Student Solutions Manual Set for Calculus Early Transcendentals Single Variable 8th Edition

676 Pages·1992·7.87 MB·English

by  Howard Anton

CONTENTS
Introduction .............................................................. 1
Chapter 1. Functions ..................................................... 2
Chapter 2. LimitsandContinuity ......................................... 43
Chapter 3. TheDerivative ................................................ 65
Chapter 4. LogarithmicandExponentialFunctions ......................... 99
Chapter 5. AnalysisofFunctionsandTheirGraphs ........................ 139
Chapter 6. ApplicationsoftheDerivative ................................. 177
Chapter 7. Integration ...................................................209
Chapter 8. ApplicationsoftheDefiniteIntegral
inGeometry,Science,andEngineering ......................... 256
Chapter 9. PrinciplesofIntegralEvaluation ............................... 292
Chapter10. MathematicalModelingwithDifferentialEquations ..............343
Chapter11. InfiniteSeries ................................................361
Chapter12. AnalyticGeometryinCalculus .................................408
Chapter13. Three-DimensionalSpace;Vectors .............................448
Chapter14. Vector-ValuedFunctions ...................................... 490
Chapter15. PartialDerivatives ............................................524
Chapter16. MultipleIntegrals .............................................573
Chapter17. TopicsinVectorCalculus ..................................... 608
AppendixA. RealNumbers,Intervals,andInequalities .......................640
AppendixB. AbsoluteValue ...............................................647
AppendixC. CoordinatePlanesandLines ...................................650
AppendixD. Distance,Circles,andQuadraticEquations ......................658
AppendixE. TrigonometryReview .........................................668
AppendixF. SolvingPolynomialEquations .................................674
CALCULUS:
A New Horizon from Ancient Roots
EXERCISE SET FOR INTRODUCTION
123 41
1. (a) x=0:123123123:::; 1000x=123:123123123:::=123+x; 999x=123; x= =
999 333
115
(b) x=12:7777:::; 10x=127:7777:::, so 9x=10x¡x=115; x=
9
(c) x=38:07818181:::; 100x=3807:818181:::; 99x=3769:74;
3769:74 376974 41886 20943
x= = = =
99 9900 1100 550
4296 537
(d) 0:4296000:::=0:4296= =
10000 1250
repeats
z }| {
22
2. (a) … is irrational, and thus has a nonrepeating decimal expansion, whereas =3:142857:::
7
22
(b) >…
7
ˆ p ! ˆ p !
223 333 63 17+15 5 355 22 63 17+15 5
3. (a) < < p < < (b) p
71 106 25 7+15 5 113 7 25 7+15 5
ˆ p !
333 63 17+15 5
(c) (d) p
106 25 7+15 5
(cid:181) ¶ (cid:181) ¶
8 2 16 2 256
4. (a) If r is the radius, then D =2r so D = r = r2. The area of a circle of radius r is
9 9 81
…r2 so 256=81 was the approximation used for ….
(b) 256=81…3:16049, 22=7…3:14268, and … …3:14159 so 256=81 is worse than 22=7.
5. The flrst series, taken to ten terms, adds to 3:0418; the second, as printed, adds to 3:1416.
1 1 1 1 1 1 1
6. (a) =0:111111:::= + + + + + +:::
9 10 100 1000 10000 100000 1000000
2 1 8 5 1 8 5
(b) =0:185185:::= + + + + + +:::
27 10 100 1000 10000 100000 1000000
14 3 1 1 1 1 1
(c) =0:311111:::= + + + + + +:::
45 10 100 1000 10000 100000 1000000
7 6 3 6 3 6 3
7. (a) =0:636363:::= + + + + + +:::
11 10 100 1000 10000 100000 1000000
8 2 4 2 4 2 4
(b) =0:242424:::= + + + + + +:::
33 10 100 1000 10000 100000 1000000
5 4 1 6 6 6 6
(c) =0:416666:::= + + + + + +:::
12 10 100 1000 10000 100000 1000000
8. (a) 1, 2, 1.75, 1.7321 (b) 1, 3, 2.33, 2.238, 2.2361
9. (a) 1, 4, 2.875, 2.6549, 2.6458 (b) 1, 25.5, 13.7, 8.69, 7.22, 7.0726, 7.0711
10. (a) Let x1 = 12(a+b), x2 = 12(a+x1), x3 = 12(a+x2), etc. Then b>x1 >x2 >¢¢¢>xn¡1 >xn >a
so all the x’s are distinct, there are inflnitely many of them and they all lie between a and b.
i
(b) x=0:99999:::, 10x=9:99999:::, 9x=9, x=1
(c) (1:999999:::)=2=0:999999:::=1; yes it is consistent, as all three are equal.
(d) 10x=9+x, so x=9=9=1. They are equal.
1
CHAPTER 1
Functions
EXERCISE SET 1.1
1. (a) around 1943 (b) 1960; 4200
(c) no; you need the year’s population (d) war; marketing techniques
(e) news of health risk; social pressure, antismoking campaigns, increased taxation
2. (a) 1989; $35,600 (b) 1983; $32,000 (c) the flrst two years; the curve is steeper (downhill)
3. (a) ¡2:9;¡2:0;2:35;2:9 (b) none (c) y =0
(d) ¡1:75•x•2:15 (e) y =2:8 at x=¡2:6; y =¡2:2 at x=1:2
max min
4. (a) x=¡1;4 (b) none (c) y =¡1
(d) x=0;3;5 (e) y =9 at x=6; y =¡2 at x=0
max min
5. (a) x=2;4 (b) none (c) x•2; 4•x (d) y =¡1; no maximum value
min
6. (a) x=9 (b) none (c) x‚25 (d) y =1; no maximum value
min
7. (a) Breaks could be caused by war, pestilence, (cid:176)ood, earthquakes, for example.
(b) C decreases for eight hours, takes a jump upwards, and then repeats.
8. (a) Yes, if the thermometer is not near a window or door or other source of sudden temperature
change.
(b) The number is always an integer, so the changes are in movements (jumps) of at least one unit.
9. (a) If the side adjacent to the building has length x then L = x + 2y. Since A = xy = 1000,
L=x+2000=x.
(b) x>0 and x must be smaller than the width of the building, which was not given.
(c) 120 (d) Lmin …89:44
20 80
80
10. (a) V =lwh=(6¡2x)(6¡2x)x (b) From the flgure it is clear that 0<x<3.
(c) 20 (d) Vmax …16
0 3
0
2
3 Chapter1
500
11. (a) V =500=…r2h so h= . Then 7
…r2
500
C=(0:02)(2)…r2+(0:01)2…rh=0:04…r2+0:02…r
…r2
10
=0:04…r2+ ; C …4:39 at r …3:4; h…13:8:
min
r
1.5 6
4
10
(b) C =(0:02)(2)(2r)2+(0:01)2…rh=0:16r2+ . Since 7
r
0:04… <0:16, the top and bottom now get more weight.
Since they cost more, we diminish their sizes in the solution,
and the cans become taller.
1.5 5.5
4
(c) r …3:1, h…16:0, C …4:76
12. (a) The length of a track with straightaways of length L and semicircles of radius r is
P =(2)L+(2)(…r) ft. Let L=360 and r =80 to get P =720+160… =1222:65 ft.
Since this is less than 1320 ft (a quarter-mile), a solution is possible.
(b) P =2L+2…r =1320 and 2r =2x+160, so
450
L= 1(1320¡2…r)= 1(1320¡2…(80+x))=660¡80…¡…x:
2 2
0 100
0
(c) The shortest straightaway is L=360, so x=15:49 ft.
(d) The longest straightaway occurs when x=0, so L=660¡80… =408:67 ft.
EXERCISE SET 1.2
1. (a) f(0) = 3(0)2¡2 = ¡2;f(2) = 3(2)2¡2 = 10;f(¡2) = 3(¡2)2¡2 = 10;f(3) = 3(3)2¡2 = 25;
p p
f( 2)=3( 2)2¡2=4;f(3t)=3(3t)2¡2=27t2¡2
p p
(b) f(0) = 2(0) = 0;f(2) = 2(2) = 4;f(¡2) = 2(¡2) = ¡4;f(3) = 2(3) = 6;f( 2) = 2 2;
f(3t)=1=3t for t>1 and f(3t)=6t for t•1.
3+1 ¡1+1 …+1 ¡1:1+1 ¡0:1 1
2. (a) g(3) = = 2; g(¡1) = = 0;g(…) = ; g(¡1:1) = = = ;
3¡1 ¡1¡1 …¡1 ¡1:1¡1 ¡2:1 21
t2¡1+1 t2
g(t2¡1)= =
t2¡1¡1 t2¡2
p p
(b) g(3) = 3+1 = 2;g(¡1) = 3;g(…) = …+1;g(¡1:1) = 3; g(t2 ¡ 1) = 3 if t2 < 2 and
p
g(t2¡1)= t2¡1+1=jtj if t2 ‚2.
p p
3. (a) x6=3 (b) x•¡ 3 or x‚ 3
ExerciseSet1.2 4
(c) x2 ¡2x+5 = 0 has no real solutions so x2 ¡2x+5 is always positive or always negative. If
x=0, then x2¡2x+5=5>0; domain: (¡1;+1).
(d) x6=0 (e) sinx6=1, so x6=(2n+ 1)…,
2
n=0;§1;§2;:::
7
4. (a) x6=¡
5
1
(b) x¡3x2 must be nonnegative; y =x¡3x2 is a parabola that crosses the x-axis at x=0; and
3
1
opens downward, thus 0•x•
3
x2¡4
(c) > 0, so x2 ¡4 > 0 and x¡4 > 0, thus x > 4; or x2 ¡4 < 0 and x¡4 < 0, thus
x¡4
¡2<x<2
(d) x6=¡1
(e) cosx•1<2, 2¡cosx>0, all x
5. (a) x•3 (b) ¡2•x•2 (c) x‚0 (d) all x (e) all x
2 3 3
6. (a) x‚ (b) ¡ •x• (c) x‚0 (d) x6=0 (e) x‚0
3 2 2
7. (a) yes (b) yes
(c) no (vertical line test fails) (d) no (vertical line test fails)
8. The sine of (cid:181)=2 is (L=2)=10 (side opposite over hypotenuse), so that L=20sin((cid:181)=2).
9. The cosine of (cid:181) is (L¡h)=L (side adjacent over hypotenuse), so h=L(1¡cos(cid:181)).
10. T 11. h 12. w
t
t
t
5 10 15
13. (a) If x < 0, then jxj = ¡x so f(x) = ¡x + 3x + 1 = 2x + 1. If x ‚ 0, then jxj = x so
f(x)=x+3x+1=4x+1;
‰
2x+1; x<0
f(x)=
4x+1; x‚0
(b) If x<0, then jxj=¡x and jx¡1j=1¡x so g(x)=¡x+1¡x=1¡2x. If 0•x<1, then
jxj=x and jx¡1j=1¡x so g(x)=x+1¡x=1. If x‚1, then jxj=x and jx¡1j=x¡1
so g(x)=x+x¡1=2x¡1;
8
< 1¡2x; x<0
g(x)= 1; 0•x<1
:
2x¡1; x‚1
14. (a) Ifx<5=2,thenj2x¡5j=5¡2xsof(x)=3+(5¡2x)=8¡2x. Ifx‚5=2,thenj2x¡5j=2x¡5
so f(x)=3+(2x¡5)=2x¡2;
‰
8¡2x; x<5=2
f(x)=
2x¡2; x‚5=2
5 Chapter1
(b) If x<¡1, then jx¡2j=2¡x and jx+1j=¡x¡1 so g(x)=3(2¡x)¡(¡x¡1)=7¡2x. If
¡1•x<2, then jx¡2j=2¡x and jx+1j=x+1 so g(x)=3(2¡x)¡(x+1)=5¡4x. If
x‚2, then jx¡2j=x¡2 and jx+1j=x+1 so g(x)=3(x¡2)¡(x+1)=2x¡7;
8
< 7¡2x; x<¡1
g(x)= 5¡4x; ¡1•x<2
:
2x¡7; x‚2
15. (a) V =(8¡2x)(15¡2x)x (b) ¡1<x<+1;¡1<V <+1 (c) 0<x<4
(d) minimum value at x = 0 or at x = 4; maximum value somewhere in between (can be approxi-
mated by zooming with graphing calculator)
16. (a) x=3000tan(cid:181) (b) (cid:181) 6=n…+…=2fornaninteger,¡1<n<1
(c) 0•(cid:181) <…=2, 0•x<+1 (d) 3000 ft
6000
0 6
0
17. (i) x=1;¡2 causes division by zero (ii) g(x)=x+1, all x
p
18. (i) x=0 causes division by zero (ii) g(x)= x+1 for x‚0
19. (a) 25–F (b) 2–F (c) ¡15–F
20. If v =48 then ¡60=WCI=1:6T ¡55; thus T =(¡60+55)=1:6…¡3–F.
p
21. If v =8 then ¡10=WCI=91:4+(91:4¡T)(0:0203(8)¡0:304 8¡0:474); thus
p
T =91:4+(10+91:4)=(0:0203(8)¡0:304 8¡0:474) and T =5–F
22. The WCI is given by three formulae, but the flprst and third don’tpwork with the data. Hence
¡15=WCI=91:4+(91:4¡20)(0:0203v¡0:304 v¡0:474); set x= v so that v =x2 and obtain
0:0203x2¡0:304x¡0:474+(15+91:4)=(91:4¡20) = 0. Use the quadratic formula to flnd the two
roots. Square them to get v and discard the spurious solution, leaving v …25.
23. Let t denote time in minutes after 9:23 AM. Then D(t)=1000¡20t ft.
EXERCISE SET 1.3
1. (e) seems best, though only (a) is bad. 2. (e) seems best, though only (a) is bad and (b) is
not good.
y
0.5 y
0.5
x
-1 1
x
-1 1
-0.5
-0.5
ExerciseSet1.3 6
3. (b) and (c) are good; (a) is very bad. 4. (b) and (c) are good; (a) is very bad.
y y
15 -11
14 -12
13 -13
12 -14
x x
-2 -1 0 1 2 -2 -1 0 1 2
5. [¡3;3]£[0;5] 6. [¡4;2]£[0;3]
y y
2
4
1
2 -3 -2 -1 1 x
x
-3 -2 -1 1 2 3
7. (a) window too narrow, too short (b) window wide enough, but too short
(c) good window, good spacing (d) window too narrow, too short
y
x
-5 5 10 15 20
-100
-200
-300
-400
-500
(e) window too narrow, too short
8. (a) window too narrow (b) window too short
(c) good window, good tick spacing (d) window too narrow, too short
y
50
x
-16 -12 -8 -4 4
-50
-100
-150
-200
-250
(e) shows one local minimum only, window too narrow, too short
7 Chapter1
9. [¡5;14]£[¡60;40] 10. [6;12]£[¡100;100]
y y
40 100
20 50
x
-5 5 10 x
8 10 12
-20
-50
-40
-100
-60
11. [¡0:1;0:1]£[¡3;3] 12. [¡1000;1000]£[¡13;13]
y y
3
10
2
1 5
x x
-0.1 0.1 -1000 1000
-1 -5
-2
-10
-3
13. [¡250;1050]£[¡1500000;600000] 14. [¡3;20]£[¡3500;3000]
y
y
-1000 1000
x
1000
-500000 x
5 10 15
-1000
-2000
15. [¡2;2]£[¡20;20] 16. [1:6;2]£[0;2]
y y
20 2
10 1.5
x 1
-2 -1 1 2
-10 0.5
-20 x
1.6 1.7 1.8 1.9 2
17. depends on graphing utility 18. depends on graphing utility
y y
6 6
4 4
2 2
x x
-4 -2 2 4 -4 -2 2 4
-2 -2
-4 -4
-6 -6
ExerciseSet1.3 8
p p
19. (a) f(x)= 16¡x2 (b) f(x)=¡ 16¡x2
y y
4 -4 -2 2 4 x
3 -1
2 -2
1 -3
x -4
-4 -2 2 4
(c) y (d) y
4
4
3
2
2
x
-4 -2 2 4
1
-2
-4 1 2 3 4
(e) No; the vertical line test fails.
p p
20. (a) y =§3 1¡x2=4 (b) y =§ x2+1
y
4
2
x
-4 -2 2 4
-2
-4
21. (a) y (b) y (c) y
1 1
x
-1 1
x x
-1 1 1 2 -1
(d) y (e) y (f) y
1 1
1
x
-p p x
x
p 2p -1 1
—1 -1
22. y
1
x
-1 1
-1
23. The portions of the graph of y =f(x) which lie below the x-axis are re(cid:176)ected over the x-axis to give the
graph of y =jf(x)j.
9 Chapter1
24. Erasetheportionofthegraphofy =f(x)whichliesintheleft-halfplaneandreplaceitwiththere(cid:176)ection
overthey-axisoftheportionintheright-halfplane(symmetryoverthey-axis)andyouobtainthegraph
of y =f(jxj).
25. (a) for example, let a=1:1 (b)
y
y
3
2
m 1
x
1 2 3
26. They are identical. 27.
y
y
15
10
x
5
x
-1 1 2 3
28. This graph is very complex. We show three views, small (near the origin), medium and large:
(a) (b) (c)
y y y
10
x
-1 2
x
-50
-1 1000
x
40
29. (a) (b)
y y
1.5 1
0.5
1
x
-3 -2 -1 1 2 3
0.5 -0.5
-1
x
-3 -2 -1 1 2 3
(c) (d) y
y
1.5 1.5
1
0.5 1
x
-1 1 2 3
0.5
x
-2 -1 1