhich

only happens once on (0,π]. Since lim

θ→0+y= 0 and y=0atθ=π,yhas a maximum when

tan θ=2θ.

(b) θ≈1.16556

Chapter 12 Supplementary Exercises 444

25. The width is twice the maximum value of yfor 0 ≤θ≤π/4:

y=rsin θ= sin θcos 2θ= sin θ−2 sin3θ, dy/dθ = cos θ−6 sin2θcos θ= 0 when cos θ=0or

sin θ=1/

√

6,y =1/

√

6−2/(6√6) = √6/9, so the width of the petal is 2√6/9.

26. (a) x2

225 −y2

1521 =1,soV=2Zh/2

0225π1+ y

1521dy =25

2028πh3+ 225πh ft3.

(b) S=2Zh/2

02πxp1+(dx/dy)2dy =4πZh/2

t225 + y2 225

1521 +225

15212!dy

26πhp6084 + h2+ 1170πln "h+√6084 + h2

78 #ft2

27. (a) The end of the inner arm traces out the circle x1= cos t, y1= sin t. Relative to the end of

the inner arm, the outer arm traces out the circle x2= cos 2t, y2=−sin 2t. Add to get the

motion of the center of the rider cage relative to the center of the inner arm:

x= cos t+ cos 2t, y = sin t−sin 2t.

(b) Same as part (a), except x2= cos 2t, y2= sin 2t,sox= cos t+ cos 2t, y = sin t+ sin 2t

0"dx

dt 2

+dy

dt 2#1/2

dt =Z2π

√5−4 cos 3tdt≈13.36489321,

L2=Z2π

√5+4costdt≈13.36489322; L1and L2appear to be equal, and indeed, with the

substitution u=3t−πand the periodicity of cos u,

L1=1

3Z5π

−πp5−4 cos(u+π)du =Z2π

√5+4cosudu=L

29. C=4Zπ/2

0"dx

dt 2

+dy

dt 2#1/2

dt =4Zπ/2

0(a2sin2t+b2cos2t)1/2dt

=4Zπ/2

0(a2sin2t+(a

2−c

2) cos2t)1/2dt =4aZπ/2

0(1 −e2cos2t)1/2dt

30. a=3,b =2,c =√5,C= 4(3) Zπ/2

0p1−(5/9) cos2udu ≈15.86543959

31. (a) r0

r1=59

61 =1−e

1+e,e=1

(b) a=93×106,r

0=a(1 −e)=59

60 93 ×106=91,450,000 mi

0"1−cos θ

60 2#1/2

dθ ≈584,295,652.5mi

32. (a) y=y0+(v

0sin α)x

v0cos α−g

2x

v0cos α2

=y0+xtan α−g

2v2

0cos2αx2

(b) dy

dx = tan α−g

0cos2αx, dy/dx =0atx=v

gsin αcos α,

y=y0+v2

gsin2α−g

2v2

0cos2αv2

0sin αcos α

g2

=y0+v2

2gsin2α

445 Chapter 12

33. α=π/4,y

0=3,x =v

0t/√2,y =3+v

t/√2−16t2

(a) Assume the ball passes through x= 391,y = 50, then 391 = v0t/√2,50 = 3 + 391 −16t2,

16t2= 344,t=√21.5,v

0=√2x/t ≈119.3 ft/s

(b) dy

dt =v0

√2−32t= 0 when t=v0

32√2,y

max =3+ v

√

32√2−16 v2

211 =3+ v

128 ≈114.2ft

0/2 + 192

−32 ,t≈−0.04 (discard) and 5.31, dist = 447.9ft

34. (a)

-1 1

-1

−1cos2πt2

2+ sin2πt2

2dt =2

35. tan ψ= tan(φ−θ)= tan φ−tan θ

1 + tan φtan θ=

dx −y

1+y

rcos θ+(dr/dθ) sin θ

−rsin θ+(dr/dθ) cos θ−sin θ

cos θ

1+rcos θ+(dr/dθ) sin θ)

−rsin θ+(dr/dθ) cos θ)sin θ

cos θ=r

dr/dθ

36. (a) From Exercise 35,

tan ψ=r

dr/dθ =1−cos θ

sin θ= tan θ

so ψ=θ/2.

(b)

p/2

37. tan ψ=r

dr/dθ =aebθ

abebθ =1

bis constant, so ψis constant.

Chapter 12 Horizon Module 446

CHAPTER 12 HORIZON MODULE

1. For the Earth, aE(1 −e2

E) = 1(1 −0.0172)=0.999711, so the polar equation is

r=aE(1 −e2

1−eEcos θ=0.999711

1−0.017 cos θ.

For Rogue 2000, aR(1 −e2

R) = 5(1 −0.982)=0.198, so the polar equation is

r=aR(1 −e2

1−eRcos θ=0.198

1−0.98 cos θ.

2. 1

-1

-1 10

3. At the intersection point A,kE

1−eEcos θ=kR

1−eRcos θ,sok

E−k

Rcos θ=kR−kReEcos θ.

Solving for cos θgives cos θ=kE−kR

kEeR−kReE

4. From Exercise 1, kE=0.999711 and kR=0.198, so

cos θ=kE−kR

kEeR−kReE

=0.999711 −0.198

0.999711(0.98) −0.198(0.017) ≈0.821130

and θ= cos−10.821130 ≈0.607408 radian.

5. Substituting cos θ≈0.821130 into the polar equation for the Earth gives

r≈0.999711

1−0.017(0.821130) ≈1.013864,

so the polar coordinates of intersection Aare approximately (1.013864,0.607408).

6. By Theorem 12.3.2 the area of the elliptic sector is ZθF

θI

2r2dθ. By Exercise 12.4.53 the area of

the entire ellipse is πab, where ais the semimajor axis and bis the semiminor axis. But

b=√a2−c2=pa2−(ea)2=a√1−e2,

so Formula (1) becomes t

T=ZθF

θI

r2dθ

2πa2√1−e2, which implies Formula (2).

7. In Formula (2) substitute T=1,θ

I= 0, and θF≈0.607408, and use the polar equation of the

Earth’s orbit found in Exercise 1:

t=ZθF

0kE

1−eEcos θ2

dθ

2πp1−e2

E≈Z0.607408

00.999711

1−0.017 cos θ2

dθ

2π√0.999711 ≈0.099792 yr.

447 Chapter 12

Note: This calculation can be done either by numerical integration or by using the integration

formula

Zdθ

(1 −ecos θ)2=

2 tan−1r1+e

1−etan θ

2

(1 −e2)3/2+esin θ

(1 −e2)(1 −ecos θ)+C,

obtained by using a CAS or by the substitution u= tan(θ/2).

8. In Formula (2) we substitute T=5

√

5 and θI=0.45, and use the polar equation of Rogue 2000’s

orbit found in Exercise 1:

TZθF

θIaR(1 −e2

1−eRcos θ2

dθ

2πa2

Rp1−e2

5√5ZθF

0.45 aR(1 −e2

1−eRcos θ2

dθ

2πa2

Rp1−e2

ZθF

0.45 aR(1 −e2

1−eRcos θ2

dθ =2tπa2

Rp1−e2

5√5.

9. (a) A CAS shows that

ZaR(1 −e2

1−eRcos θ2

dθ =a2

R2q1−e2

Rtan−1r1+e

1−e

tan θ

2+eR(1 −e2

R) sin θ

1−eRcos θ+C

(b) Evaluating the integral above from θ=0.45 to θ=θF, setting the result equal to the right

side of (3), and simplifying gives

tan−1r1+R

1−eR

tan θ

2+eRp1−e2

Rsin θ

2(1 −eRcos θ)θF

0.45

=tπ

5√5.

Using the known values of eRand t, and solving numerically, θF≈0.611346.

10. Substituting θF≈0.611346 in the equation for Rogue 2000’s orbit gives r≈1.002525 AU. So the

polar coordinates of Rogue 2000 when the Earth is at intersection Aare about (1.002525,0.611346).

11. Substituting the values found in Exercises 5 and 10 into the distance formula in Supplementary

Exercise 22 gives d=pr2

1+r2

2−2r1r2cos(θ1−θ2)≈0.012013 AU ≈1.797201 ×106km.

Since this is less than 4 million kilometers, a notiﬁcation should be issued. (Incidentally, Rogue

2000’s closest approach to the Earth does not occur when the Earth is at A, but about 9 hours

earlier, at t≈0.098768 yr, at which time the distance is about 1.219435 million kilometers.)

448

CHAPTER 13

Three-Dimensional Space; Vectors

EXERCISE SET 13.1

1. (a) (0,0,0),(3,0,0),(3,5,0),(0,5,0),(0,0,4),(3,0,4),(3,5,4),(0,5,4)

(b) (0,1,0),(4,1,0),(4,6,0),(0,6,0),(0,1,−2),(4,1,−2),(4,6,−2),(0,6,−2)

2. corners: (2,2,±2), (2,−2,±2),

(−2,2,±2), (−2,−2,±2)

(–2, –2, 2) (–2, 2, 2)

(–2, 2, –2)(–2, –2, –2)

(2, 2, –2)(2, –2, –2)

(2, –2, 2) (2, 2, 2)

3. corners: (4,2,−2), (4,2,1), (4,1,1), (4,1,−2),

(−6,1,1), (−6,2,1), (−6,2,−2), (−6,1,−2)

(–6, 2, 1)

(–6, 2, –2)

(–6, 1, –2)

(4, 1, 1)

(4, 1, –2)

(4, 2, 1)

4. (a) (x2,y

),(x

)(x1,y

),(x

)

(b) The midpoint of the diagonal has coordinates which are the coordinates of the midpoints

of the edges. The midpoint of the edge (x1,y

) and (x2,y

)is1

1+x

),y

;

the midpoint of the edge (x2,y

) and (x2,y

)is

1+y

),z

; the midpoint

of the edge (x2,y

) and (x2,y

)) is x2,y

2(z

1+z

)

. Thus the coordinates of the

midpoint of the diagonal are 1

2(x1+x2),1

2(y1+y2),1

2(z1+z2).

5. The diameter is d=p(1 −3)2+(−2−4)2+(4+12)

2=√

296, so the radius is √296/2=√

74.

The midpoint (2,1,−4) of the endpoints of the diameter is the center of the sphere.

6. Each side has length √14 so the triangle is equilateral.

7. (a) The sides have lengths 7, 14, and 7√5; it is a right triangle because the sides satisfy the

Pythagorean theorem, (7√5)2=7

2+14

(b) (2,1,6) is the vertex of the 90◦angle because it is opposite the longest side

(the hypotenuse).

8. (a) 3(b) 2(c) 5

(d) p(2)2+(−3)2=√13 (e) p(−5)2+(−3)2=√34 (f ) p(−5)2+ (2)2=√29

9. (a) (x−1)2+y2+(z+1)

2=16

(b) r=p(−1−0)2+(3−0)2+(2−0)2=√14, (x+1)

2+(y−3)2+(z−2)2=14

449 Chapter 13

2p(−1−0)2+(2−2)2+(1−3)2=1

2√5, center (−1/2,2,2),

(x+1/2)2+(y−2)2+(z−2)2=5/4

10. r=|[distance between (0,0,0) and (3,−2,4)] ±1|=√29 ±1,

x2+y2+z2=r2=√29 ±12=30±2

√

11. (x−2)2+(y+1)

2+(z+3)

2=r

(a) r2=3

2=9 (b) r2=1

2=1 (c) r2=2

2=4

12. (a) The sides have length 1, so the radius is 1

2; hence (x+2)

2+(y−1)2+(z−3)2=1

(b) The diagonal has length √1+1+1=√

3 and is a diameter, so (x+2)2+(y−1)2+(z−3)2=3

13. (x+5)

2+(y+2)

2+(z+1)

2= 49; sphere, C(−5,−2,−1), r=7

14. x2+(y−1/2)2+z2=1/4; sphere, C(0,1/2,0), r=1/2

15. (x−1/2)2+(y−3/4)2+(z+5/4)2=54/16; sphere, C(1/2,3/4,−5/4), r=3

√

6/4

16. (x+1)

2+(y−1)2+(z+1)

2= 0; the point (−1,1,−1)

17. (x−3/2)2+(y+2)

2+(z−4)2=−11/4; no graph

18. (x−1)2+(y−3)2+(z−4)2= 25; sphere, C(1,3,4), r=5

19. (a)

z(b)

z(c)

20. (a)

x = 1

(b)

zy = 1

(c)

z = 1

21. (a)

z(b)

z(c)

Exercise Set 13.1 450

22. (a)

z(b)

z(c)

23. (a) −2y+z=0 (b) −2x+z=0

24. (a) (x−a)2+(z−a)

2=a

(b) (x−a)2+(y−a)

2=a

25.

26.

z27.

28.

z29.

30.

z31.

2-3

z32.

√3

451 Chapter 13

33.

-2 2

z34.

35. (a)

1.4-1.4

-1.4

1.4 (b) z

36. (a) 1

-1

-2 2

(b)

37. Complete the square to get (x+1)

2+(y−1)2+(z−2)2= 9; center (−1,1,2), radius 3. The

distance between the origin and the center is √6<3 so the origin is inside the sphere. The largest

distance is 3 + √6, the smallest is 3 −√6.

38. (x−1)2+y2+(z+4)

2≤25; all points on and inside the sphere of radius 5 with center

at (1,0,−4).

39. (y+3)

2+(z−2)2>16; all points outside the circular cylinder (y+3)

2+(z−2)2= 16.

40. p(x−1)2+(y+2)

2+z

2=2

2+(y−1)2+(z−1)2, square and simplify to get

3x2+3y

2+3z

2+2x−12y−8z+ 3 = 0, then complete the square to get

(x+1/3)2+(y−2)2+(z−4/3)2=44/9; center (−1/3,2,4/3), radius 2√11/3.

41. Let rbe the radius of a styrofoam sphere. The distance from the origin to the center of the bowling

ball is equal to the sum of the distance from the origin to the center of the styrofoam sphere nearest

the origin and the distance between the center of this sphere and the center of the bowling ball so

√3R=√3r+r+R,(

√

3+1)r=(

√

3−1)R,r=√3−1

√3+1R=(2−√

3)R.

42. (a) Complete the square to get (x+G/2)2+(y+H/2)2+(z+I/2)2=K/4, so the equation

represents a sphere when K>0, a point when K= 0, and no graph when K<0.

(b) C(−G/2,−H/2,−I/2), r=√K/2

Exercise Set 13.2 452

43. (asin φcos θ)2+(asin φsin θ)2+(acos φ)2=a2sin2φcos2θ+a2sin2φsin2θ+a2cos2φ

=a2sin2φ(cos2θ+ sin2θ)+a

2cos2φ

=a2sin2φ+a2cos2φ=a2(sin2φ+ cos2φ)=a

EXERCISE SET 13.2

1. (a–c)

〈2, 5〉

〈–5, –4〉

〈2, 0〉

y(d–f)

-5i + 3j

3i - 2j

-6j

2. (a–c)

〈0, -8〉

〈6, -2〉

〈-3, 7〉

y(d–f)

-2 i - j

4i + 2 j

3. (a–b)

〈1, -2, 2〉

〈2, 2, –1〉

z(c–d)

z–i + 2j + 3k

2i + 3j – k

4. (a–b)

〈-1, 3, 2 〉

〈3, 4, 2 〉

z(c–d)

i - j + 2 k

2j - k

453 Chapter 13

5. (a) h4−1,1−5i=h3,−4i

3i – 4j

(b) h0−2,0−3,4−0i=h−2,−3,4i

–2i – 3j + 4kz

6. (a) h−3−2,3−3i=h−5,0i

–5i

(b) h0−3,4−0,4−4i=h−3,4,0i

– 3i + 4j

7. (a) h2−3, 8 −5i=h−1,3i

(b) h0−7, 0 −(−2)i=h−7,2i

8. (a) h−4−(−6), −1−(−2)i=h2,1i

(b) h−1,6,1i

9. (a) Let (x, y) be the terminal point, then x−1=3,x= 4 and y−(−2) = −2, y=−4.

The terminal point is (4,−4).

(b) Let (x, y, z) be the initial point, then 5 −x=−3, −y= 1, and −1−z=2sox=8,

y=−1, and z=−3. The initial point is (8,−1,−3).

10. (a) Let (x, y) be the terminal point, then x−2=7,x= 9 and y−(−1)=6,y=5.

The terminal point is (9,5).

(b) Let (x, y, z) be the terminal point, then x+2=1,y−1=2,andz−4=−3sox=−1,

y= 3, and z= 1. The terminal point is (−1,3,1).

11. (a) −i+4j−2k (b) 18i+12j−6k (c) −i−5j−2k

(d) 40i−4j−4k (e) −2i−16j−18k(f)−i+13j−2k

12. (a) h1,−2,0i(b) h28,0,−14i+h3,3,9i=h31,3,−5i

(e) h−12,0,6i−h8,8,24i=h−20,−8,−18i

(f) h8,0,−4i−h3,0,6i=h5,0,−10i

13. (a) kvk=√1+1=√

2(b) kvk=√1+49=5

√

14. (a) kvk=√9+16=5 (b) kvk=√2+7=3

Exercise Set 13.2 454

15. (a) ku+vk=k2i−2j+2kk=2

√

3(b) kuk+kvk=√14+√2

√

14+2

√

2(d) k3u−5v+wk=k−12j+2kk=2

√

(e) (1/√6)i+(1/

√

6)j−(2/√6)k(f)1

16. If one vector is a positive multiple of the other, say u=αvwith α>0, then u,vand u+vare

parallel and ku+vk=(1+α)kvk=kuk+kvk.

17. (a) k−i+4jk=√

17 so the required vector is −1/√17i+4/√17j

(b) k6i−4j+2kk=2

√

14 so the required vector is (−3i+2j−k)/

√

AB=4i+j−k,k−→

AB k=3

√

2 so the required vector is (4i+j−k)/3√2

18. (a) k3i−4jk= 5 so the required vector is −1

5(3i−4j)=−3

i+4

(b) k2i−j−2kk= 3 so the required vector is 2

3i−1

3j−2

AB =4i−3j,k−→

AB k= 5 so the required vector is 4

5i−3

19. (a) −1

2v=h−3/2,2i(b) kvk=√85, so √17

√85v=1

√5h7,0,−6ihas length √17

20. (a) 3v=−6i+9j (b) −2

kvkv=6

√26i−8

√26j−2

√26k

21. (a) v =kvkhcos π/4,sin π/4i=h3√2/2,3√2/2i

(b) v =kvkhcos 90◦,sin 90◦i=h0,2i

(d) v =kvkhcos π, sin πi=h−1,0i

22. From (12), v=hcos π/6,sin π/6i=h√3/2,1/2iand w=hcos 3π/4,sin 3π/4i=h−√2/2,√2/2i,

so v+w=((

√

3−√

2)/2,(1+√2)/2,v−w=((

√

3+√

2)/2,(1 −√2)/2)

23. From (12), v=hcos 30◦,sin 30◦i=h√3/2,1/2iand w=hcos 135◦,sin 135◦i=h−√2/2,√2/2i,so

v+w=((

√

3−√

2)/2,(1+√2)/2)

24. w =h1,0i, and from (12), v=hcos 120◦,sin 120◦i=h−1/2,√3/2i,sov+w=h1/2,

√

3/2i

25. (a) The initial point of u+v+w

is the origin and the endpoint

is (−2,5), so u+v+w=h−2,5i.

-5 5

-5

–2i + 5j

(b) The initial point of u+v+w

is (−5,4) and the endpoint

is (−2,−4), so u+v+w=h3,−8i.

-5 5

-8

3i – 8j

455 Chapter 13

26. (a) v =h−10,2iby inspection, so u−v+w=u+v+w−2v=h−2,5i+h20,−4i=h18,1i.

(b) v =h−3,8iby inspection, so u−v+w=u+v+w−2v=h3,−8i+h6,−16i=h9,−24i.

27. 6x=2u−v−w=h−4,6i,x=h−2/3,1i

28. u −2x=x−w+3v,3x=u+w−3v,x=1

(u+w−3v)=h2/3,2/3i

29. u =5

7i+2

7j+1

7k,v=8

7i−1

7j−4

7k 30. u =h−5,8i,v=h7,−11i

31. k(i+j)+(i−2j)k=k2i−jk=√

5,k(i+j−(i−2j)k=k3jk=3

32. Let A, B,Cbe the vertices (0,0), (1,3), (2,4) and Dthe fourth vertex (x, y). For the parallelogram

ABCD, −→

AD =−→

BC,hx, yi=h1,1iso x=1,y= 1 and Dis at (1,1). For the parallelogram ACBD,

−→

AD =−→

CB,hx, yi=h−1,−1iso x=−1, y=−1 and Dis at (−1,−1). For the parallelogram

ABDC, −→

AC=−→

BD,hx−1,y−3i=h2,4i,sox=3,y = 7 and Dis at (3,7).

33. (a) 5=kkvk=|k|kvk=±3k,sok=±5/3

(b) 6=kkvk=|k|kvk=2kvk,sokvk=3

34. If kkvk= 0 then |k|kvk= 0 so either k=0orkvk= 0; in the latter case, by (9) or (10), v=0.

35. (a) Choose two points on the line, for example P1(0,2) and P2(1,5); then −→

P 1P 2=h1 ,3 iis

parallel to the line, kh1,3ik =√10, so h1/√10,3/√10iand h−1/√10,−3/√10iare unit

vectors parallel to the line.

(b) Choose two points on the line, for example P1(0,4) and P2(1,3); then −→

P 1P 2=h1 ,−1 iis

parallel to the line, kh1,−1ik =√2soh1/

√

2,−1/

√

2iand h−1/√2,1/√2iare unit vectors

parallel to the line.

and P2(5,1) are on the line, so −→

P 1P 2=h5 ,1 iis perpendicular to the line, so ±1

√26h5,1iare

unit vectors perpendicular to the line.

36. (a) ±k (b) ±j (c) ±i

37. (a) the circle of radius 1 about the origin

(b) the solid disk of radius 1 about the origin

38. (a) the circle of radius 1 about the tip of r0

(b) the solid disk of radius 1 about the tip of r0

39. (a) the (hollow) sphere of radius 1 about the origin

(b) the solid ball of radius 1 about the origin

40. The sum of the distances between (x, y) and the points (x1,y

), (x2,y

) is the constant k,so

the set consists of all points on the ellipse with foci at (x1,y

) and (x2,y

), and major axis of

length k.

Exercise Set 13.2 456

41. Since φ=π/2, from (13) we get kF1+F2k2=kF1k2+kF2k2= 3600 + 900,

so kF1+F2k=30

√

5 lb, and sin α=kF2k

kF1+F2ksin φ=30

30√5,α≈26.57◦,θ =α≈26.57◦.

42. kF1+F2k2=kF1k2+kF2k2+2kF

kkF2kcos φ=14,400+10,000 + 2(120)(100)1

2=36,400, so

kF1+F2k=20

√

91 N, sin α=kF2k

kF1+F2ksin φ=100

20√91 sin 60◦=5√3

2√91,α ≈27.16◦,

θ=α≈27.16◦.

43. kF1+F2k2=kF1k2+kF2k2+2kF

kkF2kcos φ= 160,000 + 160,000 −2(400)(400)√3

so kF1+F2k≈207.06 N, and sin α=kF2k

kF1+F2ksin φ≈400

207.06 1

2,α=75.00◦,

θ=α−30◦=45.00◦.

44. kF1+F2k2=kF1k2+kF2k2+2kF

kkF2kcos φ=16+4+2(4)(2) cos 77◦,so

k≈4.86 lb, and sin α=kF2k

kF1+F2ksin φ=2

4.86 sin 77◦,α≈23.64◦,θ =α−27◦≈−3.36◦.

45. Let F1,F2,F3be the forces in the diagram with magnitudes 40,50,75 respectively. Then

F1+F2+F3=(F

1+F

)+F

. Following the examples, F1+F2has magnitude 45.83 N and

makes an angle 79.11◦with the positive x-axis. Then

k(F1+F2)+F3k2≈45.832+752+2(45.83)(75) cos 79.11◦,soF

1+F

2+F

3has magnitude ≈94.995

N and makes an angle θ=α≈28.28◦with the positive x-axis.

46. Let F1,F2,F3be the forces in the diagram with magnitudes 150,200,100 respectively. Then

F1+F2+F3=(F

1+F

)+F

. Following the examples, F1+F2has magnitude 279.34 N and

makes an angle 91.24◦with the positive x-axis. Then

kF1+F2+F3k2≈279.342+ 1002+ 2(279.34)(100) cos(270 −91.24)◦, and F1+F2+F3has

magnitude ≈179.37 N and makes an angle 91.94◦with the positive x-axis.

47. Let F1,F2be the forces in the diagram with magnitudes 8,10 respectively. Then kF1+F2khas

magnitude √82+10

2+2·8·10 cos 120◦=2

√

21 ≈9.165 lb, and makes an angle

60◦+sin−1kF1k

kF1+F2ksin 120 ≈109.11◦with the positive x-axis, so Fhas magnitude 9.165 lb and

makes an angle −70.89◦with the positive x-axis.

48. kF1+F2k=√1202+ 1502+2·120 ·150 cos 75◦= 214.98 N and makes an angle 92.63◦with the

positive x-axis, and kF1+F2+F3k= 232.90 N and makes an angle 67.23◦with the positive x-axis,

hence Fhas magnitude 232.90 N and makes an angle −112.77◦with the positive x-axis.

49. F1+F2+F=0, where Fhas magnitude 250 and makes an angle −90◦with the positive x-axis.

Thus kF1+F2k2=kF1k2+kF2k2+2kF

kkF2kcos 105◦= 2502and

45◦=α= sin−1kF2k

250 sin 105◦,so√

2≈kF

250 0.9659,kF2k≈183.02 lb,

kF1k2+ 2(183.02)(−0.2588)kF1k+ (183.02)2=62,500,kF1k= 224.13 lb.

50. Similar to Exercise 49, kF1k= 100√3N,kF

k= 100 N

51. (a) c1v1+c2v2=(2c

1+4c

)i+(−c

1+2c

)j=−4j,so2c

1+4c

2= 0 and −c1+2c

2=−4

which gives c1=2,c

2=−1.

(b) c1v1+c2v2=hc1−2c2,−3c1+6c

i=h3,5i,so c1−2c2= 3 and −3c1+6c

2=5

which has no solution.

457 Chapter 13

52. (a) Equate corresponding components to get the system of equations c1+3c

2=−1,

2c2+c3= 1, and c1+c3= 5. Solve to get c1=2,c

2=−1, and c3=3.

(b) Equate corresponding components to get the system of equations c1+3c

2+4c

3=2,

−c

1−c

3= 1, and c2+c3=−1. From the second and third equations, c1=−1−c3and

c2=−1−c3; substitute these into the ﬁrst equation to get −4 = 2, which is nonsense so the

system has no solution.

53. Place uand vtip to tail so that u+vis the vector from the initial point of uto the terminal

point of v. The shortest distance between two points is along the line joining these points so

ku+vk≤kuk+kvk.

54. (a): u+v=(u

i+u

2j)+(v

i+v

2j)=(v

i+v

2j)+u

i+u

2j=v+u

(c): u+0=u1i+u2j+0i+0j=u

i+u

2j=u

(e): k(lu)=k(l(u

i+u

2j)) = k(lu1i+lu2j)=klu1i+klu2j=(kl)u

55. (d): u+(−u)=u

i+u

2j+(−u

i−u

2j)=(u

1−u

)i+(u

1−u

)j=0

(g): (k+l)u=(k+l)(u1i+u2j)=ku1i+ku2j+lu1i+lu2j=ku+lu

(h): 1u=1(u

i+u

2j)=1u

i+1u

2j=u

i+u

2j=u

56. Draw the triangles with sides formed by the vectors u,v,u+vand ku,kv,ku+kv. By similar

triangles, k(u+v)=ku+kv.

57. Let a,b,cbe vectors along the sides of the triangle and A,B the

midpoints of aand b, then u=1

2a−1

2b=1

2(a−b)=1

cso uis

parallel to cand half as long.

58. Let a,b,c,dbe vectors along the sides of the quadrilateral

and A, B, C, D the corresponding midpoints, then

u=1

2b+1

2cand v=1

2d−1

2abut d=a+b+cso

v=1

2(a+b+c)−1

2a=1

2b+1

2c=uthus ABCD

is a parallelogram because sides AD and BC are equal

and parallel.

EXERCISE SET 13.3

1. (a) (1)(6) + (2)(−8) = −10; cos θ=(−10)/[(√5)(10)] = −1/√5

(b) (−7)(0) + (−3)(1) = −3; cos θ=(−3)/[(√58)(1)] = −3/√58

(d) (−3)(4) + (1)(2) + (2)(−5) = −20; cos θ=(−20)/[(√14)(√45)] = −20/(3√70)

2. (a) u ·v= 1(2) cos(π/6) = √3(b) u ·v= 2(3) cos 135◦=−3√2

3. (a) u ·v=−34 <0, obtuse (b) u ·v=6>0, acute

Exercise Set 13.3 458

4. Let the points be P, Q, R in order, then −→

PQ=h2−(−1),−2−2,0−3i=h3,−4,−3i,

−→

QR=h3−2,1−(−2),−4−0i=h1,3,−4i,−→

RP =h−1−3,2−1,3−(−4)i=h−4,1,7i;

since −→

QP ·

−→

QR=−3(1) + 4(3) + 3(−4) = −3<0,6PQR is obtuse;

since −→

RP ·

−→

RQ=−4(−1)+(−3) + 7(4) = 29 >0,6PRQ is acute;

since −→

PR ·

−→

PQ= 4(3) −1(−4) −7(−3)=37>0,6RP Q is acute

5. Since v1·vi= cos φi, the answers are, in order, √2/2,0,−√2/2,−1,−√2/2,0,√2/2

6. Proceed as in Exercise 5; 25/2,−25/2,−25,−25/2,25/2

7. (a) −→

AB =h1,3,−2i,−→

BC =h4,−2,−1i,−→

AB ·

−→

BC =0so −→

AB and −→

BC are orthogonal; it is a right

triangle with the right angle at vertex B.

(b) Let A, B, and Cbe the vertices (−1,0), (2,−1), and (1,4) with corresponding interior angles

α,β, and γ, then

cos α=

−→

AB ·

−→

k−→

AB kk−→

AC k

=h3,−1i·h2,4i

√10√20 =1/(5√2),α≈82◦

cos β=

−→

BA ·

−→

k−→

BAkk−→

BC k

=h−3,1i·h−1,5i

√10√26 =4/

√

65,β≈60◦

cos γ=

−→

CA ·

−→

k−→

CAkk−→

CBk

=h−2,−4i·h1,−5i

√20√26 =9/

√

130,γ≈38◦

8. −→

AB ·

−→

AP =[2i+j+2k]·[(k−1)i+(k+1)j+(k−3)k]

=2(k−1)+(k+1)+2(k−3)=5k−7=0,k =7/5.

9. (a) v ·v1=−ab +ba =0;v·v

2=ab +b(−a)=0

(b) Let v1=2i+3j,v

2=−2i−3j;

take u1=v1

kv1k=2

√13i+3

√13j,u2=−u1.

-3 3

-3

10. By inspection, 3i−4jis orthogonal to and has the same length as 4i+3j

so u1=(4i+3j)+(3i−4j)=7i−jand u2=(4i+3j)+(−1)(3i−4j)=i+7jeach make an angle

of 45◦with 4i+3j; unit vectors in the directions of u1and u2are (7i−j)/√50 and (i+7j)/

√50.

11. (a) The dot product of a vector uand a scalar v·wis not deﬁned.

(b) The sum of a scalar u·vand a vector wis not deﬁned.

(d) The dot product of a scalar kand a vector u+vis not deﬁned.

459 Chapter 13

12. (b): u·(v+w)=(6i−j+2k)·((2i+7j+4k)+(i+j−3k))=(6i−j+2k)·(3i+8j+k) = 12;

u·v+u·w=(6i−j+2k)·(2i+7j+4k)+(6i−j+2k)·(i+j−3k)=13−1=12

(c): k(u·v)=−5(13) = −65; (ku)·v=(−30i+5j−10k)·(2i+7j+4k)=−65;

u·(kv)=(6i−j+2k)·(−10i−35j−20k)=−65

13. (a) h1,2i·(h28,−14i+h6,0i)=h1,2i·h34,−14i=6

(b) k6wk=6kwk=36 (c) 24√5(d) 24√5

14. false, for example a=h1,2i,b=h−1,0i,c=h5,−3i

15. (a) kvk=√3socosα= cos β=1/

√

3, cos γ=−1/√3, α=β≈55◦,γ≈125◦

(b) kvk= 3 so cos α=2/3, cos β=−2/3, cos γ=1/3, α≈48◦,β≈132◦,γ≈71◦

16. (a) kvk= 7 so cos α=3/7, cos β=−2/7, cos γ=−6/7, α≈65◦,β≈107◦,γ≈149◦

(b) kvk= 5, cos α=3/5, cos β= 0, cos γ=−4/5, α≈53◦,β≈90◦,γ≈143◦

17. cos2α+ cos2β+ cos2γ=v2

kvk2+v2

kvk2=v2

1+v2

2+v2

3/kvk2=kvk2/kvk2=1

18. Let v=hx, y, zi, then x=px2+y2cos θ, y =px2+y2sin θ, px2+y2=kvkcos λ, and

z=kvksin λ,sox/kvk= cos θcos λ, y/kvk= sin θcos λ, and z/kvk= sin λ.

19. cos α=√3

2=√3

4,cos β=√3

√3

2=3

4,cos γ=1

2;α≈64◦,β ≈41◦,γ =60

◦

20. Let u1=ku1khcos α1,cos β1,cos γ1i,u2=ku2khcos α2,cos β2,cos γ2i,u1and u2are perpendicular

if and only if u1·u2=0soku

kku

k(cos α1cos α2+ cos β1cos β2+ cos γ1cos γ2)=0,

cos α1cos α2+ cos β1cos β2+ cos γ1cos γ2=0.

21. (a) b

kbk=h3/5,4/5i, so projbv=h6/25,8/25i

and v−projbv=h44/25,−33/25i

-2

v – projbv

projbv

(b) b

kbk=h1/√5,−2/√5i, so projbv=h−6/5,12/5i

and v−projbv=h26/5,13/5i

-5 5

-5

v – projbv

projbv

Exercise Set 13.3 460

kbk=h2/√5,1/√5i, so projbv=h−16/5,−8/5i

and v−projbv=h1/5,−2/5ix

-4

v – projbv

projbv

22. (a) b

kbk=h1/3,2/3,2/3i, so projbv=h2/3,4/3,4/3iand v−projbv=h4/3,−7/3,5/3i

(b) b

kbk=h2/7,3/7,−6/7i, so projbv=h−74/49,−111/49,222/49i

and v−projbv=h270/49,62/49,121/49i

23. (a) projbv=h−1,−1i,sov=h−1,−1i+h3,−3i

(b) projbv=h16/5,0,−8/5i,sov=h16/5,0,−8/5i+h−1/5,1,−2/5i

24. (a) projbv=h1,1i,sov=h1,1i+h−4,4i

(b) projbv=h0,−8/5,4/5i,sov=h0,−8/5,4/5i+h−2,13/5,26/5i

25. −→

AP =−i+3j,−→

AB=3i+4j,kproj−→

−→

AP k=|−→

AP ·

−→

AB |/k−→

AB k=9/5

k−→

AP k=√10,p10 −81/25=13/5

26. −→

AP =−4i+2k,−→

AB =−3i+2j−4k,kproj−→

−→

AP k=|−→

AP ·

−→

AB |/k−→

AB k=4/

√

29.

k−→

AP k=√20, p20 −16/29 = p564/29

27. Let Fbe the downward force of gravity on the block, then kFk= 10(9.8) = 98 N, and if F1and

F2are the forces parallel to and perpendicular to the ramp, then kF1k=kF2k=49

√

2N.Thus

the block exerts a force of 49√2 N against the ramp and it requires a force of 49√2 N to prevent

the block from sliding down the ramp.

28. Let xdenote the magnitude of the force in the direction of Q. Then the force Facting on the

block is F=xi−10j. Let u=−1

√2(i+j) and v=1

√2(i−j) be the unit vectors in the directions

along and against the ramp. Then Fdecomposes as F=−x−10

√2u+x+10

√

2v, and thus the block

will not slide down the ramp provided x≥10 N.

29. Three forces act on the block: its weight −300j; the tension in cable A, which has the form

a(−i+j); and the tension in cable B, which has the form b(√3i−j), where a, b are positive

constants. The sum of these forces is zero, which yields a= 450 + 150√3,b= 150 + 150√3. Thus

the forces along cables A and B are, respectively,

k150(3 + √3)(i−j)k= 450√2 + 150√6 lb, and k150(√3 + 1)(√3i−j)k= 300 + 300√3 lb.

461 Chapter 13

30. (a) Let T

Aand T

Bbe the forces exerted on the block by cables Aand B. Then

A=a(−10i+dj) and T

B=b(20i+dj) for some positive a, b. Since T

A+T

B−100j=0,we

ﬁnd a=200

3d,b=100

3d,T

A=−2000

3di+200

3j, and T

B=2000

3di+100

3j.

500

-100

-20 100

(b) An increase in dwill decrease both forces. 500

-100

-20 100

Ak≤150 is equivalent to d≥40

√65, and kT

Bk≤150 is equivalent to

d≥40

√77. Hence we must have d≥40

65.

31. Let Pand Qbe the points (1,3) and (4,7) then −→

PQ=3i+4jso W=F·

−→

PQ=−12 ft ·lb.

32. W=F·

−→

PQ=kFkk−→

PQkcos 45◦= (500)(100) √2/2=25,000√2N·m

33. W=F·15i=15·50 cos 60◦= 375 ft ·lb.

34. W=F·(15/√3)(i+j+k)=−15/√3N·m

35. With the cube as shown in the diagram,

and athe length of each edge,

d1=ai+aj+ak,d2=ai+aj−ak,

cos θ=(d

1·d

)/(kd

kkd

k)=1/3,θ ≈71◦

36. Take i,j, and kalong adjacent edges of the box, then 10i+15j+25kis along a diagonal, and a

unit vector in this direction is 2

√38i+3

√38j+5

√38k. The direction cosines are cos α=2/

√

38,

cos β=3/

√

38, and cos γ=5/

√

38 so α≈71◦,β≈61◦, and γ≈36◦.

Exercise Set 13.3 462

37. u +vand u−vare vectors along the diagonals,

(u+v)·(u−v)=u·u−u·v+v·u−v·v=kuk

2−kvk

2so (u+v)·(u−v)=0

if and only if kuk=kvk.

38. The diagonals have lengths ku+vkand ku−vkbut

ku+vk2=(u+v)·(u+v)=kuk

2+2u·v+kvk

, and

ku−vk2=(u−v)·(u−v)=kuk

2−2u·v+kvk

. If the parallelogram is a rectangle then

u·v=0soku+vk

2=ku−vk

; the diagonals are equal. If the diagonals are equal, then

4u·v=0,u·v=0souis perpendicular to vand hence the parallelogram is a rectangle.

39. ku+vk2=(u+v)·(u+v)=kuk

2+2u·v+kvk

2and

ku−vk2=(u−v)·(u−v)=kuk

2−2u·v+kvk

, add to get

ku+vk2+ku−vk2=2kuk

2+2kvk

The sum of the squares of the lengths of the diagonals of a parallelogram is equal to twice the sum

of the squares of the lengths of the sides.

40. ku+vk2=(u+v)·(u+v)=kuk

2+2u·v+kvk

2and

ku−vk2=(u−v)·(u−v)=kuk

2−2u·v+kvk

, subtract to get

ku+vk2−ku−vk

2=4u·v, the result follows by dividing both sides by 4.

41. v =c1v1+c2v2+c3v3so v·vi=civi·vibecause vi·vj=0ifi6=j,

thus v·vi=cikvik2,c

i=v·v

i/kv

ik2for i=1,2,3.

42. v1·v2=v1·v3=v2·v3= 0 so they are mutually perpendicular. Let v=i−j+k, then

c1=v·v1

kv1k2=3

7,c2=v·v2

kv2k2=−1

3, and c3=v·v3

kv3k2=1

21.

43. (a) u =xi+(x

2+1)j,v=xi−(x+1)j,θ = cos−1[(u·v)/(kukkvk)].

Solve dθ/dx = 0 to ﬁnd that the minimum value of θoccurs when x≈−3.136742 so the

minimum angle is about 40◦.

(b) Solve u·v= 0 for xto get x≈−0.682328.

44. (a) u = cos θ1i±sin θ1j,v=±sin θ2j+ cos θ2k,cos θ=u·v=±sin θ1sin θ2

(b) cos θ=±sin245◦=±1/2,θ=60

◦

is acute) when t≈2.186276 and that θmin ≈40◦when t≈0.955317; for θmax check the

endpoints t=0,π/2 to obtain θmax = cos−1(0) = π/2.

45. Let u=hu1,u

i,v=hv

i,w=hw

i. Then

u·(v+w)=hu

1+w

),u

2+w

),u

3+w

)i=hu

1+u

2+u

3+u

=hu

i+hu

i=u·v+u·w

0·v=0·v

1+0·v

2+0·v

3=0

463 Chapter 13

EXERCISE SET 13.4

1. (a) i ×(i+j+k)=



ijk

100

111



=−j+k

(b) i ×(i+j+k)=(i×i)+(i×j)+(i×k)=−j+k

2. (a) j ×(i+j+k)=



ijk

010

111



=i−k

j×(i+j+k)=(j×i)+(j×j)+(j×k)=i−k

(b) k ×(i+j+k)=



ijk

001

111



=−i+j

k×(i+j+k)=(k×i)+(k×j)+(k×k)=j−i+0=−i+j

3. h7,10,9i4. −i−2j−7k5.h−4,−6,−3i6. i +2j−4k

7. (a) v ×w=h−23,7,−1i,u×(v×w)=h−20,−67,−9i

(b) u ×v=h−10,−14,2i,(u×v)×w=h−78,52,−26i

(d) (v×w)×(u×v)=h0,56,392i

9. u ×v=(i+j)×(i+j+k)=k−j−k+i=i−j, the direction cosines are 1

√2,−1

√2,0

10. u ×v=12i+30j−6k,so± 2

√

30i+√5

√6j−1

√30k!

11. n =−→

AB ×−→

AC =h1,1,−3i×h−1,3,−1i=h8,4,4i, unit vectors are ±1

√6h2,1,1i

12. A vector parallel to the yz-plane must be perpendicular to i;

i×(3i −j+2k)=−2j−k,k−2j−kk=√

5, the unit vectors are ±(2j+k)/√5.

13. A=ku×vk=k−7i−j+3kk=√

59 14. A=ku×vk=k−6i+4j+7kk=√

101

15. A=1

2k−→

PQ ×−→

PRk=1

2kh−1,−5,2i×h2,0,3ik =1

2kh−15,7,10ik =√374/2

16. A=1

2k−→

PQ ×−→

PRk=1

2kh−1,4,8i×h5,2,12ik =1

2kh32,52,−22ik =9

√

17. 80 18. 29 19. −320. 1

21. V=|u·(v×w)|=|−16|=16 22. V=|u·(v×w)|=|45|=45

23. (a) u ·(v×w)=0,yes (b) u ·(v×w)=0,yes (c) u ·(v×w) = 245, no

Exercise Set 13.4 464

24. (a) u ·(w×v)=−u·(v×w)=−3(b) (v×w)·u=u·(v×w)=3

(e) (u×w)·v=u·(w×v)=−3(f ) v ·(w×w) = 0 because w×w=0

25. (a) V=|u·(v×w)|=|−9|=9 (b) A=ku×wk=k3i−8j+7kk=√

122

angle between uand v×wthen

cos θ=u·(v×w)

kukkv×wk=−9

√

14√14 =−9/14

so the acute angle φthat umakes with the plane determined by vand wis

φ=θ−π/2 = sin−1(9/14).

26. From the diagram,

d=kuksin θ=kukkvksin θ

kvk=ku×vk

kvku

27. (a) u =−→

AP =−4i+2k,v=−→

AB =−3i+2j−4k,u×v=−4i−22j−8k;

distance = ku×vk/kvk=2

p141/29

(b) u =−→

AP =2i+2j,v=−→

AB =−2i+j,u×v=6k; distance = ku×vk/kvk=6/

√

28. Take vand was sides of the (triangular) base, then area of base = 1

2kv×wkand

height = kprojv×wuk=|u·(v×w)|

kv×wkso V=1

3(area of base) (height) = 1

6|u·(v×w)|

29. −→

PQ=h3,−1,−3i,−→

PR=h2,−2,1i,−→

PS=h4,−4,3i,

V=1

6|−→

PQ ·(−→

PR ×−→

PS)|=1

6|−4|=2/3

30. (a) cos θ=u·v

kukkvk=−23

49 (b) sin θ=ku×vk

kukkvk=k36i−24jk

49 =12√13

492+144 ·13

492=2401

492=1

31. From Theorems 13.3.3 and 13.4.5a it follows that sin θ= cos θ,soθ=π/4.

32. ku×vk2=kuk2kvk2sin2θ=kuk2kvk2(1 −cos2θ)=kuk

kvk

2−(u·v)

33. (a) F =10jand −→

PQ=i+j+k, so the vector moment of Fabout Pis

−→

PQ ×F=



ijk

111

0100



=−10i+10k, and the scalar moment is 10√2lb·ft.

The direction of rotation of the cube about Pis counterclockwise looking along

−→

PQ ×F=−10i+10ktoward its initial point.

465 Chapter 13

(b) F =10jand −→

PQ=j+k, so the vector moment of Fabout Pis

−→

PQ ×F=



ijk

011

0100



=−10i, and the scalar moment is 10 lb·ft. The direction of rotation

of the cube about Pis counterclockwise looking along −10itoward its initial point.

PQ=j, so the vector moment of Fabout Pis

−→

PQ ×F=



ijk

010

0100



=0, and the scalar moment is 0 lb·ft. Since the force is parallel to

the direction of motion, there is no rotation about P.

34. (a) F =1000

√2(−i+k) and −→

PQ=2j−k, so the vector moment of Fabout Pis

−→

PQ ×F= 500√2

ijk

02−1

−10 1



= 500√2(2i+j+2k), and the scalar moment is

1500√2N·m.

(b) The direction angles of the vector moment of Fabout the point Pare

cos−1(2/3) ≈48◦,cos−1(1/3) ≈71◦, and cos−1(2/3) ≈48◦.

35. F makes an angle 72◦with the positive x-axis, and ~

PQ =0.2i+0.03jmakes an angle

α= tan−1(0.03/0.2) with the x-axis, so φ≈72◦−8.53◦=63.47◦and kFk= 200 N, so

k−→

PQ ×Fk=k~

PQkkFk|sin 63.47◦|= 200k(0.2i+0.030j)k|sin 63.47◦|≈36.19 N·m.

36. Part (b) : let u=hu1,u

i,v=hv

i,and w=hw1,w

i; show that

u×(v+w) and (u×v)+(u×w) are the same.

Part (c) : (u+v)×w=−[w×(u+v)] from part (a)

=−[(w×u)+(w×v)] from part (b)

=(u×w)+(v×w) from part (a)

37. Let u=hu1,u

iand v=hv1,v

i; show that k(u×v), (ku)×v, and u×(kv) are all the

same; Part (e) is proved in a similar fashion.

38. Suppose the ﬁrst two rows are interchanged. Then by deﬁnition,



b1b2b3

a1a2a3

c1c2c3

=b1

a2a3

c2c3−b2

a1a3

c1c3

+b3

a1a2

c1c2

=b1(a2c3−a3c2)−b2(a1c3−a3c1)+b

2−a

which is the negative of the right hand side of (2) after expansion. If two other rows were to be

exchanged, a similar proof would hold. Finally, suppose ∆ were a determinant with two identical

rows. Then the value is unchanged if we interchange those two rows, yet ∆ = −∆ by Part (b) of

Theorem 13.4.1. Hence ∆ = −∆,∆=0.

39. −8i−20j+2k,−8i−8k

40. (a) From the ﬁrst formula in Exercise 39 it follows that u×(v×w) is a linear combination of

vand wand hence lies in the plane determined by them.

(b) Similar to (a).

Exercise Set 13.5 466

41. If a,b,c, and dlie in the same plane then a×band c×dare parallel so (a×b)×(c×d)=0

42. Let uand vbe the vectors from a point on the curve to the points (2,−1,0) and (3,2,2), respec-

tively. Then u=(2−x)i+(−1−lnx)jand v=(3−x)i+(2−lnx)j+2k.The area of the triangle

is given by A=(1/2)ku×vk; solve dA/dx = 0 for xto get x=2.091581. The minimum area is

1.887850.

43.

−→

0×F=−→

PQ ×F+−→

QQ0×F=−→

PQ ×F, since Fand −→

QQ0are parallel.

EXERCISE SET 13.5

In many of the Exercises in this section other answers are also possible.

1. (a) L1:P(1,0),v=j,x=1,y =t

:P(0,1),v=i,x=t, y =1

:P(0,0),v=i+j,x=t, y =t

(b) L1:P(1,1,0),v=k,x=1,y =1,z =t

:P(0,1,1),v=i,x=t, y =1,z =1

:P(1,0,1),v=j,x=1,y =t, z =1

:P(0,0,0),v=i+j+k,x=t,

y=t, z =t

2. (a) L1:x=t, y =1,0≤t≤1

:x=1,y =t, 0≤t≤1

L3:x=t, y =t, 0≤t≤1

(b) L1:x=1,y =1,z =t, 0≤t≤1

L2:x=t, y =1,z =1,0≤t≤1

:x=1,y =t, z =1,0≤t≤1

:x=t, y =t, z =t, 0≤t≤1

3. (a) −→

P 1P 2=h2 ,3 iso x=3+2t,y=−2+3tfor the line; for the line segment add the condition

0≤t≤1.

(b) −→

P 1P 2=h−3,6,1iso x=5−3t,y=−2+6t,z=1+tfor the line; for the line segment add

the condition 0 ≤t≤1.

4. (a) −→

P 1P 2=h−3,−5iso x=−3t,y=1−5tfor the line; for the line segment add the condition

0≤t≤1.

(b) −→

P 1P 2=h0 ,0 ,−3 iso x=−1, y=3,z=5−3tfor the line; for the line segment add the

condition 0 ≤t≤1.

5. (a) x=2+t, y =−3−4t(b) x=t, y =−t, z =1+t

6. (a) x=3+2t, y =−4+t(b) x=−1−t, y =3t, z =2

7. (a) r0=2i−jso P(2,−1) is on the line, and v=4i−jis parallel to the line.

(b) At t=0,P(−1,2,4) is on the line, and v=5i+7j−8kis parallel to the line.

8. (a) At t=0,P(−1,5) is on the line, and v=2i+3jis parallel to the line.

(b) r0=i+j−2kso P(1,1,−2) is on the line, and v=jis parallel to the line.

9. (a) hx, yi=h−3,4i+th1,5i;r=−3i+4j+t(i+5j)

(b) hx, y, zi=h2,−3,0i+th−1,5,1i;r=2i−3j+t(−i+5j+k)

10. (a) hx, yi=h0,−2i+th1,1i;r=−2j+t(i+j)

(b) hx, y, zi=h1,−7,4i+th1,3,−5i;r=i−7j+4k+t(i+3j−5k)

11. x=−5+2t,y=2−3t12. x=t,y=3−2t

467 Chapter 13

13. 2x+2yy0=0,y

0=−x/y =−(3)/(−4)=3/4, v=4i+3j;x=3+4t,y=−4+3t

14. y0=2x=2(−2) = −4, v=i−4j;x=−2+t,y=4−4t

15. x=−1+3t,y=2−4t,z=4+t16. x=2−t,y=−1+2t,z=5+7t

17. The line is parallel to the vector h2,−1,2iso x=−2+2t,y=−t,z=5+2t.

18. The line is parallel to the vector h1,1,0iso x=t,y=t,z=0.

19. (a) y=0,2−t=0,t =2,x=7 (b) x=0,1+3t=0,t=−1/3,y =7/3

,9t

2+7t−1=0,t=−7±√85

18 ,x=−1±√85

6,y =43 ∓√85

20. (4t)2+(3t)

2=25,25t2=25,t=±1, the line intersects the circle at ±h4,3i

21. (a) z= 0 when t= 3 so the point is (−2,10,0)

(b) y= 0 when t=−2sothepointis(−2,0,−5)

22. (a) z= 0 when t= 4 so the point is (7,7,0)

(b) y= 0 when t=−3sothepointis(−7,0,7)

23. (1 + t)2+(3−t)

2= 16, t2−2t−3=0,(t+ 1)(t−3)=0;t=−1, 3. The points of intersection

are (0,4,−2) and (4,0,6).

24. 2(3t)+3(−1+2t)=6,12t=9;t=3/4. The point of intersection is (5/4,9/4,1/2).

25. The lines intersect if we can ﬁnd values of t1and t2that satisfy the equations 2 + t1=2+t

2+3t

1=3+4t

, and 3 + t1=4+2t

. Solutions of the ﬁrst two of these equations are t1=−1,

t2=−1 which also satisfy the third equation so the lines intersect at (1,−1,2).

26. Solve the equations −1+4t

1=−13+12t

,3+t

1=1+6t

, and 1 = 2 + 3t2. The third equation

yields t2=−1/3 which when substituted into the ﬁrst and second equations gives t1=−4inboth

cases; the lines intersect at (−17,−1,1).

27. The lines are parallel, respectively, to the vectors h7,1,−3iand h−1,0,2i. These vectors are not

parallel so the lines are not parallel. The system of equations 1 + 7t1=4−t

,3+t

1= 6, and

5−3t1=7+2t

2has no solution so the lines do not intersect.

28. The vectors h8,−8,10iand h8,−3,1iare not parallel so the lines are not parallel. The lines do not

intersect because the system of equations 2 + 8t1=3+8t

,6−8t

1=5−3t

,10t

1=6+t

2has no

solution.

29. The lines are parallel, respectively, to the vectors v1=h−2,1,−1iand v2=h−4,2,−2i;

v2=2v

1and v2are parallel so the lines are parallel.

30. The lines are not parallel because the vectors h3,−2,3iand h9,−6,8iare not parallel.

31. −→

P 1P2=h3 ,−7 ,−7 i ,−→

P 2P 3=h−9,−7,−3i; these vectors are not parallel so the points do not lie on

the same line.

Exercise Set 13.5 468

32. −→

P 1P 2=h2 ,−4 ,−4 i ,−→

P2P 3=h1 ,−2 ,−2 i ;−→

P 1P 2=2 −→

P2P 3so the vectors are parallel and the points

lie on the same line.

33. If t2gives the point h−1+3t

,9−6t

ion the second line, then t1=4−3t

2yields the point

h3−(4 −3t2),1 + 2(4 −3t2)i=h−1+3t

,9−6t

ion the ﬁrst line, so each point of L2is a point

of L1; the converse is shown with t2=(4−t

)/3.

34. If t1gives the point h1+3t

,−2+t

,2t

ion L1, then t2=(1−t

)/2 gives the point

h4−6(1 −t1)/2,−1−2(1 −t1)/2,2−4(1 −t1)/2i=h1+3t

,−2+t

,2t

ion L2, so each point of

L1isapointofL

; the converse is shown with t1=1−2t

35. The line segment joining the points (1,0) and (−3,6).

36. The line segment joining the points (−2,1,4) and (7,1,1).

37. A(3,0,1) and B(2,1,3) are on the line, and (method of Exercise 25)

−→

AP =−5i+j,−→

AB=−i+j+2k,kproj−→

−→

AP k=|−→

AP ·

−→

AB |/k−→

AB k=√6 and k−→

AP k=√26,

so distance = √26 −6=4

√

5. Using the method of Exercise 26, distance = k−→

AP ×−→

AB k

k−→

AB k

√

38. A(2,−1,0) and B(3,−2,3) are on the line, and (method of Exercise 25)

−→

AP =−i+5j−3k,−→

AB=i−j+3k,kproj−→

−→

AP k=|−→

AP ·

−→

AB |/k−→

AB k=15

√11 and

k−→

AP k=√35, so distance = p35 −225/11=4

p10/11. Using the method of Exercise 26,

distance = k−→

AP ×−→

AB k

k−→

AB k

p10/11.

39. The vectors v1=−i+2j+kand v2=2i−4j−2kare parallel to the lines, v2=−2v1so v1and

v2are parallel. Let t= 0 to get the points P(2,0,1) and Q(1,3,5) on the ﬁrst and second lines,

respectively. Let u=−→

PQ=−i+3j+4k,v=1

2=i−2j−k;u×v=5i+3j−k; by the method

of Exercise 26 of Section 13.4, distance = ku×vk/kvk=p35/6.

40. The vectors v1=2i+4j−6kand v2=3i+6j−9kare parallel to the lines, v2=(3/2)v1so v1

and v2are parallel. Let t= 0 to get the points P(0,3,2) and Q(1,0,0) on the ﬁrst and second

lines, respectively. Let u=−→

PQ=i−3j−2k,v=1

1=i+2j−3k;u×v=13i+j+5k,

distance = ku×vk/kvk=p195/14 (Exer. 26, Section 13.4).

41. (a) The line is parallel to the vector hx1−x0,y

1−y

1−z

iso

x=x0+(x

1−x

)t,y=y

0+(y

1−y

)t,z=z

0+(z

1−z

(b) The line is parallel to the vector ha, b, ciso x=x1+at,y=y1+bt,z=z1+ct

42. Solve each of the given parametric equations (2) for tto get t=(x−x

)/a,t=(y−y

)/b,

t=(z−z

)/c,so(x, y, z) is on the line if and only if (x−x0)/a =(y−y

)/b =(z−z

)/c.

43. (a) It passes through the point (1,−3,5) and is parallel to v=2i+4j+k

(b) hx, y, zi=h1+2t, −3+4t, 5+ti

44. Let the desired point be P(x0,y

), then −→

P 1P=(2/3) −→

P1P 2,

hx0−1 ,y

0−4,z

0+3i=(2/3)h0,1,2i=h0,2/3,4/3i; equate corresponding components to get

x0=1,y

0=14/3, z0=−5/3.

469 Chapter 13

45. (a) Let t= 3 and t=−2, respectively, in the equations for L1and L2.

(b) u =2i−j−2kand v=i+3j−kare parallel to L1and L2,

cos θ=u·v/(kukkvk)=1/(3√11),θ ≈84◦.

y=−1, z=−2+t.

46. (a) Let t=1/2 and t= 1, respectively, in the equations for L1and L2.

(b) u =4i−2j+2kand v=i−j+4kare parallel to L1and L2,

cos θ=u·v/(kukkvk)=14/

√

432,θ ≈48◦.

thus x=2+3t,y=7t,z=3+t.

47. (0,1,2) is on the given line (t=0)sou=j−kis a vector from this point to the point (0,2,1),

v=2i−j+kis parallel to the given line. u×v=−2j−2k, and hence w=j+k, is perpendicular

to both lines so v×w=−2i−2j+2k, and hence i+j−k, is parallel to the line we seek. Thus

x=t,y=2+t,z=1−tare parametric equations of the line.

48. (−2,4,2) is on the given line (t=0)sou=5i−3j−4kis a vector from this point to the point

(3,1,−2), v=2i+2j+kis parallel to the given line. u×v=5i−13j+16kis perpendicular to

both lines so v×(u×v)=45i−27j−36k, and hence 5i−3j−4kis parallel to the line we seek.

Thus x=3+5t,y=1−3t,z=−2−4tare parametric equations of the line.

49. (a) When t= 0 the bugs are at (4,1,2) and (0,1,1) so the distance between them is

√42+0

2+1

2=√

17 cm.

(b)

10 (c) The distance has a minimum value.

(d) Minimize D2instead of D(the distance between the bugs).

D2=[t−(4 −t)]2+[(1+t)−(1+2t)]2+[(1+2t)−(2+t)]2=6t

2−18t+ 17,

d(D2)/dt =12t−18 = 0 when t=3/2; the minimum

distance is p6(3/2)2−18(3/2)+17=√

14/2 cm.

50. The line intersects the xz-plane when t=−1, the xy-plane when t=3/2. Along the line,

T=25t

(1 + t)(3 −2t) for −1≤t≤3/2. Solve dT/dt = 0 for tto ﬁnd that the maximum value

of Tis about 50.96 when t≈1.073590.

EXERCISE SET 13.6

1. x=3,y =4,z =5 2. x=x0,y =y

,z =z

3. (x−2)+4(y−6)+2(z−1)=0,x+4y+2z=28

4. −(x+1)+7(y+1)+6(z−2)=0,−x+7y+6z=6

Exercise Set 13.6 470

5. z=0 6. 2x−3y−4z=0 7. n =i−j,x−y=0

8. n =i+j,P(1,0,0),(x−1)+y=0,x+y=1

9. n =j+k,P(0,1,0),(y−1)+z=0,y+z=1 10. n =j−k,y−z=0

11. −→

P 1P 2×−→

P 1P 3=h2 ,1 ,2 i×h3,−1,−2i=h0,10,−5i,for convenience choose h0,2,−1iwhich is also

normal to the plane. Use any of the given points to get 2y−z=1

12. −→

P 1P 2×−→

P 1P 3=h−1,−1,−2i×h−4,1,1i=h1,9,−5i,x+9y−5z=16

13. (a) parallel, because h2,−8,−6iand h−1,4,3iare parallel

(b) perpendicular, because h3,−2,1iand h4,5,−2iare orthogonal

14. (a) neither, because h3,−2,1iand h6,−4,3iare neither parallel nor orthogonal

(b) parallel, because h4,−1,−2iand h1,−1/4,−1/2iare parallel

15. (a) parallel, because h2,−1,−4iand h3,2,1iare orthogonal

(b) neither, because h1,2,3iand h1,−1,2iare neither parallel nor orthogonal

16. (a) parallel, because h−1,1,−3iand h2,2,0iare orthogonal

(b) perpendicular, because h−2,1,−1iand h6,−3,3iare parallel

17. (a) 3t−2t+t−5=0,t=5/2sox=y=z=5/2, the point of intersection is (5/2,5/2,5/2)

(b) 2(2 −t)+(3+t)+t= 1 has no solution so the line and plane do not intersect

18. (a) 2(3t)−5t+(−t)+1=0,1 = 0 has no solution so the line and the plane do not intersect.

(b) (1+t)−(−1+3t)+4(2+4t)=7,t=−3/14 so x=1−3/14=11/14,

y=−1−9/14 = −23/14, z=2−12/14=8/7, the point is (11/14,−23/14,8/7)

19. n1=h1,0,0i,n2=h2,−1,1i,n1·n2=2so

cos θ=n1·n2

kn1kkn

k=2

√

6=2/

√

6,θ = cos−1(2/√6) ≈35◦

20. n1=h1,2,−2i,n2=h6,−3,2i,n1·n2=−4so

cos θ=(−n1)·n2

k−n

kkn

k=4

(3)(7) =4/21,θ = cos−1(4/21) ≈79◦

(Note: −n1is used instead of n1to get a value of θin the range [0,π/2])

21. h4,−2,7iis normal to the desired plane and (0,0,0) is a point on it; 4x−2y+7z=0

22. v =h3,2,−1iis parallel to the line and n=h1,−2,1iis normal to the given plane so

v×n=h0,−4,−8iis normal to the desired plane. Let t= 0 in the line to get (−2,4,3) which is

also a point on the desired plane, use this point and (for convenience) the normal h0,1,2ito ﬁnd

that y+2z= 10.

471 Chapter 13

23. Find two points P1and P2on the line of intersection of the given planes and then ﬁnd an equation

of the plane that contains P1,P2, and the given point P0(−1,4,2). Let (x0,y

)beonthe

line of intersection of the given planes; then 4x0−y0+z0−2 = 0 and 2x0+y0−2z0−3=0,

eliminate y0by addition of the equations to get 6x0−z0−5=0;ifx

0= 0 then z0=−5, if x0=1

then z0= 1. Substitution of these values of x0and z0into either of the equations of the planes

gives the corresponding values y0=−7 and y0=3soP

(0,−7,−5) and P2(1,3,1) are on the

line of intersection of the planes. −→

P 0P 1×−→

P 0P 2=h4 ,−13,21iis normal to the desired plane whose

equation is 4x−13y+21z=−14.

24. h1,2,−1iis parallel to the line and hence normal to the plane x+2y−z=10

25. n1=h2,1,1iand n2=h1,2,1iare normals to the given planes, n1×n2=h−1,−1,3iso h1,1,−3i

is normal to the desired plane whose equation is x+y−3z=6.

26. n =h4,−1,3iis normal to the given plane, −→

P 1P 2=h3 ,−1 ,−1 iis parallel to the line,

n×−→

P 1P 2=h4 ,13,−1iis normal to the desired plane whose equation is 4x+13y−z=1.

27. n1=h2,−1,1iand n2=h1,1,−2iare normals to the given planes,

n1×n2=h1,5,3iis normal to the desired plane whose equation is x+5y+3z=−6.

28. Let t= 0 and t= 1 to get the points P1(−1,0,−4) and P2(0,1,−2) that lie on the line. Denote the

given point by P0, then −→

P 0P 1×−→

P 0P 2=h7 ,−1 ,−3 iis normal to the desired plane whose equation

is 7x−y−3z=5.

29. The plane is the perpendicular bisector of the line segment that joins P1(2,−1,1) and P2(3,1,5).

The midpoint of the line segment is (5/2,0,3) and −→

P 1P 2=h1 ,2 ,4 iis normal to the plane so an

equation is x+2y+4z=29/2.

30. n1=h2,−1,1iand n2=h0,1,1iare normals to the given planes, n1×n2=h−2,−2,2iso

n=h1,1,−1iis parallel to the line of intersection of the planes. v=h3,1,2iis parallel to the

given line, v×n=h−3,5,2iso h3,−5,−2iis normal to the desired plane. Let t= 0 to ﬁnd the

point (0,1,0) that lies on the given line and hence on the desired plane. An equation of the plane

is 3x−5y−2z=−5.

31. The line is parallel to the line of intersection of the planes if it is parallel to both planes. Normals

to the given planes are n1=h1,−4,2iand n2=h2,3,−1iso n1×n2=h−2,5,11iis parallel to

the line of intersection of the planes and hence parallel to the desired line whose equations are

x=5−2t,y=5t,z=−2+11t.

32. Denote the points by A,B,C, and D, respectively. The points lie in the same plane if −→

AB ×−→

and −→

AB ×−→

AD are parallel (method 1). −→

AB ×−→

AC =h0,−10,5i,−→

AB ×−→

AD =h0,16,−8i, these

vectors are parallel because h0,−10,5i=(−10/16)h0,16,−8i. The points lie in the same plane

if Dlies in the plane determined by A, B, C (method 2), and since −→

AB ×−→

AC =h0,−10,5i,an

equation of the plane is −2y+z+1=0,2y−z= 1 which is satisﬁed by the coordinates of D.

33. v =h0,1,1iis parallel to the line.

(a) For any t, 6·0+4t−4t=0,so(0,t,t) is in the plane.

(b) n =h5,−3,3iis normal to the plane, v·n= 0 so the line is parallel to the plane. (0,0,0) is

on the line, (0,0,1/3) is on the plane. The line is below the plane because (0,0,0) is below

(0,0,1/3).

Exercise Set 13.6 472

is on the plane. The line is above the plane because (0,0,0) is above (0,0,−3/2).

34. The intercepts correspond to the points A(a, 0,0),B(0,b,0), and C(0,0,c). −→

AB ×−→

AC =hbc, ac, abi

is normal to the plane so bcx +acy +abz =abc or x/a +y/b +z/c =1.

35. v1=h1,2,−1iand v2=h−1,−2,1iare parallel, respectively, to the given lines and to each

other so the lines are parallel. Let t= 0 to ﬁnd the points P1(−2,3,4) and P2(3,4,0) that lie,

respectively, on the given lines. v1×−→

P 1P 2=h−7,−1,−9iso h7,1,9iis normal to the desired plane

whose equation is 7x+y+9z= 25.

36. The system 4t1−1=12t

2−13, t1+3=6t

2+1,1=3t

2+ 2 has the solution (Exercise 26,

Section 13.5) t1=−4, t2=−1/3so(−17,−1,1) is the point of intersection. v1=h4,1,0iand

v2=h12,6,3iare (respectively) parallel to the lines, v1×v2=h3,−12,12iso h1,−4,4iis normal

to the desired plane whose equation is x−4y+4z=−9.

37. n1=h−2,3,7iand n2=h1,2,−3iare normals to the planes, n1×n2=h−23,1,−7iis parallel

to the line of intersection. Let z= 0 in both equations and solve for xand yto get x=−11/7,

y=−12/7so(−11/7,−12/7,0) is on the line, a parametrization of which is

x=−11/7−23t,y=−12/7+t,z=−7t.

38. Similar to Exercise 37 with n1=h3,−5,2i,n2=h0,0,1i,n1×n2=h−5,−3,0i.z=0so

3x−5y= 0, let x= 0 then y= 0 and (0,0,0) is on the line, a parametrization of which is

x=−5t,y=−3t,z=0.

39. D=|2(1) −2(−2) + (3) −4|/√4+4+1=5/3

40. D=|3(0) + 6(1) −2(5) −5|/√9+36+4=9/7

41. (0,0,0) is on the ﬁrst plane so D=|6(0) −3(0) −3(0) −5|/√36+9+9=5/

√

54.

42. (0,0,1) is on the ﬁrst plane so D=|(0)+(0)+(1)+1|/

√

1+1+1=2/

√

43. (1,3,5) and (4,6,7) are on L1and L2, respectively. v1=h7,1,−3iand v2=h−1,0,2iare,

respectively, parallel to L1and L2,v1×v2=h2,−11,1iso the plane 2x−11y+z+51=0

contains L2and is parallel to L1,D=|2(1) −11(3) + (5) + 51|/√4+121+1=25/

√

126.

44. (3,4,1) and (0,3,0) are on L1and L2, respectively. v1=h−1,4,2iand v2=h1,0,2iare parallel to

L1and L2,v1×v2=h8,4,−4i=4h2,1,−1iso 2x+y−z−3 = 0 contains L2and is parallel to

L1,D=|2(3) + (4) −(1) −3|/√4+1+1=√

45. The distance between (2,1,−3) and the plane is |2−3(1)+2(−3)−4|/√1+9+4=11/

√

14 which

is the radius of the sphere; an equation is (x−2)2+(y−1)2+(z+3)

2= 121/14.

46. The vector 2i+j−kis normal to the plane and hence parallel to the line so parametric equations

of the line are x=3+2t,y=1+t,z=−t. Substitution into the equation of the plane yields

2(3+2t)+(1+t)−(−t)=0,t=−7/6; the point of intersection is (2/3,−1/6,7/6).

47. v =h1,2,−1iis parallel to the line, n=h2,−2,−2iis normal to the plane, v·n=0sovis

parallel to the plane because vand nare perpendicular. (−1,3,0) is on the line so

D=|2(−1) −2(3) −2(0) + 3|/√4+4+4=5/

√

473 Chapter 13

48. (a)

P(x0, y0)P(x, y)

r – r0

(b) n ·(r−r0)=a(x−x

)+b(y−y

)=0

(x0,y

) that satisﬁes ax+by +d=0,soax0+bx0+d=0. If(x, y) also satisﬁes ax+by+d=0

then, subtracting, a(x−x0)+b(y−y

) = 0, which is the equation of a line with n=ha, bi

as normal.

(d) Let Q(x1,y

) be a point on the line, and position the normal n=ha, bi, with length √a2+b2,

so that its initial point is at Q. The distance is the orthogonal projection of

−→

QP0=hx0−x1,y

0−y

ionto n. Then

D=kprojn

−→

QP 0k=



−→

QP0·n

knk2n



=|ax0+by0+d|

√a2+b2.

49. D=|2(−3) + (5) −1|/√4+1=2/

√

50. (a) If hx0,y

ilies on the second plane, so that ax0+by0+cz0+d2= 0, then by Theorem

13.6.2, the distance between the planes is D=|ax0+by0+cz0+d1|

√a2+b2+c2=|−d

2+d

√

2+b

2+c

(b) The distance between the planes −2x+y+z= 0 and −2x+y+z+5

3=0is

D=|0−5/3|

√

4+1+1 =5

√

EXERCISE SET 13.7

1. (a) elliptic paraboloid, a=2,b=3

(b) hyperbolic paraboloid, a=1,b =5

(d) circular cone, a=b=1

(e) elliptic paraboloid, a=2,b=1

(f) hyperboloid of two sheets, a=b=c=1

2. (a) ellipsoid, a=√2/2,b=1/2,e =√3/3

(b) hyperbolic paraboloid, a=b=1

(d) hyperboloid of two sheets, a=1,b=2,c=1

(e) elliptic paraboloid, a=√2,b=√2/2

(f) elliptic cone, a=2,b=√3

Exercise Set 13.7 474

3. (a) −z=x2+y2, circular paraboloid

opening down the negative z-axis

(b) z=x2+y2, circular paraboloid, no change

(d) z=x2+y2, circular paraboloid, no change

(e) x=y2+z2, circular paraboloid

opening along the positive x-axis

(f) y=x2+z2, circular paraboloid

opening along the positive y-axis

4. (a) x2+y2−z2= 1, no change

(b) x2+y2−z2= 1, no change

(d) x2+y2−z2= 1, no change

475 Chapter 13

(e) −x2+y2+z2= 1, hyperboloid

of one sheet with x-axis as axis

(f) x2−y2+z2= 1, hyperboloid

of one sheet with y-axis as axis

5. (a) hyperboloid of one sheet, axis is y-axis

(b) hyperboloid of two sheets separated by yz-plane

(d) elliptic cone with x-axis as axis

(e) hyperbolic paraboloid straddling x- and z-axes

(f) paraboloid opening along the negative y-axis

6. (a) same (b) same (c) same

(d) same (e) y=x2

a2−z2

c2(f) y=x2

a2+z2

7. (a) x=0:y

25 +z2

4=1;y=0:x

9+z

4=1;

z=0:x

9+y

25 =1

+= 1

(b) x=0:z=4y

;y=0:z=x

;

z=0:x=y=0

z = x2

x2 + 4y2 = 0

z = 4y2

(0, 0, 0)

Exercise Set 13.7 476

16 −z2

4=1;y=0:x

9−z

4=1;

z=0:x

9+y

16 =1

–= 1

+= 1

8. (a) x=0:y=z=0;y=0:x=9z

;z=0:x=y

x = 9z2

x = y2

(b) x=0:−y

2+4z

2=4;y=0:x

2+z

2=1;

z=0:4x

2−y

2=4 z

y = 0

x = 0

z = 0

2;y=0:z=±x;z=0:x=y=0 z

x = 0

y = 0

z = 0

9. (a) 4x2+z2= 3; ellipse (b) y2+z2= 3; circle (c) y2+z2= 20; circle

(d) 9x2−y2= 20; hyperbola (e) z=9x

2+ 16; parabola (f) 9x2+4y

2= 4; ellipse

10. (a) y2−4z2= 27; hyperbola (b) 9x2+4z

2= 25; ellipse (c) 9z2−x2= 4; hyperbola

(d) x2+4y

2= 9; ellipse (e) z=1−4y

; parabola (f) x2−4y2= 4; hyperbola

477 Chapter 13

11.

(0, 2, 0)

(0, 0, 3)

(1, 0, 0)

Ellipsoid

z12.

(0, 3, 0)

(0, 0, 2)

(6, 0, 0)

Ellipsoid

z13.

(0, 3, 0)

(2, 0, 0)

Hyperboloid

of one sheet

14.

(0, 3, 0)

(3, 0, 0)

Hyperboloid

of one sheet

z15.

Elliptic cone

z16.

Elliptic cone

17.

(0, 0, 2) (0, 0, –2)

Hyperboloid

of two sheets

z18. z

Hyperboloid

of two sheets

19.

Hyperbolic paraboloid

20. z

xHyperbolic

paraboloid

21.

Elliptic paraboloid

z22.

Circular paraboloid

Exercise Set 13.7 478

23.

Circular cone

z24.

Elliptic paraboloid

z25.

(0, 0, 2)

(0, 2, 0)

Hyperboloid

of one sheet

26.

(3, 0, 0)

Hyperboloid

of two sheets

(-3, 0, 0)

z27.

Hyperbolic

paraboloid

z28.

(0, 0, 2)

(2, 0, 0)

Hyperboloid

of one sheet

29. z

30.

(0, 1, 0)

(0, 0, 1)

(1, 0, 0)

31.

(1, 0, 0)

(0, 1, 0)

32.

(0, 0, 1)

479 Chapter 13

33.

Circular paraboloid

(–2, 3, –9)

34.

Hyperboloid

of one sheet

(0, 0, 2)

35.

Ellipsoid

(1, –1, –2)

36.

Hyperboloid

of one sheet

(-1, 1, 2)

37. (a) x2

9+y2

4=1 (b) 6,4(c) (±√5,0,√2)

(d) The focal axis is parallel to the x-axis.

38. (a) y2

4+z2

2=1 (b) 4,2√2(c) (3,±√2,0)

(d) The focal axis is parallel to the y-axis.

39. (a) y2

4−x2

4=1 (b) (0,±2,4) (c) (0,±2√2,4)

(d) The focal axis is parallel to the y-axis.

40. (a) x2

4−y2

4=1 (b) (±2,0,−4) (c) (±2√2,0,−4)

(e) The focal axis is parallel to the x-axis.

41. (a) z+4=y

2(b) (2,0,−4) (c) (2,0,−15/4)

(d) The focal axis is parallel to the z-axis.

42. (a) z−4=−x

2(b) (0,2,4) (c) (0,2,15/4)

(d) The focal axis is parallel to the z-axis.

Exercise Set 13.7 480

43. x2+y2=4−x

2−y

2+y

2= 2; circle of radius √2

in the plane z= 2, centered at (0,0,1)

x2 + y2 = 2

(z = 2)

44. y2+z=4−2(y2+z),y

2+z=4/3;

parabolas in the planes x=±2/√3

which open in direction of the negative y-axis

z = – y2, x =

45. y=4(x

2+z

)46. y2=4(x

2+z

)

47. |z−(−1)|=px2+y2+(z−1)2,z2+2z+1=x

2+y

2+z

2−2z+1,z=(x

2+y

)/4; circular

paraboloid

48. |z+1|=2

2+y

2+(z−1)2,z2+2z+1=4x

2+y

2+z

2−2z+1

,

2+4y

2+3z

2−10z+3=0, x

4/3+y

4/3+(z−5/3)2

16/9= 1; ellipsoid, center at (0,0,5/3).

49. If z=0, x

2+y

2=1;ify= 0 then x2

a2+z2

c2= 1; since c<athe major axis has length 2a, the

minor axis length 2c.

50. x2

a2+y2

a2+z2

b2= 1, where a= 6378.1370,b = 6356.5231.

51. Each slice perpendicular to the z-axis for |z|<cis an ellipse whose equation is

a2+y2

b2=c2−z2

c2,or x

/c2)(c2−z2)+y2

(b2/c2)(c2−z2)= 1, the area of which is

πa

cpc2−z2b

cpc2−z2=πab

c2c2−z2so V=2Zc

πab

c2c2−z2dz =4

3πabc.

481 Chapter 13

EXERCISE SET 13.8

1. (a) (8,π/6,−4) (b) 5√2,3π/4,6(c) (2,π/2,0) (d) (8,5π/3,6)

2. (a) (2,7π/4,1) (b) (1,π/2,1) (c) (4√2,3π/4,−7) (d) (2√2,7π/4,−2)

3. (a) 2√3,2,3(b) −4√2,4√2,−2(c) (5,0,4) (d) (−7,0,−9)

4. (a) 3,−3√3,7(b) (0,1,0) (c) (0,3,5) (d) (0,4,−1)

5. (a) 2√2,π/3,3π/4(b) (2,7π/4,π/4) (c) (6, π/2, π/3) (d) (10,5π/6, π/2)

6. (a) 8√2,π/4,π/6

(b) 2√2,5π/3,3π/4(c) (2,0,π/2) (d) (4,π/6,π/6)

7. (a) (5√6/4,5√2/4,5√2/2) (b) (7,0,0)

8. (a) −√2/4,√6/4,−√2/2(b) 3√2/4,−3√2/4,−3√3/2

9. (a) 2√3,π/6,π/6

(b) √2,π/4,3π/4

10. (a) 4√2,5π/6,π/4

(b) 2√2,0,3π/4

11. (a) 5√3/2,π/4,−5/2

(b) (0,7π/6,−1)

12. (a) (0,π/2,−5/2) (b) (3√2,0,−3√2)

15.

(3, 0, 0)

x2 + y2 = 9

16.

y = x, x ≥ 0

z17.

z = x2 + y2

Exercise Set 13.8 482

18.

z = x

z19.

(0, 4, 0)

x2 + (y – 2)2 = 4

20.

x = 2

(2, 0, 0)

21.

(1, 0, 0)

x2 + y2 + z2 = 1

22. z

x2 – y2 = z

23.

(3, 0, 0)

x2 + y2 + z2 = 9

24.

y = √3x

25. z

z = √x2 + y2

26.

(0, 0, 2)

z = 2

483 Chapter 13

27. (0, 0, 2)

x2 + y2 + (z – 2)2 = 4

28.

(1, 0, 0)

x2 + y2 = 1

29.

(1, 0, 0)

(x – 1)2 + y2 = 1

30.

(1, 0, 0)

(x – 1)2 + y2 + z2 = 1

31. (a) z=3 (b) ρcos φ=3,ρ= 3 sec φ

32. (a) rsin θ=2,r = 2 csc θ(b) ρsin φsin θ=2,ρ= 2 csc φcsc θ

33. (a) z=3r

2(b) ρcos φ=3ρ

2sin2φ, ρ =1

3csc φcot φ

34. (a) z=√3r(b) ρcos φ=√3ρsin φ, tan φ=1

√3,φ=π

35. (a) r=2 (b) ρsin φ=2,ρ= 2 csc φ

36. (a) r2−6rsin θ=0,r= 6 sin θ(b) ρsin φ= 6 sin θ, ρ = 6 sin θcsc φ

37. (a) r2+z2=9 (b) ρ=3

38. (a) z2=r2cos2θ−r2sin2θ=r2(cos2θ−sin2θ), z2=r2cos 2θ

(b) Use the result in part (a) with r=ρsin φ,z=ρcos φto get ρ2cos2φ=ρ2sin2φcos 2θ,

cot2φ= cos 2θ

39. (a) 2rcos θ+3rsin θ+4z=1

(b) 2ρsin φcos θ+3ρsin φsin θ+4ρcos φ=1

40. (a) r2−z2=1

(b) Use the result of part (a) with r=ρsin φ,z=ρcos φto get ρ2sin2φ−ρ2cos2φ=1,

2cos 2φ=−1

Exercise Set 13.8 484

41. (a) r2cos2θ=16−z

(b) x2=16−z

2+y

2+z

2=16+y

,ρ

2=16+ρ

2sin2φsin2θ,ρ21−sin2φsin2θ=16

42. (a) r2+z2=2z(b) ρ2=2ρcos φ,ρ= 2 cos φ

43. all points on or above the paraboloid z=x2+y2, that are also on or below the plane z=4

44. a right circular cylindrical solid of height 3 and radius 1 whose axis is the line x=0,y =1

45. all points on or between concentric spheres of radii 1 and 3

46. all points on or above the cone φ=π/6, that are also on or below the sphere ρ=2

47. θ=π/6, φ=π/6, spherical (4000,π/6, π/6), rectangular 1000√3,1000,2000√3

48. (a) y=rsin θ=asin θbut az =asin θso y=az, which is a plane that contains the curve of

intersection of z= sin θand the circular cylinder r=a. From Exercise 60, Section 12.4, the

curve of intersection of a plane and a circular cylinder is an ellipse.

(b) z = sin

49. (a) (10,π/2,1) (b) (0,10,1) (c) (√101,π/2,tan−110)

50. 20

030

51. Using spherical coordinates: for point A,θA= 360◦−60◦= 300◦,φ

A=90

◦−40◦=50

◦

; for point

B,θB= 360◦−40◦= 320◦,φ

B=90

◦−20◦=70

◦

. Unit vectors directed from the origin to the

points Aand B, respectively, are

uA= sin 50◦cos 300◦i+ sin 50◦sin 300◦j+ cos 50◦k,

uB= sin 70◦cos 320◦i+ sin 70◦sin 320◦j+ cos 70◦k

The angle αbetween uAand uBis α= cos−1(uA·uB)≈0.459486 so the shortest distance is

6370α≈2,927 km.

485 Chapter 13

CHAPTER 13 SUPPLEMENTARY EXERCISES

2. (c) F =−i−j

(d) kh1,−2,2ik =3,sokr−h1,−2,2ik =3,or(x−1)2+(y+2)

2+(z−2)2=9

3. (b) x= cos 120◦=−1/2,y =±sin 120◦=±√3/2

(d) true: ku×vk=kukkvk|sin(θ)|=1

4. (d) x+2y−z=0

5. (b) (y, x, z),(x, z, y),(z, y, x)

(d) the two half-lines z=±x,x≥0inthexz-plane

6. (x+3)

2+(y−5)2+(z+4)

2=r

(a) r2=4

2=16 (b) r2=5

2=25 (c) r2=3

2=9

7. (a) −→

AB =−i+2j+2k,−→

AC =i+j−k,−→

AB ×−→

AC=−4i+j−3k, area = 1

2k−→

AB ×−→

AC k=√26/2

(b) area = 1

2hk−→

AB k=3

2h=1

2√26, h=√26/3

8. The sphere x2+(y−1)2+(z+3)

2= 16 has center Q(0,1,−3) and radius 4, and

k−→

PQ k=√12+4

2=√

17, so minimum distance is √17 −4, maximum distance is √17+4.

9. (a) a ·b=0,4c+3=0,c=−3/4

(b) Use a·b=kakkbkcos θto get 4c+3=pc

2+ 1(5) cos(π/4), 4c+3=5

2+1/

√

Square both sides and rearrange to get 7c2+48c−7=0,(7c−1)(c+7)=0soc=−7

(invalid) or c=1/7.

get c=48 ±25√3/11.

(d) a must be a scalar multiple of b,soci+j=k(4i+3j),k =1/3,c=4/3.

10. −→

OS =−→

OP +−→

PS=3i+4j+−→

QR =3i+4j+(4i+j)=7i+5j

11. (a) the plane through the origin which is perpendicular to r0

(b) the plane through the tip of r0which is perpendicular to r0

12. The normals to the planes are given by ha1,b

iand ha2,b

i, so the condition is

a1a2+b1b2+c1c2=0.

13. Since −→

AC ·(−→

AB ×−→

AD)=

−→

AC ·(−→

AB ×−→

CD)+−→

AC ·(−→

AB ×−→

AC)=0+0=0, the volume of the

parallelopiped determined by −→

AB, −→

AC, and −→

AD is zero, thus A, B, C, and Dare coplanar (lie in

the same plane). Since −→

AB ×−→

CD6=0, the lines are not parallel. Hence they must intersect.

14. The points Plie on the plane determined by A, B and C.

15. (a) false, for example i·j=0 (b) false, for example i×i=0

impossible.

Chapter 13 Supplementary Exercises 486

16. (a) Replace uwith a×b,vwith c, and wwith din the ﬁrst formula of Exercise 39.

(b) From the second formula of Exercise 39,

(a×b)×c+(b×c)×a+(c×a)×b

=(c·a)b−(c·b)a+(a·b)c−(a·c)b+(b·c)a−(b·a)c=0

17. ku−vk2=(u−v)·(u−v)=kuk

2+kvk

2−2kukkvkcos θ= 2(1 −cos θ)=4sin

(θ/2), so

ku−vk= 2 sin(θ/2)

18. −→

AB=i−2j−2k,−→

AC=−2i−j−2k,−→

AD=i+2j−3k

(a) From Theorem 13.4.6 and formula (9) of Section 13.4, 

1−2−2

−2−1−2

12−3



= 29, so V= 29.

(b) The plane containing A, B, and Chas normal −→

AB ×−→

AC=2i+6j−5k, so the equation of

the plane is 2(x−1)+6(y+1)−5(z−2)=0,2x+6y−5z=−14. From Theorem 13.6.2,

D=|2(2) + 6(1) −5(−1)|

√65 =15

√65.

19. (a) h2,1,−1i×h1,2,1i=h3,−3,3i, so the line is parallel to i−j+k. By inspection, (0,2,−1)

lies on both planes, so the line has an equation r=2j−k+t(i−j+k), that is,

x=t, y =2−t, z =−1+t.

(b) cos θ=h2,1,−1i·h1,2,1i

kh2,1,−1ikkh1,2,1ik =1/2, so θ=π/3

20. Let α=50

◦

,β =70

◦

, then γ= cos−1p1−cos2α−cos2β≈47◦.

21. 5hcos 60◦,cos 120◦,cos 135◦i=h5/2,−5/2,−5√2/2i

22. (a) Let kbe the length of an edge and introduce a coordinate system as shown in the ﬁgure,

then d=hk, k, ki,u=hk, k, 0i, cos θ=d·u

kdkkuk=2k

k

√

k

√2

=2/

√

so θ= cos−1(2/√6) ≈35◦

(b) v =h−k, 0,ki,cos θ=d·v

kdkkvk=0soθ=π/2 radians.

23. (a) (x−3)2+4(y+1)

2−(z−2)2= 9, hyperboloid of one sheet

(b) (x+3)

2+(y−2)2+(z+6)

2= 49, sphere

2−z

2= 0, circular cone

24. (a) perpendicular, since h2,1,2i·h−1,−2,2i=0

(b) L1:hx, y, zi=h1+2t, −3

2+t, −1+2ti;L

:hx, y, zi=h4−t, 3−2t, −4+2ti

487 Chapter 13

,−3

2+t

1=3−2t

,−1+2t

1=−4+2t

, solution

t1=1

2,t

2=2,x =2,y =−1,z =0

25. (a) r2=z;ρ2sin2φ=ρcos φ, ρ = cot φcsc φ

(b) r2(cos2θ−sin2θ)−z2=0,z2=r

2cos 2θ;

ρ2sin2φcos2θ−ρ2sin2φsin2θ−ρ2cos2φ=0,cos 2θ= cot2φ

26. (a) z=r2cos2θ−r2sin2θ=x2−y2(b) (ρsin φcos θ)(ρcos φ)=1,xz =1

27. (a) z

(2, 0, 0)

(0, 0, 2)

(0, 2, 0)

(b) z

p/6

(c)

p/6

z(0, 0, 2)

28. (a) z

(b)

z = 2

Chapter 13 Supplementary Exercises 488

29. (a) z

r = 1

r = 2 (b)

z = 3

z = 2

u = p/6 u = p/3

(d) z

u = p/6 u = p/3

r = 1

z = 2

z = 3

r = 2

30. (a)

(b)

1(c)

31. (a) At x=cthe trace of the surface is the circle y2+z2=[f(c)]2, so the surface is given by

y2+z2=[f(x)]2

(b) y2+z2=e2x(c) y2+z2=4−3

,soletf(x)=r4−3

32. (a) Permute xand yin Exercise 31a: x2+z2=[f(y)]2

(b) Permute xand zin Exercise 31a: x2+y2=[f(z)]2

33. z

34. −→

PQ=h1,−1,6i, and W=F·

−→

PQ =13lb·ft

489 Chapter 13

35. F =F1+F2=2i−j+3k,−→

PQ=i+4j−3k,W =F·−→

PQ=−11 N·m

36. F1= 250 cos 38◦i+ 250 sin 38◦j,F= 1000i,F2=F−F1= (1000 −250 cos 38◦)i−250 sin 38◦j;

kF2k= 1000r17

16 −1

2cos 38◦≈817.62 N·m, θ= tan−1250 sin 38◦

250 cos 38◦−1000 ≈−11◦

37. (a) F =−6i+3j−6k

(b) −→

OA=h5,0,2i, so the vector moment is −→

OA ×F=−6i+18j+15k

490

CHAPTER 14

Vector-Valued Functions

EXERCISE SET 14.1

1. (−∞,+∞); r(π)=−i−3πj2.[−1/3,+∞); r(1) = h2,1i

3. [2,+∞); r(3) = −i−ln 3j+k4.[−1,1); r(0) = h2,0,0i

5. r = 3 cos ti+(t+ sin t)j6.r=(t

2+1)i+e

−2t

7. r =2ti+ 2 sin 3tj+ 5 cos 3tk8.r=tsin ti+lntj+ cos2tk

9. x=3t

,y=−2, z=0 10. x= sin2t,y=1−cos 2t,z=0

11. x=2t−1, y=−3√t,z= sin 3t12. x=te−t,y=0,z=−5t

13. the line in 2-space through the point (2,0) and parallel to the vector −3i−4j

14. the circle of radius 3 in the xy-plane, with center at the origin

15. the line in 3-space through the point (0,−3,1) and parallel to the vector 2i+3k

16. the circle of radius 2 in the plane x= 3, with center at (3,0,0)

17. an ellipse in the plane z=−1, center at (0,0,−1), major axis of length 6 parallel to x-axis, minor

axis of length 4 parallel to y-axis

18. a parabola in the plane x=−2, vertex at (−2,0,−1), opening upward

19. (a) The line is parallel to the vector −2i+3j; the slope is −3/2.

(b) y= 0 in the xz-plane so 1 −2t=0,t=1/2thusx=2+1/2=5/2 and z= 3(1/2)=3/2;

the coordinates are (5/2,0,3/2).

20. (a) x=3+2t=0,t=−3/2soy=5(−3/2) = −15/2

(b) x=t,y=1+2t,z=−3tso 3(t)−(1+2t)−(−3t)=2,t=3/4; the point of intersection is

(3/4,5/2,−9/4).

21. (a)

(1, 0)

(0, 1) (b)

(1, -1)

(1, 1)

491 Chapter 14

22. (a)

(0, 0, 1)

(1, 1, 0)

(b)

(1, 1, 0)

(1, 1, 1)

23. r =(1−t)(3i+4j),0≤t≤124. r =(1−t)4k+t(2i+3j),0≤t≤1

25. x=2

26. y=2x+10

-5

27. (x−1)2+(y−3)2=1

28. x2/4+y

/25=1

29. x2−y2=1,x≥1

30. y=2x

2+4,x≥0

Exercise Set 14.1 492

31.

(0, 2,

/2)

(2, 0, 0)

z32.

(9, 0, 0)

(0, 4,

/2)

33.

z34.

35. x=t, y =t, z =2t

36. x=t, y =−t, z =√2√1−t2

zy + x = 0

z = √2 – x2 – y2

37. r =ti+t2j±1

3p81 −9t2−t4k

38. r =ti+tj+(1−2t)ks

x + y + z = 1

y = x

493 Chapter 14

39. x2+y2=(tsin t)2+(tcos t)2=t2(sin2t+ cos2t)=t

2=z

40. x−y+z+1=t−(1+t)/t +(1−t

)/t +1=[t

2−(1+t)+(1−t

)+t]/t =0

41. x= sin t,y= 2 cos t,z=√3 sin tso x2+y2+z2= sin2t+ 4 cos2t+ 3 sin2t= 4 and z=√3x;it

is the curve of intersection of the sphere x2+y2+z2= 4 and the plane z=√3x, which is a circle

with center at (0,0,0) and radius 2.

42. x= 3 cos t,y= 3 sin t,z= 3 sin tso x2+y2= 9 cos2t+ 9 sin2t= 9 and z=y; it is the curve

of intersection of the circular cylinder x2+y2= 9 and the plane z=y, which is an ellipse with

major axis of length 6√2 and minor axis of length 6.

43. The helix makes one turn as tvaries from 0 to 2πso z=c(2π)=3,c=3/(2π).

44. 0.2t= 10, t= 50; the helix has made one revolution when t=2πso when t= 50 it has made

50/(2π)=25/π ≈7.96 revolutions.

45. x2+y2=t2cos2t+t2sin2t=t2,px2+y2=t=z; a conical helix.

46. The curve wraps around an elliptic cylinder with axis along the z-axis; an elliptical helix.

47. (a) III, since the curve is a subset of the plane y=−x

(b) IV, since only xis periodic in t, and y, z increase without bound

(d) I, since the projection onto the yz-plane is a circle and the curve increases without bound in

the x-direction

49. (a) Let x= 3 cos tand y= 3 sin t, then z= 9 cos2t.(b) z

50. The plane is parallel to a line on the surface of

the cone and does not go through the vertex so

the curve of intersection is a parabola. Eliminate

zto get y+2=px

2+y

,(y+2)

2=x

2+y

y=x

/4−1; let x=t, then y=t2/4−1

and z=t2/4+1.

Exercise Set 14.2 494

EXERCISE SET 14.2

-2 2

-2

y2.

r'(p/4)

r''(p)

r(2p) – r(3p/2)

3. r0(t)=5i+(1−2t)j4.r

(t) = sin tj5.r

(t)=−1

2i+sec2tj+2e2tk

6. r0(t)= 1

1+t

2i+ (cos t−tsin t)j−1

2√tk

7. r0(t)=h1,2ti,

(2) = h1,4i,

r(2) = h2,4i

〈1, 4 〉

8. r0(t)=3t

i+2tj,

(1)=3i+2j

r(1) = i+j

1 2 3 4

9. r0(t) = sec ttan ti+ sec2tj,

r0(0) = j

r(0) = i

1.5

-1

10. r0(t)=2costi−3 sin tj,

r0π

6=√3i−3

rπ

6=i+3√3

-2 -1 1 2

-3

-2

-1

495 Chapter 14

11. r0(t)=2costi−2 sin tk,

r0(π/2) = −2k,

r(π/2) = 2i+j

r′(

) = -2k

(2, 1, 0)

12. r0(t)=−sin ti+ cos tj+k,

r0(π/4) = −1

√2i+1

√2j+k,

r(π/4) = 1

√2i+1

√2j+π

√

−

i + j + k

r′(

) =

√2

( , ,

13. 9i+6j 14. h√2/2,√2/2i15. h1/3,0i

16. j 17. 2i−3j+4k 18. h3,1/2,sin 2i

19. (a) continuous, lim

t→0r(t)=0=r(0) (b) not continuous, lim

t→0r(t) does not exist

20. (a) not continuous, lim

t→0r(t) does not exist. (b) continuous, lim

t→0r(t)=5i−j+k=r(0)

21. (a) lim

t→0(r(t)−r0(t)) = i−j+k

(b) lim

t→0(r(t)×r0(t)) = lim

t→0(−cos ti−sin tj+k)=−i+k

t→0(r(t)·r0(t))=0

22. r(t)·(r0(t)×r00 (t)) = 

12t3t

026t



=2t

, so lim

t→1r(t)·(r0(t)×r00 (t))=2

23. r0(t)=2ti−1

tj,r

(1)=2i−j,r(1) = i+2j;x=1+2t,y=2−t,z=0

24. r0(t)=2e

i+ 6 sin 3tj,r0(0)=2i,r(0) = i−2j;x=1+2t,y=−2, z=0

25. r0(t)=−2πsin πti+2πcos πtj+3k,r

(1/3) = −√3πi+πj+3k,

r(1/3) = i+√3j+k;x=1−√

3πt,y=√3+πt,z=1+3t

26. r0(t)=1

ti−e

−t

j+3t

k,r

(2) = 1

2i−e−2j+12k,

r(2)=ln2i+e

−2

j+8k;x=ln2+1

t,y=e

−2−e

−2

t,z=8+12t

Exercise Set 14.2 496

27. r0(t)=2i+3

√

3t+4j,t=0atP

0so r0(0)=2i+3

r(0) = −i+2j;r=(−i+2j)+t2i+3



28. r0(t)=−4 sin ti−3j,t=π/3atP

0so r0(π/3) = −2√3i−3j,

r(π/3)=2i−πj;r=(2i−πj)+t(−2

√

3i−3j)

29. r0(t)=2ti+1

(t+1)

2j−2tk,t=−2atP

0so r0(−2) = −4i+j+4k,

r(−2)=4i+j;r=(4i+j)+t(−4i+j+4k)

30. r0(t) = cos ti+ sinh tj+1

1+t

2k,t=0atP

0so r0(0) = i+k,r(0) = j;r=ti+j+tk

31. 3ti+2t

j+C 32. (sin t)i−(cos t)j+C 33. (−tcos t+ sin t)i+tj+C

34. h(t−1)et,t(ln t−1)i+C 35. (t3/3)i−t2j+ln|t|k+C 36. h−e−t,e

t,t

3i+C

37. 1

3sin 3t, 1

3cos 3tπ/3

=h0,−2/3i38. 1

3t3i+1

4t4j1

3i+1

39. Z2

0pt2+t4dt =Z2

t(1+t2)1/2dt =1

31+t

3/22

=(5

√

5−1)/3

40. −2

5(3 −t)5/2,2

5(3+t)5/2,t

3

−3

=h72√6/5,72√6/5,6i

41. 2

3t3/2i+2t

1/2

9

=52

3i+4j 42. 1

2(e2−1)i+(1−e

−1

)j+1

43. y(t)=Zy

(t)dt =1

3t3i+t2j+C,y(0) = C=i+j,y(t)=(

3+1)i+(t

2+1)j

44. y(t)=Zy

(t)dt = (sin t)i−(cos t)j+C,

y(0) = −j+C=i−jso C=iand y(t)=(1+sint)i−(cos t)j.

45. y0(t)=Zy

00 (t)dt =ti+etj+C1,y0(0) = j+C1=jso C1=0and y0(t)=ti+e

y(t)=Zy

(t)dt =1

2t2i+etj+C2,y(0) = j+C2=2iso C2=2i−jand

y(t)=1

2+2

i+(e

t−1)j

497 Chapter 14

46. y0(t)=Zy

00 (t)dt =4t

i−t

j+C

(0) = C1=0,y0(t)=4t

i−t

y(t)=Zy

(t)dt =t4i−1

3t3j+C2,y(0) = C2=2i−4j,y(t)=(t

4+2)i−(1

3+4)j

47.

1.50

1.5 48. 6

49. r0(t)=−4 sin ti+ 3 cos tj,kr(t)k=p16 cos2t+ 9 sin2t, kr0(t)k=p16 sin2t+ 9 cos2t,

krkkr0k=p144 + 49 sin2tcos2t, θ = cos−1−7 sin tcos t

p144 + 49 sin2tcos2t

rand r0are parallel when θ=0,sot=0,π/2,π,3π/2,2π; the angle between rand r0is greatest at

θmax =0.28 (t≈2.23,5.53) and θmin =−0.28, (t≈0.77,3.95), so rand r0are never perpendicular.

50. r0(t)=2ti+3t

j,kr(t)k=t

√

1+t

,kr

(t)k=t

√

4+9t

,cos θ=2+3t

√

1+t

√

4+9t

cos θ= 1 when t=0,±2

3/4

√

3 and rand r0are parallel; cos θ>0, so they are never perpendicular.

0.3

51. (a) 2t−t2−3t=−2, t2+t−2=0,(t+ 2)(t−1)=0sot=−2,1. The points of intersection

are (−2,4,6) and (1,1,−3).

(b) r0=i+2tj−3k;r

(−2) = i−4j−3k,r0(1) = i+2j−3k, and n=2i−j+kis normal to

the plane. Let θbe the acute angle, then

for t=−2: cos θ=|n·r0|/(knkkr

k)=3/

√

156, θ≈76◦;

for t= 1: cos θ=|n·r0|/(knkkr

k)=3/

√

84, θ≈71◦.

Exercise Set 14.2 498

52. r0=−2e−2ti−sin tj+3 cos tk,t= 0 at the point (1,1,0) so r0(0) = −2i+3kand hence the tangent

line is x=1−2t,y=1,z=3t. But x= 0 in the yz-plane so 1 −2t=0,t=1/2. The point of

intersection is (0,1,3/2).

53. r1(1) = r2(2) = i+j+3kso the graphs intersect at P; r0

1(t)=2ti+j+9t

kand

2(t)=i+1

tj−kso r0

1(1)=2i+j+9kand r0

2(2) = i+j−kare tangent to the graphs at P,

thus cos θ=r0

1(1) ·r0

2(2)

kr0

1(1)kkr

(2)k=−6

√86√3,θ= cos−1(6/√258) ≈68◦.

54. r1(0) = r2(−1)=2i+j+3kso the graphs intersect at P; r0

1(t)=−2e

−t

i−(sin t)j+2tkand

2(t)=−i+2tj+3t

kso r0

1(0) = −2iand r0

2(−1) = −i−2j+3kare tangent to the graphs at P,

thus cos θ=r0

1(0) ·r0

2(−1)

kr0

1(0)kkr

(−1)k=1

√14,θ≈74◦.

55. (a) r0

1=2i+6tj+3t

k,r

2=4t

k,r

1·r

2=t

dt(r1·r2)=7t

6=r

1·r

2+r

1·r

(b) r1×r2=3t

i−2t

j,d

dt(r1×r2)=18t

i−10t4j=r1×r0

2+r0

1×r2

56. (a) r0

1=−sin ti+cos tj+k,r0

2=k,r1·r2= cos t+t2;d

dt(r1·r2)=−sin t+2t=r1·r0

2+r0

1·r2

(b) r1×r2=tsin ti+t(1 −cos t)j−sin tk,

dt(r1×r2) = (sin t+tcos t)i+(1+tsin t−cos t)j−cos tk=r1×r0

2+r0

1×r2

57. d

dt[r(t)×r0(t)] = r(t)×r00 (t)+r

(t)×r

(t)=r(t)×r

00 (t)+0=r(t)×r

00 (t)

58. d

dt[u·(v×w)] = u·

dt[v×w]+du

·[v×w]=u·v×dw

dt +dv

dt ×w+du

·[v×w]

=u·v×dw

dt +u·dv

dt ×w+du

·[v×w]

59. In Exercise 60, write each scalar triple product as a determinant.

60. Let c=c1i+c2j,r(t)=x(t)i+y(t)j,r

(t)=x

(t)i+y

(t)j,r

(t)=x

(t)i+y

(t)jand use

properties of derivatives.

61. Let r1(t)=x

(t)i+y

(t)j+z

(t)kand r2(t)=x

(t)i+y

(t)j+z

(t)k, in both (8) and (9); show

that the left and right members of the equalities are the same.

62. (a) Zkr(t)dt =Zk(x(t)i+y(t)j+z(t)k)dt

=kZx(t)dt i+kZy(t)dt j+kZz(t)dt k=kZr(t)dt

(b) Similar to Part (a) (c) Use Part (a) on Part (b) with k=−1

499 Chapter 14

EXERCISE SET 14.3

1. (a) The tangent vector reverses direction at the four cusps.

(b) r0(t)=−3 cos2tsin ti+ 3 sin2tcos tj=0when t=0,π/2,π,3π/2,2π.

2. r0(t) = cos ti+ 2 sin tcos tj=0when t=π/2,3π/2. The tangent vector reverses direction at (1,1)

and (−1,1).

3. r0(t)=3t

i+(6t−2)j+2tk; smooth

4. r0(t)=−2tsin(t2)i+2tcos(t2)j−e−tk; smooth

5. r0(t)=(1−t)e

−t

i+(2t−2)j−πsin(πt)k; not smooth, r0(1) = 0

6. r0(t)=πcos(πt)i+(2−1/t)j+(2t−1)k; not smooth, r0(1/2) = 0

7. (dx/dt)2+(dy/dt)2+(dz/dt)2=(−3 cos2tsin t)2+ (3 sin2tcos t)2+0

2= 9 sin2tcos2t,

L=Zπ/2

3 sin tcos tdt=3/2

8. (dx/dt)2+(dy/dt)2+(dz/dt)2=(−3 sin t)2+ (3 cos t)2+16=25,L=Zπ

5dt =5π

9. r0(t)=he

,−e

−t

,√

2i,kr

(t)k=e

t+e

−t

,L=Z1

t+e

−t

)dt =e−e−1

10. (dx/dt)2+(dy/dt)2+(dz/dt)2=1/4+(1−t)/4+(1+t)/4=3/4, L=Z1

−1

(√3/2)dt =√3

11. r0(t)=3t

i+j+√

6tk,kr

(t)k=3t

2+1,L=Z3

(3t2+1)dt =28

12. r0(t)=3i−2j+k,kr

(t)k=√

14, L=Z4

√14 dt =√14

13. r0(t)=−3 sin ti+ 3 cos tj+k,kr0(t)k=√10, L=Z2π

√10 dt =2π

√

14. r0(t)=2ti+tcos tj+tsin tk,kr0(t)k=√5t, L =Zπ

√5tdt=π

2√5/2

15. (dr/dt)(dt/dτ)=(i+2tj)(4) = 4i+8tj=4i+ 8(4τ+1)j;

r(τ)=(4τ+1)i+(4τ+1)

j,r

(τ)=4i+ 2(4)(4τ+1)

16. (dr/dt)(dt/dτ)=h−3 sin t, 3 cos ti(π)=h−3πsin πτ,3πcos πτi;

r(τ)=h3 cos πτ, 3 sin πτi,r0(τ)=h−3πsin πτ, 3πcos πτi

17. (dr/dt)(dt/dτ)=(e

i−4e

−t

j)(2τ)=2τeτ2i−8τe−τ2j;

r(τ)=e

2i+4e

−τ2j,r

(τ)=2τeτ2i−4(2)τe−τ2j

Exercise Set 14.3 500

18. (dr/dt)(dt/dτ)=9

1/2

j+k

(−1/τ 2)=−9

2τ

5/2j−1

2k;

r(τ)=i+3τ−3/2

j+1

τk,r

(τ)=−9

−5/2

j−1

19. (a) kr0(t)k=√2,s=Zt

√2dt =√2t;r=s

√2i+s

√2j,x=s

√2,y =s

√2

(b) Similar to Part (a), x=y=z=s

√3

20. (a) x=−s

√2,y =−s

√2(b) x=−s

√3,y =−s

√3,z =−s

√3

21. (a) r(t)=h1,3,4iwhen t=0,

so s=Zt

√1+4+4du =3t, x =1+s/3,y =3−2s/3,z =4+2s/3

(b) rs=25

=h28/3,−41/3,62/3i

22. (a) r(t)=h−5,0,1iwhen t=0,sos=Zt

√

9+4+1du =√14t,

x=−5+3s/√14,y =2s/√14,z =5+s/√14

(b) r(s)s=10

=h−5+30/

√

14,20/√14,5+10/

√

14i

23. x= 3 + cos t,y= 2 + sin t,(dx/dt)2+(dy/dt)2=1,

s=Zt

du =tso t=s,x= 3 + cos s,y= 2 + sin sfor 0 ≤s≤2π.

24. x= cos3t,y= sin3t,(dx/dt)2+(dy/dt)2= 9 sin2tcos2t,

s=Zt

3 sin ucos udu=3

2sin2tso sin t=(2s/3)1/2, cos t=(1−2s/3)1/2,

x=(1−2s/3)3/2,y=(2s/3)3/2for 0 ≤s≤3/2

25. x=t3/3, y=t2/2, (dx/dt)2+(dy/dt)2=t2(t2+ 1),

s=Zt

u(u2+1)

1/2

du =1

3[(t2+1)

3/2−1] so t= [(3s+1)

2/3−1]1/2,

x=1

3[(3s+1)

2/3−1]3/2,y=1

2[(3s+1)

2/3−1] for s≥0

26. x=(1+t)

,y=(1+t)

,(dx/dt)2+(dy/dt)2=(1+t)

[4+9(1+t)2],

s=Zt

(1+u)[4 + 9(1 + u)2]1/2du =1

27([4 + 9(1 + t)2]3/2−13√13) so

1+t=1

3[(27s+13

√

13)2/3−4]1/2,x=1

9[(27s+13

√

13)2/3−4],

y=1

27[(27s+13

√

13)2/3−4]3/2for 0 ≤s≤(80√10 −13√13)/27

501 Chapter 14

27. x=etcos t,y=etsin t,(dx/dt)2+(dy/dt)2=2e

,s=Zt

√

udu =√2(et−1) so

t= ln(s/√2 + 1), x=(s/√2 + 1) cos[ln(s/√2 + 1)], y=(s/√2 + 1) sin[ln(s/√2 + 1)]

for 0 ≤s≤√2(eπ/2−1)

28. x= sin(et), y= cos(et), z=√3et,

(dx/dt)2+(dy/dt)2+(dz/dt)2=4e

,s=Zt

udu =2(e

t−1) so

et=1+s/2; x= sin(1 + s/2), y= cos(1 + s/2), z=√3(1 + s/2) for s≥0

29. dx/dt =−asin t,dy/dt =acos t,dz/dt =c,

L=Zt0

0pa2sin2t+a2cos2t+c2dt =Zt0

0pa2+c2dt =t0pa2+c2

30. x=acos t,y=asin t,z=ct,(dx/dt)2+(dy/dt)2+(dz/dt)2=a2+c2=w2,

s=Zt

wdu=wt so t=s/w;x=acos(s/w), y=asin(s/w), z=cs/w for s≥0

31. x=at −asin t,y=a−acos t,(dx/dt)2+(dy/dt)2=4a

2sin2(t/2),

s=Zt

2asin(u/2)du =4a[1 −cos(t/2)] so cos(t/2)=1−s/(4a), t= 2 cos−1[1 −s/(4a)],

cos t= 2 cos2(t/2) −1 = 2[1 −s/(4a)]2−1,

sin t= 2 sin(t/2) cos(t/2) = 2(1 −[1 −s/(4a)]2)1/2(2[1 −s/(4a)]2−1),

x=2acos−1[1 −s/(4a)] −2a(1 −[1 −s/(4a)]2)1/2(2[1 −s/(4a)]2−1),

y=s(8a−s)

8afor 0 ≤s≤8a

32. dx

dt = cos θdr

dt −rsin θdθ

dt ,dy

dt = sin θdr

dt +rcos θdθ

dt ,

dx

dt 

+dy

dt 

+dz

dt 

=dr

dt 

+r2dθ

dt 

+dz

dt 

33. (a) (dr/dt)2+r2(dθ/dt)2+(dz/dt)2=9e

,L=Zln 2

3e2tdt =3

2e2tln 2

=9/2

(b) (dr/dt)2+r2(dθ/dt)2+(dz/dt)2=5t

2+t

4=t

(5+t2),

L=Z2

t(5+t2)1/2dt =9−2

√

34. dx

dt = sin φcos θdρ

dt +ρcos φcos θdφ

dt −ρsin φsin θdθ

dt ,

dt = sin φsin θdρ

dt +ρcos φsin θdφ

dt +ρsin φcos θdθ

dt ,dz

dt = cos φdρ

dt −ρsin φdφ

dt ,

dx

dt 

+dy

dt 

+dz

dt 

=dρ

dt 

+ρ2sin2φdθ

dt 

+ρ2dφ

dt 

Exercise Set 14.4 502

35. (a) (dρ/dt)2+ρ2sin2φ(dθ/dt)2+ρ2(dφ/dt)2=3e

−2t

,L=Z2

√

−t

dt =√3(1 −e−2)

(b) (dρ/dt)2+ρ2sin2φ(dθ/dt)2+ρ2(dφ/dt)2=5,L=Z5

√

5dt =4

√

36. (a) d

dtr(t)=i+2tjis never zero, but d

dτ r3(τ)= d

dτ (τ3i+τ6j)=3τ

i+6τ5

jis zero at τ=0.

(b) dr

dτ =dr

dτ , and since t=τ3,dt

dτ = 0 when τ=0.

37. (a) g(τ)=πτ (b) g(τ)=π(1 −τ)38. t=1−τ

39. Represent the helix by x=acos t,y=asin t,z=ct with a=6.25 and c=10/π, so that the

radius of the helix is the distance from the axis of the cylinder to the center of the copper cable,

and the helix makes one turn in a distance of 20 in. (t=2π). From Exercise 29 the length of the

helix is 2πp6.252+ (10/π)2≈44 in.

40. r(t) = cos ti+ sin tj+t3/2k,r0(t)=−sin ti+ cos tj+3

2t1/2k

(a) kr0(t)k=qsin2t+ cos2t+9t/4=1

√

4+9t

(b) ds

dt =1

2√4+9t(c) Z2

2√4+9tdt=2

27(11√22 −4)

41. r0(t)=(1/t)i+2j+2tk

(a) kr0(t)k=p1/t2+4+4t

2=p(2t+1/t)2=2t+1/t

(b) ds

dt =2t+1/t (c) Z3

(2t+1/t)dt =8+ln3

42. If r(t)=x(t)i+y(t)j+z(t)kis smooth, then kr0(t)kis continuous and nonzero. Thus the angle

between r0(t) and i, given by cos−1(x0(t)/kr0(t)k), is a continuous function of t. Similarly, the

angles between r0(t) and the vectors jand kare continuous functions of t.

43. Let r(t)=x(t)i+y(t)jand use the chain rule.

EXERCISE SET 14.4

1. (a)

y(b)

503 Chapter 14

3. r0(t)=2ti+j,kr

(t)k=√

2+1,T(t)=(4t

2+1)

−1/2

(2ti+j),

T0(t)=(4t

2+1)

−1/2

(2i)−4t(4t2+1)

−3/2

(2ti+j);

T(1) = 2

√5i+1

√5j,T0(1) = 2

5√5(i−2j), N(1) = 1

√5i−2

√5j.

4. r0(t)=ti+t

j,T(t)=(t

2+t

)

−1/2

(ti+t

j),

T0(t)=(t

2+t

)

−1/2

(i+2tj)−(t+2t

)(t2+t4)−3/2(ti+t2j);

T(1) = 1

√2i+1

√2j,T0(1) = 1

2√2(−i+j), N(1) = −1

√2i+1

√2j

5. r0(t)=−5 sin ti+ 5 cos tj,kr0(t)k=5,T(t)=−sin ti+ cos tj,T0(t)=−cos ti−sin tj;

T(π/3) = −√3

2i+1

2j,T0(π/3) = −1

2i−√3

2j,N(π/3) = −1

2i−√3

6. r0(t)=1

ti+j,kr

(t)k=√

1+t

t,T(t)=(1+t

)

−1/2

(i+tj),

T0(t)=(1+t

)

−1/2

(j)−t(1+t2)−3/2(i+tj); T(e)= 1

√

1+e

2i+e

√

1+e

2j,

(e)= 1

(1+e2)3/2(−ei+j), N(e)=−e

√

1+e

2i+1

√

1+e

7. r0(t)=−4 sin ti+ 4 cos tj+k,T(t)= 1

√

17(−4 sin ti+ 4 cos tj+k),

T0(t)= 1

√

17(−4 cos ti−4 sin tj), T(π/2) = −4

√17i+1

√17k

T0(π/2) = −4

√17j,N(π/2) = −j

8. r0(t)=i+tj+t

k,T(t)=(1+t

2+t

)

−1/2

(i+tj+t

k),

T0(t)=(1+t

2+t

)

−1/2

(j+2tk)−(t+2t

)(1 + t2+t4)−3/2(i+tj+t2k),

T(0) = i,T0(0) = j=N(0)

9. r0(t)=e

[(cos t−sin t)i+ (cos t+ sin t)j+k], T(t)= 1

√

[(cos t−sin t)i+ (cos t+ sin t)j+k],

T0(t)= 1

√

[(−sin t−cos t)i+(−sin t+ cos t)j],

T(0) = 1

√3i+1

√3j+1

√3k,T0(0) = 1

√3(−i+j), N(0) = −1

√2i+1

√2j

Exercise Set 14.4 504

10. r0(t) = sinh ti+ cosh tj+k,kr0(t)k=psinh2t+ cosh2t+1=√

2 cosh t,

T(t)= 1

√

(tanh ti+j+ sech tk), T0(t)= 1

√

(sech2ti−sech ttanh tk), at t=ln2,

tanh(ln 2) = 3

5and sech(ln 2) = 4

5so T(ln 2) = 3

5√2i+1

√2j+4

5√2k,

T0(ln 2) = 4

25√2(4i−3k), N(ln 2) = 4

5i−3

11. From the remark, the line is parametrized by normalizing v, but T(t0)=v/kvk,sor=r(t

)+tv

becomes r=r(t0)+sT(t

12. r0(t)it=1 =h1,2tiit=1 =h1,2i, and T(1) = h1

√5,2

√5i, so the tangent line can be parametrized as

r=h1,1i+s1

√5,2

√5,sox=1+ s

√

,y =1+ 2s

√

13. r0(t) = cos ti−sin tj+tk,r0(0) = i,r(0) = j,T(0) = i, so the tangent line has the parametrization

x=s, y =1.

14. r(1) = i+j+√8k,r0(t)=i+j−t

√

9−t

2k,r

(1) = i+j−1

√8k,kr0(1)k=√17

√8, so the tangent

line has parametrizations r=i+j+√8k+ti+j−1

√8k=i+j+√8k+s√8

√17 i+j−1

√8k.

15. T =3

5cos ti−3

5sin tj+4

5k,N=−sin ti−cos tj,B=T×N=4

5cos ti−4

5sin tj−3

16. T0(t)= 1

√

[(cos t+ sin t)i+(−sin t+ cos t)j],N=1

√2[(−sin t+ cos t)i−(cos t+ sin t)j],

B=T×N=−k

17. r0(t)=tsin ti+tcos tj,kr0k=t, T= sin ti+ cos tj,N= cos ti−sin tj,B=T×N=−k

18. T =(−asin ti+acos tj+bk)/√a2+b2,N=−cos ti−sin tj,

B=T×N=(bsin ti−bcos tj+ak)/√a2+b2

19. r(π/4) = √2

2i+√2

2j+k,T=−sin ti+ cos tj=√2

2(−i+j),N=−(cos ti+ sin tj)=−√

2(i+j),

B=k; the rectifying, osculating, and normal planes are given (respectively) by x+y=√2,

z=1,−x+y=0.

20. r(0) = i+j,T=1

√3(i+j+k),N=1

√2(−j+k),B=1

√6(2i−j−k); the rectifying, osculating,

and normal planes are given (respectively) by −y+z=−1,2x−y−z=1,x+y+z=2.

21. (a) By formulae (1) and (11), N(t)=B(t)×T(t)= r

(t)×r

00 (t)

kr0(t)×r00 (t)k×r0(t)

kr0(t)k.

(b) Since r0is perpendicular to r0×r00 it follows from Lagrange’s Identity (Exercise 32 of Section

13.4) that k(r0(t)×r00 (t)) ×r0(t)k=kr0(t)×r00 (t)kkr0(t)k, and the result follows.

(t)k

00 (t)−(r0(t)·r00 (t))r0(t)=

u(t), so N(t)=u(t)/ku(t)k

505 Chapter 14

22. (a) r0(t)=2ti+j,r

(1)=2i+j,r

00 (t)=2i,u=2i−4j,N=1

√

i−2

√

(b) r0(t)=−4 sin ti+ 4 cos tj+k,r0(π

2)=−4i+k,r

00 (t)=−4 cos ti−4 sin tj,

r00 (π

2)=−4j,u= 17(−4j),N=−j

23. r0(t) = cos ti−sin tj+k,r00 (t)=−sin ti−cos tj,u=−2(sin ti+cos tj),kuk=2,N=−sin ti−cos tj

24. r0(t)=i+2tj+3t

k,r

00 (t)=2j+6tk,u(t)=−(4t+18t

)i+(2−18t4)j+(6t+12t

)k,

N=1

√

81t8+ 117t6+54t

4+13t

2+1−(4t+18t

)i+(2−18t4)j+(6t+12t



EXERCISE SET 14.5

1. κ≈1

0.5=2 2. κ≈1

4/3=3

3. r0(t)=2ti+3t

j,r

00 (t)=2i+6tj,κ=kr

(t)×r

00 (t)k/kr0(t)k3=6

t(4+9t

)

3/2

4. r0(t)=−4 sin ti+cos tj,r00 (t)=−4 cos ti−sin tj,κ=kr0(t)×r00 (t)k/kr0(t)k3=4

(16 sin2t+ cos2t)3/2

5. r0(t)=3e

i−e

−t

j,r

00 (t)=9e

i+e

−t

j,κ=kr

(t)×r

00 (t)k/kr0(t)k3=12e2t

(9e6t+e−2t)3/2

6. r0(t)=−3t

i+(1−2t)j,r

00 (t)=−6ti−2j,κ=kr

(t)×r

00 (t)k/kr0(t)k3=6pt2(t−1)2

(9t4+4t

2−4t+1)

3/2

7. r0(t)=−4 sin ti+ 4 cos tj+k,r00 (t)=−4 cos ti−4 sin tj,

κ=kr0(t)×r00 (t)k/kr0(t)k3=4/17

8. r0(t)=i+tj+t

k,r

00 (t)=j+2tk,κ=kr

(t)×r

00 (t)k/kr0(t)k3=√t4+4t

2+1

4+t

2+1)

3/2

9. r0(t) = sinh ti+ cosh tj+k,r00 (t) = cosh ti+ sinh tj,κ=kr0(t)×r00 (t)k/kr0(t)k3=1

2 cosh2t

10. r0(t)=j+2tk,r

00 (t)=2k,κ=kr

(t)×r

00 (t)k/kr0(t)k3=2

(4t2+1)

3/2

11. r0(t)=−3 sin ti+ 4 cos tj+k,r00 (t)=−3 cos ti−4 sin tj,

r0(π/2) = −3i+k,r00 (π/2) = −4j;κ=k4i+12kk/k−3i+kk

3=2/5,ρ =5/2

12. r0(t)=e

i−e

−t

j+k,r

00 (t)=e

i+e

−t

(0) = i−j+k,r00 (0) = i+j;κ=k−i+j+2kk/ki−j+kk

3=√

2/3,ρ =3/

√

13. r0(t)=e

(cos t−sin t)i+et(cos t+ sin t)j+etk,

r00 (t)=−2e

tsin ti+2e

tcos tj+etk,r0(0) = i+j+k,

r00 (0)=2j+k;κ=k−i−j+2kk/ki+j+kk

3=√

2/3,ρ =3

√

2/2

Exercise Set 14.5 506

14. r0(t) = cos ti−sin tj+tk,r00 (t)=−sin ti−cos tj+k,

r0(0) = i,r00 (0) = −j+k;κ=k−j−kk/kik

3=√

2,ρ=√2/2

15. r0(s)=1

2cos 1+s

2i−1

2sin 1+s

2j+√

2k,kr

(s)k=1,so

ds =−1

4sin 1+s

2i−1

4cos 1+s

2j,κ=



ds 



16. r0(s)=−

r3−2s

3si+r2s

3j,kr

(s)k=1,so

ds =1

√9−6si+1

√6sj,κ=



ds 



=r1

9−6s+1

6s=s3

2s(9 −6s)

17. (a) r0=x0i+y0j,r00 =x00 i+y00 j,kr0×r00 k=|x0y00 −x00 y0|,κ=|x

00 −y0x00 |

(x02+y02)3/2

(b) r =xi+y(x)j,r0(x)=i+(dy/dx)j,r00 (x)=(d

y/dx2)j,so

κ(x)=k(d

y/dx2)kk/ki+(dy/dx)jk3=|d2y/dx2|/[1+(dy/dx)2]3/2.

18. dy

dx = tan φ, (1 + tan2φ)3/2= (sec2φ)3/2=|sec φ|3,κ(x)= |y

00 |

|sec φ|3=|y00 cos3φ|

19. κ(x)= |sin x|

(1 + cos2x)3/2,κ(π/2)=1 20. κ(x)= 2|x|

(1+x4)3/2,κ(0)=0

21. κ(x)= 2|x|

4+1)

3/2,κ(1)=1/

√

222. κ(x)= e

−x

(1+e−2x)3/2,κ(1) = e−1

(1+e−2)3/2

23. κ(x)= 2 sec2x|tan x|

(1 + sec4x)3/2,κ(π/4)=4/(5√5)

24. By implicit diﬀerentiation, dy/dx =4x/y,d2y/dx2=36/y3so κ=36/|y|3

(1+16x

/y2)3/2;

if (x, y)=(2,5) then κ=36/125

(1+64/25)3/2=36

89√89

25. x0(t)=2t,y

(t)=3t

00 (t)=2,y

00 (t)=6t,

(1/2)=1,y

(1/2)=3/4, x00 (1/2)=2,y

00 (1/2)=3;κ=96/125

26. x0(t)=−4 sin t,y0(t) = cos t,x00 (t)=−4 cos t,y00 (t)=−sin t,

x0(π/2) = −4, y0(π/2)=0,x

00 (π/2)=0,y

00 (π/2) = −1; κ=1/16

27. x0(t)=3e

(t)=−e

−t

00 (t)=9e

00 (t)=e

−t

(0)=3,y

(0) = −1, x00 (0)=9,y

00 (0)=1;κ=6/(5√10)

28. x0(t)=−3t

(t)=1−2t,x

00 (t)=−6t,y

00 (t)=−2,

x0(1) = −3, y0(1) = −1, x00 (1) = −6, y00 (1) = −2; κ=0

507 Chapter 14

29. x0(t)=1,y

(t)=−1/t2,x

00 (t)=0,y

00 (t)=2/t3

x0(1)=1,y

(1) = −1,x

00 (1)=0,y

00 (1)=2;κ=1/

√

30. x0(t) = 4 cos 2t, y0(t)=3cost, x00 (t)=−8 sin 2t, y00 (t)=−3 sin t,

x0(π/2) = −4,y

(π/2)=0,x

00 (π/2)=0,y

00 (π/2) = −3,κ=12/4

3/2=3/2

31. (a) κ(x)= |cos x|

(1 + sin2x)3/2,

ρ(x)=(1 + sin2x)3/2

|cos x|

ρ(0) = ρ(π)=1.

(0) = ( ) = 1

(b) κ(t)= 2

(4 sin2t+ cos2t)3/2,

ρ(t)=1

(4 sin2t+ cos2t)3/2,

ρ(0)=1/2, ρ(π/2)=4

(

)

= 4

(0) =

32. x0(t)=−e

−t

(cos t+ sin t),

y0(t)=e

−t

(cos t−sin t),

x00 (t)=2e

−tsin t,

y00 (t)=−2e

−tcos t;

using the formula of Exercise 17(a),

κ=1

√2et.

3-3

33. (a) At x= 0 the curvature of I has a large value, yet the value of II there is zero, so II is not the

curvature of I; hence I is the curvature of II.

(b) I has points of inﬂection where the curvature is zero, but II is not zero there, and hence is

not the curvature of I; so I is the curvature of II.

34. (a) II takes the value zero at x= 0, yet the curvature of I is large there; hence I is the curvature

of II.

(b) I has constant zero curvature; II has constant, positive curvature; hence I is the curvature

of II.

35. (a)

1(b) 1

Exercise Set 14.5 508

36. (a) 4

-4

-1 1

(b) 4

-4

-1 1

37. (a) κ=|12x2−4|

(1+(4x

3−4x)

)

3/2(b)

f(x)

-2 2

3−4x=0atx=0,±1,f00 (x)=12x

2−4, so extrema at x=0,±1, and ρ=1/4

for x= 0 and ρ=1/8 when x=±1.

38. (a)

-30 30

-30

y(c) κ(t)= t

2+2

2+1)

3/2(d) lim

t→+∞κ(t)=0

39. r0(θ)=−rsin θ+ cos θdr

dθ i+rcos θ+ sin θdr

dθ j;

r00 (θ)=−rcos θ−2 sin θdr

dθ + cos θd2r

dθ2i+−rsin θ+ 2 cos θdr

dθ + sin θd2r

dθ2j;

κ=

r2+2dr

dθ 

−rd2r

dθ2

"r2+dr

dθ 

2#3/2.

40. Let r=abe the circle, so that dr/dθ = 0, and κ(θ)=1

r=1

41. κ(θ)= 3

√

2(1 + cos θ)1/2,κ(π/2) = 3

2√242. κ(θ)= 1

√

2θ,κ(1) = 1

√5e2

43. κ(θ)= 10 + 8 cos23θ

(1 + 8 cos2θ)3/2,κ(0) = 2

344. κ(θ)= θ

2+2

(θ

2+1)

3/2,κ(1) = 3

2√2

509 Chapter 14

45. The radius of curvature is zero when θ=π, so there is a cusp there.

46. dr

dθ =−sin θ,d2r

dθ2=−cos θ,κ(θ)= 3

3/2

√

1 + cos θ

47. Let y=t, then x=t2

4pand κ(t)= 1/|2p|

/(4p2)+1]

3/2;

t= 0 when (x, y)=(0,0) so κ(0)=1/|2p|,ρ=2|p|.

48. κ(x)= e

(1+e2x)3/2,κ0(x)=e

(1 −2e2x)

(1+e2x)5/2;κ0(x) = 0 when e2x=1/2, x=−(ln 2)/2. By the ﬁrst

derivative test, κ(−1

2ln 2) is maximum so the point is (−1

2ln 2,1/√2).

49. Let x= 3 cos t,y= 2 sin tfor 0 ≤t<2π,κ(t)= 6

(9 sin2t+ 4 cos2t)3/2so

ρ(t)=1

(9 sin2t+ 4 cos2t)3/2=1

6(5 sin2t+4)

3/2which, by inspection, is minimum when

t=0orπ. The radius of curvature is minimum at (3,0) and (−3,0).

50. κ(x)= 6x

(1+9x

)

3/2for x>0, κ0(x)= 6(1 −45x4)

(1+9x

)

5/2;κ

(x) = 0 when x=45

−1/4which, by the

ﬁrst derivative test, yields the maximum.

51. r0(t)=−sin ti+ cos tj−sin tk,r00 (t)=−cos ti−sin tj−cos tk,

kr0(t)×r00 (t)k=k−i+kk=√

2, kr0(t)k= (1 + sin2t)1/2;κ(t)=√

2/(1 + sin2t)3/2,

ρ(t)=(1+sin

2t)

3/2

√

2. The minimum value of ρis 1/√2; the maximum value is 2.

52. r0(t)=e

i−e

−t

j+√

2k,r

00 (t)=e

i+e

−t

κ(t)= √

2t+e

−2t+2,ρ(t)= 1

√

t+e

−t

)

2=2

√

2 cosh2t. The minimum value of ρis 2√2.

53. From Exercise 39: dr/dθ =aeaθ =ar,d2r/dθ2=a2eaθ =a2r;κ=1/[

p1+a

2r].

54. Use implicit diﬀerentiation on r2=a2cos 2θto get 2rdr

dθ =−2a2sin 2θ,rdr

dθ =−a2sin 2θ, and

again to get rd2r

dθ2+dr

dθ 

=−2a2cos 2θso rd2r

dθ2=−dr

dθ 

−2a2cos 2θ=−dr

dθ 

−2r2,thus



2+2dr

dθ 

−rd2r

dθ2

=3"r

2+dr

dθ 

2#,κ=3

[r2+(dr/dθ)2]1/2;dr

dθ =−a2sin 2θ

rso

r2+dr

dθ 

=r2+a4sin22θ

r2=r4+a4sin22θ

r2=a4cos22θ+a4sin22θ

r2=a4

r2, hence κ=3r

a2.

Exercise Set 14.5 510

55. (a) d2y/dx2=2,κ(φ)=|2 cos3φ|

(b) dy/dx = tan φ=1,φ=π/4,κ(π/4) = |2 cos3(π/4)|=1/

√

2,ρ=

√

(c)

-2 1

56. (a) 5

3,0,0,−5

2(b) clockwise (c) it is a point, namely the center of the circle

57. κ= 0 along y= 0; along y=x2,κ(x)=2/(1+4x

)

3/2

,κ(0) = 2. Along y=x3,

κ(x)=6|x|/(1+9x

)

3/2

,κ(0)=0.

58. (a)

2-2

y(b) For y=x2,κ(x)= 2

(1+4x

)

3/2

so κ(0) = 2; for y=x4,

κ(x)= 12x2

(1+16x

)

3/2so κ(0)=0.

κis not continuous at x=0.

59. κ=1/r along the circle; along y=ax2,κ(x)=2a/(1+4a

)

3/2

,κ(0)=2aso 2a=1/r,

a=1/(2r).

60. κ(x)= |y

00 |

(1+y02)3/2so the transition will be smooth if the values of yare equal, the values of y0

are equal, and the values of y00 are equal at x=0.Ify=e

, then y0=y00 =ex;ify=ax2+bx +c,

then y0=2ax +band y00 =2a. Equate y,y0, and y00 at x= 0 to get c=1,b= 1, and a=1/2.

61. The result follows from the deﬁnitions N=T0(s)

kT0(s)kand κ=kT0(s)k.

62. (a) B ·

ds = 0 because kB(s)k=1so dB

ds is perpendicular to B(s).

(b) B(s)·T(s)=0,B(s)·

ds +dB

·T(s)=0,butdT

ds =κN(s)soκB(s)·N(s)+dB

T(s)=0, dB

·T(s) = 0 because B(s)·N(s)=0;thus dB

ds is perpendicular to T(s).

ds is perpendicular to both B(s) and T(s) but so is N(s), thus dB

ds is parallel to N(s) and

hence a scalar multiple of N(s).

(d) If Clies in a plane, then T(s) and N(s) also lie in the plane; B(s)=T(s)×N(s)soB(s)is

always perpendicular to the plane and hence dB/ds =0,thusτ=0.

511 Chapter 14

63. dN

ds =B×dT

ds +dB

ds ×T=B×(κN)+(−τN)×T=κB×N−τN×T, but B×N=−Tand

N×T=−Bso dN

ds =−κT+τB

64. r00 (s)=dT/ds =κNso r000 (s)=κdN/ds +(dκ/ds)Nbut dN/ds =−κT+τBso

r000 (s)=−κ

T+(dκ/ds)N+κτB,r0(s)×r00 (s)=T×(κN)=κT×N=κB,

(s)×r

00 (s)] ·r000 (s)=−κ

B·T+κ(dκ/ds)B·N+κ2τB·B=κ2τ,

τ=[r

(s)×r

00 (s)] ·r000 (s)/κ2=[r

(s)×r

00 (s)] ·r000 (s)/kr00 (s)k2and

B=T×N=[r

(s)×r

00 (s)]/kr0(s)k

65. r =acos(s/w)i+asin(s/w)j+(cs/w)k,r0=−(a/w) sin(s/w)i+(a/w) cos(s/w)j+(c/w)k,

r00 =−(a/w2) cos(s/w)i−(a/w2) sin(s/w)j,r000 =(a/w3) sin(s/w)i−(a/w3) cos(s/w)j,

r0×r00 =(ac/w3) sin(s/w)i−(ac/w3) cos(s/w)j+(a

/w3)k,(r

0×r

00 )·r000 =a2c/w6,

kr00 (s)k=a/w2,soτ=c/w2and B=(c/w) sin(s/w)i−(c/w) cos(s/w)j+(a/w)k

66. (a) T0=dT

dt =dT

dt =(κN)s

0=κs0N,

N0=dN

dt =dN

dt =(−κT+τB)s

0=−κs0T+τs0B.

(b) kr0(t)k=s0so r0(t)=s

Tand r00 (t)=s

00 T+s0T0=s00 T+s0(κs0N)=s

00 T+κ(s0)2N.

00 T0+s000 T+κ(s0)2N0+[2κs0s00 +κ0(s0)2]N

=s00 (κs0N)+s

000 T+κ(s0)2(−κs0T+τs0B)+[2κs0s00 +κ0(s0)2]N

=[s

000 −κ2(s0)3]T+[3κs0s00 +κ0(s0)2]N+κτ(s0)3B.

(d) r0(t)×r00 (t)=s

00 T×T+κ(s0)3T×N=κ(s0)3B,[r

(t)×r

00 (t)] ·r000 (t)=κ

τ(s

)

6so

τ=[r0(t)×r00 (t)] ·r000 (t)

κ2(s0)6=[r0(t)×r00 (t)] ·r000 (t)

kr0(t)×r00 (t)k2

67. r0=2i+2tj+t

k,r

00 =2j+2tk,r

000 =2k,r

0×r

00 =2t

i−4tj+4k,kr

0×r

00 k=2(t

2+ 2),

τ=8/[2(t2+ 2)]2=2/(t

2+2)

68. r0=−asin ti+acos tj+bk,r00 =−acos ti−asin tj,r000 =asin ti−acos tj,

r0×r00 =ab sin ti−ab cos tj+a2k,kr0×r00 k=pa2(a2+b2),

τ=a2b/[a2(a2+b2)] = b/(a2+b2)

69. r0=eti−e−tj+√2k,r00 =eti+e−tj,r000 =eti−e−tj,r0×r00 =−√2e−ti+√2etj+2k,

0×r

00 k=√2(et+e−t), τ=(−2

√

2)/[2(et+e−t)2]=−

√

2/(e

t+e

−t

)

70. r0=(1−cos t)i+ sin tj+k,r00 = sin ti+ cos tj,r000 = cos ti−sin tj,

r0×r00 =−cos ti+ sin tj+ (cos t−1)k,

kr0×r00 k=qcos2t+ sin2t+ (cos t−1)2=q1+4sin

(t/2), τ=−1/[1 + 4 sin4(t/2)]

Exercise Set 14.6 512

EXERCISE SET 14.6

1. v(t)=−3 sin ti+ 3 cos tj

a(t)=−3 cos ti−3 sin tj

kv(t)k=p9 sin2t+ 9 cos2t=3

r(π/3)=(3/2)i+(3

√

3/2)j

v(π/3) = −(3√3/2)i+(3/2)j

a(π/3) = −(3/2)i−(3√3/2)j

23√3

()

23√3

a = − i − j

3√3

v = − i + j

2. v(t)=i+2tj

a(t)=2j

kv(t)k=√

1+4t

r(2)=2i+4j

v(2) = i+4j

a(2)=2j

(2, 4)

a = 2j

v = i + 4j

3. v(t)=e

i−e

−t

a(t)=e

i+e

−t

kv(t)k=√

2t+e

−2t

r(0) = i+j

v(0) = i−j

a(0) = i+j

(1, 1)

v = i − j

a = i + j

4. v(t)=4i−j

a(t)=0

kv(t)k=√

r(1)=6i

v(1)=4i−j

a(1) = 0

(6, 0)

v = 4i − j

a = 0

5. v =i+tj+t2k,a=j+2tk;att=1,v=i+j+k,kvk=√

3, a=j+2k

6. r =(1+3t)i+(2−4t)j+(7+t)k,v=3i−4j+k,

a=0;att=2,v=3i−4j+k,kvk=√

26, a=0

7. v =−2 sin ti+ 2 cos tj+k,a=−2 cos ti−2 sin tj;

at t=π/4, v=−√2i+√2j+k,kvk=√5, a=−√2i−√2j

8. v =et(cos t+ sin t)i+et(cos t−sin t)j+k,a=2e

tcos ti−2etsin tj;att=π/2,

v=eπ/2i−eπ/2j+k,kvk=(1+2e

π)

1/2

,a=−2e

π/2j

513 Chapter 14

9. (a) v =−aω sin ωti+bω cos ωtj,a=−aω2cos ωti−bω2sin ωtj=−ω2r

(b) From Part (a), kak=ω2krk

10. (a) v =16πcos πti−8πsin 2πtj,a=−16π2sin πti−16π2cos 2πtj;

at t=1,v=−16πi,kvk=16π,a=−16π2j

(b) x= 16 sin πt, y = 4 cos 2πt = 4 cos2πt −4 sin2πt =4−8 sin2πt, y =4−x

/32

11. v =(6/

√

t)i+(3/2)t1/2j,kvk=p36/t +9t/4,dkvk/dt =(−36/t2+9/4)/(2p36/t +9t/4)=0

if t= 4 which yields a minimum by the ﬁrst derivative test. The minimum speed is 3√2 when

r=24i+8j.

12. v =(1−2t)i−2tj,kvk=p(1 −2t)2+4t

2=√

2−4t+1,

dtkvk=8t−2

√8t2−4t+1 =0ift=1

4which yields a minimum by the ﬁrst derivative test. The

minimum speed is 1/√2 when the particle is at r=3

16i−1

16j.

13. (a)

803

(b) v = 3 cos 3ti+ 6 sin 3tj,kvk=p9 cos23t+ 36 sin23t=3

p1+3sin

23t; by inspection, maxi-

mum speed is 6 and minimum speed is 4

(d) d

dtkvk=9 sin 6t

2p1+3sin

23t= 0 when t=0,π/6,π/3,π/2,2π/3; the maximum speed is 6 which

occurs ﬁrst when sin3t=1,t=π/6.

14. (a) 8

(d) v =−6 sin 2ti+2 cos 2tj+4k,kvk=p36 sin22t+ 4 cos22t+16=2

p8 sin2t+ 5; by inspec-

tion the maximum speed is 2√13 when t=π, the minimum speed is 2√5 when t=π/2.

15. v(t)=−sin ti+ cos tj+C1,v(0) = j+C1=i,C1=i−j,v(t)=(1−sin t)i+ (cos t−1)j;

r(t)=(t+ cos t)i+ (sin t−t)j+C2,r(0) = i+C2=j,

C2=−i+jso r(t)=(t+ cos t−1)i+ (sin t−t+1)j

Exercise Set 14.6 514

16. v(t)=ti−e

−t

j+C

,v(0) = −j+C1=2i+j;C

1=2i+2jso

v(t)=(t+2)i+(2−e

−t

)j;r(t)=(t

/2+2t)i+(2t+e

−t

)j+C

r(0) = j+C2=i−j,C2=i−2jso r(t)=(t

/2+2t+1)i+(2t+e

−t−2)j

17. v(t)=−cos ti+ sin tj+etk+C1,v(0) = −i+k+C1=kso

C1=i,v(t)=(1−cos t)i+ sin tj+etk;r(t)=(t−sin t)i−cos tj+etk+C2,

r(0) = −j+k+C2=−i+kso C2=−i+j,r(t)=(t−sin t−1)i+(1−cos t)j+etk.

18. v(t)=−1

t+1j+1

−2t

k+C

,v(0) = −j+1

2k+C1=3i−jso

C1=3i−1

k,v(t)=3i−1

t+1j+1

−2t−1

2k;

r(t)=3ti−ln(t+1)j−1

−2t+1

k+C

r(0) = −1

4k+C2=2kso C2=9

4k,r(t)=3ti−ln(t+1)j+9

4−1

−2t−1

k.

19. If a=0then x00 (t)=y

00 (t)=z

00 (t)=0,sox(t)=x

t+x

,y(t)=y

t+y

,z(t)=z

t+z

, the

motion is along a straight line and has constant speed.

20. (a) If krkis constant then so is krk2, but then x2+y2=c2(2-space) or x2+y2+z2=c2

(3-space), so the motion is along a circle or a sphere of radius ccentered at the origin, and

the velocity vector is always perpendicular to the locating vector.

(b) If kvkis constant then by the Theorem, v(t)·a(t) = 0, so the velocity is always perpendicular

to the acceleration.

21. v =3t

i+2tj,a=6ti+2j;v=3i+2jand a=6i+2jwhen t=1so

cos θ=(v·a)/(kvkkak)=11/

√

130, θ≈15◦.

22. v =et(cos t−sin t)i+et(cos t+ sin t)j,a=−2etsin ti+2e

tcos tj,v·a=2e

,kvk=√

kak=2e

, cos θ=(v·a)/(kvkkak)=1/

√

2, θ=45

◦

23. (a) displacement = r1−r0=0.7i+2.7j−3.4k

(b) ∆r=r1−r0,sor

0=r

1−∆r=−0.7i−2.9j+4.8k.

24. (a)

-4 -2 2 4

-4

-2

y(b) distance = p50(1 −cos 2πt)

25. ∆r=r(3) −r(1)=8i+26/3j;v=2ti+t

j,s=Z3

p4+t

dt = (13√13 −5√5)/3.

26. ∆r=r(3π/2) −r(0)=3i−3j;v= 3 cos ti−3 sin tj,s=Z3π/2

3dt =9π/2.

515 Chapter 14

27. ∆r=r(ln 3) −r(0)=2i−2/3j+√

2ln3k;v=e

ti−e

−tj+√2k,s=Zln 3

(et+e−t)dt =8/3.

28. ∆r=r(π)−r(0) = 0;v=−2 sin 2ti+ 2 sin 2tj−sin 2tk,

kvk=3|sin 2t|,s=Zπ

3|sin 2t|dt =6Zπ/2

sin 2tdt=6.

29. In both cases, the equation of the path in rectangular coordinates is x2+y2= 4, the particles

move counterclockwise around this circle; v1=−6 sin 3ti+ 6 cos 3tjand

v2=−4tsin(t2)i+4tcos(t2)jso kv1k= 6 and kv2k=4t.

30. Let u=1−t

3in r2to get r2(u)=(3+2u)i+uj+(1−u)kso both particles move along the same

line; v1=2i+j−kand v2=−6t2i−3t2j+3t

kso kv1k=√6 and kv2k=3

√

31. (a) v =−e−ti+etj,a=e−ti+etj; when t=0,v=−i+j,a=i+j,kvk=√

2, v·a=0,

v×a=−2kso aT=0,a

N=√

(b) aTT=0,a

NN=a−a

TT=i+j (c) κ=1/

√

32. (a) v =−2tsin(t2)i+2tcos(t2)j,a=[−4t

2cos(t2)−2 sin(t2)]i+[−4t

2sin(t2) + 2 cos(t2)]j; when

t=√π/2, v=−pπ/2i+pπ/2j,a=(−π/√2−√2)i+(−π/√2+√

2)j,kvk=√π,

v·a=2

√

π,v×a=π

3/2

kso aT=2,a

N=π

(b) aTT=−√2(i−j),a

N=a−a

T=−(π/√2)(i+j)

33. (a) v =(3t

2−2)i+2tj,a=6ti+2j; when t=1,v=i+2j,a=6i+2j,kvk=√

5, v·a= 10,

v×a=−10kso aT=2

√

5, aN=2

√

(b) aTT=2√5

√5(i+2j)=2i+4j,a

N=a−a

T=4i−2j

√

34. (a) v =et(−sin t+cos t)i+et(cos t+sin t)j,a=−2etsin ti+2etcos tj; when t=π/4, v=√2eπ/4j,

a=−√2eπ/4i+√2eπ/4j,kvk=√2eπ/4,v·a=2e

π/2,v×a=2e

π/2kso aT=√2eπ/4,

aN=√2eπ/4

(b) aTT=√2eπ/4j,a

N=a−a

T=−

√

π/4i

√2eπ/4

35. (a) v =i+2tj+3t

k,a=2j+6tk; when t=1,v=i+2j+3k,a=2j+6k,kvk=√

14,

v·a= 22, v×a=6i−6j+2kso aT=22/

√

14, aN=√76/√14 = p38/7

(b) aTT=11

7i+22

7j+33

7k,a

N=a−a

T=−

7i−8

7j+9

7√14

36. (a) v =eti−2e−2tj+k,a=eti+4e

−2t

j; when t=0,v=i−2j+k,a=i+4j,kvk=√

v·a=−7, v×a=−4i+j+6kso aT=−7/√6, aN=p53/6

Exercise Set 14.6 516

(b) aTT=−7

6(i−2j+k),a

NN=a−a

TT=13

6i+19

3j+7

6√6

37. (a) v = 3 cos ti−2 sin tj−2 cos 2tk,a=−3 sin ti−2 cos tj+4 sin 2tk; when t=π/2, v=−2j+2k,

a=−3i,kvk=2

√

2, v·a=0,v×a=−6j−6kso aT=0,a

N=3

(b) aTT=0,a

NN=a=−3i

38. (a) v =3t

j−(16/t)k,a=6tj+ (16/t2)k; when t=1,v=3j−16k,a=6j+16k,kvk=√265,

v·a=−238, v×a= 144iso aT=−238/√265, aN= 144/√265

(b) aTT=−714

265j+3808

265 k,a

N=a−a

T=2304

265 j+432

265k

2653/2

39. kvk=4,v·a=−12, v×a=8kso aT=−3, aN=2,T=−j,N=(a−a

TT)/aN=i

40. kvk=√5, v·a=3,v×a=−6kso aT=3/

√

5, aN=6/

√

5, T=(1/

√

5)(i+2j),

N=(a−a

TT)/aN=(1/

√

5)(2i−j)

41. kvk=3,v·a=4,v×a=4i−3j−2kso aT=4/3, aN=√29/3, T=(1/3)(2i+2j+k),

N=(a−a

TT)/aN=(i−8j+14k)/(3√29)

42. kvk=5,v·a=−5, v×a=−4i−10j−3kso aT=−1, aN=√5, T=(1/5)(3i−4k),

N=(a−a

TT)/aN=(8i−5j+6k)/(5√5)

43. aT=d2s

dt2=d

dtp3t2+4=3t/p3t2+ 4 so when t=2,a

T=3/2.

44. aT=d2s

dt2=d

dtpt2+e−3t=(2t−3e

−3t

)/[2pt2+e−3t] so when t=0,a

T=−3/2.

45. aT=d2s

dt2=d

dtp(4t−1)2+ cos2πt =[4t−1−πcos πt sin πt]/p(4t−1)2+ cos2πt so when

t=1/4, aT=−π/√2.

46. aT=d2s

dt2=d

dtpt4+5t

2+3=(2t

3+5t)/

4+5t

2+ 3 so when t=1,a

T=7/3.

47. aN=κ(ds/dt)2=(1/ρ)(ds/dt)2=(1/1)(2.9×105)2=8.41 ×1010 km/s2

48. a =(d

s/dt2)T+κ(ds/dt)2Nwhere κ=|d2y/dx2|

[1+(dy/dx)2]3/2.Ifd

y/dx2= 0, then κ= 0 and

a=(d

s/dt2)Tso ais tangent to the curve.

49. aN=κ(ds/dt)2=[2/(1+4x

)

3/2

](3)2=18/(1+4x

)

3/2

50. y=ex,a

N=κ(ds/dt)2=[e

/(1+e2x)3/2](2)2=4e

/(1+e2x)3/2

517 Chapter 14

51. a =aTT+aNN; by the Pythagorean Theorem aN=pkak2−a2

T=√9−9=0

52. As in Exercise 51, kak2=a2

T+a2

N,81=9+a

N,a

N=√72=6

√

53. Let c=ds/dt, aN=κds

dt 2

N=1

1000c2,soc

2= 1000aN,c≤10√10√1.5≈38.73 m/s.

54. 10 km/h is the same as 100

36 m/s, so kFk= 500 1

15 100

36 2

≈257.20 N.

55. (a) v0= 320, α=60

◦

0=0sox= 160t,y= 160√3t−16t2.

(b) dy/dt = 160√3−32t,dy/dt = 0 when t=5

√

3so

max = 160√3(5√3) −16(5√3)2= 1200 ft.

max = 160(10√3) = 1600√3 ft.

(d) v(t) = 160i+ (160√3−32t)j,v(10√3) = 160(i−√3j), kv(10√3)k= 320 ft/s.

56. (a) v0= 980, α=45

◦

0=0sox= 490√2t,y= 490√2t−4.9t2

(b) dy/dt = 490√2−9.8t,dy/dt = 0 when t=50

√

2so

max = 490√2(50√2) −4.9(50√2)2=24,500 m.

max = 490√2(100√2)=98,000 m.

(d) v(t) = 490√2i+ (490√2−9.8t)j,v(100√2) = 490√2(i−j), kv(100√2)k= 980 m/s.

57. v0= 80, α=−60◦,s0= 168 so x=40t,y= 168 −40√3t−16t2;y= 0 when

t=−7√3/2 (invalid) or t=√3sox(

√

3)=40

√

3 ft.

58. v0= 80, α=0

◦

0= 168 so x=80t,y= 168 −16t2;y= 0 when t=−√42/2 (invalid) or

t=√42/2sox(

√

42/2)=40

√

42 ft.

59. α=30

◦

0=0sox=√

t/2, y=v0t/2−16t2;dy/dt =v0/2−32t,dy/dt = 0 when t=v0/64

so ymax =v2

0/256 = 2500, v0= 800 ft/s.

60. α=45

◦

0=0sox=√

t/2, y=√2v0t/2−4.9t2;y= 0 when t=0or√

/9.8so

max =v2

0/9.8=24,500, v0= 490 m/s.

61. v0= 800, s0=0sox= (800 cos α)t,y= (800 sin α)t−16t2=16t(50 sin α−t); y= 0 when t=0

or 50 sin αso xmax =40,000 sin αcos α=20,000 sin 2α=10,000, 2α=30

◦or 150◦,α=15

◦

or 75◦.

62. (a) v0=5,α=0

◦

0=4sox=5t,y=4−16t2;y= 0 when t=−1/2 (invalid) or 1/2soit

takes the ball 1/2 s to hit the ﬂoor.

(b) v(t)=5i−32tj,v(1/2)=5i−16j,kv(1/2)k=√281 so the ball hits the ﬂoor with a speed

of √281 ft/s.

the ground at the same instant.

Exercise Set 14.6 518

63. (a) v0=40,α=60,s

0=4,sox=20t, y =4+20

√

3t−16t2; when x=15,t=3

y=4+20

√

4−16 3

42

≈20.98 ft, so the water clears the corner point Awith 0.98 ft to

spare.

(b) y= 20 when −16t2+25

√

3t−16=0,t =0.668 (reject) or 1.500,x(1.500) ≈30 ft, so the

water hits the roof.

64. x=(v

/2)t, y =4+(v

√

3/2)t−16t2, solve x=15,y = 20 simultaneously for v0and t,

v0/2=15/t, t2=15

16√3−1,t ≈0.7898,v

0≈30/0.7898 ≈37.98 ft/s.

65. (a) x= (35√2/2)t, y = (35√2/2)t−4.9t2, from Exercise 17a in Section 14.5

κ=|x0y00 −x00 y0|

[(x0)2+(y

)

]

3/2,κ(0) = 9.8

352√2=0.004√2≈0.00566

(b) y0(t) = 0 when t=25

14√2,y =125

66. (a) a =aTT+aNN,a

T=d

dt2=−7.5 ft/s2,aN=κds

dt 2

ρ(132)2=1322

3000 ft/s2,

kak=pa2

T+a2

N=s(7.5)2+1322

30002

≈9.49 ft/s2

(b) cos θ=a·T

kakkTk=aT

kak≈−7.5

9.49 ≈−0.79,θ ≈2.48 radians ≈142◦

67. s0=0sox=(v

0cos α)t,y=(v

0sin α)t−gt2/2

(a) dy/dt =v0sin α−gt so dy/dt = 0 when t=(v

0sin α)/g,ymax =(v

0sin α)2/(2g)

(b) y= 0 when t= 0 or (2v0sin α)/g,sox=R=(2v

0sin αcos α)/g =(v

0sin 2α)/g when

t=(2v

0sin α)/g;Ris maximum when 2α=90

◦

,α=45

◦

, and the maximum value of R

is v2

0/g.

68. The range is (v2

0sin 2α)/g and the maximum range is v2

0/g so (v2

0sin 2α)/g =(3/4)v2

0/g,

sin 2α=3/4, α=(1/2) sin−1(3/4) ≈24.3◦or α=(1/2)[180◦−sin−1(3/4)] ≈65.7◦.

69. v0= 80, α=30

◦

0=5sox=40

√

3t,y=5+40t−16t2

(a) y= 0 when t=(−40±p(40)2−4(−16)(5))/(−32)=(5±√

30)/4, reject (5 −√30)/4toget

t=(5+√

30)/4≈2.62 s.

(b) x≈40√3(2.62) ≈181.5 ft.

70. (a) v0=v,s0=hso x=(vcos α)t,y=h+(vsin α)t−1

2gt2.Ifx=R, then (vcos α)t=R,

t=R

vcos αbut y= 0 for this value of tso h+(vsin α)[R/(vcos α)] −1

2g[R/(vcos α)]2=0,

h+ (tan α)R−g(sec2α)R2/(2v2)=0,g(sec2α)R2−2v2(tan α)R−2v2h=0.

519 Chapter 14

(b) 2gsec2αtan αR2+2gsec2αRdR

dα −2v2sec2αR −2v2tan αdR

dα =0;ifdR

dα = 0 and α=α0

when R=R0, then 2gsec2α0tan α0R2

0−2v2sec2α0R0=0,gtan α0R0−v2=0,

tan α0=v2/(gR0).

0−2v2(tan α0)R0−2v2h= 0, but from

part (b) tan α0=v2/(gR0)sosec

0= 1 + tan2α0=1+v

/(gR0)2thus

g[1+v4/(gR0)2]R2

0−2v2[v2/(gR0)]R0−2v2h=0,gR2

0−v4/g −2v2h=0,

0=v

2+2gh)/g2,R0=(v/g)pv2+2gh and

tan α0=v2/(vpv2+2gh)=v/pv2+2gh,α0= tan−1(v/pv2+2gh).

71. (a) v0(cos α)(2.9) = 259 cos 23◦so v0cos α≈82.21061, v0(sin α)(2.9) −16(2.9)2=−259 sin 23◦

so v0sin α≈11.50367; divide v0sin αby v0cos αto get tanα≈0.139929, thus α≈8◦

and v0≈82.21061/cos 8◦≈83 ft/s.

(b) From part (a), x≈82.21061tand y≈11.50367t−16t2for 0 ≤t≤2.9; the distance traveled

is Z2.9

0p(dx/dt)2+(dy/dt)2dt ≈268.76 ft.

EXERCISE SET 14.7

1. The results follow from formulae (1) and (7) of Section 12.5.

2. (a) (rmax −rmin)/(rmax +rmin)=2ae/(2a)=e

(b) rmax/rmin =(1+e)/(1 −e), and the result follows.

3. (a) From (15) and (6), at t=0,

C=v

0×b

0−GMu=v0j×r0v0k−GMu=r0v2

0i−GMi=(r

0−GM)i

(b) From (22), r0v2

0−GM =GMe, so from (7) and (17), v×b=GM(cos θi+ sin θj)+GMei,

and the result follows.

(d) From Part (c) and (10), kv×bk=kvkkbk=vr0v0. From Part (b),

kv×bk=GMq(e+ cos θ)2+ sin2θ=GM√e2+2ecos θ+1.By (10) and

Part (d), kv×bk=kvkkbk=v(r0v0)thusv=GM

r0v0pe2+2ecos θ+ 1. From (22),

r0v2

0/(GM)=1+e,GM/(r0v0)=v

/(1+e)sov=v

1+epe

2+2ecos θ+1.

4. At the end of the minor axis, cos θ=−c/a =−eso

v=v0

1+epe

2+2e(−e)+1= v

1+ep1−e

2=v

r1−e

1+e.

5. vmax occurs when θ=0sov

max =v0;vmin occurs when θ=πso

vmin =v0

1+epe

2−2e+1=v

max

1−e

1+e,thusv

max =vmin

1+e

1−e.

6. If the orbit is a circle then e= 0 so from Part (e) of Exercise 3, v=v0at all points on the orbit.

Use (22) with e= 0 to get v0=pGM/r0so v=pGM/r0.

Chapter 14 Supplementary Exercises 520

7. r0= 6440 + 200 = 6640 km so v=p3.99 ×105/6640 ≈7.75 km/s.

8. From Example 1, the orbit is 22,352 mi above the Earth, thus v≈s1.24 ×1012

26,352 ≈6859.68 mi/h.

9. From (23) with r0= 6440 + 300 = 6740 km, vesc =r2(3.99) ×105

6740 ≈10.88 km/s.

10. From (29), T=2π

√GM a3/2. But T= 1 yr = 365 ·24 ·3600 s, thus M=4π2a3

GT 2≈1.99 ×1030 kg.

11. (a) At perigee, r=rmin =a(1 −e) = 238,900 (1 −0.055) ≈225,760 mi; at apogee,

r=rmax =a(1+e) = 238,900(1 + 0.055) ≈252,040 mi. Subtract the sum

of the radius of the moon and the radius of the Earth to get

minimum distance = 225,760 −5080 = 220,680 mi,

and maximum distance = 252,040 −5080 = 246,960 mi.

(b) T=2π

/(GM)=2π

p(238,900)3/(1.24 ×1012)≈659 hr ≈27.5 days.

12. (a) rmin = 6440 + 649 = 7,089 km, rmax = 6440 + 4,340 = 10,780 km so

a=(r

min +rmax)/2 = 8934.5 km.

(b) e= (10,780 −7,089)/(10,780 + 7,089) ≈0.207.

/(GM)=2π

p(8934.5)3/(3.99 ×105)≈8400 s ≈140 min

13. (a) r0= 4000 + 180 = 4180 mi, v=p1.24 ×1012/4180 ≈17,224 mph

(b) r0= 4180 mi, v0=17,224+600 = 17,824 mi/h; e=r0v2

GM −1=(4180)(17,824)2

1.24 ×1012 −1≈0.07094.

rmax = 4180(1 + 0.07094)/(1 −0.07094) ≈4818 mi; the apogee altitude

is 4818 −4000 = 818 mi.

14. By equation (20), r=k

1+ecos θ, where k>0. By assumption, ris minimal when θ=0,

hence e≥0.

CHAPTER 14 SUPPLEMENTARY EXERCISES

2. (a) the line through the tips of r0and r1

(b) the line segment connecting the tips of r0and r1

4. (a) speed (b) distance traveled (c) distance of the particle from the origin

7. (a) r(t)=Zt

cos πu2

2du i+Zt

sin πu2

2du j;



dt 



=x0(t)2+y0(t)2= cos2πt2

2+ sin2πt2

2= 1 and r(0) = 0

521 Chapter 14

(b) r0(s) = cos πs2

2i+ sin πs2

2j,r00 (s)=−πs sin πs2

2i+πs cos πs2

2j,

κ=kr00 (s)k=π|s|

8. (a) The tangent vector to the curve is always tangent to the sphere.

(b) kvk= const, so v·a= 0; the acceleration vector is always perpendicular to the velocity

vector

4cos2t(cos2t+ sin2t)+1

4cos2t=1

9. (a) kr(t)k= 1, so by Theorem 14.2.7, r0(t) is always perpendicular to the vector r(t). Then

v(t)=Rω(−sin ωti+ cos ωtj),v =kv(t)k=Rω

(b) a =−Rω2(cos ωti+ sin ωtj),a =kak=Rω2, and a=−ω2ris directed toward the origin.

ω.

10. (a) F=kFk=mkak=mRω2=mR v2

R2=mv2

(b) R= 6440 + 3200 = 9600 km, 6.5=v=Rω = 9600ω, ω =6.5

9600 =13

19,200,

a=−ar

krk=−Rω2r

R=−Rω2(cos ωti+sin ωtj)=−13

2cos 13t

19,200i+ sin 13t

19,200j

13 kg ·km/s2≈10.77 N

11. (a) Let r=xi+yj+zk, then x2+z2=t2(sin2πt + cos2πt)=t

2=y

(b) Let x=t, then y=t2,z =±p4−t

2/3−t

4/6

Chapter 14 Supplementary Exercises 522

12.

t = 0

t = 1 t = 1

t = 2

13. (a) ker(t)k2= cos2θ+sin2θ=1,soe

(t) is a unit vector; r(t)=r(t)e(t), so they have the same

direction if r(t)>0, opposite if r(t)<0. eθ(t) is perpendicular to er(t) since er(t)·eθ(t)=0,

and it will result from a counterclockwise rotation of er(t) provided e(t)×eθ(t)=k, which

is true.

(b) d

dter(t)=dθ

dt (−sin θi+ cos θj)=dθ

dt eθ(t) and d

dteθ(t)=−dθ

dt (cos θi+ sin θj)=−dθ

dt er(t), so

v(t)= d

dtr(t)= d

dt(r(t)er(t)) = r0(t)er(t)+r(t)dθ

dt eθ(t)

dtv(t)

=r00 (t)er(t)+r

(t)dθ

dt eθ(t)+r

(t)dθ

dt eθ(t)+r(t)d

dt2eθ(t)−r(t)dθ

dt 2

er(t)

="d2r

dt2−rdθ

dt 2#er(t)+rd

dt2+2dr

dθ

dt eθ(t)

14. The height y(t) of the rocket satisﬁes tan θ=y/b, y =btan θ, v =dy

dt =dy

dθ

dt =bsec2θdθ

dt .

15. r =r0+t

−→

PQ=(t−1)i+(4−2t)j+(3+2t)k;



dt 



=3,r(s)=s−3

3i+12 −2s

3j+9+2s

16. By equation (26) of Section 14.6, r(t) = (60 cos α)ti+ ((60 sin α)t−16t2+4)j, and the maximum

height of the baseball occurs when y0(t)=0,60 sin α=32t, t =15

8sin α, so the ball clears the

ceiling if ymax = (60 sin α)15

8sin α−16152

82sin2α+4≤25,152sin2α

4≤21,sin2α≤28

75. The ball

hits the wall when x=60,t = sec α, and y(sec α)=60sinαsec α−16 sec2α+ 4. Maximize the

height h(α)=y(sec α)=60tanα−16 sec2α+ 4, subject to the constraint sin2α≤28

75. Then

h0(α)=60sec

α−32 sec2αtan α=0,tan α=60

32 =15

8, so sin α=15

√82+15

2=15

17, but for

this value of αthe constraint is not satisﬁed (the ball hits the ceiling). Hence the maximum

value of hoccurs at one of the endpoints of the α-interval on which the ball clears the ceiling, i.e.

h0,sin−1p28/75i. Since h0(0) = 60, it follows that his increasing throughout the interval, since

h0>0 inside the interval. Thus hmax occurs when sin2α=28

75,hmax = 60 tan α−16 sec2α+4=

60√28

√47 −1675

47 +4= 120√329 −1012

47 ≈24.78 ft. Note: the possibility that the baseball keeps

climbing until it hits the wall can be rejected as follows: if so, then y0(t) = 0 after the ball hits

the wall, i.e. t=15

8sin αoccurs after t= sec α, hence 15

8sin α≥sec α, 15 sin αcos α≥8,

15 sin 2α≥16, impossible.

523 Chapter 14

17. r0(1)=3i+10j+10k,soifr

(t)=3t

i+10j+10tkis perpendicular to r0(1), then

9t2+ 100 + 100t=0,t =−10,−10/9,

so r=−1000i−100j+ 500k,−(1000/729)i−(100/9)j+ (500/81)k.

18. Let r(t)=x(t)i+y(t)j, then dx

dt =x(t),dy

dt =y(t),x(0) = x0,y(0) = y0,so

x(t)=x

,y(t)=y

,r(t)=e

19. (a) dv

dt =2t

i+j+ cos 2tk,v0=i+2j−k,sox

(t)=2

3+1,y

(t)=t+2,z0

(t)=1

2sin 2t−1,

x(t)=1

4+t, y(t)=1

2+2t, z(t)=−1

4cos 2t−t+1

4, since r(0) = 0. Hence

r(t)=1

4+t

i+1

2+2t

j−1

4cos 2t+t−1

4k

(b) ds

dt it=1 =kr0(t)kit=1p(5/3)2+9+(1−(sin 2)/2)2≈3.475

20. kvk2=v(t)·v(t),2kvkd

dtkvk=2v·a,d

dt (kvk)= 1

kvk(v·a)

524

CHAPTER 15

Partial Derivatives

EXERCISE SET 15.1

1. (a) f(2,1) = (2)2(1)+1=5 (b) f(1,2) = (1)2(2)+1=3

(e) f(3a, a)=(3a)

(a)+1=9a

3+1 (f) f(ab,a−b)=(ab)2(a−b)+1=a

2−a

3+1

2. (a) 2t(b) 2x(c) 2y2+2y

3. (a) f(x+y, x −y)=(x+y)(x−y)+3=x

2−y

2+3

(b) fxy, 3x2y3=(xy)3x2y3+3=3x

4+3

4. (a) (x/y) sin(x/y)(b) xy sin(xy)(c) (x−y) sin(x−y)

5. F(g(x),h(y)) = Fx3,3y+1

=x

(3y+1)

6. g(u(x, y),v(x, y)) = gx2y3,πxy

=πxy sin hx2y32(πxy)i=πxy sin πx5y7

7. (a) t2+3t

10 (b) 0(c) 3076

8. √te−3ln

(

2+1)=√t

(t2+1)

9. (a) 19 (b) −9(c) 3

(d) a6+3 (e) −t8+3 (f) (a+b)(a−b)2b3+3

10. (a) x2(x+y)(x−y)+(x+y)=x

2x

2−y

+(x+y)=x

4−x

2+x+y

(b) (xz)(xy)(y/x)+xy =xy2z+xy

11. Fx2,y+1,z2=(y+1)e

(y+1)z2

12. gx2z3, πxyz, xy/z=(xy/z) sin πx3yz4

13. (a) f(√5,2,π,−3π)=80

√

π(b) f(1,1,...,1) =

k=1

k=n(n+1)/2

14. (a) f(−2,2,0,π/4)=1

(b) f(1,2,...,n)=n(n+ 1)(2n+1)/6, see (Theorem 2(b), Section 7.4)

15.

y16.

525 Chapter 15

17.

y18.

19. (a) all points above or on the line y=−2

(b) all points on or within the sphere x2+y2+z2=25

20. (a) all points on or between the vertical lines x=±2.

(b) all points above the line y=2x

21.

z22.

(0, 3, 0)

23. z

24. z

25. z

26.

(2, 0, 0)

(0, 2, 0)

(0, 0, 4)

Exercise Set 15.1 526

27. z

(0, 0, 1)

28.

(0, 1, 0)

(1, 0, 0)

29. z

(0, 0, 1)

(0, –1, 0)

30. z

31.

1234k = 0

y32.

k = -1

k = -2 k = 2

k = 1

k = 0x

33.

k = 2

k = 1

k = 0

k = –1

k = –2

34.

1234

k = 0

527 Chapter 15

35.

k = 2

k = 1

k = 0

k = –1

k = –2

-2

36.

k = -2

k = 2

k = -1

k = 0

k = 1

37.

(0, 4, 0)

(0, 0, 2)

(2, 0, 0)

38. z

39.

(0, 0, 3)

40.

(0, 0, 1)

( , 0, 0)

(0, − , 0)

41. concentric spheres, common center at (2,0,0)

42. parallel planes, common normal 3i−j+2k

43. concentric cylinders, common axis the y-axis

44. circular paraboloids, common axis the z-axis, all the same shape but with diﬀerent vertices along

z-axis.

45. (a) f(−1,1)=0;x

2−2x

3+3xy =0 (b) f(0,0)=0;x

2−2x

3+3xy =0

Exercise Set 15.1 528

46. (a) f(ln 2, 1) = 2; yex=2 (b) f(0,3)=3;yex=3

47. (a) f(1,−2,0)=5;x

2+y

2−z=5 (b) f(1,0,3) = −2; x2+y2−z=−2

2+y

2−z=0

48. (a) f(1,0,2)=3;xyz +3=3,xyz =0 (b) f(−2,4,1) = −5; xyz +3=−5, xyz =−8

49. (a)

T = 1

T = 2

T = 3

(b) At (1,4) the temperature is

T(1,4) = 4 so the temperature

will remain constant along

the path xy =4.

50. V=8/

p16+x2+y2p16+x2+y2

=8/V x2+y2=64/V 2−16

the equipotential curves are circles.

10 20

20 V = 2.0

V = 1.0

V = 0.5x

51. (a) f(x, y)=1−x

2−y

, because f=cis a circle of radius √1−c(provided c≤1), and the

radii in (a) decrease as cincreases.

(b) f(x, y)=p

2+y

2because f=cis a circle of radius c, and the radii increase uniformly.

(a) is the contour plot of f(x, y)=1−x

2−y

, because f=cis a circle of radius √1−c

(provided c≤1), and the radii in (a) decrease as cincreases.

2+y

2because f=cis a circle of radius √cand the radii in the plot grow like

the square root function.

52. (a) III (b) IV (c) I(d) II

53. (a) A(b) B(c) increase

(d) decrease (e) increase (f) decrease

54. (a) Calgary, since the contour lines are closer together near Calgary than they are near Chicago.

(b) The change in atmospheric pressure is about ∆p≈1012 −999 = 13, so the average rate of

change is ∆p/1600 ≈0.0081.

529 Chapter 15

55. (a)

5-5

-3

2(b)

-4

-3

56. (a) 10

-10

-10 10

(b) 10

-10

-10 10

57. (a) z

(b)

-2 -1 0 1 2

-2

-1

58. (a)

4 x

-5

6cio

(b)

0 1 2 3 4

59. (a) The graph of gis the graph of fshifted one unit in the positive x-direction.

(b) The graph of gis the graph of fshifted one unit up the z-axis.

respect to the plane z=0.

Exercise Set 15.2 530

60. (a) z

(b) If ais positive and increasing then the graph of gis more pointed, and in the limit as a→+∞

the graph approaches a ’spike’ on the z-axis of height 1. As adecreases to zero the graph of

ggets ﬂatter until it ﬁnally approaches the plane z=1.

EXERCISE SET 15.2

-1

y2.

y = x

y3.

y = 2x + 1

y5.

y6.

xy = –1

xy = 1

xy = –1

9. all of 3-space

531 Chapter 15

10. all points inside the sphere with radius 2 and center at the origin

11. all points not on the cylinder x2+z2=1 12. all of 3-space

13. 35 14. π2/215. −8

16. e−717. 018. 0

19. (a) Along x= 0 lim

(x,y)→(0,0)

x2+2y

2= lim

y→0

2y2does not exist.

(b) Along x=0,lim

(x,y)→(0,0)

x+y

x+y2= lim

y→0

ydoes not exist.

20. (a) Along y=0:lim

x→0

2= lim

x→0

xdoes not exist because 

x→+∞as x→0 so the original

limit does not exist.

(b) Along y=0:lim

x→0

xdoes not exist, so the original limit does not exist.

21. Let z=x2+y2, then lim

(x,y)→(0,0)

sin x2+y2

x2+y2= lim

z→0+

sin z

z=1

22. Let z=x2+y2, then lim

(x,y)→(0,0)

1−cos x2+y2

x2+y2= lim

z→0+

1−cos z

z= lim

z→0+

sin z

1=0

23. Let z=x2+y2, then lim

(x,y)→(0,0) e−1/(x2+y2)= lim

z→0+e−1/z =0

24. With z=1

x2+y2, lim

z→+∞

√ze−1/√z; let w=1

√z,lim

w→+∞

ew=0

25. lim

(x,y)→(0,0) x2+y2x

2−y

2

2+y

2= lim

(x,y)→(0,0) x2−y2=0

26. lim

(x,y)→(0,0) x2+4y

x

2−4y

2

2+4y

2= lim

(x,y)→(0,0) x2−4y2=0

27. along y=0:lim

x→0

2= lim

x→00 = 0; along y=x: lim

x→0

5x2= lim

x→01/5=1/5

so the limit does not exist.

28. Let z=x2+y2, then lim

(x,y)→(0,0)

1−x2−y2

x2+y2= lim

z→0+

1−z2

z2=+∞so the limit does not exist.

29. 8/330. ln 5

31. Let t=px2+y2+z2, then lim

(x,y,z)→(0,0,0)

sin x2+y2+z2

px2+y2+z2= lim

t→0+

sin t2

t=0

Exercise Set 15.2 532

32. With t=px2+y2+z2, lim

t→0+

sin t

t2= lim

t→0+

cos t

2t=+∞so the limit does not exist.

33. yln(x2+y2)=rsin θln r2=2r(ln r) sin θ, so lim

(x,y)→(0,0) yln(x2+y2) = lim

r→0+2r(ln r) sin θ=0

34. x2y2

px2+y2=(r2cos2θ)(r2sin2θ)

r=r3cos2θsin2θ, so lim

(x,y)→(0,0)

x2y2

px2+y2=0

35. e√x2+y2+z2

px2+y2+z2=eρ

ρ, so lim

(x,y,z)→(0,0,0)

e√x2+y2+z2

px2+y2+z2= lim

ρ→0+

eρ

ρdoes not exist.

36. lim

(x,y,z)→(0,0,0) tan−11

x2+y2+z2= lim

ρ→0+tan−11

ρ2=π

37. (a) No, since there seem to be points near (0,0) with z= 0 and other points near (0,0) with

z≈1/2.

(b) lim

x→0

mx3

x4+m2x2= lim

x→0

x2+m2=0 (c) lim

x→0

2x4= lim

x→01/2=1/2

(d) A limit must be unique if it exists, so f(x, y) cannot have a limit as (x, y)→(0,0).

38. (a) Along y=mx : lim

x→0

mx4

2x6+m2x2= lim

x→0

mx2

2x4+m2=0;

along y=kx2: lim

x→0

kx5

2x6+k2x4= lim

x→0

2x2+k2=0.

(b) lim

x→0

2x6+x6= lim

x→0

3=1

36=0

39. (a) lim

t→0

abct3

a2t2+b4t4+c4t4= lim

t→0

abct

a2+b4t2+c4t2=0

(b) lim

t→0

t4+t4+t4= lim

t→01/3=1/3

40. π/2 because x2+1

2+(y−1)2→+∞as (x, y)→(0,1)

41. −π/2 because x2−1

x2+(y−1)2→−∞as (x, y)→(0,1)

42. with z=x2+y2, lim

z→0+

sin z

z=1=f(0,0)

43. No, because lim

(x,y)→(0,0)

x2+y2does not exist.

Along x=0:lim

y→00/y2= lim

y→00 = 0; along y=0:lim

x→0x

/x2= lim

x→01=1.

44. Using polar coordinates with r>0, xy =r2sin θcos θand x2+y2=r2so

|xy ln x2+y2|=|r2sin θcos θln r2|≤|2r

2ln r|, but lim

r→0+2r2ln r= 0 thus

lim

(x,y)→(0,0) xy ln x2+y2=0;f(x, y) will be continuous at (0,0) if we deﬁne f(0,0)=0.

533 Chapter 15

EXERCISE SET 15.3

1. (a) 9x2y2(b) 6x3y(c) 9y2(d) 9x2

(e) 6y(f) 6x3(g) 36 (h) 12

2. (a) 2e2xsin y(b) e2xcos y(c) 2 sin y(d) 0

(e) cos y(f) e2x(g) 0(h) 4

3. (a) −1

4x3/2cos y(b) −√xcos y(c) −sin y

2√x(d) −sin y

2√x

4. (a) 8+84x

5(b) 140x4y3(c) 140x3y4(d) 140x3y4

5. (a) ∂z

∂x =3

2√3x+2y; slope = 3

8(b) ∂z

∂y =1

√3x+2y; slope = 1

6. (a) ∂z

∂x =e−y; slope = 1 (b) ∂z

∂y =−xe−y+ 5; slope = 2

7. (a) ∂z

∂x =−4 cos(y2−4x); rate of change = −4 cos 7

(b) ∂z

∂y =2ycos(y2−4x); rate of change = 2 cos 7

8. (a) ∂z

∂x =−1

(x+y)2; rate of change = −1

4(b) ∂z

∂y =−1

(x+y)2; rate of change = −1

9. ∂z/∂x = slope of line parallel to xz-plane = −4; ∂z/∂y = slope of line parallel to yz-plane = 1/2

10. The slope at Pin the positive x-direction is negative, the slope in the positive y-direction is

negative, thus ∂z/∂x < 0,∂z/∂y < 0; the curve through Pwhich is parallel to the x-axis is

concave down, so ∂2z/∂x2<0; the curve parallel to the y-axis is concave down, so ∂2z/∂y2<0.

11. ∂z/∂x =8xy3ex2y3,∂z/∂y =12x

12. ∂z/∂x =−5x4y4sin(x5y4), ∂z/∂y =−4x5y3sin x5y4

13. ∂z/∂x =x3/(y3/5+x)+3x

2ln(1 + xy−3/5),∂z/∂y=−(3/5)x4/(y8/5+xy)

14. ∂z/∂x =yexy sin(4y2),∂z/∂y=8yexy cos(4y2)+xexy sin(4y2)

15. ∂z

∂x =−y(x2−y2)

(x2+y2)2,∂z

∂y =x(x2−y2)

(x2+y2)216. ∂z

∂x =xy3(3x+4y)

2(x+y)3/2,∂z

∂y =x2y2(6x+5y)

2(x+y)3/2

17. fx(x, y)=(3/2)x2y5x2−73x

5y−7x

−1/2

y(x, y)=(1/2)x33x2−73x

5y−7x

−1/2

18. fx(x, y)=−2y/(x−y)2,fy(x, y)=2x/(x−y)2

19. fx(x, y)= y

−1/2

2+x

2,f

(x, y)=−xy−3/2

y2+x2−3

2y−5/2tan−1(x/y)

Exercise Set 15.3 534

20. fx(x, y)=3x

−y+(1/2)x−1/2y3sec √xtan √x,fy(x, y)=−x

−y+3y

2sec √x

21. fx(x, y)=−(4/3)y2sec2xy2tan x−7/3,fy(x, y)=−(8/3)ytan xy2tan x−7/3

22. fx(x, y)=2y

2cosh √xsinh xy2cosh xy2+1

2x−1/2sinh √xsinh2xy2

fy(x, y)=4xy cosh √xsinh xy2cosh xy2

23. fx(x, y)=−2x, fx(3,1) = −6; fy(x, y)=−21y2,f

(3,1) = −21

24. ∂f/∂x =x2y2exy +2xyexy,∂f/∂x



(1,1) =3e;∂f/∂y =x3yexy +x2exy,∂f/∂y



(1,1) =2e

25. ∂z/∂x =x(x2+4y2)−1/2,∂z/∂x



(1,2) =1/

√

17 ; ∂z/∂y =4y(x

2+4y2)−1/2,∂z/∂y



(1,2) =8/

√

26. ∂w/∂x =−x2ysin xy +2xcos xy, ∂w

∂x (1/2,π)=−π/4; ∂w/∂y =−x3sin x, ∂w

∂y (1/2,π)=−1/8

27. fx=8x−8y

y=−32xy3+35y

xy =fyx =−32y3

28. fx=x/px2+y2,f

y=y/px2+y2,f

xy =fyx =−xy(x2+y2)−3/2

29. fx=excos y, fy=−exsin y, fxy =fyx =−exsin y

30. fx=ex−y2,f

y=−2yex−y2,f

xy =fyx =−2yex−y2

31. fx=4/(4x−5y),f

y=−5/(4x−5y),f

xy =fyx =20/(4x−5y)2

32. fx=4xy2/(x2+y2)2,f

y=−4x

y/(x2+y2)2,f

xy =fyx =8xy(x2−y2)/(x2+y2)3

35. (a) 2x−2z(∂z/∂x)=0,∂z/∂x=x/z =±3/(2√6) = ±√6/4,∂z/∂x=±

√

6/4

(b) z=±px2+y2−1,∂z/∂x=±x/px2+y2−1=±

√

6/4

36. (a) 2y−2z(∂z/∂y)=0,∂z/∂y=y/z =±4/(2√6) = ±√6/3

(b) z=±px2+y2−1,∂z/∂y=±y/px2+y2−1=±

√

6/3

37. 3

2x2+y2+z21/22x+2z∂z

∂x=0,∂z/∂x =−x/z; similarly, ∂z/∂y =−y/z

38. 4x−3z2(∂z/∂x)

2x2+y−z3=1, ∂z

∂x =4x−2x2−y+z3

3z2;1−3z2(∂z/∂y)

2x2+y−z3=1, ∂z

∂y =1−2x2−y+z3

3z2

39. 2x+zxy ∂z

∂x +yzcos xyz +∂z

∂x sin xyz =0, ∂z

∂x =−2x+yz2cos xyz

xyz cos xyz + sin xyz ;

zxy ∂z

∂y +xzcos xyz +∂z

∂y sin xyz =0, ∂z

∂y =−xz2cos xyz

xyz cos xyz + sin xyz

40. exy(cosh z)∂z

∂x +yexy sinh z−z2−2xz ∂z

∂x =0, ∂z

∂x =z2−yexy sinh z

exy cosh z−2xz ;

exy(cosh z)∂z

∂y +xexy sinh z−2xz ∂z

∂y =0, ∂z

∂y =−xexy sinh z

exy cosh z−2xz

535 Chapter 15

41. III is a plane, and its partial derivatives are constants, so III cannot be f(x, y). If I is the graph

of z=f(x, y) then (by inspection) fyis constant as yvaries, but neither II nor III is constant as

yvaries. Hence z=f(x, y) has II as its graph, and as II seems to be an odd function of xand an

even function of y,fxhas I as its graph and fyhas III as its graph.

42. Moving to the right from (x0,y

0) decreases f(x, y), so fx<0; moving up increases f,sof

y>0.

43. (a) 30xy4−4(b) 60x2y3(c) 60x3y2

44. (a) 120(2x−y)2(b) −240(2x−y)2(c) 480(2x−y)

45. (a) fxyy(0,1) = −30 (b) fxxx(0,1) = −125 (c) fyyxx(0,1) = 150

46. (a) ∂3w

∂y2∂x =−eysin x,∂3w

∂y2∂x



(π/4,0)

=−1/√2

(b) ∂3w

∂x2∂y =−eycos x,∂3w

∂x2∂y



(π/4,0)

=−1/√2

47. (a) ∂3f

∂x3(b) ∂3f

∂y2∂x (c) ∂4f

∂x2∂y2(d) ∂4f

∂y3∂x

48. (a) fxyy (b) fxxxx (c) fxxyy (d) fyyyxx

49. (a) 2xy4z3+y(b) 4x2y3z3+x(c) 3x2y4z2+2z

(d) 2y4z3+y(e) 32z3+1 (f) 438

50. (a) 2xy cos z(b) x2cos z(c) −x2ysin z

(d) 4ycos z(e) 4 cos z(f) 0

51. fx=2z/x,fy=z/y,fz= ln(x2ycos z)−ztan z

52. fx=y−5/2zsec(xz/y) tan(xz/y), fy=−xy−7/2zsec(xz/y) tan(xz/y)−(3/2)y−5/2sec(xz/y),

fz=xy−5/2sec(xz/y) tan(xz/y)

53. fx=−y2z3/1+x

,f

y=−2xyz3/1+x

,f

z=−3xy2z2/1+x



54. fx=4xyz cosh √zsinh x2yzcosh x2yz,fy=2x

zcosh √zsinh x2yzcosh x2yz,

fz=2x

ycosh √zsinh x2yzcosh x2yz+(1/2)z−1/2sinh √zsinh2x2yz

55. ∂w/∂x =yzezcos xz,∂w/∂y =ezsin xz,∂w/∂z =yez(sin xz +xcos xz)

56. ∂w/∂x =2x/ y2+z2,∂w/∂y =−2yx2+z2/y2+z22,∂w/∂z =2zy

2−x

/y

2+z

2

57. ∂w/∂x =x/px2+y2+z2,∂w/∂y =y/px2+y2+z2,∂w/∂z =z/px2+y2+z2

58. ∂w/∂x =2y

2x+3z,∂w/∂y =3y

2x+3z,∂w/∂z =3y

2x+3z

59. (a) e(b) 2e(c) e

Exercise Set 15.3 536

60. (a) 2/√7(b) 4/√7(c) 1/√7

62.

-2

-1

-2

-1

-2

-1

-2

-1

63. (3/2) x2+y2+z2+w21/22x+2w∂w

∂x =0,∂w/∂x =−x/w; similarly, ∂w/∂y =−y/w

and ∂w/∂z =−z/w

64. ∂w/∂x =−4x/3, ∂w/∂y =−1/3, ∂w/∂z =(2x

2+y−z

3+3z

2+3w)/3

65. ∂w

∂x =−yzw cos xyz

2w+ sin xyz ,∂w

∂y =−xzw cos xyz

2w+ sin xyz ,∂w

∂z =−xyw cos xyz

2w+ sin xyz

66. ∂w

∂x =yexy sinh w

z2−exy cosh w,∂w

∂y =xexy sinh w

z2−exy cosh w,∂w

∂z =2zw

exy cosh w−z2

67. (a) fxy =15x

7+2y(b) fyz =35x

6+3y

6(d) fzz =42x

(e) fzyy = 140x3y3z6+6y(f) fxxy =30xy4z7

(g) fzyx = 105x2y4z6(h) fxxyz = 210xy4z6

68. (a) 160(4x−3y+2z)

3(b) −1440(4x−3y+2z)

2(c) −5760(4x−3y+2z)

69. fx=ex2,f

y=−e

270. fx=yex2y2,f

y=xex2y2

71. ∂w/∂xi=−isin(x1+2x

2+...+nxn)72. ∂w/∂xi=1

n n

k=1

xk!(1/n)−1

73. (a) fx=2x+2y, fxx =2,f

y=−2y+2x, fyy =−2; fxx +fyy =2−2=0

(b) zx=exsin y−eysin x, zxx =exsin y−eycos x, zy=excos y+eycos x,

zyy =−exsin y+eycos x;zxx +zyy =exsin y−eycos x−exsin y+eycos x=0

x2+y2−2y

1+(y/x)2=2x−2y

x2+y2,z

xx =−2x2−y2−2xy

(x2+y2)2,

zy=2y

x2+y2+21

1+(y/x)2=2y+2x

2+y

2,z

yy =−2y2−x2+2xy

(x2+y2)2;

zxx +zyy =−2x2−y2−2xy

(x2+y2)2−2y2−x2+2xy

(x2+y2)2=0

537 Chapter 15

74. (a) zt=−e−tsin(x/c),z

x=(1/c)e−tcos(x/c),z

xx =−(1/c2)e−tsin(x/c);

zt−c2zxx =−e−tsin(x/c)−c2(−(1/c2)e−tsin(x/c))=0

(b) zt=−e−tcos(x/c),z

x=−(1/c)e−tsin(x/c),z

xx =−(1/c2)e−tcos(x/c);

zt−c2zxx =−e−tcos(x/c)−c2(−(1/c2)e−tcos(x/c))=0

75. ux=ωsin cωtcos ωx, uxx =−ω2sin cωtsin ωx, ut=cωcos cωtsin ωx, utt =−c2ω2sin cωtsin ωx;

uxx −1

c2utt =−ω2sin cωtsin ωx −1

c2(−c2)ω2sin cωtsin ωx =0

76. (a) ∂u/∂x =∂v/∂y =2x,∂u/∂y =−∂v/∂x =−2y

(b) ∂u/∂x =∂v/∂y =excos y,∂u/∂y =−∂v/∂x =−exsin y

77. ∂u/∂x =∂v/∂y and ∂u/∂y =−∂v/∂x so ∂2u/∂x2=∂2v/∂x∂y, and ∂2u/∂y2=−∂2v/∂y∂x,

∂2u/∂x2+∂2u/∂y2=∂2v/∂x∂y −∂2v/∂y∂x,if∂

v/∂x∂y =∂2v/∂y∂x then

∂2u/∂x2+∂2u/∂y2=0;thususatisﬁes Laplace’s equation. The proof that vsatisﬁes Laplace’s

equation is similar. Adding Laplace’s equations for uand vgives Laplaces’ equation for u+v.

78. ∂z/∂y =6y,∂z/∂y (2,1) =6

79. ∂z/∂x =−x29 −x2−y2−1/2,∂z/∂x](4,3) =−2

80. (a) ∂z/∂y =8y,∂z/∂y](−1,1) =8 (b) ∂z/∂x =2x,∂z/∂x](−1,1) =−2

81. (a) ∂V/∂r =2πrh (b) ∂V/∂h =πr2

82. (a) ∂V/∂s =πsd2

6√4s2−d2(b) ∂V/∂d =πd(8s2−3d2)

24√4s2−d2

83. (a) P=10T/V,∂P/∂T =10/V ,∂P/∂T ]T=80,V=50 =1/5 lb/(in2K)

(b) V=10T/P, ∂V/∂P =−10T/P2,if V= 50 and T= 80 then

P= 10(80)/(50) = 16,∂V/∂P]

T=80,P=16 =−25/8(in5/lb3)

84. (a) ∂z/∂y =x2,∂z/∂y](1,3) =1,j+kis parallel to the tangent line so x=1,y=3+t,

z=3+t

(b) ∂z/∂x =2xy,∂z/∂x](1,3) =6,i+6kis parallel to the tangent line so x=1+t,y=3,

z=3+6t

85. 1+∂z

∂xcos(x+z) + cos(x−y)=0, ∂z

∂x =−1−cos(x−y)

cos(x+z);∂z

∂y cos(x+z)−cos(x−y)=0,

∂z

∂y =cos(x−y)

cos(x+z);∂2z

∂x∂y =−cos(x+z) sin(x−y) + cos(x−y) sin(x+z)(1 + ∂z/∂x)

cos2(x+z),

substitute for ∂z/∂x and simplify to get ∂2z

∂x∂y =−cos2(x+z) sin(x−y) + cos2(x−y) sin(x+z)

cos3(x+z).

Exercise Set 15.4 538

86. ∂V/∂r =2

3πrh =2

r(1

3πr2h)=2V/r

87. (a) ∂T/∂x =3x

2+1, ∂T /∂x](1,2) =4 (b) ∂T /∂y =4y,∂T/∂y](1,2) =8

88. ∂2R/∂R2

1=−2R2

2/(R1+R2)3,∂

R/∂R2

2=−2R2

1/(R1+R2)3,

∂2R/∂R2

1∂2R/∂R2

2=4R

/(R

1+R

)

6=h4/(R

1+R

)

i[R

/(R

1+R

)]2

=4R

/(R

1+R

)

89. f(x+∆x, y)−f(x, y)

∆x=2(x+∆x)

2−3(x+∆x)y+y

2−(2x2−3xy +y2)

∆x=4x+2∆x−3y,

fx= lim

∆x→0

f(x+∆x, y)−f(x, y)

∆x= lim

∆x→0(4x+2∆x−3y)=4x−3y;f

(2,−1)=11

f(x, y +∆y)−f(x, y)

∆y=2x2−3x(y+∆y)+(y+∆y)

2−(2x2−3xy +y2)

∆y=−3x+2y+∆y,

fy= lim

∆y→0

f(x, y +∆y)−f(x, y)

∆y=−3x+2y;f

y(2,−1) = −8

90. fx(x, y)=2

2+y

)

−1/3

(2x)= 4x

3(x2+y2)1/3,(x, y)6=(0,0);

fx(0,0) = d

dx[f(x, 0)]x=0

dx[x4/3]x=0

3x1/3x=0

=0.

91. (a) fy(0,0) = d

dy [f(0,y)]y=0

dy [y]y=0

(b) If (x, y)6=(0,0), then fy(x, y)=1

3+y

)

−2/3

(3y2)= y

3+y

)

2/3;

(x, y) does not exist where y=−x,x6=0.

EXERCISE SET 15.4

1. 42t13 2. 2(3 + t−1/3)

3(2t+t2/3)

3. 3t−2sin(1/t)4. 1−2t4−8t4ln t

2t√1+lnt−2t

4ln t

5. −10

3t7/3e1−t10/36. (1+t)etcosh (tet/2) sinh (tet/2)

7. ∂z/∂u =24u

2−16uv3−2v+3,∂z/∂v=16u

v−24u2v2−2u−3

8. ∂z/∂u =2u/v2−u2vsec2(u/v)−2uv2tan(u/v)

∂z/∂v =−2u2/v3+u3sec2(u/v)−2u2vtan(u/v)

9. ∂z/∂u =−2 sin u

3 sin v,∂z/∂v=−

2 cos ucos v

3 sin2v

10. ∂z/∂u =3+3v/u −4u, ∂z/∂v =2+3lnu+2lnv

539 Chapter 15

11. ∂z/∂u =eu,∂z/∂v =0

12. ∂z/∂u =−sin(u−v) sin u2+v2+2ucos(u−v) cos u2+v2

∂z/∂v = sin(u−v) sin u2+v2+2vcos(u−v) cos u2+v2

13. ∂T/∂r =3r

2sin θcos2θ−4r3sin3θcos θ

∂T/∂θ =−2r3sin2θcos θ+r4sin4θ+r3cos3θ−3r4sin2θcos2θ

14. dR/dφ =5e

5φ

15. ∂t/∂x =x2+y2/4x2y3,∂t/∂y =y2−3x2/4xy4

16. ∂w/∂u =2v2u2v2−(u−2v)2

[u2v2+(u−2v)

]

2,∂w/∂v =u2(u−2v)2−u2v2

[u2v2+(u−2v)

]

17. −π18. 351/2, −168

19. √3e√3,2−4√3e√320. 1161

21. F(x, y)=x

3+ cos y,dy

dx =−2xy3

3x2y2−sin y

22. F(x, y)=x

3−3xy2+y3−5, dy

dx =−3x2−3y2

−6xy +3y

2=x

2−y

2xy −y2

23. F(x, y)=e

xy +yey−1, dy

dx =−yexy

xexy +yey+ey

24. F(x, y)=x−(xy)1/2+3y−4, dy

dx =−1−(1/2)(xy)−1/2y

−(1/2)(xy)−1/2x+3 =2

√

xy −y

x−6√xy

25. D=x2+y21/2where xand yare the distances of cars A and B, respectively, from the

intersection and Dis the distance between them.

dD/dt =hx/ x2+y21/2i(dx/dt)+hy/ x2+y21/2i(dy/dt), dx/dt =−25 and dy/dt =−30

when x=0.3 and y=0.4sodD/dt =(0.3/0.5)(−25)+(0.4/0.5)(−30) = −39 mph.

26. T=(1/10)PV,dT/dt =(V/10)(dP/dt)+(P/10)(dV /dt), dV/dt = 4 and dP/dt =−1 when

V= 200 and P=5sodT/dt = (20)(−1)+(1/2)(4) = −18 K/s.

27. A=1

2ab sin θbut θ=π/6 when a= 4 and b=3soA=1

(4)(3) sin(π/6)=3.

Solve 1

2ab sin θ= 3 for θto get θ= sin−16

ab,0≤θ≤π/2.

dθ

dt =∂θ

∂a

dt +∂θ

∂b

dt =1

r1−36

a2b2−6

a2bda

dt +1

r1−36

a2b2−6

ab2db

=−6

√a2b2−36 1

dt +1

dt ,da

dt = 1 and db

dt =1

when a= 4 and b=3so dθ

dt =−6

√144 −36 1

4+1

3=−7

12√3=−7

36√3 radians/s

Exercise Set 15.4 540

28. From the law of cosines, c=√a2+b2−2ab cos θwhere cis the length of the third side.

θ=π/3soc=√

2+b

2−ab,

dt =∂c

∂a

dt +∂c

∂b

dt =1

2(a2+b2−ab)−1/2(2a−b)da

dt +1

2a2+b2−ab−1/2(2b−a)db

2√a2+b2−ab (2a−b)da

dt +(2b−a)db

dt ,da

dt = 2 and db

dt = 1 when a= 5 and b=10

so dc

dt =1

2√75[(0)(2) + (15)(1)] = √3/2 cm/s. The third side is increasing.

29. V=(π/4)D2hwhere Dis the diameter and his the height, both measured in inches,

dV/dt =(π/2)Dh(dD/dt)+(π/4)D2(dh/dt), dD/dt = 3 and dh/dt = 24 when D= 30 and

h= 240, so dV /dt =(π/2)(30)(240)(3) + (π/4)(30)2(24) = 16,200πin3/year.

30. dT

dt =∂T

∂x

dt +∂T

∂y

dt =y2

dt +2yln xdy

dt ,dx/dt = 1 and dy/dt =−4 at (3,2) so

dT/dt =(4/3)(1) + (4 ln 3)(−4)=4/3−16 ln 3◦C/s.

31. (a) xy-plane, fx=12x

y+6xy, fy=4x

3+3x

xy =fyx =12x

2+6x

(b) y6=0,f

x=3x

/y, fy=−x3/y2,f

xy =fyx =−3x2/y2

32. (a) x2+y2>1, (the exterior of the circle of radius 1 about the origin);

fx=x/px2+y2−1,f

y=y/px2+y2−1,f

xy =fyx =−xy x2+y2−1−3/2

(b) xy-plane, fx=2xcos(x2+y3),f

y=3y

2cos(x2+y3),f

xy =fyx =−6xy2sin x2+y3

33. (a) 4: f

xxx,f

xxy =fxyx =fyxx,f

xyy =fyxy =fyyx,f

yyy

(b) 5: f

xxxx,f

xxxy =fxxyx =fxyxx =fyxxx,

fxxyy =fxyxy =fxyyx =fyxyx =fyyxx =fyxxy,

fxyyy =fyxyy =fyyxy =fyyyx,f

yyyy

34. (a) Since ewhas inﬁnitely many continuous derivatives, as does xy2, by the Chain Rule the

function exy2has inﬁnitely many continuous derivatives, hence by Theorem 15.4.6,

fxyx =fxxy ; since fxy =fyx if follows that fxyx =fyxx.

(b) fxyx =fxxy =fyxx =2xy5exy2+4y

xy2

35. (a) f(tx, ty)=3t

2+t

2=t

f(x, y); n=2

(b) f(tx, ty)=pt

2+t

2=tf(x, y); n=1

y−2t

3=t

f(x, y); n=3

(d) f(tx, ty)=5/t

2+2t

2=t

−4

f(x, y); n=−4

36. (a) If f(u, v)=t

f(x, y), then ∂f

∂u

dt +∂f

∂v

dt =ntn−1f(x, y), x∂f

∂u +y∂f

∂v =ntn−1f(x, y);

let t= 1 to get x∂f

∂x +y∂f

∂y =nf(x, y).

541 Chapter 15

(b) If f(x, y)=3x

2+y

2then xfx+yfy=6x

2+2y

2=2f(x, y);

If f(x, y)=px

2+y

2then xfx+yfy=x2/px2+y2+y2/px2+y2=px2+y2=f(x, y);

If f(x, y)=x

y−2y

3then xfx+yfy=3x

y−6y

3=3f(x, y);

If f(x, y)= 5

2+2y

)

2then xfx+yfy=x5(−2)2x

(x2+2y

)

3+y5(−2)4y

(x2+2y

)

3=−4f(x, y)

37. (a) ∂z

∂x =dz

∂u

∂x,∂z

∂y =dz

∂u

∂y

(b) ∂2z

∂x2=dz

∂2u

∂x2+∂

∂x dz

du∂u

∂x =dz

∂2u

∂x2+d2z

du2∂u

∂x2

;

∂2z

∂y∂x =dz

∂2u

∂y∂x +∂

∂y dz

du∂u

∂x =dz

∂2u

∂y∂x +d2z

du2

∂u

∂x

∂u

∂y

∂2z

∂y2=∂z

∂u

∂2u

∂y2+∂

∂y ∂z

∂u∂u

∂y =dz

∂2u

∂y2+d2z

du2∂u

∂z 2

38. (a) z=f(u), u=x2−y2;∂z/∂x =(dz/du)(∂u/∂x)=2xdz/du

∂z/∂y =(dz/du)(∂u/∂y)=−2ydz/du, y∂z/∂x +x∂z/∂y =2xydz/du −2xydz/du =0

(b) z=f(u), u=xy;∂z

∂x =dz

∂u

∂x =ydz

du,∂z

∂y =dz

∂u

∂y =xdz

du,

x∂z

∂x −y∂z

∂y =xy dz

du −xy dz

du =0.

(d) xzx−yzy=xyexy −yxexy =0

39. (a) 1=−rsin θ∂θ

∂x + cos θ∂r

∂x and 0 = rcos θ∂θ

∂x + sin θ∂r

∂x; solve for ∂r/∂x and ∂θ/∂x.

(b) 0=−rsin θ∂θ

∂y + cos θ∂r

∂y and 1 = rcos θ∂θ

∂y + sin θ∂r

∂y; solve for ∂r/∂y and ∂θ/∂y.

∂x =∂z

∂r

∂x +∂z

∂θ

∂x =∂z

∂r cos θ−1

∂z

∂θ sin θ.

∂z

∂y =∂z

∂r

∂y +∂z

∂θ

∂y =∂z

∂r sin θ+1

∂z

∂θ cos θ.

(d) Square and add the results of parts (a) and (b).

(e) From Part (c),

∂2z

∂x2=∂

∂r ∂z

∂r cos θ−1

∂z

∂θ sin θ∂r

∂x +∂

∂θ ∂z

∂r cos θ−1

∂z

∂θ sin θ∂θ

∂x

=∂2z

∂r2cos θ+1

∂z

∂θ sin θ−1

∂2z

∂r∂θ sin θcos θ

+∂2z

∂θ∂r cos θ−∂z

∂r sin θ−1

∂2z

∂θ2sin θ−1

∂z

∂θ cos θ−sin θ

r

=∂2z

∂r2cos2θ+2

∂z

∂θ sin θcos θ−2

∂2z

∂θ∂r sin θcos θ+1

∂2z

∂θ2sin2θ+1

∂z

∂r sin2θ.

Exercise Set 15.4 542

Similarly, from Part (c),

∂2z

∂y2=∂2z

∂r2sin2θ−2

∂z

∂θ sin θcos θ+2

∂2z

∂θ∂r sin θcos θ+1

∂2z

∂θ2cos2θ+1

∂z

∂r cos2θ.

Add to get ∂2z

∂x2+∂2z

∂y2=∂2z

∂r2+1

∂2z

∂θ2+1

∂z

∂r.

40. zx=−2y

x2+y2,z

xx =4xy

(x2+y2)2,z

y=2x

2+y

2,z

yy =−4xy

(x2+y2)2,z

xx +zyy =0;

z= tan−12r2cos θsin θ

r2(cos2θ−sin2θ)= tan−1tan 2θ=2θ, zr=0,z

θθ =0

41. (a) By the chain rule, ∂u

∂r =∂u

∂x cos θ+∂u

∂y sin θand ∂v

∂θ =−∂v

∂xrsin θ+∂v

∂yrcos θ, use the

Cauchy-Riemann conditions ∂u

∂x =∂v

∂y and ∂u

∂y =−∂v

∂x in the equation for ∂u

∂r to get

∂u

∂r =∂v

∂y cos θ−∂v

∂x sin θand compare to ∂v

∂θ to see that ∂u

∂r =1

∂v

∂θ. The result ∂v

∂r =−1

∂u

∂θ

can be obtained by considering ∂v

∂r and ∂u

∂θ .

(b) ux=2x

x2+y2,v

y=21

1+(y/x)2=2x

x2+y2=ux;

uy=2y

x2+y2,v

x=−2y

1+(y/x)2=−2y

x2+y2=−uy;

u=lnr

,v =2θ, ur=2/r, vθ=2,sou

r=1

θ=0,v

r=0,sov

r=−1

42. z=f(u, v) where u=x−yand v=y−x,

∂z

∂x =∂z

∂u

∂x +∂z

∂v

∂x =∂z

∂u −∂z

∂v and ∂z

∂y =∂z

∂u

∂y +∂z

∂v

∂y =−∂z

∂u +∂z

∂v so ∂z

∂x +∂z

∂y =0

43. (a) ux=f0(x+ct),u

xx =f00(x+ct),u

t=cf0(x+ct),u

tt =c2f00(x+ct); utt =c2uxx

(b) Substitute gfor fand −cfor cin Part (a).

(a) and (b).

(d) sin tsin x=1

2(−cos(x+t) + cos(x−t))

44. fx(x0,y

0)=y

0)=x

∆f=(x

0+∆x)(y0+∆y)−x

0=y

∆x+x

∆y+∆x∆y

0)∆x+fy(x0,y

0)∆y+ (0)∆x+(∆x)∆ywhere 1= 0 and 2=∆x.

45. fx(x0,y

0)=2x

0)=2y

∆f=(x

0+∆x)

2+(y

0+∆y)

2−x

0+y

0=2x

∆x+(∆x)

2+2y

∆y+(∆y)

0)∆x+fy(x0,y

0)∆y+(∆x)∆x+(∆y)∆ywhere 1=∆xand 2=∆y.

543 Chapter 15

46. fx(x0,y

0)=2x

0)=x

∆f=(x

0+∆x)

0+∆y)−x

0=2x

∆x+x

∆y+y

(∆x)2+(∆x)

∆y+2x

∆x∆y

0)∆x+fy(x0,y

0)∆y+(y

∆x+∆x∆y)∆x+(2x

∆x)∆y

where 1=y0∆x+∆x∆yand 2=2x

∆x.

47. fx(x0,y

0)=3,f

0)=2y

∆f=3(x

0+∆x)+(y

0+∆y)

2−3x

0−y

0=3∆x+2y

∆y+(∆y)

0)∆x+fy(x0,y

0)∆y+ (0)∆x+(∆y)∆ywhere 1= 0 and 2=∆y.

48. (a) lim

(x,y)→(0,0) f(x, y)=0=f(0,0)

(b) lim

∆x→0

f(0+∆x, 0) −f(0,0)

∆x= lim

∆x→0

f(∆x, 0)

∆x= lim

∆x→0|∆x|

∆x, which does not exist because

lim

∆x→0+|∆x|/∆x= 1 and lim

∆x→0−|∆x|/∆x=−1.

49. fx(0,0) = lim

∆x→0

f(∆x, 0) −f(0,0)

∆x= lim

∆x→0−3∆x

∆x=−3,

fy(0,0) = lim

∆y→0

f(0,∆y)−f(0,0)

∆y= lim

∆y→0−2∆y

∆y=−2;

lim

(x,y)→(0,0) f(x, y) does not exist because f(x, y)→5if(x, y)→(0,0) where x≥0ory≥0, but

f(x, y)→0if(x, y)→(0,0) where x<0 and y<0, so fis not continuous at (0,0).

50. fx(0,0) = lim

∆x→0

f(∆x, 0) −f(0,0)

∆x= lim

∆x→0

0−0

∆x= lim

∆x→00=0

(0,0) = lim

∆y→0

f(0,∆y)−f(0,0)

∆y= lim

∆y→0

0−0

∆y= 0; along y= 0, lim

x→0

x2= lim

x→00=0,

along y=x, lim

x→0

2x2= lim

x→01/2=1/2 so lim

(x,y)→(0,0) f(x, y) does not exist.

51. (a) fx(0,0) = lim

∆x→0

f(∆x, 0) −f(0,0)

∆x= 0; similarly, fy(0,0)=0

(b) fx(0,y) = lim

∆x→0

f(∆x, y)−f(0,y)

∆x= lim

∆x→0

y(∆x2−y2)

∆x2+y2=−y

fy(x, 0) = lim

∆y→0

f(x, ∆y)−f(x, 0)

∆y= lim

∆y→0

x(x2−∆y2)

x2+∆y

2=x

∆y→0

fx(0,∆y)−fx(0,0)

∆y= lim

∆y→0−∆y

∆y=−1

fyx(0,0) = lim

∆x→0

fy(∆x, 0) −fy(0,0)

∆x= lim

∆x→0

∆x

∆x=1

(d) No, since fxy and fyx are not continuous.

52. Represent the line segment Cthat joins Aand Bby x=x0+(x

1−x

)t,y=y

0+(y

1−y

for 0 ≤t≤1. Let F(t)=f(x

0+(x

1−x

)t, y0+(y

1−y

)t) for 0 ≤t≤1; then

f(x1,y

1)−f(x

0,y

0)=F(1) −F(0). Apply the Mean Value Theorem to F(t) on the interval [0,1]

to get [F(1) −F(0)]/(1 −0) = F0(t∗), F(1) −F(0) = F0(t∗) for some t∗in (0,1) so

f(x1,y

1)−f(x

0,y

0)=F

0(t

∗

). By the chain rule, F0(t)=f

(x, y)(dx/dt)+f

(x, y)(dy/dt)=

(x, y)(x1−x0)+f

y(x, y)(y1−y0). Let (x∗,y

∗) be the point on Cfor t=t∗then

f(x1,y

1)−f(x

0,y

0)=F

0(t

∗

)=f

x(x

∗

∗)(x

1−x

0)+f

y(x

∗,y

∗)(y

1−y

0).

Exercise Set 15.5 544

53. Let (a, b) be any point in the region, if (x, y) is in the region then by the result of Exercise 52

f(x, y)−f(a, b)=f

∗

∗)(x−a)+f

y(x

∗,y

∗)(y−b) where (x∗,y

∗) is on the line segment joining

(a, b) and (x, y). If fx(x, y)=f

(x, y) = 0 throughout the region then

f(x, y)−f(a, b) = (0)(x−a)+ (0)(y−b)=0,f(x, y)=f(a, b)sof(x, y) is constant on the region.

EXERCISE SET 15.5

1. At P,∂z/∂x = 48 and ∂z/∂y =−14, tangent plane 48x−14y−z= 64, normal line x=1+48t,

y=−2−14t,z=12−t.

2. At P,∂z/∂x = 14 and ∂z/∂y =−2, tangent plane 14x−2y−z= 16, normal line x=2+14t,

y=4−2t,z=4−t.

3. At P,∂z/∂x = 1 and ∂z/∂y =−1, tangent plane x−y−z= 0, normal line x=1+t,y=−t,

z=1−t.

4. At P,∂z/∂x =−1 and ∂z/∂y = 0, tangent plane x+z=−1, normal line x=−1−t,y=0,

z=−t.

5. At P,∂z/∂x = 0 and ∂z/∂y = 3, tangent plane 3y−z=−1, normal line x=π/6, y=3t,

z=1−t.

6. At P,∂z/∂x =1/4 and ∂z/∂y =1/6, tangent plane 3x+2y−12z=−30, normal line x=4+t/4,

y=9+t/6, z=5−t.

7. By implicit diﬀerentiation ∂z/∂x =−x/z,∂z/∂y =−y/z so at P,∂z/∂x =3/4 and

∂z/∂y = 0, tangent plane 3x−4z=−25, normal line x=−3+3t/4, y=0,z=4−t.

8. By implicit diﬀerentiation ∂z/∂x =(xy)/(4z), ∂z/∂y =x2/(8z)soatP,∂z/∂x =3/8 and

∂z/∂y =−9/16, tangent plane 6x−9y−16z= 5, normal line x=−3+3t/8, y=1−9t/16,

z=−2−t.

9. The tangent plane is horizontal if the normal ∂z/∂xi+∂z/∂yj−kis parallel to kwhich occurs

when ∂z/∂x =∂z/∂y =0.

(a) ∂z/∂x =3x

,∂z/∂y =2x

y;3x

2= 0 and 2x3y= 0 for all (x, y)onthex-axis or y-axis,

and z= 0 for these points, the tangent plane is horizontal at all points on the x-axis or

y-axis.

(b) ∂z/∂x =2x−y−2, ∂z/∂y =−x+2y+ 4; solve the system 2x−y−2=0,−x+2y+4=0,

to get x=0,y=−2. z=−4at(0,−2), the tangent plane is horizontal at (0,−2,−4).

10. ∂z/∂x =6x,∂z/∂y =−2y,so6x

i−2y

j−kis normal to the surface at a point (x0,y

0,z

0)on

the surface. 6i+4j−kis normal to the given plane. The tangent plane and the given plane are

parallel if their normals are parallel so 6x0=6,x

0= 1 and −2y0=4,y

0=−2. z=−1at(1,−2),

the point on the surface is (1,−2,−1).

11. ∂z/∂x =−6x,∂z/∂y =−4yso −6x0i−4y0j−kis normal to the surface at a point (x0,y

0,z

0)on

the surface. This normal must be parallel to the given line and hence to the vector

−3i+8j−kwhich is parallel to the line so −6x0=−3, x0=1/2 and −4y0=8,y

0=−2.

z=−3/4at(1/2,−2). The point on the surface is (1/2,−2,−3/4).

12. (3,4,5) is a point of intersection because it satisﬁes both equations. Both surfaces have

(3/5)i+(4/5)j−kas a normal so they have a common tangent plane at (3,4,5).

545 Chapter 15

13. df =2xydx +x2dy =0.6+0.2=0.8,∆f=(x+∆x)

2(y+∆y)−x

2y=(1.1)2(3.2) −12·3=0.872

14. dz =6xdx −2dy =−12(0.02) −2(−0.03) = −0.18,

∆z=3(−2+0.02)2−2(4 −0.03) −(3(−2)2−2(4)) = −0.1788

15. 3/1−(−1)/2=7/216. 2(4)(−2) −(−2)3−(0 −13)=−7

17. dz =3x

dx +2x

ydy, ∆z=(x+∆x)

(y+∆y)

2−x

18. dz =yexydx +xexy dy, ∆z=e(x+∆x)(y+∆y)−exy

19. dz =7dx −2dy 20. dz = (10xy5−2)dx + (25x2y4+4)dy

21. dz =y/ 1+x

dx +x/ 1+x

dy

22. dz = 2 sec2(x−3y) tan(x−3y)dx −6 sec2(x−3y) tan(x−3y)dy

23. (a) Let f(x, y)=e

xsin y;f(0,0)=0,f

x(0,0)=0,f

y(0,0)=1,soe

xsin y≈y

(b) Let f(x, y)=2x+1

y+1 ;f(0,0)=1,f

x(0,0)=2,f

y(0,0) = −1, so 2x+1

y+1 ≈1+2x−y

24. f(1,1)=1,f

x(x, y)=αxα−1yβ,f

x(1,1) = α, fy(x, y)=βxαyβ−1,f

y(1,1) = β,so

β≈1+α(x−1)+β(y−1)

25. dT =Txdx +Tydy ≈2(−0.02) −(0.02) = −0.06,T≈T(1,3)+dT ≈93 −0.06=92.94◦

26. p(104,103) ≈p(100,98) −px(100,98)(104 −100) −py(100,98)(103 −98)

= 1008 + (−2)4 + (1)5 = 1005 mb

27. f(x, y)=(x

2+y

)

−1/2

(4,3) = −x

(x2+y2)3/2=−4

125,f

(4,3) = −y

(x2+y2)3/2=−3

125,

p(3.92)2+(3.01)2≈1

√42+3

2−4

125(−0.08) −3

125(0.01)=0.20232; actual value ≈0.202334.

28. From Exercise 24, x0.5y0.3≈1+0.5(x−1)+0.3(y−1), so

(1.05)0.5(0.97)0.3≈1+0.5(0.05)+0.3(−0.03)=1.016, actual value ≈1.01537

29. df =(2x+2y−4)dx +2xdy;x=1,y=2,dx =0.01, dy =0.04 so df =0.10

30. df =(1/3)x−2/3y1/2dx +(1/2)x1/3y−1/2dy;x=8,y=9,dx =−0.02, dy =0.03 so df =0.005

31. df =−x −2dx −y−2dy;x=−1, y=−2, dx =−0.02, dy =−0.04 so df =0.03

32. df =y

2(1 + xy)dx +x

2(1 + xy)dy;x=0,y=2,dx =−0.09, dy =−0.02 so df =−0 . 09

33. z=px2+y2,dz =xx2+y2−1/2dx +yx2+y2−1/2dy;x=3,y=4,dx =0.2,

dy =−0.04 so dz =0.088 cm.

34. dV =(2/3)πrhdr +(1/3)πr2dh;r=4,h= 20, dr =0.05, dh =−0.05 so dV =2.4π≈7.54 in3.

Exercise Set 15.5 546

35. A=xy,dA =ydx +xdy,dA/A =dx/x +dy/y,|dx/x|≤0.03 and |dy/y|≤0.05,

|dA/A|≤|dx/x|+|dy/y|≤0.08=8%

36. V=(1/3)πr2h,dV =(2/3)πrhdr +(1/3)πr2dh,dV/V =2(dr/r)+dh/h,|dr/r|≤0.01 and

|dh/h|≤0.04, |dV/V |≤2|dr/r|+|dh/h|≤0.06 = 6%.

37. z=px2+y2,dz =x

px2+y2dx +y

px2+y2dy,

z=x

x2+y2dx +y

x2+y2dy =x2

x2+y2dx

x+y2

x2+y2dy

y,



z≤x2

x2+y2

x

+y2

x2+y2

y

,if



x≤r/100 and 

y≤r/100 then



z≤x2

x2+y2(r/100) + y2

x2+y2(r/100) = r

100 so the percentage error in zis at most about r%.

38. (a) z=px2+y2,dz =xx2+y2−1/2dx +yx2+y2−1/2dy,

|dz|≤xx

2+y

−1/2|dx|+yx2+y2−1/2|dy|;ifx=3,y=4,|dx|≤0.05, and

|dy|≤0.05 then |dz|≤(3/5)(0.05)+(4/5)(0.05)=0.07 cm

(b) A=(1/2)xy,dA =(1/2)ydx +(1/2)xdy,

|dA|≤(1/2)y|dx|+(1/2)x|dy|≤2(0.05)+(3/2)(0.05)=0.175 cm2.

39. dR =R2

(R1+R2)2dR1+R2

(R1+R2)2dR2,dR

R=R2

R1+R2dR1

R1+R1

R1+R2dR2

R2,



R≤R2

R1+R2

dR1

R1

+R1

R1+R2

dR2

R2

;ifR

1= 200, R2= 400, |dR1/R1|≤0.02, and

|dR2/R2|≤0.02 then |dR/R|≤(400/600)(0.02) + (200/600)(0.02)=0.02 = 2%.

40. dP =(k/V )dT −(kT/V 2)dV ,dP/P =dT/T −dV /V ;ifdT/T =0.03 and dV/V =0.05 then

dP/P =−0.02 so there is about a 2% decrease in pressure.

41. dθ =1

√c2−a2da −a

c√c2−a2dc;ifa=3,c=5,|da|≤0.01, and |dc|≤0.01 then

|dθ|≤(1/4)(0.01)+(3/20)(0.01)=0.004 radians.

42. V=πr2h,dV =2πrhdr +πr2dh;r=2,h=5,dr =0.01, and dh =0.01 so

dV = (20π)(0.01)+(4π)(0.01)=0.24π, or about 0.754 cm3.

43. dT =π

gpL/g dL −πL

g2pL/g dg,dT

T=1

L−1

g;|dL/L|≤0.005 and |dg/g|≤0.001 so

|dT/T |≤(1/2)(0.005) + (1/2)(0.001) = 0.003=0.3%

44. Let hbe the height of the building, xthe distance to the building, and θthe angle of elevation,

then h=xtan θ,dh = tan θdx +xsec2θdθ;ifx= 100, θ=60

◦

,|dx|≤1/6 ft, and

|dθ|≤(0.2)(π/180) = π/900 radians, then |dh|≤√

(1/6) + (100)(4)(π/900) <1.7 ft.

45. (a) z=xy,dz =ydx +xdy,dz/z =dx/x +dy/y;(r+s)%.

(b) z=x/y,dz =dx/y −xdy/y2,dz/z =dx/x −dy/y;(r+s)%.

547 Chapter 15

dy,dz/z =2dx/x +3dy/y;(2r+3s)%.

(d) z=x3y1/2,dz =3x

1/2

dx +x3dy/ 2y1/2,dz/z =3dx/x +(1/2)dy/y;(3r+s/2)%.

46. z=k

xy ; at a point a, b, k

abon the surface, −k

a2b,−k

ab2,−1and hence bk, ak, a2b2is

normal to the surface so the tangent plane is bkx +aky +a2b2z=3abk. The plane cuts the x,

y, and z-axes at the points 3a, 3b, and 3k

ab , respectively, so the volume of the tetrahedron that is

formed is V=1

33k

ab 1

2(3a)(3b)=9

2k, which does not depend on a and b.

47. (a) 2t+7=(−1+t)

2+(2+t)

2=1,t=±1 so the points of intersection are (−2,1,5) and

(0,3,9).

(b) ∂z/∂x =2x,∂z/∂y =2yso at (−2,1,5) the vector n=−4i+2j−kis normal to the surface.

v=i+j+2kis parallel to the line; n·v=−4 so the cosine of the acute angle is

[n·(−v)]/(knkk−vk)=4/√

21√6=4/3

√

14. Similarly, at (0,3,9) the vector

n=6j−kis normal to the surface, n·v= 4 so the cosine of the acute angle is

4/√37√6=4/

√

222.

48. z=xf(u) where u=x/y,∂z/∂x =xf 0(u)∂u/∂x +f(u)=(x/y)f0(u)+f(u)=uf0(u)+f(u),

∂z/∂y =xf 0(u)∂u/∂y =−(x2/y2)f0(u)=−u

(u). If (x0,y

0,z

0) is on the surface then, with

u0=x0/y0,[u

0(u

)+f(u

)] i−u2

0f0(u0)j−kis normal to the surface so the tangent plane is

[u0f0(u0)+f(u

)] x−u2

0f0(u0)y−z=[u

)+f(u

)]x0−u2

0f0(u0)y0−z0

=x0

f0(u0)+f(u

)

x

0−x

f0(u

0−z

f(u

)−z

0=0

so all tangent planes pass through the origin.

49. Use implicit diﬀerentiation to get ∂z/∂x =−c2x/ a2z,∂z/∂y =−c2y/ b2z.At(x

0,z

0),

z06= 0, a normal to the surface is −c2x0/a2z0i−c2y0/b2z0j−kso the tangent plane is

−c2x0

a2z0

x−c2y0

b2z0

y−z=−c2x2

a2z0−c2y2

b2z0−z0,x0x

a2+y0y

b2+z0z

c2=x2

a2+y2

b2+z2

c2=1

50. ∂z/∂x =2x/a2,∂z/∂y =2y/b2.At(x

0,z

0) the vector 2x0/a2i+2y0/b2j−kis normal

to the surface so the tangent plane is 2x0/a2x+2y0/b2y−z=2x

/a2+2y

0/b2−z0, but

z0=x2

0/a2+y2

0/b2so 2x0/a2x+2y0/b2y−z=2z

0−z

0=z

,2x

x/a2+2y

y/b2=z+z0

51. n1=fx(x0,y

0)i+f

y(x

0,y

0)j−kand n2=gx(x0,y

0)i+g

y(x

0,y

0)j−kare normal, respectively,

to z=f(x, y) and z=g(x, y)atP;n

1and n2are perpendicular if and only if n1·n2=0,

0)g

x(x

0,y

0)+f

y(x

0)g

y(x

0,y

0)+1=0,

x(x

0)g

x(x

0,y

0)+f

y(x

0)g

y(x

0,y

0)=−1.

52. n1=fxi+fyj−k=x0

px2

0+y2

i+y0

px2

0+y2

j−k; similarly n2=−x0

px2

0+y2

i−y0

px2

0+y2

j−k;

since a normal to the sphere is N=x0i+y0j+z0k, and n1·N=px2

0+y2

0−z0=0,

2·N=−

0+y

0−z

0= 0, the result follows.

Exercise Set 15.6 548

EXERCISE SET 15.6

1. fincreases the most in the direction of III.

2. The contour lines are closer at P, so the function is increasing more rapidly there, hence ∇fis

larger at P.

3. ∇z=4i−8j4.∇z=−4e

−3y

sin 4xi−3e−3ycos 4xj

5. ∇z=x

x2+y2i+y

x2+y2j

6. ∇z=e−5xsec x2y2xy tan x2y−5i+x2tan x2yj

7. ∇f(x, y) = 3(2x+y)x2+xy2i+3xx

2+xy2j,∇f(−1,−1) = −36i−12j

8. ∇f(x, y)=−xx

2+y

−3/2i−yx

2+y

−3/2j,∇f(3,4) = −(3/125)i−(4/125)j

9. ∇f(x, y)=[y/(x+y)]i+[y/(x+y) + ln(x+y)]j,∇f(−3,4)=4i+4j

10. ∇f(x, y)=3y

2tan2xsec2xi+2ytan3xj,∇f(π/4,−3)=54i−6j

11. ∇f(x, y)=(3y/2)(1 + xy)1/2i+(3x/2)(1 + xy)1/2j,∇f(3,1)=3i+9j,

f=∇f·u=12/

√

2=6

√

12. ∇f(x, y)=2ye2xyi+2xe2xy j,∇f(4,0)=8j,D

f=∇f·u=32/5

13. ∇f(x, y)=2x/ 1+x

2+y

i+1/1+x

2+y

j,∇f(0,0) = j,Duf=−3/√10

14. ∇f(x, y)=−(c+d)y/(x−y)2i+(c+d)x/(x−y)2j,

∇f(3,4) = −4(c+d)i+3(c+d)j,D

f=−(7/5)(c+d)

15. ∇f(x, y)=12x

i+8x

yj,∇f(2,1)=48i+64j,u=(4/5)i−(3/5)j,Duf=∇f·u=0

16. ∇f(x, y)=(2x−3y)i+−3x+12y

j,∇f(−2,0) = −4i+6j,u=(i+2j)/

√

5, Duf=8/

√

17. ∇f(x, y)=y

/xi+2yln xj,∇f(1,4)=16i,u=(−i+j)/

√

2, Duf=−8√2

18. ∇f(x, y)=e

xcos yi−exsin yj,∇f(0,π/4)=(i−j)/

√

2, u=(5i−2j)/

√

29, Duf=7/

√

19. ∇f(x, y)=−y/ x2+y2i+x/ x2+y2j,

∇f(−2,2) = −(i+j)/4, u=−(i+j)/√2, Duf=√2/4

20. ∇f(x, y)=(e

y−yex)i+(xey−ex)j,∇f(0,0) = i−j,u=(5i−2j)/

√

29, Duf=7/

√

21. ∇f(x, y)=(y/2)(xy)−1/2i+(x/2)(xy)−1/2j,∇f(1,4) = i+(1/4)j,

u= cos θi+ sin θj=(1/2)i+√3/2j,Duf=1/2+√

3/8

22. ∇f(x, y)=[2y/(x+y)2]i−[2x/(x+y)2]j,∇f(−1,−2) = −(4/9)i+(2/9)j,u=j,Duf=2/9

549 Chapter 15

23. ∇f(x, y)=2sec

(2x+y)i+ sec2(2x+y)j,∇f(π/6,π/3)=8i+4j,u=(i−j)/

√

2, Duf=2

√

24. ∇f(x, y) = cosh xcosh yi+ sinh xsinh yj,∇f(0,0) = i,u=−i,Duf=−1

25. f(1,2) = 3, level curve 4x−2y+3=3,2x−y=0;

∇f(x, y)=4i−2j

∇f(1,2)=4i−2j(1, 2)

4i – 2jx

26. f(−2,2)=1/2, level curve y/x2=1/2,y=x

/2 for x6=0.

∇f(x, y)=−2y/x3i+1/x2j

∇f(−2,2)=(1/2)i+(1/4)j

i + j

(−2, 2)

27. f(−2,0) = 4, level curve x2+4y

2=4,x

/4+y

2=1.

∇f(x, y)=2xi+8yj

∇f(−2,0) = −4i

-4i

28. f(2,−1) = 3, level curve x2−y2=3.

∇f(x, y)=2xi−2yj

∇f(2,−1)=4i+2j

(2, −1)

4i + 2j

29. ∇f(x, y)=12x

i+8x

yj,∇f(−1,1)=12i−8j,u=(3i−2j)/

√

13, k∇f(−1,1)k=4

√

Exercise Set 15.6 550

30. ∇f(x, y)=3i−(1/y)j,∇f(2,4)=3i−(1/4)j,u= (12i−j)/√145, k∇f(2,4)k=√145/4

31. ∇f(x, y)=xx

2+y

−1/2i+yx

2+y

−1/2j,

∇f(4,−3)=(4i−3j)/5, u=(4i−3j)/5, k∇f(4,−3)k=1

32. ∇f(x, y)=y(x+y)

−2

i−x(x+y)

−2

j,∇f(0,2)=(1/2)i,u=i,k∇f(0,2)k=1/2

33. ∇f(x, y)=−2xi−2yj,∇f(−1,−3)=2i+6j,u=−(i+3j)/

√

10, −k∇f(−1,−3)k=−2√10

34. ∇f(x, y)=yexyi+xexy j;∇f(2,3) = e6(3i+2j), u=−(3i+2j)/

√

13, −k∇f(2,3)k=−√13e6

35. ∇f(x, y)=−3 sin(3x−y)i+ sin(3x−y)j,

∇f(π/6,π/4)=(−3i+j)/

√

2, u=(3i−j)/

√

10, −k∇f(π/6,π/4)k=−√5

36. ∇f(x, y)= y

(x+y)

2rx+y

x−yi−x

(x+y)

2rx+y

x−yj,∇f(3,1)=(

√

2/16)(i−3j),

u=−(i−3j)/√10, −k∇f(3,1)k=−√5/8

37. ∇f(x, y)=y(x+y)

−2

i−x(x+y)

−2

j,∇f(1,0) = −j,−→

PQ=−2i−j,u=(−2i−j)/

√

Duf=1/

√

38. ∇f(x, y)=−e

−xsec yi+e−xsec ytan yj,

∇f(0,π/4) = √2(−i+j), −→

PO=−(π/4)j,u=−j,Duf=−√2

39. ∇f(x, y)= yey

2√xy i+√xyey+xey

2√xy j,∇f(1,1)=(e/2)(i+3j), u=−j,Duf=−3e/2

40. ∇f(x, y)=−y(x+y)

−2

i+x(x+y)

−2

j,∇f(2,3)=(−3i+2j)/25, if Duf= 0 then uand ∇fare

orthogonal, by inspection 2i+3jis orthogonal to ∇f(2,3) so u=±(2i+3j)/

√

13.

41. ∇f(x, y)=8xyi+4x

j,∇f(1,−2) = −16i+4jis normal to the level curve through Pso

u=±(−4i+j)/√17.

42. ∇f(x, y)=(6xy −y)i+3x2−xj,∇f(2,−3)=−33i+10jis normal to the level curve through

Pso u=±(−33i+10j)/

√

1189.

43. Solve the system (3/5)fx(1,2) −(4/5)fy(1,2) = −5, (4/5)fx(1,2)+(3/5)fy(1,2) = 10 for

fx(1,2) and fy(1,2) to get fx(1,2)=5,f

(1,2) = 10. For (c), ∇f(1,2)=5i+10j,

u=(−i−2j)/

√

5, Duf=−5√5.

44. ∇f(−5,1) = −3i+2j,−→

PQ=i+2j,u=(i+2j)/

√

5, Duf=1/

√

45. ∇f(4,−5)=2i−j,u=(5i+2j)/

√

29, Duf=8/

√

46. Let u=u1i+u2jwhere u2

1+u2

2= 1, but Duf=∇f·u=u1−2u2=−2sou

1=2u

2−2,

(2u2−2)2+u2

2=1,5u

2−8u

2+3=0,u

2=1oru

2=3/5thusu

1=0oru

1=−4/5; u=jor

u=−4

5i+3

5j.

551 Chapter 15

47. (a) At (1,2) the steepest ascent seems to be in the direction i+jand the slope in that direction

seems to be 0.5/(√2/2)=1/

√

2, so ∇f≈1

2i+1

2j, which has the required direction and

magnitude.

(b)

−∇f(4, 4)

48. (a)

500

0 ft

100200 300 400

Depart from each contour line in a direction orthogonal to that contour line, as an approxi-

mation to the optimal path.

(b)

500

0 ft

100200 300 400

At the top there is no contour line, so head for the nearest contour line. From then on depart

from each contour line in a direction orthogonal to that countour line, as in Part (a).

49. ∇z=6xi−2yj,k∇zk=p36x2+4y

2=6if36x

2+4y2= 36; all points on the ellipse 9x2+y2=9.

50. ∇z=3i+2yj,k∇zk=p9+4y

,so∇k∇zk=4y

p9+4y

2=8

√

9+16 =8

51. r =ti−t2j,dr/dt =i−2tj=i−4jat the point (2,−4), u=(i−4j)/

√

17;

∇z=2xi+2yj=4i−8jat (2,−4), hence dz/ds =Duz=∇z·u=36/

√

17.

52. (a) ∇T(x, y)=y1−x

2+y



(1+x2+y2)2i+x1+x

2−y



(1+x2+y2)2j,∇T(1,1)=(i+j)/9, u=(2i−j)/

√

DuT=1/9

√



(b) u =−(i+j)/√2, opposite to ∇T(1,1)

Exercise Set 15.6 552

53. (a) ∇V(x, y)=−2e

−2xcos 2yi−2e−2xsin 2yj,E=−∇V(π/4,0)=2e

−π/2i

(b) V(x, y) decreases most rapidly in the direction of −∇V(x, y) which is E.

54. ∇z=−4xi−8yj,ifx=−20 and y= 5 then ∇z=80i−40j.

(a) u =−ipoints due west, Duz=−80, the climber will descend because zis decreasing.

(b) u =(i+j)/

√

2 points northeast, Duz=20

√

2, the climber will ascend at the rate of 20√2

ftperftoftravelinthexy−plane.

inspection ±(i+2j)/

√

5 are unit vectors in these directions; (i+2j)/

√

5 makes an angle of

tan−1(1/2) ≈27◦with the positive y-axis so −(i+2j)/√5 makes the same angle with the

negative y-axis. The compass direction should be N 27◦EorS27

◦W.

55. (a) ∇r=x

px2+y2i+y

px2+y2j=r/r

(b) ∇f(r)=∂f(r)

∂x i+∂f(r)

∂y j=f0(r)∂r

∂xi+f0(r)∂r

∂yj=f0(r)∇r

56. (a) ∇re−3r=(1 −3r)

re−3rr

(b) 3r2r=f0(r)

rrso f0(r)=3r

,f(r)=3

4+C,f(2)=12+C=1,C=−11; f(r)=3

4−11

57. ur= cos θi+ sin θj,uθ=−sin θi+ cos θj,

∇z=∂z

∂xi+∂z

∂yj=∂z

∂r cos θ−1

∂z

∂θ sin θi+∂z

∂r sin θ+1

∂z

∂θ cos θj

=∂z

∂r(cos θi+ sin θj)+1

∂z

∂θ(−sin θi+ cos θj)=∂z

∂rur+1

∂z

∂θuθ

58. (a) ∇(f+g)=(f

x+g

)i+(f

y+g

y)j=(f

i+f

j)+(g

i+g

yj)=∇f+∇g

(b) ∇(cf)=(cfx)i+(cfy)j=c(fxi+fyj)=c∇f

i+f

yj)=f∇g+g∇f

(d) ∇(f/g)=gfx−fgx

g2i+gfy−fgy

g2j=g(fxi+fyj)−f(gxi+gyj)

g2=g∇f−f∇g

(e) ∇(fn)=nf n−1fxi+nfn−1fyj=nfn−1(fxi+fyj)=nfn−1∇f

59. r0(t)=v(t)=k(x, y)∇T=−8k(x, y)xi−2k(x, y)yj;dx

dt =−8kx, dy

dt =−2ky. Divide and solve

to get y4= 256x; one parametrization is x(t)=e

−8t

,y(t)=4e

−2t

60. r0(t)=v(t)=k∇T=−2k(x, y)xi−4k(x, y)yj. Divide and solve to get y=3

25x2; one parametriza-

tion is x(t)=5e

−2t

,y(t)=3e

−4t

553 Chapter 15

61.

-3 3

-5

C = 0

C = –5

C = –10

C = –15

62. 4

-4

-6 6

T = 80

T = 95

T = 97

(5, 3)

T = 90

63. (a) z

2+3y

)]e−(x2+y2)i+[6y−2y(x

2+3y

)]e−(x2+y2)j

(d) ∇f=0if x=y=0orx=0,y =±1orx=±1,y =0.

64. dz/dt =(∂z/∂x)(dx/dt)+(∂z/∂y)(dy/dt)

=(∂z/∂xi+∂z/∂yj)·(dx/dti+dy/dtj)=∇z·r

(t)

65. ∇f(x, y)=f

(x, y)i+fy(x, y)j,if∇f(x, y) = 0 throughout the region then

fx(x, y)=f

(x, y) = 0 throughout the region, the result follows from Exercise 51, Section 15.4.

66. Let u1and u2be nonparallel unit vectors for which the directional derivative is zero. Let ube

any other unit vector, then u=c1u1+c2u2for some choice of scalars c1and c2,

Duf(x, y)=∇f(x, y)·u=c1∇f(x, y)·u1+c2∇f(x, y)·u2

=c1Du1f(x, y)+c

2f(x, y)=0.

EXERCISE SET 15.7

1. 165t32 2. 3−(4/3)t−1/3−24t−7

3t−2t2/3+4t

−63. −2tcos t2

4. 1−512t5−2560t5ln t

2t√1+lnt−512t5ln t5. 3264 6. 0

7. ∇f(x, y, z)=20x

i+8x

yz3j+12x

k,∇f(2,−1,1) = 320i−256j+ 384k,Duf=−320

Exercise Set 15.7 554

8. ∇f(x, y, z)=yzexzi+exzj+(xyexz +2z)k,∇f(0,2,3)=6i+j+6k,D

f=45/7

9. ∇f(x, y, z)= 2x

2+2y

2+3z

2i+4y

2+2y

2+3z

2j+6z

2+2y

2+3z

2k,

∇f(−1,2,4)=(−2/57)i+(8/57)j+ (24/57)k,Duf=−314/741

10. ∇f(x, y, z)=yz cos xyzi+xz cos xyzj+xy cos xyzk,

∇f(1/2,1/3,π)=(π

√

3/6)i+(π

√

3/4)j+(

√

3/12)k,Duf=(1−π)/12

11. ∇f(x, y, z)=3x

z−2xyi−x2j+x3+2z

k,∇f(2,−1,1)=16i−4j+10k,

u=(3i−j+2k)/

√

14, Duf=72/

√

12. ∇f(x, y, z)=−xx

2+z

−1/2i+j−zx

2+z

−1/2k,∇f(−3,1,4)=(3/5)i+j−(4/5)k,

u=(2i−2j−k)/3, Duf=0

13. ∇f(x, y, z)=−1

z+yi−z−x

(z+y)

2j+y+x

(z+y)

2k,∇f(1,0,−3)=(1/3)i+(4/9)j+(1/9)k,

u=(−6i+3j−2k)/7, Duf=−8/63

14. ∇f(x, y, z)=e

x+y+3z(i+j+3k), ∇f(−2,2,−1) = e−3(i+j+3k), u= (20i−4j+5k)/21,

Duf= (31/21)e−3

15. ∇f(1,1,−1)=3i−3j,u=(i−j)/

√

2, k∇f(1,1,−1)k=3

√

16. ∇f(0,−3,0)=(i−3j+4k)/6, u=(i−3j+4k)/√26, k∇f(0,−3,0)k=√26/6

17. ∇f(1,2,−2)=(−i+j)/2, u=(−i+j)/

√

2, k∇f(1,2,−2)k=1/

√

18. ∇f(4,2,2)=(i−j−k)/8, u=(i−j−k)/

√

3, k∇f(4,2,2)k=√3/8

19. ∇f(5,7,6) = −i+11j−12k,u=(i−11j+12k)/

√

266, −k∇f(5,7,6)k=−√266

20. ∇f(0,1,π/4)=2

√

2(i−k), u=−(i−k)/√2, −k∇f(0,1,π/4)k=−4

21. ∇f(2,1,−1) = −i+j−k.−→

PQ=−3i+j+k,u=(−3i+j+k)/

√

11, Duf=3/

√

22. ∇f(−1,−2,1)=13i+5j−20k,u=−k,Duf=20

23. Let ube the unit vector in the direction of a, then

Duf(3,−2,1)=∇f(3,−2,1) ·u=k∇f(3,−2,1)kcos θ= 5 cos θ=−5, cos θ=−1, θ=πso

∇f(3,−2,1) is oppositely directed to u;∇f(3,−2,1) = −5u=−10/3i+5/3j+10/3k.

24. (a) ∇T(1,1,1)=(i+j+k)/8, u=−(i+j+k)/√3, DuT=−√3/8

(b) (i+j+k)/√3(c) √3/8

25. (a) f(x, y, z)=x

2+y

2+4z

,∇f=2xi+2yj+8zk,∇f(2,2,1)=4i+4j+8k,

n=i+j+2k,x+y+2z=6

(b) r(t)=2i+2j+k+t(i+j+2k),x(t)=2+t, y(t)=2+t, z(t)=1+2t

knk=√2

√3,θ ≈35.26◦

555 Chapter 15

26. (a) f(x, y, z)=xz −yz3+yz2,n=∇f(2,−1,1) = i+3k; tangent plane x+3z=5

(b) normal line x=2+t,y=−1, z=1+3t

knk=3

√10,θ ≈18.43◦

27. Set f(x, y)=z+x−z

(y−1), then f(x, y, z)=0,n=±∇f(3,5,1) = ±(i−j−19k),

unit vectors ±1

363(i−j−19k)

28. f(x, y, z) = sin xz −4 cos yz,∇f(π, π, 1) = −i−πk; unit vectors ±1

√1+π

2(i+πk)

29. f(x, y, z)=x

2+y

2+z

,if(x

0,z

0) is on the sphere then ∇f(x0,y

0,z

0)=2(x

i+y

j+z

is normal to the sphere at (x0,y

0,z

0), the normal line is x=x0+x0t,y=y0+y0t,z=z0+z0t

which passes through the origin when t=−1.

30. f(x, y, z)=2x

2+3y

2+4z

,if(x

0,z

0) is on the ellipsoid then

∇f(x0,y

0,z

0)=2(2x

i+3y

j+4z

k) is normal there and hence so is n1=2x

i+3y

j+4z

1must be parallel to n2=i−2j+3kwhich is normal to the given plane so n1=cn2

for some constant c. Equate corresponding components to get x0=c/2, y0=−2c/3, and

z0=3c/4; substitute into the equation of the ellipsoid yields 2 c2/4+34c

2/9+49c

2/16=9,

2= 108/49, c=±6√3/7. The points on the ellipsoid are 3√3/7,−4√3/7,9√3/14and

−3√3/7,4√3/7,−9√3/14.

31. f(x, y, z)=x

2+y

2−z

,if(x

0,z

0) is on the surface then ∇f(x0,y

0,z

0)=2(x

i+y

j−z

is normal there and hence so is n1=x0i+y0j−z0k;n1must be parallel to −→

PQ=3i+2j−2kso

n1=c−→

PQ for some constant c. Equate components to get x0=3c,y

0=2cand z0=2cwhich

when substituted into the equation of the surface yields 9c2+4c

2−4c

2=1,c

2=1/9, c=±1/3

so the points are (1,2/3,2/3) and (−1,−2/3,−2/3).

32. f1(x, y, z)=2x

2+3y

2+z

(x, y, z)=x

2+y

2+z

2−6x−8y−8z+ 24,

n1=∇f1(1,1,2)=4i+6j+4k,n

2=∇f

(1,1,2) = −4i−6j−4k,n1=−n2so n1and n2are

parallel.

33. n1=2i−2j−k,n

2=2i−8j+4k,n

1×n

2=16i−10j−12kis tangent to the line, so

x(t)=1+8t, y(t)=−1+5t, z(t)=2+6t

34. f(x, y, z)=px

2+y

−z,n1=∇f(4,3,5) = 4

5i+3

5j−k,n2=i+2j+2k,n1×n2= (16i−13j+5k)/5

is tangent to the line, x(t)=4+16t, y(t)=3−13t, z(t)=5+5t

35. f(x, y, z)=x

2+z

2−25,g(x, y, z)=y

2+z

2−25,n1=∇f(3,−3,4)=6i+8k,

2=∇g(3,−3,4) = −6j+8k,n

1×n

2=48i−48j−36kis tangent to the line,

x(t)=3+4t, y(t)=−3−4t, z(t)=4−3t

36. (a) f(x, y, z)=z−8+x

2+y

,g(x, y, z)=4x+2y−z, n1=4j+k,n

2=4i+2j−k,

1×n

2=−6i+4j−16kis tangent to the line, x(t)=3t, y(t)=2−2t, z(t)=4+8t

37. dw =3x

zdx +2x

yzdy +x3y2dz, ∆w=(x+∆x)

(y+∆y)

(z+∆z)−x

38. dw =yzexyzdx +xzexyzdy +xyexyz dz, ∆w=e(x+∆x)(y+∆y)(z+∆z)−exyz

Exercise Set 15.7 556

39. dw =8dx −3dy +4dz

40. dw =8xy3z7−3ydx +12x2y2z7−3xdy +28x2y3z6+1

dz

41. dw =yz

1+x

2dx +xz

1+x

2dy +xy

1+x

2dz

42. dw =1

2√xdx +1

2√ydy +1

2√zdz

43. df =2y

dx +4xyz3dy +6xy2z2dz

=2(−1)2(2)3(−0.01) + 4(1)(−1)(2)3(−0.02) + 6(1)(−1)2(2)2(0.02)=0.96

44. df =yz(y+z)

(x+y+z)2dx +xz(x+z)

(x+y+z)2dy +xy(x+y)

(x+y+z)2dz

=(−16)(−0.04)+(−12)(0.02)+(−6)(−0.03)=0.58

45. V=`wh,dV =wh d` +`h dw +`w dh,

|dV |≤wh|d`|+`h|dw|+`w|dh|≤(4)(5)(0.05) + (3)(5)(0.05) + (3)(4)(0.05)=2.35 cm3

46. R=1/(1/R1+1/R2+1/R3), ∂R/∂R1=1

1(1/R1+1/R2+1/R3)2=R2/R2

1, similarly

∂R/∂R2=R2/R2

2and ∂R/∂R3=R2/R2

3so dR

R=(R/R1)dR1

+(R/R2)dR2

+(R/R3)dR3



R≤(R/R1)

dR1

R1

+(R/R2)

dR2

R2

+(R/R3)

dR3

R3

≤(R/R1)(0.10)+(R/R2)(0.10)+(R/R3)(0.10)

=R(1/R1+1/R2+1/R3)(0.10) = (1)(0.10)=0.10 = 10%

47. dA =1

2bsin θda +1

2asin θdb +1

2ab cos θdθ,

|dA|≤ 1

2bsin θ|da|+1

2asin θ|db|+1

2ab cos θ|dθ|

≤1

2(50)(1/2)(1/2) + 1

2(40)(1/2)(1/4) + 1

2(40)(50) √3/2(π/90)

=35/4+50π

√

3/9≈39 ft2

48. V=`wh,dV =whd` +`hdw +`wdh,|dV/V |≤|d`/`|+|dw/w|+|dh/h|≤3(r/100) = 3r%

49. ∂f/∂v =8vw3x4y5,∂f/∂w =12v

,∂f/∂x =16v

,∂f/∂y =20v

50. ∂w/∂r = cos st +ueucos ur,∂w/∂s =−rt sin st,

∂w/∂t =−rs sin st,∂w/∂u =reucos ur +eusin ur

51. ∂f/∂v1=2v

/v

3+v

,∂f/∂v2=−2v2/v2

3+v2

4,∂f/∂v3=−2v3v2

1−v2

2/v2

3+v2

42,

∂f/∂v4=−2v4v2

1−v2

2/v2

3+v2

42

52. ∂V

∂x =2xe2x−y+e2x−y,∂V

∂y =−xe2x−y+w,∂V

∂z =w2ezw,∂V

∂w =wzezw +ezw +y

557 Chapter 15

53. (a) 0(b) 0(c) 0(d) 0

(e) 2(yw +1)e

yw sin zcos z(f) 2xw(yw +2)e

yw sin zcos z

54. 128, −512, 32, 64/3

55. ∂z/∂r =(dz/dx)(∂x/∂r)=2rcos2θ/ r2cos2θ+1

,

∂z/∂θ =(dz/dx)(∂x/∂θ)=−2r

2sin θcos θ/ r2cos2θ+1



56. ∂u/∂x =(∂u/∂r)(dr/dx)+(∂u/∂t)(∂t/∂x)

=s2ln t(2x)+rs2/ty

3=x(4y+1)

21+2lnxy3

∂u/∂y =(∂u/∂s)(ds/dy)+(∂u/∂t)(∂t/∂y)

=(2rs ln t)(4) + rs2/t3xy2=8x

(4y+1)lnxy3+3x

(4y+1)

57. ∂w/∂ρ =2ρ4 sin2φ+ cos2φ,∂w/∂φ =6ρ

2sin φcos φ,∂w/∂θ =0

58. dw

dx =∂w

∂x +∂w

∂y

dx +∂w

∂z

dx =3y

3+(6xyz3)(6x)+9xy2z21

2√x−1

= 3(3x2+2)

(x−1)3/2+36x

(3x2+ 2)(x−1)3/2+9

2x(3x2+2)

√

x−1

2(3x2+ 2)(39x3−30x2+10x−4)√x−1

59. (a) V=`wh, dV

dt =∂V

∂`

dt +∂V

∂w

dt +∂V

∂h

dt =whd`

dt +`hdw

dt +`w dh

= (3)(6)(1) + (2)(6)(2) + (2)(3)(3) = 60 in3/s

(b) D=p`2+w2+h2;dD/dt =(`/D)d`/dt +(w/D)dw/dt +(h/D)dh/dt

=(2/7)(1) + (3/7)(2) + (6/7)(3) = 26/7 in/s

60. (a) ∂A/∂a =(1/2)bsin θ=(1/2)(10) √3/2=5

√

3/2

(b) ∂A/∂θ =(1/2)ab cos θ=(1/2)(5)(10)(1/2)=25/2

61. Let z=f(u) where u=x+2y; then ∂z/∂x =(dz/du)(∂u/∂x)=dz/du,

∂z/∂y =(dz/du)(∂u/∂y)=2dz/du so 2∂z/∂x −∂z/∂y =2dz/du −2dz/du =0

62. Let z=f(u) where u=x2+y2; then ∂z/∂x =(dz/du)(∂u/∂x)=2x dz/du,

∂z/∂y =(dz/du)(∂u/∂y)=2ydz/du so y∂z/∂x−x∂z/∂y =2xydz/du −2xydz/du =0

63. ∂w/∂x =(dw/dρ)(∂ρ/∂x)=(x/ρ)dw/dρ, similarly ∂w/∂y =(y/ρ)dw/dρ and

∂w/∂z =(z/ρ)dw/dρ so (∂w/∂x)2+(∂w/∂y)2+(∂w/∂z)2=(dw/dρ)2

64. Let w=f(r, s, t) where r=x−y,s=y−z,t=z−x;

∂w/∂x =(∂w/∂r)(∂r/∂x)+(∂w/∂t)(∂t/∂x)=∂w/∂r −∂w/∂t, similarly

∂w/∂y =−∂w/∂r +∂w/∂s and ∂w/∂z =−∂w/∂s +∂w/∂t so ∂w/∂x +∂w/∂y +∂w/∂z =0

65. ∂w/∂ρ = sin φcos θ∂w/∂x + sin φsin θ∂w/∂y + cos φ∂w/∂z

∂w/∂φ =ρcos φcos θ∂w/∂x +ρcos φsin θ∂w/∂y −ρsin φ∂w/∂z

∂w/∂θ =−ρsin φsin θ∂w/∂x +ρsin φcos θ∂w/∂y

Exercise Set 15.7 558

66. ∂F

∂x +∂F

∂z

∂x =0so ∂z

∂x =−∂F/∂x

∂F/∂z ,∂F

∂y +∂F

∂z

∂y =0so ∂z

∂y =−∂F/∂y

∂F/∂z .

67. ∂z

∂x =2x+yz

6yz −xy ,∂z

∂y =xz −3z2

6yz −xy

68. ln(1 + z)+xy2+z−1=0; ∂z

∂x =−y2(1+z)

2+z,∂z

∂y =−2xy(1+z)

2+z

69. yex−5 sin 3z−3z=0; ∂z

∂x =−yex

−15 cos 3z−3=yex

15 cos 3z+3,∂z

∂y =ex

15 cos 3z+3

70. ∂z

∂x =−zeyz cos xz −yexy cos yz

yexy sin yz +xeyz cos xz +yeyz sin xz ,∂z

∂y =−zexy sin yz −xexy cos yz +zeyz sin xz

yexy sin yz +xeyz cos xz +yeyz sin xz

71. ∇f(u, v, w)= ∂f

∂xi+∂f

∂y j+∂f

∂z k

=∂f

∂u

∂x +∂f

∂v

∂x +∂f

∂w

∂x i+∂f

∂u

∂y +∂f

∂v

∂y +∂f

∂w

∂y j

+∂f

∂u

∂z +∂f

∂v

∂z +∂f

∂w

∂z k=∂f

∂u∇u+∂f

∂v ∇v+∂f

∂w∇w

72. (a) ∂w

∂x =∂f

∂x +∂f

∂z

∂x (b) ∂w

∂y =∂f

∂y +∂f

∂z

∂y

73. wr=er/(er+es+et+eu), wrs =−eres/(er+es+et+eu)2,

wrst =2e

/(e

r+e

s+e

t+e

)

rstu =−6ereseteu/(er+es+et+eu)4=−6er+s+t+u/e4w=−6er+s+t+u−4w

74. ∂w/∂y1=a1∂w/∂x1+a2∂w/∂x2+a3∂w/∂x3,

∂w/∂y2=b1∂w/∂x1+b2∂w/∂x2+b3∂w/∂x3

75. (a) dw/dt =

i=1

(∂w/∂xi)(dxi/dt)

(b) ∂w/∂vj=

i=1

(∂w/∂xi)(∂xi/∂vj) for j=1,2,3

76. Let u=x2

1+x2

2+...+x

; then w=uk,∂w/∂xi=kuk−1(2xi)=2kx

k−1

∂

w/∂x2

i=2k(k−1)xiuk−2(2xi)+2kuk−1=4k(k−1)x2

iuk−2+2kuk−1for i=1,2,...,n

i=1

∂2w/∂x2

i=4k(k−1) uk−2

i=1

i+2kn uk−1

=4k(k−1)uk−2u+2kn uk−1=2kuk−1[2(k−1)+n]

which is 0 if k= 0 or if 2(k−1)+n=0,k=1−n/2.

77. dF/dx =(∂F/∂u)(du/dx)+(∂F/∂v)(dv/dx)

=f(u)g0(x)−f(v)h0(x)=f(g(x))g0(x)−f(h(x))h0(x)

559 Chapter 15

78. ∇f=fxi+fyj+fzkand ∇g=gxi+gyj+gzkevaluated at (x0,y

0,z

0) are normal, respectively,

to the surfaces f(x, y, z)=0andg(x, y, z)=0at(x

0,z

0). The surfaces are orthogonal at

(x0,y

0,z

0) if and only if ∇f·∇g=0sof

x+f

y+f

z=0.

79. f(x, y, z)=x

2+y

2+z

2−a

2=0,g(x, y, z)=z

2−x

2−y

2=0,

x+f

y+f

z=−4x

2−4y

2+4z

2=4g(x, y, z)=0

EXERCISE SET 15.8

1. (a) minimum at (2,−1), no maxima (b) maximum at (0,0), no minima

2. (a) maximum at (−1,5), no minima (b) no maxima or minima

3. f(x, y)=(x−3)2+(y+2)

, minimum at (3,−2), no maxima

4. f(x, y)=−(x+1)

2−2(y−1)2+ 4, maximum at (−1,1), no minima

5. fx=6x+2y=0,f

y=2x+2y= 0; critical point (0,0); D=8>0 and fxx =6>0 at (0,0),

relative minimum.

6. fx=3x

2−3y=0,f

y=−3x−3y

2= 0; critical points (0,0) and (−1,1); D=−9<0 at (0,0),

saddle point; D=27>0 and fxx =−6<0at(−1,1), relative maximum.

7. fx=2x−2xy =0,f

y=4y−x

2= 0; critical points (0,0) and (±2,1); D=8>0 and fxx =2>0

at (0,0), relative minimum; D=−16 <0at(±2,1), saddle points.

8. fx=3x

2−3=0,f

y=3y

2−3 = 0; critical points (−1,±1) and (1,±1); D=−36 <0at(−1,1)

and (1,−1), saddle points; D=36>0 and fxx =6>0 at (1,1), relative minimum; D=36>0

and fxx =−36 <0at(−1,−1), relative maximum.

9. fx=y+2=0,f

y=2y+x+ 3 = 0; critical point (1,−2); D=−1<0at(1,−2), saddle point.

10. fx=2x+y−2=0,f

y=x−2 = 0; critical point (2,−2); D=−1<0at(2,−2), saddle point.

11. fx=2x+y−3=0,f

y=x+2y= 0; critical point (2,−1); D=3>0 and fxx =2>0at(2,−1),

relative minimum.

12. fx=y−3x2=0,f

y=x−2y= 0; critical points (0,0) and (1/6,1/12); D=−1<0 at (0,0),

saddle point; D=1>0 and fxx =−1<0at(1/6,1/12), relative maximum.

13. fx=2x−2/

x

=0,f

y=2y−2/

xy2= 0; critical points (−1,−1) and (1,1); D=32>0

and fxx =6>0at(−1,−1) and (1,1), relative minima.

14. fx=ey= 0 is impossible, no critical points.

15. fx=2x=0,f

y=1−e

y= 0; critical point (0,0); D=−2<0at(0,0), saddle point.

16. fx=y−2/x2=0,f

y=x−4/y2= 0; critical point (1,2); D=3>0 and fxx =−4>0at(1,2),

relative minimum.

Exercise Set 15.8 560

17. fx=exsin y=0,f

y=e

xcos y= 0, sin y= cos y= 0 is impossible, no critical points.

18. fx=ycos x=0,f

y= sin x= 0; sin x=0ifx=nπ for n=0,±1,±2,... and cos x6= 0 for these

values of xso y= 0; critical points (nπ, 0) for n=0,±1,±2,...;D=−1<0at(nπ, 0), saddle

points.

19. fx=−2(x+1)e

−

(

+2x)=0,f

y=−2ye−(x2+y2+2x)= 0; critical point (−1,0); D=4e

2>0

and fxx =−2e<0at(−1,0), relative maximum.

20. fx=y−a3/x2=0,f

y=x−b

/y2= 0; critical point a2/b, b2/a;ifab>0 then D=3>0 and

fxx =2b

/a3>0ata

/b, b2/a, relative minimum; if ab < 0 then D=3>0 and

fxx =2b

/a3<0ata

/b, b2/a, relative maximum.

21.

-2 -1 0 1 2

-2

-1

∇f=(4x−4y)i−(4x−4y3)j=0when x=y, x =y3,sox=y=0orx=y=±1. At

(0,0),D =−16, a saddle point; at (1,1) and (−1,−1),D =32>0,f

xx = 4, a relative minimum.

22.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

5140

100

140

100

40 20

-20

-40

-60

-100

∇f=(2y

2−2xy +4y)i+(4xy −x2+4x)j=0when 2y2−2xy +4y=0,4xy −x2+4x= 0, with

solutions (0,0),(0,−2),(4,0),(4/3,−2/3). At (0,0), D=−16, a saddle point. At (0,−2),

D=−16, a saddle point. At (4,0),D =−16, a saddle point. At (4/3,−2/3),D =16/3,

fxx =4/3>0, a relative minimum.

23. (a) critical point (0,0); D=0

(b) f(0,0)=0,x

4+y

4≥0sof(x, y)≥f(0,0), relative minimum.

24. (a) critical point (0,0); D=0

(b) f(0,0) = 0, inside any circle centered at (0,0) there are points where f(x, y)>0 (along the

x-axis) and points where f(x, y)<0 (along the y-axis) so (0,0) is a saddle point.

561 Chapter 15

25. (a) fx=3e

y−3x

2=3e

y−x

=0,f

y=3xey−3e3y=3e

yx−e

=0,e

y=x

2and

e2y=x,x4=x,xx3−1=0sox=0,1; critical point (1,0); D=27>0 and fxx =−6<0

at (1,0), relative maximum.

(b) lim

x→−∞ f(x, 0) = lim

x→−∞ 3x−x3−1=+∞so no absolute maximum.

26. fx=8xey−8x3=8x(e

y−x

)=0,f

y=4x

y−4e

4y=4e

2−e

)=0,x

2=e

yand

x2=e3y,e3y=ey,e2y=1,soy= 0 and x=±1; critical points (1,0) and (−1,0). D= 128 >0

and fxx =−16 <0 at both points so a relative maximum occurs at each one.

27. fx=y−1=0,f

y=x−3 = 0; critical point (3,1).

Along y=0: u(x)=−x; no critical points,

along x=0: v(y)=−3y; no critical points,

along y=−4

5x+4: w(x)=−4

2+27

5x−12; critical point (27/8,13/10).

(x, y) (3,1) (0,0) (5,0) (0,4) (27/8,13/10)

f(x, y)−3 0 −5−12 −231/80

Absolute maximum value is 0, absolute minimum value is −12.

28. fx=y−2=0,f

y=x= 0; critical point (0,2), but (0,2) is not in the interior of R.

Along y=0: u(x)=−2x; no critical points,

along x=0: v(y) = 0; no critical points,

along y=4−x:w(x)=2x−x

; critical point (1,3).

(x, y) (0,0) (0,4) (4,0) (1,3)

f(x, y) 0 0 −8 1

Absolute maximum value is 1, absolute minimum value is −8.

29. fx=2x−2=0,f

y=−6y+ 6 = 0; critical point (1,1).

Along y=0: u

(x)=x

2−2x; critical point (1,0),

along y=2: u

(x)=x

2−2x; critical point (1,2)

along x=0: v

(y)=−3y

2+6y; critical point (0,1),

along x=2: v

(y)=−3y

2+6y; critical point (2,1)

(x, y) (1,1) (1,0) (1,2) (0,1) (2,1) (0,0) (0,2) (2,0) (2,2)

f(x, y) 2 −1−1330000

Absolute maximum value is 3, absolute minimum value is −1.

30. fx=ey−2x=0,f

y=xey−ey=ey(x−1) = 0; critical point (1,ln 2).

Along y=0: u

(x)=x−x

2−1; critical point (1/2,0),

along y=1: u

(x)=ex −x2−e; critical point (e/2,1),

along x=0: v

(y)=−e

; no critical points,

along x=2: v

(y)=e

y−4; no critical points (for 0 <y<1).

(x, y) (0,0) (0,1) (2,1) (2,0) (1,ln 2) (1/2,0) (e/2,1)

f(x, y)−1−e e −4−3−1−3/4e(e−4)/4≈−0.87

Absolute maximum value is −3/4, absolute minimum value is −3.

Exercise Set 15.8 562

31. fx=2x−1=0,f

y=4y= 0; critical point (1/2,0).

Along x2+y2=4:y

2=4−x

,u(x)=8−x−x

2for −2≤x≤2; critical points (−1/2,±√15/2).

(x, y) (1/2,0) −1/2,√15/2−1/2,−√15/2(−2,0) (2,0)

f(x, y)−1/4 33/4 33/4 6 2

Absolute maximum value is 33/4, absolute minimum value is −1/4.

32. fx=y2=0,f

y=2xy = 0; no critical points in the interior of R.

Along y=0: u(x) = 0; no critical points,

along x=0: v(y) = 0; no critical points

along x2+y2=1: w(x)=x−x

3for 0 ≤x≤1; critical point 1/√3,p2/3.

(x, y) (0,0) (0,1) (1,0) 1/√3,p2/3

f(x, y) 0 0 0 2√3/9

Absolute maximum value is 2

9√3, absolute minimum value is 0.

33. Maximize P=xyz subject to x+y+z= 48, x>0, y>0, z>0. z=48−x−yso

P=xy(48 −x−y)=48xy −x2y−xy2,Px=48y−2xy −y2=0,P

y=48x−x

2−2xy = 0. But

x6= 0 and y6= 0 so 48 −2x−y= 0 and 48 −x−2y= 0; critical point (16,16). PxxPyy −P2

xy >0

and Pxx <0 at (16,16), relative maximum. z= 16 when x=y= 16, the product is maximum for

the numbers 16,16,16.

34. Minimize S=x2+y2+z2subject to x+y+z= 27, x>0, y>0, z>0. z=27−x−yso

S=x2+y2+ (27 −x−y)2,Sx=4x+2y−54=0,S

y=2x+4y−54 = 0; critical point (9,9);

SxxSyy −S2

xy =12>0 and Sxx =4>0 at (9,9), relative minimum. z= 9 when x=y= 9, the

sum of the squares is minimum for the numbers 9,9,9.

35. Maximize w=xy2z2subject to x+y+z=5,x>0, y>0, z>0. x=5−y−zso

w=(5−y−z)y

2=5y

2−y

y=10yz2−3y2z2−2yz3=yz2(10 −3y−2z)=0,

z=10y

z−2y

z−3y

2=y

z(10−2y−3z)=0,10−3y−2z= 0 and 10−2y−3z= 0; critical

point when y=z=2;w

yywzz −w2

yz = 320 >0 and wyy =−24 <0 when y=z= 2, relative

maximum. x= 1 when y=z=2,xy2z2is maximum at (1,2,2).

36. Minimize w=D2=x2+y2+z2subject to x2−yz =5. x

2=5+yz so w=5+yz +y2+z2,

wy=z+2y=0,w

z=y+2z= 0; critical point when y=z=0;w

yywzz −w2

yz =3>0 and

wyy =2>0 when y=z= 0, relative minimum. x2=5,x=±

√

5 when y=z= 0. The points

±√5,0,0are closest to the origin.

37. The diagonal of the box must equal the diameter of the sphere, thus we maximize V=xyz or, for

convenience, w=V2=x2y2z2subject to x2+y2+z2=4a

,x>0,y>0,z>0; z2=4a

−x

−y

hence w=4a

2−x

x=2xy2(4a2−2x2−y2)=0,w

y=2x

y4a

2−x

2−2y

=0,

2−2x

2−y

2= 0 and 4a2−x2−2y2= 0; critical point 2a/√3,2a/√3;

wxxwyy −w2

xy =4096

27 a8>0 and wxx =−128

9a4<0at

2a/√3,2a/√3, relative maximum.

z=2a/√3 when x=y=2a/√3, the dimensions of the box of maximum volume are

2a/√3,2a/√3,2a/√3.

38. Maximize V=xyz subject to x+y+z=1,x>0, y>0, z>0. z=1−x−yso V=xy−x2y−xy2,

Vx=y(1 −2x−y)=0,V

y=x(1 −x−2y)=0,1−2x−y= 0 and 1 −x−2y= 0; critical point

(1/3,1/3); VxxVyy −V2

xy =1/3>0 and Vxx =−2/3<0at(1/3,1/3), relative maximum. The

maximum volume is V=(1/3)(1/3)(1/3)=1/27.

563 Chapter 15

39. Let x,y, and zbe, respectively, the length, width, and height of the box. Minimize

C= 10(2xy) + 5(2xz +2yz) = 10(2xy +xz +yz) subject to xyz = 16. z=16/(xy)

so C= 20(xy +8/y +8/x), Cx= 20(y−8/x2)=0,C

y= 20(x−8/y2)=0;

critical point (2,2); CxxCyy −C2

xy = 1200 >0

and Cxx =40>0 at (2,2), relative minimum. z= 4 when x=y= 2. The cost of materials is

minimum if the length and width are 2 ft and the height is 4 ft.

40. Maximize the proﬁt P= 500(y−x)(x−40) + [45,000 + 500(x−2y)](y−60)

= 500(−x2−2y2+2xy −20x+ 170y−5400).

Px= 1000(−x+y−10)=0,P

y= 1000(−2y+x+ 85) = 0; critical point (65,75);

PxxPyy −P2

xy =1,000,000 >0 and Pxx =−1000 <0 at (65,75), relative maximum. The proﬁt

will be maximum when x= 65 and y= 75.

41. (a) x=0:f(0,y)=−3y

, minimum −3, maximum 0;

x=1,f(1,y)=4−3y

2+2y, ∂f

∂y (1,y)=−6y+2=0aty=1/3, minimum 3,

maximum 13/3;

y=0,f(x, 0)=4x

, minimum 0, maximum 4;

y=1,f(x, 1)=4x

2+2x−3,∂f

∂x(x, 1)=8x+2 6= 0 for 0 <x<1, minimum −3, maximum 3

(b) f(x, x)=3x

, minimum 0, maximum 3; f(x, 1−x)=−x

+8x−3,d

dxf(x, 1−x)=−2x+8 6=0

for 0 <x<1, maximum 4, minimum −3

y(x, y)=−3y+2x= 0, solution is (0,0), which is not an interior

point of the square, so check the sides: minimum −3, maximum 14/3.

42. Maximize A=ab sin αsubject to 2a+2b=`,a>0, b>0, 0 <α<π.b=(`−2a)/2so

A=(1/2)(`a −2a2) sin α,Aa=(1/2)(`−4a) sin α,Aα=(a/2)(`−2a) cos α; sin α6= 0 so from

Aa= 0 we get a=`/4 and then from Aα= 0 we get cosα=0,α=π/2. AaaAαα −A2

aα =`2/8>0

and Aaa =−2<0 when a=`/4 and α=π/2, the area is maximum.

43. Minimize S=xy +2xz +2yz subject to xyz =V,x>0, y>0, z>0 where x,y, and zare,

respectively, the length, width, and height of the box. z=V/(xy)soS=xy +2V/y +2V/x,

Sx=y−2V/x2=0,S

y=x−2V/y2= 0; critical point ( 3

√2V, 3

√

2V); SxxSyy −S2

xy =3>0 and

Sxx =2>0 at this point so there is a relative minimum there. The length and width are each

√2V, the height is z=3

√2V/2.

44. The altitude of the trapezoid is xsin φand the lengths of the lower and upper bases are, respectively,

27 −2xand 27 −2x+2xcos φso we want to maximize

A=(1/2)(xsin φ)[(27 −2x)+(27−2x+2xcos φ)]=27xsin φ−2x2sin φ+x2sin φcos φ.

Ax= sin φ(27 −4x+2xcos φ),

Aφ=x(27 cos φ−2xcos φ−xsin2φ+xcos2φ)=x(27 cos φ−2xcos φ+2xcos2φ−x).

sin φ6= 0 so from Ax= 0 we get cosφ=(4x−27)/(2x), x6= 0 so from Aφ= 0 we get

(27 −2x+2xcos φ) cos φ−x= 0 which, for cos φ=(4x−27)/(2x), yields 4x−27 −x=0,

x=9. Ifx= 9 then cosφ=1/2, φ=π/3. The critical point occurs when x= 9 and φ=π/3;

AxxAφφ −A2

xφ = 729/2>0 and Axx =−3√3/2<0 there, the area is maximum when x= 9 and

φ=π/3.

Exercise Set 15.8 564

45. (a) ∂g

∂m =

i=1

2(mxi+b−yi)xi=2 m

i=1

i+b

i=1

xi−

i=1

xiyi!=0if

i=1

i!m+ n

i=1

xi!b=

i=1

xiyi,

∂g

∂b =

i=1

2(mxi+b−yi)=2 m

i=1

xi+bn −

i=1

yi!=0if n

i=1

xi!m+nb =

i=1

(b) The function z=g(m, b), as a function of mand b, has only one critical point, found in part

(a), and tends to +∞as either |m|or |b|tends to inﬁnity. Thus the only critical point must

be a minimum.

46. (a)

i=1

(xi−¯x)2=

i=1 x2

i−2¯xxi+¯x

=

i=1

i−2¯x

i=1

xi+n¯x2

i=1

i−2

n n

i=1

xi!2

n n

i=1

xi!2

i=1

i−1

n n

i=1

xi!2

>0son

i=1

i− n

i=1

xi!2

(b) gmm =2

i=1

i,g

bb =2n, gmb =2

i=1

xi,

D=gmmgbb −g2

mb =4

n

i=1

i− n

i=1

xi!2

>0 and gmm >0

(d) The only quadric surface of this form having a relative minimum is a paraboloid that opens

upward where the relative minimum is also the absolute minimum.

47. n=3,

i=1

xi=3,

i=1

yi=7,

i=1

xiyi=13,

i=1

i=11,y =3

4x+19

48. n=4,

i=1

xi=7,

i=1

yi=4,

i=1

i=21,

i=1

xiyi=−2,y =−36

35x+14

49.

i=1

xi=10,

i=1

yi=8.2,

i=1

i=30,

i=1

xiyi=23,n=4; m=0.5,b=0.8,y=0.5x+0.8.

50.

i=1

xi=15,

i=1

yi=15.1,

i=1

i=55,

i=1

xiyi=39.8,n=5;m=−0.55,b =4.67,y =4.67−0.55x

565 Chapter 15

51. (a) y=8843

140 +57

200t≈63.1643 + 0.285t(b)

600(1930)

35 ≈83.1143

52. (a) y≈119.84 −1.13x(b) 90

35 50

53. (a) P=2798

21 +171

350T≈133.2381 + 0.4886T(b)

1200

130

190

513 =≈−272.7096◦C

54. (a) for example, z=y

(b) For example, on 0 ≤x≤1,0≤y≤1 let z=(yif 0 <x<1,0<y<1

1/2ifx=0,1ory=0,1

55. f(x0,y

0)≥f(x, y) for all (x, y) inside a circle centered at (x0,y

0) by virtue of Deﬁnition 15.8.1.

If ris the radius of the circle, then in particular f(x0,y

0)≥f(x, y0) for all xsatisfying

|x−x0|<rso f(x, y0) has a relative maximum at x0. The proof is similar for the function

f(x0,y).

EXERCISE SET 15.9

1. (a) xy = 4 is tangent to the line, so the maximum value of fis 4.

(b) xy = 2 intersects the curve and so gives a smaller value of f.

yi+xj=λ(i+j), so solve the equations y=λ, x =λwith solution x=y=λ, but x+y=4,

so x=y= 2, and the maximum value of fis f=xy =4.

Exercise Set 15.9 566

2. (a) x2+y2= 25 is tangent to the line at (3,4), so the minimum value of fis 25.

(b) A smaller value of fyields a circle of a smaller radius, and hence does not intersect the line.

2+y

2subject to the constraint g(x, y)=3x+4y−25=0,∇f=λ∇g,

2xi+2yj=3λi+4λj, so solve 2x=3λ, 2y=4λ; solution is x=3,y = 4, minimum = 25.

3. y=8xλ,x=16yλ;y/(8x)=x/(16y), x2=2y

2so 4 2y2+8y

2= 16, y2=1,y=±1. Test

±√2,−1and (±√2,1). f−√2,−1=f√2,1=√2, f−√2,1=f√2,−1=−√2.

Maximum √2at−

√

2,−1

and √2,1, minimum −√2at−

√

2,1

and √2,−1.

4. 2x=2xλ, −1=2yλ.Ifx6= 0 then λ= 1 and y=−1/2sox

2+(−1/2)2= 25,

x2=99/4, x=±3√11/2. If x= 0 then 02+y2= 25, y=±5. Test ±3√11/2,−1/2and (0,±5).

f±3√11/2,−1/2= 101/4, f(0,−5)=5,f(0,5) = −5. Maximum 101/4at±3

√

11/2,−1/2,

minimum −5 at (0,5).

5. 12x2=4xλ,2y=2yλ.Ify6= 0 then λ= 1 and 12x2=4x,12x(x−1/3)=0,x=0orx=1/3

so from 2x2+y2= 1 we ﬁnd that y=±1 when x=0,y=±

√

7/3 when x=1/3. If y=0

then 2x2+ (0)2=1,x=±1/

√

2. Test (0,±1), 1/3,±√7/3, and ±1/√2,0.f(0,±1)=1,



1/3,±

√

7/3

=25/27, f1/√2,0=√2, f−1/√2,0=−√2. Maximum √2at

√

2,0



minimum −√2at−1/

√

2,0

.

6. 1=2xλ,−3=6yλ;1/(2x)=−1/(2y), y=−xso x2+3(−x)

2= 16, x=±2. Test (−2,2) and

(2,−2). f(−2,2) = −9, f(2,−2) = 7. Maximum 7 at (2,−2), minimum −9at(−2,2).

7. 2=2xλ, 1=2yλ, −2=2zλ;1/x =1/(2y)=−1/z thus x=2y,z=−2yso

(2y)2+y2+(−2y)

2=4,y

2=4/9, y=±2/3. Test (−4/3,−2/3,4/3) and (4/3,2/3,−4/3).

f(−4/3,−2/3,4/3) = −6, f(4/3,2/3,−4/3) = 6. Maximum 6 at (4/3,2/3,−4/3), minimum −6

at (−4/3,−2/3,4/3).

8. 3=4xλ,6=8yλ,2=2zλ;3/(4x)=3/(4y)=1/z thus y=x,z=4x/3, so

2x2+4(x)

2+(4x/3)2= 70, x2=9,x=±3. Test (−3,−3,−4) and (3,3,4).

f(−3,−3,−4) = −35, f(3,3,4) = 35. Maximum 35 at (3, 3, 4), minimum −35 at (−3,−3,−4).

9. yz =2xλ,xz =2yλ,xy =2zλ;yz/(2x)=xz/(2y)=xy/(2z)thusy

2=x

2so

x2+x2+x2=1,x=±1/

√

3. Test the eight possibilities with x=±1/√3, y=±1/√3, and

z=±1/√3 to ﬁnd the maximum is 1/3√3at 1/√3,1/√3,1/√3,1/√3,−1/√3,−1/√3,

−1/√3,1/√3,−1/√3, and −1/√3,−1/√3,1/√3; the minimum is −1/3√3at

1/√3,1/√3,−1/√3,1/√3,−1/√3,1/√3,−1/√3,1/√3,1/√3, and −1/√3,−1/√3,−1/√3.

10. 4x3=λ, 4y3=λ, 4z3=λ,sox=y=z;x+y+z=3x=1,x =y=z=1/3, minimum value

1/27, no maximum

11. f(x, y)=x

2+y

;2x=2λ,2y=−4λ;y=−2xso 2x−4(−2x)=3,x=3/10. The point is

(3/10,−3/5).

12. f(x, y)=(x−4)2+(y−2)2,g(x, y)=y−2x−3; 2(x−4) = −2λ,2(y−2) = λ;x−4=−2(y−2),

x=−2y+8soy=2(−2y+8)+3,y=19/5. The point is (2/5,19/5).

13. f(x, y, z)=x

2+y

2+z

;2x=λ,2y=2λ,2z=λ;y=2x,z=xso x+ 2(2x)+x=1,x=1/6.

The point is (1/6,1/3,1/6).

14. f(x, y, z)=(x−1)2+(y+1)

2+(z−1)2;2(x−1)=4λ,2(y+1)=3λ,2(z−1) = λ;x=4z−3,

y=3z−4so4(4z−3) + 3(3z−4)+z=2,z= 1. The point is (1,−1,1).

567 Chapter 15

15. f(x, y)=(x−1)2+(y−2)2;2(x−1)=2xλ,2(y−2)=2yλ;(x−1)/x =(y−2)/y,y=2x

so x2+(2x)

2= 45, x=±3. f(−3,−6) = 80 and f(3,6) = 20 so (3,6) is closest and (−3,−6) is

farthest.

16. f(x, y, z)=x

2+y

2+z

;2x=yλ,2y=xλ,2z=−2zλ.Ifz6= 0 then λ=−1so2x=−yand

2y=−x,x=y= 0; substitute into xy −z2= 1 to get z2=−1 which has no real solution. If

z= 0 then xy −(0)2=1,y=1/x, and also (from 2x=yλ and 2y=xλ), 2x/y =2y/x,y2=x2

so (1/x)2=x2,x4=1,x=±1. Test (1,1,0) and (−1,−1,0) to see that they are both closest to

the origin.

17. f(x, y, z)=x+y+z,x

2+y

2+z

2= 25 where x,y, and zare the components of the vector;

1=2xλ,1=2yλ,1=2zλ;1/(2x)=1/(2y)=1/(2z); y=x,z=xso x2+x2+x2= 25,

x=±5/√3. f−5/√3,−5/√3,−5/√3=−5√3 and f5/√3,5/√3,5/√3=5

√

3 so the vector

is 5(i+j+k)/√3.

18. x2+y2= 25 is the constraint; 8x−4y=2xλ,−4x+2y=2yλ;(4x−2y)/x =(−2x+y)/y,

2x2+3xy −2y2=0,(2x−y)(x+2y)=0,y=2xor x=−2y.Ify=2xthen x2+(2x)

2= 25,

x=±√5. If x=−2ythen −2y2+y2= 25, y=±√5. T−√5,−2√5=T√5,2√5= 0 and

T2√5,−√5=T−2√5,√5= 125. The highest temperature is 125 and the lowest is 0.

19. (a)

31.5

-31.5

-27

15 (b) minimum value −5,

maximum value 101/4

2−ysubject to the constraint

g(x, y)=x

2+y

2−25=0,∇f=λ∇g, 2xi−j=2λxi+2λyj, so solve 2x=2λx, −1=2λy.

If x= 0 then y=±5,f =∓5, and if x6= 0 then λ=1,y =−1/2,x

2=25−1/4=99/4,

f=99/4+1/2 = 101/4, so the maximum value of fis 101/4at(±3

√

11/2,−1/2) and the

minimum value of fis −5at(0,5).

20. (a)

0 1 2 3 4 5 6

6(b) f≈15

(d) Set f(x, y)=x

3+y

3−3xy, g(x, y)=(x−4)2+(y−4)2−25; minimize fsubject to the

constraint g=0: ∇f=λg, (3x2−3y)i+(3y

2−3x)j=2λ(x−4)i+2λ(y−4)j, so solve

(use a CAS) 3x2−3y=2λ(x−4),3y2−3x=2λ(y−4); minimum value f=14.52 at

(2.5858,2.5858)

21. Minimize f=x2+y2+z2subject to g(x, y, z)=x+y+z−27=0. ∇f=λ∇g,

2xi+2yj+2zk=λi+λj+λk, solution x=y=z= 9, minimum value 243

Exercise Set 15.9 568

22. Maximize f(x, y, z)=xy2z2subject to g(x, y, z)=x+y+z−5=0,∇f=λ∇g=λ(i+j+k),

λ=y

2=2xyz2=2xy2z, λ = 0 is impossible, hence x, y, z 6= 0, and z=y=2x, 5x−5=0,

x=1,y =z= 2, maximum value 8 at (1,2,2)

23. Minimize f=x2+y2+z2subject to x2−yz =5,∇f=λ∇g, 2x=2xλ, 2y=−zλ,2z=−yλ.

If λ6=±2, then y=z=0,x =±√5,f =5;ifλ=±2 then x= 0, and since −yz =5,

y=−z=±

√

5,f = 10, thus the minimum value is 5 at (±√5,0,0).

24. The diagonal of the box must equal the diameter of the sphere so maximize V=xyz or, for

convenience, maximize f=V2=x2y2z2subject to g(x, y, z)=x

2+y

2+z

2−4a

2=0,∇f=λ∇g,

2xy2z2=2λx, 2x2yz2=2λy, 2x2y2z=2λz. Since V6= 0 it follows that x, y, z 6= 0, hence

x=±y=±z,3x2=4a

,x=±2a/√3, maximum volume 8a3/(3√3).

25. Let x,y, and zbe, respectively, the length, width, and height of the box. Minimize

f(x, y, z) = 10(2xy) + 5(2xz +2yz) = 10(2xy +xz +yz) subject to g(x, y, z)=xyz −16=0,

∇f=λ∇g, 20y+10z=λyz, 20x+10z=λxz, 10x+10y=λz. Since V=xyz =16,x,y,z 6=0,

thus λz = 20 + 10(z/y)=20+10(z/x)=10x+10y,sox=y. Then z=16/x2,thus

20 + 10(16/x3)=20x, x3+8=x

, the only real solution of which is x= 2,thus x=y=2,z =4,

minimum value 240.

26. Minimize f(p, q, r)=2pq +2pr+2qr, subject to g(p, q, r)=p+q+r−1=0,∇

(p,q,r)f=λ∇(p,q,r)g,

2(q+r)=λ, 2(p+r)=λ, 2(p+q)=λ, solution p=q=r=1/3, minimum value 2/3.

27. Maximize A(a, b, α)=ab sin αsubject to g(a, b, α)=2a+2b−`=0,∇

(a,b,α)f=λ∇(a,b,α)g,

bsin α=2λ, a sin α=2λ, ab cos α= 0 with solution a=b, α =π/2 maximum value if parallelogram

is a square.

28. Minimize f(x, y, z)=xy +2xz +2yz subject to g(x, y, z)=xyz −V=0,∇f=λ∇g,

y+2z=λyz, x +2z=λxz, 2x+2y=λxy;λ= 0 leads to x=y=z= 0, impossible, so solve for

λ=1/z +2/x =1/z +2/y =2/y +2/x,sox=y=2z, x3=2V, minimum value 3(2V)2/3

29. (a) Maximize f(α, β, γ) = cos αcos βcos γsubject to g(α, β, γ)=α+β+γ−π=0,

∇f=λ∇g, −sin αcos βcos γ=λ, −cos αsin βcos γ=λ, −cos αcos βsin γ=λwith solution

α=β=γ=π/3, maximum value 3√3/8

(b) for example, f(α, β) = cos αcos βcos(π−α−β)

30. Find maxima and minima z=x2+4y

2subject to the constraint g(x, y)=x

2+y

2−1=0,

∇z=λ∇g, 2xi+8yj=2λxi+2λyj, solve 2x=2λx, 8y=2λy.Ify6= 0 then λ=4,x =0,y

2=1

and z=x2+4y

2=4. Ify= 0 then x2= 1 and z= 1, so the maximum height is obtained for

(x, y)=(0,±1),z = 4 and the minimum height is z= 1 at (±1,0).

569 Chapter 15

CHAPTER 15 SUPPLEMENTARY EXERCISES

1. (a) They approximate the proﬁt per unit of any additional sales of the standard or high-resolution

monitors, respectively.

(b) The rates of change with respect to the two directions xand y, and with respect to time.

3. z=px2+y2=cimplies x2+y2=c2, which is the equation of a circle; x2+y2=cis also the

equation of a circle (for c>0).

-3 3

-3

z = x2 + y2

-3 3

-3

z = √x2 + y2

5. (b) f(x, y, z)=z−x

2−y

7. (a) f(ln y, ex)=e

ln yln ex=xy (b) er+sln rs

8. (a)

y = 1

(b)

–1 1

9. wx=2xsec2(x2+y2)+√

y, wxy =8xy sec2(x2+y2) tan(x2+y2)+1

−1/2

y=2ysec2(x2+y2)+1

2xy−1/2,w

yx =8xy sec2(x2+y2) tan(x2+y2)+1

−1/2

10. ∂w/∂x =1

x−y−sin(x+y),∂2w/∂x2=−1

(x−y)2−cos(x+y),

∂w/∂y =−1

x−y−sin(x+y),∂

w/∂y2=−1

(x−y)2−cos(x+y)=∂

w/∂x2

11. Fx=−6xz, Fxx =−6z,Fy=−6yz, Fyy =−6z,Fz=6z

2−3x

2−3y

zz =12z,Fxx +Fyy +Fzz =−6z−6z+12z=0

12. fx=yz +2x, fxy =z, fxyz =1,f

xyzx =0; f

z=xy −(1/z),f

zx =y, fzxx =0,f

zxxy =0

Chapter 15 Supplementary Exercises 570

13. (a) P=10T

V,dP

dT =10

dt = 12 K/(m2min) (b) dP

dt =−10 P

dt = 240 K/(m2min)

14. (a) z=1−y

, slope = ∂z

∂y =−2y=4 (b) z=1−4x

,∂z

∂x =−8x=−8

15. x4−x+y−x3y=(x

3−1)(x−y), limit = −1, continuous

16. x4−y4

x2+y2=x2−y2, limit = lim

(x,y)→(0,0)(x2−y2) = 0, continuous

17. Use the unit vectors u=h1

√2,1

√2i,v=h0,−1i,w=h− 1

√5,−2

√5i=−√2

√5u+1

√5v, so that

Dwf=−√2

√5Duf+1

√5Dvf=−√2

√52√2+ 1

√

5(−3) = −7

√5

18. (a) n =zxi+zyj−k=8i+8j−k, tangent plane 8x+8y−z= 4 + 8 ln 2, normal line

x(t)=1+8t, y(t)=ln2+8t, z(t)=4−t

(b) n =3i+10j−14k, tangent plane 3x+10y−14z= 30, normal line

x(t)=2+3t, y(t)=1+10t, z(t)=−1−14t

19. The origin is not such a point, so assume that the normal line at (x0,y

0,z

0)6=(0,0,0) passes

through the origin, then n=zxi+zyj−k=−y0i−x0j−k; the line passes through the origin and

is normal to the surface if it has the form r(t)=−y

ti−x

tj−tkand (x0,y

0,z

0)=(x

0,2−x

lies on the line if −y0t=x0,−x0t=y0,−t=2−x

, with solutions x0=y0=−1,

x0=y0=1,x

0=y

0= 0; thus the points are (0,0,2),(1,1,1),(−1,−1,1).

20. n =2

3x−1/3

0i+2

3y−1/3

0j+2

3z−1/3

0k, tangent plane x−1/3

0x+y−1/3

0y+z−1/3

0z=x2/3

0+y2/3

0+z2/3

0=1;

intercepts are x=x1/3

0,y =y

1/3

0,z =z1/3

0, sum of squares of intercepts is x2/3

0+y2/3

0+z2/3

0=1.

21. A tangent to the line is 6i+4j+k, a normal to the surface is n=18xi+8yj−k, so solve

18x=6k, 8y=4k, −1=k;k=−1,x =−1/3,y =−1/2,z =2

22. ∆w=(1.1)2(−0.1) −2(1.1)(−0.1)+(−0.1)2(1.1) −0=0.11,

dw =(2xy −2y+y2)dx +(x

2−2x+2yx)dy =−(−0.1)=0.1

23. dV =2

3xhdx +1

3x2dh =2

32(−0.1) + 1

3(0.2) = −0.06667 m3;∆V=−0.07267 m3

24. ∇f=(2x+3y−6)i+(3x+6y+3)j=0if 2x+3y=6,x+2y=−1,x=15,y =−8,D =3>0,

xx =2>0, so fhas a relative minimum at (15,−8).

25. ∇f=(2xy −6x)i+(x

2−12y)j=0if 2xy −6x=0,x

2−12y=0;ifx= 0 then y= 0, and

if x6= 0 then y=3,x =±6, thus the gradient vanishes at (0,0),(−6,3),(6,3); fxx = 0 at all

three points, fyy =−12 <0,D =−4x

2,so(±6,3) are saddle points, and near the origin we write

f(x, y)=(y−3)x2−6y2; since y−3<0 when |y|<3, fhas a maximum by inspection.

26. ∇f=(3x

2−3y)i−(3x−y)j=0if y=x2,3x=y,sox=y=0orx=3,y =9;at

x=y=0,D =−9, saddle point; at x=3,y =9,D =9,f

xx =18>0, relative minimum

571 Chapter 15

27. (a)

P = 1

P = 3

P = 2

12345

K(b)

012345

28. (a) ∂P/∂L =cαLα−1Kβ,∂P/∂K =cβLαKβ−1

(b) the rates of change of output with respect to labor and capital equipment, respectively

29. (a) L+K= 200,000,P = 1000L0.6(200,000 −L)0.4,

dP/dL = 600(200,000 −L)0.4

L4−400 L6

(200,000 −L)0.6= 0 when L= 120,000,

P= 102,033,960.1, which is a maximum because P=0atL=0,200,000, P>0 in between,

and dP/dL = 0 has only the one solution.

(b) Since L+K= 200,000 and L= 120,000,K =80,000

30. (a) y2=8−4x

, ﬁnd extrema of f(x)=x

(8 −4x2)=−4x

4+8x

2deﬁned for −√2≤x≤√2.

Then f0(x)=−16x3+16x= 0 when x=0,±1,f00(x)=−48x2+ 16, so fhas a relative

maximum at x=±1,y =±2 and a relative minimum at x=0,y =±2

√2. At the endpoints

x=±√2,y = 0 we obtain the minimum f= 0 again.

(b) f(x, y)=x

,g(x, y)=4x

2+y

2−8=0,∇f=2xy2i+2x

yj=λ∇g=8λxi+2λyj,so

solve 2xy2=λ8x, 2x2y=λ2y.Ifx= 0 then y=±2√2, and if y= 0 then x=±√2. In

either case fhas a relative and absolute minimum. Assume x, y 6= 0, then y2=4λ, x2=λ,

use g= 0 to obtain x2=1,x =±1,y =±2, and f= 4 is a relative and absolute maximum

at (±1,±2).

31. Let a corner of the box be at (x, y, z), so that (x/a)2+(y/b)2+(z/c)2= 1. Maximize V=xyz

subject to g(x, y, z)=(x/a)2+(y/b)2+(z/c)2=1,solve ∇V=λ∇g,or

yzi+xzj+xyk=(2λx/a2)i+(2λy/b2)j+(2λz/c2)k,a

2yz =2λx, b2xz =2λy, c2xy =2λz. For the

maximum volume, x, y, z 6= 0; divide the ﬁrst and second equations to obtain a2y2=b2x2; the ﬁrst

and third to obtain a2z2=c2x2, and ﬁnally b2z2=c2y2. From g= 0 get 3(x/a)2=1,x =±a/√3,

and then y=±b/√3,z =±c/√3. The dimensions of the box are 2a

√3×2b

√3×2c

√3, and the

maximum volume is 8abc/(3√3).

32. (a) Let f(x, y)=3x

2−5xy + tan xy = 0. Then

dx =6x−5y+ysec2xy +(−5x+xsec2xy)dy

dx =0,so dy

dx =6x−5y+ysec2xy

5x−xsec2xy .

(b) Let g(x, y)=xln y+ sin(x−y)=π, dg

dx =lny+ cos(x−y)+x

y−cos(x−y)dy

dx =0,

dx =ln y+ cos(x−y)

−x/y + cos(x−y)

Chapter 15 Supplementary Exercises 572

33. F(x, y)=0,F

x+F

dx =0,F

xx +Fxy

dx +Fyx

dx +Fy

d2y

dx2=0,thus

dx =−Fx

,d2y

dx2=−Fxx +2F

xy(dy/dx)

=−FxxFy−2FxFxy

34. Denote the currents I1,I

2,I

3by x, y, z respectively. Then minimize F(x, y, z)=x

subject to g(x, y, z)=x+y+z−I= 0, so solve ∇F=λ∇g, 2xR1i+2yR2j+2zR3k=λ(i+j+k),

λ=2xR1=2yR2=2zR3, so the minimum value of Foccurs when I1:I2:I3=1

35. Solve (t−1)2/4+16e

−2t+(2−√

2= 1 for tto get t=1.8332,2.83984; the particle strikes

the surface at the points P1(0.8332,0.63959,0.64603),P

2(1.83984,0.23374,0.31482). The velocity

vectors are given by v=dx

dt i+dy

dt j+dz

dt k=i−4e−tj−1/(2√t)k, and a normal to the surface is

n=∇(x2/4+y

2+z

)=x/2i+2yj+2zk. At the points Pithese are

v1=i−0.639589j−0.369286k,v2=i−0.23374j+0.29670k;

n1=0.41661i+1.27918j+1.29207kand n2=0.91992i+0.46748j+0.62963kso

cos−1[(vi·ni)/(kvikkn

ik)] = 112.3◦,61.1◦; the acute angles are 67.7◦,61.1◦.

36. (a) F0(x)=Z1

ycos(xey)dy =sin(ex)−sin x

(b) Use a CAS to get x=π

e+1 so the maximum value of F(x)is

Fπ

e+1=Z1

sin π

e+1e

ydy ≈0.909026.

37. Let x, y, z be the lengths of the sides opposite angles α, β, γ, located at A, B, C respectively. Then

x2=y2+z2−2yz cos αand x2= 100 + 400 −2(10)(20)/2 = 300,x =10

√

3 and

2xdx

dt =2ydy

dt +2zdz

dt −2ydz

dt cos α+zdy

dt cos α−yz(sin α)dα

dt 

= 2(10)(4) + 2(20)(2) −2 10(2)1

2+ 20(4)1

2−10(20)√3

60!=60+10π

√3

so dx

dt =√3+π

6, the length of BC is increasing.

38. (a) d

dt ∂z

∂x=∂

∂x ∂z

∂xdx

dt +∂

∂y ∂z

∂xdy

dt =∂2z

∂x2

dt +∂2z

∂y∂x

dt by the Chain Rule, and

dt ∂z

∂y=∂

∂x ∂z

∂ydx

dt +∂

∂y ∂z

∂ydy

dt =∂2z

∂x∂y

dt +∂2z

∂y2

(b) dz

dt =∂z

∂x

dt +∂z

∂y

dt ,

d2z

dt2=dx

dt ∂2z

∂x2

dt +∂2z

∂y∂x

dt +∂z

∂x

d2x

dt2+dy

dt ∂2z

∂x∂y

dt +∂2z

∂y2

dt +∂z

∂y

d2y

dt2

573

CHAPTER 16

Multiple Integrals

EXERCISE SET 16.1

1. Z1

0Z2

(x+3)dy dx =Z1

(2x+6)dx =7 2. Z3

1Z1

−1

(2x−4y)dy dx =Z3

4xdx=16

3. Z4

2Z1

x2ydxdy=Z4

3ydy =2 4. Z0

−2Z2

−1

(x2+y2)dx dy =Z0

−2

(3+3y

)dy =14

5. Zln 3

0Zln 2

ex+ydy dx =Zln 3

exdx =2

6. Z2

0Z1

ysin xdydx=Z2

2sin xdx=(1−cos 2)/2

7. Z0

−1Z5

dx dy =Z0

−1

3dy =3 8. Z6

4Z7

−3

dy dx =Z6

10dx =20

9. Z1

0Z1

(xy +1)

2dy dx =Z1

01−1

x+1dx =1−ln 2

10. Zπ

π/2Z2

xcos xy dy dx =Zπ

π/2

(sin 2x−sin x)dx =−2

11. Zln 2

0Z1

xy ey2xdy dx =Zln 2

2(ex−1)dx =(1−ln 2)/2

12. Z4

3Z2

(x+y)2dy dx =Z4

31

x+1−1

x+2dx = ln(25/24)

13. Z1

−1Z2

−2

4xy3dy dx =Z1

−1

0dx =0

14. Z1

0Z1

px2+y2+1dy dx =Z1

[x(x2+2)

1/2−x(x

2+1)

1/2

]dx =(3

√

3−4

√

2+1)/3

15. Z1

0Z3

xp1−x2dy dx =Z1

x(1 −x2)1/2dx =1/3

16. Zπ/2

0Zπ/3

(xsin y−ysin x)dy dx =Zπ/2

0x

2−π2

18 sin xdx =π2/144

19. V=Z5

3Z2

(2x+y)dy dx =Z5

(2x+3/2)dx =19

20. V=Z3

1Z2

(3x3+3x

y)dy dx =Z3

(6x3+6x

)dx = 172

Exercise Set 16.1 574

21. V=Z2

0Z3

x2dy dx =Z2

3x2dx =8

22. V=Z3

0Z4

5(1 −x/3)dy dx =Z3

5(4 −4x/3)dx =30

23. (a)

(1, 0, 4)

(2, 5, 0)

z(b)

(0, 0, 5)

(3, 4, 0)

24. (a)

(1, 1, 0)

(0, 0, 2)

z(b)

(2, 2, 0)

(2, 2, 8)

25. Z1/2

0Zπ

xcos(xy) cos2πx dy dx =Z1/2

cos2πx sin(xy)iπ

0dx

=Z1/2

cos2πx sin πx dx =−1

3cos3πxi1/2

0=1

3π

26. (a)

(b) The projection onto the xy-plane consists of R1:[0,5] ×[0,1], over which lie each of the two

skew planes, and R2:[0,5] ×[1,3], over which is only the plane z=−2y+6,so

V=Z5

0Z1

((−2y+6)−y)dy dx +Z5

0Z3

(−2y+6)dy dx =45

2+20= 85

27. fave =2

πZπ/2

0Z1

ysin xy dy dx =2

πZπ/2

0−cos xyi1

0dx =2

πZπ/2

(1 −cos x)dx =1−2

575 Chapter 16

28. average = 1

3Z3

0Z1

x(x2+y)1/2dx dy =Z3

9[(1 + y)3/2−y3/2]dy = 2(31 −9√3)/45

29. Tave =1

2Z1

0Z2

010 −8x2−2y2dy dx =1

2Z1

044

3−16x2dx =14

3◦

30. fave =1

A(R)Zb

aZd

kdydx=1

A(R)(b−a)(d−c)k=k

31. 0.6211310829 32. 2.230985141

33. ZZ

f(x, y)dA =Zb

a"Zd

g(x)h(y)dy#dx =Zb

g(x)"Zd

h(y)dy#dx

="Zb

g(x)dx#"Zd

h(y)dy#

34. The integral of tan x(an odd function) over the interval [−1,1] is zero.

35. The ﬁrst integral equals 1/2, the second equals −1/2. No, because the integrand is not continuous.

36. (a)

k=1

f(x∗

k,y

∗

k)∆A∗

i=1

j=1 i

2−1

4−2j

2−1

41

22

=−4

(b) Z2

0Z2

(x−2y)dy dx =Z2

(2x−4) dx =−4

EXERCISE SET 16.2

1. Z1

0Zx

xy2dy dx =Z1

3(x4−x7)dx =1/40

2. Z3/2

1Z3−y

ydxdy=Z3/2

(3y−2y2)dy =7/24

3. Z3

0Z√9−y2

ydxdy=Z3

p9−y

2dy =9

4. Z1

1/4Zx

x2px/y dy dx =Z1

1/4Zx

x1/2y−1/2dy dx =Z1

1/4

2(x−x3/2)dx =13/80

5. Z√2π

√πZx3

sin(y/x)dy dx =Z√2π

√π

[−xcos(x2)+x]dx =π/2

6. Z1

−1Zx2

−x2

(x2−y)dy dx =Z1

−1

2x4dx =4/57. Zπ

π/2Zx2

xcos(y/x)dy dx =Zπ

π/2

sin xdx=1

Exercise Set 16.2 576

8. Z1

0Zx

ex2dy dx =Z1

xex2dx =(e−1)/29. Z1

0Zx

ypx2−y2dy dx =Z1

3x3dx =1/12

10. Z2

1Zy2

ex/y2dx dy =Z2

(e−1)y2dy =7(e−1)/3

11. (a) Z2

0Zx2

xy dy dx =Z2

2x5dx =16

(b) Z3

1Z(y+7)/2

−(y−5)/2

xy dx dy =Z3

(3y2+3y)dy =38

12. (a) Z1

0Z√x

(x+y)dy dx =Z1

(x3/2+x/2−x3−x4/2)dx =3/10

(b) Z1

−1Z√1−x2

−√1−x2

xdydx+Z1

−1Z√1−x

−√1−x

ydydx=Z1

−1

p1−x

2dx +0=0

13. Z8

4Zx

16/x

x2dy dx =Z8

(x3−16x)dx = 576 14. Z2

1Zy

xy2dx dy =Z2

2y4dy =31/10

15. Z4

0Z√y

x(1+y2)−1/2dx dy =Z4

2y(1+y2)−1/2dy =(

√

17 −1)/2

16. Zπ

0Zx

xcos ydydx=Zπ

xsin xdx=π

17. Z1

−1Z√1−x2

−√1−x2

(3x−2y)dy dx =Z1

−1

6xp1−x2dx =0

18. Z5

0Z√25−x2

5−x

ydydx=Z5

(5x−x2)dx = 125/6

19. Z2

0Z6−y

xy dx dy =Z2

2(36y−12y2+y3−y5)dy =50/3

20. Zπ/4

0Z1/√2

sin y

xdxdy =Zπ/4

4cos 2ydy=1/8

21. Z0

−1Zx3

(x−1)dy dx +Z1

0Zx

(x−1)dy dx

=Z0

−1

(x4−x3−x2+x)dx +Z1

(−x4+x3+x2−x)dx =−1/2

22. Z1/√2

0Z2x

x2dy dx +Z1

1/√2Z1/x

x2dy dx =Z1/√2

x3dx +Z1

1/√2

(x−x3)dx =1/8

577 Chapter 16

24. (a)

1 2 3

(b) (1,3),(3,27)

1Z4x3−x4

3−4x+4x2

xdydx=Z3

x[(4x3−x4)−(3 −4x+4x

)] dx =224

25. A=Zπ/4

0Zcos x

sin x

dy dx =Zπ/4

(cos x−sin x)dx =√2−1

26. A=Z1

−4Z−y2

3y−4

dx dy =Z1

−4

(−y2−3y+4)dy = 125/6

27. A=Z3

−3Z9−y2

1−y2/9

dx dy =Z3

−3

8(1 −y2/9)dy =32

28. A=Z1

0Zcosh x

sinh x

dy dx =Z1

(cosh x−sinh x)dx =1−e

−1

29. Z4

0Z6−3x/2

(3 −3x/4−y/2) dy dx =Z4

[(3 −3x/4)(6 −3x/2) −(6 −3x/2)2/4] dx =12

30. Z2

0Z√4−x2

0p4−x2dy dx =Z2

(4 −x2)dx =16/3

31. V=Z3

−3Z√9−x2

−√9−x2

(3 −x)dy dx =Z3

−3

(6p9−x2−2xp9−x2)dx =27π

32. V=Z1

0Zx

(x2+3y

)dy dx =Z1

(2x3−x4−x6)dx =11/70

33. V=Z3

0Z2

(9x2+y2)dy dx =Z3

(18x2+8/3)dx = 170

34. V=Z1

−1Z1

(1 −x)dx dy =Z1

−1

(1/2−y2+y4/2)dy =8/15

35. V=Z3/2

−3/2Z√9−4x2

−√9−4x2

(y+3)dy dx =Z3/2

−3/2

6p9−4x2dx =27π/2

36. V=Z3

0Z3

y2/3

(9 −x2)dx dy =Z3

(18 −3y2+y6/81)dy = 216/7

Exercise Set 16.2 578

37. V=8Z5

0Z√

25−x2

0p25 −x2dy dx =8Z5

(25 −x2)dx = 2000/3

38. V=2Z2

0Z√

1−(y−1)2

(x2+y2)dx dy =2Z2

01

[1 −(y−1)2]3/2+y2[1 −(y−1)2]1/2dy,

let y−1 = sin θto get V=2

Zπ/2

−π/21

3cos3θ+(1+sinθ)

2cos θcos θdθ which eventually yields

V=3π/2

39. V=4Z1

0Z√

1−x

(1 −x2−y2)dy dx =8

3Z1

(1 −x2)3/2dx =π/2

40. V=Z2

0Z√4−x2

(x2+y2)dy dx =Z2

0x2p4−x2+1

3(4 −x2)3/2dx =2π

41. Z√2

0Z2

f(x, y)dx dy 42. Z8

0Zx/2

f(x, y)dy dx 43. Ze2

1Z2

ln x

f(x, y)dy dx

44. Z1

0Ze

f(x, y)dx dy 45. Zπ/2

0Zsin x

f(x, y)dy dx 46. Z1

0Z√x

f(x, y)dy dx

47. Z4

0Zy/4

e−y2dx dy =Z4

4ye−y2dy =(1−e

−16)/8

48. Z1

0Z2x

cos(x2)dy dx =Z1

2xcos(x2)dx = sin 1

49. Z2

0Zx2

ex3dy dx =Z2

x2ex3dx =(e

8−1)/3

50. Zln 3

0Z3

xdxdy =1

2Zln 3

(9 −e2y)dy =1

2(9 ln 3 −4)

51. Z2

0Zy2

sin(y3)dx dy =Z2

y2sin(y3)dy =(1−cos 8)/3

52. Z1

0Ze

xdydx=Z1

(ex −xex)dx =e/2−1

53. (a) Z4

0Z2

√x

sin πy3dy dx; the inner integral is non-elementary.

0Zy2

sin πy3dx dy =Z2

y2sin πy3dy =−1

3πcos πy32

579 Chapter 16

(b) Z1

0Zπ/2

sin−1y

sec2(cos x)dx dy ; the inner integral is non-elementary.

Zπ/2

0Zsin x

sec2(cos x)dy dx =Zπ/2

sec2(cos x) sin xdx= tan 1

54. V=4Z2

0Z√

4−x

2+y

)dy dx =4Z2

0x

p4−x

2+1

(4 −x2)3/2dx (x= 2 sin θ)

=Zπ/2

064

3+64

3sin2θ−128

3sin4θdθ =64

2+64

4−128

1·3

2·4=8π

55. Z−1

−2Z2

xy2dy dx +Z1

−1Z2

xy2dy dx +Z2

1Z2

xy2dy dx

=Z−1

−2

3xdx+Z1

−1

3xdx+Z2

3xdx=0

56. This is the volume in the ﬁrst octant under the surface z=p1−x2−y2, so 1/8 of the volume of

the sphere of radius 1, thus π

57. Area of triangle is 1/2, so ¯

f=2Z1

0Z1

1+x

2dy dx =2Z1

01

1+x

2−x

1+x

2dx =π

2−ln 2

58. Area = Z2

(3x−x2−x)dx =4/3, so

f=3

4Z2

0Z3x−x2

(x2−xy)dy dx =3

4Z2

(−2x3+2x

4−x

/2)dx =−3

15 =−2

59. y= sin xand y=x/2 intersect at x= 0 and x=a=1.895494, so

V=Za

0Zsin x

x/2p1+x+ydydx=0.676089

EXERCISE SET 16.3

1. Zπ/2

0Zsin θ

rcos θdr dθ =Zπ/2

2sin2θcos θdθ=1/6

2. Zπ

0Z1+cos θ

rdrdθ=Zπ

2(1 + cos θ)2dθ =3π/4

3. Zπ/2

0Zasin θ

r2dr dθ =Zπ/2

3sin3θdθ=2

4. Zπ/3

0Zcos 3θ

rdrdθ=Zπ/3

2cos23θdθ=π/12

5. Zπ

0Z1−sin θ

r2cos θdrdθ=Zπ

3(1 −sin θ)3cos θdθ=0

Exercise Set 16.3 580

6. Zπ

0Zcos θ

r3dr dθ =Zπ

4cos4θdθ=3π/32

7. A=Z2π

0Z1−cos θ

rdrdθ=Z2π

2(1 −cos θ)2dθ =3π/2

8. A=4Zπ/2

0Zsin 2θ

rdrdθ=2Zπ/2

sin22θdθ=π/2

9. A=Zπ/2

π/4Z1

sin 2θ

rdrdθ=Zπ/2

π/4

2(1 −sin22θ)dθ =π/16

10. A=2Zπ/3

0Z2

sec θ

rdrdθ=Zπ/3

(4 −sec2θ)dθ =4π/3−√3

11. A=2Zπ/2

π/6Z4 sin θ

rdrdθ=Zπ/2

π/6

(16 sin2θ−4)dθ =4π/3+2

√

12. A=2Zπ

π/2Z1

1+cos θ

rdrdθ=Zπ

π/2

(−2 cos θ−cos2θ)dθ =2−π/4

13. V=8Zπ/2

0Z3

rp9−r2dr dθ =128

3√2Zπ/2

dθ =64

3√2π

14. V=2Zπ/2

0Z2 sin θ

r2dr dθ =16

3Zπ/2

sin3θdθ=32/9

15. V=2Zπ/2

0Zcos θ

(1 −r2)rdrdθ=1

2Zπ/2

(1 −sin4θ)dθ =5π/32

16. V=4Zπ/2

0Z3

dr dθ =8Zπ/2

dθ =4π

17. V=Zπ/2

0Z3 sin θ

r2sin θdrdθ=9Zπ/2

sin4θdθ=27π/16

18. V=4Zπ/2

0Z2 cos θ

rp4−r2dr dθ =32

3Zπ/2

(1 −sin3θ)dθ =16

9(3π−4)

19. Z2π

0Z1

e−r2rdrdθ=1

2(1 −e−1)Z2π

dθ =(1−e

−1

)π

20. Zπ/2

0Z3

rp9−r2dr dθ =9Zπ/2

dθ =9π/2

21. Zπ/4

0Z2

1+r

2rdrdθ=1

2ln 5 Zπ/4

dθ =π

8ln 5

22. Zπ/2

π/4Z2 cos θ

2r2sin θdrdθ=16

3Zπ/2

π/4

cos3θsin θdθ=1/3

581 Chapter 16

23. Zπ/2

0Z1

r3dr dθ =1

4Zπ/2

dθ =π/8

24. Z2π

0Z2

e−r2rdrdθ=1

2(1 −e−4)Z2π

dθ =(1−e

−4

)π

25. Zπ/2

0Z2 cos θ

r2dr dθ =8

3Zπ/2

cos3θdθ=16/9

26. Zπ/2

0Z1

cos(r2)rdrdθ=1

2sin 1 Zπ/2

dθ =π

4sin 1

27. Zπ/2

0Za

(1+r2)3/2dr dθ =π

21−1/p1+a



28. Zπ/4

0Zsec θtan θ

r2dr dθ =1

3Zπ/4

sec3θtan3θdθ=2(

√

2+1)/45

29. Zπ/4

0Z2

√1+r

2dr dθ =π

4(√5−1)

30. Zπ/2

tan−1(3/4) Z5

3 csc θ

rdrdθ=1

2Zπ/2

tan−1(3/4)

(25 −9 csc2θ)dθ

=25

2hπ

2−tan−1(3/4)i−6=25

2tan−1(4/3) −6

31. V=Z2π

0Za

hr dr dθ =Z2π

ha2

2dθ =πa2h

32. (a) V=8Zπ/2

0Za

a(a2−r2)1/2rdrdθ =−4c

3aπ(a

2−r

3/2a

3πa2c

(b) V≈4

3π(6378.1370)26356.5231 ≈1,083,168,200,000 km3

33. V=2Zπ/2

0Zasin θ

a(a2−r2)1/2rdrdθ=2

2cZπ/2

(1 −cos3θ)dθ =(3π−4)a2c/9

34. A=4Zπ/4

0Za√2 cos 2θ

rdrdθ=4a

2Zπ/4

cos 2θdθ=2a

35. A=Zπ/4

π/6Z4 sin θ

√8 cos 2θ

rdrdθ+Zπ/2

π/4Z4 sin θ

rdrdθ

=Zπ/4

π/6

(8 sin2θ−4 cos 2θ)dθ +Zπ/2

π/4

8 sin2θdθ=4π/3+2

√

3−2

36. A=Zφ

0Z2asin θ

rdrdθ=2a

2Zφ

sin2θdθ=a

2φ−1

2sin 2φ

Exercise Set 16.4 582

37. (a) I2=Z+∞

e−x2dxZ+∞

−y

2dy=Z+∞

0Z+∞

e−x2dxe−y2dy

=Z+∞

0Z+∞

e−x2e−y2dx dy =Z+∞

0Z+∞

e−(x2+y2)dx dy

(b) I2=Zπ/2

0Z+∞

e−r2rdrdθ=1

2Zπ/2

dθ =π/4(c) I=√π/2

38. (a) 1.173108605 (b) Zπ

0Z1

re−r4dr dθ =πZ1

re−r4dr ≈1.173108605

39. V=Z2π

0ZR

D(r)rdrdθ=Z2π

0ZR

ke−rrdrdθ =−2πk(1+r)e−rR

=2πk[1 −(R+1)e

−R

]

40. Ztan−1(2)

tan−1(1/3) Z2

r3cos2θdrdθ=4Ztan−1(2)

tan−1(1/3)

cos2θdθ=1

5+ 2[tan−1(2) −tan−1(1/3)] = 1

5+π

EXERCISE SET 16.4

1. (a) z

(b)

(c)

2. (a)

z(b) z

583 Chapter 16

3. (a) x=u, y =v, z =5

2+3

2u−2v(b) x=u, y =v, z =u2

4. (a) x=u, y =v, z =v

1+u

2(b) x=u, y =v, z =1

3v2−5

5. (a) x= 5 cos u, y = 5 sin u, z =v;0≤u≤2π, 0≤v≤1

(b) x= 2 cos u, y =v, z = 2 sin u;0≤u≤2π, 1≤v≤3

6. (a) x=u, y =1−u, z =v;−1≤v≤1(b) x=u, y =5+2v, z =v;0≤u≤3

7. x=u, y = sin ucos v, z = sin usin v8. x=u, y =eucos v, z =eusin v

9. x=rcos θ, y =rsin θ, z =1

1+r

210. x=rcos θ, y =rsin θ, z =e−r2

11. x=rcos θ, y =rsin θ, z =2r

2cos θsin θ

12. x=rcos θ, y =rsin θ, z =r2(cos2θ−sin2θ)

13. x=rcos θ, y =rsin θ, z =√9−r2;r≤√5

14. x=rcos θ, y =rsin θ, z =r;r≤315. x=1

2ρcos θ, y =1

2ρsin θ, z =√3

2ρ

16. x= 3 cos θ, y = 3 sin θ, z = 3 cot φ17. z=x−2y; a plane

18. y=x2+z2,0≤y≤4; part of a circular paraboloid

19. (x/3)2+(y/2)2=1;2≤z≤4; part of an elliptic cylinder

20. z=x2+y2;0≤z≤4; part of a circular paraboloid

21. (x/3)2+(y/4)2=z2;0≤z≤1; part of an elliptic cone

22. x2+(y/2)2+(z/3)2= 1; part of an ellipsoid

23. (a) x=rcos θ, y =rsin θ, z =r;x=u, y =v, z =√u2+v2;0≤z≤2

24. (a) I: x=rcos θ, y =rsin θ, z =r2;II:x=u, y =v, z =u2+v2;u2+v2≤2

25. (a) 0≤u≤3,0≤v≤π(b) 0≤u≤4,−π/2≤v≤π/2

Exercise Set 16.4 584

26. (a) 0≤u≤6,−π≤v≤0(b) 0≤u≤5,π/2≤v≤3π/2

27. (a) 0≤φ≤π/2,0≤θ≤2π(b) 0≤φ≤π, 0≤θ≤π

28. (a) π/2≤φ≤π(b) 0≤θ≤π/2,0≤φ≤π/2

29. u=1,v =2,r

u×r

v=−2i−4j+k;2x+4y−z=5

30. u=1,v =2,r

u×r

v=−4i−2j+8k;2x+y−4z=−6

31. u=0,v =1,r

u×r

v=6k;z=0 32. ru×rv=2i−j−3k;2x−y−3z=−4

33. ru×rv=(

√

2/2)i−(√2/2)j+(1/2)k;x−y+√2

2z=π√2

34. ru×rv=2i−ln 2k;2x−(ln 2)z=0

35. z=p9−y2,zx=0,z

y=−y/p9−y2,z2

x+z2

y+1=9/(9 −y2),

S=Z2

0Z3

−3

p9−y2dy dx =Z2

3πdx=6π

36. z=8−2x−2y,z

x+z

y+1=4+4+1=9,S=Z4

0Z4−x

3dy dx =Z4

3(4 −x)dx =24

37. z2=4x

2+4y

,2zzx=8xso zx=4x/z, similarly zy=4y/z thus

x+z2

y+ 1 = (16x2+16y

)/z2+1=5,S=Z1

0Zx

√

5dy dx =√5Z1

(x−x2)dx =√5/6

38. z2=x2+y2,zx=x/z,zy=y/z,z2

x+z2

y+1=(z

2+y

)/z2+1=2,

S=ZZ

√2dA =2Zπ/2

0Z2 cos θ

√2rdrdθ=4

√

2Zπ/2

cos2θdθ=√2π

39. zx=−2x,zy=−2y,z2

x+z2

y+1=4x

2+4y

2+1,

S=ZZ

Rp4x2+4y

2+1dA =Z2π

0Z1

rp4r2+1dr dθ

12(5√5−1) Z2π

dθ =(5

√

5−1)π/6

40. zx=2,z

y=2y,z

x+z

y+1=5+4y

S=Z1

0Zy

0p5+4y

2dx dy =Z1

yp5+4y

2dy = (27 −5√5)/12

41. ∂r/∂u = cos vi+ sin vj+2uk,∂r/∂v =−usin vi+ucos vj,

k∂r/∂u ×∂r/∂vk=u√4u2+1; S=Z2π

0Z2

p4u

2+1du dv = (17√17 −5√5)π/6

42. ∂r/∂u = cos vi+ sin vj+k,∂r/∂v =−usin vi+ucos vj,

k∂r/∂u ×∂r/∂vk=√2u;S=Zπ/2

0Z2v

√2ududv=√

12 π3

585 Chapter 16

43. zx=y,zy=x,z2

x+z2

y+1=x

2+y

2+1,

S=ZZ

Rpx2+y2+1dA =Zπ/6

0Z3

rpr2+1dr dθ =1

3(10√10−1) Zπ/6

dθ = (10√10−1)π/18

44. zx=x,zy=y,z2

x+z2

y+1=x

2+y

2+1,

S=ZZ

Rpx2+y2+1dA =Z2π

0Z√8

rpr2+1dr dθ =26

3Z2π

dθ =52π/3

45. On the sphere, zx=−x/z and zy=−y/z so z2

x+z2

y+1=(x

2+y

2+z

)/z2=16/(16 −x2−y2);

the planes z= 1 and z= 2 intersect the sphere along the circles x2+y2= 15 and x2+y2= 12;

S=ZZ

p16 −x2−y2dA =Z2π

0Z√15

√12

√16 −r2dr dθ =4Z2π

dθ =8π

46. On the sphere, zx=−x/z and zy=−y/z so z2

x+z2

y+1=(x

2+y

2+z

)/z2=8/(8 −x2−y2);

the cone cuts the sphere in the circle x2+y2=4;

S=Z2π

0Z2

√

8−r

2dr dθ =(8−4

√

2) Z2π

dθ = 8(2 −√2)π

47. r(u, v)=acos usin vi+asin usin vj+acos vk,kru×rvk=a2sin v,

S=Zπ

0Z2π

a2sin vdudv=2πa2Zπ

sin vdv =4πa2

48. r =rcos ui+rsin uj+vk,kru×rvk=r;S=Zh

0Z2π

rdudv =2πrh

49. zx=h

px2+y2,zy=h

px2+y2,z2

x+z2

y+1= h

2+h

2+y

)+1=(a

2+h

)/a2,

S=Z2π

0Za

√a2+h2

ardrdθ=1

2+h

2Z2π

dθ =πapa2+h2

50. Revolving a point (a0,0,b

0)ofthexz-plane around the z-axis generates a circle, an equation of

which is r=a0cos ui+a0sin uj+b0k,0≤u≤2π. A point on the circle (x−a)2+z2=b2which

generates the torus can be written r=(a+bcos v)i+bsin vk,0≤v≤2π. Set a0=a+bcos v

and b0=a+bsin vand use the ﬁrst result: any point on the torus can thus be written in the form

r=(a+bcos v) cos ui+(a+bcos v) sin uj+bsin vk, which yields the result.

51. ∂r/∂u =−(a+bcos v) sin ui+(a+bcos v) cos uj,

∂r/∂v =−bsin vcos ui−bsin vsin uj+bcos vk,k∂r/∂u ×∂r/∂vk=b(a+bcos v);

S=Z2π

0Z2π

b(a+bcos v)du dv =4π

52. kru×rvk=√u2+1;S=Z4π

0Z5

0pu

2+1du dv =4πZ5

0pu

2+1du = 174.7199011

53. z=−1 when v≈0.27955,z = 1 when v≈2.86204,kru×rvk=|cos v|;

S=Z2π

0Z2.86204

0.27955 |cos v|dv du ≈9.099

Exercise Set 16.5 586

54. (a) x=vcos u, y =vsin u, z =f(v), for example (b) x=vcos u, y =vsin u, z =1/v2

55. (x/a)2+(y/b)2+(z/c)2= cos2v(cos2u+ sin2u) + sin2v= 1, ellipsoid

56. (x/a)2+(y/b)2−(z/c)2= cos2ucosh2v+ sin2ucosh2v−sinh2v= 1, hyperboloid of one sheet

57. (x/a)2+(y/b)2−(z/c)2= sinh2v+ cosh2v(sinh2u−cosh2u)=−1, hyperboloid of two sheets

EXERCISE SET 16.5

1. Z1

−1Z2

0Z1

(x2+y2+z2)dx dy dz =Z1

−1Z2

(1/3+y

2+z

)dy dz =Z1

−1

(10/3+2z

)dz =8

2. Z1/2

1/3Zπ

0Z1

zxsin xy dz dy dx =Z1/2

1/3Zπ

2xsin xy dy dx =Z1/2

1/3

2(1−cos πx)dx =1

12 +√3−2

4π

3. Z2

0Zy2

−1Zz

−1

yz dx dz dy =Z2

0Zy2

−1

(yz2+yz)dz dy =Z2

01

3y7+1

2y5−1

6ydy =47

4. Zπ/4

0Z1

0Zx2

xcos ydzdxdy=Zπ/4

0Z1

x3cos ydxdy=Zπ/4

4cos ydy =√2/8

5. Z3

0Z√9−z2

0Zx

xy dy dx dz =Z3

0Z√9−z2

2x3dx dz =Z3

8(81 −18z2+z4)dz =81/5

6. Z3

1Zx2

xZln z

xeydy dz dx =Z3

1Zx2

(xz −x)dz dx =Z3

11

2x5−3

2x3+x2dx = 118/3

7. Z2

0Z√4−x2

0Z3−x2−y2

−5+x2+y2

xdzdydx=Z2

0Z√4−x

[2x(4 −x2)−2xy2]dy dx

=Z2

3x(4 −x2)3/2dx = 128/15

8. Z2

1Z2

zZ√3y

x2+y2dx dy dz =Z2

1Z2

3dy dz =Z2

3(2 −z)dz =π/6

587 Chapter 16

9. Zπ

0Z1

0Zπ/6

xy sin yz dz dy dx =Zπ

0Z1

x[1 −cos(πy/6)]dy dx =Zπ

(1 −3/π)xdx =π(π−3)/2

10. Z1

−1Z1−x2

0Zy

ydzdydx=Z1

−1Z1−x

dy dx =Z1

−1

3(1 −x2)3dx =32/105

11. Z√2

0Zx

0Z2−x2

xyzdzdydx=Z√

0Zx

2xy(2 −x2)2dy dx =Z√2

4x3(2 −x2)2dx =1/6

12. Zπ/2

π/6Zπ/2

yZxy

cos(z/y)dz dx dy =Zπ/2

π/6Zπ/2

ysin xdxdy =Zπ/2

π/6

ycos ydy =(5π−6

√

3)/12

14. 8Z1

0Z√1−x2

0Z√1−x2−y2

e−x2−y2−z2dz dy dx ≈2.381

15. V=Z4

0Z(4−x)/2

0Z(12−3x−6y)/4

dz dy dx =Z4

0Z(4−x)/2

4(12 −3x−6y)dy dx

=Z4

16(4 −x)2dx =4

16. V=Z1

0Z1−x

0Z√y

dz dy dx =Z1

0Z1−x

√ydydx=Z1

3(1 −x)3/2dx =4/15

17. V=2Z2

0Z4

2Z4−y

dz dy dx =2Z2

0Z4

(4 −y)dy dx =2Z2

08−4x

2+1

dx = 256/15

18. V=Z1

0Zy

0Z√1−y2

dz dx dy =Z1

0Zy

0p1−y2dx dy =Z1

yp1−y2dy =1/3

19. The projection of the curve of intersection onto the xy-plane is x2+y2=1,

V=4Z1

0Z√

1−x

0Z4−3y

dz dy dx

20. The projection of the curve of intersection onto the xy-plane is 2x2+y2=4,

V=4Z√

0Z√

4−2x

0Z8−x

−y

dz dy dx

21. V=2Z3

−3Z√

9−x

0Zx+3

dz dy dx 22. V=8Z1

0Z√

1−x

0Z√

1−x

dz dy dx

Exercise Set 16.5 588

23. (a)

(0, 0, 1)

(1, 0, 0)

(b)

(0, 9, 9)

(3, 9, 0)

(c)

(0, 0, 1)

(1, 2, 0)

24. (a)

(3, 9, 0)

(0, 0, 2)

z(b)

(0, 0, 2)

(0, 2, 0)

(2, 0, 0)

(c)

(2, 2, 0)

(0, 0, 4)

25. V=Z1

0Z1−x

0Z1−x−y

dz dy dx =1/6,f

ave =6Z1

0Z1−x

0Z1−x−y

(x+y+z)dz dy dx =3

26. The integrand is an odd function of each of x, y, and z, so the answer is zero.

589 Chapter 16

27. (a) Za

0Zb(1−x/a)

0Zc(1−x/a−y/b)

dz dy dx,Zb

0Za(1−y/b)

0Zc(1−x/a−y/b)

dz dx dy,

0Za(1−z/c)

0Zb(1−x/a−z/c)

dy dx dz,Za

0Zc(1−x/a)

0Zb(1−x/a−z/c)

dy dz dx,

0Zb(1−z/c)

0Za(1−y/b−z/c)

dx dy dz,Zb

0Zc(1−y/b)

0Za(1−y/b−z/c)

dx dz dy

(b) Use the ﬁrst integral in part (a) to get

0Zb(1−x/a)

c1−x

a−y

bdy dx =Za

2bc 1−x

a2dx =1

6abc

28. (a) Z3

0Z√9−x2

0Z√9−x2−y2

f(x, y, z)dz dy dx

(b) Z4

0Zx/2

0Z2

f(x, y, z)dz dy dx (c) Z2

0Z4−x2

0Z4−y

f(x, y, z)dz dy dx

29. V=8Za

0Zb

√

1−x

/a2

0Zc√1−x2/a2−y2/b2

dz dy dx

30. Zb

aZd

cZ`

f(x)g(y)h(z)dz dy dx =Zb

aZd

f(x)g(y)"Z`

h(z)dz#dy dx

="Zb

f(x)"Zd

g(y)dy#dx#"Z`

h(z)dz#

="Zb

f(x)dx#"Zd

g(y)dy#"Z`

h(z)dz#

31. (a) Z1

−1

xdx

Z1

2dy"Zπ/2

sin zdz

#= (0)(1/3)(1) = 0

(b) Z1

e2xdx"Zln 3

eydy#"Zln 2

e−zdz#=[(e

2−1)/2](2)(1/2)=(e

2−1)/2

32. (a) At any point outside the closed sphere {x2+y2+z2≤1}the integrand is negative, so to

maximize the integral it suﬃces to include all points inside the sphere; hence the maximum

value is taken on the region G={x2+y2+z2≤1}.

(b) 8π

EXERCISE SET 16.6

1. Let abe the unknown coordinate of the fulcrum; then the total moment about the fulcrum is

5(0 −a) + 10(5 −a) + 20(10 −a) = 0 for equilibrium, so 250 −35a=0,a=50/7. The fulcrum

should be placed 50/7 ft to the right of m1.

Exercise Set 16.6 590

2. At equilibrium, 10(0 −4) + 3(2 −4) + 4(3 −4)+m(6 −4)=0,m=25

3. A=1,¯x=Z1

0Z1

xdydx=1

2,¯y=Z1

0Z1

ydydx=1

4. A=2,¯x=1

2Z0

−1Z1+x

−1−x

xdydx+Z1

0Z1−x

−1+x

xdydx= 0, similarly ¯y=0.

5. A=1/2, ZZ

xdA=Z1

0Zx

xdydx=1/3, ZZ

ydA=Z1

0Zx

ydydx=1/6;

centroid (2/3,1/3)

6. A=Z1

0Zx2

dy dx =1/3, ZZ

xdA=Z1

0Zx

xdydx=1/4,

ydA=Z1

0Zx

ydydx=1/10; centroid (3/4,3/10)

7. A=Z1

0Z2−x2

dy dx =7/6, ZZ

xdA=Z1

0Z2−x

xdydx=5/12,

ydA=Z1

0Z2−x

ydydx=19/15; centroid (5/14,38/35)

8. A=π

4,ZZ

xdA=Z1

0Z√1−x

xdydx=1

3,¯x=4

3π

,¯y=4

3πby symmetry

9. ¯x= 0 from the symmetry of the region,

A=1

2π(b2−a2), ZZ

ydA=Zπ

0Zb

2sin θdrdθ=2

3(b

3−a

3); centroid ¯x=0,¯y=4(b3−a3)

3π(b2−a2).

10. ¯y= 0 from the symmetry of the region, A=πa2/2,

xdA=Zπ/2

−π/2Za

r2cos θdrdθ=2a

/3; centroid 4a

3π,0

11. M=Z1

0Z√x

(x+y)dy dx =13/20, Mx=Z1

0Z√x

(x+y)ydydx=3/10,

My=Z1

0Z√x

(x+y)xdydx=19/42, ¯x=My/M = 190/273, ¯y=Mx/M =6/13;

the mass is 13/20 and the center of gravity is at (190/273,6/13).

12. M=Zπ

0Zsin x

ydydx=π/4, ¯x=π/2 from the symmetry of the density and the region,

Mx=Zπ

0Zsin x

y2dy dx =4/9, ¯y=Mx/M =16

9π; mass π/4, center of gravity π

2,16

9π.

591 Chapter 16

13. M=Zπ/2

0Za

r3sin θcos θdrdθ =a

4/8, ¯x=¯yfrom the symmetry of the density and the

region, My=Zπ/2

0Za

r4sin θcos2θdrdθ =a

5/15, ¯x=8a/15; mass a4/8, center of gravity

(8a/15,8a/15).

14. M=Zπ

0Z1

r3dr dθ =π/4, ¯x= 0 from the symmetry of density and region,

Mx=Zπ

0Z1

r4sin θdrdθ=2/5, ¯y=8

5π; mass π/4, center of gravity 0,8

5π.

15. V=1,¯x=Z1

0Z1

xdzdydx=1

2, similarly ¯y=¯z=1

; centroid 1

2,1

2

16. V=πr2h=2π,¯x=¯y= 0 by symmetry, ZZZ

zdzdydx=Z2

0Z2π

0Z1

rz dr dθ dz =2π,

centroid = (0,0,1)

17. ¯x=¯y=¯zfrom the symmetry of the region, V=1/6,

¯x=1

VZ1

0Z1−x

0Z1−x−y

xdzdydx= (6)(1/24)=1/4; centroid (1/4,1/4,1/4)

18. The solid is described by −1≤y≤1,0≤z≤1−y2,0≤x≤1−z;

V=Z1

−1Z1−y2

0Z1−z

dx dz dy =4

5,¯x=1

VZ1

−1Z1−y2

0Z1−z

xdxdzdy =5

14,¯y= 0 by symmetry,

¯z=1

VZ1

−1Z1−y2

0Z1−z

zdxdzdy=2

7; the centroid is 5

14,0,2

7.

19. ¯x=1/2 and ¯y= 0 from the symmetry of the region,

V=Z1

0Z1

−1Z1

dz dy dx =4/3, ¯z=1

VZZZ

zdV =(3/4)(4/5)=3/5; centroid (1/2,0,3/5)

20. ¯x=¯yfrom the symmetry of the region,

V=Z2

0Z2

0Zxy

dz dy dx =4,¯x=1

VZZZ

xdV =(1/4)(16/3)=4/3,

¯z=1

VZZZ

zdV =(1/4)(32/9)=8/9; centroid (4/3,4/3,8/9)

21. ¯x=¯y=¯zfrom the symmetry of the region, V=πa3/6,

¯x=1

VZa

0Z√a2−x2

0Z√a2−x2−y2

xdzdydx=1

VZa

0Z√a

2−x

2−y

2dy dx

VZπ/2

0Za

r2pa2−r2cos θdrdθ =6

πa3(πa4/16)=3a/8; centroid (3a/8,3a/8,3a/8)

Exercise Set 16.6 592

22. ¯x=¯y= 0 from the symmetry of the region, V=2πa3/3

¯z=1

VZa

−aZ√a2−x2

−√a2−x2Z√a2−x2−y2

zdzdydx=1

VZa

−aZ√

−x

−

√

−x

2(a

2−x

2−y

)dy dx

VZ2π

0Za

2(a2−r2)rdrdθ=3

2πa3(πa4/4)=3a/8; centroid (0,0,3a/8)

23. M=Za

0Za

(a−x)dz dy dx =a4/2, ¯y=¯z=a/2 from the symmetry of density and

region, ¯x=1

MZa

0Za

x(a−x)dz dy dx =(2/a4)(a5/6) = a/3;

mass a4/2, center of gravity (a/3,a/2,a/2)

24. M=Za

−aZ√a2−x2

−√a2−x2Zh

(h−z)dz dy dx =1

2πa2h2,¯x=¯y= 0 from the symmetry of density

and region, ¯z=1

MZZZ

z(h−z)dV =2

πa2h2(πa2h3/6) = h/3;

mass πa2h2/2, center of gravity (0,0,h/3)

25. M=Z1

−1Z1

0Z1−y2

yz dz dy dx =1/6, ¯x= 0 by the symmetry of density and region,

¯y=1

MZZZ

y2zdV = (6)(8/105) = 16/35, ¯z=1

MZZZ

yz2dV = (6)(1/12)=1/2;

mass 1/6, center of gravity (0,16/35,1/2)

26. M=Z3

0Z9−x2

0Z1

xz dz dy dx =81/8, ¯x=1

MZZZ

x2zdV =(8/81)(81/5)=8/5,

¯y=1

MZZZ

xyz dV =(8/81)(243/8)=3,¯z=1

MZZZ

xz2dV =(8/81)(27/4)=2/3;

mass 81/8, center of gravity (8/5,3,2/3)

27. (a) M=Z1

0Z1

k(x2+y2)dy dx =2k/3, ¯x=¯yfrom the symmetry of density and region,

¯x=1

MZZ

kx(x2+y2)dA =3

2k(5k/12)=5/8; center of gravity (5/8,5/8)

(b) ¯y=1/2 from the symmetry of density and region,

M=Z1

0Z1

kx dy dx =k/2, ¯x=1

MZZ

kx2dA =(2/k)(k/3)=2/3,

center of gravity (2/3,1/2)

593 Chapter 16

28. (a) ¯x=¯y=¯zfrom the symmetry of density and region,

M=Z1

0Z1

k(x2+y2+z2)dz dy dx =k,

¯x=1

MZZZ

kx(x2+y2+z2)dV =(1/k)(7k/12)=7/12; center of gravity (7/12,7/12,7/12)

(b) ¯x=¯y=¯zfrom the symmetry of density and region,

M=Z1

0Z1

k(x+y+z)dz dy dx =3k/2,

¯x=1

MZZZ

kx(x+y+z)dV =2

3k(5k/6)=5/9; center of gravity (5/9,5/9,5/9)

29. V=ZZZ

dV =Zπ

0Zsin x

0Z1/(1+x2+y2)

dz dy dx =0.666633,

x=1

VZZZ

xdV =1.177406, y=1

VZZZ

ydV =0.353554, z=1

VZZZ

zdV =0.231557

30. (b) Use cylindrical coordinates to get

V=ZZZ

dV =Z2π

0Za

0Z1/(1+r2)

rdzdrdθ=πln(1 + a2),

z=1

VZZZ

zdV =a2

2(1 + a2) ln(1 + a2)

a→0+¯z=1

2; lim

a→+∞¯z= 0; solve z=1/4 for ato obtain a≈1.980291.

31. Let x=rcos θ,y=rsin θ, and dA =rdrdθ in formulas (10) and (11).

32. ¯x= 0 from the symmetry of the region, A=Z2π

0Za(1+sin θ)

rdrdθ=3πa2/2,

¯y=1

AZ2π

0Za(1+sin θ)

r2sin θdrdθ=2

3πa2(5πa3/4)=5a/6; centroid (0,5a/6)

33. ¯x=¯yfrom the symmetry of the region, A=Zπ/2

0Zsin 2θ

rdrdθ=π/8,

¯x=1

AZπ/2

0Zsin 2θ

r2cos θdrdθ=(8/π)(16/105) = 128

105π; centroid 128

105π,128

105π

34. ¯x=3/2 and ¯y= 1 from the symmetry of the region,

xdA=¯xA =(3/2)(6) = 9, ZZ

ydA=¯yA = (1)(6) = 6

35. ¯x= 0 from the symmetry of the region, πa2/2 is the area of the semicircle, 2π¯yis the distance

traveled by the centroid to generate the sphere so 4πa3/3=(πa2/2)(2π¯y), ¯y=4a/(3π)

Exercise Set 16.7 594

36. (a) V=1

2πa22πa+4a

3π=1

3π(3π+4)a

(b) the distance between the centroid and the line is √2

2a+4a

3πso

V=1

2πa2"2π√2

2a+4a

3π#=1

6√2π(3π+4)a

37. ¯x=kso V=(πab)(2πk)=2π

abk

38. ¯y= 4 from the symmetry of the region,

A=Z2

−2Z8−x2

dy dx =64/3soV= (64/3)[2π(4)] = 512π/3

39. The region generates a cone of volume 1

3πab2when it is revolved about the x-axis, the area of the

region is 1

2ab so 1

3πab2=1

2ab(2π¯y), ¯y=b/3. A cone of volume 1

3πa2bis generated when the

region is revolved about the y-axis so 1

3πa2b=1

2ab(2π¯x), ¯x=a/3. The centroid is (a/3,b/3).

40. Ix=Za

0Zb

y2δdydx=1

3δab3,I

0Zb

δdydx=1

3δa3b,

Iz=Za

0Zb

(x2+y2)δdydx=1

3δab(a2+b2)

41. Ix=Z2π

0Za

r3sin2θ δ dr dθ =δπa4/4; Iy=Z2π

0Za

r3cos2θ δ dr dθ =δπa4/4=I

;

z=I

x+I

y=δπa4/2

EXERCISE SET 16.7

1. Z2π

0Z1

0Z√1−r2

zr dz dr dθ =Z2π

0Z1

2(1 −r2)rdrdθ=Z2π

8dθ =π/4

2. Zπ/2

0Zcos θ

0Zr2

rsin θdzdrdθ =Zπ/2

0Zcos θ

r3sin θdrdθ=Zπ/2

4cos4θsin θdθ=1/20

3. Zπ/2

0Zπ/2

0Z1

ρ3sin φcos φdρdφdθ =Zπ/2

0Zπ/2

4sin φcos φdφdθ =Zπ/2

8dθ =π/16

4. Z2π

0Zπ/4

0Zasec φ

ρ2sin φdρdφdθ =Z2π

0Zπ/4

3a3sec3φsin φdφdθ =Z2π

3dθ =πa3/3

5. V=Z2π

0Z3

0Z9

rdzdrdθ =Z2π

0Z3

r(9 −r2)dr dθ =Z2π

4dθ =81π/2

595 Chapter 16

6. V=2Z2π

0Z2

0Z√

9−r

rdzdrdθ =2Z2π

0Z2

p9−r

dr dθ

3(27 −5√5) Z2π

dθ = 4(27 −5√5)π/3

7. r2+z2= 20 intersects z=r2in a circle of radius 2,

V=Z2π

0Z2

0Z√20−r2

rdzdrdθ=Z2π

0Z2

p20 −r2−r3)dr dθ

3(10√5−19) Z2π

dθ = 8(10√5−19)π/3

8. z=hr/a intersects z=hin a circle of radius a,

V=Z2π

0Za

0Zh

hr/a

rdzdrdθ =Z2π

0Za

a(ar −r2)dr dθ =Z2π

6a2hdθ =πa2h/3

9. V=Z2π

0Zπ/3

0Z4

ρ2sin φdρdφdθ =Z2π

0Zπ/3

3sin φdφdθ =32

3Z2π

dθ =64π/3

10. V=Z2π

0Zπ/4

0Z2

ρ2sin φdρdφdθ =Z2π

0Zπ/4

3sin φdφdθ =7

6(2−√2) Z2π

dθ = 7(2−√2)π/3

11. In spherical coordinates the sphere and the plane z=aare ρ=2aand ρ=asec φ, respectively.

They intersect at φ=π/3,

V=Z2π

0Zπ/3

0Zasec φ

ρ2sin φdρdφdθ+Z2π

0Zπ/2

π/3Z2a

ρ2sin φdρdφdθ

=Z2π

0Zπ/3

3a3sec3φsin φdφdθ+Z2π

0Zπ/2

π/3

3a3sin φdφdθ

3Z2π

dθ +4

3a3Z2π

dθ =11πa3/3

12. V=Z2π

0Zπ/2

π/4Z3

ρ2sin φdρdφdθ =Z2π

0Zπ/2

π/4

9 sin φdφdθ =9

√2

2Z2π

dθ =9

√

2π

13. Zπ/2

0Za

0Za2−r2

r3cos2θdzdrdθ =Zπ/2

0Za

(a2r3−r5) cos2θdrdθ

12a6Zπ/2

cos2θdθ=πa6/48

14. Zπ

0Zπ/2

0Z1

e−ρ3ρ2sin φdρdφdθ =1

3(1 −e−1)Zπ

0Zπ/2

sin φdφdθ =(1−e

−1

)π/3

15. Zπ/2

0Zπ/4

0Z√8

ρ4cos2φsin φdρdφdθ = 32(2√2−1)π/15

16. Z2π

0Zπ

0Z3

ρ3sin φdρdφdθ =81π

Exercise Set 16.7 596

18. Zπ/2

0Zπ/4

18 cos37 θcos φdφdθ =√2

36 Zπ/2

cos37 θdθ=4,294,967,296

755,505,013,725√2≈0.008040

19. (a) V=2Z2π

0Za

0Z√

−r

rdzdrdθ =4πa3/3

(b) V=Z2π

0Zπ

0Za

ρ2sin φdρdφdθ =4πa3/3

20. (a) Z2

0Z√4−x2

0Z√4−x2−y2

xyzdzdydx

=Z2

0Z√

4−x

2xy(4 −x2−y2)dy dx =1

8Z2

x(4 −x2)2dx =4/3

(b) Zπ/2

0Z2

0Z√4−r2

r3zsin θcos θdzdrdθ

=Zπ/2

0Z2

2(4r3−r5) sin θcos θdrdθ=8

3Zπ/2

sin θcos θdθ=4/3

0Zπ/2

0Z2

ρ5sin3φcos φsin θcos θdρdφdθ

=Zπ/2

0Zπ/2

3sin3φcos φsin θcos θdφdθ=8

3Zπ/2

sin θcos θdθ=4/3

21. M=Z2π

0Z3

(3 −z)rdzdrdθ =Z2π

0Z3

2r(3 −r)2dr dθ =27

8Z2π

dθ =27π/4

22. M=Z2π

0Za

0Zh

kzrdzdrdθ =Z2π

0Za

2kh2rdrdθ=1

4ka2h2Z2π

dθ =πka2h2/2

23. M=Z2π

0Zπ

0Za

kρ3sin φdρdφdθ =Z2π

0Zπ

4ka4sin φdφdθ =1

2ka4Z2π

dθ =πka4

24. M=Z2π

0Zπ

0Z2

ρsin φdρdφdθ =Z2π

0Zπ

2sin φdφdθ =3Z2π

dθ =6π

25. ¯x=¯y= 0 from the symmetry of the region,

V=Z2π

0Z1

0Z√2−r2

rdzdrdθ =Z2π

0Z1

p2−r

2−r

3)dr dθ =(8

√

2−7)π/6,

¯z=1

VZ2π

0Z1

0Z√2−r2

zr dz dr dθ =6

(8√2−7)π(7π/12)=7/(16√2−14);

centroid 0,0,7

16√2−14

26. ¯x=¯y= 0 from the symmetry of the region, V=8π/3,

¯z=1

VZ2π

0Z2

zr dz dr dθ =3

8π(4π)=3/2; centroid (0,0,3/2)

597 Chapter 16

27. ¯x=¯y=¯zfrom the symmetry of the region, V=πa3/6,

¯z=1

VZπ/2

0Zπ/2

0Za

ρ3cos φsin φdρdφdθ =6

πa3(πa4/16)=3a/8;

centroid (3a/8,3a/8,3a/8)

28. ¯x=¯y= 0 from the symmetry of the region, V=Z2π

0Zπ/3

0Z4

ρ2sin φdρdφdθ =64π/3,

¯z=1

VZ2π

0Zπ/3

0Z4

ρ3cos φsin φdρdφdθ =3

64π(48π)=9/4; centroid (0,0,9/4)

29. ¯y= 0 from the symmetry of the region, V=2Zπ/2

0Z2 cos θ

0Zr2

rdzdrdθ =3π/2,

¯x=2

VZπ/2

0Z2 cos θ

0Zr2

r2cos θdzdrdθ =4

3π(π)=4/3,

¯z=2

VZπ/2

0Z2 cos θ

0Zr2

rz dz dr dθ =4

3π(5π/6)=10/9; centroid (4/3,0,10/9)

30. M=Zπ/2

0Z2 cos θ

0Z4−r2

zr dz dr dθ =Zπ/2

0Z2 cos θ

2r(4 −r2)2dr dθ

=16

3Zπ/2

(1 −sin6θ)dθ = (16/3)(11π/32)=11π/6

31. V=Zπ/2

0Zπ/3

π/6Z2

ρ2sin φdρdφdθ =Zπ/2

0Zπ/3

π/6

3sin φdφdθ =4

√3−1) Zπ/2

dθ

=2(

√

3−1)π/3

32. M=Z2π

0Zπ/4

0Z1

ρ3sin φdρdφdθ =Z2π

0Zπ/4

4sin φdφdθ =1

8(2 −√2) Z2π

dθ =(2−√

2)π/4

33. ¯x=¯y= 0 from the symmetry of density and region,

M=Z2π

0Z1

0Z1−r2

(r2+z2)rdzdrdθ =π/4,

¯z=1

MZ2π

0Z1

0Z1−r2

z(r2+z2)rdzdrdθ =(4/π)(11π/120) = 11/30; center of gravity (0,0,11/30)

34. ¯x=¯y= 0 from the symmetry of density and region, M=Z2π

0Z1

0Zr

zr dz dr dθ =π/4,

¯z=1

MZ2π

0Z1

0Zr

z2rdzdrdθ =(4/π)(2π/15)=8/15; center of gravity (0,0,8/15)

35. ¯x=¯y= 0 from the symmetry of density and region,

M=Z2π

0Zπ/2

0Za

kρ3sin φdρdφdθ =πka4/2,

¯z=1

MZ2π

0Zπ/2

0Za

kρ4sin φcos φdρdφdθ =2

πka4(πka5/5)=2a/5; center of gravity (0,0,2a/5)

Exercise Set 16.7 598

36. ¯x=¯z= 0 from the symmetry of the region, V=54π/3−16π/3=38π/3,

¯y=1

VZπ

0Zπ

0Z3

ρ3sin2φsin θdρdφdθ =1

VZπ

0Zπ

4sin2φsin θdφdθ

VZπ

65π

8sin θdθ=3

38π(65π/4) = 195/152; centroid (0,195/152,0)

37. M=Z2π

0Zπ

0ZR

δ0e(ρ/R)3ρ2sin φdρdφdθ =Z2π

0Zπ

3(e−1)R3δ0sin φdφdθ =4

3π(e−1)δ0R3

38. (a) The sphere and cone intersect in a circle of radius ρ0sin φ0,

V=Zθ2

θ1Zρ0sin φ0

0Z√ρ2

0−r2

rcot φ0

rdzdrdθ =Zθ

1Zρ

0sin φ0

0rqρ2

0−r2−r2cot φ0dr dθ

=Zθ2

θ1

3ρ3

0(1 −cos3φ0−sin3φ0cot φ0)dθ =1

3ρ3

0(1 −cos3φ0−sin2φ0cos φ0)(θ2−θ1)

3ρ3

0(1 −cos φ0)(θ2−θ1).

(b) From part (a), the volume of the solid bounded by θ=θ1,θ=θ2,φ=φ1,φ=φ2, and

ρ=ρ0is 1

3ρ3

0(1 −cos φ2)(θ2−θ1)−1

3ρ3

0(1 −cos φ1)(θ2−θ1)=1

(cos φ1−cos φ2)(θ2−θ1)

so the volume of the spherical wedge between ρ=ρ1and ρ=ρ2is

∆V=1

3ρ3

2(cos φ1−cos φ2)(θ2−θ1)−1

3ρ3

1(cos φ1−cos φ2)(θ2−θ1)

3(ρ3

2−ρ3

1)(cos φ1−cos φ2)(θ2−θ1)

dφ cos φ=−sin φso from the Mean-Value Theorem cosφ2−cos φ1=−(φ2−φ1) sin φ∗where

φ∗is between φ1and φ2. Similarly d

dρρ3=3ρ

2so ρ3

2−ρ3

1=3ρ

∗2

(ρ

2−ρ

) where ρ∗is between

ρ1and ρ2. Thus cos φ1−cos φ2= sin φ∗∆φand ρ3

2−ρ3

1=3ρ

∗2

∆ρso ∆V=ρ∗2sin φ∗∆ρ∆φ∆θ.

39. Iz=Z2π

0Za

0Zh

r2δ r dz dr dθ =δZ2π

0Za

0Zh

r3dz dr dθ =1

2δπa4h

40. Iy=Z2π

0Za

0Zh

(r2cos2θ+z2)δr dz dr dθ =δZ2π

0Za

(hr3cos2θ+1

3h3r)dr dθ

=δZ2π

01

4a4hcos2θ+1

6a2h3dθ =δπ

4a4h+π

3a2h3

41. Iz=Z2π

0Za2

a1Zh

r2δ r dz dr dθ =δZ2π

0Za2

a1Zh

r3dz dr dθ =1

2δπh(a4

2−a4

42. Iz=Z2π

0Zπ

0Za

(ρ2sin2φ)δρ

2sin φdρdφdθ =δZ2π

0Zπ

0Za

4sin3φdρdφdθ =8

15δπa5

599 Chapter 16

EXERCISE SET 16.8

1. ∂(x, y)

∂(u, v)=

3−5



=−17 2. ∂(x, y)

∂(u, v)=

14v

4u−1



=−1−16uv

3. ∂(x, y)

∂(u, v)=

cos u−sin v

sin ucos v

= cos ucos v+ sin usin v= cos(u−v)

4. ∂(x, y)

∂(u, v)=

2(v2−u2)

(u2+v2)2−4uv

(u2+v2)2

4uv

(u2+v2)2

2(v2−u2)

(u2+v2)2



=4/(u

2+v

)

5. x=2

9u+5

9v,y=−1

9u+2

9v;∂(x, y)

∂(u, v)=

2/95/9

−1/92/9



6. x=lnu,y=uv;∂(x, y)

∂(u, v)=

1/u 0



7. x=√u+v/√2, y=√v−u/√2; ∂(x, y)

∂(u, v)=

2√2√u+v

−1

2√2√v−u



4√v2−u2

8. x=u3/2/v1/2,y=v1/2/u1/2;∂(x, y)

∂(u, v)=

3u1/2

2v1/2−u3/2

2v3/2

−v1/2

2u3/2

2u1/2v1/2



9. ∂(x, y, z)

∂(u, v, w)=

31 0

10−2

01 1



10. ∂(x, y, z)

∂(u, v, w)=

1−v−u0

v−vw u −uw −uv

vw uw uv 

=u2v

11. y=v, x =u/y =u/v, z =w−x=w−u/v;∂(x, y, z)

∂(u, v, w)=

1/v −u/v20

010

−1/v u/v21

=1/v

12. x=(v+w)/2,y =(u−w)/2,z =(u−v)/2,∂(x, y, z)

∂(u, v, w)=

01/21/2

1/20−1/2

1/2−1/20



=−

Exercise Set 16.8 600

13.

(0, 2)

(–1, 0) (1, 0)

(0, 0)

14.

1 2 3

(0, 0) (4, 0)

(3, 4)

15.

-3 3

-3

(2, 0)

(0, 3) 16.

1 2

17. x=1

5u+2

5v,y=−2

5u+1

5v,∂(x, y)

∂(u, v)=1

5;1

5ZZ

vdAuv =1

5Z3

1Z4

vdu dv =3

2ln 3

18. x=1

2u+1

2v,y=1

2u−1

2v,∂(x, y)

∂(u, v)=−1

2;1

2ZZ

veuvdAuv =1

2Z4

1Z1

veuvdu dv =1

2(e4−e−3)

19. x=u+v,y=u−v,∂(x, y)

∂(u, v)=−2; the boundary curves of the region Sin the uv-plane are

v=0,v =u, and u= 1 so 2 ZZ

sin ucos vdAuv =2Z1

0Zu

sin ucos vdvdu=1−1

2sin 2

20. x=pv/u,y=√uv so, from Example 3, ∂(x, y)

∂(u, v)=−1

2u; the boundary curves of the region Sin

the uv-plane are u=1,u=3,v = 1, and v=4soZZ

uv21

2udAuv =1

2Z4

1Z3

v2du dv =21

21. x=3u, y =4v, ∂(x, y)

∂(u, v)= 12; Sis the region in the uv-plane enclosed by the circle u2+v2=1.

Use polar coordinates to obtain ZZ

12pu2+v2(12) dAuv = 144 Z2π

0Z1

r2dr dθ =96π

22. x=2u, y =v, ∂(x, y)

∂(u, v)=2;Sis the region in the uv-plane enclosed by the circle u2+v2= 1. Use

polar coordinates to obtain ZZ

e−(4u2+4v2)(2) dAuv =2Z2π

0Z1

re−4r2dr dθ =(1−e

−4

)π/2

601 Chapter 16

23. Let Sbe the region in the uv-plane bounded by u2+v2=1,sou=2x, v =3y,

x=u/2,y =v/3,∂(x, y)

∂(u, v)=

1/20

01/3



=1/6, use polar coordinates to get

6ZZ

sin(u2+v2)du dv =1

6Zπ/2

0Z1

rsin r2dr dθ =π

24(−cos r2)i1

0=π

24(1 −cos 1)

24. u=x/a, v =y/b, x =au, y =bv;∂(x, y)

∂(u, v)=ab;A=ab Z2π

0Z1

rdrdθ=πab

25. x=u/3,y =v/2,z =w, ∂(x, y, z)

∂(u, v, w)=1/6; Sis the region in uvw-space enclosed by the sphere

u2+v2+w2=36so

ZZZ

6dVuvw =1

54 Z2π

0Zπ

0Z6

(ρsin φcos θ)2ρ2sin φdρdφdθ

54 Z2π

0Zπ

0Z6

ρ4sin3φcos2θdρdφdθ=192

5π

26. Let G1be the region u2+v2+w2≤1, let x=au, y =bv, z =cw, ∂(x, y, z)

∂(u, v, w)=abc;

Ix=ZZZ

(y2+z2)dx dy dz =ZZZ

(b2v2+c2w2)du dv dw

=Z2π

0Zπ

0Z1

abc(b2sin2φsin2θ+c2cos2φ)ρ4sin φdρdφdθ

=Z2π

abc

15 (4b2sin2θ+2c

)dθ =4

15πabc(b2+c2)

27. Let u=y−4x, v =y+4x, then x=1

8(v−u),y =1

2(v+u)so∂(x, y)

∂(u, v)=−1

8ZZ

vdAuv =1

8Z5

2Z2

vdu dv =1

4ln 5

28. Let u=y+x, v =y−x, then x=1

2(u−v),y =1

2(u+v)so∂(x, y)

∂(u, v)=1

−1

2ZZ

uv dAuv =−1

2Z2

0Z1

uv du dv =−1

29. Let u=x−y, v =x+y, then x=1

2(v+u),y =1

2(v−u)so∂(x, y)

∂(u, v)=1

2; the boundary curves of

the region Sin the uv-plane are u=0,v =u, and v=π/4; thus

2ZZ

sin u

cos vdAuv =1

2Zπ/4

0Zv

sin u

cos vdu dv =1

2[ln(√2+1)−π/4]

Exercise Set 16.8 602

30. Let u=y−x, v =y+x, then x=1

2(v−u),y =1

2(u+v)so∂(x, y)

∂(u, v)=−1

2; the boundary

curves of the region Sin the uv-plane are v=−u, v =u, v = 1, and v=4;thus

2ZZ

eu/v dAuv =1

2Z4

1Zv

−v

eu/vdu dv =15

4(e−e−1)

31. Let u=y/x, v =x/y2, then x=1/(u

v),y =1/(uv)so∂(x, y)

∂(u, v)=1

u4v3;

u4v3dAuv =Z4

1Z2

u4v3du dv =35/256

32. Let x=3u, y =2v,∂(x, y)

∂(u, v)=6;Sis the region in the uv-plane enclosed by the circle u2+v2=1

so ZZ

(9 −x−y)dA =ZZ

6(9 −3u−2v)dAuv =6Z2π

0Z1

(9 −3rcos θ−2rsin θ)rdrdθ=54π

33. x=u, y =w/u, z =v+w/u, ∂(x, y, z)

∂(u, v, w)=−1

ZZZ

v2w

udVuvw =Z4

2Z1

0Z3

v2w

udu dv dw = 2 ln 3

34. u=xy, v =yz, w =xz, 1≤u≤2,1≤v≤3,1≤w≤4,

x=puw/v, y =puv/w, z =pvw/u, ∂(x, y, z)

∂(u, v, w)=1

2√uvw

V=ZZZ

dV =Z2

1Z3

1Z4

2√uvw dw dv du =4(

√

2−1)(√3−1)

35. (b) If x=x(u, v),y =y(u, v) where u=u(x, y),v =v(x, y), then by the chain rule

∂x

∂u

∂x +∂x

∂v

∂x =∂x

∂x =1,∂x

∂u

∂y +∂x

∂v

∂y =∂x

∂y =0

∂y

∂u

∂x +∂y

∂v

∂x =∂y

∂x =0,∂y

∂u

∂y +∂y

∂v

∂y =∂y

∂y =1

36. (a) ∂(x, y)

∂(u, v)=

1−v−u



=u;u=x+y, v =y

x+y,

∂(u, v)

∂(x, y)=

−y/(x+y)2x/(x+y)2

(x+y)2+y

(x+y)2=1

x+y=1

∂(u, v)

∂(x, y)

∂(u, v)=1

(b) ∂(x, y)

∂(u, v)=

02v



=2v

;u=x/√y, v =√y,

∂(u, v)

∂(x, y)=

1/√y−x/(2y−3/2)

01/(2√y)

2y=1

2v2;∂(u, v)

∂(x, y)

∂(u, v)=1

603 Chapter 16

∂(u, v)=

u−v



=−2uv;u=√x+y, v =√x−y,

∂(u, v)

∂(x, y)=

1/(2√x+y)1/(2√x+y)

1/(2√x−y)−1/(2√x−y)

=−1

2px2−y2=−1

2uv ;∂(u, v)

∂(x, y)

∂(u, v)=1

37. ∂(u, v)

∂(x, y)=3xy4=3vso ∂(x, y)

∂(u, v)=1

3v;1

3ZZ

sin u

vdAuv =1

3Z2

1Z2π

sin u

vdu dv =−2

3ln 2

38. ∂(u, v)

∂(x, y)=8xy so ∂(x, y)

∂(u, v)=1

8xy ;xy 

∂(x, y)

∂(u, v)

=xy 1

8xy =1

8so

8ZZ

dAuv =1

8Z16

9Z4

du dv =21/8

39. ∂(u, v)

∂(x, y)=−2(x2+y2)so∂(x, y)

∂(u, v)=−1

2(x2+y2);

(x4−y4)exy 

∂(x, y)

∂(u, v)

=x4−y4

2(x2+y2)exy =1

2(x2−y2)exy =1

2veuso

2ZZ

veudAuv =1

2Z4

3Z3

veudu dv =7

4(e3−e)

40. Set u=x+y+2z, v =x−2y+z,w =4x+y+z, then ∂(u, v, w)

∂(x, y, z)=

112

1−21

411



= 18, and

V=ZZZ

dx dy dz =Z6

−6Z2

−2Z3

−3

∂(x, y, z)

∂(u, v, w)du dv dw = 6(4)(12) 1

18 =16

41. (a) Let u=x+y, v =y, then the triangle Rwith vertices (0,0),(1,0) and (0,1) becomes the

triangle in the uv-plane with vertices (0,0),(1,0),(1,1), and

f(x+y)dA =Z1

0Zu

f(u)∂(x, y)

∂(u, v)dv du =Z1

uf(u)du

(b) Z1

ueudu =(u−1)eui1

0=1

42. (a) ∂(x, y)

∂(r, θ)=

cos θ−rsin θ

sin θrcos θ

=r,

∂(x, y)

∂(r, θ)

(b) ∂(x, y, z)

∂(ρ, θ, φ)=

sin φcos θ−ρsin φsin θρcos φcos θ

sin φsin θρsin φcos θρcos φsin θ

cos φ0−ρsin φ

=−ρ2sin φ;

∂(x, y, z)

∂(ρ, θ, φ)

=ρ2sin φ

Chapter 16 Supplementary Exercises 604

CHAPTER 16 SUPPLEMENTARY EXERCISES

3. (a) ZZ

dA (b) ZZZ

dV (c) ZZ

Rs∂z

∂x2

+∂z

∂y2

4. (a) x=asin φcos θ, y =asin φsin θ, z =ρcos φ, 0≤θ≤2π, 0≤φ≤π

(b) x=acos θ, y =asin θ, z =z,0≤θ≤2π, 0≤z≤h

7. Z1

0Z1+√1−y2

1−√1−y2

f(x, y)dx dy 8. Z2

0Z2x

f(x, y)dy dx +Z3

2Z6−x

f(x, y)dy dx

9. (a) (1,2)=(b, d),(2,1)=(a, c), so a=2,b =1,c=1,d=2

(b) ZZ

dA =Z1

0Z1

∂(x, y)

∂(u, v)du dv =Z1

0Z1

3du dv =3

10. 0<sin √xy < 1 for 0 < x,y < π,so

0=Zπ

0Zπ

0dy dx < Zπ

0Zπ

sin √xy dy dx < Zπ

0Zπ

1dy dx =π2

11. Z1

1/2

2xcos(πx2)dx =1

πsin(πx2)i1

1/2=−1/(√2π)

12. Z2

2ey3x=2y

x=−y

dy =3

2Z2

y2ey3dy =1

2ey32

2e8−1

13. Z1

0Z2

exeydx dy 14. Zπ

0Zx

sin x

xdy dx

15.

y = sin x

y = tan (x/2)

16.

p/2

p/6

r = a

r = a(1 + cos u)

17. 2Z8

0Zy1/3

x2sin y2dx dy =2

3Z8

ysin y2dy =−1

3cos y28

3(1 −cos 64) ≈0.20271

18. Zπ/2

0Z2

(4 −r2)rdrdθ=2π

605 Chapter 16

19. sin 2θ= 2 sin θcos θ=2xy

x2+y2, and r=2asin θis the circle x2+(y−a)

2=a

,so

0Za+

√

−x

a−

√

−x

2xy

x2+y2dy dx =Za

xhln a+pa2−x2−ln a−pa2−x2idx =a2

20. Zπ/2

π/4Z2

4r2(cos θsin θ)rdrdθ=−4 cos 2θπ/2

π/4

21. Z2π

0Z2

0Z16

r2cos2θrdzdrdθ =Z2π

cos2θdθZ2

3(16 −r4)dr =32π

22. Zπ/2

0Zπ/2

0Z1

1+ρ

2ρ

2sin φdρdφdθ =1−π

4π

2Zπ/2

sin φdφ

=1−π

4π

2(−cos φ)iπ/2

0=1−π

4π

23. (a) Z2π

0Zπ/3

0Za

(ρ2sin2φ)ρ2sin φdρdφdθ =Z2π

0Zπ/3

0Za

ρ4sin3φdρdφdθ

(b) Z2π

0Z√3a/2

0Z√a2−r2

r/√3

r2dz rdr dθ =Z2π

0Z√3a/2

0Z√a2−r2

r/√3

r3dz dr dθ

−√3a/2Z√(3a2/4)−x2

−√(3a2/4)−x2Z√a2−x2−y2

√x2+y2/√3

(x2+y2)dz dy dx

24. (a) Z4

0Z√4x−x2

−√4x−x2Z4x

x2+y2

dz dy dx

(b) Zπ/2

−π/2Z4 cos θ

0Z4rcos θ

rdzdrdθ

25. Z2

0Z2−y/2

(y/2)1/3

dx dy =Z2

02−y

2−y

21/3dy =2y−y2

4−3

2y

24/32

26. A=6Zπ/6

0Zcos 3θ

rdrdθ=3Zπ/6

cos23θ=π/4

27. V=Z2π

0Za/√3

0Za

√3r

rdzdrdθ =2πZa/√3

r(a−√3r)dr =πa3

28. The intersection of the two surfaces projects onto the yz-plane as 2y2+z2=1,so

V=4Z1/

√

0Z√

1−2y

0Z1−y

dx dz dy

=4Z1/

√

0Z√

1−2y

(1 −2y2−z2)dz dy =4Z1/

√

(1 −2x2)3/2dx =√2π

29. kru×rvk=√2u2+2v

2+4,

S=ZZ

u2+v2≤4p2u2+2v

2+4dA =Z2π

0Z2

√2pr2+2rdrdθ =8π

3(3√3−1)

Chapter 16 Supplementary Exercises 606

30. kru×rvk=√1+u

,S=

0Z3u

1+u

dv du =Z2

3up1+u

du =5

3/2−1

31. (ru×rv)iu=1

v=2

=h−2,−4,1i, tangent plane 2x+4y−z=5

32. u=−3,v =0,(r

u×r

)

iu=−3

v=0

=h−18,0,−3i, tangent plane 6x+z=−9

33. A=Z4

−4Z2+y2/8

y2/4

dx dy =Z4

42−y2

8dy =32

3;¯y= 0 by symmetry;

−4Z2+y2/8

y2/4

xdxdy =Z4

−42+1

2−3

128y4dy =256

15 ,¯x=3

256

15 =8

5; centroid 8

5,0

34. A=πab/2,¯x= 0 by symmetry,

−aZb√1−x2/a2

ydydx=1

2Za

−a

(1 −x2/a2)dx =2ab2/3, centroid 0,4b

3π

35. V=1

3πa2h, ¯x=¯y= 0 by symmetry,

Z2π

0Za

0Zh−rh/a

rz dz dr dθ =πZa

rh21−r

a2dr =πa2h2/12, centroid (0,0,h/4)

36. V=Z2

−2Z4

x2Z4−y

dz dy dx =Z2

−2Z4

(4 −y)dy dx =Z2

−28−4x2+1

2x4dx =256

15 ,

−2Z4

x2Z4−y

ydzdydx=Z2

−2Z4

(4y−y2)dy dx =Z2

−21

3x6−2x4+32

3dx =1024

−2Z4

x2Z4−y

zdzdydx=Z2

−2Z4

2(4 −y)2dy dx =Z2

−2−x6

6+2x

4−8x

2+32

3dx =2048

105

¯x= 0 by symmetry, centroid 0,8

7,4

37. The two quarter-circles with center at the origin and of radius Aand √2Alie inside and outside

of the square with corners (0,0),(A, 0),(A, A),(0,A), so the following inequalities hold:

Zπ/2

0ZA

(1+r2)2rdr dθ ≤ZA

0ZA

(1+x2+y2)2dx dy ≤Zπ/2

0Z√2A

(1+r2)2rdr dθ

The integral on the left can be evaluated as πA2

4(1 + A2)and the integral on the right equals

2πA2

4(1+2A

). Since both of these quantities tend to π

4, it follows by sandwiching that

Z+∞

0Z+∞

(1+x2+y2)2dx dy =π

38. The centroid of the circle which generates the tube travels a distance

s=Z4π

0qsin2t+ cos2t+1/16 dt =√17π,soV=π(1/2)2√17π=√17π2/4.

607 Chapter 16

39. (a) The values of xin the formula of the astroidal sphere lie between −aand a; and the same is

true for x=(acos ucos v)3; similarly for yand z. Moreover,

x2/3+y2/3+z2/3=a2/3cos2ucos2v+a2/3sin2ucos2v+a2/3sin2v

=a2/3cos2v+a2/3sin2v=a2/3

(b) S=8Zπ/2

0Zπ/2

0kru×rvkdu dv

=72Zπ/2

0Zπ/2

sin ucos usin vcos4vpsin2v+ sin2ucos2ucos2vdu dv ≈4.451

0Z1

0p1−x2/3−y2/3dy dx ≈0.3590

(d) Let x=tcos3u, y =tsin3u, 0≤t≤1,0≤u≤π/2, then J=∂(x, y)

∂(t, u)=3tsin2ucos2u,

V=8Z1

0Zπ/2

(1 −t2/3)3/23tsin2ucos2ududt=3π

2Z1

t(1 −t2/3)3/2dt =4π

40. V=4

3πa3,¯

d=3

4πa3ZZZ

ρ≤a

ρdV =3

4πa3Zπ

0Z2π

0Za

ρ3sin φdρdθdφ =3

4πa32π(2)a4

4=3

41. (a) (x/a)2+(y/b)2+(z/c)2= sin2φcos2θ+sin2φsin2θ+cos2φ= sin2φ+cos2φ= 1, an ellipsoid

(b) r(φ, θ)=h2 sin φcos θ, 3 sin φsin θ, 4 cos φi;rφ×rθ=2h6 sin2φcos θ, 4 sin2φsin θ, 3 cos φsin φi,

krφ×rθk=2

p16 sin4φ+ 20 sin4φcos2θ+ 9 sin2φcos2φ,

S=Z2π

0Zπ

2q16 sin4φ+ 20 sin4φcos2θ+ 9 sin2φcos2φdφdθ ≈111.5457699

608

CHAPTER 17

Topics in Vector Calculus

EXERCISE SET 17.1

1. (a) III because the vector ﬁeld is independent of yand the direction is that of the negative x-axis

for negative x, and positive for positive

(b) IV, because the y-component is always positive, and the x-component is positive for positive

x, negative for negative x

2. (a) I, since the vector ﬁeld is constant

(b) II, since the vector ﬁeld points away from the origin

3. (a) true (b) true (c) true

4. (a) false, the lengths are equal to 1 (b) false, the y-component is then zero

y6.

y7.

y9.

y10.

11. (a) ∇φ=φxi+φyj=y

1+x

2i+x

1+x

2j=F,soFis conservative for all x, y

(b) ∇φ=φxi+φyj=2xi−6yj+8zk=Fso Fis conservative for all x, y

12. (a) ∇φ=φxi+φyj=(6xy −y3)i+(4y+3x

2−3xy2)j=F,soFis conservative for all x, y

(b) ∇φ=φxi+φyj+φzk= (sin z+ycos x)i+ (sin x+zcos y)j+(xcos z+ sin y)k=F,soFis

conservative for all x, y

609 Chapter 17

13. div F=2x+y, curl F=zi

14. div F=z3+8y

2+10zy, curl F=5z

i+3xz2j+4xy4k

15. div F= 0, curl F= (40x2z4−12xy3)i+ (14y3z+3y

)j−(16xz5+21y

16. div F=yexy + sin y+ 2 sin zcos z, curl F=−xexyk

17. div F=2

px2+y2+z2, curl F=0

18. div F=1

x+xzexyz +x

x2+z2, curl F=−xyexyz i+z

x2+z2j+yzexyzk

19. ∇·(F×G)=∇·(−(z+4y

)i+(4xy +2xz)j+(2xy −x)k)=4x

20. ∇·(F×G)=∇·((x2yz2−x2y2)i−xy2z2j+xy2zk)=−xy2

21. ∇·(∇×F)=∇·(−sin(x−y)k)=0

22. ∇·(∇×F)=∇·(−zeyzi+xexzj+3e

yk)=0

23. ∇×(∇×F)=∇×(xzi−yzj+yk)=(1+y)i+xj

24. ∇×(∇×F)=∇×((x+3y)i−yj−2xyk)=−2xi+2yj−3k

25. div (kF)=k

∂f

∂x +k∂q

∂y +k∂h

∂z =kdiv F

26. curl (kF)=k∂h

∂y −∂g

∂zi+k∂f

∂z −∂h

∂xj+k∂g

∂x −∂f

∂y k=kcurl F

27. Let F=f(x, y, z)i+g(x, y, z)j+h(x, y, z)kand G=P(x, y, z)i+Q(x, y, z)j+R(x, y, z)k, then

div (F+G)=∂f

∂x +∂P

∂x +∂g

∂y +∂Q

∂y +∂h

∂z +∂R

∂z 

=∂f

∂x +∂g

∂y +∂h

∂z +∂P

∂x +∂Q

∂y +∂R

∂z = div F+ div G

28. Let F=f(x, y, z)i+g(x, y, z)j+h(x, y, z)kand G=P(x, y, z)i+Q(x, y, z)j+R(x, y, z)k, then

curl (F+G)=∂

∂y(h+R)−∂

∂z(g+Q)i+∂

∂z(f+P)−∂

∂x(h+R)j

+∂

∂x(g+Q)−∂

∂y(f+P)k;

expand and rearrange terms to get curl F+ curl G.

29. div (φF)=φ

∂f

∂x +∂φ

∂xf+φ∂g

∂y +∂φ

∂y g+φ∂h

∂z +∂φ

∂z h

=φ∂f

∂x +∂g

∂y +∂h

∂z +∂φ

∂xf+∂φ

∂y g+∂φ

∂z h

=φdiv F+∇φ·F

Exercise Set 17.1 610

30. curl (φF)=∂

∂y(φh)−∂

∂z(φg)i+∂

∂z(φf)−∂

∂x(φh)j+∂

∂x(φg)−∂

∂y(φf )k; use the product

rule to expand each of the partial derivatives, rearrange to get φcurl F+∇φ×F

31. div(curl F)= ∂

∂x ∂h

∂y −∂g

∂z+∂

∂y ∂f

∂z −∂h

∂x+∂

∂z ∂g

∂x −∂f

∂y 

=∂2h

∂x∂y −∂2g

∂x∂z +∂2f

∂y∂z −∂2h

∂y∂x +∂2g

∂z∂x −∂2f

∂z∂y =0,

assuming equality of mixed second partial derivatives

32. curl (∇φ)=∂

∂y∂z −∂2φ

∂z∂yi+∂2φ

∂z∂x −∂2φ

∂x∂z j+∂2φ

∂x∂y −∂2φ

∂y∂xk=0, assuming equality

of mixed second partial derivatives

33. ∇·(kF)=k∇·F,∇·(F+G)=∇·F+∇·G,∇·(φF)=φ∇·F+∇φ·F,∇·(∇×F)=0

34. ∇×(kF)=k∇×F,∇×(F+G)=∇×F+∇×G,∇×(φF)=φ∇×F+∇φ×F,∇×(∇φ)=0

37. (a) curl r=0i+0j+0k=0

(b) ∇krk=∇px2+y2+z2=x

px2+y2+z2i+y

px2+y2+z2j+z

px2+y2+z2k=r

krk

38. (a) div r=1+1+1=3

(b) ∇1

krk=∇(x2+y2+z2)−1/2=−xi+yj+zk

(x2+y2+z2)3/2=−r

krk3

39. (a) ∇f(r)=f

(r)∂r

∂xi+f0(r)∂r

∂yj+f0(r)∂r

∂zk=f0(r)∇r=f0(r)

(b) div[f(r)r]=f(r)div r+∇f(r)·r=3f(r)+f0(r)

rr·r=3f(r)+rf0(r)

40. (a) curl[f(r)r]=f(r)curl r+∇f(r)×r=f(r)0+f0(r)

rr×r=0+0=0

(b) ∇2f(r) = div[∇f(r)] = div f0(r)

rr=f0(r)

rdiv r+∇f0(r)

·r

(r)

r+rf00(r)−f0(r)

r3r·r=2

(r)

r+f

00(r)

41. f(r)=1/r3,f0(r)=−3/r4, div(r/r3) = 3(1/r3)+r(−3/r4)=0

42. Multiply 3f(r)+rf0(r) = 0 through by r2to obtain 3r2f(r)+r

f0(r)=0,

d[r

f(r)]/dr =0,r

f(r)=C, f(r)=C/r3,soF=Cr/r3(an inverse-square ﬁeld).

43. (a) At the point (x, y) the slope of the line along which the vector −yi+xjlies is −x/y; the

slope of the tangent line to Cat (x, y)isdy/dx,sody/dx =−x/y.

(b) ydy =−xdx, y2/2=−x

/2+K

2+y

2=K

611 Chapter 17

44. dy/dx =x, y =x2/2+K

45. dy/dx =1/x, y =lnx+K

46. dy/dx =−y/x, (1/y)dy =(−1/x)dx, ln y=−ln x+K1,

y=eK1e−ln x=K/x

EXERCISE SET 17.2

1. (a) Z1

dy = 1 because s=yis arclength measured from (0,0)

(b) 0, because sinxy = 0 along C

2. (a) Z

ds = length of line segment = 2 (b) 0, because xis constant and dx =0

3. (a) ds =sdx

dt 2

+dy

dt 2

dt,soZ1

(2t−3t2)p4+36t

2dt =−11

108√10 −1

36 ln(√10−3) −4

(b) Z1

(2t−3t2)2 dt =0 (c) Z1

(2t−3t2)6tdt=−1

4. (a) Z1

t(3t2)(6t3)2p1+36t

2+ 324t4dt =864

5(b) Z1

t(3t2)(6t3)2dt =54

t(3t2)(6t3)26tdt=648

11 (d) Z1

t(3t2)(6t3)218t2dt = 162

5. (a) C:x=t,y=t,0≤t≤1; Z1

6tdt=3

(b) C:x=t,y=t2,0≤t≤1; Z1

(3t+6t

2−2t

)dt =3

Exercise Set 17.2 612

[3t+ 2 sin(πt/2)+πt cos(πt/2) −(π/2) sin(πt/2) cos(πt/2)]dt =3

(d) C:x=t3,y=t,0≤t≤1; Z1

(9t5+8t

3−t)dt =3

6. (a) C:x=t, y =t, z =t, 0≤t≤1; Z1

(t+t−t)dt =1

(b) C:x=t, y =t2,z =t

3,0≤t≤1; Z1

(t2+t3(2t)−t(3t2)) dt =−1

(−πsin2πt +πt cos πt −cos πt)dt =−π

2−2

7. Z3

√1+t

1+tdt =Z3

(1+t)−1/2dt =2 8. √5Z1

1+2t

1+t

2dt =√5(π/4+ln2)

9. Z1

3(t2)(t2)(2t3/3)(1 + 2t2)dt =2Z1

(1+2t

)dt =13/20

10. √5

4Z2π

e−tdt =√5(1 −e−2π)/411. Zπ/4

(8 cos2t−16 sin2t−20 sin tcos t)dt =1−π

12. Z1

−12

3t−2

3t5/3+t2/3dt =6/5

13. C:x=(3−t)

/3, y=3−t,0≤t≤3; Z3

3(3 −t)2dt =3

14. C:x=t2/3,y=t,−1≤t≤1; Z1

−12

3t2/3−2

3t1/3+t7/3dt =4/5

15. C:x= cos t,y= sin t,0≤t≤π/2; Zπ/2

(−sin t−cos2t)dt =−1−π/4

16. C:x=3−t,y=4−3t,0≤t≤1; Z1

(−37+41t−9t

)dt =−39/2

17. Z1

(−3)e3tdt =1−e

18. Zπ/2

(sin2tcos t−sin2tcos t+t4(2t)) dt =π6

192

20. (a) Zπ/2

cos21 tsin9tq(−3 cos2tsin t)2+ (3 sin2tcos t)2dt

Zπ/2

cos22 tsin10 tdt=61,047

4,294,967,296π

(b) Ze

1t5ln t+7t

(2t)+t

(ln t)1

tdt =5

36e6+59

16e4−491

144

613 Chapter 17

21. (a) C1:(0,0) to (1,0); x=t, y =0,0≤t≤1

2:(1,0) to (0,1); x=1−t, y =t, 0≤t≤1

C3:(0,1) to (0,0); x=0,y =1−t, 0≤t≤1

(0)dt +Z1

(−1)dt +Z1

(0)dt =−1

(b) C1:(0,0) to (1,0); x=t, y =0,0≤t≤1

2:(1,0) to (1,1); x=1,y =t, 0≤t≤1

C3:(1,1) to (0,1); x=1−t, y =1,0≤t≤1

4:(0,1) to (0,0); x=0,y =1−t, 0≤t≤1

(0)dt +Z1

(−1)dt +Z1

(0)dt =−2

22. (a) C1:(0,0) to (1,1); x=t, y =t, 0≤t≤1

C2:(1,1) to (2,0); x=1+t, y =1−t, 0≤t≤1

C3:(2,0) to (0,0); x=2−2t, y =0,0≤t≤1

(0)dt +Z1

2dt +Z1

(0)dt =2

(b) C1:(−5,0) to (5,0); x=t, y =0,−5≤t≤5

2:x= 5 cos t, y = 5 sin t, 0≤t≤π

−5

(0)dt +Zπ

(−25)dt =−25π

23. C1:x=t, y =z=0,0≤t≤1,Z1

0dt =0; C

2:x=1,y =t, z =0,0≤t≤1,Z1

(−t)dt =−1

C3:x=1,y =1,z =t, 0≤t≤1,Z1

3dt =3; ZC

zdx−yx2dy +3dz =0−1

2+3= 5

24. C1:(0,0,0) to (1,1,0); x=t, y =t, z =0,0≤t≤1

2:(1,1,0) to (1,1,1); x=1,y =1,z =t, 0≤t≤1

C3:(1,1,1) to (0,0,0); x=1−t, y =1−t, z =1−t, 0≤t≤1

(−t3)dt +Z1

3dt +Z1

0−3dt =−1/4

25. Zπ

(0)dt =0 26. Z1

(e2t−4e−t)dt =e2/2+4e

−1−9/2

27. Z1

e−tdt =1−e

−128. Zπ/2

(7 sin2tcos t+ 3 sin tcos t)dt =23/6

29. Represent the circular arc by x= 3 cos t, y = 3 sin t, 0≤t≤π/2.

x√yds =9

√

3Zπ/2

√sin tcos tdt=6

√

30. δ(x, y)=k

2+y

2where kis the constant of proportionality,

kpx2+y2ds =kZ1

et(√2et)dt =√2kZ1

e2tdt =(e

2−1)k/√2

Exercise Set 17.2 614

31. ZC

1+y

2ds =15kZπ/2

cos t

1+9sin

2tdt =5ktan−13

32. δ(x, y, z)=kz where kis the constant of proportionality,

kzds =Z4

k(4√t)(2+1/t)dt = 136k/3

33. C:x=t2,y=t,0≤t≤1; W=Z1

3t4dt =3/5

34. W=Z3

(t2+1−1/t3+1/t)dt =92/9+ln3 35. W=Z1

(t3+5t

)dt =27/28

36. C1:(0,0,0) to (1,3,1); x=t,y=3t,z=t,0≤t≤1

2:(1,3,1) to (2,−1,4); x=1+t,y=3−4t,z=1+3t,0≤t≤1

W=Z1

(4t+8t

)dt +Z1

(−11 −17t−11t2)dt =−37/2

37. Since Fand rare parallel, F·r=kFkkrk, and since Fis constant,

ZF·dr=Z

d(F·r)=Z

d(kFkkrk)=√

2Z4

−4

√

2dt =16

F·r=√

2(4√2)

38. Z

F·r= 0, since Fis perpendicular to the curve

39. C:x= 4 cos t, y = 4 sin t, 0≤t≤π/2

Zπ/2

0−1

4sin t+ cos tdt =3/4

40. C1:(0,3) to (6,3); x=6t, y =3,0≤t≤1

2:(6,3) to (6,0); x=6,y =3−3t, 0≤t≤1

36t2+9dt +Z1

−12

36 + 9(1 −t)2dt =1

3tan−12−2

3tan−1(1/2)

41. Represent the parabola by x=t, y =t2,0≤t≤2.

3xds =Z2

3tp1+4t

2dt = (17√17 −1)/4

42. Represent the semicircle by x= 2 cos t, y = 2 sin t, 0≤t≤π.

x2yds =Zπ

16 cos2tsin tdt=32/3

43. (a) 2πrh =2π(1)2 = 4π(b) S=ZC

z(t)dt

(2+(1/2) sin 3t)dt =4π

615 Chapter 17

44. C:x=acos t, y =asin t, 0≤t≤2π,Z−C

dt =−ZC

dt =−Z2π

dt =−2π

45. W=Z1

λ[(1 −λ)t+(3λ−1)t2−(1+2λ)t

]dt =−λ/12, W= 1 when λ=−12

46. The force exerted by the farmer is F=150+20−1

10zk=170 −3

4πtk,so

F·dr=170 −1

10zdz, and W=Z60

0170 −1

10zdz =10,020. Note that the functions

x(z),y(z) are irrelevant.

EXERCISE SET 17.3

1. ∂x/∂y =0=∂y/∂x, conservative so ∂φ/∂x =xand ∂φ/∂y =y,φ=x2/2+k(y), k0(y)=y,

k(y)=y

/2+K,φ=x

/2+y

/2+K

2. ∂(3y2)/∂y =6y=∂(6xy)/∂x, conservative so ∂φ/∂x =3y

2and ∂φ/∂y =6xy,

φ=3xy2+k(y), 6xy +k0(y)=6xy,k0(y)=0,k(y)=K,φ=3xy2+K

3. ∂(x2y)/∂y =x2and ∂(5xy2)/∂x =5y

, not conservative

4. ∂(excos y)/∂y =−exsin y=∂(−exsin y)/∂x, conservative so ∂φ/∂x =excos yand

∂φ/∂y =−exsin y,φ=excos y+k(y), −exsin y+k0(y)=−e

xsin y,

k0(y)=0,k(y)=K,φ=e

xcos y+K

5. ∂(cos y+ycos x)/∂y =−sin y+ cos x=∂(sin x−xsin y)/∂x, conservative so

∂φ/∂x = cos y+ycos xand ∂φ/∂y = sin x−xsin y,φ=xcos y+ysin x+k(y),

−xsin y+ sin x+k0(y) = sin x−xsin y,k0(y)=0,k(y)=K,φ=xcos y+ysin x+K

6. ∂(xln y)/∂y =x/y and ∂(yln x)/∂x =y/x, not conservative

7. (a) ∂(y2)/∂y =2y=∂(2xy)/∂x, independent of path

(b) C:x=−1+2t,y=2+t,0≤t≤1; Z1

(4+14t+6t

)dt =13

φ=xy2+K. Let K= 0 to get φ(1,3) −φ(−1,2)=9−(−4)=13

8. (a) ∂(ysin x)/∂y = sin x=∂(−cos x)/∂x, independent of path

(b) C1:x=πt,y=1−2t,0≤t≤1; Z1

(πsin πt −2πt sin πt + 2 cos πt)dt =0

k0(y)=0,k(y)=K,φ=−ycos x+K. Let K= 0 to get φ(π, −1)−φ(0,1)=(−1)−(−1)=0

9. ∂(3y)/∂y =3=∂(3x)/∂x,φ=3xy,φ(4,0) −φ(1,2) = −6

10. ∂(exsin y)/∂y =excos y=∂(excos y)/∂x,φ=exsin y,φ(1,π/2) −φ(0,0) = e

11. ∂(2xey)/∂y =2xey=∂(x2ey)/∂x,φ=x2ey,φ(3,2) −φ(0,0)=9e

Exercise Set 17.3 616

12. ∂(3x−y+1)/∂y =−1=∂[−(x+4y+ 2)]/∂x,

φ=3x

/2−xy +x−2y2−2y,φ(0,1) −φ(−1,2)=11/2

13. ∂(2xy3)/∂y =6xy2=∂(3x2y2)/∂x,φ=x2y3,φ(−1,0) −φ(2,−2)=32

14. ∂(exln y−ey/x)/∂y =ex/y −ey/x =∂(ex/y −eyln x)/∂x,

φ=exln y−eyln x,φ(3,3) −φ(1,1)=0

15. φ=x2y2/2, W=φ(0,0) −φ(1,1) = −1/216. φ=x2y3,W =φ(4,1) −φ(−3,0)=16

17. φ=exy,W=φ(2,0) −φ(−1,1)=1−e

−1

18. φ=e−ysin x, W =φ(−π/2,0) −φ(π/2,1) = −1−1/e

19. ∂(ey+yex)/∂y =ey+ex=∂(xey+ex)/∂x so Fis conservative, φ(x, y)=xey+yexso

F·dr=φ(0,ln 2) −φ(1,0)=ln2−1

20. ∂(2xy)/∂y =2x=∂(x

2+ cos y)/∂x so Fis conservative, φ(x, y)=x

y+ sin yso

F·dr =φ(π, π/2) −φ(0,0) = π3/2+1

21. F ·dr=[(e

y+yex)i+(xey+ex)j]·[(π/2) cos(πt/2)i+(1/t)j]dt

=π

2cos(πt/2)(ey+yex)+(xey+ex)/tdt,

so ZC

F·dr=Z2

1π

2cos(πt/2) eln t+ (ln t)esin(πt/2)+sin(πt/2)eln t+esin(πt/2)dt ≈−0.307

22. F ·dr=2t2cos(t/3)+[t

2+ cos(tcos(t/3))](cos(t/3) −(t/3) sin(t/3))dt,so

F·dr=Zπ

02t

2cos(t/3)+[t

2+ cos(tcos(t/3))](cos(t/3) −(t/3) sin(t/3))dt ≈16.503

23. No; a closed loop can be found whose tangent everywhere makes an angle <πwith the vector

ﬁeld, so the line integral ZC

F·dr>0, and by Theorem 17.3.2 the vector ﬁeld is not conservative.

24. The vector ﬁeld is constant, say F=ai+bj,soletφ(x, y)=ax +by and Fis conservative.

25. If Fis conservative, then F=∇φ=∂φ

∂xi+∂φ

∂y j+∂φ

∂z kand hence f=∂φ

∂x,g =∂φ

∂y , and h=∂φ

∂z .

Thus ∂f

∂y =∂2φ

∂y∂x and ∂g

∂x =∂2φ

∂x∂y,∂f

∂z =∂2φ

∂z∂x and ∂h

∂x =∂2φ

∂x∂z ,∂g

∂z =∂2φ

∂z∂y and ∂h

∂y =∂2φ

∂y∂z.

The result follows from the equality of mixed second partial derivatives.

26. Let f(x, y, z)=yz, g(x, y, z)=xz, h(x, y, z)=yx2, then ∂f/∂z =y, ∂h/∂x =2xy 6=∂f/∂z,thus

by Exercise 25, F=fi+gj+hkis not conservative, and by Theorem 17.3.2, ZC

yz dx+xz dy+yx2dz

is not independent of the path.

617 Chapter 17

27. ∂

∂y(h(x)[xsin y+ycos y]) = h(x)[xcos y−ysin y+ cos y]

∂

∂x(h(x)[xcos y−ysin y]) = h(x) cos y+h0(x)[xcos y−ysin y],

equate these two partial derivatives to get (xcos y−ysin y)(h0(x)−h(x)) = 0 which holds for all

xand yif h0(x)=h(x), h(x)=Cexwhere Cis an arbitrary constant.

28. (a) ∂

∂y

(x2+y2)3/2=−3cxy

(x2+y2)−5/2=∂

∂x

(x2+y2)3/2when (x, y)6=(0,0),

so by Theorem 17.3.3, Fis conservative. Set ∂φ/∂x =cx/(x2+y2)−3/2,

then φ(x, y)=−c(x

2+y

)

−1/2+k(y), ∂φ/∂y =cy/(x2+y2)−3/2+k0(y), so k0(y)=0.

Thus φ(x, y)=−c

2+y

)

1/2is a potential function.

(b) curl F=0is similar to part (a), so Fis conservative. Let

φ(x, y, z)=Zcx

(x2+y2+z2)3/2dx =−c(x2+y2+z2)−1/2+k(y, z). As in part (a),

∂k/∂y =∂k/∂z =0,soφ(x, y, z)=−c/(x2+y2+z2)1/2is a potential function for F.

29. (a) See Exercise 28, c=1;W=ZQ

F·dr=φ(3,2,1) −φ(1,1,2) = −1

√14 +1

√6

(b) Cbegins at P(1,1,2) and ends at Q(3,2,1) so the answer is again W=−1

√14 +1

√6.

30. (a) F ·dr=ydx

dt −xdy

dt dt for points on the circle x2+y2=1,so

1:x= cos t, y = sin t, 0≤t≤π, ZC1

F·dr=Zπ

(−sin2t−cos2t)dt =−π

C2:x= cos t, y =−sin t, 0≤t≤π, ZC2

F·dr=Zπ

(sin2t+ cos2t)dt =π

(b) ∂f

∂y =x2−y2

(x2+y2)2,∂g

∂x =−y2−x2

(x2+y2)2=∂f

∂y

C2in the reverse direction, does not lie in an open simply connected region inside which F

is continuous, since Fis not deﬁned at the origin, nor can it be deﬁned there in such a way

as to make the resulting function continuous there.

31. If Cis composed of smooth curves C1,C

2,...,C

nand curve Ciextends from (xi−1,y

i−1)to(x

then ZC

F·dr=

i=1 ZCi

F·dr=

i=1

[φ(xi,y

i)−φ(x

i−1,y

i−1)] = φ(xn,y

n)−φ(x

0,y

where (x0,y

0) and (xn,y

n) are the endpoints of C.

32. ZC1

F·dr+Z−C2

F·dr= 0, but Z−C2

F·dr=−ZC2

F·drso ZC1

F·dr=ZC2

F·dr,thus

F·dris independent of path.

Exercise Set 17.4 618

33. Let C1be an arbitrary piecewise smooth curve from (a, b)toapoint(x, y1) in the disk, and C2

the vertical line segment from (x, y1)to(x, y). Then

φ(x, y)=ZC

F·dr+ZC

F·dr=Z(x,y1)

(a,b)

F·dr+ZC2

F·dr.

The ﬁrst term does not depend on y;

hence ∂φ

∂y =∂

∂y ZC2

F·dr=∂

∂y ZC2

f(x, y)dx +g(x, y)dy.

However, the line integral with respect to xis zero along C2,so∂φ

∂y =∂

∂y ZC2

g(x, y)dy.

Express C2as x=x, y =twhere tvaries from y1to y, then ∂φ

∂y =∂

∂y Zy

g(x, t)dt =g(x, y).

EXERCISE SET 17.4

1. ZZ

(2x−2y)dA =Z1

0Z1

(2x−2y)dy dx = 0; for the line integral, on x=0,y

2dx =0,x

2dy =0;

on y=0,y

2dx =x2dy =0;onx=1,y

2dx +x2dy =dy; and on y=1,y

2dx +x2dy =dx,

hence I

y2dx +x2dy =Z1

dy +Z0

dx =1−1=0

2. ZZ

(1 −1)dA = 0; for the line integral let x= cos t, y = sin t,

ydx+xdy =Z2π

(−sin2t+ cos2t)dt =0

3. Z4

−2Z2

(2y−3x)dy dx =0 4. Z2π

0Z3

(1+2rsin θ)rdrdθ =9π

5. Zπ/2

0Zπ/2

(−ycos x+xsin y)dy dx =0 6. ZZ

(sec2x−tan2x)dA =ZZ

dA =π

7. ZZ

[1 −(−1)]dA =2ZZ

dA =8π8. Z1

0Zx

(2x−2y)dy dx =1/30

9. ZZ

R−y

1+y−1

1+ydA =−ZZ

dA =−4

10. Zπ/2

0Z4

(−r2)rdrdθ =−32π

11. ZZ

R−y2

1+y

2−1

1+y

2dA =−ZZ

dA =−1

619 Chapter 17

12. ZZ

(cos xcos y−cos xcos y)dA =0 13. Z1

0Z√x

(y2−x2)dy dx =0

15. (a) C:x= cos t, y = sin t, 0≤t≤2π;

=Z2π

0esin t(−sin t) + sin tcos tecos tdt ≈−3.550999378;

R∂

∂x(yex)−∂

∂yeydA =ZZ

[yex−ey]dA

=Z2π

0Z1

0rsin θercos θ−ersin θrdrdθ≈−3.550999378

(b) C1:x=t, y =t2,0≤t≤1; Z

[eydx +yexdy]=Z1

0he

2+2t

idt ≈2.589524432

C2:x=t2,y =t, 0≤t≤1; Z

[eydx +yexdy]=Z1

0h2tet+tet2idt =e+3

2≈2.859140914

−Z

≈−0.269616482; ZZ

=Z1

0Z√x

[yex−ey]dy dx ≈−0.269616482

16. (a) IC

xdy =Z2π

ab cos2tdt=πab (b) IC−ydx=Z2π

ab sin2tdt=πab

17. A=1

2IC−ydx+xdy=1

2Z2π

(3a2sin4φcos2φ+3a

2cos4φsin2φ)dφ

2a2Z2π

sin2φcos2φdφ=3

2Z2π

sin22φdφ=3πa2/8

18. C1:(0,0) to (a, 0); x=at, y =0,0≤t≤1

2:(a, 0) to (0,b); x=a−at, y =bt, 0≤t≤1

C3:(0,b)to(0,0); x=0,y=b−bt, 0≤t≤1

A=IC

xdy =Z1

(0)dt +Z1

ab(1 −t)dt +Z1

(0)dt =1

2ab

19. C1:(0,0) to (a, 0); x=at,y=0,0≤t≤1

2:(a, 0) to (acos t0,bsin t0); x=acos t,y=bsin t,0≤t≤t

3:(acos t0,bsin t0)to(0,0); x=−a(cos t0)t,y=−b(sin t0)t,−1≤t≤0

A=1

2IC−ydx+xdy =1

2Z1

(0) dt +1

2Zt0

ab dt +1

2Z0

−1

(0) dt =1

2ab t0

20. C1:(0,0) to (a, 0); x=at,y=0,0≤t≤1

2:(a, 0) to (acosh t0,bsinh t0); x=acosh t,y=bsinh t,0≤t≤t

3:(acosh t0,bsinh t0)to(0,0); x=−a(cosh t0)t,y=−b(sinh t0)t,−1≤t≤0

A=1

2IC−ydx+xdy =1

2Z1

(0) dt +1

2Zt0

ab dt +1

2Z0

−1

(0) dt =1

2ab t0

Exercise Set 17.4 620

21. W=ZZ

ydA=Zπ

0Z5

2sin θdrdθ= 250/3

22. We cannot apply Green’s Theorem on the region enclosed by the closed curve C, since Fdoes not

have ﬁrst order partial derivatives at the origin. However, the curve C, consisting of

y=x3

0/4,x

0≤x≤2; x=2,x

0≤y≤2; and y=x3/4,x

0≤x≤2 encloses a region Rin which

Green’s Theorem does hold, and

W=I

F·dr= lim

→0+I

C

F·dr= lim

→0+ZZ

R

∇·FdA =Z2

x0Zx3/4

0/41

2x−1/2−1

2y−1/2dy dx

= lim

→0+ −18

35√2−√2

4x3

0+x3/2

0+3

14x7/2

0−3

10x5/2

0!=−18

35√2

23. IC

ydx−xdy =ZZ

(−2)dA =−2Z2π

0Za(1+cos θ)

rdrdθ=−3πa2

24. ¯x=1

AZZ

xdA, but IC

2x2dy =ZZ

xdA from Green’s Theorem so

¯x=1

AIC

2x2dy =1

2AIC

x2dy. Similarly, ¯y=−1

2AIC

y2dx.

25. A=Z1

0Zx

dy dx =1

4;C1:x=t, y =t3,0≤t≤1,ZC1

x2dy =Z1

t2(3t2)dt =3

C2:x=t, y =t, 0≤t≤1; ZC2

x2dy =Z1

t2dt =1

3,IC

x2dy =ZC1−ZC2

5−1

3=4

15,¯x=8

y2dx =Z1

t6dt −Z1

t2dt =1

7−1

3=−4

21,¯y=8

21,centroid 8

15,8

21

26. A=a2

2;C1:x=t, y =0,0≤t≤a, C2:x=a−t, y =t, 0≤t≤a;C3:x=0,y =a−t, 0≤t≤a;

x2dy =0,Z

2dy =Za

(a−t)2dt =a3

3,Z

x2dy =0,I

2dy =Z

=a3

3,¯x=a

y2dx =0−Za

2dt +0=−a

3,¯y=a

3,centroid a

3,a

3

27. ¯x= 0 from the symmetry of the region,

C1:(a, 0) to (−a, 0) along y=√a2−x2;x=acos t,y=asin t,0≤t≤π

2:(−a, 0) to (a, 0); x=t,y=0,−a≤t≤a

A=πa2/2,¯y=−1

2AZπ

0−a3sin3tdt+Za

−a

(0)dt

=−1

πa2−4a3

3=4a

3π; centroid 0,4a

3π

621 Chapter 17

28. A=ab

2;C1:x=t, y =0,0≤t≤a, C2:x=a, y =t, 0≤t≤b;

C3:x=a−t, y =b−bt/a, 0≤t≤a;

ZC1

x2dy =0,ZC

2dy =Zb

a2dt =ba2,ZC3

x2dy =Za

(a−t)2(−b/a)dt =−ba2

x2dy =ZC1

+ZC2

+ZC3

=2ba2

3,¯x=2a

y2dx =0+0−Za

2a−t

adt =−ab2

3,¯y=b

3,centroid 2a

3,b

3

29. From Green’s Theorem, the given integral equals ZZ

(1−x2−y2)dA where Ris the region enclosed

by C. The value of this integral is maximum if the integration extends over the largest region for

which the integrand 1 −x2−y2is nonnegative so we want 1 −x2−y2≥0, x2+y2≤1. The

largest region is that bounded by the circle x2+y2= 1 which is the desired curve C.

30. (a) C:x=a+(c−a)t,y=b+(d−b)t,0≤t≤1

ZC−ydx+xdy =Z1

(ad −bc)dt =ad −bc

(b) Let C1,C2, and C3be the line segments from (x1,y

1)to(x

2), (x2,y

2)to(x

3), and

(x3,y

3)to(x

1), then if Cis the entire boundary consisting of C1,C

2,and C3

A=1

2ZC−ydx+xdy =1

i=1 ZCi−ydx+xdy

2[(x1y2−x2y1)+(x

3−x

)+(x

1−x

)]

2[(x1y2−x2y1)+(x

3−x

)+···+(x

1−x

)]

(d) A=1

2[(0 −0)+(6+8)+(0+2)+(0−0)]=8

31. ZC

F·dr=ZC

(x2+y)dx +(4x−cos y)dy =3ZZ

dA = 3(25 −2)=69

32. ZC

F·dr=ZC

(e−x+3y)dx +xdy =−2ZZ

dA =−2[π(4)2−π(2)2]=−24π

EXERCISE SET 17.5

1. Ris the annular region between x2+y2= 1 and x2+y2=4;

z2dS =ZZ

(x2+y2)sx2

x2+y2+y2

x2+y2+1dA

=√2ZZ

(x2+y2)dA =√2Z2π

0Z2

r3dr dθ =15

2π√2.

Exercise Set 17.5 622

2. z=1−x−y,Ris the triangular region enclosed by x+y=1,x= 0 and y=0;

xy dS =ZZ

xy√3dA =√3Z1

0Z1−x

xy dy dx =√3

24 .

3. Let r(u, v) = cos ui+vj+ sin uk,0≤u≤π, 0≤v≤1. Then ru=−sin ui+ cos uk,rv=j,

ru×rv=−cos ui+ sin uk,kru×rvk=1,ZZσ

x2ydS =Z1

0Zπ

vcos2ududv =π/4

4. z=p4−x2−y2,Ris the circular region enclosed by x2+y2=3;

(x2+y2)zdS=ZZ

(x2+y2)p4−x2−y2sx2

4−x2−y2+y2

4−x2−y2+1dA

=ZZ

2(x2+y2)dA =2Z2π

0Z√

dr dθ =9π.

5. If we use the projection of σonto the xz-plane then y=1−xand Ris the rectangular region in

the xz-plane enclosed by x=0,x=1,z= 0 and z=1;

(x−y−z)dS =ZZ

(2x−1−z)√2dA =√2Z1

0Z1

(2x−1−z)dz dx =−√2/2

6. Ris the triangular region enclosed by 2x+3y=6,x= 0, and y=0;

(x+y)dS =ZZ

(x+y)√14 dA =√14 Z3

0Z(6−2x)/3

(x+y)dy dx =5

√

14.

7. There are six surfaces, parametrized by projecting onto planes:

σ1:z=0;0≤x≤1, 0 ≤y≤1 (onto xy-plane), σ2:x=0;0≤y≤1, 0 ≤z≤1 (onto yz-plane),

σ3:y=0;0≤x≤1, 0 ≤z≤1 (onto xz-plane), σ4:z=1;0≤x≤1, 0 ≤y≤1 (onto xy-plane),

σ5:x=1;0≤y≤1, 0 ≤z≤1 (onto yz-plane), σ6:y=1;0≤x≤1, 0 ≤z≤1 (onto xz-plane).

By symmetry the integrals over σ1,σ

2and σ3are equal, as are those over σ4,σ

5and σ6, and

σ1

(x+y+z)dS =Z1

0Z1

(x+y)dx dy =1; ZZ

σ4

(x+y+z)dS =Z1

0Z1

(x+y+1)dx dy =2,

thus, ZZ

(x+y+z)dS =3·1+3·2=9.

8. Let r(φ, θ) = sinφcos θi+ sin φsin θj+ cos φk,

0≤θ≤2π, 0≤φ≤π/2; krφ×rθk=p1−cos2φ= sin φ,

(1+p1−x2−y2)dS =Z2π

0Zπ/2

(1+q1−sin2φ) sin φdφdθ

=2πZπ/2

(1 + cos φ) sin φdφ=3π

623 Chapter 17

9. Ris the circular region enclosed by x2+y2=1;

σpx2+y2+z2dS =ZZ

Rp2(x2+y2)sx2

x2+y2+y2

x2+y2+1dA

= lim

r0→0+2ZZ

R0px2+y2dA

where R0is the annular region enclosed by x2+y2= 1 and x2+y2=r2

0with r0slightly larger

than 0 because sx2

x2+y2+y2

x2+y2+ 1 is not deﬁned for x2+y2=0,so

σpx2+y2+z2dS = lim

r0→0+2Z2π

0Z1

r2dr dθ = lim

r0→0+

4π

3(1 −r3

0)=4π

10. Let r(φ, θ)=asin φcos θi+asin φsin θj+acos φk,

0≤θ≤2π, 0≤φ≤π/2; krφ×rθk=a2p1−cos2φ,

f(x, y, z)=a

2Z2π

0Zπ

2sin2φp1−cos2φdφdθ =8

3πa4

11. (a) √29

16 Z6

0Z(12−2x)/3

xy(12 −2x−3y)dy dx

(b) √29

4Z3

0Z(12−4z)/3

yz(12 −3y−4z)dy dz

9Z3

0Z6−2z

xz(12 −2x−4z)dx dz

12. (a) aZa

0Z√a2−x2

xdydx (b) aZa

0Z√a2−z2

zdydz

0Z√a2−z2

√a2−x2−z2dx dz

13. 18√29/514. a4/3

15. Z4

0Z2

y3zp4y2+1dy dz;1

2Z4

0Z4

xz√1+4xdxdz

16. aZ9

0Za/√2

a/√5

x2y

pa2−y2dy dx,aZ2a/√5

a/√2Z9

x2dx dz 17. 391√17/15 −5√5/3

18. The region R:3x

2+2y

2= 5 is symmetric in y. The integrand is

x2yz dS =x2y(5 −3x2−2y2)p1+36x

2+16y

2dy dx, which is odd in y, henceZZZ

x2yz dS =0.

Exercise Set 17.5 624

19. z=√4−x2,∂z

∂x =−x

√4−x2,∂z

∂y =0;

δ0dS =δ0ZZ

Rrx2

4−x2+1dA =2δ

0Z4

0Z1

√

4−x

2dx dy =4

3πδ0.

20. z=1

2(x2+y2), Ris the circular region enclosed by x2+y2=8;

δ0dS =δ0ZZ

Rpx2+y2+1dA =δ0Z2π

0Z√8

0pr2+1rdrdθ =52

3πδ0.

21. z=4−y

,Ris the rectangular region enclosed by x=0,x=3,y= 0 and y=3;

ydS =ZZ

yp4y2+1dA =Z3

0Z3

yp4y2+1dy dx =1

4(37√37 −1).

22. Ris the annular region enclosed by x2+y2= 1 and x2+y2= 16;

x2zdS=ZZ

x2px2+y2sx2

x2+y2+y2

x2+y2+1dA

=√2ZZ

x2px2+y2dA =√2Z2π

0Z4

r4cos2θdrdθ=1023√2

5π.

23. M=ZZ

δ(x, y, z)dS =ZZ

δ0dS =δ0ZZ

dS =δ0S

24. δ(x, y, z)=|z|; use z=pa2−x2−y2, let Rbe the circular region enclosed by x2+y2=a2, and

σthe hemisphere above R. By the symmetry of both the surface and the density function with

respect to the xy-plane we have

M=2ZZ

zdS =2ZZ

Rpa2−x2−y2sx2

a2−x2−y2+y2

a2−x2−y2+1dA = lim

r0→a−2aZZ

where Rr0is the circular region with radius r0that is slightly less than a. But ZZ

dA is simply

the area of the circle with radius r0so M= lim

r0→a−2a(πr2

0)=2πa3.

25. By symmetry ¯x=¯y=0.

dS =ZZ

Rpx2+y2+1dA =Z2π

0Z√8

0pr2+1rdrdθ =52π

zdS=ZZ

zpx2+y2+1dA =1

2ZZ

(x2+y2)px2+y2+1dA

2Z2π

0Z√8

r3pr2+1dr dθ =596π

so ¯z=596π/15

52π/3=149

65 . The centroid is (¯x, ¯y, ¯z)=(0,0,149/65).

625 Chapter 17

26. By symmetry ¯x=¯y=0.

dS =ZZ

p4−x2−y2dA =2Z2π

0Z√

√

4−r

2dr dθ =4π,

zdS =ZZ

2dA = (2)(area of circle of radius √3)=6π

so ¯z=6π

4π=3

2. The centroid is (¯x, ¯y, ¯z)=(0,0,3/2).

27. ∂r/∂u = cos vi+ sin vj+3k,∂r/∂v =−usin vi+ucos vj,k∂r/∂u ×∂r/∂vk=√10u;

3√10 ZZ

u4sin vcos vdA=3

√

10 Zπ/2

0Z2

u4sin vcos vdudv=93/

√

28. ∂r/∂u =j,∂r/∂v =−2 sin vi+ 2 cos vk,k∂r/∂u ×∂r/∂vk=2;

8ZZ

udA =8Z2π

0Z3

du dv =16πln 3

29. ∂r/∂u = cos vi+ sin vj+2uk,∂r/∂v =−usin vi+ucos vj,k∂r/∂u ×∂r/∂vk=u√4u2+1;

udA=Zπ

0Zsin v

ududv =π/4

30. ∂r/∂u = 2 cos ucos vi+ 2 cos usin vj−2 sin uk,∂r/∂v =−2 sin usin vi+ 2 sin ucos vj;

k∂r/∂u ×∂r/∂vk= 4 sin u;

4ZZ

e−2 cos usin udA=4Z2π

0Zπ/2

e−2 cos usin ududv =4π(1 −e−2)

31. ∂z/∂x =−2xe−x2−y2,∂z/∂y =−2ye−x2−y2,

(∂z/∂x)2+(∂z/∂y)2+1=4(x

2+y

−2(x2+y2)+ 1; use polar coordinates to get

M=Z2π

0Z3

r2p4r2e−2r2+1dr dθ ≈57.895751

32. (b) A=ZZ

dS =Z2π

0Z1

−1

2p40ucos(v/2)+u2+4u

2cos2(v/2) + 100du dv ≈62.93768644;

¯x≈0.01663836266; ¯y=¯z= 0 by symmetry

EXERCISE SET 17.6

1. (a) zero (b) zero (c) positive

(d) negative (e) zero (f) zero

2. (a) positive (b) zero (c) zero

(d) zero (e) negative (f) zero

Exercise Set 17.6 626

3. (a) positive (b) zero (c) positive

(d) zero (e) positive (f) zero

4. 0; the ﬂux is zero on the faces y=0,1 and z=0,1; it is 1 on x= 1 and −1onx=0

5. (a) n =−cos vi−sin vj (b) inward, by inspection

6. (a) −rcos θi−rsin θj+rk (b) inward, by inspection

7. n =−zxi−zyj+k,ZZ

F·ndS =ZZ

(2x2+2y

2+ 2(1 −x2−y2)) dS =Z2π

0Z1

2rdrdθ=2π

8. With z=1−x−y,Ris the triangular region enclosed by x+y=1,x= 0 and y= 0; use upward

normals to get

F·ndS =2ZZ

(x+y+z)dA =2ZZ

dA = (2)(area of R)=1.

9. Ris the annular region enclosed by x2+y2= 1 and x2+y2=4;

F·ndS =ZZ

R −x2

px2+y2−y2

px2+y2+2z

!dA

=ZZ

Rpx2+y2dA =Z2π

0Z2

r2dr dθ =14π

10. Ris the circular region enclosed by x2+y2=4;

F·ndS =ZZ

(2y2−1)dA =Z2π

0Z2

(2r2sin2θ−1)rdrdθ=4π.

11. Ris the circular region enclosed by x2+y2−y=0;ZZ

F·ndS =ZZ

(−x)dA = 0 since the

region Ris symmetric across the y-axis.

12. With z=1

2(6 −6x−3y), Ris the triangular region enclosed by 2x+y=2,x= 0, and y=0;

F·ndS =ZZ

R3x2+3

2yx +zxdA =3ZZ

xdA=3Z1

0Z2−2x

xdydx=1.

13. ∂r/∂u = cos vi+ sin vj−2uk,∂r/∂v =−usin vi+ucos vj,

∂r/∂u ×∂r/∂v =2u

2cos vi+2u

2sin vj+uk;

(2u3+u)dA =Z2π

0Z2

(2u3+u)du dv =18π

14. ∂r/∂u =k,∂r/∂v =−2 sin vi+ cos vj,∂r/∂u ×∂r/∂v =−cos vi−2 sin vj;

(2 sin2v−e−sin vcos v)dA =Z2π

0Z5

(2 sin2v−e−sin vcos v)du dv =10π

627 Chapter 17

15. ∂r/∂u = cos vi+ sin vj+2k,∂r/∂v =−usin vi+ucos vj,

∂r/∂u ×∂r/∂v =−2ucos vi−2usin vj+uk;

u2dA =Zπ

0Zsin v

u2du dv =4/9

16. ∂r/∂u = 2 cos ucos vi+ 2 cos usin vj−2 sin uk,∂r/∂v =−2 sin usin vi+ 2 sin ucos vj;

∂r/∂u ×∂r/∂v = 4 sin2ucos vi+ 4 sin2usin vj+ 4 sin ucos uk;

8 sin udA=8Z2π

0Zπ/3

sin ududv =8π

17. In each part, divide σinto the six surfaces

σ1:x=−1 with |y|≤1, |z|≤1, and n=−i,σ2:x= 1 with |y|≤1, |z|≤1, and n=i,

σ3:y=−1 with |x|≤1, |z|≤1, and n=−j,σ4:y= 1 with |x|≤1, |z|≤1, and n=j,

σ5:z=−1 with |x|≤1, |y|≤1, and n=−k,σ6:z= 1 with |x|≤1, |y|≤1, and n=k,

(a) ZZ

σ1

F·ndS =ZZ

σ1

dS =4,ZZ

σ2

F·ndS =ZZ

σ2

dS = 4, and ZZ

σi

F·ndS = 0 for

i=3,4,5,6soZZ

F·ndS =4+4+0+0+0+0=8.

(b) ZZ

σ1

F·ndS =ZZ

σ1

dS = 4, similarly ZZ

σi

F·ndS = 4 for i=2,3,4,5,6so

F·ndS =4+4+4+4+4+4=24.

σ1

F·ndS =−ZZ

σ1

dS =−4, ZZ

σ2

F·ndS = 4, similarly ZZ

σi

F·ndS =−4 for i=3,5

and ZZ

σi

F·ndS = 4 for i=4,6soZZ

F·ndS =−4+4−4+4−4+4=0.

18. Decompose σinto a top σ1(the disk) and a bottom σ2(the portion of the paraboloid). Then

n1=k,ZZ

σ1

F·n1dS =−ZZ

σ1

ydS =−Z2π

0Z1

2sin θdrdθ=0,

2=(2xi+2yj−k)/

p1+4x

2+4y

,ZZ

σ2

F·n2dS =ZZ

σ2

y(2x2+2y

2+1)

p1+4x

2+4y

2dS =0,

because the surface σ2is symmetric with respect to the xy-plane and the integrand is an odd

function of y. Thus the ﬂux is 0.

19. Ris the circular region enclosed by x2+y2=1;

F·ndS =ZZ

R x

px2+y2+y

px2+y2−1!dA =ZZ

x+y−px2+y2

px2+y2dA

= lim

r0→0+Z2π

0Z1

(rcos θ+rsin θ−r)dr dθ = lim

r0→0+π(r2

0−1) = −π.

Exercise Set 17.6 628

20. Let r= cos vi+uj+ sin vk,−2≤u≤1,0≤v≤2π;ru×rv= cos vi+ sin vk,

F·ndS =ZZ

(cos2v+ sin2v)dA = area of R=3·2π=6π

21. (a) ZZ

F·ndS =ZZ

F·i−∂x

∂yj−∂x

∂z kdA,ifσis oriented by front normals, and

F·ndS =ZZ

F·−i+∂x

∂yj+∂x

∂z kdA,ifσis oriented by back normals,

where Ris the projection of σonto the yz-plane.

(b) Ris the semicircular region in the yz-plane enclosed by z=p1−y2and z=0;

F·ndS =ZZ

(−y−2yz +16z)dA =Z1

−1Z√1−y2

(−y−2yz +16z)dz dy =32

22. (a) ZZ

F·∂y

∂xi−j+∂y

∂zkdA,σoriented by right normals,

and ZZ

F·−∂y

∂xi+j−∂y

∂zkdA,σoriented by left normals,

where Ris the projection of σonto the xz-plane.

(b) Ris the semicircular region in the xz-plane enclosed by z=√1−x2and z=0;

F·ndS =ZZ

(2x2−(x2+z2)+z

)dA =Z1

−1Z√1−x2

(x2+z2)dz dx =π

23. (a) On the sphere, krk=aso F=akrand F·n=akr·(r/a)=a

k−1

krk

2=a

k−1

2=a

k+1,

hence ZZ

F·ndS =ak+1 ZZ

dS =ak+1(4πa2)=4πak+3.

(b) If k=−3, then ZZ

F·ndS =4π.

24. Let r= sin ucos vi+ sin usin vj+ cos uk,ru×rv= sin2ucos vi+ sin2usin vj+ sin ucos uk,

F·(ru×rv)=a

2sin3ucos2v+1

9sin3usin2v+asin ucos3u,

F·ndS =Z2π

0Zπ

(a2sin3ucos2v+1

9sin3usin2v+asin ucos3u)du dv

=a2Zπ

sin3uduZ2π

cos2vdv+1

aZπ

sin3udu

+Z2π

sin2vdv+2πa Zπ

sin ucos3udu

=4π

3a

2+1

a=10ifa≈−1.722730,0.459525,1.263205

629 Chapter 17

EXERCISE SET 17.7

1. σ1:x=0,F·n=−x=0,ZZ

σ1

(0)dA =0 σ

2:x=1,F·n=x=1,ZZ

σ2

(1)dA =1

3:y=0,F·n=−y=0,ZZ

σ3

(0)dA =0 σ

4:y=1,F·n=y=1,ZZ

σ4

(1)dA =1

5:z=0,F·n=−z=0,ZZ

σ5

(0)dA =0 σ

6:z=1,F·n=z=1,ZZ

σ6

(1)dA =1

ZZσ

F·n=3; ZZZ

div FdV =ZZZ

3dV =3

2. For any point r=xi+yj+zkon σlet n=xi+yj+zk; then F·n=x2+y2+z2=1,so

F·ndS =ZZ

dS =4π; also ZZZ

div FdV =ZZZ

3dV = 3(4π/3)=4π

3. σ1:z=1,n=k,F·n=z

2=1,ZZ

σ1

(1)dS =π,

σ2:n=2xi+2yj−k,F·n=4x

2−4x

2−x

4−3y

σ2

F·ndS =Z2π

0Z1

04r2cos2θ−4r4cos2θsin2θ−r4cos4θ−3r4sin4θrdrdθ=π

=4π

ZZZ

div FdV =ZZZ

(2+z)dV =Z2π

0Z1

(2+z)dzrdrdθ =4π/3

4. σ1:x=0,F·n=−xy =0,ZZ

σ1

(0)dA =0 σ

2:x=2,F·n=xy =2y, ZZ

σ2

(2y)dA =8

3:y=0,F·n=−yz =0,ZZ

σ3

(0)dA =0 σ

4:y=2,F·n=yz =2z,ZZ

σ4

(2z)dA =8

5:z=0,F·n=−xz =0,ZZ

σ5

(0)dA =0 σ

6:z=2,F·n=xz =2x, ZZ

σ6

(2x)dA =8

F·n= 24; also ZZZ

div FdV =ZZZ

(y+z+x)dV =24

5. Gis the rectangular solid; ZZZ

div FdV =Z2

0Z1

0Z3

(2x−1) dx dy dz = 12.

6. Gis the spherical solid enclosed by σ;ZZZ

div FdV =ZZZ

0dV =0ZZZ

dV =0.

Exercise Set 17.7 630

7. Gis the cylindrical solid;

ZZZ

div FdV =3ZZZ

dV = (3)(volume of cylinder) = (3)[πa2(1)] = 3πa2.

8. Gis the solid bounded by z=1−x

2−y

2and the xy-plane;

ZZZ

div FdV =3ZZZ

dV =3Z2π

0Z1

0Z1−r

rdzdrdθ =3π

9. Gis the cylindrical solid;

ZZZ

div FdV =3ZZZ

(x2+y2+z2)dV =3Z2π

0Z2

0Z3

2+z

)rdzdrdθ= 180π.

10. Gis the tetrahedron; ZZZ

div FdV =ZZZ

xdV =Z1

0Z1−x

0Z1−x−y

xdzdydx=1

24.

11. Gis the hemispherical solid bounded by z=p4−x2−y2and the xy-plane;

ZZZ

div FdV =3ZZZ

(x2+y2+z2)dV =3Z2π

0Zπ/2

0Z2

ρ4sin φdρdφdθ =192π

12. Gis the hemispherical solid;

ZZZ

div FdV =5ZZZ

zdV =5Z2π

0Zπ/2

0Za

ρ3sin φcos φdρdφdθ =5πa4

13. Gis the conical solid;

ZZZ

div FdV =2ZZZ

(x+y+z)dV =2Z2π

0Z1

(rcos θ+rsin θ+z)rdzdrdθ=π

14. Gis the solid bounded by z=2xand z=x2+y2;

ZZZ

div FdV =ZZZ

dV =2Zπ/2

0Z2 cos θ

0Z2rcos θ

rdzdrdθ =π

15. Gis the solid bounded by z=4−x

,y+z= 5, and the coordinate planes;

ZZZ

div FdV =4ZZZ

x2dV =4Z2

−2Z4−x

0Z5−z

dy dz dx =4608

35 .

16. ZZ

F·ndS =ZZZ

div FdV =ZZZ

0dV =0;

since the vector ﬁeld is constant, the same amount enters as leaves.

17. ZZ

r·ndS =ZZZ

div rdV =3ZZZ

dV = 3vol(G)

631 Chapter 17

18. ZZ

F·ndS =3[π(32)(5)] = 135π

19. ZZ

curl F·ndS =ZZZ

div(curl F)dV =ZZZ

(0)dV =0

20. ZZ

σ∇f·ndS =ZZZ

div (∇f)dV =ZZZ

∇2fdV

21. ZZ

(f∇g)·n=ZZZ

div (f∇g)dV =ZZZ

(f∇2g+∇f·∇g)dV by Exercise 29, Section 17.1.

22. ZZ

(f∇g)·ndS =ZZZ

(f∇2g+∇f·∇g)dV by Exercise 29, Section 17.1;

(g∇f)·ndS =ZZZ

(g∇2f+∇g·∇f)dV by interchanging fand g;

subtract to obtain the result.

23. Let the normal n=n1i+n2j+n3k, then we have to show that

(fn1i+fn2j+fn3k)dS =ZZZ

(fxi+fyj+fzk)dV . We shall show that

fn1idS =ZZZ

fxidV , i.e. that ZZ

fn1dS =ZZZ

fxdV ; the other parts will follow in a

similar manner. Let F(x, y, z)=f(x, y, z)i, then by the Divergence Theorem

fn1dS =ZZ

F·ndS =ZZZ

∇·FdV =ZZZ

fxdV .

24. div r=c∂

∂x

(x2+y2+z2)3/2+∂

∂y

(x2+y2+z2)3/2+∂

∂z

(x2+y2+z2)3/2

=c(−2x2+y2+z2)+(−2y

2+x

2+z

)+(−2z

2+x

2+y

)

2+y

2+z

)

5/2=0

25. (a) The ﬂux through any cylinder whose axis is the z-axis is positive by inspection; by the

Divergence Theorem, this says that the divergence cannot be negative at the origin, else

the ﬂux through a small enough cylinder would also be negative (impossible), hence the

divergence at the origin must be ≥0.

(b) Similar to Part (a), ≤0.

26. (a) F =xi+yj+zk, div F=3 (b) F =−xi−yj−zk, div F=−3

27. div F= 0; no sources or sinks.

28. div F=y−x; sources where y>x, sinks where y<x.

29. div F=3x

2+3y

2+3z

; sources at all points except the origin, no sinks.

Exercise Set 17.8 632

30. div F=3(x

2+y

2+z

2−1); sources outside the sphere x2+y2+z2= 1, sinks inside the sphere

x2+y2+z2=1.

31. Let σ1be the portion of the paraboloid z=1−x

2−y

2for z≥0, and σ2the portion of the plane

z= 0 for x2+y2≤1. Then

σ1

F·ndS =ZZ

F·(2xi+2yj+k)dA

=Z1

−1Z√1−x2

−√1−x2

(2x[x2y−(1 −x2−y2)2]+2y(y

3−x)+(2x+2−3x

2−3y

)) dy dx

=3π/4;

z= 0 and n=−kon σ2so F·n=1−2x,ZZ

σ2

F·ndS =ZZ

σ2

(1 −2x)dS =π.Thus

F·ndS =3π/4+π=7π/4. But div F=2xy +3y

2+3so

ZZZ

div FdV =Z1

−1Z√1−x2

−√1−x2Z1−x2−y2

(2xy +3y

2+3)dz dy dx =7π/4.

EXERCISE SET 17.8

1. (a) The ﬂow is independent of zand has no component in the direction of k, and so by inspection

the only nonzero component of the curl is in the direction of k. However both sides of (9)

are zero, as the ﬂow is orthogonal to the curve Ca. Thus the curl is zero.

(b) Since the ﬂow appears to be tangential to the curve Ca, it seems that the right hand side of

(9) is nonzero, and thus the curl is nonzero, and points in the positive z-direction.

2. (a) The only nonzero vector component of the vector ﬁeld is in the direction of i, and it increases

with yand is independent of x. Thus the curl of Fis nonzero, and points in the negative

z-direction.

(b) By inspection the vector ﬁeld is constant, and thus its curl is zero.

3. If σis oriented with upward normals then Cconsists of three parts parametrized as

C1:r(t)=(1−t)i+tjfor 0 ≤t≤1, C2:r(t)=(1−t)j+tkfor 0 ≤t≤1,

C3:r(t)=ti+(1−t)kfor 0 ≤t≤1.

ZC1

F·dr=ZC2

F·dr=ZC3

F·dr=Z1

(3t−1)dt =1

2so

F·dr=1

2+1

2=3

2. curl F=i+j+k,z=1−x−y,Ris the triangular region in

the xy-plane enclosed by x+y=1,x= 0, and y=0;

(curl F)·ndS =3ZZ

dA = (3)(area of R) = (3) 1

2(1)(1)=3

633 Chapter 17

4. If σis oriented with upward normals then Ccan be parametrized as r(t) = cos ti+ sin tj+kfor

0≤t≤2π.

F·dr=Z2π

(sin2tcos t−cos2tsin t)dt =0;

curl F=0so ZZ

(curl F)·ndS =ZZ

0dS =0.

5. If σis oriented with upward normals then Ccan be parametrized as r(t)=acos ti+asin tjfor

0≤t≤2π.

F·dr=Z2π

0dt = 0; curl F=0so ZZ

(curl F)·ndS =ZZ

0dS =0.

6. If σis oriented with upward normals then Ccan be parametrized as r(t)=3costi+ 3 sin tjfor

0≤t≤2π.

F·dr=Z2π

(9 sin2t+ 9 cos2t)dt =9Z2π

dt =18π.

curl F=−2i+2j+2k,Ris the circular region in the xy-plane enclosed by x2+y2=9;

(curl F)·ndS =ZZ

(−4x+4y+2)dA =Z2π

0Z3

(−4rcos θ+4rsin θ+2)rdrdθ =18π.

7. Take σas the part of the plane z= 0 for x2+y2≤1 with n=k; curl F=−3y2i+2zj+2k,

(curl F)·ndS =2ZZ

dS = (2)(area of circle) = (2)[π(1)2]=2π.

8. curl F=xi+(x−y)j+6xy2k;

(curl F)·ndS =ZZ

(x−y−6xy2)dA =Z1

0Z3

(x−y−6xy2)dy dx =−30.

9. Cis the boundary of Rand curl F=2i+3j+4k,so

IF·r=ZZR

curl F·ndS =ZZR

4dA = 4(area of R)=16π

10. curl F=−4i−6j+6yk,z=y/2 oriented with upward normals, Ris the triangular region in the

xy-plane enclosed by x+y=2,x= 0, and y=0;

(curl F)·ndS =ZZ

(3+6y)dA =Z2

0Z2−x

(3+6y)dy dx = 14.

11. curl F=xk, take σas part of the plane z=yoriented with upward normals, Ris the circular

region in the xy-plane enclosed by x2+y2−y=0;

(curl F)·ndS =ZZ

xdA=Zπ

0Zsin θ

r2cos θdrdθ=0.

12. curl F=−yi−zj−xk,z=1−x−yoriented with upward normals, Ris the triangular region in

the xy-plane enclosed by x+y=1,x= 0 and y=0;

(curl F)·ndS =ZZ

(−y−z−x)dA =−ZZ

dA =−1

2(1)(1) = −1

Exercise Set 17.8 634

13. curl F=i+j+k, take σas the part of the plane z= 0 with x2+y2≤a2and n=k;

(curl F)·ndS =ZZ

dS = area of circle = πa2.

14. curl F=i+j+k, take σas the part of the plane z=1/

√

2 with x2+y2≤1/2 and n=k.

(curl F)·ndS =ZZ

dS = area of circle = π

15. (a) Take σas the part of the plane 2x+y+2z= 2 in the ﬁrst octant, oriented with downward

normals; curl F=−xi+(y−1)j−k,

F·Tds =ZZ

(curl F)·ndS

=ZZ

Rx−1

2y+3

2dA =Z1

0Z2−2x

0x−1

2y+3

2dy dx =3

(b) At the origin curl F=−j−kand with n=k, curl F(0,0,0) ·n=(−j−k)·k=−1.

direction as curl F(0,0,0) so n=−1

√2j−1

√2k.

16. (a) div(curl F)= ∂

∂x ∂h

∂y −∂g

∂z+∂

∂y ∂f

∂z −∂h

∂x+∂

∂z ∂g

∂x −∂f

∂y 

=∂2h

∂x∂y −∂2g

∂x∂z +∂2f

∂y∂z −∂2h

∂y∂x +∂2g

∂z∂x −∂2f

∂z∂y =0,

assuming equality of mixed second partial derivatives

(b) By the Divergence Theorem

(curl F)·ndS =ZZZ

div(curl F)dV =ZZZ

0dV =0.

17. Since I

E·rdr=ZZ

curl E·ndS, it follows that ZZ

curl E·ndS =−ZZ

∂B

∂t

·ndS. This

relationship holds for any volume σ, hence curl E=−∂B

∂t .

18. Parametrize Cby x= cos t, y = sin t, 0≤t≤2π. But F=x2yi+(y

3−x)j+(2x−1)kalong C

so IC

F·dr=−5π/4. Since curl F=(−2z−2)j+(−1−x

)k,

(curl F)·ndS =ZZ

(curl F)·(2xi+2yj+k)dA

=Z1

−1Z√1−x2

−√1−x2

[2y(2x2+2y

2−4) −1−x2]dy dx =−5π/4

635 Chapter 17

CHAPTER 17 SUPPLEMENTARY EXERCISES

2. (b) c

kr−r0k3(r−r0)(c) cxi+yj+zk

px2+y2+z2

3. (a) Zb

af(x(t),y(t))dx

dt +g(x(t),y(t))dy

dt dt

(b) Zb

f(x(t),y(t))px0(t)2+y0(t)2dt

4. (a) M=Z

δ(x, y, z)ds (b) L=Z

ds (c) S=ZZ

(d) A=I

xdy =−I

ydx=1

−ydx+xdy

11. ZZ

f(x, y, z)dS =ZZ

f(x(u, v),y(u, v),z(u, v))kru×rvkdu dv

13. C1:(0,0) to (1,0); x=t,y=0,0≤t≤1

2:(1,0) to (cosh t0,sinh t0); x= cosh t,y= sinh t,0≤t≤t

3: (cosh t0,sinh t0)to(0,0); x=−(cosh t0)t,y=−(sinh t0)t,−1≤t≤0

A=1

2IC−ydx+xdy =1

2Z1

(0) dt +1

2Zt0

dt +1

2Z0

−1

(0) dt =1

2t0

14. (a) F(x, y, z)= qQ(xi+yj+zk)

4π0(x2+y2+z2)3/2

(b) F =∇φ, where φ=−qQ

4π0(x2+y2+z2)1/2,soW=φ(3,1,5)−φ(3,0,0) = qQ

4π01

3−1

√35.

C:x=3,y =t, z =5t, 0≤t≤1; F·dr=qQ[0+t+25t]dt

4π0(9+t2+25t

)

3/2

W=Z1

26qQt dt

4π0(26t2+9)

3/2=qQ

4π01

√35 −1

3

15. (a) Assume the mass Mis located at the origin and the mass mat (x, y, z), then

r=xi+yj+zk,F(x, y, z)=−GmM

(x2+y2+z2)3/2r,

W=−Zt2

GmM

(x2+y2+z2)3/2xdx

dt +ydy

dt +zdz

dt dt

=GmM(x2+y2+z2)−1/2it2

=GmM 1

r2−1

r1

(b) W=3.99 ×105×1031

7200 −1

7000≈−1.597 ×109J

Chapter 17 Supplementary Exercises 636

16. Let g(x, y)=x

, then Z

gdy =ZZ

2xdA,so¯x=1

AZZ

xdA=1

2AZ

gdy =1

2AZ

2dy,

similarly for ¯y.

17. ¯x= 0 by symmetry; by Exercise 16, ¯y=1

2AZ

y2dx;C1:y=0,−a≤x≤a, y2dx =0;

2:x=acos θ, y =asin θ, 0≤θ≤π,so

¯y=−1

2(πa2/2) Zπ

a2sin2θ(−asin θ)dθ =4a

3π

18. ¯y=¯xby symmetry; by Exercise 16, ¯x=1

2AZ

x2dy;C1:y=0,0≤x≤a, x2dy =0;

2:x=acos θ, y =asin θ, 0≤θ≤π/2; C3:x=0,x

2dy =0;

¯x=1

2(πa2/4) Zπ/2

a2(cos2θ)acos θdθ=4a

3π

19. ¯y= 0 by symmetry; ¯x=1

2AZ

x2dy;A=αa2;C1:x=tcos α, y =−tsin α, 0≤t≤a;

C2:x=acos θ, y =asin θ, −α≤θ≤α;C3:x=tcos α,

y=tsin α, 0≤t≤a(reverse orientation);

2A¯x=−Za

t2cos2αsin αdt+Zα

−α

3cos3θdθ−Za

2cos2αsin α dt,

=−2a3

3cos2αsin α+2a

3Zα

cos3θdθ=−2a

3cos2αsin α+2a3

3cos2αsin α=4

3a3sin α;

¯x=2a

sin α

20. A=Za

0b−b

a2x2dx =2ab

3,C

1:x=t, y =bt2/a2,0≤t≤a;

C2:x=a−t, y =b, 0≤t≤a, x2dy =0;C

3:x=0,y =b−t, 0≤t≤b, x2dy =y2dx =0;

2A¯x=Za

(2bt/a2)dt =a2b

2,¯x=3a

2A¯y=−Za

(bt2/a2)2dt +Za

b2dt =−ab2

5+ab2=4ab2

5,¯y=3b

21. (a) Z

f(x)dx +g(y)dy =ZZ

R∂

∂xg(y)−∂

∂yf(x)dA =0

(b) W=Z

F·dr=Z

f(x)dx +g(y)dy = 0, so the work done by the vector ﬁeld around any

simple closed curve is zero. The ﬁeld is conservative.

637 Chapter 17

22. (a) Let r=dcos θi+dsin θj+zkin cylindrical coordinates, so

dt =dr

dθ

dt =ω(−dsin θi+dcos θj),v=dr

dt =ωk×r=ω×r.

(b) From Part (a), v=ωd(−sin θi+ cos θj)=−ωyi+ωxj

(d) No; from Exercise 32 in Section 17.1, if φwere a potential function for v, then

curl (∇φ) = curl v= 0, contradicting Part (c) above.

23. Yes; by imagining a normal vector sliding around the surface it is evident that the surface has two

sides.

24. Dnφ=n·∇φ,soZZ

DnφdS =ZZ

n·∇φdS =ZZZ

∇·(∇φ)dV

=ZZZ

G∂2φ

∂x2+∂2φ

∂y2+∂2φ

∂z2dV

25. By Exercise 24, ZZ

DnfdS=−ZZZ

[fxx +fyy +fzz]dV =−6ZZZ

dV =−6vol(G)=−8π

26. (a) fy−gx=exy +xyexy −exy −xyexy = 0 so the vector ﬁeld is conservative.

(b) φx=yexy −1,φ=e

xy −x+k(x),φ

y=xexy, let k(x)=0;φ(x, y)=e

xy −x

F·dr=φ(x(8π),y(8π)) −φ(x(0),y(0)) = φ(8π, 0) −φ(0,0) = −8π

27. (a) If h(x)Fis conservative, then ∂

∂y(yh(x)) = ∂

∂x(−2xh(x)), or h(x)=−2h(x)−2xh0(x) which

has the general solution x3h(x)2=C1,h(x)=Cx−3/2,soCy

3/2i−C2

1/2jis conservative,

with potential function φ=−2Cy/√x.

(b) If g(y)F(x, y) is conservative then ∂

∂y(yg(y)) = ∂

∂x(−2xg(y)), or g(y)+yg0(y)=−2g(y),

with general solution g(y)=C/y3,soF=C1

i−C

jis conservative, with potential

function Cx/y2.

28. A computation of curl Fshows that curl F=0if and only if the three given equations hold.

Moreover the equations hold if Fis conservative, so it remains to show that Fis conservative if

curl F=0. Let Cby any simple closed curve in the region. Since the region is simply connected,

there is a piecewis smooth, oriented surface σin the region with boundary C. By Stokes’ Theorem,

F·r=ZZ

(curl F)·ndS =ZZ

0dS =0.

By the 3-space analog of Theorem 17.3.2, Fis conservative.

29. (a) conservative, φ(x, y, z)=xz2−e−y(b) not conservative, fy6=gx(for example)

30. (a) conservative, φ(x, y, z)=−cos x+yz (b) not conservative, fz6=hx

Chapter 17 Horizon Module 638

CHAPTER 17 HORIZON MODULE

1. (a) If r=xi+yjdenotes the position vector, then F1·r= 0 by inspection, so the velocity ﬁeld

is tangent to the circle. The relationship F1×r=−k

2πkindicates that r,F1,kis a right-

handed system, so the ﬂow is counterclockwise. The polar form F1=−k

2πr(sin θi−cos θj)

shows that the speed is the constant k

2πr on a circle of radius r; and it also shows that the

speed is proportional to 1

rwith constant of proportionality 2kπ.

(b) Since kF1k=k

2πr, when r= 1 we get k=2πkF

–1

3. (a) F2=−q

2πkrk2rso F2is directed toward the origin, and kF2k=q

2πr is constant for constant

r, and the speed is proportional to the distance from the origin (constant of proportionality

2π). Since the velocity vector is directed toward the origin, the ﬂuid ﬂows towards the origin,

which must therefore be a sink.

(b) From Part (a) when r=1,q=2πkF

–1

5. (b) The magnitudes of the ﬁeld vectors increase, and their directions become more tangent to

circles about the origin.

origin.

639 Chapter 17

6. (a) The inward component is F2,soatr=20,15 = kF2k=q

2π(20),soq= 600π; the tangential

component is F1,soatr=20,45 = kF1k=k

2π(20),sok= 1800π.

(b) F =−1

x2+y2[(300x+ 900y)i+ (300y−900x)j]

r≤5 km/hr if r≥60√10 ≈189.7 km.

7. F =−1

2πr[(qcos θ+ksin θ)i+(qsin θ−kcos θ)j]=−q

2πr ur+k

2πruθ=−1

2πr(qur−kuθ)

8. F ·∇ψ=−1

2πr(qur−kuθ)·(−1

2π(kur+quθ)) = 1

(2πr)2(qk −kq) = 0, since urand uθare unit

orthogonal vectors.

9. From the hypotheses of Exercise 8, ψ=−1

2πkln r+α(θ),∂

∂θψ=α0(θ)=−q

2π,

α=−q

2πθ, ψ =−1

2π(kln r+qθ)

10. The streamline ψ=cbecomes kln r+qθ =−2πc, ln r=−qθ/k −2πc/k,

r=e−qθ/ke−2πc/k =κe−qθ, where κ>0.

11.

321

12. q= 600π, k = 1800π, r =κe−θ/3;atr=20,θ =π/4,κ =reθ/3=20e

π/12 ≈25.985; the desired

streamline has the polar equation r=25.985e−θ/3.

640

APPENDIX A

Real Numbers, Intervals, and Inequalities

EXERCISE SET A

1. (a) rational (b) integer, rational (c) integer, rational

(d) rational (e) integer, rational (f) irrational

(g) rational (h) integer, rational

2. (a) irrational (b) rational (c) rational (d) rational

3. (a) x=0.123123123 ..., 1000x= 123 + x,x= 123/999=41/333

(b) x=12.7777 ..., 10(x−12)=7+(x−12), 9x= 115, x= 115/9

x=3769.74

99 =376974

9900 =20943

550

(d) 4296

10000

4. x=0.99999 ...,10x=9+x,9x=9,x=1

5. (a) If ris the radius, then D=2rso 8

9D2

=16

9r2

=256

81 r2. The area of a circle of radius

ris πr2so 256/81 was the approximation used for π.

(b) 22/7≈3.1429 is better than 256/81 ≈3.1605.

6. (a) 223

71 <333

106 <63

25 17+15

√

7+15

√

5!<355

113 <22

(b) Ramanujan’s (c) Athoniszoon’s (d) Ramanujan’s

7. Line 2 3 4 5 6 7

Blocks 3,4 1,2 3,4 2,4,5 1,2 3,4

8. Line 1 2 3 4 5

Blocks all blocks none 2,4 2 2,3

9. (a) always correct (add −3 to both sides of a≤b)

(b) not always correct (correct only if a=b)

(d) always correct (multiply both sides of a≤bby 6)

(e) not always correct (correct only if a≥0)

(f) always correct (multiply both sides of a≤bby the nonnegative quantity a2)

10. (a) always correct

(b) not always correct (for example let a=b=0,c=1,d=2)

11. (a) all values because a=ais always valid (b) none

12. a=b, because if a6=bthen a<band b<aare contradictory

641 Appendix A

13. (a) yes, because a≤bis true if a<b (b) no, because a<bis false if a=bis true

14. (a) x2−5x=0,x(x−5)=0sox=0orx=5

(b) −1,0,1,2 are the only integers that satisfy −2<x<3

15. (a) {x:xis a positive odd integer}(b) {x:xis an even integer}

16. (a) not equal to Abecause 0 is not in A(b) equal to A

17. (a) false, there are points inside the triangle that are not inside the circle

(b) true, all points inside the triangle are also inside the square

(f) true, ais inside the circle (g) true

18. (a) ∅,{a1},{a2},{a3},{a1,a

},{a

}(b) ∅

19. (a) 4(b) −3

(d) −33

(e) −33

(f) −33

20. (a)

(b)

(c)

(d) none

21. (a) [−2,2] (b) (−∞,−2) ∪(2,+∞)

22. (a) −34

(b)

41168

(d)

274

(e)

(f)

1 2.3

(g) (h)

Exercise Set A 642

23. 3x<10; (−∞,10/3)

24. 1

5x≥8; [40,+∞)

25. 2x≤−11; (−∞,−11/2]

−

26. 9x<−10; (−∞,−10/9)

−

27. 2x≤1 and 2x>−3; (−3/2,1/2]

−

28. 8x≥5 and 8x≤14; [5

8,7

29. x

x−3−4<0,12 −3x

x−3<0,4−x

x−3<0;

(−∞,3) ∪(4,+∞)4

−−−

+++++

+++

+++++++

−−−−−−

−−−

4 − x

x − 3

4 − x

30. x

8−x+2= 16 −x

8−x≥0;

(−∞,8) ∪[16,+∞)16

816

−−−

−−−−−

−−−

−−

+++++++

++++++

+++

16 − x

8 − x

16 − x

8 − x

31. 3x+1

x−2−1=2x+3

x−2<0,x+3/2

x−2<0;

(−3

2,2)

+++

+−−−−

−−−

+−−−+++

++++++

−−− x − 2

−

x + 3

x − 2

x + 3

−

643 Appendix A

32. x/2−3

4+x−1>0,x−6

4+x−2>0,x+14

x+4 <0;

(−14,−4)

−4

+++

+−−−−

−−−

+−−−+++

++++++

−−−

x + 14

x + 4

−4

x + 14

−14

33. 4

2−x−1=x+2

2−x≤0; (−∞,−2] ∪(2,+∞)0

+++

−++++

+++

−−

+−−−+++

−−−−−−

+++

x + 2

2 − x

x + 2

−2

34. 3

x−5−2=13 −2x

x−5≤0,13/2−x

x−5≤0;

(−∞,5) ∪[13

2,+∞)

−−−

+++++

+++

+++++++

−−−−−−

−−− x − 5

513

− x

x − 5

− x

35. x2−9=(x+ 3)(x−3) >0;

(−∞,−3) ∪(3,+∞)

3−3

−3

+++

+−−−−

−−−

+−−−+++

++++++

−−−

x + 3

x − 3

(x + 3)(x − 3)

Exercise Set A 644

36. x2−5=(x−√

5)(x+√5) ≤0; [−√5,√5] 0

+++

+−−−−

−−−

+−−−+++

++++++

−−−

−

5x + 5

x − 5

()()

x + 5 x − 5

−

− 5

37. (x−4)(x+2)>0; (−∞,−2) ∪(4,+∞)

4−2

−2

++++++

−2

++++−−−+++x + 2

0+−−−−

−−−

++−−− x − 4

(x − 4)(x + 2)

38. (x−3)(x+4)<0; (−4,3)

−4

−4 3

−43

+++

+++++

−−−

−−−−−−−

++++++

−−−

x − 3

x + 4

(x − 3)(x + 4)

39. (x−4)(x−5) ≤0; [4,5]

4 5

−−−++++++

0+++−−−−−−− x − 5

0+++++++−−− x − 4

(x − 4)(x − 5)

40. (x−2)(x−1) ≥0; (−∞,1] ∪[2,+∞)

1 2

−−−++++++

0+++−−−−−−− x − 2

0+++++++−−− x − 1

(x − 1)(x − 2)

645 Appendix A

41. 3

x−4−2

x=x+8

x(x−4) >0; (−8,0) ∪(4,+∞)

−−+++++−−

++−−−−−−−−x − 4

++−−−−−−++

−8

40−8

+++++++−− + x + 8

x(x − 4)

(x + 8)

42. 1

x+1−3

x−2=−2x−5

(x+ 1)(x−2) ≥0,

x+5/2

(x+ 1)(x−2) ≤0;

(−∞,−5

2]∪(−1,2)

+++

−−−−−

−−−

++−−+++

+++−−

−−− x − 2

−

++++−

−−−

++−−− x + 1

−1

x + 5

(x + 1)(x − 2)

x + 5

−

43. By trial-and-error we ﬁnd that x= 2 is a root of the equation x3−x2−x−2=0sox−2

is a factor of x3−x2−x−2. By long division we ﬁnd that x2+x+ 1 is another factor so

x3−x2−x−2=(x−2)(x2+x+ 1). The linear factors of x2+x+ 1 can be determined by ﬁrst

ﬁnding the roots of x2+x+ 1 = 0 by the quadratic formula. These roots are complex numbers

so x2+x+1 6= 0 for all real x;thusx

2+x+ 1 must be always positive or always negative.

Since x2+x+ 1 is positive when x= 0, it follows that x2+x+1 >0 for all real x. Hence

x3−x2−x−2>0, (x−2)(x2+x+1)>0, x−2>0, x>2, so S=(2,+∞).

44. By trial-and-error we ﬁnd that x= 1 is a root of the equation x3−3x+2=0sox−1

is a factor of x3−3x+ 2. By long division we ﬁnd that x2+x−2 is another factor so

x3−3x+2=(x−1)(x2+x−2)=(x−1)(x−1)(x+2)=(x−1)2(x+ 2). Therefore we

want to solve (x−1)2(x+2)≤0. Now if x6= 1, then (x−1)2>0 and so x+2≤0, x≤−2. By

inspection, x= 1 is also a solution so S=(−∞,−2] ∪{1}.

45. √x2+x−6 is real if x2+x−6≥0. Factor to get (x+ 3)(x−2) ≥0 which has as its solution

x≤−3orx≥2.

46. x+2

x−1≥0; (−∞,−2] ∪(1,+∞)

47. 25 ≤5

9(F−32) ≤40, 45 ≤F−32 ≤72, 77 ≤F≤104

48. (a) n=2k,n

2=4k

2= 2(2k2) where 2k2is an integer.

(b) n=2k+1,n

2=4k

2+4k+ 1 = 2(2k2+2k) + 1 where 2k2+2kis an integer.

Exercise Set A 646

49. (a) Assume mand nare rational, then m=p

qand n=r

swhere p,q,r, and sare integers so

m+n=p

q+r

s=ps +rq

qs which is rational because ps +rq and qs are integers.

(b) (proof by contradiction) Assume mis rational and nis irrational, then m=p

qwhere pand

qare integers. Suppose that m+nis rational, then m+n=r

swhere rand sare integers

so n=r

s−m=r

s−p

q=rq −ps

sq . But rq −ps and sq are integers, so nis rational which

contradicts the assumption that nis irrational.

50. (a) Assume mand nare rational, then m=p

qand n=r

swhere p,q,r, and sare integers so

mn =p

q·r

s=pr

qs which is rational because pr and qs are integers.

(b) (proof by contradiction) Assume mis rational and nonzero and that nis irrational, then

m=p

qwhere pand qare integers and p6= 0. Suppose that mn is rational, then mn =r

where rand sare integers so n=r/s

m=r/s

p/q =rq

ps. But rq and ps are integers, so nis

rational which contradicts the assumption that nis irrational.

51. a=√2,b =√3,c =√6,d =−√2 are irrational, and a+d= 0, a rational; a+a=2

√

2, an

irrational; ad =−2, a rational; and ab =c, an irrational.

52. (a) irrational (Exercise 49(b)) (b) irrational (Exercise 50(b))

(d) √πmust be irrational, for if it were rational, then so would be π=(

√

π)

2by Exercise 50(a);

but πis known to be irrational.

53. The average of aand bis 1

2(a+b); if aand bare rational then so is the average, by Exercise 49(a)

and Exercise 50(a). On the other hand if a=b=√2 then the average of aand bis irrational, but

the average of aand −bis rational.

54. If 10x= 3, then x>0 because 10x≤1 for x≤0. If 10p/q = 3 with p, q integers, then 10p=3

so 3qmust be even (as well as 10podd), a contradiction to Exercise 48(b).

55. 8x3−4x2−2x+ 1 can be factored by grouping terms:

(8x3−4x2)−(2x−1)=4x

(2x−1)−(2x−1)=(2x−1)(4x2−1)=(2x−1)2(2x+1). The problem,

then, is to solve (2x−1)2(2x+1)<0. By inspection, x=1/2 is not a solution. If x6=1/2, then

(2x−1)2>0 and it follows that 2x+1<0, 2x<−1, x<−1/2, so S=(−∞,−1/2).

56. First, rewrite the inequality as 12x3−20x2+11x−2≥0. Now, if a polynomial in xwith integer

coeﬃcients has a rational zero p

q, then pwill be a factor of the constant term and qwill be a factor

of the coeﬃcient of the highest power of x. By trial-and-error we ﬁnd that x=1/2 is a zero, thus

(x−1/2) is a factor so

12x3−20x2+11x−2=(x−1/2) 12x2−14x+4



=2(x−1/2) 6x2−7x+2



=2(x−1/2)(2x−1)(3x−2)=(2x−1)2(3x−2).

Now to solve (2x−1)2(3x−2) ≥0 we ﬁrst note that x=1/2 is a solution. If x6=1/2 then

(2x−1)2>0 and 3x−2≥0, 3x≥2, x≥2/3soS=[2/3,+∞)∪{1/2}.

57. If a<b, then ac<bcbecause cis positive; if c<d, then bc<bdbecause bis positive, so ac<bd

(Theorem A.1(a)).

58. no, since the decimal representation is not repeating (the string of zeros does not have constant

length)

647

APPENDIX B

Absolute Value

EXERCISE SET B

1. (a) 7(b) √2(c) k2(d) k2

2. p(x−6)2=x−6ifx≥6, p(x−6)2=−(x−6) = −x+6ifx<6

3. |x−3|=|3−x|=3−xif 3 −x≥0, which is true if x≤3

4. |x+2|=x+2ifx+2≥0sox≥−2. 5. All real values of xbecause x2+9>0.

6. |x2+5x|=x

2+5xif x2+5x≥0sox(x+5)≥0 which is true for x≤−5orx≥0.

7. |3x2+2x|=|x(3x+2)|=|x||3x+2|.If|x||3x+2|=x|3x+2|, then |x||3x+2|−x|3x+2|=0,

(|x|−x)|3x+2|= 0, so either |x|−x=0or|3x+2|=0. If|x|−x= 0, then |x|=x, which is

true for x≥0. If |3x+2|= 0, then x=−2/3. The statement is true for x≥0orx=−2/3.

8. |6−2x|=|2(3 −x)|=|2||3−x|=2|x−3|for all real values of x.

9. p(x+5)

2=|x+5|=x+5ifx+5≥0, which is true if x≥−5.

10. p(3x−2)2=|3x−2|=|2−3x|=2−3xif 2 −3x≥0sox≤2/3.

13. (a) |7−9|=|−2|=2 (b) |3−2|=|1|=1

2|=|−(3+√2)|=3+√

(e) |−4−(−11)|=|7|=7 (f) |−5−0|=|−5|=5

14. √a4=p(a2)2=|a2|, but |a2|=a2because a2≥0 so it is valid for all values of a.

15. (a) Bis 6 units to the left of A;b=a−6=−3−6=−9.

(b) Bis 9 units to the right of A;b=a+9=−2+9=7.

given that b>0, it follows that b= 12.

16. In each case we solve for ein terms of f:

(a) e=f−4; eis to the left of f.(b) e=f+4;eis to the right of f.

17. |6x−2|=7

Case 1: Case 2:

6x−2=7 6x−2=−7

6x=9 6x=−5

x=3/2x=−5/6

18. |3+2x|=11

Case 1: Case 2:

3+2x=11 3+2x=−11

2x=8 2x=−14

x=4 x=−7

19. |6x−7|=|3+2x|

Case 1: Case 2:

6x−7=3+2x6x−7=−(3+2x)

4x=10 8x=4

x=5/2x=1/2

20. |4x+5|=|8x−3|

Case 1: Case 2:

4x+5=8x−34x+5=−(8x−3)

−4x=−812x=−2

x=2 x=−1/6

Exercise Set B 648

21. |9x|−11 = x

Case 2:

9x−11 = x−9x−11 = x

8x=11 −10x=11

x=11/8x=−11/10

22. 2x−7=|x+1|

Case 1: Case 2:

2x−7=x+1 2x−7=−(x+1)

x=8 3x=6

x= 2; not a solution

because xmust also

satisfy x<−1

23. 

x+5

2−x



Case 1: Case 2:

x+5

2−x=6 x+5

2−x=−6

x+5=12−6xx+5=−12+6x

7x=7 −5x=−17

x=1 x=17/5

24. 

x−3

x+4



Case 1: Case 2:

x−3

x+4=5 x−3

x+4=−5

x−3=5x+20 x−3=−5x−20

−4x=23 6x=−17

x=−23/4x=−17/6

25. |x+6|<3

−3<x+6<3

−9<x<−3

S=(−9,−3)

26. |7−x|≤5

−5≤7−x≤5

−12 ≤−x≤−2

12 ≥x≥2

S=[2,12]

27. |2x−3|≤6

−6≤2x−3≤6

−3≤2x≤9

−3/2≤x≤9/2

S=[−3/2,9/2]

28. |3x+1|<4

−4<3x+1<4

−5<3x<3

−5/3<x<1

S=(−5/3,1)

29. |x+2|>1

Case 1: Case 2:

x+2>1x+2<−1

x>−1x<−3

S=(−∞,−3) ∪(−1,+∞)

30. 

2x−1≥2

Case 1: Case 2:

2x−1≥21

2x−1≤−2

x≥31

x≤−1

x≥6x≤−2

S=(−∞,−2] ∪[6,+∞)

31. |5−2x|≥4

Case 1: Case 2:

5−2x≥45−2x≤−4

−2x≥−1−2x≤−9

x≤1/2x≥9/2

S=(−∞,1/2] ∪[9/2,+∞)

32. |7x+1|>3

Case 1: Case 2:

7x+1>37x+1<−3

7x>27x<−4

x>2/7x< −4/7

S=(−∞,−4/7) ∪(2/7,+∞)

649 Appendix B

33. 1

|x−1|<2,x6=1

|x−1|>1/2

Case 1: Case 2:

x−1>1/2x−1<−1/2

x>3/2x< 1/2

S=(−∞,1/2) ∪(3/2,+∞)

34. 1

|3x+1|≥5,x6=−1/3

|3x+1|≤1/5

−1/5≤3x+1≤1/5

−6/5≤3x≤−4/5

−2/5≤x≤−4/15

S=[−2/5,−1/3) ∪(−1/3,−4/15]

35. 3

|2x−1|≥4,x6=1/2

|2x−1|

3≤1

|2x−1|≤3/4

−3/4≤2x−1≤3/4

1/4≤2x≤7/4

1/8≤x≤7/8

S=[1/8,1/2) ∪(1/2,7/8]

36. 2

|x+3|<1,x6=−3

|x+3|

2>1

|x+3|>2

Case 1: Case 2:

x+3>2x+3<−2

x>−1x<−5

S=(−∞,−5) ∪(−1,+∞)

37. p(x2−5x+6)

2=x

2−5x+6ifx

2−5x+6≥0 or, equivalently, if (x−2)(x−3) ≥0;

x∈(−∞,2] ∪[3,+∞).

38. If x≥2 then 3 ≤x−2≤7so5≤x≤9; if x<2 then 3 ≤2−x≤7so−5≤x≤−1.

S=[−5,−1] ∪[5,9].

39. If u=|x−3|then u2−4u= 12, u2−4u−12=0,(u−6)(u+2)=0,sou=6oru=−2. If

u= 6 then |x−3|=6,sox=9orx=−3. If u=−2 then |x−3|=−2 which is impossible. The

solutions are −3 and 9.

41. |a−b|=|a+(−b)|

≤|a|+|−b|(triangle inequality)

=|a|+|b|.

42. a=(a−b)+b

|a|=|(a−b)+b|

|a|≤|a−b|+|b|(triangle inequality)

|a|−|b|≤|a−b|.

43. From Exercise 42

(i) |a|−|b|≤|a−b|; but |b|−|a|≤|b−a|=|a−b|, so (ii) |a|−|b|≥−|a−b|.

Combining (i) and (ii): −|a−b|≤|a|−|b|≤|a−b|,so||a|−|b||≤|a−b|.

650

APPENDIX C

Coordinate Planes and Lines

EXERCISE SET C

(-4, 1)

(-4, 7) (6, 7)

(6, 1)

2. area = 1

2bh =1

2(5 −(−3))(1) = 4

(-3, 2)

(4, 3)

(5, 2)

3. (a) x=2

(b) y=−3

-3

(d) y=x

(e) y≥x

(f) |x|≥1

-1 1

651 Appendix C

4. (a) x=0

(b) y=0

(d) x≥1 and y≤2

(e) x=3

(f) |x|=5

5-5

5. y=4−x

5-5

6. y=1+x

5-5

Exercise Set C 652

7. y=√x−4

8. y=−√x+1

-5 5

9. x2−x+y=0

-3

-2

-1

-1 1 2

10. x=y3−y2

-1

-2 -1 1 2

11. x2y=2

-3

-2

-1

-1 1 2

12. xy =−1

13. (a) m=4−2

3−(−1) =1

2(b) m=1−3

7−5=−1

−3−4=0 (d) m=12 −(−6)

−2−(−2) =18

0, not deﬁned

14. m1=5−2

6−(−1) =3

7,m2=7−2

2−(−1) =5

3,m3=7−5

2−6=−1

15. (a) The line through (1,1) and (−2,−5) has slope m1=−5−1

−2−1= 2, the line through (1,1) and

(0,−1) has slope m2=−1−1

0−1= 2. The given points lie on a line because m1=m2.

653 Appendix C

(b) The line through (−2,4) and (0,2) has slope m1=2−4

0+2 =−1, the line through (−2,4) and

(1,5) has slope m2=5−4

1+2 =1

3. The given points do not lie on a line because m16=m2.

16.

(4, 2)

(a)

(b)

(c)

17.

(-1, -2)

(b)

(a)

(c)

18. The triangle is equiangular because it is equilateral. The angles of

inclination of the sides are 0◦,60

◦

, and 120◦(see ﬁgure), thus the

slopes of its sides are tan0◦= 0, tan 60◦=√3, and tan 120◦=−√3.

120°

60°

19. III <II <IV <I20. III <IV <I<II

21. Use the points (1,2) and (x, y) to calculate the slope: (y−2)/(x−1)=3

(a) if x= 5, then (y−2)/(5 −1)=3,y−2 = 12, y=14

(b) if y=−2, then (−2−2)/(x−1)=3,x−1=−4/3, x=−1/3

22. Use (7,5) and (x, y) to calculate the slope: (y−5)/(x−7) = −2

(a) if x= 9, then (y−5)/(9 −7) = −2, y−5=−4, y=1

(b) if y= 12, then (12 −5)/(x−7) = −2, x−7=−7/2, x=7/2

23. Using (3,k) and (−2,4) to calculate the slope, we ﬁnd k−4

3−(−2) =5,k−4 = 25, k= 29.

24. The slope obtained by using the points (1,5) and (k, 4) must be the same as that obtained from

the points (1,5) and (2,−3) so 4−5

k−1=−3−5

2−1,−1

k−1=−8, k−1=1/8, k=9/8.

25. 0−2

x−1=−0−5

x−4,−2x+8=5x−5, 7x= 13, x=13/7

26. Use (0,0) and (x, y)togety−0

x−0=1

,y=1

x. Use (7,5) and (x, y)togety−5

x−7=2,

y−5=2(x−7), y=2x−9. Solve the system of equations y=1

2xand y=2x−9toget

x=6,y=3.

27. Show that opposite sides are parallel by showing that they have the same slope:

using (3,−1) and (6,4), m1=5/3; using (6,4) and (−3,2), m2=2/9;

using (−3,2) and (−6,−3), m3=5/3; using (−6,−3) and (3,−1), m4=2/9.

Opposite sides are parallel because m1=m3and m2=m4.

Exercise Set C 654

28. The line through (3,1) and (6,3) has slope m1=2/3, the line through (3,1) and (2,9) has slope

m2=−8, the line through (6,3) and (2,9) has slope m3=−3/2. Because m1m3=−1, the

corresponding lines are perpendicular so the given points are vertices of a right triangle.

29. (a)

y(b)

(c)

-2

y(c)

-7

30. (a)

-4

y(b)

-8 x

(c)

y(d)

655 Appendix C

31. (a)

-1

5-5

(b)

5-5

(c)

5-5

32. (a)

5-5

(b)

5-5

(c)

5-5

33. (a) m=3,b=2 (b) m=−1

4,b=3

5x+8

5so m=−3

5,b=8

5(d) m=0,b=1

(e) y=−b

ax+bso m=−b

a,y-intercept b

34. (a) m=−4, b=2 (b) y=1

3x−2

3so m=1

3,b=−2

2x+3som=−3

2,b=3 (d) y=3som=0,b=3

(e) y=−a0

xso m=−a0

,b=0

35. (a) m=(0−(−3))/(2 −0))=3/2soy=3x/2−3

(b) m=(−3−0)/(4 −0) = −3/4soy=−3x/4

36. (a) m=(0−2)/(2 −0)) = −1soy=−x+2

(b) m=(2−0)/(3 −0)=2/3soy=2x/3

37. y=−2x+4 38. y=5x−3

39. The slope mof the line must equal the slope of y=4x−2, thus m= 4 so the equation is y=4x+7.

40. The slope of the line 3x+2y=5is−3/2 so the line through (−1,2) with this slope

is y−2=−3

(x+ 1); y=−3

2x+1

41. The slope mof the line must be the negative reciprocal of the slope of y=5x+9,thusm=−1/5

and the equation is y=−x/5+6.

Exercise Set C 656

42. The slope of the line x−4y= 7 is 1/4 so a line perpendicular to it must have a slope of −4;

y+4=−4(x−3); y=−4x+8.

43. y−4=−7−4

1−2(x−2) = 11(x−2), y=11x−18.

44. y−6= 1−6

−2−(−3)(x−(−3)), y−6=−5(x+ 3), y=−5x−9.

45. The line passes through (0,2) and (−4,0), thus m=0−2

−4−0=1

2so y=1

2x+2.

46. The line passes through (0,b) and (a, 0), thus m=0−b

a−0=−b

a, so the equation is

y=−b

ax+b.

47. y=1 48. y=−8

49. (a) m1=4,m

2= 4; parallel because m1=m2

(b) m1=2,m

2=−1/2; perpendicular because m1m2=−1

(d) If A6= 0 and B6= 0, then m1=−A/B,m2=B/A and the lines are perpendicular because

m1m2=−1. If either Aor B(but not both) is zero, then the lines are perpendicular because

one is horizontal and the other is vertical.

(e) m1=4,m

2=1/4; neither

50. (a) m1=−5, m2=−5; parallel because m1=m2

(b) m1=2,m

2=−1/2; perpendicular because m1m2=−1.

(d) If B6= 0, then m1=m2=−A/B and the lines are parallel because m1=m2.IfB= 0 (and

A6= 0), then the lines are parallel because they are both perpendicular to the x-axis.

(e) m1=1/2, m2= 2; neither

51. y=(−3/k)x+4/k,k6=0

(a) −3/k =2,k=−3/2

(b) 4/k =5,k=4/5

(d) The slope of 2x−5y= 1 is 2/5so−3/k =2/5, k=−15/2.

(e) The slope of 4x+3y=2is−4/3 so the slope of the line perpendicular to it is 3/4;

−3/k =3/4, k=−4.

52. y2=3x: the union of the graphs of y=√3xand y=−√3x

5-5

657 Appendix C

53. (x−y)(x+y) = 0: the union of the graphs of

x−y= 0 and x+y=0

5-5

54. F=9

5C+32

-30 30

-30

55. u=3v

56. Y=4X+5

-1 Y

57. Solve x=5t+ 2 for tto get t=1

5x−2

5,soy=1

x−2

−3=1

x−17

5, which is a line.

58. Solve x=1+3t

2for t2to get t2=1

3x−1

3,soy=2−1

x−1

=−1

x+7

, which is a line;

1+3t

2≥1 for all tso x≥1.

59. An equation of the line through (1,4) and (2,1) is y=−3x+7. It crosses the y-axis at y= 7, and

the x-axis at x=7/3, so the area of the triangle is 1

2(7)(7/3)=49/6.

60. (2x−3y)(2x+3y)=0,so2x−3y=0,y=2

xor 2x+3y=0,

y=−2

x. The graph consists of the lines y=±2

3x.

5-5

61. (a) yes (b) yes (c) no (d) yes (e) yes (f) yes (g) no

658

APPENDIX D

Distance, Circles, and Quadratic Equations

EXERCISE SET D

1. in the proof of Theorem D.1

2. (a) d=p(−1−2)2+(1−5)2=√9+16=√

25=5

(b) 2+(−1)

2,5+1

2=(1/2,3)

3. (a) d=p(1 −7)2+(9−1)2=√36+64=√

100=10

(b) 7+1

2,1+9

2=(4,5)

4. (a) d=p(−3−2)2+(6−0)2=√25+36=√

(b) 2+(−3)

2,0+6

2=(−1/2,3)

5. (a) d=p[−7−(−2)]2+[−4−(−6)]2=√25+4=√

(b) −2+(−7)

2,−6+(−4)

2=(−9/2,−5)

6. Let A(1,1), B(−2,−8), and C(4,10) be the given points

(see diagram). A,B, and Clie on a straight line if and only if

d1+d2=d3, where d1,d2, and d3are the lengths of the line

segments AB,AC, and BC. But

d1=p(−2−1)2+(−8−1)2=3

√

10,

d2=p(4 −1)2+ (10 −1)2=3

√

10,

d3=p(4+2)

2+(10+8)

2=6

√

10; because d1+d2=d3,

it follows that A,B, and Clie on a straight line.

C(4, 10)

A(1, 1)

B(-2, -8)

7. Let A(5,−2), B(6,5), and C(2,2) be the given vertices and a,b, and cthe lengths of the sides

opposite these vertices; then

a=p(2 −6)2+(2−5)2=√25 = 5 and b=p(2 −5)2+(2+2)

2=√

25=5.

Triangle ABC is isosceles because it has two equal sides (a=b).

8. A triangle is a right triangle if and only if the square of the longest side is equal to the sum of the

squares of the other two sides (Pythagorean theorem). With A(1,3), B(4,2), and C(−2,−6) as ver-

tices and s1,s2, and s3the lengths of the sides opposite these vertices we ﬁnd that

1=(−2−4)2+(−6−2)2= 100, s2

2=(−2−1)2+(−6−3)2= 90, s2

3=(4−1)2+(2−3)2= 10,

and that s2

1=s2

2+s2

3,soABC is a right triangle.

9. P1(0,−2), P2(−4,8), and P3(3,1) all lie on a circle whose center is C(−2,3) if the points P1,

P2and P3are equidistant from C. Denoting the distances between P1,P2,P3and Cby d1,d2

and d3we ﬁnd that d1=p(0+2)

2+(−2−3)2=√29, d2=p(−4+2)

2+(8−3)2=√29, and

d3=p(3+2)

2+(1−3)2=√29, so P1,P2and P3lie on a circle whose center is C(−2,3) because

d1=d2=d3.

659 Appendix D

10. The distance between (t, 2t−6) and (0,4) is

p(t−0)2+(2t−6−4)2=pt2+(2t−10)2=√5t2−40t+ 100;

the distance between (t, 2t−6) and (8,0) is p(t−8)2+(2t−6)2=√5t2−40t+ 100,

so (t, 2t−6) is equidistant from (0,4) and (8,0).

11. If (2,k) is equidistant from (3,7) and (9,1), then

p(2 −3)2+(k−7)2=p(2 −9)2+(k−1)2,1+(k−7)2=49+(k−1)2,

1+k

2−14k+49=49+k

2−2k+1,−12k=0,k=0.

12. (x−3)/2 = 4 and (y+2)/2=−5sox= 11 and y=−12.

13. The slope of the line segment joining (2,8) and (−4,6) is 6−8

−4−2=1

3so the slope of the perpen-

dicular bisector is −3. The midpoint of the line segment is (−1,7) so an equation of the bisector is

y−7=−3(x+ 1); y=−3x+4.

14. The slope of the line segment joining (5,−1) and (4,8) is 8−(−1)

4−5=−9 so the slope of the perpen-

dicular bisector is 1

9. The midpoint of the line segment is (9/2,7/2) so an equation of the bisector is

y−7

2=1

9x−9

2;y=1

9x+3.

15. Method (see ﬁgure): Find an equation of the perpendicular bisector of the line segment joining

A(3,3) and B(7,−3). All points on this perpendicular bisector are equidistant from Aand B,thus

ﬁnd where it intersects the given line.

The midpoint of AB is (5,0), the slope of AB is −3/2thusthe

slope of the perpendicular bisector is 2/3 so an equation is

y−0=2

(x−5)

3y=2x−10

2x−3y−10=0.

The solution of the system

(4x−2y+3=0

2x−3y−10=0

gives the point (−29/8,−23/4).

A(3, 3)

B(7, -3)

4x - 2y + 3 = 0

16. (a) y= 4 is a horizontal line, so the vertical distance is |4−(−2)|=|6|=6.

(b) x=−1 is a vertical line, so the horizontal distance is |−1−3|=|−4|=4.

17. Method (see ﬁgure): write an equation of the line that goes through

the given point and that is perpendicular to the given line; ﬁnd the

point Pwhere this line intersects the given line; ﬁnd the

distance between Pand the given point.

The slope of the given line is 4/3, so the slope of a line

perpendicular to it is −3/4. x

4x – 3y + 10 = 0

(2, 1)

Exercise Set D 660

The line through (2,1) having a slope of −3/4isy−1=−

(x−2) or, after simpliﬁcation,

3x+4y= 10 which when solved simultaneously with 4x−3y+ 10 = 0 yields (−2/5,14/5) as the

point of intersection. The distance dbetween (−2/5,14/5) and (2,1) is

d=p(2+2/5)2+(1−14/5)2=3.

18. (See the solution to Exercise 17 for a description of the method.) The slope of the line

5x+12y−36=0is−5/12. The line through (8,4) and perpendicular to the given line is

y−4=12

5(x−8) or, after simpliﬁcation, 12x−5y= 76. The point of intersection of this line with

the given line is found to be 84

13,4

13and the distance between it and (8,4) is 4.

19. If B= 0, then the line Ax +C= 0 is vertical and x=−C/A for each point on the line. The line

through (x0,y

) and perpendicular to the given line is horizontal and intersects the given line at

the point (−C/A,y0). The distance dbetween (−C/A,y0) and (x0,y

)is

d=p(x

0+C/A)2+(y

0−y

)

2=r(Ax0+C)2

A2=|Ax0+C|

√A2

which is the value of |Ax0+By0+C|

√A2+B2for B=0.

If B6= 0, then the slope of the given line is −A/B and the line through (x0,y

) and perpendicular

to the given line is

y−y0=B

A(x−x0), Ay −Ay0=Bx −Bx0,Bx −Ay =Bx0−Ay0.

The point of intersection of this line and the given line is obtained by solving

Ax +By =−Cand Bx −Ay =Bx0−Ay0.

Multiply the ﬁrst equation through by Aand the second by Band add the results to get

(A2+B2)x=B2x0−ABy0−AC so x=B2x0−ABy0−AC

A2+B2

Similarly, by multiplying by Band −A,wegety=−ABx0+A2y0−BC

A2+B2.

The square of the distance dbetween (x, y) and (x0,y

)is

2=x

0−B

0−ABy0−AC

A2+B22

+y0−−ABx0+A2y0−BC

A2+B22

=(A2x0+ABy0+AC)2

(A2+B2)2+(ABx0+B2y0+BC)2

(A2+B2)2

=A2(Ax0+By0+C)2+B2(Ax0+By0+C)2

(A2+B2)2

=(Ax0+By0+C)2(A2+B2)

(A2+B2)2=(Ax0+By0+C)2

A2+B2

so d=|Ax0+By0+C|

√A2+B2.

20. d=|4(2) −3(1) + 10|

p42+(−3)2=|15|

√25 =15

5=3. 21. d=|5(8) + 12(4) −36|

√52+12

2=|52|

√169 =52

13 =4.

661 Appendix D

22. Method (see ﬁgure): Let A(0,a), B(b, 0), and C(c, 0) be the given

vertices; ﬁnd equations for the perpendicular bisectors L1,L2, and

L3and show that they all intersect at the same point.

B(b, 0)

A(0, a)

C(c, 0)

line L1: The midpoint of BC is b+c

2,0and since L1is vertical,

an equation for L1is x=b+c

line L2: The midpoint of AB is b

2,a

2; the slope of AB is −a

(if b6= 0) so the slope of

L2is b

a(even if b= 0) and an equation of L2is y−a

2=b

ax−b

2;

line L3: The midpoint of AC is c

2,a

2; the slope of AC is −a

c(if c6= 0) so the slope of

L3is c

a(even if c= 0) and an equation of L3is y−a

2=c

ax−c

2.

For the point of intersection of L1and L2, solve x=b+c

2and y−a

2=b

ax−b

2.

The point is found to be b+c

2,a2+bc

2. The point of intersection of L1and L3is obtained by

solving the system x=b+c

2and y−a

2=c

ax−c

2, its solution yields the point b+c

2,a2+bc

2.

So L1L2, and L3all intersect at the same point.

23. (a) center (0,0), radius 5 (b) center (1,4), radius 4

24. (a) center (0,0), radius 3 (b) center (3,5), radius 6

25. (x−3)2+(y−(−2))2=4

,(x−3)2+(y+2)

2=16

26. (x−1)2+(y−0)2=(

√

8/2)2,(x−1)2+y2=2

27. r= 8 because the circle is tangent to the x-axis, so (x+4)

2+(y−8)2= 64.

28. r= 5 because the circle is tangent to the y-axis, so (x−5)2+(y−8)2= 25.

29. (0,0) is on the circle, so r=p(−3−0)2+(−4−0)2=5;(x+3)

2+(y+4)

2= 25.

30. r=p(4 −1)2+(−5−3)2=√73; (x−4)2+(y+5)

2= 73.

31. The center is the midpoint of the line segment joining (2,0) and (0,2) so the center is at (1,1).

The radius is r=p(2 −1)2+(0−1)2=√2, so (x−1)2+(y−1)2=2.

32. The center is the midpoint of the line segment joining (6,1) and (−2,3), so the center is at (2,2).

The radius is r=p(6 −2)2+(1−2)2=√17, so (x−2)2+(y−2)2= 17.

33. (x2−2x)+(y

2−4y) = 11, (x2−2x+1)+(y

2−4y+4)=11+1+4,(x−1)2+(y−2)2= 16;

center (1,2) and radius 4

Exercise Set D 662

34. (x2+8x)+y

2=−8, (x2+8x+16)+y2=−8 + 16, (x+4)

2+y

2= 8; center (−4,0) and radius

2√2

35. 2(x2+2x)+2(y

2−2y)=0,2(x

2+2x+1)+2(y

2−2y+1)=2+2,(x+1)

2+(y−1)2=2;

center (−1,1) and radius √2

36. 6(x2−x)+6(y

2+y)=3,6(x

2−x+1/4)+6(y

2+y+1/4)=3+6/4+6/4,

(x−1/2)2+(y+1/2)2= 1; center (1/2,−1/2) and radius 1

37. (x2+2x)+(y

2+2y)=−2, (x2+2x+1)+(y

2+2y+1)=−2+1+1,(x+1)

2+(y+1)

2=0;

the point (−1,−1)

38. (x2−4x)+(y

2−6y)=−13, (x2−4x+4)+(y

2−6y+9)=−13+4+9,(x−2)2+(y−3)2=0;

the point (2,3)

39. x2+y2=1/9; center (0,0) and radius 1/3

40. x2+y2= 4; center (0,0) and radius 2

41. x2+(y

2+10y)=−26, x2+(y

2+10y+ 25) = −26 + 25, x2+(y+5)

2=−1; no graph

42. (x2−10x)+(y

2−2y)=−29, (x2−10x+25)+(y

2−2y+1)=−29+25+1,

(x−5)2+(y−1)2=−3; no graph

43. 16 x2+5

2x+ 16(y2+y)=7,16x

2+5

x+25

16+16y

2+y+1

4=7+25+4,

(x+5/4)2+(y+1/2)2=9/4; center (−5/4,−1/2) and radius 3/2

44. 4(x2−4x)+4(y

2−6y)=9,4(x

2−4x+4)+4(y

2−6y+9)=9+16+36,

(x−2)2+(y−3)2=61/4; center (2,3) and radius √61/2

45. (a) y2=16−x

,soy=±

√

16 −x2. The bottom half is y=−√16 −x2.

(b) Complete the square in yto get (y−2)2=3−2x−x

,soy−2=±

√

3−2x−x

or y=2±√

3−2x−x

. The top half is y=2+√

3−2x−x

46. (a) x2=9−y

2so x=±p9−y2. The right half is x=p9−y2.

(b) Complete the square in xto get (x−2)2=1−y

2so x−2=±

p1−y

,x=2±p1−y

The left half is x=2−p1−y

47. (a)

-5 5

y(b) y=√5+4x−x

=p5−(x

2−4x)

=p5+4−(x

2−4x+4)

=p9−(x−2)2

-1 5

663 Appendix D

48. (a)

-2

y(b)

-2

49. The tangent line is perpendicular to the radius at the point. The slope of the radius is 4/3, so

the slope of the perpendicular is −3/4. An equation of the tangent line is y−4=−3

(x−3), or

y=−3

4x+25

50. (a) (x+1)

2+y

2= 10, center at C(−1,0). The slope of CP is −1/3 so the slope of the tangent

is 3; y+1=3(x−2), y=3x−7.

(b) (x−3)2+(y+2)

2= 26, center at C(3,−2). The slope of CP is 5 so the slope of the tangent

is −1

5;y−3=−1

(x−4), y=−1

5x+19

51. (a) The center of the circle is at (0,0) and its radius is √20=2

√

5. The distance between Pand

the center is p(−1)2+ (2)2=√5 which is less than 2√5, so Pis inside the circle.

(b) Draw the diameter of the circle that passes through P, then the shorter segment of the

diameter is the shortest line that can be drawn from Pto the circle, and the longer segment

is the longest line that can be drawn from Pto the circle (can you prove it?). Thus, the

smallest distance is 2√5−√5=√

5, and the largest is 2√5+√

5=3

√

52. (a) x2+(y−1)2= 5, center at C(0,1) and radius √5. The distance between Pand Cis 3√5/2

so Pis outside the circle.

(b) The smallest distance is 3

2√5−√5=1

√

5, the largest distance is 3

2√5+√

5=5

√

53. Let (a, b) be the coordinates of T(or T0). The radius from (0,0) to T(or T0) will be perpendicular

to L(or L0) so, using slopes, b/a =−(a−3)/b,a2+b2=3a. But (a, b) is on the circle so a2+b2=1,

thus 3a=1,a=1/3. Let a=1/3ina

2+b

2= 1 to get b2=8/9, b=±√8/3. The coordinates of

Tand T0are (1/3,√8/3) and (1/3,−√8/3).

54. (a) p(x−2)2+(y−0)2=√2p(x−0)2+(y−1)2; square both sides and expand to get

x2−4x+4+y

2=2(x

2+y

2−2y+ 1), x2+y2+4x−4y−2 = 0, which is a circle.

(b) (x2+4x)+(y

2−4y)=2,(x

2+4x+4)+(y

2−4y+4)=2+4+4,(x+2)

2+(y−2)2= 10;

center (−2,2), radius √10.

55. (a) [(x−4)2+(y−1)2]+[(x−2)2+(y+5)

]=45

2−8x+16+y

2−2y+1+x

2−4x+4+y

2+10y+25=45

2+2y

2−12x+8y+ 1 = 0, which is a circle.

(b) 2(x2−6x)+2(y

2+4y)=−1, 2(x2−6x+9)+2(y

2+4y+4)=−1+18+8,

(x−3)2+(y+2)

2=25/2; center (3,−2), radius 5/√2.

Exercise Set D 664

56. If x2−y2= 0, then y2=x2so y=xor y=−x. The graph of x2−y2= 0 consists of the graphs

of the two lines y=±x. The graph of (x−c)2+y2= 1 is a circle of radius 1 with center at (c, 0).

Examine the ﬁgure to see that the system cannot have just one solution, and has 0 solutions if

|c|>√2, 2 solutions if |c|=√2, 3 solutions if |c|= 1, and 4 solutions if |c|<√2, |c|6=1.

2 solutions 3 solutions

4 solutions 0 solutions

57. y=x2+2

(0, 2)

58. y=x2−3

(0, -3)

3– 3

59. y=x2+2x−3

-3 1

-3

(-1, -4)

60. y=x2−3x−4

-1 4

-4

225

()

, –

61. y=−x2+4x+5

-1 5

(2, 9)

665 Appendix D

62. y=−x2+x

()

63. y=(x−2)2

(2, 0)

64. y=(3+x)

(-3, 0)

65. x2−2x+y=0

(1, 1)

66. x2+8x+8y=0

-8 x

(-4, 2)

67. y=3x

2−2x+1

()

68. y=x2+x+2

()

, –

69. x=−y2+2y+2

(3, 1)

1 – 3

1 + 3

70. x=y2−4y+5

(1, 2)

71. (a) x2=3−y,x=±

√

3−y. The right half is x=√3−y.

(b) Complete the square in xto get (x−1)2=y+1, x=1±

√

y+ 1. The left half is x=1−

√

y+1.

72. (a) y2=x+5,y=±

√

x+ 5. The upper half is y=√x+5.

(b) Complete the square in yto get (y−1/2)2=x+9/4, y−1/2=±

px+9/4,

y=1/2±px+9/4. The lower half is y=1/2−px+9/4.

Exercise Set D 666

73. (a)

-5

(b)

74. (a)

(b)

75. (a)

s(b) The ball will be at its highest point

when t= 1 sec; it will rise 16 ft.

76. (a) 2x+y= 500, y= 500 −2x.(b) A=xy =x(500 −2x) = 500x−2x2.

x=−b/(2a)=−500/(−4) = 125, so the maximum value of Ais

A= 500(125) −2(125)2= 31,250 ft2.

77. (a) (3)(2x) + (2)(2y) = 600, 6x+4y= 600, y= 150 −3x/2

(b) A=xy =x(150 −3x/2) = 150x−3x2/2

x=−b/(2a)=−150/(−3) = 50, so the maximum value of Ais

A= 150(50) −3(50)2/2 = 3,750 ft2.

667 Appendix D

78. (a) y=ax2+bx +c=ax2+b

ax+c

=ax2+b

ax+b2

4a2+c−b2

4a=ax+b

2a2

+c−b2

4a

(b) If a<0 then yis always less than c−b2

4aexcept when x=−b

2a, so the graph has its high

point there. If a>0 then yis always greater than c−b2

4aexcept when x=−b

2a, so the

graph has its low point there.

79. (a) The parabola y=2x

2+5x−1 opens upward and has x-intercepts of x=(−5±√

33)/4, so

2x2+5x−1<0if(−5−√

33)/4<x<(−5+√

33)/4.

(b) The parabola y=x2−2x+ 3 opens upward and has no x-intercepts, so x2−2x+3>0if

−∞ <x<+∞.

80. (a) The parabola y=x2+x−1 opens upward and has x-intercepts of x=(−1±√

5)/2, so

x2+x−1>0ifx<(−1−√

5)/2orx>(−1+√

5)/2.

(b) The parabola y=x2−4x+ 6 opens upward and has no x-intercepts, so x2−4x+6<0 has

no solution.

81. (a) The t-coordinate of the vertex is t=−40/[(2)(−16)] = 5/4, so the maximum height is

s= 5 + 40(5/4) −16(5/4)2= 30 ft.

(b) s=5+40t−16t2=0ift≈2.6s

√2)/4.

The length of this interval is (5 + 3√2)/4−(5 −3√2)/4=3

√

2/2≈2.1s.

82. x+3−x

2>0, x2−x−3<0, (1 −√13)/2<x<(1+√13)/2

668

APPENDIX E

Trigonometry Review

EXERCISE SET E

1. (a) 5π/12 (b) 13π/6(c) π/9(d) 23π/30

2. (a) 7π/3(b) π/12 (c) 5π/4(d) 11π/12

3. (a) 12◦(b) (270/π)◦(c) 288◦(d) 540◦

4. (a) 18◦(b) (360/π)◦(c) 72◦(d) 210◦

5. sin θcos θtan θcsc θsec θcot θ

(a) √21/5 2/5√21/2 5/√21 5/2 2/√21

(b) 3/4√7/4 3/√7 4/3 4/√7√7/3

6. sin θcos θtan θcsc θsec θcot θ

(a) 1/√2 1/√2 1 √2√2 1

(b) 3/5 4/5 3/4 5/3 5/4 4/3

7. sin θ=3/

√

10, cos θ=1/

√

10 8. sin θ=√5/3, tan θ=√5/2

9. tan θ=√21/2, csc θ=5/

√

21 10. cot θ=√15, sec θ=4/

√

11. Let xbe the length of the side adjacent to θ, then cos θ=x/6=0.3, x=1.8.

12. Let xbe the length of the hypotenuse, then sinθ=2.4/x =0.8, x=2.4/0.8=3.

13. θsin θcos θtan θcsc θsec θcot θ

(a) 225◦−1/√2−1/√2 1 −√2−√2 1

(b) −210◦1/2−√3/2−1/√32−2/√3−√3

(d) −3π/2 1 0 — 1 — 0

14. θsin θcos θtan θcsc θsec θcot θ

(a) 330◦−1/2√3/2−1/√3−2 2/√3−√3

(b) −120◦−√3/2−1/2√3−2/√3−2 1/√3

(d) −3π0−1 0 — −1 —

669 Appendix E

15. sin θcos θtan θcsc θsec θcot θ

(a) 4/5 3/5 4/3 5/4 5/3 3/4

(b) −4/5 3/5−4/3−5/4 5/3−3/4

(d) −1/2√3/2−1/√3−2 2/√3−√3

(e) 1/√21/√21√2√21

(f) 1/√2−1/√2−1√2−√2−1

16. sin θcos θtan θcsc θsec θcot θ

(a) 1/4√15/4 1/√15 4 4/√15 √15

(b) 1/4−√15/4−1/√15 4 −4/√15 −√15

(d) −3/√10 −1/√10 3 −√10/3−√10 1/3

(e) √21/5−2/5−√21/2 5/√21 −5/2−2/√21

(f) −√21/5−2/5√21/2−5/√21 −5/2 2/√21

17. (a) x= 3 sin 25◦≈1.2679 (b) x=3/tan(2π/9) ≈3.5753

18. (a) x=2/sin 20◦≈5.8476 (b) x=3/cos(3π/11) ≈4.5811

19. sin θcos θtan θcsc θsec θcot θ

(a) a/3√9−a2/3a/√9−a23/a 3/√9−a2√9−a2/a

(b) a/√a2+25 5/

√

2+25 a/5√a2+25/a √a2+25/5 5/a

20. (a) θ=3π/4±2nπ and θ=5π/4±2nπ,n=0,1,2,...

(b) θ=5π/4±2nπ and θ=7π/4±2nπ,n=0,1,2,...

21. (a) θ=3π/4±nπ,n=0,1,2,...

(b) θ=π/3±2nπ and θ=5π/3±2nπ,n=0,1,2,...

22. (a) θ=7π/6±2nπ and θ=11π/6±2nπ,n=0,1,2,...

(b) θ=π/3±nπ,n=0,1,2,...

23. (a) θ=π/6±nπ,n=0,1,2,...

(b) θ=4π/3±2nπ and θ=5π/3±2nπ,n=0,1,2,...

24. (a) θ=3π/2±nπ,n=0,1,2,... (b) θ=π±2nπ,n=0,1,2,...

25. (a) θ=3π/4±nπ,n=0,1,2,... (b) θ=π/6±nπ,n=0,1,2,...

26. (a) θ=2π/3±2nπ and θ=4π/3±2nπ,n=0,1,2,...

(b) θ=7π/6±2nπ and θ=11π/6±2nπ,n=0,1,2,...

Exercise Set E 670

27. (a) θ=π/3±2nπ and θ=2π/3±2nπ,n=0,1,2,...

(b) θ=π/6±2nπ and θ=11π/6±2nπ,n=0,1,2,...

28. sin θ=−3/5, cos θ=−4/5, tan θ=3/4, csc θ=−5/3, sec θ=−5/4, cot θ=4/3

29. sin θ=2/5, cos θ=−√21/5, tan θ=−2/√21, csc θ=5/2, sec θ=−5/√21, cot θ=−√21/2

30. (a) θ=π/2±2nπ,n=0,1,2,... (b) θ=±2nπ,n=0,1,2,...

(e) θ=±2nπ,n=0,1,2,... (f) θ=π/4±nπ,n=0,1,2,...

31. (a) θ=±nπ,n=0,1,2,... (b) θ=π/2±nπ,n=0,1,2,...

(e) θ=π/2±nπ,n=0,1,2,... (f) θ=±nπ,n=0,1,2,...

32. Construct a right triangle with one angle equal to 17◦, measure the lengths of the sides and

hypotenuse and use formula (6) for sin θand cos θto approximate sin 17◦and cos 17◦.

33. (a) s=rθ =4(π/6)=2π/3cm (b) s=rθ = 4(5π/6)=10π/3cm

34. r=s/θ =7/(π/3)=21/π 35. θ=s/r =2/5

36. θ=s/r so A=1

2r2θ=1

2r2(s/r)=1

37. (a) 2πr =R(2π−θ), r=2π−θ

2πR

(b) h=pR2−r2=pR2−(2π−θ)2R2/(4π2)=√

4πθ −θ2

2πR

38. The circumference of the circular base is 2πr. When cut and ﬂattened, the cone becomes a circular

sector of radius L.Ifθis the central angle that subtends the arc of length 2πr, then θ=(2πr)/L

so the area Sof the sector is S=(1/2)L2(2πr/L)=πrL which is the lateral surface area of the

cone.

39. Let hbe the altitude as shown in the ﬁgure, then

h= 3 sin 60◦=3

√

3/2soA=1

(3√3/2)(7) = 21√3/4.

60°

40. Draw the perpendicular from vertex Cas shown in the ﬁgure, then

h= 9 sin 30◦=9/2, a=h/ sin 45◦=9

√

2/2,

c1= 9 cos 30◦=9

√

3/2, c2=acos 45◦=9/2,

c1+c2=9(

√

3+1)/2,angle C= 180◦−(30◦+45

◦

) = 105◦

30°45°

c1c2B

41. Let xbe the distance above the ground, then x= 10 sin 67◦≈9.2 ft.

42. Let xbe the height of the building, then x= 120 tan 76◦≈481 ft.

671 Appendix E

43. From the ﬁgure, h=x−ybut x=dtan β,

y=dtan αso h=d(tan β−tan α).

44. From the ﬁgure, d=x−ybut x=hcot α,

y=hcot βso d=h(cot α−cot β),

h=d

cot α−cot β.

αβ

45. (a) sin 2θ= 2 sin θcos θ=2(

√

5/3)(2/3)=4

√

5/9

(b) cos 2θ= 2 cos2θ−1 = 2(2/3)2−1=−1/9

46. (a) sin(α−β) = sin αcos β−cos αsin β=(3/5)(1/√5) −(4/5)(2/√5) = −1/√5

(b) cos(α+β) = cos αcos β−sin αsin β=(4/5)(1/√5) −(3/5)(2/√5) = −2/(5√5)

47. sin 3θ= sin(2θ+θ) = sin 2θcos θ+ cos 2θsin θ= (2 sin θcos θ) cos θ+ (cos2θ−sin2θ) sin θ

= 2 sin θcos2θ+ sin θcos2θ−sin3θ= 3 sin θcos2θ−sin3θ; similarly, cos 3θ= cos3θ−3 sin2θcos θ

48. cos θsec θ

1 + tan2θ=cos θsec θ

sec2θ=cos θ

sec θ=cos θ

(1/cos θ)= cos2θ

49. cos θtan θ+ sin θ

tan θ=cos θ(sin θ/ cos θ) + sin θ

sin θ/ cos θ= 2 cos θ

50. 2 csc 2θ=2

sin 2θ=2

2 sin θcos θ=1

sin θ 1

cos θ= csc θsec θ

51. tan θ+ cot θ=sin θ

cos θ+cos θ

sin θ=sin2θ+ cos2θ

sin θcos θ=1

sin θcos θ=2

2 sin θcos θ=2

sin 2θ= 2 csc 2θ

52. sin 2θ

sin θ−cos 2θ

cos θ=sin 2θcos θ−cos 2θsin θ

sin θcos θ=sin θ

sin θcos θ= sec θ

53. sin θ+ cos 2θ−1

cos θ−sin 2θ=sin θ+(1−2 sin2θ)−1

cos θ−2 sin θcos θ=sin θ(1 −2 sin θ)

cos θ(1 −2 sin θ)= tan θ

54. Using (47), 2 sin 2θcos θ= 2(1/2)(sin θ+ sin 3θ) = sin θ+ sin 3θ

55. Using (47), 2 cos 2θsin θ= 2(1/2)[sin(−θ) + sin 3θ] = sin 3θ−sin θ

Exercise Set E 672

56. tan(θ/2) = sin(θ/2)

cos(θ/2) =2 sin2(θ/2)

2 sin(θ/2) cos(θ/2) =1−cos θ

sin θ

57. tan(θ/2) = sin(θ/2)

cos(θ/2) =2 sin(θ/2) cos(θ/2)

2 cos2(θ/2) =sin θ

1 + cos θ

58. From (52), cos(π/3+θ) + cos(π/3−θ) = 2cos(π/3) cos θ= 2(1/2) cos θ= cos θ

59. From the ﬁgure, area = 1

2hc but h=bsin A

so area = 1

2bc sin A. The formulas

area = 1

2ac sin Band area = 1

2ab sin C

follow by drawing altitudes from vertices Band C, respectively. A

60. From right triangles ADC and BDC,

h1=bsin A=asin Bso a/ sin A=b/ sin B.

From right triangles AEB and CEB,

h2=csin A=asin Cso a/ sin A=c/ sin C

thus a/ sin A=b/ sin B=c/ sin C.

61. (a) sin(π/2+θ) = sin(π/2) cos θ+ cos(π/2) sin θ= (1) cos θ+ (0) sin θ= cos θ

(b) cos(π/2+θ) = cos(π/2) cos θ−sin(π/2) sin θ= (0) cos θ−(1) sin θ=−sin θ

(d) cos(3π/2+θ) = cos(3π/2) cos θ−sin(3π/2) sin θ= (0) cos θ−(−1) sin θ= sin θ

62. tan(α+β)= sin(α+β)

cos(α+β)=sin αcos β+ cos αsin β

cos αcos β−sin αsin β, divide numerator and denominator by

cos αcos βand use tan α=sin α

cos αand tan β=sin β

cos βto get (38);

tan(α−β) = tan(α+(−β)) = tan α+ tan(−β)

1−tan αtan(−β)=tan α−tan β

1 + tan αtan βbecause

tan(−β)=−tan β.

63. (a) Add (34) and (36) to get sin(α−β) + sin(α+β)=2sinαcos βso

sin αcos β=(1/2)[sin(α−β) + sin(α+β)].

(b) Subtract (35) from (37).

64. (a) From (47), sin A+B

2cos A−B

2=1

2(sin B+ sin A)so

sin A+ sin B= 2 sin A+B

2cos A−B

(b) Use (49) (c) Use (48)

673 Appendix E

65. sin α+ sin(−β)=2sinα−β

2cos α+β

2, but sin(−β)=−sin βso

sin α−sin β= 2 cos α+β

2sin α−β

66. (a) From (34), Csin(α+φ)=Csin αcos φ+Ccos αsin φso Ccos φ= 3 and Csin φ=5,

square and add to get C2(cos2φ+sin2φ) = 9+ 25, C2= 34. If C=√34 then cos φ=3/

√

and sin φ=5/

√

34 so φis the ﬁrst-quadrant angle for which tanφ=5/3.

3 sin α+ 5 cos α=√34 sin(α+φ).

(b) Follow the procedure of part (a) to get Ccos φ=Aand Csin φ=B,C=√A2+B2,

tan φ=B/A where the quadrant in which φlies is determined by the signs of Aand Bbecause

cos φ=A/C and sin φ=B/C,soAsin α+Bcos α=√A2+B2sin(α+φ).

67. Consider the triangle having a,b, and das sides. The angle formed by sides aand bis π−θso

from the law of cosines, d2=a2+b2−2ab cos(π−θ)=a

2+b

2+2ab cos θ,d=√a2+b2+2ab cos θ.

674

APPENDIX F

Solving Polynomial Equations

EXERCISE SET F

1. (a) q(x)=x

2+4x+2,r(x)=−11x+6 (b) q(x)=2x

2+4,r(x)=9

3−x

2+2x−2,r(x)=2x+1

2. (a) q(x)=2x

2−x+2,r(x)=5x+5 (b) q(x)=x

3+3x

2−x+2,r(x)=3x−1

3−5,r(x)=4x

2+10

3. (a) q(x)=3x

2+6x+8,r(x)=15 (b) q(x)=x

3−5x

2+20x−100,r(x) = 504

4+x

3+x

2+x+1,r(x)=0

4. (a) q(x)=2x

2+x−1,r(x)=0 (b) q(x)=2x

3−5x

2+3x−39,r(x) = 147

6+x

5+x

4+x

3+x

2+x+1,r(x)=2

5. x0 1 −3 7

p(x)−4−3 101 5001

6. x1−1 3 −3 7 −7 21 −21

p(x)−24 −12 12 0 420 −168 10416 −7812

7. (a) q(x)=x

2+6x+13,r =20 (b) q(x)=x

2+3x−2,r =−4

8. (a) q(x)=x

4−x

3+x

2−x+1,r =−2(b) q(x)=x

4+x

3+x

2+x+1,r =0

9. Assume r=a/b a and bintegers with a>0:

(a) bdivides 1, b=±1; adivides 24, a=1,2,3,4,6,8,12,24;

the possible candidates are {±1,±2,±3,±4,±6,±8,±12,±24}

(b) bdivides 3 so b=±1,±3; adivides −10 so a=1,2,5,10;

the possible candidates are {±1,±2,±5,±10,±1/3,±2/3,±5/3,±10/3}

the possible candidates are {±1,±17}

10. An integer zero cdivides −21, so c=±1,±3,±7,±21 are the only possibilities; substitution of

these candidates shows that the integer zeros are −7,−1,3

11. (x+ 1)(x−1)(x−2) 12. (x+ 2)(3x+ 1)(x−2)

13. (x+3)

(x+1) 14. 2x4+x3−19x2+9

15. (x+ 3)(x+ 2)(x+1)

(x−3) 17. −3 is the only real root.

18. x=−3/2 is the only real root. 19. x=−2,−2/3 are the only real roots.

675 Appendix F

20. −2,−1,1/2,321. −2,2,3 are the only real roots.

23. If x−1 is a factor then p(1)=0,sok

2−7k+10=0,k

2−7k+10=(k−2)(k−5), so k=2,5.

24. (−3)7=−2187, so −3 is a root and thus by Theorem F.4(a), x+ 3 is a factor of x7+ 2187.

25. If the side of the cube is xthen x2(x−3) = 196; the only real root of this equation is x= 7 cm.

26. (a) Try to solve a

b>a

b3+ 1. The polynomial p(x)=x

3−x+ 1 has only one real root

c≈−1.325, and p(0) = 1 so p(x)>0 for all x>c; hence there is no positive rational

solution of a

b>a

b3+1.

(b) From part (a), any real x<cis a solution.

27. Use the Factor Theorem with xas the variable and yas the constant c.

(a) For any positive integer nthe polynomial xn−ynhas x=yas a root.

(b) For any positive even integer nthe polynomial xn−ynhas x=−yas a root.

Howard Anton Student Solutions Manual Set For Calculus Early Transcendentals Single Variable 8th E

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