calculus calculus foerster solutions to-textbook

C o n c e p t s a n d A p p l i c a t i o n s Second Edition

P a u l A . F o e r s t e r

Solutions Manual

Calculus

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© 2005 by Key Curriculum Press. All rights reserved.

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Printed in the United States of America

10 9 8 7 6 5 4 12 11 10 09 08

ISBN: 978-1-55953-657-8

Contents

Chapter 1 Limits, Derivatives, Integrals, and Integrals .................................................... 1

Chapter 2 Properties of Limits ................................................................................................... 9

Chapter 3 Derivatives, Antiderivatives, and Indefinite Integrals .............................. 28

Chapter 4 Products, Quotients, and Parametric Functions ......................................... 51

Chapter 5 Definite and Indefinite Integrals ....................................................................... 82

Chapter 6 The Calculus of Exponential and Logarithmic Functions ..................... 118

Chapter 7 The Calculus of Growth and Decay ............................................................ 139

Chapter 8 The Calculus of Plane and Solid Figures ................................................... 168

Chapter 9 Algebraic Calculus Techniques for the Elementary Functions .......... 213

Chapter 10 The Calculus of Motion—Averages, Extremes, and Vectors ............. 266

Chapter 11 The Calculus of Variable-Factor Products ................................................. 292

Chapter 12 The Calculus of Functions Defined by Power Series ............................. 313

iii

Overview

This Solutions Manual contains the answers to all problems in Calculus: Concepts andApplications. Solutions or key steps in the solutions are presented for all but the simplestproblems.

In most cases the solutions are presented in the form your students would be expected touse. For instance, decimal approximations are displayed as exact answers using ellipsisformat for a mathematical-world answer, then rounded to an appropriate number ofdecimal places with units of measurement applied for the corresponding real-worldanswer. An answer such as f(3) = 13.7569... ≈ 13.8 cm indicates that the precise answer,13.7569... , has been retained in memory in the student’s calculator without round-off forpossible use in subsequent computations. The ellipses indicate that the student chooses notto write all the digits on his or her paper.

Because the problems applying to the real world may be somewhat unfamiliar to both youand your students, fairly complete solutions are presented for these. Often commentary isincluded over and above what the student would be expected to write to further guide yourevaluation of students’ solutions, and in some cases reference is provided to later sectionsin which more sophisticated solutions appear. Later in the text, the details of computingdefinite integrals by the fundamental theorem are omitted because students are usuallyexpected to do these numerically. However, exact answers such as V = 8π/3 are presentedwhere possible in case you choose to have your students do the algebraic integration.

Solutions are not presented for journal entries because these are highly individual for eachstudent. The “prompts” in most problems calling for journal entries should be sufficient toguide students in making their own responses.

Where programs are called for, you may use as a model the programs in the Instructor’sResource Book. Check the publisher’s Web page (see the address on the copyright page ofthis manual) for further information on programs for specific models of the graphingcalculator.

If you or your students find any mistakes, please report them to Key Curriculum Press bysending in the Correction/Comment Form in the back of this book.

Paul A. Foerster

v

Calculus Solutions Manual Problem Set 1-2 1© 2005 Key Curriculum Press

Chapter 1—Limits, Derivatives, Integrals, and Integrals

Problem Set 1-11. a. 95 cm

b. From 5 to 5.1: average rate ≈ 26 34. cm/sFrom 5 to 5.01: average rate ≈ 27 12. cm/sFrom 5 to 5.001: average rate ≈ 27 20. cm/s So the instantaneous rate of change of d att = 5 is about 27.20 cm/s.

c. Instantaneous rate would involve divisionby zero.

d. For t = 1.5 to 1.501, rate ≈ −31.42 cm/s.The pendulum is approaching the wall: Therate of change is negative, so the distance isdecreasing.

e. The instantaneous rate of change is the limitof the average rates as the time intervalapproaches zero. It is called the derivative.

f. Before t = 0, the pendulum was not yetmoving. For large values of t, the pendulum’smotion will die out because of friction.

2. a. x = 5: y = 305, price is $3.05x = 10: y = 520, price is $5.20x = 20: y = 1280, price is $12.80

b. x = 5.1, rate ≈ 46 822. ¢/ftx = 5.01, rate ≈ 46.9820… ¢/ftx = 5.001, rate ≈ 46.9982… ¢/ft

c. 47 ¢/ft. It is called the derivative.

d. x = 10: 44 ¢/ft. x = 20: 128 ¢/ft

e. The 20-ft board costs more per foot than the10-ft board. The reason is that longer boardsrequire taller trees, which are harder to find.

Problem Set 1-2Q1. Power function, or polynomial function

Q2. f (2) = 8 Q3. Exponential function

Q4. g (2) = 9 Q5.

1

1x

h(x)

Q6. h (5) = 25 Q7. y = ax2 + bx + c, a ≠ 0

Q8. y = x Q9. y = |x|

Q10. Derivative

1. a. Increasing slowly b. Increasing fast

2. a. Increasing fast b. Decreasing slowly

3. a. Decreasing fast b. Decreasing slowly

4. a. Decreasing slowly b. Increasing slowly

5. a. Increasing fast b. Increasing slowly

c. Decreasing slowly d. Increasing fast

6. a. Decreasing fast b. Increasing slowly

c. Increasing fast d. Decreasing fast

7. a. Increasing slowly b. Increasing slowly

c. Increasing slowly

8. a. Decreasing fast b. Decreasing fast

c. Decreasing fast

9. a. Increasing fast b. Neither increasingnor decreasing

c. Increasing fast d. Increasing slowly

10. a. Decreasing slowly b. Decreasing fast

c. Decreasing fast d. Neither increasingnor decreasing

11. a.

100 200

50

100

T(x) (°C)

x (s)

x = 40: rate ≈ °1 1. /sx = 100: rate = 0°/sx = 140: rate ≈ − °0 8. /s

b. • Between 0 and 80 s the water is warmingup, but at a decreasing rate.

• Between 80 and 120 s the water is boiling,thus staying at a constant temperature.

• Beyond 120 s the water is cooling down,rapidly at first, then more slowly.

12. a.

1 2 3 4 5 6 7 8

10

20

30

40

50

60

70

v(x) (ft/s)

x (s)

2 Problem Set 1-2 Calculus Solutions Manual © 2005 Key Curriculum Press

x = 2: rate 18 (ft/s)/s x = 5: rate = 0 (ft/s)/s x = 6: rate 11 (ft/s)/s

b. Units are (ft/s)/s, sometimes written as ft/s2. The physical quantity is acceleration.

13. a.

3 4 7

2

18

h(x)

x

• Increasing at x = 3

• Decreasing at x = 7

b. h (3) = 17, h(3.1) = 17.19

Average rate =0.19

0.1= 1.9 ft/s

c. From 3 to 3.01:

average rate =0.0199

0.01= 1.99 ft/s

From 3 to 3.001:

average rate =0.001999

0.001= 1.99 ft/s

The limit appears to be 2 ft/s.

d. h (7) = 9, h(7.001) = 8.993999

Average rate =0.006001

0.001= 6.001 ft /s

The derivative at x = 7 appears to be 6 ft/s. The derivative is negative because h(x) is decreasing at x = 7.

14. a.

10

100

300

500f(t)

t

DecreaseIncrease

Not much

b. Enter y2 =y1 (x ) y1(1)

x 1

t r(t) = y2 (foxes/year)

0.97 110.5684…

0.98 109.7361…

0.99 108.9001…

1 undefined

1.01 107.2171…

1.02 106.3703…

1.03 105.5200…

c. Substituting 1 for t causes division by zero, so r(1) is undefined. Estimate: r approaches the average of r(0.99) and r(1.01), 108.0586… foxes/year. (Actual is 108.0604… .) The instantaneous rate is called the derivative.

d.

f (4.01) f (4)

0.01= 129.9697…

f (4) f (3.99)

0.01= 131.4833…

Instantaneous rate = ( 129.9697… 131.4833…)/2 = 130.7265… foxes/year (actual: 130.7287…) The answer is negative because the number of foxes is decreasing.

15. a. Average rate =a(2.1) a(2)

0.1=

52.9902… mm2/h

b. r (t) =200(1.2 t ) 200(1.22 )

t 2

2

20

40

60

r(t ) (mm2/hr)

t (mm)

r(2) is undefined.

c. r(2.01) = 52.556504… 52.556504… 52.508608… = 0.04789… Use the solver to find t when r(t) = 52.508608… + 0.01 = 52.518608… . t = 2.002088… , so keep t within 0.002 unit of 2.

16. a. v(x ) =4

3x 3 v(6) = 288

b. 6 to 6.1: average rate =

43 (6.13 63 )

0.1=

146.4133…

5.9 to 6: average rate =43 (63 5.93)

0.1=

141.6133…

Estimate of instantaneous rate is (146.4133… + 141.6133… )/2 = 144.0133… = 452.4312… cm3/cm.

c. r (x ) =43 x3 4

3 63

x 6

) (cm /cm)

6

r(x

144π�

48π�

3

x

r(6) is undefined.

Calculus Solutions Manual Problem Set 1-3 3 © 2005 Key Curriculum Press

d. r(6.1) = 146.4133… = 459.9710… r(6.1) is 7.5817… units from the derivative. Use the solver feature to find x if r(x) = 144 + 0.1. x = 6.001326… , so keep x within 0.00132… unit of 6.

17. a. i. 1.0 in./s ii. 0.0 in./s iii. 1.15 in./s

b. 1.7 s, because y = 0 at that time

18. a. i. 0.395 in./min ii. 0.14 in./min

iii. 0.105 in./min

b. The rate is negative, because y is decreasing as the tire goes down.

19. a. Quadratic (or polynomial)

b. f(3) = 30

c. Increasing at about 11.0 (2.99 to 3.01)

20. a. Quadratic (or polynomial)

b. f(1) = 12

c. Increasing at about 6.0 (0.99 to 1.01)

21. a. Exponential

b. Increasing, because the rate of change from 1.99 to 2.01 is positive.

22. a. Exponential


23. a. Rational algebraic

b. Decreasing, because the rate of change from 3.99 to 4.01 is negative.

24. a. Rational algebraic


25. a. Linear (or polynomial)


26. a. Linear (or polynomial)


27. a. Circular (or trigonometric)


28. a. Circular (or trigonometric)


29. • Physical meaning of a derivative: instantaneous rate of change

• To estimate a derivative graphically: Draw a tangent line at the point on the graph and measure its slope.

• To estimate a derivative numerically: Take a small change in x, find the corresponding

change in f(x), then divide. Repeat, using a smaller change in x. See what number these average rates approach as the change in x approaches zero.

• The numerical method illustrates the fact that the derivative is a limit.

30. Problems 13 and 14 involve estimating the value of a limit.

Problem Set 1-3 Q1. 72 ft2 Q2. y = cos x

Q3. y = 2x Q4. y = 1/x

Q5. y = x2 Q6. f(5) = 4

Q7. Q8.

x

y

x

y

Q9. Q10. x = 3

x

y

1. f(x) = 0.1x2 + 7 2. f(x) = 0.2x2 + 8

a. Approximately 30.8 a. Approximately 22.2

b. Approximately 41.8 b. Approximately 47.1

x

f (x)7

–1 5 6

f(x)

x

–2 3 5

8

3. h(x) = sin x 4. g( x) = 2 x+ 5

a. Approximately 2.0 a. Approximately 7.9

b. Approximately 1.0 b. Approximately 12.2

1

3

x

h(x)

–1 1 2

6

x

g(x)

4 Problem Set 1-4 Calculus Solutions Manual© 2005 Key Curriculum Press

5. There are approximately 6.8 squares between thecurve and the x-axis. Each square represents(5)(20) = 100 feet. So the distance is about(6.8)(100) = 680 feet.

6. There are approximately 53.3 squares between thecurve and the x-axis. Each square represents(0.5)(10) = 5 miles. So the distance is about(53.3)(5) = 266.5 miles.

7. Derivative

≈ =tan . – tan .

. – ..

1 01 0 99

1 01 0 993 42K

8. Derivative = −7 (exactly, because that is theslope of the linear function)

9. a.

100

5 108.7

t

v(t)

60

The range is 0 ≤ y ≤ 32.5660… .

b. Using the solver, x = 8.6967… ≈ 8.7 s.

c. By counting squares, distance ≈ 150 ft.The concept used is the definite integral.

d. Rate ≈ −−

=v v( . ) ( . )

. ..

5 01 4 99

5 01 4 993 1107K

About 3.1 (ft/s)/sThe concept is the derivative.The rate of change of velocity is calledacceleration.

10. a.

2

10

5

1 3 4 5

t

v(t)

b. v(4) = 9.3203… ≈ 9.3 ft/s

Domain: 0 ≤ t ≤ 4Range: 0 ≤ v(t) ≤ 9.3203…

c. By counting squares, the integral from t = 0to t = 4 is about 21.3 ft. The units of theintegral are (ft/s) · s = ft. The integral tellsthe length of the slide.

d. Rate ≈

−=v v( . ) – ( . )

. ..

3 01 2 99

3 01 2 991 8648K

About 1.86 (ft/s)/sThe derivative represents the acceleration.

11. From t = 0 to t = 5, the object travels about11.4 cm. From t = 5 to t = 9, the object travelsback about 4.3 cm. So the object is located about11.4 − 4.3 = 7.1 cm from its starting point.

12. See the text for the meaning of derivative.

13. See the text for the meaning of definite integral.

14. See the text for the meaning of limit.

Problem Set 1-4Q1. y changes at 30 Q2. Derivative ≈ −500

Q3. Q4. f (3) = 9

x

y

Q5. 100 Q6. sin (π/2) = 1

Q7. 366 days Q8. Derivative

Q9. Definite integral Q10. f (x) = 0 at x = 4

1. a.

30

20,000

t

v(t)

b. Integral ≈ + + + +5 0 5 0 5 10 15( . ( ) ( ) ( ) ( )v v v vv(20) + v(25) + 0.5v(30)) = 5(56269.45…) =281347.26… ≈ 281 000, ftThe sum overestimates the integral becausethe trapezoids are circumscribed about theregion and thus include more area.

c. The units are (ft/s)(s), which equals feet, sothe integral represents the distance thespaceship has traveled.

d. Yes, it will be going fast enough, becausev(30) = 27,919.04… , which is greater than27,000.

2. a. v(t) = 4 + sin 1.4t

5

t

3

v(t)


b. A definite integral has the units of thex-variable times the y-variable. Distance =rate × time. Because v(t) is distance/time and t is time, their product is expressed inunits of distance.

c. See graph in part a.Distance ≈ + + +0 5 0 5 0 0 5 1. ( . ( ) ( . ) ( )v v vv(1.5) + v(2) + v(2.5) + 0.5v(3)) =0.5(26.041…) = 13.02064… ≈ 13.0 ft

d. v(3) = 3.128… ≈ 3.1 mi/hMaximum speed was 5 mi/h at about 1.12 h.

3. Distance ≈ + + + + + =0 6 150 230 150 90 40 0. ( )396 ft

4. Volume ≈ 3(2500 + 8000 + 12000 + 13000 +11000 + 7000 + 4000 + 6000 + 4500) =204,000 ft3

5. Programs will vary depending on calculator. Seethe program TRAPRULE in the Instructor’sResource Book for an example. The programgives T20 = 23.819625.

6. See the program TRAPDATA in the Instructor’sResource Book for an example. The programgives T7 = 33, as in Example 2.

7. a.

1 4

7

f(x)

x

b. T10 = 18.8955T20 = 18.898875T50 = 18.89982These values underestimate the integral,because the trapezoids are inscribed in theregion.

c. T10: 0.0045 unit from the exact answerT20: 0.001125 unit from the exact answerT50: 0.00018 unit from the exact answerTn is first within 0.01 unit of 18.9 whenn = 7.T7 = 18.8908… , which is 0.0091… unitfrom 18.9.Because Tn is getting closer to 18.9 as nincreases, Tn is within 0.01 unit of 18.9 forall n ≥ 7.

8. a.

1 3

1

g(x)

x

b. T10 = 8.6700…T20 = 8.6596…Τ50 = 8.65672475…These values overestimate the integral,because the trapezoids are circumscribedabout the region.

c. T10: 0.01385… unit from answerT20: 0.003465… unit from answerT50: 0.0005545… unit from answerTn is first within 0.01 of 8.65617024… whenn = 12.T12 = 8.665795… , which is 0.009624… unitfrom 8.65617024… .Because Tn is getting closer to the exactanswer as n increases, Tn is within 0.01 unitof the answer for all n ≥ 12.

9. From the given equation,

y x= ±( / ) – .40 110 1102 2 Using the trapezoidal

rule program on the positive branch with n = 100increments gives 6904.190… for the top half ofthe ellipse. Doubling this gives an area of13,808.38… cm2. The estimate is too lowbecause the trapezoids are inscribed within theellipse. The area of an ellipse is πab, where aand b are the x- and y-radii, respectively. Sothe exact area is π (110)(40) = 4400π =13,823.007… cm2, which agrees both with theanswer and with the conclusion that thetrapezoidal rule underestimates the area.

10. Integral = 1(0.0 + 2.1 + 7.9 + 15.9 + 23.8 +29.7 + 31.8 + 29.7 + 23.8 + 15.9 + 7.9 +2.1 + 0) = 190.6The integral will have the units (in.2)(in.) = in.3,representing the volume of the football.

11. n = 10: integral ≈ 21.045n = 100: integral ≈ 21.00045n = 1000: integral ≈ 21.0000045Conjecture: integral = 21The word is limit.

12. The trapezoidal rule with n = 100 givesintegral ≈ 156.0096.Conjecture: integral = 156

13. If the trapezoids are inscribed (graph concavedown), the rule underestimates the integral.If the trapezoids are circumscribed (graph concaveup), the rule overestimates the integral.

Concave downInscribed trapezoids

Underestimates integral

Concave upCircumscribed trapezoids

Overestimates integral


Problem Set 1-51. Answers will vary.

Problem Set 1-6

Review Problems

R1. a. When t = 4, d = 90 − 80 sin [1, 2(4 − 3)] ≈15.4 ft.

b. From 3.9 to 4: average rate ≈ –40.1 ft/sFrom 4 to 4.1: average rate ≈ −29 3. ft/sInstantaneous rate ≈ −34.7 ft/sThe distance from water is decreasing, so he isgoing down.

c. Instantaneous rate ≈ − ≈d d( . ) ( . )

..

5 01 4 99

0 0270 8

d. Going up at about 70.8 ft/s

e. Derivative

R2. a. Physical meaning: instantaneous rate ofchange of a functionGraphical meaning: slope of a tangent line toa function at a given point

b. x = − 4: decreasing fastx = 1: increasing slowlyx = 3: increasing fastx = 5: neither increasing nor decreasing

c. From 2 to 2.1:

average rate = − =5 5

0 143 6547

2 1 2.

.. K

From 2 to 2.01:


0 0140

2 01 2.

..5614K

From 2 to 2.001:


0 00140 2683

2 001 2.

.. K

Differences between average rates andinstantaneous rates, respectively:43.6547… − 40.235947… = 3.4187…40.5617… − 40.235947… = 0.3255…40.2683… − 40.235947… = 0.03239…The average rates are approaching theinstantaneous rate as x approaches 2.The concept is the derivative.The concept used is the limit.

d. t = 2: 3.25 m/st = 18: 8.75 m/st = 24: 11.5 m/sHer velocity stays constant, 7 m/s, from 6 sto 16 s. At t = 24, Mary is in her final sprinttoward the finish line.

R3. By counting squares, the integral isapproximately 23.2.Distance ≈ 23 2. ft (exact answer: 23.2422…)Concept: definite integral

R4. a.

1 4

x

5

f (x)

The graph agrees with Figure 1-6c.

b. By counting squares, integral ≈ 15.0.(Exact answer is 15.)

c. T6 = 0.5(2.65 + 5.575 + 5.6 + 5.375 + 4.9 +4.175 + 1.6) = 14.9375The trapezoidal sum underestimates theintegral because the trapezoids are inscribed inthe region.

d. T50 = 14.9991; Difference = 0.0009T100 = 14.999775; Difference = 0.000225The trapezoidal sums are getting closer to 15.Concept: limit

R5. Answers will vary.

Concept Problems

C1. a. f (3) = 32 − 7.3 + 11 − 1

b. f (x) − f (3) = x2 − 7x + 11 + 1 = x2 − 7x + 12

c. f x f

x

x x

x

x x

x

( ) – ( )

–

–

–

( – )( – )

–

3

3

7 12

3

4 3

3

2

= + = =

x − 4, if x ≠ 3

d. The limit is found by substituting 3 for xin (x − 4).Limit = exact rate = 3 − 4 = −1

C2. The line through (3, f (3)) with slope −1 isy = −x + 2.

3

2 x

f(x)

The line is tangent to the graph. Zooming in bya factor of 10 on the point (3, 2) shows that thegraph becomes straighter and looks almost likethe tangent line. (Soon students will learn thatthis property is called local linearity.)


2

3

C3. a. f xx x

x

x x

x( )

( )( )= − +−

= − −−

=4 19 21

3

4 7 3

3

2

4x − 7, x ≠ 3When x = 3, 4x − 7 = 4 ⋅ 3 − 7 = 5.

b.

1 2 3 4 5 6

1

2

3

4

5

6f(x) (ft)

x (s)

5.8

4.2

2.8 3.2

c. 5.8 = 4(3 + δ ) − 7 4.2 = 4(3 − δ ) − 75.8 = 12 + 4δ − 7 4.2 = 12 − 4δ − 74δ = 0.8 −4δ = −0.8δ = 0.2 δ = 0.2

d. 4(3 + δ ) − 7 = 5 + ε12 + 4δ − 7 = 5 + ε

4δ = εδ ε= 1

4

There is a positive value of δ, namely 14 ε, for

each positive value of ε, no matter how smallε is.

e. L = 5, c = 3. “. . . but not equal to 3” isneeded so that you can cancel the (x − 3)factors without dividing by zero.

Chapter Test

T1. Limit, derivative, definite integral, indefiniteintegral

T2. See the text for the definition of limit.

T3. Physical meaning: instantaneous rate

T4.

2 5

3

6

x

y

T5. Concept: definite integralBy counting squares, distance ≈ 466.(Exact answer is 466.3496… .)

T6.

5 10 15 20 25 30 35 40

5

10

15

20

25

Speed (ft/s)

Time (s)

T7 = 5(2.5 + 5 + 5 + 10 + 20 + 25 + 20 + 5) =462.5Trapezoidal rule probably underestimates theintegral, but some trapezoids are inscribed andsome circumscribed.

T7. Concept: derivative

5 10 15 20 25 30 35 40

5

10

15

20

25

Speed (ft/s)

Time (s)

Slope ≈ −1.8 (ft/s)/s(Exact answer is −1.8137… .)Name: acceleration

T8. The roller coaster is at the bottom of the hill at25 s because that’s where it is going the fastest.The graph is horizontal between 0 and 10 secondsbecause the velocity stays constant, 5 ft/s, as theroller coaster climbs the ramp.

T9. Distance = (rate)(time) = 5(10) = 50 ft

T10. T5 = 412.5; T50 = 416.3118… ;T100 = 416.340219…

T11. The differences between the trapezoidal sum andthe exact sum are:For T5: difference = 3.8496…For T50: difference = 0.03779…For T100: difference = 0.009447…The differences are getting smaller, so Tn isgetting closer to 416.349667… .


T12. From 30 to 31:

average rate = − = −y y( ) ( )

.31 30

11 9098K

From 30 to 30.1:

average rate = − = −y y( . ) ( )

..

30 1 30

0 11 8246K

From 30 to 30.01:

average rate

= − = −y y( . ) ( )

..

30 01 30

0 011 8148K

T13. The rates are negative because the roller coaster isslowing down.

T14. The differences between the average rates andinstantaneous rate are:For 30 to 31: difference = 0.096030…For 30 to 31.1: difference = 0.010833…For 30 to 30.01: difference = 0.001095…

The differences are getting smaller, so the averagerates are getting closer to the instantaneous rate.

T15. Solve . gettingy x y

x

( ) ( ),

−−

= − +30

301 81379936 1

x = 30.092220… . So keep x within 0.092…unit of 30, on the positive side.

T16. Concept: derivative

T17. ′ ≈ −−

= − =ff f

( )( . ) ( . )

. . .4

4 3 3 7

4 3 3 7

35 29

0 610

T18. Answers will vary.


Chapter 2—Properties of Limits

Problem Set 2-1

1. a. f ( )28 10 2

2 2

0

0= − +

−=

No value for f (2) because of division by zero.

b.

x f (x)

1.997 2.994

1.998 2.996

1.999 2.998

2 undefined

2.001 3.002

2.002 3.004

2.003 3.006

Yes, f (x) stays close to 3 when x is keptclose to 2, but not equal to 2.

c. To keep f (x) within 0.0001 unit of 3, keepx within 0.00005 unit of 2. To keep f (x)within 0.00001 unit of 3, keep x within0.000005 unit of 2. To keep f (x) arbitrarilyclose to 3, keep x within 1

2 that distanceof 2.

d. The discontinuity can be “removed” bydefining f (2) to equal 3.

2.

3

2

3

x

g(x)

3

2

g(x)

x

The limit seems to be 2.

3.

3

2

x

h(x)

2

32.7

h(x)

x

There appears to be no limit, because the graphcycles infinitely as it approaches x = 3.

Problem Set 2-2Q1. Q2.

3

8

x

y

π

–1

x

y

Q3. Q4.

6

4

x

y

2–2

–4

x

y

Q5. Q6. Trapezoidal rule

1

4

x

y

Q7. Counting squares

Q8. Slope of the tangent line

Q9. Instantaneous rate of change

Q10. B

1. See the text for the definition of limit.

2. f (x) might be undefined at x = c, or might have avalue at x = c that is different from the limit.

3. Has a limit, 3 4. Has a limit, 2


7. Has no limit 8. Has no limit


11. Has no limit 12. Has no limit

13. lim ( ) .x

f x→

=3

5 For ε = 0.5, δ ≈ 0.2 or 0.3.


14. lim ( ) .x

f x→

=2

3 For ε = 0.5, δ ≈ 0.8.

15. lim ( ) .x

f x→

=6

4 For ε = 0.7, δ ≈ 0.5 or 0.6.

(The right side is more restrictive.)

16. lim ( ) .x

f x→

=4

2 For ε = 0.8, δ ≈ 0.7 or 0.8.

(The left side is more restrictive.)

17. lim ( ) .x

f x→

=5

2 For ε = 0.3, δ ≈ 0.5 or 0.6.

(The right side is more restrictive.)

18. lim ( ) .x

f x→

=3

6 For ε = 0.4, δ ≈ 0.1.

19. a. The graph should match Problem 13.

b. lim ( )x

f x→

=3

5

c. Graph is symmetrical about x = 3.Let 5 − 2 sin (x − 3) = 5 + 0.5 = 5.5.∴ sin (x − 3) = −0.25x = 3 + sin− 1 (−0.25)Max. δ = 3 − [3 + sin− 1 (−0.25)] = 0.25268…

d. Let 5 − 2 sin (x − 3) = 5 + ε.∴ sin (x − 3) = −ε/2x = 3 + sin− 1 (−ε/2)Max. δ = 3 − [3 + sin− 1 (−ε/2)] =−sin− 1 (−ε/2) = sin− 1 (ε/2), which is positivefor any positive value of ε.


b. lim ( )x

f x→

=2

3

c. The graph is symmetrical about x = 2.Let (x − 2)3 + 3 = 3 + 0.5 = 3.5.

∴ = + .x 2 0 53

Max. . . .δ = + − = =2 0 5 2 0 5 0 79373 3 K

d. Let (x − 2)3 + 3 = 3 + ε.∴ x = 2 + ε1/3

Max. δ = 2 + ε1/3 − 2 = ε1/3, which ispositive for any positive value of ε.


b. lim ( )x

f x→

=6

4

c. The right side is more restrictive.Let 1 + 3(7 − x)1/3 = 4 − 0.7 = 3.3.∴ x = 7 − (2.3/3)3

Max. δ = [7 − (2.3/3)3] − 6 = 0.5493…

d. Because the right side is more restrictive, set1 3 7 43+ − = − x ε.∴ x = 7 − [(3 − ε)/3]3

Max. δ = 7 − [(3 − ε)/3)3] − 6 = 1 − [(3 − ε)/3]3,which is positive for all positive values of ε.


b. lim ( )x

f x→

=4

2

c. The left side is more restrictive.Let 1 + 24− x = 2 + 0.8 = 2.8.

∴ 24− x = 1.8

x = −41 8

2

log .

log

Max .

log .

log.δ = − −

=4 41 8

20 84799K

d. Because the left side is more restrictive, set1 + 24− x = 2 + ε.∴ 24− x = 1 + ε

x = − +4

1

2

log( )

log

ε

Max. log(1 ) log(1 )δ ε ε= − − +

= +

4 42 2log log

,

which is positive for all ε > 0.


b. lim ( )x

f x→

=5

2

c. The right side is more restrictive.Let (x − 5)2 + 2 = 2 + 0.3 = 2.3.

∴ = + .x 5 0 3

Max = ( . ) .. δ 5 0 3 5 0 54772+ − = K

d. Because the right side is more restrictive, set(x − 5)2 + 2 = 2 + ε.∴ = + x 5 εMax. ( ) ,δ ε ε= + − =5 5 which ispositive for all ε > 0.


b. lim ( )x

f x→

=3

6

c. The graph is symmetrical about x = 3.Let 6 − 2(x − 3)2/3 = 6 − 0.4 = 5.6.∴ x = 3 + 0.23/2

Max. δ = (3 + 0.23/2) − 3 = 0.08944…

d. Let 6 − 2(x − 3)2/3 = 6 − ε.∴ x = 3 + (ε/2)3/2

Max. δ = [3 + (ε/2)3/2] − 3 = (ε/2)3/2, whichis positive for all ε > 0.

25. a. f ( )( )( ) ( )( )

25 6 5 13 5 2

5 2

5 0

0

0

0

2

= − ⋅ + −−

= =

The graph has a removable discontinuity atx = 2.Limit = 22 − 6(2) + 13 = 5

b. When f (x) = 5.1, x = 1.951191… .δ1 = 2 − 1.951191… = 0.048808…When f (x) = 4.9, x = 2.051316… .δ2 = 2.051316… − 2 = 0.051316…∴ max. δ = 0.048808…


c.f(x)

x

c = 2

L = 5

δδ

26. a.

2

8

x

y

The graph is linear.There is a removable discontinuity at x = 2.The limit appears to be 9.

b. f ( )( ) ( )

–2

4 2 7 2 2

2 2

0

0

2

= − − =

Indeterminate form

c. f xx x

xx x( )

( )( ),= + −

−= + ≠4 1 2

24 1 2

Limit = 4(2) + 1 = 9If x ≠ 2, then (x − 2) ≠ 0. Canceling is adivision process, but because (x − 2) ≠ 0,you do not risk dividing by zero.

d. If f (x) = 9.001, x = 2.00025.If f (x) = 8.999, x = 1.99975.δ1 = 2.00025 − 2 = 0.00025δ2 = 2 − 1.99975 = 0.00025Largest number is 0.00025.

e. L = 9, c = 2, ε = 0.001, δ = 0.00025

27. a. m td t d

t

t

t( )

( ) ( ) –= −−

=−

4

4

3 48

4

2

b. Removable discontinuity at x = 4.

30

4

m(t)

t

c. Limit = 24 ft/s

d. m tt t

tt t( )

( )( )

–,= − + = + ≠3 4 4

43 12 4 if

3t + 12 = 24.12 ⇒ t = 4.043t + 12 = 23.88 ⇒ t = 3.96Keep t within 0.04 s of 4 s.

e. The limit of the average velocity is theinstantaneous velocity.

Problem Set 2-3Q1. 13

Q2. Q3.

2

3

y

x

–4

y

x

Q4. Q5.

x

y

x

y

1

1

Q6. (x − 10)(x + 10) Q7. 75%Q8. Product of x and y, where x varies and y may

varyQ9.

3 1 –8 22 –21

3 –15 21

1 –5 7 0

x2 − 5x + 7

Q10. D

1.

10

2

x

y

g h

g + h

lim ( ) , lim ( ) , lim ( )x x x

f x g x h x→ → →

= = =2 2 2

10 4 6 and

∴ = +→ → →

lim ( ) lim ( ) lim ( ),x x x

f x g x h x2 2 2

Q.E.D.

x f (x)

1.96 9.9640…

1.97 9.9722…

1.98 9.9810…

1.99 9.9902…

2.00 10

2.01 10.0102…

2.02 10.0209…

2.03 10.0322…

2.04 10.0439…

All these f (x) values are close to 10.


2.

9

1.8 f

g

3

x

y

lim ( ) . lim ( )x x

f x g x→ →

= =3 3

1 8 9 and

∴ =→ →

lim ( ) . lim ( ),x x

f x g x3 3

0 2 Q.E.D.

x f (x)

2.96 1.75232

2.97 1.76418

2.98 1.77608

2.99 1.78802

3.00 1.8

3.01 1.81202

3.02 1.82408

3.03 1.83618

3.04 1.84832

All these f (x) values are close to 1.8.

3.

3

7Limit = 7

x

f(x)

The limit is 7 because f (x) is always close to 7,no matter what value x takes on. (It shouldn’tbother you that f (x) = 7 for x ≠ 3 if you think ofthe definition of limit for a while.)

4.

x

f(x) = x

6

Limit = 6

lim ( ) .x

f x→

=6

6 The y-value equals the x-value.

5.

1

5y

1

y1

y2 y

2

y

x

lim , lim . , limx x x

y y y y→ → →

= = ⋅ =1

11

21

1 22 1 5 3 and

2 1 5 31

11

21

1 2( . ) , lim lim lim= ∴ ⋅ = ⋅→ → →

x x x

y y y y

x y3 = f (x)

0.997 2.9739…

0.998 2.9825…

0.999 2.9912…

1 3

1.001 3.0087…

1.002 3.0174…

1.003 3.0262…

All these f (x) values are close to 2(1.5) = 3.

6. 2 83

3 60 53 =

= and sin

..

π

r( ).

38

0 516= =

x r (x)

2.9997 15.9894…

2.9998 15.9929…

2.9999 15.9964…

3 16

3.0001 16.0035…

3.0002 16.0070…

3.0003 16.0105…

All these r (x) values are close to 16.

lim ( ).

.

xf x

→→

3 6

3 62

0 , so the limit of a quotient

cannot be applied because of division by zero.

7. lim ( ) limx x

f x x x→ →

= − +3 3

2 9 5

= − +→ → →

lim lim limx x x

x x3

2

3 39 5 Limit of a sum

(or difference)

= ⋅ − +→ → →

lim lim limx x x

x x x3 3 3

9 5

Limit of a product,limit of a constant

= (3)(3) − 9(3) + 5 Limit of x= 9 − 27 + 5 = −13

8. lim ( ) limx x

f x x x→− →−

= + −1 1

2 3 6

= lim lim limx x x

x x→− →− →−

+ −1

2

1 13 6

Limit of a sum= lim lim lim

x x xx x x

→− →− →−⋅ + −

1 1 13 6


= (−1)(−1) + 3(−1) − 6 Limit of x= 1 − 3 − 6 = −8


9.

–8

–2 x

r(x)

r(– ) =22 4 2 12

2 2

4 8 12

0

0

0

2( ) ( )

( )

− − − −− +

= + − =

r xx x

xx x( )

( – )( ),= +

+= − ≠ −6 2

26 2

lim ( )x

r x→−

= − − = −2

2 6 8

Proof:

lim ( ) lim ( )x x

r x x→− →−

= −2 2

6 Because x ≠ −2

= +→− →−lim lim

2 2x xx (– )6 Limit of a sum

= −2 − 6 = −8, Q.E.D. Limit of x, limit of aconstant

10.13

5

x

f(x)

f ( )( )

55 3 5 40

5 5

25 15 40

0

0

0

2

= + −−

= + − =

f xx x

xx x( )

( )( – )

–,= + = + ≠8 5

58 5

lim ( )x

f x→

= + =5

5 8 13

Proof:

lim ( ) lim ( )x x

f x x→ →

= +5 5

8 Because x ≠ 5

= lim limx x

x→ →

+5 5

8 Limit of a sum

= 5 + 8 = 13, Q.E.D. Limit of x, limit of aconstant

11.41

10

5

x

f(x)

f ( )( ) ( )− = − − −

−

= − − − =

55 3 5 4 5 30

5 5125 75 20 30

0

0

0

3 2

f xx x x

xx x x( )

( )( – )

–,= + + = + + ≠

222 6 5

52 6 5

lim ( ) ( )x

f x→

= + + =5

25 2 5 6 41

Proof:

lim ( ) lim ( )x x

f x x x→ →

= + +5 5

2 2 6 Because x ≠ 5

= lim lim ( ) limx x x

x x→ → →

+ +5

2


= lim lim limx x x

x x x→ → →

⋅ + +5 5 5

2 6 Limit of a product,limit of a constanttimes a function,limit of a constant

= 5 ⋅ 5 + 2 · 5 + 6 = 41, Q.E.D.Limit of x

12.

28

3

x

f(x)

f ( )( )

33 3 5 3 21

3 3

27 9 15 21

0

0

0

3 2

= + − −−

= + − − =

f xx x x

xx x x( )

( )( – )

–,= + + = + + ≠

224 7 3

34 7 3

lim ( ) ( )x

f x→

= + + =3

23 4 3 7 28

Proof:

lim ( ) lim )x x

f x x x→ →

= + +3 3

2 4 7 (

Because x ≠ 3= + +

→ → →lim lim limx x x

x x3

2


= ⋅ + +→ → →

lim lim limx x x

x x x3 3 3

4 7 Limit of a product,limit of a constanttimes a function,limit of a constant

= 3 ⋅ 3 + 4 ⋅ 3 + 7 = 28, Q.E.D.Limit of x

13.

9

f(x)

x

–1

f ( )( ) ( ) ( )

( )− = − − − − − +

− +1

1 4 1 2 1 3

1 1

3 2

= − − + + =1 4 2 3

0

0

0


f xx x x

xx x x( )

( – )( ),= + +

+= − + ≠ −

225 3 1

15 3 1

lim ( ) ( ) ( )x

f x→−

= − − − + =1

21 5 1 3 9

Proof:

lim ( ) lim ( )x x

f x x x→− →−

= − +1 1

2 5 3

Because x ≠ −1

= + +→ → →lim lim (– ) lim

– – –x x xx x

1

2

1 15 3

Limit of a sum= ⋅ +

→ → →lim lim – lim

– – –x x xx x x

1 1 15 3

Limit of a product,limit of a constanttimes a function,limit of a constant

= (−1)(−1) + (−5)(–1) + 3 = 9, Q.E.D.Limit of x

14.

–17

2 xf(x)

f ( )( ) ( )

22 11 2 21 2 2 10

2 216 88 84 2 10

0

0

0

4 3 2

= − + − −−

= − + − − =

f xx x x x

x( )

( – )( – )

–= + +3 29 3 5 2

2

= − + + ≠x x x x3 29 3 5 2,

limx

f x→

= − + + = −2

3 22 9 2 3 2 5 17( ) ( ) ( )

Proof:

lim ( ) lim ( – )x x

f x x x x→ →

= + +2 2

3 29 3 5

Because x ≠ 2= + − + +

→ → → →lim lim ( ) lim limx x x x

x x x2

3

2

2

2 29 3 5

Limit of a sum= ⋅ ⋅ + − ⋅

→ → → → →lim lim lim ( ) lim limx x x x x

x x x x x2 2 2 2 2

9

+ +→

3 52

limx

x


= 2 ⋅ 2 ⋅ 2 + (−9)(2 ⋅ 2) + 3 ⋅ 2 + 5 = −17,Q.E.D. Limit of x

15.

x f (x)

4.990 40.8801

4.991 40.8921…

4.992 40.9040…

4.993 40.9160…

4.994 40.9280…

4.995 40.9400…

4.996 40.9520…

4.997 40.9640…

4.998 40.9760…

4.999 40.9880…

5 undefined

5.001 41.0120…

5.002 41.0240…

5.003 41.0360…

5.004 41.0480…

5.005 41.0600…

5.006 41.0720…

5.007 41.0840…

5.008 41.0960…

5.009 41.1080…

The table shows that f (x) will be within 0.1 unitof lim ( )

xf x

→=

541 if we keep x within 0.008 unit

of 5.

16.f (x)

x

9

–1

When x is close to –1, f (x) is close to 9.

17. f xx x

x x

x x

x x

x

x( )

( )( )

( )( )= − +

− += − −

− −= −

−

2

2

5 6

6 9

2 3

3 3

2

3You cannot find the limit by substituting intothe simplified form because the denominator stillbecomes zero.

18. f xx

x x

x x x

x x

x x

x

( )( )( )

( )( )= −

− += − + +

− −

= + +−

3

2

2

2

8

4 4

2 2 4

2 2

2 4

2

You cannot find the limit by substituting intothe simplified form because the denominator stillgoes to zero.


19. a. 5(0)1/2 = 0 = v(0)5(1)1/2 = 5 = v(1)5(4)1/2 = 10 = v(4)5(9)1/2 = 15 = v(9)5(16)1/2 = 20 = v(16)

b. a

v v( ) .9

9 001 9

9 001 90 8333101≈ =( . ) – ( )

. –K

Conjecture: a( ) . /9 0 83 5 6= =Units of a(t): (mi/h)/s

c. av t v

t

t

tt t( ) lim

( ) – ( )

–lim

–

–

/

99

9

5 15

99 9

1 2

= =→ →

=+

=+

→

→

lim( – )

( – )( )

lim

/

/ /

/

t

t

t

t t

t

9

1 2

1 2 1 2

9 1 2

5 3

3 3

5

3

= 5

6, which agrees with the conjecture.

d. Distance = integral of v(t) from 1 to 9. By thetrapezoidal rule with n = 100 increments,integral ≈ 86.6657… . The units are(mi/h) · s. To convert to ft, multiply by 5280and divide by 3600, getting 127.1111…(exact: 127 1

9 ) . The truck went about 127 ft.

20. a. Derivative ≈ =2 1 2

2 1 212 61

3 3. –

. –.

b.x

x

x x x

x

3 28

2

2 2 4

2

–

–

( – )( )

–= + + =

x x2 2 4+ + , provided x ≠ 2. This expressionapproaches 12 as x approaches 2.

Proof:

lim–

–lim ( )

x x

x

xx x

→ →= + +

2

3

2

28

22 4

Because x ≠ 2= + +


x x2

2

2 22 4

Limit of a sum= ⋅ + +


x x x2 2 2

2 4


= 2 · 2 + 2 · 2 + 4 = 12, Q.E.D.Limit of x

c. The line through point (2, 8) with slope 12 isy = 12x − 16. The line appears to be tangentto the graph of f at point (2, 8).

12

1

2

x

f(x)

8

21. By the symmetric difference quotient,

derivative ≈ = −0 7 0 7

2 0 010 05994

5 01 4 99. – .

( . ). .

. .

K

22. By the trapezoidal rule with n = 100,integral ≈ 11 8235. K .

23. Prove that limx c

n nx c→

= for any positive integer n.

Proof:

Anchor:

If n = 1, limx c

x c c→

=1 1= by the limit of x.

Induction Hypothesis:

Assume that the property is true for n = k.∴ =

→limx c

k kx c

Verification for n = k + 1:lim lim ( )x c

k

x c

kx x x→

+

→= ⋅1

= ⋅ = ⋅→ →

lim limx c

k

x c

kx x c c By the inductionhypothesis

= +ck 1

Conclusion:

∴ =→

limx c

n nx c for all integers n ≥ 1, Q.E.D.

24. Answers will vary.

Problem Set 2-4Q1. Instantaneous rate of change

Q2. Product of x and y, where x varies and ycan vary

Q3. 0.0005

Q4.

Q5. Exponential function

Q6.y = cos x

x

Q7. (x + 6)(x − 1) Q8. 53

Q9. 120 Q10. 103

1. a. Has left and right limits

b. Has no limit

c. Discontinuous. Has no limit



b. Has a limit

c. Discontinuous. No f (3)


b. Has a limit

c. Continuous


b. Has a limit

c. Continuous

5. a. Has no left or right limit

b. Has no limit

c. Discontinuous. No limit or f (2)


b. Has a limit

c. Continuous (Note that the x-value 5 is not atthe discontinuity.)


b. Has a limit

c. Discontinuous. f (1) ≠ limit


b. Has no limit

c. Discontinuous. No limit


b. Has a limit

c. Discontinuous. No f (c)


b. Has no limit

c. Discontinuous. No limit, no f (c)

11. Answers may vary. 12. Answers may vary.

3

x

f(x)

4

f(x)

x


5

x

f(x)

–2

x

f(x)

f(–2)


x

f(x)

6

2

x

f(x)


10

f(x)

x

–2

–2

5 x

f(x)


x

f(x)

6

4

1

x

f(x)

5

3

21. Discontinuous at x = −3

22. Discontinuous at x = 11

23. Discontinuous at x = π/2 + π n, where n is aninteger

24. Nowhere discontinuous

25.

x

f(x)

2

1

2

3

Discontinuous because limx

f x→

=2

2( ) and f (2) = 3

26.

2

1

2

x

g(x)

Discontinuous because g(x) has no limit as xapproaches 2


27.

2

3

x

s(x)

Discontinuous because s(x) has no limit as xapproaches 2 from the left (no real functionvalues to the left of x = 2)

28.

2

x

p(x)

1

Discontinuous because p(x) has no limit as xapproaches 2

29.

2

1

x

h(x)

Discontinuous because there is no value of h(2)

30.

2

3

x

f(x)

Discontinuous because f (x) has no limit as xapproaches 2

31.

c f c( )lim ( )

x cf x

→ −lim ( )

x cf x

→ +lim ( )x c

f x→ Continuous?

1 4 2 2 2 removable

2 1 1 1 1 continuous

4 5 5 2 none step

5 none none none none infinite

32.

c f c( )lim ( )

x cf x

→ −lim ( )

x cf x

→ +lim ( )x c

f x→ Continuous?

1 3 2 3 none step

2 1 4 4 4 removable


5 5 5 none none infinite

33. a.

x

d(x)

2

3

b. lim ( ) , lim ( ) .x x

d x d x→ →− +

= =2 2

3 3 Limit = 3.

Continuous.

34. a.

3

2

1

x

h(x)

b. lim ( ) , lim ( ) .x x

h x h x→ →− +

= =1 1

3 2 No limit.

Not continuous.

35. a.

2

x

m(x)

9

7

b. lim ( ) , lim ( ) .x x

m x m x→ →− +

= =2 2

9 7 No limit.

Not continuous.

36. a.

–1

2

q(x)

x

b. lim ( ) , lim ( ) .x x

q x q x→− →−− +

= =1 1

2 2 Limit = 2.

Continuous.

37. 9 – 22 = 2k∴ k = 2.5

g(x)

x

2

5


38. 0.4(1) + 1 = k(1) + 2∴ k = −0.6

x

f(x)

1

1.4

39. (32)k = 3k − 3∴ k = −1/2.

3

x

u(x)

1

40. −k + 5 = (−1)2k∴ k = 5/2

x

v(x)

–1

5

41. a. b − 1 = a(1 − 2)2 ⇒ b − 1 = a

b. a = −1 ⇒ b = 0. Continuous at x = 1.

x

f(x)

1 1

a = –1, b = 0

c. For example, a = 1 ⇒ b = 2. Continuousat x = 1.

1

1 x

f(x)

e.g., a = 1, b = 2

42. lim ( ) limx x

f x k x k→ →− −

= − = −2 2

2 2 2 4

lim ( ) lim . . ( )x x

f x kx k k→ →+ +

= = =2 2

1 5 1 5 2 3

For f (x) to be continuous at x = 2, these twolimits must be equal, so find k such thatk2 − 4 = 3kk2 − 3k − 4 = 0(k − 4)(k + 1) = 0so k = 4 and k = −1 are the two values of k thatwill make f (x) continuous at x = 2.

43. Let T(θ ) = the number of seconds it takes tocross.

T( )if

40

sin, if 0 or 90θ

θ

θθ θ=

= °° < < ° ° < < °

24 90

90 180

,

90

40

θ

θ

44. a.

x

f(x)

4

1

b. f (x) seems to approach 4 as x approaches 1.

c. f (1.0000001) = 1.0000001 + 3 + 10− 13 ≈4.0000001, which is close to 4.

d. There is a vertical asymptote at x = 0. Youmust get x much closer to 1 than x =1.0000001 for the discontinuity to show up.

45. For any value of c, P(c) is determined by additionand multiplication. Because the set of realnumbers is closed under multiplication andaddition, P(c) will be a unique, real number forany real value x = c. P(c) is the limit of P(x) asx approaches c by the properties of the limit of aproduct of functions (for powers of x), the limitof a constant times a function (for multiplicationby the coefficients), and the limit of a sum (forthe individual terms). Therefore, P is continuousfor all values of x.

46. a. limx→0

|sgn x| = 1 but f (0) = 0

lim ( ) ( ),x

f x f→

≠0

0 so discontinuous

b.

x

g(x)

2

3

c.

x

h(x)

1

–1 1


d. For x > 0, a(x) = x/x = 1 = sgn x.For x < 0, a(x) = (−x)/x = −1 = sgn x.For x = 0, a(0) is not defined.∴ a(x) = sgn x for all x ≠ 0, Q.E.D.

e.

2

–2

–2x

f(x)

π

Problem Set 2-5Q1. No limit Q2. 3

Q3. 4 Q4. 3

Q5. 2 Q6. No

Q7. No Q8. Yes

Q9. No Q10. Yes

1. • limx

f x→−∞

∞ ( ) = • limx

f x→− −3

4( ) = –

• limx

f x→− +

=3

3( ) • limx

f x→

= ∞1

( ) –

• limx

f x→

=2

1( ) • limx

f x→ −

= ∞3

( )

• limx

f x→ +

=3

2( ) • limx

f x→∞

( ) does not

exist.

2. • limx

g x→−∞

=( ) 2 • limx

g x→− −

=2

4( )

• limx

g x→− +

=2

3( ) – • limx

g x→ −

= ∞1

( )

• limx

g x→

=2

3( ) • limx

g x→ −

=3

4( )

• lim ( )x

g x→∞

= −2


x

f(x)

2

2

x

f(x)


7

–5

x

f(x)

x

f(x)

7. a.

3

2

x

f(x)

b. limx

f x→ +

∞3

( ) = , limx

f x→ −

∞3

( ) = – ,

lim ( ),x

f x→3

none, limx

f x→∞

( ) = ,2

limx

f x→−∞

( ) = 2

c. 21

3100+

−=

x1

398

x −=

x – 31

98=

x = =3

1

983 0102. K

x f (x)

3.01 102

3.001 1002

3.0001 10002

All of these f (x) values are greater than 100.lim ( )x

f x→ +

= ∞3

means that f (x) can be kept

arbitrarily far from zero just by keeping xclose enough to 3 on the positive side.There is a vertical asymptote at x = 3.

d. 21

32 001+

−=

x.

1

30 001

3 1000

1003

xx

x

−=

− ==

.

x f (x)

1004 2.00099…

1005 2.00099…

1006 2.00099…

All of these f (x) values are within 0.001 unitof 2. lim

xf x

→∞=( ) 2 means that you can keep

f (x) arbitrarily close to 2 by making the valueof x arbitrarily large. y = 2 is a horizontalasymptote.

8. a.

π/2

1x

g(x)

b. lim ( ) , lim ( )/ /x x

g x g x→ →− +

= ∞ = −∞π π2 2

The limit is infinite because |g(x)| can be keptarbitrarily far from zero. You can’t saylim ( )

/xg x

→= ∞

π 2 because the left and right


limits are not the same (one is positive and theother is negative).

c. sec x = −1000∴ cos x = −0.001x = arccos (−0.001) = 1.57179…

x g(x)

1.5717 –1106.5…

1.5716 –1244.2…

1.5715 –1421.1…

All of these f (x) values are less than −1000.lim ( )

/xg x

→ += −∞

π 2 means that arbitrarily far

g(x) can be kept arbitrarily far from zero in thenegative direction by keeping x close enoughto π

2 on the positive side.The line x = π

2 is a vertical asymptote.

9. a.

5

3r(x)

x

2

b. lim ( )x

r x→∞

= 2 because (sin x)/x approaches zero.

c. r( )

sin( ). ,28 2

28

282 00967= + = K which is

within 0.01 unit of 2.

r( )

sin( )32 2

32

32= + = 2.01723 ,K which is

more than 0.01 unit away from 2.

y

x

1.99

2.01

28 32

r

Keeping r(x) within 0.01 unit of 2 means you

want to keep sin . ,xx < 0 01 or |sin x| <

0.01 |x|. You are looking for a large value ofx, so you know x will be positive, so youwant |sin x| < 0.01x. You can’t get rid of theabsolute value symbol on the sine becausesine will keep alternating as x gets larger.You know |sin x| ≤ 1 for all valuesof x, so you need to make 0.01x > 1, orx > 100. So D = 100.

d. The line y = 2 is an asymptote. Even thoughr (x) oscillates back and forth across this line,the limit of r(x) is 2 as x approaches infinity,satisfying the definition of asymptote.

e. The graph suggests that lim ( ) .x

r x→

=0

3

(The exact value is e, 2.7182… .)

10. a. h x x x( ) ( / )= +1 1

10

1

2

3

x

h(x)

b. There is a compromise number (bigger than1, but finite) that wins. (The exact limit is e.)

11. The limit is infinite. y is unbounded as xapproaches infinity. If there were a number Esuch that log x < E for all x > 0, then you couldlet x = 102E so that log x = log 102E = 2E, whichis greater than E, which was assumed to be anupper bound.

12. “Wanda, here’s what happens to a fraction whenthe denominator gets close to zero: 1

0 1 10. = ,1

0.00011

0.00001, , ,= =10 000 100 000. The answers

just keep getting bigger and bigger. When thedenominators get bigger and bigger, the fractiongets closer and closer to zero, like this:

110

1100

110000 1 01 001= = =. , 0. , 0. .”

13. a. The definite integral is the product of theindependent and dependent variables. Becausedistance = (rate)(time), the integral representsdistance in this case.

b. T9 = 17.8060052… T45 = 17.9819616… T90 = 17.9935649…T450 = 17.9994175…

c. The exact answer is 18. It is a limit becausethe sums can be made as close to it as youlike, just by making the number of trapezoidslarge enough (and thus keeping their widthsclose to zero). The sums are smaller than theintegral because each trapezoid is inscribedunder the graph and thus leaves out a part ofits respective strip of the region.

d. Tn is 0.01 unit from 18 when it equals 17.99.From part b, this occurs between n = 45 and n= 90. By experimentation,T66 = 17.9897900… and T67 = 17.9900158… .Therefore, the approximation is within 0.01unit of 18 for any value of n ≥ 67.An alternative solution is to plot the graph ofthe difference between 18 and Tn as a functionof the number of increments, n, or to do aregression analysis to find an equation. Thebest-fitting elementary function is an inversepower variation function, y = (5.01004…)


(x− 1.48482…). The graph of this function and threeof the four data points are shown here. UseTRACE or the solver feature of your grapher tofind n ≈ 67.

0.01

0.1

y

n

67

90

14. a. Work = force × distance. Because a definiteintegral measures the y-variable times thex-variable, it represents work in this case.

b. By the trapezoidal rule, T10 = 24.147775…and T100 = 24.004889… . The units arefoot-pounds.

c. The integer is 24.

d. By experimentation, T289 = 24.001003… andT290 = 24.000998… .∴ D = 290

15. Length = 100 sec x = 100/cos xLength > 1000 ⇒ 100/cos x > 1000cos x < 0.1 (because cos x is positive)x > cos–1 0.1 (because cos is decreasing)x > 1.4706289…π/2 − 1.4706289… = 0.100167…x must be within 0.100167… radian of π/2.The limit is (positive) infinity.

16. a. f (2) = 5 · 2 · 0 · (1/0), which has theform 0 · ∞.g(2) = 5 · 2 · 0 · (1/0)2, which has theform 0 · ∞.h(2) = 5 · 2 · 02 · (1/0), which has theform 0 · ∞.

b. f x x xx

x x( ) ( )= − ⋅ = ≠5 21

25 2

–,

∴ =→

( )limx

f x2

10

g x x xx

x

xx( ) ( )= − ⋅ = ≠5 2

1

2

5

222( – ) –

,

∴→

( )limx

g x2

is infinite.

h x x xx

x x x( ) ( ) = ( – ), = − ⋅ ≠5 21

25 2 22

–∴

→ ( ) =lim

xh x

20

c. The indeterminate form 0 · ∞ could approachzero, infinity, or some finite number.

Problem Set 2-6

Q1. 53 Q2. 53

Q3. Undefined Q4. 5

Q5. Undefined Q6. Does not exist

Q7. 1 Q8. +

Q9. Indeterminate Q10. C

1. IVT applies on [1, 4] because f is a polynomialfunction, and polynomial functions arecontinuous for all x.f (1) = 18, f (4) = 3∴ There is a value x = c in (1, 4) for whichf (c) = 8.Using the intersect or solver feature,c = 1.4349… , which is between 1 and 4.

f (x)

x

f(1)

f(4)

8

1 4c

2. IVT applies on [0, 6] because f is a polynomialfunction, and polynomial functions arecontinuous for all x.f (0) = −8, f (6) = −0.224∴ There is a value x = c in (0, 6) for whichf (c) = −1.Using the intersect or solver feature,c = 5.8751… , which is between 0 and 6.

–0.224

–8

0 6c xf(x)

3. a. For 1 ≤ y < 2 or for 5 < y ≤ 8, the conclusionwould be true. But for 2 ≤ y ≤ 5, it would befalse because there are no values of x in [1, 5]that give these values for f (x).

b. The conclusion of the theorem is true becauseevery number y in [4, 6] is a value of g(x) forsome value of x in [1, 5].

4. a. f f f( ) , ( ) , ( . ) . ,2 4 3 8 0 5 2 1 414= = = = K

f ( )5 8=b. f is continuous at x = 3 because it has a limit

and a function value and they both equal 8.

c. f is continuous nowhere else. Because thesets of rational and irrational numbers aredense, there is a rational number between anytwo irrational numbers, and vice versa. Sothere is no limit of f (x) as x approaches anynumber other than 3.

d. The conclusion is not true for all values of ybetween 1 and 4. For instance, if y = 3, thenc would have to equal log2 3. But log2 3 isirrational, so f (c) = 8, which is not between1 and 4.


5. Let f (x) = x2. f is a polynomial function, so it iscontinuous and thus the intermediate valuetheorem applies. f (1) = 1 and f (2) = 4, so thereis a number c between 1 and 2 such that f (c) = 3.By the definition of square root, c = 3 , Q.E.D.

6. Prove that if f is continuous, and if f (a) ispositive and f (b) is negative, then there is at leastone zero of f (x) between x = a and x = b.

Proof:

f is continuous, so the intermediate valuetheorem applies. f (a) is positive and f (b) isnegative, so there is a number x = c between aand b for which f (c) = 0. Therefore, f has at leastone zero between x = a and x = b, Q.E.D.

7. The intermediate value theorem is called anexistence theorem because it tells you that anumber such as 3 exists. It does not tell youhow to calculate that number.

8. Telephone your sweetheart’s house. An answer tothe call tells you the “existence” of thesweetheart at home. The call doesn’t tell suchthings as how to get there, and so on. Also,getting no answer does not necessarily mean thatyour sweetheart is out.

9. Let f (t) = Jesse’s speed − Kay’s speed. f (1) =20 − 15 = 5, which is positive. f (3) = 17 − 19 =−2, which is negative. The speeds are assumed tobe continuous (because of laws of physics), so fis also continuous and the intermediate valuetheorem applies.So there is a value of t between 1 and 3 forwhich f (t) = 0, meaning that Jesse and Kay aregoing at exactly the same speed at that time.The existence of the time tells you neither whatthat time is nor what the speed is. An existencetheorem, such as the intermediate value theorem,does not tell these things.

10. Let f (x) = number of dollars for x-ounce letter.f does not meet the hypothesis of the IVT on theinterval [1, 9] because there is a stepdiscontinuity at each integer value of x. There isno value of c for which f (c) = 2 because f (x)jumps from 1.98 to 2.21 at x = 8.

11. You must assume that the cosine is functioncontinuous. Techniques:

• c = cos− 1 0.6 = 0.9272…

• Using the solver feature, c = 0.9272...

• Using the intersect feature, c = 0.9272...

12. You must assume that 2x is continuous.f (0) = 20 = 1, because any positive number to the0 power equals 1.

• c = = =loglog

log2 33

21 5849. ...

• Using the solver feature, c = 1.5849...

• Using the intersect feature, c = 1.5849...

13. This means that a function graph has a highpoint and a low point on any interval in whichthe function is continuous.

x

f(x)

a bc1 c2

If the function is not continuous, there may be apoint missing where the maximum or minimumwould have been.

x

f(x)

a b

Another possibility would be a graph with avertical asymptote somewhere between a and b.

14. Prove that if f is continuous on [a, b], the imageof [a, b] under f is all real numbers between theminimum and maximum values of f (x),inclusive.

Proof:

By the extreme value theorem, there are numbersx1 and x2 in [a, b] such that f (x1) and f (x2) are theminimum and maximum values of f (x) on [a, b].Because x1 and x2 are in [a, b], f is continuous onthe interval whose endpoints are x1 and x2. Thus,the intermediate value theorem applies on thelatter interval. Thus, for any number y betweenf (x1) and f (x2), there is a number x = c betweenx1 and x2 for which f (c) = y, implying that theimage of [a, b] under f is all real numbersbetween the minimum and maximum values off (x), inclusive, Q.E.D.

Problem Set 2-7Review Problems


R1. a. f ( )336 51 15

3 3

0

0= − +

−=

Indeterminate form


b. f x x x( ) = – ,4 5 3≠

9–9

9

–9

y

x

At x = 3 there is a removable discontinuity.

c. For 0.01, keep x within 0.0025 unit of 3. For0.0001, keep x within 0.000025 unit of 3. Tokeep f (x) within ε unit of 7, keep x within14 ε unit of 3.

R2. a. L f xx c

=→

lim ( ) if and only if for any number

ε > 0, no matter how small,there is a number δ > 0 such that if x iswithin δ units of c, but x ≠ c, then f (x) iswithin ε units of L.

b. limx

f x→

=1

2( )

limx

f x→2

( ) does not exist.

limx

f x→

=3

4( )

limx

f x→4

( ) does not exist.

limx

f x→5

( ) = 3

c. limx

f x→

=2

3( )

Maximum δ: 0.6 or 0.7

d. The left side of x = 2 is the more restrictive.

Let 2 + x – 1 = 3 − 0.4 = 2.6.∴ x = 1 + 0.62 = 1.36∴ maximum value of δ is 2 − 1.36 = 0.64.

e. Let f x( ) = 3 − ε.

2 1 3+ − = −x εx = (1 − ε)2 + 1Let δ = 2 − [(1 − ε)2 + 1] = 1 − (1 − ε)2,which is positive for all positive ε < 1. Ifε ≥ 1, simply take δ = 1. Then δ will bepositive for all ε > 0.

R3. a. See the limit property statements in the text.

b. •

20

3

x

g(x)

–19

• The limit of a quotient property does notapply because the limit of the denominatoris zero.

• g xx x x

x( ) = − − +

−( )( )3 10 2

3

2

g(x) = x2 − 10x + 2, x ≠ 3

You can cancel the (x − 3) because thedefinition of limit says “but not equal to 3.”

• lim ( )x

g x→3

= + +→ → →

lim lim (– ) limx x x

x x3

2

3 310 2

Limit of a sum= ⋅ +

→ → →lim lim – limx x x

x x x3 3 3

10 2


= 3 · 3 − 10(3) + 2 Limit of x= −19, which agrees with the graph.

c. • f (x) = 2x,

g xx x

x

x x

x( ) = − +

−= − −

−

2 8 15

3

3 5

3

( )( )

= −x + 5, x ≠ 3lim ( ) , limx x

f x g x→ →

= =3 3

8 2 ( )

• p(x) = f (x) · g(x)limx

p x→

= ⋅ =3

8 2 16( )

x p(x)

2.997 15.9907…

2.998 15.9938…

2.999 15.9969…

3 undefined

3.001 16.0030…

3.002 16.0061…

3.003 16.0092…

All these p(x) values are close to 16.

• r xf x

g x( ) = ( )

( )

lim ( )x

r x→

= =3

8

24

3

5

10

15

20y

x

f

g

r

d. For 5 to 5.1 s: average velocity = −15.5 m/s.

Average velocity = =f t f

t

( ) – ( )

–

5

535 5 50

5

5 2 5

5

2t t

t

t t

t

– –

–

– ( – )( – )

–= =


−5(t − 2), for t ≠ 5. Instantaneous velocity =limit = −5(5 − 2) = −15 m/s.The rate is negative, so the distance above thestarting point is getting smaller, whichmeans the rock is going down.Instantaneous velocity is a derivative.

R4. a. • f is continuous at x = c if and only if

1. f (c) exists

2. lim ( )x c

f x→

exists

3. limx c

f x f c→

=( ) ( )

• f is continuous on [a, b] if and only if f iscontinuous at every point in (a, b), andlim ( ) ( )x a

f x f a→ +

= and limx b

f x f b→ −

=( ) ( ).

b.

c f c( )lim

x cf x

→ −( ) lim ( )

x cf x

→ +lim ( )x c

f x→ Continuous?

1 none none none none infinite

2 1 3 3 3 removable

3 5 2 5 none step



c. i. ii.

x

y

1

x

y

2

iii. iv.

x

y

3

x

y

4

v. vi.

5

x

y

L

f(6)

x

y

6

vii.

1

5

–2

x

y

d.

2

4

x

f(x)

2

The left limit is 4 and the right limit is 2, sof is discontinuous at x = 2, Q.E.D.Let 22 = 22 − 6(2) + k.∴ k = 12

R5. a. limx

f x→

= ∞4

( ) means that f (x) can be kept

arbitrarily far from 0 on the positive side justby keeping x close enough to 4, but not equalto 4.limx

f x→∞

=( ) 5 means that f (x) can be made to

stay arbitrarily close to 5 just by keeping xlarge enough in the positive direction.

b. • lim ( )x

f x→−∞

does not exist.

• limx

f x→−

=2

1( )

• limx

f x→ −

= ∞2

( )

• limx

f x→ +

= −∞2

( )

• limx

f x→∞

=( ) 2

c. f (x) = 6 − 2− x

limx

f x→∞

=( ) 6

f (x) = 5.999 = 6 − 2− x

2− x = 0.001

x = − log

log

0 001

2

.

x = 9.965...

x f (x)

10 5.999023…

20 5.999999046…

30 5.99999999907…

All of these f (x) values are within 0.001 of 6.

d. g(x) = x− 2

lim ( )x

g x→

= ∞0

g(x) = 106 = x− 2

x2 = 10− 6

x = 10− 3

x g(x)

0.0009 1.2345… ⋅ 106

0.0005 4,000,000

0.0001 1 ⋅ 108


All of these g(x) values are larger than1,000,000.

e. v(t) = 40 + 6 t

n Trapezoidal Rule

50 467.9074…

100 467.9669…

200 467.9882…

400 467.9958…

The limit of these sums seems to be 468.

By exploration,T222 = 467.98995…T223 = 467.99002…∴ D = 223

R6. a. See the text statement of the intermediatevalue theorem.The basis is the completeness axiom.See the text statement of the extreme valuetheorem.The word is corollary.

b. f (x) = −x3 + 5x2 − 10x + 20f (3) = 8, f (4) = −4So f (x) = 0 for some x between 3 and 4 bythe intermediate value theorem.The property is continuity.The value of x is approximately 3.7553.

c.

–4

3

x

f(x)

f (−6) = 1 and f (−2) = 5 by tracing on thegraph or by simplifying the fraction to getf (x) = x + 7, then substituting. You will notalways get a value of x if y is between 1 and5. If you pick y = 3, there is no value of x.This fact does not contradict the intermediatevalue theorem. Function f does not meet thecontinuity hypothesis of the theorem.

Concept Problems

C1.

hh

f

g

y

x

7

4

Conjecture: lim ( )x

f x→

=4

7

C2. f (1) = 12 − 6 ⋅ 1 + 9 = 4As x → 1 from the left, f (x) → 12 + 3 = 4.As x → 1 from the right, f (x) → 12 − 6 + 9 = 4.∴ lim

xf x f

→14 1( ) = = ( )

∴ f is continuous at x = 4, Q.E.D.For the derivative, from the left side,f x f

x

x

x

x x

x

( ) – ( )

–

–

–

( – )( )

–

1

1

3 4

1

1 1

1

2

= + = + =

x + 1, x ≠ 1∴ ′ = + =

→ −limx

f x1

1 1 2( )

For the derivative, from the right side,f x f

x

x x

x

x x

x

( ) – ( )

–

– –

–

( – )( – )

–

1

1

6 9 4

1

1 5

1

2

= + = =

x − 5, x ≠ 1∴ lim

xf x

→ +′ −

11 5 4( ) = – =

So f is continuous at x = 1, but does not have avalue for the derivative there because the rate ofchange jumps abruptly from 2 to −4 at x = 1. Ingeneral, if a function has a cusp at a point, thenthe derivative does not exist, but the function isstill continuous.

C3. The graph is a y = x2 parabola with a stepdiscontinuity at x = 1. (Use the “rise-run”property. Start at the vertex. Then run 1, rise 1;run 1, rise 3; run 1, rise 5; . . . . Ignore thediscontinuity at first.) To create thediscontinuity, use the signum function withargument (x − 1). Because there is no value forf (1), the absolute value form of the signumfunction can be used.

y xx

x= 2 2

1

1+ − | – |

–

C4. The quantity | ( ) – |f x L is the distance betweenf (x) and L. If this distance is less than ε, thenf (x) is within ε units of L. The quantity |x − c|is the distance between x and c. The right part ofthe inequality, |x – c| < δ, says that x is within δunits of c. The left part, 0 < |x − c|, says that xdoes not equal c. Thus, this definition of limit isequivalent to the other definition.

Chapter Test

T1. f is continuous at x = c if and only if

1. f (c) exists

2. lim ( )x c

f x→

exists

3. limx c

f x f c→

=( ) ( )

f is continuous on [a, b] if and only if f iscontinuous at all points in (a, b), andlim ( ) ( )x a

f x f a→ +

= and lim ( ) ( )x b

f x f b→ −

= .


T2. a. • limx

f x→ −

=2

3( ) • limx

f x→ +

=2

4( )

• lim ( )x

f x→2

does not exist. • lim ( )x

f x→ −

=6

2

• lim ( )x

f x→ +

=6

2 • limx

f x→

=6

2( )

b. f is continuous on [2, 6] because it iscontinuous for all values in (2, 6) andlim ( ) ( )x

f x f→ +

=2

2 and lim ( ) ( )x

f x f→ −

=6

6 .

T3. See the text statement of the quotient property.

T4. a. Left: −4; right: −4

b. Limit: −4

c. Discontinuous

T5. a. Left: none; right: none

b. Limit: none

c. Discontinuous

T6. a. Left: 6; right: 6

b. Limit: 6

c. Continuous

T7. a. Left: −2; right: 3

b. Limit: none

c. Discontinuous

T8.

4

4

x

y

T9. a. b.

x

f(x)

x

g(x)

1

c. d.

1

–1

h(x)

x

1

1

x

s(x)

T10. a. f ( )( )( )

,30 5 0 8 0 3

0 3

0

0

2

= − ⋅ + −−

=

an indeterminate form

b. lim ( ) lim ( ),x x

f x x x x→ →

= − + ≠3 3

2 5 8 3

Definition of limit“x ≠ c”

= + − +→ → →

lim lim ( ) limx x x

x x3

2

3 35 8

Limit of a sum= ⋅ + − ⋅ +

→ → →lim lim ( ) limx x x

x x x3 3 3

5 8


= 3 · 3 + (−5) · 3 + 8 Limit of x= 2, Q.E.D.

T11. If k f xx x

x x= =

≤+ >

12

1 2

2

, ( ),

,

lim limx x

f x f x→ →− +

= =2 2

4 3( ) , ( )

∴ f is discontinuous at x = 3.

2

5f(x)

x

T12. limx

f x k→ −

= ⋅2

22( )

limx

f x k→ +

+2

2 ( ) =

∴ 4k = 2 + kk = 2/3

T13. See graph in T11.

T14. a. limx

T x→∞

=( ) 20

From the graph, it appears that if x > 63 ft,then T(x) is within 1° of the limit.The graph of T has a horizontal asymptote atT = 20.

b. T = 20 + 8(0.97x) cos 0.5x. The amplitude ofthe cosine factor is 8 0 97( . ).x Make thisamplitude < 0.1.8(0.97c) = 0.1

0.97c = 0.0125

c = log .

log .

0 0125

0 97

c = 143.8654…∴ T is within 0.1 unit of 20 wheneverx > 143.8654… .

c. The time of day would be mid-afternoon,when the temperature of the surface ishighest.

T15. a. Use either TRACE or TABLE to show:d(0) = 0, d(10) = 6, d(20) = 14, d(30) = 24,d(40) = 36, and d(50) = 50.

b. Average rate = −−

=d d( . ) ( )

.

20 1 20

20 1 2014 0901 14

20 1 200 901

. –

. –= . cm/day


c. Average rate = −−

=d t d

t

( ) ( )20

200 01 0 5 14

20

0 01 70 20

20

2. . –

–

. ( )( – )

–

t t

t

t t

t

+ = + =

0.01t + 0.7, t ≠ 20. The limit as t approaches20 is 0.01(20) + 0.7, which equals0.9 cm/day. This instantaneous rate is calledthe derivative.

d. The glacier seems to be speeding up becauseeach 10-day period it moved farther than it hadin the preceding 10-day period.

T16. c(0) = p(0) = 10, so each has the same speed att = 0. lim ( ) . lim ( ) .

t tc t p t

→∞ →∞= = ∞16 Surprise for

Phoebe!

T17. f xkx x

kx x( )

,

– ,=

≤>

2 2

10 2

if

if

limx

f x k k→ −

= ⋅ =2

22 4( )

limx

f x k→ +

= −2

10 2( )

Make 4k = 10 − 2k ⇒ k = 5/3. There is a cuspat x = 2.

2

10f(x)

x

T18. h(x) = x3. h(1) = 1 and h(2) = 8, so 7 is betweenh(1) and h(2). The intermediate value theoremallows you to conclude that there is a realnumber between 1 and 8 equal to the cube rootof 7.



Chapter 3—Derivatives, Antiderivatives,and Indefinite Integrals

Problem Set 3-11. The graph is correct.

2. Average rate = = =f f( . ) – ( )

.

. –

.

5 1 5

0 1

3 21 3

0 12.1 km/min

3.

5

2

x

y

4. The graph of r has a removable discontinuityat x = 5.

rf f

( )( ) ( )

55 5

5 5

0

0= −

−=

5. r xf x f

x

x x

x( )

( ) ( )= −−

= − + −−

5

5

8 18 3

5

2

= − −−

= − ≠( )( ),

x x

xx x

5 3

53 5

′ =→

f r xx

( ) lim ( )55

= − −−

= − =→ →

lim( )( )

lim( )x x

x x

xx

5 5

5 3

53 2

The derivative is the velocity of the spaceship,in km/min.

6. Find the equation of the line through (5, f (5)),or (5, 3), with slope 2.y − 3 = 2(x − 5) ⇒ y = 2x − 7

5

3

x

y

This line is tangent to the graph of f (x) at (5, 3).

7. As you zoom in, the line and the graph appear tobe the same.


Problem Set 3-2Q1. Instantaneous rate of change

Q2. x + 9 Q3. 18

Q4.

x

y

1

Q5. 9x2 − 42x + 49

Q6. “−” sign

Q7. Q8.

x

y

3

x

y

5

2

Q9. Newton and Leibniz

Q10. D

1. See the text for the definition of derivative.

2. Physical: Instantaneous rate of change of thedependent variable with respect to theindependent variableGraphical: Slope of the tangent line to the graphof the function at that point

3. a. fx

xx′

→( ) =3

0 6 5 4

33

2

lim. – .

–

= lim. ( – )( )

–.

x

x x

x→

+ =3

0 6 3 3

33 6

b. Graph of the difference quotient m(x)

3

3.6

x

m(x)

c., d. Tangent line: y = 3.6x − 5.4

3

1 x

f(x)

4. a. f x x fx

xx( ) = – . , ( ) =0 2 6

0 2 7 2

62

6

2

′ +→

lim– . .

–

= + = −→

lim– . ( – )( )

–.

x

x x

x6

0 2 6 6

62 4


b.

6

–2.4

x

m(x)

c., d. Tangent line: y = −2.4x + 7.2

6

–7.2

xf(x)

5. fx x

xx′ + + +

+→−(– ) =2

5 1 5

22

2

lim

= lim( )( )

x

x x

x→−

+ ++

=2

2 3

21

6. (– ) =fx x

xx′ + +

+→−4

6 2 10

44

2

lim–

= + ++→−

lim( )( )

x

x x

x4

4 2

42= –

7. ( ) =fx x x

xx′ + +

→1

4 8 6

11

3 2

lim– –

–

= = –lim( – )( – – )

–x

x x x

x→1

21 3 2

14

8. fx x x

xx′ +

+→−(– ) =1

4 6 8

11

3 2

lim– – –

= ++

=→−lim

( )( – – )x

x x x

x1

21 2 2

11

9. fx

xx′ + +

→( ) =3

0 7 2 0 1

33lim

– . .

–

= = −→

lim– . ( – )

–.

x

x

x3

0 7 3

30 7

10. fx

xx′

→( ) =4

1 3 3 2 2

44lim

. – – .

–

= =→

lim. ( – )

–.

x

x

x4

1 3 4

41 3

11. fxx

′+

=→−

(– ) =15 5

10

1lim

–

12. fxx

′ + =→

( ) =32 2

30

3lim

–

–

13. The derivative of a linear function equals theslope. The tangent line coincides with the graphof a linear function.

14. The derivative of a constant function is zero.Constant functions are horizontal and don’tchange! The tangent line coincides with thegraph.

15. a. Find f ′ ( 1) = 2, then plot a line through point(1, f (1)) using f ′ ( 1) as the slope. The line isy = 2x − 1.

b. Near the point (1, 1), the tangent line and thecurve appear nearly the same.

c. The curve appears to get closer and closer tothe line.

d. Near point (1, 1) the curve looks linear.

e. If a graph has local linearity, the graph nearthat point looks like the tangent line.Therefore, the derivative at that point could besaid to equal the slope of the graph at thatpoint.

16. a. f ( x) = x2 + 0.1 (x − 1)2/3

f ( 1) = 12 + 0.1(1 − 1)2/3 = 1 + 0 = 1, Q.E.D.The graph appears to be locally linear at(1, 1), because it looks smooth there.

b. Zoom in by a factor of 10,000.

1

1

c. The graph has a cusp at x = 1. It changesdirection abruptly, not smoothly.

d. If you draw a secant line through (1, 1) from apoint just to the left of x = 1, it has a largenegative slope. If you draw one from a pointjust to the right, it has a large positive slope.In both cases, the secant line becomes verticalas x approaches 1 and a vertical line hasinfinite slope. So there is no real numberequal to the derivative.

17. a.

3

5

x

f(x)7

b. First simplify the equation.

f xx x

x( )

,

,=

+ ≠=

2 3

7 3

if

if


The difference quotient is

m xx

x

x

x( ) =

( ) –

–

–

–

+ =2 7

3

5

3

3

5

x

m(x)

c.

x f (x)

2.997 667.66…

2.998 1001

2.999 2001

3.000 undefined

3.001 −1999

3.002 −999

3.003 −665.66…

The difference quotients are all large positivenumbers on the left side of 3. On the rightside, they are large negative numbers. For aderivative to exist, the difference quotientmust approach the same number as x getscloser to 3.

18. a.

1

2

x

s(x)

b. m xx

x( ) =

| sin ( – ) |

–

1

1

1

1

x

m(x)

c. As x approaches 1 from the left, m(x)approaches −1. As x approaches 1 from theright, m(x) approaches 1. Because the left andright limits are unequal, there is no derivativeat x = 1.

19. a. ′ +→

f xx x

xx( ) = lim

. – . . –

–3

20 25 2 5 7 25 2

3

=→

lim. ( – )( – )

–x

x x

x3

0 25 3 7

3

=→

lim ,x

x3

0 25 7 1 . ( – ) = – Q.E.D.

The tangent line on the graph has slope −1.

b.

Draw secant linesfrom here.

f (x)

3

2

x

As the x-distance between the point and 3decreases, the secant lines (solid) approach thetangent line (dashed).

c. The same thing happens with secant linesfrom the left of x = 3. See the graph in part b.

d.

3

4

x

Draw secant linesfrom here.

g(x)

e. A derivative is a limit. Because the left andright limits are unequal, there is no derivativeat x = 3.

f. m xx

x( ) =

– cos

–

66

3

π

. By table,

x m(x)

2.9 3.1401…

2.99 3.1415…

3 undefined

3.01 −3.1415…

3.1 −3.1401…

Conjecture: The numbers are π and −π.

20. From Problem 19, parts b and c, the tangent lineis the limit of the secant lines as x approaches c.Because the slope of the secant line is the average


rate of change of f (x) for the interval from x to c(or from c to x) and the derivative, f ′(c), is thelimit of this average rate, the slope of the tangentline equals f ′(c).

Problem Set 3-3Q1. 3

Q2.

–2

y

x

5

Q3.y

x

Q4. 20%

Q5. 3x2 − 2x − 8

Q6. 25x2 − 70x + 49

Q7. log 6

Q8.

Q9.

Q10. “lim”x c→

is missing.

1. a.40

f´

–2 2

f

x

y

b. f ′(x) is positive for −2 < x < 2. The graph off is increasing for these x-values.

c. f ( x) is decreasing for x satisfying |x| > 2.f ′(x) < 0 for these values of x.

d. Where the f ′ graph crosses the x-axis, thef graph has a high point or a low point.

e. See the graph in part a.

f. Conjecture: f ′ is quadratic.

2.15

3

x

y

g

g'

The graph does not have the high and lowpoints that are typical of a cubic function. As xincreases, the graph starts to roll over and forma high point, but it starts going back up againbefore that happens. This behavior is revealed bythe fact that the derivative is positive everywhere.Between x = 0 and x = 1, the derivative reachesa low point, indicating that the slope is aminimum, but the slope is still positive andthe graph of g is still going up.

3. a.

200h

h´

x

y

–2 1 2.5

b. The h′ graph looks like a cubic functiongraph. Conjecture: Seventh-degree functionhas a sixth-degree function for its derivative.

c. By plotting the graph using a friendly window,then tracing, the zeros of h′ are −2, 1, 2.5.

d. If h′(x) = 0, the h graph has a high point ora low point. This is reasonable because ifh′(x) = 0, the rate of change of h(x) is zero,which would happen when the graph stopsgoing up and starts going down, or viceversa.

e. See the graph in part a.

4.

102 x

y

q

q'

The graph does not have the expected shape for aquartic function. The two high points and the lowpoint all appear to occur as a high point atx = 2. The derivative graph crosses the x-axis justonce, at x = 2, indicating that this is the onlyplace where the function graph is horizontal.


5. a.

4

5

x

y

f

f´

b. Amplitude = 1, period = 2π = 6.283… c. The graph of f ′ also has amplitude 1 and

period 2π.d.

f

g

f´ and g´5

4

x

y

The graphs of f and g are the same shape,spaced 1 unit apart vertically. The graphs off ′ and g′ are identical! This is to be expectedbecause the shapes of the f and g graphs arethe same.

6.

4

5

x

y

f

f´

The function available on the grapher is y = cos x.The amplitude is 1, the period is 2π, and theshape is sinusoidal. cos 0 = 1, and the graph is ata high point, y = 1, when x = 0.

7. 8.

x

y

4

3f

f´

x

y

4

3f'

f

9. 10.

x

f

f´

y

1

5

x

y

2

3f'

f

11. The derivative for f ( x) = 2x is consistently belowthat of the function itself. This fact implies thatf ( x) does not increase rapidly enough to make the

derivative equal the function value. So the basemust be greater than 2. By experimenting, 3 istoo large, but not by much. You can use trialand error with bases between 2 and 3, checkingthe results either by plotting the graph and thenumerical derivative or by constructing tables.An ingenious method that some students comeup with uses the numerical derivative andnumerical solver features to solvenDeriv(bx, x, 1) = b1 at x = 1. The answer isabout 2.718281… . (In Section 3-9, students willlearn that this number is e, the base of naturallogarithms.) The graph of f ( x) = 2.781…x

and its numerical derivative are shown here.

f and f '

x

y

1

1

12. Answer will vary depending on calculator.

13. a. Maximum area = (12.01)2 = 144.2401 in.2

Minimum area = (11.99)2 = 143.7601 in.2

Range is 143.7601 ≤ area ≤ 144.2401.Area is within 0.2401 in.2 of the ideal.

b. Let x be the number of inches.Area = x2.The right side of 12 is more restrictive, so setx2 = 144.02.∴ x = 144.021/2 = 12.000833…Keep the tile dimensions within 0.0008 in. of12 in.

c. The 0.02 in part b corresponds to ε, and the0.0008 corresponds to δ.

14. The average of the forward and backwardsdifference quotients equals

1

2

f x h f x

h

f x f x h

h

( ) – ( ) ( ) – ( – )+ +

= +

1

2

f x h f x h

h

( ) – ( – )

= +f x h f x h

h

( ) – ( – )

2, Q.E.D.

15. a. f ( x) = x3 − x + 1 ⇒ f ( 1) = 1

fx x

xx′ +

→( ) =1

1 1

11

3

lim( – ) –

–

= = +→ →

lim–

–lim

( )( – )

–x x

x x

x

x x x

x1

3

11

1 1

1=

→limx

x x1

1 2( + ) =


b. Forward: f f( . ) – ( )

.

. –

..

1 1 1

0 1

1 231 1

0 12 31= =

Backwards: f f( ) – ( . )

.

– .

..

1 0 9

0 1

1 0 829

0 11 71= =

Symmetric:

f f( . ) – ( . )

( . )

. – .

..

1 1 0 9

2 0 1

1 231 0 829

0 22 01= =

The symmetric difference quotient is closer tothe actual derivative because it is the averageof the other two, and the other two span theactual derivative.

c. f ( 0) = 1

fx x

xx′ +

→( ) =0

1 1

00

3

lim( – ) –

–

= lim–

lim( – )x x

x x

xx

→ →= = −

0

3

0

2 1 1

d. Forward: f f( . ) – ( )

.

. –

.

0 1 0

0 1

0 901 1

0 1= = –0.99

Backwards:

f f(– . ) – ( )

.

.

.

0 1 0

0 1

1 099 1

0 1= − = –0.99

Symmetric:

f f( . ) – (– . )

( . )

. – .

..

0 1 0 1

2 0 1

0 901 1 099

0 20 99= = −

All three difference quotients are equal becausef ( x) changes just as much from −0.1 to 0 as itdoes from 0 to 0.1.

16.

h Backwards Forward Symmetric

0.1 −1.1544… 3.1544… 1

0.01 −3.6415… 5.6415… 1

0.001 −9 11 1

The backwards difference quotients are becominglarge and negative, while the forward differencequotients are becoming large and positive. Theiraverage, the symmetric difference quotient, isalways equal to 1.


Problem Set 3-4Q1. 9x2 − 24x + 16

Q2. a3 + 3a2b + 3ab2 + b3

Q3. See the text definition of derivative.

Q4.f x h f x h

h

( ) – ( – )+2

Q5. No limit (infinite) Q6. log 73

Q7. 3 Q8. Pythagorean theorem

Q9. 10 Q10. C

1. f ( x) = 5x4 ⇒ f ′ ( x) = 20x3

2. y = 11x8 ⇒ dy/dx = 88x7

3. v = 0.007t− 83 ⇒ dv/dt = −0.581t− 84

4. v xx

v x x( ) = ( ) = – ––9

10

18

1

2⇒ ′

5. M(x) = 1215 ⇒ M′ (x) = 0 (Derivative of aconstant)

6. f (x) = 4.7723 ⇒ f ′ (x) = 0 (Derivative of aconstant)

7. y = 0.3x2 − 8x + 4 ⇒ dy/dx = 0.6x − 8

8. r = 0.2x2 + 6x − 1 ⇒ dr /dx = 0.4x + 6

9.d

dxx( – )13 1= −

10. f (x) = 4.5x2 − x ⇒ f′ (x) = 9x − 1

11. y = x2.3 + 5x− 2 − 100x + 4 ⇒dy/dx = 2.3x1.3 − 10x− 3 − 100

12.d

dxx x x x x x( – – )/ –2 5 2 1 3 5 24 3 14

2

58 3+ = − +– / –

13. v = (3x − 4)2 = 9x2 − 24x + 16 ⇒ dv/dx = 18x − 24

14. u = (5x − 7)2 = 25x2 − 70x + 49 ⇒du/dx = 50x − 70

15. f (x) = (2x + 5)3 = 8x3 + 60x2 + 150x + 125 ⇒f ′ (x) = 24x2 + 120x + 150

16. f (x) = (4x − 1)3 = 64x3 − 48x2 + 12x − 1 ⇒f ′ (x) = 192x2 − 96x + 12

17. P xx

x P x x( ) – ( ) –= + ⇒ ′ =2

24 1

18. Q xx x

x Q x x x( ) = – + ( ) = + –3 2

2

3 21 1+ ⇒ ′

19. f (x) = 7x4

′ = +→

f xx h x

hh( ) lim

( ) –0

4 47 7

= + + +→

lim ( )h

x x h xh h x0

3 2 2 3 328 42 28 7 28=

By formula, f ′(x) = 7 ⋅ 4x3 = 28x3, which checks.

20. g(x) = 5x3

g xx h x

hh′ +

→( ) = lim

0

3 35 5( ) –

= + + =→

limh

x xh h x0

2 2 215 15 5 15( )

By formula, g ′ (x) = 5 ⋅ 3x2 = 15x2, which checks.

21. v(t) = 10t2 − 5t + 7

v tt h t h t t

hh′ + − + + − − +

→( ) = lim

[ ( ) ( ) ] ( )0

2 210 5 7 10 5 7

= +→

lim–

h

th h h

h0

220 10 5

= +→

lim ( – )h

t h t0

20 10 5 20 5= –

By formula, v ′ (t) = 10 ⋅ 2t − 5 = 20t − 5, whichchecks.


22. s t t t( ) = – + .4 26 3 7

′ = + − + + − − +→

s tt h t h t t

hh( ) lim

[( ) ( ) . ] [ . ]0

4 2 4 26 3 7 6 3 7

= + + +→

lim– –

h

t h t h th h th h

h0

3 2 2 3 4 24 6 4 12 6

= + + +→

lim ( – – )h

t t h th h t h0

3 2 2 34 6 4 12 6

= 4 123t t– By formula, s t t t t t′ ⋅( ) = – = –4 6 2 4 123 3 , which checks.

23. Mae should realize that you differentiatefunctions, not values of functions. If yousubstitute a value for x into f (x) = x4, you getf (3) = 34 = 81, which is a new function, g(x) =81. It is the derivative of g that equals zero.Moral: Differentiate before you substitute for x.

24. a. v(x) = h′(x) = −10x + 20

b. The book was going down at 10 m/s.The velocity is −10, so h(x) is decreasing.

c. The book was 15 m above where he threw it.

d. 2 s. The book is at its highest point when thevelocity is zero. v(x) = 0 if and only if x = 2.

25.

x

f´

f

y

9

7

26.

6

x

g

g'

y3

27. a.

f´

f

y

x

10

1

b. The graph of f ′ is shown dashed in part a.

c. There appear to be only two graphs becausethe exact and the numerical derivative graphsalmost coincide.

d. f (3) = −6.2

f ′(3) = 3.8 (by formula)

f ′(3) ≈ 3.8000004 (depending on grapher)

The two values of f ′( )3 are almost identical!

28. a. g(x) = x− 1. Conjecture: g′(x) = −1 ⋅ x− 2.Conjecture is confirmed.

y1

y 2 and y3

1

1

x

y

b. h(x) = x1/ 2. Conjecture: g′(x) = 0.5x− 1/ 2.Conjecture is confirmed.

y1

y2 and y3

1

2

x

y

c. e(x) = 2x. Conjecture: e′ (x) = x ⋅ 2x− 1.Conjecture is refuted!

y1

y2

y3

111

x

y

29. f x x x( ) = –/1 2 2 13+f x x f′ ′ =( ) = + , ( ) 1

2– / 9

41 2 2 4

Increasing by 9/4 y-units per x-unit at x = 4

30. f (x) = x− 2 − 3x + 11f ′(x) = −2x− 3 − 3, f ′(1) = −5Decreasing by 5 y-units per x-unit at x = 1

31. f (x) = x1.5 − 6x + 30f ′(x) = 1.5x0.5 − 6, f ′(9) = −1.5Decreasing by 1.5 y-units per x-unit at x = 9

32. f x x x( ) = – + +3 1

f x x f′ = ′ =( ) – + , ( ) – .32

– /1 2 1 2 0 0606K

Decreasing by approximately 0.0607 y-unit perx-unit at x = 2

33. f xx

x x f x x x( ) = ′3

2 2

33 5 2 3– – + , ( ) = – –


High and low points of the f graph are at thex-intercepts of the f ′ graph.

–1 3

x

y

5 f´

f

34. f xx

x x f x x x( ) = – + + , ( ) = – +3

2 2

32 3 9 4 3′

High and low points of the f graph are at thex-intercepts of the f ′ graph.

1 3

15f

f ' x

y

35. If f (x) = k ⋅ g(x), then f ′(x) = k ⋅ g′(x).

Proof:

f xf x h f x

hk g x h k g x

h

kg x h g x

h

kg x h g x

h

h

h

h

h

′ +

= ⋅ + ⋅

= ⋅ +

= ⋅ +

→

→

→

→

( ) = lim( ) – ( )

lim( ) – ( )

lim( ) – ( )

lim( ) – ( )

0

0

0

0

= ⋅ ′k g x( ) , Q.E.D.

Dilating a function f (x) vertically by a constant kresults in the new function g(x) = k ⋅ f (x). Whathas been shown is that

d

dxk f x k

d

dxf x( ( )) ( )⋅ = ⋅

That is, dilating a function vertically by aconstant k dilates the derivative function by aconstant factor of k.

36. If f x x( ) = 5 , then f c c′ =( ) 5 4 .

Proof:

f cf x f c

x c

x c

x c

x c x x c x c xc c

x c

x x c x c xc c

c c c c c

x c

x c

x c

x c

′ =

=

= − + + + +−

= + + + +

=

→

→

→

→

( )

+ + + +

lim( ) – ( )

–

lim–

–

lim( )( )

lim ( )

5 5

4 3 2 2 3 4

4 3 2 2 3 4

4 4 4 4 4

= 5c4, Q.E.D.

37. If f (x) = xn, then f′(x) = nxn− 1.

Proof:

f xx h x

hh

n n

′ +→

( ) = lim( ) –

0

= + + − + + −

→

− −

lim( )

h

n n n n nx nx h n n x h h x

h0

1 12

2 21 L

= + − + +

→

− − −lim ( )h

n n nnx n n x h h0

1 2 11

21 L

= + + + +nxn–1 0 0 0L = ,–nxn 1 which is from the second term in the

binomial expansion of (x + h)n, Q.E.D.

38. If y u u u un n= + + + + ,1 2 3 L where the ui aredifferentiable functions of x, prove that

′ ′ ′ ′ ′y u u u un n= + + + +1 2 3 L for all integers n ≥ 2.

Proof:

Anchor: For n = 2, y u u2 1 2= + .∴ ′ ′ ′y u u2 1 2= + by the derivative of a sum of thetwo functions property, thus anchoring theinduction.Induction hypothesis:Suppose that for n = k > 2,

′ ′ ′ ′ ′y u u u uk k= + + + +1 2 3 L .Verification for n = k + 1:Let y u u u u uk k k+ + .1 1 2 3 1= + + + + +LThen y u u u u uk k k+ += ( + + + + ) +1 1 2 3 1L , whichis a sum of two terms.∴ ′ ′ ′y u u u u uk k k+ += ( + + + + ) +1 1 2 3 1L ,which, by the anchor,= ( + + + +′ ′ ′ ′ + ′+u u u u uk k1 2 3 1L )= ′ + ′ + ′ + + ′ + ′+u u u u uk k1 2 3 1L ,which completes the induction.

Conclusion:∴ ′ ′ ′ ′ ′y u u u un n= + + + +1 2 3 L for all integersn ≥ 2, Q.E.D.

39. a. f x x x f x x x x′ ⇒( ) = – + ( ) = – +3 10 5 5 52 3 2

b. g(x) = f (x) + 13 is also an answer to part abecause it has the same derivative as f (x). Thederivative of a constant is zero.

c. The name antiderivative is chosen because itis an inverse operation of taking the derivative.

d.d

dxg x

d

dxf x C[ ( )] [ ( ) ]= + =

d

dxf x

d

dxC

d

dxf x( ) ( )+ =

The word indefinite is used because of theunspecified constant C.

40. a. f ′(x) = 5x4 ⇒ f (x) = x5 + Cf(2) = 23 ⇒ (2)5 + C = 23C = −9∴ f (x) = x5 − 9

b. f ′(x) = 0.12x2 ⇒ f (x) = 0.04x3 + Cf (1) = 500 ⇒ 0.04(1)3 + C = 500


C = 499.96∴ f (x) = 0.04x3 + 499.96

c. ′ = ⇒ = +f x x f x x C( ) ( )3 14

4

f C( ) ( )5 2 5 214

4= ⇒ + =C = −154.25

∴ = − f x x( ) .14

4 154 25

Problem Set 3-5Q1. No values of t Q2. dy dx x/ = 10

Q3. ′ = − −y x51 4 Q4. ′ =f x x( ) . .1 7 0 7

Q5. ( / )( )d dx x3 5 3+ = Q6. f ( )3 45=

Q7. f ′( ) =3 30 Q8. 45

Q9. ε Q10. C

1. y t t t= – +.5 3 74 2 4

vdy

dtt t= = – . + ,.20 7 2 73 1 4

adv

dtt t= = – . .60 10 082 0 4

2. y t t= . ––0 3 54

vdy

dtt a

dv

dtt= = – . – , = =– –1 2 5 65 6

3. x = −t3 + 13t2 − 35t + 27. The object starts outat x = 27 ft when t = 0 s. It moves to the left tox ≈ 0.15 ft when t ≈ 1.7 s. It turns there andgoes to the right to x = 70 ft when t = 7 s. Itturns there and speeds up, going to the left for allhigher values of t.

10

x

y

Starts at t = 0, x = 27

Turns at t = 1.7, x = 0.15

Turns at t = 7, x = 76

4. x t t t t= – + – +4 3 211 38 48 50. The object startsat x = 50 ft when t = 0 s. It moves to the leftto x ≈ 30 ft when t ≈ 1.0 s. Then it moves to theright to x ≈ 34.8 ft when t ≈ 2.4 s. The objectmoves to the left again, turning at x ≈ 9.4 ftwhen t ≈ 4.8 s and then moving back to the rightfor higher values of t.

Starts at t ≈ 0, x ≈ 50

x

y

10

Turns at t ≈ 4.8, x ≈ 9.4

Turns at t ≈ 2.4, x ≈ 34.8

Turns at t ≈ 1.0, x ≈ 30.0

5. a. x t t t= – + – +3 213 35 27 (See Problem 3.)v t t a t= – + – , = – +3 26 35 6 262

b. v( ) = – + – = –1 3 26 35 12So x is decreasing at 12 ft/s at t = 1.a( ) = – + =1 6 26 20So the object is slowing down at 20 (ft/s)/sbecause the velocity and acceleration are inopposite directions when t = 1.v( ) ( ) ( )6 3 6 26 6 35 132= − + − =So x is increasing at 13 ft/s at t = 6.a( ) ( )6 6 6 26 10= − + = −So the object is slowing down at 10 (ft/s)/sbecause the velocity and acceleration are inopposite directions when t = 6.v( ) = – ( ) + ( ) – = –8 3 8 26 8 35 192

So x is decreasing at 19 ft/s at t = 8.a( ) = – ( ) + = –1 6 8 26 22So the object is speeding up at 22 (ft/s)/sbecause the velocity and acceleration are in thesame directions when t = 8.

c. At t = 7, x has a relative maximum becausev( ) =7 0 at that point and is positive justbefore t = 7 and negative just after. No, x isnever negative for t in [0, 9]. It starts out at27 ft, decreases to just above zero aroundt = 1.7, and does not become negative untilsome time between t = 9.6 and 9.7.

6. a. x t t t t= – + – +4 3 211 38 48 50 (See Problem 4.)v t t t a t t= – + – , = – +4 33 76 48 12 66 763 2 2

b. At t = 1, v( ) = –1 1 and a( ) =1 22. At t = 3,v( ) = –3 9 and a( ) = –3 14 . At t = 5, v( ) =5 7and a( ) =5 46 . The object is slowing down att = 1 because the velocity and acceleration arein opposite directions. The object is speedingup at t = 3 and t = 5 because velocity andacceleration are in the same direction.

c. v = 0 when t = 1.0475… , 2.3708… ,or 4.8315… .

d. The displacement is at a maximum or aminimum whenever v = 0.

1 2 3 4 5

10 t

y

x

v

e. a = 0 when t = 1.6413… or 3.8586… .When a = 0, v is at a maximum or minimum


point and the graph of x is at its steepest fortimes around these values.

1 2 3 4 5

10 t

y

x

va

7. a.

300

10

t

y

d

d´

b. v = d′ = 30 − 2t. Velocity is positive for0 ≤ t < 15. Calvin is going up the hill forthe first 15 s.

c. At 15 seconds his car stopped. d(15) = 324,so distance is 324 feet.

d. 99 30 0 33 3 0 332+ – = ( – )( + ) = =t t t t t⇒ ⇒or t = −3. He’ll be back at the bottom whent = 33 s.

e. d( ) =0 99. The car runs out of gas 99 ft fromthe bottom.

8. a.

30

200

v = 251y

ta = v'

v

b. Trace the v′ graph to find a( )0 32≈ . Theacceleration decreases because the velocity isapproaching a constant. In the real world, thisoccurs because the wind resistance increases asthe velocity increases.

c. The limit is 251 ft/s as t approaches infinity.The term 0.88t approaches zero as t gets verylarge, leaving only 1 inside the parentheses.

d. 90% of terminal velocity is 0.9(251) =225.9 ft/s.Algebraic solution:251 1 0 88 225 9 0 88 0 1( – . ) = . – . = – .t t⇒

t = = ≈log .

log .

0 1

0 8818 012394 18 0. ... . s

Numerical solution gives the same answer.Graphical solution: Trace to v(t) = 225.9.T is between 18 and 18.5.

e. Find the numerical derivative.v′(18.0123…) ≈ 3.2086… , which isapproximately 10% of the initial acceleration.

9. a. d t t t d t t( ) = – . ( ) = – .18 4 9 18 9 82 ⇒ ′d′( ) = – . = .1 18 9 8 8 2d′ ⋅( ) = – . = – .3 18 9 8 3 11 4d′ is called velocity in physics.

b. At t = 1 the football is going up at 8.2 m/s.At t = 3 the football is going down at11.4 m/s. The ball is going up when thederivative is positive and coming down whenthe derivative is negative. The ball is goingup when the graph slopes up and comingdown when the graph slopes down.

c. d′( ) = – .4 21 2, which suggests that theball is going down at 21.2 m/s. However,d( ) = – .4 6 4, which reveals that the ballhas gone underground. The function givesmeaningful answers in the real world only ifthe domain of t is restricted to values thatmake d(t) nonnegative.

10. a.

5

2

t

y

v

a

b. The acceleration at the bottom of the swingis 0. The acceleration is greatest at either endof its swing.

11.

5

2

t

y

a

v

12. v(t) = 15t0.6 . Because v t x t( ) = ( )′ , x(t) musthave had t1 6. in it. The derivative of t1.6 can beassumed to be 1.6t0.6 . So the coefficient of t1 6.

must be 15 1 6/ . , or 9.375. But x(0) was 50.Thus, x(t) = 9.375t1.6 + 50. The derivative x′(t)really does equal v(t). Using this equation,x( ) = . ( ) + = . ..10 9 375 10 50 423 2251 6 KSo the distance traveled is 423.225… − 50 =373.225… , or about 373 ft.

13. The average rate is defined to be the change inthe dependent variable divided by the change inthe independent variable (such as total distancedivided by total time). Thus, the difference quotient is an average rate. The instantaneousrate is the limit of this average rate as the changein the independent variable approaches zero.


14. a.

10

y

t

3

v

b. y is a relative maximum when t ≈ 0, 4,8, … .y is a relative minimum when t ≈ 2, 6, 10, … .

c. The velocity is a relative maximum whent ≈ 3 or 7. The displacement graph at thesetimes appears to be increasing the fastest.

d. The equation used in the text is

y tt= +2 0 852

. cosπ

The student could observe that the period is 4,leading to the coefficient π/2. The amplitudedecreases in a way that suggests an exponentialfunction with base close to, but less than, 1.The additive 2 raises the graph up two units,as can be ascertained by the fact that the graphseems to converge to 2 as t gets larger. Thenumerical derivative of the function shown inpart a agrees with the graph of the velocity.Note that the actual maximum and minimumvalues occur slightly before the values of tread from the graph in part a. For instance,the maximum near t = 4 is actually att = 3.9343… .

15. y xdy

dxx

d y

dxx= ⇒ = ⇒ =5 15 303 2

2

2

16. y xdy

dxx

d y

dxx= ⇒ = ⇒ =7 28 844 3

2

22

17. y x xdy

dxx x= + ⇒ = + ⇒9 18 52 5 4

d y

dxx

2

2318 20= +

18. y x x= − + ⇒10 15 422

dy

dxx

d y

dx= − ⇒ =20 15 20

2

2

19. m′( ) = .5 153 4979K

m′( ) = .10 247 2100K

These numbers represent the instantaneous rate ofchange of the amount of money in the account.The second quantity is larger because the moneygrows at a rate proportional to the amount ofmoney in the account. Because there is moremoney after 10 years, the rate of increase shouldalso be larger.

′′ =m ( ) .5 14 6299K

′′ =m ( ) .10 23 5616KBoth quantities are in the units ($/yr)yr.The quantities represent the instantaneous rate ofchange of the instantaneous rate of change of theamount of money in the account. For example,at t = 5, the rate of increase of the account(153.50 $/yr) is increasing at a rate of14.63 ($/yr)/yr.

20. ′ = −p ( ) .7 4 9510K

′ = −p ( ) .14 2 4755K∴ ′ = ′p p( ) ( )14 71

2

The fact that these derivatives are negative tellsus that the amount of nitrogen 17 is decreasing.

′′ =p ( ) .7 0 4902K

′′ =p ( ) .14 0 2451KBoth quantities are in units (% of nitrogen17/s)/s. The quantities represent the rate ofchange of the rate of change of the percentage ofnitrogen 17 remaining. For example, at t = 7 s,the rate of decrease (−4.95%/s) is changing at arate of 0.49 (%/s)/s.

21.

5

x

y

2f´f

The graph of the derivative looks like the graphof y = cos (x).

22.

5

x

y

2

f' f

The graph of the derivative looks like the graphof y = −sin (x).

Problem Set 3-61.

y

x

3

10


2. The graph confirms the conjecture.

5

x

y

2y1

y2 y 3and

3. g(x) = sin 3xConjecture: g′(x) = 3 cos 3xThe graph confirms the conjecture.

3

10

x

y

g

g´

4. h x x( ) = sin 2

Conjecture: h x x x′( ) = 2 2cosThe graph confirms the conjecture.

5

x

y

1 hh'

5. t x x( ) = .sin 0 7

Conjecture: t x x x′( ) = . – . .0 7 0 3 0 7cosThe graph confirms the conjecture.

10

1x

y

t

t´

6. f (x) = sin [g(x)]f is a composite function.g is the inside function.sine is the outside function.Differentiate the outside function with respect tothe inside function. Then multiply the answer bythe derivative of the inside function with respectto x.

7. a. f (x) = sin 3x. Inside: 3x. Outside: sine.

b. h(x) = sin3 x. Inside: sine. Outside: cube.

c. g(x) = sin x3. Inside: cube. Outside: sine.

d. r(x) = 2cos x. Inside: cosine. Outside:exponential.

e. q(x) = 1/(tan x). Inside: tangent. Outside:reciprocal.

f. L(x) = log (sec x). Inside: secant. Outside:logarithm.


Problem Set 3-7

Q1 exists

Q2 exists

Q3

Any order is acceptable.

. ( )

. lim ( )

. lim ( ) ( )

( )

f c

f x

f x f cx c

x c

→

→=

Q4. No (not continuous) Q5. dy dx x/ = – /16 1 5

Q6. f (x) = −10x− 3 Q7. Antiderivative

Q8.y = sin x

x

Q9.y = cos x

x

Q10. C

1. a. Let y = f (u), u = g(x).dy

dx

dy

du

du

dx= ⋅

b. y f g x g x′ ′ ⋅ ′= ( )] ( )[

c. To differentiate a composite function,differentiate the outside function with respect tothe inside function, then multiply by the deriva-tive of the inside function with respect to x.

2. f x x( ) ( )= −2 31

a. f ′(x) = 3(x2 − 1)2(2x) = 6x(x2 − 1)2

b. (x2 − 1)3 = x6 − 3x4 + 3x2 − 1,

so f ′(x) = 6x5 − 12x3 + 6x.

c. From part a, f ′(x) = 6x(x2 − 1)2 =6x(x4 − 2x2 + 1) = 6x5 − 12x3 + 6x, so thetwo answers are equivalent.

3. f x x f x x x( ) ( )= ⇒ ′ = − ⋅ = −cos sin sin3 3 3 3 3

4. f x x f x x( ) ( )= ⇒ ′ =sin cos5 5 5

5. g x x g x x x( ) ( ) ( ) ( )= ⇒ ′ = −cos sin3 2 33

6. h x x h x x x( ) ( ) ( ) ( )= ⇒ ′ =sin cos5 4 55

7. y = (cos x)3 ⇒y′ = 3(cos x)2 ⋅ (−sin x) = −3 cos2

x sin x

8. f x x( ) ( )= ⇒sin 5

f x x x x x′ = ⋅ =( ) ( )5 54 4sin cos sin cos

9. y x y x x= ⇒ ′ =sin sin cos6 56

10. f x x( ) = ⇒cos7

f x x x x x′ = ⋅ − =( ) ( ) –7 76 6cos sin cos sin


11. y = −6 sin 3x ⇒ y′ = −18 cos 3x

12. f (x) = 4 cos (−5x) ⇒f′ (x) = 4[−sin (−5x)] ⋅ (−5) = 20 sin (−5x)

13.d

dxx x x(cos ) cos ( sin )4 37 4 7 7 7= ⋅ − ⋅

= −28 cos3 7x sin 7x

14.d

dxx x x(sin ) sin cos9 813 9 13 13 13= ⋅

= 117 sin8 13x cos 13x

15. f (x) = 24 sin5/3 4x ⇒f′ (x) = 40 sin2/3 4x ⋅ cos 4x ⋅ 4

= 160 sin2/3 4x cos 4x

16. f (x) = −100 sin6/5 (−9x) ⇒f′ (x) = −120 sin1/5 (−9x) ⋅ cos (−9x) ⋅ (−9)

= 1080 sin1/5 (−9x) cos (−9x)

17. f (x) = (5x + 3)7 ⇒f′ (x) = 7(5x + 3)6

⋅ 5 = 35(5x + 3)6

18. f (x) = (x2 + 8)9 ⇒f′ (x) = 9(x2 + 8)8

⋅ 2x = 18x(x2 + 8)8

19. y = (4x3 − 7)− 6 ⇒y′ = −6(4x3 − 7)− 7

⋅ 12x2 = −72x2(4x3 − 7)− 7

20. y = (x2 + 3x − 7)− 5 ⇒y′ = −5(x2 + 3x − 7)− 6

⋅ (2x + 3)= −5(2x + 3)(x2 + 3x − 7)− 6

21. y = [cos (x2 + 3)]100 ⇒y′ = 100 [cos (x2 + 3)]99 ⋅ [−sin (x2 + 3)] ⋅ 2x

= −200x cos99(x2 + 3) sin (x2 + 3)

22. y = [cos (5x + 3)4]5 ⇒ y′ = 5[cos (5x + 3)4]4 ⋅[−sin (5x + 3)4] ⋅ 4(5x + 3)3

⋅ 5 =−100(5x + 3)3 cos4 (5x + 3)4 sin (5x + 3)4

23. y = 4 cos 5x ⇒ dy

dx = 4(−sin 5x)5 = −20 sin 5x ⇒

d y

dx

2

2 = −20(cos 5x)5 = −100 cos 5x

24. y = 7 sin (2x + 5) ⇒dy

dx = 7 cos (2x + 5)(2) = 14 cos (2x + 5) ⇒

d y

dx

2

2 = 14[−sin (2x + 5)](2) =

−28 sin (2x + 5)

25. f′ (x) = cos 5x ⇒ f (x) = 1

5sin 5x + C

26. f′ (x) = 10 sin 2x ⇒ f (x) = −5 cos 2x + C

27. f (x) = 5 cos 0.2xf′ (x) = −5 sin 0.2x ⋅ 0.2 = −sin 0.2xf′ (3) = −sin 0.6 = −0.5646… andf (3) = 4.1266…

The line has the equationy = −0.5646…x + 5.8205… .

The line is tangent to the graph.

3

5y

x

28. y = 7 sin π t + 12t1.2

velocity = dy

dt= 7π cos π t + 14.4t0.2

Yes, there are times when the beanstalk isshrinking. The velocity graph is negative forbrief intervals, and the y-graph is decreasing inthese intervals.

10

50

t

y and v

y

v

29. a. V rdV

drr= ⇒ =4

343 2π π

dV/dr is in (cm3/cm), or cm2.

b. r = 6t + 10

c.dr

dt= 6 (not surprising!). Units: cm/min

d.dV

dt

dV

dr

dr

dt= ⋅

When t = 5, r = 40. So

dV

dr= =4 40 64002π π( ) .

∴ = ⋅ =dV

dt6400 6 38 400 3π π, min. cm /

dV/dr has units cm2, and dr/dt has units

cm/min, so dV/dt has units cmcm2 ⋅min

,

which becomes cm3/min, Q.E.D. Thismatches the commonsense answer that rate of

change of volume is volume

time

cm

min

3

= .

e. V t= +4

36 10 3π

( )

∴ = + = +dV

dtt t4 6 10 6 24 6 102 2π π( ) ( ) ( )

When t = 5, dV

dt= + =24 6 5 10 38 4002π π[ ] ,( ) .


30. a.

∆u

∆xx

u

∆u does not approach zero as ∆x approacheszero from the left side. (∆u does approach zeroas ∆x approaches zero from the left side.)

b.

∆u

∆ xx

u

∆u does approach zero as ∆x approaches zerofrom either side.

Problem Set 3-8Q1. f′(x) = 9x8 Q2. dy/dx = −3 sin x

Q3. y′ = 72x5 (5x6 + 11)1.4 Q4. s′ = 0

Q5. 12 Q6. 1

Q7. Yes (continuous) Q8. f (x) = −cos x2

Q9. Q10. E

4

4 y

x

f'

1. a.

15

Increasingx

d(x)

25

5

3

45

23

y(t) = C + A cos B(t − D)Vertical displacement = 25 = CAmplitude = 0.5(40) = 20 = APhase disp. (for cosine) = 3 = DPeriod = 60/3 = 20, so B = 2π/20 = π/10.(Note that B is the angular velocity in radiansper second.)

y t t( ) = +25 2010

3cos ( – )π

b. ′ =y t t( ) – 210

3π πsin ( – )

c.

′ = =y ( ) – 3.6931615 210

15 3π πsin ( – ) K

y(t) is increasing at about 3.7 ft/s.

d. The fastest that y(t) changes is 2π, or6.28… ft/s. The seat is at y(t) = 25 ft abovethe ground then.

2. a. y = C + A cos B(x − D).B = 2π /6 = π /3 rad/sD = phase displacement = 1.3 sA = 0.5(110 − 50) = 30 cmC = 110 − 30 or 50 + 30, which equals 80 cm.

∴ = +d t80 303

1 3cos ( – . )π

b. d t′ = –103

1 3π πsin ( – . )

c. d′ = − =( ) 5 10

35 1 3 21 02135π π

sin ( – . ) . K

d′ = =( ) –11 10

311 1 3 21 02135π π

sin ( – . ) . K

At both times, the pendulum is moving awayfrom the wall at about 21.0 cm/s. Theanswers are the same because the times areexactly one period apart.

d. d′ = = −( ) –20 10

320 1 3 21 02135π π

sin ( – . ) . K

The pendulum is moving toward the wall.Because the derivative is negative, d isdecreasing, which in this problem impliesmotion toward the wall.

e. The fastest is 10π ≈ 31.4 cm/s, when d = 80.

f. 0 103

1 33

1 3 0= ⇒ − =– π π πsin ( – . ) sin ( . )t t

⇒ = ⇒ − = +−π π π3

1 3 03

1 3 01( – . ) sin ( . )t t n

⇒ t − 1.3 = 3n ⇒ t = 1.3 + 3n.

The first positive time occurs when n = 0,that is, when t = 1.3 s. When the velocityis zero, the pendulum is at its maximumheight.

3. a. The curb has slope (3.25 − 0.75)/44 = 2.5/44.∴ equation is f (x) = 0.75 + (2.5/44)x.

b. Sinusoid has period 8 ft, so B = 2π /8 = π /4.Amplitude = 0.5(0.75 − 0.25) = 0.25 ft. Lowend of ramp is a low point on the sinusoid.∴ sinusoidal axis is at y = 0.25 when x = 0and goes up with slope 2.5/44.Sinusoid is at a low point when x = 0. Sophase displacement is zero if the cosine issubtracted.


∴ equation is

g x x x( ) .= + −0 252 5

440 25

4

.. cos

π

(There are other correct forms.)

c. ′ = +g x x( )2 5

44 16 4

.sin

π π

′ = + =g ( ) ft/ft9

2 5

44 16 49 0 1956

.sin ( ) .

π πK

Going up at about 0.2 vertical ft perhorizontal ft

′ = + = −g (15) ft/ft

2 5

44 16 415 0 0820

.sin ( ) .

π πK

Going down at about 0.08 vertical ft perhorizontal ft. A positive derivative impliesg(x) is getting larger and thus the child isgoing up. A negative derivative implies g(x)is getting smaller and thus the child is goingdown.

d. By tracing the ′g graph, maximum value ofg′ (x) is 0.2531… ft/ft (about 14.2° up).Minimum is −0.1395… ft/ft (about 7.9°down).

4. a. Let d = day number and L(d ) = number ofminutes.14 hours 3 minutes is 843 minutes. 10 hours15 minutes is 615 minutes.∴ amplitude = (1/2)(843 − 615) = 114 min.Sinusoidal axis is at L(d) = 615 + 114 =729 min.Assuming a 365-day year, B = 2π/365.Phase displacement = 172

∴ + ( ) =L d d729 1142

365172cos ( – )

π

On August 7, d = 219.

L( ) = 219 729 1142

365219 172+ =cos ( – )

π

807.67… , or about 13 hours 28 minutes.

b. L d d′ − −( ) =228

365

2

365172

π πsin ( )

On August 7, d = 219.

′ − =L ( ) =219228

365

2

365219 172

π πsin ( – )

–1.42009…Rate ≈ –1.42 min/day(Decreasing at about 1.42 min/day)

c. The greatest rate occurs when the sine is1 or −1.Rate is 228π/365 ≈ 1.96 min/day.1/4 year is about 91 days. So greatest rateoccurs at day 172 ± 91, which is day 263 orday 81 (September 20 or March 22).

5. In general, the period for a pendulum formed bya weight suspended by a string of negligiblemass is 2π L g/ , where L is the length from the

pivot point to the center of mass (actually, thecenter of percussion) of the weight, and g is the

gravitational acceleration, about 9.8 m/s2.Consequently, if the pendulum is 1 meter long,its period will be 2 1 9 8 2 007π / . . ,= K or about

2 s. This is the period for a complete back-and-forth swing. You must quadruple the length of apendulum to double its period. A pendulum hungfrom the ceiling will have a period slow enoughto measure fairly precisely. A good way to getmore accuracy is to count the total time for tenswings, then divide by 10. The period is roughlyconstant for any (moderate) amplitude, as long asthe amplitude is not too big. This fact is notobvious to the uninitiated student and is worthspending time showing. It is quite dramatic towatch a pendulum take just as long to make tenswings with amplitude 2 cm as it does withamplitude 20 or 30 cm.

6. The following data were computed from actualsunrise and sunset times for San Antonio foreach ten days. You can get similar informationfor your locality from the local weather bureau ornewspaper office, from the Nautical AlmanacOffice, U.S. Naval Observatory, Washington,D.C., 20390, or from the Internet.

Day Min Day Min Day Min

0 617 130 811 260 738

10 623 140 823 270 720

20 632 150 833 280 703

30 645 160 840 290 686

40 660 170 842 300 669

50 676 180 842 310 653

60 693 190 836 320 639

70 711 200 828 330 628

80 729 210 816 340 620

90 747 220 803 350 615

100 764 230 789 360 615

110 780 240 772

120 797 250 755

100 200 300

d

L(d )

800

700

0

600

The graph shows a good fit to the data. But thereis a noticeable deviation in the fall and winter,here the day is slightly longer than predicted.The main reason for the discrepancy, apparently,


is the fact that in the fall and winter, Earth iscloser to the Sun and hence moves slightly morerapidly through its angle with the Sun thanduring the spring and summer.

7. a.

f

hg

1

4

y

x

The limits are all equal to 4.

b. f (x) < g(x) and lim ( ) lim ( )x x

f x g x→ →

= =1 1

4

f (x) ≤ h (x) ≤ g(x)

c.

x f(x) h(x) g(x)

0.95 3.795 3.8 3.805

0.96 3.8368 3.84 3.8432

0.97 3.8782 3.88 3.8818

0.98 3.9192 3.92 3.9208

0.99 3.9598 3.96 3.9602

1.00 4 4 4

1.01 4.0398 4.04 4.0402

1.02 4.0792 4.08 4.0808

1.03 4.1182 4.12 4.1218

1.04 4.1568 4.16 4.1632

1.05 4.195 4.2 4.205

d. From the table, if x is within 0.02 unit of 2,then both f (x) and g(x) are within 0.1 unit of4. From the table, δ = 0.01 or 0.02 willwork, but 0.03 is too large. All the values ofh(x) are between the corresponding values off (x) and g(x), and the three functions allapproach 4 as a limit.

8. Prove that limsin

.x

x

x→=

00 See the text proof.

9. a. The numbers are correct.

b.

x (sin x)/x

0.05 0.99958338541…

0.04 0.99973335466…

0.03 0.99985000674…

0.02 0.99993333466…

0.01 0.99998333341…

Values are getting closer to 1.

c. Answers will vary according to calculator. Forthe TI-83 in TABLE mode, starting x at 0 andusing ∆ x = 10−7 shows that all values roundto 1 until x reaches 1.8 × 10−6, whichregisters as 0.999999999999.

d. Answer will depend on calculator. For TI-83in TABLE mode, (sin 0.001)/0.001 is0.999999833333, which agrees exactly withthe value published by NBS to 12 places.

e. If students have studied Taylor series(Chapter 12) before taking this course, theywill be able to see the reason. The Taylorseries for sin 0.001 is

0 001

0 001

3

0 001

5

0 001

7

3 5 7

..

!

.

!

.

!− + − + L

= 0.00100 00000 00000 00000 000…

− 0.00000 00001 66666 66666 666…

+ 0.00000 00000 00000 00833 333…_______________________________

= 0.00099 99998 33333 34166 666…

10. See the text proof.

11. See the text proof.

12. a. See the text statement of the theorem.

b. Proof:

Given any ε > 0, there is a δ f > 0 such that0 < |x − c | < δ f ⇒ |f (x) − L | < ε, becauselim ( ) .x c

f x L→

= Similarly, there is a δ g > 0

such that 0 < |x − c | < δ g ⇒ |g(x) − L| < ε.Let δ be the smaller of δ f and δ g. Then 0 <|x − c | < δ ⇒ 0 < |x − c | < δ f = |f (x) − L| <ε ⇒ f (x) < L + ε , and also 0 < |x − c | <δ ⇒ 0 < |x − c | < δg ⇒ |g(x) − L| < ε ⇒L − ε < g(x). Then L − ε < g(x) < h(x) < f(x)< L + ε, so |h(x) − L | < ε, so lim ( )

x ch x L

→= .

Q.E.D.

c. See Figure 3-8c or 3-8d.

13. a. The limit seems to equal 2.

b.

2

1g

g

h

h

c. See the graph in part b. The lines haveequations g(x) = x + 1 and h(x) = 3 − x.

d. Prove that limx

y→1

2= .


Proof:

lim( )x

x→

+ = + =1

1 1 1 2

lim( )x

x→

− = − =1

3 3 1 2

For x < 1, g(x) ≤ y ≤ h(x).∴ the squeeze theorem applies, and lim

xy

→ −12= .

For x > 1, h(x) ≤ y ≤ g(x).∴ the squeeze theorem applies, and lim

xy

→ +12= .

Because both left- and right-hand limits equal 2, lim

xy

→12= , Q.E.D.

e. The word envelope (a noun) is used becausethe small window formed by the two lines“envelops” (a verb) the graph of the function.

f. As |x| becomes large, (x − 1) · sin–

1

1x=

sin [ /( – )]

/( – )

1 1

1 1

x

x takes on the form

sin (argument)

(argument) as the argument approaches

zero. Thus the limit is 1 and y approaches2 + 1, which equals 3.


Problem Set 3-9Q1. 1 Q2. –sin x

Q3.d

dxx

d

dxx x

2

2 (cos ) ( sin ) cos= − = −

Q4. y = sin x + C

Q5. x8 Q6. x48

Q7. log 32 = 5 log 2 Q8.

x

y

Q9. Cube function Q10. C

1. M(x) = 1000e0.06 x

a. M′(x) = 1000(0.06)e0.06 x = 60e0.06 x

M′(1) = 63.7101… $/yrM′(10) = 109.3271… $/yrM′(20) = 199.2070… $/yr

b. M(0) = $1000M(1) = $1061.84M(2) = $1127.50M(3) = $1197.22Increase from year 0 to year 1: $61.84Increase from year 1 to year 2: $65.66Increase from year 2 to year 3: $69.72No, the amount of money in the account doesnot change by the same amount each year.

c. 1061.84/1000 = 1.06184, so the APR for0 to 1 year is approximately 6.184%.1127.50/1061.84 = 1.061836… , so the APRfor 1 to 2 years is approximately 6.184%.1197.22/1127.50 = 1.06183… , so the APRfor 2 to 3 years is approximately 6.184%.The APR is higher than the instantaneousrate. Savings institutions may prefer toadvertise the APR instead of the instantaneousrate because the APR is higher.

2. a. f (t) = 10e− 0.34 t ⇒f ′ (t) = 10(−0.34)e− 0.34 t = −3.4e− 0.34 t

f ′ (0) = −3.4f ′ (2) = −1.7224…f ′ (4) = −0.8726…f ′ (6) = −0.4420Factor of change from 0 to 2:−1.7224…/−3.4 = 0.5066Factor of change from 2 to 4:−0.8726…/−1.7224… = 0.5066Factor of change from 4 to 6:−0.4420…/−0.8726… = 0.5066

b. f (0) = 10f (2) = 5.0661…f (4) = 2.5666…f (6) = 1.3002…Factor of change from 0 to 2:5.0661…/10 = 0.5066Factor of change from 2 to 4:2.5666…/5.0661… = 0.5066Factor of change from 4 to 6:1.3002…/2.5666… = 0.5066The factors of change are the same in part aand part b.

c.

5

10

t

y

f

f '

The values of f are negative because theamount of 18-F is decreasing as time goes on.

3. a. A p p A pp

( ) . = − ⇒ ′ = −63 23 5

23 5ln ( )

.

10

50

x

y

A

A´

b. If the pressure is increasing, then the altitudeis decreasing. A′(10) = −2.35, so the altitudeis changing at −2.35 thousand feet/psi. That


is, a change of 1 psi would indicate that thealtitude had decreased by 2.35 thousand feet.The negative sign means that the altitude isdecreasing.

c. ′ = − =A ( ) . .5 4 7 4 7

′ = − =A ( ) .10 2 35 2.35

This shows that the altitude is changing fasterat 5 psi than it is at 10 psi.

d. A(p) = 063 − 23.5 ln p = 0−23.5 ln p = −63ln p = 2.6808…p = 14.5975…The fact that A(p) is negative for all values ofp greater than 14.5975… means that if the airpressure is above 14.5975 psi, then the planemust be beneath sea level.

4. x = 3000e0.05 y

a. ln x = ln (3000e0.05 y)ln x = ln 3000 + ln e0.05 y

ln x − ln 3000 = 0.05y

y x= −( )1

0 053000

.ln ln

y = 20 ln x – 20 ln 3000

b. y(3000) = 20 ln 3000 − 20 ln 3000 = 0y(4000) = 20 ln 4000 − 20 ln 3000 =5.7536…y(5000) = 20 ln 5000 − 20 ln 3000 =10.2165…y(6000) = 20 ln 6000 − 20 ln 3000 =13.8629…Number of years to get from $3000 to $4000:5.7536…Number of years to get from $4000 to $5000:4.4628…Number of years to get from $5000 to $6000:3.6464…The time intervals decrease as the amount ofmoney increases because when there is moremoney in the account, it takes less time toearn the given amount of interest.

c. y x yx

= − ⇒ ′ =20 20 300020

ln ln yr/$

y′(3000) = 0.0066…y′(4000) = 0.005y′(5000) = 0.004y′(6000) = 0.0033…This shows that the number of years it takesto earn each dollar decreases as the amount ofmoney increases.

5. f (x) = 5e3x ⇒ f ′ (x) = 15e3x

6. f (x) = 7e− 6x ⇒ f ′ (x) = −42e− 6x

7. g(x) = −4ecos x ⇒g′(x) = −4(ecos x)(−sin x) = 4(sin x) ecos x

8. h(x) = 8e− sin x ⇒h ′ (x) = 8e− sin x (−cos x) = −8(cos x)e− sin x

9. y = 2 sin (e4x) ⇒ y′ = 2 cos (e4x) 4e4x = 8e4x cos (e4x)

10. y = 6 cos (e− 0.5 x) ⇒y′ = 6[−sin (e− 0.5 x)](−0.5) e− 0.5 x = 3e− 0.5 x sin (e− 0.5 x)

11. f x x f xx x

( ) ln ( ) ( )= ⇒ ′ = ⋅ =10 710

77

10

12. g x x g xx x

( ) ln ( )= ⇒ ′ = ⋅ =9 49

44

9

13. T x Tx

xx

= ⇒ ′ = =1818

3543

32ln ( )

14. P x Px

xx

= ⇒ ′ = =−10001000

0 77000 7

0 70 3ln ..

..

15. y = 3 ln (cos 5x) ⇒

′ = − = −yx

x x3

55 5 15 5

cos( sin ) tan

16. y = 11 ln (sin 0.2x) ⇒

′ = =yx

x x11

0 20 2 0 2 2 2 0 2

sin .(cos . ) . . cot .

17. u = 6 ln (sin x0.5 ) ⇒

′ = =− −ux

x x x x6

0 5 30 50 5 0 5 0 5 0 5

sin(cos ) . cot.

. . . .

18. v x= ⇒0 09 8. ln (cos )

′ = −vx

x x0 09

888 7.

cos( sin ) = −0.72x7 tan x8

19. r x e r xe

exx

x( ) ln ( )= ⇒ ′ = ⋅ =11

Not surprising because we could have first usedthe fact that natural log and exp are inverses:r (x) = ln ex = x ⇒ r ′ (x) = 1

20. c(x) = eln x = x ⇒ c ′ (x) = 1c ′ (2) = 1, c ′ (3) = 1, c ′ (4) = 1

21. f (x) = 3x ⇒ f ′ ( x ) = ( ln 3 ) 3 x

22. g (x) = 0.007x ⇒ g′(x) = (ln 0.007) 0.007x

23. y = 1.6cos x ⇒ y′ = (ln 1.6)1.6cos x (−sin x) =−ln 1.6 sin x (1.6cos x)

24. y = sin 5x ⇒ y′ = cos 5x (ln 5)5x

25. y xdy

dx xx

xx= ⇒ = ⋅ = = ⇒−ln 5

54 11

55

5

d y

dxx

x

2

22

255= − = −−

26. y edy

dxe

d y

dxex x x= ⇒ = ⇒ =7 7

2

277 49

27. y = e− 0.7 x ⇒ y′ = −0.7e− 0.7 x ⇒ y″ = 0.49e− 0.7 x


28. y x yx x

x= ⇒ ′ = ⋅ = = ⇒−ln 81

88

1 1

′′ = − = −−y xx

1122

29. f ′ (x) = 12e2x ⇒ f ( x ) = 6 e 2 x + C

30. y′ = 5x ln 5 ⇒ y = 5x + C



R1. a.

x Average rate of change from 2 to x

1.97f f( ) ( . )

.

2 1 97

0 0311 82

− = .

1.98f f( ) ( . )

.

2 1 98

0 0211 88

− = .

1.99f f( ) ( . )

.

2 1 99

0 0111 94

− = .

2.01f f( . ) ( )

.

2 01 2

0 0112 06

− = .

2.02f f( . ) ( )

.

2 02 2

0 0212 12

− = .

2.03f f( . ) ( )

.

2 03 2

0 0312 18

− = .

The derivative of f at x = 2 is approximately 12.

b. r xf x f

x( )

( ) ( )= −−

2

2r(2) is of the form 0

0 .lim ( )x

r x→2

appears to be 12.

c. r xx

x

x x x

x( )

( )( )

( )= −

−= − + +

−

3 28

2

2 2 4

2= + + ≠x x x2 2 4 2,

lim ( ) limx x

r x x x→ →

= + +2 2

2 2 4 because x ≠ 2.

∴ =→

lim ( )x

r x2

12

d. The answers in parts a, b, and c are the same.

R2. a. f cf x f c

x cx c′ −

→( ) = lim

( ) ( )

–b. f (x) = 0.4x2 − x + 5

′ = − + −→

fx x

xx( ) lim

. .

–3

0 4 5 5 6

33

2

= +→

lim( – )( . . )

–x

x x

x3

3 0 4 0 2

3= +

→lim( . . )x

x3

0 4 0 2 1 4= .

c. m xx x

x( ) =

0 4 0 6

3

2. – .

–

−

3

1

2

x

m(x)

d. Line: y = 1.4x + 1.4

3

4

x

y

e. The line is tangent to the graph.

f. Yes, f does have local linearity at x = 3.Zooming in on the point (3, 5.6) shows thatthe graph looks more and more like the line.

R3. a.

50y1

y2

1 x

y

b. See the graph in part a.

c. The y1 graph has a high point or a low pointat each x-value where the y2 graph is zero.

d.

20

p

p´

t

y

1

Take the numerical derivative at t = 3, 6,and 0.p′(3) ≈ −2.688… . Decreasing at about2.69 psi/h when t = 3.p′(6) ≈ −1.959… . Decreasing at about1.96 psi/h when t = 6.p′(0) ≈ −3.687… . Decreasing at about3.69 psi/h when t = 0.The units are psi/h. The sign of the pressurechange is negative because the pressure isdecreasing. Yes, the rate of pressure change isgetting closer to zero.

R4. a. See the text for the definition of derivative.

b. Differentiate


c. If y = xn, then y′ = nxn− 1.

d. See solution to Problem 35 in Problem Set3-4.

e. See the proof in Section 3-4.

f.dy

dx is pronounced “d y, d x.”

d

dxy( ) is pronounced “d, d x, of y.”

Both mean the derivative of y with respect to x.

g. i. f x x f x x( ) = ( ) =/ /763

59 5 4 5⇒ ′

ii. g x xx

x( ) = –76

742

− − + ⇒

g x xx′ − − −( ) = –283

15

iii. h(x) = 73 ⇒ h′ (x) = 0

h. f ′ = =( ) ./32 32 201 6635

4 5( ) exactly. The

numerical derivative is equal to or very closeto 201.6.

i.

4

4y

f´

f

x

R5. a. vdx

dtx t= ′or ( ).

adv

dtv t a

d x

dtx t= or ( ), = or ( )′ ′′

2

2

b. d y

dx

2

2 means the second derivative of y with

respect to x.

y = 10x4 ⇒ y′ = 40x3 ⇒ y″ = 120x2

c. ′ = ⇒ = +f x x f x x C( ) ( ) .12 33 4 f (x) isthe antiderivative, or the indefinite integral,of f (x).

d. The slope of y f x= ( ) is determined by thevalue of ′f x( ). So the slope of y f x= ( ) atx = 1 is ′ = −f ( ) ,1 1 at x = 5 is ′ =f ( ) ,5 3 andat x = −1 is ′ − =f ( ) .1 0

5

5

x

y f´f

e. i. y = −0.01t3 + 0.9t2 − 25t + 250

vdy

dtt t= = − + −0 03 1 8 252. .

adv

dtt= = − +0 06 1 8. .

ii. a ( 15) = −0.06(15) + 1.8 = 0.9 (km/s)/s

v(15) = −0.03(152) + 1.8(15) − 25= −4.75 km/s

The spaceship is slowing down at t = 15because the velocity and the accelerationhave opposite signs.

iii. v = −0.03t2 + 1.8t − 25 = 0By using the quadratic formula or thesolver feature of your grapher,t = 21.835… or t = 38.164… .The spaceship is stopped at about 21.8and 38.2 seconds.

iv. y = −0.01t3 + 0.9t2 − 25t + 250 = 0By using TRACE or the solver feature of your grapher, t = 50.v(50) = −10Because the spaceship is moving at10 km/s when it reaches the surface, it isa crash landing!

R6. a.

1

1

x

cosinederivative

y

b. The graph of the derivative is the same as thesine graph but inverted in the y-direction.Thus, ( )cos sinx x′ = − is confirmed.

c. −sin 1 = −0.841470984…Numerical derivative ≈ −0.841470984…The two are very close!

d. Composite function

f′(x) = −2x sin (x2)

R7. a. i.dy

dx

dy

du

du

dx= ⋅

ii. f (x) = g(h(x)) ⇒ f ′(x) = g ′(h(x)) ⋅ h ′(x)

iii. The derivative of a composite function isthe derivative of the outside function withrespect to the inside function times thederivative of the inside function withrespect to x.

b. See the derivation in the text. This derivationconstitutes a proof. ∆u must be nonzerothroughout the interval.


c. i. f (x) = (x2 − 4)3

f ′(x) = 3(x2 − 4)2 ⋅ 2x = 6x(x2 − 4)2

ii. f (x) = x6 − 12x4 + 48x2 − 64f′(x) = 6x5 − 48x3 + 96xExpanding the answer to part i givesf′(x) = 6x5 − 48x3 + 96x, which checks.

d. i. f′(x) = −3x2 sin x3

ii. g′(x) = 5 cos 5x

iii. h′(x) = 6 cos5 x (−sin x)= −6 sin x cos5 x

iv. k′(x) = 0

e. f ′(x) = 12 cos 3x ⇒ f ″ (x) =12(−sin 3x)3 = −36 sin 3x f ′(x) = 12 cos 3x ⇒ f (x) = 4 sin 3x + C f (x) is the second derivative of f (x). f (x) is the antiderivative, or indefiniteintegral, of f (x).

f. W = 0.6x3 and dx/dt = 0.4dW

dt

dW

dx

dx

dtx x= = . . .⋅ ⋅ =1 8 0 4 0 722 2

If x = 2, W = 0.6 ⋅ 23 = 4.8 lbdW/dt = 0.72(22) = 2.88The shark is gaining about 2.88 lb/day.If x = 10, W = 0.6 ⋅ 103 = 600 lb.dW/dt = 0.72(102) = 72The shark is gaining about 72 lb/day.The chain rule is used to get dW/dt fromdW/dx by multiplying the latter by dx/dt.

R8. a. limsin

x

x

x→=

01

x (sin x)/x

−0.05 0.99958338541…

−0.04 0.99973335466…

−0.03 0.99985000674…

−0.02 0.99993333466…

–0.01 0.99998333341…

0.00 undefined

0.01 0.99998333341…

0.02 0.99993333466…

0.03 0.99985000674…

0.04 0.99973335466…

0.05 0.99958338541…

The values of (sin x)/x approach 1 as xapproaches 0.

b. See the text for the statement of the squeezetheorem. Squeeze (sin x)/x between cos x andsec x.

c. See the proof in Section 3-8 of the text.

d. cos x = sin (π/2 − x)cos′ x = cos (π/2 − x) (−1)

= −sin x, Q.E.D.

e. d(t) = C + A cos B(t − D)C = 180, A = 20D = 0 for cosine because hand starts at a highpoint.B = 2π/60 = π/30 because period is 60 s.

d t t( ) = +180 2030

cosπ

′ = −d t t( )2

3 30

π πsin

At 2, t = 10: d ′(10) ≈ −1.81 cm/sAt 3, t = 15: d ′(15) ≈ −2.09 cm/sAt 7, t = 35: d ′(35) ≈ 1.05 cm/sAt the 2 and 3, the tip is going down, so thedistance from the floor is decreasing, which isimplied by the negative derivatives. At the 7,the tip is going up, as implied by the positivederivative.

R9. a. p x e p x ex x( ) ( ) ( . ). .= ⇒ ′ = −− −100 100 0 10 1 0 1

= −10e− 0.1 x

p′(0) = −10p′(10) = −3.6787…p′(20) = −1.3533…

The rates are negative because the amount ofmedication in your body is decreasing.To find the biological half-life, find x such that

p x p( ) ( )= =1

20 50

100 500 1e x− =.

e x− =0 1 1

2.

− =

0 1

1

2. lnx

x = −101

2ln

x = 6.9314…The half-life is 6.9314… h.p(2(6.9314…)) = p(13.8629…) =100e− 0.1(13.8629…) = 25

After two half-lives have elapsed, 25% of themedicine remains in your body.

b. i. f (x) = 5e2x ⇒ f ′(x) = 5(2)e2x = 10e2x

ii. ydy

dxx x= ⇒ =7 7 7(ln )

iii. d

dxx

xx x[ln (cos )]

cos( sin ) tan= − = −1

iv. y x xdy

dx xx= = ⇒ = ⋅ = ⇒−ln ln8 18 8

18

d y

dxx

x

2

22

288= − = −−

c. ′ = ⇒ = +f x e f x e Cx x( ) ( )12 43 3


d.

x

y

3

3 y2

y3y1

y1 = ex is the inverse of y2 = ln x, so y1 is areflection of y2 across the line y = x.

Concept Problems

C1. a. f (x) = x7, g(x) = x9. So h(x) = f(x) ⋅ g(x) = x16.

b. h′(x) = 16x15

c. f ′(x) = 7x6, g′(x) = 9x8. So f ′(x) ⋅ g′(x) =63x14 ≠ h ′(x).

d. h′(x) = f ′ (x) ⋅ g(x) + f (x) ⋅ g′(x) =7x6 ⋅ x9 + x7 ⋅ 9x8 = 16x15

C2. a. f xx x

x( ) =

– sin

sin.

2 f (0) has the form 0/0,

which is indeterminate. f is discontinuous atx = 0 because f (0) does not exist.

b. By graph (below) or by TABLE , f (x) seems toapproach −1 as x approaches zero. Define f (0)to be −1.

π

5

x

f(x)

c. Conjecture: The function is differentiable atx = 0. The derivative should equal zerobecause the graph is horizontal at x = 0.

d. ′ =→

ff x f

xh( ) lim

( ) – ( )

–0

0

00

=→

lim

– sin

sin– (– )

x

x x

xx0

21

= +→

lim– sin sin

sinx

x x x

x x0

2

Using TABLE for numerator, denominator, andquotient shows that the numerator goes tozero faster than the denominator. For instance,if x = 0.001,

quotient = ×

×=1 1666 10

9 999 10

9

7

.

.

–

–

K

KK0.00116

Thus, the limit appears to be zero. (The limitcan be found algebraically to equal zero byl’Hospital’s rule after students have studiedSection 6-5.)

Chapter Test

T1. See the definition of derivative in Section 3-2or 3-4.

T2. Prove that if f (x) = 3x4, then f ′ (x) = 12x3.

Proof:

′ = + = + −→ →

f xf x h f x

h

x h x

hh h( ) lim

( ) – ( )lim

( )0 0

4 43 3

= + + + + −→

limh

x x h x h xh h x

h0

4 3 2 2 3 4 43 12 18 12 3 3

= + + + =→

lim ( ) ,h

x x h xh h x0

3 2 2 3 312 18 12 3 12

Q.E.D.

T3. If you zoom in on the point where x = 5, thegraph appears to get closer and closer to thetangent line. The name of this property is locallinearity.

Slope = 2Slope = 2

5

x

y

5

T4. Amos substituted before differentiating insteadof after. Correct solution is f (x) = 7x ⇒f ′ (x) = 7 ⇒ f ′ (5) = 7.

T5. f (x) = (7x + 3)15 ⇒ f ′ (x) = 105(7x + 3)14

T6. g(x) = cos (x5) ⇒ g ′ (x) = −5x4 sin x5

T7. d

dxx

xx x[ln (sin )]

sincos cot= ⋅ =1

T8. y = 36x ⇒ y′ = (ln 3)36x(6) = 6(ln 3)36x

T9. f (x) = cos (sin5 7x) ⇒f ′(x) = −sin (sin5 7x) ⋅ 5 sin4 7x ⋅ cos 7x ⋅ 7

= −35 sin (sin5 7x) sin4 7x cos 7x

T10. y = 60x2/3 − x + 25 ⇒ y′ = 40x− 1/3 − 1

T11. y edy

dxe

d y

dxex x x= ⇒ = ⇒ =9 9

2

299 81

T12. y′ ≈ 0.6 (Function is y = −3 + 1.5x, for whichthe numerical derivative is 0.6081… .)

T13. y = 3 + 5x− 1.6

v(x) = 5(−1.6)x− 2.6 = −8x− 2.6

a(x) = −8(−2.6)x− 3.6 = 20.8x− 3.6

Acceleration is the second derivative of thedisplacement function.

T14. f ′ (x) = 72x5/4 ⇒ f (x) = 32x9/4


T15. f ′ (x) = 5 sin x and f (0) = 13f (x) = −5 cos x + C

13 = −5 cos 0 + C ⇒ C = 18f (x) = −5 cos x + 18

T16. f (x) = cos 3x ⇒ f ′ (x) = −3 sin 3xf ′ (5) = −3 sin 15 = −1.95086…Decreasing at 1.95… y-units per x-unit.

T17. f xx

x( )

sin=

1

1

x

f(x)

As x approaches zero, f (x) approaches 1.The squeeze theorem states:

If (1) g(x) ≤ h(x) for all x in a neighborhood of c,

(2) lim ( ) lim ( ) ,x c x c

g x h x L→ →

= = and (3) f is a

function for which g(x) ≤ f (x) ≤ h(x) for all x inthat neighborhood of c, then lim ( ) .

x cf x L

→=

T18.

h5 1h

h

−

–0.0003 1.6090…

–0.0002 1.6091…

–0.0001 1.6093…

0 undefined

0.0001 1.6095…

0.0002 1.6097…

0.0003 1.6098…

ln 5 = 1.6094… . The table shows that

lim ln .h

h

h→

− =0

5 15

Proof:

d

dx hx

h

x h x

( ) lim55 5

0= −

→

+ Definition of derivative.

= −→

55 1

0

x

h

h

hlim Factor out 5x.

= 5x · ln 5 Evaluate.

T19. v(t) = 251(1 − 0.88t)a(t) = 251[− ln (0.88)] 0.88t = −251(ln 0.88)0.88ta(10) = −251(ln 0.88)(0.88)(10) = 8.9360…

Numerical derivative gives 8.9360… as well.

T20. If the velocity and the acceleration have oppositesigns for a particular value of t, then the object isslowing down at that time.

T21. a. v(t) = t1.5 + 3 ⇒ a(t) = 1.5t0.5

b. d tt

t C( ).

.

=

+ +2 5

2 53

d(1) = 20

1

2 53 1 20

2 5.

.( )+ + =C

3.4 + C = 20C = 16.6∴ d (t) = 0.4t2.5 + 3t + 16.6

c. d (9) − d (1) = 120.8

This represents the displacement between thefirst and ninth seconds.

T22. a. c t t( ) = +300 22

365cos

π ⇒

c t t′ −( ) =4

365

2

365

π πsin

b. c′ = − ⋅

( )273

4

365

2

365273

π πsin

= 0.03442… ppm/day

c. Rate is ( ).

, ,6 10

0 03442

1 000 000

1

24 60 6015× ⋅ ⋅

⋅ ⋅=K

2390.6627… , which is approximately 2390tons per second!



Chapter 4—Products, Quotients, and Parametric Functions

Problem Set 4-11. f (x) = 3 cos x ⇒ f ′ (x) = −3 sin x

g(x) = 2 sin x ⇒ g′(x) = 2 cos x

2.

10

x

p(x)6

p′(2) ≈ −3.9218…p(x) is decreasing at x = 2 because ′p ( ) < .2 0This fact corresponds with the graph, whichslopes steeply in the negative direction at x = 2.f ′ (2) ⋅ g′(2) = (−3 sin 2)(2 cos 2) = 2.2704…So p′ (2) ≠ f ′ (2) ⋅ g′ (2).

3.

x

10

6q(x)

q is the cotangent function.

q ′ (2) = −1.8141…

q x( ) is decreasing at x = 2.

f ′ (2)/g′ (2) = (−3 sin 2)/(2 cos 2) = 3.2775…

So q ′ (2) ≠ f ′ (2)/g′ (2).

4.

3

2x

y

t = 2 here

The geometric figure seems to be an ellipse.

5. See graph in Problem 4.∆x = 3 cos 2.1 − 3 cos 1.9 = −0.54466…∆y = 2 sin 2.1 − 2 sin 1.9 = −0.16618…

dy

dx

y

x≈ ∆

∆= = 0.3051

– .

– .

0 16618

0 54466

K

KK

At , . ,tdy dt

dx dt= = =2

2 2

3 20 3051

/

/

cos

– sinK

which agrees with the difference quotient.

6. You’ll see in Section 4-2 that

p′(x) = f ′ (x)g(x) + f (x)g′(x).∴ p′(2) = (–3 sin 2)(2 sin 2) + (3 cos 2)(2 cos 2) =–3.9218… , which agrees with Problem 2.

You’ll see in Section 4-3 that

′ ′ ′q x

f x g x f x g x

g x( ) =

( ) ( ) – ( ) ( )

[ ( )]2

∴ ′ = =q ( )23 2 2 2 3 2 2 2

2 2 2

(– sin )( sin ) – ( cos )( cos )

( sin )

−1.8141… , which agrees with Problem 3.

Problem Set 4-2

Q1. ′ =y x3

41 4– / Q2. y′ = 1/x

Q3.dy

dxx= − −30 5 7 7( )– Q4.

d

dxx x(sin ) cos2 2 2=

Q5. v′ = −3 cos2 t sin t Q6. L′ = 2m + 5

Q7. y = sin x3 + C Q8. ′ ≈ −y 3

Q9. 4 ft/s Q10. B

1. f ( x) = x3 cos x ⇒ f ′ (x) = 3x2 cos x – x3 sin x

2. f ( x) = x4 sin x ⇒ f ′ (x) = 4x3 sin x + x4 cos x

3. g(x) = x1.5e2x ⇒ g′ (x) = 1.5x0.5e2x + 2x1.5 e2x

4. h(x) = x− 6.3 ln 4x ⇒h′ (x) = −6.3x− 7.3 ln 4x + x− 6.3 (1/4x)4

= −6.3x− 7.3 ln 4x + x− 7.3

5. y = x7(2x + 5)10 ⇒dy/dx = 7x6(2x + 5)10 + x7(10)(2x + 5)9 ⋅ 2

= x6(2x + 5)9(34x + 35)

6. y = x8(3x + 7)9 ⇒dy/dx = 8x7(3x + 7)9 + x8(9)(3x + 7)8(3)

= x7(3x + 7)8 (51x + 56)

7. z = ln x sin 3x ⇒z′ = (1/x) sin 3x + 3 ln x cos 3x

8. v = e5x cos 2x ⇒ v′ = 5e5x cos 2x − 2e5x sin 2x

9. y = (6x + 11)4(5x − 9)7 ⇒y ′ = 4(6x + 11)3(6)(5x − 9)7

+ (6x + 11)4(7)(5x − 9)6(5)

= (6x + 11)3(5x − 9)6(330x + 169)

10. y = (7x – 3)9(6x − 1)5 ⇒y′ = 9(7x – 3)8(7)(6x − 1)5

+ (7x − 3)9(5)(6x − 1)4(6)

= (7x − 3)8(6x − 1)4(588x − 153)


11. P = (x2 − 1)10(x2 + 1)15 ⇒P′ = 10(x2 − 1)9(2x)(x2 + 1)15

+ (x2 − 1)10(15)(x2 + 1)14(2x)

= 10x(x2 − 1)9(x2 + 1)14[2(x2 + 1) + 3(x2 – 1)]

= 10x(x2 − 1)9(x2 + 1)14(5x2 − 1)

12. P(x) = (x3 + 6)4(x3 + 4)6 ⇒P′ (x) = 4(x3 + 6)3(3x2)(x3 + 4)6

+ (x3 + 6)4 ⋅ 6(x3 + 4)5

⋅ 3x2

= 6x2(x3 + 6)3(x3 + 4)5[2(x3 + 4) + 3(x3 + 6)]

= 6x2(x3 + 6)3(x3 + 4)5(5x3 + 26)

13. a ( t) = 4 sin 3t cos 5t ⇒a′(t) = 12 cos 3t cos 5t + 4 sin 3t(−5 sin 5t)

= 12 cos 3t cos 5t − 20 sin 3t sin 5t

14. v = 7 cos 2t sin 6t ⇒v′ = −14 sin 2t sin 6t + 7 cos 2t(6 cos 6t)

= −14 sin 2t sin 6t + 42 cos 2t cos 6t

15. y = cos (3 sin x) ⇒y′ = −sin (3 sin x) · 3 cos x

= −3 sin (3 sin x) cos x

16. y = sin (5 cos x) ⇒ y′ = cos (5 cos x) · (−5 sin x)

= −5 cos (5 cos x) sin x

17. y = cos e6x ⇒dy/dx = 6e6x(−sin e6x) = −6e6x sin e6x ⇒d 2y/dx2 = −36e6x sin e6x − 6e6x(6e6x cos e6x)

= −36e6x sin e6x − 36e12x cos e6x

18. y = ln (sin x) ⇒ dy/dx = (1/sin x) cos x

= (cos x)(sin x)− 1 ⇒ d 2y/dx2

= (cos x) [−(sin x)]− 2 (cos x) + (−sin x) ⋅ (sin x)− 1

= −−

+ = − + = −cos

sincot csc

2

22 21 1

x

xx x

19. z = x3(5x − 2)4 sin 6x ⇒z′ = 3x2(5x − 2)4 sin 6x + x3[(4)(5x − 2)3(5) sin 6x

+ (5x − 2)4(6 cos 6x)]

= 3x2(5x − 2)4 sin 6x + 20x3(5x − 2)3 sin 6x

+ 6x3(5x − 2)4 cos 6x

20. u = 3x5(x2 − 4) cos 10x

u′ = 15x4(x2 − 4) cos 10x + 3x5[2x cos 10x

+ (x2 − 4) ⋅ (−10 sin 10x)]

= 15x4 ⋅ (x2 − 4) cos 10x + 6x6 cos 10x

− 30x5(x2 − 4) sin 10x

21. If y = uvw, where u, v, and w are differentiablefunctions of x, then y′ = u′vw + uv′w + uvw′.

Proof:

y = uvw = (uv)w

∴ y′ = (uv)′w + (uv)w′ = (u′v + uv′)w + (uv)w′= u′vw + uv′w + uvw′, Q.E.D.

22. If y u u u un n= …1 2 3 where u un1… are differentiablefunctions of x, then

′ = ′ + ′y u u u u u u u un n n1 2 3 1 2 3K K

+ ′ + + ′u u u u u u u un n1 2 3 1 2 3K L K K .

23. z = x5 cos6 x sin 7x ⇒z′ = 5x4 cos6 x sin 7x + x5 6 cos5 x (−sin x) ⋅

sin 7x + x5 cos6 x ⋅ 7 cos 7x

= 5x4 cos6 x sin 7x − 6x5 cos5 x sin x sin 7x

+ 7x5 cos6 x cos 7x

24. y = 4x6 sin3 x cos 5x ⇒y′ = 24x5 sin3 x cos 5x + 4x6 3 sin2 x cos x ⋅

cos 5x + 4x6 sin3 x(−5 sin 5x)

= 24x5 sin3 x cos 5x + 12x6 sin2 x cos x ⋅cos 5x − 20x6 sin3 x sin 5x

25. y = x4 (ln x)5 sin x cos 2x ⇒y′ = 4x3(ln x)5 sin x cos 2x + x4 ⋅ 5(ln x)4(1/x) ⋅

sin x cos 2x + x4(ln x)5 cos x cos 2x

+ x4(ln x)5 sin x ⋅ (−2 sin 2x)

= 4x3(ln x)5 sin x cos 2x + 5x3(ln x)4 sin x ⋅cos 2x + x4(ln x)5 cos x cos 2x

− 2x4(ln x)5 sin x sin 2x

26. u = x5 e2x cos 2x sin 3x ⇒u′ = 5x4 e2x cos 2x sin 3x + x5 · 2e2x ·

cos 2x sin 3x + x5 e2x(−2 sin 2x) sin 3x

+ x5 e2x cos 2x · 3 cos 3x

= 5x4 e2x cos 2x sin 3x + 2x52e2x cos 2x sin 3x

− 2x5 e2x sin 2x sin 3x + 3x5 e2x cos 2x cos 3x

27. a. y(t) = 4 + 3e− 0.1 t cos πt

v(t) = y′(t) = 3(−0.1)e− 0.1 t cos π t+ 3e− 0.1 t(−π sin π t)

= e− 0.1 t(−0.3 cos π t − 3π sin π t )

b. v(2) = e− 0.2 (−0.3 cos 2π − 3π sin 2π)

= e−0.2(−0.3 − 0) = −0.2456…

There is not a high point at t = 2 becausev(2) ≠ 0.

v(t) = 0 ⇒ e− 0.1 t(−0.3 cos π t − 3π sin π t) = 0

⇒ −0.3 cos π t = 3π sin π t ⇒ t = 1.9898…

28. a. y(t) = t sin t ⇒ v(t) = y′(t) = sin t + t cos tGraph confirms Figure 4-2d.


b. v exceeds 25.

v

25

25

c. In 1940, wind-induced vibrations in theTacoma Narrows Bridge increased inamplitude until the bridge collapsed.

29. Prove that the derivative of an odd function is aneven function and that the derivative of an evenfunction is an odd function.

Proof:

For any function, the chain rule givesd

dxf x f x f x( ) ( ) ( ) ( ).− = ′ − ⋅ − = − ′ −1

For an odd function,d

dxf x

d

dxf x f x( ) = ( ).− = − ′[– ( )]

∴ −f ′ (−x) = −f ′ (x) or f ′ (−x) = f ′ (x),

and the derivative is an even function.For an even function,d

dxf x

d

dxf x f x( ) ( ) ( ).− = = ′

∴ −f ′ (−x) = f ′ (x) or f ′ (−x) = −f ′ (x),and the derivative is an odd function, Q.E.D.

30. f (x) = 2 sin x cos x ⇒f ′ (x) = 2 cos x · cos x + 2 sin x(−sin x)

= 2 cos2 x − 2 sin2 x = 2 cos 2xg(x) = sin 2x ⇒ g′(x) = 2 cos 2x = f ′ (x), Q.E.D.f(0) = 0 and g(0) = 0∴ f (x) = 2 sin x cos x = sin 2x = g(x), by theuniqueness theorem for derivatives, Q.E.D.f (x) = cos2 x − sin2 x ⇒f ′ (x) = 2 cos x(−sin x) − 2 sin x cos x

= −4 sin x cos x = −3 sin 2xg(x) = cos 2x ⇒g′(x) = (−2 sin 2x) = −sin 2x = f ′ (x), Q.E.D.f(0) = 1 and g(0) = 1

∴ f (x) = cos2 x − sin2 x = cos 2x = g(x) by theuniqueness theorem, Q.E.D.

31. Prove that if fn(x) = xn, then ′ = −f x nxnn( ) 1

for allintegers ≥ 1.

Proof (by induction on n):

If n = 1, then f1(x) = x1 = x, which implies that

′ = =f x x101 1( ) , which anchors the induction.

Assume that for some integer n = k > 1,′ = −f x kxk

k( ) .1

For n = k + 1, fk+ 1(x) = xk+ 1 = (xk)(x).

By the derivative of a product property,′ = ′ + ′ = ′ ++f x x x x x x x xk

k k k k1( ) ( ) ( ) ( )( ) ( ) ( ) .

Substituting for (xk)′ from the inductionhypothesis,

′ = + = + = + =+−f x kx x x kx x k xk

k k k k k1

1 1( ) ( )( ) ( )(k + 1)x( k+ 1 ) −1, completing the induction.∴ ′ = −f x nxn

n( ) 1 for all integers ≥ 1, Q.E.D.

32. Way 1: y = (x + 3)8(x − 4)8

y′ = 8(x + 3)7 · (x − 4)8 + (x + 3)8 · 8(x − 4)7

= 8(x + 3)7(x − 4)7(x + 3 + x − 4)= 8(x + 3)7(x − 4)7(2x − 1)

Way 2: y = (x2 − x − 12)8

y′ = 8(x2 − x − 12)7(2x − 1) = 8(x + 3)7(x − 4)7(2x − 1), which checks.

33. a.

f

f´

x

f(x)

5

1

b. f ′ (x) = 3x2 sin x + x3 cos xThe graph in part a is correct.

c. The numerical derivative graph duplicates thealgebraic derivative graph, as in part a, thusshowing that the algebraic derivative is right.

34. a.

1/9

1.4

500,000599,128

–500,000

–1.5

x

f(x)

b. f ′ (x) = 4(5x − 7)3(5) · (2x + 3)5

+ (5x − 7)4 · (5)(2x + 3)4(2)= 10(5x − 7)3(2x + 3)4[2(2x + 3)

+ 5x − 7]= 10(5x − 7)3(2x + 3)4(9x − 1)

c. f ′ (x) = 0 ⇔ 5x − 7 = 0 or 2x + 3 = 0or 9x − 1 = 0∴ x = 7/5 = 1.4, or x = −3/2 = −1.5,or x = 1/9See graph in part a.

d. f (1.4) = 0, f (−1.5) = 0, f (1/9) = 599,127.6… .See graph in part a.

e. False. The graph may have a point where itlevels off and then continues changing in thesame direction, as at x = −1.5 in part a.

35. a. A = L WdA

dt

dL

dtW L

dW

dt= ⋅ + ⋅


b. W tdW

dtt= + ⇒ = −2 2 2cos sin

L tdL

dtt= + ⇒ =3 2 2 4 2sin cos

dA

dtt t= +( )( )4 2 2 2cos cos

+ + −( )( )3 2 2 2sin sint tAt t = 4, dA/dt = 7.132… , so A is increasing.At t = 5, dA/dt = −4.949… , so A isdecreasing.

Problem Set 4-3Q1. 1066x1065

Q2. f (x) = 12x5 + C

Q3. y′ = 3x2 sin x + x3 cos x

Q4. dy/dx = −sin (x7) ⋅ 7x6 = −7x6 sin (x7)

Q5. f ′ (x) = 0 (derivative of a constant)

Q6. 54e9t

Q7. See the text for the definition of derivative.

Q8. Instantaneous rate of change at a given x

Q9. (x − 3)4(x − 3 + 2x) = 3(x − 3)4(x − 1)Q10.

4

4 y

x

1. f xx

xf x

x x x x

x( ) ( )= ⇒ ′ =

3 2 3

2

3

sin

sin – cos

sin

2. f xx

xf x

x x x x

x( ) ( )= ⇒ ′ = +4 3 4

2

4

cos

cos sin

cos

3. g xx

x( )

cos

ln= ⇒

3

′ = − ⋅ − ⋅g x

x x x x x

x( )

cos ( sin ) ln cos ( / )

(ln )

3 12 3

2

= − −3 2 3

2

ln sin cos (cos )/

(ln )

x x x x x

x

4. h xx

e x( )sin= ⇒

5

3

′ = − ⋅h x

x x e x e

e

x x

x( )sin cos sin

( )

5 34 3 5 3

3 2

= −5 34 5

3

sin cos sinx x x

e x

5. yx

x= ⇒sin

cos

10

20

′ =+

yx x x x

x

10 10 20 20 10 20

202

cos cos sin sin

cos

6. yx

x= ⇒cos

sin

12

18

′ =yx x x x

x

– sin sin – cos cos

sin

12 12 18 18 12 18

182

7. y3x 7

6x 5=

−

+⇒

yx x

x x′ =

+ −

+=

+

3 6 5 3 7 6

6 5

57

6 52 2

( ) – ( )( )

( ) ( )

8. yx

x= +

−⇒10 9

5 3

′ = ⋅ − − + ⋅−

=yx x

x x

10 5 3 10 9 5

5 3

75

5 32 2

( ) ( )

( )

–

( – )

9. zx

x

dz

dx= + ⇒( )

( – )

8 1

5 2

6

9

= + ⋅ − − + ⋅ −−

6 8 1 8 5 2 8 1 9 5 2 5

5 2

5 9 6 8

18

( ) ( ) ( ) ( ) ( )( ) ( )

( )

x x x x

x

= − + +( ) ( )

( – )

8 1 120 141

5 2

5

10

x x

x

10. Ax

x

dA

dx=

+⇒( – )

( )

4 1

7 2

7

4

= − ⋅ + − − ⋅ ++

7 4 1 4 7 2 4 1 4 7 2 7

7 2

6 4 7 3

8

( ) ( ) ( ) ( ) ( ) ( )

( )

x x x x

x

= + + − −+

= + ++

= ++

28 4 1 7 2 7 2 4 1

7 2

28 4 1 7 2 3 3

7 2

84 4 1 1

7 2

6 3

8

6 3

8

6

5

( – ) ( ) [( ) ( )]

( )

( – ) ( ) ( )

( )

( – ) ( )

( )

x x x x

x

x x x

x

x x

x

11. Qe

xQ

x e x e x

x

x x x

= ⇒ ′ = −3 3 3

3 2

2sin

sin cos

sin

12. rx

x= ⇒ln

cos

4

′ = − −r

x x x x x

x

4 13 4 4

2

( / )cos (ln )( sin )

cos

= + = +( cos )/ (ln ) sin

cos

cos ln sin

cos

4 44

2

4

2

x x x x

x

x x x x

x x

13.d

dxx x( )– /60 804 3 7 3= − − /

14.d

dxx x( )– /24 7 3 = − − /56 10 3

Problems 15–22 and 25–26 can be done using eitherthe power rule or the quotient rule.

15. r xx

x r x xx

( ) ( )= = ⇒ ′ = − =− −1212 36

363

3 44

–

16. t xx

x t x xx

( ) ( )= = ⇒ ′ = − =− −5151 867

86717

17 1818

–


17. v xx

x( ) = = ⇒−14

0 514 0 5 1

cos .(cos . )

v′(x) = −14(cos 0.5x)–2(−sin 0.5x)(0.5)

= 7 0 5

0 52

sin .

cos .

x

x

18. a xx

x( ) = = ⇒−20202

2

sin(sin )

a ′ (x) = −40(sin x) −3 (cos x)

= – cos

sin

403

x

x

19. r xx

x r x xx

( ) ( )= = ⇒ ′ = − = −− −1 11 22

20. s xx

x s x xx

( ) ( )= = ⇒ ′ = − =− −12

22

2 33

–

21. W xx

x( ) ( )5= = − ⇒10

110 13 5

3

( – )–

W′(x) = 150x2(x3 − 1)4

22. T xx x

x x( ) = = ⇒−1 1

cos sin(cos sin )

′ = − −−T x x x x x( ) ( ) (cos sin sin sin2

+ =cos cos )sin – cos

cos sinx x

x x

x x

2 2

2 2 ,

which transforms to

T xx

xx x′ = = −( )

– cos

sincsc cot

2

24 2 2

14

2

23. T xx

x( ) = ⇒sin

cos

T xx x x x

x

x x

x xx

′ =

= + = =

( )(cos )(cos ) – (sin )(– sin )

cos

cos sin

cos cossec

2

2 2

2 221

(T is for tangent function.)

24. C xx

x( ) = ⇒cos

sin

C xx x x x

x

x x

x xx

′ =

= = = −

( )(– sin )(sin ) – (cos )(cos )

sin

– sin – cos

sin

–

sincsc

2

2 2

2 221

(C is for cotangent function.)

25. C xx

C xx

x( ) ( )= ⇒ ′ =1 0

2sin

– cos

sin

= − ⋅ = −1

sin

cos

sincsc cot

x

x

xx x

(C is for cosecant function.)

26. S xx

x( ) ( )= = ⇒−1 1

coscos

S′(x) = −(cos x)− 2(−sin x)

= =sin

cossec tan

x

xx x2

(S is for secant function.)

27. a. v tt

( ) = 1000

3 –

v( ) mi h11000

3 1500= = /

–

v( ) mi h21000

3 21000= = /

–

v( ) .31000

3 3

1000

0= =

– No value for v(3).

b. v(t) = 100(3 − t)− 1 ⇒

a(t) = −1000(3 − t)− 2 − 1 = 1000

3 2( – )t

a( ) (mi h) h11000

3 12502= = / /

( – )

a( ) (mi h) h21000

3 210002= = / /

( – )

a( ) = .31000

3 3

1000

02( – )= No value for a (3).

c.

2000v

a

t

a or v

3

d.1000

3500 2 32( – )t

t= ⇒ = − ⇒( )2

3 2 3 2− = ± ⇒ = ± ⇒t t

t = − =3 2 1 585. K in the domain.Range is 0 ≤ t < 1.585… .

28. a. Because they are walking in the samedirection, their relative rate is the difference(x − 5).

b. t xx

( ) = 300

5–, assuming Willie’s rate is

constant.t(6) = 300 s, t(8) = 100 s, t(10) = 60 s,t(5) = 300/0, which is infinite, t(4) = −300,which is not reasonable in the real world,and t(5.1) = 3000 s. A reasonable domainis x > 5.

c. t(x) = 300(x – 5)− 1

t′(x) = −300(x − 5)− 2 = −−300

5 2( )xt′(6) = −300 s/(ft/s)

d. t′(5) does not exist because of division byzero. More fundamentally, t′(5) does not existbecause t(5) does not exist.

29. f xx

x( ) = +

+⇒3 7

2 5


f xx x

x x′ = ⋅ + + ⋅

+=

+( )

3 2 5 3 7 2

2 5

1

2 52 2

( ) – ( )

( ) ( )

f ′ = =( ) .4

1

1690 005917159K

Using 4.1, f ′(4) ≈ 0.005827505… .

Using 4.01, f ′(4) ≈ 0.005908070… .

Using 4.001, f ′(4) ≈ 0.005916249… .

f ′(4) (exact) = 0.005917159…

Difference quotients are approaching f ′(4).

30. a. Sketch. See accurate plot in part b.

b. f xx

x( ) = ⇒

2 8

3

–

–

′ = = − +−( )

f xx x x

x

x x

x( )

2 3 8 1

3

6 8

3

2

2

2

2

( – ) – ( – )( )

( – )

f

f'

x3

5

f(x)

y2 and y3 both agree with the graph of f ′.

c.

x f (x) f ′(x)

2.95 −14.05 −399

2.96 −19.04 −624

2.97 −27.363… −1110.11…

2.98 −44.02 −2499

2.99 −94.01 −9999

3.00 undefined undefined

3.01 106.01 −9999

3.02 56.02 −2499

3.03 39.363… −1110.11…

3.04 31.04 −624

3.05 26.05 −399

f ( x) changes faster and faster as x approaches3, shooting off to negative infinity as xapproaches 3 from the negative side and topositive infinity as x approaches 3 from thepositive side. Note that the rates aresymmetrical about x = 3.

d. There is a relative minimum at x = 4 and arelative maximum at x = 2.

′ = − +−

=f ( )( )

( )2

2 6 2 8

2 30

2

2

′ = − +−

=f ( )( )

( )4

4 6 4 8

4 30

2

2

31. If y = xn, where n is a negative integer, theny′ = nxn− 1.

Proof:

Let n = −p, where p is a positive integer.

∴ = =− y xx

pp

1

∴ ′ = ⋅ ⋅ y

x px

x

p p

p

0 1 1

2

– –

because p is a

positive integer.

= − = − = −− − − −px

xpx px

p

pp p p

–

.1

21 2 1

Replacing −p with n gives y′ = nxn− 1, Q.E.D.32.

1

1

x

y

y

y '

1

1

x

y

y'

y


Problem Set 4-4Q1. (sin x)/(tan x) = cos x

Q2. 1/(sec x) = cos x

Q3. sin2 3 + cos2 3 = 1

Q4. f ′ (x) = ex sin x + ex cos x

Q5. g xx x x

x′ = +( )

cos sin

cos2

Q6. h ′ (x) = −(15/7)(3x)–12/7

Q7. dy/dx = 3(cos x)− 4 sin x

Q8. Limit = −3

Q9. (Function is secant.)

1

π

x

y

yy'

Q10. C

1. f (x) = tan 5x ⇒ f ′(x) = 5 sec2 5x

2. f (x) = sec 3x ⇒ f ′(x) = 3 sec x tan x

3. y = sec x7 ⇒ y′ = 7x6 sec x7 tan x7

4. z = tan x9 ⇒ z′ = 9x8 sec2 (x9)


5. g(x) = cot e11x ⇒ g ′ (x) = −11e11x csc2 (e11x)

6. h(x) = csc e10x ⇒ h ′ (x) = −10e10x csc (e10x) cot (e10x)

7. r (x) = ln (csc x) ⇒

′ =r xx

( )1

csc(−csc x cot x) = −cot x

8. p(x) = ln (cot x) ⇒

p xx

xx x

′ = − =( ) ( )1 12

cotcsc

cos sin

9. y = tan5 4x ⇒(d/dx)(y) = 5 tan4 4x · sec2 4x · 4

= 20 tan4 4x sec2 4x

10. y = tan7 9x ⇒(d/dx)(y) = 7 tan6 9x · sec2 9x · 9

= 63 tan6 9x sec2 9x

11. (d/dx)(sec x tan x) = sec x tan x · tan x +sec x · sec2 x = sec x tan2 x + sec3 x

12. (d/dx)(csc x cot x) = −csc x cot x · cot x +csc x · (−csc2 x) = −csc x cot2 x − csc3 x

13. y = sec x csc x ⇒y′ = sec x tan x · csc x + sec x · (−csc x cot x) = sec2 x − csc2 x

14. y = tan x cot x = 1 for all x ⇒ y′ = 0

15. yx

x= tan

sin = sec x ⇒ y′ = sec x tan x

16. yx

x= cot

cos = csc x ⇒ y′ = −csc x cot x

17. yx

x= ⇒5 7

14

ln

cot

yx x x

x

x x x x

x

x x

x

x

x′ = ⋅ − −

= +

= +

5 7 14 5 7 14 14

14

5 14 70 7 14

145

14

70 7

14

17

2

2

2

2

2

( )cot ln ( csc )

cot

( cot )/ ln csc

cot

cot

ln

cos

18. yx

e x= ⇒4 1040

csc

yx x e x e

e

x x

x′ = − −4 10 10 10 4 10 4040 40

40 2

( csc cot ) csc ( )

( )

= − −40 10 10 160 1040

csc cot cscx x x

e x

19. w = tan (sin 3x) ⇒w′ = sec2 (sin 3x) · 3 cos 3x

= 3 sec2 (sin 3x) · cos 3x

20. t = sec (cos 4x) ⇒t′ = sec (cos 4x) tan (cos 4x) · (−4 sin 4x)

= −4 sec (cos 4x) tan (cos 4x) sin 4x

21. S(x) = sec2 x − tan2 x = 1 ⇒ S′(x) = 0(The differentiation formulas give the same.)

22. m(x) = cot2 x − csc2 x = −1 ⇒ m′(x) = 0(The differentiation formulas give the same.)

23. A(x) = sin x2 ⇒ A′(x) = cos x2 · 2x = 2x cos x2

24. f (x) = cos x3 ⇒ f ′ (x) = −sin x3 · 3x2

= −3x2 sin x3

25. F(x) = sin2 x ⇒ F ′ (x) = 2 sin x cos x

26. g(x) = cos3 x ⇒g′(x) = 3 cos2 x · (−sin x) = −3 cos2 x sin x

27. y = tan x ⇒ dy/dx = sec2 x ⇒d2y/dx2 = 2 sec x(sec x tan x) = 2 sec2 x tan x

28. y = sec x ⇒ y′ = sec x tan x ⇒y″ = (sec x tan x) · tan x + sec x · sec2 x

= sec x tan2 x + sec3 x

29. y xx

x= = ⇒cot

cos

sin

yx x x x

x′ = ⋅ ⋅– sin sin – cos cos

sin2

= = −–

sincsc

12

2

xx or:

yx

x= = ⇒−1

tantan( ) 1

y′ = −1 ⋅ (tan x)− 2 ⋅ sec2 x = −csc2 x

30. y xx

x= = = ⇒cscsin

sin1 1( )–

y x xx

xx x′ = − = − = −−( )sin cos

cos

sincsc cot2

2

31. a. See graph in part b.

b. f (x) = tan x ⇒ f ′(x) = sec2 x. Predicted graphshould be close to actual one.

f

f´

1

1 x

y

c.

tan . – tan .

( . )

1 01 0 99

2 0 013 42646416= . K

tan′ 1 = sec2 1 = (1/cos 1)2 = 3.42551882…Difference quotient is within 0.001 of actual.

32. a. f (x) = sec x ⇒ f ′ (x) = sec x tan x ⇒f ′ (1) = sec 1 tan 1 = 2.8824…

b.

1

5

x

yy1

y 2


The formula is confirmed by the fact that theline is tangent to the graph.

c.

f f '

x

y

π/2

1

If f ′ (x) is negative, the graph of f isdecreasing.

33. a. y/10 = tan x ⇒ y = 10 tan x, Q.E.D.

b. y′ = 10 sec2 x. At x = 1, y′ = 10 sec2 1 =34.2551… . y is increasing at about34.3 ft/radian.

( . )34 2551

180K

π = 0.5978… ft/degree

c. y = 535 ⇒ x = tan− 1 53.5 = 1.55210…∴ y′ = 10 sec2 1.55210… = 28632.5…y is increasing at about 28,632.5 ft/radian.

34. a. tan xy= =opposite side

adjacent side 500∴ y = 500 tan x, Q.E.D.

b. dy/dt = 500 sec2 x · dx/dt

c. dx/dt = 0.3 rad/sAt y = 300, x = tan− 1 (300/500) = 0.5404…∴ dy/dt = 500 (sec2 0.5404…)(0.3)

= 500(1.36)(0.3) = 204 ft/s35. a. y = sin x + C

b. y x C= − +12 2cos

c. y x C= +13 3tan

d. y x C= − +14 4cot

e. y = 5 sec x + C


Problem Set 4-5Q1. sin′ x = cos x Q2. cos′ x = −sin x

Q3. tan′ x = sec2 x Q4. cot′ x = −csc2 x

Q5. sec′ x = sec x tan x Q6. csc′ x = −csc x cot x

Q7. f ′ (1) is infinite. Q8. f ′ (3) is undefined.

Q9. f ′ (4) = −1 Q10. f ′ (6) = 0

1. See Figure 4-5d. 2. See Figure 4-5d.

3. See Figure 4-5d. 4. See Figure 4-5d.

5. The principal branch of the inverse cotangentfunction goes from zero to π so that the functionwill be continuous.

6. There are no values of the inverse secant for xbetween −1 and 1, so the inverse secant function

cannot be continuous. (Some texts restrict therange of the inverse cosecant to 0 ≤ y ≤ π/2 sothat the function will be continuous, but doingso throws away the other half of the possiblevalues.)

7. sin (sin− 1 0.3) = 0.3

8. cos− 1 (cos 0.8) = 0.8

9. y = sin− 1 x ⇒ sin y = x ⇒ cos y · y′ = 1 ⇒

yy x

′ = =1 1

1 2cos –, Q.E.D.

y

x1

[Because sin y = (opposite leg)/(hypotenuse), putx on the opposite leg and 1 on the hypotenuse.

Adjacent leg = 1 2– ,x and cos y =(adjacent)/(hypotenuse).]

10. y = cos− 1 x ⇒ cos y = x ⇒ −sin y · y′ = 1 ⇒

yy x

′ = − = −1 1

1 2sin –, Q.E.D.

y

1

x

1 – x 2√

[Because cos y = (adjacent leg)/(hypotenuse), putx on the adjacent leg and 1 on the hypotenuse.

Opposite leg = 1 2– ,x and sin y =(opposite)/(hypotenuse).]

11. y = csc− 1 x ⇒ csc y = x ⇒ −csc y cot y · y′ ⇒

yy y x x

′ = − = −1 1

12csc cot – if x > 0

If x < 0, then y is in Quadrant IV (see Fig-ure 4-5d). So both csc y and cot y are negative,and thus their product must be positive.

∴ ′ = − ,yx x

1

12| | – Q.E.D.

y

1x

x 2 – 1√

[Because csc y = (hypotenuse)/(opposite leg), putx on the hypotenuse and 1 on the opposite leg.

1 – x 2√


Adjacent leg ,= x2 1– and csc y = x and cot y =(adjacent)/(opposite).]

12. y = cot− 1 x ⇒ cot y = x ⇒ −csc2 y · y′ = 1 ⇒

yy x x

′ = − = −+( )

= −+

1 1

1

1

122

2 2csc, Q.E.D.

y

x

1

1 + x 2√

[Because cot y = (adjacent leg)/(opposite leg),put x on the adjacent leg and 1 on the opposite

leg. Hypotenuse = +1 2x , and csc y =(hypotenuse)/(opposite).]

Problems 13−18 are shown done “from scratch,” asin Example 1. If students practice doing them thisway, they will not be dependent on memorizedformulas. Problem 13 shows how an alternatesolution could be found using the formulas and thechain rule.

13. y = sin− 1 4x ⇒ sin y = 4x ⇒ cos y · y′ = 4 ⇒

′ = =yy x

4 4

1 16 2cos –

y

4x1

1 – 16x2√

Alternate solution by application of the formula:

y x yx x

= ⇒ ′ = ⋅ =−sin– ( ) –

1

2 24

1

1 44

4

1 16

14. y = cos− 1 10x ⇒ cos y = 10x ⇒

−sin y · y′ = 10 ⇒ yy x

′ = − = −10 10

1 100 2sin –

y

1

10x

1 – 100x2√

15. y e y ex x= ⇒ = ⇒−cot cot1 0 5 0 5. .

− ⋅ ′ = ⇒csc2 0 50 5y y e x. .

′ = − = −+( )

= −+

ye

y

e

e

e

e

x x

x

x

x

0 5 0 5

1

0 50 5

2

0 5

2

0 5. . .

1

. . .

csc

y

1 + ex

e

1

0.5x

√

16. y = tan− 1 (ln x) ⇒ tan y = ln x ⇒sec2 y · y′ = l/x ⇒

′ = =+( )

=+

yx y x x x x

1 1

1

1

122

2 2sec ln ( ln )

y

1

1 + ln2x√ln x

17. yx

yx

y y y= ⇒ = ⇒ ⋅ ′ = ⇒−sec sec sec tan1

3 3

1

3

′ = =⋅

y xy y

xx

1

3

1

32

9 33sec tan

( – ) /

, if > 0

If x < 0, then y is in Quadrant II, where bothsec y and tan y are negative. So their product ispositive.

∴ ′ =y

x x

3

2 9| | –

y

x

3

x 2 – 9√

18. yx

yx= ⇒ = ⇒−csc csc1

10 10

− ⋅ ′ = ⇒csc coty y y1

10

yy y x x

′ = − = −⋅

1

10

1

1010

100100

2csc cot –

If x < 0, then y is in Quadrant IV, where bothcsc y and cot y are negative. So their product ispositive.

∴ ′ = −yx x

10

1002| | –

y

10

x 2

– 100

x

√

For Problems 19−24, a solution is shown using theappropriate formula.

19. y = cos− 1 5x2

′ = − ⋅ = −yx

xx

x

1

1 510

10

1 252 2 4– ( ) –

20. f (x) = tan− 1 x3

′ =+

⋅ =+

f xx

xx

x( )

1

13

3

13 22

2

6( )


21. g(x) = (sin− 1 x)2

′ = ⋅−g x xx

( ) 21

1

1

2sin

–

22. u = (sec− 1 x)2

′ = ⋅−u xx x

21

1

1

2 sec

| | –

23. v = x sin− 1 x + (1 − x2)1/ 2

′ = ⋅ + ⋅ + ⋅ −

= + − =

− − /

− −

v x xx

x x

xx

x

x

xx

11

1

1

21 2

1 1

1

2

2 1 2

1

2 2

1

sin–

( – )

sin– –

sin

( )

The surprise is that you now have seen a formulafor the antiderivative of the inverse sine function.

24. f (x) = cot− 1 (cot x) = x ⇒ f ′ (x) = 1 (Surprise!!)Application of the formulas gives the sameresult.

25. a. tan θ = x/100, so θ = tan− 1 (x/100), Q.E.D.

b.d

dx x x

θ =+

⋅ =+

1

1 100

1

100

100

100002 2( / )

d

dt

d

dx

dx

dt x

dx

dt

θ θ= ⋅ =+

⋅100

10000 2

c. If x = 500 ft and dθ/dt = −0.04 rad/s, then

− =+

⋅0 04100

10000 5002.dx

dtdx

dt= = −(– . )( )0 04 260000

100104

The truck is going 104 ft/s.

104(3600/5280) = 70.909… ≈ 71 mi/h26. a. θ = tan− 1 (50/x) − tan− 1 (30/x) or

θ = cot− 1 (x/50) − cot− 1 (x/30)The inverse tangent equation has theadvantage that the function appears on thecalculator. The inverse cotangent equation hasthe advantage that x is in the numerator of theargument, which makes the chain rule lesscomplicated to use.)

b.d

dx

x

x

x

x

θ =+

−+

–

( / )

–

( / )

– –50

1 50

30

1 30

2

2

2

2

=+

++

–50

2500

30

9002 2x x

= ++ +–

( )( )

20 30000

2500 900

2

2 2

x

x x

c. dθ/dx = 0 ⇒ −20x2 + 30000 = 0 ⇒20x2 = 30000

x = ± = ±1500 38 729. KAbout 38.7 ft

d. Maximum is between x = 38 and 39.

0.5

100

x

40

θ

27.

x Num. Deriv.* Alg. Deriv.

−0.8 −1.666671… −1.666666…

−0.6 −1.250000… −1.25

−0.4 −1.091089… −1.091089…

−0.2 −1.020620… −1.020620…

0 −1.000000… −1

0.2 −1.020620… −1.020620…

0.4 −1.091089… −1.091089…

0.6 −1.250000… −1.25

0.8 −1.666671… −1.666666…

*The precise value for the numericalderivative will depend on the tolerance towhich the grapher is set. The values givenby numerical derivative and the formula arevery close.

28. a. y xdy

dx x x= ⇒ =−sec

| | –

1

2

1

1

At x = 2,

dy

dx= =1

2 30 288675

| |. .K

The answer is reasonable because the graphslopes up at x = 2, with slope significantlyless than 1.

b. At x = 2, y = sec− 1 2 = cos− 1 (1/2) =1.04719… .d

dyy y y(sec sec tan) =

At y = 1.047… ,

d

dyy(sec ) (sec . )(tan . )= =1 047 1 047K K

3.464101… .

c. The answer to part b is the reciprocal of theanswer to part a. That is,

13 464101. K =

0.288675… . Thus, the derivative of theinverse secant at x = c is the reciprocal ofthe derivative of the secant at y = sec− 1 c.

29. a. y = sin− 1 x ⇒ sin y = x ⇒ cos y · y′ = 1 ⇒

′ =yy

1

cos, Q.E.D.


b. ′ = = =yx

1 1

0 61 251 1cos(sin ) cos(sin . )– – .

yx

′ = = = =1

1

1

1 0 6

1

0 81 25

2 2– – . .. , Q.E.D.

c. y f x f y x f yd

dxy= ⇒ = ⇒ ′ ⋅ = ⇒−1 1( ) ( ) ( ) ( )

d

dxy

f y

d

dxf x

f f x( )

( )[ ( )]

[ ( )],–=

′⇒ =

′ −1 11

1

Q.E.D.

d. f ( x) = x3 + x = 10 ⇒ (x − 2)(x2 + 2x + 5) = 0

⇒ x = 2 (only)

∴ h(10) = 2

Because h(x) = f −1(x) and f ′(x) = 3x2 + 1,

′ =′

=′

=⋅ +

=hf h f

( ) .101

10

1

2

1

3 2 11132[ ( )] ( )/

30. The inverse trig cofunctions, cos− 1, cot− 1, andcsc− 1, are the ones whose derivatives are precededby a minus sign.

Problem Set 4-6Q1. See the text for the definition of continuity.


Q3. y′ = −6x− 2 + C Q4. cos′ x = −sin x

Q5. dy/dx = sec2 x Q6. 1 12| |x x −( )Q7. f ′ (x) = 4x3; f ″ (x) = 12x2; f ″ (2) = 48

Q8. dy/dx = 15x2(x3 + 1)4

Q9. Integral ≈ 5.4 (Function is y = 2− x.)

Q10. E

1. Continuous 2. Neither

3. Neither 4. Both

5. Neither 6. Neither

7. Both 8. Neither

9. Neither 10. Neither

11. Continuous 12. Both

For Problems 13−20, sample answers are given.Equations do not necessarily correspond to the graphsshown.

13. a.

5

f(x)

x

3

b. f ( x) = x + 2

14. a.

x

f(x)

4

–2

b. f ( x) = x2

15. a.

6

x

f(x)

b. f xx x

x( )

( )( )= − +−

6 1

616. a.

1

2

f(x)

x

b. f xx x

xx

x( )

( ) ,

,=

−−

≠

=

2 1

15

if 1

if 1

17. a.

—5

x

f(x)

b. f xx x

x x( )

,

,=

≤ −> −

if

if

5

3 5

18. a.

–1

3

f(x)

x


b. f ( x) = (x + 1)2/3 + 3

19. a.

4

x

f(x)

7

b. f xx x

x x( )

,

,=

− <− ≥

2 9 4

11 4

if

if

20. a. No such function

x

f(x)

Not possible.Differentiabilityimpliescontinuity.

b. No such function

21. Continuous 22. Both

3

x

f(x)

4

x

f(x)

2

23. Both 24. Neither

1

x

y

x

f(x)

π/2

25. f xx x

a x b x( )

,

( – ) ,=

<+ ≥

3

2

1

2 1

if

if For f to be continuous at x = 1,

lim lim[ ( ) ]x x

x a x b→ →− +

= − + ⇒1

3

1

22

1 = a(1 − 2)2 + b ⇒ a + b = 1 ⇒ b = 1 − a

For f to be differentiable at x = 1,

lim lim ( )x x

x a x a→ →− +

= − ⇒ = − ⇒1

2

12 23 3 2 (1 2)

a = −1.5

b = 1 − a = 1 − (−1.5) ⇒ b = 2.5

1

1

x

f(x)

f is differentiable at x = 1.

26. f xx x

ax b x( )

if

if =

+ ≥+ <

–( – ) ,

,

3 7 2

2

2

3

For f to be continuous at x = 2,

lim ( ) limx x

ax b x→ →− +

+ = − + ⇒2

3

2

23 7[ ( – ) ]

a ⋅ 23 + b = 6 ⇒ 8a + b = 6 ⇒ b = 6 − 8a

For f to be differentiable at x = 2,

lim lim )x x

ax x a→ →− +

= − ⇒ ⋅ = ⇒2

2

2

23 2 3 3 2 2[ ( – ]

a = 1 6/

b = 6 − 8(1/6) ⇒ b = 14/3

2

x

f(x)

6


27. f xax x

x x b x( ) =

+ <− + ≥

2

2

10 2

6 2

, if

, if For f to be continuous at x = 2,

lim ( ) lim ( )x x

ax x x a→ →− +

+ = − + ⇒ + =2

2

2

210 6 6 4 10

4 – 12 + b ⇒ b = 4a + 18For f to be differentiable at x = 2,lim lim ( )x x

ax x a→ →− +

= ⇒ ⋅ = ⋅ ⇒2 2

2 2 6 2 2 2 2 6– –

a = −0.5b = 4(–0.5) + 18 ⇒ b = 16

2

10

f(x)

x


28. f xa x x

bx x( )

if

, if =

≤

− >

/ , 1

12 2 1

For f to be continuous at x = 1,


lim lim ( )x x

a x bx a b→ →− +

= ⇒ = ⋅ ⇒1 1

2 212 1 12 1/ – / –

a + b = 12For f to be differentiable at x = 1,lim limx x

ax bx a b→

−

→

−− +− = − ⇒ − ⋅ = − ⋅ ⇒

1

2

1

22 1 2 1

a = 2b∴ 2b + b = 12 ⇒ b = 4a = 2 · 4 ⇒ a = 8

1

10

x

f(x)


29. f xe x

b x x

ax

( )ln

=≤

+

, if

, if >

1

1For f to be continuous at x = 1,lim lim ( ln )x

ax

x

ae b x e b→ →− +

= + ⇒ =1 1

For f to be differentiable at x = 1,lim limx

ax

x

aae x ae→ →− +

= ⇒ =1 1

1 1( / )

Solve by grapher: a = 0.5671… andb = 1.7632…

1

2

x

f(x)


30. f xa x x

e xbx( )sin

,=

≥

, if < /

if /

2 3

2 3

ππ

For f to be continuous at x = 2π/3,

lim sin lim( / ) ( / )x x

bxa x e→ →− +

= ⇒2 3 2 3π π

ae a

ebb3

2

2

32 3

2 3

= ⇒ =ππ

//

For f to be differentiable at x = 2π/3,lim( / )x→ −2 3π

a x bex

bxcos lim( / )

= ⇒→ +2 3π

− = ⇒ =abe a beb b

222 3 2 3π π/ /–

So 2

32

2

32

1

3

2 32 3e

be b bb

bπ

π/

= ⇒ = ⇒ = −– –/

= … = =−– . and 0 5773

2

30 34462 3 3a e π /( ) . K

3

0.5

x

f(x)

f is differentiable at x = 2π/3.

31. a. yax bx cx d x

x k x=

+ + + ≤ ≤+

3 2 0 0 5

0 5

, .

,

if

if > .

For y to contain the origin,a ⋅ 03 + b ⋅ 02 + c ⋅ 0 + d d= ⇒ =0 0For y′ = 0 at x = 0, y′ = 3ax2 + 2bx + c ⇒ 0 = 3a ⋅ 02 + 2b ⋅ 0 + c ⇒ c = 0For y′ = 1 at x = 0.5, y′ = 3ax2 + 2bx + c ⇒ 1 = 3a(0.5)2 + 2b(0.5) + c ⇒ 1 = 0.75a + b ⇒b = 1 – 0.75aFor ′′ = =y x0 0 5 at . , ′′ = + ⇒y ax b6 20 = 3a + 2bSolve for a and b:3a + 2(1 – 0.75a) = 0 ⇒ 1.5a = –2 ⇒a = – 4/3 b = 2

b. For the function to be continuous,

lim ( ) lim ( ). .x x

x x x k→ →− +

− + = + ⇒0 5

43

3 2

0 52

− + = + ⇒43

3 20 5 2 0 5 0 5( . ) ( . ) . kk = = −– . 6661

6 0 1 K

32. Equation of the linear part of the fork isy – 20 = 5(x – 10) ⇒ y = 5x – 30

∴ =+ <

− ≥

yax bx x

x x

3 if

if

,

,

10

5 30 10

For y to be continuous at x = 10,a b⋅ + ⋅ = ⋅10 10 5 10 303 –1000a + 10b = 20 ⇒ b = 2 – 100aFor y to be differentiable at x = 10,3 10 52a b⋅ + =300a + (2 – 100a) = 5200a = 3 ⇒ a = 3/200b = 2 – 100(3/200) ⇒ b = 0.5

33. f xx

x

xx

x

( ),

,=

− −−

≠

=

2 2

22

4 2

if

if

Simplifying the equation for f (x) gives

f x

x x

x x

x

( )

,

,=

+ <

−=

2

2

1 2

1

2

if

if > 2

4, if


Taking the derivative for each branch gives

′ =<>=

f x

x x

x x

x

( )

,

,

2 2

2 2

2

if

if

undefined, if

Taking the left and right limits giveslim ( ) ; lim ( ) .x x

f x f x→ →− +

′ = ⋅ = ′ = ⋅ =2 2

2 2 4 2 2 4

Using the definition of derivative, taking the

limit from the left, ′ = + −−

→→ −

f xx

xx( ) lim ,

2

2 1 4

2

1

0which is infinite. The same thing happens fromthe right. As the following graph shows, thesecant lines become vertical as x approaches 2from either side.

4

2

x

f(x) Secantslopebecomesinfinite.

Thus, f is not differentiable at x = 2, eventhough the right and left limits of f ′ (x) are equalto each other. The function must be continuousif it is to have a chance of being differentiable.

34. a. d t

t

tt

tt

( )

..

., .

, .

=

−+

≤

−

≥

60 50 5

0 50 5

150 21

0 5

if

if

′ =− + <

>

−

−d t

t t

t t( )

. ( . ) , .

, .

60 5 0 5 0 5

150 0 5

2

2

if

if

The inequality signs must be < and > becausealthough the function is defined at x = 0.5,the derivative is not.

b. d d′ = = ⇒( ) ( –1 150 1 1502) is continuous atx = 1 because it is differentiable there.

c. lim ( ) . ( . . ) ..x

d t→

−−

′ = − + = −0 5

260 5 0 5 0 5 60 5

lim ( ) ( . ).x

d t→

−+

′ = =0 5

2150 0 5 600

As the ball was about to be hit, it wasapproaching the plate at 60.5 ft/s.Just after the ball was hit, it was going awayfrom the plate at 600 ft/s.

d. Function d is not differentiable at t = 0.5because d ′ (t) approaches different limits fromboth sides of x = 0.5.Function d is continuous at t = 0.5 becauseyou get zero as the limit of d (t) as tapproaches zero from either left or right.

e. A regulation baseball diamond has thepitcher’s mound 60.5 feet from home plate.Substituting zero for t gives d (0) = 60.5,confirming that the pitcher was on the moundat that time.

35. a. y = mx + b ⇒ y′ = m, which is independentof x.∴ linear functions are differentiable for all x.∴ linear functions are continuous for all x.

b. y = ax2 + bx + c ⇒ y′ = 2ax + b, whichexists for all x by the closure axioms.∴ quadratic functions are differentiable forall x.∴ quadratic functions are continuous for all x.

c. y = 1/x = x–1 ⇒ y′ = –x–2, which exists forall x ≠ 0 by closure and multiplicative inverseaxioms.∴ the reciprocal function is differentiable forall x ≠ 0.∴ the reciprocal function is continuous for allx ≠ 0.

d. y = x ⇒ y′ = 1, which is independent of x.∴ the identity function is differentiable forall x.∴ the identity function is continuous forall x.

e. y = k ⇒ y′ = 0, which is independent of x.∴ constant functions are differentiable forall x.∴ constant functions are continuous for all x.

36. See text proof.

Problem Set 4-7Q1. y′ = 243x1214 Q2. dy/dx = 2/(x–1)2

Q3. f ′ (x) = 1 + ln x Q4. y′(x) = 5e5x cos e5x

Q5. (d /dx)(y) = 3x2, x ≠ 0; d2y/dx2 = 6x, x ≠ 0

Q6. y′ = 0 Q7. ′ = − −θ 1 1 2/ x

Q8. v(t) is decreasing at t = 5.

Q9. Q10. E

1

π/2

x

y

y'

1. x = t4, y = sin 3tdy

dx

dy dt

dx dt

t

t

d y

dx

d

dx

t

t= = ⇒ =

/

/

cos cos3 3

4

3 3

43

2

2 3


= − ⋅ −( sin )( / ) cos ( )( / )

( )

9 3 4 3 3 12

4

3 2

3 2

t dt dx t t t dt dx

t

= − − ÷36 3 36 3

16

3 2

6

t t t t

t

dx

dt

sin cos

= − −36 3 36 3

64

3 2

9

t t t t

t

sin cos

= − −9 3 9 3

16 7

t t t

t

sin cos

2. x = 6 ln t, y = t3

dy

dx

dy dt

dx dt

t

tt

d y

dx

d

dxt t dt dx t

dx

dt

t

tt

= = = ⇒

= = = ÷

= =

/

/ /

) .

.

/

3

60 5

0 5 1 5 1 5

1 5

60 25

23

2

2

3 2 2

23

.

( . ) . ( /

.

3. a. x = 2 + t, y = 3 – t2

t x y

–3 –1 –6

–2 0 –1

–1 1 2

0 2 3

1 3 2

2 4 –1

3 5 –6

b.

3

2

x

y

c.dy

dx

dy dt

dx dt

tt= = − = −/

/

2

12

If t = 1, dy dx/ –= 2 and (x, y) = (3, 2).Line through (3, 2) with slope –2 is tangentto the graph. See part b.

d. x = 2 + t ⇒ t = x – 2 ⇒ y = 3 – (x – 2)2

This is the Cartesian equation of a parabolabecause only one of the variables is squared.

e. By direct differentiation, dy dx x/ – ( – ).= 2 2At (x, y) = (3, 2), dy/dx = –2(3 – 2) = –2,which agrees with part c.dy/dx = –2(x – 2) = –2(2 + t – 2) = –2t,which agrees with part c.

4. a. x = t2, y = t3

t x y

–3 9 –27–2 4 –8–1 1 –1 0 0 0 1 1 1 2 4 8 3 9 27

b.

x

y

5

5

c.dy

dx

dy dt

dx dt

t

tt= = =/

/.

3

21 5

2

If t = 1, dy/dx = 1.5 and (x, y) = (1, 1).Line through (1, 1) with slope 1.5 is tangentto the graph. See graph in part b.

d. x t t x y x y x= ⇒ = ⇒ = ⇒ =2 1 2 1 2 3 1 5/ / .( )The name semicubical is picked because 1.5is half of 3, the exponent for a cubic function.The name parabola is used because the equationlooks similar to y = x2 for a parabola.

e. By direct differentiation, dy/dx = 1.5x0.5 .At (x, y) = (1, 1), dy/dx = 1.5 ⋅ 10.5 = 1.5,which agrees with part c.dy/dx = 1.5x0.5 = 1.5(t2)0.5 = 1.5t, whichagrees with part c.

5. a. The graph confirms the figure in the text.

b.dy

dx

t

tt=

−= −5

3

5

3

cos

sincot

c. If t = π/4, x y= =3 2 2 5 2 2/ and / .(x, y) = (2.121… , 3.535…)dy

dx= − = −5

3 45 3cot /

π

5

–5

3–3

y

x



d. False. The line from (0, 0) to (2.1… , 3.5…)does not make an angle of 45° with thex-axis. (This shows that the t in parametricfunctions is not the same as the θ in polarcoordinates.)

e. The tangent line is horizontal if dy/dx = 0.

∴ cos t = 0 and sin t ≠ 0.This happens at t = π/2, 3π/2, … .Points are (0, 5), (0, –5).Tangent line is vertical if dy/dx is infinite.

∴ sin t = 0 and cos t ≠ 0.This happens at t = 0, π, 2π, … .Points are (3, 0), (–3, 0). See graph in part c.

f. x/3 = cos t ⇒ (x/3)2 = cos2 ty/5 = sin t ⇒ (y/5)2 = sin2 tAdding left and right sides of the equationsgives (x/3)2 + (y/5)2 = cos2 t + sin2 t.∴ (x/3)2 + (y/5)2 = 1, which is a standard formof the equation of an ellipse centered at theorigin, with x-radius 3 and y-radius 5.

6. a. The graph confirms the figure in the text.

b.dy

dx

t t

t t

t

tt=

−= − =24

24

2

2

sin cos

cos ( sin )

sin

costan–

∴ dy/dx = –tan t

c. If t = 1, x = 8 cos3 1 = 1.2618… , and

y = 8 sin3 1 = 4.7665… ,

(x, y) = (1.2618… , 4.7665…).

At t = 1, dy/dx = –tan 1 = –1.5574… .

x

y

8–8

8

–8


d. dx/dt = –24 cos2 t sin tdy/dt = 24 sin2 t cos tThe cusps occur where t is a multiple of π/2.At each such value, dx/dt and dy/dt equal zero.t = 0 gives the cusp at (8, 0).lim( / ) lim(– tan ) tant t

dy dx t→ →

= = =0 0

0 0–

So the graph becomes horizontal at (8, 0).t = π/2 gives the cusp at (0, 8).

lim ( / ) lim ( tan ),/ /t t

dy dx t→ →

= −π π2 2

which is infinite.

So the graph becomes vertical at (0, 8).

e. x/8 = cos3 t ⇒ (x/8)2/3 = cos2 t

y/8 = sin3 t ⇒ (y/8)3/2 = sin2 t

∴ (x/8)2/3 + (y/8)2/3 = cos2 t + sin2 t

⇒ x2/3 + y2/3 = 4

7. a. x = 6 + 5 cos t, y = 3 + 5 sin t

6

3

dy/dx isinfinitehere.

x

y

b.dy

dx

t

tdy dx t=

−⇒ = −5

5

cos

sin/ cot

c. dy/dx = 0 if cos t = 0 and sin t ≠ 0.∴ t = 0.5π, 1.5π, 2.5π, …dy/dx is infinite if sin t = 0 and cos t ≠ 0.∴ t = 0, π, 2π, …At a point where dy/dx is infinite, dx/dt mustbe zero. This happens where t n= ±π π/ ,2so dy/dx = 5 cos t = 0 at those points. Seegraph in part a.

d.x

ty

t− = − =6

5

3

5cos sin and

x yt t

−

+ −

= +6

5

3

5

2 22 2cos sin

x y−

+ −

=6

5

3

51

2 2

This is an equation of a circle centered at(6, 3) with radius 5.

e. The 6 and 3 added in the original equationsare the x- and y-coordinates of the center.The coefficients, 5, for cosine and sine in theoriginal equations are the x- and y-radii,respectively. Because the x- and y-radii areequal, the graph is a circle.

8. x = cos2 t, y = sin2 t

dy

dx

t t

t tt t= − = ≠ ≠2

21 0 0

cos ( sin )

sin coscos sin– ( , )

1

1

x

y

The graph is a line segment with a slope of –1.x + y = cos2 t + sin2 t ⇒ x + y = 1This is the equation of a line with slope –1,confirming what was observed on the graph.The parametric equations restrict the ranges of xand y to the first quadrant, no matter what is thedomain of t. This is true because cos2 t and sin2 tare never negative.The Cartesian equation allows–∞ < x < ∞ and –∞ < y < ∞.


9. a. The grapher confirms the figure in the text.

b.dy

dx

t t

t t

t t

t t= −

− −= −

− −2 2 2

2 2 2

2

2

cos cos

sin sin

cos cos

sin sin

c. Cusps occur where both dx/dt and dy/dt = 0.A graphical solution shows that this occurs att = 0, t = 2π/3, t = 4π/3, t = 2π, … .(A cusp could also happen if dx/dt = 0 anddy/dt ≠ 0, but for this figure there is no suchplace.)

0 2 2π4π3 3__ __π

dx/dt

dy/dt

t

dx/dt or dy/dt

At t = 0, 2π, … , the tangent appears tobe horizontal. At t = 2π/3, 4π/3, 8π/3,10π/3, … , there appears to be a tangentline but not a horizontal one.A numerical solution shows the followingvalues as t approaches 2π/3:

t dy/dx

2π/3 – 0.1 –1.547849…

2π/3 – 0.01 –1.712222…

2π/3 – 0.001 –1.730052…

2π/3 indeterminate

2π/3 + 0.001 –1.734052…

2π/3 + 0.01 –1.752225…

2π/3 + 0.1 –1.951213…

dy/dx seems to be approaching about –1.732as t approaches 2π/3.

[The exact answer is − 3, which studentswill be able to prove easily with l’Hospital’srule after they have studied Section 6-5. JoanGell and Cavan Fang have shown clevertrigonometric transformations that “remove”the removable discontinuity and lead to thesame answer. These are

1. Use the sum and product properties ondy/dx:

dy

dx

t t

t t=

−2 1 5 0 5

2 1 5 0 5

sin . sin .

sin . cos .

= − ≠tan . /0 5 0t dx dt if

As t dy dx→ → =2 3 3π π/ , / – ( / ) – 3tan .

2. Use the double argument properties ondy/dx:dy

dx

t t

t t t=

+cos – ( cos – )

–(sin sin cos )

2 1

2

2

= ++

=( – cos )( cos )

–(sin )( cos )

– cos

– sin

1 1 2

1 2

1t t

t t

t

t,

which approaches − 3 as t → 2π/3.]


b.dy

dx

a t t

a tt t= =4

222

3cos (– sin )

seccos sin

–

(The answer is independent of a.)

c. x a ta t

t

a t

t2 2 2

2 2

2

2 2

244 4 1= = =tan

sin

cos

( – cos )

cosy = 2a cos2 t ⇒ cos2 t = y/(2a)

∴ = = −x

a y a

y a

a a y

y2

2 24 1 2

2

4 2[ – /( )]

/( )

( )

x2y = 8a3 – 4a2y ⇒ (x2 + 4a2)y = 8a3 ⇒

ya

x a=

+8

4

3

2 2

a yx

= ⇒ =+

3216

362

d. y a x a= + ⇒−8 43 2 2 1( )

dy

dxa x a x

a x

x a= − + ⋅ =

+−8 4 2

16

43 2 2 2

3

2 2 2( )–

( )

e. At t = π/4, x = 2a tan (π/4) = 2a.From part d,

dy

dx

a a

a a

a

a=

+= =– ( )

[( ) ]

–16 2

2 4

32

64

3

2 2 2

4

4 –1/2

From part b,dy

dx= – ( / ) ( / )2 4 43cos sinπ π

= =– ( / ) ( / ) – / ,2 2 2 2 2 1 23 which agrees.

At t = π/4, x = 2a tan (π/4) = 2a = 6 andy = 2a cos2

(π/4) = 2a(1/2) = a = 3.A line through (6, 3) with slope –1/2 istangent to the graph at that point.

5

y

x

10

–5

–5–10–15 0 5 10 15

t = π/4

11. a. x = cos t + t sin ty = sin t – t cos tThe grapher confirms the figure in the text.[Note: In the derivation of these equationsfrom the geometric definition of involute,


x = cos t + t cos (t – π/2)y = sin t + t sin (t – π/2)(cos t, sin t) is the point of tangency of thestring.Because the circle is a unit circle, the lengthof the string is also t, the central angle inradians.The string makes an angle of (t – π/2) withthe positive x-axis so that(t cos (t – π/2), t sin (t – π/2)) is a vectorrepresenting the unwound string.The cofunction properties and odd-evenproperties from trig are used to simplify theequations so that the calculus will be easier.]

b.dy

dx

t t t t

t t t t= +

+ +cos – [cos (– sin )]

– sin (sin cos )

= =t t

t tt

sin

costan

c. At t = π, dy/dt = tan π = 0. The string willbe pointing straight up from the x-axis. Thediagram shows that the tangent to the graph ishorizontal at this point.

x

1

1

y

String

(x, y)

t = π

12. a. x starts at a middle point and increases.y starts at a high point and decreases.∴ x = 25 + 15 sin Bt

y = 20 + 15 cos BtThe period is 60 seconds.So B = 2π/60 = π/30

∴ = +x t25 1530

sinπ

y t= +20 1530

cosπ

b. dx dt t/ = π π2 30

cos

dy dt t/ –= π π2 30

sin

At t = 5,

dx dt/ / .= = =π π π

2 63 4 1 3603cos K

dy dt/ – – / – .= = =π π π

2 64 0 7853sin K

c. The slope of the circular path is dy/dx.At t = 5,dy

dx= = =– /

/

ππ

4

3 41 3 0 5773– / – . K

d.x

tx

t–

sin–

sin25

15 30

25

15 30

22= ⇒

=π π

yt

yt

–cos

–cos

20

15 30

20

15 30

22= ⇒

=π π

Because sin cos2 2

30 301

π πt + = ,

x y– –.

25

15

20

151

2 2

+

=

This is an equation of a circle centered at(25, 20) with radius 15, confirming that thepath really is a circle.

13. The actual solutions will vary depending on theperiod of the pendulum, as determined by thelength of the string. The following solutionsupposes that the period turns out to be 3.1seconds.

x t y t= =302

3 120

2

3 1cos

.sin

.

π π

dy

dx

t

tt= =

( / . ) cos.

–( / . ) sin .

40 3 123 1

60 3 12

2

3

2

3 1

π π

π ππ

3.1

– cot

At t = 5, x ≈ –22.8, y ≈ –13.0,and dy/dx ≈ –0.78.

If the measurements have been accurate, thependulum will be above the coin when t = 5.

14. The graph looks like an ellipse that moves in thex-direction as t increases. Because y starts at ahigh point and varies between 5 and 1, the ellipsehas center at y = 3 and y-radius 2. Thus, anequation for y would be y = 3 + 2 cos t.x starts at 0 and increases. If the ellipse hadx-radius 0.5, an equation for x would bex = 0.5 sin t. The graph of this ellipse is

x

y5

10

The graph seems to move over 1 unit to the righteach cycle. Thus, if t increases by 2π, x increasesby 1. The equations are thusx = t/(2π) + 0.5 sin t, y = 3 + 2 cos tThe graph here duplicates the one in the text.

x

y5

10


To locate “interesting” features,dy

dx

dy dt

dx dt

t

t= =

+/

/

– sin

/( ) . cos

2

1 2 0 5π.

For horizontal tangents, dy/dt = 0 and dx/dt ≠ 0.∴ 2 sin t = 0 ⇔ t = 0 + πn (n an integer)Thus, x = 0, 0.5, 1, 1.5, … .For vertical tangents, dx/dt = 0 and dy/dt ≠ 0.∴ 1/(2π) + 0.5 cos t = 0 ⇔ cos t = –1/πSolving numerically for t givest = 1.8947… + 2π n or 4.3884… + 2πn.For crossing points, x = 0.5, 1.5, 2.5, …from symmetry on the graph. If x = 0.5, then1/(2π)t + 0.5 sin t = 0.5.Solving numerically for the value of t closest to0, t = 0.8278… .y(0.8278…) = 3 + 2 cos 0.8278… = 4.3529…A crossing point is (0.5, 4.3529…) at t =0.8278… .


b. (x = cos 4t, y = sin t)

1

1

x

y

If n is an even number, the graph comes toendpoints and retraces its path, making twocomplete cycles as t goes from 0 to 2π.If n is an odd number, the graph does notcome to endpoints. It makes one completecycle as t goes from 0 to 2π.

c. i. (x = cos 5t, y = sin t)

1

1 y

x

ii. (x = cos 6t, y = sin t)

1

1 y

x

d. n = 1. (x = cos t, y = sin t)

1

1 y

x

n = 2. (x = cos 2t, y = sin t)

1

1 y

x

If n = 1, the graph is a circle.If n = 2, the graph is a parabola.

e. Jules Lissajous (1822–1880) lived in France.Nathaniel Bowditch (1773–1838) lived inMassachusetts.

Problem Set 4-8Q1. y′ = 2001x2000 Q2. y′ = ln (2001)2001x

Q3. 5 Q4. f ′(u) = –csc2 u

Q5. product Q6. 1/(1 + 9x2)

Q7. x3 + C Q8. Instantaneous rateQ9. Q10. –2.4033… ft/s

6

4

x

y

y'

1. x3 + 7y4 = 13 ⇒ 3x2 + 28y3y′ = 0 ⇒

yx

y′ = –

3

28

2

3

2. 3x5 − y4 = 22 ⇒ 15x4 − 4y3y′ = 0 ⇒ yx

y′ = 15

4

4

3

3. x ln y = 104 ⇒ 1 · ln y + xy

y⋅ ⋅ ′1 = 0 ⇒

′ = −y

y y

x

ln

4. yex = 213 ⇒ pxy′ + yex = 0 ⇒ y′ = –y


5. x + xy + y = sin 2x ⇒1 + y + xy′ + y′ = 2 cos 2x ⇒y′(x + 1) = 2 cos 2x – 1 – y ⇒

yx y

x′ =

+2 2 1

1

cos – –

6. cos xy = x – 2y ⇒(–sin xy) ( y + xy′) = 1 – 2y′ ⇒y′(–x sin xy + 2) = 1 + y sin xy ⇒

yy xy

x xy′ = +1

2

sin

– sin

7. x0.5 – y0.5 = 13 ⇒0.5x–0.5 – 0.5y–0.5y′ = 0 ⇒ y′ = y0.5 /x0.5

8. x1.2 + y1.2 = 64 ⇒ 1.2x0.2 + 1.2y0.2y′ = 0 ⇒y′ = –x0.2 /y0.2

9. e xy = tan y ⇒ exy(1 · y + x · y′) = y′sec2 y ⇒ ye xy

+ xy′e xy = y′sec2 y ⇒ xy′e xy – y′sec2 y = –ye xy

⇒ y′(xexy – sec2 y) = –yexy ⇒ y′ = −−ye

xe y

xy

xy sec2

10. ln (xy) = tan–1 x ⇒ tan (ln xy) = x ⇒

sec2 (ln xy) · 1

xy

(1 · y + y′x) = 1 ⇒ y + y′x =

xy cos2 (ln xy) y′x = xy cos2 (ln xy) – y ⇒

′ = −y

xy xy y

x

cos (ln )2

11. (x3y4)5 = x – y ⇒5(x3y4)4(3x2y4 + x3 · 4y3y′) = 1 – y′ ⇒y′(20x15y19 + 1) = 1 – 15x14y20 ⇒

yx y

x y′ =

+1 15

1 20

14 20

15 19

–

12. (xy)6 = x + y ⇒6(xy)5(y + xy′) = 1 + y′ ⇒y′(6x6y5 – 1) = 1 – 6x5y6 ⇒

yx y

x y′ = 1 6

6 1

5 6

6 5

–

–

13. cos2 x + sin2 y = 1 ⇒2 cos x · (–sin x) + 2 sin y · cos y · y′ = 0 ⇒

yx x

y y′ = cos sin

cos sin

14. sec2 y – tan2 x = 1 ⇒2 sec y · sec y tan y · y′ – 2 tan x · sec2 x = 0 ⇒

yx x

y y′ = sec tan

sec tan

2

2

15. tan xy = xy ⇒(sec2 xy) · (y + xy′) = y + xy′ ⇒y′(x sec2 xy – x) = y – y sec2 xy ⇒

yy xy

x xyy

y

x′ = ⇒ ′ =( – sec )

(sec – )

1

1

2

2 –

16. cos xy = xy ⇒(–sin xy) · (y + xy′) = y + xy′ ⇒

y′(–x – x sin xy) = y + y sin xy ⇒

yy xy

x xyy

y

x′ = + ⇒ ′ =( sin )

(– – sin )

1

1–

17. sin y = x ⇒ cos y · y′ = 1 ⇒ y′ = sec y

18. cos y = x ⇒ –sin y · y′ = 1 ⇒ y′ = –csc y

19. csc y = x ⇒ –csc y cot y · y′ = 1 ⇒y′ = –sin y tan y

20. cot y = x ⇒ –csc2 y · y′ = 1 ⇒ y′ = –sin2 y

21. y = cos− 1 x ⇒ cos y = x ⇒ –sin y · y′ = 1 ⇒

yy x

′ = =– –1 1

1 2sin –

y

1

x

√1 – x 2

22. y = ln x ⇒ ey = x ⇒ ey · y′ = 1 ⇒ y′ = =1 1

e xy

23. y = x11/5 ⇒ y5 = x11 ⇒ 5y4 · y′ = 11x10 ⇒

yx

y

x

x

x

xx′ = = = =11

5

11

5

11

5

11

5

10

4

10

11 5 4

10

44 56 5

( )/ // ,

which is the answer obtained using the derivativeof a power formula, Q.E.D.

24. Prove that if y = xn, where n = a/b and a and bare integers, then y′ = nan− 1.

Proof:

y = xn = xa/ b ⇒ yb = xa.Because a and b are integers,byb− 1 y′ = ax a− 1

yax

by

ax

b x

ax

bx

a

bx

a

b

a

a b b

a

a a ba a a b′ = = = = − − −

–

–

–

/ –

–

– /( / )

( )

1

1

1

1

11

= =a

bx nxa b n/ – – ,1 1 Q.E.D.

25. a. x2 + y2 = 100At (–6, 8), (–6)2 + 82 = 100, which shows that (–6, 8) is on the graph, Q.E.D.

b. x2 + y2 = 100 ⇒ 2x + 2y · dy/dx = 0 ⇒ dy/dx = –x/yAt (–6, 8), dy/dx = –(–6)/8 = 0.75.A line at (–6, 8) with slope 0.75 is tangent tothe graph, showing that the answer isreasonable.

x

y10

10


c. x = 10 cos t y = 10 sin tdy

dx

t

t

t

t= =10

10

cos

– sin

cos

sin–

At x = –6, t = cos–1 (–0.6).sin [cos–1 (–0.6)] = 0.8

∴ = =dy

dx– . ,

– .

.

0 6

0 80 75

which agrees with part b, Q.E.D.

26. a. x2 – y2 = 36At (10, –8), 102 – (–8)2 = 36, which showsthat (10, –8) is on the graph, Q.E.D.

b. x2 – y2 = 36 ⇒ 2x – 2y · dy/dx = 0 ⇒dy/dx = x/yAt (10, –8), dy/dx = 10/(–8) = –1.25.A line at (10, –8) with slope –1.25 is tangentto the graph, showing that the answer isreasonable.

x

y

10

10

c. x = 6 sec t y = 6 tan tdy

dx

t t

t

t

t= =6

6 2

sec tan

sec

tan

sec

At x = 10, t = ±sec–1 (10/6).tan [±sec–1 (10/6)] = ±8/6.Choose the negative value because y < 0.

∴ = =dy

dx

– /

/

10 6

8 61 25– . ,

which agrees with part b, Q.E.D.

27. a. x3 + y3 = 64 ⇒ 3x2 + 3y2 · dy/dx = 0 ⇒dy/dx = –x2/y2

x = 0: y3 = 64 ⇒ y = 4∴ dy/dx = –0/16 = 0The tangent is horizontal (see the next graph).x = 2: 8 + y3 = 64 ⇒ y3 = 56 ⇒y = 3.8258…∴ dy/dx = –22/(3.8258…)2 = –0.2732…The tangent line has a small negative slope,which agrees with the graph.x = 4: 64 + y3 = 64 ⇒ y = 0∴ dy/dx = –42/0, which is infinite.The tangent line is vertical.

x

y

10

10

b. y = x: x3 + x3 = 64 ⇒ x3 = 32 ⇒x = 3.1748…dy/dx = –x2/y2 = –x2/x2 = –1

c. y = (64 – x3)1/3

As x becomes infinite, (64 – x3)1/3 gets closerto (–x3)1/3, which equals –x. The graph hasa diagonal asymptote at y = –x, anddy/dx → –1.

d. By analogy with the equation of a circle, suchas x2 + y2 = 64

28. a. First simplify the equation.[(x – 6)2 + y2][(x + 6)2 + y2] = 1200(x – 6)2(x + 6)2 + (x – 6)2y2 + (x + 6)2y2 + y4

= 1200(x2 – 36)2 + (x2 – 12x + 36 + x2 + 12x + 36)y2

+ y4 = 1200x4 − 72x2 + 1296 + 2x2y2 + 72y2 + y4 = 1200x4 − 72x2 + 2x2y2 + 72y2 + y4 = −96Differentiate the simplified equationimplicitly.4x3 − 144x + 4xy2 + 4x2y · dy/dx

+ 144y · dy/dx + 4y3 · dy/dx = 0(4x2 y + 144y + 4y3) · dy/dx = −4x3

+ 144x − 4xy2

dy

dx

x x xy

x y y y= +

+ +– –3 2

2 3

36

36At x = 8: (4 + y2)(196 + y2) = 1200784 + 200y2 + y4 = 1200y4 + 200y2 − 416 = 0

y2 200 41664

22 058806= ± =–. K or

−202.0…y = ±1.4348542… (No other real solutions)At (8, 1.434…), dy/dx = −1.64211… .At (8, –1.434…), dy/dx = 1.64211… .Both answers agree with the moderately steepnegative and positive slopes, respectively.

x

y

10

5

b. At the x-intercepts, y = 0.∴ (x − 6)2 (x + 6)2 = 1200(x2 − 36)2 = 1200

x = ± ± = ±36 1200 8.4048K or±1.1657…Derivative appears to be infinite at eachx-intercept.

At x = + =36 1200 8.4048K ,


dy

dx= +

+ +–( . ) ( . ) – ( . )( )

( . ) ( ) ( )

8 4 36 8 4 8 4 0

8 4 0 36 0 0

3

2 3

K K K

K

= 896 29

0

.,

K which is infinite, as conjectured.

c. From part a,x4 − 72x2 + 2x2y2 + 72y2 + y4 = −96 ⇒y4 + (2x2 + 72)y2 + (x4 − 72x2 + 96) = 0

y2 =

− + ± + − − +( ) ( ) ( )( )2 72 2 72 4 1 72 96

2

2 2 2 4 2x x x x

y x x2 2 236 144 1200= − − ± –

Only the positive part of the ambiguoussign ± gives real solutions for y.

y x x= ± +– – –2 236 144 1200

Plot the graph letting y1 equal the positivebranch and y2 equal the negative branch. Thegraph is as in the text. The two loops maynot appear to close, depending on the windowyou use for x.

d. Repeating the algebra of parts a and c with1400 in place of 1200 gives

y x x= ± +– – –2 236 144 1400

Plot the graph as in part a. The two ovals inthe original graph merge into a single closedfigure resembling an (unshelled) peanut.

x

y

10

5

e. The two factors in the equation[(x − 6)2 + y2][(x + 6)2 + y2] = 1200are the squares of the distances from (x, y) tothe points (6, 0) and (−6, 0), respectively.The product of the distances is 1200, aconstant.

Problem Set 4-9Q1. y2 + 2xyy′ Q2. Implicit differentiation

Q3. Product rule Q4. Chain rule

Q5. Speeding up Q6. smaller

Q7. cos x − x sin x Q8.1

x

dx

dt

Q9. −2e− x + xe− x Q10. E

1. Know: dA

dt= 12 mm2/h. Want:

dr

dt.

A rdA

dtr

dr

dt

dr

dt r= ⇒ = ⇒ =π π

π2 2

6

3

r

dr/dt

1

dr

dt= =2

0 6366π

. K mm/h when r = 3 mm.

dr

dt varies inversely with the radius.

2. Know: dr

dt= 2 cm/s. Want:

dV

dt.

V rdV

dtr

dr

dt= ⇒ =4

343 2π π

dV

dt= =72 226 1946 3π . /K cm s at r = 3 cm

dV

dt= =288 904 7786π . /K cm s3 at r = 6 cm

3 6

dV/dt

r

500

The graph shows that the larger the balloon gets,the faster Phil must blow air to maintain the2 cm/s rate of change of radius.

3. Know: dA

dt= −144 cm /s.2 Want:

da

dt.

A = πab and a = 2b ⇒ A a= 1

22π

dA

dta

da

dt

da

dt a

dA

dt= ⇒ =π

π1

b a

da

dt= ⇒ = ⇒ = − = −12 24

61 9098

π. K

≈ −1 91. cm/sThe length of the major axis is 2a, so the majoraxis is decreasing at 12/π cm/s.

4. Know: dK

dt= 100 000, MJ/s;

dm

dt= −20 kg/s.

Want: dV

dt. (Note: 1 megaJoule—MJ—is the

energy required to accelerate a 1-kg mass by1 km/s through a distance of 1 km; it can beexpressed 1 MJ = 1 kg · km2/s2.)

K mVdK

dtV

dm

dtmV

dV

dt= ⇒ = + ⇒1

2

1

22 2


dV

dt mV

dK

dt

V

m

dm

dt= −1

2dV

dt=

⋅−

⋅=100000

5000 10

10 20

2 50002 02

(– ). (km/s)/s

5. Let y = Milt’s distance from home plate.Let x = Milt’s displacement from third base.

Know:dx

dt= −20 ft/s. Want:

dy

dt.

y x ydy

dtx

dx

dt2 2 290 2 2= + ⇒ =

⇒ = ⋅ =+

dy

dt

x

y

dx

dt

x

x

–20

902 2

90

10

x

dy/dt

At xdy

dt= = − ≈ −45 8 944 8 9, . K . ft/s

(exact: −4 5 ).

At xdy

dt= =0 0, ft/s, which is reasonable because

Milt is moving perpendicular to his line fromhome plate.

6. Let y = displacement from stern to dock alongpier. Let x = displacement from bow to pieralong dock.

Know: dy

dt= −3 m/s. Want:

dx

dt.

x2 + y2 = 2002

2 2 03

2002 2x

dx

dty

dy

dt

dx

dt

y

x

dy

dt

y

y+ = ⇒ = − =

–

At ydx

dt= = =120

360

1602 25, . m/s.

200

10

y

dx/dt

120

7. a. Let L = length. Let W = width. Let H = depth(meters).

Know: dW

dt

dL

dt= = −0 1 0 3. m/s; . m/s.

Want: dH

dt.

LWH = ⇒20dL

dtWH L

dW

dtH LW

dH

dt⋅ + ⋅ ⋅ + ⋅ = 0

dH

dt

H

W

dW

dt

H

L

dL

dt

LW L W

= − ⋅ − ⋅

= − −200 1

200 32 2( . ) (– . )

b.dH

dt= − ⋅

⋅+ ⋅

⋅=20 0 1

5 2

20 0 3

5 20 022 2

. ..

Depth is increasing at 0.02 m/s.

8. Let L = distance between spaceships.

Know:dx

dt

dy

dt= = −80 50 km/s; km/s.

Want: dL

dt.

L x y LdL

dtx

dx

dty

dy

dt2 2 2 2 2 2= + ⇒ = +

⇒ =+

dL

dt x yx y

180 50

2 2( – )

dL

dt= ⋅ ⋅ = =1

130080 500 50 1200

200

13( – )

–

−15.3846…

Distance is decreasing at about 15.4 km/s.

9. a. Let x = distance from bottom of ladder towall. Let y = distance from top of ladder tofloor.

20 0 2 22 2 2= + ⇒ = + ⇒x y xdx

dty

dy

dtdy

dt

x

y

dx

dt= −

Note that the velocity of the weight is −dy/dt,so

vx

x

dx

dt=

400 2–

b. v = ⋅ − = − = −4

3843

6

40 6123( ) . ft/sK

c. Here xdx

dtv= = = → ∞20 2

40

0, , so (!!)

10. a.

1200 in.2W

L

D

Know: dL

dt. Want:

dW

dt.

LWdL

dtW L

dW

dt= ⇒ ⋅ + ⋅ =1200 0

⇒ = − ⋅ = − ⋅dW

dt

W

W

dL

dtW

dL

dt1200

1

12002

/

b. − = − ⇒ = ⇒21

12006 202W W( ) in.

L = 60 in.


c. D L W DdD

dtL

dL

dtW

dW

dt2 2 2 2 2 2= + ⇒ = +

⇒ =+

+

dD

dt L WL

dL

dtW

dW

dt

12 2

At L = 60 and W = 20,dD

dt=

++ −1

20 6060 6 20 2

2 2[ ( ) ( )]

= = =320

40001 6 10 5 0596. . K

Diagonal is increasing at about 5.06 in./min.

11. a. Let h = depth of water. Let r = radius of waterat surface. Let V = volume of water.

Know: dh

dt = 5 m/h. Want:

dV

dt.

V r h= 1

32π

By similar triangles, r

h =

3

5 ⇒ r =

3

5h

∴ =

=V h h h

1

3

3

5

3

25

23π π

dV

dth

dh

dt= 9

252π

At h = 3,

dV

dt= = =81

516 2 50 8938π π. . K ≈

50.9 m3/h.

b. i. Know: dV

dt = −2 m3/h. Want:

dh

dt.

dV

dth

dh

dt

dh

dt h

dV

dt= ⇒ =9

25

25

92

2ππ

dh

dt= = −–50

1440 1105

π. K ≈

−0.11 m/h at h = 4 m

ii. dh

dt → −∞ as h → 0 m

c. i. Know: dV

dtk h= .

dV

dth k= − = ⇒ = −0 5 4 0 25. at .

dV

dth= −0 25.

ii. dV

dt= − = −0 25 0 64 0 2. . m /h3.

at mh = 0 64.

iii. dV

dth= − = ⇒0 2 0 64. at . m

dh

dt= = − ≈25

9 0 640 2 0 43172π ( . )

(– . ) . K

−0.43 m/h

12. Let h = altitude. Let r = radius. Let V = volumeof cone.

Know: dh

dt = −6 ft/min;

dr

dt = 7 ft/min. Want:

dV

dt.

V r hdV

dtr

dr

dth r

dh

dt= ⇒ = +1

3

2

3

1

32 2π π π

dV

dt= + − =2

38 3 7

1

38 62π π( )( )( ) ( ) ( )

−16π ft3/min = −50.2654…Volume is decreasing at about 50.3 ft3/min.

13. a. Let ω = angular velocity in radians per day.

ω π ω πE M, = =2

365

2

687

d

dt

θ ω ω π= − = −

=E M 2

1

365

1

687

644

2507550 008068

π = . K ≈ 0.00807 rad/day

b. T = −

= ≈

−1

365

1

687778 7422

1

. K

778.7 days

The next time after 27 Aug. 2003 when thetwo planets will be closest is 779 days later,on 14 Oct. 2005 (or 15 Oct., if the planetswere aligned later than about 6:11 a.m. backon 27 Aug. 2003). Because the actual orbitsof Earth and Mars are not as simple aspreviously assumed, the actual closestdistances are not always the same. In fact, theapproach on 27 Aug. 2003 was the closestone in nearly 60,000 years! Nor is the periodbetween close approaches quite so simple.The next close approach will actually be on30 Oct. 2005, not 15 Oct.

c. By the law of cosines,

D2 = 932 + 1412 − 2 · 93 · 141 cos θD = 28530 26226– cos θ million mi

d.dD

dt

d

dt= 26226

2 28530 26226

sin

cos

θθ

θ–

=⋅

⋅26226

1365

1687

2

2 28530 26226

–

–

π θ

θ

sin

cosmillion mi/day

=⋅ ⋅

⋅

⋅

1 000 000 262261

3651

6872

24 2 28530 26226

, , sin

cos

–

–

π θ

θ

=⋅ −

⋅

−

1 092 750 0001

3651

68728530 26226

, , , sin

cos

π θ

θmi/h

To find out how fast D is changing today,first determine how many days after 27 Aug.2003 it is today, then multiply that number

by d

dt

θ π= −

1

365

1

6872 to find θ, then

substitute θ into the previous expressions.


e. To maximize dD

dt, plot the variable part of

dD

dt, y =

−sin

cos.

θθ28530 26226

y0.01

π 2π

θ

From the graph, it is clear that the maximumoccurs well before θ = π/2 (90°). Using themaximize feature, the maximum occurs atθ ≈ 0.8505… , or 48.7…°.(The exact value is cos− 1 (93/141). One canfind this by finding (d/dt)(dD/dt) and setting itequal to zero. One can also see this bydecomposing Earth’s motion vector into twocomponents—one toward/away from Mars andthe other perpendicular to the first. The rate ofchange in D is maximized when all of Earth’smotion is along the Earth-to-Mars component,which occurs when the Earth-Mars-Suntriangle has a right angle at Earth.In this case, cosθ = 93

141 .)

f. θ π= −

1

365

1

6872 t if t = days since

27 Aug. 2003.

D t=

28530 262261

365

1

6872– cos – π

1000

200

t

D

The graph is not a sinusoid. The high and lowpoints are not symmetric.

14. As B moves from negative values of x to positivevalues of x, the length of AB decreases to about0.56 unit, then begins to increase when the x-value of point B passes about −0.3.Let l = length of AB.

Know: dx

dt= 2 units/s. Want:

dl

dt.

l e xdl

dt

e x xdx

dte

dx

dt

e x

e x

x

x x

x

x

= + ⇒

= + ⋅ +

= +

+

−

0 8 2

0 8 2 1 2 0 8

0 8

0 8 2

1

22 0 8

0 8 2

.

. / .

.

.

( ) .

.

dl

dt = −1.9963… units/s at x = −5 units.

dl

dt = 2.6610… units/s at x = 2 units.

The length of AB is at a minimum when dl/dt = 0.Use your grapher to solve 0.8e0.8x + 2x = 0.At x = −0.3117… , the length of AB stopsdecreasing and starts increasing.

Problem Set 4-10

Review Problems


R1. a. x = g(t) = t3 ⇒ g′(t) = 3t2

y = h(t) = cos t ⇒ h′(t) = −sin tIf f (t) = g(t) · h(t) = t3 cos t, then, for example,f ′(1) = 0.7794… by numerical differentiation.g′(1) · h′(1) = 3(12) · (−sin 1) = −2.5244…∴ f ′(t) ≠ g′(t) · h′(t), Q.E .D.

b. If f (t) = g(t)/h(t) = t3/cos t, then, for example,f ′(1) = 8.4349… by numerical differentiation.g′(1)/h′(1) = 3(12)/(−sin 1) = 3.5651…∴ f ′(t) ≠ g′(t)/h′(t), Q.E .D.

c. y = cos tx = t3 ⇒ t = x1/3 ⇒ y = cos (x1/3 )dy

dxx x= − ⋅ −sin ( )/ /1 3 2 31

3

At x = 1,

dy

dx= − ⋅ = −sin 1 0.280490

1

3K .

If x = 1, then t = 11/3 = 1.

∴ = = = −dy dt

dx dt

t

t

/

/

– sin – sin

3

1

30 2804902 . ,K

which equals dy/dx, Q.E.D.

R2. a. If y = uv, then y′ = u′v + uv′.b. See the proof of the product formula in the

text.

c. i. f (x) = x7 ln 3x ⇒f ′(x) = 7x6 ln 3x + x

x7 3

3⋅ = 7x6 ln 3x + x6

ii. g(x) = sin x cos 2x ⇒g′(x) = cos x cos 2x − 2 sin x sin 2x

iii. h(x) = (3x − 7)5(5x + 2)3

h′(x) = 5(3x − 7)4(3) · (5x + 2)3

+ (3x − 7)5(3)(5x + 2)2(5)= 15(3x − 7)4(5x + 2)2(5x + 2

+ 3x − 7)= 15(3x − 7)4(5x + 2)2(8x − 5)

iv. s(x) = x8e− x ⇒ s′(x) = −x8e− x + 8x7e− x

d. f (x) = (3x + 8)(4x + 7)i. f ′(x) = 3(4x + 7) + (3x + 8)(4) = 24x + 53

ii. f (x) = 12x2 + 53x + 56

f ′(x) = 24x + 53, which checks.


R3. a. If y = u/v, then yu v uv

v′ = ′ ′–

2 .

b. See proof of quotient formula in text.

c. i. f xx

x( ) = ⇒sin 10

5

f xx x x x

xx x x

x

′ = ⋅ ⋅

=

( )10 10 10 5

10 10 5 10

5 4

10

6

cos – sin

cos – sin

ii. g xx

xg x( ) ( )= + ⇒ ′( )

( – )

2 3

9 5

9

4

= + ⋅ − − + ⋅ − ⋅−

9 2 3 2 9 5 2 3 4 9 5 9

9 5

8 4 9 3

8( ) ( ) ( ) ( )

( )

x x x x

x

= +18 2 3 5 11

9 5

8

5

( ) ( – )

( – )

x x

x

iii. h(x) = (100x3 − 1)− 5 ⇒h ′(x) = −5(100x3 − 1)− 6 · 300x2

= −1500x2(100x3 − 1)− 6

d. y = 1/x10

As a quotient:

yx x

x xx′ = ⋅ ⋅ = = − −0 1 10 10

1010 9

20 1111– –

As a power:

y = x− 10

y′ = −10x− 11, which checks.

e. t xx

xx( ) = =sin

costan

′ =t xx x x x

x( )

cos cos – sin (– sin )

cos2

= + = =cos sin

cos cossec

2 2

2 221x x

x xx

t ′ (1) = sec2 1 = 3.4255…

f. m xt x t

x

x

x( ) = =( ) – ( )

–

tan – tan

–

1

1

1

1

1

1

x

m(x)

3.42...

x m(x)

0.997 3.40959…

0.998 3.41488…

0.999 3.42019…

1 undefined

1.001 3.43086…

1.002 3.43622…

1.003 3.44160…

The values get closer to 3.4255… as xapproaches 1 from either side, Q.E.D.

R4. a. i. y = tan 7x ⇒ y′ = 7 sec2 7x

ii. y = cot (x4) ⇒ y′ = −4x3 csc2

(x4)

iii. y = sec e x ⇒ y′ = e x sec e x tan e x

iv. y = csc x ⇒ y′ = −csc x cot x

b. See derivation in text for tan′x = sec2 x.

c. The graph is always sloping upward, whichis connected to the fact that tan′ x equals thesquare of a function and is thus alwayspositive.

�

x

y

1

d. f (t) = 7 sec t ⇒ f ′(t) = 7 sec t tan t

f ′ (1) = 20.17…

f ′ (1.5) = 1395.44…

f ′ (1.57) = 11038634.0…

There is an asymptote in the secant graph att = π /2 = 1.57079… . As t gets closer to thisvalue, secant changes very rapidly!

R5. a. i. y x yx

= ⇒ ′ =+

−tan 123

3

1 9

ii. d

dxx

x x(sec )

| | –

–1

2

1

1=

iii. c x x c xx

x( ) ( ) ( )2= ⇒ ′ =−cos

– cos

–

–1

1

2

2

1

b. y x yx

= ⇒ ′ =−

−sin 1

2

1

1

1

�/2

x

y

y′ = =( )01

1 01

2–, which agrees with the

graph.

y′ = =( )11

1 1

1

02–, which is infinite.

The graph becomes vertical as x approaches 1from the negative side. y′(2) is undefinedbecause y(2) is not a real number.

R6. a. Differentiability implies continuity.


b. i. Answers may vary. ii. Answers may vary.

c

x

f(x)

c

x

f(x)

iii. No such function. iv. Answers may vary.

c

x

f(x)

c. i.

1

2

x

f(x)

ii. f is continuous at x = 1 because right andleft limits both equal 2, which equals f (1).

iii. f is differentiable. Left and right limitsof f ′ (x) are both equal to 2, and f iscontinuous at x = 2.

d. g xx x

x ax b x( )

if

if =

≤ ≤+ + ≤

sin ,

,

–1

2

0 1

0

g xx x

x a x′ =

− < <+ <

−

( )if

if

( ) ,

,

/1 0 1

2 0

1 2

limx

g x a b b→ −

= + + =0

0 0( )

lim ( ) sinx

g x→

−+

= =0

1 0 0

∴ b = 0lim

–xg x a a

→′ = + =

00( )

lim /

xg x

→

−+

′ = =0

1 21 1( )

∴ a = 1

10

1

x

g(x)

The graph appears to be differentiable andcontinuous at x = 0.

R7. a. x e y tdy

dx

dy dt

dx dt

t

et

t= = ⇒ = = ⇒2 32

2

3

2,

/

/

d y

dx

d

dx

t

e t

2

2

2

2

3

2=

= ⋅ − ⋅

= − ÷ = −

6 2 3 4

2

3 3 3 3

2

2 2 2

2 2

2

2

2

4

t dt dx e t e dt dx

e

t t

e

dx

dt

t t

e

t t

t

t t

( / ) ( / )

( )

b. x = (t/π) cos t

y = (t/π) sin tdy

dx

dy dt

dx dt

t t t

t t t= = +

+/

/

( / ) sin ( / )(cos )

( / ) cos ( / )(– sin )

1

1

π ππ π

= +sin cos

cos – sin

t t t

t t tWhere the graph crosses the positive x-axis,t = 0, 2π, 4π, 6π, … .If t = 6π, x = 6 and y = 0.∴ (6, 0) is on the graph.If t = 6π, thendy

dx= + = + =sin cos

cos – sin –

6 6 6

6 6 6

0 6

1 06

π π ππ π π

π π .

So the graph is not vertical where it crossesthe x-axis. It has a slope of 6π = 18.84… .

c. At a high point, y is a maximum and x is zero.Use cosine for y and sine for x.For y, the sinusoidal axis is at 25 ft.For x, the sinusoidal axis is at 0 ft.Both x and y have amplitude 20 ft, the radiusof the Ferris wheel.The phase displacement is 3 seconds.The period is 20 seconds, so the coefficientof the arguments of sine and cosine is2π/20 = π /10.

x t= 2010

3sin ( – )π

y t= +25 2010

3cos ( – )π

dx dt t/ = 210

3π πcos ( – )

dy dt t/ = −210

3π πsin ( – )

When t = 0, dy/dt = 5.0832… .The Ferris wheel is going up at about5.1 ft/s.When t = 0, dx/dt = 3.6931… .The Ferris wheel is going right at about3.7 ft/s.dy

dx

dy dt

dx dt= /

/dy/dx will be infinite if dx/dt = 0 anddy/dt ≠ 0.

dx/dt = 0 if 210

3 0π πcos ( – )t = .

π π π10

32

( – )t n= + (where n is an integer)

t = 8 + 5n

The first positive time is t = 8 s.


R8. a. y = x8/5 ⇒ y5 = x8

5 88

5

8

5

8

54 7

7

4

7

8 5 43 5y y x y

x

y

x

xx′ = ⇒ ′ = = =

( )//

Using the power rule directly:

y x y x= ⇒ ′ =8 5 85

3 5/ /

b. y3 sin xy = x4.5 ⇒

3y2y′ · sin xy + y3(cos xy)(y + xy′) = 4.5x3.5

y′[3y2 sin xy + xy3

cos xy]

= 4.5x3/5 − y4 cos xy

ydy

dx

x y xy

y xy xy xy′ = =

+4 5

3

3 5 4

2 3

. – cos

sin cos

.

c. i. 4y2 − xy2 = x3 ⇒8yy′ − y2 − x · 2yy′ = 3x2

y′(8y − 2xy) = 3x2 + y2

ydy

dx

x y

y xy′ = = +3

8 2

2 2

–At (2, 2), dy/dx = 2. At (2, −2), dy/dx = −2.Lines at these points with these slopes aretangent to the graph (see diagram).

x

y

5

5

2

ii. At (0, 0), dy/dx has the indeterminate form0/0, which is consistent with the cusp.

iii. To find the asymptote, solve for y.(4 − x)y2 = x3

yx

x2

3

4=

–As x approaches 4 from the negative side,y becomes infinite. If x > 4, y2 is negative,and thus there are no real values of y.Asymptote is at x = 4.

R9.

70

x

z

Let x = Rover’s distance from the table.Let z = slant length of tablecloth.

Know:dx

dt= 20 cm/s. Want:

dz

dt at z = 200.

z2 = x2 + 702

2 2zdz

dtx

dx

dt =

dz

dt

x

z

dx

dt

x

z= = 20

At z = 200, x = =200 70 30 392 2–

dz

dt= = = . .

20 30 39

2003 39 18 7349

⋅..

The glass moves at the same speed as thetablecloth, or about 18.7 cm/s, which is about1.3 cm/s slower than Rover.

Concept Problems

C1. a. Let (x, y) be the coordinates of a point on thetangent line.y y

x xm y m x x y

–

–0

00 0= ⇒ = − +( )

b. Substituting (x1, 0) for (x, y) gives

0 1 0 0 1 00= − + ⇒ = −m x x y x x

y

m( ) , Q.E.D.

c. The tangent line intersects the x-axis at (x2, 0).Repeating the above reasoning with x2 and x1

in place of x1 and x0 gives

x xy

m2 11= −

Because y1 = f (x1) and m = f ′(x1),

x xf x

f x2 11

1

= −′( )

( ), Q.E.D.

d. Programs will vary according to the kind ofgrapher used. The following steps are needed:

• Store f (x) in the Y= menu.

• Input a starting value of x.

• Find the new x using the numericalderivative.

• Display the new x.

• Save the new x as the old x and repeat.

For f (x) = x2 − 9x + 14, the program shouldgive x = 2, x = 7.

e. For g(x) = x3 − 9x2 + 5x + 10, first plot thegraph to get approximations for the initialvalues of x.

20

1

x

g(x)

Run the program three times with x0 = −1, 1,and 8. The values of x are

x = −0.78715388…

x = 1.54050386…

x = 8.24665002…

The answers are the same using the built-insolver feature. The same preliminary analysisis needed to find starting values of x.


f. f (x) = sec x − 1.1

Starting with x0 = 1, it takes seven iterationsto get x = 0.429699666… .

C2. a. The connecting rod, the crankshaft, and they-axis form a triangle with angle φ = θ − π /2included between sides of 6 cm and(y − 8) cm.

8

y – 820

6

θφ

By the law of cosines,

202 = (y − 8)2 + 62 − 2 · 6 ·(y − 8) cos (θ − π /2)

202 = (y − 8)2 + 62 − 12(y − 8) sin θ(y − 8)2 − 12 sin θ (y − 8) − 364 = 0

Solve for y − 8 using the quadratic formula.

y − = + ⋅8 6 36 3642sin sin .θ θ( + ) (The

solution with the negative radical gives atriangle below the origin, which has no real-life meaning.)

y = + + ⋅8 6 2 9 912sin sinθ θ +

b. vdy

dt

d

dt

d

dt= = +

⋅6

18

9 912cos

sin cos

sinθ θ θ θ

θθ

+

vd

dt

d

dt= +

⋅6

9 2

9 912cos

sin

sinθ θ θ

θθ

+

c. ad y

dt=

2

2

= −

6 sin θ θd

dt

2

+

18

91 2 9

9 91

4

2 3 2

2cos sin

sin

θ θθ

θ–

( + ) /

d

dt

= ⋅

18

91 2 9

9 916

4

2 3 2

2cos sin

sinsin

θ θθ

θ θ–

( + )–/

d

dt

(There are many other correct forms of theanswer, depending on how you use thedouble-argument properties and Pythagoreanproperties from trigonometry.)Note that the angular velocity is constant at6000π radians per minute, sod

dt

θ π= 100 rad/s.

d. See the graph. Note that a line at a = −980is so close to the x-axis that it does notshow up.

500,000

3

θ

a

Solving graphically and numerically,a < −980 for θ ∈ (0.2712… , 2.8703…). Thepiston is going down (v < 0) forθ ∈ (π /2, 3π /2).So the piston is going down with accelerationgreater than gravity for θ between π /2 and2.8703… .

Chapter Test

T1. y = uv ⇒ y′ = u′v + uv′

T2. ′ =

+ ∆+ ∆

−

∆⋅ + ∆

+ ∆

∆ →

y

u uv v

uv

x

v v v

v v vxlim

( )

( )0

= + ∆ − + ∆∆ + ∆

∆ →

lim( ) ( )

( )x

u u v u v v

x v v v0

= + ∆ − − ∆∆ + ∆

∆ →

lim( )x

uv uv uv u v

x v v v0

=+ ∆

⋅ ∆ ⋅ − ∆∆

∆ →

lim( )x v v v

u v u v

x0

1

=+ ∆

⋅ ∆∆

− ∆∆

∆ →

lim( )x v v v

u

xv u

v

x0

1

= −

= ′ − ′1

2 2v

du

dxv u

dv

dx

u v uv

v[Because as ∆x → 0, u/x and ∆ ∆v x/ become

du/dx and dv/dx and v → 0, so ( ) ]v v v v+ →∆ 2 .

T3. cotcos

sinx

x

x=

= − −sin sin cos cos

sin

x x x x

x2

= − + = − = −(sin cos )

sin sincsc

2 2

2 221x x

x xx

T4. y x y x y y= ⇒ = ⇒ ′ = ⇒−sin sin cos1 1

yy

y y y y′ = + = ⇒ = − ⇒11 12 2 2 2

coscos sin cos sin

cos siny y x yx

= − = − ⇒ =−

1 11

1

2 2

2

T5.dy

dx

dy dt

dx dt

t

tt

d y

dx

d

dxt= = = = =/

/

4

22 2

32

2

22( )

4 44

22t dt dx t

dx

dt

t

t( / ) = ÷ = =


T6. c(x) = cot 3x

c′(x) = −3 csc2 3x, which is negative for all

permissible values of x.c′(5) = −3 csc2 15 = −3/sin2 15 = −7.0943…

c(t) is decreasing at about 7.1 y-units/x-unit.

T7. f (x) = sec x ⇒ f ′ (x) = sec x tan x

f ′ (2) = sec 2 tan 2 = 5.25064633…

Use m(x) for the difference quotient.

m xx

x( ) = 1 1 2

2

/cos – /cos

–

x m(x)

1.997 5.28893631…

1.998 5.27611340…

1.999 5.26335022…

2.000 undefined

2.001 5.23800134…

2.002 5.22541482…

2.003 5.21288638…

T8. Answers may vary.

x

f(x)

2

7

1–2

T9. f (x) = mx + b

f ′ (x) = m for all x

∴ f is differentiable for all x.

∴ f is continuous for all x, Q.E.D.

T10. f (x) = sec 5x ⇒ f ′(x) = 5 sec 5x tan 5x

T11. y = tan7/3 x ⇒ y′ = 73 tan4/3 x

T12. f (x) = (2x – 5)6(5x – 1)2

f ′ (x) = 6(2x – 5)5(2) ⋅ (5x – 1)2

+ (2x – 5)6 ⋅ 2(5x – 1) ⋅ 5= 2(2x – 5)5(5x – 1)[6(5x – 1) + 5(2x – 5)]

= 2(2x – 5)5(5x – 1)(40x – 31)

T13. ( )f xe

x

x

= ⇒3

ln

′ = − = −f x

e x e x

x

xe x e

x x

x x x x

( )ln ( / )

(ln )

ln

(ln )

3 1 33 3

2

3 3

2

T14. x = sec 2t

y = tan 2t3

dy

dx

dy dt

dx dt

t t

t t

t t

t t= = ⋅

⋅=/

/

sec

sec tan

sec

sec tan

2 3 2 2 2 32 6

2 2 2

3 2

2 2

T15. y = 4 sin− 1 (5x3)

yx

xx

x′ = ⋅ ⋅ =4

1

1 515

60

1 253 2

22

6– ( ) –T16. 9x2 − 20xy + 25y2 − 16x + 10y − 50 = 0 ⇒

18x − 20y − 20xy ′ + 50yy ′ − 16 + 10y ′ = 0y ′(−20x + 50y + 10) = −18x + 20y + 16

ydy

dx

x y

x y′ = = + +

+ +–

–

18 20 16

20 50 10

= + ++ +

–

–

9 10 8

10 25 5

x y

x y

If x = −2, then

36 + 40y + 25y2 + 32 + 10y − 50 = 0

25y2 + 50y + 18 = 0

Solving numerically givesy = −0.4708… or y = −1.5291… , both ofwhich agree with the graph.

(Solving algebraically by the quadratic formula,

y = − ±1 7 5 / , which agrees with the numericalsolutions.)

At (−2, −0.4708…), dy/dx = 1.60948… .

At (−2, −1.5291…), dy/dx = −0.80948… .

The answers are reasonable, because lines ofthese slopes are tangent to the graph at therespective points, as shown here.

x

y

5

5

–2

T17. f xx x

a x b x( )

if

if=

+ ≤+ >

3

2

1 1

2 1

,

( – ) ,

f xx x

a x x′ =

<>

( )if

if

3 1

2 2 1

2,

( – ),

For equal derivatives on both sides of x = 1,

lim–x

f x→

′ = ⋅ =1

23 1 3( )

limx

f x a a→ +

′ = − = −1

2 1 2 2( ) ( )

∴− = ⇒ = −2 3 1 5a a .

For continuity at x = 1,

lim–x

f x→

= + =1

31 1 2( )

limx

f x a b a b→ +

= − + = +1

1 2( ) ( )2

∴ + =a b 2

Substituting a = −1.5 gives b = 3.5.


The graph shows differentiability at x = 1.

1

2

x

f(x)

Values of b other than 3.5 will still cause thetwo branches to have slopes approaching 4as x approaches 1 from either side as long asa = −1.5. However, f will not be continuous,and thus will not be differentiable, as shownhere for b = 4.5.

1

2

x

f(x)

T18. y = x7/3 ⇔ y3 = x7

3y2y′ = 7x6

yx

y

x

xx x′ = = = =−7

3

7

3

7

3

7

3

6

2

6

7 3 26 14 3 4 3

( )// /

This answer agrees with y′ = nxn− 1. 4/3 is7/3 − 1.

T19. cot = adjacent/opposite = x/5 = cot− 1 (x/5)

T20.d

dx x x

θ = −+

⋅ = −+

1

1 5

1

5

1

5 5 252 2( / ) ( / )

= −+

= −+

1

5 5

5

252 2( / )x x

T21.dx

dt= −420 mi/h

T22.d

dt

d

dx

dx

dt x x

θ θ= ⋅ = −+

⋅ − =+

5

25420

2100

252 2( )

T23. The plane is changing fastest when x approacheszero, when the plane is nearest the station.

30

25

x

y



Chapter 5—Definite and Indefinite Integrals

Problem Set 5-11. f (1000) = 20 + (0.000004)(10002) = 24 $/ft

f (4000) = 20 + (0.000004)(40002) = 84 $/ft

The price increases because it is harder and slowerto drill at increasing depths.

2. T6 = 500(24 + 29)/2 + 500(29 + 36)/2

+ 500(36 + 45)/2 + 500(45 + 56)/2

+ 500(56 + 69)/2 + 500(69 + 84)/2= 13,250 + 16,250 + 20,250 + 25,250

+ 31,250 + 38,250 = 144,500

About $144,500, an overestimate because thetrapezoids are circumscribed above the curve.(Note that T6 can be found more easily by firstfactoring out the 500, then adding the functionvalues.)

3. R6 = 500(26.25) + 500(32.25) + 500(40.25) +500(50.25) + 500(62.25) + 500(76.25) = 143,750

About $143,750(Note that R6 can be found more easily by firstfactoring out the 500, then adding the functionvalues.)R6 is close to T6. (They differ by less than 1%.)

4. T100 = 144,001.8, T500 = 144,000.072

Conjecture: Exact value is $144,000.

5. g (x) = 20x + 13 0 000004( . )x3 + C

g (4,000) − g (1,000) = (165,333.3333… + C) − (21,333.3333… + C) = 144,000,

which is the conjectured value of the definiteintegral!The other name for antiderivative is indefiniteintegral.

6. a. f (x) = x7 + C b. y = − cos x + C

c. u = 0.5e2x + C d. v = 132 (4x + 5)8 + C

Problem Set 5-2Q1. Answers may vary.

y

x

Area = productof x and y

a b


Q3. f ′ (x) = −(ln 2)2− x Q4. y = sin x + C

Q5. y′ = 4 m/s Q6. sec x tan x (derivative)

Q7. 1 Q8. That constant.

Q9. 0 Q10. B

1. f (x) = 0.2x4 ⇒ f ′ (x) = 0.8x3 ⇒ f ′ (3) = 21.6;f (3) = 0.2(34) = 16.2

∴ y − 16.2 = 21.6(x − 3) ⇒ y = 21.6x − 48.6

x f (x) y Error

3.1 18.47042 18.36 0.11042

3.001 16.22161… 16.2216 0.0000108…

2.999 16.17841… 16.1784 0.0000107…

2. g (x) = sec x ⇒ g ′ (x) = sec x tan x ⇒g ′ (π /3) = 2 3 = 3.464…

g (π /3) = sec (π /3) = 2

Linear function is y − 2 = 2 3 (x − π /3) ⇒y = 2 3 (x − π /3) + 2.

dx f (x) y Error

0.04 2.15068… 2.13856… 0.01212…

− 0.04 1.87184… 1.86143… 0.01041…

0.001 2.003471… 2.003464… 7.01 × 10− 6

3. a. f (x) = x2 ⇒ f ′ (x) = 2x ⇒ f ′ (1) = 2

Tangent line: y − 1 = 2(x − 1) ⇒ y = 2x − 1The graph shows a zoom by factor of 10.

1

1

graph tangent line

Local linearity describes the property of thefunction because if you keep x close to 1 (inthe “locality” of 1), the curved graph of thefunction looks like the straight graph of thetangent line.

b.

x f (x) y Error, f (x) − y

0.97 0.9409 0.94 0.0009

0.98 0.9604 0.96 0.0004

0.99 0.9801 0.98 0.0001

1 1 1 0

1.01 1.0201 1.02 0.0001

1.02 1.0404 1.04 0.0004

1.03 1.0609 1.06 0.0009


The table shows that for x-values close to 1 (thepoint of tangency), the tangent line is a closeapproximation to the function values.

4. f (x) = x2 − 0.1(x − 1)1/3

Zooming in on (1, 1) shows that the graph goesvertical at x = 1. This observation is confirmedalgebraically.f ′ (x) = 2x − (1/3)(0.1)(x − 1)−2/3

f ′ (1) = 2 − (1/3)(0.1)(0)−2/3 , which is infinite.

graphtangent line

1

1

f does not have local linearity at x = 1. Becausethe slope of the graph becomes infinite, no linearfunction can approximate the graph there. If f isdifferentiable at x = c, then f is locally linearthere. The converse is also true. If f is locallylinear at x = c, then f is differentiable there.

5. a. Let A be the number of radians in θ degrees.

By trigonometry, tan A = x

100 ⇒

A = tan− 1 x

100.

Because 1 radian is 180/π degrees,

θ = 180

π tan− 1

x

100, Q.E .D.

b. dθ = 180

π ·

1

1 100 2+ ( / )x ·

1

100 dx

= 1 8

1 100 2

. /

( / )

π+ x

dx

x = 0: dθ = 0.5729… dx

x = 10: dθ = 0.5672… dx

x = 20: dθ = 0.5509… dx

c. At x = 0, θ = 0. For x = 20, dx = 20.∴ θ ≈ (0 + 0.5729…)(20) = 11.459…°The actual value is (180/π)(tan− 1 0.2) =11.309… .The error is 0.1492…°, which is about 1.3%.

d. 0.5729… is approximately 0.5, so multiplyingby it is approximately equivalent to dividingby 2. For a 20% grade, this estimate gives 10°,compared to the actual angle of 11.309…°, anerror of about 11.6%. For a 100% grade, thisestimate gives 50°, compared to the actualangle of 45°, an error of about 11.1%.

6. dV = 4π r 2 drdr = 0.03 and r = 6, so dV = 4π (62)(0.03) =4.32π ≈ 13.57 mm3

V ≈ 43 π (63) + 4.32π = 288π + 4.32π =

292.32π ≈ 918.350 mm3

Actual volume is V = 4

3π (6.033) =

292.341636π ≈ 918.418 mm3.∆V = 4

33 4

336 03 6π π( . ) ( )− = 4.341636π ≈

13.640 mm3

Error is 292.32π − 292.341636π = − 0.021636π,or about 0.068 mm3 too low.

7. a. (6000 · 0.05)/365 = 0.8219… , or about82 cents.

b. m = 6000e (0.05/365)t ⇒dm = 6000(0.05/365)e(0.05/365)t dtSubstituting t = 0 and dt = 1 gives dm =0.8219… , the same as part a.Substituting t = 0 and dt = 30 gives dm =24.6575… ≈ $24.66.Substituting t = 0 and dt = 60 gives dm =49.3150… ≈ $49.32.

c. t = 1: ∆m = 6000e (0.05/365)(1) − 6000 =0.8219… , almost exactly equal to dm.t = 30: ∆m = 6000e (0.05/365)(30) − 6000 =24.7082… , about 5 cents higher than dm.t = 60: ∆m = 6000e (0.05/365)(60) − 6000 =49.5182… , about 20 cents higher than dm.As t increases, dm is a less accurateapproximation for ∆m.

8. a. dS = − 1.636 dtMarch 11: dS = − 1.636(10) = − 16.36minutesSunrise time ≈ 6:26 − 0:16 = 6:10 a.m.,which agrees with the tabulated value.March 21: dS = − 1.636(20) = − 32.72 minutesSunrise time ≈ 6:26 − 0:33 = 5:53 a.m.,which agrees with the tabulated value.

b. By September 1, t = 185, givingdS = − 1.636(185) = − 302.66, or 5:04 hours.

So the predicted sunrise time would be6:26 − 5:04 = 1:22 a.m. Because the sunrisereaches its earliest in mid-June, the timepredicted by dS is not reasonable.

9. y = 7x3 ⇒ dy = 21x2 dx

10. y = − 4x11 ⇒ dy = − 44x10 dx

11. y = (x4 + 1)7 ⇒ dy = 28x3(x4 + 1)6 dx

12. y = (5 − 8x)4 ⇒ dy = − 32(5 − 8x)3 dx

13. y = 3x2 + 5x − 9 ⇒ dy = (6x + 5) dx

14. y = x2 + x + 9 ⇒ dy = (2x + 1) dx

15. y = e− 1.7x ⇒ dy = − 1.7e− 1.7x dx


16. y = 15 ln x1/3 ⇒ dy = 15 1

3

51 3

2 3

xx

xdx/

/⋅ =−

17. y = sin 3x ⇒ dy = 3 cos 3x dx

18. y = cos 4x ⇒ dy = − 4 sin 4x dx

19. y = tan3 x ⇒ dy = 3 tan2 x sec2 x dx

20. y = sec3 x ⇒ dy = 3 sec3 x tan x dx

21. y = 4x cos x ⇒ dy = (4 cos x − 4x sin x) dx

22. y = 3x sin x ⇒ dy = (3 sin x + 3x cos x) dx

23. y = x2/2 − x/4 + 2 ⇒ dy = (x − 1/4) dx

24. y = x3/3 − x/5 + 6 ⇒ dy = (x2 − 1/5) dx

25. y = cos (ln x) ⇒ dy = −sin(ln )x

xdx

26. y = sin (e0.1 x) ⇒ dy = 0.1e0.1 x cos (e0.1 x) dx

27. dy = 20x3 dx ⇒ y = 5x4 + C

28. dy = 36x4 dx ⇒ y = 7.2x5 + C

29. dy = sin 4x dx ⇒ y = − (1/4) cos 4x + C

30. dy = cos 0.2x dx ⇒ y = 5 sin 0.2x + C

31. dy = (0.5x − 1)6 dx ⇒ y = (2/7)(0.5x − 1)7 + C

32. dy = (4x + 3)− 6 dx ⇒ y = (− 1/20)(4x + 3)− 5 + C

33. dy = sec2 x dx ⇒ y = tan x + C

34. dy = csc x cot x dx ⇒ y = − csc x + C

35. dy = 5 dx ⇒ y = 5x + C

36. dy = − 7 dx ⇒ y = − 7x + C

37. dy = (6x2 + 10x − 4) dx ⇒ y = 2x3 + 5x2 − 4x + C

38. dy = (10x2 − 3x + 7) dx ⇒y = (10/3)x3 − (3/2)x2 + 7x + C

39. dy = sin5 x cos x dx ⇒ y = (1/6) sin6 x + C

40. dy = sec7 x tan x dx = sec6 x(sec x tan x dx) ⇒ y = (1/7) sec7 x + C

41. a. y = (3x + 4)2(2x − 5)3 ⇒y′ = 2(3x + 4)(3)(2x − 5)3

+ (3x + 4)2 ⋅ 3(2x − 5)2 ⋅ 2= 6(3x + 4)(2x − 5)2[2x − 5 + 3x + 4]

∴ dy = 6(3x + 4)(2x − 5)2(5x − 1) dx

b. dy = 6(7)(− 3)2(4)(− 0.04) = − 60.48

c. x = 1 ⇒ y = (7)2(− 3)3 = − 1323

x = 0.96 ⇒ y = − 1383.0218…

∴ ∆y = − 1383.0218… − (− 1323)

= − 60.0218…

d. − 60.48 is close to − 60.0218… .

42. a. y = sin 5x ⇒ dy = 5 cos 5x dx

b. dy = 5 cos (5π /3) ⋅ 0.06 = 0.15

c. x = π /3 ⇒ y = sin (5π /3) = − 3 /2

= − 0.86602…

x = π /3 + 0.06 ⇒ y = − 0.679585565…

∴ ∆y = − 0.679… − (− 0.866…)

= 0.186439…

d. 0.15 is (fairly) close to 0.186439... .

Problem Set 5-3Q1. Antiderivative = x3 + C

Q2. Indefinite integral = (1/6)x6 + C

Q3. y′ = 3x2 Q4. y′ = ln 3(3x)

Q5. dy = ln 3(3x) dx Q6. ′′ =y x5 4

Q7. Integral = sin x + C Q8. y′ = − sin x

Q9. 1 Q10. E

1. x dx x C10 111

11= +∫

2. x dx x C20 211

21= +∫

3. 44

56 5x dx x C− −= − +∫

4. 9 7 6x dx x C− −= − +∫ 3

2

5. cos sinx dx x C = +∫6. sin cosx dx x C = − +∫7. 4 7

4

77cos sinx dx x C = +∫

8. 20 920

99sin cosx dx x C = − +∫

9. 55

0 30 3 0 3e dx e Cx x. .

.= +∫

10. 2 2000 01 0 01e dx e Cx x− −= − +∫ . .

11. 44

4m

m

dm C= +∫ ln

12. 8 48 4

8 4.

.rr

dr C= +∫ ln .

13. ( ) ( ) ( )4 91

44 9 42 2v dv v dv+ = +∫∫

= +1

124 9( )v 3 + C

14. ( ) ( ) ( )5 53 171

33 17 3p dp p dp+ = +∫∫

= +1

183 17( )p 6 + C

15. ( ) ( ) ( )38 51

58 5 53− = − − −∫ ∫x dx x dx

= − +1

208 5 4( – )x C

16. ( ) ( ) ( ) ( )4 420 1 20− = − − −∫∫ x dx x dx

= − 1

520( – )x 5 + C


17. ( ) 6sin cos sinx x dx x C= +∫ 1

77

18. ( ) 8cos sin cosx x dx x C= − +∫ 1

99

19. cos sin cos4 51

5θ θ θ θ d C= − +∫

20. sin cos sin5 61

6θ θ θ θ d C= +∫

21. ( )x x dx x x x C2 3 23 51

3

3

25+ − = + − +∫

22. ( ) x x dx x x x C2 3 24 11

32− + = − + +∫

23. ( ) ( ) x dx x x x dx2 3 6 4 25 15 75 125+ = + + +∫∫= 1

7x7 + 3x5 + 25x3 + 125x + C

24. ( ) ( )x dx x x dx3 2 6 36 12 36− = − +∫∫= 1

7x7 − 3x4 + 36x + C

25. e x x dx e Cx xsec secsec tan = +∫26. e x dx e Cx xtan tansec2 = +∫27. sec tan2 x dx x C = +∫28. csc cot2 x dx x C = − +∫29. tan sec tan7 2 81

8x x dx x C = +∫

30. cot csc cot8 2 91

9x x dx x C = − +∫

31. csc cot csc csc cot9 8x x dx x x x dx ( )= ∫∫= − 1

9csc9 x + C

32. sec tan sec sec tan7 6x x dx x x x dx ( )= ∫∫= 1

7sec7 x + C

33. v (t) = 40 + 5 t = 40 + 5t1/ 2

D t t dt t t C( ) ( )= + = + +∫ 40 5 4010

31 2 3 2/ /

D (0) = 0 ⇒ 0 = 40 ⋅ 0 + 10

303/ 2 + C ⇒ C = 0

∴ D (t) = 40t + 10

3t3/ 2

D (10) = 505.4092... ≈ 505 ft

34. a. f (x) = 0.3x2 + 1T100 = 9.300135

b. g x x dx x x C( ) ( . ) .= + = + +∫ 0 3 1 0 12 3

c. g (4) − g (1) = 6.4 + 4 + C − 0.1 − 1 − C =9.3, which is about equal to the definiteintegral! It is also interesting that the constantC drops out.

35. Prove that if f and g are functions that can beintegrated, then

[ ( ) ( )] ( ) ( ) .f x g x dx f x dx g x dx+ = + ∫∫∫Proof:

Let h x f x dx g x dx( ) ( ) ( ) .= + ∫∫By the derivative of a sum property,

h xd

dxf x dx

d

dxg x dx′ = + ∫∫( ) ( ) ( ) .

By the definition of indefinite integral appliedtwice to the right side of the equation,h ′ (x) = f (x) + g (x).By the definition of indefinite integral applied inthe other direction,

h x f x g x dx( ) [ ( ) ( )] = +∫By the transitive property, then,

[ ( ) ( )] ( ) ( ) , f x g x dx f x dx g x dx+ = +∫ ∫ ∫Q.E.D.

36. Calvin says x x dx x x x Ccos sin cos= + +∫ .

Phoebe checked this by differentiating:d

dxx x x C( sin cos )+ +

= 1 ⋅ sin x + x ⋅ (cos x) − sin x + 0 = x cos x

By the definition of indefinite integral, she knewthat Calvin was right.

37. a.

C v (t)

1.5 12.25

2.5 16.25

3.5 22.25

Sum: 50.75

Integral ≈ 50.75

b.

c v (t)

1.25 11.5625

1.75 13.0625

2.25 15.0625

2.75 17.5625

3.25 20.5625

3.75 24.0625

Sum: 101.8750

Integral ≈ (101.8750)(0.5) = 50.9375


c. As shown in Figures 5-3c and 5-3d, theRiemann sum with six increments hassmaller regions included above the graph andsmaller regions excluded below the graph, sothe Riemann sum should be closer to theintegral.

d. Conjecture: Exact value is 51.By the trapezoidal rule with n = 100,integral ≈ 51.00045, which agrees with theconjecture.

e. The integral is the product of v (t) and t, andthus has the units (ft/min)(min), or ft. So theobject went 51 ft. Average velocity = 51/3 =17 ft/min.


Problem Set 5-4Q1. y′ = sin x + x cos x

Q2. tan x + C

Q3. f ′(x) = sec2 x

Q4. (1/4)x4 + C

Q5. z′ = − 7 sin 7x

Q6. − cos u + C

Q7. Limit = 8

Q8.

x

y

4

7

Q9. If a + b = 5, then a = 2 and b = 3.

Q10. No

1. x dx2

1

4

∫ 2. x dx3

2

6

∫c f (c) c f (c)

1.25 1.5625 2.25 11.390625

1.75 3.0625 2.75 20.796875

2.25 5.0625 3.25 34.328125

2.75 7.5625 3.75 52.734375

3.25 10.5625 4.25 76.765625

3.75 14.0625 4.75 107.171875

Sum = 41.8750 5.25 144.703125

R6 = (0.5)(41.875) 5.75 190.109375

= 20.9375 Sum = 638.000000

R8 = (0.5)(638) = 319

3. 31

3x dx

−∫ 4. 21

2x dx

−∫c f (c) c f (c)

− 0.75 0.43869… − 0.75 0.59460…

− 0.25 0.75983… −0.25 0.84089…

0.25 1.31607… 0.25 1.18920…

0.75 2.27950… 0.75 1.68179…

1.25 3.94822… 1.25 2.37841…

1.75 6.83852… 1.75 3.36358…

2.25 11.84466… Sum = 10.04849…

2.75 20.51556… R6 = (0.5)(10.04…)

Sum = 47.94108… = 5.024249…

R8 = (0.5)(47.94…)

= 23.97054…

5. sin x dx1

2

∫ 6. cos x dx 0

1

∫ c f (c) c f (c)

1.1 0.891207… 0.1 0.995004…

1.3 0.963558… 0.3 0.955336…

1.5 0.997494… 0.5 0.877582…

1.7 0.991664… 0.7 0.764842…

1.9 0.946300… 0.9 0.621609…

Sum = 4.790225… Sum = 4.214375…

R5 = (0.2)(4.79…) R5 = (0.2)(4.21…)

= 0.958045… = 0.842875…

7. tan.

.

x dx0 4

1 2

∫L4 = 0.73879… , U4 = 1.16866…M4 = 0.92270… , T4 = 0.95373…∴ M4 and T4 are between L4 and U4, Q.E.D.

8. 101

3

∫ / :x dx

L4 = 9.5, U4 = 12.8333…M4 = 10.89754… , T4 = 11.1666…∴ M4 and T4 are between L4 and U4, Q.E.D.

9. ln x dx1

5

∫ is underestimated by the trapezoidal

rule and overestimated by the midpoint rule.

5

2

y

x

5

2

y

x


10. e dxx 0

2

∫ is overestimated by the trapezoidal rule

and underestimated by the midpoint rule.

1

4

y

x

1

4

y

x

11. a. h (x) = 3 + 2 sin xFor an upper sum, take sample points at xequals 1, π/2, 2, 3, 4, and 6.

b. For a lower sum, take sample points at xequals 0, 1, 3, 4, 3π/2, and 5.

c. U6 = 1[h (1) + h (π/2) + h (2) + h (3) + h (4)+ h (6)] = 21.71134…

L6 = 1[h (0) + h (1) + h (3) + h (4) + h (3π/2)+ h (5)] = 14.53372…

12. Programs will vary depending on the type ofgrapher used. See the program in the Programsfor Graphing Calculators section of theInstructor’s Resource Book.

13. a. For x dx2

1

4

,∫ the program should give the

values listed in the text.

b. L100 = 20.77545, L500 = 20.955018.Ln seems to be approaching 21.

c. U100 = 21.22545, U500 = 21.045018.Un also seems to be approaching 21.f is integrable on [1, 4] if Ln and Un have thesame limit as n approaches infinity.

d. The trapezoids are circumscribed around theregion under the graph and thus contain morearea (see left diagram). For rectangles, the“triangular” part of the region that is left outhas more area than the “triangular” part that isadded, because the “triangles” have equal basesbut unequal altitudes (see right diagram).

Trapezoidincludesmore area.

x

y

Rectangleleaves outmore area.

x

y

14. a. x dx2

0

3

∫U100 = 9.13545, L100 = 8.86545

Conjecture: Integral equals 9 exactly.

b. The sample points will be at the right of eachinterval, 1 ⋅ 3/n, 2 ⋅ 3/n, 3 ⋅ 3/n, . . . ,n ⋅ 3/n.

c. Un = (3/n)(1 ⋅ 3/n)2 + (3/n)(2 ⋅ 3/n)2

+ (3/n)(3 ⋅ 3/n)2 + ⋅ ⋅ ⋅ + (3/n)(n ⋅ 3/n)2

d. Un = (3/n)3(12 + 22 + 32 + ⋅ ⋅ ⋅ + n2)

= (3/n)3(n/6)(n + 1)(2n + 1)

= (4.5/n2)(n + 1)(2n + 1)

U100 = (4.5/1002)(101)(201) = 9.13545, whichis correct.

e. Using the formula, U1000 = 9.013504… ,which does seem to be approaching 9

f. Un

n

n

nn = ⋅ + ⋅ +4 5

1 2 1.

= 4.5(1 + 1/n)(2 + 1/n)

As n approaches infinity, 1/n approaches zero.∴ Un approaches 4.5(1 + 0)(2 + 0),which equals 9, exactly!

15. x dx3

0

2

∫Find an upper sum using the sample points

1 ⋅ 2/n, 2 ⋅ 2/n, 3 ⋅ 2/n, . . . , n ⋅ 2/n.

Un = (2/n)(1 ⋅ 2/n)3 + (2/n)(2 ⋅ 2/n)3

+ (2/n)(3 ⋅ 2/n)3 + ⋅ ⋅ ⋅ + (2/n)(n ⋅ 2/n)3

= (2/n)4(13 + 23 + 33 + ⋅ ⋅ ⋅ + n3)

= (2/n)4[(n/2)(n + 1)]2

= 4/n2 ⋅ (n + 1)2 = 4(1 + 1/n)2

lim ( )n

nU→∞

= + =4 1 0 42

Problem Set 5-5

Q1. x2/2 + 2x + C Q2.10

10

t

Cln

+

Q3. − cot x + C Q4. − csc x cot x

Q5.5 4(ln )x

xC+ Q6.

1 2

x

y

Q7. Answers may vary. Q8. Answers may vary.

x

y5

3

x

y

1

2

Q9. No limit (infinite) Q10. D

1. See the text for the statement of the mean valuetheorem.

2. See the text for the statement of Rolle’stheorem.


3. g (x) = 6/x; [1, 4]

1 4

6

2

g(x)

x

c

m = = −6 4 6

4 13 2

/ –

–/

g ′ (x) = − 6x− 2

∴ − 6c− 2 = − 3/2 ⇒ c = 2

Tangent at x = 2 parallels the secant line.

4. f (x) = x4; [− 1, 2]

f(x)

–1 2

10

x

c

m = =16 1

2 15

–

– (– )

g ′ (x) = 4x3

∴ = ⇒ = =4 5 5 4 1 0773 3c c / . K

Tangent at x = 1.077… parallels the secant line.

5. c x x( ) ; , = +

2 02

cosπ

1

x

c(x)

/2c

m = = − =cos ( / ) – cos

/ –

ππ

π2 0

2 02 0 6366/ . K

c′(x) = − sin x

∴ − sin c = − 2/π ⇒ c = 0.69010…

Tangent at x = 0.690… parallels the secant line.

6. h x x( ) ; [ , ]= −5 1 9

c1 9

5h(x)

x

m = = −2 4

9 11 4

–

–/

h ′ (x) = − (1/2)x− 1/2

∴ (− 1/2)c− 1/2 = − 1/4 ⇒ c = 4

Tangent at x = 4 parallels the secant line.

7. f (x) = x cos x on [0, π/2]

1

1

2_0

f(x)

x

c

f ′ (x) = cos x − x sin x∴ f is differentiable for all x.x cos x = 0 ⇔ x = 0 or cos x = 0cos x = 0 at ±π/2 + 2π n, where n is an integer∴ hypotheses are met on [0, π/2].Using the solver feature, f ′ (c) = 0 atc = 0.86033… .Horizontal line at x = 0.86033… is tangent.

8. f (x) = x2 sin x

π0

1

f(x)

x

c

f ′ (x) = 2x sin x + x2 cos x∴ f is differentiable for all x.x2 sin x = 0 ⇔ x = 0 or sin x = 0sin x = 0 at x = 0 + π n , where n is an integerInterval: [0, π]Using the solver feature, f ′ (c) = 0 atc = 2.28892… .Horizontal line at x = 2.288… is tangent.

9. f (x) = (6x − x2)1/2

60

3

f(x)

x

c

f x x x x′ = ⋅ −−( ) ( )/12

2 1 26 6 2( – )

∴ f is differentiable on (0, 6).f is continuous at x = 0 and x = 6.(6x − x2)1/2 = 0x(6 − x) = 0 ⇒ x = 0 or 6Interval: [0, 6](6c − c2)− 1/2(3 − c) = 0 ⇒ c = 3Horizontal line at x = 3 is tangent.

π

π


10. f (x) = x4/3 − 4x1/3

c

4

1

f(x)

x

0

f x x x′ = − −( ) / /43

1 3 43

2 3

∴ f is differentiable for all x ≠ 0.f is continuous at x = 0.f (x) = 0 ⇒ x4/3 − 4x1/3 = 0 ⇒x1/3(x − 4) = 0 ⇒ x = 0 or 4Interval: [0, 4]43

1 3 43

2 3 0c c/ /− =−

43

2 3 1 0 1c c c− − = ⇒ =/ ( )

Horizontal line at x = 1 is tangent.

11. a. d (t) = 1000(1.09t )

d (50) = 1000(1.0950) = 74,357.520…

= $74,357.52 (Surprising!)

b. Average rate is

74357 5 1000

50

. –K =

1467.150… ≈ $1,467.15 per year.

c. d ′ (t) = ln (1.09)1000(1.09)t

d ′ (0) ≈ $86.18 per yeard ′ (50) ≈ $6,407.96 per yearThe average of these is $3,247.07 per year,which does not equal the average in part b.

d. Solving 1000(1.90)t ln 1.09 =1000 1 09 1000

50

50( . ) − algebraically gives

( . )1 091 09 1

50 1 09

50t = −.

ln .

⇒ = −t ln ln

.

ln .1 09

1 09 1

50 1 09

50

.

= ln (1.0950 − 1) − ln 50 − ln (ln 1.09) ⇒

t = − −ln ( . ) ln – ln (ln . )

ln .

1 09 1 50 1 09

1 09

50

= 32.893 years.K

This time is not halfway between 0 and 50.

12. d t

t

tt

tt

( )if

if = +

≤

≥

431

11

200 11

1

–,

– ,

The hypotheses of the mean value theorem do notapply on any interval that contains t = 1 as aninterior point, such as [0, 2] and [0.5, 2], becaused is not differentiable there. The hypotheses doapply on any interval not containing 1 and on

intervals for which 1 is an endpoint, suchas [1, 2].The conclusion is true if the instantaneousvelocity, d ′ (t), ever equals the average velocity.The average velocity equals

md d= =( ) – ( )2 0

228 5 0 2. ft/s for [ , ],

m

d d= =( ) – ( . )

..

2 0 5

1 557 111 0 5 2K ft/s for [ . , ],

md d= =( ) – ( )2 1

1100 1 2 ft/s for [ , ].

Between t = 0 and t = 1, d ′ (t) is negative.Above t = 1, d ′ (t) = 200t− 2.For d ′(c) = 28.5 ft/s,200c− 2 = 28.5 ⇒ c = 2.649… .But 2.649… is outside (0, 2), so the conclusionis not true. See the left graph.For d ′ (c) = 57.111… ft/s,200c− 2 = 57.111… ⇒ c = 1.871… .Because 1.871… is in (0.5, 2), the conclusion istrue. See the right graph.

100

200

0.5 1 2

t

d(t)

c?

c is outside(0, 2).

100

200

0.5 1 2

t

d(t)

c?

c is in(0.5, 2).

For d ′ (c) = 100, 200c− 2 = 100 ⇒ c = 1.414… .Because 1.414… is in (1, 2), the conclusion istrue, as is guaranteed by the mean value theorem.The fact that the conclusion is true when thehypotheses are met illustrates the fact that thehypotheses are sufficient. The fact that theconclusion can be true even if the hypotheses arenot met proves that the hypotheses are notnecessary.

13. See Figure 5-5d. 14. Answers may vary.

a b

f(x)

x


f(x)

xa

d

b

a b

x

f(x)


17. Answers may vary.

x

a c b

f(x)

18. Michel Rolle (1652− 1719) lived in France.Sources will vary.

19. f (x) = x2 − 4x 20. f (x) = x2 − 6x + 5

f (1) = − 3 ≠ 0 f (2) = − 3 ≠ 0Conclusion is not true. Conclusion is not true.f ′ (2) = 0, but 2 is not f ′ (3) = 0, but 3 is notin the interval (0, 1). in the interval (1, 2).

10

x

f(x)

–3

x

f(x)

1 2

–3

21. f (x) = x2 − 4x 22. f (x) = x2 − 6x + 5

f (2) = − 4 ≠ 0 f (4) = − 3 ≠ 0Conclusion is not true. Conclusion is true.f ′ (2) = 0, but 2 is not in f ′ (3) = 0, and 3 is inthe open interval (0, 2). the interval (1, 4).

0 2

1x

f(x)

1 4

xc

f(x)1

23. f (x) = x2 − 4x 24. f (x) = |x − 2| − 1f (3) = − 3 ≠ 0 f is not differentiable

at x = 2.Conclusion is true. Conclusion is not

true.f ′ (2) = 0, and 2 is in f ′ (x) never equals 0.the interval (0, 3).

30

–3

x

f(x)

c

1 3

1x

f(x)

25. f (x) = 1/x 26. f (x) = x − [x]

f (0) does not exist. f is discontinuous at1 and 2.

Conclusion is Conclusion isnot true. not true.

f ′ (x) never equals 0. f ′ (x) never equals 0.

5

5

x

f(x)

1 2

1x

f(x)

27. f (x) = 1 − (x − 3)2/3 28. f x( ) =

x x x

x

3 26 11 6

2

– –

–

+

f is not differentiable f is not continuous orat x = 3. differentiable at 2.Conclusion is Conclusion isnot true. not true.f ′ (x) never equals 0. There is no point at x = 2

to draw the tangent line.

f(x)

x

1

2 4

1 3

x

f(x)

1

No point oftangency

29. g xx x x

x( ) = +3 27 13 6

2

– –

–

= + = − + ≠( – )( – )

–

x x x

xx x x

2 5 3

25 3 2

22 ,

Thus, g is discontinuous at x = 2, and thehypotheses of the mean value theorem are notmet. The conclusion is not true for [1, 3],because the tangent line would have to contain(2, g (2)), as shown in the left graph. Theconclusion is true for (1, 5), because the slope ofthe secant line is 1, and g ′ (x) = 1 at x = 3, whichis in the interval (1, 5). See the right graph.

2 x

g(x)

3

1 3 2 x

g(x)

1

3

3

5

30. h (x) = x2/3 ⇒ h ′ (x) = (2/3)x− 1/3

∴ h is differentiable for all x ≠ 0.h ′ (0) would be 0− 1/3 = 1/(01/3) = 1/0, which isinfinite. The hypotheses of the mean valuetheorem are met on the interval [0, 8], becausethe function need not be differentiable at anendpoint. The hypotheses are not met on [− 1, 8],because the point x = 0 where h is notdifferentiable is in the open interval (− 1, 8). Tosee if the conclusion of the mean value theorem


is true anywhere, find the slope of the secant line(see next graph).

m = =4 1

8 11 3

–

– (– )/

The tangent line has slope h ′ (c) = 1/3. Therefore,(2/3)c− 1/3 = 1/3 ⇒ c− 1/3 = 1/2 ⇒ c1/3 = 2 ⇒ c = 8.So the conclusion of the mean value theorem isnot true because 8 is at the endpoint of theinterval, not in the open interval (− 1, 8).

–1 8

4

x

h(x)

31. a. f xx x

x x( )

if

if=

≥+ <

3 3 3

3 3

– ,

,

3

6

x

f(x)

b. f is continuous at x = 3 because the right andleft limits both equal 6. f is not differentiableat x = 3 because the left limit of f ′ (x) is 1and the right limit is 3.

c. f is not differentiable at x = 3, which is in (1, 6).The secant line has slope 11/5. The tangent linehas slope either 1 or 3, and thus is never 11/5.

d. f is integrable on [1, 6]. The integral equals41.5, the sum of the areas of the two trapezoidsshown in this diagram.

3

6

x

f(x)

1 6

4

15

32. a. f (t) = number of miles in t hourst = number of hours drivenFor the mean value theorem to apply on[a, b], f must be differentiable on (a, b) andcontinuous at t = a and t = b.

b. The 60 mi/h equals the slope of the secantline. Therefore, there must be a tangent lineat some value t = c in (a, b) with slope equalto 60. This tangent line’s slope is theinstantaneous speed at t = c. Therefore, thespeed was exactly 60 at some time betweent = a and t = b, Q.E.D.

33. a. f (x) = 25 − (x − 5)2 + 4 cos 2π (x − 5)The graph agrees with Figure 5-5l.

b. f ′ (x) = − 2(x − 5) − 4 sin 2π (x − 5) ⋅ 2π =− 2x + 10 − 8π sin 2π (x − 5) f ′ (5) = 0Because the derivative at x = 5 is 0, thetangent line at x = 5 is horizontal. This isconsistent with x = 5 being a high point onthe graph.

c.

20

4

x

y1

y2

m(x)

x m (x) x m (x)

3.0 2 5.5 − 16.5

3.5 6.8333… 6.0 − 14.0 1 6.5 − 6.833…

4.5 16.5 7.0 − 25.0 no value

The difference quotient is positive when x isless than 5 and negative when x is greaterthan 5.

d. In the proof of Rolle’s theorem, the left limitof the difference quotient was shown to bepositive or zero and the right limit wasshown to be negative or zero.The unmentioned hypothesis isdifferentiability on the interval (a, b). Thefunction f is differentiable. Because there is avalue of f ′(5), both the left and right limitsof the difference quotient must be equal. Thisnumber can only be zero, which establishesthe conclusion of the theorem. Theconclusion of Rolle’s theorem can be trueeven if the hypotheses aren’t met. Forinstance, f (x) = 2 + cos x has zero derivativesevery π units of x, although f (x) is neverequal to zero.

34. a. mf f= = =( . ) – ( )

.

. –

.

4 5 2

2 5

5 5 4

2 50 6.

g (x) − 4 = 0.6(x − 2) ⇒g (x) = 0.6x + 2.8

Your graph should agree with Figure 5-5m.

b. f ′ (x) = 1 − π sin π xUsing the solver feature, f ′(c) = 0.6 atc = 2.0406… , 2.9593… , and 4.0406… ,all of which are in (2, 4.5).


c. h (x) = f (x) − g (x)

2

5

x

y

y1

y2

y3

For c1 = 2.0406… :

x h(x)

1.7906… –0.2925…

1.8406… –0.1865…

1.8906… –0.1022…

1.9406… –0.0411…

1.9906… –0.0041…

2.0406… 0.0081…

2.0906… –0.0039…

2.1406… –0.0397…

2.1906… –0.0977…

2.2406… –0.1760…

2.2906… –0.2723…

For c = 2.9593… : For c = 4.0406… :

x h(x) x h(x)

2.7093… –1.3274… 3.7906… 0.5075…

2.7593… –1.4237… 3.8406… 0.6134…

2.8093… –1.5021… 3.8906… 0.6977…

2.8593… –1.5601… 3.9406… 0.7588…

2.9093… –1.5959… 3.9906… 0.7958…

2.9593… –1.6081… 4.0406… 0.8081…

3.0093… –1.5958… 4.0906… 0.7960…

3.0593… –1.5589… 4.1406… 0.7602…

3.1093… –1.4979… 4.1906… 0.7022…

3.1593… –1.4136… 4.2406… 0.6239…

3.2093… –1.3077… 4.2906… 0.5276…

h(c1) = h(2.0406…) = 0.0081…So h(c1) is an upper bound for h(x).h (c2) = h (2.9593…) = − 1.60811669…So h (c2) is a lower bound for h (x).h (c3) = h (4.0406…) = 0.808116698…So h (c3) is an upper bound for h (x).

d. f meets the hypotheses of the mean valuetheorem, because f is differentiable for all x.h (x) = f (x) − g (x)∴ h ′(x) = f ′(x) − g ′(x)∴ h ′(c) = f ′(c) − g ′(c)For each of the values of c in part b, h ′(c) = 0.

∴ f ′(c) − g ′(c) = 0∴ f ′(c) = g ′(c)∴ f ′(c) = the slope of the secant line, Q.E.D.

35. The hypotheses of the mean value theoremstate that f should be differentiable on the openinterval (a, b) and continuous at x = a and x = b.If f is differentiable on the closed interval [a, b],it is automatically continuous at x = a and x = b,because differentiability implies continuity.

36. a. h (x) = f (x) − g (x)The mean value theorem applies to h becauseboth f and g are given to be differentiable, anda linear combination of differentiablefunctions is also differentiable.

b. By the mean value theorem, there is a numberc in (a, b) for which

′ = −−

h ch b h a

b a( )

( ) ( ).

If f (a) = g (a) + D1 and f (b) = g (b) + D2,then h (a) = D1 and h (b) = D2.

∴ ′ =h cD D

b a( )

–

–2 1

c. If D1 ≠ D2, then h ′ (c) ≠ 0.But h ′ (x) = f ′ (x) − g ′ (x) by the derivative ofa sum, and thus h ′(x) = 0 for all x in thedomain.∴ ′ =h c( ) ,0 which contradicts ′ ≠h c( ) .0So the supposition that D1 ≠ D2 is false,meaning that D1 and D2 are equal, Q.E.D.

37. By the definition of antiderivative (indefinite

integral), g x dx( ) = ∫ 0 if and only if g ′ (x) = 0.

Any other function f for which f ′ (x) = 0 differsfrom g (x) by a constant. Thus the antiderivativeof zero is a constant function, Q.E.D.

38. f (x) = (cos x + sin x)2, and g (x) = sin 2x

y

x

1

1

g

f

x f (x) g (x)

0 1 0

1 1.9092… 0.9092…

2 0.2431… − 0.7568…

3 0.7205… − 0.2794…

4 1.9893… 0.9893…

In each case, f (x) = g (x) + 1.


Proof:

(cos x + sin x)2 = cos2 x + 2 cos x sin x + sin2 x

= 2 cos x sin x + 1 = sin 2x + 1, Q.E.D.

39. The hypotheses of Rolle’s theorem say that fis differentiable on the open interval (a, b).Because differentiability implies continuity,f is also continuous on the interval (a, b).Combining this fact with the hypothesis ofcontinuity at a and at b allows you to concludethat the function is continuous on the closedinterval [a, b].

40. The intermediate value theorem applies tocontinuous functions, whereas the mean valuetheorem applies to differentiable functions. Bothare existence theorems, concluding that thereis a value x = c in the open interval (a, b). Forthe intermediate value theorem, f (c) equals apre-selected number v between f ( a) and f ( b).For the mean value theorem, f ′ (c) equals theslope of the secant line connecting (a, f ( a))and (b, f ( b)).


Problem Set 5-6Q1. r′(x) = m(x)


Q3. Increasing at 6 units/unit

Q4. dy = sec x tan x dx

Q5. y′ = 8x(x2 + 3)3

Q6. d 2z/dz2 = −25 sin u

Q7. f ′(x) = 0

Q8. 4.5

Q9. See Figure 5-5b.

Q10. E

1. a. I x dx= −∫ 10 1 5

4

9.

= (−10/0.5)(9− 0.5 ) − (−10/0.5)(4− 0.5 )

= −20/3 + 20/2 = 10/3 = 3.33333…

The +C and −C add up to zero.

b.

4 9

x

y

1

c. Pick sample points at left ends of subintervalsfor U5 and at right ends for L5.

k x f ( x)

1 4 1.25

2 5 0.89442719…

3 6 0.68041381…

4 7 0.53994924…

5 8 0.44194173…

Sum = 3.80673199…

U5 = (1)(3.80673199…) = 3.80673199…

k x f ( x)

1 5 0.89442719…

2 6 0.68041381…

3 7 0.53994924…

4 8 0.44194173…

5 9 0.37037037…

Sum = 2.92710236…

L5 = (1)(2.92710236…) = 2.92710236…

Average = (U5 + L5)/2 = 3.36691717… .Average overestimates the integral,3.33333… .This fact is consistent with the fact that thegraph is concave up.

d. Use sample points at the midpoints.

M10 = 3.32911229…

M100 = 3.33329093…

M1000 = 3.33333290…

Sums are converging toward 10/3.

2. I x dx= = − − −∫ sin cos . ( cos ).

1 5 00

1 5

= 0.92926279…

Using sample points at the midpoints,

M10 = 0.93013455…

M100 = 0.92927151…

M1000 = 0.92926288…

Integral = 0.92926279…

The sums are converging toward the integral.The rectangle and the region differ by the two“triangular” regions. Because the sample pointis at the midpoint of the subinterval, the“triangles” have equal bases. Because the graphis concave down, the “triangle” below thehorizontal line has a larger altitude, andthus a larger area, than the one above the line.So the rectangle includes more area on the left


than it leaves out on the right, and thusoverestimates the integral.

Rectangleincludesmore area.

y

x

3. See the statement of the fundamental theorem inthe text.

4. See the work in the text preceding the proof ofthe fundamental theorem for a derivation of aRiemann sum that is independent of the numberof subintervals.

5. See the text proof of the fundamental theorem.

6. If c is picked as the point in (a, b) where the

mean value theorem is true for g x f x dx( ) ( ) ,= ∫then the exact integral equalsg b g a

b ab a

( ) – ( )

( – )( ),⋅ − which equals g(b) − g(a).

7. v(t) = 100 − 20(t + 1)1/2

Distance = +∫ [ – ( ) ] /

0

81 2100 20 1t dt

= − +10040

31 3 2

0

8

t t( ) /

= − − +80040

39 0

40

313 2 3 2( ) ( )/ /

= 4531

3ft

8. a. h(x) = x1/2

k x f ( x)

1 4.25 2.0615528…

2 4.75 2.1794494…

3 5.25 2.2912878…

4 5.75 2.3979157…

5 6.25 2.5

6 6.75 2.5980762…

7 7.25 2.6925824…

8 7.75 2.7838821…

9 8.25 2.8722813…

10 8.75 2.9580398…

Sum = 25.3350679…

M10 = (0.5)(25.3350679…) = 12.66753…

b. h(u)∆u and h(u + ∆u)∆u are terms in a lowersum and an upper sum, respectively, becauseh(x) is increasing.

∴ h(u)∆u < A(u + ∆u) – A(u) < h(u + ∆u)∆u

c. h uA u u A u

uh u u( )

( ) ( )( )< + − < +∆

∆∆

But the limits of h(u) and h(u + ∆u) both equalh(u) because h is continuous and h(u) isindependent of ∆u. Therefore, by the squeezetheorem,

lim( ) – ( )

( ).∆u

A u u A u

uh u

→

+ ∆∆

=0

But the limit on

the left is defined to be dA/du.∴ dA/du = h(u), Q.E.D.

d. dA = h(u) du

∴ A(u) = ∫ h(u) du = ∫ u1/2 du = (2/3)u3/2 + C∴ A(u) = (2/3)u3/2 − 16/3

e. A( )9 12 23= , which agrees with M10 = 12.667… .

(Note also that A(9) < M10, which is expectedbecause the graph of h is concave down.)

9. a. Answers may vary. b. Answers may vary.

x

f(x)

x

f(x)

c. Answers may vary. d. Answers may vary.

x

f(x)

x

f(x)

e. Answers may vary. f. Answers may vary.

x

f(x)

x

f(x)


Problem Set 5-7

Q1. 16

6x C+ Q2. 118

63 7( )x C+ +Q3. − +−1

33x C Q4. 1

66sin x C+

Q5. 15 5sin x C+ Q6. x + C

Q7. tan x + C Q8. y″ = −1/x2

Q9. definite Q10. indefinite

1. x dx x2 3

1

4

1

41

3

1

364

1

31 21= = − =∫ ( ) ( )

2. x dx x3 4

2

5

2

51

4

1

4625

1

416

609

4152

1

4= = − = =∫ ( ) ( )


3. ( ) ( )1 31

91 32 3

2

3

2

3

+ = +− −∫ x dx x

= − − =1

91000

1

9125 125( ) ( )

4. ( – ) ( – )5 21

155 22

1

43

1

4

x dx x− −∫ =

= = = =1

155832 343

6175

15

1235

3411

2

3[ – (– )]

5. 60 36 36 32 1 11162 3 5 3

1

8

1

8

x dx x/ /= = − =∫ ( )

6. 24 9 6 9 6 32 1 297 63 2 5 2

1

4

1

4

x dx x/ /. . .= = − =∫ ( )

7. 5 5 40 10 302

8

2

8

dx x= = − =∫8. dx x= = − =∫20

50

20

50

50 20 30

9. ( )x x dx x x x2

2

03 2

2

0

3 71

3

3

27+ + = + +

− −∫= − − + −

= =0

8

36 14

32

310

2

3

10. ( )x x dx x x x2

3

03 2

3

0

4 101

32 10+ + = + +

− −∫= 0 − (−9 + 18 − 30) = 21

11. 4 51

44 5 4

1

11 2

1

1

x dx x dx+ = +− −∫ ∫ ( ) / ( )

= ⋅ + = = =−

1

4

2

34 5

1

627 1

13

34

1

33 2

1

1

( ) ( – )/x

12. 2 101

22 10 2

3

31 2

3

3

x dx x dx+ = +− −∫ ∫ ( ) / ( )

= ⋅ + = = =−

1

2

2

32 10

1

364 8

56

318

2

33 2

3

3

( ) ( – )/x

13. 4 4 4 1 4 1 80 0

sin cosx dx x= − = − − + =∫π π

( ) ( )

14. 6 6 6 1 6 1 122

2

2

2

cos sin/

/

/

/

x dx x= = − − =− −∫ π

π

π

π

( ) ( )

or: Integral of an even function betweensymmetric limits.

2 6 12 12 1 0 120

2

0

2

cos sin/ /

x dx x= = − =∫π π

( )

15. (sec cos )/

/2

6

3

x x dx+∫π

π

= + = + − −tan sin / //

/x x

π

π

6

33 3 2 1 3 2/ 1

= − =( / ) / .7 6 3 1 2 1 52072K

16. (sec tan sin ) sec cos/ /

x x x dx x x+ = −∫0

3

0

3π π

= − − + =21

21 1 1 5.

17. e dx e e ex x2

0

42

0

42 4 01

2

1

2

1

27 5

ln lnln .∫ = = − =

18. e dx e e ex x− − −= − = − +∫0

3

0

3 3 0ln ln ln

= − + =1

31

2

3

19. sin cos sin3 4

1

2

1

21

4x x dx x =∫

= =1

42 1 0 0455664 4(sin – sin ) . K

20. ( cos ) sin ( cos )11

514 5

3

3

3

3

+ = − +− −∫ x x dx x

= − + + + =1

51 3

1

51 3 05 5( cos ) [ cos(– )]

or: Integral equals zero because an odd function isintegrated between symmetric limits.

21. cos sin.

.

.

.

31

33

0 1

0 2

0 1

0 2

x dx x=∫= =1

30 6 0 3 0 0897074(sin . – sin . ) . K

22. sin cos. .

21

22

0

0 4

0

0 4

x dx x = −∫= − =1

20 8 0(cos . – cos ) 0.1516466K

23. ( – sin )x x x dx dx7 3

5

5

6 4 2 2 2+ + =−∫ ∫

0

5

= = − =2 2 20 0 20( )0

5x

24. (cos – tan ) cosx x x dx x dx+ =−∫ ∫10 23

1

1

0

1

= = − =2 2 1 2 0 1 682941

0

1sin sin sinx . K

25. x dx−

−∫2

1

1

has no value because y = x− 2 has a

vertical asymptote at x = 0, which is within theinterval.

26. x dx−∫ 2

2

has no value because the integrand is

not a real number for negative values of x.

27. Integral = −(area) 28. Integral = area

x3 6

-5

f(x)

5 7

1

f(x)

x


29. Integral ≠ area 30. Integral ≠ area

7 x

0

f(x)

1

x8

2

1 f(x)

1

31. f x dx f x dxb

a

a

b

( ) ( ) = − = −∫ ∫ 7

32. 4 4 4 7 28 ( ) ( ) ( )f x dx f x dxa

b

a

b

= = =∫ ∫33. g x dx g x dx g x dx

a

c

a

b

b

c

( ) ( ) ( )= +∫ ∫ ∫= 12 + 13 = 25

34. f x dxa

c

( ) ∫ cannot be determined.

35. f x dx g x dxa

c

a

c

( ) ( )∫ ∫+ cannot be determined.

36. [ ( ) ( )]f x g x dxa

b

+ =∫f x dx g x dx

a

b

a

b

( ) ( ) 7 12 19∫ ∫+ = + =

37.

5 10

7

y

x

y = f' (x)

f(3) = 7

y = f(x)

38.

7

8

y

x

y = f ' (x)

f(1) = 8

y = f(x )

39. Statement:“If f ( x) < g(x) for all x in [a, b],

then f x dx g x dxa

b

a

b

( ) ( ) .∫ ∫< ”

Converse:

“If f x dx g x dxa

b

a

b

( ) ( ) ,∫ ∫<

then f ( x) < g(x) for all x in [a, b].”The converse can be shown to be false by anycounterexample in which the area of the regionunder the g graph is greater than the area underthe f graph, but the g graph touches or crossesthe f graph somewhere in [a, b]. Onecounterexample isf ( x) = 1.5 and g(x) = 2 + cos x on [0, 2π].

g

f

y3

20

x

40. x dx x C2

1

43

1

41

3∫ = +

= + − − =1

34

1

31 213 3( ) ( )C C

The two C’s will always cancel, so it is notnecessary to write them.

Problem Set 5-8Q1. 30x2.4 + C

Q2. 30(42.4 − 1) = 805.72…

Q3. y x′ = −1/ –1 2

Q4. f ′ (x) = 3x2 sin x + x3

cos x

Q5.

x1

1

f(x) and f'(x)

f

f'

Q6. Yes, continuous

Q7. Increasing at x = 7

Q8. f ( a) = f ( b) = 0

Q9. v(9) = 450 ft/s

Q10. a(9) = 25 (ft/s)/s

π


1. a.

t

50

100

v

t = a t = b1

(t, v)

dt

dy = v dt = (55 + 12t0.6 ) dt

b. Displacement = +∫ ( ).55 12 0 6t dta

b

( ). .55 12 55 7 50 6 1 6

0

1

0

1

+ = +∫ t dt t t.

= 55 + 7.5 − 0 − 0 = 62.5 mi

( ). .55 12 55 7 50 6 1 6

1

2

1

2

+ = +∫ t dt t t.

= 110 + 22.735… − 55 − 7.5 ≈ 70.2 mi

c. ( ). .55 12 55 7 50 6 1 6

0

2

0

2

+ = +∫ t dt t t.

= 110 + 22.735… − 0 − 0 = 132.735… ,which equals the sum of the two integralsabove.

d. ( ).55 12 0 6

0+ =∫ t dt

b

300

55 7 5 3001 6

0t t

b+ =. .

55b + 7.5b1.6 = 300b ≈ 4.13372… ≈ 4.134 h

300

x

y

4.134

e. v(4.13372…) = 55 + 12(4.13372…)0.6 =83.1181… . At the end of the trip, you weregoing about 83 mi/h.

2.

10

10

v

t

dv

(t, v)

dy v dt e dtt= = −15 0 1.

15 150 150 1500 1 0 1

0

20

0

202e dt e et t− − −= − = − +∫ . .

= 129.6997… ft

3. a.

2

5

y

x

dx

(x, y)

b. dA = y dx = 10e0. 2x dx

c. 10 50 50 500 2 0 2

0

2

0

20 4e dx e ex x. . .= = −∫

d.

10 0 2

0

2

e dxx.∫ = 24.59123K

The region is approximately a trapezoid withheight 2 and bases 10 and y(2). y(2) =14.9182… , so the area of the trapezoid is2/2(10 + 14.9182…) = 24.9182… .

4.

3

4

y

x

dx

(x, y )

dA = (6x − x2 ) dx

( )6 31

33 6

1

36 362

0

62 3

0

62 3x x dx x x− = − = ⋅ − ⋅ =∫

The area of the circumscribed rectangle is 6 ⋅ 9 =54. The area of the parabolic region is two-thirdsthis area.

5. a. dA = [x + 2 − (x2 − 2x − 2)] dx

The top and bottom of the strip are nothorizontal, so the area of the strip is slightlydifferent from dA. As dx approaches zero, thedifferences in height at different values of x inthe strip become smaller, so the differencebetween dA and the area of the strip getssmaller.

b. y1 = y2 ⇒ x + 2 = x2 − 2x − 2 ⇒0 = x2 − 3x − 4 ⇒ 0 = (x − 4)(x + 1) ⇒x = 4 or x = −1

( )− + + = − + +− −∫ x x dx x x x2

1

43 2

1

4

3 41

3

3

24

= − ⋅ + ⋅ + ⋅

1

34

3

24 4 43 2

− − ⋅ − + ⋅ − + ⋅ −

1

31

3

21 4 13 2( ) ( ) ( )

= =125

620 83333. K

c. R100 = 20.834375 (Checks)


6.

π⁄4

1y

x

dx

(x, y )1

(x, y )2

The curves intersect at x = π /4.

dA = (cos x − sin x) dx

(cos sin ) (sin cos )/ /

x x dx x x− = +∫0

4

0

4π π

= −2 1

7. a.

90

6

x

F

dx

(x, F )

b. dW = F dx = 0.6x dx

W x dx x= =∫ 0 6 0 3 2

0

9

0

9

. .

= 24.3 − 0 = 24.3 inch-pounds

c. The region under F from x = 0 to x = 9 is atriangle with base 9 and height F(9) = 5.4. Sothe area is 1/2 ⋅ 9 ⋅ 5.4 = 24.3.

d. dW is found by multiplying F by dx. F ismeasured in pounds, and dx is measured ininches, so the units of dW are(pounds)(inches), or inch-pounds.

8.

5

30

F

x

dx

(x, F )

dW x dx= 5020

cosπ

5020

1000

200

10

0

10

cos sinπ

ππ

x dx x=∫= − =

= …

1000

2

10000

1000

318 3098π

ππ π

sin sin

.

The midpoint Riemann sum R100 gives318.313… , which is close to the answer foundusing integration.

9. a.

dx

(x, T )

0.5 1

x

T

20

b. dD = T dx = [20 − 12 cos 2π(x − 0.1)] dx

c. D x dx= ∫ [ – cos ( – . )].

20 12 2 0 10

0 5

π

= − −20 6 2 0 1x x( ) ( . )0

0.5/ sinπ π

= 10 − (6 /π) sin 0.8π − 0 + (6 /π) sin (−0.2π)= 7.75482… ≈ 7.75 degree-days

d. From noon to midnight,

D x dx= ∫ [ – cos ( – . )].

20 12 2 0 10 5

1

π

= − −20 6 2 0 1x x( ) ( . )0.5

1/ sinπ π

= 20 − (6 /π) sin 1.8π − 10 + (6 /π) sin 0.8π= 12.24517… ≈ 12.25 degree-days

The total number of degree-days isD = 7.75482… + 12.24517… =20 degree-days.Note that this answer can be found moreeasily by observing that in one full cycle of asinusoid, there is just as much area above thesinusoidal axis, T = 20, as there is below it.So the average temperature difference for theday is 20 degrees, making the number ofdegree-days for one day equal to 20.

10. a.

10 30

T

C

1000

dT

dH = C dT= (−0.016T 3 + 0.678T 2 + 7.45T + 796) dT

H T T= +∫ (– . .0 016 0 6783 2

10

30

+ +7 45 796. )T dT

= − + +0 004 0 226 3 7254 3 2. . .T T T

+ 796T 10

30= −3240 + 6102 + 3352.5

+ 23880 + 40 − 226 − 372.5 − 7960= 21,576 Btu

b. (2000)(21576) = 43,152,000 BtuThe property is the integral of a constant


times a function. That is, 200010

30

C dT ∫ =

200010

30

C dT .∫11. a.

xdx

P

1000

1000 200

(x, P)

b. dC = P dx = (100 + 0.06x2) dx

C x dx x xb b

= + = +∫ ( . )100 0 06 100 0 022 3

0.

0

= 100b + 0.02b3 − 0 − 0

∴ C = 100b + 0.02b3

c. b = 100: C = 100(100) + 0.02(1003) =$30,000

b = 200: C = 100(200) + 0.02(2003) =$180,000

For 100 m to 200 m, the cost should be180,000 − 30,000 = $150,000.

As a check,

( . )100 0 06 100 0 022 3

100

200

100

200

+ = +∫ x dx x x.

= 100(200) + 0.02(2003) − 100(100)− 0.02(1003) = 150000

Thus, P dx P dx P dx0

200

0

100

100

200

∫ ∫ ∫= + ,

which shows that the sum of integrals withthe same integrand applies.

12. Using trapezoids, the area is approximately10(0/2 + 38 + 50 + 62 + 60 + 55 + 51 + 30 +3/2) = 3475 ft2. The fundamental theorem cannotbe used because the function is specified only bydata, not by an equation whose antiderivative canbe found.

Plan of attack for area problems:

• Do geometry to get dA in terms of samplepoint (x, y).

• Do algebra to get dA in terms of one variable.

• Do calculus to sum the dA’s and take the limit(i.e., integrate).

13. y

x

(x, y)

(x, 0)

4

1 5

The graph intersects the x-axis at x = 1 and x = 5.y = −x2 + 6x − 5 = −(x − 1) ( x − 5) = 0 ⇒ x = 1,5, which confirms the graphical solution.

dA = (y − 0) dx = (−x2 + 6x − 5) dx

A x x dx x x x= + = − + −∫ (– – )2 3 2

1

5

1

5

6 51

33 5

= − + − + − + =125

375 25

1

33 5 10

2

3

14.–2 3

–6

xy

(x, 0)

(x, y)

The graph intersects the x-axis at x = −2and x = 3.y = x2 − x − 6 = (x + 2)(x − 3) = 0 ⇒ x = −2,3, which confirms the graphical solution.dA = (0 − y) dx = (−x2 + x + 6) dx

A x x dx x x x= + + = − + +− −∫ (– )2 3 2

2

3

2

3

61

3

1

26

= − + + − − + =99

218

8

32 12 20

5

6

15.(x, y)

y

x

–2

(0, y)

4

The graph intersects the y-axis at y = 1and y = 4.x = (y − 1) ( y − 4) = 0 ⇒ y = 1, 4, whichconfirms the graphical solution.

dA = (0 − x) dy = −(y − 1) ( y − 4) =(−y2 + 5y − 4) dy

A y y dy y y y= + = − + −∫ (– – )2 3 2

1

4

1

4

5 41

3

5

24

= − + − + − + =64

340 16

1

3

5

24 4

1

2

16. y

x

–1

5

5

(x, y)(0, y)


The graph intersects the y-axis at y = −1and y = 5.

x = 5 + 4y − y2 = (1 + y) ( 5 − y) = 0 ⇒ y =−1, 5, which confirms the graphical solution.

dA = (x − 0) dy = (5 + 4y − y2) dy

A y y dy y y y= + = + −− −∫ ( – )5 4 5 2

1

32 2 3

1

5

1

5

= + − + − − =25 50125

35 2

1

336

17.y

x

–1 4

2

(x, y1)

(x, y2)

The graphs intersect at x = −1 and x = 4.x2 − 2x − 2 = x + 2 ⇔ x2 − 3x − 4 = 0 ⇔ x =−1, 4, which confirms the graphical solution.dA = (y2 − y1) dx = (−x2 + 3x + 4) dx

A x x dx x x x= − + + = − + +− −∫ ( )2 3 2

1

4

1

4

3 41

3

3

24

= − + + − − + =64

3

1

3

3

2

5

624 16 4 20

18.

y

x

10

–2

4(x, y1)

(x, y2)

The graphs intersect at x = −2 and x = 4.−2x + 7 = x2 − 4x − 1 ⇔ x2 − 2x − 8 = 0 ⇔(x + 2)(x − 4) = 0 ⇔ x = −2, 4, which confirmsthe graphical solution.

dA = (y1 − y2) dx = (−x2 + 2x + 8) dx

A x x dx x x x= + + = − + +− −∫ (– )2 3 2

2

4

2

4

2 81

38

= − + + − − + =64

316 32

8

34 16 36

19.y

x

(x, y2)

(x, y1)

2–2

6

The graphs intersect at x = −2 and x = 2.0.5x2 + 2x = −x2 + 2x + 6 ⇔ 1.5x2 = 6 ⇔ x =−2, 2, which confirms the graphical solution.dA = (y2 − y1) dx = (−1.5x2 + 6) dx

A x dx x x= + = − +− −∫ (– . )1 5 6 0 5 62 3

2

2

2

2

.

= −4 + 12 − 4 + 12 = 16

20.

0 5

5

x

y

(x, y2)

(x, y1)

The graphs intersect at x = 0 and x = 5.0.2x2 + 3 = x2 − 4x + 3 ⇔ 0.8x2 − 4x = 0 ⇔0.8x(x − 4) = 0 ⇔ x = 0, 5, which confirms thegraphical solution.

dA = (y1 − y2) dx = (−0.8x2 + 4x) dx

A x x dx x x= + = − +∫ (– . )0 8 44

1522 3 2

0

5

0

5

= − + + − =500

15

2

350 0 0 16

21.

(x, y2)

x

2

0

(x, y1)

5

y

The graphs intersect at x = 0 and x = 5.dA = (y1 − y2) dx = (2e0. 2x − cos x) dx

A e x dx e xx x= − = −∫ ( cos ) sin. .2 100 2 0 2

0

5

0

5

= 10e − sin 5 − 10 + 0 = 18.1417…

22.

x

y

1

(x, y2)

(x, y1)

dA = (y2 − y1) dx = (e2x − sec2 x) dx

A e x dx e xx x= = −∫ ( – sec ) tan2 2 2

0

1

0

1

0 5.

= 0.5e2 − tan 1 − 0.5 + 0 = 1.6371…


23.

4

(x1, y)(x2, y)

–2 2

x

y

1

The graphs intersect at y = 1 and y = 4.Write y = x + 3 as x = y − 3.y − 3 = −y2 + 6y − 7 ⇒ y2 − 5y + 4 = 0 ⇒(y − 1)(y − 4) = 0 ⇒ y = 1, 4, which confirmsthe graphical solution.

dA = (x2 − x1) dy = (−y2 + 5y − 4) dy

A y y dy y y y= + = − + −∫ (– – )2 3 2

1

4

1

4

5 41

3

5

24

= − + − + − + =64

340 16

1

3

5

24 4

1

224.

(x2, y) (x1, y)

y

x

5

–5

8

The graphs intersect at y = −5 and y = 5.Write y = −2x1 + 11 as x1 = 5.5 − 0.5y.

5.5 − 0.5y = 0.25y2 − 0.5y − 0.75 ⇒0.25y2 = 6.25 ⇒ y = −5, 5, which confirms thegraphical solution.

dA = (x1 − x2) dy = (−0.25y2 + 6.25) dy

A y dy y y= + = − +− −∫ (– . . )0 25 6 25

1

12

25

42 3

5

5

5

5

= − + − + =125

12

125

4

125

12

125

4

2

341

25.

(x, y1)y

x

(x, y2)

–1 2

4

The graphs intersect at x = −1 and x = 2.x3 − 4x = 3x2 − 4x − 4 ⇒ x3 − 3x2 + 4 =(x + 1) ( x − 2)2 = 0 ⇒ x = −1, 2, whichconfirms the graphical solution.

dA = (y1 − y2) dx = (x3 − 3x2 + 4) dx

A x x dx x x x= + = − +− −∫ ( – )3 2 4 3

1

2

1

2

3 41

44

= − + − − + =4 8 8 1 4 61

4

3

426.

(x, y2)

(x, y1)

y

x

–1 8

2

The graphs intersect at x = −1 and x = 8.x2/3 = (x + 1)1/2 + 1 ⇒ x = −1, 8 numerically,which confirms the graphical solution.Or x2/ 3 − 1 = (x + 1)1/ 2 ⇒ (x2/ 3 − 1)2 = x + 1.Write t = x1/ 3, so (t2 − 1)2 = t3 + 1 ⇒t4 − t3 − 2t2 = t2(t + 1)(t − 2) = 0 ⇒t = 0, −1, 2 ⇒ x = t3 = 0, −1, 8.

But x = 0 is extraneous from the irreversible stepof squaring both sides. So x = −1, 8.

dA = (y2 − y1) dx = [(x + 1)1/2 + 1 − x2/3] dx

A x x dx= + +−∫ [( ) – ]/ /1 11 2 2 3

1

8

= + + −−

2

31

3

53 2 5 3

1

8

( ) /x x x /

= + − − + − =18 8 0 1 796

5

3

5

1

5

27. Wanda: You can always tell the right waybecause the altitude of the strip should bepositive. This will happen if you take(larger value) minus (smaller value). In this case,if you slice vertically, it’s line minus curve(see graph).

curve

line

x

y

For curve minus line, you’d get the opposite ofthe right answer. Note that if you slicehorizontally, it would be curve minus line.

28. a. Peter: Horizontal slicing would be awkwardbecause for some values of y the length of thestrip would be given by line minus curve, butin others it would be boundary minus curve,and yet elsewhere it would be curve minuscurve. If you use vertical slices, the length


of the strip will always be line minus curve.(See graphs.)

x

y

x

y

(x, y1)

(x, y2)

b. Peter: In the graph on the right, y1 − y2 willbe positive. Because y2 is negative, you willget (pos.) − (neg.), which is equivalent to(pos.) + (pos.). Thus, the altitude for the stripis positive.

29.

x

y

h(x, y)

(x, ah2)

–h

The graph shows the parabolic region fromx = −h to x = h and a strip from the graph to ahorizontal line at y = ah2.

dA = (ah2 − y) dx = (ah2 − ax2) dx

A ah ax dx ah x axh

h h

= = −

−∫ ( – )2 2 2 3

0

21

3

= −

=2

1

3

4

33 3 3a h h ah

Area of rectangle = 2h(ah2) = 2ah3

∴ = =area of region

area of rectangle,

( / )4 3

2

2

3

3

3

ah

ah Q.E.D.

The graph shows y = 67 − 0.6x2 and the liney = 7, with a circumscribed rectangle.

67

y = 7

y

x

7 = 67 − 0.6x2 ⇒ 0.6x2 = 60 ⇒ x = ±10Rectangle has width 10 − (−10) = 20 and length67 − 7 = 60. Area of region = 2

3 20 60 800( )( ) = .

30. dA = sin x dx

A x dx x= = − = − − + =∫ sin cos0 0

1 1 2π π

( ) , which

is a rational number.

1

π

x

y

(x, y)y = cos x

–π/10 π/10

x

y = 7 cos 5x

y7

For y = 7 cos 5x, width is 1/5 as much andaltitude is 7 times as much.∴ A = (2)(1/5)(7) = 2.8

31.

–2 3

8

x

y(x, y

2)

(x, y1)

The graphs intersect at x = −2 and x = 3.dA = (y2 − y1) dx = (−x2 + x + 6) dx

A x x dx x x x= + + = − + +− −∫ (– )2 3 2

2

3

2

3

61

3

1

26

= − + + − − + = =9

9

218

8

32 12 20

5

620 8333. K

R10 = 20.9375R100 = 20.834375R1000 = 20.83334375The Riemann sums seem to be approaching theexact answer.

32.

10

π 2π 3π 4π

x

t(x)

t′(x) = 1 + cos xt′(x) = 0 ⇒ cos x = −1 ⇒x = π + 2π n = … , π, 3π, 5π, …t′(x) is never negative, so t′(x) does not changesigns. These points are plateau points.

Problem Set 5-9

Q1.1

3

1

23 2x x x C+ + + Q2.

4

77 4x C/ +

Q3. y x′ = −2

31 3/ Q4. − +−1

33e Cx

Q5. −csc x + C Q6. x− 1

Q7. See Section 5-4. Q8. mean value

Q9. Q10.

4

x

y

x

y

1

1


Plan of attack for volume problems:

• Do geometry to get dV in terms of samplepoint (x, y).

• Do algebra to get dV in terms of one variable.

• Do calculus to add up the dV’s and take the limit(i.e., integrate).

1. a. dV = π x 2 dyy = 9 − x2 ⇒ x2 = 9 − ydV = π (9 − y) dy

b. V y dy y y= − = −∫ π π( ) ( . ) 9 9 0 5 2

0

9

0

9

= π (81 − 40.5) − π (0 − 0) = 40.5π= 127.2345…

c. R100 = 127.2345… (Checks.)

d. Volume of circumscribed cylinder is 9(π ⋅ 32)= 254.4690… . Half of this is 127.2345… ,equal to the volume of the paraboloid.

2. dV = π x2 dyy = 10 − 2x ⇒ x = 5 − 0.5y∴ dV = π (5 − 0.5y)2 dy

V y dy y= − = − −∫ π π( . ) ( . )2 3

0

10

5 0 52

35 0 5

0

10

= − + = = …2

30

2

3125

250

3261 7993π π π

( ) ( ) .

Vcylinder = πr2h, so 1

3 of that is V r h= 1

32π .

Here, r = 5 and h = 10, so V = =1

35 102π ( )( )

250

3

π, as found by integrating.

3. a. dV = π y2 dx = π (3e− 0.2 x)2 dx = 9π e−0.4 x dx

b. 9 22 50 4 0 4

0

5

0

5

π πe dx ex x− −= −∫ . ..

= −22.5π e−2 + 22.5π e0 = 61.1195…The midpoint Riemann sum R100 gives61.1185… , which is close to the answerfound using integration.

c. Slice perpendicular to the axis of rotation, soslice vertically if rotating about the x-axis andhorizontally if rotating about the y-axis.

4. y = 4x − x2 is rotated about the x-axis.

41

x

y(x, y)

dV = π y2 dx = π (4x − x2)2 dx

V x x dx x x x dx= − = +∫ ∫π π( )4 16 82 2

1

42 3 4

1

4

( – )

= − +

= = …π π16

32

1

530 6 96 1323 4 5

1

4

x x x . .


5. y = x1.5 is rotated about the x-axis.

(x, y)

19

27

x

y

dV = π y2 dx = π x3 dx

V x dx x= = = = …∫ π π π3 4

1

9

1

9 1

41640 5152 2119.

6. y = ln x ⇒ x = ey is rotated about the y-axis.

(x, y) x

y

1

1

dV = π x2 dy = π e2y dy

V e dy e ey y= = = − =∫ π π π2 2

0

12

0

1

2 21 10 0359( ) . K

7. y = x3/4 ⇒ x = y4/ 3 is rotated about the y-axis.

(x, y)

x

y

1

8

1 16

dV = π x2 dy = π y8/3 dy

V y dy y= = ⋅ =∫ π π π8 3 11 3

1

83

11

6141

11/ /

1

8

= 1753.8654…

8. y x= 14

and y = 8x2, intersecting at (0, 0) and(2, 16), are rotated about the y-axis. Area of crosssection is π πx x1

222− .


x1 = y1/ 4, and x y21

8=

∴ = − = −

dV x x dy y y dyπ π( ) /

12

22 1 2 21

64

V y y dy= −

∫ π 1 2 2

0

16 1

64/

= −

= =π π2

3

1

192

64

367 02063 2 3

0

16

y y/ . K

The midpoint Riemann sum R100 gives V ≈67.0341… , which is close to the answer foundusing integration.

9. y1 = e0.4 x and y2 = x + 1, from x = 0 to x = 3, arerotated about the x-axis.Area of cross section is π πy y2

212− .

dV y y dx x e dxx= − = + −π π( ) [( ) ]222

12 0 81 .

V x e dx

x e

e

e

x

x

= + −

= + −

= − − +

= − = …

∫ π

π

π

π

[( ) ]

.

1.25 1.25

( . . ) . ft

2

.

1

1

31 1 25

64

3

1

3

22 25 1 25 26 6125

0 8

0

3

3 0 8

0

3

2 4

2 4 3

.

.

.

( )

The midpoint Riemann sum R100 givesV = 26.6127… , which is close to the answerfound using integration.

10. y1 = x1/3 and y2 = 10e− 0.1 x are rotated about thex-axis. Only the back half of the solid is shown.

80

10y

x

(x, y2)

(x, y1)

dV y y dx e x dxx= − = −−π π( ) ( )22

12 0 2 2 3100 . /

V e x dx

e x

e

x

x

= −

= − −

= − + = …

−

−

−

∫ π

π

π

( )

( . )

( . ) .

0

8

.

100

500 0 6

500 480 8 1193 3394

0 2 2 3

0

8

0 2 5 3

1 6

. /

. /

11. y = 4 − x1 ⇒ x1 = 4 − y, and y = 4 − x22

⇒ x2 =4 – ,y intersecting at x = 0 and x = 1, are

rotated about the y-axis. Only the back half ofthe solid is shown.

x

y4

3

1

(x1, y)

(x2, y)

dV x x dy y y dy= − = − − −π π( ) [( ) ( ) ]222

12 4 4

= π (−y2 + 7y − 12) dy

V y y dy= − + −∫ π ( ) 4

2

37 12

= − + −

π 1

3

7

2123 2

3

4

y y y

= − + − + − +

π 64

356 48 9

63

236

= = …1

60 523598π .

12. y = ax2 ⇒ x = (y/a)1/2, from (0, 0) to (r, h), isrotated about the y-axis.

r

h

(x, y)x

y

dV = π x2 dy = π (y/a) dy = (π /a)y dy

V a y dy a y a hh h

= = ⋅ = −∫ ( / ) ( / )( )π π π( / )1

2

1

202

0 0

2

Because y = ax2, h = ar2.

∴ = =V a ar ar1

2

1

22 2 4( / )( )π π

Volume of circumscribed cylinder isV c = π r2h = π r2(ar2) = π ar4.Thus, the volume of the paraboloid is half thevolume of the circumscribed cylinder, Q.E.D.

13. a. y = 0.3x1.5 is rotated about the x-axis.

2

4

(x, y)

x

y

dV = π y2 dx = π (0.3x1.5 )2 dx = π (0.09x3) dx

V x dx x= =

= = …

∫ π π

π

( . ) .

5.76 18.09557

0 09 0 02253 4

0

4

0

4


b. R10 = 5.7312πR100 = 5.75971…πR1000 = 5.7599971…πValues are getting closer to V = 5.76π.

14. y = 4 − x2 ⇒ x = (4 − y)1/2 dyInner radius is 3 − x; outer radius is 3.

dV = π[32 − (3 − x)2] dy

= π{9 − [3 − (4 − y)1/2]2} dy

= π [6(4 − y)1/2 − 4 + y] dy

V y y dy= − − +∫ π[ ( ) ]1/26 4 40

4

= − − − +π[ ( ) . ]3/2

0

44 4 4 0 5 2y y y

= π (0 − 16 + 8 + 32 + 0 − 0) = 24π= 75.3982…

15. y = 4 − x2 is rotated about the line y = −5. Onlythe back half of the solid is shown.

(x, y)4

y = –5

x

y

2

dV = π[(y + 5)2 − 52] dx

= π[(9 − x2)2 − 52] dx = π (56 − 18x2 + x4) dx

V x x dx

x x x

= − +

= − +

= − + − + −= = …

∫ π

π

ππ

( )

( . )

( . )

. .

0

2

56 18

56 6 0 2

112 48 6 4 0 0 0

70 4 221 168

2 4

0

2

3 5

16. Cross sections perpendicular to the x-axis aresquares with side length (y2 − y1). The curvesintersect at (0, 0) and (1, 1).

x1

y

1

dV = (y2 − y1)2 dx = (x1/5 − x2)2 dx

V x x dx

x x x dx

= −

= − +

∫∫

( )

( )

/

/ /

1 5 2

0

12

2 5 11 5 4

0

1

2

= − +

= =5

7

5

8

1

5

81

2800 28927 5 16 5 5

0

1

x x x/ / . K

17. Cross sections perpendicular to y-axis are squaresof edge 2x, where (x, y) is a sample point on theline in the xy-plane.

y = –(15/4)x + 15

4

15

(x, y)

x

y

Equation of line is

y x x y= − + ⇒ = −15

415 4

4

15.

dV x dy y dy

y dy

= = −

= −

( )22

2

2 4 44

15

64 11

15

64 11

15320 1

1

150

15

0

15

−

= − −

∫ y dy y

2 3

= 320 3 cm

The circumscribed rectangular box has volumel · w · h = 8 · 8 · 15 = 960 = 3V, so the pyramidis 1/3 the volume of the circumscribed rectangularsolid, Q.E.D.

The volume of a pyramid is one-third the volumeof the circumscribed rectangular box, just as thevolume of a cone is one-third the volume of thecircumscribed cylinder.

18. Center line: y = 0.2x2

Upper bound: y = 0.16x2 + 1Radius of circular cross section is 1 − 0.04x2.The tip of the “horn” is where 0.2x2 = 0.16x2 + 1with x ≥ 0, which is at x = 5.

dV = π (1 − 0.04x2)2 dx

= π (1 − 0.08x2 + 0.0016x4) dx

V x x dx

x x x

= − +

= − +

= − + − + −

= = … ≈

∫ π

π

π

π

( . . )

. cm

0

5

3

1 0 08 0 0016

0 08

3

0 0016

5

510

31 0 0 0

8

38 3775 8 4

2 4

0

5

3 5. .

.


19. a. y = x0.6

Pick sample point (x, y) on the curve withinthe slice. One leg of the isosceles triangle isy, so the other leg is also equal to y.

dV y dx x dx= =1

2

1

22 1 2.

b. V x dx x= = = ⋅ −∫ 1

2

1

4 4

1

4 44 01 2 2 2

0

4

0

42 2. .

. ..

= 4.7982…


c. If the cross sections were squares, they wouldhave twice the area of the triangles, so dV

would be twice as much and V = ⋅ =1

2 242 2

..

9.5964… .

20. y = ex, y = 3, and x = 0. Cross sectionsperpendicular to the x-axis are rectangles withheight equal to 4 times the base. Each base haslength (3 − y).

x

y

12

dV y y dx

e e dx e dxx x x

= − −

= − − = −

( )[ ( )]

( )[ ( )] ( )

3 4 3

3 4 3 4 3 2

V e dx e e dx

x e e

x x x

x x

= − = − +

= − +

= − ⋅ + ⋅ − + −

= …

∫ ∫4 3 4 9 6

4 9 61

2

4 9 3 6 31

29 0 6

1

2

7 5500

2

0

32

0

3

2

0

3

( ) ( )

ln

ln ln

ln

.

21. y = x2 and y = 2 − x2, intersecting at x = 1.Cross sections perpendicular to the x-axis areequilateral triangles. Each base has length (y2 − y1).

x

y

1

1

Using properties of special right triangles, youcan find that an equilateral triangle with

base b has height3

2b.

bb

b √ 312

60°12

dV bh dx y y y y dx

y y dx x x dx

x dx

= = − −

= − = − −

= −

1

2

1

2( )

3

2( )

( ) ( )

( )

2 1 2 1

2 12 2 2 2

2 2

3

4

3

42

3

42 2

V x dx

x x dx

x x x

= −

= − +

= − +

= − + − + −

= =

∫∫

3

42 2

3

44 8 4

3

44

8

3

4

5

3

44

8

3

4

50 0 0

8 3

15

0

12 2

2 4

0

1

3 5

0

1

( )

( )

0.9237...

22. y = ln x and y = 1. Cross sections perpendicularto the y-axis are rectangles with height equal to1/2 the base. Each base has length x.

x

y

1

5

dV x x dy x dy= ⋅ =12

12

2

Solve y = ln x for x, to get x = ey.

dV e dy e dyy y= =12

2 12

2( )

V e dy e ey y= = = − =∫ 1

2

1

4

1

4

1

41 59722 2

0

12

0

1

. K

23. a. Line has equation y x= 12 , 0 ≤ x ≤ 6.

b. The log has radius = 6, so the circle is

x2 + z2 = 36, or z x x= = −36 362 2– ( ) .1/2

c. dV = y ⋅ 2z ⋅ dx = 1

2x · 2(36 − x2)1/2 ⋅ dx

= (36 − x2)1/2 (x dx)

V x x dx

x x dx

=

= − −

∫∫

( – )

( – )

36

1

236 2

2 1 2

0

6

2 1 2

0

6

/

/

( )

( )

= − ⋅ =1

2

2

336 722 3 2

0

6

( – )x / 3 in.


24. The points (0, 0) and (r, h) in xy-coordinates areon the line running up the top surface, so the

line is yh

rx= . The circle forming the boundary

for the bottom surface has radius = r and center(0, 0) in xz-coordinates, so the circle is x2 + z2 =r2, or z r x= 2 2– . The slab at x = x0 is

rectangular of height yhx

r= 0 , width

2 2 202z r x= – , and thickness dx, so

dVhx

rr x dx= −2 2 2 , and

Vh

rx r x dx

h

rr x x dx

r

r

= −

= − − −

∫∫

2

2

2 2

0

2 2 1 2

0( ) ( )/

= − ⋅ − = − ⋅ −

=

2

3

2

30

2

3

2 2 3 2

0

3 2 3

2

h

rr x

h

rr

r h

r

( ) ( )/ /

25. A cone of radius r and altitude h can be generatedby rotating about the x-axis the line

yr

hx= from x = 0 to h.

(x, y)

rh x

y

dV y dxr

hx dx= =π π2

2

22

Vr

hx dx

r

hx r h

h h

= = ⋅ =∫π π π2

22

2

23

0 0

21

3

1

3, Q.E.D.

26. a. Equation of circle in xy-plane is x2 + y2 = 100.

dV = πx2 dy = π(100 − y2) dy

V y dy= −−∫ π ( ) 100 2

10

10

= −

−

π 1001

33

10

10

y y

= − + −

π 10001

31000 1000

1

31000( ) ( )

= 4

3π (1000) cm3

b. Formula: V r= = =43

3 43

3 4310 1000π π π ( ) cm3,

which agrees with the answer by calculus.

27. Sphere can be generated by rotating about they-axis the circle x2 + y2 = r2.

Slicing perpendicular to the y-axis as in Problem26 gives dV = πx2 dy = π(r2 − y2) dy.

V r y dy r y yr

r

r

r

= − = −

− −

∫ π π( )2 2 2 31

3

= −

− − +

=π π πr r r r r3 ,

1

3

1

3

4

33 3 3 3

Q.E.D.

28. The graph shows slices perpendicular to x-axiswith sample points (x, y) and (x, z).

a

by

x

z

(x, y)

(x, z)c

Equation of ellipsoid is x

a

y

b

z

c

+

+

=

2

.2 2

1

For a fixed value of x, the x-term will beconstant. Subtracting this term from both sidesof the equation gives an equation of the form

y

b

z

ck

+

=

2 22, where k2 = 1 − (x/a)2.

Dividing both sides by k2 givesy

kb

z

kc

+

=

2 2

1. Thus, the y- and z-radii are

kb and kc, which have the original ratio b/c.Therefore, each elliptical cross section is similarto the ellipse at the yz-plane, Q.E.D.dV = πyz dxBecause z = (c/b)y, dV = π (c/b)y2 dx.The ellipse in the xy-plane (z = 0) has equation

x

a

y

b

+

=

2 2

1, from which y2 = (b/a)2(a2 − x2).

∴ dV = π (c/b)(b/a)2(a2 − x2) dx

V c b b a a x dxa

a

= −−∫ π ( / )( / ) ( )2 2 2

= −

−

π ( / )( / )2c b b a a x xa

a2 31

3

= ⋅ =π π( / )( / )2c b b a a abc4

3

4

33

Note that the volume formula for a sphere is aspecial case of the volume formula for anellipsoid in which a = b = c = r, the radius of thesphere.

29.50 + 2L

L

y

L

50


Note that the top of each isosceles trapezoidalcross section has length 50 + 2L yards, wherey

LL y y= ° ⇒ = ° =tan cot cot( ) ( ) .52 52

52

180

π

So each slab is dV L y dx= + +12 50 2 50( ) ;

dV y y dx= +5052

1802 cot

π, and

V y yk k

k

= +

⋅

= ≈=

∑ 5052

18030

1 649 443 6 1 649 443

2

0

19

cot

, , . , , .

π

K yd3

Cost = 12 ⋅ 1,649,443.6… ≈ $19,793,324

Problem Set 5-10

Q1.1

33x x C+ + Q2. 24

Q3. sec2 x dx x C= +∫ tan Q4. 2 sec2 x tan x

Q5. Answers may vary. Q6. See graph in Q5.

dt t

v

(t, v)

Q7. See graph in Q5. Q8. d (disp) = v dt

Q9. v dt a

b

∫ Q10. A

1. cos x dx .0.3

1.4

≈∫ 0 6899295233K

2. ( – )x x dx2

1

4

3 5 13 5+ =∫ .

3. 2 10 0988652x dx ≈∫ .0

3

K

4. tan 0.1

1.4

x dx∫ ≈ 1 76714178. K

5. cos sin sin sin.

.

.

.

x dx x0 3

1 4

0 3

1 4

1 4 0 3∫ = = − =. .

0.6899295233…

For the ten digits of the answer shown bycalculator, there is no difference between thissolution and the solution to Problem 1.

6. ( )x x dx x x x2

1

43 2

1

4

3 51

3

3

25− + = − +

∫

= − +

1

34

3

24 5 43 2( ) ( ) ( )

− − +

=1

31

3

21 5 1 13 53 2( ) ( ) ( ) .

There is no difference between this answer andthe solution to Problem 2.

7.

x

y

2

π

πsin2

04 9348∫ =x dx . K

We cannot compute this integral algebraicallybecause we do not know an antiderivative forsin2 x.

8.

x

y

5

2

π (ln ) .x dx2

2

5

14 6673∫ = K

We cannot compute this integral algebraicallybecause we do not know an antiderivative for(ln x)2.

9. a.

–20 20

–2

2 Si x

x

b. (sin x)/(x) approaches 1 as x approaches zero.

c. Answers will vary depending on the grapherused. The TI-83 gives Si 0.6 = 0.58812881using TRACE or 0.588128809608 using TABLE,both of which are correct to as many decimalplaces as the NBS values.

d. By TABLE, Si x seems to be oscillating betweenabout 1.53 and 1.61 when x is between 20and 30. The limit is somewhere betweenthese two numbers, say about 1.57. Theactual limit is π /2, which equals 1.570796… .

e.

–20 20

–2

2 Si x

x

The f graph is positive when x is between −πand π (as well as elsewhere), and has itsgreatest values there, which agrees with the


large positive slope of the Si x graph in thisregion. Each place where the Si x graph has ahigh or low point, the f ( x) graph has a zero,corresponding to the zero slope of the Si xgraph. So f ( x) is the derivative of Si x.

10. a. erf x e dttx

= −∫2 2

0πThe integrand e t− 2

is an even functionintegrated between symmetrical limits. Thus,rather than using the entire interval [−x, x],one may find the integral on [0, x] and doublethe result.

b.y = erf x

x

1

2

c. By TABLE, values of erf x are

x erf x

1 0.842700792…

2 0.995322265…

3 0.999977909…

4 0.999999984…

The values approach 1, meaning that thefraction of the data between −x and x isvirtually 100% when x is beyond 4.

d. Answers will vary depending on thegrapher used. The TI-83 gives erf 0.5 =0.52049987781… using TABLE, which iscorrect to as many decimal places as the NBSvalue.

e.y = erf x

x

1

2

The slope of y = erf x appears to equal about1 when x = 0 and decreases toward zero as xincreases, which agrees with the graph of f.

11. a. (speed) [( / )/ ](dt ≈ + ⋅ + ⋅∫ 2 60 3 33 4 25 2 270

12

+ 4 ⋅ 13 + 2 ⋅ 21 + 4 ⋅ 5 + 9)= (1/90)(310) = 3.444… ≈ 3.4 nautical miles

b. T6 = (1/30)(33 ⋅ 0.5 + 25 + 27 + 13 + 21 + 5 + 9 ⋅ 0.5)= (1/30)(112) = 3.7333… ≈ 3.7 nautical miles

c. The answer by Simpson’s rule should becloser, because the graph is represented bycurved segments instead of straight ones.

12. (Data and CAT scans for this problem wereprovided by Dr. James Stewart of San Antonio.)

a. A dD ( . / )( .≈ + ⋅ + ⋅∫ 0 8 3 6 8 4 6 8 2 20 10

8

. .

+ 4 ⋅ 25.3 + 2 ⋅ 29.5 + 4 ⋅ 34.6 + 2 ⋅ 38.4+ 4 ⋅ 33.9 + 2 ⋅ 15.8 + 4 ⋅ 6.1 + 2.3)

= (0.8/3)(643.5) = 171.6 cm3

b. The mass will be 171.6 g, which is withinthe normal range of 150 to 200 g.

13. a.

0.5

Distance (in.)

Force (lb)

300

b. Let F = force, x = displacement, W = work.

W F dx= ∫ 0

0 5.

≈ (0.05/3)(0 + 4 ⋅ 120 + 2 ⋅ 240 + 4 ⋅ 360+ 2 ⋅ 370 + 4 ⋅ 330 + 2 ⋅ 290 + 4 ⋅ 280+ 2 ⋅ 270 + 4 ⋅ 270 + 190)

= (0.05/3)(7970) = 132.8333…≈ 132.8 inch-pounds

14. Let C = heat capacity (Btu/lb mole)/ºF,T = temperature (ºF), H = heat (Btu/lb-mole).Approximate values of C to the nearest 0.02 arefrom the given figure.

T C Simpson’s factor

500 8.44 1

1000 9.24 4

1500 10.08 2

2000 10.84 4

2500 11.48 2

3000 11.98 4

3500 12.36 2

4000 12.68 4

4500 12.94 1

H C dT= ≈ +∫ 500

38 44 9 24 4

500

4500

[ . ( . )( )

+ (10.08)(2) + (10.84)(4) + (11.48)(2)+ (11.98)(4) + (12.36)(2)+ (12.68)(4) + 12.94]

= = ≈500

3268 18 44696 6666( . ) . K 44,697 Btu

The answers students get will vary slightly.


15. a. Simpson’s rule will give a more accurateanswer because the function y = sin x isapproximated better by quadratic functionsthan by straight lines.

b. S4 = (1/3)(π /4)[sin 0 + 4 sin (π /4) +2 sin (π/2) + 4 sin (3π /4) + sin π] = 2.0045…

T4 = (1/2)(π /4)[sin 0 + 2 sin (π/4) +2 sin (π /2) + 2 sin (3π /4) + sin π] = 1.8961…

sin cos0

x dx xπ π

π∫ = − = − =0

0 2cos cos

Simpson’s rule does give a betterapproximation of the integral because S4 iscloser to 2 than is T4.

16. Programs will vary depending on the type ofgrapher used. See the program in the Programsfor Graphing Calculators section of theInstructor’s Resource Book.

17. Using a Simpson’s rule program, the mass of thespleen is 171.6 cm3.

18. Enter Y12 2

= −

πe x . A Simpson’s rule program

gives S50 = 0.5204998781… and S100 =0.5204998778… . There is little differencebetween the two estimations, and both are closeto the tabulated value.

19. a.

1

2

y

tx

As x varies, the area beneath the curve y = 1/tfrom t = 1 to t = x varies also.

b. Using the power formula on t dt−∫ 1 gives

t0

0.

Division by zero is undefined, so thisapproach does not work.

c. Graph Y1 = fnint(x− 1, x, 1, x). (Entries maybe different for different calculators.) Thegraph looks like y = ln x. The value of f (x)is negative for x < 1 because for these valuesthe lower limit of integration is larger than theupper limit, resulting in negative values for dx.

5

1

y

x

d. f (2) = 0.6931…

f ( 3) = 1.0986…

f ( 6) = 1.7917…

f ( 2) + f ( 3) = f ( 2 ⋅ 3). This is a property oflogarithmic functions.



R1. a. The width of each region is 4. So

T3 = (4/2)[v(4) + 2v(8) + 2v(12) + v(16)] =2[22 + 2(26.9705…) + 2(30.7846…) + 34] =343.0206… . T3 underestimates the integralbecause v(t) is concave down, so trapezoidsare inscribed under the curve.

b. R3 = 4[v(6) + v(10) + v(14)] = 4(24.6969… +28.9736… + 32.4499…) = 344.4821…

This Riemann sum is close to the trapezoidal-rule sum.

c. T50 = 343.9964… , and T100 = 343.9991…Conjecture: The exact value of the integralis 344.

d . g ( t) = 10t + 4t1.5

g(16) − g(4) = 344

This is the value the trapezoidal-rule sums areapproaching.

R2. a. The slope of the linear function is the sameas the slope of the curve at x = 1. So theslope is

f ( x) = sin π x ⇒ f ′(x) = π cos π x ⇒f′( 1) = π cos π = −πAt x = 1, y = sin π = 0

y − 0 = −π (x − 1) ⇒ l(x) = −π x + π

1

1y

xf(x)

l (x)

0.8 1 1.2

x

f(x)l (x)

0.2

As you zoom in, you see that f ( x) is veryclose to the line l(x) for values near x = 1.


For x = 1.1, the error is sin [π ( 1.1)] −[−π ( 1.1) + π] = 0.0051…For x = 1.001, the error is sin [π ( 1.001)] −[−π(1.001) + π] = 5.1677… × 10− 9.

b. i. y = csc5 2x ⇒ dy = −10 csc5

2x cot 2x dx

ii. y = x5/5 − x− 3/3 ⇒ dy = (x4 + x− 4) dx

iii. y = (7 − 3x)4 ⇒ dy = −12(7 − 3x)3 dx

iv. y = 5e− 0. 3x ⇒ dy = −1.5e− 0. 3x dx

v. y x dyx

x dx= ⇒ = ⋅ ⋅ln( )

( )( )421

24 2 24

3

= 4/x dx; or y = ln ( 2x)4 = 4 ln ( 2x) ⇒

dy = 41

22⋅ ⋅

xdx = 4/x dx

c. i. dy = sec x tan x dx ⇒ y = sec x + C

ii. dy x dx y x C= + ⇒ = + +( )53 71

183 7 6( )

iii. dy = 5 dx ⇒ y = 5x + C

iv. dy = 0.2e− 0.2 x ⇒ y = −e− 0.2 x + C

v. dy dx y Cxx

= ⇒ = +66

6

ln

d. i. y = (2x + 5)1/2 ⇒ dy = (2x + 5)− 1/2 dx

ii. x = 10 and dx = 0.3 ⇒dy = 25− 1/2 ⋅ 0.3 = 0.06

iii. ∆y = (2 ⋅ 10.3 + 5)1/2 − (2 · 10 + 5)1/2

= 0.059644…

iv. 0.06 is close to 0.059644… .

R3. a. See the text for the definition of indefiniteintegral.

b. i. 12 7 22 3 5 3x dx x C/ /= +∫ .

ii. sin cos sin6 71

7x x dx x C= +∫

iii. ( )x x dx x x x C2 3 28 31

34 3− + = − + +∫

iv. 12 43 3e dx e Cx x∫ = +

v. 77

7x

x

dx C∫ = +ln

R4. a. See the text for the definition of integrability.

b. See the text for the definition of definiteintegral.

c. sec.

.

x dx0 2

1 4

∫i. U6 = 2.845333…

ii. L6 = 1.872703…

iii. M6 = 2.209073…

iv. T6 = 2.359018…

d. U6 L6

0.2 1.4

x

5y

0.2 1.4

x

5y

M6 T6

0.2 1.4

x

5y

0.2 1.4

x

5y

e.

y

x

f(x)

R5. a. The hypothesis is the “if ” part of a theorem,and the conclusion is the “then” part. (Hypo-means “under,” and -thesis means “theme.”)

b. d t t( ) = +20 34

sinπ

Average velocity 1.5 m/s= =d d( ) – ( )

–

2 0

2 0

Instantaneous velocity = d ′(t) = 0.75π cos π4

t

d c c′ = =( ) . .0 754

1 5π πcos

π4

c = cos− 1 (2/π) = 0.880689…

c = 1.12132… ≈ 1.12 s

c. g(x) = x4/3 − 4x1/3 = x1/3(x − 4)

g(x) = 0 ⇒ x = 0 or x = 4. Interval is [0, 4].

g′(x) = (4/3)x1/3 − (4/3)x− 2/3 = (4/3)x− 2/3(x − 1)

g′(c) = 0 ⇒ c = 1

At x = 0, g′(0) takes the form 1/0, which isinfinite.Thus, g is not differentiable at x = 0.However, the function need not bedifferentiable at the endpoints of the interval,just continuous at the endpoints anddifferentiable at interior points.

d. For a function to be continuous on a closedinterval, the limit needs to equal the functionvalue only as x approaches an endpoint fromwithin the interval. This is true for function f


at both endpoints, but not true for function gat x = 2. The graphs show that the conclusionof the mean value theorem is true for f butnot for g.

2 7

4

f(x)

x

Tangent parallels secant

c

Secant

2 7

4

g(x)

x

Secant

No tangentparallels secant

(Middle branch has the equation y = 1.4( x−2 ) .Point c = 4.4825… .)

e. g is the linear function containing the points(a, f (a)) and (b, f (b)). h is the function h (x) =f ( x) − g(x). Thus, h(a) = h ( b) = 0, satisfyingone hypothesis of Rolle’s theorem. The othertwo hypotheses are satisfied because f and gare differentiable and continuous at theappropriate places, and a difference ofdifferentiable and continuous functions alsohas these properties. The c in (a, b) for whichh′(c) = 0 turns out to be the c in (a, b) forwhich f ′ (c) equals the slope of the secant line,g′(c), which equals [f (b) − f (a)]/(b − a).

f.

x

f(x)8

1

Points are 18

14

38

12

58

34

78, , , , , , and .

g. If r′(x) = s′(x) for all x in an interval, thenr(x) = s(x) + C for some constant C.

R6. a. g x x dx x C( ) .= = +∫ 1 5 2 50 4. .

Without loss of generality, let C = 0.

g c

g g′ = =( ) .12 1

2 11 862741

( ) – ( )

–K

∴ = …⇒c11 5 1 862741. .

c1 = (1.862741…)1/1.5 = 1.513915927…

Similarly, c2 = 2.50833898… .

c3 = 3.505954424…

For x dx1 5

1

.4

,∫

R3 = (1.513…)1.5 + (2.508…)1.5 + (3.505…)1.5

= 12.4, which is the exact value of theintegral.

b. ( – )10 10 1 32 3

1

3

1

3

x dx x x= −− −∫ ( / )

= 30 − 9 − (−10) + (−1/3) = 92/3 = 30.6666…

c. T100 = 30.6656, which is close to 92/3.

d. M10 = 30.72

M100 = 30.6672

M1000 = 30.666672

These Riemann sums are approaching 92/3.

R7. a. i. x dx x− − − −= − = − + =∫ 2 1

1

5

1

51 15 1 4 5 /

ii. ( )x x dx2 5

3

4

3+∫ ( )

= +∫( / ) ( )1 2 3 22 5

3

4

( )x x dx

= +( / )( )112 32 63

4x

= (1/12)(19)6 − (1/12)(12)6

= 3,671,658.08…

iii. (sin – ) cosx dx x x5 50 0

= − −∫π π

= −cos π − 5π + cos 0 + 0 = 2 − 5π

iv. 4 2 2 2 482 2

0

5

0

52 5 0e dx e e ex x= = − =∫

ln lnln

v. 33

3

3

3

3

3

78

31

4

1

4 4 1x

x

dx∫ = = − =ln ln ln ln

= 70.9986…

b.

x

y

–5

Integral is negative, because each y-value inthe Riemann sum is negative.

c. ( sin – )4 6 8 4 2 47 3

10

10

x x x dx dx+ + =−∫ ∫0

10

= =8 80x 0

10

d. Total area = sum of two areas

x

f(x)

a c b


R8. a.

dt t

v

(t, v)

dy = v dt = 150t0.5 dt

y t dt t= = =∫ 150 100 27000 5 1 5

0

9

0

9. . ft

For [ , ], ..0 4 150 8000 5

0

4

y t dt= =∫For [ , ], .4 9 150 2700 8000 5

4

9

y t dt= = −∫= 1900.

So v t dt( ) = = +∫ 2700 800 19000

9

= + ∫∫ v t dt v t dt( ) ( ) .4

9

0

4

b. dA = x dyy = ln x ⇒ x = ey

dA = ey dy

e dy e e ey y

0

4

0

44 0 4 1 3

ln lnln∫ = = − = − =

R9. a. y = e0.2 x, from x = 0 to x = 4, is rotated aboutthe x-axis.

0 4

1

(x, y)

x

y

dV = π y2 dx = πe0.4 x dx

V e dx ex x= =∫ π π0 4 0 4

0

4

0

4

2 5. . .

= 2.5π (e1.6 − 1) = 31.0470…

b. y x= 10 25. and y = x2, intersecting at (0, 0) and

(1, 1) in Quadrant I, is rotated about they-axis. Only the back half of the solid isshown.

1

1

x

y

(x2, y)

(x1, y)

y x x y= ⇒ =10 25

14.

y = x2 ⇒ x2 = y

dV x x dy y y dy= − = −π π( ) ( ) 22

12 2 8

V y y dy y y= − = −

= =

∫ π π

π

( )

.

2 8 3 9

0

1

0

11

3

1

9

2

90 6981K

c. y = x1 + 2 ⇒ x1 = y − 2

y = 3x2 − 6 ⇒ x2 = 1

3y + 2

Graphs intersect at y = 6. Diameter of circularcross section is (x2 − x1).

dV = π [0.5(x2 − x1)]2 dy

= +

− −

= −

π π4

1

32 2

44

2

3

2 2

y y dy y dy( )

V = dV0

6

∫ ≈ 25.1327… (exactly 8π)

The right circular cone of altitude 6 andradius 2 also has volume 1

3 π π⋅ ⋅ =2 6 82 .

R10. a.

log x dx =∫ 6 09131

10

. K

The integral is reasonable because countingsquares gives approximately 6.

1 10

1t

v(t)

b. dW = v · y · dx = (1000 + 50x)(4 − 0.2x2) dx

( )( . ) , .1000 50 4 0 2 10 897 52

0

3

+ − =∫ x x dx

c. v t dt( )3

5

∫ = 1/3(0.2)[29 + 4(41) + 2(50) + 4(51)

+ 2(44) + 4(33) + 2(28) + 4(20)+ 2(11) + 4(25) + 39] = 67.6

Values of velocity are more likely to beconnected by smooth curves than by straightlines, so the quadratic curves given bySimpson’s rule will be a better fit than thestraight lines given by the trapezoidal rule.

Concept Problems

C1. a.

5b

f(b)

π/2

5–

5

–5

b.

3

1 b

f (b)


The two lines are horizontal asymptotes. y4 isthe inverse of y1.

c. The graph of f in part a is a reflection ofthe graph of y = tan x in part b across the liney = x. Function f seems to be the inversetangent function, f (b) = tan− 1 b. (In Chapter 9,students will learn that this is actually true.)

C2. f (x) = ax2 + bx + cy

x

d k e

f (d ) = ad 2 + bd + cf (e) = ae2 + be + c

∴ = + + − + +−

mae be c ad bd c

e d

2 2( )

= − + −−

= + +a e d b e d

e da e d b

( ) ( )( )

2 2

f ′(x) = 2ax + b ⇒ f ′(k) = 2ak + b∴ 2ak + b = a(e + d) + b

2ak = a(e + d)

k = (1/2)(e + d)∴ k is at the midpoint of [d, e], Q.E.D.

C3. S n n( ) = + + + + +0 1 2 32 2 2 2 2L

a. S(0) = 0, S(1) = 1, S(2) = 5, S(3) = 14S(n) = an3 + bn2 + cn + d0 = 0 + 0 + 0 + d1 = a + b + c + d5 = 8a + 4b + 2c + d14 = 27a + 9b + 3c + dSolving this system gives a = 1/3, b = 1/2,c = 1/6, d = 0.∴ S(n) = (1/3)n3 + (1/2)n2 + (1/6)n

= (n/6)(n + 1)(2n + 1)

b. By equation,S(4) = (4/6)(5)(9) = 30S(5) = (5/6)(6)(11) = 55By addition,S(4) = 0 + 1 + 4 + 9 + 16 = 30, which checks.S(5) = 0 + 1 + 4 + 9 + 16 + 25 = 55, whichchecks.S(1000) = (1000/6)(1001)(2001) =333,833,500

c. Prove that S(n) = (n/6)(n + 1)(2n + 1) for anypositive integer n.

Proof (by induction on n):

Anchor:

For n = 1, S(n) = (1/6)(2)(3) = 1, the correctanswer, which anchors the induction.

Induction hypothesis:

Assume that for some integer n = k > 1,S(k) = (k/6)(k + 1)(2k + 1).Verification for n = k + 1:

S k k k( ) ( )2+ = + + + + +1 0 1 12 2 2L

= + + + + +( ) ( )20 1 12 2 2L k k

= (k/6)(k + 1)(2k + 1) + (k + 1)2

= [(k + 1)/6][k(2k + 1) + 6(k + 1)]= [(k + 1)/6][2k2 + 7k + 6]= [(k + 1)/6][(k + 2)(2k + 3)]= [(k + 1)/6][(k + 1) + 1] ⋅

[2(k + 1) + 1],which is the formula with (k + 1) in place ofk, thus completing the induction.∴ S(n) = (n/6)(n + 1)(2n + 1) for any positiveinteger n, Q.E.D.

d. S n n( ) = + + + + +0 1 2 33 3 3 3 3LS(0) = 0S(1) = 0 + 1 = 1S(2) = 0 + 1 + 8 = 9S(3) = 0 + 1 + 8 + 27 = 36S(4) = 0 + 1 + 8 + 27 + 64 = 100(The answers are perfect squares!)Assume that S(n) = an4 + bn3 + cn2 + dn + e.0 = 0 + 0 + 0 + 0 + e1 = a + b + c + d + e9 = 16a + 8b + 4c + 2d + e36 = 81a + 27b + 9c + 3d + e100 = 256a + 64b + 16c + 4d + eSolving this system givesa = 1/4, b = 1/2, c = 1/4, d = 0, e = 0.∴ S(n) = (1/4)n4 + (1/2)n3 + (1/4)n2

= (1/4)n2(n2 + 2n + 1)= [(n/2)(n + 1)]2,

which agrees with the observation that S(n) isa perfect square.By equation,S(5) = [(5/2)(6)]2 = 225S(6) = [(6/2)(7)]2 = 441By addition,S(5) = 03 + 13 + 23 + 33 + 43 + 53 = 225,which checks.S(6) = 03 + 13 + 23 + 33 + 43 + 53 + 63 =441, which checks.

C4. a. 4 100

sin sinx x dxπ

∫= +∫ [– cos cos (– )]2 11 2 9

0x x dx

π

= − − −2

1111

2

99

0sin sinx x( )

π

= −0 − 0 + 0 + 0 = 0, Q.E.D.There is just as much area below the x-axis asthere is above it, so the integral is 0.


b. 40

sin sinx nx dxπ

∫= − + + −∫ { cos [( ) ] cos [( ) ]}2 1 2 1

0n x n x dx

π

=+

+ + −–sin ]

–sin

2

11

2

11

0nn x

nn x[( ) [( ) ]

π

=+

+ + −–sin

–sin

2

11

2

11

nn

nn( ) ( )π π

−+

−–sin

–sin

2

10

2

10

n n

If n is an integer, the first two terms willinvolve sines of integer multiples of π, andare thus equal to 0. The last two terms are 0unless n = ±1. Thus, the integral equals 0 forany integer n > 1, Q.E.D.

C5. a. Algebraic solution:

Pick sample points ck at the right end of eachsubinterval. Because f (x) is increasing on theinterval [1, 9], the high points of f (x) arelocated at the right ends of the subintervalsand the low points are at the left ends.

∴ ==

∑U f c xn k k

k

n

( )∆1

and L f c xn k k

k

n

= −=

∑ ( )1

1

∆

Note that c0 = 1 and cn = 9. Subtracting gives

U L f c f c xn n k k k

k

n

− = − −=

∑[ ( ) ( )]1

1

∆

= − + − +[ ( ) ( )] [ ( ) ( )]f c f c x f c f c x1 0 1 2 1 2∆ ∆ L+ [ f ( cn) − f ( cn− 1)]∆xn

≤ [ f ( c1) − f ( c0)] ||P|| + [ f ( c2) − f ( c1)] ||P||

+ + − −L [ ( ) ( )] || ||f c f c Pn n 1

= − + − + +[ ( ) ( ) ( ) ( )f c f c f c f c f cn1 0 2 1 L ( ) − f ( cn− 1)] ||P||= [ f ( cn) − f ( c0)] ||P|| = ||P||(1.29 − 1.21)∴ Un − Ln ≤ ||P||(1.29 − 1.21), Q.E.D.

Graphical solution:

The difference Un − Ln is equal to the area ofthe spaces between the lower and upperrectangles in Figure 5-11g. Imagine thesespaces moved over to the left so that theyalign at x = 1 (graph). The spaces can becircumscribed with a rectangle of base ||P||and altitude (1.29 − 1.21). Thus, Un − Ln ≤||P|| (1.29 − 1.21), Q.E.D.

1 9

x

f(x)

Norm = largest ∆ x

Slide them over.

Norm

1.21

1.29

b. From part a, 0 ≤ Un − Ln ≤ ||P|| (1.29 − 1.21).As ||P|| approaches zero, the rightmostmember of the inequality goes to zero. By the

squeeze theorem, lim ( ) ,P

n nU L→

− =0

0 which

implies lim lim .P

nP

nU L→ →

=0 0

So f is integrable

on [1, 9] by the definition of integrability,Q.E.D.

c. Prove that g(x) = 1/x is integrable on [1, 4].

Proof:

Partition the interval [1, 4] into nsubintervals whose widths are not necessarilyequal. Let ||P|| be the norm of the partition.Pick sample points ck at the left end of eachsubinterval. Because g (x) is decreasing on[1, 4], the high points are located at the leftends of the subintervals and the low pointsare at the right ends (graph).

1 4

x

y1

By algebraic or graphical reasoning as inpart a, Un − Ln ≤ ||P||(1 − 1/4). As ||P||approaches zero, Un − Ln is squeezed to zero.Thus, Un and Ln approach the same limit,which implies that g is integrable on [1, 4],Q.E.D.

d. This reasoning cannot be applied directly toh (x) = sin x on the interval [0, 3] becauseh (x) is both increasing and decreasing ondifferent parts of the interval. Thus, the highpoints are not always at the same end of thesubinterval and the high point at π/2 may notbe at either end of a subinterval (graph).

x

y = sin x

0 3

1

π/2

The reasoning could be applied indirectly byfirst splitting the interval [0, 3] into [0, π /2]and [π /2, 3] so that h(x) is increasing on oneand decreasing on the other.

Chapter Test

T1. Indefinite integral:

g x f x dx( ) ( ) = ∫ if and only if g′(x) = f (x).


T2. Definite integral:Let Ln and Un be lower and upper sums of f (x)on the interval [a, b]. Then

f x dx L Un

na

b

nn( ) lim lim ,= =

→∞ →∞∫provided the two limits are equal.

T3. Fundamental theorem:If f is an integrable function, and

g x f x dx( ) ( ) , = ∫ then

f x dx g b g aa

b

( ) ( ) ( ). = −∫T4. The function f is continuous on the interval

[3, 8] and differentiable on (3, 8). It does notmatter that it is not differentiable at theendpoints.

T5.

3

x

f (x)

8c

T6. Hypotheses: f (a) = f (b) = 0Differentiable on (a, b).Continuous at x = a and x = b.Conclusion: There is a c in (a, b) such thatf ′ (c) = 0.

a

c

b

x

f(x)

T7.

3

x

f (x)

8

T8. y = esin x ⇒ dy = cos x esin x dx

T9. 0 10 1

0 1.

.

ln .x

x

dx C= +∫T10. ( ) ( ) ( )4 13

1

724 133 5 2 3 6x x dx x C+ = + +∫

T11. x dxx2

3

1

4

1

4 3 3

3

4

3

1

321= = − =∫

T12. ( )12 10 310

33 2 4

2

23

2

2

x x dx x x+ = +− −

∫= − − − = −48

80

348

80

353

1

3

T13. π [( ) ( ) ]x x dyc

d

12

22−∫

T14. The slope of the linear function is the same asthe slope of the curve at x = 1. So the slope isfound by y = x3 ⇒ y′ = 3x2 ⇒ y′(1) = 3.At x = 1, y = 1.y − 1 = 3(x − 1) ⇒ y = 3x − 2

1

1

x

y

(1, 1)

1

1

1.2

1.2

As you zoom in on (1, 1), you see that the graphof y = x3 is locally linear.

T15. a. 12 480 25 0 25

0

3

0

3

e dx ex x. .=∫= − =48 48 53 61600 75 0e e. . K

b. M50 = 53.6154…T50 = 53.6170…S50 = 53.61600081…The midpoint Riemann sum error is0.000502646712… .The trapezoidal-rule error is 0.0010052962… .The midpoint Riemann sum error is half ofthe trapezoidal-rule error, because2(0.000502646712…) = 0.0010052962… .The Simpson’s rule error is0.000000015077… , which is much smallerthan the error for the other two methods.

T16. a.

y = cos x

x

(x, y)

y

π

1

b. dV y dx dV x dx= ⇒ =1

2

1

22 2cos

V x dx= ∫ 1

22

0

2

cos/π


c. This integral cannot be evaluated algebraicallybecause we do not know an antiderivative forcos2 x.

V x dx= =∫ 1

20 39262

0

2

cos ./

Kπ

T17. a. g x x dx x C x( ) . . .= = + =∫ 0 3 0 1 0 12 3 3

b.

1 2.64... 4

x

f(x)

5

1 2.64... 4

x

g(x)

5

c. mg g= = =( ) – ( )

–

. – .4 1

4 1

6 4 0 1

32 1.

g′(x) = f ( x) = 0.3x2

∴ 0.3c2 = 2.1 ⇒ c = 7 = 2.6457513…

In the right graph in part b, the tangent line atx = 2.64… is parallel to the secant line fromx = 1 to x = 4.

d. f ( 2.645…) = 0.3(2.645…)2 = 2.1 (exactly)The point is (2.645… , 2.1).Area of region under graph equals area ofrectangle, as shown in the graph on the left inpart b.



Chapter 6—The Calculus of Exponentialand Logarithmic Functions

Problem Set 6-11. The integral would be 1

00P , which involves

division by zero.

N Integral

1000 01500 0.4054…2000 0.6931…2500 0.9162…500 −0.6931…100 −2.3205…

1000 2000

1

2

–1

–2

Integral

N

The graph resembles a logarithmic function.

2. 0 05 0 05 0 5 0 0 50

10

0

10

.∫ = = − =dt t. . . , Q.E.D.

0.5 is between 0.4054… and 0.6931… , thevalues of the left integral for N = 1500 andN = 2000.

By solver, 1

1000 P

N

∫ dP = 0.5 when

N = 1648.7212… , or about 1649 people.

3. At 20 years, the integral on the right equals 1. At0 years, the integral equals 0. Solving for N atthese times gives N = 2718.2818… for 20 years,and N = 1000 (as expected!) for 0 years.

5 10 15 20

1000

2000

3000Population

Time (yr)

The graph resembles an exponential function.

4. ln 1648.7212… − ln 1000 = 0.5, exactly. Thisis the value of the integral on the left!

Problem Set 6-2

Q1.1

0 70 7

.x C. + Q2. 9

Q3. f ′ ( x) = −2 cos x sin x Q4. continuous

Q5. differentiable Q6. yx

′ = 1

1 2–

Q7. y′ = − csc x cot x Q8. Riemann sum

Q9. indefinite integral, or antiderivative

Q10. log 12


1. y = ln 7x ⇒ y′ = 1/(7x) · 7 = 1/x2. y = ln 4x ⇒ y′ = 1/(4x) · 4 = 1/x3. f (x) = ln x5 ⇒ f ′ (x) = 1/(x5) · 5x4 = 5/x4. f (x) = ln x3 ⇒ f ′ (x) = 1/(x3) · 3x2 = 3/x5. h (x) = 6 ln x− 2 ⇒ h′ (x) = 6/(x− 2) · (−2x− 3) = −12/x6. g (x) = 13 ln x− 5 ⇒

g′ (x) = 13/(x− 5) · (−5x− 6) = −65/x7. r (t) = ln 3t + ln 4t + ln 5t ⇒

r ′ (t) = 1/(3t) · 3 + 1/(4t) · 4 + 1/(5t) · 5 = 3/t8. v (z) = ln 6z + ln 7z + ln 8z ⇒

v′ (z) = 1/(6z) · 6 + 1/(7z) · 7 + 1/(8z) · 8 = 3/z9. y = (ln 6x)(ln 4x) ⇒

y′ = 1/(6x) · 6 · (ln 4x) + (ln 6x)[1/(4x) · 4]

= (1/x)(ln 4x + ln 6x) or ln 24 2x

x

10. z = (ln 2x)(ln 9x) ⇒z′ = 1/(2x) · 2 · (ln 9x) + (ln 2x)[1/(9x) · 9]

= (1/x)(ln 9x + ln 2x) or ln 18 2x

x

11. yx

x = ln

ln

11

3 ⇒

yx x x x

x′ = ⋅ ⋅ ⋅1 11 11 3 11 1 3 3

3 2

/( ) (ln ) – (ln ) /( )

(ln )

= ln – ln

(ln )

3 11

3 2

x x

x x or

ln( / )

(ln )

3 11

3 2x x

12. yx

x= ⇒ln

ln

9

6

yx x x x

x′ = ⋅ ⋅ ⋅ ⋅1 9 9 6 9 1 6 6

6 2

/( ) (ln ) – (ln ) /( )

(ln )

= ln – ln

(ln )

6 9

6 2

x x

x x or

ln( / )

(ln )

2 3

6 2x x

13. p = (sin x)(ln x) ⇒ p′ = (cos x)(ln x) + (sin x)(1/x)

14. m = (cos x)(ln x) ⇒m′ = (−sin x)(ln x) + (cos x)(1/x)

15. y = cos (ln x) ⇒ y′ = −sin (ln x) · (1/x)

16. y = sin (ln x) ⇒ y′ = cos (ln x) · (1/x)


17. y = ln (cos x), where cos x > 0 ⇒ y′ = (1/cos x) · (−sin x) = −tan x (Surprise!)

18. y = ln (sin x), where sin x > 0 ⇒ y′ = (1/sin x) · (cos x) = cot x (Surprise!)

19. T (x) = tan (ln x) ⇒ T ′ (x) = sec2 (ln x) · (1/x)

20. S (x) = sec (ln x) ⇒ S ′ (x) = sec (ln x) · tan (ln x) · (1/x)

21. y = (3x + 5)−1 ⇒ y′ = −(3x + 5)−2 · 3 = −3(3x + 5)−2

22. y = (x3 − 2)−1 ⇒y′ = −(x3 − 2)−2 · 3x2 = −3x2(x3 − 2)

23. y = x4 ln 3x ⇒ y′ = 4x3 ln 3x + x4 · 1/(3x) · 3 = 4x3 ln 3x + x3

24. y = x7 ln 5x ⇒ y′ = 7x6 ln 5x + x7 · 1/(5x) · 5 = 7x6 ln 5x + x6

25. y = ln (1/x) ⇒ y′ = 1/(1/x) · (−x−2) = −1/x26. y = ln (1/x4) ⇒ y′ = 1/(1/x)4 · (−4x−5) = −4/x

27. 7 7/ | | x dx x C= +∫ ln

28. 5 5/ x dx x C= +∫ ln | |

29. 1 31

3/( ) x dx x C= +∫ ln | |

30. 1 81

8/( ) | | x dx x C= +∫ ln

31.x

xdx

xx dx

2

3 32

5

1

3

1

53

+=

+∫∫ ( )

= + +1

353ln || x C

32.x

xdx

xx dx

5

6 65

4

1

6

1

46

– –( )= ∫∫

= +1

646ln –| |x C

33.x

xdx

xx dx

5

6 65

9

1

6

1

96

– –(– )= −∫ ∫

= − +1

69 6ln –| |x C

34.x

xdx

xx dx

3

4 43

10

1

4

1

104

– –(– )= −∫ ∫

= − +1

410 4ln –| |x C

35.sec tan

secln sec

x x dx

xx C

11

+= + +∫ | |

36.sec

tanln | tan |

2

11

x dx

xx C

+= + +∫

37.cos

sinln | sin |

x dx

xx C= +∫

38.sin

cos

– sin

cosln | cos |

x dx

x

x dx

xx C= − = − +∫∫

39. ( / ) ln ln ln. .

1 4 0 50 5

4

0 5

4

w dw w= = −∫ | | .

= ln 8 = 2.079441…

40. ( / ) ln ln ln. .

1 10 0 10 1

10

0 1

10

v dv v= = −∫ | | .

= ln 100 = 4.605170…

41. ( / ) ln ln ln. .

1 3 0 10 1

3

0 1

3

x dx x | | | | | . |= = − − −−

−

−

−

∫= ln 3 − ln 0.1 = ln 30 = 3.401197…

42. ( / ) ln ln ln. .

1 4 0 20 2

4

0 2

4

x dx x= = − − −−

−

−

−

∫ | | | | | . |

= ln 4 − ln 0.2 = ln 20 = 2.995732…

43.x dx

x xx dx

1 2

3 24

9

3 24

91 2

1

2

3

1

1

3

2

/

/ /+=

+⋅∫ ∫ /

= + = =2

31

2

328 9 0 7566533 2

4

9

ln | | (ln – ln )/x . K

44.x dx

x xx dx

−−

+=

+⋅∫ ∫

1 3

2 31

8

2 31

81 3

2

3

2

1

2

2

3

/

/ //

= + = =3

22

3

26 3 1 5 22 3

1

8

ln | | (ln – ln ) ln/x .

= 1.039720…

45. ( )5ln (ln )xdx

xx C= +∫ 1

66

46.ln

ln (ln )x

xdx x

dx

xx C= = +∫∫ ( ) 1 21

2

47. f x t dt f x xx

( ) ( )= ⇒ ′ =∫ cos cos3 32

48. f x t t dtx

( ) = + ⇒∫ ( – )2

510 17

′ = + −f x x x( ) 2 10 17

49.d

dxt dt x

x

tan tan3

2

3∫

=

50.d

dxdtt

xx2

1−∫

= 2

51. f x dt f x xt xx

( ) ( )= ⇒ ′ = ⋅∫ 3 2 32

2

1

52. g x t dtx

( ) = ⇒∫0

cos

g x x x′ = ⋅ −( ) ( )cos sin

53. h x t dtx

( ) = + ⇒−

∫ 1 2

0

3 5

h x x′ = +( ) 3 1 3 5 2( – )

54. p x t dt p x x xx

( ) ( ) ( )= + ⇒ ′ = + ⋅∫ ( )–

4 7 12 7 2

11 1 3

3

55. ( / ) ln ln ln5 5 5 3 5 11

3

1

3

x dx x= = −∫ | | = 5 ln 3

= 5.493061…


Midpoint Riemann sum: M100 = 5.492987…Trapezoidal rule: T100 = 5.493209…Numerical integration: 5.493061…


57. a. By finding areas, g (0) ≈ −2.7, g (1) = 0,g(2) = 1, g(3) ≈ 0.3, g(4) ≈ −0.3, g(5) ≈ 0.7,g (6) ≈ 3.3, g (7) = 6, and g (8) = 7.

1 2 3 5 6 7 8

1

2

3

4

5

6

7

–1

–2

–3

g

x

y

b. h x f t dt h x f xx

( ) ( )= ⇒ ′ = − ⋅−

∫ ( ) ( )1

12

2

1 2x ⇒

h′(2) = f (3) ⋅ 4 = −1 ⋅ 4 = −4

58. a. By finding areas, g (0) = −6, g (1) = −2.5,

g (2) = 0, g (3) = 1.5, g (4) = 2, g (5) = 1.5,

g (6) = 0.75, g (7) = 0.5, g (8) = 0.75,

g (9) = 1.5, and g (10) = 2.75.

1 2 3 4 5 7 8 9 10

1

2

3

4

–1

2–

3–

4–

5–

6–

g

x

y


( ) ( ) ( ) ( )= ⇒ ′ = ⋅ ⇒∫ 2 22

2

h′(4) = f (8) ⋅ 2 = (0.5)(2) = 1

59. ( / ) ln ln ln1 10001000 1000

P dP P NN N

= = −∫ | |

0 05 0 05 0 50

10

0

10

. . dt t= =∫ .

ln N − ln 1000 = 0.5

lnN

10000 5= .

Ne

10000 5= .

N = 1000e0.5 ≈ 1648.721…≈ 1649 people

60. a. F + 30 = k/h0 + 30 = k/20 ⇒ k = 600∴ F + 30 = 600/h ⇒ F = 600/h − 30

b.

h

F

30

10 20

dh

(h, F)

c. Work equals force times displacement.Because the force varies, a definite integralmust be used.

d. The work done compressing the air a smallamount, dh, is approximately equal to theforce at the sample point (h, F ) times dh(see part b).

dW = F dh = (600/h − 30) dh

∴ = ∫W h dh( / – )600 3020

10

= −600 30ln | | 20

10h h

= 600 ln 10 − 300 − 600 ln 20 + 600= −115.8883… ≈ −116 inch-pounds

This number is negative because each value ofdh is negative and F is positive, making theirproduct negative.

e. Distance is measured in inches, force ismeasured in pounds, and we are finding theirproduct.

61. a. d (f ) = a + b ln f0 = a + b ln 53, 10 = a + b ln 16010 = b ln 160 − b ln 53 ⇒

b = =10

160 539 050741

ln – ln. ...

a = −9.050741… ln 53 = −35.934084…d ( f ) = –35.934084… + 9.050741… ln f

b. f d cm d ′ (part c)

53 0 0.1707…

60 1.1227… 0.1508…

70 2.5197… 0.1292…

80 3.7265… 0.1131…

100 5.7461… 0.0905…

120 7.3962… 0.0754…

140 8.7914… 0.0646…

160 10.0 0.0565…

The measured distances will vary. Theyshould be close to the calculated distances.

c. d ′ ( f ) = b/f = 9.050…/f. See table in part b.d. d ′ ( f ) is in cm/10 kHz.e. d ′ ( f ) decreases as f gets larger; this is

consistent with the spaces between thenumbers getting smaller as f increases.


62. a. ln 2 = 0.693147…

ln 3 = 1.098612…

ln 6 = 1.791759…

ln 2 + ln 3 = ln 6

Conjecture: ln (ab) = ln a + ln b

b. ln (10/2) = ln 5 = 1.609437…

ln 10 = 2.302585…

ln 2 = 0.693147…

ln (10/2) = ln 10 − ln 2

Conjecture: ln (a/b) = ln a − ln b

c. ln (210) = ln 1024 = 6.931471…

ln 2 = 0.6931471…

ln (210) = 10 ln 2

Conjecture: ln (a b ) = b ln a

d.ln

log

2

2= …2.30258

ln

log

3

32 30258= ….

They seem to be the same.ln 10 = 2.30258…

12 30258

loge= ….

log 4 = 0.60205… and

ln

ln

.

.

4

10

1 3862

2 302580 60205= =K

KK.


Problem Set 6-3

Q1. y′ = 1/(1 + x2) Q2.1

44 1ln | |x C+ +

Q3. 1 Q4. 1/4

Q5. 35 Q6. 8

Q7.

1x

y

2

Q8. f is differentiable on (a, b).

Q9. f is continuous at x = a and x = b.

Q10. B

1. ln 6 + ln 4 = 1.79175… + 1.38629… =3.17805…ln 24 = 3.17805…

2. ln 5 + ln 7 = 1.60943… + 1.94591… =3.55534…ln 35 = 3.55534…

3. ln 2001 − ln 667 = 7.60140… − 6.50279… =1.09861…ln (2001/667) = ln 3 = 1.09861…

4. ln 1001 − ln 77 = 6.90875… − 4.34380… =2.56494…ln (1001/77) = ln 13 = 2.56494…

5. 3 ln 1776 = 3(7.48211…) = 22.44635…ln (17763) = ln 5,601,816,576 = 22.44635…

6. 4 ln 1066 = 4(6.97166…) = 27.88667…ln (10664) = ln 1,291,304,958,736 = 27.88667…

7. See the text for the proof of the uniquenesstheorem.

8. See the text for the proof.

9. Prove that ln (a/b) = ln a − ln b forall a > 0, b > 0.

Proof:

Let f (x) = ln (x/b), g(x) = ln x − ln b for x, b > 0Then f ′ (x) = (b/x)(1/b) = 1/x, andg′ (x) = (1/x) − 0 = 1/x.∴ f ′ (x) = g′ (x) for all x > 0.f (b) = ln (b/b) = ln 1 = 0g(b) = ln b − ln b = 0∴ f (b) = g (b).∴ f (x) = g (x) for all x > 0 by the uniquenesstheorem.∴ ln (x/b) = ln x − ln b for all x > 0.∴ ln (a/b) = ln a − ln b for all a > 0 and b > 0,Q.E.D.

10. Prove that ln (a b) = b ln a for all a > 0 and all b.

Proof:

Let f (x) = ln (xb); g (x) = b ln x for x > 0.Then f ′ (x) = 1/(xb) · bxb− 1 = b/x andg′ (x) = b (1/x) = b/x.∴ f ′ (x) = g′ (x) for all x > 0.f (1) = ln (1b) = ln 1 = 0g(1) = b ln 1 = b · 0 = 0∴ f (1) = g (1).∴ f (x) = g (x) for all x > 0 by the uniquenesstheorem.∴ ln (xb) = b ln x for all x > 0.∴ ln (ab) = b ln a for all a > 0 and all b, Q.E.D.

11. Prove that ln (a/b) = ln a − ln b forall a > 0, b > 0.

Proof:

ln (a/b) = ln (a · b− 1) = ln a + ln b− 1 =ln a + (−1) ln b = ln a − ln b∴ ln (a/b) = ln a − ln b, Q.E.D.

12. Example: ln (2 + 3) = ln 5 = 1.60943…ln 2 + ln 3 = 0.69314… + 1.09861… =1.79175…∴ ln (2 + 3) ≠ ln 2 + ln 3.∴ ln (a + b) = ln a + ln b is false, Q.E.D.


13. See the text definition of ln x.

14. logln

lnlog

logb

a

a

xx

bx

b= =

15. f (x) = log3 x ⇒ f ′ (x) = 1/(x ln 3)f ′ (5) = 0.182047…The graph shows a tangent line with a smallpositive slope.

5

1x

f(x)

16. f (x) = log0.8 x ⇒ f ′ (x) = 1/(x ln 0.8)f ′ (4) = −1.120355…The graph shows a tangent line with slope ≈ −1.

x

f(x)

4

10

5

17. g (x) = 8 ln (x5) = 40 ln x ⇒ g′ (x) = 40/x

18. h (x) = 10 ln (x0.4 ) = 4 ln x ⇒ h′ (x) = 4/x

19. T(x) = log5 (sin x) ⇒ ′ =⋅

⋅ T xx

x( )sec ln

cos1

5T′ (x) = (cot x)/(ln 5)

20. R (x) = log4 (sec x) ⇒

R xx

x xx′ =

⋅⋅ =( )

1

4 4sec lnsec tan

tan

ln

21. p (x) = (ln x)(log5 x) ⇒p′ (x) = (1/x) · (log5 x) + (ln x) · [1/(x ln 5)]

= ⋅ + =1

5 5

2

5

ln

ln

ln

ln

ln

ln

x

x

x

x

x

x

22. q (x) = (log9 x)/(log3 x) = ⋅ = =ln

ln

ln

ln

ln

ln

x

x9

3 3

2 3

1

2∴ q ′ (x) = 0 because q (x) is constant.

23. f xx

xx x( ) =

= −lnsin

ln ln sin3

3

= 3 ln x − ln sin x ⇒ f ′ (x) = 3/x − cot x

24. f (x) = ln (x4 tan x) = ln x4 + ln (tan x)= 4 ln x + ln (tan x) ⇒

f xx

x

x′ = +( )

4 2sec

tan= +4 1

x x xsin cos

25.d

dxx

d

dxx xx( ) ( )ln ln3 3= =

33

3 3ln lnxx

xx+ = +

26.d

dx

d

dxxx(ln ) (sec ln )sec5 5=

= ln 5 sec x tan x

27. a. y = 7 · (2 − 0.9x)dy/dx = 7(−0.9x)(ln 0.9)dy/dx = 0.737523…(0.9x)x = 0: dy/dx = 0.737… mi/hx = 1: dy/dx = 0.663… mi/hx = 5: dy/dx = 0.435… mi/hx = 10: dy/dx = 0.257… mi/hThe lava is slowing down.

b. y/7 = 2 − 0.9x

0.9x = 2 − y/7x ln 0.9 = ln (2 − y/7)x = (1/ln 0.9)[ln (2 − y/7)]

c.dx

dy y= ⋅ ⋅ −

( / . )1 0 9

1

2 7

1

7ln

– /dx

dy y= 9 491221

14

.

–

K

y = 10: dx/dy = 2.372… h/mi

d. If x = 10, then y = 7(2 − 0.910), so

dx

dy= = …9 491221

14 7 2 0 93 88865110

.

– ( – . )

K. .

e. 3.88… is the reciprocal of 0.257… , thevalue of dy/dx when x = 10, not when y = 10.

28. a. 1000(1.06t) = M ⇒ 1.06t = M/1000 ⇒log1.06 1.06t = log1.06 (M/1000) ⇒t = log1.06 (M/1000)

b. dt dMM

/ = ⋅1

1 06 1000

1

1000(ln . )( / )

= 1

1 06M ln .

c. If , /M dt dM= = =10001

1000 1 06ln .0.01716… yr/$. At this rate, with $1000 inthe account, it would take 0.017… year, orabout 6 days, to earn a dollar of interest.

d. dt/dM gets smaller as M increases; moreinterest is earned when M is larger, so it takesless time to accumulate $1000.

29. The intersection point is at x = 2.7182818… ,which is approximately e.


Problem Set 6-4Q1. y′ = 3/x Q2. (−1/10)(5x)− 2 + C

Q3. 15 5ln x C+ Q4. y x′ = −1 1 2/ –


Q5. 5 tan (5x) Q6.ln

ln

23

17

Q7. 36 Q8. 8

Q9. B Q10. E


1. a. R(t) = aekt

60,000 = aek·0 ⇒ a = 60,0002,400,000 = aek·2 = 60,000e2k

40 = e2k ⇒ 2k = ln 40 ⇒k = (ln 40)/2 = 1.844…(Store 1.844… without round-off as k.)∴ R(t) = 60,000e1.844…t

b. R ( 5 ) = 60,000e1.844…(5) = 607,157,310.7…About 607 million rabbits.

c. 2 = 60,000e1.844…t ⇒ 1/30,000 = e1.844…t ⇒−ln 30,000 = 1.844…t ⇒ t = −5.589…So the first pair of rabbits was introducedabout 5.6 years earlier, or in about 1859.

2. a. v (t) = 20,000e− 0.1 t

v(0) = 20,000e0 = 20,000$20,000 when built.

b. v (10) = 20,000e− 1 = 7357.588…v(11) = 20,000e− 1.1 = 6657.421…At 10 years, value is $7357.59.At 11 years, value is $6657.42.So depreciation is 7357.59 − 6657.42 =$700.17.

c. v′ (t) = −2000e− 0.1 t

v′ (10) = −2000e− 1 = −735.758… ,or about $736 per year.This rate is higher than the actual depreciationin part b because the latter rate is an averagefor the year. The rate at the end will be lowerthan 736 to give the average of 700.

d. 5,000 = 20,000e− 0.1 t

0.25 = e− 0.1 t

ln (0.25) = −0.1tt = (ln 0.25)/(−0.1) = 13.8629… ≈14 yr.

3. a. m(t) = 1000(1.06)t

ln m(t) = ln 1000 + t ln 1.061/m(t) · m′ (t) = 0 + ln 1.06m ′ (t) = m (t) · ln 1.06m ′ (t) = 1000(1.06)t (ln 1.06)m ′ (0) = 58.27 $/yrm ′ (5) = 77.98 $/yrm ′ (10) = 104.35 $/yr

b. m(0) = $1000.00m(5) = $1338.23m(10) = $1790.85The rates are increasing. $338.23 is earnedbetween 0 and 5 years; $452.62 is earned

between 5 and 10 years, which agrees withthe increasing derivatives shown in part a.

c.m t

m t

t

t

′ = =( )

( )

( . ) (ln . )

( . )ln

1000 1 06 1 06

1000 1 061 06.

∴ m′ (t)/m(t) = ln 1.06, a constantd. m(1) = 1060.00. So you earn $60.00.

The rate starts out at only $58.27/year buthas increased enough by year’s end to makethe total for the year equal to $60.00.

4. d (t) = 200t · 2− t ⇒ ln d (t) = ln 200t − t ln 21

1 2d t

d t t( )

ln⋅ ′ = − ⇒( ) /

′ = ⋅ / −−d t t tt( ) ( )(1 ln 2)200 2

d ′ (1) = (200 · 2− 1)(1 − ln 2) = 30.685…d ′ (2) = (400 · 2− 2)(1/2 − ln 2) = −19.314…So the door is opening at about 30.7°/s at1 second and closing at about 19.3°/s at2 seconds, which agrees with the graph.

1 2

100

t

d(t)

The widest opening occurs when d ′ (t) = 0.Solving numerically for t in(200t · 2− t)(1/t − ln 2) = 0,t = 1.44269… .d (1.44269…) = 106.147…So the widest is about 106° at t ≈ 1.4 s.

5. enn

n

= +

→∞

lim 11

and e nn

n= +→

lim( ) /

0

11

When you substitute ′ for n in the first equation,you get the indeterminate form 1∞ . When yousubstitute 0 for n in the second equation, youalso get the indeterminate form 1∞ .

n (1 + 1/n)n

100 2.70481…

1000 2.71692…

10000 2.71814…

n (1 + n)1/ n

0.01 2.70481…

0.001 2.71692…

0.00001 2.71826…

6. y = 17e− 5x ⇒ y′ = −85e− 5x

7. y = 667e− 3x ⇒ y′ = −2001e− 3x

8. h(x) = x3ex ⇒ h′ (x) = 3x2ex + x3ex = x2ex (3 + x)

9. g (x) = x− 6ex ⇒ g′ (x) = −6x− 7ex + x− 6ex = x− 7ex (−6 + x)


10. r (t) = et sin t ⇒ r ′ (t) = et sin t + et cos t

11. s (t) = et tan t ⇒ s′ (t) = et tan t + et sec2 t

12. u = 3exe−x = 3 ⇒ u′ = 0

13. v = e4xe− 4x = 1 ⇒ v′ = 0

14. ye

xy

e x e x

x

x x x

= ⇒ ′ = −ln

ln ( / )

(ln )

12

15. yx

ey

x e x e

ex

x x

x= ⇒ ′ = − ⋅ln ( / ) ln12

16. y = 4esec

x ⇒ y′ = 4esec

x sec x tan x

17. y = 7ecos

x ⇒ y′ = −7ecos

x sin x

18. y = 3 ln e2x = 6x ln e = 6x ⇒ y′ = 6

19. y = 4 ln e5x = 4 · 5x = 20x ⇒ y′ = 20

20. y = (ln e3x)(ln e4x) = 3x · 4x = 12x2 ⇒ y′ = 24x

21. y = (ln e− 2x)(ln e5x) = −2x · 5x = −10x2 ⇒y′ = −20x

22. g(x) = 4eln 3x = 4 · 3x = 12x ⇒ g′ (x) = 12

23. h (x) = 6eln 7x = 6 · 7x = 42x ⇒ h′ (x) = 42

24. y = ex + e− x ⇒ y′ = ex − e− x

25. y = ex − e− x ⇒ y′ = ex + e− x

26. y e y e x x ex x x= ⇒ ′ = ⋅ =5 5 2 2 53 3 3

15 15

27. y e y e x x ex x x= ⇒ ′ = ⋅ =8 8 5 405 5 54 4

28. f (x) = 0.42x ⇒ ln f (x) = 2x ln 0.4 ⇒1

2 0 4 0 4 2 0 42

f xf x f x x

( )ln ln′ = ⇒ ′ = ⋅( ) . ( ) . .

29. f (x) = 10− 0.2 x ⇒ ln f (x) = −0.2x ln 10 ⇒1

0 2 10f x

f x( )

ln′ = − ⇒( ) .

′ = −−f x x( ) ( ..10 0 2 100 2 ln )

30. g (x) = 4(7x) ⇒ ln g (x) = ln 4 + x ln 7 ⇒1

7 4 7 7g x

g x g x x

( )ln ln′ = ⇒ ′ =( ) ( ) ( )

31. h (x) = 1000(1.03x) ⇒ ln h (x) = ln 1000 +

x ln( )

ln1 031

1 03. ( ) .⇒ ′ = ⇒h x

h x

′h x( ) = 1000(1.03x) ln 1.03

32. c (x) = x5 · 3x ⇒ ln c (x) = 5 ln x + x ln 3 ⇒1

5 3c x

c x x( )

ln′ = + ⇒( ) /

′c x( ) = x5 · 3x(5/x + ln 3)

33. m(x) = 5x · x7 ⇒ ln m(x) = x ln 5 + 7 ln x ⇒1

5 7m x

m x x( )

ln′ = + ⇒( ) /

m ′ (x) = 5x · x7(ln 5 + 7/x)

34. y = (ln x)0.7 x ⇒ ln y = 0.7x ln (ln x) ⇒1

0 7 0 71

yy x x

x x′ = + ⋅ ⇒. ln (ln ) .

ln

′ = +

⋅y x

xx x0 7

0 7 0 7. ( )ln ln.

ln(ln ) .

35. y = xln x ⇒ ln y = ln x · ln x ⇒ ln y = (ln x)2 ⇒1

21

yy x

x′ = ⋅ ⇒ln ′ =

⋅y

x

xx x2 ln ln = 2 ln x · xln x−1

36. y = (csc 5x)2x ⇒ ln y = 2x ln (csc 5x) ⇒ 1

yy′ =

2 5 21

5ln csc

csc( )

x x

x+ (– csc cot )5 5 5x x ⇒

y′ = (csc 5x)2x [2 ln (csc 5x) − 10x cot 5x]37. y = (cos 2x)3x ⇒ ln y = 3x ln (cos 2x) ⇒

13 2 3

1

22 2

yy x x

xx′ = +ln cos

cos(– sin )( ) ⇒

y′ = (cos 2x)3x [3 ln (cos 2x) − 6x tan 2x]38. Two solution methods are possible.

Differentiate directly:

yx

x= + ⇒ln

–

5 2

7 8

′ =+

+ ⋅

yx

x

x x

x

7 8

5 2

5 7 8 5 2 7

7 8 2

– ( – ) – ( )

( – )

=+

–

( )( – )

54

5 2 7 8x xOr simplify using properties of logarithms first:

y = ln (5x + 2) − ln (7x − 8) ⇒

′ =+

− =+

yx x x x

5

5 2

7

7 8

54

5 2 7 8–

–

( )( – )39. Two solution methods are possible.

Differentiate directly:y = ln [(4x − 7)(x + 10)]

′ =+

⋅ + + − ⋅yx x

x x1

4 7 104 10 4 7 1

( – )( )[ ( ) ( ) ]

= ++

8 33

4 7 10

x

x x( – )( )Or simplify using properties of logarithms first:y = ln (4x − 7) + ln (x + 10)

′ =−

++

= ++

yx x

x

x x

4

4 7

1

10

8 33

4 7 10( – )( )

40. y = (2x + 5)3 4 1x − ⇒

ln y = 3 ln (2x + 5) +

1

24 1ln ( )x − ⇒

1 6

2 5

2

4 1yy

x x′ =

++

−⇒

′ =+

+−

+ −y

x xx x

6

2 5

2

4 12 5 4 13[( ) ]

= + +−

( )( )28 4 2 5

4 1

2x x

x

41. yx

xy x= +

−⇒ = + −( )

( )ln ln ( )

10 3

4 510 10 3

10

3

3 4 51 30

10 3

15

4 5ln ( )− ⇒ ′ =

++

−⇒x

yy

x x

′ =+

+−

+−

yx x

x

x

30

10 3

15

4 5

10 3

4 5

10

3

( )

( )

= − +−

( )( )

( )

270 105 10 3

4 5

9

4

x x

x


42.d

dxdtt

xx10 10

3∫

=

43.d

dxt dt x

x

ln ln3∫

=

44.d

dxt dt x

x

log log ( )25

4

24 4∫

=

45.d

dxt dt x x

x

ln (cos ) (ln cos ).6 3

22

2∫

=

46.d

dxx

d

dxx

d

dx x x

2

25

2

2 255 5

(ln ) ( ln )= =

= −

47.d

dxe

d

dxe ex x x

2

27 7 77 49( ) ( )= =

48. e dx e Cx x5 51

5= +∫

49. e dx e Cx x7 71

7= +∫

50. 77

2 72

2x

x

dx C= +∫ ln

51. 1 051 05

1 05.

.

ln .x

x

dx C= +∫52. 6 6e dx e Cx x= +∫53. e dx e Cx x0 2 0 25. .= +∫54. e x dx e Cx xsin sincos = +∫55. e x dx e Cx xtan tansec2 = +∫56. e dx x dx x Cx3 3 41

4ln∫ ∫= = +

57. 60 60 5 1505 2e dx x dx x Cxln∫ ∫= = +

58. ( ) ( )11

10212 50 2 2 51+ = + +∫ e e dx e Cx x x

59. ( ) ( )11

40414 100 4 4 101− = − − +∫ e e dx e Cx x x

60. ( )e e dx e ex x x x− = +− −∫0

2

0

2

= + − − =−e e2 2 1 1 5 524391. ...

Numerically: integral ≈ 5.524391... (Checks.)

61. ( ) ( )– –e e dx e ex x x x+ = −− −∫ 1

2

1

2

= e2 − e−2 − e−1 + e1 = 9.604123…Numerically: integral ≈ 9.604123... (Checks.)

62. Step 2: Definition of derivative.

Step 3: Logarithm of a quotient, applied inreverse.

Step 4: Write division as multiplication by thereciprocal; distribute division over addition.

Step 6: 1/x does not depend on h, so it is a“constant” with respect to h.

Step 7: Logarithm of a power, applied in reverse.

Step 9: The expression in parentheses has the form(1 + n)1/n, whose limit is e as n approaches zero.



Problem Set 6-5

Q1. e ≈ 2.71828 Q2. enn

n

= +

→∞

lim 11

or

e nn

n= +→

lim( ) /

0

11

Q3. 1 Q4. x

Q5. x Q6. e

Q7. (ln x)/(ln b) Q8. ex

Q9. −e−x + C Q10. E

1. limsin

x

x

x→→

0

2 5

3

0

0

= =→

limcos

x

x0

10 5

3

10

3

11

x

y

2. limtan

x

x

x→→

0

4 3

5

0

0

= =→

limsec

x

x0

212 3

5

12

5

11 x

y

3. limtan

x

x

x→→

0

0

0

= =→

limsec

x

x0

2

11

4. limsin

x

x

x→→

0

0

0

= =→

limcos

,x

x0 1

1 a “well-known” limit.


5. limcos

x

x

x→

− →0 2

1 0

0

= →→

limsin

x

x

x0 2

0

0

= =→

limcos

x

x0 2

1

2

6. limcosx

x

x→ −→

0

2

3 1

0

0

=−

→→

limsinx

x

x0

2

3 3

0

0

=−

= −→

limcosx x0

2

9 3

2

9

7. limsin

x

x

x→ +→

02

0

0

= = ∞→ +lim

cosx

x

x0 2

8. limcos

x

x

x x→

−+

→0 2

1 0

0

=+

=→

limsin

x

x

x0 1 20

9. limln

/x

x

x→ +→ −∞

∞0 1

= −= − =

→

−

− →+ +lim lim ( )x x

x

xx

0

1

20

0

10. lim .x

xe

x→= ∞

0

3

2 Form is 1

0

11. limlnx

xe e

x→

− →1 5

0

0

= =→ −lim

x

xe

x

e1 15 5

12. limln

x

x x

x x→

− +− +

→1 2

1

2 1

0

0

= −−

→→

−

limx

x

x1

1 1

2 2

0

0

= − = −→

−

limx

x1

2

2

1

2

13. lim

cos cos.

x

x

x→

+ = = −2

3 5 11

226 43297K

14. limtan

.x

x

x→ −= ∞

2 2

Form is tan 2

0

15. limx

xe

x→∞→ ∞

∞2

= → ∞∞→∞

limx

xe

x2

= = ∞→∞

limx

xe

2

16. limx x

x

e→∞→ ∞

∞

3

= → ∞∞→∞

limx x

x

e

3 2

= → ∞∞→∞

limx x

x

e

6

= =∞

→∞

lim .x xe

60

6 Form:

17. lim–

limx x

x

x→∞ →∞

+ = =3 17

4 11

3

4

3

4

18. lim limx x

x

x→∞ →∞+= = −2 – 7

3 5

–7

5

7

5

19. lim– –

– –x

x x x

x x x→∞

++

→ ∞∞

3 2

3 2

5 13 21

4 9 11 17

= ++

→ ∞∞→∞

lim–

–x

x x

x x

3 10 13

12 18 11

2

2

=+

→ ∞∞→∞

lim–

x

x

x

6 10

24 18

= =→∞

limx

6

24

1

4

20. limx

x

x→∞

+ → ∞∞

3 2

7 8

5

5 –

= = =→∞ →∞

lim limx x

x

x

15

35

15

35

3

7

4

4

21. L xx

x= →→ +lim

0

00

ln lim ( ln ) limln

–L x xx

xx x= = → −∞

∞→ →+ +0 01

=−

= − =→

−

− →+ +lim lim ( )x x

x

xx

0

1

20

0

∴ = =L e0 1

22. L xx

x= →→ +lim (sin )sin

0

00

ln lim sin (ln sin ) limlnsin

cscL x x

x

xx x= = → −∞

∞→ →+ +0 0

= ⋅−

=−→ →+ +

lim/(sin ) cos

csc cotlim

cscx x

x x

x x x0 0

1 1

= − =→ +lim ( sin )x

x0

0

∴ = =L e0 1

23. L xx

x= →→

∞−

lim (sin )/

tan

π 21

ln lim tan ln sin/

L x xx

=→ −π 2

( )

= →

= ⋅−

= −

→

→

→

−

−

−

limln sin

cot

lim( /sin ) cos

csc

limcos sin

sin

/

/

/

x

x

x

x

x

x x

x

x x

x

π

π

π

2

22

2

2

0

0

1


= − =→ −lim ( cos sin )

/xx x

π 20

∴ = =L e0 1

24. L xx

x= →→

−+

∞lim /( )

1

1 1 1

ln lim [ /( ) ln ]L x xx

= − ⋅→ +1

1 1

=−

→ = =→ →+ +lim

lnlim

/x x

x

x

x1 11

0

0

1

11

∴ = =L e e1

25. L ax ax

x= + → ∞ ≥→∞

lim( ) ( .)/1 01 0 Note:

ln lim [ / ln ( )] limln ( )

L x axax

xx x= ⋅ + = + → ∞

∞→∞ →∞1 1

1

= + ⋅ =+

=→∞ →∞

lim/( )

limx x

ax a a

ax

1 1

1 10

∴ = =L e0 1

26. L axx

x= + →→

∞lim ( ) /

0

11 1

ln lim [ / ln ( )] limln ( )

L x axax

xx x= ⋅ + = + →

→ →0 01 1

1 0

0

= + ⋅ =+

=→ →

lim/( )

limx x

ax a a

axa

0 0

1 1

1 1

∴ =L ea

27. L xx

x= →→ +lim /(ln )

0

3 00

ln lim ] limL x xx x

= ⋅ = =→ →+ +0 0

3 3[3/(ln ) ln

∴ = =L e3 20 08553. ...

28. L xx

x= →→ +lim ( ) /(ln )

0

5 07 0

ln lim [ /(ln ) ln( )]L x xx

= ⋅→ +0

5 7

= → −∞−∞

= ⋅ ⋅ =

→

→

+

+

limln ( )

ln

lim[ /( )]

/

x

x

x

x

x

x

0

0

5 7

5 1 7 7

15

∴ = =L e5 148 4131. ...

29. limx xx e→

−−

→ ∞ − ∞

0

1 1

1

= − →→

lim–

( – )x

x

x

e x

x e0

1

1

0

0

=+ ⋅

→→

lim–

( – )x

x

x x

e

e x e0

1

1 1

0

0

=+

→→

lim–

–x

x

x x

e

e xe0

1

1

0

0

=+ +

=→

limx

x

x x x

e

e e xe0

1

2

30. limsinx x x→

−

→ ∞ − ∞0

1 1

= − →→

limsin

sinx

x x

x x0

0

0

= −+

→→

limcos

cos sinx

x

x x x0

1 0

0

= −− +

=→

limsin

sin cosx

x

x x x0 20

31. f x x x( ) .= −sec tan2 2

2 2

π π Where secant and

tangent are defined, the Pythagorean propertiestell us that f (x) = 1.

–1 1 3 5

1

x

f(x)

32. Using l’Hospital’s rule leads to

limsec

tanlim

sec tan

sec/ /x x

x

x

x x

x→ →=

π π2 2 2

= =→ →lim

tan

seclim

sec

sec tan/ /x x

x

x

x

x xπ π2 2

2

=→lim

sec

tan,

/x

x

xπ 2 the original expression!

Using tan x = (sec x)/(csc x), the expressionreduces to

limsec

(sec )/(csc )lim csc

/ /x x

x

x xx

→ →= =

π π2 21

33. L xx

k x= →→ +lim /(ln )

0

00

ln L k x x k kx x

= ⋅ = =→ →+ +lim ln ln lim

0 0[ /( ) ]

∴ L = ek

The graph turns out to be a horizontal line,y = ek, defined for x > 0.

x

yy = e k

By the definition of a power,

x x e ek x k x k x x k/ / /= = =( ) ( ) ( )ln ln ln ln1 1

34. a. f xg x

h x

x

x x( ) = =

+( )

( )

. – .

. – .

0 3 2 7

0 2 2 4 2

2

2

g(3) = 0.3(9) − 2.7 = 0,h(3) = 0.2(9) − 2(3) + 4.2 = 0, Q.E.D.

b. g′ (x) = 0.6x ⇒ g′ (3) = 1.8h′ (x) = 0.4x − 2 ⇒ h′ (3) = −0.8


Tangent lines at (3, 0) have these equations.

For g: y1 = 1.8(x − 3)For h: y2 = −0.8(x − 3)

c.y

y

x

xx1

2

1 8 3

0 8 32 25= = − ≠. ( – )

– . ( – ). , for 3.

g

h

′′

= = −( )

( )

.

– .

3

3

1 8

0 82 25. , which equals y1/y2,

Q.E.D.

d. Because the ratio g (x)/h(x) approaches theratio y1/y2 as x approaches 3, and becausey1/y2 equals g′ (3)/h′ (3) for all x ≠ 3, the ratiog (x)/h(x) also approaches g′ (3)/h′ (3) as xapproaches 3. This is what l’Hospital’s ruleconcludes.

If g (3) or h (3) were any number other than 0,the canceling of the (x − 3)’s in part c couldnot be done, and the ratio would almostcertainly not equal 1.8/(−0.8).

e. The graph shows a removable discontinuity at(3, −2.25):

3

1 x

f(x)

35. a. For yearly compounding, m(t) =1000(1 + 0.06)t. For semiannual compound-ing, m ( t) = 1000(1 + 0.06/2)2t because thereare two compounding periods per year, eachof which gets half the interest rate.

b. m(t) = 1000(1 + 0.06/n)nt

lim limn n

ntm t n→∞ →∞

= +( ) ( . / )1000 1 0 06

= +→∞

1000 1 0 06lim ( . / )n

ntn

Let L nn

nt= +→∞

lim ( . / )1 0 06 .

ln L nt nn

= +→∞

lim [ ln ( . / )]1 0 06

= + →→∞

limln ( . / )

/( )n

n

nt

1 0 06

1

0

0

= + ⋅ −−→∞

−

−lim/( . / ) ( . )

/n

n n

n t

1 1 0 06 0 06 2

2

=+

=→∞

lim.

( . / )n

t

nt

0 06

1 0 060 06.

∴ = ⇒ =→∞

L e m t et

n

t0 06 0 061000. .lim ( )

When interest is compounded continuously,m(t) = 1000e0. 06t.

c.

t m(t), Annual m(t),Continuous Difference

5 1,338.23 1,349.86 11.63

20 3,207.14 3,320.12 112.98

50 18,420.15 20,085.54 1,665.38

d. For 7% interest, compounded continuously,m (t) = 1000e0.0 7t.

36. a. f (x) = xn, g (x) = ln x, h (x) = ex

lim( )

( )lim

lnx x

nf x

g x

x

x→∞ →∞= → ∞

∞

= = = ∞→∞ →∞

lim/

lim ,–

x

n

x

nnx

xnx n

1

10if >

∴ a power function is higher-order than thenatural log function.

lim( )

( )lim

x x

n

x

f x

h x

x

e→∞ →∞= → ∞

∞

= → ∞∞→∞

lim ,–

x

n

x

nx

e

1

if n − 1 > 0

Eventually, the exponent of the power willbecome zero, in which case the limit takes theform constant/∞, which is 0.∴ a power function is lower-order than anexponential function.Using “<” to represent “is lower-order than,”natural log < power < exponential.

b. i. limln

x

x

x→∞=3

05 ii. lim .x x

x

e→∞=

100

0 01 0

iii. limln

.

x

xe

x→∞= ∞

0 3

100

iv. lim limx x

x

x x→∞ →∞= =1

0

v. lim lim.x

x

x x

xe

ee

→∞ →∞= = ∞0 2

0 8.


Problem Set 6-61. y = ln (3x + 4) ⇒ y′ = 3/(3x + 4)

2. y = ln (3x5) = ln 3 + 5 ln x ⇒ y′ = 5/x3. y = ln (e3x) = 3x ⇒ y′ = 3

4. y = ln (sin 4x) ⇒ y′ = =4 4

4

cos

sin

x

x 4 cot 4x

5. y = ln (cos5 x) = 5 ln (cos x) ⇒

′ = − = −yx

xx

55

sin

costan

6. y = ln (e5) = 5 ⇒y′ = 0 (Derivative of a constant!)


7. y = ln [cos (tan x)] ⇒

y ′ = ⋅– sin (tan )

cos (tan )

x

x sec2 x = −tan (tan x) sec2 x

8. y x x x x= + = − +ln – ln2 22 31

22 3( ) ⇒

′ =

−− +

= −− +

yx

x x

x

x x

1

2

2 2

2 3

1

2 32 2

9. y = cos (ln x) ⇒ y′ = −(1/x) sin (ln x)

10. y = sin x · ln x ⇒ y′ = cos x · ln x + (1/x) sin x

11. y = e7x ⇒ y′ = 7e7x

12. y e y x ex x= ⇒ ′ =3 3

3 2

13. y e e x y xx x= = = ⇒ ′ =5 5 45

5ln ln

14. y = ecos x ⇒y′ = ecos x · (−sin x) = −ecos x sin x

15. y = cos (ex) ⇒ y′ −sin (ex) · ex = −ex sin ex

16. y = (cos3 x)(e3x) ⇒y′ = 3 cos2 x (−sin x) · e3x + cos3 x · e3x · 3

= −3e3x cos2 x sin x + 3e3x cos3 x= 3e3x cos2 x (−sin x + cos x) (Factoringoptional)

17. y e y x ex x= ⇒ ′ =5 5

5 4

18. y e y e ee e xx x

= ⇒ ′ = ⋅19. sin y = ex ⇒ cos y · y′ = ex ⇒

y′ = e

y

e

e

x x

xcos –=

1 2(See sketch.)

1sin y = e x

y

(Showing cos – y e x= 1 2 )

20. y = ex · ln x ⇒y′ = ex · ln x + ex · (1/x) = ex (ln x + 1/x)

21. y = 11

/tx

∫ dt ⇒ y′ = 1/x

22. tan y = ex ⇒ sec2 y · y′ = ex ⇒

y′ = e

y

e

e

x x

xsec2 21=

+(See sketch.)

1

ex

y

√1 + e 2x

23. y = ln (eln x) = ln x ⇒ y′ = 1/x24. y = 2x ⇒ ln y = x ln 2 ⇒ (1/y)y′ = ln 2 ⇒

y′ = y ln 2 = 2x ln 2

25. y e ex xx

= = =ln ln2 2 2 ⇒y′ = 2x ln 2 (See Problem 24.)

26. y = e2 ln x = e xln 2

= x2 ⇒ y′ = 2x

27. y = x2 ⇒ y′ = 2x

28. y = ex ln x (which equals xx) ⇒y′ = ex ln x [ln x + x (1/x)] = xx (ln x + 1)

29. y = xx = (eln x)x = ex ln x ⇒y′ = xx (ln x + 1) (See Problem 28.)

30. y = x ln x − x ⇒ y′ = ln x + x · (1/x) − 1 = ln x(Note: This answer reveals that the integral ofln x is x ln x − x.)

31. y = ex(x − 1) ⇒ y′ = ex(x − 1) + ex · 1 = xex

32. y = 1

2( )–e ex x+ ⇒ y′ =

1

2( – )–e ex x

33. y = 1

2( – )–e ex x

⇒ y′ = 1

2( )–e ex x+

(Problems 32 and 33 are the hyperbolic cosineand sine functions, respectively. See Chapter 8.)

34. y = e

e

x

x1+ ⇒

y′ = e e e e

e

x x x x

x

⋅ + ⋅+

( ) – ( )

( )

1

1 2 =+e

e

x

x( )1 2

35. y = 5x ⇒ ln y = x ln 5 ⇒ (1/y)y′ = ln 5 ⇒ y′ = y ln 5 = 5x ln 5

36. y = log5 x = ln

ln

x

5 ⇒ y′ =

1

5

/

ln

x = 1

5x ln

37. y = x− 7 log2 x = x− 7 · ln

ln

x

2 ⇒

y′ = 1

27

18 7

ln– ln– –x x x

x⋅ + ⋅

= xx

–

ln(– ln )

8

27 1+

38. y = 2− x cos x ⇒y′ = 2− x (−ln 2) · cos x + 2− x (−sin x)

= −2− x(ln 2 · cos x + sin x) (Factoring optional)

39. y = e− 2x ln 5x ⇒y′ = −2e− 2x · ln 5x + e− 2x · (1/x)

= e− 2x (−2 ln 5x + 1/x)

40. y yx

x x= ⇒ ′ = ⋅ =7

7

1

77 7 7

ln lnln

41. yx

ex x y

xe= = = ⇒ ′ =log

loglog ln3

3

1

42. yx

ex x y

xe= = = ⇒ ′ =log

loglog ln10

10

1

43. y xx= = ⋅( )( )log ln

ln

lnln8 8

88 = ln x ⇒

y′ = 1

x

44. y = (log4 x)10 ⇒

y′ = 10(log4 x)9 · 1

4x ln=

10

44

9(log )

ln

x

x

45. y = log5 x7 = 7 log5 x =

7

5

ln

ln

x ⇒ y′ =

7

5x ln


46. y = tan ex ⇒ y′ = sec2 ex · ex = ex sec2 ex

47. y = esin x ⇒ y′ = esin x cos x

48. y = ln csc x ⇒y′ = (1/csc x) · (−csc x cot x) = −cot x

49. y = 35 ⇒ y′ = 0 (Derivative of a constant!)

50. y = ln (cos2 x + sin2 x) = ln 1 = 0 ⇒ y′ = 0

51. y = sin x ⇒ y′ = cos x

52. y = sin− 1 x ⇒ y′ = 1

1 2– x

53. y = csc x ⇒ y′ = −csc x cot x

54. y = tan− 1 x ⇒ y′ = 1

1 2+ x

55. y = tan x ⇒ y′ = sec2 x

56. y = cot x ⇒ y′ = −csc2 x

57. e dx e Cx x4 41

4 = +∫

58. e dx e x C4 4= +∫59. x e dx e x dx e Cx x x3 34 4 41

44

1

4∫ ∫= = +( )

60. cos cossin sinx e dx e x dxx x⋅ =∫ ∫ ( )

= +e Cxsin

61.(ln )

(ln ) (ln )x

xdx x

xdx x C

55 61 1

6= = +∫∫

62. 5 5x xdx e dx=∫ ∫ ln

= ∫( / ) 1 5 55ln lnlne dxx

= + = +1

5

5

55

ln lnlne C Cx

x

63. e dx dx Cx xx

ln

ln5 5

5

5 = = +∫ ∫

(See Problem 62.)

64.1

2

1

2( ) ( – )– –e e dx e e Cx x x x+ = +∫

65.1

1 t

x

∫ dt = ln x (By definition!)

66. e dx e Cx x− −= − +∫

67. 22

2x

x

dx C= +∫ ln

68. ( ) .x dx x Cxx

− + = + +∫ 0 2 0 83 1 253

3. .

ln

69. ( / ) | | 3 3x dx x C= +∫ ln

70.

44

4

12

48 656170

1

2

1

2x

x

dx = = =∫ ln ln. K

71. (ln ) (ln )xx

dx x C9 101 1

10= +∫

72. cos sin x dx x C= +∫73. e dx x dx x Cxln = = +∫ ∫ 1

22

74. ln ln( ) e dx x e dx x dx x Cx3 23 33

2= = = +∫ ∫∫

75. 0 dx C=∫ (Integral of zero is a constant.)

76. cos secx x dx dx x C = = +∫ ∫1

77. sec sec21

22 2x dx x dx=∫ ∫ ( )

= + +1

22 2ln | sec tan |x x C

78. tan tan31

33 3x dx x dx∫ ∫= ( )

= +1

33ln | sec |x C

79. cot cot41

44 4x dx x dx=∫ ∫ ( )

= +1

44ln | sin |x C

80. csc csc51

55 5x dx x dx=∫ ∫ ( )

= − + +1

55 5ln | csc cot |x x C

81. limcos

x

x

x→

− →0

1 0

0

= limsin

x

x→

=0 1

0

82. lim– cosx

x

x→→

0 1

0

0

= = ∞→

lim0x x

1

sin (Reciprocal of Problem 81.)

83. lim– cos

/

– cos ( / )/x

x

x→= =

π

ππ

π2 1

2

1 2 2

84. limcosx

x

x→ += ∞

π

π1 0

Form:

85. lim– sin

x

x x

x→ →

0 3

5 5 0

0

= →→

lim– cos

x

x

x0 2

5 5 5

3

0

0


= →→

limsin

x

x

x0

25 5

6

0

0

= = =

→lim

cosx

x0

125 5

6

125

620 8333. K

86. lim( . / )x

xx→∞

∞+ →1 0 03 1

Let L xx

x= +→∞

lim( . / )1 0 03 .

ln L = + = +→∞ →∞

ln lim ( . / ) lim lnx

x

x

xx x1 0 03 1 0 03( . / )

= +→∞

lim [ lnx

x x( . / )] 1 0 03

= + →→∞

limln ( . )–

–x

x

x

1 0 03 0

0

1

1

= + ⋅ =→∞

lim/( . / ) (– . )

–

–

–x

x x

x

1 1 0 03 0 030 03

2

2 .

∴ L = e0.03 = 1.03045…

87. lim ( . ) /

x

xx→∞

+ → ∞1 0 03 1 0

Let L xx

x= +→∞

lim( . )1 0 03 1/ .

ln lim lnL x xx

= + → ⋅ ∞→∞

[( / ) ( . )]1 1 0 03 0

= + → ∞∞→∞

limln ( . )

x

x

x

1 0 03

= + ⋅ =→∞

lim/( . ) .

x

x1 1 0 03 0 03

10

∴ L = e0 = 1

88. limx

x

x→∞→ ∞

∞2

2

= → ∞∞→∞

limln

x

x

x

2 2

2

= = ∞→∞

lim(ln )

x

x2 2

2

2

or: limx

x

x→∞= ∞2

2 by (exponential)/(power)

89. lim ( . ) /( )

x

xx→

− ∞→2

3 20 5 1

Let L = ln (0.5x)3/(2 −x) .

ln lim–

ln .Lx

xx

= ⋅

→ ∞ ⋅

→2

3

20 5 0

= →→

limx

x

x2

3 0 5

2

0

0

ln .

–

= ⋅ = −→

lim/( . ) .

–x

x2

3 0 5 0 5

1

3

2

∴ L = e− 3/2 = 0.22313…

90. lim–

–x xe x→

⇒ ∞ − ∞

0 3

1

1

1

3

= →→

lim– ( – )

( – )x

x

x

x e

x e0

3

3

3 1

3 1

0

0

=+ ⋅

→→

lim–

( – )x

x

x x

e

e x e0

3

3 3

3 3

3 1 3 3

0

0

=+ +

= −→

lim–

x

x

x x x

e

e e xe0

3

3 3 3

9

9 9 27

1

2



R1. a. dM/dt = 0.06M ⇒ M− 1 dM = 0.06 dt

∴ =− ∫∫ M dM dtx

1

0

5

1000 06. , Q.E.D.

0.06 0.06 0.30

5dt t= =∫0

5

b. Solving numerically for x in

M dMx

− =∫ 1

1000 3.

gives x ≈ 134.9858… .

c. There will be $134.99 in the account, so theinterest will be $34.99.

R2. a. Integrating x− 1 by the power rule results in

division by zero:x

C–1 1

1 1

+

− ++ .

b. If g x f t dta

x

( ) = ∫ ( ) and f (x) is continuous in

a neighborhood of a, then g′ (x) = f (x).

ln xt

dtx

= ∫ 11

d

dxx

d

dx tdt

x

x

(ln ) =

=∫ 1 1

1

c. i. y = (ln 5x)3 ⇒ y′ = (3/x)(ln 5x)2

ii. f (x) = ln x9 = 9 ln x ⇒ f ′ (x) = 9/x

iii. y = csc (ln x) ⇒y′ = −csc (ln x) cot (ln x) · (1/x)

iv. g x t dt g x x xx

( ) ( )= ⇒ ′ =∫ csc csc2 2

1

2

d. i.sec tan

sec secsec tan

x x

xdx

xx x dx= ∫∫ 1

= +ln sec || x C

ii.10

102

3

2

3

xdx x=∫ ln

–

–

–

–

| |

= − −10 3 10 2ln | – | ln | |= 10(ln 3 − ln 2) = 4.054651…

iii. x x dx x x dx2 3 1 3 1 241

34 3( ) ( ) ( )− = −− −∫∫

= +1

343ln | – |x C


e. By finding areas, h (1) = −2.5, h (2) = 0,h (6) = 2.7, h (10) = 0, and h (11) = −2.5.

1 2 6 11

4

–4

y

x or t

y = f (t )

y = h(x)

f. i. y (100) ≈ 70 names; 70% rememberedy(1) = 1 name; 100% remembered

ii. yx

′ =+

101

100y′ (100) = 101/(200) = 0.505 names/persony′ (1) = 101/101 = 1 name/person

iii. Paula has probably not forgotten anynames as long as x − y < 0.5. Aftermeeting 11 people, she remembers about10.53… ≈ 11 names, but after meeting12 people, she remembers about 11.44… ≈11 names.

R3. a. i. See the text for the definition of logarithm.

ii. See the text for the definition of ln x.

iii. See the text for the statement of theuniqueness theorem.

iv. See the text for the proof.

v. See the solution to Problem 10 inLesson 6-3.

b. i. e n e nn

n

n

n= + = +→ →∞

lim ( ) lim ( / )/

0

11 1 1or

ii. logln

lnb xx

b=

c. i. y xx

yx

= = ⇒ ′ =logln

ln ln4 4

1

4

ii. f x xx

( ) ( )= = ⇒log cosln(cos )

ln2 2

′ = ⋅ −f xx

x( ) ( )1

2(cos )(ln )sin

= − tan

ln

x

2

iii. y x yx= = ⇒ ′ =log log5 59 9 9log5

R4. a. i.

2

10

y

x

ii.

–2

10

y

x

iii.

5

1

y

x

b. i. f (x) = x1.4e5x ⇒f ′ (x) = 1.4x0.4e5x + 5x1.4 e5x

ii. g (x) = sin (e− 2x) ⇒ g′ (x) = −2e− 2x cos(e− 2x)

iii.d

dxe

d

dxxx( ) ( )ln = = 1

iv. y = 100x ⇒ y′ = (ln 100)100 x

v. f (x) = 3.7 · 100.2 x ⇒ f ′ (x) = 0.74 ln 10 · 100.2 x

v i . r (t) = t tan t ⇒ ln r = tan t ln t ⇒1 2

rr t t

t

t′ = + ⇒sec ln

tan

′ = +

r t t t

t

tttan sec ln

tan2

c. y = (5x − 7)3 (3x + 1)5 ⇒ln y = 3 ln (5x − 7) + 5 ln (3x + 1) ⇒1 15

5 7

15

3 1yy

x x′ =

−+

+⇒

′ =−

++

− +y

x xx x

15

5 7

15

3 15 7 3 13 5( ) ( )

= (120x − 90)(5x − 7)2 (3x + 1)4

d. i. 10 52 2e dx e Cx x− −= − +∫ii. e x dx e Cx xcos cossin = − +∫

iii. e dx ex x− −

− −= −∫ 0 1 0 1

2

2

2

2

10. .

= − + =−10 10 4 026720 2 0 2e e. . . K

iv. 10

10

0 2 100 2

0 2.

.

. lnx

x

dx C= +∫e. i. The exposure is the product of C (t) and t,

where C (t) varies. Thus, a definite integralmust be used.


ii. E x e dt et xx

( ) . ( ). .= = − +− −∫ 150 937 5 10 16 0 16

0

E(5) = 937.5(−e− 0.8 + 1) =516.25… ppm · days E (10) =937.5(−e− 1.6 + 1) = 748.22… ppm · daysAs x grows very large, E (x) seems toapproach 937.5.

iii. E ′ (x) = 150e− 0.16 x = C (x)E ′ (5) = 67.39… ppm (or ppm · daysper day)E′ (10) = 30.28… ppm

f. i. From Figure 6-7d, the maximumconcentration is about 150 ppm at about2 hours. (These values can be found moreprecisely by setting the numerical oralgebraic derivative equal to zero, solvingto get t = −1/ln 0.6 = 1.9576… . ThenC (1.9576…) = −200/(e ln 0.6) =144.0332… .)

ii. C (t) = 200t · 0.6t

C′ (t) = 200t · 0.6t ln 0.6 + 200 · 0.6t

C′ (1) = 200 · 0.61(ln 0.6 + 1) = 58.70…

C′(5) = 200 · 0.65(5 ln 0.6 + 1)

= −24.16… < 0

C(t) is increasing at about 58.7 ppm/hwhen t = 1 and decreasing at about24.2 ppm/h when t = 5. The concentrationis increasing if C′ (t) is positive anddecreasing if it is negative.

iii. Solving 50 = 200t · 0.6t numerically for tgives t ≈ 0.2899… and t ≈ 6.3245… .So C(t) > 50 for 6.3245… − 0.2899… =6.03… , or about 6 hours.

iv. C1(t) = 200t · 0.3t

100

C(t)

1

50t

From the graph, the maximum is about60 ppm around t = 1. (Exactly, t = −1/ln 0.3 =0.8305… , for which C (0.8305…) =−200/(e ln 0.3) = 61.11092… ≈ 61.1 ppm.)Repeating the computations of part iiigives C (t) > 50 for 0.409… < t < 1.473… ,or for about 1.06 hours.In conclusion, the concentration peakssooner at a lower concentration and staysabove 50 ppm for a much shorter time.ln 0.5 = −0.025t

R5. a. limx

x

x→∞

−−

→ ∞−∞

2 3

7 5

2

2

=−

= −→∞

limx

x

x

4

10

2

5

b. limcos

x x

x x

e x→

− +− −

→0

2 1

1

0

0

= +−

→→

limsin

x x

x x

e0

2

1

0

0

= + = + =→

limcos

x x

x

e0

2 2 1

13

c. limx

xx e→∞

− → ∞ ⋅3 0

= → ∞∞→∞

limx x

x

e

3

= → ∞∞→∞

limx x

x

e

3 2

= → ∞∞→∞

limx x

x

e

6

= = ∞→∞

limx xe

60 (Form: 6/ )

d. L xx

x= →→

∞lim tan( / )

1

2 1π

ln lim [tan( / ) ln ]L x xx

= ⋅→1

2π

= →→

limln

cot ( / )x

x

x1 2

0

0π

=−

=−

= −→

lim/

( / )csc / /x

x

x1 2

1

2 2

1

2

2

π π π π

∴ L = e− 2/ π = 0.529077…

e. limx

x→

=2

43 48

f. lim (tan sec ) lim ( )/ /x x

x x→ →

− = =π π2

2 2

21 1

g. Examples of indeterminate forms:

0/0, ∞/∞, 0 · ∞, 00, 1∞ , ∞0, ∞ − ∞

R6. a. i. y = ln (sin4 7x) = 4 ln sin 7x ⇒y′ = 4(1/sin 7x) · cos 7x · 7 = 28 cot 7x

ii. y = x− 3e2x ⇒y′ = −3x− 4 · e2x + x− 3 · 2e2x

= x− 4e2x (2x − 3)

iii. y = cos (2x) ⇒ y′ = −sin (2x) · 2x ln 2

iv. y xx

yx

= = ⇒ ′ =logln

ln ln34 4

3

4

3

b. i. e dx e Cx x− −= − / +∫ 1 7 1 71 1 7. .( . )

ii. 2sec sec tanx x x dx∫= ∫ e x x dxxln sec sec tan2


ii. E x e dt et xx

( ) . ( ). .= = − +− −∫ 150 937 5 10 16 0 16

0

E(5) = 937.5(−e− 0.8 + 1) =516.25… ppm · days E(10) =937.5(−e− 1.6 + 1) = 748.22… ppm · daysAs x grows very large, E(x) seems toapproach 937.5.

iii. E ′ (x) = 150e− 0.16x = C (x)E ′ (5) = 67.39… ppm (or ppm · daysper day)E′ (10) = 30.28… ppm

f. i. From Figure 6-7d, the maximumconcentration is about 150 ppm at about2 hours. (These values can be found moreprecisely by setting the numerical oralgebraic derivative equal to zero, solvingto get t = −1/ln 0.6 = 1.9576… . ThenC (1.9576…) = −200/(e ln 0.6) =144.0332… .)

ii. C (t) = 200t · 0.6t

C′ (t) = 200t · 0.6t ln 0.6 + 200 · 0.6t

C′ (1) = 200 · 0.61(ln 0.6 + 1) = 58.70…

C′(5) = 200 · 0.65(5 ln 0.6 + 1)

= −24.16… < 0

C(t) is increasing at about 58.7 ppm/hwhen t = 1 and decreasing at about24.2 ppm/h when t = 5. The concentrationis increasing if C′ (t) is positive anddecreasing if it is negative.

iii. Solving 50 = 200t · 0.6t numerically for tgives t ≈ 0.2899… and t ≈ 6.3245… .So C(t) > 50 for 6.3245… − 0.2899… =6.03… , or about 6 hours.

iv. C1(t) = 200t · 0.3t

100

C(t)

1

50t

From the graph, the maximum is about60 ppm around t = 1. (Exactly, t = −1/ln 0.3= 0.8305… , for which C (0.8305…) =−200/(e ln 0.3) = 61.11092… ≈ 61.1ppm.)Repeating the computations of part iiigives C (t) > 50 for 0.409… < t < 1.473…, or for about 1.06 hours.In conclusion, the concentration peakssooner at a lower concentration and staysabove 50 ppm for a much shorter time.ln 0.5 = −0.025t

R5. a. limx

x

x→∞

−−

→ ∞−∞

2 3

7 5

2

2

=−

= −→∞

limx

x

x

4

10

2

5

b. limcos

x x

x x

e x→

− +− −

→0

2 1

1

0

0

= +−

→→

limsin

x x

x x

e0

2

1

0

0

= + = + =→

limcos

x x

x

e0

2 2 1

13

c. limx

xx e→∞

− → ∞ ⋅3 0

= → ∞∞→∞

limx x

x

e

3

= → ∞∞→∞

limx x

x

e

3 2

= → ∞∞→∞

limx x

x

e

6

= = ∞→∞

limx xe

60 (Form: 6/ )

d. L xx

x= →→

∞lim tan( / )

1

2 1π

ln lim [tan( / ) ln ]L x xx

= ⋅→1

2π

= →→

limln

cot ( / )x

x

x1 2

0

0π

=−

=−

= −→

lim/

( / )csc / /x

x

x1 2

1

2 2

1

2

2

π π π π

∴ L = e− 2/ π = 0.529077…

e. limx

x→

=2

43 48

f. lim (tan sec ) lim ( )/ /x x

x x→ →

− = − = −π π2

2 2

2

1 1

g. Examples of indeterminate forms:

0/0, ∞/∞, 0 · ∞, 00, 1∞ , ∞0, ∞ − ∞

R6. a. i. y = ln (sin4 7x) = 4 ln sin 7x ⇒y′ = 4(1/sin 7x) · cos 7x · 7 = 28 cot 7x

ii. y = x− 3e2x ⇒y′ = −3x− 4 · e2x + x− 3 · 2e2x

= x− 4e2x (2x − 3)

iii. y = cos (2x ) ⇒ y′ = −sin (2x ) · 2x ln 2

iv. y xx

yx

= = ⇒ ′ =logln

ln ln34 4

3

4

3

b. i. e dx e Cx x− −= − / +∫ 1 7 1 71 1 7. .( . )

ii. 2sec sec tanx x x dx∫= ∫ e x x dxxln sec sec tan2


C4. a. Suppose there is a number M > 0 such thatln x ≤ M for all x > 0. Let x = eM +1. Thenln x = ln eM +1 = (M + 1) ln e = M + 1 > M.This contradicts ln x ≤ M for all x > 0. Thusthe supposition is false, and there can be nosuch number M that is an upper bound forln x, Q.E.D.

b. If M were a lower bound for ln x, then −Mwould be an upper bound for ln (1/x), butpart a shows no such number can exist.

c. ln′ x = 1/x, which shows that ln isdifferentiable for all x > 0. Thus, ln iscontinuous for all x > 0 becausedifferentiability implies continuity.

d. Because ln is continuous for all x > 0, theintermediate value theorem applies. Thus, if kis between ln a and ln b, there is a number cbetween a and b such that ln c = k.

ln a

ln b

a bc

k

x

y = ln x

e. Part a shows k cannot be an upper bound forln, so there must be some b > 0 such thatln b > k. Similarly, part b shows k is nota lower bound, so some a > 0 exists forwhich ln a < k. By part d there is somenumber c between a and b such that ln c = k,Q.E.D.

f. The domain of ln is the positive reals, and therange is all reals; the domain of the inverse toln (i.e., exp) is the range of ln (i.e., all reals),and the range of the inverse is the domain ofln (i.e., positive reals).

C5. a. g x t dt t dtx

x( ) = = − ⇒∫∫ sin sin

4

4 2

2

g′ (x) = −2x sin x2

b. g x t dtx

x

( ) = ∫ sintan

2

= + ⇒∫∫ sin sintan

t dt t dtx

x 4

4

2

g′ (x) = −2x sin x2 + sin (tan x) sec2 x

c. g x f t dtu x

v x

( ) = ⇒∫ ( )( )

( )

g′ (x) = f (v(x)) · v′ (x) − f ( u ( x ) ) · u′ (x)

C6. log cabin(or log cabin + C, which equals “houseboat”)

Chapter Test

T1. ln xt

dtx

= ∫ 11

T2. enn

n

= +

→∞

lim 11

or e nn

n= +→

lim ( ) /

0

11

T3. If g (x) = f t dta

x

( )∫ and f ( t ) is continuous in a

neighborhood of a, then g′ (x) = f ( x ) .

T4. If (1) f ′ ( x ) = g′ (x) for all x in the domain and(2) f ( a ) = g ( a ) for some a in the domain, thenf ( x ) = g ( x ) for all x in the domain.

T5. Prove that ln x = loge x for all x > 0.

Proof:

Let f ( x ) = ln x, and g (x) = loge x.f ′ ( x ) = 1/x andg′ (x) = (1/x) · loge e = (1/x) · 1 = 1/x∴ f ′ ( x ) = g′ (x) for all x > 0f ( 1 ) = ln 1 = 0 and g (1) = loge 1 = 0∴ f ( 1 ) = g (1)∴ by the uniqueness theorem,f ( x ) = g (x) for all x > 0.∴ ln x = loge x for all x > 0, Q.E.D.

T6. f ( x ) = ln (x3ex )

a. ′ = ⋅ + = +f xx e

x e x e xxx x( ) ( ) /

13 3 13

2 3

b. f ( x ) = 3 ln x + x ln e = 3 ln x + x ⇒ f ′ ( x ) = 3/x + 1 (Checks.)

T7. y = e2x ln x3 = 3e2x ln x ⇒y e x e xx x′ = ⋅ + ⋅ /6 3 12 2ln ( )

= + /3 2 12e x xx ( )ln

T8. v = ln (cos 10x) ⇒v′ = 1/(cos 10x) · (−10 sin 10x) = −10 tan 10x

T9. f ( x ) = (log2 4x)7 = [(ln 4x)/(ln 2)]7 ⇒f ′ ( x ) = 7[(ln 4x)/(ln 2)]6 · [(1/4x) · 4 · (1/ln 2)]

= 7 4

22

6(log )

ln

x

x

T10. t ( x ) = ln (cos2 x + sin2 x) = ln 1 = 0 ⇒t′ (x) = 0

T11. p x etx

( )ln

= ∫1sin t dt ⇒

p′ (x) = eln x sin ln x · 1/x = sin ln x

T12. e dx e Cx x5 51

5= +∫

T13. ( ) ( )6ln (ln )x dx x x C/ = +∫ 1

77

T14. sec ln sec tan51

55 5x dx x x C= + +∫ | |


T15. 51

55

1

525 1

0

2

0

2x xdx = =∫ ln ln

( – )

= 14.9120…

T16. lim–

lnx

x

x→∞→ −∞

∞5 3

4

=⋅

= = −∞→∞ →∞

lim–

[ /( )]lim(– )

x xxx

3

1 4 43

T17. L xx

x= → ∞→ −lim (tan )

/

cot

π 2

0

ln lim cot ln tan/

L x xx

= ⋅ → ⋅∞→ −π 2

0[ ( )]

= → ∞∞→ −

limln (tan )

tan/x

x

xπ 2

= ⋅ = =→ →− −lim

( / tan ) sec

seclim cot

/ /x x

x x

xx

π π2

2

22

10

∴ L = e0 = 1

T18. a.

1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

-1

-2

-3

-4

-5

y

t or x

f

g


( ) ( ) ( ) ( ) ,= ⇒ ′ = − ⋅−

∫ 3 5 32

3 5

h f f′ = − ⋅ = ⋅ = ⋅ =( ) ( ) ( )3 9 5 3 4 3 1 3 3

T19. ln ,xt

dtx

= ∫ 11

so ln ..

1 81

1

1 8

= ∫ tdt

M4 0 2

1

1 3

1

1 5

1

1 70 58664= + + +

=.

1

1.1.

. . .K

From calculator, ln 1.8 = 0.58778… .

T20. g x t dt tx x

( ) sin cos= = −∫2 2

2 2

= −cos x2 + cos 2 ⇒ g′(x) = 2x sin x2

′ = =∫g xd

dxt dt x x

x

( ) sin sin2 2

2

2

T21. Let h ( x ) = f ( x ) − g (x).

Then h ( a ) = f ( a ) − g (a) = 0 andh ( b ) = f ( b ) − g (b) ≠ 0.

∴ ≠h b h a

b a

( ) – ( )

–0

By the mean value theorem, there is a number cbetween a and b such that

′ =h ch b h a

b a( )

( ) – ( )

–.

∴ h′ (c) ≠ 0

But h ′ ( x ) = f ′ ( x ) − g′ (x), which equals 0 forall values of x.

∴ h ′ ( c ) = 0

This result thus contradicts the mean valuetheorem, Q.E.D.

T22. a. F ( x ) = 60e0.1 x ⇒ F ′ ( x ) = 6e0.1 x so

F ′ ( 5 ) = 6e0.5 = 9.8923… lb/ftF ′ ( 1 0 ) = 6e = 16.3096… lb/ft

b. Work equals force times displacement. But theforce varies at different displacements. Thus, adefinite integral has to be used.

c.

dx

F

x

5

100 (x, F)

dW = F dx = 60 e0.1 x dx

W e dxx= ∫ 60 0 1

0

5.

= = −600 600 10 1

0

5 0 5e ex. .( )

W ≈ 389.23 ft-lb


Problem Set 6-8Cumulative Review, Chapters 1–6

1. f ( x ) = 2x

f ′ ≈ =( ) .3

2 2

0 25 549618

3 1 2 9. .–

.K

2. There are about 10.0 squares, each 20 units.

∴ ≈∫ g x dx( ) 20010

50

(Function is g x x x( ) .= + +2 0 12

15sin ,

π so exact

answer is 200.)

3. L f xx c

= →lim ( ) if and only if for any ε > 0, there is

a δ > 0 such that if x is within δ units of c butnot equal to c, f ( x ) is within ε units of L.

4. Answers may vary.

3

4

x

f(x)


5. f ′ (x) = +→

lim( ) – ( )

h

f x h f x

h0

or ′ = −−→

f cf x f c

x cx c( ) lim

( ) ( )

6. f (x) = x3

f xx h x

hh′ = + −

→( ) lim

( )0

3 3

= + + +→

lim–

h

x x h xh h x

h0

3 2 2 3 33 3

= + + =→

lim( )h

x xh h x0

2 2 23 3 3 , Q.E.D.

7. f (x) = x3 ⇒ f ′ (x) = 3x2

f ′ (5) = 3·52 = 75

f ′ (5) ≈ 5 01 4 99

0 0275 0001

3 3. – .

..=

f ′ (5) ≈ 5 001 4 999

0 00275 000001

3 3. – .

..=

The symmetric differences are getting closer to75 as ∆ x gets closer to zero.

8. f (x) = 3 5 3 5 1 2x x– ( ) /= −

f ′ (x) = 1

23 5 3 1 5 3 51 2 1 2( – )x x− −⋅ = −/ /. ( )

f ′ (7) = 1.5(21 − 5)−1/2 = 1.5/4 = 0.375 = 3/8

9. Line with slope of 3/8 is tangent to the graph atx = 7.

5

5

x

f(x)

8

3

10. a. y = e2x cos 3x ⇒y′ = 2 e2x cos 3x − 3e2x sin 3x

b. q xx

xq x( )

ln

tan( )= ⇒ ′ =

(tan )/ ln sec

tan tan sin

x x x x

x x x

x

x

−= −

2

2 2

1 ln

c. d

dx

d

dxx x x

2

225 5 5 5 5( ) [(ln ) ] (ln )= =

11. For the function to be differentiable,lim ( ) lim ( )x x

ax x x b→ →− +

+ = − + +2

2

2

21 6 and

lim lim ( )x x

ax x→ →− +

= − +2 2

2 2 6 .

4a + 1 = 8 + b and 4a = 2 ⇒ a = 1

2 and b = −5

2

2

y

x

12. Optional graph showing upper sum:

1 4

10

x

y = x 2

x dx2

1

4

∫U6 = 0.5(1.52 + 22 + 2.52 + 32 + 3.52 + 42)

= 24.875

13. M10 = 20.9775M100 = 20.999775Sums seem to be approaching 21.

14. a. cos sin cos5 61

6x x dx x C= − +∫

b. ( / ) | | 1 x dx x C= +∫ ln

c. tan ln sec ln cosx dx x C x C= + = − +∫ | | | |

d. sec ln sec tanx dx x x C= + +∫ | |

e. ( ) ( ) ( )/ /3 51

33 5 31 2 1 2x dx x dx− = −∫∫

= ⋅ + = +1

3

2

33 5

2

93 53 2 3 2( – ) ( – )x C x C/ /

15. x dx x2 3

1

4

1

4 1

3

64

3

1

321= = − =∫ ,

which agrees with the conjecture in Problem 13.

16. The graph shows a tangent line at x = c parallelto the secant line.

a c b

x

f(x)

Statement:

If f is differentiable on (a, b) and continuous atx = a and x = b, then there is a number x = c in

(a, b) such that f xf b f a

b a′ =( )

( ) – ( )

–.

17. y = x9/7

y7 = x9

7y6 y′ = 9x8

yx

y

x

xx x x′ = = = = =− −9

7

9

7

9

7

9

7

9

7

8

6

8

9 7 68 54 7 2 7 9 7 1

( )// / /

as from the derivative of a power formula


18. If x− 1 were the derivative of a power, then thepower would have to be x0. But x0 = 1, so itsderivative equals 0, not x− 1. Thus, x− 1 is not thederivative of a power, Q.E.D.

19. f x t dtx

( ) = ⇒∫ costan

31

f ′ (x) = cos (3 tan x) · sec2 x

20. f x t dt f x xx

( ) ,1

= ⇒ ′ =∫ ( / ) ( ) /1 1 Q.E.D.

21. Prove ln xa = a ln x for any constant a and allx > 0.

Proof:

Let f (x) = ln xa and g (x) = a ln x.

Then f xx

ax ax

a

xaa′ = ⋅ = ⋅ =−( )

1 11 and

g x ax

a

x′ = ⋅ =( ) .

1

∴ f ′ (x) = g′ (x) for all x > 0f (1) = ln (1a) = ln 1 = 0 and g(1) = a ln 1 = 0∴ f (1) = g (1)∴ f (x) = g (x) for all x > 0, and thusln xa = a ln x for all x ≥ 0, Q.E.D.

22. x = 5 cos t, y = 3 sin t

∴ = = = −dy

dx

dy dt

dx dt

t

tt

/

/

cos

– sincos

3

5

3

5

23. At t = 2, (x, y) = (5 cos 2, 3 sin 2) = (−2.08… , 2.72…).

At t = 2, dy

dx= − = …3

52 0 2745cot . .

The graph shows that a line of slope 0.27…at point (−2.08… , 2.72…) is tangent to thecurve.

5

3

x

y

24. y = tan− 1 t

vdy

dt tt= =

+= + −1

112

2( ) 1

adv

dtt t

t

t= = − + ⋅ = −

+−1 1 2

2

12 2

2 2( )( )

25. lim–

sinx

xe

x→→

0

3 1

5

0

0

= =→

limx

xe

x0

33

5 5

3

5cos

26. L nn

n= + →→

∞lim ( ) /

0

11 1

ln lim ln ( )Ln

nn

= +

→ ∞ ⋅→0

11 0

= + →→

limln ( )

n

n

n0

1 0

0

= + =→

lim/( )

n

n0

1 1

11

∴ L = e1 = e, Q.E.D.

27. Know: dx

dt

dy

dt= − =30 40 ft/s, ft/s

Want: dz

dt when x = 200 and y = 100

x2 + y2 = z2

2 2 2xdx

dty

dy

dtz

dz

dt+ =

When x = 200 and y = 100, z = =50 000 100 5, .

2 200 30 2 100 40 2 100 5( )( ) ( )( )− + = ⋅ dz

dt

dz

dt= − = −20

58 94427. K ft/s

The distance z is decreasing.

28. f x( )2

5

∫ dx ≈ (1/3)(0.5)(100 + 4 · 150 + 2 · 170 +

4 · 185 + 2 · 190 + 4 · 220 + 300)= (1/3)(0.5)(3340) = 556 2

3

29. Area of cross section = πy2

Because the end of the radius is on a line throughthe origin with slope r/h, y = (r/h)x.

∴ = =Area [( / ) ]2π πr h x

r

hx

2

22

dx

(x, Area)

Area

x

h

dV = (Area) dx

∴ = = ∫∫V dxr

hx dx

hh

( )Areaπ 2

22

00

= ⋅ = − =π π πr

hx

r

hh r h

h2

23

0

2

23 3 21

3

1

30

1

3( ) , Q.E.D.



Chapter 7—The Calculus of Growth and Decay

Problem Set 7-1

1. D (0) = 500D (10) = 895.4238482…D (20) = 1603.567736…

2. D ′ (t) = 500(ln 1.06)(1.06t ) $/yrD ′ (0) = 29.13445406…D ′ (10) = 52.17536994…D ′ (20) = 93.43814108…The rate of change, in $/yr, increases as theamount in the account increases.

3. R tD t

D t

t

t( )( )

( )

(ln . ) ( . )

( . )= ′ = ⋅ ⋅

⋅500 1 06 1 06

500 1 06= ln 1.06 = 0.0582689081…

R(0) = ln 1.06R(10) = ln 1.06R(20) = ln 1.06

4. The percent interest rate stays the same:approximately 5.83%.

5. f (x) = a ⋅ bx ⇒f ′(x) = a ⋅ (ln b) ⋅ bx = (ln b)(a ⋅ bx )

= (ln b) ⋅ f (x)

So f ′(x) is directly proportional to f (x).

6. See Problem 11 in Section 7-2.


1

y

x1

y

x

Q3. Q4.

x

y

1

y

x

Q5. Q6.

x

y y

x

Q7. Q8.y

x

3

x

y

Q9. Q10.

4

x

y

3

3

x

y

1. a. B = number of millions of bacteria;t = number of hours

dB dt kB dB B k dt B kt C/ = ⇒ = ⇒ = +∫ ∫/ ln | |

| |B e e e B C ekt C kt C kt= = ⋅ ⇒ =+1

b. 5 510

1= ⇒ =⋅C e Ck

7 = 5e3k ⇒ ln (7/5) = 3k

⇒ = ⋅ =k

1

3

7

50 112157ln . K

∴ = =

=B e et

tt5 5

7

551 3 7 5

30 112157( / ) ln( / )

/. K

c.B

t5

10

d. B = 5(7/5)24/3 = 73.78945…About 74 million

e. 1000 = 5(7/5)t/3 ⇒ ln (1000/5) = t/3 ⋅ ln (7/5)

t = =3 200

7 547 24001

ln

ln ( / ). K

About 47 hours after start, so in a little lessthan 2 days

2. a. N = number of units of radiation from N17;t = number of seconds

dN dt kN dN N k dt/ / = ⇒ = ∫∫⇒ ln |N| = kt + C| | N e N C ekt C kt= ⇒ =+

1

b. 3 × 1017 = C1ek·0 ⇒ C1 = 3 × 1017

5.6 × 1013 = 3 × 1017e60k

⇒ × =−ln )( .1 866 10 604K k⇒ k = −0.143103…

∴ = × −N e t3 1017 0 143103. K

c.N

t

3 × 1017


d. t = 5(60) = 300 s

N e= × =−3 10 0 06799117 0 143103 300( . )( ) .K KIt will not be safe because 0.067… > 0.007.

3. a. F = number of mg; t = number of minutes

dF dt kF dF F k dt F kt C/ /= ⇒ = ⇒ = +∫∫ ln | |

| |F e F C ekt C kt= ⇒ =+1

50 = C1ek·0 ⇒ C1 = 50

30 = 50e20k ⇒ ln (30/50) = 20k⇒ k = (1/20) ⋅ ln (0.6) = −0.025541…

( . )( . ) / / .F e et t t= = = −50 50 0 6 500 6 20 20 0 025541ln K

b.F

t

50

50

c. F = 50(0.6)(60/20) = 10.8 mg (exactly)

d. 0.007 = 50(0.6)t/20

⇒ ln (0.007/50) = ln (0.6)t/20⇒ t = 347.4323…

About 5 h 47 min

4. a. V = number of dollars trade-in value;t = number of months from the present

dV dt kV dV V k dt V kt C/ /= ⇒ = ⇒ = +∫ ∫ ln | |

| |V e V C ekt C kt= ⇒ =+1

b. 4200 420010

1= ⇒ =⋅C e Ck

4700 4200 4700 4200 33= ⇒ = −−e kk( )( ) ( / )ln

k = (−1/3) ln (4700/4200) = −0.037492…

∴ = −V e t4200 0 037492. K

c.

–30 30

t

V

4200

d. At 1 year after V = 4700, t = 9 months.V = 4200e( −0.037492…)(9) = 2997.116…About $3000

e. 1200 = 4200e− 0.037492…t

⇒ ln (1200/4200) = −0.037492…t⇒ t = (−1/0.037492…) ⋅ ln (1200/4200)= 33.4135…About 33 months from the present

f. 31 months before V = 4700, t = −34.

∴ = =− −V e4200 15026 7950 037492 34( . )( ) .K KAbout $15,000

g. The difference between $16,000 and $15,000is the dealer’s profit.

5. a. dC/dt = kC

b. dC C k dt C kt D/ = ⇒ = +∫ ∫ ln | |

⇒ = ⇒ =+| |C e C D ekt D kt1

0 00372 0 0037210

1. .= ⇒ =⋅D e Dk

0.00219 = 0.00372e8k

⇒ ln (0.00219/0.00372) = 8k⇒ k = (1/8) ⋅ ln (219/372) = −0.0662277…∴ C = 0.00372e− 0.0662277…t

c. Either: C = 0.015⇒ 0.015 = 0.00372e− 0.0662277…t

ln 4.0322… = −0.0662277…tt = −21.05… , which is before the poison wasinhaled,or: t = −20 ⇒ C = 0.00372e− 0.0662277…( −20)

C = 0.0139… , which is less than 0.015∴ the concentration never was that high.

d.

20,000

t

P

50

100

e. (1/2)(0.00372) = 0.00372e− 0.0662277…t

ln (1/2) = −0.0662277…t ⇒ t = 10.4661…About 10.5 hours

6. a. dP/dt = kP

b. dP P k dt P kt C/ = ⇒ | | = +∫ ∫ ln

⇒ = ⇒ =+| |P e P C ekt C kt1

100 10010

1= ⇒ =⋅C e Ck

50 = 100e5750k ⇒ ln 0.5 = 5750k⇒ k = −0.0001205473…∴ P = 100e− 0.0001205473…t

c. P = 100e( −0.0001205473…)(4000) = 61.74301…About 61.7%

d. 48 37 100 0 0001205473. .= −e tK

ln 0.4837 = −0.0001205473…tt = 6024.939…The wood is about 6025 years old. For 1996,the flood would have been 1996 − (−4004) =6000 years ago, so the wood is old enough.

e.

20,000

t

P

5061.7

4000 5750

100


7. dM/dt = kM ⇒ M = Cekt by the techniques inProblems 1–6, where C is the initial investment.∴ M varies exponentially with t.Let i = the interest rate as a decimal.dM/dt = Ck ⋅ ekt

At t = 0, dM/dt = Ci.∴ Ci = Ck ⋅ e0 ⇒ k = i ⇒ M = Ceit

Examples:

$1000 at 7% for 5 yr: $1419.07$1000 at 7% for 10 yr: $2013.75$1000 at 14% for 5 yr: $2013.75$1000 at 14% for 10 yr: $4055.20Leaving the money in the account twice as longhas the same effect as doubling the interest rate.Doubling the amount invested obviously doublesthe money at any particular time, but that doesn’ttell us how that compares with doubling the timeor the interest rate.Algebraically, Cei ·2t = Ce2i ·t shows that doublingthe time is equivalent to doubling the interestrate. Solving Ce2it > 2Ceit gives Ce2it − 2Ceit > 0⇒ Ceit(eit − 2) > 0 ⇒ eit > 2 (because Ceit > 0, soC > 0, being an investment) ⇒ it > ln 2. Sodoubling either the time or the interest rate willalways eventually yield more than doubling theinvestment, once t is high enough. For example,at 7%, 0.07t > ln 2 ⇔ t > (ln 2)/0.07 =9.9021… ⇒ t, so by 9 years 11 months,doubling the time or interest rate will yield morethan doubling the investment.

8. Assume an investment of $1000 at 7% per year.For 5 years, as in Problem 7:Annually: M = 1000(1.07)5 = $1402.55Quarterly: M = 1000(1.0175)20 = $1414.78Monthly: M = 1000(1.00583…)60 = $1417.63Daily: M = 1000(1.0001917808…)1825 =$1419.02Continuously (Problem 7): $1419.07Note that compounding continuously is only5 cents better than compounding daily for a$1000 investment in 5 years!

M = M0(1 + k/n)nt

Let L k nn

nt= + →→∞

∞lim ( / ) .1 1

ln lim [ lnL nt k nn

= ⋅ + / → ∞ ⋅→∞

( )] 01

= ⋅ + → ∞∞→∞

limln( / )

–n

t k n

n

11

=⋅

+⋅ −

−→∞

−

−lim/

( )

n

tk n

kn

n

11

2

2

=+

=→∞

lim/n

kt

k nkt

1∴ L = ekt

∴ =→∞

limn

ktM M e0 , which is the continuous

compounding equation.

9.dy

dxy= 0 3.

dy

ydx=∫ ∫0 3.

ln |y| = 0.3x + C

| | = = ⋅+y e e ex C x C0 3 0 3. .

y = ±eC ⋅ e0.3 x = C1e0.3 x

−4 = C1e0 ⇒ C1 = −4, showing that C1 can be

negative.∴ y = −4e0.3 x

5

10

20

–10

–20

y

x

10.dy

dxy= −0 2.

dy

yx dx= − ∫∫ 0 2.

ln |y| = −0.2x + C

| | = = ⋅− + −y e e ex C x C0 2 0 2. .

y = ±eC ⋅ e− 0.2 x = C1e− 0.2 x

30 = C1e− 1.4

C

e1 1 4

307 3979= =. . K

∴ y = 7.3979…e− 0.2 x

11. dy dx ky dy y k dx y kx C/ = ⇒ / = ⇒ | | = +∫ ∫ ln 1

| |y e y Cekx C kx= ⇒ =+ 1

y Ce C y y y ek kx( ) ( ) ,0 000= ⇒ = ⇒ =⋅

Q.E.D.

Problem Set 7-3Q1. Cekx Q2. (kx2)/2 + C

Q3. kx + C Q4. −cos x + C

Q5. 1 1 2/ – x Q6. 5 cos x

Q7. tan x

Q8.

x1

1

y' or y

y

y'

Q9. lim lim∆ ∆x

nx

nL U→ →

=0 0

Q10. B

1. a. dM/dt = 100 − S

b. S = kM ⇒ dM/dt = 100 − kM


c.dM

kMdt

k

k dM

kMdt

100

1

100−= ⇒ − −

−= ∫∫∫∫

⇒ − | − | = +1100

kkM t Cln

⇒ |100 − kM| = e− kte− kC

⇒ 100 − km = C1e− kt ⇒ kM = 100 − C1e

− k t

⇒ =Mk

C e kt1100 1( – )–

Substitute M = 0 when t = 0.

01

100 10010

1= ⇒ =k

C e C( – )

∴ =Mk

e kt1001( – )–

d. k = 0.02 ⇒ M = 5000(1 – e–0.02 t)

e.

M

t

5000

30 60 90

f. t = 30: $2255.94 ($3000 in, $744.06 spent)t = 60: $3494.03 ($6000 in, $2505.97 spent)t = 90: $4173.51 ($9000 in, $4826.49 spent)

g. t = 365: (365.23 or 366 could be used.)M = 5000(1 – e− 7.30 ) = 4996.622…≈ $4996.62 in the accountdM/dt = 100 − 0.02(4996.622…) = 0.06755…Increasing at about $0.07 per day

h. lim lim .

t t

tM e→∞ →∞

−= 5000 1 0 02( – )

= 5000(1 − 0) = 5000

2. dM/dt = 100 + kM (k = daily interest rate)dM

kMdt

k

k dM

kMdt

100

1

100+= ⇒

+= ⇒∫∫∫∫

1100 100

kkM t C kM e ekt kCln | || |+ = + ⇒ + = ⇒

100 + km = C1ekt ⇒ kM = −100 + C1e

kt ⇒

Mk

C ekt= −11001( )

Substitute M = 0 when t = 0.

01

100 10010

1= ⇒ =k

C e C( – )

∴ =Mk

ekt1001( – )

Let k = 0.0002 (0.02% per day).∴ M = (500000)(e0.0002 t – 1)

The graph is almost straight. The $100/daydeposits far exceed the interest for the first fewyears.

t

500

50,000

M

Make a table of M and dM/dt for variousnumbers of years. Neglect leap years.

Years M dM/dt

0 0 100.00

1 37865 107.57

10 537540 207.51

20 1652980 430.60

After 1 year, the $100/day is putting more intothe account. After 10 years, the interest hasstarted putting in more than the $100/day. After20 years, the interest puts in about $331 a day,while the winnings still put in only $100 a day.As t approaches infinity, the amount in theaccount becomes infinite!

3. a. E = RI + L(dI/dt)

b. L dI/dt = E – RIL dI

E RIdt

L

R

RdI

E RIdt

–= ⇒ − −

−=∫ ∫∫∫

⇒ = +– | – |L

RE RI t Cln

⇒ = − −| – |E RI e eR L t R L C( / ) ( / )

⇒ − = ⇒ =−E RI C e IR

E C eR L t R L t1 1

1( / ) –( / )[ – ]

Substitute I = 0 when t = 0.

01

10

1= ⇒ =R

E C e C E( – )

∴ =IE

Re R L t[ – ]–( / )1

c. I e t= 110

101 10 20[ – ]–( / )

I = 11(1 – e− 0.5 t)

I

t

5 10

11


d. i. I = 11(1 – e− 0.5) = 4.3281… ≈ 4.33 amps

ii. I = 11(1 – e–5) = 10.9258… ≈ 10.93 amps

iii. lim lim .

t t

tI e→∞ →∞

−= =11 1 11 1 00 5( – ) ( – )

= 11 amps

e. I = 0.95(11) = 10.45

10 45 11 1

0 95 1

0 05

0 5 0 05

0 5

0 5

0 5

. ( – )

. –

.

. .

.

.

.

=

=

=− =

−

−

−

e

e

e

t

t

t

t

ln

t = −2 ln 0.05 = 5.9914…About 6 seconds

4. a. R = C(dT/dt) + hT

b. C dT/dt = R – hT

C dT R hT dt/( )− = ∫∫− −

−= ∫∫C

h

h dT

R hTdt

− − = +C

hR hT t Dln | |

| | ( / ) ( / )R hT e eh C t h C D− = − −

R hT D e h C t− = /1

–( )

T h R D e h C t= / /( )[ – ]–( )1 1

Substitute T = 0 when t = 0.

01

10

1= ⇒ =h

R D e D R( – )

∴ = TR

he h C t[ – ]–( / )1

c. T = (50/0.04)[1 – e− (0.04/ 2)t]T = 1250(1 – e− 0.02t)

d.

1250T

t

100 200

e. Use TRACE or TABLE.

t = 10: T = 226.586… ≈ 227°t = 20: T = 412.099… ≈ 412°t = 50: T = 790.150… ≈ 790°t = 100: T = 1080.830… ≈ 1081°t = 200: T = 1227.105… ≈ 1227°

f. lim lim .

t t

tT e→∞ →∞

−= =1250 1 1250 1 00 02( – ) ( – )

= 1250°g. T = 0.99(1250) = 1237.5

1237.5 = 1250(1 – e− 0.02t)

0.99 = 1 – e− 0.02t

e− 0.02t = 0.01

−0.02t = ln 0.01

t = −50 ln 0.01 = 230.258…

About 230 seconds

5. a.dV

dtkV= 1 2/

b. V dV k dt− = ⇒∫∫ 1 2/

22

1 22

V kt C Vkt C/ = + ⇒ = +

V varies quadratically with t.

c. Initial conditions t = 0; V = 196;dV/dt = −28:

1960

2281 2/ = ⋅ + ⇒ =k C

C

and −28 = k ⋅ 1961/ 2 ⇒ k = −2

∴ =− +

⇒ = −

V

tV t

2 28

214

2

2( )

d. False. dV/dt = 2t − 28, so the water flowsout at 28 ft3/min only when t = 0. Forinstance, at t = 5, dV/dt = −18, whichmeans water flows out at only 18 ft3/min.So it takes longer than 7 min to emptythe tub.

e. 0 = (t − 14)2 ⇒ the tub is empty att = 14 min.

f.

14

100

V

t

g. See the solution to Problem C4 in ProblemSet 7-7.

6. The following data were gathered in the author’sclass in December 1994. Times t are in secondsand volumes V are in mL. Note that a burettereads the amount of fluid delivered, so youmust subtract the reading from 50 to find thevolume remaining. Use food coloring in thewater to make the liquid level easier to read.Read from the bottom of the meniscus (thecurved surface of the liquid).


Seconds Reading Volume

0 0 50

10 2.4 47.6

20 4.4 45.6

30 6.4 43.6

40 8.5 41.5

50 10.5 39.5

60 12.4 37.6

70 14.3 35.7

80 16.1 33.9

90 17.8 32.2

100 19.9 30.1

110 21.2 28.8

120 22.8 27.2

130 24.5 25.6

140 25.6 24.4

150 27.4 22.6

160 28.6 21.4

170 30.0 20.0

180 31.3 18.7

190 32.6 17.4

200 33.8 16.2

210 35.1 14.9

220 36.4 13.6

230 37.4 12.6

240 38.5 11.5

250 39.5 10.5

260 40.6 9.4

M M M

320 46.1 3.9

360 49.3 0.7

Using quadratic regression with these data,V = 0.000209255…t2 − 0.20964…t + 49.54… .The data and the equation can be plotted on thegrapher, as shown.

t

V

100

50

The volume does seem to vary quadratically withtime. Because there is still fluid in the burettewhen V = 0, the graph crosses the t-axis, unlikethe graphs in Problem 5 and Example 1. The

position of the vertex can be used to predict theposition of the stopcock and the time when thefluid would all be gone if the burette were ofuniform diameter all the way down to thestopcock. For the preceding data, the vertex is at

t = − − ≈0 20964

2 0 000209255500

.

( )( . )

K

K s

V ≈ −3.0 mL

So the stopcock should be found at a pointcorresponding to about 3 mL below the bottommark.

7. a. n = 1, k = 1, C = −3:

∴ = ⇒ = ⇒ = −∫ ∫ dy dx y dy y dx y x/ / ln | | 3

⇒ |y| = e x−3 = exe− 3 ⇒ y = ±0.04978…ex

1

1

x

y

b. n = 0.5, k = 1, C = −3:

∴ = ⇒ =−∫ ∫ dy dx y y dy dx/ . .0 5 0 5

⇒ = − ⇒ = −2 31

430 5y x y x. ( )2

Note: x ≥ 3 because y0.5 is a positive number.

3

1

x

y

c. n dy dx ky y dy k dx= − ⇒ = ⇒ =− ∫∫1 1/

⇒ = + ⇒ = ± +1

22 22y kx C y kx C

k C y x= = − ⇒ = ± −1 3 2 6,

5

3

x

y

n dy dx ky y dy k dx= − ⇒ = ⇒ =− ∫∫2 2 2/

⇒ = + ⇒ = +1

33 33 3y kx C y kx C

k C y x= = − ⇒ = −1 3 3 93,


5

3

x

y

d. For ndy

dxky y dy k dxn n> = ⇒ =− ∫∫1,

⇒ −−

= + >− −y

nkx C n

n( )

,1

11 because

so yn kx Cn

= −− ⋅ +−

1

11 ( ) ( )

which has a vertical asymptote at x = −C/kbecause the denominator equals zero for thispoint.

Note that the radical will involve a ± signwhen the root index is even (for example,when n is odd).

For , , : ( )n k C y x= = = − = − − −2 1 3 3 1

2

3

x

y

For n k C yx

= = = − = −± −

3 1 31

2 6, , :

2

3

x

y

Note that the graph shows two branches.

e. For so ndy

dxky k y kx C= = = = +0 0, , ,

a linear function. For k = 1, C = −3, y = x − 3.

2

3

x

y

8. dB/dt = c ( M − kB), where k and c are constants.dB

M kBc dt

k

k dB

M kBc dt

−= ⇒ − −

−=∫ ∫ ∫∫1

⇒ − − = +1

kM kB ct Cln | |

⇒ − = − −| |M kB e ekct kcC

⇒ − =

⇒ = −

−

−

M kB C e

Bk

M C e

kct

kct

1

11

( )

Use the initial condition B = 0 when t = 0.

0 = (1/k) ( M − C1e0) ⇒ C1 = M

∴ =BM

ke kct( – )–1

Use the initial condition kB = 80 whenB = 1000.

80 = k(1000) ⇒ k = 0.08

Use the initial condition dB/dt = 500 when t = 0.From dB/dt = c (M − kB), 500 = c (M − 0) ⇒c = 500/M.∴ particular equation is

B M e

B M e

M t

M t

= −= −

−

−( / . )[ ]

. [ ]

. /( / )

( / )

0 08 1

12 5 1

0 08 500

40

Assume various values of M:M = 1000: B = 12500(1 − e− 0.04 t)M = 5000: B = 62500(1 − e− 0.008 t )M = 10000: B = 125000(1 − e− 0.004 t )

250

100,000

M = 1000

M = 5000

M = 10000

500

B

t

As shown on the graph, the sales start outincreasing at the same rate (500 bottles/day). Ast increases, the number of bottles/day increases,approaching a steady state equal to 12.5M.

To find the break-even time, first find the totalnumber of bottles sold as a function of time. B isin bottles per day, so the total sales in x days,T(x), is

T x B dtx

( ) .= ∫0

Use, for example, M = $10,000/day.

T x e dt

t e

x e

x e

tx

t x

x

x

( ) ( )

( / . )

( )

[ ( )]

.

.

.

.

= −

= +

= + − −

= − −

−

−

−

−

∫ 125000 1

125000 1 0 004

125000 250 0 250

125000 250 1

0 004

0

0 004

0

0 004

0 004

[ ]

For selling prices of $0.25 and $0.50/bottle, thetotal numbers of dollars are

D25(x) = 31250[x − 250(1 − e− 0.004 x)]D50(x) = 62500[x − 250(1 − e− 0.004 x)]


The total amount spent on advertising is M ⋅ x,or A(x) = 10000x.The three graphs can be plotted by grapher. For$0.25/bottle, the break-even time is 207 days.For $0.50/bottle, the break-even time is 90 days(less than half!).

Dollars (millions)

x

2

1

20790

$0.50/bottle

$0.25/bottle

9. The differential equation is dT/dt = k(1200 − L),and L = h (T − 70), where h is a proportionalityconstant.

∴ dT/dt = k(1200 + 70h − hT )

dT h hT k dt/( ) 1200 70+ − = ∫ ∫(−1/h) ln |1200 + 70h − hT | = kt + C

ln |1200 + 70h − hT | = −kht − hC

|1200 + 70h − hT | = e− kht ⋅ e− hC

1200 + 70h − hT = C1e− kht ⇒

hT = 1200 + 70h − C1e− kht

T = 1200/h + 70 − (C1/h) ⋅ e− kht

Use T = 70 when t = 0.

70 = 1200/h + 70 − C1/h ⋅ e− kh⋅ 0

⇒ C1 = 1200

∴ T = 1200/h + 70 − 1200/h ⋅ e− kht

T = 70 + (1200/h)(1 − e− kht)

Substitute t = 0, L = 0, and dT/dt = 3 into theoriginal differential equation.3 = k(1200 − 0) ⇒ k = 0.0025

∴ T = 70 + (1200/h)(1 − e− 0.0025 ht)Using T = 96 when t = 10,26 = (1200/h)(1 − e− 0.025 h).Solving numerically gives h ≈ 11.7347… .

∴ equation is T ≈ 70 + 102.26…(1 − e− 0.02933…t ).

Time data for various temperatures can be foundby grapher or by substituting for T andsolving for t.

100

180°

170°160°140°

Never reaches

39 72 130

t

T

T t

140° 39 min

155° 61 min

160° 72 min

170° 130 min

180° Never!

The limit of T as t increases is 70 +102.26…(1 + 0), which equals 172.26…°. Thus,the temperature never reaches 180°. When theheater turns off, the differential equation becomesdT

dtkh T T C e kht= − − ⇒ = + −( ) .70 70 2

Using T = 160 at time t = 0 when the heaterturns off, T = 70 + 90e− 0.02933…t.To find the time taken to drop to 155°, substitute155 70 90 0 02933= + −e t. .K

Solving numerically or algebraically givest = 1.9… . Thus, it takes only 2 minutes for thetemperature to drop 5°! By contrast, from thepreceding table, it takes 11 minutes(t = 61 to t = 72 in the table) to warm back upfrom 155° to 160°.

The design of the heater is inadequate because ittakes much longer to warm up by a certainamount than it does to cool back down again.Near 172°, a slight increase in the thermostatsetting for the heater makes a great increase inthe time taken to reach that setting. For instance,it takes an hour (72 minutes to 130 minutes) towarm the 10 degrees from 160° to 170°. Theseinadequacies could be corrected most easily byadding more insulation. The resulting decrease inh would make the heater cool more slowly, heatup faster, and reach the 180 degrees it currentlycannot reach. Decreasing h would also reduce thepower consumption.

10. a. dP/dT = kP/T 2

dP P k dT T P kT C/ = / ⇒ = − +∫ ∫ −2 1ln | |

⇒ = ⇒ =− / + −| |P e P C ek T C k T1

/

b. 0 054 1293. = − /C e k

3 95 1343. = − /C e k

( . . ) ( )3 95 0 054 343 293/ = − / + /e k k

ln (3.95/0.054) = −k/343 + k/293k = / / − / + /

=[ ( . . )] ( )

.ln 3 95 0 054 1 343 1 293

8627 812641KFrom ln |P| = −kT −1 + C,C = ln 0.054 + 8627.812641…/293.C = 26.52768829… ⇒ C1 = e26.52768829…

P e e T= … − …/26 52768829 8627 812641. .

P e T= − /( . . )26 52768829 8627 812641K K


c.

Temperature T P Actual*

10 283 0.0190… 0.021

20 293 0.054 0.054

30 303 0.142… 0.133

40 313 0.354… 0.320

50 323 0.832… 0.815

60 333 1.85… 1.83

70 343 3.95 3.95

80 353 8.05… 7.4(meltingpoint)

90 363 15.7… 12.6

100 373 29.8… 18.5

110 383 54.6… 27.3

200 473 3972.1… 496.5*Source: Lange’s Handbook of Chemistry, 1952, p. 1476.

The function models the data well up to themelting point, but not above it. Thedifferences between the predicted and actualanswers are most likely due to the fact thatnaphthalene changes from solid to liquid at80°C; the constants for solid and liquidnaphthalene differ.Use initial conditions for T = 90, 110 as inpart b to get a better equation for the liquid:12.6 = C1e

− k/ 363

496.5 = C1e− k/ 473

k = [ln (496.5/12.6)]/(−1/473 + 1/363)= 5734.569702…

C = ln 12.6 + 5734.569702…/363= 18.33140949…

⇒ C1 = e18.33140949…

∴ P = e(18.33140949… −5734.569702…/ T )

With the new equation,

Temperature T P Actual

10 283 0.144… 0.021

20 293 0.289… 0.054

30 303 0.551… 0.133

40 313 1.01… 0.320

50 323 1.78… 0.815

60 333 3.03… 1.83

70 343 5.01… 3.95

80 353 8.05… 7.4(meltingpoint)

90 363 12.6 12.6

100 373 19.2… 18.5

110 383 28.7… 27.3

200 473 496.5 496.5

So the new equation models the data abovethe melting point, but not below it.

d. Using the equation for liquid naphthalene,760 = e(18.33140949… −5734.569702…/ T ) ,

ln 760 = 18.33140949… − 5734.569702…/T

T .= =5734 569702

18 33140949 760490 214

.

. – ln

K

KK

About 490 K, or 217°C (actual: 218°C)

e. Answers will vary.

Problem Set 7-4Q1. y ′ = 5x4 Q2. y ′ = 5x ln 5

Q3. (1/8)x 8 + C Q4. 7 x/ln 7 + C

Q5. y ′ = −y/x or y ′ = −3x− 2

Q6. 87.5

Q7. Q8.

1

1

y

x1

1

x

y

Q9. g(5) − g(1) Q10. E

1. a. dy/dx = x/(2y)At (3, 5), dy/dx = 3/10 = 0.3.At (−5, 1), dy/dx = −5/2 = −2.5.On the graph, the line at (3, 5) slopes upwardwith a slope less than 1. At (−5, 1) the lineslopes downward with a slope much steeperthan −1.

b. The figure looks like one branch of ahyperbola opening in the y-direction. (Thelower branch shown on the graph is also partof the solution, but students would not beexpected to find this graphically.)

y

5

5

x(–5, 1) (1, 2) (5, 1)

(3, 5)

c. See graph in part b. The figure looks like theright branch of a hyperbola opening in thex-direction. (The left branch is also part of the


solution, but students would not be expectedto find this graphically.)

d.dy

dx

x

yy dy x dx y x C= ⇒ = ⇒ = +∫ ∫2

21

22 2

x = 5, y = 1 ⇒ C = 1 − 12.5 = −11.5

By algebra, x2 − 2y2 = 23. This is theparticular equation of a hyperbola opening inthe x-direction, which confirms theobservations in part c.

2. At (3, 3), dy/dx = 0.1(3) = 0.3, which isreasonable because the slope is positive and lessthan 1. At (0, −2), dy/dx = 0.1(−2) = −0.2, whichis reasonable because the slope is negative andless than 1 in absolute value. The next graphshows the two particular solutions. For the first,y (6) ≈ 4.0. For the second, y (6) ≈ −3.6.dy

dxy y C e x= ⇒ =0 1 1

0 1. .

For (3, 3), the particular solution isy = 2.2224…e0.1 x, giving y(6) = 4.0495… .

For (0, −2), the particular solution isy = −2e0.1 x, giving y(6) = −3.6442… .

Both graphical answers are close to these actualanswers.

1 2 3 4 5 6–1–2

1

2

3

4

–1

–2

–3

–4

y

x

3. a. At ( , ), . .3 23

2 20 75

dy

dx= − = −

( )( )

At ( , ), ,1 01

2 0

dy

dx= −

( )( ) which is infinite.

1 2 3–1–2–3

1

2

–1

–2

y

x

b. See the graph from part a. The figuresresemble half-ellipses.

c.dy

dx

x

y= −

2

2y dy = −x dx

2 y dy x dx = −∫ ∫

y x C2 21

2= − +

( ) .− = − + ⇒ =11

21 1 52 2( ) C C

y2 = 1.5 − 0.5x2

y x . . = − −1 5 0 5 2

(Use the negative square root because of theinitial condition.)

The graph agrees with part b.

From the next-to-last line, add 0.5x2 to bothsides, getting 0.5x2 + y2 = 1.5, which is theequation of an ellipse because x2 and y2 havethe same sign but unequal coefficients.

4. a. At ( , ), ( ) .3 1 3 1 1 0dy

dx= − =

At ( , ), ( ) .1 2 1 1 2 1dy

dx= − = −

At ( , ), ( ) .0 1 0 1 1 0− = + =dy

dx

1 2 3–1–2–3

1

2

3

–1

y

x

b. See the graph from part a. Both graphs have ahorizontal asymptote at y = 1.

c.dy

dxx y= −( )1

dy

yx dx

1–=

dy

yx dx

1–=∫ ∫

−ln |1 − y| = 0.5x2 + C

1 0 5 2

− = ± ⋅− −y e ex C.

y C e x= + −1 10 5 2. (C1 can be positive or

negative.)

−1 = 1 + C1e0 ⇒ C1 = −2

∴ = − − .y e x1 2 0 5 2

The grapher confirms the graph in part b.

As | | , ..x e x→ ∞ →−0 5 2

0 So y → 1,which agrees with the horizontal asymptoteat y = 1.


5.

0.5 1 1.5 2–0.5–1–1.5–2

0.5

1

1.5

2

–0.5

–1

–1.5

–2

y

x

6.

0.5 1 1.5 2–0.5–1–1.5–2

0.5

1

1.5

2

–0.5

–1

–1.5

–2

y

x

7.

0.5 1 1.5 2–0.5–1–1.5–2

0.5

1

1.5

2

–0.5

–1

–1.5

–2

y

x

8.

0.5 1 1.5 2–0.5–1–1.5–2

0.5

1

1.5

2

–0.5

–1

–1.5

–2

y

x

9. a.

x

y

(3, 2)

(1, –2)

b.dy

dxxy= −0 2.

Evidence: At (1, 1) the slope was given tobe −0.2, which is true for this differentialequation. As x or y increases from thispoint, the slope gets steeper in the negativedirection, which is also true for thisdifferential equation. In Quadrants I and III theslopes are all negative, and in Quadrants IIand IV they are all positive. (Note: The

algebraic solution is y Ce x= −0 1 2. .)

10. a. Initial condition (0, 2)

(0, –5)

(0,–2.5)

y

(0, 2)

x5

5

b. See the graph in part a with initial condition(0, −5). The graph goes toward −∞ in they-direction instead of toward +∞.

c. If a ruler is aligned with the slope lines, thelines that form a straight line are the onescrossing the y-axis at −2.5 with slope −1/2.(In courses on differential equations, studentswill learn that the given equation is a first-order linear equation that can be solved usingan integrating factor. The general solution isy = Ce0.2 x − 0.5x − 2.5. For C = 0, the curveis the line y = −0.5x − 2.5, which intersectsthe y-axis at (0, −2.5).)


11. a. Initial condition (0, 2)

P

t

(0, 18)

(4, 2)(0, 2)

10.5

b. See the graph in part a with initial condition(4, 2). The graph is the same as that in part abut shifted over 4 months. This behavior isto be expected because dP/dt depends only onP, not on t, and both initial conditions havethe same value of P.

c. See the graph in part a with initial condition(0, 18). The population is decreasing to thesame asymptote, P = 10.5, as in parts aand b.

d. The asymptote at P = 10.5 indicates that theisland can sustain only 1050 rabbits. If thepopulation is lower than that, it increases. Ifthe population is higher than that, itdecreases. The number 10.5 is a value of Pthat makes dP/dt equal zero. Note that there isanother asymptote at P = 0, which alsomakes dP/dt equal zero.

12. a. dv/dt = 32.16 − 0.0015v2

The slope at (5, 120) appears to be about 1,but dv/dt actually equals 32.16 −0.0015(120)2 = 10.56. The answers aredifferent because the graph is scaled by afactor of 10.

b. Initial condition (0, 0)

t

v

(5,120)

(0,180)

146.4...

(0, 0) (5, 0)

50

c. Terminal velocity occurs when dv/dt = 0.0 = 32.16 − 0.0015v2

v = (32.16/0.0015)1/2 = 146.424… ≈ 146 ft/s

The graph shows this velocity for timesabove about 15 seconds.

d. See the graph in part b with initial condition(5, 0). The graph is identical to the one inpart b except shifted 5 seconds to the right.This behavior is to be expected because thedifferential equation is independent of t.

e. See the graph in part b. This graph decreasesto the terminal velocity because the diverstarts out going faster.

f. Similarities include: Both models have ahorizontal asymptote that the particularsolutions approach from above or below.Both models decrease rapidly and graduallylevel off for values above the asymptoticlimit.Differences include: For values below theasymptotic limit, one model starts with rapidincrease and gradually slows its growth,whereas the other starts with a slow increasethat becomes more rapid growth beforeslowing toward the asymptote.

13. a. mamg

r= 2 By hypothesis

dv

dt

g

r= 2 Divide by m; a

dv

dt= .

dv

dr

dr

dt

g

r⋅ = 2 Chain rule

dv

drv

g

r⋅ = 2 v

dr

dtr= =( )distance

dv

dr

g

r v= 2 Divide by v.

b.dv

dr( , )5 2 1 2488 .= −

dv

dr( , )1 10 6 244 .= −

dv

dr( , )10 4 0 1561 .= −

These slopes agree with those shown.c. Initial condition (r, v) = (1, 10)

From the graph, the velocity is zero at r ≈ 5.So the spaceship is about 4 Earth radii, orabout 25,000 km, above the surface.

r

v(1,18)

5

14.11...

(1,12)

(2,10)(1,10)

6.12...

4.37...(10,4)

(5,2)


The precise value of r can be foundalgebraically.dv

dr r vv dv

rdr= ⇒ = ∫∫– . – .62 44 62 44

2 2

⇒ = +v

rC

2

2

62 44.

For the solution through (1, 10), C = 50 −62.44 = −12.44, so the ship starts fallingwhen v = 0 at r = 62.44/12.44 ≈ 5.

d. See the graph in part c with initial condition(r, v) = (1, 12). The graph levels off between4 and 5 km/s. The precise value of v can befound algebraically.

Cv

r= − = ⇒ = +72 62 44 9 56

2

62 449 56

2

. . ..

Because r > 0, v is never zero, so thespaceship never stops and falls back.As r approaches infinity, v2/2 approaches9.56, and thus v approaches

( )( . )2 9 56 4 37= …. km/s.

e. See the graph in part c with initial condition(r, v) = (1, 18). The graph levels off atv ≈ 14 km/s. Here the spaceship loses about4 km/s of velocity, whereas it loses 7 or8 km/s when starting at 12 km/s. Both caseslose the same amount of kinetic energy,which is proportional to v2 (the change in v2

is the same in both cases). The precise valueof v can be found algebraically as in part d.For the solution through (1, 18),C = 162 − 62.44 = 99.56. As r → ∞,v → = …( )( . )2 99 56 14 11. km/s.

f. See the graph in part c with initial condition(r, v) = (2, 10). The graph levels off atabout 6 km/s, so the spaceship does escape.Alternatively, note that the solution through(2, 10) lies above the solution through(1, 12). The precise value of v can be foundalgebraically as in parts d and e. For thesolution through (2, 10), C = 50 − 31.22 =18.78. As r → ∞, v → =( )( . )2 18 786.12… km/s.

14. See the Programs for Graphing Calculatorssection of the Instructor’s Resource Book.

Problem Set 7-5Q1. ky Q2. y = Ce3x

Q3. 4.8 Q4. 100

Q5. −ln |1 − v| + C Q6. sec x tan x

Q7.

1

1x

y

y' y'

Q8. 3x2y5 + 5x3y4y′ = 1 + y′Q9. continuous Q10. A

1. a. dy = −(x/y) dyFor (1, 3), dy = −(1/3)(0.5) = −0.1666… ,so new y ≈ 3 − 0.1666… = 2.8333… atx = 1.5.For (1.5, 2.8333…), dy =−(1.5/2.8333…)(0.5) = −0.2647… ,so new y ≈ 2.8333… − 0.2647… = 2.5686…at x = 2.

x y

0 3.2456…

0.5 3.1666…

1 3

1.5 2.8333…

2 2.5686…

2.5 2.1793…

3 1.6057…

The Euler’s y-values overestimate the actualvalues because the tangent lines are on theconvex side of the graph and the convex sideis upward.

b. dy = −(x/y) dy

y dy x dx= −∫∫0.5y2 = −0.5x2 + C0.5(32) = −0.5(12) + C ⇒ C = 50.5y2 = −0.5x2 + 5

y x= −10 2 (Use the positive square root.)

At x y= = =3 10 3 12, – .

The particular solution stops at the x-axisbecause points on the circle below the x-axiswould lead to two values of y for the samevalue of x, making the solution not afunction.

The Euler’s value of 1.6057… overestimatesthe actual value by 0.6057… .

2. a. dy = (x/y) dyFor (1, 2), dy = (1/2)(0.5) = 0.25, so newy ≈ 2 + 0.25 = 2.25 at x = 1.5.For (1.5, 2.25), dy = (1.5/2.25)(0.5) =0.3333… , so new y ≈ 2.25 + 0.3333… =2.5833… at x = 2.


x y

0 1.6071…

0.5 1.75

1 2

1.5 2.25

2 2.5833…

2.5 2.9704…

3 3.3912…

The Euler’s y-values underestimate the actualvalues because the tangent lines are on theconvex side of the graph and the convex sideis downward. The error is greater at x = 0because the graph is more sharply curvedbetween x = 0 and x = 1 than it is betweenx = 1 and x = 3.

b. dy = x/y dy

y dy x dx= ∫∫0.5y2 = 0.5x2 + C0.5(22) = 0.5(12) + C ⇒ C = 1.50.5y2 = 0.5x2 + 1.5

y x= +2 3 (Use the positive square root.)

At x y= = + = = …0 0 3 3 1 7320, . .

The particular solution stops at the x-axisbecause points on the circle below the x-axiswould lead to two values of y for the samevalue of x, making the solution not afunction.

The Euler’s value of 1.6071… underestimatesthe actual value by 0.1249… unit.

3. dx = 0.2. Make a table showing values of dy =0.2(dy/dx) and new y = old y + dy.

x dy/dx dy y

2 3 0.6 1

2.2 5 1.0 1.6

2.4 4 0.8 2.6

2.6 1 0.2 3.4

2.8 −3 −0.6 3.6

3 −6 −1.2 3.0

3.2 −5 −1.0 1.8

3.4 −3 −0.6 0.8

3.6 −1 −0.2 0.2

3.8 1 0.2 0.0

4 2 0.4 0.2

1 2 3 4

1

2

3

4

y

x

You cannot tell whether the last value of y is anoverestimate or an underestimate because theconvex side of the graph is downward in someplaces and upward in other places.

4. dx = 0.3. Make a table showing values of dy =0.3(dy/dx) and new y = old y + dy.

x dy/dx dy y

1 −3 −0.9 2

1.3 −2 −0.6 1.1

1.6 −1 −0.3 0.5

1.9 0 0 0.2

2.2 1 0.3 0.2

2.5 2 0.6 0.5

2.8 3 0.9 1.1

3.1 4 1.2 2

3.4 5 1.5 3.2

3.7 6 1.8 4.7

3.9 7 2.1 6.5

1 2 3 4

1

2

3

4

5y

x

The approximate values of y underestimate theactual values of y because the convex side of thegraph is down.



7. a. and b. dy

dxxy= −0 2.

x

y

(3, 2)

(1, –2)


8. a. and b. dy

dxx y= − +0 1 0 2. .

x

5

5

y

(0, 2)

(0, 4)

c. When the graph is observed, the slope linesseem to follow a straight path using (0, 2.5)as an initial condition. Euler’s methodconfirms this.(In differential equations, students will learnhow to solve such first-order linears bymultiplying both sides by the integratingfactor e− 0.2x . The general solution isy = Ce0.2x + 0.5x − 2.5. For C = 0, theparticular solution is y = 0.5x + 2.5.)

9. a. Using dr = 0.6, v(13.6) ≈ 0.1414… andv(14.2) ≈ −1.2900… , so the spacecraft seemsto reverse direction somewhere between thesetwo values of r, as shown in the graph inpart b.

b. Using dr = 0.1, v(20) ≈ 4.5098… , and thevalues are leveling off, as shown in thegraph.

dr = 0.1

dr = 0.6

10 20

10

20 v

r

Actual

c.dv

dr r v= – .62 44

2

v dv r dr= − −∫∫ 62 44 2.

0 5 62 442 1. .v r C= +−

0 5 12 62 44 1 9 562 1. ( ) . ( )= + ⇒ =− C C .0 5 62 4 9 562 1. . .v r= +−

v r= +124 88 19 121. .–

When r = 20, v = 5.0362… .Because the graph is concave up (convex sidedown), the Euler’s solution underestimatesthe actual velocity. The first increment, wherethe graph is so steep, makes a large error thataccumulates as the iterations continue,

putting the graph into a region of the slopefield from which the spacecraft would notescape Earth’s gravity.

d. Let v1 be the initial velocity at r = 1.Solving for C gives0 5 62 441

2. .v C= +C v= −0 5 62 441

2. .

If v1 2 62 44< ( . ), then C is negative,

making v r C= +128 88 1. – an imaginarynumber when r is large enough. If

v1 > 2 62 44( . ), then C is positive, making va positive real number for all positive

values of r. (The asymptote is v C= .)

10. a. v(2) = 61.6831… , v(4) = 106.2850… ,v(6) = 129.7139… , v(8) = 139.9323… ,v(10) = 143.9730… , v(20) = 146.4066…These values will be overestimates becausethe graph is concave down (convex side up),so the Euler’s tangent lines will be above theactual graph, as in the next graph.

v

t

Actual

Euler

100

10 20

200

b. v(2) = 157.7979… , v(4) = 150.5128… ,v(6) = 147.9234… , v(8) = 146.9777… ,v(10) = 146.6290… , v(20) = 146.4254…These values will be underestimates becausethe graph is concave up (convex side down),so the Euler’s tangent lines will be below theactual graph.

c.dv

dtv= ⇔ = ⇔0 0 0015 32 162. .

v = =32 16

0 0015146 42404

.

.. (store)K

The terminal velocity is about 146.4 ft/s.

d. The table shows the values of v and theerrors, Euler minus actual. The errors increaseonly for a while, then approach zero becauseboth the Euler’s solution and the actualsolution approach the same asymptote. (It isnot always true that values farther from the


initial condition have a greater error in theirEuler’s approximation.)

t Euler’s v Actual v Error

2 61.6831… 60.4791… 1.2040…

4 106.2850… 103.3298… 2.9552…

6 129.7139… 126.8383… 2.8756…

8 139.9323… 137.9573… 1.9749…

10 143.9730… 142.8466… 1.1264…

20 146.4066… 146.3792… 0.0274…

The graph in part a shows the Euler’ssolutions from parts a and b, and the actualsolution from part c, thus confirminggraphically the numerical answers to thisproblem.

11. a. For x ≤ 5, the radicand 25 − x2 is non-negative, giving a real-number answer for y.For x > 5, the radicand is negative, giving noreal solution.

b. The slope field shows that the graph will beconcave up (convex side down), making theEuler’s tangent lines lie below the graph,leading to an underestimate.

At x y= = − = − …4 9 0 6 25 4 9 0 59692. , . . .– .

The Euler’s solution at x = 4.9 is−0.8390… , which is an underestimatebecause −0.8390… < −0.5969… but isreasonably close to the actual value.

c. The Euler’s solutions for the given points are

x y

5.1 −0.3425…

5.2 0.1935…

5.3 −0.7736…

6.6 26.9706…

From 5.1 to 5.2,

dy = − ( )( . )

( )(– . )( . )

9 5 1

25 0 34250 1

K

= 0.5360… , indicating that the graph is stilltaking upward steps.

From 5.2 to 5.3,

dy = − ( )( . )

( )( . )( . )

9 5 2

25 0 19350 1

K

= −0.9672… , indicating that the graph takesa relatively large downward step. The signchange in dy happens whenever the priorEuler’s y-value changes sign. The graph startsover on another ellipse representing a differentparticular solution.

d. Euler’s method can predict values that areoutside the domain, which are inaccurate.


Problem Set 7-6Q1. definition of definite integral

Q2. fundamental theorem of calculus

Q3. definition of indefinite integral

Q4. the intermediate value theorem

Q5. Rolle’s theorem

Q6. the mean value theorem

Q7. the chain rule Q8. general

Q9. particular Q10. initial

1. a. dB/dt is proportional to B, which means thatthe larger the population is, the faster itgrows. But dB/dt is also proportional to(30 − B)/30, which means that the closer B isto 30, the slower it grows. dB/dt > 0 when0 < B < 30 because when the population isless than 30 million the population willincrease until it reaches the carrying capacity.dB/dt < 0 when B > 30 because when thepopulation is greater than 30 million, thepopulation will decrease until it reaches thecarrying capacity.

b.B

t

Euler

Actual

10 20 30 40

10

20

30

40

For the initial condition (0, 3), the populationgrows, leveling off at B = 30. For the initialcondition (10, 40), the population dropsbecause it is starting out above the maximumsustainable value (carrying capacity).

c.

t B

0 3

10 13.8721…

20 26.4049…

30 29.5565…

40 29.9510…

See the graph in part b. The graph shows thatthe Euler’s points and graphical solution areclose to each other.


d.dB

dtB

B= ⋅0 2130

30.

–

30

300 21

B BdB dt

( – )= .

Separate the variables.

1 1

300 21

B BdB dt+

=

–.

By partial fractions(see Example 1).

ln |B| − ln |30 − B| = 0.21t + CWhy the “−” sign?

−ln |B| + ln |30 − B| = −0.21t − CTo simplify latersteps.

30

27

39

10 21

1

10

– B

BC e

C e

C e C

t

C

=

= ±

= ⇒ =

−

−

.

1

Substitute the initialcondition (0, 3) tofind C1.

309

30 9

0 21

0 21

– B

Be

B Be

t

t

=

− =

−

−

.

.

Be

Btt Solve for explicitly

in terms of .=

+ −30

1 9 0 21.

At , 26.4326 .t B

e= =

+=20

30

1 9 4 2– . K

The Euler’s value, B ≈ 26.4049… , is veryclose to this precise value.

e.d

dB

dB

dtB

= − +0 014 30 0 007. ( . )

Derivative = 0 if −0.014B + 30(0.007) = 0,which is true if and only if B = 15. This value is halfway between B = 0 andB = 30.

1530

1 9 0 21=+ e t– . ⇒ t ≈ 10.4629…

The point of inflection is (10.4629… , 15).

2. a. A logistic function is reasonable because thenumber of houses grows at an increasing ratefor a while, then slows down as the numberapproaches 120, the “carrying capacity” of thesubdivision.

b.dy

dxy

y= ⋅0 9120

120.

–

120

1200 9

y ydy dx

( – )= .

1 1

120y y+

–

dy = 0.9 dx

ln |y| − ln |120 − y| = 0.9x + C−ln |y| + ln |120 − y| = −0.9x − C120

10 9

1– y

yC e C ex C= = ±− −.

115

5= C1e

0 ⇒ C1 = 23

Substitute the initialcondition (0, 5) tofind C1.

12023

120 23

0 9

0 9

– y

ye

y ye

x

x

=

− =

−

−

.

.

ye x=

+120

1 23 0 9– .

The graph confirms that the particularsolution follows the slope lines.

y (houses)

50

5 10

100

x (years)

c. 70% of 120 is 84.

84

120

1 234 42530 9=

+⇒ ≈

exx– . . ,K or about

4 years 5 months Solve numericallyfor x.

1 lot left means 119 lots built on.

119120

1 23 0 9=+

⇒ ≈ …e

xx– . 8.7940 , or about

8 years 10 months.

d.dy

dxy

yy y= ⋅ =0 9

120

120

0 9

120120 2.

– .( – )

d

dy

dy

dxy

= 0 9

120120 2

.( – )

The derivative is zero if 120 − 2y = 0, whichis true if and only if y = 60. This value ishalfway between y = 0 and y = 120.If y < 60, the derivative is positive, so dy/dxis increasing. If y > 60, the derivative isnegative, so dy/dx is decreasing. Therefore,dy/dx is a maximum when y = 60, and thenumber of houses is increasing the mostrapidly at this point of inflection.


3. a.dy

dxky

M y

M= ⋅ –

0 5 10 100 5

10 10. = ⇒ =k

MM

k

M M( )( – )

.

( – )

1 1 24 241 1

24 24. = ⇒ =k

MM

k

M M( )( – )

.

( – )

∴−

=−

0 5

10 10

1 1

24 24

.

( )

.

( )M MEliminate k byequating the twovalues of k/M.

12(M − 24) = 11.0(M − 10)12M − 11M = 288 − 110 ⇒ M = 178

Solve for M.

kk

178

0 5

10 178 10

89

1680=

−⇒ = =.

( )0.05297… (Store this.)Ajax expects to sell 178,000 CDs based onthis mathematical model.

b.dy

dxy

y= ⋅ ⋅ −89

1680

178

178

y (thousand CDs)

50 100

100

178

200

x (days)

The slope field has horizontal slope lines atabout y = 178, thus confirming M = 178.

c. The general solution is yM

ae kx=+ −1

.

Substitute M = 178 and k = 89/1680 =0.05297… and the initial condition y = 10 atx = 0.

10178

116 80=

+⇒ =

aea .

The equation is y

e x=+ −

178

1 16 8 0 05297. . K .

See the graph in part b. The graph follows theslope lines.

d. At x = 50, y = 81.3396… .At x = 51, y = 83.6844… .83.6844… − 81.3396… = 2.35447…They expect to sell about 2354 CDs on the51st day.

e. The point of inflection is halfway betweeny = 0 and y = 178, that is, at y = 89.

89

178

1 16 8 0 05297=+ −. .e xK

Solving numerically gives x ≈ 53.2574… ,or on the 54th day.

4.dy

dxky

M y

M= ⋅ −

M dy

y M yk dx

( )−= ∫∫

1 1

y M ydy k dx+

−

= ∫∫See Section 9-7 for aquick way to resolveinto partial fractions.

ln |y| − ln |M − y| = kx + CThe differential of thesecond denominatoris −dy.

−ln |y| + ln |M − y| = −kx − C

lnM y

ykx C

− = − −

M y

ye e ekx C C kx− = = ⋅− − − −

M y

ye e C e C eC kx kx C− = ± ⋅ = = ±− − − −

1 1

M − y = C1ye− kx

y + C1ye− kx = M

yM

C e

M

aekx kx=+

=+− −1 11

a = C1, Q.E.D.

5. a. At t = 5.5, F ≈ 1.7869… ≈ 2 fish left.At t = 5.6, F ≈ −11.0738… , meaning nofish are left.

The fish are predicted to become extinct injust over 5.5 years.

Part a, dt = 0.1

F (fish)

t (years)

Part b

Parts c and d

5 10

200

500

1000

b. See the graph in part a with initial condition(3, 1200), showing that the fish populationwill decrease because the initial condition isabove the 1000 maximum sustainable.

c. See the graph in part a with initial condition(0, 300), showing that the population risesslowly at first, then faster, eventually


slowing down as the population approachesthe 1000 maximum sustainable (carryingcapacity).

d. Let F = y + 200.dy

dt

y y= ⋅ ⋅ −130

200

800

1000200000

800130

y ydy dt

( )−= ∫∫

250 250

800130

y ydy dt+

−

= ∫∫See Section 9-7 forquick partial fractions.

250 ln |y| − 250 ln |800 − y| = 130t + CWhy “−” ?

800 13 25− = ⋅− −y

ye et C( / )

8001

13 25− = −y

yC e t( / )

Substitute y = 100 (F = 300) when t = 0.800 100

10071

01

− = ⇒ =C e C

8007 13 25− = −y

ye t( / )

ye t=

+ −800

1 7 13 25( / )

Fe t=

++−

800

1 720013 25( / )

See the graph in part a, showing that thesketch from part c reasonably approximatesthis precise algebraic solution.

6. Answers will vary. Here is a typical run with aclass of 25 people.

x N

0 1

1 3

2 6

3 13

4 21

5 25

6 25

7 25

8 25

Logistic regression gives

N

e x=+ −

25 5083

1 43 1120 1 3032

.

. .

K

K K

The logistic function fits reasonably well (asshown in this graph), especially if you useseveral values of the maximum number ofpeople as shown in the table.

1 2 3 4 5 6 7 8

5

10

15

20

25N

x

7. a. and b.

Year P ∆P/∆t (∆P/∆t)/P

1940 131.7

1950 151.4 2.38 0.01571…

1960 179.3 2.59 0.01444…

1970 203.2 2.36 0.01161…

1980 226.5 2.275 0.01004…

1990 248.7

You can’t find ∆P/∆t for 1940 and 1990because you don’t know values of P bothbefore and after these values.

c. Using linear regression on the values of(∆P/∆t)/P without round-off gives

10 02802596 0 0000792747

P

P

tP

∆∆

≈ …− …. . .

The correlation coefficient is r = −0.98535… .For the other types of regression:r = −0.978… for logarithmicr = −0.981… for exponentialr = −0.971… for powerThus, a linear function fits best because r isclosest to −1.

d.1

0 02802596 0 0000792747P

dP

dtP≈ …− …. .

⇒ = …− …dP

dtP P( . .0 02802596 0 0000792747 )

e.

100

500P

0

t

–50 50

Stable population at 353.5 million


f.

Year t Euler Actual* Euler**

1890 −50 44.6… 62.9 46.1…

1900 −40 56.9… 76.0 58.3…

1910 −30 71.7… 92.0 72.9…

1920 −20 89.2… 105.7 90.1…

1930 −10 109.3… 122.8 109.8…

1940 0 131.7 131.7 131.7

1950 10 155.4… 151.4 155.0…

1960 20 180.1… 179.3 179.2…

1970 30 204.7… 203.2 203.5…

1980 40 228.2… 226.5 226.9…

1990 50 249.9… 248.7 248.8…

2000 60 269.3… 281.4 268.6

2010 70 286.1…

2020 80 300.2…

2030 90 311.8…

2040 100 321.1…*Data from The World Book Encyclopedia .

**Note that although linear regression gives the“best” fit for (∆ P/∆ t ) /P versus P , actually plottingthe graph shows that the data point for 1960 isconsiderably out of line.

0.015

0.014

0.013

0.012

0.011

150 200

(∆P/∆t)/P

P

1950

1960

1980

1970

Using the two endpoints, 1950 and 1980,gives (∆P/∆t)/P = 0.002716… −0.00007557…P. Using this equation givespopulations much closer to the actual onesfor the given years, as shown in therightmost column of the table in part f. Thisis, of course, no guarantee that the latermodel fits any better in the future than theformer one.

g. The population growth rate is zeroif dP/dt = 0.LetP(0.02802596… − 0.0000792747…P) = 0.P = 0 or P = (0.02802596…)/(0.0000792747…) = 353.5…Predicted ultimate population ≈ 353.5 millionDifferential equation: P = 353.5… makesdP/dt = 0.Graph: P = 353.5… is a horizontalasymptote.

h. See the graph in part e. Data do follow thesolution.

i. Sample answer: The predicted populationsagree fairly well with the data for the sixgiven years. The fit is exact for 1940 becausethis point was used as an initial condition.For the other five years, the predictedpopulations are a bit higher than the actualpopulation.

j. Actual data are given in the table in part f.

k. The predicted population for 2010 from part fis 286.1… million. Using 486.1 million asan initial condition in 2010 gives thefollowing predictions:

Year t Euler

2010 70 486.1…

2020 80 444.5…

2030 90 417.7…

2040 100 399.7…

2050 110 387.1…

The logistic model predicts that thepopulation will drop, approaching theultimate value of 353.5 million from above.This behavior shows up in the slope field ofpart e because the slopes are negative forpopulations above 353.5.

8. a.1

10 10

10

10y y

A

y

B

y

A y By

y y( – ) –

( – )

( – )= + = +

The numerator of the first fraction mustequal that of the last fraction for all valuesof y. That is, 1 = 10A − Ay + By. Theconstant and linear coefficients on the leftmust equal the corresponding ones on theright. Thus, 1 = 10A and 0 = −Ay + By.So A = B = 0.1.

b.1

10

0 1 0 1

10y ydy

y ydy

( – )

. .

–= +

∫∫

= −

∫0 1

1 1

10. ,

y ydy

– which equals 3 dx.∫

∴ 0.1(ln |y| − ln |y − 10|) = 3x + C

ln–

ln–y

y

y

yx C

10

1030 10= − = +

y

y ye x C–10

110 30 10= − = − +( )

101 10 30

ye eC x= ± − −

yke

k exC=

+= ± −10

1 3010

– , where , Q.E.D.


c.dP

dtP P= …− …( . . )0 02802 0 00007927

= 0.00007927…P(353.5… − P)

1

353 50 00007927

P PdP dt

( . – )KK= ∫∫ .

= −

∫1

353 5

1 1

353 5. – .K KP PdP

= ∫0 00007927. K dt

ln |P| − ln |P − 353.5…|= 353.5…(0.00007927…t + C)

ln

– .ln

– .P

P

P

P353 5

353 5

K

K= −

= 0.02802…t + 353.5…C

P

Pe t C– .353 5 0 02802 353 5K = − … + …( . . )

353 51 0 02802. K

Pke t= + ….

P

ke t=+

353 5

1 0 02802

.– . ...

K

For the initial condition t = 0, P = 131.7,

k = − = …353 5

131 71 1 684

.

.

K. .

d.

Year t Algebraic Euler Actual

1940 0 131.7 131.7 131.7

1950 10 155.5… 155.4… 151.4

1960 20 180.2… 180.1… 179.3

1970 30 204.7… 204.7… 203.2

1980 40 228.2… 228.2… 226.5

1990 50 249.8… 249.9… 248.7

The two methods of evaluating themathematical model agree almost perfectly.However, the fact that they agree with eachother is no guarantee that they will fit the realworld as closely as they match each other.

9.dR

dtk R

dR

Rk dt R k t C= ⇒ = ⇒ = +1 1 1ln | |

⇒ = ⇒ =| |R e e R C eC k t k t1 11

R is increasing because k1 > 0.

10.dF

dtk F

dF

Fk dt= − ⇒ = −2 2

⇒ = − +ln | |F k t C2

⇒ = ⇒ =− −| |F e e F C eC k t k t2 22

F is decreasing because −k2 < 0.

11.dR

dtk R k RF= −1 3

dF

dtk F k RF= − +2 4

12.dF

dR

dF dt

dR dt

k F k RF

k R k RF= = +/

/

–

–2 4

1 3

The dt cancels out.

13. R = 70, F = 15

⇒ = + ⋅

⋅= =dF

dR

– .

– .

..

15 0 025 1050

70 0 04 1050

11 25

280 4017K

14. The slope at (70, 15) is about 0.4.

100

50

F

R(70, 15)

At R FdR

dt

dF

dt= = = =70 15 28 11 25, , , and . ,

which are both positive. So both populations areincreasing and the graph starts up and to theright.

15. The populations vary periodically and the graphis cyclical. The fox population reaches itsmaximum 1/4 cycle after the rabbit populationreaches its maximum.

16. Neither population changes when dR/dt =dF/dt = 0.dF/dt = 0 ⇔ F = 0 or R = 1/0.025 =40 (4000 rabbits)dR/dt = 0 ⇔ R = 0 or F = 1/0.04 = 25 foxes

17. Assume that dF/dt still equals −F + 0.025RF.dF

dR

dF dt

dR dt

F RF

R RF R= = +/

/

– .

– . – .

0 025

0 04 0 01 2

R FdF

dR= = ⇒ =

−= −70 15

11 25

210 5357 and

.. K

18.

100

50

F

R(70, 15)

Note that the slope at (70, 15) is now negative.

19. The populations now spiral to a fixed point. Therabbit population stabilizes at the same value asin Problem 16, R = 40 (4000 rabbits), which issurprising. The stable fox population decreasesfrom 25 to 15.


20. Assume that dF/dt still equals = −F + 0.025RF.

dF

dR

dF dt

dR dt

F RF

R RF R= = +/

/

– .

– . – . –

0 025

0 04 0 01 102

R F

dF

dR= = ⇒ =

−= −70 15

11 25

310 3629 and

.. K

21.

100

50

F

R(70, 15)

(70, 30)

Note that the slope at (70, 15) is about −0.4.

22. The fox and rabbit populations spiral toward afixed point. Again, and even more surprisingly,the rabbits stabilize at R = 40 (4000). But thestable fox population is reduced to 8 or 9. Alongthe way, the model shows that the foxes arereduced to about 1, thus becoming in danger ofextinction!

23. See the graph in Problem 21 with initialcondition (70, 30). With this many foxes andhunters chasing rabbits, the rabbits becomeextinct. At this point, the foxes have beenreduced to just 5. After the rabbits becomeextinct, the foxes decrease exponentiallywith time, eventually becoming extinctthemselves.



R1. P(t) = 35(0.98 t )P′(t) = 35(0.98t) ln 0.98

t P(t) P′(t) P′(t)/P(t)

0 35 −0.7070… −0.2020…

10 28.5975… −0.5777… −0.2020…

20 23.3662… −0.4720… −0.2020…

P t

P t

t

t

′ = =( )

( )

( . ) ln .

( . )ln

35 0 98 0 98

35 0 980 98.

= −0.2020… , which is a constant, Q.E.D.

R2. a. V = speed in mi/h; t = time in sdV

dtkV=

b.dV

Vk dt= ∫∫

ln |V | = kt + C

| | V e e ekt C C kt= = ⋅+

V = C1ekt

C1 can be positive or negative, so theabsolute value sign is not needed for V. In thereal world, V is positive, which also makesthe absolute value sign unnecessary.

c. 400 = Cek·0 ⇒ C = 400

500 400

1 25

400 005578

40=

⇒ = = …

⋅e

k

k

ln ..

V = 400e0.005578…t

d. 750 = 400e0.005578…t

⇒ = = … ≈tln .

. ...

1 875

0 005578112 68 113. s

R3. a. y dy dx y x C− = ⇒ = +∫∫ 1 2 6 3/ 2( )

b. y = (3x − 4)2 (y = (3x − 14)2 does not workbecause at (3, 5), dy/dx = −30 but 6y1/2 = 30.)

c.

(3, 25)y

x

10

1 2 3

d. At x = 2, y′ = 12 and y = 4.See graph in part c.A line through (2, 4) with slope 12 is tangentto the graph.

e. i. dN/dt = 100 − kNdN

kNdt

100 –∫ ∫=

−(1/k) ln |100 − kN| = t + CUsing (0, 0) gives −(1/k) ln 100 = C.Substituting this value for C gives−(1/k) ln |100 − kN| = t − (1/k) ln 100.ln |100 − kN| − ln 100 = −ktln |1 − (k/100)N| = −kt1 100− = −( / )k N e kt

N k e kt= − −( / )( )100 1Using (7, 600) and solving numericallygives k ≈ 0.045236.∴ N = 2210.6…(1 − e− 0.045236 t)

ii. t = 30: About 1642 names


iii. limt

N→∞

= … − = …2210.6 (1 0) 2210.6

The brain saturates at about 2211 names.

iv. Let dN/dt = 30.

30 10070

1547 4= − ⇒ = = …kN Nk

.

names. Substituting this for N gives

1547 4 2210 6 1 0 045236. . ( )..K K= − −e t

e t− = − =0 045236 1

1547 4

2210 60 3. .

.

.

K

K (exactly)

t = = … ≈ln .

– .

0 3

0 0452326 6 27

K. days

or:

30 = N(t) − N(t − 1)

= − +− − −2210 6 0 045236 0 045236 1. [ ]. . ( )K e et t

= − +− − −2210 6 10 045236 1 0 045236. ( ). ( ) .Ke et t

⇒ t ≈ 27 days

R4. a.dy

dx xyy= − +20

0 05.

At (2, 5), dy/dx = −1.75.At (10, 16), dy/dx = 0.675.The slopes at (2, 5) and (10, 16) agree withthese numbers.

b. Initial conditions (1, 8) and (1, 12)

5

5

y

x

(10, 16)

(2, 5)

(1, 8)

(1, 10)

(1, 12)

The solution containing (1, 8) crosses thex-axis near x = 7, converges asymptoticallyto the y-axis as x approaches zero, and issymmetric across the x-axis. The solutioncontaining (1, 12) goes to infinity as x goesto infinity.

c. See the graph in part b with initial condition(1, 10). The solution containing (1, 10)behaves more like the one containing(1, 12), although a slight discrepancy inplotting may make it seem to go theother way.

R5. a.dy

dx xyy= − +20

0 05.

Table with initial condition (1, 9), ∆x = 1:

x y (∆x = 1) y (∆x = 0.1)

1 9 9

2 7.227… 7.707…

3 6.205… 6.949…

4 5.441… 6.413…

5 4.794… 5.999…

6 4.200… 5.662…

7 3.616… 5.377…

8 3.007… 5.130…

9 2.326… 4.910…

10 1.488… 4.712…

11 0.2185… 4.529…

12 −8.091… 4.359…

13 4.199…

14 4.045…

15 3.896…

16 3.750…

17 3.604…

18 3.457…

19 3.306…

20 3.150…

M M

28.9 0.1344…

29 −0.3810…

5

5

y

x

(1, 9)

∆x = 1

∆x = 0.1

For ∆x = 1, the graph crosses the x-axis atabout x = 11.

b. See the table in part a for ∆x = 0.1.See the graph in part a.

c. The accuracy far away from the initialcondition is very sensitive to the size of theincrement. For instance, in part a the firststep takes the graph so far down that itcrosses the x-axis before running off theedge of the grid. The greater accuracy with∆x = 0.1 shows that the graph actually doesnot cross the x-axis before x = 20.


d. Continuing the computations in part c, thegraph crosses the x-axis close to x = 28.9.See the table in part a.

R6. a.

y (hundred beavers)

x (years)

5 10

5

10

The population is decreasing because it isabove the maximum sustainable, 900 beavers(y = 9). By Euler’s method, y ≈ 9.3598… , orabout 936 beavers, at x = 3 years.

b. See the graph in part a with initial condition(3, 100), showing that the population isexpected to increase slowly, then morerapidly, then more slowly again, leveling offasymptotically toward 900. This happensbecause the initial population of 100 is belowthe maximum sustainable.

c.dy

dxy

y= ⋅0 69

9.

–

yae x=

+9

1 0 6– .

Substitute into thegeneral equation.

19

1 1 8=+ ae– .

Substitute the initialcondition (3, 1).

a = 8e1.8 = 48.3971… Solve for a.

y

e e ex x=+

=+−

9

1 8

9

1 48 39711 8 0 6 0 6. . – .. K

The point ofinflection is halfwaybetween theasymptotes at y = 0and y = 9.

4 59

1 8 1 8 0 6. =+ −e e x. .

Substitute 4.5 for y.

x = ln (8e1.8 )/0.6 = 6.4657… ≈ 6.5 yr

d.dy

dx

x

y= – . ( – )

( – )

0 5 6

7dy = 0 when x = 6, and dx = 0 when y = 7. Sothe stable point is (6, 7), corresponding to thepresent population of 600 Xaltos natives and7000 yaks.

(Note that the general solution to the differ-ential equation is (x − 6)2 + 2(y − 7)2 = C,and the specific solution for the given initialcondition is (x − 6)2 + 2(y − 7)2 = 0, whosegraph is a single point.)

e. Initial condition (9, 7)

5

5

x

y

(6, 7) (9, 7) (15, 7) (19, 7)

Suddenly there are too many predators for thenumber of prey, so the yak populationdeclines. Because y is decreasing from (9, 7),the graph follows a clockwise path.

f. See the graph in part e with initial condition(19, 7). The graph crosses the x-axis atx ≈ 14.4, indicating that the yaks are huntedto extinction. (The Xaltos would then starveor become vegetarian!)

g. See the graph in part e with initial condition(15, 7). The graph never crosses the x-axis,but crosses the y-axis at y ≈ 2.3, indicatingthat the yak population becomes so sparsethat the predators become extinct. (The yakpopulation would then explode!)

Concept Problems

C1. a.dy

dxk y y dy k dx= ⋅ ⇒ = ⇒/ − /1 2 1 2

2 0 51 2 2y kx C y kx C/ = + = +, so [ . ( )] .

b. The differential equation would have tobecome y1 3/

after it is integrated. So theoriginal equation would have to containy− /2 3

after the variables have been separated.

Conjecture: dy

dxky= 2 3/

c. Confirmation:dy

dxky y dy k dx= ⇒ = ⇒/ − /2 3 2 3

3y1/ 3 = kx + C ⇒ y = [(1/3)(kx + C)]3, acubic function, Q.E.D.

d. For n ≠ 0, dy

dxk y n n= ⋅ ⇒− /( )1

y dy k dx ny kx Cn n n− − / /= ⇒ = + ⇒( ) 1 1

y = [(1/n)(kx + C)]n


For example: dy

dxy y dy dx= ⇒ =/ − /7 8 7 8

⇒ = + ⇒ = / +/8 1 81 8 8y x C y x C[( )( )]

C2. a.

Ticket Price People

2.00 460

2.50 360

3.00 320

4.00 260

4.50 140

5.50 120

6.00 80

b.

500

6

P

N

Function behaves (more or less) linearly.Let N = number of tickets and P = number of$/ticket.By linear regression, N ≈ −90.83P + 605.4,with correlation coefficient r = −0.9747… .

c. Let M = total number of dollars.M ≈ P ⋅ N = P(−90.83P + 605.4)M ≈ −90.83P2 + 605.4P

d. Maximize M: M′ ≈ −181.66P + 605.4

M ′ = 0 ⇔ P ≈ 605 4

181 66

.

. = 3.332…

Maximum M at P ≈ 3.332… because M′changes from positive to negative there (orbecause the graph of M is a parabola openingdownward).Charge $3.30 or $3.35.

e. M has a local maximum at this price becausecharging more than the optimum pricereduces attendance enough to reduce the totalamount made, whereas charging less than theoptimum price increases attendance, but notenough to make up for the lower price perticket.

C3. a. g t e e t

( ) . .

= − −10 0 8 0 5

The graph does look like Figure 7-7e.

lim . lim. .

t

e ee et

tt

→∞

− −→∞

−=10 100 8 0 5 0 5

= =− ⋅10 100 8 0e .

Limit is 10, indicating maximum possiblepopulation.

b. a = 421.3692… , c = 0.7303036… , andk = 0.01589546… , either by twice takinglogarithms as suggested, or by this method:Taking ln once ⇒ ln a − ce− kt = ln P, soln a − ce10k = ln 179ln a − c = ln 203ln lna ce k− =−10 226Then substituting ln a = c + ln 203 into thefirst and third equations gives

c(1 − e10k) = ln 179 − ln 203c e k( )1 226 20310− = −− ln ln

Substituting c e c e ek k k( ) ( )1 110 10 10− = − =− −

− −−e k10 179 203( )ln ln into the previousequation yields

e k− = − −−

= −−

10 226 203

179 203

226 203

203 179

ln ln

ln ln

ln ln

ln ln

so k = − −

−

=1

10

226 203

203 1790 01589ln

ln ln

ln ln. .K

Then find c using c e k( )1 22610− = −− lnln 203 and find a using 203 = −ae c .c = 0.7303… and a = 421.3692…

431.3...

t

g(t)

100

100

Note that this model predicts an ultimatepopulation of lim ( )

tP t

→∞≈ 421 million.

c. Now a = 551.1655… , c = 0.9988291… ,k = 0.01186428… , and the ultimatepopulation is lim ( )

tP t

→∞≈ 551 million. Thus,

the Gompertz model is quite sensitive to asmall change in initial conditions. Thepredicted ultimate population increased by130 million with only a 1 million change inone data point!

C4. dV/dt = −2V1/ 2 + F, where F is a constant.dV

F Vdt

−=∫ ∫2 1 2/

The integral on the right is not the integral ofthe reciprocal function because the numeratorcannot be made the differential of thedenominator. A slope field gives informationabout the solutions. The following graph is forF = 20 ft3/min flowing in. (The dashed lineshows the solution with F = 0, the originalcondition.) Starting with 196 ft3 in the tub, thevolume levels off near 100 ft3. Starting below100 ft3, the volume would increase toward 100.


14

196 F = 20

F = 0

t

V

If the inflow rate is too high, the tub willoverflow. The next graph is for F = 40 ft3/min.In this case, the stable volume is above theinitial 196 ft3.

14

196

F = 0

t

VF = 40

It is possible to antidifferentiate the left side bythe algebraic substitution method of ProblemSet 9-11, Problems 101–106. The generalsolution is

t CF

F V V+ = − − −2

2 1 2 1 2ln ( )/ /

and the particular solution for V = 196 at t = 0 is

tF F

F VV− = − −

−−14

2

28

2 1 21 2ln //

Unfortunately, it is difficult or impossible tosolve for V. The volume will asymptoticallyapproach F2/4, overflowing the tub if F2/4 > tubcapacity.

Chapter Test

T1.dy

dxky=

T2. Solving a differential equation means finding theequation of the function whose derivative appearsin the differential equation.

T3. The general solution involves an arbitraryconstant of integration, C. A particular solutionhas C evaluated at a given initial condition.

T4.

5

5

y

x

(0, –4)

T5. The concave side of the graph is up, so theactual graph curves up from the Euler’s tangentlines, making the Euler’s method values an

underestimate. (Or: The convex side of the graphis down, so the Euler’s tangent lines are belowthe actual graph.)

T6. General logistic differential equation:dy

dxky

M y

M= ⋅ −

T7.dy

dxy= 0 4.

dy

ydx∫ ∫= 0 4.

ln |y| = 0.4x + C|y| = eCe0.4 x

y = C1e0.4 x

−5 = C1e0.4(0) = C1

y = −5e0.4 x

T8.dy

dxy y dy dx= ⇒ = ⇒− ∫∫12 121 2 1 2/ /

2 121 2y x C/ = +

T9. a.dP

dtkP P Cekt= ⇒ =

P = 3000 at t = 0 ⇒ P = 3000ekt

b. P = 2300 at t = 5 ⇒

k = = − ⇒1

5

2300

30000 05314ln . K

P(25) = 794.6…Phoebe will not quite make it because thepressure has dropped just below 800 psi bytime t = 25.or:800 3000 0 05314= −e t. K

t =−

=1

0 05314

800

300024 87

.ln .

KK

Phoebe will not quite make it because thepressure has dropped to 800 just before t = 25.

T10. a. y = number of grams of chlorine dissolvedt = number of hours since chlorinator wasstarteddy

dtky= −30

dy

kydt

30 –∫ ∫=

− − = +130

kky t Cln | |

ln |30 − ky| = −kt + C1

30 − ky = C2e− k t

y = 0 when t = 0 ⇒ C2 = 30∴ ky = 30(1 − e− kt)

yk

e kt= 301( – )–

The rate of escape is ky = 13 when y = 100.So k = 0.13.

∴ = = − −y e et t30

0 131 230 7 10 13 0 13

.( – ) .– . K( ).


b. 200 230 7 1 0 13= … − −. ( e t. )

e t− = − =0 13 1

200

230 70 1333. .

. KK

t = =ln .

– .

0 1333

0 1315 499

KK. ≈ 15.5 hr

T11. a.dy

dxy

y= ⋅0 516

16.

–

yae ae

ax=+

⇒ =+

⇒ =16

12

16

170 5 0– .

ye x=

+16

1 7 0 5– .

b. At x = 0, y = 2:dy = 0.5(2)(16 − 2)/(16)(0.1) = 0.0875At x = 0.1, y ≈ 2 + 0.0875 = 2.0875,so dy = 0.5(2.0875)(16 − 2.0875)/(16)(0.1) =0.09075… .At x = 0.2, y ≈ 2.0875 + 0.09075… =2.17825… .

The precise solution is ye

=+

16

1 7 0 1– . =

2.18166… , which is greater than2.17825… , as expected because thegraph is concave up (convex side downward).

c. 416

1 7 0 5=+ e x– . ⇒ x = [ln (3/7)]/−0.5 =

1.6945…About 1 month 21 days

d.y (hundred lilies)

x (months)

The graph shows that the number of lilies isexpected to decrease toward 1600 (y = 16)because of overcrowding.

T12. a.

R (roadrunners)

C (coyotes)

(80, 700)

The graph starts going downward and to theright from (80, 700) because the coyotepopulation is relatively high, thus decreasingthe number of roadrunners.

b. There can be two different values for theroadrunner population for a particular coyotepopulation because the two events happen attwo different times. For example, coyotes areincreasing from 80 when there are 700 road-runners, but later they are decreasing from 80when there are about 200 roadrunners.


Problem Set 7-8Cumulative Review, Chapters 1–7

1.

200

8

t

v(t)

(t, v(t))

v(t) dt represents the distance traveled in time dt.

2. Definite integral

3. ( – )t t t dt3 2

0

8

21 100 80+ +∫= − + +1

47 50 804 3 2

0

8t t t t

= 1280 mi

4. M100 = 1280.0384

M1000 = 1280.000384

The Riemann sums seem to be approaching 1280as n increases. Thus, the 1280 that was found bypurely algebraic methods seems to give thecorrect value of the limit of the Riemann sum.

5.

200

8

t

v(t)

(t, v(t))

6. Any Riemann sum is bounded by thecorresponding lower and upper sums. That is,Ln ≤ Rn ≤ Un.

By the definition of integrability, the limits of Ln

and Un are equal to each other and to the definiteintegral. By the squeeze theorem, then, the limitof Rn is also equal to the definite integral.


7. Definition:

f x dx L Ua

b

xn

xn( ) lim lim

0 0= =∫ → →∆ ∆

provided that the two limits are equal.Fundamental theorem: If f is integrable on

[ , ] and ( ) ( ) , a b g x f x dx= ∫then f x dx g b g a

a

b

( ) ( ) ( ). = −∫Or: If F x f t dt

a

x

( ) ( ) ,= ∫ then ′ =F x f x( ) ( ).

8. Numerically, the integral equals 1280. Bycounting, there are approximately 52 squares.Thus, the integral ≈ 52(25)(1) = 1300.

9. vv v′ ≈ = − /

( ) . (mi )

44 1 3 9

0 219 9

( . ) – ( . )

.

min

min

vv v′ ≈ = −( ) .

(mi/min)

min4

4 01 3 99

0 0219 9999

( . ) – ( . )

.

10. f cf x f c

x cx c′ =

→( ) lim

( ) – ( )

– or

f xf x x f x

xx′ = + ∆

∆→( ) lim

( ) – ( )∆ 0

11. v′(t) = 3t2 − 42t + 100 ⇒ v′(4) = −20

12. Slowing down. v′(4) < 0 and v(4) = 208 > 0 ⇒velocity is positive but decreasing ⇒ speed isslowing down.

13. The line has slope −20, and passes through(4, 208). The line is tangent to the graph.

100

200

100 5

v(t)

t

5

–100

Slope= –20

14. Acceleration

15. At a maximum of v(t), v′(t) will equal zero.

3t2 − 42t + 100 = 0 ⇔ t =± ⋅ ⋅42 42 4 3 100

6

2 –

t = 3.041… or 10.958…

So the maximum is not at exactly t = 3.16. v″(t) = 6t − 42

17. Know: dx

dt= 0 3. . Want:

dy

dt and

dz

dt.

y edy

dte

dx

dtex x x= ⇒ = − ⇒ = −− − −6 3 0 90 5 0 5 0 5. . ..

At x = 2, dy

dte= − = −−0 9 0 33101. . .K

y is decreasing at about 0.33 unit per second.

18. z x y zdz

dtx

dx

dty

dy

dt2 2 2 2 2 2= + ⇒ = +

dz

dt zx e ex x= + ⋅1

0 3 6 0 90 5 0 5[ . (– . )]– . – .

At x z= = + = …2 2 2 2072 2 97862 2, . . ,K

so

dz

dt= = −

1

2 97860 6 0 7308 0 04391

.( . – . )

KK K. .

∴ z is decreasing at about 0.044 unit per second.

19.dm

dtkm=

20.dm

mk dt m kt C= ⇒ = + ⇒∫∫ ln | |

| |m e m C ekt C kt= ⇒ =+1

21. Exponentially

22. General

23. 10000 = C1e0 ⇒ C1 = 10000

10900 10000 1= ⋅ek

⇒ k = ln 1.09 ⇒ m = 10000eln(1.09) t

= 10000(1.09)t

24. False. The rate of increase changes as the amountin the account increases. At t = 10,m = 10000(1.09)10 ≈ 23673.64.The amount of money would grow by$13,673.64, not just $9,000.

25. By Simpson’s rule, y dx30

42

∫≈ + ⋅ + ⋅ + ⋅ + ⋅2

374 4 77 2 83 4 88 2 90(

+ ⋅ + =4 91 89 1022) .

26. By symmetric difference quotient, at x = 36

y′ ≈ =90 83

2 21 75

–

( ). .

27. If f is differentiable on (a, b) and continuous atx = a and x = b, and if f ( a) = f ( b) = 0, then thereis a number x = c in (a, b) such that f ′ (c) = 0.

28.

a c b

f(b)

f(a)

f(x)

x

tangent

secant

29.f(x) and f´(x)

x

5

0 5 10

f f´

f´


30.

x

f(x)

1

2

Step discontinuity at x = 1.

31. g(x) = x1/ 3(x − 1)

g x x x x x x′ = + − = −/ − / − /( ) ( ) ( )13 2 3 2 31

31

1

34 1

g′(0) is undefined because 0 2 3− / takes on the form

1 02 3/ / or 1/0.

g(x)

x

1

–1 1

32. e0.2 x = 0.6x ⇒ x ≈ 3.0953… (Store as a.)Or x ≈ 7.5606… (Store as b.)dA = (0.6x − e0.2 x) dx

A x e dxx

a

b

= ≈∫ ( . – ).0 6 0 87870 2 . K

(Integrate numerically.)

33. dV = π [(e0.2 x)2 − (0.6x)2] dx

V e x dxx

a

= ≈∫π .( – . ).0 4 2

00 36 8 0554K

(Integrate numerically.)

34.dy

dx

x

y= 0 25.

Initial conditions: (0, 3) and (10, 4)

5

5

x

y

(0, 3)(10, 4)

35. See the graph in Problem 34. Any initialcondition for which y = 0.5x, such as (2, 1),gives the asymptote.

36. y dy x dx y x C= ⇒ = +∫∫ 0 25 0 5 0 1252 21. . .

⇒ x2 − 4y2 = CInitial condition: (10, 4)100 − 64 = C ⇒ C = 36

∴ − = ⇒ = ±x y y x2 2 24 36 0 5 36. –

37. x y= = = …10 5 0 5 10 5 36 4 308422. : . .. –

38. At ( , ), . . .10 4 0 2510

40 625

dy

dx= ⋅ =

Using ∆x = 0.5, y(10.5) ≈ 4 + (0.625)(0.5) =4.3125, which is close to the exact value of4.30842… .

39.d

dxx

x

x(sin )

–

–1 32

6

3

1=

40. x = ln (cos t) and y = sec t

dx dtt

t t/ = ⋅ − = −1

cossin tan( )

dy/dt = sec t tan tdy

dx

dy dt

dx dt

t t

tt y= = = − = −/

/

sec tan

– tansec

41.dx

xx C

4 3

1

34 3

–ln | |= − − +∫

42. h x e h x ex x x( ) ( ) 5= = ⇒ ′ = =⋅ ⋅5 5 55 5 5ln lnln ln

43. limsin cos – –

x

x x x

x→

+ →0 2

5 3 5 1 0

0

= →→

limcos – sin –

x

x x

x0

5 5 3 3 5

2

0

0

= = −→

lim– sin – cos

x

x x0

25 5 9 3

24 5.

44. yx x x

x= +sin cos – –5 3 5 1

2 , showing a

removable discontinuity at (0, −4.5).

5

–5

–1 1

x

y

(0, –4.5)




Chapter 8—The Calculus of Plane and Solid Figures

Problem Set 8-1

1. f (x) = x3 − 6x2 + 9x + 3f ′ (x) = 3x2 − 12x + 9

ff´

x

y

1 32 4

g (x) = x3 − 6x2 + 15x − 9g ′ (x) = 3x2 − 12x + 15

x

y

1 32 4

ff´

h (x) = x3 − 6x2 + 12x − 3h ′ (x) = 3x2 − 12x + 12

f´

x

y

1 32 4

f

Positive derivative ⇒ increasing functionNegative derivative ⇒ decreasing functionZero derivative ⇒ function could be at a highpoint or a low point, but not always.

2. The functions have vertex points at values of xwhere their derivatives change sign. If thederivative is never zero, as for function g, thefunction graph has no vertex points. If thederivative is zero but does not change sign, asfor function h, the function graph just levelsoff, then continues in the same direction, withno vertex.

3. g ′′ (x) = (d/dx)(3x2 − 12x + 15) = 6x − 12h ′′ (x) = (d/dx)(3x2 − 12x + 12) = 6x − 12All the second derivatives are the same!

4. The curves are concave up where the secondderivative is positive and concave down where thesecond derivative is negative.

5. Points of inflection occur where the firstderivative graph reaches a minimum.Points of inflection occur where the secondderivative graph crosses the x-axis.


1

1x

y

1

1x

y

Q3. Q4.

1

1x

y

1

1x

y

Q5. Q6.

1

1 x

y

1

1x

y

Q7. Q8.

1

1x

y

1

1x

y

Q9. Q10.

1

1 x

y

2

1x

y

1.

x

f´(x)

f (x)

2

+ –0

max.

x 2

– ––

no p.i.

f´´(x)

f (x)

2.

x

f'(x)

f(x)

2

– +0

min.

x

f"(x)

f(x)

2

+ +0

no p.i.


3.

x 2

+ 0 +

plateau

f´(x)

f (x)

x 2

– 0 +

p.i.

f´´(x)

f (x)

4.

x

f'(x)

f(x)

2

undef.+ –

max.

x

f"(x)

f(x)

2

undef.+ +

no p.i.

5.

x 2

– +undef.

min.

f´(x)

f (x)

x 2

0 0undef.

no p.i.

f´´(x)

f (x)

6.

x

f'(x)

f(x)

2

– –undef.

no max./min.

x

f"(x)

f(x)

2

– +undef.

p.i.

7.

x 2

no max./min.

–f´(x)

f (x)

x 2

p.i.

+ –f´´(x)

f (x)

8.

x

f'(x)

f(x)

2

–

no max./.min.

x

f"(x)

f(x)

2

0– +

p.i.

9.

x 2

– –undef.

no max./min.

f´(x)

f (x)

x 2

– +undef.

no p.i.

f´´(x)

f (x)

10.

x

f'(x)

f(x)

2

undef.+ –

no max.min.

x

f"(x)

f(x)

2

undef.+ +

no p.i.

11.

x –2 1 3

0 0 0+ – –+

max. min. max.

f´(x)

f (x)

x –1 2

00– + –

p.i. p.i.

f´´(x)

f (x)

x

f(x)

–2 –1 1 2 3

12.

–3 –1 3

0 0 0– + + –

min. plateau max.

f'(x)

f(x)

x

x –2 –1 2

0 0 0+ – + –

p.i. p.i. p.i.

f(x)

f"(x)

x

f(x)

–3 –2 –1 2 3


13.

x 1

0+ – –

–2

∞

max. plateau

f´(x)

f (x)

x 1

0+ + –

–2

∞

no p.i. p.i.

f´´(x)

f (x)

x

f(x)

–2 1

14.

2 3 4

0 ∞ 0– + + –

f(x)

max. max.none

f'(x)

x

3

∞+ –

f(x)

p.i.

f"(x)

x

x

f(x)

2 3 4

15.

3

zero+ –

–1 1 5

00 e.p.e.p.

min. max. min.

x

f´(x)

f (x)

3

zero– –

–1 1 5

00 e.p.e.p.

no p.i.

x

f´´(x)

f (x)

x

f(x)

–1 1 3 5

16.

6

–+ +

2 7

00 e.p.e.p.

1

f(x)

min. max.max. min.

x

f'(x)

x 5

– +

3 7

00 e.p.e.p.

1

zero

f(x)

no p.i.

f"(x)

f(x)

1 2 3 5 6 7

x

17.

3 6

3

–3

y

xf´

f´

f´f

18.

4 8

4

–4

y

x

f'

f' f' f

19.

4 8

4

y

x

f´ f´

f

20.

4

4

y

x

2

f'

f


21. f (x) = 3ex − xex

f ′ (x) = 3ex − ex − xex = ex (2 − x)f ′ (2) = e2 (2 − 2) = 0 ⇒ critical point at x = 2f ′′ (x) = 2ex − ex − xex = ex (1 − x)f ′′ (2) = e2 (1 − 2) = −7.3890... < 0∴ local maximum at x = 2

1 2 3

5

f(x)

x

The graph confirms a maximum at x = 2.

22. f x x( ) = −sinπ4

f x x′ = −( )π π4 4

cos

f ′ = − = ⇒( )24 4

2 0π π

cos ( ) critical point

at x = 2

′′ =f x x( )π π2

16 4sin

′′ = = >f ( ) .216 4

2 0 6168 02π π

sin ( ) K

∴ local minimum at x = 2

2

1

–1

f (x )

x

The graph confirms a minimum at x = 2.

23. f (x) = (2 − x)2 + 1f ′ (x) = −2(2 − x)f ′ (2) = −2(2 − 2) = 0 ⇒ critical point at x = 2f ″ (x) = 2 ⇒ f ″ (2) = 2 > 0∴ local minimum at x = 2

2

1

f(x)

x


24. f (x) = −(x − 2)2 + 1f ′ (x) = −2(x − 2)f ′ (2) = −2(2 − 2) = 0 ⇒ critical point at x = 2f ″ (x) = −2 ⇒ f ″ (2) = −2 < 0∴ local maximum at x = 2

2

1

f(x)

x

The graph confirms a maximum at x = 2.

25. f (x) = (x − 2)3 + 1f ′ (x) = 3(x − 2)2

f ′ (2) = 3(2 − 2)2 = 0 ⇒ critical point at x = 2f ″ (x) = 6(x − 2)f ″ (2) = 6(2 − 2) = 0, so the test fails.f ′ (x) goes from positive to positive asx increases through 2, so there is a plateauat x = 2.

2

1

f(x)

x

The graph confirms a plateau at x = 2.

26. f (x) = (2 − x)4 + 1f ′ (x) = −4(2 − x)3

f ′ (2) = −4(2 − 2)3 = 0 ⇒ critical point at x = 2f ″ (x) = 12(2 − x)2

f ″ (2) = 12(2 − 2)2 = 0, so the test fails.f ′ (x) changes from negative to positive asx increases through 2, so there is a localminimum at x = 2.

2

1

f(x)

x


27. a. f (x) = 6x5 − 10x3

f ′ (x) = 30x4 − 30x2 = 30x2(x + 1)(x − 1)f ′ (x) = 0 ⇔ x = −1, 0, or 1 (critical pointsfor f (x))

′′ = − = + −f x x x x x x( ) 120 603 60 2 1 2 1( )( )

f ″ (x) = 0 ⇔ x = 0, ± 1 2/ (critical points for

f ′ (x))

b. The graph begins after the f-critical point atx = −1; the f ′ -critical point at x = − 1 2/ is

shown, but is hard to see.

c. f ′ (x) is negative for both x < 0 and x > 0.


28. a. f (x) = 0.1x4 − 3.2x + 7f ′ (x) = 0.4x3 − 3.2 = 0.4(x − 2)(x2 + 2x + 4)x2 + 2x + 4 has discriminant = 22 − 4 · 4 < 0,so f ′ (x) = 0 ⇔ x = 2 (critical point for f (x)).f ″ (x) = 1.2x2

f ″ (x) = 0 ⇔ x = 0 (critical point for f ′ (x))

b. f ″ (x) does not change sign at x = 0.( f ″ (x) ≥ 0 for all x)

c. f ″ (c) = 0, but f ′ (c) ≠ 0

29. a. f x xe x( ) = −

f x xe e e xx x x′ = − + = −− − −( ) ( )1

f ′ (x) = 0 ⇔ x = 1 (critical point for f (x))′′ = − = −− − −f x xe e e xx x x( ) 2 2( )

f ″ (x) = 0 ⇔ x = 2 (critical point for f ′ (x))

b. Because f (x) approaches its horizontalasymptote (y = 0) from above, the graph mustbe concave up for large x; but the graph isconcave down near x = 1, and the graph issmooth; somewhere the concavity mustchange from down to up.

c. No. e x− ≠ 0 for all x, so xe xx− = ⇔ =0 0.

30. a. f (x) = x2 ln xf ′ (x) = x + 2x ln x = x (1 + 2 ln x)f (x) and f ′ (x) are undefined at x = 0, sof x x x e′ = ⇔ = − ⇔ = =−( ) .0 0 5 0 5ln .

0.6065… (critical point for f (x)).f ″ (x) = 3 + 2 ln x

′′ = ⇔ = − ⇔ = =−f x x x e( ) ln .0 1 5 1 5.0.2231… (critical point for f ′ (x)).

b. lim ln limln

lim/

–– –x x x

x xx

x

x

x→ → →+ + += =

0

2

02

03

1

2

= − =→ +lim .x

x0

20 5 0 by L’Hospital’s rule.

limx

x→ −0

2 ln x does not exist because x2 ln x is

undefined for x < 0.

c. All critical points from part a appear,although the inflection point at x e= −1 5.

ishard to see on the graph.

31. a. f (x) = x5/3 + 5x2/3

′ = + = +− −f x x x x x( ) ( )/ / /5

3

10

3

5

322 3 1 3 1 3

f ′ (x) = 0 ⇔ x = −2, and f ′ (x) is undefinedat x = 0 (critical points for f (x)).

′′ = − = −− − −f x x x x x( ) ( )/ / /10

9

10

9

10

911 3 4 3 4 3

f ″ (x) = 0 ⇔ x = 1 (critical point for f ′(x);f ′ (0) is undefined, so f ′ has no critical pointat x = 0).

b. The y-axis (x = 0) is a tangent line becausethe slope approaches −∞ from both sides.

c. There is no inflection point at x = 0 becauseconcavity is down for both sides, but there isan inflection point at x = 1.

32. a. f x x x( ) = −1 2 0 23. .

f x x x x x′ = − = −− −( ) . . . ( )1 2 0 6 0 6 2 10 2 0 8 0 8. . .

f ′ (x) = 0 ⇔ x = 0.5, and f ′ (x) is undefinedat x = 0 (critical points for f (x)).′′ = + = +− − −f x x x x x( ) . . .0 24 0 48 0 24 20 8 1 8 1 8. . . ( )

f ″ (x) = 0 ⇔ x = −2 (critical point for f ′ (x);f ′ (0) is undefined, so f ′ has no critical pointat x = 0).

b. f (0) = 01.2 − 3 ⋅ 00.2 = 0 has only one value.

c. Curved concave up because f ′′(x) > 0 forx < −2

33. a. f (x) = −x3 + 5x2 − 6x + 7

7

1

x

f(x)

Maximum (2.5, 7.6), minimum (0.8, 4.9),points of inflection (1.7, 6.3)No global maximum or minimum

b. f ′(x) = −3x2 + 10x − 6

′ = ⇔ = ± =f x x( ) .01

35 7 2 5485( ) K or

0.7847…

′′ = − + ′′ = ⇔ = =f x x f x x( ) ; ( )6 10 05

31.666…

c. f ′′(0.7847…) = −6(0.7847…) + 10 =5.2915… > 0, confirming local minimum.

d. Critical and inflection points occur onlywhere f, f ′, or f ′′ is undefined (no such pointsexist) or is zero (all such points are foundabove).

34. a. f (x) = x3 − 7x2 + 9x + 10

f(x)

10

1

x

Maximum (0.8, 13.2), minimum (3.9, −2.1),points of inflection (2.3, 5.6)No global maximum or minimum


b. f ′(x) = 3x2 − 14x + 9

′ = = ± =f x x( ) at .0

1

37 22 3 896( ) K or

0.769…

′′ = − ′′ = = =f x x f x x( ) ; ( ) at .6 14 0

7

32 333K

c. f ′′(0.769…) = 6(0.769…) − 14 =−9.3808… < 0, confirming local maximum.

d. Critical and inflection points occur onlywhere f, f ′, or f ′′ is undefined (no such pointsexist) or is zero (all such points are foundabove).

35. a. f (x) = 3x4 + 8x3 − 6x2 − 24x + 37,x ∈ [−3, 2]

–3 2

80

x

f(x)

Maximum (−3, 82), (−1, 50), (2, 77),minimum (−2, 45), (1, 18), points ofinflection (−1.5, 45.7), (0.2, 32.0)Global maximum at (−3, 82) and globalminimum at (1, 18)

b. f ′(x) = 12x3 + 24x2 − 12x − 24= 12(x + 2)(x − 1)(x + 1)

f ′(x) = 0 ⇔ x = −2, −1, 1f ′(x) is undefined ⇔ x = −3, 2.f ′′(x) = 36x2 + 48x − 12 = 12(3x2 + 4x − 1);

′′ = ⇔ = − ± = …f x x( ) 0.215201

32 7( )

or −1.5485…f ′′(x) is undefined ⇔ x = −3, 2.

c. f ′′(−2) = 12[3(4) + 4(−2) − 1] = 36 > 0,confirming local minimum.

d. Critical and inflection points occur onlywhere f, f ′, or f ′′ is undefined (only atendpoints) or is zero (all such points are foundabove).

36. a. f (x) = (x − 1)5 + 4, x ∈ [−1, 3]

1

10

3–1

x

f(x)

Maximum (3, 36), minimum (−1, −28), plateau and points of inflection (1, 4)Global maximum at (3, 36) and globalminimum at (−1, −28)

b. f ′(x) = 5(x − 1)4

f ′(x) = 0 ⇔ x = 1; f ′(x) is undefined ⇔x = −1, 3.f ′′(x) = 20(x − 1)3;f ′′(x) = 0 ⇔ x = 1; f ′′(x) is undefined ⇔x = −1, 3.

c. f ′′(1) = 20(1 − 1)3 = 0, so the test fails.

d. Critical and inflection points occur onlywhere f, f ′, or f ′′ is undefined (only atendpoints) or is zero (all such points are foundabove).

37. f (x) = ax3 + bx2 + cx + d; f ′(x) = 3ax2 + 2bx + c;f ′′(x) = 6ax + 2b ⇒ f ′′(x) = 0 at x = −b/(3a)Because the equation for f ′′(x) is a line withnonzero slope, f ′′(x) changes sign at x = −b/(3a),so there is a point of inflection at x = −b/(3a).

38. f (x) may not have a local maximum orminimum (if f ′(x) is never zero); if this is notthe case, then the maximum and minimum occurwhere f ′(x) = 3ax2 + 2bx + c = 0, at

xb b a c

a

b

a

b ac

a= ± ⋅ ⋅ = ±– – – –

,2 4 4 3

6 3

3

3

2 2

and the maximum and minimum occur at

b ac a2 3 3– /( )units on either side of the

inflection point −b/(3a) (see Problem 33).

39. f (x) = ax3 + bx2 + cx + df ′(x) = 3ax2 + 2bx + c; f ′′(x) = 6ax + 2bPoints of inflection at (2, 3) ⇒ f ′′(3) = 0 ⇒18a + 2b = 0Maximum at (5, 10) ⇒ f ′(5) = 0 ⇒ 75a + 10b +c = 0(3, 2) and (5, 10) are on the graph ⇒27a + 9b + 3c + d = 2.125a + 25b + 5c + d = 10Solving this system of equations yields

f x x x x( ) = − + − −1

2

9

2

15

2

5

23 2 .

3 5

5

(5, 10)

x

f(x)

(3, 2)

The graph confirms maximum (5, 10) and pointsof inflection (3, 2).

40. f (x) = ax3 + bx2 + cx + df ′(x) = 3ax2 + 2bx + c; f ′′(x) = 6ax + 2bPoints of inflection at (2, 7) ⇒ f ′′(2) = 0 ⇒12a + 2b = 0Maximum at (−1, 61) ⇒ f ′(−1) = 0 ⇒3a − 2b + c = 0


(2, 7) and (−1, 61) are on the graph ⇒8a + 4b + 2c + d = 7.−a + b − c + d = 61Solving this system of equations yieldsf (x) = x3 − 6x2 − 15x + 53.

2–1

80f(x)

x

The graph confirms maximum (−1, 61) andpoints of inflection (2, 7).

41. a. f (x) = x3 ⇒ f ′(x) = 3x2

f ′(−0.8) = 1.92f ′(−0.5) = 0.75f ′(0.5) = 0.75f ′(0.8) = 1.92

b. The slope seems to be decreasing from −0.8to −0.5; f ″(x) = 6x < 0 on −0.8 ≤ x ≤ −0.5,which confirms that the slope decreases. Theslope seems to be increasing from 0.5 to 0.8;f ″(x) = 6x > 0 on 0.5 ≤ x ≤ 0.8, whichconfirms that the slope increases.

c. The curve lies above the tangent line.

42. Ima could notice that y ′ = 0 at x = 0(or y ′ = 3 at x = ±1), so the graph could notpossibly be a straight line with slope = 1.

43. a.

cx

f(x)

b.

cx

f(x)

c.

cx

f(x)

d.

cx

f(x)

e.

(Locallyconstant)

cx

f(x)

44. f (x) = 10(x − 1)4/3 + 2f (1) = 2, so f (1) is defined.

f x x′ =( ) /40

31 1 3( – )

f ′(1) = 0, so f is differentiable at x = 1.

f x x″ = −( ) /40

91 2 3( – )

f ″(1) has the form40

90

40

91 02 3( ) ( / )– / or , so

f ″(1) is infinite.There seems to be a cusp at (1, 2), but zoomingin on this point reveals that the tangent isactually horizontal there.

f(x)

2

1

x

See Problem 20 in Problem Set 10-6 forcalculation of curvature.

45. f (x) = e0.06 x, f ′(x) = 0.06e0.06 x,f ″(x) = 0.0036e0.06 x

g (x) = 1 + 0.06x + 0.0018x2 + 0.000036x3

g ′(x) = 0.06 + 0.0036x + 0.000108x2

g ″(x) = 0.0036 + 0.000216xf (0) = 1 and g (0) = 1f ′(0) = 0.06 and g ′(0) = 0.06f ″(0) = 0.0036 and g ″(0) = 0.0036(In fact, f ′″(0) = g ′″(0).)But f (10) = e0.6 = 1.822… ≠ g (10) = 1.816;f ′(10) = 0.109… ≠ g ′(10) = 0.1068.Because f (x) > 0 for all x, f has no x-intercept.But g (0) = 1 and g (−100) = −23.By the intermediate value theorem, g (x) = 0somewhere between x = −100 and x = 0, meaningthat g does have an x-intercept.


46. f x xx

x

x( ) if

if= + ≠

=

( – ) sin

–,

,

11

12 1

2 1

3

2.01

2

1.991

f(x)

lim lim ( – )–

( )x x

f x xx

f→ →

= ⋅ + = =1 1

311

12 2 1( ) sin

(The limit of the first term is zero because(x − 1)3 approaches zero and the sine factor isbounded.)∴ f is continuous at x = 1.

ff x f

x

x x

x

xx

x

x

x

′ =

= +

= =

→

→

→

( )

sin 0

11

1

1 1 1 2 2

1

11

1

1

1

3

1

2

lim( ) – ( )

–

lim[( – ) sin( /( – ))] –

–

lim ( – )–

(x − 1)2 → 0 and the sine factor is bounded.∴ f ′(1) = 0

1.999

2

2.001

1 1.10.9

The graph is zoomed in by a factor of 10 bothways. The graph does appear to be locally linearat x = 2. Although the sine factor makes aninfinite number of cycles in any neighborhood ofx = 1, the (x − 1)3 factor approaches zero sorapidly that the graph is “flattened out.” Thename pathological is used to describe the fact thatthe graph makes an infinite number of cycles in abounded neighborhood of x = 1.


Problem Set 8-3Q1. y x′ = − + −3 3 5( ) 2 Q2. ln |x + 6| + C

Q3. − −2

35 3x / Q4. 3x1/3 + C

Q5. − +−x C1 Q6. x + CQ7. ln |sin x| + CQ8. Q9.

x

y

1

1

2

1x

y"

Q10. D

1. Let x = total width of pen, y = length of pen.Domains: 0 ≤ x ≤ 300, 0 ≤ y ≤ 200Maximize A(x) = xy.

2 3 600 2002

3x y y x+ = ⇒ = −

∴ = −A x x x( ) 2002

32

300150

max.A(x)

x

The graph shows a maximum at x ≈ 150.

Algebraically, A x x′ = −( ) .2004

3A′(x) = 0 ⇔ x = 150, confirming the graph.

x y= ⇒ = − ⋅ =150 2002

3150 100

Make the total width 150 ft and length 100 ft.(Note: The maximum area was not asked for.)

2. a. Let x = width of a room across the front,y = depth of a room from front to back.Domains: x ≥ 0, y ≥ 0Minimize P (x) = 12x + 7y.xy = 350 ⇒ y = 350x− 1

∴ P (x) = 12x + 2450x− 1

14.28...

x

P(x)

The graph shows a minimum at x ≈ 14.Algebraically, P ′ (x) = 12 − 2450x− 2.P ′ (x) = 0 ⇔ 2450x− 2 = 12 ⇔x = ± = ± = ± …2450 12 35 6 14 288/ / .

Minimum is at x y= = =35 6 10 6/ , 24.49… .Make rooms 14.3 ft across and 24.5 ft deep.

b. For 10 rooms, P (x) = 20x + 11y =20x + 3850x− 1.

P ′ (x) = 20 − 3850x− 2 = 0 at x = 192 5.

Minimum at x = = …192 5 13 87. . ,

y = = …350 192 5 25 22/ ..

Make rooms 13.9 ft across and 25.2 ft deep.For 3 rooms, P (x) = 6x + 4y = 6x + 1400x− 1.

P ′(x) = 6 − 1400x− 2 = 0 at x = =1400 6/

10 7 3/


Minimum at x = = …10 7 3 15 27/ . ,

y = = …5 21 22 91.

Make rooms 15.3 ft across and 22.9 ft deep.

3. a. Let x = width of rectangle, 2x = length ofrectangle, y = width of square.A rect = 2x2, Asq = y2

For minimal rectangle, 2x2 ≥ 800 ⇒ x ≥ 20.For minimal square, y2 ≥ 100 ⇒ y ≥ 10.Perimeter P = 6x + 4y = 600 ⇒y = 150 − 1.5x∴ 150 − 1.5x ≥ 10 ⇒ x ≤ 140/1.5 =93.3333…Domain: 20 ≤ x ≤ 93.3333…

b. Total area A(x) = 2x2 + y2

= 2x2 + (150 − 1.5x)2

= 22500 − 450x + 4.25x2

20,000

93.3

x

A(x)

20

c. The graph shows a maximum at endpointx = 93.3333… .A′(x) = −450 + 8.5xA′(x) = 0 ⇔ x = 450/8.5 = 52.9411…Because A(52.9…) is a minimum, themaximum occurs at an endpoint.A(20) = 15200, A(93.3333…) =17522.2222…Greatest area ≈ 17,522 ft2

4. a. Let r = radius of circle, s = width of squareDiameter ≥ 50 ⇒ r ≥ 25Circumference ≤ 1000 ⇒ 2πr ≤ 1000 ⇒r ≤ 500/πDomain of r : 25 ≤ r ≤ 500/π = 159.154…Minimize A(r) = πr2 + s2.Perimeter 2πr + 4s = 1000 ⇒ s = 250 − πr/2∴ A(r) = πr2 + (250 − πr/2)2

7025 160

r

Amax.

min.

The graph shows minimum area at x ≈ 70.A′(r) = 2πr + 2(250 − πr/2)(−π/2)A′(r) = 0 ⇔ 2πr − π(250 − πr/2) = 0 ⇒r = 500/(4 + π) = 70.012…A(25) = 46370.667…A(70.012…) = 35006.197…

A(159.154…) = 79577.471…Minimum area at r = 70.012… ,s = 1000/(4 + π) = 140.024…For square, 4(140.024…) ≈ 560.For circle, 2π(70.012…) ≈ 440.Use 440 yd for square and 560 yd for circle.(You could build a square corral with side 140around the circular fence of radius 70 toenclose a total area of only 19,607 yd 2 , butBig Bill might not like your solution!)

b. The graph of A versus r shows that themaximum area occurs at the largest possiblecircle. Big Bill should use all 1000 yards forthe circular fence and not build a corral.

5. a. Let x = length of square base, z = heightof box.Domain of x x: .0 120 10 954≤ ≤ = …Maximize V(x) = x2z.Area = x2 + 4xz = 120 ⇒ z = 30/x − x/4∴ V(x) = 30x − x3/4

6.32... 10.95...0

max.V

x

The graph shows a maximum at x ≈ 6.3.

V′(x) = 30 − 3x2/4 = 0 at x = ± 40

x = − 40 is out of the domain.

Critical points at x = 0, x x= =40 120,

V V( ) , ( )0 0 120 0= =V( ) .40 20 40 126 49= = …Maximum at x = = …40 6 324. ,

z = = …40 2 3 162/ .

Make the box 6.32 cm square by 3.16 cmdeep.

b. Conjecture: An open box with square base ofside length x and fixed surface area A willhave maximal volume when the base lengthis twice the height, which occurs when

x A= /3 (see the solution to Problem 8b).

6. a. Domain of x is 0 ≤ x ≤ 6.

b. V(0) = 0 cm2

V(1) = 180 cm2

V(2) = 256 cm2 (largest volume for an integervalue of x)V(3) = 252 cm2

V(4) = 192 cm2

V(5) = 100 cm2

V(6) = 0 cm2


c. V(x) = (20 − 2x)(12 − 2x)x= 240x − 64x2 + 4x3

200

6

x

V(x)

The graph shows a maximum at x ≈ 2.4.V′(x) = 240 − 128x + 12x2 = 0 at

x = ±( )128 4864 24/ = 2.427… or 8.239…

x = 8.239… is out of the domain.V(2.427…) = 262.68… is a maximumbecause it is positive and V(0) = V(6) = 0.Maximum volume ≈ 262.7 cm2 atx ≈ 2.43 cm

7. Let x = length, y = depth, C(x) = total cost.Domains: x > 0, y > 0Area of bottom = 5xTotal area of sides is (10 + 2x)y.Minimize C(x) = 10(5x) + 5(10 + 2x)y.Volume = 72 ⇒ 5xy = 72 ⇒y x x= = −72 5 14 4 1/( ) .

∴ = + + −C x x x x( ) ( )( . )50 5 10 2 14 4 1

C x x x( ) = + +−50 720 1441

3.794...

x

C

The graph shows a minimum at x ≈ 3.8.C′(x) = 50 − 720x− 2 = 0 ⇔ x = ± 72 5/ =±3.7947…x = −3.7947… is out of the domain.Minimum is at x = 3.7947 because C′(x)changes from negative to positive there.C(3.7947…) = 120 10 + 144 ≈ 523.47Minimum cost is $523.47.

8. a. Maximize V = xyz.Fixed area A = xy + 2xz + 2yz

⇒ y = (A − 2xz)/(x + 2z)

∴ =+

VAxz x z

x z

– 2

2

2 2

dV

dx

z x z x Az

x z= +

+– –

( )

2 8 2

2

2 2 3 2

2

dV

dxx z z A= = − + +0 2 4 2 at

yA z z z A

z z A z

z z A

= + +

+ + +

= − + +

– (– )

–

2 2 4

2 4 2

2 4

2

2

2

Therefore, x = y for maximum volume,Q.E.D.

b. Let x = y. Maximize V = xyz = x2z.Fixed area A = xy + 2xz + 2yz = x2 + 4xz⇒ z = A/(4x) − x/4∴ V = (A/4)x − x3/4

dV

dxA x x A= − = = ±( / ) at 4 3 4 0 32/ /

dV/dx goes from positive to negative atx A= /3 ⇒ maximum at x A= /3.

z A A A A x= − = =/( ) /4 3 3 41

23

1

2/ / /

c. For the maximal box in part b, the depth ishalf the length of the base. Thus, the box isshort and fat. This makes sense because theproblem is equivalent to maximizing thevolume of two open boxes with the secondbox placed upside-down on the first. Theresulting single closed box will havemaximum volume when it is a cube, whichwill happen if each open box is half a cube.

9. For y = ex, minimize D x x y( ) = + =2 2

x e x2 2+ .

–0.4263...

1

D(x)

x

The graph shows a minimum at x ≈ −0.43.

D x x e x ex x′ = + +−( ) /1

22 22 2 1 2 2( ) ( )

D′(x) = 0 ⇔ 2x + 2e2x = 0 ⇔ x = −e2x

Because x appears both algebraically andexponentially, there is no analytic solution.Solving numerically gives x ≈ −0.4263. Bygraphing D(x), D(−0.4263) is a minimum.Closest point to the origin is(x, y) = (−0.4263… , 0.6529…).

10. Minimize A(r) = πr2 + 2rx, r ≥ 20.2π r + 2x = 400 ⇒ x = 200 − π rx ≥ 100 ⇒ r ≤ 100/π∴ domain is 20 ≤ r ≤ 100/π.A(r) = π r2 + 2r(200 − π r) = 400r − π r2


10,000

20 31.83...

r

A(r)

The graph shows a minimum at endpoint x = 20.A′ = 400 − 2πrA′ = 0 ⇔ r = 200/π = 63.6… (out of domain)A′ > 0 for all r in the domain.∴ minimum occurs at left end of domain, r = 20.x = 200 − 20π = 137.168…Make radius of semicircles 20 m and straightsections 137.17 m.

11.

y

x

x – 1

L

1

8

L x x y( ) = +2 2 .

Domains: x ≥ 1, y ≥ 8Minimize L2(x) = x2 + y2.

Using similar triangles, y

x x= 8

1–⇒ y

x

x= 8

1–.

∴ L2(x) = x2 + 64

1

2

2

x

x( – )

10

1 5

x

L

The graph shows a minimum of L(x) at x ≈ 5.

(L2)′(x) = 2x − 128

1 3

x

x( – )

(L2)′(x) = 0 ⇔ 2128

1 3xx

x=

( – )⇔

x = 0 (out of domain) or (x − 1)3 = 64 ⇔ x = 5

By graph, L(x) is a minimum at x = 5.Shortest ladder has length L( )5 5 5= ≈ 11.18 ft.

12. Let x and y be the segments shown.

x

yL

7

5

L x x y( ) = + + +( ) ( )7 52 2

Maximize L2(x) = (x + 7)2 + (y + 5)2.Using similar triangles, y/7 = 5/x ⇒ y =35/x.∴ L2(x) = (x + 7)2 + (35/x + 5)2

L2(x) = x2 + 14x + 49 + 1225/x2 + 350/x + 25

5.59...

x

L

20

The graph shows a minimum of L(x) at x ≈ 5.6.( ( ))L x x x x2 2 32 14 350 2450′ = + − −− −

By numerical solution, (L2)′ = 0 at x ≈ 5.5934… .(Exact answer is x = 1753 .)But a minimum distance L in the hall impliesthat the maximal ladder that will go through thehall is at x = 5.5934… .L2(5.5934…) = 285.3222…L(5.5934…) = 16.8914…No ladder longer than 16.8 ft (rounded down) canpass through the hall.

13. Let r = radius, h = height.V = πr2h2r + 2h = 1200 ⇒ h = 600 − r∴ V = πr2(600 − r) = π(600r2 − r3)

V

400

r

The graph shows a maximum at r ≈ 400.V′ = π(1200r − 3r2)V′ = 0 ⇔ r = 0 or r = 400From graph, maximum is at r = 400.h = 600 − 400 = 200Maximum volume occurs with rectangle400 mm wide (radius), 200 mm high.

14. Rotating a square does not give the maximumvolume. The solution to Problem 13 gives acounterexample. Repeating the calculations withperimeter P instead of 1200 gives r = (1/3)P andh = (1/6)P, showing that the proportions formaximum volume are with radius twice theheight.

15. a. Let r = radius, h = height.V = πr2h = π(3.652)(10.6) = 141.2185π

= 443.6510… cm3


b. A = 2πrh + 2πr2

V = πr2h = 141.2185π ⇒ h = 141.2185/r2

∴ A = 2πr(141.2185/r2) + 2πr2

A r r= +−2 141 2185 1 2π ( . )

c. A

r

4.13...

500

The graph shows a minimum at x ≈ 4.1.A r r′ = − +−2 141 2185 22π ( . )A′ = 2π/r2(−141.2185 + 2r3)A′ = 0 ⇔ r3 = 70.60925 ⇒r = = …70 60925 4 13323 . .Minimum at r = 4.1… because A′ goes fromnegative to positive.

h = =141 2815 70 60925 2 70 609253 3. /( )2. .

= 8.2664…Radius ≈ 4.1 cm, height ≈ 8.3 cmBecause height = 2 × radius, height = diameter.So minimal can is neither tall and narrow norshort and wide.

d. Normally proportioned can is taller andnarrower than minimal can. For normal can,A = 2π(3.65)(10.6) + 2π(3.65)2 =326.8041… .For minimal can, A = 2π(4.13…)(8.26…) +2π(4.13…)2 = 322.014… .Difference is 4.78… cm2.Percent: (4.789…)(100)/326.80… = 1.465…≈ 1.5% of metal in normal can

e. Savings = (0.06)(20 × 106)(0.01465…)(365) =6.419… × 106, or about $6.4 million!

16. a. C r r k rh r k r( ) .= + = + −2 2 2 282 4372 2 1π π π πC r rk r′ = − −( ) .4 282 437 2π π

= −−4 70 609252 3πr kr( . )

C′(r) = 0 at r k= 70 609253 . /

C r k r″ = + >−( ) .4 564 874 03π π for all r > 0,so this is a local minimum.If the normal can is the cheapest to make,then 3 65 70 609253. = ⇒. /k

k = = …−70 60925 3 65 1 45203. ( . ) . .

This is reasonable because metal for the endsis cut into circles, so some must be wasted.

b. Now it takes (2r)2 cm2 of metal to make eachend of the can, so the function to minimize isC r r k rh r k r( ) . .= + = + −8 2 8 282 4372 2 1π πC r rk r′ = − −( ) .16 282 437 2π

C r rk

′ = =( ) at0282 437

163

. π

C r k r″ = + >−( ) .16 564 874 03π for all r > 0,so this is a local minimum.If the normal can is the cheapest to make,

then 3 65282 437

16

282 437

16 3 653

3. = ⇒ =. .

( . )

π πk

k

= 1.1404… .To minimize the area (not the cost) of thecan, minimize 8 2 8 282 4372 2 1r rh r r+ = + −π π. .

C r r r′ = = ⇒−( ) – .16 282 437 02π

r = =282 437

163

. π3.8126 cm

h =( )

=141 2185

282 437 169 7099

32

.

. /.

πK cm.

The proportions of this can are closer to thoseof the normal can.

c. If the metal for the ends can be cut withoutwaste, then it takes π(r + 0.6)2 to make eachend and (2πr + 0.5)h to make the sides, sominimizeC(r) = 2π(r + 0.6)2 + (2πr + 0.5)h

= + + + −2 0 6 141 2185 2 0 5 2π π( . ) . ( . )2r r rC r r r′ = + − −( ) ( . ) .4 0 6 282 437 2π π

− − .141 2185 3rC′(r) = 0 at r ≈ 3.9966 by graphing calculator.C r r r″ = + + >− −( ) . .4 564 874 423 6555 03 4π πfor all r > 0, so this is a minimum point.Minimal can has r ≈ 3.9966… ,h ≈ 8.8411… cm.But if the metal for the ends is cut fromsquares, then it takes 4(r + 0.6)2 to make eachend and (2πr + 0.5)h to make the sides, sominimize:C(r) = 8(r + 0.6)2 + (2πr + 0.5)h

= + + + −8 0 6 141 2185 2 0 5 2( . ) . ( . )2r r rπC r r r′ = + − −( ) ( . ) .16 0 6 282 437 2π

− − .141 2185 3rC′(r) = 0 at r ≈ 3.6776… by graphingcalculator.

′′ = + + >− −C r r r( ) 16 564 874 423 6555 03 4. .πfor all r > 0, so this is a minimum point.Minimal can has r ≈ 3.6776… ,h ≈ 10.4411… .This is close to the normal can!

17. a. Volume of cup = π(2.5)2 · 7 = 43.75πLet r = radius of cup, h = height of cup.Minimize A(r) = πr2 + 2πrh.π πr h h r2 243 75 43 75= ⇒ = −. .∴ = + −A r r r( ) .π π2 187 5

100

3.52...

r

A


The graph shows a minimum at r ≈ 3.5 cm.A r r r r r′ = − = −− −( ) . ( . )2 87 5 2 43 752 2 3π π πA′(r) = 0 at r = 43 753 . = 3.5236… .There is a minimum at x = 3.5236… becauseA(r) goes from decreasing to increasing.(See graph.)h r= = =−43 75 43 75 43 752 3 3. ( . ) / .Minimal cup has r ≈ 3.52 cm, h ≈ 3.52 cm.

b. Ratio is d : h = 2r : h = 2 : 1.

c. Current cup design uses π(2.5)2 + π · 5 · 7 =41.25π = 129.59… cm2 = 0.012959… m2 percup, which costs(300,000,000)(0.012959…)(2.00)≈ $7,775,441.82 per year.Minimal cup design uses 3π(43.75)2/3 =117.01… cm2 = 0.011701… m2 per cup,which costs (300,000,000)(0.011701…)(2.00)≈ $7,021,141.88 per year.Switching to minimal cup design wouldsave 754,299.93 ≈ $754,000 per year inpaper costs (about 10% of the current annualpaper bill), but would likely result in loss ofsales because a cup of that shape is hard todrink from.

d. Let r = radius of cup, h = height of cup.π πr h V h V r2 2= ⇒ = −( / )

Minimize ( ) .A r r rh r Vr= + = + −π π π2 2 12 2

A r r Vr r V′ = − = =−( ) at2 2 02 3π π/′′ = + >−A r Vr( ) 2 4 03π for all r > 0, so this is

a minimum.Minimal cup has r = V/π3 ,

h V V V r= = =−( / )( / )π π π2 3 3/ / .

18. a. A = yz = (30 + 0.2x)(40 − 0.2x)A(x) = 1200 + 2x − 0.04x2

Left rectangle: A(0) = 1200 in.2

Right rectangle: A(100) = 1000 in.2

b. A(80) = 1104 in.2

c.

25 100

1000

x

A(x)

The graph shows a maximum at x ≈ 25.A′(x) = 2 − 0.08x = 0 at x = 25.Critical points at x = 0, 25, 100A(25) = 1225 in.2; A(0) = 1200 in.2;A(100) = 1000 in.2 (from part a)Maximum area at x = 25 in., minimum area forx = 100 in.

19. Maximize A(x) = 2xy = 2x cos x.Use 0 ≤ x ≤ π/2 for the domain of x.

1

1

x

A(x)

The graph shows a maximum at x ≈ 0.86.A′(x) = 2 cos x − 2x sin xA′(x) = 0 when x = cot x.Solving numerically gives x ≈ 0.8603… .A(0) = A(π/2) = 0; A(0.8603…) = 1.1221…Maximum area = 1.1221…

20.

50

200

Street

x y

Let x = width of store, y = length of store.Minimize C(x) = 100x + 80(x + 2y).xy y x= ⇒ = −4000 4000 1

C x x x( ) = + −180 640000 1

y ≤ 200 ⇒ x ≥ 20, so domain of x is20 ≤ x ≤ 50.Graph shows minimum at x endpoint x = 50.

20 50

50,000C(x)

x

C x x′ = − =−( ) 180 640000 02

at x = 80 5

3= 59.628… , outside the domain.

C(20) = $35,600.00; C(50) = $21,800.00Minimum cost is at x = 50, y = 4000/50 = 80.Bill should build the store 50 ft × 80 ft.

21. a. A = 0.5xy = 0.5x cot x

lim limtanx x

Ax

x→ →= →

0 0 2

0

0

= =→

limsecx x0 2

1

2

1

2b. Domain of x is 0 < x ≤ π/2.

π/2

0.5

A(x)

x

The graph shows that the area approaches amaximum as x approaches the endpoint x = 0from the positive side.


A x x x x′ =( )1

22(cot – csc )

A′(x) = 0 when x = cos x sin x or2x = 2 sin x cos x = sin 2x,which happens at x = 0.A(π/2) = 0, so the “maximum” occurs at x = 0.But x = 0 is not in the domain; A(x) can getarbitrarily close to 1/2, but never achieve it.

22.y

x

(x, y)

3x

Domain of x is 0 ≤ x ≤ 3.Maximize A = 0.5(3 − x) ( y) = 0.5(3 − x)ex =1.5ex − 0.5xex.

4

2 30

x

A(x)

The graph shows a maximum at x ≈ 2.A′(x) = 1.5ex − 0.5ex − 0.5xex = 0.5ex(2 − x)A′(x) = 0 at x = 2, confirming the graph.A′(x) > 0 for x < 2, and A′(x) < 0 for x > 2,confirming maximum point at x = 2.Maximum area A(2) = e2/2 = 3.69452… .

23. a. Maximize A(x) = 2xy = 2x(9 − x2) =18x − 2x3.Domain: 0 ≤ x ≤ 3

1.7320 3

20

x

A(x)

The graph shows a maximum at x ≈ 1.7.′ = − = = ± =A x x x( ) at 18 6 0 32 ±1.732…

−1.732 is out of the domain.A A A( ) ( ) ; ( 3) .0 3 0 12 3 20 7846= = = = K

Maximal rectangle has width = 2 3,

length = 9 − 3 = 6.

b. Maximize P(x) = 4x + 2y = 4x + 18 − 2x2.

10 3

20

x

P(x)

The graph shows a maximum at x ≈ 1.P′(x) = 4 − 4x = 0 at x = 1P(0) = 18; P(1) = 20; P(3) = 12Maximal rectangle has width = 2,length = 9 − 1 = 8.

c. No. The maximum-area rectangle is 2 3 by 6.The maximum-perimeter rectangle is 2 by 8.

24. a. Maximize V(x) = πx2y = πx2(9 − x2) =9πx2 − πx4.Domain: 0 ≤ x ≤ 3

2.121...0 3

50

x

V(x)

The graph shows a maximum at x ≈ 2.1.V x x x x′ = − = = ±( ) at ,18 4 0 0 4 53π π . .

− 4 5. is out of the domain.

V V V( ) ( ) , ( ) .0 3 0 4 5 20 25= = = =. π63.6172…Maximum is at x y= = − =4 5 9 4 5 4 5. , . . .

Maximal cylinder has radius = =4 5.2.12132… and height = 4.5.

b. Maximize L(x) = 2πxy = 2πx(9 − x2) =18π x − 2π x3.

1.732...0 3

50

x

L(x)

The graph shows a maximum at x ≈ 1.7.L x x x′ = − = = ±( ) at 18 6 0 32π π .

− 3 is out of the domain.

L L L( ) ( ) ; ( ) .0 3 0 3 12 3 65 2967= = = =π K

Maximum is at x y= = − =3 9 3 6, .

Maximal cylinder has radius = =31.7320… and height = 6.

c. Maximize A(x) = 2πx2 + 2π xy = 2π x2 +2π x(9 − x2) = 2π x2 + 18π x − 2π x3.

2.097...0 3

50

x

A(x)

The graph shows a maximum at x ≈ 2.1.A′(x) = 18π + 4π x − 6π x2


A′(x) = 0 at x = ± =1 2 7

32 0971. K or

−1.430…−1.430… is out of the domain.A(0) = 0; A(2.0971…) = 88.2727… ;A(3) = 18π = 56.5486…

Maximal cylinder has radius = + =1 2 7

3

2.0971… and height = =52 4 7

9

–

4.6018… .

d. No. The maximum-volume cylinder hasdimensions different from both of themaximum-area cylinders in parts b and c.

e. No. Rotating the maximum-area rectangledoes not produce the maximum-volumecylinder. But it produces the cylinder withmaximum lateral area.

f. If y = a2 − x2, the paraboloid has radius = a.V = πx2(a2 − x2) = π(a2x2 − x4)V′ = π(2a2x − 4x3)V x x a′ = ⇔ = = ± /0 0 2 or .

V is maximum at x a= / 2.

For the cylinder of maximum volume,(cylinder radius):(paraboloid radius) = /1 2,a constant.Note: This ratio is also constant (1 3/ ) for the

cylinder of maximum lateral area, but is notconstant for the cylinder of maximum totalarea.

25. a. x2 + y2 = 100, 0 ≤ x ≤ 10

Maximize V x x y x x( ) .= ⋅ =π π2 2 22 100 –

b.

100

2000

x

V(x)

8.16...

The graph shows a maximum volume atx ≈ 8.2.

V xx

xx x′ = +( )

–

––

2

1004 100

3

2

2π π

= +–

–

6 400

100

3

2

π πx x

x

V x x′ = = = =( ) at , .0 0

200

3

10 6

38 1649K

V(0) = V(10) = 0

V10 6

3

4000 3

92418 399

= =π. K

Maximal cylinder has radius = 8.1649… ,

height

= =20 3

311 5470. ,K and volume =

2418.39… .

c. Height radius= ⋅ 2

Volume of sphere Vs = ⋅ =4

31000

4000

3π π

Volume of maximal cylinder Vc = 4000 3

9

π

∴ = / V Vc s 3

26. Let r = radius of cone, h = height.

Lateral area A(r) = π r · (slant height) = +πr r h2 2

V r h h r= = ⇒ = −1

35 152 2π π

∴ = +A r r r r( ) π 2 4225 –

h ≥ 2r ⇒ 2r ≤ 15r− 2

Domain of r is 0 7 5 1 95743< ≤ =r . .. K

1

20

r

A(r)

1.957...

The graph shows a minimum of A(r) at endpointr = 1.957… .Minimize ( ) ( ).A r r r2 2 4 2225= + −π( ( )) ( ) atA r r r r2 2 3 3 64 450 0 112 5′ = − = = =−π .

2.1971… , which is out of the domain.A(1.9574…) = 26.915… , lim

rA r

→ += ∞

0( ) .

Minimal cone has radius = = …7 5 1 95743 . .

and height = = =2 2 7 5 3 91483r . .. KMake r ≈ 1.96 ft and h ≈ 3.91 ft.

27. a. Lateral area L(x) = 2π xyDomains: 0 ≤ x ≤ 5 and 0 ≤ y ≤ 7Equation of element of cone is

y x y x= − + ⇒ = − +7

57 1 4 7. .

∴ L(x) = 2πx(−1.4x + 7) = 2π(−1.4x2 + 7x)

5

50

x

L(x)

2.50

The graph shows a maximum of L(x) atx ≈ 2.5.L′(x) = 2π(−2.8x + 7)L′(x) = 0 at x = 2.5.


L′(x) goes from positive to negative atx = 2.5.∴ maximum lateral area at radius x = 2.5 cm.

b. Total area A(x) = 2πxy + 2πx2

= 2πx(−1.4x + 7) + 2πx2

A(x) = 2π(7x − 0.4x2)

150

A(x)

x

50

The graph shows a maximum at endpointx = 5.A′(x) = 2π(7 − 0.8x) = 0 at x = 8.75, out ofdomain.∴ maximum is at an endpoint, x = 5.A(0) = 0; A(5) = 2π(52) = 50π = 157.07…Maximum area is with the degenerate cylinderconsisting only of the top and bottom, radius5 and height 0.

28. a. Let r = radius of cone, h = height of cone(constants).Let (x, y) be a sample point on cone element.Domain of x is 0 ≤ x ≤ r.L(x) = 2πxy.Equation of element of cone isy = (−h/r)x + h.∴ L(x) = 2πx[(−h/r)x + h] = 2πh(−x2/r + x)L′(x) = 2πh(−2x/r + 1)L′(x) = 0 at x = r/2.L′(x) goes from positive to negative atx = r/2.∴ maximum lateral area at radius x = r/2.

b. A(x) = 2πxy + 2πx2

= 2πx[h − (h/r)x] + 2πx2

A(x) = 2π[(1 − h/r)x2 + hx]A′(x) = 2π[2(1 − h/r)x + h] = 0 at

xh

h r= –

( – / )2 1

A′(x) = 0 at xh

h r

rh

h r= =–

– / ( – )2 2 2If h ≤ 2r, then A′(x) ≠ 0 for all x ≤ r, so inthis case the critical points are the endpoints,x = 0, r.A(0) = 0; A(r) = 2πr2

If h ≥ 2r, then 02

≤ ≤rh

h rr

( – ), so this is a

critical point; Arh

h r

rh

h r2 2

2

( – ) ( – )

= π.

A′(x) goes from positive to negative at

xrh

h r=

2( – ).

Maximum area at xrh

h r=

2( – ) if h ≥ 2r;

x = r otherwise.

c. From part b, the maximal cylinder degeneratesto two circular bases if the radius of the coneis at least half the height.

29. Maximize V = π y2x.Ellipse equation is (x/9)2 + (y/4)2 = 1, fromwhich y2 = (16/81)(81 − x2).∴ V = (16π/81)(81x − x3)Domain: 0 ≤ x ≤ 9

150

0 5.196... 10

x

V

The graph shows a maximum V at x ≈ 5.2.V′ = (16π/81)(81 − 3x2) = (16π/27)(27 − x2)V′ = 0 at x = ± = ± …27 5 196.−5.196… is out of the domain.V V V( ) ( ) ; ( ) .0 9 0 27 32 3 174 1= = = = …πAt x = 5.196… , y2 = (16/81)(81 − 27) =32/3 ⇒ y = 32 3/ = 3.2659…

∴ maximum volume ≈ 174.1 cm3 at radius ≈3.27 m and height ≈ 5.20 m.

30. Maximize C(y) = πy2x, the volume of the cylinder.The parabola has an equation of the formx = ay2 + 16.0 = a ⋅ 16 + 16 ⇒ a = −1 ⇒ x = 16 − y2

V(y) = πy2(16 − y2) = π(16y2 − y4)Domain: 0 ≤ y ≤ 4

0 4

300

2.828...

y

F(y)

The graph shows a maximum V(y) at y ≈ 2.8.C′(y) = π(32y − 4y3) = 4πy(8 − y2) = 0 aty = ±0 8, .

y = − 8 is out of the domain.

C C C( ) ( ) , ( ) .0 4 0 8 64 201 0619= = = = …πMaximum C(y) at y = 8.

At y x= =8 8, .

Maximal cylinder has radius = ≈8 2 83. m,height = 8 m, and volume = 64π ≈ 201.1 m3.Maximize F(y), the volume of the frustum.Note that Vf = (1/3)πh(R2 + r2 + Rr), where


Vf = volume of frustum, h = height of frustum,R = larger radius, and r = smaller radius.

∴ = + +F y x y y( ) ( )1

316 42π

= − + +1

316 4 162 2π ( )( )y y y

F y y y y( ) ( )= + − −1

3256 64 4 3 4π

0 4

300

1.821...

y

F(y)

The graph shows a maximum F(y) at y ≈ 1.8.

′ = − −F y y y( ) ( )1

364 12 42 3π

F ′(y) = 0 ⇔ 64 − 12y 2 − 4y 3 = 0Solving numerically for y close to 1.8 givesy ≈ 1.8216… .Substituting y = 1.8216… givesx = 16 − y 2 ≈ 12.6816… .

F x y y( . ) ( )1 8216

1

316 42K = + + ≈π

353.318… .Maximal frustum has radii = 4 m and ≈1.82 m,height ≈ 12.68 m, and volume ≈ 353.3 m3.The maximal frustum contains ≈ 152.3 m3 morethan the maximal cylinder, about 75.7% more.

31. a. If f (c) is a local maximum, thenf (x) − f (c) ≤ 0 for x in a neighborhood of c.For x to the left of c, x − c < 0.

Thus, f x f c

x c

( ) – ( )

–≥ 0 (neg./neg.) and

′ = ≥→ −

f cf x f c

x cx( ) lim

( ) – ( )

–.

00

For x to the right of c, x − c > 0.

Thus, f x f c

x c

( ) – ( )

–≤ 0 (neg./pos.) and

′ = ≤→ +

f cf x f c

x cx( ) lim

( ) – ( )

–.

00

Therefore, 0 ≤ f ′(c) ≤ 0.Because f ′(c) exists, f ′(c) = 0 by the squeezetheorem, Q.E.D.

b. If f is not differentiable at x = c, then f ′(c)does not exist and thus cannot equal zero.Without this hypothesis, the reasoning inpart a shows only that f ′(x) changes sign atx = c. There could be a cusp, a removablediscontinuity, or a step discontinuity at x = c.

c. The converse would say that if f ′(c) = 0, thenf (c) is a local maximum. This statement is

false because f (c) could be a local minimumor a plateau point.

32. a. Let x = length of corral (parallel to wall), y =width of corral (perpendicular to wall).A = xyIf x ≤ 600, then 1000 = x + 2y ⇔ y = 500 − 0.5x.If x ≥ 600, then 1000 = x + 2y + (x − 600) ⇔y = 800 − x.

∴ =≤>

Ax x x

x x x

500 0 5 600

800 600

2

2

– . ,

– ,

500

x

150,000A

The graph shows a maximum A at x ≈ 500.

Ax x

x x′ =

<>

500 600

800 2 600

– ,

– ,

For x < 600, A′ = 0 ⇔ x = 500.For x > 600, A′ = 0 ⇔ x = 400 (out of thedomain).A′ is undefined at the cusp, x = 600.Maximum at x = 500 because graph isparabola opening downward.Or: Check the critical points.A(500) = 500(500) − 0.5(500)2 = 125,000A(600) = 500(600) – 0.5(600)2 = 120,000 ft2

Maximum area is 125,000 f t 2 at x = 500 ft.

b. If x ≤ 400, then 1000 = x + 2y ⇔y = 500 − 0.5x.If x ≥ 400, then 1000 = x + 2y + (x − 400)⇔ y = 700 − x.

∴ =≤>

Ax x x

x x x

500 0 5 400

700 400

2

2

– . ,

– ,

400

x

150,000A

The graph shows a maximum A at the cusp,x = 400.

′ =<>

Ax x

x x

500 400

700 2 400

– ,

– ,

For x < 400, A′ = 0 ⇔ x = 500 (out of thedomain).For x > 400, A′ = 0 ⇔ x = 350 (out of thedomain).


Maximum area is at the cusp, x = 400.A = 700(400) − 4002 = 120,000Maximum area is 120,000 f t 2.

c. If x ≤ 200, then 1000 = x + 2y ⇔y = 500 − 0.5x.If x ≥ 200, then 1000 = x + 2y + (x − 200) ⇔y = 600 − x.

∴ =≤>

Ax x x

x x x

500 0 5 200

600 200

2

2

– . ,

– ,

300

x

150,000A

The graph shows a maximum A at x ≈ 300.

′ =<>

Ax x

x x

500 200

600 2 200

– ,

– ,

For x < 200, A′ = 0 ⇔ x = 500 (out of thedomain).For x > 200, A′ = 0 ⇔ x = 300.A′ is undefined at the cusp, x = 200.Maximum area is at x = 300 because graph isa parabola opening downward.Or: Check critical points.A(300) = 600(300) − 3002 = 90,000A(200) = 500(200) − 0.5(200)2 = 80,000 ft2

Maximum area is 90,000 ft2 at x = 300 ft.



x

y

x

y

Q3. Q4.

x

y

x

y

Q5. Q6.

x

y

x

y

Q7. Q8. Sample answer:

x

y

2

1

y

x

Q9. tan x C+ Q10. B

1. a. y = 4 − x2

dV = 2πxy ⋅ dx = 2π(4x − x3) dx

b. 0 = 4 − x2 = (2 − x)(2 + x) at x = ±2

V x x dx x x= − = −

∫ 2 4 2 2

1

43

0

22 4

0

2

π π( )

= 8π = 25.1327…

c. y = 4 − x2 ⇒ x2 = 4 − yUpper bound of solid is at y = 4.dV = πx2 dy = π(4 − y) dy

V y dy y y= − = − =∫ π π( ) ( . )4 4 0 50

42

0

4

8π = 25.1327… , which is the same answeras by cylindrical shells in part b.

2. a. Height of cylinder = 8 − x

b. y = x2/3 ⇒ x = y3/2

dV = 2π(8 − x)y dy = 2π(8 − y3/2)y dy= 2π(8y − y5/2) dy

c. At x = 8, y = 82/3 = 4.

V y y dy y y= − = −

∫ 2 8 2 4

2

75 2 2 7 2

0

4

0

4

π π( )/ /

= = …384

7172 3387π .

d. dV = π y2 dx = π x4/3 dx

V x dx x= = = =∫ π π π4 3 7 3

0

8

0

83

7

384

7/ /

172.3387… , which is the same as thevolume by cylindrical shells in part c.

3. The graph shows y = −x2 + 4x + 3, from x = 1 tox = 4, sliced parallel to the y-axis, with samplepoint (x, y), rotated about the y-axis, showingback half of solid only.

(x, y)

1 4

1 x

y

dV = 2πxy ⋅ dx = 2π(−x3 + 4x2 + 3x) dx

V x x x dx= − + +∫ 2 4 33 2

1

4

π ( )

≈ 268.6061… (exactly 85.5π)


Circumscribed hollow cylinder of radii 1 and4 and height 7 has volume π(42 − 12) ⋅ 7 =329.8… , which is a reasonable upper bound forthe calculated volume.

4. The graph shows y = x2 − 8x + 17, from x = 2 tox = 5, sliced parallel to the y-axis, with samplepoint (x, y), rotated about the y-axis, showingback half of solid only.

(x, y)

x

y

2 5

5

dV = 2πxy ⋅ dx = 2π(x3 − 8x2 + 17x) dx

V x x x dx= − +∫ 2 8 173 2

2

5

π ( )

≈ 117.8097… (exactly 37.5π)Circumscribed hollow cylinder of radii 2 and 5and height 5 has volume π(52 − 22) ⋅ 5 =329.8… , which is a reasonable upper boundfor the calculated volume. Assuming that thepart of the solid above y = 2 could be fit intothe “trough,” the volume is approximatelyπ(52 − 22) ⋅ 2 = 131.9… , which is close to thecalculated volume.

5. The graph shows x = −y2 + 6y − 5, intersectingy-axis at y = 1 and y = 5, rotated about thex-axis, showing back half of solid only.

(x, y)

x

y

1

5

4

(0, y)

dV = 2πy(x − 0) ⋅ dy = 2π(−y3 + 6y2 − 5y) dy

V y y y dy= − + −∫ 2 6 53 2

1

5

π ( )

≈ 201.0619… (exactly 64π)Circumscribed hollow cylinder of radii 1 and 5and height 4 has volume π(52 − 12) ⋅ 4 =301.5… , which is a reasonable upper bound forthe calculated volume.

6. The graph shows x = y2 − 10y + 24, intersectingy-axis at y = 4 and 6, rotated about the x-axis,showing back half of solid only.

x

y6

–1

(x, y)(0, y)

dV = 2πy(0 − x) ⋅ dy = 2π(−y3 + 10y2 − 24y) dy

V y y y dy= − + −∫ 2 10 243 2

4

6

π ( )

≈ 41.8879… exactly 40

3π

Circumscribed hollow cylinder of radii 4 and 6and height 1 has volume π(62 − 42) ⋅ 1 =62.83… , which is a reasonable upper bound forthe calculated volume.

7. Figure 8-4h shows y = x3, intersecting the liney = 8 at x = 2 and the line x = 1. Rotate aboutthe y-axis. Slice parallel to the y-axis. Picksample points (x, y) on the graph and (x, 8) onthe line y = 8.dV = 2πx(8 − y) ⋅ dx = 2π(8x − x4) dx

V x x dx= −∫ 2 8 4

1

2

π ( )

≈ 36.4424… (exactly 11.6π)Circumscribed hollow cylinder of radii 2 and 1and height 7 has volume π(22 − 12) ⋅ 7 = 65.9… ,which is a reasonable upper bound for thecalculated volume.

8. The graph shows y = 1/x, intersecting line y = 4at x = 0.25 and the line x = 3, rotated about they-axis, showing back half of solid only.

3

x(x, y)

y = 4

0.25

y(x, 4)

dV = 2πx ⋅ (4 − y) · dx = 2π(4x − 1) dx

V x dx= −∫ 2 4 10 25

3

π ( ).

≈ 95.0331… (exactly 30.25π)Circumscribed hollow cylinder of radii 0.25 and 4and height 3.7 has volume π(32 − 0.252)(3.7) =103.8… , which is a reasonable upper bound forthe calculated volume.

9. Figure 8-4i shows y = 1/x2, intersecting the linex = 5 at y = 0.04 and the line y = 4. Rotate


about the x-axis. Slice parallel to the x-axis. Picksample points (x, y) on the graph and (5, y) onthe line x = 5.dV = 2πy(5 − x) ⋅ dy = 2π(5y − y1/2 ) dy

V y y dy= −

≈ …∫ 2 5 1 2

0 04

4

π

π

( )

217.8254 (exactly 69.336 )

/

.

Circumscribed cylinder of radius 4 and height 4.5has volume π ⋅ 42 ⋅ 4.5 = 226.1… , which is areasonable upper bound for the calculatedvolume.

10. The graph shows y = x2/3 , intersecting the liney = 1 and intersecting the line x = 8 at y = 4,rotated about the x-axis, showing back half ofsolid only.

(8, y)(x, y)

x

y

81

1

4

dV = 2πy(8 − x) ⋅ dy = 2π(8y − y5/2 ) dy

V y y dy= −

≈ …

∫ 2 8

473

7

5 2

1

4

π

π

( )

149.0012 exactly

/


11. Figure 8-4j shows y1 = x2 − 6x + 7 and y2 =x + 1, intersecting at (1, 2) and (6, 7). Rotateabout the y-axis. Slice parallel to y-axis. Picksample points (x, y1) and (x, y2).dV = 2πx ⋅ (y2 − y1) ⋅ dx = 2π(−x3 + 7x2 − 6x) dx

V x x x dx= − + −

≈ …

∫ 2 7 6

1455

6

3 2

1

6

π

π

( )

458.1489 exactly


12. The graph shows y x x y= ⇒ =11 3

13/ and y =

0.5x2 − 2 ⇒ x2 = 2y + 4, intersecting at (8, 2)in Quadrant I and bounded by the x-axis, rotatedabout the x-axis, showing back half of solidonly.

2

84

(x1, y)

(x2 , y)

x

y

dV = 2πy(x2 − x1) dy = 2π(2y2 + 4y − y4) dy

V y y y dy= + −

≈

∫ 2 2 4

1313

15

2 4

0

2

π

π

( )

43.5634 exactly K

Circumscribed cylinder of radius 2 and height 8has volume π ⋅ 22 ⋅ 8 = 100.5… , which is areasonable upper bound for the calculatedvolume.

13. Figure 8-4k shows y = x3/2 , from x = 1 to x = 4.Rotate about the line x = 5. Slice parallel to they-axis. Pick sample point (x, y).dV = 2π(5 − x)y ⋅ dx = 2π(5x3/2 − x5/2 ) dx

V x x dx= −

≈ …

∫ 2 5

161 5676 513

7

3 2 5 2

1

4

π

π

( )

. exactly

/ /

Circumscribed cylinder of radius 4 and height 8has volume π(42) ⋅ 8 = 402.1… , which is areasonable upper bound for the calculatedvolume.

14. The graph shows y = x− 2, from x = 1 to x = 2,rotated about the line x = 3, showing back half ofsolid only.

(x, y)

1

1 2 3

x

y

dV x y dx x x dx= − ⋅ = −− −2 3 2 3 2 1π π( ) ( )

V x x dx= −

≈ … −

− −∫ 2 3 2 1

1

2

π

π

( )

5.0696 (exactly (3 2 ln 2))Circumscribed hollow cylinder of radii 1 and 2and height 1 has volume π(22 − 12) ⋅ 1 = 9.4… ,which is a reasonable upper bound for thecalculated volume.

15. The graph shows y1 = x4 and y2 = 5x + 6,intersecting at x = −1 and x = 2, rotated aboutthe line x = 4, showing back half of solidonly.


(x, y2)

(x, y1)

–1 2 4

16

x

y

dV = 2π(4 − x)(y2 − y1) dx= 2π(4 − x)(5x + 6 − x4) dx

V x x x dx= − + −

≈ …−∫ 2 4 5 6 4

1

2

π

π

( )( )

390.1858 (exactly 124.2 )Circumscribed hollow cylinder of radii 2 and 5and height 16 has volume π(52 − 22) ⋅ 16 =1055.5… , which is a reasonable upper bound forthe calculated volume.

16. The graph shows y1 = x = x1/2 and y2 = 6 − x,intersecting at x = 4 in Quadrant I and boundedby the line x = 1, rotated about the line x = −1,showing back half of solid only.

(x, y2)

(x, y1) x

41

1

5

y

–1

dV = 2π(x + 1)(y2 − y1) dx= 2π(x + 1)(6 − x − x1/2) dx

V x x x dx= + − −

≈

∫ 2 1 6

109 5368 3413

15

1 2

1

4

π

π

( )( )

. exactly

/

K


17. Figure 8-4l shows y1 = −x2 + 4x + 1 and y2 =1.4x, intersecting at x = 0 and x = 3.3740…(store as b). Rotate about the line x = −2. Sliceparallel to the y-axis. Pick sample points (x, y1)and (x, y2).dV = 2π(x + 2) ⋅ (y1 − y2) dx

= 2π(x + 2)(−x2 + 4x + 1 − 1.4x) dx

V x x x dxxb

= + − + + −∫ 2 2 4 1 1 42

0π ( )( . )

≈ 163.8592…Circumscribed hollow cylinder with radii 2 and5.4 and height 4 has volume π(5.42 − 22)4 =316.1… , a reasonable upper bound for calculatedvolume.

18. The graph shows y1 and y2 as described inProblem 17, but rotated about the line y = −1,showing back half of solid only. Slicingperpendicular to the x-axis is appropriatebecause slicing parallel to it would give stripsof length (curve) minus (curve) at some valuesof y and (curve) minus (other curve) at othervalues of y.

0 3

1

(x, y1)

(x, y2)

x

y

y = –1

dV = π [(y1 + 1)2 − (y2 + 1)2] dx= π [(−x2 + 4x + 2)2 − (1.4x + 1)2] dx

Limits of integration are 0 to b, whereb = 3.3740… , as in Problem 17.

V x x dxxb

= − + + − +∫ π[( ) ( . ) ] 22 2

04 2 1 4 1

≈ 181.0655…Circumscribed hollow cylinder of radii 2 and 6and height 3.4 has volume π(62 − 22) · 3.4 =341.8… , a reasonable upper bound for thecalculated volume.

19. Slice perpendicular to the y-axis. Pick samplepoints (x, y) on the graph of y = x3 and (1, y)on the line x = 1.y = x3 ⇒ x = y1/3; y1/3 = 1 at y = 1dV = π ( x2 − 12) dy = π (y2/3 − 1) dy

V y dy= − ≈ …∫ π ( ) ./2 3

1

8

1 36 4424

(exactly 11.6π ), which agrees with the answer toProblem 7.

20. See the graph for Problem 8. Sliceperpendicular to the y-axis. Pick sample points(x, y) on the graph of y = 1/x and (3, y) on theline x = 3.dV x dy y dy= − = − −π π( ) ( ) 3 92 2 2

V y dy= − ≈ …−∫ π ( ) .9 95 03312

1 3

4

/(exactly

30.25π ) , which agrees with the answer toProblem 8.

21. The graph shows y = x1/3, from x = 0 to x = 8,rotated about the x-axis, showing back half ofsolid only.


(8, y)(x, y)

8

2

x

y

y = x1/3 ⇒ x = y3

dV = 2πy ( 8 − x) ⋅ dy = 2π ( 8y − y4) dy

V y y dy y y= − = −

∫ 2 8 2 4

1

54 2 5

0

2

0

2

π π( )

= 2π ( 16 − 6.4) = 19.2π = 60.3185789…R8 = 19.3662109… π = 60.8407460…R100 = 19.2010666… π = 60.3219299…R1000 = 19.2000106… π = 60.3186124…Rn is approaching 19.2π as n increases.

22. a. y = sin x from x = 0 to x = 2, rotated aboutthe y-axis, as in Figure 8-4m. Slice parallelto the y-axis. Pick sample point (x, y) onthe graph.dV = 2πxy ⋅ dx = 2πx sin x dx

V x x dx= ≈ …∫ 2 10 94270

π sin .2

numerically (exactly 2π ( sin 2 − 2 cos 2),integrating by parts).

b. The integrand, x sin x, is a product of twofunctions, for which the antiderivative cannotbe found using techniques known so far.

23. a. x = 5 cos t, dx = −5 sin t dty = 3 sin t, dy = 3 cos t dtSlice parallel to the x-axis, then rotate aboutthe x-axis. Pick sample points (−x, y) at theleft end of the strip and (x, y) at the right end.dV = 2πy [ x − (−x)] ⋅ dy = 4πxy dy

= 4π ( 5 cos t)(3 sin t)(3 cos t dt)= 180π cos2 t sin t dt

Limits of integration are y = 0 to y = 3.At y = 0, t = 0. At y = 3, t = π/2.

V t t dt= ∫ 180 2

0

2

ππ

cos sin/

= −60 30

2π πcos

/t

= −60π ( 0 − 1) = 60π = 188.4955…b. Slice the region in Quadrant I perpendicular to

the x-axis, then rotate about the x-axis. Picksample point (x, y) on the graph.dV = πy2 dx = π ( 3 sin t)2(−5 sin t dt)

= −45π sin3 t dtLimits of integration are from x = −5 tox = 5.At x = −5, t = π. At x = 5, t = 0.

V t dt= − ≈ …∫ 45 188 495530

ππ

sin .

(exactly 60π ) , which agrees with the volumefound in part a.

The integral can be found algebraicallyusing the Pythagorean properties fromtrigonometry.sin3 t = (1 − cos2 t) sin t = sin t − cos2 t sin t

V t dt t t dt= − − −∫∫ 45 45 20

π πππ

sin cos sin0

( )

= −45 15 3 0π π πcos cost t

= 45π − 15π − (−45π ) + (−15π ) = 60πc. Slice the region parallel to the line x = 7 and

rotate about that line. Pick sample points(x, y) and (x, −y) on the upper and lowerbranches.dV = 2π ( 7 − x)[y − (−y)] dx

= 4π ( 7 − 5 cos t)(3 sin t)(−5 sin t dt)= −60π ( 7 − 5 cos t)(sin2 t) dt

Limits of integration are t = π to t = 0, as inpart b.V ≈ 2072.6169… (exactly 210π 2, using thehalf-argument properties for sin2 t, as inProblem 16 of Problem Set 5-9, or by usingintegration by parts as in Chapter 9).



4

16

x

y

0 1 4

16

x

y

Q3. A x dx= ∫ 2

1

4

Q4. A x= 1

33

1

4

Q5. A = 21

Q6.

4

16

x

y

Q7. V x dx= ∫ 2 3

1

4

π Q8. V x= π2

4

1

4

Q9. V = 127.5π Q10. E

1. a.y

x

2

1

0


b. L e en n

n

≈ +

= …=

∑ ( . ) [ – ]. . ( – )0 4 2 0 4 0 4 1 2

1

5

6.7848c. dy = e x dx

dL dx dy e dxx= + = +2 2 21

L e dxx= + ≈ …∫ 1 6 78862

0.

2

numerically

2. a.y

x

3

1

0

b. L n n

n

≈ +

= …=

∑ ( . ) [ – ]. . ( – )0 6 2 22 0 6 0 6 1 2

1

5

7.7853c. dy = (2x ln 2) dx

dL dx dy dxx= + = +2 2 21 2 2( ln )

L dxx= + ≈∫ 1 2 2 7 79202

0

3

( ln ) . K

numerically

3. a.

1.50

10

x

y

b. L n nn

≈ +=

∑ ( . ) [tan . – tan . ( – )]0 3 0 3 0 3 12 2

1

5

= 14.4394…

c. dy = sec2 x dx

dL dx dy x dx= + = +2 2 41 sec

L x dx= + ≈∫ 1 14 44884

0

1 5

sec.

. K

numerically

4. a.

1.50

10

x

y

1

b. L n nn

≈ +=

∑ ( . ) [sec . – sec . ( – )]0 3 0 3 0 3 12 2

1

5

= 13.7141…

c. dy = sec x tan x dx

dL dx dy x x dx= + = +2 2 2 21 tan sec

L x x dx= + ≈ …∫ 1 2 2

0

1 5

tan sec/

13.7304

numerically

5. a.

6–1

9

1 x

y

b. dy = (2x − 5) dx

dL dx dy x dx= + = +2 2 21 2 5( – )

L x dx= + ≈∫ 1 2 5 15 86172

1

6

( – ) . K

c. Low point is (2.5, −3.25). Chords from(1, −1) to (2.5, −3.25) and from (2.5, −3.25)to (6, 9) have combined length 7 3125. +

162 3125 15 4. = …. , which is a reasonablelower bound for L.

6. a.

0 4

4

x

y

b. dy = (4 − 2x) dx

dL dx dy x dx= + = +2 2 21 4 2( – )

L x dx= + ≈∫ 1 4 2 9 29352

0

4

( – ) . K

c. Chords from (0, 0) to (2, 4) and from(2, 4) to (4, 0) have combined length2 20 8 9442= …. , which is a reasonablelower bound for L.

7. a.

2

16

–1

x

y


b. dy = −4x3 dx

dL dx dy x dx= + = +2 2 61 16

L x dx= + ≈

−∫ 1 16 18 24706

1

2

. K

c. Chords from (−1, 15) to (0, 16) and (0, 16) to

(2, 0) have combined length 2 260+ =17.5… , which is a reasonable lower boundfor L.

8. a.

–1 9

50

x

y

b. dy = (3x2 − 18x + 5) dx

dL dx dy= +2 2

= + +1 3 18 52 2( – )x x dx

L x x dx= + + ≈−∫ 1 3 18 5 219 48732 2

1

9

( – ) . K

c. Using five chords with ∆x = 2, L ≈204.4605… , which is a reasonable lowerbound for L.

9. a.

1 e

1x

y

b. dy x x dx x x dx= ⋅ =− −2 21 1ln ln

dL dx dy x x dx= + = +2 2 1 21 2( ln )–

L x x dxe

= + ≈ …∫ 1 2 7 60431 2

0 1( ln )–

..

c. Chords from x = 0.1 to x = 1 and from x = 1to x = e have combined length 7.3658… ,which is a reasonable lower bound for L.

10. a.

5

5 4π0

x

y

b. dy = (sin x + x cos x) dx

dL dx dy= +2 2

= + +1 2(sin cos )x x x dx

L x x x dx= + + ≈∫ 1 2

0

4

(sin cos )π

54.1699K

c. Eight chords of ∆x = π/2 extend from middleto high to middle to low points on the graph.Lengths sum to 52.6109… , a reasonablelower bound for L.

11. a.

10

1.50

x

y

b. dy = sec2 x dx


L x dx= + ≈∫ 1 14 44884

0

1 5

sec.

. K

c. Distance between the endpoints is14.1809… , which is a reasonable lowerbound for L.

12. a.

10

1.50

x

y

b. dy = sec x tan x dx

dL x x dx= +1 2(sec tan )

L x x dx= + ≈∫ 1 2

0

1 5

(sec tan ).

13.7304K

c. The distance between the endpoints is13.2221… , which is a reasonable lowerbound for L.

13. a.

5

5y

x


b. dx = −15 cos2 t sin t dt, dy = 15 sin2 t cos t dt

dL dx dy= +2 2

= +(– cos sin ) ( sin cos )15 152 2 2 2t t t t dt

L t t t t dt= +∫ (– cos sin ) ( sin cos )15 152 2 2 2

0

2π

≈ 30

To see why the answer is so simple,transform the radicand and use thefundamental theorem.

L t t t t dt= +∫ 225 2 2 2

0

2

(sin cos ) (cos sin )π

= ∫7 5 2 2. 0

2

( sin cos )t t dtπ

= ∫7 5 22

0

2

. sin t dtπ

= = ⋅∫ ∫7 5 2 7 5 4 20

2

0

2

. | | ./

sin sint dt t dtπ π

=

− =30

1

22 30

0

2

( ) (exactly!)cos/

tπ

c. Circle of radius 5 (i.e., x = 5 sin t, y =5 cos t) has circumference 10π = 31.4152… ,which is close to the calculated value of L.

14. a.

5

10

x

y

b. dx = 5(−2 sin t + 2 sin 2t) dtdy = 5(2 cos t − 2 cos 2t) dt

dL dx dy= + =2 2

[ (– sin sin )] [ ( cos – cos )]5 2 2 2 5 2 2 22 2t t t t+ + dt

L =

[ (– sin sin )] [ ( cos – cos )]5 2 2 2 5 2 2 22 2

0

2

t t t t dt+ +∫π

≈ 80To see why the answer is so simple,transform the radicand algebraically anduse the fundamental theorem.

L t t t t dt= ∫10 2 2 2 2 2– sin sin – cos cos0

2π

= −∫10 2 10

2

– cos cost dt A Bπ

(using ( ))

=+∫10 2

1

1

2

0

2 – cos

cos

t

tdt

π

=+∫10 2

10

2 |sin |

cos

t

tdt

π

= + −∫20 2 1 1 2

0( cos ) sint t dt/ ( )

π

= − +40 2 1 1 2

0( cos )t / π

= − + + =40 2 1 1 40 2 1 1 801 2 1 2( – ) ( )/ /

c. Maximum/minimum values of y are±7 5 3. . Circle of radius 7.5 3 has

circumference 15 3 81 6209π = . .K

15. a.

4

4

x

y

b. dx = (−5 sin t + 5 sin 5t) dtdy = (5 cos t − 5 cos 5t) dt

dL dx dy= +2 2 =

(– sin sin ) ( cos – cos )5 5 5 5 5 52 2t t t t dt+ +

L t t t t dt= + +∫ (– sin sin ) ( cos – cos )5 5 5 5 5 52 2

0

2π

≈ 40To see why the answer is so simple,transform the radicand and use thefundamental theorem.

L t t t t dt= ∫5 2 2 5 2 5 0

2

– sin sin – cos cosπ

= −∫5 2 1 40

2

– cos cos ( )t dt A B(using )π

=+∫5 2

1 4

1 4

2

0

2 – cos

cos

t

tdt

π

=+∫5 2

4

1 40

2 |sin |

cos

t

tdt

π

= + −∫40 2 1 4 41 2

0

4

( cos ) sin/

t t dt/ ( )π

= − +20 2 1 4 1 20

4( cos )

/t / π

= − + ⋅ =0 20 2 2 40

c. Maximum/minimum values of x, y are±3 3. Circle of radius 3 3 has circumference32.6483… , which is close.


16. a.

t = 4π

5

5

x

y

b. dx = (−sin t + sin t + t cos t) dt = t cos t dtdy = (cos t − cos t + t sin t) dt = t sin t dt

dL dx dy t t t t dt

t dt t dt t

= + = += = ≥

2 2 2 2( cos ) ( sin )

| | (because 0)

L t dt t= = = =∫ 0 5 8 78 95682

0

42. .

0

4π ππ K

c. Circle of radius 4π = 12.5663… would havecircumference = 8π 2.

17. a.

40

30

x

y

b. dy = 6x1/2 dx

dL dx dy x dx= + = +2 2 1 36

L x dx x dx= + = +∫ ∫1 361

361 36 36

0

41 2

4

( ) /

0 ( )

= + =1

541 36

1

54145 13 2

0

4 3 2( ) ( – )/x /

= 32.3153…

c. The chord connecting the endpoints has length32.2490… , which is a reasonable lowerbound for L.

18. a.

1

1 2

x

y

b. dy x x dx= − −( / ) 2 24

dL dx dy x x dx= + = +2 2 2 2 21 4( / – )–

= + +1 16 1 24 4x x dx/ – / –

= + = + −( / )–x x dx x x dx2 2 2 2 24 4| / |

L x x= +∫ ( / )–2 2

1

2

4 dx (because integrand > 0)

= − = =−x x3 1

1

212 1

1

121 0833/ . K

c. Distance between endpoints is

1 006944 6. K K= 1.0034 , which is areasonable lower bound for L.

19. a.

5

1 8

x

y

b. dy x dx= −2 1 3/

dL dx dy x dx= + = +2 2 2 31 4 – /

L x dx= +∫ 1 4 2 3

1

8– /

= + −∫ ( )/ /x x dx2 3 1 2 1 3

1

8

4 / ( )

= +

∫ −3

24

2

32 3 1 2

1

81 3( )/ /x x dx/

= ⋅ +3

2

2

342 3 3 2

1

8

( )/x /

= − =8 8 5 5 11 4470. K

c. Distance between endpoints is 130 =11.4017… , which is a reasonable lowerbound for L.

20. a.

0 3

5

x

y

b. dy x x dx x x dx= + = +1

22 2 22 1 2 2 1 2( ) / /( )

dL dx dy x x dx= + = + +2 2 2 21 2( )

= + + = +1 2 14 2 2x x dx x dx( )

L x dx x x= + = + =∫ ( )11

3122 3

0

3

0

3

c. Distance between endpoints is 11.6123… ,which is a reasonable lower bound for L.

21. Construct an x-axis at water level and a y-axisthrough the vertex of the parabola.

–2100 2100

x

y

750

220

0


General equation is y − 220 = ax2.Substitute (2100, 750) for (x, y).

750 220 210053

4410002− = ⋅ ⇒ =a a

Equation of parabola is y x= +53

4410002202 .

dy x dx= 106

441000

dL dx dy x dx= + = +2 2 2 21 106 441000( / )

L x dx= +∫ 1 106 441000 2 2

2100

2100

( / )–

≈ 4372.0861… numerically ≈ 4372 feet.The answer is reasonable because the 4200 feetbetween supports is a lower bound for L.

22. y e e dy e e dxx x x x= + = −− −0 2 0 2. ( ), . ( )

dL dx dy

e e dxx x

= +

= +

2 2

21 0 04. ( – )–

L e e dxx x= +

≈ … ≈−∫ 1 0 04 2

4

4

. ( – )–

24.1722 24.2 ftThe parabola with vertex (0, 0.4) and endpoints(±4, 0.2(e4 + e−4)) = (±4, 10.9232…) has equationy = ax2 + 0.4. Substituting (4, 10.9232…) gives10.9232… = 16a + 0.4 ⇒ a = 0.6577… .y = 0.6577… x2 + 0.4 ⇒ dy = 1.3154… x dx

dL dx dy x dx= + = +2 2 21 1 7303. ...

L x dx= + ≈ … ≈−∫ 1 1 7303 23 21932

4. K . 23.2 ft,

4

which is about a foot shorter than the catenary,as shown by graph:

4–4

10

x

y

23. Outer ellipse:x = 120 cos t, dx = −120 sin t dty = 100 sin t, dy = 100 cos t dt

dL dx dy

t t dt

= +

= +

2 2

2 2120 100(– sin ) ( cos )

L t t dt= +

≈ … ≈∫ (– sin ) ( cos )120 1002 2

0

2π

692.5791 692.6 mInner ellipse:x = 100 cos t, dx = −100 sin t dty = 50 sin t, dy = 50 cos t dt

dL dx dy

t t dt

= +

= +

2 2

2 2100 50(– sin ) ( cos )

L t t dt= +

≈ … ≈∫ (– sin ) ( cos )100 502 2

0

2π

484.4224 484.4 m

24.

8

x

y5

–8

–5

dx = −16 sin 2t, dy = 5 cos t

dL dx dy

t t dt

= +

= +

2 2

2 216 2 5(– sin ) ( cos )

Curve appears to have length

L t t dt= +∫ (– sin ) ( cos )16 2 52 2

0

2π

= 68.7694…Length should be less than the lengths of threecircumscribing segments, 16 + 16 + 10 = 42.The discrepancy is explained by the fact that theparabola is traced twice as t goes from 0 to 2π.Actual length ≈ (0.5)(68.7694…) = 34.384… ,for which 42 is a reasonable upper bound.

25. 9 42

32 3 3 2x y x y= ⇔ = ± / .

3

0 3 4

x

y

dx = y1/2 dy

dL dx dy y dy= + = +2 2 1 21( ) /

L y dy y= + = +

= = …

∫ ( ) ( )12

31

42

34 6666

1 2 3 2

0

3

0

3/ /

.

26. x2 = y3 ⇔ x = ±y1.5

2x dx = 3y2 dy ⇒ 4x2 dx2 = 9y4 dy2

⇒ = ⇒ =4 99

43 2 4 2 2 2y dx y dy dx y dy

Note that dy < 0 between (−1, 1) and (0, 0):

–1 8

x

y

1

4


For x in [−1, 0], x = −y1.5 , dx = −1.5y0.5 dy,

dL dx dy y dy= − + = − +2 2 2 25 1. .

For x in [0, 8], x = y1.5 , dx = 1.5y0.5 dy,

dL dx dy y dy= + = +2 2 2 25 1. .

L y dy y dy= − + + +∫∫ 2 25 1 2 25 10

4

. .1

0

= − + + +8

272 25 1

8

272 25 13 2 3 2( . ) ( . )y y/

1

0/

0

4

= + + = …8

271 3 25 10 1 10 51313 2 3 2(– . – )/ / .

27. xt

t dx t t t dt= =π π

cos (cos – sin ), 1

yt

t dy t t t dt= = +π π

sin (sin cos ), 1

dL dx dy= +2 2

= + +1 2 2

π(cos – sin ) (sin cos )t t t t t t dt

= +11 2

πt dt

The curve crosses the x-axis exactly when sin t= 0, when t is a multiple of π. There are sevencrossings after the beginning, so t should runbetween 0 and 7π. To check this, note that thecurve ends at (−7, 0), so solve (t/π) cos t = −7with t = nπ ⇒ (nπ/π ) cos nπ = −7 ⇒n cos nπ = −7 ⇒ n = 7 ⇒ 0 ≤ t ≤ 7π.

L t dt= + ≈ …∫11 77 65082

0

7

π

π.

The integral can be evaluated algebraically bytrigonometric substitution as in Section 9-6,giving

11

21 12 2 2+ = + + + +( )

+∫ t dt t t t t Cln .

28. x = r cos t, dx = −r sin t dty = r sin t, dy = r cos t dt

dL dx dy r t r t dt= + = +2 2 2 2 2 2sin cos

= r dt (for r ≥ 0)The range 0 ≤ t ≤ 2π generates the entire circle.

Circumference = = =∫ r dt rt r0

2

0

2

2π π

π , Q.E.D.

29. y = A sin x, dy = A cos x dx

dL dx dy A x dx= + = +2 2 2 21 cos

Pick a convenient interval for x such as [0, 2π].

L A x dx= +∫ 1 2 2cos0

2π

A L

0 6.283185… (= 2π)

1 7.640395…

2 10.540734…

3 13.974417…

Doubling A doubles the amplitude of thesinusoid. However, it less than doubles thelength of the sinusoid for much the same reasonthat doubling one leg of a right triangle does notdouble the hypotenuse. In the limit as Aapproaches infinity, doubling A approachesdoubling the length.

30. x = cos t, dx = −sin t dty = A sin t, dy = A cos t dt

dL dx dy t A t dt= + = +2 2 2 2 2sin cos

The entire ellipse is generated as t increases from0 to 2π.

L t A t dt= +∫ sin cos2 2 2

0

2π

A L

0 4 (a double line segment)

1 6.283185… (= 2π)

2 9.688448…

3 13.364893…

Doubling A doubles one axis of the ellipsewithout changing the other axis. That is why thelength does not double when A doubles. Thereasoning is similar to that in the solution toProblem 29.

31. The function y x= − −( )2 1 has a vertical

asymptote at x = 2, which is in the interval[1, 3]. So the length is infinite. Mae’s partitionof the interval skips over the discontinuity, asshown in the graph.

1 2 3

x

y25

Mae'serror

32. The sample points are all of the form (n/2,sin nπ), which all lie on the x-axis and thereforefail to measure the wiggly bits.

100

x

y

1

Amos's sample points

The length of the curve is 40 times the length ofthe part from x = 0 to x = 0.25 (by symmetry),so Amos could use five subintervals of [0, 0.25]to estimate the length of half of one arch, then


multiply his answer by 40 to find the totallength.


Problem Set 8-6

Q1. 1 9 4+ x dx Q2. 1 4+ sec x dx

Q3.1

66sin x C+ Q4. 156

Q5. xex + ex

Q6. Maximum y = 7 (at x = 1)

Q7. f xf x x f x

xx′ = + ∆

∆→( ) lim

( ) – ( )∆ 0

or ( )f cf x f c

x cx c′ =

→lim

( ) – ( )

–


Q9.1

22 2ln sec tan| |x x C+ +

Q10. D

1. a. The graph shows y = 0.5x2, from x = 0 tox = 3, rotated about the y-axis.

(x, y)

30

x1

y

dy = x dx

dL dx dy x dx= + = +2 2 21

dS x dL x x dx= ⋅ = +2 2 1 2π π

S x x dx= + ≈ …∫ 2 1 64 13612

0

3

π .

b. The inscribed cone of height 4.5 and radius3 has lateral surface area = πrL =π · 3 · 3 4 5 50 97222 2+ = …. . , which is

a reasonable lower bound for S.

c. S x x dx= +∫ π ( ) ( )1/21 22

0

3

= + = −2

31

2

310 10 12

0

3

π π( ) ( )3/2x

= 64.1361… , agreeing with the answerfound numerically.

2. a. The graph shows y = sin x, from x = 0 tox = π, rotated about the x-axis.

0 π

(x, y)

x

y

1

b. dy = cos x dx

dL dx dy x dx= + = +2 2 21 cos

dS y dL x x dx= ⋅ = +2 2 1 2π π sin cos

S x x dx= + ≈ …∫ 2 1 14 42352

0π

πsin cos .

c. The circumscribed cylinder of length πand radius 1 has lateral area = 2π 2 =19.7392… , which is a reasonable upperbound for S.

3. The graph shows y = ln x, from x = 1 to x = 3,rotated about the x-axis.

31

1

x

y

(x, y)

dy x dx= −1

dL dx dy x dx= + = +2 2 21 –

dS y dL x x dx= ⋅ = +2 2 1 2π π ln –

S x x dx= + ≈ …∫ 2 1 9 02422

1

3

π ln – .

4. The graph shows y = ln x, from x = 1 to x = 3,rotated about the y-axis, showing back half ofsurface only.

1 3

1

x

y

(x, y)

dL x dx= + −1 2 , from Problem 3.

dS x dL x x dx= ⋅ = +2 2 1 2π π –

S x x dx= + ≈−∫ 2 1 28 30472

1

3

π . K

5. The graph shows y x x= = −1 1/ ,from x = 0.5 to

x = 2, rotated about the y-axis.

2

2

x

y

(x, y)

dy x dx= − −2

dL dx dy x dx= + = +2 2 41 –


dS x dL x x dx= ⋅ = +2 2 1 4π π –

S x x dx= + ≈∫ 2 1 15 51814

0 5

2

π –

.. K

6. The graph shows y = 1/x = x− 1, from x = 0.5 tox = 2, rotated about the x-axis.

2

2

x

y

(x, y)

dL x dx= +1 4– , from Problem 5.

dS y dL x x dx= ⋅ = +2 2 11 4π π – –

S x x dx= + ≈−∫ 2 1 15 51811 4

0 5

2

π –

.. K

(Note that surfaces 5 and 6 are congruent.)

7. The graph shows y = x3, from x = 0 to x = 2,rotated about the y-axis.

(x, y)

2

8

0

x

y

dy = 3x2 dx

dL dx dy x dx= + = +2 2 41 9

dS x dL x x dx= ⋅ = +2 2 1 9 4π π

S x x dx= + ≈∫ 2 1 9 77 32454

0

2

π . K

8. The graph shows y = −x3 + 5x2 − 8x + 6, fromx = 0 to x = 3, rotated about the y-axis.

(x, y)

x

y

3

6

dy = (−3x2 + 10x − 8) dx

dL dx dy= +2 2

= + +1 3 10 82 2(– – )x x dx

dS = 2π x ⋅ dL

= + +2 1 3 10 82 2π x x x dx(– – )

Graph intersects x-axis where y = 0.−x3 + 5x2 − 8x + 6 = −(x − 3)(x2 − 2x + 2) = 0at x = 3.

S x x x dx= + +∫ 2 1 3 10 82 2

0

3

π (– – )

≈ 58.7946…9. The graph shows y x= = x1/2, from x = 0 to

x = 1, rotated about the x-axis.

(x, y)

0 1

1

x

y

dy x dx

dL dx dy x dx

=

= + = +

−0 5

1 0 25

1 2

2 2 1

. /

. –

dS y dL x x dx= ⋅ = +2 2 1 0 251 2 1π π / –.

= + = +2 0 25 2 0 25 1 2π πx dx x dx. ( . ) /

S x dx x= + = +∫ 2 ( 0.25)4

3( 0.25) 1/2 3/2

0

1

0

1

π π

= − =4

31 25 0 125 5 33043 2π

( . . ) ./ K

10. The graph shows y = x3, from x = 1 to x = 2,rotated about the x-axis, showing back half ofsurface only.

(x, y)

1 2

1

8 y

x

dy = 3x2 dx

dL dx dy x dx= + = +2 2 41 9

dS = 2πy ⋅ dL = 2πx3 (1 + 9x4)1/2 dx

S x x dx= +∫ 2 (1 9 )1/2π 3 4

1

2

= +∫π18

( ) ( )/1 9 364 1 2

1

2

x x dx

= ⋅ +π18

2

31 9 4 3 2

1

2

( ) /x

= =π

27145 10 199 48043 2 3 2( – )/ / . K

11. The graph shows y x x= + −4 28 4/ / , from x = 1

to x = 2, rotated about the x-axis, showing backside of surface only.


1 2

2

x

y

dy = (x3/2 − x− 3/2) dx = 0.5(x3 − x− 3) dx

dL dx dy x x dx= + = +2 2 3 3 21 0 25. ( – )–

= + +1 0 25 0 5 0 256 6. – . . –x x dx

= + = + −0 25 0 53 3 2 3 3. ( )–x x dx x x dx. ( )

dS = 2π y ⋅ dL= 2π ( x4/8 + x− 2/4)[0.5(x3 + x− 3)] dx

= + +π8

3 27 5( )–x x x dx

S x x x dx= + +∫π8

3 27 5

1

2

( )–

= +

π8

1

8

3

2

1

28 2 4

1

2

x x x– –

= + +

=

= …

π π8

32 61

32

1

8

3

2

1

24

155

256

14 4685

– – –

.12. The graph shows y = x2, from x = 0 to x = 2,

rotated about the y-axis.

(x, y)

x

y

4

0 2

dy x dx

dL dx dy x dx

=

= + = +

2

1 42 2 2

dS x dL x x dx= ⋅ = +2 2 1 4 2π π

S x x dx= +∫2 1 4 2

0

2

π

= + = +∫π π4

1 4 86

1 42 1 2

0

22 3 2

0

2

( ) ( ( )/ /x x dx x)

= − =π

617 1 36 17693 2( ) ./ K

13. The graph shows y x= +1

322 3 2( ) ,/ from x = 0 to

x = 3, rotated about the y-axis.

30

1

12

(x, y)

x

y

dL = (1 + x2) dx, from Problem 20 in Section 8-5

dS = 2πx ⋅ dL = 2π ( x + x3) dx

S x x dx x x= + = +

∫ 2

1

23 2 4

0

3

0

3

π π( )

= 49.5π = 155.5088…

14. The graph shows y = 2x1/3, from x = 1 tox = 8, rotated about the y-axis, showing backhalf of surface only.

4

2

1 8

x

y

(x, y)

dy x dx= −2

32 3/

dL dx dy x dx= + = +2 2 4 314

9– /

dS x dL x x dx= ⋅ = +2 2 14

94 3π π – /

S x x dx= +∫2 14

94 3

1

8

π – /

= +

∫2

4

91 3 4 3

1

8 1 2

π x x dx/ //

= +

⋅∫3

2

4

9

4

34 3

1 21 3

1

8

π x x dx//

/

= +

π x 4 3

3 2

1

84

9/

/

= − =π27

148 13 204 04353 2 3 2( ) ./ / K

15. The graph shows y x x= + −1 1

43 1

3, from x = 1 to

x = 3, rotated about the line y = −1, showingback half of surface only.

(x, y)

1

9

(x, –1)

3

x

y


dx x x dx= −

−2 21

4

dL dx dy x x dx= + = + −

−2 2 2 22

11

4

= + − + −11

2

1

164 4x x dx

= +

= +

− −x x dx x x dx2 22

2 21

4

1

4dS = 2π ( y + 1) ⋅ dL

= + +

+

− −21

3

1

41

1

43 1 2 2π x x x x dx

= + + + +

− −21

3

1

3

1

4

1

165 2 2 3π x x x x x dx

S x x x x x dx

x x x x x

= + + + +

= + + − −

= =

− −

− −

∫21

3

1

3

1

4

1

16

21

18

1

3

1

6

1

4

1

32

1015

18318 1735

5 2 2 3

1

3

6 3 2 1 2

1

3

π

π

π . K

16. The graph shows y x x= + −1

3

1

43 1, from x = 1 to

x = 3, rotated about line x = 4.

(x, y)

(4, y)

1 3 4

9

x

y

dL x x dx= +

−2 21

4, from Problem 15

dS x dL x x x dx= − ⋅ = − +

−2 4 2 41

42 2π π( ) ( )

= − + −

− −2 41

42 3 2 1π x x x x dx

S x x x x dx= − + −

− −∫2 41

42 3 2 1

1

3

π

= − − −

−24

3

1

4

1

43 4 1

1

3

π x x x xln | |

= −

=2 15

1

3

1

43 94 6164π ln . K

17. a. x y y x2 2 225 25+ = ⇒ = −dy x x dx= − − −( )25 2 1 2/

dL dx dy x x dx= + = + − −2 2 2 2 11 25( )

dS = 2πy ⋅ dL

= − + − −2 25 1 252 2 2 1π x x x dx( )

= − + =2 25 102 2π πx x dx dx

b. i. S dx x0 10

1

0

1

10 10 10, = = =∫ π π π

ii. S dx x1 21

2

1

2

10 10 10, = = =∫ π π π

iii. S dx x2 32

3

2

3

10 10 10, = = =∫ π π π

iv. S dx x3 43

4

3

4

10 10 10, = = =∫ π π π

v. S dx x4 54

5

4

5

10 10 10, = = =∫ π π π

c. The two features exactly balance each other.The area of a zone of a sphere is a functionof the height of the zone only, and isindependent of where the zone is locatedon the sphere.

18. Suppose that the sphere is centered at the origin,as in Problem 17. The equation of a greatcircle in the xy-plane is x2 + y2 = r2, from

which y r x r x= − = −2 2 2 2 1 2( ) ./

dy = −x(r2 − x2)− 1/2dx

dL dx dy x r x dx= + = + − −2 2 2 2 2 11 ( )dS y dL

r x x r x dx

r x x dx r dx r

= ⋅

= − + −

= − + = >

−

2

2 1

2 2 0

2 2 1 2 2 2 2 1

2 2 2

π

π

π π

( ) ( )

( )

/

if

S r dx rx rr

r

r

r

= = =− −∫ 2 2 4 2π π π , Q.E.D.

19. Pick a sample point in the spherical shell atradius r from the center. Surface area at thesample point is 4π r2. Volume of shell isapproximately (surface area)(thickness).dV r dr= ⋅4 2π

V r dr r RR R

= = =∫ 44

3

4

32

0

3

0

3π π π , Q.E.D.

20. V rdV

drr S= ⇒ = =4

343 2π π , Q.E.D.

or: V S drdV

drS= ⇒ =∫ by the definition of

indefinite integral.

21. y = ax2, dy = 2ax dx

dL dx dy a x dx= + = +2 2 2 2 1 21 4( ) /

dS x dL x a x dx= ⋅ = +2 2 1 4 2 2 1 2π π ( ) /

S x a x dxr

= +∫2 1 4 2 2 1 2

0π ( ) /

= +∫π4

1 4 822 2 1 2 2

0aa x a x dx

r

( ) ( )/

= + = + −π π6

1 46

1 4 122 2 3 2

02

2 2 3 2

aa x

aa r

r

( ) [( ) ]/ /


22. Let h be the height of the paraboloid from thevertex to the center of the base. Because h is thevalue of y when x = r, h = a r 2. Substituting intothe formula for S from Problem 21 gives

Sa

ah= + −π6

1 4 123 2[( ) ]/

Let a = 1 and evaluate S for various h. Find thezone areas by subtracting. Use the TABLE feature.

h S Zone

0π6

0( ) N.A.

1

π6

10 1803( . )K

π6

10 1803( . )K

2π6

26( )π6

15 8196( . )K

3

π6

45 8721( . )K

π6

19 8721( . )K

4

π6

69 0927( . )K

π6

23 2206( . )K

5π6

95 2340( . )Kπ6

26 1412( . )K

6 π6

124( )

π6

28 7659( . )K

The property is not true for paraboloids. Theareas of zones of equal height are greater if thezone is farther away from the vertex.

23. x = 5 cos t, dx = −5 sin t dty = 3 sin t, dy = 3 cos t dt

5

3(x, y)

x

y

dL dx dy t t dt= + = − +2 2 2 25 3( sin ) ( cos )dS y dL= ⋅2π

= − +2 3 5 32 2π ( sin ) ( sin ) ( cos )t t t dt

S t t t dt= − +

≈ …∫ 6 5 32 2

0π

πsin ( sin ) ( cos )

165.7930

From ( / ) ( / ) , . .x y y x5 3 1 0 6 252 2 2+ = = ± − Using the upper branch of the graph,dy x x dx= − − −0 6 25 2 1 2. ( ) ./

dL dx dy x x dx= + = + − −2 2 2 2 11 0 36 25. ( )

At x = ±5, dL involves division by zero, whichis awkward, and makes the Cartesian equation

inappropriate for finding the arc length of anellipse.For the surface area, however, the offendingdenominator cancels out, giving

dS x dx= 0 24 25 162 2. ,π – which is definedat x = ±5.

24. a. x = 35 sec t, dx = 35 sec t tan t dty = 100 + 80 tan t, dy = 80 sec2 t dty = 0 ⇔ 100 + 80 tan t = 0 ⇒ tan t = −5/4t = −−tan 1 5 4 ( / )Radius at base is x = −5/4 =−35 1sec tan[ ( )]56.0273… ≈ 56.0 ft.

b. At top, t = 0.5.Radius: x = 35 sec 0.5 = 39.8822… ≈ 39.9 ftHeight: y = 100 + 80 tan 0.5 = 143.7041…≈ 143.7 ft

c. From the information given in parts a and b,it can be assumed that −π /2 < t < π /2.

Minimize xdx

dtt t t: at= = =35 0 0sec tan

(or, because cos t has a max at t = 0,sec t = 1/cos t has a minimum there).Minimum radius = 35 ftHeight = y = 100 + tan 0 = 100 ft

d. dL dx dy= +2 2

= ⋅ +35 802 2 2 2 4sec tan sect t t dt

dS = 2πx ⋅ dL

= ⋅ +2 35 35 802 2 2 2 4π ( )sec sec tan sect t t t dt

S dS= ≈ … ≈− −∫ 37 756 5934 37 757 2

5 4

0 5

1, . , ft

tan ( / )

.

e. Volume , . ft3≈ ⋅ = … ≈S4

1212 585 5311

466.1307… yd3

25. From Figure 8-6m, a circle of radius L has areaπL2 and circumference 2πL. The circumference ofthe cone’s base is 2πR. The arc length of thesector of the circle of radius L must be equal tothis, so the sector is (2πR)/(2πL) = R/L of thecircle and has surface area S = πL2(R/L) = πRL,Q.E.D.

26. S = πRL − πrlThe objective is to get the lateral area in terms ofthe slant height of the frustum, L − l.

S R Lr

Rl= − ⋅

π

= − ⋅

πR L

l

Ll , because

r

R

l

L= .

= πR

LL l( – )2 2


= +πR

LL l L l( )( – )

= + ⋅

−π R R

l

LL l( )

= + ⋅

−π R R

r

RL l( )

= π ( R + r)(L − l)

= +

−2

2π R r

L l( ), Q.E.D.

Problem Set 8-7Q1. 15x2 − 14x + 4 Q2. 12(4x − 9)2

Q3. 3 sin2 x cos x Q4. 3 sec 3x tan 3x

Q5. − −e x Q6. −1/x2

Q7. ln |x| + C Q8.1

22x C+

Q9. 3x + C Q10. x + C

1. a. r = 10 sin θ ⇒ dA = 50 sin2 θ dθ

A d= ≈ …∫ 50 157 07962

0

2

sin θ θπ

.

(exactly 50π )b. The area of the circle is π ⋅ 52 = 25π.

The calculated area is twice this because thecircle is traced out twice as θ increases from 0to 2π. Although r is negative for π < θ < 2π,dA is positive because r is squared.

2. a. r = 10 sin θ ⇒ dr = 10 cos θ dθ

dL dr r d

d d

= +

= + =

2 2

2 2100 100 10

( )

cos sin

θ

θ θ θ θ

L d= = =∫ 10 10 200

2

0

2

θ θ ππ π

The circumference is 2π ⋅ 5 = 10π. Thecalculated length is twice this value becausethe circle is traced out twice as θ increasesfrom 0 to 2π. The calculus of this sectionalways gives the dynamic answer as thedistance traveled by a point on the curve asθ increases from one value to another. Thispath length does not necessarily equal thelength of the curve.

3. a. r = 4 + 3 sin θ. The calculator graph confirmsthat the text figure is traced out once as θincreases from 0 to 2π.

b. dA d= +1

24 3 2( sin )θ θ

A dA= ≈ …∫ 64 40260

2

.π

(exactly 20.5π)

c. dr = 3 cos θ dθ

dL dr r d= +2 2( )θ

= + +( cos ) ( sin )3 4 32 2θ θ θd

L dL= ≈∫ 28 8141

0

2

.π

K

4. a. r = 5 − 3 cos θ. The calculator graph confirmsthat the text figure is traced out once as θincreases from 0 to 2π.

b. dA d= 1

25 3 2( – cos )θ θ

A dA= ≈ …∫ 92 6769 29 50

2

. (exactly . )ππ

c. dr = 3 sin θ dθdL dr r d= +2 2( )θ

= +( sin ) ( – cos )3 5 32 2θ θ θd

L dL= ≈ …∫ 34 31360

2

.π

5. a. r = 7 + 3 cos 2θ. The calculator graphconfirms that the text figure is traced out onceas θ increases from 0 to 2π.

b. dA d= +1

27 3 2 2( cos )θ θ

A dA= ≈ …∫ 168 07520

2

.π

(exactly 53.5π)

c. dr = −6 sin 2θ dθ

dL dr r d= +2 2( )θ

= + +(– sin ) ( cos )6 2 7 3 22 2θ θ θd

L dL= ≈ …∫ 51 45110

2

.π

6. a. r = 8 cos 2θ. The calculator graph confirmsthat the text figure is traced out once asθ increases from 0 to 2π.

b. dA d= 1

28 2 2( cos )θ θ

A dA= ≈∫ 100 5309 32

0

2

. (exactly )K ππ

c. dr = −16 sin θ dθ

dL dr r d= +2 2( )θ

= +(– sin ) ( cos )16 2 8 22 2θ θ θd

L dL= ≈∫ 77 5075

0

2

. Kπ

7. a. 5 = 5 + 5 cos θ. The calculator graphconfirms that the text figure is traced out onceas θ increases from 0 to 2π.

b. dA d= +1

25 5 2( cos )θ θ

A dA= ≈∫ 117 8097 37 5

0

2

. (exactly . )K ππ


c. dr = −5 sin θ dθ

dL dr r d= +2 2( )θ

= + +(– sin ) ( cos )5 5 52 2θ θ θd

L dL= =∫ 400

2

(exactly)π

8. a. r = 10

3 2– cos θ. The calculator graph

confirms that the text figure is traced out onceas θ increases from 0 to 2π.

b. dA =

1

2

10

3 2

2

– cosθ

A dA= ≈∫ 84 2977 12 5

0

2

. (exactly )K ππ

c. dr d= – sin

( – cos )

20

3 2 2

θθ

θ

dL dr r d= +2 2( )θ

=

+

– sin

( – cos ) – cos

20

3 2

10

3 22

2 2θθ θ

θd

L dL= ≈∫ 33 0744

0

2

. Kπ

9. a.

1

1

r = sin 3θ makes one complete cycle as θincreases from 0 to π.

b. dA d= 1

23 2(sin )θ θ

A dA= ≈∫ 0 7853 0 25

0. (exactly .K π

π)

c. dr = 3 cos 3θ dθdL dr r d= +2 2( )θ

= +( cos ) (sin )3 3 32 2θ θ θd

L dL= ≈∫ 6 68240

. Kπ

10. a.

4

4

b. dA d= 1

24 4 2( sec – cos )θ θ θ

A dA= ≈

−∫ 4 55571

1

. K

(exactly 16 tan 1 − 24 + 4 sin 2)

c. dr = 4(sec θ tan θ + sin θ) dθ

dL dr r d= +2 2( )θ

= + +4 2 2(sec tan sin ) (sec – cos )θ θ θ θ θ θd

L dL= ≈

−∫ 10 95341

1

. K

11. r = 49 2cos θr = 0 ⇔ 2θ = cos−1 0 = ±π /2 + 2π n (n an integer)θ = ±π /4 + π nThe right-hand loop corresponds tononnegative values of the integrand,−π /4 ≤ θ ≤ π /4.

dA d= 1

249 2( cos )θ θ

A d= =− −∫ 1

249 2 12 25 2

4

4

4

4

( cos ) . sin/

/

/

/

θ θ θπ

π

π

π

= 24 5.

Area of both loops is 49.

12. The graph of r = csc θ + 4 shows a closed loopfrom θ ≈ 3.4 to θ ≈ 6.

5

5

The graph passes through the pole where r = 0.csc 4 0 csc ( 4)θ θ+ = ⇔ = − =−1

sin− − = − … +1 0 25 0 2526 2( . ) . orπ n[π − (−0.2526…)] + 2π nDesired range is 3.3942… ≤ θ ≤ 6.0305… .

dA d= +1

24 2(csc )θ θ

A dA= ≈ …∫ 8 4553

3 3942

6 0305

. .

.

K

K

13. r1 = 4 + 4 cos θ and r2 = 10 cos θ intersect where4 + 4 cos θ = 10 cos θθ π= = ± …+−cos 1 2 3 0 8410 2( / ) . .n

(The graphs also touch at the pole, but not forthe same value of θ. For the cardioid,θ = π + 2πn. For the circle, θ = π /2 + 2π n.)Region outside the cardioid and inside


the circle is generated as θ goes from−0.841… to 0.841… .

dA r r d= 1

2 22

12( – ) θ

= − +1

210 4 4[( ) ( ) ] 2 2cos cosθ θ θd

A dA= ≈

−∫ 18 88630 841

0 841

. KK

K

.

.

(exactly 26 cos (2/3) (4/3)− −1 5)

14. r1 = 5 and r2 = 5 − 5 cos θ intersect at θ = −π /2and π /2.

5

5

dA r r d= 1

2 12

22( – ) θ

= − −1

25 5 52[ ( ) ] 2cos θ θd

Integrate from −π /2 to π /2, because in QuadrantsII and III the cardioid lies outside the circle.

A dA= ≈ −

−∫ 30 3650 50 6 252

2

. (exactly .K ππ

π)

/

/

15. a. r = 0.5θ. The graph starts at θ = 0 and makesthree revolutions, so θ increases from 0 to 6π.

dr = 0.5 dθ

dL dr r d d= + = +2 2 2 20 5 0 5( ) . ( . )θ θ θ

L dL= ≈ …∫ 89 85890

6

.π

b. dA d d= =1

20 5

1

82 2( . )θ θ θ θ

Area swept out for third revolution inQuadrant I is

A d32 3

4

4 53

4

4 5 1

8

1

24

217

192= = =∫ θ θ θ π

π

π

π

π ..

Area swept out for second revolution inQuadrant I is

A d22 3

2

2 53

2

2 5 1

8

1

24

61

192= = =∫ θ θ θ π

π

π

π

π ..

Area of region between second and thirdrevolution in Quadrant I is A3 = A2 =13

1625 19253π = …. .

16. The graph of r = 4 + 6 cos θ shows a closed loopfrom θ ≈ 2.3 to θ ≈ 4.0.

10

4

r = 4 + 6 cos θ = 0 ⇔ cos θ = −2/3θ π= − = ± … +−cos 1 2 3 2 3005 2( / ) . n

dA d= +1

24 6 2( cos )θ θ

The outer loop is swept out as θ increases from−2.3005… to 2.3005.

A dA1

2 3005

2 3005

105 0506= ≈∫ . KK

K

– .

.

The inner loop is generated as θ increases from2.3005… to 3.9826… .

A dA2

2 3005

3 926

1 7635= ≈∫ . KK

K

.

.

Area of the region between the loops isA1 − A2 ≈ 103.2871… .

17. a.

4

2

dr = −2.5θ −1.5 dθdL dr r d d= + = +2 2 3 16 25 25( ) . – –θ θ θ θ

dL ≈ …∫ 31 08722

6

.π

π

/

b. The graph shows sectors of central angles 1,2, and 3 radians.

Area of sector is A r( ) .θ θ= 1

22

A( ) ( ) .11

25 1 12 52= =( )

A( ) ( ) .21

23 5355 2 12 52= =( . ...)

A( ) ( ) .31

22 8867 3 12 52= =( . )K

In general, A( ) ( ) . ,θ θ θ= =1

25 12 51 2 2( )– / which

is independent of the value of θ.


18. The graph shows r = sec θ and a segment fromθ = 0 to 1.5.

1

10

The point with polar coordinates (r, θ) has xy-coordinates x = r cos θ, y = r sin θ. The graphgiven by r = sec θ can be writtenx = r cos θ = sec θ cos θ = 1y = r sin θ = sec θ sin θ = tan θ(i.e., −∞ < y < ∞). Thus, this graph is theline x = 1.By calculus, the segment from θ = 0 to θ = 1.5has length as follows:dr = sec θ tan θ dθdL dr r d= +2 2( )θ

= +(sec tan ) secθ θ θ θ2 2 d

= + =sec tan secθ θ θ θ θ2 21 d d

L d= =∫ sec tan.

2

0

1 5

0

1 5

θ θ θ.

= tan 1.5 − 0 = 14.1014…As shown above, y = tan θ.At θ = 1.5, y = tan 1.5, confirming the calculus.

19. A typical record has grooves of inner radius6.6 cm and outer radius 14.6 cm, and takesabout 24 minutes to play. There are thus(33.333…)(24) or about 800 grooves in a spaceof (14.6 − 6.6) or 8.0 cm. Thus, the groovesdecrease in radius by about 8.0/800 = 0.01 cmper revolution. A simple equation of the spiral is

r = =0 01

2

0 005. .

πθ

πθ

which assumes that the grooves start at the centerand have a pitch of 0.01 cm. The innermostactual groove is at θ = 6.6π/0.005 = 1320π,and the outermost groove is at θ = 14.6π/0.005= 2920π.dr = −(0.005/π) dθdL dr r d= +2 2( )θ

= +( . / ) [( . / ) ]0 005 0 0052 2π π θ θd

= +0 0051 2.

πθ θd

L dL= ≈ …∫ 53 281 41201320

2920

, . cmπ

π

= 16,960.0002…π cmRough check: Average radius = 10.6 cmL should equal approximately the sum of 800

circles of radius 10.6 cm. L ≈ 800(2π · 10.6) =16,960π cm, which is very close to the calculated16,960.0002…π cm.(The integral can be evaluated algebraically bythe tangent trigonometry substitution fromChapter 9. The result, 16,960.00021…π, isremarkably close both to the numerical answerand to the sum of the lengths of the 800 circlesof average radius 10.6 cm.)

20. a. r = = − −100

3 2100 3 2 1

– coscos

θθ( )

dA d=

= − −

1

2100 3 2 1 2

2

[ ( – cos ) ]–θ θ

θ θ5000(3 2 cos ) d

A dA= ≈ … ≈∫ 974 3071 9740

0 2

. (kilo-mi)2.

b. Solving ( )5000 3 2 2

0 8− =−∫ cos

.t dt

θ

974.3071… gives θ ≈ 1.88976… .

c. P = ka1.5

(27.3)(24) = k(240)1.5

k = 0.17622…

d. The major axis of the spaceship’s orbit is120 thousand miles, so a = 60.P = k · 601.5 = 81.9 hours (precise answer)

e. The total area of the ellipse is

A d= − −∫ 5000 3 2 2

0

2

( )cos θ θπ

= 8429.7776… (kilo-mi)2

Fraction of area from θ = 0 to θ = 0.2 is(974.3071…)/(8429.7776…) = 0.1155… .This fraction is the same as the fraction of theperiod. Thus, the time is 0.1155…(81.9) =9.4659… hours to go from θ = 0 to θ = 0.2,and the same for θ to go from 0.8 to1.88976… .

f. dr d

d

= − − ⋅

= − −

−

−

100 3 2 2

200 3 2

2

2

( )

( )

cos ( sin )

sin cos

θ θ θ

θ θ θ

dL dr r d= +

= − − −

2 2

2

( )θ

θ θ[( 200 sin (3 2 cos ) )2

+ − −( ( ) ) ]2 1/2100 3 2 1cos θ θd

From θ = 0 to θ = 0.2,

L dL= ≈∫ 20 22280

0 2

. K.

≈ 20.2 kilo-mi.

From θ = 0.8 to θ = 1.88976… ,

L dL= ≈∫ 56 78960 8

1 88

. KK

.

.

≈ 56.8 kilo-mi.

g. Average speed from θ = 0 to θ = 0.2 is

20 2228

9 46592 1363

.

.

K

KK= . , or about 2136 mi/h.


Average speed from θ = 0.8 to

θ = 1.88976… is

56 7896

9 46595 9993

.

.

K

KK= . ,

or about 5999 mi/h.

h. When the spaceship is farthest from Earth, itsradial velocity (toward the Earth) is zero. Asit proceeds in its orbit, it can be thought of asfalling toward the Earth, thus picking upspeed. The reverse is true on the other side ofthe Earth, where it is moving away and isthus being slowed by gravity.

21. a. Count 5 spaces to the right and about 7.5spaces down from the given point.Slope ≈ −1.5.

b. r = θx = θ cos θ ⇒ dx = dθ · cos θ − θ sin θ dθy = θ sin θ ⇒ dy = dθ · sin θ + θ cos θ dθdy

dx

dy d

dx d= = +/

/

sin cos

cos – sin

θθ

θ θ θθ θ θ

At θ = 7, dy/dx = −1.54338… , thusconfirming the answer found graphically.

22. a. x = r cos θ, y = r sin θ⇒ y/x = sin θ/cos θ = tan θ

r

r cos

r sin

slope = r sin

= tan

r cos

b. The slope of any line is tan φ, where φ is theangle between the x-axis and the line.And, because the tangent line has slopedy

dx

dy d

dx d= /

/( )

θθ

by the chain rule ,

tan/

/φ θ

θ .= dy d

dx d

c. tan ψ = tan (φ − θ) =+

tan – tan

tan tan

φ θφ θ1

=+ ⋅

⋅

dy ddx d

yx

dy ddx d

yx

x dx d

x dx d

//

–

//

/

/

θθ

θθ

θθ1

=+

xdyd

ydxd

xdxd

ydyd

θ θ

θ θ

–

d. dx/dθ = −r sin θ;

dy/dθ = r cos θ xdy

dy

dx

d

θ θ−

= ⋅ − ⋅ −r r r r ( ) ( )cos cos sin sinθ θ θ θ= r2 cos2 θ + r2 sin2 θ = r2

e. r x y rdr

dx

dx

dy

dy

d2 2 2 2 2 2= + ⇒ = +

θ θ θ

⇒ = +rdr

dx

dx

dy

dy

d

θ θ θSubstitute these expressions in parts d and einto the top and bottom of the expression inpart c to show the property.

f. tan/

– cos

sin

– cos

sinψ

θθ

θθ

θ = = = =r

dr d

a a

a

1

tan θ/2, using the half-angle formula. Thenψ = θ/2 + nπ. But 0 ≤ θ ≤ 2π, and 0 ≤ ψ ≤ π,which implies n = 0, so ψ = θ/2.

g. tan/

ψθ θ

= ⇒ = ⋅r

dr d

dr

drconst ⇒ r = Cekθ

Note that dr

dkCe kr k

dr d

rk

θθθ= = ⇒ = =/

1

tancot

ψψ= .

Equations for the spiral will vary.



R1. a.

3–10

50

x

yf

g

h

b. f ′ (x) = 3x2 − 18x + 30; f ″(x) = 6x − 18g′ (x) = 3x2 − 18x + 27; g″(x) = 6x − 18h′ (x) = 3x2 − 18x + 24; h″(x) = 6x − 18

c. h′ (x) = 3(x − 2)(x − 4) = 0 at x = 2 and 4h″(2) = −6 < 0, so h has a local maximumat x = 2.h″(4) = 6 > 0, so h has a local minimumat x = 4.

d. g′ (x) = 3(x − 3)2 = 0 only at x = 3.g′ (x) > 0 on both sides of x = 3, so this isneither a maximum nor a minimum point.

e. From the graphs, each point of inflectionappears at x = 3. Because each secondderivative equals 6x − 18, each one equalszero when x = 3.


R2. a.

x 2

+ undef. +

no max.or min.

f´(x)

f (x)

x 2

+ undef. –

p.i.

f´´(x)

f (x)

b.f (x)

x

–2 1 3 5

c. i. f x x f x x′ = − ′′ =− −( ) , ( )/ /2

31

2

91 3 4 3–

ii. Zooming in shows that there is a localminimum cusp at (0, 0) and a localmaximum with zero derivative at x ≈ 0.3.

1

0.5

0

x

f(x)

Algebraically, f ′ (x) = 0 at x = (2/3)3 =8/27, and f ′ (x) is undefined at x = 0, thuslocating precisely the minimum andmaximum found by graphing.Because there are no other critical valuesof x, there are no other maximum orminimum points.

iii. f ″(x) is undefined at x = 0, and f ″(x) < 0everywhere else; f ″ never changes sign, sothere are no inflection points.

iv. f (0) = 0, f (8/27) = 4/27, f (5) = −2.0759…Global maximum at (8/27, 4/27).Global minimum at (5, −2.0759…).

d. f (x)

x

2

1

The graph shows that f x x e x( ) = −2 has local

minimum at x = 0, local maximum at x ≈ 2,and points of inflection at x ≈ 3.4 and atx ≈ 0.6.f x xe x e x x ex x x′ = − = −− − −( ) ( )2 22

f x e xe x e x x ex x x x″ = − + = − +− − − −( ) ( )2 4 2 42 2

f ′ (x) = 0 at x = 0, 2

Minimum at (0, 0). Maximum at (2, 0.5413…).f ″(x) = 0 at x = + = …2 2 3 4142. and atx = − = …2 2 0 5857.

f ″(x) changes sign at each of these x-values,which implies points of inflection at(0.5857… , 0.1910…), (3.4142… ,0.3835…).

R3. a. Let x = width of a cell, y = length of the cell.xy y x x= ⇒ = <−10 10 01;

Minimize ( ) .L x x y x x= + = + −12 7 12 70 1

The graph shows minimum L (x) at x ≈ 2.4.

100

2

x

L(x)

L x x′ = − −( ) 12 70 2

L x x′ = = = …( ) at .0 70 12 2 4152/

At , . .x y= = = …70 12 10 12 70 4 1403/ /

Overall length of battery is 6(2.4152…) =14.4913… .Optimal battery is about 14.5″ by 4.1″,which is longer and narrower than the typicalbattery, 9″ by 6.7″. Thus, minimal walllength does not seem to be a majorconsideration in battery design.

b. The graph shows y = 8 − x3, from x = 0 tox = 2, with rectangle touching sample point(x, y) on the graph, rotated about the y-axis,generating cubic paraboloid and inscribedcylinder.

2

8

x

y

(x, y)

Domain of x is 0 ≤ x ≤ 2.Maximize V (x) = π r 2h = π x 2y = 8π x2 − π x5.The graph shows that V (x) has a maximum atx ≈ 1.5.

2

30

0

V(x)

x


V ′ (x) = 16π x − 5π x4 = π x(16 − 5x3)V x x x′ = = = = …( ) at and .0 0 16 5 1 47363 /

Maximal rectangle has x = = …16 5 1 47363 / . ,

y = 8 − 16/5 = 4.8.R4. a. The graph shows y x y x= =1

1 322/ ,and

intersecting at (0, 0) and (1, 1), rotated aboutthe x-axis, sliced parallel to the x-axis,showing back half of solid only.

1

0 1

x

y

(x2, y)

(x1, y)

x1 = y3, x2 = y1/2

dV = 2π y (x2 − x1) · dy = 2π y (y1/2 − y3) dy

V dV= ≈ …∫ 1 25660

1

. (exactly 0.4π )

b. The graph shows y1 and y2 as in part a, butsliced perpendicular to the x-axis, generatingplane washer slices.

10

1

x

y

(x, y1 )

(x, y2 )

dV y y dx x x dx= − = −π π( ) ( )12

22 2 3 4/

V dV= ≈ …∫ 1 2566.0

1

(exactly 0.4π), which is

the same answer as in part a, Q.E.D.

c. i. The graph shows y = x2 and y = 4,intersecting at (2, 4) and (−2, 4), rotatedabout the y-axis, showing back half ofsolid only.

(x, y)

x

y

4

2–2

(x, 4)

dV = 2π x (4 − y) · dx = 2π x (4 − x2) dxV ≈ 25.1327… (exactly 8π)(Disks can also be used.)

ii. The graph shows the region described inpart i, rotated about the x-axis, showingback half of solid only.

x

y

4

2–2

(x, 4)

(x, y)

dV = π (42 − y2) dx = π (16 − x4) dx

V dV= ≈ …∫ 160 84952

.2

– (exactly 51.2π)

(Cylindrical shells can also be used.)

iii. The graph shows the region described inpart i, rotated about the line y = 5,showing back half of solid only.

x

y

4

2–2

(x, y)

(x, 4)5

dV = π [(5 − y)2 − 12] dx= π [(5 − x2)2 − 1] dx

V dV= ≈ …

∫ 174 2536

7

152. exactly 55

2

–π

(Cylindrical shells can also be used.)

iv. The graph shows the region described inpart i, rotated about the line x = 3,showing back half of solid only.

(x, y)x

y

4

2–2

(x, 4)

3

(3, y)

dV = 2π (3 − x) · (4 − y) · dx = 2π (3 − x)(4 − x2) dx

V dV= ≈ …∫ 201 06192

2

.–

(exactly 64π)

(Washers can also be used.)R5. a. y = x2 from x = −1 to x = 2.

dy = 2x dx

dL dx dy x dx= + = +2 2 21 2( )

L dL= ≈ …∫ 6 12571

.2

–


b. y = x3/2 from x = 0 to x = 9.dy = 1.5x1/2 dx

dL dx dy x dx= + = +2 2 1 2 21 1 5( . )/

L x dx= +∫ ( . )1 2 25 1 2

0

9/

= +∫1

2 251 2 25 2 251 2

0

9

.( . )x dx/ ( . )

= +2

6 751 2 25 3 2

0

9

.( . )x /

= = …2

6 7521 25 1 28 72813 2

.( . – )/ .

Distance between the endpoints is

10 26 27 85672 2+ = …. , so the answer

is reasonable.

c. x = t cos π t ⇒ dx = (cos π t − π t sin π t ) dty = t sin π t ⇒ dy = (sin π t + π t cos π t ) dtThe graph shows t increases from 0 to 4.

4

4

x

y

dL dx dy= +2 2

= + +(cos – sin ) (sin cos )π π π π π πt t t t t t dt2 2

= +1 2( )π t dt

L t dt= + ≈∫ 1 25 72552 2

0

4

π . K

R6. a. The graph shows y = x1/3 , from x = 0 tox = 8, rotated about the y-axis, showing theback half of the solid only.

(0, y)(x, y)

2

80

x

y

dy x dx= −1

32 3/

dL dx dy x dx= + = +

2 2 2 32

11

3– /

dS x dL x x dx= ⋅ = +

2 2 1

1

32 3

2

π π – /

S x x dx= +

∫2 1

1

32 3

2

π 0

8– /

= +

∫2

1

91 3 4 3

1 2

π / //

0

8

x x dx

= +

∫3

2

1

9

4

34 3

1 21 3π

//

0

8

x x dx/

= +

π x 4 3

3 2

0

81

9/

/

= =π

27145 1 203 04363 2( – )/ . K

The disk of radius 8 has area 64π =201.0619… , so the answer is reasonable.

b. The graph shows y = tan x, from x = 0 tox = 1, rotated about the line y = −1, showingthe back half of the solid only.

(x, y)

(x,–1)

1

1

0

x

y

dy = sec2 x dx


dS = 2π (y + 1) · dL

= + +2 1 1 4π ( )tan secx x dx

S dS= ≈∫ 20 4199

0

1

. K

R7. a. r = θ ⇒ dr = dθdL dr rd d= + = +2 2 21( )θ θ θ

L dL= ≈∫ 32 4706

0

5 2

. Kπ /

b. dA r d d= = .1

2

1

22 2θ θ θ Area of the region

between the curves equals the area traced outfrom t = 2π to t = 5π /2 minus the area tracedout from t = 0 to t = π /2.

A d d= −∫ ∫1

2

1

22

2

5 22

0

2

θ θ θ θπ

π π/ /

= −1

6

1

63

2

5 23

0

2

θ θπ

π π/ /

= − − + =

= …

1

62 5 2 0 5 0

7 5

63 3 3 3 3 3π π( . . )

38.7578

.


Concept Problems

C1. a. The graph of µ(t) = 130 − 12T + 15T 2 − 4T 3

from T = 0 to T = 3 shows maxima at T = 0and T ≈ 2.0 and minima at T ≈ 0.5 andT = 3.

130

30

T

µ(T)

To maximize µ(T):

µ′ (T ) = −12 + 30T − 12T 2

= −6(2T − 1)(T − 2)

µ′ (T ) = 0 at T = 1

22,

µ(0) = 130; µ 1

2127 25

= . ;

µ(2) = 134; µ(3) = 121

Maximum viscosity occurs at T = 2, or 200°.

b. Minimum viscosity occurs at endpoint,T = 3, or 300°.

c.

C2. The graph of f (x) = (x − 1)4 + x shows that thegraph straightens out at x = 1 but does notchange concavity.

x

y

1

1

f ′ (x) = 4(x − 1)3 + 1; f″(x) = 12(x − 1)2,so f ′ (1) = 1 and f″(1) = 0.f″(x) > 0 for all x ≠ 1. In particular, f″(x) does notchange sign at x = 1. Thus, the graph is straightat x = 1, but not horizontal. Zooming in on (1,1) shows that the graph resembles y = x when xis close to 1, although it is actually concave upslightly.

C3. The graphs of f (x) = x2/3 and g x x( ) = −1 3/ show a

cusp at x = 0 for function f and a verticalasymptote at x = 0 for function g.

y

x

f

g

0 1

1

C4. a. y x= + +

3 1 25 13

52

. cos ( – )π

y′ =

2 5 13

53

53

. +

−

cos ( – ) sin ( – )π π π

x x

= +

– .cos ( – ) sin ( – )

2 5

31

35

35

π π πx x

dL dx= + +

12 5

31

2– .( cos )sin

π χ χ

where temporarily stands forχ π3

5( – )x

L dL= ≈ …∫ 5 77265

7 5

..

b. ′′ =y

– .cos cos sin sin

2 5

31

π χ χ χ χ χ[( ) ( ) ]+ + − d

dx

= − + ⋅– .cos cos

2 5

32 1

32π χ χ π

( )

= +– .(cos )( cos – )

2 5

91 2 1

2π χ χ

y ″ = 0 ⇔ cos χ = −1 or cos χ = 0.5χ = π + 2π n or χ = ±π /3 + 2π nx = 8 + 6n, 4 + 6n, or 6 + 6nThe only zero of y ′′ within the domain isx = 6, so the point of inflection must beat x = 6.

c. dS = 2π (x − 4) dL, where dL is as in part a.

S dS= ≈ …∫ 78 23735

7 5

..

d. x y= ⇒ = + +

7 5 3 1 25 1

5

6

2

. . cosπ

= + −3 1 25 1 3 2 2. ( / )

= + − = …3 1 25 1 75 3 3 0224. . .( )dV = 2π (x − 4) · (y − 3.0224…) · dx

V x dx= − + − −

= …∫ 2 4 1 25 1 1 75 3

58 8652

2

5

7 5

π χ( ) . [( ) ( . )]

.

cos.

C5. The 2000 World Almanac and Book of Factslists the area of Brazil as 3,286,478 square miles.Individual answers will vary.


C6. Let the cylinder lie on the x-axis and the hole lieon the y-axis so that the z-axis is perpendicularto both the cylinder and the hole. The cylinder isthus described by y2 + z2 ≤ 25, and the hole byx2 + z2 ≤ 9.Slice the hole with planes perpendicular to thez-axis. Then for −3 ≤ z ≤ 3, the cross section atz of the hole is a rectangle with height

2 and width .y z x z= =2 25 2 2 92 2– –

Area of cross-section rectangle is

4 225 34 2 4– z z+ ,

so . dV z z dz= +4 225 34 2 4–Thus, the volume of the hole (and thus of theuranium that once filled the hole) is

V z z dz= +−∫ 4 225 34 2 4

3

3

–

= 269.3703… cm3

According to the CRC Handbook, the densityof uranium is 19.1 g/cm3. So the mass of theuranium drilled out ism = (269.3703…)(19.1) = 5144.97… g.Value is 200(5144.97…) ≈ $1,029,000.

C7. Draw x- and y-axes with origin at the center ofthe circle on one face of the cube.

x

y

1

1

(x, y)

The solid remaining consists of eight identicalcorner pieces. Each corner piece consists of a cubeand three identical spikes. The spikes have squarecross sections when sliced perpendicular to theappropriate axes. The hole perpendicular to thexy-plane cuts a circle in that plane with equationx2 + y2 = 1. The cube shown in the diagrambegins at x = y so that 2x2 = 1,from which x = 2 2/ . Each cube is thus

( / )1 2 2− cm on a side, and thus has volume

Vc ( / ) . cm= − = …1 2 2 0 02512623 3.

Consider the leftmost spike in the precedingdiagram. Pick a sample point (x, y) on the partof the circle in that spike. The cross sectionperpendicular to the x-axis for this spike is a

square of side ( ) . 1 1 1 2− = −y x– Thus,

dV x dxs2( ) .= −1 1 2–

Because the spike goes from x = 0 to x = 2 2/ ,

V dVs s . .= ≈ …∫ 0 01096420

2 2/

(The integral can be evaluated algebraically usingtrigonometry substitution, as in Chapter 9. The

exact value is Vs .= − −11 2

12 4

1

2

π)

The 24 spikes (3 for each of the eight corners) areidentical.Thus, the total volume remaining isV V V= +

= − +

= + − = …

8 24

8 1 2 2 2411 2

12 4

1

2

8 8 2 6

3

c s

3

( / )

0.46415 cm

– –π

π

Chapter Test

T1.

x

y

3

0 1 2 3 4

T2.

1 2 3 4 5 6 7

1

2

3

4

5

6y

x

Derivative

Function

T3. A = xy = x(500 − 0.5x) = 500x − 0.5x2

A′ = 500 − xA′ = 0 ⇔ x = 500A′ goes from positive to negative at x = 500⇒ local maximum at x = 500.A(0) = A(1000) = 0 ⇒ global maximumat x = 500.Maximum at x = 500, y = 250.


T4. V x y y dxa

b

= −∫2 1 2π ( )

T5. a. dA r dr= 1

22

b. dL dr r d= +( ) ( )2 2θ

c. dL dx dy= +2 2

d. dS x dL= −2 1π ( )

T6. f (x) = x3 − 7.8x2 + 20.25x − 13

f ′ (x) = 3x2 − 15.6x + 20.25

= 3(x − 2.5)(x − 2.7)

f ′ (x) changes from positive to negative atx = 2.5 and from negative back to positive atx = 2.7. So there is a local maximum at x = 2.5and a local minimum at x = 2.7.f ″(x) = 6x − 15.6 = 6(x − 2.6)f ″(x) = 0 at x = 2.6f ′ (2.6) = −0.03, so the graph is not horizontal atthe inflection point.

T7. y = x3 ⇒ dy = 3x2 dx

dL dx dy x dx= + = +2 2 2 21 3( )

L x dx= + = …∫ 1 9 8 63034

0

2

.

T8. dS = 2πx · dL = 2πx 1 9 4+ x dx

S x x dx= + = …∫ 2 1 9 77 32454

0

2

π .

T9. V(x) = π x2(8 − y) = 8π x2 − π x5

The graph shows a maximum V (x) at x ≈ 1.5.

0 1 2

40

x

y

V′(x) = 16π x − 5π x4 = 0 at x = 0 or 3.21/3

V(0) = V(2) = 0V(3.21/3) = 4.8 · 3.22/3π > 0, so this is amaximum.Maximal cylinder has V = 4.8 · 3.22/3π cm3 =32.7459… cm3.

T10. Slicing parallel to the y-axis generates cylindricalshells of radius x extending from the samplepoint (x, y) to the line y = 8.dV = 2πx · (8 − y) · dx = 2πx (8 − x3) dx

V x x dx x x= − = −∫ 2 8 2 4 0 24 2 5

0

2

π π( ) ( . )0

2

= 19.2π = 60.3185…

T11. Vcyl = π · 22 · 8 = 32πV

Vsolid

cyl

.= =19 2

320 6

. ππ

So, Vsolid = 0.6Vcyl .

T12. a.

5

2

x

y

b. dx = −5 sin t dt, dy = 2 cos t dt

dL dx dy= +2 2

= +(– cos ) ( sin )5 22 2t t dt

L dL= ≈ …∫ 23 01310

2

.π

c. Slicing perpendicular to the x-axis generatescircular slices of radius y, where sample point(x, y) is on the upper branch of the ellipse.dV = πy2 dx = 4π sin2 t (−5 sin t dt)

= −20π sin3 t dtLeftmost slice is at t = π, and rightmost sliceis at t = 0.

V t dt= − = …

= …∫ 20 83 7758

26 6666

3π

ππ

sin .0

. (numerically)V can be evaluated algebraically bytransforming two of the three sin t factorsinto cosines.

V t t dt= − −∫ 20 1 2ππ

( )0

cos sin

= − + ∫∫ 20 20 200

π πππ

sin cos sint dt t t dt

= −

=20

20

326

2

33

0

π π ππ

cos cost t

The x-radius is 5, and the y-radius is 2.4

3π (x-radius)(y-radius)2 = 4

3π (5)(2)2 =

262

3π , Q.E.D.

(In general, if a = x-radius and b = y-radius,the parametric functions are x = a cos t, y =b sin t. Repeating the preceding algebraic

solution gives V ab= 4

32π .)

T13. r = 5e0.1 θ

dr = 0.5e0.1 θ dθ


dL dr rd= +2 2( )θ

= +( . ) ( ). .0 5 50 1 2 0 1 2e e dθ θ θ= e d0 1 25 25. θ θ.

The spiral starts at r = 5 = 5e0.1·0 and makes threecomplete revolutions, so 0 ≤ θ ≤ 6π.

L e d e= = ∫ 0 1 0 1

0

6

0

6

25 25 10 25 25. .θ θππ

θ. .

= − = …10 25 25 1 280 69610 6. ( ) ..e π

T14. dA r d e d= =1

212 52 0 2θ θθ. .

The area between the second and third revolutionsequals the area swept out for the third revolution

minus the area swept out for the secondrevolution. In Quadrant I, the third revolutionextends from θ = 4π to θ = 4.5π and the secondrevolution extends from θ = 2π to θ = 2.5π.

A e d e d= − ∫∫ 12 5 12 50 2 0 2

2

2 5

4

4 5

. .. .θ θ

π

π

π

πθ θ

..

= −62 5 62 50 24

4 5 0 22

2 5. .. .e eθ

ππ θ

ππ. .

= 62.5(e0.9 π − e0.8 π − e0.5 π + e0.4 π )= 203.7405…



Chapter 9—Algebraic Calculus Techniquesfor the Elementary Functions

Problem Set 9-1

1. V x x dx= ⋅ ≈ …∫2 3 58640

2

ππ

cos/

.

2. f (x) = x sin x ⇒ f ′(x) = x cos x + sin x

3. f x dx x x dx x dx′ = + ∫∫∫ ( ) cos sin

4. x x dx f x dx x dxcos sin= ′ − ∫∫∫ ( )

= f (x) + cos x + C (by definition of indefiniteintegral)

= x sin x + cos x + C

5. V x x dx= ∫20

2

ππ

cos/

= +2 2 02π π πx x xsin cos /

= π2 − 2π6. V = π2 − 2π = 3.5864… , which is the same as

the approximation, to the accuracy shown.

7. The method involves working separately with thedifferent “parts” of the integrand. The functionf (x) = x sin x was chosen because one of theterms in its derivative is x cos x, which is theoriginal integrand. See Section 9-2.

Problem Set 9-2

Q1. y′ = x sec2 x + tan x Q2.1

1111x C+

Q3. Q4.1

33sin x C+

x

y

Q5. 5 cos2 5x − 5 sin2 5x Q6.

x

y

Q7. r′(x) = t (x) Q8. lim( ) – ( )

h

f x h f x

h→

+0

Q9. ≈ 110/6 Q10. C

1. x x dxsin∫ u = x dv = sin x dx

du = dx v = −cos x

= − − −∫x x x dxcos cos( )

= −x cos x + sin x + C

2. x x dxcos 3∫ u = x dv = cos 3x dx

du = dx v x= 1

33sin

= − ∫1

33 3x x x dxsin sin

1

3

= + +1

33

1

93x x x Csin cos

3. xe dxx4∫ u = x dv = e4x dx

du = dx v e x= 1

44

= − ∫1

4

1

44 4xe e dxx x

= − +1

4

1

164 4xe e Cx x

4. 6 3xe dxx−∫ u = 6x dv = e− 3x dx

du = 6 dx v e x= − −1

33

= −

− −

− −∫( ) ( )61

36

1

33 3x e e dxx x

= − − +− −22

33 3xe e Cx x

5. ( )x e dxx+ −∫ 4 5 u = x + 4 dv = e− 5x dx

du = dx v e x= − −1

55

= − + ⋅ +− −∫( )x e e dxx x41

5

1

55 5

= − − − +− − −4

5

1

5

1

255 5 5e xe e Cx x x

= − − +− −21

25

1

55 5e xe Cx x

6. ( )x e dxx+∫ 7 2 u = x + 7 dv = e2x dx

du = dx v e x= 1

22

= + ⋅ − ∫( )x e e dxx x71

2

1

22 2

= + − +7

2

1

2

1

42 2 2e xe e Cx x x

= + +13

4

1

22 2e xe Cx x

7. x x dx3 ln ∫ u = ln x dv = x3 dx

du = x− 1 dx v x= 1

44

= − ∫1

4

1

44 3x x x dxln

= − +1

4

1

164 4x x x Cln


8. x x dx5 3ln∫ u = ln 3x dv = x5 dx

du = x− 1 dx v x= 1

66

= − ∫1

63

1

66 5x x x dxln

= − +1

63

1

366 6x x x Cln

9. x e dxx2∫ u = x2 dv = ex dx

du = 2x dx v = ex

= − ∫x e xe dxx x2 2

u = 2x dv = ex dx

du = 2 dx v = ex

= − −

∫x e xe e dxx x x2 2 2

= x2ex − 2xex + 2ex + C

10. x x dx2 sin∫ u = x2 dv = sin x dx

du = 2x dx v = −cos x

= − − −∫x x x x dx2 2cos cos( )

u = 2x dv = −cos x dx

du = 2 dx v = −sin x

= − − − − −

∫x x x x x dx2 2 2cos sin sin( )

= −x2 cos x + 2x sin x + 2 cos x + C

11. ln x dx∫ u = ln x dv = dx

du = x− 1 dx v = x

= − ⋅ −∫x x x x dxln 1

= x ln x − x + C

Problem Set 9-3

Q1.1

66r C+ Q2. 2m cos 2m + sin 2m

Q3. tan x + C Q4.1

18113 6( )x C+ +

Q5.1

4114x x C+ + Q6. 1

Q7. 1/2

Q8. V f x g x dxa

b

= −∫π [ ( ) ( ) ]2 2

Q9. Q10. B

2

4

x

y

1. x e dxx3 2∫

e

e

u dv

x3 e2x

3x2 12 e

2x

6x14

2x

618

2x

01

16 e2x

+

+

–

–

+

= − + − +1

2

3

4

3

4

3

83 2 2 2 2 2x e x e xe e Cx x x x

2. x e dxx5 −∫ u dv

x 5 e –x

5x 4 –e –x

20x 3 e –x

60x 2 –e –x

120x e –x

120 –e –x

0 e –x

+

+

+

–

–

–

+

= −x5e− x − 5x4e− x − 20x3e− x − 60x2e− x − 120xe− x

− 120e− x + C

3. x x dx4 sin∫ u dvx 4 sin x

4x 3 –cos x12x 2 –sin x24x cos x24 sin x0 –cos x

+

+

+

–

–

–

= −x4 cos x + 4x3

sin x + 12x2 cos x

− 24x sin x − 24 cos x + C

4. x x dx2 cos∫ u dvx 2 cos x2x sin x2 –cos x0 –sin x

+

+

–

–

= x2 sin x + 2x cos x − 2 sin x + C

5. x x dx5 2cos∫ u dvx 5 cos 2x

5x 4 12 sin 2x

20x 3 –14 cos 2x

60x 2 –18 sin 2x

120x1

16 cos 2x

1201

32 sin 2x

0 –164 cos 2x

+

+

+

–

–

–

+


= + −1

22

5

42

5

225 4 3x x x x x xsin cos sin

− + + +15

42

15

42

15

822x x x x x Ccos sin cos

6. x x dx3 5sin∫ u dvx3 sin 5x

3x2 –15 cos 5x

6x –125 sin 5x

61

125 cos 5x

01

625 sin 5x

+

+

–

–

+

= − + +1

55

3

255

6

12553 2x x x x x xcos sin cos

− +6

6255sin x C

7. e x dxx sin∫ u dvex sin xex –cos xex –sin x

+

–+

= − + −∫e x e x e x dxx x xcos sin sin

⇒ ∫2 e x dxx sin

= − + +e x e x Cx xcos sin 1

⇒ ∫ e x dxx sin

= − + +1

2

1

2e x e x Cx xcos sin

8. e x dxx cos∫ u dve x cos xe x sin xe x –cos x

+

–+

= + − ∫e x e x e x dxx x xsin cos cos

⇒ ∫2 e x dxx cos

= ex sin x + ex cos x + C1

⇒ ∫ e x dxx cos

= + +1

2

1

2e x e x Cx xsin cos

9. e x dxx3 5cos∫ u dve3x cos 5x

3e3x 15 sin 5x

9e 3x 125 cos 5x

+

–

+ –

= +1

55

3

2553 3e x e xx xsin cos

− ∫9

2553e x dxx cos

⇒ ∫34

2553e x dxx cos

= + +1

55

3

2553 3


⇒ ∫ e x dxx3 5cos

= + +5

345

3

3453 3e x e x Cx xsin cos

10. e x dxx4 2sin∫ u dve 4x sin 2x

4e 4x –12 cos 2x

16e 4x –14 sin 2x

+

–

+

= − + − ∫1

22 2 4 24 4 4e x e x e x dxx x xcos sin sin

⇒ ∫5 24e x dxx sin

= − + +1

22 24 4


⇒ ∫ e x dxx4 2sin

= − + +1

102

1

524 4e x e x Cx xcos sin

11. x x dx7 3ln∫-----------------

u dvln 3x x7

1/x18 x8

118 x7

0164 x8

+

–

+

= − +1

83

1

648 8x x x Cln

12. x x dx5 6ln∫-------------------------

u dvln 6x x 5

1/x16 x 6

116 x 5

0136 x 6

+

–

+

= − +1

66

1

366 6x x x Cln

13. x dx x C4 577

5ln

ln= +∫ (ln 7 is a constant!)

14. e dx e Cx x7 755

7cos

cos= +∫15. sin cos sin5 61

6x x dx x C= +∫

16. x x dx x x dx( ) ( ) ( ) /31

23 22 2 3 2 2 3− = − − −∫∫ /

= − +3

103 2 5 3( – )x C/


17. x x dx3 5( )1/2+∫ u dvx 3 (x + 5)1/2

3x 2 23(x + 5)3/2

6x415(x + 5)5/2

68

105(x + 5)7/2

016945(x + 5)9/2

+

+

–

–

+

= + − +2

35

4

553 3 2 2 5 2x x x x( ) ( )/ /

+ + − + +16

355

32

31557 2 9 2x x x C( ) / /( )

18. x x dx x x dx2 2 1 22 2∫ ∫− = −( ) /

u dvx2 (2 – x)1/2

2x –23(2 – x)3/2

2415(2 – x)5/2

08

105(2 – x)7/2+

+

–

– –

= − − − −2

32

8

1522 3 2 5 2x x x x( ) ( )/ /

− +16

1052 7 2( – )x C/

19. ln ln lnx dx x dx x x x C5 5 5 5= = − +∫ ∫

20. e dx x dx x Cxln 7 277

2= = +∫ ∫

21. x e dxx5 2

∫--------------------------

--------------------------

2

2

u dvx 4 xex2

4x 3 1 ex2

2x2 xex2

4x1 ex2

2 xex2

012 e x2

+

+

–

–

= − + +1

24 22 2 2

x e x e e Cx x x

22. x e dxx5 3

∫---------------------------

u dv

x 3 x 2 ex3

3x2 13 ex3

1 x 2 ex3

0 ex3

+

–

+ 13

= − +1

3

1

33 3 3

x e e Cx x

23. x x dx( )3ln∫--------------------------

--------------------------

--------------------------

2

8

8

16

u dv(ln x)3 x

3 (ln x)2/x12 x2

3 (ln x)2 1 x

6 (ln x)/x14 x2

6 ln x14 x

6/x1 x2

61 x

01 x2

+

+

–

–

+

= −1

2

3

42 3 2x x x x( ) ( )2ln ln

+ − +3

4

3

82 2x x x Cln

24. x x dx3( )2ln∫--------------------------

--------------------------

u dv(ln x)2 x 3

2(ln x)/x14 x4

2 ln x14 x3

2/x116 x 4

2116 x3

0164 x 4

+

+

–

–

= − + +1

4

1

8

1

324 4 4x x x x x C( )2ln ln

25. x x dx3 2 1( )4+∫----------------------------------

u dvx2 x(x2 + 1)4

2x110 (x2 + 1)5

15 x(x2 + 1)5

0112 (x2 + 1)6

+

–

+

= + − + +1

101

1

6012 2 5 2 6x x x C( ) ( )

26. x x dx x x3 2 3 2 1 23 3− = −∫ ∫ ( ) /

-----------------------------------

u dvx 2 x(x2 – 3)1/2

2x13(x2 – 3)3/2

23 x(x2 – 3)3/2

015(x2 – 3)5/2

+

–

+

= − − +1

33

2

1532 2 3 2 2 5 2x x x C( ) / /( – )

27. cos2 x dx∫ u dvcos x cos x–sin x sin x

+–


= − −∫cos sin sinx x x dx( ) 2

= + −∫cos sin cosx x x dx( )1 2

= + − ∫cos sin cosx x x x dx2

⇒ = + +∫2 21cos cos sinx dx x x x C

⇒ = + +∫ cos cos sin2 1

2

1

2x dx x x x C

28. sin .2 0 4x dx∫ u dvsin 0.4x sin 0.4x

0.4 cos 0.4x –2.5 cos 0.4x+

–

= − + ∫2 5 0 4 0 4 0 42. . . .sin cos cosx x x dx

= − + −∫2 5 0 4 0 4 1 0 42. . . ( . ) sin cos sinx x x dx

= − + −∫ ∫2 5 0 4 0 4 0 42. . . . sin cos sinx x dx x dx

⇒ ∫2 0 42sin . x dx

= −2.5 sin 0.4x cos 0.4x + x + C1

⇒ ∫ sin2 0 4. x dx

= −1.25 sin 0.4x cos 0.4x + 0.5x + C

29. sec3 x dx∫ u dvsec x sec2 x

sec x tan x tan x+

–

= − ∫sec tan sec tanx x x x dx2

= − −∫sec tan sec secx x x x dx( ) 2 1

= − + +∫sec tan sec ln sec tanx x x dx x x3 | |

⇒ ∫2 3sec x dx

= sec x tan x + ln | sec x + tan x | + C1

⇒ ∫ sec3 x dx

= + + +1

2

1

2sec tan ln sec tanx x x x C| |

30. sec tan2 x x dx∫= ⋅ = +∫ ( ) ( )1sec sec tan secx x x dx x C

1

22

31. logln

ln31

3x dx x dx= ∫∫

= +1

3ln( ln – )x x x C

= − +x x x Clogln31

3

32. logln

ln101

10x dx x dx =∫ ∫

= +1

10ln( ln – )x x x C

= − +x x x Clogln10

1

10

33. sin cosx dx x C= − +∫34. cos sinx dx x C= +∫35. csc ln csc cotx dx x x C= − + +∫ | |

36. sec ln sec tanx dx x x C= + +∫ | |

37. tan ln cosx dx x C= − +∫ | |

38. cot ln sinx dx x C= +∫ | |

39. x2∫ cos x dx

For the first integral, Wanda integrated cos x anddifferentiated x2, but in the second integral she

plans to differentiate cos∫ x dx and integrate 2x,

effectively canceling out what she did in the firstpart. She will get

x x dx x x x x x x dx2 2 2 2cos sin sin cos ,= − + ∫∫which is true but not very useful!

40. x x dx2 cos∫Amos’s choice of u and dv transforms

x x dx x x x x dx2 3 31

3

1

3∫ ∫+cos cos sininto ,

which is more complicated than the originalexpression.

41. After two integrations by parts,

e x dxx sin ∫= − + − ∫e x e x e x dxx x xcos sin sin

but after two more integrations,

e x dx e x e x e xx x x xsin cos sin cos= − + +∫− + ∫e x e x dxx xsin sin

Two integrations produced the original integralwith the opposite sign (which is useful), and twomore integrations reversed the sign again to givethe original integral with the same sign (whichis not useful).

42. cos cos2 1

21 2x dx x dx= +∫ ∫ ( )

= +

+1

2

1

22x x Csin


By the double-argument properties fromtrigonometry,1

2

1

22

1

2x x C x x x C+

+ = + +sin ( sin cos )

which is equivalent to the answer in Problem 27found using integrating by parts.

43.

x

y

(1,1/e)

3

1

y′ = −xe− x + e− x = e− x(1 − x)Critical points at x = 0, 1, 3; maximum atx = 1.

A xe dxx= −∫0

3

= − −− −( )xe ex x

0

3

= −3e− 3 − e− 3 + 1 = −4e− 3 + 1 = 0.8008…

44. y = 12x2e− x

Area from x = 0 to x = b is

A b x e dxxb

( ) = −∫ 12 2

0

= − − −− − −12 24 242

0x e xe ex x x b

= −12b2e− b − 24be− b − 24e− b + 24

The first two terms approach zero as b → ∞by L’Hospital’s rule. The third term alsoapproaches 0.∴ =

→∞limb

bA 24

45. y = ln xdV = πy2

dx = π (ln x)2 dx

V x dx= ∫ π (ln )2

1

5

-----------------------

-----------------------

u dv(ln x)2 1

2 (ln x)/x x2 ln x 1

2/x x

2 10 x

+

–

–

+

= − +π π πx x x x x( )2ln ln2 21

5

= 5π ( ln 5)2 − 10π ln 5 + 10π − 0 + 0 − 2π= 15.2589…

46. Consider u dv ,∫ and write dv v C= +∫ . Then

u dv u v C v C du= + − +∫ ∫( ) ( )

= + − − ∫∫uv Cu v du C du

= + − − = −∫ ∫uv Cu v du Cu uv v du

Thus, the constant cancels out later, Q.E.D.

47. For integration by parts, u dv uv v du .= −∫ ∫Applying limits of integration gives

u dv uv v duc

d

u a

u b

a

b

= −∫ ∫=

=

= − − ∫( ) bd ac v dua

b

The quantity (bd − ac) is the area of the“L-shaped” region, which is the area of the largerrectangle minus the area of the smaller one.Thus, the integral of u dv equals the area of theL-shaped region minus the area represented by theintegral of v du.

48. ln ln lnax dx a x dx= +∫ ∫ ( )

= x ln a + x ln x − x + C= x ln ax − x + C

49. sin7 x dx∫ u dvsin6 x sin x

6 sin5 x cos x –cos x+

–

= − + ∫sin cos sin cos6 5 26x x x x dx

= − + −∫sin cos sin sin6 5 26 1x x x x dx( )

= − + − ∫∫sin cos sin sin6 5 76 6x x x dx x dx

7 67 6 5sin sin cos sinx dx x x x dx= − + ∫∫sin sin cos sin7 6 51

7

6

7x dx x x x dx = − + ∫∫

The fractions are 1/(old exponent) and(old exponent − 1)/(old exponent). The newexponent is 2 less than the old exponent. So

sin sin cos7 61

7x dx x x= −∫

+ +

∫6

7

1

5

4

54 3– sin cos sinx x x dx

= − −1

7

6

356 4sin cos sin cosx x x x

+ +

∫24

35

1

3

2

32– sin cos sinx x x dx

= − −1

7

6


− − +8

35

16

352sin cos cosx x x C


Problem Set 9-4

Q1. uv v du− ∫


Q2. Q3.

x

y

2π

3

x

y

2π

1

Q4. y′ = 1 + ln 5x Q5.1

66sin x C+

Q6. ln |x| + C

Q7.

x

y

3

1

Q8.1

1 2+ xQ9. ln | sec x + tan x | + C

Q10. D

1. sin9 x dx∫ u dvsin8 x sin x

8 sin7 x cos x –cos x+

–

= − + ∫sin cos sin cos8 7 28x x x x dx

= − + −∫sin cos sin sin8 7 28 1x x x x dx ( )

= − + − ∫∫sin cos sin sin8 7 98 8x x x dx x dx

9 89 8 7 = − + ∫ ∫sin sin cos sinx dx x x x dx

sin sin cos sin9 8 71

9

8

9x dx x x x dx= − +∫ ∫

2. cos10 x dx∫ u dvcos 9 x cos x

–9 cos 8 x sin x sin x–+

= + ∫cos sin cos sin9 8 29x x x x dx

= + −∫cos sin cos cos9 8 29 1x x x x dx ( )

= + − ∫ ∫cos sin cos cos9 8 109 9x x x dx x dx

10 910 9 8cos cos sin cosx dx x x x dx= +∫ ∫cos cos sin cos10 9 81

10

9

10x dx x x x dx= + ∫∫

3. cot cot cot12 10 2x dx x x dx= ∫ ∫= −∫ cot csc10 2 1x x dx ( )

= − ∫ ∫cot csc cot10 2 10x x dx x dx

= − − ∫1

1111 10cot cotx x dx

4. tan tan tan20 18 2x dx x x dx = ∫ ∫= −∫ tan sec18 2 1x x dx( )

= − ∫ ∫tan sec tan18 2 18x x dx x dx

= − ∫1

1919 18tan tanx x dx

5. sec13 x dx∫ u dvsec 11 x sec 2 x

11 sec 10 x sec x tan x tan x+

–

= − ∫sec tan sec tan11 11 211x x x x dx

= − − ∫sec tan sec sec11 11 211 1x x x x dx( )

= − + ∫∫sec tan sec sec11 13 1111 11x x x dx x dx

12 1113 11 11sec sec tan secx dx x x x dx= + ∫∫sec sec tan sec13 11 111

12

11

12x dx x x x dx= + ∫∫

6. csc100 x dx∫u dv

csc 98 x csc 2 x–98 csc 97 x csc x cot x –cot x–

+

= − − ∫csc cot csc cot98 98 298x x x x dx

= − − −∫csc cot csc csc98 98 298 1x x x x dx( )

= − − + ∫∫csc cot csc csc98 100 9898 98x x x dx x dx

99 98100 98 98csc csc cot cscx dx x x x dx= − + ∫∫csc csc cot csc100 98 981

99

98

99x dx x x x dx= − + ∫∫

7. cosn x dx∫ u dvcos n – 1 x cos x

–(n –1) cosn – 2 x sin x sin x–+

= + −− −∫cos sin cos sinn nx x n x x dx1 2 21( )

= + − −− −∫cos sin cos cosn nx x n x x dx1 2 21 1( ) ( )

= + −− −∫cos sin cosn nx x n x dx1 21( )

− − ∫( )n x dxn1 cos

n x dx x x n x dxn n ncos cos sin cos= + −− −∫∫ 1 21( )

cos cos sin–

cosn n nx dxn

x xn

nx dx= +− −∫∫ 1 11 2


8. sinn x dx∫ u dvsinn –1 x sin x

(n –1) sinn –2 x cos x –cos x+

–

= − + −− −∫sin cos sin cosn nx x n x x dx1 2 21( )

= − + − −− −∫sin cos sin sinn nx x n x x dx1 2 21 1( ) ( )

= − + −− −∫sin cos sinn nx x n x dx1 21( )

− − ∫( )n x dxn1 sin

n x dx x x n x dxn n nsin sin cos sin= − + −− −∫∫ 1 21( )

sin sin cos–

sinn n nx dxn

x xn

nx dx= − +− −∫∫ 1 11 2

9. tan tan tann nx dx x x dx= −∫∫ 2 2

= −−∫ tan secn x x dx2 2 1( )

= −− −∫∫ tan sec tann nx x dx x dx2 2 2

= −− −∫1

11 2

nx x dxn n

–tan tan

10. cot cot cotn nx dx x x dx= −∫∫ 2 2

= −−∫ cot cscn x x dx2 2 1( )

= −− −∫∫ cot csc cotn nx x dx x dx2 2 2

= − −− −∫1

11 2

nx x dxn n

–cot cot

11. cscn x dx∫u dv

csc n – 2 x csc 2x–(n – 2) cscn – 3 x csc x cot x –cot x–

+

= − − −− −∫csc cot csc cotn nx x n x x dx2 2 22( )

= − − − −− −∫csc cot csc cscn nx x n x x dx2 2 22 1( ) ( )

= − − −− ∫csc cot cscn nx x n x dx2 2( )

+ − −∫( )n x dxn2 2csc

( )n x dxn− ∫1 csc

= − + −− −∫csc cot cscn nx x n x dx2 22( )

cscn x dx∫= − +− −∫1

1

2

12 2

nx x

n

nx dxn n

–csc cot

–

–csc

12. secn x dx∫u dv

sec n – 2 x sec 2 x(n – 2) secn – 3 x sec x tan x tan x–

+

= − −− −∫sec tan sec tann nx x n x x dx2 2 22( )

= − − −− −∫sec tan sec secn nx x n x x dx2 2 22 1( ) ( )

= − −− ∫sec tan secn nx x n x dx2 2( )

+ − −∫( )n x dxn2 2sec

( )n x dxn− ∫1 sec

= + −− −∫sec tan secn nx x n x dx2 22( )

secn x dx∫= +− −∫1

1

2

12 2

nx x

n

nx dxn n

–sec tan

–

–sec

13. sin5 x dx∫= − + − −

+1

5

4

5

1

3

2

34 2sin cos sin cos cosx x x x x C

= − − − +1

5

4

15

8

154 2sin cos sin cos cosx x x x x C

14. cos5 x dx∫= + +

+1

5

4

5

1

3

2

34 2cos sin cos sin sinx x x x x C

= + + +1

5

4

15

8

154 2cos sin cos sin sinx x x x x C

15. cot6 x dx∫= − − − − − −

+1

5

1

35 3cot cot cotx x x x C( )

= − + − − +1

5

1

35 3cot cot cotx x x x C

16. tan7 x dx∫= − − +

+1

6

1

4

1

26 4 2tan tan tan ln cosx x x x C| |

= − + + +1

6

1

4

1

26 4 2tan tan tan ln cosx x x x C| |

17. sec sec tan tan4 21

3

2

3x dx x x x C= + +∫

18. csc csc cot cot4 21

3

2

3x dx x x x C= − − +∫

19. a. y = cos x is on top; y = cos3 x is in the

middle; y = cos5 x is on the bottom.

b. For y = cos x, area ≈ 2.0000… .For y = cos3 x, area ≈ 1.3333… .For y = cos5 x, area ≈ 1.06666… .

c. A x dx x12

2

2

2

= =∫ −cos sin

– /

/

/

/

π

π

π

π

sin (π/2) − sin (−π/2) = 2


A x dx33

2

2

= ∫ cos– /

/

π

π

= +−

1

3

2

32

2

2

cos sin sin/

/

x x xπ

π

= +1

32 2

2

322cos / ) sin / ) sin / )( ( (π π π

− − − − −1

32 2

2

32cos ( sin ( sin (2 π π π/ ) / ) / )

= + − + = = …02

30

2

3

4

31 3333.

Observe that A A3 12

3= .

A x dx

x x x dx

A

55

2

2

4

2

23

2

21

5

4

5

4

5

4

5

4

3

4

5

2

3

16

151 066666

=

= +

= + = ⋅ = ⋅ ⋅ =

= …

∫

∫− −

cos

cos sin cos

– /

/

/

/

/

/

π

π

π

π

π

π

0 2

.

3

Observe that A A5 34

5= .

d. Based on the graphs, the area under cos xshould be greater than that under cos3 x,which in turn is greater than the area undercos5 x. This is exactly what happens with thecalculated answers: A1 > A3 > A5.

e.

x

y

1

y = cos100 x

–0.5π 0.5π

f. Yes, lim cos– /

/

n

n x dx→∞

=∫ 02

2

.π

π

Following the pattern in part c, for odd n,

An n n

n n nn = ⋅( – )( – )( – ) ( )( )( )

( )( – )( – ) ( )( )

1 3 5 4 2 1

2 4 5 32

K

K,

the denominator gets large faster than thenumerator. However, because both go toinfinity, this observation is not decisive.The following epsilon proof by Cavan Fangestablishes the fact rigorously, using thedefinition of limit in the form “For anyε > 0, there is an N > 0 such that whenevern > N, An < ε.”Proof:

Pick 0 < ε < 2π.

Then 04

1< <cos ,ε

so there exists N > 0

such that cosN ε επ4 2

< .

Note that if ε π4 2

< < ,x then cos cosx <ε4

⇒ cosN x < .επ2

Now, for any n > N,

cos/

/n x dx

−∫ π

π

2

2

= +∫ ∫2 2cos cosn nx dx x dx0

/4

/4

/2

.ε

ε

π

But 2 < ( < ) . /

cos cos/

n nx dx dx x22

10

4

0

4 εε ε∫∫ =

And 2 < ( > )/4

/2

cos cos/

/n Nx dx x dx n N2

4

2

ε

π

ε

π

∫ ∫< < < .

/

22 2 24

2

0

2επ

επ

ε επε

π πdx dx xN

/

/

cos∫ ∫ =

So/2

/2

cosn x dx−∫ π

π

= + <∫ ∫2 2cos cos .n nx dx x dx0

/4

/4

/2ε

ε

πε

∴ =→∞ −∫lim cos ,

/

/

n

n x dxπ

π

2

2

0 Q.E.D.

20. cos cos cos5 4x dx x x dx=∫ ∫= −∫ ( )21 2sin cosx x dx

= − +∫ ( )1 2 2 4sin sin cosx x x dx

= −∫ ∫cos sin cosx dx x x dx2 2

+ ∫ sin cos4 x x dx

= − + +sin sin sinx x x C2

3

1

53 5

∴ =−∫A x dx5

5

2

2

sin/

/

π

π

= − +

2

2

3

1

53 5

0

2

sin sin sin/

x x xπ

= − + = = …24

3

2

5

16

151 0666. , which agrees with

the answer from Problem 19.

21. sec3 x dx∫ u dvsec x sec 2 x

sec x tan x tan x+

–


= − −∫sec tan sec secx x x x dx ( )2 1

= − + ∫∫sec tan sec secx x x dx x dx3


2 3sec x dx ∫= sec x tan x + ln |sec x + tan x| + C1

sec3 x dx∫= +1

2

1

2sec tanx x ln |sec x + tan x| + C

Note that the answer is half the derivative ofsecant plus half the integral of secant.

22. sinn ax dx∫u dv

sinn –1 ax sin axa(n – 1) sinn – 2 ax cos ax –

1a cos ax–

+

= − −1 1

aax axnsin cos

+ − −∫( )n ax ax dxn1 2 2sin cos

= − −1 1

aax axnsin cos

+ − −−∫( ) ( )n ax ax dxn1 12 2sin sin

= − + −− −∫111 2

aax ax n ax dxn nsin cos sin( )

− − ∫( )n ax dxn1 sin

n ax dxnsin∫= − + −− −∫1

11 2

aax ax n ax dxn nsin cos sin( )

sinn ax dx∫= − +− −∫1 11 2

anax ax

n

nax dxn nsin cos

–sin

sin sin cos5 431

153 3x dx x x= −∫

− − +4

453 3

8

4532sin cos cosx x x C

23. sin sin cos3 21

3ax dx

aax ax= −∫

+ ∫2

322sin ax dx (From Problem )

= − − +1

3

2

32

aax ax

aax Csin cos cos

= − + +1

322

aax ax Ccos sin( ) , Q.E.D.

Or: d

dx aax ax– (cos )(sin )

1

322 +

= − +1

322

aa ax ax(– sin )(sin )

− 1

32

aax a ax ax(cos )( sin cos )

= +1

32 23 2(sin sin – sin cos )ax ax ax ax

= +1

32 13 2[sin sin ( – cos )]ax ax ax

= +1

323 2[sin sin (sin )]ax ax ax

= sin3 ax

∴ ∫ sin3 ax dx

= − + +1

322

aax ax C(cos )(sin ) , Q.E.D.

24. Use integration by parts, or use the technique ofProblem 20, as shown here.

cos cos cos3 2ax dx ax ax dx= ∫∫= −∫ ( )1 2sin cosax ax dx

= − ∫∫ cos sin cosax dx ax ax dx2

= − +1 1

33

aax

aax Csin sin

= +1

33 2

aax ax C(sin )( – sin )

= +1

32 2

aax ax(sin )( cos ) + C, Q.E.D.

Or: Differentiate, as in the alternate solution forProblem 23.

Problem Set 9-5Q1. f ′(1) = −4 Q2. g′(2) = 1/2

Q3. h′(3) = −12 Q4. t′(4) = π/24

Q5. p′(5) = 6e5 Q6. x = 83 = 512

Q7. Q8. integration by parts

x

y

3

5

Q9. Q10. E

1 3 5 6 8

x

–1

1

f'(x) and f(x)

f

f'

1. sin cos sin5 21x dx x x dx= −∫ ∫ ( )2

= − +∫ ( )1 2 2 4cos cos sinx x x dx

= − + − +cos cos cosx x x C2

3

1

53 5


2. cos sin cos7 21x dx x x dx= −∫ ∫ ( )3

= − + −∫ ( )1 3 32 4 6sin sin sin cosx x x x dx

= − + − +sin sin sin sinx x x x C3 5 73

5

1

7

3. cos sin cos7 29 1 9 9x dx x x dx ( )3= −∫ ∫= − +∫ (1 3 9 3 92 4sin sinx x

− sin6 9x) cos 9x dx

= −1

99

1

993sin sinx x

+ − +1

159

1

6395 7sin sinx x C

4. sin cos sin3 210 1 10 10x dx x x dx= −∫∫ ( )

= − + +1

1010

1

30103cos cosx x C

5. sin cos sin4 53 31

153x x dx x C= +∫

6. cos sin cos8 97 71

637x x dx x C= − +∫

7. cos sin6 38 8x x dx∫= −∫ cos cos sin6 28 1 8 8x x x dx( )

= −∫ ( )cos cos sin6 88 8 8x x x dx

= − + +1

568

1

7287 9cos cosx x C

8. sin cos4 32 2x x dx∫= −∫ sin sin cos4 22 1 2 2x x x dx( )

= −∫ ( )sin sin cos4 62 2 2x x x dx

= − +1

102

1

1425 7sin sinx x C

9. sin cos sin cos cos5 2 2 21x x dx x x x dx= −∫ ∫ ( )2

= − +∫ ( )cos cos cos sin2 4 62x x x x dx

= − + − +1

3

2

5

1

73 5 7cos cos cosx x x C

10. cos sin cos sin sin3 2 2 21x x dx x x x dx∫ ∫= −( )

= −∫ ( )sin sin cos2 4x x x dx

= − +1

3

1

53 5sin sinx x C

11. cos cos2 1

21 2x dx x dx= +∫∫ ( )

= + +1

2

1

42x x Csin

12. sin cos2 1

21 2x dx x dx= −∫∫ ( )

= − +1

2

1

42x x Csin

13. sin cos2 51

21 10x dx x dx= −∫∫ ( )

= − +1

2

1

2010x x Csin

14. cos cos2 61

21 12x dx x dx= +∫∫ ( )

= + +1

2

1

2412x x Csin

15. sec tan sec4 2 21x dx x x dx= +∫∫ ( )

= + +1

33tan tanx x C

16. csc cot csc6 2 21x dx x x dx= +∫ ∫ ( )2

= + +∫ ( )cot cot csc4 2 22 1x x x dx

= − − − +1

5

2

35 3cot cot cotx x x C

17. csc cot csc8 2 26 6 1 6x dx x x dx= +∫ ∫ ( )3

= + + +∫ ( )cot cot cot csc6 4 2 26 3 6 3 6 1 6x x x x dx

= − − −1

426

1

106

1

667 5 3cot cot cotx x x

− +1

66cot x C

18. sec tan sec4 2 2100 100 1 100x dx x x dx∫ ∫= +( )

= + +1

300100

1

1001003tan tanx x C

19. tan sec tan10 2 111

11x x dx x C= +∫

20. cot csc cot8 2 91

9x x dx x C= − +∫

21. sec tan sec sec tan10 9x x dx x x x dx∫ ∫= ( )

= +1

1010sec x C

22. csc cot csc csc cot8 7x x dx x x x dx∫ ∫= ( )

= − +1

88csc x C

23. sec sec (sec )10 10 1020 20 20dx dx x C= = +∫ ∫24. csc csc csc8 8 812 12 12dx dx x C= = +∫ ∫ ( )

25. ( )cos sin cos sin2 2 21

22x x dx x dx x C− = = +∫ ∫


26. ( )cos sin2 2x x dx dx x C+ = = + ∫ ∫27. ( ) 2sin csc cotx dx x dx x C− = = − +∫ ∫ 2

28. ( ) 2cos sec tan3 31

332x dx x dx x C−∫ ∫= = +

29. sec3 x dx∫= + + +1

2

1


30. csc3 x dx∫= − − + +1

2

1

2csc cot ln csc cotx x x x C| |

31. a. cos sin5 3x x dx∫

u dvsin 3x cos 5x

3 cos 3x15 sin 5x

–9 sin 3x – 125 cos 5x

+

–

+

= +1

55 3

3

255 3sin sin cos cosx x x x

+ ∫9

255 3cos sinx x dx

16

255 3cos sinx x dx∫

= + +1

55 3

3

255 3 1sin sin cos cosx x x x C

cos sin5 3x x dx∫= + +5

165 3

3

165 3sin sin cos cosx x x x C

b. cos sin5 30

2

x x dxπ

∫= + =5

165 3

3

163 5 0

0

2

sin sin cos cosx x x xπ

Because the integral finds the area aboveminus the area below, this calculation showsthe two areas are equal.

32. a.y

πx

1

b. A x dx x x= = − =∫ sin cos cos3 3

0 0

1

3

4

3

π π

c. Numerically: A ≈ 1.3333… (Checks.)

d. A = 0 because sin3 x is an odd function[sin3 (−x) = −sin3 x] and the integral of an

odd function between symmetrical limits isequal to zero.

33. dV = π y2 dx = π sin2 x dx

V x dx x dx= =∫ ∫π ππ πsin ( – cos )2

0 021 2

= − =π π ππ

2 42 2

0

2x xsin /

34. a. y = sec2 x

dV = π[(y + 3)2 − 32] dx

= π (sec4 x + 6 sec2 x) dx

V x x dx

x x x dx

x x x dx

x x

= +

= + +

= +

= +

= + =

∫∫∫

π

π

π

π π

π π

(sec sec )

tan sec sec

(tan sec sec )

tan tan

tan tan ( . )

4 2

2 2 2

2 2 2

0

1

3

0

1

3

6

1 6

7

37

31 7 1 38 2049

0

1

0

1

[( ) ]

K

b. dV = 2π (x + 3) y dx = 2π (x + 3)(sec2 x) dx

V x x dx x dx= +∫ ∫2 60

12 2

0

1

π πsec sec

= − +∫2 2 60

1

0

1

0

1

π π πx x x dx xtan tan tan

= +8 1 20

1π πtan ln cos| |x

= 8π tan 1 + 2π ln (cos 1) (= 35.2738…)

35. dA r d d= = +1

2

1

25 42 2θ θ θ( cos )

A d= + +∫1

216 40 252

0

4

( cos cos )/

θ θ θπ

= + + +4 2 2 2025

2 0

4

θ θ θ θπ

sin sin/

= + + + = …π π2 10 225

829 1012. ,

which agrees with the numerical answer.

36. dA r d a d= = +1

2

1

212 2θ θ θ( )2cos

A a d

a

= (1+ 2 +

= + + +

∫1

2

1

22

1

2

1

42

2 2

0

2

2

0

2

cos cos

sin sin

θ θ θ

θ θ θ θ

π

π

)

= 3

22πa , which is 1.5 times Acircle.



Problem Set 9-6

Q1.1

33sin x C+ Q2. − +1

44cos x C

Q3. − +1

55ln cos| |x C Q4.

1

66ln sin| |x C+

Q5.1

77 7ln sec tan| |x x C+ +

Q6. 5 sec2 5x

Q7. y′ = 4 cos 4x

Q8. d

Q9. See the text for the statement of fundamentaltheorem of calculus.

Q10. See the text for the definition of indefiniteintegral, Section 5-3.

Note: A radical without a sign in front of it means thepositive root. Because trigonometric functions can bepositive or negative, the radical should technically bereplaced by the absolute value of the appropriatetrigonometric function. Fortunately, this turns out to beunnecessary. If x has been replaced by a sin θ, a tan θ,or a sec θ, it is assumed that θ is the correspondinginverse trigonometric function. So θ is restricted to therange of that inverse trigonometric function. Thus,respectively,

a x a2 2– cos= ∈| |, and θ θ Quadrant I or IV

a x a2 2+ = ∈| |, and sec θ θ Quadrant I or IV

x a a2 2– tan= ∈| |, and θ θ Quadrant I or II

For the first two, the absolute value is unnecessarybecause cos θ ≥ 0 and sec θ ≥ 0 in the respectivequadrants. For the secant substitution, if x is negative,then θ is in Quadrant II, where tan θ < 0. Thus, theradical equals the opposite of a tan θ, and one shouldwrite

x a a2 2– tan= − θWhere the integral of sec θ occurs, one gets

ln x x a x+ +2 2 0for >

− − + <ln x x a x2 2 0for

The second form can be transformed into the first bytaking advantage of the property −ln n = ln (1/n). Thus,

− − =ln – ln– –

x a xx a x

2 2

2 2

1

= +ln –x a x2 2

which can be shown by rationalizing the denominatorof the fraction and incorporating the constant ln a2

(or 2 ln a) into the constant of integration. Becausethe major focus of this section is on the correct

substitution to use and the ensuing calculus, andbecause algebraic techniques are of less importance nowthat technology is used for evaluating integrals, thestudent is not expected to carry along the absolute valuejust to eliminate it later.

1. 49 2– x dx∫

u

v

θ

√ 49 – x2

7x

Let . , ,x

x dx d7

7 7= = =sin sin cosθ θ θ θ

49 77

2 1– cos sinxx= = −θ θ,

∴ = ∫ ∫49 7 72– cos cosx dx dθ θ θ( )

= = +∫ ∫4949

21 22cos cosθ θ θ θd d( )

= + +49

2

49

42θ θsin C

= + +49

2

49

2θ θ θsin cos C

= + ⋅ ⋅ +−49

2 7

49

2

1

7

1

7491 2sin –

xx x C

= + − +−49

2 7

1

2491 2sin

xx x C

2. 100 2– x dx∫

u

v

θ

√ 100 – x2

10x

Let . , ,x

x dx d10


100 1010


∴ = ∫ ∫100 10 102– cos cosx dx dθ θ θ( )

= = + )∫ ∫100100

21 22cos cosθ θ θ θd d(

= 50θ + 25 sin 2θ + C= 50θ + 50 sin θ cos θ + C

= + ⋅ ⋅ +−5010

501

10

1

101001 2sin –

xx x C

= + +−5010

1

21001 2sin –

xx x C


3. x dx2 16+∫

u

v

θ

4

x

√x2 + 16

Let . , ,x

x dx d4

4 4 2= = =tan tan secθ θ θ θ

xx2 116 44

+ = = −sec tanθ θ,

∴ + =∫ ∫x dx d2 216 4 4sec secθ θ θ( )

= ∫16 3sec θ θd (Compare Problem 21 in

Problem Set 9-4.)

= + + +16

2

16

2 1sec tan ln sec tanθ θ θ θ| | C

= + + + + +1

216 8

16

4 42

2

1x xx x

Cln

= + + + + − +1

216 8 16 8 42 2

1x x x x Cln ln

= + + + + +1

216 8 162 2x x x x Cln

4. 81 2+∫ x dx

u

v

θ

9

x

√81 + x2

Let . , ,x

x dx d9


81 99

2 1+ = = −xx

sec tanθ θ,

∴ + =∫ ∫81 9 92 2x dx dsec secθ θ θ( )

= ∫81 3sec θ θd (Compare Problem 21 in

Problem Set 9-4.)

= + + +81

2

81


= + + + + +1

281

81

2

81

9 92

2

1x xx x

Cln

= + + + + − +1

281

81

281

81

292 2

1x x x x Cln ln

= + + + + +1

281

81

2812 2x x x x Cln

5. 9 12x dx–∫

u

v

θ

1

3x √9x2 – 1

Let . ,3

1

1

3

xx= =sec secθ θ

dx d= 1

3sec tanθ θ θ,

9 1 32 1x x– tan sec= = −θ θ,

∴ =

∫∫ 9 1

1

32x dx d– tan sec tanθ θ θ θ

= ∫1

32sec tanθ θ θd

= − )∫1

33(sec secθ θ θd

= + +1

6

1

6sec tan ln sec tanθ θ θ θ| |

− + +1

3ln sec tan| |θ θ C

= − + +1

6

1

6sec tan ln sec tanθ θ θ θ| | C

= − + +1

29 1

1

63 9 12 2x x x x C– ln –

6. 16 12x dx–∫

u

v

θ

1

4x √16x2 – 1

Let . ,4

1

1

4

xx= =sec secθ θ

dx d= 1

4sec tanθ θ θ,

16 1 42 1x x– tan sec= = −θ θ,

∴ =

∫ ∫16 1

1

42x dx d– tan sec tanθ θ θ θ

= ∫1

42sec tanθ θ θd

= −∫1

43( )sec secθ θ θd

= + +1

8

1


− + +1

4ln sec tan| | θ θ C


= − + +1

8

1


= − + +1

216 1

1

84 16 12 2x x x x C– ln –

7.dx

x17 2–∫

u

v

θ

x√17

√17 – x 2

Let / .x 17 = sin θx dx d= =17 17sin cosθ θ θ, ,

17 1717


∴ =∫ ∫dx

x

d

17

17

172–

cos

cos

θ θθ

= = + = +−∫ d Cx

Cθ θ sin 1

17

8.dx

x13 2–∫

u

v

θ

x√ 13

√13 – x 2

Let / . ,x x13 13= =sin sinθ θdx d= ,13 cos θ θ

13 1313


∴ =∫ ∫dx

x

d

13

13

132–

cos

cos

θ θθ

= = + = +−∫ d Cx

Cθ θ sin 1

13

9.dx

x2 1+∫

u

v

θ

x

1

√ x2 + 1

Let . ,x

dx d1

2= =tan secθ θ θ

x x2 11+ = = −sec tanθ θ,

∴+

= =∫ ∫ ∫dx

x

dd

2

2

1

sec

secsec

θ θθ

θ θ

= + + = + + +ln sec tan ln| | θ θ C x x C2 1

10.dx

x2 121–∫

u

v

θ

x

11

√x 2 – 121

Let . ,x

x11

11= =sec secθ θ

dx d= 11sec tanθ θ θ,

xx2 1121 11

11– tan sec= , = −θ θ

∴ = ∫∫ dx

x

d2 121

11

11–

sec tan

tan

θ θ θθ

= = + +∫ sec ln sec tanθ θ θ θ | |d C1

= + +ln–x x

C11

121

11

2

1

= + − +ln – lnx x C21121 11

= + +ln –x x C2 121

11. x x dx2 2 9–∫

u

v

θ

x

3

√x 2 – 9

Let . ,x

x3

3= =sec secθ θ

dx = 3 sec θ tan θ dθ,

xx2 19 33

– tan sec= = −θ θ,

∴∫ x x dx2 2 9–

= ∫ ( ) ( ) ( ) 9 3 32sec tan sec tanθ θ θ θ θd

= ∫81 3 2sec tanθ θ θd

= −

∫ ∫81 5 3sec secθ θ θ θd d

= + −

∫ ∫81

1

4

3

43 3 3sec tan sec secθ θ θ θ θ θd d

= − ∫81

4

81

4sec tan sec3 3θ θ θ θd


= −81

4

81

83sec tan sec tanθ θ θ θ

− + +81

8 1ln sec tan| |θ θ C

= ⋅ ⋅ − ⋅ ⋅81

4 27

9

3

81

8 3

9

3

3 2 2x x x x– –

− + +81

8 3

9

3

2

1ln–x x

C

= −1

49

9

893 2 2x x x x– –

− + + +81

89 32

1ln – lnx x C81

8

= −1

49

9

893 2 2x x x x– –

− + +81

892ln –x x C

12. x x dx2 29 –∫

u

v

θ

x3

√9 – x2

Let . , ,x

x dx d3


9 33


∴∫ x x dx2 29 –

= ∫ ( sin ) cos cos9 3 32 θ θ θ θ( ) ( )d

= ∫81 2 2sin cosθ θ θd

= −∫81 2 4( )cos cosθ θ θd

= −∫8181

42 3cos cos sinθ θ θ θd

− ⋅ ∫3 81

42cos θ θd

= + −∫81

81 2

81

43( )cos cos sinθ θ θ θd

= + − +81

8

81

162

81

43θ θ θ θ sin cos sin C

= + − +81

8

81

8

81

43θ θ θ θ θsin cos cos sin C

= + − +81

8

81

81 2 2θ θ θ θ ( )sin cos cos C

= −81

8 31sin

x

− ⋅ ⋅ ⋅

+81

8 3

9

31

2 9

9

2 2x x xC

––

( – )

= − − +−81

8 3

1

82 9 91 2 2sin –

xx x x C( )

13. ( ) /1 2 3 2−∫ x dx

u

v

θ

x1

√1 – x2

Let x = sin θ. dx = cos θ dθ,

1 2 1– cos sinx x= = −θ θ,

∴ − = ∫∫ ( ) ( )/1 2 3 2 3x dx dcos cosθ θ θ

= ∫ cos4 θ θd

= + ∫1

4

3

43 2cos sin cosθ θ θ θd

= + +∫1

4

3

81 23cos sin cosθ θ θ θ( ) d

= + + +1

4

3

8

3

1623cos sin sinθ θ θ θ C

= + + +1

4

3

8

3

83cos sin sin cosθ θ θ θ θ C

= − + + +−1

41

3

8

3

812 3 2 1 2x x x x x C( ) / sin –

14. ( ) /x dx2 3 281− −∫

u

v

θ

9

x√x 2 – 81

Let . , ,x

x dx d9

9 9= = =sec sec sec tanθ θ θ θ θ

x x2 181 91

9– tan sec= = −θ θ,

∴ − −∫ ( ) /x dx2 3 281

= ∫ −( tan ) sec tan9 93θ θ θ θ( )d

= ∫1

81 2

sec

tan

θ θθd

= ∫1

81cot cscθ θ θd

= − +1

81csc θ C

= +–

–

x

xC

81 812


15.dx

x81 2+∫

u

v

θ

9

x

√81 + x 2

Let . , ,x

x dx d9


81 99

2 1+ = = −xx

sec tanθ θ,

∴+

= = = +∫ ∫∫dx

x

dd C

81

9

81

1

9

1

92

2

2

sec

sec

θ θθ

θ θ

= +−1

9 91tan

xC

16.dx

x25 12 +∫

u

v

θ

1

5x

√25x2 + 1

Let . , ,5

1

1

5

1

52x

x dx d= = =tan tan secθ θ θ θ

25 1 52 1x x+ = = −sec tanθ θ,

∴+

=⋅

= = +∫∫∫ dx

x

dd C

25 1 5

1

5

1

52

2

2

sec

sec

θ θθ

θ θ

= +−1

551tan x C

17. a.x dx

x2 25+∫

u

v

θ

5

x√x 2 + 25

Let . , ,x

x dx d5


x x2 125 51

5+ = = −sec tanθ θ,

∴+

=⋅∫∫ x dx

x

d2

2

25

5 5

5

tan ( sec )

sec

θ θ θθ

= = +∫5 5tan sec secθ θ θ θd C

= + +x C2 25

b.x dx

xx x dx

2

2 1 2

25

1

225 2

+= + − /∫∫ ( ) ( )

= + +x C2 25 , which agrees with part a.Moral: Always check for an easy way tointegrate before trying a more sophisticatedtechnique!

18. a.x dx

x2 49–∫

u

v

θ

7

x√x 2 – 49

Let .x

7= sec θ x = 7 sec θ,


xx2 149 77


∴ = ∫∫ x dx

x

d2 49

7 7

7–

sec ( sec tan )

tan

θ θ θ θθ

= = + = +∫7 7 492 2sec tan –θ θ θd C x C

b.x dx

xx x dx

2

2 12

49

1

249 2

–= − − /∫∫ ( ) ( )

= +x C2 49– ,which agrees with part a.

19.dx

x9 5 2– ( – )∫

u

v

θ

3 x – 5

√ 9 – (x – 5)2

Let . , ,x

x dx d–

sin sin cos5

35 3 3= = + =θ θ θ θ

9 5 35

32 1– ( – ) cos sin

–x

x= = −θ θ,

∴ = ∫∫ dx

x

d

9 5

3

32– ( – )

cos

cos

θ θθ

= = + = +−∫ d Cx

Cθ θ sin–1 5

3

20.dx

x36 2 2– ( )+∫

u

v

θ

6 x + 2

√ 36 – (x + 2)2


Let . , ,x

x dx d+ = = − =2

66 2 6sin sin cosθ θ θ θ

36 2 62

62 1– ( ) cos sinx

x+ = = +−θ θ,

∴+

= ∫∫ dx

x

d

36 2

6

62– ( )

cos

cos

θ θθ

= = + = + +−∫ d Cx

Cθ θ sin 1 2

6

21.dx

x x

dx

x2 28 20 4 36+=

+∫∫ – ( ) –

u

v

6

x + 4 √ (x + 4)2 –36

Let . ,x

x+ = = −4

66 4sec secθ θ


x x x2 28 20 4 36 6+ = + =– ( ) – tan θ,

θ = +−sec 1 4

6

x

∴+

= ∫∫ dx

x x

d2 8 20

6

6–

sec tan

tan

θ θ θθ

= ∫ sec θ θd

= ln | sec θ + tan θ | + C1

= + + + +ln–x x x

C4

6

8 20

6

2

1

= + + + − +ln – lnx x x C4 8 20 621

= + + + +ln –x x x C4 8 202

22.dx

x x

dx

x2 214 50 7 1– ( – )+=

+∫∫

u

v

θ

1

x – 7

√(x – 7)2 + 1

Let . , ,x

x dx d–

tan tan sec7

17 2= = + =θ θ θ θ

x x x2 214 50 7 1– ( – ) sec+ = + = θ,

θ = tan− 1 (x − 7)

∴+

= = ∫∫∫ dx

x x

dd

2

2

14 50–

sec

secsec

θ θθ

θ θ

= ln | sec θ + tan θ | + C

= + + − +ln –x x x C2 14 50 7

23. 100 2

3

8

– x dx−∫

u

v

θ

10 x

√100 – x 2

Let . , ,x

x dx d10


100 1010


∴−∫ 100 2

3

8

– x dx

= ⋅∫ 10 101

1

0 3

0 8

cos cossin (– . )

sin .

–

–

θ θ θd

= ∫100 2

0 3

0 8

1

1

cossin (– . )

sin .

–

–

θ θd

= +∫50 ( cos )sin (– . )

sin .

–

–

1 21

1

0 3

0 8

θ θd

= +50 25 sin 2θ θsin (– . )

sin .–

–

1

1

0 3

0 8

= +− −50 sin 0.8 25 sin (2 sin 0.8)11

− 50 sin− 1 (−0.3) − 25 sin [2 sin− 1 (−0.3)]= 99.9084…Numerical integration: 99.9084… (Checks.)

24. x dx2

1

4

25+∫–

u

v

θ

5

x√x 2 + 25

Let . , ,x

x dx d5


x x2 125 5 0 2+ = = −sec tanθ θ, .

∴ +∫ x dx2

1

4

25–

= ⋅∫ 5 5 2

0 2

0 8

1

1

sec sectan (– . )

tan .

–

–

θ θ θd

= ∫25 3

0 2

0 8

1

1

sectan (– . )

tan .

–

–

θ θd

= + +−

−

−

25

2

25

2 1

1

0 2

0 8sec tan

tan ( . )

tan .θ θ θ θln |sec tan |


= ⋅−25

20 8 0 81sec tan( . ) .

+ +−25

20 8 0 81ln sec tan| ( . ) . |

− − ⋅ −−25

20 2 0 21sec [tan ]( . ) ( . )

− − −−25

20 2 0 21ln sec [tan ]| ( . ) . |

= 26.9977…Numerical integration: 26.9977… (Checks.)

25. y = 3x2

dL y dx x dx= + ′ = +1 1 362 2( )

L x dx= +∫ 1 36 2

0

5

u

v

θ

1

6x√ 1 + 36x2

Let . , ,6

1

1

6

1

62x

x dx d= = =tan tan secθ θ θ θ

1 36 62 1+ = = −x xsec tanθ θ,

∴ L dx

x

= ⋅=

=

∫ sec secθ θ θ1

62

0

5

==

=

∫1

63

0

5

sec θ θdx

x

= + +=

=1

12

1

12 0

5

sec tan ln |sec tan |θ θ θ θx

x

= + + + +1

21 36

1

121 36 62 2

0

5

x x x xln

= + + =5

2901

1

12901 30 75 3828ln . K

Numerical integration: L = 75.3828… (Checks.)

26. a. 9 25 2253

5252 2 2x y y x+ = ⇒ = ± –

Slice the region vertically. Pick a samplepoint (x, y) on the positive branch of thegraph, within the strip.

dA y dx x dx= =26

525 2–

A x dx= ∫6

525 2

3

4

––

u

v

θ

5x

√ 25 – x2

Let , ,x

x dx d5

5 5= . = =sin sin cosθ θ θ θ

25 55


∴ A dx

x

= ⋅=

=

∫6

55 5

3

4

cos cos–

θ θ θ

==

=

∫30 2

3

4

cos–

θ θdx

x

= +=

=

∫15 1 23

4

( cos )–

θ θdx

x

= +=−

=15

15

22

3

4

θ θsinx

x

= + ==15 15 3

4θ θ θsin cos –xx

= +−

=−

=15

5

3

5251 2

3

4sin –

xx x

x

x

= +−15 0 83

54 91sin ( ).

− − −−15 0 63

53 161sin (– )( . )

= − − +− −15 0 8 0 6 14 41 1[sin sin ]. ( . ) .

= + = …15

214 4 37 9619

π. .

Numerical integration: A = 37.9619…(Checks.)

b. A x dx= ∫6

525 2

5

5

––

x = 5 ⇒ θ = π/2, x = −5 ⇒ θ = −π/2

∴ A d= ∫30 2

2

2

cos– /

/

θ θπ

π

= +−

1515

22

2

2

θ θπ

πsin

/

/

= + + − −15

2

15

2

15

2

15

2

π π π πsin sin ( )

= 15π = 47.1238…The area is π (x-radius)(y-radius).

27. x y r y r x x y r2 2 2 2 2 0+ = ⇒ = ± = = ±– , at

Slice the region inside the circle perpendicular tothe x-axis. Pick sample point (x, y) on thepositive branch of the circle, within the strip.

dA y dx r x dx= = −2 2 2 2

A r x dxr

r

= ∫2 2 2––

u

v

θ

rx

√r 2 – x 2

Let x

r= sin θ. x = r sin θ, dx = r cos θ dθ,

r x rx

r2 2 1– cos sin= = −θ θ,

x = r ⇒ θ = π/2, x = −r ⇒ θ = −π/2

∴ = ⋅∫A r r d22

2

cos cos– /

/

θ θ θπ

π

=− /

/

∫2 2 2

2

2

r dcos θ θπ

π


= +− /

/

∫r d2

2

2

1 2( cos )θ θπ

π

= +−

r r2 2

2

21

22θ θ

π

πsin

/

/

= ⋅ + + ⋅r r r2 2 2

2

1

2 2

π π πsin

− − =1

22 2r rsin ( )π π

∴ A = πr2, Q.E.D.

28.x

a

y

by

b

aa x

+

= ⇒ = ±

2 22 21 –

Slice the region inside the ellipse perpendicularto the x-axis. Pick sample point (x, y) on thepositive branch of the ellipse, within the strip.

dA y dxb

aa x dx= =2

2 2 2–

Ab

aa x dx

a

a

= ∫2 2 2––

u

v

θ

ax

√a 2 – x 2

Let x

a= sin θ. x = a sin θ, dx = a cos θ dθ,

a x ax

a2 2 1– cos sin= = −θ θ,

x = a ⇒ θ = π/2, x = −a ⇒ θ = −π/2

∴ = ⋅∫Ab

aa a d

22

2

cos cos– /

/

θ θ θπ

π

= ∫2 2

2

2

ab dcos– /

/

θ θπ

π

= +∫ab d( cos )– /

/

1 22

2

θ θπ

π

= +−

ababθ θ

π

π

22

2

2

sin/

/

= + + − − =ab ab ab abab

π π π π π2 2 2 2

sin sin ( )

∴ A = πab

Note that if a = b = r, then πab = πr2, the area ofa circle.

29. dV x dya

bb y dy b y b= = − ≤ ≤π π2

2

22 2( – ) ,

Va

bb y dy

b

b

= ⋅ ∫π2

22 2( – )

–

= ⋅

=−

π πa

bb y

ya b

b

b2

22

32

3

4

3–

∴ =V a b4

32π

Rotating instead about the x-axis is equivalent tointerchanging the a and b, giving V ab= 4

32π .

30. x y y x2 2 29 9− = ⇒ = ± –

Slice the region perpendicular to the x-axis. Picka sample point (x, y) on the positive branch ofthe hyperbola, within the strip.

dA y dx x dx= =2 2 92 –

A x dx= ∫2 92

3

5

–

u

v

θ

3

x√x 2 – 9

Let x

3= secθ. x = 3 sec θ, dx = 3 sec θ tan θ dθ,

xx2 19 33


∴ = ⋅=

=

∫A dx

x

2 3 33

5

tan sec tanθ θ θ θ

==

=

∫18 2

3

5

tan secθ θ θdx

x

==

=

∫18 3

3

5

(sec – sec )θ θ θdx

x

= 9 sec θ tan θ + 9 ln | sec θ + tan θ |− + =

= |18

3

5ln sec tan |θ θ

x

x

= − + ==9 9 3

5sec tan ln sec tanθ θ θ θ| | xx

= − +x x x x 2 2

3

5

9 91

3

1

39– ln –

= 20 − 9 ln 3 = 10.1124…Numerical integration: A = 10.1124… (Checks.)

31. dV x y dx x x dx= =2 2 4 92π π( ) –

V x x dx= ∫4 92

3

5

π –

= ∫2 9 22

3

5

π x x dx– ( )

= ⋅ /2 3 2π 2

392

3

5

( – )x

= ⋅ = = …4

364

256

3268 0825π π .

32. From Problems 30 and 31, A = 20 − 9 ln 3,

V = 256

3π .

V x A x= ⋅ ⇒ = =2

128

3 20 9 34 2192π

( – ln ). K

x is a little more than halfway through theregion.


33. x = a cos t ⇒ dx = −a sin t dty = b sin tdA = 2y dx = 2(b sin t)(−a sin t dt)

= −2ab sin2 t dtx = −a ⇒ t = π, x = a ⇒ t = 0

∴ = − = − ∫∫A ab t dt ab t dt2 1 2200

sin ( – cos )ππ

= − + = + + − =abtab

t ab ab2

2 0 0 00

sinπ

π π( )

∴ A = πab, as in Problem 28.

With this method, you get sin2 t dt,∫ directly.

With trigonometric substitution in Problem 28,

you get cos2 t dt,∫ indirectly.

34. r = 0.5θ ⇒ dr/dθ = 0.5

dL r dr d d d= + = +2 2 20 25 0 25( / ) . .θ θ θ θ

= +0 5 12. θ θd

L d= +∫0 5 12

0

6

. θ θπ

u

v

θ

1

φ

√ θ2 + 1

Let θ = tan φ ⇒ dθ = sec2 φ dφ.

θ φ φ θ2 11+ = = −sec tan,

∴ = ⋅=

=

∫L d0 5 2

0

6

. sec secφ φ φθ

θ π

==

=

∫0 5 3

0

6

. sec φ φθ

θ πd

= +0 25 0 25. . |sec tan ln secφ φ φ+ =

=tan φ θ

θ π|

0

6

= + + + +0 25 1 0 25 12 2

0

6

. .θ θ θ θπ

ln

= + + + +1 5 36 1 0 25 36 1 62 2. .π π π πln

= 89.8589… , same as numericalintegration.

35. See the note preceding the solutions for thissection. For the sine and tangent substitution,the range of the inverse sine and inverse tangentmake the corresponding radical positive. For thesecant substitution, the situation is morecomplicated but still gives an answer of the samealgebraic form as if x had been only positive.

Problem Set 9-7Q1. (x + 5)(x − 5) Q2. x2 + 2x − 15

Q3. (x + 2)(x − 6) Q4. x2 + 14x + 49

Q5. (x + 4)2 Q6. x2 − 64

Q7. e x

Q8. b2 − 4ac = −1500, so x2 + 50x + 1000 is prime.

Q9. b2 − 4ac = −144, so x2 + 36 is prime.

Q10. B

1.11 15

3 2

4

1

7

22

x

x xdx

x xdx

–

– – –+= +

∫∫

= 4 ln |x − 1| + 7 ln |x − 2| + C

2.7 25

7 8

2

1

9

82

x

x xdx

x xdx

+ =+

+

∫∫ – –

–

–= −2 ln |x + 1| + 9 ln |x − 8| + C

3.( – )

– –

/ /

–

5 11

2 8

7 2

2

3 2

42

x dx

x x x xdx=

++

∫∫

= + + − +7

22

3

24ln ln| | | |x x C

4.( – )

– –

/ /

–

3 12

5 50

9 5

5

6 5

102

x dx

x x x xdx=

++

∫∫

= + + − +9

55

6

510ln ln| | | |x x C

5.21

7 10

7

5

7

22

dx

x x x xdx

+ +=

++

+

∫∫ –

= −7 ln |x + 5| + 7 ln |x + 2| + C

6.10

9 36

2

3

8

122

x dx

x x x xdx

– – –=

++

∫∫

= 2 ln |x + 3| + 8 ln |x − 12| + C

7.9 25 50

1 7 2

2x x

x x xdx

– –

( )( – )( )+ +∫=

++ +

+

∫ 2

1

3

7

4

2x x xdx

–= 2 ln |x + 1| + 3 ln |x − 7| + 4 ln |x + 2| + C

8.7 22 54

2 4 1

2x x

x x xdx

++∫ –

( – )( )( – )

= ++

+

∫ 3

2

1

4

5

1x x xdx

–

–

–= 3 ln |x − 2| − ln |x + 4| + 5 ln |x − 1| + C

9.4 15 1

2 5 6

2

3 2

x x

x x xdx

++∫ –

– –

=+

++

+

∫ –

–

1

3

2

1

3

2x x xdx

= −ln |x + 3| + 2 ln |x + 1| + 3 ln |x − 2| + C

10.– –

– –

3 22 31

8 19 12

2

3 2

x x

x x xdx

++∫

= + +

∫ –

–

–

– –

2

1

4

3

3

4x x xdx

= −2 ln |x − 1| − 4 ln |x − 3| + 3 ln |x − 4| + C

11.3 2 12 9

1

3 2x x x

xdx

+ +∫ –

–

= + − +

∫ 3 5 7

2

12x x

xdx

–

= + − + − +x x x x C3 25

27 2 1ln | |


12.x x x

x xdx

3 2

2

7 5 40

2 8

–

– –

+ +∫= − +

∫ x

x

x xdx5

3

2 82 – –

= − ++

+

∫ x

x xdx5

1

2

2

4–

= − + + + − +1

25 2 2 42x x x x Cln ln| | | |

13.4 6 11

1 4

2

2

x x

x xdx

+ ++ +∫ ( )( )

= ++

++

∫ x

x xdx

2

1

3

42

=+

++

++∫∫∫1

2

2

12

1

3

42 2

x dx

x

dx

x

dx

x

= + + + + +−1

21 2 3 42 1ln tan ln| | | |x x x C

14.4 15 1

5 3 1

2

3 2

x x

x x xdx

– –

– + +∫= +

∫ 3

1

2

4 12x

x

x xdx

–

–

– –

= − + − − +3 11

24 12ln ln| | | |x x x C

Note thatx

x x x x

–

– –

/

–

/

– –,

2

4 1

1 2

2 5

1 2

2 52 =+

+

but1

22 5

1

22 5ln lnx x− + + − −

= − −1

24 12ln | |,x x so the answer comes out

the same.

15.4 18 6

5 1

2

2

x x

x xdx

+ ++ +∫ ( )( )

=+

++

++

∫ 1

5

3

1

2

1 2x x xdx

–

( )

= ln |x + 5| + 3 ln |x + 1| + 2(x + 1)− 1 + C

16.

= + +

∫ 5 2

7

3

7 2x x xdx

–

– ( – )

= 5 ln |x| − 2 ln |x − 7| − 3(x − 7)− 1 + C

17.dx

x x x

dx

x3 2 36 12 8 2– – ( – )+=∫ ∫

= − +−1

22 2( – )x C

18.1

4 6 4 1 14 3 2 4x x x xdx

dx

x+ + + +=

+∫ ∫ ( )

= − + +−1

31 3( )x C

19. a.dy

dty

y dy

y ydt= ⇒ =2

1000

1000

1000

10002

–

( – )

1000

10002

dy

y ydt

( – )=∫ ∫

1 1

10002

y ydy dt+

=∫ ∫–

ln |y| − ln |1000 − y| = 2t + C

ln–

y

yt C

10002= +

y

ye t C

10002

–= + (Note that 0 ≤ y < 1000.)

1000 100012 2− = ⇒ − =− − − −y

ye

yet C t C

10001 12 2

ye ke k et C t C= + = + =− − − − ( )

yke t=

+1000

1 2–

Initial condition y = 10 when t = 0 ⇒ k = 99.

ye t=

+1000

1 99 2–

b. ye

( ) . students11000

1 9969 4531 692=

+= … ≈–

have heard the rumor after one hour.

ye

( ) .41000

1 99967 8567 9688=

+= … ≈–

students have heard by lunchtime.

ye

( ) .–81000

1 99999 9888 100016=

+= … ≈

students—everyone knows by the end ofthe day!

c. It is quicker to analyze the original differentialequation, which already refers to the derivative,than to analyze the equation found in part a.

Maximize y yy′ = 2

1000

1000

–

= 1

5001000 2( – )y y .

′′ = ′ ′ = =y y yy y1

5001000 2 0 500( – ) when .

This is the maximum point because y″ > 0for y < 500 and y″ < 0 for y > 500 (andy′ > 0 for all t).So the rate of spreading (y′) is greatest when500 students have heard the news. This occurswhen

5001000

1 99 2=+ e t–

99 1 22e t− + =

e t− =2 1

99

t = = …1

299 2 2975ln . hr

3 53 245

14 49

2

3 2

x – x

x – x xdx

++∫


d. The graph follows the slope-field pattern.

t

y

2 4 6

1000

20. a. Assume that an infected person and anuninfected person have about the same chanceof meeting any other infected person (i.e.,infected people are not quarantined). Aninfected person can meet N − P uninfectedpeople out of the total population, so thechance of meeting an uninfected person will be(N – 1)/N, so of an average infected person’sthree contacts per day, 3(N − 1)/N of themwill be with uninfected persons. (Actually(N − P)/(N − 1) because the total populationthat someone can meet is N − 1—people don’tmeet themselves outside the Twilight Zone—but (N − P)/N is reasonably close enough fornow.) So there are P infected people, eachmeeting an average of 3(N − P)/N uninfectedpeople per day, for a grand total of 3P(N − P)/Ncontacts between infected and uninfectedpeople per day.

b. If 10% of the contacts with infected peopleper day result in infection, then the numberof new infections per day should be 0.1 timesthe number of contacts between infected anduninfected people, that is,dP

dt

P N P

NP

N P

N= ⋅ =0 1

30 3. . .

( – ) –

c.dP

dtP

N P

N

N dP

P N Pdt= ⇒ =0 3 0 3. .

–

( – )

N dP

P N Pdt

( – )∫ ∫= 0.3

1 10 3

P N PdP dt+

=∫ ∫–

.

ln |P| − ln |N − P| = 0.3t + C

ln–

P

N Pt C= +0 3.

P

N Pe t C

–= +0.3 (Note that 0 ≤ P < N.)

N

Pe ke k et C t C= + = + =− − − −1 10 3 0 3. . ( )

P tN

ke t( ) =+1 0 3– .

Initial condition P(0) = P0

⇒ =+

⇒ = −PN

kk

N

P001

1.

∴ =+

( )P tN

N P e t1 100 3( / – ) – .

d. N = 1000 and P0 = 10

⇒ =+

P te t( )

1000

1 99 0 3– .

Pe

( )71000

1 99 2 1=+ – . = 76.2010… ≈ 76 people

infected after 1 week

e. Solve P(t) = 990.

9901000

1 99 0 3=+ e t– .

1 99100

990 3+ =−e t.

e0.3 t = 992

t = = ≈2

0 399 30 6341 31

.ln . daysK

21. Ax x

dxx x

dxb b

=+

= ++

∫ ∫25

3 4

5

1

5

422 2– –

–

= − − + =+

5 1 5 4 51

422

ln ln ln–

| | | |x xx

xb

b

=+

− =+

+51

45

1

65

1

45 6ln

–ln ln

–ln

b

b

b

b

A( ) .7 56

115 6 5 9281= + = …ln ln

lim lim ln lnb b

A b→∞ →∞

= +( ) 51

15 6

(l’Hospital’s rule)

= 5 ln 6 = 8.9587…So the area does approach a finite limit.

22. dV x y dxx

x xdx= =

+2

50

3 42π π–

Vx

x xdx

x xdx

b b

=+

= ++

∫ ∫50

3 4

10

1

40

422

π π π– –2

= − + +10 1 40 4 2π πln ln| | | |x x b

= 10π ln |b − 1| + 40π ln |b + 4| − 40π ln 6V(7) = 40π ln 11 − 30π ln 6 = 132.4590…

limb

V b→∞

= ∞( ) because both ln terms become

infinite and are added.(Note that if the region were rotated about thex-axis, the limit of the volume would be

finite. The answer would be 535

62 6π – ln

= 35.3400… .

23. a.x

x xdx

x xdx

–

–

/

–

/

–

3

6 8

1 2

2

1 2

42 +∫ ∫= +

= + +1

22

1

24ln ln| – | | – |x x C

b. x2 – 6x + 8 = (x – 3)2 – 1

u

v

θ

1

x – 3 √ (x – 3)2 – 1


Let x – 3 = sec θ. dx = sec θ tan θ dθ,

( – ) – tan secx x3 1 32 1= =θ θ, ( – )–

∴+

=∫ ∫x

x xdx

x

xdx

–

–

–

( – ) –

3

6 8

3

3 12 2

= = ∫∫ sec

tan(sec tan )

sec

tan

θθ

θ θ θ θθ

θ2

2

d d

= ln | tan θ | + C

= +ln ( – ) –x C3 12

= + +ln –x x C2 6 8

c.x

x xdx

x

x xdx

–

–

–

–

3

6 8

1

2

2 6

6 82 2+=

+∫ ∫= + +1

26 82ln | – |x x C

d. From part a,1

22

1

24ln ln| – | | – |x x C+ +

= +1

22 4ln |( – )( – )|x x C

= + +1

26 82ln | – |x x C

which is the answer in part c. This equals

ln ln –| – | ,/x x C x x C2 1 2 26 8 6 8+ + = + +which is the answer from part b. So all threeanswers are equivalent, Q.E.D.

24. a. When the population is very much smallerthan the maximum, (m – p) behaves like aconstant, and dp/dt = k(m – p) · p isapproximately proportional to p. But when pis approaching m, then (m – p) goes to zero,so dp/dt = kp(m – p) goes to zero.

b. dp/dt = kp(m – p) = k(mp – p2). So dp/dt is aquadratic function of p. Thus, the turningpoint is at

pm

m= =– / .2 1

2(– )

If k > 0, the graph of dp/dt versus p opensdownward and the turning point is amaximum.So the population grows fastest whenp = m/2.

c.dp

dtkp m p

dp

p m pk dt= ⇒ =( – )

( – )

dp

p m pk dt

( – )= ∫∫

1 1/ /

–

m

p

m

m pdp k dt+

=∫ ∫

1 11m

pm

m p kt Cln ln| | – | – | = +

d. ln–

p

m pkmt C= + 2 (C2 = mC1)

p

m pe C ekmt C kmt

–= =+ 2

3 ( )C eC3

2=

Note that > > > .m pp

m p0 0⇒

–

m p

pbe

m

pbe b Ckmt kmt– = ⇒ = =– –– ( / )1 1 3

m

pbe p

m

bekmt

kmt= + ⇒ =+

11

––

At time t = 0, p = p0.

∴ =+

⇒ = +pm

bm p b0 01

1( )

∴ = ++

pp b

be kmt0 1

1

( )–

Letting K = km, p pb

be Kt= ++ −01

1, Q.E.D.

e. Let p denote millions of people. Thenp0 = 179.3.Substitute p(10) = 203.2.

203 2 179 31

1 10. .= ++

b

be K–

⇒ 203.2 + 203.2be–10K = 179.3 + 179.3b⇒ b(203.2e–10K – 179.3) = –23.9

⇒ =be K

– .

. – .–

23 9

203 2 179 310

By substituting p(20) = 226.5 andtransforming,

be K= – .

. – ..–

47 2

226 5 179 320

Equating the two values of b and solvingnumerically for K gives K = 0.0259109… .

∴ = = …be

– .

. – .– . ...

23 9

203 2 179 31 06304360 259109 .

∴ =+

pe

179 32 063036

1 1 063036 0 0259109.. ...

. ... – . ...

Check that this equation gives a goodapproximation for 1990.

pe

( ) .30 179 32 0630

1 1 0630 30 0 0259= ⋅+ ⋅ ⋅

. ...

. ... – . ...

= 248.4892… ≈ 248.5 million people,which is close to the actual population,248.7 million.

f. pe

( ) .40 179 32 0630

1 1 0630 40 0 0259= ⋅+ ⋅ ⋅

. ...

. ... – . ...

= 268.6144… ≈ 268.6 million people, whichis lower than the actual population by about13 million people.

g. k p pb

bep b

t t kt> ( )01

110 0⇒ = +

+= +

→∞ →∞lim lim –

= 179.3 · (1 + 1.0630…)= 369.9024… ≈ 369.9 million people


h. If p(10) had been 204.2, then K would havebeen given by

– .

. – .

– .

. – .– –

24 9

204 2 179 3

47 2

226 5 179 310 20e eK K=

⇒ K = 0.0343965…

⇒ =be

– .

. – .– . ...

24 9

204 2 179 30 0343965

= 0.721075…

So the ultimate population would have beenlimt

p p b→∞

= + = + …0 1 179 3 1 0 7210( ) . ( . )

= 308.5888… ≈ 308.6 million people.An increase of 1 million in one of the initialconditions causes a decrease of over61 million in the predicted maximumpopulation! So this model does have a fairlysensitive dependence on the initial conditions.

Problem Set 9-8Q1. integration by parts Q2. partial fractions

Q3. x = tan θ or θ = tan–1 x

Q4. x = sec θ or θ = sec–1 x

Q5. x = sin θ or θ = sin–1 x

Q6.1

1612 8( )x C+ +

Q7. 7 (at f (1)). Q8. 3 (at f (5)).

Q9. undefined Q10. B

1. tan−∫ 1 x dx u dvtan–1 x 1

dx1 + x2 x–

+

= −+

− ∫x xx dx

xtan 1

2 1

= −+

− ∫x xx dx

xtan 1

2

1

2

2

1

= − + +−x x x Ctan ln | |1 21

21 (Checks.)

2. cot−∫ 1 x dx u dvcot–1 x 1

dx1 + x2 x

+

–

= ++

− ∫x xx dx

xcot 1

2 1

= ++

− ∫x xx dx

xcot 1

2

1

2

2

1

= + + +−x x x Ccot ln | |1 21

21 (Checks.)

3. cos−∫ 1 x dx

√

u dvcos –1 x 1dx

1 – x2x–

+

= +−

− ∫x xx dx

xcos 1

21

= − − −− −∫x x x x dxcos ( ) ( )/1 2 1 21

21 2

= − ⋅ − +−x x x Ccos ( ) /1 2 1 21

22 1

= − − +−x x x Ccos 1 21 (Checks.)

4. sin−∫ 1 x dx

√

u dvsin–1 x 1dx

1 – x2x

+

–

= −−

− ∫x xx dx

xsin 1

21

= + − −− −∫x x x x dxsin ( ) ( )/1 2 1 21

21 2

= + ⋅ − +−x x x Csin ( ) /1 2 1 21

22 1

= + − +−x x x Csin 1 21

5. sec−∫ 1 x dx

√

sec –1 x 1dx

|x| x2 – 1x–

+

u dv

= −−

− ∫x xx dx

x xsec

| |

1

2 1

= −−

=− ∫x x xdx

xx x xsec sgn ( /| | sgn )1

2 1

u

v

θ

1

x √x 2 – 1

Let .x

1= sec θ

dx = sec θ tan θ dθx2 1– = tan θ

θ = sec–1 x

∴ ∫ sec–1 x dx

= ∫x x xd

sec sgnsec tan

tan– –1 θ θ θ

θ= ∫x x x dsec sgn sec– –1 θ θ


= x sec–1 x – sgn x ln |sec θ + tan θ| + C

= x sec–1 x – sgn x ln –x x C+ +2 1

(Checks.)Note: This answer can be transformed to

x x x x Csec ln ( – ) .– – | |1 2 1+ +

6. csc−∫ 1 x dx

√

u dvcsc –1 x 1dx

|x| x2 – 1x

+

–

= + ∫x xx dx

x xcsc

| | –

– 1

2 1

= + ∫x x xdx

xcsc sgn

–

–1

2 1

u

v

θ

1x

√ x2 – 1

Let x = csc θ. dx = –csc θ cot θ dθ,

x x2 11– cot csc= =θ θ, –

∴∫ csc– 1 x dx

= ∫x x xd

csc sgncsc cot

cot– –1 θ θ θ

θ= ∫x x x dcsc sgn csc– –1 θ θ

= x csc–1 x + sgn x ln | csc θ + cot θ | + C

= + + +x x x x x Ccsc sgn ln ––1 2 1

(Checks.)Note: This answer can be transformed to

x x x x Csec ln –− + + +1 2 1(| | ) .

7. tan tan ln− −= − +∫ 1 1 2

1

4

1

41

21x dx x x x| |

= − − +− −4 41

217 1

1

221 1tan ln tan ln

= − − =−4 44

1

2

17

23 44781tan ln

π. K

Numerically, tan− =∫ 1

1

4

3 4478x dx . .K

8.

π /2

1 3

x

y

Simpson’s rule for y = sec− 1 x: n = 10 ⇒∆x = 0.2

A y y y y y y≈ + + + + + +0 2

34 2 4 41 1 2 1 4 1 6 2 8 3

.( ). . . .L

= 1.919692K

A x dx= −∫ sec 1

1

3

= − +−x x x x xsec sgn ln –1 2

1

31( )

= − ⋅ + − + ⋅− −3 3 1 3 8 1 1 11 1sec ln sec ln( )

= 1.930131… .

The Simpson’s rule answer differs from thisby 0.0104… , or about 0.5%.

9. By vertical slices,

A x dx=

∫ π

21

0

1

– sin–

= − −−π2

11 2

0

1

x x x xsin –

= − − − + + =−π2

1 0 0 0 1 11sin

By horizontal slices, A y dy= ∫ sin/

0

2π

= − = − + =cos cos cos/y 02

20 1π π

, which is the

same answer as by vertical slices.

10. By cylindrical shells, dV = 2π x tan− 1 x dx.

V x x dx= −∫2 1

0

1

π tan

x x dxtan−∫ 1

12 x

u dvtan–1 x x

dx1 + x 2 2 –

+

= −+

− ∫1

2

1

2 12 1

2

2x xx

xdxtan

= − −+

− ∫1

2

1

21

1

12 1

2x xx

dxtan

= − + +− −1

2

1

2

1

22 1 1x x x x Ctan tan

∴ = − +− −V x x x xπ π π2 1 10

1tan tan

= − + − + −− −π π πtan tan1 11 1 0 0 0

= − = ⋅ −−2 1 24

1π π π π πtan

= − =1

21 79322π π . K

Compare this with a cylinder (π r2h) minus acone (π r 2h/3), both of radius 1 and altitude π / 4,which has volume 2π ( π / 4)/3 = π2/6 = 1.6449… ;the volume is slightly less than V, which isexpected because the cylinder minus the cone isgenerated by rotating a line that lies below thegraph.


Problem Set 9-9

Q1. xx= = −55

1tan tanθ θ or

Q2. xex − ex + C Q3.1

33tan x C+

Q4. 2x1/ 2 + C Q5. ln |x| + C

Q6. reduction formula Q7. False

Q8. dx dy dr rd2 2 2 2+ +or ( )θ

Q9. (1 − x2)−1/ 2 Q10. C

1.

x

y

cosh

sinh

1

1x

y

tanh

coth

1

1

x

y

sech

csch

1

1

2.

x

y

1

1cosh–1

sinh

–1

x

y

–1coth

tanh–1 1

1

x

y

1

1 csch–1–1sech

–1

3.d

dxx x xtanh tanh sec3 23= h2

4.d

dxx x x5 3 15 3sec sec tanhh h 3= −

5. cosh sinh cosh5 61

6x x dx x C= +∫

6. ( )sinh cosh (sinh )x x dx x C− −= − +∫ 3 21

2

= − +1

2csch2 x C

Or: ( ) h2sinh cosh csc cothx x dx x x dx− =∫ ∫3

= − + = − + +1

2

1

212

1coth (csc )x C x Ch21

= − +1

2csch2 x C

7.d

dxx x(csc sin )h

= −csch x coth x sin x + csch x cos x

8.d

dxx x x x x x(tan tanh ) sec tanh tan sec= +2 2h

9. sec tanhh2 41

44x dx x C= +∫

10. sec tanh sech h7 71

77x x dx x C= − +∫

11.d

dxx x x x x x( coth ) coth csc3 2 3 23= − h

12.d

dxx x( csc ).2 5 4h

= 2.5x1.5csch 4x − 4x2.5 csch 4x coth 4x

13. tanh ln coshx dx x=∫ ( )1

3

1

3

= ln (cosh 3) − ln (cosh 1) = 1.875547…

14. sinh coshx dx x= =− −∫ 4

4

4

4

0

(Note that sinh is an odd function.)

15.d

dx

x

x

sinh

ln

5

3

= 5 5 3 5

3

1

2

cosh ln – sinh

(ln )

–x x x x

x

16.d

dx

x

x

cosh

cos

6

3

= +6 6 3 3 6 3

32

sinh cos cosh sin

cos

x x x x

x

17. x x dxsinh∫ u dvx sinh x1 cosh x0 sinh x

+

–+

= x cosh x − sinh x + C

∴∫ x x dxsinh0

1

= − = −x x xcosh sinh cosh sinh0

11 1

= e− 1 = 0.36787…

18. x x dx2 cosh∫ u dvx 2 cosh x2x sinh x

2 cosh x0 sinh x–

–

+

+

= x2 sinh x – 2x cosh x + 2 sinh x + C

∴∫ x x dxa

b2 cosh

= − +x x x x xa

b2 2 2sinh cosh sinh


= b2 sinh b – 2b cosh b + 2 sinh b– a2 sinh a + 2a cosh a – 2 sinh a

19.d

dxx

x( sinh )–3 4

12

16 1

1

2=

+

20.d

dxx

x

x( tanh )

––5

15

11 3

2

6=

21. tanh−∫ 1 5x dx

= + − +−1

55

1

101 51x x x Ctanh ln | ( ) |2

22. 4 61cosh−∫ x dx

= − +−4

66

4

66 11 2 1 2x x x Ccosh [( ) – ] /

= − +−2

36

2

336 11 2x x x Ccosh –

23. Let x = 3 sinh t, dx = 3 cosh t dt,

x t t2 29 9 9 3+ = ⋅ + =sinh cosh ,

tx= −sinh .1

3

∴ + = ⋅∫ ∫x dx t t dt2 9 3 3cosh cosh

= ∫9 2cosh t dt u dvcosh t cosh tsinh t sinh t

+–

= − ∫9 9 2cosh sinh sinht t t dt

= − −∫9 9 12cosh sinh (cosh )t t t dt

= − + ∫∫9 9 92cosh sinh cosht t t dt

∴ = + +∫18 9 921cosh cosh sinht dt t t t C

∴ = + +∫9 4 5 4 52cosh . cosh sinh .t dt t t t C

= ⋅ + ⋅ + +−4 59

3 34 5

3

21. . sinh

x x xC

= + + +−0 5 9 4 53

2 1. . sinhx xx

C

24. Let x = 5 cosh t, dx = 5 sinh t dt,

x t t2 225 25 25 5– cosh – sinh= ⋅ = ,

tx= −cosh 1

5.

∴ = ⋅∫∫ x dx t t dt2 25 5 5– sinh sinh

= ∫25 2sinh t dt u dvsinh t sinh t

cosh t cosh t–+

= − ∫25 25 2sinh cosh cosht t t dt

= − +∫25 25 12sinh cosh sinht t t dt( )

= − − ∫∫25 25 252sinh cosh sinht t t dt dt

∴ = − +∫50 25 2521sinh sinh cosht dt t t t C

∴ = − +∫25 12 5 12 52sinh sinh cosht dt t t t C. .

= ⋅ ⋅ − −12 525

5 512 5

5

21. .

x x x–cosh

= − +−0 5 25 12 55

2 1. .x xx

C– cosh

25. a. Figure 9-9g shows that the horizontal forceis given by the vector (h, 0) and the verticalforce is the vector (0, v), so their sum, thetension vector, is the vector (h, v), which has

slope v

h. Because the tension vector points

along the graph, the graph’s slope, y ′, also

equals v

h.

b. v = weight of chain below (x, y) = s ⋅ w

⇒ ′ = = ⋅ = ⋅yv

h

s w

h

w

hs

c. ds dx dy dx dy dx= + = +2 2 2 21[ ( / ) ]

= + ′1 2( )y dx

⇒ ′ = = + ′d yw

hds

w

hy dx( ) 1 2( )

d. [ ( ) ] ( )1 2 12+ ′ ′− /∫ y d y

= + − /∫ ( ) ( )1 2 12sinh sinht d t

= − /∫ ( ) ( )cosh cosh2 12t t dt

= = + = ′ +−∫ dt t C y Csinh 11

w

hdx

w

hx C= +∫ 2

⇒ ′ = +−sinh 1 yw

hx C

e. At x = 0, y ′ = 0, so

sinh− = + ⇒ =1 0 0 0w

hC C .

f. sinh sinh− ′ = ⇒ ′ =1 yw

hx y

w

hx

g.dy

dx

w

hx dy

w

hx dx= ⇒ = ∫∫sinh sinh

⇒ = +yh

w

w

hx Ccosh

26. a. y x kk

C= = ⇒ = +2 0 21

0when cosh

⇒ 2 = k + C ⇒ C = 2 − k

y x kk

k= = ⇒ = + −5 4 54

2when cosh

Using the solver feature of your grapher,k ≈ 3.0668… .


y x= ……

+ − …3 06681

3 06682 3 0668. .cosh

.

y x= ……

− …3 06681

3 06681 0668. .cosh

.b. y (20) = 1040.9739…

c. y = 4:

4 3 06681

3 06681 0668= …

…− …. .cosh

.x

cosh

.

.

.

1

3 0668

5 0668

3 0668K

K

Kx =

x = … ……

= …−3 06685 0668

3 06683 33551. .cosh

.

.

By symmetry, x = ±3.3355… .The answer can be found numerically usingthe solver feature of your grapher.

d. yk

x y′ = ′ = …sinh1

3 1 1418; ( ) .

e. A kk

x k dx= +

∫ cosh –

–

12

1

3

= + −−

kk

x k x2

1

312sinh ( )

= ……

−…

( . )3 0668

3

3 0668

1

3 06682 sinh

.sinh

–

.

+ 4(2 − 3.0668…)= 9.5937…

f. L y dx= + ′−∫ 1 2

1

3

( )

= +−∫ 1 2

1

3

sinh ( / )x k dx

= =− −∫ cosh sinh

1 11

3

1

3

kx dx k

kx

= +

= …k

k ksinh sinh

3 14 5196.

27. a. The vertex is midway between the poles,so y = 110 ft when x = 150 ft.

yh

w

w

hx C= +cosh

= +400

0 8

0 8

400

lb

lb ft. /cosh

.x C

110 5001

500150= ⋅

+cosh C

⇒ C = 110 − 500 cosh 0.3

y x= + −5001

500110 500 0 3cosh cosh .

The cable comes closest to the ground atx = 0.y (0) = 500 cosh 0 + 110 − 500 cosh 0.3= 610 − 500 cosh 0.3 = 87.3307… ≈ 87.3 ft

b. y x′ = sinh1

500

L x dx

x dx

= +

=

∫∫−

1 500

1

500

2

150

150

150

150

sinh ( / )

cosh

–

=−

5001

500 150

150

sinh x

= 500 sinh 0.3 − 500 sinh (−0.3)= 1000 sinh 0.3 = 304.5202… ≈ 304.5 ft

A faster method is:Half weight of cable = vertical tension at(150, 110) = h ⋅ y ′ (150) (CompareProblem 25.)

Weight .= ⋅ =2 4001

500150 800 0 3sinh sinh

= 243.6162… ≈ 243.6 lb(Note: Because w ⋅ L = weight, either of thesemethods could give both the weight and thelength.)

c. T h v= +2 2 ; h is constant and v is greatestat the ends, so the maximum tension is atx = 150 ft.

T h hy( )150 1502 2= + ′[ ( )]

= + =400 1 0 3 400 0 32sinh . cosh .

= 418.1354… ≈ 418.1 lb

d. The general equation is yh

w

w

hx C= +cosh .

If y (0) = 100 and y (150) = 110, find h suchthat y (150) − y (0) = 10. Solve:h

w

w

h

h

wcosh 150 10− = , or

cosh120

18

h h− =

By grapher, h = 901.3301... ≈ 901.3 lb.

28. The answers will depend on the dimensionsof the chain used. Note that the answer isindependent of the kind of chain. You mightshow students how a heavy chain and a lightchain of equal length will hang in the samecatenary if they are suspended from the samepoints.Assume that the dimensions are the same as inExample 5.

a. Vertex: (0, 20). Supports: (±90, 120).

b. y = 51.78… cosh

.

1

51 7831 78

KKx – .

c. Note: To conserve class time, you might havestudents plot only each 20 cm for x, as shownhere for Example 5. Use the TABLE feature.


x y

0 20.0

±20 23.9

±40 36.2

±60 58.8

±80 95.1

d. A clever way to make sure the measurementsare vertical is to hold a book against the boardwith its bottom edge along the chalk tray.Then hold the meterstick against the verticaledge of the book. It is crucial that the pointsbe plotted accurately to get the dramaticimpact of “perfect fit.”

e. For a quadratic function with vertex on they-axis, y = ax2 + c. Using the data forExample 5,20 = a (0) + c ⇒ c = 20

120 90 201

812= + ⇒ =a a( )

y x= +1

81202

20

10

x

yparabola

catenary

The parabola is more curved at the vertex.

f. For Example 5,

dL y dx x dx= + ′ = +1 11

51 782 2( ) sinh

. K

= cosh

.

1

51 78Kx dx

L x dx=

−∫ cosh.

1

51 7890

90

K

= =

−51 78

1

51 78285 349

90

90

. .KK

Ksinh.

x

≈ 285.3 cmThe actual length should be close to this.

29. a. y = sinh x

dS x dL x x dx= = +2 2 1 2π π cosh

S x x dx= +∫2 1 2

0

1

π cosh

= 5.07327… by numerical integration≈ 5.07 ft2

b. Cost = 2(57)(5.07327…) = 578.3532… ≈$578.35

c. Slice perpendicular to the y-axis.dV = π x2 dy = π (sinh–1 y)2 dyTop of bowl is at

y = − =sinh . .1

1

241 133534K

∴ = ∫ V y dyπ (sinh )–

.1 2

0

1 133K

= 1.25317… by numerical integration≈ 1.253 ft3

30. a. y kk

x C= − +cosh1

Inner catenary: yinner(0) = 612, yinner(260) = 0

6120

612= − + ⇒ = +kk

C C kii

i i icosh

0260

612= − + +kk

kii

icosh

⇒ ki = 97.1522… (numerically)

yinner = 97.1522… − +

+cosh

.

1

97 15221 612

Kx

Outer catenary: youter (0) = 630, youter (315) = 0

630 = −ko cosh0

ko

+ Co ⇒ Co = 630 + ko

0 = −ko cosh315

ko

+ 630 + ko

⇒ ko = 127.7114… (numerically)

y xouter .= − +

+127 7114

1

127 71141 630K

Kcosh

.

b. The graphs are the same as in Figure 9-9k:

100

100 x

y

c. A y dx y dx= −− −∫ ∫outer inner

315

315

260

260

= − + +−

kx

kk x xo

oo

2

315

315

630sinh

+ − −−

kx

kk x xi

ii

2

260

260

612sinh

= 54323.2729… ≈ 54,323 ft2

d.dy

dx kxouter

o

so= −sinh ,1

Lk

x dx= +−∫ 1

12

315sinh

o

315

= =− −∫ cosh sinh

1 1

315

315

kx dx k

kx

o315

315

oo


= ⋅ ⋅2 127 7114

315

127 7114. K

Ksinh

.= 1493.7422… ≈ 1494 ft.

e.

′ − = − − =ykouter

o

( ) sinh sinh.

315315 315

127 7144K

= 5.8481… is the spider’s starting slope.

f. José can fly through at altitude yinner(x) ifx ≥ 50 + 120/2 = 110.yinner(110) = 542.7829… , so the plane canfly through at heights between 0 and 542 feet.(Because of the curvature of the arch and thevertical thickness of the plane, the closestdistance is slightly less than 50 feet when thehorizontal distance is 50 feet. The plane canfly through at slightly higher altitudes bybanking slightly.)

31. a. H(x) = csch x ⇒ H′(x) = −csch x coth xH′(1) = −csch 1 coth 1 = −1.1172855…

b. H′ ≈( )h h

11 01 0 99

0 02

csc ( . ) – csc ( . )

.

= −1.11738505… . The answers differ by0.0000995… , which is about 0.0089% ofthe actual answer.

32. sec sin tanhh ( )x dx x= −∫ 1

1

2

1

2

= sin–1 (tanh 2) − sin–1 (tanh 1)= 0.435990…

Numerically,

sech .x dx =∫ 0 4359901

2

K

(Checks.)

33. By parts:

e x dxx sinh 2∫ u dvsinh 2x ex

2 cosh 2x ex 4 sinh 2x ex +

–

+

= − + ∫e x e x e x dxx x xsinh cosh sinh2 2 2 4 2

⇒ − ∫3 2e x dxx sinh

= − + ⇒ ∫e x e x C e x dxx x xsin cosh sinhh 2 2 2 21

= − +2

32

1

32e x e x Cx xcosh sinh

By transforming to exponential form:

e x dx e e e dxx x x xsinh 21

22 2= − −∫∫ ( )

= − = + +− −∫1

2

1

6

1

23 3( )e e dx e e Cx x x x

Transforming to exponential form is easier!(Note that the two answers can be shown to beequivalent either by transforming the first to

exponential form or by transforming the secondto hyperbolic form, as shown here.)

1

6

1

2

1

6

1

23 2 2e e C e e e Cx x x x x+ + = +

+− −

= + − +

+− −e e e e e Cx x x x x1

3

1

3

1

6

1

62 2 2 2

= + − +2

3 2

1

3 2

2 2 2 2

ee e

ee e

Cxx x

xx x– ––

= − +2

32

1

32e x e x Cx xcosh sinh

34. e x dxx sinh∫ u dve x sinh xe x cosh xe x sinh x

+

–+

= − + ∫e x e x e x dxx x xcosh sinh sinh

⇒ = − +∫0 e x dx e x x Cx xsinh cosh sinh )(

The original integral reappeared with the samecoefficient, so when it was added again to the leftside, it exactly canceled out the desired integral.Use the exponential form of sinh x.

e x dx e e e dxx x x xsinh )= − −∫∫ 1

2(

= −∫1

212( )e dxx

= − +1

4

1

22e x Cx

35. a. cosh2 x − sinh2 x

= +

−

e e e ex x x x– ––

2 2

2 2

= + + − + =e e e ex x x x2 2 2 22

4

2

41

– ––

∴ cosh2 x − sinh2 x = 1, Q.E.D.

b.1 1

22 2

2cosh(cosh – sinh )

coshxx x

x=

⇒ 1 − tanh2 x = sech2 x

c.1 1

22 2

2sinh(cosh – sinh )

sinhxx x

x=

⇒ coth2 x − 1 = csch2 x

36. a. Substitute 2x for x in the definition of sinh x.

b. sinh ( – )–21

22 2x e ex x=

= ⋅ ⋅ ⋅ +

=

−21

2

1

22

( – )

sinh cosh

–e e e e

x x

x x x x( )


c. cosh ( )–21

22 2x e ex x= +

= + + + +1

41

1

412 2 2 2( ) ( – )– –e e e ex x x x

= +

+

1

2

1

2

2 2

( ) ( – )– –e e e ex x x x

= cosh2 x + sinh2 x

d. cosh2 x − sinh2 x = 1 ⇒ cosh2 x = 1 + sinh2 x⇒ cosh 2x = cosh2 x + sinh2 x= (1 + sinh2 x) + sinh2 x = 1 + 2 sinh2 x

e. 1 + 2 sinh2 x = cosh 2x⇒ 2 sinh2 x = cosh 2x − 1

⇒ =sinh (cosh – )2 1

22 1x x

37. a. On the circle, u2 + v2 = 1 ⇒2u du + 2v dv = 0 ⇒ dv = (−u/v) du.

dL du dv du u v du

u v

vdu

vdu

udu

Ldu

u

= + = +

= + = =

= ∫

2 2 2 2 2

2 2

2 2

22

1

1 1

1

1

( / )

–

–cos

= − = − + =− −cos coscos

12

1 11 2 2u

The curve along the hyperbola from u = 1 tou = cosh 2 has length greater than the linesegment along the horizontal axis from (1, 0)to (cosh 2, 0). This segment has lengthL = cosh 2 − 1 = 2.762… . So the length ofthe curve is greater than 2, Q.E.D.

b. The area of the triangle that circumscribes thesector is 0.5(2 sinh 2 cosh 2) = sinh 2 cosh 2.The area of the sector is the area of thistriangle minus the area of the region betweenthe upper and lower branches of the hyperbolafrom u = 1 to u = cosh 2.Slice this region vertically. Pick sample point(u, v) on the upper branch, within the strip.Let t be the argument of sinh and cosh at thesample point. 0 ≤ t ≤ 2.dA = 2v du = 2 sinh t d(cosh t) = 2 sinh2 t dt

A t dt= ≈∫2 11 6449582

0

2

sinh . K

Thus, the area of the sector iscosh 2 sinh 2 − 11.644958… = 2, Q.E.D.

c. By definition of the circular functions,x is the length of the arc from (1, 0) to(cos x sin x). So the total arc has length 2x.The circumference of a unit circle is 2π, andits area is π. Thus,

Ax

xsector ,= =2

2ππ Q.E.D.

d. Slice as in part b.

Let l t dt= ∫ sinh2 u dvsinh t sinh t

cosh t cosh t–+

= − ∫sinh cosh cosht t t dt2

= − +∫sinh cosh ( sinh )t t t dt1 2

= − − ∫sinh cosh sinht t t t dt2

∴ = − +∫2 2sinh sinh cosht dt t t t C

sinh sinh cosh210 5 0 5t dt t t t C= − +∫ . .

Slicing as in part b, the area A between theupper and lower branches of the hyperbola is

A t dtx

= ∫2 2

0sinh

= − −2 0 5 0 5 0( . . )sinh coshx x x = sinh x cosh x − x

Thus, the area of the sector iscosh x sinh x − (sinh x cosh x − x) = x,Q.E.D.

38. a. y = sinh− 1 x ⇒ sinh y = x ⇒ cosh y y′ = 1

′ = =+

=+

yy y x

1 1

1

1

12 2cosh sinh, Q.E.D.

b. y = tanh− 1 x ⇒ tanh y = x ⇒ sech2 y y′ = 1

′ = = =yy y x

1 1

1

1

12 2 2sec – tan –h h, Q.E.D.

c. y = coth− 1 x ⇒ coth y = x⇒ −csch2 y y′ = 1

yy y x

′ = = =1 1

1

1

12 2 2– –(coth – ) –,

cschQ.E.D.

d. y = sech− 1 x ⇒ sech y = x⇒ −sech y tanh y y′ = 1

yy y

′ = 1

– tanhsech

= 1

1 2– –sech sechy y

= − 1

1 2x x–, Q.E.D.

e. y = csch− 1 x ⇒ csch y = x⇒ −csch y coth y y′ = 1

yy y y y

′ = =+

1 1

1 2– coth –csch csch csch

= −+

1

1 2| |x x, Q.E.D.


Problem Set 9-10Q1. Q2. sinh x + C

1

1 x

y

Q3. sinh x Q4. −sin x

Q5. sin x + C Q6. y = x3

Q7. y = tan x Q8. y = sinh x or x3 + x

Q9. y = ex Q10. A

1. a.1

2

x

y

It might converge because the integrandapproaches zero as x approaches infinity.

b. ( / ) lim ( )1 2 2

22x dx x dx

b

b

=→∞

−∞

∫∫= − = − + =

→∞

−

→∞lim lim( / / )b

b

bx b1

21 1 2

1

2

Integral converges to 1

2.

2. a.

3

1

x

y


b. 1 4 4

33/ lim

bx dx x dx

b

=→∞

−∞

∫∫= − = +

=

→∞

−

→∞lim lim –b

b

bx

b

1

3

1

3

1

81

1

813

33

The integral converges to 1

81.

3. a.

x

y

1

1


b. ( / ) ( / )1 11 1

x dx x dxb

=→∞

∞

∫ ∫limb

= = = ∞→∞ →∞

lim ln lim (ln – )b

b

bx b| |

10

The integral diverges.

4. a.

1

1x

y

It might converge because the integrandbecomes infinite only as x approaches zero.

b. ( / ) lim ( / )1 10

1

0

1

x dx x dxa a

=→ + ∫∫

= = = ∞→ →+ +lim ln lim ( – ln )

a a ax a

0

1

00| |


5. a. y

x1

1


b. 1 0 2 0 2

11/ . .x dx x dx

b

b

=→∞

−∞

∫∫ lim

= = = ∞→∞ →∞

lim lim ( . – . ).

b

b

bx b1 25 1 25 1 250 8

1

0 8. .


6. a.

x

y

1

1


b. 1 1 2

1

1 2

1/ . .x dx x dx

b

b

=∞

→∞

−∫ ∫lim

= − = + =→∞

−

→∞lim lim (– )– .

b

b

bx b5 5 5 50 2

1

0 2.

The integral converges to 5.

7. a.

1

1

x

y



b. 1 0 2

0

0 21

0

1

/ .x dx x dxa a

=→

−+ ∫∫ lim .

= =→ →+ +lim lim ( . – . ).

a a ax a

0

0 81

0

0 81 25 1 25 1 25. .

= 1.25The integral converges to 1.25.

8. a. y

x

1

1


b. 1 1 2

0

1

0

1 21

/ . .x dx x dxa a

=∫ ∫→

−+

lim

= − = + = ∞→

−

→+ +lim lim (– )– .

a a ax a

0

0 21

0

0 25 5 5.


9. a.

0 1

1

x

y


b. 1 1 1 12 2

00/( ) /( )+ = +

→∞

∞

∫∫ x dx x dxb

b

lim

= = =→∞

−

→∞lim tan lim (tan – )–

b

b

bx b1

0

1 02

π

The integral converges to π /2.

10. a.

0 x1

1


b. 1 1 1 10 0

/( ) /( )+ = +∞

→∞∫ ∫x dx x dxb

b

lim

= + = + = ∞→∞ →∞

lim ln lim [ln ( ) – ]b

b

bx b| |1 1 0

0


11. a.

0

y

x

1

1

It might converge because the integrandbecomes infinite only as x approaches 0 or 1.

b. To determine whether this converges, split theintegral into two pieces. Each piece mustconverge in order for the integral to converge.The integral can be written

10

1

/( ) x x dxln∫= +∫ ∫1 1

0/( ) /( )

1

x x dx x x dxc

cln ln

= +→ →+ − ∫∫lim ln lim ln

a b c

b

a

c

x x dx x x dx0 1

1 1/( ) /( )

= +→ →+ −lim ln ln lim ln ln

a a

c

b c

b

x x0 1

| | | |

=

+→

→

+

−

lim (ln |ln | – ln |ln |)

lim (ln |ln | – ln |ln |)a

b

c a

b c0

1

= ∞ + ∞For the integral to converge, both limits mustexist. Because neither exists, the integraldiverges.

12. a.

0 3

y

x

1

1


b. The indefinite integral can be written

( ) ( / ) .ln (ln )x dx x x C− −= − +∫ 2 1

1 12 2

33/[ ( ) ] /[ ( ) ]x x dx x x dx

b

b

ln lim ln=→∞

∞

∫∫= −

→∞

−lim lnb

b

x( ) 1

3

= + =→∞

− −lim [–(ln ) (ln ) ]–

bb 1 13 (ln 3) 1

The integral converges to(ln 3)− 1 = 0.910239… .

13. a.

2

x

1

y



b. e dx e dxx

b

xb

−

→∞

−∞

= ∫∫ 0 4 0 4

22

. .lim

= −→∞

−limb

xb

e2 5 0 4

2. .

= + =→∞

− −lim(– . . ). – .

b

be e e2 5 2 5 2 50 4 0 8 0 8. .

The integral converges to 2.5e− 0.8

= 1.1233… .

14. a.

0 1

1

x

y

It diverges because the integrand does notapproach zero as x approaches infinity.

b. (Not applicable)

15. a.

0–1 2

1

x

y

It does not converge because the integrand isundefined for x < 0.

b. (Not applicable)

16. a.

0 3 7

1

1

x

y

It might converge because the integrandbecomes infinite only as x approaches 3.

b. ( – )x dx3 2 3

1

7−∫ /

= +→

−

→

−− + ∫∫lim ( – ) lim ( – )

b a a

b

x dx x dx3

2 3

3

2 37

13 3/ /

= − + −→ →− +lim limb

b

a ax x

3

1 3

1 3

1 37

3 3 3 3( ) ( )/ /

= − −→ −

lim [ ( – ) ( ) ]/ /

bb

3

1 3 1 33 3 3 2

+ − −→ +lim [ ( ) ( ) ]/ /

aa

3

1 3 1 33 4 3 3

= 3 ⋅ 21/3 + 3 ⋅ 41/3

The integral converges to 3 ⋅ 21/3 + 3 ⋅ 41/3

= 8.5419… .

17. a.

0 1

x

y1

It might converge because the integrand seemsto approach zero as x approaches infinity.

b. Integrate by parts:

xe dx e x Cx x− −= − + +∫ ( 1)

xe dx xe dxx

b

xb

−

→∞

−∞

= ∫∫ lim00

= − + = + + =→∞

−

→∞lim lim [– ( ) ]–

b

xb

b

be x e b( )1 1 1 10

(The first term is zero by l’Hospital’s rule.)Integral converges to 1.

18. a.

0 3

x

y

1

1

It might converge because the integrandbecomes infinite only as x approaches 1.

b. ( – )x dx1 2

0

3−∫

= +→

−

→

−+ ∫∫lim ( – ) lim ( – )

–b a a

b

x dx x dx1

2

1

23

01 1

= − − + − −→

−

→

−+

lim lim–b

b

a ax x

1

1

0 1

13

1 1( ) ( )

= +→lim [–( – ) (– ) ]

–

– –

bb

1

1 11 1

+ +→ +lim [– ( – ) ]– –

aa

1

1 12 1

= ∞ + ∞For the integral to converge, both limits mustexist. Because neither exists, the integraldiverges.

19. a.

20

1

x

y


b. (Not applicable)


20. a.

20

1

x

y


b. (Not applicable)

21. As b x dxb

→ ∞ ∫, cos0

oscillates between −1 and

1 and never approaches a limit. Similarly,

sin x dxb

0∫ oscillates between 0 and 2.

22. a. I /.. .

1 0011 001 1 001

111= =

→∞

−∞

∫∫ x dx x dxb

b

lim

= −→∞

−limb

bx1000 0 001

1

.

= +→∞

lim (– )– .

bb1000 10000 001

= 1000 (converges), Q.E.D.

I /.. .

0 9990 999 0 999

111= =

→∞

−∞

∫∫ x dx x dxb

b

lim

=→∞

limb

bx1000 0 001

1

.

=→∞

lim ( – ).

bb1000 10000 001

= ∞ (diverges), Q.E.D.

b. I /11

1= = ∞∞

∫ x dx (see Problem 3), so I1

diverges.

c. “Ip converges if p > 1 and diverges if p ≤ 1.”

23. a. y = 1/x = x− 1

dA = y dx = x− 1 dx

A x dx= = ∞−∞

∫ 1

1 (See Problem 3)

The area does not approach a finite limit.

b. By plane slices, dV = πy2 dx = πx− 2 dx.

V x dx xb b

bb

= = −→∞

−

→∞

−∫lim limπ π2 1

11

= + =→∞

lim (– )–

bbπ π π1

The volume converges to π.c. By cylindrical shells, dV = 2π xy dx

= 2π x(x− 1) dx = 2π dx.

V dx xb

b

b

b

= =→∞ →∞∫lim lim2 2

1 1π π

= = ∞→∞

lim ( – )b

b2 2π π

Volume diverges.

d. False. The volume could approach a constantas in part b or become infinite as in part c.

24. y = −1/x ⇒ x = −y− 1

Slice the vertical cross section horizontally.dA = x dy = −y− 1 dy

A y dy ya a a a

= − = −→−∞

−

→ ∞

−

∫lim lim ln–

–

11 1

| |

= + = ∞→−∞lim (– ln |– | ln | |)

aa1

The area of the bucket’s surface is greater thanthe area of the cross section, and the cross-sectional area diverges. Thus, the bucket hasinfinite surface area. The bucket is congruent tothe solid in Problem 23b, which has volumeapproaching π. Thus, π cubic units of paintwould fill the bucket but could not coat thewhole surface!

25. a. f x t e dtx t( ) = −∞

∫0

f te dt te eb

tb

b

t tb

( )10 0

= =→∞

−

→∞∫lim lim (– – )– –

= + + =→∞

lim (– – )– –

b

b bbe e 0 1 1

(Using l’Hospital’s rule on be− b gives

lim lim limb

b

b b b bbeb

e e→∞

−

→∞ →∞= = =1

0.)

f t e dtb

t( )2 2

0=

→∞

−∞

∫lim

= − +→∞

− −∞

∫limb

tb

tt e te dt2

0 02

= + + =→∞

lim (– )–

b

bb e2 0 2 1 2( )

(Using l’Hospital’s rule on b2e− b gives

lim lim limb

b

b b b bb eb

e e→∞

−

→∞ →∞= = =2 2 2

0.)

f t e dtb

t( )3 3

0=

→∞

−∞

∫lim

= − +→∞

− −∞

∫limb

tb

tt e t e dt3

0

2

03

= + + =→∞

lim (– )–

b

bb e3 0 3 2 6( )

(Using l’Hospital’s rule on b3e− b gives

lim lim lim limb

b

b b b b b bb eb

e

b

e e→∞

−

→∞ →∞ →∞= = = =3

23 6 60.)

b. Conjecture:

f ( 4) = 4 f (3) = 24 = 4!

f ( 5) = 5 f (4) = 120 = 5!

f ( 6) = 6 f (5) = 720 = 6!

c. f x t e dtx t( ) = −∞

∫0u dvtx e–t

xtx–1 –e–t +

–


= − +→∞

− − −∞

∫limb

x tb

x tt e x t e dt0

1

0

= + + −→∞

lim (– )–

b

x bb e x f x0 1 ( )

= 0 + 0 + x f (x − 1)= x f (x − 1), Q.E.D.

( lim (– )–

b

x bb e→∞

= 0 can be proved by

mathematical induction usingl’Hospital’s rule.)

d. Part a shows that f (1) = 1 = 1!.Part c shows that f (n) = nf (n − 1) =n(n − 1) f (n − 2) = n(n − 1)(n − 2)…(2)(1)= n!, Q.E.D.

e. t e dtt3

0

1000

6− ≈∫The value of b that makes the integral comewithin 0.000001 of 6 can be found numerically(though it will be slow), or algebraically:

t e dttb

3

0

−∫= − − − − +− − − −b e b e be eb b b b3 23 6 6 6

|b3e− b + 3b2e− b + 6be− b + 6e− b| < 0.000001for b > 23.4050… , say, b ≈ 24.

f. 0 5 0 5

0

0 5

0

24

. ! .= ≈ ≈−∞

−∫ ∫t e dt t e dtt t.

0.886227311…

From the graphs, t0.5e− t < t3e− t for x ≥ 24. Theerror in 0.5! from stopping at b = 24 is thearea under the “tail” of the graph from b = 24.

Error < < .= −∞

−∞

∫ ∫t e dt t e dtt t0 5

24

3

240 000001.

The difference between the tabulated valueof 0.5! and the value calculated here is

0.8862269255 − 0.866227311…

= −0.000000386

which is less in absolute value than 0.000001.Note, however, that the difference is negativebecause the calculated value is larger thanthe tabulated value. This observation meansthat either the tabulated value is incorrect orthere is more inaccuracy in the numericalintegration algorithm than there is in the errorcaused by dropping the tail of the integral.(Using a smaller tolerance in the numericalintegrator gives a value of 0.8862269252… .)

g. Using the tabulated value of 0.5!,1.5! = 1.5(0.5!) = 1.3293…2.5! = 2.5(1.5!) = 3.3233…3.5! = 3.5(2.5!) = 11.6317…

h. 0 0

0 0! = =

−∞

→∞∫ t e dt et

b

t blim – –

= + =→∞

lim(– )–

b

be 1 1, Q.E.D.

i. (−1)! = 0!/0, which is infinite. So (−2)! and(−3)!, which equal (−1)!/(−1) and (−2)!/(−2),are also infinite. However,(−0.5)! = 0.5!/(0.5) = 1.77245…(−1.5)! = (−0.5)!/(−0.5) = −3.54490…(−2.5)! = (−1.5)!/(−1.5) = 2.36327…all of which are finite.

j. 0 52

0 886226925. ! . ,= =πK which agrees

with the tabulated value.26. dW = F dr = 1000r− 2 dr

At the earth’s surface, r = 1.

W r dr rb

b= =

−

→∞

∞−∫ 1000 10002

1

11

lim –

= +→∞

lim (– )–

bb1000 10001

= 1000 radius-poundsThus, the amount of work does not increasewithout bound as r goes to infinity.

27. a. 22

21

3x x

xdx− −

−

∫

= − −−

+ − −−

∫ ∫2

2

22

2

21

2

2

3x xx

xdx

x

xdx

b. 22

22

2

21

2

2

3x xx

xdx

x

xdx− −

−

+ − −−

∫ ∫

= − −−

+ − −−

→

→

−

+

∫∫

lim

lim

b

xb

a

x

a

x

xdx

x

xdx

2 1

2

3

22

2

22

2

c. limb

xb x

xdx

→ −− −

−

∫2 1

22

2

+ − −−

→ + ∫lim

a

x

a

x

xdx

2

3

22

2

= + + −→ →− + ∫∫lim ( ) lim ( )

b

x

a

x

a

b

dx dx2 2

3

12 1 2 1

= + +→ →− +lim ( /ln ) lim ( /ln – )b

xb

a

x

ax x

2 1 2

3

2 2 2 2

= + −→ −lim ( /ln – /ln )b

b b2

2 2 2 2 1

+ +→ +lim ( /ln – – /ln )a

a a2

32 2 3 2 2

= + − − + −4 2 2 2 2 1 8 2 3/ / /ln ln ln− + = / /4 2 2 6 2ln ln

The integral converges to 6/ln 2 = 8.6561… .d. The integral is defined by dividing the

interval into Riemann partitions and summingthe subintervals. But the Riemann partitionsmay be chosen so that the discontinuitiesare at endpoints of subintervals. Then thesubintervals corresponding to each continuouspiece may be summed separately.

e. False. Some discontinuous functions (notably,piecewise continuous functions) are integrable.



Problem Set 9-11

1. y = sec 3x tan 3x ⇒y′ = (3 sec 3x tan 3x) tan 3x + sec 3x (3 sec2 3x)

= 3 sec 3x tan2 3x + 3 sec3 3x2. y = sinh 5x tanh 5x ⇒

y′ = (5 cosh 5x) tanh 5x + sinh 5x (5 sech2 5x)= 5 sinh 5x + 5 sinh 5x sech2 5x or5 sinh 5x + 5 tanh 5x sech 5x

3. x x dxcosh 4∫

16

u dvx cosh 4x1

14 sinh 4x

01

cosh 4x

+

–

+

= − +1

44

1

164x x x Csinh cosh

4. x x dxcos∫ u dvx cos x1 sin x0 –cos x+

–

+

= + +x x x Csin cos5. f (x) = (3x + 5)− 1 ⇒ f ′(x) = −3(3x + 5)− 2

6. f (x) = (5 − 2x− 1) ⇒ f ′(x) = 2(5 − 2x)− 2

7. ( )3 51

33 51x dx x C+ = + +−∫ ln | |

8. ( ) | |15 21

25 2− = − +−∫ x dx x Cln –

9. t(x) = tan5 4x ⇒t′(x) = 5 tan4 4x (4 sec2 4x) = 20 tan4 4x sec2 4x

10. h(x) = sech3 7x ⇒h′(x) = 3 sech2 7x (−7 sech 7x tanh 7x)

= −21 sech3 7x tanh 7x

11. sin cos2 1

21 2x dx x dx ( )= −∫ ∫

= − +1

2

1

42x x Csin

= − +1

2

1

2x x x Csin cos (or integrate by parts)

12. cos cos2 1

21 2x dx x dx= +∫∫ ( )

= + +1

2

1

42x x Csin

= + +1

2

1

2x x x Csin cos (or integrate by parts)

13. yx

x=

+⇒6 11

2

–

yx x

x x′ = +

+=

+6 2 6 11 1

2

23

22 2

( ) – ( – )( )

( ) ( )

14. yx

x= + ⇒5 9

4–

yx x

x x′ = + =5 4 5 9 1

4

29

42 2

( – ) – ( )( )

( – )

–

( – )

15.6 11

26

23

2

x

xdx

xdx

–

+= −

+

∫ ∫

= 6x – 23 ln | x + 2 | + C

16.5 9

45

29

4

x

xdx

xdx

+ = +

∫ ∫– –

= 5x + 29 ln | x − 4 | + C

17. f t t t( ) ( )1/2= + = + ⇒1 12 2

f t t tt

t′ = + =

+−( ) ( )

1

21 2

1

2 1 2

2( ) /

18. g t t t( ) ( )1/2= = − ⇒2 21 1–

g t t tt

t′ = =−( ) ( )

1

21 2

1

2 1 2

2( – )

–

/

19. 1 12 2+ = +∫ ∫t dt dtan tanθ θ( )

= =∫ sec sec tan3 1

2θ θ θ θd

+ + +1

2ln sec tan| |θ θ C

= + + + + +1

21

1

212 2t t t t Cln

20. t dt d2 21 1– sec – sec=∫ ∫ θ θ( )

= = −∫∫ sec tan sec secθ θ θ θ θ θ2 3d d( )

= + +1

2

1


− ln | sec θ + tan θ | + C

= − + +1

2

1


= − + +1

21

1

212 2t t t t C– ln –

21. y = x 3 e x ⇒y′ = 3x2ex + x3ex = x2ex(3 + x)

22. y = x4e− x ⇒y′ = 4x3e− x − x4e− x = x3e− x(4 − x)

23. x e dxx3∫ u dvx 3 e x

3x 2 e x

6x e x

6 e x

0 e x

+

+

–

–+

= x3ex − 3x2 ex + 6xex − 6ex + C

24. x e dxx4 −∫ u dvx 4 e –x

4x 3 –e –x

12x 2 e –x

24x –e –x

24 e –x

0 –e –x–

–

–

+

+

+

= −x4 e− x − 4x3

e− x −12x2 e− x − 24xe− x − 24e− x + C

25. f x x f xx

x( ) ( ) ( ) /= ⇒ ′ = = −− −sin–

1

2

2 1 21

11


26. g x x g xx

( ) ( )= ⇒ ′ =+

−tan 12

1

1

27. sin−∫ 1 x dx u dvsin–1 x 1

(1 – x2)–1/2 x+

–

= − −− −∫x x x x dxsin 1 2 1 21( ) ( )/

= x sin− 1 x − (−0.5)(2)(1 − x2)1/2 + C

= + +−x x x Csin –1 21

28. tan−∫ 1 x dx u dvtan–1 x 1

11 + x2 x–

+

= −+

− ∫x xx

x dxtan ( )12

1

1

= − + +−x x x Ctan ln1 21

21| |

29.1

4 5

1 6

5

1 6

12x xdx

x xdx

+=

++

∫∫ –

– / /

–

= − + + − +1

65

1

61ln ln| | | |x x C

30.1

6 7

1 8

1

1 8

72x xdx

x xdx

– –

– / /

–=

++

∫∫

= − + + − +1

81

1

87ln ln| | | |x x C

31.1

4 5

1

2 92 2x xdx

xdx

+=

+∫ ∫– ( ) –

= ∫ 1

3 93

2( sec ) –( sec tan )

θθ θ θd

= =∫ ∫1

33

tansec tan sec

θθ θ θ θ θ( )d d

= ln | sec θ + tan θ | + C

= + + + − +ln ( ) ( )1

32

1

32 92

1x x C

= + + + − +ln x x x C2 4 52

32.1

6 7

1

3 162 2x xdx

xdx

− −=

− −∫ ∫ ( )

= ∫ 1

4 164

2( sec ) –( sec tan )

θθ θ θd

= = ∫∫ 1

44

tan( sec tan ) sec

θθ θ θ θ θd d

= + +ln |sec tan |θ θ C

= − + +ln ( ) ( – ) –1

43

1

43 162

1x x C

= − + +ln – –x x x C3 6 72

33. f (x) = tanh x ⇒ f ′(x) = sech2 x

34. f (x) = coth x ⇒ f ′(x) = −csch2 x

35. tanhsinh

coshln coshx dx

x dx

xx C∫ ∫= = +| |

(Absolute value is optional.)

36. cothcosh

sinhln sinhx dx

x dx

xx C= = +∫ ∫ | |

(Absolute value is necessary.)

37. y = e2x cos 3x

y′ = (2e2x) cos 3x + e2x(−3 sin 3x)

= e2x(2 cos 3x − 3 sin 3x)

38. y = e−3x cos 4x

y′ = (−3e−3x) cos 4x + e−3x(−4 sin 4x)

= −e−3x(3 cos 4x + 4 sin 4x)

39. e x dxx2 3cos∫

9

u dve 2x cos 3x

2e 2x 13 sin 3x

4e 2x –1 cos 3x

+

–

+

= +1

33

2


− ∫4

932e x dxx cos

13

932e x dxx cos∫

= + +1

33

2

932 2


e x dxx2 3cos∫= + +3

133

2


40. e x dxx−∫ 3 4cos u dve –3x cos 4x

–3e –3x 14 sin 4x

9e –3x –116 cos 4x+

–

+

= −− −1

44

3


− −∫9

1643e x dxx cos

25

1643e x dxx−∫ cos

= − +− −1

44

3

1643 3


e x dxx−∫ 3 4cos

= − +− −4

254

3



Note: As a check for integrals such asProblems 39 and 40, the numerators of thecoefficients equal the 3 and 4 in the argumentsof e−3x and cos 4x. The denominators equal32 + 42, or 25.

41. g (x) = x3 ln 5x ⇒g ′ (x) = (3x2) ln 5x + x3 (5/5x)

= x 2 (3 ln 5x + 1)

42. h (x) = x2 ln 8x ⇒h ′(x) = (2x) ln 8x + x2 (8/8x)

= x(2 ln 8x + 1)

43. x x dx3 5ln∫------------------------

u dvln 5x x3

1x

14 x4

114 x 3

0116x 4

+

–

+

= − +1

45

1

164 4x x x Cln

44. x x dx2 8ln∫------------------------

9

3

3

u dvln 8x x2

1x

1x3

11x2

01x3+

–

+

= − +1

38

1

93 3x x x Cln

45. yx

x x x=

+ + +⇒

( )( )( )2 3 4ln y = ln x − ln (x + 2)

− ln (x + 3) − ln (x + 4) ⇒y′ = y[x− 1 − (x + 2)− 1 − (x + 3)− 1 − (x + 4)− 1]

=+ + +

⋅ − +− −x

x x xx x

( )( )( )2 3 421 1[ ( )

− (x + 3)− 1 − (x + 4)− 1]

46. yx

x x x= ⇒

( – )( – )( – )1 2 3ln y = ln x − ln (x − 1)

− ln (x − 2) − ln (x − 3) ⇒y′ = y[x− 1 − (x − 1)− 1 − (x − 2)− 1 − (x − 3)− 1]

= ⋅ − −− −x

x x xx x

( – )( – )( – )1 2 311 1[ ( )

− (x − 2)− 1 − (x − 3)− 1]

47.x

x x xdx

( )( )( )+ + +∫ 2 3 4

=+

++

−+

∫ –1

2

3

3

2

4x x xdx

= −ln |x + 2| + 3 ln |x + 3| − 2 ln |x + 4| + C

48.x

x x xdx

( – )( – )( – )1 2 3∫= − +

∫ 1 2

1

2

2

3 2

3

/

– –

/

–x x xdx

= − − − + − +1

21 2 2

3

23ln ln ln| | | | | |x x x C

49. y = cos3 x sin x ⇒y′ = (−3 cos2 x sin x) sin x + cos3 x (cos x)

= −3 cos2 x sin2 x + cos4 x

50. y = sin5 x cos x ⇒y′ = (5 sin4

x cos x) cos x + sin5 x (−sin x)= 5 sin4 x cos2 x − sin6 x

51. cos sin cos3 41

4x x dx x C( ) = − +∫

52. sin cos sin5 61

6x x dx x C( ) = +∫

53. cos sin cos3 21x dx x x dx= −∫ ∫ ( )

= −∫ ∫cos sin cosx dx x x dx2 ( )

= − +sin sinx x C1

33

Or: cos cos sin cos3 21

3

2

3x dx x x x dx= +∫ ∫

= + +1

3

2

32cos sin sinx x x C

54. sin cos sin5 21x dx x x dx= −∫ ∫ ( ) ( )2

= − +∫ ( )( )1 2 2 4cos cos sinx x x dx

= −∫ ∫sin cos sinx dx x x dx2 2

+ ∫ cos sin4 x x dx

= − + − +cos cos cosx x x C2

3

1

53 5

Or: sin sin cos sin5 4 31

5

4

5x dx x x x dx= − + ∫∫

= − − + ∫1

5

4

15

8

154 2sin cos sin cos sinx x x x x dx

= − −1

5

4


− +8

15cos x C

55. cos cos sin cos4 3 21

4

3

4x dx x x x dx= + ∫∫

= + + ∫1

4

3

8

3

83cos sin cos sinx x x x dx

= + + +1

4

3

8

3

83cos sin cos sinx x x x x C


56. sin sin cos sin6 5 41

6

5

6x dx x x x dx= − +∫ ∫

= − −1

6

5


+ ∫15

242sin x dx

= − −1

6

5


− + ∫15

48

15

48sin cosx x dx

= − −1

6

5


− + +5

16

5

16sin cosx x x C

57. g (x) = (x4 + 3)3 ⇒g ′(x) = 3(x4 + 3)2(4x3) = 12x3(x4 + 3)2

58. f (x) = (x3 − 1)4 ⇒f ′(x) = 4(x3 − 1)3(3x2) = 12x2(x3 − 1)3

59. ( ) ( )3x dx x x x dx4 12 8 43 9 27 27+ = + + +∫∫= + + + +1

13

27

52713 9 5x x x x C

60. ( )4x dx3 1−∫= − + − +∫ ( )x x x x dx12 9 6 34 6 4 1

= − + − + +1

13

2

5

6

713 10 7 4x x x x x C

61. ( )3x x dx x C4 3 4 431

163+ = + +∫ ( )

62. ( ) ( )x x dx x C3 4 2 3 511

151− = − +∫

63. ( )x dx x x C4 531

53+ = + +∫

64. ( )x dx x x C3 411

4− = − +∫

65. f x t dt f x xx

( ) ( ) ( ) ( )= + ⇒ ′ = +∫ 4 3

1

4 33 3

66. h x t dt h x xx

( ) ( ) ( ) ( )= − ⇒ ′ = −∫ 3

5

4 3 41 1

67. xe dxx

1

2

∫ u dvx ex

1 ex

0 ex

+

–+

= − = − − + = =xe e e e e e ex x

1

2 2 2 22 7 3890. K

68. xe dxx−∫0

2 u dvx e–x 1 –e–x 0 e–x

+–+

= − − = − − + +− − − −xe e e ex x0

2 2 22 0 1

= −3e− 2 + 1 = 0.59399…

69. r(x) = xex ⇒ r ′(x) = xex + ex

70. s(x) = xe− x ⇒ s′(x) = −xe− x + e− x

71. q xx

x( )

ln= + ⇒2

′ = ⋅ − + ⋅ = − −q x

x x x

x

x

x( )

( / ) (ln ) ln1 2 1 12 2

72. r xx

x( )

(ln )= + ⇒3 4

′ = ⋅ − + ⋅r x

x x x x

x( )

(ln ) ( / ) [(ln ) ]3 1 4 12 3

2

= − −3 42 3

2

(ln ) (ln )x x

x

73.ln

(ln )x

xdx x

dx

x

+ = +∫ ∫22

= + +1

22 2(ln )x C

74.(ln )

(ln )x

xdx x

dx

x xdx

334 4+ = +∫ ∫ ∫

= + +1

444(ln ) ln | |x x C

(The absolute values are optional because ln xappears in the original integrand, so onlypositive values of x can be used.)

75. f x e f x xex x( ) ( )= ⇒ ′ =2 2

2

76. f x e f x x ex x( ) ( )= ⇒ ′ =3 3

3 2

77. xe dx e Cx x2 21

2∫ = +

78. x e dx e Cx x2 3 31

3∫ = +

79. x e dxx3 2

∫---------------------

2

u dvx 2 xe x2

2x12e x2

21xe x2

014ex2 +

–

+

= − +1

2

1

22 2 2

x e e Cx x


80. x e dxx5 3

∫---------------------

u dvx3 x2 ex 3

3x2 13ex3

1 x2 ex3

013e x3

+

–

+

= − +1

3

1

33 3 3

x e e Cx x

81. e bx dxax∫ cos u dveax cos bx

aeax 1b sin bx

a2 e ax –1b2 cos bx+

–

+

= + − ∫12

2

2be bx

a

be bx

a

be bx dxax ax axsin cos cos

a b

be bx dxax

2 2

2

+ ∫ cos

= + +12 1b

e bxa

be bx Cax axsin cos

e bx dxax∫ cos

=+

++

+b

a be bx

a

a be bx Cax ax

2 2 2 2sin cos

(for a, b not both 0)

e bx dx x Cax cos = +∫ (for a = b = 0)

82. e bx dxax sin∫

b2

u dve ax sinbx

ae ax –1b cosbx

a2 eax –1

sinbx

+

–

+

= − +12b

e bxa

be bxax axcos sin

− ∫a

be bx dxax

2

2 sin

a b

be bx dxax

2 2

2

+ ∫ sin

= − + +12 1b

e bxa

be bx Cax axcos sin

e bx dxax sin∫=

+−

++a

a be bx

b

a be bx Cax ax

2 2 2 2sin cos

(for a, b not both 0)

e bx dx Cax sin =∫ (for a = b = 0)

83. sin ( cos )2 1

21 2cx dx cx dx∫ ∫= −

= − +1

2

1

42x

ccx Csin (for c ≠ 0)

sin2 cx dx C=∫ (for c = 0)

84. cos cos2 1

21 2cx dx cx dx= +∫ ∫ ( )

= + +1

2

1

42x

ccx Csin (for c ≠ 0)

cos2 cx dx x C= +∫ (for c = 0)

85. f xax b

cx d( ) = +

+

f xa cx d c ax b

cx d

ad bc

cx d′ = + +

+=

+( )

( ) – ( )

( )

–

( )2 2

(for c, d not both 0)(undefined for c = d = 0)

86. f (x) = (ax + b)n

f x na ax b n′ = + −( ) ( ) 1

(for a, b not both 0, or n ≥ 1)f ′(x) = 0 (for a = b = 0 and 0 ≤ n ≤ 1)

(undefined for a = b = 0 and n < 0)

87.ax b

cx ddx

a

c

b a c d

cx ddx

++

= + −+

∫ ∫ ( / )

= + + +ax

c

bc ad

ccx d C

–ln | |2 (for c ≠ 0)

ax b

cx ddx

a

dx

b

dx C

++

= + +∫ 22

(for c = 0, d ≠ 0)(undefined for c = d = 0)

88. ( )ax b dxax b

a nCn

n

+ = ++

++

∫ ( )

( )

1

1(for n ≠ −1, a ≠ 0)

( )ax b dxa

ax b Cn+ = + +∫ 1ln | |

(for n = −1, a ≠ 0)

( )ax b dx b x Cn n+ = +∫ (for a = 0)

89.x dx

x ax a x dx

2 2

2 2 1 21

22

+= +∫ ∫ −( ) ( )/

= ⋅ + + = + +1

22 2 2 1 2 2 2( ) /x a C x a C

90.x dx

a xa x x dx

2 2

2 2 1 21

22

−= − − −−∫∫ ( ) ( )/

= − ⋅ − + = − +1

22 2 2 1 2 2 2( ) /a x C a x C–

(for a ≠ 0)(undefined for a = 0)

91.dx

x a

d a

a a2 2 2 2 2+=

+∫∫ ( tan )

tan

θθ

= =∫ ∫a d

ad

sec

secsec

2 θ θθ

θ θ

= ln |sec θ + tan θ | + C1

= + + +ln1 12 2

1ax a

ax C

= + + +ln x a x C2 2


92.dx

a x

d a

a a2 2 2 2 2−=

−∫ ∫ ( sin )

sin

θθ

= = = + = +∫ ∫ −a d

ad C

x

aC

cos

cossin

θ θθ

θ θ 1

(for a ≠ 0)(undefined for a = 0)

93. f (x) = x2 sin ax ⇒ f ′(x) = 2x sin ax + ax2 cos ax

94. f (x) = x2 cos ax ⇒ f ′(x) = 2x cos ax − ax2 sin ax

95. x ax dx2 sin∫ u dvx 2 sinax2x –

1a cosax

2 –1a2 sinax

01a3 cosax–

–

+

+

= − + + +1 2 222 3a

x axa

x axa

ax Ccos sin cos

(for a ≠ 0)

x ax dx C2 sin =∫ (for a = 0)

96. x ax dx2 cos∫ u dvx 2 cosax2x 1

a sinax

2 –1a2 cosax

0 –1a3 sinax

+

+

–

–

= + − +1 2 222 3a

x axa

x axa

ax Csin cos sin

(for a ≠ 0)

x ax dx x C2 31

3cos = +∫ (for a = 0)

97. sinh coshax dxa

ax C= +∫ 1 (for a ≠ 0)

sinh ax dx C=∫ (for a = 0)

98. cosh sinhax dxa

ax C= +∫ 1(for a ≠ 0)

cosh ax dx x C= +∫ (for a = 0)

99. cos−∫ 1 ax dx u dvcos –1 ax 1–a

√1 – (ax)2x–

+

= +−

− ∫x axax dx

axcos

( )

1

21

= − − −− −∫x axa

ax a x dxcos 1 2 1 2 21

21 2[ ( ) ] ( )/

= − +−x axa

ax Ccos – ( )1 211

(for a ≠ 0)

cos− = +∫ 1

2ax dx x C

π(for a = 0)

100. sin−∫ 1 ax dx u dvsin–1 ax 1a

1 – (ax)2x

+

–

√

= −− ∫x axax dx

axsin

– ( )

1

21

= + − −− −∫x axa

ax a x dxsin 1 2 1 2 21

21 2[ ( ) ] ( )/

= + +−x axa

ax Csin – ( )1 211 (for a ≠ 0)

sin− =∫ 1 ax dx C (for a = 0)

101.1

1+∫ xdx Let u x= +1 .

x = (u − 1)2

dx = 2(u − 1) du

= = − ∫∫∫ 2 12 2

( – )u du

udu u du( / )

= 2u − 2 ln |u| + C= + − + +2 1 2 1( ) |x x Cln |

Or: 2 2 1 1x x C− + +ln | |Absolute values are optional because1 0+ x > .

102.1

1– xdx∫ Let u x= −1 .

x = (1 − u)2

dx = 2(u − 1) du

= = − ∫∫∫ 2 12 2

( – )u du

udu u du ( / )

= − +2 2u u Cln | |

= − − − +2 1 2 1( ) | |x x Cln

Or: − − − +2 2 1 1x x Cln | |

103.1

1 4+∫ xdx Let u x= +1 4 .

x = (u − 1)4

dx = 4(u − 1)3 du

= = − + −∫∫ 4 14 12 12 4

32( – )u du

uu u u du( / )

= − + − +4

36 12 43 2u u u u Cln | |

= + − + + +4

31 6 1 12 14 3 4 2 4( )x x x( ) ( )

− + +4 1 4ln x C

Or: 4

32 4 4 14 3 4 2 4 4

1( ) ln ,x x x x C− + − + +( ) | |


by expanding the powers or by starting withu x= 4 .

Absolute values are optional because 1 04+ x > .

104.1

3x xdx

+∫ Let u x= 1 6/ .

x = u6

dx = 6u5 du

=+

=+∫∫ 6 6

1

5

3 2

3u du

u u

u du

u

= − + −+

∫ 6 6 6

6

12u u

udu

(by long division)= − + − + +2 3 6 6 13 2u u u u Cln | |

= − + − + +2 3 6 6 13 6 6x x x x Cln ( )

105.1

1edx

x +∫ Let u ex= + 1.

ex = u2 − 1 x = ln (u2 − 1)

dxu du

u=

−2

12

= = −+

∫∫ 2

1

1

1

1

12

du

u u udu

– –(by partial fractions)

= − − + +ln ln| | | |u u C1 1

= + − − + + +ln ln( ) ( )e e Cx x1 1 1 1

106.1

1edx

x –∫ Let u ex= – .1

ex = u2 + 1 x = ln (u2 + 1)

dxu du

u=

+2

12

=+

= + = +− −∫ 2

12 2 12

1 1du

uu C e Cxtan tan –

107. a. Let t = x/2 and substitute, gettingcos x = 2 cos2 (x/2) − 1 andsin x = 2 sin (x/2) cos (x/2).

b. cossec ( / )

xx

= −2

212

= 2 2

2

2

2

– sec ( / )

sec ( / )

x

x

= ++

2 1 2

1 2

2

2

– [ tan ( / )]

tan ( / )

x

x

=+

1 2

1 2

2

2

– tan ( / )

tan ( / )

x

x, Q.E.D.

sinsin ( / )

cos ( / )cosx

x

xx= 2

2

222 ( / )

= 2 21

22tansec ( / )

( / )xx

=+2 2

1 22

tan ( / )

tan ( / )

x

x, Q.E.D.

c. u x u x dxdu

u= ⇒ = ⇒ =

+−tan tan( / )2 2

2

11

2

cos–

xu

u=

+1

1

2

2 and sin xu

u=

+2

1 2 from part b.

d.1

1+∫ cos xdx

=+

+

⋅+∫ 1

111

2

12

2

2– uu

du

u

=+ +

= ∫∫ 2

1 12 2

du

u udu

( ) ( – )

e.1

12

+= = + = +∫∫ cos

tanx

dx du u C x C( / )

108. a. seccos

x dxx

dx= ∫∫ 1

= + ⋅+

=∫ ∫1

1

2

1

2

1

2

2 2 2

u

u

du

u udu

– –

b. sec–

x dxu u

du= ++

∫∫ 1

1

1

1 = −ln | 1 − u | + ln |1 + u| + C

= + + = + +ln–

lntan ( / )

– tan ( / )

1

1

1 2

1 2

u

uC

x

xC

c. sec lntan ( / )

– tan ( / )x dx

x

xC= +

⋅+∫ 1 2

1 1 2

= + +lntan ( / ) tan ( / )

– tan ( / ) tan ( / )

ππ4 2

1 4 2

x

xC

= ln |tan (π/4 + x/2)| + C

d. i. sec ln tan |x dx x= +∫ | ( / / )π 4 20

1

0

1

= ln |tan (π/4 + 1/2)| − ln |tan π/4|= ln |tan (π/4 + 1/2)| = 1.226191…

ii. sec ln sec tan |x dx x x= +∫ |0

1

0

1

= ln |sec 1 + tan 1| − ln |1 + 0| =ln |sec 1 + tan 1| = 1.226191… , whichagrees with the answer in part i.

109.1

1

1

111

2

12

2

2– cos–

–xdx

uu

du

u=

+

⋅+∫∫

=+

= = +∫∫ 2

1 1

12 2 2

du

u u

du

u uC

( ) – ( – )

–

= −cot (x/2) + C

110.1

1

1

12

1

2

12

2+=

++

⋅+∫ ∫sin x

dx uu

du

u

=+ +

=+

=+

+∫∫ 2

1 2 1

1

12 2

du

u u

du

u uC

( ) ( )

–

=+

+–

tan ( / )

1

2 1xC


111.cos

– cos

–

––

x

xdx

uu

uu

du

u1

11

111

2

1

2

2

2

2

2= +

+

⋅+∫ ∫

=+

= −+

∫ ∫1

1

1 2

1

2

2 2 2 2

–

( )

u

u udu

u udu

= − − +−12 1

uu Ctan

= − − +−1

22 21

tan ( / )tan [tan ( / )]

xx C

= −cot (x/2) − x + C

Or: cos

– cos – cos

x

xdx

xdx

11

1

1= − +

∫ ∫

= − +∫ ∫dxx

dx1

1– cos= −x − cot (x/2) + C (using Problem 109)

Problem Set 9-121. Answers will vary.

Problem Set 9-13

Review ProblemsR0. Answers will vary.

R1. f ( x ) = x cos x ⇒f ′( x ) = x(−sin x) + (1) cos x = cos x − x sin x

⇒ + = ′∫x x dx C f x dxcos ( )

= −∫ (cos sin )x x x dx

= − ∫sin sinx x x dx

⇒ = − +∫ x x dx x x x Csin sin cos

x x dx x x xsin sin cos= −∫1

4

1

4

= sin 4 − 4 cos 4 − sin 1 + cos 1 = 1.5566…

Numerically,

x x dxsin . .≈∫ 1 55661

4

K

R2. 5 2x x dxsin∫ u = 5x dv = sin 2x dx

du = 5 dx v x= − 1

22cos

= −

+ ∫5

1

22

1

22 5x x x dxcos cos ( )

= − + +5

22

5

42x x x Ccos sin

R3. a. x x dx3 2cos∫ u dvx3 cos 2x

3x2 12 sin 2x

6x –14 cos 2x

6 –18 sin 2x

0116 cos 2x

+

+

–

–

+

= +1

22

3

423 2x x x xsin cos

− − +3

42

3

82x x x Csin cos

b. e x dxx4 3sin∫13

9

u dve4x sin 3x

4e 4x – cos 3x

16e –1 sin 3x+

–

+

4x

= +–1

33

4

934 4e x e xx xcos sin

–16

934e x dxx∫ sin

⇒ ∫25

934e x dxx sin

= − + +1

33

4

934 4


⇒ ∫ e x dxx4 3sin

= − + +3

253

4

2534 4e x e x Cx xcos sin

c. x x dx(ln )2∫------------------------2

------------------------

u dv(ln x)2 x

2 ln x · 1x

1x2

ln x x1x

12x2

112x

014x2

+

+

–

–

= − + +1

2

1

2

1

42 2 2 2x x x x x C( )ln ln

d. Slice parallel to the y-axis. Pick a samplepoint (x, y) on the graph, within the slice.dV = 2πx · y · dx = 2πx(x ln x) dx

= 2πx2 ln x dx

V x x dx= ∫2 2

1

2

π ln

----------------------

u dvln x x2

1x

13 x3

113 x2

019 x3 +

–

+


= −

2

1

3

1

93 3

1

2

π x x xln

= − + = −16

32

16

9

2

9

16

32

14

9π π π π πln ln

= 6 7268. K

R4. a. cos30 dx∫ u dvcos 29x cosx

29 cos 28x sinx sinx+

––

= + ∫cos sin cos sin29 28 229x x x x dx

= + −∫cos sin cos )29 2829 1x x x x dx( cos2

⇒ ∫30 30cos dx

= + ∫cos sin cos29 2829x x x dx

⇒ ∫ cos30 dx

= + ∫1

30

29

3029 28cos sin cosx x x dx

b. sec sec tan sec6 4 41

5

4


= +1

5

4

154 2sec tan sec tanx x x x

+ ∫8

152sec x dx

= +1

5

4

154 2sec tan sec tanx x x x

+ +8

15tan x C

c. tan tan tann nx dx x x dx=∫ ∫ − ( )2 2

= −−∫ tan secn x x dx2 2 1( )

= −− −∫ ∫tan sec tann nx x dx x dx2 2 2

=−

−− −∫1

11 2

nx x dxn ntan tan

R5. a. cos ( sin ) (cos )5 2 21x dx x x dx= −∫ ∫= − +∫ ( )( )1 2 2 4sin sin cosx x x dx

= − + +sin sin sinx x x C2

3

1

53 5

b. sec (tan ) (sec )6 2 2 21x dx x x dx= +∫∫= + +∫ (tan tan )(sec )4 2 22 1x x x dx

= + + +1

5

2

35 3tan tan tanx x x C

c. sin ( cos )2 71

21 14x dx x dx= −∫ ∫

= − +1

2

1

2814x x Csin

d. sec3 x dx∫ u dvsecx sec 2x

secx tanx tanx–+

= − ∫sec tan tan secx x x x dx2

= − −∫sec tan sec secx x x dx( )2 1

= + −∫ ∫sec tan sec secx x x dx x dx3

2 3sec x dx∫= + + +sec tan |sec tan |x x x x Cln

sec3 x dx∫= + + +1

2

1

2sec tan | tan |x x x x Cln sec

e. tan tan tan9 9 932 32 32dx dx x C= = +∫ ∫( ) ( )

f. r = 9 + 8 sin θ

dA r d d= = +1

2

1

29 82 2θ θ θ( sin )

A d= + +∫1

264 144 812

0

4

( sin sin )/

θ θ θπ

= − + +∫1

232 1 2 144 81

0

4

[ ( cos ) sin ]/

θ θ θπ

d

= −

− +16

1

22 72

81

2 0

4

θ θ θ θπ

sin cos/

= − − + +4 8 36 281

872π π

= + − =113

864 36 2 57 4633π . K

R6. a. x dx2 49−∫

u

v

θ

7

x √x 2 – 49

Let x

x7

7= =sec . secθ θ,


xx2 149 77

− = = −tan , secθ θ

∴ −∫ x dx2 49

= ∫ ( tan )( sec tan )7 7θ θ θ θd

= ∫49 2sec tanθ θ θd

= −

∫ ∫49 3sec secθ θ θ θd d


= + +49

1

2

1

2sec tan |sec tan |θ θ θ θln

− +

+ln |sec tan |θ θ C1

= − + +49

2

49

2 1sec tan | tan |θ θ θ θln sec C

= ⋅ ⋅ −49

2 7

49

7

2x x

− + − +49

2 7

49

7

2

1lnx x

C

= − − + −1

249

49

2492 2x x x xln

+ +49

27 1ln C

= − − + − +1

249

49

2492 2x x x x Cln

b. x x dx x dx2 210 34 5 9− + = − +∫∫ ( )

u

v

θ

3

x – 5

√ (x – 5)2 + 9

Let x − =5

3tan .θ

x = 5 + 3 tan θ, dx = 3 sec2 θ dθ,

( ) sec , tanxx− + = = −−5 9 3

5

32 1θ θ

∴ +∫ ( – )x dx5 92

= =∫ ∫( sec )( sec ) sec3 3 92 3θ θ θ θ θd d

= + + +9

2

9

2 1sec tan ln |sec tan |θ θ θ θ C

=− + −9

2

5 9

3

5

3

2( )x x

+− +

+ − +9

2

5 9

3

5

3

2

1ln( )x x

C

= − + ⋅ −1

25 9 52( ) ( )x x

+ − + + − − +9

25 9 5

9

232

1ln ( ) lnx x C

= +1

25 10 342( – ) –x x x

+ + + − +9

210 34 52ln –x x x C

c. 1 0 25 2– . x dx∫

u

v

θ

10.5x

√ 1 – 0.25x2

Let 0 5

1

. x = sin θ.

x = 2 sin θ, dx = 2 cos θ dθ,

1 0 252

2 1– . cos , sinxx= = −θ θ

1 0 25 22– . (cos )( cos )x dx d∫ ∫= θ θ θ

= = +∫ ∫2 1 22cos ( cos )θ θ θ θd d

= + + = + +θ θ θ θ θ1

22sin sin cosC C

= + − +−sin .1 2

2

1

21 0 25

xx x C

d. Slice region vertically. Pick sample point(x, y) on the upper branch of the circle,within the strip.

dA y dx x dx= =2 2 25 2–

u

v

θ

5x

√ 25 – x 2

Let x

x dx d5

5 5= = =sin sin cosθ θ θ θ. , ,

25 55

2 1– cos , sinxx= = −θ θ

A x dx= −∫ 2 25 2

3

4

==

=

∫2 5 53

4

cos ( cos )θ θ θdx

x

= +=

=

∫25 1 23

4

( cos )θ θdx

x

= + ==25 12 5 2 3

4θ θ. sin xx

= + ==25 25 3

4θ θ θsin cos xx

= + ⋅ ⋅ −−255

255

1

5251 2

3

4

sinx x

x

= + − −− −25 0 8 4 9 25 0 6 3 161 1sin . sin . = 25(sin− 1 0.8 − sin− 1 0.6) = 7.0948…

R7. a.( )

– –

( )

( )( – )

6 1

3 4

6 1

1 42

x dx

x x

x dx

x x

+ = ++∫ ∫

=+

+−

∫ 1

1

5

4x xdx

= ln |x + 1| + 5 ln |x − 4| + C


b.5 21 2

1 2 3

2x x

x x xdx

– –

( – )( )( – )+∫= +

+−

∫ 3

1

4

2

2

3x x xdx

– –

= 3 ln |x − 1| + 4 ln |x + 2| − 2 ln |x − 3| + C

c.5 3 45

9

5 3 45

9

2

3

2

2

x x

x xdx

x x

x xdx

+ ++

= + ++∫ ∫ ( )

= ++

= + +∫ −5 3

95

321

x xdx x

xCln | | tan

(The second integral may be found byinspection or by trigonometric substitution.)

d.5 27 32

4

2

2

x x

x xdx

+ ++∫ ( )

= ++

−+

∫ 2 3

4

1

4 2x x xdx

( )

= 2 ln | x | + 3 ln | x + 4 | + (x + 4)− 1 + C

= + ++

+ln | ( ) |x xx

C2 341

4

e.dy

dxy y= 0 1 3 8. ( – )( – )

dy

y ydx

( – )( – ).

3 80 1=∫ ∫

−−

+−

=∫ ∫1 5

3

1 5

80 1

/ /.

y ydy dx

− − + − = +1

53

1

58 0 1 1ln | | ln | | .y y x C

−ln | y − 3 | + ln | y − 8 | = 0.5x + C

Substituting (0, 7) gives

C = –ln 4 + ln 1 = –ln 4.

ln–

. lny

yx

8

30 5 4

−= −

y

ye ex x–

.. ln .8

30 250 5 4 0 5

−= =−

− =y

ye x–

–. .8

30 25 0 5

((y – 8)/(y – 3) < 0 because (0, 7) is on thegraph)

ye x= +

+3

5

1 0 25 0 5. .

The graph shows that solution fits slope field.

x

y

7

R8. a.

2

1

x

y

b. f ( x ) = sec− 1 3x

′ = =f xx x x x

( )| | ( ) – | | –

3

3 3 1

1

9 12 2

c. tan−∫ 1 5x dx u dvtan–1 5x 1

51 + 25x2 x

+

–

= −+

− ∫x xx

xdxtan 1

255

1 25

= −+

− ∫x xx

x dxtan ( )125

1

10

1

1 2550

= − + +−x x x Ctan ln | |1 251

101 25

(Absolute values are optional because1 + 25x2 > 0.)

d. “Obvious” way: Slice the region vertically.Pick a sample point (x, y) on the graph,within the strip.dA = y dx = cos− 1 x dx

A x dx x x x= = − −− −∫ cos cos1 1 2

0

1

0

1

1

= − − + =−cos 11 0 0 1 1Easier way: Slice horizontally. Pick a samplepoint (x, y) on the graph within the strip.dA = x dy = cos y dy

A y dy y= = = − =∫ cos sin/ /

0

2

0

2

1 0 1π π

R9. a.

x

y

1

1

b.

x

y

1

1


c. h(x) = x2 sech xh′(x) = –x2 sech x tanh x + 2x sech x

d. f ( x ) = sinh− 1 5x

′ =+

f xx

( )5

25 12

e. tanhcosh

sinh31

33x dx

xx dx=∫ ∫

= +1

33ln |cosh |x C

(Absolute values are optional becausecosh 3x > 0.)

f. cosh−∫ 1 7x dx u dvcosh –1 7x 1

749x2 – 1

x+

–

√

= −−

− ∫x xx

xdxcosh 1

27

7

49 1

= − −− −∫x x x x dxcosh ( ) ( )/1 2 1 271

1449 1 98

= − ⋅ − +−x x x Ccosh ( ) /1 2 1 271

142 49 1

= − − +−x x x Ccosh 1 271

749 1

g. cosh2 x – sinh2 x

= + − −1

2

1

22 2( ) ( )– –e e e ex x x x

= + + − − +1

42

1

422 2 2 2( ) ( )– –e e e ex x x x

= 1, Q.E.D.

h. The general equation is y kk

x C= +cosh .1

y = 5 at x = 0 ⇒ 5 = k cosh 0 + C⇒ C = 5 – k

y x kk

k= = ⇒ = + −7 3 73

5at cosh

⇒ = −23

kk

kcosh

⇒ k = 2.5269… (solving numerically)

yt= + −2 5269

2 52695 2 5269. cosh

..K

KK

y( ) .10 68 5961 20= …

= + −2 52692 5269

5 2 5269. cosh.

.KK

Kx

⇒ x = ±6.6324… (solving numerically)

R10. a. ( ) lim ( ). .x dx x dxb

b

− = −−

→∞

∞−∫ ∫2 21 2

3

1 2

3

= − −→∞

−lim ( ) .

b

b

x5 2 0 2

3

= + =→∞

lim [– ( – ) ]– .

bb5 2 5 50 2


b. tan lim tan//

x dx x dxa a

=→ −∫ ∫ππ 22

0 0

=→ −lim ln |sec |

/a a

xπ 2

0

= − = −∞→ −lim (ln |sec | ln |sec |)

/aa

π 20


c. x dx−

−∫2 3

1

1/

= +→

−

− →

−− +∫ ∫lim lim/ /

b

b

a ax dx x dx

0

2 3

1 0

2 31

= +→ − →− +lim lim/ /

b

b

a ax x

0

1 3

1 0

1 31

3 3

= − − + − =→ →− +lim [ ( )] lim ( )/ /

b ab a

0

1 3

0

1 33 3 3 3 6


d. xx

xdx− −

−

∫ | |1

10

4

= + + −→ →− +∫ ∫lim ( ) lim ( )

b

b

a ax dx x dx

1 0 1

4

1 1

= +

+ −

→ →− +

lim lim/ /

b

b

a a

x x x x1

3 2

0 1

3 242

3

2

3

= + −

→ −

lim /

bb b

1

3 22

30

+ ⋅ − − +

→ +

lim / /

aa a

1

3 2 3 22

34 4

2

3

= ⋅ + + ⋅ − − ⋅ +2

31 1

2

34 4

2

31 13 2 3 2 3 2/ / /

= + − + =116

34 1

10

3

The integral converges to 10

33 333= . .K

e. x dxp−∞

∫1 converges if p > 1 and diverges

otherwise.

R11. a. f x x x f x xx

x( ) sin ( ) sin= ⇒ ′ = +

−− −1 1

21

b. I = −∫ x x dxsin 1 u dvsin–1 x x

11 – x2

12x2 –

+

√

= −− ∫1

2

1

2 1

2 12

2x x

x dx

xsin

–

Let I1

2

21= ∫ x dx

x– and x = sin θ.

∴ = − =dx d xcos , cos ,θ θ θ 1 2

θ = sin− 1 x


I1

22= =∫ ∫sin cos

cossin

θ θ θθ

θ θdd

= − = − +∫1

21 2

1

2

1

42( cos ) sinθ θ θ θd C

= − +1

2

1

2θ θ θsin cos C

= − − +−1

2

1

211 2sin x x x C

∴ = − + − +− − I1

2

1

4

1

412 1 1 2x x x x x Csin sin

c.d

dxe e ex x xtanh sec= h2

d. ( )x x dxx x

dx3 13

1− =−

−∫ ∫=

+∫ 1

1 1x x xdx

( – )( )

= − +−

++

∫ 1 1 2

1

1 2

1x x xdx

/ /

= − + − + + +ln | | ln | | ln | |x x x C1

21

1

21

e. f ( x ) = (1 − x2)1/2

′ = − − = − −− −f x x x x x( ) ( ) ( ) ( )/ /1

21 2 12 1 2 2 1 2

f. I = −∫ ( ) /1 2 1 2x dx

Let x = sin θ.∴ = − = dx d xcos , ( ) cos ,/θ θ θ1 2 1 2

θ = sin− 1 x

∴ = ⋅ =∫ ∫ I cos cos cosθ θ θ θ θd d2

= + = + +∫1

21 2

1

2

1

42( cos ) sinθ θ θ θd C

= + +1

2

1

2θ θ θsin cos C

= + − +−1

2

1

211 2sin x x x C

g. g x x g x xx

( ) ( ) ( )2= ⇒ ′ = ⋅ln ln21

h. x x dxln∫--------------------

u dvln x x

1x

12x2

112x

014x2

+

–

+

= − +1

2

1

42 2x x x Cln

R12. For ( ) ,/9 2 1 2− −∫ x x dx the x dx can be

transformed to the differential of the insidefunction by multiplying by a constant,

− − − = − − +−∫1

29 2 92 1 2 2 1 2( ) ( ) ( ) ,/ /x x dx x C

and thus has no inverse sine.

For ( ) ,/9 2 1 2− −∫ x dx transforming the dx to the

differential of the inside function, −2x dx,requires multiplying by a variable. Because theintegral of a product does not equal the product ofthe two integrals, you can’t divide on the outsideof the integral by −2x. So a more sophisticatedtechnique must be used, in this case,trigonometric substitution. As a result, aninverse sine appears in the answer:

( ) sin/93

2 1 2 1− = +− −∫ x dxx

C

Concept Problems

C1. sech x dx x dx∫ ∫= −1 2tanh

u

v

θ

1tanh x

√ 1 – tanh2x

Let tanh x = sin θ.∴ x = tanh− 1 (sin θ) and θ = sin− 1 (tanh x)

dx d d=−

⋅ =1

1

12sin

coscosθ

θ θθ

θ

1 12 2– tanh – sin cosx = =θ θ

∴ = ⋅ =∫ ∫ ∫sec coscos

h x dx d dθθ

θ θ1

= + = +−θ C x Csin (tanh ) ,1 Q.E.D.

sec sin (tanh )h0

11

0

1

∫ = −x dx x

= sin− 1 (tanh 1) − sin− 1 (tanh 0)= sin− 1 (tanh 1) = 0.86576948…Numerical integration gives 0.86576948… ,which agrees with the exact answer.

C2. From sinh 2A = 2 sinh A cosh A,let A = x/2, sosinh x = 2 sinh (x/2) cosh (x/2) ⇒ csch x

= =⋅ ⋅

1 1

2 2 2sinh sinh ( / ) cosh ( / )x x x

cscsinh ( / ) cosh ( / )

sec ( / )

sec ( / )h

h

hx

x x

x

x=

⋅ ⋅⋅1

2 2 2

2

2

2

2

=⋅

=sec ( / )

tanh ( / )

sec ( / )

tanh ( / )

h h22

2

2 2

12

2

2

x

x

x

x

∴ = ⋅

∫ ∫csc

tanh ( / )sec ( / )h h2x dx

xx dx

1

2

1

22

= ln | tanh (x/2) | + C, Q.E.D.

csc ln | tanh ( / )|h1

2

1

2

2∫ =x dx x

= =ln | tanh | ln | tanh |1 1 2 0 49959536– ( / ) . K

Numerical integration gives 0.49959536… .


C3. From sin 2A = 2 sin A cos A, let A = x/2, sosin x = 2 sin (x/2) cos (x/2)

cscsin sin ( / ) cos ( / )

xx x x

= =1 1

2 2 2

= ⋅1

2 2 2

2

2

2

2sin ( / ) cos ( / )

sec ( / )

sec ( / )x x

x

x

= =sec ( / )

tan ( / )

sec ( / )

tan ( / )

22

2

2 2

12

2

2

x

x

x

x

∴ = ⋅

∫ ∫csc

tan ( / )sec ( / )x dx

xx dx

1

2

1

222

= ln | tan (x/2) | + C, Q.E.D.Or:Let u = tan (x/2), as in Problem 107 of ProblemSet 9-11.

Then anddxdu

ux

u

u=

+= +2

1

1

22

2

csc

∴ = + ⋅+∫ ∫csc x dx

u

u

du

u

1

2

2

1

2

2

= = + = +∫ ( / ) ln | | ln | tan ( / )| ,1 2u du u C x C

Q.E.D.Confirmation:

csc ln | tan ( / )|. .0 5

1

0 5

1

2∫ =x dx x

= − =ln tan ln tan .1

2

1

40 7605K

Numerical integration gives 0.7605… .

Note that tan ( / )sin ( / ) cos ( / )

cos ( / )x

x x

x2

2 2 2

2 22=

=+

=+

sin

cos csc cot,

x

x x x1

1 so

ln | tan ( / )| ln |csc cot | .x x x2 = − +

C4. Ax

dx=+−∞

∞

∫ 1

1 2

=+

++→−∞ →∞∫ ∫lim lim

a a b

b

xdx

xdx

1

1

1

12

0

20

= +→−∞

−

→∞

−lim tan lim tana a b

bx x1 0 1

0

= − + −→−∞

−

→∞

−lim ( tan ) lim (tan )a b

a b0 01 1

= −(−π/2) + (π/2) = πC5. Prove that f (x) = ln x is unbounded above.

Proof:Assume f (x) = ln x is not unbounded above.Then there is a number M > 0 such thatln x < M for all x > 0.Let x = eM+ 1.Then ln x = ln eM+ 1 = M + 1.∴ ln x > M, which is a contradiction.∴ the assumption is false, and ln x is unboundedabove, Q.E.D.

Chapter Test

T1. sin cos sin5 61

6x x dx x C= +∫

T2. x x dx3 6sinh∫ u dvx3 sinh 6x

3x2 16 cosh 6x

6x 136 sinh 6x

61

216 cosh 6x

01

1296 sinh 6x+

–

–

+

+

= −1

66

1

1263 2x x x xcosh sinh

+ − +1

366

1

2166x x x Ccosh sinh

T3. cos−∫ 1 x dx

√

u dvcos –1 x 1

–1

1 – x2x

+

–

= +− ∫x xx

xdxcos

–

1

21

= − − −− −∫x x x x dxcos ( ) ( )/1 2 1 21

21 2

= − − +−x x x Ccos ( )( ) /1 2 1 21

22 1

= − − +−x x x Ccos 1 21

T4. sec3∫ x dx

= + + +1

2

1

2sec tan ln |sec tan |x x x x C

T5. e x dxx2 5cos∫ u dve 2x cos 5x

2e 2x 15 sin 5x

4e 2x –125 cos 5x

+

–

+

= +1

55

2


− ∫4

2552e x dxx cos

29

2552e x dxx∫ cos

= + +1

55

2



295

2



T6. ln 3x dx∫ u dvln 3x 1

1/x x–+

= − = − +∫x x dx x x x Cln ln3 3

T7. f (x) = sech3 (e5x) ⇒f ′(x) = 3 sech2 (e5x) ⋅ [−sech (e5x)

tanh (e5x)] ⋅ 5e5x

= −15e5x sech3 (e5x) tanh (e5x)

T8. g x x g xx

( ) sin ( )–

= ⇒ ′ =−1

2

1

1

T9. f (x) = tanh− 1 x ⇒ tanh f (x) = x, |x| ≤ 1sech2 f(x) ⋅ f ′(x) = 1[1 − tanh2 f (x)] ⋅ f ′(x) = 1(1 – x2) ⋅ f ′(x) = 1

′ = <f xx

x( )–

, | |1

112

′ = = =f ( . )– . .

.0 61

1 0 36

1

0 641 5625

Numerically, f ′(0.6) ≈ 1.5625… (depending onthe tolerance of the calculator).

T10. General equation is y kk

x C= +cosh .1

y = 1 at x = 0 ⇒ 1 = k cosh 0 + C ⇒C = 1 − k

y = 3 at x kk

k= ⇒ = +5 35

1cosh –

Solving numerically, k ≈ 6.5586… .

y x= + −6 5586

1

6 55861 6 5586. K

KKcosh

..

T11. a. i. I =+

=∫ ∫x

x xdx

x

xdx

–

–

–

( – ) –

3

6 5

3

3 42 2

u

v

θ

x – 3

2

√ (x – 3)2 – 4

Let . – ,x

x–

sec sec3

23 2= =θ θ


( – ) – tan sec–

xx

3 4 23

22 1= =θ θ, –

∴ = ∫ I( sec )( sec tan )

tan

2 2

4 2

θ θ θ θθ

d

= ⋅ = +∫ 1 21tan

sec ln | tan |θ

θ θ θd C

= − − +ln ( )1

23 42

1x C

= − − +ln ( )x C3 42

= − + +1

26 52ln | |x x C

ii. x

x xdx

x xdx

−− +

=−

+−

∫ ∫3

6 5

1 2

1

1 2

52

/ /

= + +1

21

1

25ln | | ln | |x x C– –

= +1

21 5ln | |( – )( – )x x C

= + +1

26 52ln | | ,x x C–

which agrees with part a.

iii.x

x xdx

–

–

3

6 52 +∫

=+

⋅ −∫1

2

1

6 52 62x x

x dx–

( )

= + +1

26 52ln | |x x C– , as in parts a and b.

b. See parts i, ii, and iii.

T12. cos ( cos )2 1

21 2x dx x dx= +∫ ∫

= + +1

2

1

42x x Csin

T13. a. i. cos sin cos5 2 21x dx x x dx=∫ ∫ ( – )

= +∫ ( – sin sin ) cos1 2 2 4x x x dx

= + +sin sin sinx x x C–2

3

1

53 5

ii. cos cos sin cos5 4 31

5

4


= +1

5

4

154 2cos sin cos sinx x x x

+ ∫8

15cos x dx

= +1

5

4

154 2cos sin cos sinx x x x

+ +8

15sin x C

b.1

5

4

15

8

154 2cos sin cos sin sinx x x x x+ +

= 1

51 2 2( – sin ) sinx x

+ +4

151

8

152( – sin ) sin sinx x x

= +1

5

2

5

1

53 5sin sin sinx x x–

+ +4

15

4

15

8

153sin sin sinx x x–


= + +

− +

1

5

4

15

8

15

2

5

4

153sin sinx x

+ 1

55sin x

= +sin sin sinx x x– 2

3

1

53 5

T14. xe dxx−∞

∫ 0 1

0

. u dvx e –0.1x 1 –10e –0.1x 0 100e –0.1x

+

––

=→∞

lim (– – )– . – .

b

x xb

xe e10 1000 1 0 1

0

= + +→∞

lim (– – )– . – .

b

b bbe e10 100 0 1000 1 0 1

= + +

→∞

lim .b b

b

e–

10 1001000 1

= +

→∞

lim. .b be

–10

0 11000 1 (by l’Hospital’s rule)

= 100



Chapter 10—The Calculus of Motion—Averages,Extremes, and Vectors

Problem Set 10-11. v(t) = 100(0.8)t − 30 = 100et ln 0.8 − 30 = 0

⇒ e tt ln . ln .

ln ..0 8 0 3

0 3

0 85 3955= ⇒ = =. K

v becomes negative after t0 ≈ 5.40 min.

2. s v dt e dtttt

up = = ∫∫ ( – )ln .100 300 8

00

00

= 151.8341… (numerically) ≈ 151.8 ft

s v dt e dtt

ttdown = − = −∫∫ ( – )ln .100 300 8

1010

00

= 51.8110… (numerically) ≈ 51.8 ftDistance = sup + sdown = 203.6452… ≈ 203.6 ft

3. Displacement = sup − sdown = 100.0231… ≈100.0 ftThe displacement is positive, so Calvin isupstream of his starting point.

4. Displacement = ∫ ( – )ln .100 300 8

0

10

e dtt

= 100.0231… (numerically) ≈ 100.0 ft

5. Distance = = −∫∫ | | | | .v dt e dtt100 300 8

0

10

0

10ln

= 203.6452… (numerically) ≈ 203.6 ft

Problem Set 10-2

Q1. 120 mi Q2. 25 mi/hQ3. 1.25 h Q4. f ′(x) = 1/xQ5. x ln x − x + C Q6. f ′(t) = sec2 t

Q7. g′(t) = sech2 t Q8.1

33x C+

Q9.1

22

lnx C+ Q10. ln lnln2 2 22ex x=

1. a. v(t) = t2 − 10t + 16 on [0, 6]v(t) = (t − 2)(t − 8) = 0 ⇔ t = 2 or 8 sv(t) > 0 for t in [0, 2). v(t) < 0 for t in (2, 6].

b. For [0, 2), displacement

= + =∫ ( – )t t dt2

0

2

10 16 142

3

Distance = 142

3ft

For (2, 6], displacement

= + = −∫ ( – )t t dt2

2

6

10 16 262

3

Distance = 2 ft62

3

c. Displacement = + = −∫ ( – )t t dt2

0

6

10 16 12 ft

Distance = − + =∫ | | ftt t dt2

0

6

10 16 411

3

d. Displacement = + −

= −14

2

326

2

312 ft

Distance = + =142

326

2

341

1

3ft

e. a(t) = v′(t) = 2t − 10a(3) = 2(3) − 10 = −4 (ft/s)/s

2. a. v(t) = tan 0.2t on [10, 20]v(t) = 0 ⇔ t = … 0, 5π, 10π, … = 5π in[10, 20]v(t) is infinite ⇔ t = … 2.5π, 7.5π, … ,none of which is in [10, 20].v(t) < 0 for t in [10, 5π). v(t) > 0 for t in(5π, 20].

b. For [10, 5π), displacement = ∫ tan 0 210

5

. t dtπ

= 5 ln | sec π | − 5 ln | sec 2 | = −4.3835…Distance = 4.3835… ≈ 4.38 cm

For (5π, 20], displacement = ∫ tan 0 25

20

. t dtπ

= 5 ln |sec 4| − 5 ln | sec π | = 2.1259…Distance = 2.1259… ≈ 2.13 cm

c. Displacement = = −∫ tan 0 2 2 257610

20

. .t dt K ≈

−2.26 cm

Distance = = ≈∫ | . | .tan 0 2 6 509510

20

t dt K

6.51 cm

d. Displacement = −4.3835… + 2.1259… =−2.2576… ≈ −2.26 cmDistance = −(−4.3835…) + 2.1259… =6.5095… ≈ 6.51 cm

e. a ( t) = v′(t) = 0.2 sec2 ta ( 15) = 0.2 sec2 3 = 0.2040… ≈ 0.20 (cm/s)/s

3. a. v t t( ) = −secπ24

2 on [1, 11]

v(t) = 0 when

cos .π24

0 5 8t t= ⇔ = in [1, 11]

v(t) < 0 for t in [1, 8). v(t) > 0 for t in(8, 11].


b. For [1, 8), displacement =

∫ sec –

π24

21

8

t dt

= + −24

3 316

ππ π

ln sec tan

− + +24

24 242

ππ π

ln sec tan

= −4.9420…Distance ≈ 4.94 kmFor (8, 11], displacement

=

∫ sec –

π24

28

11

t dt

= + −24 11

24

11

2422

ππ πln sec tan

− + +24

3 316

ππ π

ln sec tan

= 4.7569…Distance ≈ 4.76 km

c. Displacement =

=∫ sec –

π24

21

11

t dt

− ≈ −0 1850 0 19. kmK .

Distance = − = ≈∫ secπ24

2 9 69901

11

t dt . K

9.70 km

d. Displacement = −4.9420… + 4.7569… =−0.1850… ≈ −0.19 kmDistance = −(−4.9420…) + 4.7569… =9.6990… ≈ 9.70 km

e. a t v t t t( ) ( ) = ′ = π π π24 24 24

sec tan

a ( 6) = 0.1851… ≈ 0.19 (km/h)/h

exactly π24

2

4. a. v(t) = t3 − 5t2 + 8t − 6 on [0, 5]v(t) = (t − 3)(t2 − 2t + 2) = 0 ⇔ t = 3 in [0, 5]v < 0 for t in [0, 3). v > 0 for t in (3, 5].

b. For [0, 3), displacement =

( – – )t t t dt3 2

0

3

5 8 6 63

4+ = −∫

Distance = 63

4mi

For (3, 5], displacement =

( – – )t t t dt3 2

3

5

5 8 6 242

3+ =∫

Distance = 242

3mi

c. Displacement =

( – – )t t t dt3 2

0

5

5 8 6 1711

12+ =∫ mi

Distance = − + − =∫ | | mit t t dt3 2

0

5

5 8 6 315

12

d. Displacement = − + =63

424

2

317

11

12mi

Distance = − −

+ =6

3

424

2

331

5

12mi

e. a ( t) = v′(t) = 3t2 − 10t + 8a(2.5) = 1.75 (mi/min)/min

5. a t t v( ) , ( ) , on [ , ]/= = −1 2 0 18 0 16

v t t dt t C v C( ) ; ( ) 8/ /= = + = − ⇒ = −∫ 1 2 3 22

30 18 1

v t t( ) /= −2

3183 2

Displacement ft= −

= −∫ 2

318 14

14

153 2

0

16

t dt/

Distance ft/= − =∫ 2

318 179

7

153 2

0

16

t dt

6. a t t v( ) , ( ) , on [ . , . ]= =−1 1 0 0 4 1 6

v t t dt t C t v C( ) ( ); ( )= = + > = ⇒ =−∫ 1 0 1 0 0ln

v t t( ) = ln

Displacement .= = − …∫ ln.

.

t dt 0 08140 4

1 6

≈ −0 081. cm

Distance | | .= = …∫ ln.

.

t dt 0 38540 4

1 6

≈ 0 385. cm

7. a t t v( ) , ( ) , on [ , ]= = −6 0 9 0sin πv t t dt t C v( ) ; ( )= = − +∫ 6 6 0sin cos

= − ⇒ = −9 3Cv(t) = −6 cos t − 3

Displacement = =∫ (– cos – )6 30

t dtπ

−9.4247… ≈ −9.42 km (exact: −3π km)

Distance | | .= − − = …∫ 6 3 13 53380

cos t dtπ

≈ 13.53 km (exact: 6 3 + π )

8. a(t) = sinh t, v(0) = −2, on [0, 5]

v t t dt t C( ) = = +∫ sinh cosh

v(0) = −2 ⇒ C = −3v(t) = cosh t − 3

Displacement = =∫ (cosh – )t dt30

5

59.2032… ≈ 59.20 mi (exact: sinh 5 − 15)

Distance | |

. . mi

= −

= ≈∫ cosh t dt3

64 1230 64 120

5

K

9. a. v = t1/2 − 2 = 0 ⇔ t = 4 s;v < 0 if t < 4, v > 0 if t > 4

b. Displacement ft= =∫ ( – )/t dt1 2

1

9

2 11

3

c. Distance | ft= − =∫ | /t dt1 2

1

9

2 4


10. a. v t t= = = … − −sin 2 01

20

1

2at , , , , ,π π π

π π, , 3

2…

sin ,2 0 01

2t ≥

on , soπ

Distance cm= =∫ sin/

2 10

2

t dtπ

b. Displacement cm= =∫ sin.

2 10

4 5

t dtπ

Distance cm= =∫ sin.

2 90

4 5

t dtπ

Or: The regions where the graph is belowthe x-axis cancel out the regions where thegraph is above the axis, leaving only oneuncancelled region above the graph, soDisplacement = area from part a = 1 cm. Theabsolute values of the regions above andbelow the graph are the same, so Distance =9 times the area from part a = 9 cm.

11. a. v = 60 − 2t

Displacement ft= =∫ ( – )60 2 30010

40

t dt

b. Distance | | ft= − =∫ 60 2 50010

40

t dt

12. a. a tt t

t( )

in =

>

40 0 015 9 8 0 100

9 8 100

cos . – . , [ , ]

– . ,

For t in [0, 100],

v t t dt

t t C

( ) ( . . )

. .

= −

= − +

∫ 40 0 015 9 8

40

0 0150 015 9 8

cos

.sin

v(0) = 0 ⇒ C = 0

For , ( ) . .t v t dt t C> = − = − +∫100 9 8 9 8

v( ) . .10040

0 0151 5 980 1679 986= − = …

.sin

⇒ C = 1679.986… + 980 = 2659.986…

v t t t t

t t( ) .

sin . . , [ ,

. . ,= −

− + >

40

0 0150 015 9 8 0

9 8 2659 986 100

in 100]

K

3030

100

t

a(t)

100

1000

t

v(t)

b. a t= = =−01

0 015

9 8

401at

.cos

.

88 2184 88 2. . sK ≈v = 0 at t = 2,659.986…/9.8 = 271.4272… ≈271.4 s

c. Displacement ( )= ∫ v t dt0

300

=

∫ 40

0 0150 015 9 8

0

100

.sin . – .t t dt

+ +∫ (– . . )9 8 2659 986

300

t dtK100

≈ 116202.27… + 139997.32… ≈ 256,200 m

Distance = 116202.27… +

| . . |− +∫ 9 8 2659 986

100

300

t dtK

= 116202.27… + 147998.09… ≈ 264,200 mThe distance is greater than the displacement,which agrees with the fact that the velocitybecomes negative at t = 271.4… s.

d. v(300) = −9.8(300) + 2,659.986… =−280.0133… , so the rocket is movingdownward (falling) at about 280 m/s.

13. a.tend

saav

(mi/h)/svend

mi/hvav

mi/hsend

mi

0 — 0 — 0

5 2.95 14.75 7.375 0.0102…

1 0 3.8 33.75 24.25 0.0439…

1 5 1.75 42.5 38.125 0.0968…

2 0 0.3 4 4 43.25 0.1569…

2 5 0 4 4 4 4 0.2180…

3 0 0 4 4 4 4 0.2791…

3 5 0 4 4 4 4 0.3402…

4 0 −0.2 4 3 43.5 0.4006…

4 5 −0.9 38.5 40.75 0.4572…

5 0 −2.6 25.5 3 2 0.5017…

5 5 −3.5 8 16.75 0.525

6 0 −1.6 0 4 0.5305…

b. At t = 60, vend = 0, ∴ the train is at rest.

c. The train is just starting at t = 0; itsacceleration must be greater than zero to get itmoving, even though it is stopped at t = 0.Acceleration and velocity are differentquantities; the velocity can be zero butchanging, which means the acceleration isnonzero.

d. Zero acceleration means the velocity isconstant, but not necessarily zero.

e. The last entry in the last column is thedisplacement at time t = 60. Thus, it is0.5305… ≈ 0.53 mi between stations.


14. a.

tend

saav

(mi/h)/svend

mi/hvav

mi/hsend

mi

0 — 6000 — 400

10 8.5 6085 6042.5 416.7847…

20 22 6305 6195 433.9930…

30 33 6635 6470 451.9652…

40 39.5 7030 6832.5 470.9444…

50 42.5 7455 7242.5 491.0625…

60 53 7985 7720 512.5069…

70 71 8695 8340 535.6736…

80 83.5 9530 9112.5 560.9861…

90 47.5 10005 9767.5 588.1180…

100 3 10035 10020 615.9513…

b. According to these calculations, the spaceshipis only about 620 mi from the launchpad andmoving at only about 10,000 mi/h. So thespecifications are definitely not met, and theproject should be sent back to the drawingboard.

15. a. adv

dtv a dt at C= ⇒ = = +∫ ;

v = v0 when t = 0 ⇒ C = v0 ⇒ v = v0 + at

b. vds

dts v dt v at dt= ⇒ = = + =∫ ∫ ( )0

v t at C021

2+ +

s = s0 when t = 0 ⇒ C = s0 ⇒

s v t at s= + +02

01

2

16. Use s(t) for displacement. Assume v(0) =s(0) = 0.

a. a tt

t( )

,

,=

≤ <≥

2 0 6

0 6

if

if

v tt t

t( )

,

,=

≤ <>

2 0 6

12 6

if

if

s tt t

t t( )

,

– ,=

≤ <>

2 0 6

12 36 6

if

if

−36 comes from the initial condition,s(6) = 36.

t

a(t)

2

0 6 10

10

0 6 10

t

v(t)

0 6 10

50

t

s(t)

b. The acceleration suddenly jumps from 0 to 2at t = 0 and drops back to 0 at t = 6. (Thevelocity graph has cusps in both places.)

c. a t t( ) = −2 23

cosπ

lim lim – cos

cos

t ta t t

→ →+ +=

= − =0 0

2 23

2 2 0 0

( )π

lim lim – cos

cos

– –t ta t t

→ →=

= − =6 6

2 23

2 2 2 0

( )π

π

Because a(t) is continuous at t = 0 and 6,there are no sudden changes in acceleration.

d. a t t t

t( ) – cos ,

,= ≤ ≤

≥

2 2

30 6

0 6

πif

if

v t t t t

t( ) – sin ,

,= ≤ ≤

≥

2

6

30 6

12 6π

πif

if

e.

10

0 6 10

t

v(t)

There are no step discontinuities in a(t), andthus the graph of v(t) is smooth.

f. 26

3

18

30

62

20

6

t t dt t t−

= +

∫ π

ππ

πsin cos

= + − − =3618

018

362 2π πElevator goes 36 ft.


g. The elevator will take another 36 ft to slowdown and stop. So the deceleration shouldstart where the elevator is 564 ft up, aboutthe 47th floor (from part h, one floor = 12 ft).

h. The elevator takes a total of 12 s to accelerateand decelerate. During these intervals ittravels a total of 72 ft, leaving 528 ft for theconstant velocity portion. At 12 ft/s, this partof the trip will take 44 s. Thus, the total triptakes 56 s.

i. The elevator must start to decelerate halfwaythrough the trip, where s(t) = 6 ft. Solving

26

36

0t t dt

b

– sinπ

π

=∫

numerically for b gives b ≈ 3.1043… ≈3.1 s.a(3.1043…) = 3.9880… ≈ 4.0 ft/s2

By symmetry, the deceleration process muststart at this time, meaning the accelerationjumps to −3.9880… ft/s2. The graph lookslike this:

6.2

5

0

t

a(t)

Thus, the passengers get a large jerk at themidpoint of the trip.One way to remedy the problem is to reducethe acceleration so that the elevator goes only6 ft instead of 36 ft in the first 6 seconds.That is,

a t t( ) = −1

3

1

3 3cos

π

You may think of other ways.

Problem Set 10-3

Q1. 50 mi/h Q2. 30 mi

Q3. 20 min Q4. 2πQ5. No local maximum Q6. 1.5

Q7. f (x) = 16 (at x = 1) Q8. infinite

Q9. Mean value theorem Q10. D

1. a. y x x dxav = + = =∫1

45

1

4164 413

1

5

( – ) ( )

b. The rectangle has the same area as the shadedregion.

50

1 5

y = 41

x

f(x)

c. 41 = c3 − c + 5c = 3.4028… , which is in [1, 5].

2. a. y x x dxav = + =∫1

87 4

1

61 2

1

9

( – )/


y = 4.1666...5

1 9

x

f(x)

c. 41

671 2= − +c c/

c = 5.0892… , which is in [1, 9].

3. a. y x dxav . .= = …∫1

63 0 2 2 0252

1

7

sin


1 7

3

x

g(x)

y = 2.0252...

c. 2.0252… = 3 sin 0.2cc = 3.7053… , which is in [1, 7].

4. a. y x dxav .= = …∫1

12 5181

0 5

1 5

tan.

.


0.5 1.5

10

x

h(x)

y = 2.5181...


c. 2.5181… = tan cc = 1.1927… , which is in [0.5, 1.5].

5. a. y t dtav = =∫1

82

1

61

9


1 9

t

v(t)

3y = 2.1666...

c. 21

6= c

c = 425

36, which is in [1, 9].

6. a. y e dt etav ( )= − = +−∫1

3100 1

100

32 3

0

3

( )–

= 68.3262…


t

v(t)

y = 68.32...

100

0 3

c. 68.3262… = 100(1 − e− c)c = 1.1496… , which is in [0, 3].

7. yk

ax dx akk

av = =∫1 1

32 2

0

8. yk

ax dx akk

av = =∫1 1

43 3

0

9. yk

ae dxk

a ex kk

av ( )= = −∫1 11

0

10. yk

x dxk

kk

av | |= =∫1 10

tan ln sec

11. a(t) = 6t− 1/2

v(t) = 12t1/2 + C; v(0) = 60 ⇒ C = 60

v(t) = 12t1/2 + 60

s(t) = 8t3/2 + 60t + s0

v(25) = 120 ft/s

Displacement = s(25) − s(0) = 2500 ft

vav = 2500/25 = 100 ft/s

12. The general equation of a parabola with vertex(h, k) is v − k = a(t − h)2. Vertex is at(t, v) = (2, 50), sov − 50 = a(t − 2)2

. v = 30 when t = 0, so−20 = a(−2)2 ⇒ a = −5.

v = 50 − 5(t − 2)2

v t dtav mi/h= =∫1

450 5 2 43

1

32

0

4

[ – ( – ) ]

This is just 131

3mi/h above the speed limit.

If Ida wins her appeal, her fine will be

7 131

393

1

393 33⋅ = ≈$ $ . , which is $46.67 less

than what she now faces.

13. Consider an object with constant acceleration a,for a time interval [t0, t1].

v t a dt at C( ) = = +∫At t = t0, v(t) = v0 ⇒ v0 = at0 + C ⇒C = v0 − at0.∴ v(t) = at + v0 − at0 = v0 + a(t − t0)

vv a t t dt

t tt

t

av =+∫ [ ( – )]

–

0 0

1 0

0

1

= +

1 1

2

1

21 00 1 1 0

20 0 0 0

2

t tv t a t t v t a t t

–( – ) – – ( – )

= + −v a t t0 ( )1

2 1 0

The average of v0 and v1 is1

2

1

21

2

0 1 0 0 1 0

0 1 0

( ) [ ( – )]v v v v a t t

v a t t

+ = + +

= + −( )

∴ vav = the average of v0 and v1, Q.E.D.

14. Counterexample: In Problem 11, the car’sacceleration is a t= 6/ .The initial velocity is

v(0) = 60 ft/s; the final velocity after 25 secondsis v(25) = 120 ft/s; and the average velocity isvav = 100 ft/s. But the average of the initial and

final velocities is 1

20 25 90[ ( ) ( )]v v v+ = ≠ft/s .av

15. a. Integral = area = 12(100 + 70)/2 + 6(40) +12(40 + 10)/2 = 1560yav = 1560/30 = 52, or $52,000Cost of inventory = 0.50(52000)/100 =$260.00

b. At x = 12, they may have had a single, largesale, dropping the inventory from $70,000 to$40,000. There is no day on which the inventoryis worth $52,000.

10 20 30

50

100

y (thousand dollars)

x (days)

No x where y = 52


16.

–5

–10

5 10 15 20 25 30x (ft)

yWater surface

(ft)

y = – 6.5142av

Integral = −(area of 4 rectangles, 2 trapezoids,and 2 quarter-circles)2(−8) + 8(−10) + 7(−3) + 1(−2) +7[−10 + (−5)]/2 + 5[−5 + (−3)]/2 − π(22) −π(1)2/4 = −195.4269…yav = −195.4269…/30 = −6.5142… , or about6.51 feet deep.The volume would equal 6.5142… times the areaof the horizontal cross section times the numberof gallons in a cubic foot.

17. Integral ≈ 3(16/2 + 15 + 15 + 17/2) +2(17 + 20)/2 + 1(20 + 14)/2 + 3(14/2 + 10 +9 + 8 + 9/2) = 139.5 + 37 + 17 + 115.5 = 309yav = 309/24 = 12.875 ≈ 12.9°C

The average of the high and low temperatures is(20 + 8)/2 = 14°C, which is higher than theactual average. Averaging high and lowtemperatures is easier than finding the average bycalculus, but the latter is more realistic for suchapplications as determining heating and airconditioning needs.

18. a. At x = 3, y = 81.3139… ≈ 81.3 mg.

y e dxxav

. mg

= =

= … ≈

−∫1

3200

1

3395 6202

131 8734 131 9

0 3

0

3. ( . )

.

K

b. k = 81.3139… , so the equation isy = 281.3139…e− 0.3(x− 3 ) .

y e dxx

av = +

∫1

6395 6202 281 3139 0 3 3

3

6

. . – . ( – )K K

= + =1

6395 6202 556 4674 158 6812( . . )K K K.

≈ 158 7. mg

c. As the graph shows, there are two times in[0, 6] at which there are 158.7 mg. So theconclusion of the mean value theorem is true,in spite of the discontinuity.

1 2 3 4 5 6 7

100

200

300

y (mg)

x (h)Two times

y = 158.68...av

19. v = A sin 120π t and y = | A sin 120π t |

y A t dtav | |= ∫1

1 60120

0

1 60

/sin

/

π

= −∫ ∫60 120 60 1200

1120

1120

1 60

A t dt A t dtsin sin/

/

/

π π

= − +At

At

2120

2120

1120

1 60

ππ

ππcos cos

/

/

0

1/120

= + + =A A

20 2

2

ππ π π

π(– cos cos cos – cos )

If , thenavyA

A= = ⇒ =1102

110 55π

π

= 172.78… V.The average value of one arc of

y x x dx=−

=∫sin sin , is and1

0

20π π

π

y = sin x has a maximum value of 1. Ahorizontal stretch does not affect the averagevalue. Write a proportion to find the maximumof a sinusoidal curve with an average value

of 110. 2

1

110/,

π =m

so m = 55π.

20. a. d = k sin x

d k x dx

kx x

k k

av2 2 2

0

2

2

0

2

2 2

1

2

2

1

2

1

42

20 0 0

2

=

=

= + =

∫π

π

ππ

π

π

sin

– sin

( – – )

∴ = = rms / .k k2 0 7071K

b. cos sin sin cos2 1 21

2

1

222 2x x x x= − ⇒ = −

Thus, sin2 x is a sinusoid.

π 2π

1x

y

c. By symmetry across the line y = 1

2, the

average of y x= −1

2

1

22cos (and hence

y = sin2 x) over [0, 2π] is1

2. Thus, the

average of is .y k x k= 2 2 21

2sin

∴ = rms k/ ,2 as in part a.


d. By symmetry, it suffices to find the averageand rms for one arch of the graph, that is,over [0, π].

y x dxav | |= ∫10π

πsin

= ≥∫10 0

0ππ

πsin sinx dx x (because in [ , ])

= − =1 2

0π π

π

cos x

d x dxav .2 2

0

12 0 094715= ≈∫π

ππ

(|sin | – / ) K

∴ rms ≈ 0.094715…1/2 = 0.3077…

The maximum distance between high and lowpoints for this curve is 1; a sinusoidal curvewith maximum distance 1 between high and

low points has equation y x= 1

2sin , with

rms /= =2 4 0 3535. K(using part a). This

number is greater than the rms for |sin x|,so |sin x| is smoother.

Problem Set 10-4Q1. x = 81 Q2. y′ = −x(100 − x2)− 1/ 2

Q3. − +1

3100 2 3 2( – ) /x C Q4. y′ = 3 ⋅ (1 − 9x2)− 1/ 2

Q5.1

2

1

42 2xe e Cx x− + Q6. y′ = sech2 x

Q7. 1.5 Q8. t = 1 and t = 4

Q9. t = 4 Q10. A

1.

x

50

100

100 – x

T x x= + +1

250

1

51002 2 ( – )

The graph shows a minimum T at x ≈ 22 m.

100

100

x

T

Algebraic solution:

′ = + ⋅ −−T x x1

450 2

1

52 2 1 2( ) /

′ = ⇔ + =−T x x01

250

1

52 2 1 2( ) /

5x = 2(502 + x2)1/2

25x2 = 4 · 502 + 4x2

x = ± = ±100 21 21 8217/ . K

Ann should swim toward a point about 21.8 mdownstream.

2.100

100 – x x

30

T x x = + +1

13100

1

12302 2( – )

The graph shows a minimum T at x ≈ 72.

100

10

x

T

Algebraic solution:

′ = − + + ⋅−T x x1

13

1

2430 22 2 1 2( ) /

′ = ⇔ = + −T x x01

13

1

12302 2 1 2( ) /

13x = 12(302 + x2)1/2

169x2 = 144 · 302 + 144x2

x = ±72The diver should swim for 100 − 72 = 28 m,then dive.

3.

x 1000 – x1000

300

C x x= − + +40 1000 50 3002 2( )The graph shows a minimum C at x ≈ 400(exactly x = 400).

100,000C

x

1000

The pipeline should be laid 600 m along the roadfrom the storage tanks, then straight across thefield to meet the well.

4.

120

400

400 – x x


W x x= − + +3000 400 4000 1202 2( )The graph shows a minimum W at x ≈ 136(exactly x = =360 7 136 067/ . K).

2,000,000

400

x

W

The walkway should go 400 −136.067…≈ 263.9 m parallel to the street, then crossthe street.

5. a. For minimal path, x = 100 21/ .

∴ =+

= = . / ,sin θ x

x500 4 2 5

2 2 Q.E.D.

b. For minimal path, x = 400.

sin θ =+

= =x

x3000 8 40 50

2 2. / , Q.E.D.

6. Distance swimming = +p x2 2 . Distance

walking = k − x.

Ts

p xw

k x= + +1 12 2 ( – )

Tx

s p x w s w′ =

+− = −

2 2

1 1 1sin θ

Ts

w′ = ⇔ =0 sin θ , Q.E.D.

7. sin θ = 12

13

x =

=−30

12

13721tan sin

The diver should swim 100 − 72 = 28 m, thendive. The algebraic solution is easier than beforebecause no algebraic calculus needs to be done.Mathematicians find general solutions to gaininsight, and to find patterns and methods to alloweasier solution of similar problems.

8. sin θ =+

= =x

x120

3000

4000

3

42 2

16x2 = 9(1202 + x2)7 9 120 360 7 136 0672 2x x= ⋅ ⇔ = ± = …/ .

The walkway should go 400 − 136.067… ≈263.9 m parallel to the street, then cross thestreet.The algebraic solution is easier than beforebecause no algebraic calculus needs to be done.Mathematicians find general solutions to gaininsight, and to find patterns and methods to alloweasier solution of similar problems.

9.

x C(x), approximate

300 49,213

390 49,002

400 49,000

410 49,002

500 49,155

The table shows that a near miss will havevirtually no effect on the minimal cost. Forinstance, missing the optimal value of x by10 m will make about a $2 difference in cost,and missing by 100 m makes only a $150 to$200 difference.

10. T x x x( ) = + +1

5500

1

312002 2( – )

The graph shows a local minimum at x ≈ 900 ft(exact: 900 ft), which is out of the domain.

500

500 900

x

T(x)

The minimum occurs at an endpoint of thedomain. Because Calvin can walk entirely alongpavement when x = 0, there is a removablediscontinuity in the above function andT(0) = 100 + 240 = 340 s. Because T(500) =433.333… , which is greater than 340, theminimum time is at x = 0. Calvin’s time isminimized by staying on the sidewalks. If roadconstruction (for instance) prevented Calvin fromwalking on Heights Street, his time would beminimized by walking directly to Phoebe’shouse.

11.

x300 – x

70

120

T x x= + + +1

50120

1

13070 3002 2 2 2( – )

The graph shows a minimum T at x ≈ 48 yd.

300

5

x

T


Algebraically:

Tx

x

x

x′ =

+−

+50 120

300

130 70 3002 2 2 2

–

( – )

Setting T ′ = 0 and simplifying leads to afourth-degree equation, which must be solvednumerically. Minimum is at x = 47.8809…≈ 47.9 yd.

12.

x300 – x

70

120θ

θ

1

2

θ

θ1

2

From Problem 11,

′ =+

−+

Tx

x

x

x50 120

300

130 70 3002 2 2 2

–

( – )By trigonometry,

sin θ1 2 2120=

+

x

x,

sin–

( – )θ2 2 2

300

70 300=

+

x

x

∴ ′ = − T1

50

1

1301 2sin sinθ θ

For minimal path, T ′ = 0. Thus,

1

50

1

1301 2sin sinθ θ=

sin

sin

θθ

1

2

50

130= , Q.E.D.

13. Tv

a xv

b k x= + + +1 1

1

2 2

2

2 2( – )

′ =+

−+

Tx

v a x

k x

v b k x12 2

22 2

–

( – )

sin θ1 2 2=

+

x

a x,

sin–

( – )θ2 2 2

=+

k x

b k x

∴ ′ = − Tv v

1 1

11

22sin sinθ θ

For minimal path, T ′ = 0. Thus,

1 1

11

22v v

sin sinθ θ=

sin

sin

θθ

1

2

1

2

= v

v, Q.E.D.

14. a. The light rays take the minimal time to getfrom one point to another, just as RobinsonCrusoe wanted to take the minimal time toget from hut to wreck.

b. Light always takes the path requiring the leasttime between two points.

c. When you look at the object, your mind tellsyou that the light rays go straight. Actually,they are bent, as shown in the diagram. Sothe object is deeper than it appears to be.Because θwater < θair, vwater < vair.

Apparentdepth

Actualdepth

θAir

θ Water

Actual path of light rays

Apparent path of light rays


Problem Set 10-5Q1. f ′(x) = sin x + x cos x Q2. g″(x) = x− 1

Q3. xex − ex + C Q4. Snell’s law

Q5. x

Q6. Q7.

x

y

1

1

Q8. Total distance

Q9. Newton and Leibniz Q10. C

1. D tt

D t= + ′ = − −11 2

The graphs show zero derivative and localminimum of D at t = 1, and maximum of D att = 3.

D

D'

3

3

t

D or D'

D t t′ = ⇔ = ⇔ = ±0 1 12 , confirming the graph.Minimum is D(1) = 2, or 2000 mi.

Maximum is D( ) ,3 31

3= or about 3333 mi.

2. Fuel cost per mile = k · v2.At v = 30, cost = 0.18.

0 18 301

50002. = ⋅ ⇒ =k k

Driver cost is 20 20100 2000

tv v

= ⋅ = .

∴ = + ⋅ = + Cv

v

v

v2000

5000100

2000

50

2 2

Cv

v′ = − +2000

252


The graphs show minimum C at v ≈ 37 mi/h;C′ is multiplied by 10 so that it is easier to seeits behavior around C′ = 0.C v′ = = = …0 10 50 36 83 at . 403

C or C' times 100

C

C' times 1037

100

85 100

v

3. Maximize f(x) = x − x2.f ′(x) = 1 − 2x; f ′(x) = 0 at x = 0.5;f ″(x) = −2, so graph is concave downeverywhere.Maximum of f (x) is at x = 0.5.

4. Maximize f (x) = x − x2 for x ≥ 2.The graph shows maximum at endpoint x = 2.

2

1 xf(x)

Because f (2) is the maximum and it is negative,there is no number greater than 2 that exceeds itssquare.

5. a. St

tF

tG

t

t t=

+=

+=

+ +100

1

9

9

900

1 9; ;

( )( )

The graph shows a maximum of G at t = 3hours.

S

G

F times 20

10

50

t

y

b. Gt

t t

t

t t=

+ +=

+ +900

1 9

900

10 92( )( )

Gt t t t

t t′ = + + +

+ +900 10 9 900 2 10

10 9

2

2 2

( ) – ( )

( )

=+ +

900 9

10 9

2

2 2

( – )

( )

t

t tG t′ = ⇔ = ±0 3

Because G′ changes from positive to negativeat t = 3, there is a local maximum there, asin the graph.Fran should study for 3 hours.

c. Optimum grade = G(3) = 56.25 ≈ 56 (Notgood!)

i. G(4) = 55.3846… ≈ 55, about 1 pointless.

ii. G(2) = 54.5454… , ≈ 55, about 1 pointless.

6. a. µ = 130 − 12T + 15T 2 − 4T 3, 0 ≤ T ≤ 3d

dTT T T T

µ = − + − = − − −12 30 12 6 2 1 22 ( )( )

d

dTT T

µ = = =0 0 5 2 at . or

µ(0) = 130µ(0.5) = 127.25µ(2) = 134µ(3) = 121

Maximum viscosity occurs at T = 2, or 200°.

b. Minimum viscosity = 121 centipoise atT = 3, or 300°.

c.d

dt

d

dT

dT

dt

µ µ= ⋅

Because , .T tdT

dt t T= = =1

2

1

2

When , . and .TdT

dt

d

dT= = =1 0 5 6

µ

∴ = = . ( )d

dt

µ0 5 6 3

Viscosity is increasing at 3 centipoise/min.

7. a. Put a coordinate system with origin at thecenter of the cone’s base. Pick a sample point(x, y) where the cylinder touches the elementof the cone. Thus, x is the radius of thecylinder and y is its altitude. The volume andsurface area areV = π x 2yA = 2π x 2 + 2π xyThe cone element has equation y = −0.6x + 6.V = π x 2(−0.6x + 6) = π(−0.6x3 + 6x2)A = 2π x2 + 2π x(−0.6x + 6)

= π(0.8x2 + 12x)

10

600y

A

V

x

b. From the graphs, the maximum volumeoccurs where the radius x ≈ 6.7 in. Themaximum area occurs at x = 10, where all ofthe area is in the two bases of the cone.Algebraically,

V ′ = π ( −1.8x2 + 12x)

V x x′ = ⇔ = =0 0 62

3 or


Maximum V is at x = 62

3 in., as shown on

the graph, and y = 2 in.

A ′ = π (1.6x + 12)

A′ = 0 ⇔ x = −7.5, which is out of thedomain.

A(0) = 0 and A(10) = 200π, so maximum Ais at x = 10 in., as shown on the graph, andy = 0 in.

The maximum volume and maximum area donot occur at the same radius.Note that the radius of the cone is largecompared to its altitude. Thus, the increase inareas of the two bases of the cylinder offsetsthe decrease in its lateral area as x increases,making the maximum area that of thedegenerate cylinder of altitude zero.

8. a. Put a coordinate system with origin at thecenter of the cone’s base. Pick a sample point(x, y) where the cylinder touches the elementof the cone. Thus, x is the radius of thecylinder and y is its altitude.

Know: dy

dt= 2 in./min. Want:

dV

dt.

V = πx2yThe cone element has equation y = −3x + 18,

from which x y= −61

3.

V y y y y y= −

⋅ = − +

π π6

1

336 4

1

9

22 3

dV

dyy y y y= − +

= − −π π36 8

1

3

1

36 182 ( )( )

dV

dt

dV

dy

dy

dty y= ⋅ = − −2

36 18π ( )( )

When ydV

dt= = −12 24, . π

V is decreasing at 24π = 75.3982… ≈75.4 in.3/min.

b. If t ∈ [0, 9], then y ∈ [0, 18].dV

dty y= ⇔ = =0 6 18 or

V(0) = 0; V(6) = 96π; V(18) = 0Maximum V is 96π in.3 at t = 3 min.

c. If t ∈ [4, 6], then y ∈ [8, 12].No critical points for V are in [8, 12].V(8) = 88.8888…π; V(12) = 48πMaximum V is 88.8888…π ≈ 279.3 in.3 att = 4 min.

9. Know: . m /min. Want: .dV

dt

dy

dt= −0 7 3

dV = πx2 dy

∴ = ⇒ =

dV

dtx

dy

dt

dy

dt xπ

π2

2

0 7– .

When the water is 3 m deep, y = 8.

Because y x x= + =4 45 3, .

dy

dt= = − ≈– .0 7

30 1286 0 129

π. . m/minK

10. a. Pick a sample point (x, y) where the cylindertouches the parabola. Thus, the radius of thecylinder is x and its altitude is y.

Know: . . Want: .dx

dt

dV

dt= 0 3

V = π x2y = πx2(4 − x2) = π(4x2 − x4)dV

dtx x

dx

dtx x= − ⋅ = −π π( ) . ( )8 4 1 2 23 3

When . , . .xdV

dt= = −1 5 0 45π

b.dV

dxx x= −4 2 3π ( )

dV

dxx= ⇔ = ±0 0 2 2, is out of domain)(–

V V V( ) ( ) ; ( )0 2 0 2 4= = = πMaximum volume = 4π ≈ 12.6 units3 atradius = 2 units.

11. a. w = 1000 + 15t (lb); p = 0.90 − 0.01t ($/lb)A = (1000 + 15t)(0.90 − 0.01t)

= 900 + 3.5t − 0.15t2 ($)

b.dA

dtt t= − = = =3 5 0 3 0

35

311

2

3. . at days

Maximum A at t = 112

3, not a minimum,

because dA

dtgoes from positive to negative

there.

c. A 112

3920

5

12920 42

= ≈ $ .

12. a. 0 ≤ D ≤ 130 ⇒ 0 ≤ 20x + 10 ≤ 130 ⇒

− ≤ ≤1

26x

0 ≤ W ≤ 310 ⇒ 0 ≤ 10(x2 − 8x + 22) ≤310 ⇒ −1 ≤ x ≤ 9Given x ≥ 1, the domain of x is [1, 6].

b. Minimize/maximize W on x ∈ [1, 6].dW

dxx x= − = =10 2 8 0 4( ) at .

W(1) = 150; W(4) = 60; W(6) = 100Minimum: W = 60 ft (at x = 4 mi)Maximum: W = 150 ft (at x = 1 mi)

c. C = k · D · W= k · (20x + 10) · 10(x2 − 8x + 22)= 100k(2x3 − 15x2 + 36x + 22) (k > 0)dC

dxk x x

k x xdC

dxx x

= − +

=

= ⇔ = =

100 6 30 36

600 2 3

0 2 3

2( )

( – )( – )

or


C(1) = 4500k; C(2) = 5000k; C(3) = 4900k;C(6) = 13,000kCheapest bridge at x = 1 mi.

d. No. The shortest bridge at x = 4 mi wouldcost C(4) = 5400k, which is 900k more thanthe cheapest bridge at x = 1.

Problem Set 10-6Q1. −x cos x + sin x + C Q2. 2xe3x + 3x2e3x

Q3.2

2

x

Cln

+ Q4. 53 = 125

Q5.1

474x + Q6.

6 6

3

22

32 2sec

sec lnt

e xxt =

Q7. parametric Q8. x ln x − x + C

Q9. x2 Q10. E

1. The velocity is tangent to the path, theacceleration is toward the concave side of thepath, and there is an obtuse angle betweenacceleration and velocity.

y

x

r

v

a→

→

→

2. The velocity is tangent to the path, theacceleration is toward the concave side of thepath, and there is an acute angle between theacceleration vector and the velocity vector.

y

x

r

v

a

→

→

→

3. a.r r rr e t i e t jt t= +( ) ( )cos sin

r r rv e t e t i e t e t jt t t t= − + +( ) ( )cos sin sin cos

r

Kr r

v i j( )1 0 8186 3 7560= − + …. .

x is decreasing at t = 1 because dx/dt isnegative.

Speed = + =0 8186 3 75602 2. . ...K

3.8442… ≈ 3.84 cm/s

b. L =

( cos sin ) ( sin cos )e t e t e t e t dtt t t t− + +∫ 2 2

0

1

= = = ≈∫ 2 2 1 2 4300 2 432

0

1

e dt et ( – ) . .K

Distance from origin

= +( cos ) ( sin )e e1 2 1 21 1

= e = 2.7812… ≈ 2.78 cm

(Note that the distance traveled is less than thedistance from the origin because the particlestarted at (1, 0), not at (0, 0).)

c. r r ra e t i e t jt t= − +( ) ( )2 2sin cos .

r r ra i j( ) . ... . ...1 4 5747 2 9373= − +( ) ( )

4. a. r r rr t i t j1 2 2= − + −( ) ( ) and2

r r rr t i t j2 1 5 4 1 5 2= − + −( . ) ( . )

r r r r r rv i t j v i j1 21 2 2 1 5 1 5= + − = +( ) and . .

r r r r r ra i j a i j1 20 2 0 0= + = +and

r r r r r rv i j v i j1 21 2 1 1 5 1 5( ) and ( )= − = +. .

r r r r r ra i j a i j1 21 0 2 1 0 0( ) and ( )= + = +

s12 21 1 2 2 2( ) . 4 cm/s and= + ≈

s22 21 1 5 1 5 2 12( ) . cm/s= + ≈. .

b. Distance1

4

= + ≈∫ 1 2 22 2[ ( – )]t dt

6.1257… ≈ 6.13 m

c. The paths cross at (x, y) = (−1, 1) and (2, 4).By tracing on the grapher,

rr1 is at (−1, 1)

when t = 1, but rr2 is not at (−1, 1) until

t = 2.By further tracing, both paths are at (2, 4)when t = 4.So the particles collide only at (x, y) = (2, 4)when t = 4.

5. a.r r rr t t i t j( ) sin cos= +( . ) ( . )10 0 6 4 1 2r r rv t t i t j( ) cos sin= + −( . ) ( . . )6 0 6 4 8 1 2r r ra t t i t j( ) sin cos= − + −( . . ) ( . . )3 6 0 6 5 76 1 2

b.r r rr i j( . ) sin cos0 5 10 0 3 4 0 6= +( . ) ( . )

= … + …2 9552 3 3013. .r ri j

r r rv i j( . ) cos sin0 5 6 0 3 4 8 0 6= + −( . ) ( . . )

= … − …5 7320 2 7102. .r ri j

r r ra i j( . ) sin cos0 5 3 6 0 3 5 76 0 6= − + −( . . ) ( . . )

= − … − …1 0638 4 7539. .r ri j

The graph shows r rr v, , and

ra at t = 0.5.

x

y

t = 7

t = 0.5at

an a

va

v

These vectors make sense because the head ofrr is on the graph,

rv is tangent to the graph,

and ra points to the concave side of the graph.

c. The object is speeding up because the anglebetween

ra and

rv is acute.


d. | |rv( . ) ( cos . ) (– . sin . )0 5 6 0 3 4 8 0 62 2= +

= 6.3404…

r ra v( . ) ( . ) sin cos0 5 0 5 3 6 0 3 6 0 3⋅ = −( . . )( . )

+ (−5.76 cos 0.6)(−4.8 sin 0.6)= 6.7863… , so the angle is acute.

P

a v

v= ⋅ = …

r r

r( . ) ( . )

| ( . )|

0 5 0 5

0 51 0703.

rr

ra Pv

vt ( . )0 50 5

0 5= ( . )

| ( . )|

= +

Pi j

v

( cos . ) (– . sin . )

| ( . )|

6 0 3 4 8 0 6

0 5

r r

r

= … − …0 9676 0 4575. .r ri j

r r ra a an t( . ) ( . )0 5 0 5 0 5= −( . )

= − … − …2 0314 4 2964. .r ri j

See the graph in part b.

e. The object is speeding up at | ( . )|ra Pt 0 5 =

= 1.0703… ≈ 1.07 (ft/s)/s.

f.r r rr i j( ) sin cos7 10 4 2 4 8 4= +( . ) ( . )

= − … − …8 7157 2 0771. .r ri j

r r r

r rv i j

i j

( ) cos sin7 6 4 2 4 8 8 4

2 9415 4 1020

= + −

= − … − …

( . ) ( . . )

. .r r ra i j( ) sin cos7 3 6 4 2 5 76 8 4= − + −( . . ) ( . . )

= … + …3 1376 2 9911. .r ri j

See the graph in part b.The object is slowing down because theangle between

ra and

rv is obtuse.

(Note that Pa v

v= ⋅ = − …

r r

r( ) ( )

| ( )|

7 7

74 2592. , so

the object is slowing down at 4.2592…≈ 4.26 (ft/s)/s.)

g. r r r r rr i j i j( ) sin cos0 10 0 4 0 0 4= + = +( ) ( )

r r r r rv i j i j( ) cos sin0 6 0 4 8 0 6 0= + − = +( ) ( . )

r r ra i j( ) sin cos0 3 6 0 5 76 0= − + −( . ) ( . )

= −0 5 76r ri j.

r ra v( ) ( )0 0 0 6 5 76 0 0⋅ = + − =( )( ) ( . )( )∴

ra( )0 and

rv( )0 are perpendicular, Q.E.D.

This means the object is neither slowingdown nor speeding up at t = 0.

6. a. r r rr t t i t j( ) cos sin= +( . ) ( . )8 0 8 6 0 4

r r rv t t i t j( ) sin cos= − +( . . ) ( . . )6 4 0 8 2 4 0 4

r r ra t t i t j( ) cos sin= − + −( . . ) ( . . )5 12 0 8 0 96 0 4

b. r r rr i j( ) cos sin1 8 0 8 6 0 4= +( . ) ( . )

= … + …5 5736 2 3365. .r ri j

r r rv i j( ) sin cos1 6 4 0 8 2 4 0 4= − +( . . ) ( . . )

= − … + …4 5910 2 2105. .r ri j

r r ra i j( ) cos sin1 5 12 0 8 0 96 0 4= − + −( . . ) ( . . )

= − … − …3 5671 0 3738. .r ri j

The graph shows r rr v, , and

ra at t = 1.

x

y

t = 10.5

t = 1

r

v

a

v

a

r

at

an

t = 1.25πa

These vectors make sense because the head of rr is on the graph,

rv is tangent to the graph,

and ra points to the concave side of the graph.

c. The object is speeding up because the anglebetween

ra and

rv is acute.

d. | |rv( ) (– . sin . ) ( . cos . )1 6 4 0 8 2 4 0 42 2= +

= 5.0955…r ra v( ) ( ) cos sin1 1 5 12 0 8 6 4 0 8⋅ = − −( . . )( . . )

+ (−0.96 sin 0.4)(2.4 cos 0.4)= 15.5506… , so the angle is acute.

Pa v

v= ⋅ = …

r r

r( ) ( )

| ( )|

1 1

13 0518.

rr

ra Pv

vt ( )11

1= ( )

| ( )|

= +P

i j

v

(– . sin . ) ( . cos . )

| ( )|

6 4 0 8 2 4 0 4

1

r r

r

= − … + …2 7496 1 3239. .r ri j

r r r

r ra a a

i j

n t( ) ( )

. .

1 1 1

0 8174 1 6977

= −

= − … − …

( )

See the graph in part b, showing rat and

ran

at t = 1.

e. The object is speeding up at

| ( )| ( ) . . (ft/s)s.ra Pt 1 1 3 0518 3 05= = … ≈

f. r r rr i j( . ) cos sin10 5 8 8 4 6 4 2= +( . ) ( . )

= − … − …4 1543 5 2294. .r ri j

r r r

r rv i j

i j

( . )10 5

5 4694 1 1766

= − +

= − … − …

( 6.4 sin 8.4) (2.4 cos 4.2)

. .

r r r

r ra i j

i j

( . ) cos sin10 5 5 12 8 4 0 96 4 2

2 6587 0 8367

= − + −

= … + …

( . . ) ( . . )

. .See the graph in part b, showing

r rr v, , and

ra

at t = 10.5.

The object is slowing down at t = 10.5because the angle between

ra and

rv is obtuse

at that time. (Note that

Pa v

v= ⋅ = − …

r r

r( . ) ( . )

| ( . )|

10 5 10 5

10 52 7552. ,

so the object is slowing down at about2.8 (ft/s)/s.)


g. The object is stopped when

r r r rv t t i t j( ) sin cos= − + =( . . ) ( . . )6 4 0 8 2 4 0 4 0⇔ −6.4 sin 0.8t = 0 and 2.4 cos 0.4t = 0.Using −6.4 sin 0.8t = −1.28 sin 0.4tcos 0.4t, you see that

r rv t( ) = 0 exactly when

cos 0.4t = 0; the first time this happens is att = 1.25π s.

r r rr i j( . ) cos sin1 25 8 6 0 5π π π= +( ) ( . )

= − +8 6r ri j

r r r

r ra i j

i j

( . ) cos sin1 25 5 12 0 96 0 5

5 12 0 96

π π π= − + −

= −

( . ) ( . . )

. .See the graph in part b, showing

ra at

t = 1.25π.The acceleration vector points along the pathat t = 1.25π. So the object is stopped, but ithas a nonzero acceleration. At first glance,this fact may be surprising to you!

7. a.r r rr t t i t j( ) cos sin=

+

10

66

6

π π

r r rv t t i t j( ) sin cos= −

+

10

6 6

6

6 6

π π π π

r r rr t v t t t i( ) ( ) cos sin+ = −

10

6

10

6 6

π π π

+ +

6

6

6

6 6sin cos

π π πt t j

r

The graph shows the path of r rr v+ .

x

y

0

1

23

4

5

6

7

89

10

11

12

b. See the graph in part a, showing vectorsrv .

c. For r rr v+ ,

xt t

10 6 6 6= −cos sin

π π π

yt t

6 6 6 6= +sin cos

π π π

xt t t

10 6 3 6 6

22

= −cos cos sin

π π π π

+

π π6 6

22sin t

yt t t

6 6 3 6 6

22

= +sin sin cos

π π π π

+

π π6 6

2

cos2 t

∴

+

= +

x yt t

10 6 6 6

2 22 2cos sin

π π

+

+

π π π π6 6 6 6

22 2sin cost t

2

x y

10 6 6

2

+

= +

2 2

1π

x y2

2

2

2100 1 36 36 1 36( / ) ( / )++

+=

π π1

This is the equation of an ellipse centered atthe origin with x-radius 11.2878… andy-radius 6.7727… .

d.

r r ra t t i t j( ) cos sin= −

+ −

10

36 6

6

36 6

2 2π π π π

r rr t a t( ) ( )+

= −

+ −

1010

36 66

6

36 6

2 2π π π πcos sint i t j

r r

See the graph in part a, showing an ellipticalpath followed by the heads of the accelerationvectors.

e. The direction of each acceleration vector is theopposite of the corresponding position vectorand is thus directed toward the origin.

Note that r ra t r t( ) ( )= − π 2

36.

8. a.r r rr t t t i t t j( ) cos sin= +( . ) ( . )0 5 0 5r rv t t t t i( ) cos sin= −( . . )0 5 0 5

+ +( . . )0 5 0 5sin cost t t jr

r ra t t t t i( ) sin cos= − −( . )0 5

+ −( . )cos sint t t j0 5r

b.r r rr i j( . ) cos sin8 5 4 25 8 5 4 25 8 5= +. . . .

= − … + …2 5585 3 3935. .r ri j

r rv i( . ) cos sin8 5 0 5 8 5 4 25 8 5= −( . . . . )

+ +( . . . . )0 5 8 5 4 25 8 5sin cosrj

= − … − …3 6945 2 1593. .r ri j

r ra i( . ) sin cos8 5 8 5 4 25 8 5= − −( . . . )

+ −( . . . )cos sin8 5 4 25 8 5rj

= … − …1 7600 3 9955. .r ri j

r r rr i j( )12 = +6 cos12 6 sin 12

= … − …5 0631 3 2194. .r ri j

r rv i( ) cos sin12 0 5 12 6 12= −( . )

+ +( . )0 5 12 6 12sin cosrj

= … + …3 6413 4 7948. .r ri j

r ra i( ) sin cos12 12 6 12= − −( )

+ −( )cos sin12 6 12rj

= − … + …4 5265 4 0632. .r ri j


The graph shows that rr ( . )8 5 and

rr ( )12 really

do terminate on the path.

x

y

t = 8.5

t = 12at

va

r

r

a

v

c. See the graph in part b, showing rv( . ),8 5

r rv a( ), ( . ),12 8 5 and

ra( )12 . The velocity

vectors point along the path as it spiralsoutward, and the acceleration vectors pointinward to the concave side of the graph.

d. In both cases, the angle between ra and

rv

appears to be acute. Check using dotproducts.

r ra v( . ) ( . )8 5 8 5 2 125⋅ = . , which is positive.

r ra v( ) ( )12 12 3⋅ = , which is also positive.Thus, the angles are acute, and the object isspeeding up at both times.

e. At h, .t a v= ⋅ =12 12 12 3r r( ) ( )

| | .rv( ) .12 36 25 6 0207= = …r

r r

r

r

r

r r

r r

aa v

v

v

v

i j

i j

t ( )

. .

1212 12

12

12

12

3

36 253 6413 4 7948

0 3013 0 3968

= ⋅ ⋅

= +

= … + …

( ) ( )

| ( )|

( )

| ( )|

.( . ... . ... )

r r r

r ra a a

i j

n t( ) ( )

. .

12 12 12

4 8279 3 6664

= −

= − … + …

( )

See the graph in part b, showing ran and r

at att = 12.

f. Speed | | .= = = …rv( ) .12 36 25 6 0207

≈ 6.02 mi/h

Speed is increasing by P( )123

36 25= =

.0.4982… ≈ 0.498 (mi/h)/h.

g. See the graph in part b, showing r rr t a t( ) ( ).+

The heads seem to lie on a unit circle.Algebraic verification:r r r rr t a t t i t j( ) ( ) sin cos+ = − + , which is acircle.

9. a. r r r r rr x xi yj xi x j( ) = + = + 2

r r r r rv x

dx

dti

dy

dtj

dx

dti x

dx

dtj( ) = + = + 2

b.

dx

dtv x i xj= − ⇒ = − −3 3 6r r r( )

r r rv i j( )2 3 12= − −

At , speed | |x v= = = =2 2 153r( )

12.3693… ≈ 12.4 cm/s.

c. The graph shows rr ( )2 and

rv( )2 .

This is reasonable because rv( )2 points along

the curve to the left, indicating that x isdecreasing.

x = 2

at

a

an

v

5

20

x

y

d. From part b, r r rv x i xj( ) ,= − −3 6

∴ = − =

r r r ra x i

dx

dtj j( ) 0 6 18

r ra j( )2 18= . See the graph in part c.

e.r

r r

r

r

raa v

v

v

vt ( )22 2

2

2

2= ⋅ ⋅( ) ( )

| ( )|

( )

| ( )|

= – (– – )216 3 12

153

r ri j

= + = … + …72

17

288

174 2352 16 9411

r r r ri j i j. .

r r ra a an t( ) ( )2 2 2= −( )

= − + = − … + …72

17

18

174 2352 1 0588

r r r ri j i j. .

rat ( )2 is parallel to the curve.

ran ( )2 is

normal to the curve and points inward to theconcave side.

f. The object is slowing down when x = 2because the angle between

ra( )2 and

rv( )2 is

obtuse, as shown by the graph and by the factthat the dot product is negative. Also,

rat ( )2

points in the opposite direction of rv( )2 .

g. dL dy dx dx x dx= + = +1 1 42 2( / )dL

dtx

dx

dt= + =1 4 52

At ,xdx

dt

dx

dt= = + =2 5 1 4 2 172( )

⇒ = = … ≈dx

dt

5

171 2126 1 21. . cm/s.

10. r r rr i j( )1 8 8615 4 8410= … + …. .

r r rq i j( )2 2 9659 4 3406= − … + …. .

r r rq i j( . )1 5 0 2065 6 6362= − … + …. .

r r rq i j( . )1 1 2 5579 7 4880= … + …. .

r rv t t t t t i( ) cos cos sin sin= −( . . )12 0 5 6 0 5

+ +( . . )12 0 5 6 0 5cos sin sin cost t t t jr

r r rv i j( )1 3 2693 7 5391= … + …. .


The graph shows average velocity vectorsapproaching the instantaneous velocity vector

rv( )1 as t approaches 1. The instantaneousvelocity vector is tangent to the graph and pointsin the direction of motion.

10

10

x

y

vt = 1.1t = 1.5

t = 2

11. r r rr t t i t j( ) sin cos= +( . ) ( . )10 0 6 4 1 2

dL dx dt dy dt dt

t t dt

= +

= +

( / ) ( / )

( cos . ) (– . sin . )

2 2

2 26 0 6 4 8 1 2

L dL= ≈ …∫ 12 0858. ft (numerically)0

2

12.r r rr t t i t j( ) cos sin=

+

10

66

6

π π

dL dx dt dy dt dt

t t dt

= +

= ⋅

+ ⋅

( / ) ( / )

–sin cos

2 2

2 210

6 6

6

6 6

π π π π

One complete cycle of the curve is 0 ≤ t ≤ 12, so

L dL= = … ≈∫ 51 0539 51 1. (numerically) . ft.0

12

13. a.r r ra t i j( ) = −0 32r r r r rv t i j dt C i t C j( ) = − = + − +∫ ( ) ( )0 32 321 2

r r rv i j C C( )0 130 0 130 01 2= − + ⇔ = − =and

∴ = − − r r rv t i tj( ) 130 32

b.

r r rr t i tj dt( ) = − −∫ ( )130 32

= − + + − +( ) ( )130 1632

4t C i t C jr r

r r rr i j C C( )0 60 5 8 60 5 83 4= + ⇔ = =. . and

∴ = − + + − + ( . ) ( )r r rr t t i t j( ) 130 60 5 16 82

c. When the ball passes over the plate, x(t) = 0,so t = 60.5/130 = 0.4653… . At that time,y(t) = 4.5346… , which is slightly above thestrike zone.

d. At t = 0, dx/dt = 200 cos 15°,dy/dt = 200 sin 15°.As in part a,

r r rv t C i t C j( ) = + − + =1 232( )

( ) ( )200 15 32 200 15cos sin° + − + °r ri t j

As in part b, r rr t t i( ) cos= ° +( )200 15

( ) .− + ° +16 200 15 32t t jsinr

e. When x = 400, t = 2 sec 15° = 2.0705… sy(2.0705…) = 41.5846… .Phyllis makes the home run because41.5… > 10.

100

400

x

y

14. a. r r ra t i y t j( ) = + ′′3 ( )

r r rv t t C i y t j( ) = + + ′( ) ( ( ))3 1

r r rv i j C( )0 0 0 01= + ⇒ =

r r rr t t C i y t j( ) = + +( . ) ( ( ))1 5 2

2

r r rr i j C( )0 0 0 02= + ⇒ =

∴ = + = + ( . ) ( ( )) ( . ) ( ( ))r r r r rr t t i y t j t i x t j( ) sin1 5 1 52 2

r r rr t t i t j( ) sin= +( . ) ( . )1 5 1 52 2

b.r r rv t t i t t j( ) cos= +( ) ( . )3 3 1 5 2

If x = 6, t = 2.r r

Kr

v i j( )2 = +6 5.7610

Speed | | .

. m/

= = + = …≈

rKv( ) .

min

2 6 5 76 8 3180

8 32

2 2

15. a. d = a + b cos tt = 0: 240 = a + b cos 0 = a + bt = π: −60 = a + b cos π = a − b2a = 180 ⇒ a = 902b = 300 ⇒ b = 150∴ d = 90 + 150 cos t

b. r rr t t t i( ) cos cos= +( )90 150 2

+ +( )90 150sin sin cost t t jr

r rv t t t t i( ) sin cos sin= − −( )90 300

+ + −(90 cos t t t j150 1502 2cos sin )r

r rv t t t i( ) sin sin= − −( )90 150 2

+ +( )90 150 2cos cost t jr

r

Kr

Kr

v i j( )1 212 1270 13 7948= − −. .

Speed

. . cm/s

= += ≈

(– . ) (– . )212 1 13 7

212 5750 212 6

2 2K K

K

c.r ra t t t i( ) cos cos= − −( )90 300 2

+ − −( )90 300 2sin sint t jr

r

Kr

Kr

a i j( )1 76 2168 348 5216= −. .

P

a v

v( ) .1

1 1

153 4392= ⋅ = −

r r

r K( ) ( )

| ( )|

rr

r

Kr

Kr

a Pv

v

i j

t ( ) ( )

. .

1 11

1

53 3266 3 4678

=

= +

( )

| ( )|


r r r

Kr

Kr

a a a

i j

n t( ) ( )

. .

1 1 1

22 8902 351 9894

= −

= −

( )

Annie is slowing down. The angle betweenthe acceleration and velocity vector is obtuse,as revealed by the negative dot product. She isslowing down at 53.4392… ≈ 53.4 cm/s.

16. a. r r rr t t i t j= + +( . ) ( . )0 5 4 0 5sin cosr r rv t t i t j( ) cos sin= + + −( . ) ( . )0 5 2 0 5

r r ra t t i t j( ) sin cos= − + −( ) ( . )0 5

r r rv i j( )14 0 6367 1 3139= … − …. .

r r ra i j( )14 0 9906 0 7539= − … − …. .

Speed | | . mi/h= = …rv( )14 1 4601

P

a v

v( )14

14 14

14= ⋅

r r

r( ) ( )

| ( )|

= =0 3598

1 46010 2464

.

.

K

KK.

rr

r

r r

a Pv

v

i j

t ( ) ( )

. .

14 1414

14

0 1074 0 2217

=

= … − …

( )

| ( )|

r r r

r ra a a

i j

n t( ) ( )

. .

14 14 14

1 0980 0 5312

= −

= − … − …

( )

The log is speeding up at t = 14. You can tellby the fact that P(14) is positive, and thus theangle between

ra( )14 and

rv( )14 is acute, which

means that rat ( )14 points in the same direction

as rv( )14 . It is speeding up at 0.2464… ≈

0.246 (mi/h)/h.

b. dL dx dy= +2 2

= + +( . cos ) (– sin . )0 5 2 0 52 2t t dt

L dL= ≈ …

≈∫ 22 7185

0

14

. (numerically)

22.7 mi

Average speed ≈ 1

1422 7185( . )K

= … ≈1 6227 1 62. . mi/h

17. a. r r rr t t t i t j( ) sin cos= − + +( ) ( )5 12 15 12

r r rv t t i t j( ) cos sin= − + −( ) ( )5 12 12r r ra t t i t j( ) sin cos= + −( ) ( )12 12

b. r r rv i j( . )2 5 14 6137 7 1816= … − …. .

r

Kr

Kr

a i j( . )2 5 7 1816 9 6137= +. .

The graph shows and .r rv a( . ) ( . )2 5 2 5

5

5

x

y

v

a

c. P

a v

v( . )2 5

2 5 2 5

2 5= ⋅

r r

r( . ) ( . )

| ( . )|

= = …60 2 5

169 120 2 52 2052

sin .

– cos ..

rr

r

r r

a Pv

v

i j

t ( . ) ( . )

. .

2 5 2 52 5

2 5

1 9791 0 9726

=

= … − …

( . )

| ( . )|

r r r

r ra a a

i j

n t( . ) ( . )

. .

2 5 2 5 2 5

5 2024 10 5863

= −

= … + …

( . )

d. rv( . )2 5 is reasonable because its graphpoints along the path in the direction ofmotion.

ra( . )2 5 is reasonable because it points

toward the concave side of the path. Theroller coaster is traveling at | |

rv( . )2 5 =

16.2830… ft/s. Its speed is increasing atP(2.5) = 2.2052… ft/s2, as shown by the factthat P(2.5) is positive, meaning that the anglebetween

ra( . )2 5 and

rv( . )2 5 is acute.

e. The path is at a high point when they-component of

rr is a maximum. This

happens when cos t = 1, or t = 0 + 2π n.r r ra n i j( )0 2 0 12+ = −π , pointing straightdown.Similarly, the path is at a low point whencos t = −1, or t = π + 2π n.r r ra n i j( )π π+ = +2 0 12 , pointing straight up,Q.E.D.

f. dL dx dy= +2 2

= +( – cos ) (– sin )5 12 122 2t t dt

L dL= ≈ …

≈∫ 78 7078

78 70

2

. (numerically)

. ft

π

18. Recall that | | | | .r ri j= = 1

The angle between r ri iand is 0, so

r r r ri i i i⋅ = =| || | .cos 0 1

Similarly, 1.r rj j⋅ =


The angle between

r ri jand is ,

π2

so

r r r ri j i j⋅ = = | | | | .cos

π2

0

∴ ⋅ = + ⋅ + = ( ) ( )r r r r r rv v x i y j x i y j1 2 1 1 2 2

x x i i x y i j y x j i y y j j1 2 1 2 1 2 1 2

r r r r r r r r⋅ + ⋅ + ⋅ + ⋅

= ( ) + ( ) + ( ) + ( )x x x y y x y y1 2 1 2 1 2 1 21 0 0 1= ,x x y y1 2 1 2+ Q.E.D.

19. r r rr t t i t j( ) sin cos= +( . ) ( . )10 0 8 10 0 6

+ ( ).6 0 5t kr

r r rv t t i t j( ) cos sin= + −( . ) ( . )8 0 8 6 0 6

+−( ).3 0 5t k

r

r ra t t i( ) sin= −( . . )6 4 0 8

+ − + − −( . cos . )3 6 0 6 1 5 1 5t j t kr r

( . ).

r r r rv i j k( ) ( . ) ( . )1 8 0 8 6 0 6 3= + − +cos sinr r r ra i j k( ) sin cos .1 6 4 0 8 3 6 0 6 1 5= − + − −( . . ) ( . . )

To determine whether the object is speeding upor slowing down, find the dot product.r ra v( ) ( ) sin cos1 1 6 4 0 8 8 0 8⋅ = − +( . . )( . )( . . )( . ) ( . )( )− − + −3 6 0 6 6 0 6 1 5 3cos sin

= −20.0230…∴ the object is slowing down because the anglebetween

ra( )1 and

rv( )1 is obtuse.

20. a. This is an example of the chain rule.

b. dy/dx equals the slope of the velocity vector,and tan φ also equals the slope of this vector.Thus, tan φ = dy/dx.

By the chain rule, dy/dx = (dy/dt)/(dx/dx),Q.E.D.

c. tan/

/

( )

( )φ = = ′

′dy dt

dx dt

y t

x t

⇒ = ′′

d

ds

d

ds

y t

x t(tan )

( )

( )φ

⇒ = ′ ′′ ′′ ′′

sec–2

2φ φd

ds

x y x y

x

dt

ds

∴ = ′ ′′ ′′ ′′

d

ds

x y x y

x ds dt

φφ–

(sec )( / )2 2

But sec2 φ = 1 + tan2 φ = 1 + y′2/x′2, so

ds dt v/ | |.=r

∴ = ′ ′′ ′′ ′

′ + ′ ′

d

ds

x y x y

x y x v

φ –

( / ) | |2 2 21r

= ′ ′′ ′′ ′

′ + ′= ′ ′′ ′′ ′x y x y

x y v

x y x y

v

–

( ) | |

–

| |2 2 3r r , Q.E.D.

d. x = 5 cos t ⇒ x′ = −5 sin t ⇒ x″ = −5 cos t

y = 3 sin t ⇒ y′ = 3 cos t ⇒ y′′ = −3 sin t

| |rv t t= +25 92 2sin cos

∴ = ++

d

dt

t t

t t

φ 15 15

25 9

2 2

2 2 3 2

sin cos

( sin cos ) /

=+

15

16 92 3 2( sin ) /t

which is maximized for sin2 t = 0 at (± 5 , 0),the ends of the major axis, and is minimizedfor sin2 t = 1 at (0, ±3), the ends of the minoraxis, Q.E.D.

e. x = r cos t ⇒ x′ = −r sin t ⇒ x′′ = −r cos ty = r sin t ⇒ y′= r cos t ⇒ y′′ = −r sin t

| |rv r t r t r= + =2 2 2 2cos sin

∴ = + = d

ds

r t r t

r r

φ 2 2 2 2

3

1sin cos

| | | |a constant equal to the reciprocal of the radius.

f. x = 5 cos2 t ⇒ x′ = −10 cos t sin t = −5 sin 2t⇒ x′′ = −10 cos 2ty = 3 sin2 t ⇒ y′ = 6 sin t cos t = 3 sin 2t⇒ y′′ = 6 cos 2t

| | | |rv t t t= + =25 2 9 2 34 22 2sin sin sin

∴ = +d

ds

t t t t

t

φ – sin cos sin cos

|sin |

30 2 2 30 2 2

34 2 = 0, Q.E.D.

g. At (5, 0), sin t = 0, so d

ds

φ = =15

9

5

93 2/ .

Radius of curvature = =9

51 8.

h. The osculating circle has radius 1.8 and centeron the x-axis at x = 4 − 1.8 = 3.2. Equationsare

x = 3.2 + 1.8 cos t

y = 1.8 sin t

The graph shows the osculating circle. Thename is appropriate because the circle “kisses”the ellipse at the point (5, 0).

3.2 5

3

x

y



R1. v t= − =3 0 at t = 9 s

v > 0 for t > 9 s


Displacement from t = 0 to t = 9 is

( – )t dt3 90

9

= −∫ .

They have moved 9 ft closer to the sawmill.

From t = 0 to t = 25:

Displacement = =∫ ( – )t dt3 81

30

25

ft

Distance = − =∫ | | ftt dt3 261

30

25

R2. a. i.

3

5

t

v(t)

ii. Displacement = ≈ −∫ ( – )2 8 3 80221

4t dt . K

≈ −3.8 cm (exactly 14/ln 2 − 24)

iii. Distance = −∫ | |2 81

4t dt ≈ 10.8853…

≈ 10.9 cm (exactly 2/ln 2 + 8)

b.

tend a aav

0 2 — 30 speeding up

5 8 5 55 speeding up

10 1 4.5 77.5 speeding up

15 0 0.5 80 neither

20 −10 −5 55 slowing down

25 −20 −15 −20 slowing down

(Note that the object is speeding up, slowingdown, or neither, exactly when aend > 0,aend < 0, or aend = 0, respectively, in theoriginal table.)

R3. a. i. v t dtav ( / ) / .= = =∫1

36 2 0 6366

0

3

sin π π K

ii. v t dtav ( / )= =∫1

66 0

3

9

sin π

iii. v t dtav ( / )= =∫1

126 0

0

12

sin π

b. i. f x x x x x x( ) ( ) at= − = − = =6 6 0 0 62 3 2 ,

Average = =∫1

66 182 3

0

6

( – )x x dx

ii. The rectangle has the same area as theshaded region.

5

18

6

x

f(x)

iii. The average of the two values of f(x) at theendpoints is 0, not 18.

R4. a.

θ θ

x

200

700

700 – x

Let x = distance from intersection to cutoff.0 ≤ x ≤ 700Let T = total time taken.

T x x= + +1

5 7200

1

6 27002 2

. .( – )

700

100

x

T

T x x

x x

′ = ⋅ + −

= ⋅ + −

−

−

1

5 7

1

2200 2

1

6 21

5 7200

1

6 2

2 2 1 2

2 2 1 2

.( )

.

. .

/

/

( )

( )

T′ = 0 ⇔ 6.2x = 5.7(2002 + x2)1/2

38.44x2 = 32.49(2002) + 32.49x2

x x2

232 49 200

5 95467 3544= ⇔ = ±. ( )

.. K

Or: Let θ = angle of incidence.Minimal path occurs for θ = sin− 1 (5.7/6.2).x = 200 tan θ = 467.3544…Note that at x = 0, Juana goes entirely alongthe sidewalk.

T( ) .0

1

6 2900 145 1612= ⋅ =

.K

T(467.3544…) = 126.7077…T(700) = 127.7212…

Turning at a point about 467 ft from theintersection of the two sidewalks takes theminimum time, although it takes only asecond longer to head straight for the Englishbuilding.


b.

θ

θx

6

10

10 – x

Let x = distance from closest point on thebeach to the cutoff point, 0 ≤ x ≤ 10.Let C = total cost of the road.

C x x= − + +5 10 13 36 2( )The graph shows a minimum C at x ≈ 2.5.

10

150

x

C

Let θ be the angle of incidence.By the minimal path property, the cost isminimized when

sin θ = =x

bridge length

5

13x = 6 tan [sin− 1 (5/13)] = 2.5C(2.5) = 122C(10) = 151.6047…The minimum cost is $122,000, obtained bygoing 7.5 km along the beach, then cuttingacross to the island. This path saves about$29,600 over the path straight to the island.

R5. a. i. a(t) = 6t − t2, t in [0, 10]a′(t) = 6 − 2ta′(t) = 0 ⇔ t = 3a(0) = 0; a(3) = 9; a(10) = −40Maximum acceleration = 9 at t = 3.Minimum acceleration = −40 at t = 10.

ii. v t t t dt t t C( ) ( )= − = − +∫ 6 31

32 2 3

v(0) = 0 ⇒ C = 0

∴ = − ( )v t t t31

32 3

v′(t) = a(t) = t(6 − t)

v′(t) = 0 ⇔ t = 0 or t = 6

v v v( ) ; ( ) ; ( )0 0 6 36 10 331

3= = = −

Maximum velocity = 36 at t = 6.

Minimum velocity at .= − =331

310t

iii. s t v t dt t t C( ) ( )= = − +∫ 3 41

12

Because s(t) measures distance from thestarting point, s(0) = 0, which implies thatC = 0.

∴ = − ( )s t t t3 41

12

s t v t t t′ = = −( ) ( ) ( )1

392

s′(t) = 0 ⇔ t = 0 or t = 9

s s s( ) ; ( ) ; ( )0 0 9 1821

410 166

2

3= = =

Maximum displacement at .= =1821

49t

Minimum displacement = 0 at t = 0.

b. i. Let t = number of days Dagmar has beensaving, P(t) = number of pillars inDagmar’s account, and V(t) = real value(in constant day-zero pillars) of money inaccount after t days.P(t) = 50t (assuming continuousdepositing)V(t) = P(t)(0.50.005 t ) = 50t(0.50.005 t )

ii. The graph shows a maximum V(t) at t ≈289 days.

6000

500

t

V(t)

V′(t) =50(0.50.005 t ) + 50t[0.005(0.50.005 t )] ln 0.5V′(t) = 0 ⇔ 1 = −0.005t ln 0.5

t = − =200

0 5288 5390

ln .. K

Dagmar’s greatest purchasing power willbe after about 289 days.

R6. a. i. and ii.

v

a

a

Speeding up

Slowing down

b. i. r r rr t t i t j( ) cosh sinh= +( ) ( )5 3r r rv t t i t j( ) sinh cosh= +( ) ( )5 3r r ra t t i t j( ) cosh sinh= +( ) ( )5 3

r r r

Kr

Kr

r i j

i j

( ) cosh sinh1 5 1 3 1

7 7154 3 5256

= +

= +

( ) ( )

. .

r

Kr

Kr

v i j( )1 5 8760 4 6292= +. .

r

Kr

Kr

a i j( )1 7 7154 3 5256= +. .


ii. The graph shows r rr v( ) ( )1 1, , and

ra( )1 .

Note that r ra r( ) ( )1 1= so that the

acceleration vector points directly awayfrom the origin when it is drawn from thehead of

rr ( ).1 (For an elliptical path, the

acceleration vector points directly towardthe origin.)

5 10

5

x

y

r

v

aAsymptote

iii. Speed | |= = + =rv ( ) sinh cosh1 25 1 9 12 2

7.4804… ≈ 7.48 units/minr r

Ka v( ) ( ) sinh cosh1 1 34 1 1 61 6566⋅ = = .The object is speeding up at t = 1, asshown by the positive dot product and bythe acute angle between

ra( )1 and

rv( )1 .

rr r

r

r

r

r

r

r r

aa v

v

v

v

v

v

i j

t ( )

.

. .

11 1

1

1

1

34 1 1

25 1 9 11

1 1018 1

6 4744 5 1007

2 2

= ⋅

=+

= …

= … + …

( ) ( )

| ( )|

( )

| ( )|

sinh cosh

sinh cosh( )

( )

( ( ) . . )r r ra i jn 1 1 2409 1 5751= … − …

| ( )|r

r r

raa v

vt 11 1

1

61 6566

7 4804= ⋅ = …

…| ( ) ( )|

| ( )|

.

.= …8 2423.

rat ( )1 points in the same direction as

rv( ),1

as indicated by the positive dot product andby the acute angle between

ra( )1 and

rv( ),1

so the object is speeding up at about8.24 units/min2.

iv. Distance = ∫ ds0

1

= +∫ 25 92 2

0

1

sinh cosht t dt

= 4.5841… (numerically) ≈ 4.58 units

v. r r rr t v t t t i( ) ( ) cosh sinh+ = +( )5 5

+ +( )3 3sinh cosht t jr

Note that the y-coordinate is 0.6 times thex-coordinate, so the head lies on y = 0.6x,one asymptote of the hyperbola.

Concept Problems

C1. a.

–5

0 4

t

v(t)5

b. v = t3 − 7t2 + 15t − 9 = (t − 1)(t − 3)2

v = 0 ⇔ t = 1, 3

Particle is stopped at 1 s and 3 s.

c. v′ = 3t2 − 14t + 15 = (3t − 5)(t − 3)v′ = 0 at t = 5/3, 3

v v( ) ; ;0 95

3

32

271

5

27= −

= =

v v( ) ; ( )3 0 4 3= = Maximum velocity at t = 4, minimumvelocity at t = 0.

d. v″(t) = 6t − 14v″(t) = 0 ⇔ t = 7/3v″(t) changes from negative to positive att = 7/3, so there is a point of inflection atthat point.

e. At t = 7/3, the particle’s acceleration stopsdecreasing and starts increasing. Thus, theminimum acceleration is at that time.

f. y v t dt t t t t C= = − + − +∫ ( )1

4

7

3

15

294 3 2

y(0) = 4 ⇒ C = 4

∴ = − + − + y t t t t1

4

7

3

15

29 44 3 2

g.

4

0 4

t

y(t)

h. y′(t) = v(t) = 0 when t = 1, 3.

y y y y( ) ; ( ) ; ( ) ; ( )0 4 15

123

7

44

8

3= = = =

Maximum y at t = 0, minimum y at t = 1.

i. ′′ = ′ = − − = =y v t t t( )( ) at ,3 5 3 05

33

y″ changes sign at t = 5

3and at t = 3, so there

are points of inflection at these values of t.


j. At t = 5/3, the particle stops accelerating andstarts decelerating, so the velocity at that timeis a local maximum. At t = 3, the particlestops decelerating and starts accelerating, sothe velocity is a local minimum.

k. y is never negative because its minimumvalue is 5/12 at t = 1.

l. Displacement ( ) ( ) ( )= = −∫ v t y y4 00

4

= − = −8

34

4

3ft

m. Distance | ( )| ft= =∫ v t dt 55

60

4

n. v v dtav (displacement) ft/s= = ⋅ = −∫1

4

1

4

1

30

4

o. | | | | (distance) ft/sav0

4

v v dt= = ⋅ =∫1

4

1

4

35

24

C2. Assume that the maximum g a human canwithstand is A and that the distance fromNew York to Los Angeles is D km.Recall that 1 g = 9.81 (m/s)/s.For the fastest trip, the passenger accelerates atA g for the first D/2 km, then decelerates at −A gfor the last D/2 km.Starting at rest, the velocity t seconds afterleaving New York, when accelerating atthe maximum rate, is v(t) = A · 9.81 tand the distance from New York is

s t A t( ) . .= ⋅1

29 81 2

The passenger reaches the halfway point of

the trip when s t D( ) = ⋅10001

2 (because D is km

and s is m), so the first half of the trip takes

tD

A= 1000

9 81.seconds. By symmetry, the second

half takes exactly as long, so the minimum time

for the trip is tD

A= 2

1000

9 81.seconds.

For example, suppose that it is 4000 km fromNew York to Los Angeles and that the humanbody can withstand A = 5 g. Then the minimal

time is t = = …21000 4000

9 81 5571 1372

( )

. ( ). , or

about 9.5 min.

C3. Let x = distance from center along clock hand,L = length of web, and θ = central angle.

Know: dx

dt

d

dt= − =0 7

30. cm/s; rad/s.

θ π

Want:dL

dttat s.= 10

By the law of cosines,

L2 = x2 + 252 − 2 · x · 25 · cos θ

2 2 50 50LdL

dtx

dx

dt

dx

dtx

d

dt = − +cos sinθ θ θ

At , cm, , ,t x= = = =10 183

1

2θ π θcos

sin θ = 3

2.

So L = + ⋅ =18 25 25 18 4992 2 –

dL

dt= − ⋅ + ⋅ ⋅

1

49918 0 7 25

1

20 7. .

+ ⋅ ⋅ ⋅

25 183

2 30

π

= 1.6545… cm/s

C4. a. Let t = time since vertex of cone touchedwater, y = distance from vertex of cone tobottom of cylinder (0 ≤ y ≤ 15), h = altitudeof submerged part of cone, r = radius ofsubmerged part of cone, and D = depth ofwater in cylinder.

Know: dy

dt

dD

dt= −2 cm/ . Want: .min

Volume of water is 15 · 72π = 735π cm3.

Volume of submerged part of cone is 1

32πr h.

Volume of submerged part of cone plus wateris π · 72 · D.

∴ = + 49 7351

32π π πD r h

49 7351

32D r h= +

But , and , soD h y r h= + = 5

12

49 7351

3

25

1443D D y= + ⋅ ( – )

4925

1442dD

dtD y

dD

dt

dy

dt= −

( – )

Find D when y = 10.

49 73525

43210 3D D= + ( – )

Solving numerically gives D ≈ 15.1624… .

Substitute this for D, 10 for y, and −2 fordy/dt.

49

25

1445 1624 22dD

dt

dD

dt= +

( . )K


Solving algebraically or numerically givesdD

dt= … ≈0 2085 0 21. . cm/ .min

b. When the cone is completely submerged, thetotal volume is

7351

35 12 8352π π π+ ⋅ ⋅ = .

In this case, D = = …835

4917 0408

ππ

. .

When the cone first becomes completelysubmerged,y = 17.0408… − 12 = 5.0408… .Thus when y = 1 the cone is alreadycompletely submerged and the depth D is notchanging.

C5. Let h(t) = f(t) − g(t), so h′(t) = f ′(t) − g′(t). Thenh(a) = h(b) = 0 because f(a) = g(a), f(b) = g(b).

By the mean value theorem (or Rolle’s theorem),there exists an x = c in (a, b) such that

h ch b h a

b a′ = =( ) .

( ) – ( )

–0

But because h′(c) = f ′(c) − g′(c) at time c,f ′(c) = g′(c), so the knights have the samevelocity at this time.

C6. Let L = length of track, z = vertical coordinate ofa point on the track, and T( θ ) = number ofminutes to reach the top.v = 30 − 60 sin θDomain of θ is 0 6≤ ≤θ π / because v would be

negative for acute angles θ > π/6.Because θ is constant, v is constant. Thus,

TL

v

L( ) .θ

θ= =

30 60– sin

To find L for any value of θ, consider z to be anindependent variable. By trigonometry,dL

dzdL dz= =csc cscθ θ, and thus .

L dzz

= ==∫ csc cscθ θ θ1000

0

1000

( is constant)

(Another way to find L is to “unroll” the trackinto a vertical plane. Because the track alwaysmakes an angle of θ with the horizontal, thiswill result in a right triangle with hypotenuse =L, altitude = 1000, and base angle θ. Thensin θ = 1000/L so that L = 1000 csc θ.)

∴ = ( )T θ θθ

1000

30 60

csc

– sin

= −100

32 2 1(sin – sin )θ θ

T ′ = − − ⋅−( )θ θ θ100

32 2 2(sin sin )

(cos )θ θ θ− 4 sin cos

T′(θ ) = 0 ⇔ cos θ ( 1 − 4 sin θ ) = 0cos θ = 0 only for values of θ outside thedomain.∴ 1 − 4 sin θ = 0 ⇔ θ = sin− 1 0.25T(θ ) approaches positive infinity as θ approacheseither end of the domain. So T(θ ) is a minimumfor θ = sin− 1 0.25.Optimal trip takes T(sin− 1 0.25)

= =−100

30 25 2 0 25 266

2

32 1[ . – ( . ) ] s,

or 4 minutes 262

3seconds.

Chapter Test

T1. If the acceleration and velocity have the samesign, the object is speeding up. If the accelerationand velocity have opposite signs, the object isslowing down.

T2. Displacement = ∫ y dx0

60

= ⋅ + + ⋅ − ⋅ +

= + − = −

820 10

215 0 10

25 15

2120 0 200 80 ft

Distance | | ft= = + + =∫ y dx 120 0 200 3200

60

T3. Average value of f ( x) ≈ 2.8. The graph showsequal areas above and below y = 2.8, and thepoint x = c where f ( c) ≈ 2.8 by the mean valuetheorem for integrals.

5 10

5

f(x)

x

c

Average ¯ 2.8

Equal areas

T4. r r rr i j= +7 3The velocity vector points in the negativex-direction and the acceleration vector makes anobtuse angle with the velocity vector, indicatingthat the object is slowing down.

5 10

5

y

x

v

a

→

→

T5. v t= + 60

Displacement ft= + =∫ ( )t dt60 15831

30

25


T6.

θθ

x 7 – x

3

7

By the minimal path property,

sin θ = =360

8000 45. .

∴ θ = sin− 1 0.45

x = 3 tan θ = 3 tan (sin− 1 0.45) = 1.5117…

So the cheapest path is 7 − x = 5.4882… mialong the road, then turning toward Ima’s house.

This path costs

360 5 4882 800 9 1 5117 2⋅ + ⋅ +. K K.= 4663.2685… ≈ $4663, which is about $257cheaper than the proposal.

T7. f(x) = x3 − 4x + 5, x ∈ [1, 3]

f ′(x) = 3x2 − 4

′ = ⇔ = ± = ±f x x( ) .0 4 3 1 1547/ K

(The negative value is out of the domain.)

f(1) = 2

f(1.1547…) = 1.9207… , the minimum.

f(3) = 20, the maximum.

Average = + =∫1

24 5 73

1

3

( – )x x dx

The graph shows a minimum at x = 1.1547… ,a maximum at x = 3, and an average of 7. Thearea of the rectangle of altitude 7 equals the areaof the region under the graph.

7

20

1 3

x

f(x)

T8. a. v(t) = 10(0.5 − 2− t)

Distance = = ≈∫ | ( ) | . . ftv t dt 3 6067 3 610

2

K

(exactly 2.5/ln 2)

Displacement ( )

. . ft

=

= − ≈ −∫ v t dt

0

2

0 8202 0 82K

(exactly 10 − 7.5/ln 2)

b. a(0) = v′(0) = 10 ⋅ 2− 0 ln 2 = 10 ln 2

= 6.9314… ≈ 6.93 (ft/s)/s

c. v(0) = −5

Because v(0) is negative and a(0) is positive,the object is slowing down.

T9.

t a aav vend vav displend

0 4 — 50 — 0

7 6 5 85 67.5 472.5

14 10 8 141 113 1263.5

21 13 11.5 221.5 181.25 2532.25

The object traveled about 2532.25 cm.

Average velocity was about

1

212532 25 120 583 120 6⋅ = ≈( . ) . . cm/s.K

T10. r r rr t t i t j( ) cos sin= +( . ) ( . )10 0 4 10 0 6

r r rv t t i t j( ) sin cos= − +( . ) ( . )4 0 4 6 0 6

T11. r r ra t t i t j( ) cos sin= − + −( . . ) ( . . )1 6 0 4 3 6 0 6

T12. r r rr i j( ) cos sin2 10 0 8 10 1 2= +( . ) ( . )

= +6 9670 9 3203. .K Ki j

The graph shows rr ( )2 .

5

5

0x

y

r

a

v

t = 2

T13.r r rv i j( ) sin cos2 4 0 8 6 1 2= − +( . ) ( . )

= − +2 8694 2 1741. .Kr

Kr

i j

See the graph in Problem T12.

The velocity vector is tangent to the path,pointing in the direction of motion.

T14. r r ra i j( ) cos sin2 1 6 0 8 3 6 1 2= − + −( . . ) ( . . )

= − −1 1147 3 3553. .Kr

Kr

i j

See the graph in Problem T12.

T15. r ra v( ) ( )2 2⋅

= 6.4 cos 0.8 sin 0.8 − 21.6 cos 1.2 sin 1.2= −4.0963… ≈ −4.10 (mi/h)2/h

| ( ) sin . cos .rv 2 16 0 8 36 1 22 2| = +

= 3.6000… mi/h

Pa v

v= ⋅ = − ≈ −

r r

r K( ) ( )

| ( ) |

2 2

21 1378 1 14. . (mi/h)/h

rr

rK

rK

ra

P

vv i jt ( ) . .2

2

22 0 9069 0 6871= = −( )

| ( ) |( )


r r ra a an t( ) ( )2 2 2= −( )

= − −2 0216 2 6681. .Kr

Kr

i j

t = 2

an

at

a

v

T16. The tangential component rat ( )2 has direction the

opposite of rv( ),2 so

rv is decreasing and the object

is slowing down at t = 2.

T17. Object is slowing down at |ra Pt ( )| | ( )|2 2= =

1 1378. (mi/h)/h.K

T18. ran ( )2 points inward to the concave side because

ran is the component of acceleration that pulls theobject out of the straight path into a curve.

T19. dL dx dy v t dt= + =2 2 | |r( )

= ⋅ + ⋅16 0 4 36 0 62 2sin . cos .t t dt

L dL= =∫ 10 0932

0

2

. K (numerically) ≈ 10.09 mi



Chapter 11—The Calculus of Variable-Factor Products

Problem Set 11-11.

4

10

x

F

(x, F)

∆x = 0.2

F ≈ F(4) = 80e− 2 = 10.8268… ≈ 10.83 lbin the strip.W ≈ F(4) · ∆x = 16e− 2 = 2.1653… ≈ 2.17 ft-lb

2. dW = 20xe− 0.5 x dx

3. W xe dxx= =−∫ 20 69 12890 5

0

7. . K

(exactly 80 − 360e− 3.5 )

4. W ≈ 69.13 ft-lb

5. The amount of work done from x = 0 to x = b is

W x dx

be e

xb

b b

=

= − − +

−

− −

∫ 20

40 80 80

0 5

0

0 5 0 5

.

. .

limb

W→∞

= + + =0 0 80 80 ft-lb

(Use l’Hospital’s rule for be− 0.5 b.)

Problem Set 11-2

Q1. 2 Q2. 102

3

Q3. v(t) = ln | sec t | + C Q4. a ( t) = t− 1

Q5. Fundamental theorem of calculus

Q6. Riemann

Q7. Integration by parts

Q8. Implicit differentiation

Q9. Heaviside method Q10. A

1. Ignore the weight of the rope.Let y = the distance from the bottom of the well.Slope of linear function is −8/50 = −0.16.Weight = 20 − 0.16ydW = (weight) dy = (20 − 0.16y) dy

W y dy= =∫ ( – . )20 0 16 8000

50

ft-lb

2. a. Let y = number of miles up.

Slope of linear function is − = −20 702

7/ .

Weight (tons)= −302

7y

dW y dy= −

30

2

7

W y dy=

=∫ 30

2

71400

0

70

– mi-tons

b. W total = 90 tons · 70 mi = 6300 mi-tonsExcess energy becomes kinetic energy ofrocket and spent fuel.

3. Hooke’s law: F = k · sdW = ks ds

W ks ds k= =∫ 500

10

4. a. F = −x3 + 6x2 − 12x + 16

The graph starts at the high force of 16 lb,levels off, then drops to F = 0 at x = 4.

4

15

x

F

b. dW = F dx = (−x3 + 6x2 − 12x + 16) dx

W x x x dx= + + =∫ (– – )3 2

0

4

6 12 16 32 ft-lb

5.

10

(x, y)

17

15

dV = π x2 dy∴ dW = (17 − y)(62.4)(π x2 dy)

By similar triangles, x = 1.5y.∴ dW = (17 − y)(62.4)(π · 2.25y2 dy)

= 140.4π (17y2 − y3) dy

W dW= =∫ 1 396 752 09370

10

, , . ...

≈ 1.4 million ft-lb (exactly 444,600π)

6. dV = π x2 dyAt x = 4, y = 16.∴ dW = (26 − y)(54.8)(π x2 dy)

Because y = x2,dW = (26 − y)(54.8)(π y dy)

= 54.8π (26y − y2) dy

W dW= ∫0

16

= 337,891.2751… ≈ 337,891 ft-lb


7. a. Draw x- and y-axes with the origin at thecenter of the sphere. To fill the sphere halffull, the water must be pumped from −120 toy, where y is negative. Integration is fromy = −20 to y = 0.

dV = π x2 dyx2 + y2 = 202 ⇒ x2 = 400 − y2

∴ dW = [y − (−120)](62.4)[π (400 − y2) dy]= 62.4π (y + 120)(400 − y2) dy

W dW= =−∫ 117 621 229

20

0

, ,

≈ 117.6 million ft-lb(exactly 62.4π · 600000)

b. For filling the tank, the limits ofintegration are from −20 to 20.

dW =

−∫ 250 925 288 420

20

, , . K

≈ 250.9 million ft-lb

(exactly 62.4π · 1,280,000)This answer can be found quickly by liftingthe entire weight of the water through thedistance the center of the sphere moves,namely 120 ft.

W =

⋅

= ⋅

( . )

. , ,

62 44

320 120

62 4 1 280 000

3( )( )π

π

(Note that the amount of work to fill theentire tank is more than twice the amountneeded to half-fill it. The work to fill the tophemisphere is greater than that to fill thebottom hemisphere because the same amountof water has to be lifted through a greaterdisplacement.)

8. Slice the water horizontally. Pick sample point(x, y) on the curve y = 0.0002x4 within theslice.dV = 15 · 2x · dy = 30(5000y)1/4 dy

= 300(0.5y)1/4 dy∴ dW = (30 − y)(67)[300(0.5y)1/4 dy]

= 20100(30 − y)(0.5y)1/4 dy

W dW= ≈∫ 9 134 6020

16

, , ft-lb

exactly ( )( . ) /20100 0 5 5404

91 4

9. a. If x is the distance between the piston and thecylinder head and F is the force exerted by thehot gases, then dW = F dx.F = pA, where p is the pressure and A is thearea of the piston.∴ dW = pA dxA dx = dV∴ dW = p dVp = k1V

− 1.4

Initial condition V = 1 at p = 1000 ⇒k1 = 1000.∴ dW = 1000V −1.4 dV

W V dV= ≈−∫ 1000 1504 73201 4

1

10. . K

≈ 1504.7 in.-lb (exactly 2500(1 − 10− 0.4 ))

b. Initial condition p = 15 at V = 10 ⇒k2 = 15 ⋅ 101.4

dW = 15 ⋅ 101.4V −1.4 dV

W V dV= ⋅ ≈ −−∫ 15 10 566 95741 4 1 4

10

1. . . K

So about 567 in.-lb of work is done incompressing the gases (exactly37.5 ⋅ 101.4 (10− 0.4 − 1)).(Mathematically, the work is negative becausethe force is positive and dx is negative.Physically, the work is negative becauseenergy is taken out of the surroundings to putinto the gases. Positive work indicates thatenergy is put into the surroundings by theexpanding gases.)

c. Net amount of work ≈ 1504.7320… −566.9574 = 937.7746… ≈ 937.8 in.-lb

d. Carnot (kar-NO), Nicolas Léonard Sadi,1796−1832, was a French physicist and apioneer in the field of thermodynamics.

Problem Set 11-3Q1. 2 cm3 Q2. 3 cm3

Q3. Q4.

π

1 x

y

1

x

y

Q5. Q6.

1

x

y

x

y

Q7. (mass)/(volume) Q8. (force)(displacement)

Q9.1

1 2– xQ10. B

1. a. The graph shows y = ln x, rotated aboutx = 0, showing back half of solid only.

(x, y)

1 3

1x

y


Slice the region parallel to the axis ofrotation, generating cylindrical shells. Picksample point (x, y) on the curve, within theslice.ρ = kx− 1

dm = ρ dV = (kx− 1)(2π x ln x dx)= 2π k ln x dx

m k x dx k= ≈∫ 2 8 14191

3

π .ln K

(exactly 2π k(3 ln 3 − 2))

b. Slice perpendicular to the axis of rotation,generating plane washers.ρ = 5 + 2ydm = ρ dV = (5 + 2y) ⋅ π ( 32 − x2) dy

= π ( 5 + 2y)(9 − e2y) dy

m dm= ≈∫ 108 1103

0

3

. Kln

(exactly π [36 ln 3 + 9 (ln 3)2 − 16])

2. The graph shows y = sin x, rotated about they-axis, showing back half of solid only.

π

1

y

x

Slice the region parallel to the axis of rotation,generating cylindrical shells.ρ = kxdm = ρ dV = kx ⋅ 2π xy dx = 2π kx2 sin x dx

m dm k k= ≈ −∫ 36 8798 2 42

0. (exactly ( ) )K π π

π

3. a. The graph shows y = 9 − x2, rotated about they-axis.

3

9

x

y

(x, y)

Slice the region perpendicular to the axis ofrotation, generating plane disks.ρ = k,dm = k dV = k ⋅ π x2 dy = k ⋅ π ( 9 − y) dy

m dm k= =∫ 40 50

9

. π

Or: Slice parallel to the axis of rotation.dm = k ⋅ 2π xy dx = 2π kx(9 − x2) dx

m dm k= =∫ 40 50

3

. π

Or: Volume of paraboloid is half the volumeof the circumscribed cylinder, or0.5(π ⋅ 32)(9) = 40.5π, so m = 40.5π k.

b. Slice perpendicular to the axis of rotation.ρ = ky2

dm = ρ dV = ky2 ⋅ π ( 9 − y) dy

m dm k= =∫ 546 750

9

. π

c. Slice parallel to the axis of rotation.ρ = k(1 + x)dm = ρ ⋅ 2 π xy dx = 2π kx(1 + x)(9 − x2) dx

m dm k= =∫ 105 30

3

. π

d. The solid in part b has the largest mass.

4. a. The graph shows y x1 = and y2 = 0.5x,intersecting at (0, 0) and (4, 2), rotated aboutthe x-axis, showing back half of solid only.

4

2

x

y

y1

y2

Slice perpendicular to the axis of rotation,generating plane washers.Pick sample points (x, y1) and (x, y2).ρ = kxdm dV kx y y dx

kx x x dx

= = ⋅ −

= −

ρ π

π

( )

( . )

12

22

20 25

m k k= =16

316 7551π . K

b. Slice parallel to the axis of rotation,generating cylindrical shells.

Pick sample points (x1, y) and (x2, y).ρ = ky2

dm = ρ dV = ky2 ⋅ 2π y(x2 − x1) dy= 2 π ky3(2y − y2) dy

m dm k k= = =∫ 64

1513 4041

0

2

π . K

5. a. Prediction: The cone on the left, with higherdensity at its base, has greater mass becausehigher density is in the larger part of the cone.

b. Set up a coordinate system with the originat the center of the base. Slice each coneperpendicular to its axis, generating planedisks.Pick sample point (x, y) on the element ofthe cone, y = 6 − 2x.dV = π x 2 dy = π ( 3 − 0.5y)2 dyFor the cone on the left, ρ = 80 − 5y.dm = (80 − 5y) · π(3 − 0.5y)2 dy

m dm= =∫ 13050

6

π oz


For the cone on the right, ρ = 50 + 5y.dm = (50 + 5y) · π ( 3 − 0.5y)2 dy

m dm= =∫ 10350

6

π oz

∴ the cone on the left has the greater mass, aspredicted in part a.

6. a. Prediction: The cylinder on the left, withhigher density at walls, has greater massbecause higher density is in the larger partof the cylinder.

b. Set up a coordinate system with the origin atthe center of the bottom base. Slice eachcylinder parallel to its axis, generatingcylindrical shells. Pick sample point (x, 6).dV = 2π x · 6 · dx = 12π x dxFor the cylinder on the left, ρ = 50 + 10x.dm = (50 + 10x) · 12π x dx

= 12π (50x + 10x2) dx

m dm= =∫ 37800

3

π oz

For the cylinder on the right, ρ = 80 − 10x.dm = (80 − 10x) · 12π x dx

= 12π ( 80x − 10x2) dx

m dm= =∫ 32400

3

π oz

∴ the cylinder on the left has the greatermass, as predicted in part a.

7. y1 = 4 − 2x2 and y2 = 3 − x2, rotated aboutthe x-axis.The graphs intersect at (1, 2) in Quadrant I.Slice perpendicular to the axis of rotation,generating plane washers.

Pick sample points (x, y1) and (x, y2).

ρ π= = −kx dV y y dx212

22, ( )

dm = ρ dV = π kx2(7 − 10x2 + 3x4) dx

m dm k k= = =∫ 16

212 3935

0

1

π . K

8. Rotate the region in Problem 7 about the y-axis.Slice the region parallel to the axis of rotation,generating cylindrical shells.Pick sample points (x, y1) and (x, y2).

ρ = −e x , dV = 2π x(y1 − y2) dx = 2π x (1 − x2) dx

dm dV xe x dxx= = −−ρ π ( ) 2 1 2

m dm e= = −−∫ 0 9444 2 14 51

0

1

. (exactly ( ))K π

9. Set up axes with the origin at the center of thelower base and the y-axis coaxial with thecylinder’s axis.Slice perpendicular to the axis of the cylinder,generating plane disks of constant radius 0.5.

ρ is given in the table in the text.

dV dy dy= =π π0 5

42.

dm dV dy= =ρ π ρ4

m dy= ∫π ρ4 0

2

Simpson’s rule cannot be used because there isan odd number of increments. Use the trapezoidalrule.

m ≈ + + + + +

π4

0 41

210 9 9 9 8 9 6 9 4

1

29 0( . ) ( ) . . . . ( . )

= 4.82π ≈ 15.14 g

10. a. The graph shows y1 = 4 − x2 and y2 =4x − x2, intersecting at (1, 3), rotated aboutthe y-axis, showing back half of solid only.

(x, y )1

(x, y )2

x

y4

1

Slice parallel to the axis of rotation,generating cylindrical shells.Pick sample points (x, y1) and (x, y2).ρ = kx, dV = 2π x (y1 − y2) dx

= 2π x (4 − 4x) dxdm = ρ dV = 2π k x2(4 − 4x) dx

m dm k k= = =∫ 2

32 0943

0

1

π . K

b. The graph shows the curves in part a, rotatedabout the x-axis, showing back half of solidonly.

(x, y )1

(x, y )2

4

1

x

y

Slice perpendicular to the axis of rotation,generating plane washers.

Pick sample points (x, y1) and (x, y2).

ρ π= = −kx dV y y dx , ( )12

22

dm = ρ dV = π kx(16 − 24x2 + 8x3) dx

m dm k k= = =∫ 3 6 11 3097

0

1

. . π K


c. The region is rotated about the y-axis asin part a.Slice perpendicular to the axis of rotation,generating plane disks.Pick sample points (x1, y) and (x2, y).Below y = 3, disks have radius x2.Above y = 3, disks have radius x1.

ρ = ky

For in [ , ], x dV x dx0 3 22= π

= −π ( ) .2 4 2– y dy

For x in [3, 4], dV x dx y dy= = −π π ( ) .12 4

m ky y dy= ⋅ −∫ π ( )2 4 2

0

3

–

+ ⋅ −∫ ky y dyπ ( )43

4

= + = =114

151

2

33 6 11 3097π π π . .k k k kK

(Coincidentally, this answer equals the answerto part b.)

11. a. The graph shows a sphere with origin at itscenter.

(x, y)

r

r y

x

Slice the upper semicircular regionperpendicular to the x-axis and rotate it to getplane disks.Pick a sample point (x, y).Equation of the circle in the xy-plane isx2 + y2 = r2.ρ = k|x|, dV = π y2 dx = π (r2 − x2) dxdm = ρ dV = k|x| π (r2 – x2) dx

m k x r x dx

k r x x dx

k r x x r k

r

r

r

r

= −

=

= −

=

∫∫

π

π

π π

| | ( )2 2

2 3

0

2 2 4

0

4

2

21

2

1

4

1

2

–

( – )

b. Slice the right semicircular region parallel tothe y-axis, and rotate it to get cylindricalshells coaxial to the y-axis.

ρ π π = = ⋅ ⋅ =kx dV x y dx x r x dx , 2 2 4 2 2–

. dm dV kx r x dx= =ρ π4 2 2 2 –

m k x r x dxr

= ∫4 2 2 2

0π –

Let x = r sin θ.

Then , .dx r d r x r= =cos – cosθ θ θ2 2

x r= ⇒ = =−θ πsin 1 1 2/

∴ = ⋅ ⋅∫ m k r r r d4 2 2

0

2

π θ θ θ θπ

sin cos cos/

= ∫4 4 2 2

0

2

π θ θ θπ

r k dsin cos/

= ∫π θ θπ

r k d4 2

0

2

2sin/

(half-argument property)

= ∫1

21 44

0

2

π θ θπ

r k d( – cos )/

(half-argument property)

= −

=1

2

1

44

1

44

0

22 4π θ θ π

π

r k r ksin/

c. Slice into spherical shells. Pick a samplepoint on the x-axis within the shell.Then x is the radius of the shell and 4π x2

is the area of the shell at the sample point.

∴ dV = 4π x2 dx

ρ = kx

dm = ρ dV = 4π kx3 dx

m k x dx kx krr r

= = =∫4 3

0

40

4π π π

12. Assume Earth is spherical, with radius3960 mi = 3960 ⋅ 5280 ⋅ 12 ⋅ 2.54 cm =637,300,224 cm, and slice into sphericalshells with radius x and dV = 4π x2 dx.

ρ = −128

6373002243x

g cm/

dm dV xx

dx= = −

ρ π π48

32

6373002242

3

m dV= ∫0

637300224

= −

168

6373002243

4

0

637300224

π π

xx

= 8π ⋅ 6373002243 ≈ 6.505 × 1027 g

Mass is about 6.505 × 1021 metric tons!

13. The graph shows y = ex, from x = 0 to π/2,rotated about the y-axis, showing back half ofsolid only.

(x, y)

1

1x

y


Slice parallel to the axis of rotation, generatingcylindrical shells.Pick a sample point (x, y).ρ = cos x, dV = 2π x ⋅ y ⋅ dx = 2π xex dxdm = ρ dV = cos x 2π xex dx

m dm e= ≈ −

∫ 8 6261

212

0

2

. exactly /K π πππ /

14. a = 4 mi, b = 1 mi, c = 0.5 mi

a.x

a

y

b

z

c

+

+

=

2 2 2

1,

where a = 4, b = 1, c = 0.5.

The cross section at z = z0 < c has equation

x

a

y

b

z

c

+

= −

2 20

2

1 .

This is the equation of an ellipse with

x a z c-radius and1 02– ( / )

y b z c-radius .1 02– ( / )

b. Slice horizontally into plane elliptical disks.The area of the cross section isπ (x-radius)(y-radius) = π ab(1 − (z/c)2)= 4π (1 − (z/0.5)2) = 4π (1 − 4z2).

ρ = 0.08(5280)3 e− 0.2 z lb/mi3

dm = ρ dV = 0.08 ⋅ 52803e− 0.2 z ⋅ 4π (1 – 4z2) dz

= 0.32 ⋅ 52803π e−0.2 z(1 – 4z2) dz

m dm= ≈ ⋅ ⋅∫ 0 32 5280 1 0089533

0

0 5

. . K.

≈ 4.7525… × 1010 lb, or about23,762,540 tons

(exactly 0.32 · 52803π (1100e− 0.1 − 995))

c. Volume of semi-ellipsoid = =2

3

4

33π π miabc

Weight

49,326,507,160 lb,

= ⋅ ⋅

≈

0 08 52804

33. π

1,801,427,783 lb more (≈ 3.8% more thanactual)

d. V ab z c dzc

= −∫2 1 2

0 [ ( / ) ]π

= − ⋅

21

33

0

πab z c z cc

( / )

=

=2

2

3

4

3π πab c abc, Q.E.D.

Problem Set 11-4

Q1. −52 = −25 Q2. (−11)2 = 121

Q3. sin 2x = 2 sin x cos x

Q4. cos ( cos )2 1

21 2x x= +

Q5.– /

( – )

/

( – )

1 3

2

1 3

5x x+

Q6.1

35

1

32ln ln| | | |x x C− − − +

Q7.1 3

2

1 3

52 2

/

( – )

– /

( – )x x+

Q8. g x f x dx g x f x( ) ( ) ( )= ⇔ ′ =∫ ( )

Q9. f ′(2) > 0: increasing Q10. E

1. a. The graph shows y = 9 − x2, rotated aboutthe y-axis.

(x, y)

x

y

3

9

Slice the region perpendicular to the axis ofrotation, generating plane disks.

dV = π x2 dy = π (9 − y) dy

V y dy= − =∫ π π( ) .9 40 50

9

b. Each point in a disk is about y units from thexz-plane, where y is at the sample point (x, y).

dMxz = y dV = π (9y − y2) dy

M y y dyxz = − =∫ π π( ) .9 121 52

0

9

c. y V M yxz⋅ = ⇒ = =121 5

40 53

.

.

ππ

x z= = 0 by symmetry.The centroid is at (0, 3, 0).

2. a.x y

y x12 5

1 25 11

144

2 22 2

+

= ⇒ = −

Slice the ellipsoidal region above the x-axisperpendicular to the x-axis, generating planedisks as the region rotates.Pick sample point (x, y).

dV y dx x dx= = −

π π2 225 1

1

144

V x dx=

∫25 1

1

1442

0

12

π –

= −

=25

1

4322003

0

12

π πx x

This answer equals 2

312 52⋅ ⋅ ⋅π , which is

expected because the volume of a (whole)

ellipsoid is V abc= 4

3π .


b. Each point in a disk is about x units from theyz-plane, where x is at the sample point(x, y).

dM x dV x x dxyz = = −

25 1

1

1442π

M dMyz yz= =∫ 9000

12

π

c. x V M xyz⋅ = ⇒ = =900

2004 5

ππ

.

y z= = 0 by symmetry.

The centroid is at (4.5, 0, 0).

3. a. See the graph in Problem 1.Each point in a disk is about y units from thexz-plane, where y is at the sample point(x, y), so each point has about the samedensity.

ρ = ky1/3, dm = ρ dV = kπ(9y1/3 − y4/3) dy

m dm k= =∫ 170 1375

0

9

. K

exactly =

9

289

33π k

b. Each point in a disk is about y units from thexz-plane, where y is at the sample point(x, y).dMxz = y dm = kπ (9y4/3 − y7/3) dy

M dM

k k

xz xz=

=

∫0

9

43612 4952

9

289. exactlyK π

c. y m M yk

kxz⋅ = ⇒ = 612 4952

170 1375

.

.

K

K

= 3 6. (exactly)x z= = 0 by symmetry.The center of mass is at (0, 3.6, 0).

d. False. The centroid is at (0, 3, 0), but thecenter of mass is at (0, 3.6, 0).

4. a. Slice the ellipsoid as in Problem 2.Each point in a disk is about x units fromthe yz-plane, where x is at the sample point(x, y), so each point has about the samedensity as at the sample point.

ρ π π= = = −

kx dV y dx x dx, 2 225 1

1

144

dm dV kx x dx= = −

ρ π 25 1

1

1442

m dm k= =∫ 2827 4333

0

12

. K

(exactly 900π k)

b. Each point in a disk is about x units from theyz-plane, where x is at the sample point(x, y).

dM x dm kx x dxyz = = −

25 1

1

1442 2π

M dM kyz yz= =∫ 57600

12

π

c. x m M xk

kyz⋅ = ⇒ = =5760

9006 4

ππ

.

y z= = 0 by symmetry.Center of mass is at (6.4, 0, 0).

d. False. The centroid is at (4.5, 0, 0), but thecenter of mass is at (6.4, 0, 0).

5. a. y = ex

Slice the region parallel to the y-axis.dA = y dx = ex dx

A e dx ex= = − =∫ 2

0

2

1 6 3890. K

Each point in a strip is about x units from they-axis, where x is at the sample point (x, y).∴ dMy = x dA = xex dx

M xe dx ey

x= = + =∫ 2

0

2

1 8 3890. K

x A My⋅ = ⇒

xe

e= +2

2

1

1– = 1.3130… (= coth 1)

b. Strips in part a generate plane disks. Eachpoint in a disk is about x units from theyz-plane, where x is at the samplepoint (x, y).dV = πy2 dx = πe2x dx

V e dx ex= = =∫ π π2 4

0

2

21 84 1917 . ( – ) K

dMyz = x dV = πxe2x dx

M dM eyz yz= = + =∫ 3

4

1

4129 42924

0

2

π π . K

x V M xe

eyz⋅ = ⇒ = + =3 1

2 11 5373

4

4( – ). K

c. False. For the solid, x is farther from theyz-plane.

6. a. Slice the region parallel to the y-axis.dA = sec x dx

A x dx= = +∫ sec ln ( )/

2 30

3

π

= 1.3169…dMy = x dA = x sec x dx

M x x dxy = ∫ sec/

0

3π

= 0.7684… (numerically)

x A M xy⋅ = ⇒ = =0 7684

1 3169

.

.

K

KK0.5835


b. Strips in part a generate plane disks. Eachpoint in a disk is about x units from theyz-plane, where x is at the samplepoint (x, y).

dV = π y2 dx = π sec2 x dx

V x dx= = =∫ π π

πsec

/2

0

3

3 5 4413. K

dMyz = x dV = π x sec2 x dx

M x x dxyz = =∫ π

πsec

/2

0

3

3 5206. K

exactlyπ π

2 3

3

1

2+

ln

x V M xyz⋅ = ⇒ = 0 6470. K

exactlyπ3

2

3−

ln

c. False. For the solid, x is farther from theyz-plane.

7. Construct axes with the origin at a vertex and thex-axis along the base, b.Slice the triangle parallel to the x-axis.

The width of a strip is bb

hy− .

dA bb

hy dy= −

dM y dA byb

hy dyx = = −

2

M byb

hy dyx

h

=

∫ – 2

0

= − =1

2 3

1

62 3

0

2byb

hy bh

h

y A M ybh

bhhx⋅ = ⇒ = =

1612

1

3

2

, Q.E.D.

8. a. y = x2/3 from x = 0 to x = 8.Slice the region vertically. Pick a samplepoint (x, y) on the graph within the strip.(See the graph in part e.)

dA = y dx = x2/3 dx

A x dx= =∫ 2 3

0

8

19 2/ .

b. Slice the region parallel to the x-axis so thateach point in a strip is about y units from thex-axis, where y is at the sample point (x, y).

dMx = y(8 − x) dy = (8y − y5/2) dy

M y y dyx = =∫ ( – )/8 27 42855 2

0

4

. K

exactly 273

7

c. Each point in a strip of part a isapproximately x units from the y-axis, wherex is the value in the sample point (x, y).

dMy = x dA = x5/3 dx

M x dxy = =∫ 5 3

0

8

96/

d. x A M xy⋅ = ⇒ = =96

19 25

.

y A M yx⋅ = ⇒ = = …27 4285

19 21 4285

.

.

K.

exactly 13

7

Centroid is at (5, 1.4285…).

e. The balance point is shown on the graph.

8

4

x

y

9. a. Slice the region parallel to the y-axis so thateach point in a strip will be about x unitsfrom the y-axis, where x is at the samplepoint (x, y).

dA = y dx = sin x dx

A x dx= =∫ sin 20

(exactly)π

(This may be “well-known” by now.)dMy = x dA = x sin x dx

M x x dxy = = … =∫ sin0

3 1415π

π. (exactly)

x A M xy⋅ = ⇒ = π2

, Q.E.D.

(Or just note the symmetry.)

b. dM2y = x2 dA = x2 sin x dx

M x x dxy22

05 8696= = …∫ sin

π.

(exactly )π 2 4−

c. x A M xy2

2

2 4

2⋅ = ⇒ = = …π –

1.7131

10. a. Set up axes with the x-axis along the base, B.

dM2B = y2 dA = y2B dy

M y B dy BHB

H

22 3

0

1

3= =∫

b. Set up axes with the x-axis through thecentroid.

dM2c = y2 dA = y2B dy

M y B dy BHcH

H

22

0 5

0 531

12= =∫– .

.


c. Set up axes with the x-axis along the base, B.

dM y dA y BB

Hy dyB2

2 2= = −

M ByB

Hy dy BHB

H

22 3 3

0

1

12=

=∫ –

d. Use the axes in part c. The distance from thecentroidal axis to a sample point (x, y) is

y H− 1

3.

dM y H dA y H BB

Hy dyc2

2 21

3

1

3= −

= −

−

M BH

y y Hy H dycH

H

2 = + +

∫ – –

– /

/ 1 5

3

7

9

1

93 2 2

3

2 3

= − + − +B

Hy By BHy BH y

H

4

5

9

7

18

1

94 3 2 2

0

= 1

363BH

11. a. Slice into cylindrical shells so that each pointin a shell will be about r units from the axis.

The altitude of a shell is a constant, H.dM = r2 dV = r2 2πrH dr

M Hr dr HRR

= =∫ 21

23 4

0π π

r V M rHR

R Hr R2 2

4

2

12 1

2⋅ = ⇒ = ⇒ =

π

πb. Slice the cone into cylindrical shells so that

each point in a shell will be about r unitsfrom the axis.

The altitude of a shell is HH

Rr− .

dM r dV r r HH

Rr dr= = −

2 2 2π

M H rR

r dr HRh

R

=

=∫2

1 1

103 4

0

4π π–

r V M rHR

R Hr R2 2

4

2

11013

0 3⋅ = ⇒ = ⇒ =π

π.

c. Slice the sphere into cylindrical shells so thateach point in a shell is about r units from theaxis.The equation of the sphere is r2 + y2 = R2.The altitude of a shell is 2y.

dM r dV r r R r dr= = ⋅2 2 2 22 2π –

M r R r drR

= ∫4 3 2 2

0π –

Let r = R sin θ.

dr R d R r R= =cos – cosθ θ θ, 2 2

r = 0 ⇒ θ = 0, r = R ⇒ θ π=2

M R R R d= ⋅ ⋅∫4 3 3

0

2

π θ θ θ θπ

sin cos cos/

= ∫4 5 2 4

0

2

π θ θ θ θπ

R d(cos – cos ) sin/

= − +

4

1

3

1

55 3 5

0

2

π θ θπ

R cos cos/

= − + + −

=4 0 0

1

3

1

5

8

155 5π πR R

r V M rR

Rr R2 2

5

3

81543

0 4⋅ = ⇒ = ⇒ =π

π.

12. Assume the clay has uniform density ρ.

Cylinder: H R V R H R= = = =2 2 10002 3C C C, π π

RC

/

cm=

500 1 3

π

Second moment of volume = =1

22 4 5π π( )C C CR R R

(from Problem 11a)

Second moment of mass

= =

= …ρπ ρ

πρRC

/

, .52 3

500500

14 684 1932

Sphere: V R R= = ⇒ =

4

31000

75031 3

ππS S

/

cm

Second moment of volume = 8

155πRS

(from Problem 11c)Second moment of mass

= =

ρ π ρ

π8

15400

75052 3

RS

/

= 15,393.3892…ρThe sphere has higher moment of mass.

13. a. Set up axes with the x-axis through thecentroid.

dM2 = y dA = y2 · B dy

M B y dy yH

H

H

H

22 3

0 5

0 5

0 5

0 51

3= =

− −∫ .

.

.

.

= 1

123BH , Q.E.D.

(Same answer as in Problem 10b)

b. i. B = 2, H = 12; M2 = 288;stiffness = 288k

ii. B = 12, H = 2; M2 = 8; stiffness = 8kA board on its edge is 36 times stiffer.

c. i. Set up axes with the x-axis through thecentroid.From y = 0 to y = 2, dM2 = y2

· 2 dy.From y = 2 to y = 4, dM2 = y2

· 4 dy.


By symmetry, M dM2 20

4

2= ∫= + =∫ ∫2 2 2 4 1602

0

22

2

4

y dy y dy .

Stiffness = 160k

ii. From y = 0 to y = 4, dM2 = y2 · 1 dy.

From y = 4 to y = 6, dM2 = y2 · 4 dy.

By symmetry, M dM2 20

6

2= ∫= + =∫∫2 2 4 4482 2

4

6

0

4

y dy y dy .

Stiffness = 448k (2.8 times stiffer!)

d. Increasing the depth does seem to increasestiffness greatly, but making the beam verytall would also make the web very thin,perhaps too thin to withstand much force.

14. a. dA = y dx = x3 dx

A x dx= =∫ 3

0

2

4

b. dV = 2π x · y · dx = 2π x4 dx

V x dx= =∫ 2 12 84

0

2

π π.

c. dMy = x dA = x4 dx

M x dxy = =∫ 4

0

2

6 4.

The volume integral is 2π times the momentintegral.

d. x A M xy⋅ = ⇒ = =6 4

41 6

..

e. The centroid travels 2 3 2π π . .x =(Area)(Distance traveled by centroid) =(4)(3.2π) = 12.8π, which equals the volume.Thus, the theorem of Pappus is confirmed.

15. a. Area of a small circle = π r2

The centroid of the small circle is its center,so the distance from the axis of rotation tothe centroid is R. Thus, the theorem ofPappus implies

V = 2π RA = 2π R(π r 2) = 2π 2r 2R

b. Area of a semicircle = 1

22πr

Volume of a sphere = 4

33πr

2

43

212

4

3

3

2π

π

π π πr A V r

r

rr⋅ = ⇒ =

⋅=

16. Pick a closed region that does not lie on bothsides of the y-axis.Slice the region parallel to the y-axis so that eachpoint in the strip will be about r units from they-axis (see graph).

f(r)

r

r

y

a b

Let f(r) be the length of the strip or the sum ofthe lengths if the region has S-shaped parts.Let A be the area of the region.dMy = r dA = r f (r) d r

M r f r drya

b

= ∫ ( )

Rotate the region about the y-axis. The stripsgenerate cylindrical shells.

dV = 2π r f (r) d r

V r f r dr r f r dra

b

a

b

= = ∫∫ 2 2π π( ) ( )

= 2π My

But My also equals r A⋅ .∴ = = ( )( )V rA r A2 2π π= (distance traveled by centroid)(area of region),Q.E.D.

Problem Set 11-5Q1. centroid Q2. center of mass

Q3. radius of gyration Q4. definite integration

Q5. indefinite integration (or antidifferentiation)

Q6. ρ = (mass) ÷ (volume)

Q7. x1/ 2 Q8. ln | sec x + tan x | + C

Q9. y′ = (x2 + 1)− 1 Q10. A

1. a. Slice the trough face horizontally so that eachpoint in a strip is about the same distancebelow the surface as at the sample point(x, y).y = 2x4 ⇒ x = (0.5y)1/4

p = k(2 − y), dA = 2x dy = 2(0.5y)1/4 dydF = p dA = 2k(2 − y)(0.5y)1/4 dy

F dF k k= =

∫ 2 8444

128

450

2

. exactlyK

b. dMx = y dF = y · 2k(2 − y)(0.5y)1/4 dy

M dM k kx x= =

∫ 2 1880

256

1170

2

. exactlyK

c. y F M y

k

kx⋅ = ⇒ = 2 1880

2 8444

.

.

K

K

=

0 7692

10

13. exactlyK

x = 0 by symmetry.

The center of pressure is at 010

13,

.

2. a. The graph shows y = x2, between y = 0 andy = 100.


10

100

x

y

(x, y)

Width at y = 100 ft is 2 y = 20 ft, Q.E.D.

b. Slice the dam face horizontally so that eachpoint in a strip is the same distance below thesurface as the sample point (x, y).dA = 2x dy = 2y1/2 dx

A y dx= =∫ 2 13331

31 2 2

0

100/ ft

(2/3 the area of the circumscribed rectangle)

c. p = k(100 − y) with k = 62.4 lb/ft3

dF = p dA = 2k(100 − y)(y1/2) dy

F dF k= = =∫ 53 3331

33 328 000

0

100

, , ,

Force is 3,328,000 lb, or 1664 tons.

d. dMx = y dF = 2ky(100 − y)(y1/2) dy

M dM kx x= =

=∫ 16000000

7142 628 571 4285

0

100

, , . K

≈ 142.6… million lb-fte. y F Mx⋅ = ⇒

y = =300

742

7

8 = 42.8571… ≈ 42.86 ft

3. a. Slice the bulkhead horizontally so that eachpoint in a strip is the same distance belowthe surface as the sample point (x, y).

x y

20

32

321

4 4

+

=–

x y= − −

20 1

1

321

4 1/4

dA x dy y dy= = − −

2 40 1

1

321

4 1 4

/

A dA= ≈ ≈∫ 1186 6077 1186 6 2

0

32

. . ftK

b. p = 67(32 − y)

dF = p dA

= − ⋅ − −

67 32 40 1

1

321

4 1 4

( )

/

y y dy

F dF= ≈∫ 1 199 294 1645

0

32

, , . K

≈ 1 199. million lbc. dMx = y dF

= ⋅ − ⋅ − −

y y y dy67 32 40 1

1

321

4 1 4

( )

/

M dMx x= ≈∫ 13 992 028 2564

0

32

, , . K

≈ 13.992 million lb-ft

d. y F M yx⋅ = ⇒ ≈ 11 6668. ftK

x = 0 by symmetry.Center of pressure is at about (0, 11.67) ft.

e. Moment of area:

dM y dA y y dy= = ⋅ − −

40 1

1

321

4 1 4/

M dM= ≈∫ 20 071 5364

0

32

, . K

≈ 20.07 thousand ft3

y A M y⋅ = ⇒ ≈ 16 9150. ftK

x = 0 by symmetry.The centroid is at about (0, 16.92) ft.The centroid is different from center ofpressure.

f. Area below waterline:

A dAw = ≈ ≈∫ 548 6345 548 6 2

0

16

. . ftK

First moment of area below waterline:

M dMw = ≈ ≈∫ 4749 3398 4749 3 3

0

16

. . ftK

y A M yw w⋅ = ⇒ ≈ 8 6566. ftK

x = 0 by symmetry.The center of buoyancy is at about(0, 8.66) ft.

4. a. Equation of ellipse is x y

6 31

2 2

+

= .

b. Slice the ellipse horizontally so that eachpoint in a strip is y units from the surfacewhere y is at the sample point and y isnegative.Surface of oil is at y = 0 ⇒ p = −50y.

x y dA x dy y dy= = =2 9 2 4 92 2– –,

dF p dA y y dy= = − ⋅50 4 9 2–

= −200 9 2y y dy–

F dF= =−∫ 1800

3

0

lb (exactly)

5. a.

10

60y

x

(x, y)


Slice the wing parallel to the y-axis. Picksample point (x, y) within the strip.

dA y dx x dx= = 6020

cosπ

A dA= = =

−∫2400

763 943710

10

π. K

≈ 763.9 ft2

b. dF p dA k x x dx= = − ⋅( | |)10 6020

cosπ

F dF k= =

−∫ 10

10

4 863 4168, . K

exactly48000

2πk

c. Make 4863.4168…k ≥ 96.k ≥ 0.0197… tons/ft2 (exactly 0.002π 2 )

6. a. y = 100 − x2 intersects the x-axis at x = ±10.Slice the wing parallel to the y-axis. Picksample point (x, y) within the strip.

p = 90 − 7x

dA = y dx = (100 − x2) dx

dF = p dA = (90 − 7x)(100 − x2) dx

F dF= =−∫ 120 000

10

10

, lb (exactly)

b. dMy = x dF = x(90 − 7x)(100 − x2) dx

M dMy y= = −−∫

560000

310

10

lb-ft

c. x F M xy⋅ = ⇒ = −15

9ft

d. p ky y p y= = = =6

560 50 (because at )

dA x dy y dy= =2 2 100 –

dF p dA y y dy= = 12

5100 –

F dF= =∫ 64 0000

100

, lb (exactly)

e. dM y dF y y dyx = = 12

51002 –

M dMx x= =∫ 3 657 142 80

100

, , . K

≈ 3.657 million lb-ft exactly25600000

7

f. y F Mx⋅ = ⇒

y = ≈

57 1428 57 14

400

7. . ft exactlyK

7. a. Slice the region as shown in Figure 11-5f.At a sample point (x, t), d(dM2x) = t2 dx dt.

dM t dx dt t dt dxxt

t y

t

t y

22 2

00= =

=

=

=

=

∫∫= =

=

=1

3

1

33

0

3t dx y dxt

t y

= 1

30 25 4 4 1 3 3[ . ( – ) – ( – ) ]/x x dx

b. M dMx xx

x

2 24

4

0 53338

15= =

=

=

∫ . exactlyK–

8. a. Slice the region parallel to the y-axis so thateach point in a strip will have about the samepressure as at the sample point (x, y).y = e− x

p = kx2, dA = (1 − e− x) dxdF = p dA = kx2(1 − e− x) dx

F dF k= =∫ 0 95140

5

. Kln

exactly1

35

1

55

2

55

8

53 2(ln ) (ln ) ln+ + −

k

b. Slice the region parallel to the x-axis so thateach point in a strip will have about the samepressure as at the sample point (x, y).x = −ln y, p = ky− 1

dA = (ln 5 − x) dy = (ln 5 + ln y) dy= ln (5y) dy

dF = p dA = ky− 1ln (5y) dyx = ln 5 ⇒ y = e− l n 5 = 0.2

F dF k k= =

∫ 1 2951

1

25 2. exactly

0.2

1

K (ln )

c. Slice the region parallel to the y-axis. Thenslice a strip parallel to the x-axis as shown inFigure 11-5g.At sample point (x, t), p = kx2t− 1.d(dF) = p dA = kx2t− 1 dx dt

dF kx t dx dt kt dt x dxt y

t

t y

t

= =

− −

=

=

=

=

∫∫ 2 1 11

21

= = − ===( ) ( )k t x dx k y x dx kx dxt y

tln ln1 2 2 30

F kx dx k k= = =∫ 3 4

0

5 1

45 1 6774( ) .ln

ln

K

9. The integrals in Problems 7 and 8 can be writtenin the form

f x t dt dxt c

t d

x a

x b

( , )=

=

=

=

∫∫The result is called a double integral because twointegrals appear. (Hiding inside each integral is asecond integral!)

10. a. y x= 58

2tanπ

y x x= ⇒ = ⇒ = ±58

1 22tanπ


Slice the floodgate parallel to the y-axis.

dA x dx= −

5 5

82tan

π

A dA= = ≈

−∫ 14 5352 14 54 2

2

2

. .K ft

exactly 4080−

π

b. Slice the floodgate parallel to the x-axis sothat each point in a strip has about the samepressure as at the sample point (x, y).p = k(20 − y) with k = 62.4 lb/ft2

y x x y= ⇒ = −58

80 22 1tan tan .

ππ

dA x dy y dy= = −216

0 21

πtan .

dF p dA k y y dy= = − ⋅ −( )2016

0 21

πtan .

F dF k= =∫ 248 2628

0

5

. K

= 15491.6027… ≈ 15.49 thousand lb

exactly 8005200

3−

π

k

(The force can also be found by slicingparallel to the y-axis as in part a, then slicingthe strip horizontally and using a doubleintegral. In this case, the pressure at asample point (x, t) isp = k(20 − t)d(dF) = p dA = k(20 − t) dt dxThe first integration is from t = y to t = 5.The second integration is from x = −2 tox = 2.)

c. Let µ (Greek letter mu) = coefficient offriction.

µ µ⋅ = ⇒ =F 10000

10000

15491 6027. K= 0 6455. K

Problem Set 11-6

Q1.1

101101x C+ Q2. 0

Q3. x ln x − x + C Q4. 2 sin x cos x = sin 2x

Q5. (force)(displacement) Q6. y′ = 3(1 + 9x2)− 1

Q7. x = −2 Q8. 2 sec2 x tan x

Q9. ′′ = −y x9 3cos Q10. D

1. Partition the interval into small subintervalsof width dT so that C is about the same at anypoint in a subinterval. The amount of heat,dH, to raise the temperature by dT isdH = C dT = (10 + 0.3T1/2) dT.

H dH= =∫ 13 200100

,900

calories (exactly)

2. a. v(t) = 55 + 6 t − t2

v(0) = 55 + 6 · 0 − 02 = 55 mi/hv(3) = 55 + 6 · 3 − 32 = 64 mi/hv(6) = 55 + 6 · 6 − 62 = 55 mi/h, Q.E.D.

b. Cost of a short time, dt, at speed v isdC = 3(v − 55) dt = 18t − 3t2 dt.Total ticket cost is

C dC= =∫ 108 (exactly).0

6

Fine should be $108.00.

3. a. Cost per foot, P, = ax2 + bx + ca · 02 + b · 0 + c = 500 ⇒ c = 500

a b

a b

a

b

⋅ + ⋅ + =⋅ + ⋅ + =

⇒==

100 100 500 820

200 200 500 1180

0 002

3

2

2

.

P(x) = 0.002x2 + 3x + 500

b. P(700) = 0.002 · 7002 + 3 · 700 + 500= $3580/ft

c. Cost to dig a short distance, dx, isdC = P dx = (0.002x2 + 3x + 500) dx.Cost to dig 1000 feet is

( .0 002 3 5008

310002

0

10002x x dx+ + = ⋅∫ ) .

Cost is about $2,666,667.

d. Cost to dig 500 feet twice (once fromeach end) is

C x x dx= + + = ⋅∫2 0 002 3 50017

35002 2

500

( . ) .0

Cost is about $1,416,667.Savings is about $1,250,000!

4. a. velocity · area has the units in.

sin. ,2⋅

which is in.3/s, correct for flow rate.

b. v = 4 − x2 ⇒ v′ = −2xv′ changes from positive to negative at x = 0.∴ there is a maximum flow rate at the centerof the pipe where x = 0.(Or simply observe that the graph of v isa parabola opening downward with vertexat x = −b/(2a) = 0.)v(2) = 4 − 22 = 0, Q.E.D.

c. Slice the water in the pipe into cylindricalshells.Each point in a shell has about the samewater velocity as at the sample point x unitsfrom the axis.Let F = flow rate in in.3/s.dF = v dA = (4 − x2) · 2πx · dx

= 2π(4x − x3) dx

F x x dx= − = = …∫ 2 4 8 25 13273π π( ) .0

2

≈ 25.13 in.3/s

d. 25.1327… in.3/s · 60 s/min · 1 gal/231 in.3 =6.5279… ≈ 6.53 gal/min


e. 4 in./s · π · 22 in.2 = 16π = 13.0559…≈ 13.06 gal/min (exactly twice the actualrate)

f. The problem is equivalent to finding thevolume of a solid of rotation by cylindricalshells. The velocity takes the place of thealtitude of a shell.

5. a.

600

5

x

F

b. F has a step discontinuity at x = 2.

c. dW = F dxBecause the graph is linear on [0, 2], the workequals the area of the triangle.

W = ⋅ ⋅ =1

22 600 600 in.-lb

d. W F dx= ∫2

5

By Simpson’s rule,

W ≈ + ⋅ + ⋅ + ⋅1

30 5 450 4 470 2 440 4 420( . )(

+ ⋅ + ⋅ +2 410 4 390 330)

= 12662

3in.-lb

e. Total work ≈ + =600 12662

31866

2

3 in.-lb

f. Yes, a piecewise continuous function such asthis one can be integrable. See Problem 27 inProblem Set 9-10.

6. a. Slice the solid into disks parallel to thexz-plane so that each point in a disk has aboutthe same density as at the sample point (x, y).

y = 4 − x2 ⇒ x2 = 4 − y

dm = ρ dV = k · π x 2 dy = kπ(4 − y) dy

m k y dy k= − =∫ π π( ) g4 80

4

b. Each point in a disk of part a is also about thesame distance from the xz-plane as the samplepoint (x, y).Let K stand for the constant.

dF = K · dm · y− 1/2 = K · kπ (4 − y) dy · y− 1/2

= Kkπ (4y− 1/2 − y1/2) dy

F dF Kk Kk= = =∫ 32

333 5103

0

4

π . K

7. Slice the solid into cylindrical shells so that eachpoint in a shell is about the same distance fromthe y-axis as the sample point (x, y).

dM2y = x2 dm = x2 · ρ dV = x2 · k · 2π xy dx= 2π kx3(4 − x2) dx

M dM ky y2 2

0

2 32

333 5103= = =∫ π . kg-cm2K

8. a. T(D) = 20 sin 2π DT(0) = 20 sin 0 = 0T(1/4) = 20 sin π /2 = 20, which checks.

b. Partition the time interval into shortincrements of width dD so that T is about thesame at any time in the increment as it is atthe sample point (D, T ) .Let H = number of degree-days.dH = T dD = 20 sin 2π D dD

H dH= = =∫ 10

3 18300

1 4

π. K

/

≈ 3.18 degree-days

9. a. m = 2000 − 5t (mass in kilograms, time inseconds)

b. a = F/m = 7000(2000 − 5t)− 1

= 1400(400 − t)− 1

c. adv

dt t= = 1400

400 –

dvt

dt=∫ ∫ 1400

400 –v = −1400 ln | 400 − t | + CAssume the car starts at rest at t = 0.0 = −1400 ln 400 + C ⇒ C = 1400 ln 400

v tt

( ) = 1400400

400ln

| – |

d. v( ) . . m/s20 1400

20

1971 8106 71 81= = ≈ln K

vds

dt t= = 1400

400

400ln

–

st

dt= =∫ 1400400

400711 9673

0

20

ln–

. K

≈ −

712 0 28000 1 1920

19. m exactly ln

10. Slice the tract parallel to the tracks so that eachpoint in the strip will have about the samevalue per square kilometer as at the samplepoint (x, y).Let v = thousands of dollars per square kilometerand W = thousands of dollars the land is worth.v = kx = 200x (v = 200 at x = 1)dW = v dA = 200x[(4 − x2) − (4x − x2)] dx

= 800(x − x2) dxThe curves intersect at x = 1.

W x x dx= − =∫ 800 1331

32

0

1

( )


The land is worth about $133,333.If all the land were worth $200,000 per km2,

W A x dx= = =∫200 200 4 4 4000

1

( – ) .

The land would be worth $400,000.Actual value is about $267,000 less.

11. a. Slice the tract parallel to the y-axis so thateach point in a strip will be about the samevalue per square unit as at the samplepoint (x, y).y = cos xLet v = value of land per square unit andW = worth of the land.v = kx, dA = y dx = cos x dxdW = v dA = kx cos x dx

W dA k k= = −

=∫ ππ

21 0 5707

0

2

. K/

b. Slice the tract parallel to the x-axis so thateach point in a strip will be about the samevalue per square unit as at the samplepoint (x, y).v = kydW = v dA = v · x dy = ky · cos− 1 y dy

W dW k k= = =∫ π8

0 39260

1

. K

12.

(x, y)

x

3

9y

Slice the wall parallel to the ground so thateach point in the slice will cost about thesame to paint per square meter.Let r = rate in dollars per square meter andC = cost in dollars to paint the wall.r = ky2 = 3y2 (r = 12 when y = 2)dA x dy y dy= =2 2 9 –

dC r dA y y dy= = ⋅3 2 92 –

C dC= ≈

∫ $ . exactly 1999 54 1999

19

350

9

13. a. Let v = value of land per square kilometer,W = worth of the land in dollars, andr = distance from center of town.Slice the city into circular rings of width dr sothat each point in a ring will be about r unitsfrom the center.v r dA r dr= − =10 3 2, πdW v dA r r dr= = − ⋅ ( ) 10 3 2π

W dW= = =∫ 36 113 0973

0

3

π . K

≈ 113.1 million dollars

b. v e v kkr= = ⇒ = − ⇒10 3 11

310, ( ) ln

v e r= −10 10 3(ln ) /

dW v dA e r drr= = ⋅−10 210 3( ) /ln π

W dW= = ≈∫ 71.4328 71.4 millionK

0

3

dollars exactly ( )18

109 102

π(ln )

– ln

c. By Simpson’s rule,

W v r dr= ⋅ ≈∫ 21

30 3 2

0

3

π π( . )( ) (10 + 4 ⋅ 12

+ 2 ⋅ 15 + 4 ⋅ 14 + 2 ⋅ 13 + 4 ⋅ 10+ 2 ⋅ 8 + 4 ⋅ 5 + 2 ⋅ 3 + 4 ⋅ 2 + 1)

= 52.2π = 163.9911… ≈ 164.0 milliondollars

d. This problem is equivalent to volume bycylindrical shells, where the value of the landper square unit takes the place of the altitudeof the cylinder. It is also equivalent to thewater flow in Problem 4 of this problem set.

e. Answers will vary.

14. a. p = 100[(x − 8)1/2 − 0.5(x − 8)]dF = p dA = 2p dx

= 200[(x − 8)1/2 − 0.5(x − 8)] dx

F dF= = ≈∫ 177 1236 177

8

10

. lbK

exactly800 2

3200−

b. Average pressure = total force/total area

= = ≈177 1236

444 2809 44 3 2. K

K. . lb/ft

exactly200 2

350−

c. dMyz = x dF = 2px dx= 200x[(x − 8)1/2 − 0.5(x − 8)] dx

M dMyz yz= = ≈∫ 1602 8706 1603

8

10

. lb-ftK

exactly7360 2

3

5600

3−

d. x F M xyz⋅ = ⇒ = 1602 8706

177 1236

.

.

K

K= 9.0494… ft

Calvin should stand about 10 − 9.0494 ft

≈ 111

2in. from the end.

15. a. f (x) = 9 − x2 = (3 − x)(3 + x) = 0 only atx = ±3.

g x x x x

x x x

( )

only at .

= − − + +

= − + = = ±

1

33 9

1

33 3 0 3

3 2

2( – )( )


b. A x dxf = =∫ ( – )–

9 362

3

3

A x x x dxg = + +

=∫ – –

–

1

33 9 363 2

3

3

To simplify algebraic integration, you coulduse

A x dxf = ∫2 9 2

0

3

( – )

A x dxg = ∫2 9 2

0

3

( – ) , where the odd terms

integrate to zero between symmetrical limits.Thus, the two integrals are identical.

c. The high point of f comes at x = 0.The high point of g comes where g′(x) = 0.g′(x) = −x2 − 2x + 3 = −(x + 3)(x − 1)g′(x) = 0 ⇔ x = −3 or x = 1The high point is at x = 1.

d. Slice the region under the g graph parallel tothe y-axis so that each point in a strip will beabout the same distance from the y-axis as thesample point (x, y).

dMy = x dA = x g(x) dx

= − − + +

1

33 94 3 2x x x x dx

M dMy y= =∫ 21 63

3

.–

x A M xy⋅ = ⇒ = =21 6

360 6

..

e. False. For the symmetrical region under thegraph of f, the centroid is on the line throughthe high point. But for the asymmetricalregion under the graph of g, the high point isat x = 1 and the centroid is at x = 0.6.

f. False

Area to left = =∫ g x dx( ) .17 10723

0 6

–

.

(exactly)

Area to right = =∫ g x dx( ) .18 89280 6

3

. (exactly)

(or 36 − 17.1072 = 18.8928)

g. Let S stand for skewness.

dS x dA x g x dx= −

= −

3

5

3

5

3 3

( )

S x x x x dx=

− − + +

= − ⋅⋅

∫ –

–

–

3

5

1

33 9

17 773764 3

7 125

33 2

3

3

5

. exactlyK

h. By symmetry, the centroid of the area under fis on the y-axis, so x = 0. ThendS = x3 dA = x3(9 − x2) dx

S x x dx= − =∫ 3 2

3

3

9 0( )–

(odd function

integrated between symmetrical limits)

The “skewness” being zero reflects thesymmetry of this region. It is not skewedat all.

i. For example, graph

g x x x x( ) .− = − − +1

33 93 2

gNewgraph

y

x

3–3

16. a. y = x2

dL dx dy x dx= + = +2 2 21 4

dM x dL x x dxy = = + 1 4 2

b. M dMy y= =∫ 5 7577

0

2

. K

exactly1

1217 17 1( – )

c. L x dx= + =∫ 1 4 4 64672

0

2

. K

exactly1

417 4 17ln ( )+ +

d. x L M xy⋅ = ⇒ = =5 7577

4 64671 2390

.

.

K

KK.

e. dS x dL x x dx= = +2 2 1 4 2π π

S dS= =

∫ 36 1769

617 17 1

0

2

. exactlyKπ

( – )

f. Integral for S is 2π times the integral for My!

17. In Problem 16 1 2390, .R x= = K andL = 4.6467… .2π RL = 2π (1.2390…)(4.6467…) = 36.1769… ,which equals S, Q.E.D.

18. The centroid of the small circle is at its center,R units away from the axis.The arc length L of the small circle is 2πr.Surface area S = 2π R(2πr) = 4π 2rR



R1. Slice the region parallel to the force axis so thateach point in a strip has about the same force asat the sample point (x, F ) .

dW = F dx = 30e− 0.2 x dx

W e dx ex= = −− −∫ 30 150 10 2 2

0

10. ( )

= 129.6997… ≈ 129.7 ft-lb


R2. a. dW = F dx = kx− 2 dx

W kx dx k= = −−∫ 2

3

1 2

3ft-lb

(Mathematically, the answer is negativebecause dx is negative. Physically, the answeris negative because the magnets absorbenergy from their surroundings rather thanreleasing energy to their surroundings.)

b. Construct axes with the origin at the vertex ofthe cone. An element of the cone in the xy-

plane has the equation y x x y= =7

3

3

7 or .

Slice the water horizontally into disks so thateach point in a disk is lifted about the samedistance as the sample point (x, y) on theelement of the cone.

F = 0.036 dV = 0.036 ⋅ πx2 dy

= ⋅0 0369

492. π y dy

Each disk is lifted (10 − y) cm.dW = (10 − y) dF

= −( )( . )10 0 0369

492y y dyπ

W dW= = = …∫0

7

3.591 11.2814π

≈ 11.28 in.-lb

R3. a. The graph shows the region in Quadrant Iunder the graph of y = 8 − x3 rotated aboutthe y-axis.

2

8

(x, y)

x

y

Slice the region parallel to the x-axis,generating disks, so that each point in a diskis about the same distance from the xz-planeas the sample point (x, y).

ρ = ky, dV = π x2 dy = π(8 − y)2/3 dy

m k y y dy k= − =∫ π π ( ) ./8 57 62 3

0

8

b. Slice the region parallel to the y-axis,generating cylindrical shells, so that eachpoint in a shell is about the same distancefrom the y-axis as the sample point (x, y).

ρ = ex, dV = 2π xy dx = 2π(8x − x4) dx

m e x x dxx= − =∫ 2 8 644

0

2

π π ( )

R4. a. The width of a strip at the samplepoint (x, y) is

w bb

hy dA b

b

hy dy= − ⇒ = −

dM y dA byb

hy dyx = = −

2

M byb

hy dy by

b

hyx

h h

=

= −∫ – 2

0

2 3

0

1

2 3

= − − + =1

2 30 0

1

62 3 2bh

b

hh bh

y A y bh Mx⋅ = ⋅ =1

2

⇒ = =ybh

bhh

1612

1

3

2

, Q.E.D.

b. The graph shows the region under y = ex

rotated about the y-axis, showing back half ofsolid only.

(x, y)

x

y

1

1

Slice the region parallel to the y-axis,generating cylindrical shells, so that eachpoint in a shell will be about the samedistance from the y-axis as the samplepoint (x, y).

dV = 2πx ⋅ y ⋅ dx = 2πxex dx

dM2y = x2 dV = 2πx3ex dx

M x e dxyx

23

0

1

2 3 5401= =∫ π . K

(exactly 12π − 4πe)

R5. Draw axes with the x-axis at ground level and they-axis through the upper vertex of the triangle.Slice the face of the building horizontally so thatthe wind pressure at any point in a strip is aboutequal to the pressure at the sample point (x, y).

dA y dy= −

150

150

400

dF p dA e y dyy= = ⋅ − −

− ( ).200 150 1 11

4000 01

F dF= = …∫0

400

3736263.2708

≈ 3.736 million lb(exactly 30000(125 − 25e− 4))


R6. a. Let x = number of feet at which drill isoperating, and r(x) = number of dollars perfoot to drill at x feet.r(x) = a · bx

r (0) = 30 ⇒ a = 30

50 305

310000

110000

= ⋅ ⇒ =

b b

/

∴ =

( )

/

r xx

305

3

10000

(or ( ) . / .r x e ex x) ln= =− ⋅30 300 6 10000 0 00005108256K

b. dC r x dx dxx

= =

( )

/

305

3

10000

C dx

x

=

=∫ 30

5

36965243 17

10000

0

50000 /

. K

≈ 6.965 million dollars

exactly − ⋅

30

10000

0 6

5

31

5

ln .–

Concept ProblemsC1. a. Either slice the region parallel to the y-axis,

dA = (8 − y) dx = (8 − x3) dx

A x dx= =∫ ( – )8 123

0

2

or slice parallel to the x-axis,

A y dy= =∫ 1 3

0

8

12/

b. i. Use slices parallel to the x-axis so thateach point in a strip will be about thesame distance from the x-axis as thesample point (x, y).

dMx = y dA = y(y1/3 dy)

M y dyx = = =∫ 4 3

0

8 384

754 8571/ . K

ii. Use slices parallel to the y-axis so thateach point in a strip will be about thesame distance from the y-axis as thesample point (x, y).

dMy = x dA = x(8 − x3) dx

M x x dxy = =∫ ( – )8 9 64 .0

2

c. x A M xy⋅ = ⇒ = =9 6

120 8

..

y A M yx⋅ = ⇒ = = =384 7

12

32

74 5714

/. K

Centroid is at (0.8, 4.5714…).

d. i. With slices parallel to the x-axis,

dV = 2π y · x · dy = 2π y4/3 dy

V y dy= = =∫ 2

768

7344 67754 3

0

8

π π/ . K

With slices perpendicular to the x-axis,

dV = π (82 − y2) dx = π (64 − x6) dx

V x dx= − = =∫ π π( ) .64

768

7344 67756

0

2

K

ii. With slices parallel to the y-axis,

dV = 2π x · (8 − y) · dx = 2π x(8 − x3) dx

V x x dx= − =∫ 2 8 19 24

0

2

π π( ) .

= …60 3185.

With slices perpendicular to the y-axis,

dV = π x2 dy = π y2/3 dy

V y dy= = =∫ π π2 3

0

8

19 2 60 3185/ . . K

iii. With slices parallel to the line x = 3,

dV = 2π (3 − x) · (8 − y) · dx= 2π (3 − x)(8 − x3) dx

V x x dx= − − =∫ 2 3 8 52 83

0

2

π π( )( ) .

= 165.8760…

With slices perpendicular to the linex = 3,

dV = π [32 − (3 − x)2] dy

= π [9 − (3 − y1/3)2] dy

V y dy= − − =∫ π π[9 3 52 81 3

0

8

( ) ] . / 2

= 165.8760…

e. i. The centroid is 32/7 units from the x-axis.

∴ = ⋅ ⋅ = V 232

712

768

7π π

= 344.6775… (Checks.)

ii. The centroid is 0.8 unit from the y-axis.

V = 2π · 0.8 · 12 = 19.2π= 60.3185… (Checks.)

iii. The centroid is 3 − 0.8 = 2.2 units fromthe line x = 3.V = 2π · 2.2 · 12 = 52.8π = 165.8760…(Checks.)

f. Use horizontal slices so that each point in adisk will be about the same distance from thexz-plane as the sample point (x, y).

dMxz = y dV = y(π x2 dy) = y · π y2/3 dy

M y dyxz = = =∫ π π5 3

0

8

96 301 5928/ . K

g. y V M yxz⋅ = ⇒ = =96

19 25

ππ.

x z= = 0 by symmetry.

Centroid is at (0, 5, 0).


h. No. For the solid, y = 5, but for the region,y = …4.5714 .

i. Use slices of the region parallel to the y-axisso that each point in a resulting cylindricalshell will be about the same distance from they-axis as the sample point (x, y).

ρ = kx2,dV = 2πx(8 − y) dx = 2π(8x − x4) dx

dm = ρ dV = kx2 · 2π (8x − x4) dx

= 2π k(8x3 − x6) dx

m k x x dx k= − =∫ 2 8192

73 6

0

2

π π( )

= 86.1693…k

j. Use cylindrical shells as in part i so that eachpoint in a shell will be about the samedistance from the y-axis as the samplepoint (x, y).

dM2 = x2 · dm = 2π k(8x5 − x8) dx

M k x x dx k25 8

0

2

2 8512

9= − =∫ π π( )

= 178.7217…

k. Use vertical slices of the region so that eachpoint in a strip will have about the samepressure acting on it as at the samplepoint (x, y).

p = 3 − x, dA = (8 − y) dx = (8 − x3) dx

dF = p dA = (3 − x)(8 − x3) dx

F x x dx= =∫ .0

2

( – )( – )3 8 26 43

(Note the similarity to the integral inpart d.iii.)

l. F = kz− 2

F = 26.4 at z = 1 ⇒ k = 26.4 ⇒ F = 26.4z− 2

dW = F dz = 26.4z− 2 dz

W z dz= =−∫ 26 4 17 62

1

3

. .

m. Use horizontal slices so that each point in aresulting disk will be at about the sametemperature as the sample point (x, y).dH = CT dm = 0.3(10 − y)(5.8π y2/3 dy)

H y y dy= − =∫ 1 74 10 167 042 3. ( )( ) .

/

0

8

π π

= 524.7716… ≈ 524.8 cal

C2. Let f (x) be the height of a vertical strip at x (orcombined heights if the region being rotated isnot convex). Let x = a and x = b be the left andright boundaries of the region.

dV x dA x f x dx V x f x dxa

b

= = ⇒ = ∫2 2 2π π π ( ) ( )

dM x dA x f x dx M x f x dxy ya

b

= = ⇒ = ∫ ( ) ( )

Note that V = 2π My, thus showing that the twoproblems are mathematically equivalent, Q.E.D.

C3. dMxz = y dVxz = yπ x2 dy = π y(9 − y) dy

M y y dyxz = − =∫π π π ( ) .0

9

9 121 5

dM2y = x2 dVy = x2 2π xy dx = 2π x3(9 − x2) dx

M x x dxy23 22 9 121 5= − =∫ π π ( ) . ,

0

3

Q.E.D.

This is not true in general. Counterexample:Rotate the region under y = 2 − 2x2.

dM y dV y x dy y y dyxz xz= = = −

π π

2 11

2

M y y dyxz = −

=∫ π π

0

2

11

2

2

3dM2y = x2 dVy = x2 2π xy dx = 2π x3(2 − 2x2) dx

M x x dxy23 22 2 2

1

3

2

3= − =∫ π π π ( ) , not .

0

1

General proof: For any paraboloid of height Hand base radius R, let h = distance (along theaxis) from the base and r = radius. Then a

generating parabola is given by h HH

Rr= − ⋅2

2.

dM h dV h r dh h H hR

Hdhbase ( )= = = −π π2

2

MR

HHh h dh

h

h H

base ==

=

∫2

2

0π ( – )

= −

=

=

=R

H

Hh h R H

h

h H22 3

0

2 2

2

1

3

1

6π π

dM r dV r rh dr2axis = =2 2 2π

= −

r r H

H

Rr dr2

222π

M H rR

r drr

r R

23

25

02

1axis =

=

=

∫π –

= −

=

=

=

21

4

1

6

1

64

26

0

4π πH rR

r R Hr

r R

In the original example, H = R2, so the twomoments turned out to be equal.

C4. a. Assume m ≠ 0.The area of the trapezoid is

Ab b

hma mb

b a= + ⋅ = +1 2

2 2( – )

= −1

22 2m b a( )

Integrating, y dx mx dx mxa

b

a

b

a

b

≈ =∫∫ 1

22

= − =1

22 2m b a A( ) , Q.E.D.

The length is L b a m= − +( ) .1 2

Integrating, dL dx b a La

b

a

b

≈ = − ≠∫∫ ( ) ,

Q.E.D.


b. Note that r = mh.The volume of the cone is

V r h m h= =1

3

1

32 2 3π π .

Integrating dV ≈ π y2 dx = π m2x2 dx,

π π ,m x dx m h Vh

2 2 2 3

0

1

3= =∫ Q.E.D.

The surface area is S r r h= + =π

2 2

π .mh m2 21+Integrating dS ≈ 2π y dx = 2π mx dx,

2 2

0π π ,mx dx mh S

h

= ≠∫ Q.E.D.

c. Exact area of a strip:

∆ ∆ ∆ ∆ ∆A mx m x x x y x y x= + + ∆ = +1

2

1

2( ( ))

Exact volume of frustum:

∆ ∆

∆ ∆

∆ ∆ ∆ ∆

V m x x m x x x m x x

m x x x x x

y x y y x y x

= + ∆ + + ∆ +

= + + ∆

= + + ∆

π

π

π

3

33 3

1

3

2 2 2 2 2

2 2 2

2 2

( ( ) ( ) )

( )

( )

( ( ) )

d. dA y dx y dx y dx y dx− = +

−1

2∆

= 1

2∆y dx

dV − π y2 dx

= + +

−π πy dx y y dx y dx y dx2 2 21

3∆ ∆

= +π y y dx y dx∆ ∆1

32

Both differences contain only higher-orderinfinitesimals.

e. If dQ = ∆Q leaves out only infinitesimals of

higher order, then dQa

b

∫ is exactly equal to Q.

f. Reasons:

i. 0.5 and ∆y are constant with respect to thesummation, so they can be pulled out.

ii. The sum of all the subsegments ∆x of[a, b] must be b − a, the whole interval.

iii. ∆y has limit zero as ∆x goes to zero.

Chapter TestT1. a. force · displacement

b. mass

c. force

d. area · displacement

e. second moment of volume

f. x

T2. dW = (40x − 10x2) dx

W x x dx= − =∫ ( )40 10 902

1

4

T3. y m M yxz⋅ = ⇒ ⋅ =200 3000∴ = = 3000/200 15 cmy

T4. The center of the circle is (8, 9) and the radiusis 7, so the circle is on just one side of the axisof rotation (the y-axis). So the solid satisfies thehypothesis of the theorem of Pappus.The centroid of the circle is (8, 9), thedisplacement from the y-axis is R = 8, and thearea of the circle is 49π.∴ = = = = ( )( )( ) V RA2 2 8 49 784π π π π2463.0086…

T5. Using exponential regression,F ≈ 29.9829… (1.0626…)x

dW = F dx

W F dx= ≈ … ≈∫ 412 4652 412 5. . ft-lb0

10

(By the trapezoidal rule, W ≈ 413 ft-lb.Simpson’s rule cannot be used because there isan odd number of increments.)

T6. a.

(x, y)

x

y

2

1

Slice the region parallel to the y-axis so that eachpoint in a strip will be about the same distancefrom the y-axis as the sample point (x, y).dMy = x dA = xex dx

M xe dx eyx= = + = … ≈∫ 2 3

0

2

1 8 3890 8 39. . in.

b. dM2y = x2 dA = x2ex dx

M x e dx eyx

22 2

0

2

2 2 12 7781

12 78

= = − = …

≈

∫ .

. in.4

c. A e dx ex= = − = … ≈∫ 2 1 6 3890 6 39. . in.20

2

x A M xe

ey⋅ = ⇒ = + = … ≈2

2

1

11 3130 1 31

–. . in.

T7. The graph shows y = x1/2 from x = 0 to x = 16,rotated about the x-axis.

4

16x

y(x, y)


Slice the region parallel to the x-axis so that eachpoint in a resulting cylindrical shell will beabout the same distance from the x-axis as thesample point (x, y).ρ ρ π= = = ⋅ −3 3 2 16y dm dV y x y dy, ( )= 6π y2(16 − y2) dy

m y y dy= − =∫ 6 16 819 22 4π π( ) . 0

4

= 2573.5927… ≈ 2573.6 g

T8. a. Slice the end of the trough parallel to thex-axis so that each point in a strip has aboutthe same pressure acting on it as the samplepoint (x, y), where x ≥ 0.p = 62.4(8 − y), dA = 2x dy = 2y1/3 dydF = p dA = 62.4(8 − y) · 2y1/3 dy

F y y dy= − ⋅

= ⋅

= … ≈

∫ 62 4 8 2

62 464 9

75134 6285 5134 6

1 3. ( )

. . . lb

/

0

8

b. dMx = y dF = 62.4(8 − y) · 2y4/3 dy

F y y dy= − ⋅ = ⋅ ⋅∫ 62 4 8 29 2

3562 44 3

10

. ( ) ./

0

8

= 16430.8114… ≈ 16.43 thousand lb-ft

y F M yx⋅ = ⇒ = =16430 8114

5134 62853 2

.

.

K

K. ft

x = 0 by symmetry.Center of pressure is at (0, 3.2).

T9. a. Slice the seating area into concentric rings ofwidth dr. Each point in a ring will be aboutthe same distance from the center as thesample point.Let W = worth of the seating and v = valueper square foot.dW = v dA = 150r− 1 · 2π r dr = 300π dr

W dr bb

= = −∫ 300 300 3030

π π ( ) dollars

b. 300π (b − 30) = 60000 ⇒

b = + = … ≈30200

93 6619 93 7π

. . ft


Chapter 12—The Calculus of Functions Definedby Power Series

Problem Set 12-1

1. f xx

( ) ,= 6

1–P x x x x x x5

2 3 4 56 6 6 6 6 6( ) = + + + + +

1–1

100

–100

x

yf P5

f

P5

The graph of P5 fits the graph of f reasonablywell for about −0.8 < x < 0.6.The graph of P5 bears no resemblance to thegraph of f at x = 2 and at x = −2, for example.

2. P6(x) = P5(x) + 6x6

1–1

100

–100

x

yf P5

f

P5

P6

The graph of P5 fits the graph of f slightly better,perhaps for −0.9 < x < 0.7.

3. P5(0.5) = 11.8125, P6(0.5) = 11.90625,f (0.5) = 12∴ P6(0.5) is closer to f (0.5) than P5(0.5) is.P5(2) = 378, P6(2) = 762, f (2) = −6∴ P6(2) is not closer to f (2) than P5(2) is.

4. Possible conjecture: P(x) converges to f (x) for−1 < x < 1, or perhaps for −1 ≤ x ≤ 1.

5. P0(1) = 6 P0(−1) = 6P1(1) = 12 P1(−1) = 0P2(1) = 18 P2(−1) = 6P3(1) = 24 P3(−1) = 0P4(1) = 30 P4(−1) = 6

For x = 1, the sums just keep getting larger andlarger as more terms are added. For x = −1, thesums oscillate between 0 and 6. In neither casedoes the series converge. If the answer toProblem 4 includes x = 1 or x = −1, theconjecture will have to be modified.

6. P5(0.5) = 11.8125 and f (0.5) = 12The values differ by 0.1875, and6(0.5)6 = 0.09375.P5(−0.5) = 3.9375 and f (−0.5) = 4

The values differ by 0.0625, and6(−0.5)6 = −0.09375.For P5(0.5), the difference is greater than thevalue of the next term of the series. This result isto be expected because the rest of the seriesis formed by adding more positive terms.For P5(−0.5), the difference is less in absolutevalue than the absolute value of the next term.This result is reasonable because the termsalternate in sign so that you are adding andsubtracting ever smaller quantities.

7. A geometric series; the common ratio

Problem Set 12-2Q1. . . . for any ε > 0 there is a D > 0 such that

if x > D, then f (x) is within ε units of L.

Q2. the fundamental theorem of calculus

Q3. the fundamental theorem of calculus

Q4. the mean value theoremQ5. derivative Q6. cos x − x sin x

Q7. x sin x + cos x + C Q8. dA r d= 1

22 θ

Q9. f (x) = e− x Q10. D

1. Series: 200 − 120 + 72 − 43.2 + 25.92 −15 552. + LSums: 200, 80, 152, 108.8, 134.72, 119,168, …

100

10

125

Sn

n

S = ⋅+

=2001

1 0 6125

.The series converges to 125.

| |125 125 2001 0 6

1 0 6− = − ⋅

+Sn

n– (– . )

.

= − ⋅125 1 1 61 0 6

1 6.

– (– . )

.

n

= 125|1 − (1 − (−0.6)n)|

= 125|−0.6n| = 125(0.6n)

125 0 6 0 0001

0 0001125

0 627 48( . ) . .n n= ⇔ = =ln . /

ln .K

Make n ≥ 28.Sn will be within 0.0001 unit of 125 for allvalues of n ≥ 28.


2. Series: 30 + 33 + 36.3 + 39.93 + 43.923 +48.3153 + K

Sums: 30, 63, 99.3, 139.23, 183.153,231.4683, …

10

100

Sn

n

The graph shows divergence.

S 30 4,133,883.70 (Wow!)100 = ⋅ =1 1 1

1 1 1

100– .

– .K

The formula S = t1/(1 − r) gives −300 for S, butit has no meaning because the series does notconverge.

3. a. Series:

7(0.8 ) 7 5.6 4.48 3.584n

n

−

=

∞

= + + + +∑ 1

1

L

Sums: 7, 12.6, 17.08, 20.664, 23.5312, …S4 = 20.664, so the amount first exceeds20 µg at the fourth dose.

S = ⋅ =71

1 0 835

– ., so the total amount

never exceeds 40 µg.The graph confirms that the partial sums ofthe series approach 35 asymptotically andfirst exceed 20 µg at the fourth dose.

1 2 3 4 5 6 7 8 9 10

10

20

30

40

Asymptote

n

Goes > 20

Stays > 20

b. tn = Sn − 7, so the sequence is 0, 5.6, 10.08,13.664, 16.5312, … .See the graph in part a. The open circles showthe partial sums just before a dose.t7 = 20.6599… . The amount remainsabove 20 µg for n ≥ 7.

c. See the graph in part a.

4. a. Perimeters are 16, 16 0 5. , 16(0.5), … ,which is a geometric sequence with t1 = 16and .r = 0 5.

b. S10

5

1 2161 0 5

1 0 552 9203= ⋅ =– .

– . / . cmK

c. The total perimeter converges to

16

1

1 0 554 62741 2⋅ =

– . / . cm.K

d. The sum of the areas is 16 8 4 2+ + + +L ,which is a convergent geometric series withr = 0.5.

S = ⋅ =161

1 0 532

– . cm2

5. a. The interest rate for one month is0.09/12 = 0.0075.

Months Dollars

0 1,000,000.00

1 1,007,500.00

2 1,015,056.25

3 1,022,669.17

b. Worth is (1,000,000)(1.007512) =$1,093,806.90; interest is $93,806.90.

c. The first deposit is made at time t = 0, thesecond at time t = 1, and so forth, so at timet = 12, the term index is 13.

d. Meg earned $93,806.90 the first year.

APR . %= ⋅ =93806 90

1000000100 9 3806

.K

e. (1,000,000)(1.0075n) = 2,000,000

n = =ln

ln .

2

1 007592 7657. K

After 93 months

6. a. The interest rate for one month is0.108/12 = 0.009.S5 = 100 + 100(1.009) + 100(1.009)2

+ 100(1.009)3 + 100(1.009)4

+ 100(1.009)5

= ⋅ =1001 1 009

1 1 009613 66

6– .

– .$ .

b. There are six terms because the term index ofthe first term is zero.

c. 10 years equals 120 months. There will havebeen 121 deposits after 10 years because theinitial deposit was made at time 0. So thereare 121 terms.

S120

121

1001 1 009

1 1 00921 742 92= ⋅ =– .

– .$ , .

The principal is 121(100) = $12,100.The interest is 21,742.92 − 12,100 =$9,642.92.

7. a. Sequence: 20, 18, 16.2, 14.58, 13.122, …

b. S4 = 20 + 18 + 16.2 + 14.58 = 68.78 ft

c. S = ⋅ =201

1 0 9200

– .So the ball travels 200 ft before it comes torest.


d. For the 10-ft first drop, 10 = 0.5(32.2)t2.t = (10/16.1)1/2 = 0.7881…The total time for the 20-ft first cycle is2(0.7881…) = 1.5762… s.For the 18-ft second cycle,t = 2(9/16.1)1/2 = 1.4953… s.

e. The times form a geometric series with firstterm 1.5762… and common ratio equal to0.91/2 = 0.9486… . So the series of timesconverges to

S = ⋅ =1 5762

1

1 0 930 71551 2. .K K

– . /

The model predicts that the ball comes to restafter about 30.7 s.

8. a.

Iteration Total Length

0 27

1 36

2 48

3 64

Because each segment is divided into fourpieces, each of which is 1/3 of the originallength, the length at the next iteration can becalculated by multiplying the previous lengthby 4/3.

b. The sequence of lengths diverges because thecommon ratio, 4/3, is greater than 1. Thus,the total length of the snowflake curve isinfinite!

c. From geometry, the area of an equilateral

triangle of side s is A s= 3

42.

The number of triangles added is 3, 12, 48,192, … .The side of each added triangle is 3, 1, 1/3,1/9, … .The added areas form the series

3

43 3 12 1 48 1 3 192 1 9[ ( ) ( ) ( / ) ( / ) ]2 2 2 2+ + + + L

= ⋅ ⋅ + +3

43 9 4 1 3 4 1 3 4 1 32 0 1 2[ ( / ) ( / ) ( / )2 4 6

+ +4 1 33 8( / ) ]L

= ⋅ ⋅ + + + +3

163 9 4 9 4 9 4 9 4 92[ / ( / ) ( / ) ( / ) ]2 3 4 L

= ⋅ ⋅ ⋅ =3

163 9 4 9

1

1 4 912 15 32 ( / ) .

– /The area of the pre-image is

3

49 20 25 32⋅ = . .

The total area is 32 4 3 56 1184. . cm .2= K

9. f xx

( ) = 6

1–

P x x x x x x( ) = + + + + + +6 6 6 6 6 62 3 4 5 L

′ = + + + + +P x x x x x( ) 6 12 18 24 302 3 4 L

′′ = + + + +P x x x x( ) 12 36 72 1202 3 L

′′′ = + + +P x x x( ) 36 144 360 2 L

f ′(x) = 6(1 − x)− 2

f ″(x) = 12(1 − x)− 3

f ″′(x) = 36(1 − x)− 4

P′(0) = 6 and f ′(0) = 6P″(0) = 12 and f ″(0) = 12P″′(0) = 36 and f ″′(0) = 36Conjecture: P fn n( ) ( )( ) ( )0 0= for all values of n.

Problem Set 12-3Q1. 0.3333… Q2. 0.4444…

Q3.2

3Q4.

4

9Q5. 13 Q6. 125

Q7. mass ⋅ displacement Q8. centroid

Q9. ln x + C Q10. D

1. f (x) = 5e2x

f ′(x) = 10e2x

f ″(x) = 20e2x

f ′′′(x) = 40e2x

f (4)(x) = 80e2x

2. P1(x) = c0 + c1x and ′ =P x c1 1( )P1(0) = c0 and f (0) = 5 ⇒ c0 = 5

′ =P c1 10( ) and f ′(0) = 10 ⇒ c1 = 10

∴ P1(x) = 5 + 10x

3. P2(x) = c0 + c1x + c2x2

′ = + ′′ =P x c c x P x c2 1 2 2 22 2( ) and ( )

P2(0) = c0 and f (0) = 5 ⇒ c0 = 5′ =P c2 10( ) and f ′(0) = 10 ⇒ c1 = 10′′ =P c2 20 2( ) and f ′′(0) = 20 ⇒ 2c2 = 20

⇒ c2 = 10∴ P2(x) = P1(x) = 5 + 10x + 10x2

c0 and c1 are the same as for P1(x).

4. P3(x) = c0 + c1x + c2x2 + c3x

3

P4(x) = c0 + c1x + c2x2 + c3x

3 + c4x4

′′′ = + =P x c c x P x c4 3 4 44

46 24 24( ) and ( )( )

′′′ = ′′′ =P c f4 30 6 0 40( ) and ( )

⇒ = ⇒ =6 4020

33 3c c

P c f44

440 24 0 80( ) ( )( ) and ( )= =

⇒ = ⇒ =24 8010

34 4c c

c0, c1, and c2 are the same as before.

5. P x x x x32 35 10 10

20

3( ) = + + +

P x x x x x42 3 45 10 10

20

3

10

3( ) = + + + +


f P4

P3

P4

P31

100y

x

6. P4 is indistinguishable from f for about−1 < x < 0.9.

7. P3(1) = 31.6666666…P4(1) = 35.0000000…f (1) = 5e2 = 36.9452804…∴ P4(1) is closer to f (1) than P3(1), Q.E.D.

8. c4

480

24

5 2

4= = ⋅

!The 5 is the coefficient in 5e2x.The 2 is the exponential constant.The 4 is the exponent of x in the last term.

9. c c3

3

2

220

6

5 2

3

20

2

5 2

2= = ⋅ = = ⋅

! !, ,

c c1

1

0

010

1

5 2

15

5 2

00 1= = ⋅ = = ⋅ =

! !( ! ),

10. Conjecture:

c c5

5

6

65 2

5

160

120

4

3

5 2

6

320

720

4

9= ⋅ = = = ⋅ = =

! !,

11. P xn

xn

n

n( ) = ⋅

=

∞

∑ 5 2

0!


x

y

x

y

Q3. Q4.

x

y

x

y

Q5. Q6.

x

y

y

x

Q7. exponent Q8. coefficient

Q9. power Q10. D

1. a.

n f xn( ) ( ) f n( ) ( )0 P n( ) ( )0 cn

0 e x 1 c0 1

1 e x 1 c1 1

2 e x 1 2!c21

2!

3 e x 1 3!c31

3!

∴ = + + + +P x x x x( ) ,1

1

2

1

32 3

! !L Q.E.D.

b. Next two terms: L L+ + +1

4

1

54 5

! !x x

c.1

0n

xn

n!=

∞

∑d.

S3

ex

5

y

x

3

e. The two graphs are indistinguishable forapproximately −1 < x < 1.

f. Solve ex − S3(x) = 0.0001 for x close to 1.x ≈ 0.2188…Solve ex − S3(x) = 0.0001 for x close to −1.x = −0.2237…The interval is −0.2237… < x < 0.2188… .

g. The ninth partial sum is S8(x).Solve ex − S8(x) = 0.0001 for x close to 1.x ≈ 1.4648…Solve S8(x) − ex = 0.0001 for x close to −1.x = −1.5142…The interval is −1.5142… < x < 1.4648… .

2. a. By equating derivatives:

n f xn( ) ( ) f n( ) ( )0 P n( ) ( )0 cn

0 cos x 1 c0 1

1 −sin x 0 c1 0

2 −cos x −1 2!c2 − 1

2!

3 sin x 0 3!c3 0

4 cos x 1 4!c41

4!

5 −sin x 0 5!c5 0

6 −cos x −1 6!c6 − 1

6!

7 sin x 0 7!c7 0

8 cos x 1 8!c81

8!


∴ = − + − + −P x x x x x( )

! ! ! !1

1

2

1

4

1

6

1

82 4 6 8 L ,

Q.E.D.

b. L L− + − +1

10

1

12

1

1410 12 14

! ! !x x x

c. (– )( )!

11

22

0

n n

nn

x=

∞

∑d. y = cos x

cos

S4

S7

x

y

e. See the graph in part d, showing S7(x) (eighthpartial sum).The graphs are indistinguishable forapproximately −5.5 < x < 5.5.

f. Solve S7(x) − cos x = 0.0001 for x closeto 5.5.x ≈ 4.5414…(Note that some solvers may give an errormessage. In this case, zoom in by table, starting at x = 5 and using increments of 0.1;then x = 4.5, and increments of 0.01, and soforth.)By symmetry, the interval is−4.5414… < x < 4.5414… .

g. Both functions are even. P(x) is even becauseit has only even powers of x.

3. a. S33 5 70 6 0 6

1

30 6

1

50 6

1

70 6( . ) .= − + −

!( . )

!( . )

!( . )

= 0.564642445…sin 0.6 = 0.564642473…∴ S3(0.6) ≈ sin 0.6, Q.E.D.

b. sin 0.6 = 0.564642473…Tail = sin 0.6 − Sn(0.6)First term of the tail is tn+1.

sin 0.6 − S1(0.6) = 0.0006424733…t2 = 0.000648sin 0.6 − S2(0.6) = −0.00000552660…t3 = −0.00000555428…sin 0.6 − S3(0.6) = 0.0000000276807…t4 = 0.0000000277714…In each case, the tail is less in magnitude thanthe absolute value of the first term of the tail,Q.E.D.

c. Make | | .tn+−< ×1

200 5 10 .1

2 30 6 5 102 3 21

( )!( . )

nn

+< ×+ −

Inequality is first true for n = 8.Use at least nine terms (n = 8).

4. a. P x x x x x( ) = + + + +1

3

1

5

1

73 5 7

! ! !L

b. By equating derivatives:

n f xn( )( ) f n( )( )0 P n( ) ( )0 cn

0 sinh x 0 c0 0

1 cosh x 1 c1 1

2 sinh x 0 2!c2 0

3 cosh x 1 3!c3

1

3!

4 sinh x 0 4!c4 0

5 cosh x 1 5!c5

1

5!

6 sinh x 0 6!c6 0

7 cosh x 1 7!c7

1

7!

∴ = + + + +P x x x x x( ) ,1

3

1

5

1

73 5 7

! ! !L Q.E.D.

c. S3(0.6) = 0.636653554…sinh 0.6 = 0.636653582…

∴ S3(0.6) ≈ sinh 0.6, Q.E.D.

d. Solve S3(x) − sinh x = 0.0001 for xclose to 1.x ≈ 1.4870…By symmetry, the interval is−1.4870… < x < 1.4870… .

e. P x x x x′ = + ⋅ + ⋅ + ⋅ +( ) 1

1

33

1

55

1

772 4 6

! ! !L

= + + + +11

2

1

4

1

62 4 6

! ! !x x x L

f. Find S3(0.6) for the P′ series.S3(0.6) = 1.1854648cosh 0.6 = 1.18546521…∴ S3(0.6) ≈ cosh 0.6, and thus the P′(x)series seems to represent cosh x, Q.E.D.

g. P x dx( )∫ = + ⋅ + ⋅ + +1

2

1

3

1

4

1

5

1

62 4 6x x x C

! !L

Simplifying and letting C = 1 gives

1

1

2

1

4

1

62 4 6+ + + +

! ! !x x x L ,

which is the series for cosh x, Q.E.D.

5. a. f (x) = ln x f (1) = 0f ′(x) = x− 1 f ′(1) = 1

′′ = − −f x x( ) 2 ′′ = −f ( )1 1

f x x′′′ = −( ) 2 3 f ′′′ =( )1 2

P x x x x( ) ( )= − − +11

21

1

312 3( – ) ( – )

− +1

41 4( – )x L


P x x x x′ = − − + − − − +( ) ( ) ( ) ( )1 1 1 12 3 K

′′ = − + − − − +P x x x( ) ( ) ( )1 2 1 3 1 2 K

′′′ = − − +P x x( ) ( )2 6 1 K

P(1) = 0 = f (1)P′(1) = 1 = f ′(1)P″(1) = −1 = f ″(1)P′′′(1) = 2 = f ′′′(2), Q.E.D.

b. L L+ − +1

51

1

615 6( – ) ( – )x x

c. P xn

xn n

n

( ) = ⋅+

=

∞

∑(– ) ( – )11

11

1

d.

1

1x

y

ln x

S10

e. S10(1.2) = 0.182321555…ln 1.2 = 0.182321556…S10(1.95) = 0.640144911…ln 1.95 = 0.667829372…S10(3) = −64.8253968…ln 3 = 1.0986122…S10(x) fits ln x in about 0 < x < 2.

This is a wider interval of agreement than thatfor the fourth partial sum, which looks likeabout 0.3 < x < 1.7.S10(1.2) and ln 1.2 agree through the eighthdecimal place. The values of S10(1.95) andln 1.95 agree only to one decimal place. Thevalues of S10(3) and ln 3 bear no resemblanceto each other.

6. a. P x x x x( ) ( )= − − +11

21

1

312 3( – ) ( – )

− +1

41 4( – )x K

n tn(3)

1 2

2 −2

3 2.6666…

4 −4

5 6.4

6 −10.6666…

The absolute values of the terms are gettinglarger as n increases.

b. lim limx

nx

n

tn→∞ →∞

= → ∞∞

| |2

= → ∞→∞

limln

x

n2 2

1 1= ∞

The series cannot possibly converge becausethe terms do not approach zero as napproaches infinity.

c.

n tn(1.2)

1 0.22 −0.023 0.0026666…4 −0.00045 0.0000646 −0.00001066…

The absolute values of the terms areapproaching zero as n increases.

d. Tail = ln 1.2 − Sn(1.2)First term of the tail is tn+1.

ln 1.2 − S1(1.2) = −0.01767…t2 = −0.02ln 1.2 − S2(1.2) = 0.002321…t3 = 0.002666…ln 1.2 − S3(1.2) = −0.0003451…t4 = −0.0004In each case, the tail is less in magnitude thanthe absolute value of the first term of the tail.

7. a. f (x) = tan− 1 x

P xn

xn n

n

( ) =+

+

=

∞

∑(– )11

2 12 1

0

= − + − +x x x x1

3

1

5

1

73 5 7 L

b. y = tan− 1 x, y = S5(x), and y = S6(x) (sixthand seventh partial sums)

1

1

y

x

f(x)

S6

S5

S5

S6

Both partial sums fit the graph of f very wellfor about −0.9 < x < 0.9. For x > 1 andx < −1, the partial sums bear no resemblanceto the graph of f.

Problem Set 12-5Q1. 4! = 24 Q2. 3! = 6

Q3. 4!/4 = 6 Q4. n = 3


Q5. n = m − 1 Q6. m = 1

Q7. 0! = 1!/1 = 1 Q8. (−1)! = 0!/0 = 1/0 = ∞

Q9. x x/ 2 7– Q10. A

1. f u eu( ) =

= + + + + + +1

1

2

1

3

1

4

1

52 3 4 5u u u u u

! ! ! !K

2. f u u( ) = ln

= − − + − +( )u u u u1

1

21

1

31

1

412 3 4( – ) ( – ) ( – ) K

3. f u u( ) = sin

= − + − + − +u u u u u u

1

3

1

5

1

7

1

9

1

113 5 7 9 11

! ! ! ! !L

4. f u u( ) = cos

= − + − + − +1

1

2

1

4

1

6

1

8

1

102 4 6 8 10

! ! ! ! !u u u u u L

5. f u u( ) = cosh

= + + + + + +11

2

1

4

1

6

1

8

1

102 6 8 10

! ! ! ! !u u u u u4 L

6. f u u( ) = sinh

= + + + + + +u u u u u u

1

3

1

5

1

7

1

9

1

113 5 7 9 11

! ! ! ! !L

7. f u u u u u u u( ) ( )= − = + + + + + +−1 11 2 3 4 5 L

8. f u u( ) = −tan 1

= − + − + − +u u u u u u1

3

1

5

1

7

1

9

1

113 5 7 9 11 L

9. x x x x x x x xsin

! ! ! != − + − + −

1

3

1

5

1

7

1

93 5 7 9 L

= − + − + −x x x x x2 4 6 8 101

3

1

5

1

7

1

9! ! ! !L

10. x x x x x x x xsinh! ! ! !

= + + + + +

1

3

1

5

1

7

1

93 5 7 9 L

= + + + + +x x x x x2 4 6 8 101

3

1

5

1

7

1

9! ! ! !L

11. cosh x3

= + + + + +1

1

2

1

4

1

6

1

83 2 3 4 3 6 3 8

!( )

!( )

!( )

!( )x x x x L

= + + + + +1

1

2

1

4

1

6

1

86 12 18 24

! ! ! !x x x x L

12. cos x2

= − + − + −1

1

2

1

4

1

6

1

82 2 2 2

!( )

!( )

!( )

!( )x x x x2 4 6 8 L

= − + − + −11

2

1

4

1

6

1

84 8 12 16

! ! ! !x x x x L

13. ln ( ) ( – ) ( – )x x x x2 2 2 2 2 31

1

21

1

31= − − + −L

(Or: ln x2 = 2 ln x = 2(x − 1) − (x − 1)2

+ −2

31 3( – ) )x L

14. e x x xx− = + + +2

11

2

1

32 2 2 2 3(– )

!(– )

!(– )

+ +1

42 4

!(– )x L

= − + − + −1

1

2

1

3

1

42 4 6 8x x x x

! ! !L

15.

e dt t t t t dttx x

− = + +

∫ ∫

2

0

2 4 6 8

01

1

2

1

3

1

4–

!–

! !–L

= − + ⋅ − ⋅ + ⋅ −x x x x x

1

3

1

5

1

2

1

7

1

3

1

9

1

43 5 9

! ! !7 L

16. ln ( – )//

( ) ( ) 3 3 11

23 1 2

1 31 3t dt t t

xx

= − −∫∫

+ − +…

1

33 1

1

43 13 4( – ) ( – )t t dt

=⋅ ⋅

−⋅ ⋅

1

3 2 13 1

1

3 3 23 12 3( – ) ( – )t t

+⋅ ⋅

−⋅ ⋅

+1

3 4 33 1

1

3 5 43 14 5

1 3

( – ) ( – )/

t tx

L

= − +1

63 1

1

183 1

1

363 12 3 4( – ) ( – ) ( – )x x x

− +1

603 1 5( – )x L

17.

1

114

4 8 12 16

xx x x x

+= − + − + −L

18.9

3

3

1 32 2x x+=

+ ( / )

= − + − +

3 1

1

3

1

3

1

32

24

36x x x L

= − + − +3

1

3

1

32 4

26x x x L

19.

1

114

0

4 8 12 16

0tdt t t t t dt

x x

+= + +∫ ∫ ( – – – )K

= − + − + −x x x x x

1

5

1

9

1

13

1

175 9 13 17 K

20.

9

33

1

3

1

322 4

26

00 tdt t t t dt

xx

+= + +

∫∫ – – K

= − +

⋅−

⋅+

⋅−3

1

3

1

3 5

1

3 7

1

3 93 5

27

39x x x x x K


21.d

dxx(sinh )2

= + + + +

d

dxx x x x2 6 10 141

3

1

5

1

7! ! !K

= + + + +2

6

3

10

5

14

75 9 13x x x x

! ! !K

= + + + +2

2

2

2

4

2

65 9 13x x x x

! ! !K

= + + + +…21

12

1

3605 9 13x x x x

Alternate solution:

d

dxx x x(sinh ) cosh2 22=

= + + + +

2 1

1

2

1

4

1

64 8 12x x x x

! ! !K

= + + + +…22

2

2

4

2

65 9 13x x x x

! ! !

22. d

dxx(cos ).0 5

= + +

d

dxx x x x1

1

2

1

4

1

6

1

82 3 4–

! !–

! !–K

= − + − + −1

2

2

4

3

6

4

82 3

! ! ! !x x x K

= − + − + −…1

2

1

12

1

240

1

100802 3x x x

Alternate solution:

d

dxx x x(cos ) sin.0 5 0 5 0 51

2= − − . .

= − − + − +

−1

2

1

3

1

5

1

70 5 0 5 1 5 2 5 3 5x x x x x. . . . .

! ! !L

= − +⋅

−⋅

+⋅

−1

2

1

2 3

1

2 5

1

2 72 3

! ! ! !x x x L

= − + − + −1

2

2

4

3

6

4

82 3

! ! ! !x x x L

Multiply by 1/1, 2/2, 3/3, 4/4, … and simplify.

23. P x x x428 3 2

0 7

22( ) ( )

.

!( )= − + − + −

+ − − −0 51

32

0 048

423 4.

!( )

.

!( )x x

= −8 + 3(x − 2) + 0.35(x − 2)2

+ 0.085(x − 2)3 − 0.002(x − 2)4

24. P x x x x52 37 2 1

0 48

21

0

31( ) ( )

.

!( )

!( )= + + − + + +

+ + − +0 36

41

0 084

514 5.

!( )

.

!( )x x

= 7 + 2(x + 1) − 0.24(x + 1)2

+ 0.015(x + 1)4 − 0.0007(x + 1)5

25.a. f (0.4) ≈ 2 + 0.5(1.4) − 0.3(1.4)2 − 0.18(1.4)3

+ 0.02(1.4)4 = 1.694912We must assume that the series converges forx = 0.4.

b. f (−1) = c0 = 2f ′(−1) = c1 = 0.5f″(−1) = 2!c2 = 2(−0.3) = −0.6f″′(−1) = 3!c3 = 6(−0.18) = −1.08f (4)(−1) = 4!c4 = 24(0.02) = 0.48

c. g(x) = f (x2 − 1) ≈ P4(x2 − 1)= 2 + 0.5(x2 − 1 + 1) − 0.3(x2 − 1 + 1)2

− 0.18(x2 − 1 + 1)3 + 0.02(x2 − 1 + 1)4

= 2 + 0.5x2 − 0.3x4 − 0.18x6 + 0.02x9

Sixth-degree polynomial:

g(x) ≈ 2 + 0.5x2 − 0.3x6

g(1) ≈ 2.2

d. g ′(x) = x − 1.2x3 + terms in higher powersof x.∴ g ′(0) = 0g″(x) = 1 − 3.2x2 + terms in higher powersof x.∴ g″(0) = 1 > 0. ∴ (0, g(0)) is a localminimum.

e. g t dt t t dtx x

( ) ( . . )0

2 4

02 0 5 0 3∫ ∫≈ + −

= + −

= + −

20 5

3

0 3

5 0

21

60 06

3 5

3 5

t t tx

x x x

. .

.

26. a. f (1) ≈ P4(1)= −4 + 3(1 − 2) + 0.5(1 − 2)2

− 0.09(1 − 2)3 − 0.06(1 − 2)4

= −6.47We must assume that the series converges forx = 1.

b. f (2) = c0 = −4f ′(2) = c1 = 3f″(2) = 2!c2 = 2(0.5) = 1f ′″(2) = 3!c3 = 6(−0.09) = −0.54f (4)(2) = 4!c4 = 24(−0.06) = −1.44

c. g(x) = f (x2 + 2) ≈ P(x2 + 2)= −4 + 3(x2 + 2 − 2) + 0.5(x2 + 2 − 2)2

−0.09(x2 + 2 − 2)3 − 0.06(x2 + 2 − 2)4

= −4 + 3x2 + 0.5x4 − 0.09x6 − 0.06x8

Fourth-degree polynomial:

g(x) ≈ −4 + 3x2 + 0.5x4


d. g′(x) = 6x + 2x3 + terms in higher powersof x.∴ g′(0) = 0g″(x) = 6 + 6x2 + terms in higher powersof x.∴ g″(0) = 6 > 0∴ (0, g(0)) = (0, 4) is a local minimum.

e. h x g t dt t t dtx x

( ) ( ) ( . )= ≈ − + +∫ ∫0

2 4

04 3 0 5

= − + + = − + +4 0 1 4 0 13 50

3 5t t t x x xx

. .

27. f (x) = sin x, about x = π/4:

f x x x( ) = +

−

⋅

2

2

2

2 4

2

2 2 4

2

–!

–π π

−⋅

+

⋅

2

2 3 4

2

2 4 4

3

!–

!–x x

π π 4

+

⋅

−2

2 5 4!–x

π 5

L

28. f (x) = cos x, about x = π/4:

f x x x( ) = −

−

⋅

2

2

2

2 4

2

2 2 4

2

–!

–π π

+⋅

+

⋅

2

2 3 4

2

2 4 4

3 4

!–

!–x x

π π

−

⋅

−2

2 5 4

5

!–x

πL

29. f (x) = ln x, about x = 1:

f x x x x( ) ( )= − − +11

21

1

312 3( – ) ( – )

− +1

41 4( – )x L

30. f (x) = log x, about x = 10:

f xx x

( ) = + ⋅ − ⋅11

10

10

10

1

2 10

10

10

2

2ln

( – )

ln

( – )

+ ⋅ − ⋅ +1

3 10

10

10

1

4 10

10

10

3

3

4

4ln

( – )

ln

( – )x xL

31. f (x) = (x − 5)7/3 , about x = 4:

f x x x( ) = − + − ⋅1

7

34

7 4

3 242

2( – )!( – )

+ ⋅ ⋅ − ⋅ ⋅ ⋅7 4 1

3 34

7 4 1 2

3 443

34

4

!( – )

(– )

!( – )x x

+ ⋅ ⋅ ⋅ −7 4 1 2 5

3 545

5(– )(– )

!( – )x L

32. f (x) = (x + 6)4.2, about x = −5:

f x x x( ) . ( )= + + + ⋅ +1 4 2 54 2 3 2

25 2. .

!( )

+ ⋅ ⋅ + 4 2 3 2 2 2

35 3. . .

!( )x

+ ⋅ ⋅ ⋅ + +4 2 3 2 2 2 1 2

45 4. . . .

!( )x L

33. By equating derivatives:

n f xn( )( ) f n( )( )0 P n( )( )0 cn

0 cos 3x 1 c0 1

1 −3 sin 3x 0 c1 0

2 −9 cos 3x −9 2!c2 − 9

2!

3 27 sin 3x 0 3!c3 0

4 81 cos 3x 81 4!c4

81

4!

5 −243 sin 3x 0 5!c5 0

6 −729 cos 3x −729 6!c6 − 729

6!

∴ = − + − + cos

! ! !3 1

9

2

81

4

729

62 4 6x x x x L

By substitution:

cos

!( )

!( )

!( )3 1

1

23

1

43

1

632 4 6x x x x= − + − +L

= − + − +1

9

2

81

4

729

62 4 6

! ! !x x x L

The two answers are equivalent. Substitutiongives the answer much more easily in this case.

34. By equating derivatives:

n f xn( )( ) f n( )( )0 P n( )( )0 cn

0 ln (1 + x) 0 c0 01 (1 + x)− 1 1 c1 12 −(1 + x)− 2 −1 2!c2 − 1

2!3 2(1 + x)− 3 2 3!c3 2

3!4 −6(1 + x)− 4 −6 4!c4 − 6

4!

∴ + = − + − + ( )ln 1

1

2

1

3

1

42 3 4x x x x x L

By substitution, substitute (1 + x) for u in

ln ( – ) ( – ) ( – )u u u u u= − − + − +( )1

1

21

1

31

1

412 3 4 L

ln ( ) .1

1

2

1

3

1

42 3 4+ = − + − +x x x x x L

The two answers are equivalent. Substitutiongives the answer much more easily in this case.

35. S4(1.5) = 0.40104166… ;ln 1.5 = 0.40546510…Error = 0.00442344…

Fifth term = =1

51 5 1 0 006255( . – ) .

The error is smaller in absolute value than thefirst term of the tail.


36. Solve numerically for x close to 2:S4(x) − ln x = 0.0001x ≈ 1.2263…Solve numerically for x close to 0.1:ln x − S4(x) = 0.0001x ≈ 0.7896…Interval is about 0.7896… < x < 1.2263… .

37. a. tan− = − + −1 3 5 71

3

1

5

1

7x x x x x

+ − +1

9

1

119 11x x L

∴ = − + − + − +− tan 11 1

1

3

1

5

1

7

1

9

1

11L

The tenth partial sum is S9(1).

Sn

n

n

9

0

9

1 11

2 10 760459904( ) .=

+=

=∑(– )

( )K

4S9(1) = 3.04183961…π = 3.14159265…The error is about 3%.

b. The fiftieth partial sum is S49(1).4S49(1) = 3.12159465…π = 3.14159265…The error is about 0.6%.(It is merely an interesting coincidence thatalthough 4S49(1) differs from π in the seconddecimal place, several other decimal placeslater on do match up!)

c. By the composite argument propertiesfrom trig,

tan tan tan− −+

1 11

2

1

3

=

+

⋅

tan tan tan tan

– tan tan tan tan

– –

– –

1 1

1 1

12

13

112

13

=+

⋅=

12

13

112

13

1–

∴ = +− − − ,tan tan tan1 1 111

2

1

3 Q.E.D.

p Sn

n

n

n

= =+

=

+

∑4 4 11

2 1

1

29

0

9 2 1

(– )

++

=

+

∑(– )11

2 1

1

30

9 2 1n

n

n

n

= −( )+

+

= …

+ +

=∑4

1

2 1

1

2

1

3

3 14159257

2 1 2 1

0

9 n n n

nn

.π = 3.14159265…

The answer differs from π by only 1 in theseventh decimal place. The improvement inaccuracy is accounted for by the fact that theinverse tangent series converges much morerapidly for x = 1/2 and x = 1/3 than it does forx = 1. In Problem 17 of Problem Set 12-6,you will see that the interval of convergencefor the inverse tangent series is −1 ≤ x ≤ 1.In general, power series converge slowly atthe endpoints of the convergence interval.

38. sin

! ! !x x x x x= − + − +1

3

1

5

1

73 5 7 L

cos

! ! !x x x x= − + − +1

1

2

1

4

1

62 4 6 L

x x x x+ + + +

1

3

2

15

17

3153 5 7 L

1

1

2

1

4

1

6

1

3

1

5

1

72 4 6 3 5 7− + − + + +

! ! !–

! !–

!x x x x x x xL L

x x x x–! !

–!

1

2

1

4

1

63 5 7+ +L

1

3

4

5

6

73 5 7x x x− + −

! !L

1

3

1

6

1

723 5 7x x x– –+ L

2

15

64

75 7x x− −

!L

2

15

1

155 7x x– +L

17

3157x −L

∴ = + + + + tan x x x x x

1

3

2

15

17

3153 5 7 L

S4(0.2) = 0.202710024…tan 0.2 = 0.202710035…

39. Define a xf a

ix ai

ii( ) = −

( )( )

!( ) , the ith term of the

general Taylor series. So, f x a xi

i

( ) ==

∞

∑ ( ).0

We must assume d

dxa x

d

dxa x

n

n i

i

n

n i

i

( )=

∞

=

∞

∑ ∑=0 0

( );

that is, the nth derivative of an infinite series isthe infinite sum of the nth derivatives of theindividual terms.

For , ( )i nd

dxa x

f a

i

d

dxx a

n

n i

i n

ni< = ⋅ −

( ) ( )

!( )

= ⋅ =f a

i

i( ) ( )

!0 0


For ,i nd

dxa x

d

dxa x

n

n i

n

n n= =( ) ( )

= ⋅ −f a

n

d

dxx a

n n

nn

( ) ( )

!( )

= ⋅ −f a

nn x a

n( ) ( )

!!( )0

For , ( )i nd

dxa x

f a

i

d

dxx a

n

n i

i n

ni> = ⋅ −

( ) ( )

!( )

= ⋅ − − − + − =−f x

ii i i i n x a

ii n

( ) ( )

!( )( ) ( )( )1 2 1 0K

for x = a.

So ( ) for and , andd

dxa a i n i n

n

n i = < >0

d

dxa a f a

n

n nn( ) ( ).( )=

Thus, d

dxa x

n

n i

i

( )=

∞

∑0

evaluated at x = a

is d

dxa a f a

n

n nn( ) ( ).( )=

40. Brook Taylor: 1685−1731Colin Maclaurin: 1698−1746Sir Isaac Newton: 1642−1727Gottfried Wilhelm von Leibniz: 1646−1716

41. a.t

tn

x

nx

n

nxn

n

n n

n n

+

+ +

+= + =

+−1

2 1

1

11

11

11

1 11

(– ) ( – )

(– ) ( – )| |

b. r x102

111 2= =for .

r x109 5

111 95= =.

for .

r x1020

113= =for

c. rn

nx x

n

nx

n n=

+− = −

+= −

→∞ →∞lim lim

11 1

11| | | | | |

d. r = 1.1 for x = −0.1r = 1 for x = 0r = 0.9 for x = 0.1r = 0.9 for x = 1.9r = 1 for x = 2

e. Possible conjecture: The series converges toln x whenever the value of x makes r < 1, anddiverges whenever the value of x makes r > 1.

f. The series should converge for r < 1.r = |x − 1| < 1 ⇒ −1 < (x − 1) < 1 ⇒0 < x < 2


Problem Set 12-6Q1. sin x Q2. sinh x

Q3. e−x Q4. ex

Q5. (1 − x)− 1 (−1 < x ≤ 1)

Q6.1

22sin x C+

Q7. 3 sec2 3x Q8. 1

Q9. e Q10. B

1. a.n

x x x x xnn

n4

1

4

2

16

3

64

4

2562 3 4

1

= + + +=

∞

∑

+ +5

10245x L

b. Lt

t

n x

nxn

n

nn

n

n

n

n= = + ⋅→∞

+→∞

+

+lim lim11

1

1

4

4( )

= + =→∞

x n

n

xn4

1

4lim

Lx x

x< ⇔ < ⇔ − < < ⇔ − < <14

1 14

1 4 4

Open interval of convergence is (−4, 4).

c. Radius of convergence = 4.

2. a.x

nx x x x

n

nn

⋅= +

⋅+

⋅+

⋅=

∞

∑ 2

1

2

1

2 4

1

3 8

1

4 162 3 4

1

+⋅

+1

5 325x L

b. Lt

t

x

n

n

xn

n

nn

n

n

n

n= =+ ⋅

⋅ ⋅→∞

+→∞

+

+lim lim( )

11

11 2

2

=+

=→∞

x n

n

xn2 1 2lim

Lx x

x< ⇔ < ⇔ − < < ⇔ − < <12

1 12

1 2 2

Open interval of convergence is (−2, 2).


3. a.( ) ( )2 3

2 32 3

21

2x

nx

xn

n

+ = + + +

=

∞

∑ ( )

+ + + + +( ) ( )2 3

3

2 3

4

3 4x xL

b. Lx

n

n

xn

n

n= ++

⋅+→∞

+

lim( )

( )

2 3

1 2 3

1

= ++

= +→∞

| | | |2 31

2 3xn

nx

xlim

L < 1 ⇔ |2x + 3| < 1 ⇔ −1 < 2x + 3 < 1⇔ −2 < x < −1Open interval of convergence is (−2, −1).

c. Radius of convergence = 1

2.


4. a.( – )5 7

21

x

n

n

n=

∞

∑= + +( – ) ( – ) ( – )5 7

2

5 7

4

5 7

6

2 3x x x

+ +( – )5 7

8

4xL

b. Lx

n

n

xn

n

n=+

⋅→∞

+

lim( – )

( ) ( – )

5 7

2 1

2

5 7

1

= −+

= −→∞

| | | |5 71

5 7xn

nx

nlim

L < 1 ⇔ |5x − 7| < 1 ⇔ −1 < 5x − 7 < 1⇔ 1.2 < x < 1.6

Open interval of convergence is (1.2, 1.6).

c. Radius of convergence = 0.2.

5. a.n

nx n

n

3

1

8!

( – )=

∞

∑= − + +( )x x x8

8

28

27

682 3( – ) ( – )

+ +64

248 4( – )x L

b. Ln x

n

n

n xn

n

n= ++

⋅−→∞

+

lim( ) ( – )

( )!

!

( )

1 8

1 8

3 1

3

= − +

⋅

+

→∞| | lim

nx

n

n n8

1 1

1

3

= − ⋅ ⋅ =| |x 8 1 0 0

L < 1 for all values of x.

Series converges for all values of x.

c. Radius of convergence is infinite.

6. a.n

nx n

n

!( )4

1

2+=

∞

∑= + + + + +( )x x x2

2

162

6

8122 3( ) ( )

+ + +24

2562 4( )x L

b. Ln x

n

n

n xn

n

n= + ⋅ ++

⋅⋅ +→∞

+

lim( )! ( )

( ) ! ( )

1 2

1 2

1

4

4

= + + ⋅+

→∞| | ( ) x n

n

nn2 1

1

4

lim

= + + ⋅ = ∞→∞

| | [( ) ]x nn

2 1 1lim

The series converges only for |x + 2| = 0⇔ x = −2.


7. sin(– )

( )!x

nx

nn

n

=+

+

=

∞

∑ 1

2 12 1

0

Note that |(−1)n| can be left out of the ratio.

Lx

n

n

xn

n

n=+

⋅ +→∞

+

+lim( )!

( )!2 3

2 12 3

2 1

=+ +

= ⋅→∞

xn n

xn

2 21

2 3 2 20lim

( )( )∴ L < 1 for all x and the series converges forall x.

8. cos(– )

( )!x

nx

nn

n

==

∞

∑ 1

22

0

Note that |(−1)n| can be left out of the ratio.

Lx

n

n

xn

n

n=+

⋅→∞

+

lim( )!

( )!2 2

22 2

2

=+ +

= ⋅→∞

xn n

xn

2 21

2 2 2 10lim


9. sinh( )!

xn

x n

n

=+

+

=

∞

∑ 1

2 12 1

0

Lx

n

n

xn

n

n=+

⋅ +→∞

+

+lim( )!

( )!2 3

2 12 3

2 1

=+ +

= ⋅→∞

xn n

xn

2 21

2 3 2 20lim


10. cosh( )!

xn

x n

n

==

∞

∑ 1

22

0

Lx

n

n

xn

n

n=+

⋅→∞

+

lim( )!

( )!2 2

22 2

2

=+ +

= ⋅→∞

xn n

xn

2 21

2 2 2 10lim


11. en

xx n

n

==

∞

∑ 1

0!

Lx

n

n

xx

nx

n

n

n n=

+⋅ =

+= ⋅

→∞

+

→∞lim

( )!

!lim

1

1

1

10| | | |

∴ L < 1 for all x and the series converges forall x.

12. en

xxn

n

n

−

=

∞

= ∑ (– )

!

1

0

Lx

n

n

xx

nx

n

n

n n=

+⋅ =

+= ⋅

→∞

+

→∞lim

( )!

!lim

1

1

1

10| | | |

∴ L < 1 for all x and the series converges for all x.

13. tn = xnn!

Lx n

x nx n x

n

n

n n= + = + = ⋅ ∞

→∞

+

→∞lim

( )!

!| | lim ( ) | |

1 11

L = ∞ for all x ≠ 0; L = 0 at x = 0.∴ the series converges only for x = 0.


14. tn

xn nn= !

100

Ln x

n xn

n

n

n

n= + ⋅→∞

+

+lim( )!

!

1

100

1001

1

= + → ⋅ ∞→∞

| | | |xn

xnlim

1

100L = ∞ for all x ≠ 0; L = 0 at x = 0.∴ the series converges only for x = 0.

15. cosh( )!

101

2102

0

==

∞

∑ nn

n

Ln

nn

n

n=+

⋅→∞

+

lim( )!

( )!10

2 2

2

10

2 2

2

=+ +

=→∞

101

2 2 2 102 lim

( )( )n n nL = 0 < 1 ⇒ series converges.

16. ln (– ) (– . )0 1 11

0 91

1

. = +

=

∞

∑ n n

nn

tn = −(0.9)n/n

n tn t tn n+1/

1 −0.9 0.45

2 −0.405 0.6

3 −0.243 0.675

4 −0.164025 0.72

5 −0.118098 0.75

9 −0.043046721 0.81

15 −0.0137260754… 0.84375

35 −0.0007151872… 0.875

Ratio seems to approach 0.9.

Proof:

Ln

n

n

n

n

n

n

n

=+

⋅

=+

→∞

+

→∞

lim(– . )

(– . )

lim

0 9

1 0 9

0 91

1

.

= 0.9(1) = 0.9, Q.E.D.

17. a. P xn

xn n

n

( ) =+

+

=

∞

∑(– )11

2 12 1

0

Note that |(−1)n| = 1 for all n.

Lx

n

n

xn

n

n=+

⋅ +→∞

+

+lim2 3

2 12 3

2 1

= ++

= ⋅→∞

xn

nx

n

2 22 1

2 31lim by l’Hospital’s rule

L < 1 ⇔ x2 < 1 ⇔ −1 < x < 1Open interval of convergence is (−1, 1).

b.

tan–1

S4

S3S4

S3

x

y

1

1

The graphs of the partial sums of P(x) andtan− 1 x fit very well for −1 < x < 1. Thepartial sums diverge from tan− 1 x for x outsidethis interval.

c. S3(0.1) = 0.09966865238095…

d. tan− 1 0.1 = 0.09966865249116… ;Tail = 0.00000000011021…

e. First term of tail = =1

90 1 9( . )

0.00000000011111… ,which is larger than the tail.

18. a. y = x2 sin 2x, from x = 0 to x = 1.5, rotated

about the y-axis.A slice of the region parallel to the axis ofrotation generates a cylindrical shell.dV = 2π x · y · dx = 2πx3

sin 2x dx

V x x dx= ∫ 2 23

0

1 5

π sin.

Integrate by parts.

u dvx3 sin 2x

3x 2 –12 cos 2x

6x –14 sin 2x

618 cos 2x

0116 sin 2x

+

+

–

–

+

V x x x x= − +2

1

22

3

423 2π cos sin

+ −

3

42

3

82

0

1 5

x x xcos sin.

= − +

2

9

163

21

163π cos sin

= …4 662693947.

V x x dx= ≈ …∫ 2 2 4 6626939473π sin .0

1.5

The answers are the same to at least ninedecimal places.

b. Omitting the 2 23π , x x dxsin∫= − +

∫ x x x x3 3 521

32

1

52( )

!( )

!( )

− +

1

72 7

!( )x dxL


= − + − +

∫ 2

2

3

2

5

2

74

36

58

710x x x x dx

! ! !L

= −⋅

+⋅

2

5

2

7 3

2

9 55

37

59x x x

! !

−

⋅+ +2

11 7

711

!x CL

=+ ⋅ +

++

+

=

∞

∑(– )( ) ( )!

12

2 5 2 1

2 12 5

0

nn

n

nn n

x C

For this series,

Lx

n n

n n

xn

n n

n n= ⋅+ ⋅ +

⋅ + ⋅ +⋅→∞

+ +

+ +lim( ) ( )!

( ) ( )!2

2 7 2 3

2 5 2 1

2

2 3 2 7

2 1 2 5

= ++

⋅+ +→∞ →∞

42 5

2 7

1

2 3 2 22x

n

n n nn nlim lim

( )( )= 4x2

⋅ 1 ⋅ 0∴ the series converges for all x and thusconverges for x = 1.5.So,

V ≈ −⋅

+

⋅2

2

51 5

2

7 31 5

2

9 51 55

37

59π ( . )

!( . )

!( . )

−⋅

+⋅

2

11 71 5

2

13 91 5

711

913

!( . )

!( . )

= 4.67164363…The answer is within 0.01 of the answerfound in part a.

c. tn nn

nn

n+

++= − ⋅

+ ⋅ +1

2 32 71

2 2

2 7 2 31 5( )

π( ) ( )!

( . )

By table search, | tn+ 1 | < 0.5 × 10− 10 forn ≥ 10.Because n starts at zero for the first term, youwould need 11 terms to estimate the volumeto 10 decimal places.

19. a. Assume this series can be integrated term byterm.

f x e dttx

( ) = −∫2

0

= − + − + − +

∫ 1

1

2

1

3

1

4

1

52 4 6 8 10

0t t t t t dt

x

! ! ! !K

= − +⋅

−⋅

+⋅

x x x x x1

3

1

5 2

1

7 3

1

9 43 5 7 9

! ! !

−⋅

+1

11 511

!x K

b.

1

1

x

y

f(x)

S5

The partial sum is reasonably close forapproximately −1.5 < x < 1.5.

c. tn n

xn

nn=

++(– )

( ) !

1

2 12 1

Note that | (−1)n | = 1.

Lx

n n

n n

xn

n

n=+ +

⋅ +→∞

+

+lim( )( )!

( ) !2 3

2 12 3 1

2 1

= ++

⋅+→∞ →∞

xn

n nn n

2 2 1

2 3

1

1lim lim

= x2 ⋅ 1 ⋅ 0 < 1 for all x.

d. Erf x does seem to be approaching 1 as xincreases, as shown by the following tablegenerated by numerical integration.

x erf x

1 0.8427007929…

2 0.9953222650…

3 0.9999779095…

4 0.9999999845…

5 0.9999999999…

20. a. Assume this series can be integrated term byterm.

sin

! ! !

t

t tt t t t= ⋅ − + −

1 1

3

1

5

1

73 5 7

+ − +

1

9

1

119 11

! !t t K

= − + − + − +11

3

1

5

1

7

1

9

1

112 4 6 8 10

! ! ! ! !t t t t t K

=+=

∞

∑(– )( )!

12 1

2

0

nn

n

t

n

Si x =

11

3

1

5

1

7

1

9

1

112 4 6 8 10

0− + − + − +

∫ ! ! ! ! !

t t t t t dtx

K

= −⋅

+⋅

−⋅

x x x x1

3 3

1

5 5

1

7 75

! ! !3 7

+⋅

−⋅

+1

9 9

1

11 119

! !x x11 K

=+ +

+

=

∞

∑(– )( )( )!

11

2 1 2 12 1

0

n n

nn n

x

b. Ratio for is sin t

t

Lt

n

n

tn

n

n=+

⋅ +→∞

+

lim( )!

( )!2 2

22 3

2 1

=+ +

= ⋅→∞

tn n

tn

2 lim( )( )

1

2 3 2 202

∴ L < 1 for all values of t.Series for (sin t)/t converges for all valuesof t.


Ratio for Si x is

Lx

n n

n n

x

xn

n n n

x

n

n

n

n n

=+ +

⋅ + +

= ++

⋅+ +

= ⋅ ⋅

→∞

+

+

→∞ →∞

lim( )( )!

( )( )!

lim lim( )( )

2 3

2 1

2

2

2 3 2 3

2 1 2 1

2 1

2 3

1

2 3 2 2

1 0∴ L < 1 for all values of x.Series for Si x converges for all values of x.The radii of convergence for both series areinfinite.

c. The third partial sum is S2(x).

S23 50 6 0 6

1

3 30 6

1

5 50 6( . ) .= −

⋅+

⋅!( . )

!( . )

= 0.5881296Si 0.6 = 0.5881288…The answers are quite close!

d.

x

y

Si (x)

TenthPartial Sum

π 2π 3π−π−2π−3π

1

−1

S9(x) is reasonably close to Si x for−3π < x < 3π .

21. a. Given L tn

nn=

→∞lim where L < 1.

By the definition of limit as n → ∞, thereis a number k > 0 for any ε > 0 such thatif n > k, then tn

n is within ε units of L.

Thus, tnn < L + ε, Q.E.D.

b. L < 1 ⇒ 1 − L > 0So take any ε < 1 − L⇒ L + ε < L + 1 − L⇒ L + ε < 1.

c. For all integers n > k,

0 ≤ < + ⇒ ≤ < +t L t Lnn

nnε ε0 ( ) and

(L + ε)n < (L + ε)n− k for all n > k becauseL + ε < 1, so 0 ≤ tn < (L + ε)n− k, Q.E.D.

d. Because 0 ≤ tn < (L + ε)n− k for all n > k, itfollows that the tail after tn satisfies0 1 2 3≤ + + ++ + +t t tn n n L

< (L + ε)n+ 1− k + (L + ε)n+ 2− k + (L + ε)n+ 3− k + K

= (L + ε)n+ 1− k[1 + (L + ε) + (L + ε)2 + K ],

which converges because L + ε < 1.

e. The tail of the series is increasing and isbounded above by

( ) [ ( ) ( ) ]L L Ln k+ + + + + + =+ −ε ε ε1 21 K

( )

– ( )

–L

L

n k++

+εε

1

1So the series converges, Q.E.D.

22. L nn

n=→∞

lim

Because ln is a continuous function,ln ln lim lim lnL n n

n

n

n

n= =→∞ →∞

( )

= = = → ∞∞→∞ →∞ →∞

lim ln limln

limln

n n xnn

n

n

x

x

1

=→∞

lim/

x

x1

1 by l’Hospital’s rule

= 0∴ L = e0 = 1, Q.E.D.

23. ln(– )

( – )xn

xn

n

n

=+

=

∞

∑ 11

1

1

L tx

nnn

nn

nn= =

→∞ →∞lim | | lim

( – )1

= = = −→∞

| – |

lim

| – |x

n

xx

n

n

1 1

11| |

L < 1 ⇔ |x − 1| < 1 ⇔ 0 < x < 2Open interval of convergence is (0, 2).

24.1

1n

xnn

n=

∞

∑L

x

n

x

nx

n

n

nn

n= = = ⋅

→∞ →∞lim

| |lim

| || | 0

∴ L < 1 for all values of x, and thus the seriesconverges for all values of x, Q.E.D.

25. n xn n

n=

∞

∑1

L n x nx xn

n nn

n= = = ⋅ ∞

→∞ →∞lim | | lim | | | |

L = 0 if x = 0 and is infinite if x ≠ 0.∴ the series converges only if x = 0.

26.n

nxn

n

n

!

=

∞

∑1

Use the ratio technique.

Ln x

n

n

n xn

n

n

n

n= + ⋅+

⋅⋅→∞

+

+lim( )!

( ) !

1

1

1

1

=+

=+

= ⋅→∞ →∞

| | | | | |xn

nx

nx

en

n

n n nlim( )

lim( / )1

1

1 1

1

L < 1 ⇔ | |xe

⋅ 1 < 1 ⇔ −e < x < e

Open interval of convergence is (−e, e).

Problem Set 12-7Q1. geometric Q2. multiplying by 2

Q3. common ratio Q4. ln x


Q5. 14

2

16

42 4− +

! !x x Q6. −1/3!

Q7. 2.5 Q8. 1 < x < 7

Q9. ( ) ( )2 3 32e i t jtr r

+ cos

Q10. D

1. a. Sn

n

n

51

1

5

16

66

2

6

3

6

4

6

5= = − + − ++

=∑(– )

! ! ! ! !

= − + − + =6 3 11

4

1

203 8.

b.

Tail = = = − + − ++

=

∞

∑Rn

n

n

51

6

16 6

6

6

7

6

8(– )

! ! ! !K

c. Hypotheses: (1) signs are strictly alternating,(2) |tn| are strictly decreasing, and(3) lim .

nnt→∞

= 0

| | | | / ! /R t5 6 6 6 1120< = =

d. Absolute convergence means that | |tn

n=

∞

∑1

converges.If the convergent series were not absolutelyconvergent, it would be called “conditionallyconvergent.”

e. When you show absolute convergence, youfind the partial sums of |tn|. The partial sumsmust be increasing because |tn| is positive.|tn| is decreasing because the series isconvergent.

2. a. Comparison test (direct)

b. Integral test

c. nth term test

d. Geometric series test

e. Ratio test (or ratio technique)

f. Limit comparison test

g. p-series test

3. a. Sn

n

5 21

51

11

4

1

9

1

16

1

25= = + + + +

=∑

= = …11669

36001 463611.

b.

Tail = = = + + +=

∞

∑Rn

n

5 26

1 1

36

1

49

1

64L

The graph shows the tail bounded above by

( / ) .1 0 22

5x dx =

∞

∫ .

5 6 7 8 9

0.03

tn

n

The series converges because the sequence ofpartial sums is increasing and bounded aboveby 0.2.

c. ( / ) ( / )1 121000

2

1001x dx R x< <

∞∞

∫∫ 1000

1/1001 < R1000 < 1/1000R1000 ≈ 0.5(1/1001 + 1/1000)

= 0.00099950049…S = S1000 + R1000

≈ 1.643934… + 0.00099950049…= 1.644934…

The answer is correct to at least three decimalplaces.(The exact answer is π 2/6 = 1.64493406… .The estimate is actually correct to ninedecimal places.)

d. 0.5[1/(n + 1) + 1/n] = 0.0000005 (sixdecimal places)n ≈ 1,999,999.5, so use about 2 millionterms.

4. a. Sn

n

5

1

51

11

2

1

3

1

4

1

5= = + + + +

=∑

= = …217

602 2833.

p = 1, a harmonic series.

b. Tail = = = + + +=

∞

∑Rn

n

5

6

1 1

6

1

7

1

8L . The

graph shows the tail for R5 bounded below by

( / )16

x dx∞

∫ .

5

0.1

0.2

6 7 8 9

t

n

n

( / ) lim (ln – ln )1 66

x dx bb

= = ∞→∞

∞

∫∴ the series diverges because a lower bound isinfinite.

c. Graphing the rectangles to the left of then-values leads to R5 < ∞, which does notimply that the tail is finite.

d. S x dx xn

n n

> > ⇔ =+ +

∫1000 1 10001

1

1

1

if ( / ) ln

ln (n + 1) > 1000 ⇔ n > e1000 − 1≈ 1.97 · 10434 s. It would take approximately6.24 · 10420 yr.

5. a. 1

1

2

1

3

1

4− + − + L

The series converges because(1) strictly alternating signs, (2) strictlydecreasing |tn|, (3) tn → 0.


b. ( / ) lim (ln – ln )1 11

x dx bb

= = ∞→∞

∞

∫Τhe series | |tn

n=

∞

∑1

diverges, so the given series

does not converge absolutely.

c. S1000 = 0.692647… , S1001 = 0.693646… ,ln 2 = 0.693147…

| S1000 − ln 2| = 0.0004997… , |S1001 − ln 2| =0.0004992… , |t1001| = 1/1001 =0.00009900…∴ both partial sums are within |t1001| of ln 2.

d. No term is left out. No term appears morethan once.

Series is

1

2

1

4

1

6

1

8

1

10

1

12− + − + − +K

= + +

1

21

1

2

1

3

1

4– – K .

∴ the series converges to 1

22ln .

Conditional convergence means that whetherthe series converges, and, if so, what value itconverges to, depends on the condition thatyou do not rearrange the terms.

6. a. 11

4

1

9

1

16− + − +K

The series converges because(1) strictly alternating signs, (2) strictlydecreasing |tn|, (3) tn → 0.

b. | |tn

n=

∞

∑1

converges by the p-series test

because p > 1.

c. Ln

nn

=+

⋅ =→∞

lim( )

1

1 112

2

, so the ratio test

fails because L is not less than 1.

7. a. sin (– )( )!

xn

xn n

n

=+

+

=

∞

∑ 11

2 12 1

0

t32 3 11

1

2 3 10 6= − ⋅

⋅ +⋅ ⋅ +( ) .3

( )!

= − = −1

70 6 0 000005554285717

!. . K

b. S130 6 0 6

1

30 6 0 564( . ) . . .= − =

!

S23 50 6 0 6

1

30 6

1

50 6 0 564648( . ) . . . .= − + =

! !

c. R1 = sin 0.6 − S1(0.6) = 0.0006424…R2 = sin 0.6 − S2(0.6) = −0.0000055266…|R1| = 0.0006424…|t2| = 0.000648∴ |R1| < |t2||R2| = 0.0000055266…|t3| = 0.0000055542…∴ |R2| < |t3|

d. The terms are strictly alternating in sign, theterms are strictly decreasing in absolute value,and the terms approach zero for a limit asn → ∞. Thus, the series converges by thealternating series test.

Or:| | | |R tn n< +1 for all n ≥ 1, as shown by examplein part c.

limn

nt→∞ + =| |1 0 because it takes the form 0

∞.

∴ =→∞

| | ,limn

nR 0 and thus the series converges.

Or: Use the ratio technique.

Ln

n

n n

n

n

n

n

=+

⋅ +

=+ +

=

→∞

+

+

→∞

lim.

( )!

( )!

.

lim( )( )

0 6

2 3

2 1

0 6

0 361

2 3 2 20

2 3

2 1

.

Because L < 1, the series converges.

8. The sequence converges because limn

nt→∞= 2, a

(finite) real number.The series does not converge because lim

nnt→∞

≠ 0.

9. a.

1

3

1

8

1

15

1

24+ + + +K

Compare with the p-series with p = 2:

lim/( )

/lim ,

n n

n

n

n

n→∞ →∞

− =−

=1 1

1 11

2

2

2

2 a positive

real number.∴ the series converges by the limitcomparison test.

b. The p-series with p = 2 begins1

4

1

9

1

16

1

25+ + + . These terms form a lower

bound, not an upper bound, so the directcomparison test fails.

c. If n started at 1, the first term would be 1/0,which is infinite.

10. a. The seventh term of 1

0 60

nn

n!

.=

∞

∑ is

t661

60 6 0 0000648= =

!. . .

b. Sn

n

n

4

0

41

0 6 1 8214= ==

∑ !. .

e0.6 = 1.8221188…S4 differs from e0.6 by 0.00071880… , whichis greater than t5 = 0.000648, but not muchgreater. The difference is greater than t5

because all subsequent terms are added, notsubtracted. It is not much greater than t5

because the subsequent terms are very small.


c.

n 5 6 7

Tail: 0.000648 0.0000648 0.000005554…

Geometricseries: 0.000648 0.0000648 0.00000648

Terms of the e0.6 series are formed bymultiplying the previous term:

tn

tn n= −0 6

1.

Terms of the geometric series are formed bymultiplying the previous term by 0.1:t tn n= −0 1 1.For n ≥ 7, 0.6/n is smaller than 0.1, so theterms of the e0.6 series are smaller than thecorresponding terms of the geometric series.Thus, the geometric series forms an upperbound for the tail of the e0.6 series afterterm t6.

d. Geometric series converges to

0 00006481

1 0 10 000072. . .⋅ =

– .

e. The tail of the series after t6 is bounded by0.000072.The entire series is bounded byS6(0.6) + 0.000072 = 1.8221128 + 0.000072= 1.8221848.e0.6 = 1.8221188…So the upper bound is just above e0.6 , Q.E.D.

11. a.

1

1

2

1

3

2

4

6

5

24+ + + + + K

Ln

n

n

n

n

n nn n= + ⋅ = + ⋅

=

→∞ →∞lim

!

( – )!lim

1 1 1 10

∴ the series converges because L < 1.

b.

1

1

2

2

3

6

4

241 1

1

2

1

3+ + + + = + + + +K K

! !This is the Maclaurin series that convergesto e1.

c. Ln n

n n

n

nn n= =

→∞ →∞lim

/( – )!

/ !lim

!

( – )!

1

1= = ∞

→∞limn

n

∴ the test fails because the limit of the ratiois infinite.

12. a. U:

2

4

4

10

8

28

16

82+ + + + K

G:

2

3

4

9

8

27

16

81+ + + + K

V:

2

2

4

8

8

26

16

80+ + + + K

The terms of U are bounded above by thecorresponding terms of G, and so U convergesby the direct comparison test.

The terms of V are bounded below by thecorresponding terms of G, and so the directcomparison test fails in this case.

b. Ln

n n

n n n

n

n= =→∞ →∞

lim/( – )

/lim

–

2 3 1

2 3

3

3 1

= =→∞

limln

lnn

n

n

3 3

3 31

(using l’Hospital’s rule)∴ the V series converges because the G seriesconverges and L is a (finite) positive number.

13. Divergent harmonic series

14. Convergent p-series, p > 115. Convergent alternating series meeting the three

hypotheses

16. Divergent p-series, p ≤ 1

17.3

43

3

4

3

16

3

640

nn

= + + + +=

∞

∑ K

Converges because it is a geometric series withcommon ratio 1/4, which is less than 1 inabsolute value

18.

3

41

3

4

9

16

27

640

n

nn

= + + + +=

∞

∑ L

Converges because it is a geometric series withcommon ratio 3/4, which is less than 1 inabsolute value

19.

1

2 1

1

1

1

3

1

5

1

70

( )! ! ! ! !nn

+= + + + +

=

∞

∑ K

= + + + +1

1

6

1

120

1

5040K

Converges by comparison with geometricseries with t0 = 1 and r = 1/6

20.

1

31

1

3

1

9

1

270

(– )nn

= − + − +=

∞

∑ K

Converges by the alternating series test. (Termsare strictly alternating. Terms are strictlydecreasing in absolute value. tn approaches zeroas n approaches infinity.)

21.

n

nn

3

42

1

8

15

27

80

64

255

125

624–= + + + +

=

∞

∑ K

Diverges. Use the integral test.

x

xdx x

b

b3

44

2 21

1

41

–lim ln= −→∞

∞

∫ | |

= − −

= ∞→∞

lim lnb

b1

41 04( )

Or: Compare with a harmonic series.n

n nn n

3

42 2

1

1

–=

∞

=

∞

∑ ∑> → ∞


22.

1

2

2

3

3

4

4

5+ + + + K , divergent because tn does

not approach zero

23.

– –1

1

1

2

1

3

1

4+ + + + K , a convergent alternating

series meeting the three hypotheses

24.

sin sin sin sin sinnn

= + + + +=

∞

∑ 0 1 2 30

K

= + …+ …+ …+0 0 8414 0 9092 0 1411. . . K

Diverges. tn does not approach 0 as n → ∞.

25. Lt

t

n

nn

n

nn

n

n= = + = >→∞

+→∞

+

lim lim( / ) /( )

( / ) /1

14 3 1

4 34 3 1/ .

Diverges by the ratio test

26. Convergent geometric series with | r | = 7/11 < 1

27. Diverges because tn does not approach zero

28. Lt

tn

n

n

=→∞

+lim 1

= +→∞

+

lim[( ) – ]/

[ – ]/n

n

n

n

n

1 1 2

1 2

2 1

2

= + ++

= <→∞

1

2

1 1

1

1

21

2

2lim( )

n

n

nConverges by the ratio test

29.1 1

22 x xdx x dx x

b

b

lnlim (ln )=→∞

−∞

∫∫ ( / )

= − = ∞→∞

lim ln ln ln lnb

b[ ( ) ( )]2

Diverges by the integral test

30. Converges by direct comparison with the

convergent p-series 22

3n

n=

∞

∑31.

1 2 6 241 2 3 4e e e e

+ + + + K

Diverges because tn does not approach zero

32. Converges to e by the definition of e

33. x = − + − + − +1 1 1 1 1 1: K

Diverges by the nth term testx = + + + + +9 1 1 1 1 1: K

Diverges by the nth term testComplete interval is (1, 9).

34. x = − − + − + −1

1

3

1

18

1

81

1

324: K

Converges by the alternating series test

x = + + + +51

3

1

18

1

81

1

324: K

Converges by comparison with the geometric

series

1

3

1

9

1

27

1

81+ + + + K

Complete interval is [−1, 5].

35. x = − − + − + −41

1

1

2

1

3

1

4: K

Converges by the alternating series test

x = − + + + +2

1

1

1

2

1

3

1

4: K

Diverges. p series with p = 0.5, which is lessthan 1.Complete interval is [−4, −2).

36. x = − − + − + − +2

1

31 1 1 1 1: K

Diverges by the nth term test

x = − + + + + +12

31 1 1 1 1: K

Diverges by the nth term test

Complete interval is −

2

1

31

2

3, – .

37. n x n

n

( )−=

∞

∑ 31

Ln x

n xn

n

n= +→∞

+

lim( )( – )

( – )

1 3

3

1

= − + = − ⋅→∞

| | lim | |xn

nx

n3

13 1

L < 1 ⇔ | x − 3 | < 1 ⇔ 2 < x < 4At x = 2 the series is − + − + −1 2 3 4 L , whichdiverges because the terms do not approach zero.At x = 4 the series is 1 2 3 4+ + + +L , whichdiverges because the terms do not approach zero.Interval of convergence is (2, 4).

38.5

21

n n

n

x

n

⋅

=

∞

∑L

x

n

n

xn

n n

n n= ⋅+

⋅⋅→∞

+ +

lim( )

5

1 5

1 1

2

2

=+

= ⋅

→∞5

15 1

2

| | lim | |xn

nx

n

L < 1 ⇔ 5| x | < 1 ⇔ −0.2 < x < 0.2

At x = −0.2 the series is − + − + −1

1

4

1

9

1

16L ,

which is a convergent alternating series.

At x = 0.2 the series is 1

1

4

1

9

1

16+ + + +K ,

which is a convergent p-series with p = 2.Interval of convergence is [−0.2, 0.2].

39.x

n

n

n=

∞

∑1

Lx

n

n

xn

n

n=+

⋅→∞

+

lim1

1

=+

= ⋅→∞

| | | |xn

nx

nlim

11

L < 1 ⇔ | x | < 1 ⇔ −1 < x < 1

At x = −1 the series is − + − + −1

1

2

1

3

1

4K ,



At x = 1 the series is 1

1

2

1

3

1

4+ + + +K ,

which is a divergent harmonic series (p-serieswith p = 1).Interval of convergence is [−1, 1).

40.(– ) ( – )1 6

24

n n

nn

x

n ⋅=

∞

∑

Lx

n

n

xn

n

n

n

n=+ ⋅

⋅ ⋅→∞

+

+lim( – )

( ) ( – )

6

1 2

2

6

1

1

= −+

= − ⋅→∞

1

26

1

1

26 1| | limx

n

nx

n| |

L x x< ⇔ − < ⇔ < <11

26 1 4 8| |

At x = 4 the series is

1

4

1

5

1

6

1

7+ + + +K ,

which is a divergent harmonic series (p-serieswith p = 1).


1

4

1

5

1

6

1

7− + − +K ,

which is a convergent alternating series.Interval of convergence is (4, 8].

41. (– ) ( )1 5

2

1 2

1

n n

n

x

n

+

=

∞ ⋅ +∑L

x

n

n

xn

n

n= ++

⋅+→∞

+

lim( )

( ) ( )

5

2 1

2

5

2 2

2

= ++

= + ⋅→∞

( ) ( )xn

nx

n5

15 12 2lim

L < 1 ⇔ (x + 5)2 < 1 ⇔ −6 < x < −4

At x = −6 the series is 1

2

1

4

1

6

1

8− + − +K ,


At x = −4 the series is

1

2

1

4

1

6

1

8− + − +K ,

which is a convergent alternating series.Interval of convergence is [−6, −4].

42. ( )x

n

n

n

+

=

∞

∑ 12

1

Lx

n

n

xn

n

n= ++

⋅+→∞

+

lim( )

( ) ( )

1

1 1

1

2

2

= ++

= + ⋅

→∞| | | |x

n

nx

n1

11 1

2

lim

L < 1 ⇔ | x + 1 | < 1 ⇔ −2 < x < 0

At x = −2 the series is − + − + −1

1

4

1

9

1

16K ,



4

1

9

1

16+ + + +K , which

is a convergent p-series with p = 2.Interval of convergence is [−2, 0].

43.ln ( )n

nxn

n

++=

∞

∑ 1

10

Ln x

n

n

n xn

n

n= + ⋅+

⋅ ++ ⋅→∞

+

limln ( )

ln ( )

2

2

1

1

1

= ++

⋅ ++→∞ →∞

| |xn

n

n

nn nlim

ln ( )

ln ( )lim

2

1

1

2

= ++

⋅ ++→∞ →∞

| |xn

n

n

nn nlim

/( )

/( )lim

1 2

1 1

1

2

(by l’Hospital’s rule)

= ++

⋅ ++

= ⋅ ⋅→∞ →∞

| | | |xn

n

n

nx

n nlim lim

1

2

1

21 1

L < 1 ⇔ | x | < 1 ⇔ −1 < x < 1

At x = −1 the series isln ln ln ln

.1

1

2

2

3

3

4

4− + − +K

By l’Hospital’s rule,

limln ( )

lim/

n n

n

n

n→∞ →∞

= =1

10.

Because the terms decrease in absolute value andapproach zero for a limit, the series converges bythe alternating series test.At x = 1 the series is

ln ln ln ln1

1

2

2

3

3

4

4+ + + +L .

ln ( )lim (ln )

x

xdx x

b

b

=

→∞

∞

∫ 1

22

11

= −

= ∞→∞

lim (ln )b

b1

202

Thus, the series diverges by the integral test.Interval of convergence is [−1, 1].

44. 5 31

( )x n

n

−=

∞

∑L

x

xx

n

n

n= = −→∞

+

lim( – )

( – )

5 3

5 33

1

| |

L < 1 ⇔ | x − 3 | < 1 ⇔ 2 < x < 4At x = 2 the series is − + − + −5 5 5 5 L , whichdiverges because the terms do not approach zero.At x = 4 the series is 5 5 5 5+ + + +L , whichdiverges because the terms do not approach zero.Interval of convergence is (2, 4).

45.4

0

n

nn

x=

∞

∑

Lx

x

xn

n

n

n

n= ⋅ =→∞

+

+lim| |

4

4

41

1

Lx

xx x< ⇔ < ⇔ > ⇔ < − >1

41

41 4 4

| |

| | or


At x = −4 the series is 1 1 1 1− + − +L , whichdiverges because the terms do not approach zero.At x = 4 the series is 1 1 1 1+ + + +L , whichdiverges because the terms do not approach zero.Intervals of convergence are (−∞, −4) and (4, ∞).(Note that the series in Problems 45 and 46 havenegative powers of x and are called Laurent seriesrather than Taylor series.)

46.1

1xn

n=

∞

∑

Lx

x xn

n

n= =

→∞ +lim

| |1

1

Lx

x< ⇔ < ⇔ >11

11| |

| |1 ⇔ x < −1 or x > 1

At x = −1 the series is − + − + −1 1 1 1 L ,which diverges because the terms do not approachzero.At x = 1 the series is 1 1 1 1+ + + +L ,which diverges because the terms do not approachzero.Intervals of convergence are (−∞, −1) and (1, ∞).

47. a. Assume all the blocks have equal mass = m,with the center of mass at the center of theblock, and equal length = L.Write Hn = the distance the nth blockoverhangs the (n + 1)th block. (n = 1 forthe top block.)Note that according to the rule, Hn = thedistance between the rightmost edge of thenth block and the center of mass of the pileof the top n blocks.

Now, the center of mass of the nth block is12 L units from its rightmost edge, and the

center of mass of the pile of the top n − 1blocks is 0 units from (i.e., right on top of )the edge of the nth block according to therule.

Therefore, the center of mass of the pile of the

top n blocks is 1 1

20 1

nmL m n m⋅ + ⋅ −

( )

units from the edge of the nth block; that is,

Hn

Ln = 1

2, Q.E.D.

b. The total distance the top (first) blockoverhangs the nth block is H H1 2+ + +LHn−1. So for a pile of n blocks, the top block

will project entirely beyond the bottom blockif

L H Hn< + + =−1 1L

1

2

1

4

1

6

1

2 1L L L

nL+ + + +L

( – )

The first n for which

1

1

2

1

4

1

6

1

2 1< + + + + =L

( – )nnis 5.

c. To make a pile with overhang H, find an n

such that 1

1

2

1

3

1

1

2+ + + + >Ln

H

L– (this is

possible because the harmonic series divergesto infinity). Then a stack of n blocks willhave total overhang

H Hn1 1+ + −L

= + + + +1

2

1

4

1

6

1

2 1L L L

nLK

( – )

= + + + +

+

< ⋅ =1

21

1

2

1

3

1

1

1

2

2L

nL

H

LHK

(The achieved overhang is greater than H, soone may pull blocks slightly back—movingblocks back can only make the pile morestable—until the overhang equals H exactly.)

d. The theoretical overhang for a stack of 52objects is

H H H

L L L L

L

L

= + +

= + + + +

= + + + +

=

1 51

1

2

1

4

1

6

1

102

1

2

1

4

1

6

1

102

2 2594

L

L

K

K.

slightly more than two-and-a-quarter cardlengths.

48. The least upper bound postulate says that anynon-empty set of real numbers that has an upperbound has a least upper bound. In particular, anynumber less than this least upper bound cannotbe an upper bound for the set.The set of real numbers {t1, t2, t3, …} is non-empty and is bounded above (given). Therefore,this set has a least upper bound L. Any numberless than L is also less than some tD in the set.

Claim: L tb

n=→ ∞lim

Proof:

Pick a number ε > 0.Because L is an upper bound for tn, L + ε is alsoan upper bound.Because L is the least upper bound for tn,L − ε is not an upper bound.∴ there exists an integer D > 0 such thattD > L − ε.But the values of tn are increasing.∴ tn > tD > L − ε for all n > D.


Keep n > D.Then L − ε < tn < L + ε.Thus, tn is within ε units of L for all n > D.∴ L t

nn=

→∞lim by the definition of limit as

n → ∞, Q.E.D.

Problem Set 12-8Q1. ratio Q2. |common ratio| < 1Q3. for all values of x Q4. radius = 1

Q5. x x x x− + − +1

3

1

5

1

73 5 7 K

Q6.S4

sin

Q7. ln |sec x + tan x| + CQ8. y′ = sec2 xQ9. tan x + C Q10. Newton and Leibniz

1. a. cosh xn

x n

n

==

∞

∑ 1

22

0( )!

S5 4( ) = 27.2699118K

b. Rf c

5

2 5 22 5 24

2 5 24( )

=⋅ +

⋅+

⋅ +( . ) ( )

( )!

f x x( ) ( ) =12 cosh

∴ = < + M cosh ( )–41

23 24 4

= 40.5312… < 41

| ( ) | .R5

12441

124 1 4360≤ ⋅ =

!K

S5(4) is within 2 of cosh 4 in the units digit.

c. cosh 4 = 27.3082328…S5(4) = 27.2699118…cosh 4 − S5(4) = 0.0383… , which is wellwithin the 1.4360… upper bound found byLagrange form.

2. a. sinh( )!

xn

x n

n

=+

+

=

∞

∑ 1

2 12 1

0

S9(5) = 74.2032007…

b. Rf c

9

2 9 32 9 35

2 9 35( ) =

⋅ +⋅

⋅ +⋅ +

( ) ( )

( )!f (21)(x) = cosh x

∴ = < + M cosh ( )–51

23 25 5

= 121.5156… < 122

| ( )| .R9

215122

215 0 001138≤ ⋅ =

!K

S9(5) is within 2 units of sinh 5 in the thirddecimal place.

c. sinh 5 = 74.2032105…S9(5) = 74.2032007…sinh 5 − S9(5) = 0.00000981… , which iswell within the 0.001138… upper boundfound by Lagrange form.

3. a. ex

nx

n

n

==

∞

∑ !0

The fifteenth partial sum isS14(3) = 20.0855234… .

b. Rf c

14

15153

153( ) = ⋅

( ) ( )

!f (15)(x) = ex

∴ M = e3 < 33 = 27

| ( )| .R14

15327

153 0 0002962≤ ⋅ =

!K

S14(3) is within 3 units of e3 in the fourthdecimal place.

c. e3 = 20.085536923…S14(3) = 20.085523458…e3 − S14(3) = 0.00001346… , which is withinthe 0.0002962… found by Lagrange form.

4. a. ln (– ) ( – )xn

xn n

n

= ⋅+

=

∞

∑ 11

11

1

S8(0.7) = −0.356671944…

b. Rf c

8

990 7

90 7 1( . ) = ⋅ −

( ) ( )

!( . )

f (9)(x) = 8!x− 9

∴ M = 8!(0.7)− 9

| ( . )| ( . ) 9R8

990 7

8 0 7

90 3

1

93 7≤ ⋅ =!( . )

!( / )

–

= 5.4195… × 10− 5

S8(0.7) is within 6 units of ln 0.7 in the fifthdecimal place.(Note that for ln x, the Lagrange form of theremainder simplifies to

| ( )|R xn

x

xn

n

≤+

⋅

+1

1

1 1| – |

For x < 0.5, the fraction |x − 1|/x is greaterthan 1.The Lagrange form of the remainder becomesinfinite as n → ∞ and is thus not useful.)

c. ln 0.7 = −0.356674943…S8(0.7) = −0.356671944…|ln 0.7 − S8(0.3)| = 2.9998 × 10− 6, which iswithin the 5.4195… × 10− 5 found byLagrange form.

5. For sinh 2, all derivatives are bounded by cosh 2.

cosh ( )–21

23 2 4 6252 2< + = .

The general term is tnn

n=+

⋅ +1

2 122 1

( )!.


For six-place accuracy,

| ( )| . .Rnn

n24 625

2 32 0 5 102 3 6≤

+⋅ < ×+ −.

( )!The second inequality is first true for n = 6.Use at least seven terms (n = 6).

6. For cosh 3, all derivatives are bounded by cosh 3.

cosh ( )–31

23 2 13 56253 3< + = .


n= ⋅1

232

( )!.

For eight-place accuracy,

| ( )| . .Rnn

n313 5625

2 23 0 5 102 2 8≤

+⋅ < ×+ −.

( )!The second inequality is first true for n = 10.Use at least 11 terms (n = 10).

7. For ln x, the nth derivative (n ≥ 1) on [x, 1]is bounded by M = (−1)n+ 1(n − 1)!(0.6)− n.

| ( . )| | . |Rn

nn

nn0 6

0 6

10 6 1

11≤

+⋅ −

++!( . )

( )!

–( )

=+

⋅

+1

1

2

3

1

n

n

For seven-place accuracy,

1

1

2

30 5 10

17

n

n

+⋅

< ×

+−. .

This inequality is first true for n = 32.Use at least 32 terms.

8. For e10, all derivatives are bounded by e10.e10 < 310 = 59049

For five-place accuracy,

| ( )| . .Rnn

n1059049

110 0 5 101 5≤

+⋅ < ×+ −

( )!The second inequality is first true for n = 43.Use 44 terms (n = 43).

9. cosh 2 = 3.76219569…S4(2) = 3.76190476…cosh 2 − S4(2) = 0.000290929…


n= ⋅1

222

( )!.

Rf c

c4

2 4 22 4 2

10

22 4 2

22

10( ) =

⋅ +⋅ = ⋅

⋅ +⋅ +

( ) ( )

( )!cosh

!

∴ ⋅ = .cosh!

c2

100 000290929

10

K

cosh c = 1.03098027…c = cosh− 11.0309… = 0.2482… ,which is between 0 and 2.

10. e5 = 148.413159…S19(5) = 148.413107…e5 − S19(5) = 5.1234… × 10− 5

The general term is 1

5n

n

!.

Rf c

ec19

2020

20

520

55

20( ) = ⋅ = ⋅

( ) ( )

! !

∴ ⋅ = × − .ec 5

205 1234 10

205

!K

ec = 1.30702776…c = ln 1.3070… = 0.2677… ,which is between 0 and 5.

11. cos!( . )

!( . )

!( . )2 4 1

1

22 4

1

42 4

1

62 42 4. 6= − + −

+ − +1

82 4

1

102 48 10

!( . )

!( . ) L

= 1 − 2.88 + 1.3824 − 0.2654208

+ −0 0273004. L LThe terms are strictly alternating. They aredecreasing in absolute value after t1, and theyapproach zero for a limit as n → ∞.Therefore, the hypotheses of the alternating seriestest apply, and | Rn(2.4) | < | tn+ 1| =

1

2 22 4 2 2

( )!( . ) .

nn

++

For six-place accuracy, make| Rn(2.4) | < 0.5 × 10− 6.The inequality is first true for n = 7.Use 8 terms (n = 7).

12. e− = + − + + +2 2 3 41 21

22

1

32

1

42( )

!(– )

!(– )

!(– )

+ + +1

52

1

62 6

!(– )

!(– )5 L

= − + − …+ …− …+1 2 2 1 3333 0 6666 0 2666. . . LThe terms are strictly alternating. They aredecreasing in absolute value after t2, and theyapproach zero for a limit as n → ∞.Therefore, the hypotheses of the alternating series

test apply, and | ( )| | | .R tnn n

n− < =+

⋅++2

1

121

1

( )!For seven-place accuracy, make| Rn(−2) | < 0.5 × 10− 7.The inequality is first true for n = 14.Use 15 terms (n = 14).

13. a. Sn

n

10 31

101

1 19753198= = …=

∑ .

R x dx xb

b10

3 21010

0 5< =

=−

→∞

∞

∫ lim – . –

0.5(10− 2) = 0.005

R x dx xb

b10

3 21111

0 5> =

=−

→∞

∞

∫ lim – . –

0.5(11− 2) = 0.00413223…R10 ≈ 0.5(0.005 + 0.00413223…) =0.00456611…


S ≈ 1.19753198… + 0.00456611… =1.20209810…Error < 0.5(0.005 − 0.00413223…) =0.00043388… (about three decimal places)

b. Using both the upper and lower bounds,

error < −

+

∞∞

∫∫0 5 3 3

1. x dx x dx

nn

= − +− −0 25 0 25 12 2. . ( ) .n nSolve 0.25n− 2 − 0.25(n + 1)− 2 = 0.000005 toget n = 45.9194… .Use 46 terms.Using only the upper bound, Rn < 0.000005

if . .x dxn

3 0 000005<∞

∫Set 0.5n− 2 = 0.000005n = (0.000005/0.5)− 1/2

= 316.2277… , which rounds up to 317terms, considerably more that the 46 terms togive this accuracy by comparing upper andlower bounds of Rn.

14. a. Sn

n

100 1 051

1001

4 698244= = …=

∑ . .

R x dx xb

b100

1 05 0 05100100

20< =

=−

→∞

∞

∫ . lim – – .

20(100− 0.05 ) = 15.886564…

R x dx xb

b100

1 05 0 05101101

20> =

=−

→∞

∞

∫ . lim – – .

20(101− 0.05 ) = 15.878662…R100 ≈ 0.5(15.886564… + 15.878662…) =15.882613…S ≈ 4.698244… + 15.882613… =20.580858…Error < 0.5(15.886564… − 15.878662…) =0.003950… (about two decimal places)

b. Error . . .< −

+

∞∞

∫∫0 5 1 05 1 05

1x dx x dx

nn

= 20n− 0.05 − 20(n + 1)− 0.05

Solve 20n− 0.05 − 20(n + 1)− 0. 05 = 0.000005 toget n = 111840.2309… .Use 111,841 terms.With a value of p such as 1.05, which iscloser to 1 than 3 is, it takes more termsbecause the terms approach zero more slowly.

15. a. Sn

n

10 20

101

11 9817928=

+= …

=∑ .

The series converges because the terms of thetail starting at t1 are bounded above by theconvergent p-series with p = 2.

b. Rx

dx xb

bb

10 21

1010

1

1<

+=

=

→∞∫ lim tan–

π /2 − tan− 110 = 0.0996686…

Rx

dx xb

bb

10 21

1111

1

1>

+=

=

→∞∫ lim tan–

π /2 − tan− 111 = 0.0906598…

R10 ≈ 0.5(0.0996686… + 0.0906598…) =0.0951642…S ≈ 1.9817928… + 0.0951642… =2.0769570…Error < 0.5(0.0996686… − 0.0906598…) =0.004504…S ≈ 2.0769570… is correct to at least twodecimal places.Make Rn < 0.00005.Rn ≈ 0.5[(π/2 − tan− 1 n) − (π /2 − tan− 1 (n + 1))]= 0.5(tan− 1 (n + 1) − tan− 1 n)Solve 0.5(tan− 1 (n + 1) − tan− 1 n) = 0.00005to get n = 99.4962… .Use 100 terms.

16. In this p-series, p = 0.5, which is not greaterthan or equal to 1. Thus, the series diverges andthe remainder is infinite.

17. en

n

n

2 = ⋅=

∞

∑ 12

0!

From Example 1, S10 = 7.38899470… .By Lagrange form, | R10 | < 0.0004617… .Use a geometric series as an upper bound.

t t1111

12121

112

1

122= ⋅ = ⋅

! !and

Common ratio rt

t= =12

11

1

6

∴ < ⋅ ⋅ = … | | .R10111

112

1

116

0 00006156! –

The geometric series gives a better estimate ofthe remainder than does the Lagrange form.

18. en

n n

n

−

=

∞

= ⋅∑2

0

11

2(– )!

S10 = 0.135379188…

| | | | .R t10 11

112

110 000051306< = = …

!This number appears to be a better estimate ofthe error. However, it represents an error of ≈| R10 |/S(10) = 0.03789…%.e2 ≈ 1/S10 = 7.38665971…A 0.037…% error for this value would be0.002799… , which is a worse estimate of theerror than that by Lagrange or by geometricseries.(In general, an error of ε% in1/f (x) gives a

maximum error of εε1 100– /

in the value of f (x).

So an error of 0.03789…% in 1/e2 means an

error of 0 03789

1 0 00037890 03788 2.

– .

K

KK= . % in .)e


19. a.

250

239 7887357

π= . K

Thus, 250 radians is 39 complete cycles plus0.7887… additional cycle, orb = (2π)(0.7887…) = 4.9557730… radians.sin b = −0.970528019…sin 250 = −0.970528019… (Checks.)The value of b can be calculated efficientlyusing the fraction part command. For atypical grapher, b = f Part(250/(2π))2π.

b. From Figure 12-8c, you can tell that the valueof c is one cycle back from the value of b.c = b − 2π = −1.32741228…Check:sin c = −0.970528019…sin 250 = −0.970528019… (Checks.)In general:If b is in [0, π/2], then c = b.If b is in (π/2, 3π/2], then c = π − b.If b is in (3π/2, 2π], then c = 2π − b.

c. From Figure 12-8c, you can tell that thevalue of d is a quarter-cycle ahead of the valueof c. The value of the sine is the opposite ofthe corresponding value of cos d.

d c= + =π2

0 243384039. K

Check:−cos d = −0.970528019…sin 250 = −0.970528019… (Checks.)In general:

c d c x d∈ − −

= + = −π π π

2 4 2, : and sin cos

c d c x d∈ −

= − = −π

40, and : sin sin

c d c x d∈

= =0

4, : sin sin and

π

c d c x d∈

= − =π π π4 2 2

, : and sin cos

d. For x in [0, π/4], both the sine and cosineseries meet the hypotheses of the alternatingseries test. Thus, the error in S5(x) is boundedby | t6 |, the first term of the tail. | t6 | is greaterfor the cosine series than for the sine series.The maximum of | t6 | in the interval is atx = π/4.

∴ < =⋅ +

⋅ + | ( )| | ( / )|R x t5 62 6 24

1

2 6 24π π

( )!( / )

= 3.8980… × 10− 13,which is small enough to guarantee that sin xwill be correct to ten decimal places.For direct calculation,

| ( )| .Rnn

n2501

2 3250 0 5 102 3 10<

+⋅ < ⋅+ −

( )!

The second inequality is first true when n is348. Because both numerator and denominatormay overflow most computers, you cancalculate values of ln | Rn(250) | as follows:

ln ln ln| ( )| ( )R n in

i

n

250 2 3 2502

2 3

= + −=

+

∑Then compare the values with ln (0.5 ·10− 10).So you would need to use 349 terms.Unfortunately, even this procedure would notbe practical because the terms themselveswould have to be calculated to ten or moredecimal places, and they are so large that eachterm overflows most computers’ capacities.

e. The program will have the following steps.The particular commands will depend on thegrapher or computer used.

• Put in a value of x.

• Find b, as shown in part a.

• Find c, as shown in part b.

• Find d, as shown in part c.

• Choose the function and sign, as shownin part c.

• Calculate and display the answer.

20. For sin 1, | ( )| | |R tnn n

n11

2 311

2 3< =+

⋅ =++

( )!1

2 3( )!n +.

Set . .1

2 30 5 10 23

( )!n +< × −

This inequality is first true for n = 11.Use at least 12 terms (n = 11).Using the technique in Problem 19,

| ( )| . .Rnn

n11

2 24 0 5 102 2 23<

+< ⋅+ −

( )!( / )π

The second inequality is first true for n = 10.You would save only one term by the method ofProblem 19.

21. a. Apply the mean value theorem to f ′(x) on[a, x]. There is a number x = c in (x, a) suchthat

′′ = ′ ′f c

f x f a

x a( )

( ) – ( )

–⇒ f ′(x) = f ′(a) + f ″(c)(x − a), Q.E.D.

b. f x dx f a dx f c x a dx′ = ′ + ′′ −∫ ∫ ∫( ) ( ) ( )( )

f x f a x f c x a C( ) ( )= ′ + ′′ ⋅ +( ) ( – )1

22

Substituting the initial condition (a, f (a))gives

f (a) = f ′(a)a + f ″(c)(0) + C ⇒C = f (a) − f ′(a)a


f x( ) =

′ + ′′ ⋅ + − ′f a x f c x a f a f a a( ) ( ) ( )( ) ( – )1

22

f x f a f a x a f c x a( ) ( ) ( )( ) ( ) ,= + ′ − + ′′ −1

22( )

Q.E.D.

c. Apply the mean value theorem to f ″(x)on [a, x].There is a number x = c in (a, x) such that

′′′ = ′′ ′′f c

f x f a

x a( )

( ) – ( )

–⇒ ′′ = ′′ + ′′′ −f x f a f c x a( ) ( ) ( )( )

Integrate once to get f ′(a).

′′ = ′′ + ′′′ −∫∫∫ f x dx f a dx f c x a dx( ) ( ) ( )( )

′ = ′′ + ′′′ ⋅ +f x f a x f c x a C( ) ( ) ( )1

22( – )

Use (a, f ′(a)) as an initial condition.

f a f a a f c C′ = ′′ + ′′′ + ⇒( ) ( ) ( )( )1

20

C f a f a a= ′ − ′′( ) ( )

f x f a x f c x a′ = ′′ + ′′′ −( ) ( ) ( )( )21

2+ ′ − ′′f a f a a( ) ( )

f x f a f a x a f c x a′ = ′ + ′′ − + ′′′ −( ) ( ) ( )( ) ( )( )21

2

Integrate again to get f(x).

f x dx f a dx f a x a dx′ = ′ + ′′ −∫ ∫ ∫( ) ( ) ( )( )

+ ′′′ −∫ 1

2f c x a dx( )( )2

f x f a x f a x a( ) ( ) ( )( )2= ′ + ′′ −1

2

+ ′′′ − +1

6f c x a C( )( )3

Use (a, f ″(a)) as an initial condition.

f a f a a f a f c C( ) ( ) ( )( ) ( )( )= ′ + ′′ + ′′′ +1

20

1

60

⇒ = − ′C f a f a a( ) ( )

f x f a x f a x a( ) ( ) ( )( )2= ′ + ′′ −1

2!

+ ′′′ − + − ′1

33

!f c x a f a f a a( )( ) ( ) ( )

f x f a f a x a f a x a( ) ( ) ( )( ) ( )( )2= + ′ − + ′′ −1

2!

+ ′′′ −1

3!,f c x a( )( )3 Q.E.D.

d. The technique is mathematical induction.

22. a. f xe x

x

x

( )if

if = ≠

=

– –,

,

20

0 0

It is given that f n( )( )0 0= for all n > 0.c0 = f (0) = 0c1 = f ′(0) = 0

2!c2 = f ″(0) = 0 ⇒ c2 = 03!c3 = f ″′(0) = 0 ⇒ c3 = 0…∴ series is 0 + 0x + 0x2 + 0x3 + … , Q.E.D.

b. Each partial sum of the Maclaurin seriesequals zero for any value of x. Thus, thesequence of partial sums converges to zerofor all x. But f (x) does not equal zero exceptat x = 0. Thus, the series converges to f (x)only at x = 0.

c. e x x xx− −−

= + − + + +2

11

2

1

32 2 2 2 3( )

!(– )

!(– )– – L

= − + − +− − −1

1

2

1

32 4 6x x x

! !L

d. The fourth partial sum, S3(2) =0.7786458333… .

f e e( ) .2 0 77880078302 0 252

= = =− −− . K

The partial sum is close to f (2), so it isreasonable to make the conjecture that theLaurent series converges to f (2).

23. Using the Lagrange form of the remainder, thevalue of ex is given exactly by

en

x R xx nk

n

k

= +=

∑ 1

0!

( ), where

R xf c

kxk

kk( ) =

+

++

( ) ( )

( )!

11

1 and c is between 0 and x.

| ( )| | |R xM

kxk

k≤+

+

( )!11

Because all derivatives of e x equal e x, the value ofM for any particular value of x is also e x, whichis less than 3x, if x ≥ 0; or 1, if x < 0.

lim lim( )!k

kk

xkR x

kx

→∞ →∞

+<+

| ( )| | |3

11

which approaches 0 as k → ∞ by the ratiotechnique.Because the remainder approaches zero as napproaches infinity, ex is given exactly by

en

xx n

n

==

∞

∑ 1

0!

, Q.E.D.



R1. f x

xP x x x x( ) and ( )= = + + + +9

19 9 9 92 3

–L

1

20

x

yP6

P5

f


The graph shows that P5(x) and P6(x) are close tof (x) for x between about −0.7 and 0.6, and bearlittle resemblance to f (x) beyond ±1.P5(0.4) = 14.93856P6(0.4) = 14.975424f (0.4) = 15∴ P6(0.4) is closer to f (0.4) than P5(0.4) is,Q.E.D.P5(x) = 9 + 9x + 9x2 + 9x3 + 9x4 + 9x5;P5(0) = 9

′ = + + + + ′ =P x x x x x P52 3 4

59 18 27 36 45 0 9( ) ; ( )

′′ = + + + ′′ =P x x x x P52 3

518 54 108 180 0 18( ) ; ( )

′′′ = + + ′′′ =P x x x P52

554 216 540 0 54( ) ; ( )

f (x) = 9(1 − x)− 1; f (0) = 9f ′(x) = 9(1 − x)− 2; f ′(0) = 9f ″(x) = 18(1 − x)− 3; f ″(0) = 18

′′′ = − ′′′ =−f x x f( ) ( ) ; ( )54 1 0 544

∴ ′ = ′ ′′ = ′ ( ) ( ), ( ) ( ),P f P f5 50 0 0 0and ( ) ( )′′′ = ′′′P f5 0 0

Pn(x) is a subseries of a geometric series.

R2. a. Series is 3 + 2.7 + 2.43 + 2.187 + L .

After 10 days, S10

10

31 0 9

1 0 9= ⋅ =– .

– .19.5396… .About 19.5 mm increase in 10 days

S = ⋅ =31

1 0 930

– .About 30 mm increase eventually

b. Let x be the amount invested to have0.5 million dollars at the end of 19 years.Interest rate is 10% per year, so the amountat the end of a year is 1.1 times the amount atthe beginning of the year.x(1.119) = 0.5x = 0.5(1.1− 19) = 0.081753995…They must invest $81,754.00 now in orderto make the last payment.The total to invest is the sum0 5 1 1 0 5 1 1 0 5 1 11 2 19. ( . ) . ( . ) . ( . ).− − −+ + +L

This is the nineteenth partial sum of thegeometric series with first term 0.5(1.1)− 1

and common ratio 1.1− 1.

S191

19

10 5 1 11 1 1

1 1 14 182460045= ⋅ = …−. ( . ) .

– .

– .

–

–

They must invest $4,182,460.05 now tomake all 19 payments.

R3. P(x) = c0 + c1x + c2x2 + c3x

3 + c4x4 + L

f (x) = 7e3x ⇒ f (0) = 7 ⇒ c0 = 7f ′(x) = 21e3x ⇒ f ′(0) = 21 ⇒ c1 = 21f ″(x) = 63e3x ⇒ f ″(0) = 63 ⇒ 2!c2 = 63 ⇒ c2 = 31.5

′′′ = ⇒ ′′′ = ⇒f x e fx( ) ( )189 0 1893

c3 = 189/3! = 31.5

R4. a. e0.12 = 1.127496851…S3(0.12) = 1.127488, which is close to e0.12 .

b. cos 0.12 = 0.9928086358538…S3(0.12) = 0.9928086358528, which is close.

c. sinh (0.12) = 0.1202882074311…S3(0.12) = 0.1202882074310… , which isclose.

d. ln 1.7 = 0.530628251…S20(1.7) = 0.530612301… , which is close.ln 2.3 = 0.83290912…S20(2.3) = −4.42067878… , which is notclose.

R5. a. A Maclaurin series is a Taylor series expandedabout x = 0.

b. Substitute t = x + 1 into

ln ( – ) ( – )t t t t= − − +( )11

21

1

312 3

− +1

41 4( – )t L .

ln ( )x x x x x+ = − + − +1

1

2

1

3

1

42 3 4 L

=+

=

∞

∑ (– )1 1

1

nn

nn

x

c. Assume one may integrate this function termby term.

ln ( )x dx x x x C+ = −

⋅+

⋅− +∫ 1

1

2

1

3 2

1

4 32 3 4 L

=+

++

+

=

∞

∑ (– )

( )

1

1

11

1

nn

nn n

x C

d. ln ln( ) ( ) ( ) ( )x dx x x x C+ = + + − + +∫ 1 1 1 1 1

= + + + − + = −x x x x C C Cln ln( ) ( ) ( )1 1 11

= − + − +x x x x2 3 4 51

2

1

3

1

4L

+ − + − + − +x x x x x C

1

2

1

3

1

42 3 4 L

= −⋅

+⋅

− +1

2

1

3 2

1

4 32 3 4x x x CL ,

which is the same as the series in part c.

e.

t t dt t t t dtxx

cos!( )

!( )2 2 2 2 4

001

1

2

1

4= − + −

∫∫ L

= + +

∫ t t t t dt

x

–! !

–!

1

2

1

4

1

65 9 13

0L

= −⋅

+⋅

−⋅

+1

2

1

6 2

1

10 4

1

14 62 6 10 14x x x x

! ! !L

(Note that the series can be transformed to

= − + − +

1

2

1

3

1

5

1

72 2 3 2 5 2 7x x x x

!( )

!( )

!( ) L

= =∫1

2

1

22 2 2

0sin cos sin )x t t dt x

x

, and .


f. tan− =+∫1

20

1

1x

tdt

x

= − + − + −∫ [ ( ) ( ) ( ) ]1 2 2 2 2 3 2 4

0t t t t dt

x

L

(| | )t ≤ 1

= − + − + −x x x x x

1

3

1

5

1

7

1

93 5 7 9 L

g. f(3) = 5 ⇒ c0 = 5f ′(3) = 7 ⇒ c1 = 7f ″(3) = −6 ⇒ c2 = −6/2! = −3

′′′ = ⇒ = =f c( ) . . / ! .3 0 9 0 9 3 0 153

∴ f (x) = 5 + 7(x − 3) − 3(x − 3)2 +0.15(x − 3)3 + L

R6. a. = − −−

=

∞

∑ ( ) ( )3 51

n n

n

x

= − + − +1

35

1

95

1

2752 3( – ) ( – ) ( – )x x x L

b. Lx

xx

n

n n

n n= = −→∞

+ +

lim(– ) ( – )

(– ) ( – )| |

–( )

–

3 5

3 5

1

35

1 1

L < 1 ⇔ |x − 5| < 3 ⇔ 2 < x < 8Open interval of convergence is (2, 8).Radius of convergence = 3

c. cosh( )!

xn

x n

n

==

∞

∑ 1

22

0

Lx

n

n

xn

n

n=+

⋅→∞

+

lim( )!

( )!2 2

22 2

2

=+ +

= ⋅→∞

xn n

xn

2 21

2 2 2 10lim

( )( )L < 1 for all x.Series converges for all x, Q.E.D.

d. e1 2 2 31 1 21

21 2

1

31 2. .= + + +

!( . )

!( . )

+ +1

41 2 4

!( . ) L

S4(1.2) = 3.2944 (the fifth partial sum)e1.2 = 3.32011692…Error = e1.2 − S4(1.2) = 0.02571692…

The first term of the tail is t551

51 2= =

!( . )

0.020736.The error is greater than t5, but not muchgreater.

e.

x

ln

S11

S10

y

1

1 2

The open interval of convergence is (0, 2).Both partial sums fit ln well within this

interval. Above x = 2 the partial sumsdiverge rapidly to ±∞. Below x = 0 the partialsums give answers, but there are no realvalues for ln x.

R7. a. S10

10

10001 0 8

1 0 84463 129088= ⋅ =– .

– .. (exactly)

b. S n

n

= = ==

∞

∑1000 0 81000

1 0 85000

0

.– .

S − S10 = 536.870912, which differs from thelimit by about 10.7%.

c. “Tail”

d. “Remainder”

e. R x dx bb

1010

< = − +

−

→∞

−∞

∫ 3 2 21

2

1

210lim ( )–

= 0.005

R x dx bb

103 2 21

2

1

211> = − +

−

→∞

−∞

∫ lim ( )–

11

= 0.004132…Series converges because the tail is boundedabove by 0.005.S = S10 + R10 ≈ 1.197531… + 0.5(0.005 +0.004132…) = 1.202098…R10 is approximately 0.5(0.005 − 0.004132…)= 0.0004338… , so S10 is correct to aboutthree decimal places.

f. − + + + +1

4

1

3

1

22

1

59L

Ln

n

n

nn n= = =

→∞ →∞lim

/( – )

/lim

–

1 5

1 51

3

3

3

3

(Apply l’Hospital’s rule three times.)∴ the series converges because L is a positivereal number.The terms of the F series begin1

1

1

8

1

27

1

64+ + + + L .

Although the F series converges, its terms(after t1) are less, not greater, than thecorresponding terms of the S series, sothe comparison test is inconclusive.

g. 2/1! + 4/2! + 8/3! + 16/4! + 32/5! + L= 2 + 2 + 1.3333… + 0.6666…

+ …+ ==

∞

∑0 2666 21

. / !L n

n

n

The terms are decreasing starting at t2, whichcan be seen numerically, above, oralgebraically by the fact that the next term isformed by multiplying the numerator by 2and the denominator by more than 2.R1 is bounded by the geometric series withfirst term 2 and common ratio1.3333…/2 = 2/3.


Because | common ratio | is less than 1,the geometric series converges(to 2/(1 − 2/3) = 6).

Thus, the tail after the first partial sum isbounded above by a convergent geometricseries, Q.E.D.

h. The given series converges because, aswritten, it meets the three hypotheses of thealternating series test. It does not convergeabsolutely because replacing all minus signswith plus signs gives the divergent harmonicseries.

The given series is the Taylor series for ln xexpanded about x = 1 and evaluated at x = 2.The remainders approach zero, so the seriesconverges to ln 2.

Rearrange the series this way:

11

2

1

4

1

3

1

6

1

8

1

5

1

10– – –

− +

− +

− +1

12L

Each term of the series appears exactly once.Simplifying inside the parentheses andfactoring out 1/2 gives1

2

1

4

1

6

1

8

1

10

1

12− + − + − + L

= + + +

1

2

1

1

1

2

1

3

1

4

1

5

1

6– – – L

The series in parentheses is the original seriesthat converges to ln 2. So the series asrearranged converges to 0.5 ln 2, Q.E.D.

i. |R10000| < |t10001| = 1/10001

= 0.0000999900…

Upper bound is 1/10001.

j. i. 10 3

21

n n

n

x

n

( – )

=

∞

∑

Lx

n

n

x

xn

nx

n

n n

n n

n

=+

⋅−

= −+

= − ⋅

→∞

+ +

→∞

lim( – )

( ) ( )

10 3

1 10 3

10 31

10 3 1

1 1

2

2

2

| | lim | |

L < 1 ⇔ 10 |x − 3| < 1 ⇔ 2.9 < x < 3.1

At x = 2.9 the series is

− + − + −1

1

4

1

9

1

16L ,

which converges by the alternating seriestest.

At x = 3.1 the series is

1

1

4

1

9

1

16+ + + + L ,

which converges because it is a p-serieswith p = 2.

Interval of convergence is [2.9, 3.1].

ii. (– ) ( )1 1

21

n

n

n

n

x

n

+⋅=

∞

∑

Lx

n

n

x

xn

nx

n

n

n

n

n

n

= ++ ⋅

⋅ ⋅+

= ++

= + ⋅

→∞

+

+

→∞

lim( )

( ) ( )

lim

1

1 2

2

1

1

21

1

1

21 1

1

1

| | | |

L x x< ⇔ + < ⇔ + <11

21 1 1 2| | | |

⇔ − < <3 1x


1

1

2

1

3

1

4+ + + + L , which is a divergent

harmonic series.


− + − + −1

1

2

1

3

1

4L , which converges

by the alternating series test.

Interval of convergence is (−3, 1].

k. i. 10

10 10 5 1 66660

nn

!= + + +

=

∞

∑ . K

+ +0 4166. K L

The tail after S0 is bounded above by theconvergent geometric series with firstterm 10 and common ratio 0.5. Thus, theseries converges.

(Other justifications are possible.)

ii. ( )– –nn

3 1

1

5+=

∞

∑ = 1.2 + 0.325 + 0.2370… +

0.215625 + 0.208 + 0.2046… +

0 2029. ... + L

Diverges because tn → 0.2, not 0,as n → ∞

iii. Converges. The general term can berewritten 3(2/5)n, so the series is aconvergent geometric series with commonratio r = 2/5.

iv. Diverges. p-series with p = 1/3, which isnot greater than 1.

v. Converges by the ratio technique

lim( )!

! ( )!

! !

( )!n n

nn

n

n

n→∞ ++

⋅ + ⋅⋅ ⋅ ⋅

+4

3 1 3

3 3

31

= ++

= ⋅ <→∞

1

3

4

1

1

31 1lim

n

n

n


R8. a. cosh( )!

21

222

0

= ⋅=

∞

∑ nn

n

Fourth partial sum is S3(2).The (2n)th derivative of cosh x is cosh x,so all derivatives are bounded by

cosh ( )–21

23 2 4 6252 2< + = . .

| ( )| .R3

824 625

82 0 02936≤ ⋅ =.

!K

Error is less than 0.03.

b. en

n

n

3

0

13= ⋅

=

∞

∑ !

All derivatives of ex are equal to ex, so allderivatives are bounded by e3 < 33 = 27.For 20-place accuracy,

| ( )| . .Rnn

n327

13 0 5 101 20≤

+⋅ < ×+ −

( )!

The second inequality is first true for n = 33.Use at least 34 terms (n = 33).

c. Using the Lagrange form of the remainder, thevalue of cosh 4 is given exactly by

cosh( )!

41

24 42

0

= ⋅ +=

∑ nRn

k

n

k

( ), where

Rf c

kk

kk( )4

2 24

2 22 2=

+⋅

++

( ) ( )

( )! and c is between

0 and 4.

| ( )|RM

kkk4

2 242 2≤

+⋅ +

( )!Because all even derivatives of cosh x equalcosh x, for any value of x between 0 and 4 wecan use cosh 4 for M, and cosh 4 is less than1

23 2 40 531254 4( )–+ = . .

Use M = 41.

lim lim( )!k

kk

kRk→∞ →∞

+<+

⋅| ( )|441

2 242 2

=+

=→∞

+

414

2 20

2 2

lim( )!k

k

k

By the ratio technique, this fractionapproaches zero as k approaches infinity.Therefore, because the remainder approacheszero as k approaches infinity, cosh 4 is given

exactly by cosh( )!

,41

242

0

= ⋅=

∞

∑ nn

n

Q.E.D.

d. sinh( )!

0 61

2 10 62 1

0

. .=+

⋅ +

=

∞

∑ nn

n

S3(0.6) = 0.636653554…sinh 0.6 = 0.636653582…

sinh 0.6 − S3(0.6) = 2.7862… × 10− 8

Rf c c

3

2 3 32 3 3 90 6

2 3 30 6

90 6( . ) =

⋅ +⋅ = ⋅

⋅ +⋅ +

( ) ( ).

cosh

!.

= 2.7862… × 10− 8

cosh c = 1.00328…c = cosh− 1

1.00328… = 0.0809… , which isin the interval (0, 0.6).

e. ln (– ) ( – )xn

xn n

n

= +

=

∞

∑ 11

11

1

| Rn(1.3) | < | tn+ 1 |For 20-place accuracy, make

1

11 3 1 0 5 101 20

nn

+< ×+ −( . – ) . .

This inequality is first true for n = 35.Use at least 35 terms.

f. Sn

n

50 41

501

1 08232064= ==

∑ . K

R x dx xb

b50

4 3

501 3< = −

=−

→∞

−∞

∫ lim ( / )50

(1/3)(50− 3) = 0.000002666…

R x dx xb

b50

4 3

511 3> = −

=−

→∞

−∞

∫ lim ( / )51

(1/3)(51− 3) = 0.000002512…

The series converges because the sequence ofpartial sums is increasing, and the tail afterS50 is bounded above by 0.000002512… .R50 ≈ 0.5(0.000002666… + 0.000002512…)

= 0.000002589…S ≈ 1.082232064… + 0.000002589…

= 1.082323235…Error < 0.5(0.000002666… − 0.000002512…)= 0.0000000769… (about seven decimalplaces)

Concept Problems

C1. Recall that i i i i i= = − = − =–1 1 12 3 4, , , ,

so i4n = 1 and i4n+ 2 = −1 for all n.

a. cos!( )

!( )

!( )ix ix ix ix= − + −1

1

2

1

4

1

62 4 6

+ −1

88

!( )ix L

= − + − + −1

2 4 6 8

22

44

66

88i

xi

xi

xi

x! ! ! !

L

= − + − + −1

1

2

1

4

1

6

1

82 4 6 8–

! !

–

! !x x x x L

= + + + + +11

2

1

4

1

6

1

82 4 6 8

! ! ! !x x x x L

= cosh x, Q.E.D.


b. sin

!( )

!( )

!( )ix ix ix ix ix= − + − +1

3

1

5

1

73 5 7 L

= − + − +ix i

ix i

ix i

ix

23

45

67

3 5 7! ! ! L

= − + − +ix i x i x i x

–

! !

–

!

1

3

1

5

1

73 5 7 L

= + + + +

i x x x x

1

3

1

5

1

73 5 7

! ! !L

= i sinh x, Q.E.D.

c. e ix ix ix ixix = + + + +11

2

1

3

1

42 3 4

!( )

!( )

!( )

+ +1

55

!( )ix L

= + + + + + +1

2 3 4 5

22

33

44

55ix

ix

ix

ix

ix

! ! ! !L

= + + + + + +11

2 3

1

4 52 3 4 5ix x

ix x

ix

–

!

–

! ! !L

= − + −1

1

2

1

42 4

! !x x L

+ − + −

i x x x

1

3

1

53 5

! !L

= cos x + i sin x, Q.E.D.

d. Using the formula in part c (Euler’s formula):

e iπ = cos π + i sin π = −1 + i ⋅ 0 = −1,Q.E.D.

C2. tan tan tan41

5

1

2391 1− −−

=

+

⋅

tan tan – tan tan

tan tan tan tan

– –

– –

415

1239

1 415

1239

1 1

1 1

=

+

⋅

tan tan –

tan tan

–

–

415

1239

1 415

1239

1

1

To evaluate tan 41

51tan−

, recall that

tantan

– tan2

2

1 2AA

A= .

Therefore,

tan tan

–

21

5

25

115

5

121

2−

=

=

and tan tan

–

2 21

5

1012

15

12

120

1191

2⋅

=

=−

Substituting this value gives

tan tan tan–

41

5

1

239

120119

1239

1120119

1239

11 1− −−

=

+ ⋅=

Thus, 41

5

1

2391

41 1 1tan tan tan− − −− = = π

, Q.E.D.

The two series are

41

54

1

5

4

3

1

5

4

5

1

5

4

7

1

51

3 5 7

tan− = ⋅ −

+

−

+K

tan− = ⋅ −

+

−1

3 51

2394

1

239

4

3

1

239

4

5

1

239L

Rnn

n1

5

4

2 3

1

5

2 3

<

+

+

Rnn

n1

239

1

2 3

1

239

2 3

<

+

+

| Total remainder |

<+

⋅

+

+ +1

2 34

1

5

1

239

2 3 2 3

n

n n

To get π accurate to 50 places, as shown, theremainder for π must be less than 0.5 × 10− 50.For π/4, the remainder must be less than0.125 × 10− 50.The inequality

1

2 34

1

5

1

2390 125 10

2 3 2 350

n

n n

+⋅

+

< ×

+ +−.

is first true for n = 34.Use at least 35 terms.

C3. a. y″ + 9xy = 0; y = 5 and y′ = 7 when x = 0.y = c0 + c1x + c2x

2 + c3x3 + c4x

4 + c5x5

+ +c x66 L

y′ = c1 + 2c2x + 3c3x2 + 4c4x

3 + 5c5x4

+ +6 65c x L

y″ = 2c2 + 6c3x + 12c4x2 + 20c5x

3

+ +30 64c x L

b. Substitute y = 5 into the y-equation: c0 = 5Substitute y′ = 7 into the y′-equation: c1 = 7

c. Constant term: 2c2 = 0 ⇒ c2 = 0

x c c c-term: .6 9 01

69 5 7 53 0 3+ = ⇒ = ⋅ = −(– )

x c c24 112 9 0-term: + = ⇒

c41

129 7 5 25= ⋅ = −(– ) .

x c c c35 2 520 9 0

1

209 0 0-term: + = ⇒ = ⋅ =(– )

x c c46 330 9 0-term: + = ⇒

c61

309 7 5 2 25= =(– )(– . ) .

d. y = 5 + 7x + 0x2 − 7.5x3 − 5.25x4 + 0x5

+ +2 25 6. x L

S6(0.3) = 5 + 7(0.3) − 7.5(0.3)3 − 5.25(0.3)4

+ 2.25(0.3)6 = 6.85661525


e. To ascertain convergence or divergence, noticethat y can be written as three separate series.

y c c x c x= +

⋅+

⋅ ⋅ ⋅+0 0

32

069

2 3

9

2 3 5 6

(– ) (– )L

+ +

⋅+

⋅ ⋅ ⋅+c x c x c x1 1

42

179

3 4

9

3 4 6 7

(– ) (– )L

+ +

⋅+

⋅ ⋅ ⋅+c x c x c x2

22

52

289

4 5

9

4 5 7 8

(– ) (– )L

If 0.3 is substituted for x, the first and secondseries have terms that are strictly alternating,decreasing in absolute value, and approachingzero for a limit as n approaches infinity.Thus, these series converge by the alternatingseries test. The third series is zero becausec2 = 0. Thus, the entire series for y convergeswhen x = 0.3.

Chapter Test

T1. e x x x

nxx n n− = − + − + + − ⋅ +1

1

2

1

31

12 3

! ! !L L( )

Note: The last ellipsis mark is necessary or thiswould stand for a Taylor polynomial (finitenumber of terms), not a Maclaurin series(infinite number of terms).

T2.

5

15 5 5 52 3

+= − + − +

xx x x L

Geometric series, common ratio r = −x

T3. cos!( – )

!( – )x x x= − + −1

1

2

1

42 4π π

+ −1

66

!( – )x π L

T4. R xf c

x5

66

6( ) =

( ) ( )

!, where c is between 0 and x.

f x n x f c cn n n( ) ( )( ) ( ) ( !)( ) ! = − + ⇒ =− + −1 1 61 6 7( ) ( )

∴ = =R xc

xx

c5

76

6

7

6

6( ) ,

!

!

–

where 0 < c ≤ x.

T5. sin

!( )

!( )

!( )( )x x x x x2 2 2 3 2 5 2 71

3

1

5

1

7= − + − +L

= − + − +x x x x2 6 10 141

3

1

5

1

7! ! !L

=+

+

=

∞

∑(– )( )!

11

2 14 2

0

n n

nn

x

T6. The alternating harmonic series

1

1

2

1

3

1

4− + − + L converges conditionally, but

not absolutely. The condition is that the termsremain in the order presented and not berearranged.

T7. f(x) = ln x, f(1) = 0, c0 = 0f ′(x) = x− 1, f ′(1) = 1, c1 = 1

′′ = − ′′ = − = − = −−f x x f c c ( ) , ( ) , ! ,22 21 1 2 1

1

2

′′′ = ′′′ = = =−f x x f c c( ) 2 1 2 3 21

33

3 3, ( ) , ! ,

f x x f c( ) ( )( ) ! , ( ) !, ! !,4 4 443 1 3 4 3= − = − = −−

c4

1

4= − L

∴ = − − + ( )ln ( – ) ( – )x x x x11

21

1

312 3

− +1

41 4( – )x L , Q.E.D.

T8. 1000 999+ +L converges becauser = 0.999 < 1.(It converges to 1000/(1 − 0.999) = 1,000,000.) 0 0001 0 0002. .+ +L diverges because r = 2 ≥ 1.

T9.( – )2 5

31

x

n

n

n=

∞

∑

Lx

n

n

xn

n

n=+

⋅→∞

+

lim( – )

( – )

2 5

3 3

3

2 5

1

= −+

= − ⋅→∞

| | | |2 51

2 5 1xn

nx

nlim

L < 1 ⇔ |2x − 5| < 1 ⇔ −1 < 2x − 5 < 1⇔ 4 < 2x < 6 ⇔ 2 < x < 3Open interval of convergence is (2, 3).Radius of convergence is 0.5.

T10. At x = 2 the series is

− + − + −1

3

1

6

1

9

1

12L , which converges by

the alternating series test.At x = 3 the series is

1

3

1

6

1

9

1

12+ + + +L , which is a divergent

harmonic series (or 1/3 of p-series withp = 1).Converges at x = 2, diverges at x = 3

T11. f xt

dtx

( ) =+∫ 1

1 20

= + +∫ ( – – )1 2 4 6

0t t t dt

x

L

= − + − +x x x x

1

3

1

5

1

73 5 7 L

=+

+

=

∞

∑(– )11

2 12 1

0

n n

nn

x

(The same as tan− 1 x)

T12. Lx

n

n

xn

n

n=+

⋅ +→∞

+

+lim2 3

2 12 3

2 1

= ++

= ⋅→∞

xn

nx

n

2 22 1

2 31lim

L < 1 ⇔ x2 < 1 ⇔ −1 < x < 1



− + − + −1

1

3

1

5

1

7L , which converges by the

alternating series test.


1

3

1

5

1

7− + − +L , which

converges by the alternating series test.Interval of convergence is [−1, 1].

T13. f (0.6) ≈ S19(0.6) (the 20th partial sum)

= 0.540419500…

T14. f (0.6) ≈ 0.540419500… numerically

T15. f (0.6) = tan− 1 0.6 = 0.540419500… exactlyThe answers to Problems T13 and T14 are correctto at least ten decimal places.

T16. The first term of the tail for S19(0.6) is

t2041 111

410 6 1 9562 10= = … × −( . ) . , which agrees

with the observation that S19(0.6) is correct to atleast ten decimal places.

T17. cosh! ! !

x x x x= + + + +11

2

1

4

1

62 4 6 L

==

∞

∑ 1

22

0( )!n

x n

n

T18. All even derivatives of cosh x equal cosh x.

Derivatives are bounded by

cosh . .31

2

1

23 2 13 56253 3 3 3= + < + = =( ) ( )– –e e M

For ten-place accuracy,

| ( )| . .Rnn

n313 5625

2 23 0 5 102 2 10≤

+⋅ < ×+ −.

( )!

The second inequality is first true for n = 11.Use at least 12 terms (n = 11).

T19. a. p-series, with p = 1.5

b. x dx bb

−

→∞

∞− −= − + =∫ 1 5

1

0 5 0 52 2 1 2. lim [ ( )]. .

∴ the series converges because the integralconverges, Q.E.D.(As additional information, this calculationalso proves that R1 is bounded above by 2and thus that S is bounded above by 3.)

c. S100 = 2.41287409…

d. R x dx1001 5

100< −

∞

∫ .

= − + =→∞

− −lim [ ( )] .. .

bb2 2 100 0 20 5 0 5

R x dx1001 5

101> −

∞

∫ .

= +→∞

lim [– ( )]– . – .

bb2 2 1010 5 0 5

= 0.19900743…

R100 ≈ 0.5(0.2 + 0.19900743…) =0.19950371…

S = S100 + R100 ≈ 2.41287409… +0.19950371… = 2.61237781…

(As additional information, the error in R100

is less than 0.5(0.2 − 0.1900743…) =0.0004962… , making S correct to about twodecimal places.)


Problem Set 12-10Cumulative Review Number 1

1. Limit: See Sections 2-2 and 2-5.Derivative: See Sections 3-2 and 3-4.Indefinite integral: See Section 5-3.Definite integral: See Section 5-4.

2. a. Continuity at a point: See Section 2-4.

b. Continuity on an interval: See Section 2-4.

c. Convergence of a sequence: A sequenceconverges if and only if lim

nnt→∞ exists.

d. Convergence of a series: A series convergesif and only if the sequence of partial sumsconverges.

e. Natural logarithm: See Section 3-9.

f. Exponential: ax = ex ln a

3. a. Mean value theorem: See Section 5-5.

b. Intermediate value theorem: See Section 2-6.

c. Squeeze theorem: See Section 3-8.

d. Uniqueness theorem for derivatives: SeeSection 6-3.

e. Limit of a product property: See Section 2-3.

f. Integration by parts formula: See Section 9-2.

g. Fundamental theorem of calculus: SeeSection 5-6.

h. Lagrange form of the remainder: SeeSection 12-8.

i. Parametric chain rule: See Section 4-7.

j. Polar differential of arc length: SeeSection 8-7.

4. a. f x t dtx

( ) sech= + ⇒∫ 13

f x x′ = +( ) sech1

b. f (x) = ax ⇒ f ′(x) = ax ln a

c. f (x) = xa ⇒ f ′(x) = axa− 1

d. f (x) = xx ⇒ ln f (x) = x ln x1/f (x) · f ′(x) = ln x + x · (1/x)f (x) = (ln x + 1) – f (x)f ′(x) = xx ln x + xx


e. e6x cos 3x dx dv ue6x cos 3x

6e6x 13sin 3x

36e6x – 19cos 3x

+

+

–

= +1

33

2


− ∫4 36e x dxx cos

5 36e x dxx cos ∫= + +1

33

2

336 6



153

2


f. cosh sinh cosh5 61

6x x dx x C= +∫

g. sec3 x dx∫= + + +1

2

1


h. ( ) | |sin cos ln sin5 51

551x x dx x C− = +∫

i. limcos –

lim– sin

x x

x

x

x

x→ →=

0 2 0

7 1

13

7 7

26

= = −→

lim– cos

x

x0

49 7

26

49

26

j. L xx

x=→

lim ( – ) /

0

31

ln limln ( – )

lim–

–L

x

x xx x= = = −

→ →0 0

3 1 3

13

L = e− 3 = 0.0497…

5. a.dy

dxx y= − +0 2 0 3 0 3 1 8. . . , ( , )

x

y10

10

b. If x = 9, y ≈ 5.413… , which agrees with thegraph.

6. a. p = k(40 − y)

b. y = 0.1x2 ⇒ x = (10y)1/2

dA = 2x dy = 2(10y)1/2 dy

A y dy y= = ⋅∫ 2 102

10

2

310 3 2

0

40

( )1/2

0

40

( ) /

= 3200/3 = 1066.6666… yd2

(Or: Area = 2/3 of circumscribed rectangle =(2/3)(1600) = 3200/3, etc.)

c. dF = p dA = k(40 − y) ⋅ 2(10y)1/2 dy

F dF k= = …∫ 17066 60

40

. lb (exactly

256,000k/15)

d. dM = y dF = y ⋅ k(40 − y) ⋅ 2(10y)1/2 dy

M dM k= = …∫ 292 571 4, . lb-yd0

40

(exactly 10,240,000k/35)

e. F y M yM

F

k

k⋅ = = = =, yd

10240000 35

256000 1517

1

7

/

/By symmetry, x = 0.

Center of pressure is at 0 171

7,

.

7. a. z = 30 − 0.5y

b. For a cross section,A = 2xz = 2(10y)1/2(30 − 0.5y).A = 101/2(60y1/2 − y3/2)A′ = 101/ 2(30y− 1/ 2 − 1.5y1/ 2)

= (101/ 2)(y− 1/ 2)(30 − 1.5y)A′ = 0 ⇔ 30 − 1.5y = 0 ⇔ y = 20A′ is infinite ⇔ y = 0.A(0) = 0A(20) = 565.6854… (exactly 400 2 )A(40) = 400Maximum at y = 20; minimum at y = 0

c. dV = 2xz dy = 101/2(60y1/2 − y3/2) dy

V dV= =∫ 192000

40

(exactly)

Use 19200/5 = 3840 truckloads.

d. dL dx dy x dx= + = +2 2 21 0 04.

L dL= = … ≈−∫ 92 9356 92 9

20. . yd

20

8. r r rr t i t j= +( . ) ( . )100 0 03 50 0 03cos sin

r r rv t i t j= − +( . ) ( . . )3 0 03 1 5 0 03sin cos

Speed | |= = +rv (– sin . ) ( . cos . )3 1 5 1 5 1 52 2

= 2.9943… ≈ 2.99 ft/s

9. Si 0

tu

udu

t

= ∫ sin

= + +

∫ 1 1

3

1

5

1

73 5 7

uu u u u du

t

–! !

–!

L0

= + +

∫ 1

1

3

1

5

1

72 4 6

0–

! !–

!u u u du

t

L


= −

⋅+

⋅−

⋅+t t t t

1

3 3

1

5 5

1

7 73 5 7

! ! !L

+

+ +++(– )

( )( )!

1

2 1 2 12 1

nn

n nt L

Lt

n n

n n

tn

n

n=+ +

⋅ + +→∞

+

+lim( )( )!

( )( )!2 3

2 12 3 2 3

2 1 2 1

= ++ + +

= ⋅→∞

tn

n n nt

n

2 22 1

2 3 2 3 2 20lim

( )

( )( )( )∴ L < 1 for all values of t, and the seriesconverges for all values of t.Third partial sum is

S23 50 6 0 6

1

3 30 6

1

5 50 6( . ) .= −

⋅+

⋅!( . )

!( . )

= 0.5881296

| | | | .R t2 371

7 70 6 0 0000007934< =

⋅= …

!( . )

The answer is correct to within ±1 in the sixthdecimal place.

Si . .0 6 0 5881288090

0 6

= ≈ …∫ sin. u

udu

Note that this answer agrees with the third partialsum to within 1 in the sixth decimal place.

10. r = 5 + 4 cos θ

dA d= +1

25 4 2( cos )θ θ

A dA= ≈ … ≈∫ 103 6725 103 7 332

0

2

. . ft (exactly )ππ

11.dV

dtkV

dV

Vk dt= ⇒ =

ln | V | = kt + CV = C1e

kt

At t = 0, V = 300.300 = C1

dV

dt= −5 when V = 300

− = ⇒ = −5 3001

60k k

∴ V = 300e− (1 /60) t

At t = 10, V = 300e− 1/6 = 253.9445… ≈ 253.9 million gal.

Cumulative Review Number 21. Derivative: See Sections 3-2 and 3-4.

2. Definite integral: See Section 5-4.

3. Mean value theorem: See Section 5-5.

4. f x g t dt f x g xx

( ) ( ) ( )= ⇒ ′ =∫3( )

5. tanh sec tanh5 61

6x x dx x Ch2 = +∫

6. x x dxsinh 2∫ dv u x sinh 2x1

12 cosh 2x

014 sinh 2x

+

–

+

= − +1

22

1

42x x x Ccosh sinh

7.3 14

3 2

1

3

4

2

x

x xdx

x xdx

++

=+

+

∫ ∫( )( – )

–

–= −ln | x + 3 | + 4 ln | x − 2 | + C

8.

sinh

! ! !

x

xdx

xx x x x dx∫ ∫= + + + +

1 1

3

1

5

1

73 5 7 L

= + + + +

∫ 1

1

3

1

5

1

72 5 6

! ! !x x x dxK

= +⋅

+⋅

+⋅

+…+x x x x C1

3 3

1

5 5

1

7 73 5 7

! ! !

9.n x n

nn

( )−

=

∞

∑ 5

31

Ln x

n xn

n

n

n

n= + ⋅→∞

+

+lim( )( – )

( – )

1 5

3

3

5

1

1

= − + = − ⋅ = −→∞

1

35

1 1

35 1

1

35| | | | | |x

n

nx x

nlim

L x x< ⇔ − < ⇔ − < − <11

35 1 3 5 3| |

Open interval of convergence is 2 < x < 8.

10. x dx x dxa a

−

→

−∫ ∫=+

0 998

0

1

0

0 9981

. .lim

=→ +lim

..

a ax

0

0 00211

0 002

= −→ +lim ( ).

aa

0

0 002500 500

= 500

11. y x= 2

yx dx

x= = ⋅ =∫ 2

3

9

3

3

9

9 3

1

6

1

339

–12. f ( x) = x2

f ( 4) = 16f ( 3.99) = 15.9201, which is within 0.08 unitof 16.f ( 4.01) = 16.0801, which is not within 0.08unit of 16.Thus, δ = 0.01 is not small enough to keepf (x) within 0.08 unit of 4.

13. V A dx= ∫2

10

≈ + + + +2

3153 4 217 2 285 4 319 343[ ( ) ( ) ( ) ]

= 2140 ft3


14. r = 4 sin 2θ

dA r d d= =1

28 22 2θ θ θsin

A d= ≈ ≈∫ 8 2 6 2831 6 282 2

0

2

sin/

θ θπ

. . ft

(exactly 2π)

15. (x/5)2 + (y/3)2 = 1

y x= ± +0 6 25 2. Use– ( .)

dA y dx x dx= =2 1 2 25 2. –

A dA= ≈ ≈∫ 17 6021 17 6

1

5

. .K square units

( sin sin )exactly ( . ) .15 1 0 2 1 2 61 1− −− −

16. y = 0.0016x4

dA = (16 − 0.0016x4) dx

A x dx= =−∫ ( – . )16 0 0016 2564 2

10

10

ft

17. dL dx dy x dx= + = +2 2 3 21 0 0064( . )

L dL= ≈ … ≈−∫ 42 5483 42 55

10

10

. . ft

18. p = 62.4(16 − y)dA = 2x dy = 10y1/4 dy

dF = p dA = 62.4(16 − y) ⋅ 10y1/4 dy

F dF= ≈ … ≈∫ 113595 73 113 6000

16

. , lb

exactly 11359511

15

19. lim limln

x xy

x

x→∞ →∞= → ∞

∞

= =→∞

lim/

x

x1

10

20. yx x x

x

x

x′ = =( / )( ) – (ln )( ) – ln1 1 1

2 2

y′ = 0 ⇔ ln x = 1 ⇔ x = e = 2.718… ftThere is a maximum at x = e because y′ goesfrom positive to negative there.

21. yx x x x

x

x

x′′ = = +(– / )( ) – ( – ln )( ) – ln1 1 2 3 22

4 3

y″ = 0 ⇔ ln x = 1.5 ⇔ x = e1.5 = 4.4816…≈ 4.48 ftThere is a point of inflection at x ≈ 4.48 ftbecause y″ changes sign there.

22.

1

40

x

y

23. ln ( – ) ( – )x x x x= − − + −( )1

1

21

1

312 3 L

+ +

+(– )( – )

11

1nn

nx L

Lx

n

n

xn

n

n=+

⋅→∞

+

lim( – )

( ) ( – )

1

1 1

1

= −+

= − ⋅ = −→∞

| | | | | |xn

nx x

n1

11 1 1lim

L < 1 ⇔ | x − 1 | < 1 ⇔ −1 < x − 1 < 1

⇔ 0 < x < 2

At x = 0 the series is − − − − −1

1

2

1

3

1

4L ,

which is a divergent harmonic series.


2

1

3

1

4− + − +L , which

converges because it meets the three hypothesesof the alternating series test.∴ interval of convergence is 0 < x ≤ 2, Q.E.D.

24. | | | |R tn

xn nn< =

+++

111

11( – )

For ln 1.4 to 20 places, make

0 4

10 5 10

120. n

n

+−

+< ×. .

Solving numerically gives n > 45.817… .Use 46 terms.

25. If the velocity is 0 ft/s at time t = 0, the shipspeeds up, approaching approximately 34 ft/sasymptotically as t increases.

If the velocity is 50 ft/s at time t = 0, the shipslows down, again approaching 34 ft/sasymptotically as t increases.(The graphs are shown here. The differentialequation is dv/dt = 0.7(34 − v).)

t

v

26.r r rr t i t j= +( ) ( )ln sin 2r r rv t i t j= +( / ) ( )1 2 2cosr r ra t i t j= − + −( / ) ( )1 4 22 sin


Cumulative Review Number 31. δ is clearly smaller than necessary.

L+ε

L–ε

L

c – δ c c+δ

x

f(x)

2. See Sections 3-2 and 3-4 for definitions ofderivative.Graphical meaning: slope of tangent linePhysical meaning: instantaneous rate of change

3. g x f x dx g x f x( ) ( ) if and only if ( ) ( ). = ′ =∫4. f t dt L U

tn

tn

r

s

( ) ,= =→ →∫ lim lim

∆ ∆0 0 where Ln and Un are

lower and upper Riemann sums, respectively,provided the two limits are equal.

5. l’Hospital’s rule

limcos

–x x

x x

e→→

0 51

0

0

= →→

limcos – sin

– –x x

x x x

e0 55

1

5= −0.2

6. y = tan (sin 5x)y′ = sec2 (sin 5x) ⋅ 5 cos 5xChain rule

7. y = (5x − 3)(2x + 7)4(x − 9)ln y = ln (5x − 3) + 4 ln (2x + 7) + ln (x − 9)

y yx x x

′ = ++

+

5

5 3

8

2 7

1

9– –

8. y = tan− 1 xtan y = x, sec2 y y′ = 1

yy y

′ = =+

1 1

12 1sec tan–

yx

′ =+1

1 2

9. sin cos sin7 81

8x x dx x C= +∫

10. x dx2 9+∫ x = 3 tan θ

dx = 3 sec2 θ dθx2 9 3+ = sec θ

= ∫ 9 3sec θ θd

= + + +9

2

9


= + ⋅ + + + +9

2

9

3 3

9

2

9

3 3

2 2

1x x x x

Cln

= + + + + +1

29

9

292 2x x x x Cln

11.3 11

2 3

5

3

2

12

x

x xdx

x xdx

–

–

–

–+=

++

∫∫

= 5 ln | x + 3 | − 2 ln | x − 1| + C

12. sin−∫ 1 x dx u dvsin–1 x 1

11 – x2

x–

+

√

= − −− −∫x x x x dxsin ( ) ( )/1 2 1 21

= + − −− −∫x x x x dxsin ( ) ( )/1 2 1 21

21 2

= + − +−x x x Csin 1 21

13. Fundamental theorem of calculusSee Section 5-6 for statement.

14. See Figure 5-5b.

15. f x h t dt f x h xx

( ) ( ) ( ) ( )= ⇒ ′ =∫3

16. f (x) = xe− x

f ′(x) = e− x − xe− x

f ″(x) = −e− x − e− x + xe− x = e− x(x − 2)f ″(x) = 0 ⇔ x = 2f ″(x) changes sign at x = 2.∴ the only point of inflection is at x = 2.

17. y = sin x from x = 0 to x = 2.

dL dx dy x dx= + = +2 2 21 cos

L dL= ≈∫ 2 3516

0

2

. K

18. a. x dx xa a

−

→=

+∫ 3 4

0

1 4

0

16 16

4/ /lim

= =→ +lim ( – )/

aa

0

1 48 4 8

b. Average value = =8

16 0

1

2–

19. r = 10 cos θdA = 50 cos2 θ dθ

A d= ≈ …∫ 50 13 34782cos θ θ .0.5

1

(exactly 12.5(1 + sin 2 − sin 1))

20.r r rr t i t j= + −2 13r r rv ti t j= − −2 3 2

r r rv i j( )1 2 3= −Speed .= = …13 3 6055

Distance from origin is | | .rr t t= +4 29 –

d r

dtt t t t

| |( )–

r

= + −− −1

29 4 184 2 1 2 3 3/ ( )

= −7 10/ at t = 1

Distance is decreasing at 2.2135… .


21. y = cos xdV = 2πx ⋅ y ⋅ dx

V x x dx= ≈ …∫ 2 3 58640

2

ππ

cos/

.

(exactly 2π (π/2 − 1))

22. a. y = −1.5x + 6A = xy = −1.5x2 + 6xA′ = −3x + 6A′ = 0 ⇔ −3x + 6 = 0 ⇔ x = 2A(0) = 0, A(4) = 0, A(2) > 0Thus, maximum area is at x = 2, Q.E.D.

b. V = πx2y = π(−1.5x3 + 6x2)V′ = π(−4.5x2 + 12x)

V x x′ = ⇔ = =0 0 22

3or

V V V( ) ( ) and0 4 0 22

30= =

>

Thus, maximum volume is at x = 22

3.

23. V ≈ + + + +1

32 51 4 37 2 41 4 63 59( )[ ( ) ( ) ( ) ]

= 3942

33ft

24. a. erf ( )/x e dttx

= − −∫2 1 2

0

2

π

f x e dttx

( ) = −∫2

0

= + +

∫ 1

1

2

1

3

1

42 4 6 8

0–

!–

! !–t t t t dt

x

L

= − +

⋅−

⋅+

⋅−x x x x x

1

3

1

5 2

1

7 3

1

9 43 5 7 9

! ! !L

b. f xn n

xn n

n

( ) =+

+

=

∞

∑(– )( ) !

11

2 12 1

0

Lx

n n

n n

xn

n

n=+ +

⋅ +→∞

+

+lim( )( )!

( ) !2 3

2 12 3 1

2 1

= ++ +

= ⋅→∞

xn

n nx

n

2 22 1

2 3 10lim

( )

( )( )L < 1 for all values of x, and thus the seriesconverges for all values of x, Q.E.D.

Final Examination

1.sin . – sin

.

1 1 1

0 1= 0.497363752…

sin . – sin

.

1 01 1

0 01= 0.536085981…

sin . – sin

.

1 001 1

0 001= 0.539881480…

2. f ′(1) = cos 1 = 0.540302305…The quotients in Problem 1 are converging tocos 1.

3. f xf x h f x

hh′ = +

→( ) lim

( ) – ( )0

f cf x f c

x cx c′ =

→( ) lim

( ) – ( )

–

4. f (x) = ex

limx

f x e→

=2

2( )

If f (x) = e2 + 0.1, x = ln (e2 + 0.1) =2.01344… .If f (x) = e2 − 0.1, x = ln (e2 − 0.1) =1.98637… .On the left, keep x within 0.01362… unit of 2.On the right, keep x within 0.01344… unit of 2.So you must keep x within 0.01344… unit of 2.

5. L f xx c

=→

lim ( ) if and only if for any ε > 0 there is

a δ > 0 such that if x is within δ units of c butnot equal to c, then f (x) is within ε units of L.

6. ε = 0.1, δ = 0.01344…

7. See Figure 1-3a.

8. Distance ≈ 17.4 m

10

1 1.3 1.6 1.9 2.2 2.5 2.8

t

r

9. Distance ≈ + + + +1

30 3 7 4 9 2 13 4 12( . )[ ( ) ( ) ( )

2 10 4 8 5 17 4( ) ( ) ] . ,+ + = which agrees withProblem 8.

10. v(t) = te− t

Distance ≈ 0.4[v(0.2) + v(0.6) + v(1) + v(1.4) +v(1.8)] = 0.601474…

11. Distance = = − −− − −∫ te dt te et t t

0

2

0

2

= −2e− 2 − e− 2 + 0 + 1 = 1 − 3e− 2 = 0.593994…

The difference is 0.00748… , which is about1.26%.

12. If f is integrable on [a, b] and g x f x dx( ) ( ) , = ∫then f x dx g b g a

a

b

( ) ( ) ( ).= −∫13. f (x) = x2/3

f x x′ = −( ) /2

31 3

f is differentiable everywhere except at x = 0. Butf is continuous at x = 0 because the limit of f (x)as x → 0 is zero, the same as f (0). Thus,


f meets the hypotheses of the mean valuetheorem because it is differentiable on (0, 1) andcontinuous at 0 and 1.Slope of the secant line from (0, 0) to (1, 1) is 1.

f c c c′ = = ⇔ =−( )2

31

8

271 3/

Tangent at x = 8/27 is parallel to secant.

1

18/27

x

f(x)

14. a. Example: 5 3

2 3

x

x xdx

–

( – )( )+∫= +

+

∫ 1

2

4

3x xdx

–

= ln | x − 2 | + 4 ln | x + 3 | + C

b. Example: 9 2– x dx∫x = 3 sin θdx = 3 cos θ dθ

9 32– cosx = θ

= = +∫ ∫99

21 22cos cosθ θ θ θd d( )

= + +

= + +

9

2

9

42

9

2

9

2

θ θ

θ θ θ

sin

sin cos

C

C

= + +−9

2 3

1

291 2sin –

xx x C

15. sec3 x dx∫ dvusec x sec 2 x

sec x tan x tan x+

–


= − +∫ ∫sec tan sec secx x x dx x dx3

2 3sec sec tan secx dx x x x dx= +∫ ∫sec sec tan ln sec tan3 1

2

1

2x dx x x x x C∫ = + + +| |

16.dy

dxky

dy

yk dx y kx C= ⇒ = ⇒ = + ⇒∫ ∫ ln | |

| y | = ekx+ C = ekxeC ⇒ y = C1ekx

17. Cross section of solid at any point in the slice isessentially the same as at the sample point.

2

4

x

y

(x, y)

18. Height at any point in the slice is essentially thesame as at the sample point.

2

4

x

y

(x, y)

19. dMy = x dA = x(4 − x2) dx

M x x dxy = − =∫ ( )4 42

0

2

A x dx= =∫ ( – )416

32

0

2

xA M xy= ⇒ = =4

16 3

3

4/

20. Let H = number of calories added.dH = C dT = (10 + 0.3T1/2) dT

H T dT= + =∫ ( . )/10 0 3 13 2001 2

100

900

, cal

21. a. Si Sixu

udu x

x

x

x

= ⇒ ′ =∫ sin sin0

b. Si x u u u dut

= + +

∫ 1

1

3

1

5

1

72 4 6

0–

! !–

!K

= −

⋅+

⋅−

⋅+t t t t

1

3 3

1

5 5

1

7 73 5 7

! ! !K

c. Si . ( . ) .0 7 0 7 0 71

3 30 71

3≈ = −⋅

=S!( . )

0.68094444…

d. | ( . )| | ( . )|R t1 250 7 0 7

1

5 50 7< =

⋅=

!( . )

0.0002801…S1(0.7) equals Si 0.7 correct to three decimalplaces and is within ±0.3 in the fourthdecimal place.

e. See Cumulative Review Number 1,Problem 9.

22. r r rr t i t j= +( ) ( )3 2

r r r r r rv t i t j v i j= + ⇒ = +( ) ( ) .3 2 0 5 0 75 12 ( . )r r r r r ra t i j a i j= + ⇒ = +( ) ( )6 2 0 5 3 2( . )


1

1

y

x

a

v→

→

The object is speeding up at t = 0.5 because theangle between the velocity and accelerationvectors is acute, indicating that the tangentialcomponent of acceleration acts in the samedirection as the velocity.(Algebraically,

r rv a⋅ = 4 25. , which is positive,

again indicating an acute angle.)

23. r = cos θ

dA r d d= =1

2

1

22 2θ θ θcos

A d= ≈∫ 1

20 23912

0

6

cos/

θ θπ

. K

exactlyπ24

3

16+

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