3000 Solved Problems in Calculus

465 Pages·2011·20.86 MB·English

SCHAUM'S
OUTLINE OF
3000 SOLVED
PROBLEMS IN
Calculus
Elliot Mendelson, Ph.D.
Professor of Mathematics
Queens College
City University of New York
Schaum's Outline Series
MC
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CONTENTS
Chapter 1 INEQUALITIES 1
Chapter 2 ABSOLUTE VALUE 5
Chapter 3 LINES 9
Chapter 4 CIRCLES 19
Chapter 5 FUNCTIONS AND THEIR GRAPHS 23
Chapter 6 LIMITS 35
Chapter 7 CONTINUITY 43
Chapter 8 THE DERIVATIVE 49
Chapter 9 THE CHAIN RULE 56
Chapter 10 TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES 62
Chapter 11 ROLLE'S THEOREM, THE MEAN VALUE THEOREM, AND THE SIGN
OF THE DERIVATIVE 69
Chapter 12 HIGHER-ORDER DERIVATIVES AND IMPLICIT DIFFERENTIATION 75
Chapter 13 MAXIMA AND MINIMA 81
Chapter 14 RELATED RATES 88
Chapter 15 CURVE SKETCHING (GRAPHS) 100
Chapter 16 APPLIED MAXIMUM AND MINIMUM PROBLEMS 118
Chapter 17 RECTILINEAR MOTION 133
Chapter 18 APPROXIMATION BY DIFFERENTIALS 138
Chapter 19 ANTIDERIVATIVES (INDEFINITE INTEGRALS) 142
Chapter 20 THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF
CALCULUS 152
Chapter 21 AREA AND ARC LENGTH 163
Chapter 22 VOLUME 173
Chapter 23 THE NATURAL LOGARITHM 185
Chapter 24 EXPONENTIAL FUNCTIONS 195
Chapter 25 L'HOPITAL'S RULE 208
Chapter 26 EXPONENTIAL GROWTH AND DECAY 215
iii
iv CONTENTS
Chapter 27 INVERSE TRIGONOMETRIC FUNCTIONS 220
Chapter 28 INTEGRATION BY PARTS 232
Chapter 29 TRIGONOMETRIC INTEGRANDS AND SUBSTITUTIONS 238
Chapter 30 INTEGRATION OF RATIONAL FUNCTIONS: THE METHOD
OF PARTIAL FRACTIONS 245
Chapter 31 INTEGRALS FOR SURFACE AREA, WORK, CENTROIDS 253
Surface Area of a Solid of Revolution / Work / Centroid of a Planar Region /
Chapter 32 IMPROPER INTEGRALS 260
Chapter 33 PLANAR VECTORS 268
Chapter 34 PARAMETRIC EQUATIONS, VECTOR FUNCTIONS, CURVILINEAR
MOTION 274
Parametric Equations of Plane Curves / Vector-Valued Functions /
Chapter 35 POLAR COORDINATES 289
Chapter 36 INFINITE SEQUENCES 305
Chapter 37 INFINITE SERIES 312
Chapter 38 POWER SERIES 326
Chapter 39 TAYLOR AND MACLAURIN SERIES 340
Chapter 40 VECTORS IN SPACE. LINES AND PLANES 347
Chapter 41 FUNCTIONS OF SEVERAL VARIABLES 361
Multivariate Functions and Their Graphs / Cylindrical and Spherical Coordinates /
Chapter 42 PARTIAL DERIVATIVES 376
Chapter 43 DIRECTIONAL DERIVATIVES AND THE GRADIENT.
EXTREME VALUES 392
Chapter 44 MULTIPLE INTEGRALS AND THEIR APPLICATIONS 405
Chapter 45 VECTOR FUNCTIONS IN SPACE. DIVERGENCE AND CURL.
LINE INTEGRALS 425
Chapter 46 DIFFERENTIAL EQUATIONS 431
INDEX 443
To the Student
This collection of solved problems covers elementary and intermediate calculus, and much of advanced
calculus. We have aimed at presenting the broadest range of problems that you are likely to encounter—the
old chestnuts, all the current standard types, and some not so standard.
Each chapter begins with very elementary problems. Their difficulty usually increases as the chapter pro-
gresses, but there is no uniform pattern.
It is assumed that you have available a calculus textbook, including tables for the trigonometric, logarith-
mic, and exponential functions. Our ordering of the chapters follows the customary order found in many
textbooks, but as no two textbooks have exactly the same sequence of topics, you must expect an occasional
discrepancy from the order followed in your course.
The printed solution that immediately follows a problem statement gives you all the details of one way to
solve the problem. You might wish to delay consulting that solution until you have outlined an attack in your
own mind. You might even disdain to read it until, with pencil and paper, you have solved the problem
yourself (or failed gloriously). Used thus, 3000 Solved Problems in Calculus can almost serve as a supple-
ment to any course in calculus, or even as an independent refresher course.
V
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HAPTER 1
nequalities
1.1 Solve 3 + 2*<7.
2x < 4 [Subtract 3 from both sides. This is equivalent to adding -3 to both sides.]
Answer x<2 [Divide both sides by 2. This is equivalent to multiplying by 5.] In interval notation, the
solution is the set (—°°, 2).
1.2 Solve 5 - 3* < 5x + 2.
5-3>x<5x + 2, 5<8* + 2 [Add 3x to both sides.], 3<8* [Subtract 2 from both sides.]
Answer 1 <x [Divide both sides by 8.] In interval notation, the solution is the set (|,°°).
1.3 Solve -7<2x + 5<9.
-7 < 2* + 5 < 9, -12 < 2x < 4 [Subtract 5 from all terms.]
Answer — 6 < x < 2 [Divide by 2.] In interval notation, the solution is the set (—6,2).
1.4 Solve 3<4x-l<5.
3<4x-l<5, 4<4x<6 [Add 1 to all terms.]
Answer 1 s x < \ [Divide by 4.] In interval notation, the solution is the set [1, |).
1.5 Solve 4<-2x + 5<7.
4<-2x + 5<7, -K-2jc<2 [Subtracts.]
Answer \ >*>-! [Divide by -2. Since -2 is negative, we must reverse the inequalities.] In interval
notation, the solution is the set [-1, |).
1.6 Solve 5 < \x. + 1 s 6.
5<|x + l<6, 4<|*s5 [Subtract 1.]
Answer 12<^sl5 [Multiply by 3.] In interval notation, the solution is the set [12,15].
1.7 Solve 2/jc<3.
x may be positive or negative. Case 1. x>0. 2/x<3. 2<3x [Multiply by AC.], |<jc [Divide by 3.]
Case 2. x<0. 2/x<3. 2>3x [Multiply by jr. Reverse the inequality.], |>jc [Divide by 3.] Notice
that this condition |>x is satisfied whenever jc<0. Hence, in the case where x<0, the inequality is
satisfied by all such x.
Answer f < x or x < 0. As shown in Fig. 1-1, the solution is the union of the intervals (1,«) and (—°°, 0).
Fig. 1-1
1.8 Solve
We cannot simply multiply both sides by x - 3, because we do not know whether x - 3 is positive or
negative. Case 1. x-3>0 [This is equivalent to x>3.] Multiplying the given inequality (1) by the
positive quantity x-3 preserves the inequality: * + 4<2;t-6, 4<x-6 [Subtract jr.], 10<x [Add
6.] Thus, when x>3, the given inequality holds when and only when x>10. Case 2. x-3<0 [This
is equivalent to x<3]. Multiplying the given inequality (1) by the negative quantity x — 3 reverses the
inequality: * + 4>2*-6, 4>x-6 [Subtract*.], 10>x [Add 6.] Thus, when x<3, the inequality
1
2 CHAPTER 1
(1) holds when and only when x < 10. But x < 3 implies x < 10, and, therefore, the inequality (1) holds
for all x<3.
Answer *>10 or x<3. As shown in Fig. 1-2, the solution is the union of the intervals (10, oo) and
(~»,3).
Fig. 1-2
1.9 Solve

1. x + 5>0 [This is equivalent to x>-5.]. We multiply the inequality (1) by x + 5. x<I Case 1. x + 5>0 [This is equivalent to x>-5.]. We multiply the inequality (1) by x + 5. x<
x + 5, 0<5 [Subtract x.] This is always true. So, (1) holds throughout this case, that is, wheneverx + 5, 0<5 [Subtract x.] This is always true. So, (1) holds throughout this case, that is, whenever
x>-5. Case 2. x + 5<0 [This is equivalent to x<-5.]. We multiply the inequality (1) by x + 5. The
inequality is reversed, since we are multiplying by a negative number. x>x + 5, 0>5 [Subtract*.] Butinequality is reversed, since we are multiplying by a negative number. x>x + 5, 0>5 [Subtract*.] But
0 > 5 is false. Hence, the inequality (1) does not hold at all in this case.
Answer x > -5. In interval notation, the solution is the set (-5, °°).
1.10 Solve

Case 1. x + 3>0 [This is equivalent to jc>-3.]. Multiply the inequality (1) by x + 3. x-7>
2x + 6, -7>x+6 [Subtract x.], -13>x [Subtract 6.] But x<-13 is always false when *>-3.
Hence, this case yields no solutions. Case 2. x + 3<0 [This is equivalent to x<— 3.]. Multiply the
inequality (1) by x + 3. Since x + 3 is negative, the inequality is reversed. x-7<2x + 6, —7<x + 6
[Subtract x.] ~\3<x [Subtract 6.] Thus, when x < —3, the inequality (1) holds when and only when
*>-13.
Answer —13 < x < —3. In interval notation, the solution is the set (—13, —3).
1.11 Solve (2jt-3)/(3;t-5)>3.
Case 1. 3A.-5>0 [This is equivalent to *>§.]. 2x-3>9x-l5 [Multiply by 3jf-5.], -3>
7x-15 [Subtract 2x.], I2>7x [Add 15.], T a* [Divide by 7.] So, when x>f, the solutions must
satisfy x<". Case 2. 3x-5<0 [This is equivalent to x<|.]. 2* - 3 < 9* - 15 [Multiply by 3*-5.
Reverse the inequality.], -3<7jr-15 [Subtract 2*.], 12 < 7x [Add 15.], ^ s x [Divide by 7.] Thus,
when x< f, the solutions must satisfy x^ !f. This is impossible. Hence, this case yields no solutions.
Answer f < x s -y. In interval notation, the solution is the set (§, ^].
1.12 Solve (2*-3)/(3*-5)>3.
Remember that a product is positive when and only when both factors have the same sign. Casel. Jt-2>0
and x + 3>0. Then x>2 and jt>—3. But these are equivalent to x>2 alone, since x>2 im-and x + 3>0. Then x>2 and jt>—3. But these are equivalent to x>2 alone, since x>2 im-
plies x>-3. Case 2. *-2<0 and A: + 3<0. Then x<2 and jc<—3, which are equivalent to
x<—3, since x<-3 implies x<2.
Answer x > 2 or x < -3. In interval notation, this is the union of (2, °°) and (—<», —3).
1.13 Solve Problem 1.12 by considering the sign of the function f(x) = (x — 2)(x + 3).
Refer to Fig. 1-3. To the left of x = — 3, both x-2 and x + 3 are negative and /(*) is positive. As
one passes through x - — 3, the factor x - 3 changes sign and, therefore, f(x) becomes negative. f(x)
remains negative until we pass through x = 2, where the factor x — 2 changes sign and f(x) becomes and
then remains positive. Thus, f(x) is positive for x < — 3 and for x > 2. Answer
Fig. 1-3
INEQUALITIES 3
1.14 Solve (x-l)(x + 4)<0.
The key points of the function g(x) = (x - l)(x + 4) are x = — 4 and x = l (see Fig. 1-4). To the
left of x = -4, both x — 1 and x + 4 are negative and, therefore, g(x) is positive. As we pass through
x = — 4, jr + 4 changes sign and g(x) becomes negative. When we pass through * = 1, A: - 1 changes sign
and g(x) becomes and then remains positive. Thus, (x - \)(x + 4) is negative for -4 < x < 1. Answer
Fig. 1-4
Fig. 1-5
1.15 Solve x2 - 6x + 5 > 0.
Factor: x2 -6x + 5 = (x - l)(x - 5). Let h(x) = (x - \)(x - 5). To the left of x = 1 (see Fig. 1-5),
both .* - 1 and jc - 5 are negative and, therefore, h(x) is positive. When we pass through x = \, x-\
changes sign and h(x) becomes negative. When we move further to the right and pass through x = 5, x — 5
changes sign and h(x) becomes positive again. Thus, h(x) is positive for x < 1 and for x>5.
Answer x > 5 or x < 1. This is the union of the intervals (5, °°) and (—°°, 1).
1.16 Solve x2 + Ix - 8 < 0.
Factor: x2 + Ix - 8 = (x + &)(x - 1), and refer to Fig. 1-6. For jc<-8, both x + 8 and x-l
are negative and, therefore, F(x) = (x + 8)(x - 1) is positive. When we pass through x = -8, x + 8
changes sign and, therefore, so does F(x). But when we later pass through x = l, x-l changes sign and
F(x) changes back to being positive. Thus, F(x) is negative for -8 < x < 1. Answer
Fig. 1-6
Fig. 1-7
1.17 Solve 5x - 2x2 > 0.
Factor: 5x - 2x2 = x(5 - 2x), and refer to Fig. 1-7. The key points for the function G(x) = x(5 - 2x)
are x = 0 and *=|. For x<Q, 5-2x is positive and, therefore, G(x) is negative. As we pass
through x = 0, x changes sign and. therefore, G(x) becomes positive. When we pass through x= |,
5 — 2x changes sign and, therefore, G(x) changes back to being negative. Thus, G(x) is positive when and only
when 0 < x < |. Answer
1.18 Solve (Jt-l)2(* + 4)<0.
(x — I)2 is always positive except when x = 1 (when it is 0). So, the only solutions occur when
* + 4<0 and jc^l.
Answer x<— 4 [In interval notation, (—=°, — 4).]
1.19 Solve x(x-l)(x + l)>0.
The key points for H(x) = x(x - l)(x + 1) are x = 0, x = l, and jc=-l (see Fig. 1-8). For x to
the left of — 1, x, x — 1, and x + 1 all are negative and, therefore, H(x) is negative. As we pass
through x = — 1, x + 1 changes sign and, therefore, so does H(x). When we later pass through x = 0, x
changes sign and, therefore, H(x) becomes negative again. Finally, when we pass through x = l, x-\
changes sign and H(x) becomes and remains positive. Therefore, H(x) is positive when and only when
— 1 < A: < 0 or x>\. Answer