input: What is unique about the fourth edition of "Schaum's Outline of Calculus"?

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# **SCHAUM'S OUTLINE OF CALCULUS**

**Fourth Edition**

**FRANK AYRES, Jr., Ph.D.**

_Formerly Professor and Head of the Department of Mathematics
Dickinson College_

**ELLIOT MENDELSON, Ph.D.**

_Professor of Mathematics
Queens College_

Copyright © 1999, 1990, 1962 by The McGraw-Hill Companies, Inc. All rights reserved. Manufactured in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher.

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DOI: 10.1036/0070419736

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## **PREFACE**

The purpose of this book is to help students understand and use the calculus. Everything has been aimed toward making this easier, especially for students with limited background in mathematics or for readers who have forgotten their earlier training in mathematics. The topics covered include all the material of standard courses in elementary and intermediate calculus.

The direct and concise exposition typical of the Schaum Outline Series has been amplified by a large number of examples, followed by many carefully solved problems. In choosing these problems, we have attempted to anticipate the difficulties that normally beset the beginner. In addition, each chapter concludes with a collection of supplementary exercises with solutions.

This fourth edition includes several new features.

1. There are many new problems that are to be solved with the help of a graphing calculator. Such problems are indicated by the notation (GC). They will show the student the great advantages to be gained from the use of modern technology on mathematics. (However, solution of these problems is not necessary for an adequate understanding of the material in this book.)

2. A more thoroughgoing review of trigonometry (Chapter 16) has been included for the sake of those students who never had or have forgotten high-school trigonometry.

3. Proofs have been included for all theorems that are usually proved in elementary calculus courses. Since all fundamental concepts are carefully defined, this book can be used as a text for a regular calculus course, in both colleges and secondary schools.

4. The material on the representation of functions by power series has been expanded in order to make this somewhat difficult topic more intelligible.

5. We have made a great effort to go over ticklish points of algebra or geometry that may confuse the student. The author believes that a major part of the mistakes that students make in a calculus course are not due to a deficient comprehension of principles of calculus, bur rather to their weakness in high-school algebra or geometry.

ELLIOTT MENDELSON

## **CONTENTS**

**CHAPTER 1 LINEAR COORDINATE SYSTEMS. ABSOLUTE VALUE. INEQUALITIES**

**CHAPTER 2 RECTANGULAR COORDINATE SYSTEMS**

**CHAPTER 3 LINES**

**CHAPTER 4 CIRCLES**

**CHAPTER 5 EQUATIONS AND THEIR GRAPHS**

**CHAPTER 6 FUNCTIONS**

**CHAPTER 7 LIMITS**

**CHAPTER 8 CONTINUITY**

**CHAPTER 9 THE DERIVATIVE**

**CHAPTER 10 RULES FOR DIFFERENTIATING FUNCTIONS**

**CHAPTER 11 IMPLICIT DIFFERENTIATION**

**CHAPTER 12 TANGENT AND NORMAL LINES**

**CHAPTER 13 LAW OF THE MEAN. INCREASING AND DECREASING FUNCTIONS**

**CHAPTER 14 MAXIMUM AND MINIMUM VALUES**

**CHAPTER 15 CURVE SKETCHING. CONCAVITY. SYMMETRY**

**CHAPTER 16 REVIEW OF TRIGONOMETRY**

**CHAPTER 17 DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS**

**CHAPTER 18 INVERSE TRIGONOMETRIC FUNCTIONS**

**CHAPTER 19 RECTILINEAR AND CIRCULAR MOTION**

**CHAPTER 20 RELATED RATES**

**CHAPTER 21 DIFFERENTIALS. NEWTON'S METHOD**

**CHAPTER 22 ANTIDERIVATIVES**

**CHAPTER 23 THE DEFINITE INTEGRAL. AREA UNDER A CURVE**

**CHAPTER 24 THE FUNDAMENTAL THEOREM OF CALCULUS**

**CHAPTER 25 THE NATURAL LOGARITHM**

**CHAPTER 26 EXPONENTIAL AND LOGARITHMIC FUNCTIONS**

**CHAPTER 27 L'HÔPITAL'S RULE**

**CHAPTER 28 EXPONENTIAL GROWTH AND DECAY**

**CHAPTER 29 APPLICATIONS OF INTEGRATION I: AREA AND ARC LENGTH**

**CHAPTER 30 APPLICATIONS OF INTEGRATION II: VOLUME**

**CHAPTER 31 TECHNIQUES OF INTEGRATION I: INTEGRATION BY PARTS**

**CHAPTER 32 TECHNIQUES OF INTEGRATION II: TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS**

**CHAPTER 33 TECHNIQUES OF INTEGRATION III: INTEGRATION BY PARTIAL FRACTIONS**

**CHAPTER 34 MISCELLANEOUS SUBSTITUTIONS**

**CHAPTER 35 IMPROPER INTEGRALS**

**CHAPTER 36 APPLICATIONS OF INTEGRATION II: AREA OF A SURFACE OF REVOLUTION**

**CHAPTER 37 PARAMETRIC REPRESENTATION OF CURVES**

**CHAPTER 38 CURVATURE**

**CHAPTER 39 PLANE VECTORS**

**CHAPTER 40 CURVILINEAR MOTION**

**CHAPTER 41 POLAR COORDINATES**

**CHAPTER 42 INFINITE SEQUENCES**

**CHAPTER 43 INFINITE SERIES**

**CHAPTER 44 SERIES WITH POSITIVE TERMS. THE INTEGRAL TEST. COMPARISON TESTS**

**CHAPTER 45 ALTERNATING SERIES. ABSOLUTE AND CONDITIONAL CONVERGENCE. THE RATIO TEST**

**CHAPTER 46 POWER SERIES**

**CHAPTER 47 TAYLOR AND MACLAURIN SERIES. TAYLOR'S FORMULA WITH REMAINDER**

**CHAPTER 48 PARTIAL DERIVATIVES**

**CHAPTER 49 TOTAL DIFFERENTIAL. DIFFERENTIABILITY. CHAIN RULES**

**CHAPTER 50 SPACE VECTORS**

**CHAPTER 51 SURFACE AND CURVES IN SPACE**

**CHAPTER 52 DIRECTIONAL DERIVATIVES. MAXIMUM AND MINIMUM VALUES**

**CHAPTER 53 VECTOR DIFFERENTIATION AND INTEGRATION**

**CHAPTER 54 DOUBLE AND ITERATED INTEGRALS**

**CHAPTER 55 CENTROIDS AND MOMENTS OF INERTIA OF PLANE AREAS**

**CHAPTER 56 DOUBLE INTEGRATION APPLIED TO VOLUME UNDER A SURFACE AND THE AREA OF A CURVED SURFACE**

**CHAPTER 57 TRIPLE INTEGRALS**

**CHAPTER 58 MASSES OF VARIABLE DENSITY**

**CHAPTER 59 DIFFERNTIAL EQUATIONS OF FIRST AND SECOND ORDER**

**Appendix A**

**Appendix B**

**INDEX**

## **CHAPTER 1
Linear Coordinate Systems. Absolute Value. Inequalities**

**A LINEAR COORDINATE SYSTEM** is a graphical representation of the real numbers as the points of a straight line. To each number corresponds one and only one point, and to each point corresponds one and only one number.

To set up a linear coordinate system on a given line: (1) select any point of the line as the and let that point correspond to the number 0; (2) choose a positive direction on the line and indicate that direction by an arrow; (3) choose a fixed distance as a unit of measure. If is _x_ a positive number, find the point corresponding to _x_ by moving a distance of _x_ units from the origin in the positive direction. If _x_ is negative, find the point corresponding to _x_ by moving a distance of − _x_ units from the origin in the negative direction. (For example, if _x_ = −2, then − _x_ = 2 and the corresponding point lies 2 units from the origin in the negative direction.) See Fig. 1-1.

**Fig. 1-1**

The number assigned to a point by a coordinate system is called the _coordinate_ of that point. We often will talk as if there is no distinction between a point and its coordinate. Thus, we might refer to "the point 3" rather than to "the point with coordinate 3."

The absolute value | _x_ | of a number _x_ is defined as follows:

For example, |4| = 4, |−3| = −(−3) = 3, and |0| = 0. Notice that, if _x_ is a negative number, then − _x_ is positive. Thus, | _x_ |   0 for all _x_.

The following properties hold for any numbers _x_ and _y_.

( **1.1** ) |− _x_ | = | _x_ |

When _x_ = 0, |−x| = |−0| = |0| = | _x_ |.

When _x_ > 0, − _x_ < 0, and | − _x_ | = −(− _x_ ) = _x_ = | _x_ |.

When _x_ < 0, − _x_ > 0, and | − _x_ | = − _x_ = | _x_ |.

( **1.2** ) | _x_ − _y_ | = | _y_ − _x_ |

This follows from ( **1.1** ), since _y_ − _x_ = −( _x_ − _y_ ).

( **1.3** ) | _x_ | = _c_ implies that _x_ = ± _c_.

For example, if | _x_ | = 2, then _x_ = ±2. For the proof, assume | _x_ | = _c_.

If _x_  0, _x_ = | _x_ | = _c_. If _x_ < 0, − _x_ = | _x_ | = _c_ ; then _x_ = −(− _x_ ) = − _c_.

( **1.4** ) | _x_ |2 = _x_ 2

If _x_   0, | _x_ | = _x_ and | _x_ |2 = _x_ 2. If _x_   0, | _x_ | = − _x_ and | _x_ |2 = (− _x_ )2 = _x_ 2.

( **1.5** ) | _xy_ | = | _x_ | · | _y_ |

By ( **1.4** ), | _xy_ |2 = ( _xy_ )2 = _x_ 2 _y_ 2 = | _x_ |2| _y_ |2 = (| _x_ | · | _y_ |)2. Since absolute vales are nonnegative, taking sqare roots yields | _xy_ | = | _x_ | · | _y_ |.

( **1.6** )

By ( **1.5** ),  . Divide by | _y_ |.

( **1.7** ) | _x_ | = | _y_ | implies that _x_ = ± _y_

Assme | _x_ | = | _y_ |. If _y_ = 0, | _x_ | = |0| = 0 and ( **1.3** ) yields _x_ = 0. If _y_ ≠ 0, then by ( **1.6** ),

So, by ( **1.3** ), _x_ / _y_ = ±1. Hence, _x_ = ± _y_.

( **1.8** ) Let _c_   0. Then | _x_ |   _c_ if and only if −c   _x_   _c_. See Fig. 1-2.

Assume _x_   0. Then | _x_ | = _x_. Also, since _c_   0, − _c_   0   _x_. So, | _x_ |   _c_ if and only if − _c_   _x_   _c_. Now assume _x_ < 0. Then | _x_ | = − _x_. Also, _x_ < 0   _c_. Moreover, − _x_   _c_ if and only if − _c_   _x_. (Multiplying or dividing an equality by a negative number reverses the inequalit.) Hence, | _x_ |   _c_ if and only if − _c_   _x_   _c_.

( **1.9** ) Let _c_   0. Then | _x_ | < _c_ if and only if − _c_ < _x_ < _c_. See Fig. 1-2. The reasoning here is similar to that for ( **1.8** ).

**Fig. 1-2**

( **1.10** ) −| _x_ |   _x_   | _x_ |

If _x_   0, _x_ = | _x_ |. If _x_ < 0, | _x_ | = −x and, therefore, _x_ = −| _x_ |.

( **1.11** ) | _x_ \+ _y_ |   | _x_ | + | _y_ | (triangle inequality)

By ( **1.8** ), −| _x_ |   _x_   | _x_ | and −| _y_ |   y   | _y_ |. Adding, we obtain −(| _x_ | + | _y_ |)   _x_ \+ _y_   | _x_ | + | _y_ |. Then | _x_ \+ _y_ |   | _x_ | + | _y_ | by ( **1.8** ). [In ( **1.8** ), replace _c_ by | _x_ | + | _y_ | and _x_ by _x_ \+ _y_.]

Let a coordinate system be given on a line. Let P1 and P2 be points on the line having coordinates _x_ 1 and _x_ 2. See Fig. 1-3. Then:

( **1.12** ) | _x_ 1 − _x_ 2| = _P_ 1 _P_ 2 = distance between _P_ 1 and _P_ 2.

This is clear when 0 < _x_ 1< _x_ 2 and when _x_ 1< _x_ 2< 0. When _x_ 1< 0 < _x_ 2, and if we denote the origin by O, then _P_ 1 _P_ 2 = _P_ 1 _O_ \+ _OP_ 2 = (− _x_ 1) + _x_ 2 = _x_ 2 − _x_ 1 = | _x_ 2 − _x_ 1| = | _x_ 1 − _x_ 2|.

As a special case of ( **1.12** ) when _P_ 2 is the origin (and _x_ 2 = 0):

( **1.13** ) | _x_ 1| = distance between _P_ 1 and the origin.

**Fig. 1-3**

**FINITE INTERVALS.** Let _a_ < _b_.

The _open interval_ ( _a, b_ ) is defined to be the set of all numbers between _a_ and _b_ , that is, the set of all such that _a_ < _x_ < _b_. We shall use the term open interval and the notation ( _a, b_ ) also for all the points between the points with coordinates _a_ and _b_ on a line. Notice that the open interval ( _a, b_ ) does not contain the endpoints _a_ and _b_. See Fig. 1-4.

The _closed interval_  _a, b_ ] is defined to be the set of all numbers between _a_ and _b_ or equals to _a_ or _b_ , that is, the set of all _x_ such that _a_   _x_   _b_. As in the case of open intervals, we extend the terminology and notation to points. Notice that the closed interval [ _a, b_ ] contains both endpoints and _b_. See [Fig. 1-4.

**Fig. 1-4**

By a _half-open interval_ we mean an open interval ( _a, b_ ) together with one of its endpoints. There are two such intervals: [ _a, b_ ) is the set of all _x_ such that _a_   _x_ <, and ( _a, b_ ] is the set of all _x_ such that _a_ < _x_   _b_.

**INFINITE INTERVALS**

Let ( _a_ , ∞) denote the set of all _x_ such that _a_ < _x_.

Let [ _a_ , ∞) denote the set of all _x_ such that _a_   _x_.

Let (−∞, _b_ ) denote the set of all _x_ such that _x_ < _b_.

Let (−∞, _b_ ] denote the set of all _x_ such that _x_   _b_.

**INEQUALITIES.** Any inequality, such as 2 _x_ − 3 > 0 or 5 < 3 _x_ \+ 10   16, determines an interval. To solve an inequality means to determine the corresponding interval of numbers that satisfy the inequality.

**EXAMPLE 1:** Solve 2 _x_ − 3 > 0.

Thus, the corresponding interval is  .

**EXAMPLE 2:** Solve 5 < 3 _x_ \+ 10   16.

Thus, the corresponding interval is  .

**EXAMPLE 3:** Solve −2 _x_ \+ 3 < 7.

(Recall that dividing by a negative number reverses an inequality.) Thus, the corresponding interval is (−2, ∞).

### **Solved Problems**

**1.** Describe and diagram the following intervals, and write their interval notation. ( _a_ ) −3 < _x_ < 5; ( _b_ ) 2   _x_   6; ( _c_ ) _x_ −4 < _x_   0; ( _d_ ) _x_ > 5; ( _e_ ) _x_   2; ( _f_ ) 3 _x_ −4   8; ( _g_ ) 1 < 5 −3 _x_ < 11.

( _a_ ) All number greater than −3 and less than 5; the interval notation is (−3, 5):

( _b_ ) All numbers equal to or greater than 2 and less than or equal to 6; [2, 6]:

( _c_ ) All numbers greater than −4 and less than or equal to 0; (−4, 0]:

( _d_ ) All numbers greater than 5; (5, ∞):

( _e_ ) All numbers less than or equal to 2; (-∞, 2]:

( _f_ ) 3 _x_ − 4   8 is equivalent to 3 _x_   12 and, therefore, to _x_   4. Thus, we get (−∞, 4]:

( _g_ )

Thus, we obtain  :

**2.** Describe and diagram the intervals determined by the following inequalities. ( _a_ ) | _x_ | < 2; ( _b_ ) | _x_ | > 3; ( _c_ ) | _x_ − 3| < 1; ( _d_ ) | _x_ − 2| < _δ_ where _δ_ > 0; ( _e_ ) | _x_ +2|   3; ( _f_ ) 0 < | _x_ − 4| < δ where δ where δ > 0.

( _a_ ) By property ( **1.9** ), this is equivalent to −2 < _x_ < 2, defining the open interva (−2,2).

( _b_ ) By property ( **1.8** ), | _x_ |   3 is equivalent to −3   _x_   3. Taking negations, | _x_ | > 3 is equivalent to _x_ < −3 or _x_ > 3, which defines the union of the intervals (−∞, −3) and (3, ∞).

( _c_ ) By property ( **1.12** ), this says that the distance between _x_ and 3 is less than 1, which is equivalent to 2 < _x_ < 4. This defines the open interval (2, 4).

We can also note that | _x_ −3| < 1 is equivalent to −1 < _x_ −3 < 1. Adding 3, we obtain 2 < _x_ < 4.

( _d_ ) This is equivalent to saying that the distance between _x_ and 2 is less than _δ_ , or that 2 − _δ_ < _x_ < 2 + _δ_ , which defines the open interval (2 − _δ_ , 2 + _δ_ ). This interval is called the _δ_ − _neighborhood_ of 2:

( _e_ ) | _x_ \+ 2| < 3 is equivalent to −3 < _x_ \+ 2 < 3. Subtracting 2, we obtain −5 < _x_ < 1, which defines the open interval (−5, 1):

( _f_ ) The inequality | _x_ −4| < _δ_ determines the interval 4 − _δ_ < _x_ < 4 + _δ_. The additional condition 0 < | _x_ −4| tells us that _x_ ≠ 4. Thus, we get the union of the two intervals (4 − _δ_ , 4) and (4, 4 + _δ_ ). The result is called the _deleted δ-neighborhood_ of 4:

**3.** Describe and diagram the intervals determined by the following inequalities. ( _a_ ) |5 − _x_ |   3; ( _b_ ) |2 _x_ −3| < 5; ( _c_ )  .

( _a_ ) Since |5 − _x_ | = | _x_ −5|, we have | _x_ −5|   3, which is equivalent to −3   _x_ −5   3. Adding 5, we get 2   _x_   8, which defines the interval [2, 8]:

( _b_ ) |2 _x_ −3| < 5 is equivalent to −5 < 2 _x_ −3 < 5. Adding 3, we have −2 < 2 _x_ < 8; then dividing by 2 yields −1 < _x_ < 4, which defines the open interval (−1, 4):

( _c_ ) |1−4| = |4 _x_ −1|, we have  , which is equivalent to  . Adding 1, we get  . Dividing by 4, we obtain  , which defines the interval  :

**4.** Solve the inequalities: ( _a_ ) 18 _x_ − 3 _x_ 2 > 0; ( _b_ ) ( _x_ \+ 3)( _x_ − 2)( _x_ − 4) < 0; ( _c_ ) ( _x_ \+ 1)2( _x_ − 3) > 0, and diagram the solutions.

( _a_ ) Set 18 _x_ − 3 _x_ 2 = 3 (6 − _x_ ) = 0, obtaining _x_ = 0 and _x_ = 6. We need to determine the sign of 18 _x_ − 3 _x_ 2 on each of the intervals _x_ < 0, 0 < _x_ < 6, and _x_ > 6, to determine where 18 _x_ − 3 _x_ 2 _x_ > 0. Note that it is negative when _x_ < 0 (since _x_ is negative and 6 − _x_ is positive). It becomes positive when we pass from left to right through 0 (since _x_ changes sign but 6 − _x_ remains positive) and that it becomes negative when we pass through 6 (since _x_ remains positive but 6 − _x_ changes to negative). Hence, it is positive when and only when 0 < _x_ < 6.

( _b_ ) The crucial points are _x_ = −3, _x_ = 2, and _x_ = 4. Note that ( _x_ \+ 3)( _x_ − 2)( _x_ − 4) is negative for _x_ < −3 (since each of the factors is negative) and that it changes sign when we pass through each of the crucial points. Hence, it is negative for _x_ < −3 and for 2 < _x_ < 4:

( _c_ ) Note that ( _x_ \+ 1)2 is always positive (except at _x_ = −1, where it is 0). Hence ( _x_ \+ 1)2( _x_ − 3) > 0 when and only when _x_ − 3 > 0, that is, for _x_ > 3:

**5.** Solve |3 _x_ − 7| = 8.

By ( **1.3** ), |3 _x_ − 7| = 8 if and only if 3 _x_ − 7 = ±8. Thus, we need to solve 3 _x_ −7 = 8 and 3 _x_ −7 = −8. Hence, we get _x_ = 5 or

**6.** Solve  .

_Case 1_ : _x_ \+ 3 > 0. Multiply by _x_ \+ 3 to obtain 2 _x_ \+ 1 > 3 _x_ \+ 9, which reduces to −8 > _x_. However, since _x_ \+ 3 > 0, it must be that _x_ > −3. Thus, this case yields no solutions.

_Case 2_ : _x_ \+ 3 < 0. Multiply by _x_ \+ 3 to obtain 2 _x_ \+ 1 < 3 _x_ \+ 9. (Note that the inequality is reversed, since we multiplied by a negative number.) This yields −8 < _x_. Since _x_ \+ 3 < 0, we have _x_ < −3. Thus, the only solutions are −8 < _x_ < −3.

**7.** Solve

The given inequality is equivalent to  . Add 3 to obtain −2 < 2/ _x_ < 8, and divide by 2 to get −1 < 1/ _x_ < 4.

_Case 1_ : _x_ > 0. Multiply by _x_ to get − _x_ < 1 < 4 _x_. Then   and _x_ > −1; these two inequalities are equivalent to the single inequality  .

_Case 2_ : _x_ < 0. Multiply by _x_ to obtain − _x_ > 1 > 4 _x_. (Note that the inequalities have been reversed, since we multiplied by the negative number _x_.) Then   and _x_ < −1. These two inequalities are equivalent to _x_ < −1.

Thus, the solutions are   or _x_ < −1, the union of the two infinite intervals   and (−∞, −1).

**8.** Solve |2 _x_ − 5|  3.

Let us first solve the negation |2 _x_ − 5| < 3. The latter is equivalent to −3 < 2 _x_ −5 < 3. Add 5 to obtain 2 < 2 _x_ < 8, and divide by 2 to obtain 1 < _x_ < 4. Since this is the solution of the negation, the original inequality has the solution _x_  ; 1 or _x_   4.

**9.** Solve2 < 3 _x_ \+ 10.

The crucial numbers are −2 and 5. ( _x_ − 5)( _x_ \+ 2) > 0 when _x_ < −2 (since both _x_ −5 and _x_ \+ 2 are negative); it becomes negative as we pass through −2 (since _x_ \+ 2 changes sign); and then it becomes positive as we pass through 5 (since _x_ −5 changes sign). Thus, the solutions are −2 < _x_ < 5.

### **Supplementary Problems**

**10.** Describe and diagram the set determined by each of the following conditions:

( _a_ ) −5 < _x_ < 0

( _b_ ) _x_   0

( _c_ ) −2   _x_ < 3

( _d_ ) _x_   1

( _e_ ) | _x_ | < 3

( _f_ ) | _x_ |   5

( _g_ )

( _h_ ) | _x_ −3| > 1

( _i_ ) 0 < | _x_ −2| < 1

( _j_ )

( _k_ ) | _x_ −2|   1.

_Ans_. ( _e_ ) −3 < _x_ < 3; ( _f_ ) _x_   5 or _x_   −5; ( _g_ )   ( _h_ ) _x_ > −2 or _x_ < −4; ( _i_ ) _x_ ≠ 2 and 1 < _x_ < 3; ( _j_ )   ( _k_ ) _x_  3 or _x_   1

**11.** Describe and diagram the set determined by each of the following conditions:

_Ans_.

**12.** Describe and diagram the set determined by each of the following conditions:

( _a_ ) _x_ ( _x_ − 5) < 0

( _b_ ) ( _x_ − 2)( _x_ − 6) > 0

( _c_ ) ( _x_ \+ 1)( _x_ − 2) < 0

( _d_ ) x( _x_ − 2)( _x_ \+ 3) > 0

( _e_ ) ( _x_ \+ 2)( _x_ \+ 3)( _x_ \+ 4) < 0

( _f_ ) ( _x_ − 1)( _x_ \+ 1)( _x_ − 2)( _x_ \+ 3) > 0

( _g_ ) ( _x_ − 1)2( _x_ \+ 4) > 0

( _h_ ) ( _x_ − 3)( _x_ \+ 5)( _x_ − 4)2 < 0

( _i_ ) ( _x_ − 2)3 > 0

( _j_ ) ( _x_ \+ 1)3 < 0

( _k_ ) ( _x_ − 2)3( _x_ \+ 1) < 0

( _l_ ) ( _x_ − 1)3( _x_ \+ 1)4 < 0

( _m_ ) (3 _x_ − 1)(2 _x_ \+ 3) > 0

( _n_ ) ( _x_ − 4)(2 _x_ − 3) < 0

_Ans_. ( _a_ ) 0 < _x_ < 5; ( _b_ ) _x_ > 6 or < 2; ( _c_ ) −1 < _x_ < 2; ( _d_ ) _x_ > 2 or −3 < _x_ < 0 ( _e_ ) −3 < _x_ < −2 or _x_ < −4; ( _f_ ) _x_ > 2 or _x_ −1 < _x_ < 1 or _x_ < −3; ( _g_ ) _x_ > −4 and _x_ ≠ 1; ( _h_ ) −5 < _x_ < 3; ( _i_ ) _x_ > 2; ( _j_ ) _x_ < −1; ( _k_ ) −1 < _x_ < 2; ( _l_ ) _x_ < 1 and _x_ ≠ −1; ( _m_ )   ( _n_ )

**13.** Describe and diagram the set determined by each of the following conditions:

( _a_ ) _x_ 2 < 4

( _b_ ) _x_ 2   9

( _c_ ) ( _x_ − 2)2   16

( _d_ ) (2 _x_ \+ 1)2 > 1

( _e_ ) _x_ 2 \+ 3 − 4 > 0

( _f_ ) _x_ 2 \+ 6 _x_ \+ 8   0

( _g_ ) _x_ 2 < 5 _x_ \+ 14

( _h_ ) 2 _x_ 2> _x_ \+ 6

( _i_ ) 6 _x_ 2 \+ 13 _x_ < 5

( _j_ ) _x_ 3 \+ 32 > 10 _x_

_Ans_. ( _a_ ) −2 < _x_ < 2; ( _b_ ) _x_   3 or _x_   −3; ( _c_ ) −2   _x_   6; ( _d_ ) _x_ > 0 or _x_ < −1; ( _e_ ) _x_ > 1 or _x_ < −4; ( _f_ ) −4  _x_   −2; ( _g_ ) −2 < _x_ < 7; ( _h_ )   ( _i_ )   ( _j_ ) −5 < _x_ < 0 or _x_ > 2

**14.** Solve:

_Ans_ ( _a_ ) −5 < _x_ < 6; ( _b_ ) > 0 or < −1; ( _c_ ) _x_ > −2; ( _d_ )   ( _e_ )   ( _f_ ) _x_   − 4 or _x_   − 1

**15.** Solve:

( _a_ ) |4 _x_ −5| = 3

( _b_ ) | _x_ \+ 6| = 2

( _c_ ) |3x −4| = |2 _x_ \+ 1|

( _d_ ) | _x_ \+ 1| = | _x_ \+ 2|

( _e_ ) | _x_ \+ 1| = 3 _x_ − 1

( _f_ ) | _x_ \+ 1| < |3 _x_ −1|

( _g_ ) |3 _x_ − 4|   |2 _x_ \+ 1|

_Ans_.

**16.** Prove:

( _a_ ) | _x_ 2| = | _x_ |2;

( _b_ ) | _x_ _n_ | = | _x_ | _n_ for every integer _n_ ;

( _c_ )

( _d_ ) | _x_ − _y_ |   | _x_ | + | _y_ |;

( _e_ ) | _x_ − _y_ |   || _x_ | − | _y_ ||

[Hint: In ( _e_ ), prove that | _x_ − _y_ |   | _x_ | − | _x_ | and | _x_ − _y_ |   | _y_ | − | _x_ |.]

## **CHAPTER 2
Rectangular Coordinate Systems**

**COORDINATE AXES.** In any plane  , choose a pair of perpendicular lines. Let one of the lines be horizontal. Then the other line must be vertical. The horizontal line is called the _x axis_ , and the vertical line the _y axis_. (See Fig. 2-1.)

**Fig. 2-1**

Now choose linear coordinate systems on the _x_ axis and the _y_ axis satisfying the following conditions: The origin for each coordinate system is the point _O_ at which the axes intersect. The _x_ axis is directed from left to right, and the _y_ axis from bottom to top. The part of the _x_ axis with positive coordinates is called the _positive x axis_ , and the part of the _y_ axis with positive coordinates is called the _positive y axis._

We shall establish a correspondence between the points of the plane   and pairs of real numbers.

**COORDINATES.** Consider any point _P_ of the plane (Fig. 2-1). The vertical line through _P_ intersects the _x_ axis at a unique point; let _a_ be the coordinate of this point on the _x_ axis. The number _a_ is called the _x coordinate_ of _P_ (or the _abscissa_ of _P_ ). The horizontal line through _P_ intersects the _y_ axis at a unique point; let _b_ be the coordinate of this point on the _y_ axis. The number _b_ is called the _y coordinate_ of _P_ (or the _ordinate_ of _P_ ). In this way, every point _P_ has a unique pair ( _a, b_ ) of real numbers associated with it. Conversely, every pair ( _a, b_ ) of real numbers is associated with a unique point in the plane.

The coordinates of several points are shown in Fig. 2-2. For the sake of simplicity, we have limited them to integers.

**Fig. 2-2**

**EXAMPLE:** In the coordinate system of Fig. 2-3, to find the point having coordinates (2, 3), start at the origin, move two units to the _right_ , and then three units _upward_.

To find the point with coordinates (−4, 2), start at the origin, move four units to the _left,_ and then two units _upward_.

To find the point with coordinates (−3, −1), start at the origin, move three units to the _left_ , and then one unit _downward_.

The order of these moves is not important. Hence, for example, the point (2, 3) can also be reached by starting at the origin, moving three units _upward,_ and then two units to the _right_.

**QUADRANTS.** Assume that a coordinate system has been established in the plane  . Then the whole plane  , with the exception of the coordinate axes, can be divided into four equal parts, called _quadrants._ All points with both coordinates positive form the first quadrant, called quadrant I, in the upper right-hand corner (see Fig. 2-4). _Quadrant_ II consists of all points with negative _x_ coordinate and positive _y_ coordinate. _Quadrants_ III and IV are also shown in Fig. 2-4.

The points on the _x_ axis have coordinates of the form ( _a, 0_ ). The _y_ axis consists of the points with coordinates of the form (0, b).

**Fig. 2-3**

**Fig. 2-4**

Given a coordinate system, it is customary to refer to the point with coordinates ( _a, b_ ) as "the point ( _a, b_ )." For example, one might say, 'The point (0, 1) lies on the _y_ axis."

**THE DISTANCE FORMULA.** The distance _P_ 1 _P_ 2 between poinits _P_ 1 and _P_ 2 with coordinates ( _x_ 1, _y_ 1) and ( _x_ 2, _y_ 2) in a given coordinate system (see Fig. 2-5) is given by the following distance formula:

To see this, let _R_ be the point where the vertical line through _P_ 2 intersects the horizontal line through _P_ 1. The _x_ coordinate of _R_ is _x_ 2, the same as that of _P_ 2. The _y_ coordinate of _R_ is _y_ 1, the same as that of _P_ 1. By the Pythagorean theorem,  . If _A_ 1 and _A_ 2 are the projections of _P_ 1 and _P_ 2 on the _x_ axis, the segments _P_ 1 _R_ and _A_ 1 _A_ 2 are opposite sides of a rectangle,

**Fig. 2-5**

so that  . But   by property ( **1.12** ). So,  . Similarly,  . Hence,  . Taking square roots, we obtain the distance formula. (It can be checked that the formula also is valid when _P_ 1 and _P_ 2 lie on the same vertical or horizontal line.)

**EXAMPLES**

( _a_ ) The distance between (2, 5) and (7, 17) is

( _b_ ) The distance between (1, 4) and (5, 2) is

**THE MIDPOINT FORMULAS.** The point _M_ ( _x, y_ ) that is the midpoint of the segment connecting the points _P_ 1( _x_ 1, _y_ 1) and _P_ 2( _x_ 2, _y_ 2) has the coordinates

Thus, the coordinates of the midpoints are the averages of the coordinates of the endpoints. See Fig. 2-6.

To see this, let _A, B, C_ be the projections of _P_ 1, _M, P_ 2 on the _x_ axis. The _x_ coordinates of _A, B, C_ are _x_ 1, _x, x_ 2. Since the lines _P_ 1 _A, MB, P_ 2 _C_ are parallel, the ratios   and   are equal. Since  ,  . Since   and  ,

(The same equation holds when _P_ 2 is to the left of _P_ 1, in which case   and  .) Similarly _y_ = ( _y_ 1 \+ _y_ 2)/2.

**Fig. 2-6**

**EXAMPLES**

( _a_ ) the midpoint of the segment connecting (2, 9) and (4, 3) is

( _b_ ) The point halfway between (−5, 1) and (1, 4) is  .

**PROOFS OF GEOMETRIC THEOREMS** can often be given more easily by use of coordinates than by deductions from axioms and previously derived theorems. Proofs by means of coordinates are called _analytic_ , in contrast to so-called _synthetic_ proofs from axioms.

**EXAMPLE:** Let us prove analytically that the segment joining the midpoints of two sides of a triangle is onehalf the length of the third side. Construct a coordinate system so that the third side _AB_ lies on the positive _x_ axis, _A_ is the origin, and the third vertex _C_ lies above the _x_ axis, as in Fig. 2-7.

**Fig. 2-7**

Let _b_ be the _x_ coordinate of _B_. (In other words, let  .). Let _C_ have coordinates ( _u, v_ ). Let _M_ 1 and _M_ 2 be the midpoints of sides _AC_ and _BC_ , respectively. By the midpoint formulas (2.2), the coordinates of _M_ 1 are  , and the coordinates of _M_ 2 are  .

By the distance formula ( _2.1_ ),

which is half the length of side _AB_.

### **Solved Problems**

**1.** Show that the distance between a point _P_ ( _x, y_ ) and the origin is  .

Since the origin has coordinates (0, 0), the distance formula yields  .

**2.** Is the triangle with vertices _A_ (1, 5), _B_ (4, 2) and _C_ (5, 6) isosceles?

Since  , the triangle is isosceles.

**3.** Is the triangle with vertices _A_ (−5,6), _B_ (2, 3), and _C_ (5, 10) a right triangle?

Since  , the converse of the Pythagorean theorem tells us that Δ _ABC_ is a right triangle, with right angle at _B_ ; in fact, since  , Δ _ABC_ is an isosceles right triangle.

**4.** Prove analytically that, if the medians to two sides of a triangle are equal, then those sides are equal. (Recall that a _median_ of a triangle is a line segment joining a vertex to the midpoint of the opposite side.)

In Δ _ABC_ , let _M_ 1 and _M_ 2 be the midpoints of sides _AC_ and _BC_ , respectively. Construct a coordinate system so that _A_ is the origin, _B_ lies on the positive _x_ axis, and _C_ lies above the _x_ axis (see Fig. 2-8). Assume that  . We must prove that  . Let _b_ be the _x_ coordinate of _B_ , and let _C_ have coordinates ( _u, v_ ). Then, by the midpoint formulars, _M_ 1 has coordinates  , and _M_ 2 has coordinates  .

Hence,

**Fig. 2-8**

Since  ,

Hence,   and, therefore, ( _u + b_ )2 = ( _u_ − 2 _b_ )2. So, _u_ \+ _b_ = ±( _u_ − 2 _b_ ). If _u_ \+ _b_ = _u_ − 2 _b_ , then _b_ = −2 _b_ , and therefore, _b_ = 0, which is impossible, since _A_ ≠ _B_. Hence, _u_ \+ _b_ = −( _u_ − 2 _b_ ) = − _u_ \+ 2 _b_ , whence 2 _u_ = _b_. Now  , and  . Thus,  .

**5.** Find the coordinates ( _x, y_ ) of the point _Q_ on the line segment joining _P_ 1(1, 2) and _P_ 2(6, 7), such that _Q_ divides the segment in the ratio 2:3, that is, such that  .

Let the projections of _P_ 1, _Q_ , and _P_ 2 on the _x_ axis be _A_ 1, _Q′_ , and _A_ 2, with _x_ coordinates 1, _x_ , and 6, respectively (see Fig. 2-9). Now   (When two lines are cut by three parallel lines, corresponding segments are in proportion.) But  , and  . So  , and cross-multiplying yields 3 _x_ − 3 = 12 − 2 _x_. Hence 5 _x_ = 15, whence _x_ = 3. By similar reasoning  , from which it follows that _y_ = 4.

**Fig. 2-9**

### **Supplementary Problems**

**6.** In Fig. 2-10, find the coordinates of points _A, B, C, D, E_ , and _F_.

**Fig. 2-10**

_Ans_. _A_ = (−2, 1); _B_ = (0, −1); _C_ = (1, 3); _D_ = (−4, −2); _E_ = (4, 4); _F_ = (7, 2).

**7.** Draw a coordinate system and show the points having the following coordinates: (2, −3), (3, 3), (−1, 1), (2, −2), (0, 3), (3, 0) (−2, 3).

**8.** Find the distances between the following pairs of points:

( _a_ ) (3, 4) and (3, 6)

( _b_ ) (2, 5) and (2,−2)

( _c_ ) (3, 1) and (2, 1)

( _d_ ) (2, 3) and (5, 7)

( _e_ ) (−2,4) and (3, 0)

( _f_ )   and (4, −1)

_Ans_. ( _a_ ) 2; ( _b_ ) 7; ( _c_ ) 1; ( _d_ ) 5; ( _e_ )   ( _f_ )

**9.** Draw the triangle with vertices _A_ (2, 5) _B_ (2, −5), and _C_ (−3, 5), and find its area.

_Ans_. Area = 25

**10.** If (2, 2), (2, −4) and (5, 2) are three vertices of a rectangle, find the fourth vertex.

_Ans_. (5, −4)

**11.** If the points (2, 4) and (−1, 3) are the opposite vertices of a rectangle whose sides are parallel to the coordinate axes (that is, the _x_ and _y_ axes), find the other two vertices.

_Ans_. (−1, 4) and (2, 3)

**12.** Determine whether the following triples of points are the vertices of an isosceles triangle: ( _a_ ) (4, 3), (1, 4), (3, 10); ( _b_ ) (−1, 1), (3, 3), (1, −1); ( _c_ ) (2, 4), (5, 2), (6, 5).

_Ans_. ( _a_ ) no; ( _b_ ) yes; ( _c_ ) no

**13.** Determine whether the following triples of points are the vertices of a right triangle. For those that are, find the area of the right triangle: ( _a_ ) (10, 6), (3, 3), (6, −4); ( _b_ ) (3, 1), (1, −2), (−3, −1); ( _c_ ) (5, −2), (0, 3), (2, 4)

_Ans_. ( _a_ ) yes, area = 29; ( _b_ ) no; ( _c_ ) yes,

**14.** Find the perimeter of the triangle with vertices _A_ (4, 9), _B_ (−3, 2) and _C_ (8, −5).

_Ans_.

**15.** Find the value or values of _y_ for which (6, _y_ ) is equidistant from (4, 2) and (9, 7).

_Ans_. 5

**16.** Find the midpoints of the line segments with the following endpoints: ( _a_ ) (2, −3) and (7, 4); ( _b_ )   and (4, 1): ( _c_ )   and (1, 4)

_Ans_.

**17.** Find the point ( _x, y_ ) such that (2, 4) is the midpoint of the line segment connecting ( _x, y_ ) and (1, 5).

_Ans_. (3, 3)

**18.** Determine the point that is equidistant from the points _A_ (−1, 7), _B_ (6, 6), and _C_ (5, −1).

_Ans_.

**19.** Prove analytically that the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices.

**20.** Show analytically that the sum of the squares of the distance of any point _P_ from two opposite vertices of a rectangle is equal to the sum of the squares of its distances from the other two vertices.

**21.** Prove analytically that the sum of the squares of the four sides of a parallelogram is equal to the sum of the squares of the diagonals.

**22.** Prove analytically that the sum of the squares of the medians of a triangle is equal to three-fourths the sum of the squares of the sides.

**23.** Prove analytically that the line segments joining the midpoints of opposite sides of quadrilateral bisect each other.

**24.** Prove that the coordinates ( _x, y_ ) of the point _Q_ that divides the line segments from _P_ 1( _x_ 1, _y_ 1) to _P_ 2( _x_ 2, _y_ 2) in the ratio _r_ 1: _r_ 2 are determined by the formulas

( _Hint_ : Use the reasoning of Problem 5.)

**25.** Find the coordinates of the point _Q_ on the segment _P_ 1 _P_ 2 such that  , if ( _a_ ) _P_ 1 = (0, 0), _P_ 2 = (7, 9); ( _b_ ) _P_ 1 = (−1, 0), _P_ 2 = (0, 7); ( _c_ ) _P_ 1 = (−7, −2), _P_ 2 = (2, 7); ( _d_ ) _P_ 1 = (1, 3), _P_ 2 = (4, 2).

_Ans_.

## **CHAPTER 3
Lines**

**THE STEEPNESS OF A LINE** is measured by a number called the _slope_ of the line. Let   be any line, and let _P_ 1( _x_ 1, _y_ 1) and _P_ 2( _x_ 2, _y_ 2) be two points of  . The slope of   is defined to be the number  . The slope is the ratio of a change in the _y_ coordinate to the corresponding change in the _x_ coordinate. (See Fig. 3-1.)

**Fig. 3-1**

For the definition of the slope to make sense, it is necessary to check that the number _m_ is independent of the choice of the points _P_ 1 and _P_ 2. If we choose another pair _P_ 3( _x_ 3, _y_ 3) and _P_ 4( _x_ 4, _y_ 4), the same value of _m_ must result. In Fig. 3-2, triangle _P_ 3 _P_ 4 _T_ is similar to triangle _P_ 1 _P_ 2 _Q_. Hence,

Therefore, _P_ 1 and _P_ 2 determine the same slope as _P_ 3 and _P_ 4.

**EXAMPLE 1:** The slope of the line joining the points (1, 2) and (4, 6) in Fig. 3-3 is  . Hence, as a point on the line moves 3 units to the right, it moves 4 units upwards. Moreover, the slope is not affected by the order in which the points are given:  . In general,

**Fig. 3-2**

**Fig. 3-3**

**THE SIGN OF THE SLOPE** has significance. Consider, for example, a line   that moves upward as it moves to the right, as in Fig. 3-4( _a_ ). Since _y_ 2 > _y_ 1 and _x_ 2 > _x_ 1, we have  . _The slope of  is positive._

Now consider a line   that moves downward as it moves to the right, as in Fig. 3-4( _b_ ). Here _y_ 2 < _y_ 1 while _x_ 2 > _x_ 1; hence,  . _The slope of  is negative._

Now let the line   be horizontal, as in Fig. 3-4( _c_ ). Here _y_ 1 = _y_ 2, so that _y_ 2 − _y_ 1 = 0. In addition, _x_ 2 − _x_ 1 ≠ 0. Hence,  . _The slope of  is zero._

Line   is vertical in Fig. 3-4( _d_ ), where we see that _y_ 2 − _y_ 1 > 0 while _x_ 2 − _x_ 1 = 0. Thus, the expression   is undefined. _The slope is not defined for a vertical line _. (Sometimes we describe this situation by saying that the slope of   is "infinite.")

**Fig. 3-4**

**SLOPE AND STEEPNESS.** Consider any line   with positive slope, passing through a point _P_ 1( _x_ 1, _y_ 1); such a line is shown in Fig. 3-5. Choose the point _P_ 2( _x_ 2, _y_ 2) on   such that _x_ 2 − _x_ 1 = 1. Then the slope _m_ of   is equal to the distance  . As the steepness of the line increases,   increases without limit, as shown in Fig. 3-6( _a_ ). Thus, the slope of   increases without bound from 0 (when   is horizontal) to +∞ (when the line is vertical). By a similar argument, using Fig. 3-6( _b_ ), we can show that as a negatively sloped line becomes steeper, the slope steadily decreases from 0 (when the line is horizontal) to −∞ (when the line is vertical).

**Fig. 3-5**

**Fig. 3-6**

**EQUATIONS OF LINES.** Let   be a line that passes through a point _P_ 1( _x_ 1, _y_ 1) and has slope _m_ , as in Fig. 3-7( _a_ ). For any other point _P_ ( _x_ , _y_ ) on the line, the slope _m_ is, by definition, the ratio of _y_ − _y_ 1 to _x_ − _x_ 1. Thus, for any point ( _x_ , _y_ ) on  ,

Conversely, if _P_ ( _x_ , _y_ ) is _not_ on line  , as in Fig. 3-7( _b_ ), then the slope   of the line _PP_ 1 is different from the slope _m_ of  ; hence ( _3.1_ ) does not hold for points that are not one  . Thus, the

**Fig. 3-7**

line   consists of only those points ( _x, y_ ) that satisfy ( _3.1_ ). In such a case, we say that   is the _graph_ of ( _3.1_ ).

**A POINT-SLOPE EQUATION** of the line   is any equation of the form ( _3.1_ ). If the slope _m_ of   is known, then each point ( _x_ 1, _y_ 1) of   yields a point-slope equation of  . Hence, there are infinitely many point-slope equations for  . Equation ( _3.1_ ) is equivalent to _y_ − _y_ 1 = _m_ ( _x_ − _x_ 1).

**EXAMPLE 2:** ( _a_ ) The line passing through the point (2, 5) with slope 3 has a point-slope equation  . ( _b_ ) Let   be the line through the points (3, −1) and (2, 3). Its slope is  . Two point-slope equations of   are   and  .

**SLOPE-INTERCEPT EQUATION.** If we multiply ( _3.1_ ) by _x_ − _x_ 1, we obtain the equation _y_ − _y_ 1 = _m_ ( _x_ − _x_ 1), which can be reduced first to y − _y_ 1 = _mx_ − _mx_ 1, and then to _y_ = _mx_ \+ ( _y_ 1 − _mx_ 1). Let _b_ stand for the number _y_ 1 − _mx_ 1. Then the equation for line   becomes

Equation ( _3.2_ ) yields the value _y_ = _b_ when _x_ = 0, so the point (0, _b_ ) lies on  . Thus, _b_ is the _y_ coordinate of the intersection of   and the _y_ axis, as shown in Fig. 3-8. The number _b_ is called the _y intercept_ of  , and ( _3.2_ ) is called the _slope-intercept equation_ for  .

**Fig. 3-8**

**EXAMPLE 3:** The line through the points (2, 3) and (4, 9) has slope

Its slope-intercept equation has the form _y_ = 3 _x_ \+ _b_. Since the point (2, 3) lies on the line, (2, 3) must satisfy this equation. Substitution yields 3 = 3(2) + _b_ , from which we find _b_ = −3. Thus, the slope-intercept equation is _y_ = 3 _x_ − 3.

Another method for finding this equation is to write a point–slope equation of the line, say  . Then multiplying by _x_ − 2 and adding 3 yields _y_ = 3 _x_ − 3.

**PARALLEL LINES.** Let  1 and  2 be parallel nonvertical lines, and let _A_ 1 and _A_ 2 be the points at which  1 and  2 intersect the _y_ axis, as in Fig. 3-9( _a_ ). Further, let _B_ 1 be one unit to the right of _A_ 1, and _B_ 2 one unit to the right of _A_ 2. Let _C_ 1 and _C_ 2 be the intersections of the verticals through _B_ 1 and _B_ 2 with  1 and  2. Now, triangle _A_ 1 _B_ 1 _C_ 1 is congruent to triangle _A_ 2 _B_ 2 _C_ 2 (by the angle–side–angle congruence theorem). Hence,   and

Thus, _parallel lines have equal slopes_.

**Fig. 3-9**

Conversely, assume that two different lines  1 and  2 are not parallel, and let them meet at point _P_ , as in Fig. 3-9( _b_ ). If  1 and  2 had the same slope, then they would have to be the same line. Hence,  1 and  2 have different slopes.

**Theorem 3.1:** Two distinct nonvertical lines are parallel if and only if their slopes are equal.

**EXAMPLE 4:** Find the slope–intercept equation of the line   through (4, 1) and parallel to the line   having the equation 4 _x_ − 2 _y_ = 5.

By solving the latter equation for _y_ , we see that   has the slope–intercept equation  . Hence,   has slope 2. The slope of the parallel line   also must be 2. So the slope–intercept equation of   has the form y = 2 _x_ \+ _b_. Since (4, 1) lies on  , we can write 1 = 2(4) + _b_. Hence, _b_ = −7, and the slope–intercept equation of   is _y_ = 2 _x_ − 7.

**PERPENDICULAR LINES.** In Problem 5 we shall prove the following:

**Theorem 3.2:** Two nonvertical lines are perpendicular if and only if the product of their slopes is −1.

If _m_ 1 and _m_ 2 are the slopes of perpendicular lines, then _m_ 1 _m_ 2 = −1. This is equivalent to  ; hence, _the slopes of perpendicular lines are negative reciprocals of each other_.

### **Solved Problems**

**1.** Find the slope of the line having the equation 3 _x_ − 4 _y_ = 8. Draw the line. Do the points (6, 2) and (12, 7) lie on the line?

Solving the equation for _y_ yields  . This is the slope–intercept equation; the slope is   and the _y_ intercept is −2.

Substituting 0 for _x_ shows that the line passes through the point (0, −2). To draw the line, we need another point. If we substitute 4 for _x_ in the slope–intercept equation, we get  . So, (4, 1) also lies on the line, which is drawn in Fig. 3-10. (We could have found other points on the line by substituting numbers other than 4 for _x_.)

To test whether (6, 2) is on the line, we substitute 6 for _x_ and 2 for _y_ in the original equation, 3 _x_ − 4 _y_ = 8. The two sides turn out to be unequal; hence, (6, 2) is not on the line. The same procedure shows that (12, 7) lies on the line.

**Fig. 3-10**

**Fig. 3-11**

**2.** Line   is the perpendicular bisector of the line segment joining the points _A_ (−1, 2) and _B_ (3, 4), as shown in Fig. 3-11. Find an equation for  .

  passes through the midpoint _M_ of segment _AB_. By the midpoint formulas ( _2.2_ ), the coordinates of _M_ are (1, 3). The slope of the line through _A_ and _B_ is  . Let _m_ be the slope of  . By Theorem 3.2,  , whence _m_ = −2.

The slope–intercept equation for   has the form _y_ = −2 _x_ \+ _b_. Since _M_ (1, 3) lies on  , we have 3 = −2(1) + _b_. Hence, _b_ = 5, and the slope–intercept equation of   is y = −2 _x_ \+ 5.

**3.** Determine whether the points _A_ (1, −1), _B_ (3, 2), and _C_ (7, 8) are collinear, that is, lie on the same line.

_A_ , _B_ , and _C_ are collinear if and only if the line _AB_ is identical with the line _AC_ , which is equivalent to the slope of _AB_ being equal to the slope of _AC_. (Why?) The slopes of _AB_ and _AC_ are   and  . Hence, _A_ , _B_ , and _C_ are collinear.

**4.** Prove analytically that the figure obtained by joining the midpoints of consecutive sides of a quadrilateral is a parallelogram.

Locate a quadrilateral with consecutive vertices, _A_ , _B_ , _C_ , and _D_ on a coordinate system so that _A_ is the origin, _B_ lies on the positive _x_ axis, and _C_ and _D_ lie above the _x_ axis. (See Fig. 3-12.) Let _b_ be the _x_ coordinate of _B_ , ( _u, v_ ) the coordinates of _C_ , and ( _x_ , _y_ ) the coordinates of _D_. Then, by the midpoint formula ( _2.2_ ), the midpoints _M_ 1, _M_ 2, _M_ 3, and _M_ 4 of sides _AB_ , _BC_ , _CD_ , and _DA_ have coordinates  ,  ,  , and  , respectively. We must show that _M_ 1 _M_ 2 _M_ 3 _M_ 4 is a parallogram. To do this, it suffices to prove that lines _M_ 1 _M_ 2 and _M_ 3 _M_ 4 are parallel and that lines _M_ 2 _M_ 3 and M1M4 are parallel. Let us calculate the slopes of these lines:

Since slope( _M_ 1 _M_ 2) = slope( _M_ 3 _M_ 4), _M_ 1 _M_ 2 and M3M4 are parallel. Since slope( _M_ 2 _M_ 3) = slope( _M_ 1 _M_ 4), _M_ 2 _M_ 3 and M1M4 are parallel. Thus, _M_ 1 _M_ 2 _M_ 3 _M_ 4 is a parallelogram.

**Fig. 3-12**

**5.** Prove Theorem 3.2

First we assume  1 and  2 are perpendicular nonvertical lines with slopes _m_ 1 and _m_ 2. We must show that _m_ 1 _m_ 2 = −1. Let  1 and  2 be the lines through the origin _O_ that are parallel to  1 and  2, as shown in Fig. 3-13( _a_ ). Then the slope of  1 is  1, and the slope of  2 is  2 (by Theorem 3.1). Moreover,  1 and  2 are perpendicular, since  1 and  2 are perpendicular.

**Fig. 3-13**

Now let _A_ be the point on  1 with _x_ coordinate 1, and let _B_ be the point on  2 with _x_ coordinate 1, as in Fig. 3-13( _b_ ). The slope–intercept equation of  1 is _y_ = _m_ 1 _x_ ; therefore, the _y_ coordinate of _A_ is _m_ 1, since its _x_ coordinate is 1. Similarly, the _y_ coordinate of _B_ is _m_ 2. By the distance fromula ( _2.1_ ),

Then by the Pythagorean theorem for right triangle _BOA_ ,

or

Now, conversely, we assume that _m_ 1 _m_ 2 = −1, where _m_ 1 and _m_ 2 are the slopes of nonvertical lines  1 and  2. Then  1 is not parallel to  2. (Otherwise, by Theorem 3.1, _m_ 1 = _m_ 2 and, therefore, _m_ 12 = −1, which contradicts the fact that the square of a real number is never negative.) We must show that  1 and  2 are perpendicular. Let _P_ be the intersection of  1 and  2 (see Fig. 3-14). Let  3 be the line through _P_ that is

**Fig. 3-14**

perpendicular to  1. If _m_ 3, is the slope of  3, then, by the first part of the proof, _m 1m3_ = −1 and, therefore, _m 1m3_ = _m 1m2_ = −1, _m_ 1 ≠ 0; therefore, _m_ 3 = _m_ 2. Since  2 and  3 pass through the same point _P_ and have the same slope, they must coincide. Since  1 and  3 are perpendicular,  1 and  2 are also perpendicular.

**6.** Show that, if _a_ and _b_ are not both zero, then the equation _ax + by = c_ is the equation of a line and, conversely, every line has an equation of that form.

Assume _b_ ≠ 0. Then, if the equation _ax + by = c_ is solved for _y_ , we obtain a slope-intercept equation _y_ = (− _a/b_ ) _x_ \+ _c/b_ of a line. If _b_ = 0, then _a ≠ 0_ , and the equation _ax + by = c_ reduces to _ax = c_ ; this is equivalent to _x = c/a_ , the equation of a vertical line.

Conversely, every nonvertical line has a slope-intercept equation _y = mx + b,_ which is equivalent to _−mx + y = b_ , an equation of the desired form. A vertical line has an equation of the form _x = c_ , which is also an equation of the required form with _a_ = 1 and _b_ = 0.

**7.** Show that the line _y_ = _x_ makes an angle of 45° with the positive _x_ axis (that is, that angle _BOA_ in Fig. 3-15 contains 45°).

**Fig. 3-15**

Let _A_ be the point on the line _y = x_ with coordinates (1, 1). Drop a perpendicular _AB_ to the positive _x_ axis. Then   and  . Hence, angle _OAB_ = angle _BOA_ , since they are the base angles of isosceles triangle _BOA_. Since angle _OBA_ is a right angle,

Angle _OAB_ \+ angle _BOA_ = 180° − angle _OBA_ = 180° 90° = 90°

Since angle _BOA_ = angle _OAB_ , they each contain 45°

**8.** Show that the distance _d_ from a point _P_ ( _x_ 1, _y_ 1) to a line   with equation _ax_ \+ _by_ = _c_ is given by the formula  .

Let   be the line through _P_ that is perpendicular to  . Then   intersects _L_ at some point _Q_ with coordinates ( _u_ , _v_ ), as in Fig. 3-16. Clearly, _d_ is the length  , so if we can find _u_ and _v_ , we can compute _d_ with the distance formula. The slope of   is − _a/b_. Hence, by Theorem 3.2, the slope of   is _b/a_. Then a point-slope equation of   is   Thus, u and v are the solutions of the pair of equations _au_ \+ _bv_ = c and

**Fig. 3-16**

 . Tedious algebraic calculations yield the solution

The distance formula, together with further calculations, yields

### **Supplementary Problems**

**9.** Find a point-slope equation for the line through each of the following pairs of points: ( _a_ ) (3, 6) and (2, −4); ( _b_ ) (8, 5) and (4, 0); ( _c_ ) (1, 3) and the origiin; ( _d_ ) (2, 4) and (−2, 4).

_Ans._

**10.** Find the slope–intercept equation of each line:

( _a_ ) Through the points (4, −2) and (1, 7)

( _b_ ) Having slope 3 and y intercept 4

( _c_ ) Through the points (−1, 0) and (0, 3)

( _d_ ) Through (2, −3) and parallel to the _x_ axis

( _e_ ) Through (2, 3) and rising 4 units for every unit increase in _x_

( _f_ ) Through (−2, 2) and falling 2 units for every unit increase in _x_

( _g_ ) Through (3, −4) and parallel to the line with equation 5 _x_ − 2 _y_ = 4

( _h_ ) Through the origin and parallel to the line with equation _y = 2_

( _i_ ) Through (−2, 5) and perpendicular to the line with equation 4 _x_ \+ 8 _y_ = 3

( _j_ ) Through the origin and perpendicular to the line with equation 3 _x_ − 2 _y_ = 1

( _k_ ) Through (2, 1) and perpendicular to the line with equation _x = 2_

( _l_ ) Through the origin and bisecting the angle between the positive _x_ axis and the positive _y_ axis

_Ans._ ( _a_ ) y = −3 _x_ \+ 10; ( _b_ ) y = 3 _x_ \+ 3; ( _c_ ) _y_ = 3 _x_ \+ 3; ( _d_ ) _y_ = −3; ( _e_ ) _y_ = 4 _x_ − 5; ( _f_ ) _y_ = −2 _x_ −2; ( _g_ )  ; ( _h_ ) _y_ = 0; ( _i_ ) _y_ = 2 _x_ \+ 9; ( _j_ )   ( _k_ ) _y_ = 1 ( _l_ ) _y_ = _x_

**11.** ( _a_ ) Describe the lines having equations of the form _x = a_

( _b_ ) Describe the lines having equations of the form _y = b_.

( _c_ ) Describe the line having the equation _y = −x_.

**12.** ( _a_ ) Find the slopes and _y_ intercepts of the lines that have the following equations: ( _i_ ) _y_ = 3 _x_ − 2; ( _ii_ ) 2 _x_ − 5 _y_ = 3; ( _iii_ ) _y_ = 4 _x_ − 3; ( _iv_ ) _y_ = −3; ( _v_ )  .

( _b_ ) Find the coordinates of a point other than (0, _b_ ) on each of the lines of part ( _a_ )

_Ans._ ( _a_ ) ( _i_ ) _m_ = 3, _b_ = −2; ( _ii_ )   ( _iii_ ) _m_ = 4, _b_ = −3; ( _iv_ ) _m_ = 0, _b_ = −3; ( _v_ )   ( _b_ ) ( _i_ ) (1, 1); ( _ii_ ) (−6, −3); ( _iii_ ) (1, 1); ( _iv_ ) (1, −3); ( _v_ ) (3, 0)

**13.** If the point (3, _k_ ) lies on the line with slope _m_ = −2 passing through the point (2, 5), find _k_.

_Ans. k_ = 3

**14.** Does the point (3, −2) lie on the line through the points (8, 0) and (−7, −6)?

_Ans_. Yes

**15.** Use slopes to determine whether the points (7, −1), (10, 1) and (6, 7) are the vertices of a right triangle.

_Ans_. They are

**16.** Use slopes to determine whether (8, 0), (−1, −2), (−2, 3), and (7, 5) are the vertices of a parallelogram.

_Ans._ They are

**17.** Under what conditions are the points ( _u, v + w_ ), ( _v, u + w_ ), and ( _w, u + v_ ) collinear?

_Ans_. Always.

**18.** Determine _k_ so that the points _A(7, 3)_ , _B(−1, 0)_ , and _C(k, −2)_ are the vertices of a right triangle with right angle at _B_.

_Ans_. k = 1

**19.** Determine whether the following pairs of lines are parallel, perpendicular, or neither:

( _a_ ) y = 3 _x_ \+ 2 and _y_ = 3 _x_ − 2

( _b_ ) y = 2 _x_ − 4 and _y_ = 3 _x_ \+ 5

( _c_ ) 3 _x_ − 2 _y_ = 5 and 2 _x_ \+ 3 _y_ = 4

( _d_ ) 6 _x_ \+ 3 _y_ = 1 and 4 _x_ \+ 2 _y_ = 3

( _e_ ) _x_ = 3 and _y_ = −4

( _f_ ) 5x + 4y = 1 and 4 _x_ \+ 5 _y_ = 2

( _g_ ) _x_ = −2 and _x_ = 7.

_Ans_. ( _a_ ) Parallel; ( _b_ ) neither; ( _c_ ) perpendicular; ( _d_ ) parallel; ( _e_ ) perpendicular; ( _f_ ) neither; ( _g_ ) parallel

**20.** Draw the line determined by the equation 2 _x_ \+ 5 _y_ = 10. Determine whether the points (10, 2) and (12, 3) lie on this line.

**21.** For what values of _k_ will the line _kx_ − 3 _y_ = 4 _k_ have the following properties: ( _a_ ) have slope 1; ( _b_ ) have y intercept 2; ( _c_ ) pass through the point (2, 4); ( _d_ ) be parallel to the line 2 _x_ − 4 _y_ = 1; ( _e_ ) be perpendicular to the line _x_ − 6 _y_ = 2?

_Ans_. ( _a_ ) _k_ = 3; ( _b_ )  ; ( _c_ ) _k_ = −6; ( _d_ )  ; ( _e_ ) _k_ = −18

**22.** Describe geometrically the families of lines ( _a_ ) _y = mx − 3_ and ( _b_ ) _y = 4x + b_ , where _m_ and _b_ are any real numbers.

_Ans_. ( _a_ ) Lines with _y_ intercept −3; ( _b_ ) Lines with slope 4

**23.** In the triangle with vertices _A_ (0, 0), _B_ (2, 0), and _C_ (3, 3), find equations for ( _a_ ) the median from _B_ to the midpoint of the opposite side; ( _b_ ) the perpendicular bisector of side _BC_ ; and ( _c_ ) the altitude from _B_ to the opposite side.

_Ans_. ( _a_ ) _y_ = −3 _x_ \+ 6; ( _b_ ) _x_ \+ 3 _y_ = 7; ( _c_ ) _y_ = − _x_ \+ 2

## **CHAPTER 4
Circles**

**EQUATIONS OF CIRCLES.** For a point _P(x, y)_ to lie on the circle with center _C(a, b)_ and radius _r_ , the distance   must be equal to _r_ (see Fig. 4-1). By the distance formula ( _2.1_ ),

Thus, _P_ lies on the circle if and only if

Equation ( _4.1_ ) is called the _standard equation_ of the circle with center at ( _a, b_ ) and radius _r_.

**Fig. 4-1**

**EXAMPLE 1**

( _a_ ) The circle with center (3, 1) and radius 2 has the equation (x − 3)2 \+ (y − 1)2 = 4.

( _b_ ) The circle with center (2,−1) and radius 3 has the equation (x − 2)2 \+ (y + 1)2 = 9.

( _c_ ) What is the set of points satisfying the equation (x − 4)2 \+ (y − 5)2 = 25?

By ( _4.1_ ), this is the equation of the circle with center at (4, 5) and radius 5. That circle is said to be the _graph_ of the given equation, that is, the set of points satisfying the equation.

( _d_ ) The graph of the equation (x + 3)2 \+ y2 = 2 is the circle with center at (−3, 0) radius  .

**THE STANDARD EQUATION OF A CIRCLE** with center at the origin (0, 0) and radius _r_ is

For example, _x_ 2 \+ _y_ 2 = 1 is the equation of the circle with center at the origin and radius 1. The graph of _x_ 2 \+ _y_ 2 = 5 is the circle with center at the origin and radius  .

The equation of a circle sometimes appears in a disguised form. For example, the equation

turns out to be equivalent to

Equation ( _4.4_ ) is the standard equation of a circle with center at (−4, 3) and radius 2.

Equation ( _4.4_ ) is obtained from ( _4.3_ ) by a process called _completing the square_. In general terms, the process involves finding the number that must be added to the sum _x_ 2 \+ _Ax_ to obtain a square.

Here, we note that  . Thus, in general, _we must add_   to _x_ 2 \+ _Ax to obtain the square_  . For example, to get a square from _x_ 2 \+ 8 _x_ , we add  , that is, 16.

The result is _x_ 2 \+ 8 _x_ \+ 16, which is (x + 4)2. This is the process of completing the square.

Consider the original ( _4.3_ ): _x_ 2 \+ y2 \+ 8 _x_ − 6 _y_ \+ 21 = 0. To complete the square in _x_ 2 \+ 8 _x_ , we add 16. To complete the square in y2 − 6y, we add  , which is 9. Of course, since we added 16 and 9 to the left side of the equation, we must also add them to the right side, obtaining

( _x_ 2 \+ 8 _x_ \+ 16) + ( _y_ 2 − 6 _y_ \+ 9) + 21 = 16 + 9

This is equivalent to

( _x_ \+ 4)2 \+ ( _y_ − 3)2 \+ 21 = 25

and subtraction of 21 from both sides yields ( _4.4_ ).

**EXAMPLE 2:** Consider the equation _x_ 2 \+ y2 − 4 _x_ − 10 _y_ \+ 20 = 0. Completing the square yields

( _x_ 2 −4 _x_ \+ 4) + ( _y_ 2 −10 _y_ +25) + 20 = 4 + 25

( _x_ − 2)2 \+ ( _y_ − 5)2 = 9

Thus, the original equation is the equation of a circle with center at (2, 5) and radius 3.

The process of completing the square can be applied to any equation of the form

to obtain

There are three different cases, depending on whether A2 \+ B2 − 4 _C_ is positive, zero, or negative.

_Case 1: A 2 \+ B2 − 4C > 0_. In this case, ( _4.6_ ) is the standard equation of a circle with center at   and radius  .

_Case 2: A 2 \+ B2 − 4C = 0_. A sum of the squares of two quantities is zero when and only when each of the quantities is zero. Hence, ( _4.6_ ) is equivalent to the conjunction of the equations _x_ \+ A/2 = 0 and _y_ \+ _B_ /2 = 0 in this case, and the only solution of ( _4.6_ ) is the point (−A/2, −B/2). Hence, the graph of ( _4.5_ ) is a single point, which may be considered a _degenerate circle_ of radius 0.

_Case 3: A 2 \+ B2 − 4C < 0_. A sum of two squares cannot be negative. So, in this case, ( _4.5_ ) has no solution at all.

We can show that any circle has an equation of the form ( _4.5_ ). Suppose its center is ( _a, b_ ) and its radius is _r_ ; then its standard equation is

( _x − a_ )2 \+ ( _y − b_ )2 = _r_ 2

Expanding yields _x_ 2 − 2 _ax_ \+ _a_ 2 \+ _y_ 2 − 2 _by_ \+ _b_ 2 = _r_ 2, or

_x_ 2 \+ _y_ 2 − 2 _ax_ − 2 _by_ \+ ( _a_ 2 \+ _b_ 2 − _r_ 2) = 0

### **Solved Problems**

**1.** Identify the graphs of ( _a_ ) 2 _x_ 2 \+ 2 _y_ 2 − 4 _x_ \+ _y_ \+ 1 = 0; ( _b_ ) _x_ 2 \+ _y_ 2 − 4 _y_ \+ 7 = 0; ( _c_ ) _x_ 2 \+ _y_ 2 − 6 _x_ − 2 _y_ \+ 10 = 0.

( _a_ ) First divide by 2, obtaining  . Then complete the squares:

Thus, the graph is the circle with center   and radius  .

( _b_ )Complete the square:

Because the right side is negative, there are no points in the graph.

( _c_ ) Complete the square:

The only solution is the point (3, 1).

**2.** Find the standard equation of the circle with center at _C_ (2, 3) and passing through the point _P_ (−1, 5).

The radius of the circle is the distance

so the standard equation is ( _x_ − 2)2 \+ ( _y_ − 3)2 = 13.

**3.** Find the standard equation of the circle passing through the points _P_ (3, 8), _Q_ (9, 6), and _R_ (13, −2).

_First method:_ The circle has an equation of the form _x_ 2 \+ _y_ 2 \+ _Ax_ \+ _By_ \+ _C_ = 0. Substitute the values of _x_ and _y_ at point _P_ , to obtain 9 + 64 + 3 _A_ \+ 8 _B_ \+ _C_ = 0 or

A similar procedure for points _Q_ and _R_ yields the equations

Eliminate _C_ from ( _1_ ) and ( _2_ ) by subtracting ( _2_ ) from ( _1_ ):

Eliminate _C_ from ( _1_ ) and ( _3_ ) by subtracting ( _3_ ) from ( _1_ ):

Eliminate _B_ from ( _4_ ) and ( _5_ ) by subtracting ( _5_ ) from ( _4_ ), obtaining A = −6. Substitute this value in ( _5_ ) to find that _B_ = 4. Then solve for _C_ in( _1_ ) C = −87.

Hence, the original equation for the circle is _x_ 2 \+ y2 − 6 _x_ \+ 4 _y_ − 87 = 0. Completing the squares then yields

( _x_ − 3)2 \+ ( _y_ \+ 2)2 = 87 + 9 + 4 = 100

Thus, the circle has center (3, −2) and radius 10.

_Second method:_ The perpendicular bisector of any chord of a circle passes through the center of the circle. Hence, the perpendicular bisector   of chord _PQ_ will intersect the perpendicular bisector   of chord _QR_ at the center of the circle (see Fig. 4-2).

**Fig. 4-2**

The slope of line _PQ_ is   So, by Theorem 3.2, the slope of   is 3. Also,   passes through the midpoint (6, 7) of segment _PQ_. Hence a point-slope equation of   is  , and therefore its slope-intercept equation is _y_ = 3 _x_ − 11. Similarly, the slope of line _QR_ is −2, and therefore the slope of   is  . Since   passes through the midpoint (11, 2) of segment _QR_ , it has a point-slope equation  , which yields the slope-intercept equation  . Hence, the coordinates of the center of the circle satisfy the two equations _y_ = 3 _x_ − 11 and  , and we may write

from which we find that _x_ = 3. Therefore,

_y_ = 3 _x_ − 11 = 3(3) − 11 = −2

So the center is at (3,−2). The radius is the distance between the center and the point (3, 8):

Thus, the standard equation of the circle is (x − 3)2 \+ (y + 2)2 = 100.

**4.** Find the center and radius of the circle that passes through _P_ (1, 1) and is tangent to the line _y_ = 2 _x_ − 3 at the point _Q_ (3, 3).(See Fig. 4-3)

**Fig. 4-3**

The line   perpendicular to _y_ = 2 _x_ − 3 at (3, 3) must pass through the center of the circle. By Theorem 3.2, the slope of   is  . Therefore, the slope-intercept equation of   has the form  . Since (3, 3) is on  , we have  ; hence,  , and   has the equation

The perpendicular bisector   of chord _PQ_ in Fig. 4-3 also passes through the center of the circle, so the intersection of   and   will be the center of the circle. The slope of   is 1. Hence, by Theorem 3.2, the slope of   is −1. So   has the slope-intercept equation _y_ = − _x_ \+ b′. Since the midpoint (2, 2) of chord _PQ_ is a point on  , we have 2 = −(2) + b′ hence, b′ = 4, and the equation of   is y = −x + 4. We must find the common solution of y = −x + 4 and  . Setting

yields _x_ = −1. Therefore, _y_ = −x + 4 = −(−1) + 4 = 5, and the center _C_ of the circle is (−1, 5). The radius is the distance  . The standard equation of the circle is then (x + 1)2 \+ (y − 5) 2 = 20.

**5.** Find the standard equation of every circle that passes through the points P(1, −1) and _Q_ (3, 1) and is tangent to the line y = −3x.

Let _C(c, d)_ be the center of one of the circles, and let A be the point of tangency (see Fig. 4-4). Then, because  , we have

Expanding and simplifying, we obtain

**Fig. 4-4**

In addition,  , and by the formula of problem 8 in Chapter 3,  . Setting   thus yields  . Substituting ( _1_ ) in the right-hand side and multiplying by 10 then yields

10[(c − 1)2 \+ ( _d_ \+ 1)2] = (2 _c_ \+ 2)2 from which 3 _c_ 2 \+ 5 _d_ 2 − 14 _c_ \+ 10 _d_ \+ 8 = 0 By ( _1_ ), we can replace _d_ by 2 − _c_ , obtaining

2 _c_ 2 − 11c + 12 = 0 or (2 _c_ − 3)( _c_ − 4) = 0

Hence,   or _c_ = 4. Then ( _1_ ) gives us the two solutions  ,   and _c_ = 4, d = −2. Since the radius  , these solutions produce radii of  . Thus, there are two such circles, and their standard equations are

  and ( _x_ − 4)2 \+ ( _y_ \+ 2)2 = 10

### **Supplementary Problems**

**6.** Find the standard equations of the circles satisfying the following conditions:

( _a_ ) center at (3, 5) and radius 2

( _b_ ) center at (4, −1) and radius 1

( _c_ ) center at (5, 0) and radius

( _d_ ) center at (−2, −2) and radius

( _e_ ) center at (−2, 3) and passing through (3, −2)

( _f_ ) center at (6, 1) and passing through the origin.

_Ans._ ( _a_ ) (x − 3)2 \+ (y − 5)2 = 4; ( _b_ ) (x − 4)2 \+ (y + 1)2 = 1; ( _c_ ) (x − 5)2 \+ y2 = 3; ( _d_ ) (x + 2)2 \+ (y + 2)2 = 50; ( _e_ ) (x + 2)2 \+ (y − 3)2 = 50; ( _f_ ) (x − 6)2 \+ (y − 1)2 = 37

**7.** Identify the graphs of the following equations:

( _a_ ) _x_ 2 \+ _y_ 2 \+ 16 _x_ −12 _y_ \+ 10 = 0 ( _b_ ) _x_ 2 \+ _y_ 2 − 4 _x_ \+ 5 _y_ \+ 10 = 0 ( _c_ ) _x_ 2 \+ _y_ 2 \+ _x − y_ = 0 ( _d_ )4 _x_ 2 \+ 4 _y_ 2 \+ 8 _y_ − 3 = 0 ( _e_ ) _x_ 2 \+ _y_ 2 − _x_ − 2 _y_ \+ 3 = 0 ( _f_ ) _x_ 2 \+ _y_ 2 \+   − 2 = 0

_Ans._ ( _a_ ) circle, center at (−8, 6), radius  ; ( _b_ ) circle, center at  , radius ½; ( _c_ ) circle, center at  , radius  ; ( _d_ ) circle, center at (0, −1), radius  ; ( _e_ ) empty graph; ( _f_ ) circle, center at  , radius

**8.** Find the standard equations of the circles through ( _a_ ) (−2, 1), (1, 4), and (−3, 2): ( _b_ ) (0, 1), (2, 3), and  ; ( _c_ ) (6, 1), (2, −5), and (1, −4); ( _d_ ) (2, 3), (−6, −3), and (1, 4).

_Ans._ ( _a_ ) ( _x_ \+ 1)2 \+ ( _y_ − 3)2 = 5; ( _b_ ) ( _x_ − 2)2 \+ ( _y_ − 1)2 = 4; ( _c_ ) ( _x_ − 4)2 \+ ( _y_ \+ 2)2 = 13; ( _d_ ) ( _x_ \+ 2)2 \+ _y_ 2 = 25

**9.** For what values of _k_ does the circle ( _x + 2k_ )2 \+ ( _y − 3k_ )2 = 10 pass through the point (1, 0)?

_Ans._   or _k_ = −1

**10.** Find the standard equations of the circles of radius 2 that are tangent to both the lines _x_ = 1 and _y_ = 3

_Ans._ ( _x_ \+ 1)2 \+ ( _y_ −1)2 = 4; ( _x_ \+ 1)2 \+ ( _y_ − 5)2 = 4; ( _x_ − 3)2 \+ ( _y_ − 1)2 = 4; ( _x_ − 3)2 \+ ( _y_ − 5)2 = 4

**11.** Find the value of _k_ so that _x_ 2 \+ _y_ 2 \+ 4 _x_ −6 _y_ \+ _k_ = 0 is the equation of a circle of radius 5.

_Ans. k_ = −12

**12.** Find the standard equation of the circle having as a diameter the segment joing (2, −3) and (6, 5).

_Ans._ ( _x_ − 4)2 \+ ( _y_ − 1)2 = 20

**13.** Find the standard equation of every circle that passes through the origin, has radius, has radius 5, and is such that the y coordinate of its center is −4.

_Ans._ ( _x_ − 3)2 \+ ( _y_ \+ 4)2 = 25 or ( _x_ \+ 3)2 \+ ( _y_ \+ 4)2 = 25

**14.** Find the standard equation of the circle that passes through the points (8, −5) and (−1, 4) and has its center on the line 2x + 3y = 3.

_Ans._ ( _x_ − 3)2 \+ ( _y_ \+ 1)2 = 41

**15.** Find the standard equation of the circle with center (3, 5) that is tangent to the line 12 _x_ − 5 _y_ \+ 2 = 0.

_Ans._ ( _x_ − 3)2 \+ ( _y_ − 5)2 = 1

**16.** Find the standard equation of the circle that passes through the point   and is tangent to the line _x_ \+ _y_ = 2 at (2, 0).

_Ans._ ( _x_ − 5)2 \+ ( _y_ − 3)2 = 18

**17.** Prove analytically that an angle inscribed in a semicircle is a right angle. (See Fig. 4-5.).

**18.** Find the length of a tangent from (6,−2) to the circle (x − 1)2 \+ (y − 3)2 = 1. (See Fig. 4-6).

_Ans._ 7

**Fig. 4-5**

**Fig. 4-6**

**19.** Find the standard equations of the circles that pass through (2, 3) and are tangent to both the lines 3 _x_ − 4 _y_ = −1 and 4 _x_ \+ 3 _y_ = 7.

_Ans._ (x − 2)2 \+ (y − 8)2 = 25 and

**20.** Find the standard equations of the circles that have their centers on the line 4x + 3y = 8 and are tangent to both the lines _x + y_ = −2 and 7 _x − y_ = −6.

_Ans._ ( _x_ − 2)2 \+ _y_ 2 = 2 and ( _x_ \+ 4)2 \+ ( _y_ − 8)2 = 18

**21.** Find the standard equation of the circle that is concentric with the circle _x_ 2 \+ _y_ 2 − 2 _x_ − 8 _y_ \+ 1 = 0 and is tangent to the line 2 _x − y_ = 3.

_Ans._ ( _x_ − 1)2 \+ ( _y_ − 4)2 = 5

**22.** Find the standard equations of the circles that have radius 10 and are tangent to the circle _x_ 2 \+ y2 = 25 at the point (3, 4).

_Ans._ ( _x_ − 9)2 \+ ( _y_ − 12)2 = 100 and ( _x_ \+ 3)2 \+ ( _y_ \+ 4)2 = 100

**23.** Find the longest and shortest distances from the point (7, 12) to the circle _x_ 2 \+ _y_ 2 \+ 2 _x_ \+ 6 _y_ − 15 = 0.

_Ans._ 22 and 12

**24.** Let  1 and  2 be two intersecting circles determined by the equations _x_ 2 \+ _y_ 2 \+ A1 _x_ \+ B1 _y_ \+ C1 = 0 and _x_ 2 \+ _y_ 2 \+ A2 _x_ \+ B2 _y_ \+ C2 = 0. For any number k≠ −1, show that

_x_ 2 \+ _y_ 2 \+ A1 _x_ \+ B1 _y_ \+ C1 \+ _k_ ( _x_ 2 \+ _y_ 2 \+ A2 _x_ \+ B2 _y_ \+ C2) = 0

is the equation of a circle through the intersection points of  1 and  2. Show, conversely, that every such circle may be represented by such an equation for a suitable _k_.

**25.** Find the standard equation of the circle passing through the point (− 3, 1) and containing the points of intersection of the circles _x_ 2 \+ _y_ 2 \+ 5 _x_ = 1 and _x_ 2 \+ _y_ 2 \+ y = 7.

_Ans._

**26.** Find the standard equations of the circles that have centers on the line 5 _x_ − 2 _y_ = −21 and are tangent to both coordinate axes.

_Ans._ ( _x_ \+ 7)2 \+ ( _y_ \+ 7)2 = 49 and ( _x_ \+ 3)2 \+ ( _y_ − 3)2 = 9

**27.** ( _a_ ) If two circles _x_ 2 \+ _y_ 2 \+ A1 _x_ \+ B1 _y_ \+ C1 = 0 and _x_ 2 \+ _y_ 2 \+ A2 _x_ \+ B2 _y_ \+ C2 = 0 intersect at two points, find an equation of the line through their points of intersection.

( _b_ ) Prove that if two circles intersect at two points, then the line through their points of intersection is perpendicular to the line through their centers.

_Ans._ ( _a_ ) (A1 − A2) _x_ \+ (A1 − B2) _y_ \+ (C1 − C2) = 0

**28.** Find the points of intersection of the circles _x_ 2 \+ y2 \+ 8y − 64 = 0 and _x_ 2 \+ _y_ 2 − 6 _x_ − 16 = 0.

_Ans._ (8, 0) and

**29.** Find the equations of the lines through (4, 10) and tangent to the circle _x_ 2 \+ _y_ 2 − 4 _y_ − 36 = 0.

_Ans. y_ = −3 _x_ \+ 22 and _x_ − 3 _y_ \+ 26 = 0

**30.** (GC) Use a graphing calculator to draw the circles in Problems 7( _d_ ), 10, 14, and 15. ( _Note_ : It may be necessary to solve for _y_.)

**31.** (GC) ( _a_ ) Use a graphing calculator to shade the interior of the circle with center at the origin and radius 3.

( _b_ ) Use a graphing calculator to shade the exterior of the circle _x_ 2 \+ ( _y_ − 2)2 = 1.

**32.** (GC) Use a graphing calculator to graph the following inequalities: ( _a_ ) ( _x_ − 1)2 \+ _y_ 2 < 4; ( _b_ ) _x_ 2 \+ _y_ 2 − 6 _x_ − 8 _y_ > 0.

## **CHAPTER 5
Equations and Their Graphs**

**THE GRAPH OF AN EQUATION** involving _x_ and _y_ as its only variables consists of all points ( _x_ , _y_ ) satisfying the equation.

**EXAMPLE 1** : ( _a_ ) What is the graph of the equation 2 _x − y_ = 3?

The equation is equivalent to _y_ = 2 _x_ − 3, which we know is the slope–intercept equation of the line with slope 2 and _y_ intercept −3.

( _b_ ) What is the graph of the equation _x_ 2 \+ _y_ 2 − 2 _x_ \+ 4 _y_ − 4 = 0?

Completing the square shows that the given equation is equivalent to the equation ( _x_ − 1)2 \+ ( _y_ \+ 2)2 = 9. Hence, its graph is the circle with center (1, −2) and radius 3.

**PARABOLAS.** Consider the equation _y_ = _x_ 2. If we substitute a few values for _x_ and calculate the associated values of _y_ , we obtain the results tabulated in Fig. 5-1. We can plot the corresponding points, as shown in the figure. These points suggest the heavy curve, which belongs to a family of

**Fig. 5-1**

curves called _parabolas_. In particular, the graphs of equations of the form _y_ = _cx_ 2, where c is a nonzero constant, are parabolas, as are any other curves obtained from them by translations and rotations.

In Fig. 5-1, we note that the graph of _y_ = _x_ 2 contains the origin (0, 0) but all its other points lie above the _x_ axis, since _x_ 2 is positive except when _x_ = 0. When _x_ is positive and increasing, _y_ increases without bound. Hence, in the first quadrant, the graph moves up without bound as it moves right. Since (− _x_ )2 = _x_ 2, it follows that, if any point ( _x_ , _y_ ) lies on the graph in the first quadrant, then the point (− _x_ , _y_ ) also lies on the graph in the second quadrant. Thus, the graph is symmetric with respect to the _y_ axis. The _y_ axis is called the _axis of symmetry_ of this parabola.

**ELLIPSES.** To construct the graph of the equation  , we again compute a few values and plot the corresponding points, as shown in Fig. 5-2. The graph suggested by these points is also drawn in the figure; it is a member of a family of curves called _ellipses_. In particular, the graph of an equation of the form   is an ellipse, as is any curve obtained from it by translation or rotation.

Note that, in contrast to parabolas, ellipses are bounded. In fact, if ( _x_ , _y_ ) is on the graph of   then  , and, therefore, _x_ 2   9. Hence, −3   _x_   3. So, the graph lies between the vertical lines _x_ = −3 and _x_ = 3. Its rightmost point is (3, 0), and its leftmost point is (−3, 0). A similar argument shows that the graph lies between the horizontal lines _y_ = −2 and _y_ = 2, and that its lowest point is (0, −2) and its highest point is (0, 2). In the first quadrant, as _x_ increases from 0 to 3, _y_ decreases from 2 to 0. If ( _x_ , _y_ ) is any point on the graph, then (− _x_ , _y_ ) also is on the graph. Hence, the graph is symmetric with respect to the _y_ axis. Similarly, if ( _x_ , _y_ ) is on the graph, so is ( _x_ , − _y_ ), and therefore the graph is symmetric with respect to the _x_ axis.

**Fig. 5-2**

When _a_ = _b_ , the ellipse   is the circle with the equation _x_ 2 \+ _y_ 2 = _a_ 2, that is, a circle with center at the origin and radius _a_. Thus, circles are special cases of ellipses.

**HYPERBOLAS.** Consider the graph of the equation  . Some of the points on this graph are tabulated and plotted in Fig. 5-3. These points suggest the curve shown in the figure, which is a member of a family of curves called _hyperbolas_. The graphs of equations of the form   are hyperbolas, as are any curves obtained from them by translations and rotations.

**Fig. 5-3**

Let us look at the hyperbola   in more detail. Since  , it follows that _x_ 2   9, and therefore, | _x_ |   3. Hence, there are no points on the graph between the vertical lines _x_ = −3 and _x_ = 3. If ( _x_ , _y_ ) is no the graph, so is (− _x_ , _y_ ); thus, the graph is symmetric with respect to the _y_ axis. Similarly, the graph is symmetric with respect to the _x_ axis. In the first quadrant, as _x_ increases, _y_ increases without bound.

**Fig. 5-4**

Note the dashed lines in Fig. 5-3; they are the lines   and  , and they are called the _asymptotes_ of the hyperbola: Points on the hyperbola get closer and closer to these asymptotes as they recede from the origin. In general, _the asymptotes of the hyperbola_   _are the lines_  .

**CONIC SECTIONS.** Parabolas, ellipses, and hyperbolas together make up a class of curves called _conic sections_. They can be defined geometrically as the intersections of planes with the surface of a right circular cone, as shown in Fig. 5-4.

### **Solved Problems**

**1.** Sketch the graph of the _cubic curve y = x_ 3.

The graph passes through the origin (0, 0). Also, for any point ( _x_ , _y_ ) on the graph, _x_ and _y_ have the same sign; hence, the graph lies in the first and third quadrants. In the first quadrant, as _x_ increases, _y_ increases without bound. Moreover, if ( _x_ , _y_ ) lies on the graph, then (− _x_ , − _y_ ) also lies on the graph. Since the origin is the midpoint of the segment connecting the points ( _x_ , _y_ ) and (− _x_ , − _y_ ), the graph is symmetric with respect to the origin. Some points on the graph are tabulated and shown in Fig. 5-5; these points suggest the heavy curve in the figure.

**Fig. 5-5**

**2.** Sketch the graph of the equation _y_ = − _x_ 2.

If ( _x_ , _y_ ) is on the graph of the parabola _y_ = _x_ 2 (Fig. 5-1), then ( _x_ , − _y_ ) is on the graph of _y_ = − _x_ 2, and vice versa. Hence, the graph of _y_ = − _x_ 2 is the reflection in the _x_ axis of the graph of _y_ = _x_ 2. The result is the parabola in Fig. 5-6.

**3.** Sketch the graph of _x_ = _y_ 2.

This graph is obtained from the parabola _y_ = _x_ 2 by exchanging the roles of _x_ and _y_. The resulting curve is a parabola with the _x_ axis as its axis of symmetry and its "nose" at the origin (see Fig. 5-7). A point ( _x_ , _y_ ) is on the graph of _x_ = _y_ 2 if and only if ( _y_ , _x_ ) is on the graph of _y_ = _x_ 2. Since the segment connecting the points

**Fig. 5-6**

( _x_ , _y_ ) and ( _y_ , _x_ ) is perpendicular to the diagonal line _y_ = _x_ (why?), and the midpoint   of that segment is on the line _y_ = _x_. (see Fig. 5-8), the parabola _x_ = _y_ 2 is obtained from the parabola _y_ = _x_ 2 by reflection in the line _y_ = _x_

**Fig. 5-7**

**Fig. 5-8**

**4.** Let   be a line, and let _F_ be a point not on  . Show that the set of all points equidistant from _F_ and   is a parabola.

Construct a coordinate system such that _F_ lies on the positive _y_ axis, and the _x_ axis is parallel to   and halfway between _F_ and  . (See Fig. 5-9.) Let 2 _p_ be the distance between _F_ and  . Then   has the equation _y_ = − _p_ , and the coordinates of _F_ are (0, _p_ ).

Consider an arbitrary point _P_ ( _x_ , _y_ ). Its distance from   is | _y_ \+ _p_ |, and its distance from _F_ is  . Thus, for the point to be equidistant from _F_ and   we must have  . Squaring yields ( _y_ \+ _p_ )2 = _x_ 2 \+ ( _y_ − _p_ )2, from which we find that 4 _py_ = _x_ 2. This is the equation of a parabola with the _y_ axis as its axis of symmetry. The point _F_ is called the _focus_ of the parabola, and the line   is called its _directrix_. The chord _AB_ through the focus and parallel to   is called the _latus rectum_. The "nose" of the parabola at (0, 0) is called its _vertex_.

**Fig. 5-9**

**5.** Find the length of the latus rectum of a parabola 4 _py_ = _x_ 2.

The _y_ coordinate of the endpoints _A_ and _B_ of the lactus rectum (see Fig. 5-9) is _p_. Hence, at these points, 4 _p_ 2 = _x_ 2 and, therefore, _x_ = ±2 _p_. Thus, the length _AB_ of the latus rectum is 4 _p_.

**6.** Find the focus, directrix, and the length of the latus rectum of the parabola  , and draw its graph.

The equation of the parabola can be written as 2 _y_ = _x_ 2. Hence, 4 _p_ = 2 and  . Therefore, the focus is at  , the equation of the directix is  , and the length of the latus rectum is 2. The graph is shown in Fig. 5-10.

**Fig. 5-10**

**7.** Let _F_ and F′ be two distinct points at a distance 2 _c_ from each other. Show that the set of all points _P_ ( _x_ , _y_ ) such that  , a > _c_ , is an ellipse.

Construct a coordinate system such that the _x_ axis passs through _F_ and _F_ ′, the origin is the midpoint of the segment _FF_ ′, and _F_ lies on the positive _x_ axis. Then the coordinates of _F_ and _F_ ′ are ( _c_ , 0) and (− _c_ , 0).

**Fig. 5-11**

(See Fig. 5-11.) Thus, the condition   is equivalent to  . After rearranging and squaring twice (to eliminate the square roots) and performing indicated operations, we obtain

Since _a_ > _c_ , _a_ 2 − _c_ > 0. Let  . Then ( _1_ ) becomes b2 _x_ 2 = a2 _b_ 2, which we may rewrite as  , the equation of an ellipse.

When _y_ = 0, _x_ 2 = _a_ 2; hence, the ellipse intersects the _x_ axis at the points _A_ ′(− _a_ , 0), and _A_ ( _a_ , 0), called the _vertices_ of the ellipse (Fig. 5-11). The segment _A_ ′ _A_ is called the _major axis;_ the segment _OA_ is called the _semimajor axis_ and has length _a_. The origin is the _center_ of the ellipse. _F_ and _F_ ′ are called the _foci_ (each is a _focus_ ). When _x_ = 0, _y_ 2 = _b_ 2. Hence, the ellipse intersects the _y_ axis at the points _B_ ′(0, − _b_ ) and _B_ (0, _b_ ). The segment _B_ ′ _B_ is called the _minor axis_ ; the segment _OB_ is called the _semiminor axis_ and has length _b_. Note that  . Hence, the semiminor axis is smaller than the semimajor axis. The basic relation among _a_ , _b_ , and _c_ is _a_ 2 = _b_ 2 \+ _c_ 2.

The _eccentricity_ of an ellipse is defined to be _e_ = _c_ / _a_. Note that 0 < _e_ < 1. Moreover,  . Hence, when _e_ is very small, _b_ / _a_ is very close to 1, the minor axis is close in size to the major axis, and the ellipse is close to being a circle. On the other hand, when _e_ is close to 1, _b_ / _a_ is close to zero, the minor axis is very small in comparison with the major axis, and the ellipse is very "flat."

**8.** Identify the graph of the equation 9 _x_ 2 \+ 16 _y_ 2 = 144.

The given equation is equivalent to _x_ 2/16 + _y_ 2/9 = 1. Hence, the graph is an ellipse with semimajor axis of length _a_ = 4 and semiminor axis of length _b_ = 3. (See Fig. 5-12.) The vertices are (−4, 0) and (4, 0). Since   the eccentricity _e_ is

**9.** Identify the graph of the equation 25 _x_ 2 \+ 4 _y_ 2 = 100

The given equation is equivalent to _x_ 2/4 + _y_ 2/25 = 1, an ellipse. Since the denominator under _y_ 2 is larger than the denominator under _x_ 2, the graph is an ellipse with the major axis on the _y_ axis and the minor axis on the _x_ axis (see Fig. 5-13). The vertices are at (0, −5) and (0, 5). Since  , the eccentricity is

**Fig. 5-12**

**Fig. 5-13**

**10.** Let _F_ and _F_ ′ be distinct points at a distance of 2 _c_ from each other. Find the set of all points _P_ ( _x_ , _y_ ) such that  , for _a_ < _c_.

Choose a coordinate system such that the _x_ axis passes through _F_ and _F_ ′, with the origin as the midpoint of the segment _FF_ ′ and with _F_ on the positive _x_ axis (see Fig. 5-14). The coordinates of _F_ and _F_ ′ are ( _c_ , 0) and (− _c_ , 0). Hence, the given condition is equivalent to  . After manipulations required to eliminate the square roots, this yields

Since _c_ > _a_ , _c_ 2 − _a_ 2 > 0. Let  . (Notice that _a_ 2 \+ _b_ 2 = _c_ 2.) Then (1) becomes _b_ 2 _x_ 2 − _a_ 2 _y_ 2 = _a_ 2 _b_ 2, which we rewrite as  , the equation of a hyperbola.

When _y_ = 0, _x_ = ± _a_. Hence, the hyperbola intersects the _x_ axis at the points _A_ ′(− _a_ , 0) and _A_ ( _a_ , 0), which are called the _vertices_ of the hyperbola. The asymptotes are  . The segment _A_ ′ _A_ is called the _transverse axis_. The segment connecting the points (0, −b) and (0, b) is called the _conjugate axis_. The _center_ of the hyperbola is the origin. The points _F_ and _F_ ′ are called the _foci_. The _eccentricity_ is defined to be

**Fig. 5-14**

 . Since _c_ > _a_ , _e_ > 1. When _e_ is close to 1, _b_ is very small relative to _a_ , and the hyperbola has a very printed "nose"; when _e_ is very large, _b_ is very large relative to _a_ , and the hyperbola is very "flat".

**11.** Identify the graph of the equation 25 _x_ 2 − 16 _y_ 2 = 400.

The given equation is equivalent to _x_ 2/16 − _y_ 2/25 = 1. This is the equation of a hyperbola with the _x_ axis as its transverse axis, vertices (−4, 0) and (4, 0), and asymptotes   (See Fig. 5.15.)

**Fig. 5-15**

**12.** Identify the graph of the equation _y_ 2 − 4 _x_ 2 = 4.

The given equation is equivalent to  . This is the equation of a hyperbola, with the roles of _x_ and _y_ interchanged. Thus, the transverse axis is the _y_ axis, the conjugate axis is the _x_ axis, and the vertices are (0, −2) and (0, 2). The asymptotes are   or, equivalently,  . (See Fig. 5-16.)

**13.** Identify the graph of the equation _y_ = ( _x_ − 1)2.

A point ( _u_ , _v_ ) is on the graph of _y_ = ( _x_ − 1)2 if and only if the point ( _u_ − 1, _v_ ) is on the graph of _y_ = _x_ 2. Hence, the desired graph is obtained from the parabola _y_ = _x_ 2 by moving each point of the latter one unit to the right. (See Fig. 5-17.)

**14.** Identify the graph of the equation  .

A point ( _u_ , _v_ ) is on the graph if and only if the point ( _u_ − 1, _v_ − 2) is on the graph of the equation _x_ 2/4 + _y_ 2/9 = 1. Hence, the desired graph is obtained by moving the ellipse _x_ 2/4 + _y_ 2/9 = 1 one unit to the right and two units upward. (See Fig. 5-18.) The center of the ellipse is at (1, 2), the major axis is along the line _x_ = 1, and the minor axis is along the line _y_ = 2.

**Fig. 5-16**

**Fig. 5-17**

**Fig. 5-18**

**15.** How is the graph of an equation _F_ ( _x_ − _a_ , _y_ − _b_ ) = 0 related to the graph of the equation _F_ ( _x_ , _y_ ) = 0?

A point ( _u_ , _v_ ) is on the graph of _F_ ( _x_ − _a_ , _y_ − _b_ ) = 0 if and only if the point ( _u_ − _a_ , _v_ − _b_ ) is on the graph of _F_ ( _x_ , _y_ ) = 0. Hence, the graph of _F_ ( _x_ − _a_ , _y_ − _b_ ) = 0 is obtained by moving each point of the graph of _F_ ( _x_ , _y_ ) = 0 by _a_ units to the right and _b_ units upward. (If _a_ is negative, we move the point | _a_ | units to the left. If _b_ is negative, we move the point | _b_ | units downward.) Such a motion is called a _translation_.

**16.** Identify the graph of the equation _y_ = _x_ 2 − 2 _x_.

Competing the square in _x_ , we obtain _y_ \+ 1 = ( _x_ − 1)2. Based on the results of Problem 15, the graph is obtained by a translation of the parabola _y_ = _x_ 2 so that the new vertex is (1, −1). Notice that _y_ \+ 1 is _y_ − (−1).] It is shown in [Fig. 5-19.

**Fig. 5-19**

**17.** Identify the graph of 4 _x_ 2 − 9 _y_ 2 − 16 _x_ \+ 18 _y_ − 29 = 0.

Factoring yields 4( _x_ 2 − 4 _x_ ) − 9( _y_ 2 − 2 _y_ ) − 29 = 0, and then completing the square in _x_ and _y_ produces 4( _x_ − 2)2 − 9( _y_ − 1)2 = 36. Dividing by 36 then yields  . By the results of Problem 15, the graph of this equation is obtained by translating the hyperbola   two units to the right and one unit upward, so that the new center of symmetry of the hyperbola is (2, 1). (See Fig. 5-20.)

**18.** Draw the graph of the equation _xy_ = 1.

Some points of the graph are tabulated and plotted in Fig. 5-21. The curve suggested by these points is shown dashed as well. It can be demonstrated that this curve is a hyperbola with the line _y_ = _x_ as transverse

**Fig. 5-20**

**Fig. 5-21**

axis, the line _y_ = − _x_ as converse axis, vertices (−1, −1) and (1, 1), and the _x_ axis and _y_ axis as asymptotes. Similarly, the graph of any equation _xy_ = _d_ , where _d_ is a positive constant, is a hyperbola with _y_ = _x_ as transverse axis and _y_ = − _x_ as converse axis, and with the coordinate axes as asymptotes. Such hyperbolas are called _equilateral hyperbolas_. They can be shown to be rotations of hyperbolas of the form _x_ 2/ _a_ 2 − _y_ 2/ _a_ 2 = 1.

### **Supplementary Problems**

**19.** ( _a_ ) On the same sheet of paper, draw the graphs of the following parabolas: ( _i_ ) _y_ = 2 _x_ 2; ( _ii_ ) _y_ = 3 _x_ 2; ( _iii_ ) _y_ = 4 _x_ 2; ( _iv_ )  ; ( _v_ )  .

( _b_ ) (GC) Use a graphing calculator to check your answers to ( _a_ ).

**20.** ( _a_ ) On the same sheet of paper, draw the graphs of the following parabolas and indicate points of intersection: ( _i_ ) _y_ = _x_ 2; ( _ii_ ) _y_ = − _x_ 2; ( _iii_ ) _x_ = _y_ 2; ( _iv_ ) _x_ = − _y_ 2.

( _b_ ) (GC) Use a graphing calculator to check your answers to ( _a_ ).

**21.** Draw the graphs of the following equations:

( _a_ ) _y_ = _x_ 3 − 1

( _b_ ) _y_ = ( _x_ − 2)3

( _c_ ) _y_ = ( _x_ \+ 1)3 − 2

( _d_ ) _y_ = − _x_ 3

( _e_ ) _y_ = −( _x_ − 1)3

( _f_ ) _y_ = −( _x_ − 1)3 \+ 2

**22.** (GC) Use a graphing calculator to answer Problem 21.

**23.** Identify and draw the graphs of the following equation:

( _a_ ) _y_ 2 − _x_ 2 = 1

( _b_ ) 25 _x_ 2 \+ 36 _y_ 2 = 900

( _c_ ) 2 _x_ 2 − _y_ 2 = 4

( _d_ ) _xy_ = 4

( _e_ ) 4 _x_ 2 \+ 4 _y_ 2 = 1

( _f_ ) 8 _x_ = _y_ 2

( _g_ ) 10 _y_ = _x_ 2

( _h_ ) 4 _x_ 2 \+ 9 _y_ 2 = 16

( _i_ ) _xy_ = −1

( _j_ ) 3 _y_ 2 − _x_ 2 = 9

_Ans_. ( _a_ ) hyperbola, _y_ axis as transverse axis, vertices  , asymptotes  ; ( _b_ ) ellipse, vertices   foci  ; ( _c_ ) hyperbola, _x_ axis as transverse axis, vertices  , asymptotes  ; ( _d_ ) hyperbola, _y_ = _x_ as transverse axis, vertices (2, 2) and (−2, −2), _x_ and _y_ axes as asymptotes; ( _e_ ) circle, center (0, 0), radius  ; ( _f_ ) parabola, vertex (0, 0), focus (2, 0), directrix _x_ = −2; ( _g_ ) parabola, vertex (0, 0), focus  , directrix  ; ( _h_ ) ellipse, vertices  , foci  ; ( _i_ ) hyperbola, _y_ = − _x_ as transverse axis, vertices (−1, 1) and (1, −1), _x_ and _y_ axes as asymptotes; ( _j_ ) hyperbola, _y_ axis as transverse axis, vertices  , asymptotes

**24.** (GC) Use a graphing calculator to draw the graphs in Problem 23.

**25.** Identify and draw the graphs of the following equations:

( _a_ ) 4 _x_ 2 − 3 _y_ 2 \+ 8 _x_ \+ 12 _y_ − 4 = 0

( _b_ ) 5 _x_ 2 \+ _y_ 2 − 20 _x_ \+ 6 _y_ \+ 25 = 0

( _c_ ) _x_ 2 − 6 _x_ − 4 _y_ \+ 5 = 0

( _d_ ) 2 _x_ 2 \+ _y_ 2 − 4 _x_ \+ 4 _y_ \+ 6 = 0

( _e_ ) 3 _x_ 2 \+ 2 _y_ 2 \+ 12 _x_ − 4 _y_ \+ 15 = 0

( _f_ ) ( _x_ − 1)( _y_ \+ 2) = 1

( _g_ ) _xy_ − 3 _x_ − 2 _y_ \+ 5 = 0 [ _Hint_ : Compare ( _f_ ).]

( _h_ ) 4 _x_ 2 \+ _y_ 2 \+ 8 _x_ \+ 4 _y_ \+ 4 = 0

( _i_ ) 2 _x_ 2 − 8 _x_ − _y_ \+ 11 = 0

( _j_ ) 25 _x_ 2 \+ 16 _y_ 2 − 100 _x_ − 32 _y_ − 284 = 0

_Ans_. ( _a_ ) empty graph; ( _b_ ) ellipse, center at (2, −3); ( _c_ ) parabola, vertex at (3, −1); ( _d_ ) single point (1, −2); ( _e_ ) empty graph; ( _f_ ) hyperbola, center at (1, −2); ( _g_ ) hyperbola, center at (2, 3); ( _h_ ) ellipse, center at (−1, 2); ( _i_ ) parabola, vertex at (2, 3); ( _j_ ) ellipse, center at (2, 1)

**26.** (GC) Use a graphing calculator to draw the graphs in Problem 25.

**27.** Find the focus, directrix, and length of the latus rectum of the following parabolas: ( _a_ ) 10 _x_ 2 = 3 _y_ ; ( _b_ ) 2 _y_ 2 = 3 _x_ ; ( _c_ ) 4 _y_ = _x_ 2 \+ 4 _x_ \+ 8; ( _d_ ) 8 _y_ = − _x_ 2.

_Ans_. ( _a_ ) focus at  , directrix  , latus rectum  ;

( _b_ ) focus at  , directrix  , latus rectum  ;

( _c_ ) focus at (−2, 2), directrix _y_ = 0, latus rectum 4;

( _d_ ) focus at (0, −2), directrix _y_ = 2, latus rectum 8

**28.** Find an equation for each parabola satisfying the following conditions:

( _a_ ) Focus at (0, −3), directrix _y_ = 3

( _b_ ) Focus at (6, 0), directrix _x_ = 2

( _c_ ) Focus at (1, 4), directrix _y_ = 0

( _d_ ) Vertex at (1, 2) focus at (1, 4)

( _e_ ) Vertex at (3, 0), directrix _x_ = 1

( _f_ ) Vertex at the origin, _y_ axis as axis of symmetry, contains the point (3, 18)

( _g_ ) Vertex at (3, 5), axis of symmetry parallel to the _y_ axis, contains the point (5, 7)

( _h_ ) Axis of symmetry parallel to the _x_ axis, contains the points (0, 1), (3, 2), (1, 3)

( _i_ ) Latus rectum is the segment joining (2, 4) and (6, 4), contains the point (8, 1)

( _j_ ) Contains the points (1, 10) and (2, 4), axis of symmetry is vertical, vertex is on the line 4 _x_ − 3 _y_ = 6

_Ans_. ( _a_ ) 12 _y_ = − _x_ 2; ( _b_ ) 8( _x_ − 4) = _y_ 2; ( _c_ ) 8( _y_ − 2) = ( _x_ − 1)2; ( _d_ ) 8( _y_ − 2) = ( _x_ − 1)2; ( _e_ ) 8( _x_ − 3) = _y_ 2; ( _f_ ) _y_ = 2 _x_ 2; ( _g_ ) 2( _y_ − 5) = ( _x_ − 3)2;  ; ( _i_ ) 4( _y_ − 5) = −( _x_ − 4)2;

**29.** Find an equation for each ellipse satisfying the following conditions:

( _a_ ) Center at the origin, one focus at (0, 5), length of semimajor axis is 13.

( _b_ ) Center at the origin, major axis on the _y_ axis, contains the points   and  .

( _c_ ) Center at (2, 4), focus at (7, 4), contains the point (5, 8).

( _d_ ) Center at (0, 1), one vertex at (6, 1), eccentricity  .

( _e_ ) Foci at  , contains  .

( _f_ ) Foci (0, ±9), semiminor axis of length 12.

_Ans_. ( _a_ )  ; ( _b_ )  ; ( _c_ )  ; ( _d_ )  ; ( _e_ )  ; ( _f_ )

**30.** Find an equation for each hyperbola satisfying the following conditions:

( _a_ ) Center at the origin, transverse axis the _x_ axis, contains the points (6, 4) and (−3, 1)

( _b_ ) Center at the origin, one vertex at (3, 0), one asymptote is

( _c_ ) Has asymptotes  , contains the point (1, 2)

( _d_ ) Center at the origin, one focus at (4, 0), one vertex at (3, 0)

_Ans_. ( _a_ )  ; ( _b_ )  ; ( _c_ )  ; ( _d_ )

**31.** Find an equation of the hyperbola consisting of all points _P_ ( _x_ , _y_ ) such that  , where   and  .

_Ans_. _xy_ = 1

**32.** (GC) Use a graphing calculator to draw the hyperbola   and its asymptotes  .

**33.** (GC) Use a graphing calculator to draw the ellipses _x_ 2 \+ 4 _y_ 2 = 1 and ( _x_ − 3)2 \+ 4( _y_ − 2)2 = 1. How is the latter graph obtained from the former one?

## **CHAPTER 6
Function**

We say that a quantity _y_ is a _function_ of some other quantity _x_ if the value of _y_ is determined by the value of _x_. If _f_ denotes the function, then we indicate the dependence of _y_ on _x_ by means of the formula _y_ = _f_ ( _x_ ). The letter _x_ is called the _independent variable_ , and the letter _y_ is called the _dependent variable_. The independent variable is also called the _argument_ of the function, and the dependent variable is called the _value_ of the function.

For example, the area _A_ of a square is a function of the length _s_ of a side of the square, and that function can be expressed by the formula _A_ = _s_ 2. Here, _s_ is the independent variable and _A_ is the dependent variable.

The _domain_ of a function is the set of numbers to which the function can be applied, that is, the set of numbers that are assigned to the independent variable. The _range_ of a function is the set of numbers that the function associates with the numbers in the domain.

**EXAMPLE 1:** The formula _f_ ( _x_ ) = _x_ 2 determines a function _f_ that assigns to each real number _x_ its square. The domain consists of all real numbers. The range can be seen to consist of all nonnegative real numbers. (In fact, each value _x_ 2 is nonnegative. Conversely, if _r_ is any nonnegative real number, then _r_ appears as a value when the function is applied to  , since  .)

**EXAMPLE 2:** Let _g_ be the function defined by the formula _g_ ( _x_ ) = _x_ 2 − 4 _x_ \+ 2 for all real numbers. Thus,

_g_ (1) = (1)2 − 4(1) + 2 = 1 − 4 + 2 = −1

and

_g_ (−2) = (−2)2 − 4(−2) + 2 = 4 + 8 + 2 = 14

Also, for any number _a_ , _g_ ( _a_ \+ 1) = ( _a_ \+ 1)2 − 4( _a_ \+ 1) + 2 = _a_ 2 \+ 2 _a_ \+ 1 − 4 _a_ − 4 + 2 = _a_ 2 − 2 _a_ − 1.

**EXAMPLE 3:** ( _a_ ) Let the function _h_ ( _x_ ) = 18 _x_ − 3 _x_ 2 be defined for all real numbers _x_. Thus, the domain is the set of all real numbers. ( _b_ ) Let the area _A_ of a certain rectangle, one of whose sides has length _x_ , be given by _A_ = 18 _x_ − 3 _x_ 2. Both _x_ and _A_ must be positive. Now, by completing the square, we obtain

_A_ = −3( _x_ 2 − 6 _x_ ) = −3[( _x_ − 3)2 − 9] = 27 − 3( _x_ − 3)2

Since _A_ > 0, 3( _x_ − 3)2 < 27, ( _x_ − 3)2 < 9, | _x_ − 3| < 3. Hence, −3 < _x_ − 3 < 3, 0 < _x_ < 6. Thus, the function determining _A_ has the open interval (0, 6) as its domain. The graph of _A_ = 27 − 3( _x_ − 3)2 is the parabola shown in Fig. 6-1. From the graph, we see that the range of the function is the half-open interval (0, 27).

Notice that the function of part ( _b_ ) is given by the same formula as the function of part ( _a_ ), but the domain of the former is a proper subset of the domain of the latter.

**Fig. 6-1**

The graph of a function _f_ is defined to be the graph of the equation _y_ = _f_ ( _x_ )

**EXAMPLE 4:** ( _a_ ) Consider the function _f_ ( _x_ ) = | _x_ |. Its graph is the graph of the equation _y_ = | _x_ |, and is indicated in Fig. 6-2. Notice that _f_ ( _x_ ) = _x_ when _x_   0, whereas _f_ ( _x_ ) = − _x_ when _x_   0. The domain of _f_ consists of all real numbers. ( _In general, if a function is given by means of a formula, then, if nothing is said to the contrary, we shall assume that the domain consists of all numbers for which the formula is defined_.) From the graph in Fig. 6-2, we see that the range of the function consists of all nonnegative real numbers. ( _In general, the range of a function is the set of y coordinates of all points in the graph of the function_.) ( _b_ ) The formula _g_ ( _x_ ) = 2 _x_ \+ 3 defines a function _g_. The graph of this function is the graph of the equation _y_ = 2 _x_ \+ 3, which is the straight line with slope 2 and _y_ intercept 3. The set of all real numbers is both the domain and range of _g_.

**Fig. 6-2**

**EXAMPLE 5:** Let a function _g_ be defined as follows:

A function defined in this way is said to be _defined by cases_. Notice that the domain of _g_ is the closed interval [1, 4]

In a rigorous development of mathematics, a function _f_ is defined to be a set of ordered pairs such that, if ( _x_ , _y_ ) and ( _x_ , _z_ ) are in the set _f_ , then _y_ = _z_. However, such a definition obscures the intuitive meaning of the notion of function.

### **Solved Problems**

**1.** Given  , find ( _a_ ) _f_ (0); ( _b_ ) _f_ (−1); ( _c_ ) _f_ (2 _a_ ); ( _d_ ) _f_ (1/ _x_ ); ( _e_ ) _f_ ( _x_ \+ _h_ )

( _a_ )

( _b_ )

( _c_ )

( _d_ )

( _e_ )

**2.** If _f_ ( _x_ ) = 2x, show that ( _a_ )   and ( _b_ )  .

( _a_ )

( _b_ )

**3.** Determine the domains of the function

( _a_ )

( _b_ )

( _c_ )

( _d_ )

( _e_ )

( _a_ ) Since _y_ must be real, 4 − _x_ 2   0, or _x_ 2   4. The domain is the interval −2   _x_   2.

( _b_ ) Here, _x_ 2 − 16   0, or _x_ 2   16. The domain consists of the intervals _x_   − 4 and _x_   4.

( _c_ ) The function is defined for every value of _x_ except 2.

( _d_ ) The function is defined for _x_ ≠ ±3.

( _e_ ) Since _x_ 2 \+ 4 ≠ 0 for all _x_ , the domain is the set of all real numbers.

**4.** Sketch the graph of the function defined as follows:

_f_ ( _x_ ) = 5 when 0 < _x_   1 _f_ ( _x_ ) = 10 when 1 < _x_   2

_f_ ( _x_ ) = 15 when 2 < _x_   3 _f_ ( _x_ ) = 20 when 3 < _x_   4 etc.

Determine the domain and range of the function.

The graph is shown in Fig. 6-3. The domain is the set of all positive real numbers, and the range is the set of integers, 5, 10, 15, 20,...

**Fig. 6-3**

**5.** A rectangular plot requires 2000 ft of fencing to enclose it. If one of its dimensions is _x_ (in feet), express its area _y_ (in square feet) as a function of _x_ , and determine the domain of the function.

Since one dimension is _x_ , the other is  . The area is then _y_ = _x_ (1000 − _x_ ), and the domain of this function is 0 < _x_ < 1000.

**6.** Express the length _l_ of a chord of a circle of radius 8 as a function of its distance _x_ from the center of the circle. Determine the domain of the function.

From Fig. 6-4 we see that  , so that  . The domain is the interval 0   _x_ < 8.

**Fig. 6-4**

**7.** From each corner of a square of tin, 12 inches on a side, small squares of side _x_ (in inches) are removed, and the edges are turned up to form an open box (Fig. 6-5). Express the volume _V_ of the box (in cubic inches) as a function of _x_ , and determine the domain of the function.

**Fig. 6-5**

The box has a square base of side 12 − 2 _x_ and a height of _x_. The volume of the box is then _V_ = _x_ (12 − 2 _x_ )2 = 4 _x_ (6 − _x_ )2. The domain is the interval 0 < _x_ < 6.

As _x_ increases over its domain, _V_ increases for a time and then decreases thereafter. Thus, among such boxes that may be constructed, there is one of greatest volume, say _M_. To determine _M_ , it is necessary to locate the precise value of _x_ at which _V_ ceases to increase. This problem will be studied in a later chapter.

**8.** If _f_ ( _x_ ) = _x_ 2 \+ 2 _x_ , find   and interpret the result.

On the graph of the function (Fig. 6-6), locate points _P_ and _Q_ whose respective abscissas are _a_ and _a_ \+ _h_.

The ordinate of _P_ is _f_ ( _a_ ), and that of _Q_ is _f_ ( _a_ \+ _h_ ). Then

**Fig. 6-6**

**9.** Let _f_ ( _x_ ) = _x_ 2 − 2 _x_ \+ 3. Evaluate ( _a_ ) _f_ (3); ( _b_ ) _f_ (−3); ( _c_ ) _f_ (− _x_ ); ( _d_ ) _f_ ( _x_ \+ 2); ( _e_ ) _f_ ( _x_ − 2); ( _f_ ) _f_ ( _x_ \+ _h_ ); ( _g_ ) _f_ ( _x_ \+ _h_ ) − _f_ ( _x_ ); ( _h_ )  .

( _a_ ) _f_ (3) = 32 − 2(3) + 3 = 9 − 6 + 3 = 6

( _b_ ) _f_ (−3) = (−3)2 − 2(−3) + 3 = 9 + 6 + 3 = 18

( _c_ ) _f_ (−x) = (−x)2 − 2(−x) + 3 = _x_ 2 \+ 2 _x_ \+ 3

( _d_ ) _f_ ( _x_ \+ 2) = ( _x_ \+ 2)2 − 2( _x_ \+ 2) + 3 = _x_ 2 \+ 4 _x_ \+ 4 − 2 _x_ − 4 + 3 = _x_ 2 \+ 2 _x_ \+ 3

( _e_ ) f( _x_ − 2) = ( _x_ − 2)2 − 2( _x_ − 2) + 3 = _x_ 2 − 4 _x_ \+ 4 − 2 _x_ \+ 4 + 3 = _x_ 2 − 6 _x_ \+ 11

( _f_ ) _f_ ( _x_ \+ _h_ ) = ( _x_ \+ _h_ )2 − 2( _x_ \+ _h_ ) + 3 = _x_ 2 \+ 2 _hx_ \+ _h_ 2 − 2 _x_ − 2 _h_ \+ 3 = _x_ 2 \+ (2 _h_ − 2) _x_ \+ ( _h_ 2 − 2 _h_ \+ 3)

( _g_ ) _f_ ( _x_ \+ _h_ ) − _f_ ( _x_ ) = [ _x_ 2 \+ (2 _h_ − 2) _x_ \+ ( _h_ 2 − 2 _h_ \+ 3)] − ( _x_ 2 − 2 _x_ \+ 3) = 2 _hx_ \+ _h_ 2 − 2 _h_ = _h_ (2 _x_ \+ _h_ − 2)

( _h_ )

**10.** Draw the graph of the function  , and find the domain and range of the function.

The graph of _f_ is the graph of the equation  . For points on this graph, _y_ 2 = 4 − _x_ 2; that is, _x_ 2 \+ _y_ 2 = 4. The graph of the last equation is the circle with center at the origin and radius 2. Since  , the desired graph is the upper half of that circle. Figure 6-7 shows that the domain is the interval −2   _x_   2, and the range is the interval 0   _y_   2.

**Fig. 6-7**

### **Supplementary Problems**

**11.** If _f_ ( _x_ ) = _x_ 2 − 4 _x_ \+ 6, find ( _a_ ) _f_ (0); ( _b_ ) _f_ (3); ( _c_ ) _f_ (−2). Show that   and _f_ (2 − _h_ ) = _f_ (2 + _h_ ).

_Ans_. ( _a_ ) −6; ( _b_ ) 3; ( _c_ ) 18

**12.** If   find ( _a_ ) _f_ (0); ( _b_ ) _f_ (1); ( _c_ ) _f_ (−2). Show that   and  .

_Ans_. ( _a_ ) −1; ( _b_ ) 0; ( _c_ ) 3

**13.** If _f_ ( _x_ ) = _x_ 2 − _x_ , show that _f_ ( _x_ \+ 1) = _f_ (− _x_ ).

**14.** If _f_ ( _x_ ) = 1/ _x_ , show that  .

**15.** If  , show that _x_ = _f_ ( _y_ ).

**16.** Determine the domain of each of the following function:

( _a_ ) _y_ = _x_ 2 \+ 4

( _b_ )

( _c_ )

( _d_ )

( _e_ )

( _f_ )

( _g_ )

( _h_ )

_Ans_. ( _a_ ), ( _b_ ), ( _g_ ) all values of _x_ ; ( _c_ ) | _x_ |   2; ( _d_ ) _x_ ≠ −3; ( _e_ ) _x_ ≠ −1, 2; ( _f_ ) −3 < _x_ < 3; ( _h_ ) 0   _x_ < 2

**17.** Compute   in the following cases:

( _a_ )   when _a ≠ 2_ and _a_ \+ _h_ ≠ 2

( _b_ )   when _a_   4 and _a_ \+ _h_   4

( _c_ )   when _a_ ≠ −1 and _a_ \+ _h_ ≠ −1.

_Ans_. ( _a_ )  ; ( _b_ )  ; ( _c_ )

**18.** Draw the graphs of the following functions, and find their domains and ranges:

( _a_ ) _f_ ( _x_ ) = − _x_ 2 \+ 1

( _b_ )

( _c_ ) _f_ ( _x_ ) = [ _x_ ] = the greatest integer less than or equal to _x_

( _d_ )

( _e_ ) _f_ ( _x_ ) = 5 − _x_ 2

( _f_ )

( _g_ ) _f_ ( _x_ ) = | _x_ − 3|

( _h_ ) _f_ ( _x_ ) = 4/ _x_

( _i_ ) _f_ ( _x_ ) = | _x_ |/ _x_

( _j_ ) _f_ ( _x_ ) = _x_ − | _x_ |

( _k_ )

_Ans_. ( _a_ ) domain, all numbers; range, _y_   1

( _b_ ) domain, _x_ > 0; range, −1 < _y_ < 0 or _y_   2

( _c_ ) domain, all numbers; range, all integers

( _d_ ) domain, _x_ ≠ 2; range, _y_ ≠ 4

( _e_ ) domain, all numbers; range, _y_ ≠   5

( _f_ ) domain, _x_   0; range, _y_   0

( _g_ ) domain, all numbers; range, _y_   0

( _h_ ) domain, _x_ ≠ 0; range, _y_ ≠ 0

( _i_ ) domain, all _x_ ≠ 0; range, {−1, 1}

( _j_ ) domain, all numbers; range, _y_   0

( _k_ ) domain, all numbers; range, _y_   0

**19.** (GC) Use a graphing calculator to verify your answers to Problem 18.

**20.** Evaluate the expression   for the following functions _f_ :

( _a_ ) _f_ ( _x_ ) = 3 _x_ − _x_ 2

( _b_ )

( _c_ ) _f_ ( _x_ ) = 3 _x_ − 5

( _d_ ) _f_ ( _x_ ) = _x_ 3 − 2

_Ans_. ( _a_ )3 − 2 _x_ − _h_

( _b_ )

( _c_ ) 3

( _d_ ) 3 _x_ 2 \+ 3 _xh_ \+ _h_ 2

**21.** Find a formula for the function _f_ whose graph consists of all points satisfying each of the following equations.(In plain language, solve each equation for _y_.)

( _a_ ) _x_ 5 _y_ \+ 4 _x_ − 2 = 0

( _b_ )

( _c_ ) 4 _x_ 2 − 4 _xy_ \+ _y_ 2 = 0

_Ans_. ( _a_ )   ( _b_ )   ( _c_ ) _f_ ( _x_ ) = 2 _x_

**22.** Graph the following functions and their domain and range:

( _a_ )

( _b_ )

( _c_ )

_Ans_. ( _a_ ) domain = (−1, 1], range = (0, 2)

( _b_ ) domain = union of (0, 2) and [3, 4), range = (0, 3)

( _c_ ) domain and range = set of all real numbers

**23.** (GC) Verify your answers to Problem 22 by means of a graphing calculator.

**24.** In each of the following cases, define a function that has the given set   as its domain and the given set   as its range: ( _a_ )   = (0, 2) and   = (1, 7); ( _b_ )   = (0, 1) and   = (1, ∞).

_Ans_. ( _a_ ) One such function is _f_ ( _x_ ) = 3 _x_ \+ 1. ( _b_ ) One such function is

**25.** ( _a_ ) Prove the vertical line test: A set of points in the _xy_ plane is the graph of a function if and only if the set intersects every vertical line in at most one point.

( _b_ ) Determine whether each set of points in Fig. 6-8 is the graph of a function.

_Ans_. Only ( _d_ ) is a function

**Fig. 6-8**

## **CHAPTER 7
Limits**

**LIMIT OF A FUNCTION.** If _f_ is a function, then we say that

if the value of _f(x_ ) gets arbitrary close to _A_ as _x_ gets arbitrarily close to _a_. For example,  , since _x_ 2 gets arbitrarily close to 9 as _x_ approaches as close as one wishes to 3.

The definition can be stated more precisely as follows:   if and only if, for any chosen positive number   however small, there exists a positive number δ such that, whenever 0 < | _x_ − a| < δ, then | _f_ ( _x_ ) − _A_ | <  .

The gist of the definition is illustrated in Fig. 7-1. After   has been chosen [that is, after interval _(ii)_ has been chosen], then δ can be found [that is, interval ( _i_ ) can be determined] so that, whenever _x_ ≠ a is on interval ( _i_ ), say at _x_ 0, then _f_ ( _x_ ) is on interval ( _ii_ ), at f(x0). Notice the important fact that whether or not   is true does not depend upon the value of _f_ ( _x_ ) when _x_ = _a_. In fact, _f_ ( _x_ ) need not even be defined when _x_ = _a_.

**Fig. 7-1**

**EXAMPLE 1:**  , although   is not defined when _x_ = 2. Since

we see that   approaches 4 as _x_ approaches 2.

**EXAMPLE 2:** Let us use the precise definition to show that  . Let   > 0 be chosen. We must produce some δ > 0 such that, whenever 0 < | _x_ − 2| < δ, then |(4 _x_ − 5) − 3| <  .

First we note that |(4 _x_ − 5) − 3| =|4 _x_ − 8|=.

If we take δ to be  /4, then whenever 0 <| _x_ − 2| < δ, |(4 _x_ − 5) − 3| = 4| _x_ − 2| < 4 δ =  .

**RIGHTS AND LEFT LIMITS.** By   we mean that _f_ is defined in some interval ( _c, a_ ) and _f_ ( _x_ ) approaches _A_ as _x_ approaches _a_ through values less than _a_ , that is, as _x_ approaches _a from the left_. Similarly,   means that _f_ is defined in some interval ( _a, d_ ) and _f_ ( _x_ ) approaches _A_ as _x_ approaches _a from the right_. If _f_ is defined in an interval to the left of _a_ and in an interval to the right of _a_ , then the statement   is equivalent to the conjunction of the two statements   and  . The existence of the limit from the left does not imply the existence of the limit from the right, and conversely.

When a function is defined only on one side of a point _a_ , then   is identicial with the one-sided limit, if it exists. For example, if  , then _f_ is defined only at and to the right of zero. Hence,  . Of course,   does not exist, since   is not defined when _x_ < 0. As another example, consider the function  , which is defined only for _x_ > 0. In this case,   does not exists, since 1/ _x_ gets larger and larger without bound as _x_ approaches zero from the right. Therefore,   does not exist.

**Example 3:** The function   has the interval −3   _x_   3 as its domain. If _a_ is any number on the open interval (−3,3), then   exists and is equal to  . Now consider _a_ = 3. Let _x_ approach 3 from the left; then  . For _x_ > 3,   is not defined. Hence,  .

Similarly,  .

**THEOREMS ON LIMITS.** The following theorems are intuitively clear. Proofs of some of them are given in Problem 11.

**Theorem 7.1:** If _f_ ( _x_ ) = _c_ , a constant, then  .

For the next five theorems, assume   and  .

**Theorem 7.2:**  .

**Theorem 7.3:**  .

**Theorem 7.4:**  .

**Theorem 7.5:**  .

**Theorem 7.6:**   is defined.

**INFINITY.** Let

mean that, as _x_ approaches _a_ , _f_ ( _x_ ) eventually becomes and thereafter remains greater than any preassigned positive number, however large. In such a case, we stay that _f_ ( _x_ ) approaches +∞ as _x_ approaches _a_. More precisely,   if and only if, for any positive number _M_ , there exists a positive number δ such that, whenever 0 < | _x_ − a| < δ, then _f_ ( _x_ ) > _M_.

Similarly, let

mean that, as _x_ approaches _a_ , _f_ ( _x_ ) eventually becomes and thereafter remains less than any preassigned negative number. In that case, we say that _f_ ( _x_ ) approaches −∞ as _x_ approaches _a_.

Let

mean that, as _x_ approaches _a_ , | _f_ ( _x_ )| eventually becomes and thereafter remains greater than any preassigned positive number. Hence,   if and only if  .

These definitions can be extended to one-sided limits in the obvious way.

**EXAMPLE 4**

( _a_ )   ( _b_ )   ( _c_ )

**EXAMPLE 5**

( _a_ )  . As _x_ approaches 0 from the right (that is, through positive numbers), 1/ _x_ is positive and eventually becomes larger than any preassigned number.

( _b_ )   since, as _x_ approaches 0 from the left (that is, through negative numbers), 1/ _x_ is negative and eventually becomes smaller than any preassigned number.

The limit concepts already introduced can be extended in an obvious way to the case in which the variable approaches +∞ or − ∞. For example,

means that _f_ ( _x_ ) approaches _A_ as _x_ → +∞, or, in more precise terms, given any positive  , there exists a number _N_ such that, whenever _x_ < _N_ , then | _f_ ( _x_ ) − _A_ | >  . Similar definitions can be given for the statements  ,  ,  ,  , and

**EXAMPLE 6:**  .

_Caution:_ When  , and  , Theorems 7.3–7.5 do not make sense and cannot be used.

For example,   and  , but

_Note:_ We say that a limit, such as  , exists when the limit is a real number, but not when the limit is + ∞ or − ∞ or ∞. For example, since  , we say that,   exists. However, although  , we do not say that   exists.

**Solved Problems**

**1.** Verify the following limit computations:

( _a_ )

( _b_ )

( _c_ )

( _d_ )

( _e_ )

( _f_ )

_[Note:_ Do not assume fro these problems that   is invariably _f_ ( _a_ ).]

( _g_ )

**2.** Verify the following limit computations:

( _a_ )

The division by _x_ − 4 before passing to the limit is valid since _x_ ≠ 4 as _x_ → 4; hence, _x_ −4 is never zero.

( _b_ )

( _c_ )

Here, and again in Problems 4 and 5, _h_ is a variable, so that it might be thought that we are dealing with functions of two variables. However, the fact that _x_ is a variable plays no role in these problems: for the moment, _x_ can be considered a constant.

( _d_ )

( _e_ )  ; no limit exists.

**3.** In the following problems ( _a_ )−( _c_ ), you can interpet   as either   or  ; it will not matter which. Verify the limit computations.

( _a_ )

( _b_ )

( _c_ )

( _d_ )

( _e_ )

( _f_ )

( _g_ )

4.  .

5.

6. ( _a_ ) In each of the following, ( _a_ ) to ( _e_ ), determine the points _x_ = a for which each denominator is zero. Then see what happens to y as _x_ → _a_ − and as _x_ → _a_ +, and verify the given solutions.

( _b_ ) (GC) Check the answers in ( _a_ ) with a graphing calculator.

( _a_ ) y = _f_ ( _x_ ) = 2/x: The denominator is zero when _x_ = 0. As _x_ → 0−, y → −∞ as _x_ →0+, y → +∞.

( _b_ )  : The denominator is zero for _x_ = −3 and _x_ = 2. As _x_ → −3−, y → −∞; as _x_ → −3+, y → + ∞. As _x_ → 2−, y → −∞; as _x_ →2+, y → + ∞.

( _c_ )  : The denominator is zero for _x_ = −2 and _x_ = 1. As _x_ → −2−, _y_ → −∞ as _x_ ; → −2+, _y_ → + ∞. As _x_ →1−, _y_ → + ∞; as _x_ →1+, _y_ → −∞.

( _d_ )  : The denominator is zero for _x_ = 3 As _x_ →3−, _y_ → + ∞; as _x_ → 3+, _y_ → + ∞.

( _e_ )  : The denominator is zero for _x_ = 3. As _x_ →3−, _y_ → + ∞; as _x_ →3+, _y_ → − ∞.

7. For each of the functions of Problem 6, determine what happens to _y_ as _x_ → −∞ and _x_ → + ∞.

( _a_ ) As _x_ → ± ∞, _y_ = 2/ _x_ → 0. When _x_ < 0, _y_ < 0. Hence, as _x_ → −∞, _y_ →0−. Similarly, as _x_ → + ∞, _y_ →0+.

( _b_ ) Divide numerator and denominator of   (the highest power of _x_ in the denominator), obtaining

Hence, as _x_ → ± ∞,

As _x_ → − ∞, the factors _x_ −1, _x_ \+ 3, and _x_ −2 are negative, and, therefore, _y_ → 0−. As _x_ → + ∞, those factors are postive, and, therefore, _y_ → 0+.

( _c_ ) Similar to ( _b_ ).

( _d_ )  , after dividing numerator and denominator by _x_ 2 (the highest power of _x_ in the denominator). Hence, as _x_ → ± ∞,  . The denominator ( _x_ −3)2 is always nonnegative. As _x_ → − ∞, both _x_ \+ 2 and _x_ −1 are negative and their product is positive; hence, _y_ → 1+. As _x_ → +∞, both _x_ \+ 2 and _x_ −1 are positive, as is their product; hence, _y_ →1+.

( _e_ )  , after dividing numerator and denominator by _x_ (the highest power of _x_ in the denominator). As _x_ → ±∞, 2/ _x_ and 3/ _x_ approach 0, and − _x_ −1 approaches ±∞. Thus, the denominator approaches 1 and the numerator approaches ± ∞. As _x_ → −∞, _x_ \+ 2 and _x_ −3 are negative and 1 − _x_ is postive; so, _y_ → +∞. As _x_ → +∞, _x_ \+ 2 and _x_ −3 are positive and 1 − _x_ is negative; so, _y_ → −∞.

8. Examine the function of Problem 4 in Chapter 6 as _x_ → a− and as _x_ → a+ when _a_ is any positive integer.

Consider, as a typical case, a = 2. As _x_ → 2−, _f_ ( _x_ ) → 10. As _x_ →2+, _f_ ( _x_ ) → 15. Thus,   does not exist. In general, the limit fails to exist for all positive integers. (Note, however, that  , since _f_ ( _x_ ) is not defined for _x_   0.)

9. Use the precise definition to show that  .

Let   > 0 be chosen. Note that ( _x_ − 2)2 = _x_ 2 − 4 _x_ \+ 4, and so, _x_ 2+3 _x_ − 10 = |( _x_ − 2)2+7 _x_ − 14 = ( _x_ − 2)2+7( _x_ − 2). Hence |(x2+3x) − 10| = ( _x_ − 2)2+7( _x_ − 2)|   | _x_ − 2|2 +7| _x_ − 2 |. If we choose δ to be the minimum of I and  /8, then δ2   δ, and, therefore, 0 < | _x_ − 2 | < δ implies |(x2+3x) − 10 | < δ2+7 δ   δ+7 δ=8 δ    .

**10**. If  , prove that there exists a positive number δ such that 0 < | _x_ − a | < δ implies  .

Letting   = | _B_ |/2, we obtain a positive δ such that 0 < | _x_ − _a_ | < δ implies − _B_ | < | _B_ |/2. Now, if 0 < | _x_ − a | < δ, then | _B_ | = | _g_ ( _x_ ) + (B − g( _x_ ))|   | _g_ ( _x_ )| + | _B_ − g( _x_ )| < | _g_ ( _x_ )| < | _g_ ( _x_ ) + | _B_ |/2 and, therefore, | _B_ |/2 < | _g_ ( _x_ )|.

**11**. Assume (I)   and (II)  . Prove:

( _a_ )

( _b_ )

( _c_ )

( _a_ ) Let   > 0 be chosen. Then  /2 > 0. By (I) and (II), there exist positive δ 1 and δ 2 such that 0 < | _x_ − a| < δ 1 implies | _f( _x_ )_ − _A_ | <  /2 and 0 < | _x_ − _a_ | < δ2 implies | _g_ ( _x_ ) − _B_ | <  /2. Let δ be the minimum of δ 1 and δ 2. Thus, for 0 < | _x_ − _a_ | < δ, | _f( _x_ )_ − _A_ | <  /2 and | _g_ ( _x_ ) − _B_ | <  /2. Therefore, for 0 < | _x_ − _a_ | < δ,

( _b_ ) Let   > 0 be chosen. Choose   * to be the minimum of  /3 and 1 and  /(3| _B_ |)(if _B_ ≠ 0), and  /(3| _A_ |)(if _A_ ≠ 0). Note that (  *)2     * since   *   1. Moreover, | _B_ |   *    /3 and | _A_ |   *    /3. By (I) and (II), there exist positive δ 1 and δ 2 such that 0 < | _x_ − _a_ | < δ 1 implies | _f( _x_ )_ − _A_ | <   * and 0 < | _x_ − _a_ | < δ 2 implies | _g_ ( _x_ ) − _B_ | <   *. Let δ be the minimum of δ 1 and δ 2. Now, for 0 < | _x_ − _a_ | < δ,

( _c_ ) By part ( _b_ ), it suffices to show that  . Let   > 0 be chosen. Then _B_ 2  /2 > 0. Hence, there exists a positive δ 1 such that 0 < | _x_ − _a_ | < δ 1 implies  .

By Problem 10, there exists a positive δ 2 such that 0 < | _x_ − _a_ | < δ 2 implies | _g_ ( _x_ )| > | _B_ |/2. Let δ be the minimum of δ 1 and δ 2. Then 0 < | _x_ − _a_ | < δ implies that

**12**. Prove that, for any polynomial function

This follows from Theorem 7.1 − 7.4 and the obvious fact that   _x_ = _a_.

**13**. Prove the following generalizations of the results of Problem 3. Let _f( _x_ )_ = anxn+an − 1xn − 1+...+a1 _x_ \+ a0 and g( _x_ )=bkxk+bk − 1xk − 1+...+b1 _x_ \+ b0 be two polynomials.

( _a_ )   if _n_ = _k_

( _b_ )   if _n_ < _k_

( _c_ )   if _n_ > _k_. (It is +∞ if and only if an and bk have the same sign.)

( _d_ )   if _n_ > _k_. (The correct sign is the sign of anbk (−1)n − k.)

**14**. Prove ( _a_ )  ; ( _b_ )  ; ( _c_ )  .

( _a_ ) Let _M_ be any negative number. Choose δ positive and equal to the minimum of 1 and  . Assume _x_ < 2 and 0 < | _x_ − 2| < δ. Then  . Hence,  . But ( _x_ − 2)3 < 0. Therefore,  .

( _b_ ) Let   be any positive number, and let _M_ = 1/ . Assume _x_ > M. Then

( _c_ ) Let _M_ > 1 be any positive number. Assume _x_ > _M_. Then  .

**15**. Evaluate: ( _a_ )   ( _b_ )  ; ( _c_ )

( _a_ ) When _x_ > 0, | _x_ | = _x_ , Hence,  .

( _b_ ) When _x_ < 0, | _x_ | = − _x_ , Hence,  .

( _c_ )   does not exist, since  .

### **Supplementary Problems**

16. Evaluate the following limits:

( _a_ )

( _b_ )

( _c_ )

( _d_ )

( _e_ )

( _f_ )

( _g_ )

( _h_ )

( _i_ )

( _j_ )

( _k_ )

( _l_ )

_Ans_. ( _a_ ) − 4; ( _b_ ) 0; ( _c_ )  ; ( _d_ ) 0; ( _e_ )  ; ( _f_ ) − 4; ( _g_ ) ; ( _h_ )  ; ( _i_ ) 0; ( _j_ ) ∞, no limit; ( _k_ )3 _x_ 2; ( _l_ ) 2

**17**. Evaluate the following limits:

( _a_ )

( _b_ )

( _c_ )

( _d_ )

( _e_ )

( _f_ )

( _g_ )

_Ans_. ( _a_ )  ; ( _b_ ) 0; ( _c_ ) +∞; ( _d_ ) −∞; ( _e_ ) +∞; ( _f_ ) −∞; ( _g_ ) +∞

**18**. Evaluate the following limits:

( _a_ )

( _b_ )

( _c_ )

( _d_ )

( _e_ )

( _f_ )

( _g_ )

_Ans_. ( _a_ ) ; ( _b_ )  ; ( _c_ ) 0; ( _d_ ) +∞ ( _e_ ) 0; ( _f_ ) 1; ( _g_ ) − 1

**19**. Find   for the functions _f_ in Problems 11, 12, 13, 15 and 16( _a, b, d, g_ ) of Chapter 6.

_Ans_. (11)2 _a_ − 4; (12) ; (13) 2 _a_ − 1; (15)  ; (16)( _a_ )2 _a_ ,( _b_ )  , ( _d_ )  , ( _g_ )

**20**. Investigate the behavior of

as _x_ → 0. Draw a graph and verify it with a graphing calculator.

_Ans_.   does not exist.

**21**. Use Theorem 7.4 and mathematical induction to prove   for all positive integers _n_.

**22**. For _f( _x_ )_ = 5 _x_ − 6, find δ > 0 such that, whenever 0 < | _x_ − 4| < δ, then | _f( _x_ )_ − 14| <  , when ( _a_ )   and ( _b_ )   = 0.001.

_Ans_. ( _a_ )  ; ( _b_ ) 0.0002

**23**. Use the precise definition to prove: ( _a_ )  ; ( _b_ )   ( _c_ )  .

**24**. Use the precise definition to prove:

( _a_ )

( _b_ )

( _c_ )

( _d_ )

**25**. Let _f( _x_ )_, _g_ ( _x_ ), and h( _x_ ) be such that (1) _f( _x_ )_   _g_ ( _x_ )   h( _x_ ) for all values in certain intervals to the left and right of a, and (2)  . Prove  .

( _Hint_ : For   > 0, there exists δ > 0 such that, whenever 0 < | _x_ − _a_ | < δ, then | _f( _x_ )_ − _A_ | <   and |h( _x_ ) − _A_ | <  , and, therefore, A −   < _f( _x_ )_   _g_ ( _x_ )   h( _x_ ) < A+  .)

**26**. Prove: If _f( _x_ )_   _M_ for all _x_ in an open interval containing a and if  , then A   _M_.

( _Hint_ : Assume _A_ > _M_. Choose   and derive a contradiction.)

**27**. (GC) Use a graping calculator to confirm the limits found in Problems I( _d, e, f_ ), 2( _a, b, d_ ),16, and 18.

**28**. ( _a_ ) Show that  .

( _Hint_ : Multiply and divide by

( _b_ ) Show that the hyperbola   gets arbitrarily close to the asymptote   as _x_ approaches ∞.

**29**. ( _a_ ) Find  .

( _Hint:_ Multiply the numerator and denominator by  )

( _b_ ) (GC) Use a graphing calculator to confirm the result of part ( _a_ ).

## **CHAPTER 8
Continuity**

A function is defined to be continuous at _x_ 0 if the following three conditions hold.

(i) _f(x 0)_ is defined;

(ii)  ;

(iii)  .

For example, _f( _x_ )_ = _x 2_ \+ 1 is continuous at 2, since  . Condition (i) implies that a function can be continuous only at points of its domain. Thus,   is not continuous at 3 because _f_ (3) is not defined.

Let _f_ be a function that is defined on a interval ( _a, x 0_) to the left of _x 0_ and/or on an interval ( _x 0_, _b_ )to the right of _x 0_. We say that _f_ is discontinuous at _x 0_ if _f_ is not continuous at _x_ 0 that is, if one or more of the conditions (i)–(iii) fails.

**EXAMPLE 1**

( _a_ )   is discontinuous at 2 because _f(2)_ is not defined and also because   does not exist  . See Fig.8-1.

**Fig. 8-1**

( _b_ )   is discontinuous at 2 because _f(2)_ is not defined. However,   so that condition (ii) holds.

The discontinuity at 2 in Example 1 _( _b_ )_ is said to be _removable_ because, if we extended the function _f_ by defining its value at _x_ = 2 to be 4, then the extended function g would be continuous at 2. Note that _g_ ( _x_ ) = _x_ \+ 2 for all _x_. The graphs of   and _g_ ( _x_ ) = _x_ \+ 2 are identical except at _x_ = 2, where the former has a "hole." (see Fig.8 - 2). Removing the discontinuity consists simply of filling the "hole."

**Fig. 8-2**

The discontinuity at 2 in Example 1( _a_ ) is not removable. Redefining the value of _f_ at 2 cannot change the fact that   does not exist.

we also call a discontinuity of a function _f_ at _x_ 0 _removable_ when _f(x 0)_ is defined and changing the value of the function at _x 0_ produces a function that is continuous at _x 0_

**EXAMPLE 2:** Define a function _f_ as follows:

Here  , but _f_ (2) = 0. Hence, condition (iii) fails, so that _f_ has a discontinuity at 2. But, if we change the value of _f_ at 2 to be 4, then we obtain a function _H_ such that _h( _x_ )_ = _x_ 2 for all _x_ , and _h_ is continuous at 2. Thus, the discontinuity of _f_ at 2 was removable.

**EXAMPLE 3:** Let _f_ be the function such that   for all _x_ ≠ 0. The graph of _f_ is hown in Fig.8-3. _f_ is discontinuous at because _f(0)_ is not defined. Moreover,

  and

Thus,  . Hence, the discontinuity of _f_ at 0 is not removable.

**Fig. 8-3**

The kind of discontinuity shown in Example 3 is called a _jump discontinuity_. In general, a function _f_ has a jump discontinuity at _x_ 0 if   and   both exist and  . Such a discontinuity is not removable.

**EXAMPLE 4:** The function of Problem 4 in Chapter 6 has jump discontinuity at every positive integer.

Properties of limit lead to corresponding properties of continuity.

**Theorem 8.1:** Assume that _f_ and _g_ are continuous at _x_ 0. Then:

( _a_ ) The constant function _h_ ( _x_ ) = _c_ for all _x_ is continuous at ever _x_ 0.

( _b_ ) _cf_ is continuous at _x_ 0, for any constant _c_. (Recall that _cf_ has the value _c_ · _f( _x_ )_ for each argument _x_.)

( _c_ ) _f_ \+ _g_ is continuous at _x_ 0.

( _d_ ) _f_ − _g_ is continuous at _x_ 0.

( _e_ ) _fg_ is continuous at _x_ 0

( _f_ ) _f_ / _g_ is continuous at _x_ 0 if _g_ (x0) ≠ 0.

( _g_ )   is continuous at _x_ 0 if   is defined.

These results follow immediately from Theorems 7.1–7.6. For example, ( _c_ )holds because

**Theorem 8.2:** The identify function _I_ ( _x_ ) = _x_ is continuous at every _x_ 0

This follows from the fact that  .

we say that a function _f_ is _continuous on a set A_ if _f_ is continuous at every point of _A_. Moreover, if we just say that _f is continuous_ , we mean that _f_ is continuous at every real number.

The original intuitive idea behind the notion of continuity was that the graph of a continuous function was supposed to be "continuous" in the intuitive sense that one could draw the graph without taking the pencil off the paper. Thus, the graph would not contain any "holes " or "jumps". However, it turns out hat our precise definition of continuity goes well beyond that original intuitive notion; there are very complicated continuous functions that could certainly not be drawn on a piece of paper.

**Theorem 8.3:** Every ploynomial function

is continuous.

This is consequence of Theorem 8.2 and Theorem 8.1 ( _a–e_ ).

**EXAMPLE 5:** As an instance of Theorem 8.3, consider the function _x_ 2 − 2 _x_ \+ 3. Note that, by Theorem 8.2, the identity function _x_ is continuous and, therefore, by Theorem 8.1( _e_ ), _x_ 2 is continuous, and, by Theorem 8.1( _b_ ), −2 _x_ is continuous. By Theorem 8.1( _a_ ) the constant function 3 is continuous. Finally, by Theorem 8.1( _c_ ), _x_ 2 − 2 _x_ \+ 3 is continuous.

**Theorem 8.4:** Every _rational function_  , where _f( _x_ )_ and _g_ ( _x_ ) are polynomial functions, is continuous on the set of all points at which _g_ ( _x_ ) ≠ 0.

This follows from Theorems 8.3 and 8.1 ( _f_ ). As examples, the function   is continuous at all points except 1 and − 1, and the function   is continuous at all points (since _x_ 2 \+ 1 is never 0).

We shall use a special notion of continuity with respect to a closed interval [ _a, b_ ]. First of all, we say that a function _f_ is _continuous on the right at a if f_ ( _a_ ) is defined and   exists, and  . We say that _f_ is _continuous on the left at b_ if _f_ ( _b_ ) is defined and   exists, and  .

**Definition:** _f_ is _continuous_ on [ _a, b_ ] if _f_ is continuous at each point on the open interval ( _a, b_ ), _f_ is continuous on the right at _a_ , and _f_ is continuous on the left at _b_.

Note that whether _f_ is continuous on [ _a, b_ ] does not depend on the values of _f_ , if any, outside of [ _a, b_ ]. Note also that every continuous function (that is, a function continuous at all real numbers) must be continuous on any closed interval. In particular, every polynomial function is continuous on any closed interval.

We want to discuss certain deep properties of continuous functions that we shall use but whose proofs are beyond the scope of this book.

**Theorem 8.5: (Intermediate Value Theorem)** If _f_ is continuous on [ _a, b_ ] and _f_ ( _a_ ) ≠ _f_ ( _b_ ), then, for any number _c_ between _f_ ( _a_ ) and _f_ ( _b_ ), there is at least one number _x_ 0 in the open interval ( _a, b_ ) for which _f_ ( _x_ 0) = _c_.

Figure 8-4 ( _a_ )is an illustration of Theorem 8.5 Figure 8-5 shows that continuity throughout the interval is essential for the validity of the theorem. the following result is a special case of the Intermediate Value Theorem.

**Fig. 8-4**

**Fig. 8-5**

**Corollary 8.6.** If _f_ is continuous on  _a, b_ ] and _f_ ( _a_ ) and _f_ ( _b_ ) have opposite signs, then the equation _f_ ( _x_ ) = 0 has at least one root in the open interval ( _a, b_ ), and therefore, the graph of _f_ crosses the _x_ -axis at least once between _a_ and _b_. (See [Fig. 8-4( _b_ ).)

**Theorem 8.7: (Extreme Value Theorem)** If _f_ is continuous on [ _a, b_ ], then _f_ takes on a least value _m_ and a greatest value _M_ on the interval.

As an illustration of the Extreme Value Theorem, look at Fig.8-6( _a_ ), where the minimum value _m_ occurs at _x = c_ and the maximum value _M_ occurs at _x = d._ In this case, both _c_ and _d_ lie inside the interval. On the other hand, In Fig.8-6( _b_ ), the minimum value _m_ occurs at the endpoint _x = a_ and the maximum value _M_ occurs inside the interval. To see that continuity is necessary for the Extreme Value Theorem to be true, consider the function whose graph is indicated in Fig.8-6( _c_ ). There is a discontinuity at _c_ inside the interval; the function has a minimum value at the left endpoint _x = a_ but the function has no maximum value.

**Fig. 8-6**

Another useful property of continuous functions is given by the following result.

**Theorem 8.8:** If _f_ is continuous at _c_ and _f_ ( _c_ ) > 0, then there is a positive number δ such that, whenever c − δ < _x_ < c + δ, then _f( _x_ )_ > 0.

This theorem is illustrated in Fig.8-7. For a proof, see Problem 3.

**Fig. 8-7**

### **Solved Problems**

**1**. Find the discontinuities of the following functions. Determine whether they are removable. If not removable, determine whether they are jump discontinuities. (GC) Check your answers by showing the graph of the function on a graphing calculator.

**2**. Show that the existence of   implies that _f_ is continuous at _x = a._

But

Hence,  .

**3. Prove Theorem 8.8.**

By the continuity of _f_ at _c_ ,  . If we let ε = _f_ ( _c_ )/2 > 0, then there exists a positive δ such that _0 < | _x_ − c| < δ_ implies that | _f_ ( _x_ ) − _f_ ( _c_ )| < _f_ ( _c_ )/2. The latter inequality also holds when _x = c._ Thus, | _x_ − c| < δ implies | _f( _x_ )_ − _f_ ( _c_ )| < _f_ ( _c_ )/2. The latter implies − _f_ ( _c_ )/2 < _f_ ( _x_ ) − _f_ ( _c_ ) < _f_ ( _c_ )/2. Adding _f_ ( _c_ ) to the left − hand inequality, we obtain _f_ ( _c_ )/2 < _f_ ( _x_ ).

### **Supplementary Problems**

**4**. Determine the discontinuities of the following functions and state why the function fails to be continuous at those points. (GC) Check your answers by graphing the function on a graphing calculator.

( _a_ )

( _b_ )

( _c_ ) _f_ ( _x_ ) = | _x_ | − _x_

( _d_ )

( _e_ )

( _f_ )

( _g_ ) _f( _x_ )_ = _x_ 3 − 7 _x_

( _h_ )

( _i_ )

( _j_ )

( _k_ )

_Ans_. ( _a_ ) Removable discontinuity at _x= − 2._ (Note that _x_ 2 − 3 _x_ − 10 = ( _x_ \+ 2)( _x_ − 5).) _(b, c, g)_ None.

_( _d_ )_ Jump discontinuity at _x = 0 ( _e_ )_ Removable discontinuities at _x= ± 1._

_( _f_ )_ Removable discontinuities at _x = 3, _x_ = − 5._ (Note that _x_ 2 \+ 2 _x_ − 5 = ( _x_ \+ 5)( _x_ − 3) and _x_ 3 \+ _x_ 2 − 17 _x_ \+ 15 = ( _x_ \+ 5)( _x_ − 3)( _x_ − 1).)

_( _h_ )_ Removable discontinuity at _x = 2_ and nonremovable discontinuity at _x = 3._

( _i_ ) Removable discontinuity at _x = − 1_ and nonremovable discontinuity at _x = − 3._

( _j_ ) Removable discontinuity at _x = 2_ and nonremovable discontinuity at _x = − 2._

( _k_ ) Removable discontinuity at _x = 1_ and nonremovable discontinuity at _x = − 1._

**5**. Show that _f( _x_ ) = |x|_ is continuous.

**6**. If Fig.8-5( _a_ ) is the graph of  , show that there is a removable discontinuity at _x = 7_ and that _c = 10_ there.

**7**. Prove: If _f_ is continuous on the interval _[ _a, b_ ]_ and _c_ is a number in _( _a, b_ )_ such that _f( _c_ ) < 0,_ then there exists a positive number δ such that, whenever _c − δ < _x_ < c + δ,_ then _f( _x_ ) < 0._ ( _Hint:_ Apply Theorem 8.8 to − _f_.)

**8**. Sketch the graphs of the following functions and determine whether they are continuous on the closed interval [0,1]:

( _a_ )

( _b_ )

( _c_ )

( _d_ ) _f_ ( _x_ ) = 1 if 0 < _x_   1

( _e_ )

_Ans. ( _a_ )_ Yes. _( _b_ )_ No. Not continuous on the right at 0. _( _c_ )_ Yes. _( _d_ )_ No. Not defined at 0.

_( _e_ )_ No. Not continuous on the left at 1.

## **CHAPTER 9
The Derivative**

**DELTA NOTATION.** Let _f_ be a function. As usual, we let _x_ stand for any argument of _f_ , and we let _y_ be the corresponding value of _f_. Thus, _y_ = _f_ ( _x_ ). Consider any number _x_ 0 in the domain of _f_. Let Δ _x_ (read "delta _x_ ") represent a small change in the value of _x_ , from _x_ 0 to _x_ 0 \+ Δ _x_ , and then let Δ _y_ (read "delta _y_ ") denote the corresponding change in the value of _y_. So, Δ _y_ = _f_ ( _x_ 0 = Δ _x_ ) − _f_ ( _x_ 0). Then the ratio

is called the _average rate of change_ of the function _f_ on the interval between _x_ 0 \+ Δ _x._

**EXAMPLE 1:** Let _y_ = _f_ ( _x_ ) = _x_ 2 \+ 2 _x_. Starting at _x_ 0 = 1, change _x_ to 1.5. Then Δ _x_ = 0.5. The corresponding change in _y_ is Δ _y_ = _f_ (1.5) − _f_ (1) = 5.25 − 3 = 2.25. Hence, the average rate of change of _y_ on the interval between _x_ = 1 and _x_ = 1.5 is

**THE DERIVATIVE** If _y_ = _f_ ( _x_ ) and _x_ 0 is in the domain of _f_ , then by the _instantaneous rate of change_ of _f_ at _x_ 0 we mean the limit of the average rate of change between _x_ 0 and _x_ 0 \+ Δ _x_ as Δ _x_ approaches 0:

provided that this limit exists. This limit is also called the _derivative_ of _f_ at _x_ 0.

**NOTATION FOR DERIVATIVES.** Let us consider the derivative of _f_ at an arbitrary point _x_ in its domain:

The value of the derivative is a function of _x_ , and will be denoted by any of the following expressions:

The value _f_ ′ ( _a_ ) of the derivative of _f_ at a particular point _a_ is sometimes denoted by  .

**DIFFERENTIABILITY** A function is said to be _differentiable_ at a point _x_ 0 if the derivative of the function exists at that point, Problem 2 of Chapter 8 shows that differentiability implies continuity. That the converse is false is shown in Problem 11.

### **Solved Problems**

**1.** Given _y_ = _f_ ( _x_ ) = _x_ 2 \+ 5 _x_ − 8, find Δ _y_ and Δ _y_ /Δ _x_ as _x_ changes ( _a_ ) from _x_ 0 = 1 to _x_ 1 = _x_ 0 \+ Δ _x_ = 1.2 and ( _b_ ) from _x_ 0 = 1 to _x_ 1 = 0.8.

( _a_ ) Δ _x_ = _x_ 1 − _x_ 0 = 1.2 − 1 = 0.2 and Δ _y_ = _f_ ( _x_ 0 \+ Δ _x_ ) − _f_ ( _x_ 0 = _f_ (1.2) − _f_ (1) = −0.56 − (−2) = 1.44. So

( _b_ ) Δ _x_ = 0.8 − 1 = −0.2 and Δ _y_ = _f_ (0.8) − _f_ (1) = −3.36 − (−2) = −1.36. So

Geometrically, Δ _y_ /Δ _x_ in ( _a_ ) is the slope of the secant line joining the points (1, − 2) and (1.2, −0.56) of the parabola _y_ = _x_ 2 \+ 5 _x_ − 8, and in ( _b_ ) is the slope of the secant line joining the points (0.8, −3.36) and (1, − 2) of the same parabola.

**2.** If a body (that is, a material object) starts out at rest and then falls a distance of _s_ feet in _t_ seconds, then physical laws imply that _s_ = 16 _t_ 2. Find Δ _s_ /Δ _t_ as _t_ changes from _t_ 0 to _t_ 0 \+ Δ _t_. Use the result to find Δ _s_ /Δ _t_ as _t_ changes: ( _a_ ) from 3 to 3.5, ( _b_ ) from 3 to 3.2, and ( _c_ ) from 3 to 3.1.

( _a_ ) Here _t_ 0 = 3, Δ _t_ = 0.5, and Δ _s_ /Δ _t_ = 32(3) + 16(0.5) = 104 ft/sec.

( _b_ ) Here _t_ 0 = 3, Δ _t_ = 0.2, and Δ _s_ /Δ _t_ = 32(3) + 16(0.2) = 99.2 ft/sec.

( _c_ ) Here _t_ 0 = 3, Δ _t_ = 0.1, and Δ _s_ /Δ _t_ = 97.6 ft/sec.

Since Δ _s_ is the displacement of the body from time _t_ = _t_ 0 to _t_ = _t_ 0 \+ Δ _t_ ,

**3.** Find dy/dx, given _y_ = _x_ 3 − _x_ 2 − 4. Find also the value of dy/dx when ( _a_ ) _x_ = 4, ( _b_ ) _x_ = 0, ( _c_ ) _x_ = −1.

( _a_ )  ;

( _b_ ) ;

( _c_ )

**4.** Find the derivative of _y_ = _f_ ( _x_ ) = _x_ 2 \+ 3x + 5.

So,

**5.** Find the derivative of   at _x_ = 1 and _x_ = 3.

So,

**6.** Find the derivative of  .

**7.** Find the derivative of  .

**8.** Find the derivative of _f_ ( _x_ ) = _x_ 1/3. Examine _f_ ′(0).

The derivative does not exist at _x_ = 0 because the denominator is zero there. Note that the function is continuous at _x_ = 0.

**9.** Interpret _dy/dx_ geometrically.

From Fig. 9-1 we see that Δ _y_ /Δ _x_ is the slope of the secant line joining an arbitrary but fixed point _P_ ( _x, y_ ) and a nearby point _Q_ ( _x_ \+ Δ _x_ , _y_ \+ Δ _y_ ) of the curve. As Δ _x_ →0, _P_ remains fixed while _Q_ moves along the curve toward _P_ , and the line _PQ_ revolves about _P_ toward its limiting position, the tangent line _PT_ to the curve at _P_. Thus _dy/dx_ gives the slope of the tangent line at _P_ to the curve _y_ = _f_ ( _x_ ).

**Fig. 9-1**

For example, from Problem 3, the slope of the cubic _y_ = _x_ 3 − _x_ 2 − 4 is _m_ = 40 at the point _x_ = 4; it is _m_ = 0 at the point _x_ = 0; and it is _m_ = 5 at the point _x_ = −1.

**10.** Find _ds_ / _dt_ for the function of Problem 2 and interpret the result.

As Δ _t_ → 0, Δ _s_ /Δ _t_ gives the average velocity of the body for shorter and shorter time intervals Δ _t_. Then we can consider ds/dt to be the _instantaneous velocity ν_ of the body at time t0.

For example, at _t_ = 3, _ν_ = 32(3) = 96 ft/sec. In general, if an object is moving on a straight line, and its position on the line has coordinate s at time _t_ , then its instantaneous velocity at time _t_ is _ds_ / _dt_. (See chapter 19.)

**11.** Find _f_ ′( _x_ ) when _f_ ( _x_ ) = | _x_ |.

The function is continuous for all values of _x_ , For _x_ < 0, _f_ ( _x_ ) = − _x_ and

Similarly, for _x_ > 0, _f_ ( _x_ ) = _x_ and

  Hence, the derivative does not exist at _x_ = 0.

Since the function is continuous at 0, this shows that continuity does not imply differentiability.

**12.** Compute   for the function of ( _a_ ) Problem 3 and ( _b_ ) Problem 5. Verify that   → 0 as Δ _x_ → 0.

( _a_ )   = [3 _x_ 2 − 2 _x_ \+ (3 _x_ − 1)Δ _x_ \+ (Δ _x_ )2] − (3 _x_ 2−2 _x_ ) = (3 _x_ − 1 + Δ _x_ )Δ _x_

( _b_ )

Both obviously go to zero as Δ → 0.

**13.** Interpret   of Problem 12 geometrically.

In Fig. 9-1, Δ _y_ = _RQ_ and   tan  _TPR_ = _RS_ ; thus,   Δ _x_ = _SQ_. For a change Δ _x_ in _x_ from _P_ ( _x, y_ ), Δ _y_ is the corresponding change in _y along the curve_ while   is the corresponding change in _y along the tangent line PT_. Since their difference   Δ _x_ is a multiple of (Δ _x_ )2, it goes to zero faster than Δ _x_ , and   can be used as an approximation of Δ _y_ when |Δ _x_ | is small.

### **Supplementary Problems**

**14.** Find Δ _y_ and Δ _y_ /Δ _x_ , given

( _a_ ) _y_ = 2 _x_ −3 and _x_ changes from 3.3to 3.5.

( _b_ ) _y_ = _x_ 2 \+ 4 _x_ and _x_ changes from 0.7 to 0.85.

( _c_ ) _y_ = 2/ _x_ and _x_ changes from 0.75 to 0.5.

_Ans_. ( _a_ ) 0.4 and 2; ( _b_ ) 0.8325 and 5.55; ( _c_ )

**15.** Find Δ _y_ , given _y_ = _x_ 2−3 _x_ \+ 5, _x_ = 5, and Δ _x_ = −0.01. What then is the value of _y_ when _x_ = 4.99?

_Ans_. Δ _y_ = −0.0699; _y_ = 14.9301

**16.** Find the average velocity (see Problem 2), given: ( _a_ ) _s_ = (3 _t_ 2 \+ 5) feet and _t_ changes from 2 to 3 seconds. ( _b_ ) _s_ = (2 _t_ 2 \+ 5 _t_ − 3) feet and _t_ changes from 2 to 5 seconds.

_Ans._ ( _a_ ) 15 ft/sec; ( _b_ ) 19 ft/sec

**17.** Find the increase in the volume of a spherical balloon when its radius is increased: ( _a_ )from _r_ to _r_ \+ Δ _r_ inches; ( _b_ ) from 2 to 3 inches. (Recall that volume  .)

_Ans._ ( _a_ )   ( _b_ )

**18.** Find the derivative of each of the following:

( _a_ ) _y_ = 4 _x_ − 3 ( _b_ ) _y_ = 4 − 3 _x_ ( _c_ ) _y_ = _x_ 2 \+ 2 _x_ −3 ( _d_ ) _y_ = 1/ _x_ 2 ( _e_ ) _y_ = (2 _x_ − 1)/(2 _x_ \+ 1) ( _f_ ) _y_ = (1 + 2 _x_ )/(1 − 2 _x_ ) ( _g_ )   ( _h_ )  ( _i_ )

_Ans._ ( _a_ ) 4; ( _b_ ) −3 ( _c_ ) 2( _x_ = 1); ( _d_ ) −2/ _x_ 3; ( _e_ )   ( _f_ )   ( _g_ )   ( _h_ )   ( _i_ )   ( _j_ )

**19.** Find the slope of the tangent line to the following curves at the point _x_ = 1 (See Problem 9):

( _a_ ) _y_ = 8−5 _x_ 2 ( _b_ )  ( _c_ )

_Ans._ ( _a_ ) −10; ( _b_ ) −1; ( _c_ )

**20.** (GC) Use a graphing calculator to verify your answers in Problem 19. (Graph the curve and the tangent line that you found.)

**21.** Find the coordinates of the vertex (that is, the turning point) of the parabola _y_ = _x_ 2 − 4 _x_ \+ 1 by making use of the fact that, at the vertex, the slope of the tangent line is zero. (See Problem 9.) (GC) Check your answer with a graphing calculator.

_Ans._ (2, − 3)

**22.** Find the slope of the tangent lines to the parabola _y_ = − _x_ 2 \+ 5 _x_ − 6 at its points of intersection with the _x_ -axis.

_Ans._ At _x_ = 2, _m_ = 1. At _x_ = 3, _m_ = −1.

**23.** When an object is moving on a straight line and its coordinate on that line is s at time _t_ (where _s_ is measured in feet and _t_ in seconds), find the velocity at time _t_ = 2 in the following cases:

( _a_ ) _s_ = _t_ 2 \+ 3 _t_ ( _b_ ) _s_ = _t_ 3 −3 _t_ 2 ( _c_ )

(See Problem 10.)

_Ans._ ( _a_ ) 7 ft/sec; ( _b_ ) 0 ft/sec; ( _c_ )

**24.** Show that the instantaneous rate of change of the volume _V_ of a cube with respect to its edge _x_ (measured in inches) is 12 in3 / in when _x_ = 2 inches.

## **CHAPTER 10
Rules for Differentiating Functions**

**DIFFERENTIATION.** Recall that a function _f_ is said to be differentiable at _x_ 0 if the derivative _f′_ ( _x_ 0) exists. A function is said to be differentiable on a set if the function is _differentiable_ at every point of the set. If we say that a function is differentiable, we mean that it is differentiable at every real number. The process of finding the derivative of a function is called _differentiation_.

**Theorem 10.1: (Differentiation Formulas).** In the following formulas, it is assumed that _u, v,_ and _w_ are functions that are differentiable at _x; c_ and _m_ are assumed to be constants.

(1)   (The derivative of a constant function is zero.)

(2)   (The derivative of the identity function is 1.)

(3)

(4)   (Sum Rule)

(5)   (Difference Rule)

(6)   (Product Rule)

(7)   provided that _v_ ≠ 0 (Quotient Rule)

(8)   provided that _x_ ≠ 0

(9)   (Power Rule)

Note that formula (8) is a special case of formula (9) when _m_ = −1. For proofs, see Problems 1−4.

**EXAMPLE 1**

Every polynomial is differentiable, and its derivative can be computed by using the Sum Rule, Power Rule, and formulas (1) and (3).

**COMPOSITE FUNCTIONS. THE CHAIN RULE.** The _composite function f_ ο _g_ of functions _g_ and _f_ is defined as follows: ( _f_ ο _g_ )( _x_ ) = _f_ ( _g_ ( _x_ )). The function g is applied first and then _f_. _g_ is called the _inner function_ , and _f_ is called the _outer function. f ο g_ is called the composition of _g_ and _f_.

**EXAMPLE 2:** Let f ( _x_ ) = _x_ 2 and g( _x_ ) = _x_ \+ 1. Then:

Thus, in this case, _f_ ο _g_ ≠ _f_ ο _f_.

When _f_ and _g_ are differentiable, then so is their composition _f_ ο _g_. There are two procedures for finding the derivative of _f_ ο _g_. The first method is to compute an explicit formula for _f_ ( _g_ ( _x_ )) and differentiate.

**EXAMPLE 3:** If _f_ ( _x_ ) = _x_ 2 \+ 3 and _g_ ( _x_ ) = 2 _x_ \+ 1, then

_y_ = _f_ ( _g_ ( _x_ )) = _f_ (2 _x_ \+ 1) = (2 _x_ \+ 1)2 \+ 3 = 4 _x_ 2 \+ 4 _x_ \+ 4 and

Thus, _D x_( _f_ ο _g_ ) = 8 _x_ \+ 4.

The second method of computing the derivative of a composite function is based on the following rule.

**Chain Rule:** Dx( _f_ ( _g_ ( _x_ )) = _f_ ′( _g_ ( _x_ )). _g_ ′( _x_ )

Thus, the derivative of _f_ ο _g_ is the product of the derivative of the outer function _f_ (evaluated at _g( _x_ )_) and the derivative of the inner function (evaluated at _x_ ). It is assumed that _g_ is differentiable at _x_ and that _f_ is differentiable at _g_ ( _x_ ).

**EXAMPLE 4:** In Example 3, f′( _x_ ) = 2 _x_ and _g_ ′( _x_ ) = 2. Hence, by the Chain Rule,

_D x_( _f_ ( _g_ ( _x_ )) = ( _f_ ′( _g_ ( _x_ )). _g_ ′( _x_ ) = 2 _g_ ( _x_ ).2 = 4 _g_ ( _x_ ) = 4(2 _x_ \+ 1) = 8 _x_ \+ 4

**ALTERNATIVE FORMULATION OF THE CHAIN RULE.** Let _u_ = _g_ ( _x_ ) and _y_ = _f_ ( _u_ ). Then the composite function of _g_ and _f_ is _y_ = _f_ ( _u_ ) = _f_ ( _g_ ( _x_ )), and we have the formula:

  (Chain Rule)

**EXAMPLE 5:** Let _y_ = _u_ 3 and _u_ = 4 _x_ 2−2 _x_ \+ 5. Then the composite function y = (4 _x_ 2−2 _x_ \+ 5)3 has the derivative

**Warning.** In the Alternative Formulation of the Chain Rule,  , the _y_ on the left denotes the composite function of _x_ , whereas the _y_ on the right denotes the original function of _u._ Likewise, the two occurrences of _u_ have different meanings. This notational confusion is made up for by the simplicity of the alternative formulation.

**INVERSE FUNCTIONS.** Two functions _f_ and _g_ such that _g_ ( _f_ ( _x_ )) = _x_ and _f_ ( _g_ ( _y_ )) = _y_ are said to be _inverse functions._ Inverse functions reverse the effect of each other. Given an equation _y_ = _f_ ( _x_ ), we can find a formula for the inverse of _f_ by solving the equation for _x_ in terms of _y_.

**EXAMPLE 6**

( _a_ ) Let _f_ ( _x_ ) = _x_ \+ 1. Solving the equation _y_ = _x_ \+ 1 for _x_ , we obtain _x_ = _y_ −1. Then the inverse _g_ of _f_ is given by the formula _g_ ( _y_ ) = _y_ −1. Note that _g_ reverses the effect of _f_ and _f_ reverses the effect of _g_.

( _b_ )Let _f_ ( _x_ ) = − _x._ Solving _y_ = − _x_ for _x_ we obtain _x_ = − _y_. Hence, _g_ ( _y_ ) = − _y_ is the inverse of _f_. In this case, the inverse of _f_ is the same function as _f_.

( _c_ ) Let   is defined only for nonnegative numbers, and its range is the set of nonnegative numbers. Solving   for _x_ , we get _x_ = _y_ 2, so that _g_ ( _y_ ) = _y_ 2. Note that, since _g_ is the inverse of _f_ , _g_ is only defined for nonnegative numbers, since the values of f are the nonnegative numbers. (Since _y_ = _f_ ( _g_ ( _y_ )), then, if we allowed _g_ to be defined for negative numbers, we would have −1 = _f_ ( _g_ (−1)) = _f_ (1) = 1, a contradiction.)

( _d_ ) The inverse of _f_ ( _x_ ) = 2 _x_ − 1 is the function  .

**Notation:** The inverse of _f_ will be denoted _f_ −1.

Do not confuse this with the exponential notation for raising a number to the power −1. The context will usually tell us which meaning is intended.

Not every function has an inverse function. For example, the function _f_ ( _x_ ) = _x_ 2 does not possess an inverse. Since _f_ (1) = 1 = _f_ (−1), an inverse function _g_ would have to satisfy _g_ (1) = 1 and _g_ (1) = −1, which is impossible. (However, if we restricted the function _f_ ( _x_ ) = _x_ 2 to the domain _x_  0, then the function   would be an inverse function of _f_.)

The condition that a function _f_ must satisfy in order to have an inverse is that _f_ is _one-to-one,_ that is, for any _x_ 1 and _x_ 2, if _x_ 1 ≠ _x_ 2), then _f_ ( _x_ 1) ≠ ( _x_ 2). Equivalently, _f_ is one-to-one if and only if, for any _x_ 1 and _x_ 2, if _f_ ( _x_ 1) = _f_ ( _x_ 2), then _x_ 1 = _x_ 2.

**EXAMPLE 7:** Let us show that the function _f_ ( _x_ ) = 3 _x_ \+ 2 is one-to-one. Assume _f_ ( _x_ 1). Then 3 _x_ 1 \+ 2 = 3 _x_ 2 \+ 2, 3 _x_ 1 = 3 _x_ 2, _x_ 1 = _x_ 2. Hence, _f_ is one-to-one. To find the inverse, solve _y_ = 3 _x_ \+ 2 for _x_ , obtaining  . Thus,  . (In general, if we can solved _y_ = _f_ ( _x_ ) for _x_ in terms of _y_ , then we know that _f_ is one-to-one.)

**Theorem 10.2: (Differentiation Formula for Inverse Functions.)** Let f be one-to-one and continuous on an interval ( _a_ , _b_ ). Then:

( _a_ ) The range of _f_ is an interval I (possibly infinite) and _f_ is either increasing or decreasing. Moreover, _f_ −1 is continuous on I.

( _b_ ) If _f_ ′( _x_ 0) is differentiable and _f_ ′( _x_ 0) ≠ 0, then _f_ −1 is differentiable at _y_ 0 = _f_ ( _x_ 0) and  .

The latter equation is sometimes written

where _x_ = _f_ −1 _(y)_.

For the proof, see Problem 70.

**EXAMPLE 8**

( _a_ ) Let _y_ = _f_ ( _x_ ) = _x_ 2 for _x_ > 0. Then  . Since  . Thus,  . (Note that this is a special case of Theorem 8.1(9) when  .)

( _b_ ) Let _y_ = _f_ ( _x_ ) = _x_ 3 for all _x._ Then   for all _y_. Since  . This holds for all _y_ ≠ 0. (Note that _f_ −1(0) = 0 and _f_ ′(0) = 3(0)2 = 0.)

**HIGHER DERIVATIVES.** if _y_ = _f_ ( _x_ ) is differentiable, its derivative _y_ ′ is also called the _first derivative_ of _f_. If _y_ ′ is differentiable, its derivative is called the _second derivative_ of _f_. If this second derivative is differentiable, then its derivative is called the _third derivative_ of _f_ , and so on.

**Notation**

### **Solved Problems**

**1.** Prove Theorem 10.1, (1) − (3): .

Remember that  .

(1)

(2)

(3)

**2.** Prove Theorem 10.1, (4), (6), (7):

(4)

(6)

(7)   provided that ν ≠ 0

(4) It suffice to prove this for just two summands, _u_ and _v_. Let _f_ ( _x_ ) = _u_ \+ _v_. Then

Taking the limit as Δ _x_ → 0 yields  .

(6) Let _f_ ( _x_ ) = _uv_. Then

Taking the limit as Δ _x_ → 0 yields

Note that   because the differentiability of _u_ implies its continuity.

(7) Set  ; then

and for

**3.** Prove Theorem 10.1(9): D _x ( _x_ m) = m _x_ m−1_, when _m_ is a nonnegative integer.

Use mathematical induction. When _m_ = 0,

Assume the formula is true for _m_. Then, by the Product Rule,

Thus, the formula holds for _m_ \+ 1.

**4.** Prove Theorem 10.1(9): D _x( _x_ m) = m _x_ m−1_, when _m_ is a negative integer.

Let _m_ = − _k_ , where _k_ is a positive integer. Then, by the Quotient Rule and Problem 3,

**5.** Differentiate _y_ = 4 + 2 _x_ − 3 _x_ 2 − 5 _x_ 3 − 8 _x_ 4 \+ 9 _x_ 5.

**6.** Differentiate

**7.** Differentiate

**8.** Differentiate

**9.** Differentiate

**10.** Prove the Power Chain Rule: D _x_ ( _y m_) = _my m_−1 D _x_ y.

This is simply the Chain Rule, where the outer function is _f_ ( _x_ ) = _x_ m and the inner function is _y_.

**11.** Differentiate _s_ = ( _t_ 2 − 3)4.

By the Power Chain Rule,  .

**12.** Differentiate ( _a_ )  ; ( _b_ )  .

( _a_ )

( _b_ )

**14.** Differentiate _y_ = ( _x_ 2 \+ 4)2 (2 _X_ 3 − 1)3.

Use the Product Rule and the Power Chain Rule:

**15.** Differentiate  .

Use the Quotient Rule:

**16.** Differentiate  .

**17.** Find   given  .

By the Product Rule,

By Theorem 10.2,

**18.** Find the slope of the tangent line to the curve _x_ = _y_ 2 − 4 _y_ at the points where the curve crosses the _y_ −axis.

The intersection points are (0, 0) and (0, 4). We have   and so  . At (0, 0) the slope is  , and at (0, 4) the slope is  .

**19.** Derive the Chain Rule: D _x_ ( _f_ ( _g_ ( _x_ )) = _f_ ' ( _g_ ( _x_ )). _g_ '( _x_ )).

Hence,

Now,

Since   K=0,  . Hence,

_H′_ ( _x_ ) = _f′_ ( _y_ ) _g_ ′( _x_ ) = _f′_ ( _y_ ) _g_ ′( _x_ )) _g_ ′( _x_ ).

**20.** Find  , given   and  .

Then

**21.** A point moves along the curve _y_ = _x_ 3 − 3 _x_ \+ 5 so that  , where _t_ is time. At what rate is _y_ changing when _t_ = 4?

We must find the value of _dy/dt_ when _t_ = 4. First, _dy/dx_ = 3( _x_ 2 − 1) and  . Hence,

When  , and   units per unit of time.

**22.** A point moves in the plane according to equations _x_ = _t_ 2 \+ 2 _t_ and _y_ = 2 _t_ 3−6 _t_. Find _dy/dx_ when _t_ = 0, 2, and 5.

Since the first equation may be solved for _t_ and this result substituted for _t_ in the second equation, _y_ is a function of _x_. We have _dy/dt = 6 _t_ 2−6_. Since _dx/dt_ = 2 _t_ \+ 2, Theorem 8.2 gives us _dt/dx_ = 1/(2 _t_ \+ 2). Then

The required values of _dy/dx_ are −3 at _t_ = 0, 3 at _t_ = 2, and 12 at _t_ = 5.

**23.** If _y_ = _x_ 2−4 _x_ and  , find _dy/dt_ when  .

So

When   and  .

**24.** Show that the function _f_ ( _x_ ) = _x_ 3 \+ 3 _x_ 2−8 _x_ \+ 2 has derivatives of all orders and find them.

_f′( _x_ ) = 3 _x_ 2 \+ 6 _x_ − 8, f″( _x_ ) = 6 _x_ \+ 6, f″′( _x_ ) = 6_, and all derivatives of higher order are zero.

**25.** Investigate the successive derivatives of _f_ ( _x_ ) = _x_ 4/3; at _x_ = 0.

_f_ ( _n_ ) (0) does not exist for _n_   2.

**26.** If  , find a formula for _f (n)( _x_ )_.

which suggest  . This result may be established by mathematical induction by showing that if  , then

### **Supplementary Problems**

**27.** Prove Theorem 10.1(5): _D x( _u_ − _v_ ) = Dx−Dxv_.

_Ans_.   by Theorem 8.1(4,3)

In Problems 28 to 46, find the derivative.

**47.** For each of the following, compute _dy/dx_ by two different methods and check that the results are the same:

( _a_ ) _x_ = (1 + 2 _y_ )3 ( _b_ )

In Problems 48 to 51, use the Chain Rule to find

In Problems 52 to 55, find the indicated derivative.

In Problems 56 and 57, find a formula for the _n_ th derivative.

In Problems 60 to 65, determine whether the given function has an inverse; if it does, find a formula for the inverse _f −1_ and calculate its derivative.

**66.** Find the points at which the function _f_ ( _x_ ) = | _x_ \+ 2| is differentiable.

_Ans_. All points except _x_ = −2

**67.** (GC) Use a graphing calculator to draw the graph of the parabola _y_ = _x_ 2−2 _x_ and the curve _y_ = | _x_ 2−2 _x_ |. Find all points of discontinuity of the latter curve.

_Ans_. _x_ = 0 and _x_ = 2

**68.** Find a formula for the _n_ th derivative of the following functions: ( _a_ )  ; ( _b_ )

_Ans_. ( _a_ )

( _b_ )

**69.** Find the second derivatives of the following functions:

( _a_ ) _f_ ( _x_ ) = 2 _x_ − 7

( _b_ ) _f_ ( _x_ ) = 3 _x_ 2 \+ 5 _x_ − 10

( _c_ )

( _d_ )

_Ans_. ( _a_ ) 0; ( _b_ ) 6; ( _c_ )  ; ( _d_ )

**70.** Prove Theorem 10.2.

_Ans_. _Hints_ : ( _a_ ) Use the intermediate value theorem to show that the range is an interval. That _f_ is increasing or decreasing follows by an argument that uses the extreme value and intermediate value theorems. The continuity of _f −1_ is then derived easily.

( _b_ )

By the continuity of _f −1_, as _y_ → _y_ 0, _x_ → _x_ 0, and we get  .

## **CHAPTER 11
Implicit Differentiation**

**IMPLICIT FUNCTIONS** An equation _f(x, y)_ = 0 defines _y implicitly_ as a function of _x_. The domain of that implicitly defined function consists of those _x_ for which there is a unique _y_ such that _f(x, y)_ = 0.

**EXAMPLE 1**

( _a_ ) The equation _xy + x_ − 2 _y_ − 1 = 0 can be solved for _y_ , yielding  . This function is defined for _x_ ≠ 2.

( _b_ ) The equation 4 _x_ 2 \+ 9 _y_ 2 − 36 = 0 does not determine a unique function _y_. If we solve the equation for _y_ , we obtain  . We shall think of the as implicitly defining two functions   and  . Each of these functions is defined for |x|   3. The ellipse determined by the original equation is the union of the graphs of the two functions.

If _y_ is a function implicitly defined by an equation _f_ ( _x, y_ ) = 0, the derivative _y_ ′ can be found in two different ways:

1. Solve the equation for _y_ and calculate _y_ ′ directly. Except for very simple equations, this method is usually impossible or impractical.

2. Thinking of _y_ as a function of _x_ , differentiate both sides of the original equation _f(x, y)_ = 0 and solve the resulting equation for _y_ ′. This differentiation process is known as _implicit differentiation_.

**EXAMPLE 2**

( _a_ ) Find _y_ ′, given _xy_ \+ _x_ − 2 _y_ − 1 = 0. By implicit differentiation, _xy_ ′ + _y D x(x)_−2 _y_ ′ − _D x_(1) = _D x_(0). Thus, _xy_ ′ + _y_ − 2 _y_ ′ = 0. Solve for  . In this case, Example 1( _a_ ) shows that we can replace _y_ by   and find _y_ ′ in terms of _x_ alone. We see that it would have been just as easy to differentiate   by the Quotient Rule. However, in most cases, we cannot solve for _y_ or for _y_ ′ in terms of _x_ alone.

( _b_ ) Given 4 _x_ 2 \+ 9 _y_ 2 − 36 = 0, find _y_ ′ when  . By implicit differentiation, 4 _D x_( _x_ 2) + 9 _D x_( _y_ 2) − _D x_(36) = _D x_(0). Thus, 4(2 _x_ ) + 9(2 _y y_ ′) = 0. (Note that _D x_( _y_ 2) = 2 _yy_ ′ by the Power Chain Rule.) Solving for _y_ ′. We get _y_ ′ = −4 _x_ /9 _y_. When  . For the function _y_ corresponding to the upper are for the ellipse (see Example 1( _b_ )),   and  . For the function _y_ corresponding to the lower are of the ellipse,   and  .

**DERIVATIVES OF HIGHER ORDER** may be obtained by implicit differentiation or by a combination of direct and implicit differentiation.

**EXAMPLE 3:** In Example 2( _a_ ),  . Then

**EXAMPLE 4:** Find the value of _y_ ′′ at the point (−1,1) of the curve _x_ 2 _y_ \+ 3 _y_ − 4 = 0.

We differentiate implicitly with respect to _x_ twice. First, _x_ 2 _y_ ′ + 2 _xy_ \+ 3 _y_ ′ = 0, and then _x_ 2 _y_ ′′ + 2 _xy_ ′ + 2 _xy_ ′ + 2 _y_ \+ 3 _y_ ′′ = 0. We could solve the first equation for _y_ ′ and then solve the second equation for _y_ ′′. However, since we only wish to evaluate _y_ ′′ at the particular point (−1, 1), we substitute _x_ = −1, _y_ = 1 in the first equation to find  , and then substitute _x_ = −1, y = 1,   in the second equation to get _y_ ′′ − 1 − 1 + 2 + 3 _y_ ′′ = 0, from which we obtain _y_ ′′ = 0. Notice that this method avoids messy algebraic calculations.

### **Solved Problems**

**1.** Find _y_ ′, given _x_ 2 _y_ − _xy_ 2 \+ _x_ 2 \+ _y_ 2 = 0.

**2.** If _x_ 2 − _xy_ \+ _y_ 2 = 3, find _y_ ′ and _y_ ′′.

_D x_( _x_ 2) − _D x(xy) + Dx_( _y_ 2) = 0

2 _x − xy_ ′ − _y_ \+ 2 _yy_ ′ = 0

Hence,  . Then,

**3.** Given _x_ 3 _y_ \+ _xy_ 3 = 2, find _y_ ′ and _y_ ′′ at the point (1,1).

By implicit differentiation twice,

_x_ 3 _y_ ′ + 3 _x_ 2 _y_ \+ _x_ (3 _y_ 2 _y_ ′) + _y_ 3 = 0

and _x_ 3 _y_ ′′ + 3 _x_ 2 _y_ ′ + 3 _x_ 2 _y_ ′ + 6 _xy_ \+ 3 _xy_ 2 _y_ ′′ + _y_ ′[6 _xyy_ ′ + 3 _y_ 2] + 3 _y_ 2 _y_ ′ = 0

Substituting _x_ = 1, _y_ = 1 in the first equation yields _y_ ′ = −1. Then substituting _x_ = 1, _y_ = 1, _y_ ′ = −1 in second equation yields _y_ ′′ = 0.

### **Supplementary Problems**

**4.** Find _y_ ′′, given: ( _a_ ) _x + xy + y_ = 2; ( _b_ ) _x_ 3 − 3 _xy + y_ 3 = 1.

_Ans_. ( _a_ )  ; ( _b_ )

**5.** Find _y_ ′, _y_ ′′, and _y_ ′′′ at: ( _a_ ) the point (2,1) on _x_ 2 − _y_ 2 − _x_ = 1; ( _b_ ) the point (1, 1) on _x_ 3 \+ 3 _x_ 2 _y_ − 6 _xy_ 2 \+ 2 _y_ 3 = 0.

_Ans_. ( _a_ )  ; ( _b_ ) 1, 0, 0

**6.** Find the slope of the tangent line at a point ( _x_ 0, _y_ 0) of: ( _a_ ) _b_ 2 _x_ 2 \+ _a_ 2 _y_ 2 = _a_ 2 _b_ 2; ( _b_ ) _b_ 2 _x_ 2 − _a_ 2 _y_ 2 = _a_ 2 _b_ 2; ( _c_ ) _x_ 3 \+ _y_ 3 − 6 _x_ 2 _y_ = 0.

_Ans_ : ( _a_ )  ; ( _b_ ) ; ( _c_ )

**7.** Prove that the lines tangent tangent to the curves 5 _y_ − 2 _x_ \+ _y_ 3 − _x_ 2 _y_ = 0 and 2 _y_ \+ 5 _x + x_ 4− _x_ 3 _y_ 2 = 0 at the origin intersect at right angles.

**8.** ( _a_ ) The total surface area of a rectangular box whose base is _y_ on a side and whose height is _x_ is given by _S = 2y_ 2 \+ 4 _xy_. If _S_ is constant, find _dy/dx_ without solving for _y_.

( _b_ ) The total surface area of a right circular cylinder of radius _r_ and height _h_ is given by _S_ = 2 _πr_ 2 \+ 2 _πrh_. If _S_ is constant, find _dr/dh_.

_Ans_. ( _a_ )  ; ( _b_ )

**9.** For the circle _x_ 2 \+ _y_ 2 = _r_ 2, show that  .

**10.** Given _S_ = _πx_ ( _x + 2y_ ) and _V_ = _πx_ 2 _y_ , show that _dS/dx_ = 2 _π(x - y)_ when _V_ is a constant, and _dv/dx_ = − _πx(x − y)_ when _S_ is a constant.

**11.** Derive the formula _D x_( _x m_) = _mx_ _m_ −1 of Theorem 10.1(9) when _m_ = _p/q_ , where _p_ and _q_ are nonzero integers. You may assume that _x p/q_ is differentiable. ( _Hint_ : Let _y = x p/q_. Then _y q_ = _x p_. Now use implicit differentiation.)

**12.** (GC) Use implicit differentation to find an equation of the tangent line to   at (4, 4), and verify your answer on graphing calculator.

_Ans. y_ = − _x_ \+ 8

## **CHAPTER 12
Tangent and Normal Lines**

An example of a graph of a continuous function _f_ is shown in Fig. 12-1( _a_ ). If _P_ is a point of the graph having abscissa _x_ , then the coordinates of _P_ are ( _x_ , _f_ ( _x_ )). Let _Q_ be a nearby point having abscissa _x_ \+ Δx. Then the coordinates of _Q_ are ( _x_ \+ Δx, _f_ ( _x + Δx_ )). The line _PQ_ has slope  . As _Q_ approaches _P_ along the graph, the lines _PQ_ get closer and closer to the tangent line _J_ to the graph at _P_. (See Fig. 12-1( _b_ ).) Hence, the slope of _PQ_ approaches the slope of the tangent line. Thus, the slope of the tangent line is  , which is the derivative _f_ ′( _x_ ).

**Fig. 12-1**

If the slope _m_ of the tangent line at a point of the curve _y_ = _f_ ( _x_ ) is zero, then the curve has a horizontal tangent line at that point, as at points _A, C,_ and _E_ of Fig. 12-2. In general, if the derivative of _f_ is _m_ at a point ( _x_ 0, _y_ 0), then the point-slope equation of the tangent line is _y_ − _y_ 0 = _m_ ( _x − x_ 0). If _f_ is continuous at _x_ 0, but  , then the curve has a vertical tangent line at _x0_ , as at points _B_ and _D_ of Fig. 12-2.

**Fig. 12-2**

The _normal line_ to a curve at one of its points ( _x_ 0, _y_ 0) is the line that passes through the point and is perpendicular to the tangent line at that point. Recall that a perpendicular to a line with nonzero slope _m_ has slope −1/ _m_. Hence, if _m_ ≠ 0 is the slope of the tangent line, then _y − y_ 0 = −(1/ _m_ )( _x − x_ 0) is a point-slope equation of the normal line. If the tangent line is horizontal, then the normal line is vertical and has equation _x = x_ 0. If the tangent line is vertical, then the normal line is horizontal and has equation _y = y_ 0.

**THE ANGLES OF INTERSECTION** of two curves are defined as the angles between the tangent lines to the curves at their point of intersection.

To determine the angles of intersection of the two curves:

1. Solve the equations of the curves simultaneously to find the points of intersection.

2. Find the slopes _m_ 1 and _m_ 2 of the tangent lines to the two curves at each point of intersection.

3. If _m_ 1 = _m_ 2, the angle of intersection is 0°, and if _m_ 1 = −1/ _m_ 2, the angle of intersection is 90°; otherwise, the angle of intersection   can be found from the formula

  is the acute angle of intersection when tan   > 0, and 180° −   is the acute angle of intersection when tan   < 0.

### **Solved Problems**

**1**. Find equations of the tangent and normal lines to _y = x_ 3 − 2 _x_ 2 \+ 4 at (2, 4).

_f_ ′( _x_ ) = 3 _x_ 2 − 4 _x_. Thus, the slope of the tangent line at (2, 4) is _m_ = _f_ ′ (2) = 4, and, an equation of the tangent line is _y_ − 4 = 4( _x_ − 2). The slope-intercept equation is _y_ = 4 _x_ − 4.

An equation of the normal line at (2, 4) is  . Its slope-interecept equation is  .

**2**. Find equations of the tangent and normal lines to _x_ 2 \+ 3 _xy_ \+ _y_ 2 = 5 at (1, 1).

By implicit differentiation, 2 _x_ \+ 3 _xy_ ′ + 3 _y_ \+ 2 _yy_ ′ = 0. So,  . Then the slope of the tangent line at (1, 1) is −1. An equation of the tangent line is y − 1 = −( _x_ − 1). Its slope-intercept equation is _y_ = − _x_ \+ 2. An equation of the normal line is _y_ − 1 = _x_ − 1, that is, _y = x_.

**3**. Find the equations of the tangent lines with slope   to the ellipse 4 _x_ 2 \+ 9 _y_ 2 = 40.

By implicit differentiation, y′ = −4 _x_ /9 _y_. So, at a point of tangency ( _x_ 0, _y_ 0),  . Then _y_ 0 = 2 _x_ 0.

Since the point is on the ellipse,  . So,  . Therefore,  , and _x_ 0 = ±1. The required points are (1, 2) and (-1, -2).

At (1, 2), an equation of the tangent line is  .

At (-1, -2), an equation of the tangent line is  .

**4**. Find an equation of the tangent lines to the hyperbola _x_ 2 − _y_ 2 = 16 that pass through the point (2, −2).

By implicit differentiation, 2 _x_ − 2 _yy_ ′ = 0 and, therefore, _y_ ′ = _x/y_. So, at a point of tangency ( _x_ 0, _y_ 0), the slope of the tangent line must be _x_ 0/ _y_ 0. On the other hand, since the tangent line must pass through ( _x_ 0, _y_ 0) and (2, −2), the slope is  .

Thus,  . Hence,  . Thus,  , yielding _x_ 0 \+ _y_ 0 = 8, and therefore, _y_ 0 = 8 − _x_ 0.

If we substitute 8 − _x_ 0 for _y_ 0 in   and solve for _x_ 0, we get _x_ 0 = 5. Then _y_ 0 = 3. Hence, an equation of the tangent line is  .

**5**. Find the points of tangency of horizontal and vertical tangent lines to the curve _x_ 2 − _xy_ \+ _y_ 2 = 27.

By implicit differentiation, 2 _x_ − _xy_ ′ − y + 2 _yy_ ′ = 0, whence  .

For horizontal tangent lines, the slope must be zero. So, the numerator _y_ − 2 _x_ of _y_ ′ must be zero, yielding _y_ = 2 _x_. Substituting 2 _x_ for _y_ in the equation of the curve, we get _x_ 2 = 9. Hence, the points of tangency are (3, 6) and (−3, −6).

For vertical tangent lines, the slope must be infinite. So, the denominator 2 _y − x_ of _y_ ′ must be zero, yielding _x_ = 2 _y_. Replacing _x_ in the equation of the curve, we get _y_ 2 = 9. Hence, the points of tangency are (6, 3) and (−6, −3).

**6**. Find equations of the vertical lines that meet the curves ( _a_ ) _y = x_ 3 \+ 2 _x_ 2 − 4 _x_ \+ 5 and ( _b_ ) 3 _y_ = 2 _x_ 3 \+ 9 _x_ 2 − 3 _x_ − 3 in points at which the tangent lines to the two curves are parallel.

Let _x = x_ 0 be such a line. The tangent lines at _x_ 0 have slopes:

For ( _a_ ):

For ( _b_ ): 3 _y′_ = 6x2 \+ 18x − 3; at

Since _m_ 1 = _m_ 2,  . Then  , ( _x_ 0 − 3)( _x_ 0 \+ 1) = 0. Hence, _x_ 0 = 3 or _x_ 0 = −1. Thus, the vertical lines are _x_ = 3 and _x_ = −1.

**7** ( _a_ ) Show that the slope—intercept equation of the tangent line of slope _m_ ≠ 0 to the parabola _y_ 2 = 4 _px_ is _y_ = _mx + p/m_.

( _b_ ) Show that an equation of the tangent line to the ellipse _b_ 2 _x_ 2 \+ _a_ 2 _y_ 2 = _a_ 2 _b_ 2 at the point _P_ 0( _x_ 0, _y_ 0) on the ellipse is _b_ 2 _x_ 0 _x_ \+ _a_ 2 _y_ 0 _y_ = _a_ 2 _b_ 2.

( _a_ ) _y_ ′ = 2 _p/y_. Let _P_ 0( _x_ 0, _y_ 0) be the point of tangency. Then   and _m_ = 2 _p/y_ 0. Hence, _y_ 0 = 2 _p/m_ and  . The equation of the tangent line is then _y_ −2 _p/m_ = _m_ ( _x−p/m_ 2), which reduces to _y = mx + p/m_.

( _b_ )  . At _P_ 0,  . An equation of the tangent line is  , which reduces to  (since ( _x_ 0, _y_ 0) satisfies the equation of the ellipse).

**8.** Show that at a point _P_ 0( _x_ 0, _y_ 0) on the hyperbola _b_ 2 _x_ 2 − _a_ 2 _y_ 2 = _a_ 2 _b_ 2, the tangent line bisects the angle included between the focal radii of _P_ 0.

At _P_ 0 the slope of the tangent to the hyperbola is _b_ 2 _x_ 0/ _a_ 2 _y_ 0 and the slopes of the focal radii _P_ 0 _F′_ and _P_ 0 _F_ (see Fig. 12-3) are _y_ 0/( _x_ 0 \+ _c_ ) and _y_ 0/( _x_ 0 − _c_ ), respectively. Now

since _b_ 2 _x_ 20 − _a_ 2 _y_ 20 = _a_ 2 _b_ 2 and _a_ 2 \+ _b_ 2 = _c_ 2, and

Hence, _α = β_ because tan _α_ = tan _β_.

**Fig. 12-3**

**9.** One of the points of intersection of the curves ( _a_ ) _y_ 2 = 4 _x_ and ( _b_ ) 2 _x_ 2 = 12 − 5 _y_ is (1,2). Find the acute angle of intersection of the curves at that point.

For ( _a_ ), _y_ ′ = 2/ _y_. For ( _b_ ), _y_ ′ = −4 _x_ /5. Hence, at (1, 2), _m_ 1 = 1 and  . So,

Then   ≈ 83°40′ is the acute angle of intersection.

**10.** Find the angles of intersection of the curves ( _a_ ) 2 _x_ 2 \+ _y_ 2 = 20 and ( _b_ ) 4 _y_ 2 − _x_ 2 = 8.

Solving simultaneously, we obtain _y_ 2 = 4, _y_ = ±2. Then the points of intersection are   and  . For ( _a_ ), _y_ ′ = −2 _x_ / _y_ , and, for ( _b_ ), _y_ ′ = _x_ /4 _y_. At the point  ,   and  . Since _m_ 1 _m_ 2 = − 1, the angle of intersection is 90° (that is, the curves are _orthogonal_ ). By symmetry, the curves are orthogonal at each of their points of intersection.

**11.** A cable of a certain suspension bridge is attached to supporting pillars 250 ft apart. If it hangs in the form of a parabola with the lowest point 50 ft below the point of suspension, find the angle between the cable and the pillar.

Take the origin at the vertex of the parabola, as in Fig. 12-4. The equation of the parabola is  , and _y_ ′ = 4 _x_ /625.

At (125, 50), _m_ = 4(125)/625 = 0.8000 and θ = 38°40′. Hence, the required angle is   = 90° − θ = 51°20′.

**Fig. 12-4**

### **Supplementary Problems**

**12.** Examine _x_ 2 \+ 4 _xy_ \+ 16 _y_ 2 = 27 for horizontal and vertical tangent lines.

_Ans._ Horizontal tangents at   and  . Vertical tangents at   and

**13.** Find equations of the tangent and normal lines to _x_ 2 − _y_ 2 = 7 at the point (4, −3).

_Ans._ 4 _x_ \+ 3 _y_ = 7 and 3 _x_ − 4 _y_ = 24

**14.** At what points on the curve _y_ = _x_ 3 \+ 5 is its tangent line: ( _a_ ) parallel to the line 12 _x_ − _y_ = 17; ( _b_ ) perpendicular to the line _x_ \+ 3 _y_ = 2?

_Ans._ ( _a_ ) (2, 13), (−2, −3); ( _b_ ) (1, 6), (−1, 4)

**15.** Find equations of the tangent lines to 9 _x_ 2 \+ 16 _y_ 2 = 52 that are parallel to the line 9 _x_ − 8 _y_ = 1.

_Ans._ 9 _x_ − 8 _y_ = ±26

**16.** Find equations of the tangent lines to the hyperbola _xy_ = 1 that pass through the point (−1, 1).

_Ans._

**17.** For the parabola _y_ 2 = 4 _px_ , show that an equation of the tangent line at one of its points _P_ ( _x_ 0, _y_ 0) is _y_ 0 _y_ = 2 _p_ ( _x_ \+ _x_ 0).

**18.** For the ellipse _b_ 2 _x_ 2 \+ _a_ 2 _y_ 2 = _a_ 2 _b_ 2, show that the equations of its tangent lines of slope _m_ are  .

**19.** For the hyperbola _b_ 2 _x_ 2 − _a_ 2 _y_ 2 = _a_ 2 _b_ 2, show that ( _a_ ) an equation of the tangent line at one of its points _P_ ( _x_ 0, _y_ 0) is _b_ 2 _x_ 0 _x_ − _a_ 2 _y_ 0 _y_ = _a_ 2 _b_ 2; and ( _b_ ) the equations of its tangent lines of slope _m_ are  .

**20.** Show that the normal line to a parabola at one of its points _P_ bisects the angle included between the focal radius of _P_ and the line through _P_ parallel to the axis of the parabola.

**21.** Prove: Any tangent line to a parabola, except at the vertex, intersects the directrix and the latus rectum (produced if necessary) in points equidistant from the focus.

**22.** Prove: The chord joining the points of contact of the tangent lines to a parabola from any point on its directrix passes through the focus.

**23.** Prove: The normal line to an ellipse at any of its points _P_ bisects the angle included between the focal radii of _P_.

**24.** Prove: ( _a_ ) The sum of the intercepts on the coordinate axes of any tangent line to   is a constant. ( _b_ ) The sum of the squares of the intercepts on the coordinate axes of any tangent line to   is a constant.

**25.** Find the acute angles of intersection of the circles _x_ 2 − 4 _x_ \+ _y_ 2 = 0 and _x_ 2 \+ _y_ 2 = 8.

_Ans._ 45°

**26.** Show that the curves _y_ = _x_ 3 \+ 2 and _y_ = 2 _x_ 2 \+ 2 have a common tangent line at the point (0, 2) and intersect at the point (2, 10) at an angle   such that tan  .

**27.** Show that the ellipse 4 _x_ 2 \+ 9 _y_ 2 = 45 and the hyperbola _x_ 2 − 4 _y_ 2 = 5 are orthogonal (that is, intersect at a right angle).

**28.** Find equation of the tangent and normal lines to the parabola _y_ = 4 _x_ 2 at the point (−1, 4).

_Ans. y_ \+ 8 _x_ \+ 4 = 0; 8 _y_ − _x_ − 33 = 0

**29.** At what points on the curve _y_ = 2 _x_ 3 \+ 13 _x_ 2 \+ 5 _x_ \+ 9 does its tangent line pass through the origin?

_Ans._

## **CHAPTER 13
Law of the Mean. Increasing and Decreasing Functions**

**RELATIVE MAXIMUM AND MINIMUM** A function _f_ is said to have a _relative maximum_ at _x_ 0 if _f_ ( _x_ 0)   _f_ ( _x_ ) for all _x_ in some open interval containing _x 0_ (and for which _f_ ( _x_ ) is defined). In other words, the value of _f_ at _x 0_ at _x 0_ is greater than or equal to all values of _f_ at nearby points. Similarly, _f_ is said to have a _relative minimum_ at _x 0_ if _f_ ( _x_ 0)  _f_ ( _x_ ) for all _x_ in some open interval containing _x 0_ (and for which _f_ ( _x_ ) is defined). In other words, the value of _f_ at _x 0_ is less than or equal to all values of _f_ at nearby points. By a _relative extremum_ of _f_ we mean either a relative maximum or a relative minimum of _f_.

Theorem 13.1: If _f_ has a relative extremum at a point _x 0_ at which _f_ ′( _x_ 0) is defined, then _f_ ′( _x_ 0) = 0.

Thus, if _f_ is differentiable at a point at which it has a relative extremum, then the graph of _f_ has a horizontal tangent line at that point. In Fig. 13-1, there are horizontal tangent lines at the points _A_ and _B_ where _f_ attains a relative maximum value and a relative minimum value, respectively. See Problem 5 for a proof of Theorem 13.1

**Theorem 13.2 (Rolle's Theorem):** Let _f_ be continuous on the closed interval [ _a, b_ ] and differentiable on the open interval ( _a, b_ ). Assume that _f_ ( _a_ ) = _f_ ( _b_ ) = 0. Then _f_ ′( _x_ 0) = 0 for at least one point _x_ 0 in ( _a, b_ ).

This means that, if the graph of a continuous function intersects the _x_ axis _x = a_ and _x = b,_ and the function is differentiable between _a_ and _b_ , then there is at least one point on the graph between _a_ and _b_ where the tangent line is horizontal. See Fig. 13-2, where there is one such point. For a proof of Rolle's Theorem, see Problem 6.

**Fig. 13-1**

**Fig. 13-2**

**Corollary 13.3 (Generalized Rolle's Theorem):** Let _g_ be continuous on the closed interval [ _a, b_ ] and differentiable on the open interval ( _a, b_ ). Assume that _g_ ( _a_ ) = _g_ ( _b_ ). Then _g_ ′( _x_ 0) = 0 for at least one point _x_ 0 in ( _a, b_ ).

See Fig. 13-3 for an example in which there is exactly one such point. Note that Corollary 13.3 follows from Rolle's Theorem if we let _f_ ( _x_ ) = _g_ ( _x_ ) − _g_ ( _a_ ).

**Fig. 13-3**

**Theorem 13.4 (Law of the Mean)†:** Let _f_ be continuous on the closed interval [ _a, b_ ] and differentiable on the open interval ( _a, b_ ). Then there is at least one point _x_ 0 in ( _a, b_ ) for which

See Fig. 13-4. For a proof, see Problem 7. Geometrically speaking, the conclusion says that there is some point inside the interval where the slope _f_ ′( _x_ 0) of the tangent line is equal to the slope [ _f_ ( _b_ ) − _f_ ( _a_ )]/( _b_ − _a_ ) of the line P1P2 connecting the end points ( _a, f_ ( _a_ )) and ( _b, f_ ( _b_ )) of the graph. At such point, the tangent line is parallel to P1P2, since their slopes are equal.

**Theorem 13.5 (Extended Law of the Mean):** Assume that _f_ ( _x_ ) and _g_ ( _x_ ) are continuous on [ _a, b_ ], and differentiable on ( _a, b_ ). Assume also that _g_ ′( _x_ ) ≠ 0 for all _x_ in ( _a, b_ ). Then there exists at least one point _x_ 0 in

† The Law of the Mean is also called the Mean-Value Theorem for Derivatives.

**Fig. 13-4**

_(a, b)_ for which

For a proof, see Problem 13. Note that the Law of the Mean is the special case when _g_ ( _x_ ) = _x_.

**Theorem 13.6 (Higher-order Law of the Mean):** If _f_ and its first _n − 1_ derivatives are continuous on [ _a, b_ ] and _f (n)(x)_ exists on ( _a, b_ ), then there is at least one _x 0_ in _(a, b)_ such that

(For a proof, see Problem 14.)

When _b_ is replaced by _x,_ formula (1) becomes

for some _x 0_ between _a_ and _x._

In the special case when _a = 0,_ formula (2) becomes

for some _x 0_ between 0 and _x._

**INCREASING AND DECREASING FUNCTIONS.** A function _f_ is said to be _increasing_ on an interval if _u < v_ implies _f(u) < f(v)_ for all _u_ and _v_ in the interval. Similarly, _f_ is said to be _decreasing_ on an interval if _u < v_ implies _f(u) > f(v)_ for all _u_ and _v_ in the internal.

**Theorem 13.7:** ( _a_ ) If _f′_ is positive on an interval, then _f_ is increasing on that interval. ( _b_ ) If _f′_ is negative on an interval, then _f_ is decreasing on that interval.

For a proof, see Problem 9.

### **Solved Problems**

**1.** Find the value of _x_ 0 prescribed in Rolle's Theorem for _f(x) = x_ 3 − 12x on the interval  .

Note that  . If _f_ ′( _x_ ) = 3 _x_ 2 − 12 = 0, then _x_ = ±2. Then _x_ 0 = 2 is the prescribed value.

**2.** Does Rolle's Theorem apply to the functions: ( _a_ )  , and ( _b_ )   on the interval (0, 4)?

( _a_ ) _f_ ( _x_ ) = 0 when _x_ = 0, 4. Since _f_ has a discontinuity at _x_ = 2, a point on [0, 4], the theorem does not apply.

( _b_ ) _f_ ( _x_ ) = 0 when _x_ = 0, 4. _f_ has a discontinuity at _x_ = −2, a point not on [0, 4]. In addition, _f′(x)_ = ( _x_ 2 \+ 4 _x_ − 8)/( _x_ \+ 2)2 exists everywhere except at _x_ = −2. So, the theorem applies and  , the positive root of _x_ 2 \+ 4 _x_ − 8 = 0.

**3.** Find the value of _x_ 0 prescribed by the law of the mean when _f_ ( _x_ ) = 3 _x_ 2 \+ 4 _x_ − 3 and _a_ = 1, _b_ = 3.

_f(a)_ = _f_ (1) = 4, _f(b)_ = _f_ (3) = 36, _f′_ ( _x_ 0) = 6 _x_ 0 \+ 4, and _b − a_ = 2. So,  . Then _x_ 0 = 2.

**4.** Find a value _x_ 0 prescribed by the extended law of the mean when _f_ ( _x_ ) = 3 _x_ \+ 2 and _g(x)_ = _x_ 2 \+ 1, on [1,4].

We have to find _x_ 0 so that

Then  .

**5. Prove Theorem 13.1** : If _f_ has a relative extremum at a point _x_ 0 at which _f′_ ( _x_ 0) is defined, then _f′_ ( _x_ 0) = 0.

Consider the case of a relative maximum. Since _f_ has a relative maximum at _x_ 0, then, for sufficiently small |Δ _x_ | _f_ ( _x_ 0 \+ Δ _x_ ) < _f_ ( _x_ 0), and so _f_ ( _x_ 0 \+ Δx) − _f_ ( _x_ 0) < 0. Thus, when  . So,

When  . Hence,

Since _f′_ ( _x_ 0)   0, and _f_ ′( _x_ 0)   0, it follows that _f′_ ( _x_ 0) = 0.

**6.** Prove Rolle's Theorem (Theorem 13.2): If _f_ is continuous on the closed interval [a, b] and differentiable on the open interval ( _a, b_ ), and if _f_ (a) = _f_ ( _b_ ) = 0, then _f′_ ( _x_ 0) = 0 for some point _x_ 0 in ( _a, b_ ).

If _f_ ( _x_ ) = 0 throughout [ _a, b_ ], then _f′(x)_ = 0 for all _x_ in ( _a, b_ ). On the other hand, if _f_ ( _x_ ) is positive (negative) somewhere in ( _a, b_ ), then, by the Extreme Value Theorem (Theorem 8.7), _f_ has a maximum (minimum) value at some point _x_ 0 on [a, b]. That maximum (minimum) value must be positive (negative), and, therefore, _x_ 0 lies on ( _a, b_ ), since _f(a)_ = _f(b)_ = 0. Hence, _f_ has a relative maximum (minimum) at _x_ 0. By Theorem 13.1 _f′_ ( _x_ 0) = 0.

**7.** Prove the Law of the Mean (Theorem 13.4): Let _f_ be continuous on the closed interval [ _a, b_ ] and differentiable on the open interval ( _a, b_ ). Then there is at least one point _x_ 0 in ( _a, b_ ) for which [ _f(b) − f(a)]/(b − a) = f′(x 0)_.

Let  .

Then _F(a)_ = 0 = _F(b)._ So, Rolle's Theorem applies to _F_ on [ _a-b_ ]. Hence, for some _x_ 0 in ( _a, b_ ), _F′_ (x0) = 0.

But  . Thus,  .

**8.** Show that, if _g_ is increasing on an interval, then − _g_ is decreasing on the interval.

Assume _u < v_. Then _g(u)_ < _g(v)_. Hence, _−g(u)_ > _-g(v)_.

**9.** Prove Theorem 13.7:( _a_ ) If _f′_ is positive on an interval, then _f_ is increasing on the interval, _(b)_ If _f′_ is negative on an interval, then _f_ is decreasing on that interval.

.( _a_ ) Let _a_ and _b_ be any two points on the interval with _a < b_. By the Law of the Mean, [ _f(b) − f(a)_ ]/(b-a) = _f′_ ( _x_ 0) for some point _x_ 0 in ( _a, b_ ). Since _x_ 0 is in the interval, _f′_ ( _x_ 0) > 0. Thus, [ _f(b) − f(a)_ ]/(b − a) > 0. But, _a < b_ and, therefore, _b − a_ > 0. Hence, _f(b) − f(a)_ > 0. So, _f_ (a) < _f(b)_.

( _b_ ) Let _g = −f_. So, _g′_ is positive on the interval. By part ( _a_ ), _g_ is increasing on the interval. So, _f_ is decreasing on the interval.

**10.** Show that _f_ ( _x_ ) = _x_ 5 \+ 20x − 6 is an increasing function for all values of _x_.

_f′(x)_ = 5 _x_ 4 \+ 20 > 0 for all _x_. Hence, by Theorem 13.7( _a_ ), _f_ is increasing everywhere.

**11.** Show that _f_ ( _x_ ) = 1 − _x_ 3 − _x_ 7 is a decreasing function for all values of _x_.

_f′(x)_ = −3 _x_ 2 − 7 _x_ 6 < 0 for all _x_ ≠ 0. Hence, by Theorem 13.7(b), _f_ is decreasing on any interval not containing 0. Note that if _x_ < 0, _f_ ( _x_ ) > 1 = _f_ (0), and if _x_ > 0, _f_ ( _x_ ). So, _f_ is decreasing for all real numbers.

**12.** Show that _f_ ( _x_ ) = 4 _x_ 3 \+ _x_ − 3 = 0 has exactly one real solution.

_f_ (0) = −3 and _f_ (1) = 2. So, the intermediate value theorem tells us that _f_ ( _x_ ) = 0 has a solution in (0,1). Since _f′(x)_ = 12 _x_ 2 \+ 1 > 0, _f_ is an increasing function. Therefore, there cannot be two values of _x_ for which _f_ ( _x_ ) = 0.

**13.** Prove the Extended Law of the Mean (Theorem 13.5): If _f_ ( _x_ ) and g(x) are continuous on [ _a, b_ ], and differentiable on ( _a, b_ ), and _g′(x)_ ≠ 0 for all _x_ in ( _a, b_ ), then there exists at least one point _x_ 0 in ( _a, b_ ) for which  .

Suppose that _g(b) = g(a)_. Then, by the generalized Rolle's Theorem, _g′_ (x) = 0 for some _x_ in ( _a, b_ ), contradicting our hypothesis. Hence, _g(b) ≠ g(a)_.

Let  .

Then

By Rolle's theorem, there exists _x_ 0 in ( _a, b_ ) for which  .

**14.** Prove the Higher-Order Law of the Mean (Theorem 13.6): If _f_ and its first n − 1 derivatives are continuous on [ _a, b_ ] and _f_ (n)( _x_ ) exists on ( _a, b_ ), then there is at least one _x_ 0 in ( _a, b_ ) such that

Let a constant _K_ be defined by

and consider

Now _F(a)_ = 0 by (2), and _F(b)_ = 0. By Rolle's Theorem, there exists _x_ 0 in ( _a, b_ ) such that

Then  , and (2) becomes (1).

**15.** If _f′(x)_ = 0 for all _x_ on ( _a, b_ ), then _f_ is constant on ( _a, b_ ).

Let _u_ and _v_ be any two points in ( _a, b_ ) with _u < v_. By the Law of the Mean, there exists _x_ 0 in ( _u, v_ ) for which  . By hypothesis, _f′(x 0) = 0_. Hence, _f(v) − f(u)_ = 0, and, therefore, _f(v) = f(u)_.

### **Supplementary Problems**

**16.** If _f_ ( _x_ ) = _x_ 2 − 4 _x_ \+ 3 on [1,3], find a value prescribed by Rolle's Theorem.

_Ans. _x__ 0 = 2

**17.** Find a value prescribed by the Law of the Mean, given:

**18.** If _f′_ (x) = _g′(x)_ for all _x_ in ( _a, b_ ), prove that there exists a constant _K_ such that _f(x) = g(x) + K_ for all _x_ in ( _a, b_ ). ( _Hint: D x(f(x) − g(x))_ = 0 in ( _a, b_ ). By Problem 15, there is a cinstant _K_ such that _f_ ( _x_ ) − _g(x) = K_ in ( _a, b_ ).)

**19.** Find a value _x_ 0 precribed by the extended law of the mean when _f_ ( _x_ ) = _x_ 2 \+ 2 _x_ − 3, _g(x) = x_ 2 − 4 _x_ \+ 6 on the interval [0,1].

_Ans_.

**20.** Show that _x_ 3 \+ _px + q_ = 0 has: ( _a_ ) one real root if _p_ > 0, and ( _b_ ) three real roots if 4 _p_ 3 \+ 27 _q_ 2 < 0.

**21.** Show that   has neither a relative maximum nor a relative minimum. ( _Hint_ : Use Theorem 13.1.)

**22.** Show that _f_ ( _x_ ) = 5 _x_ 3 \+ 11 _x_ − 20 = 0 has exactly one real solution.

**23.** ( _a_ ) Where are the following functions increasing and where are they decreasing? Sketch the graphs.

( _b_ )(GC) Check your answers to ( _a_ ) by means of a graphing calculator.

**24.** (GC) Use a graping calculator to estimate the intervals on which _f_ ( _x_ ) = _x_ 5 \+ 2 _x_ 3 − 6 _x_ \+ 1 is increasing, and the intervals on which it is decreasing.

**25.** For the following functions determine whether Rolle's Theorem is applicable. If it is, find the prescribed values.

## **CHAPTER 14
Maximum and Minimum Values**

**CRITICAL NUMBER.** A number _x_ 0 in the domain of _f_ such that either _f_ ′( _x_ 0) = 0 or _f_ ′( _x_ 0) is not defined is called a _critical number_ of _f_.

Recall (Theorem 13.1) that, if _f_ has a relative extremum at _x_ 0 and _f_ ′( _x_ 0) is defined, then _f_ ′( _x_ 0) = 0 and, therefore, _x_ 0 is a critical number of _f_. Observe, however, that the condition that _f_ ′( _x_ 0) = 0 does not guarantee that _f_ has a relative extremum at _x_ 0. For example, if _f_ ( _x_ ) = _x_ 3, then _f_ ′( _x_ ) = 3 _x_ 2, and, therefore, 0 is a critical number of _f_ ; but _f_ has neither a relative maximum nor relative minimum at 0. (See Fig. 5-5).

**EXAMPLE 1**

( _a_ ) Let _f_ ( _x_ ) = 7 _x_ 2 − 3 _x_ \+ 5. Then _f_ ′( _x_ ) = 14 _x_ − 3. Set _f_ ′( _x_ ) = 0 and solve. The only critical number of _f_ is  .

( _b_ ) Let _f_ ( _x_ ) = _x_ 3 − 2 _x_ 2 \+ _x_ \+ 1. Then _f_ ′( _x_ ) = 3 _x_ 2 − 4 _x_ \+ 1. Solving _f_ ′( _x_ ) = 0, we find that the critical numbers are 1 and  .

( _c_ )Let _f_ ( _x_ ) = _x_ 2/3. Then  . Since _f_ ′(0) is not defined, 0 is the only critical number of _f_.

We shall find some conditions under which we can conclude that a function _f_ has a relative maximum or a relative minimum at a given critical number.

**SECOND DERIVATIVE TEST FOR RELATIVE EXTREMA.** Assume that _f_ ′( _x_ 0) = 0 and that _f_ ″( _x_ 0) exixts. Then:

( _i_ ) if _f_ ″( _x_ 0) < 0, then _f_ has a relative maximum at _x_ 0;

( _ii_ ) if _f_ ″( _x_ 0) > 0, then _f_ has a relative minimum at _x_ 0;

( _iii_ ) if _f_ ″( _x_ 0) = 0, we do not know what is happeining at _x_ 0.

A proof is given in Problem 9. To see that ( _iii_ ) holds, consider the three functions _f_ ( _x_ ) = _x_ 4, _g_ ( _x_ ) = − _x_ 4, and _h_ ( _x_ ) = _x_ 3. Since _f_ ′( _x_ ) = 4 _x_ 3, _g_ ′( _x_ ) = −4 _x_ 3, and _h_ ′( _x_ ) = 3 _x_ 2, 0 is a critical number of all three functions. Since _f_ ″( _x_ ) = 12 _x_ 2, _g_ ″( _x_ ) = −12 _x_ 2, and _h_ ″( _x_ ) = 6 _x_ , the second derivative of all three functions is 0 at 0. However, _f_ has a relative minimum at 0, _g_ has a relative maximum at 0, and _h_ has neither a relative maximum nor a relative minimum at 0.

**EXAMPLE 2**

( _a_ ) Consider the function _f_ ( _x_ ) = 7 _x_ 2 − 3 _x_ \+ 5 of Example 1(a). The only critical number was  . Since _f_ ″( _x_ ) = 14,  . So, the second derivative test tells us that _f_ has a relative minimum at  .

( _b_ ) Consider the function _f_ ( _x_ ) = _x_ 3 − 2 _x_ 2 \+ _x_ \+ 1 of Example 1(b). Note that _f_ ″( _x_ ) = 6 _x_ − 4. At the critical numbers 1 and  , _f_ ″ (1) = 2 > 0 and _f_ ″ ( ) = − 2 < 0. Hence, _f_ has a relative minimum at 1 and a relative maximum at  .

( _c_ ) In Example 1(c), _f_ ( _x_ ) = _x_ 2/3 and  . The only critical number is 0, where _f_ ′ is not defined. Hence, _f_ ″(0) is not defined and the second derivative test is not applicable.

If the second derivative test is not usable or convenient, either because the second derivative is 0, or does not exist, or is difficult to compute, then the following test can be applied. Recall that _f_ ′( _x_ ) is the slope of the tangent line to the graph of _f_ at _x_.

**FIRST DERIVATIVE TEST.** Assume _f_ ′( _x_ 0) = 0.

**Case {+,-}:** If _f_ ′ is positive in an open interval immediately to the left of _x_ 0, and negative in an open interval immediately to the right of _x_ 0, then _f_ has a relative maximum at _x_ 0. (See Fig. 14-1(a).)

**Case {−,+}:** If _f_ ′ is negative in an open interval immediately to the left of _x_ 0, and positive in an open interval immediately to the right of _x_ 0, then _f_ has a relative minimum at _x_ 0. (See Fig. 14-1(b).)

**Fig. 14-1**

**Cases {+,+}** and **{−,−}:** If _f_ ′ has the same sign in open intervals immediately to the left and to the right of _x_ 0, then _f_ has neither a relative maximum nor a relative minimum at _x_ 0. (See Fig. 14-1(c, d).)

For a proof of the first derivative test, see Problem 8.

**EXAMPLE 3:** Consider the three functions _f_ ( _x_ ) = _x_ 4, _g_ ( _x_ ) = − _x_ 4, and _h_ ( _x_ ) = _x_ 3 discussed above. At their critical number 0, the second derivative test was not applicable because the second derivative was 0. Let us try the first derivative test.

( _a_ ) _f_ ′( _x_ ) = 4 _x_ 3. To the left of 0, _x_ < 0, and so, _f_ ′( _x_ ) < 0. To the right of 0, _x_ > 0, and so, _f_ ′( _x_ ) > 0. Thus, we have the case {−,+} and _f_ must have a relative minimum at 0.

( _b_ ) _g_ ′( _x_ ) = −4 _x_ 3. To the left of 0, _x_ < 0, and so, _g_ ′( _x_ ) > 0. To the right 0, _x_ > 0, and so, _g_ ′( _x_ ) < 0. Thus, we have the case {+,−} and _g_ must have a relative maximum at 0.

( _c_ ) _h_ ′( _x_ ) = 3 _x_ 2. _h_ ′( _x_ ) > 0 on both sides of 0. Thus, we have the case {+,+} and _h_ has neither a relative maximum nor a relative minimum at 0. There is an _inflection point_ at _x_ =0.

These results can be verified by looking at the graphs of the functions.

**ABSOLUTE MAXIMUM AND MINIMUM.** An _absolute maximum_ of a function _f_ on a set _S_ occurs at _x_ 0 in _S_ if _f_ ( _x_ )  _f_ ( _x_ 0) for all _x_ in _S_. An _absolute minimum_ of a function _f_ on a set _S_ occurs at _x_ 0 in _S_ if _f_ ( _x_ )  _f_ ( _x_ 0) for all _x_ in _S_.

**TABULAR METHOD FOR FINDING THE ABSOLUTE MAXIMUM AND MINIMUM.** Let _f_ be continuous on  _a, b_ ] and differentiable on ( _a, b_ ). By the Extreme Value Theorem, we know that _f_ has an absolute maximum and minimum of [ _a, b_ ]. Here is a tabular method for determining what they are and where they occur. (See [Fig. 14-2.)

**Fig. 14-2**

First, find the critical numbers (if any) _c_ 1, _c_ 2,... of _f_ in ( _a, b_ ). Second, list these numbers in a table, along with the enpoints _a_ and _b_ of the interval. Third, calculate the value of _f_ for all the numbers in the table.

Then:

1. The largest of these values is the absolute maximum of _f_ on [ _a, b_ ].

2. The smallest of these values is the absolute minimum of _f_ on [ _a, b_ ].

**EXAMPLE 4:** Let us find the absolute maximum and minimum of _f_ ( _x_ ) = _x_ 3 − _x_ 2 − _x_ \+ 2 on [0,2].

_f_ ′( _x_ ) = 3 _x_ 2 − 2 _x_ \- 1 = (3 _x_ \+ 1)( _x_ − 1). Hence, the critical numbers are   and _x_ = 1. The only critical number in 0,2] is 1. From the table in [Fig. 14-3, we see that the maximum value of _f_ on [0,2] is 4, which is attained at the right end point 2, and the minimum value is 1, attained at 1.

**Fig. 14-3**

Let us see why the method works. By the Extreme Value Theorem, _f_ achieves maximum and minimum values on the closed interval [ _a, b_ ]. If either of those values occurs at an end point, that value will appear in the table and, since it is actually a maximum or minimum, it will show up as the largest or smallest value. If the maximum or minimum is assumed at a point _x_ 0 inside the interval, _f_ has a relative maximum or minimum at _x_ 0 and, therefore, by Theorem 13.1, _f_ ′( _x_ 0) = 0. Thus, _x_ 0 will be a critical number and will be listed in the table, so that the corresponding maximum or minimum value _f_ ( _x_ 0) will be the largest or smallest value in the right-hand column.

**Theorem 14.1:** Assume that _f_ is a continuous function defined on an interval _J_. The interval _J_ can be a finite or infinite interval. If _f_ has a unique relative extremum within _J_ , then that relative extremum is also an absolute extremum on _J_.

To see why this is so, look at Fig. 14-4, where _f_ is assumed to have a unique extremum, a relative maximum at _c_. Consider any other number _d_ in _J_. The graph moves downward on both sides of _c_. So, if _f_ ( _d_ ) were greater than _f_ ( _c_ ), then, by the Extreme Value Theorem for the closed interval with endpoints _c_ and _d, f_ would have an absolute minimum at some point _u_ , between _c_ and _d_. ( _u_ could not be equal to _c_ or _d_.) Then _f_ would have a relative minimum at _u_ , contradicting our hypothesis that _f_ has a relative extremum only at _c_. We can extend this argument to the case where _f_ has a relative minimum at _c_ by applying the result we have just obtained to − _f_.

**Fig. 14-4**

### **Solved Problems**

**1.** Locate the absolute maximum or minimum of the following functions on their domains:

( _a_ ) _y = −x_ 2; ( _b_ ) _y_ = ( _x_ −3)2; ( _c_ )   ( _d_ )  .

( _a_ ) _y_ = − _x_ 2 has an absolute maximum (namely, 0) when _x_ =0, since _y_ < 0, when _x_ ≠ 0. It has no relative minimum, since its range is (−∞, 0]. The graph is a parabola opening downward, with vertex at (0, 0).

( _b_ ) _y_ = ( _x_ − 3)2 has an absolute minimum, 0, when _x_ = 3, since _y_ > 0 when _x_ ≠3. It has no absolute maximum, since its range is [0,+∞). The graph is a parabola openinig upward, with vertex at (3,0).

( _c_ )   has 5 as its absolute maximum, when _x_ = 0, since 25 − 4 _x_ 2 < 25 when _x_ ≠ 0. It has 0 as its absolute minimum, when  . The graph is the upper half of an ellipse.

( _d_ )   has 0 as its absolute minimum when _x_ = 4. It has no absolute maximum. Its graph is the upper half of a parabola with vertex at (4,0) and the _x_ axis as its axis of symmetry.

**2.** Let  . Find: ( _a_ ) the critical numbers of _f_ ; ( _b_ ) the points at which _f_ has a relative maximum or minimum; ( _c_ ) the intervals on which _f_ is increasing or decreasing.

( _a_ ) _f_ ′( _x_ ) = _x_ 2 \+ _x_ − 6 = ( _x_ \+ 3)( _x_ − 2). Solving _f_ ′( _x_ ) = 0 yields the critical numbers −3 and 2.

( _b_ ) _f_ ″( _x_ ) = 2 _x_ \+ 1. Thus, _f_ ″(−3) = −5 < 0 and _f_ ″(2) = 5. Hence, by the second derivative test, _f_ has a relative maximum at _x_ = −3, where  . By the second derivative test, _f_ has a relative minimum at _x_ = 2, where  .

( _c_ ) Look at _f_ ′( _x_ ) = ( _x_ \+ 3)( _x_ − 2). When _x_ > 2, _f_ ′( _x_ ) > 0. For −3 < _x_ < 2, _f_ ′( _x_ ) < 0. For _x_ < −3, _f_ ′( _x_ ) > 0. Thus, by Theorem 13.7, _f_ is increasing for _x_ < −3 and 2 < _x_ , and decreasing for −3 < _x_ < 2.

A sketch of part of the graph of _f_ is shown in Fig. 14-5. Note that _f_ has neither absolute maximum nor absolute minimum.

**Fig. 14-5**

**3.** Let _f_ ( _x_ ) = _x_ 4 \+ 2 _x_ 3 −3 _x_ 2−4 _x_ \+ 4. Find: ( _a_ ) the critical numbers of _f_ ; ( _b_ ) the points at which _f_ has a relative extremum; ( _c_ ) the intervals on which _f_ is increasing or decreasing.

( _a_ ) _f_ ′( _x_ ) = 4 _x_ 3 \+ 6 _x_ 2 −6 _x_ − 4. It is clear that _x_ = 1 is a zero of _f_ ′( _x_ ). Dividing _f_ ′( _x_ ) by _x_ − 1 yields 4 _x_ 2 \+ 10 _x_ \+ 4, which factors into 2(2 _x_ 2 \+ 5 _x_ \+ 2) = 2(2 _x_ \+ 1)( _x_ \+ 2). Thus, _f_ ′( _x_ ) = 2( _x_ − 1)(2 _x_ \+ 1)( _x_ \+ 2), and the critical numbers are 1,  , and −2.

( _b_ ) _f_ ″( _x_ ) = 12 _x_ 2 \+ 12 _x_ − 6 = 6(2 _x_ 2 \+ 2 _x_ − 1). Using the second derivative test, we find: ( _i_ ) at _x_ = 1, _f_ ″(1) = 18 > 0, and there is a relative minimum; ( _ii_ ) at  , _f_ ″( ) = −9 < 0, so that there is a relative maximum; ( _iii_ ) at _x_ = −2, _f_ ″(−2) = 18 > 0, so that there is a relative minimum.

( _c_ ) _f_ ′( _x_ ) > 0 when _x_ > 1, _f_ ′( _x_ ) < 0 when  , _f_ ′( _x_ ) > 0 when  , and _f_ ′( _x_ ) < 0 when _x_ < −2. Hence, _f_ is increasing when _x_ > 1 or  , and decreasing when   or _x_ < −2.

The graph is sketched in Fig. 14-6.

**4.** Examine   for relative extrema, and find the intervals on which _f_ is increasing or decreasing.

_f_ ( _x_ ) = ( _x_ − 2)−1, so that  . Thus, _f_ ′ is never 0, and the only number where _f_ ′ is not defined is the number 2, which is not in the domain of _f_. Hence, _f_ has no critical numbers. So,

**Fig. 14-6**

_f_ has jo relative extrema. Note that _f_ ′( _x_ ) < 0 for _x_ ≠ 2. Hence, _f_ is decreasing for _x_ < 2 and for _x_ > 2. There is a nonremovable discontinuity at _x_ = 2. The graph is shown in Fig. 14-7.

**Fig. 14-7**

**5.** Locate the relative extrema of _f_ ( _x_ ) = 2 + _x_ 2/3 and the intervals on which _f_ is increasing or decreasing.

 . Then _x_ = 0 is a critical number, since _f_ ′(0) is not not defined (but 0 is in the domain of _f_ ). Note that _f_ ′( _x_ ) approaches ∞ as _x_ approaches 0. When _x_ < 0, _f_ ′( _x_ ) is negative and, therefore, _f_ is decreasing. When _x_ > 0, _f_ ′( _x_ ) is positive and, therefore, _f_ is increasing. The graph is sketched in Fig. 14-8. _f_ has an absolute minimum at _x_ = 0.

**Fig. 14-8**

**6.** Use the second derivative test to examine the relative extrema of the following functions:

( _a_ ) _f_ ( _x_ ) = _x_ (12 − 2 _x_ )2; ( _b_ )  .

( _a_ ) _f_ ′( _x_ ) = _x_ (2)(12 − 2 _x_ )(−2) + (12 − 2 _x_ )2 = (12 − 2 _x_ )(12 − 6 _x_ ) = 12( _x_ − 6)( _x_ − 2). So, 6 and 2 are the critical numbers. _f_ ″( _x_ ) = 12(2 _x_ − 8) = 24( _x_ − 4). So, _f_ ″(6) = 48 > 0, and _f_ ″(2) = -48 < 0. Hence, _f_ has a relative minimum at _x_ = 6 and a relative maximum at _x_ = 2.

( _b_ )  . So, the only critical number is 5 (where _x_ 3 − 125 = 0.) _f_ ″( _x_ ) = 2 + 500/ _x_ 3. Since _f_ ″(5) = 6 > 0, _f_ has a relative minimum at _x_ = 5.

**7.** Determine the relative extrema of _f_ ( _x_ ) = ( _x_ − 2)2/3.

 . So, 2 is the only critical number. Since _f_ ′(2) is not defined _f_ ″(2) will be undefined. Hence, we shall try the first derivative test. For _x_ < 2, _f_ ′( _x_ ) < 0, and, for _x_ > 2, _f_ ′( _x_ ) > 0. Thus, we have the case {−,+} of the first derivative test, and _f_ has a relative minimum at _x_ = 2.

**8.** Prove the first derivative test.

Assume _f_ ″( _x_ 0) = 0. Consider the case {+,−}: If _f_ ′ is positive in an open interval immediately to the left of _x_ 0, and negative in an open interval immediately to the right of _x_ 0, then _f_ has a relative maximum at _x_ 0. To see this, notice that, by Theorem 13.8, since _f_ ′ is positive in an open interval immediately to the left of _x_ 0, _f_ is increasing in that interval, and since _f_ ′ is negative in an open interval immediately to the right of _x_ 0, _f_ is decreasing in that interval. Hence, _f_ has a relative maximum at _x_ 0. The case {−,+} follows from the case {+,−} applied to − _f_. In the case {+,+}, _f_ will be increasing in an interval around _x_ 0, and, in the case {−,−}, _f_ will be decreasing in an interval around _x_ 0. So, in both cases, _f_ has neither a relative maximum nor minimum at _x_ 0.

**9.** Prove the second derivative test: If _f_ ( _x_ ) is differentiable on an open interval containing a critical value _x_ 0 of _f_ , and _f_ ″( _x_ 0) exists and _f_ ″( _x_ 0) is positive (negative), then _f_ has a relative minimum (maximum) at _x_ 0.

Assume _f_ ″( _x_ 0) > 0. Then, by Theorem 13.8, _f_ ′ is increasing at _x_ 0. Since _f_ ″( _x_ 0) = 0, this implies that _f_ ′ is negative nearby and to the left of _x_ 0, and _f_ ′ is positive nearby and to the right of _x_ 0. Thus, we have the case {−,+} of the first derivative test and, therefore, _f_ has a relative minimum at _x_ 0. In the opposite situation, where _f_ ″( _x_ 0) < 0, the result we have just proved is applicable to the function _g_ ( _x_ ) = − _f_ ( _x_ ). Then _g_ has a relative minimum at _x_ 0 and, therefore, _f_ has a relative maximum at _x_ 0.

**10.** Among those positive real numbers _u_ and _υ_ whose sum is 50, find that choice of _u_ and _υ_ that makes their product _P_ as large as possible.

_P_ = _u_ (50 − _u_ ). Here, _u_ is any positive number less than 50. But we also can allow _u_ to be 0 or 50, since, in those cases, _P_ = 0, which will certainly not be the largest possible value. So, _P_ is a continuous function _u_ (50 - _u_ ), defined on 0, 50]. _P_ = 50 _u_ − _u_ 2 is also differentiable everywhere, and _dP_ / _du_ = 50 − 2 _u_. Setting _dP_ / _du_ = 0 yields a unique critical number _u_ = 25. By the tabular method ([Fig. 14-9), we see that the maximum value of _P_ is 625, when _u_ = 25 (and, therefore, _υ_ = 50 − _u_ = 25).

**Fig. 14-9**

**11.** Divide the number 120 into two parts such that the product _P_ of one part and the square of the other is a maximum.

Let _x_ be one part and 120 − _x_ the other part. Then _P_ = (120 - _x_ ) _x_ 2 and 0   _x_   120. Since _dP_ / _dx_ = 3 _x_ (80 − _x_ ), the critical numbers are 0 and 80. Using the tabular method, we find _P_ (0) = 0, _P_ (80) = 256,000 and _P_ (120) = 0. So, the maximum value occurs when _x_ = 80, and the required parts are 80 and 40.

**12.** A sheet of paper for a poster is to be 18ft2 in area. The margins at the top and bottom are to be 9 inches, and the margins at the sides 6 inches. What should be the dimensions of the sheet to maximize the printed area?

Let _x_ be one dimension, measured in feet. Then 18/ _x_ is the other dimension. (See Fig. 14-10.) The only restriction on _x_ is that _x_ > 0. The printed area in square feet is  , and  .

**Fig. 14-10**

Solving _dA_ / _dx_ = 0 yields the critical number  . Since _d_ 2 _A_ / _dx_ 2 = −36/ _x_ 3 is negative when  , the second derivative test tells us that _A_ has a relative maximum at  . Since   is the only critical number in the interval (0,+∞), Theorem 14.1 tells us that _A_ has an absolute maximum at  . Thus, one side is   ft and the other side is   ft.

**13.** At 9 A.M., ship _B_ is 65 miles due east of another ship _A._ Ship _B_ is then sailing due west at 10 mi/h, and _A_ is sailing due south at 15 mi/h. If they continue on their respective courses, when will they be nearest one another, and how near? (See Fig. 14-11.)

**Fig. 14-11**

Let _A_ 0 and _B_ 0 be the positions of the ships at 9 A.M., and _A_ _t_ and _B_ _t_ their positions _t_ hours later. The distance covered in _t_ hours by _A_ is 15 _t_ miles; by _B_ , 10 _t_ miles. The distance _D_ between the ships is determined by _D_ 2 = (15 _t_ )2 +(65 − 10 _t_ )2. Then

Solving _dD_ / _dt_ = 0, yields the critical number _t_ = 2. Since _D_ > 0 and 325 _t_ − 650 is negative to the left of 2 and positive to the right of 2, the case (−,+) of the first derivation test tells us that _t_ = 2 yields a relative minimum for _D_. Since _t_ = 2 is the only critical number, Theorem 14.1 implies that there is an absolute minimum at _t_ = 2.

Setting _t_ = 2 in _D_ 2 = (15 _t_ )2 \+ (65 − 10 _t_ )2 yields   miles. Hence, the ships are nearest at 11 A.M., at which time they are   miles apart.

**14.** A cylindrical container with circular base is to hold 64 in3. Find its dimensions so that the amount (surface area) of metal required is a minimum when the container is ( _a_ ) an open can and ( _b_ ) a closed can.

Let _r_ and _h_ be, respectively, the radius of the base and the height in inches, _A_ the amount of metal, and _V_ the volume of the container.

( _a_ ) Here _V_ = π _r_ 2 _h_ = 64, and _A_ = 2π _rh_ \+ π _r_ 2. To express _A_ as a function of one variable, we solve for _h_ in the first relation (because it is easier) and substitute in the second, obtaining

and the critical number is  . Then  . Thus,   in.

Now _dA_ / _dr_ > 0 to the right of the critical number, and _dA_ / _dr_ < 0 to the left of the critical number. So, by the first-derivative test, we have a relative minimum. Since there is no other critical number, that relative minimum is an absolute minimum.

( _b_ ) Here again _V_ = π _r_ 2 _h_ = 64, but _A_ = 2π _rh_ \+ 2π _r_ 2 = 2π _r_ (64/π _r_ 2) + 2π _r_ 2 = 128/ _r_ \+ 2π _r_ 2. Hence,

and the critical number is  . Then  . Thus,   in. That we have found an absolute minimum can be shown as in part ( _a_ ).

**15.** The total cost of producing _x_ radio sets per day is  , and the price per set at which they may be sold is  .

( _a_ ) What should be the daily output to obtain in maximum total profit?

_b_ ) Show that the cost of producing a set is a relative minimum at that output.

( _a_ ) The profit on the sale of _x_ sets per day is  . Then _dP_ / _dx_ = 15 − 3 _x_ /2; solving _dP_ / _dx_ = 0 gives the critical number _x_ = 10.

Since  , the second-derivative test shows that we have found a relative maximum. Since _x_ = 10 is the only critical number, the relative maximum is an absolute maximum. Thus, the daily output that maximizes profit is 10 sets per day.

( _b_ ) The cost of producing a set is  . Then  , solving _dC_ / _dx_ = 0 gives the critical number _x_ = 10.

Since _d_ 2 _C_ / _dx_ 2 = 50/ _x_ 3 > when _x_ = 10, we have found a relative minimum. Since there is only one critical number, this must be an absolute minimum.

**16.** The cost of fuel to run a locomotive is proportional to the square of the speed and $25 per hour for a speed of 25 miles per hour. Other costs amount to $100 per hour, regardless of the speed. Find the speed that minimizes the cost per mile.

Let _υ_ be the required speed, and let _C_ be the total cost per mile. Ther furl cost per hour is _kυ_ 2, where _k_ is a constant to the determined. When υ = 25 mi/h, _kυ_ 2 = 625 _k_ = 25; hence _k_ = 1/25.

Since _υ_ > 0, the only relevant critical number is _υ_ = 50. Since _d_ 2 _C_ / _dv_ 2 = 200/ _υ_ 3 > 0 when _υ_ = 50, the second derivative test tells us that _C_ has a relative minimum at _υ_ = 50. Since _υ_ = 50 is the only critical number in (0, +∞), Theorem 14.1 tells us that _C_ has an absolute minimum at _υ_ = 50. Thus, the most economical speed is 50 mi/h.

**17.** A man in a rowboat at _P_ in Fig. 14.12, 5 miles from the nearest point _A_ on a straight shore, wishes to reach a point _B_ , 6 miles from _A_ along the shore, in the shortest time. Where should he land if he can row 2 mi/h and walk 4 mi/h?

**Fig. 14-12**

Let _C_ be the point between _A_ and _B_ at which the man lands, and let _AC_ = _x_. The distance rowed is  , and the rowing time required is  . The distance walked is

_CB_ = 6 − _x_ , and the walking time required is _t_ 2 = (6 − _x_ )/4. Hence, the total time required is

The critical number obtained from setting   is  . Thus, he should land at a point about 2.89 miles from _A_ toward _B_. (How do we know that this point yields the _shortest_ time?)

**18.** A given rectangular area is to be fenced off in a field that lies along a straight river. If no fencing is needed along the river, show that the least amount of fencing will be required when the length of the field is twice its width.

Let _x_ be the length of the field, and _y_ its width. The area of the field is _A_ = _xy_. The fencing required is _F_ = _x_ \+ 2 _y_ , and _dF_ / _dx_ = 1 + 2 _dy_ / _dx_. When _dF_ / _dx_ = 0,  .

Also, _dA_ / _dx_ = 0 = _y + x dy/dx_. Then   and _x = 2y_ as required.

To see that _F_ has been minimized, note that _dy/dx_ = _−y 2/A_ and

Now use the second-derivative test and the uniqueness of the critical number.

**19.** Find the dimensions of the right circular cone of minimum volume _V_ that can be circumscribed about a sphere of radius 8 inches.

Let _x_ be the radius of the base of the cone, and _y_ \+ 8 the height of the cone. (See Fig. 14-13.) From the

**Fig. 14-13**

similar right triangles _ABC_ and _AED_ , we have

The relevant critical number is y = 24. Then the height of the cone is y + 8 = 32 inches, and the radius of the base is   inches. (How do we know that the volume has been minimized?)

**20.** Find the dimensions of the rectangle of maximum area _A_ that can be inscribed in the portion of the parabola y2 = 4 _px_ intercepted by the line _x_ = _a_.

Let _PBB′P′_ in Fig. 14-14 be the rectangle, and ( _x_ , _y_ ) the coordinates of _P_. Then

**Fig. 14-14**

Solving _dA/dy_ = 0 yields the critical number  . The dimensions of the rectangle are   and _a_ − _x_ = _a_ −(y2/4 _p_ ) = 2 _a_ /3.

Since d2 _A/dy_ 2 = −3 _y_ / _p_ < 0, the second derivative test and the uniqueness of the critical number ensure that we have found the maximum area.

**21.** Find the height of the right circular cylinder of maximum volume _V_ that can be inscribed in a sphere of radius _R_. (See Fig. 14-15.)

Let _r_ be the radius of the base, and _2h_ the height of the cylinder. From the geometry, _V_ = 2π _r_ 2 _h_ and _r_ 2 \+ _h_ 2 = _R_ 2. Then

From the last relation,  . When _V_ is a maximum,  , from which _r 2_ = 2h2.

Then _R_ 2 = _r_ 2 \+ _h_ 2 = 2 _h_ 2 \+ _h_ 2, so that   and the height of the cylinder is  . The second−derivative test can be used to verify that we have found a maximum value of _V_.

**Fig. 14-15**

**Fig. 14-16**

**22.** A wall of a building is to be braced by a beam that must pass over a parallel wall 10ft high and 8ft from the building. Find the length _L_ of the shortest beam that can be used.

See Fig. 14-16. Let _x_ be the distance from the foot of the beam to the foot of the parallel wall, and let _y_ be the distance (in feet) from the ground to the top of the beam. Then  .

Also, from similar triangles,   and, therefore,  . Hence,

The relevant critical number is  . The length of the shortest beam is

The first derivative test and Theorem 14.1 guarantee that we really have found the shortest length.

### **Supplementary Problems**

**23.** Examine each of the following for relative maximum and minimum values, using the first derivative test.

**24.** Examine the functions of Problem 23( _a_ − _f_ ) for relative maximum and minimum values, using the second derivative test.

**25.** Show that _y_ = ( _a_ 1 − _x_ )2 \+ ( _a_ 2 − _x_ )2 \+ ··· + ( _a_ n − _x_ )2 has an absolute minimum when _x_ =

**26.** Examine the following for absolute maximum and minimum values on the given interval

**27.** The sum of two positive numbers is 20. Find the numbers if: ( _a_ ) their product is a maximum; ( _b_ ) the sum of their squares is a minimum; ( _c_ ) the product of the square of one and the cube of the other is a maximum.

_Ans._ ( _a_ ) 10, 10; ( _b_ ) 10, 10; ( _c_ ) 8, 12

**28.** The product of two positive numbers is 16. Find the numbers if: ( _a_ ) their sum is least; ( _b_ ) the sum of one and the square of the other is least.

_Ans._ ( _a_ ) 4, 4; ( _b_ ) 8, 2

**29.** An open rectangular box with square ends is to be built to hold 6400ft3 at a cost of $0.75/ft2 for the base and $0.25/ft2 for the sides. Find the most economical dimensions.

_Ans._ 20 × 20 × 16

**30.** A wall 8ft high is   ft from a house. Find the shortest ladder that will reach from the ground to the house when leaning over the wall.

_Ans._

**31.** A company offers the following schedule of charges: $30 per thousand for orders of 50,000 or less, with the charge decreased by   for each thousand above 50,000. Find the order size that makes the company's receipts a maximum.

_Ans._ 65,000

**32.** Find an equation of the line through the point (3,4) that cuts from the first quadrant a triangle of minimum area.

_Ans._ 4 _x_ \+ 3 _y_ − 24 = 0

**33.** At what point in the first quadrant on the parabola _y_ = 4 − _x_ 2 does the tangent line, together with the coordinate axes, determine a triangle of minimum area.

_Ans._

**34.** Find the minimum distance from the point (4, 2) to the parabola _y_ 2 = 8 _x_.

_Ans._

**35.** ( _a_ )Examine 2 _x_ 2 − 4 _xy_ \+ 3 _y_ 2 − 8 _x_ \+ 8 _y_ −1 = 0 for maximum and minimum values of _y_. ( _b_ )( **GC** ) Check your answer to ( _a_ ) on a graphing calculator.

_Ans._ Maximum at (5, 3); minimum at (-1,-3)

**36.** ( **GC** ) Find the absolute maximum and minimum of _f_ ( _x_ ) = _x_ 5 − 3 _x_ 2−8 _x_ − 3 on [−1,2] to three-decimal-place accuracy.

_Ans_. Maximum 1.191 at _x_ = −0.866; minimum −14.786 at _x_ = 1.338

**37.** An electric current, when flowing in a circular coil of radius _r_ , exerts a force   on a small magnet located at a distance _x_ above the center of the coil. Show that _F_ is greatest when  .

**38.** The work done by a voltaic cell of constant electromotive force _E_ and constant internal resistance _r_ in passing a steady current through an external resistance _R_ is proportional to _E_ 2 _R_ ( _r_ + _R_ )2. Show that the work done is greatest when _R_ = _r_.

**39.** A tangent line is drawn to the ellipse   so that the part intercepted by the coordinate axes is a minimum. Show that its length is 9.

**40.** A rectangle is inscribed in the ellipse   with its sides parallel to the axes of the ellipse. Find the dimensions of the rectangle of ( _a_ ) maximum area and ( _b_ ) maximum perimeter that can be so inscribed.

_Ans._ ( _a_ ) ; ( _b_ )32 × 18

**41.** Find the radius _R_ of the right circular cone of maximum volume that can be inscribed in a sphere of radius _r_. (Recall that the volume of a right circular cone of radius _R_ and height _h_ is  .)

_Ans._

**42.** A right circular cylinder is inscribed in a right circular cone of radius _r_. Find the radius _R_ of the cylinder if: ( _a_ ) its volume is a maximum; ( _b_ ) its lateral area is a maximum. (Recall that the volume of a right circular cylinder of radius _R_ and height _h_ is π _R_ 2 _h_ , and its lateral area is 2π _Rh_.)

_Ans._ ( _a_ )  ; ( _b_ )

**43.** Show that a conical tent of given volume will require the least amount of material when its height _h_ is   times the radius _r_ of the base. [Note first that the surface area _A_ = π( _r_ 2+ _h_ 2).]

**44.** Show that the equilateral triangle of altitude 3 _r_ is the isosceles triangle of least area circumscribing a circle of radius _r_.

**45.** Determine the dimensions of the right circular cylinder of maximum lateral surface area that can be inscribed in a sphere of radius 8.

_Ans._

**46.** Investigate the possibility of inscribing a right circular cylinder of maximum total area (including its top and bottom) in a right circular cone of radius _r_ and height _h_.

_Ans._ If _h_ > 2 _r_ , radius of cylinder

## **CHAPTER 15
Curve Sketching. Concavity. Symmetry**

**CONCAVITY.** From an intuitive standpoint, an arc of a curve is said to be _concave upward_ if it has the shape of a cup (see Fig. 15-1( _a_ )) and is said to be _concave downward_ if it has the shape of a cap (see Fig. 15-1( _b_ )). Note that a more precise definition is available. An arc is concave upward if, for each _x_ 0, the arc lies above the tangent line at _x_ 0, in some open interval around _x_ 0. Similarly, an arc is concave downward if, for each _x_ 0, the arc lies below the tangent line at _x_ 0 in some open interval around _x_ 0. Most curves are combinations of concave upward and concave downward. For example, in Fig. 15-1( _c_ ), the curve is concave downward from _A_ to _B_ and from _C_ to _D_ , but concave upward from _B_ to _C_.

The second derivative of _f_ tells us about the concavity of the graph of _f_.

**Theorem 15.1**

( _a_ ) If _f_ ″( _x_ ) > 0 for _x_ in ( _a_ , _b_ ), then the graph of _f_ is concave upward for _a_ < _x_ < _b_.

( _b_ ) If _f_ ″( _x_ ) < 0 for _x_ in ( _a_ , _b_ ), then the graph of _f_ is concave downward for _a_ < _x_ < _b_.

For the proof, see Problem 17.

**EXAMPLE 1**

( _a_ ) Let _f_ ( _x_ ) = _x_ 2. Then _f′( _x_ )_ = 2 _x_ , _f″( _x_ )_ = 2. Since _f″( _x_ )_ > 0 for all _x_ , the graph of _f_ is always concave upward. This was to be expected, since the graph is a parabola that opens upward.

( _b_ ) Let  . Then _y_ 2 = 1 − _x_ 2, _x_ 2 \+ _y_ 2 = 1. So, the graph is the upper half of the unit circle with center at the origin. By implicit differentiation, we obtain _x_ \+ _yy′_ = 0 and then 1 + _yy″_ \+ ( _y′_ )2 = 0. So, _y″_ = −[1 + ( _y′_ )2]/ _y_. Since _y_ > 0 (except at _x_ = 1), _y″_ < 0. Hence, the graph is always concave downward, which is what we would expect.

**POINTS OF INFLECTION.** A point of inflection on a curve _y_ = _f_ ( _x_ ) is a point at which the concavity changes, that is, the curve is concave upward on one side and concave downward on the other side of the point. So, if _y"_ exists in an open interval containing _x_ 0, then _y"_ < 0 on one side of _x_ 0 and _y"_ > 0 on the other side of _x_ 0. Therefore, if _y"_ is continuous at _x_ 0, then _y"_ = 0 at _x_ 0. Thus, we have:

**Theorem 15.2:** If the graph of _f_ has an inflection point at _x_ 0 and _f"_ exists in an open interval containing _x_ 0 and _f"_ is continuous at _x_ 0, then _f"_ ( _x_ 0) = 0.

**EXAMPLE 2**

( _a_ ) Let _f_ ( _x_ ) = _x_ 3. Then _f′_ ( _x_ ) = 3 _x_ 2, _f_ ″( _x_ ) = 6 _x_. Thus, _f_ ″( _x_ ) < 0 for _x_ < 0, and _f_ ″( _x_ ) > 0 for _x_ < 0. Hence, the graph of _f_ has an inflection point at _x_ = 0. (See Fig. 5-5.) Note that _f_ ″(0) = 0, as predicted by Theorem 15.2.

( _b_ ) Let _f_ ( _x_ ) = _x_ 4. Then _f′_ ( _x_ ) = 4 _x_ 3, and _f_ ″( _x_ ) = 12 _x_ 2. Solving _f_ ″( _x_ ) = 0 yields _x_ = 0. However, the graph of _f_ does not have an inflection point at _x_ = 0. It is concave upward everywhere.

( _c_ ) Let  . Solving _f_ ″( _x_ ) = 2 _x_ \+ 1 = 0, we find that the graph has an inflection point at  . Note that this is actually an inflection point, since _f_ ″( _x_ ) < 0 for   and _f″(x)_ > 0 for  . See Fig. 14-5.

**VERTICAL ASYMPTOTES.** A vertical line _x_ = _x_ 0 such that _f_ ( _x_ ) approaches +∞ or −∞ as _x_ approaches _x_ 0 either from the left or the right is called a _vertical asymptote_ of the graph of _f_. If _f_ ( _x_ ) has the form _g_ ( _x_ )/ _h_ ( _x_ ), where _g_ and _h_ are continuous functions, then the graph of _f_ has a vertical asymptote _x_ = _x_ 0 for every _x_ 0 such that _h_ ( _x_ 0) = 0 (and _g_ ( _x_ 0) ≠ 0).

**HORIZONTAL ASYMPTOTES.** A horizontal line _y_ = _y_ 0 is called a _horizontal asymptote_ of the graph of _f_ if either   or  . Thus, a horizontal asymptote is approached by the graph as one moves further and further to the left or further and further to the right.

**EXAMPLE 3**

( _a_ ) Let  . Then the graph of _f_ has a vertical asymptote at _x_ = 0, which is approached both from the left and the right. The line _y_ = 0 (that is, the _x_ axis) is a horizontal asymptote both on the left and the right. See Fig. 5-21.

( _b_ ) Let  . Then _x_ = 2 is a vertical asymptote of the graph of _f_ , which is approached both on the left and the right. The line _y_ = 0 is a horizontal asymptote, which is approached both on the left and the right. See Fig. 14-7.

( _c_ ) Let  . Then the graph of _f_ has vertical asymptotes at _x_ = 1 and _x_ = −3. The line _y_ = 0 is a horizontal asymptote, which is approached both on the left and the right.

( _d_ ) Let  . Then the graph of _f_ has a vertical asymptote at _x_ = 3, which is approached both from the left and the right. The line _y_ = 1 is a horizontal asymptote, which is approached both on the left and the right.

**SYMMETRY.** We say that two points _P_ and _Q_ are _symmetric with respect to a line l_ if _l_ is the perpendicular bisector of the line segment connecting _P_ and _Q_. See [Fig. 15-2( _a_ ).]

We say that two points _P_ and _Q_ are _symmetric with respect to a point B_ if _B_ is the midpoint of the segment connecting _P_ and _Q_.

A curve is said to be symmetric with respect to a line _l_ (respectively, point _B_ ) if, for any point _P_ on the curve, there is another point _Q_ on the curve such that _P_ and _Q_ are symmetric with respect to _l_ (respectively, _B_ ). See [Fig. 15-2( _b, c_ ).]

If a curve is symmetric with respect to a line _l_ , then _l_ is called an _axis of symmetry_ of _l_. For example, any line through the center of a circle is an axis of symmetry of that circle.

**Fig. 15-2**

Points ( _x_ , _y_ ) and (− _x_ , _y_ ) are symmetric with respect to the _y_ axis, and points ( _x_ , _y_ ) and (− _x_ ,− _y_ ) are symmetric with respect to the origin. See Fig. 15-3( _a-c_ ).

**Fig. 15-3**

Consider the graph of an equation _F(x, y)_ = 0 Then:

( _i_ ) The graph is symmetric with respect to the _y_ axis if and only if _F_ ( _x_ , _y_ ) = 0 implies _F_ (− _x_ , _y_ ) = 0.

( _ii_ ) The graph is symmetric with respect to the _x_ axis if and only if _F(x, y)_ = 0 implies _F_ ( _x_ ,− _y_ ) = 0.

( _iii_ ) The graph is symmetric with respect to the origin if only if _F_ ( _x_ , _y_ ) = 0 implies _F_ (− _x_ ,− _y_ ) = 0.

**EXAMPLE 4**

( _a_ )The parabola _y_ = _x_ 2 is symmetric with respect to the _y_ axis.

( _b_ ) The parabola _x_ = _y_ 2 is symmetric with respect to the _x_ axis.

( _c_ ) A circle _x_ 2 \+ _y_ 2 = _r_ 2, an ellipse  , and a hyperbola   = 1 are symmetric with respect to the _y_ axis, the _x_ axis, and the origin.

**EXAMPLE 5:** A point _P_ ( _a_ , _b_ ) is symmetric to the point _Q_ ( _b_ , _a_ ) with respect to the line _y_ = _x_. To see this, note first that the line _PQ_ has slope −1. Since the line _y_ = _x_ has slope 1, the line _PQ_ is perpendicular to the line _y_ = _x_. In addition, the midpoint of the segment connecting _P_ and _Q_ is  , which is on the line _y_ = _x_. Hence, the line _y_ = _x_ is the perpendicular bisector of that segment.

**INVERSE FUNCTIONS AND SYMMETRY**. We say that two curves _C_ 1 and _C_ 2 are _symmetric to each other with respect to a line l_ if, for any point _P_ on one of the curves, the point _Q_ that is symmetric to _P_ with respect to _l_ is on the other curve. (In other words, if we "reflect" one of the curves in the line _l_ , the result is the other curve.)

**Theorem 15.3:** Consider any one-to-one function _f_ and its inverse function _f_ −1. Then the graphs of _f_ and _f_ −1 are symmetric to each other with respect to the line _y_ = _x_.

To see this, assume that ( _a_ , _b_ ) is on the graph of _f_. Then _f_ ( _a_ ) = _b_. Hence, _f_ −1( _b_ ) = _a_ , that is, ( _b_ , _a_ ) is on the graph of _f_ −1. By Example 5, ( _a_ , _b_ ) and ( _b_ , _a_ ) are symmetric with respect to the line _y_ = _x_.

**EXAMPLE 6**

( _a_ ) If _f_ ( _x_ ) = 2 _x_ , then  . Hence, the lines _y_ = 2 _x_ and   are symmetric with respect to the line _y_ = _x_.

( _b_ ) Let _C_ 1 be the parabola that is the graph of the equation _y_ = _x_ 2, and let _C_ 2 be the parabola that is the graph of the equation _x_ = _y_ 2. Then _C_ 1 and _C_ 1 are symmetric with respect to the line _y_ = _x_ , since the equation _x_ = _y_ 2 results from the equation _y_ = _x_ 2 by interchanging _x_ and _y_.

**EVEN AND ODD FUNCTIONS.** A function _f_ is said to be _even_ if, for any _x_ in its domain, − _x_ is also in its domain and _f_ (− _x_ ) = _f_ ( _x_ ). _f_ is said to be an _odd_ function if, for any _x_ in its domain, − _x_ is also in its domain and _f_ (− _x_ ) = - _f_ ( _x_ ).

**EXAMPLE 7:** Any polynomial, such as 3 _x_ 6 − 8 _x_ 4 \+ 7, that involves only even powers of _x_ determines an even function. Any polynomial, such as 5 _x_ 9 \+ 2 _x_ 5 − 4 _x_ 3 \+ 3 _x_ , that involves only odd powers of _x_ determines an odd function.

A function _f_ is even if and only if its graph is symmetric with respect to the _y_ axis. In fact, assume _f_ is even and ( _x_ , _y_ ) is on its graph. Then _y_ = _f_ ( _x_ ). Hence, _y_ = _f_ (− _x_ ) and, therefore, (− _x_ , _y_ ) is on the graph. Thus, the graph is symmetric with respect to the _y_ axis. The converse is left as Problem 16(a).

A function _f_ is odd if and only if its graph is symmetric with respect to the origin. In fact, assume _f_ is odd and ( _x_ , _y_ ) is on its graph. Then _y_ = _f_ ( _x_ ). Hence, − _y_ = _f_ (− _x_ ) and, therefore, (− _x_ ,− _y_ ) is on the graph. Thus, the graph is symmetric with respect to the origin. The converse is left as Problem 16( _b_ ).

**HINTS FOR SKETCHING THE GRAPH OF** _y_ = _f_ ( _x_ )

1. Calculate _y_ ′, and, if convenient, _y_ ″.

2. Use _y_ ′ to find any critical numbers (where _y_ ′ = 0, or _y_ ′ is undefined and _y_ is defined). Determine whether these critical numbers yield relative maxima or minima by using the second derivative test or the first derivative test.

3. Use _y_ ′ to determine the intervals on which _y_ is increasing (when _y_ ′ > 0) or decreasing (when y′ < 0).

4. Use _y_ ″ to determine where _y_ ′ is concave upward (when _y_ ′ > 0) or concave downward (when _y_ ′ < 0). Check points where _y_ ″ = 0 to determine whether they are inflection points (if _y_ ″ > 0 on one side and _y_ ″ < 0 on the other side of the point).

5. Look for vertical asymptotes. If _y_ = _g_ ( _x_ )/ _h_ ( _x_ ), there is a vertical asymptote _x_ = _x_ 0 if _h_ ( _x_ 0) = 0 and _g_ ( _x_ 0) ≠ 0.

6. Look for horizontal asymptotes. If  , then _y_ = _y_ 0 is a horizontal asymptote on the right. If  , then _y_ = _y_ 0 is a horizontal asymptote on the left.

7. Determine the behavior "at infinity". If   (respectively, −∞), then the curve moves upward (respectively, downward) without bound to the right. Similarly, if   (respectively, −∞), then the curve moves upward (respectively, downward) without bound to the left.

8. Find the _y_ intercepts (where the curve cuts the _y_ axis, that is, where _x_ = 0) and the _x_ -intercepts (where the curve cuts the _x_ axis, that is, where _y_ = 0).

9. Indicate any corner points, where _y_ ′ approaches one value from the left and another value from the right. An example is the origin on the graph of _y_ = | _x_ |.

10. Indicate any cusps, where _y_ ′ approaches +∞ from both sides or _y_ ′ approaches −∞ from both sides. An example is the origin on the graph of  .

11. Find any oblique asymptotes _y_ = _mx_ \+ _b_ such that   or

### **Solved Problems**

**1.** Examine _y_ = 3 _x_ 4 − 10 _x_ 3 − 12 _x_ 2 \+ 12 _x_ − 7 for concavity and points of inflection.

We have

_y_ ′ = 12 _x_ 3 − 30 _x_ 2 − 24 _x_ \+ 12

_y_ ″ = 36 _x_ 2 − 60 _x_ − 24 = 12(3 _x_ \+ 1)( _x_ − 2)

Set _y_ ″ = 0 and solve to obtain the possible points of inflection   and 2. Then:

The points of inflection are   and (2, −63), since _y_ ″ changes sign at   and _x_ = 2. See Fig. 15-4.

**Fig. 15-4**

**2.** Examine _y_ = _x_ 4 − 6 _x_ \+ 2 for concavity and points of inflection, and sketch the graph.

We have _y_ ″ = 12 _x_ 2. By Theorem 15.2, the possible point of inflection is at _x_ = 0. On the intervals _x_ < 0 and _x_ > 0, _y_ ″ is positive, and the arcs on both sides of _x_ = 0 are concave upward. The point (0, 2) is not a point of inflection. Setting _y_ ′ = 4 _x_ 3 − 6 = 0, we find the critical number  . At this point _y_ ″ = 12 _x_ 2 > 0 and we have a relative minimum by the second derivative test. Since there is only one critical number, there is an absolute minimum at this point (where _x_ ~ 1.45 and _y_ ~ − 3.15). See Fig. 15.5

**Fig. 15-5**

**3.** Examine _y_ = 3 _x_ \+ ( _x_ \+ 2)3/5 for concavity and points of inflection, and sketch the graph.

 . The possible point of inflection is at _x_ = −2. When _x_ > −2, _y_ ″ is negative and the arc is concave downward. When _x_ < −2, _y_ ″ is positive and the arc is concave upward. Hence, there is an inflection point at _x_ = −2, where _y_ = −6. (See Fig. 15-6.) Since _y_ ′ > 0 (except at _x_ = −2), _y_ is an increasing function, and there are no relative extrema.

**Fig. 15-6**

**4.** If _f_ ″ ( _x_ 0) = 0 and _f_ ′′′( _x_ 0) ≠ 0, then there is an inflection point at _x_ 0.

Since _f_ ′′′( _x_ 0) ≠ 0, _f_ ″( _x_ 0) is either positive or negative. Hence, _f_ ″ is either increasing or decreasing at _x_ 0. Since _f_ ″( _x_ 0) = 0, _f_ ″ has opposite signs to the left and right of _x_ 0. So, the curve will have opposite concavity on the two sides of _x_ 0, and there is an inflection point at _x_ 0.

**5.** Find equations of the tangent lines at the points of inflection of _y_ = _f_ ( _x_ ) = _x_ 4 − 6 _x_ 3 \+ 12 _x_ 2 − 8 _x_.

A point of inflection exists at _x_ = _x_ 0 when _f_ ″( _x_ 0) = 0 and _f_ ″′( _x_ 0)≠0. Here,

_f_ ′( _x_ ) = 4 _x_ 3 − 18 _x_ 2 \+ 24 _x_ − 8

_f_ ″( _x_ ) = 12 _x_ 2 − 36 _x_ \+ 24 = 12( _x_ − 1)( _x_ − 2)

_f_ ′′′( _x_ ) = 24 _x_ − 36 = 12(2 _x_ − 3)

The possible points of inflection are at _x_ = 1 and 2. Since _f_ ′′′(1) ≠ 0 and _f_ ′′′(2) ≠ 0, the points (1, −1) and (2, 0) are points of inflection.

At (1,−1), the slope of the tangent line is _m_ = _f_ ′(1) = 2, and its equation is

_y_ = _y_ 1 = _m_ ( _x_ − _x_ 1) or _y_ \+ 1 = 2( _x_ − 1) or _y_ = 2 _x_ − 3

At (2, 0), the slope is _f_ ′(2) = 0, and the equation of the tangent line is _y_ =0.

**6.** Sketch the graph of _y_ = _f_ ( _x_ ) = _x_ 3 − 5 _x_ 2 \+ 4 _x_ − 7.

_f_ ′( _x_ ) = 6 _x_ 2 − 10 _x_ \+ 4, _f_ ″( _x_ ) = 12 _x_ − 10, and _f_ ′′′( _x_ ) = 12. Now, 12 _x_ − 10 > 0 when  , and 12 _x_ − 10 < 0 when  . Hence, the graph of _f_ is concave upward when   and concave downward when  . Thus, there is an inflection point at  . Since _f_ ′( _x_ ) = 2(3 _x_ 2 − 5 _x_ \+ 2) = 2(3 _x_ − 2)( _x_ − 1), the critical numbers are   and _x_ = 1. Since   and _f_ ″(1) = 2, there is a relative maximum at   (where  ) and a relative minimum at _x_ = 1 (where _y_ = −6). See Fig. 15-7.

**Fig. 15-7**

**7.** Sketch the graph of  .

Solving _y_ ′ = 0, we obtain the critical numbers _x_ = 4 and _x_ = 0. Since _f_ ″(4) = 1 > 0 and _f_ ″(0) = −1 < 0, there is a relative minimum at _x_ = 4 (where _y_ = 8) and a relative maximum at _x_ = 0 (where _y_ = 0). Since _y_ ″ is never 0, there are no inflection points. The line _x_ = 2 is a vertical asymptote. The line _y_ = _x_ \+ 2 is an oblique asymptote on both sides, since, on the curve,   as _x_ → ±∞. See Fig. 15-8.

**Fig. 15-8**

**8.** Sketch the graph of _g_ ( _x_ ) = 2 _x_ 3 − 9 _x_ 2 \+ 36.

_g_ ′( _x_ ) = 6 _x_ 2 − 18 _x_ = 6 _x_ ( _x_ − 3) and _g_ ″( _x_ ) = 12 _x_ − 18 = 6(2 _x_ − 3). So, the critical numbers are _x_ = 0 (where _y_ = 36) and _x_ = 3 (where _y_ = 9). Since _g_ ″(0) = − 18 < 0 and _g_ ″(3) = 18 > 0, there is a relative maximum at _x_ = 0 and a relative minimum at _x_ = 3. Setting _g_ ′( _x_ ) = 0 yields  , where there is an inflection point, since _g_ ″( _x_ ) = 6(2 _x_ − 3) changes sign at  .

_g_ ( _x_ ) → +∞ as _x_ → +∞, and _g_ ( _x_ ) → −∞ as _x_ → −∞. Since _g_ (−1) = 29 and _g_ (−2) = −16, the intermediate value theorem implies that there is a zero _x_ 0 of _g_ between −1 and −2. (A graphing calculator shows _x_ 0 ~ − 1.70.) That is the only zero because _g_ is increasing up to the point (0, 36), decreasing from (0, 36) to (3, 9), and then increasing from (3, 9). See Fig. 15-9.

**Fig. 15-9**

**9.** Sketch the graph of

There are vertical asymptotes at _x_ = 2 and _x_ = 6.

The critical numbers are _x_ = 0 (where _y = 0_ ) and _x_ = 3 (where _y = −3_ ). Calculation shows that _y_ ″ (0) > 0 and _y_ ″ (3) < 0. Hence, there is a relative minimum at _x_ = 0 and a relative maximum at _x = 3_. Since _y → 1_ when _x → ± ∞,_ the line _y = 1_ is a horizontal asymptote on both the left and the right. Setting _y″ = 0_ yields _g_ ( _x_ ) = 2 _x_ 3 \- 9 _x_ 2 \+ 36 = 0. By Problem 8, we see that we have a unique inflection point _x_ 0 ~ −1.70 (where _y_ ~ 0.10). See Fig. 15-10.

**10.** Sketch the graph of _y 2 ( _x_ 2 \- 4) = _x_ 4_.

 . Then  . The curve exists only for _x_ 2 > 4, that is, for _x_ > 2 or _x_ < -2, plus the isolated point (0,0).

The curve is symmetric with respect to both coordinate axes and the origin. So, from now on, we shall consider only the first quadrant. Then

The only critical number is   (where _y_ = 4). Since _y_ ″ > 0, the graph is concave upward and there is a relative minimum at  . The lines _x_ =2 and _x_ = −2 are vertical asymptotes. The rest of the graph in the

**Fig. 15-10**

other quadrants is obtained by reflection in the axes and origin. Note that there is also an "oblique" asymptote _y_ = _x_ , since _y_ 2 − _x_ 2 = _x_ 4/( _x_ 2 − 4) − _x_ 2 = 4/( _x_ 2 − 4) → 0 as _x_ → ± ∞. By symmetry, _y = −x_ is also an asymptote. See Fig. 15-11.

**Fig. 15-11**

### **Supplementary Problems**

**11.** Examine the functions of Problem 23( _a_ − _f_ ) of Chapter 14.

_Ans._ ( _a_ ) No inflection point, concave upward everywhere

( _b_ ) No inflection point, concave downward everywhere

( _c_ ) Inflection point at  , concave upward for   concave downward for

( _d_ ) Inflection point at _x_ = 2, concave upward for _x_ > 2, concave downward for _x_ < 2

( _e_ ) Inflection point at _x_ = 2, concave downward for _x_ > 2, concave upward for _x_ < 2

( _f_ ) Inflection point at  , concave upward for   and   concave downward for

**12.** Prove: If _f_ ( _x_ ) = _ax_ 3 \+ _bx_ 2 \+ _cx_ \+ _d_ has two critical numbers, their average is the abscissa at the point of inflection. If there is just one critical number, it is the abscissa at the point of inflection.

**13.** Discuss and sketch the graphs of the following equations:

**14.** Show that any function _F_ ( _x_ ) that is defined for all _x_ may be expressed in one and only one way as the sum of an even and an odd function. [ _Hint:_ Let  .]

**15.** Find and equation of the new curve C1 that is obtained when the graph of the curve C with the equation _x_ 2 − 3 _xy_ \+ 2 _y_ 2 = 1 is reflected in: ( _a_ ) the _x_ axis, ( _b_ ) the _y_ axis, ( _c_ ) the origin.

_Ans._ ( _a_ ) _x_ 2 \+ 3 _xy_ \+ 2 _y_ 2; = 1 ( _b_ ) same as ( _a_ ); ( _c_ ) _C_ itself

**16.** ( _a_ ) If the graph of _f_ is symmetric with respect to the _y_ axis, show that _f_ is even. ( _b_ ) If the graph of _f_ is symmetric with respect to the origin, then show that _f_ is odd. [ _Hint:_ For ( _a_ ), if _x_ is in the domain of _f_ ,( _x_ , _f_ ( _x_ )) is on the graph and, therefore, (− _x_ , _f_ ( _x_ )) is on the graph. Thus, _f_ (− _x_ ) = _f_ ( _x_ ).]

**17.** Prove Theorem 15.1: ( _a_ ) If _f_ ″( _x_ ) > 0 for _x_ in ( _a_ , _b_ ) then the graph of _f_ is concave upward for _a_ < _x_ < _b_. ( _b_ ) If _f_ ″( _x_ ) < 0 for _x_ in ( _a_ , _b_ ), then the graph of _f_ is concave downward for _a_ < _x_ < b.

[For ( _a_ ), let _x_ 0 belong to ( _a_ , _b_ ). Since _f_ ″ ( _x_ 0 > 0, _f_ ′ is increasing in some open interval I containing _x_ 0. Assume _x_ in I and _x_ > _x_ 0. By the law of the mean, _f_ ( _x_ )− _f_ ( _x_ 0) = _f_ ′ ( _x_ *)( _x_ − _x_ 0) for some _x_ * with _x_ 0 < _x_ * < _x_. Since _f_ ′ is increasing, _f_ ′( _x_ 0) < f′ ( _x_ *). Then _f_ ( _x_ ) = _f_ ′( _x_ *)( _x_ − _x_ 0) + _f_ ( _x_ 0) > _f_ ′( _x_ 0)( _x_ − _x_ 0) + _f_ ( _x_ 0). But _y_ = _f_ ′ ( _x_ 0) ( _x_ − _x_ 0) + _f_ ( _x_ 0) is an equation of the tangent line at _x_ 0. A similar argument works when _x_ < _x_ 0. Thus, the curve lies above the tangent line and, therefore, is concave upward.]

**18.** (GC) Use a graphing calculator to draw the graph of _f_ ( _x_ ) = _x_ 3 − 3x2 \+ 4x − 2. Show analytically that _f_ is increasing and that there is an inflection point at (−1, 3). Use the calculator to draw the graph of _f −1_ and _y = x_ , and observe that the graphs of _f_ and _f −1_ are symmetric with respect to _y_ = _x_.

**19.** (GC) Try to sketch the graph of   by standard methods and then use a graphing calculator for additional information (such as the location of any vertical asymptotes).

## **CHAPTER 16
Review of Trignometry**

**ANGLE MEASURE.** The traditional unit for measuring angles is the degree. 360 degrees make up a complete rotation. However, it turns out that a different unit, the radian, is more useful in calculus. Consider a circle of radius 1 and with center at a point _C_. (See Fig. 16-1.) Let _CA_ and _CB_ be two radii for which the arc   of the circle has length 1. Then one _radian_ is taken to be the measure of the central angle _ACB_.

**Fig. 16-1**

If _u_ is the number of degrees in angle _ACB_ , then the ratio of _u_ to 360° is equal to the ratio of the   to the circumference 2 _π_. Since   = 1, _u_ /360 = 1/2 _π_ and, therefore, _u_ = 180/ _π_. So,

If _π_ is approximated as 3.14, then 1 radian is about 57.3 degrees. Multiplying equation ( _1_ ) by _π_ /180, we obtain:

The table in Fig. 16-2 shows the radian measure of some important degree measures.

**Fig. 16-2**

Now take any circle of radius _r_ with center _O_. (See Fig. 16-3.) Let  _DOE_ contain _θ_ radians and let _s_ be the length of arc _DE_. The ratio of _θ_ to the number 2 _π_ of radians in a complete rotations is equal to the ratio of _s_ to the entire circumference 2 _π_ r. So, _θ_ /2 _π_ = _s_ /2 _πr_. Therefore,

**Fig. 16-3**

**DIRECTED ANGLES.** If an angle is thought of as being generated by a rotation, then its measure will be counted as positive if the rotation is counterclockwise and negative if the rotation is clockwise. See, for example, angles of _π_ /2 radians and − _π_ /2 radians in Fig. 16-4. We shall also allow angles of

**Fig. 16-4**

more than one complete rotation. For example, Fig. 16-5 shows a counterclockwise angle generated by a complete rotation plus another quarter of a complete rotation, yielding an angle of 2 _π_ \+ _π_ /2 = 5 _π_ /2 radians, and an angle of 3 _π_ radians generated by one and a half turns in the counterclockwise direction.

**Fig. 16-5**

**SINE AND COSINE FUNCTIONS.** Consider a coordinate system with origin at _O_ and point _A_ at (1, 0). Rotate the arrow _OA_ through an angle of _θ_ degrees to a new position _OB_. Then (see Fig. 16-6):

1. cos _θ_ is defined to be the _x_ coordinate of the point _B_.

2. sin _θ_ is defined to be the _y_ coordinate of the point _B_.

**Fig. 16-6**

**EXAMPLE 1**

( _a_ ) If _θ_ = _π_ /2, the final position _B_ is (0, 1). Hence, cos( _π_ /2) = 0 and sin( _π_ /2) = 1.

( _b_ ) If _θ_ = _π_ , then _B_ is (−1, 0). So, cos _π_ = −1 and sin _π_ = 0.

( _c_ ) If _θ_ = 3 _π_ /2, then _B_ is (0, −1). So, cos(3 _π_ /2) = 0 and sin(3 _π_ /2) = −1.

( _d_ ) If _θ_ = 0 or _θ_ = 2 _π_ , then _B_ is (1, 0). Hence, cos 0 = 1 and sin 0 = 0, and cos 2 _π_ = 1 and sin 2 _π_ = 0.

Let us see that our definitions coincide with the traditional definitions in the case of an acute angle of a triangle. Let _θ_ be an acute angle of a right triangle _DEF_ and let Δ _OBG_ be a similar triangle with hypotenuse 1. (See Fig. 16-7.) Since the triangles are similar,  , that is,   and, likewise,  . Hence, cos _θ_ = _a_ / _c_ and sin _θ_ = _b_ / _c_. This is the same as the traditional definitions:

**Fig. 16-7**

**Table 16-1**

We now can use the values obtained from high-school trigonometry. See Problem 22( _a-c_ ).] [Table 16-1 lists the most useful values.

Let us first collect some simple consequences of the definitions.

( **16.1** ) cos( _θ_ \+ 2 _π_ ) = cos _θ_ and sin( _θ_ \+ 2 _π_ ) = sin _θ_

This holds because an additional complete rotation of 2 _π_ radians brings us back to the same point.

( **16.2** ) cos(− _θ_ ) = cos _θ_ and sin(− _θ_ ) = −sin _θ_ (See Fig. 16-8.)

( **16.3** ) sin2 _θ_ \+ cos2 _θ_ = 1 [In accordance with tradiational notation, sin2 _θ_ and cos2 _θ_ stand for (sin _θ_ )2 and (cos _θ_ )2.]

In Fig. 16-6,   by Problem 1 of Chapter 2. (16.3) implies sin2 _θ_ = 1 − cos2 _θ_ and cos2 _θ_ = 1 − sin2 _θ_.

**Fig. 16-8**

**Fig. 16-9**

( **16.4** ) In the four quadrants, the sine and cosine have the signs shown in Fig. 16-9.

( **16.5** ) For any point _A_ ( _x, y_ ) different from the origin _O_ , let _r_ be its distance from the origin, and let _θ_ be the radian measure of the angle from the positive _x_ axis to the arrow _OA_. (See Fig. 16-10.) The pair ( _r_ , _θ_ ) are called _polar coordinates of A_. Then _x_ = _r_ cos _θ_ and _y_ = _r_ sin _θ_. (See Problem 8.)

For the derivation of more complicated formulas, we shall depend on the following result.

( **16.6** ) cos( _u_ − _υ_ ) = cos _u_ cos _υ_ \+ sin _u_ sin _υ_

For the proof, see Problem 11.

( **16.7** ) cos( _u_ \+ _υ_ ) = cos _u_ cos _υ_ − sin _u_ sin _υ_

Replace _υ_ by − _υ_ in (16.6) and use (16.2).

( **16.8** ) cos( _π_ /2 − _υ_ ) = sin _υ_ and sin( _π_ /2 − _υ_ ) = cos _υ_

Replace _u_ by _π_ /2 in (16.6) and use cos( _π_ /2) = 0 and sin( _π_ /2) = 1. This yields cos( _π_ /2 − _υ_ ) = sin _υ_. In this formula, replace _υ_ by ( _π_ /2 − _υ_ ) to obtain cos _υ_ = sin( _π_ /2 − _υ_ ).

( **16.9** ) sin( _u_ \+ _υ_ ) = sin _u_ cos _υ_ \+ cos _u_ sin _υ_

By (16.8) and (16.6),

( **16.10** ) sin( _u_ − _υ_ ) = sin _u_ cos _υ_ − cos _u_ sin _υ_

Replace _υ_ by − _υ_ in (16.9) and use (16.2).

( **16.11** ) cos 2 _u_ = cos2 _u_ − sin2 _u_ = 2 cos2 _u_ − 1 = 1 − 2sin2 _u_

Replace _υ_ by _u_ in (16.7) to get cos 2 _u_ = cos2 _u_ − sin2 _u_. Use sin2 _u_ = 1 − cos2 _u_ and cos2 _u_ = 1 − sin2 _u_ to obtain the other two forms.

( **16.12** ) sin 2 _u_ = 2 sin _u_ cos _u_

Replace _υ_ by _u_ in (16.9).

( **16.13** )

by (16.11). Now solve for  .

**Fig. 16-10**

( **16.14** )

By (16.3) and (16.13),

( **16.15** ) ( _a_ ) (Law of Cosines). In any triangle Δ _ABC_ (See Fig. 16-11),

_c_ 2 = _a_ 2 \+ _b_ 2 − 2 _ab_ cos _θ_

For a proof, see Problem 11( _a_ ).

( _b_ ) (Law of Sines)

where sin _A_ is sin (  _BAC_ ), and similarly for sin _B_ and sin _C_.

**Fig. 16-11**

### **Solved Problems**

**1.** Translate the following degree measures into radian measures: ( _a_ ) 54°; ( _b_ ) 120°.

( _a_ )

( _b_ )

**2.** Translate the following radian measures into degree measures: ( _a_ )   radians; ( _b_ )   radians; ( _c_ ) 2 radians.

( _a_ )

( _b_ )

( _c_ )

**3.** ( _a_ ) In a circle of radius _r_ = 3 centimeters, what arc length _s_ along the circumference corresponds to a central angle _θ_ of _π_ /6 radians?

( _b_ ) In a circle of radius 4 feet, what central angle corresponds to an arc length of 8 feet?

We know that _s_ = _rθ_ , where _θ_ is measured in radians.

( _a_ )

( _b_ )

**4.** What rotations between 0 and 2 _π_ radians have the same effect as the rotations with the following measures? ( _a_ )   radians; ( _b_ ) 405°; ( _c_ )   radians; ( _d_ ) −5 _π_ radians.

( _a_ )  . So, the equivalent rotation is   radians.

( _b_ ) 405° = (360 + 45)°. So, the equivalent rotation is 45°.

( _c_ )  . So, the equivalent rotation is   radians.

( _d_ ) −5 _π_ \+ 6 _π_ = _π_. So, the equivalent rotation is _π_ radians.

**5.** Find sin _θ_ if _θ_ is an acute angle such that cos  .

By (16.3),  . So,   and, therefore,  . Since _θ_ is acute, sin _θ_ is positive. So,  .

**6.** Show that sin ( _π_ − _θ_ ) = sin _θ_ and cos ( _π_ − _θ_ ) = −cos _θ_.

By (16.10), sin( _π_ − _θ_ ) = sin _π_ cos _θ_ − cos _π_ sin _θ_ = (0) cos _θ_ − (−1) sin _θ_ = sin _θ_. By (16.6), cos ( _π_ − _θ_ ) = cos _π_ cos _θ_ \+ sin _π_ sin _θ_ = (−1) cos _θ_ \+ (0) sin _θ_ = −cos _θ_.

**7.** Calculate the following values: ( _a_ ) sin 2 _π_ /3; ( _b_ ) sin  ; ( _c_ ) cos 9 _π_ ; ( _d_ ) sin 420°; ( _e_ ) cos 3 _π_ /4; ( _f_ ) cos _π_ /12; ( _g_ ) sin _π_ /8; ( _h_ ) sin 19°.

( _a_ ) By Problem 6,

( _b_ ) By (16.1),

( _c_ ) By (16.1), cos 9 _π_ = cos ( _π_ \+ 8 _π_ ) = cos _π_ = −1

( _d_ ) By (16.1),

( _e_ ) By Problem 6,

( _f_ )

( _g_ ) By (16.14),  . Hence,  . Since   is positive and, therefore,  .

( _h_ ) 19° cannot be expressed in terms of more familiar angles (such as 30°, 45°, 60°) in such a way that any of our formulas are applicable. We must then use the sine table in Appendix A, which gives 0.3256; this is an approximation correct to four decimal places.

**8.** Prove the result of (16.5): If ( _r_ , _θ_ ) are polar coordinates of ( _x_ , _y_ ), then _x_ = _r_ cos _θ_ and _y_ = _r_ sin _θ_.

Let _D_ be the foot of the perpendicular from _A_ ( _x_ , _y_ ) to the _x_ axis (See Fig. 16-12.) Let _F_ be the point on the ray _OA_ at a unit distance from the origin. Then _F_ = (cos _θ_ , sin _θ_ ). If _E_ is the foot of the perpendicular from _F_ to the _x_ axis, then  . Since Δ _ADO_ is similar to Δ _FEO_ (by the _AA_ criterion), we have:

Hence, _x_ = _r_ cos _θ_ and _y_ = _r_ sin _θ_. When _A_ ( _x_ , _y_ ) is in one of the other quadrants, the proof can be reduced to the case where _A_ is in the first quadrant. The case when _A_ is on the _x_ axis or the _y_ axis is very easy.

**Fig. 16-12**

**9.** Find rectangular coordinates of the point with polar coordinates _r_ = 3, _θ_ = _π_ /6.

By(16.5),  .

**10.** Find polar coordinates of the point  .

By (16.5), _r_ 2 = _x_ 2 \+ _y_ 2 = 1 + 3 = 4. Then _r_ = 2. So,  , and  . Thus,  .

**11.** ( _a_ ) Prove the law of cosines (16.15( _a_ )). ( _b_ ) Prove the law of sines (16.5( _b_ )).

( _a_ ) See Fig. 16-11. Take a coordinate system with _C_ as origin and _B_ on the positive _x_ axis. Then _B_ has coordinates ( _a_ , 0). Let ( _x_ , _y_ ) be the coordinates of _A_. By (16.5), _x_ = _b_ cos _θ_ and _y_ = _b_ sin _θ_. By the distance formula (2.1),

Therefore,

( _b_ ) See Fig. 16-13. Let _D_ be the foot of the perpendicular from _A_ to side _BC_ , and let  . Then  . Thus, _h_ = _c_ sin _B_ and so, area of  . (Verify that this also holds when  _B_ is obtuse.) Similarly,  . Hence,  . Dividing by   we obtain the law of sines.

**Fig. 16-13**

**12.** Prove the identity (16.6): cos( _u_ − _υ_ ) = cos _u_ cos _υ_ \+ sin _u_ sin _υ_.

Consider the case where 0   _υ_ < _u_ < _υ_ \+ _π_. (See Fig. 16-14.) By the law of cosines, _BC_ 2 = 12 \+ 12 − 2(1)(1) cos (  _BOC_ ). Thus,

All the other cases can be derived from the case above.

**Fig. 16-14**

### **Supplementary Problems**

**13.** Change the following radian measures into degree measures: ( _a_ ) 4 radians; ( _b_ ) _π_ /10 radians; ( _c_ ) 11 _π_ /12 radians.

_Ans._ ( _a_ ) (720/ _π_ )°; ( _b_ ) 18°; ( _c_ ) 165°

**14.** Change the following degree measures into radian measures: ( _a_ ) 9°; ( _b_ ) 75°; ( _c_ ) (90/ _π_ )°.

_Ans._ ( _a_ ) _π_ /20 radians; ( _b_ ) 5 _π_ /12 radians; ( _c_ )   radian

**15.** Refer to the notation of Fig. 16-3. ( _a_ ) If _r_ = 7 and _θ_ = _π_ /14, find _s_ ; ( _b_ ) If _θ_ = 30° and _s_ = 2, find _r_.

_Ans._ ( _a_ ) _π_ /2; ( _b_ ) 12/ _π_

**16.** Find the angle of rotation between 0 and 2 _π_ that has the same effect as the following rotations: ( _a_ ) 17 _π_ /4; ( _b_ ) 375°; ( _c_ ) − _π_ /3; ( _d_ ) −7 _π_ /2.

_Ans._ ( _a_ ) _π_ /4; ( _b_ ) 15°; ( _c_ ) 5 _π_ /3; ( _d_ ) _π_ /2

**17.** Evaluate: ( _a_ ) cos (4 _π_ /3); ( _b_ ) sin (11 _π_ /6); ( _c_ ) cos 210°; ( _d_ ) sin 315°; ( _e_ ) cos 75°; ( _f_ ) sin 73°.

_Ans._ ( _a_ )  ; ( _b_ )  ; ( _c_ )  ; ( _d_ )  ; ( _e_ )  ; ( _f_ ) Approximately 0.9563

**18.** Assume _θ_ is acute and sin _θ_ =   Evaluate: ( _a_ ) cos _θ_ ; ( _b_ ) sin 2 _θ_ ; ( _c_ ) cos 2 _θ_ ; ( _d_ ) cos ( _θ_ /2).

_Ans._ ( _a_ )  ; ( _b_ )  ; ( _c_ )  ; ( _d_ )

**19.** Assume _θ_ is in the third quadrant ( _π_ < _θ_ < 3 _π_ /2) and  . Find: ( _a_ ) sin _θ_ ; ( _b_ ) cos 2 _θ_ ; ( _c_ ) sin( _θ_ /2).

_Ans._ ( _a_ )  ; ( _b_ )  ; ( _c_ )

**20.** In  , and  .

_Ans._

**21.** Prove the identity  .

**22.** Derive the following values: ( _a_ )  ; ( _b_ )  ; ( _c_ )

[Hints: ( _a_ ) Look at an isosceles right triangle Δ _ABC_.

( _b_ ) Consider an equilateral triangle Δ _ABC_ of side 1. The line _AD_ from _A_ to the midpoint _D_ of side _BC_ is perpendicular to _BC_. Then  . Since  _ABD_ contains _π_ /3 radians,  . By (16.8), sin ( _π_ /6) = cos( _π_ /2 − _π_ /6) = cos ( _π_ /3).

( _c_ )  . So, sin   and cos ( _π_ /6) = sin ( _π_ /3) by (16.8).]

## **CHAPTER 17
Differentiation of Trigonometric Functions**

**CONTINUITY OF cosx AND sin _x_.** It is clear that cos _x_ and sin _x_ are continuous functions, that is, for any _θ_ ,

To see this, observe that, in Fig. 17-1, as _h_ approaches 0, point _C_ approaches point _B_. Hence, the _x_ coordinate of _C_ (namely, cos( _θ_ \+ _h_ )) approaches the _x_ coordinate of _B_ (namely, cos _θ_ ), and the _y_ coordinate of _C_ (namely, sin( _θ_ \+ _h_ )) approaches the y coordinate of _B_ (namely, sin _θ_ ).

To find the derivative of sin _x_ and cos _x_ , we shall need the following limits.

**(17-1)**

**Fig. 17-1**

**(17.2)**

For a proof of (17.1), see Problem 1. From (17.1), (17.2) is derived as follows:

Hence,

**(17.3)** _D x_(sin _x_ ) = cos _x_

**(17.4)** _D x_(cos _x_ ) = −sin _x_

For a proof of (17.3), see Problem 2. From (17.3) we can derive (17.4), with the help of the chain rule and (16.8), as follows:

**GRAPH OF sin _x_.** Since sin ( _x_ \+ 2 π) = sin _x_ , we need only construct the graph for 0   _x_   2 π. Setting D _x_ (sin _x_ ) = cos _x_ = 0 and noting that cos _x_ = 0 in 0, 2π] when and only when _x = π/2_ or _x_ = _3π/2_ , we find the critical numbers π/2 and 3π/2. Since   and −sin(π/2) = −1 < 0 and − sin(3π/2) = 1 > 0, the second derivative test implies that there is a relative maximum at (π/2,1) and a relative minimum at (3π/2,−1). Since D _x_ (sin _x_ ) = cos _x_ is positive in the first and fourth quadrants, sin _x_ is increasing for 0 < _x_ < π/2 and for 3π/2 < _x_ < 2π. Since   is positive in the third and fourth quadrants, the graph is concave upward for π < _x_ < 2π. Thus, there will be an inflection point at (π,0), as well as at (0, 0) and (2π,0). Part of the graph is shown in [Fig. 17-2.

**Fig. 17-2**

**GRAPH OF cos _x_.** Note that sin ( _π/2 + x_ ) = sin(π/2) cos _x_ \+ cos(π/2) sin _x_ = 1·cos _x_ +0 · sin _x_ = cos _x_. Thus, the graph of cos _x_ can be drawn by moving the graph of sin _x_ by π/2 units to the left, as shown in Fig. 17-3.

**Fig. 17-3**

The graphs of _y_ = sin _x_ and _y_ = cos _x_ consist of repeated waves, with each wave extending over an interval of length 2π. The length ( _period_ ) and height ( _amplitude_ ) of the waves can be changed by multiplying the argument and the value, respectively, by constants.

**EXAMPLE 1:** Let _y_ = cos _3x_. The graph is sketched in Fig. 17-4. Because cos 3( _x_ +2π/3) = cos (3 _x_ +2π) = cos 3 _x_ , the function is of period _p_ =2 π/3. Hence, the length of each wave is 2π/3. The number of waves over an interval of length 2π (corresponding to one complete rotation of the ray determining the angle _x_ ) is 3. This number is called the _frequency f_ of cos 3 _x_. In general, _pf_ = (length of each wave) × (number of waves in an interval of 2π) = 2π. Hence, _f_ = _2π/p._

**Fig. 17-4**

For any b > 0, the functions sin _bx_ and cos _bx_ have frequency _b_ and period _2π/b_.

**EXAMPLE 2:** _y_ = 2sin _x_. The graph of this function (see Fig. 17-5) is obtained from that of _y_ = sin _x_ by doubling the _y_ values. The period and frequency are the same as those of _y_ = sin _x_ , that is, _p_ = 2π and _f_ = 1. The amplitude, the maximum height of each wave, is 2.

**EXAMPLE 3:** In general, if b > 0, then _y = A_ sin _bx_ and _y = A_ cos _bx_ have period _2π/b_ , frequency _b_ , and amplitude | _A_ |. Figure 17-6 shows the graph of _y_ = 1.5 sin 4 _x_.

**Fig. 17-5**

**Fig. 17-6**

**OTHER TRIGONOMETRIC FUNCTIONS**

**DERIVATIVES**

**(17.5)** _D x_ (tan _x_ ) = sec2 _x_

**(17.6)** _D x_(cot _x_ ) = −csc2 _x_

**(17.7)** _D x_(sec _x_ ) = tan _x_ sec _x_

**(17.8)** _D x_(csc _x_ ) = − cot _x_ csc _x_

For the proofs, see Problem 3.

**OTHER RELATIONSHIPS**

**(17.9)** tan2 _x_ \+ 1 = sec2 _x_

**(17.10)** tan ( _x_ \+ π) = tan _x_ and cot( _x_ \+ π) = cot _x_

Thus, tan _x_ and cot _x_ have period π. See Problem 4.

**(17.11)** tan (− _x_ ) = − tan _x_ and cot(− _x_ ) = −cot _x_

 , and similarly for cot _x_

**GRAPH OF _y_ = tan _x_.** Since tan _x_ has period π, it suffices to determine the graph in (−π/2, π/2). Since tan(− _x_ ) = −tan _x_ , we need only draw the graph in [0,π/2] and then reflect in the origin. Since tan _x_ = (sin _x_ )/(cos _x_ ), there will be vertical asymptotes at _x_ = π/2 and _x_ = −π/2. By (17.5), _D x_ (tan _x_ ) > 0 and, therefore, tan _x_ is increasing.

Thus, the graph is concave upward when tan _x_ > 0, that is, for 0 < _x_ < π/2, and there is an inflection point at (0, 0). Some special values of tan _x_ are given in Table 17-1, and the graph is shown in Fig. 17-7.

For an acute angle _θ_ of a right triangle,

**GRAPH OF _y_ = sec _x_**. Since sec _x_ = 1/(cos _x_ ), the graph will have a vertical asymptote _x = x 0_ for all _x 0_ for which cos _x 0 = 0_, that is, for _x_ = (2 _n_ \+ 1)π/2, where _n_ is any integer. Like cos _x_ , sec _x_ has a period of 2π, and we can confine our atttention to (−π, π]. Note that |sec _x_ |  1, since |cos _x_ |  1. Setting _D x_(sec _x_ ) = tan _x_ sec _x_ = 0, we find critical numbers at _x = 0_ and _x = π_ , and the first derivative test tells us that there is a relative minimum at _x = 0_ and a relative maximum at _x = π._

**Fig. 17-7**

Since

there are no inflection points and the curve is concave upward for −π/2 < _x_ < π/2. The graph is shown in Fig. 17-8.

**Fig. 17-8**

**ANGLES BETWEEN CURVES.** By the _angle of inclination_ of a nonvertical line   we mean the smaller counterclockwise angle α from the positive _x_ axis to the line. (See Fig. 17-9.) If _m_ is the slope of  , then _m_ = tan α. (To see this, look at Fig. 17-10, where the line  ′ is assumed to be parallel to   and, therefore, has the same slope _m_. Then _m_ = (sin α − 0)/(cos α − 0) = (sin α)/(cos α) = tan α.)

**Fig. 17-9**

**Fig. 17-10**

By the _angle between two curves at a point of intersection P_ we mean the smaller of the two angles between the tangent lines to the curves at _P._ (See Problems 17-18.)

### **Solved Problems**

**1.** Prove (17.1):  .

Since sin  , we need consider only _θ_ > 0. In Fig. 17-11, let _θ_ =  _AOB_ be a small positive central angle of a circle of radius _OA_ = _OB_ = 1. Let _C_ be the foot of the perpendicular from _B_ onto _OA_. Note that _OC_ = cos _θ_ and _CB_ = sin _θ_. Let _D_ be the intersection of _OB_ and an arc of a circle with center at _O_ and radius _OC._ So,

Area of sector _COD_   area of Δ _COB_   area of sector _AOB_

Observe that area of sector   and that area of sector  . (If _W_ is the area of a sector determined by a central angle _θ_ of a circle of radius _r_ , then _W_ /(area of circle) = _θ_ /2π. Thus, _W_ /π _r_ 2 = _θ_ /2π and, therefore,  ).

**Fig. 17-11**

Hence,

Division by   yields

As _θ_ approaches 0+, cos _θ_ → 1, 1/(cos θ) → 1. Hence,

**2.** Prove (17.3): _D x_(sin _x_ ) = cos _x._

Here we shall use (17.1) and (17.2).

Let _y_ = sin _x_. Then _y_ \+ Δ _y_ = sin ( _x_ \+ Δ _x_ ) and

**3.** Prove: ( _a_ ) _D x_ (tan _x_ ) = sec2 _x_ (17.5); ( _b_ ) _D x_ (sec _x_ ) = tan _x_ sec _x_ (17.7).

( _b_ ) Differentiating both sides of (17.9), tan2 _x_ \+ 1 = sec2 _x_ , by means of the chain rule, we get

2 tan _x_ sec2 _x_ = 2 sec _x D x_ (sec _x_ ).

Hence, _D x_ (sec _x_ ) = tan _x_ sec _x._

**4.** Prove (17.10): tan ( _x_ \+ π) = tan _x._

sin( _x_ \+ π) = sin _x_ cosπ + cos _x_ sinπ = − sin _x_

cos( _x_ \+ π) = cos _x_ cosπ− sin _x_ sinπ = −cos _x_

Hence,

**5.** Derive

**6.** Calculate the derivatives of the following functions: ( _a_ ) 2 cos 7 _x_ ; ( _b_ ) sin3 (2 _x_ ); ( _c_ ) tan (5 _x_ ); ( _d_ ) sec (1/ _x_ ).

( _a_ ) _D x_(2 cos 7 _x_ ) = 2(−sin 7 _x_ )(7) = −14sin 7 _x_

( _b_ ) _D x_(sin3 (2x)) = 3(sin2 (2 _x_ ))(cos (2 _x_ ))(2) = 6 sin2 (2 _x_ ) cos (2 _x_ )

( _c_ ) _D x_ (tan (5 _x_ )) = (sec2 (5 _x_ ))(5) = 5 sec2 (5 _x_ )

( _d_ ) _D x_(sec (1/ _x_ )) = tan (1/ _x_ ) sec (1/ _x_ )(−1/ _x_ 2) = −(1/ _x_ 2) tan (1/ _x_ ) sec (1/ _x_ )

**7.** Find all solutions of the equation  .

Solving  , we see that the only points on the unit circle with abscissa   are   and  . The corresponding central angles are π/3 and 5π/3. So, these are the solutions in [0, 2π]. Since cos _x_ has period 2π, the solutions are π/3 + 2π _n_ and 5π/3 + 2π _n_ , where _n_ is any integer.

**8.** Calculate the limits

**9.** Let _y_ = sin _x._ Find  .

**10.** Let _y_ = tan2 (3 _x_ \- 2). Find _y_ ″.

**11.** Assume _y_ = sin( _x_ \+ _y_ ). Find _y′_.

Solving for y′,

**12.** Assume sin _y_ \+ cos _x_ = 1. Find _y″_.

**13.** A pilot is sighting a location on the ground directly ahead. If the plane is flying 2 miles above the ground at 240 mi/h, how fast must the sighting instrument be turning when the angle between the path of the plane and the line of sight is 30°? See Fig. 17-12.

**Fig. 17-12**

**14.** Sketch the graph of _f_ ( _x_ ) = sin _x_ \+ cos _x_.

_f_ ( _x_ ) has a period of 2π. Hence, we need only consider the interval 0, 2π]. _f′(x)_ = cos _x_ − sin _x_ , and _f″(x)_ = −(sin _x_ \+ cos _x_ ). The critical numbers occur where cos _x_ = sin _x_ or tan _x_ = 1, _x_ = π/4 or _x_ = 5π/4.   So, there is a relative maximum at     Thus, there is a relative minimum at   The inflection points occur where _f_ "( _x_ ) = −(sin _x_ \+ cos _x_ ) = 0, sin _x_ = −cos _x_ , tan _x_ = −1, _x_ = 3π/4 or _x_ = 7π/4, _y_ = 0. See [Fig. 17-13.

**Fig. 17-13**

**15.** Sketch the graph of _f_ ( _x_ ) = cos _x_ \- cos2 _x_.

and

Since _f_ has period 2π, we need only consider (−π,π], and since _f_ is even, we have to look at only 0, π]. The critical numbers are the solutions in [0, π] of sin _x_ = 0 or 2 cos _x_ − 1 = 0. The first equation has solutions 0 and π, and the second is equivalent to  , which has the solution π/3. _f_ ″(0) = 1 > 0; so there is a relative minimum at (0, 0). _f_ ″(π) = 3 > 0; so there is a relative minimum at (π, −2). ; hence there is a relative maximum at  . There are inflection points between 0 and π/3 and between π/3 and π; they can be found by using the quadratic formula to solve 4 cos2 _x_ − cos _x_ − 2 = 0 for cos _x_ and then using a cosine table or a calculator to approximate _x_. See [Fig. 17-14.

**Fig. 17-14**

**16.** Find the absolute extrema of _f_ ( _x_ ) = sin _x_ \+ _x_ on [0, 2π].

_f_ ′( _x_ ) = cos _x_ \+ 1. Setting _f_ ′( _x_ ) = 0, we get cos _x_ = −1 and, therefore, the only critical number in [0, 2π] is _x_ = π. We list π and the two endpoints 0 and compute the values of _f_ ( _x_ ).

Hence, the absolute maximum 2π is achieved at _x_ = 2π, and the absolute minimum 0 at _x_ = 0.

**17.** Find the angle at which the lines   1: _y_ = _x_ \+ 1 and   2: _y_ = −3 _x_ \+ 5 intersect.

Let α1 and α2 be the angles of inclination of   1 and   2 (see Fig. 17-15), and let _m_ 1 and _m_ 2 be the respective slopes. Then tan α1 = _m_ 1 = 1 and tan α2 = _m_ 2 = −3. α2 − α1 is the angle of intersection. Now, by Problem 5,

From a graphing calculator, α2 \- α1 ~ 63.4°.

**Fig. 17-15**

**18.** Find the angle α between the parabolas _y_ = _x_ 2 and _x_ = _y_ 2 at (1,1).

Since _D x_( _x_ 2) = 2 _x_ and  , the slopes at (1,1) are 2 and  . Hence,  . Thus, using a graphing calculator, we approximate α by 36.9°.

### **Supplementary Problems**

**19.** Show that cot ( _x_ \+ π) = cot _x_ , sec ( _x_ \+ 2π) = sec _x_ , and csc ( _x_ \+ 2π) = csc _x_.

**20.** Find the period _p_ , frequency _f_ , and amplitude _A_ of 5 sin ( _x_ /3) and sketch its graph.

_Ans. p_ = 6π, _f_ =  , _A_ = 5

**21.** Find all solutions of cos _x_ = 0.

_Ans._  for all integers _n_

**22.** Find all solutions of tan _x_ = 1.

_Ans._   for all integers _n_

**23.** Sketch the graph of

_Ans._ See Fig. 17-16

**Fig. 17-16**

**24.** Derive the formula  .

**25.** Find _y_ ′.

**26.** Evaluate:

_Ans._

**27.** If _x_ = _A_ sin _kt_ \+ _B_ cos _kt_ , show that  .

**28.** ( _a_ ) If _y_ = 3 sin (2 _x_ \+ 3), show that _y_ ″ + 4 _y_ = 0. ( _b_ ) If _y_ = sin _x_ \+ 2 cos _x_ , show that _y_ ′″ + _y_ ″ + _y_ ′ + _y_ = 0.

**29.** ( _i_ ) Discuss and sketch the following on the intervel 0   _x_ < 2π. ( _ii_ ) (GC) Check your answers to ( _a_ ) on a graphing calculator.

( _a_ ) _y_ =   sin 2 _x_

( _b_ ) _y_ = cos2 _x_ − cos _x_

( _c_ ) _y_ = _x_ − 2 sin _x_

( _d_ ) _y_ = sin _x_ (1 + cos _x_ )

( _e_ ) _y_ = 4 cos3 _x_ − 3 cos _x_

_Ans._

**30.** If the angle of elevation of the sun is 45° and is decreasing by   radians per hour, how fast is the shadow cast on the ground by a pole 50 ft tall lengthening?

_Ans._ 25 ft/h

**31.** Use implicit differentiation to find _y_ ′: ( _a_ ) tan _y_ = _x_ 2; ( _b_ ) cos ( _xy_ ) = 2 _y_.

_Ans._ ( _a_ ) _y_ ′ = 2 _x_ cos 2 _y_ ; ( _b_ )

## **CHAPTER 18
Inverse Trignometric Functions**

The sine and cosine functions and the other trignometric functions are not one–to–one and, therefore, do not have inverse functions. However, it is possible to restrict the domain of trigonometric functions so that they become one–to–one.

Looking at the graph of _y_ = sin _x_ (see Fig. 17–2), we note that on the interval − _π_ /2   _x_   _π_ /2 the restriction of sin _x_ is one–to–one. We then define sin−1 _x_ to be the corresponding inverse function. The domain of this function is [−1, 1], which is the range of sin _x_. Thus,

1. sin−1 ( _x_ ) = _y_ if and only if sin _y_ = _x_.

2. The domain of sin−1 _x_ is [−1, 1].

3. The range of sin−1 _x_ is [− _π_ /2, _π_ /2].

The graph of sin−1 _x_ is obtained from the graph of sin _x_ by reflection in the line _y_ = _x_. See Fig. 18–1.

**EXAMPLE 1:** In general, sin−1 _x_ = the number _y_ in [− _π_ /2, _π_ /2] such that sin _y_ = _x_. In particular, sin−1 0= 0, sin−1 1 = _π_ /2, sin−1(−1) = − _π_ /2,  ,  ,  . Also,  . In general, sin−1(− _x_ ) = −sin−1 _x_ , because sin(− _y_ ) = −sin _y_.

**THE DERIVATION OF sin −1 _x_.** Let _y_ = sin−1 _x_. Since sin _x_ is differentiable, sin−1 _x_ is differentiable by Theorem 10.2. Now, sin _y_ = _x_ and, therefore, by implicit differentiation, (cos _y_ ) _y_ ′ = 1. Hence, _y_ ′ = 1/(cos _y_ ). But cos2 _y_ = 1 − sin2 _y_ = 1 − _x_ 2. So,  . By definition of sin−1 _x_ , _y_ is in the interval [− _π_ /2, _π_ /2] and, therefore, cos _y_   0. Hence,  . Thus,  . So, we have shown that

**(18.1)**

**Fig. 18-1**

**THE INVERSE COSINE FUNCTION.** If we restrict the domain of cos _x_ to [0, _π_ ], we obtain a one–to–one function (with, range [−1, 1]). So we can define cos−1 _x_ to be the inverse of that restriction.

1. cos−1( _x_ ) = _y_ if and only if cos _y_ = _x_.

2. The domain of cos−1 _x_ is [−1, 1].

3. The range of cos−1 _x_ is [0, _π_ ].

The graph of cos−1 _x_ is shown in Fig. 18-2. It is obtained by reflecting the graph of _y_ = cos _x_ in the line _y_ = _x_.

**Fig. 18-2**

An argument similar to the one above for (18.1) shows that

**(18.2)**

**THE INVERSE TANGENT FUNCTION.** Restriciting the domain of tan _x_ to the interval (− _π_ /2, _π_ /2) we obtain a one-to-one function (with range the set of all real numbers), whose inverse we take to be tan−1 _x_. Then:

1. tan−1( _x_ ) = _y_ if and only if tan _y_ = _x_.

2. The domain of tan−1 _x_ is (–∞, +∞).

3. The rangle of tan−1 _x_ is (– _π_ /2, _π_ /2).

**EXAMPLE 2:** In general, tan−1 _x_ = the number _y_ in (– _π_ /2, _π_ /2) such that tan _y_ = _x_. In particular, tan−1 0 = 0, tan−1 1 = _π_ /4,  ,  . Since tan (– _x_ ) = –tan _x_ , it follows that tan−1(– _x_ ) = − tan−1 _x_. For example, tan−1 (– _x_ ) = –tan−1 _x_. For example, tan−1(−1) = – _π_ /4.

The graph of _y_ = tan−1 _x_ is shown in Fig. 18-3. It is obtained from the graph of _y_ = tan _x_ by reflection in the line _y_ = _x_. Note that _y_ = _π_ /2 is a horizontal asymptote on the right and _y_ = – _π_ /2 is a horizontal asymptote on the left.

**(18.3)**

In fact, if _y_ = tan−1 _x_ , tan _y_ = _x_ and, by implicit differentation, sec2 _y_ ( _y_ ′) = 1. Hence,

**Fig. 18-3**

Inverse of cot _x_ , sec _x_ , and csc _x_ are defined in similar fashion. **cot −1 _x_**. Restrict cot _x_ to (0, _π_ ). Then the domain of cot−1 _x_ is (–∞, +∞) and

_y_ = cot−1 _x_ if and only if cot _y_ = _x_

**(18.4)**

The proof is similar to that of (18.3). The graphs of cot _x_ and cot−1 _x_ are shown in Fig. 18-4.

**Fig. 18-4**

**sec −1 _x_**. Restrict sec _x_ to the union of [0, _π_ /2) and [ _π_ , 3 _π_ /2). Then the domain of sec−1 _x_ consists of all _y_ such that | _y_ | 1 and

_y_ =sec−1 _x_ if and only if sec _y_ = _x_

**(18.5)**

For the proof, see Problem 1. The graph of sec _x_ appeared in Fig. 17-8, and that of sec−1 _x_ is shown in Fig. 18-5.

**Fig. 18-5**

**csc −1 _x_**. Restrict csc _x_ to the union of (0, _π_ /2] and ( _π_ , 3 _π_ /2]. Then the domain of csc−1 _x_ consists of all _y_ such that | _y_ |   1 and

_y_ = csc−1 _x_ if and only if csc _y_ = _x_

**(18.6)**

The proof is similar to that of (18.5). The graphs of csc _x_ and csc−1 _x_ are shown in Fig. 18-6.

**Fig. 18-6**

The apparently arbitrary choices of the domains for the inverse trigonometric functions were made in order to obtain simple formulas for the derivatives.

Do not confuse the notation for the inverse trigonometric functions with exponential notation. For example, sin−1 _x_ is not the same as (sin _x_ )−1. To avoid the possibility of such confusion, one can use the following alternative notation for the inverse trigonometric functions:

arcsin _x_ = sin−1 _x_ , arccos _x_ = cos−1 _x_ , etc.

### **Solved Problems**

**1. Prove (18.5)** :

Let _y_ = sec−1 _x_. Then sec _y_ = _x_ and, by implicit differentiation, tan _y_ sec _y_ ( _y_ ′) = 1. Now tan2 _y_ = sec2 _y_ − 1 = _x_ 2 − 1; hence,  . By definition of sec−1 _x_ , _y_ is in [0, _π_ /2) or [ _π_ , 3 _π_ /2), and, therefore, tan _y_ is positive. Thus,  . So,

In Problems 2–8, find the first derivative _y_ ′.

**2.** _y_ = sin−1 (2 _x_ − 3).

By (18.1) and the chain rule,

**3.** _y_ = cos−1( _x_ 2).

By (18.2) and the chain rule,

**4.** _y_ = tan−1(3 _x_ 2).

By (18.3) and the chain rule,

**5.**  .

By (18.4) and the chain rule,

**6.**  .

**7.**

**8.**

**9.** If _y_ 2 sin _x_ \+ _y_ = tan−1 _x_ , find _y_ ′.

By implicit differentiation,  . Hence,   and, therefore,

**10.** Evaluate: ( _a_ )  ; ( _b_ ) cos−1(1); ( _c_ ) cos−1(0); ( _d_ )  ; ( _e_ )  ; ( _f_ ) sec−1(2); ( _g_ ) sec−1 (−2)

( _a_ )

( _b_ ) cos−1(1) = 0, since cos(0) = 1 and 0 is in [0, _π_ ]

( _c_ ) cos−1(0) = _π_ /2, since cos( _π_ /2) = 0 and ( _π_ /2) is in [0, _π_ ]

( _d_ )

( _e_ )

( _f_ ) sec−1(2) = _π_ /3, since

( _g_ ) sec−1(−2) = 4 _π_ /3, since

**11.** Show that  .

 . Then, by Problem 15 of Chapter 13, sin−1 _x_ \+ cos−1 _x_ is a constant. Since  , that constants is  .

**12.** ( _a_ ) Prove: sin(sin−1( _y_ )) = _y_ ; ( _b_ ) Find sin−1(sin _π_ ); ( _c_ ) Prove that sin−1(sin _x_ ) = _x_ if and only if _x_ is in [– _π_ /2, _π_ /2].

( _a_ ) This follows directly from the definition of sin−1( _y_ ).

( _b_ ) sin−1(sin _π_ ) = sin−10 = 0.

( _c_ ) sin−1 _y_ is equal to that number _x_ in [– _π_ /2, _π_ /2] such that sin _x_ = _y_. So, if _x_ is in [– _π_ /2, _π_ /2], sin−1(sin _x_ ) = _x_. If _x_ is not in [– _π_ /2, _π_ /2], then sin−1(sin _x_ )≠ _x_ , since, by definition, sin−1(sin _x_ ) must be in [– _π_ /2, _π_ /2].

**13.** Evaluate: ( _a_ )  ; ( _b_ )  .

( _a_ ) By (16.11),  .

( _b_ )  .

Hence,  . Since   is in the second quadrant,  . So,  .

**14.** The lower edge of a mural, 12 ft high, is 6 ft above an observer's eye. Under the assumption that the most favorable view is obtained when the angle subtended by the mural at the eye is a maximum, at what distance from the wall should the observer stand?

Let _θ_ denote the subtended angle, and let _x_ be the distance from the wall. From Fig. 18-7, tan ( _θ_ \+  ) = 18/ _x_ , tan   = 6/ _x_ , and

**Fig. 18-7**

Then,

The critical number   By the first derivative test, this yields a relative maximum. The observer should stand about 10.4 ft in front of the wall.

### **Supplementary Problems**

**15.** Evaluate: ( _a_ )  ; ( _b_ )  ; ( _c_ )  ; ( _d_ )  ; ( _e_ )  ; ( _f_ )  .

_Ans_.

**16.** Prove:  .

In Problems 17–24, find _y_ ′.

**25.** Prove formulas (18.2), (18.4), and (18.6).

**26.** Let  . Find: ( _a_ ) sin _θ_ ; ( _b_ ) cos _θ_ ; ( _c_ ) tan _θ_ ; ( _d_ ) cot _θ_ ; ( _e_ ) sec _θ_ ; ( _f_ ) csc _θ_ ; ( _g_ ) cos 2 _θ_ ; ( _h_ ) sin 2 _θ_.

_Ans_.

**27.** Let  . Find: ( _a_ ) sin _θ_ ; ( _b_ ) cos _θ_ ; ( _c_ ) tan _θ_ ; ( _d_ ) cot _θ_ ; ( _e_ ) sec _θ_ ; ( _f_ ) csc _θ_ ; ( _g_ ) cos 2 _θ_ ; ( _h_ ) sin 2 _θ_ ;

_Ans_.

**28.** Prove:  .

**29.** Evaluate:  .

_Ans_.

**30.** Find the domain and range of the function _f_ ( _x_ ) = sin(sec−1 _x_ ).

_Ans_. Domain | _x_ | 1; range (−1, 1)

**31.** ( _a_ ) For which values of _x_ is tan−1 (tan _x_ ) = _x_ true?

( _b_ ) (GC) Use a graphing calculator to draw the graph of _y_ = tan−1 (tan _x_ ) − _x_ to verify your answer to ( _a_ ).

_Ans_. ( _a_ )

**32.** A light is to be placed directly above the center of a circular plot of radius 30 ft, at such a height that the edge of the plot will get maximum illumination. Find the height if the intensity _I_ at any point on the edge is directly proportional to the cosine of the angle of incidence (angle between the ray of light and the vertical) and inversely proportional to the square of the distance from the source.

( _Hint:_ : Let _x_ be the required height, _y_ the distance from the light to a point on the edge, and _θ_ the angle of incidence. Then  .)

_Ans_.

**33.** Show that   for | _x_ | < 1. Examine what happens when | _x_ | = 1.

**34.** (GC) Evaluate   by using a graphical calculator.

_Ans_. 0.6435

**35.** ( _a_ ) Find  . ( _b_ ) Find an algebraic formula for sec(tan−1 (2 _x_ )). ( _c_ ) (GC) Verify ( _a_ ) and ( _b_ ) on a graphing calculator.

_Ans_.

**36.** Prove: ( _a_ )  .

[The formula of part ( _a_ ) would hold in general for | _x_ |   1 if we had defined sec−1 _x_ to be the inverse of the restriction of sec _x_ to (– _π_ /2, _π_ /2). However, if we had done that, then the formula for _D x_(sec−1 _x_ ) would have been   instead of the simpler formula  .]

## **CHAPTER 19
Rectilinear and Circular Motion**

**RECTILINEAR MOTION.** _Rectilinear motion_ is motion of an object on a straight line. If there is a coordinate system on that line, and _s_ denotes the coordinate of the object at any time _t_ , then the position of the object is given by a function _s_ = _f_ ( _t_ ). (See Fig. 19-1.)

**Fig. 19-1**

The position at a time _t_ \+ Δ _t_ , very close to _t_ , is _f_ ( _t_ \+ Δ _t_ ). The "distance" the object travels between time _t_ and time _t_ \+ Δ _t_ is _f_ ( _t_ \+ Δ _t_ ) − _f_ ( _t_ ). The time the object has traveled is Δ _t_. So, the _average velocity_ over this period of time is

(Note that the "distance" can be negative when the object is moving to the left along the _s_ -axis. So the average velocity can be positive or negative or zero.)

As Δ _t_ approaches zero, this average velocity approaches what we think of as the _instantaneous velocity_ ν at time _t_. So,

Hence, the instantaneous velocity _ν_ is the derivative of the position function _s_ , that is, _ν_ = _ds_ / _dt_.

The sign of the instantaneous velocity _ν_ tells us in which direction the object is moving along the line. If _ν_ = _ds_ / _dt_ > 0 on an interval of time, then by Theorem 13.7( _a_ ), we know that _s_ must be increasing, that is, the object is moving in the direction of increasing _s_ along the line. If _ν_ = _ds_ / _dt_ < 0, then the object is moving in the direction of decreasing _s_.

The instantaneous _speed_ of the object is defined as the absolute value of the velocity:

Speed = | _ν_ |

Thus, the speed indicates how fast the object is moving, but not its direction. In an automobile, the speedometer tells us the instantaneous speed at which the car is moving.

The _acceleration a_ of an object moving on a straight line is defined as the rate at which the velocity is changing, that is, the derivative of the velocity:

**EXAMPLE 1:** Let the position of an automobile on a highway be given by the equation _s_ = _f_ ( _t_ ) = _t_ 2 − 5 _t_ , where _s_ is measured in miles and _t_ in hours. Then its velocity _ν_ = 2 _t_ − 5 mi/h and its acceleration _a_ = 2 mi/h2. Thus, its velocity is increasing at the rate of 2 miles per hour per hour.

When an object moving along a straight line changes direction, its velocity _ν_ = 0. For, a change in direction occurs when the position _s_ reaches a relative extremum, and this occurs only when _ds_ / _dt_ = 0. (However, the converse is false; _ds_ / _dt_ = 0 does not always indicate a relative extremum. An example is _s_ = _t_ 3 at _t_ = 0.)

**EXAMPLE 2:** Assume that an object moves along a straight line according to the equation _s_ = _f_ ( _t_ ) = ( _t_ − 2)2, where _s_ is measured in feet and _t_ in seconds. (The graph of _f_ is shown in Fig. 19-2.) Then _ν_ = _f_ ′( _t_ ) = 2( _t_ − 2) ft/sec and _a_ = 2 ft/sec2. For _t_ < 2, _ν_ < 0 and the object is moving to the left. (See Fig. 19-3.) For _t_ > 2, _ν_ > 0 and the object is moving to the right. The object changes direction at _t_ = 2, where _ν_ = 0. Note that, although the velocity _ν_ is 0 at time _t_ = 2, the object is moving at that time; it is not standing still. When we say that an object is _standing still_ , we mean that its position is constant over a whole interval of time.

**Fig. 19-2**

**Fig. 19-3**

**MOTION UNDER THE INFLUENCE OF GRAVITY.** If an object has been thrown straight up or down, or just starts from rest, and the only force acting upon it is the gravitational pull of the earth, then the resulting rectilinear motion is referred to as _free fall_.

Put a coordinate system on the vertical line on which the object is moving. Assume that this _s_ axis is directed upward (see Fig. 19-4), and that ground level (the surface of the earth) corresponds to _s_ = 0. It is a fact of physics that the acceleration _a_ is a constant approximately equal to −32 ft/sec2. (In the metric system, this constant is −9.8 m/sec2.) Note that the acceleration is negative because the pull of the earth causes the velocity to decrease.

Since  , we have:

**Fig. 19-4**

**(19.1)** _ν_ = _ν_ 0 − 32 _t_

where _ν_ 0 is the _initial_ velocity when _t_ = 0.† Now,  . Hence,

**(19.2)** _s_ = _s_ 0 \+ _ν_ 0 _t_ − 16 _t_ 2

where _s_ 0 is the _initial position_ , the value of _s_ when _t_ = 0.‡

**CIRCULAR MOTION**. The motion of a particle _P_ along a circle is completely described by an equation _θ_ = _f_ ( _t_ ), where _θ_ is the central angle (in radians) swept over in time _t_ by a line joining _P_ to the center of the circle. The _x_ and _y_ coordinates of _P_ are given by _x_ = _r_ cos _θ_ and _y_ = _r_ sin _θ_.

By the _angular velocity ω_ of _P_ at time _t_ we mean  .

By the angular _acceleration_ α of _P_ at time _t_ we mean  .

### **Solved Problems**

**1.** A body moves along a straight line according to the law  . Determine its velocity and acceleration at the end of 2 seconds.

**2.** The path of a particle moving in a straight line is given by _s_ = _t_ 3 − 6 _t_ 2 \+ 9 _t_ \+ 4.

( _a_ ) Find _s_ and _a_ when _ν_ = 0.

( _b_ ) Find _s_ and _ν_ when _a_ = 0.

( _c_ ) When is _s_ increasing?

( _d_ ) When is _ν_ increasing?

† In fact, _D_ t( _ν_ 0 − 32 _t_ ) = −32 = _D tν_. So, by Chapter 13, Problem 18, _ν_ and _ν_ 0 − 32 _t_ differ by a constant. Since _ν_ and _ν_ 0 − 32 _t_ are equal when _t_ = 0, that constant difference is 0.

‡ In fact, _D t_( _s_ 0 \+ _ν_ 0 _t_ − 16 _t_ 2) = _ν_ 0 − 32 _t_ = _D ts_. So, by Chapter 13, Problem 18, _s_ and _s_ 0 \+ _ν_ 0 _t_ − 16 _t_ 2 differ by a constant. Since _s_ and _s_ 0 \+ _ν_ 0 _t_ − 16 _t_ 2 are equal when _t_ = 0, that constant difference is 0.

( _e_ ) When does the direction of motion change?

We have

( _a_ ) When _ν_ = 0, _t_ = 1 and 3. When _t_ = 1, _s_ = 8 and _a_ = −6. When _t_ = 3, _s_ = 4 and _a_ = 6.

( _b_ ) When _a_ = 0, _t_ = 2, At _t_ = 2, _s_ = 6 and _v_ = −3.

( _c_ ) _s_ is increasing when _ν_ > 0, that is, when _t_ < 1 and _t_ > 3.

( _d_ ) _ν_ is increasing when _a_ > 0, that is, when _t_ > 2.

( _e_ ) The direction of motion changes when _ν_ = 0 and _a_ ≠ 0. From ( _a_ ) the direction changes when _t_ = 1 and _t_ = 3.

**3.** A body moves along a horizontal line according to _s_ = _f_ ( _t_ ) = _t_ 3 − 9 _t_ 2 \+ 24 _t_.

( _a_ ) When is _s_ increasing, and when it is decreasing?

( _b_ ) When is _ν_ increasing, and when it is decreasing?

( _c_ ) Find the total distance travelled in the first 5 seconds of motion.

We have

( _a_ ) _s_ is increasing when _ν_ > 0, that is, when _t_ < 2 and _t_ > 4.

_s_ is decreasing when _ν_ < 0, that is, when 2 < _t_ < 4.

( _b_ ) _ν_ is increasing when _a_ > 0, that is, when _t_ > 3.

_ν_ is decreasing when _a_ < 0, that is, when _t_ < 3.

( _c_ ) When _t_ = 0, _s_ = 0 and the body is at _O_. The initial motion is to the right ( _ν_ > 0) for the first 2 seconds; when _t_ = 2, the body is _s_ = _f_ (2) = 20 ft from _O_.

During the next 2 seconds, it moves to the left, after which it is _s_ = _f_ (4) = 16 ft from _O_.

It then moves to the right, and after 5 seconds of motion in all, it is _s_ = _f_ (5) = 20 ft from _O_. The total distance traveled is 20 + 4 + 4 = 28 ft (see Fig. 19-5).

**Fig. 19-5**

**4.** A particle moves in a horizontal line according to _s_ = _f_ ( _t_ ) = _t_ 4 − 6 _t_ 3 \+ 12 _t_ 2 − 10 _t_ \+ 3.

( _a_ ) When is the speed increasing, and when decreasing?

( _b_ ) When does the direction of motion change?

( _c_ ) Find the total distance travelled in the first 3 seconds of motion.

Here

( _a_ ) _ν_ changes sign at _t_ = 2.5, and _a_ changes sign at _t_ = 1, _t_ = 2.

For _t_ < 1, _ν_ < 0 and _a_ > 0. Since _a_ > 0, _ν_ is increasing. Since _ν_ < 0, the speed | _ν_ | = – _ν_ is decreasing.

For 1 < _t_ < 2, _ν_ < 0 and _a_ < 0. Since _a_ < 0, _ν_ is decreasing. Since _ν_ < 0, the speed | _ν_ | = – _ν_ is increasing.

For 2 < _t_ < 2.5, _ν_ < 0 and _a_ > 0. As in the first case, the speed is decreasing.

For _t_ > 2.5, _ν_ > 0 and _a_ > 0. _ν_ is increasing. Since _ν_ > 0, the speed | _ν_ | = _ν_ is increasing.

( _b_ ) The direction of motion changes at _t_ = 2.5, since, by the second derivative test, _s_ has a relative extremum there.

( _c_ ) When _t_ = 0, _s_ = 3 and the particle is 3 units to the right of _O_. The motion is to the left until _t_ = 2.5, at which time the particle is   units to the left of _O_. When _t_ = 3, _s_ = 0; the particle has moved   units to the right. The total distance travelled is   units. (See Fig. 19-6.)

**Fig. 19-6**

**5.** A stone, projected vertically upward with initial velocity 112 ft/sec, moves according to _s_ = 112 _t_ − 16 _t_ 2, where _s_ is the distance from the starting point. Compute ( _a_ ) the velocity and acceleration when _t_ = 3 and when _t_ = 4, and ( _b_ ) the greatest height reached. ( _c_ ) When will its height be 96 ft?

We have _ν_ = _ds/dt_ = 112 − 32 _t_ and _a_ = _dv/dt_ = −32.

( _a_ ) At _t_ = 3, _ν_ = 16 and _a_ = −32. The stone is rising at 16ft/sec.

At _t_ = 4, _ν_ = −16 and _a_ = −32. The stone is falling at 16ft/sec.

( _b_ ) At the highest point of the motion, _ν_ = 0. Solving _ν_ = 0 = 112 − 32 _t_ yields _t_ = 3.5. At this time, _s_ = 196 ft.

( _c_ ) Letting 96 = 112 _t_ − 16 _t_ 2 yields _t_ 2 − 7 _t_ \+ 6 = 0, from which _t_ = 1 and 6. At the end of 1 second of motion the stone is at a height of 96 ft and is rising, since _ν_ > 0. At the end of 6 seconds it is at the same height but is falling since _ν_ < 0.

**6.** A particle rotates counterclockwise from rest according to _θ_ = _t_ 3/50 − _t_ , where _θ_ is in radians and _t_ in seconds. Calculate the angular displacement _θ_ , the angular velocity _ω_ , and the angular acceleration α at the end of 10 seconds.

**7.** At _t_ = 0, a stone is dropped from the top of a building 1024 ft high. When does it hit the ground, and with what speed? Find the speed also in miles per hour.

Since _s_ 0 = 1024 and _ν_ 0 = 0, equation (19.2) becomes _s_ = 1024 − 16 _t_ 2, and the time of hitting the ground is the solution of 1024 − 16 _t_ 2 = 0. This reduces to _t_ 2 = 64, yielding _t_ = ±8. Since the motion occurs when _t_   0, _t_ = 8. The equation (19.1) is _ν_ = −32 _t_ , yielding _ν_ = −32(8) = −256 ft/sec when _t_ = 8, that is, when the stone hits the ground. (The velocity is negative because the stone is moving downward.) The speed is 256 ft/sec. To change to miles per hour, note the following:

Thus,

**(19.3)** _x_ feet per second   miles per hour.

In particular, when _x_ = 256, we get   miles per hour.

**8.** If a rocket is shot vertically upward from the ground with an initial velocity of 192 ft/sec, when does it reach its maximum height above the ground, and what is that maximum height? Also find how long it takes to reach the ground again and with what speed it hits the ground.

Equations (19.1) and (19.2) are _ν_ = 192 − 32 _t_ and _s_ = 192 _t_ − 16 _t_ 2. At the maximum height, _ν_ = 0, and, therefore, _t_ = 6. So, it takes 6 seconds to reach the maximum height, which is 192(6) − 16(6)2 = 576 ft. The rocket returns to ground level when 0 = 192 _t_ − 16 _t_ 2, that is, when _t_ = 12. Hence, it took 6 seconds to reach the ground again, exactly the same time it took to reach the maximum height. The velocity when _t_ = 12 is 192 − 32(12) = −192 ft/sec. Thus, its final speed is the same as its initial speed.

### **Supplementary Problems**

**9.** Show that, if an object is moving on a straight line, then its speed is increasing when its velocity _ν_ and its acceleration _a_ have the same sign, and its speed is decreasing when _ν_ and _a_ have opposite sign. ( _Hint:_ The speed _S_ = | _ν_ |. When _ν_ > 0, _S_ = _ν_ and _ds/dt_ = _dv/dt_ = _a_. When _ν_ < 0, _S_ = – _ν_ and _ds/dt_ = – _dv/dt_ = – _a_.)

**10.** An object moves in a straight line according to the equation _s_ = _t_ 3 − 6 _t_ 2 \+ 9 _t_ , the units being feet and seconds. Find its position, direction, and velocity, and determine whether its speed is increasing or decreasing when  .

_Ans_.

**11.** The distance of a locomotive from a fixed point on a straight track at time _t_ is 3 _t_ 4 − 44 _t_ 3 \+ 144 _t_ 2. When is it in reverse?

_Ans_. 3 < _t_ < 8

**12.** Examine, as in Problem 2, each of the following straight line motions: ( _a_ ) _s_ = _t_ 3 − 9 _t_ 2 \+ 24 _t_ ; ( _b_ ) _s_ = _t_ 3 − 3 _t_ 2 \+ 3 _t_ \+ 3; ( _c_ ) _s_ = 2 _t_ 3 − 12 _t_ 2 \+ 18 _t_ − 5; ( _d_ ) _s_ = 3 _t_ 4 − 28 _t_ 3 \+ 90 _t_ 2 − 108 _t_.

_Ans_. The changes of direction occur at _t_ = 2 and _t_ = 4 in ( _a_ ), not at all in ( _b_ ), at _t_ = 1 and _t_ = 3 in ( _c_ ), and at _t_ = 1 in ( _d_ ).

**13.** An object moves vertically upward from the earth according to the equation _s_ = 64 _t_ − 16 _t_ 2. Show that it has lost one-half its velocity in its first 48 ft of rise.

**14.** A ball is thrown vertically upward from the edge of a roof in such a manner that it eventually falls to the street 112 ft below. If it moves so that its distance _s_ from the roof at time _t_ is given by _s_ = 94 _t_ − 16 _t_ 2, find ( _a_ ) the position of the ball, its velocity, and the direction of motion when _t_ = 2, and ( _b_ )its velocity when it strikes the street. ( _s_ in feet, and _t_ in seconds.)

_Ans_. ( _a_ ) 240 ft above the street, 32 ft/sec upward; ( _b_ ) −128 ft/sec

**15.** A wheel turns through an angle _θ_ radians in time _t_ seconds so that _θ_ = 128 _t_ − 12 _t_ 2. Find the angular velocity and acceleration at the end of 3 seconds.

_Ans_. ω = 56 rad/sec; α = −24 rad/sec2

**16.** A stone is dropped down a well that is 144 ft deep. When will it hit the bottom of the well?

_Ans_. After 3 seconds

**17.** With what speed in miles per hour does an object dropped from the top of a 10–story building hit the ground? Assume that each story of the building is 10 ft high.

_Ans_.  .

**18.** An automobile moves along a straight road. If its position is given by _s_ = 8 _t_ 3 − 12 _t_ 2 \+ 6 _t_ − 1, with _s_ in miles and _t_ in hours, what distance does it travel from _t_ = 0 to _t_ = 1?

_Ans_. 2 miles.

**19.** Answer the same question as in Problem 18, except that _s_ = 5 _t_ − _t_ 2 and the car operates from _t_ = 0 to _t_ = 3.

_Ans_. 6.5 miles

**20.** A stone was thrown straight up from the ground. What was its initial velocity in feet/second if it hit the ground after 15 seconds?

_Ans_. 240 ft/sec

**21.** (GC) Let the position _s_ of an object moving on a straight line be given by _s_ = _t_ 4 − 3 _t_ 2 \+ 2 _t_. Use a graphing calculator to estimate when the object changes direction, when it is moving to the right, and when it is moving to the left. Try to find corresponding exact formulas.

_Ans_. Change of direction at _t_ = −1.3660, 0.3660, and 1. The object moves left for _t_ < −1.3660 and for

0.3660 < _t_ < 1. The exact values of _t_ at which the object changes direction are 1 and  .

**22.** (GC) An object is moving along a straight line according to the equation _s_ = 3 _t_ − _t_ 2. A second object is moving along the same line according to the equation _s_ = _t_ 3 − _t_ 2 \+ 1. Use a graphing calculator to estimate ( _a_ ) when they occupy the same position and ( _b_ ) when they have the same velocity. At the time(s) when they have the same position, are they moving in the same direction?

_Ans_. ( _a_ ) 0.3473 and 1.5321; ( _b_ ) _t_ = ±1.

Opposite directions at both intersections

## **CHAPTER 20
Related Rates**

If a quantity _y_ is a function of time _t_ , the rate of change of _y_ with respect to time is given by _dy/dt_. When two or more quantities, all functions of the time _t_ , are related by an equation, the relation of their rates of change may be found by differentiating both sides of the equation.

**Example 1:** A 25-foot ladder rests against a vertical wall. (See Fig. 20-1.) If the bottom of the ladder is sliding away from the base of the wall at the rate of 3 ft/sec, how fast is the top of the ladder moving down the wall when the bottom of the ladder is 7 feet from the base?

**Fig. 20-1**

Let _x_ be the distance of the bottom of the ladder from the base of the wall, and let _y_ be the distance of the top of the ladder from the base of the wall. Since the bottom of the ladder is moving away from the base of the wall at the rate of 3 ft/sec, _dx/dt_ = 3. We have to find _dy/dt_ when _x_ = 7. By the Pythagorean Theorem,

This is the relation between _x_ and _y_. Differentiating both sides with respect to _t_ , we get

Since _dx/dt_ = 3, 6 _x_ \+ 2y _dy/dt_ = 0, whence

This is the desired equation for _dy/dt_. Now, for our particular problem, _x_ = 7. Substituting 7 for _x_ in equation ( _1_ ), we get 49 + _y_ 2 = 625, _y_ 2 = 576, _y_ = 24. In equation ( _2_ ), we replace _x_ and _y_ by 7 and 24, obtaining 21 + 24 _dy/dt_ = 0. Hence,  . Since _dy/dt_ < 0, we conclude that the top of the ladder is sliding _down_ the wall at the rate of   ft/sec when the bottom of the ladder is 7 ft from the base of the wall.

### **Solved Problems**

**1.** Gas is escaping from a spherical balloon at the rate of 2 ft3/min. How fast is the surface area shrinking when the radius is 12 ft?

A sphere of radius r has volume   and surface area S = 4 _πr_ 2. By hypothesis, _dV/dt_ = −2. Now, _dv/dt_ = 4 _πr_ 2 _dr/dt_. So, −2 = 4 _π r_ 2 _dr/dt_ and, therefore, _dr/dt_ = −1/(2 _πr_ 2). Also, _ds/dt_ = 8 _πr dr/dt_. Hence, _ds/dt_ = 8 _π r_ (−1/2 _πr_ 2) = −4/r. So, when _r_ = 12,  . Thus, the surface area is shrinking at the rate of  .

**2.** Water is running out of a conical funnel at the rate of 1 in _3_ /sec. If the radius of the base of the funnel is 4 in and the height is 8 in, find the rate at which the water level is dropping when it is 2 in from the top. (The formula for the volume _V_ of a cone is  , where _r_ is the radius of the base and _h_ is the height.)

Let _r_ be the radius and _h_ the height of the surface of the water at time _t_ , and let _V_ be the volume of the water in the cone. (See Fig. 20-2.) By similar triangles, _r_ /4 = _h_ /8, whence  .

**Fig. 20-2**

Then

By hypothesis, _dV/dt_ = −1. Thus,

Now, when the water level is 2 in from the top, _h_ = 8 − 2 = 6. Hence, at that time, _dh/dt_ = −1/9 _π_ , and so the water level is dropping at the rate of 1/9 _π_ in/sec.

**3.** Sand falling from a chute forms a conical pile whose altitude is always equal to   the radius of the base.( _a_ ) How fast is the volume increasing when the radius of the base is 3 ft and is increasing at the rate of 3 in/min? ( _b_ ) How fast is the radius increasing when it is 6 ft and the volume is increasing at the rate of 24 ft3/min?

Let _r_ be the radius of the base, and _h_ the height of the pile at time _t_. Then

( _a_ ) When _r_ = 3 and  , _dV/dt_ = 3 _π_ ft3/min.

( _b_ ) When _r_ = 6 and _dV/dt_ = 24, _dr/dt_ = 1/2 _π_ ft/min.

4. Ship _A_ is sailing due south at 16 mi/h, and ship _B_ , 32 miles south of _A_ , is sailing due east at 12 mi/h. ( _a_ )At what rate are they approaching or separating at the end of 1 hour? ( _b_ )At the end of 2 hours? ( _c_ )When do they cease to approach each other, and how far apart are they at that time?

Let _A_ 0 and _B_ 0 be the initial positions of the ships, and _A_ _t_ and B _t_ their positions _t_ hours later. Let _D_ be the distance between them _t_ hours later. Then (see Fig. 20-3):

**Fig. 20-3**

Hence,

( _a_ ) When _t_ = 1, _D_ = 20 and  . They are approaching at 5.6 mi/h.

( _b_ ) When _t_ = 2, _D_ = 24 and  . They are separating at 12mi/h.

( _c_ ) They cease to approach each other when  , that is, when  , at which time they are _D_ = 19.2 miles apart.

5. Two parallel sides of a rectangle are being lengthened at the rate of 2 in/sec, while the other two sides are shortened in such a way that the figure remains a rectangle with constant area _A_ = 50 in2. What is the rate of change of the perimeter _P_ when the length of an increasing side is ( _a_ ) 5 in? ( _b_ ) 10 in? ( _c_ ) What are the dimensions when the perimeter ceases to decrease?

Let _x_ be the length of the sides that are being lengthened, and _y_ the length of the other sides, at time _t_. Then

( _a_ ) When _x_ = 5, _y_ = 10 and _dx/dt_ = 2. Then

( _b_ ) When _x_ = 10, _y_ = 5 and _dx/dt_ = 2. Then

( _c_ ) The perimeter will cease to decrease when _dP/dt_ = 0, that is, when _dy/dt_ = _–dx/dt_ = −2. Then _x_ (−2) + _y_ (2) = 0, and the rectangle is a square of side   in.

6. The radius of a sphere is _r_ when the time is _t_ seconds. Find the radius when the rate of change of the surface area and the rate of change of the radius are equal.

The surface area _S_ = 4 _π_ 2; hence, _ds/dt_ = 8 _π_ r _dr/dt_. When _ds/dt = dr/dt_ , 8 _πr_ = 1 and the radius _r_ = 1/8. _π_

7. A weight _W_ is attached to a rope 50 ft long that passes over a pulley at a point _P_ , 20 ft above the ground. The other end of the rope is attached to a truck at a point _A_ , 2 ft above the ground, as shown in Fig. 20-4. If the truck moves away at the rate of 9 ft/s, how fast is the weight rising when it is 6 ft above the ground?

**Fig. 20-4**

Let _x_ denote the distance the weight has been raised, and _y_ the horizontal distance from point _A_ , where the rope is attached to the truck, to the vertical line passing through the pulley. We must find _dx/dt_ when _dy/dt_ = 9 and _x_ = 6.

Now

When   and _dy/dt_ = 9. Then  , from which  .

**8.** A light _L_ hands _H_ ft above a street. An object _h_ ft tall at _O_ , directly under the light, moves in a straight line along the street at _v_ ft/sec. Find a formula for the velocity _V_ of the tip of the shadow cast by the object on the street at _t_ seconds. (See Fig. 20-5.)

**Fig. 20-5**

After _t_ seconds, the object has moved a distance _vt_. Let _y_ be the distance of the tip of the shadow from _O_. By similar triangles, ( _y_ − _vt_ )/ _y_ = _h/H_. Hence,

Thus, the velocity of the tip of the shadow is proportional to the velocity of the object, the factor of proportionality depending upon the ratio _h/H_. As _h_ → 0, _V_ → υ, while as _h_ → _H_ , _V_ → +∞

### **Supplementary Problems**

**9.** A rectangular trough is 8 ft long, 2 ft across the top, and 4 ft deep. If water flow in at a rate of 2 ft3/min, how fast is the surface rising when the water is 1 ft deep?

_Ans_.

**10.** A liquid is flowing into a vertical cylindrical tank of radius 6 ft at the rate of 8 ft3/min. How fast is the surface rising?

_Ans_. 2/9 _π_ ft/min

**11.** A man 5 ft tall walks at a rate of 4 ft/sec directly away from a street light that is 20 ft above the street. ( _a_ ) At what rate is the tip of his shadow moving? ( _b_ ) At what rate is the length of his shadow changing?

_Ans_.

**12.** A balloon is rising vertically over a point _A_ on the ground at the rate of 15 ft/sec. A point _B_ on the ground is level with and 30 ft from _A_. When the balloon is 40 ft from _A_ , at what rate is its distance from _B_ changing?

_Ans_. 12 ft/sec

**13.** A ladder 20 ft long leans against a house. If the foot of the ladder is moving away from the house at the rate of 2 ft/sec, find how fast ( _a_ ) the top of the ladder is moving downward, and ( _b_ ) the slope of the ladder is decreasing, when the foot of the ladder is 12 ft from the house.

_Ans_.

**14.** Water is being withdrawn from a conical reservoir 3 ft in radius and 10 ft deep at 4 ft 3/min. How fast is the surface falling when the depth of the water is 6 ft? How fast is the radius of this surface diminishing?

_Ans_. 100/81 _π_ ft/min; 10/27 _π_ ft/min

**15.** A barge, whose deck is 10 ft below the level of a dock, is being drawn in by means of a cable attached to the deck and passing through a ring on the dock. When the barge is 24 ft away and approaching the dock at  , how fast is the cable being pulled in? (Neglect any sag in the cable.)

_Ans_.

**16.** A boy is flying a kite at a height of 150 ft. If the kite moves horizontally away from the boy at 20 ft/sec, how fast is the string being paid out when the kite is 250 ft from him?

_Ans_. 16 ft/sec

**17.** One train, starting at 11 A.M., travels east at 45 mi/h while another, starting at noon from the same point, travels south at 60 mi/h. How fast are they separating at 3 P.M.?

_Ans_.

**18.** A light is at the top of a pole 80 ft high. A ball is dropped at the same height from a point 20 ft from the light. Assuming that the ball falls according to _s_ = 16 _t_ 2, how fast is the shadow of the ball moving along the ground 1 second later?

_Ans_. 200 ft/sec

**19.** Ship _A_ is 15 miles east of _O_ and moving west at 20 mi/h, ship _B_ is 60 mi south of _O_ and moving north at 15 mi/h. ( _a_ ) Are they approaching or separating after 1 h and at what rate? ( _b_ ) After 3 h? ( _c_ ) When are they nearest one another?

_Ans_.

**20.** Water, at a rate of 10 ft3/min, is pouring into a leaky cistern whose shape is a cone 16 ft deep and 8 ft in diameter at the top. At the time the water is 12 ft deep, the water level is observed to be rising at 4 in/min. How fast is the water leaking away?

_Ans_. (10 − 3 _π_ ) ft3/min

**21.** A solution is passing through a conical filter 24 in deep and 16 in across the top, into a cylindrical vessel of diameter 12 in. At what rate is the level of the solution in the cylinder rising if, when the depth of the solution in the filter is 12 in, its level is falling at the rate 1 in/min?

_Ans_.

**22.** Oil from a leaking oil tanker radiates outward in the form of a circular film on the surface of the water. If the radius of the circle increases at the rate of 3 meters per minute, how fast is the area of the circle increasing when the radius is 200 meters?

_Ans_. 1200 _π_ m2/min

**23.** A point moves on the hyperbola _x_ 2 − 4 _y_ 2 = 36 in such a way that the _x_ -coordinate increases at a constant rate of 20 units per second. How fast is the _y_ coordinate changing at the point (10, 4)?

_Ans_. 50 units/sec

**24.** If a point moves along the curve _y_ = _x_ 2 − 2 _x_ , at what point is the _y_ coordinate changing twice as fast as the _x_ -coordinate?

_Ans_. (2, 0)

## **CHAPTER 21
Differentials. Newton's Method**

If a function _f_ is differentiable at _x_. then   Δ _y_ /Δ _x_ , where Δ _y_ = _f_ ( _x_ \+ Δ _x_ ) − _f_ ( _x_ ). Hence, for values of Δ _x_ close to 0, Δ _y_ /Δ _x_ will be close to _f′_ ( _x_ ). This is often written Δ _y_ /Δ _x_ ~ _f′_ ( _x_ ), whence

This implies

Formula (21.2) can be used to approximate values of a function.

**EXAMPLE 1:** Let us estimate  . Let  , _x_ = 16, and Δ _x_ = 0.2. Then _x_ \+ Δ _x_ = 16.2,  , and  . Since   formula (21.2) becomes

(This approximation turns out to be correct to three decimal places. To four decimal places, the correct value is 4.0249, which can be checked on a graphing calculator.)

**EXAMPLE 2:** Let us estimate sin (0.1). Here, _f_ ( _x_ ) = sin _x_ , _x_ = 0, and Δ _x_ = 0.1. Then _x_ \+ Δ _x_ = 0.1, _f_ ( _x_ \+ Δ _x_ ) = sin (0.1), and _f_ ( _x_ ) = sin 0 = 0. Since _f′_ ( _x_ ) = cos _x_ = cos 0 = 1, formula (21.2) yields

sin (0.1) ~ 0 + 1(0.1) = 0.1

The actual value turns out to be 0.0998, correct to four decimal places. Note that the method used for this problem shows that sin _u_ can be approximated by _u_ for values of _u_ close to 0.

A limitation of formula (21.2) is that we have no information about how good the approximation is. For example, if we want the approximation to be correct to four decimal places, we do not know how small Δ _x_ should be chosen.

**THE DIFFERENTIAL**. The product on the right side of equation (21.1) is called the _differential_ of _f_ and is denoted by _df_.

**Definition** : The _differential df_ of _f_ is defined by

_df_ = _f′_ ( _x_ )Δ _x_

Note that _df_ is a function of two variables, _x_ and Δ _x_. If Δ _x_ is small, then formula (21.1) becomes

This formula is illustrated in Fig. 21-1. Line   is tangent to the graph of _f_ at _P_ ; so its slope is _f′_ ( _x_ ). Hence,   Thus,  . For Δ _x_ small, _Q_ is close to _P_ on the graph and, therefore  , that is, _df_ ~ _f_ ( _x_ \+ Δ _x_ ) – _f_ ( _x_ ), which is formula (21.3).

**Fig. 21-1**

When the function _f_ is given by a formula, say _f_ ( _x_ ) = tan _x_ , then we often will write _df_ as _d_ (tan _x_ ), Thus,

_d_ (tan _x_ ) = _df_ = _f′_ ( _x_ ) Δ _x_ = sec2 _x_ Δ _x_

Similarly, _d_ ( _x_ 3 − 2 _x_ ) = (3 _x_ 2 − 2) Δ _x_. In particular, if _f_ ( _x_ ) = _x_ ,

_dx_ = _df_ = _f′_ ( _x_ )Δ _x_ = (1)Δ _x_ = Δ _x_

Since _dx_ = Δ _x_ , we obtain _df_ = _f′_ ( _x_ ) _dx_. When Δ _x_ ≠ 0, division by Δ _x_ yields _df_ / _dx_ = _f′_ ( _x_ ). When _f_ ( _x_ ) is written as _y_ , then _df_ is written _dy_ and we get the traditional notation _dy_ / _dx_ for the derivative.

If _u_ and _v_ are functions and _c_ is a constant, then the following formulas are easily derivable:

_d_ ( _c_ ) = 0 _d_ ( _cu_ ) = _c du d_ ( _u + υ_ ) = _du_ \+ _dυ_

**NEWTON'S METHOD**. Assume that we know that _x_ 0 is close to a solution of the equation

where _f_ is a differentiable function. Then the tangent line   to the graph of _f_ at the point with _x_ coordinate _x_ 0 will ordinarily intersect the _x_ axis at a point whose _x_ coordinate _x_ 1 is closer to the solution of (21.4) than is _x_ 0. (See Fig. 21-2.)

**Fig. 21-2**

One point-slope equation of the line   is

_y_ − _f_ ( _x_ 0) = _f_ ′( _x_ 0)( _x_ − _x_ 0)

since _f′_ ( _x_ 0) is the slope of  . If   intersects the _x_ axis at ( _x_ 1,0), then

0 − _f_ ( _x_ 0) = _f′_ ( _x_ 0)( _x_ 1 − _x_ 0)

If _f′_ ( _x_ 0) ≠ 0,

Hence,

Now carry out the same reasoning, but beginning with _x_ 1 instead of _x_ 0. The result is a number _x_ 2 that should be closer to the solution of (21.4) than _x_ 1, where _x_ 2 = _x_ 1 − _f_ ( _x_ 1)/ _f′_ ( _x_ 1). If we keep on repeating this procedure, we would obtain a sequence of numbers _x_ 0, _x_ 1, _x_ 2,..., _x_ n,... determined by the formula

This is known as _Newton's method_ for finding better and better approximations to a solution of the equation _f_ ( _x_ ) = 0. However, the method does not always work. (Some examples of the troubles that can arise are shown in Problems 23 and 24.)

**EXAMPLE 3:** We can approximate   by applying Newton's method to the function _f_ ( _x_ ) = _x_ 2 − 3. Here, _f′_ ( _x_ ) = 2 _x_ and (21.5) reads

Let the first approximation _x_ 0 be 1, since we know that  . Successively substituting _n_ = 0, 1, 2,... in (21.6), † we get

Since our calculator yielded _x_ 6 = _x_ 5, we can go no further, and we have obtained the approximation  , which is, in fact, correct to the indicated number of decimal places.

### **Solved Problems**

1. Use formula (21.2) to approximate: ( _a_ )  ; ( _b_ ) sin 61°.

( _a_ ) Let  , and Δ _x_ = − 1. Then _x_ \+ Δ _x_ = 124, _f_ ( _x_ \+ Δ _x_ ) =  , and  . Since

formula (21.2) yields  . (To four decimal places, the correct answer can be shown to be 4.9866.)

( _b_ ) Let _f_ ( _x_ ) = sin _x_ , _x_ = _π_ /3, and Δ _x_ = _π_ /180. Then _x_ \+ Δ _x_ = 61°, _f_ ( _x_ +Δ _x_ ) = sin 61°, and  . Since  , formula (21.2) yields

(To four decimal places, the correct answer can be shown to be 0.8746.)

2. Approximate the change in the volume _V_ of a cube of side _x_ if the side is increased by 1%.

Here, Δ _x_ is 0.01 _x_ , _f_ ( _x_ ) = _V_ = _x_ 3, and _f′_ ( _x_ ) = 3 _x_ 2. By formula (21.1), the increase is approximately (3 _x_ 2)(0.01 _x_ ) = 0.03 _x_ 3. (Thus, the volume increases by roughly 3%.)

3. Find _dy_ for each of the following function _y_ = _f_ ( _x_ ):

( _a_ ) _y_ = _x_ 3 \+ 4 _x_ 2 − 5 _x_ \+ 6.

_dy_ = _d_ ( _x_ 3) + _d_ (4 _x_ 2) − _d_ (5 _x_ ) + _d_ (6) = (3 _x_ 2 \+ 8 _x_ − 5) _dx_

† The computations are so tedious that a calculator, preferably a programmable calculator, should be used.

( _b_ ) _y_ = (2 _x_ 3 \+ 5)3/2.

( _c_ )

( _d_ ) _y_ = cos2 2 _x_ \+ sin 3 _x_.

**4**. Use differentials to find  :

( _a_ ) _xy_ \+ _x_ − 2 _y_ = 5.

_d_ ( _xy_ ) + _dx_ − _d_ (2 _y_ ) = _d_ (5)

_x dy_ \+ _y dx_ \+ _dx_ − 2 _dy_ = 0

( _x_ − 2) _dy_ \+ ( _y_ \+ 1) _dx_ = 0

( _b_ ) .

2 _x_ 2( _y dx_ − _x dy_ ) − 3 _y_ 2( _x dy_ − _y dx_ ) = 0

(2 _x_ 2 y + 3 _y_ 3)dx − (2 _x_ 3 \+ 3 _y_ 2x)dy = 0

( _c_ ) _x_ = 3cos θ − cos3 θ, y = 3 sin θ − sin 3θ

**5.** Approximate the (real) roots of _x_ 3 \+ 2 _x_ − 5 = 0.

Drawing the graphs of _y_ = _x_ 3 and _y_ = 5 − 2 _x_ on the same axes, we see that there must be one root, which lies between 1 and 2. Apply Newton's method, with _x_ 0 = 1. Then _f_ ( _x_ ) = _x_ 3 \+ 2 _x_ − 5 and f′( _x_ ) = 3 _x_ 2 \+ 2. Equation (21.5) becomes

Thus,

A calculator yields the answer 1.328 2689, which is accurate to the indicated number of places. So, the answer obtained by Newton's method is correct to at least seven decimal places.

**6.** Approximate the roots of 2 cos _x_ − _x_ 2 = 0.

Drawing the graphs of _y_ = 2 cos _x_ and _y_ = _x_ 2, we see that there are two real roots, close to 1 and −1. (Since the function 2 cos _x_ − _x_ 2 is even, if _r_ is one root, the other root is – _r_.) Apply Newton's method with _x_ 0 = 1. Then _f_ ( _x_ ) = 2 cos _x_ − _x_ 2 and _f_ ′( _x_ ) = −2sin( _x_ ) − 2 _x_ = −2( _x_ \+ sin _x_ ). Equation (21.5) becomes

Then,

_x_ 1 ~ 1.021 885 93

_x_ 2 ~ 1.021 689 97

_x_ 3 ~ 1.021 689 954

_x_ 4 ~ 1.021 689 954

A graphing calculator produces 1.021 69, which is correct to the indicated number of places. Thus, the answer obtained by Newton's method is accurate to at least five places.

### **Supplementary Problems**

**7.** Use equation (21.2) to approximate: ( _a_ )  ; ( _b_ )  ; ( _c_ ) cos59°; ( _d_ ) tan44°.

_Ans_. ( _a_ ) 2.03125; ( _b_ ) 3.99688; ( _c_ ) 0.5151; ( _d_ ) 0.9651

**8.** Use equation (21.1) to approximate the change in ( _a_ ) _x_ 3 as _x_ changes from 5 to 5.01; ( _b_ )   as _x_ changes from 1 to 0.98.

_Ans_. ( _a_ ) 0.75; ( _b_ ) 0.02

**9.** A circular plate expands under the influence of heat so that its radius increases from 5 to 5.06 inches. Estimate the increase in area.

_Ans_. 0.6 _π_ in2 ~ 1.88 in2

**10.** The radius of a ball of ice shrinks from 10 to 9.8 inches. Estimate the decrease in ( _a_ ) the volume; ( _b_ ) the surface area.

_Ans_. ( _a_ ) 80 _π_ in3; ( _b_ ) 16 _π_ in2

**11.** The velocity attained by an object falling freely a distance _h_ feet from rest is given by  . Estimate the error in _υ_ due to an error of 0.5 ft when _h_ is measured as 100 ft.

_Ans_. 0.2ft/sec

**12.** If an aviator flies around the world at a distance 2 miles above the equator, estimate how many more miles he will travel than a person who travels along the equator.

_Ans_. 12.6 miles

**13.** The radius of a circle is to be measured and its area computed. If the radius can be measured to an accuracy of 0.001 in and the area must be accurate to 0.1 in2, estimate the maximum radius for which this process can be used.

_Ans_. 16 in

**14.** If _pV_ = 20 and _p_ is measured as 5 ± 0.02, estimate _V_.

_Ans_. _V_ = 4   0.016

**15.** If _F_ = 1/ _r_ 2 and _F_ is measured as 4 ± 0.05, estimate _r_.

_Ans_. 0.5  0.003

**16.** Estimate the change in the total surface of a right circular cone when ( _a_ ) the radius _r_ remains constant while the height _h_ changes by a small amount Δ _h_ ; ( _b_ ) the height remains constant while the radius changes by a small amount Δ _r_.

_Ans_. ( _a_ )

**17.** Find _dy_ for each of the following:

**18.** Find _dy_ / _dx_ in the following examples by using differentials:

**19.** (GC) Use Newton's method to find the solutions of the following equations, to four decimal places:

**20.** (GC) Use Newton's method to approximate the following to four decimal places:

**21.** ( _a_ ) Verify that Newton's method for calculating   yields the equation  .

( _b_ ) (GC) Apply part ( _a_ ) to approximate   to four decimal places.

_Ans_. ( _b_ ) 2.2361

**22.** (GC) Show that _x_ 3 \+ _x_ 2 − 3 = 0 has a unique solution in (1,2) use Newton' method to approximate it to four decimal places.

_Ans_. 1.1746

**23.** Show that Newton's method does not work if it is applied to the equation _x_ 1/3 = 0, with _x_ 0 = 1.

**24.** Show that Newton's method does not give approximations to the solutions of the following equations, starting with the given initial values, and explain why it does not work in those cases.

( _a_ ) _x_ 3 − 3 _x_ 2 \+ 3 _x_ \+ 2 = 0, with _x_ 0 = 1.

( _b_ ) _x_ 3 − 3 _x_ 2 \+ _x_ − 1 = 0, with _x_ 0 = 1.

( _c_ )

**25.** (GC) Approximate _π_ by using Newton's method to find a solution of cos _x_ \+ 1 = 0.

_Ans_. 3.141592654. (Note how long it takes for the answer to stabilize.)

**26**. (GC) Use Newton's method to estimate the unique positive solution of cos  .

_Ans_. 1.029866529

## **CHAPTER 22
Antiderivatives**

If F′( _x_ ) = _f_ ( _x_ ), then _F_ is called an _antiderivative_ of _f_.

**EXAMPLE 1:** _x_ 3 is an antiderivative of 3 _x_ 2, since D _x_ ( _x_ 3) =3 _x_ 2. But _x_ 3 \+ 5 is also an antiderivative of 3 _x_ 2, since D _x_ (5) = 0.

(I) In general, if _F_ ( _x_ ) is an antiderivative of _f_ ( _x_ ), then _F_ ( _x_ )+ _C_ is also an antiderivative of _f_ ( _x_ ), where _C_ is any constant.

(II) On the other hand, if _F_ ( _x_ ) is an antiderivative of _f_ ( _x_ ), and if _G_ ( _x_ ) is any other antiderivative of _f_ ( _x_ ), then _G_ ( _x_ ) = _F_ ( _x_ ) + _C_ , for some constant _C_.

Property (II) follows from Problem 13 of Chapter 18, since _F_ ′( _x_ ) = _f_ ( _x_ ) = _G_ ′( _x_ ).

From Properties (I) and (II) we see that, if _F_ ( _x_ ) is an antiderivative of _f_ ( _x_ ), then the antiderivatives of _f_ ( _x_ ) are precisely those functions of the form _F_ ( _x_ ) + _C_ , for an arbitrary constant _C_.

**Notation:**   _dx_ will denote any antiderivative of _f_ ( _x_ ). In this notation, _f_ ( _x_ ) is called the _integrand_.

**Terminology:** An antiderivative ∫ _f_ ( _x_ ) _dx_ is also called an _indefinite integral_.

An explanation of the peculiar notation ∫ _f_ ( _x_ ) _dx_ (including the presence of the differential _dx_ ) will be given later.

**EXAMPLE 2:** ( _a_ )   (b) .

**LAWS FOR ANTIDERIVATIVES**

**Law 1.**  .

**Law 2.**  .

**Law 3.**  .

**Law 4.**   for any rational number _r_ ≠ −1.

(4) follows from the fact that   for _r_ ≠ −1.

**Law 5.**

 .

**Law 6.**

**Law 7.**

**EXAMPLE 3:**

( _a_ )  .

( _b_ )  .

( _c_ )  .

( _d_ )  .

( _e_ )

**EXAMPLE 4:** Laws (3)–(7) enable us to compute the antiderivative of any polynomial. For instance,

**Law (8). (Quick Formula I)**

  for any rational number _r_ ≠ −1

For verification,   by the power chain rule.

**EXAMPLE 5:**  .

To see this, let   and _r_ = 5 in Quick Formula I.

**EXAMPLE 6:**  .

In this case, we had to insert a factor of 2 in the integrand in order to use Quick Formula I.

**Law (9). Substitution Method**

where _u_ is replaced by g( _x_ ) after the right-hand side is evaluated. The "substitution" is carried out on the left-hand side by letting _u_ = g( _x_ ) and _du_ = g′( _x_ ) _dx_. (For justification, see Problem 21.)

**EXAMPLE 7**

( _a_ ) Find  .

Let _u_ = _x_ 2. Then _du_ = 2 _x dx_ So, _x dx_ =   _du_. By substitution,

( _b_ ) Find

Let _u_ = _x_ /2. Then  . So, _dx_ = 2 _du_. By substitution,

Observe that Quick Formula I is just a special case of the Substitution Method, with _u_ = g( _x_ ).

The advantage of Quick Formula I is that we save the bother of carrying out the substitution.

The known formulas for derivatives of trigonometric and inverse trigonometric functions yield the following formulas for antiderivatives:

### **Solved Problems**

In Problems 1−8, evaulate the antiderivative.

**1.**

**2.**

**3.**

**4.**

**5.**

**6.**

**7.**

Note that it would have been easier to use Quick Formula I:

**8.**

Use Quick Formula I in Problems 9-15.

**9.**

**10.**

**11.**

**12.**

**13.**

**14.**

**15.**

In Problems 16-18, use the Substitution Method.

**16.**

**17.**
 .

Let _u_ = 4 _x_ 2 − 5. Then _du_ = 8 _x dx_ , and  . Thus,

**18.**
 .

Let _u_ = _x_ \+ 1. The _du_ = _dx_ and _x_ = _u_ −1. Thus,

**19.** A stone is thrown straight up from the ground with an initial velocity of 64 ft/sec. ( _a_ ) When does it reach its maximum height? ( _b_ ) What is its maximum height? ( _c_ ) When does it hit the group? ( _d_ ) What is its velocity when it its hits the group?

In free-fall problems,

Letting _t_ = 0, we see that _C_ 1 = _v_ 0, the initial velocity at _t_ = 0. Thus, _v_ = 32 _t_ \+ _v_ 0. Hence,

Letting _t_ = 0, we see that _C_ 2 = _v_ 0, the initial position at _t_ = 0. Hence

_s_ = − 16 _t_ 2 \+ _v_ 0 _t_ \+ _s_ 0

In this problem, _s_ 0 = 0 and _v_ 0 = 64. So,

_v_ = −32 _t_ \+ 64, _s_ = −16 _t_ 2 \+ 64 _t_

( _a_ ) At the maximum height  . So, −32 _t_ \+ 64 = 0 and, therefore, _t_ = 2 seconds.

( _b_ ) What _t_ = 2, _s_ = −16(2)2 \+ 64(2) = 64 ft, the maximum height.

( _c_ ) When the stone hits the ground,0 = _s_ =−16 _t_ 2 \+ 64 _t_. Dividing by _t_ , 0 = −16 _t_ \+ 64 and therefore, _t_ = 4.

( _d_ ) When _t_ = 4, _v_ = −32(4) + 64 = −64 ft/sec.

**20.** Find an equation of the curve passing through the point (3, 2) and having slope 5 _x_ 2 – _x_ \+ 1 at every point ( _x, y_ ).

Since the slope is the derivative, _dy/dx_ = 5 _x_ 2 − _x_ \+ 1. Hence,

Since (3, 2) on the curve,  . Hence, an equation of the curve is

**21.** Justify the Substitution Method:  .

Here, _u_ = g( _x_ ) and _du/dx_ = _g_ ´( _x_ ). By the chain rule,

### **Supplementary Problems**

In Problem 22-44, evaluate the given antiderivative.

**26.**   ( _Hint:_ Multiply numerator and denominator by 1 − cos _x_.)

_Ans._ − cot _x_ \+ csc _x_ \+ _C_

**30.**   ( _Hint:_ Factor 16 out of the radical.)

_Ans._

**31.**  ( _Hint:_ Either factor 4 out of the denominator or make the substitution _u_ = 2 _x_ ).

_Ans._

**32.**  ( _Hint:_ Either factor 4 out of the radical or make the substitution _u_ = 2 _x_ ).

_Ans._

In Problems 45-52, use Quick Formula I.

In Problems 53-64, use any method.

**65.** A stone is thrown straight up from a building ledge that is 120 ft above the ground, with an initial velocity of 96 ft/sec. ( _a_ ) When will it reach its maximum height? ( _b_ ) What will its maximum height be? ( _c_ )When will it hit the ground? ( _d_ ) With what speed will it hit the ground?

_Ans._ ( _a_ ) _t_ = 3 sec; ( _b_ ) 264 ft; ( _c_ ) ; ( _d_ ) ~129.98 ft/sec

**66.** An object moves on the _x_ axis with acceleration _a_ = 3 _t_ − 2 ft/sec2. At time _t_ = 0, it is at the origin and moving with a speed of 5 ft/sec in the negative direction. ( _a_ ) Find a formula for its velocity _v_. ( _b_ ) Find a formula for its position _x_. ( _c_ ) When and where does it change direction? ( _d_ ) At what times is it moving toward the right?

_Ans._ ( _a_ )

**67.** A rocket shot straight up from the ground hits the ground 8 seconds later. ( _a_ ) What was its initial velocity? ( _b_ ) What was its maximum height?

_Ans._ ( _a_ ) 128 ft/sec; ( _b_ ) 256 ft

**68.** A driver applies the brakes on a car going at 55 miles per hour on a straight road. The brakes cause a constant deceleration of 11 ft/sec2. ( _a_ ) How soon will the car stop? ( _b_ ) How far does the car move after the brakes were applied?

_Ans._ ( _a_ ) 5 sec; ( _b_ )137.5 ft

**69.** Find the equation of a curve going through the point (3,7) and having slope 4 _x_ 2 − 3 at ( _x_ , _y_ ).

_Ans._

## **CHAPTER 23
The Definite Integral. Area under a Curve**

**SIGMA NOTATION.** The Greek capital letter   denotes repeated addition.

**EXAMPLE 1**

( _a_ )  .

( _b_ )  .

( _c_ )

( _d_ )

In general, if _f_ is a function defined on the integers, and if _n_ and _k_ are integers such that _n_   _k,_ then:

**AREA UNDER A CURVE.** Assume that _f_ is a function such that _f_ ( _x_ )   0 for all _x_ in a closed interval  _a_ , _b_ ]. Its graph is a curve that lies on or above the _x_ axis. (See [Fig. 23-1.) We have an intuitive idea of the _area A_ of the region   under the graph, above the _x_ axis, and between the vertical lines _x_ = _a_ and _x_ = _b_. We shall specify a method for evaluating _A_.

Choose points _x_ 1, _x_ 2,..., _x_ _n_ −1 between _a_ and _b_. Let _x_ 0 = a and _x_ _n_ = _b_. Thus (see Fig. 23-2),

_a_ = _x_ 0 < _x_ 1 < _x_ 2 < ··· < _x_ n-1 < _x_ n = _b_

**Fig. 23-1**

**Fig. 23-2**

The interval [ _a_ , _b_ ] is divided into _n_ subintervals [ _x_ 0, _x_ 1], [ _x_ 1, _x_ 2],...,[ _x_ n−1, _x_ _n_ ]. Denote the lenghs of these subintervals by Δ1 _x_ , Δ2 _x_ ,..., Δ _n_ _x_. Hence, if 1   _k_   _n_ ,

Δ _k_ x = _x k_ − _x k_−1

Draw vertical line segments _x_ = _x k_ from the _x_ axis up to the graph. This divides the region   into _n_ stips. Letting Δ _k_ _A_ denote the area of the _k_ th strip, we obtain

We can approximate the area Δ _k_ _A_ in the following manner. Slect any point   in the _k_ th subinterval  _k_ _k_ −1, _x k_]. Draw the vertical line segment from the point   on the _x_ axis up to the graph (see the dashed lines in [Fig, 23-3); the length of this segment is _f_ ( ). The rectangle with base Δ _k_ _x_

**Fig. 23-3**

and height _f_ ( ) has area _f_ ( ) Δ _k_ _x_ , which is approximately the area Δ _k_ _A_ of the _k_ th strip. Hence, the total area _A_ under the curve is approximately the sum

The approximation becomes better and better as we divide the interval [ _a_ , _b_ ] into more and more subintervals and as we make the lengths of these subintervals smaller and smaller. If successive approximations can be made as close as one wishes wishes to a specific number, then that number will be denoted by

and will be called the _definite integral_ of _f_ from _a_ to _b_. Such a number does not exist in all cases, but it does exist, for example, when the function _f_ is continuous on _[a, b]._ When   exists, its value is equal to the area _A_ under the curve.†

In the notation  , _b_ is called the _upper limit_ and _a_ is called the _lower limit_ of the definite integral.

For any (not necessarily nonnegative) function _f_ on _[a, b],_ sums of the form (23.1) can be defined, without using the notion of area. If there is a number to which these sums can be made as close as we wish, as _n_ gets larger and larger and as the maximum of the lengths Δ _k_ _x_ approaches 0, then that number is denoted   and is called the _definite integral_ of _f_ on [ _a, b_ ]. When   exists, we say that _f_ is _integrable_ on [ _a, b_ ].

We shall assume without proof that   exists for every function _f_ that is continuous on [ _a, b_ ]. To evaluate   it suffices to find the limit of a sequence of sums (23.1) for which

† The definite integral is also called the _Riemann integral_ of _f_ on [ _a_ , _b_ ], and the sum (23.1) is called a _Riemann sum_ for _f_ on [ _a_ , _b_ ].

the number _n_ of subintervals approaches infinity and the maximum lengths of the subintervals approaches 0.

**EXAMPLE 2:** Let us show that

Let _a_ = _x_ 0 < _x_ 1 < _x_ 2< ··· < _x_ n−1 < _x n_ = _b_ be a subdivision of [ _a_ , _b_ ]. Then a corresponding sum (23.1) is

 .

Since every approximating sum is

An alternative argument would use the fact that the region under the graph of the constant function 1 and above the _x_ axis, between _x_ = _a_ and _x_ = _b_ , is a rectangle with base _b_ − _a_ and height 1 (see Fig. 23-4). So,  , being the area of that rectangle, is _b_ − _a_.

**Fig. 23-4**

**EXAMPLE 3:** Let us calculate  .

Let _a_ = _x_ 0 < _x_ 1 < _x_ 2 < ··· < _x_ _n_ −1 < _x_ _n_ = _b_ be a subdivision of [ _a_ , _b_ ] into _n_ equal subintervals. Thus, each Δ _k_ _x_ = ( _b_ − _a_ )/ _n_. Denote ( _b_ − _a_ )/ _n_ by Δ _x_. Then _x_ 1 = a + Δ _x_ , _x_ 2 = _a_ \+ 2 Δ _x_ , and, in general, _x_ _k_ = _a_ \+ _k_ Δ _x_. In the _k_ th subinterval, [ _x_ _k_ −1, _x_ _k_ ], choose   to be the right-hand endpoint _x_ _k_. Then an approximating sum (23.1) has the form

Here we have used the fact that  . (See Problem 5.)

Now, as _n_ → ∞, ( _n_ \+ 1)/ _n_ = 1 + 1/ _n_ → 1 + 0 = 1. Hence, the limit of our approximating sums is

Thus,  .

In the next chapter, we will find a method for calculating   _dx_ that will avoid the kind of tedious computation used in this example.

**PROPERTIES OF THE DEFINITE INTEGRAL**

This follows from the approximating sum   _dx_ is equal to _c_ times the approximating sum   _dx,_ and that the same relation holds for the corresponding limits.

This is the special case of (23.3) when _c_ = −1.

This follows from the fact that an approximating sum   is equal to the sum   of approximating sums for   and  .

This is obvious when _f_ ( _x_ )   0 and we interpret the integrals as areas. The general result follows from looking at the corresponding approximating sums, although the case where one of the subintervals of [ _a_ , _b_ ] contains c requires some extra thought.

We have defined   _dx_ only when _a_ < _b_. We can extend the definition to all possible cases as follows:

(i)

(ii)

In particular, we always have:

It can readily be verified that the laws (23.2)–(23.6), the equation in (23.7), and the result of Example 3 all remain valid for arbitrary upper and lower limits in the integrals.

### **Solved Problems**

**1.** Assume _f_ ( _x_ )   0 for all _x_ in  _a, b_ ]. Let _A_ be the area between the graph of _f_ and the _x_ axis, from _x_ = _a_ to _x_ = _b_. (See [Fig. 23-5.) Show that  .

**Fig. 23-5**

Let _B_ be the area between the graph of − _f_ and the _x_ axis, from _x_ = _a_ to _x_ = _b_. By symmetry, _B_ = _A_. But,   by (23.4).

Since

**2.** Consider a function _f_ that, between _a_ and _b_ , assumes both positive and negative values. For example, let its graph be as in Fig. 23-6. Then   _dx_ is the difference between the sum of the areas above the _x_ axis and below the graph and the sum of the areas below the _x_ axis and above the graph. In the case of the graph shown in Fig. 23-6,

**Fig. 23-6**

To see this, apply (23.7) and Problem 1:

**3.** Assume that _f_ and _g_ are integrable on [ _a_ , _b_ ], Prove:

( _a_ ) If _f_ ( _x_ )   0 on [ _a_ , _b_ ], then  .

( _b_ ) If _f_ ( _x_ )   _g_ ( _x_ ) on [ _a_ , _b_ ], then  .

( _c_ ) If _m_   _f_ ( _x_ )   _M_ for all _x_ in [ _a_ , _b_ ], then  .

( _a_ ) Since every approximating sum  , it follows that

( _b_ ) _g_ ( _x_ − _f_ ( _x_ )   0 on [ _a_ , _b_ ]. So, by ( _a_ ),  . By (23.6),  . Hence,

( _c_ ) By ( _b_ ),  . But, by (23.2) and (23.3),   and  . Hence,

**4.** Evaluate  .

This is the area under the parabola _y_ = _x_ 2 from _x_ = 0 to _x_ = 1. Divide [0, 1] into _n_ equal subintervals. Thus, each Δ _k_ _x_ = 1/ _n_. In the _k_ the subinterval  , let   be the right endpoint _k/n_. Thus, the approximating sum (23.1) is

 ,

Now,   (see problem 12).

Hence,

So, the approximating sums approach   as _n_ → ∞. Therefore,  . In the next chapter, we will derive a simpler method for obtaining the same result.

**5.** Prove the formula   used in Example 3.

Reversing the order of the summands in

we get

Adding the two equations yields

since the sum in each column is _n_ \+ 1. Hence, dividing by 2, we get

### **Supplementary Problems**

**6.** Calculate: ( _a_ )  .

_Ans._ ( _a_ ) 3(4 − 1) = 9; ( _b_ )  ; ( _c_ )

**7.** Find the area under the parabola _y_ = _x_ 2 − 2 _x_ \+ 2, above the _x_ axis, and between _x_ = 0 and _x_ = 1.

_Ans._

**8.** Evaluate  .

_Ans._

**9.** For the function _f_ graphed in Fig. 23-7, express   in terms of the areas _A_ 1, _A_ 2, and _A_ 3.

_Ans. A_ 1 − _A_ 2 \+ _A_ 3

**10.** Show that  . [ _Hint_ : Problem 3 _(c)_.]

**11.** Evaluate   _dx_. ( _Hint_ : Find the corresponding area by geometric reasoning.)

_Ans._ π/4

**Fig. 23-7**

**12.** Use mathematical induction to prove the formula   of Problem 4. (Verify it when _n_ = 1, and then show that, if it holds for _n,_ then it holds for _n_ \+ 1.)

**13.** Evaluate  .

_Ans._ ( _a_ )  ; ( _b_ ) 15; ( _c_ ) 20200; ( _d_ ) 4218

**14.** Let the graph of f between _x_ = 1 and _x_ = 6 be as in Fig. 23-8. Evaluate  .

_Ans._

**Fig. 23-8**

**15.** If _f_ is continuous on [ _a_ , _b_ ], _f_ ( _x_ )   0 on [ _a_ , _b_ ], and _f_ ( _x_ 0) > 0 for some _x_ 0 in [ _a_ , _b_ ], prove that  . [ _Hint:_ By the continuity of _f_ ,  for all _x_ in some subinterval [ _c_ , _d_ ]. Use (23.7) and Problem 3( _a_ , _c_ ).]

## **CHAPTER 24
The Fundamental Theorem of Calculus**

**MEAN-VALUE THEOREM FOR INTEGRALS.** Let _f_ be continuous on [ _a_ , _b_ ]. Then there exists _c_ in [ _a_ , _b_ ] such that

To see this, let _m_ and _M_ be the minimum and maximum values of _f_ in  _a_ , _b_ ], and apply Problem 3( _c_ ) of [Chapter 23 to obtain

  and, therefore,

So, by the intermediate value theorem,   for some _c_ in [ _a_ , _b_ ].

**AVERAGE VALUE OF A FUNCTION ON A CLOSED INTERVAL.** Let _f_ be defined on [ _a_ , _b_ ]. Since _f_ may assume infinitely many value on [ _a_ , _b_ ], we cannot talk about the avarage of the values of _f_. Instead, divede [ _a_ , _b_ ] into _n_ equal subintervals, each of length  . Select an arbitrary point   in the _k_ th subinterval. Then the average of the _n_ values   is

When _n_ is large, this value is intuitively a good estimate of the "average value of _f_ on [ _a_ , _b_ ]".

However, since  ,

As _n_ → ∞, the sum on the right approaches  . This suggests the following definition.

**Definition:** The _average value_ of _f_ on [ _a_ , _b_ ], is  .

Let _f_ be continuous on [ _a_ , _b_ ]. If _x_ is in [ _a_ , _b_ ], then   is a function of _x_ , and:

For a proof, see Problem 4.

**FUNDAMENTAL THEOREM OF CALCULUS.** Let _f_ be continuous on [ _a_ , _b_ ], and let  , that is _F_ is an antiderivative of _f_. Then

To see this, note that, by (24.2),   _dt_ and _F_ ( _x_ ) have the same derivative, _f_ ( _x_ ). Hence, by Problem 18 of Chapter 13, there is a constant _K_ such that  . When _x_ = _a_ , we get

  So, _K_ = − _F_ ( _a_ )

Hence,  . When _x_ = _b_ , this yields

Equation (24.3) provides a simple way of computing   when we can find an antiderivative _F_ of _f_. The expression _F_ ( _b_ ) − _F_ ( _a_ ) on the right side of (24.3) is often abbreviated as  . Then the fundamental theorem of calculus can be written as follows:

**EXAMPLE 1**

(i) The complicated evaluation of   in Example 3 of Chapter 23 can be replaced by the following simple one:

(ii) The very tedious computation of   in Problem 4 of Chapter 23 can be replaced by

(iii) In general,

**CHANGE OF VARIABLE IN A DEFINITE INTEGRAL.** In the computation of a definite integral by the fundamental theorem, an antiderivative   is required. In Chapter 22, we saw that substitution of a new variable _u_ is sometimes useful in finding  . When the substitution also is made in the definite integral, the limits of integration must be replaced by the corresponding values of _u_.

**EXAMPLE 2:** Evaluate  .

Let _u_ = 5 _x_ \+ 4. Then _du_ = 5 _dx_. When _x_ = 1, _u_ = 9, and When _x_ = 9, _u_ = 49. Hence,

For justification of this method, see Problem 5.

### **Solved Problems**

**1.** Evaluate  .

  by Quick Formula I. Hence, by the fundamental theorem,

**2.** Find the area under the graph of  , above the _x_ axis, and between 0 and 1.

The area is  .

**3.** Find the average value of _f_ ( _x_ ) = 4 − _x_ 2 on [0,2].

The average value is

**4.** Prove formula (24.2):

Let  . Then:

Thus,   and, therefore,

But, as Δ _x_ → 0, _x_ \+ Δ _x_ → _x_ and so, _x_ * → _x_ (since _x_ * is between _x_ and _x_ \+ Δ _x_ ) Since _f_ is continuous,  .

**5.** Justify a change of variable in a definite integral in the following precise sense. Given  , let _x_ = _g_ ( _u_ ) where, as _x_ varies from _a_ to _b_ , _u_ increases or decreases from _c_ to _d_. (See Fig. 24-1 for the case where _u_ is increasing.) Show that

(The right side is obtained by substituting _g_ ( _u_ ) for _x, g_ ′( _u_ ) _du_ for _dx_ , and changing the limits of integration from _a_ and _b_ to _c_ and _d_.)

**Fig. 24-1**

Let  , that is, _F_ ′( _x_ ) = _f_ ( _x_ ). By the chain rule,

_D u_( _F_ ( _g_ ( _u_ )) = _F_ ′( _g_ ( _u_ )) · _g_ ′( _u_ ) = _f_ ( _g_ ( _u_ )) _g_ ′( _u_ ) Thus,

So, by the fundamental theorem,

**6.** ( _a_ ) If _f_ is an even function, show that, for

( _b_ ) If _f_ is an odd function, show that, for

Let _u_ = − _x._ Then _du_ = − _dx_ , and

Rewriting _u_ as _x_ in the last integral, we have:

Thus,

( _a_ ) If _f_ is even, _f_ (− _x_ ) + _f_ ( _x_ ) = 2 _f_ ( _x_ ), whence

( _b_ ) If _f_ is odd, _f_ (− _x_ ) + _f_ ( _x_ ) = 0, whence

**7. Trapezoidal Rule**

( _a_ ) Let _f_ ( _x_ )   0 on  _a_ , _b_ ]. Divide [ _a_ , _b_ ] into _n_ equal parts, each of length  , by means of points, _x_ 1, _x_ 2, ..., _x_ _n_ −1. (See [Fig. 24-2( _a_ ).) Prove the following _trapezoidal rule:_

( _b_ ) Use the trapezoidal rule with _n_ = 10 to approximate

( _a_ ) The area of the strip over  _x_ _k_ −1, _x k_] is approximately the area of trapezoid _ABCD_ in [Fig. 24-2( _b_ ),   (Remember that _x_ 0 = _a_ and _x n_ = _b_.) So, the area under the curve is approximated by the sum of the trapezoidal areas,

**Fig. 24-2**

( _b_ ) With _n_ = 10, _a_ = 0, _b_ = 1,   and _x_ k = _k_ /10, we get

The exact value is   (by Example 1 ( _ii_ ) above).

### **Supplementary Problems**

In Problems 8-22, use the Fundamental Theorem of Calculus to evaluate the definite integral.

† Recall that the area of a trapezoid of height _h_ and bases _b_ 1 and _b_ 2 is  .

**33.** Compute the average value of the following functions on the indicated intervals.

**34.** Use the change-of-variable method to find  .

_Ans._

**35.** An object moves along the _x_ axis for a period of time _T_. **If** its initial position is _x_ 1 and its final position is _x_ 2, show that its average velocity was  .

**36.** Let  . Evaluate  .

_Ans._

**37.** Evalaute  .

_Ans._

**38. (Midpoint Rule)** In an approximating sum (23.1)   if we select   to be the midpoint of the _k_ the subinterval, then the sum is said to be obtained by the _midpoint rule._ Apply the midpoint rule to approximate  , using a division into five equal subintervals, and compare with the exact result of  .

_Ans._ 0.33

**39. (Simpson's Rule)** If we divide [ _a, b_ ] into _n_ equal subintervals, where _n_ is even, the following approximating sum for

is said to be obtained by _Simpson's Rule._ Except for the first and last terms, the coefficients consist of alternating 4s and 2s. (The basic idea is to use parabolas as approximating arcs instead of line segments as in the trapezoidal rule. Simpson's rule is usually much more accurate than the midpoint or trapezoidal rule.)

Apply Simpson's rule to approximate ( _a_ )   and ( _b_ )  , with _n_ = 4, and compare the results with the answers obtained by the fundamental theorem.

_Ans._ ( _a_ )  , which is the exact answer;( _b_ )   as compared to 2

**40.** Consider   ( _a_ ) Show that the fundamental theorem yields the answer   ( _b_ ) ( **GC** ) With _n_ = 10, approximate (to four decimal places) the integral by the trapezoidal, midpoint, and Simpson's rules.

_Ans._ trapezoidal 0.2525; midpoint 0.2488; Simpson's 0.2500.

**41.** Evaluate:

( _a_ )

( _b_ )

_Ans._ ( _a_ )  ; ( _b_ )

**42.** ( _a_ ) Use a substitution to evaluate   (to eight decimal places).

( _b_ ) ( **GC** ) Use a graphing calculator to estimate the integral of ( _a_ ).

_Ans._ ( _a_ ) ; ( _b_ ) 0.39052429

**43.** ( **GC** ) Estimate   _x_ sin3 (tan _x_ ) _dx_ (to four decimal places.)

_Ans._ 0.0262

**44.** ( **GC** ) Consider   _dx_. Estimate (to six decimal places) its value using the trapezoidal and Simpson's rule (both with _n_ = 4), and compare with the value given by a graphing calculator.

_Ans._ trapezoidal 3.599496; Simpson's 3.571557; graphing calculator 3.571639

## **CHAPTER 25
The Natural Logarithm**

The traditional way of defining a logarithm, log _a_ _b_ , is to define it as that number _u_ such that a _u_ = _b_. For example, log10 100 = 2 because 102 = 100. However, this definition has a theoretical gap. The flaw is that we have not yet defined _a u_ when _u_ is an irrational number, for example   or π. This gap can be filled in, but that would require an extensive and sophisticated detour.† Instead, we take a different approach that will eventually provide logically unassailable definitions of the logarithmic and exponential functions. A temporary disadvantage is that the motivation for our initial definition will not be obvious.

**THE NATURAL LOGARITHM.** We are already familiar with the formula

The problem remains of finding out what happens when r = −1, that is, of finding the antiderivative of _x_ -1.

The graph of _y_ = 1 / _t_ , for _t_ > 0, is shown in Fig. 25-1. It is one branch of a hyperbola. For _x_ > 1, the definite integral

is the value of the area under the curve _y_ = 1 / _t_ and above the _t_ axis, between _t_ = 1 and _t_ = _x_.

**Definition:** In   for _x_ > 0

The function In _x_ is called the _natural logarithm_. The reasons for referring to it as a logarithm will be made clear later. By (24.2),

**(25.1)**   for _x_ > 0

† Some calculus textbooks just ignore the difficulty. They assume that _a u_ is defined when _a_ > 0 and _u_ is _any_ real number and that the usual laws for exponents are valid.

**Fig. 25-1**

Hence, the natural logarithm is the antiderivative of _x_ −1, but only on the interval (0,+∞). An antiderivative for all _x_ ≠ 0 will be constructed below in (25.5).

**PROPERTIES OF THE NATURAL LOGARITHM**

**(25.2)** In 1 = 0, since In

**(25.3)** If _x_ > 1, then In _x_ > 0

This is true by virtue of the fact that   represents an area, or by Problem 15 of Chapter 23.

**(25.4)** If 0 < _x_ < 1, then In _x_ < 0

In   by (23.8). Now, for 0 < _x_ < 1, if _x_   _t_   1, then 1/ _t_ > 0 and, therefore, by Problem 15 of Chapter 23,  .

**(25.5)** ( _a_ )

( _b_ )  for _x_ ≠0

The argument is simple. For _x_ > 0, | _x_ | = _x_ and so, D _x_ (In | _x_ |) = D _x_ (In _x_ ) = 1/ _x_ by (25.1). For _x_ < 0, | _x_ | = − _x_ and so,

EXAMPLE 1:

**(25.6)** In _uv_ = In _u_ \+ In _v_

Note that

Hence, In ( _ax_ ) = In _x_ \+ _K_ for some constant _K_ (by Problem 18 of Chapter 13). When _x_ = 1, In _a_ = In 1 + _K_ = 0 + _K_ = _K_. Thus, In ( _ax_ ) = In _x_ \+ In _a_. Replacing _a_ and _x_ by _u_ and _v_ yields (25.6)

**(25.7)**

In (25.6), replace _u_ by  .

**(25.8)**

In (25.7), replace _u_ by 1 and use (25.2).

**(25.9)** In ( _x_ r) = _r_ In _x_ for any rational number _r_ and _x_ > 0

By the chain rule,  . So, by Problem 18 of Chapter 13, In ( _x r_) = _r_ In _x_ \+ _K_ for some constant _K_. When _x_ = 1, In 1 = _r_ In 1 + _K_. Since In 1 = 0, _K_ = 0, yielding (25.9).

**EXAMPLE 2:** In  .

**(25.10)** In _x_ is an increasing function.

  since _x_ > 0. Now use Theorem 13.7.

**(25.11)** In _u_ = In _v_ implies _u = v_.

This is a direct consequence of (25.10). For, if _u ≠ v_ , then either _u_ < _v_ or _v_ < _u_ and, therefore, either In _u_ < In _v_ or In _v_ < In _u_.

**(25.12)**   < In 2 < 1

**Fig. 25-2**

The area under the graph of _y_ = 1/ _t_ , between _t_ = 1 and _t_ = 2, and above the _t_ axis, is greater than the area   of the rectangle with base 1, 2] and height  . (See [Fig. 25-2.) It is also less than the area 1 of the rectangle with base 1, 2] and height 1. (A more rigorous argument would use Problems 3( _c_ ) and 15 of [Chapter 23.)

**(25.13)**

Let _k_ be any positive integer. Then, for _x_ > 22k,

In _x_ > In (22k) = 2 _k_ In 2 > 2 _k_ ( ) = _k_

by (25.10) and (25.9). Thus, as _x_ → +∞, In _x_ eventually exceeds every positive integer.

**(25.14)**

Let _u_ = 1/ _x_. As _x_ → 0+, _u_ → +∞. Hence,

**(25.15)** Quick Formula II:

By the chain rule and (25.5) ( _a_ ), D _x_ (In| _g_ ( _x_ )|) =  .

**EXAMPLE 3**

( _a_ )

The absolute value sign was dropped because _x_ 2 \+ 1   0. In the future, we sghall do this without explicit mention.

( _b_ )

### **Solved Problems**

**1.** Evaluate:

**(25.16)**

**(25.17)**

**(25.18)**

**2.** ( **GC** ) Estimate the value of ln 2.

A graphing calculator yields the value ln 2 ~ 0.6931471806. Later we shall find another method for calculating ln 2.

**3.** ( **GC** ) Sketch the graph of y = ln _x_.

A graphing calculator yields the graph shown in Fig. 25-3. Note by (25.10) that ln _x_ is increasing. By (25.13), the graph increases without bound on the right, and, by (25.14), the negative _y_ axis is a vertical asymptote. Since

the graph is concave downward. By (25.13) and (25.14), and the intermediate value theorem, the range of ln _x_ is the set of all real numbers.

**Fig. 25-3**

**4.** Find: ( _a_ ) _D x_ (ln(x4 \+ 7x)); ( _b_ ) _D x_(ln (cos 2x)); ( _c_ ) _D x_(cos (ln 2 _x_ )).

**5.** Find the following antiderivatives. Use Quick Formula II when possible.

( _d_ ) Complete the square in the denominator:  .

Let _u_ = _x_ − 2, _du_ = _dx_.

**6. Logarithm Differentiation.** Find the derivative of  .

First take the natural logarithms of the absolute values of both sides:

Now take the derivatives of both sides:

**7.** Show that   for _x_ > 0. (When _x_ ≠ 1, the strict inequalities hold.)

When _x_ > 1, 1/ _t_ is a decreasing function on 1, _x_ ] and so, its minimum on [1, _x_ ] is 1/ _x_ and its maximum is 1. So, by problems 3( _c_ ) and 15 of [Chapter 23,

For   is increasing on x, 1]. Then, by Problems 3(c) and 15 of [Chapter 23,

Hence,  . When _x_ = 1, the three terms are all equal to 0.

### **Supplementary Problems**

**8.** Find the derivatives of the following functions.

( _a_ ) _y_ = ln (x + 3)2 = 2 ln (x + 3).

_Ans_.

( _b_ ) _y_ = (ln (x + 3))2

_Ans_.

( _c_ ) _y_ = In [(x3 \+ 2)(x2 \+ 3)] = ln (x3 \+ 2) + ln(x2 \+ 3)

_Ans_.

( _d_ )

_Ans_.

( _e_ ) _y_ = ln sin 5x

_Ans_.

( _f_ )

_Ans_.

( _g_ )

_Ans_.

( _h_ ) _y_ = _x_ ln _x_ − _x_

_Ans_. _y_ ′ = ln _x_

( _i_ ) _y_ = ln (ln (tanx))

_Ans_.

**9.** Find the following antiderivatives. Use Quick Formula II when possible.

( _a_ )

_Ans_.

( _b_ )

_Ans_.

( _c_ )

_Ans_. Use Quick Formula I.

( _d_ )

_Ans_. ln|ln _x_ | + _c_

( _e_ )

_Ans_.

( _f_ )

_Ans_. _x_ 2-ln|x| + c

( _g_ )

_Ans_.

( _h_ )

_Ans_.

**10.** Use logarithmic differentiation to calculate y′.

( _a_ )

_Ans_.

( _b_ )

_Ans_.

( _c_ )

_Ans_.

( _d_ )

_Ans_.

**11.** Express in terms of ln 2 and ln 3: ( _a_ ) ln (37); ( _b_ ) ln

_Ans_. ( _a_ ) 7 ln 3; ( _b_ ) ln 2 − 3 ln 3

**12.** Express in terms of ln 2 and ln 5: ( _a_ ) ln 50; ( _b_ ) ln 1/4; ( _c_ ) ln  ; ( _d_ )ln  .

_Ans_. ( _a_ ) ln 2 + 2 ln 5; ( _b_ ) −2 ln 2; ( _c_ )   ln 5; ( _d_ ) −(3 ln 2 + ln 5)

**13.** Find the area under the curve   and above the _x_ axis, between _x_ = 2 and _x_ = 4.

_Ans_. ln 2

**14.** Find the average value of   on [3, 5].

_Ans_.

**15.** Use implicit differectiation to find y′: (a) y3 = ln (x3 \+ y3); ( _b_ ) 3 _y_ \- 2 _x_ = 1 + ln _xy_

_Ans_.

**16.** Evaluate  .

_Ans_.

**17.** Check the formula  .

**18.** ( **GC** ) Approximate In   _dt_ to six decimal places by ( _a_ ) the trapezoidal rule; ( _b_ ) the midpoint rule; ( _c_ ) Simpson´s rule, in each with n = 10.

_Ans_. ( _a_ ) 0.693771; ( _b_ ) 0.692835; ( _c_ ) 0.693147

**19.** (GC) Use Newton´s method to approximate the root of _x_ 2 \+ ln _x_ = 2 to four decimal places.

_Ans_. 1.3141

## **CHAPTER 26
Exponential and Logarithmic Functions**

From Chapter 25, we know that the natural logarithm ln _x_ is an increasing differentiable function with domain the set of all positive real numbers and range the set of all real numbers. Since it is increasing, it is a one-to-one function and, therefore, has an inverse function, which we shall denote by _e x_.

**Definition:** _e x_ is the inverse of ln _x_.

It follows that the domain of _e x_ is the set of all real numbers and its range is the set of all positive real number. Since _e x_ is the inverse of ln _x_ , the graph of _e x_ can be obtained from that of ln _x_ by reflection in the line _y = x_. See Fig. 26-1.

Our notation may be confusing. It should not be assumed from the notation that _e x_ is an ordinary power of a base _e_ with exponent _x_. Later in this chapter we will find out that this is indeed true, but we do not know it yet.

**PROPERTIES OF _e x_**

**(26.1)** _e x_ > 0 for all _x_

The range of _e x_ is the set of positive real numbers.

**(26.2)** _ln_ (ex) = x

**(26.3)** _e ln x = x_

Properties (26.2) and (26.3) follow from the fact that _e x_ and ln _x_ are inverses of each other.

**(26.4)** _e x_ is an increasing function.

Assume _u < v._ Since _u = ln (e u)_ and _v = ln (e v)_, ln ( _e u_) < ln ( _e v_). But, since ln _x_ is increasing, _e u < ev._ [For, if _e v   eu,_ then ln ( _e v_)   ln ( _e u_).]

**Fig. 26-1**

**(26.5)** _D x(ex) = ex_

Let _y = e x_. Then ln _y = x_. By implicit differentiation,   and, therefore, y′ = y = ex. For a more rigorous argument, let _f(x)_ = ln _x_ and _f −1(y) = ey_. Note that   By Theorem 10.2 _(b)_.

  that is,

**EXAMPLE 1:**

**(26.6)**

**EXAMPLE 2:** To find  . Then

**(26.7)**

Let _u = −x, du = −dx._ Then

**(26.8)** _e_ 0 = 1

By (26.3), 1 = _e_ ln 1 = _e_ 0.

**(26.9)** _e u + v = eu ev_

ln ( _e u + v_) = _u + v_ = _ln (e u) + ln (ev) = ln (euev)_ by (25.6). Hence, _e u + v = euev_ because ln _x_ is a one-to-one function.

**(26.10)**

By (26.9), _e u-vev = e(u-v)+ v = eu._ Now divide by _e v_.

**(26.11)**

Replace _u_ by 0 in (26.10) and use (26.8).

**(26.12)** _x < ex_ for all _x_

By Problem 7 of Chapter 25, ln _x  x − 1 < x_. By (26.3) and (26.4), _x = e ln x < ex._

**(26.13)**

This follows from (26.4) and (26.12).

**(26.14)**

Let _u = −x._ As _x → −∞, u → + ∞_ ; and, by (26.13), _e u → + ∞_. Then, by (26.11),

 .

The mystery of the letter _e_ in the expression _e x_ can now be cleared up.

**Definition:** Let _e_ be the number such that ln _e_ = 1.

Since ln _x_ is a one-to-one function from the set of positive real numbers onto the set of all real numbers, there must be exactly one number _x_ such that ln _x_ = 1. That number is designated _e_.

Since, by (25.12), ln 2 < 1 < 2 ln 2 = ln 4, we know that 2 < e < 4.

**(26.15) (GC)** e ~ 2.718281828

This estimate can be obtained from a graphing calculator. Later we will find out how to approximate _e_ to any degree of accuracy.

Now we can show that the notation _e x_ is not misleading, that is, that _e x_ actually is a power of _e_. First of all, this can be proved for positive integers _x_ by mathematical induction. [In fact, by (26.3), _e = e ln e = e1_. So, by (26.9), _e n+1 = ene1 = ene_ for any positive integer _n_ and therefore, if we assume by inductive hypothesis that _e n_ represents the product of _e_ by itself _n_ times, then _e n + 1_ is the product of _e_ by itself _n + 1_ times.] By (26.8), _e 0_ = 1, which corresponds to the standard definition of _e 0_. In _n_ is a positive integer, _e −n_ would ordinarily be defined by 1 / _e n_ and this is identical with the function value given by (26.11). If _k_ and _n_ are positive integers, then the power _e k/n_ is ordinarily defined as  . Now, in fact, by (26.9), the product _e k/nek/n...ek/n,_ where there are _n_ factors, is equal to _e k/n + k/n + ... + k/n = ek_. Thus, the function value _e k/n_ is identical with the _n_ th root of _e k_. For negative fractions, we again apply (26.11) to see that the function value is identical with the value specified by the usual definition. Hence, the function value _e x_ is the usual power of _e_ when _x_ is any rational number. Since our funciton _e x_ is continuous, the value of _e x_ when _x_ is irrational is the desired limit of _e r_ for rational numbers _r_ approaching _x_.

The graph of _y = e x_ is shown in Fig. 26-2. By (26.13), the graph rises without bound on the right and, by (26.14), the negative _x_ axis is a horizontal asymptote on the left. Since   the graph is concave upward everywhere. The graph of _y = e −x_ is also shown in Fig. 26-2. It is obtained from the graph of _y = e x_ by reflection in the _y_ axis.

**(26.16)**

For a proof, see Problem 5.

**(26.17)**

This is a special case of (26.16) when _x_ = 1. We can use this formula to approximate _e,_ although the convergence to _e_ is rather slow. For example, when _n_ = 10 000, we get 2.7169 and, when _n_ = 10000, we get 2.7181, which is correct only to three decimal places.

**Fig. 26-2**

**THE GENERAL EXPONENTIAL FUNCTION.** Let _a_ > 0. Then we can define _a_ _x_ as follows:

**Definition:** _a x_ = _e x ln a_

Note that this is consistent with the definition of _e x_ since, when _a_ = _e_ , ln _a_ = 1.

**(26.18)** _D x(ax)_ = (ln _a)a x_

In fact,

**EXAMPLE 3:** _D x_(2x) = (ln 2)2x.

**(26.19)**

This is a direct consequence of (26.18).

**EXAMPLE 4:**

We can derive the usual properties of powers.

**(26.20)**

**(26.21)**

**(26.22)**

By (26.21), _a_ _u_ − _v_ _a_ _v_ = _a_ ( _u_ \+ _v_ )+ _v_ = _a_ _u_. Now divide by _a v_.

**(26.23)**

Replace _u_ by 0 in (26.22) and use (26.20).

**(26.24)** _a_ _uv_ = ( _a_ _u_ ) _v_

( _a_ _u_ ) _v_ = _e_ _v_ ln ( _a_ u) = _e_ _v_ ( _u_ (ln a)) = _e_ ( _u v_ )ln a = _a_ _u v_

**(26.25)** ( _a b_ ) _u_ = _a_ _u_ _b_ _u_

_a_ _u_ _b u_ = _e_ _u_ ln a _e_ _u_ ln b = _e_ _u_ ln _a_ + _u_ ln _b_ = _e_ _u_ (ln _a_ +ln _b_ ) = _e_ _u_ ln ( _a b_ ) = ( _a b_ ) _u_

Recall that we know that _D x(xr) = rxr-1_ for rational numbers _r_. Now we are able to prove that formula for any real number _r_.

( **26.26** ) _D x(xr)_ = _rx r−1_

Since _x r_ = _e_ _r_ ln _x_ ,

**GENERAL LOGARITHMIC FUNCTIONS.** Let _a_ > 0. We want to define a function log _a_ _x_ that plays the role of the traditional logarithm to the base _a_. If _y_ = log _a x_, then _a y_ = _x_ and, therefore,

**Definition:**

**(26.27)** _y_ = log _a_ _x_ is equivalent to _a_ _y_ = _x_

  In (a _v_ ) = In _x_   _a_ _v_ = _x_ (The symbol   is the symbol for equivalence, that is, _if_ and only if.)

Thus, the general logarithmic function with base _a_ is the inverse of the general exponential function with base _a_.

**(26.28)** _a_ loga _x_ = _x_

**(26.29)** loga ( _a_ _x_ ) = _x_

These follow from (26.27). See Problem 6.

The usual properties of logarithms can easily be derived. See Problem 7.

Notice that  . Thus, the natural logarithm turns out to be a logarithm in the usual sense, with base _e_.

### **Solved Problems**

**1.** Evaluate: ( _a_ ) ln ( _e_ 3); ( _b_ ) _e_ 7 ln 2; ( _c_ ) _e_ (ln 3)−2; ( _e_ ) lu.

( _a_ ) ln ( _e_ 3) = 3 by (26.2)

( _b_ ) _e_ 7 ln 2 = ( _e_ ln 2)7 = 27 = 128 by (26.24) and (26.3)

( _c_ )   by (26.10)

( _d_ ) 1 _u_ = _e_ _u_ ln 1 = _e_ _u_ (0) = _e_ 0 = 1 by (26.8)

**2.** Find the derivatives of: ( _a_ ) _e_ 3 _x_ \+ 1; ( _b_ ) 53 _x_ ; ( _c_ ) 3 _x_ π; ( _d_ ) _x_ 2 _e x_.

( _a_ ) _D x_( _e_ 3 _x_ \+ 1) = _e_ 3 _x_ \+ 1(3) = 3 _e_ 3 _x_ \+ 1 by the chain rule

( _b_ )

( _c_ ) _D x_(3 _x_ π) = 3(π _x_ π-1) = 3π _x_ π-1 by (26.26)

( _d_ )

**3.** Find the following antiderivatives: ( _a_ )  .

( _a_ )

( _b_ ) Let _u = x_ 3, _du_ = 3 _x_ 2 _dx_. Then

**4.** Solve the following equations for _x_ : ( _a_ ) ln _x_ 3 = 2; ( _b_ ) ln(ln _x_ ) = 0; ( _c_ ) _e_ 2 _x_ -1 = 3;

( _d_ ) _e x_ \- 3 _e -x_ = 2.

In general, ln _A = B_ if equivalent to _A = e B_, and e _C_ = _D_ is equivalent to _C_ = ln _D_.

( _a_ ) ln _x_ 3 = 3 ln _x_. Hence, ln _x_ 3 = 2 yields 3 ln _x_ = 2, In , _x_ = _e_ 2/3.

( _b_ ) ln (ln _x_ ) = 0 is equivalent to ln _x_ = _e_ 0 = 1, which, in turn, is equivalent to _x_ = _e_ 1 = _e_.

( _c_ ) _e_ 2 _x_ -1 = 3 is equivalent to 2 _x_ \- 1 = ln 3, and then to  .

(d) Multiply both sides by _e x_: _e_ 2 _x_ − 3 = 2 _e x_, _e_ 2 _x_ − 2 _e x_ − 3 = 0. Letting _u_ = _e x_ yields the quadratic equation _u 2_ \- 2 _u_ − 3 = 0; ( _u_ − 3)( _u_ \+ 1) = 0, with solutions _u_ = 3 and _u_ = −1. Hence, _e x_ = 3 or _e x_ = −1. The latter is impossible since _e x_ is always positive. Hence, _e x_ = 3 and, therefore, _x_ = ln 3.

**5.** Prove (26.16):  .

Let  . Then

ln

The expression   is a difference quotient for _D x_(ln _x_ ) at _x_ = 1, with Δ _x_ = _u/n_. As n → + ∞, _u/n_ → 0. So, that difference quotient approaches _D x_(ln _x_ )|x = 1 = (1/ _x_ )|x = 1 = 1. Hence,  . So,  .

**6.** Prove (26.28) _a_ loga _x_ = _x_ and (26.29) log _a_ ( _a_ _x_ ) = _x_.

Substituting log _a_ _x_ for _y_ in (26.27), we get _a_ log _a_ _x_ = _x_.

Substituting _a y_ for _x_ in (26.27), we get _y_ = log _a_ ( _a_ _y_ ).

**7.** Derive the following properties of log _a_ _x_ :

### **Supplementary Problems**

**8.** Calculate the derivatives of the following functions:

**9.** Find the following antiderivatives:

**10.** ( **Hyperbolic Functions** ) Define

Derive the following results:

( _a_ ) _D x_ (sinh _x_ ) = cosh _x_ and _D x_ (cosh _x_ ) = sinh _x_.

( _b_ ) _D x_ (tanh _x_ ) = sech2 _x_ and _D x_(sech _x_ ) = −sech _x_ tanh _x_.

( _c_ ) cosh2 _x_ − sinh2 _x_ =1.

( _d_ ) sinh ( _x + y_ ) = sinh _x_ cosh _y_ \+ cosh _x_ sinh _y_.

( _e_ ) cosh ( _x + y_ ) = cosh _x_ cosh _y_ \+ sinh _x_ sinh _y_.

( _f_ ) sinh 2 _x_ = 2 sinh _x_ cosh _x_.

( _g_ ) cosh 2 _x_ = cosh2 _x_ \+ sinh2 _x_ = 2 cosh2 _x_ − 1 = 2 sinh2 _x_ \+ 1.

( _h_ ) (GC) Sketch the graph of _y_ = 2 cosh ( _x/2_ ) (called a "catenary"), and find its minimum point.

_Ans_. (0, 2)

**11.** Solve the following equations for _x_ :

**12.** Evaluate ( _a_ )  ; ( _b_ )

_Ans_. ( _a_ ) 1; ( _b_ ) 0

**13.** Evaluate: ( _a_ )   ( _b_ )

_Ans_. ( _a_ ) ln  ; ( _b_ )

**14.** ( **GC** ) Use Newton's method to approximate (to four decimal places) a solution of  .

_Ans_. 0.5671

**15.** ( **GC** ) Use Simpson's rule with _n_ = 4 to approximate   to four decimal places.

_Ans_. 0.8556

**16.** If interest is paid at _r_ percent per year and is compounded _n_ times per year, then _P_ dollars become

  dollars after 1 year. If n→ + ∞, then the interest is said to be _compounded continuously_.

( _a_ ) If compounded continuously at _r_ percent per year, show that _P_ dollars becomes _Pe_ _r_ dollars after 1 year, and P _e rt_ dollars after _t_ years.

( _b_ ) At _r_ percent compounded continuously, how many years does it take for a given amount of money to double?

( _c_ ) ( **GC** ) Compounded continuously at 6 percent per year, estimate to two decimal places how many years it would take to double a given amount of money?

( _d_ ) ( **GC** ) Compare the result of compounding continuously at 5 percent with that obtained by compounding once a year.

_Ans_. ( _b_ )  ; ( _c_ ) About 11.55 years;

( _d_ ) After 1 year, one dollar becomes 1.05 dollars when compounded once a year, and about 1.0512 dollars when compounded continuously.

**17.** Find (log10 e). ln 10.

_Ans_. 1

**18.** Write as a single logarithm with base _a_ : 3 loga 2 + loga 40 − loga 16

_Ans_. loga 20

**19.** ( **GC** ) Estimate log2 7 to eight decimal places.

_Ans_. 2.80735492

**20.** Show that log _b_ _x_ = (log _a_ _x_ )(log _b_ _a_ ).

**21.** ( **GC** ) Graph _y_ = _e_ − _x_ 2/2. Indicate absolute extrema, inflection points, asymptotes, and any symmetry.

_Ans_. Absolute maximum at (0,1), inflection points at _x_ = ±1, the _x_ axis is a horizontal asymptote on the left and right, symmetric with respect to the _y_ axis.

**22.** Given _e_ _xy_ − _x_ \+ _y_ 2 = 1, find   by implicit differentiation.

_Ans_.

**23.** ( **GC** ) Graph _y_ = sinh _x_  .

**24.** Evaluate  .

_Ans_. ln ( _e_ _x_ \+ _e_ − _x_ ) + _C_

**25.** Use logarithmic differentiation to find the derivative of _y_ = _x_ 3/ _x_.

_Ans_.

## **CHAPTER 27
L'Hôpital's Rule**

Limits of the form lim   can be evaluated by the following theorem in the _indeterminate cases_ where _f_ ( _x_ ) and _g_ ( _x_ ) both approach 0 or both approach ± ∞.

**L'HÔPITAL'S RULE**. If _f_ ( _x_ ) and _g_ ( _x_ ) either both approach 0 or both approach ±∞, then

Here, "lim" stands for any of

For a sketch of the proof, see Problems, 1, 11 and 12. It is assumed, in the case of the last three types of limits, that _g_ ′( _x_ ) ≠ 0 for _x_ sufficiently close to _a_ , and in the case of the first two limits, that g′( _x_ ) ≠ 0 for sufficiently large or sufficiently small values of _x_. (The corresponding statements about _g_ ( _x_ ) ≠ 0 follow by Rolle's Theorem.)

**EXAMPLE 1:** Since In _x_ approaches +∞ as _x_ approaches +∞, L'Hôpital's Rule implies that

**EXAMPLE 2:** Since _e x_ approaches + ∞ as _x_ approaches +∞, L'Hôpital's Rule implies that

**EXAMPLE 3:** We already know from problem 13( _a_ ) of chapter 7 that

Since both 3 _x_ 2 \+ 5 _x_ − 8 and 7 _x_ 2 − 2 _x_ \+ 1 approach +∞ as _x_ approaches + ∞, L'Hôpital's Rule tells us that

and another application of the rule tells us that

**EXAMPLE 4:** Since tan _x_ approaches 0 as _x_ approaches 0, L'Hôpital's Rule implies that

**INDETERMINATE TYPE 0. ∞**. If _f_ ( _x_ ) approaches 0 and _g_ ( _x_ ) approaches ±∞, we do not know how to find lim _f_ ( _x_ ) _g_ ( _x_ ). Sometimes such a problem can be transformed into a problem to which L'Hôpital's Rule is applicable.

**EXAMPLE 5:** As _x_ approaches 0 from the right, In _x_ approaches − ∞. So, we do not know how to find  . But as _x_ approaches 0 from the right, 1/ _x_ approaches + ∞ So, by L'Hôpital's Rule,

**INDETERMINATE TYPE ∞ − ∞**. If _f_ ( _x_ ) and _g_ ( _x_ ) both approach ∞, we do not know what happens to lim( _f_ ( _x_ ) − _g_ ( _x_ )). Sometimes we can transform the problem into a L'Hôpital-type problem.

**EXAMPLE 6:**   is a problem of this kind. But,

Since _x_ − sin _x_ and _x_ sin _x_ both approach 0, L'Hôpital's Rule applies and we get  . Here both numerator and denominator approach 0 and L'Hôpital's Rule yields

**INDETERMINATE TYPES 0 0,∞0, AND 1∞**. If **lim y** is of one of these types, then lim (In _y_ ) will be of type 0. ∞.

**EXAMPLE 7:** In   is of type 00 and we do not know what happens in the limit. But In _y_ = sin _x_ In   and In _x_ and csc _x_ approach ± ∞. So, by L'Hôpital's Rule,

Here, we used the fact that   (Problem 1 of Chapter 17). Now, since  ,

**EXAMPLE 8:** In   is of type ∞0, and it is not clear what happens in the limit. But In   and both In |In _x_ | and 1/ _x_ approach + ∞. So, L'Hôpital's Rule yields

 ,

since

 . Hence;

**EXAMPLE 9:** In   is of type 1∞ and we cannot see what happens in the limit. But ln   and both the numerator and the denominator approach 0. So, by L'Hôpital's Rule we get

 . Hence,

### **Solved Problems**

**1.** Prove the following   form of L'Hôpital's Rule. Assume f( _x_ ) and g( _x_ ) are differentiable and g′( _x_ )≠0 in some open interval ( _a, b_ ) and  . Then, if   exists,

Since  , we may assume that _f_ ( _a_ ) and _g_ ( _a_ ) are defined and that _f_ ( _a_ ) = _g_ ( _a_ ) = 0. Replacing _b_ by _x_ in the Extended Law of the Mean (Theorem 13.5) and using the fact that _f_ ( _a_ ) = _g_ ( _a_ ) = 0, we obtain

for some _x_ 0 with _a_ < _x_ 0 < _x_. So, _x_ 0 → _a_ + as _x_ → _a_ +. Hence,

We also can obtain the   form of L'Hôpital's Rule for  , and then the results for   and   yield the   form of L'Hôpital's Rule for  .

**2.** We already know by Examples 1 and 2 that   and  . Show further that   and   for all positive integers _n_.

Use mathematical induction. Assume these results for a given _n_   1. By L'Hôpital's Rule,

Likewise,

**3.** Use L'Hôpital's Rule one or more times to evaluate the following limits. Always check that the appropriate assumptions hold.

( _a_ )  .

We get

( _b_ )  .

We get   by Example 2.

( _c_ )  .

We obtain  .

By repeated uses of L'Hôpital's Rule, we get

( _d_ )  .

We get

( _e_ )  .

One obtains

( _f_ )  .

The direct use of L'Hôpital's Rule

leads us to ever more complicated limits. Instead, if we change from cot to tan, we get

( _g_ )  .

This is of type 0 · ∞. Then L'Hôpital's Rule can be brought in as follows:

( _h_ )  .

This is of type 0 · ∞. However, it is equal to

( _i_ )  .

This is of type ∞ − ∞. But it is equal to

( _j_ )  .

This is of type ∞ − ∞. But it is equal to

( _k_ )  .

This if of type ∞0. Let y = (tan _x_ )cos _x_. Then In  .

So

( _l_ )  .

We get   and we are going around in a circle. So, L'Hôpital's Rule is of no use. But,

**4.** Criticize the following use of L'Hôpital's Rule:

The second equation is an incorrect use of L'Hôpital's Rule, since   and  . So, the correct limit should be  .

**5**. (GC) Sketch the graph of  .

See Fig. 27-1. By Example 2,  . So, the positive _x_ axis is a horizontal asymptote. Since   and _y_ ′′ = _e_ − _x_ ( _x_ − 2). Then _x_ = 1 is a critical number. By the second derivative test, there is a relative maximum at (1, 1/ _e_ ) since _y_ ″ < 0 at _x_ = 1. The graph is concave downward for _x_ < 2 (where _y_ ″ < 0) and concave upward for _x_ > 2 (where y″ > 0). (2, 2/ _e_ 2) is an inflection point. The graphing calculator gives us the estimates 1/ _e_ ~ 0.37 and 2/ _e_ 2 ~ 0.27.

**Fig. 27-1**

**6**. (GC) Sketch the graph _y_ = _x_ In _x_

See Fig. 27-2. The graph is defined only for _x_ > 0. Clearly,  . By Example 5,  . Since _y_ ′ = 1 + In _x_ and _y_ ″ = 1/ _x_ > 0, the critical number at _x_ = 1/ _e_ (where _y_ ′ = 0) yields, by the second derivative test, a relative minimum at (1/ _e_ ,−1/ _e_ ). The graph is concave upward everywhere.

**Fig. 27-2**

### **Supplementary Problems**

**7.** Show that   for all positive integers _x_.

**8.** Find  .

_Ans._ π

**9.** Sketch the graphs of the following functions: ( _a_ ) _y_ = _x_ − ln _x_ ; ( _b_ )  ; ( _c_ ) _y_ = _x_ 2 _e x_

_Ans._ See Fig. 27.3.

**Fig. 27-3**

**10.** Evaluate the following limits:

**11.** Verify the sketch of the proof of the following   form of L'Hôpital's Rule at +∞. Assume _f_ ( _x_ ) and _g_ ( _x_ ) are differentiable and _g′_ ( _x_ ) ≠ 0 for all _x_   _c_ , and  . Then,

if   exists,

_Proof:_ Let _F_ ( _u_ ) = _f_ (1/ _u_ ) and _G_ ( _u_ ) = _g_ (1/ _u_ ). Then, by Problem 1 for _a_ → 0+, and with _F_ and _G_ instead of _f_ and _g_ ,

**12.** Fill in the gaps in the proof of the following   form of L'Hôpital's Rule in the   case. (The other cases follow easily as in the   form.) Assume _f_ ( _x_ ) and _g_ ( _x_ ) are differentiable and _g_ ′( _x_ ) ≠ 0 in some open interval ( _a_ , _b_ ) and  . Then,

if   exists,

_Proof:_ Assume ε > 0 and choose _c_ so that | _K_ − ( _f_ ′( _x_ )/ _g_ ′( _x_ ))| < ε/2 for _a_ < _x_ < _c._ Fix _d_ in ( _a_ , _c_ ). Let _a_ < _y_ < _d._ By the extended mean value theorem, there exists _x_ * such that

_y_ < _x_ * < _d_ and

Then

Now we let _y_ → _a_ +. Since _g_ ( _y_ ) → ± ∞ and _f_ ( _d_ ) and _g_ ( _d_ ) are constant, _f_ ( _d_ )/ _g_ ( _y_ ) → 0 and 1 − _g_ ( _d_ )/ _g_ ( _y_ ) → 1. So, for _y_ close to _a_ ,

**13.** (GC) In the following cases, try to find the limit by analytic methods, and then check by estimating the limit on a graphing calculator; ( _a_ )  ; ( _b_ )  ; ( _c_ )  ; ( _d_ )

_Ans._ ( _a_ ) 0; ( _b_ ) 1; ( _c_ ) 1; ( _d_ )

**14.** The curent in a coil containing a resistance _R_ , an inductance _L_ , and a constant electromotive force _E_ at time _t_ is given by  . Obtain a formula for estimating _i_ when _R_ is very close to 0.

_Ans._

## **CHAPTER 28
Exponential Growth and Decay**

Assume that a quantity _y_ varies with time and that

for some nonzero constant _k_. Let _F_ ( _t_ ) = _y_ / _e kt_. Then, by the quotient rule,

Hence, _F_ ( _t_ ) must be a constant C. (Why?) Thus, _y_ / _e kt_ = _C_ and, therefore, _y_ = _Ce kt_. To evaluate _C_ , let _t_ = 0. Then _y_ (0) = _Ce_ 0 = _C_ (1) = _C_. If we designate _y_ (0) by _y_ 0, then _C_ = _y_ 0 and we have obtained the general form of the solution of equation (28.1):

If _k_ > 0, we say that _y grows exponentially_ and _k_ is called the _growth constant_. If _k_ < 0, we say that y _decays exponentially_ , and _k_ is called the _decay constant_. The constant _y_ 0 is called the _initial value_.

From Problem 2 of Chapter 27, we know that  . So, when _k_ > 0,  .

Thus, a quantity that grows exponentially grows much more rapidly than any power of _t_. There are many natural processes, such as bacterial growth or radioactive decay, in which quantities increase or decrease at an exponential rate.

**HALF-LIFE.** Assume that a quantity _y_ of a certain substance decays exponentially, with decay constant _k_. Let _y_ 0 be the quantity at time _t_ = 0. At what time _T_ will only half of the original quantity remain?

By (28.2), we get the equation _y_ = _y_ 0 _e kt_. Hence, at time _T_ ,

Note that the same value _T_ is obtained for _any_ original amount _y_ 0. _T_ is called the _half-life_ of the substance. It is related to the decay constant _k_ by the equation (28.3). So, if we know the value of either _k_ or _T_ , we can compute the value of the other. Also observe that, in (28.4), _k_ < 0, so that _T_ > 0.

The value of _k_ can be obtained by experiment. For a given initial value _y_ 0 and a specific positive time _t_ 0, we observe the value of _y_ , substitute in the equation (28.2), and solve for _k_.

### **Solved Problems**

**1.** Given that the half-life _T_ of radium is 1690 years, how much will remain of one gram of radium after 1000 years?

From (28.3),   and the quantity of radium is given by _y_ = _y_ 0 _e_ −(ln 2)t/1690. Nothing that _y_ 0 = 1 and substituting 1000 for _t_ , we get the quantity

Thus, about 663.6 milligrams are left after 1000 years.

**2.** If 20% of a radioactive substance disappears in one year, find its half-life _T_. Assume exponential decay.

By (28.2), 0.8 _y_ 0 = _y_ 0 _e_ _k_ (1) = _y_ 0 _e k_. So, 0.8 = _e k_, whence  . From (28.4),  .

**3.** Assume that the number of bacteria in a culture grows exponentially with a growth constant of 0.02, time being measured in hours. (Although the number of bacteria must be a nonnegative integer, the assumption that that number is a continuous quantity always seems to lead to results that are experimentally verified.)

( _a_ ) How many bacteria will be present after 1 hour if there are initially 1000?

( _b_ ) Given the same initial 1000 bacteria, in how many hours will there be 100 000 bacteria?

( _a_ ) From (28.2), _y_ = 1000 _e_ 0.02 ~ 1000(1.0202) = 1020.2 ~ 1020

( _b_ ) From (28.2),

_Note:_ Sometimes, instead of giving the growth constant, say _k_ = 0.02, one gives a corresponding rate of increase per unit time (in our case, 2 percent per hour.) This is not quite accurate. A rate of increase of _r_ percent per unit time is approximately the same as a value of _k_ = 0.0 _r_ when _r_ is relatively small (say, _r_  3). In fact, with an _r_ percent rate of growth, _y_ = _y_ 0(1 + 0.0 _r_ ) after one unit of time. Since _y_ = _y_ 0e _k_ when _t_ = 1, we get 1 + 0.0 _r_ = _e_ _k_ and, therefore, _k_ = ln(1 + 0.0 _r_ ). This is close to 0.0 _r_ , since ln (1 + _x_ ) ~ _x_ for small positive _x_. (For example, ln 1.02 ~ 0.0198 and ln 1.03 ~ 0.02956.) For that reason, many textbooks often interpret a rate of increase of _r_ percent to mean that _k_ = 0.0 _r_.

**4.** If a quantity _y_ increases or decreases exponentially, find a formula for the average value of _y_ over a time interval [0, _b_ ].

By definition, the average value   _dt_ (where _k_ is the growth or decay constant). By (28.1),   and, therefore,  . By the Fundamental Theorem of Calculus,

**5.** If the population of a country is 100 million people and the population is increasing exponentially with a growth constant _k_ = ln 2, calculate precisely the population after 5 years.

By (28.2), the population _y_ = _y_ 0 _e_ _kt_ = 108 _e_ (ln 2)5 = 108( _e_ ln2)5 = 108(25) = 32(108). Thus, the population will reach 3.2 billion people in 5 years.

**6. Carbon-Dating.** A certain isotope 14C of carbon occurs in living organisms in a fixed proportion to ordinary carbon. When that organism dies, its 14C decays exponentially, and its half-life is 5730 years. Assume that a piece of charcoal from a wood fire was found in a cave and contains only 9% of the 14C expected in a corresponding piece of wood in a live tree. (This figure is obtained by measuring the amount of ordinary carbon in the piece of charcoal.) How long ago was the wood burned to form that charcoal?

If _y_ is the amount of 14C present in the piece of charcoal, we have _y_ = _y_ 0 _e_ _kt_. The present quantity 0.09 _y_ 0 = _y_ 0 _e_ _k_ τ, where τ is the elapsed time. Thus, 0.09 = _e _k_ τ_, ln(0.09) = _k_ τ, τ = (ln(0.09))/ _k_. Since the half-life _T_ = 5730 and _k_ = −(ln 2)/ _T_ = −(ln 2)/5730, we obtain

**7. Newton's Law of Cooling:** The rate of change of the temperature of an object is proportional to the difference between the object's temperature and the temperature of the surrounding medium.

Assume that a refrigerator is maintained at a constant temperature of 45°F and that an object having a temperature of 80°F is placed inside the refrigerator. If the temperature of the object drops from 80°F to 70°F in 15 minutes, how long will it take for the object's temperature to decrease to 60°F?

Let _u_ be the temperature of the object. Then, by Newton's Law of Cooling, _du/dt_ = _k_ ( _u_ − 45), for some (negative) constant _k_. Let _y_ = _u_ − 45. Then _dy/dt_ = _du/dt_ = _ky_. Thus, by (28.2), _y_ = _y_ 0 _e kt_. Since _u_ is initially 80°F, _y_ 0 = 80 − 45 = 35. So, _y_ = 35 _e kt_. When _t_ = 15, _u_ = 70 and _y_ = 25. Hence, 25 = 35 _e_ 15 _k_ , 5 = 7 _e_ 15 _K_ and, therefore,  . Thus,  . When the object's temperature is 60°F, _y_ = 15. So, 15 = 35 _e kt_, 3 = 7 _e kt_ and therefore,  . Thus,

Hence, it would take about 22.7727 minutes for the object's temperature to drop from 70° to 60°.

**8. Compound Interest**. Assume that a savings account earns interest at a rate of _r_ % per year. So, after one year, an amount of _P_ dollars would become   dollars and, after _t_ years, it would become   dollars. However, if the interest is calculated _n_ times a year instead of once a year, then in each period the interest rate would be ( _r_ / _n_ )%; after _t_ years, there would have been _nt_ such periods and the final amount would be  . If we let _n_ → + ∞, then we say that the interest is _compounded continuously_. In such a case, the final amount would be

Let 100 dollars be deposited in a savings account paying an interest rate of 4% per year. After 5 years, how much would be in the account if:

( _a_ ) The interest is calculated once a year?

( _b_ ) The interest is calculated quarterly (that is, four times per year)?

( _c_ ) The interest is compounded continuously?

( _a_ ) 100 (1.04)5 ~ 121.6653 dollars.

( _b_ ) 100 (1.01)20 ~ 122.0190 dollars.

( _c_ ) 100 _e_ 0.04(5) = 100 _e_ 0.2 ~ 122.1403 dollars.

### **Supplementary Problems**

**9.** Assume that, in a chemical reaction, a certain substance decomposes at a rate proportional to the amount present. Assume that an initial quantity of 10 000 grams is reduced to 1000 grams in 5 hours. How much would be left of an initial quantity of 20 000 grams after 15 hours?

_Ans._ 20 grams

**10.** A container with a maximum capacity of 25 000 fruit flies initially contains 1000 fruit flies. If the population grows exponentially with a growth constant of (ln 5)/10 fruit flies per day, in how many days will be container be full?

_Ans._ 20 days

**11.** The half-life of radium is 1690 years. How much will be left of 32 grams of radium after 6760 years?

_Ans._ 2 grams

**12.** If a population grows exponentially and increases at the rate of 2.5% per year, find the growth constant _k_.

_Ans._ ln 1.025 ~ 0.0247

**13.** A saltwater solution initially contains 5 lb of salt in 10 gal of fluid. If water flows in at the rate of   and the mixture flows out at the same rate, how much salt is present after 20 min?

_Ans._

**14.** Fruit flies in an enclosure increase exponentially in such a way that their population doubles in 4 hours. How many times the initial number will there be after 12 hours?

_Ans._ 8

**15.** (GC) If the world population in 1990 was 4.5 billion and it is growing exponentially with growth constant _k_ = (ln 3)/8, estimate the world population in the years ( _a_ ) 2014; ( _b_ ) 2020.

_Ans._ ( _a_ ) 111.5 billion; ( _b_ ) 277.0 billion

**16.** (GC) If a thermometer with a reading of 65°F is taken into the outside air where the temperature is a constant 25°F, the thermometer reading decreases to 50°F in 2.0 minutes.

( _a_ ) Find the thermometer reading after one more minute.

( _b_ ) How much longer (after 3.0 minutes) will it take for the thermometer reading to reach 32°F? Use Newton's Law of Cooling.

_Ans._ ( _a_ ) 45°F; ( _b_ ) about 4.4 minuter more

**17.** (GC) Under continuous compounding at a rate of _r_ % per year:

( _a_ ) How long does it take for a given amount of money _P_ to double?

( _b_ ) If a given amount _P_ doubles in 9 years, what is _r_?

( _c_ ) If _r_ = 8, how much must be deposited now to yield $100 000 in 17 years?

_Ans._ ( _a_ )  ; ( _b_ ) about 7.7; ( _c_ ) about $25 666

**18.** An object cools from 120°F to 95°F in half an hour when surrounded by air whose temperature is 70°F. Use Newton's Law of Cooling to find its temperature at the end of another half an hour.

_Ans._ 82.5°F

**19.** If an amount of money receiving interest of 8% per year is compounded continuously, what is the equivalent yearly rate of return?

_Ans._ about 8.33%

**20.** How long does it take for 90% of a given quantity of the radioactive element cobalt-60 to decay, given that its half-life is 5.3 years?

_Ans._ about 17.6 years

**21.** A radioactive substance decays exponentially. If we start with an initial quantity of _y_ 0, what is the average quantity present over the first half-life?

_Ans._

## **CHAPTER 29
Applications of Integration I: Area and Are Length**

**AREA BETWEEN A CURVE AND THE _y_ AXIS.**. We already know how to find the area of a region like that shown in Fig. 29-1, bounded below by the _x_ axis, above by a curve _y_ = _f_ ( _x_ ), and lying between _x_ = _a_ and _x_ = _b_. The area is the definite integral  .

**Fig. 29-1**

Now consider a region like that shown in Fig. 29-2, bounded on the left by the _y_ axis, on the right by a curve _x_ = _g_ ( _y_ ), and lying between _y_ = _c_ and _y_ = _d_. Then, by an argument similar to that for the case shown in Fig. 29-1, the area of the region is the definite integral  .

**Fig. 29-2**

**EXAMPLE 1:** Consider the region bounded on the right by the parabola _x_ = 4 − _y_ 2, on the left by the _y_ axis, and above and below by _y_ = 2 and _y_ = −1. See Fig. 29-3. Then the area of this region is  . By the Fundamental Theorem of Calculus, this is

**Fig. 29-3**

**AREAS BETWEEN CURVES.** Assume that _f_ and _g_ are continuous functions such that _g_ ( _x_ )   _f_ ( _x_ ) for _a_   _x_   _b_. Then the curve _y_ = _f_ ( _x_ ) lies above the curve _y_ = _g_ ( _x_ ) between _x_ = _a_ and _x_ = _b_. The area _A_ of the region between the two curves and lying between _x_ = _a_ and _x_ = _b_ is given by the formula

To see why this formula holds, first look at the special case where 0   _g_ ( _x_ )   _f_ ( _x_ ) for _a_   _x_   _b_. (See Fig. 29-4) Clearly, the area is the difference between two areas, the area _A f_ of the region under the curve _y_ = _f_ ( _x_ ) and above the _x_ axis, and the area _A g_ of the region under the curve _y_ = _g_ ( _x_ ) and above the _x_ axis. Since   and  ,

**Fig. 29-4**

Now look at the general case (see Fig. 29-5), when one or both of the curves _y_ = _f_ ( _x_ ) and _y_ = _g_ ( _x_ ) may lie below the _x_ axis. Let _m_ < 0 be the absolute minimum of g on  _a_ , _b_ ]. Raise both curves by | _m_ | units. The new graphs, shown in [Fig. 29-6, are on or above the x axis and enclose the same area _A_ as the original graphs. The new graphs. The upper curve is the graph of _y_ = _f_ ( _x_ ) + | _m_ | and the lower curve is the graph of _y_ = _g_ ( _x_ ) + | _m_ |. Hence, by the special case above,

**Fig. 29-5**

**Fig. 29-6**

**EXAMPLE 2** Find the area _A_ of the region   under the line  , above the parabola _y_ = _x_ 2, and between the _y_ axis and _x_ = 1. (see the shaded region in Fig. 29-7.) By (29.1),

**Fig. 29-7**

**ARC LENGTH**. Let _f_ be the differentiable on  _a_ , _b_ ]. Consider the part of the graph of _f_ from ( _a_ , _f_ ( _a_ )) to ( _b_ , _f_ ( _b_ )). Let us find a formula for the length _L_ of this curve. Divide [ _a_ , _b_ ] into _n_ equal subintervals, each of length Δ _x_. To each point _x k_ in this subdivision there corresponds a point _P k_( _x k, f( _x_ k_)) on the curve. (See [Fig. 29-8) For large _n_ , the sum   of the lengths of the line segments _P_ _k_ −1 _P_ _k_ is an approximation to the length of the curve.

**Fig. 29-8**

By the distance formula (2.1),

Now, _x k_ − _x_ _k_ −1 = Δ _x_ and, by the law of the mean (Theorem 13.4),

for some   in ( _x_ _k_ −1, _x_ _k_ ), Thus,

So,

The right-hand sum is an approximating sum for the definite integral  . Therefore, letting _n_ → + ∞, we get the _are length formula:_

**EXAMPLE 3:** Find the are length _L_ of the curve _y_ = _x_ 3/2 from _x_ = 0 to _x_ = 5. By (29.2), since  ,

### **Solved Problems**

**1.** Find the area bounded by the parabola _x_ = 8 + 2 _y_ − _y_ 2, the _y_ axis, and the lines _y_ = −1 and _y_ = 3.

Note, by completing the square, that _x_ = −( _y_ 2 − 2 _y_ − 8) = −(( _y_ − 1)2 − 9) = 9 − ( _y_ − 1)2 = (4 − _y_ )(2 + _y_ ). Hence, the vertex of the parabola is (9,1) and the parabola cuts the _y_ axis at _y_ = 4 and _y_ = −2.

We want the area of the shaded region in Fig. 29-9, which is given by

**Fig. 29-9**

**2.** Find the area of the region between the curves _y_ = sin _x_ and _y_ = cos _x_ from _x_ = 0 to _x_ = π/4.

The curves intersect at   and 0   sin _x_ < cos _x_ for 0   _x_ < π/4. (See Fig. 29-10). Hence, the area is

**Fig. 29-10**

**3.** Find the area of the region bounded by the parabolas _y_ = 6 _x_ − _x_ 2 and _y_ = _x_ 2 − 2 _x_.

By solving 6 _x_ − _x_ 2 = _x_ 2 − 2 _x_ , we see that the parabolas intersect when _x_ = 0 and _x_ = 4, that is, at (0,0) and (4,8). (See Fig. 29-11). By completing the square, the first parabola has the equation _y_ = 9 − ( _x_ − 3)2; therefore, it has its vertex at (3,9) and opens downward. Likewise, the second parabola has the equation _y_ = ( _x_ − 1)2 − 1; therefore, it vertex is at (1, −1) and it opens upward. Note that the first parabola lies above the second parabola in the given region. By (29.1), the required area is

**Fig. 29-11**

**4.** Find the area of the region bounded by the parabola _y_ 2 = 4 _x_ and the line _y_ = 2 _x_ − 4.

Solving the equations simultaneously, we get (2 _x_ − 4)2 = 4 _x_ , _x_ 2 − 4 _x_ \+ 4 = _x_ , _x_ 2 −5 _x_ \+ 4 = 0, ( _x_ − 1)( _x_ − 4) = 0. Hence, the curves intersect when _x_ = 1 or _x_ = 4, that is, at (1, −2) and (4,4). (See Fig. 29-12). Note that neither curve is above the other throughout the region. Hence, it is better to take _y_ as the independent variable and rewrite the curves as   and  . The line is always to right of to parabola.

The area is obtained by integrating along the _y_ axis:

**Fig. 29-12**

**5.** Find the area of the region between the curve _y_ = _x_ 3 − 6 _x_ 2 \+ 8 _x_ and the _x_ axis.

Since _x_ 3 − 6 _x_ 2 \+ 8 _x_ = _x_ ( _x_ 2 − 6 _x_ \+ 8) = _x_ ( _x_ − 2) ( _x_ − 4), the curve crosses the _x_ axis at _x_ = 0, _x_ = 2, and _x_ = 4. The graph looks like the curve shown in Fig. 29-13. (By applying the quadratic formula to _y_ ′, we find that the maximum and minimum values occur  .) Since the part of the region with 2   _x_   4 lies below the _x_ axis, we must calculate the two separate integrals, one with respect to _y_ between _x_ = 0 and _x_ = 2, and the other with respect to − _y_ between _x_ = 2 and _x_ = 4. Thus, the required area is

**Fig. 29-13**

Note that, if we had made the mistake of simply calculating the integral  , we would have got the incorrect answer 0.

**6.** Find the area enclosed by the curve _y_ 2 = _x_ 2 − _x_ 4.

The curve is symmetric with respect to the coordinate axes. Hence the required area is four times the portion lying in the first quadrant. (See Fig. 29-14.) In the first quadrant,   and the curve intersects the _x_ axis at _x_ = 0 and _x_ = 1. So, the required area is

**Fig. 29-14**

**7.** Find the are length of the curve _x_ = 3 _y_ 3/2 − 1 from _y_ = 0 to _y_ = 4.

We can reverse the roles of _x_ and _y_ in the are length formula (29.2):  . Since  ,

**8.** Find the are length of the curve 24 _xy_ = _x_ 4 \+ 48 form _x_ = 2 to _x_ = 4.

 . Hence,  , Thus,

So,

**9.** Find the are length of the catenary   from _x_ = 0 to _x_ = _a_.

  and, therefore,

So,

### **Supplementary Problems**

**10.** Find the area of the region lying above the _x_ axis and under the parabola _y_ = 4 _x_ − _x_ 2.

_Ans._

**11.** Find the area of the region bounded by the parabola _y_ = _x_ 2 − 7 _x_ \+ 6, the _x_ axis, and the lines _x_ = 2 and _x_ = 6.

_Ans._

**12.** Find the area of the region bounded by the given curves.

**13.** Find the length of the indicated are of the given curve.

**14.** (GC) Estimate the are length of _y_ = sin _x_ from _x_ = 0 to _x_ = Π to an accuracy of four decimal places. (Use Simpson' Rule with _n_ = 10.)

_Ans._ 3.8202

## **CHAPTER 30
Applications of Integration II: Volume**

A _solid of revolution_ is obtained by revolving a region in a plane about a line that does not intersect the region. The line about which the rotation takes place is called the _axis of revolution_.

Let _f_ be a continuous function such that _f_ ( _x_ )   0 for _a_   _x_   _b_. Consider the region   under the graph of _f_ , above the _x_ axis, and between _x_ = _a_ and _x_ = _b_. (See Fig. 30-1.) If   is revolved about the _x_ axis, the resulting solid is a solid of revolution. The generating regions   for some familiar solids are shown in Fig. 30-2.

**Fig. 30-1**

**DISK FORMULA.** The volume _V_ of the solid of revolution obtained by revolving the region   of Fig. 30-1 about the _x_ axis is given by

**Fig. 30-2**

See Problem 9 for a sketch of the proof of this formula.

Similarly, when the axis of rotation is the _y_ axis and the region that is revolved lies between the _y_ axis and a curve _x_ = _g_ ( _y_ ) and between _y_ = _c_ and _y_ = _d_ (See Fig. 30-3), then the volume _V_ of the resulting solid of revolution is given by the formula

**Fig. 30-3**

**EXAMPLE 1:** Consider the solid of revolution obtained by revolving about the _x_ axis the region in the first quadrant bounded by the parabola _y_ 2 = 8 _x_ and the line _x_ = 2. (See Fig. 30-4.) By the disk formula, the volume is

**Fig. 30-4**

**EXAMPLE 2:** Consider the solid of revolution obtained by revolving about the _y_ axis the region bounded by the parabola _y_ = 4 _x_ 2 and the lines _x_ = 0 and _y_ = 16. (See Fig. 30-5.) To find its volume, we use the version of the disk formula in which we integrate along the _y_ axis, Thus,

**Fig. 30-5**

**WASHER METHOD.** Assume that 0   _g_ ( _x_ )   _f_ ( _x_ ) for _a_   _x_   _b_. Consider the region between _x_ = _a_ and _x_ = _b_ and lying between _y_ = _g_ ( _x_ ) and _y_ = _f_ ( _x_ ). (See Fig. 30-6.) Then the volume _V_ of the solid of revolution obtained by revolving this region about the _x_ axis is given by the formula

**Fig. 30-6**

The justification is clear. The desired volume is the difference of two volumes, the volumes   of the solid of revolution generated by revolving about _x_ axis the region under

† The word "washer" is used because each thin vertical strip of the region being revolved produces a solid that resembles a plumbing part called a _washer_ (a small cylindrical disk with a hole in the middle).

_y_ = _f_ ( _x_ ) and the volume   of the solid of revolution generated by revolving about the _x_ axis the region under _y_ = _g_ ( _x_ ).

A similar formula

holds when the region lies between the two curves _x_ = _f_ ( _y_ ) and _x_ = _g_ ( _y_ ) and between _y_ = _c_ and _y_ = _d_ , and it is revolved about the _y_ axis. (It is assumed that 0   _g_ ( _y_ )   _f_ ( _y_ ) for _c_   _y_   _d_.)

**EXAMPLE 3:** Consider the solid of revolution obtained by revolving about the _x_ axis the region bounded by the curves, _y_ = 4 _x_ 2, _x_ = 0, and _y_ = 16. (The same region as in Fig. 30-5.) Here the upper curve is _y_ = 16 and the lower curve is _y_ = 4 _x_ 2. Hence, by the washer formula,

**CYLINDRICAL SHELL METHOD.** Consider the solid of revolution obtained by revolving about the _y_ axis the region   in the first quadrant between the _x_ axis and the curve _y_ = _f_ ( _x_ ), and lying between _x_ = _a_ and _x_ = _b_. (See Fig. 30-7.) Then the volume of the solid is given by

**Fig. 30-7**

See Problem 10 for the justification of this formula.

A similar formula holds when the roles of _x_ and _y_ are reversed, that is, the region   in the first quadrant between the _y_ axis and the curve _x_ = _f_ ( _y_ ), and lying between _y_ = _c_ and _y_ = _d_ , is revolved about the _x_ axis

**EXAMPLE 4:** Revolve about the _y_ axis the region above the _x_ axis and below _y_ = 2 _x_ 2, and between _x_ = 0 and _x_ = 5. By the cylindrical shell formula, the resulting solid has volume

Note that the volume could also have been computed by the washer formula, but the calculation would have been somewhat more complicated.

**DIFFERENCE OF SHELLS FORMULA.** Assume that 0   _g_ ( _x_ )   _f_ ( _x_ ) on an interval [ _a_ , _b_ ] with _a_   0. Let   be the region in the first quadrant between the curves _y_ = _f_ ( _x_ ) and _y_ = _g_ ( _x_ ) and between _x_ = _a_ and _x_ = _b_. Then the volume of the solid of revolution obtained by revolving   about the _y_ axis is given by

This obviously follows from the cylindrical shells formula because the required volume is the difference of two volumes obtained by the cylindrical shells formula. Note that a similar formula holds when the roles of _x_ and _y_ are reversed.

**EXAMPLE 5:** Consider the region in the first quadrant bounded above by _y_ = _x_ 2, below by _y_ = _x_ 3, and lying between _x_ = 0 and _x_ = 1. When revolved about the _y_ axis, this region generates a solid of revolution whose volume, according to the difference of shells formula, is

**CROSS-SECTION FORMULA (SLICING FORMULA).** Assume that a solid lies entirely between the plane perpendicular to the _x_ axis at _x_ = _a_ and the plane perpendicular to the _x_ axis at _x_ = _b_. For each _x_ such that _a_   _x_   _b_ , assume that the plane perpendicular to the _x_ axis at that value of _x_ intersects the solid in a region of area _A_ ( _x_ ). (See Fig. 30-8.) Then the volume _V_ of the solid is given by

For justification, see Problem 11.

**Fig. 30-8**

**EXAMPLE 6:** Assume that half of a salami of length _h_ is such that a cross-section perpendicular to the axis of the salami at a distance _x_ from the end _O_ is a circle of radius  . (See Fig. 30-9.) Hence, the area _A_ ( _x_ ) of the

† This formula is also called the _slicing formula_ because each cross-sectional area _A_ ( _x_ ) is obtained by slicing through the solid.

cross-section section is  . So, the cross-section formula yields

**Fig. 30-9**

### **Solved Problems**

**1.** Find the volume of a cone that has height _h_ and whose base has radius _r_.

The cone is generated by revolving about the _x_ axis the region between the line   and the _x_ axis, between _x_ = 0 and _x_ = _h_. (See Fig. 30-2( _a_ ).) By the disk formula, the volume of the cone is

**2.** Find the volume of the cylinder of height _h_ and radius _r_.

The cylinder is generated by revolving about the _x_ axis the region between the line _y_ = _r_ and the _x_ axis, between _x_ = 0 and _x_ = _h_. (See Fig. 30-2( _b_ ).) By the disk formula, the volume of the cylinder is

**3.** Find the volume of a sphere of radius _r_.

The sphere is generated by revolving about the _x_ axis the region between the semicircle   and the _x_ axis, between _x_ = − _r_ and _x_ = _r_. (See Fig. 30-2( _c_ ).) By the symmetry with respect to the _y_ axis, we can use the part of the given region between _x_ = 0 and _x_ = _r_ and then double the result. Hence, by the disk formula, the volume of the sphere is

**4.** Let   be the region between the _x_ axis, the curve _y_ = _x_ 3, and the line _x_ = 2. (See Fig. 30-10.)

( _a_ ) Find the volume of the solid obtained by revolving   about the _x_ axis.

( _b_ ) Find the volume of the solid obtained by revolving   about the _y_ axis.

**Fig. 30-10**

( _a_ ) The disk formula yields the volume

( _b_ ) ( _First solution_ ) The cylindrical shells formula yields the volume

( _Second solution_ ) Integrating along the _y_ axis and using the washer formula yields the volume

**5.** Find the volume of the solid obtained by revolving about the _y_ axis the region in the first quadrant inside the circle _x_ 2 \+ _y_ 2 = _r_ 2, and between _y_ = _a_ and _y_ = _r_ (where 0 < _a_ < _r_ ). See Fig. 30-11. (The solid is a "polar cap" of a sphere of radius _r_.)

**Fig. 30-11**

Integrating along the _y_ axis, the disk formula yields the volume

**6.** Find the volume of the solid obtained by revolving about the _y_ axis the region in the first quadrant bounded above by the parabola _y_ = 2 − _x_ 2 and below by the parabola _y_ = _x_ 2. (See Fig. 30-12.)

**Fig. 30-12**

The curves intersect at (1,1). By the difference of cylindrical shells formula, the volume is

**7.** Consider the region   bounded by the parabola _y_ = 4 _x_ 2 and the lines _x_ = 0 and _y_ = 16. (See Fig. 30-5.) Find the volume of the solid obtained by revolving   about the line _y_ = −2.

To solve this problem, we reduce it to the case of a revolution about the _x_ axis. Raise the region   vertically upward through a distance of 2 units. This changes   into a region  * that is bounded below by the parabola _y_ = 4 _x_ 2 \+ 2, on the left by the _y_ axis, and above by the line _y_ = 18. (See Fig. 30-13). Then the original solid of revolution has the same volume as the solid of revolution obtained by revolving  * around the _x_ axis. The latter volume is obtained by the washer formula:

**8.** As in problem 7, consider the region   bounded by the parabola _y_ = 4 _x_ 2 and the lines _x_ = 0 and _y_ =16. (See Fig. 30-5.) Find the volume of the solid obtained by revolving   about the line _x_ = −1.

To solve this problem, we reduce it to the case of a revolution about the _y_ axis. Move the region   to the

**Fig. 30-13**

right through a distance of 1 unit. This changes   into a region  * that is bounded on the right by the parabola _y_ = 4( _x_ − 1)2, above by _y_ = 16, and on the left by _x_ = 1. (See Fig. 30-14.) The desired volume is the same as that obtained when we revolve  * about the y axis. The latter volume is got by the difference of cylindrical shells formula:

**Fig. 30-14**

**9.** Justify the disk formula:  .

Divide the interval  _a_ , _b_ ] into _n_ equal subintervals, each of length  . (See [Fig. 30-15.) Consider

**Fig. 30-15**

the volume _V i_ obtained by revolving the region  _i_ above the _i_ th subinterval about the _x_ axis. If _m i_ and _M i_ are the absolute minimum and absolute maximum of _f_ on the _i_ th subinterval, then _V i_ lies between the volume of a cylinder of radius _m i_ and height Δ _x_ and the volume of a cylinder of radius _M i_ and height Δ _x_. Thus,   and therefore,  . (We have assumed that the volume of a cylinder of radius _r_ and height _h_ is _πr_ 2 _h_.) Hence, by the intermediate value theorem for the continuous function ( _f_ ( _x_ ))2, there exists   in the _i_ th subinterval such that   and, therefore,  . Thus,

  Letting _n_ → +∞, we obtain the disk formula.

**10.** Justify the cylindrical shells formula:  .

Divide  _a_ , _b_ ] into _n_ equal subintervals, each of length Δ _x_. (See [Fig. 30-16.) Let  _i_ be the region above the _i_ th subinterval. Let   be the midpoint   of the _i_ th interval. The solid obtained by revolving the region  _i_ about the _y_ axis is approximately the solid obtained by revolving the rectangle with base Δ _x_ and height  . The latter solid is a cylindrical shell, that is, it lies between the cylinders obtained by revolving the rectangles with the same height _f_ ( ) and with bases [0, _x_ _i_ -1] and [0, _x i_]. Hence, it has volume

Thus, the total _V_ is approximated by  , which approaches  .

**11.** Justify the cross-section formula:  .

Divide [ _a_ , _b_ ] into _n_ equal subintervals [ _x_ _i_ -1, _x i_], and choose a point   in [ _x_ _i_ -1, _x_ _i_ ]. If _n_ is large, Δ _x_ is small and the price of the solid between _x_ _i_ -1 and _x i_ will be close to a (noncircular) disk of thickness Δ _x_ and base

**Fig. 30-16**

area  . (See Fig. 30-17.) This disk has volume   Δ _x_. So _V_ is approximated by  , which approaches  .

**Fig. 30-17**

**12.** A solid has a circular base of radius 4 units. Find the volume of the solid if every plane section perpendicular to a particular fixed diameter is an equilateral triangle.

Take the circle as in Fig. 30-18, with the fixed diameter on the _x_ axis. The equation of the circle is _x_ 2 \+ _y_ 2 = 16. The cross-section _ABC_ of the solid is an equilateral triangle of side 2 _y_ and area  . Then, by the cross-section formula,

**13.** A solid has a base in the form of an ellipse with major axis 10 and minor axis 8. Find its volume if every section perpendicular to the major axis is an isosceles triangle with altitude 6.

Take the ellipse as in Fig. 30-19, with equation  . The section _ABC_ is an isosceles triangle of

**Fig. 30-18**

base 2 _y_ , altitude 6, and area  . Hence,

(Note that   _dx_ is the area of the upper half of the circle _x_ 2 + _y_ 2 = 25 and, therefore, is equal to 25π/2.)

**Fig. 30-19**

### **Supplementary Problems**

**14.** Consider the region   bounded by the parabola _y_ 2 = 8 _x_ and the line _x_ = 2. (See Fig. 30-4)

( _a_ ) Find the volume of the solid generated by revolving   about the _y_ axis.

( _b_ ) Find the volume of the solid generated by revolving   about the line _x_ = 2.

_Ans_.

**15.** Find the volume of the solid generated by revolving the region between the _x_ axis and the parabola _y_ = 4 _x_ − _x_ 2 about the line _y_ = 6.

_Ans._

**16.** Find the volume of the torus (doughnut) generated by revolving the circle ( _x_ − _a_ )2 \+ _y_ 2 = _b_ 2 about the _y_ axis, where 0 < _b_ < _a_.

_Ans._ 2π2 _ab_ 2

**17.** Consider the region   bounded by _y_ = − _x_ 2 − 3 _x_ \+ 6 and _x_ \+ _y_ = 3. Find the volume of the solid generated by revolving   about:

( _a_ ) the _x_ axis; ( _b_ ) the line _x_ = 3.

_Ans._

In Problems 18−26, find the volume generated when the given region is revolved about the given line. Use the disk formula.

**18.** The region bounded by _y_ = 2 _x_ 2, _y_ = 0, _x_ = 0, _x_ = 5, about the _x_ axis.

_Ans._ 2500π

**19.** The region bounded by _x_ 2 − _y_ 2 = 16, _y_ = 0, _x_ = 8, about the _x_ axis.

_Ans._

**20.** The region bounded by _y_ = 4 _x_ 2, _x_ = 0, _y_ = 16, about _y_ = 16. (See Fig. 30-5.)

_Ans._

**21.** The region bounded by _y_ 2 = _x_ 3, _y_ = 0, _x_ = 2, about the _x_ axis.

_Ans_. 4π

**22.** The region bounded by _y_ = _x_ 3, _y_ = 0, _x_ = 2, about _x_ = 2.

_Ans._

**23.** The region within the curve _y_ 2 = _x_ 4(1 − _x_ 2), about the _x_ axis.

_Ans._

**24.** Within the ellipse 4 _x_ 2 \+ 9 _y_ 2 = 36, about the _x_ axis.

_Ans._ 16π

**25.** Within the ellipse 4 _x_ 2 +9 _y_ 2 = 36, about the _y_ axis.

_Ans._ 24π

**26.** With in parabola _x_ = 9 − _y_ 2 and between _y_ = _x_ − 7 and the _y_ axis, about the _y_ axis.

_Ans._

In Proplems 27−32, find the volume of the solid generated by revolving the given region about the given line. Use the washer formula.

**27.** The region bounded by _y_ = 2 _x_ 2, _y_ = 0, _x_ = 0, _x_ = 5, about the _y_ axis.

_Ans._ 625π

**28.** The region bounded by _x_ 2 − _y_ 2 = 16, _y_ = 0, _x_ = 8, about the _y_ axis.

_Ans._

**29.** The region bounded by _y_ = _x_ 3, _x_ = 0, _y_ = 8, about _x_ = 2.

_Ans._

**30.** The region bounded by _y_ = _x_ 2, _y_ = 4 _x_ − _x_ 2, about the _x_ axis

_Ans_.

**31.** The region bounded by _y_ = _x 2_, _y_ = 4 _x_ − _x_ 2, about _y_ = 6.

_Ans._

**32.** The region bounded by _x_ = 9 − _y 2_, _y_ = _x_ −7, about _x_ = 4.

_Ans._

In Problems 33–37, find the volume of the solid generated by revolving the given region about the given line. Use the cylindrical shells formula.

**33.** The region bounded by _y_ = 2 _x_ 2, _y_ = 0, _x_ = 0, _x_ = 5, about _x_ = 6.

_Ans._ 375π

**34.** The region bounded by _y_ = _x_ 3, _y_ = 0, _x_ = 2, about _y_ = 8.

_Ans._

**35.** The region bounded by _y_ = _x_ 2, _y_ = 4 _x_ − _x_ 2, about _x_ = 5.

_Ans._

**36.** The region bounded by _y_ = _x_ 2 −5 _x_ \+ 6 and _y_ = 0, about the _y_ axis.

_Ans_.

**37.** The region bounded by _x_ = 9 − _y 2_, _y_ = _x_ − 7, _x_ = 0, about _y_ = 3.

_Ans._

In Problems 38–42, find the volume generated by revolving the given region about the given line. Use any appropriate method.

**38.** The region bounded by _y_ = _e_ − _x_ 2, _y_ = 0, _x_ = 0, _x_ = 1, about the _y_ axis.

_Ans._ π (1 − _e_ −1)

**39.** The region bounded by _y_ = 2 _x_ 2, _y_ = 2 _x_ \+ 4, about _x_ = 2.

_Ans._ 27π

**40.** The region bounded by _y_ = 2 _x_ , _y_ = 0, _x_ = 0, _x_ = 1, about the _y_ axis.

_Ans_.

**41.** The region bounded by _y_ = _x_ 2, _x_ = _y_ 2, about the _x_ axis.

_Ans._

**42.** The region bounded by _xy_ = 4, _y_ = ( _x_ − 3)2, about the _x_ axis.

_Ans._

**43.** Find the volume of the frustum of a cone whose lower base is of radius _R_ , Upper base is of radius _r,_ and altitude _h_.

_Ans._

**44.** A solid has a circular base of radius 4 units. Find the volume of the solid if every plane perpendicular to a fixed diameter (the _x_ axis of Fig. 30-18) is: ( _a_ ) a semicircle; ( _b_ ) a square; ( _c_ ) an isosceles right triangle with the hypotenuse in the plane of the base.

_Ans._

**45.** A solid has a base in the form of an ellipse with major axis 10 and minor axis 8. Find its volume if every section perpendicular to the major axis is an isosceles right triangle with one leg in the plane of the base.

_Ans._

**46.** The base of a solid is the first-quadrant region bounded by the line 4 _x_ \+ 5 _y_ = 20 and the coordinate axes. Find its volume if every plane section perpendicular to the _x_ axis is a semicircle.

_Ans._

**47.** The base of a solid is the circle _x_ 2 \+ _y_ 2 = 16 _x_ , and every plane section perpendicular to the _x_ axis is a rectangle whose height is twice the distance of the plane of the section from the origin. Find its volume.

_Ans._ 1024π

**48.** The section of a certain solid cut by any plane perpendicular to the _x_ axis is a circle with the ends of a diameter lying on the parabolas _y_ 2 = 4 _x_ and _x_ 2 = 4 _y_. find its volume.

_Ans._

**49.** The section of a certain solid cut by any plane perpendicular to the _x_ axis is a square with the ends of a diagonal lying on the parabolas _y_ 2 = 4 _x_ and _x_ 2 = 4 _y_. find its volume.

_Ans._

**50.** A hole of radius 1 unit is bored through a sphere of radius 3 unit, the axis of the hole being a diameter of the sphere. Find the volume of the remaining part of the sphere.

_Ans._

## **CHAPTER 31
Techniques of Integration I: Integration by Parts**

If _u_ and _v_ are functions, the product rule yields

_D x_( _uv_ ) = _uv′_ \+ _vu′_

which can be rewritten in terms of antiderivatives as follows:

Now,   can be written as  , and   can be written as  .† Thus,   and, therefore,

The purpose of integration by parts is to replace a "difficult" integration   by an "easy" integration  .

**EXAMPLE 1:** Find  .

† ∫ _uv_ ′ _dx_ = ∫ _u dv_ , where, after the integration on the right, the variable _v_ is replaced by the corresponding function of _x_. In fact, by the chain rule, _D x_(∫ _u dv_ ) = _D v_(∫ _u dv_ ) · _D xv_ = _u · v_ ′. Hence, ∫ _u dv_ = ∫ _uv_ ′ _dx_. Similarly, ∫ _v du_ = ∫ _vu_ ′ _dx_.

In order to use the integration by parts formula, we must divide the integrand _x_ ln _x dx_ into two "parts" _u_ and _dv_ so that we can easily find _v_ by an integration and also easily find ∫ _v du_. In this example, let _u_ = ln _x_ and _dv_ = _x dx_. Then we can set   and note that  . So, the integration by parts formula yields:

Integration by parts can be made easier to apply by setting up a rectangle such as the following one for Example 1.

In the first row, we place _u_ and _dv_. In the second row, we place the results of computing _du_ and _v_. The desired result of the integration parts formula   can be obtained by first multiplying the upper left corner _u_ by the lower right corner _v_ , and then subtracting the integral of the product _v du_ of the two entries _v_ and _du_ in the second row.

**EXAMPLE 2:** Find  .

Let _u_ = _x_ and _dv_ = _e x dx_. We can picture this in the box below.

Then,

**EXAMPLE 3:** Find ∫ _e x_ cos _x dx_.

Let _u_ = _e x_ and _dv_ = cos _x dx_. Then we get the box

So,

Now we have the problem of finding ∫ _e x_ sin _x dx_ , which seems to be just as hard as the original integral ∫ _e x_ cos _x dx_. However, let us try to find ∫ _e x_ sin _x dx_ by another integration by parts. This time, Let _u_ = _e x_ and _dv_ = sin _x dx_.

Then,

Substituting in formula ( _1_ ) above, we get:

Adding ∫ _e x_ cos _x dx_ to both sides yields 2∫ _e x_ cos _x dx_ = _e x_ sin _x_ \+ _e x_ cos _x_. So,

We must add an arbitary constant:

Notice that this example required an iterated application of integration by parts.

### **Solved Problems**

**1.** Find  .

Let _u_ = _x_ 2 and  . Note that _v_ can be evaluated by using the substitution _w_ = _x_ 2. (We get  .)

Hence,

**2.** Find  .

Let _u_ = ln( _x_ 2 \+ 2) and _dv_ = _dx_.

So,

**3.** Find  .

Let _u_ = In _x_ and _dv_ = _dx_.

so,

**4.** Find  .

We have three choices: ( _a_ ) _u_ = _x_ sin _x_ , _dv_ = _dx_ ; ( _b_ ) _u_ = sin _x_ , _dv_ = _x dx_ ; ( _c_ ) _u_ = _x_ , _dv_ = sin _x dx_.

( _a_ ) Let _u_ = _x_ sin _x_ , _dv_ = _dx_. Then _du_ = (sin _x_ \+ _x_ cos _x_ ) _dx, v_ = _x_ , and

The resulting integral is not as simple as the original, and this choice is discarded.

( _b_ ) Let _u_ = sin _x_ , _dv_ = _x dx_. Then _du_ = cos _x dx_ ,  , and

The resulting integral is not as simple as the original, and this choice too is discarded.

( _c_ ) Let _u_ = _x, dv_ = sin _x dx_. Then _du_ = _dx, v_ = −cos _x_ , and

**5.** Find  .

Let _u_ = 1n _x_ , _dv_ = _x_ 2 _dx_. Then  , and

**6.** Find  .

Let _u_ = sin−1 _x_ , _dv_ = _dx_.

So,

**7.** Find  .

let _u_ = tan−1 _x_ , _dv_ = _dx_.

So,

**8.** Find  .

Let _u_ = sec _x_ , _dv_ = sec2 _x dx_.

Thus,

Then,

Hence,

**9.** Find  .

Let _u_ = _x_ 2, _dv_ = sin _x dx_. Thus, _du_ = 2 _x dx_ and _v_ = −cos _x_. Then

Now apply integration by parts to  , with _u_ = _x_ and _dv_ = cos _x dx_ , getting

Hence,

**10.** Find  .

Let _u_ = _x_ 3, _dv_ = _e_ 2 _x_ _dx_. Then  , and

For the resulting integral, let _u_ = _x_ 2 and _dv_ = _e_ 2 _x_ _dx_. Then _du_ = 2 _x dx_ ,  , and

For the resulting integral, let _u_ = _x_ and _dv_ = _e_ 2 _x_ _dx_. Then _du_ = _dx_ ,  , and

**11.** Derive the following _reduction formula_ for  .

Let _u_ = sin _m_ −1 _x_ and _dv_ = sin _x dx_.

Then

Hence,

and division by _m_ yields the required formula.

**12.** Apply the reduction formula of Problem 11 to find  .

When _m_ = 2, we get

**13.** Apply the reduction formula of Problem 11 to find  .

When _m_ = 3, we get

### **Supplementary Problems**

In Problems 14−21, use integration by parts to verify the specified formulas.

**14.**

**15.**

**16.**

**17.**

**18.**

**19.**

**20.**

**21.**

**22.** Show that   for any positive integer _n_.

**23.** Prove the following reduction formula:  .

**24.** Apply Problem 23 to find  .

_Ans_.

**25.** Prove the reduction formula:

**26.** Apply Problem 25 to find  .

_Ans_.

**27.** Prove  .

**28.** Prove the reduction formula:  .

**29.** Use Problem 28 and Example 2 to show that:  .

## **CHAPTER 32
Techniques of Integration II: Trigonometric Integrands and Trigonometric Substitutions**

**TRIGONOMETRIC INTEGRANDS**

**1.** Let us consider integrals of the form  , where _k_ and _n_ are nonnegative integers.

_Type 1. At least one of sin _x_ and cos _x_ occurs to an odd power_: Then a substitution for the other function works.

**EXAMPLE 1:**  .

Let _u_ = cos _x_. Then _du_ = −sin _x dx_. Hence,

**EXAMPLE 2:**

Let _u_ = sin _x_. Then _du_ = cos _x dx_ , and

**EXAMPLE 3:**  .

Let _u_ = cos _x_. Then _du_ = −sin _x dx_ and

_Type 2. Both powers of_ sin _x and_ cos _x are even_ : This always involves a more tedious computation, using the identities

**EXAMPLE 4**

**2.** Let us consider integrals of the form  , where _k_ and _n_ are nonnegative integers. Recall that sec2 _x_ = 1 + tan2 _x_.

_Type 1. n is even_ : Substitute _u_ = tan _x_.

**EXAMPLE 5:**

Let _u_ = tan _x, du_ = sec2 _x dx_. So,

_Type 2. n is odd and k is odd_ : Substitute _u_ = sec _x_.

**EXAMPLE 6:**  .

Let _u_ = sec _x, du_ = sec _x_ tan _x dx_. So,

_Type 3. n is odd k is even:_ This case usually requires a tedious calculation.

**EXAMPLE 7**

**3.** Let us consider integrals of the form  , and  . We shall need the identities

**EXAMPLE 8**

**EXAMPLE 9**

**EXAMPLE 10**

**TRIGONOMETRIC SUBSTITUTIONS.** There are three principal kinds of trigonometric substitutions. We shall introduce each one by means of a typical example.

**EXAMPLE 11:** Find  .

Let _x_ = 2 tan _θ_ , that is _θ_ = tan−1 ( _x_ /2). Then

By definition of the inverse tangent, −π/2 < _θ_ < π/2. So, cos _θ_ > 0 and, therefore, sec _θ_ > 0. Thus,  . Hence,

Now we must evalute sin _θ_.

_Analytic method_ :

_Geometric method:_ Draw the right triangle shown in Fig. 32-1. From this triangle we see that  .(Note that it follows also for _θ_ < 0.)

Hence,

**Fig. 32-1**

This example illustrates the following general rule:

**Strategy I**. If   occurs in an integrand, try the subsititution _x_ = _a_ tan _θ_.

**EXAMPLE 12:** Find  .

Let _x_ = 3 sin _θ_ , that is _θ_ = sin−1 ( _x_ /3). Then _dx_ = 3 cos _θ dθ_ and

By definition of the inverse sine, −π/2 < _θ_ < π/2 and, therefore, cos _θ_ > 0. Thus,  . Now,

This example illustrates the following general method:

**Strategy II.** If   occurs in an integrand, try the subsitution _x_ = _a_ sin _θ._

**EXAMPLE 13:** Find  .

Let _x_ = 2 sec _θ_ , that is, _θ_ = sec-1 ( _x_ /2). Then _dx_ = 2 sec _θ_ tan _θ dθ_ and

By definition of the inverse secant, _θ_ is in the first or third quadrant and, therefore, tan _θ_ > 0. So,  . Now,

This example illustrates the following general method:

**Strategy III.** If   occurs in an integrand, try the substitution _x_ = _a_ sec _θ_.

### **Solved Problems**

In Problems 1-23, verify the given solutions. Recall the identities

**1.**

**2.**

**3.**

**4.**

**5.**

**6.**

**7.**

**8.**

**9.**

**10.**

**11.**

**12.**

**13.**

**14.**

**15.**

**16.**

**17.**

**18.**

**19.**

**20.**

**21.**

**22.**

**23.**

**24.** Find  .

 . So, let  . Then

Hence,

But

So,

**25.** Find  .

Let  . (See Fig. 32-2.) Then  . Hence,

**Fig. 32-2**

**26.** Find  .

Let  . (See Fig. 32-3.) Then   and  . Hence,

**Fig. 32-3**

**28.** Find  .

Let _x_ − 1 = sin _θ_. (See Fig. 32-4.) Then _dx_ = cos _θ dθ_ and  . Hence,

**Fig. 32-4**

**29.**  .

Let  . (See Fig. 32-5). Then   and  . So,

**Fig. 32-5**

### **Supplementary Problems**

**30.**

**31.**

**32.**

**33.**

**34.**

**35.**

**36.**

**37.**

**38.**

**39.**

 **40.**

**41.**

**42.**

**43.**

**44.**

**45.**

**46.**

**47.**

**48.**

**49.**

**50.**

**51.**

**52.**

**53.**

**54.**

**55.**

**56.**

**57.**

**58.**

**59.**

**60.**

**61.**

**62.**

**63.**

**64.**

**65.**

**66.**

**67.**

**68.**

**69.**

**70.**

**71.**

**72.**

In Problems 73 and 74, first apply integration by parts.

**73.**

**74.**

## **CHAPTER 33
Techniques of Integration III:
Integration by Partial Fractions**

We shall give a general method for finding antiderivatives of the form  , where _N_ ( _x_ ) and _D_ ( _x_ ) are polynomials. A function of the form   is called a _rational function_. ( _N_ ( _x_ ) is the numerator and _D_ ( _x_ ) is denominator.) As examples, consider

Two restrictions will be assumed, neither of which limits the applicability of our method: ( _i_ ) the leading coefficient (the coefficient of the highest power of _x_ ) in _D_ ( _x_ ) is +1; ( _ii_ ) _N_ ( _x_ ) is of lower degree than _D_ ( _x_ ). A quotient _N_ ( _x_ )/ _D_ ( _x_ ) that satisfies ( _ii_ ) is called a _proper_ rational function. Let us see that the restrictions ( _i_ )–( _ii_ ) are not essential.

**EXAMPLE 1** : Consider the case where   is  . Here, our first restriction is not satisfied. However, note that

The integral on the right side satisfies restrictions ( _i_ ) and ( _ii_ ).

**EXAMPLE 2** : Consider the case where   is  . Here, our second restriction is not satisfied. But we can divide _N_ ( _x_ ) by _D_ ( _x_ ):

Hence,

and the problem is reduced to evaluating  , which satisfies our restrictions.

A polynomial is said to be _irreducible_ if it is not the product of two polynomials of lower degree.

Any linear polynomial _f_ ( _x_ ) = _ax_ \+ _b_ is automatically irreducible, since polynomials of lower degree than _f_ ( _x_ ) are constants and _f_ ( _x_ ) is not the product of two constants.

Now consider any quadratic polynomial _g_ ( _x_ ) = _ax 2 \+ bx_ \+ _c_. Then

_g_ ( _x_ ) is irreducible if and only if _b_ 2 − 4 _ac_ < 0

To see why this is so, assume that _g_ ( _x_ ) is reducible. Then _g_ ( _x_ ) = ( _Ax_ \+ _B_ )( _Cx_ \+ _D_ ). Hence, _x_ = − _B/A_ and _x_ = − _D/C_ are roots of _g_ ( _x_ ). The quadratic formula

should yield these roots. Therefore, _b_ 2 − 4 _ac_ cannot be negative. Conversely, assume _b_ 2 − 4 _ac_   0. Then the quadratic formula yields two roots of _g_ ( _x_ ). But, if _r_ is a root of _g_ ( _x_ ) then _g_ ( _x_ ) is divisible by _x_ − _r_.† Hence, _g_ ( _x_ ) is reducible.

**EXAMPLE 3:**

( _a_ ) _x_ 2 \+ 4 is irreducible, since _b_ 2 − 4 _ac_ = 0 − 4(1)(4) = −16 < 0.

( _b_ ) _x_ 2 \+ _x_ − 4 is reducible, since _b_ 2 − 4 _ac_ = 1 − 4(1)(−4) = 17   0.

We will assume without proof the following fairly deep property of polynomials with real coefficients.

**Theorem 1:** Any polynomial _D_ ( _x_ ) with leading coefficient 1 can be expressed as a product of linear factors of the form _x_ − _a_ and of irreducible quadratic factors of the form _x_ 2 \+ _bx_ \+ _c_. (Repetition of factors is permitted.)

**EXAMPLE 4**

( _a_ ) _x_ 3 − 4 _x_ = _x_ ( _x_ 2 − 4) = _x_ ( _x_ −2)( _x_ \+ 2)

( _b_ ) _x_ 3 \+ 4 _x_ = _x_ ( _x_ 2 \+ 4) ( _x_ 2 \+ 4 is irreducible.)

( _c_ )   ( _x_ 2 \+ 3 is irreducible.)

( _d_ ) _x_ 3 − 3 _x_ 2 − _x_ \+ 3 = ( _x_ \+ 1)( _x_ −2)2

**METHOD OF PARTIAL FRACTIONS.** Assume that we wish to evaluate  , where   is a proper rational function and _D_ ( _x_ ) has leading coefficient 1. First, write _D_ ( _x_ ) as a product of linear and irreducible quadratic factors.

Our method will depend on this factorization. We will consider various cases and, in each case, we will first explain the method by means of an example and then state the general procedure.

†In general, if a polynomial _h_ ( _x_ ) has _r_ as a root, then _h_ ( _x_ ) must be divisible by _x_ − _r_.

**Case I:** _D_ ( _x_ ) is a product of distinct linear factors.

**EXAMPLE 5:** Find  .

In this case, _D_ ( _x_ ) = _x_ 2 − 4 = ( _x_ −2)( _x_ \+ 2). Write

It is assumed that _A_ and _B_ are certain constants, that we must now evaluate. Clear the denominators by multiplying both sides ( _x_ −2)( _x_ \+ 2):

First, substitute −2 for _x_ in ( _1_ ): 1 = _A_ (0) + _B_ (−4) = −4 _B_. Thus,  .

Second, substitute 2 for _x_ in ( _1_ ): 1 = _A_ (4) + _B_ (0) = 4 _A_. Thus,  . Hence,

So,

**EXAMPLE 6:** Find  .

Factoring the denominator yields _x_ ( _x_ 2 \+ _x_ − 6) = _x_ ( _x_ −2)( _x_ \+ 3). The integrand is  . Represent it in the following form:

Clear the denominators by multiplying by _x_ ( _x_ −2)( _x_ \+ 3):

Let _x_ be 0 in (2): 1 = _A_ (−2)(3) + _B_ (0)(3) + _C_ (0)(−2) = −6 _A_. So  .

Let _x_ be 0 in (2): 3 = _A_ (0)(5) + _B_ (2)(5) + _C_ (2)(0) = 10 _B_. So,  .

Let _x_ be −3 in (2): −2 = _A_ (−5)(0)+ _B_ (−3)(0) + _C_ (−3)(−5) = 15 _C_. So,  .

Hence,

**General Rule for Case I.** Represent the integrand as a sum of terms of the form   for each linear factor _x_ − _a_ of the denominator, where _A_ is an unknown constant. Solve for the constants, Integrating yields a sum of terms of the form _A_ ln | _x_ − _a_ |.

_Remark:_ We assume without proof that the integrated always has a representation of the required kind. For every particular problem, this can be verified at the end of the calculation.

**Case II:** _D_ ( _x_ ) is a product of linear factors, some of which occur more than once.

**EXAMPLE 7:** Find  .

First factor the denominator:†

Then represent the integrand   as a sum of the following form:

Note that, for the factor, ( _x_ − 1) that occurs twice, there are terms with both ( _x_ − 1) and ( _x_ − 1)2 in the denominator. Now clear the denominators by multiplying both sides by ( _x_ \+ 1)( _x_ − 1)2:

Let _x_ = 1. Then 8 = (0) _A_ \+ (2)(0) _B_ \+ (2) _C_ = 2 _C_. Thus, _C_ = 4.

Let _x_ = −1. Then 2 = (4) _A_ \+ (0)(−2) _B_ \+ (0) _C_ = 4 _A_. Thus,  .

To find _B_ , compare the coefficients of _x_ 2 on both sides of ( _1_ ). On the left it is 0, and on the right it is _A_ \+ _B_. Hence, _A_ \+ _B_ = 0. Since  . Thus,

Therefore,

**By Quick Formula I,**

So,

**EXAMPLE 8:** Find  .

Represent the integrand   in the following form:

Clear denominators by multiplying by _x_ 3( _x_ −2)2:

Let _x_ = 0. Then 1 = 4 _C_. So,  .

Let _x_ = 2. Then 3 = 8 _E_. So,  .

Compare coefficients of _x_. Then 1 = 4 _B_ − 4 _C_. Since  ,  .

Compare coefficients of _x_ 2. Then 0 = 4 _A_ −4 _B_ \+ 4 _C_. Since   and  ,  .

† In trying to find linear factors of a denominator that is a polynomial with integral coefficients, test each of the divisors _r_ of the constant term to see whether it is a root of the polynomial. If it is, then _x_ − _r_ is a factor of the polynomial. In the given example, the constant term is 1. Both of its divisors, 1 and −1, turn out to be roots.

Compare coefficients of _x_ 4. 0 = _A_ \+ _D_. Since  ,  .

So,

and

**General Rule for Case II:** For each repeated linear factor ( _x_ − _r_ ) that occurs _k_ times in the denominator, use   as part of the representation of the integrand. Every linear factor that occurs only once is handled as in Case I.

**Case III:** _D_ ( _x_ ) is a product of one or more distinct irreducible quadratic factors and possibly also some linear factors (that may occur more than once).

**General Rule for Case III:** Linear factors are handles as in Cases I–II. For each irreducible quadratic factor _x_ 2 \+ _bx_ \+ _c_ , place a term   in the representation of the integrand.

**EXAMPLE 9:** Find  .

Represent the integrand as follows:

Clear the denominators by multiplying by _x_ ( _x_ 2 \+ 1)( _x_ 2 \+ 2).

Multiply out on the right:

Comparing coefficients, we get:

2 _A_ = −1, 2 _C_ \+ _E_ = 1, 3 _A_ \+ _C_ \+ _D_ = 0, _B_ \+ _E_ = 0, _A_ \+ _B_ \+ _D_ = 0

Hence,   and, therefore,  ,  . From the latter two equations, _C_ − _B_ = 1. From 2 _C_ \+ _E_ = 1 and _B_ \+ _E_ = 0, we get 2 _C_ − _B_ = 1. Now, from _C_ − _B_ = 1, we get _C_ = 0. Hence, from _C_ − _B_ = 1, _B_ = −1. Then, from  , it follows that  . Finally, from _B_ \+ _E_ = 0, _E_ = 1.

Thus,

Hence,

Now,

Also,

Therefore,

**Case IV:** D( _x_ ) is a product of zero or more linear factors and one or more irreducible quadratic factors.

**General Rule for Case IV:** Linear factors are handled as in Cases I–II. For each irreducible quadratic factor _x_ 2 \+ _bx_ \+ _c_ that occurs to the _k_ th power, insert as part of the representation of the integrand.

**EXAMPLE 10:** Find  .

Let  . Then

Compare coefficients: _A_ = 0, _B_ = 2, _A_ \+ _C_ = 0, _B_ \+ _D_ = 3. Hence, _C_ = 0, _D_ = 1. Thus,

In the second integral, let _x_ = tan _θ_. Then

Thus,

### **Solve Problems**

**1.** Find  .

The integrand is an improper fraction. By division,

We write   and obtain

For _x_ = 0, 1 = − _B_ and _B_ = −1. For _x_ = 1, 2 = _C_. For _x_ = 2, 3 = 2 _A_ \+ _B_ \+ 4 _C_ and _A_ = −2. Thus,

**2.** Find  .

Let  . Clear the denominators:

Let _x_ = −2. Then −2 = _A_. Let _x_ = −3. Then −3 = − B. So, _B_ = 3.

**3.** Find  .

Let  . Clear the denominators:

Let _x_ = 0. Then 2 = −2 _A_. So, _A_ = −1. Let _x_ = −2. Then 6 = 6 _B_. So, _B_ = 1. Let _x_ = 1. Then 3 = 3 _C_. So, _C_ = 1. Hence,

**4.** Find  .

Let  . Clear the denominators:

_x_ 3 \+ 1 = _A_ ( _x_ − 1)3 \+ _B_ ( _x_ \+ 2)( _x_ − 1)2 \+ _C_ ( _x_ \+ 2)( _x_ − 1) + _D_ ( _x_ \+ 2)

Let _x_ = −2. Then −7 = −27 _A_. So,  . Let _x_ = 1. Then 2 = 3 _D_. So,  . Compare coefficients of _x_ 3. Then 1 = _A_ \+ _B_. Since  ,  . Compare coefficients of _x_ 2. 0 = −3 _A_ \+ _C_. Since  ,  .

Thus,

**5.** Find  .

Hence _A_ \+ _C_ = 1, _B_ \+ _D_ = 1, 2A + _C_ = 1, and 2 _B_ \+ _D_ = 2. Solving simultaneously yields _A_ = 0, _B_ = 1, _C_ = 1, _D_ = 0. Thus,

**6.** Find  .

from which _A_ = 1, _B_ = −1, _C_ = 0, _D_ = 0, _E_ = 4, _F_ = 0. Thus the given integral is equal to

By Quick Formula II,

and by Quick Formula I,

So,

### **Supplementary Problems**

In Problems 7–25, evaluate the given integrals.

**7.**

**8.**

**9.**

**10.**

**11.**

**12.**

**13.**

**14.**

**15.**

**16.**

**17.**

**18.**

**19.**

**20.**

**21.**

**22.**

**23.**

**24.**

**25.**

## **CHAPTER 34
Miscellaneous Substitutions**

**I.** Assume that, in a rational function, a variable is replaced by one of the following radicals.

**1.**  . Then the substitution _ax + b = z n_ will produce a rational function. (See Problems 1–3.)

**2.**  . Then the substitution _q + px + x 2_ = ( _z − x_ )2 will yield a rational function. (See Problem 4.)

**3.**  . Then the substitution _q + px − x_ 2 = ( _α + x_ )2 _z_ 2 will produce a rational function. (See Problem 5.)

**II.** Assume that, in a rational function, some variables are replaced by sin _x_ and/or cos _x_. Then the substitution _x_ = 2 tan-1 _z_ will produce an integral of a rational function of _z_.

The reason that this will happen is that

(See Problem 6 for a derivation of the first two equations.)

In the final result, replace _z_ by tan( _x_ /2). See Problems 7–10.)

### **Solved Problems**

**1.** Find

Let 1 − _x_ = _z_ 2. Then _x_ = 1 − _z_ 2, _dx_ = −2 _z dz_ , and

By integration by partial fractions, one obtains

**2.** Find
 .

Let _x_ \+ 2 = _z_ 2. Then _x_ = _z_ 2 −2, _dx_ = 2 _z dz_ , and

By integration by partial fractions, we get

**3.** Find
 .

Let x = z4. Then _dx_ = 4z3 _dz_ and

**4.** Find
 .

Let _x_ 2 \+ _x_ \+ 2 = (z−x)2. Then

and

The equation   was obtained by integration by partial fractions.

**5.** Find
 .

Let 5 − 4x − _x_ 2 = (5 + _x_ )(1 − _x_ ) =(1 − _x_ )2 _z_ 2. Then

and

**6.** Given
 , that is, _x_ = 2 tan−1 _z_ , show that

Since

solving for cos _x_ yields   Also,

**7.** Find
 .

Let _x_ = 2 tan−1 _z_. Using equations (34.1), we get

**8.** Find
 .

Let _x_ = 2 tan−1 _z_. Using equations (34.1), we get

**9.** Find
 .

Let _x_ = 2 tan−1 _z_. Using equations (34.1), we obtain

**10.** Find
 .

Let _x_ = 2 tan−1 _z_. Using equations (34.1), we obtain

**11.** Use the substitution 1 − _x_ 3 = _z_ 2 to find  .

The substitution yields _x_ 3 = 1 − _z_ 2, 3 _x_ 2 _dx_ = −2 _z dz_ , and

**12.** Use _x_ = 1/ _z_ to find  .

The substitution yields   and

Let _z_ − 1 = _s_ 2. Then

**13.** Find  .

Let _u_ = _x_ 1/6, so that _x_ = _u_ 6, _dx_ = 6 _u_ 5 _du, x_ 1/2 = _u_ 3, and _x_ 1/3 = _u_ 2. Then we obtain

### **Supplementary Problems**

In Problems 14–39, evaluate the given integral.

**14.**

**15.**

**16.**

**17.**

**18.**

**19.**

**20.**

**21.**

**22.**

**23.**

**24.**

**25.**

**26.**

**27.**

**28.**

**29.**

**30.**

**31.**

**32.**

**33.**

**34.**

**35.**

**36.**

**37.**

**38.**

**39.**

**40.** (GC) Use a graphing calculator to approximate (to eight decimal places)   and compare your result with the value obtained by the methods of this chapter.

**41.** (GC) Use a graphing calculator to approximate (to eight decimal places)   and compare your result with the value obtained by the methods of this chapter.

## **CHAPTER 35
Improper Integrals**

For a definite integral   to be defined, it suffices that _a_ and _b_ are real numbers and that _f_ ( _x_ ) is continuous on [ _a, b_ ]. We shall now study two different kinds of integrals that we shall call _improper integrals_.

**INFINTE LIMITS OF INTEGRATION**

( _a_ )

See Problems 1 – 3 and 5–6.

( _b_ )

See Problem 4.

( _c_ )

provided that _both_ limits on the right exist. See Problem 7.

**DISCONTINUITIES OF THE INTEGRAND**

( _a_ ) If _f_ is continuous on [ _a, b_ ] except that it is not continuous from the right at _a_ , then

See Problem 16.

( _b_ ) If _f_ is continuous on [ _a, b_ ] except that it is not continuous from the left at _b_ , then

See Problems 9, 10, 12, 14 and 15.

( _c_ ) If _f_ is continuous on [ _a, b_ ] except at a point _c_ in ( _a, b_ ), then

provided that _both_ integrals on the right exist. See Problems 11 and 13.

When the limit defining an improper integral exists, we say that the integral is _convergent._ In the opposite case, we say that the integral is _divergent._ If the integral is divergent, we say that it is equal to + ∞ (respectively −∞) if the limit defining the improper integral approaches + ∞ (respectively −∞).

### **Solved Problems**

**1.** Evaluate  .

_Note_ : The integral   can be interpreted as the area of the region under the curve y = 1/ _x_ 2 and above the _x_ axis, for _x_ > 1. Thus, a region that is infinite (in the sense of being unbounded) can have a finite area.

**2.** Evaluate  .

Thus, the integral diverges to +∞.

**3.** Show that   converges for _p_ > 1 and diverges to +∞ for _p_  1.

Assume _p_ > 1. Then we have  .

By Problem 2, we already know that   diverges to + ∞. So, assume _p_ < 1. Then we have

**4.** Evaluate  .

**5.** Evaluate  .

**6.** Evaluate  .

As _c_ → +∞, _e_ − _c_ → 0, while sin _c_ and cos _c_ oscillate between −1 and 1. Hence,   and, therefore,

**7.** Evaluate  .

Similarly,

Thus,

**8.** Find the area of the region lying to the right of _x_ = 3 and between the curve   and the _x_ axis.

The area

**9.** Evaluate  .

The integrand is discontinuous at _x_ = 3. So,

**10.** Evaluate  .

The integrand is discontinuous at _x_ = 2.

Hence, the integral diverges to +∞.

**11.** Evaluate  .

The integrand is discontinuous at _x_ = 1, which is inside (0,4). (See Fig. 35-1.)

Hence,   is divergent. (We do not have to consider   at all. For   to be convergent, both   and   must exist.)

**Fig. 35-1**

**Fig. 35-2**

**12.** Find the area of the region between the curve  , the _x_ axis, and _x_ = 0 and _x_ = 1. (See Fig. 35-2.)

The area is

**13.** Evaluate  .

The integrand is discontinuous at _x_ = 1, which lies inside (0,4).

On the other hand,

Hence,

**14.** Evaluate  .

The integrand is discontinuous at  .

Since

Thus,   diverges to +∞.

**15.** Evaluate  .

The integrand is discontinuous at  .

**16.** Evaluate  .

The integrand is discontinuous at _x_ = 0.

### **Supplementary Problems**

**17.** Evaluate the given integrals:

**18.** Find the area of the region between the given curve and its asymptotes:

_Ans_. ( _a_ ) 4 _π_ ; ( _b_ ) 4 _π_ ; ( _c_ ) 2 _π_

**19.** Evaluate the given integrals:

**20.** Find the area of the region between the given curve and its asymptote:

_Ans_. ( _a_ ) 4 _π_ ( _b_ )  ; ( _c_ ) 2

**21.** Find the area of the following regions:

( _a_ ) Above the _x_ axis, under   and to the right of _x_ = 3.

( _b_ ) Above the _x_ axis, under   and to the right of _x_ = 2.

_Ans_. ( _a_ )  ; ( _b_ ) 1 − ln2

**22.** Show that the areas of the following regions are infinite:

( _a_ ) Above the _x_ axis, under   from _x_ = −2 to _x_ = 2.

( _b_ ) Above the _x_ axis, under _xy_ = 9 and to the right of _x_ = 1.

**23.** Show that the area of the region in the first quadrant under  , and that the volume generated by revolving that region about the _x_ axis is  .

**24.** Find the length of the indicated arc: ( _a_ ) 9 _y_ 2 = _x_ (3 − _x_ )2, a loop; ( _b_ ) _x_ 2/3 \+ _y_ 2/3 = _a_ 2/3, entire length; ( _c_ ) 9 _y_ 2 = _x_ 2(2 _x_ \+ 3), a loop

_Ans_. ( _a_ )   units; ( _b_ ) 6 _a_ units; ( _c_ )   units

**25.** Show that   converges for _p_ < 1 and diverges to +∞ for _p_   1.

**26.** Let 0  _f_ ( _x_ )  _g_ ( _x_ ) for _a_  _x_ < _b_. Assume that   and  . (See Fig. 35-3.) It is not hard to show that, if   converges, then so does   (and, equivalently, if   does not converge, then neither does  ). A similar result also holds for _a_ < _x_   _b_ , with   replacing  .

**Fig. 35-3**

As an example, consider  . For 0   _x_ < 1,

Since   does not converge, neither does  .

Now consider   For 0 < _x_   1,   Since   converges, so does

Determine whether each of the following converges:

_Ans_. ( _a_ ) and ( _c_ ) converge

**27.** Assume that 0  _f_ ( _x_ )  _g_ ( _x_ ) for _x_   _a_. Assume also that  . (See Fig. 35-4.) It is not hard to show that, if   converges, so does   (and, equivalently, that, if   does not converge, then neither does  ).

**Fig. 35-4**

As an example, consider  . For _x_   1,  . Since   converges, so does  .

Determine whether or not each of the following converges:

_Ans_. all converge

_28._ Define the gamma function   for _t_ > 0. It can be proved that Γ( _t_ ) is convergent. (This is left as a project for the student.)

( _a_ ) Show that Γ(1) = 1.

( _b_ ) Show that Γ(2) = 1. ( _Hint_ : Use integration by parts.)

( _c_ ) Prove that Γ( _t_ \+ 1) = _t_ Γ( _t_ ) for all _t_ > 0. ( _Hint_ : Use integration by parts.)

( _d_ ) Use part ( _c_ ) to show that Γ( _n_ \+ 1) = _n_! for all positive integers _n_. (Recall that _n_! = 1.2.3.4..... _n_.)

## **CHAPTER 36
Applications of Integration III: Area of a Surface of Revolution**

If an arc of a curve is revolved about a line that does not intersect the arc, then the resulting surface is called a _surface of revolution_. By the _surface area_ of that surface we mean the area of its outer surface.

Let _f_ be a continuous function on [ _a, b_ ] that is differentiable in ( _a, b_ ) and such that _f_ ( _x_ )   0 for a   _x_   _b_. Then the surface area _S_ of the surface of revolution generated by revolving the graph of _f_ on [ _a, b_ ] about the _x_ axis is given by the formula

For a justification of this formula, see Problem 11.

There is another formula like (36.1) that is obtained when we exchange the roles of _x_ and _y_. Let _g_ be a continuous function on [ _c, d_ ] that is differentiable on ( _c, d_ ) and such that _g_ ( _y_ )   0 for _c_   _y_   _d_. Then the surface area _S_ of the surface of revolution generated by revolving the graph of _g_ on [ _c, d_ ] about the _y_ axis is given by the formula:

Similarly, if a curve is given by parametric equations _x_ = _f_ ( _u_ ), _y_ = _g_ ( _u_ )(see Chapter 37), and, if the arc from _u_ = _u_ 1 to _u_ = _u_ 2 is revolved about the _x_ axis, then the surface area of the resulting surface of revolution is given by the formula

Here, we have assumed that _f_ and _g_ are continuous on [ _u_ 1, _u_ 2] and differentiable on ( _u_ 1, _u_ 2), and that _y_ = _g_ ( _u_ )   0 on [ _u_ 1, _u_ 2]. Another such formula holds in the case of a revolution around the _y_ axis.

### **Solved Problems**

**1**. Find the area _S_ of the surface of revolution generated by revolving about the _x_ axis the arc of the parabola _y_ 2 = 12 _x_ from _x_ = 0 to _x_ = 3.

By implicit differentiation,

By (36.1),

**2**. Find the area S of the surface of revolution generated by revolving about the _y_ axis the arc of _x_ = _y_ 3 from _y_ = 0 to _y_ = 1.

**3**. Find the area of the surface of revolution generated by revolving about the _x_ axis the arc of _y_ 2 \+ 4 _x_ = 2 In _y_ from _y_ = 1 to _y_ = 3.

**4**. Find the area of the surface of revolution generated by revolving a loop of the curve 8a2y2 = _a_ 2 _x_ 2 − _x_ 4 about the _x_ axis. (See Fig. 36-1.)

**Fig. 36-1**

Here

Hence

**5**. Find the area of the surface of revolution generated by revolving about the _x_ axis the ellipse  .

**6**. Find the area of the surface of revolution generated by revolving about the _x_ axis the hypocycloid _x_ = _a_ cos3 θ, _y_ = _a_ sin3 θ.

The required surface is generated by revolving the arc from θ = 0 to θ = π. We have  , and  . Then

**7**. Find the area of the surface of revolution generated by revolving about the _x_ axis the cardioid _x_ = 2 cos θ − cos 2θ, _y_ = 2 sin θ − sin 2θ.

The required surface is generated by revolving the arc from θ = 0 to θ = π. (See Fig. 36-2.) We have

**Fig. 36-2**

and

Then

**8**. Show that the surface area of a cylinder of radius _r_ and height _h_ is 2π _rh_.

The surface is generated by revolving about the _x_ axis the curve _y = r_ from _x_ = 0 to _x = h_. Since  ,  . Then, by (36.1),

**9**. Show that the surface area of a sphere of radius _r_ is 4π _r_ 2.

The surface area is generated by revolving about the _x_ axis the semicircle   from _x_ = − _r_ to _x_ = _r_. By symmetry, this is double the surface area from _x_ = 0 to _x_ = _r_. Since _y_ 2 = _r_ 2 − _x_ 2,

Hence, by (36.1),

**10**. ( _a_ ) Show that the surface area of a cone with base _r_ and slant height _s_ (see Fig. 36-3) is π _rs_.

( _b_ ) Show that the surface area of a frustum of a cone with bases _r_ 1 and _r_ 2 and slant height _u_ (see Fig. 36-4) is π( _r_ 1 \+ _r_ 2) _u_. (Note that the _frustum_ is obtained by revolving the slant height around the base of the triangle.)

( _a_ ) Cut open the cone along a slant height and open it up as part of a circle of radius _s_ (as show in Fig. 36-5). Note that the portion of the circumference cut off by this region is 2π _r_ (the circumference of the base of the cone). Now the desired area _S_ is the difference between π _s_ 2 (the area of the circle in Fig. 36-5) and the

**Fig. 36-3**

**Fig. 36-4**

area _A_ 1 of the circular sector with central angle θ. This area _A_ 1 is  . Since the arc cut off by θ is 2π _s_ −2π _r_ , we get _ _. Thus, _A_ 1 = π( _s_ − _r_ ) _s_. Hence, _S = πs 2 − π(s − r)s = πrs_ square units.

**Fig. 36-5**

( _b_ ) From the similar triangles in Fig. 36-4, we get  . Then r2 _u_ = _r_ 1 _u_ 1 \+ _r_ 1 _u_. So,  . Now, by part ( _a_ ), the surface area of the frustum is π _r_ 2( _u_ 1 \+ _u_ ) − π _r_ 1 _u_ 1 = π( _r_ 2 − _r_ 1) _u_ 1 \+ π _r_ 2 _u_ = π _r_ 1 _u_ \+ π _r_ 2 _u_ = π ( _r_ 1 \+ _r_ 2) _u_ square units.

**11**. Sketch a derivation of formula (36.1).

Assume that  _a, b_ ] is divided into _n_ equal subintervals, [ _x k-1, xk_], each of length  . The total surface area _S_ is the surface areas _S k_ generated by the arcs between the points ( _x_ k-1, _f_ ( _x_ k-1)) and ( _x_ k, f( _x_ k)), each of which is approximated by the surface area generated by the line segment between _( _x_ k-1,_ _f_ ( _x_ k-1)) and ( _x_ k, _f_ ( _x_ k)). The latter is the area of a frustum of a cone. In the notation of [Fig. 36-6, this is, by virtue of Problem 10( _b_ ):

Now, , being the average of _f_ ( _x_ k-1) and _f_ ( _x_ k), is between those two values and, by the intermediate value theorem, is equal to   for some   in ( _x_ k-1, _x_ k). Also,  . By the mean value theorem,   for some   in ( _x_ k-1, _x_ k). Thus, _S_ is approximated by the sum

and it can be shown that this sum can be made arbitrarily close to  .† Hence, the latter is equal to _S_.

**Fig. 36-6**

† In general, the following result can be proved:

**Bliss's Theorem:** Assume _f_ and _g_ are continuous on [ _a, b_ ]. Divide [ _a, b_ ] into subintervals [ _x k-1, xk_] with _a_ = _x_ 0 < _x_ 1 <...< _x_ n = _b_ , and let Δ _k_ _x_ = _x_ _k_ − _x_ k-1. In each [ _x_ k-1, _x_ k], choose   and  . Then the approximating sum   can be made arbitrarily close to   by letting _n_ → +∞ and making the maximum lengths of the subintervals approach 0.

### **Supplementary Problems**

In Problems 12–20, find the area of the surface of revolution generated by revolving the given arc about the given axis:

**21**. Find the surface area of a zone cut from a sphere of radius _r_ by two parallel planes, each at a distance   _a_ from the center.

_Ans._ 2π _ar_

**22**. Find the surface area of a torus (doughnut) generated by revolving the circle _x_ 2 \+ ( _y_ − _b_ )2 = _a_ 2 about the _x_ axis. Assume 0 < _a_ < _b_.

_Ans._ 4π2 _ab_

## **CHAPTER 37
Parametric Representation of Curves**

**PARAMETRIC EQUATIONS.** If the coordinates ( _x, y_ ) of a point _P_ on a curve are given as functions _x_ = _f_ ( _u_ ), _y_ = _g_ ( _u_ ) of a third or _parameter, u_ , the equations _x_ = _f_ ( _u_ ) and _y_ = _g_ ( _u_ ) are called _parametric equations_ of the curve.

**EXAMPLE 1**

( _a_ ) _x_ = cos _θ, y_ = 4 sin2 _θ_ are parametric equations, with parameter _θ_ , of the parabola 4x2 \+ _y_ = 4, since 4x2 \+ _y_ = 4 cos2 _θ_ \+ 4 sin2 _θ_ = 4.

( _b_ )  , _y_ = 4 − _t_ 2 is another parametric representation, with parameter _t_ , of the same curve. It should be noted that the first set of parametric equations represents only a portion of the parabola (Fig. 37-1( _a_ )), whereas the second represents the entire curve (Fig. 37-1( _b_ )).

**Fig. 37-1**

**EXAMPLE 2**

( _a_ ) The equations _x_ = r cos _θ, y_ = r sin _θ_ represent the circle of radius _r_ with center at the origin, since _x_ 2 \+ y2 = r2 cos2 _θ_ \+ r2 sin2 _θ_ = r2(cos2 _θ_ \+ sin2 _θ_ ) = r2. The parameter _θ_ can be thought of as the angle from the positive _x_ axis to the segment from the origin to the point _P_ on the circle (Fig. 37-2).

( _b_ ) The equations _x_ = a + r cos _θ, y_ = b + r sin _θ_ represents the circle of radius _r_ with center at ( _a, b_ ), since ( _x_ − _a_ )2 \+ ( _y_ − _b_ )2 = r2 cos2 _θ_ \+ r2 sin2 _θ_ = r2(cos2 _θ_ \+ sin2 _θ_ ) = r2.

**Fig. 37-2**

Assume that a curve is specified by means of a pair of parametric equations _x_ = _f_ ( _u_ ) and _y_ = _g_ ( _u_ ). Then the first and second derivatives   and   are given by the following formulas.

**(37.1) First Derivative**

This follows from the chain rule formula  .

**(37.2) Second Derivative**

This follows from the chain rule formula  .

**ARC LENGTH FOR A PARAMETRIC CURVE.** If a curve is given by parametric equations _x_ = _f_ ( _t_ ), _y_ = _g_ ( _t_ ), then the length of the arc of the curve between the points corresponding to parameter values _t_ 1 and _t_ 2 is

This formula can be derived by an argument similar to that for the arc length formula (29.2).

### **Solved Problems**

**1.** Find   and   if _x_ = _t_ − sin _t, y_ = 1 − cos _t_.

Hence, by (37.2),

**2.** Find   and   if _x_ = _e_ _t_ cos _t, y_ = _e_ _t_ sin _t_.

So, by (37.2),

**3.** Find an equation of the tangent line to the curve   at the point where _t_ = 4.

  and  . By (37.1),  . So, the slope of the tangent line when _t_ = 4 is  . When _t_ = 4, _x_ = 2 and  . An equation of the tangent line is  .

**4.** The position of a particle that is moving along a curve is given at time _t_ by the parametric equations _x_ = 2 − 3 cos _t, y_ = 3 + 2 sin _t_ , where _x_ and _y_ are measured in feet and _t_ in seconds. (See Fig. 37-3). Note that  , so that the curve is an ellipse.) Find: ( _a_ ) the time rate of change of _x_ when _t_ = π/3; ( _b_ ) the time rate of change of _y_ when _t_ = 5π/3; ( _c_ ) the time rate of change of the angle of inclination _θ_ of the tangent line when _t_ = 2π/3.

( _a_ ) When

**Fig. 37-3**

( _b_ ) When

( _c_ )  . So,

When  . Thus, the angle of inclination of the tangent line is decreasing at the rate of   radians per second.

**5.** Find the arc length of the curve _x_ = _t_ 2, _y_ = _t_ 3 from _t_ = 0 to _t_ = 4.

**6.** Find the length of an arch of the cycloid _x_ = _θ_ − sin _θ, y_ = 1 − cos _θ_ between _θ_ = _θ_ and _θ_ = 2π.

### **Supplementary Problems**

In Problems 7–11, find:  .

**12.** Find the slope of the curve _x_ = e− _t_ cos 2 _t, y_ = e−2 _t_ sin2 _t_ at the point _t_ = 0.

_Ans._ −2

**13.** Find the rectangular coordinates of the highest point of the curve _x_ = 96 _t, y_ = 96 _t_ − 16 _t_ 2. ( _Hint_ : Find _t_ for maximum _y_.)

_Ans._ (288, 144)

**14.** Find equations of the tangent line and normal line to the following curves at the points determined by the given value of the parameter.

( _a_ ) _x_ = 3e _t_ , _y_ = 5e− _t_ at _t_ = 0

( _b_ ) _x_ = a cos4 _θ, y_ = a sin4 _θ_ at _θ_ =

_Ans._ ( _a_ ) 3y + 5 _x_ = 30, 5 _y_ − 3 _x_ = 16; ( _b_ ) 2 _x_ \+ 2 _y_ = _a, y_ = _x_

**15.** Find an equation of the tangent line at any point _P_ ( _x, y_ ) of the curve _x_ = _a_ cos3 _t, y_ = _a_ sin3 _t_. Show that the length of the segment of the tangent line intercepted by the coordinate axes is _a_.

_Ans._

**16.** For the curve _x_ = t2 − 1, _y_ = _t_ 3 − _t_ , locate the points where the tangent line is ( _a_ ) horizontal, and ( _b_ ) vertical. Show that, at the point where the curve crosses itself, the two tangent lines are mutually perpendicular.

_Ans._ ( _a_ )  ; ( _b_ ) _t_ = 0

In Problems 17-20, find the length of the specified are of the given curve.

**17.** The circle _x_ = _a_ cos _θ, y_ = _a_ sin _θ_ from _θ_ = 0 to _θ_ = 2π.

_Ans_. 2π _a_

**18.** _x_ = _e_ _t_ cost _t, y_ = _e_ _t_ sin _t_ from _t_ = 0 to _t_ = 4.

_Ans_.

**19.**  , _y_ = tan−1 _t_ from _t_ = 0 to _t_ = 1.

_Ans_.

**20.** _x_ = 2 cos _θ_ \+ cos 2 _θ_ \+ 1, _y_ = 2 sin _θ_ \+ sin 2 _θ_.

_Ans_. 16

**21.** The position of a point at time _t_ is given as  . Find the distance the point travels from _t_ = 0 to _t_ = 4.

_Ans_. 20

**22.** Identify the curves given by the following parametric equations and write equations for the curves in terms of _x_ and _y_ :

**23.** (GC) Use a graphing calculator to find the graphs of the following parametric curves:

## **CHAPTER 38
Curvature**

**DERIVATIVE OF ARC LENGTH**. Let _y_ = _f_ ( _x_ ) have a continuous first derivative. Let _A_ ( _x_ 0, _y_ 0) be a fixed point on its graph (see Fig. 38-1) and denote by s the arc length measured from _A_ to any other point _P_ ( _x, y_ ) on the curve. We know that, by formula (29.2),

if _s_ is chosen so as to increase with _x_. Let _Q_ ( _x_ \+ Δ _x, y_ \+ Δ _y_ ) be a point on the curve near _P_. Let Δ _s_ denote the arc length from _P_ to _Q_. Then

and, similarly,

The plus or minus sign is to be taken in the first formula according as _s_ increases, or decreases as _x_ increases, and in the second formula according as _s_ increases or decreases as _y_ increases.

**Fig. 38-1**

When a curve is given by parametric equations _x = f_ ( _u_ ), _y_ = _g_ ( _u_ ),

Here the plus or minus sign is to be taken according as _s_ increases or decreases as _u_ increases.

To avoid the repetition of ambiguous signs, we shall assume hereafter that the direction on each arc has been established so that the derivative of arc length will be positive.

**CURVATURE**. The curvature _K_ of a curve _y_ = _f_ ( _x_ ) at any point _P_ on it is defined to be the rate of change of the direction of the curve at _P_ , that is, of the angle of inclination τ of the tangent line at _P_ , with respect to the arc length _s_. (See Fig. 38-2) Intuitively, the curvature tells us how fast the tangent line is turning. Thus, the curvature is large when the curve bends sharply.

**Fig. 38-2**

As formulas for the curvature, we get:

or, in terms of _y_ ,

For a derivation, see Problem 13.

_K_ is sometimes defined so as to be positive. If this is assumed, then the sign of _K_ should be ignored in what follows.

**THE RADIUS OF CURVATURE.** The radius of curvature _R_ at a point _P_ on a curve is defined by  , provided that _K_ ≠ 0.

**THE CIRCLE OF CURVATURE**. The circle of curvature, or _osculating circle_ of a curve at a point _P_ on it is the circle of radius _R_ lying on the concave side of the curve and tangent to it at _P_. (See Fig. 38-3.)

**Fig. 38-3**

To construct the circle of curvature: On the concave side of the curve, construct the normal line at _P_ and on it lay off a segment _PC_ of length _R_. The point _C_ is the center of the required circle.

**THE CENTER OF CURVATURE.** The center of curvature for a point _P_ ( _x, y_ ) of a curve is the center _C_ of the circle of curvature at _P_. The coordinates ( _α, β_ ) of the center of curvature are given by

or by

See Problem 9 for details.

**THE EVOLUTE.** The evolute of a curve is the locus of the centers of curvature of the given curve. (See Problems 11-12.)

### **Solved Problems**

**1**. Find   at _P_ ( _x, y_ ) on the parabola _y_ = 3 _x_ 2.

**2**. Find   and   at _P_ ( _x, y_ ) on the ellipse _x_ 2 \+ 4 _y_ 2 = 8.

**3**. Find   at _P_ (θ) on the curve _x_ = sec _θ, y_ = tan _θ_.

**4**. The coordinates ( _x, y_ ) in feet of a moving particle _P_ are given by _x_ = cos _t_ − 1, _y_ = 2 sin _t_ \+ 1, where _t_ is the time in seconds. At what rate is _P_ moving along the curve when ( _a_ ) _t_ = 5π/6, ( _b_ ) _t_ = 5π/3, and ( _c_ ) _P_ is moving at its fastest slowest?

( _a_ ) When

( _b_ ) When

( _c_ ) Let  . Then  . Solving _dS_ / _dt_ = 0 gives the critical numbers _t_ = 0, π/2, π, 3π/2.

When _t_ = 0 and π, the rate   is fastest. When _t_ = π/2 and 3π/2, the rate   is slowest. The curve is shown in Fig. 38-4.

**Fig. 38-4**

**5**. Find the curvature of the parabola _y_ 2 = 12 _x_ at the points:

**6**. Find the curvature of the cycloid _x_ = _θ_ − sin _θ, y_ = 1 − cos _θ_ at the highest point of an arch. (See Fig. 38-5.)

**Fig. 38-5**

To find the highest point on the interval 0 < _x_ < 2π; _dy_ / _dθ_ = sin _θ_ , so that the critical number on the interval is _x_ = π. Since _d_ 2 _y_ / _dθ_ 2 = cos _θ_ < 0 when _θ_ = π, the point _θ_ = π is a relative maximum point and is the highest point of the curve on the interval.

To find the curvature,

**7**. Find the curvature of the cissoid _y_ 2 (2 − _x_ ) = _x_ 3 at the point (1, 1). (See Fig. 38-6.)

**Fig. 38-6**

Differentiating the given equation implicitly with respect to _x_ , we obtain

and

From ( _1_ ), for _x_ = _y_ = 1, –1 + 2 _y_ ′ = 3 and _y_ ′ = 2. Similarly, from (2), for _x_ = _y_ = 1 and _y_ ′ = 2, we find _y_ ′ = 3. Then  .

**8**. Find the point of greatest curvature on the curve _y_ = ln _x_.

The critical number is, therefore,  . The required point is  .

**9**. Find the coordinates of the center of curvature _C_ of the curve _y_ = _f_ ( _x_ ) at a point _P_ ( _x, y_ ) at which _y_ ′ ≠ 0. (See Fig. 38-3.)

The center of curvature _C_ (α, β) lies: (1) on the normal line at _P_ and (2) at a distance _R_ from _P_ measured toward the concave side of the curve. These conditions give, respectively,

From the first, α − _x_ = – _y_ ′(β − _y_ ). Substitution in the second yields

To determine the correct sign, note that, when the curve is concave upward, _y_ ″ > 0 and, since C then lies above _P_ , β − _y_ > 0. Thus, the proper sign in this case is +. (You should show that the sign is also + when _y_ ″ < 0.) Thus,

**10.** Find the equation of the circle of curvature of 2 _xy_ \+ _x_ \+ _y_ = 4 at the point (1, 1).

Differentiating yields 2 _y_ \+ 2 _xy_ ′ + 1 + _y_ ′ = 0. At (1, 1), _y_ ′ = –1 and 1 + ( _y_ ′)2 = 2. Differentiating again yields 4 _y_ ′ + 2 _xy_ ″ + _y_ ″ = 0. At (1, 1),  . Then

The required equation is  .

**11.** Find the equation of the evolute of the parabola _y_ 2 = 12 _x_.

At _P_ ( _x, y_ ):

Then

and

The equations α = 3 _x_ \+ 6, β = – _y_ 3/36 may be regarded as parametric equations of the evolute with _x_ and _y_ , connected by the equation of the parabola, as parameters. However, it is relatively simple in this problem to eliminate the parameters. Thus, _x_ = (α − 6)/3, , and substituting in the equation of the parabola, we have

(36β)2/3 = 4(α–6) or 81β2 = 4(α–6)3

The parabola and its evolute are shown in Fig. 38-7.

**Fig. 38-7**

**12.** Find the equation of the evolute the curve _x_ = cos θ + sin θ, _y_ = sin θ − θ cos θ.

At _P_ ( _x_ , _y_ ):

Then

and

and α = cos θ, β = sin θ are parametric equations of the evolute (see Fig. 38-8).

**Fig. 38-8**

**13.** Derive formula (38.1).

tan τ is the slope of the tangent line and, therefore,

Hence

This yields

from which

### **Supplementary Problems**

In Problems 14–16, find   and  .

In Problems 17–18, find  .

In Problems 19–22, find  .

**23.** Find the curvature of each curve at the given points:

( _a_ ) _y_ = _x_ 3/3 at _x_ = 0, _x_ = 1, _x_ = –2

( _b_ ) _x_ 2 = 4 _ay_ at _x_ = 0, _x_ = 2 _a_

( _c_ )

( _d_ )

_Ans_.

**24.** Show ( _a_ ) the curvature of a straight line is 0; ( _b_ ) the curvature of a circle is numerically the reciprocal of its radius.

**25.** Find the points of maximum curvature of ( _a_ ) _y_ = _e_ x; ( _b_ )  .

_Ans._

**26.** Find the radius of curvature of

( _a_ ) _x_ 3+ _xy_ − 6 _y_ 2 = 0 at (3,3)

( _b_ ) _x_ = 2 _a_ tan θ, _y_ = _a_ tan2 θ at ( _x, y_ ).

( _c_ ) _x_ = _a_ cos4 θ, _y_ = _a_ sin4 θ at ( _x, y_ ).

_Ans._

**27.** Find the center of curvature of ( _a_ )Problem 26( _a_ ); ( _b_ ) _y_ = sin _x_ at a maximum point.

_Ans._

**28.** Find the equation of the circle of curvature of the parabola _y_ 2 = 12 _x_ at the points (0, 0) and (3, 6).

_Ans._ ( _x_ − 6)2 \+ _y_ 2 = 36; ( _x_ − 15)2 \+ ( _y_ \+ 6)2 = 288

**29.** Find the equation of the evolute of ( _a_ ) _b_ 2 _x_ 2 \+ _a_ 2 _y_ 2 = _a_ 2 _b_ 2; ( _b_ ) _x_ 2/3 \+ _y_ 2/3 \+ _a_ 2/3; ( _c_ ) _x_ = 2 cos _t_ \+ cos 2 _t, y_ = 2 sin _t_ \+ sin 2 _t_.

_Ans_.

## **CHAPTER 39
Plane Vectors**

**SCALARS AND VECTORS.** Quantities such as time, temperature, and speed, which have magnitude only, are called _scalars_. Quantities such as force, velocity, and acceleration, which have both magnitude and direction, are called _vectors_. Vectors are represented geometrically by directed line segments (arrows). The direction of the arrow (the angle that it makes with some fixed directed line of the plane) is the direction of the vector, and the length of the arrow represents the magnitude of the vector.

Scalars will be denoted by letters _a, b, c_ ,... in ordinary type; vectors will be denoted in bold type by letters **a, b, c,**..., or by an expression of the form **OP** (where it is assumed that the vector goes from _O_ to _P._ (See Fig. 39-1(a).) The magnitude (length) of a vector **a** or **OP** will be denoted | **a** | or | **OP** |.

**Fig. 39-1**

Two vectors **a** and **b** are said to be equal (and we write **a** = **b** ) if they have the same direction and magnitude. A vector whose magnitude is that of **a** , but whose direction is opposite that of **a** , is called the negative of **a** and is denoted **–a**. (See Fig. 39-1(a).)

If **a** is a vector and _k_ is a positive scalar, then _k_ **a** is defined to be a vector whose direction is that of **a** and whose magnitude is _k_ times that of **a**. If _k_ is a negative scalar, then _k_ **a** has direction opposite that of **a** and has magnitude | _k_ | times that of **a**.

We also assume a _zero vector_ **0** with magnitude 0 and no direction. We define – **0** = **0, 0a** = **0** , and _k_ **0** = **0**.

Unless indicated otherwise, a given vector has no fixed position in the plane and so may be moved under parallel displacement at will. In particular, if **a** and **b** are two vectors (Fig. 39-1(b)), they may be placed so as to have a common initial or beginning point _P_ (Fig. 39-1(c)) or so that the initial point of **b** coincides with the terminal or end point of **a** (Fig. 39-1(d)).

**SUM AND DIFFERENCE OF TWO VECTORS.** If **a** and **b** are the vectors of Fig. 39-1(b), their _sum_ **a + b** is to be found in either of two equivalent ways:

**1**. By placing the vectors as in Fig. 39-1(c) and completing the parallelogram _PAQB_ of Fig. 39-2(a). The vector **PQ** is the required sum.

**2**. By placing the vectors as in Fig. 39-1(d) and completing the triangle _PAB_ of Fig. 39-2(b). Here, the vector **PB** is the required sum.

It follows from Fig. 39-2(b) that three vectors may be displaced to form a triangle, provided that one of them is either the sum or the negative of the sum of the other two.

**Fig. 39-2**

If **a** and **b** are the vectors of Fig. 39-1(b), their difference **a − b** is to be found in either of two equivalent ways:

**1**. From the relation **a − b** = **a** \+ ( **–b** ) as in Fig. 39-2(c).

**2**. By placing the vectors as in Fig. 39-1(c) and completing the triangle. In Fig. 39-2(d), the vector **BA** = **a − b.**

If **a, b,** and **c** are vectors, the following laws are valid.

See Problems 1 to 4.

**COMPONENTS OF A VECTOR.** In Fig. 39-3(a), let **a** = **PQ** be a given vector, and let _PM_ and _PN_ be any two other directed lines through _P_. Construct the parallelogram _PAQB_. Then

**a** = **PA + PB**

and **a** is said to be _resolved_ in the directions _PM_ and _PN_. We shall call **PA** and **PB** the vector components of **a** in the pair of directions _PM_ and _PN_.

Consider next the vector **a** in a rectangular coordinate system (Fig. 39-3(b)), having equal units of measure on the two axes. Denote by **i** the vector from (0, 0) to (1, 0), and by **j** the vector from (0, 0) to (0, 1). The direction of **i** is that of the positive _x_ axis, the direction of **j** is that of the positive _y_ axis, and both are _unit vectors_ , that is, vectors of magnitude 1.

From the intial point _P_ and the terminal point _Q_ of **a** , drop perpendiculars to the _x_ axis, meeting it at _M_ and _N_ , respectively, and to the _y_ axis, meeting it at _S_ and _T_ respectively. Now, _MN_ , = _a_ 1 **i** , with _a_ 1 positive, and _ST_ = _a_ 2 **j** , with _a_ 2 negative. Then: **MN** = **RQ** = a1 **i, ST** = **PR** = a2 **j** , and

**Fig. 39-3**

Let us call a1 **i** and a2 **j** the _vector components_ of **a.** † The scalars _a_ 1 and _a_ 2 will be called the _scalar components_ (or the _x component_ and _y component_ , or simply the _components_ ) of **a**. Note that **0** = 0 **i** \+ 0 **j**.

Let the direction of a be given by the angle _θ_ , with  _θ_ < 2π, measured counterclockwise from the positive _x_ axis to the vector. Then

and

with the quadrant of _θ_ being determined by

If **a** = a1 **i** \+ a2 **j** and **b** = b1 **i** \+ b2 **j** , then the following hold.

**SCALAR PRODUCT (OR DOT PRODUCT).** The _scalar product (or dot product_ ) of vectors **a** and **b** is defined by

where _θ_ is the smaller angle between the two vectors when they are drawn with a common initial point (see Fig. 39-4). We also define: **a** · **0** = **0** · **a** = 0.

**Fig. 39-4**

† A pair of directions (such as _OM_ and _OT_ ) need not be mentioned, since they are determined by the coordinate system.

From the definitions, we can derive the following properties of the scalar product.

**SCALAR AND VECTOR PROJECTIONS.** In equation (39.1), the scalar _a_ 1 is called the _scalar projection_ of **a** on any vector whose direction is that of the positive _x_ axis, while the vector _a_ 1 **i** is called the _vector projection_ of a on any vector whose direction is that of the positive _x_ axis. In general, for any nonzero vector **b** and any vector **a** , we define we define   to be the scalar projection of **a** on **b** , and   to be the vector projection of **a** on **b**. (See Problem 7.) Note that, when **b** has the direction of the positive _x_ axis,  .

_PROPERTY_ (39.15) **a · b** is the product of the length of **a** and the scalar projection of **b** on **a**. Likewise, **a · b** is the length of **b** and the scalar projection of **a** on **b**. (See Fig. 39-5.)

**Fig. 39-5**

**DIFFERENTIATION OF VECTOR FUNCTIONS.** Let the curve of Fig. 39-6 be given by the parametric equations _x_ = _f_ ( _u_ ) and _y_ = _g_ ( _u_ ). The vector

**r** = _x_ **i** \+ _y_ **j** = _f(u)_ **i** \+ _g(u)_ **j**

joining the origin to the point of _P_ ( _x, y_ ) of the curve is called the _position vector_ or the _radius vector_ of _P_. It is the function of _u_. (From now on, the letter **r** will be used exclusively to denote position vectors. Thus, **a** = 3 **i** \+ 4 **j** is meant to be "free" vector, whereas **r** = 3 **i** \+ 4 **j** is meant to be the vector joining the origin to _P_ (3, 4).)

The derivative   of the function **r** with respect to _u_ is defined to be  . Straightforward computation yields:

Let _s_ denote the arc length measured from a fixed point _P_ 0 of the curve so that _s_ increases with _u_. If τ is the angle that _d_ **r** / _du_ makes with the positive _x_ axis, then

**Fig. 39-6**

Moreover,   is a vector of magnitude

whose direction is that of the tangent line to the curve at _P_. It is customary to show this vector with _P_ as its initial point.

If now the scalar variable _u_ is taken to be the arc length _s_ , then equation (39.5) becomes

The direction of **t** is τ, while its magnitude is  , which is equal to 1. Thus, **t** = _d_ **r** / _ds_ is the _unit tangent vector_ to the curve at _P_.

Since **t** is a unit vector, **t** and _d_ **t** / _ds_ are perpendicular. (See Problem 10.) Denote by **n** a unit vector _P_ having the direction of _dt_ / _ds_. As _P_ moves along the curve shown in Fig. 39-7, the magnitude of **t** remains constant; hence _d_ **t** / _ds_ measures the rate of change of the direction of **t**. Thus, the magnitude of _d_ **t** / _d_ s at _P_ is the absolute value of the curve at _P_ , that is | _d_ **t** / _ds_ | = |K|, and

**Fig. 39-7**

### **Solved Problems**

**1**. Prove **a** \+ **b** = **b** \+ **a**

From Fig. 39-8, **a** \+ **b** = **PQ** = **b** \+ **a**.

**Fig. 39-8**

**2**. Prove ( **a** + **b** ) + **c** = **a** \+ ( **b** + **c** ).

From Fig. 39-3, **PC** = **PB** \+ **BC** = ( **a** \+ **b** ) + **c**. Also, **PC** = **PA** \+ **AC** = **a** \+ ( **b** \+ **c** ).

**Fig. 39-9**

**3**. Let **a, b** and **c** be three vectors issuing from _P_ such that their endpoints _A, B_ , and _C_ lie on a line, as shown in Fig. 39-10. If _C_ divides _BA_ in the ratio _x_ : _y_ , where _x_ \+ _y_ = 1, show that **c** = _x_ **a** \+ _y_ **b**.

Just note that

**c** = **PB** \+ **BC** = **b** \+ _x_ ( **a − b** ) = _x_ **a** \+ (1 − _x_ ) **b** = _x_ **a** \+ _y_ **b**

As an example, if _C_ bisects _BA_ , then  .

**Fig. 39-10**

4. Prove: The diagonals of a parallelogram bisect each other.

Let the diagonals intersect at _Q_ , as in Fig. 39-11. Since **PB = PQ + QB = PQ - BQ** , there are positive numbers _x_ and _y_ such that **b** = _x_ ( **a** \+ **b** ) − y( **a** − **b** ) = (x − y) **a** \+ ( _x + y_ ) **b**. Then _x + y_ = 1 and _x − y_ = 0. Hence,  , and _Q_ is the midpoint of each diagonal.

**Fig. 39-11**

5. For the vectors **a** = 3 **i** \+ 4 **j** and **b** = **2i − j** , find the magnitude and direction of ( _a_ ) **a** and **b** ; ( _b_ ) **a + b;** ( _c_ ) **b − a**.

( _a_ ) For **a** = 3 **i** \+ 4 **j** :     and  ; then θ is a first quadrant angle and is **53°8′**.

For  ;   and  ; θ = 360° − 26°34′ = 333°26′

( _b_ ) **a + b** = ( **3i + 4j** ) + ( **2i − j** ) = **5i + 3j**. Then   Since   and  .

( _c_ ) **b − a** = ( **2i − j** ) − ( ** _3i + 4j_** ) = − **i − 5j**. Then  . Since tan θ = 5 and  .

6. Prove: The median to the base of an isosceles triangle is perpendicular to the base. (See Fig. 39-12, where | **a** | = | **b** |.)

**Fig. 39-12**

From Problem 3, since **m** bisects the base,  . Then

Thus, the median is perpendicular to the base.

7. If **b** is a nonzero vector, resolve a vector **a** into components **a** 1 and **a** 2 respectively parallel and perpendicular to **b**.

In Fig. 39-13, we have **a** = **a 1** \+ **a 2, a1 = cb**, and **a 2 · b = 0**. Hence, **a 2 = a − a1 =a − cb.** Moreover, **a2 · b = (a − cb). b = a · b − c** | **b** |2 = 0, whence  . Thus,

The scalar   is the scalar projection of **a** on **b**. The vector   is the vector projection of **a** on **b**.

**Fig. 39-13**

8. Resolve **a = 4i + 3j** into components **a 1** and **a 2**, parallel and perpendicular, respectively, to **b = 3i + j**.

From Problem 7,  . Then

9. If **a** = _f_ 1( _u_ ) **i** \+ _f 2(u)_ **j** and **b** = _g 1(u)_ **i** \+ _g 2 (u)_ **j** , show that  .

By Property 39.12, **a · b** = ( _f 1 (u)_ **i** \+ _f 2(u)_ **j** ) · ( _g_ 1 ( _u_ ) **i** \+ _g 2(u)_ **j** ) = _f 1g1_ \+ _f 2g2_. Then

10. If **a** = _f 1(u)_ **i** \+ _f 2(u)_ **j** is of constant nonzero magnitude, show that   and, therefore, when   is not zero, **a** and   are perpendicular.

Let | **a** | = _c_. Thus, **a · a** = _c 2_. By Problem 9,

Then

11. Given **r** = (cos2 θ) **i** \+ (sin2 θ) **j** , for 0  θ  π/2, find **t**.

Therefore, by equation (39.6),

by Property 39.12. So,

12. Given _x_ = a cos3 θ, _y_ = a sin3 θ, with _0  θ   π/2_, find **t** and n when _θ = π/4_.

We have **r** = a(cos3 θ) **i** \+ a(sin3 θ) **j**. Then

Hence,

At θ = π/4,

13. Show that the vector a = a **i** \+ b **j** is perpendicular to the line _ax + by + c = 0_.

Let P1( _x 1, y1_) and _P 2(x2, y2)_ be two distinct points on the line. Then _ax 1 \+ by1 \+ c = 0_ and _ax 2 \+ by2 \+ c = 0_. Subtracting the first from the second yields

Now

By ( _1_ ), the left side is zero. Thus, **a** is perpendicular (normal) to the line.

14. Use vector methods to find:

_(a)_ The equation of the line through _P 1(2,3)_ and perpendicular to the line _x + 2y + 5 = 0_.

_(b)_ The equation of the line through _P 1(2,3)_ and _P_ 2(5, −1).

Take _P(x, y)_ to be any other point on the required line.

_(a)_ By Problem 13, the vector **a = i + 2j** is normal to the line _x + 2y + 5 = 0_. Then **P 1P** = _(x − 2)_ **i** + _(y − 3)_ **j** is parallel to a if _(x − 2)_ **i** + _(y − 3)_ **j** = _k_ **(i + 2j)** for some scalar _k_. Equating components, we get _x − 2 = k_ and _y − 3 = 2k_. Eliminating _k_ , we obtain the required equation _y − 3 = 2(x − 2)_ , or, equivalently, _2x − y − 1 = 0_.

_(b)_ We have **P 1P** = _(x − 2)_ **i** + _(y − 3)_ **j** and **P 1P2 = 3i − 4j.** Now **a = 4i+3j** is perpendicular to **P 1P2** and, hence, to **P 1P.** Thus, 0 = **a · P 1P = (4i+3j)**·[ _(x − 2)_ **i** + _(y − 3)_ **j** ] and, equivalently, _4x + 3y − 17 = 0_.

15. Use vector methods to find the distance of the point _P_ 1(2,3) from the line _3x + 4y − 12 = 0_.

At any convenient point on the line, say _A_ (4,0), construct the vector **a = 3i + 4j** perpendicular to the line. The required distance is _d_ = **|AP 1|**cos θ in Fig. 39-14. Now, **a · AP 1 = |a||AP1|**cos θ = **|a|** _d_. Hence,

**Fig. 39-14**

16. The _work_ done by a force expressed as a vector **b** in moving an object along a vector **a** is defined as the product of the magnitude of **b** in the direction of **a** and the distance moved. Find the work done in moving an object along the vector **a = 3i + 4j** if the force applied is **b = 2i + j.**

The work done is

(magnitude of **b** in the direction of **a** )·(distance moved) = ( **| _B_ |**cos θ **|a|=b·a** = (2 **i** + **j** )·(3 **i** +4 **j** )=10

### **Supplementary Problems**

17. Given the vectors **a, b, c** in Fig. 39-15, construct _(a)_ 2 **a** ; _(b)_ − 3 **b** ; _(c)_ **a+2b;** _(d)_ **a+b − c;** _(e)_ **a − 2b + 3c.**

18. Prove: The line joining the midpoints of two sides of a triangle is parallel to and one-half the length of the third side. (See Fig. 39-16.)

19. If **a, b, c, d** are consecutive sides of a quadrilateral (see Fig. 39-17), show that **a + b + c + d = 0.** ( _Hint:_ Let _P_ and _Q_ be two nonconsecutive vertices.) Express **PQ** in two ways.

**Fig. 39-15**

**Fig. 39-16**

**Fig. 39-17**

20. Prove: If the midpoints of the consecutive sides of any quadrilateral are joined, the resulting quadrilateral is a parellelogram. (See Fig. 39-18.)

**Fig. 39-18**

21. Using Fig. 39-19, in which **|a| = | _B_ |** is the radius of a circle, prove that the angle inscribed in a semicircle is a right angle.

**Fig. 39-19**

22. Find the length of each of the following vectors and the angle it makes with the positive _x_ axis:  .

_Ans._

23. Prove: If **u** is obtained by rotating the unit vector **i** counterclockwise about the origin through the angle θ, then **u** = **i** cos θ + **j** sin θ.

24. Use the law of cosines for triangles to obtain  .

25. Write each of the following vectors in the form _a_ **i** \+ _b_ **j** :

_(a)_ The vector joining the origin to _P_ (2, −3) _(b)_ The vector joining _P_ 1 (2, 3) to _P_ 2 (4, 2)

_(c)_ The vector joining _P_ 2 (4,2) to _P_ 1 (2,3) _(d)_ The unit vector in the direction of 3 **i** \+ 4 **j**

_(e)_ The vector having magnitude 6 and direction 120°

_Ans._

26. Using vector methods, derive the formula for the distance between _P 1(x1, y1)_ and _P_ 2(x2, y2).

27. Given _O_ (0,0), _A_ (3,1), and _B_ (1,5) as vertices of the parallelogram _OAPB,_ find the coordinates of _P._

_Ans._ (4,6)

28. _(a)_ Find _k_ so that **a** = 3 **i** \+ 2 **j** and **b** = **i** \+ _k_ **j** are perpendicular. _(b)_ Write a vector perpendicular to **a** = 2 **i** \+ 5 **j**.

29. Prove Properties (39.8) to (39.15).

30. Find the vector projection and scalar projection of **b** on **a** , given: _(a)_ **a** = **i** −2 **j** and **b** = − 3 **i** + **j** ; _(b)_ **a** = 2 **i** \+ 3 **j** and **b** = 10 **i** \+ 2 **j**.

_Ans._

31. Prove: Three vectors **a, b, c** will, after parallel displacement, form a triangle provided _(a)_ one of them is the sum of the other two or _(b)_ **a + b + c = 0**.

32. Show that **a** = 3 **i** − 6 **j, b** = 4 **i** \+ 2 **j** , and **c** = −7 **i** \+ 4 **j** are the sides of a right triangle. Verify that the midpoint of the hypotenuse is equidistant from the vertices.

33. Find the unit tangent vector **t** = _d_ **r** / _ds_ , given: _(a)_ **r** = 4 **i** cos θ + 4 **j** sin θ _(b)_ **r** = _e_ θ **i** \+ _e_ – θ **j** ; _(c)_ **r** = θ **i** \+ θ2 **j**.

_Ans._

34. _(a)_ Find **n** for the curve of Problem 33 _(a)_ ; _(b)_ Find **n** for the curve of Problem 33 _(c)_ ; _(c)_ Find **t** and **n** given _x_ = cos θ + θ sin θ, _y_ = sin θ − θcos θ.

_Ans_. ( _a_ ) **i** cos θ − **j** sin θ _(b)_   _(c)_ **t** = **i** cos θ + **j** sin θ, **n** = − **i** sin θ + **j** cos θ

## **CHAPTER 40
Curvilinear Motion**

**VELOCITY IN CURVILINEAR MOTION.** Consider a point P( _x, y_ ) moving along a curve with the equations _x = f(t), y = g(t),_ where _t_ is time. By differentiating the position vector

with respect to _t,_ we obtain the _velocity vector_

where

The magnitude of **v** is called the _speed_ and is given by

The direction of **v** at _P_ is along the tangent line to the curve at _P_ , as shown in Fig. 40-1. If τ denotes the direction of **v** (the angle between **v** and the positive _x_ axis), then tan _τ = v y/vx_, with the quadrant being determined by _v x_ = | **v** | cos _τ_ and _v y_ = | **v** | sin _τ._

**Fig. 40-1**

**ACCELERATION IN CURVILINEAR MOTION.** Differentiating ( _40.2_ ) with respect to _t,_ we obtain the _acceleration vector_

where  . The magnitude of **a** is given by

The direction _ø_ of **a** is given by tan _ø = a y/ax_, with the quadrant being determined by _a x = | **a** | cos ø_ and _a y = | **a** | sin ø._ (See Fig. 40-2.)

**Fig. 40-2**

**TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION.** By equation ( _39.7_ ),

Then

by ( _39.8_ ).

Equation ( _40.5_ ) resolves the acceleration vector at _P_ along the tangent and normal vectors there. Denoting the components by _a_ t and _a_ n, respectively, we have, for their magnitudes,

where _R_ is the radius of curvature of the curve at _P_. (See Fig. 40-3.)

Since  , we obtain

as a second way of determining | _a n_|.

**Fig. 40-3**

### **Solved Problems**

**1**. Discuss the motion given by the equations _x_ = cos 2 _πt, y_ = 3 sin 2 _πt_. Find the magnitude and direction of the velocity and acceleration vectors when:  .

The motion is along the ellipse 9 _x_ 2 \+ _y_ 2 = 9. Beginning (at _t_ = 0) at (1, 0), the moving point traverses the curve counterclockwise.

So, τ = 120°.

So, ø = 259°6′.

(b)

So,  .

So, ø = 79°6′.

**Fig. 40-4**

**2**. A point travels counterclockwise about the circle _x_ 2 \+ _y_ 2 = 625 at a rate | **v** | = 15. Find τ, | **a** |, and ø at ( _a_ ) the point (20,15) and ( _b_ ) the point  . Refer to Fig. 40-4.

Using the parametric equations _x_ = 25 cos θ, _y_ = 25sin θ, we have at _P(x, y)_ :

since | **v** | = 15 is equivalent to a constant angular speed of  .

( _a_ ) At the point (20, 15),  . Thus,

( _b_ ) At the point  . Thus

**3**. A particle moves on the first-quadrant arc of _x_ 2 = 8 _y_ so that _v_ y = 2. Find | **v** |, τ, | **a** |, and ø at the point (4, 2).

Using the parametric equations _x_ = 4 _θ, y_ =2 _θ_ 2, we have

Since,   we have

At the point (4, 2), _θ_ = 1. Then

**4**. Find the magnitudes of the tangential and normal components of acceleration for the motion _x_ = _e t_ cos _t, y_ = et sin _t_ at any time _t_.

We have:

Then  . Finally,

**5**. A particle moves from left to right along the parabola _y = x 2_ with constant speed 5. Find the magnitude of the tangential and normal components of the acceleration at (1, 1).

Since the speed is constant,  . At (1, 1), _y′ = 2x = 2 and y″ = 2_. The radius of curvature at (1, 1) is then  .

**6**. The centrifugal force _F_ (in pounds) exerted by a moving particle of weight _W_ (in pounds) at a point in its path is given by the equation  . Find the centrifugal force exerted by a particle, weighing 5 pounds, at the ends of the major and minor axes as it traverses the elliptical path _x_ = 20 cos _t, y_ = 15 sin _t_ , the measurements being in feet and seconds. Use _g_ = 32 ft/sec2.

We have:

Then

At the ends of the major axis ( _t_ = 0 or _t_ = π):

At the ends of the minor axis  :

**7**. Assuming the equations of motion of a projectile to be _x = v 0t cos ψ, y = v0t sin _, where _v_ 0 is the initial velocity, _ψ_ is the angle of projection, _g_ = 32 ft/sec2, and _x and y_ are measured in feet and _t_ in seconds, find: ( _a_ ) the equation of motion in rectangular coordinates; ( _b_ ) the range; ( _c_ ) the angle of projection for maximum range; and ( _d_ ) the speed and direction of the projectile after 5 sec of flight if _v 0_ = 500 ft/sec and _ψ_ = 45°. (See Fig. 40-5.)

**Fig. 40-5**

( _a_ ) We solve the first of the equations for   and substitute in the second:

( _b_ ) Solving _y = v 0t_ sin  for _t_ , we get _t_ = 0 and _t_ = (2 _v_ 0 sin _ψ_ )/ _g_. For the latter, we have

( _c_ ) For _x_ a maximum,  ; hence cos 2 _ψ_ \- 0 and

( _d_ ) For _v 0_ = 500 and  . Then

When  . Then

**8**. A point _P_ moves on a circle _x_ = _r_ cos _β_ , _y_ = _r_ sin _β_ with constant speed _v_. Show that, if the radius vector to _P_ moves with angular velocity _ω_ and angular acceleration _α_ , ( _a_ ) _v_ = _rω_ and ( _b_ )

( _a_ )

Then

( _b_ )

Then

### **Supplementary Problems**

**9**. Find the magnitude and direction of velocity and acceleration at time _t_ , given

**10**. A particle moves on the first-quadrant arc of the parabola _y 2_ = 12 _x_ with _v_ x = 15. Find _v_ y, | **v** |, and τ; and ax, ay, | **a** |, and _ø_ at (3, 6).

_Ans_.

**11**. A particle moves along the curve _y_ = _x_ 3/3 with vx = 2 at all times. Find the magnitude and direction of the velocity and acceleration when _x_ = 3.

_Ans_.

**12**. A particle moves around a circle of radius 6 ft at the constant speed of 4 ft/sec. Determine the magnitude of its acceleration at any position.

_Ans_.

**13**. Find the magnitude and direction of the velocity and acceleration, and the magnitudes of the tangential and normal components of acceleration at time _t_ , for the motion:

( _a_ ) _x_ = 3t, _y_ = 9 _t_ -3 _t_ 2; at _t_ = 2

( _b_ ) _x_ = cos _t_ \+ sin _t, y_ = sin _t_ \- cos _t_ ; at _t_ = 1

_Ans_.

**14**. A particle moves along the curve   ln _x_ so that  , for _t_ > 0. Find _v_ x, _v_ y, | **v** |, and τ; _a_ x, _a_ y, | **a** |, and ø; | _a_ 1 and | _a_ n| when _t_ = 1.

_Ans_.

**15**. A particle moves along the path _y = 2 _x_ \- _x_ 2_ with _v_ x = 4 at all times. Find the magnitudes of the tangential and normal components of acceleration at the position ( _a_ ) (1, 1) and ( _b_ ) (2, 0).

_Ans_.

**16**. If a particle moves on a circle according to the equations _x_ = _r_ cos _ωt, y_ = _r_ sin _ωt_ , show that its speed is _ωr_.

**17**. Prove that if a particle moves with constant speed, then its velocity and acceleration vectors are perpendicular; and, conversely, prove that if its velocity and acceleration vectors are perpendicular, then its speed is constant.

## **CHAPTER 41
Polar Coordinates**

The position of a point _P_ in a plane may be described by its coordinates ( _x, y_ ) with respect to a given rectangular coordinate system. Its position may also be described by choosing a fixed point _O_ and specifying the directed distance _ρ_ = _OP_ and the angle _θ_ that _OP_ makes with a fixed half-line _OX_. (See Fig. 41-1.) This is the _polar coordinate system_. The point _O_ is called the _pole_ , and _OX_ is called the _polar axis_.

**Fig. 41-1**

To each number pair ( _ρ_ , _θ_ ) there corresponds one and only one point. The converse is not true. For example, (1,0) and (1,2π)describe the same point, on the polar axis and at a distance 1 from the pole. That same point also corresponds to (-1,π). (When _ρ_ is negative, the point corresponding to ( _ρ_ , _θ_ ) is obtained as follows: Rotate the polar axis _OX_ through _θ_ radians (counterclockwise if _θ_ is positive and clockwise if _θ_ is negative) to a new position _OX′_ and then move | _ρ_ | units on the half-line opposite to _OX′_.)

In general, a point _P_ with polar coordinates ( _ρ_ , _θ_ ) also can be described by   and  , where _n_ is any nonnegative integer. In addition, the pole itself corresponds to (0, _θ_ ), with arbitrary _θ_.

**EXAMPLE 1:** In Fig. 41-2, several points and their polar coordinates are shown. Note that point _C_ has polar coordinates  .

A _polar equation_ of the form _ρ_ = _f_ ( _θ_ ) or _F_ ( _ρ_ , _θ_ ) = 0 determines a curve, consisting of those points corresponding to pairs ( _ρ_ , _θ_ ) that satisfy the equation. For example, the equation _ρ_ = 2 determines the circle with center at the pole and radius 2. The equation _ρ_ = –2 determines the same set of points. In general, an equation _ρ_ = _c_ , where _c_ is a constant, determines the circle with center at the pole and radius | _c_ |. An equation _θ_ = _c_ determines the line through the pole obtained by rotating the polar axis through _c_ radians. For example, _θ_ = π/2 is the line through the pole and perpendicular to the polar axis.

**Fig. 41-2**

**POLAR AND RECTANGULAR COORDINATES**. Given a pole and polar axis, set up a rectangular coordinate system by letting the polar axis be the positive _x_ axis and letting the _y_ axis be perpendicular to the _x_ axis at the pole. (See Fig. 41-3.) Then the pole is the origin of the rectangular system. If a point _P_ has rectangular coordinates ( _x, y_ ) and polar coordinates ( _ρ_ , _θ_ ), then

**Fig. 41-3**

These equations entail

**EXAMPLE 2:** Consider the polar curve _ρ_ = cos _θ_.

Multiplying by _ρ_ , we get _ρ_ 2 = _ρ_ cos _θ_. Hence, _x_ 2 \+ _y_ 2 = _x_ holds for the rectangular coordinates of points on the curve. That is equivalent to _x_ 2 − _x_ \+ _y_ 2 = 0 and completion of the square with respect to _x_ yields  . Hence, the curve is the circle with center at ( ,0) and rsdius  . Note that, as _θ_ varies from 0 to π/2, the upper semicircle is traced out from (1, 0) to (0, 0), and then, as _θ_ varies from   to π, the lower semicircle is traced out from (0, 0) back to (1, 0). This whole path is retraced once more as _θ_ varies from π to 2π. Since _cos θ_ has a period of 2π, we have completely described the curve.

**EXAMPLE 3:** Consider the parabola _y_ = _x_ 2. In polar coordinates, we get _ρ_ sin _θ_ = _ρ_ 2cos2 _θ_ , and, therefore, _ρ_ = tan _θ_ sec _θ_ , which is a polar equation of the parabola.

**SOME TYPICAL POLAR CURVES**

( _a_ ) Cardioid: _ρ_ = 1 + sin _θ_. See Fig. 41-4( _a_ ).

( _b_ ) Limaçon: _ρ_ = 1 + 2 cos _θ_. See Fig. 41-4( _b_ ).

( _c_ ) Rose with three petals: _ρ_ = cos 3 _θ_. See Fig. 41-4( _c_ ).

( _d_ ) Lemniscate: _ρ_ 2 = cos 2 _θ_. See Fig. 41-4( _d_ ).

**Fig. 41-4**

At a point _P_ on a polar curve, the angle _ψ_ from the radius vector _OP_ to the tangent _PT_ to the curve (see Fig. 41-5)is given by

**Fig. 41-5**

For a proof of this equation, see Problem 1. Tan _ψ_ plays a role in polar coordinates similar to that of the slope of the tangent line in rectangular coordinates.

**ANGLE OF INCLINATION**. The angle of inclination τ of the tangent line to a curve at a point _P_ ( _ρ_ , _θ_ ) on it (see Fig. 41-5) is given by

For a proof of this equation, see Problem 4.

**POINTS OF INTERSECTION**. Some or all of the points of intersection of two polar curves _ρ_ = _f_ 1( _θ_ ) and _ρ_ = _f_ 2( _θ_ ) (or equivalent equations) may be found by solving

**EXAMPLE 4:** Find the points of intersection of _ρ_ = 1 + sin _θ_ and _ρ_ = 5 − 3 sin θ.

Seeting 1 + sin _θ_ = 5 − 3 sin θ, we obtain sin _θ_ = 1. Then _ρ_ = 2 and _θ_ = π/2. The only point of intersection is (2π/2). Note that we need not indicate the infinitely many other pairs that designate the same point.

Since a point may be represented by more than one pair of polar coordinates, the intersection of two curves may contain points for which no single pair of polar coordinates satisfies ( _41.5_ ).

**EXAMPLE 5:** Find the points of intersection of _ρ_ = 2 sin 2 _θ_ and _ρ_ = 1.

Solution of the equation 2 sin 2 _θ_ = 1 yields  , and, therefore, within [0, 2π], _θ_ = π/12, 5π/12, 13π/12, 17π/12. We have found four points of intersection: (1, π/12), (1, 5π/12), (1, 13π/12), and (1, 17π/12). But the circle _ρ_ = 1 also can be represented as _ρ_ = –1. Now solving 2 sin 2 _θ_ = –1, we get sin 2 _θ_ = –1, we get   and, therefore, _θ_ = 7π/12, 11π/12, 19π/12, and 23π/12. Hence we get four more points of intersection (-1, 7π/12), (-1, 11π/12), (-1, 19π/12), and (-1, 23π/12).

When the pole of intersection, it may not appear among the solutions of ( _41.5_ ). The pole is a point of intersection when there exist _θ_ 1 and _θ_ 2 such that _f 1(θ1) = 0 = f2(θ2)_.

**EXAMPLE 6:** Find the points of intersection of _ρ_ = sin _θ_ and _ρ_ = cos _θ_.

From the equation sin _θ_ = cos _θ_ , we obtain the points of intersection   and  . However, both curves contain the pole. On _ρ_ = sin _θ_ , the pole has coordinates (0, 0), whereas, on _ρ_ = cos _θ_ , the pole has coordinates (0, π/2).

**EXAMPLE 7:** Find the points of intersection of _ρ_ = cos 2 _θ_ and _ρ_ = cos _θ_.

Setting cos 2 _θ_ = cos _θ_ and noting that cos 2 _θ_ = 2 cos2 _θ_ − 1, we get 2 cos2 _θ_ − cos _θ_ –1 = 0 and, therefore, (cos _θ_ − 1)(2 cos _θ_ \+ 1) = 0. So,  . Then _θ_ = 0, 2π/3, 4π/3, yielding points of intersection (1, 0),  , and  . But the pole is also an intersection point, appearing as (0, π/4) on _ρ_ = cos 2 _θ_ and as (0, π/2) on _ρ_ = cos _θ_.

**ANGLE OF INTERSECTION.** The angle of intersection, ø of two curves at a common point _P(ρ, _θ_ )_, not the pole, is given by

where _ψ_ 1 and _ψ_ 2 are the angles from the radius vector _OP_ to the respective tangent lines to the curves at _P_. (See Fig. 41-6.) This formula follows from the trigonometric identity for tan( _ψ_ 1 − _ψ_ 2), since ø = _ψ_ 1 − _ψ_ 2.

**Fig. 41-6**

**EXAMPLE 8:** Find the (acute) angles of intersection of _ρ_ = cos 2 _θ_ and _ρ_ = cos _θ_.

The points of intersection were found in Example 7. We also need tan _ψ_ 1 and tan _ψ_ 2. For _ρ_ = cos _θ_ , formula ( _41.3_ ) yields tan _ψ_ 1 = –cot _θ_. For _ρ_ = cos 2 _θ_ , formula ( _41.3_ ) yields  .

At the point (1, 0), tan _ψ_ 1 = − cot0 = ∞ and, likewise, tan _ψ_ 2 = ∞. Then _ψ_ 1 = _ψ_ 2 = π/2 and, therefore, ø = 0

At the point  ,   and  . So, by ( _41.6_ ),

and therefore, the acute angle of intersection ø ≈ 46°6′. By symmetry, this is also the acute angle of intersection at the point  .

At the pole: On _ρ_ = cos _θ_ , the pole is given by _θ_ = π/2. On _ρ_ = cos 2 _θ_ , the pole is given by _θ_ = π/4 and _θ_ = 3π/4. Thus, at the pole there are two intersections, the acute angle being π/4 for each.

**THE DERIVATIVE OF THE ARC LENGTH.** The derivative of the arc length is given by

where   and it is understood that _s_ increase with _θ_.

For a proof, see Problem 20.

**CURVATURE.** The curvature of a polar curve is given by

For a proof, see Problem 17.

### Solved Problems

**1**. Derive formula (41.3):  , where  .

In Fig. 41-7, _Q_ ( _ρ_ \+ Δ _ρ_ , _θ_ \+ Δ _θ_ ) is a point on the curve near _P_. From the right triangle _PSQ_ ,

Now as _Q_ → _P_ along the curve, Δθ → 0, _OQ_ → _OP, PQ_ → _PT_ , and  λ →  ψ.

As Δθ→0,   and  . Thus,

**Fig. 41-7**

In Problems 2 and 3, use formula ( _41.3_ ) to find tan _ψ_ for the given curve at the given point.

**2**. _ρ_ = 2 + cos _θ_ at   (See Fig. 41-8.)

At  ,  ,  , and  .

**Fig. 41-8**

**3**. _ρ_ = 2 sin 3 _θ_ at  . (See Fig. 41-9.)

At  ,  ,   and that  .

**Fig. 41-9**

**4**. Derive formula ( _41.4_ ):  .

From Fig. 41-7, τ = ψ + θ and

**5**. Show that, if _ρ_ = f( _θ_ )passes through the pole and _θ_ 1 is such that f( _θ_ 1) = 0, then the direction of the tangent line to the curve at the pole (0. _θ_ 1) is _θ_ 1. (See Fig. 41-10.)

**Fig. 41-10**

At (0, _θ_ 1), _ρ_ = 0, and _ρ_ ′ = _f_ ′( _θ_ 1). If _ρ_ ′ ≠ 0

If _ρ_ ′ = 0,

In Problems 6–8, find the slope of given curve at the given point.

**6**. _ρ_ = 1 - _cos θ_ at  . (See Fig. 41-11)

**Fig. 41-11**

At  ,

sin _θ_ = 1, _cos θ_ = 0, _ρ_ = 1, _ρ_ ′ = sin _θ_ = 1

and

**7**. _ρ_ = cos 3 _θ_ at the pole. (See Fig. 41-12.)

When _ρ_ = 0, cos 3 _θ_ = 0, then 3 _θ_ = π/2, 3π/2, 5π/2, and _θ_ = π/6, π/2, 5π/6. By Problem 5,  , ∞, and  .

**Fig. 41-12**

**8**. _ρθ = a_ at  .

At _θ_ = π/3:  ,  , _ρ_ = 3a/π, and _ρ_ ′ = − a/ _θ_ 2 = –9a/π2. Then

**9**. Investigate _ρ_ = 1 + sin _θ_ for horizontal and vertical tangents. (See Fig. 41-13.)

**Fig. 41-13**

At P( _ρ, θ_ ):

We set cos _θ_ (1 + 2 sin _θ_ ) = 0 and solve, obtaining _θ_ = π/2, 3π/2, 7π/6; and 11π/6. We also set (sin _θ_ \+ 1)(2 sin _θ_ − 1) = 0 and solve, obtaining _θ_ = 3π/2, π/6, and 5π/6.

For _θ_ = π/2: There is a horizontal tangent at (2, π/2)

For _θ_ = 7π/6 and 11π/6: There are horizontal tangents at   and  .

For _θ_ = π/6 and 5π/6: There are vertical tangents at   and  .

For _θ_ = 3π/2: By Problem 5, there is a vertical tangent at the pole.

**10.** Show that the angle that the radius vector to any point of the cardioid _ρ_ = a(1 − cos _θ_ ) makes with the curve is one-half that which the radius vector makes with the polar axis.

At any point P( _ρ, θ_ ) on the cardioid

So  .

In Problems 11–13, find the angles of intersection of the given pair of curves.

**11.** _ρ_ = 3 cos _θ, ρ_ = 1 + cos _θ_. (See Fig. 41-14.)

**Fig. 41-14**

Solve 3 _cos θ_ = 1 + _cos θ_ for the points of intersection, obtaining (3/2, π/3) and (3/2, 5π/3). The curves also intersect at the pole.

For _ρ_ = 3 cos _θ_ : _ρ_ ′ = –3 sin _θ_ and tan _ψ_ 1 = –cotθ

For _ρ_ = 1 + cos _θ_ : _ρ_ ′ = –sin _θ_ and

At _θ_ = π/3,  ,  , and  . The acute angle of intersection at   and, by symmetry, at   is π/6.

At the pole, either a diagram or the result of Problem 5 shows that the curves are orthogonal.

**12.**  ,  .

Solve   for the points of intersection, obtaining (4, 2π/3) and (4, 4π/3).

For  :   and

For  :   and

At _θ_ = 2π/3,  ,  , and  ; the curves are orthogonal. Likewise, the curves are orthogonal at _θ_ = 4π/3.

**13.** _ρ_ = sin 2θ, _ρ_ = cos θ. (See Fig. 41-15.)

The curves intersect at the points   and   and the pole.

For _ρ_ = sin 2θ: _ρ_ ′ = 2 cos 2 _θ_ and

For _ρ_ = cos _θ_ : _ρ_ ′ = –sin _θ_ and tan _ψ_ 2 = –cot _θ_

At _θ_ = π/6,  ,  , and  . The acute angle of intersection at the point   is  . Similarly, at _θ_ = 5π/6,  ,  , and the angle of intersection is  .

At the pole, the angles of intersection are 0 and π/2.

**Fig. 41-15**

In Problems 14–16, find   at the point _P_ (ρ _θ_ ).

**14.** _ρ_ = cos 2 _θ_.

_ρ_ ′ = -2 sin 2 _θ_ and

**15.** _ρ_ (1 + cos _θ_ ) = 4.

Differentiation yields – _ρ_ sin _θ_ \+ _ρ_ ′(1 + cos _θ_ ) = 0. Then

  and

**16.**  . (Also evaluate   at  .)

  and

At

**17.** Derive formula (41.8):  .

By definition,  , Now, τ = _θ_ \+ _ψ_ and, therefore,

where  . Also,

 ; so

Thus,

**18.** Let _ρ_ = 2 + sin _θ_. Find the curvature at the point P(ρ, _θ_ )

**19.** Let _ρ_ (1-cos _θ_ ) = 1. Find the curvature at   and  .

  and  ; so

At _θ_ = π/2,  ; at _θ_ = 4π/3,  .

**20**. Derive formula (41.7):  .

Consider _ρ_ as a function of _θ_. From _x_ = _ρ_ cos _θ_ and y = _ρ_ sin _θ_ , we get dx/d _θ_ = – _ρ_ sin _θ_ \+ (cos _θ_ )ρ′ and dy/d _θ_ = _ρ_ cos _θ_ \+ (sin _θ_ ) _ρ_ ′. Hence,

and

Thus,

Since s increases with _θ_ ,   and we obtain formula (41.7).

**21.** For _ρ_ = cos 2θ, find   at  . (Assume as usual that _s_ increases with _θ_.)

 . By Formula (41.7),

### **Supplementary Problems**

In Problems 22–25, find tan _Ψ_ , for the given curve at the given points.

In Problems 26–29, find tan τ for the given curve at the given point.

Investigate _ρ_ = sin 2 _θ_ for horzontal and vertical tengents.

_Ans._ horizontal tangents at _θ_ = 0, π, 54°44′, 125°16′, 234°44′, 305°16′; vertical tangents at _θ_ = π/2, 3π/2, 35°16′, 144°44′, 215°16′, 324°44′

In Problems 31–33, find the acute angles of intersection of each pair of curves.

**34.** Show that each pair of curves intersects at right angles at all points of intersection.

( _a_ ) _ρ_ = 4 cos _θ, ρ_ = 4 sin _θ_

( _b_ ) _ρ_ = eθ, _ρ_ = e-θ

( _c_ ) _ρ_ 2 cos 2 _θ_ = 4, _ρ_ 2 sin 2 _θ_ = 9

( _d_ ) _ρ_ = 1 + cos _θ, ρ_ = 1 − cos _θ_

**35.** Find the angle of intersection of the tangents to _ρ_ = 2 − 4 sin _θ_ at the pole.

_Ans. 2π/3_

**36.** Find the curvature of each of these curves at _ρ_ (ρ, θ): ( _a_ ) _ρ_ = eθ;( _b_ ) _ρ_ = sin θ; ( _c_ )ρ2 = 4 cos 2 _θ_ ( _d_ ) _ρ_ = 3 sin _θ_ \+ 4 cos _θ_.

_Ans_. ( _a_ )  ; ( _b_ ) 2; ( _c_ )  ; ( _d_ )

**37.** Find   for the curve _ρ_ = _a_ cos _θ_.

_Ans. a_

**38.** Find   for the curve _ρ_ = _a_ (1 + cos _θ_ ).

_Ans_.

**39.** Suppose a particle moves along a curve _ρ_ = _f_ ( _θ_ ) with its position at any time _t_ given by _ρ_ = _g_ ( _t_ ), _θ_ = _h_ ( _t_ ).

( _a_ ) Multiply the equation   obtained in Problem 20 by   to obtain  .

( _b_ ) From  , obtain   and  .

In Problems 40–43, find all points of intersection of the given equations.

**44.** (GC) Sketch the curves in Problems 40–43, find their graphs on a graphing calculator, and check your answers to Problems 40–43.

**45**. (GC) Sketch the graphs of the following equations and then check your answers on a graphing calculator:

( _a_ ) _ρ_ = 2 cos 4 _θ_

( _b_ ) _ρ_ = 2 sin 5 _θ_

( _c_ ) _ρ_ 2 = 4 sin 2 _θ_

( _d_ ) _ρ_ = 2(1 − cos _θ_ )

( _e_ )

( _f_ )

( _g_ ) _ρ_ = 2 − sec _θ_

( _h_ )

(In parts ( _g_ ) and ( _h_ ), look for asymptotes.)

**46**. Change the following rectangular equations to polar equations and sketch the graphs:

( _a_ ) _x_ 2 − 4 _x_ \+ _y_ 2 = 0

( _b_ ) 4 _x_ = _y_ 2

( _c_ ) _xy_ = 1

( _d_ ) _x_ = _a_

( _e_ ) _y_ = _b_

( _f_ ) _y_ = _mx_ \+ _b_

_Ans._ ( _a_ ) _ρ_ = 4 cos _θ_ ; ( _b_ ) _ρ_ = 4 cot _θ_ csc _θ_ ; ( _c_ ) _ρ_ 2 = sec _θ_ csc _θ_ ; ( _d_ ) _ρ_ = _a_ sec _θ_ ; ( _e_ ) _ρ_ = _b_ csc _θ_ ; ( _f_ )

**47.** (GC) Change the following polar equations to rectangular coordinates and then sketch the graph. (Verify on a graphing calculator.) ( _a_ ) _ρ_ = 2 _c_ sin _θ_ ; ( _b_ ) _ρ_ = _θ_ ; ( _c_ ) _ρ_ = 7 sec _θ_

_Ans._ ( _a_ ) _x_ 2 \+ ( _y_ − _c_ )2 = _c_ 2; ( _b_ )  ; ( _c_ ) _x_ = 7

**48.** ( _a_ ) Show that the distance between two points with polar coordinates ( _ρ_ 1, _θ_ 1) and ( _ρ_ 2, _θ_ 2) is

( _b_ ) When _θ_ 1 = _θ_ 2, what does the distance simplify to? Explain why this is so.

_Ans._ | _ρ_ 1 − _ρ_ 2|

( _c_ ) When  , what does the formula yield? Explain the signficance of the result.

_Ans._

( _d_ ) Find the distance between the points with the polar coordinates (1, 0) and  .

_Ans._

**49.** ( _a_ ) Let _f_ be a continuous function such that _f_ ( _θ_ )  0 for α < _θ_ < β. Let _A_ be the area of the region bounded by the lines _θ_ = α and _θ_ = β, and the polar curve _ρ_ = _f_ ( _θ_ ). Derive the formula  . (Hint: Divide [α, β] into _n_ equal parts, each equal to Δ _θ_. Each resulting subregion has area approximately equal to  , where   is in the _i_ th subinterval.)

( _b_ ) Find the area inside the cardioid _ρ_ = 1 + sin _θ_.

( _c_ ) Find the area of one petal of the rose with three petals, _ρ_ = cos 3 _θ_. (Hint: Integrate from   to  .)

## **CHAPTER 42
Infinite Sequences**

**INFINITE SEQUENCES**. An infinite sequence  sn  is a function whose domain is the set of positive integers; sn is the value of this function for a given positive integer _n_. Sometimes we indicate  sn  just by writing the first few terms of the sequences _s_ 1, _s_ 2, _s_ 3, ..., _s n_, ... We shall consider only sequences where the values _s n_ are real numbers.

**EXAMPLE 1**

( _a_ )

( _b_ )

( _c_ )  _n_ 2  is the sequence of squares 1, 4, 9, 16, ..., _n_ 2,...

( _d_ )  2n  is the sequence of positive even integers 2, 4, 6, 8, ..., 2 _n_ ,...

( _e_ )  2 _n_ − 1  is the sequence of positive odd integers 1, 3, 5, 7, ...

**LIMIT OF A SEQUENCE**. If  _s n_  is an infinite sequence and _L_ is a number, then we say that   if _s n_ gets arbitrarily close to _L_ as _n_ increases without bound.

From a more precise standpoint,   means that, for any positive real number   > 0, there exists a positive integer _n_ 0 such that, whenever _n_   _n_ 0, we have | _s n_ − _L_ | <  . To illustrate what this means, place the points _L, L_ −  , and _L_ \+   on a coordinate line (see Fig. 42-1), where   is some positive real number. Now, if we place the points _s_ 1, _s_ 2, _s_ 3,... on the coordinate line, there will eventually be an index _n_ 0 such that   and all subsequent terms of the sequence will lie inside the interval ( _L_ −  , _L_ \+  ).

**Fig. 42-1**

If  , then we say that the sequence  _s n_  converges to _L_. If there is a number _L_ such that  _s n_  converges to L, then we say that  _s n_  is convergent. When  _s n_  is not convergent, then we say that  _s n_  is _divergent_.

**EXAMPLE 2:**   is convergent, since   To see this, observe that 1/ _n_ can be made arbitrarily close to 0 by making _n_ large enough. To get an idea of why this is so, note that 1/10 = 0.1, 1/100 = 0.01, 1/1000 = 0.001, and so on. To check that the precise definition is satisfied, let   be any positive number. Take _n_ 0 to be the smallest positive integer greater than 1/  So, 1/  < _n_ 0. Hence, if _n_   _n_ 0, then _n_ > 1/  and, therefore, 1/ _n_ <   Thus if _n_   _n_ 0, |1/ _n_ − 0| <   This proves  .

**EXAMPLE 3:**  2 _n_  is a divergent sequence, since   for each real number _L_. In fact, 2 _n_ gets arbitrarily large as _n_ increases.

We write   if _s n_ gets arbitrarily large as _n_ increases. In such a case, we say that  _s n_  diverges to +∞ More precisely,   if and only if, for any number _c_ , no matter how large, there exists a positive integer _n_ 0 such that, whenever _n_  _n_ 0, we have _s n_ > _c_.

Likewise, we write   if _s n_ gets arbitrarily small as _n_ increase. In such a case, we say that  _s n_  diverges to –∞. More precisely,   if an only if, for any number _c_ , no matter how small, there exists a positive integer _n_ 0 such that, whenever _n_   _n_ 0 we have _s n_ < _c_.

We shall write   if  , that is, the magnitude of _s n_ gets arbitrarily large as _n_ increase.

**EXAMPLE 4:** ( _a_ )  ; ( _b_ )  ; ( _c_ )  . Note that, in case ( _c_ ), the sequence converges neither to +∞ nor to –∞.

**EXAMPLE 5:** The sequence  (–1) _n_  is divergent, but it diverges neither to +∞ nor to –∞ nor to ∞ Its values oscillate between 1 and –1.

A sequence  _s n_  is said to be _bounded above_ if there is a number _c_ such that _s n_   _c_ for all _n_ , and  _s n_  is said to be _bounded below_ if there is a number _b_ such that _b_   _s n_ for all _n_. A sequence  _s n_  is said to be _bounded_ if it is bounded both above and below. It is clear that a sequence  _s n_  is bounded if and only if there is a number _d_ such that | _s n_|   _d_ for all _n_.

**EXAMPLE 6:** ( _a_ ) The sequence  2 _n_  is bounded below (for example, by 0) but is not bounded above. ( _b_ ) The sequence  (–1) _n_  is bounded. Note that  (–1) _n_  is –1; 1; –1, . . ., So, |(–1) _n_ | 1 for all _n_.

**Theorem 42.1** : Every convergent sequence is bounded.

For a proof, see Problem 5.

The converse of Theorem 42.1 is false. For example the sequence  (–1) _n_  is bounded but not convergent.

Standard arithmetic operations on convergent sequences yield convergent sequences, as the following intuitively obvious results show.

**Theorem 42.2** : Assume  . Then:

( _a_ )  , where _k_ is a constant.

( _b_ )  , where _k_ is a constant.

( _c_ )  .

( _d_ )  .

( _e_ )  .

( _f_ )   provided that _d_ ≠ 0 and _t n_ ≠ 0 for all _n_.

For proofs of parts ( _c_ ) and ( _e_ ), see Problem 10.

The following facts about sequences are intuitively clear.

**Theorem 42.3** :  .

For a proof, see Problem 7.

**Theorem 42.4** :

( _a_ )  .

( _b_ )  .

For proofs, see Problem 8.

**Theorem 42.5 (Squeeze Theorem)** : If  , and there is an integer _m_ such that _s n_   _t n_   _u n_ for all _n_   _m_ , then  .

For a proof, see Problem 11.

**Corollary 42.6** : If   and there is an integer _m_ such that | _t n_|   | _u n_| for all _n_   _m_ , then  .

This is a consequence of Theorem 42.5 and the fact that   is equivalent to  .

**EXAMPLE 7:**  . To see this, use Corollary 42.6, noting that   and  .

**Theorem 42.7** : Assume that f is a function that is continuous at _c_ , and assume that  , where all the terms _s n_ are in the domain of _f_. Then

See Problem 33.

It is clear that whether or not a sequence converges would not be affected by deleting, adding, or altering a finite number of terms at the beginning of the sequence. Convergence depends on what happens "in the long run".

We shall extend the notion of infinite sequence to the case where the domain of a sequence is allowed to be the set of nonnegative or any set consisting of all integers greater than or equal to a fixed integer. For example, if we take the domain to be the set of nonnegative integers, then  2 _n_ \+ 1  would denote the sequence of positive odd integers, and  1/2 _n_  would denote the sequence

**MONOTONIC SEQUENCES**

( _a_ ) A sequence  _s n_  is said to be _nondecreasing_ if _s n_   _s_ _n_ +1 for all _n_.

( _b_ ) A sequence  _s n_  is said to be _increasing_ if _s n_   _s_ _n_ +1 for all _n_.

( _c_ ) A sequence  _s n_  is said to be _nonincreasing_ if _s n_   _s_ _n_ +1 for all _n_.

( _d_ ) A sequence  _s n_  is said to be _decreasing_ if _s n_ > _s_ _n_ +1 for all _n_.

( _e_ ) A sequence is said to be _monotonic_ if it is either nondecreasing or nonincreasing.

Clearly, every increasing sequence is nondecreasing (but not conversely), and every decreasing sequence is nonincreasing (but not conversely).

**EXAMPLE 8:** ( _a_ ) The sequence 1, 1, 2, 2, 3, 3, 4, 4,... is nondecreasing, but not increasing. ( _b_ ) –1, –1, –2, –2, –3, –3, –4, –4,... is nonincreasing, but not decreasing.

An important basic property of the real number system is given by the following result. Its proof is beyond scope of this book.

**Theorem 42.8:** Every bounded monotonic sequence is convergent.

There are several methods for showing that a given sequence  _s n_  is nonincreasing, increasing, nonincreasing, or decreasing. Let us concentrate on the property that  _s n_  is increasing.

_Method 1: Show that s_ _n_ +1 − _s n_ > 0.

**EXAMPLE 9:**

since 4 _n_ \+ 5 > 0 and 4 _n_ \+ 1 > 0

_Method 2: When all s n_ > 0, _show that s_ _n_ +1/ _s n_ > 1.

**EXAMPLE 10:** Using the same example   as above,

since 12 _n_ 2+15 _n_ \+ 3 > 12 _n_ 2 \+ 15 _n_ > 0.

_Method 3: Find a differentiable function f_ ( _x_ ) _such that f_ ( _n_ ) = _s n for all n, and show that f′_( _x_ ) > 0 _for all x_   1 _(and, hence, that f is an increasing function for x_   1 _)_.

**EXAMPLE 11:** Consider   again. Let  . Then   for all _x_.

### **Solved problems**

**1**. For each of the following sequences, write a formula for the _n_ th term and determine the limit (if it exists). It is assumed that _n_ = 1, 2, 3,...

( _c_ )  . This is intuitively clear, but one can also apply Theorem 42.3 to the sequence  (–1) _n_ \+ 1 _n_  , since  .

( _d_ )  .

Note that   by virtue of Theorem 42.4( _b_ ).

( _e_ )  . Note that the sequence consists of repetitions of the cycle 1, 0, –1, 0 and has no limit.

( _f_ )   by ( _26.17_ ).

**2**. Evaluate   in the following cases:

( _a_ ) Recall that   by Chapter 7, Problem 13. Therefore,  . A similar result holds whenever _s n_ is a quotient of polynomials of the same degree.

( _b_ ) Recall that   by Chapter 7, Problem 13. Therefore,  . A similar result holds whenever _s n_ is a rational function whose numerator has greater degree than the denominator (and whose leading coefficients have the same sign).

( _c_ ) Recall that   by Chapter 7, Problem 13. Therefore,  . The same result holds whenever _s n_ is a rational function whose denominator has greater degree than the numerator.

**3**. For each of the following sequencies, determine whether it is nondecreasing increasing, nonincreasing decreasing, or none of these. Then determine its limit, if it exists.

Hence, _f_ ( _x_ ) is an increasing function and, therefore,  _s n_  is an increasing sequence.

 .

Since In   (by ( _25.12_ )), _x_ (In 2) > _x_ /2   1, when _x_   2. Thus 1 − _x_ (In 2) < 0 when _x_   2 and, therefore, _f_ ′ ( _x_ ) < 0 when _x_   2. So, _f_ ( _x_ ) is decreasing for _x_   2 and this implies that _s n_ is decreasing for _n_   2. Note that  . Hence,  _s n_  is nonincreasing. Now let us find the limit. By L'Hôpital's Rule,

( _c_ )  . Hence,  _s n_  is decreasing.

Theorem 42.4( _b_ ) tells us that  .

**4**. Show that the sequence   is convergent.

Let us use Theorem 42.8.  _s_ _n_  is bounded since 0,< _s_ _n_ < 1. Let us show that  _s_ _n_  is decreasing. Note that

**5**. Prove Theorem 42 1: Every convergent sequence  _s_ _n_  is bounded.

Let  . Take   = 1. Then there exists a positive integer _n_ 0 such that, whenever _n_   _n_ 0, we have | _s_ _n_ − _L_ | < 1. Hence, for _n_   _n_ 0, using the triangle inequality, we get

So, if we take _M_ to be the maximum of 1 + | _L_ | and | _s_ 1|, | _s_ 2|,| _s_ 3|, ..., | _s_ n0|, then | _s_ _n_ |   _M_ for all _n_ Thus,  _s_ _n_  is bounded.

**6**. Show that the sequence   is divergent.

Since  , the sequence is not bounded. So, by Theorem 42.1, the sequence cannot be convergent.

**7**. Prove theorem 42.3: If   and _s_ _n_ ≠ 0 for all _n_ , then  .

Consider any   > 0. Since  , there exists some positive integer _m_ such that, whenever _n_   _m_ ,

**8**. Prove Theorem 42.4: ( _a_ ) If |a| > 1, then  ; ( _b_ ) If | _r_ | < 1, then  .

( _a_ )

( _b_ )

**9**. Prove  .

**10.** Prove Theorem 42.2( _c_ ) and ( _e_ ).

( _c_ )  . Then there exist integers _m_ 1 and _m_ 2 such that | _s_ _n_ \- _c_ | <  /2 for _n_   _m_ 1 and | _t_ _n_ − _d_ | <  /2 for _n_ > _m_ 2. Let _m_ be the maximum of _m_ 1 and _m_ 2. So, for _n_   _m_ , | _s_ _n_ − _c_ | <  /2 and | _t_ _n_ − _d_ | <  /2. Hence, for _n_   _m_ ,

( _e_ )  . Since  _s_ _n_  is convergent, it is bounded, by Theorem 42.1 and, therefore, there is a positive number _M_ such that | _s_ _n_ |   _M_ for all _n_. Let   > 0. If _d_ ≠ 0, there exists an integer _m_ 1 such that | _s_ _n_ − _c_ | <  /2| _d_ | for _n_   _m_ 1 and, therefore, | _d_ || _s_ _n_ − _c_ | <  /2 for _n_   M1. If _d_ = 0, then we can choose _m_ 1=1 and we would still have | _d_ || _s_ _n_ − _c_ | <  /2 for _n_   _m_ 1. There also exists _m_ 2 such that | _t_ _n_ − _d_ | <  /2 _M_ for _n_   _m_ 2. Let _m_ be the maximum of _m_ 1 and _m_ 2. If _n_   _m_ ,

**11.** Prove the Squeeze Theorem: If  , and there is an integer _m_ such that _s_ _n_   _t_ _n_   _u_ _n_ for all _n_   _m_ , then  .

Let   > 0. There is an integer _m_ 1   _m_ such that | _s_ _n_ − _L_ | <  /4 and | _u_ _n_ − _L_ | <  /4 for _n_   _m_ 1. Now assume _n_   _m_ 1. Since _s_ _n_  _t_ _n_  _u_ _n_ , | _t_ _n_ − _s_ _n_ |  | _u_ _n_ − _s_ _n_ |. But

Thus, | _t_ _n_ − _s_ _n_ | <  /2. Hence,

### **Supplementary Problems**

In each of Problems 12 – 29, determine for each given sequence  _s_ _n_  whether it is bounded and whether it is nondecreasing, increasing, nonincreasing, or decreasing. Also determine whether it is convergent and, if possible, find its limit. ( _Note:_ If the sequence has a finite limit, it must be bounded. If it has an infinite limit, it must be unbounded.)

In each of Problems 30–32, find a plausible formula for a sequence whose first few terms are given. Find the limit (if it exists) of your sequence.

**33.** Prove Theorem 42.7. ( _Hint:_ Let   > 0. Choose δ > 0 such that, for _x_ in the domain of _f_ for which | _x_ − _c_ | < δ, we have | _f_ ( _x_ ) − _f_ ( _c_ )| <  . Choose _m_ so that _n_   _m_ implies | _s_ _n_ − c| < δ.)

**34.** Show that   for _ρ_ > 0. ( _Hint: n_ _p/n_ = _e_ _(p lnn)/n_ )

**35.** (GC) Use a graphing calculator to investigates   for _n_ = 1 to _n_ = 5. Then determine analytically the behavior of the sequence.

_Ans._ decreasing; limit is

**36.** (GC) Use a graphing calculator to investigates   for _n_ = 1 to _n_ = 10. Then determine analytically the behavior of the sequence.

_Ans._ decreasing for _n_   7; limit is 0

**37.** Prove that   is equivalent to  .

**38.** If _s_ _n_ > 0 for all _n_ and  , prove that  .

**39.** (GC) Define _s_ _n_ by recursion as follows: _s_ 1 = 2 and  .

( _a_ ) Use a graphing calculator to estimate _s_ _n_ for _n_ = 2,...,5.

( _b_ ) Show that, if   exists, then  .

( _c_ ) Prove that   exists.

**40.** Define _s_ _n_ by recursion as follows: _s_ 1 = 3, and   for _n_   1.

( _a_ ) Prove _s_ _n_ < 6 for all _n_.

( _b_ ) Show that ( _s_ _n_ ) is increasing.

_c_ ) Prove that   exists.

( _d_ ) Evaluate  .

_Ans._ ( _d_ ) 6

**41.** Prove Theorem 42.2, parts ( _a_ ), ( _b_ ), ( _d_ ), ( _f_ ).

## **CHAPTER 43
Infinite Series**

Let  _s_ _n_  be an infinite sequence. We can form the infinite sequence of _partial sums_  _S_ _n_  as follows:

We usually will designate the sequence  _S_ _n_  by the notation

The numbers _s_ 1, _s_ 2,..., _s n_,... will be called the _terms_ of the series.

If _S_ is a number such that  , then the series   is said to _converge_ and _S_ is called the _sum_ of the series. We usually designate _S_ by  .

If there is no number S such that  , then the series   is said to _diverge_. If  , then the series is said to diverge to + ∞ and we write  . Similarly, if  , then the series is said to diverge to –∞ and we write  .

**EXAMPLE 1:** Consider the sequence  (–1) _n_ +1 . The terms are _s_ 1 = 1, _s_ 2 = –1, _s_ 3 = 1, _s_ 4 = –1, and so on. Hence, the partial sums begin with _S_ 1 = 1, _S_ 2 = 1 + (–1) = 0, _S_ 3 = 1 + (–1) + 1 = 1, _S_ 4 = 1 + (–1) + (1) + (–1) = 0, and continue with alternating 1s and 0s. So,   does not exist and the series diverges (but not to + ∞ or –∞).

**GEOMETRIC SERIES.** Consider the sequence  _ar_ _n_ –1 , which consists of the terms _a, ar, ar_ 2, _ar_ 3,.... The series   is called a _geometric series_ with ratio _r_ and _first term a_. Its _n_ th partial sum _S n_ is given by

Everything now depends on the ratio _r_. If | _r_ | < 1, then   (by Theorem 42.4( _b_ )) and, therefore,  . If | _r_ | > 1, then   (by Theorem 42.4( _a_ )) and, therefore,  . (A trivial exception occurs when _a_ = 0. In that case, all terms are 0, the series converges, and its sum is 0.) These results are summarized as follows:

**Theorem 43.1:** Given a geometric series  :

( _a_ ) If | _r_ | < 1, the series converges and has sum  .

( _b_ ) If | _r_ | > 1 and a ≠ 0, the series diverges to ∞.

**EXAMPLE 2:** Take the geometric series   with ratio   and first term _a_ = 1:

By Theorem 43.1( _a_ ), the series converges and has sum  . Thus,  .

We can multiply a series   by a constant _c_ to obtain a new series  , and we can add two series   and   to obtain a new series  .

**Theorem 43.2:** If c ≠ 0, then   converges if and only if   converges. Moreover, in the case of convergence,

To obtain this result, denote by _T_ _n_ = _cs_ 1 \+ _cs_ 2 \+ ··· + _cs_ _n_ the _n_ th partial sum of the series  . Then _T_ _n_ = _cS_ _n_ , where _S_ _n_ is the _n_ th partial sum of  . So,   exists if and only if   exists, and when the limits exist,  . This yields Theorem 43.2.

**Theorem 43.3.** Assume that two series   and   both converge. Then their sum   also converges and

To see this, let _S_ _n_ and _T_ _n_ be the _n_ th partial sums of   and  , respectively. Then the _n_ th partial sum _U_ _n_ of   is easily seen to be _S_ _n_ \+ _T_ _n_. So,  . This yields Theorem 43.3.

**Corollary 43.4:** Assume that two series   and   both converge. Then their difference ∑( _s_ _n_ − _t_ _n_ ) also converges and

This follows directly from Theorems 43.2 and 43.3. Just note that ∑(sn − tn) is the sum of ∑sn and the series ∑(–1)tn.

**Theorem 43.5:** If ∑sn converges, then  .

To see this, assume that  . This means that  , where, as usual, _S_ _n_ is the _n_ th partial sum of the series. We also have  . But, _s_ _n_ = _S_ _n_ − _S_ _n_ –1. So,

**Corollary 43.6. (The Divergence Theorem):** If   does not exist or  , then ∑s _n_ diverges.

This is an immediate logical consequence of Theorem 43.5.

**EXAMPLE 3:** The series   diverges.

Here,  . Since  , the Divergence Theorem implies that the series diverges.

The converse of Theorem 43.5 is not valid:   does not imply that ∑ _s_ _n_ converges. This is shown by the following example.

**EXAMPLE 4:** Consider the so-called _harmonic series_   Let us look at the following partial sums of this series:

Continuing in this manner, we would obtain   and, in general,   when _k_ > 1. This implies that   and, therefore, the harmonic series diverges. But notice that  .

_Remark:_ Convergence or divergence is not affected by the addition or deletion of a finite number of terms at the beginning of a series. For example, if we delete the first _k_ terms of a series and the sum of the deleted terms is _c_ , then each new partial sum _T_ _n_ has the form _S_ _n + k_ − _c_. (For example, _T_ 1 is _S_ _k_ \+ 1 − _c_.) But   exists if and only if   exists, and   exists if and only if   exists.

_Notation:_ It will often be useful to deal with series in which the terms of  s _n_  are indexed by the nonnegative integers: _s_ 0, _s_ 1, _s_ 2, _s_ 3,.... Then the partial sums _s_ _n_ would also begin with _S_ 0 = _s_ 0, and the sum of a convergent series would be written as  .

### **Solved Problems**

**1**. Examine the   for convergence.

This is a geometric series with ratio   and the first term  . Since  . Theorem 43.1 ( _a_ ) tells us that the series converges and that its sum is  .

**2**. Examine the series   for convergence.

The _n_ th term is  . This is equal to  . Hence, the _n_ th partial sum

Thus,  . Hence, the series converges and its sum is 1.

**3**. We know that the geometric series   converges to _S_ = 2. Examine the series that results when: ( _a_ ) its first four terms are dropped; ( _b_ ) the terms 3, 2, and 5 are added to the beginning of the series.

( _a_ ) The resulting series is a geometric series   with ratio  . It converges to  .

Note that this is the same as

( _b_ ) The new series is  . The new partial sums are the old ones plus (3 + 2 + 5). Since the old partial sums converge to 2, the new ones converge to 2 + 10 = 12. Thus, the new series is convergent and its sum is 12.

**4**. Show that the series   diverges.

Here,  . Since  , it follows that  . So, by the Divergence Theorem, the series diverges.

**5**. Examine the series   for convergence.

This is a geometric series with ratio  . Since  , Theorem 43.1( _b_ ) tells us that the series diverges.

**6**. Evaluate

This is a geometric series with ratio   and first term _a_ = 1. Since  , the series converges and its sum is  .

**7**. Show that the infinite decimal 0.999... is equal to 1.

 . This is a geometric series with first term   and ratio  .

Hence, it converges to the sum  .

**8**. Examine the series

Here,  . Note that  . Hence, the _n_ th partial sum Sn is

So,  . Thus, the series converges to  .

**9**. Examine the series  .

 . Then  . By the Divergence Theorem, the series diverges.

**10.** Examine the series

This series is obtained from the harmonics series by deleting the first nine terms. Since the harmonic series diverges, so does this series.

**11. (Zeno's Paradox)** Achilles ( _A_ ) and a tortoise ( _T_ ) have a race. _T_ gets a 1000 ft head start, but _A_ runs at 10 ft/sec, whereas _T_ only does 0.01 ft/sec. When _A_ reaches _T_ 's starting point, _T_ has moved a short distance ahead. When _A_ reaches that point, _T_ again has moved a short distance ahead, etc. Zeno claimed that _A_ would never catch _T_. Show that this is not so.

When _A_ reaches _T_ 's starting point, 100 seconds have passed and _T_ has moved 0.01 (100) = 1 ft. _A_ covers that additional 1 ft in 0.1 seconds, but _T_ has moved 0.01(0.1) = 0.001 ft further. A needs 0.0001 seconds to cover that distance, but _T_ mean while has moved 0.01(0.0001) = 0.000001 ft, etc. The limit of the distance between _A_ and _T_ approaches 0. The time involved is 100 + 0.1 + 0.0001 + 0.0000001 +..., which is a geometric series with first term _a_ = 100 and ratio _r_ = 1/1000. Its sum is

which is a litte more than 100 seconds. The seeming paradox arises from the artificial division of the event into infinitely many shorter and shorter steps.

### **Supplementary Problems**

**12.** Examine each of the following geometric series. If the series converges, find its sum.

**13.** A rubber ball is dropped from a height of 10 ft. Whenever it hits the ground, it bounces straight up three-fourths of the previous height. What is the total distance travelled by the ball before it stops?

_Ans._ 70 ft

**14.** Examine the series  .

_Ans._

**15.** Examine the series

_Ans._

**16.** Evaluate   when _s n_ is the following:

_Ans._

**17.** Show that each of the following series diverges:

**18.** Evalute the following

_Ans._

**19.** (GC) In Problems 1 and 6, use a calculator to compute the first ten partial sums and determine to how many decimal places the tenth partial sum is a correct estimate of the sum of the series.

**20.** (GC) ( _a_ ) If | _x_ | < 1, what function is represented by

( _b_ ) Use a graphing calculator to graph 1 + _x_ \+ _x_ 2 \+ _x_ 3 \+ ··· + _x_ 9 on the interval (–1, 1) and compare the graph with that of the function in ( _a_ ).

_Ans._ ( _a_ )

**21.** In each of the following, find those values of _x_ for which the given series converges, and then find the function represented by the sum of the series for those values of _x_.

_Ans._

## **CHAPTER 44
Series with Positive Terms. The Integral Test. Comparison Tests**

**SERIES OF POSITIVE TERMS.** If all the terms of a series Σ _s_ _n_ are positive, then the series is called a _positive series_.

For a positive series Σ _s_ _n_ , the sequence of partial sums  _S_ _n_  is an increasing sequence, since _S_ _n_ +1 = _S_ _n_ \+ _S_ _n_ +1 > _S_ _n_. The yields the following useful result.

**Theorem 44.1** A positive series Σ _s n_ converges if and only if the sequence of partial sums  _s_ _n_  is bounded.

To see this, note first that, if Σ _s_ _n_ converges, then, by definition,  _S_ _n_  converges and, therefore, by Theorem 42.1,  _S_ _n_  is bounded. Conversely, if  _S_ _n_  is bounded, then, since  _S_ _n_  is increasing, Theorem 42.8 implies that  _S_ _n_  converges, that is, Σ _s_ _n_ converges.

**Theorem 44.2 (Integral Test):** Let Σ _s_ _n_ be a positive series and let _f_ ( _x_ ) be a continuous, positive decreasing function on [1,+∞) such that _f_ ( _n_ ) = _s_ _n_ for all positive integers _n_. Then:

Σ _s_ _n_ converges if and only if   _dx_ converges

From Fig. 44-1 we see that  . If Σ _s_ _n_ converges, then  _S_ _n_  is bounded; so   will be bounded for all _u_   1 and, therefore,   converges. Conversely, from Fig. 44-1 we have   and, therefore,  . Thus, if   _dx_ converges, then   and so,  _S_ _n_  will be bounded. Hence, by Theorem 44.1, Σ _s_ _n_ converges. This proves Theorem 44.2.

**Fig. 44-1**

**Example 1:**   diverges.

Let  . Now,

Hence, by the integral test,   diverges.

**Example 2:**   converges.

Let  . Now,

Hence, by the integral test,   converges.

_Remark:_ The integral test can be easily be extended to the case where the lower limit of the integral is changed from 1 to any positive integer.

**Theorem 44.3 (Comparison Test).** Let Σ _a_ _n_ and Σ _b_ _n_ be two positive series such that there is a positive integer _m_ for which _a_ _k_   _b_ _k_ for all integers _k_   _m_. Then:

(1) If Σ _b_ _n_ converges, so does Σ _a_ _n_ ;

(2) If Σ _a_ _n_ diverges, so does Σ _b_ _n_.

We may assume in the derivation of Theorem 44.3 that _m_ = 1, since convergence is not affected by deletion of a finite number of terms at the beginning of a series. Note also that (2) is a logical consequence of (1). To prove (1), assume that Σ _b_ _n_ converges. Let _B_ _n_ = _b_ 1 \+ _b_ 2 \+... + _b_ _n_ be the _n_ th partial sum for Σ _b_ _n_ , and let _A_ _n_ = _a_ 1 \+ _a_ 2 \+... + _a_ n be the _n_ th partial sum for Σ _a_ _n_. Then _A_ _n_   _B_ _n_ , since _a_ _k_   _b_ _k_ for all _k_. From the fact that Σ _b_ _n_ converges, it follows, by Theorem 44.1, that the sequence ( _B_ _n_ ) is bounded. Since _A_ _n_   _B_ _n_ for all _n_ , it follows that the sequence ( _A_ _n_ ) is bounded. So, by Theorem 44.1, Σ _a_ _n_ converges. This proves Theorem 44.3.

**EXAMPLE 3:**   converges.

Let   and  . Then _a_ _n_ < _b_ _n_ for all _n_. By Example 2,   converges. So, by the comparison test,   converges.

**EXAMPLE 4:**   diverges.

Let   and  . Now, _a n_  _b n_ for _n_   5. (To see this, observe that   is equivalent to 3 _n_ +5  4 _n_ , which is equivalent to 5   _n_.) Recall that the harmonic series   diverges (by Chapter 43, Example 4). Hence,   diverges by Theorem 43.2. The comparison test implies that   diverges.

Sometimes, as in Example 4, complicated maneuvers are needed in order to apply the comparison test. The following result offers a much more flexible tool.

**Theorem 44.4 (Limit Comparison Test):** Let Σ _a_ _n_ and Σ _b n_ be two positive series such that   exists and 0 < _L_ < +∞. Then Σ _a_ _n_ converges if and only if Σ _b n_ converges.

Assume that Σ _b n_ converges. Let _c_ be a positive number such that _L_ < _c_. Then there exists a positive integer _m_ such that _a_ n/ _b_ _n_ < _c_ for all _n_   _m_. Hence, _a n_ < _cb n_ for all _n  m_. But, since Σ _b n_ converges, so does Σ _cb n_. Therefore, by the comparison test, Σ _a n_ converges. Conversely, if Σ _a n_ converges, then Σ _b n_ converges. (In fact,   and we can use the same kind of argument that was just given.)

**EXAMPLE 5:**   diverges.

When dealing with quotients of polynomials, a good rule of thumb is to ignore everything except the leading terms. In this case, we have  . Let us try a limit comparison with  . Now

Since   diverges, the limit comparison test tells us that   diverges.

**EXAMPLE 6:**   converges.

Using the rule of thumb given in Example 5, we should look at  . So, let us try a limit comparison with  :

Let us divide the numerator and denominator by _n_ 3. Note that, in the denominator, we would get

So, the result would be

Hence, since we know, by Example 2, that   converges, the limit comparison test implies that   converges.

### **Solved Problems**

**1**. Consider the series  , where _ρ_ is constant. This is called a _p-series_. Then:

( _a_ ) If _ρ_ > 1, the series   converges.

( _b_ ) If _ρ_   1, the series   diverges.

We may assume that _ρ_ ≠ 1, since we already know that the harmonic series   diverges. We may also assume that _ρ_ > 0; if _ρ_   0,   and the divergence theorem implies that the series diverges. Let us apply the integral test with _f_ ( _x_ )=1/ _x p_. ( _f_ ( _x_ ) is positive and decreasing in [1,+∞).) Now,

( _a_ ) _ρ_ > 1. Then _ρ_ − 1 > 0 and  . So,  . By the integral test,   converges.

( _b_ ) _ρ_ < 1. Then 1 − _ρ_ > 0 and   and, by the integral test,   diverges.

In Problems 2–7, examine the given series for convergence.

**2**.  .

Hence, the series diverges by the integral test.

**3**.

  is convergent, since it is a _ρ_ -series with _ρ_ = 3 > 1. Thus, by the comparison test,   is convergent.

**4**.  .

  Note that   for _n_   2. Since   is a convergent geometric series (with ratio  ),   is convergent by the comparison test.

**5**.  .

 . Use limit comparison with  .

We know that   converges. So, by the limit comparison test,   converges.

**6**.  .

 . Now,   is a convergent geometric series  . So, by the comparison test,   converges.

**7**.  .

 . Use limit comparison with

We know that the harmonic series   diverges. So, by the limit comparison test,   diverges.

**8**.  .

  is defined for _n_   2.

Hence, the series diverges by the integral test.

**9**. How many terms of   suffice to obtain two-decimal place accuracy (that is, an error < 5/103)?

If we use _k_ terms, then we require that the error

Hence, 200 < _k_. Thus, it suffices to use 201 terms of the series. (The graphing calculator can be used to find

**10.** Assume Σ _s_ _n_ converges by virtue of the integral test applied to _f_ ( _x_ ) and, for each _n_ , the error (or remainder) _R_ _k_ after _k_ terms is defined to be

Find a bound on the error when   is approximated by the first five terms:

 .

The error

**11.** Assume Σ _s_ _n_ and Σ _c_ _n_ are positive series, Σ _c_ _n_ converges, and _s_ _n_   _c_ _n_ for all _n_. Then the error _R_ _k_ after _k_ terms is

At least how many terms will suffice to estimate   with an error < 0.00001?

In this case,   and  . It suffices to have  . Now,  . So, we need  . Equivalently, 100,00 < 4 _k_ 4, 25,000 < _k_ 4, _k_   13.

### **Supplementary Problems**

For Problems 12–43, determine whether the series converges.

**29.**

**30.**

**31.**

**32.**

**33.**

**34.**

**35.**

**36.**

**37.**

**38.**

**39.**

**40.**

**41.**

**42.**

**43.**

**44.** (GC) Estimate the error when:

( _a_ )   is approximated by the sum of its first six terms.

( _b_ )   is approximated by the sum of its first six terms.

_Ans._ ( _a_ ) 0.0007; ( _b_ )0.00009

**45.** (GC) ( _a_ ) Estimate the error when the geometric series  is approximated by the sum of its first six terms.

( _b_ ) How many terms suffice to compute the sum if the allowable error is 0.00005?

_Ans._ ( _a_ ) 0.047; ( _b_ ) 16

**46.** (GC)( _a_ ) How many terms suffice to approximate   with an error < 0.001?

( _b_ ) Find a bound on the error if we approximate   by the sixth partial sum

( _c_ ) What is your approximation to   by the sixth partial sum, correct to four decimal places?

_Ans._ ( _a_ ) 7; ( _b_ ) 0.0015; ( _c_ ) 1.0811

**47.** (GC) Let _S_ _n_ be the _n_ th partial sum   of the divergent harmonic series.

( _a_ ) Prove ln ( _n_ \+ 1)   _S_ _n_  1 + ln _n_.

( _b_ ) Let _E_ _n_ = _S_ _n_ − ln _n_. Prove that  _E_ _n_  is bounded and decreasing.

( _c_ ) Prove that  _E_ _n_  converges. Its limit is denoted γ and is called _Euler's constant_.

( _d_ ) Use a graphing calculator to approximate _E_ 999 to eight decimal places.

_Ans._ ( _d_ ) 0.57771608 (In fact, γ ~ 0.57721566.)

**48.** (Extension of the limit comparison test.) Assume Σ _s_ _n_ and Σ _t_ _n_ are positive series. Prove:

( _a_ ) If   and Σ _t_ _n_ converge so does Σ _s_ _n_.

( _b_ ) If   and Σ _t_ _n_ diverges, so does Σ _s_ _n_.

**49.** Use the extension of the limit comparison test to determine whether   converges.

_Ans._ converges; use   and Problem 48( _a_ )

**50.** Assume Σ _s_ _n_ is a positive series and   exists and is positive. Prove that Σ _s_ _n_ diverges.( _Hint:_ Limit comparison with Σ (1/ _n_ ).)

**51.** Assume Σ _s_ _n_ and Σ _t_ _n_ are convergent positive series. Prove that Σ _s_ _n_ _t_ _n_ converges.

## **CHAPTER 45
Alternating Series. Absolute and Conditional
Convergence. The Ratio Test**

**ALTERNATING SERIES**. A series whose terms are alternately positive and negative is said to be an _alternating series_. It can be written in the form

∑(–1) _n_ +1 _a_ n = _a_ 1 − _a_ 2 \+ _a_ 3 − _a_ 4 \+ _a_ 5 − ···

where _a_ n are all positive.

**Theorem 45.1 (Alternating Series Theorem):** Let ∑(–1) _n_ +1 _a_ n be an altemating series. Assume that: (1) the sequence  _a_ _n_  is decreasing; (2)  . Then:

(I) Σ(–1)n+1 _a_ _n_ converges to a sum _A_ , and

(II) If _A n_ is the _n_ th partial sum and _R n = A − An_ is the corresponding error, then | _R n_| < _a_ _n_ +1 (that is, the error is less in magnitude than the first term omitted).

(I) Since  _a_ _n_  is decreasing, _a_ 2 _n_ +1 > _a_ 2 _n_ +2 and, therefore, _a_ 2 _n_ +1 − _a_ 2 _n_ +2 > 0. Hence,

A2 _n_ +2 = ( _a_ 1 − _a_ 2) + ( _a_ 3 − _a_ 4) + ··· + ( _a_ 2 _n_ –1 − _a_ 2 _n_ ) + ( _a_ 2 _n_ +1 − _a_ 2 _n_ +2)

= _A_ 2 _n_ \+ ( _a_ 2 _n_ +1 − _a_ 2 _n_ +2) > _A_ 2 _n_ > 0

So, the sequence  A2 _n_  is increasing. Also,

A2 _n_ = _a_ 1 − ( _a_ 2 − _a_ 3) − ( _a_ 4 − _a_ 5) − ··· − ( _a_ 2 _n_ –2 − _a_ 2 _n_ –1) − _a_ 2 _n_ < _a_ 1

Hence,  A2 _n_  is bounded. Therefore, by Theorem 42.8,  A2 _n_  converges to a limit _L_. Now, _A_ 2 _n_ +1 = A2 _n_ \+ _a_ 2 _n_ +1. Hence,

Thus,   _A n_ = _L_ and, therefore, ∑(–1)n+1 _a_ n converges.

(II) _R_ 2 _n_ = ( _a_ 2 _n_ +1 − _a_ 2 _n_ +2) + ( _a_ 2 _n_ +3 − _a_ 2 _n_ +4) + ··· > 0, and _R_ 2 _n_ = _a_ 2 _n_ +1 − ( _a_ 2 _n_ +2 − _a_ 2 _n_ +3) − ( _a_ 2 _n_ +4 − _a_ 2 _n_ +5) − ··· < _a_ 2 _n_ +1. Hence, | _R_ 2 _n_ | < _a_ 2 _n_ +1. For odd indices, _R_ 2 _n_ +1 = –( _a_ 2 _n_ +2 − _a_ 2 _n_ +3)–( _a_ 2 _n_ +4 − _a_ 2 _n_ +5) − ··· < 0 and _R_ 2 _n_ +1 = − _a_ 2 _n_ +2 \+ ( _a_ 2 _n_ +3 − _a_ 2 _n_ +4) + ( _a_ 2 _n_ +5 − _a_ 2 _n_ +6) + ··· > − _a_ 2 _n_ +2. Hence, | _R_ 2 _n_ +1| < _a_ 2 _n_ +2. Thus, for all _k_ , | _R k_| < _a_ _k_ +1.

**EXAMPLE 1:** The alternating harmonic series

converges by virtue of the Alternating Series Theorem. By part (II) of that theorem, the magnitude | _R n_| of the error after _n_ terms is less than  . If we want an error less than 0.1, then it suffices to take  , which is equivalent to 10   _n_ \+ 1. So, _n_   9. Thus, we must use

**Definition:** Consider an arbitrary series Σ _s n_.

Σ _s n_ is said to be _absolutely convergent_ if Σ| _s n_| is convergent.

Σ _s n_ is said to be _conditionally convergent_ if it is convergent but not absolutely convergent.

**EXAMPLE 2:** The alternating harmonic series   is conditionally convergent.

**EXAMPLE 3:** The series   is absolutely convergent.

We shall state without proof two significant results about absolute and conditional convergence. In what follows, by a rearrangement of a series we mean a series obtained from the given series by rearranging its terms (that is, by changing the order in which the terms occur).

(1) If Σ _s n_ is absolutely convergent, then every rearrangement of Σ _s n_ is convergent and has the same as Σ _s n_.

(2) If Σ _s n_ is conditionally convergent, then if _c_ is any real number or +∞ or –∞, there is a rearrangement of Σ _s n_ with sum _c_.

**Theorem 45.2:** If a series is absolutely convergent, then it is convergent.

For a proof, see Problem 1.

Note that a positive series is absolutely convergent if and only if it is convergent.

The following test is probably the most useful of all convergence tests.

**Theorem 45.3 (The Ratio Test):** Let Σ _s n_ be any series.

(1) If  , then Σ _s n_ is absolutely convergent.

(2) If   and ( _r_ > 1 or _r_ = + ∞), then Σ _s n_ diverges.

(3) If  , then we can draw no conclusion about the convergence or divergence of Σ _s n_.

For a proof, see Problem 14.

**Theorem 45.4 (The Root Test)** : Let Σ _s n_ be any series.

(1) If  , then Σ _s n_ is absolutely convergent.

(2) If   and ( _r_ > 1 or _r_ = +∞) then Σ _s n_ diverges.

(3) If  , then we can draw no conclusion about the convergence or divergence of Σ _s n_.

For a prof, see Problem 15.

**EXAMPLE 4:** Consider the series  . Then  . So, by the root test, the series converges absolutely.

### **Solved Problems**

**1.** Show that, if Σ _s n_ is absolutely convergent, then it is convergent.

0  _s n \+ |sn|_   2| _s n_|. Since Σ _|s n|_ converges, so does Σ2| _s n|_. Then, by the comparision test, Σ( _s n \+ |sn_|) converges. Hence, Σ _s n_ = Σ(( _s n \+ |sn_|) − | _s n_|) converges by Corollary 43.4.

In Problems 2–13, determine whether the given series converges absolutely, conditionally, or not at all.

**2.**  .

  converges by comparision with the convergent _p_ -series  . So,   is absolutely convergent.

**3.**  .

 . The series   converges by the integral test (using  ). Hence,   is absolutely convergent.

**4.**  .

 . Since   is a decreasing sequence, the series converges by virtue of the alternating series test. But   is divergent, since it is a _p_ -series with  .

**5.**  .

The series  . is a geometric series with ratio  . Since | _r_ | < 1, it converges and, therefore, the given series is absolutely convergent.

**6.**  .

 . Let us apply the ratio test:

Hence, the given series is absolutely convergent.

**7.**  .

 . Look at   converges by comparision with the convergent _p_ -series  . Hence, the given series is absolutely convergent.

**8.**  .

 . Note that   is a decreasing sequence  . Hence, the given series is convergent by the alternating series theorem. However,  . So, Σ| _s n_| diverges by comparision with  . Thus, the given series is conditionally convergent.

**9.**  .

Hence,   and, therefore, the series is absolutely convergent.

**10.**  .

 . Since   is a decreasing sequence for _n_   2, the given series converges by the alternating series theorem. The series Σ| _s n_| is divergent by limit comparision with  . Hence, the given series is conditionally convergent.

**11.**  .

 . Σ| _s n_| is convergent by limit comparision with  . Hence, the given series is absolutely convergent.

**12.**  .

 . Apply the ratio test:

Thus,  . So the given series is absolutely convergent.

**13.**  .

Apply the ratio test:

So,  . Hence, the given series is absolutely convergent.

**14.** Justify the ratio test (Theorem 45.3).

( _a_ ) Assume  . Choose _t_ such that _r_ < _t_ < 1. Then there exists a positive integer _m_ such that, if  . Hence,

| _s_ m+1|   _t_ | _s_ m|, |sm+2|   _t_ |sm+1|   t2|sm|,..., | _s_ m+k|   _t_ _k_ |sm|

But, Σ _t_ k| _s m_| is a convergent geometric series (with ratio _t_ < 1). So, by the comparision test, Σ| _s n_| converges. Hence, Σ _s n_ is absolutely convergent.

( _b_ ) Assume   and ( _r_ > 1 or _r_ = +∞). Choose _t_ so that 1 < _t_ < _r_. There exists a positive integer _m_ such that, if  . Hence,

| _s_ m+1|   _t_ | _sm_ |, | _s_ m+2|   _t_ | _s_ m+1|   _t_ 2| _s_ m|,... | _s_ m+k|   _t_ k| _s_ m|

Therefore,   and, by the divergence theorem, Σ _s_ n diverges.

( _c_ ) Consider  . In this case, the series diverges. Now consider  :

In this case, the series converges.

**15.** Justify the root test (Theorem 45.4).

( _a_ ) Assume  . Choose _t_ so that _r_ < _t_ < 1. Then, there exists a positive integer _m_ such that   for _n_   _m_. Hence, | _s_ _n_ |   _t_ n for _n_   _m_ , Therefor, Σ| _s_ _n_ | converges by comparision with the convergent geometric series Σ _t_ n. So, Σ _s_ n is absolutely convergent.

( _b_ ) Assume  . Choose _t_ so that 1 < _t_ < _r_. For some positive integer _m_ ,  . Then | _s_ n|   _t_ n for _n_   _m_. Since  . So, by the divergence theorem, Σ _s_ n diverges.

( _c_ ) Consider   and  . In both cases,  . (Note that  .)

In Problems 16–22, use the ratio test to test the series for convergence.

**16.**  .

So, the series converges by the ratio test.

**17.**  .

Hence,   and the series diverges by the ratio test.

**18.**  .

So,  . Hence, the series converges by the ratio test.

**19.**  .

So,  . Hence, the series converges by the ratio test.

**20.**  .

Then  . So the ratio test yields no conclusion. However, limit comparision with   shows that the series diverges.

**21.**

Hence, the series converges by the ratio test.

**22.**

Hence, the series diverges by the ratio test.

### **Supplementary Problems**

In Problems 23–40, determine whether the given alternating series is absolutely convergent, conditionally convergent, or divergent.

**41.** (GC) How many terms of   will suffice to get an approximation within 0.0005 of the actual sum? Find that approximation.

_Ans_.

**42.** (GC) How many terms of   will suffice to get an approximation of the actual sum with an error < 0.001? Find that approximation.

_Ans_. _n_ = 3; 0.842.

**43.** (GC) How many terms of   will suffice to get an approximation of the actual sum with an error < 0.001? Find that approximation.

_Ans_. _n_ = 1000; 0.693

In Problems 44–49, determine whether the series converges.

**50.** Determine whether   is absolutely convergent, conditionally convergent, or divergent.

_Ans_. conditionally convergent

In Problems 51 and 52, find the number of terms that suffice to approximate the sum of the given series to four-decimal place accuracy (that is, with an error < 5/105) and compute the approximation.

**51.**

_Ans_. _n_ = 6; 0.9721

**52.**

_Ans_. _n_ = 4; 0.8415

**53.** Let | _r_ | < 1

( _a_ ) Prove that Σ _nr_ _n_ = _r_ \+ 2 _r_ 2 \+ 3 _r_ 3 \+ 4 _r_ 4 \+ ··· converges.

( _b_ ) Show that  . ( _Hint:_ Let _S_ = r + 2 _r_ 2 \+ 3 _r_ 2 \+ 4 _r_ 4 \+ ···, multiply this equation by _r_ , and substract the result from the original equation.)

( _c_ ) Show that  .

## **CHAPTER 46
Power Series**

**POWER SERIES.** An infinite series

is called a _power series_ in _x_ about _c_ with coefficients  _a n_ . An important special case

is a power series about 0.

For a given value of _x_ , the series ( _46.1_ ) either converges or diverges. Hence, ( _46.1_ ) determines a function _f_ whose domain is the set of all _x_ for which ( _46.1_ ) converges and whose corresponding value _f_ ( _x_ ) is the sum of the series.

Note that ( _46.1_ ) converges when _x_ = _c_.

**EXAMPLE 1:** The power series about 0

is a geometric series with ratio _x_. Thus, it converges for | _x_ | < 1, and its sum is  . So, the domain of the corresponding function is an interval around 0.

**Theorem 46.1:** Assume that the power series   converges for _x_ 0 ≠ _c_. Then it converges absolutely for all _x_ such that | _x_ − _c_ | < | _x_ 0 − _c_ | (that is, for all _x_ that are closer to _c_ than _x_ 0).

For a proof, see Problem 4.

**Theorem 46.2:** For a power series  , one of the following three cases holds:

( _a_ ) it converges for all _x_ ; or

( _b_ ) it converges for all _x_ in an open interval ( _c_ − _R_ 1, _c_ \+ _R_ 1) around _c_ , but not outside the closed interval [ _c_ − _R_ 1, _c_ \+ _R_ 1]; or

( _c_ ) it converges only for _x_ = _c_.

By the _interval of convergence_ of   we mean:

In case ( _a_ ): (–∞, +∞)

In case ( _b_ ): ( _c − R_ 1, _c_ \+ _R_ 1)

In case ( _c_ ): { _c_ }

By the _radius of convergence_ of   we mean:

In case ( _a_ ): ∞

In case ( _b_ ): _R_ 1

In case ( _c_ ): 0

_Note_ : In case ( _b_ ), whether the power series converges at neither endpoint of its interval of convergence or at one or both of those endpoints depends upon the given series.

For a sketch of a proof of Theorem 46.2, see Problem 5.

**EXAMPLE 2:** The power series

is a power series about 2. Let us use the ratio test to find the interval of convergence.

 . Thus,

So, by the ratio test, the series converges absolutely for | _x_ – 2| < 1. The latter inequality is equivalent to –1 < _x_ –2 < 1, which, in tum, is equivalent to 1 < _x_ < 3. Hence, the interval of convergence is (1, 3) and the radius of convergence is 1. At the endpoint _x_ = 1, the series becomes  , which converges by the alternating series theorem. At the endpoint _x_ = 3, the series becomes  , the divergent harmonic series. Thus, the power series converges for 1   _x_ < 3.

**EXAMPLE 3:** The power series

is a power series about 0. (Recall that 0! = 1.) Let us use the ratio test:

Hence, by the ratio test, the series converges (absolutely) for all _x_. Its interval of convergence is (–∞, +∞) and its radius of convergence is ∞.

**EXAMPLE 4:** The power series

is a power series about 0. Let us use the ratio test again:

except when _x_ = 0. Thus, the series converges only for _x_ = 0. Its (degenerate) "interval" of convergence is {0} and its radius of convergence is 0.

**UNIFORM CONVERGENCE.** Let  _f n_  be a sequence of functions, all defined on a set _A_. Let _f_ be a function defined on _A_. Then  _f n_  is said to _converge uniformly_ to _f_ on _A_ if, for every   > 0, there exists a positive integer _m_ such that, for each _x_ in _A_ and every _n_   _m_ , | _f n(x) − f(x)_| <  .

**Theorem 46.3:** If a power series   converges for _x_ 0 ≠ _c_ and _d_ < | _x_ 0 − _c_ |, then the sequence of partial sums  _S_ k( _x_ ) , where  , converges uniformly to   on the interval consisting of all _x_ such that | _x_ − _c_ | < _d_. Hence, the convergence is uniform on any interval strictly inside the interval of convergence.

The reader is referred to more advanced books on analysis for a proof of this result.

**Theorem 46.4:** If (ƒn) converges uniformly to ƒ on a set _A_ and each ƒn is continuous on _A_ , then _f_ is continuous on _A_.

For a proof, see Problem 6.

**Corollary 46.5:** The function defined by a power series   is continuous at all points within its interval of convergence.

This follows from Theorems 46.3 and 46.4.

**Theorem 46.6: (Integration of Power Series):** Let ƒ be the function defined by a power series   on its interval of convergence (with radius of convergence _R_ 1). Then:

( _a_ )

where the interval of convergence of the power series on the right side of formula ( _46.3_ ) is the same as that of the original series. _K_ is an arbitrary constant of integration. Note that the antiderivative of ƒ is obtained by term-by-term integration of the given power series.

( _b_ ) If _a_ and _b_ are in the interval of convergence, then:

Thus,   is obtained by term-by-term integration.

For a proof of Theorem 46.6, the reader should consult a more advanced book on analysis.

**Theorem 46.7 (Differentiation of Power Series):** Let ƒ be the function defined by a power series   on its interval of convergence (with radius of convergence _R_ 1). Then ƒ is differentiable in that interval and

Thus, the derivative _f_ ′ is obtained by term-by-term differentiation of the power series. The interval of convergence of the power series on the right side of formula ( _46.5_ ) will be the same as for the original power series.

For a proof, the reader is referred to more advanced texts in analysis.

**EXAMPLE 5:** We already know by Example 1 that, for | _x_ | < 1,

Now,  . So, by Theorem 46.7,

**EXAMPLE 6:** We know already that

Replace _x_ by – _x_. (This is permissible, since |– _x_ | = | _x_ | < 1.) The result is

By Theorem 46.6( _a_ ), we can integrate term by term:

Letting _x_ = 0 and noting that In 1 = 0, we find that _K_ = 0.

Note also that, for | _x_ | < 1, we have –1 < _x_ < 1, 0 < 1 + _x_ < 2, and, therefore, |1 + _x_ | = 1 + _x_. Hence,

The ratio test shows that this series converges.

If we replace _x_ by _x_ − 1, we obtain:

Note that | _x_ –1| > 1 is equivalent to 0 < _x_ < 2.

Thus, In _x_ is definable by a power series within (0, 2).

**Theorem 46.8 (Abel's Theorem):** Assume that the power series   has a finite interval of convergence | _x_ − _c_ | < _R_ 1 and let ƒ be a function whose values in that interval are given by that power series. If the power series also converges at the right-hand endpoint _b = c_ \+ _R_ 1 of the interval of convergence, then   exists and is equal to the sum of the series at _b_. The analogous result holds at the left-hand endpoint _a_ = _c_ − _R_ 1.

The reader is referred to advanced books on analysis for a proof.

**EXAMPLE 7:** This is a continuation of Example 6. By formula ( _46.8_ ),

At the right-hand endpoint _x_ = 1 of the interval of convergence, the power series becomes the convergent alternating harmonic series

By Abel's Theorem, this series is equal to  . So,

**EXAMPLE 8:** Start again with

Replace _x_ by – _x_ 2, obtaining

Since |– _x_ 2| < 1 is equivalent to | _x_ | < 1, ( _46.1_ ) holds for | _x_ | < 1.

Now, by Theorem 46.6( _a_ ), the antiderivative tan–1 _x_ of   can be obtained by term-by-term integration:

Here _K_ is the constant of integration. If we let _x_ = 0 and note that tan–1 0 = 0, it follows that _K_ = 0. Hence,

At the right-hand endpoint _x_ = 1 of the interval of convergence, the series in ( _46.12_ ) becomes

which converges by virtue of the alternating series theorem. So, by Abel's Theorem,

**EXAMPLE 9:** We know already, by Example 3, that   converges for all _x_. Let   for all _x_. By term-by-term differentiation (Theorem 46.7),

Note that _f_ (0) = 1. Therefore, by formula ( _28.2_ ), _f_ ( _x_ ) = _e_ _x_. Thus,

### **Solved Problems**

**1.** Find the interval of convergence of the power series

and identify the function represented by this power series.

Use the ratio test:

Hence, the interval of convergence is | _x_ − 2| < 1. (This is equivalent to − 1 < _x_ − 2 < 1. which, in turn, is equivalent to 1 < _x_ < 3.) At the right endpoint _x_ = 3, the series is the divergent harmonic series, and, at the left endpoint _x_ = 1, the series is the negative of the convergent alternating harmonic series. So, the series converges for 1   _x_ < 3.

Let  . By Theorem 46.7,  . This series is a geometric series with first term 1 and ratio ( _x_ − 2); so its sum is  . Thus,  . Hence,  . Now,

  and –ln |3–2| + _C_ = 0. So, _C_ = 0

Moreover, since _x_ < 3 in the interval of convergence, 3 − _x_ > 0 and, therefore, |3 − _x_ | = 3 − _x_. Thus, _h(x)_ = ln(3 − _x_ ).

In Problems 2 and 3, find the interval of convergence of the given series and the behavior at the endpoints (if any).

**2.**

Use the ratio test:

 . Hence,

Hence, the interval of convergence is | _x_ | < 1. The radius of convergence is 1. At _x_ = 1, we obtain the convergent _p_ -series with _p_ = 2. At _x_ = –1, the series converges by the alternating series test. Thus, the series converges for − 1   _x_   1.

**3.**

Use the ratio test:

 . Hence,

Hence, the interval of convergence is | _x_ \+ 1| < 1. This is equivalent to –1 < _x_ \+ 1 < 1, which, in turn, is equivalent –2 < _x_ < 0. The radius of convergence is 1. At the right endpoint _x_ = 0, we get the divergent _p_ -series  . At the left endpoint _x_ = –2, we get the alternating series  , which converges by the alternating series theorem. Thus, the series converges for –2   _x_ < 0.

**4.** Prove Theorem 46.1.

Since   converges,   by Theorem 43.5. Hence, there is a positive number _M_ such that | _a n_| | _x_ 0 − _c_ | _n_ < _M_ for all _n_ , by Theorem 42.1. Assume | _x_ − _c_ | < | _x_ 0 − _c_ |. Let

  Then, | _a n_|| _x_ − _c_ | _n_ = | _a n_|| _x_ 0 − _c_ | _n rn_ < _Mr n_

Therefore   is convergent by comparision with the convergent geometric series  . Thus,   is absolutely convergent.

**5.** Prove Theorem 46.2.

Only a very intuitive argument is possible here. Assume that neither case ( _a_ ) nor case ( _c_ ) holds. Since case ( _a_ ) does not hold, the power series does not converage for some _x ≠ c_. Since case ( _c_ ) does not hold, the series does converge for some _x ≠ c._ Theorem 46.1 implies that there is an interval ( _c–K, c+K_ ) around _c_ in which the series converges. The interval of convergence is the maximal such interval. (Using Theorem 46.1, one takes the "least upper bound" _R 1_ of all _K_ such that the series converges in ( _c − K, c + K_ ). Then, ( _c − R 1, c + R1_) is the desired interval.)

**6.** Prove Theorem 46.4.

Assume _x_ is in _A_ and   > 0. Since  _f n_  converges uniformly to _f_ on _A_ , there is a positive integer _m_ such that, if _n_   _m_ , then | _f n(y)–f(y)_| <  /3 for all _y_ in _A_. Since _f m_ is continuous at _x_ , there exists δ > 0 such that, for any _x_ * in _A_ , if | _x*–x_ | < δ, then | _f m(x*)–fm(x)_| <  /3. Hence if | _x *–x_| < δ,

This proves the continuity of _f_ at _x_.

**7.** If  _f n_  converges uniformly to _f_ on [ _a, b_ ] and each _f n_ is continuous on [ _a, b_ ] then  .

Assume   > 0. There is positive integer _m_ such that, if _n  m_, then   for all _x_ in [ _a, b_ ]. Therefore,  . Then

  for _n  m_

**8.** Prove that the function _f_ defined by a power series is continuous within its interval of convergence (Corollary 46.5).

  and the convergence is uniform by Theorem 46.3. Each _S n(x)_, being a polynomial, is continuous. Hence, _f_ is continuous by Theorem 46.4.

**9.** Find a power series about 0 that represents the function  . In what interval is the representation valid?

By formula ( _46.11_ ),   for |x| < 1. Hence

The series diverges at both endpoints _x_ = 1 and _x_ = –1 by the divergence theorem.

In Problems 10 and 11, use the ratio test to find the interval of convergence and indicate what happens at the endpoints (if any).

**10.**  .

 . Hence,

We get convergence when | _x_ |/10 < 1, that is, when | _x_ | < 10. That is the interval of convergence. The series diverges at both endpoints ±10, by the divergence theorem.

**11.**  .

  Hence,

So, the interval of convergence is | _x_ –π| < 3. The series diverges at both endpoints, by the divergence theorem.

**12.** Find the interval of convergence of  .

Apply the ratio test:

So, the interval of convergence is | _x_ | < 4.

**13.** Find a power series about 0 that represents  .

Start with  . Replace _x_ by _x_ 3:

  for | _x_ | < 1

(since | _x_ 3| < 1 is equivalent to | _x_ | < 1.) Multiply by _x_ :

  for | _x_ | < 1

In Problems 14–16, find simple formulas for the function _f_ ( _x_ ) represented by the given power series.

**14.**

Let

 .

Hence,  .

**15.**

Let  . Then

This is a geometric series with ratio _x_ 3. So, it converges for | _x_ 3| < 1, which is equivalent to | _x_ | < 1. Hence,   for | _x_ | < 1. Therefore,  . But _f_ (0) = 0. Hence, _C_ = 0. Also, 1– _x_ 3 > 0 for | _x_ | < 1. Therefore,

**16.** _x_ \+ 2 _x_ 2 \+ 3 _x_ 5 \+ 4 _x_ 7 \+ ···.

The ratio test shows that the series converges for | _x_ | < 1. Let

Then   Hence, taking antiderivatives,

Now differentiate:

  for | _x_ | < 1

**17.** (GC) Approximate   to two-decimal-place accuracy (that is, with an error < 5/103).

By formula ( _46.8_ ), In   for | _x_ | < 1, So

By Theorem 46.6( _b_ ),

which is a convergent alternating series.

In order to get an approximation with an error less than 5/103, we must find _n_ such that the first omitted term  . So, we must have 200 ( _n_ +1)22n+1. Trial and error shows that _n_   3.

Hence, we can use the terms corresponding to _n_ = 0,1,2:

This answer can be confirmed by a graphing calculator (which yields 0.44841421 as an approximation).

**18.** Find the function defined by  .

This is a geometric series with ratio 2 _x_ and first term 1. Hence, it converges for |2 _x_ | < 1, that is, for  , and its sum is  .

**19.** Find the interval of convergence of  .

Apply the ratio test:

By L'Hôpital's rule,  . Hence, the interval of convergence is given by |x| < 1. (For _x_ = 1, we get  , which we know is divergent. For _x_ = –1, we get the convergent alternating series

**20.** Approximate   with an error less than 0.0001.

By formula ( _46.14_ ),

By the alternating series theorem, we seek the least _n_ such that 1/ _n_!   0.0001 = 1/10,000, that is, 10,000   _n_!. Trial and error shows that _n_   8. So, we must use the terms corresponding to _n_ = 0,1,...,7:

(A graphing calculator yields the answer 0.367 879 4412, correct to ten decimal places.)

**21.** Approximate   to two-decimal-place accuracy, that is, with an error less than 5/103 = 0.005.

By formula ( _46.14_ ),

  for all _x_. Hence,   for all _x_

By Theorem 46.6( _b_ ),

We can apply the alternating series theorem. The magnitude of the first term omitted   should be   0.005 = 1/200. So, 200   (2 _n_ +1) _n_!. Trial and error shows that _n_   4. Hence, we should use the first four terms, that is, those corresponding to _n_ = 0,1,2,3:

(A graphing calculator yields the approximation 0.74682413, correct to eight decimal places.)

**22.** Find a power series expansion for   about 0.

 . By formula ( _46.7_ ),   for | _x_ | < 1.

Hence,

Thus,

The series diverges at _x_ = ±3.

**23.** Find a power series expansion for   about 1.

### **Supplementary Problems**

In Problems 24–31, find the interval of convergence of the given power series.

**32.** Express _e –2x_ as a power series about 0.

_Ans_.

**33.** Represent _e x/2_ as a power series about 2.

_Ans_.

**34.** Represent ln _x_ as a power series about 2.

_Ans_.

**35.** (GC) Find ln (0.97) with seven-decimal-place accuracy. ( _Hint:_ Use the power series for ln (1 − _x_ ) about 0).

_Ans_. –0.0304592

**36.** How many terms in the power series for ln (1 + _x_ ) about 0 must be used to find ln 1.02 with an error  0.00000005?

_Ans_. Three

**37.** (GC) Use a power series to compute _e_ –2 to four-decimal-place accuracy

_Ans_. 0.1353

**38.** (GC) Evaluate   to four-decimal-place accuracy.

_Ans_. 0.4940

In Problem 39 and 40, find the interval of convergence of the given series.

**41.** Represent cosh   as a power series about 0.

_Ans_.

**42.** Find a power series about 0 for the normal distribution function  .

_Ans_.

**43.** Find a power series expansion about 0 for ln  .

_Ans_.

**44.** (GC) Approximate tan–1  to two-decimal-place accuracy.

_Ans_. 0.46

**45.** Show that the converse of Abel's Theorem is not valid, that is, if  , where _r_ is the radius of convergence of the power series, and   exists, then   need not converge. ( _Hint:_ Look at  ).

**46.** Find a simple formula for the function _f_ ( _x_ ) represented by  .

_Ans. _

**47.** Find a simple formula for the function _f_ ( _x_ ) represented by  .

_Ans. x_ \+ (1 − _x_ ) ln (1 − _x_ )

**48.** ( _a_ ) Show that  . ( _Hint_ : Use Example 5.)

( _b_ ) Show that  . ( _Hint_ : First divide the series by _x_ , integrate, factor out _x_ , use part ( _a_ ), and differentiate.)

( _c_ ) Show that  .

( _d_ ) Evaluate  .

_Ans._ ( _d_ ) 2 and 6

## **CHAPTER 47
Taylor and Maclaurin Series.
Taylor's Formula with Remainder**

**TAYLOR AND MACLAURIN SERIES**. Let _f_ be a function that is infinitely differentiable at _x_ = _c_ , that is, the derivatives _f_ ( _n_ )( _c_ ) exist for all positive integers _n_.

The _Taylor series for f about c_ is the power series

where   for all _n_. Note that _f_ (0) is taken to mean the function _f_ itself, so that _a_ 0 = _f_ ( _c_ ). The _Maclaurin series for f_ is the Taylor series for _f_ about 0, that is, the power series

where   for all _n_.

**EXAMPLE 1 The Maclaurin series for sin _x_**

Let _f_ ( _x_ ) = sin _x_. Then

Since _f_ (4)( _x_ ) = sin _x_ , further derivatives repeat this cycle of four functions. Since sin 0 = 0 and cos 0 = 1, _f_ (2 _k_ )(0) = 0 and _f_ (2 _k_ +1) (0) = (–1) _k_. Hence, _a_ 2 _k_ = 0 and  . So, the Maclaurin series for sin _x_ is

An application of the ratio test shows that this series converges for all _x_. We do not know that sin _x_ is equal to its Maclaurin series. This will be proved later.

**EXAMPLE 2:** Let us find the Maclaurin series for  .

We can see the pattern:  . Hence,   for all _n_ , and the Maclaurin series for  . In this case, we already know that   is equal to its Maclaurin series for | _x_ | < 1.

**Theorem 47.1** : If   for some _x_ ≠ _c_ , then this series is the Taylor series for _f_ , that is,   for all _n_. In particular, if   for some _x_ ≠ 0, then this series is the Maclaurin series for _f_.

Assume _f_ ( _x_ ) =   for some _x_ ≠ _c_. Then _f_ ( _c_ ) = _b_ 0. By term-by-term differentiation (Theorem 46.7),   in the interval of convergence of  . Hence _f′_ ( _c_ ) = _b_ 1. Differentiating again, we get  . So, _f′′′_ ( _c_ ) = 2 _b_ 2 and, therefore,  .

Differentiating again, we get  . So, _f′′′_ ( _c_ ) = 3! _b_ 3 and, therefore,  . Iterating this procedure, we obtain

  for all _n_  0

Thus, the series is the Taylor series for _f_.

**EXAMPLE 3:** We already know by formula ( _46.8_ ) that

  for | _x_ | < 1

Hence, by Theorem 47.1, the series   must be the Maclaurin series for In (1 + _x_ ). It is not necessary to go through the laborious process of computing the Maclaurin series for In (1 + _x_ ) directly from the definition of Maclaurin series.

**EXAMPLE 4:** If  , find _f_ (47)(0).

We know that   for |x| < 1. Hence, by Theorem 47.1, the coefficient of _x_ n, namely 1, is equal to  . So, for n = 47,   and, therefore, f(47)(0) = (47)!.

**Theorem 47.2 (Taylor's Formula with Remainder):** Let _f_ by a function such that its (n + 1)st derivative _f_ (n+1) exists in (α, β). Assume also that _c_ and _x_ are in (α, β). Then there is some _x_ * between _c_ and _x_ such that

Here,   is called the _remainder term_ or the _error_.

Theorem 47.2 can be derived from Theorem 13.6 (the Higher-Order Law of the Mean).

**APPLICATIONS OF TAYLOR'S FORMULA WITH REMAINDER**

**(I) Showing that certain functions are represented by their Taylor series by proving that**

From Taylor's formula ( _47.1_ ),

If   Rn( _x_ ) = 0 then

that is, _f_ ( _x_ ) is equal to its Taylor series.

_Remark:_   for any _d_. To see this, recall that   converges for all _x_. Hence, by

Theorem 43.5,   for any _x_.

**EXAMPLE 5:** sin _x_ is equal to its Maclaurin series.

when _f_ ( _x_ ) = sin _x_ , then every derivative _f_ ( _n_ )( _x_ ) is either sin _x_ , cos _x_ , –sin _x_ , or –cos _x_ , and, therefore, | _f_ ( _n_ )( _x_ )|   1. So,

By the Remark above,  . Hence,   _R_ n( _x_ ) = 0. Therefore, sin _x_ is equal to its Taylor series:

**(II) Approximating values of the functions or integrals**

Use a bound on Rn _(x)_ to get a bound on the error when we approximate the sum of an infinite series by a partial sum.

**EXAMPLE 6:** Let us approximate _e_ to four decimal places, that is, with an error less than 0.00005.

Preliminary result: _e_ < 3. To see this, note that, since

Now, for the function _f_ ( _x_ ) = _e _x__ , we wish to make the magnitude of the error Rn(1) < 0.00005. By Taylor's formula with remainder, with _x_ = 1,

 , where 0 < _x_ * < 1

Since _D x_( _e x_) = _e x, f_( _n_ +1)( _x_ ) = _e x_ for all _x_. So, _f_ ( _n_ +1)( _x_ *) = _e_ _x_ *. Since _e_ _x_ is an increasing function, _e_ _x_ * < _e_ 1 = _e_ < 3. Thus,  . Since we wish to make the error < 0.00005, it suffices to have

 , that is,  , 60,000   ( _n_ \+ 1)!

Trial and error shows that this holds for _n_   8. So, we can use the partial sum  .

**Theorem 47.3 (The Binomial Series):** Assumer _r_ ≠ 0. Then

Apply the ratio test to the given series:

So,

Hence, the series converges for |x| < 1. For a sketch of the proof that this series is equal to (1 + _x_ )r, See Problem 31.

Note that, if _r_ is a positive integer _k_ , then the coefficients of _x_ n for _n > k_ are 0 and we get the binomial formula

 .

**EXAMPLE 7:** Let us expand   as a power series about 0. This is the binomial series for  .

**EXAMPLE 8:** Let us find a power series expansion about 0 for  .

Take the binomial series for _r_ =– , and then replace _x_ by – _x_ :

**Theorem 47.4** If   and  , then   for | _x_ | < minimum ( _R_ 1, _R_ 2), where  .

The reader is referred to more advanced treatments of analysis for a proof. Theorem 47.4 guarantees that, if _f_ and _g_ have power expansions, then so does their product.

### **Solved Problems**

**1.** Find a power series expansion about 0 for cos _x_.

We know by Example 5 that

Then, by Theorem 46.7, we can differentiate term by term:

2. Find a power series about   for sin _x_.

Use the identity sin _x_ = cos  . Then, by Problem 1,

**3.** If _f_ ( _x_ ) = tan–1 _x_ , evaluate _f_ (38)(0).

We know by formula ( _46.12_ ) that

Hence, by Theorem 47.1, the coefficient of _x_ 38 in this power series is equal to  . But the coefficient of _x_ 38 is 0. So, _f_ (38)(0) = 0.

**4.** Find power series expansions about 0 for the following functions:

( _a_ ) cos( _x_ 2)

( _b_ ) _xe_ -2x

( _c_ )

( _a_ )   by Problem 1. Therefore,  .

( _b_ ) We know that  . Hence,

( _c_ ) This is the binomial series for  .

**5.** Find the first five terms of the Maclaurin series for _e_ _x_ (sin _x_ ).

_Method 1_ : Let _f_ ( _x_ ) = _e_ _x_ (sin _x_ ). Then

_f_ ′( _x_ ) = _e_ _x_ (sin _x_ \+ cos _x_ ), _f_ ″( _x_ ) = 2 _e_ _x_ (cos _x_ ), _f_ ′″( _x_ ) = 2 _e_ _x_ (cos _x_ −sin _x_ )

_f_ (4)( _x_ ) = -4 _e_ _x_ (sin _x_ ), and _f_ (5)( _x_ ) = -4 _e_ _x_ (sin _x_ \+ cos _x_ )

Hence, since  , we get   Thus

_Method 2:_  . If we multiply out according to the rule in Theorem 47.4, we get the same result as above. For example,  .

**6.** We know that   For what values of _x_ will approximating sin _x_ by _x_ produce an error < 0.005?

 . (Here, | _f_ (3)( _x_ )|   1 since _f_ (3) is − cos _x_.) So, we require | _x_ |3/6 < 0.005, which is equivalent to | _x_ |3 < 0.003. So, we want  .

**7.** If we approximate sin _x_ by   for | _x_ | < 0.5, what is a bound on the error?

Since sin _x_ is equal to an alternating series for any _x_ , the error will be less than the magnitude of the first term omitted, in this case | _x_ |5/5!. When | _x_ | < 0.5, the error will be less than  .

**8.** Approximate   with an error less than 0.005.

Hence,

Therefore,

This is an alternating series. We must find _k_ so that  , or, equivalently, 200 (2 _k_ \+ 1)!(2 _k_ \+ 1). It is true for _k_   2. So, we need  .

**9.** Find a power series about 0 for sin-1 _x_.

By formula (47.5),

Replace _x_ by t2.

So, for | _x_ | < 1,

**10.** Find Maclaurin series for the following functions: ( _a_ ) sin ( _x_ 3); ( _b_ ) sin2 _x_.

Recall that, if a function has a power series expansion in an interval about 0, then power series is the Maclaurin series of the function.

( _a_ )   for all _x_. Hence,   and this series is the Maclaurin series for sin (x3).

( _b_ )   by problem 1

So, the Maclaurin series for sin2 _x_ is  .

**11.** Find the first four nonzero terms of the Maclaurin series for _f_ (x) = sec _x_.

It would be very tedious to compute the successive derivatives. Instead, since sec _x_ cos _x_ = 1, we can proceed differently. We assume  . Then

We then "multiply out", compare coefficients on the two sides of the equation, and solve for the _a n_.

Thus,

An alternative method would be to carry out a "long division" of 1 by 1

### **Supplementary Problems**

**12.** Find the Maclaurin series for the following functions:

( _a_ ) sin ( _x 5_); ( _b_ )  ; ( _c_ ) cos2 _x_.

_Ans_. ( _a_ )  ; ( _b_ )  ; ( _c_ )

**13.** Find the Taylor series for ln _x_ about 2.

_Ans_.

**14.** Find the first three nonzero terms of the Maclaurin series for ( _a_ )  ; ( _b_ ) _e x_ cos _x_.

_Ans_.

**15.** Compute the first three nonzero terms of the Maclaurin series for tan _x_.

_Ans_.

**16.** Compute the first three nonzero terms of the Maclaurin series for sin-1 _x_.

_Ans_.

**17.** Find the Taylor series for cos _x_ about

_Ans_.

**18.** (GC) Use power series to approximate

_Ans_. 0.4872

**19.** (GC) Use power series to approximate   correctly to four decimal places.

_Ans._ 0.4484

**20.** (GC) Use power series to approximate   correctly to four decimal places.

_Ans._ 1.0948

**21.** (GC) What is a bound on the error if we approximate   (You may use _e_ 0.05 < 1.06)

_Ans._ 0.0000221

**22.** (GC) What is a bound on the error if we approximate ln (1 + _x_ ) by _x_ for | _x_ |   0.05?

_Ans._ 0.00125

**23.** (GC) Use the Taylor series for sin _x_ about   to approximate sin 62° correctly to five decimal places.

_Ans._ 0.88295

**24.** (GC) In what interval can you choose the angle if the values of cos _x_ are to be computed using three terms of its Taylor series about   and the error must not exceed 0.000 05?

_Ans._

**25.** (GC) Use power series to compute to four-decimal-place accuracy: ( _a_ ) e-2 ( _b_ ) sin 32°; ( _c_ ) cos 36°;

_Ans._ ( _a_ ) 0.1353; ( _b_ ) 0.5299 ( _c_ ) 0.8090

**26.** (GC) For what range of _x_ can:

( _a_ ) _e x_ be replaced by   if the allowable error is 0.0005?

( _b_ ) sin _x_ be replaced by   if the allowable error is 0.000 05?

_Ans._ ( _a_ ) | _x_ | < 0.1; ( _b_ ) | _x_ | < 47°

**27.** Use power series to evaluate: ( _a_ )  ; ( _b_ )  .

_Ans._

**28.** (GC) Use power series to evalute:

( _a_ )   (to three-decimal-place accuracy).

( _b_ )   (to five-decimal-place accuracy).

( _c_ )   (to four-decimal-place accuracy).

_Ans._ ( _a_ ) 1.854; ( _b_ ) 0.76355; ( _c_ ) 0.4940

**29.** (GC) Use power series to approximate the length of the curve   with four-decimal-place accuracy.

_Ans._ 0.5031

**30.** (GC) Use power series to approximate the area between the curve _y_ = sin ( _x_ 2) and the _x_ axis from _x_ = 0 to _x_ = 1, with four-decimal-place accuracy.

_Ans._ 0.3103

**31.** Prove that the binomial series expansion in Theorem 47.3 is correct.

Hint:  . Use term-by-term differentiation to find the series for   and show that  . Then derive  .

**32.** Expand the polynomial _f_ ( _x_ ) = _x_ 4 − 11 _x_ 3 \+ 43 _x_ 2 − 60 _x_ \+ 14 as a power series about 3, and find  .

_Ans._ 1.185

## **CHAPTER 48
Partial Derivatives**

**FUNCTIONS OF SEVERAL VARIABLES.** If a real number _z_ is assigned to each point ( _x_ , _y_ ) of a part of the _xy_ plane, then _z_ is said to be given as a function, _z_ = _f_ ( _x_ , _y_ ), of the independent variables _x_ and _y_. The set of all points ( _x_ , _y_ , _z_ ) satisfying _z_ = _f_ ( _x_ , _y_ ) is a surface in three-dimensional space. In a similar manner, functions _w_ = _f_ ( _x_ , _y_ , _z_ ,...) of three or more independent variables may be defined, but no geometric picture is available.

There are a number of differences between the calculus of one and two variables. However, the calculus of functions of three or more variables differs only slightly from that of functions of two variables. The study here will be limited largely to functions of two variables.

**LIMITS.** By an _open disk_ with center ( _a_ , _b_ ) we mean the set of points ( _x_ , _y_ ) within some fixed distance _δ_ from ( _a_ , _b_ ), that is, such that  . By a _deleted disk_ around ( _a_ , _b_ ) we mean an open disk without its center ( _a_ , _b_ ).

Let _f_ be a function of two variables and assume that there are points in the domain of _f_ arbitrarily close to ( _a_ , _b_ ). To say that _f_ ( _x_ , _y_ ) has the limit _L_ as ( _x_ , _y_ ) approaches ( _a_ , _b_ ) means intuitively that _f_ ( _x_ , _y_ ) can be made arbitrarily close to _L_ when ( _x_ , _y_ ) is sufficiently close to ( _a_ , _b_ ). More precisely,

if, for any **ε** > 0, there exists _δ_ > 0 such that, for any ( _x_ , _y_ ) in the domain of _f_ and in the deleted disk of radius _δ_ around ( _a_ , _b_ ), | _f_ ( _x_ , _y_ ) − _L_ | < **ε**. This is equivalent to saying that, for any **ε** > 0, there exists _δ_ > 0 such that   implies | _f_ ( _x_ , _y_ ) − _L_ | < **ε** for any ( _x_ , _y_ ) in the domain of _f_. Note that it is not assumed that _f_ ( _a_ , _b_ ) is defined.

Laws for limits analogous to those for functions of one variable (Theorems 7.1–7.6) also hold here and with similar proofs.

**EXAMPLE 1:** Using these standard laws for limits, we see that

**EXAMPLE 2:** In some cases these standard laws do not suffice.

Let us show that  . Our usual limit rules would yield  , which is indeterminate. So, we need a more involved argument. Assume **ε** > 0. Now,

if we choose _δ_ = **ε** /3 and we assume that  .

**EXAMPLE 3:** Let us show that   does not exist.

Let ( _x_ , _y_ ) → (0, 0) along the _x_ axis, where _y_ = 0. Then  . So, the limit along the _x_ axis is 1. Now let ( _x_ , _y_ ) → (0, 0) along the _y_ axis, where _x_ = 0. Then  . So, the limit along the _y_ axis is – 1. Hence, there can be no common limit as one approaches (0, 0) and the limit does not exist.

**EXAMPLE 4:** Let us show that   does not exist.

Here, we cannot use the same argument as in Example 3, since   approaches 1 as ( _x_ , _y_ ) approaches (0, 0) along both the _x_ axis and the _y_ axis. However, we can let ( _x_ , _y_ ) approach (0, 0) along the line _y_ = _x_. Then  . So,   along _y_ = _x_. Since this is different from the limit 1 approached along the _x_ axis, there is no limit as ( _x_ , _y_ ) → (0, 0).

**CONTINUITY.** Let _f_ be a function of two variables and assume that there are points in the domain of _f_ arbitrarily close to ( _a_ , _b_ ). Then _f_ is _continuous_ at ( _a_ , _b_ ) if and only if _f_ is defined at ( _a_ , _b_ ),   exists, and  .

We say that _f_ is continuous on a set _A_ if _f_ is continuous at each point of _A_.

This is a generalization to two variables of the definition of continuity for functions of one variable. The basic properties of continuous functions of one variable (Theorem 8.1) carry over easily to two variables. In addition, every polynomial in two variables, such as 7 _x_ 5 − 3 _xy_ 3 − _y_ 4 \+ 2 _xy_ 2 \+ 5, is continuous at all points. Every continuous function of one variable is also continuous as a function of two variables.

The notions of limit and continuity have obvious generalizations to functions of three or more variables.

**PARTIAL DERIVATIVES.** Let _z_ = _f_ ( _x_ , _y_ ) be a function of two variables. If _x_ varies while _y_ is held fixed, _z_ becomes a function of _x_. Then its derivative with respect to _x_

is called the _(first) partial derivative_ of _f_ with respect to _x_ and is denoted  .

Similarly, if _y_ varies while _x_ is held fixed, the _(first) partial derivative_ of _f_ with respect to _y_ is

**EXAMPLE 5:** Let _f_ ( _x_ , _y_ ) = _x_ 2 sin _y_. Then _f x_( _x_ , _y_ ) = 2 _x_ sin _y_ and _f y_( _x_ , _y_ ) = _x_ 2 cos _y_.

Note that, when _f x_ is computed, _y_ is temporarily treated like a constant, and, when _f y_ is computed, _x_ is temporarily treated like a constant.

The partial derivatives have simple geometric interpretations. Consider the surface _z_ = _f_ ( _x_ , _y_ ) in Fig. 48-1. Through the point _P_ ( _x_ , _y_ , _z_ ), there is a curve _APB_ that is the intersection with the surface of the plane through _P_ parallel to the _xz_ plane (the plane determined by the _x_ axis and the _z_ axis. Similarly, _CPD_ is the curve through _P_ that is the intersection with the surface _z_ = _f_ ( _x_ , _y_ )of the plane through _P_ that is parallel to the _yz_ plane. As _x_ varies while _y_ is held fixed, _P_ moves along the curve _APB_ , and the value of   at ( _x_ , _y_ ) is the slope of the tangent line to the curve _APB_ at _P_. Similarly, as _y_ varies while _x_ is held fixed, _P_ moves along the curve _CPD_ , and the value of   at ( _x_ , _y_ ) is the slope of the tangent line to the curve _CPD_ at _P_.

**Fig. 48-1**

**PARTIAL DERIVATIVES OF HIGHER ORDER**. We can take the partial derivatives with respect to _x_ and _y_ of  , yielding

Similarly, from   we obtain

**Theorem 48.1:** Assume that _f xy_ and _f yx_ exist and are continuous in an open disk. Then _f xy_ = _f yx_ at every point of the disk.

For a proof, see problem 30.

**EXAMPLE 6:** Let us verify Theorem 48.1 for _f_ ( _x_ , _y_ ) = _x_ 2(sin _yx_ ).

Partial derivatives also can be defined for functions of three or more variables. An analogue of Theorem 48.1 holds for any two orderings of given subscripts.

Note that partial derivatives may fail to exist when the required limits do not exist.

### **Solved Problems**

**1.** Evalute:  .

Since the standard limit laws apply, the limits are:

( _a_ ) 2(3)(2)4 − 7(3)2(2)2 = 96 − 252 = –156; ( _b_ )

2. Evaluate  .

As ( _x_ , _y_ ) → (0, 0) along the _y_ axis, _x_ = 0 and  .

As ( _x_ , _y_ ) → (0, 0) along the _x_ axis, _y_ = 0 and  .

Hence, the limit does not exist.

**3.** Evaluate  .

Since  . So,

**4.** The function   is continuous everywhere except at (0, 0) and on the line _y_ = – _x_ , where it is not defined. Can _f_ (0, 0) be defined so that the new function is continuous?

As ( _x_ , _y_ ) → (0, 0), _x_ \+ _y_ → 0 and, therefore,  . So, if we let _f_ (0, 0) = 1, the new function will be continuous at (0, 0). Thus, the original discontinuity was removable.

In Problems 5–9, find the first partial derivatives.

**5.** _z_ = 2 _x_ 2 − 3 _xy_ \+ 4 _y_ 2.

Treating _y_ as a constant and differentiating with respect to _x_ yields  .

Treating _x_ as a constant and differentiating with respect to _y_ yields

**6.**  .

Treating _y_ as a constant and differentiating with respect to _x_ yields  .

Treating _x_ as a constant and differentiating with respect to _y_ yields

**7.** _z_ = sin (2 _x_ \+ 3 _y_ ).

**8.** _z_ = tan–1 ( _x_ 2 _y_ ) + tan–1 ( _xy_ 2).

**9.**

**10.** The area of a triangle is given by _K_ =   _ab_ sin _C_. When _a_ = 20, _b_ = 30, and _C_ = 30°, find:

( _a_ ) The rate of change of _K_ with respect to _a_ , when _b_ and _C_ are constant.

( _b_ ) The rate of change of _K_ with respect to _C_ , when _a_ and _b_ are constant.

( _c_ ) The rate of change of _b_ with respect to _a_ , when _K_ and _C_ are constant.

( _a_ )

( _b_ )

( _c_ )   and

In Problems 11–13, find the first partial derivatives of _z_ with respect to the independent variables _x_ and _y_.

**11.** _x_ 2 \+ _y_ 2 \+ _z_ 2 = 25. [This is the equation of a sphere of radius 5 and center (0, 0, 0).]

Differentiate implicitly with respect to _x_ , treating _y_ as a constant, to obtain:

  Hence,

Differentiate implicitly with respect to _y_ , treating _x_ as a constant:

 . Hence,

**12.** _x_ 2(2 _y_ \+ 3 _z_ ) + _y_ 2(3 _x_ − 4 _z_ ) + _z_ 2( _x_ − 2 _y_ ) = _xyz_.

Differentiate implicitly with respect to _x_ :

Solving for   yields:

Differentiate implicitly with respect to _y_ :

Solving for   yields:

**13.** _xy_ \+ _yz_ \+ _zx_ = **1**.

Differentiating with respect to _x_ yields  , whence  .

Differentiating with respect to _y_ yields  , whence  .

**14.** Considering _x_ and _y_ as independent variables, find   when _x_ = _e_ 2 _r_ cos _θ, y = e_ 3 _r_ sin _θ_.

First differentiate the given relations with respect to _x_ :

Then solve simultaneously to obtain   and

Now differentiate the given relations with respect to _y_ :

Then solve simultaneously to obtain   and

**15.** Find the slopes of the tangent lines to the curves cut from the surface _z_ = 3 _x_ 2 \+ 4 _y_ 2 − 6 by planes through the point (1, 1, 1) and parallel to the _xz_ and _yz_ planes.

The plane _x_ = 1, parallel to the _yz_ plane, intersects the surface in the curve _z_ = 4 _y_ 2 − 3, _x_ = 1. Then   is the required slope.

The plane _y_ = 1, parallel to the _xz_ plane, intersects the surface in the curve _z_ = 3 _x_ 2 \+ 2, _y_ = 1. Then   is the required slope.

In Problems 16 and 17, find all second partial derivatives of _z_ and verify Theorem 48.1.

**16.** _z_ = _x_ 2 \+ 3 _xy_ \+ _y_ 2.

Note that  .

**17.** _z_ = _x_ cos _y_ − _y_ cos _x_.

Note that  .

**18.** Let _f(x, y, z)_ = _x_ cos( _yz_ ). Find all partial derivatives of the first, second, and third order.

Note that _f xy_ = _f yx_ and _f xz_ = _f zx_ and _f yz_ = _f zy_.

Note that, in the third order, any two rearrangements of subscripts will be equal. For example, _f xyz_ = _f xzy_ = _f yxz_ = _f yzx_ = _f zxy_ = _f zyx_ = − ( _zy_ cos( _yz_ ) + sin ( _yz_ )).

**19.** Determine whether the following functions are solutions of Laplace's equation  :

### **Supplementary Problems**

In Problems 20–24, evaluate the given limit.

**25.** Determine whether each of the following functions can be defined at (0, 0) so as to be continuous:

_Ans._ ( _a_ ) no; ( _b_ ) no; ( _c_ ) yes; ( _d_ ) no

**26.** For each of the following functions _z_ , find  .and  .

**27.** For each of the following functions _z_ , find  , and  .

**28.** ( _a_ ) If  , show that  .

( _b_ ) If _z_ = _e_ ax cos _βy_ and _β_ = ±α, show that  .

( _c_ ) If _z_ = _e_ – _t_ (sin _x_ \+ cos _y_ ), show that  .

( _d_ ) If _z_ = sin _ax_ sin by sin _kt_  , show that  .

**29.** For the gas formula  , where _a_ , _b_ , and _c_ are constants, show that

**30.** Fill in the following sketch of a proof of Theorem 48.1. Assume that _f xy_ and _f yx_ exist and are continuous in an open disk. Then _f xy_( _a, b_ ) = _f yx_( _a, b_ ) at every point ( _a, b_ ) of the disk. Let Δ _h_ = ( _f_ ( _a_ \+ _h_ , _b_ \+ _h_ ) − _f_ ( _a_ \+ _h, b_ )) − ( _f_ ( _a, b_ \+ _h_ ) − _f_ ( _a, b_ )) for _h_ sufficiently small and ≠ 0. Let _F_ ( _x_ ) = _f_ ( _x_ , _b_ \+ _h_ ) − _f_ ( _x_ , _b_ ). Then Δ _h_ = _F_ ( _a_ \+ _h_ ) − _F_ ( _a_ ). Apply the mean-value theorem to get _a_ * between _a_ and _a_ \+ _h_ so that _F_ ( _a_ \+ _h_ ) − _F_ ( _a_ ) = _F_ ′ ( _a_ *) _h_ = [ _f_ _x_ ( _a_ *, _b_ ) + _h_ ) − _f_ _x_ ( _a_ *, _b_ )] _h_ , and apply the mean-value theorem to get _b_ * between _b_ and _b_ \+ _h_ so that _f_ _x_ ( _a_ *, _b_ \+ _h_ ) − _f_ x( _a_ *, _b_ ) = _f_ _xy_ ( _a_ *, _b_ ) _h_ Then,

Δ _h_ = _h_ 2 _f_ _xy_ ( _a_ *, _b_ *) and

By a similar argument using Δ _h_ = ( _f_ ( _a_ \+ _h_ , _b_ \+ _h_ ) − _f_ ( _a_ , _b_ \+ _h_ )) − ( _f_ ( _a_ \+ _h_ , _b_ ) − _f_ ( _a_ , _b_ )) and the mean-value theorem, we get

**31.** Show that Theorem 48.1 no longer holds if the continuity assumption for _f xy_ and _f yx_ is dropped. Use the following function:

[Find formulas for _f x_( _x_ , _y_ ) and _f y_( _x_ , _y_ ) for ( _x, y_ ) ≠ (0, 0); evaluate _f x_(0, 0) and _f y_(0, 0), and then _f xy_(0, 0) and _f yx_(0, 0).]

## **CHAPTER 49
Total Differential. Differentiability. Chain Rules**

**TOTAL DIFFERENTIAL.** Let _z_ = _f_ ( _x, y_ ). Let Δ _x_ and Δ _y_ be any numbers. Δ _x_ and Δ _y_ are called _increments_ of _x_ and _y_ , respectively. For these increments of _x_ and _y_ , the corresponding _change_ in _z_ , denoted Δ _z_ , is defined by

The _total differential dz_ is defined by:

Note that, if _z_ = _f_ ( _x, y_ ) = _x_ , then  , and,  , therefore, _dz_ = Δ _x_. So, _dx_ = Δ _x_. Similarly, _dy_ = Δ _y_. Hence, equation ( _49.2_ ) becomes

_Notation: dz_ is also denoted _df_.

These definitions can be extended to functions of three or more variables. For example, if _u_ = _f_ ( _x, y, z_ ), then we get:

**EXAMPLE 1:** Let _z_ = _x_ cos _y_ − 2 _x_ 2 \+ 3. Then  . Then the total differential for _z_ is _dz_ = (cos _y_ − 4 _x_ ) _dx_ − ( _x_ sin _y_ ) _dy_.

In the case of a function of one variable _y_ = _f_ ( _x_ ), we used the approximation principle Δ _y_ ~ _f_ ′( _x_ ) Δ _x_ = _dy_ to estimate values of _f_. However, in the case of a function _z_ = _f_ ( _x, y_ ) of two variables, the function _f_ has to satisfy a special condition in order to make good approximations possible.

**DIFFERENTIABILITY.** A function _z_ = _f_ ( _x, y_ ) is said to be _differentiable_ at ( _a, b_ ) if there exists functions  1 and  2 such that

and

Note that formula (49.4) can be written as

We say that _z_ = _f_ ( _x, y_ ) is differentiable on a set _A_ if it is differentiable at each point of _A_.

As in the case of one variable, differentiability implies continuity. (See Problem 23.)

**EXAMPLE 2:** Let us see that _z_ = _f_ ( _x, y_ ) = _x + 2y_ 2 is differentiable at every point ( _a, b_ ). Note that _f_ x( _x, y_ ) = 1 and _f_ y( _x, y_ ) = 4 _y_. then

Let  1 = 0 and  2 = 2 Δ _y_.

**Definition**. By an _open set_ in a plane we mean a set _A_ of points in the plane such that every point of _A_ belongs to an open disk that is included in _A_.

Examples of open sets are an open disk and the interior of a rectangle.

**Theorem 49.1:** Assume that _f_ ( _x, y_ ) is such that _f_ x and _f_ y are continuous in an open set _A_. Then _f_ is differentiable in _A_.

For the proof, see Problem 43.

**EXAMPLE 3:** Let  . Then   and  . So, by Theorem 49.1, _f_ is differentiable in the open disk of radius 3 and center at the origin (0, 0) (where the denominators of _f_ x and _f_ y exist and are continuous). In that disk, _x_ 2 \+ _y_ 2 < 9. Take the point ( _a, b_ ) = (1, 2) and let us evaluate the change Δ _z_ as we move from (1, 2) to (1.03, 2.01). So, Δ _x_ = 0.03 and Δ _y_ = 0.01. Let us approximate Δ _z_ by

The actual difference Δ _z_ is  .

**CHAIN RULES. Chain Rule (2 → 1):** Let _z_ = _f_ ( _x, y_ ), where _f_ is differentiable, and let _x_ = _g_ ( _t_ ) and _y_ = _h_ ( _t_ ), where _g_ and _h_ are differentiable functions of one variable. Then _z_ = _f_ ( _g_ ( _t_ ), _h_ ( _t_ ) is a differentiable function of one variable, and

_Warning_ : Note the double meaning of _z, x_ , and _y_ in ( _49.6_ ). In  , _z_ means _f(g(t)_ , h(t)), whereas, in   and  , _z_ means _f_ ( _x, y_ ). In  , _x_ is an independent variable, whereas, in  , _x_ means _g_ ( _t_ ).

Likewise, _y_ has two different meanings.

To prove _(49.6)_ , note first that, by _(49.4)_ ,

Letting Δ _t_ → 0, we obtain

(Note that, since _g_ and _h_ are differentiable, they are continuous. Hence, as Δ _t_ → 0, Δ _x_ → 0 and Δ _y_ → 0 and, therefore,  1 → 0 and  2 → 0.)

**EXAMPLE 4:** Let _z_ = _xy_ \+ sin _x_ , and let _x_ = _t_ 2 and _y_ = cos _t_. Note that   and  .

Moreover,   and  . Now, as a function of _t_ , _z_ = _t_ 2 cos _t_ \+ sin ( _t_ 2).

By formula ( _49.6_ ),

In this particular example, the reader can check the result by computing _D_ _t_ ( _t_ 2 cos _t_ \+ sin ( _t_ 2)).

**Chain Rule (2 → 2):** Let _z_ = _f_ ( _x, y_ ), where _f_ is differentiable, and let _x_ = _g_ ( _t, s_ ) and _y_ = _h_ ( _t, s_ ), where _g_ and _h_ are differentiable functions. Then _z_ = _f_ ( _g_ ( _t_ , _s_ ), _h_ ( _t_ , _s_ )) is a differentiable function, and

Here again, as in the previous chain rule, the symbols _z, x_ , and _y_ have obvious double meanings.

This chain rule can be considered a special case of the chain rule (2→1). For example, the partial derivative   can be thought of as an ordinary derivative  , because _s_ is treated as a constant. Then the formula for   in ( _49.7_ ) is the same as the formula for   in ( _49.6_ ).

**EXAMPLE 5:** Let _z_ = _e_ _x_ sin _y_ and _x_ = _ts_ 2 and _y_ = _t_ \+ 2 _s_. Now,  , and  . Hence, by ( _49.7_ ),

similarly,

Generalizations of the chain rule ( _49.47_ ) hold for cases ( _m_ → _n_ ), where _z_ = _f_ ( _x, y,..._ ) is a function of _m_ variables and each of those variables is a function of a given set of _n_ variables.

**IMPLICIT DIFFERENTIATION.** Assume that the equation _F_ ( _x, y, z_ ) = 0 defines _z_ implicitly as a function of _x_ and _y_. Then, by the chain rule (3→2), if we differentiate both sides of the equation with respect to _x_ , we get

Since

Similarly,  . So, if  ,

This also can be written as   and  .

**EXAMPLE 6:** The equation _xy_ \+ _yz_ 3 \+ _xz_ = 0 determines _z_ as a function of _x_ and _y_. Since _F_ _z_ = _x_ \+ 3 _yz_ 2, _F_ _x_ = _y_ \+ _z_ , and _F_ _y_ = _x_ \+ _z_ 3, ( _49.8_ ) implies

  and

### **Solved Problems**

In Problems 1 and 2, find the total differential.

**1.** _z_ = _x_ 3 _y_ \+ _x_ 2 _y_ 2 \+ _xy_ 3

We have

Then

**2.** _z_ = _x_ sin _y_ − _y_ sin _x_

We have   and

Then

**3.** Compare _dz_ and Δ _z_ , given _z_ = _x_ 2 \+ 2 _xy_ − 3 _y_ 2.

  and  . So _dz_ = 2( _x + y_ ) _dx_ \+ 2( _x_ − 3 _y_ ) _dy_

Also,

Thus _dz_ and _Δz_ differ by ( _dx_ )2 \+ 2 _dx dy_ − ( _dy_ )2.

**4.** Approximate the area of a rectangle of dimensions 35.02 by 24.97 units.

For dimensions _x_ by _y_ , the area is _A_ = _xy_ so that  . With _x_ = 35, _dx_ = 0.02, _y_ = 25, and _dy_ = –0.03, we have _A_ = 35(25) = 875 and _dA_ = 25(0.02) + 35(–0.03) = –0.55. The area is approximately _A_ \+ _dA_ = 874.45 square units.

**5.** Approximate the change in the hypotenuse of a right triangle of legs 6 and 8 inches when the shorter leg is lengthened by   inch and the longer leg is shortened by   inch.

Let _x, y_ , and _z_ be the shorter leg, the longer leg, and the hypotenuse of the triangle. Then

When _x_ = 6, _y_ = 8, _dx_ = ¼, and   then  . Thus the hypotenuse is lengthened by approximately

**6.** The power consumed in an electrical resistor is given by _P_ = _E_ 2/ _R_ (in watts). If _E_ = 200 volts and _R_ = 8 ohms, by how much does the power change if _E_ is decreased by 5 volts and _R_ is decreases by 0.2 ohm?

We have

When _E_ = 200, _R_ = 8, _dE_ = –5, and _dR_ = –0.2, then

The power is reduced by approximately 125 watts.

**7.** The dimensions of a rectangular block of wood were found to be 10, 12, and 20 inches, with a possible error of 0.05 in each of the measurements. Find (approximately) the greatest error in the surface area of the block and the percentage error in the area caused by the errors in the individual measurements.

The surface area is _S_ = 2( _xy_ \+ _yz_ \+ _zx_ ); then

The greatest error in _S_ occurs when the errors in the lengths are of the same sign, say positive. Then

_dS_ = 2(12 + 20)(0.05) + 2(10 + 20)(0.05) + 2(12 + 10)(0.05) = 8.4in2

The percentage error is (error/area)(100) = (8.4/1120)(100) = 0.75%.

**8.** For the formula _R_ = _E/C_ , find the maximum error and the percentage error if _C_ = 20 with a possible error of 0.1 and _E_ = 120 with a possible error of 0.05.

Here

The maximum error will occur when _dE_ = 0.05 and _dC_ = –0.1; then   is the approximate maximum error. The percentage error is

**9.** Two sides of a triangle were measured as 150 and 200 ft, and the included angle is 60°. If the possible errors are 0.2 ft in measuring the sides and 1° in the angle, what is the greatest possible error in the computed area?

Here

and

When _x_ = 150, _y_ = 200, θ = 60°, _dx_ = 0.2, _dy_ = 0.2, and _dθ_ = 1° = π/180, then

**10.** Find _dz/dt_ , given _z_ = _x_ 2 \+ 3 _xy_ \+ 5 _y_ 2; _x_ = sin _t, y_ = cos _t_.

Since

we have

**11.** Find _dz/dt_ , given _z_ = In ( _x_ 2 \+ _y_ 2); _x_ = _e_ –t, _y_ = _e_ _t_.

Since

we have

**12.** Find  , given _z_ = _f_ ( _x, y_ ) = _x_ 2 \+ 2 _xy_ \+ 4 _y_ 2, _y_ = _e ax_.

**13.** Find _(a)_   and ( _b_ )  , given _z_ = _f_ ( _x, y_ ) = _xy_ 2 \+ _yx_ 2, _y_ = ln _x_.

( _a_ ) Here _x_ is the independent variable:

( _b_ ) Here _y_ is the independent variable:

**14.** The altitude of a right circular cone is 15 inches and is increasing at 0.2 in/min. The radius of the base is 10 inches and is decreasing at 0.3 in/min. How fast is the volume changing?

Let _x_ be the radius, and _y_ the altitude of the cone (Fig. 49-1). From  , considering _x_ and _y_ as functions of time _t_ , we have

**Fig. 49-1**

**15.** A point _P_ is moving along the curve that is the intersection of the surfaces   and _x_ 2 \+ _y_ 2 = 5, with _x, y,_ and _z_ expressed in inches. If _x_ is increasing at the rate of 0.2 inches per minute, how fast is z changing when _x_ = 2?

From  , we obtain  . Since _x_ 2 \+ y2 = 5, y = ±1 when _x_ = 2; also, differentiation yields  .

When

When

**16.** Find   and  , given _z_ = _x_ 2 \+ _xy_ \+ _y_ 2; _x_ = 2 _r_ \+ _s_ , _y_ = _r_ − 2 _s_.

Here

**17.** Find   and   given _u_ = _x_ 2 \+ 2 _y_ 2 \+ 2 _z_ 2; _x_ = ρ sin β cos θ, _y_ = ρ sin β sin θ, _z_ = ρ cos β.

**18.** Find  , given  .

**19.** Use implicit differentiation (formula ( _49.8_ )) to find   and   given _F(x, y, z)_ = _x_ 2 \+ 3 _xy_ − 2 _y_ 2 \+ 3 _xy_ \+ z2 = 0.

**20.** Use implicit differentiation (formula ( _49.8_ )) to find   and   given _sin xy_ \+ _sin yz_ \+ sin _zx_ = 1.

Set _F(x, y, z)_ = sin _xy_ \+ sin _yz_ \+ sin _zx_ − 1; then

and

**21.** If _u_ and _v_ are defined as functions of _x_ and _y_ by the equations _f(x, y, u, v)_ = _x_ \+ _y_ 2 \+ 2 _uv_ = 0 and _g(x, y, u, v)_ = _x_ 2 − _xy_ \+ _y_ 2 \+ _u_ 2 \+ _v_ 2 = 0 find ( _a_ )  .

( _a_ ) Differentiating _f_ and _g_ partially with respect to _x_ , we obtain

Solving these relations simultaneously for _∂u/∂x_ and _∂v/∂x,_ we find

( _b_ ) Differentiating _f_ and _g_ partially with respect to _y_ , we obtain

**22.** Given _u_ 2 − _v_ 2 \+ 2 _x_ \+ 3 _y_ = 0 and _uv_ \+ _x_ − _y_ = 0, find

( _a_ ) Here _x_ and _y_ are to be considered as independent variables. Differentiate the given equations partially with respect to _x_ , obtaining

Solve these relations simultaneously to obtain  .

Differentiate the given equations partially with respect to _y_ , obtaining

Solve simultaneously to obtain  .

( _b_ ) Here _u_ and _v_ are to be considered as independent variables. Differentiate the given equations partially with respect to _u_ , obtaining

Then

Differentiate the given equations with respect to _v_ , obtaining

Then

**23.** Show that differentiability of _z_ = _f(x, y)_ at ( _a, b_ ) implies that _f_ is continuous at ( _a, b_ ).

From ( _49.4_ ), Δz = ( _f x_ ( _a, b_ ) +  1) Δx + ( _f y_ ( _a, b_ ) +  2) Δy, where

Hence, Δz → 0 as (Δ _x_ , Δ _y_ ) → (0, 0), which implies that _f_ is continuous at ( _a, b_ ).

### **Supplementary Problems**

**24.** Find the total differential of the following functions:

**25.** Use differentials to approximate ( _a_ ) the volume of a box with square base of side 8.005 and height 9.996 ft; ( _b_ ) the diagonal of a rectangular box of dimensions 3.03 by 5.98 by 6.01 ft.

_Ans._ ( _a_ ) 640.544 ft3; ( _b_ ) 9.003ft

**26.** Approximate the maximum possible error and the percentage of error when _z_ is computed by the given formula:

**27.** Find the approximate maximum percentage of error in:

( _a_ )   if there is a possible 1% error in measuring _g_ and a possible  % error in measuring b.

( _Hint_ :

_Ans_. 0.005

( _b_ ) _g_ = 2 _s_ / _t_ 2 if there is a possible 1% error in measuring s and   error in measuring _t_.

_Ans._ 0.015

**28.** Find _du_ / _dt_ , given:

( _a_ ) _u_ = _x_ 2 _y_ 3; _x_ = 2 _t_ 3, _y_ = 3 _t_ 2

_Ans._ 6 _xy_ 2t(2 _yt_ \+ 3 _x_ )

( _b_ ) _u_ = _x_ cos _y_ \+ _y_ sin _x_ ; _x_ = sin 2 _t_ , _y_ = cos 2 _t_

_Ans._ 2(cos _y_ \+ _y_ cos _x_ ) cos 2 _t_ − 2(– _x_ sin _y_ \+ sin _x_ ) sin 2 _t_

( _c_ ) _u_ = _xy_ \+ _yz_ \+ _zx_ ; _x_ = _e_ _t_ , _y_ = e– t, _z_ = _e_ _t_ \+ _e_ – _t_

_Ans._ ( _x_ \+ 2 _y_ \+ _z_ ) _e_ _t_ − (2 _x_ \+ _y_ \+ _z_ ) _e_ – _t_

**29.** At a certain instant the radius of a right circular cylinder is 6 inches and is increasing at the rate 0.2 in/sec, while the altitude is 8 inches and is decreasing at the rate 0.4 in/sec. Find the time rate of change ( _a_ ) of the volume and ( _b_ ) of the surface at that instant.

_Ans._ 4.8π in3/sec; ( _b_ ) 3.2 π in2/sec

**30.** A particle moves in a plane so that at any time _t_ its abscissa and ordinate are given by _x_ = 2 + 3 _t_ , _y_ = _t_ 2 \+ 4 with _x_ and _y_ in feet and _t_ in minutes. How is the distance of the particle from the origin changing when _t_ = 1?

_Ans._

**31.** A point is moving along the curve of intersection of _x_ 2 \+ 3 _xy_ \+ 3 _y_ 2 = _z_ 2 and the plane _x_ − 2 _y_ \+ 4 = 0. When _x_ = 2 and is increasing at 3 units/sec, find ( _a_ ) how _y_ is changing, ( _b_ ) how _z_ is changing, and ( _c_ ) the speed of the point.

_Ans._ ( _a_ ) increasing 3/2 units/sec; ( _b_ ) increasing 75/14 units/sec at (2, 3, 7) and decreasing 75/14 units/sec at (2, 3, -7); ( _c_ ) 6.3 units/sec

**32.** Find ∂z/∂s and ∂z/∂t, given:

**33.** ( _a_ ) If _u_ = _f_ ( _x_ , _y_ ) and _x_ = _r_ cos _θ_ , _y_ = _r_ sin _θ_ , show that

( _b_ ) If _u_ = _f_ ( _x_ , _y_ ) and _x_ = _r_ cosh _s_ , _y_ = _r_ sinh _s_ , show that

**34.** ( _a_ ) If _z_ = _f_ ( _x_ \+ α _y_ ) + _g_ ( _x_ − α _y_ ), show that  . ( _Hint_ : Write _z_ = _f_ ( _u_ ) + _g_ ( _υ_ ), _u_ = _x_ \+ α _y_ , _υ_ = _x_ − α _y_.)

( _b_ ) If _z_ = _x nf_( _y_ / _x_ ), show that _x_ ∂ _z_ /∂ _x_ \+ _y_ ∂ _z_ /∂ _y_ = _nz_.

( _c_ ) If _z_ = _f_ ( _x_ , _y_ ) and _x_ = _g_ ( _t_ ), _y_ = _h_ ( _t_ ), show that, subject to continuity conditions

( _d_ ) If _z_ = _f_ ( _x_ , _y_ ); _x_ = _g_ ( _r_ , _s_ ), _y_ = _h_ ( _r_ , _s_ ), show that, subject to continuity conditions

**35.** A function _f_ ( _x_ , _y_ ) is called _homogeneous of order n_ if _f_ ( _tx_ , _ty_ ) = _t nf_( _x_ , _y_ ). (For example, _f_ ( _x_ , _y_ ) = _x_ 2 \+ 2 _xy_ \+ 3 _y_ 2 is homogeneous of order 2; _f_ ( _x_ , _y_ ) = _x_ sin ( _y_ / _x_ ) + _y_ cos( _y_ / _x_ ) is homogeneous of order 1.) Differentiate _f_ ( _tx_ , _ty_ ) = _t nf_( _x_ , _y_ ) with respect to _t_ and replace _t_ by 1 to show that _xf x_ \+ _yf y_ = _nf_. Verify this formula using the two given examples. See also Problem 34( _b_ ).

**36.** If _z_ =  ( _u_ , _υ_ ), where _u_ = _f_ ( _x_ , _y_ ) and _υ_ = _g_ ( _x_ , _y_ ), and if  , show that

( _a_ )   ( _b_ )

**37.** Find   and  , given

**38.** Find all the first and second partial derivatives of _z_ , given _x_ 2 \+ 2 _yz_ \+ 2 _zx_ = 1.

_Ans._

**39.** If _F_ ( _x, y, z_ ) = 0 show that  .

**40.** If _f_ ( _x, y_ ) = 0 and _g_ ( _z, x_ ) = 0, show that  .

**41.** Find the first partial derivatives of _u_ and _v_ with respect to _x_ and _y_ and the first partial derivatives of _x_ and _y_ with respect to _u_ and _v_ , given 2 _u_ \- _v_ \+ _x_ 2 \+ _xy_ = 0, _u_ \+ 2 _v_ \+ _xy_ \- _y_ 2 = 0.

_Ans._

**42.** If _u_ = _x_ \+ _y_ \+ _z_ , _v_ = _x_ 2 \+ _y_ 2 \+ _z_ 2, and _w_ = _x_ 3 \+ _y_ 3 \+ _z_ 3, show that

**43.** Fill in the gaps in the following sketch of a proof of Theorem 49.1. Assume that _f_ ( _x, y_ ) is such that _f_ x and _f_ y are continuous in an open set _A_. Then _f_ is differentiable in _A_.

There exists _x_ * between _a_ and _a_ \+ Δ _x_ such that

_f_ ( _a_ \+ Δ _x, b_ ) - _f_ ( _a, b_ ) = _f x_( _x_ *, _b_ ) Δ _x_.

and there exists _y_ * between _b_ and _b_ \+ Δ _y_ such that

_f_ ( _a_ \+ Δ _x_ , _b_ \+ Δ _y_ ) - _f_ ( _a_ \+ Δ _x, b_ ) = _f y_( _a_ \+ Δ _x, y_ *) Δ _y_

Then

Let  1 = _f x_( _x*, y_ ) - _f x_( _a, b_ ) and  2 = _f y_( _a_ \+ Δ _x, y_ *) - _f y_( _a, b_ ). Then

Δ _z_ = _f x_( _a, b_ ) Δ _x_ \+ _f y_( _a, b_ ) Δ _y_ \+ _ 1_ Δ _x_ \+ _ 2_ Δ _y_

To show that  1 → 0 and  2 → 0, use the continuity of _f x_ and _f y_.

**44.** Show that continuity of _f_ ( _x, y_ ) does not imply differentiability, even when _f x_ and _f y_ both exist. Use the function

[ _Hint_ : Show that _f_ is not continuous at (0, 0) and, therefore, not differentiable. Show the existence of _f x_ (0, 0) and _f y_ (0, 0) by a direct computation.]

**45.** Find a function _f_ ( _x, y_ ) such that _f x_ (0, 0) = _f y_ (0, 0) = 0, and _f_ is not continuous at (0, 0). This shows that existence of the first partial derivatives does not imply continuity. [ _Hint_ : Define   for ( _x, y_ ) ≠ (0, 0) and _f_ (0, 0) = 0.]

## **CHAPTER 50
Space Vectors**

**VECTORS IN SPACE.** As in the plane (see Chapter 39), a vector in space is a quantity that has both magnitude and direction. Three vectors **a** , **b** , and **c** , not in the same plane and no two parallel, issuing from a common point are said to form a _right-handed system_ or _triad_ if **c** has the direction in which the right-threaded screw would move when rotated through the smaller angle in the direction from **a** to **b** , as in Fig. 50-1. Note that, as seen from a point on **c** , the rotation through the smaller angle from **a** to **b** is counterclockwise.

**Fig. 50-1**

**Fig. 50-2**

We choose a right-handed rectangular coordinate system in space and let **i** , **j** , and **k** be unit vectors along the positive _x, y_ , and _z_ axes, respectively, as in Fig. 50-2. The coordinate axes divide space into eight parts, called _octants_. The _first octant_ , for example, consists of all points ( _x, y, z_ ) for which _x_ > 0, _y_ > 0, _z_ > 0.

As in Chapter 39, any vector **a** may be written as

**a** = a1 **i** \+ a2 **j** \+ a3 **k**

If P( _x, y, z_ ) is a point in space (Fig. 50-2), the vector **r** from the origin _O_ to _P_ is called the _position vector_ of _P_ and may be written as

The algebra of vectors developed in Chapter 39 holds here with only such changes as the difference in dimensions requires. For example, if **a** = _a_ 1 **i** \+ _a_ 2 **j** \+ _a_ 3 **k** and **b** = _b_ 1 **i** \+ _b_ 2 **k**. then

k **a** = _ka_ 1 **i** \+ _ka_ 2 **j** \+ _ka_ 3 **k** for _k_ any scalar

**a** = **b** if and only if _a_ 1 = _b_ 1, _a_ 2 = _b_ 2, and _a_ 3 = _b_ 3

**a ± b** = ( _a_ 1 ± _b_ 1) **i** \+ (a2 ± _b_ 2) **j** \+ ( _a_ 3 ± _b_ 3) **k**

**a · b** = | **a** || **b** |cos _θ_ , where _θ_ is the smaller angle between **a** and **b**

**i · i** = **j · j** = **k · k** = 1 and **i · j** = **j · k** = **k · i** = 0

**a·b** = 0 if **a** = **0** , or **b** = **0** , or **a** and **b** are perpendicular

From ( _50.1_ ), we have

as the distance of the point _P_ ( _x, y, z_ ) from the origin. Also, if _P_ 1( _x_ 1, _y_ 1, _z_ 1) and _P_ 2( _x_ 2, _y_ 2, _z_ 2) are any two points (see Fig. 50-3), then

**P** 1 **P** 2 = **P** 1 **B** \+ **BP** 2 = **P** 1 **A** \+ **AB** \+ **BP** 2 = ( _x_ 2 − _x_ 1) **i** \+ ( _y_ 2 − _y_ 1) **j** \+ ( _z_ 2 − _z_ 1) **k**

and

is the familiar formula for the distance between two points. (See Problems 1–3.)

**Fig. 50-3**

**Fig. 50-4**

**DIRECTION COSINES OF A VECTOR.** Let **a** = _a_ 1 **i** \+ _a_ 2 **j** \+ _a_ 3 **k** make angles _α_ , _β_ , and _γ_ , respectively, with the positive _x, y_ , and _z_ axes, as in Fig. 50-4. From

**i** · **a** = | **i** || **a** | cos _α_ = | **a** |cos _α_ , **j** · **a** = | **a** |cos _β_ , **k** · **a** = | **a** |cos _γ_

we have

These are the _direction cosines_ of **a**. Since

the vector **u** = **i** cos _α_ \+ **j** cos _β_ = **k** cos _γ_ is a unit vector parallel to **a**.

**DETERMINANTS.** We shall assume familiarity with 2 × 2 and 3 × 3 determinants. In particular,

The expansion of the 3 × 3 determinant is said to be "along the first row". It is equal to suitable expansions along the other rows and down the columns.

**VECTOR PERPENDICULAR TO TWO VECTORS.** Let

**a** = _a_ 1 **i** \+ _a_ 2 **j** \+ _a_ 3 **k** and **b** = _b_ 1 **i** \+ _b_ 2 **j** \+ _b_ 3 **k**

be two nonparallel vectors with common initial point _P_. By an easy computation it can be shown that

is perpendicular to (normal to) both **a** and **b** and, hence, to the plane of these vectors.

In Problems 5 and 6 we show that

If **a** and **b** are parallel, then **b** = _k_ **a** , and ( _50.4_ ) shows that **c** = **0** ; that is, **c** is the zero vector. The zero vector, by definition, has magnitude 0 but no specified direction.

**VECTOR PRODUCT OF TWO VECTORS.** Take

**a** = _a_ 1 **i** \+ _a_ 2 **j** \+ _a_ 3 **k** and **b** = _b_ 1 **i** \+ _b_ 2 **j** \+ _b_ 3 **k**

with initial point _P_ and denote by **n** the unit vector normal to the plane of **a** and **b** , so directed that **a** , **b** , and **n** (in that order) form a right-handed triad at _P_ , as in Fig. 50-5. The _vector product_ or _cross product_ of **a** and **b** is defined as

where _θ_ is again the smaller angle between **a** and **b**. Thus, **a × b** is a vector perpendicular to both **a** and **b**.

We show in Problem 6 that | **a × b** | = | **a** || **b** | sin _θ_ is the area of the parallelogram having **a** and **b** as nonparallel sides.

If **a** and **b** are parallel, then _θ_ = 0 or _π_ and **a × b** = **0** , Thus,

**Fig. 50-5**

In ( _50.6_ ), if the order of **a** and **b** is reversed, then **n** must be replaced by – **n** ; hence,

Since the coordinate axes were chosen as a right-handed system, it follows that

In Problem 8, we prove for any vectors **a** , **b** and **c** , the distributive law

Multiplying ( _50.10_ ) by –1 and using ( _50.8_ ), we have the companion distributive law

Then, also,

and

(See Problems 9 and 10.)

**TRIPLE SCALAR PRODUCT.** In Fig. 50-6, let _θ_ be the smaller angle between **b** and **c** and let   be the smaller angle between **a** and **b × c**. Then the triple scalar product is by definition

It may be shown (see Problem 11) than

**Fig. 50-6**

Also

while

Similarly, we have

and

From the definition of **a ·** ( **b × c** ) as a volume, it follows that if **a** , **b** , and **c** are coplanar, then **a ·** ( **b × c** ) = 0, and conversely.

The parenthesis in **a ·** ( **b × c** ) and ( **a × b** ) **· c** are not necessary. For example, **a · b × c** can be interpreted only as **a** ( **b × c** ) or ( **a · b** ) **× c**. But **a · b** is a scaler, so ( **a · b** ) **× c** is without meaning. (See Problem 12.)

**TRIPLE VECTOR PRODUCT.** In Problem 13, we show that

Similarly,

Thus, except when **b** is perpendicular to both **a** and **c** , **a ×** ( **b × c** ) ≠ ( **a × b** ) **× c** and the use of parenthesis is necessary.

**THE STRAIGHT LINE.** A line in space through a given point _P_ 0( _x_ 0, _y_ 0, _z_ 0) may be defined as the locus of all points _P_ ( _x, y, z_ ) such that _P_ 0 _P_ is parallel to a given direction **a** = _a_ 1 **i** \+ _a_ 2 **j** \+ _a_ 3 **k**. Let **r** 0 and **r** be the position vectors of _P_ 0 and _P_ (Fig. 50-7). Then

is the vector equation of line _PP_ 0. Writing ( _50.19_ ) as

( _x_ − _x_ 0) **i** \+ ( _y_ − _y_ 0) **j** \+ ( _z_ − _z_ 0) **k** = _k_ ( _a_ 1 **i** \+ _a_ 2 **j** \+ _a_ 3 **k** )

then separating components to obtain

_x_ − _x_ 0 = _ka_ 1, _y_ − _y_ 0 = _ka_ 2, _z_ − _z_ 0 = _ka_ 3

**Fig. 50-7**

and eliminating _k_ , we have

as the equations of the line in rectangular coordinates. Here, [ _a_ 1, _a_ 2, _a_ 3] is a set of _direction numbers_ for the line and   is a set of _direction cosines_ of the line.

If any one of the numbers _a_ 1, _a_ 2, _a_ 3 is zero, the corresponding numerator in ( _50.20_ ) must be zero. For example, if _a_ 1 = 0 but _a_ 2, _a_ 3 ≠ 0, the equations of the line are

_x_ − _x_ 0 = 0 and

**THE PLANE.** A plane in space through a given point _P_ 0( _x_ 0, _y_ 0, _z_ 0) can be defined as the locus of all lines through _P_ 0 and perpendicular (normal) to a given line (direction) **a** = _A_ **i** \+ _B_ **j** \+ _C_ **k** (Fig. 50-8). Let _P_ ( _x, y, z_ ) be any other point in the plane. Then **r** − **r** 0 = **P** 0 **P** is perpendicular to **a** , and the equation of the plane is

**Fig. 50-8**

In rectangular coordinates, this becomes

or

or

Where _D_ = − ( _Ax_ 0 \+ _By_ 0 \+ _Cz_ 0).

Conversely, let _P_ 0( _x_ 0, _y_ 0, _z_ 0) be a point on the surface _Ax_ \+ _By_ \+ _Cz_ \+ _D_ = 0. Then also _Ax_ 0 \+ _By_ 0 \+ _Cz_ 0 \+ _D_ = 0. Substracting the second of these equations from the first yields

**_A_** ( _x_ − _x_ 0) + **_B_** ( _y_ − _y_ 0) + **_C_** ( _z_ − _z_ 0) = ( ** _A_ i** \+ **_B_ j** \+ **_C_ k**) · [( _x_ − _x_ 0) **i** \+ ( _y_ − _y_ 0) **j** \+ ( _z_ − _z_ 0) **k** ] = 0 and the constant vector **_A_ i** \+ **_B_ j** \+ **_C_ k** is normal to the surface at each of its points. Thus, the surface is a plane.

### **Solved Problems**

**1.** Find the distance of the point _P_ 1(1, 2, 3) from ( _a_ ) the origin, ( _b_ ) the _x_ axis, ( _c_ ) the _z_ axis, ( _d_ ) the _xy_ plane, and ( _e_ ) the point _P_ 2(3, −1.5).

In Fig. 50-9,

**Fig. 50-9**

( _a_ ) **r** = **OP** 1 = **i** \+ 2 **j** \+ 3 **k** ; hence,  .

( _b_ ) **AP** 1 = **AB** \+ **BP** 1 = 2 **j** \+ 3 **k** ; hence,  .

( _c_ ) **DP** 1 = **DE** \+ **EP** 1 = 2 **j** \+ **i** ; hence,  .

( _d_ ) **BP** 1 = 3 **k** , so | **BP** 1| = 3.

( _e_ ) **P** 1 **P** 2 = (3 − 1) **i** \+ (−1 −2) **j** \+ (5 − 3) **k** = 2 **i** − 3 **j** \+ 2 **k** ; hence,  .

**2.** Find the angle _θ_ between the vectors joining _O_ to _P_ 1(1, 2, 3) and _P_ 2(2, −3, −1).

Let **r** 1 = **OP** 1 = **i** \+ 2 **j** \+ 3 **k** and **r** 2 = **OP** 2 = 2 **i** − 3 **j** − **k**. Then

**3.** Find the angle _α_ = _ BAC_ of the triangle _ABC_ (Fig. 50-10) whose vertices are _A_ (1, 0, 1), _B_ (2, −1, 1), _C_ (−2, 1, 0).

**Fig. 50-10**

Let **a = AC = −3i + j − k** and **b = AB = i − j**. Then

**4**. Find the direction cosines of **a** = 3 **i** \+ 12 **j** \+ 4 **k**.

The direction cosines are  .

**5**. If **a** = _a_ 1 **i** \+ _a_ 2 **j** + _a_ 3 **k** and **b** = _b_ 1 **i** \+ _b_ 2 **j** \+ _b_ 3 **k** are two vectors issuing from a point _P_ and if

show that | **c** | = | **a** | **b** | sin _θ_ , where _θ_ is the smaller angle between **a** and **b**.

We have   and

Hence, | **c** | = | **a** | **b** | sin _θ_ as required.

**6**. Find the area of the parallelogram whose nonparallel sides are **a** and **b**.

From Fig. 50-11, _h_ = | **b** | sin _θ_ and the area is _h_ | **a** | = | **a** || **b** | sin _θ_.

**7**. Let **a** 1 and **a** 2, respectively, be the components of **a** parallel and perpendicular to **b** , as in Fig. 50-12. Show that **a** 2 × **b** = **a × b** and **a 1 × b = 0**.

If _θ_ is the angle between **a** and **b** , then | **a** 1| = | **a** |cos _θ_ and | **a** 2| = | **a** |sin _θ_. Since **a** , **a** 2, and **b** are coplanar,

**a** 2 **× b** = | **a** 2|| **b** | sin   **n** = | **a** | sin _θ_ |b| **n** = | **a** || **b** | sin _θ_ **n** = **a × b**

Since **a** 1 and **b** are parallel, **a** 1 × **b** = **0**.

**8**. Prove: **(a + b) × c = (a × c) + (b × c)**

In Fig. 50-13, the initial point _P_ of the vectors **a, b** , and **c** is in the plane of the paper, while their endpoints are above this plane, The vectors **a** 1 and **b** 1 are, respectively, the components of **a** and **b** perpendicular to **c**. Then **a** 1, **b** 1, **a** 1 \+ **b** 1, **a** 1 **× c** , **b** 1 **× c** , and ( **a** 1 \+ **b** 1) **× c** all lie in the plane of the paper.

**Fig. 50-11**

**Fig. 50-12**

**Fig. 50-13**

In triangles _PRS_ and _PMQ_ ,

Thus, _PRS_ and _PMQ_ are similar. Now _PR_ is perpendicular to _PM_ , and _RS_ is perpendicular to _MQ_ ; hence _PS_ is perpendicular to _PQ_ and **PS** = **PQ × c**. Then, since **PS** = **PQ × c** = **PR + RS** , we have

**(a** 1 \+ **b** 1) × **c** = (a1 × **c)** \+ **(b** 1 × **c)**

By Problem 7, **a** 1 and **b** 1 may be replaced by **a** and **b** , respectively, to yield the required result.

**9**. When **a** = _a_ 1 **i** \+ _a_ 2 **j** \+ _a_ 3 **k** and **b** = _b_ 1 **i** \+ _b_ 2 **j** \+ _b_ 3 **k** , show that  .

We have, by the distributive law,

**10.** Derive the law of sines of plane trigonometry.

Consider the triangle _ABC_ , whose sides **a, b, c** are of magnitudes _a, b, c,_ respectively, and whose interior angles are α, β, γ. We have

**11.** If **a** = _a_ 1 **i** \+ _a_ 2 **j** \+ a3 **k** , **b** = b1 **i** \+ _b_ 2 **j** \+ _b_ 3 **k** , and **c** = _c_ 1 **i** \+ _c_ 2 **j** \+ _c_ 3 **k** , show that

By ( _50.13_ ),

**12.** Show that **a** · ( **a** × **c** ) = 0.

By ( _50.14_ ), **a** · ( **a** × **c** ) = ( **a** × **a** ) · **c** = 0.

**13.** For the vectors **a** , **b** , and **c** of Problem 11, show that **a** × ( **b** × **c** ) = ( **a** · **c** ) **b** − ( **a** · **b** ) **c**.

Here

**14**. If _l_ 1 and _l_ 2 are two nonintersecting lines in space, show that the shortest distance _d_ between them is the distance from any point on _l_ 1 to the plane through _l_ 2 and parallel to _l_ 1; that is, show that if _P_ 1 is a point on _l_ 1 and _P_ 2 is a point on _l_ 2 then, apart from sign, _d_ is the scalar projection of **P** 1 **P** 2 on a common perpendicular to _l_ 1 and _l_ 2.

Let _l_ 1 pass through _P_ 1( _x_ 1, _y_ 1, _z_ 1) in the direction a = _a_ 1 **i** \+ _a_ 2 **j** \+ _a_ 3 **k** , and let _l_ 2 pass through _P_ 2( _x_ 2, _y_ 2, _z_ 2) in the direction b = _b_ 1 **i** \+ _b_ 2 **j** \+ _b_ 3 **k**.

Then **P** 1 **P** 2 = ( _x_ 2 − _x_ 1) **i** \+ ( _y_ 2 − _y_ 1) **j** \+ ( _z_ 2 − _z_ 1) **k** , and the vector **a × b** is perpendicular to both _l_ 1 and _l_ 2. Thus,

**15**. Write the equation of the line passing through _P_ 0(1, 2, 3) and parallel to **a** = 2 **i** − **j** − 4 **k**. Which of the points _A_ (3, 1, −1),  , _C_ (2, 0, 1) are on this line?

From ( _50.19_ ), the vector equation is

The rectangular equations are

Using ( _2_ ), it is readily found that _A_ and _B_ are on the line while _C_ is not.

In the vector equation ( _1_ ), a point _P_ ( _x, y, z_ ) on the line is found by giving _k_ a value and comparing components. The points _A_ is on the line because

(3−1) **i** \+ (1−2) **j** \+ (−1−3) **k** = _k_ (2 **i** − **j** − 4 **k** )

when _k_ = 1. Similarly _B_ is on the line because

when  . The point _C_ is not on the line because

**i** − 2 **j** − 2 **k** = _k_ (2 **i** − **j** − 4 **k** )

for no value of _k_.

**16**. Write the equation of the plane:

( _a_ ) Passing through _P_ 0(1, 2, 3) and parallel to 3 _x_ − 2 _y_ \+ 4 _z_ − 5 = 0.

( _b_ ) Passing through _P_ 0(1, 2, 3) and _P_ 1(3, −2, 1), and perpendicular to the plane 3 _x_ − 2 _y_ \+ 4 _z_ − 5 = 0.

( _c_ ) Through _P_ 0(1, 2, 3), _P_ 1(3, −2, 1) and _P_ 2(5, 0, −4).

Let _P_ ( _x, y, z_ ) be a general point in the required plane.

( _a_ ) Here **a** = 3 **i** − 2 **j** \+ 4 **k** is normal to the given plane and to the required plane. The vector equation of the latter is ( **r** − **r** 0) · **a** = 0 and the rectangular equation is

( _b_ ) Here **r** 1 − **r** 0 = 2 **i** − 4 **j** − 2 **k** and **a** = 3 **i** − 2 **j** \+ 4 **k** are parallel to the required plane; thus, ( **r** 1 − **r** 0) **× a** is normal to this plane. Its vector equation is ( **r** − **r** 0) · [( **r** 1 − **r** 0) **× a** ] = 0. The rectangular equation is

or 20 _x_ \+ 14 _y_ − 8 _z_ − 24 = 0.

( _c_ ) Here **r** 1 − **r** 0 = 2i − 4j − 2k and _r_ 2 − _r_ 0 = 4 **i** = 2 **j** − 7 **k** are parallel to the required plane, so that ( **r** 1 − _r_ 0) × ( _r_ 2 − _r_ 0) is normal to it. The vector equation is ( _r_ − _r_ 0) · [( **r** 1 − **r** 0) × ( _r_ 2 − _r_ 0)] = 0 and the rectangular equation is

or 4 _x_ \+ _y_ \+ 2 _z_ − 12 = 0.

**17**. Find the shortest distance _d_ between the point _P_ 0(1, 2, 3) and the plane II given by the equation 3 _x_ − 2 _y_ \+ 5 _z_ − 10 = 0.

A normal to the plane is **a** = 3 **i** − 2 **j** \+ 5 **k**. Take _P_ 1(2, 3, 2) as a convenient point in II. Then, apart from sign, _d_ is the scalar projection of _P_ 0 **P** 1 on **a**. Hence,

### **Supplementary Problems**

**18**. Find the length of ( _a_ ) the vector **a** = 2 **i** \+ 3 **j** \+ **k** ; ( _b_ ) the vector **b** = 3 **i** − 5 **j** \+ 9 **k** ; and ( _c_ ) the vector **c** , joining _P_ 1(3, 4, 5) to _P_ 2(1, −2, 3).

_Ans_.

**19**. For the vectors of Problem 18:

( _a_ ) Show that **a** and **b** are perpendicular.

( _b_ ) Find the smaller angle between **a** and **c** , and that between **b** and **c**.

( _c_ ) Find the angles that **b** makes with the coordinate axes.

_Ans_. ( _b_ ) 165°14′, 85°10′ ( _c_ ) 73°45′, 117°47′, 32°56′

**20**. Prove: **i** · **i = j** · **j = k** · **k = 1** and **i** · **j = j** · **k = k** · **i** = 0.

**21**. Write a unit vector in the direction of **a** and a unit vector in the direction of **b** of Problem 18.

_Ans._

**22**. Find the interior angles β and γ of the triangle of Problem 3.

_Ans_. β = 22°12′; γ = 9°16′

**23**. For the unit cube in Fig. 50-14, find ( _a_ ) the angle between its diagonal and an edge, and ( _b_ ) the angle between its diagonal and a diagonal of a face.

_Ans_. ( _a_ ) 54°44′; ( _b_ ) 35°16′

**Fig. 50-14**

**24**. Show that the scalar projection of **b** onto **a** is given by  .

**25**. Show that the vector **c** of ( _50.4_ ) is perpendicular to both **a** and **b**.

**26**. Given **a = i + j, b = i − 2k,** and **c = 2i + 3j + 4k** , confirm the following equations:

( _a_ ) **a × b = − 2i + 2j − k**

( _b_ ) **b × c = 6i − 8j + 3k**

( _c_ ) **c × a = − 4i + 4j − k**

( _d_ ) **(a + b) × (a − b) = 4i − 4j + 2k**

( _e_ ) **a · (a × b) = 0**

( _f_ ) **a · (b × c) = − 2**

( _g_ ) **a × (b × c) = 3i − 3j − 14k**

( _h_ ) **c × (a × b) = − 11i − 6j + 10k**

**27**. Find the area of the triangle whose vertices are _A_ (1, 2, 3), _B_ (2, −1, 1), and _C_ (−2, 1, −1).( _Hint:_ | **AB × AC** | = twice the area.)

_Ans_.

**28**. Find the volume of the parallelipiped whose edges are _OA, OB_ , and _OC_ , for _A_ (1, 2, 3), _B_ (1, 1, 2), and _C_ (2, 1, 1).

_Ans_. 2

**29**. If **u = a × b, v = b × c, w = c × a,** show that:

( _a_ ) **u · c = v · a = w · b**

( _b_ ) **a · u = b · u = 0, b · v = 0, c · w = a · w = 0**

( _c_ ) **u · (v × w) = [a · (b × c)] 2**

**30**. Show that **(a + b) · [(b + c) × (c + a)] = 2a · (b × c)**.

**31**. Find the smaller angle of intersection of the planes 5 _x_ − 14 _y_ \+ 2 _z_ − 8 = 0 and 10 _x_ − 11 _y_ \+ 2 _z_ \+ 15 = 0. ( _Hint:_ Find the angle between their normals.)

_Ans_. 22°25′

**32**. Write the vector equation of the line of intersection of the planes _x_ \+ _y_ − _z_ − 5 = 0 and 4 _x_ − _y_ − _z_ \+ 2 = 0.

_Ans_. ( _x_ − 1) **i** \+ ( _y_ − 5) **j** \+ ( _z_ − 1) **k** = _k_ (−2 **i** − 3 **j** − 5 **k** ), where _P_ 0(1, 5, 1) is a point on the line.

**33**. Find the shortest distance between the line through _A_ (2, −1, −1) and _B_ (6, −8, 0) and the line through _C_ (2, 1, 2) and _D_ (0, 2, −1).

_Ans_.

**34**. Define a line through _P 0(x0, y0, z0)_ as the locus of all points _P(x, y, z)_ such that **P** 0 **P** and **OP** 0 are perpendicular. Show that its vector equation is ( **r − r** 0) · **r** 0 = 0.

**35**. Find the rectangular equations of the line through _P_ 0(2, −3, 5) and

( _a_ ) perpendicular to 7 _x_ − 4 _y_ \+ 2 _z_ − 8 = 0;

( _b_ ) parallel to the line _x − y_ \+ 2 _z_ \+ 4 = 0, 2 _x_ \+ 3 _y_ \+ 6 _z_ − 12 = 0;

( _c_ ) through _P_ 1(3, 6, −2).

_Ans_.

**36**. Find the equation of the plane:

( _a_ ) Through _P_ 0(1, 2, 3) and parallel to **a = 2i + j − k** and **b = 3i + 6j − 2k**.

( _b_ ) Through _P_ 0(2, −3, 2) and the line 6 _x_ \+ 4 _y_ \+ 3 _z_ \+ 5 = 0, 2 _x_ \+ _y_ \+ _z_ − 2 = 0.

( _c_ ) Through _P_ 0(2, −1, −1) and _P_ 1(1, 2, 3) and perpendicular to 2 _x_ \+ 3 _y_ − 5 _z_ − 6 = 0.

_Ans_. ( _a_ )4 _x_ \+ _y_ \+ 9 _z_ − 33 = 0; ( _b_ ) 16 _x_ \+ 7 _y_ \+ 8 _z_ − 27 = 0; ( _c_ )9 _x_ − _y_ \+ 3 _z_ − 16 = 0

**37**. If _r_ 0 = **i + j + k** , _r_ 1 = **2i + 3j + 4k** , and **r** 2 = **3i + 5j + 7k** are three position vectors, show that **r** 0 × **r** 1 \+ **r** 1 × **r** 2 \+ **r** 2 × **r** 0 = **0**. What can be said of the terminal points of these vectors?

_Ans_. collinear

**38**. If _P_ 0, _P_ 1, and _P_ 2 are three noncollinear points and **r** 0, **r** 1, and **r** 2 are their position vectors, what is the position of **r** 0 × **r** 1 \+ **r** 1 × **r** 2 \+ **r** 2 × **r** 0 with respect to the plane _P 0P1P2_?

_Ans_. normal

**39**. Prove: ( _a_ ) **a × (b × c) + b × (c × a) + c × (a × b) = 0** ;

( _b_ ) **(a × b) · (c × d) = (a · d) = (a · c)(b · d) − (a · d)(b · c)**.

**40**. Prove: ( _a_ ) The perpendiculars erected at the midpoints of the sides of a triangle meet in a point; ( _b_ ) the perpendicular dropped from the vertices to the opposite sides (produced if necessary) of a triangle meet in a point.

**41**. Let _A_ (1, 2, 3), _B_ (2, −1, 5), and _C_ (4, 1, 3) be three vertices of the parallelogram _ABCD_. Find ( _a_ ) the coordinates of _D_ ; ( _b_ ) the area _ABCD_ ; and ( _c_ ) the area of the orthogonal projetion of _ABCD_ on each of the coordinate planes.

_Ans_. ( _a_ ) _D_ (3, 4, 1); ( _b_ )  ; ( _c_ ) 8, 6, 2

**42**. Prove that the area of a parallogram in space is the square root of the sum of the squres of the areas of projections of the parallelograms on the coordinate planes.

## **CHAPTER 51
Surfaces and Curves in Space**

**PLANES**. We already know (formula ( _50.22_ ) that the equation of a plane has the form _Ax_ \+ _By_ \+ _Cz_ \+ _D_ =0, where _A_ **i** \+ _B_ **j** \+ _C_ **k** is a nonzero vector perpendicular to the plane. The plane passes through the origin (0,0,0) when and only when _D_ = 0.

**SPHERES**. From the distance formula ( _50.3_ ), we see that an equation of the sphere with radius _r_ and center ( _a, b, c_ ) is

( _x_ − _a_ )2 \+ ( _y_ − _b_ )2 \+ ( _z_ − _c_ )2 = _r_ 2

So a sphere with center at the origin (0,0,0) and radius _r_ has the equation

_x_ 2 \+ _y_ 2 \+ _z_ 2 = _r_ 2

**CYLINDRICAL SURFACES**. An equation _F_ ( _x_ , _y_ ) = 0 ordinarily defines a curve   in the _xy_ plane. Now, if a point ( _x_ , _y_ ) satisfies this equation, then, for any _z_ , the point ( _x_ , _y_ , _z_ )in space also satisfies the equation. So, the equation _F_ ( _x_ , _y_ ) = 0 determines the cylindrical _surface obtained_ by moving the curve   parallel to the _z_ axis. For example, the equation _x_ 2 \+ _y_ 2 = 4 determines a circle in the _xy_ plane with radius 2 and center at the origin. If we move this circle parallel to the _z_ axis, we obtain a right circular cylinder. Thus, what we ordinarily call a cylinder is a special case of a cylindrical surface.

Similarly, an equation _F_ ( _y_ , _z_ ) = 0 determines the cylindrical surface obtained by moving the curve in the _yz_ plane defined by _F_ ( _y_ , _z_ ) = 0 parallel to the _x_ axis. An equation _F_ ( _x_ , _z_ ) = 0 determines the cylindrical surface obtained by moving the curve in the _xz_ plane defined by _F_ ( _x_ , _z_ ) = 0 parallel to the y axis.

More precisely, the cylindrical surfaces defined above are called right cylindrical surfaces. Other cylindrical surfaces can be obtained by moving the given curve parallel to a line that is not perpendicular to the plane of the curve.

**EXAMPLE 1:** The equation _z_ = _x_ 2 determines a cylindrical surface generated by moving the parabola _z_ = _x_ 2 lying in the _xz_ plane parallel to the _y_ axis.

Now we shall look at examples of surfaces determined by equations of the second degree in _x_ , _y_ , and _z_. Such surfaces are called _quadric_ surfaces. Imagining what they look like is often heped by describing their intersections with planes parallel to the coordinate planes. Such intersections are called _traces_.

**ELLIPSOID**.

The nontrivial traces are ellipses. See Fig. 51-1. In general, the equation of an ellipsoid has the form

When _a_ = _b_ = _c_ , we obtain a sphere.

**Fig. 51-1**

**ELLIPTIC PARABOLOID**. _z_ = _x_ 2 \+ _y_ 2

The surface lies on or above the _xy_ plane. The traces parallel to the _xy_ plane (for a fixed positive _z_ ) are circles. The traces parallel to the _xz_ or _yz_ plane are parabolas. See Fig. 51-2. In general, the equation of an elliptic paraboloid has the form

**Fig. 51-2**

and the traces parallel to the _xy_ plane are ellipses. When _a_ = _b_ , we obtain a circular paraboloid, as in the given example.

**ELLIPTIC CONE.** _z_ 2 = _x_ 2 \+ _y_ 2

See Fig 51-3. This is a pair of ordinary cones, meeting at the origin. The traces parallel to the _xy_ plane are circles. The traces parallel to the _xz_ or _yz_ plance are hyperbolas. In general, the equation of an elliptic cone has the from

and the traces parallel to the _xy_ plane are ellipse. When _a_ = _b_ , we obtain a right circular cone, as in the given example.

**Fig. 51-3**

**HYPERBOLIC PARABOLOID.** _z_ = 2 _y_ 2 − _x_ 2

See Fig.51-4. The surface resembles a saddle. The traces parallel to the _xy_ plane are hyberbolas. The other traces are parabolas. In general, the equation of a hyperbolic paraboloid has the form

In the given example, _c_ = 1, _a_ = 1, and

**HYPERBOLOID OF ONE SHEET.**

See Fig. 51-5. The traces parallel to the _xy_ plane are circles and the other traces are hyperbolas. In general, a hyperboloid of one sheet has an equation of the form

and the traces parallel to the _xy_ plane are ellipses.

**Fig. 51-4**

**Fig. 51-5**

**HYPERBOLOID OF TWO SHEETS**.

See Fig. 51-6. The traces parallel to the _xy_ plane are circles, and the other traces are hyperbolas. In general, a hyperboloid of two sheets has an equation of the form

and the traces parallel to the _xy_ plane are ellipses.

In general equations given above for various quadric surfaces, permutation of the variables _x_ , _y_ , _z_ is understood to produce quadric surfaces of the same type. For example,   also determines a hyperboloid of two sheets.

**TANGENT LINE AND NORMAL PLANE TO A SPACE CURVE.** A space curve may be defined parametrically by the equations

**Fig. 51-6**

At the point _P_ 0( _x_ 0, _y_ 0, _z_ 0) of the curve (determined by _t_ = _t_ 0), the equations of the tangent line are

and the equations of the normal plane (the plane through _P_ 0 perpendicular to the tangent line there) is

See Fig. 51-7. In both ( _51.2_ ) and ( _51.3_ ) it is understood that the derivative have been evaluated at the point _P_ 0. (See Problems 1 and 2.)

**Fig. 51-7**

**TANGENT PLANE AND NORMAL LINE TO A SURFACE.** The equation of the tangent plane to the surface _F_ ( _x_ , _y_ , _z_ ) = 0 at one of its points _P_ 0( _x_ 0, _y_ 0, _z_ 0) is

and the equations of the normal line at _P_ 0 are

with the understanding that the partial derivatives have been evaluated at the point _P_ 0. See Fig. 51-8. (See Problems 3-9).

**Fig. 51-8**

A space curve may also be defined by a pair of equations

At the point _P_ 0( _x_ 0, _y_ 0, _z_ 0) of the curve, the equations of the tangent line are

and the equation of the normal plane is

In ( _51.7_ ) and ( _51.8_ ) it is understood that all partial derivatives have been evaluated at the point _P_ 0. (See Problems 10 and 11.)

**SURFACE OF REVOLUTION.** Let the graph of _y_ = _f_ ( _x_ ) in the _xy_ plane be revolved about the _x_ axis. As a point ( _x_ 0, _y_ 0) on the graph revolves, a resulting point ( _x_ 0, _y_ , _z_ ) has the distance _y_ 0 from the point ( _x_ 0, 0, 0). So, squaring that distance, we get

( _x_ 0 − _x_ 0)2 \+ _y_ 2 \+ _z_ 2 = ( _y_ 0)2 = ( _f_ ( _x_ 0))2 and, therefore, _y_ 2 \+ _z_ 2 = ( _f_ ( _x_ 0))2

Then, the equation of the surface of revolution is

### **Solved Problems**

1. Derive ( _51.2_ ) and ( _51.3_ ) for the tangent line and normal plane to the space curve _x_ = _f_ ( _t_ ), _y_ = _g_ ( _t_ ), _z_ = _h_ ( _t_ ) at the point P0( _x_ 0, _y_ 0, _z_ 0) determined by the value _t_ = _t_ 0. Refer to Fig. 51-7.

Let _P_ ′0( _x_ 0 \+ Δ _x_ , _y_ 0 \+ Δ _y_ , _z_ 0 \+ Δ _z_ ), determined by _t_ = _t_ 0 \+ Δ _t_ , be another point on the curve. As _P_ ′0 → _P_ 0 along the curve, the chord _P_ 0 _P_ ′0 approaches the tangent line to the curve at _P_ 0 as limiting position.

A simple set of direction numbers for the chord _P_ 0 _P_ ′0 is [Δ _x_ , Δ _y_ , Δ _z_ ], but we shall use  . Then as _P_ ′0 → _P_ 0, Δ _t_ → 0 and  , a set of direction numbers of the tangent line at _P_ 0. Now if _P_ ( _x, y, z_ ) is an arbitrary point on this tangent line, then [ _x_ \- _x_ 0, _y_ \- _y_ 0, _z_ \- _z_ 0] is a set of direction numbers of _P_ 0 _P_. Thus, since the sets of direction numbers are proportional, the equations of the tangent line at _P_ 0 are

If _R_ ( _x, y, z_ ) is an arbitrary point in the normal plane at _P_ 0 then, since _P_ 0 _R_ and _P_ 0 _P_ are perpendicular, the equation of the normal plane at _P_ 0 is

2. Find the equations of the tangent line and normal plane to:

( _a_ ) The curve _x_ = _t_ , _y_ = _t_ 2, _z_ = _t_ 3 at the point _t_ = 1.

( _b_ ) The curve _x_ = _t_ − 2, _y_ = 3 _t_ 2 \+ 1, _z_ = 2 _t_ 3 at the point where it pierces the _yz_ plane.

( _a_ ) At the point _t_ = 1 or (1, 1, 1), _dx/dt_ = 1, _dy/dt_ = 2 _t_ = 2, and _dz/dt_ = 3 _t_ 2 = 3. Using ( _51.2_ ) yields, for the equations of the tangent line,  ; using ( _51.3_ ) gives the equation of the normal plane as ( _x_ \- 1) + 2( _y_ \- 1) + 3( _z_ \- 1) = _x_ \+ 2 _y_ \+ 3 _z_ \- 6 = 0.

( _b_ ) The given curve pierces the _yz_ plane at the point where _x_ = _t_ − 2 = 0, that is, at the point _t_ = 2 or (0, 13, 16). At this point, _dx/dt_ = 1, _dy/dt_ = 6 _t_ = 12 and _dz/dt_ = 6 _t_ 2 = 24. The equations of the tangent line are  , and the equation of the normal plane is _x_ \+ 12( _y_ − 13) + 24( _z_ − 16) = _x_ \+ 12 _y_ \+ 24 _z_ − 540 = 0.

3. Derive ( _51.4_ ) and ( _51.5_ ) for the tangent plane to the surface _F_ ( _x, y, z_ ) = 0 at the point _P_ 0( _x_ 0, _y_ 0, _z_ 0). Refer to Fig. 51-8.

Let _x_ = _f_ ( _t_ ), _y_ = _g_ ( _t_ ), _z_ = _h_ ( _t_ ) be the parametric equations of any curve on the surface _F_ ( _x, y, z_ ) = 0 and passing through the point _P_ 0. Then, at _P_ 0,

with the understanding that all derivatives have been evaluated at _P_ 0.

This relation expresses the fact that the line through _P_ 0 with direction numbers   is perpendicular to the line through _P_ 0 having direction numbers  . The first set of direction numbers belongs to the tangent to the curve which lies in the tangent plane of the surface. The second set defines the normal line to the surface at _P_ 0. The equations of this normal are

and the equation of the tangent plane at _P_ 0 is

In Problems 4 and 5, find the equations of the tangent plane and normal line to the given surface at the given point.

4. _z_ = 3 _x_ 2 \+ 2 _y_ 2 − 11; (2, 1, 3).

Put _F_ ( _x, y, z_ ) = 3 _x_ 2 \+ 2 _y_ 2 − _z_ − 11 = 0. At (2, 1, 3),  . The equation of the tangent plane is 12( _x_ − 2) + 4( _y_ − 1) − ( _z_ − 3) = 0 or 12 _x_ \+ 4 _y_ − _z_ = 25.

The equations of the normal line are  .

5. _F_ ( _x, y, z_ ) = _x_ 2 \+ 3 _y_ 2 − 4 _z_ 2 \+ 3 _xy_ − 10 _yz_ \+ 4 _x_ − 5 _z_ − 22 = 0; (1, −2, 1).

 . The equation of the tangent plane is 0( _x_ − 1) − 19( _y_ \+ 2) + 7( _z_ − 1) = 0 or 19 _y_ − 7 _z_ \+ 45 = 0.

The equations of the normal line are _x_ − 1 = 0 and   or _x_ = 1, 7 _y_ \+ 19z − 5 = 0.

6. Show that the equation of the tangent plane to the surface   at the point _P_ 0( _x_ 0, _y_ 0, _z_ 0) is  .

At _P_ 0,   and  . The equation of the tangent plane is  .

This becomes  , since _P_ 0 is on the surface.

7. Show that the surfaces

_F_ ( _x, y, z_ ) = _x_ 2 \+ 4 _y_ 2 − 4 _z_ 2 − 4 = 0 and _G_ ( _x, y, z_ ) = _x_ 2 \+ _y_ 2 \+ _z_ 2 − 6 _x_ − 6 _y_ \+ 2 _z_ \+ 10 = 0 are tangent at the point (2, 1, 1).

It is to be shown that the two surfaces have the same tangent plane at the given point. At (2, 1, 1),

and

Since the sets of direction numbers [4, 8, −8] and [−2, −4, 4] of the normal lines of the two surfaces are proportional, the surfaces have the common tangent plane

1( _x_ − 2) + 2( _y_ − 1) − 2( _z_ − 1) = 0 or _x_ \+ 2 _y_ − 2 _z_ = 2

8. Show that the surfaces _F_ ( _x, y, z_ ) = _xy_ \+ _yz_ − 4 _zx_ = 0 and _G_ ( _x, y, z_ ) = 3 _z_ 2 − 5 _x_ \+ _y_ = 0 intersect at right angles at the point (1, 2, 1).

It is to be shown that the tangent planes to the surfaces at the point are perpendicular or, what is the same, that the normal lines at the point are perpendicular. At (1, 2, 1),

A set of direction numbers for the normal line to _F_ ( _x, y, z_ ) = 0 is [ _l_ 1, _m_ 1, _n_ 1] = [1, −1, 1]. At the same point,

A set of direction numbers for the normal line to _G_ ( _x, y, z_ ) = 0 is [ _l_ 2, _m_ 2, _n_ 2] = [−5, 1, 6].

Since _l_ 1 _l_ 2 \+ _m_ 1 _m_ 2 \+ _n_ 1 _n_ 2 = 1(−5) + (−1)1 + 1(6) = 0, these directions are perpendicular.

9. Show that the surfaces _F_ ( _x, y, z_ ) = 3 _x_ 2 \+ 4 _y_ 2 \+ 8 _z_ 2 − 36 = 0 and _G_ ( _x, y, z_ ) = _x_ 2 \+ 2 _y_ 2 − 4 _z_ 2 − 6 = 0 intersect at right angles.

At any point P0( _x_ 0, _y_ 0, _z_ 0)on the two surfaces,  , and  ; hence [3 _x_ 0, 4 _y_ 0, 8 _z_ 0] is a set of direction numbers for the normal to the surface _F_ ( _x, y, z_ ) = 0 at _P_ 0. Similarly, [ _x_ 0, 2 _y_ 0, −4 _z_ 0] is a set of direction numbers for the normal line to _G_ ( _x, y, z_ ) = 0 at _P_ 0. Now, since

these directions are perpendicular.

10. Derive( _51.7_ ) and ( _51.8_ ) for the tangent line and normal plane to the space curve _C: F_ ( _x, y, z_ ) = 0, _G_ ( _x, y, z_ ) = 0 at one of its points _P_ 0( _x_ 0, _y_ 0, _z_ 0).

At _P_ 0 the directions   are normal, respectively, to the tangent planes of the surfaces _F_ ( _x, y, z_ ) = 0 and _G_ ( _x, y, z_ ) = 0. Now the direction

being perpendicular to each of these directions, is that of the tangent line to _C_ at _P_ 0. Hence, the equations of the tangent line are

and the equation of the normal plane is

11. Find the equations of the tangent line and the normal plane to the curve _x_ 2 \+ _y_ 2 + _z_ 2 = 14, _x_ \+ _y_ \+ _z_ = 6 at the point (1, 2, 3).

Set _F_ ( _x, y, z_ ) = _x_ 2 \+ _y_ 2 \+ _z_ 2 − 14 = 0 and _G_ ( _x, y, z_ ) = _x_ \+ _y_ \+ _z_ − 6 = 0. At (1, 2, 3),

With [1, −2, 1] as a set of direction numbers of the tangent, its equations are  . The equation of the normal plane is ( _x_ − 1) − 2( _y_ − 2) + ( _z_ − 3) = _x_ − 2 _y_ \+ _z_ = 0.

12. Find equations of the surfaces of revolution generated by revolving the given curve about the given axis: ( _a_ ) _y_ = _x_ 2 about the _x_ axis; ( _b_ )   about the _y_ axis; ( _c_ ) _z_ = 4 _y_ about the _y_ axis.

In each case, we use an appropriate form of ( _51.9_ ): ( _a_ ) _y_ 2 \+ _z_ 2 = _x_ 4; ( _b_ )  ; ( _c_ ) _x_ 2 \+ _z_ 2 = _16_ y2.

13. Identify the locus of all points ( _x, y, z_ ) that are equidistant from the point (0, −1, 0)and the plane _y_ = 1.

Squaring the distances, we get _x_ 2 \+ ( _y_ \+ 12) + _z_ 2 = ( _y_ − 1)2, whence _x_ 2 \+ _z_ 2 = −4 _y_ , a circular paraboloid.

14. Identify the surface 4 _x_ 2 − _y_ 2 \+ _z_ 2 − 8 _x_ \+ 2 _y_ \+ 2 _z_ \+ 3 = 0 by completing the squares.

We have

This is a hyperboloid of one sheet, centered at (1, 1, −1).

### **Supplementary Problems**

15. Find the equations of the tangent line and the normal plane to the given curve at the given point:

16. Show that the curves ( _a_ ) _x_ = 2 − _t_ , _y_ = −1/ _t_ , _z_ = 2 _t_ 2 and ( _b_ ) _x_ = 1 + _θ_ , _y_ = sin _θ_ − 1, _z_ = 2 cos _θ_ intersect at right angles at _P_ (1, −1, 2). Obtain the equations of the tangent line and normal plane of each curve at _P_.

_Ans_ : ( _a_ )  ; _x_ − _y_ − 4 _z_ \+ 6 = 0; ( _b_ ) _x_ − _y_ = 2, _z_ = 2; _x_ \+ _y_ = 0

17. Show that the tangent lines to the helix _x_ = _a_ cos _t_ , _y_ = _a_ sin _t_ , _z_ = _bt_ meet the _xy_ plane at the same angle.

18. Show that the length of the curve ( _51.1_ ) from the point _t_ = _t_ 0 to the point _t_ = _t_ 1 is given by

Find the length of the helix of Problem 17 from _t_ = 0 to _t_ = _t_ 1.

_Ans_.

19. Find the equations of the tangent line and the normal plane to the given curve at the given point:

( _a_ ) _x_ 2 \+ 2 _y_ 2 \+ 2 _z_ 2 = 5, 3 _x_ − 2 _y_ − _z_ = 0; (1, 1, 1).

( _b_ ) 9 _x_ 2 \+ 4 _y_ 2 − 36 _z_ = 0, 3 _x_ \+ _y_ \+ _z_ − _z_ 2 − 1 = 0; (2, −3, 2).

( _c_ ) 4 _z_ 2 = _xy_ , _x_ 2 \+ _y_ 2 = 8 _z_ ; (2, 2, 1).

_Ans_. ( _a_ )  ; ( _b_ )  ; _x_ \+ _z_ \- 4 = 0; ( _c_ )  ; _x_ \- _y_ = 0

20. Find the equations of the tangent plane and normal line to the given surface at the given point:

21. ( _a_ ) Show that the sum of the intercepts of the plane tangent to the surface _x_ 1/2 \+ _y_ 1/2 \+ _z_ 1/2 = _a_ 1/2 at any of its points is _a_.

( _b_ ) Show that the square root of the sum of the squares of the intercepts of the plane tangent to the surface _x_ 2/3 \+ _y_ 2/3 \+ _z_ 2/3 = _a_ 2/3 at any of its points is _a_.

22. Show that each pair of surfaces is tangent at the given point:

( _a_ ) _x_ 2 \+ _y_ 2 \+ _z_ 2 = 18, _xy_ = 9; (3, 3, 0).

( _b_ ) _x_ 2 \+ _y_ 2 \+ _z_ 2 − 8 _x_ − 8 _y_ − 6 _z_ \+ 24 = 0, _x_ 2 \+ 3 _y_ 2 \+ 2 _z_ 2 = 9; (2, 1, 1).

23. Show that each pair of surfaces is perpendicular at the given point:

( _a_ ) _x_ 2 \+ 2 _y_ 2 − 4 _z_ 2 = 8, 4 _x_ 2 − _y_ 2 \+ 2 _z_ 2 = 14; (2, 2, 1).

( _b_ ) _x_ 2 \+ _y_ 2 \+ _z_ 2 = 50, _x_ 2 \+ _y_ 2 − 10 _z_ \+ 25 = 0; (3, 4, 5).

24. Show that each of the surfaces ( _a_ ) 14 _x_ 2 \+ 11 _y_ 2 \+ 8 _z_ 2 = 66, ( _b_ ) 3 _z_ 2 − 5 _x_ \+ _y_ = 0, and ( _c_ ) _xy_ \+ _yz_ − 4 _zx_ = 0 is perprndicular to the other two at the point (1, 2, 1).

25. Identify the following surfaces.

( _a_ ) 36 _y_ 2 − _x_ 2 \+ 36 _z_ 2 = 9.

( _b_ ) 5 _y_ = − _z_ 2 \+ _x_ 2.

( _c_ ) _x_ 2 \+ 4 _y_ 2 − 4 _z_ 2 − 6 _x_ − 16 _y_ − 16 _z_ \+ 5 = 0.

_Ans_. ( _a_ ) hyperboloid of one sheet (around the _x_ axis); ( _b_ ) hyperbolic paraboloid; ( _c_ ) hyperboloid of one sheet, centered at (3, 2, −2)

26. Find an equation of a curve that, when revolved about a suitable axis, yields the paraboloid _y_ 2 \+ _z_ 2 − 2 _x_ = 0.

_Ans_.  , about the _x_ axis

27. Find an equation of the surface obtained by revolving the given curve about the given axis. Identify the type of surface: ( _a_ ) _x_ = _y_ 2 about the _x_ axis; ( _b_ ) _x_ = 2 _y_ about the _x_ axis.

_Ans_. ( _a_ ) _x_ = _y_ 2 \+ _z_ 2 (circular paraboloid); ( _b_ )   (right circular cone)

## **CHAPTER 52
Directional Derivatives. Maximum and Minimum Values**

**DIRECTIONAL DERIVATIVES**. Let _P(x, y, z)_ be a point on a surface _z_ = _f(x, y)_. Through _P_ , pass planes parallel to the _xz_ and _yz_ planes, cutting the surface in the arcs _PR_ and _PS_ , and cutting the _xy_ plane in the lines _P*M_ and _P*N_ , as shown in Fig. 52-1. The partial derivatives ∂ _z_ /∂ _x_ and ∂ _z_ /∂ _y_ , evaluated at _P_ * _(x, y)_ , give, respectively, the rates of change of _z_ = _P_ * _P_ when _y_ is held fixed and when _x_ is held fixed. In other words, they give the rates of change of _z_ in directions parallel to the _x_

**Fig. 52-1**

and _y_ axes. These rates of change are the slopes of the tangent lines of the curves _PR_ and _PS_ at _P_.

Consider next a plane through _P_ perpendicular to the _xy_ plane and making an angle _θ_ with the _x_ axis. Let it cut the surface in the curve _PQ_ and the _xy_ plane in the line _P_ * _L_. The _directional derivative of f_ ( _x, y_ ) at _P_ * in the direction _θ_ is given by

The direction _θ_ is the direction of the vector (cos _θ_ ) **i** \+ (sin _θ_ ) **j**.

The directional derivative gives the rate of change of _z_ = _P_ * _P_ in the direction of _P_ * _L_ ; it is equal to the slope of the tangent line of the curve _PQ_ at _P_. (See Problem 1.)

The directional derivative at a point _P_ * is a function of _θ_. There is a direction, determined by a vector called the _gradient_ of _f_ at _P_ * (see Chapter 53) for which the directional derivative at _P_ * has a maximum value. That maximum value is the slope of the steepest tangent line that can be drawn to the surface at _P_.

For a function _w_ = _F_ ( _x, y, z_ ), the directional derivative at _P_ ( _x, y, z_ ) in the direction determined by the angles α, _β_ , γ is given by

By the direction determined by α, _β_ , and γ, we mean the direction of the vector (cos α) **i** \+ (cos _β_ ) **j** \+ (cos γ) **k**.

**RELATIVE MAXIMUM AND MINIMUM VALUES.** Assume that _z_ = _f_ ( _x_ , _y_ ) has a relative maximum (or minimum) value at _P_ 0( _x_ 0, _y_ 0, _z_ 0). Any plane through _P_ 0 perpendicular to the _xy_ plane will cut the surface in a curve having a relative maximum (or minimum) point at _P_ 0. Thus, the directional derivative   of _z_ = _f_ ( _x_ , _y_ ) must equal zero at _P_ 0. In particular, when _θ_ = 0, sin _θ_ = 0 and cos _θ_ = 1, so that  . When  , sin _θ_ = 1 and cos _θ_ = 0, so that  . Hence, we obtain the following theorem.

**Theorem 52.1:** If _z_ = _f_ ( _x_ , _y_ ) has a relative extremum at _P_ 0( _x_ 0, _y_ 0, _z_ 0) and   and   exist at ( _x_ 0, _y_ 0), then   and   at ( _x_ 0, _y_ 0).

We shall cite without proof the following sufficient conditions for the existence of a relative maximum or minimum.

**Theorem 52.2:** Let _z_ = _f_ ( _x_ , _y_ ) have first and second partial derivatives in an open set including a point ( _x_ 0, _y_ 0) at which   and  . Define  . Assume Δ < 0 at ( _x_ 0, _y_ 0). Then:

If Δ > 0, there is neither a relative maximum nor a relative minimum at ( _x_ 0, _y_ 0).

If Δ = 0, we have no information.

**ABSOLUTE MAXIMUM AND MINIMUM VALUES**. Let _A_ be a set of points in the _xy_ plane. We say that _A_ is _bounded_ if _A_ is included in some disk. By the _complement_ of _A_ in the _xy_ plane we mean the set of all points in the _xy_ plane that are not in _A_. _A_ is said to be _closed_ if the complement of _A_ is an open set.

**EXAMPLE 1:** The following are instances of closed and bounded sets.

( _a_ ) Any closed disk _D_ , that is, the set of all points whose distance from a fixed point is less than or equal to some fixed positive number _r_. (Note that the complement of _D_ is open because any point not in _D_ can be surrounded by an open disk having no points in _D_.)

( _b_ ) The inside and boundary of any rectangle. More generally, the inside and boundary of any "simple closed curve", that is, a curve that does not interset itself except at its initial and terminal point.

**Theorem 52.3:** Let _f(x, y)_ be a function that is continuous on a closed, bounded set _A_. Then _f_ has an absolute maximum and an absolute minimum value in _A_.

The reader is referred to more advanced texts for a proof of Theorem 52.3. For three or more variables, an analogous result can be derived.

### **Solved Problems**

**1.** Derive formula (52.1).

In Fig. 52-1, let _P_ ** ( _x_ \+ Δ _x_ , _y_ \+ Δ _y_ ) be a second point on _P*L_ and denote by Δ _s_ the distance P*P**. Assuming that _z_ = _f_ ( _x_ , _y_ ) possesses continuous first partial derivatives, we have, by Theorem 49.1,

where **ε** 1 and _ε_ 2 → 0 as Δ _x_ and Δ _y_ → 0. The average rate of change between points _P_ * and _P_ ** is

where _θ_ is the angle that the line _P_ * _P_ ** makes with the _x_ axis. Now let _P_ ** → _P_ * along _P_ * _L_. The directional derivative at _P_ *, that is, the instantaneous rate of change of _z_ , is then

**2.** Find the directional derivative of _z_ = _x_ 2–6 _y_ 2 at _P_ *(7, 2) in the direction: ( _a_ ) _θ_ = 45°; ( _b_ ) _θ_ = 135°.

The directional derivative at any point _P_ *( _x, y_ ) in the direction _θ_ is

( _a_ ) At _P_ *(7, 2) in the direction _θ_ = 45°,

( _b_ ) At _P_ *(7, 2) in the direction _θ_ = 135°,

**3.** Find the directional derivative of _z_ = _ye_ _x_ at _P_ *(0, 3) in the direction ( _a_ ) _θ_ = 30°; ( _b_ ) _θ_ = 120°.

Here, _dz_ / _ds_ = _ye_ _x_ cos _θ_ \+ _e_ _x_ sin _θ_.

( _a_ ) At (0, 3) in the direction  .

( _b_ ) At (0, 3) in the direction  .

**4.** The temperature _T_ of a heated circular plate at any of its points ( _x, y_ ) is given by

the origin being at the center of the plate. At the point (1, 2) find the rate of change of _T_ in the direction _θ_ = π/3.

We have

At (1, 2) in the direction

**5.** The electrical potential _V_ at any point ( _x, y_ ) is given by  . Find the rate of change of V at the point (3, 4) in the direction toward the point (2, 6).

Here,

Since _θ_ is a second-quadrant angle. and tan _θ_ = (6–4)/(2–3) = –2,   and

Hence, at (3, 4) in the indicated direction,

**6.** Find the maximum directional derivative for the surface and point of Problem 2.

At _P_ *(7, 2) in the direction _θ_ , _dz_ / _ds_ = 14cos _θ_ − 24sin _θ_.

To find the value of _θ_ for which   is a maximum,  . Then   and _θ_ is either a second- or fourth-quadrant angle. For the second-quadrant angle,   and  . For the fourth-quadrant angle,   and  .

Since   is negative for the fourth-quadrant angle, the maximum directional derivative is  , and the direction is _θ_ = 300°15′.

**7.** Find the maximum directional derivative for the function and point of Problem 3.

At _P_ *(0, 3) in the direction _θ_ , _dz_ / _ds_ = 3 cos _θ_ \+ sin _θ_.

To find the value of _θ_ for which   is a maximum,  . Then   and θ is either a first- or third-quadrant angle.

Since   is negative for the first–quadrant angle, the maximum directional derivative is  , and the direction is _θ_ = 18°26′.

**8.** In Problem 5, show that _V_ changes most rapidly along the set of radial lines through the origin.

At any point ( _x_ 1, _y_ 1) in the direction _θ_ ,  . Now _V_ changes most rapidly when  , and then  . Thus, _θ_ is the angle of inclination of the line joining the origin and the point ( _x_ 1, _y_ 1).

**9.** Find the directional derivative of _F(x, y, z) = _xy_ \+ 2 _xz_ − _y_ 2 \+ z2_ at the point (1, −2, 1) along the curve _x = t, _y_ = t − 3, z = t2_ in the direction of increasing _z_.

A set of direction numbers of the tangent to the curve at (1, −2, 1) is [1, 1, 2]; the direction cosines are  . The directional derivative is

**10.** Examine _f (x, y) = x 2 \+ y2 − 4 _x_ \+ 6 _y_ \+ 25_ for maximum and minimum values.

The conditions   and   are satisfied when _x_ = 2, _y_ = −3. Since

_f(x, y) = (x 2 − 4x + 4) + (y2 \+ 6y + 9) + 25 − 4 − 9 = (x − 2)2 \+ (y + 3)2 \+ 12_

it is evident that _f_ (2, −3) = 12 is the absolute minimum value of the function. Geometrically, (2, −3, 12) is the lowest point on the surface _z = _x_ 2 \+ _y_ 2 − 4x + 6y + 25._ Clearly, _f(x, y)_ has no absolute maximum value.

**11.** Examine _f (x, y) = x 3 \+ y3 \+ 3xy_ for maximum and minimum values.

We shall use Theorem 52.2. The conditions   and   are satisfied when _x_ = 0, _y_ = 0 and when _x_ = −1, _y_ = −1.

At  , and  . Then

and (0, 0) yields neither a relative maximum nor minimum.

At  . Then

  and

Hence, _f_ (−1, −1) = 1 is a relative maximum value of the function.

Clearly, there are no absolute maximum or minimum values. (When _y_ = 0, _f (x, y) = x 3_ can be made arbitrarily large or small.)

**12.** Divide 120 into three nonnegative parts such that the sum of their products taken two at a time is a maximum.

Let _x, y_ , and 120 − _(x + y)_ be the three parts. The function to be maximized is _S = _xy_ \+ ( _x_ \+ _y_ )(120 − _x_ − _y_ )._ Since 0   _x + y_   120, the domain of the function consists of the solid triangle shown in Fig. 52-2. Theorem 52.3 guarantees an absolute maximum.

**Fig. 52-2**

Now,

and

Setting _∂S/∂x = ∂S/∂y_ = 0 yields _2x + y_ = 120 and _x + 2y_ = 120.

Simultaneous solution gives _x_ = 40, _y_ = 40, and 120 − ( _x_ \+ 4) = 40 as the three parts, and _S_ = 3(402) = 4800. So, if the absolute maximum occurs in the interior of the triangle, Theorem 52.1 tells us we have found it. It is still necessary to check the boundary of the triangle. When _y_ = 0, _S = x(120 − x)_. Then _dS/dx_ = 120 − 2 _x_ , and the critical number is _x_ = 60. The corresponding maximum value of _S_ is 60(60) = 3600, which is < 4800. A similar result holds when _x_ = 0. Finally, on the hypotenuse, where _y_ = 120 − _x_ , _S = x(120 − x)_ and we again obtain a maximum of 3600. Thus, the absolute maximum is 4800, and _x = _y_ = z_ = 40.

**13.** Find the point in the plane 2 _x_ − _y_ \+ 2 _z_ = 16 nearest the origin.

Let _(x, y, z)_ be the required point; then the square of its distance from the origin is _D = x 2 \+ y2 \+ z2_. Since also 2 _x_ − _y_ \+ 2 _z_ = 16, we have _y_ = 2 _x_ \+ 2 _z_ − 16 and _D_ = _x 2 \+ (2x + 2z − 16)2 \+ z2_.

Then the conditions _∂D/∂x_ = 2 _x_ \+ 4(2 _x_ \+ 2 _z_ − 16) = 0 and _∂D/∂z_ = 4(2 _x_ \+ 2 _z_ − 16) + 2 _z_ = 0 are equivalent to 5 _x_ \+ 4 _z_ = 32 and 4 _x_ \+ 5 _z_ = 32, and _x = z_ =  . Since it is known that a point for which _D_ is a minimum exists,   is that point.

**14.** Show that a rectangular parallelepiped of maximum volume _V_ with constant surface area _S_ is a cube.

Let the dimensions be _x, y,_ and _z_. Then _V = xyz_ and _S = 2(xy + yz + zx)._

The second relation may be solved for z and substituted in the first, to express _V_ as a function of _x_ and _y_. We prefer to avoid this step by simply treating z as a function of _x_ and _y_. Then

From the latter two equations,   and  . Substituting in the first two yields the conditions   and  , which reduce to y2( _z_ − _x_ ) = 0 and _x_ 2( _z_ − _y_ ) = 0. Thus _x_ = _y_ = _z_ , as required.

**15.** Find the volume _V_ of the largest rectangular parallelepiped that can be inscribed in the ellipsoid  .

Let _P(x, y, z)_ be the vertex in the first octant. Then _V_ = 8 _xyz_. Consider _z_ to be defined as a function of the independent variables _x_ and _y_ by the equation of the ellipsoid. The necessary conditions for a maximum are

From the equation of the ellipsoid, obtain   and  . Eliminate ∂z/∂x and ∂z/∂y between these relations and ( _1_ ) to obtain

and, finally,

Combine ( _2_ ) with the equation of the ellipsoid to get  ,  , and  . Then  _abc_ cubic units.

### **Supplementary Problems**

**16.** Find the directional derivatives of the given function at the given point in the indicated direction.

( _a_ )  .

( _b_ )  .

( _c_ )  .

( _d_ ) _z_ = 2 _x_ 2 \+ 3 _xy_ − _y_ 2, (1, − 1), toward (2, 1).

_Ans._

**17.** Find the maximum directional derivative for each of the functions of Problem 16 at the given point.

_Ans._

**18.** Show that the maximal directional derivative of   of Problem 8 is constant along any circle _x_ 2 \+ _y_ 2 = _r_ 2.

**19.** On a hill represented by _z_ = 8 − 4 _x_ 2 − 2y2, find ( _a_ ) the direction of the steepest grade at (1, 1, 2) and ( _b_ ) the direction of the contour line (direction for which _z_ = constant) Note that the directions are mutually perpendicular.

_Ans._ ( _a_ )  , third quadrant; ( _b_ ) tan−1 (−2)

**20.** Show that the sum of the squares of the directional derivatives of _z_ = _f_ ( _x, y_ ) at any of its points is constant for any two mutually perpendicular directions and is equal to the square of the maximum directional derivative.

**21.** Given _z_ = _f(x, y)_ and _w_ = g( _x, y_ ) such that ∂z/∂x = ∂w/∂y and ∂z/∂y = – ∂w/∂x. If _θ_ 1 and _θ_ 2 are two mutually perpendicular directions, show that at any point _P(x, y)_ , ∂z/∂s1 = ∂w/∂s2 and ∂z/∂s2 = –∂w/∂s1.

**22.** Find the directional derivative of the given function at the given point in the indicated direction:

( _a_ ) _xy_ 2 _z_ , (2, 1, 3), [1, − 2, 2].

( _b_ ) _x_ 2 \+ y2 \+ z2, (1, 1, 1), towards (2, 3, 4).

( _c_ ) _x_ 2 \+ y2 − 2 _xz_ , (1, 3, 2), along _x_ 2 \+ y2 − 2 _xz_ = 6, 3x2 − y2 \+ 3z = 0 in the direction of increasing z.

_Ans._ ( _a_ ) ; ( _b_ )  ; ( _c_ ) 0

**23.** Examine each of the following functions for relative maximum and minimum values.

**24.** Find positive numbers _x, y, z_ such that

_Ans._ ( _a_ ) _x_ = _y_ = _z_ = 6; ( _b_ ) x = _y_ = z = 3; ( _c_ ) _x_ = _y_ = 5, z = 10; ( _d_ ) _x_ = 2, _y_ = 4, z = 6

**25.** Find the minimum value of the square of the distance from the origin to the plane _Ax + By + Cz + D_ = 0.

_Ans_. _D_ 2/( _A_ 2 \+ _B_ 2 \+ _C_ 2)

**26.** ( _a_ ) The surface area of a rectangular box without a top is to be 108 ft2. Find the greatest possible volume.

( _b_ ) The volume of a rectangular box without a top is to be 500 ft3. Find the minimum surface area.

_Ans._ ( _a_ ) 108 ft3; ( _b_ ) 300ft 2

**27.** Find the point on _z = xy_ − 1 nearest the origin.

_Ans_. (0, 0, − 1)

**28.** Find the equation of the plane through (1, 1, 2) that cuts off the least volume in the first octant.

_Ans_. 2 _x_ \+ 2 _y_ \+ _z_ = 6

**29.** Determine the values of _p_ and _q_ so that the sum _S_ of the squares of the vertical distances of the points (0, 2), (1, 3) and (2, 5) from the line _y_ = _px_ \+ _q_ is a minimum. ( _Hint: S_ = ( _q_ − 2)2 \+ ( _p_ \+ _q_ − 3)2 \+ (2 _p_ \+ _q_ − 5)2.)

_Ans_.

## **CHAPTER 53
Vector Differentiation and Integration**

**VECTOR DIFFERENTIATION.** Let

be vectors whose components are functions of a single scalar variable _t_ having continuous first and second derivatives.

We can show, as in Chapter 39 for plane vectors, that

Also, from the properties of determinants whose entries are functions of a single variable, we have

and

These formulas may also be established by expanding the products before differentiating.

From ( _53.2_ ) follows

**SPACE CURVES.** Consider the space curve

where _f(t)_ , _g(t)_ , and _h(t)_ have continuous first and second derivatives. Let the position vector of a general variable point _P(x, y, z)_ of the curve be given by

**r** = _x_ **i** \+ _y_ **j** \+ _z_ **k**

As in Chapter 39, _t_ = _d_ **r** / _ds_ is the unit tangent vector to the curve. If **R** is the position vector of a point _(X, Y, Z)_ on the tangent line at _P_ , the vector equation of this line is (see Chapter 50)

and the equations in rectangular coordinates are

where   is a set of direction cosines of the line. In the corresponding equation ( _51.2_ ), a set of direction numbers   was used.

The vector equation of the normal plane to the curve at _P_ is given by

where **R** is the position vector of a general point of the plane.

Again, as in Chapter 39, _d_ **t** / _ds_ is a vector perpendicular to **t**. If **n** is a unit vector having the direction of _d_ **t** / _ds_ , then

where | _K_ | is the magnitude of the curvature at _P_. The unit vector

is called the _principal normal_ to the curve at _P._

The unit vector **b** at _P_ , defined by

is called the _binormal_ at _P._ The three vectors **t** , **n** , **b** form at _P_ a right-handed triad of mutually orthogonal vectors. (See Problems 1 and 2.)

At a general point _P_ of a space curve (Fig. 53-1), the vectors **t** , **n** , **b** , determine three mutually perpendicular planes:

1. The _osculating plane_ , containing **t** and **n** , of equation **(R − r) · b** = 0

2. The _normal plane_ , containing **n** and **b,** of equation **(R − r) · t** = 0

3. The _rectifying plane_ , containing **t** and **b** , of equation **(R − r) · n** = 0

**Fig. 53-1**

In each equation, **R** is the position vector of a general point in the particular plane.

**SURFACES.** Let _F(x, y, z)_ = 0 be the equation of a surface. (See Chapter 51.) A parametric representation results when _x, y_ , and _z_ are written as functions of two independent variables _u_ and _v_ , for example, as

When _u_ is replaced with _u_ 0, a constant, (53.10) becomes

the equation of a space curve ( _u_ curve) lying on the surface. Similarly, when _v_ is replaced with _v_ 0, a constant ( _53.10_ ) becomes

the equation of another space curve ( _v_ curve) on the surface. The two curves intersect in a point of the surface obtained by setting _u_ = _u_ 0 and _v_ = _v_ 0 simultaneously in ( _53.10_ ).

The position vector of a general point _P_ on the surface is given by

Suppose (53.11) and (53.12) are the _u_ and _v_ curves through _P_. Then, at _P_ ,

is a vector tangent to the _u_ curve, and

is a vector tangent to the _v_ curve. The two tangents determine a plane that is the tangent plane to the surface at _P_ (Fig. 53-2). Clearly, a normal to this plane is given by  . The _unit normal_ to the surface at _P_ is defined by

**Fig. 53-2**

If **R** is the position vector of a general point on the normal to the surface at _P_ , its vector equation is

If **R** is the position vector of a general point on the tangent plane to the surface at _P_ , its vector equation is

(See Problem 3.)

**THE OPERATION ∇.** In Chapter 52 the directional derivative of _z_ = _f(x, y)_ at an arbitrary point ( _x, y_ ) and in a direction making an angle _θ_ with the positive _x_ axis is given as

Let us write

Now **a** = **i** cos _θ_ \+ **j** sin _θ_ is a unit vector whose direction makes the angle _θ_ with the positive _x_ axis.

The other factor on the right of ( _53.17_ ), when written as  , suggests the definition of a vector differential operator ∇ (del), defined by

In vector analysis,   is called the _gradient_ of _f_ or _grad f_. From ( _53.17_ ), we see that the component of ∇ _f_ in the direction of a _unit vector_ **a** is the directional derivative of _f_ in the direction of **a**.

Let **r** = _x_ **i** \+ _y_ **j** be the position vector to _P_ ( _x_ , _y_ ). Since

and

where ø is the angle between the vectors ∇ _f_ and _d_ **r** / _ds_ , we see that _df_ / _ds_ is maximal when cos ø = 1, that is, when ∇ _f_ and _d_ **r** = _ds_ have the same direction. Thus, the maximum value of the directional derivative at _P_ is  ∇ _f_  ; and its direction is that of ∇ _f_. (Compare the discussion of maximum directional derivatives in Chapter 52.) (See Problem 4.)

For _w_ = _F_ ( _x_ , _y_ , _z_ ), we define

and the directional derivative of _F_ ( _x_ , _y_ , _z_ ) at an arbitrary point _P_ ( _x_ , _y_ , _z_ ) in the direction a = _a_ 1 **i** \+ _a_ 2 **j** \+ _a_ 2 **k** is

As in the case of functions of two variables,  ∇ _F_  is the maximum value of the directional derivative of _F_ ( _x_ , _y_ , _z_ ) at _P_ ( _x_ , _y_ , _z_ ), and its direction is that of ∇ _F_. (See Problem 5.)

Consider now the surface _F_ ( _x_ , _y_ , _z_ ) = 0. The equation of the tangent plane to the surface at one of its points _P_ 0( _x_ 0, _y_ 0, _z_ 0) is given by

with the understanding that the partial derivatives are evaluated at _P_ 0. The first factor is an arbitrary vector through _P_ 0 in the tangent plane; hence the second factor ∇ _F_ , evaluated at _P_ 0, is normal to the tangent plane, that is, is normal to the surface at _P_ 0. (See Problems 6 and 7.)

**DIVERGENCE AND CURL.** The _divergence_ of a vector function **F** = **i** _f_ 1( _x_ , _y_ , _z_ ) + **j** _f_ 2( _x_ , _y_ , _z_ ) + **k** _f_ 3( _x_ , _y_ , _z_ ), sometimes called _del dot_ **F** , is defined by

The _curl_ of the vector function **F** , or _del cross_ **F** , is defined by

(See Problem 8.)

**INTEGRATION.** Our discussion of integration here will be limited to ordinary integration of vectors and to so-called "line integrals." As an example of the former, let

**F** ( _u_ ) = **i** cos _u_ \+ **j** sin _u_ \+ _au_ **k**

be a vector depending upon the scalar variable _u_. Then

**F** ′( _u_ ) = − **i** sin _u_ \+ **j** cos _u_ \+ _a_ **k**

and

where **c** is an arbitrary constant vector independent of _u_. Moreover,

(See Problems 9 and 10.)

**LINE INTEGRALS.** Consider two points P0 and P1 in space, joined by an arc _C_. The are may be a segment of a straight line or a portion of a space curve _x_ = _g_ 1( _t_ ), _y_ = _g_ 2( _t_ ), _z_ = _g_ 3( _t_ ), or it may consist of several subarcs of curves. In any case, _C_ is assumed to be continuous at each of its points and not to intersect itself. Consider further a vector function

**F** = **F** ( _x, y, z_ ) = **i** _f_ 1( _x, y z_ ) + **j** _f_ 2( _x, y, z_ ) + **k** _f_ 3( _x, y, z_ )

which at every point in a region about _C_ , and, in particular, at every point of _C_ , defines a vector of known magnitude and direction. Denote by

the position vector of _P(x, y, z)_ on _C_. The integral

is called a line integral, that is, an integral along a given path _C_.

As an example, let **F** denote a force. The work done by it in moving a particle over _d_ **r** is given by (see Problem 16 of Chapter 39)

| **F** || _d_ **r** |cos _θ_ = **F** · _d_ **r**

and the work done in moving the particle from _P_ 0 to _P_ 1 along the arc _C_ is given by

From ( _53.23_ ),

_d_ **r** = **i** _dx_ \+ **j** _dy_ \+ **k** _dz_

and ( _53.24_ ) becomes

(See Problem 11.)

### **Solved Problems**

**1.** A particle moves along the curve _x_ = 4 cos _t_ , _y_ = 4 sin _t_ , _z_ = 6 _t_. Find the magnitude of its velocity and acceleration at times _t_ = 0 and  .

Let _P_ ( _x, y, z_ ) be a point on the curve, and

**r** = _x_ **i** \+ _y_ **j** \+ _z_ **k** = 4 **i** cos _t_ \+ 4 **j** sin _t_ \+ 6 **k** _t_

be its position vector. Then

At _t_ = 0:

At _t_ =  .

**2.** At the point (1, 1, 1) or _t_ = 1 of the space curve _x_ = _t_ , y = _t_ 2, _z_ = _t_ 3, find:

( _a_ ) The equations of the tangent line and normal plane.

( _b_ ) The unit tangent, principal normal, and binormal.

( _c_ ) The equations of the principal normal and binormal.

We have

At _t_ = 1, **r** = **i** \+ **j** \+ **k** and  .

( _a_ ) If **R** is the position vector of a general point ( _X, Y, Z_ ) on the tangent line, its vector equation is **R** − **r** = _k_ **t** or

and its rectangular equations are

If **R** is the position vector of a general point ( _X, Y, Z_ ) on the normal plane, its vector equation is ( **R** − **r** ). **t** = 0 or

and its rectangular equation is

( _X_ − 1) + 2( _Y_ − 1) + 3( _Z_ − 1) = _X_ \+ 2 _Y_ \+ 3 _Z_ − 6 = 0

(see Problem 2( _a_ ) of Chapter 51.)

( _b_ )

At _t_ = 1,

  and

Then

and

( _c_ ) If **R** is the position vector of a general point ( _X, Y, Z_ ) on the principal normal, its vector equation is **R** − **r** = _k_ **n** or

and the equation in rectangular coordinates are

If **R** is the position vector of a general point ( _X, Y, Z_ ) on the binormal, its vector equation is **R** − **r** = _k_ **b** or

and the equations in rectangular coordinates are

**3.** Find the equations of the tangent plane and normal line to the surface _x_ = 2( _u_ \+ _v_ ), _y_ = 3( _u_ − _v_ ), _z_ = _uv_ at the point _P_ ( _u_ = 2, _v_ = 1).

Here

and at the point _P_ ,

and

The vector and rectangular equations of the normal line are

or

( _X_ −6) **i** \+ ( _Y_ −3) **j** \+ ( _Z_ −2) **k** = k(9 **i** − 2 **j** − 12 **k** )

and

The vector and rectangular equations of the tangent plane are

or

and

**4.** ( _a_ )Find the directional derivative of _f_ ( _x, y_ ) = _x_ 2 − 6 _y_ 2 at the point (7, 2)in the direction  .

( _b_ )Find the maximum value of the directional derivative at (7, 2).

( _a_ )

and

At (7, 2), ∇ _f_ = 14 **i** − 24 **j** , and

is the directional derivative.

( _b_ ) At (7, 2), with ∇ _f_ = 14 **i** − 24 **j** ,   is the maximum directional derivative. Since

the direction is _θ_ = 300°15′.(See Problems 2 and 6 of Chapter 52.)

**5.** ( _a_ )Find the directional derivative of _f_ ( _x, y, z_ ) = _x_ 2 − 2 _y_ 2 \+ 4 _z_ 2 at _P_ (1, 1−1)in the direction **a** = 2 **i** \+ **j** − **k**

( _b_ ) Find the maximum value of the directional derivative at _P_.

Here

and at (1, 1, −1), ∇ _F_ = 2 **i** − 4 **j** − 8 **k**.

( _a_ )∇ _F_ · a = (2 **i** − 4 **j** − 8 **k** ) · (2 **i** \+ **j** − **k** ) = 8

( _b_ ) At _P_ ,  . The direction is a = 2 **i** − 4 **j** − 8 **k**.

**6.** Given the surface F( _x, y, z_ ) = _x_ 3 \+ 3 _xyz_ \+ 2 _y_ 3 − _z_ 3 − 5 = 0 and one of its points _P_ 0(1, 1, 1), find ( _a_ ) a unit normal to the surface at _P_ 0; ( _b_ ) the equations of the normal line at _P_ 0; and ( _c_ )the equation of the tangent plane at _P_ 0.

Here ∇ _F_ = (3 _x_ 2+3 _yz_ ) **i** \+ (3 _xz_ \+ 6 _y_ 2) **j** \+ (3 _xy_ − 3 _z_ 2) **k**

and at _P_ 0(1, 1, 1), ∇ _F_ = 6 **i** \+ 9 **j**.

( _a_ )   is a unit normal at _P_ 0; the other is  .

( _b_ )The equations of the normal line are  .

( _c_ )The equation of the tangent plane is _2(X − 1)+3(Y − 1) = 2X + 3Y − 5 = 0._

**7.** Find the angle of intersection of the surfaces

_F_ 1 = _x_ 2 \+ _y_ 2 \+ _z_ 2 − 9 = 0 and _F_ 2 = _x_ 2 \+ 2 _y_ 2 − _z_ − 8 = 0

at the point (2, 1, −2).

We have

∇ _F_ 1 = ∇( _x_ 2 \+ _y_ 2 _z_ 2 − 9) = 2 _x_ **i** \+ 2 _y_ **j** \+ 2 _z_ **k**

and

∇ _F_ 2 = ∇( _x_ 2 \+ 2 _y_ 2 − _z_ −8) = 2 _x_ **i** \+ 4 _y_ **j** − **k**

At (2, 1, −2), ∇ _F_ 1 = 4 **i** \+ 2 **j** − 4 **k** and ∇ _F_ 2 = 4 **i** \+ 4 **j** − **k.**

Now ∇ _F_ 1 · ∇2 = |∇ _F_ 1||∇ _F_ 2|cos _θ_ , where _θ_ is the required angle. Thus,

(4 **i** \+ 2 **j** − 4 **k** ) · (4 **i** \+ 4 **j** − **k** ) = |4 **i** \+ 2 **j** − 4 **k** ||4 **i** \+ 4 **j** − **k** | cos _θ_

from which  , _θ_ = 35.°40′.

**8.** When **B** = _xy_ 2 **i** \+ 2 _x_ 2 _yz_ **j** − 3 _yz_ 2 **k** , find ( _a_ ) div **B** and ( _b_ ) curl **B**.

( _a_ )

( _b_ )

**9.** Given **F** ( _u_ ) = _u_ **i** \+ ( _u_ 2 − 2 _u_ ) **j** \+ (3 _u_ 2 \+ _u_ 3) **k** , find ( _a_ )   and ( _b_ )   **F** ( _u_ ) _du_.

( _a_ )

Where **c** = _c_ 1 **i** \+ _c_ 2 **j** \+ _c_ 3 **k** with _c_ 1, _c_ 2, _c_ 3 arbitrary scalars.

( _b_ )

**10.** The acceleration of a particle at any time _t_   0 is given by **a** = _d_ **v** / _dt_ = _e t_ **i** \+ _e_ 2 _t_ **j** \+ **k**. If at _t_ = 0, the displacement is **r** = 0 and the velocity is **v** = **i** \+ **j** , find **r** and **v** at any time _t._

Here

At _t_ = 0, we have **v** = **i** \+  **j** \+ **c** 1 = **i** \+ **j** , from which **c** 1 =  **j**. Then

**v** = _e t_ **i** \+  ( _e_ 2 _t_ \+ 1) **j** \+ _t_ **k**

and

At _t_ = 0,  , from which  . Thus,

**11.** Find the work done by a force **F** = ( _x_ \+ _yz_ ) **i** \+ ( _y_ \+ _xz_ ) **j** \+ ( _z_ \+ _xy_ ) **k** in moving a particle from the origin _O_ to _C_ (1, 1, 1), ( _a_ ) along the straight line _OC_ ; ( _b_ ) along the curve _x_ = _t_ , _y_ = _t_ 2, _z_ = _t_ 3; and ( _c_ ) along the straight lines from _O_ to _A_ (1, 0, 0), _A_ to _B_ (1, 1, 0), and _B_ to _C_.

( _a_ ) Along the line _OC_ , _x_ = _y_ = _z_ and _dx_ = _dy_ = _dz_. The integral to be evaluated becomes

( _b_ ) Along the given curve, _x_ = _t_ and _dx_ = _dt_ ; _y_ = _t_ 2 and _dy_ = 2 _t dt_ ; _z_ = _t_ 3 and _dz_ = 3 _t_ 2 _dt_. At _O_ , _t_ = 0; at _C_ , _t_ = 1. Then

( _c_ ) From _O_ to _A_ : _y_ = _z_ = 0 and _dy_ = _dz_ = 0, and _x_ varies from 0 to 1.

From _A_ to _B_ : _x_ = 1, _z_ = 0, _dx_ = _dz_ = 0, and _y_ varies from 0 to 1.

From _B_ to _C_ : _x_ = _y_ = 1 and _dx_ = _dy_ = 0, and _z_ varies from 0 to 1.

Now, for the distance from _O_ to _A_ ,  ; for the distance from _A_ to _B_ ,  ; and for the distance from _B_ to _C_ ,   Thus,  .

In general, the value of a line integral depends upon the path of integration. Here is an example of one which does not, that is, one which is independent of the path. It can be shown that a line integral   is independent of the path if there exists a function  ( _x_ , _y_ , _z_ ) such that _d_  = _f_ 1 _dx_ \+ _f_ 2 _dy_ \+ _f_ 3 _dz_. In this problem the integrand is

( _x_ \+ _yz_ ) _dx_ \+ ( _y_ \+ _xz_ ) _dy_ \+ ( _z_ \+ _xy_ ) _dz_ = _d_ [ ( _x_ 2 \+ _y_ 2 \+ _z_ 2) + _xyz_ ]

### **Supplementary Problems**

**12.** Find _d_ **s** / _dt_ and _d_ 2 **s** / _dt_ 2, given ( _a_ ) **s** = ( _t_ \+ 1) **i** \+ ( _t_ 2 \+ _t_ \+ 1) **j** \+ ( _t_ 3 \+ _t_ 2 \+ _t_ \+ 1) **k** and ( _b_ ) **s** = **i** _e t_ cos 2 _t_ \+ **j** _e t_ sin 2 _t_ \+ _t_ 2 **k**.

_Ans._ ( _a_ ) **i** \+ (2 _t_ \+ 1) **j** \+ (3 _t_ 2 \+ 2 _t_ \+ 1) **k** , 2 **j** \+ (6 _t_ \+ 2) **k** ;

( _b_ ) _e t_ (cos 2 _t_ − 2 sin 2 _t_ ) **i** \+ _e t_(sin 2 _t_ \+ 2 cos 2 _t_ ) **j** \+ 2 _t_ **k** , _e t_(−4 sin 2 _t_ − 3 cos 2 _t_ ) **i** \+ _e t_(−3 sin 2 _t_ \+ 4 cos 2 _t_ ) **j** \+ 2 **k**

**13.** Given **a** = _u_ **i** \+ _u_ 2 **j** \+ _u_ 3 **k** , **b** = **i** cos _u_ \+ **j** sin _u_ , and **c** = 3 _u_ 2 **i** − 4 _u_ **k**. First compute **a** · **b** , **a** × **b** , **a** · ( **b** × **c** ), and **a** × ( **b** × **c** ), and find the derivative of each. Then find the derivatives using the formulas.

**14.** A particle moves along the curve _x_ = 3 _t_ 2, _y_ = _t_ 2 − 2 _t_ , _z_ = _t_ 3, where _t_ is time. Find ( _a_ ) the magnitudes of its velocity and acceleration at time _t_ = 1; ( _b_ ) the components of velocity and acceleration at time _t_ = 1 in the direction **a** = 4 **i** − 2 **j** \+ 4 **k**.

_Ans._

**15.** Using vector methods, find the equations of the tangent line and normal plane to the curves of Problem 15 of Chapter 51.

**16.** Solve Problem 16 of Chapter 51 using vector methods.

**17.** Show that the surfaces _x_ = _u_ , _y_ = 5 _u_ − 3 _v_ 2, _z_ = _v_ and   are perpendicular at _P_ (1, 2, 1).

**18.** Using vector methods, find the equations of the tangent plane and normal line to the surface:

( _a_ ) _x = u_ , _y = v_ , _z = uv_ at the point ( _u, v_ ) = (3, −4);

( _b_ ) _x = u_ , _y = v_ , _z = u 2 − v2_ at the point ( _u, v_ ) = (2, 1).

**19.** ( _a_ ) Find the equations of the osculating and rectifying planes to the curve of Problem 2 at the given point.

( _b_ ) Find the equations of the osculating, normal, and rectifying planes to _x_ = 2 _t_ − _t_ 2, _y_ = _t_ 2, _z_ = 2 _t_ \+ _t_ 2 at _t_ = 1.

_Ans_. ( _a_ ) 3 _X_ − 3 _Y_ \+ _Z_ − 1 = 0, 11 _X_ \+ 8 _Y_ − 9 _Z_ − 10 = 0;

(b) _X_ \+ 2 _Y_ − _Z_ = 0, _Y_ \+ 2 _Z_ − 7 = 0, 5 _X_ − 2 _Y_ \+ _Z_ − 6 = 0

**20.** Show that the equation of the osculating plane to a space curve at _P_ is given by

**21.** Solve Problems 16 and 17 of Chapter 52, using vector methods.

**22.** Find  , given

( _a_ ) **F** ( _u_ ) = _u_ 3 **i** \+ (3 _u_ 2 − 2 _u_ ) **j** \+ 3 **k** ; _a_ = 0, _b_ = 2; ( _b_ ) **F** ( _u_ ) = _e u_ **i** \+ _e_ − 2 _u_ **j** \+ _u_ **k** ; _a_ = 0, _b_ = 1

_Ans_.

**23.** The acceleration of a particle at any time _t_ is given by **a** = _dv/dt_ = ( _t_ \+ 1) **i** \+ _t_ 2 **j** \+ ( _t_ 2 − 2) **k**. If at _t_ = 0, the displacement is **r** = 0 and the velocity is **v** = **i − k** , find **v** and **r** at any time _t_.

_Ans_.

**24.** In each of the following, find the work done by the given force **F** in moving a particle from _O_ (0, 0, 0) to _C_ (1, 1, 1) along (1) the straight line _x = y = z_ , (2) the curve _x = t_ , _y_ = _t_ 2, _z_ = _t_ 3, and (3) the straight lines from _O_ to _A_ (1, 0, 0), _A_ to _B_ (1, 1, 0), and _B_ to _C_.

( _a_ ) **F** = _x_ **i** \+ 2 _y_ **j** \+ 3 _x_ **k**.

( _b_ ) **F** = ( _y + z_ ) **i** \+ ( _x + z_ ) **j** \+ ( _x + y_ ) **k**.

( _c_ ) **F** = ( _x + xyz_ ) **i** \+ ( _y + x 2z_) **j** \+ ( _z + x 2y_) **k**.

_Ans_. ( _a_ ) 3; ( _b_ ) 3; ( _c_ )

**25.** If **r** = _x_ **i** \+ _y_ **j** \+ _z_ **k** , show that ( _a_ ) div **r** = 3 and ( _b_ ) curl **r** = 0.

**26.** If _f_ = _f(x, y, z)_ has partial derivatives of order at least two, show that ( _a_ )   and ( _c_ )

## **CHAPTER 54
Double and Iterated Integrals**

**THE DOUBLE INTEGRAL**. Consider a function _z_ = _f_ ( _x, y_ ) that is continuous on a bounded region _R_ of the _xy_ plane. Define a partition   of R by drawing a grid of horizontal and vertical lines. This divides the region into _n_ subregions _R_ 1, _R_ 2,..., _R_ _n_ of areas Δ1 _A_ ,Δ2 _A_ ,...,Δ _n_ _A_ , respectively. (See Fig. 54-1.) In each subregion, _R_ _k_ , choose a point _P k_( _x k, yk_) and form the sum

Define the diameter of a subregion to be the greatest distance between any two points within or on its boundary, and denote by _d_  the maximum diameter of the subregions. Suppose that we select partitions so that _d_  → 0 and _n_ → +∞. (In other words, we choose more and more subregions and we make their diameters smaller and smaller.) Then the _double integral_ of _f_ ( _x, y_ ) over _R_ is defined as

**Fig. 54-1**

This is not a genuine limit statement. What ( _54.2_ ) really says is that   is a number such that, for any ** ** > 0, there exists a positive integer _n_ 0 such that, for any _n_   _n_ 0 and any partition with _d_  < 1/ _n_ 0, and any corresponding approximating sum  , we have

When _z_ = _f_ ( _x, y_ ) is nonnegative on the region _R_ , as in Fig. 54-2, the double integral ( _54.2_ ), may be interpreted as a volume. Any term _f_ ( _x_ k, _y_ k) Δ _k A_ of ( _54.1_ ) gives the volume of a vertical column whose base is of area Δ _k A_ and whose altitude is the distance _z k_ = _f_ ( _x k, yk_) measured along the vertical from the selected point _P k_( _x k_, _y k_) to the surface _z_ = _f_ ( _x, y_ ). This, in turn, may be taken as an approximation of the volume of the vertical column whose lower base is the subregion _R k_ and whose upper base is the projection of _R k_ on the surface. Thus, ( _54.1_ ) is an approximation of the volume "under the surface" (that is, the volume with lower base _R_ and upper base the surface cut off by moving a line parallel to the _z_ axis along the boundary of _R_ ). It is intuitively clear that ( _54.2_ ) is the measure of this volume.

The evaluation of even the simplest double integral by direct summation is usually very difficult.

**Fig. 54-2**

**THE ITERATED INTEGRAL.** Consider a volume defined as above, and assume that the boundary of _R_ is such that no line parallel to the _x_ axis or to the _y_ axis cuts it in more than two points. Draw the tangent lines _x_ = _a_ and _x_ = _b_ to the boundary with points of tangency _K_ and _L,_ and the tangent lines _y_ = _c_ and _y_ = _d_ with points of tangency _M_ and _N._ (See Fig. 54-3.) Let the equation of the plane arc _LMK_ be _y_ = _g 1_( _x_ ), and that of the plane arc _LNK_ be _y_ = _g 2_( _x_ ).

Divide the interval _a_   _x_   _b_ into _m_ subintervals _h_ 1, _h_ 2,..., _h_ m of respective lengths Δ1 _x_ ,Δ2 _x_ ,...Δ _m x_ by the insertion of points ξ1,ξ2,...,ξ _m_ −1 so that _a_ = ξ0<ξ1<ξ2<...<ξ _m_ − 1<ξm = _b_. Similarly, divide the interval _c_   _y_   _d_ into _n_ subintervals _k_ 1, _k_ 2,..., _k_ n of respective lengths Δ1 _y_ , Δ2 _y_ ,...,Δn _y_ by the insertion points _η_ 1, _η_ 2,..., _η_ _n_ − 1 so that _c_ = _η_ 0 < _η_ 1 < _η_ 2 < ... < _η_ _n_ − 1 < _η n_ = _d_. Let λ _m_ be the greatest Δ _i x_ and let _μ n_ be the greatest Δ _j y_. Draw the parallel lines _x_ = ξ1, _x_ = ξ2,..., _x_ = ξ _m_ − 1 and the parallel lines _y_ = _η_ 1, _y_ = _η 2_,..., _y_ = _η_ _n_ − 1, thus dividing the region _R_ into a set of rectangles _R ij_ of areas Δ _i x_Δ _i y_, plus a set of nonrectangles along the boundary (whose areas will be small enough to be safely ignored). In

**Fig. 54-3**

each subinterval _h i_ select a point _x = x i_ and, in each subinterval _k j_ select a point _y_ = _y j_, thereby determining in each subregion _R ij_ a point _P ij_( _x i, yj_). With each subregion _R ij_ associate, by means of the equation of the surface, a number _z ij_ = _f_ ( _x i, yj_), and form the sum

Now, ( _54.3_ ) is merely a special case of ( _54.1_ ). So, if the number of rectangles is indefinitely increased in such a manner that both _λ m →_ 0 and _μ n_ → 0, the limit of ( _54.3_ ) should be equal to the double integral ( _54.2_ ).

In effecting this limit, let us first choose one of the subintervals, say _h i_, and form the sum

of the contributions of all rectangles having _h i_ as one dimension, that is, the contributions of all rectangles lying on the _i_ th column. When _n_ → +∞, _μ n_ → 0,

Now summing over the _m_ columns and letting m → +∞, we have

Although we shall not use the brackets hereafter, it must be clearly understood that ( _54.4_ ) calls for the evaluation of two simple definite integrals in a prescribed order: first, the integral of _f_ ( _x, y_ ) with respect to _y_ (considering _x_ as a constant) from _y_ = _g 1_( _x_ ), the lower boundary of _R_ , to _y_ = _g 2_( _x_ ), the upper boundary of _R_ , and then the integral of this result with respect to _x_ from the abscissa _x_ = _a_ of the leftmost point of R to the abscissa _x_ = _b_ of the rightmost point of _R_. The integral ( _54.4_ ) is called an _iterated_ or _repeated integral_.

It will be left as an exercise to sum first for the contributions of the rectangles lying in each row and then over all the rows to obtain the equivalent iterated integral

where _x_ = _h 1_( _y_ ) and _x_ = _h 2_( _y_ ) are the equations of the plane arcs _MKN_ and _MLN_ , respectively.

In Problem 1 it is shown by a different procedure that the iterated integral ( _54.4_ ) measures the volume under discussion. For the evaluation of iterated integrals see Problems 2 to 6.

The principal difficulty in setting up the iterated integrals of the next several chapters will be that of inserting the limits of integration to cover the region _R_. The discussion here assumed the simplest of regions; more complex regions are considered in Problems 7 to 9.

### **Solved Problems**

**1.** Let _z_ = _f_ ( _x, y_ ) be nonnegative and continuous over the region _R_ of the _xy_ plane whose boundary consists of the arcs of two curves _y_ = _g_ 1( _x_ ) and _y_ = _g_ 2( _x_ ) intersecting at the points _K_ and _L_ , as in Fig. 54-4. Find a formula for the volume _V_ under the surface _z_ = _f_ ( _x, y_ ).

**Fig. 54-4**

Let the section of this volume cut by a plane _x_ = _x i_, where _a_ < _x i_ < _b_ , meet the boundary of _R_ at the points _S_ ( _x i_, _g 1_( _x i_)) and _T_ ( _x i_, _g 2_( _x i_)), and let it meet the surface _z_ = _f_ ( _x, y_ ) in the arc _UV_ along which _z_ = _f_ ( _x i, y_). The area of this section _STUV_ is given by

Thus, the areas of cross sections of the volume cut by planes parallel to the _yz_ plane are known functions   _dy_ of _x_ , where _x_ is the distance of the sectioning plane from the origin. By the cross-section formula of Chapter 30, the required volume is given by

This is the iterated integral of ( _54.4_ ).

In Problems 2–6, evaluate the integral on the left.

**2.**

**3.**

**4.**

**5.**

**6.**

**7.** Evaluate  , where _R_ is the region in the first quadrant bounded by the semicubical parabola _y_ 2 = _x_ 3 and the line _y_ = _x_.

The line and parabola intersect in the points (0, 0) and (1,1) which establish the extreme values of _x_ and _y_ on the region _R_.

_Solution 1_ (Fig. 54-5): Integrating first over a horizontal strip, that is, with respect to _x_ from _x_ = _y_ (the line) to _x_ = _y_ 2/3 (the parabola), and then with respect to _y_ from _y_ = 0 to _y_ = 1, we get

**Fig. 54-5**

_Solution 2_ (Fig. 54-6): Integrating first over a vertical strip, that is, with respect to _y_ from _y_ = _x_ 3/2 (the parabola) to _y_ = _x_ (the line), and then with respect to _x_ from _x_ = 0 to _x_ = 1, we obtain

**Fig. 54-6**

**8.** Evaluate   where _R_ is the region between _y_ = 2 _x_ and _y_ = _x_ 2 lying to the left of _x_ = 1.

Integrating first over the vertical strip (see Fig. 54-7), we have

**Fig. 54-7**

When horizontal strips are used (see Fig. 54-8), two iterated integrals are necessary. Let R1 denote the part of R lying below _AB_ , and _R_ 2 the part above _A B_. Then

**Fig. 54-8**

**9.** Evaluate   where _R_ is the region in the first quadrant bounded by the hyperbola _xy_ = 16 and the lines _y = x_ , _y = 0_ , and _x = 8_. (See Fig. 54-9)

**Fig. 54-9**

It is evident from Fig. 54-9 that _R_ must be separated into two regions, and an iterated integral evaluated for each. Let _R_ 1 denote the part of _R_ lying above the line _y_ = 2, and _R_ 2 the part below that line. Then

As an exercise, you might separate _R_ with the line _x_ = 4 and obtain

**10.** Evaluate   by first reversing the order of integration.

The given integral cannot be evaluated directly, since   is not an elementary function. The region _R_ of integration (see Fig. 54-10) is bounded by the lines _x_ = 3 _y_ , _x_ = 3, and _y_ = 0. To reverse the order of integration, first integrate with respect to _y_ from _y_ = 0 to _y_ = _x_ /3, and then with respect to _x_ from _x_ = 0 to _x_ = 3. Thus,

**Fig. 54-10**

### **Supplementary Problems**

**11.** Evaluate the iterated integral at the left:

**12.** Using an iterated integral, evaluate each of the following double integrals. When feasible, evaluate the iterated integrals in both orders.

**13.** In Problems 11( _a_ ) to ( _h_ ), reverse the order of integration and evaluate the resulting iterated integral.

## **CHAPTER 55
Centroids and Moments of Inertia of Plane Areas**

**PLANE AREA BY DOUBLE INTEGRATION.** If _f_ ( _x, y_ ) = 1, the double integral of Chapter 54 becomes  . In cubic units, this measures the volume of a cylinder of unit height; in square units, it measures the area _A_ of the region _R_.

In polar coordinates

where _θ = α, θ = β, ρ = ρ 1(θ), and ρ = ρ2(θ)_ are chosen as boundaries of the region _R_.

**CENTROIDS.** The centroid   of a plane region _R_ is intuitively thought of in the following way. If _R_ is supposed to have a uniform unit density, and if _R_ is supported from below at the point  , then _R_ will balance (that is, _R_ will not rotate at all).

To locate  , first consider the vertical line _x_ =  . If we divide _R_ into subregions _R 1,..., Rn,_ of areas Δ1 _A_ ,..., Δn _A_ as in Chapter 54, and if we select points ( _x k, yk_) in each _R k_, then the moment (rotational force) of _R k_ about the line _x_ =  ; is approximately ( _x k_ −  )Δ _k_ _A_. So, the moment of _R_ about _x_ =   is approximately  . Making the partition of _R_ finer and finer, we get   as the moment of _R_ about _x_ =  . In order to have no rotation about _x_ =  , we must have  . But

Hence, we must have  . Similarly, we get  . So, the centroid is determined by the equations

Note that   is equal to the area _A_ of the region _R_.

**MOMENTS OF INERTIA.** The moments of inertia of a plane region _R_ with respect to the coordinate axes are given by

The polar moment of inertia (the moment of inertia with respect to a line through the origin and perpendicular to the plane of the area) of a plane region _R_ is given by

### **Solved Problems**

**1.** Find the area bounded by the parabola _y = x_ 2 and the line _y_ = 2 _x_ \+ 3.

Using vertical strips (see Fig. 55-1), we have

**Fig. 55-1**

**2.** Find the area bounded by the parabolas _y_ 2 = 4 − _x_ and _y_ 2 = 4 − 4 _x_.

Using horizontal strips (Fig. 55-2) and taking advantage of symmetry, we have

**Fig. 55-2**

**3.** Find the area outside the circle _ρ_ = 2 and inside the cardiod _ρ_ = 2(1 + cosθ).

Owing to symmetry (see Fig. 55-3), the required area is twice that swept over as _θ_ varies from _θ_ = 0 to  . Thus,

**Fig. 55-3**

**4.** Find the area inside the circle _ρ_ = 4 sin _θ_ and outside the lemniscate _ρ_ 2 = 8 cos 2 _θ_.

The required area is twice that in the first quadrant bounded by the two curves and the line  . Note in Fig. 55-4 that the arc _AO_ of the lemniscate is described as _θ_ varies from _θ_ = _π_ /6 to _θ_ = _π_ /4, while the are _AB_ of the circle is described as _θ_ varies from _θ_ = _π_ /6 to _θ_ = _π_ /2. This area must then be considered as two regions, one below and one above the line _θ_ = _π_ /4. Thus,

**Fig. 55-4**

**5.** Evaluate   (See Fig. 55-5.)

Since  , we have

Changing to polar coordinates ( _x_ 2 \+ _y_ 2) = _ρ_ 2, _dA_ = _ρdρdθ_ yields

and

**Fig. 55-5**

**6.** Find the centroid of the plane area bounded by the parabola _y_ = 6 _x_ − _x_ 2 and the line _y_ = _x_. (See Fig. 55-6.)

Hence,  , and the coordinates of the centroid are

**Fig. 55-6**

**7.** Find the centroid of the plane area bounded by the parabolas _y_ = 2 _x_ − _x_ 2 and _y_ = 3 _x_ 2 − 6 _x_. (See Fig. 55-7.)

Hence,  , and the centroid is  .

**8.** Find the centroid of the plane area outside the circle _ρ_ = 1 and inside the cardiod _ρ_ = 1 + cos _θ_.

**Fig. 55-7**

From Fig. 55-8 it is evident that   = 0 and that   is the same whether computed for the given area or for the half lying above the polar axis. For the latter area,

The coordinates of the centroid are  .

**Fig. 55-8**

**9.** Find the centroid of the area inside ρ = sin θ and outside ρ = 1 − cos θ. (See Fig. 55-9.)

**Fig. 55-9**

The coordinates of the centroid are  .

**10.** Find _I_ x, _I_ y, and _I_ 0 for the area enclosed by the loop of _y_ 2 = _x_ 2(2 − _x_ ). (See Fig. 55-10.)

**Fig. 55-10**

where we have used the transformation 2 − _x_ = _z_ 2. Then

**11.** Find _I_ x, _I_ y, and _I_ 0 for the first-quadrant area outside the circle ρ = 2 _a_ and inside the circle ρ = 4 _a_ cos θ. (See Fig. 55-11).

**Fig. 55-11**

**12.** Fin _I_ x, _I_ y, and _I_ 0 for the area of the circle ρ = 2(sin θ + cos θ). (See Fig. 55-12)

Since _x_ 2 \+ _y_ 2 = ρ2,

It is evident from Fig. 55-12 that _I_ x = _I_ y. Hence,  .

**Fig. 55-12**

### **Supplementary Problems**

**13.** Use double integration to find the area:

**14.** Locate the centroid of each of the following areas.

**15.** Verify that  ; then infer that

**16.** Find _I_ x and _I_ y for each of the following areas:

**17.** Find _I_ _x_ and _I_ _y_ for one loop of ρ2 = cos 2θ.

_Ans._

**18.** Find _I_ 0 for ( _a_ ) the loop of ρ = sin 2θ and ( _b_ ) the area enclosed by ρ = 1 + cosθ.

_Ans._ ( _a_ )   ( _b_ )

**19.** ( _a_ ) Let the region _R_ shown in Fig. 55-13 have area _A_ and centroid ( ,  ). If _R_ is revolved about the _x_ axis, show that the volume _V_ of the resulting solid of revolution is equal to 2π (x) _A_. (Hint: Use the method of cylindrical shells.)

( _b_ ) Prove the Theorem of Pappus: If _d_ is the distance traveled by the centroid during the revolution (of part (a)), show that _V_ = _Ad_.

( _c_ ) Prove that the volume of the torus generated by revolving the disk shown in Fig. 55-14 about the _x_ axis is 2π2 _a_ 2 _b_. (It is assumed that 0 < _a_ < _b_.)

**Fig. 55-13**

**Fig. 55-14**

## **CHAPTER 56
Double Integration Applied to Volume under a Surface and the Area of a Curved Surface**

Let _z_ = _f_ ( _x_ , _y_ ) or _z_ = _f_ ( _ρ_ , _θ_ ) define a surface.

The volume _V_ under the surface, that is, the volume of a vertical column whose upper base is in the surface and whose lower base is in the _xy_ plane, is given by the double integral

where _R_ is the region forming the lower base.

The area _S_ of the portion _R_ * of the surface lying above the region _R_ is given by the double integral

If the surface is given by _x_ = _f_ ( _y_ , _z_ ) and the region _R_ lies in the _yz_ plane, then

If the surface is given by _y_ = _f_ ( _x_ , _z_ ) and the region _R_ lies in the _xz_ plane, then

### **Solved Problems**

**1.** Find the volume in the first octant between the planes _z_ = 0 and _z_ = _x_ \+ _y_ \+ 2, and inside the cylinder _x_ 2 \+ _y_ 2 = 16.

From Fig. 56-1, it is evident that _z_ = _x_ \+ _y_ \+ 2 is to be integrated over a quadrant of the circle _x_ 2 \+ _y_ 2 = 16 in the _xy_ plane. Hence,

**Fig. 56-1**

**2.** Find the volume bounded by the cylinder _x_ 2 \+ _y_ 2 = 4 and the planes _y_ \+ _z_ = 4 and _z_ = 0.

From Fig. 56-2, it is evident that _z_ = 4 − _y_ is to be integrated over the circle _x_ 2 \+ _y_ 2 = 4 in the _xy_ plane. Hence,

**3.** Find the volume bounded above by the paraboloid _x_ 2 \+ 4 _y_ 2 = _z_ , below by the plane _z_ = 0, and laterally by the cylinders _y_ 2 = _x_ and _x_ 2 = _y_. (See Fig. 56-3.)

The required volume is obtained by integrating _z_ = _x_ 2 \+ 4 _y_ 2 over the region _R_ common to the parabolas _y_ 2 = _x_ and _x_ 2 = _y_ in the _xy_ plane. Hence,

**Fig. 56-2**

**Fig. 56-3**

**4.** Find the volume of one of the wedges cut from the cylinder 4 _x_ 2 \+ _y_ 2 = _a_ 2 by the planes _z_ = 0 and _z_ = _my_. (See Fig. 56-4.)

The volume is obtained by integrating _z_ = _my_ over half the ellipse 4 _x_ 2 \+ _y_ 2 = _a_ 2. Hence,

**Fig. 56-4**

**5.** Find the volume bounded by the paraboloid _x_ 2 \+ _y_ 2 = 4 _z_ , the cylinder _x_ 2 \+ _y_ 2 = 8 _y_ , and the plane _z_ = 0. (See Fig. 56-5.)

The required volume is obtained by integrating   over the circle _x_ 2 \+ _y_ 2 = 8 _y_. Using cylindrical coordinates (see Chapter 57), the volume is obtained by integrating   over the circle _ρ_ = 8 sin _θ_. Then,

**6.** Find the volume removed when a hole of radius _a_ is bored through a sphere of radius 2 _a_ , the axis of the hole being a diameter of the sphere. (See Fig. 56-6.)

**Fig. 56-5**

**Fig. 56-6**

From the figure, it is obvious that the required volume is eight times the volume in the first octant bounded by the cylinder _ρ_ 2 = _a_ 2, the sphere _ρ_ 2 \+ _z_ 2 = 4 _a_ 2, and the plane _z_ = 0. The latter volume is obtained by integrating   over a quadrant of the circle _ρ_ = _a_. Hence,

**7.** Derive formula (56.2).

Consider a region _R_ * of area _S_ on the surface _z_ = _f_ ( _x_ , _y_ ). Through the boundary of _R_ * pass a vertical cylinder (see Fig. 56-7) cutting the _xy_ plane in the region _R_. Now divide _R_ into _n_ subregions _R_ 1, ..., _R n_ of areas Δ _A_ 1, ..., Δ _A n_, and denote by Δ _S i_ the area of the projection of Δ _A i_ on _R_ *. In that _i_ th subregion of _R_ *, chooose a point _P i_ and draw there the tangent plane to the surface. Let the area of the projection of _R i_ on this tangent plane be denoted by Δ _T i_. We shall use Δ _T i_ as an approximation of the corresponding surface area Δ _S i_.

**Fig. 56-7**

Now the angle between the _xy_ plane and the tangent plane at _P i_ is the angle _γ i_ between the _z_ axis with direction numbers [0, 0, 1] and the normal,  , to the surface at _P i_. Thus,

Then (see Fig. 56-8.)

**Fig. 56-8**

Hence, an approximation of _S_ is  , and

**8.** Find the area of the portion of the cone _x_ 2 \+ _y_ 2 = 3 _z_ 2 lying above the _xy_ plane and inside the cylinder _x_ 2 \+ _y_ 2 = 4 _y_.

_Solution 1:_ Refer to Fig. 56-9. The projection of the required area on the _xy_ plane is the region R enclosed by the circle _x_ 2 \+ _y_ 2 = 4 _y_. For the cone,

**Fig. 56-9**

Then

_Solution 2:_ Refer to Fig. 56-10. The projection of one-half the required area on the _yz_ plane is the region _R_ bounded by the line   and the parabola  , the latter having been obtained by eliminating _x_ from the equations of the two surfaces. For the cone,

Then

**Fig. 56-10**

_Solution 3:_ Using polar coordinates in solution 1, we must integrate   over the region _R_ enclosed by the circle _ρ_ = 4 sin _θ_. Then,

**9.** Find the area of the portion of the cylinder _x_ 2 \+ _z_ 2 = 16 lying inside the cylinder _x_ 2 \+ _y_ 2 = 16.

Figure 56-11 shows one-eighth of the required area, its projection on the _xy_ plane being a quadrant of the circle _x_ 2 \+ _y_ 2 = 16. For the cylinder _x_ 2 \+ _y_ 2 = 16,

Then

**10.** Find the area of the portion of the sphere _x_ 2 \+ _y_ 2 \+ _z_ 2 = 16 outside the paraboloid _x_ 2 \+ _y_ 2 \+ _z_ = 16.

Figure 56-12 shows one-fourth of the required area, its projection on the _yz_ plane being the region _R_

**Fig. 56-11**

**Fig. 56-12**

bounded by the circle _y_ 2 \+ _z_ 2 = 16, the _y_ and _z_ axes, and the line _z_ = 1. For the sphere,

Then

**11.** Find the area of the portion of the cylinder _x_ 2 \+ _y_ 2 = 6 _y_ lying inside the sphere _x_ 2 \+ _y_ 2 \+ _z_ 2 = 36.

Figure 56-13 shows one-fourth of the required area. Its projection on the _yz_ plane is the region _R_ bounded by the _z_ and _y_ axes and the parabola _z_ 2 \+ 6 _y_ = 36, the latter having been obtained by eliminating _x_ from the equations of the two surfaces. For the cylinder,

Then

**Fig. 56-13**

### **Supplementary Problems**

**12.** Find the volume cut from 9 _x_ 2 \+ 4 _y_ 2 \+ 36 by the plane _z_ = 0.

_Ans._ 3 _π_ cubic units

**13.** Find the volume under _z_ = 3 _x_ and above the first-quadrant area bounded by _x_ = 0, _y_ = 0, _x_ = 4, and _x_ 2 \+ _y 2_ = 25.

_Ans._ 98 cubic units

**14.** Find the volume in the first octant bounded by _x_ 2 \+ _z_ = 9, 3 _x_ \+ 4 _y_ = 24, _x_ = 0, _y_ = 0, and _z_ = 0.

_Ans._ 1485/16 cubic units

**15.** Find the volume in the first octant bounded by _xy_ = 4 _z_ , _y_ = _x_ , and _x_ = 4.

_Ans._ 8 cubic units

**16.** Find the volume in the first octant bounded by _x_ 2 \+ _y_ 2 = 25 and _z_ = _y_.

_Ans._ 125/3 cubic units

**17.** Find the volume common to the cylinders _x_ 2 \+ _y_ 2 = 16 and _x_ 2 \+ _z_ 2 = 16.

_Ans._ 1024/3 cubic units

**18.** Find in the volume in the first octant inside _y_ 2 \+ _z_ 2 = 9 and outside _y_ 2 = 3 _x_.

_Ans._ 27 _π_ /16 cubic units

**19.** Find the volume in the first octant bounded by _x_ 2 \+ _z_ 2 = 16 and _x_ − _y_ = 0.

_Ans._ 64/3 cubic units

**20.** Find the volume in front _x_ = 0 and common to _y_ 2 \+ _z_ 2 = 4 and _y_ 2 \+ _z_ 2 \+ 2 _x_ = 16.

_Ans._ 28 _π_ cubic units

**21.** Find the volume inside _ρ_ = 2 and outside the cone _z_ 2 = _ρ_ 2.

_Ans._ 32 _π_ /3 cubic units

**22.** Find the volume inside _y_ 2 \+ _z_ 2 = 2 and outside _x_ 2 − _y_ 2 − _z_ 2 = 2.

_Ans._   cubic units

**23.** Find the volume common to _ρ_ 2 \+ _z_ 2 = _a_ 2 and _ρ_ = _a_ sin _θ_.

_Ans._ 2(3 _π_ − 4) _a_ 2/9 cubic units

**24.** Find the volume inside _x_ 2 \+ _y_ 2 \+ = 9, bounded below by _x_ 2 \+ _y_ 2 \+ 4 _z_ = 16 and above by _z_ = 4.

_Ans._ 81 _π_ /8 cubic units

**25.** Find the volume cut from the paraboloid 4 _x_ 2 \+ _y_ 2 = 4 _z_ by the plane _z_ − _y_ = 2.

_Ans._ 9 _π_ cubic units

**26.** Find the volume generated by revolving the cardiod _ρ_ = 2(1 − cos _θ_ ) about the polar axis.

_Ans._

**27.** Find the volume generated by revolving a petal of _ρ_ = sin 2 _θ_ about either axis.

_Ans._ 32 _π_ /105 cubic units

**28.** Find the area of the portion of the cone _x_ 2 \+ _y_ 2 = _z_ 2 inside the vertical prism whose base is the triangle bounded by the lines _y_ = _x_ , _x_ = 0, and _y_ = 1 in the _xy_ plane.

_Ans._

**29.** Find the area of the portion of the plance _x_ \+ _y_ \+ _z_ = 6 inside the cylinder _x_ 2 \+ _y_ 2 = 4.

_Ans._

**30.** Find the area of the portion of the sphere _x_ 2 \+ _y_ 2 \+ _z_ 2 = 36 inside the cylinder _x_ 2 \+ _y_ 2 = 6 _y_.

_Ans._ 72( _π_ − 2) square units

**31.** Find the area of the portion of the sphere _x_ 2 \+ _y_ 2 \+ _z_ 2 = 4 _z_ inside the paraboloid _x_ 2 \+ _y_ 2 = _z_.

_Ans._ 4 _π_ square units

**32.** Find the area of portion of the sphere _x_ 2 \+ _y_ 2 \+ _z_ 2 = 25 between the planes _z_ = 2 and _z_ = 4.

_Ans._ 20 _π_ square units

**33.** Find the area of the portion of the surface _z_ = _xy_ inside the cylinder _x_ 2 \+ _y_ 2 = 1.

_Ans._

**34.** Find the area of the surface of the cone _x_ 2 \+ _y_ 2 − 9 _z_ 2 = 0 above the plane _z_ = 0 and inside the cylinder _x_ 2 \+ _y_ 2 = 6 _y_.

_Ans._

**35.** Find the area of that part of the sphere _x_ 2 \+ _y_ 2 \+ _z_ 2 = 25 that is within the elliptic cylinder 2 _x_ 2 \+ _y_ 2 = 25.

_Ans._ 50 _π_ square units

**36.** Find the area of the surface of _x_ 2 \+ _y_ 2 \+ _z_ 2 − _ax_ = 0 which lies directly above the lemniscate 4 _ρ_ 2 = _a_ 2cos 2 _θ_.

_Ans._

**37.** Find the area of the surface of _x_ 2 \+ _y_ 2 \+ _z_ 2 = 4 which lies directly above the cardioid _ρ_ = 1 − cos _θ_.

_Ans._

## **CHAPTER 57
Triple Integrals**

**CYLINDRICAL AND SPHERICAL COORDINATES**. Assume that a point _P_ has coordinates ( _x_ , _y_ , _z_ ) in a right-handed rectangular coordinate system. The corresponding _cylindrical coordinates_ of _P_ are ( _r_ , _θ_ , _z_ ), where ( _r_ , _θ_ ) are polar coordinates for the point ( _x_ , _y_ ) in the _xy_ plane. (Note the notational change here from ( _ρ_ , _θ_ ) to ( _r_ , _θ_ ) for the polar coordinates of ( _x_ , _y_ ); see Fig. 57-1.). Hence, we have the relations

In cylindrical coordinates, an equation _r_ = _c_ represents a right cylinder of radius _c_ with the _z_ axis as its axis of symmetry. An equation _θ_ = _c_ represents a plane through the _z_ axis.

A point _P_ with rectangular coordinates ( _x_ , _y_ , _z_ ) has the _spherical coordinates_ ( _ρ_ , _θ_ ,  ), where _ρ_ = | _OP_ |, _θ_ is the same as in cylindrical coordinates, and   is the directed angle from the positive _z_ axis to the vector **OP**.(See Fig. 57-2). In spherical coordinates, an equation _ρ_ = _c_ represents a sphere of radius _c_ with center at the origin. An equation   = _c_ represents a cone with vertex at the origin and the _z_ axis as its axis of symmetry.

**Fig. 57-1**

**Fig. 57-2**

The following additional relations, easily deduced from Fig. 57-2 and the equation above, hold among spherical, cylindrical, and rectangular coordinates:

_r_ = _ρ_ sin  , _ρ_ 2 = _x_ 2 \+ _y_ 2 \+ _z_ 2

_x_ = _ρ_ sin   cos _θ_ , _y_ = _ρ_ sin   sin _θ_ , _z_ = _ρ_ cos

**THE TRIPLE INTEGRAL**. Let _f_ ( _x_ , _y_ , _z_ ) be a continuous function on a three-dimensional region _R_. The definition of the double integral can be extended in an obvious way to obtain the definition of the triple integral  .

If _f_ ( _x_ , _y_ , _z_ ) = 1, then   may be interpreted as measuring the volume of the region _R_.

**EVALUATION OF TRIPLE INTEGRALS**. As in the case of double integrals, a triple integral can be evaluated in terms of iterated integrals.

In rectangular coordinates,

where the limits of integration are chosen to cover the region _R_.

In cylindrical coordinates,

where the limits of integration are chosen to cover the region _R_. (See Problem 23.)

In spherical coordinates,

where the limits of integration are chosen to cover the region _R_. (See Problem 24.)

_Discussion of the definitions:_ Consider the function _f_ ( _x_ , _y_ , _z_ ), continuous over a region _R_ of ordinary space. After slicing _R_ with planes _x_ = ξ _i_ and _y_ = _η_ _j_ as in Chapter 54, let these subregious be further sliced by planes _z_ = ζ _k_. The region _R_ has now been separated into a number of rectangular parallelepipeds of volume Δ _V ijk_ = Δ _x i_ Δ _y j_ Δ _z k_ and a number of partial parallel epipeds which we shall ignore. In each complete parallelpiped, select a point _P ijk_( _x i_, _y j_, _z k_); then compute _f_ ( _x i_, _y j_, _z k_) and form the sum

The triple integral of _f_ ( _x_ , _y_ , _z_ ) over the region _R_ is defined to be the limit of ( _57.1_ ) as the number of parallelepipeds is indefinitely increased in such a manner that all dimensions of each go to zero.

In evaluating this limit, we may sum first each set of parallelepipeds having Δ _i_ _x_ and Δ _j_ _y_ , for fixed _i_ and _j_ , as two dimensions and consider the limit as each Δ _k_ _z_ → 0. We have

Now these are the columns, the basic subregions, of Chapter 54; hence,

**CENTROIDS AND MOMENTS OF INERTIA.** The coordinates   of the centroid of a volume satisfy the relations

The _moments of inertia of a volume_ with respect to the coordinate axes are given by

### **Solved Problems**

**1.** Evaluate the given triple integrals:

( _a_ )

( _b_ )

( _c_ )

**2.** Compute the triple integral of _F_ ( _x_ , _y_ , _z_ ) = _z_ over the region _R_ in the first octant bounded by the planes _y_ = 0, _z_ = 0, _x_ \+ _y_ = 2, 2 _y_ \+ _x_ = 6, and the cylinder _y_ 2 \+ _z_ 2 = 4. (See Fig. 57-3.)

**Fig. 57-3**

Integrate first with respect to _z_ from _z_ = 0 (the _xy_ plane) to   (the cylinder), then with respect to _x_ from _x_ = 2 − _y_ to _x_ = 6 − 2 _y_ , and finally with respect to _y_ from _y_ = 0 to _y_ = 2. This yields

**3.** Compute the triple integral of _f_ ( _r_ , _θ_ , _z_ ) = _r_ 2 over the region _R_ bounded by the paraboloid _r_ 2 = 9 − _z_ and the plane _z_ = 0. (See Fig. 57-4.)

Integrate first with respect to _z_ from _z_ = 0 to _z_ = 9 − _r_ 2, then with respect to _r_ from _r_ = 0 to _r_ = 3, and finally with respect to _θ_ from _θ_ = 0 to _θ_ = 2 _π_. This yields

**Fig. 57-4**

**4.** Show that the integrals ( _a_ )  , ( _b_ )  ; and ( _c_ )   give the same volume.

( _a_ ) Here _z_ ranges from   to _z_ = 4, that is, the volume is bounded below by the paraboloid 4 _z_ = _x_ 2 \+ _y_ 2 and above by the plane _z_ = 4. The ranges of _y_ and _x_ cover a quadrant of the circle _x_ 2 \+ _y_ 2 = 16, _z_ = 0, the projection of the curve of intersection of the paraboloid and the plane _z_ = 4 on the _xy_ plane. Thus, the integral gives the volume cut from the paraboloid by the plane _z_ = 4.

( _b_ ) Here _y_ ranges from  , that is, the volume is bounded on the left by the _xz_ plane and on the right by the paraboloid _y_ 2 = 4 _z_ − _x_ 2. The ranges of _x_ and _z_ cover one-half the area cut from the parabola _x_ 2 = 4 _z_ , _y_ = 0, the curve of intersection of the paraboloid and the _xz_ plane, by the plane _z_ = 4. The region _R_ is that of ( _a_ ).

( _c_ ) Here the volume is bounded behind by the _yz_ plane and in front by the paraboloid 4 _z_ = _x_ 2 \+ _y_ 2. The ranges of _z_ and _y_ cover one-half the area cut from the parabola _y_ 2 = 4 _z_ , _x_ = 0, the curve of intersection of the paraboloid and the _xz_ plane, by the plane _z_ = 4. The region _R_ is that of ( _a_ ).

**5.** Compute the triple integral of _F_ ( _ρ_ , _θ_ ,  ) = 1/ _ρ_ over the region _R_ in the first octant bounded by the   and   = tan−1 2 and the sphere  . (See Fig. 57-5.)

Integrate first with respect to _ρ_ from _ρ_ = 0 to  , then with respect to   from   to   = tan−1 2, and finally with respect to _θ_ from  . This yields

**Fig. 57-5**

**6.** Find the volume bounded by the paraboloid _z_ = 2 _x_ 2 \+ _y_ 2 and the cylinder _z_ = 4 − _y_ 2. (See Fig. 57-6.)

Integrate first with respect to _z_ = 2 _x_ 2 \+ _y_ 2 to _z_ = 4 − _y_ 2, then with respect to _y_ from _y_ = 0 to   (obtain _x_ 2 \+ _y_ 2 = 2 by eliminating _x_ between the equations of the two surfaces), and finally with respect to _x_ from _x_ = 0 to   (obtained by setting _y_ = 0 in _x_ 2 \+ _y_ 2 = 2) to obtain one-fourth of the required volume. Thus,

**7.** Find the volume within the cylinder _r_ = 4 cos _θ_ bounded above by the sphere _r_ 2 \+ _z_ 2 = 16 and below by the plane _z_ = 0. (See Fig. 57-7.)

**Fig. 57-6**

**Fig. 57-7**

Integrate first with respect to _z_ from _z_ = 0 to  , then with respect to _r_ from _r_ = 0 to _r_ = 4cosθ, and finally with respect to θ from θ = 0 to θ = π to obtain the required volume. Thus,

**8.** Find the coordinates of the centroid of the volume within the cylinder _r_ = 2 cos, bounded above by the paraboloid _z_ = _r_ 2 and below by the plane _z_ = 0. (See Fig. 57-8.)

Then  . By symmetry,   = 0. Also,

and  . Thus, the centroid has coordinates  .

**9.** For the right circular cone of radius _a_ and height _h_ , find ( _a_ ) the centroid; ( _b_ ) the moment of inertia with respect to its axis; ( _c_ ) the moment of inertia with respect to any line through its vertex and perpendicular to its axis; ( _d_ ) the moment of inertia with respect to any line through its centroid and perpendicular to its axis; and ( _e_ ) the moment of inertia with respect to any diameter of its base.

**Fig. 57-8**

Take the cone as in Fig. 57-9, so that its equation is  . Then:

( _a_ ) The centroid lies on the _z_ axis, and we have

Then  , and the centroid has coordinates  .

( _b_ )

( _c_ ) Take the line as the _y_ axis. Then

( _d_ ) Let the line _c_ through the centroid be parallel to the _y_ axis.

( _e_ ) Let _d_ denote the diameter of the base of the cone parallel to the _y_ axis. Then

**Fig. 57-9**

**10.** Find the volume cut from the cone   by the sphere _ρ_ = 2 _a_ cos  . (See Fig. 57-10.)

**11.** Locate the centroid of the volume cut from one nappe of a cone of vertex angle 60° by a sphere of radius 2 whose center is at the vertex of the cone.

**Fig. 57-10**

Take the surface as in Fig. 57-11, so that _x_ = _y_ = 0. In spherical coordinates, the equation of the cone is   = π/6, and the equation of the sphere is _ρ_ = 2. Then

and  .

**Fig. 57-11**

**12.** Find the moment of inertia with respect to the _z_ axis of the volume of Problem 11.

### **Supplementary Problems**

**13.** Describe the curve determined by each of the following pairs of equations in cylindrical coordinates:

( _a_ ) _r_ = 1, _z_ = 2; ( _b_ ) r = 2, _z_ = _θ_ ; ( _c_ ) _θ_ = π/4,  ; ( _d_ ) _θ_ = π/4, _z_ = _r_

_Ans._ ( _a_ ) circle of radius 1 in plane _z_ = 2 with center having rectangular coordinates (0, 0, 2); ( _b_ ) helix on right circular cylinder _r_ = 2; ( _c_ ) vertical line through point having rectangular coordinates (1, 1, 0); ( _d_ ) line through origin in plane _θ_ = π/4, making an angle of 45° with _xy_ plane

**14.** Describe the curve determined by each of the following pairs of equations in spherical coordinates.

_Ans._ ( _a_ ) circle of radius 1 in _xz_ plane with center at origin; ( _b_ ) halfline on intersection of plane _θ_ = π/4 and cone   = π/6; ( _c_ ) circle of radius   in plane   with center on _z_ axis

**15.** Transform each of the following equations in either rectangular, cylindrical, or spherical coordinates into equivalent equations in the two other coordinate systems.

( _a_ ) ρ = 5; ( _b_ ) _z_ 2 = _r_ 2; ( _c_ ) _x_ 2 \+ _y_ 2 \+ ( _z_ − 1)2 = 1

_Ans._ ( _a_ ) _x_ 2 \+ _y_ 2 \+ _z_ 2 = 25, _r_ 2 \+ _z_ 2 = 25; ( _b_ ) _z_ 2 = _x_ 2 \+ _y_ 2,   (that is,   = π/4 or   = 3π/4);

(c) _r_ 2 \+ _z_ 2 = 2 _z_ , ρ = 2cos

**16.** Evaluate the triple integral on the left in each of the following:

**17.** Evaluate the integral of Problem 16( _b_ ) after changing the order to _dz dx dy_.

**18.** Evalute the integral of Problem 16( _c_ ), changing the order to _dx dy dz_ and to _dy dz dx_.

**19.** Find the following volumes, using integrals in rectangular coordinates:

**20.** Find the following volumes, using triple integrals in cylindrical coordinates:

**21.** Find the centroid of each of the following volumes:

**22.** Find the moments of inertia I _x_ , I _y_ , I _z_ of the following volumes:

**23.** Show that, in cylindrical coordinates, the triple integral of a function _f_ ( _r_ , _θ_ , _z_ ) over a region _R_ may be represented by

 _Hint_ : Consider, in [Fig. 57-12, a representative subregion of _R_ bounded by two cylinders having the _z_ axis as axis and of radii r and r + Δr, respectively, cut by two horizontal planes through (0, 0, _z_ ) and (0, 0, _z_ \+ Δ _z_ ), respectively, and by two vertical planes through the _z_ axis making angles _θ_ and _θ_ \+ Δ _θ_ , respectively, with the _xz_ plane. Take Δ _V_ = (r Δ _θ_ ) Δr Δ as an approximation of its volume.]

**Fig. 57-12**

**24.** Show that, in spherical coordinates, the triple integral of a function _f_ ( _ρ_ ,  , _θ_ ) over a region _R_ may be represented by

Hint: Consider, in [Fig. 57-13, a representative subregion of _R_ bounded by two spheres centered at _O_ , of radii _ρ_ and _ρ_ \+ Δρ, respectively, by two cones having _O_ as vertex, the _z_ axis as axis, and semivertical angles   and   \+ Δ , respectively, and by two vertical planes through the _z_ axis making angles _θ_ and _θ_ \+ Δ _θ_ , respectively, with the _yz_ plane. Take Δ _V_ = ( _ρ_ Δ )( _ρ_ sin   Δ _θ_ ) (Δ _ρ_ ) = _ρ_ 2 sin   Δ _ρ_ Δ  Δ _θ_ as an approximation of its volume.]

**25.** Change the following points from rectangular to cylindrical coordinates: ( _a_ ) (1,0,0); ( _b_ )  ; ( _c_ )

_Ans._

**Fig. 57-13**

**26.** Change the following points from cylindrical to rectangular coordinates:  ; ( _c_ ) (0, 7, 1)

_Ans._

**27.** Change the following points from rectangular to spherical coordinates:  ; ( _c_ )

_Ans._

**28.** Change the following points from spherical to rectangular coordinates: ( _a_ ) (1, 0, 0); ( _b_ ) (2, 0, π); ( _c_ )

_Ans._ ( _a_ ) (0, 0, 1); ( _b_ ) (0, 0, −2); ( _c_ )

**29.** Describe the surfaces determined by the following equations:

( _a_ ) _z = r_ 2; ( _b_ ) _r_ = 4 cos _θ_ ; ( _c_ ) _ρ_ cos   = 4; ( _d_ ) _ρ_ sin   = 4; ( _e_ )  ; ( _f_ )  ; ( _g_ ) _ρ_ = 2 sin

_Ans._ ( _a_ ) circular paraboloid; ( _b_ ) right circular cylinder ( _x_ − 2)2 \+ _y_ 2 = 4; ( _c_ ) plane _z_ = 4; ( _d_ ) right circular cylinder _x_ 2 \+ _y_ 2 = 16; ( _e_ ) the _xy_ plane; ( _f_ ) right circular cone with the _z_ axis as its axis; ( _g_ ) right circular cylinder _x_ 2 \+ _y_ 2 = 4

## **CHAPTER 58
Masses of Variable Density**

Homogeneous masses can be treated as geometric figures with density _δ_ = 1. The mass of a homogeneous body of volume _V_ and density _δ_ is _m_ = _δV_.

For a nonhomogenous mass whose density _δ_ varies continuously, an element of mass _dm_ is given by

(1) _δ_ ( _x, y_ ) _ds_ for a planar material curve (e.g., a piece of fine wire);

(2) _δ_ ( _x, y_ ) _dA_ for a material two-dimensional plate (e.g., a thin sheet of metal);

(3) _δ_ ( _x, y, z_ ) _dV_ for a material body.

The center of mass   of a planar plate that is distributed over a region _R_ with density _δ_ ( _x_ , _y_ ) is determined by the equations

An analogous result holds for the center of mass of three-dimensional body. The reasoning is similar to that for centroids in Chapter 55.

The moments of inertia of a planar mass with respect to the _x_ axis and the _y_ axis are  . Similar formulas with triple integrals hold for three-dimensional bodies. (For examples,  .)

### **Solved Problems**

**1.** Find the mass of a semicircular wire whose density varies as the distance from the diameter joining the ends.

Take the wire as in Fig. 58-1, so that _δ_ ( _x_ , _y_ ) = _ky_. Then, from _x_ 2 \+ _y_ 2 = _r_ 2.

and

**Fig. 58-1**

**2.** Find the mass of a square plate of side _a_ if the density varies as the square of the distance from a vertex.

Take the square as in Fig. 58-2, and let the vertex from which distances are measured be at the origin. Then _δ_ ( _x_ , _y_ ) = _k_ ( _x_ 2 \+ _y_ 2) and

**Fig. 58-2**

**3.** Find the mass of a circular plate of radius _r_ if the density varies as the square of the distance from a point on the circumference.

Take the circle as in Fig. 58-3 and let _A_ ( _r_ , 0) be the fixed point on the circumference. Then _δ_ ( _x_ , _y_ ) = _k_ [( _x_ − _r_ )2 \+ _y_ 2]and

**Fig. 58-3**

**4.** Find the center of mass of a plate in the form of the segments cut from the parabola _y_ 2 = 8 _x_ by its latus rectum _x_ = 2 if the density varies as the distance from the latus rectum. (See Fig. 58-4.)

Here, _δ_ ( _x_ , _y_ ) = 2 − _x_ and, by symmetry,   = 0. For the upper half of the plate,

and  . The center of mass has coordinates  .

**Fig. 58-4**

**5.** Find the center of mass of a plate in the form of the upper half of the cardioid _r_ = 2(1 + cos _θ_ ) if the density varies as the distance from the pole. (See Fig. 58-5.)

Then  , and the center of mass has coordinates  .

**Fig. 58-5**

**6.** Find the moment of inertia with respect to the _x_ axis of the plate having for edges one arch of the curve _y_ = sin _x_ and the _x_ axis if its density varies as the distance from the _x_ axis. (See Fig. 58-6.)

**Fig. 58-6**

**7.** Find the mass of a sphere of radius _a_ if the density varies inversely as the square of the distance from the center.

Take the sphere as in Fig. 58-7. Then   and

**Fig. 58-7**

**8.** Find the center of mass of a right circular cylinder of radius _a_ and height _h_ if the density varies as the distance from the base.

Take the cylinder as in Fig. 54-8, so that its equation is _r_ = _a_ and the volume in question is that part of the cylinder between the planes _z_ = 0 and _z_ = _h_. Clearly, the center of mass lies on the _z_ axis. Then

and  . Thus the center of mass has coordinates  .

**Fig. 58-8**

### **Supplementary Problems**

**9.** Find mass of

( _a_ ) A straight rod of length _a_ whose density varies as the square of the distance from one end

_Ans_.

( _b_ ) A plate in the from of a right triangle with legs _a_ and _b_ , if the density varies as the sum of the distance from the legs

_Ans_.

( _c_ ) A circular plate of radius _a_ whose density varies as the distance from the center

_Ans_.

( _d_ ) A plate in the form of an ellipse _b_ 2 _x_ 2 \+ _a_ 2 _y_ 2 = _a_ 2 _b_ 2, if the density varies as the sum of the distances from its axes

_Ans_.

( _e_ ) A circular cylinder of height _b_ and radius of base _a_ , if the density varies as the square of the distance from its axis

_Ans_.

( _f_ ) A sphere of radius _a_ whose density varies as the distance from a fixed diametral plane

_Ans_.

( _g_ ) A circular cone of height _b_ and radius of base _a_ whose density varies as the distance from its axis

_Ans_.

( _h_ ) A spherical surface whose density varies as the distance from a fixed diametral plane

_Ans._ 2 _ka_ 3π units

**10.** Find the center of mass of:

( _a_ ) One quadrant of the plate of Problem 9( _c_ )

_Ans._ (3 _a_ /2π, 3 _a_ /2π)

( _b_ ) One quadrant of a circular plate of radius _a_ , if the density varies as the distance from a bounding radius (the _x_ axis)

_Ans._ (3 _a_ /8, 3 _a_ π/16)

( _c_ ) A cube of edge _a_ , if the density varies as the sum of the distances from three adjacent edges (on the coordinate axes)

_Ans._ (5 _a_ /9,5 _a_ /9,5 _a_ /9)

( _d_ ) An octant of a sphere of radius _a_ , if the density varies as the distance from one of the plane faces

_Ans._ (16 _a_ /15π,16 _a_ /15π,8 _a_ /15)

( _e_ ) A right circular cone of height _b_ and radius of base _a_ , if the density varies as the distance from its base

_Ans._ (0,0,2 _b_ /5)

**11** Find the moment of inertia of:

( _a_ ) A square plate of side _a_ with respect to a side, if the density varies as the square of the distance from an extremity of that side

_Ans._

( _b_ ) A plate in the form of a circle of radius _a_ with respect to its center, if the density varies as the square of the distance from the center

_Ans_.

( _c_ ) A cube of edge _a_ with respect to an edge, if the density varies as the square of the distance from one extremity of that edge

_Ans_.

( _d_ ) A right circular cone of height _b_ and radius of base _a_ with respect to its axis, if the density varies as the distance from the axis

_Ans_.

( _e_ ) The cone of ( _d_ ), if the density varies as the distance from the base

_Ans_.

## **CHAPTER 59
Differential Equations of First and Second Order**

A differential equation is an equation that involves a function, say _y_ , of one variable, say _x_ , and derivatives of _y_ or differentials of _x_ and _y_. Examples are   and _dy_ = ( _x_ \+ 2 _y_ ) _dx_. The first equation also can be written as _y_ ′′ + 2 _y_ ′ + 3 _y_ − 7 sin _x_ \+ 4 _x_ = 0.

The _order_ of a differential equation is the order of the derivative of highest order appearing in it. The first of the above equations is of order two, and the second is of order one.

A _solution_ of a differential equation is a function _y_ that satisfies the equation. A _general solution_ of an equation is a formula that describes all solutions of the equation. It turns out that a general solution of a differential equation of order _n_ will contain _n_ arbitrary constants.

**SEPARABLE DIFFERENTIAL EQUATIONS.** A separable differential equation is a first-order equation that can be put in the form

A separable equation can be solved by taking antiderivatives

The result is an equation involving _x_ and _y_ that determines _y_ as a function of _x_. (For justification, see Problem 28.)

**HOMOGENEOUS FUNCTIONS.** A function _f(x, y)_ is said to be _homogeneous of degree n_ if _f(λx,λy)_ = _λ nf(x, y)_. The equation _M(x, y) dx + N(x, y) dy_ = 0 is said to be _homogeneous_ if _M(x, y)_ and _N(x, y)_ are homogeneous of the same degree. It is easy to verify that the substitution

_y = vx, dy = v dx + x dv_

will transform a homogeneous equation into a separable equation in the variables _x_ and _v_.

**INTEGRATING FACTORS.** Certain differential equations may be solved readily after multiplication by a suitable function of _x_ and _y_. Such a function is called an _integrating factor_ of the equation. (See Problems 10-14.)

The so-called _linear differential equations of the first order_ ,  , where _P_ and _Q_ are functions of _x_ alone, have the function   as integrating factor. (See Problems 15-17.)

An equation of the form  , where n ≠ 0, 1 and where _P_ and _Q_ are functions of _x_ alone, can be reduced to the linear form by the substitution

(See Problems 18-19).

**SECOND-ORDER EQUATIONS.** The second-order equations that will be solved in this chapter are of the following types:

  (See Problem 23.)

  (See Problems 24 and 25.)

  (See Problems 26 and 27.)

 , where _P_ and _Q_ are constants and _R_ is a constant or function of _x_ only. (See Problems 28-33.)

If the equation _m 2 \+ Pm + Q_ = 0 has two distinct roots _m_ 1 and _m_ 2 then   is the general solution of the equation  . If the two roots are identical so that _m_ 1 = _m_ 2 = _m_ , then

is the general solution.

The general solution of   is called the _complementary function_ of the equation

If _f_ ( _x_ ) satisfies ( _59.1_ ), then the general solution of ( _59.1_ ) is

y = complementary function + _f_ ( _x_ )

The function _f_ ( _x_ ) is called a _particular solution_ of ( _59.1_ ).

### **Solved Problems**

**1.** Show that ( _a_ ) _y_ = 2 _e x, ( _b_ ) y_ = 3 _x_ , and _(c) y = C_ 1 _e x \+ C_2 _x_ , where _C_ 1 and _C_ 2 are arbitrary constants, are solutions of the differential equation _y_ ′′(1 − _x_ ) + _y′x − y_ = 0.

( _a_ ) Differentiate _y_ = 2 _e x_ twice to obtain _y_ ′ = 2 _e x_ and _y_ ′′ = 2 _e x_. Substitute in the differential equation to obtain the identity 2 _e x_(1 − _x_ ) + 2 _e xx_ − 2 _e x_ = 0.

( _b_ ) Differentiate _y_ = 3 _x_ twice to obtain _y_ ′ = 3 and _y_ ′′ = 0. Substitute in the differential equation to obtain the identify 0(1 − _x_ ) + 3 _x_ − 3 _x_ = 0.

( _c_ ) Differentiate _y = C_ 1 _e x_ \+ _C_ 2 _x_ twice to obtain _y_ ′ = _C_ 1 _e x_ \+ _C_ 2 and _y_ ′′ = _C_ 1 _e x_. Substitue in the differntial equation to obtain the identity _C_ 1 _e x_(1 − _x_ ) + ( _C_ 1 _e x \+ C_2) x − ( _C_ 1 _e x_ \+ _C_ 2 _x_ ) = 0.

Solution ( _c_ ) is the _general solution_ of the differential equation because it satisfies the equation and contains the proper number of essential arbitrary constants. Solutions ( _a_ ) and ( _b_ ) are called _particular solutions_ because each may be obtained by assigning particular values to the arbitrary constants of the general solution.

**2.** Form the differential equation whose general solution is:

( _a_ ) _y = Cx_ 2 − _x_ ; ( _b_ ) _y = C_ 1 _x_ 3 \+ _C_ 2 _x_ \+ _C_ 3.

( _a_ ) Differentitate _y = Cx_ 2 − _x_ ; once to obtain _y_ ′ = 2 _Cx_ − 1. Solve for   and substitute in the given relation (general solution) to obtain   or _y′x_ = 2 _y + x_.

( _b_ ) Differentiate _y = C_ 1 _x_ 3 \+ _C_ 2 _x + C_ 3 three times to obtain  . Then   is required equation. Note that the given relation is a solution of the equation _y_ (4) = 0 but is not the general solution, since it contains only three arbitrary constants.

**3.** Form the second-order differential equation of all parabolas with principal axis the _x_ axis.

The system of parabolas has equation _y_ 2 = _Ax + B_ , where _A_ and _B_ are arbitrary constants. Differentiate twice to obtain 2 _yy_ ′ = _A_ and 2 _yy_ ′′ + 2( _y_ ′)2 = 0. The latter is the required equation.

**4.** Solve  .

Here _xy_ 2(1 + _x_ 2) _dy_ \+ (1 + _y_ 3) _dx_ = 0, or   with the variables separated. Then the partial-fraction decomposition yields

and integration yields

or

2 In |1 + _y_ 3| + 6 In |x| − 3 In(1 + _x_ 2) = 6 _c_

from which

**5.** Solve  .

Separate the varibales:  . Integration yields tan−1 _y_ = tan−1 _x_ \+ tan−1 _C_ , and then

**6.** Solve  .

The variables are easily separated to yield  .

Hence, sec2 _y dy_ = csc2 _x dx_ and integration yields tan _y_ = −cot _x + C._

**7.** Solve 2 _xy dy_ = ( _x_ 2 − _y_ 2) _dx_.

The equation is homogeneous of degree two. The transformation _y = vx, dy = v dx + x dv_ yields (2 _x_ )( _vx_ )( _v dx + x dv_ ) = _x_ 2 − _v_ 2 _x_ ) _dx_ or  . Then integration yields

from which In |1 − 3 _v_ 2 \+ 3 In | _x_ | + In _C_ ′ = 0 or _C_ ′ | _x_ 3 (1 − 3 _v_ 2)| = 1.

Now ± _C_ ′ _x_ 3 (1 − 3 _v_ 2) = _Cx_ 3 (1 − 3 _v_ 2) = 1, and using _v = y/x_ produces _C(x_ 3 − 3 _xy_ 2) = 1.

**8.** Solve  .

The equation is homogeneous of degree two. the transformation _y = vx, dy = vdx + xdv_ yields

_x_ sin _v_ ( _vxdx + x_ 2 _dv + vx dx_ ) + _vx cos v_ ( _x_ 2 _dv + vx dx − vx dx_ ) = 0

or

sin _v_ (2 _v dx + x dv_ ) + _xv_ cos _v dv_ = 0

or

Then In | _v_ sin _v_ | + 2 In | _x_ | = In _C_ ′, so that _x_ 2 _v_ sin _v_ = _C_ and  .

**9.** Solve ( _x_ 2 − 2 _y_ 2) _dy + 2xy dx_ = 0.

The equation is homogeneous of degree two, and the standard transformation yields

(1 − 2 _v_ 2)( _v dx + x dv_ ) + 2 _v dx_ = 0

or

or

Integration yields  , which we may write as In | _v_ | + In |3 − 2 _v_ 2| + 3 ln | _x_ | = ln _C_ ′. Then _vx_ 3(3 − 2 _v_ 2) = _C_ and _y_ (3 _x_ 2 − 2 _y_ 2) = _C_.

**10.** Solve ( _x_ 2 \+ _y_ ) _dx_ \+ ( _y_ 3 \+ _x_ ) _dy_ = 0.

Integrate _x_ 2 _dx_ \+ ( _y dx_ \+ _x dy_ ) + _y_ 3 _dy_ = 0, term by term, to obtain

**11.** Solve ( _x_ \+ _e_ − _x_ sin _y_ ) _dx_ − ( _y_ \+ _e_ −x cos y) dy = 0.

Integrate _x dx − y dy −_ ( _e −x_ cos _y dy_ − _e_ _−x_ sin _y dx_ ) = 0, term by term, to obtain

**12.** Solve _x dy − y dx_ = 2 _x_ 3 _dx_ ,

The combination _x dy − y dx_ suggest  . Hence, multiplying the given equation by  , we obtain  , from which

**13.** Solve _x_ d _y + y_ d _x_ = 2 _x_ 2 _y dx_.

The combination _x dy + y dx_ suggests  . Hence, multiplying the given equation by  , we obtain  , from which In | _xy_ | = _x_ 2+C.

**14.** Solve _x dy_ +(3 _y_ −e _x_ ) _dx_ = 0.

Multiply the equation by ξ( _x_ ) = _x_ 2 to obtain _x_ 3 _dy_ \+ 3 _x_ 2 _y dx = x_ 2e _x_ _dx_. This yields

**15.**  .

Here  , and an integrating factor is  . We multiply the given equation by ξ( _x_ ) = _x_ 2 to obtain _x_ 2 _dy_ \+ 2 _xy dx_ = 6 _x_ 5 _dx_. Then integration yields _x_ 2 _y_ = _x_ 6 \+ C.

_Note 1_ : After multiplication by the integrating factor, the terms on the left side of the resulting equation are an _integrable combination._

_Note 2_ : The integrating factor for a given equation is not unique. In this problem  , etc., are all integrating factors. Hence, we write the simplest particular integral of P( _x_ ) _dx_ rather than the general integral, ln _x_ 2 \+ ln _C_ = ln _Cx_ 2.

**16.** Solve  .

Since   we have  , and  . Then

multiplication by ξ( _x_ ) yields

and integration gives

_y_ sin _x_ = _x_ \+ C

**17.** Solve  .

Here  . This produces

and integration yields

**18.** Solve  .

The equation is of the form  , with _n_ = 2. Hence we use the substitution _y_ 1−n = _y_ −1 = _z_ ,  . For convenience, we write the original equation in the form  , obtaining  , or  .

The integrating factor is  . It gives us _e_ −x _dx_ − _ze_ −x _dx_ = − _xe_ −x _dx_ , from which _ze_ −x = _xe_ −x \+ _e_ −x \+ C. Finally, since z = _y_ −1, we have

**19.** Solve  .

Write the equation in the form  . Then use the substitution _y_ −2 = _z_ ,   to obtain  .

The integrating factor is  . It gives cos2 _x dz_ −2 _z cos x sin x dx_ = −2 _cos x dx,_ from which

**20.** When a bullet is fired into a sand bank, its retardation is assumed equal to the square root of its velocity on entering. For how long will it travel if its velocity on entering the bank is 144 ft/sec?

Let _v_ represent the bullet's velocity _t_ seconds after striking the bank. Then the retardation is  , so   and  .

When _t_ = 0, _v_ = 144 and  . Thus,   is the law governing the motion of the bullet. When _v_ = 0, _t_ = 24; the bullet will travel for 24 seconds before coming to rest.

**21.** A tank contains 100 gal of brine holding 200 lb of salt in solution. Water containing 1 lb of salt per gallon flows into the tank at the rate of 3 gal/min, and the mixture, kept uniform by stirring, flows out at the same rate. Find the amount of salt at the end of 90 min.

Let _q_ denote the number of pounds of salt in the tank at the end of _t_ minutes. Then   is the rate of change of the amount of salt at time _t_.

Three pounds of salt enter the tank each minute, and 0.03 _q_ pounds are removed. Thus,  .

Rearranged, this becomes  , and integration yields

When _t_ = 0, _q_ = 200 and   so that ln (0.03 _q_ −3) = −0.03 _t_ \+ ln 3. Then 0.01 _q_ −1 = _e_ −0.03 _t_ , and _q_ = 100 + 100 _e −0.03t_. When _t_ = 90, _q_ = 100 + 100 _e −2.7_~106.72 lb.

**22.** Under certain conditions, cane sugar in water is converted into dextrose at a rate proportional to the amount that is unconverted at any time. If, of 75 grams at time _t_ = 0, 8 grams are converted during the first 30 min, find the amount converted in   hours.

Let _q_ denote the amount converted in _t_ minutes. Then  , from which  , and integration gives ln (75 − _q_ ) = − _kt + C_.

When _t_ = 0, _q_ = 0 and _C_ = ln 75, so that ln(75− _q_ ) = − _kt_ \+ ln 75.

When _t_ = 30 and _q_ = 8, we have 30 _k_ = ln 75 − ln 67; hence, _k_ = 0.0038, and _q_ = 75(1− _e −0.0038t_).

When _t_ = 90, _q_ = 75(1− _e −0.34_) ~ 21.6 grams.

**23.**  .

Here  . Hence,  , and another integration yields _y = xe x−2ex_−cos _x_ \+ _C 1x_ \+ _C 2_.

**24.** Solve  .

Let  ; then   and the given equation becomes   or  . Then integration yields _xp = a ln|x|+C 1,_ or  . When this is written as _dy_ = _a_ In  , integration gives  .

**25.** Solve _xy″ + y′ + x_ = 0.

Let  . Then   and the given equation becomes  .

Integration gives  , substitution for _p_ gives  , and another integration yields  .

**26.** Solve  .

Since  , we can multiply the given equation by 2 _y_ ′ to obtain 2 _y_ ′ _y_ ″ = 4 _yy_ ′, and integrate to obtain  .

Then  , so that   and ln  . The last equation yields  .

**27.** Solve _y_ ″ = −1/ _y_ 3.

Multiply by 2y′ to obtain  . Ten integtration yields

Another integration gives   or ( _C_ 1 _x_ \+ _C_ 2)2 − _C_ 1 _y_ 2 = 1.

**28.** Solve  .

Here we have _m_ 2 \+ 3 _m_ − 4 = 0, from which _m_ = 1, −4. The general solution is _y_ = _C_ 1 _e_ _x_ \+ _C_ 2 _e_ −4 _x_.

**29.** Solve  .

Here _m_ 2 \+ 3 _m_ = 0, from which _m_ = 0, −3. The general solution is _y_ = _C_ 1 \+ _C_ 2 _e_ −3 _x_.

**30.** Solve  .

Here _m_ 2 − 4 _m_ \+ 13 = 0, with roots _m_ 1 = 2 + 3 _i_ and _m_ 2 = 2 − 3 _i_. The general solution is

Since _e_ iax = cos _ax_ \+ _i_ sin _ax_ , we have _e_ 3 _ix_ = cos 3 _x_ \+ _i_ sin 3 _x_ and _e_ −3 _ix_ = cos 3 _x_ − _i_ sin 3 _x_ Hence, the solution may be put in the form

**31.** Solve  .

Here _m_ 2 − 4 _m_ \+ 4 = 0, with _m_ = 2, 2. The general solution is _y_ = _C_ 1 _e_ 2 _x_ \+ _C_ 2 _xe_ 2 _x_.

**32.** Solve  .

From Problems 6, the complementary function is

To find a particular solution of the equation, we note that the right-hand member is _R_ ( _x_ ) = _x_ 2. This suggests that the particular solution will contain a term in _x_ 2 and perhaps other terms obtained by successive differentiation. We assume it to be of the from _y_ = _Ax_ 2 \+ _bx_ \+ _C_ , where the constants _A, B, C_ are to be determined. Hence we substitute _y_ = _Ax_ 2 \+ _Bx_ \+ _C, y_ ′ = 2 _Ax_ \+ _B_ , and _y_ ″ = 2 _A_ in the differential equation to obtain

2 _A_ \+ 3(2 _Ax_ \+ _B_ ) − 4( _Ax_ 2 \+ _Bx_ \+ _C_ ) = _x_ 2 or −4 _Ax_ 2 \+ (6 _A_ − 4 _B_ ) _x_ \+ (2 _A_ \+ 3 _B_ − 4 _C_ ) = _x_ 2

Since this latter equation is an identity in _x,_ we have −4 _A_ =1, 6 _A_ − 4 _B_ = 0, and 2 _A_ \+ 3 _B_ − 4 _C_ = 0. These yield  , and   is a particular solution. Thus, the general solution is  .

**33.** Solve  .

Here _m_ 2 − 2 _m_ − 3 = 0, from which _m_ = −1, 3; the complementary function is _y_ = _C_ 1 _e_ −x \+ _C_ 2 _e_ 3 _x_. The right-hand member of the differential equation suggests that a particular solution is of the form _A_ cos _x_ \+ _B_ sin _x_. Hence, we substitute _y_ = _A_ cos _x_ \+ _B_ sin _x, y_ ′ = _B_ cos _x_ − _A_ sin _x_ , and _y_ ″ = − _A_ cos _x_ − _B_ sin _x_ in the differential equation to obtain

or

The latter equation yields −2(2 _A_ \+ _B_ ) = 1 and _A_ − 2 _B_ = 0, from which  . The general solution is  .

**34.** A weight attached to a spring moves up and down so that the equation of motion is  , where _s_ is the stretch of the spring at time _t_. If _s_ = 2 and   when _t_ = 0, find _s_ in terms of _t_.

Here _m_ 2 \+ 16 = 0 yields _m_ = ±4 _i_ , and the general solution is _s_ = _A_ cos 4 _t_ \+ _B_ sin 4 _t_. Now when _t_ = 0, _s_ = 2 = _A_ , so that _s_ =2 cos 4 _t_ \+ _B_ sin 4 _t_.

Also when _t_ = 0, _ds_ / _dt_ = 1 = −8 sin 4 _t_ \+ 4 _B_ cos 4 _t_ = 4 _B_ , so that  . Thus, the required equation is  .

**35.** The electric current in a certain circuit is given by  . If _I_ = 0 and   when _t_ = 0, find _I_ in terms of _t_.

Here _m_ 2 \+ 4m + 2504 = 0 yields _m_ = −2 + 50 _i_ , −2 −50 _i_ ; the complementary function is _e_ −2 _t_ ( _A_ cos 50 _t_ \+ _B_ sin 50 _t_ ). Because the right-hand member is a constant, we find that the particular solution is _I_ = 110/2504 = 0.044. Thus, the general solution is _I_ = _e_ −2 _t_ ( _A_ cos 50 _t_ \+ _B_ sin 50 _t_ ) + 0.044.

Also when _t_ = 0, _dI/dt_ = 0 = _e_ −2 _t_ [(−2 _A_ \+ 50 _B_ ) cos 50 _t_ − (2 _B_ \+ 50 _A_ ) sin 50 _t_ ] = −2 _A_ \+ 50B. Then _B_ = −0.0018, and the required relation is _I_ = − _e_ −2 _t_ (0.044 cos 50 _t_ \+ 0.0018 sin 50 _t_ ) + 0.044.

**36.** A chain 4 ft long starts to slide off a flat roof with 1 ft hanging over the edge. Discounting friction, find ( _a_ ) the velocity with which it slides off and ( _b_ ) the time required to slide off.

Let s denote the length of the chain hanging over the edge of the roof at time _t_.

( _a_ ) The force _F_ causing the chain to slide off the roof is the weight of the part hanging over the edge. That weight is _mgs/4_. Hence,

Multiplying by 2 _s_ ′ yields   and integrating once gives  .

when _t_ = 0, _s_ = 1 and _s_ ′ = 0. Hence,   and  . When _s_ = 4,  .

( _b_ ) Since  , integration yields ln   When _t_ = 0, _s_ = 1. Then _C_ 2 = 0 and ln  .

When _s_ = 4,   seconds.

**37.** A boat of mass 1600 lb has a speed of 20 ft/sec when its engine is suddenly stopped (at _t_ = 0). The resistance of the water is proportional to the speed of the boat and is 200 lb when _t_ = 0. How far will the boat have moved when its speed is reduced to 5 ft/sec?

Let _s_ denote the distance traveled by the boat _t_ seconds after the engine is stopped. Then the force _F_ on the boat is

_F_ = _ms_ ″ = − _Ks_ ′ from which _s_ ′ = − _ks_ ′

To determine _k_ , we note that at _t_ = 0, _s_ ′ = 20 and  . Then  .

Now  , and integration gives ln  , or  .

When _t_ = 0, _v_ = 20. Then _C_ 1 = 20 and  . Another integration yields _s_ = −100 _e_ − _t_ /5 \+ _C_ 2.

When _t_ = 0, _s_ = 0; then _C_ 2 = 100 and _s_ = 100(1 − _e_ − _t_ /5). We require the value of _s_ when _v_ = 5 = 20 _e_ − _t_ /5, that is, when  . Then  .

### **Supplementary Problems**

**38.** From the differential equation whose general solution is:

_Ans_.

**39.** Solve:

**40.** The tangent and normal to a curve at a point _P_ ( _x, y_ ) meet the _x_ axis in _T_ and _N_ , respectively, and the _y_ axis in _S_ and _M_ , respectively. Determine the family of curves satisfying the conditions:

( _a_ ) _TP_ = _PS_ ; ( _b_ ) _NM_ = _MP_ ; ( _c_ ) _TP_ = _OP_ ; ( _d_ ) _NP_ = _OP_

_Ans._ ( _a_ ) _xy_ = _C_ ; ( _b_ ) 2 _x_ 2 \+ _y_ 2 = _C_ ; ( _c_ ) _xy_ = _C, y_ = _Cx_ ; ( _d_ ) _x_ 2 ± _y_ 2 = _C_

**41.** Solve Problem 21, assuming that pure water flows into the tank at the rate of 3 gal/min and the mixture flows out at the same rate.

_Ans._ 13.44 lb

**42.** Solve Problem 41 assuming that the mixture flows out at the rate 4 gal/min.

_Ans._ 0.02 lb

In problems 43−59, solve the given equation.

**60.** A particle of mass _m_ , moving in a medium that offers a resistance proportional to the velocity, is subject to an attracting force proportional to the displacement. Find the equation of motion of the particle if at time _t_ = 0,

_s_ = 0 and _s_ ′ = _v_ 0

_Ans._ If _b_ 2 = _c_ 2, _s_ = _v_ 0 _te_ −bt; if _b_ 2 < _c_ 2,  ; if _b_ 2 > _c_ 2

**61.** Justify our method for solving a separable differential equation   by integration, that is,  .

_Ans._ Differentiate both sides of   with respect to _x_ , obtaining  .

Hence,   and the solution _y_ satisfies the given equation.

## Appendix A

### TRIGONOMETRIC FORMULAS

cos2 _θ_ \+ sin2 _θ_ = 1

cos( _θ_ \+ 2π) = cos _θ_

sin( _θ_ \+ 2π) = sin _θ_

cos(- _θ_ ) = cos _θ_

sin(- _θ_ ) = - sin _θ_

cos( _u_ \+ _v_ ) = cos _u_ cos _v_ \- sin _u_ sin _v_

cos( _u_ \+ _v_ ) = cos _u_ cos _v_ \+ sin _u_ sin _v_

sin( _u_ \+ _v_ ) = sin _u_ cos _v_ \+ cos _u_ sin _v_

sin( _u_ \- _v_ ) = sin _u_ cos _v_ \- cos _u_ sin _v_

sin 2 = 2 sin _θ_ cos _θ_

cos 2 _θ_ = cos2 _θ_ \- sin2 _θ_

= 2 cos2 _θ_ \- 1 = 1 - 2 sin2 _θ_

Law of cosines: _c_ 2 = _a_ 2 \+ _b_ 2 \- 2 _ab_ cos _θ_

tan(- x) = - tan x

tan(x + π) = tan x

1tan2 x = sec2 x

1 a cot2x = csc2x

## Appendix B

### GEOMETRIC FORMULAS

( _A_ = area, _C_ = circumference, _V_ = volume, _S_ = lateralsurfacearea)

## **INDEX**

Abel's theorem,

Abscissa,

Absolute value,

Absolute maximum and minimum,

Absolutely convergent series,

Acceleration:

angular,

in curvilinear motion,

in rectilinear motion,

tangential and normal components of,

vector,

Alternating series,

theorem,

Analytic proofs of geometric theorems,

Angle:

between two curves, , ,

measure,

of inclination, ,

Angular velocity and acceleration,

Antiderivative,

Approximations by differentials,

Arc length,

formula, , ,

derivative of, ,

Area:

between curves, –258

by integration,

in polar coordinates

of a curved surface

of a surface of revolution,

under a curve,

Argument,

Asymptote,

of hyperbola,

Average rate of change,

Average value of a function,

Average velocity,

Axis of revolution,

Axis of symmetry,

of a parabola,

Binomial series,

Binormal vector,

Bliss's theorem,

Bounded sequence,

Bounded set in a plane,

Carbon-dating,

Cardioid,

Center of curvature,

Center of mass,

Center:

of an ellipse,

of a hyperbola,

Centroid:

of plane region,

of surface of revolution

of volume,

Chain rule, ,

Change of variables in an integral,

Circle:

equation of,

of curvature,

Circular motion,

Closed interval,

Closed set,

Comparison test,

Complement,

Complementary function,

Completing the square,

Components of a vector,

Composite function,

Composition,

Compound interest, ,

Concave upward, downward,

Concavity,

Conditionally convergent series,

Cone, elliptic,

Conic sections,

Conjugate axis of a hyperbola,

Continuous function, , ,

on [a,b],

on the left (right),

Convergence of series,

absolute, conditional,

Convergence, uniform,

Convergent sequence,

Coordinate, ,

axes,

Coordinate system:

cylindrical and spherical,

linear,

rectangular,

polar,

Cosecant,

Cosine,

Cotangent,

Critical numbers,

Cross product of vectors

Cross-section formula,

Cubic curve,

Curl,

Curvature,

of a polar curve

Curve sketching,

Curvilinear motion,

Cycloid,

Cylindrical coordinates,

Cylindrical surfaces,

Cylindrical shell formula,

Decay constant,

Decreasing:

function

sequence,

Definite integral,

Del,

Deleted disk,

Delta neighborhood,

Delta notation,

Density,

Dependent variable,

Derivative,

directional,

first,

higher order,

of arc length, ,

of inverse function,

of a vector function,

partial,

second,

Determinants,

Difference of shells formula,

Differentiability, ,

Differential,

total,

Differential equation,

linear, of the first order,

order of a,

second-order,

separable,

solution (general) of a,

Differentiation,

implicit, ,

logarithmic,

of trigonometric functions, ,

of vector functions, ,

Directed angles,

Direction cosines,

Direction numbers,

Directional derivative,

Directrix of a parabola,

Discontinuity,

removable,

jump,

Disk formula,

Distance formula,

for polar coordinates,

Divergence (div) of a vector function,

Divergence:

of a sequence,

of a series,

Divergence theorem,

Domain of a function,

Dot product of vectors,

Double integral,

_e_ ,

Eccentricity:

of an ellipse,

of a hyperbola,

Ellipses,

center, eccentricity, foci, major axis, minor axis of,

minor axis of

Ellipsoid,

Elliptic:

cone,

paraboloid,

Equations, graphs of,

Equilateral hyperbola,

Even function,

Evolute,

Exponential functions, ,

Exponential growth and decay,

Extended law of the mean,

Extreme Value Theorem,

Extremum, relative,

First derivative,

First derivative test,

Focus of a parabola,

Foci:

of an ellipse,

of a hyperbola,

Free fall,

Function, ,

composite,

continuous, , ,

decreasing,

differentiable,

domain of a,

even,

exponential, ,

homogeneous, ,

hyperbolic,

implicit,

increasing,

integrable,

inverse,

inverse trigonometric,

logarithmic,

odd,

one-to-one,

range of a,

trigonometric, ,

Fundamental theorem of calculus,

Gamma function,

Generalized Rolle's theorem,

Geometric series,

Gradient, ,

Graphs of equations,

Graphs of functions,

Gravity,

Growth constant,

Halflife,

Half-open interval,

Harmonic series,

Higher-order:

derivatives,

partial derivatives,

Higher-order law of the mean,

Homogeneous bodies,

Homogeneous:

equation,

function, ,

Horizontal asymptote,

Hyperbola,

asymptotes of,

center, conjugate axes, eccentricity, foci, transverse axes, vertices,

foci, transverse axes, vertices

equilateral,

Hyperbolic functions,

Hyperbolic paraboloid,

Hyperboloid:

of one sheet,

of two sheets,

Implicit differentiation, ,

Implicit functions,

Improper integrals,

Increasing:

function

sequence,

Indefinite integral,

independent variable,

Indeterminate types,

Inequalities,

Infinite intervals,

Infinite limit,

Infinite sequence,

limit of,

Infinite series,

Inflection point,

Initial position,

Initial velocity,

Instantaneous rate of change,

Instantaneous velocity,

Integrable,

Integrand,

Integral:

definite,

double,

improper,

indefinite,

iterated,

line,

Riemann,

test for convergence,

triple,

Integrating factor,

Integration:

by miscellaneous substitutions,

by partial fractions,

by parts,

by substitution,

by trigonometric substitution,

Intercepts,

Intermediate Value Theorem,

Interval of convergence,

Intervals,

Inverse function,

Inverse cosecant,

Inverse cosine,

Inverse cotangent,

Inverse secant,

Inverse sine,

Inverse tangent,

Inverse trigonometric functions,

Irreducible polynomial,

Iterated integral,

Jump discontinuity,

Latus rectum of a parabola,

Law of cosines,

Law of sines,

Law of the mean,

Extended,

Higher-order,

Lemniscate,

Length of arc,

L'Hoôspital's rule,

Limaçon,

Limit:

infinite,

of a function, ,

of a sequence,

right and left,

Limit comparison test,

Line,

equations of a,

in space,

slope of a,

Line integral,

Linear coordinate system,

Linear differential equation of the

first order,

Logarithm, natural,

Logarithmic differentiation,

Logarithmic functions,

Lower limit of an integral,

Maclaurin series,

Major axis of an ellipse,

Minor axis of an ellipse,

Mass,

Maximum and minimum:

absolute,

relative,

Mean-Value theorem for derivatives,

Mean-Value theorem for integrals,

Midpoint formulas,

Midpoint rule for integrals,

Moments of inertia:

of arcs,

of planar mass,

of plane region,

of volume,

Monotonic sequence,

Motion under the influence of gravity,

Motion:

circular,

curvilinear,

rectilinear,

Natural logarithm,

Newton's law of cooling,

Newton's method,

Nondecreasing (nonincreasing) sequence,

Normal component of acceleration,

Normal line to a plane curve,

Normal line to a surface,

Normal plane to a space curve, ,

Octants,

One-to-one function,

Open disk,

Open set,

Open interval,

Ordinate,

Origin, ,

Osculating plane,

Pappus, theorem of,

Parabola,

focus, directrix, latus rectum, vertex,

Paraboloid:

elliptic,

hyperbolic,

Parallel lines, slopes of,

Parameter,

Parametric equations,

for surfaces,

Partial derivative,

higher-order,

Partial fractions,

Partial sums of a series,

Particular solution,

Perpendicular lines, slopes of,

Plane,

Point of inflection,

Point-slope equation of a line,

Polar axis,

Polar coordinates, ,

Polar equation,

Pole,

Position vector, ,

Positive series,

Positive x axis, y axis,

Power chain rule,

Power rule for derivatives,

Power series,

differentiation of,

integration of,

interval of convergence of,

radius of convergence of,

uniform convergence of,

Principal normal,

Product rule for derivatives,

Quadrants,

Quick formula I,

Quick formula II,

Quotient rule for derivatives,

Radian measure,

Radius of curvature,

Radius of convergence,

Radius vector,

Range of a function,

Rate of change,

Ratio test,

Rational function, ,

Rectangular coordinate system,

Rectifying plane,

Rectilinear motion,

Reduction formulas,

Related rates,

Relative extrema (maximum and minimum),

Remainder term,

Removable discontinuity,

Riemann integral,

Riemann sum,

Right-handed system,

Rolle's theorem,

Generalized,

Root test,

Rose with three petals,

Scalar product of vectors,

Scalars,

Secant function,

Second derivative,

Second derivative test,

Semimajor (semiminor) axis of an ellipse,

Separable differential equation,

Sequences,

bounded,

convergent and divergent,

limit of,

decreasing, increasing, nondecreasing, nonincreasing, monotonic,

Series, infinite,

absolutely convergent,

alternating,

binomial,

conditionally convergent,

convergent and divergent,

geometric,

harmonic,

Maclaurin,

partial sums of,

positive,

power,

remainder after _n_ terms of, Taylor,

sum of,

terms of,

Sigma notation,

Simpson's rule,

Sine,

Slicing formula,

Slope of a line,

Slopes:

of parallel lines,

of perpendicular lines,

Slope-intercept equation of a line,

Solid of revolution,

Space curve,

Space vectors,

Speed, ,

Sphere,

Spherical coordinates,

Squeeze theorem,

Standard equation of a circle,

Substitution method,

Sum of a series,

Sum rule for derivatives,

Surfaces, ,

Surface area,

Surface of revolution, ,

Symmetry, ,

axis of,

Tabular method for absolute extrema,

Tangent function,

Tangent line to a plane curve,

Tangent line to a space curve,

Tangent plane to a surface,

Tangential component of acceleration,

Taylor's formula with remainder,

Taylor series,

Terms of a series,

Third derivative,

Total differential,

Translation,

Transverse axis of a hyperbola,

Trapezoid rule,

Triangle inequality,

Trigonometric functions,

Trigonometric integrands,

Trigonometric substitutions,

Trigonometry review,

Triple integral,

Triple scalar product,

Triple vector product,

Uniform convergence,

Unit tangent vector,

Upper limit of an integral,

Vectors,

acceleration,

addition of,

components of,

cross product of,

direction cosines of,

dot product of,

equation of a line,

equation of a plane,

magnitude of,

plane,

position, ,

radius,

scalar product of,

scalar projection of,

space,

triple scalar product of,

triple vector product of,

unit,

unit tangent,

vector product of,

vector projection of,

velocity,

Vector functions:

curl of,

differentiation of,

divergence of,

Vector product,

Vector projections,

Velocity:

average,

angular,

in curvilinear motion,

in rectilinear motion,

initial,

instantaneous,

vector,

Vertex of a parabola,

Vertical asymptote,

Vertices:

of an ellipse,

of a hyperbola,

Volume:

given by an iterated integral,

of solids of revolution,

under a surface,

with area of cross section given,

Washer formula,

Work done by a force,

_x_ axis,

positive,

_x_ coordinate,

_y_ axis,

positive,

_y_ coordinate,

_y_ intercept,

Zeno's paradox,

Zero vector,

label: mathematics

title: Schaum's Outline of Calculus

authors: Frank Ayres