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Full text of "Differential calculus"

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WORKS OF H. B. PHILLIPS, PH.D. 

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JOHN WILEY & SONS, Inc. 



Differential Equations. Second Eklition, Rewritten. 

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,) ,oA^ ^^j^ J-^i^ >'? 



cyK 






k) 



IFFERENTIAL CALCULUS 



BY 



H. B. PHILLIPS, Ph.D. 

Associate Professor of Mathematics in the Massachusetts 
Institute of Technology 



NEW YORK 
JOHN WILEY & SONS, Inc. 

London: CHAPMAN & HALL, Lixitbd 






Copyright, 1916^ 

BY 

H. B. PHILLIPS 




79996^ 



Stanbope ipress 

K H. GILSON COMPANY 
BOSTON, U.S.A. 



PREFACE 



In this text on differential calculus I have continued the 
)lan adopted for my Analytic Geometry, wherein a few cen- 
ral methods are expounded and appUed to a large variety 
»f examples to the end that the student may learn principles 
nd gain power. In this way the differential calculus makes 
nly a brief text suitable for a term's work and leaves for the 
ntegral calculus, which in many respects is far more impor- 
ant, a greater proportion of time than is ordinarily devoted 
o it. 

As material for review and to provide problems for which 
nswers are not given, a supplementary list, containing about 
alf as many exercises as occur in the text, is placed at the 
nd of the book. 

I wish to acknowledge my indebtedness to Professor H. W. 
'yler and Professor E. B. Wilson for advice and criticism 
nd to Dr. Joseph Lipka for valuable assistance in preparing 
tie manuscript and revising the proof. 

H. B. PHILLIPS. 

Boston, Mass., August, 1916. 



CONTENTS 



^APTEB Pages 

I. Introduction 1- 9 

II. Derivative and Differential 10- 18 

III. Differentiation of Algebraic Functions 19- 31 

IV. Rates 32-38 

V. Maxima and Minima 39- 48 

VI. Differentiation of Transcendental Functions. ' 49- 62 

VII. Geometrical Applications 63- 84 

/'III. Velocity and Acceleration in a Curved Path . 85- 93 

IX. Rolle's Theorem and Indeterminate Forms 94-100 

X. Series and Approximations 101-112 

XI. Partl\l Differentiation 113-139 

Supplementary Exercises 140-153 

Answers 154-160 

Index 161-162 



DIFFERENTIAL CALCULUS 



CHAPTER I 
INTRODUCTION 

1. Definition of Function. — A quantity y is called a 
function of a quantity x if values of y are determined by values 
of X. 

Thus, U y = 1 — x^, y is a function of x; for a value of x^ 
determines a value of y. Similarly, the area of a circle is a 
function of its radius; for, the radius being given, the area is 
determined. 

It is not necessary that only one value of the function 
correspond to a value of the variable. Several values may be 
determined. Thus, if x and y satisfy the equation 
x^ — 2xy-\-y^ = x, 

then y is a function of x. To each value of x correspond two 
values of y found by solving the equation for y. 

A quantity u is called a function of several variables if u is 
determined when values are assigned to those variables. 

Thus, if z = x- + y^, then 2 is a function of x and y; for, 
values being given to x and y, a value of 2 is determined. 
Similarly, the volume of a cone is a function of its altitude 
and radius of base; for the radms and altitude being assigned, 
the volume is determined. 

2. Kinds of Functions. — An expression containing 
variables is called an explicit function of those variables. 
Thus Vx + y is an explicit function of x and y. Similarly, if 

y = Vx-^ 1, 
y is an explicit function of x. 



DIFFERENTIAL CALCULUS Chap. I. 

A quantity determined by an equation not solved for that 
quantity is called an implicit function. Thus, if 
x^ — 2 xy -\- y^ = X, 

\y is an implicit function of x. Also x is an implicit function 
of y. 

Explicit and implicit do not denote properties of the func- 
tion but of the way it is expressed. An implicit function is 
rendered explicit by solving. For example, the above equa- 
tion is equivalent to 

y = X ± Vx, 
in which y appears as an explicit function of x. 

A rational function is one representable by an algebraic 
expression containing no fractional powers of variable quanti- 
ties. For example, 

xV5 + 3 
x'^ -\-2x 
is a rational function of x. 

An irrational function is one represented by an algebraic 
expression which cannot be reduced to rational form. Thus 
V X + y is an irrational function of x and y. 

A function is called algebraic if it can be represented by an 
algebraic expression or is the solution of an algebraic equa- 
tion. All the functions previously mentioned are algebraic. 

Functions that are not algebraic are called transcendental. 
For example, sin x and log x are transcendental functions of x. 

3. Independent and Dependent Variables. — In most 
problems there occur a number of variable quantities con- 
nected by equations. Arbitrary values can be assigned to 
some of these quantities and the others are then determined. 
Those taking arbitrary values are called independent vari- 
ables; those determined are called dependent variables. 
Which variables are taken as independent and which as de- 
pendent is usually a matter of convenience. The number of 
independent variables is, however, determined by the equa- 
tions. 



Chap. I. INTRODUCTION 3 

For example, in plotting the curve 

y = 3^ + X, 

values are assigned to x and values of y are calculated. The 
independent variable is x and the dependent variable y. We 
might assign values to y and calculate values of x but that 
would be much more difficult. 

4. Notation. — A particular function of x is often repre- 
sented by the notation / (x) , which should be read, function 
of X, or / of X, not / times x. For example, 

fix) = V^TT 
means that/ (ar) is a symbol for Vx^ -\. i. Similarly, 

y=f(x) 

means that y is some definite (though perhaps unknown) 
function of x. 

If it is necessarj' to consider several functions in the same 
discussion, they are distinguished by subscripts or accents or 
by the use of different letters. Thus, /i (x), f-z (x), /' (x), 
/" (x) , g (x) (read /-one of x, /-two of x, /-prime of x, /-second 
of X, g of x) represent (presumably) different functions of x. 

Functions of several variables are expressed by WTiting 
commas between the variables. For example, 

v=f{r,h) 

expresses that t; is a function of r and h and 

V = f{a, 6, c) 

expresses that y is a function of a, 6, c 

The / in the symbol of a function should be considered as 
representing an operation to be performed on the variable or 
variables. Thus, if 

f(x) = V^^~+l, 

f represents the operation of squaring the variable, adding 1, 
and extracting the square root of the result. If x is replaced 



4 DIFFERENTIAL CALCULUS Chap. I. 

by any other quantity, the same operation is to be performed 
on that quantity. For example, 

/ (2) = V22+ 1 = Vs. 
f{y+l) = V(y +1)2+1 = \/i/2 + 2y + 2. 
Similarly, if 

/ {x, y) = x^ + xy - y\ 
then / (1, 2) = 12 + 1 . 2 - 22 = -1. 

If / (x, y, 0) = a:2 + i/2 + ^^ 

then f (2, -3, 1) = 22 + (-3)2 + 1 = 14. 

EXERCISES 

3 3 3 

1. Given x + y = a , express y as an explicit function of x. 

2. Given logio (x) = sin y, express x as an explicit function of y. 
Also express y as an explicit function of x. 

3. If / (x) = x2 - 3 X + 2, show that / (1) = / (2) = 0. 

4. If F (x) = x* + 2 x2 + 3, show that F (-a) = F (a). 

5. li F (x) = x+ -, find F {x + 1). Also find F (x) + 1. 

6. If <!> (x) = Vx2 - 1, find <> (2 X). Also find 2 </. (x). 

7. If^W=2T^3,find^(^). Also find ^. 

8. If /i (x) = 2% /2 (X) = xS find /i !/2 (y)]. Also find /^ [/i (y)]. 

9. If / (x, ?/) = X - -, show that/ (2, 1) = 2/ (1, 2) = 1. 

10. Given / (x, y) = x^ + xy, find / (y, x). 

11. On how many independent variables docs the volume of a right 
circular cylinder depend? 

12. Three numbers x, y, z satisfy two equations 

X2 + ?/2 + 22 = 5, 

X ■\- y -\rz ==1. 
How many of these numbers can be taken as independent variables? j 

6. Limit. — If in any process a variable quantity ap- 
proaches a constant one in such a way that the difference of 
the two becomes and remains as small as you please, the con- 
stant is said to be the limit of the variable. 

The use of limits is well illustrated by the incommensurable 



Chap. I. INTRODUCTION 5 

cases of geometry and the determination of the area of a 
circle or the volume of a cone or sphere. 

6. Limit of a Function. — As a variable approaches a 
limit a function of that variable may approach a limit. Thus, 
as X approaches 1, a:^ + 1 approacnes 2. 

We shall express that a variable x approaches a limit a by 
the notation 

X = a. 

The symbol = thus means "approaches as a limit." 

Let / {x) approach the limit A as x approaches a; this is 
expressed by 

lim/(x) = A, 

which should be read, " the limit of / (x), as x approaches a, 
is A." 
Example 1. Find the value of 

lim 



sHi)- 



As X approaches 1, the quantity x + - approaches 1 + t 



or 2, Hence 



1s(^+i) = 2. 



Ex. 2. Find the value of 

sin 6 



lim 



;e=0 1 + COS d 
As 9 approaches zero, the function given approaches 

Hence 

,. sin d _ 

hm :r— = 0. 

9=0 1 + cos 

7. Properties of Limits. — In finding the limits of func- 
tions frequent use is made of certain simple properties that 
follow almost immediately from the definition. 



6 DIFFERENTIAL CALCULUS Chap. I. 

1. The limit of the sum of a finite number of functions is 
equal to the sum of their limits. 

Suppose, for example, X, Y, Z are three functions ap- 
proaching the limits A, B, C respectively. Then X-\-Y-{-Z 
is approaching A -{- B -\- C. Consequently, 

lim (X + F + Z) = ^ + 5 + C = lim Z + lim 7 + lim Z. 

2. The limit of the 'product of a finite number of functions 
is equal to the product of their limits. 

If, for example, X, Y, Z approach A, B, C respectively, 
then XYZ approaches ABC, that is, 

lim XYZ = ABC = lim X lim Y lim Z. 

3. // the limit of the denominator is not zero, the limit of the 
ratio of two functions is equal to the ratio of their limits. 

Let X, Y approach the limits A, B and suppose B is not 

X A 

zero. Then y approaches ^ , that is, 

,. X A limX 
^''^Y=B=V;^' 

A 
If B is zero and A is not zero, t, will be infinite. Then 

ts 

X A X 

=7 cannot approach ^ ^sa limit; for, however large ^ may 

become, the difference of t^ and infinity will not become small. 
8. The Form ^. — When x is replaced by a particular 

value, a function sometimes takes the form r- Although this 

symbol does not represent a definite value, the function may 
have a definite limit. This is usually made evident by writ- 
ing the function in a different form. 
Example 1. Find the value of 

,. x'-l 
lim r 



Chap. I. INTRODUCTION 7 

When X is replaced by 1, the function takes the form 
1-1 
1-10* 
Since, however, 

a^-1 , , 

^^=^ + 1' 

the function approaches 1 + 1 or 2. Therefore 

3:2—1 

Hm^ ^ = 2. 

1=5=1 a: — 1 

Ex. 2. Find the value of 

,. (VTTx - 1) 
lim -• 

x=0 X 

When X = the given function becomes 
1-1 ^0 

o" 

Multiplying numerator and denominator by Vl -f a: + 1, 
Vl + x- 1 ^ X 1 

X ~ X (Vl+x + 1) ~ Vl+x + 1 * 

As X approaches 0, the last expression approaches ^. Hence 

9. Infinitesimal, -r- A variable approaching zero as a 
limit is called an infiriiiesimal. 

Let a and /3 be two infinitesimals. If 

lim- 

is finite and not zero, a and /3 are said to be infinitesimals of 
the same order. If the limit is zero, a is of higher order than 

/3. If the ratio - approaches infinity, /3 is of higher order 

than a. Roughly speaking, the higher the order, the smaller 
the infinitesimal. 



8 DIFFERENTIAL CALCULUS Chap. I. 

For example, let x approach zero. The quantities 

X/>«2 /y^ /y^ Ck^ n 

are infinitesimals arranged in ascending order. Thus x* is of 
higher order than x^; for 

x^ 
lim — = lim x^ = 0. 

x=0X^ 1=0 

Similarly, a^ is of lower order than a^, since 

X* X 

approaches infinity when x approaches zero. 

As X approaches x , cos x and cot x are infinitesimals of the 

same order; for 

,. cos a; ,. . 
Iim — 7 — = lim sm x = 1, 
.,cota; i-fl^ 

which is finite and not zero. 

EXERCISES 
Find the values of the following limits: 

... a;*- 2 a; + 3 ... Vl - x» - Vl + x' 

1. hm ^ 4. Iim z ■ 

x=Q X — 5 x=o X* 

_ ,. sin 9 4- cosd 

, sin 2 + cos 2 9 6. hm ^-^• 

6=2 edzoi&n9 

^ ,. x^ - 3x + 2 - ,. sin » 

3. hm r-! 6. lim 



xAi X — 1 9=0 sin 29 

7. By the use of a table of natural sines find the value of ! 

,. sin X 

lim 

x=o a; 

8. Define as a limit the area within a closed curve. 

9. Define as a limit the volume within a closed surface. 

10. Define V2. 

11. On the segment PQ (Fig. 9a) construct a series of equilateral tri- 
angles reaching from P to Q. As the number of triangles is increased, 



Chap. I. 



INTRODUCTION 



their bases approaching zero, the polygonal line PABC, etc., approaches 
PQ. Does its length approach that of PQl 



\ / 



Fig. 9a. 

12. Inscribe a series of cylinders in a cone 
as shown in Fig. 9b. As the number of cyl- 
inders increases indefinitely, their altitudes 
approaching zero, does the sum of the vol- 
umes of the cylinders approach that of the 
cone? Does the sum of the lateral areas of 
the cylinders approach the lateral area of the 
cone? 




Fig. 9b. 



1 



13. Show that when x approaches zero, tan - does not approach a 
limit. 



14. As X approaches 1, which of the infinitesimals 1 — a; and Vl — x 
is of higher order? V 

16. As the radius of a sphere approaches zero, show that its volume 
is an infinitesimal of higher order than the area of its surface and of the 
same order as the volume of the circumscribing cylinder. 



CHAPTER II 
DERIVATIVE AND DIFFERENTIAL 

10. Increment. — When a variable changes value, the 
algebraic increase (new value minus old) is called its in- 
crement and is represented by the symbol A written before 
the variable. 

Thus, if X changes from 2 to 4, its increment is 

Ax = 4 - 2 = 2. 

If X changes from 2 to —1, 

Ax = -1-2= -3. 

When the increment is positive there is an increase in 
value, when negative a decrease. 

Let 2/ be a function of x. When x receives an increment 

Ax, an increment A?/ will be 
determined. The increments 
of X and y thus correspond. 
To illustrate this graphically 
let X and y be the rectangular 
coordinates of a point P. An 
equation 

y=f{x) 
represents a curve. When x 
changes, the point P changes 
to some other position Q on the curve. The increments of 
X and y are 

Ax = PR, Ay = RQ. (10) 

11. Continuous Function. — A function is called con^ 
tinuous if the increment of the function approaches zero as 
the increment of the variable approaches zero. 

10 




Chap. II. 



DERIVATIVE AND DIFFERENTIAL 



11 



In Fig. 10, y is a continuous function of x; for, as Ax 
approaches zero, Q approaches P and so Ay approaches zero. 

In Figs. 11a and lib are shown two ways that a function 
can be discontinuous. In Fig. 11a the curve has a break at 




T 


I 


X 





/ 


if 



Fig. 11a. 



Fig. lib. 



P. As Q approaches P', Ax = PR approaches zero, but 
Ay = RQ does not. In Fig. lib the ordinate at x = a is 
infinite. The increment Ay occurring in the change from 
a: = a to any neighboring value is infinite. 

12. Slope of a Curve. — As Q moves along a continuous 
curve toward P, the line PQ 



turns about P and usually 


approaches a limiting posi- 


tion PT. This line PT is 


called the tangent to the 


curve at P. 


The slope of PQ is 


RQ Ay 


PR Ax' 



As Q approaches P, Ax ap- 
proaches zero and the slope 
of PQ approaches that of PT. 




Fig. 12a. 



Therefore 



Slope of the tangent = tan = lim -r^ • (12) 



12 



DIFFERENTIAL CALCULUS 



Chap. II. 



The slope of the tangent at P is called the slope of the curve 
at P. 

Exam-pie. Find the 
slope of the parabola 
y = x'^ aX the point (1, 1). 

Let the coordinates of 
P \yQ X, y. Those of 
Q are x -\- Aa:, y + Ly, 
Since P and Q are both 
on the curve, 

= 'r2 




y =x'^ 



and 



?/ + Ay = (a; + Aa;)2 = 
a^2 + 2 a; Ax + (Ax)2. 



Subtracting these equations, we get 

^y = 2x^x-\- (Ax)2. 

Dividing by Ax, 

A?/ 



Ax 



= 2 X + Ax. 



As Ax approaches zero, this approaches 
Slope at P = 2 X. 

This is the slope at the point with abscissa x. The slope at 
(1, 1) is then 2-1 = 2. 



13. Derivative. — Let 2/ be a function of x. If 



^y 
Ax 



approaches a limit as Ax approaches zero, that limit is called 
the derivative of y with respect to x. It is represented by the 
notation D^y, that is, 

A-. 

(13a) 



D^y = lim -^• 

Ax=0 Ax 



If a function is represented by /(x), its derivative with 
respect to x is often represented by/' (x). Thus 

A/(x) 



/' (x) = lim 

Az=0 



Ax 



= D^(x). 



(13b) 



Chap. II. DERIVATIVE AXD DIFFERENTIAL 



13 




Fig. 13. 



In Art. 12 we found that this Umit represents the slope of 
the curve y = Six). The derivative is, in fact, a function 
of X whose value is the slope of 
the curve at the point with ab- 
scissa X. 

The derivative, being the limit 

Aw 
of ^, is approximately equal 

to a small change in y divided 
by the corresponding small 
change in x. It is then large or 
small according as the small in- 
crement of y is large or small in 
comparison with that of x. 
If small increments of x and 

Aw 
y have the same sign -r^ and 
Ax 

its limit Dxy are positive. If they have opposite signs D^y 

is negative. Therefore D^y is positive when x and y increase 

and decrease together and negative when one increases as the 

other decreases. 

Example. y = 3? — Zx-\-2. 

Let X receive an increment Ax, The new value of x is 
x + Ax. The new value of j/ is y -f Ay. Since these satisfy 
the equation, 

2/ + Ay = (x + Ax)3 - 3 (x + Ax) + 2. 

Subtracting the equation 

y = x3-3x-t-2, 
we get 

Ay = 3 x2 Ax + 3 X (Ax)^ -J- (Ax)' - 3 Ax. 

Dividing by Ax, 

Aw 

^=3x2 + 3xAx4- (Ax)2 - 3. 

As Ax approaches zero this approaches the limit 
D^y = 3 x2 - 3. 



14 DIFFERENTIAL CALCULUS Chap. II. 

The graph is shown in Fig. 13. At A (where x = \) y = 
and DxV = 3 • 1 — 3 = 0. The curve is thus tangent to the 
a;-axis at A. The slope is also zero at B (where x = —1). 
This is the highest point on the arc AC. On the right of A 
and on the left of B, the slope DxV is positive and x and y in- 
crease and decrease together. Between A and B the slope 
is negative and y decreases as x increases. 

EXERCISES 

1. Given y = Vx, find the increment of y when x changes from 
X = 2 to X = 1.9. Show that the increments approximately satisfy 
the equation 

Ay ^ 1 
Ax 2 Vx 

2. Given y = logio x, find the increments of y when x changes from 
50 to 51 and from 100 to 101. Show that the second increment is ap- 
proximately half the first. 

3. The equation of a certain line is y = 2 x + Z. Find its slope by 

Au 
calculating the limit of — • .- — ^ 

4. Construct the parabola y = x^ — 2 x. Show that its slope at 
the point with abscissa x is 2 (x — 1). Find its slope at (4, 8). At 
what point is the slope equal to 2? 

6. Construct the curve represented by the equation y = x* — 2 x*. 
Show that its slope at the point with abscissa x is 4 x (x" — 1). At what 
points are the tangents parallel to the x-axis? Indicate where the slope 
is positive and where negative. 

In each of the following exercises show that the derivative has the 
value given. Also find the slope of the corresponding curve at x = — 1. 

6. y = (x + 1) (x + 2), Dx2/ = 2 X + 3. 

7. y = X*, Dzy = 4x'. 

8. 2/ = x» - x2, Dj^y = 3x^ -2 x. 

9. t/ = -. ^'y=-^^- 

10. If X is an acute angle, is Dx cos x positive or negative? 

11. For what angles is Dx sin x positive and for what angles negative? 

14. Approximate Value of the Increment of a Function. — 
Let y be a function of x and represent by e a quantity such 
that ^y 



Chap. II. DERIVATIVE AND DIFFERENTIAL 15 

As Ax approaches zero, -r^ approaches D^y and so « ap- 
proaches zero. 
The increment of y is 

Ay = Bxy Ax -f cAx. 
The part 

D,y Ax (14) 

is called the -principal part of Ay. It differs from Ay by an 
amount cAx. As Ax approaches zero, e approaches zero, and 
so eAx becomes an indefinitely small fraction of Ax. It is an 
infinitesimal of higher order than Ax. If then the principal 
part is used as an approximation for Ay, the error will be 
only a small fraction of Ax when Ax is sufficiently small. 

Example. When x changes from 2 to 2.1 find an approxi- 
mate value for the change in y = -• 

X 

In exercise 9, page 14, the derivative of - was found to 

X 

be 1. Hence the principal part of Ay is 

-Aax= -j(.l) = -0.0250. 

-J.2 4 

The exact increment is 

^^ = (21)5 - ^^ = -0-«232. 

The principal part represents Ay with an error less than 002 
which is 2% of Ax. 

15. Differentials. — Let x be the independent variable 
and let y be a function of x. The principal part of Ay is 
called the differential of y and is denoted by dy; that is, 

dy = D^y Ax. (15a) 

This equation defines the differential of any ftmction y of x. 
In particular, if y = x, Dzy = 1, and so 

dx = Ax, (15b) 

that is, the differential of the independent variable is equal to 



16 



DIFFERENTIAL CALCULUS 



Chap. II. 



its increment and the differential of any function y is equal to 
the product of its derivative and the increment of the independent 
variable. 

Combining 15a and 15b, we get 



whence 



dy = Dxy dx, 
dy 



dx 



= D^y, 



(15c) 
(15d) 



dy 



that is, the quotient ^ is equnl to the derivative of y with respect 

to X. 

Since D^y is the slope of the curve y =f(x), equations 15b 

and 15c express that dy and dx are the sides of the right tri- 
angle PRT (Fig. 15) with 
hypotenuse PT extending 
along the tangent at P. 
On this diagram, Ax and Ay 
are the increments 

Ax = PR, Ay = RQ, 
occurring in the change 
from P to Q. The differen- 
tials are 

dx = PR, dy = RT. 
A point describing the 
curve is moving when it 
passes through P in the direction of the tangent PT. The 
differential dy is then the amount y would increase when x 
changes to x + Ax if the direction of motion did not change. 
In general the direction of motion does change and so the 
actual increase Ay = RQ is different from dy. If the in- 
crements are small the change in direction will be small and 
so Ay and dy will be approximately equal. 

Equation 15c was obtained under the assumption that x 
was the independent variable. It is still valid if x and y are 
continuous functions of an independent variable t. For then 
dx = DtX At, dy = Dty At. 



T 




y 




X 




P^ 


J 

^ dx 


Ay 






^/[H 


-Ax---> 


R 













x! 



Fig. 15. 



Chap. II. DERIVATIVE AND DIFFERENTIAL 17 

The identity 



Ay _ Ay Ax 
At Ax ' At 
gives in the limit 

Dty = D^y • DtX. 
Hence 

Dty At = D^y • Dtx At, 
that is, 

dy = D^y dx. 




X 4- I 

Example 1. Given y = , find dy. 




In this case 




,, + Ax+l x+1 


Ax 


^ x + Ax X X 


(x + Ax) 


Consequently, 

Ay 1 




Ax X (x + Ax) 




As Ax approaches zero, this approaches 




^= -i. 

dx 3?' 
Therefore 

, dx 




Ex. 2. Given x = f*, y = <3, find ^• 





The differentials of x and y are found to be 
dx = 2tdl, dy = '^f^dt. 
Division then gives, 

dx~r" 

Ex. 3. An error of 1% is made in measuring the side of a 
square. Find approximately the error in the calculated area. 

Let X be the correct measure of the side and x + Ax the 
value found by measurement. Then dx = Ax = ±0.01 x. 



18 DIFFERENTIAL CALCULUS Chap. II, 

The error in the area is approximately 

dA =d (a;2) =2xdx= ±0.02 x^ = ±0.02 A, 
which is 2% of the area. 

EXERCISES 

1. Let n be a positive integer and y = x". Expand 

Ay = {x + Ax)" — x" 
by using the binomial theorem. Show that 

-^ = nx"-». 
ax 

What is the principal part of Ay? 

2. Using the results of Ex. 1, find an approximate value for the in- 
crement of x« when x changes from 1.1 to 1.2. Express the error as a 
percentage of Ax. 

dA 

3. If A is the area of a circle of radius r, show that -y- is equal to the 

ar 
■circumference. 

4. If the radius of a circle is measured and its area calculated by 
using the result, show that an error of 1% in the measurement of the 
radius will lead to an error of about 2% in the area. 

5. If V is the volume of a sphere with radius r, show that -p is equal 

ar 
to the area of its surface. 

6. Let V be the volume of a cylinder with radius r and altitude h. 
Show that if r is constant -rr is equal to the area of the base of cylinder 

and if A is constant -j- is equal to the lateral area. 

7. li y — f (x) and for all variations in x, dx = Ax, dy = Ay, show 
that the graph of y = / (x) is a straight line. 

8. If y is the independent variable and x = f (y), make a diagram 
showing dx, dy, Ax, and Ay. 

9. If the y-axis is vertical, the x-axis horizontal, a body thrown hori- 
zontally from the origin with a velocity of 50 ft. per second will in / 
seconds reach the point ' 

X = 50t, y = -16 fi. I 

Find the slope of its path at that point. 

10. A line turning about a fixed point P intersects the x-axis at A 
and the y-axis at B. If Ki and K2 are the areas of the triangles OP A (and 
OPB, show that 

dKi^PA^ ■ 

dK, P&' 



CHAPTER III 

DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 

16. The process of finding derivatives and differentials 
is called differentiation. Instead of applying the direct 
method of the last chapter, differentiation is usually per- 
formed by means of certain formulas derived by that method. 
In this work we use the letter d for the operation of taking the 

differential and the symbol -r- for the operation of taking 

the derivative with respect to x. Thus 

d {u -\- v) = differential of {u-\-v), 

-j-{u -j- v) = derivative of (m + v) with respect to x. 

To obtain the derivative with respect to x we proceed as in 
finding the differential except that d is everywhere replaced 

^^dx' 
^^ 17. Formulas. — Let u, v, w be continuous functions of a 
single variable x, and c, n constants.* 

I. dc = 0. 

n. d {u + v) = du + dv. 
in. d {cu) = c du. 

IV. d (vv) = udv -\- V du. 

du — u dv 



VI. d (m") = nw'*-^ du. 



* It is assumed that the functions w, v, w have derivatives. There 
exist continuous functions, 

" = / (x), 

19 



20 DIFFERENTIAL CALCULUS Chap. III. 

18. Proof of I. — The differential of a constant is zero. 

When a variable x takes an increment Ax, a constant does 

Ac 
not vary. Consequently, Ac = 0, ^ = 0, and in the limit 

dc 

~r- == 0. Clearing of fractions, 

dc = dx ' = 0. 

19. Proof of- n. — The differential of the sum of a finite 
number of functions is equal to the sum of their differentials. 

Let 

y = u-\- V. 

When X takes an increment Ax, u will change to m + Aw, v 
to y + Ay, and yioy -]- Ay. Consequently 

y + Ay = u-i- Au-i-v + Av. 

Subtraction of the two equations gives 

Ay = Au + Av, 



whence 



Ay _ Au Av 
Ax ~ Ax Ax 



As A^ approaches zero, ^. ^, ^^ approach |, |, ^° 

respectively. Therefore 

dy _du dv 
dx dx dx' 
and so 

dy = du-\- dv. 

By the same method we can prove 

I d(w±y±iy± • • ' ) = du^dv zLdw ± • ••, 

audi that 

Au 
I Ax 

does not approach a limit as Ax approaches zero. Such a function has 
no derivative DxU and therefore no differential 

du = Dzudx. 



Chap. III. ALGEBRAIC FUNCTIONS 21 

20. Proof of in. — The differential of a constant times a 
function is equal to the constant times the differential of the 
function. 

Let y = cu. H*J^^ ^'^ 

Then y + Ay = c (m + Am) m +z\ij -- ^^x ♦ Ax) (. 

and so Ay = c Am, ^'j^ fC^^^^)-fC^l 

Ay Am AV- /Yil^i^M^ 

Ai = 'Ai- ^^ ^^ 

As Ax approaches zero, -r-^ and c -r- approach 3^ and c -t- • 



Therefore ofif. /.W, f^jc+Ax)- 

^^c— ^^ A^^c' — ^^ 

dx dx* 

whence 

dy = c du. 

Fractions with a constant denominator should be differen- 
tiated by this formula. Thus 



^(c)=^(^)=^'^^- 



21. Proof of IV. — The differential of the product of two 
functions is equal to the first times the differential of the second 
plus the second times the differential of the first. 

Let y = uv. 

Then y + Ay = (m + Am) (y + Aw) 

= uv -{- V Am + (m + Am) Ap. 

Subtraction ^ves 

Ay = f Am + (m + Am) Ar, 



whence 



Ay Am , , , . . Ap 



Since m is a continuous function, Am approaches zero as Ax 
approaches zero. Therefore, in the limit, 

dy _ du, dv 
dx dx dx' 



22 DIFFERENTIAL CALCULUS Chap. III. 

and so 

dy = V du-\- u dv. 

In the same way we can show that 

d (uvw) = uv dw -\- uw dv + vw du. 

22. Proof of V. — The differential of a fraction is equal to 
the denominator times the differential of ike numerator minus 
the numerator times the differential of the denominator ^ all 
divided hy the square of the denominator. 

Let 



u 



Then 
and 



Ly = 



u + Am u V Am — m Ay 



V + Ay V V {v -\- Ay) 
Dividing by Ax, 

. Am Ay 

Am y-T M-r- 

-r^ = Ax Ax 

Ax — -, 1 — 7-T"* 

V {v -\- Av) 

Since y is a continuous function of x, Av approaches zero as Ax 
approaches zero. Therefore 

du dv 

dy _ dx dx 

dx v^ 

whence 

, V du — udv 

dy = — "— i 

23. Proof of VI. — The differential of a variable raised to a 
constant power is equal to the product of the exponent, the variable 
raised to a power one less, and the differential of the variable. 

We consider three cases depending on whether the exponent 
is a positive whole number, a positive fraction, or a negative 
number. For the case of irrational exponent, see Ex. 25, 
page 6J . 



Chap. III. AL^tEBRAIC FUNCTIONS 2^ 

(1) Let n be a positive, integer and y = w". Then 

nin — 1) 



and 



Ay = nw"-» Am + ^ ^^, ^^ W-^ (Am)^ + 



= nM"~^ 



Dividing by Au, 

Aw , , n (n — 1) , /* X , 
-r^ = nu"-^ H ^ — - M"-2 (Am) + 

As Am approaches zero, this approaches 

dy 

du 

Consequently, 

dy = nti"~^ du, 

V 

TO ~ 

(2) Let n be a positive fraction - and y = u''= u^. Then 

y« = M". 

Since p and g are both positive integers, we can differentiate 
both sides of this equation by the formula just proved^ 
Therefore 

gy«~^ dy = pw^"* du. 
p 
Solving for dy and substituting m' for y, we get 

P-i f"^ 

dy = ^- — — du = - u du = nM"~^ du. 



qu « 
(3) Let n be a negative number —m. Then 

y = M" = M""* = 

^ M" 

Since 7?i is positive, we can find d (u") by the formulas proved 
above. Therefore, by V, 

J M"'d(l) — 1 d(M'") —mu'"-^du , , _, , 

dy = ' ., — - = ^- • =v-«fnM~"*~^aM = nu'^^ du^ 




24 DIFFERENTIAL CALCULUS Chap. III. 

Therefore, whether n is an integer or fraction, positive or 
negative, 

d (w") = nw"~^ du. 

If the numerator of a fraction is constant, this formula can be 
used instead of V. Thus 



d{~\= d {cu~^) = —cu~^ du. 



Example 1. y = Aa^. 
Using formulas III and VI, 

dy = Ad{x?) = 4 • 3 a;2 dx = 12 a:2 dx. 

Ex.2, y = Vx-\--^+3. 
Vx 

This can be written 

y = x^ -}- x~^ + 3. 

Consequently, by II and VI, 

dy ^d (x^) ^ d (x'h ^ d (3) 
dx dx dx dx 

1 -idx 1 -sdx , ^ 

2 dx 2 dx^ 
1 1 



2V^ 2V^ 
Ex. 3. y=(x + a) (x^-b^). 
Using IV, with u = x + a, v = x^ — h^, 

= (x + a) (2x-0) + (x2-62) (1+0) 
= 3 x2 + 2 ax - 62. 

Ex.^.y = ^-^' 



Chap. III. ALGEBRAIC FUNCTIONS 25 

Using V, with w = a^ + 1, y = x- — 1, 

(x» - 1) d (x2 + 1) - (x* 4- 1) d (x* - 1) 



dy = 



(X^ _ 1)2 

(x^- l)2xdx- (x' + l)2xcte 
(x2 - 1)2 
4xdx 



Ex. 5. t/ = Vx2- 1. 
Using VI, with w = x^ — 1, 



= i(x«-l)-i(2x) = 



X 



Vx2-1 
Ex. 6. x^ + xy — y^ = 1. 

We can consider y a function of x determined by the equa- 
tion. Then 

d (x2) + d (xy) - d (y') = d (1) = 0, 
that is, 

2 X dx -\- X dy -^ y dx — 2 y dy = 0, 
{2x-\-y)dx + ix-2y)dy = 0. 
Consequently, 

dy ^ 2x + y 
dx 2y — X 

Ex.7. x = t-{-j, y = t--. 

In this case 

dx = dt-rp, dy = dt-{--^* 

Consequently, 

dy ^ ^ t^ ^ f-hl 
dx 1 ^ - 1 ' 



Ex. 8. Find an approximate value of y = ( 1 when 



26 DIFFERENTIAL CALCULUS Cbap. III. 

When X = 0, y = 1. Also 

, 2dx 

dy = - 



3 {I - x)i (1 + x)^ 
When X = this becomes 

dy = —I dx. 

If we assume that dy is approximately equal to Ay, the change 
in y when x changes from to 0.2 is approximately 
dy = -f (0.2) = -0.13. 

The required value is then 

y = l- 0.13 = .87. 

EXERCISES 

In the following exercises show that the differentials and derivatives 
have the values given: 

l/t^y = 3x* + 4x^ - 6x^ + 5, di ■= 12 {x^ + x^ - x) dx. 
x3 - x2 + 1 dy 3 x^ - 2 X 




^ ~ 5 ' dx 5 

4. ?/ = (x + 2 a) (x - a)2, dy = Z (x^ - a«) dx. 

5. 1/ = X (2 X - 1) (3 X + 2), ^^ = 18 x^ + 2 X - 2. 

1 , -2xdx 

2 X + 3 -22 (to rf 1 V ^ITi-g 

d 1 - 2x _ 2x d g' - 2 s'' 

^' dx (X - 1)2 - (X - 1)'* ^^' ds « ^«' - «' = v^rrr^i- 

ax \x / x^ Va^ - x* 

14 ± d't^ = ^^^7=- 16. ^xY = 2xi/^ + 2x-^j/^. 

^ ' dxVx^+1 i£^ + l)Vx*-l dx " ^ "dx 

C2+3x«)S 20 , . xdxi-'ydu 

16. d^^^^ = - :^ (2+3x<')Sdx. 17. d Vx^ + r/^ = W^FP^' 



Chap. III. ALGEBRAIC FUNCTIONS 27 

18. y = (i+l)(2-3i)»(2x-3)', 

^ = (24 + 13x -36i«) (2- 3z) (2x - 3)». 
ax 

_ a:* dy _ mcuf^~^ 

19- y- m» ^- n. • 

(a + fti")" (a + 6x»)«+* 

* 3z* ' dx x4 Vx* + 1 

(x + VTT70'^^ , (x + vrr^)"-' 
^^ ^ = — ^+1 — + — -^r^n — ' 

dj/ =2(i + v^l +xO"(ic- 



22. X* + y* = a*, 



24. 2x2-3xy + 4t/» = 3x, 



ax 
dy _ 4x - 3y - 3 

dx ~ 3x -8y 



25. - + ?^ = 1, ydx-xdy = 0. 
y X ' 

1 dx do " 

26. y=-, + , = 

27. i/=" + x*"*/" = x**", mydx = nx dy. 

t 2t + S dy , 

28. x=^^, y=-^— ^^^ = 0. 

29. x=<-Vi2-l, y = <+V<2-l, xdy + ydx=0. 

- 3a< ^ 3a<^ dy ^ 2t -t* 

1 + Z'' ^ ~ 1 4- i'' dx ~ 1 - 2t^' 

X 

31. Given y = / ^ „ , find an approximate value for y when 

X = 4.2. 

32. Find an approximate value of 



\/l 



x + 1 



x^ + x + 1 
when X = .3. 

33. Given y = a^, find dy and At/ when x changes from 3 to 3.1. Is 
d(/ a satisfactory approximation for Ay? Express the difference as a 
percentage of Ay. 

34. Find the slope of the curve 

J/ = X (x» + 31)^ 
at the point x = 1. 

35. Find the points on the parabola y^ = 4 ax where the tangent is 
inclined at an angle of 45° to the x-axis. 



28 DIFFERENTIAL CALCULUS Chap. III. 

. 36. Given y = (a + a?) Va — x, for what values of x does y increase 
as X increases and for what values does y decrease as x increases? 

37. Find the points P (x, y) on the curve 

,1 

where the tangent is perpendicular to the line joining P to the origin. 

38. Find the angle at which the circle 

x2 + 2/2 = 2x-32/ 
intersects the x-axis at the origin. 

39. A line through the point (1,2) cuts the z-a^ at (x, 0) and the 
2,-axis at (0, y). Find g- " ^ -- ^^ ( X -^O 

40. If x'* — X + 2 = 0, why is the equation 

not satisfied? ^ 

41. The distances x, x' of a point and its image from a lens are con- 
nected bv the equation 

x^x' /' 
/ being constant. If L is the length of a small object extending along 
the axis perpendicular to the lens and L' is the length of its image, show 
that 

L \x] 
approximately, x and x' being the distances of the object and its image 
from the lens. 

24. Higher Derivatives. — The first derivative -i^ is a 

function of x. Its derivative with respect to x, wntten -r-^ » 

is called the second derivative of y with respect to x. That 
is, 

dx^ dx \dx) 
Similarly, 

ty = ±('^\ 

d3? dx W/ 






Chap. III. ALGEBRAIC FUNCTIONS 29 

The derivatives of / (x) with respect to x are often written 
/' (x), /" (x), r (x), etc. Thus, if y = / (x). 

|=n.,g=r(x). g=r'(x).etc. 

Example 1. y = x^. 

Differentiation with respect to x gives 

S = #(6) = 0. 

All higher derivatives are zero. 
Ex. 2. x2 + xy + y^ = 1. 
Differentiating with respect to x, 

whence 

dy ^ 2x + y 
rfx X + 2 y 

The second derivative is 

^J ^ _± ( 2x + y \ ^^Tx' ^y 
dx^ dx\x + 2y) (x + 2 y)^ 

dy 
Replacing -^ by its value in terms of x and y and reducing, 

cPy 6 (x2 + xy + y2) 6 



dx2 (x + 2y)3 (x + 2y)' 

The last expression is obtained by using the equation of the 
curve x^ + xy + y2 = i gy differentiating this second 
derivative we could find the third derivative, etc. 



30 DIFFERENTIAL CALCULUS Chap. IH. 

25. Change of Variable. — We have represented the 
second derivative by -— . This can be regarded as the quo- 
tient obtained by dividing a second differential 

d^y = d (dy) 

by (dxy. The value of d^y will however depend on the vari- 
able with respect to which y is differentiated. 



Thus, suppose y = x"^, x = t^. Then -j^= ^ and so 



dx^ 



d'y = 2 (dxY = 2 (3 f2 dty = 18 f^ {dt)\ 
If, however, we differentiate with respect to t, since y = i^, 
g=30f^and 

d'y = 30 t^ (dty, 

which is not equal to the value obtained when we differen- 
tiated y with respect to x. 

For this reason we shall not use differentials of the second 
or higher orders except in the numerators of derivatives. 

Two derivatives like --t^ and ■— must not be combined like 

fractions because cPy does not have the same value in the two 
cases. 

If we have derivatives with respect to t and wish to find 
derivatives with respect to x, they can be found by using the 
identical relation 

du 
d _ du dt_ _ dt ,nK\ 

dx dt dx dx 

dt 

For example, 

<Py 

d_ /dy\ ^ d_ /dy\ di^^de_^ 

dx\dtj dt\dt) dx dx^ 

dt 



Chap. IV. 


ALGEBRAIC FUNCTIONS 


approxima _. ^1 . , 1 c , 
, . , e. Given x = t — - , y = t-\--:, find 
velocity a t ^ t 


case 

/ 

This e 

would m 

A^ a rule ,, 
>. . ^ luently, 


dy ^ t^ t^-l 

dx ,, , dt f -\-l 
dt + j. 


dl'y _ d (t^-\\ 


At dt At 1 



31 



dry 
dx^' 



4<» 



dx2 dx \t^ + 1/ (f + 1)2 dx {p + 1)' . , 1 {^ + If 

EXERCISES 
Find ^ and -r4, in each of the followong exercises: 

1. y = _ • 6. x^ + y^ = a^ 

2. J/ = V^"ir^. 6. x2 - 2 2/! = 1. 

3. y = {x - ly (x + 2)*. 7. xy = 2 +y. 

4. j/! = 4 X. 8. x' + y' = a'. 
9. If a and 6 are constant and y = ax'' + bx, show that 

dx* ax 
10. If a, b, c, d are constant and y = ax^ -\- bx^ + ex + d, show that 
d*y 



dx^ «• 



11. Show that 



12. Show that 



d^ I .d^ _ \_ . fPx 
di\dt ^l~ dfl' 



dxVdx^ '^'^ dx^^^^'dx ^y) ''dx* 

13. Given x = t^ + t', ?/ = f - <', find ^ and ^■ 

dx' dj^ 

14. By differentiating the equation 

dx ^ 
dy 

with respect to x, find j^ in terms of derivatives of x with respect to y. 



■ Chap. III. 

Vited the 
CHAPTi^R IV 1 the quo- 

RATES 

26. Rate of Change. — If the change in a qua'/ 
proportional to the time in which it occurs, z is sai*-' /the vari- 
at a constant rate. If Az is the change occurri 9/ in- 
terval of time A^, the rate of change of z is 

If the rate of change of z is not constant, it will be nearly 

Az 

constant if the interval At is very short. Then -^ is ap- 
proximately the rate of change, the approximation becoming 
greater as the increments become less. The exact rate of 
change at the time t is consequently defined as 

limff = |, (26) 

At=o At at 

that is, the rate of change of any quantity is its derivative with 

respect to the time. 

If the quantity is increasing, its rate of change is positive; 

if decreasing, the rate is negative. 

27. Velocity Along a Straight Line. — Let a particle P 
move along a straight line (Fig. 27). Let s = OP be con- 

8 As 

,^ ^ ■^■■■■■■^ 



p 

Fig. 27. 

sidered positive on one side of 0, negative on the other. If 
the particle moves with constant velocity the distance As in 
the time At, its velocity is 

As ) 

At' ^ 
If the velocity is not constant, it will be nearly so when At 

As 
is very short. Therefore -rr is approximately the velocity, the 

32 



Chap. IV. RATES 33 

approximation becoming greater as A< becomes less. The 
velocity at the time t is therefore defined as 

This equation shows that ds is the distance the particle 
would move in a time dt if the velocity remained constant. 
As a rule the velocity will not be constant and so ds will be 
different from the distance the particle does move in the 
time dt. 

When s is increasing, the velocity is positive; when s is 
decreasing, the velocity is negative. 

Example. A body starting from rest falls approximately 
s=mf 
feet in t seconds. Find its velocity at the end of 10 seconds. 

The velocity at any time t is 

f = ^ = 32 « ft./sec.* 
dt 

At the end of 10 seconds it is 

y = 320 ft./sec. 
28. Acceleration Along a Straight Line. — The accelera- 
tion of a particle moving along a straight line is defined as 
the rate of change of its velocity. That is 

This equation shows that dv is the amount v would increase 
in the time dt if the acceleration remained constant. 

The acceleration is positive v/hen the velocity is increasing, 
negative when it is decreasing. 

Example. At the end of t seconds the vertical height of a 
ball thrown upward with a velocity of 100 ft./sec. is 

/i = 100 < - 16 ^. 
Find its velocity and acceleration. Also find when it is 
rising, when falling, and when it reaches the highest point. 

* The notation ft./sec. means feet per second. Similarly, ft./sec.*, 
used for acceleration, means feet per second per second. 



-p 




Fig. 29. 



34 DIFFERENTIAL CALCULUS Chap. IV. 

The velocity and acceleration are 

v = j^ = (100 - 32 t) ft./sec, 

a = ^= -32ft./sec.2. 
at 

The ball will be rising while v is positive, that is, until t = 
-^ = 3i. It will be falling after ^ = 3^. It will be at the 

highest point when < = 3|. 

29. Angular Velocity and Acceleration. — Consider a 

body rotating about a fixed axis. Let 6 be the angle turned 
through at time t. The angular veloc- 
ity is defined as the rate of change of 
6, that is, 

angular velocity = w = -r- • 
at 

The angular acceleration is the rate 

of change of angular velocity, that is, 

, IX- do} cPd 

angular acceleration =« = —- = —-. 

at dt^ 

Exam-pie 1. A wheel is turning 100 revolutions per minute 
about its axis. Find its angular velocity. 

The angle turned through in one minute will be 

CO = 100 • 2 TT = 200 IT radians/min. 
Ex. 2. A v^^heel, starting from rest under the action of a 
constant moment (or tv/ist) about its axis, will turn in t 
seconds through an angle 

e = kt^, 

h being constant. Find its angular velocity and acceleration 
at time t. 
By definition 

w = 37 = 2 /e/ rad./sec, 
dt 

a = -TT = 2k rad./sec.'. 
ai ' 



Chap. IV. / RATES 35 

30. Related Rates. — In many cases the rates of change 
of certain variables are known and the rates of others are to 
be calculated. This is done by expressing the quantities 
whose rates are wanted in terms of those whose rates are 
known and taking the derivatives with respect to t. 

Example 1. The radius of a cylinder is increasing 2 ft, /sec. 
and its altitude decreasing 3 ft./sec. Find the rate of change 
of its volume. 

Let r be the radius and h the altitude. Then 
V = irr^h. 
The derivative with respect to t is 



By hypothesis 
Hence 



dv _« d^ , o u^'"' 



dr _ dh _ 
'^~^' dt~ '^' 



$ = 4 Trr/i - 3 7rr2. 
at 



This is the rate of increase when the radius is r and altitude 
h. If r = 10 ft. and /i = 6 ft., 

-77= — 60 T cu. ft./sec. 
at 

Ex. 2. A ship B sailing south at 16 miles per hour is north- 
west of a ship A sailing east at 10 miles per hour. At what 
rate are the ships approaching? 

Let X and y be the distances of the ships A and B from the 
point where their paths cross. The distance between the 

ships is then 

s = Vx2 + y\ 

This distance is changing at the rate 

dx dy 
ds J^di'^^'dU 
dt Vx2 + w2 



36 DIFFERENTIAL CALCULUS Chap. IV. 

By hypothesis, 

1=10 *= -16, ^= = ^^ = cos45° = ^. 

at at Vx^ + ?/2 Vx^ + 2/2 v2 

Therefore 

ds 10 - 16 



dt V2 



= -3V2 mi./hr. 



The negative sign shows that s is decreasing, that is, the 
ships are approaching. 

EXERCISES 

1. From the roof of a house 50 ft. above the street a ball is thrown 
upward with a speed of 100 ft. per second. Its height above the ground 
t seconds later will be 

h = 50+ loot - U fi. 

Find its velocity and acceleration when I = 2. How long df^3s it con- 
tinue to rise? What is the highest point reached? » 

2. A body moves in a straight line according to the law 

s = i /< - 4 ^3 + 16 ^2. 

Find its velocity and acceleration. During what interval is the velocity 
decreasing? When is it moving backward? 

3. If V is the velocity and a the acceleration of a particle moving 
along the a:-axis, show that 

adx = V dv. 

4. If a particle moves along a line with the velocity 

v^ = 2 gs, 

where g is constant and s the distance from a fixed point in the line, show 
that the acceleration is constant. 

5. When a particle moves with constant speed around a circle with 
center at the origin, its shadow on the a;-axi3 moves with velocity v 
satisfying the equation 

1,2 4- n^x^ = nVS 

n and r being constant. Show that the acceleration of the shadow is 
proportional to its distance from the origin. 

6. A wheel is turning 500 revolutions per minute. What is its 
angular velocity? If the wheel is 4 ft. in diameter, with what speed does 
it drive a belt? 



Chap. IV. RATES 37 

7. A rotating wheel is brought to rest by a brake. Assuming the 
friction between brake and wheel to be constant, the angle turned 
through in a time t will be 

= a + bt — cP, 

a, b, c being constants. Find the angular velocity and acceleration. 
When will the wheel come to rest? 

8. A wheel revolves according to the law w = 30 / + ^, where « b 
the speed in radians per minute and t the time since the wheel started- 
A second wheel turns according to the law 6 = iC*, where / is the time 
in seconds and the angle in degrees through which it has turned. Which 
wheel is turning faster at the end of one minute and how much? 

9. A wheel of radius r rolls along a line. If r is the velocity and a 
the acceleration ot its center, a> the angular velocity and a the angular 
acceleration about its axis, show that 

V = roj, a = Ta. 

^ 10. The depth of water in a cylindrical tank, 6 feet in diameter, is 
increasing 1 foot per minute. Find the rate at which the water is flow- 
ing in. 

11. A stone dropped into a pond sends out a series of concentric 
ripples. If the radius of the outer ripple increases steadily at the rate 
of 6 ft. /sec., how rapidly is the area of disturbed water increasing at 
the end of 2 seconds? 

12. At a certain instant the altitude of a cone is 7 ft. and the radiiis 
of its base 3 ft. If the altitude is increasing 2 ft. /sec. and the radius of 
its base decreasing 1 ft. /sec., how fast is the volume increasing or de- 
creasing? 

13. The top of a ladder 20 feet long sUdes down a vertical wall. Find 
the ratio of the speeds of the top and bottom when the ladder makes an 
angle of 30 degrees with the ground. 

14. The cross section of a trough 10 ft. long is an equilateral triangle. 
If water flows in at the rate of 10 cu. ft. /sec., find the rate at which the 
depth is increasing when the water is 18 inches deep. 

15. A man 6 feet tall walks at the rate of 5 feet per second away from 
a lamp 10 feet from the groimd. When he is 20 feet from the lamp post, 
find the rate at which the end of his shadow is moving and the rate at 
which his shadow is growing. 

16. A boat moving 8 miles per hour is laying a cable. Ass umin g 
that the water is 1000 ft. deep, the cable is attached to the bottom and 
stretches in a straight line to the stem of the boat, at what rate is the 
cable leaving the boat when 2000 ft. have been paid out? 

17. Sand when poured from a height on a level siurface forms a cone 
with constant angle /3 at the vertex, depending on the material. If the 



/ 



/ 



38 DIFFERENTIAL CALCULUS Chap. [V. 

sand is poured at the rate of c cu. ft. /sec, at what rate is the radius in- 
creasing when it equals a? 

18. Two straight railway tracks intersect at an angle of 60 degrees. 
On one a train is 8 miles from the junction and moving toward it at the 
rate of 40 miles per hour. On the other a train is 12 miles from the 
junction and moving from it at the rate of 10 miles per hour. Find 
the rate at which the trains are approaching or separating. 'K-C/t^tA-^ 

19. An elevated car nmning at a constant elevation of 50 ft. above 
the street passes over a surface car, the tracks crossing at right angles. 
If the speed of the elevated car is 16 miles per hour and that of the sur- 
face car 8 miles, at what rate are the cars separating 10 seconds after 
they meet? ^7 f ? ' j ; i_^ 

20. The rays of the sun make an angle of 30 degrees with the hori- 
zontal. A ball drops from a height of 64 feet. How fast is its shadow 
moving just before the ball hits the ground? 



I 



CHAPTER V 
MAXIMA AND MINIMA 

31. A function of a; is said to have a maximum at x = a, 
if when x = a the function is greater than for any other value 
in the immediate neighborhood of a. It has a minimum if 
when X = a the function is less than for any other value of x 
sufficiently near a. 

If we represent the function by y and plot the curve 
y = / (^)> ^ maximum occurs at the top, a minimum at the 
bottom of a wave. 

If the derivative is continuous, as in Fig. 31a, the tangent 
is horizontal at the highest and lowest points of a wave and 
the slope is zero. Hence in determining maxima and minima 
of a function / (x) we first look for values of x such that 
d 



dx 



/(x)=/'(x)=0. 



If a is a root of this equation, / (a) may be a maximum, a 
minimum, or neither. 




Fig. 31a. 



If the slope is positive on the left of the point and negative 
on the right, as at A, the curve falls on both sides and the 
ordinate is a maximum. That is, / (x) has a maximum value 

39 



40 DIFFERENTIAL CALCTijUS Chap. V. 

atx = a, iff (x) is positive for values >f xa little less and nego/- 
live for values a little greater than a. 

If the slope is negative on the left and positive on the right, 
as at B, the curve rises on both sides and the ordinate is a 
minimum. That is, / (x) has a minimum at x = a, if f (x) is 
negative for values of x a little less and positive for values a little 
greater than a. 

If the slope has the same sign on both sides, as at C, the 
curve rises on one side and falls on the other and the ordinate 
is neither a maximum nor a minimum. That is, / (re) has 
neither a maximum nor a minimum at x = a if f (x) has the 
same sign on both sides of a. 

Example 1. The sum of two numbers is 5. Find the maxi- 
mum value of their product. 

Let one of the numbers be x. The other is then 5 — x. 
The value of x is to be found such that the 
product 

y = X (5 — x) = 5 X — x^ 

is a maximum. The derivative is 

^ = 5-20:. 
ax 

This is zero when x = |. If x is less than 
f , the derivative is positive. If x is greater 
than f the derivative is negative. Near 

X = ^ the graph then has the shape shown in Fig. 31b. At 

re = f the function has its greatest value 

f (5 - f ) = \^ 

Ex. 2. Find the shape of the pint cup which requires for 
its construction the least amount of tin. 

Let the radius of base be r and the depth h. The area of 
tin used is 

A=rr^ + 2 rrh. 




Let V be the number of cubic inches in a pint. Then 

V = n^h. 



I 



Chap. V 



MAXIMA AND MINIMA 



41 



Consequently, 



and 



h = 



irr» 



A 2 . 2f 

r 



Since r and v are constants, 
dA 



dr 



= 27rr 



-■7^=K^> 



This is zero if irr^ = v. If there is a maximum or minimum 
it must then occur when 

r = v/-; 

dA 
for, if r has any other value, -j- will have the same sign on 

both sides of that value and A will be neither a maximum nor 
a minimum. Since the amount of tin used cannot be zero 
there must be a least amount. This must then be the value 
of A when v = irr^. Also v = rr^h. We therefore conclude 
that r = h. The cup requiring the 
least tin thus has a depth equal to 
the radius of its base. 

Ex. 3. The strength of a rec- 
tangular beam is proportional to 
the product of its width by the 
square of its depth. Find the 
strongest beam that can be cut 
from a circular log 24 inches in 
diameter. 

In Fig. 31c is shown a section 
of the log and beam. Let x be the breadth and y the depth 
of the beam. Then 

X2 + J/2 = (24)2. 
The strength of the beam is 

S = kxy^ = kx (242 _ 3.2)^ 




Fig. 31c. 



42 



DIFFERENTIAL CALCULUS 



Chap. V. 



k being constant. 



The derivative of *S is 
dS 



dx 



= k (242 - 3 x2). 



If this is zero, x = ±8 V3. Since x is the breadth of the 
beam, it cannot be negative. Hence 

a: = 8 a/3 

is the only solution. Since the log cannot be infinitely strong, 
there must be a strongest beam. Since no other value can 
give either a maximum or a minimum, a; = 8 Vs must be 
the width of the strongest beam. The corresponding depth 
is y = 8 Vq. 

Ex. 4. Find the dimensions of the largest right circular 
cylinder inscribed in a given right circular cone. 

Let r be the radius and h the altitude of the cone. Let 
X be the radius and y the altitude of 
an inscribed cylinder (Fig. 31d). From 
the similar triangles DEC and ABC, 
DE^AB 

EC EC ' 

that is. 




y ^h^ 
r — X r 
The volume of the cylinder is 



y = - (r - x). 



Fig. 31d. 



V = irx^y = — (rx^ — X?). 



Equating its derivative to zero, we get 
2 rx - 3 a;2 = 0. 

Hence x = or x = | 
not give the maximum, 
radius must then be x 



The value x = obviously does 
Since there is a largest cylinder, its 
I r. By substitution its altitude is 
then found to be ?/ = \h. 

32. Method of Finding Maxima and Minima. — The 
method used in solving these problems involves the following 
steps: 



Chap. V. MAXIMA AND MINIMA 43 

(1) Decide what is to be a maximum or minimum. Let 
it be y. 

(2) Express y in terms of a single variable. Let it be x. 
It may be convenient to express y temporarily in terms of 

several variable quantities. If the problem can be solved by 
our present methods, there will be relations enough to elimi- 
nate all but one of these. 

dv 

(3) Calculate -^ and find for what values of x it is zero. 

(4) It is usually easy to decide from the problem itself 

whether the corresponding values of y are maxima or minima. 

du 
If not, determine the signs of -j^ when x is a little less and 

ax 

a little greater than the values in question and apply the 

criteria given in Art. 3L 

EXERCISES 

Find the maximum and minimum values of the following functions: 

1. 2 x» - 5 X + 7. 3. X* -2x^ + 6. 

\/2. 6 + 12i -i». A. —=£=' 

V a* — X* 

Show that the following functions have no maxima or minima: 

6. x». 7. 6x5 - 15 X* + 10x3. 

\/ 6. x' + 4 X. 8. X Va» + x*. 

9. Show that x A — has a maximum and a minimum and that the 

X 

maximum is less than the minimum. 

10. The sum of the square and the reciprocal of a number is a mini- 
mum. Find the number. 

11. Show that the largest rectangle with a given perimeter is a 
square. 

?/ 12. Show that the largest rectangle that can be inscribed in a given 

circle is a square. 
, 13. Find the altitude of the largest cylinder that can be inscribed in 

a sphere of radius a. 
, ■^ 14. A rectangular box with square base and open at the top is to be 
" made out of a given amount of material. If no allowance is made for 

thickness of material or waste in construction, what are the dimensions 

of the largest box that can/be made? 



44 DIFFERENTIAL CALCULUS Chap. V. 

; ,16. A cylindrical tin can closed at both ends is to have a given 
capacity. Show that the amount of tin used will be a minimum when 
the height equals the diameter. 

16. What are the most economical proportions for an open cylindrical 
water tank if the cost of the sides per square foot is two-thirds the cost 
of the bottom per square foot? 

17. The top, bottom, and lateral surface of a closed tin can are to be 
cut from rectangles of tin, the scraps being a total loss. Find the most 
economical proportions for a can of given capacity. 

18. Find the volume of the largest right cone that can be generated 
by revolving a right triangle of hypotenuse 2 ft. about one of its sides. 

19. Four successive measurements of a distance gave Ci, 02, as, cu as 
results. By the theory of least squares the most probable value of the 
distance is that which makes the sum of the squares of the four errors a 
minimum. What is that value? 

20. If the sum of the length and girth of a parcel post package must 
not exceed 72 inches, find the dimensions of the largest cylindrical jug 
that can be sent by parcel post. 

21. A circular filter paper of radius 6 inches is to be folded into a 
conical filter. Find the radius of the base of the filter if it has the 
maximum capacity. 

22. Assuming that the intensity of light is inversely proportional to 
the square of the distance from the source, find the point on the line 
joining two sources, one of wliich is twice as intense as the other, at 
which the illumination is a minimum. 

23. The sides of a trough of triangular section are planks 12 inches 
wide. Find the width at the top if the trough has the maximum 
capacity. 

24. A fence 6 feet high runs parallel to and 5 feet from a wall. Find 
the shortest ladder that will reach from the ground over the fence to 
the wall. 

26. A log has the form of a frustum of a cone 29 ft. long, the diameters 
of its ends being 2 ft. and 1 ft. \ beam of square section is to be cut 
from the log. Find its length if the volume of the beam is a maximum. 

26. A window has the form of a rectangle surmounted by a semi- 
circle. If the perimeter is 30 ft;, find the dimensions so that the greatest 
amount of fight may be admitted. 

27. A piece of wire 6 ft. long is to be cut into 6 pieces, two of one length 
and four of another. The two former are bent into circles which are 
held in parallel planes and fastened together by the four remaining 
pieces. The whole forms e, model of a right cylinder. Calculate the 
lengths into which the wire must be divided to produce the cylinder of 
greatest volutue. 



Chap. V. MAXIMA AND MINIMA 45 

28. Among all circular sectors with a given perimeter, find the one 
which has the greatest area. 

29. A ship B is 75 miles due east of a ship A. IS B sails west at 12 
miles per hour and A south at 9 miles, find when the ships will be closest 
together. 

30. A man on one side of a river J mile wide wishes to reach a point 
on the opposite side 5 miles further along the bank. If he can walk 4 
miles an hour and swim 2 miles an hour, find the route he should take 
to make the trip in the least time. 

81. Find the length of the shortest line which will divide an equi- 
lateral triangle into parts of equal area. 

32. A triangle is inscribed in an oval curve. If the area of the tri- 
angle is a maximum, show graphically that the tangents at the vertices 
of the triangle are parallel to the opposite sides. 

33. A and C are points on the same side of a plane mirror. A ray of 
light passes from A to C by way of a point B on the mirror. Show that 
the length of the path ABC will be a minimum when the lines AB; 
CB make equal angles with the perpendicular to the mirror. 

34. Let the velocity of Ught in air be t'l and in water ij. The path of 
a raj' of Ught from a point A in the air to a point C below the surface of 
the water is bent at B where it enters the water. If di and fit are the 
angles made by AB and BC with the perpendicular to the surface, show 
that the time required for light to pass from A to C \st11 be least if £ is so 
placed that 

singi _ Vi 
sin di Vi 

35. The cost per hour of propelling a steamer is proportional to the 
cube of her speed through the water. Find the speed at which a boat 
should be run against a current of 5 miles per hour in order to make a 
given trip at least cost. 

36. If the cost per hour for fuel required to run a steamer is propor- 
tional to the cube of her speed and is S20 per hour for a speed of 10. knots, 
and if the other expenses amount to $ ICO per hour, find the most econona- 
ical speed in still water. 

33. other Types of Maxima and Minima. — The method 
given in Art. 31 is sufficient to determine maxima and minima 
if the function and its derivative are one-valued and continu- 
ous. In Figs. 33a and 33b are shown some tj'pes of maxima 
and minima that do not satisfy these conditions. 

^At B and C, Fig. 33a, the tangent is vertical and the de- 



46 



DIFFERENTIAL CALCULUS 



Chap. V. 



that on the right. The derivative is discontinuous. At A 
and E the curve ends. This happens in problems where 
values beyond a certain range are impossible. According to 




Fig. 33a. 

our definition, y has maxima &t A, B, D and minima at C 
and E. 

If more than one value of the function corresponds to a 

single value of the vari- 
able, points like A and 
B, Fig. 33b, may occur. 
At such points two values 
of y coincide. 

These figures show 
that in determining max- 
ima and minima special 
attention must be given 
to places where the de- 
FiG. 33b. rivative is discontinuous, 

the function ceases to exist, or two values of the function 
coincide. 

Example 1. Find the maximum and minimum ordinates 
on the curve y* = a^. 
In this case, y = x^ and 

dx 3^ • 
No finite value of x makes' the derivative zero, but x = 




Chap. V. 



MAXIMA AND MINIMA 



47 



makes it infinite. Since y is never negative, the value is a 
mipitrmm (Fig. 33c). 




Ex. 2. A man on qne side of a river ^ mile wide wishes to 
reach a point on the opposite side 2 miles down the river. If 
he can row 6 miles an hour and walk 4, find the route he 
should take to make the trip in the least time. 




Fig. 33d. 

Let A (Fig. 33d) be the starting point and B the destina- 
tion. Suppose he rows to C, x miles down the river. The 
time of rowing will be ^ VxM-~l and the time of walking 
i (2 — x). The total time is then 

t = i V^Ti + i (2 - x). 
Equating the derivative to zero, we get 

6\/z* + i 4 
which reduces to 5 x* + | = 0. This has no real solution. 



48 DIFFERENTIAL CALCULUS Chap. V. 

The trouble is that J (2 — x) is the time of walking only 
if C is above B. If C is below B, the time is \{x — 2). 
The complete value for t is then 

the sign being + if x < 2 and — if x > 2. The graph of 
the equation connecting x and t is shown in Fig. 33e. At 
X = 2 the derivative is discontinuous. Since he rows faster 
than he walks, the minimum obviously occurs when he rows 
all the way, that is, x = 2. 

EXERCISES 

Find the maximum and minimum values of y on the following curves: 

1.- x' 4- y' = o'. \J 3. y* = a:* - 1. 

2. ^ = x* (a; - 1). 4. X = l^ + i\ y = (^ - i*. 

6. Find the rectangle of least area having a given perimeter. 

6. Find the point on the parabola j/^ = 4x nearest the point 
(-1,0). 

7. A wire of length I is cut into two pieces, one of which is bent to 
form a circle, the other a square. Find the lengths of the pieces when 
the sum of the areas of the square and circle is greatest. 

8. Find a point P on the line segment AB such that PA^ + PB^ is 
a maximum. 

9. If the work per hour of moving a car along a horizontal track is 
proportional to the square of the velocity, what is the least work re- 
quired to move the car one mile? 

10. If 120 cells of electromotive force E volts and internal resistance 
2 ohms are arranged in parallel rows with x cells in series in each row, 
the current which the resulting battery will send through an external 
resistance of \ ohm is 

^ 60xE 

^ x* + 20* 

How many cells should be placed in each row to give the maximum 
current? 



CHAPTER VI 



DIFFERENTIATION OF TRANSCENDENTAL 
FUNCTIONS 

34. Formulas for Differentiating Trigonometric Func- 
tions. — Let u be the circular measure of an angle. 

Vn. d sin w = cos ti du. 4^1]d^ = cosv 

CK^ . , 

Vin. <2 cos u = — smwdu. dc£i^ . -^r*^-/ 

IX. d tan u = sec"* u du. j +<u>'^ , d^c '^ - 

X. d cot M = — CSC** It rfu. <*£^- 

XI. d sec u = sec u tan u di/! ^4^ ^ ^^^ >= ^" ^ 

Xn. d CSC M = —CSC u cot M da. cU^l^^ - -cos^Ci'^^ "^ 

The negative sign occurs in the differentials of all co- 
functions. 

35. The Sine of a Small Angle. — Inspection of a table 
of natural sines will show that 
the sine of a small angle is very 
nearly equal to the circular meas- 
ure of the angle. Thus 

sin l** = 0.017452, 
r 



180 



= 0.017453. 



We should then expect that 



,. sin , 
hm— -= 1. 



(35) 




Fig. 35. 



To show this graphically, let 9 = AOP (Fig. 35). Draw 
PM p)erpendicular to OA. The circular measure of the angle 
is defined by the equation 

arc arc AP 



6 = 



rad. 



OP 



49 



60 DIFFERENTIAL CALCULUS Chap. VI. 

,, . ^ MP „ 
Also sin 6 = jyp . Hence 

sing^ MP ^ chord QP 
d arc AP arc QP 

As 6 approaches zero, the ratio of the arc to the chord ap* 
proaches 1 (Art. 53). Therefore the limit of — r— is 1. 

36. Proof of VII, the Differential of the Sine. — Let 

J/ = sin w. ■ 

Then • fcj--5»n3^ 

J/ + Ay = sin (w + Aw) ' - 3(^^ 3 

and so 

Ay = sin (w + Aw) — sin w. 

It is shown in trigonometry that 

sin A - sin 5 = 2 cos § (^ + ^) sin ^ {A - B), 

If then A = w + Aw, B = u, 

^ Ay = 2 cos (w + § Aw) sin \ Aw, 
whence ^''" ^ ^ ■^**'^ -^^1^^%^^)^^ ii^^j^ 4- c 

^ = 2 cos ^ +4 Aw) j"2u ^ =1^0^ tw +% Aw) ^^1^ . 

As Aw approaches zero *^( 

sin I Aw _ sing 
~^5Au~~ B 

approaches 1 and cos (w + i Aw) approaches cos w. There- 
fore 

dy 

3^ = cos w. 

aw 

Ck)nsequently, *^L " -^ ^^ 

dy = cos w dw. 

37. Proof of Vm, the Differential of the Cosine. —By 

trigonometry 

cos w = sinf^ — wj- , 



Chap. VI. 



TRANSCENDENTAL FUNCTIONS 



53 



22. Find the points on the curve y = x + sin 2 x where the tangent 
is parallerfto the line y = 2 x + S. 

23. A weight supported by a spring hangs at rest at the origin. If 

the weight is lifted a distance A and let fall, its height at any subsequent 

time ( will be 

y = A cos (2 mt), 

n being constant. Find its velocity and acceleration as it passes the 
origin. WTiere is the velocity greatest? Where is the acceleration 
greatest? 

24. A revolving light 5 miles from a straight shore makes one com- 
plete revolution per minute. Find the velocity along the shore of the 
beam of light when it makes an angle of 60 degrees with the shore line._ 

^ 26. In Ex. 24 with what velocity would the light be rotating if the 
spot of light is moving along the shore 15 miles per hour when the 
beam makes with the shore line an angle of 60 degrees? 

26. Given that two sides and the included angle of a triangle have at 
a certain instant the values 6 ft., 10 ft., and 30 degrees resf)ectiv6ly, and 
that these quantities are changing at the rates of-S ft., —2 it., and 10 
degrees per second, how fasf is the area of the/triangle ch^ging? 

27. OA is a crank and AB a connecting nxi attached to a piston B 
moving along a line through 0. If OA re*x>l\W about with angular 
velocity w, prove that thp velocity of fi is wOC/where C is the point in 
which the line BA cuts/he line through perpendicular to OB. 



3/ 



Gcoa 



rims 
^eam that 



28. An alley 8 ft 
/What is the longest 

street into the alley? 

29. A needle rest^'ith one end 
needle will sink to a p>osition in 
If the length of them^edle eq 
the position of equmbrium? 

30. A rope with a ring at o 
horizontal line and held taut 
the rope slips freely, tUe wei 
the angle formed at th^ bott* 




dicubir to a street 27 ft. wide, 
mitved horixontally along the 

a smAOth hemispherical bowl. The 

hich tpe center is as low as possible. 

meter of the bowl, what will be 



is looped over two pegs in the same 
weight fastened to the free end. If 
descend as far as f>ossible. Find 
of the loop. 

31. Find the angle at the bottom of the loop in Ex. 30 if the rope is 
looped over a circular pulley instead of the two pegs. 

32. A gutter is to b^ made by bending into shape a strip of copper so 
that the cross section sl^all be an arc of a circle. If the width of the strip 
is a, find the radius of the cross section when the carrying capacity is a 
maximum. 

33. A spoke in the front wheel of a bicycle is at a certain instant per- 
pendicular to one in the rear wheel. If the bicycle rolls straight ahead, 
an what position will the outer ends of the two ^x)kes be closest together? 



64 DIFFERENTIAL CALCULUS Chap. VI. 

39. Inverse Trigonometric Functions. — The symbol 
sin~* X is used to represent the angle whose sine is x. Thus 
y = sin~^ X, x = sin y 

are equivalent equations. Similar definitions apply to cos~^ x 
tan~^ x, cot~* x, sec~^ x, and csc~^ x. 

Since supplementary angles and those differing by mul- 
tiples of 2 7r have the same sine, an indefinite number of 
angles are represented by the same symbol sin~^ x. The 
algebraic sign of the derivative depends on the angle dif- 
ferentiated. In the formulas given below it is assumed that 
sin-i u and csc~^ u are angles in the first or fourth quadrant, 
cos~^ u and sec"^ u angles in the first or second quadrant. 
If angles in other quadrants are differentiated, the opposite 
sign must be used. The formulas for tan~^ u and cot~^ u 
are valid in all quadrants. 
^ 40. Formulas for Differentiating Inverse Trigonometric 
Functions. — 

Xm. dsin-^u = 

XIV. d cos-^u = - 

XV. d tan-* u = 

XVI. d cot-* u = - 

XVII. dsec-^u = 



XVIII. dcsc-'u = fl' • d[a>^^^.-_l-^ 




y = sin~* u. 



41. Proof of the Formulas. — Let 

Then 

sin y = u. 
Differentiation gives 

cos y dy = du, 




Chap. VI. 



TRANSCENDENTAL FUNCTIONS 



55 



whence ' 
But 




dy = 



du 
cosy 



cosy 



= =fc Vl - sin* y = zfc Vl - u». 



If 2/ is an angle in the first or fourth quadrant, cos y is positive. 

Hence 

cos y = Vl — u* 

and so 

, du 

dy = , 

The other formulas are proved in a similar way. 
EXERCISES 



\J L y = Bin-'(3x-l), 
2. y=coe-(l-?). 

5. y = 8ec-» V4 1 + 1, 
ft 1 -. 3 . 

V 7. y = tan-i' 



dy = 



dy = 



Zdx 



v6x - 9i» 



rfx 



V2ax-x* 

dy _ -2 
dx x» + l' 



(4 1 + 1) Vi 
dy 1 



X + a 
8. X = csc~i (sec 6), 

s/.9. y = r- ^ 



10. y = sec" 



Vtf -x» 
1 



Vl -I* 



<^ V2 + 2x-4x» 

dx~ 3? +a*' 

d9 ^• 

^^ ^ 

<** (o» - x») Va» - 2 1»* 
dy 1 



11. y = §V^^^ + ^'Bin-»-, ^ = 



< o i -I 4 sin I 

12. y = tan-i r-— — , 

3 + 5 cosx 



dx Vl -z» 






Va»-x». 
4 



13. y = S€< ■» 



2x»-l' 



dx 5 + 3cosx 
^= 2 



56 DIFFERENTIAL CALCULUS Chap.VL 

\/l4. y - a8in-»5+ V^rr^, ^ - V^^. 
" o ' dx ^ a -{■ X 

16. y = 2 (3 X + 1)* + 4 cot-i ^^^-±^, 

rfy^ i 

dx (3x + l)* + 4(3x4-l)** 
1 _, 3x dy 1 -x" 

16. I/-gtan 2 + 2x»' dx 4x* + 17x2+4' 

„ _,x + 1 2 _, 2x dy X + 1 

17. « = cos ' — s :^co8 ^5 , j^ = . =L' 

2 V3 3-x dx (3_a;)V3-2x-x» 

1ft _ ^x' - g* 1_ _jX dy 1 

"• ^~ 2a^x« 2a»^'' o' dx ~ -^a v^TTT^j* 



19 



, ,x + Vi2 + 4x-4 '^I/. 
I. y = tan-i ^ 1 ^Z - 



2 <te xVx2+4x-4 



20. y = X sin-i x + Vl - x^, ^ = , ^ 

dx2 Vl -x' 

21. y = x2 sec-i I - 2 Vx^^Tl^ ^ = 2xsec-i|- 

Z uX i2 

22. Let 8 he the arc from the x-axis to the point (x, y) on the circle 
x2 + y2 = a*. Show that 

* = -?, |.|, *. = ^. + .^. 

V 23. Let A be the area bounded by the circle x* + y" = o*, the y-axis 
and the vertical line through (x, y). Show that 

A = xy + a^ tan~* - , dA = 2ydx. 

24. The end of a string wound on a pulley moves with velocity v il 
along a line perpendicular to the axis of the pulley. Find the angular 
velocity with which the pulley turns. 

^ 25. A tablet 8 ft. high is placed on a wall with its base 20 ft. above the 
level of an observer's eye. How far from the wall should the observer 
stand that the angle of vision subtended by the tablet be a maximum? 

42. Exponential and Logarithmic Functions. — If a is a 
positive constant and u a variable, a" is called an exponential 
function. If w is a fraction, it is understood that o" is the 
positive root. 

If y = a", then u is called the logarithm of i/ to base a. 
That is, 

y = a", ?' = logaV ' " 

r 






*. .^^^SCENDENTAL FUNCTIONS 57 

Elimination of u 



are by definition equivalent equations 
gives the important identity 

aio«.v = y. (42b) 

This expresses symbolically that the logarithm is the power 

to which the base must be raised to equal the number. 

43. The Curves y = a". — Let a be greater than 1. 

The graph of 

y = a' 



has the general appearance of Fig. 43. 
increment h, the increment in y 
is 
At/ = 0*+* — a' = a' (a'' — 1). 

This increases as x increases. If 
then X increases by successive 
amounts h, the increments in y 
form steps of increasing height. 
The curve is thus concave upward 
and the arc lies below its chord. 
The slope of the chord AP join- 
ing A (0, 1) and P (x, y) is 
a' — \ 



If X receives a small 




Fig. 43. 



As Pi moves toward A the slope 
of APx increases. As P2 moves toward A the slope of APj 
decreases. Furthermore the slopes of AP2 and APi approach 
equality; for 

and a~* approaches 1 when k approaches zero. Now if two 
numbers, one always increasing, the other always decreasing, 
approach equality, they approach a common limit. Call 
this limit m. Then 

lim^^ = m. (43) 



bb 



DIFFERENTUL CALCUI.Ao 



jpp. 



m^^^^Wii^-^- 



This is the slope of the curve 2/ = a* at the point where it 
crosses the ^/-axis. 

44. Definition of e. — We shall now show that there is a 
number e such that 

^-^ - (44) 

In fact, let 

•■ 1 1 



lim 



where m is the slope found in Art. 43. Then 



,. e* - 1 ,. o*" - 1 1 ,. a*" - 1 1 

hm = lim = — hm = — • m = 1. 

x=o a: 1^0 X m x^o x_ m 

m 
The curves y = a' all pass through the point A (0, 1). 
Equation (44) expresses that when a = e the slope of the 
curve at A is 1. If a > e the slope is greater than 1. If 
a < e, the slope is less than 1 . 





Y 


1 1 . 


/ 






a>e, 1 


/ 






i 

I / ^ 


/a<e ■ 




A 


I/-'-' 




— 


:::SS5^ 













X- 



Fia. 44. 

We shall find later that 

e = 2.7183 

approximately. Logarithms to base e are called natural 
logarithms. In this book we shall represent natural log- 
arithms by the abbreviation In. Thus In u means the natural 
logarithm of u. 



Chap. VI. TRANSCENDENTAL FUNCTIONS 59 

45. Differentials of Exponential and Logarithmic Func- 
tions. — 

XIX. de~ = c- dw. - e = I -*--L 4 -L-(- 

U. L? 
XX. da" = a" In a du. 

XXL d\nu = ^-^}i^'Jk cJe^.e"" 

46. Proof of XIX, the Differential of e". — Let 

Then ^ 

whence di<2^ yy^^ 

Ay = e«+^» - e« = e" (e^" - 1) ZHT ^ 
land ^ 

Au Au 

As Au approaches zero, by (44), C 

e^" - 1 
Au 
approaches 1. Consequently, \ 

3^ = e", dy = e" dw. 

47. Proof of XX, the Differential of a". — The identity 

a = e'°" 
gives a" = e"*"". 

Consequently, 

da" = e" •" ° d (w In a) = e" •" ° In a dw = a" Ina du. 

48. Proof of XXI and XXII, the Differential of a Log- 
arithm. — Let 

y = Inw. 
Then e" = u. 

Differentiating, 

^ dy = du. 



60 

Therefore 



DIFFERENTIAL CALCULUS 



du du 



Chap. VI. 



The derivative of loga u is found in a similar way. 
Example 1. y = Iri (sec^ x). 

d sec^ X 2 sec x (sec x tan x dx) 



^ sec^ X 
Ex. 2. y = 2*^'^"'^ 



sec^ X 



dy = 2^"""' In 2d (tan-^ x) = — T+l^ 



— = 2 tan X dx. 



2tan-iiin2da; 



EXERCISES 



I. 2/ = ea;, 

\/ 2. y = a*^'^^*^ 

z-l 

3. 2/ = e=^+S 

^ . 5. 1/ = x" + n*, 
\j e. y = a'sf, 

7. y = In (3 x2 + 5 X + 1), 

8. 2/ = In sec* x, 

9. y = In (x + Va;2 - a?), 
^10. y = In (sec ax + tan ax), 

II. y = In (o* + &*), 



dx X* 



dx 
dy 



= 2 a**'' ^ "^ In o sec* 2 a;. 



2 — 
dx (x + 1)2 
dy_( 2 Y 
dx Ve* + e-^/ 

— = wx*~^ + n'' In n. 
dx 

^ = a='x°-Mo + a;lno). 
dx 



dy 
dx 



6x4-5 



3x2 + 5x + l 



^ = 2 tan X. 
dx 

dy L_ 



-^ = a sec ax 
dx 



', 12. y = In sin X + 5 cos* x, 

1 
2 



IS. 2/ = 9l^*«'°2~2wH^' 



dx Vx* — o* 

a sec ax. 

_aMno+_b^4n6 
^ - a» + 6* 
dy _ cos' x ^ 
dx ~ sin X 

dy_ 1 , 
dx sin* z 



Chap. VI. TRANSCENDENTAL FUNCTIONS 61 

,. _1, X' 1_ dy ^ 8 

14. y - ^in^, _^ a:«-4' dx x(i»- 4)>* 

16. 1/ = - In , > :7- = , 

« o + Va» -I* <« X V o» - X* 



16. y = iB ( VI+3 + vT+2) + V(x + 3) (I + 2), ^ = y |^ 



17. y = In (VT+^ + V^, ^ = 



•ii^ 2 Vx» + ax 



18. y = X tan-i --% In (x«+a»), ^ = tan"' - • 
a Z ax a 

V 19. y = ef" (sin ax — cos ax), -^ = 2 ae" sin ax. 

ax 

\l 20. y = i tan* x — 5 tan* z — In cos x, -p = tan' x. 

22. X = e* + e-', y = e< - e"', = - ^• 

23. y=llnx, g=^(21nx-3). 

24. y = x^, g=(x + n)e'. 

26. By taking logarithms of both sides of the equation y = x" and 
differentiating, show that the formula 

--x" = nx"-* 
dx 

is true even when n is irrational. 

26. Find the slope of the catenary 

y = |Ve^+e'-) 

at X = 0. 

27. Find the points on the curve y = e** sin x where the tangent 
18 parallel to the i-axis. 

y 28. If y = Ae"* + Be""*, where A and B axe constant, show that 

29. If y = 2e"'*, where z is Miy function of x, show that 

di*^*'<ix^^^ * dx» 
SO. For what values of x does 

y = 5 hi (x - 2) + 3hi (x + 2) + 4» 
Bcrease as x increases? 



62 DIFFERENTIAL CALCULUS Chap. VI. 

31. From equation (44) show that 

e = lim (1 + x) *• 

32. If the space described by a point is s = ae' + &e""', show that the 
acceleration is equal to the space passed over. 

33. Assuming the resistance encountered by a body sinking in water 
to be proportional to the velocity, the distance it descends in a time t is 

g and k being constants. Show that the velocity v and acceleration a 

satisfy the equation 

a = g — kv. 

Also show that for large values of t the velocity is approximately con- 
stant. 

34. Assuming the resistance of air proportional to the square of the 
velocity, a body starting from rest will fall a distance 



s=^ln 



/e« + e-*^'^ 



in a time t. Show that the velocity and acceleration satisfy the equa- 
tion 

k^i^ 
a = g 

Also show that the velocity approaches a constant value. 




CHAPTER VII 
GEOMETRICAL APPLICATIONS 

49. Tangent Line and Normal. — Let mi be the slope 

of a given curve at Pi (xi, yi). It is shown in analytic 
geometry that a line 
through (xi, yi) with slope 
nil is represented by the 
equation 
y — yi = miix - xi). 

This equation then rep- 
resents the tangent at 
(,^1, yi) where the slope of 
the curve is mi. 

The line PiN perpen- 
dicular to the tangent at 
its point of contact is 
called the normal to the curve at Pi. Since the slope of the 

tangent is mi, the slope of a perpendicular line is and so 

TWl 

y-yi= - — (^-xi) 

7/1 1 

is the equation of the normal at (xi, yi). 

Example 1. Find the equations of the tangent and normal 
to the eUipse x^ + 2 ^^ = 9 at the point (1,2). 
The slope at any point of the curve is 
dy _ X 
dx 2y 

At (1, 2) the slope is then 

mi = —I. 
,The equation of the tangent is 

2/-2 = -i(x-l). 
63 



Fig. 49. 



64 



DIFFERENTIAL CALCULUS 



Chap. VII. 



and the equation of the normal is 

y-2 = A(x-l). 

Ex. 2. Find the equation of the tangent to x^ — y^ = a? 
at the point (xi, y\). 

Xi ' 

The slope at (xi, yi) is — . The equation of the tangent 



yi 



is then 



Xi 



y-yi = -(.^- ^i) 

2/1 



^ -r.Z — 



which reduces to 

XiX - yiy = xi^ - yi' 

Since {xi, yi) is on the curve, Xi^ — y-^ = a^. The equation 
of the tangent can therefore be reduced to the form 

XiX - yiy = a\ 

50. Angle between Two Curves. — By the angle be- 
tween two curves at a 
point of intersection we 
mean the angle between 
their tangents at that 
point. 

Let mi and nii be the 
slopes of two curves at 
a point of intersection. 
It is shown in analytic 
geometry that the angle 
/3 from a line with slope 
mi to one with slope 




Fig. 50a. 
m2 satisfies the equation 

taujS = 



m2 — mi 
1 +mim2 



(50) 



This equation thus gives the angle /3 from a curve with slope 
mi to one with slope nh, the angle being considered positive^ 
when measured in the counter-clockwise direction. 



I 



I 



Chap. VII. 



GEOMETRICAL APPLICATIONS 



65 



Example. Find the angles determined by the line y = x 
and the parabola y = x^. 

Solving the equations simultaneously, we find that the line 
and parabola intersect at 
(1,- 1) and (0, 0). The slope i r 

of the line is 1. The slope at 
any point of the parabola is 

ax 



At (1, 1) the slope of the 
parabola is then 2 and the 
angle from the line to the 
parabola is then given by 




Fig. oOb. 



tan /3i = 



2-1 



whence 



1 

3' 



1+2 

/3i = tan-4 = 18° 26'. 

At (0, 0) the slope of the parabola is and so the angle 
from the line to the parabola is given by the equation 

tan Pi = = — 1. 

^ 1+0 ' 

whence 

ft = -45°. 

The negative sign signifies that the angle is measured in the 
clockwise direction from the line to the parabola. 

EXERCISES 

Find the tangent and normal to each of the following curves at the 
point indicated: 

1. The circle 3^ + f = 5 at (-1, 2). 

2. The hj-perbola xt/ = 4 at (1, 4). 

3. The parabola y* = ax at i = a. 

4. The exponential curve y = oif at x = 0. 

6. The sine curve i/ = 3 sin x at x = ^• 

D 



X* IT 

6. The eUipse -j + ^ = 1, at (x,, y,). 



66 DIFFERENTIAL CALCULUS Chap. VIL 

/ 7. The hyperbola x^ -\- xy — y^ = 2 x, at (2, 0). 

8. The semicubical parabola y^ = x^, at ( — 8, 4). 

9. Find the equation of the normal to the cycloid 

X = a {4> — sin <^), y = a (1 — cos <^) 

at the point ^ = 4>\. Show that it passes through the point where the 
rolling circle touches the x-axis. 

Find the angles at which the following pairs of curves intersect : 
10. 2/2 = 4 x, x^ = 4 2/. 13. y = sin x, y — cos x. 

y/ 11. x^ + ?/2 = 9, x^ + y'^ — 6 x = 9. 14. y = logio x, y = In x. 
12. x2 -+ 2/2 + 2 X = 7, 2/2 = 4 X. 15. y = \ (e^ + e"^), 2/ = 2 e*. 
1^ 16. Show that for all values of the constants a and h the curves 
x^ — y^ = a*, xy = l^ 
intersect at right angles. 
17. Show that the curves 

y = 6°'", y = e"^ sin (&x + c) 
are tangent at each point of intersection. 
^ 18. Show that the part of the tangent to the hyperbola xy = a? in- 
tercepted between the coordinate axes is bisected at the point of tan- 
gency. 
^ 19. Let the normal to the parabola y^ = ax at P cut the x-axis at A'^. 
Show that the projection of PN on the x-axis has a constant length. 

20. The focus F of the parabola y^ = ax is the point (l a,0). Show 
that the tangent at any point P of the parabola makes equal angles 
with FP and the line through P parallel to the axis. 
/' 21. The foci of the ellipse 

^ + P=1' «>^ 

are the points F' ( - Va^ - b^, O) and F (Va* - 6^^ q) . Show that the^ 
tangent at any point P of the ellipse makes equal angles with FP and 
F'P. 

22. Let P be any point on the catenary y = ^\f' +e ° ) , M the 

projection of P on the x-axis, and A'^ the projection of M on the tangent 
at P. Show that MA'' is constant in length, 

23. Show that the portion of the tangent to the tractrix 

a. /a + y/a^ - xA ,-- -, 

intercepted between the 2/-axis and the point of tangency is constant in 
iength. 



ft 




Fig. 51. 



Chap. VII. GEOMETRICAL APPLICATIONS 67 

24. Show that the angle between the tangent at any point P and the 
line joining P to the origin is the same at all points of the curve 

hi Vx- + y2 = k tan-i -• 
^ X 

26. A point at a constant distance along the normal from a given 

curve generates a curve which is called parallel to the first. Find the 

parametric equations of the parallel curve generated by the point at 

distance h along the normal drawn inside of the ellipse 

X = a cos 4>, y = b sin <t>. 

61. Direction of Curvature. — A curve is said to be con- 
cave upward at a point P if 
the part of the curve near P 
lies above the tangent at P. 
It is concave downward at Q 
if the part near Q lies below 
the tangent at Q. 

A • T dhj . 
At 'points where -p^ is pos- 

d^n 
itive, the curve is concave upward; where -7-^ is negative, the 

curve is concave downward. 
For 

d^y _ d (dy\ 
dx^ dx \dx/ 

(Py dv 

If then -T-| is positive, by Art. 13, -r- • the slope, increases as x 

increases and decreases as x decreases. The curve therefore 
rises above the tangent on both sides of the point. If, 

however, -j^ is negative, the slope decreases as x increases 
dx^ 

and increases as x decreases, and so the curve falls below the 
tangent. 

52. Point of Inflection. — A point like A (Fig. 52a), on 
one side of which the curve is concave upward, on the other 
concave downward, is called a point of inflection. It is 
assumed that there is a definite tangent at the point of in- 
flection. A point like B is not called a point of inflection. 



68 



DIFFERENTIAL CALCULUS 



Chap. VII. 



The second derivative is positive on one side of a point of 
inflection and negative on the other. Ordinary functions 
change sign only by passing through zero or infinity. Hence 

to find points of inflection we find where r^ is zero or infinite. 




-t 



Fig. 52a. 



If the second derivative changes sign at such a point, it is a 
point of inflection. If the second derivative has the same 
sign on both sides, it is not a point of inflection. 





Fig. 52b. 



Fig. 52c. 



Example 1. Examine the curve 3y = x* — Qx^ for direc- 
tion of curvature and points of inflection. 
The second derivative is 

(Py 



da^ 



= 4 (x2 - 1). 



Chap. Vn. GEOMETRICAL APPLICATIONS 69 

This is zero at x = ± 1. It is positive and the curve con- 
cave upward on the left of x = —1 and on the right of 
X = -\-l. It is negative and the curve concave downward 
between x = —1 and x =-\- I. The second derivative 
changes sign at A (—1, — f) and B (+1, — §), which are 
therefore points of inflection (Fig. o2b), 

Ex. 2. Ejcamine the curve y = x* for points of inflection. 
In this case the second derivative is 

This is zero when x is zero but is positive for all other values 
of X. The second derivative does not change sign and there 
is consequently no point of inflection (Fig. o2c). 

Ex. 3. If X > 0, show that sin x > x — 5-: •* 

o! 



Let 



^3 

y = sm X - X + ^j 



We are to show that y > 0. Differentiation gives 

dy , , a^2 cPy 

-y- = cos X — 1 + ?ri ' T^ = — sm x + x. 

dx 2! dx2 

When X is positive, sin x is less than x and so -7^ is positive. 

Therefore -^ increases with x. Since -^ is zero when x is 
ax ax 

du 
zero, j^ is then positive when x > 0, and so y increases with 

X. Since y = when x = 0, y is therefore positive when 
X > 0, which was to be proved. 

* If n is any p>ositive integer n ! represents the product <rf the integers 
frwn 1 to n. Thus 

3! = l-2-3 = 6. 



70 DIFFERENTIAL CALCULUS Chap. VII. 



EXERCISES 

Examine the following curves for direction of curvature and points 
of inflection: 

1. 2/ = a;3 - 3 z + 3. 6. ?/ = xe*. 

2. 2/ = 2 x3 - 3 a;2 - 6 X + 1. 6. y = e"^. 

3. y = x^ - 4 x» + 6 x2 + 12 X. 7. x^y - 4 x + 3 y = 0. 

4. 2/3 = X - 1. 8. X = sin <, 2/ = ^ sin 3 t. 

Prove the following inequalities: 
9. X In X - X - ^' + I > 0, if < X < L 

10. (x - 1) e^ + 1 > 0, if X > 0. 

11. 6== < 1 + X + ^ e°, if < X < a. 

12. hisecx>|, if _|<a;<|. 

13. According to Van der Waal's equation, the pressure p and 
volume y of a gas at constant temperature T are connected by the equa- 
tion 

RT a 

^ m{v -h) v^' 

a, b, m, and 12 being constants. If p is taken as ordinate and v as ab- 
scissa, the curve represented by this equation has a point of inflection. 
The value of T for which the tangent at the point of inflection is hori' 
zontal is called the critical temperature. Show that the critical tem- 
perature is 

„ _ 8 am 
~ 27 lib' 

63. Length of a Curve. — The length of an arc PQ of a 
curve is defined as the hmit (if there is a hmit) approached 
by the length of a broken line with vertices on PQ as the 
number of its sides increases indefinitelj^ their lengths ap- 
proaching zero. 

We shall now show that if the slope of a curve is continu- 
ous the ratio of a chord to the arc it subtends approaches 1 as 
the chord approaches zero. 

In the arc PQ (Fig. 53) inscribe a broken line PABQ. 
Projecting on PQ, we get 

PQ = proj. PA -f proj. AB -{■ proj. BQ. 



GEOMETRICAL APPLICATIONS 



71 



The projection of a chord, such as AB, is equal to the prod- 
uct of its length by the cosine of the angle it makes with PQ. 
On the arc AB is a tangent RS parallel to AB. Let a be the 
argest angle that any tangent on the arc PQ makes with the 




Fig. 53. 

chord PQ. The angle between RS and PQ is not greater 
than a. Consequently, the angle between AB and PQ is not 
greater than a. Therefore 

proj. AB = AB cos a. 
Similarly, 

proj. PA = PA cos a, 

proj. BQ = BQ cos a. 
Adding these equations, we get 

PQ = (PA +AB + BQ) cos a. 

It is evident that this result can be extended to a broken 
ine with any number of sides. As the number of sides in- 
sreases indefinitely, the expression in parenthesis approaches 
ihe length of the arc PQ. Therefore 

PQ = arc PQ cos a, 

;liat is, 

chord PQ 

^pc- = cos a. 

arc PQ 

If the slope of the curve is continuous, the angle a ap- 
proaches zero as Q approaches P. Hence cos a approaches 



L and 



,. chord PQ , 

lim 5;^ = 1. 

Q-P arc PQ ^ 



72 



DIFFERENTIAL CALCULUS 



Chap. VII. 



Since the chord is always less than the arc, the limit cannot 
be greater than 1. Therefore, finally, 



,. chord PQ . 
lim ^^7^ = 1. 



(53) 



Q^P arc PQ 

54. Differential of Arc. — Let s be the distance measured 
along a curve from a fixed point A to a variable point P. 
Then s is a function of the coordinates of P. Let </> be the 
angle from the positive direction of the x-axis to the tangent 
PT drawn in the direction of increasing s. 







T 


r 










Q 




/ .^^^'' 






/^^ • 
























w ^ 








B 




1 




S 






\ 

A 




~o 








Fig. 54a. 



Fig. 54b. 



If P moves to a neighboring position Q, the increments in 
X, y, and s are 

Ax = PR, Ay = RQ, As = arc PQ. 

From the figure it is seen that 

Ax Ax As 
cos (RPQ) = PQ = ^ PQ' 

. /r.T>/^N Ay Ay As 

As Q approaches P, RPQ approaches and 

As _ arc PQ 

PQ ~ chord PQ 
approaches 1. The above equations then give in the limit 



dx 



dy 



cos<^ = ^. 8m</, = ^ 



(54a) 



Chap. VII. 



GEOMETRICAL APPLICATIONS 



73 



These equations express that dx and dy are the sides of a 
right triangle with hj-potenuse ds extending along the tangent 
(Fig. 54b). All the equations connecting dx, dy, ds, and <f> can 
be read off this triangle. One of particular importance is 

ds^ = dx^-\- dy\ (54b) 

55. Curvature. — If an arc is everywhere concave toward 
its chord, the amount it is bent can be measured by the angle 
/3 between the tangent-s at its ends. The ratio 
/3 <{>' - <t> _ A<l> 

~ As 



arc PP' 



As 



The 



is the average bending per unit length along PP'. 
limit as P' approaches P, 

,. A<f> d<t> 

A«^o As ds 

is called the curvature at P. It is greater where the curve 
bends more sharph% less where it is more nearly straight. 




Fig. 55a. 
In case of a circle (Fig. 55b) 



Fig. 55b. 



<i> = d-\- 



s = aB. 



Consequently, 



d<t> _ dB 1 
ds add a* 



74 DIFFERENTIAL CALCULUS Chap. VII. 

that is, the curvature of a circle is constant and equal to the 

reciprocal of its radius. 

66. Radius of Curvature. — We have j ust seen that the 

radius of a circle is the reciprocal of its curvature. The 

radius of curvature of any curve is defined as the reciprocal of 

its curvature, that is, 

ds 
radius of curvature = p = jr ' (56a) 

It is the radius of the circle which has the same curvature as 
the given curve at the given point. 

To express p in terms of x and y we note that 



Consequently, 



* = tan-.|. 



^dx 
, 1 , /dy\ _ dx^ 

Also ds = Vdx^ + dyK 

Substituting these values for ds and d4>, we get 



dPy 
dx" 



(56b) 



If the radical in the numerator is taken positive, p will have 

d^v 
the same sign as -r^ , that is, the radius will be positive when 

the curve is concave upward. If merely the numerical value 
is wanted, the sign can be omitted. 
By a similar proof we could show that 



h(l)T 

dy' 



(56c) 



Chap. VII. GEOMETRICAL APPLICATIONS 75 

Example 1. Find the radius of curvature of the parabola 
y2 = 4 x at the point (4, 4). 

At the point (4, 4) the derivatives have the values 

dy^2^1 ^^_4 1^ 

dx y 2' dx^ y^ ~ IQ 

Therefore 

P - ^ - _j_ - - 10 V5. 

dx" 16 

The negative sign shows that the curve is concave downward. 
The length of the radius is 10 Vo. 

Ex. 2. Find the radius of curvature oi the curve repre- 
sented by the polar equation r = a cos 6. 

The expressions for x and y in terms of 6 are 

X = r cos 5 = a cos 6 cos 6 = a cos^ ^, 
y = r sin ^ = a cos dsmd. 

Consequently, 

dy a (cos* B — sin* 6) a cos 2 6 . « „ 

■T- = ^ a ■ a = ^-^ = -cot 2 d, 

dx —2a cos 6 smd — asm 2d 



m 



^ = V^^/ = _ 2 CSC* 2 Odd ^ 2 
dx* dx asm2d dd a 



[1 + cot* 20]^ (csc*2 0)i 



= — a 



a 



2 ,^, 2 esc' 2 2 

esc' 2 

a 

The radius is thus constant. The curve is in fact a circle. 

67. Center and Circle of Curvature. — At each point of 
a curve is a circle on the concave side tangent at the point 
with radius equal to the radius of curvature. This circle is 
called the circle of curvature. Its center is called the center 
of curvature. 

Since the circle and curve are tangent at P, they have the 



76 



DIFFERENTIAL CALCULUS 



Chap. VII. 



dv 
same slope -^ at P. Since they have the same radius of 

curvature, the second derivatives will also be equal at P. 




Fig. 57. 



The circle of curvature is thus the circle through P such that 

dv d/^v 

~ and 74 have the same values for the circle as for the curve 

dx dx^ 

at P. 



EXERCISES 

1. The length of arc measured from a fixed point on a certain curve 
is s = x^ + X. Find the slope of the curve at x = 2. 

2. Can X = cos s,y = sin s, represent a curve on which s is the length 
of arc measured from a fixed point ? Can x = sec s, y = tan s, represent 
such a curve? 

Find the radius of curvature on each of the following curves at the 
point indicated: 



3. ^2 + p=l. at (0,6). 



5. r = e^, at = ^• 



4. x^ + xy + y^ = 3, at (1, 1). 6. r = a (1 + cos 0), at = 0. 

Find an expression for the radius of curvature at any point of each of 
the following curves: 



7. y =l\^ + e »). 



9. X =■ iy* — ihxy. 



8. X = In sec y. 10. r = a sec* f 9. 

11. Show that the radius of curvature at a point of inflection is 
infinite. 



Chap. VII. 



GEOMETRICAL APPLICATIONS 



77 



12. A point on the circumference of a circle rolling along the x-axis 
generates the cycloid 

X = a (ip — sm<f>), J/ = o (1 — COS 4>)t 

a being the radius of the roUing circle and <f> the angle through which it 
has turned. Show that the radius of the circle of curvature is bisected 
by the point where the rolling circle touches the x-axis. 

13. A string held taut is unwound from a fixed circle. The end of 
the string generates a curve with parametric equations 

X = a cos + ad sin 9, y = asind — aO cos 9, 

a being the radius of the circle and 9 the angle subtended at the center 
by the arc unwound. Show that the center of curvatiu"e corresponding 
to any point of this path is the point where the string is tangent to the 
circle. 

14. Show that the radius of curvature at any point (x, y) of the hypo- 
cycloid X* + 2/' = a' is three times the perpendicular from the origin 
to the tangent at (x, y). 

1 cos x 

58. Limit of It is shown in trigonometry 



that 



Consequently, 



1 — cos X = 2 sin' r 



1 2sin2| 

1 — cos X 2.x 

= = sin - 

X X 2 



/. x\ 
sin- 



X 

\ 2 / 



As X approaches zero, 



. X 

X 

2 



approaches 1. Therefore 



iimi^:i^^ = 0.1 = 0. 

1=0 X 



69. Derivatives of Arc in Polar Coordinates. — The 
angle from the outward drawn radius to. the tangent drawn 

Fthe direction of Increasing s is usually represented by the 
tter ^. 



78 



DIFFERENTIAL CALCULUS 



Chap. VIL 



Let r, 6 be the polar coordinates of P, and r + Ar, d + M 
those of Q (Fig. 59a). Draw QR perpendicular to PR and 
let As = arc PQ. Then 

RQ _ (y+Ar) sin A9 _ ^ , ^ >, sin ^^ ^^ As 

'As'PQ' 



sin (/2PQ) = 
cos (RPQ) = 



PQ PQ 

PR _ (r + Ar) cos A0 - r 
PQ~ 



Ad 



Af 
= cos {^9) p^ - 



PQ 

r (1 — cos A9) 

PQ 



= cos {^^) 



Ar As r (1 — cos A0) A0 As 



As PQ 



Ad 



As PQ 





Fig. 59a. Fig. 59b. 



As Ad approaches zero, 

1- /no^x , 1- sinA0 , ,. 1-cosA^ 
\im (RPQ) =rl/, lim-^^=l, hm ^ 

The above equations then give in the limit, 



= 0, lim 



As 
PQ 



1. 



sin }p = 



rdd 
ds 



cos\f/ = 



dr 
ds 



(59a) 



These equations show that dr and rdd are the sides of a 
right triangle with hypotenuse ds and base angle ^. From 
this triangle all the equations connecting dr, dd, ds, and \p 
can be obtained. The most important of these are 

. , rdB 
tan^ = 



dr 



ds^ = dr^ + r2 dd\ 



(59]j) 



Chap. VII. GEOMETRICAL APPLICATIONS 79 

Example. The logarithmic spiral r = as^. 
In this case, dr = ae^ dd and so 

The angle ^ is therefore constant and equal to 45 degrees. 
The equation 

, dr 1 
^^ = di = ^ 

dr . 
shows that -r is also constant and so r and s increase propor- 
tionally. 

EXERCISES 
Find the angle ^ at the point indicated on each of the following curves: 

1. The spiral r = ad, a.t 6 = ^ • 

o 

2. The circle r = a and at 6 =-• 

4 

3. The straight line r = a sec 6, at 9 = ^ • 

o 

4. The eUipse r (2 — cos d) = k, at = ^• 

6. The lemniscate r* = 2 a* cos 2 0, at = f x. 

6. Show that the curves r = ae^, r = ae~^ are perpendicular at each 
of their points of intersection. 

7. Find the angles at which the curves r = a cos 0, r = a sin 2 fl 
intersect. 

8. Find the points on the cardioid r = o (1 — cos 9) where the tan- 
gent is parallel to the initial line. 

9. Let P (r, 9) be a point on the hyperbola r^ sin 2 S = c. Show 
that the triangle formed by the radius OP, the tangent at P, and the 
X-axis is isosceles. 

10. Find the slope of the curve r = e** at the point where 9 =%• 

4 

60. Angle between Two Directed Lines in Space. — 

A directed line is one along which a positive direction is 
assigned. This direction is usually indicated by an arrow. 



80 



DIFFERENTIAL CALCULUS 



Chap. VII. 



An angle between two directed lines is one along the sides 
of which the arrows point away from the vertex. There are 
two such angles less than 360 degrees, their sum being 360 
degrees (Fig. 60). They have the same cosine. 

If the lines do not intersect, the angle between them is de- 
fined as that between intersecting lines respectively parallel 
to the given lines. 




Fig. 60. 



Fig. 61. 



61. Direction Cosines. — It is shown in analytic geome- 
try* that the angles a, /3, 7 between the coordinate axes and 
the line P1P2 (directed from Pi to P2) satisfy the equations 



zi — Zi 



COS a = p p , cos /3 = ^p p^ , cos 7 = -5-B- ' (61a) 
rir2 -Lii^i i^U2 

These cosines are called the direction cosines of the line. 
They satisfy the identity 

cos^ a + cos^ /3 + cos^ 7 = 1. (61b) 

If the direction cosines of two lines are cos ai, cos j8i, cos 71 
and cos 0:2, cos /32, cos 72, the angle 6 between the lines is given 
by the equation 

cos 6 = cos ai cos az + cos ^2 cos ^2 + cos 71 cos 72. (61c) 
In particular, if the lines are perpendicular, the angle d is 90 
degrees and 

= cos ai cos a2 + cos jSi cos 02 + cos 71 cos 72. (6 Id) 

* Cf. H. B. Phillips, Analytic Geometry, Art. 64, et seq. 




Chap. VII. GEOMETRICAL APPLICATIONS 81 

62. Direction of the Tangent Line to a Curve. — The 
tangent line at a point P of a curve is defined as the limiting 
position PT approached by the secant 
PQ as Q approaches P along the curve. 

Let s be the arc of the curve measured 
from some fixed point and cos a, cos /3, 
cos 7 the direction cosines of the tangent 
drawn in the direction of increasing s. 

If X, y, z are the coordinates of P, ^^" ^^^' 

X + Ax, y + At/, z + Az, those of Q, the direction cosines of 
PQare 

Arc Ay Az 
PQ' PQ' PQ' 

As Q approaches P, these approach the direction cosines of 
the tangent at P. Hence 

Ax ,. Ax As 
cos a = lim 757: = lim -r— 757. • 
Q=P PQ As PQ 

On the curve, x, y, z are functions of s. Hence 



,. Ax dx ,. As ,. arc ^ * 
^^°^A^=d^' ^^°^PQ = ^^°^c-hord = ^- 



Therefore 



Similarly, 



cos a = 3- • (62a) 

as 



cos^ = ^, cosT = |. (62a) 



These equations show that if a distance ds is measured 
along the tangent, dx, dy, dz are its projections on the coordi- 
nate axes (Fig. 62b). Since the square on the diagonal of a 

* The proof that the limit of arc/chord is 1 waa given in Art. 53 
for the case of plane curves with continuous slope. A similar proof 
can be given for any curve, plane or space, that is continuous in 
direction. 



82 



DIFFERENTIAL CALCULUS 



Chap. VII. 



rectangular parallelopiped is equal to the sum of the squares 
of its three edges, 

c?.s2 = dx^ + di/ + dz\ (62b) 




Fig. 62b. 

Example. Find the direction cosines of the tangent to the 
parabola 

X = at, y = bt, z = \ cP 

at the point where t = 2. 
At i = 2 the differentials are 

dx = a dt, dy = b dt, dz = \ctdt = c dt, 

ds = zt Vdx^ + dy^ + dz^ = ± Va^ + b^ -{- c^ dt. 

There are two algebraic signs depending on the direction s is 
measured along the curve. If we take the positive sign, the 
direction cosines are 

dx _ a dy _ b 

ds ~ Va^ + 62 _j_ c2 ' ds ~ Va^ + fe^-fc^' 

dz c 

ds ~ Va2 + 62 + c2 

63. Equations of the Tangent Line. — It is shown in 
analytic geometry that the equations of a straight line 



Chap. VII. GEOMETRICAL APPLICATIONS 83 

through a point Pi (xi, yi, Zi) with direction cosines propor- 
tional to A, B, C are 

x-xi _ y -yi _ z-zi ,^„v 

The direction cosines of the tangent line are proportional 
to dx, dy, dz. If then we replace A, B, C hy numbers pro- 
portional to the values of dx, dy, dz at Pi, (63) will represent 
the tangent line at Pi. 

Example 1. Find the equations of the tangent to the curve 
x = t, y = f, z = i^ 

at the point where t = I. 

The point of tangency is t — 1, Xi = 1, yi = 1, Zi = 1. 
At this point the differentials are 

dx : dy : dz = dt :2tdt iSe dt = I : 2 : 3. 
The equations of the tangent line are then 
X — 1 _y — 1 _ 2—1 
1 2 ~~3 

Ex. 2. Find the angle between the curve Sx -\- 2y — 2 z 
= 3, 4 x^ + y^ = 2 2^ and the line joining the origin to 
(1, 2, 2). 

The curve and line intersect at (1, 2, 2). Along the curve 
y and z can be considered functions of x. The differentials 
satisfj^ the equations 

Zdx + 2dy -2dz = 0, Sxdx-\-2y dy = 4:zdz. 
At the point of intersection these equations become 
3dx-\-2dy-2dz = 0, Sdx -{- ^dy = Sdz. 
Solving for dx and dy in terms of dz, we get 
dx = 2dz, dy = —2 dz. 
Consequently, 

ds = Vdx^ -\- dy^ + dz^ = ddz 
and 

dx 2 a dy -2 dz 1 

cos a = -5- = ;r, cos /3 = V- = -5- , COS 7 = -r = 5. 

ds 3 ds 3 ' ds 3 



84 DIFFERENTIAL CALCULUS Chap. VII. 

The line joining the origin and (1, 2, 2) has direction cosines 
equal to 

12 2 
3» 3> 3' 

The angle 6 between the line and curve satisfies the equation 

2-4 + 2 

cos e = - — ^^^^ = 0. 

The line and curve intersect at right angles. 

EXERCISES 

Find the equations of the tangent lines to the following curves at the 
points indicated: 

1. a; = sec t, y = tan t, z = at, at t = -:• 

2. X = e*, y = e~*, z = fi, at < = 1. 

3. X = e^ sin /, y = e* cos t, z = kt, at < = 5* 

4. On the circle 

X = a cos 6, y = a cos (0 + 0^), z = a cos ( ^ + 5 "" J 

show that ds is proportional to dd. 

5. Find the angle at which the helix 

X = a cos 6, y = a sin 6, z = kd 
cuts the generators of the cylinder x^ + y^ = a^ on which it lies. 

6. Find the angle at which the conical helix 

X = t cos t, y = t sin t, z = t 
cuts the generators of the cone x^ + y^ = z^ on which it lies. 

7. Find the angle between the two circles' cut from the sphere 
x'^ + y^ + z^ = 14 by the planes x — y + z = and x + y — z = 2. 



CHAPTER Vin 

VELOCITY AND ACCELERATION IN A CURVED 

PATH 

64. Speed of a Particle. — When a particle moves along 
a curve, its speed is the rate of change of distance along the 
path. 

Let a particle P move along the cm^e AB, Fig. 64. Let s 
be the arc from a fixed point A to P. The speed of the par- 
ticle is then 





Fig. 64. 



Fig. 65a. 



66. Velocity of a Particle. — The velocity of a particle 
at the point P in its path is defined as the vector* PT tangent 
to the path at P, drawn in the direction of motion with length 
equal to the speed at P. To specify the velocity we must 
then give the speed and direction of motion. 

* A vector is a quantity having length and direction. The direction 
is usually indicated by an arrow. Two vectors are called equal when 
" they extend along the same line or along parallel lines and have the 
i same length and direction. 

85 




86 DIFFERENTIAL CALCULUS Chap. VIII. 

The particle can be considered as moving instantaneously 
in the direction of the tangent. The velocity indicates in 
magnitude and direction the distance it would move in a 

unit of time if the speed and direc- 
tion of motion did not change. 

Example. A wheel 4 ft. in di- 
ameter rotates at the rate of 500 
revolutions per minute. Find the 
speed and velocity of a point on its 

Let OA be a fixed line through the 
center of the wheel and s the distance along the wheel from 
OA to a moving point P. Then 

s = 2dh. 

The speed of P is 

^ = 2^=2 (500) 2 TT = 20007r ft./mm. 
at at 

Its velocity is 2000 tt ft./min. in the direction of the tangent 
at P. The speeds of all points on the rim are the same. 
Their velocities differ in direction. 

66. Components of Velocity in a Plane. — To specify a 
velocity in a plane it is customary to give its components, 
that is, its projections on the coordinate axes. 

If PT is the velocity at P (Fig. 66), the x-component is 

„^ „^ ds , ds dx dx 

PQ = Prcos<^=^cos0 = ^^ = ^, 

and the ^/-component is 

^^ ^,r, . ds . ds dy dy 

The components are thus the rates of change of the coordinates. 
Since 

PT^ = PQ2 + QT\ 



Chap. VIII. VELOCITY AND ACCELERATION 



87 



the speed 'S expressed in terms of the components by the 
equation 

(dsV ^ (dxV (dyV 

\dt) \dtl "^ \dtj 





Fig. 66. 



Fig. 67. 



67. Components in Space. — If a particle is moving 
along a space curve, the projections of its velocity on the 
three coordinate axes are called components. 

Thus, if PT (Fig. 67) represents the velocity of a point, its 
components are 

ds dx _ dx 
dt ds dt * 



PQ = PT cos a 



dt ds dt 

RT = PT cosy = ^^^ = ^- 
dt ds dt 

Smce PT^ = PQ^ + QR^ -\- RT^, the speed and compo- 
lents are connected by the equation 



(f=(tr+(fj+(fj 



88 DIFFERENTIAL CALCULUS Chap. VIII. 

68. Notation. — In this book we shall indicate a vector 
with given components by placing the components in brack- 
ets. Thus to indicate that a velocity has an a;-component 
equal to 3 and a ^/-component equal to —2, we shall simply 
say that the velocity is [3, —2]. Similarly, a vector in space 
with x-component a, ^/-component h, and 2-component c will 
be represented by the symbol [a, h, c]. 

Example 1. Neglecting the resistance of the air a bullet 
fired with a velocity of 1000 ft. per second at an angle of 30 
degrees with the horizontal plane will move a horizontal 
distance 

x = 500t Vd 

and a vertical distance 

y = 500t- 16.1 1^ 

in t seconds. Find its velocity and speed at the end of 10 
seconds. 
The components of velocity are 

^ = 500 V3, § = 500 - 32.2 t 
at at 

At the end of 10 seconds the velocity is then 

V= [500 V3, 178] 
and the speed is 

j^ = V(500V3)' + (178)2 = 884 ft./sec." 

Ex. 2. A point on the thread of a screw which is turned 
into a fixed nut describes a helix with equations 

X = r cos 6, y = rsind, z = kd, 

S being the angle through which the screw has turned, r the 
radius, and k the pitch of the screw. Find the velocity and 
speed of the point. 



I 



Chap. VIII. VELOCITY AND ACCELERATION 



89 



The components of velocity are 

dx . ^dd dy „dd dz J dd 

Since -77 is the angular velocity « with which the screw is 

rotating, the velocity of the moving px)int is 
V = [— rw sin 6, rw cos d, ha] \ 
and its speed is 

ds 



dt 



= VrV gjn2 e^r^f^ cos" d + kW = « Vr^ + k^, 



which is constant. 




/:5i 






Fig. 69a. 



Fig. 69b. 



69. Composition of Velocities. — By the sum of two 

velocities Vi and V2 is meant the velocity Vi + V2 whose 
components are obtained by adding corresponding compo- 
nents of Vi and Fo. Similarly, the difference V2 — Vi is the 
velocity whose components are obtained by subtracting the 
components of Vi from the corresponding ones of V2. 

Thus, if 

Vi = [ai, bi], Vi = [aa, 62], 
Fi + F2= [01 + 02,61 + 62], Vz-Vt= [02-01,62-6,]. 

If Fi and F2 extend from the same point (Fig. 69a), 
Fi + F2 is one diagonal of the parallelogram with Fi and 
F2 as adjacent sides and F2 — Fi is the other. In this case 
F2 — Fi extends from the end of Fi to the end of Fj. 



90 



DIFFERENTIAL CALCULUS 



Chap. VIII. 



By the product wF of a vector by a number is meant a 
vector m times as long as V and extending in the same direc- 
tion if m is positive but the opposite direction if m is negative. 
It is evident from Fig. 69b that the components of mF are 
m times those of F. 



F 1 

The quotient — can be considered as a product — F. 
m m 



Its 



A = Hm . 

At=0 i^t 



components are obtained by dividing those of F by m. 

70. Acceleration. — The acceleration of a particle mov- 
ing along a curved path is the rate of change of its velocity 

AF dV 
dt ' 
In this equation AF is a vector 

and -TT 
At 

ing the 

byA^ 

Let the particle move from 

the point P where the velocity 

is F to an adjacent point P' 

where the velocity is F + A F. 




is obtained by divid- 
components of AF 



Fig. 70a. 



The components of velocity will change from 



dx 
dt 



|t° 



dx .dx 
dt^ di 



dy. .dy 
dt'^ dt 



Consequently, 

_ Vdx dy] 
^ ~ Idt ' Tt\ 
Subtraction and division by At give 



F + AF 



=[' 



dx ^^ f. dx 
dt'^^dt 



dy,Kdy' 
dt'^ dt 



. ,, r . dx . dyl A F 



dx . dy 



dt 
At 



dt 

At 



As At approaches zero, the last equation approaches 
._dV_VdH d^l 



Chap. VIII. VELOCITY AND ACCELERATION 



91 



In the same way the acceleration of a particle moving in 
space is found to be 






(70b) 



Equations 70a and 70b express that the components of the 
acceleration of a moving particle are the second derivatives of 
Us coordinates with respect to 
the time. 

Example. A particle 
moves with a constant 
speed V around a circle of 
radius r. Find its velocity 
and acceleration at each 
point of the path. 

Let = AOP. The co- 
ordinates of P are 

x== r cos Qy 
The velocity of P is 




-[- 



r sm -t: , r cos Q 
dt 



„. ^ ds dd 

Smce s = r^, y: = V = r -fi , 
dt dt 



The velocity can therefore be 



written 

V=[—vsmd, wcos^]. 
Since v is constant, the acceleration is 



. dV Vd , . ^s d , ,,1 
= |_-.cos0^. -.sm^^J. 



Replacing ;77 by - , this reduces to 



[t'- v^ ~\ v^ 
COS0, sua© = -[— COS0, — sin^l. 
r r J r 

Now [— COS0, — sin0] is a vector of unit length directed 



92 DIFFERENTIAL CALCULUS Chap. VIIL 

along PO toward the center. Hence the acceleration of P is 
directed toward the center of the circle and has a magnitude 

equal to — . 

EXERCISES 

1. A point P moves with constant speed v along the straight line y=a. 
Find the speed with which the line joining P to the origin rotates. 

2. A rod of length a sUdes with its ends in the x- and y-axes. If the 
end in the x-axis moves with constant speed v, find the velocity and speed 
of the middle point of the rod. 

3. A wheel of radius a rotates about its center with angular speed « 
while the center moves along the x-axis with velocity v. Find the velocity 
and speed of a point on the perimeter of the wheel. 

4. Two particles Pi (xi, yi) and Pt (xz, 2/2) move in such a way that 

xi = 1 + 2 <, r/i = 2 - 3 <2, 

X2 = 3 + 2 <2, y^= - 4 i». 

Find the two velocities and show that they are always parallel. 

5. Two particles Pi ( xi, yi, Zi) and P2 (x2, 2/2, Z2) move in such a way 
that 

Xi = o cos 6, yi = a cos (6 + i ir), Zi = a cos (^ + § jt), 
Xi = a sin d, yz = a sin {9 + i ir) , Zi = a ain {9 -{■ i it) . 

Find the two velocities and show that they are always at right angles. 

6. A man can row 3 miles per hour and walk 4. He wishes to cross 
a river and arrive at a point 6 miles further up the river. If the river is 
If miles wide and the current flows 2 miles per hour, find the course he 
shall take to reach his destination in the least time. 

7. Neglecting the resistance of the air a projectile fired with velocity 
[a, b, c] moves in t seconds to a position 

X = at, y = bt, z = ct — \ gfi. 

Find its speed, velocity, and acceleration. 

8. A particle moves along the parabola 3? = ay in such a way that 

■J- is constant. Show that its acceleration is constant. 
dt 

9. When a wheel rolls along a straight line, a point on its circum- 
ference describes a cycloid with parametric equations 

X = o (^ — sin 0), y = a (I — cos <t>), 

a being the radius of the wheel and </> the angle through which it hab 
rotated. Find the speed, velocity, and acceleration of the moving point. 



Chap. Vni. VELOCITY AND ACCELERATION 93 

10. Find the acceleration of a particle moving with constant speed v 
along the cardioid r = a {1 — cob 6). 

11. If a string is held taut while it is unwoxind from a fixed circle, its 
end describes the curve 

X = a cos d + a d an 0, y = a tan d — a 9 coe d, 

e being the angle subtended at the center by the arc unwound. Show 

that the end moves at each instant with the same velocity it would have 

dd 
if the straight part of the string rotated with angular velocity -t- about 

the point where it meets the fixed circle. 

12. A piece of mechanism consists of a rod rotating in a plane with 
constant angular velocity w about one end and a ring sUding along the 
rod with constant speed v. (1) If when t = the ring is at the center 
of rotation, find its position, velocity, and acceleration as functions of the 
time. (2) Find the velocity and acceleration immediately after t = ti, 
if at that instant the rod ceases to rotate but the ring continues sliding 
with unchanged speed along the rod. (3) Find the velocity and acceler- 
ation immediately after / = /i if at that instant the ring ceases sliding 
but the rod continues rotating. (4) How are the three velocities re- 
lated? How are the three accelerations related? 

13. Two rods AB, BC are hinged at B and lie in a plane. A is 
fixed, AB rotates with angular speed « about A, and BC rotates with 
angular speed 2 u about B. (1) If when t = 0, C lies on AB produced, 
find the path, velocity, and acceleration of C. (2) Find the velocities 
and accelerations immediately after < = /i if at that instant one of the 
rotations ceases. (3) How are the actual velocity and acceleration 
related to these partial velocities and accelerations? 

14. A hoop of radius a rolls with angular velocity on along a horizon- 
tal line, while an insect crawls along the rim with speed cuat. If when 
t = the insect is at the bottom of the hoop, find its path, velocity, and 
acceleration. The motion of the insect results from three simultaneous 
actions, the advance of the center of the hoop with speed cu^, the rota- 
tion of the hoop about its center with angular speed a>i, and the crawl 
of the insect advancing its radius with angular speed wi. Find the three 
velocities and accelerations which result if at the time t = ti two of these 
actions cease, the third continuing unchanged. How are the actual 
velocity and acceleration related to these partial velocities and accelera- 
tions? 



CHAPTER IX 



ROLLE'S THEOREM AND INDETERMINATE 
FORMS 

71. Rolle's Theorem. — // /' (x) is continuous, there is at 
least one real root of f (x) = between each pair of real roots 
off{x) =0. 

To show this consider 
the curve 

y=fix)- 

Let / (x) be zero at 
X = a and x = b. Be- 
tween a and b there 
must be one or more 
^^^- ^^^- points P at maximum 




distance from the x-axis. 
horizontal and so 



\t such a point the tangent is 



| = /'(x) = 0. 

That this theorem may not hold if /' (x) is discontinuous 
is shown in Figs. 71b and 71c. In both cases the curve 

Y 




Fig. 71b. 



Fig. 71c. 



crosses the x-axis at a and b but there is no intermediate 
point where the slope is zero. 

n 



Chap. IX. ROLLE'S THEOREM 95 

Example. Show that the equation 
a:' + 3x - 6 = 
cannot have more than one real root. 
Let 

Then 

/' (x) = 3 a;2 + 3 = 3 (x^ + 1). 

Since /' (x) does not vanish for any real value of x, / (x) =0 
cannot have more than one real root ; for if there were two 
there would be a root of /' (x) = between them. 
72. Indeterminate Forms. — The expressions 

^, -, 0.x, X - 00, 1-, 0^ 00° 

are called indeterminate forms. No definite values can be 
assigned to them. 

If when X = a a function / (x) assumes an indeterminate 
form, there may however be a definite limit 

lim/(x). 

x=a 

In such cases this limit is usually taken as the value of the 
function at x = a. 
For example, when x = the function 
2x _0 
X "0* 
It is e\'ident, however, that 

Urn — = lim (2) = 2. 

x=0 X 

This example shows that an indeterminate form can often be 

made definite b}' an algebraic change of form. 

aa 

73. The Forms -r and - . — We shall now show that, if 
00 ' 

f (x) 
for a particular value of the variable a fraction „\ \ assumes 

F(x) 

the form - or — , numerator and denominator can be replaced 



96 DIFFERENTIAL CALCULUS Chap. IX. 

by their derivatives without changing the value of the limit 
approached by the fraction as x approaches a. 

1. Let /' {x) and F' (x) be continuous between a and 6 
Iff («) = 0, F (a) = 0, and F (6) is not zero, there is a number 
Xi between a and b such that 

fib)_f(x^) 



Fib) F'(xi) 

To show this let ^^ = R. Then 
Fib) 



(73a) 



fib) -RFib)=0, 
Consider the function 

/ ix) - RF ix). 

This function vanishes when x = b. Since / (a) = 0, 
F (a) = 0, it also vanishes when x = a. By RoUe's Theorem 
there is then a value Xi between a and b such that 

/' ixO - RF' ixO = 0. 
Consequently, 

Fib) "" F'ix^)' 

which was to be proved. 

2. Let/' ix) and F' ix) be continuous near a. /// (a) = 
and F (a) = 0, then 



(73b) 



r f(x) ,. fix) 
:,=„ i^ (a:) x=« F' ix) 

For, if we replace b by x, (73a) becomes 

fix) ^fix,) 
Fix) F'ix,y 

Xi being between a and x. Since Xi approaches a as x ap- 
proaches a, 

iS F (x) i;E F' ixO S F' (x) 
3. In the neighborhood of x = a, let /' ix) and F' ix) be 



Chap. IX. ROLLE'S THEOREM 97 

continuous at all points except x = a. If f (x) and F (x) 
approach infinity as x approaches a, 

"™ F (X) "2 F' (I) 
To show this let c be near a and on the same side as x. 
Since f (x) — f (c) and F {x) — F (c) are zero when x = c, 
by Theorem 1, 

f{xr)^ f{x)-f(c) ^ /(x)' fix) 
F'(xi) F (x) - F(c) F(x) F(c) ' 

where Xi is between x and c. As x approaches a, / (x) and 
F (x) increase indefinitely. The quantities / (c)/f (x) and 
F (c)/F (x) approach zero. The right side of this equation 
therefore approaches 

Since Xi is between c and a, by taking c sufficiently near to 
a the left side of the equation can be made to approach 

.2 F' (x) 

Since the two sides are always equal, we therefore conclude 
that 

"2 F (x) 'iS f ' (x) 

sin X 
Example 1. Find the value approached by as x 

approaches zero. 

Since the numerator and denominator are zero when x = 0, 
we can apply Theorem 2 and so get 

,. sin X ,. cosx 
lim = lim — - — = 1. 

x=0 ^ 1=0 1 

Ex. 2. Find the value of lim -. r^ • 

x=, (tt — x)* 



98 DIFFERENTIAL CALCULUS Chap. IX. 

When X = T the numerator and denominator are both 
zero. Hence 

,. l + cosx ,. (— sinx) 

hm -, Tj- = hm -^^. ^ = -• 

x=^ (tt - xf x=T -2 {tt — x) 

Since this is indeterminate we apply the method a second 
time and so obtain 

,. sin X _ y cosx _ 1 

x=x ^ [tt X) x=ic ■^ ^ 

The value required is therefore \. 

+ Q fl ^ O* 

Ex. 3. Find the value approached by — as x ap- 

tan X 

proaches x • 
id 

TT 

When X approaches - the numerator and denominator of 

this fraction approach oo. Therefore, by Theorem 3, 

lim tan 3 a: ,. 3 sec^ 3 a; ,. 3 cos^ x 

. T -7 = lim ^ = lim — ^-^r- ' 

a;=2 tanx sec^x cos^Sx 

TT 0- 

When X is replaced by ^ the last expression takes the form j: • 
Therefore 

,. 3 cos^ aj ,. 6 cos x sin x 
hm — TTir- = hm 



cos^ 3 a; 6 cos 3 x sin 3 x 

cos^ X — sin^ X 1 



= lim 



3 (cos2 3 X - sin2 3 x) 3 



74. The Forms • oo and oo — oo. — By transforming 

e expression t^ 
For example, 



the expression to a fraction it will take the form r: or — 



a; In a; 



Chap. DC. ROLLE'S THEOREM 99 

has the form • oo when x = 0. It can, however, be 
written 

, In a; 

I xmx = -r— » 



00 

which has the form — 

00 

The expression 

sec X — tan x 

has the form x — oo when x = ^- It can, however, be 

written 

1 sin a; 1 — sin x 

sec X — tan x = = , 

cos X cos x cos X 

which becomes ^ when a: = ^ • 

75. The Forms 0**, 1* , oo°. — The logarithm of the given 
function has the form • x. From the limit of the log- 
arithm the limit of the function can be determined. 

1 

Example. Find the limit of (1 + x)' as x approaches 

zero. 

1 

Let i/ = (l+x)'. 



Then 



1 1 1 /I I \ In (1 + x) 
\ny =-ln(l + x) =^ '- 



TNTien x is zero this last expression becomes ^ . Therefore 

^^!n(l+£) _ ,.„ 1 ^ J 

x-0 a: 1 + X 

The Umit of In y being 1, the limit of y is e.] 



100 



DIFFERENTIAL CALCULUS 



Chap. IX. 



EXERCISES 
1. Show by RoUe's Theorem that the equation 
X* — 4x — 1 = 

cannot have more than two real roots. 

Determme the values of the following limits: 



3^-1 

'ITxio - l' 
x" - 1 

'ir X — 1 

1 — cos X 

x=o sin X 



2. Lira 

3. Lim 

a;=l 

4. Lim 

x=0 
6. Lim 



X — a 

. Lim : • 

1=0 a; — sm a; 

_ T • x^ cos X 

7. Lim -. 

x=0 cos X — 1 

8. Lim^5j£^. 

x=3 X —3 

rt T • In cos a; 

9. Lim • 

x=0 X 

10. Lim 

11. Lim 



=^2 (x - 2)2 

1 + cos X — sin a; 
cos X (2 sin x — 1) 



12. Lim ^"gio (s^" ^ ~ sin a) 
x=a logio (tan X — tan a) 

13. Li^6sinx-6x + x3^ 
a;=0 a;2 

14. Lim ^^^^ <A - 2 tan <^ 

^ 1 + cos 4 </> 

16. Limi^. 
xdro cot .r 

16. Lim^. 



17. Lim 



sec S X — X 



If sec X — X 



18. Liml±^^. 
x=^|sec(x + ^j 

19. Lim a; cot x. 

20. Lim tan x cos 3 x. 



21. Lim (x + a)lnfl+-V 

X = CI0 \ X/ 



22. 


Lim (x - 3) cot (ttx). 




a;=3 


23. 


Limn ffx + ^)~f(x)' 
n=oo L V n/ _ 




f 1 


24. 


Lim x^e**. 




x=0 


26. 


Lim ). 

a:=0 \a; ^ - 1/ 


26. 


Lim (cotx — Inx). 




x=0 


27. 


Lim tanx — : r-r— | 

.tL smx— sm*a;J 

^~9 




28. 


Lira X*. 




x=0 


29. 


Lim(sinx)**°*. 




-=^1 




1 


30. 


Lim (1 + ax)*. 




x=0 




1 


81. 


T<im (x"* — a*") in «. 

Z»ao 



CHAPTER X 

SERIES AND APPROXIMATIONS 

76. Mean Value Theorem. — // / (x) and /' (x) are can- 
tinuousfram x = a to x = b, there is a value X\ between a and b 
such thai 



fib)-f(a) 
b — a 



= /' (xi). 



(76T 



To show this consider the curve y = f(x). Since f (a) 
and / (6) are the ordinates at x = a and x = b, 
f(b)-f(a) ^ ^ ^j ^^^^^ ^^ 
b — a 
On the arc AB let Pi be a point at maximum distance from 




Fig. 76. 

the chord. The tangent at Pi will be parallel to the chord 
and so its slope/' (xi) will equal that of the chord. Therefore 

b — a 

which was to be proved. 

Replacing 6 by x and solving for/ (x), equation (76) becomes 

Six) =/(a)4-(x-a)/'(xi). 
101 



102 DIFFERENTIAL CALCULUS Chap. X. 

Xi being between a and x. This is a special case of a more 
general theorem which we shall now prove. 

77. Taylor's Theorem. — If f (x) and all its derivatives 
used are continuous from a to x, there is a value Xi between a and x 
such that 

f (x) =f{a) + {x- a)r (a) +^^^r (a) 

+ ^^^ /'" («)+••• +^^^ /" (^>). 
To prove this let 

<i>{x) =f{x)-f{a)-{x-a)f'{a) 

_ {x - aY . ^ _ . . . _ {x - ay-' . 

2! ^ ^""^ (n- 1)! ^ ^"^* 

It is easily seen that 

<t> (a) = 0, <l>' (a) = 0, <t>" (a) = 0, 

. . . r-' (a) = 0, 0" (x) = /» (x). 
When x = a the function 

<l>(x) 
{x - a)* 

therefore assumes the form ^. By Art. 73 there is then a 

value 2i between a and x such that 

<f> (x) <t>' (zi) 



(x — aY n {zi — a)"'^ 
This new expression becomes ^ when z, = a. There is con- 
sequently a value Zi between Zi and a (and so between x and a) 
such that 

<f>'(zi) ^ <i>" (z,) 

n {zi — a)"-i n (n - 1) (22 - a)""" ^ 

A continuation of this argument gives finally 

<i>{x) ^ 0" (Zn) ^ /" (Zn) V 

(x — a)" n! n! * ' 



Chap. X. SERIES AND APPROXIMATIONS 103 

2» being between x and a. If Xi = 2„ we then have 

Equating this to the original value of <t> (x) and solving for 
/ (x), we get 

f{x)=f{a) + (x-a)r(ay 

+ (^V'(a)+...+^V"(xO. 

which was to be proved. 
Example. Prove 

hix=(x-l) 2~ + ~~3 4^7"* 

where Xi is between 1 and x. 
When X = 1 the values of In x and its derivatives are 

/(x)=bi(x). /(1)=0. 

/' (X) = l^ r (1) = 1, 

r(x) = -i, r(i) = -1, 

r'(x) = |. r'(i) = 2, 

r"(x) = -^, r"(xo = -(!),• 

Taking a = 1, Taylor's Theorem gives 

hix = + l(x-l)-i(x-l)2 + ?(x-l)3-^^^^\ 

which is the result required. 

78. Approximate Values of Functions. — The last term 
in Taylor's formula 

(x — a)" , , , _ 



104 DIFFERENTIAL CALCULUS Chap. X. 

is called the remainder. If this is small, an approximate 
value of the function is 

S(x)=fia) + ix-a)ria) 

(x - ay „ , . (x - g)"-^ . , , . 

+ -^j / (a) + • • • + -(^j-JTiyr /" («)' 

the error in the approximation being equal to the remainder. 

To compute/ (x) by this formula, we must know the values 
of/ (a), /' (a), etc. We must then assign a value to a such that 
/(^)» /' (^)> ^^c., are known. Furthermore, a should be as 
close as possible to the value x at which f (x) is wanted. For, 
the smaller x — a, the fewer terms (x — a)^, (x — aY, etc., 
need be computed to give a required approximation. 

Example 1. Find tan 46° to four decimals. 

The value closest to 46° for which tan x and its derivatives 

are known is 45". Therefore we let a = t' 

4 

/(a;) = tana;, "^W^^' 

. f (x)=sec«x, ^'(S)^^' 

f" (x) = 2 sec2 X tan x, f" (|) = 4, 

f" (x) = 2 sec* X -\- A. sec^ x tan^ x. 
Using these values in Taylor's formula, we get 

tan. = l + 2(.-|) + i(x-.)%/^(.-|; 
and 

tan46« = l + 2(j|5) + 2(j|5)'= 1.0355 

approximately. Since Xi is between 45° and 46°, /'" (xi) 
does not differ much from 

f" (45°) = 8 + 8 = 16. 



Chap. X. SERIES AND APPROXIMATIONS 105 

The error in the above approximation is thus very nearly 

6" (l^) < 3750? ^ 40:000 = 0-0^25. 
It is therefore correct to 4 decimals. 

Ex. 2. Fmd the value of e to four decimals. 

The only value of x for which e* and its derivatives are 
known is a; = 0. We therefore let a be zero. 

fix) = e-, / (x) = e*, r{x)=^, , f- (x) = e*, 

/(O) = 1, /' (0) = 1, /" (0) = 1, , /» (xx) = e^. 

By Taylor's Theorem, 

^ = i + x + 2i + 3i+ • • • + (;rriy! + ^- 

Letting x = 1, this becomes 

6 = 1+1+^ + 57+ • • • + 7 TV-i + — 1- 

2! 3! (n — 1)! n! 

In particular, if n = 2, 

e = 2 + I e*'. 
Since Xi is between and 1. e is then between 2| and 2 + 
^ e, and therefore between 2^ and 4. To get a better ap- 
proximation let n = 9. Then 

e = l + l + ^ + Jr+--- +3^2-7183 

approximately, the error being 

9! ^ 9] ^ 9] ^ .00002. 
The value 2.7183 is therefore correct to four decimals. 
EXERCISES 

Determine the values of the following functions correct to four 
decimals: 

1. sin 5". 5. sec (10°). 

2. cos 32°. 6. hi (t?5). 

3. cot 43°. 7. VI. 

4. tan 58°. 8. tan"! i^). 
9. Given In 3 = 1.0986, hi 5 = 1.6094, find hi 17. 



106 DIFFERENTIAL CALCULUS Chap. X. 

79. Taylor's and Maclaurin's Series. — As w increases 
indefinitely, the remainder in Taylor's formula 

Rn = ■^^— ^/" (Xx) 

n 
often approaches zero. In that case 

S{x)=\\m\f{a) + {x-a)r{a)+ • • • + ^T'^lXj '"' («)| 

This is usually written 

Six) =/ (a) + (X ~ a)r (a) +^^^V" («) 



3! 

the dots at the end signifying the limit of the sum as the 
number of terms is indefinitely increased. Such an infinite 
sum is called an infinite series. This one is called Taylor's 
Series. 

In particular, if a = 0, Taylor's Series becomes 

/(a;)=/(0)+a;f(0)+|r(0)+|^r'(0)+ .... 

This is called Maclaurin's Series. 

Example. Show that cos x is represented by the series 

x^ , X* x* , 
cosx=l-2| + 4j-^+ • • • . 

The series given contains powers of x. This happens when 

a = 0, that is, when Taylor's Series reduces to Maclaurin's. 

/ (x) = cos X, / (0) = 1, 

/' (x)= -sinx, /(0)=0, 

r(x) = -cosx, /"(0)=-l, 

/'" (x) = sin X, f" (0) = 0, 

f" (x) = cos X, J"" (0) = 1. 

These values give 

cos X = 1 - 2j + 41 - • • • ± -, /" (a;i). 



Chap. X. SERIES AND APPROXIMATIONS 107 

The nth derivative of cos x is ±cos x or isin x, depending 
on whether n is even or odd. Since sin x and cos x are never 
greater than !,/„ (xi) is not greater than 1. Furthermore 

^ _ X X X X 

n! ~ l'2'3* ' ' ' n 
can be made as small as you please by taking n sufl&ciently 
large. Hence the remainder approaches zero and so 

, X2 , X* x^ , 

cosx = l-2! + 4!~6!"^ ' ' * ' 
which was to be proved. 

EXERCISES 

1. smx = x-3J + ^-yj + - • • . 

- - , ^ 2x» 4x« 4x6 ,8x^, 

6. (a + x)" = a» + no»-»x + " %7 ^^ """^ + • • • , if |xl*< |o|. 

8. lnx = ln3+^-^^^' + ^^^ .iflx-3I<l. 

9. ln(x + 5)-ln6+^^-^|^'+^|^' ,if|x-l|<l. 

80. Convergence and Divergence of Series. — An in- 
finite series is said to converge if the sum of the first n terms 
approaches a Umit as n increases indefinitely. If this sum 
does not approach a limit, the series is said to diverge. 

The series for sin x and cos x converge for all values of x. 
The geometrical series 

a -\- ar -{- ar^ -\- ar^ -\- ar* -\- ■ • • 

* The symbol |x| is used to represent the numerical value of x with- 
out its algebraic sign. Thus, | — 3 1 = | 3 | = 3. 



108 DIFFERENTIAL CALCULUS Chap. X 

converges when r is numerically less than 1. For the sum of 
the first n terms is 

Sn = a -]r ar -\- ar^ + ...-{- ar""-^ = a \ ~ ^" • 

1 — r 

If r is numerically less than 1, r" approaches zero and Sn 
approaches 

1 — r 

as n increases indefinitely. 
The series 

1-1 + 1-1 + 1-1+.. . 

is divergent, for the sum oscillates between and 1 and does 
not approach a limit. The geometrical series 
1 + 2 + 4 + 8+16+ • • • 

diverges because the sum increases indefinitely and so does 
not approach a limit. 

81. Tests for Convergence. — The convergence of a 
series can often be determined from the problem in which it 
occurs. Thus the series 

1 — ^ 4- ^ _ ^^ _j_ 

2!"^ 4! 6"!"^ ' ' * 

converges because the sum of n terms approaches cos aj as n 
increases indefinitely. 

The terms near the beginning of a series (if they are all 
finite) have no influence on the convergence or divergence of 
the series. This is determined by terms indefinitely far out 
in the series. 

82. General Test. — Fcrr the series 

Wl + W2 + W3 + • • • + Wn + • • • 

to converge it is necessary and sufficient that the sum of terms 
beyond Un approach zero as n increases indefinitely. 

For, if the series converges, the sum of n terms must ap- 
proach a limit and so the sum of terms beyond the nth must 
approach zero. 



Chap. X. SERIES AND APPROXIMATIONS 109 

83. Comparison Test. — A series is convergent if beyond 
a certain point its terms are in numerical value respectively less 
than those of a convergent series whose terms are all positive. 

For, if a series converges, the sum of terms beyond the nth 
will approach zero as n increases indefinitely. If then another 
series has lesser corresponding terms, their smn will approach 
zero and the series will converge. 

84. Ratio Test. — If the ratio -^ of consecutive terms 

^ Un ^ 

approaches a limit r as n increases indefinitely, the series 

^1 + 1^2 + ^3+ • • • +W„ + Un+l + • • • 

IS convergent if r is numerically less than 1 and divergent if r is 
numerically greater than 1 . 

Since the limit is r, by taking n sufficiently large the ratio 
of consecutive terms can be made as nearly r as we please. 
If r < 1, let ri be a fixed number between r and 1. We can 
take n so large that the ratio of consecutive terms is less than 
ri. Then 

Wn+l < riMn, Mn+2 < ^Mn+i < r^U,,, etC, 

Beyond w„ the terms of the given series are therefore less than 
those of the geometrical progression 

Un + riUn + rihin + • • • 

which converges since rx is numerically less than 1. Con- 
sequently the given series converges. 

If, however, r is greater than 1, the terms of the series must 
ultimately increase. The terms do not then approach zero 
and their sum cannot approach a limit. 

Example. Find for what values of x the series 
X + 2 x2 ^ 3 3^ ^ 4 3^ _^ , . . 

converges. 

The ratio of consecutive terms is 

Wn+i (n + 1) x"+i 



Un nx" 



-(}-¥■ 



110 DIFFERENTIAL CALCULUS Chap. X. 

The limit of this ratio is 

r = lim [I + -] X = X. 

n=oo \ n/ 

The series will converge if x is numerically less than 1. 

85. Power Series. — A series of powers of {x — a) of 
the form 

P (x) = oo + ai (x — a) + 02 (x — a)2 + a3 (x — a)' + • • • , 

where a, ao, ai, a^, etc., are constants, is called a power series. 
If a power series converges when x = b, it will converge for 
all values of x nearer to a than b is, that is, such that 
\x — a\ < \b — a\. 
In fact, if the series converges when x = b, each term of 
Oo + ai (6 - a) + 02 (6 - o)2 + as (6 - a)3 -I- • • . 
will be less than a maximum value M, that is, 

|a„ {b - o)"| < M. 
Consequently, 

The terms of the series 

oo + oi (x — o) + 02 (a; — 0)2 -{- as (x — a)^ + • • • 
are then respectively less than those of the geometrical series 

in which the ratio is 

\x — a\ 



\b-a\ 

If then |x — o|<|6 — a|, the progression and consequently 
the given series will converge. 

// o power series diverges when x = b, it will diverge for all 
valu£s of X further from a than b is, that is, such that 

jr — o| > |b — to|, 



Chap. X. SERIES AND APPROXIMATIONS 111 

For it could not converge beyond b, since by the proof just 
given it would then converge at b. 

This theorem shows in certain cases why a Taylor's Series 
is not convergent. Take, for example, the series 

X^ 3^ X^ 

ln(l + a;) =^-2!~^3]~4]'*" ' ' * * 

As X approaches — 1, In (1 + x) approaches infinity. Since 
a convergent series cannot have an infinite value, we should 
expect the series to diverge when x = — 1. It must then 
diverge when x is at a distance greater than 1 from a = 0. 
The series in fact converges between x = — 1 and x = 1 and 
diverges for values of x numerically greater than 1. 

86. Operations with Power Series. — It is shown in 
more advanced treatises that convergent series can be added, 
subtracted, multiplied and divided like polynomials. In 
case of division, however, the resulting series will not usually 
converge beyond a point where the denominator is zero. 

Example. Express tan x as a series in powers of x. 

We could use Maclaurin's series with / (x) = tan x. It is 
easier, however, to expand sin x and cos x and divide the one 
by the other to get tan x. Thus 

x^ x° 

smx ^~6"'~120~'** ,x3,2x«, 

tanx= = z -T =^+-5-+^rT-+ • * * • 

cos X i_^_i_^_ 315 

EXERCISES 

1. Show that 

l°(l-^) = l^(l+^)-l^(l-^)=2(x+| + | + y+ • •• y 

and that the series converges when \x\ < 1. 

2. By expanding cos 2 x, show that 

. , 1 — cos 2 X - X* _, ar» , -. !« 
sin3^=— ^ . = 22;-23^ + 2«gj . 

Prove that the series converges for all values of x. 



112 DIFFERENTIAL CALCULUS Chap. X. 

3. Show that 

and that the series converges for all values of x. 

4. Given / (x) = sm"i x, show that 

Expand this by the binomial theorem and determine /" (z), etc., by 
differentiating the result. Hence show that 

, 1 x» 1 3a:5 1 3 5a;7 
8m-x=x + -^ + -.--^+^.^.gy+... 

and that the series converges when |x| < L 

6. By a method similar to that used in Ex. 4, show that 

tan-i x = x--5-+g — y+-'« 

and that the series converges when |x| < L 
6. Prove 

sec X = = l+-^+-^x*+-««. 

cos X 2 24 

For what values of x do you think the series converges? 



I 



CHAPTER XI 

PARTIAL DIFFERENTIATION 

87. Functions of Two or More Variables. — A quantity 
u is called a function of two independent variables x and y, 

u =fix,y), 

if u is determined when arbitrarj^ values (or values arbitrary 
within certain limits) are assigned to x and y. 
For example, 

U = VI — X2 _ y2 

is a function of x and y. If w is to be real, x and y must be so 
chosen that x^ + y^ is not greater than 1. Within that 
limit, however, x and y can be chosen independently and a 
value of u will then be determined. 

In a similar way we define a function of three or more in- 
dependent variables. An illustration of a function of vari- 
ables that are not independent is furnished by the area of a 
triangle. It is a function of the sides a, b, c and angles A , B, 
C of the triangle, but is not a function of these six quantities 
considered as independent variables; for, if values not be- 
longing to the same triangle are given to them, no triangle 
and consequently no area will be determined. 

The increment of a function of several variables is its in- 
crease when all the variables change. Thus, if 

u=f{x, y), 
u + Au = f{x -\- Ax,y -h Ay) 
and so 

Aw = / (x + Ax, y + Ay) - / (x, y). 

A function is called continuous if its increment approaches 
zero when all the increments of the variables approach zero. 

113 



114 DIFFERENTIAL CALCULUS Chap. XI. 

88. Partial Derivatives. — Let 

u =f(x, y) 

be a function of two independent variables x and y. If we 
keep y constant, w is a function of x. The derivative of this 
function with respect to x is called the 'partial derivative of u 
with respect to x and is denoted by 

fx °' ^'^^^y^' 

Similarly, if we differentiate with respect to y with x con- 
stant, we get the partial derivative with respect to y denoted 

by 

— or fy{x,y). 

For example, if 

u = x^ -\- xy — y"^, 
then 

du ^ . du - 

a^=2x + y, Yy^^-^y- 

Likewise, if w is a function of any number of independent 
variables, the partial derivative with respect to one of them 
is obtained by differentiating with the others constant. 

89. Higher Derivatives. — The first partial derivatives 
are functions of the variables. By differentiating these 
functions partially, we get higher partial derivatives. 

For example, the derivatives of — with respect to x and y 



are 



Similarly, 



dx \dx/ dx^' dy \dx) dydx 

d /du\ _ d^u d /du\ _ d^u 
dx \dy) ~ dxdy' dy \dy) ~ dy^ 

It can be shown that 

dhi _ dhi 
dxdy dydx* 



Chap. XI. PARTIAL DIFFERENTIATION 115 

if both derivatives are continuous, that is, partial derivatives 
are independent of the order in which the differentiations are 
'performed* 

Example, u — xhf + xy^. 

^''^ = A (x2 + 2xy) =2x + 2y, ^ = ^(x2 + 2a^)=2x. 



dxdy dx ' ' "" "' dy^ dy 

90. Dependent Variables. — It often happens that some 
of the variables are functions of others. For example, let 

w = x^ + ?/' + z* 

and let 2 be a function of x and y. When y is constant, z will 
be a function of x and the partial derivative of u with respect 
to X will be 

--=2x4-2 z— • 
dx dx 

Similarly, the pari;ial derivative with respect to y with x con- 
stant is 

— = 2y + 2z — • 

dy '^ dy 

If, however, we consider z constant, the partial derivatives 

are 

du ^ du ^ 
— = 2 X. — = 2 V. 
dx ' dy ^ 

The value of a partial derivative thus depends on what quantities 
are kept constant during the differentiation. 

The quantities kept constant are sometimes indicated by 
subscripts. Thus, in the above example 

dxly,^ ' \dx/y dx V^x/a ^ dx 

* For a proof see Wilson, Advanced Calculus,^ 50. 



116 



DIFFERENTIAL CALCULUS 



Chap. XI. 



Then — will repre- 




It will usually be clear from the context what independent 

variables u is considered a function of. 

sent the derivative with all those variables except x constant. 
Example. If a is a side and A the opposite angle of a right 

^ triangle with hypotenuse c, find (—] • 

\dcjA 

From the triangle it is seen that 

c a = c sin ^. 

Fig. 90. Differentiating with A constant, we get 

da . . 
— = sm A, 
dc 

which is the value required. 

91. Geometrical Representation. — Let z = f (x, y) be 
the equation of a surface. The points with constant y- 
coordinate form the curve AB (Fig. 91a) in which the plane 
y = constant intersects the surface. In this plane z is the 
vertical and x the horizontal coordinate. Consequently, 

dz 
dx 

is the slope of the curve AB at P. 

Similarly, the locus of points with given x is the curve CD 
and 

dy 

is the slope of this curve at P. 

Example. Find the lowest point on the paraboloid 

2 = x2 + i/2 — 2a: — 4y + 6. 

At the lowest point, the curves AB and CD (Fig. 91b) will 
have horizontal tangents. Hence 



|^=2x-2 = 0, 
dx 



1 = 2.-4 = 0. 



Chap. XI. 



PARTIAL DIFFERENTIATION 



117 



Consequently, x = I, y = 2. These values substituted in 
the equation of the surface give z = I. The point required 
is then (1, 2, 1). That this is really the lowest point is shown 
by the graph. 




Fig. 91a. 



Fig. 91b. 



EXERCISES 

In each of the following exercises show that the partial derivatives 
satisfy the equation given: 



1. u = 



x' + y^ 
x + y' 



du , du ' 
dx dy 



2. e = (x + a)(, + 6). g | = z. 



3. z = (x' + j/»)", 



dz dz 

^^~ dy 



dx 



4. M = In (x« H- xy + ?/«), x--- + y—- = 2. 

dx ay 



z X y' 



6. u = tan"' 



7. u = 



(^)' 



1 



du . du , du _ 

dx» ^ dy^ 

S^u . cShx dhi. 



Vx* + y^ + 2*' <^* ^y* ^2* 

In each of the following exercises verify that 

dx dy dy dx 



■jo [dxjy 



118 DIFFERENTIAL CALCULUS Chap. XL 

8. u = -. 10. u ^ sin (x + y). 

9. M = In (x^ + 2/2)^ 11, ^ _ ^yg^ 
12. Given i; = V^~+^i'+^, verify that 

9^ _ dh) 
dx dy dz dz dy dx 
Prove the following relations assuming that 2 is a function of x and y: 

15. . = (. + .).^, | + |.(l+. + ,)(l+| + |)e«^. 

.. /aw aw\ / dz dz\ 

U.u = xyz, z^x--y-) = u\x~-y-y 

^ ' dxdy \dxdy^ dx dyj ^ 

.- a / dtt 3z\ d^w a^z 

16. • — l Z — — U — \ = Z — : — U • 

dx\ dx dxj dx^ dx^ 

17. If X = r cos 8, y = r sin d, show that 
fdx^ 

18. Let a and b be the sides of a right triangle with hypotenuse e and 
opposite angles A and B, Let p be the perpendicular from the vertex 
of the right angle to the hypotenuse. Show that 

') --■ 

ijA C 

19. If K is the area of a triangle, a side and two adjacent angles of 
which are c, A, B, show that 

ldK\ ^h2 ldK\ ^a^ 

20. If K is the area of a triangle with sides a, b, c, show that 

If) -loot A. 
\dajb,c 2 

21. Find the lOwest point on the surface 

z = 2x^ + y'^ + Sx -2y + 9. 

22. Find the highest point on the surface 

z = 2 y - x^ + 2 xy - 2 y^ + 1. 

92. Increment. — Let u = f (x, y) be a function of two 
independent variables x and y. When x changes to a; + Aa; 
and y to y -\- Ay, the increment of u is 

^u=f{x + ^x,y + Ay) - f (x, y). (92a) 



[dajb c^' [daj 



Chap. XI. PARTIAL DIFFERENTIATION 119 

By the mean value theorem, Art. 76, 
/ (x + Ax, y + ^y) =fix,y + Ay) + Az/, (xi, y + Ay), 
Xi lying between x and x + Ax. Similarly 

/ (x, y + Ay) = / (x, y) + lyfy (x, yi), 
yi being between y and y + Ay. Using these values in (92a), 
we get 

Aw = Ax/x (xi, y + Ay) + Ay/„ (x, yi). (92b) 

As Ax and Ay approach zero, Xi approaches x and yi ap- 
proaches y. If /x (x, y) and /„ (x, y) are continuous, 

a.. 

/. (iCi, y + Ay) = /, (x, y) + ci = — + ei, 

a-. 

/v (a;, yi) = /» (a;, y) + €2 = ^ + ^, 

ci and €2 approaching zero as Ax and Ay approach zero. 
These values substituted in (92b) give 

Aw = ^ Ax + ^ Ay + ei Ax + €2 Ay. (92c) 

The quantity 

V- Ax + — Ay 
dx dy 

is called the principal part of Aw. It differs from Au by an 
amount ei Ax + ez Ay. As Ax and Ay approach zero, ci and e» 
approach zero and so this difference becomes an indefinitely 
small fraction of the larger of the increments Ax and Ay. 
We express this by sajdng the principal part differs from Aw 
by an infinitesimal of higher order than Ax and Ay (Art. 9). 
When Ax and Ay are sufficiently small this principal part then 
gives a satisfactory approximation for Aw. 

Analogous results can be obtained for any number of in- 
dependent variables. For example, if there are three inde- 
pendent variables x, y, z, the principal part of Aw is 

dw . , du . , du . 
— Ax+— Ay + — Az. 
dx dy dz 

In each case, if the partial derivatives are continuous, the 



120 DIFFERENTIAL CALCULUS Chap. XI. 

principal part differs from Am by an amount which becomes 
indefinitely small in comparison with the largest of the in- 
crements of the independent variables as those increments 
all approach zero. 

Example. Find the change in the volume of a cylinder 
when its length increases from 6 ft. to 6 ft. 1 in. and its diam- 
eter decreases from 2 ft. to 23 in. 

Since the volume is y = irr^, the exact change is 
Av = TT (1 - ^y (6 + tV) - x. p. 6 = -0.413 TTCU. ft. 
The principal part of this increment is 

I A. + I AA = 2 . rt (- 1) + . r^ (j|) = -0.417 . cu', ft. 

93. Total Differential. — If m is a function of two inde- 
pendent variables x and y, the total differential of u is the 
principal part of Aw, that is, 

This definition applies to any function of x and y. The 
particular values u = x and u = y give 

dx = Ax, dy = Ay, (93b) 

that is, the differentials of the independent variables are equal 
lo their increments. 

Combining (93a) and (93b), we get 

<^w = ^ fix + ^ dy. (93c) 

We shall show later (Art. 97) that this equation is valid even 

if x and y are not the independent variables. 

The quantities 

, du , , du , 

dxU = r- dx, dyU = -r- dy 
dx dy ^ 

are called partial differentials. Equation (93c) expresses 
tha t the total differential of a function is equal to the sum of the 
partial differentials obtained by letting the variables change one 
at a time. 



Chap. XI. PARTIAL DIFFERENTIATION 121 

Similar results can be obtained for functions of any number 
of variables. For instance, if m is a function of three inde- 
pendent variables x, y, z, 

du = ^- Ax + ■— Ay -\- —- Az. 
dx dy dz 

The particular values u = x, u = y,u = z give 

dx = Ax, dy = Ay, dz = Az, 

The pre\aous equation can then be written 

du=pdx-\-pdy+^dz (93d) 

dx dy dz 

and in this form it can be proved valid even when x, y, z are 
not the independent variables. 

Example 1. Find the total differential of the function 
w = x^ + xy^. 

By equation (93c) 

, du . du J 
du = -r~dx + --dy 
dx dy 

= (2x2/ + y') dx+(x'-{-2xy) dy. 

Ex. 2. Find the error in the volume of a rectangular box 
due to small errors in its three edges. 

Let the edges be 5, y, z. The volume is then 
V = xyz. 

The error in v, due to small errors Ax, Ay, Az in x, y, z, is Av. 
If the increments are sufficiently small, this will be approxi- 
mately 

dv = yz dx -\- xz dy + xy dz. 

Dividing by v, we get 

dv _ yz dx -{- xz dy -\- xy dz 
V xyz 

dx , dy , dz 

= h — H 

xyz 

dx 
Now — expresses the error (2x as a fraction or percentage of x. 



122 DIFFERENTIAL CALCULUS Chap. XI. 

The equation just obtained expresses that the percentage 
error in the volume is equal to the sum of the percentage 
errors in the edges. If, for example, the error in each edge 
is not more than one per cent, the error in the volume is not 
more than three per cent. 

94. Calculation of Differentials. — In proving the formu- 
las of differentiation it was assumed that u, v, etc., were 
functions of a single variable. It is easy to show that the 
same formulas are valid when those quantities are functions 
of two or more variables and du, dv, etc., are their total 
differentials. 

Take, for example, the differential of uv. By (93c) the 
result is 

d (uv) = -r- (uv) du -\--r- (uv) dv = V du -\- u dv, 
du dv 

which is the formula IV of Art. 17. 
Example, u = ye^ -\- ze". 
Differentiating term by term, we get 

du = y^ dx -\- ^ dy -}- ze" dy + e" dz. 

We obtain the same result by using (93d) ; for that formula 
gives 

^-1/ Silt Sxt 

du = -;- dx -^ -T- dy + -r- dz = yei' dx -\- (e' + ze^) dy + e« dz. 
dx dy dz 

95. Partial Derivatives as Ratios of Differentials. — 

The equation 

dxU = -r- dx 
dx 

shows that the partial derivative — is the ratio of two dif- 
ferentials dxU and dx. Now dxU is the value of du when the 
same quantities are kept constant that are constant in the 

calculation of ^r-. Therefore, the partial derivative ^r- is the 
dx dx 



Chap. XI. PARTIAL DIFFERENTIATION 123 

value to which -r- reduces when du and dx are determined with 
dx 

the same quantities constant that are constant in the calculation 

. du 

Example. Given u = x^ -\- y^ -{- z^, v = xyz, find (^) * 

Dififerentiating the two equations with v and z constant, 

we get 

du = 2 X dx -\- 2 y dy, = yz dx -\- xz dy. 

Eliminating dy, 

du = 2xdx-2^ dx = 2 f ^' ~ ^' ) dx. 
X \ X / 

Under the given conditions the ratio of du to dx is then 

du ^ 2 (x^ - y^y 

dx X 

Since i; and z were kept constant, this ratio represents f — j ; 

that is, 

lbu\ ^ 2 (x^ - y») 

W/r,z ^ 

EXERCISES 

1. One side of a right triangle increases from 5 to 5.2 while the other 
decreases from 12 to 11.75. Find the increment of the hypotenuse and 
its principal part. 

2. A closed box, 12 in. long, S in. wide, and 6 in. deep, is made of 
material \ inch thick. Find approximately the volume of material 
used. 

3. Two sides and the included angle of a triangle are h = 20, c = 30, 
and A = 45°. By using the formula 

a* = 6^ + c^ — 2 6c cos A, 

find approximately the change in a when b increases 1 unit, c decreases 
5 unit, and A increases 1 degree. 

4. The period of a simple pendulum is 

r = 2xv'-. 

Find the error in T due to small errors in I and g. A 



124 DIFFERENTIAL CALCULUS Chap. XI. 

6. If J7 is computed by the formula, 

s = h gi\ 

find the error in g due to small errors in s and t. 

6. The area of a triangle is determined by the formula 

K = ^ ah smC. 
Find the error in K due to small errors in a, b, C. 
Find the total differentials of the following functions: 

7. xyV. 9. - + ^ + -. 

y z X 

8. xyainix + y). 10. tan-i - + tan-i -• 

X y 

11. The pressure, volume, and temperature of a perfect gas are con- 
nected by the equation -pv = kt, k being constant. Find dp in terms of 
dv and dt. 

12. If X, y are rectangular and r, d polar coordinates of the same 
point, show that 

xdy - ydx = r^djd, dx^ + dy"^ = dr^ + r^ dfi. 

13. li X = u — V, y = u^ -\- v^, find ( — I ' 

\dV jy 

14. li u = xy + yz -\- zx, x^ + z'^ = 2 yz, find I -r- ) • 

16. li yz = ux + v^, vx = uy + z^, find ( — ) 

\dzju,x 

16. A variable triangle with sides a, h, c and opposite angles A, B, C 
is inscribed in a fixed circle. Show that 



cos A cos B cos C 

96. Derivative of a Function of Several Variables. — 

Let u = f (x, y) and let x and y be functions of two variables 
^ and t. When t changes to < + Af, x and y will change to 
:c + Ax and y + Ay. The resulting increment in u will be 

Aw = — Ax + — Ai/ + d Ax + C2 Ay. 

Consequently, 

Am _ du Lx du ^y Ax Ay 
A<~dxA7"^dyAf"^''A«'^''A< 

As A< approaches zero, Ax and Ay will approach zero and so 



Chap. XI. PARTIAL DIFFERENTIATION 125 

ci and 62 will approach zero. Taking the limit of both sides, 

dt dx dt dy dt 

dx 
If a: or y is a function of t only, the partial derivative — 

at 

or ^ is replaced by a total derivative -r: or -^ . If both x 
dt at at 

and y are functions of <, w is a function of t with total deriva- 
tive 

du _du dx du ^ /OAK> 

dt~didt'^dydt' ^^^^ 

Likewise, if t( is a function of three variables x, y, z, that 
depend on t, 

du _ du dx du dy du dz (Qf^\ 

m ~ dx It '^ dy It '^ dz W ^ ^ 

As before, if a variable is a function of t only, its partial de- 
rivative is replaced by a total one. Similar results hold for 
any number of variables. 
The term 

du dx 

dx ~di 

is the result of differentiating u with respect to t, leaving all 
the variables in u except x constant. Equations (96a) and 
(96c) express that if u is a function of several variable quanti- 

dll 

ties, — can he obtained by differentiating with respect to t as if 

only one of those quantities were variable at a time and adding 
the results. 

Example 1. Given y = x', find -p- 

The function x* can be considered a function of two vari- 
ables, the lower x and the upper x. If the upper x is held 
constant and the lower allowed to vary, the derivative (as in 
case of X") is 

X . x*-i = «*, 



126 DIFFERENTIAL CALCULUS Chap. XI. 

If the lower x is held constant while the upper varies, the 
derivative (as in case of a") is 

af In a;. 

The actual derivative of y is then the sum 

-^ = 0:=" -h af In x. 

Ex. 2. Given u = f (x,y, z), y and z being functions of x, 

n J du 
find 3-- 
ax 

By equation (96c) the result is 

du _ du du dy du dz 
dx dx dy dx dz dx 

In this equation there are two derivatives of u with respect 
to X. If y and z are replaced by their values in terms of x, u 
"will be a function of x only. The derivative of that function 

is -T- . liy and z are replaced by constants, u will be a second 

du 
function of x. Its derivative is t- • 

ax 

Ex. 3. Given u = f {x,y, z), z being a function of x and y. 
Find the partial derivative of u with respect to x. 

It is understood that y is to be constant in this partial 
differentiation. Equation (96c) then gives 

du _ du du dz 
dx dx dz dx 

In this equation appear two partial derivatives of u with 
respect to x. If z is replaced by its value in terms of x and y, 
u will be expressed as a function of x and y only. Its partial 
derivative is the one on the left side of the equation. If z is 
kept constant, u is again a function of x and y. Its partial 
derivative appears on the right side of the equation. We 
must not of course use the same symbol for both of these 
derivatives. A way to avoid the confusion is to use the 



Chap. XI. PARTIAL DIFFERENTIATION 127 

letter / instead of u on the right side of the equation. It then 
becomes 

du _ df^.dl dz 
' dx dx dz dx 

It is understood that/ (x, y, z) is a definite function of x, y, z 

and that -r=^ is the derivative obtained with all the variables 
dx 

but x constant. 

97. Change of Variable. — If m is a function of x and y 
we have said that the equation 

, du , , du J 
du = -r- dx ->r -— ay 
dx dy 

is true whether x and y are the independent variables or not. 
To show this let s and t be the independent variables and x 
and y functions of them. Then, by definition, 

du = -r- ds + -^ dt. 
ds dt 

Since u is a function of x and y which are functions of s and t, 
by equation (96a), 

du _ du d3C du dy du _ du dx du dy 

ds dx ds dy ds ' dt dx dt dy dt 

Consequently, 

J /du dx , du dy\ J , fdu dx , dudy\ ,^ 

du/dXj , dx ,A , du/dy J , du jA du , , du , 
= rx{a^''' + M'^) + ry[fa'^+mV = rx''^ + Ty'''<' 

which was to be proved. 

A similar proof can be given in case of three or more 
variables. 

98. Implicit Functions. — If two or more variables are 
connected by an equation, a differential relation can be ob- 
tained by equating the total differentials of the two sides of 
the equation. 



128 



DIFFERENTIAL CALCULUS 



Chap. XI. 



Example 1. f (x, y) = 0. 
In this case 



d'fix,y) =^dx+^dy = d*0 = 0.\ 



Consequently, 



dy 
dx 



Ex. 2. / (x, y, z) = 0. 
Differentiation gives 

df 



df 



dx 



df 



^dx+^dy+^dz^^a, 
dx dy dz 

If z is considered a function of x and y, its partial derivative 
with respect to x is found by keeping y constant. Titen 
dy = and 

d£ _ _ ax 

dx ~ dj_' 
dz 

Similarly, if a; is constant, dx = and 

dz dy_ 

dy~ df 
dz 

Ex. 3. /i {x, y, z) = 0, /2 {x, y, z) = 0. 
We have two differential relations 



df: 



dfi 



dfi 



f^ax+f^dy^f^dz^O. 
fUx-^l^dy^fUz^O, 

We could eliminate y from the two equations /i = 0, fz = 0. 
We should then obtain s as a function of x. The total de- 



Chap. XI. 



PARTIAL DIFFERENTIATION 



129 



rivative of this function is found by eliminating dy and solving 

dz 
for the ratio -r- . The result is 
dx 



dz 


dfidf2_ 
dy dx 


_dhdh 
dx dy 


dx 


dfidh_ 
dz dy 


_dfidf2 
dy dz 



99. Directional Derivative. — Let u = f {x, y). At each 
point P (x, y) in the xy-plane, u has a definite value. If we 
move away from P in any definite direction PQ, x and y will 




Fig. 99. 



be functions of the distance moved. Thfc derivative of u 
with respect to s is 

du du dx , dii dy dii , du . 

ds dx ds dy ds dx dy 

This is called the derivative of u in the direction PQ. The 

partial derivatives -r- and -r- are special values of -r- which 
^ dx dy ds 

result when PQ is drawn in the direction of OX or OY. 

Similarly, if w = / (x, y, z), 

du du dx , dudy , du dz du , du ^ , du 

T- = T~T~ + ^Tr + ^j~ = "E~ cos a-f^- cos /3 + -T- cos y 

ds dx ds dy ds dz ds dx dy dz 

is the rate of change of u w4th respect to s as we move along a 
line with direction cosines cos a, cos /3, cos y. The partial 



130 DIFFERENTIAL CALCULUS Chap. XI. 

du 
derivatives of u are the values to which -r- reduces when s is 

ds 

measured in the direction of a coordinate axis. 

Example. Find the derivative of x^ + y^ in the direction 
<t> = 45° at the point (1, 2). 

The result is 

■^(x^-hy') = 2x^+2y^==2xcos<f>-\-2ysm<f> 
OS OS OS 

= 2~ + 4-4== 3V2. 

100. Exact Differentials. — If P and Q are functions of 
two independent variables x and y, 

Pdx + Qdy 

may or may not be the total differential of a function w of a; 
and y. If it is the total differential of such a function, 

P dx -\- Q dy = du = -r- dx -\- -T- dy. 
dx dy 

Since dx and dy are arbitrary, this requires 



du 
dx 


dy 


Consequently, 




dP dH 


dQ dhi 


dy ~ dydx 


dx dx dy 



Since the two second derivatives of u with respect to x and 
y are equal, 

An expression P dx -\- Qdy \s called an exact differential if 
it is the total differential of a function of x and y. We have 
just shown that (100a) must then be satisfied. Conversely, 
it can be shown that if this equation is satisfied P dx ■{• Qdy 
is an exact differential.* 

^ See Wilson, Advanced Calculus, § 92. 



(100b) 



Chap. XI. PARTIAL DIFFERENTIATION 131 

Similarly, if 

Pdx + Qdy + Rdz 

is the differential of a function u of x, y, z, 

dP^dQ dQ^dR dR^dP 

dy dx ' dz dy' dx dz 

and conversely. 

Example 1. Show that 

(x2 -h2xy)dx+ (x2 + y') dy 

is an exact differential. 
In this case 

The two partial derivatives being equal, the expression is 
exact. 

Ex. 2. In thermodynamics it is shown that 
dU = T dS — p dv, 

V being the internal energy, T the absolute temperature, S 
the entropy, p the pressure, and v the volume of a homogene- 
ous substance. Any two of these five quantities can be 
assigned independently and the others are then determined. 
Show that 



\dp)s \ds). 



The result to be proved expresses that 
TdS + v dp 

is an exact differential. That such is the case is shown by 
replacing T dS by its value dU + pdv. We thus get 
TdS + vdp = dU -\- pdv-^vdp = d(U + pv). 

EXERCISES 
1. If w = / (x, y), y = <f> (x), find ^• 



2. If « = / {x, y, z),z=4> (x), find [^ 



132 DIFFERENTIAL CALCULUS Chap. XI. 

3. l!u =f{x, y, z), z = <l> (x, y), y = ^ (x), find ^. 

4. If M = / (x, y), y = 4> {x, r), r = ^ (x, s), find f ^ j , f |^j , 



and 



5. U/ (X, J,, 2) = 0, z = F (X, y), find^- 

6. If F (x, y, z) = 0, show that 

dx dy i 
dy dZ dx 



dx dy dz^ . 



7. If M = x/ (z), z = -, show that x - — f- V t— = «• 

X dx " dy 

8. If w =/ (r, s), r = x+a<, s = y + 6<, show that -rr = a t- + & t — 

3i dx 5j/ 

9. If z = / (x + ay), show that -r- = o — ' 

dy dx 

10. li u = f (x, y), x = r cos 9, y — r sin 0, show that 

(5r+(^sy=(s)'+(s)' 

11. The position of a pair of rectangular axes moving in a plane is 
determined by the coordinates h, k of the moving origin and the angle <t> 
between the moving x-axis and a fixed one. A variable point P has co- 
ordinates x', y' with respect to the moving axes and x, y with respect to 
the fixed ones. Then 

X =f (x', y', h, k, <i>), y = F (x', y', h, k, 0). 

Find the velocity of P. Show that it is the sum of two parts, one repre- 
senting the velocity the point would have if it were rigidly connected 
with the moving axes, the other representing its velocity with respect 
to those axes conceived as fixed. 

12. Find the directional derivatives of the rectangular coordinates 
X, y and the polar coordinates r, ^ of a point in a plane. Show that they 
are identical with the derivatives with respect to s given in Arts. 54 and 
69. 

13. Find the derivative of x^ — y' 'm the direction <f> = 30° at the 
point (3, 4). 

14. At a distance r in space the potential due to an electric charge e 

is F = - . Find its directional derivative. 
r 

16. Show that the derivative of xy along the normal at any point of 

the curve x^ — y^ = a^ is zero. 



Chap. XI. 



PARTIAL DIFFERENTIATION 



133 



16. Given u = f {x, y), show that 

if «i and 82 are measvu*ed along i>erpendiciilar directions. 

Determine which of the following expressions are exact differentials: 
y dx — X dy. 

{2x + y)dx + (x-2y) dy. 
exdx+ eydy + (x + J/) asdz. 
yzdx — xzdy + y^dz. 
Under the conditions of Ex. 2, page 131, show that 

\dT)p \dp)T \dT}v KdvJT 

In case of a perfect gas, pv = kT. Using this and the equation 
dU = TdS -pdv. 



17. 
18. 
19. 
20. 
21. 



22. 



show that 



dp 



Since U is always a function of p and T, this last equation expresses 
that U is & function of T only. 

101. Direction of the Normal at a Point of a Surface. — 

Let the equation of a surface be 

F (x, y, z) = 0. 
Differentiation gives 



dF , .dF, ,dF, ^ 
dx dy " dz 



(101a) 



Let PN be the Une 
through P (x, y, z) with 
direction cosines propor- 
tional to 

dF^dF^dF 
dx' dy ' dz 
If P moves along a curve 
on the surface, the direc- 
tion cosines of its tangent 
PT are proportional to 

dx : dy : dz. 

Equation (101a) expresses that PN and PT are perpendicu- 
lar to each other (Art. 61). Consequently PN is perpendicu- 




134 



DIFFERENTIAL CALCULUS 



Chap. XI. 



lar to all the tangent lines through P. This is expressed by- 
saying PA'" is the normal to the surface at P. We conclude 
that the normal to the surface F (x, y, z) = at P {x, y, z) has 
direction cosines proportional to 

^:^:^. (101) 

dx dy dz 

102. Equations of the Normal at Pi {xi, yi, zi). — Let A, 
B, C be proportional to the direction cosines of the normal 
to a surface at Pi (xi, yi, zi). The equations of the normal 
are (Art. 63) 

x- xi _ y-yr _ z- Zi 

A ~ B ~ C ' ^^^^^ 

103. Equation of the Tangent Plane at Pi (cci, j/i, si). — 

All the tangent lines at Pi on the surface are perpendicular 




Fig. 103. 

to the normal at that point. All these lines therefore lie in a 
plane perpendicular to the normal, called the tangent plane 
at Pi. 

It is shown in analytical geometry that \i A, B, C are pro- 
portional to the direction cosines of the normal to a plane 
passing through (xi, y\, Zi), the equation of the plane is 

A{x-Xx)^B{y- y{) + C (z - z,) = 0.* (103) 
* See Phillips, Analytic Geometry, Art. 68. 



Ch.\p. XI. P.\RTL\L DIFFERENTL\TIOX 135 

If A, B, C are proportional to the direction cosines of the 
normal to a surface at Pi, this is then the equation of the 
tangent plane at Pi. 

Example. Find the equations of the normal line and tan- 
gent plane at the point (1, —1, 2) of the ellipsoid 

x2 + 2t/2 + 3z» = 3x+12. 
The equation given is equivalent to 

x2 + 2 j/2 + 3 22 _ 3 X - 12 = 0. 
The direction cosines of its normal are proportional to the 
partial derivatives 

2 z - 3 : 4 y : 6 2. 

At the point (1, —1, 2), these are proportional to 

A :B :C = -1 : -4 : 12 = 1 : 4 : -12. 
The equations of the normal are 

X - 1 _ y 4- 1 2-2 
1 ~ 4 ~ -12 ' 
The equation of the tangent plane is 

X - 1 + 4 (y + 1) - 12 (0 - 2) = 0. 

EXERCISES 
Find the equations of the normal and tangent plane to each of the 
following surfaces at the point indicated: 

1. Sphere, z» + y» + 2' = 9, at (1, 2, 2). 

2. Cylinder, x» + ly + !/» = 7, at (2, -3, 3). 

3. Cone, 2^ = x» + t/*, at (3, 4, 5). 

4. Hj'perbohc paraboloid, xy = 3 z — 4, at (5, 1, 3). 
6. EUiptic paraboloid, x = 2 y* -|- 3 2*, at (5, 1, 1). 

6. Find the locxis of points on the cylinder 

(x + 2)* + (y - 2)« = 4 
where the normal is parallel to the xy-plane. 

7. Show that the normal at anj- point P (x, y, z) of the surface 
J/* 4- 2* = 4 X makes equal angles with the x-axis and the line joining 
P and A (1, 0, 0). 

8. Show that the normal to the spheroid 

9 "^25 
at P (x, y, z) determines equal angles with the lines joining P with 
A' (0, -4, 0) and A (0, 4, 0). 



136 DIFFERENTIAL CALCULUS Chap. XI. 

104. Maxima and Minima of Functions of Several 
Variables. — A maximum value of a function w is a value 
greater than any given by neighboring values of the variables. 
In passing from a maximum to a neighboring value, the func- 
tion decreases, that is 

Aw < 0. (104a) 

A minimum value is a value less than any given by neigh- 
boring values of the variables. In passing from a minimum 
to a neighboring value 

Aw > 0. (104b) 

If the condition (104a) or (104b) is satisfied for all small 

changes of the variables, it must be satisfied when a single 

variable changes. If then all the independent variables but 

X are kept constant, u must be a maximum or minimum in x. 

du 
If — is continuous, by Art. 31, 

i = 0. (1040 

Therefore, if the first 'partial derivatives of u with respect to the 
independent variables are continuous, those derivatives must be 
zero when u is a maximum or minimum. 

When the partial derivatives are zero, the total differential 
is zero. For example, if x and y are the independent vari- 
ables, 

du =^dx +^ dy = ' dx + ' dy = 0. (104d) 

Therefore, if the first partial derivatives are continuous, the 
total differential of u is zero when u is either a maximum or a 
minimum. 

To find the maximum and minimum values of a function, 
we equate its differential or the partial derivatives with re- 
spect to the independent variables to zero and solve the 
resulting equations. It is usually possible to decide from 
the problem whether a value thus found is a maximum, 
minimum, or neither. 



Chap. XI. PARTIAL DIFFERENTIATION 137 

Example 1. Show that the maximuin rectangular paralielo- 
piped with a given area of surface is a cube. 

Let X, y, zhe the edges of the parallelepiped. If F" is the 
volume and A the area of its surface 

V = xyz, A = 2 xy ■\- 2 xz -\- 2 yz. 

Two of the variables x, y, z are independent. Let them be 
X, y. Then 

A -2xy 



z — 



Therefore 



V = 



2{x-\-y) 
xy (A -2 xy) 



2{x + y) 
&V^t P -2x^-4xy l ^ 
dx 2L {x + yy J "' 
dV ^x^T A -2y^ -4 xy 1 
dy 2L (x-\-yr J "• 

The values x = 0, y = cannot give maxima. Hence 

A - 2x^ - 4:xy = 0, A - 2y^ - 4:xy = 0. 

Solving these equations simultaneously with 

A = 2 xy -\- 2 xz •\- 2 yz, 
we get 

We know there is a maximum. Since the equations give 
only one solution it must be the maximum. 
Ex. 2. Find the point in the plane 

x + 2i/ + 30= 14 

nearest to the origin. 

The distance from any point (x, y, z) of the plane to the 
origin is 

D = a/x2 + 2/2 + 2?. 



138 DIFFERENTIAL CALCULUS Chap. XI. 

If this is a minimum 

d.j) ^ ^dx-\-ydy-\-zdz ^ ^ 
Va;2 + 7/2 + 22 
that is, 

X dx + y dy -{- z dz = 0. (104e) 

From the equation of the plane we get 

dx -\- 2 dy -{- 3 dz = 0. (104f) 

The only equation connecting x, y, z is that of the plane. 
Consequently, dx, dy, dz can have any values satisfying this 
last equation. If x, y, z are so chosen that Z) is a minimum 
(104e) must be satisfied by all of these values. If two linear 
equations have the same solutions, one is a multiple of the 
other. Corresponding coefficients are proportional. The 
coefficients of dx, dy, dz in (104e) are x, y, z. Those in (104f) 
are 1, 2, 3. Hence 

X _y _ z 

I~2~3" 

Solving these simultaneously with the equation of the plane, 
we get a; = 1, 2/ = 2, 2 = 3. There is a minimum. Since 
we get only one solution, it is the minimum. 

EXERCISES 

1. An open rectangular box is to have a given capacity. Find the 
dimensions of the box requiring the least material. 

2. A tent having the form of a cylinder surmounted by a cone is to 
contain a given volume. Find its dimensions if the canvas required is a 
minimum. 

3. When an electric current of strength / flows through a wire of 
resistance R the heat produced is proportional to PR. Two terminals 
are connected by three wires of resistances Ri, Ri, Rz respectively. A 
given current flowing between the terminals will divide between the 
wires in such a way that the heat produced is a minimum. Show that 
the currents Ii, h, h in the three wires will satisfy the equations 

IiRi = TiRi = /jRj. 

4. A particle attracted toward each of three points A, B, C with a 
force proportional to the distance will be in equilibrium when the sum 



Chap. XI. PARTUL DIFFERENTUTION 139 

of the squares of the distances from the pwints ib least. Find the posi- 
tion of equilibrium. 

6. Show that the triangle of greatest area with a given perimeter is 
equilateral. 

6. Two adjacent sides of a room are plane mirrors. A ray of light 
starting at P strikes one of the mirrors at Q, is reflected to a point R on 
the second mirror, and b there reflected to 5. If P and S are in the 
same horizontal plane find the positions of Q and R so that the path 
PQRS may be as short as possible. 

7. A table has four legs attached to the top at the comers Ai, Ai, 
A 3, Ai of a square. A weight W placed upon the table at a point of the 
diagonal A1A3, two-thirds of the way from Ai to A3, will cause the legs 
to shorten the amounts Si, sj, sj, S4, while the weight itself sinks a dis- 
tance h. The increase in potential energy due to the contraction of a 
leg is /:«*, where A; is constant and s the contraction. The decrease in 
potential energy due to the sinking of the weight is Wh. The whole 
system \s"ill settle to a position such that the potential energy is a mini- 
mum. Assuming that the top of the table remains plane, find the 
ratios of Sj, Sj, Sj, S4. 



SUPPLEMENTARY EXERCISES 
CHAPTER III 

Find the differentials of the following functions: 

^- ^ • 6. X (a2 + a;2) Va^ - x\ 

2 , "" 7 (2x+l)(2x + 7)^ 
6 Vox* + & (2x + 5)3 

3 2Vax'' + bx (x + 2)« (x + 4)^ 

6x * ' (x + 1)2 (x + 3)6" 

4, 2ax + b , (2x»-l)V^Hn: 

Vox- + &a; + c x' 

(ax + ft)'H-2 6(ax+b)"+i „_, 

^- o2 (n + 2) a^ (n + 1) ' 10- a; (x" + n) » ' 

Find -T- in each of the following cases: 

11. 2x2-4x2/4-3^2 = 6x-4y + 18. 

12. x^ + 3 x«7/ = 2/3. 

13. x = Sy^ + 2yK 

14. (x2 + 2/2)2 = 2a2(x2-2/«). 

15. X = 1 + -^, y = 2t- ^ 



i-1' ^ " (<-l)« 

< 1 

16. X = / ' y = , 

V 1 + <2 '^ Vl - /2 

17. X = < (<2 + a2)4 y = t{t^-\- a2)5. 

18. X = z2 + 2 8, 2 = 2/2 + 22/. 

19. x2 + 22 = a2, 2/2 = &2. 

20. The volume elasticity of a fluid is e = —v-j-. 

according to Boyle's law, p» = constant, show that e = p. 

21. When a gas expands without receiving or giving out heat, the 
pressure, volume, and temperature satisfy the equations 

pv = RT, pv" = C, 

R, n, and C being constants. Find -^ and ^« 

140 






SUPPLEMENTARY EXERCISES 141 



22. If 9 is the volume of a spherical segment of altitude h, show that 
jT is equal to the area of the circle forming the plane face of the segment. 

an 

23. If a polynomial equation 

/(x)=0 
has two roots equal to r, / (x) has (x — r)* as a factor, that is, 

/ (x) = (X - r)Vi (x), 
where /i (x) is a polynomial in x. Hence show that r is a root of 

f (x) = 0, 
where/' (x) is the derivative of/ (x). 

Show by the method of Ex. 23 that each of the following equations 
has a double root and find it: 

24. x^ - 3 x» + 4 = 0. 

25. x3 - x^ - 5 X - 3 = 0. 

26. 4x»-8x* -3x + 9 =0. 

27. 4x^ - 12x» + x* + 12x + 4 =0. 

Find -^ and -j-^ in each of the following cases. 

<ir ox* 

28. w = X Vo* - X*. 31. ax + by + c = Q. 

^ 32. X = 2 + 3<, y = 4-5/. 

''•'-(?+T?- 33. x = ^., „. " 



30. XT/ =a^ * + !' ^ + 1 

34. If J/ = x^, find g and 0- 

35. Given x* — y* = 1, verify that 

d^y _ _ (Pxfdy^ 
dbfi ~ dy^ \dxj ' 

36. If ra is a positive integer, show that 

-5-: x" = constant, 
dx" 

37. If « andi; are functions of x, show that 

d* , . ihi , .dhi dv , „ dhi <Pv , . du cPv , d*v 
^(u.) -=^.. + 4^.^ +6^.^ + 4^.^ + u^. 

CJompare this with the binomial expansion for (u + »)*. 

38. If / (x) = (x — r)3/i (x), where /i (x) is a polynomial, show that 

/'(r)=/"(r)=0. ^ 



142 DIFFERENTIAL CALCULUS 

CHAPTER IV 

39. A particle moves along a straight line the distance 

s = 4 <3 - 21 i2 + 36 < + 1 
feet in t seconds. Find its velocity and acceleration. When is the 
particle moving forward? When backward? When is the velocity 
increasing? When decreasing? 

40. Two trains start from different points and move along the same 
track in the same direction. If the train in front moves a distance 6 t^ 
in t hours and the rear one 12 t^, how fast will they be approaching or 
separating at the end of one hour? At the end of two hours? When 
will they be closest together? 

41. If s = Vt, show that the acceleration is negative and propor- 
tional to the cube of the velocity. 

42. The velocity of a particle moving along a straight line is 

V = 2P -3t. 
Find its acceleration when t = 2. 

k 

43. li V- = -, where k is constant, find the acceleration. 

s ' 

44. Two wheels, diameters 3 and 5 ft., are connected by a belt. 
What is the ratio of their angular velocities and which is greater? 
What is the ratio of their angular accelerations? 

45. Find the angular velocity of the earth about its axis assuming 
that there are 365| days in a year. 

46. A wheel rolls down an inclined plane, its center moving the 
distance s = 5P in t seconds. Show that the acceleration of the 
wheel about its axis is constant. 

47. An amount of money is drawing interest at 6 per cent. If the 
interest is immediately added to the principal, what is the rate of 
change of the principal? 

48. If water flows from a conical funnel at a rate proportional to 
the square root of the depth, at what rate does the depth change? 

49. A kite is 300 ft. high and there are 300 ft. of cord out. If the 
kite moves horizontally at the rate of 5 miles an hour directly away 
from the person flying it, how fast is the cord being paid out? 

50. A particle moves along the parabola 

100 y = 16x2 
in such a way that its abscissa changes at the rate of 10 ft./ sec. Find 
the velocity and acceleration of its projection on the j/-axis. 

51. The side of an equilateral triangle is increasing at the rate of 
10 ft. per minute and its area at the rate of 100 sq. ft. per minute. 
How large is the triangle? 



SUPPLEMENTARY EXERCISES 143 

CHAPTER V 
52. The velocity of waves of length X in deep water is proportional to 



V^ 






when a is a constant. Show that the velocity is a minimum when 
X = a. 

53. The sum of the surfaces of a sphere and cube is given. Show 
that the sum of the volumes is least when the diameter of the sphere 
equals the edge of the cube. 

54. A box is to be made out of a piece of cardboard, 6 inches square, 
by cutting equal squares from the comers and turning up the sides. 
Find the dimensions of the largest box that can be made in this way. 

55. A gutter of trapezoidal section is made by joining 3 pieces of 
material each 4 inches wide, the middle one being horizontal. How 
wide should the gutter be at the top to have the maximum capacity? 

56. A gutter of rectangular section is to be made by bending into 
shape a strip of copper. Show that the capacity of the gutter wOl be 
greatest if its width is twice its depth. 

57. If the top and bottom margins of a printed page are each of 
width a, the side margins of width b, and the text covers an area c, 
what should be the dimensions of the page to use the least paper? 

58. Find the dimensions of the largest cone that can be inscribed 
in a sphere of radius a. 

59. Find the dimensions of the smallest cone that can contain a 
sphere of radius a. 

60. To reduce the friction of a Uquid against the walls of a channel, 
the channel should be so designed that the area of wetted svu^ace is as 
small as possible. Show that the best form for an open rectangular 
channel with given cross section is that in which the width equals 
twice the depth. 

61. Find the dimensions of the best trapezoidal chaimel, the banks 
.iiaking an angle 6 with the vertical. 

62. Find the least area of canvas that can be used to make a conical 
tent of 1000 cu. ft. capacity. 

63. Find the maximum capacity of a conical tent made of 100 sq. ft. 
of canvas. 

64. Find the height of a light above the center of a table of radius a, 
so as best to illuminate a point at the edge of the table; assimaing that 
the illumination varies inversely as the square of the distance from the 
Ught and directly as the sine of the angle between the rays and the 
surface of the table. 



144 DIFFERENTIAL CALCULUS 

65. A weight of 100 lbs., hanging 2 ft. from one end of a lever, is to 
be raised by an upward force appUed at the other end. If the lever 
weighs 3 lbs. to the foot, find its length so that the force may be a 
minimum. 

66. A vertical telegraph pole at a bend in the line is to be supported 
from tipping over by a stay 40 ft. long fastened to the pole and to a 
stake in the grovmd. How far from the pole should the stake be 
driven to make the tension in the stay as small as possible? 

67. The lower corner of a leaf of a book is folded over so as just to 
reach the inner edge of the page. If the width of the page is 6 inches, 
find the width of the part folded over when the length of the crease is 
a minimum. 

68. If the cost of fuel for running a train is proportional to the 
square of the speed and $10 per hom* for a speed of 12 mi./hr., and 
the fixed charges on $90 per hour, find the most economical speed. 

69. If the cost of fuel for running a steamboat is proportional to 
the cube of the speed and $10 per hour for a speed of 10 mi./hr., and 
the fixed charges are $14 per hour, find the most economical speed 
against a current of 2 mi./hr. 



CHAPTER VI 

Differentiate the following fimctions: 

sin X 76. sec^ x — tan^ x. 



70. 
71. 
72. 



. „ 77, sin'- sec =• 

smd X 3 



1 — cos TO i ^ 

, , „ 78. tan- 

1 + cos B 1 — X 

sin e ,__ 2 tan x 



73. sin ax cos ax. ' 1 — tan^ x 

„, ,9 80. 5 sec^ - 7 sec* 9. 

74. COtTT — CSCtt- o, - , 

2 2 81. sec X CSC a; — 2 cot X. 

75. tan 2 x — cot 2 x. 

Differentiate both sides of each of the following equations and show 
that the resulting derivatives are equal. 

82. sec^ x + csc-x = sec'' x csc^ x. 

83. sin2x = 2 sin x cos x. 

84. sin 3 X = 3 sin X — 4 sin' x. 

85. sin (x + a) = sin x cos a + cos x sin a. 

86. sec'^ X = 1 + tan* x. 

87. sin X + sin o = 2 sin J (x + a) cos ^ (x — a). 



SUPPLEMENTARY EXERCISES 145 

88. coso - cosx = 2sm Ha; + o) sin Ha; - o). 
Find ^ and $^ in each of the following cases: 

89. x = a cos" 9, y = a sin- 6. 

90. X = a cos* e, y = a sin* d. 

91. X = tan9 -e, y = cosB. 

92. X = sec* e, y = tan* e. 

93. X =8ec9, y = tanO. 

94. X = CSC — cot ff, y = esc + cot 6. 

Differentiate the following functions: 
95. Bin-V/^- ^^2. ac8C-^ + V^^^. 



96. cos- (^). 103. pq-.-cot- 



97. tan 

tan 



(^)- 



104. Vl — X sin-i X — "n/x- 



,-. ,x + 1 , . .z — 1 

- ^ , , lOo. sec-i r+sin * — --: • 

2 2x + l x-1 x + 1 



VS"^ ^^ 106. sm-^ ;+^^^^ - 

_ + a cos X 



99. cos-^^^py- jQ. 1 os-x + ^vT:^ 

V5 ^ 

100. CSC ' 2x-l ' 108. Vi^-a* -asec-»-- 



101 . sec 



-'K^-^y- 



109. e^. lis. -tan-'-+^hi(a* + x»). 



110. Ve*. 



a a ' 2 

119. e-«cos(o+W). 



121. (. + l),„(x+l)-x-i. 



113. 7=^. 

114. a^lnx. ,-_ , V.T + a + Vx — a 
,,, 1 • n 122. In , = . 

llo. lnsin"x. Vi 4- a — Va- — a 

116. hilnx. 123. tan"! H^* + e"*). 

117. Infi^y 124. hi(Vi + V7T2). 

125. (x + l)hi(x* + 2x + 5)+|tan-i^^. 

126. sec§xtan|x + ln(sec|x + tan|x). 



146 DIFFERENTIAL CALCULUS 

127. x sec-i I /^x + -] - In (x2 + 1) . 

128. I In f I x2 + 1 ) - I a; + tan-i |x. 

CHAPTER VII 

Find the equations of the tangent and normal to each of the follow- 
ing curves at the point indicated: 

129. y^ = 2x + y,at (1, 2). 

130. x2 - 2/2 = 5, at (3, 2). 

131. x^ + y^ = x + 3y,at i-1,1). 

132. x* + ?/* = 2, at (1, 1). 

133. y = \nx, at (1, 0). 

134. x2 (x + y) = a" (x - y), at (0, 0). 

135. X = 2 cos 0, 2/ = 3 sin 0, at 6 = 5- 

136. r = o (1 + cosd), at (9 = ^• 

Find the angles at which the following pairs of curves intersect: 

137. x2 + 2/2 = 8 X, 2/2 (2 - x) = x^ 

138. 2/^ = 2 ox + a^ x2 = 2 62/ + 62. 

139. x2 = 4: ay, (x^ + 'ia^)y = 8a\ 

140. y^ = &x, x2 + ^2 = 16, 

141. 2/ = He^ + e-^ ), 2/ = 1. 

142. 2/ = sin X, 2/ = sin 2 x. 

143. Show that all the curves obtained by giving different values to 
n in the equation 

are tangent at (a, 6). 

144. Show that for all values of a and 6 the curves 

x' — 3 X2/2 = 0, 3 x^y — y^ = h, 
intersect at right angles. 

Examine each of the following curves for direction of curvature and 
points of inflection : 
i^c 1 —X 

145. y = rr^- 

146. 2/ = tan x. 

147. x = 62/2-2i/». 



1 



SUPPLEMENTARY EXERCISES 147 

148. x = 2<-i, ^/ = 2f + |• 
149. Clausius's equation connecting the pressure, volume, and tem- 
perature of a gas is 

RT c 

^ v-a r(t; + b)'' 
R, a, b, c being constants. If T is constant and p, v the coordinates 
of a point, this equation represents an isothermal. Find the value of 
T for which the tangent at the point of inflection is horizontal. 

150. If two curves y = f (x), y = F (x) intersect at x = a, and 
/' (a) = F' (a), but /" (a) is not equal to F" (a), show that the curves 
are tangent and do not cross at x = a. Apply to the curves y = 3? 
and y = x* at X = 0. 

151. If two curves y = f (x), y = F (x) intersect at x = a, and 
r (o) = F' (a), /" (a) = F" (a), but /'" (o) is not equal to F'" (a), 
show that the curves are tangent and cross at x = a. Apply to the 
ciuTres y = X* and y = x* + (x — 1)' at x = 1. 

Find the radius of ciu^rature on each of the following curves at the 
point indicated: 

152. Parabola y* = ax at its vertex. 

X* w* . 

153. Ellipse — + rr = 1 at its vertices. 

a- b- 



154. Hyperbola ^ - ^ = 1 at x = Va* + 6». 

155. 1/ = Incscx, at f ^, )• 



156. x = §siny — jln (sec y + tan y), at any point (x, y). 

157. X = a cos's, y = a sin' 6, at any point. 

158. Find the center of curvature of y = In (x — 2) at (3, 0). 

Find the angle ^ at the point indicated on each of the following 
curves: 

159. r = 2*, at fl = 0. 

160. r = a + 6 cos d, at = ^• 

161. r(l -cose) =k,ate = |- 

162. r = a sin 20, at » = |- 

Find the angles at which the following pairs of curves intersect: 

163. r (1 — cosd) = a, r = o (1 — cosd). 

164. r = asec*s> r = 6csc*=' 



148 DIFFERENTIAL CALCULUS 

165. r = acosd, r = a cos 2 d. 

166. r = asecd, r = 2 a sin d. 

Find the equations of the tangent Unes to the following curves at 
the points indicated: 

2 

167. X = 2 1, y = - z = f^, at t = 2. 

168. X = sint, y = cos t, z = sec t, at t = 0. 

169. x^ + y^+z^ = 6, x + y + z = 2, at (1, 2, -1). 

170. z=x^ + y%z'' = 2x- 2y, at (1, -1, 2). 



CHAPTER VIII 

171. A point describes a circle with constant speed. Show that 
its projection on a fixed diameter moves with a speed proportional 
to the distance of the point from that diameter. 

172. The motion of a point (x, y) is given by the equations 

x = ^ V^^rT2 4- |%in-i 1 , 

^ ^ Of 

y = \ V^2Tp72 -j. ^ In + v^rqr^). 

Show that its speed is constant. 

Find the speed, velocity, and acceleration in each of the following 

cases: 

173. x = 2 + 3<, 2/ = 4-9<. 

174. X = a cos {o)t + a), y = a sin (ost + a). 

175. X = a + at, y = b + fit, z = c -\- yt. 

176. a; = e* sin t, y = e^ cos t, z = kt. 

177. The motion of a point P (x, y) is determined by the equations 

X = a cos {nt + «), 2/ = & sin {nt + a). 
Show that its acceleration is directed toward the origin and has a 
magnitude proportional to the distance from the origin. 

178. A particle moves with constant acceleration along the parab- 
ola y^ = 2 ex. Show that the acceleration is parallel to the x-axis. 

179. A particle moves with acceleration [a, o] along the parabola 
2/2 = 2 ex. Find its velocity. 

-T-^ , -T-| extends along the normal at 

(x, y) and is in magnitude equal to the curvature at (x, y) , 



SUPPLEMENTARY EXERCISES 149 

CHAPTER IX 

181. Show that the function 

xi-1 
vanishes at x = — 1 and x = 1, but that its derivative does not vanish 
between these values. Is this an exception to Rolle's theorem? 

182. Show that the equation 

x*-5x + 4 = 

has only two distinct real roots. 

183. Show that 

x*sm- 
Lim — : = 0, 

a: =5= sm X 

but that this value cannot be found by the methods of Art. 73. Explain. 

184. Show that 

_ . 1 — cos X „ 
Lim = 0. 

x:S.O cos X 

Why cannot this result be obtained by the methods of Art. 73? 
Find the values of the following limits: 

185. Lim ,^'-,^ . 189. Lim f i - ^- 
x = ol-cos2x z = 0\X* X / 

,„^ ^. vTx - Vl2 -X 190. Limx"e-'*. 

186. Lim 



x = 32x -3 Vl9 -ox 



z = 00 

.«, _. xlnx 

TX 191. Lim — 



187. Lim 



tan^ ■ x = osin2x - xcotx 



^ 1 1 + esc (x — 1) 192. Lim (.sec x) *»• 



188. Lim^(^-^^ 



■0 



; A I cot (irx) 

193. The area of a regular polygon of n sides inscribed in a circle 
of radius a is 

na'sm - cos — 
n n 

Show that this approaches the area of the circle when n increases 
indefinitely. 

194. Show that the curve 

x^ + y^ = Zxy 
is tangent to both coordinate axes at the origin. 



150 DIFFERENTIAL CALCULUS 



CHAPTER X 

Determine the values of the following functions correct to four 
decimals: 

195. cos 62°. 198. vTl. 

196. sin 33°. 199. tan-i (t^)- 

197. hi (1.2). 200. esc (31°). 

201. Calculate tt by expanding tan~* x and using the formula 

\ = tan-i (1). 

202. Given hi 5 = 1.6094, calculate hi 24. 

203. Prove that _ 

D = V^h 
is an approximate formula for the distance of the horizon, D being the 
distance in miles and h the altitude of the observer in feet. 

Prove the following expansions indicating if possible the values of 
X for which they converge: 

204. In (1 + a;2) = hi 10 + f (x - 3) - 2^ (x - 3)2 + • • • . 

205. ln(e^ + e-)=f-g + g+ .... 

206. In (1 + siax) = X - h x^ + ^ ^ - ^ X* + ' • ' . 

207. e^secx = 1 + x + x^ + ^x^ + h x* + • . . . 

208. In (x + Vr+^0 =^-|f+^f+---- 

209. ln^ = 2p + J-3 + ^ + ...]. 

X — 1 La;3x^5x* J 

210. In tan X = hi X + ^ + ^ + • • • . 

211. e«-- = l+x + |-j-^ . 

212. eta°- = l+x + |j+^' + ^+ • • • . 

Determine the values of x for which the following series converge: 

213. 1+3^ + 1+1+5+ '■ ' • 

2H. (.-,) + (^'+fe^* + (1^111'+.... 

215. 1 +2x4-3x2 + 4x3+ ... . 

X + 2 (X + 2)' (X + 2)» _ ^ 

216. 2 + -jT2-+-2T3~ + ~3T4~+ • 



SUPPLEMENTARY EXERCISES 151 



CHAPTER XI 

In each of the following exercises show that the partial derivatives 
satisfy the equation given: 

217. u = xy + y^z\ x - + z - = y -■ 

218. 2 = x'-2xV + y*, J/?+^§=0- 

ax ay 

219.„.(x + ,)ln^. x(g-g)..f. 

221. u = xy + -, 

222. z = ln(^ + y^), g + ^ = 0. 

223. u = y-±^, 

y -z 

axay "^ dj/az "^ dz* di»* 

Prove the following relations assuming that z is a function oK x and y^ 

224. u = ix + y- zy, 

du _du _ dudz du dz 
3y dx dy dx dx dy 

225. u = z + e^, 

du du dz dz 

X V — = X — — 1/ — • 

dx ^ dy dx " dy 

226. u = 2 (a? - J/*), 

^ai + ^a^^^^-^a^a^+'^ai;) 

227. If X = I (e* + e"*), y ="^{6^ - e"*), show that 

Var/e \dxjy 

228. lixyz = a', show that 



152 DIFFERENTIAL CALCULUS 

In each of the following exercises find A2 and its principal part, 
assuming that x and y are the independent variables. When Ax and 
Ay approach zero, show that the difference of Az and its principal part 
is an infinitesimal of higher order than Ax and Ay. 

229. z = xy. 232. z = V^'+^z. 

230. z = x2 -2/2 + 2X. 

231. z = —^ 

x^ + 1 

Find the total differentials of the following functions: 

233. ax* + bxY + cy*. 

234. hi{x^ + y^+z^). 

235. x* tan-i ^ - y^ tan-i -• 

X " y 

236. yzfF + zxe" + xye'. 

237. If w = x»/ (2), z = ^, show that 

238. Ku =f(r,s), r = x + y, s = x — y, show that 

5m . du _ ^du 
dx dy ~ dr' 

239. If M =/(r, s, 0, r = -, s = ^, < = -, show that 

y z X 

du , du , du „ 

240. If a is the angle between the x-axis and the Hne OP from the ' 
origin to P (x, y, z), find the derivatives of a in the directions parallel 
to the coordinate axes. 

241. Show that 

(cot y — ysecx tan x) dx — (x esc* y + sec x) dy 
is an exact differential. 

Find the equations of the normal and tangent plane to each of the 
following surfaces at the point indicated: 

242. x2 + 2y2 - z2 = 16, at (3, 2, -1). 

243. 2x + 3?/ -4z = 4, at (1, 2, 1). 

244. z2 = 8xy, at (2, 1, -4). 

245. 2/ = z2 - x2 + 1, at (3, 1, -3). 

246. Show that the largest rectangular parallelopiped with a givai 
surface is a cube. 



SUPPLEMENTARY EXERCISES 153 

247. An open rectangular box is to be constructed of a given amount 
of material. Find the dimensions if the capacity is a maximum. 

248. A body has the shape of a hollow cylinder with conical ends. 
Find the dimensions of the lai^est body that can be constructed from 
a given amoimt of material. 

249. Find the volume of the largest rectangular parallelepiped that 
can be inscribed in the ellipsoid. 

250. Show that the triangle of greatest area inscribed in a given 
circle is equilateral. 

251. Find the p)oint so situated that the sum of its distances from 
the vertices of an acute angled triangle is a minimum. 

252. At the point (x, y, z) of space find the direction along which a 
given function F (x, y, z) has the largest directional derivative. 



I 

i 



ANSWERS TO EXERCISES 

Page 8 

1. -f 4. -1. 

2. V2. 6. 1. 

3. -1. 6. i. 

Page 14 
3. 2. 

5. The tangents are parallel to the x-axis at (—1, —1), (0, 0), and 

(1, —1). The slope is positive between ( — 1, —1) and (0, 0) 

and on the right of (1, —1). 

10. Negative. 

11. Positive in 1st and 4th quadrants, negative in 2nd and 3rd. 

Pages 27, 28 

31. When x = 4, t/ = | and dy = 0.072 dx. When x changes to 4.2, 

dx = 0.2 and an approximate value for y is y + dy = 0.814. 
This agrees to 3 decimals with the exact value. 

32. When x = 0, the function is equal to 1 and its differential is — dx. 

When x = 0.3, an approximate value is then 1 — dx = 0.7. 
The exact value is 0.754. 
34. 18. 35. (a, 2 a). 

36. Increases when x < 5, decreases when x > ■=• 

o o 

37. x=±^. 39. -(^:n)5 2 

38. tan-i|. 

Page 31 
1 2 4 9 ^ «' 



{x-iy (x-l)» • V^rr^j* (a2-x2)3 

3. (x - 1)2 (x + 2)2 (7x + 2), (x - 1) (x + 2)2 (42x2 + 24x - 12). 

4. 2. _4 7 1-^ 2 



y' y' x-r (x-l)3 

8. 



^ y^ x* 3xij/S 

X 1 



6 



2y' 42/» 

154 



ANSWERS TO EXERCISES 155 



13. 



^ 12 gj _ 12 

dl» ~ U2 + 30'' rfy* U2-30*' 



Pages 3&-38 

1. r = 100 — 32 f, a = — 32. Rises until t = 3|. Highest point 

h = 206.25. 

2. V = P ~ 12P + S2t, a = SC- - 241 + 32. Velocity decreasing 

between t = 1.691 and t = 6.309. Moving backward when t 
is negative or between 4 and 8. 

7. CO = 6 — 2 d, a = —2 c. 'SVTieel comes to rest when < = tj-- 

J c 

10. 9xcu. ft./min. 15. 12^ ft./sec., 7| ft./sec. 

11. 144 X sq. ft./sec. 16. 4 VS mi.,/hr. 

12. Decreasing 8tcu. ft./sec. ^_ ctan/3, , 

^ 17. z- it./aec. 

13. %^ : 1. TO* ' 

14. ^ V3 in. /sec. 

18. Neither approaching nor separating. 

19. 25.8 ft. sec. 20. 64 \^ ft./sec. 



Pages 43-45 

1. Minimum 3|. 2. Minimum — 10, maximum 22. 

3. Maximimi at x = 0, minima at x = — 1 and x = +1. 

4. Minimum when x = 0. 13. ' a V3. 
10. fv^2. 

14. Length of base equals twice the depth of the box. 

16. Radius of base equals two-thirds of the altitude. 

4 

17. Altitude equals - times diameter of base. 

X 

16 X V3 20. Girth equals twice length. 

27 21. Radius equals 2 Ve inches. 

19. i(ai + «i + aj + 04). 

22. The distance from the more intense source is v^2 times the dis- 

tance from the other source. 

23. 12 V2. 25. 19ift. 

24. [5' 4- 6*]'. 

26. Radius of semicircle equals height of rectangle. 

27. 4 pieces 6 inches long and 2 pieces 2 ft. long. 

28. The angle of the sector is two radians. 

29. At the end of 4 hours. 



166, DIFFERENTIAL CALCULUS 

30. He should land 4.71 miles from his destination. 

a V2 
3L — -^ , a being the length of side. 

35. 2|mi./hr. 36. 13.6 knots. 



Page 48 

1. Maximum = a, minimum = —a. 

2. Maximum = 0, minimum = — {^)K 

3. Minimum = — 1. 

4. Minimum = 0, maximum = ^. 
10. Either 4 or 5. 

Pages 62, 53 

19. A = 3. „„ IT , . 

20 ^ = _ 7 B = -y? 22. WTT ± g, n bemg any mteger. 

21. V3 -|. 

23. Velocity = —2irnA, acceleration = 0. 

24. —^ miles per minute. ^"- 2+b^tV3. 

or o V ,. 28. 13 Vl3. 

25. I radians per hour. 

29. The needle will be incUned to the horizontal at an angle of about 

32° 30' 

30. 120°. a 

31. 120°. '*'^- TT* 

33. If the spokes are extended outward, they will form the sides of an 
isosceles triangle. 

Page 56 

V 

24. w = - cos <p, r being the radius of pulley and <t> the angle formed 

T 

by the string and line along which its end moves. 

25. 4V35. 

Page 61 

27. X = rnr + cot~i 2, n being any integer. 
30. a;<-3, x>2, or -2 < x < 1. 



Pages 65, 66 

1. 2y-x = 5, y + 2x = 0. 

2. 7/ + 4x = 8, 4y-x = 15. 

3. 2y=F X = d:a, y ±2x = ±Sa. 



ANSWERS TO EXERCISES 157 

4. y = aix\nb + l), x + ayhxb = a^hab. 

a^ vr 

7. x + y = 2, 2-y = 2. 12. 90°. 

8. x + 3i/ = 4, y-3x = 28. 13. tan-»2V2. 

9. y -\-xX&a.\4>x = a<iniaiih<h- _, In 10 - 1 
10. 90°, tan-if. ^^' ^^ In 10 + 1' 
11- 45°. 15. tan-i3V3. 



Page 70 

1. Point of inflection (0, 3). Concave upward on the right of this 

point, downward on the left. 

2. Point of inflection (|, —\). Concave upward on the right of this 

point, downward on the left. 

3. The curve is everjTV'here concave upward. There is no point of 

inflection. 

4. Point of inflection (1, 0). Concave on the left of this point, down- 

ward on the right. 

6. Point of inflection! —2, — -j)- Concave upward on the right, 
downward on the left. 

6. Points of inflection at x = ± — ^^ . Concave downward between 

V2 
the points of inflection, upward outside. 

7. Points of inflection (0, 0), (±3, ±1). Concave upward when 

-3 < X < or X > 3. 

8. Point of inflection at the origin. Concave upward on the left of 

the origin. 

Pages 76, 77 
1. ±2V6. -7 t 



7. ^. 



a 



^- b' 8. aecy. 

4. 3V2. 9. ^^ + ^>'. 

4y 

^•^'^' 10. 2a8ec«^- 

6. ta. 2 



158 DIFFERENTIAL CALCULUS 



Page 79 



There are two angles \}/ depending on the direction in which s is 
measured along the curve. In the following answers only one 
of these angles is given. 

2 "L. 7. 0°, 90°, and tan-» 3 Vs. 

'4 s. d = dziT. 

3. |. 10- 3. 

o 

Page 84 



1. 



X 



,_!^ 5. tan-?. 



•v/o 1 2 — J- 5. tan"' 7 • 

— ^'^ _ y — 1 ^ 4_ k 

V2 2 a ■ ft X -1 « 

6. tan '■——' 
- X — e , 2 — 1 V2 

2. = 1 — we = — -^ — • 

e ^ 2 7. 69° 29'. 



3. 



irfc 



Pages 92, 93 

1. The angular speed is ^ ^ , where x is the abscissa of the moving 

point. 

2. If Xi is the abscissa of the end in the x-axis and yi the ordinate ct' 

the end in the y-axis, the velocity of the middle point is 

the upper signs being used if the end in the x-axis moves to the 
right, the lower signs if it moves to the left. The speed is ^ — • 

3. The velocity is 

[v — aw sin 0, aw cos 6], 

where is the angle from the x-axis to the radius through the 
moving point. The speed is 

y/i^ -\- aW — 2 omv sin 6. 

6. The boat should be pointed 30° up the river. 

7. Velocity = [o, h, c — gl], Acceleration = [0, 0, —g\, 

Speed = Va2 -\-¥-^{e- gt)K 



ANSWERS TO EXERCISES 169 



9. Velocity = [oko (1 — cos<^), aw sin <^], 

Speed = aw ^2 — 2 cos <> = 2 aw sin ^ ^, 
Acceleration = \au?€\a.<i>, aoj^cosi^]. 



^ r 3r^ sinf g ZrP- cos f g ~| 
L 4 a sin ^ ' 4 o sin 5 9 J 



12. X = vi cos w/, y = vt sin w/. The velocity is the sum of the par- 

tial velocities, but the acceleration is not. 

13. X = a cos cot +b cos 2 at, y = a sin w< + & sin 2 ut. The velocity is 

the sum of the partial velocities and the acceleration the sum 
of the partial accelerations. 

14. X = ao)it — a sin {ui + oh) t, y = a cos («i + Wi) t. The velocity is 

the sum of the partial velocities and the acceleration the sum 
of the partial accelerations. 

Page 100 



2. 


A. 


3. 


n. 


4. 


0. 


5. 


e». 


6. 


2. 


7. 


-2. 


8. 


1. 


9. 


0. 


10. 


ir2. 


11. 


1. 


12. 


1. 


13. 


0. 


14. 


-i- 


15. 


0. 


16. 


0. 


17. 


-i. 


1. 


0.0872. 


2. 


0.8480. 


3. 


1.0724. 


4. 


1.6003. 


5. 


1.0154. 


21. 


(-2,1,0) 



18. 


h 


19. 


1. 


20. 


-3. 


21. 


a. 




1 


22. 






ir 


23. 


f'{x)dx. 


24. 


00. 


25. 


1 

2- 


26. 


00. 


27. 


00. 


28. 


1. 


29. 


1. 


30. 


a. 


31. 


gm 


Page 106 




6. 


0.1054. 


7. 


1.6487. 


8. 


0.0997. 


9. 


2.833. 


Page 118 




22. 


(1, 1, 2) 



160 DIFFERENTIAL CALCULUS 

Pages 123, 124 

1. Increment = —0.151, principal part = —0.154. 
. dT l(dl dg\ ^. ,, , J ^ ., 

~T ~ 2\~l a J' '^"^^® "' ^^^ ^ ™^y "^ either positive or 

negative, the percentage error in T may be I the sum of the 
percentage errors in I and g. 
6. The percentage error in g may be as great as that in s plus twice 
that in T. 

13. -"^-tl. 15, 2z-^-uy ^ 

u zx — 2uv 

14. hx^ + y^ + xy - z-'). 

Pages 131, 132 

J du ^df_ dl_d6^ ^ /du\ ^df_ df^d^^ 

' dx dx~^ dy dx' "■ \dx/y dx "*" dz dx' 

' dx dx dy dx dz \dx dy dx*' 

dldF_dldF 13. 3V3-4. 



dx ^,dldF ' 14. --3(xcosa+2/cos^+zcoS7). 

dy "^ dz dy 



p dz dy dx dx dy p 

5. - = -^T TT-^TT- 14. _£ 

Page 135 
j_ ^l^y_-2^£-^^ (a;_i)+2(i/-2)+2(z-2)=0. 

2. Normal, y + 4 x = 5, z = 3. 
Tangent plane, x — 4 j/ — 14 = 0. 

a; - 3 2/ - 4 g - 5 01^ c n 

3. — g— = ^ = -3^ , 3x + 42/ — 5z = 0. 

a; — 5 ?/ — 1 z — 3 ,^ „ , ^ 

4. "l— = "^-5— = ^^. a: + 52/-32-l =0. 

a;-5 7/-lz-l cicn 

5. ^rr'"~^ = ~6~' a;-42/-6z + 5=0. 

6. x + z = y-z = db V2. 

Pages 138, 139 

1. The box should have a square base with side equal to twice the 

depth. 

2. The cylinder and cone have volumes in the ratio 3 : 2 and lateral 

surfaces in the ratio 2 : 3. 
4. The center of gravity of the triangle ABC, 



INDEX 



The numbers refer to the pages. 



Acceleration, along a straight line, 
33. 
angular, 34. 

in a curved path, 90, 91. 
Angle, between directed lines in 
space, 79, 80. 
between two plane curves, 64. 
Approximate value, of the incre- 
ment of a function, 14, 15, 
118-120. 
Arc, differential of, 72. 

Continuous function, 10, 113. 
Convergence of infinite series, 107- 

111. 
Curvature, 73. 

center and circle of, 75. 

direction of, 67. 

radius of, 74. 
Curve, length of, 70. 

slope of, 11. 

Dependent variables, 2, 115. 
Derivative, 12. 

directional, 129. 

higher, 28, 29, 114. 

of a function of several vari- 
ables, 124-127. 

partial, 114. 
Differential, of arc, 72. 

of a constant, 20. 

of a fraction, 22. 

of an nth power, 22. 

of a product, 21. 

of a sum, 20. 

total, 120, 121. 
Differentials, 15. 

exact, 130, 131. 

of algebraic functions, 19-31. 

of transcendental functions, 49- 
62. 

partial, 120. 
Differentiation, of algebraic func- 
tions, 19-31 



Differentiation, of transcendoital 
functions, 49-62. 
partial, 113-139. 
Directional derivative, 129. 
Direction cosines, 80, 81. 
Direction of curvatiu^, 67. 
Divergence of infinite series, 107- 
111. 

Exact differentials, 130, 131. 
Exponential fimctions, 56-62. 

Function, continuous, 10, 113. 

discontinuous, 10. 

explicit, 1. 

imphcit, 2, 127, 128. 

irrational, 2. 

of one variable, 1. 

of several variables, 113. 

rational, 2. 
Functions, algebraic, 2, 19-31. 

exfKinential, 56-62. 

inverse trigonometric, 54-56. 

logarithmic, 56-62. 

transcendental, 2, 49-62. 

trigonometric, 49-53. 
Ftmctional notation, 3. 

Geometrical apphcations, 63-84. 

ImpUcit functions, 2, 127. 
Increment, 10. 

of a function, 14, 15, 118, 119. 
Independent variable, 2. 
Indeterminate forms, 95-100. 
Infinitesimal, 7. 
Infinite series, 106-112. 

convergence and divergence of, 
107-111. 

Maclaurin's, 106. 

Taylor's, 106. 
Inflection, 67. 



161 



162 



INDEX 



Length of a curve, 70. 

Limit, of a function, 5. 

.sin 5 .„ 
of -^,49. 

Limits, 4-9. 

properties of, 5, 6. 
Logarithms, 56, 58. 

natural, 58. 

Maclaurin's series, 106. 
Maxima and minima, exceptional 
types, 45, 46. 

method of finding, 42, 43. 

one variable, 39-48. 

several variables, 136-138. 
Mean value theorem, 101. 

Natural logarithm, 58. 
Normal, to a plane curve, 63. 
to a surface, 133, 134. 

Partial derivative, 114. 

geometrical representation of, 
116, 117. 
Partial, differentiation, 113-139. 

differential, 120. 
Plane, tangent, 134. 
Point of inflection, 67, 68. 
Polar coordinates, 77-79. 
Power series, 110, 111. 

operations with, 111. 



Rate of change, 32. 
Rates, 32-38. 
related, 35. 
Related rates, 35. 
RoUe's theorem, 94. 

Series, 106-112. 

convergence and divergence of, 
107-111. 

Maclaurin's, 106. 

power, 110, 111. 

Taylor's, 106. 
Sine of a small angle, 49. 
Slope of a curve, 11. 
Speed, 85. 

Tangent plane, 134. 
Tangent, to a plane curve, 63. 

to a space curve, 81-83. 
Taylor's, theorem, 102. 

series, 106. 
Total differential, 120, 121. 

Variables, change of, 30, 127. 

dependent, 2, 115. 

independent, 2. 
Vector, 85. 

notation, 88. 
Velocities, composition of, 89, 90. 
Velocity, components of, 86, 87. 

along a curve, 85-89. 

along a straight line, 32. 

angular, 34. 






\ 



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INTEGRAL CALCULUS 



BY 



H. B. PHILLIPS, Ph.D. 

AssocicUe Professor of Malhematics in the MassachttseUt 
Institxde of Technology 



r- 



NEW YORK 
JOHN WILEY & SONS, Isc 

Lo(ax>N: CHAPMAN & HALL, LmtTBD 



Copyright, 1917, 

BY 

H. B. PHILLIPS 



Stanbope ]press 

f, H.GILSON COMPANY 
BOSTON, U.S.A. 



5-25 



PREFACE 



This text on Integral Calculus completes the course in 
mathematics begun in the Analytic Geometry and continued 
in the Differential Calculus. Throughout this course I have 
endeavored to encourage individual work and to this end 
have presented the detailed methods and formulas rather 
as suggestions than as rules necessarily to be followed. 

The book contains more exercises than are ordinarily 
needed. As material for review, however, a supplementary 
list of exercises is placed at the end of the text. 

The appendix contains a short table of integrals which 
includes most of the forms occurring in the exercises. Through 
the courtesy of Prof. R. G. Hudson I have taken a two-page 
table of natural logarithms from his Engineers' Manual. 

I am indebted to Professors H. W. Tyler, C. L. E. Moore, 
and Joseph Lipka for suggestions and assistance in preparing 
the manuscript. 

H. B. PHILLIPS. 
Cambridge, Mass. 
June, 1917. 



CONTENTS 



PAOsa 

HAFTXB 1-13 ^ 

I. Integration 

II. Formulas and Methods of Integration 14-34 

III. Definite Integrals 35-46 

IV. Simple Areas and Volumes 4/-59 

V. Other Geometrical Applications 60-69 

VI. Mechanical and Phtsical Appucations 70-89 

VII. Approximate Methods 90-96 

VIII. Double Integration 97-111 

IX. Triple Integration 112-125 

X. Differential Equations 126-lo6 

Supplementary Exercises 157-170 

Answers 171-185 

Table of Integrals 186-189 

Table of Natural Logarithms 190-191 



Index. 



193-194 



INTEGRAL CALCULUS 



CHAPTER I 

INTEGRATION 

1. Integral. — A function F ix) whose differential is 
equal to / (x) dx is called an integral of / {x) dx. Such a 

function is represented by the notation / / (x) dx. Thus 
F (x) = Jf (x) dx, dF (x) = / (x) dx, 

are by definition equivalent equations. The process of 
finding an integral of a given differential is called integration. 
For example, since d (x-) = 2 x dx, 



/^ 



2 X dx = x^. 
Similarly, 

I cos xdx = sin x, j e' dx = e'. 

The test of integration is to differentiate the integral. If 
it is correct, its differential must be the expression integrated. 
2. Constant of Integration. — If C is any constant, 

d[F{x) +C] =dF{x). 

If then F (x) is one integral of a given differential, F {x) + C 
is another. For example, 

j 2xdx = x^ -\- C, / cos X dx = sin X + C, 

"where C is any constant. 



2 Integration Chap. 1 

We shall now prove that, if two continuous junctions of one 
variable have the same differential, their difference is constant. 

Suppose Fi {x) and F^ {x) are 
functions having the same differ- 
ential. Then 

dFi {x) = dFi (x). 

I Let y = Fi (x) - Fi {x) and plot 
^ =• the locus representing y as a 



X 

function of x. The slope of this 

Fig. 2. locus is 

dy-^ dF^jx) -dF,{x) ^Q 
dx dx 

Since the slope is everywhere zero, the locus is a horizontal 
line. The equation of such a line \s, y = C. Therefore, 

Fi (x) - F, (x) = C, 

which was to be proved. 

If then F (x) is one continuous integral of / (x) dx, diTiy 
other continuous integral has the form 

Jf{x)dx = F{x) + C. (2> 

Any value can be assigned to C. It is called an arbitrary 
constant. 

3. Formulas. — Let a and n be constants, u, v, w, 
variables. 

I. I du db dv zt dw = I du ± j dv zt j dw. 
n. fa du = a I du. 



u»^' 



III. / u" da = ^ , , + C, if n is not -1. 
J n + 1 

IV. fu-^du=f^ = lnu + C. 



Art. 3 Formulas 3 

These formulas are proved by showing that the differential 
of the right member is equal to the expression under the 
integral sign. Thus to prove III we differentiate the right 
side and so obtain 

\n+ I / n + 1 

Formula I expresses that the integral of an algebraic sum 
of differentials is obtained by integrating them separately 
and adding the results. 

Formula II expresses that a constant factor can be trans- 
ferred from one side of the symbol I to the other without 

changing the result. A variable cannot be transferred in this 
way. Thus it is not correct to write 



I xdx = X I dx = xi^. 
Example 1. / x* dx. 



Apply Formula III, letting u = x and n = 5. Then dx = 
du and 



/ 






Ex.2. CzVidx. 
By Formula II we have 

fs Vx dx = 3 fxs dx = ^+C = 2x^ + C. 

Ex. 3. Ax - 1) (x + 2) dx. 

We expand and integrate term by term. 

J{x - 1) (x + 2) dx = J {7^ + X - 2) dx 

= ix3 + Ax2-2x + C. 



4 Integration Chap. 1 

Ex. 4. I dx. 

Dividing by a^ and using negative exponents, we get 



In a: + 2a;-i-^a:-2 + C 
2 1 
a; 2x2 



— '• •V.J 



/V2 



^x. 5. / V2 X + 1 dx. 



li u = 2 X -{- I, du = 2 dx. We therefore place a factor 
2 before dx and | outside the integral sign to compensate 
for it. 



rV2 X + 1 dx = ^ r(2x + l)^2dx = ^ fu^ 



du 



Apply IV with w = x^ + 1. Then du = 2xdx and 

/xdx 1 r2xdx 1 Tdw 1, , /^ i ^/ , ■ v , ^ 

By division, we find 

4x4-2 ^ , 4 



2x- 1 ' 2x-l 

Therefore 



Art. 4 Motion of a Particle fi 

EXERCISES 
Find the values of the following integrals: 



Vi (x2 + 2x + 1) dx. / ^"' J x^+ax + ft*^- 

(2 X + a) dx 



J^. /(V^- v^)'dx. l/21- /: 



Vx- + ax + 6 

7. fx (x + a) (x + b) dx. «^' f <'d/ 

•^ ^^^* J 1 — a 



a/* 

"di. 



^8. /2£±3dx. ^. jK.-,0« 

j (x» + i)y-2) ^^^ • //,;_i_x.^d^. 

^-L-^- .-./(,-i)'f. 

X^ C X dx A,y3 o 

4. Motion of a Particle. — Let the acceleration of a 
particle moving along a straight line be a, the velocity v, 
and the distance passed over s. Then, 

dv ds 

a = 

Consequently, 



" = 5' " = * 



dv = adt, ds = vdt. 



•6 Integration Chap. 1 

If then a is a known function of the time or a constant, 

v= iadt + Ci, s = iv dt + C2. (4) 

If the particle moves along a curve and the components of 
velocity or acceleration are known, each coordinate can be 
found in a similar way. 

Example 1. A body falls from rest under the constant 
acceleration of gravity g. Find its velocity and the distance 
traversed as functions of the time t. 

In this case 

dv 

Hence 



v=Jgdt = gt + C. 



Since the body starts from rest, v = when t = 0. These 

values of v and t must satisfy the equation v = gt -\- C. 

Hence 

= g-0 + C, 

ds 
whence C = and v = gt. Since v = -r., ds = gtdt and 



= Jgtdt + C = hQt' + C. 



When f = 0, s = 0. Consequently, C = and s = \ gf. 

Ex. 2. A projectile is fired with a velocity Vo in a direction 
making an angle a with the horizontal plane. Neglecting 
the resistance of the air, find its motion. 

Pass a vertical plane through the line along which the 
particle starts. In this plane take the starting point as 
origin, the horizontal line as x-axis, and the vertical line as 
y-axis. The only acceleration is that of gravity acting 
downward and equal to g. Hence 

^ = ^=-g. 
di^ ' dt^ ^ 



Art. 6 



CtJBVEs WITH A Given- Slope 



Integration gives, 
dx 
di 



= Ci 



|=-^ + c,. 



dx dii 

When ^ = 0> "j: ^^^ ~jI are the components of t'o. Hence 

C\ = t'o cos a, Cj = Vo sin a, 
and 

dx 

-J- = Vo cos a, 

at 

dy 

-^ = Vosma — gt. 

Integrating again, we 
get 

x = v4 cos a, 

y = Vot sin a — I gf-, 




Fig. 4. 



the constants being zero because x and y are zero when 
t = 0. 

5. Curves with a Given Slope. — If the slope of a curve 
Is a given function of x, 






l=/w. 


then 






dy =f (x) dx 


and 





y 



= //(x) 



dx-[-C 



is the equation of the curve. 
Since the constant can 
have any value, there are an 
infinite number of curves 
having the given slope. If 
the curve is required to pass 
through a ^ven point P, the 

value of C can be found by substituting the coordinates of P 

in the equation after integration. 



Fig. 5. 



8 Integration Chap. 1 

Example 1. Find the curve passing through (1, 2) with 
slope equal to 2 x. 
In this case 



ax 



Hence 



y = j2xdx = x'^ + C. 



Since the curve passes through (1, 2), the values x = 1, 
y = 2 must satisfy the equation, that is 

2 = 1 + C. 

Consequently, C = 1 and y = x^ -\- 1 is the equation of the 
curve. 
Ex. 2. On a certain curve 

d^y 

— - = r 

dx^ ^• 

If the curve passes through (— 2, 1) and has at that point 
the slope — 2, find its equation. 
By integration we get 

At (- 2, 1), x = -2 and ^ = - 2. Hence 

- 2 = 2 + C, 
or C = — 4. Consequently, 

y =J{h x2 - 4) dx = i rr^ - 4x + C. 

Since the curve passes through (—2, 1), 

1 = - I + 8 + C. 
Consequently, C = — 5f , and 

y = \a^ — 4:X — bl 
is the equation of the curve. 



Arte 



Separation of the VARLtBLES 



6. Separation of the Variables. — The integration formu- 
las contain only one variable. If a differential contains two 
or more variables, it must be reduced to a form in which 
each term contains a single variable. If this cannot be done, 
we cannot integrate the differential by our present methods. 

Example 1. Find the curves such that the part of the 
tangent included between the coordinate axes is bisected 
at the point of tangency. 

Let P (x, y) be the point at which AB \& tangent to the 
curve. Since P is the middle 
point oi AB, 

OA =2y, OB = 2x. 
The slope of the curve at P is 



dy 
dx 



OA 
OB 




This can be written 

y X 

Since each term contains a single variable, we can integrate 
and so get 

In y + In X = C. 
This is equivalent to 

Inxy = C. 
Hence 

xy = e^ = k. 

C, and consequently k, can have any value. The curves 
are rectangular hyperbolas with the coordinate axes as 
asymptotes. 

Ex. 2. According to Newton's law of cooling, 

when k is constant, a the temperature of the air, and 6 the 
temperature at the time t of a body cooling in the air. Find 
^ as a function of t. 



10 Integbation Chap. 1 

Multiplying by dt and dividing by 6 — a, Newton's 
equation becomes 

d — a 
Integrating both sides, we get 

\n{d- a) = -kt-\- C. 
Hence 

e - a = e-*'+<^ = e^e-*'. 

When t = 0,\ete = do. Then 

do— a = e^e" = e^, 
and so 

d - a= {do- a) e-*' 

is the equation required. 

Ex. 3. The retarding effect of fluid friction on a rotating 
disk is proportional to the angular speed co. Find co as a 
function of the time t. 

The statement means that the rate of change of w is pro- 
portional to oj, that is, 

do) J , 

It = ^'"' 

where k is constant. Separating the variables, we get 
— = kdt, 

0} 

whenco 

\n 0} = kt -\- C , 
and 

CO = e*^'+<^ = e^g*'. 

Let Wo be the value of w when t = 0. Then 

050 = e^'^e^ = e^. 
Replacing e^ by wq, the previous equation becomes 

CO = OJoC*', 

which is the resuJt required. 



Art. 6 Separation of the Variables 11 

Ex. 4. A cylindrical tank full of water has a leak at the 
bottom. Assuming that the water escapes at a rate pro- 
portional to the depth and that ^ of it escapes the first 
day, how long will it take to half empty? 

Let the radius of the tank be a, its height h and the depth 
of the water after t days x. The volume of the water at any 
time is ira^x and its rate of change 

, dx 
This is assumed to be proportional to x, that is, 

Va^ "77 =KX, 

at 
where k is constant. Separating the variables, 

irar dx , -. 
= kat. 

X 

Integration gives 

7ra2 In X = kt + C. 

When t = the tank is full and x = h. Hence 

TraHnh = C. 

Subtracting this from the preceding equation, we get 

xa^ In T = kt. 
n 

When t = 1, X = -^jj h. Consequently, 

xa' In ^ji = k. 
When X = \h, 

to,^ In -r , , 

. n, In^ « c- 1 

< = ^^ = j^ = 6.5/ days. 



12 Integration Chap. 1 

-y EXERCISES 

V 1. If the velocity of a body moving along a line ia v = 2 1 + 3 fl, 

fiiid the distance traversed between t = 2 and i = 5. 

/I. Find the distance a body started vertically downward with a 

velocity of 30 ft. /sec. will fall in the time t, 
./ 3. From a point 60 ft. above the street a ball is thrown vertically 

upward with a speed of 100 ft. /sec. Find its height as a function of 

the time. Also find the highest point reached. 
. ^ 4. A rifle ball is fired through a 3-inch plank the resistance of which 

causes a negative constant acceleration. If its velocity on entering 

the plank is 1000 ft. /sec. and on leaving it 500 ft. /sec, how long does 

it take the ball to pass through? 
\^\, 6. A particle starts at (1, 2). After i seconds the component of its 
'' velocity parallel to the x-axis is 2 < — 1 and that parallel to the t/-axis 

is 1 — <. Find its coordinates as functions of the time. Also find the. 

equation of its path. 
y' 6 A bullet is fired at a velocity of 3000 ft. /sec. at an angle of 45° from 

a point 100 ft. above the ground. Neglecting the resistance of the air, 

find where the bullet will strike the ground. 

7. Find the motion of a particle started from the origin with velocity 

Vo in the vertical direction, if its acceleration is a constant X in a direc- 
tion making 30° with the horizontal plane. 
t/ 8. Find the equation of the curve with slope 2 — x passing through 

(1, 0). 
y^"9. Find the equation of the curve with slope equal to y passing 

through (0, 1). 
^ 10. On a certain curve 

1 = ^^+3. 

If the curve passes through (1, 2), find its lowest point. 
11. On a certain curve 

■J- = X— 1. 

/ If the curve passes through (— 1, 1) and has at that point the slope 2, 
find its equation. 
/ 12. On a certain curve 

v/ 



If the slope is — 1 at x = 0, find the difference of the ordinates at x = 3 
and X = 4. 



L' 



Art. 6 Separation- of the Vablables 13 

13. The pressure of the air p and altitude above sea level h are con- 
noted by the equation 

dh '"P' 

where k is constant. Show that p = po^"**, when pjis the pressure at 
sea level. 

14. Radium decomposes at a rate proportional to the amount 
present. If half the original quantity disappears in 1800 years, what 
percentage disappears in 100 years? 

15. When bacteria grow in the presence of imlimited food, they 
increase at a rate proportional to the number present. Express that 
number as a function of the time. 

16. Cane sugar is decomposed into other substances through the 
presence of acids. The rate at which the process takes place is propor- 
tional to the mass x of sugar still unchanged. Show that x = cc~^. 
What does c represent? 

17. The rate at which water flows from a small opening at the 
bottom of a tank is proportional to the square root of the depth of the 
water. If half the water flows from a cylindrical tank in 5 minutes, 
find the time required to empty the tank. 

18. Solve Ex. 17, when the cylindrical tank is replaced by a conical 
funnel. 

19. A sum of money is placed at compound interest at 6 per cent per 
annum, the interest being added to the principal at each instant. Ilovr 
many years will be required for the sum to double? 

20. The amount of light absorbed in penetrating a thin sheet cf 
water is proportional to the amount falling on the surface and approxi- 
mately proportional to the thickness of the sheet, the approximation 
increasing as the thickness approaches zero. Show that the rate of 
change of illumination is proportional to the depth and so find the 
illumination as a function of the depth. 



cL 



CHAPTER II 

FORMULAS AND METHODS OF INTEGRATION 

7. Formulas. — The following is a short list of integra- 
tion formulas. In these u is any variable or function of a 
single variable and du is its differential. The constant is 
omitted but it should be added to each function determined 
by integration. A more extended list of formulas is given in 
the Appendix. 

u" du = — r^ , if n is not — 1. 
n + 1 

TT r*^" 1 
n. / — = In u. 

J u 
in. / cos udu = sin u. 

IV. / sin udu = — cos u. \/ ^ 

V. / sec** udu = tan u. 

VI. / esc** udu = — cot u. v/ ^ 

Vn. / sec u tan udu = sec u. 

VIII. / CSC u cot u du = — CSC u.^ €> 

IX. I tan a du =4 — In cos u. 

X. / cot u du = In sin u. v_^ \ 

XI. / sec u du = In (sec u + tan u). 
14 



.\rt 8 Integration by StTBJsTrrmox' 17 

Xn. Jcscuda = ln(cscu-'^^^™^- Letu=xV3, 

xm. r_±= = sin-^-* 



/flit 1 U y du 

u'* + a** a a — w- 

XV. / =^^= = - sec ^ — \^ I r 



XVI • ^" 



r di 

J II \ ir 

r-74i= = In (u + VII^T^. 
xvn. rT±i-^ = -Lin!l^. \ 

J u'^ - a" 2a u -{-a \ 

XVm. Tc- du = e*. 

Any one of these formulas can be proved by showing that 
the differential of the right member is equal to the expression 
under the integral sign. Thus to show that 

/ sec M dw = In (sec u + tan u), 

we note that 

J , / , ^ . (sec u tan u + sec' u) du , 

d In (sec u + tan u) = ;-— = sec m du. 

sec u -\- tan u 

8. Integration by Substitution. — When some function 
of the variable is taken as u, a given differential may assume 
the form of the differential in one of the integration formulas 
or differ from such form onlj'^ by a constant factor. Inte- 
gration accomplished in this way is called integration by 
substitution. 

Ekch differential is the product of a function of u by du. 
More errors result from failing to pay attention to the du 

* In Formulas XIII and XV it is assumed that sin-* - is an angle in 

the 1st or 4th qvAdrant, wid sec-* - an angle in the 1st or 2nd quadrant. 
In other cases the algebraic sign of the result must be changed. 



I Methods of Integration Chap. U 

4ie cause. Thus the student may 
m Formula III that the integral of a 
iO write 



/ 



cos 2 X dx = sin 2 x. ' 



sm^ xdx _, ?cf-^- 
+ cos ^ X lA. 



FORMULAS 

7. Formt'' ^^ let 2 x = u, dx is not du but \ du and so 

tion form- ^ C i . i • o 

. , , cos 2 X arc = 4 / cos m aw = ^ sm w = f sm 2 x. 
smgle Y< ''J , 

omitt*' r ^^ 

by ia'xampZe 1. / sin' x cos x dx. 

;*;;. If we let u = sin x, du = cos x dx and 

/ sin' X cos xdx — j u^du = \u* + C = j sin* x + C. 

Ex.2. Jf 

We observe that sin ^ x dx differs only by a constant 
fa tor from the differential of 1 + cos ^ x. Hence we let 

w = 1 + cos 3 X. 

Then dw = — I sin ^ X dx, sin ^ x dx = — 3 dw, 

, r sin ^ xdx ^ rdw „, , ^ 

and / -— r-^ — r— = — 3/ — = -31nw + C 

J 1 + cos f X J u 

= - 3 In (1 + cos ^ x) + C. 
Ex. 3. I (tan x + sec x) sec x dx. 

Expanding we get 

/ (tan X + sec x) sec x dx = / tan x sec x dx + / sec^ x dx 

= sec X + tan x + C 

n ^ r 3dx ■ 

JEx, 4. / , 

J V2-3x» 



Art 8 Integration by Substitutiox' 17 

This resembles the integral in formula XIII. Let u =x Vs, 
« = V2. Then du = Vsdx and 

r 3 — 

r 3dx ^ I V3 ^ /- r du ■ 

J V2-Zx^ J Va2 - u^ ^J Va2 - u^ 



= VSsin-i - + C = V3 sin-i^-^ + C. 
o V2 



Ex. 0. / * 



f V4 <2 - 9 

This suggests the integral in formula XV. Let w =2i, 
a = 3. Then 

/ rf^ _ /^ 2dt _ r du 

t V4 f 2 - 9 ~ J 2 f V4 <2 - 9 ~ J w Vw2 _ a- 



= - sec~^ - + C = 7, sec~^ tt + C'- 
a a 3 3 



J V2 2:2 + 1 



+ 

This may suggest formula XVI. If, however, we let 
u = X V2, du = V2 dx, which is not a constant times x dx. 
We should let 

M = 2x2 + 1. 

Then xdx = I du and 

r xdx 1 rdu 1 r . J 

= |V^4-C = ^V2x2 + l+C. 
Ex. 7. / e**° ' sec2 x dx. 

If M = tan X, by formula XVIII 

/ c**°' sec2 X dx = / e" du = e" + C = e**° ' 4- C. 



18 . Formulas and Methods of Integration Chap. 2 

EXERCISES 
Determine the values of the following integrals: 

1. J (sin 2 X - cos 3 x) dx. 21. J cos« x sin a; dx. 



3. Jsin (n< 4- a) d<. i 23. ^~ 

4. fsec^^edd. 24. r «ec^ (ax) dx 
•^ -^ 1 + tan (ax) 

6. ycsc ^ cot ^ dd. I 25. r ^^^ 



sec" X dx 
+ 2 tan X 
cos 2 X dx 
sin 2x' 



V3— 2x2 

6. Jcos^sin^de. 26 f— ?i^ 

J 3 X- + 



/ dx 
cos" X 27. I — 

- J "^ X 



+ 4 
dx 



V3 x" — 4 



■ J sin"x ' 29. r ^i^ — 



sin X dx -^ V 7 x" + 1 

cos* X 30. r__zE_ 



sin" X 29. 

11. J f CSC 2 — cot- j CSC -de 

12. J'cos(x2— l)xdx. 



31. 



32. 



/ dx 
VTx^ 

C- 

J X Va"x" 

/ 
/ 



dx 



3- 4x" 

dx 



13. r+ff ^ dx. 

cos"3x gg /' (3x- 2)dx 

14. r(8ecx- 1)2 dx. ' -^ ^4-x" 

/• 2x + 3 
. „ rl — sin X J 1 34. j dx. 

15. I dx. -^ V x" + 4 

^ cosx . ' 

/x -p 4 
4 X" - 5 '^* 

17 r (co8x + sinx)" , 36. f ^^~^ 

// ntj f COS . 

sin" X cos X dx. K ^ '* J ^2^ 



dx. 

cos X dx 



sin" X 
sin X cos x dx 



*n r ■, oo /•^''in X C( 

19. I tan' X sec* x da?, JttJ. I — ;== 

J ) J V 2 — sin" X 

20. r8ec"xtanxdxl , 89. r_22££i^. 

^ ^ ^ J 1 + sin" X 



Art 9 Integrals Co.vtai.vi.vg a Qctadratic Expression 19 

sec-xdx , -_ r^'dx 



40. 



/ sec-xdx ^ ^ rj 

tanx Vtan^x- l' * -^ 1 + e*' 



Vl-cosd" 60- /« '^• 

• /x[4-(lnx)^r 6L j^^ 
r sin X cos x ax 

•^ v'cos- X — siii^ X 52. /*_^_«£ 



45. r-^ 



e-»^ 



x^ CO r g°'' <^J^ 

I k2^ 63. , 



dx 



46. /e-..dx. ^^^--,g ^^ 

47. /(."+e-)«dx. 64. /e^ + e- 

9. Integrals Containing ax^ + 6x + c. — Integrals con- 
taining a quadratic expression ax- -\-hx -\- C can often be 
reduced to manageable form by completing the square of 
03? + hx. 

Example 1. / ^ ., , f — j-^- 

Completing the square, we get 

3 x2 + 6 a; + 5 = 3 (x2 + 2 X + 1) + 2 = 3(x + 1)- + 2. 

If then M = (x + 1) Vd, 

du 



f dx ^ r rf (x + 1) ^ _i_ r_( 

J 3x2 + 6x + 5 ~J 3(x + l)2 + 2 V3 J w- 



+ 2 



= 1 tan-(£±lI^ + C. 

Ve V2 



J V2 - 3 X - 



The coefficient of x- being negative, we place the terms x* 
and 3 X in a parenthesis preceded by a minus sign. Thus 

2 - 3 X - x2 = 2 - (x2 + 3 x) = V- - (a; + iY- 



20 Formulas and Methods of Integration Chap. 2: 

If then, w = x + I, we have 



r 2dx ^ 2 r 

J V2-dx-x^ J 

Ex.3. ^ (2^-1)^^- 



, = 2sin-i f-7^ + C. 



/; 



V4 x2 + 4 a: + 2 
Since the numerator contains the first power of x, we 
resolve the integral into two parts, 

dx 



r (2x-l)dx ^1 r (Sx-\-4:)dx ^ C 
J V4rr2 -4-4 3:4-2 4 J V4rr2 4-4 3:4-2 J ^ 



V4x2-f4a:+2 4«^ V 4x^+4 x-h2 ^ V4x2+4x+2 
In the first integral on the right the numerator is taken equal 
to the differential of 4 a;^ + 4 x + 2. In the second the 
numerator is dx. The outside factors \ and — 2 are chosen 
so that the two sides of the equation are equal. The first 
integral has the form 

The second integral is evaluated by completing the square.. 
The final result is 

J* {'2,x—\)dx 1 / . „ , . r-x 
>/4x2+4x + 2 2 

- In (2x -f l-f V4x2 + 4x + 2) + C. 

EXERCISES 

1 r__— ^£__ 7 r (2 X -f- 5) dx 

' Jx2-f6x4-13" J4x«— 4x— 2* 



J V9 -1- 4 -r J. tJ 



(2 X - 1) dx 



^- •^V 2-H4x-4x i' -^Vax^-ex + i 

3^ C dx ^- J 3 x2 + 2 X -h 2 



>•/ 



X dx 
j-2x 
(2x 
rf:c "" J (2x 4-1)^4x^-1- 4 x-1 



VS x2 + 4 X -h 2 f (2 X -1- 3) dx. 



/; 



VI+6X-5X-' 11. r .f'^r^V'^,. - 

J (x^— 2x 4- 3)1 
•^ (x- 3) V2x2- 12x4-15 ^2. J y- 



6 



--dx. 
dx ^^ r e'dx 



• J (i-|-o)(x-f 6)" ^^' J 



(i-|-o)(x4-6) c/ 2e*^4-3e*— 1 



•• J Integrals of Trigonometric Fdnctioks 21 

10. Integrals of Trigonometric Functions. — A power of 
a trigonometric function multiplied bj' its differential can be 
integrated by Formula I. Thus, if w = tan x, 

I tan^ X • sec- xdx = I u* du = ^ tan^ x -\- C. 

Differentials can often be reduced to the above form by 
trigonometric transformations. This is illustrated by the 
following examples. 

Example 1 . I sin* x cos* x dx. 

If we take cos xdx as du and use the relation cos^ x = 
1 — sin^ X, the other factors can be expressed in terms of 
sin X without introducing radicals. Thus 

/ sin* X cos' xdx = j sin* x cos* x • cos x dx 

= I sin* a; (1 — sin* x) dsinx = } sin* a; — | sin'^ a: + C 

Ex. 2. / tan' x sec* x dx. 

If we take sec* x dx as du and use the relation sec* x = 
1 + tan- x, the other factors can be expressed in terms rf 
u = tan x A^nthout introducing radicals. Thus 

I tan' X sec* xdx = j tan' x • sec* x • sec* x dx 

= j tan' X (1 + tan* x) d tan x 

= itan*x+itan«x + C. 

Ex. 3. I tan^xsec'xrfx. 



^2 Formulas and Methods of Integration Chap. i\ 

If we take tan x sec x dx = d sec x as du, and use the rela- 
tion tan^ X = sec^ x — 1, the integral takes the form 

/ tan^ X sec' xdx= j tan^ x • sec^ x • tan x sec x dx 

= j (sec^ a; — 1) sec^ X'dsecx 

= i sec* X — I sec' x-{- C. 

Ex. 4. I sin 2 a; cos 3 x dx. 

This is the product of the sine of one angle and the cosine 
of another. This product can be resolved into a sum or 
difference by the formula 

sin AcosB = ^ [sin (A + 5) + sin (A - B)]. 

Thus ~~^ 

sin 2 X cos Sx = ^ [sin 5 a; + sin (— x)] 
= I [sin 5 re — sin x] 
Consequently, 

/ sin 2 X cos dxdx = ^ j (sin 5 a; — sin x) dx 

= — tV cos 5 X + ^ cos x + C 
Ex. 5. / tan* x dx. 

If we replace tan^ x by sec^ a; — 1, the integral becomes 
/ tan* a: da; = I tan' a; (sec- a; — l)dx = jtan^x — / tan^xdx. 
The integral is thus made to depend on a simpler one 
/ tan' X dx. Similarly, 
/ tan' xdx = j tail x (sec^ a; — 1) da: = ^ tan^ a: + In cos x. 

Hence finally 

I tan* xdx = I tan* x — | tan^ x — In cos x + C. 



Art. 12 



Trigonometric SuBSTmrrioxs 



23 



11. Even Powers of Sines and Cosines. — Integrals of 
the f^rm 

sin" X cos" X dx, 



f^ 



where m or n is odd can be evaluated by the methods of 
Art. 10. If both m and n are even, however, those methods 
fail. In that case we can evaluate the integral by the use 
of the formulas 

1 — cos 2 U 



sin^u = 



cos'^u = 



sin u cos u = 



1 + cos 2 u 

2 ' 

sin 2u 



2 



(11) 



dx 



Example 1. / co^xdx. 

By the above formulas 

I cos^ xdx = I (cos^ xy dx = j I ^ j 

= r(i + |cos2x + icos22a;) 
= f [\ + ^ cos2x + 1(1 -\- cos^x)]dx 
— f a; + 4 sin 2 X + 3^5 sin 4 X + C. 
Ex. 2. / cos^ X sin- x dx. 

I cos* X sin* xdx = I J sin* 2xdx= j 1(1 — cos Ax) dx 

= I 2: — ^V sin 4 X + C. 

12. Trigonometric Substitutions. — Differentials con- 
taining Va- — X-, Va2 4- X-, or Vx- — a*, which are not 



24 Formulas and Methods of Integration Chap. 2 

reduced to manageable form by taking the radical as a new 
variable, can often be integrated by one of the folbwing 
substitutions: 

For Va^ — x^, let x = a sin Q. 
For Va^ + x^, let a; = a tan Q. 
For V'a:^ — a^, let x = a sec Q. 

Example 1. l Va^ — x^dx. 

Let X = a sin 6. Then 

V a^ — x"^ = a cos 0, dx = a cos c?^. 
Consequently, 

fVa''- x^dx = a2 Ccos^ddd = ^(^ + ^sin 2 ^W C. 

Since a: = a sin d, 

• , a^ 1 • « « • « /, a; Va^ — x"^ 

6 = sm~^ - > - sm 2 = sm cos 6 = — - — 5 • 

a 2 a^ . 

Hence finally 

fV^^^r^^dx = ^sin-i-+^Va2-x2 + C. 
J 2 a 2 

^^•2- J ^^2^aY 

If we let a; = a tan ^, 3:^ + 0^ = a^ sec^ 6, dx = a sec^ d^j 



and 



/(^^ = i/s"S-0 = ^/^^^'^^^ 



= iT^ (^ + sin cos e) + C. 
Since 

a; = otan0, = tan-^-, sin cos = -t-i — :.■ 
' a a^ -\- x^ 

Hence 

/ dx _ 11" _i 5 , «a; "I ^ 
(a;2 + a2)2-2a3L a "^a^ ^^^^J "^ ^' 



Art 12 



it6 : 77 
Trigonometric ScBSTrrunoNis 



1. J sin' X dx. 

2. J cos'xdz. 

3. J (cos z + sin x)» dx, 

4. J cos* z sin* X dx. 

5. J sin* § I cos* \xdx. 

6. rsin«3»coe»3ddff. 

7. JCcos* — sin* 9) sin ad0 

g /• cos^ X dx 

J 1 — sin X 
g /• cos- zdx 

J sin X 

10 r sin* g dg 
J cos g 

11. I sec*xdx. 

12. jcsc^ydy. 

13. j tan*xdx. 

14. pec3g + tan»g 
J sec + tan fl 



EXERCISES 

21. J sin* ax di. 

22. J cos* ax dx. 

23. Jcoe* X sin* X dx. 

24. Jcos* i X sin* ^ rdx. 
26. I sin'xdx. 
26. j^_^. . 



16. Jtan ^ X sec* ^ x dx. 

16. j tan* 2 x sec' 2 x dx. 

17. Jcot'xdx. 

18. I tan'xdx. 

j^g /• cos*xdx 
J sin* X 

y 20. J sec' X C8C X dx. 



sinx 



27. ( - 

•/ 1 + cos X 

28. j/l +Bined0. 

29. J*Vx*— a«di. 

80. fV^^+^'dx. 

31. f. ^'^ . 

32. f ^-^ ■ 
•^ (x*- o*)* 




(a*— x2)i 
36. fx^V3? + a*dx. 

37. r-^^=. 

-/ X* V I* + a* 
88. f^x'— 4x + 5dx, 

39. r ^-^---^^^^ 

•^ V2— 2x— 4x» 



26 Formulas and Methods of Integration Chap, 2 

13. Integration of Rational Fractions. — A fraction, such 
as 



x^ -2x-S 



whose numerator and denominator are polynomials is called 
a rational fraction. 

If the degree of the numerator is equal to or greater than 
that of the denominator, the fraction should be reduced by 
division. Thus 

= X + 2 ' 



x'-2x-Z ' ' x2-2a;+3 

A fraction with numerator of lower degree than its denomi- 
nator can be resolved into a sum of partial fractions with 
denominators that are factors of the original denominator. 
Thus 

lOx + G ^ lOx + 6 ^ 9 , 1 
x^-2x-^ (a;-3)(x+l) x-3"^x + l' 

These fractions can often be found by trial. If not, pro- 
ceed as in the following examples. 

Case 1. Factors of the denominator all of the first degree 
^id none repeated. 

„ , rx* + 2x + 6. 

Dividing numerator by denominator, we get 

x* + 2x-\-Q ^^_^ 3x2 + 6 ^ 



x^ -\- x^ — 2x 7? -\-x^ — 2x 

3x2 + 6 



= x-l + 



X (x - 1) (x + 2) 



Assume 



3 x2 + 6 ^A^. ^ JL. ^ 



X (x - 1) (x + 2) X ' X - 1 ' X + 2 
The two sides of this equation are merely different ways of 



Art. 13 Integration of Rational Fractions 27 

writing the same function. If then we clear of fractions, the 
two sides of the resulting equation 

3 x2 + 6 = A (x - 1) (x + 2) + fix (x + 2) + Cz (a; - 1) 
= {A-{-B-\-C)x^-\-{A-\-2B-C)x-2A 

are identical. That is 

A+B + C = 3, A-\-2B-C = 0, -2A = Q. 

Solving these equations, we get 

A = - 3, B = S, C = 3. 

Conversely, if A, B, C, have these values, the above equa- 
tions are identically satisfied. Therefore 

rx^ + 2x + 6 , n 1 3 , 3 , 3 \ , 
J^ +a:^-2x ^=Jl^-^-x+^^ + ^^^ 

= |x2-x- 31nx + 31n(x- 1) +31n(x + 2)+C 

z X 

The constants can often be determined more easily by 
substituting particular values for x on the two sides of the 
equation. Thus, the equation above, 

3 x2 + 6 = A (x - 1) (x + 2) + fix (x + 2) + Cx (x - 1) 

is an identity, that is, it is satisfied by all values of x. In 
particular, if x = 0, it becomes 

6= -2A, 

tihence A = — 3. Similarly, by substituting x = 1 and 
X = — 2, we get 

9 = 35, 18 = 6 C, 
whence fi = 3, C = 3. 

Case 2. Factors of the denominator all of first degree 
but some repeated. * 



„ r (8x^ + 7)dx 

J (x + l)(2x + ] 



28 Formulas and Methods of Integration Chap. 2 

Assume 

Corresponding to the repeated factor {2 x -\- ly, we thus 
introduce fractions with (2 x + 1)' and all lower powers as 
denominators. Clearing and solving as before, we find 

A = 1, B = 12, C = - C, D = 0. 
Hence 

Case 3. Denominator containing factors of the second 
degree but none repeated. 



r 



Ex.3. /"*-^^t£ + idx. 



The factors of the denominator are x — 1 and x^ -\- x -\- 1. 
Assume 

4 x^ + x+ 1 ^ A Bx-\-C 

X^ -\ X - l^ x^ + X -\- l 

With the quadratic denominator x^ + x + 1, we thus use a 
numerator that is not a single constant but a linear function 
Ex + C. Clearing fractions and solving for A, B, C, we find 

A =2, B = 2, C = 1. 
Therefore 

J x^ — 1 J \x — I x^ + X 4- 1/ 

= 21n (x - 1) + In (x2 + X 4- 1) + C. 

Case 4. Denominator containing factors of the second 
degree, some being repeated. 

Ex.4:. C-f^A'^.dx. 



(X2+ 1)2' 



Art 14 I^'TEGRAI;s 29 

Assume 

x' + 1 ^A Bx-hC Dx-\-E 



X (X^ + 1)2 X ' (X2 + 1)2 ' X- + 1 

Corresponding to the repeated second degree factor (x^ + 1)^, 
we introduce partial fractions having as denominators 
(x^ + ly and all lower powers of x^ + 1, the numerators 
being all of first degree. Clearing fractions and solving for 
A, B, C,D, E, we find 

A = l, B=-l, C=-l, D=-l, E = l. 
Hence 

J X (X2 + 1)2 "^^ J Ix (X2 +1)2 X2 + 1 '^ 

= In , + ^ tan-i x — „ , , , ,. + C 

Vx2+ 1 2 2 (x2 + 1) ' 

14. Integrals Containing {ax + b)'. — Integrals contain- 

p 
ing (ox + 6)9 can be rationalized by the substitution 

ax-\-h = z'^. 

If several fractional powers of the same linear function 
ax + 6 occur, the substitution 

ax + & = 2" 

may be used, n being so chosen that all the roots can be 
extracted. 

Examvlel. f ^^■ 
J l + Vi 

Let X = 2^. Then dx = 2 z dz and 

= 2 2 - 2 In (1 + 2) + C 

= 2 Vi - 2 In (l + Vx) + C. 

Ex.2. rJ2£r3)^^x 
J r2r- 



(2x-3)^+l 



30 



Formulas and Methods of Integration 



Chap. 2 



To rationalize both (2 a; — 3)* and (2 a; — 3)», let 
2x-Z = z\ Then 



J (2a; -3^ + 1 J 2' + 1 J\ 2'+! 



dz 



= 3(y - f + 1 - 2 + tan-i2) + C 

= f (2 a; - 3)^ - § (2 X - 3)^ + (2 a: - 3) ^ 
- 3 (2a; - 3)« + tan-^ (2a: - 3)« + C. 



1. 

2. 

3. 

4. 

6. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 



X2- 

J x' 



X3 + X2 



3a; + 2 
2x + 3 



dx. 



EXERCISES 
L4. 



x^ + 1 
{x'' - 1) 
x3- 1 



dx. 



dx. 
xdx 



r x^ + l ^^^ 
J X {x^ — "^ 

r x^— ] 
J 4x»- 

J (X + 1) (x + 3) (X + 5) 

16xdx 

(2x-l)(2x-3)(2x-5)' 

f4^.dx. 
J x^ — x^ 

x'dx 

(x + l)(x- 1)** 

J dx 

/(f^y- 

/ dx 
x^— X* 
/ x'dx 
(x*— 4)*' 
xdx 



/ xdx 
(x* - 4)> 



/ x*dx 
X*- 1' 

/dx 
x3 + 1* 

x^dx 
x^ + l' 



dx 



X* — X* + x^ — 1 
2 x2 + X — 2 



/• 2x-= + ; 
J (x2- 

/ 



20. 



21. 



(X2— 1)2 ■'^^" 

X* + 24 x^ - 8 X 
(x + D* dx 



dx. 



J X 



i_ ^i 



/x* — X' 



x» + 1 



dx. 



22. /x^/^JT6dx. 



23. 



24. 



26. 



vT+^- 1 



r vx- 

»/ X 

/ 
/ 



+ 3 
dx 



dx. 



(X* - 1) (X* + 1) 

dx 

y/x+l - Vx— 1 



Art. 15 Integration by Parts 31 

15. Integration by Parts. — From the formula 

d (uv) = udv -\- udv 
we get f^d** 

udv = d {uv) — V du, 
whence 

j udv = uv — j V du. (15) 

If j vdu is known this gives / v du. Integration by the 
use of this formula is called integration by parts. 

Example 1. jlnxdx. 

dx 
Let u = \n X, dv = dx. Then du = — , v = x, and 

X 

l]nxdx = \nx'X— j x- — ' 
= xQnx- 1) +C. 
Ex. 2. / x^ sin x dx. 

Let u = x^ and dv = sin x dx. Then du = 2 x dx, t; =» 
— cos x, and 



/ x^ sin xdx = —x^ cos x + I ! 



^ sin xdx = —x^ cos x -\- j 2x cos x dx. 

A second integration by parts with u = 2 x, dv = cosx dx^ 
gives 

I 2 X cos xdx = 2xsmx — j 2 sin xdx 

= 2 X sin a; + 2 cos x + C 
Hence finally 

/ x* sin xdx = — x^ cos x + 2 x sin x + 2 cos x + C. 



32 Formulas and Methods of Lvtegration Chap. 2 

The method of integration by parts applies particularly 
to functions that are simplified by differentiation, like In x, 
or to products of functions of different classes, like x sin x. 
In applying the method the given differential must be re- 
solved into a product u • dv. The part called dv must have 
a known integral and the part called u should usually be 
simplified by differentiation. 

Sometimes after integration by parts a multiple of the 
original differential appears on the right side of the equation. 
It can be transposed to the other side and the integral can 
be solved for algebraically. This is shown in the following 
•examples. 

Ex. 3. / Va2 - x^dx 



/v^ 



Integrating by parts with u = Va^ — x^, dv = dx, we get 

— x^dx 



/v a^ — x^dx = X Va^ — x^ — / , 
J Va2- 



Adding a^ to the numerator of the integral and subtracting 
^n equivalent integral, this becomes 

/Va^ — x^dx = x Va^ — x"^— I , dx-^-d? / . 

J Vo2-x2 J Va'-x^ 

= X Va^ - x^ - fVa'-x^dx + a^ f /^ - 
J J Va^-x^ 

Transposing / ^a^ - ^^ dx and dividing by 2, we get 



/ 



Va2 -x^dx = ^ Va2 - x^ + ^ sin"^ ^ + C. 



Ex. 4. I e"* cos hx dx. 

Integrating by parts with u = e^', dv = cos bx dx, we get 

/. , e"^ sin 6a; a C n. • v j 
e'"' cos bx dx = — r r / e"^ sm bx dx. 



Art. 16 Reduction Formulas 3S 

Integrating by parts again with u = ef', do = sin hx dx, 
this becomes 

/, , e^^sinftx a[ e"'cos6x , a /*_, • t j T 
€f"cos,hxdx= T r T hr I (f'smhxdxX 

(b sin hx -\- a cos hx\ a^ P . . , 
" ^' i P j " ^ J e"sm6x(ix. 

Transposing the last integral and dividing by 1 + -^, this 

gives 

fh sin hx -\- a cos hx\ 



j e" cos 6a; rfx = e*"' f 



a2 + 62 / 

16. Reduction Formulas. — Integration by parts is often 
used to make an integral depend on a simpler one and so to 
obtain a formula by repeated appUcation of which the given 
integral can be determined. 

To illustrate this take the integral 

/ sin" X dx, 

wnere n is a positive integer. Integrating by parts with 
u = sin*"^ X, dv = sin x dx, we get 

/sin"xdx= — sin"~^xcosx+ / (n — 1) sin*~'xcos'a;dx 

= — sin""^ x cos X + (n— 1) I sin""- x (1 — sin' x) dx 
= —sin""' X cos X -{- (n— 1) / sin""' x dx 

— (n— 1) / sin'xdx. 

Transposing the last integral and dividing by n, we get 

/. „ J sin"~' X cos X , n — I C . , , 
sin" xdx = 1 I sin"~' x dx. 
n n J 

By successive application of this formula we can make 
/ sin" X dx depend on j dx or j sin x dx according as n 



IS 

I even or odd. 

I 



34 Formulas and Methods of Integration Chap. 2 

Example. j sin^ x dx. 

By the formula just proved 

/. « J sin^ X cos X , b C • A 1 
sin^ xdx = ^ 1" fi / ^^^ ^ ^^ 

sin^ x cos x , 5 r sin^ a; cos re , 3 T . „ , "I 
= g + g|_ ^ + jjsin^a:dxj 

sin^a;cosa; 5 . , 5 . . 5 . ^ 

= a H7 sin^ X cos a; — 7^ sin a; cos X + 77; a; + (7. 

o 24 Id Id 

EXERCISES 

1. \ X cos 2 X dx. 11. i X? e-* dx. 

2. Jinx- xdx. 12. J* (x — 1)2 sin (2 x) dx. 

3. Jsin-ixdx. 13_ JVi^^T^dx. 

4. I X tan-^ X dx. >. , 

•^ 14. J Va^ + x^dx. 

5. fin (x + Va-'' + x'O dx. 

/, ^ j^ 16. I e^ sin 3 x dx. 
In X dx J 

Vx- 1 ^ 

- ^T 16. I (f cos X dx. 

7. J In (In x)^. -^ 

8. Jx^ sec- xdx. "• /^-^«i'^2xdx. 

9. re-*ln(c* + l)dx. 18. j'sec'^edd. 

10. (x^e^dx.. 19. I sin 2 X cos 3 xdx. 

20. Prove the formula 

/_ , V , sec"-'' X tan x , n — 2 /• «_» / s , 
sec" (x) dx = 1 r I sec"-2 (x) dx. 
n — 1 n— IJ 

and xise it to integrate J sec* x dx. 

21. Prove the formula 

and use it to integrate j (o* — x*)' dx. 



CHAPTER III 



DEFINITE INTEGRALS 

17. Summation. — Between x = a and a: = 6 let f{x) be 
a continuous function of x. Di\'ide the interval between a 
and h into any number of equal parts Ax and let X\, Xz, . . , 
Xn, be the points of di\'ision. Form the sum 

/(a)Aa:+/(x,)Ax+/(x2)Aa;+ • • • +/(x,)Ax. 

This sum is represented by 
the notation 

Since / (a), /(xi), /(xj), 
etc., are the ordinates of 
the curve y = f (x) at 
X = Xi, X2, etc., the terms 
/ (a) Ax, / (xi) Ax, / (xa) Ax, 




Fig. 17a. 



etc., represent the areas of the rectangles in Fig. 17a, 
and 2^^f (x) Ax is the sum of those rectangles. 

Example 1. Find the value of V' x^ Ax when Ax = |, 

The inter%'al between 1 and 2 
is divided into parts of length 
Ax = ^. The points of division 
are 1{, U, If. Therefore 

2^'x2Ax= P.Ax+(|)2Ax-h 
(1)2 Ax + (1)2 Ax 
= -«/Ax = ^3.i = 1.97. 

Ex. 2. Find approximately the area bounded by the 
X-axis, the curve y = V^, and the ordinates x = 2, x = 4. 
From Fig. 176 it appears that a fairly good approxima- 

35 




Fig. 176. 



36 



Definite Integrals 



Chap. 3 



tion will be obtained by dividing the interval between 2 and 
4 into 10 parts each of length 0.2. The value of the area 
thus obtained is 

^Wx Ax={V2-\-V2^+\/2A-\- ' • • +V3^) (0.2) =3.39. 

The area correct to two decimals (given by the method of 
Art. 20) is 3.45. 

18. Definite and Indefinite Integrals. — If we increase 
indefinitely the number of parts into which 6 — a is divided, 

f (x) Ax usually 

approaches a limit. This limit is called the definite integral 
of / (x) dx between x = a and x = b. It is represented by 

/ f{x)dx. That is 
f'f (x) dx = lim 2 V (^) Aa;. (18) 



the notation 



The number a is called the lower limit, b the upper limit of 
the integral. 

In contradistinction to the definite integral (which has a 
definite value), the integral that we have previously used 
(which contains an undetermined constant) is called an in- 
definite integral. The 
connection between the 
two integrals will be 
shown in Art. 21. 

19. Geometrical 
Representation.— If 
the curve y = f (x) lies 
above the a:-axis and 
a < 6, as in Fig. 17a, 

/ / (x) dx represents 

the limit approached by the sum of the inscribed rectangles 
and that limit is the area between x = a and x = b bounded 
by the curve and the ic-axis. 




Fig. 19o. 



Art. 19 



Geometrical Representation 



37 



At a point below the a;-axis the ordinate / {x) is negative 
and so the product / (x) Ax is the negative of the area of the 
corresponding rectangle. Therefore (Fig. 19a) 

^ / (x) Ax = (sum of rectangles above OX) 

— (sum of rectangles below OX), 
and in the limit 



i: 



f (x) dx = (area above OX) — (area below OX) (19a) 




Fig. 196. 

If, however, a > 6, as in Fig. 196, x decreases as we pass 
from o to 6, Ax is negative and instead of the above equation 
we have 



/ (x) dx = (area below OX) — (area above OX). (19b) 



Example 1. Show graphically that / sin'xdx = 0. 

The curve y = sin^ x is 
shown in Fig. 19c. Be- 
tween X = and x = 2ir the 
areas above and below the 
X-axis are equal. Hence 



27r 



r 



sin'xdx = Ai — ^2 = 0. 



Fig. 19c. 



Ex. 2. Show that 



/ €-''■ dx = 2 / e-^' dx. 




38 



Definite Integrals 



Chap. 3 



The curve y = e~^' is shown in Fig. 19a. It is symmetrical 

with respect to the ^-axis. 
The area between x = —1 
and a: = is therefore equal 
to that between x = and 
X = 1. Consequently 




Fig. 19d. 



Je'' dx = Ai-\-A2^ 2 At 

= 2 fe-^'dx. 
Jo 



2.x: 



Ax = 1. 



EXERCISES 
Find the values of the following sums: 
1. ^ xAx, Ax = \. 
10 Ax 

X 

3. V]_ VxAx, Ax = \. 

4. Show that 

V 

7 J sin X Ax = 1 — cos - 

approximately. Use a table of natural sines and take Ax = -^^ 

5. Calculate x approximately by the formula 
>i Ax 



'-s:r 



+ X2 



Ax = 0.1. 



6. Find correct to one decimal the area bounded by the parabola 
y = X*, the X-axis, and the ordinates x = 0, x = 2. The exact area 
is f . 

7. Find correct to one decimal the area of the circle x^ + y^ = 4. 
By representing the integrals as areas prove graphically the following 

equations : 

8. f'sin {2x)dx = 0. 
Jo 

J-2t 
cos^ X dx = 0. 


ir 

10. I sin' xdx = 2 \ sin* x dx. 

Jo Ja 



Art 20 



Debit ATTV£ of Ab£a 



39 



It 



+» xdx 



1 +x* 



= 0. 



dx 



•X 

" J_l 1 + X* Jo l+x* 

13. J" fix)dx =j"' fia- x)dx. 

20. Derivative of Area. — The area A bounded by a 

curve 

y=f{x), 

a fixed ordinate x = a, and a movable ordinate MP, is a 
function of the abscissa x of the movable ordinate. 

Let X change to x + Ax. y 
The increment of area is 
AA = MPQN. 

Construct the rectangle 

MP'Q'N equal in area to 

MPQN. If some of the 

points of the arc PQ are 

above P'Q', others must be 

below to make MPQN and 

MP'Q'N equal. Hence P'Q' 

intersects PQ at some point R. Let y' be the ordinate of R. 

Then y' is the altitude of MP'Q'N and so 

^A = MPQN = MP'Q'N = y' Ax. 

Consequently 




Fig. 20. 



^A 

Ax 



= y' 



When Ax approaches zero, if the curve is continuous, y' 
approaches y. Therefore in the hmit 



dA . . 

Let the indefinite integral of / (x) dx be 
f{x)dx = Fix)-\-C. 



(20a) 



I' 



40 Definite Integrals Chap. 3 

From equation (20a) we then have 

A = Cf{x)dx = F{x)+C. 

The area is zero when x = a. Consequently 

= F(a)+C, 
whence C = — F (a) and 

A=F{x) -F(a). 

This is the area from x = a to the ordinate MP with abscissa 
X. The area between a: = a&ndx = 6 is then 

A = F (b) - F (a). (20b) 

The difference F (h) — F (a) is often represented by the 



notation F {x) 



, that is, 



Fix) 



= Fib)-F (a). (20c) 



21. Relation of the Definite and Indefinite Integrals. — 

The definite integral I / (x) dx is equal to the area bounded 

by the curve y = f (x), the x-axis, and the ordinates x = a, 
x = b. If 

'*fix)dx = F(x) + C, 



f^ 



by equation (20b) this area is F (b) — F (a). We therefore 
conclude that 



£ 



f{x)dx^F{x) 



^F{b)-F{a), (21) 



that is, to find the value of the definite integral j f (x) dx, 

substitute x = a, and x = bin the indefinite integral J f (x) dx 
and subtract the former from the latter result. 



Art 22 



Properties of Definite Integrals 



41 



Example. Find the value of the integral 
'1 dx 



r 

Jo 



1 +x» 



The value required is 

n dx 

Jo 1+X2 



= tan~^ X I = tan~* 1 



tan-iO = 7- 
4 



22. Properties of Definite Integrals. — A definite inte- 
gral has the foUowing simple properties: 

I. rfix)dx= - rfix)dx. 

n. r / (x) dx = ff ix) dx-^i'f (x) dx. 

m. /V W dx^ih- a)f(xd, a = x^ = h. 

The first of these is due to the fact that if Ax is positive 
when X varies from a to 6, it is negative when x varies from 
h to a. The two integrals thus represent the same area with 
different algebraic signs. 

r 





Fig. 22a. 



Fig. 22&. 



The second property expresses that the area from a to c 
is equal to the sum of the areas from a to 6 and b to c. This 
is the ease not only when h is between a and c, as in Fig. 22a, 
but also when 6 is beyond c, as in Fig. 226. In the latter 



case 



sum 



I f (x) dx is negative and the 

rf(x)dx-\- rf{x)dx 

da t/5 

is equal to the difference of the two areas. 



42 



Definite Integrai^ 



Chap. 3 



Equation III expresses that the area PQMN is equal to 
that of a rectangle P'Q'MN with 
altitude between MP and NQ. 

23. Infinite Limits. — It has 
been assumed that the limits a 
and h were finite. If the integral 




'J a 



dx 



Fig. 22c. 

approaches a limit when b increases indefinitely, that limit 

/ (x) dx. That is, 

a 

f f (x) dx = lim fV (x) dx. (23) 

t/r. 5=30 ^a 

If the indefinite integral 

Jf{x)dx = F{x) 

approaches a limit when x increases indefinitely, 

r /(re) dx = lim [F (6) - F (a)] = F (oo) - F (a). 

»/a 6 = 00 

The value is thus obtained by equation (21) just as if the 
limits were finite. 



..00 

Example 1. / — 
»/o 1 



dx 



+ x^ 
The indefinite integral is 

f dx ^ 
J l-\-x' 



tan~^ X. 



When X approaches infinity, this approaches jr . Hence 



J/100 
r 



dx 



= tan~^a; 



J 



Art. 24 Infixtte Values op the PuNcnoN iZ 

r*aa 

Ex. 2. / cosxdx. 



£' 



The indefinite integral sin a: does not approach a limit 
when X increases indefinitely. Hence 



r 



cosxdx 



has no definite value. 

24. Infinite Values of the Function. — If the function 

/ (x) becomes infinite when x = 6, j f{x) dxis defined as 

the limit 

Jf (x) dx = lim / / (x) dx, 
a z=b *J a 

z being between a and 6. 
Similarly, if / (a) is infinite, 

J/ (x) dx = lim / / (x) dx, 
a z = a *J z 

z being between a and 6. 

If the function becomes infinite at a point c between o and 

b, I f (x) dx is defined by the equation 

rf{x)dx= rf{x)dx+ rf(x)dx. (24) 

t/a i/a »/c 



Example 1. / 



dx 



When X = 0, -77= is infinite. We therefore divide the 

vx 



integral into two parts: 



J -I V X J_i V X Jo 



dx ^ 3 3^ 

v^ 2 2 



44 Definite Integrals Cha|i. 3 

dx 



Ex.2 



£ 



If we use equation (21), we get 

'^dx^_l 
'-1 x^ X 



x: 



1 

= -2. 

-1 



Since the integral is obviously positive, the result — 2 is 

absurd. This is due to the fact that -5 becomes infinite 

x^ 

when X = Q. Resolving the integral into two parts, we get 

ri dx ndx , n dx 

25. Change of Variable. — If a change of variable is 
made in evaluating an integral, the limits can be replaced by 
the corresponding values of the new variable. To see this, 
suppose that when x is expressed in terms of t, 



s> 



fix)dx = Fix) 
is changed into 

' <i>{t)dt = ^{t). 



f< 



If to, tif are the values of t, corresponding to Xo, Xi, 

F (xo) = $ (to), F (x,) = $ (<i), 
and so 

F{xr)-F{xo)=^{U)-^{k), 
that is 



rf(x)dx= f <f>{t) 



dt. 



If more than one value of t corresponds to the same value 
of X, care should be taken to see that when t varies from k 
to ti, X varies from Xo to Xi, and that for all intermediate 
values, / (x) dx = 4> (t) dt. 



Example. j Va^ — x^ dx. 



Art. 26 Change of Variables 45 

Substituting x = a sind, we find 

When X = a, sin0 = 1, and = K' When x = —a, sind 
= — 1 and d = —-. Therefore 

T r 

J'' V^^^^dx = a^r cos^ Odd =^ld-\-^sm 2e]' = ^• 

~2 "2 

Since sin^Tr = —1, it might seem that we could use |t 
as the lower limit. We should then get 



"'I 



cos'dde= -^ 



s 

This is not correct because in passing from |t to ^x, 
crosses the third and second quadrants. There cos 6 is 
negative and 

Va^ — 7? dx = {— a cos d) cos Odd, 
and not o' cos' d d6 as assumed above. 

EXERCISES 
Find the values of the following definite integrab: 

/•4 6. I xlnxdx. 

1. 1 sec* a; dr. ■'- 

J a Va- — 



8. J tanxdx. 



3. f {x-l)^dx. 

. (^ xdx - ra\a2( £ _£\ 

^-4 Vj2 + 144 Jo >• ^ 

^ sin' do. 10. J^ ;^- 



46 Definite Integrals Chap. 3 

J 13. f e-k'x^xdx. 

11. J csc^xdx. •'<' 



2 

12 



»/i X y/x^ — 1 



Evaluate the following definite integrals by making the change of 
variable indicated: 



r* Vr — 1 
16. -^ -dx, x-\=z\ 



X 

dz _^ ^ 1^ 



18. I ^^-. —-; r-T-f Sin 0=2. 

Jo 6 — 5 sm + sin^ 9 
Jo a^+x^ ^ 



CHAPTER IV 

SIMPLE AREAS AND VOLUMES 

26. Area Bounded by a Plane Curve. Rectangular 
Coordinates. — The area bounded by the curve y = / (x), 
the X-axis and two ordinates x = a, x = b, is the limit 
approached by the sum of rectangles y Ax. That is, 





Fig. 26a. 



Fig. 266. 



A = hm V y Ax = / ydx= I f{x)dx. (26a) 

Ai=0 ^^ a *J a *J a 

Similarly, the area bounded by a curve, the abscissas 
y = a,y = b, and the y-axis is 



A = lim 2 X A?/ = / xdy. 

Ay=0 %J a 



(26b) 



Example 1. Find the area bounded by the curve x = 2 -|- 
y — y- and the i/-axis. 

47 



48 



Simple Areas and Volumes 



The curve (Fig. 26c) crosses the 2/-axis at y = 
y = 2. The area required is, therefore, 



Chap. 4 
• 1 and 




Fig. 26c 



A = J\dy=£^ {2 + y-y^)dy = 2y + ^-y^ 



= 4i 



Ex. 2. Find the area within the circle x^ -{■ y^ = 16 and 
parabola x^ = 6 ?/. 

Solving the equations simultaneously, we find that 
the parabola and circle intersect at P (— 2V3, 2) and 
Q (2\/3, 2). The area MPQN (Fig. 2M) under the circle is 

ydx= I \/l6-x2da: = ^x + 4V3. 

The area MPO + OQN under the parabola is 



x 



2v/3 3.2 S ,- 



-2V3 6 

The area between the curves is the difference 
MPQN - MPO - OQN = J/tt + J Vs. 



Ex. 3. Find the area within the hypocycloid x = a sin^ 0, 
y = a cos^ 4>. 



Art. 26 Rectangular Coordinates 

The area OAB in the first quadrant is 

Jydx= / a CDs' <^ • 3 a sin^ <f> cos <t> d<t> 

X 

= 3 a^ I cos* sin'^ 4>d<l) = g\ x a^ 



49 




Fig. 26c. 

The entire area is then 

^•OAB = lira\ 

EXERCISES 

1. Find the area bounded by the Hne 2 y— S x— 5 =0, the x-axis, 
and the ordinates x = 1, x = 3. 

.^. 2. Find the area bounded by the parabola y = 3 x-, the y-axis, and 
r the abscissas y = 2, y = 4. 

3. Find the area bounded hy y' = x, the line y = — 2, and the 
ordinates x = 0, x = 3. 

4. Find the area bounded by the parabola y = 2 x — x- and the 
X-axis. 

.y 6. Find the area bounded by y = In x, the x-axis, and the ordinates 
'z = 2, X = 8. 

6. Find the area enclosed by the ellipse 

^ + g = l. 
a- cr 

7. Find the area bounded by the coordinate axes and the curve 
x* + j/i = a*. 



50 Simple Areas and Volumes Chap. 4 

8. Find the area within a loop of the curve x^ = y^ (i— y^). 
y 9. Find the area within the loop of the curve y"^ = (x— I) {x— 2)2. 
/ 10. Show that the area bounded by an arc of the hyperbola xy = k^, 
the X-axis and the ordinates at its ends, is equal to the area bounded by 
the same arc, the y-axis and the abscissas at its ends. 
tX 11. Find the area bounded by the curves y^ = 4. ax, x^ = A ay.~\ 
y- 12. Find the area bounded by the parabola y = 2x— x"^ and the 
iline y = — X. 

/ 13. Find the areas of the two parts into which the circle x^ + r/^ = g 
is divided by the parabola y^ = 2 x. 

/ 14. Find the area within the parabola x^ = 4 y + 4 and the circle 
a;2 4-^ = 16. 

5. Find the area bounded by y* = 4 x, x^ = 4 y, and x'' + y^ = 5. 
16. Find the area of a circle by using the parametric equations 
X = a cos 0, y = a sin 6. 
fn^ Find the area bounded by the x-axis and one arch of the cycloid. 

X = a (4) — sin </>), y = a (1 — cos 0). 

</ 18. Find the area within the cardioid 

X = a cos d {1 — cos 6), y = a sin 0(1 — cos 6). 

19. Find the area bounded by an arch of the trochoid, 

X = cuj) — b sin </>, y = a — b cos <(>, 

and the tangent at the lowest points of the curve. 

20. Find the area of the ellipse x" — xy + y^ = 3. 

21. Find the area bounded by the curve y"^ = ^ — — — and its as- 
ymptote X = 2 a. 

22. Find the area within the curve 

l+(f)' = '- 

27. Area Bounded by a Plane Curve. Polar Coordi- 
nates. — To find the area of the sector POQ bounded by two 
radii OP, OQ and the arc PQ of a given curve. 

Divide the angle POQ into any number of equal parts A^ 
and construct the circular sectors shown in Fig. 27a. One 
of these sectors ORS has the area 

i OR^ A0 = ^ r2 A^. 



Art 27 



Polar Coordinates 



51 



If a and /S are the limiting values of 6, the sum of all the 
sectors is then 

As A^ approaches zero, this sum approaches the area A of 
the sector POQ. Therefore 



A= lim V §r2A5= / ^r^ 



dd. 



(27) 




Fig. 27a. 



Fig. 276. 



In this equation r must be replaced by its value in terms 
of d from the equation of the curve. 

Example. Find the area of one loop of the curve r ^ 
a sin 2 (Fig. 276). 

TT 

A loop of the curve extends from d = to 6 =-. Its 
area is 

A = r ^r^ dd = r ^sm^2e) dd 



=fra- 



cos 4:d)dd = 



xa*= 



52 Simple Areas and Volumes Chap. 4 

EXERCISES 

1. Find the area of the circle r = a. 

2. Find the area of the circle r = a cos Q. 

3. Find the area bounded by the coordinate axes and the line 

r = asecfe— -^y 

4. Find the area bounded by the initial line and the first turn of the 
spiral r = ae^- 

\b^ Find the area of one loop of the curve r^ = o^ cos 2 6. 

6. Find the area enclosed by the curve r = cos fl + 2. 
TT) Find the area within the cardioid r = a (1 + cos 0). 

8. Find the area bounded by the parabola r — a sec^ -^ and the 
y-axis. 

9. Find the area bounded by the parabola 

2a 
f = . 

1 — cos 6 

and the radii 9 = -:, 9 = ■^- 
4 2 

10. Find the area bounded by the initial line and the second and 
third tiu-ns of the spiral r = aB. 

11. Find the area of the curve r = 2 a cos 3 9 outside the circle 
r = a. 

12. Show that the area of the sector bounded by any two radii of 
the spiral rd = am proportional to the difference of those radii. 

13. Find the area common to the two circles r = a cos 9, r — 
a cos 6 + asin 9. 

n 

14. Find the entire area enclosed by the curve r = a cos' 5 • 

16. Find the area within the curve (r — aY = a? {I — ff^). 

16. Through a point within a closed curve a chord is drawn. Show 
that, if either of the areas determined by the chord and curve is a maxi- 
mum or minimum, the chord is bisected by the fixed point. 

28. Volume of a Solid of Revolution. — To find the 
volume generated by revolving the area ABCD about the 
X-axis. 

Inscribe in the area a series of rectangles as shown in 
Fig. 28a. One of these rectangles PQSR generates a circular 



Art 28 



Volume of a Solid of Revolution' 



53 



cylinder with radius y and altitude Ax. The volume of this 
cylinder is 

Ty-^x. 

D 




Fig. 28a. 
If a and b are the limiting values of x, the sum of the cylinders 



IS 



2 TZ/2AX. 

The volume generated by the area is the limit of this sum 

x?/-Ax= I iry-dx. (28) 

If the area does not reach the axis, as in Fig. 286, let ?/i 
and t/2 be the distances from the axis to the bottom and top 

r 




Fig. 286. 

of the rectangle PQRS. When revolved about the axis^ 
it generates a hollow cy Under, or washer, of volume 
T (2/2^ - Vi^) Ax. 




54 Simple Areas and Volumes Chap. 4 

The volume generated by the area is then 

b /»6 

V = lim y TT (?/2^ - yi^) Ax = I TT (yi^ - y^) dx. 

Ai=0 ^^ a tJ a 

If the area is revolved about some other axis, y in these 
formulas must be replaced by the perpendicular from a point 
of the curve to the axis and x by the distance along the axis 
to that perpendicular. 

Example 1. Find the 
volume generated by re- 
volving the ellipse 

O Ax ax 

Fig. 28c. about the x-axis. 

From the equation of the curve we get 

The volume required is, therefore, 

J-'o ^52 fa 4 

Tty^dx = -Y J {a- — x^) dx = -Tah\ 

Ex. 2. A circle of radius a is revolved about an axis in its 
plane at the distance b 
(greater than a) from its D 

center. Find the volume 
generated. 

Revolve the circle, 
Fig. 28d, about the line 
CD. The rectangle MN 
generates a washer with 
radii 

R^ = b-x=b- Va^-y^, 

R^=b-^x=b-\- Va^-y\ Fig. 28d. 

The volume of the washer is 

T {Ri" - Ri") =4x6 Va'-y-'Ay. 




Art. 28 Volume of a Solid of Revolution 55 

The volume required is then 

V = r 4x6 Va2 _ y2dy = 2Tr^a%. 

Ex. 3. Find the volume generated by revolving the circle 
r = a sin 6 about the x-axis. 
In this case 

y = rsind = a sin- 6, 

X = rcosd = acosflsin^. 

The volume required is 



V = / Ty^ dx = I ira? sin* (cos- d — sin- 6) d9 = 



x-a-' 



The reason for using x as the lower limit and as the upper is 
to make dx positive along the upper part ABC of the curve. 




Fig. 28e. 



As 6 varies from x to 0, the point P describes the path 
OABCO. Along OA and CO dx is negative. The integral 
thus gives the volume generated by MABCN minus that 
generated by 0AM and OCN. 



EXERCISES 

1. Find the volume of a sphere by integration. 

2. Find the volume of a right cone by integration. 

3. Find the volume generated by revolving about the z-axis the area 
X)unded by the i-axis and the parabola y = 2 x — x*. 



56 Simple Areas and Volumes Chap. 4 

4. Find the volume generated by revolving about OY the area 
bounded by the coordinate axes and the parabola x' + 2/ = a • 

5. Find the volume generated by revolving about the x-axis the 

area bounded by the catenary y = -x ye" -j- e"y , the x-axis and the 

lines X = ± a. 

6. Find the volume generated by revolving one arch of the sine curve 
y = sin X about OX. 

1/ 7. A cone has its vertex on the surface of a sphere and its axis 

coincides with a diameter of the sphere. Find the common volume. 
V 8. Find the volume generated by revolving about the y-axis, the 

part of the parabola y"^ = "iax cut off by the line x = a. 
iX 9. Find the volume generated by revolving about x = a the part of 

thp parabola y^ = 4tax cut off by the line x = a. 
\/ 10. Find the volume generated by revolving about y = — 2 a the 

part of the parabola r/^ = 4 ax cut off by the line x — a. 

11. Find the volume generated by revolving one arch of the cycloid 

X = a ((^ — sin <^), y = a (1 — cos <^) 

about the x-axis. 

12. Find the volume generated by revolving the curve 

X = o cos' 0, 2/ = o sin' <> 
about the y-axis. 

13. Find the volume generated by revolving the cardioid r = a (1 -|- 
cos (?) about the initial line. 

14. Find the volume generated by revolving the cardioid r = 

o (1 + cos 0) about the line x = — j* 

16. Find the volume generated by revolving the eUipse 

x^ -h xy -h 2/2 = 3 \t^^ ^* 

about the x-axis. 

16. Find the volume generated by revolving about the line y = x 

the part of the parabola x* -H 2/* = "* cut off by the line x -{■ y = a. 

29. Volume of a Solid with Given Area of Section. — 
Divide the solid into slices by parallel planes. Let X be the 
area of section at distance x from a fixed point. The plate 
PQRS with lateral surface perpendicular to PQR has the 
volume 

PQR 'Ax = X Ax. 



Art. 29 Volume of a Solid with Given Area of Section 57 
If a and h are the limiting values of x, the sum of such plates is 

Xx^x. 

The volume required is the limit of this sum 

v = \\my\XAx=rXdx. (29) 



Ai=0 




Example 1. Find the volume of the ellipsoid 



?! _L ^' 4_ !! = 1 
^2 -r ^2 "f- ^2 ^• 




Fig. 296. 

The section perpendicular to the x-axis at the distance x 
from the center is an ellipse 

^ -J- ?! = 1 _ ^ 

52-1-^2 A o2* 



58 Simple Areas and Volumes Chap. 4 

The semi-axes of this ellipse are 

MP = cv/l-|, MQ = 6\/l-^. 
By exercise 6, page 49, the area of this ellipse is 
TT . MP . MQ = Tbc ll - ~) • 

The volume of the ellipsoid is, therefore, 

I Trbc (1 ^jdx = ^ irabc. 

Ex. 2. The axes of two equal right circular cylinders 
intersect at right angles. Find the common volume. 




Fig. 29c. 



In Fig. 29c, the axes of the cylinders are OX and OZ and 
OABC is I of the common volume. The section of OABC 
by a plane perpendicular to OF is a square of side 



MP = MQ = Va^ - y\ 



I 



Art. 29 Volume of a Solid with Given Abea of Section 59 

The area of the section is therefore 

MP • MQ = a2 - y*, 

and the required volume is 

EXERCISES 

1. Find the volume of a pjTamid by integration. 

2. A wedge is cut from the base of a right circular cylinder by a 
plane passing through a diameter of the base and inclined at an angle 
a to the base. Find the volume of the wedge. 

3. Two circles have a diameter in common and lie in perpendicular 
planes. A square moves in such a way that its plane is perpendicular 
to the common diameter and its di^onals are chords of the circles. 
Find the volimae generated. 

4. The plane of a moving circle is perpendicular to that of an ellipse 
and the radius of the circle is an ordinate of the ellip)se. Find the vol- 
ume generated when the circle moves from one vertex of the eUipse to 
the other. 

5. The plane of a moving triangle is perpendicular to a fixed diam- 
eter of a circle, its base is a chord of the circle, and its vertex Ues on a 
line parallel to the fixed diameter at distance h from the plane of the 
circle. Find the volume generated by the triangle in moving from one 
end of the diameter to the other. 

6. A triangle of constant area A rotates about a line perpendicular 
to its plane while advancing along the line. Find the volume swept 
out in advancing a distance h. 

7. Show that if two soUds are so related that every plane parallel to 
a fixed plane cuts from them sections of equal area, the volumes of the 
soUds are equal. 

8. A cj-lindrical siu^ace passes through two great circles of a sphere 
■which are at right angles. Find the volume within the cylindrical 
surface and sphere. 

9. Two cylinders of equal altitude h have a common upper base and 
their lower bases are tangent. Find the volume common to the two 
cj-linders. 

10. .\ circle moves with its center on the 2-axis and its plane parallel 
to a fixed plane inclined at 45° to the 2-axis. If the radius of the circle 
is always r = Va- — z*, where 2 is the coordinate of its center, find the 
volimae described. 



CHAPTER V 

OTHER GEOMETRICAL APPLICATIONS 

30. Infinitesimals of Higher Order. — In the applica- 
tions of the definite integral that we have previously made, 
the quantity desired has in each case been a limit of the form 

lim V' fix) Ax. 

Ai=0 ^a 

We shall now consider cases involving limits of the form 
lim V Fix, Ax) 

when F ix, Ax) is only approximately expressible in the form 
/ (x) Ax. Such cases are usually handled by neglecting 
infinitesimals of higher order than Ax. That such neglect 
does not change the limit is indicated by the following 
theorem : 

// for values of x between a and b, F ix, Ax) differs from 
f ix) Ax by an infinitesimal of higher order than Ax, 

lim Y Fix, Ax) =lim V fix) Ax. 

To show this let e be a number so chosen that 
F ix. Ax) = fix) Ax -\-e Ax. 

If F ix. Ax) and / ix) Ax differ by an infinitesimal of higher 
order than Ax, e Ax is of higher order than Ax and so e ap- 
proaches zero as Ax approaches zero (Differential Calculus, 
Art. 9). The difference 

y" Fix, Ax) - V'fix) Ax = V%Ax 
60 



Art. 31 



Rectangtjlar Coordinates 



61 



Am 



kO. 



is graphically represented by a sum of rectangles (Fig. 30), 
whose altitudes are the various values of c. Since all these 
values approach zero * with Aa;, 
the total area approaches zero 
and so 

limVV(x,Aa;)=limV /(x)Ax, - 

which was to be proved. ^^^- ^^• 

31. Length of a Curve. Rectangular Coordinates. — 

In the arc AB of a curve inscribe a series of chords. The 
length of one of these chords PQ is 

VA^M^A^ = y/l + (^)'Aa:, 



Y 






Ay 


B 




A 


f" 









c 


1 




i X 



Fig. 31a. 



and the sura of their lengths is 



The length of the arc AB is defined as the limit approached 
by this sum when the number of chords is increased indefi- 
nitely, their lengths approaching zero. 

* For the discussion to be strictly accurate it vaaai be shown that 
there is a number larger than any of the e's which approaches zero. In 
the language of higher mathematics, the approach to the limit must be 
uniform. In ordinary cases that certainly would be true. A similar 
remark applies to all the applications of the above theorem. 



62 Other Geometrical Applications Chap. 5 



The quantity V 1 +(t^) is not a function of x alone. 



\^x) 
When Ao; approaches zero, however, the difference of 



es zero. If then we 



V/^ + {^ ^^^ v/l + (^J approach 

replace V1+(t^) Aa;byyi+ f -^ J Ax, the error is an 

infinitesimal of higher order than \x. Therefore the length 
of arc is 

dv 
In applying this formula ~ must be determined from the 

equation of the curve. The result can also be written 

s= r Vd^+~dy^. (31) 

In this formula, y may be expressed in terms of x, or x in 

terms of y, or both may be expressed in terms of a parameter. 

In any case the limits are the values at A and B of the 

variable that remains. 

Example 1. Find the length of the arc of the parabola 

y^ = 4:X between x = and x = 1. -\ 

dx V 
In this case t- = ^ . The limiting values of y are and 2. 
dy 2 

Hence 

s = J'\/l+(^Jdy = £lV^^^dy = V2 + \nil + V2). 

Ex. 2. Find the perimeter of the curve 

x = a cos^ </>, y = a sin^ </». 
In this case 



ds = Vdx^ + dy^ = Vo a^ cos* sin^ 0+ 9 a^ sin" <^ cos^ <f> d<l> 
= 3 a cos (f) sin <j) d<i>. 



Art. 32 Polar Coordinates 63 

One-fourth of the curve is described when <f> varies from 
to jz. Hence the perimeter is 



= 4 rSa 
Jo 



cos <f> sin <f>d<f> = 6 a. 



EXERCISES 

1. Find the circumference of a circle by integration. 

2. Find the length of y^ = x' between (0, 0) and (4, 8). 

3. Find the length of x = In sec y between y = and y = -^y 

4. Find the length ofx = \jf— |lny between y = 1 and y = 2. 

5. Find the length of y = e^ between (0, 1) and (1, e). 

6. Find the perimeter of the curve 

x' +2/ = tt • 

7. Find the length of the catenary 



y = -M + e 9 



between x = — a and x = a. 

8. Find the length of one arch of the cycloid 

x — a {<(> — sin <f>), y = a {1 — cos (^). 

9. Find the length of the involute of the circle 

X = a (cos +dsia6), y = a (sin 6—6 cos 6), 

between ^ = and 6 = 2v. 

10. Find the length of an arc of the cycloid 

X = a{d + siD.6), y = a{l—cosd). 

If s is the length of arc between the origin and any point (x, y) of the 
same arch, show that 

s2 = 8 ay. 

32. Length of a Curve. Polar Coordinates. — The 
differential of arc of a curve is (Differential Calculus, Arts.. 
54,59) 

ds = Vdx^ + di/2 = Vdr2 + r2 dd\ 



64 Other Geometrical Applications 

Equation (31) is, therefore, equivalent to 



Chap. 5 



(32) 



r 




i 








M 






^•^ 













X 



Fig. 32. 
In this case dr = add and 



In using this formula, 
r must be expressed in 
terms of or ^ in terms 
of r from the equation 
of the curve. The limits 
are the values at A and B 
of the variable that re- 
■^ mains. 

Example. Find the 
length of the first turn 
of the spiral r = ad. 



Jo 



2 dd^ + a2^2 (IQ2 



Jo 



+ 02, 



= 7ra Vl +4x-+^ln(2x + Vl+4x2). 



EXERCISES 

1. Find the circumference of the circle r — a. 

2. Find the circumference of the circle r = 2 a cos Q. 

3. Find the length of the spiral r = e"* between = and Q — -> 

a 

4. Find the distance along the straight line r = a sec ( ^ ~ 5 ) from 
= to (9 = ^• 

6. Find the arc of the parabola r = a sec* J 6 cut off by the ^-axis. 

6. Find the length of one loop of the curve 

r = a cos* 7' 
4 

7. Find the perimeter of the cardioid 

r = a (1 + cos6). 

8 

8. Find the complete perimeter of the curve r = a sin' = . 

o 



Art. 33 



Area of a Surface of Revolution 



05 



33. Area of a Surface of Revolution. — To find the area 
generated by revolving the arc AB about the x-axis. 

Join A and Z^ by a broken line with vertices on the arc. 
Let X, y be the coor- 
dinates of P and X + q_ 
Ax, y -\- Ay those of Q. 
The chord PQ generates 
a frustum of a cone 
whose area is 



7r(2y + Ay) PQ = 
ir(2y + A?/) VAx2+A2/2. 




Fig. 33a. 



The area generated by the broken line is then 



2) 7r(2 2/ + Ay)VAx2 + Ai/2. 



The area S generated by the arc AB is the limit ap- 
proached by this sum when Ax and Ay approach zero. Neg- 
lecting infinitesimals of higher order, (2 i/ + Ay) VAx^ + Ay^ 
can be replaced by 2 y Vdx^ + dy^ = 2y ds. Hence the 
area generated is 

S= \2Tryds. (33a) 

In this formula y and ds must be calculated from the 
equation of the curve. The limits are the values at A and 
B of the variable in terms of which they are expressed. 

Similarly, the area generated by revolving about the 
2/-axis is 



S 



=i>^ 



xds. 



(33b) 



Example. Find the area of the surface generated by 
revohing about the y-axis the part of the curve y = \ — x* 
above the x-axis. 



66 Other Geometrical Applications 

In this case 



Chap. 5 



ds = y 1 + (^Jdx = Vl + 4x''dx. 

The area required is generated by the part AB of the curve 
between x = and x = 1. Hence 




Fig. 336. 



S = f 2t xds= j 2txVi-\-4x'' 

D v) 



dx 



EXERCISES 

1. Find the area of the surface of a sphere. 

2. Find the area of the surface of a right circular cone. 

3. Find the area of the spheroid generated by revolving an ellipse 
about its major axis. 

4. Find the area generated by revolving the curve x' + 2/ = o 
about the 2/-axis. 

6. Find the area generated by revolving about OX, the part of the 
catenary 






between x = — a and x = a. 

6. Find the area generated by revolving one arch of the cycloid 

X = a (<t> — sin <^), y = a (1 — cos (f>) 
about OX. 



Art 34 



Unconventional Methods 



67 



7. Find the area generated by revolving the cardioid r = a (1 + cos 0) 
about the initial line. 
/ 8. The arc of the circle 

«* + y* = a' 

between (a, 0) and (0, a) is revolved about the line x -{- y = a. Find 
the area of the surface generated. 

9. The arc of the parabola ^ = 4 x between x = and x = 1 is 
I revolved about the line y = — 2. Find the area generated. 

10. Find the area of the surface generated by revolving the lemnis- 

! cate r^ = 2 a' cos 2 about the line fl = 7- 

4 

j 34. Unconventional Methods. — The methods that have 
I been given for finding lengths, areas, and volumes are the 
I ones most generally applicable. 
In particular cases other methods 
may give the results more easily. 
To solve a problem by integra- 
tion, it is merely necessary to ex- 
press the required quantity in any 
way as a limit of the form used in 
defining the definite integral. 

Example 1. When a string held 
taut is unwound from a fixed 
circle, its end describes a curve 
called the involute of the circle. 
Find the length of the part de- 
scribed when the first turn of the 
string is unwound. 

Let the string begin to unwind at A. When the end 
reaches P the part unwound QP is equal to the arc AQ. 
Hence 

QP = AQ = ad. 

When P moves to R the arc PR is approximately the arc of 
a circle with center at Q and central angle M. Hence 




Fig. 34a. 



PR = ad Id 



68 



Other Geometrical Applications 



Chap. 6 



approximately. The length of the curve described when d 
varies from to 2 tt is then 



s = lim 2i ad Ad 



Ae=o 



Jo 



addd = 2ira\ 



Ex. 2. Find the volume generated by rotating about the 
y-axis the area bounded by the parabola x^ = y — 1, the 
X-axis, and the ordinates a: = ±1. 

Resolve the area into slices by ordinates at distances Aar 
apart. When revolved about the y-axis, the rectangle PM 
between the ordinates x, x -\- Ax generates a hollow cylinder 
whose volume is 

TT (x -\- Ax)- y — TTX^y = 2 irxy Ax + Ty {AxY. 




Fig. 346. 



Fig. 34c. 



Since Try {AxY is an infinitesimal of higher order than Ax, 
the required volume is 

lim V 2 TTxy Ax = I 2 ttx (1 + x^) dx = %Tr. 

Ex. 3. Find the area of the cylinder x^ -\- y^ = ax within 
the sphere x^ -\- y^ -\- z^ = a^. 

Fig. 34c shows one-fourth of the required area. Divide 
the circle OA into equal arcs As. The generators through 



Art. 34 Unconventional Methods 6^ 

the points of division cut the surface of the cylinder into 
rips. Neglecting infinitesimals of higher order, the area 
the strip MPQ is MP • As. If r, d are the polar coordinates 
M, r = a cos d and 

As = a A0, MP = Va- — r- = a sin 6. 
The required area is therefore given by 



— = hm 

4 Ad- 



im T^a^sinflA^ = Ja-smddd. 

Consequently 

S = 4a' / sindde = Aa\ 
Jo 

EXERCISES 

1. Find the area swept over by the string in example 1, page 67. 

2. Find the area of surface cut from a right circular cyUnder by a 
plane passing through a diameter of the base and incUned 45° to the 
base. 

3. The axes of two right circular cylinders of equal radius intersect 
at right angles. Find the area of the soUd coounon to the two cylinders 
(Fig. 29c). 

4. An equilateral triangle of side a is revolved about a line parallel 
to the base at distance b below the base. Find the volume generated. 

5. The area bounded by the hyperbola i* — y^ = a- and the lines 
y = ±a is revolved about the x-axis. Find the volume generated. 

6. The vertex of a cone of vertical angle 2 o is the center of a sphere 
of radius a. Find the volume common to the cone and sphere. 

7. The axis of a cone of altitude h and radius of base 2 a is a gen- 
erator of a cylinder of radius a. Find the area of the surface of the 
cyhnder within the cone. * 

8. Find the area of the surface of the cone in Ex. 7 within the 
cylinder. 

9. Find the volume of the cylinder in Ex. 7 within the cone. 



CHAPTER VI 

MECHANICAL AND PHYSICAL APPLICATIONS 

35. Pressure. — The pressure of a liquid upon a hori- 
zontal area is equal to the weight of a vertical column of the 
liquid having the area as base and reaching to the surface. 
By the pressure at a point P in the Hquid is meant the pressure 
upon a horizontal surface of unit area at that point. The 



1 
Fig. 35a. 




Fig. 356. 



volume of a column of unit section and height h is h. 
the pressure at depth h is 

p = wh, 



Hence 



(35a) 



w being the weight of a cubic unit of the liquid. 

To find the pressure upon a vertical plane area (Fig. 356), 
we make use of the fact that the pressure at a point is the 
same in all directions. The pressure upon the strip AB 
parallel to the surface is then approximately 

pAA, 

p being the pressure at any point of the strip and AA its 
area. The reason for this not being exact is that the pressure 

70 



Art. 35 Pressitbe 71 

at the top of the strip is a little less than at the bottom. 
This difference is, however, infinitesimal, and, since it multi- 
plies ilA, the error is an infinitesimal of higher order than 
Ai4. The total pressure is, therefore, 

P = lim VplA = fpdA = w ChdA. (35b) 

Before integration dA must be expressed in terms of h. 
The limits are the values of h at the top and bottom of the 
submerged area. In case of water the value of w is about 
62.5 lbs. per cubic foot. 

Example. Find the water pressure upon a s^-nicircle of 




Fig. 3oc. 

radius 5 ft., if its plane is vertical and its diameter in the 
surface of the water. 

In this case the element of area is 

dA = 2 V25 - h^ dh. 
Hence 

P = w jhdA =2w jhV2o-h^dh 

= i5 o. ^ = 2.^0. (62.5) = 5208.3 lbs. 

EXERCISES 

1. Find the pressure sustained by a rectangular floodgate 10 ft. 
broad and 12 ft. deep, the upper edge being in the surface of the water. 

2. Find the pressure on the lower half of the floodgate in the pre- 
ceding problem. 

3. Find the pressure on a triangle of base 6 and altitude h, sub- 
merged so that its vertex is in the surface of the water, and its altitude 
vertical. 

4. Find the pressure upon a triangle of base b and altitude h, sub- 
merged so that its base is in the surface of the liquid and its altitude 
vertical. 



72 Mechanical and Physical Applications Chap. 6 

5. Find the pressure upon a semi-ellipse submerged with one axis 
in the surface of the liquid and the other vertical. 

6. A vertical masonry dam in the form of a trapezoid is 200 ft. long 
at the surface of the water, 150 ft. long at the bottom, and 60 ft. high. 
What pressure must it withstand? 

7. One end of a water main, 2 ft. in diameter, is closed by a vertical 
bulkhead. Find the pressure on the bulkhead if its center is 40 ft. 
below the surface of the water. 

8. A rectangular tank is filled with equal parts of water and oil. If 
the oil is half as heavy as water, show that the pressure on the sides is 
one-fourth greater than it would be if the tank were filled with oil. 

36. Moment. — Divide a plane area or length into small 
parts such that the points of each part differ only infinitesi- 
mally in distance from a given axis. Multiply each part by 
the distance of one of its points from the axis, the distance 
being considered positive for points on one side of the axis 
and negative for points on the other. The limit approached 
by the sum of these products when the parts are taken 
smaller and smaller is called the moment of the area or length 
with respect to the axis. 

Similarly, to find the moment of a length, area, volume, 
or mass in space with respect to a plane, we divide it into 
elements whose points differ only infinitesimally in distance 
from the plane and multiply each element by the distance of 
one of its points from the plane (considered positive for 
points on one side of the plane and negative on the other). 
The moment with respect to the plane is the limit approached 
by the sum of these products when the elements are taken 
smaller and smaller. 

Example. Find the moment of a rectangle about an axis 
parallel to one of its sides at distance c. 

Divide the rectangle into strips parallel to the axis (Fig. 
36). Let y be the distance from the axis to a strip. The 
area of the strip is b Ay. Hence the moment is 

ybAy== J bydy = ab[c-\--]. 



Art. 37 Center of Gravity of a Length or Area in a Plane 



73 



Since ab is the area of the rectangle and c + x is the distance 

from the axis to its center, the moment is equal to the product 
of the area and the distance from the axis to the center of the 

rectangle. 

b 













- 


r-*~ 






a 












; 


1 




' 
















Fig. 36. 



Fig. 37a. 



37. The Center of Gravity of a Length or Area in a 
Plane. — The center of gravity of a length or area in a 
plane is the point at which it could be concentrated without 
changing its moment with respect to any axis in the plane. 

Let C (x, y) be the center of gravity of the arc AB (Fig. 
37a), and let s be the length of the arc. The moment of AB 
with respect to the x-axis is 



/; 



yds. 



If the length s were concentrated at C, its moment would be 
sy. By the definition of center of gravity 



sy 



whence 



J yds, 



Similarly, 



- 1: 



xds 



X = 



74 



Mechanical and Physical Applications 



Chap. 6 



The limits are the values at A and B of the variable in terms 

of which the integral is expressed. 

Let C {x, y) be the center of gravity of an area (Figs. 376, 

37c). Divide the area into strips dA and let {x, y) be the 

center of gravity of the 
strip dA . The moment 
of the area with respect 
to the X-axis is 




/^ 



dA. 



Fig. 376. 



If the area were con- 
centrated at C, the 
moment would be Ay, 
where A is the total 
area. Hence 




Fig. 37c 



The strip is usually taken parallel to a coordinate axis. 
The area can, however, be divided into strips of any other 
kind if convenient. 

Example 1. Find the center of gravity of a quadrant of 
the circle x"^ -\- y^ = a^. 

In this case ~^ 

i 

ds = Vdx^ -j- dy"^ = -dx 

\J 



f\ 



Art. 37 Center OF Gravity OF A Length OR Area IN A Plane 75 
and 



j yds = j y ' -dx = a^. 



The length of the arc is 



s = - (2 7ra) = -a. 



Hence 



y = 



I 



yds 



2_a 

IT 



T 


^^ 


is 




X 


^ , 










Tig. 37d. 



It is evident from the symmetry of the figure that x has the 
same value. 

Ex. 2. Find the center of gravity of the area of a semi- 
circle. 

From symmetry it is evident that the center of gravity is 
in the y-axis (Fig. 37e). Take the element of area parallel 
to OX. Then dA = 2xdy and 



j ydA = I 2xydy = 2 j y Va^ — y^ dy = 



ia\ 



The area is A = jr a^. Hence 



CydA 4 o 




Fig. 37e. 




Fig. 37/. 



Ex. 3. Find the center of gravity of the area bounded by 
the X-axis and the parabola y = 2x — x^. 

Take the element of area perpendicular to OX. If {x, y) 



76 Mechanical and Physical Applications Chap. 6 

are the coordinates of the top of the strip, its center of gravity 
is (a^, I). Hence its moment with respect to the x-axis is 

The moment of the whole area about OX is then 

i2 



15 



The area is 



A = I ydx = I (2X — X^) dx =;: 



Hence y = %. Similarly, 

\xdA r{2x^-x^) 



dx 



X = 



= 1. 



38. Center of Gravity of a Length, Area, Volume, or 
Mass in Space. — The center of gravity is defined as the 

point at which the 
mass, area, length, or 
volume can be con- 
centrated without 
changing its moment 
with respect to any 
plane. 

Thus to find the cen- 
ter of gravity of a solid 
mass (Fig. 38a) cut it 
into slices of mass Am. 
If {x, y, z) is the cen- 
ter of gravity of the 
slice, its moment with 
respect to the a;y-plane is z Am and the moment of the whole 
mass is 

lim V z Am = I z dm. 

Am*0 ^^ t/ 




Fig. 38o. 



Art 38 Center of Gravitt 77 

If the whole mass M were concentrated at its center of 
gravity {x, y, z), the moment with respect to the xy-plane 
! would be zM. Hence 



zM 



or 



= 12 dm, 

J 



zdm 



Similarly, 



/ X dm j y dm 



^ = ^M-' y = -M-' (^^> 



The mass of a unit volume is called the density. If then 
dv is the volume of the element dm and p its density, 

dm = pdv. 

To find the center of gravity of a length, area, or volume 
it is merely necessary to replace M in these formulas by s, 
S, or V. 

Example 1. Find the center of gravity of the volume of 
an octant of a sphere of radius a. 

The volume of the slice (Fig. 38a) is 

dv = I irx^ dz = \w (a^ — z^) dz. 
Hence 

fzdv = £l.ia^-z^)zdz = f-^a\ 
The volume of an octant of a sphere is ^ ird^. Hence 

- /^^" i^^' 3 
V x , 8 

6" 

From symmetry it is evident that x and y have the same 
value. 



78 



MECHANICAL AND PHYSICAL APPLICATIONS 



Chap. 6 



Ex. 2. Find the center of gravity of a right circular cone 
whose density is proportional to the distance from its base. 
Cut the cone into slices parallel to 
the base. Let y be the distance of a 
slice from the base. Except for in- 
finitesimals of higher order, its volume 
n is TX^ dy, and its density is ky where 
K is constant. Hence its mass is 

Am = kwx^y dy. 




Fig. 38fe. 



By similar triangles x = rih — ?/). 
Hence 



M 
Therefore, finally, 



h' 



= J dm = J -i^iji-yfydy = 



h' 



12 



/ ydm 

y = —n— 



2, 
M 5 



EXERCISES 

1. The wind produces a uniform pressure upon a rectangular door. 
Find the moment tending to turn the door on its hinges. 

2. Find the moment of the pressure upon a rectangular floodgate 
about a horizontal line through its center, when the water is level with 
the top of the gate. 

3. A triangle of base h and altitude h is submerged with its base 
horizontal, altitude vertical, and vertex c feet below the surface of the 
water. Find the moment of the pressure upon the triangle about a 
horizontal line through the vertex. 

4. Find the center of gravity of the area of a triangle, 

6. Find the center of gravity of the segment of the parabola y^ = ax, 
cut ofT by the line a; = a. 



Art 38 Center of Gravity 7^ 

6. FLid the center of gravity of the area of a quadrant of the ellipse 

a* ^ 6'- 

7. Find the center of gravity of the area bounded by the coordinate 
axes and the parabola x' + y* = a'. 

8. Find the center of gravity of the area above the x-axis bounded 

by the curve x* + y = a*. 

9. Find the center of gravity of the area bounded by the x-axis and 
one arch of the curve y = sin x. 

10. Find the center of gravity of the area bounded by the two parab- 
olas y- = ax, x^ = ay. 

11. Find the center of gravity of the area of the upper half of the 
cardioid r = o (1 -f- cosfl). 

12. Find the center of gravity of the area bounded by the x-axis and 
one arch of the cycloid, 

X = a (<^— sin0), y = a (1— co8<^). 

13. Find the center of gravity of the area within a loop uf the lemnis- 
cate r- = a^ cos 2 d. 

14. Find the center of gravity of the arc of a semicircle of radius a. 
16. Find the center of gravity of the arc of the catenary 

y = ^(^ + e"°) 

between x = — a and x = a. 

16. Find the center of gravity of the arc of the curve x* + y* = a* 
in the first quadrant. 

17. Find the center of gravity of the arc of the curve 

X = \y^— h\ny 

between y = 1 and y = 2. 

18. Find the center of gravity of an arch of the cycloid 

X = a{<i>— sm<i>), y — a{\— cos4>). 

19. Find the center of gravity of a right circular cone of constant 
density. 

20. Find the center of gravity of a hemisphere of constant density. 

21. Find the center of gravity of the solid generated by revolving 
about OX the area bounded by the parabola y- = 4 x and the line x = 4. 

22. Find the center of gravity of a hemisphere whose density is 
proportional to the distance from the plane face. 

23. Find the center cf gravity of the soUd generated by rotating a 
sector of a circle about one of its bounding radii. 



so Mechanical and Physical Applications Chap. 6 

24. Find the center of gravity of the solid generated by revolving 
the cardioid r = a (1 + cos e) about the initial line. 

26. Find the center of gravity of the wedge cut from a right circular 
cylinder by a plane passing through a diameter of the base and making 
with the base the angle a. 

26. Find the center of gravity of a hemispherical surface. 

27. Show that the center of gravity of a zone of a sphere is midway 
between the bases of the zone. 

28. The segment of the parabola' i/^ = 2ax cut off by the line x = a 
is revolved about the x-axis. Find the center of gravity of the surface 
generated. 

39. Theorems of Pappus. Theorem I. — If the arc of a 
plane curve is revolved about an axis in its plane, and not 
crossing the arc, the area generated is equal to the product 
of the length of the arc and the length of the path described 
by its center of gravity. 

Theorem II. If a plane area is revolved about an axis in 
its plane and not crossing the area, the volume generated is 
equal to the product of the area and the length of the path 
described by its center of gravity. 

To prove the first theorem, let the arc be rotated about 
the X-axis. The ordinate of its center of gravity is 



I' 



^ yds 

whence 



27r / yds = 2irys. 



The left side of this equation represents the area of the 
surface generated. Also 2Try is the length of the path 
described by the center of gravity. This equation, therefore, 
expresses the result to be proved. 

To prove the second theorem let the area be revolved about 
the X-axis. From the equation 



JydA 



Art. 39 



Theorems of Pappus 



81 



we get 

2ir I ydA = 2iryA. 

Since 2t f y dA is the volume generated, this equation is 

equivalent to theorem II. 

Example 1. Find the area of the torus generated by- 
revolving a circle of radius a about an axis in its plane at 
distance b (greater than a) from its 
center. 

Since the circumference of the circle 
is 2 7ra and the length of the path de- 
scribed by its center 2 wb, the area gen- 
erated is 



S = 27ra-2 7r6 = 4:ir-ab. 




Fig. 39a. 



Ex. 2. Find the center of gravity of the area of a semi- 
circle by using Pappus's theorems. 

When a semicircle of radius a is revolved about its diameter, 
the volume of the sphere generated is J xa'. If y is the 
distance of the center of gravity of the semicircle from this 
diameter, by the second theorem of Pappus, 

^ Tra^ = 2 Try A = 2 x^ • i- xa^, 
whence 

$xa' 




y = 



TT^a- 



4a 
3x" 



Ex. 3. Find the volume gen- 
erated by revolving the cardioid 
r = a (1 + cos 6) about the initial 
line. 

The area of the triangle OPQ 
is approximately 
Fig. 396. 1 r^ Ad, 

and its center of gravity is f of the distance from the vertex 
to the base. Hence 

y = f r sin 0. 



82 Mechanical and Physical Applications Chap, ft 

By the second theorem of Pappus, the volume generated by 
OPQ is then approximately 

2 TT?/ A A = f 7rr^ sin 6 M. 
The entire volume is therefore 

V = j lTr^smddd = ^Tra^ J {1 + coseysmddd 

Jo t/o 

2 -(l + eos^)-'|- 8 , 

O 4 I o 

EXERCISES 

V 1. By using Pappus's theorems find the lateral area and the volume 
of a right circular cone. 

Y 2. Find the volume of the torus generated by revolving a circle of 
radius a about an axis in its plane at distance b (greater than a) from 
its center. 

J 3. A groove with cross-section an equilateral triangle of side i inch 
is cut around a cylindrical shaft 6 inches in diameter. Find the volume 
of material cut away. 

</ 4. A steel band is placed around a cylindrical boiler 48 inches in 
diameter. A cross-section of the band is a semi-ellipse, its axes being 6 
and Vq inches, respectively, the greater being parallel to the axis of the 
boiler. What is the volume of the band? 
l^y^. The length of an arch of the cycloid 

X = a {<j> — sin <^), y = a (I — cos <j>) 

is 8 a, and the area generated by revolving it about the x-axis is ^ iro*. 
Find the area generated by revolving the arch about the tangent at its 
highest point. 

i/^&. By the method of Ex. 3, page 81, find the volume generated by 
revolving the lemniscate r^ = 2 a^ cos 2 about the x-axis. 

7. Obtain a formula for the volume generated by revolving the 
polar element of area about the line x = — a. Apply this formula to 
obtain the volume generated by revolving about x = — a the sector of 
the circle r = a bounded by the radii d = — a, 6 = -\- a. 

8. A variable circle revolves about an axis in its plane. If tho 
distance from the center of the circle to the axis is 2 a and its radius 
is a sin 6, where d is the angle of rotation, find the volume of the horn- 
shaped solid that is generated. 

9. Can the area of the surface in Ex. 8 be found in a similar way? 



Art. 40 



Moment of Inertia 



83 



10. The vertex of a right circular cone is on the surface of a right 
circular cylinder and its axis cuts the axis of the cyhnder at right angles. 
Find the volume common to the cyhnder and cone (use sections deter- 
mined by planes through the vertex of the cone and the generators of 
the cylinder). 

40. Moment of Inertia. — The moment of inertia of a 
particle about an axis is the product of its mass and the square 
of its distance from the axis. 

To find the moment of inertia of a continuous mass, we 
divide it into parts such that the points of each differ only 
infinitesimally in distance from the axis. Let Aw be such a 
part and R the distance of one of its points from the axis. 
Ebccept for infinitesimals of higher order, the moment of 
inertia of \m about the axis is R- Aw. The moment of 
inertia of the entire mass is therefore 



7 = UmX^^^^= fR'dm. 

Am=0 ^ J 



(40) 



By the moment of inertia of a length, area, or volume, we 
mean the value obtained by using the differential of length, 
area, or volume in place of 
dm in equation (40). 

Example 1. Find the mo- 
ment of inertia of a right cir- 
cular cone of constant density 
about its axis. 

Let p be the density, h the 
altitude, and a the radius of 
the base of the cone. Di\nde 
it into hollow cyhndrical sUces 
by means of cylindrical sur- 
faces having the same axis as 
the cone. By similar triangles 
the altitude y of the cylin- 
drical surface of radius r is 




Fig. 40a. 



y=-(a-r). 



84 



Mechanical and Physical Applications 



Chap. C 



Neglecting infinitesimals of higher order, the volume between 
the cylinders of radii r and r + Ar is then 

Ay = 2 irry Ar = r (a — r) dr. 

a 



The moment of inertia is therefore 

2Trhp 
a 



I = I r^ dm = I 



r^p dv = 
The mass of the cone is 



t/O 



r^(a — r) dr = 



Tpha* 
10 




M = pv = I wpa^h. 
I = 



Hence 

Ex. 2. Find the moment of in- 
ertia of the area of a circle about 
a diameter of the circle. 

Let the radius be a and let the x- 
axis be the diameter about which 
the moment of inertia is taken. 
Divide the area into strips by lines parallel to the x-axis. 
Neglecting infinitesimals of higher order, the area of such a 
strip is 2 a; Ay and its moment of inertia 2 xy^ Ay. The 
moment of inertia of the entire area is therefore 



Fig. 406. 



/ = j2xy^dy = 2 j Va-' - y^y^ dy = 



ira* 



EXERCISES 



1. Find the moment of inertia of the area of a rectangle about one 
of its edges. 

2. Find the moment of inertia of a triangle about its base. 

3. Find the moment of inertia of a triangle about an axis through 
its vertex parallel to its base. 

4. Find the moment of inertia about the r/-axis of the area bounded 
by the parabola y^ = iax and the line x = a. 

5. Find the moment of inertia of the area in Ex. 4 about the line 
X = a. 

6. Find the moment of inertia of the area of a circle about the axis 
perpendicular to its plane at the center. (Divide the area into rings 
with centers at the center of the circle.) 



Art. 41 Work Done by a Forge 8^ 

7. Find the moment of inertia of a cylinder of mass M and radius a. 
about its axis. 

8. Find the moment of inertia of a sphere of mass M and radius a. 
about a diameter. 

9. An ellipsoid is generated by revolving the eUipse 

^-ul^= 1 

about the x-axis. Find its moment of inertia about the x-axis. 

10. Find the moment of inertia of a hemispherical shell of constant 
density about the diameter perpendicular to its plane face. 

11. Prove that the moment of inertia about any axis is equal to the 
moment of inertia about a parallel axis through the center of gravity 
plus the product of the mass and the square of the distance between the 
two axes. 

12. Use the answer to Ex. 6, and the theorem of Ex. 11 to determine 
the moment of inertia of a circular area about an axis, perpendicular to- 
ils plane at a point of the circumference. 

41. Work Done by a Force. — Let a force be applied to 
a body at a fixed point. Wtien the body moves work is done 
by the force. If the force is constant, the work is defined as 
the product of the force and the distance the point of appli- 
cation moves in the direction of the force. That is, 

W = Fs, (41a) 

where W is the work, F the force, and s the distance moved 
in the direction of the force. 

If the direction of motion does 
not coincide with that of the force, 

the work done is the product of the ^^^^ I ^j . 

force and the projection of the dis- Fig. 41a. 

placement on the force. Thus when the body moves from. 
A to 5 (Fig. 41a) the work done by the force F is 

W = Fs cos e. (41b> 

If the force is variable, we divide the path into parts As. 
'Tn moving the distance As, the force is nearly constant and. 
so the work done is approximately FcosflAs. As the 




m 



Mechanical and Physical Applications 



Chap. 6 



intervals As are taken shorter and shorter, this approximation 
becomes more and more accurate. The exact work is then 
the Hmit 

W = ]imy.FcosdAs= jFcoseds. (41c) 

As=0 ^ J 

To determine the value of W, we express F cos 6 and ds 
in terms of a single variable. The limits of integration are 
the values of this variable at the two ends of the path. If 
the displacement is in the direction of the force, 6 = 0, 
cos 6=1 and 



W 



-P 



ds. 



(41d) 




-B 



vwvvww-^- 



Fig. 416. 



Fig. 41c. 



Example 1. The amount a helical spring is stretched is 
proportional to the force applied. If a force of 100 lbs. is 
required to stretch the spring 1 inch, find the work done in 
stretching it 4 inches. 

Let s be the number of inches the spring is stretched. The 
force is then 

F = ks, 

k being constant. When s = 1, F = 100 lbs. Hence k = 
100 and 

F =lOOs. 

The work done in stretching the spring 4 inches is 

JFds = I 100 s ds = 800 inch pounds = 66§ foot pounds. 
*Jo 



Art. 41 



Work Done by a Force 



87 



Ex. 2. A gas is confined in a cylinder with a movable piston. 
Assuming Boyle's law pv = k, find the work done by the 
pressure of the gas in pushing out the piston (Fig. 41rf). 

Let V be the volume of gas in the cylinder and p the pressure 
per unit area of the piston. If A is the area of the piston, 
pA is the total pressure of the gas upon it. If s is the distance 
the piston moves, the work done is 

W = JpAds. 

But A ds = dv. Hence 

W= f^pdv= r'^dv = k\n^ 

is the work done when the volume expands from I'l to i'2. 




Fig. Aid. Fig. 41e. 

Ex. 3. The force with which an electric charge Ci repels 
a charge e^ at distance r is 

ke-iCz 



where k is constant. Find the work done by this force 
when the charge e^ moves from r = a to r = 6, Ci remaining 
fixed. 

Let the charge d move from A to B along any path AB 
(Fig. 41e). The work done by the force of repulsion is 

W = jFcosdds = ffdr = r^dr 

The work depends only on the end points A and B and not 
on the path connecting them. 



Mechanical and Physical Applications 



Chap. 6 



EXERCISES 

1. According to Hooke's law the force required to stretch a bar from 
the length a to the length a -{- x ia 

kx 



where K is constant. Find the work done in stretching the bar from the 
length a to the length b. 

2. Supposing the force of gravity to vary inversely as the square of 
the distance from the earth's center, find the work done by gravity on 
a meteor of weight w lbs., when it comes from an indefinitely great 
distance to the earth's surface. 

3. If a gas expands without change of 
temperature, according to van der Waal's 
equation, 



^ >, 



P 



V — b 




a, b, c being constant. Find the work done 
when the gas expands from the volume vi to 
the volume t'2. 

4. The work in foot pounds required 
to move a body from one altitude to 
another is equal to the product of its 
weight in pounds and the height in feet 
41J. ^j^^^ j^ jg raised. Find the work required 

to pump the water out of a cylindrical cistern of diameter 4 ft. and 

depth 8 ft. 

5. A vertical shaft is supported by a flat step bearing (Fig. 41/). 
The frictional force between a small part of the shaft and the bearing is 
HP, where p is the pressure between the two and m is a constant. If the 
pressure per unit area is the same at all points of the supporting surface, 
and the weight of the shaft and its load is P, find the work of the fric- 
tional forces during each revolution of the shaft. 

6. When an electric current flows a distance x through a homo- 
geneous conductor of cross-section A, the resistance is 

kx 
A' 

where K is a constant depending on the material. Find the resistance 
when the current flows from the inner to the outer surface of a hollow 
cylinder, the two radii being a and b. 



Art 41 



Work Doxe by a Force 



89 



7. Find the resistance when the current flows from the inner to the 
outer surface of a hollow sphere. 

8. Find the resistance when the cmrent 
flows from one base of a tnmcated cone to 
the other. 

9. When an electric current i flows an in- 
finitesimal distance AB (Fig. 41^) it produces 
at any point O a magnetic force (perpendicular 
to the paper) equal to 

ide 
r ' 

where r is the distance between AB and 0. Find the force at the center 
of a circle due to a current i flowing around it. 

10. Find the magnetic force at the distance c from an infinite straight 
line along which a current i is flowing. 




CHAPTER VII 
APPROXIMATE METHODS 

42. The Prismoidal Formula. — Let yi, ys, be two 

ordinates of a curve at distance h apart, and let 2/2 be the 

ordinate midway between 
them. The area bounded 
by the a:-axis, the curve, and 
the two ordinates is given 
approximately by the for- 
mula 

Fig. 42a. A= lh(yi-\- 4:yo+ ys). (42a) 

This is called the prismoidal formula because of its similarity 
to the formula for the volume of a prismoid. 
If the equation of the curve is 




y = a -i- bx -i- cx~ -\- dx?, 



(42b) 



where a, 6, c, d, are constants (some of which may be zero), 
the prismoidal formula gives the exact area. To prove this 
let k be the abscissa of the middle ordinate and t the dis- 
tance of any other ordinate from it (Fig. 42a). Then 

X = k -[-t. 

[f we substitute this value for x, (42b) takes the form 

y = A+Bt + Ct^-{- Dt\ 

where A, B, C, D are constants. The ordinates yi, 1/2, ys are 

Hence 



h h 

obtained by substituting t = — k, 0,^. 



90 



Art. 42 



The Prismoidal Forjiula 



91 



Also the area is 



-2 



12 



This is equivalent to 



which was to be proved. 

If the equation of the curve does not have the form (42b), 
it may be approximately equivalent to one of that type and 
so the prismoidal formula may give an approximate value 
for the area. 

While we have illustrated the prismoidal formula by the 
area under a curve, it may be used equally well to determine 
a length or volume or any other quantity represented by a 
definite integral, 

V (x) dx. 



£ 



Since such an integral represents the area under the curve 
y = / (^)> its value can be found by replacing h in (42a) by 

6 - a and t/i, y^, yz by / (a), / f ^ ^ j , / (6) respectively. 

Exam-pie 1. Find the 
area bounded by the 
X-axis, the curve y = 
e~^\ and the ordinates 
X = 0, X = 2. 

The integral 



/ 



e-'^dx 




Fig. 426. 



cannot be expressed in terms of elementary functions. 
Therefore we cannot obtain the area by the methods that we 



92 



Approximate Methods 



Chap. 7 



have previously used. The ordinates yi, y^, yz, in this case 
are 

yi = 1, y2 = e-S yz = e-^. 

The prismoidal formula, therefore, gives 
A 



= |(l +^+^1=0.869. 



The answer correct to 3 decimals (obtained from a table) is 
0.882. 

Ex. 2. Find the length of the parabola ?/- = 4 x from 
a; = 1 to X = 5. 

The length is given by the formula 



=xv^-^ 



dx. 



By integration we find s = 4.726. To apply the prismoidal 
formula, let 



y 



-^' 



Then /i = 4, 
and 



Vl = V2, 7/2 = Vl, 7/3 = Vf, 

s = 4 ( V2 + 4 v^ + Vf ) = 4.752. 

Ex. 3. Find the vol- 
ume of the spheroid 
generated by revolving 
the ellipse 

a- 0- 

about the x-axis. 

The section of the 




Fig. 42c. 



spheroid perpendicular to OX has the area 
A = Try'- = irh- ll - ^y 



Art. 43 



Simpson's Rule 



93 



Its volume is 



=£■ 



Adx. 



Since A is a polynomial of the second degree in x (a sp)ecial 
case of a third degree polynomial), the prismoidal formula 
gives the exact volume. The three cross-sections corre- 
sponding to x = —a, X = 0, X = a, are 

Ai = 0, Ao = irlr, Ai = 0. 
Hence 



y = ^[Ai + 4A2 + A3]=| 



ra62. 



43. Simpson's Rule. — Divide the area between a curve 
and the x-axis into any even number of parts by means of 
equidistant ordinates yi, y^, 1/3, ... , y„. (An odd number 
of ordinates will be needed.) Simpson's rule for determining 
approximately the area between yi and y„ is 

^l/, + 4y2 + 2i/, + 4i/4 + 2i/3+ • • • 4- 



A 



1 + -i + 2 + 4 -f 2 + 



+ 1 



jby 



(43) 



h being the distance between the ordinates yi and ?/„, In 
the numerator the end coefficients are 1. The others are 
alternately 4 and 2. The 
denominator is the sum of 
the coefficients in the num- 
erator. 

This formula is obtained 
by applying the prismoidal 
formula to the strips taken 
two at a time and adding 
the results. Thus if the area 



Y 


/ 


'Jx 


V. 


y» 


"4 


^ 

"s 







£ 















Fig. 43. 



is divided into four strips by the ordinates r/i, 1/2, ys, J/4, y 

h 



the part between r/i and 1/3 has a base equal to 
area as given by the prismoidal formula is 



2' 



Its 



94 



Approximate Methods 



Chap. 7 ' 



Similarly the area between ys and y^ is 

The sum of the two is 

A^h ( ^^' + ^ ^2 + ^ ^3 + 4 ^4 + ys N 

By using a sufficiently large number of ordinates in 
Simpson's formula, the result can be made as accurate as 
desired. 

Example. Find In 5 by Simpson's rule. Since 



in 5 



ndx 



1 . 



we take y = - m Simpson's formula. Dividing the interval 
into 4 parts we get 



In 5 = 4 /i 



+ 4»^ + 2.^ + 4-i + ^ 
12 



= 1.622. 



If we divide the interval into 8 parts, we get 

In 5 = ^\- (1 + I + I + I + f + f + I + I + i) = 1.6108. 
The value correct to 4 decimals is 

In 5 = 1.6094. 

44. Integration in Series. — In calculating integrals it 
is sometimes convenient to expand a function in infinite 
series and then integrate the series. This is particularly the 
case when the integral contains constants for which numerical 
values are not assigned. For the process to be valid all 
series used should converge. 

Example. Find the length of a quadrant of the ellipse 



^^t.= 1 
a^ "^ 62 ^' 



Art 44 Integration in Series 95 

Let a be greater than 6. Introduce a parameter <t> by the 

equation 

X = a sin <t>. 

Substituting this value in the equation of the ellipse, we find 
y = b cos 4>. 

Using these values of x and y we get 

r 

s= fVdx^ + dy- = jVa- - (a^ - b^) sin^ <f, d<t>. 

This is an eUiptic integral. It cannot be represented by an 
expression containing only a finite number of elementary 
functions. We therefore express it as an infinite series. By 
the binomial theorem 



Va- — (a- — b~) sin^ (f> 

= «L^ - 2 -^^^^^ '^-2li-^j ^^ '^ • • -J- 
Since 

we find by integrating term by term 

*~"L2 8 a' 128V a^ j ' gj 

^Trgf g^-b^ 3 /a- - ¥V 1 

~ 2 L 4a2 64V a- )''']' 

If a and 6 are nearly equal, the value of s can be calculated 
very rapidly from the series. 

EXERCISES 

1. Show that the prismoidal formula gives the correct volume in 
each of the following cases: (o) sphere, (6) cone, (r) cyhnder, (d) 
P3'ramid, (e) segment of a sphere, (/) truncated cone or pyramid. 

2. Find the error when the value of the integral j i*dx is found 
by the prismoidal formula. 



•96 Approximate Methods Chap. 7 

In each of the following cases compare the value given by the pris- 
moidal formula with the exact value determined by integration. 

3. Area bounded hy y = Vx, y = 0, x = 1, x = 3. 

4. Arc of the curve y = 3^ between x=— 2, x = +2. 

5. Volume generated by revolving about OX one arch of the sine 
■curve y = sin x. 

6. Area of the surface of a hemisphere. 

Compute each of the following by Simpson's rule using 4 intervals: 



dx 
4 Jo 1 +x^ 



dx 



'•r 



Vi+x» 
9. Length of the curve y = Inx from a; = 1 to a; = 5. 

10. Surface of the spheroid generated by rotating the ellipse x* + 
■4y^ = 4 about the x-axis. 

11. Volume of the solid generated by revolving about the x-axis the 

^rea bounded hy y = 0, y = ^ , ^ , x = — 2, x = 2. 



12. Find the value of 



by expanding in series. 
13. Express 



J cos (x*) dx, 




x 



' sin (Xx) dx 



as a series in powers of X. 

14. Find the length of a quadrant of the ellipse x* + ly^ = 2. 



CHAPTER VIII 
DOUBLE INTEGRATION 
45. Double Integrals. — The notation 
/ / f{x,y)dxdy 

is used to represent the result of integrating first with respect 
to y (leaving x constant) between the limits c, d and then 
with respect to x between the limits a, b. 

As here defined the first integration is with respect to the 
variable whose differential stands last and its limits are 
attached to the last integral sign. Some writers integrate 
in a different order. In reading an article it is therefore 
necessary to know what convention the author uses. 

Example. Find the value of the double integral 



Jo J -i 



(x- -H y^) dx dy. 



We integrate first with respect to y between the limits —x, 
X, then with respect to x between the limits 0, 1. The result 
is 

f j\x-'+y'')dxdy = j^dx{xhf + \y'y_^ = J\x'dx = l. 

46. Area as a Double Integral. — Divide the area be- 
tween two curves y = f {x), y = F (x) into strips of width 
Ax. Let P be the point {x, y) and Q the point (x + Ax, y + 
Ay). The area of the rectangle PQ is Ax Ay. The area 
of the rectangle RS (Fig. 46a) is 



Ax V Ay = Ax / dy. 

^f(z) J fix) ^ 



97 



98 Double Integration Chap. 8 

The area bounded by the ordinates x = a, x = h is then. 

Sb nF(x) nb PF(x) 

Ax j dy = j j dx dy. 
^-y, a Jf(x) J a JS(.i) 

If it is simpler to cut the area into strips parallel to the 
a;-axis, the area is 



= J I dy dx, 



i 



the limits in the first integration being the values of x at the 
ends of a variable strip; those in the second integration, the 
values of y giving the limiting strips. 

Example. Find the area bounded by the parabola y"^ = 
4 ox + 4 a^ and the straight line y = 2a — x (Fig. 466). 




Fig. 46a. 



Fig. 46&. 



Solving simultaneously, we find that the parabola and the 
line intersect at A (0, 2 a) and B (Sa, —6a). Draw the 
strips parallel to the rc-axis. The area is 

y2 



/2a f2a-y po / 

/ dydx= I 2( 



4a'\ , 64 , 



The limits in the first integration are the values of x at i? 
and S, the ends of the variable strip. The limits in the 
second integration are the values oi y at B and A, corre- 
sponding to the outside strips. 



Art 47 



Volume by Double Integratiox 



99 



47. Volume by Double Integration. — To find the 
volume under a surface z = / (x, y) and over a given region 
in the xy-plane. 

The volume of the prism TQ standing on the base Ax Ay 
(Fig. 47a) is 

z Ax Ay. 

The volume of the plate RT is then 

Ax Ay = Ax / 2 dy, 

a»=u R J fix) 

/ (x), F (x) being the values of y at R, S. The entire volume 
is the limit of the sum of such plates 

Ax / zdy = I I zdxdy, 

ui=u a Jf(.x) J a Jf(x) 

a, b being the values of x corresponding to the outside plates. 
Example. Find the volume bounded by the surface 
az = a^ — x^ — 4 y2 and the xy-plane. 
z 




Fig. 47a. 



Fig. 476. 



Fig. 476 shows one-fourth of the required volume, 
y = 0. At iS, 2 = and so 

y = ^ Vo« - x2. 



Ati2, 



100 



Double Integration 



Chap. 8 



The limiting values of x at and A are and a. Therefore 
v = ^\ I zdxdy = il / -id'-x''-4y^)dxdy 

3 a Jo 4 



48. The Double Integral as the Limit of a Double 
Summation. — Divide a plane area by lines parallel to the 
coordinate axes into rectangles with sides Ax and Ay. Let 
(a:, y) be any point within one of these rectangles. Form 
the product 

/ {x, y) Ax Ay. 

This product is equal to the volume of the prism standing 
on the rectangle as base and reaching the surface z = f {x, y) 
at some point over the base. Take the sum of such products 




Fig. 48a. 



for all the rectangles that lie entirely within the area. We 
represent this sum by the notation 

2) ^J{x,y)AxAy. 

When Ax and Ay are taken smaller and smaller, this sum 
approaches as limit the double integral 



// 



/ {x, y) dx dy, 



Art 48 



The Limit of a Double Summation 



lot 



with the limits determined by the given area; for it approaches 
the volume over the area and that volume is equal to the 
double integral. 
Whenever then a quantity is a limit of a sum of the form. 

%^f(x,y)AxAy 



its value can be found by double integration. Furthermore, 
in the formation of this sum, infinitesimals of higher order 
than Ax Ay can be neglected without 
changing the Umit. For, if e Az A?/ 
is such an infinitesimal, the sum of 
the errors thus made is 

%%€^x Ay. 

When Ax and Ay approach zero, e 
approaches zero. The sum of the 
errors approaches zero, since it is 
represented by a volume whose thick- 
ness approaches zero. 

Example 1. An area is bounded 
by the parabola y' = 4 ax and the 
line X = a. Find its moment of 
inertia about the axis perpendicular to its plane at the 
origin. 

Divide the area into rectangles Ax Ay. The distance of 
any point P (x, y) fro m the ax is perpendicular to the plane 
at is i2 = OP = Vx- + y^. If then (x, y) is a point 
within one of the rectangles, the moment of inertia of that 
rectangle is 

Rr ^x \y = (x^ -\- y«) Ax Ay, 

approximately. That the result is approximate and not 
exact is due to the fact that different points in the rectangle 
differ slightly in distance from the axis. This difference is. 




Fig. 486. 



102 



Double Integkation 



Chap. 8 



however, infinitesimal and, since R^ is multiplied by Ax Ay, 
the resulting error is of higher order than Ax A?/. Hence in 
the limit 



J-2\^z 




344 

(x^ -\- y^) dx dy = j^a*- 

Ex. 2. Find the center of grav- 
ity of the area bounded by the 
parabolas y^ = 4:X -j- 4:,y^ = — 2x 
+ 4. 

By symmetry the center of 
gravity is seen to be on the 
X-axis. Its abscissa is 



- / 



xdA 



X = 



If we wish to use double inte- 
gration we have merely to replace 
dA by dx dy or dy dx. From the 
figure it is seen that the first 
integration should be with respect to x. Hence 



Fig. 48c. 



X2 /'i(4-»2) 
I xdy dx 
_ 2 Ji(yi-4) 

11 dydx 

«/-2 t/ J (1^2-4) 



16 

8 



EXERCISES 
Find the values of the following double integrals: 
dxdy 



^' J, Ji {x + yy 

Ja 



rdedr. 



-i /•1V3 



3. I I xydxdy. 



J -2 IT /•'« 
I e-kr^rdddr. 
•'0 

6. C C {x^ + y^)dydx. 

Jo Jy 

I dy dx. 

*'0 



Art. 49 Double Integratiox. Polar Coordinates 103 

7. Find the area bounded by the parabola y^ = 1x and the line 
X = y. 

8. Find the area bounded by the parabola j/* = 4 ax, the line 
X + y = 3 a, and the x-axis. 

9. Find the area enclosed by the ellipse 

{y - xT- + x^ = l. 

j 1 10. Find the volume under the paraboloid z = -i — x^ — y^ and over 
the square bounded by the lines x = ±l,i/ = ±lin the xy-plane. 

11. Find the volume bounded by the x(/-plane, the cyhnder x- + y^ 
= 1, and the plane x + y + z = 3. 

12. Find the volume in the first octant bounded by the cylinder 
(x — 1)- + (i/ — 1)^ = 1 and the paraboloid xy = z. 

13. Find the moment of inertia of the triangle bounded by the 
coordinate axes and the line x + y = 1 about the line perpendicular to 
its plane at the origin. 

14. Find the moment of inertia of a square of side a about the axis 
perpendicular to its plane at one comer. 

15. Find the moment of inertia of the triangle bounded by the lines 
x + y = 2, X = 2, y = 2 about the x-axis. 

16. Find the moment of inertia of the area bounded by the parab- 
ola y- = ax and the line x = a about the line y = — a. 

17. Find the moment of inertia of the area bounded by the hyper- 
bola xy = 4 and the line x -{- y = 5 about the line y = x. 

18. Find the moment of inertia of a cube about an edge. 

19. A wedge is cut from a cylinder by a plane passing through a 
diameter of the base and inclined 45° to the base. Find its moment of 
inertia about the axis of the cylinder. 

20. Find the center of gravity of the triangle formed by the lines 
x = y, x + y = 4, X — 2y = 4:. 

21. Find the center of gravity of the area bounded by the parabola 
J/* = 4 ax -f 4 a* and the line y = 2a — x. 

49. Double Integration. Polar Coordinates. — Pass 

through the origin a series of lines making with each other 
equal angles Ad. Construct a series of circles with centers 
"at the origin and radii differing by Ar. The hues and circles 
di\ide the plane into curved quadrilaterals (Fig. 49a). 

Let r, 6 be the coordinates of P, r + Ar, 4- A0 those of 
Q. Since PR is the arc of a circle of radius r and subtends 
the angle Ad at the center, PR = r Ad. Also RQ = Ar. 



104 Double Integration Chap. 8 

When Ar and \d are very small PRQ will be approximately 
a rectangle with area 

PR'RQ = rAd Ar. 




Fig. 49a. 

It is very easy to show that the error is an infinitesimal of 
higher order than A^ Ar. (See Ex. 5, page 107.) Hence 
the sum 



^^rAeAr, 



taken for all the rectangles within a curve, gives in the limit 
the area of the curve in the form 



A 



= j jrdddr. (49a) 



The limits in the first integration are the values of r at 
the ends A, B of the strip across the area. The limits in the 
second integration are the values of d giving the outside 
strips. 

If it is more convenient the first integration may be with 
respect to 6. The area is then 



= // 



r dr dd. 



Art. 49 



Double Integration. Polar Coordinates 



105 



The first limits are the values of 6 at the ends of a strip 
between two concentric circles (Fig. 496). The second hmite 
are the extreme values of r. 




Fig. 496. 



Fig. 49c. 



The element of area in polar coordinates is 
dA = rdddr. 



(49b) 



We can use this in place of dA in finding moments of inertia, 
volumes, centers of gravity, or any other quantities expressed 
by integrals of the form 



/ 



fir,d)dA. 



Example 1. Change the double integral 

{x'^ + y^)dxdy 



Jo Jo 



to polar coordinates. 

The integral is taken over the area of the semicircle 
y = V2 ax — x^ (Fig. 49c). In polar coordinates the 
equation of this circle is r = 2 a cos 6. The element of area 



106 Double Integration Chap. 8 

dx dy can be replaced by r cW dr.* Also x^ -\- y^ = r^. Hence 

J^2a n V2ai-i2 ni /»2 COS 9 

/ (x^ -\- y'^) dx dy = \ I r^ • r dd dr. 

Jo Jo J 

The limits for r are the ends of the sector OP. The limits 

for 6 give the extreme sectors 6 = 0, 9 = -• 

Ex. 2. Find the moment of in- 
ertia of the area of the cardioid 
r = a (1 + cos 6) about the axis 
-X perpendicular to its plane at the 
origin. 

The distance from any point P 
(r, d) (Fig. 49d) to the axis of rota- 
tion is 

OP = r. 




Fjg. 49d. 
Hence the moment of inertia is 



na(l + cosO) „4 /V OK 

r^-rdedr= - (I -\- cosdYdd = ^ira*. 
iJo 10 

Ex. 3. Find the center of gravity of the cardioid in the 
preceding problem. 

The ordinate of the center of gravity is evidently zero. 
Its abscissa is 

xdA 2 / / r cosd ' rdddr 

Jo Jo 



X = 



J'm 



If 



rdddr 



5 



Ex. 4. Find the volume common to a sphere of radius] 
2 a and a cylinder of radius a, the center of the sphere being] 
on the surface of the cylinder. 

* This does not mean that 

dx dy = rdd dr, 

but merely that the sum of all the rectangular elements in the circle is 
equal to the sum of all the polar elements. 



Art. 49 



Double Integration. Polar Coordinates 



107' 



Fig. 49e shows one-fourth of the required volume. Take 
a system of polar coordinates in the xy-plane. On the 
element of area r dd dr stands a prism of height 
z= V4 a2 - r', 




Fig. 49e. 
The volume of the prism is z-rdddr and the entire volume is 



2acoe9 



dB 



I V^a'-r^.rdddr = 4:l ^ 

X 

= ^ r (1 - sin'^) de = ^a' (3t - 4). 



EXERCISES 

Find the values of the following integrals by changing to polar 
coordinates: 

(i2 + y2)dy(ir. 3. J J e-^'*+*''> dxdy. 

2. J^ J^ dxdy. ^ J J ^d^-x'-y^dxdy. 

6. Find the area bounded by two circles of radii a, a + £i.a and two 
lines through the origin, making with the initial line the angles a, 
■ j a + ^a, respectively. Show that when Aa and Aa approach zero, the 
' result differs from 

a Act Aa 

by an infinitesimal of higher order than Aa Ao. 



108 Double Integration Chap. 8 

6. The central angle of a circular sector is 2 a, Find the moment 
of inertia of its area about the bisector of the angle. 

7. An area is bounded by the circle r = a V2 and the straight 

line r = asec id — j] . Find its moment of inertia about the axis per- 
pendicular to its plane at the origin. 

8. Find tha center of gravity of the area in Ex. 6. 

9. The center of a circle of radius 2 a lies on a circle of radius a. 
Find the moment of inertia of the area between them about the 
common tangent. 

10. Find the moment of inertia of the area of the lemniscate r^ = 
2 a^ cos 2 about the axis perpendicular to its plane at the origin. 

11. Find the moment of inertia of the area of the circle r = 2 a 

outside the parabola, r = a sec* ^ about the axis perpendicular to its 
plane at the origin. 

12. Find the moment of inertia about the y-axis of the area within 
the cu-cle {x - aY + (y - o)* = 2 a?. 

13. The density of a square lamina is proportional to the distance 
from one corner. Find its moment of inertia about an edge passing 
through that corner. 

14. Find the moment of inertia of a cylinder about a generator. 

15. Find the moment of inertia of a cone about its axis. 

16. Find the volume under the spherical surface x^ -{- y^ ■\- z^ = a? 
and over the lemniscate r* = a^ cos 2 5 in the xy-plane. 

17. Find the volume bounded by the xy-plane, the paraboloid 
az = X* + y'^ and the cylinder x* + y* = 2 ax. 

18. Find the moment of inertia of a sphere of density p about a 
diameter. 

19. Find the volume generated by revolving one loop of the curve 1 
r = a cos 2 d about the initial line. 

50. Area of a Surface. — Let an area A in one plane 
projected upon another plane. The area of the projection 

A' = A cos 0, 

when <f) is the angle between the planes. 

To show this divide A into rectangles by two sets of lines 
respectively parallel and perpendicular to the intersection 
MN of the two planes. Let a and h be the sides of one of 



Art. 60 Area of a Sttrface 109 

these rectangles, a being parallel to MN. The projection of 
this rectangle will be a rectangle with sides 

a' = a, h' — h cos ^, 
and area 

a'h' = ab cos <{>. 

The sum of the projections of all the rectangles is 

^a'b' = ^ab cos 4>. 

As the rectangles are taken smaller and smaller this 
approaches as limit 

A' = A cos <t>, 

which was to be proved. 




Fig. 50a. 

To find the area of a curved surface, resolve it into elements 
whose projections on a coordinate plane are equal to the 
differential of area dA in that plane. The element of surface 
can be considered as lying approximately in a tangent plane. 
Its area is, therefore, approximately 

dA 

cos<f> 

where is the angle between the tangent plane and the 

coordinate plane on which the area is projected. The area 

of the surface is the limit 

J cos </) 



110 



Double Integration 



Chap. 8 



The angle between two planes is equal to that between 
the perpendiculars to the planes. Therefore ()> is equal to 
z 




Fig. 50&. 

the angle between the normal to the surface and the co- 
ordinate axis perpendicular to the plane on which we project. 
If the equation of the surface is 

F (x, y, z) = 0, 

the cosine of the angle between its normal and the 2-axis is 
(Differential Calculus, Art. 101) 

dF 

dz 



cos 7 = 



v/fj 



dxj'^\dy)'^[dzj 



py 



The cosines of the angles between the normal and the x-axis 

or w-axis are obtained by replacing ^r- by 7— or t— . In 

dz "^ dx dy 

finding areas the algebraic sign is assumed to be positive. 

Example 1. Find the area of the sphere x^ + y^ -{- z^ = a^ 
within the cylinder x^ -}- y^ = ax. 

Project on the x^z-plane. The angle <f) is then the angle 7 

between the normal to the sphere and the 2!-axis. Its cosine is 

z z 

cos 7 = 



Vx2 + 2/2 + 22 a 



J 



Art. 60 Area of a Surface 111 

Using polar coordinates in the xj/-plane, 

z = Va- — x^ — if = Va2 — j^. 
Hence the area of the surface is 

S = fj" = 4 r r°-^£iL = 2aHx - 2). 
J COS 7 Jo Jo V a^ — r^ 

Ex. 2. Find the area of the surface of the cone y^ -h 
z^ = x2 in the first octant bounded by the plane y -\- z = a. 
Project on the yz-plane. Then <f> = a and 

X X 1_ 

Vx2 +1/2+22 ~ V2x2 ~ V2 

The area on the cone is therefore 



S= r r " V2dydz = 
Jo Jo 




EXERCISES 

1. Find the area of the triangle cut from the plane 

x + 2y + 3z = 6 

by the coordinate planes. 

2. Find the area of the surface of the cylinder x- -\- y- = a* between 
the planes z = 0, 2 = mx. 

3. Find the area of the surface of the cone x- -\- y- = z* cut out by 
the cylinder i^ + t/^ = 2 ax. 

4. Find the area of the plane x + y + z = 2a in the first octant 
bounded by the cyUnder x^ + y^ = a*. 

6. Find the area of the surface ^ = 2xy above the zy-plane bounded 
by the planes j/ = 1, x = 2. 

6. Find the area of the surface of the cylinder x^ + y^ = 2ax 
between the ly-plane and the cone x- -\- y- = z-. 

7. Find the area of the surface of the paraboloid y- -\- z- = 2 ox, 
intercepted by the parabolic cyUnder rf- = ax and the plane x = o. 

8. Find the area intercepted on the cylinder in Ex. 4. 

9. A square hole of side a is cut through a sphere of radius a. If 
the axis of the hole is a diameter of the sphere, find the area of the 
surface cut out. 



CHAPTER IX 



TRIPLE INTEGRATION 



51. Triple Integrals. — The notation 

I I fix, y, z) dx dy dz 



is used to represent the result of integrating first with respect 
to z (leaving x and y constant) between the limits Zi and Za, 
then with respect to y (leaving x constant) between the 
limits 2/1 and y^, and finally with respect to x between the 
limits Xi and Xg- 




Fig. 52a. 

52. Rectangular Coordinates. — Divide a solid into 
rectangular parallelepipeds of volume Ax Ai/ A2 by planes 
parallel to the coordinate planes. To find the volume of 

112 



'Art. 62 Rectangitlar Coobdinates 113 

the solid, first take the sum of the parallelepipeds in a vertical 
column PQ. The result is 



V Ax \y A2 = Ax Ay / dz, 



Zi and Zt being the values of z at the ends of the colunm. 
Then sum these columns along a base MN and so obtain the 
volume of the plate MNR. The result is 

Um V Ax Ay / dz = \x J I dy dz, 

t/i and ya being the Umiting values of y in the plate. Finally, 
take the sum of these plates. The result is the triple integral 

V = Um X -^"^ / / dy dz = I J I dxdydz, 

Xi, X2 being the limiting values of x. 

It may be more convenient to begin by integrating with 
respect to x or y. In any case the limits can be obtained 
from the consideration that the first integration is a summa- 
tion of parallelepipeds to form a prism, the second a summa- 
tion of prisms to form a plate, and the third integration a 
summation of plates. 

Let (x, y, 2) be any point of the parallelepiped Ax Ay \z. 
Multiply Ax Ay Az by / (x, y, z) and form the sum 

2 2 2-^ ^^' ^' ^^ ^* ^^ ^^ 

taken for all parallelepipeds in the soUd. When Ax, Ay, and 
Az approach zero, this sum approaches the triple integral 



III' 



f (x, y, 2) dx dy dz 

as limit. It can be shown that terms of higher order than 
Ax Ay Az can be neglected in the sum without changing 
the limit. 



114 



Triple Integration 



Chap. 9 

The differential of volume in rectangular coordinates is 

dv = dx dy dz. 

This can be used in the formulas for moment of inertia, center 
of gravity, etc., those quantities being then determined by 
triple integration. 
Example 1. Find the volume of the ellipsoid 

c^ y^ & 

Fig. 52a shows one-eighth of the required volume. Therefore 
V = S I I I dxdydz. 

The limits in the first int egration are the values z = at P 
and z = cyl i^fl ^^ ^- '^^^ Umits in the second 

integration are the values of y a t M a nd N. At M, y = 

I x^ 
and at iV, 2 = 0, whence y = hy \ ^. Finally, the limits 

for X are and a. Therefore 



V = S I I I dx dy dz = ^ irdbc. 




Fig. 52&. 



Ex. 2. Find the center of gravity of the solid bounded 
by the paraboloid y^ -j- 2 z^ = 4 x and the plane x = 2. 



Art. 62 Rectangular Coordinates 115 

By symmetry y and z are zero. The x-coordinate is 
xdv 4 1/ / xdzdydx . 

Jo Jo Jiiyi+2^) _4 



X = 



Jd. 



111'^'^ 



dx 



The limits for x are the values x = j (y^ + 2 2*) at P and 
X = 2 at Q. At S, X = 2 and y = V4 x - 2 2^ = V8 - 2 2*. 
The Umits for y are, therefore, y = at ^ and y = 
VS - 2 2^ at S. The limits for 2 are 2 = at'A and 2 = 2 
atB. 

Ex. 3. Find the moment of inertia of a cube about an 
edge. 




Fig. 52c. 

Place the cube as shown in Fig. 52c and determine its 
moment of inertia about the 2-axis. The distance of any 
point (x, y, z) from the 2-axis is 

R = Vx2 + i/2. 

Hence the moment of inertia is 

I / (3^ + y')dxdydz=la\ 

»/o «/0 

where a is the edge of the cube. 



1 



116 



Triple Integration 



Chao. 9 



EXERCISES 

1. Find by triple integration the volume of the pyramid determined 
by the coordinate planes and the plane x + y -\- z = 1. 

2. Find the moment of inertia of the pyramid in Ex. 1 about the 
a>axis. 

3. A wedge is cut from a cylinder of radius a by a plane passing 
through a diameter of the base and inclined 45° to the base. Find its 
center of gravity. 

4. Find the volume bounded by the paraboloid f; + -; = 2 - and 

b^ c' a 

the plane x = a. 

6. Express the volume of the cone 

(z - 1)2 = a;2 + 2/2 

in the first octant as a triple integral in 6 ways by integrating with dx, 
dy, dz, arranged in all possible orders. 

6. Find the volume bounded by the surfaces j/^ = 4 a" — 3 ax, y^ = 
ax,z = ±/i. 

7. Find the volume bounded by the cylinder ^ = \ — x — y and 
the coordinate planes. 

53. Cylindrical Coordinates. — Let M be the projection 
of P on the a;?/-plane. Let r, Q be the polar coordinates of M 

in the xy-plane. The cyhndrical 
coordinates of P are r, Q, z. 

From Fig. 53a it is evident that 
x = r cos Q, y = T sin d. 

By using these equations we can 
change any rectangular into a 
cylindrical equation. 

The element of volume in cy- 
lindrical coordinates is the volume 
PQ, Fig. 536, bounded by two 
cylindrical surfaces of radii r, 
r + Ar, two horizontal planes z, z -\- Az, and two planes 
through the 2-axis making angles ^, ^ + A^ witli OX. The 
base of PQ is equal to the polar element MN in the xy~ 
plane. Its altitude PR is Az. Hence 

dv = rdd dr dz. (53) 




Fig. 53a. 



Art. 63 



Cttjxdrical Coordixates 



117 



This value of dv can be used in the formulas for volume, 
center of gravity, moment of inertia, etc. In problems con- 

z 




Fig. 536. 




Fig. 53c. 



lected with cylinders, cones, and spheres, the resulting 
itegrations are usually much easier in cylindrical than in 
jtangular coordinates. 



118 



Triple Integration 



Chap. 9 



Example 1. Find the moment of inertia of a cylinder 
about a diameter of its base. 

Let the moment of inertia be taken about the x-axis, 
Fig. 53c. The square of the distance from the element PQ 

to the X-axis is 

i^a = ^2 _|_ 22 = ^2 sin2 4- z\ 

The moment of inertia is therefore 

R^dv= I I I {r^smH-\-z^)rdddzdr 

Jo Jo Jo 



IT 



(3a2+4/i2). 



The first integration is a summation for elements in the 
wedge RS, the second a summation for wedges in the slice 
OMN, the third a summation for all such slices. 

Ex. 2. Find the volume bounded by the xy-plane, the 
cylinder x^ -\- y^ = ax, and the sphere x^ -{- y^ -\- z^ = c?. 

z 




Fig. 53d. 

In cylindrical coordinates, the equations of the cylinder 
and sphere are r = a cos B and r^ -\- z^ = a^. The volume j 
required is therefore 

v 

P2 r a cose f*^ai>-f 

v = 2 I I I rd9drdz= ^a^Zir -^). 

Jo Jo Jo 



Art. 64 



Sphekical Coordinates 



119 



54. Spherical Coordinates. — The spherical coordinates 
of the point P (Fig. 54a) are r = OP and the two angles 6 and 
<i>. From the diagram it is 
easily seen that 

a; = r sin cos 6, 
i/ = r sin sin Q, 
z = r cos (j>. 

The locus r = const, is a 
sphere with center at 0; 6 = 
const, is the plane through OZ 
making the angle 6 with OX; 
<f> = const, is the cone gener- 
ated by lines through making the angle ^ with OZ. 

The element of volume is the volume PQRS bounded 




Fig. 54a. 




Fig. 545. 

by the spheres r,r -\- Ar, the planes 6,9 + Ad, and the cones 
</>, + A0. When Ar, A^, and A0 are very small this is 



120 Triple Integration Chap. 9 

approximately a rectangular parallelepiped. Since OP = r 
and POR = A<^, 

PR = r A</). 

Also OM = OP sin <^ and the arc PS is approximately equal 
to its projection MN, whence 

PS = MN = rsm(j> Ad, 

approximately. Consequently 

Av = PR'PS'PQ = r" sin (/> A^ • A0 • Ar, 

approximately. When the increments are taken smaller 

and smaller, the result becomes more and more accurate. 

Therefore 

dv = r^ sin (j>ddd<j) dr. (54) 

Spherical coordinates work best in problems connected 
with spheres. They are also very useful in problems where 
the distance from a fixed point plays an important role. 




Fig. 54c. 

Example. If the density of a solid hemisphere varies as 
the distance from the center, find its center of gravity. 

Take the center of the sphere as origin and let the 2-axis 
be perpendicular to the plane face of the hemisphere. By 
symmetry it is evident that x and y are zero. The density 



Art. 55 Attraction 121 

is p = kr, where A; is constant. Also z = r cos <p. Hence 



z = 



fzdrn^f_ 



krzdv 



rr r2 Pa 



r* cos <t> sin <l>dd d<t) dr 



Jr»2T /»2 na 
»/0 t/0 



-5"- 



r^ sin <t> dd d<t> dr 



EXERCISES 

1. Find the volume bounded by the sphere a? + y* + 2^ = 4 and 
the paraboloid x* + y* = 3 z. 

2. A right cone is scooped out of a right cylinder of the same height 
and base. Find the distance of the center of gravity of the remainder 
from the vertex. 

3. Find the volume bounded by the surface z = e— (^H-i*) and the 
xy-plane. 

4. Find the moment of inertia of a cone about a diameter of its base. 

5. Find the volvune of the cylinder i* + i/^ = 2 ax intercepted 
between the paraboloid x^ + y^ = 2az and the xy-plane. 

6. Find the center of gravity of the volume common to a sphere of 
radius a and a cone of vertical angle 2 a, the vertex of the cone being at 
the center of the sphere. 

7. Find the center of gravity of the volume boimded by a spherical 
surface of radius a and two planes passing through its center and in- 
cluding an angle of 60°. 

8. The vertex of a cone of vertical angle - is on the surface of a 

sphere of radius a. If the axis of the cone is a diameter of the sphere, 
find the moment of inertia of the volume common to the cone and 
sphere about this axis. 

55. Attraction. — Two particles of masses mi, mz, sepa- 
rated by a distance r, attract each other with a force 

kmiiiv. 




122 v Triple Integration Chap. 9 

where fc is a constant depending on the units of mass, dis- 
tance, and force used. A similar law expresses the attraction 
or repulsion between electric charges. 

To find the attraction due to a continuous mass, resolve 
it into elements. Each of these attracts with a force given 
by the above law. Since the 
forces do not all act in the 
same direction we cannot ob- 
tain the total attraction by 
merely adding the magnitudes 
of the forces due to the sev- ^^°- ^^"• 

eral elements. The forces must be added geometrically. 
For this purpose we calculate the sum of the components 
along each coordinate axis. The force having these sums 
as components is the resultant attraction. 

If dm is the mass of an element at P, r its distance from 0, 
and 6 the angle between OX and OP, the attraction between 
this element and a unit particle at is 
, 1 • dm _ k dm 

This force acts along OP. Its component along OX is 
cos 6 • k dm 

The component along OX of the total attraction is then 

'*k cos 6 dm 



= / 



The calculation of this integral may involve single, double, 
or triple integration, depending on the form of the attracting 
mass. 

Example 1. Find the attraction of a uniform wire of 
length 2 I, and mass M on a particle of unit mass at distance | 
c along the perpendicular at the center of the wire. 

Take the origin at the unit particle and the a:-axis perpen- 
dicular to the wire. Since particles below OX attract down- 



Art. 66 



Attraction 



123 



ward just as much as those above OX attract upward, the 
vertical component of the total attraction is zero. The 
component along OX is, therefore, the total attraction. 
The mass of the length dy of the wire is 

Mdy 



Hence 



X = 



21 

kM ^cos 6 dy 
21 



r- 



For simplicity of integration it is better to use 6 as variable. 
Then y = c tan 6, dy = c sec^ 6 dd, and 




X 



~ 21 J-„ 



Fig. 556. 



cos • c sec2 d dd kM 



where a is the angle XOA 
X = 



& sec^ Q cl 

In terms of I this is 
kM 



sma, 



Ex. 2. Find the attraction of a homogeneous cylinder of 
mass M upon a particle of unit mass on the axis at distance c 
from the end of the cyUnder. 

By symmetry it is clear that the total attraction will act 
along the axis of the cyUnder. Take the origin at the attract- 
ing particle and let the y-ends be the axis of the cyhnder. 



124 



Triple Integration 



Chap. 9 



Divide the cylinder into rings generated by rotating the 
elements dx dy about the y-axis. The volume of such a 
ring is 

2 irx dx dy 

and its mass is 

M 2M 

dm = — 57 '2-Kxdxdy = —zj-xdx dy. 




Since all points of this ring are at the same distance from O 
and the joining Unes make the same angle 6 with OY, the 
vertical component of attraction is 

, fcosOJm _ , rydm _ 2 Mk A+* A xydydx 
J r" ~ J H ~ a^h Jc Jo (r^+y2)f 

= ^ Ih + V^T^ - Va'-\-(c-hh)q . 
aril 



Art. 56 Attraction 125 

EXERCISES 

1. Find the attraction of a imiform wire of mass M and length I on 
a particle of unit mass situated in the line of the wire at distance c from 
its end. 

2. Find the attraction of a wire of mass M bent in the form of a 
semicircle of radius a on a unit particle at its center. 

3. Find the attraction of a flat disk of mass M and radius a on a 
unit particle at the distance c in the perpendicular at the center of the 
disk. 

4. Find the attraction of a homogeneoxis cone upon a unit particle 
situated at its vertex. 

6. Show that, if a sphere is concentrated at its center, its attraction 
upon an outside particle will not be changed. 

6. Find the attraction of a homogeneous cube upon a particle at 
one comer. 



CHAPTER X 
DIFFERENTIAL EQUATIONS 

56. Definitions. — A differential equation is an equation 
containing differentials or derivatives. Thus 

(x^ + y^) dx -\-2xydy = 0, 

^^_dy^ 2 
dx"^ dx 

are differential equations. 

A solution of a differential equation is an equation connect- 
ing the variables such that the derivatives or differentials 
calculated from that equation satisfy the differential equa- 
tion. Thus y = x^ — 2 xisa solution of the second equation 
above; for when a;^ — 2 a; is substituted for y the equation is 
satisfied. 

A differential equation containing only a single independent 
variable, and so containing only total derivatives, is called 
an ordinary differential equation. An equation containing 
partial derivatives is called a partial differential equation. 
We shall consider only ordinary differential equations in this 
book. 

The order of a differential equation is the order of the 
highest derivative occurring in it. 

57. Illustrations of Differential Equations. — Whenever 
an equation connecting derivatives or differentials is known, 
the equation connecting the variables can be determined by 
solving the differential equation. A number of simple cases 
were treated in Chapter I. 

The fundamental problem of integral calculus is to find 
the function 



= jf{x)dx, 



y 

126 



Arts? 



Illustrations of Differential EkjUATioNS 



127 



when / (x) is given. This is equivalent to solving the differ- 
ential equation 

dy = f (x) dx. 

Often the slope of a curve is known as a function of x and y. 

The equation of the curve can be found by solving the 
differential equation. 

In mechanical problems the velocity or acceleration of a 
particle may be known in terms of the distance s the particle 
has moved and the time t. 



ds 
dt='^ 



df 



= a. 



The position s can be determined as a function of the time by 
solving the differential equation. 

In physical or chemical problems the rates of change of 
the variables may be known as functions of the variables, 
and the time. The values of those variables at any time can 
be found by solving the differential equations. 

Example. Find the curve in which the cable of a suspension 
bridge hangs. 




Let the bridge be the x-axis and let the i/-axis pass through 
the center of the cable. The portion of the cable AP is in 



I 

128 Differential Equations Chap. 10 

equilibrium under three forces, a horizontal tension fl" at A, 
a tension FT in the direction of the cable at P, and the 
weight of the portion of the bridge between A and P. The 
weight of the cable, being very small in comparison with that 
of the bridge, is neglected. 

The weight of the part of the bridge between A and P is 
proportional to x. Let it be Kx. Since the vertical com- 
ponents of force must be in equilibrium 
r sin = Kx. 

Similarly, from the equilibrium of horizontal components, 
we have 

Tcos<l> = H. 
Dividing the former equation by this, we get 



tan <f> = y^x. 

£1 



But tan <j) = -^ . Hence 

dx H ' 
The solution of this equation is 

The curve is therefore a parabola. 

58. Constants of Integration. Particular and General 
Solutions. — To solve the equation 

we integrate once and so obtain an equation with one arbi- 
trary constant, 



y = jf(x)dx + c. 



Art. 58 Constants of Integration 129 

To solve the equation 

we integrate twice. The result 

y = j jf{x)dx^+ CiX + Cj 

contains two arbitrary constants. Similarly, the integral of 
the equation 

contains n arbitrary constants. 

These illustrations belong to a special type. The rule 
indicated is, however, general. The complete, or general, 
solution of a differential equation of the nth order in two vari- 
ables contains n arbitrary constants. If particular values are 
assigned to any or all of these constants, the result is still a 
solution. Such a solution is called a particular solution. 

In most problems leading to differential equations the 
result desired is a particular solution. To find this we 
usually find the general solution and then determine the 
constants from some extra information contained in the 
statement of the problem. 

Example L Show that 

aJf-f_y2_ 2cx = 

is the general solution of the differential equation 
y^-x^-2xy^ = 0. 
Differentiating x^ -\- y- — 2cx = 0, we get 

whence 

dy _ c — X 
dx u 



130 Differential Equations Chap. 10 

Substituting this value in the differential equation, it becomes 

yi — x^ — 2xy -^ = y~ — x^ — 2x{c — x) = if -]-x^ — 2cx = 0. 

Hence x^ -{- y^ — 2 ex = Q \s, a solution. Since it contains 
one constant and the differential equation is one of the first, 
order, it is the general solution. 

Ex. 2. Find the differential equation of which y = Cie"" + 
C2e^^ is the general solution. 

Since the given equation contains two constants, the 
differential equation is one of the second order. We there- 
fore differentiate twice and so obtain 







d'y 
dx" 


= C\e- 


' + 4 cae^*. 


EUminating 


Cl, 


we get 










d:'y 

dx" 


dy _ 
dx 


= 2c2e2^, 






dy _ 
dx 


■y = 


de"'. 



Hence 

diy_dy^^/dy_\ 
dx^ dx \dx / 
or 

p^-3^ + 2y=0. 
ax- ax 

This is an equation of the second order having y = Cie" -\- 
C2fi^' as solution. It is the differential equation required. 

EXERCISES 

In each of the following exercises, show that the equation given is a 
solution of the differential equation and state whether it is the general 
or a particular solution. 



Art. 59 The Fikst Obder ix Two Variables 131 

2.^-yS=cx, ix' + y')dx-2xydy = 0. 

3^ 1/ = 06* sin X, g_2| + 2y = 0. 

4. y =ci+ctsin(x+c), ^ + df "^ ®- 

Find the differential equation of which each of the following equa* 
tions is the general solution: 

_ cj 7. y = Ci sin X + ci coe z. 

5. y-cix + -' ^ x*y = ci+c4hxz + cvr». 

6. y = cxe'. 9. i^ + Cixy + Cjy* = 0. 

59. Differential Equations of the First Order in Two 

dv 
Variables. — By solving for -p an equation of the first 

order in two variables x and y can be reduced to the form 

To solve this equation is equivalent to finding the curves 
with slope equal to f(x,y). The solution contains one 




Fig. 59. 

arbitrarj' constant. There is consequently an infinite num- 
ber of such curves, usually one through each point of the 
plane. 

We cannot always solve even this simple tj'pe of equa- 
tion. In the follo\sing articles some cases will be discussed 



132 Differential Equations Chap. 10 

which frequently occur and for which general methods of 
solution are known. 

60. Variables Separable. — A differential equation of 
the form 

Mdx + Ndy = 

is called separable if each of the coefficients M and N «;on- 
tains only one of the variables or is the product of a function 
of X and a function of y. By division the x's and dx can be 
brought together in the first term, the y's and dy in the 
second. The two terms can then be integrated separately 
and the sum of the integrals equated to a constant. 

Example 1. (I -^ x^) dy — xy dx = 0. 

Dividing by (1 + x^) y, this becomes 

dy _ xdx 
y ~ 1 +a;2' 
whence 

ln^ = iln(l+a;2)+c. 

If c = In k, this is equivalent to 

\ny = lnVT+^+ln/c = lnfc VT+i", 
and so 

y = k Vl + x^, 

where k is an arbitrary constant. 

Ex. 2. Find the curve in which the area bounded by thej 
curve, coordinate axes, and a variable ordinate is proportions 
to the arc forming part of the boundary 

Let A be the area and s the length of arc. Then 

A = ks. 
Differentiating with respect to x, 

dA _ yds 
dx dx' 



or 



''-v/'+dj 



Art. 61 



Exact Differential Equations 



[13a 



Solving for ■—, 



dy_ Vy^ - k^ 
dx~ k ' 



whence 



dy 



dx 



Vt/2 - fc2 k' 



The solution of this is 



In (2/ + Vi/2 _ A-2) =^ + c. 



-+e 



Therefore 

y + V?/2 — ^-2 = e* = e*e* = Cie* , 
where Ci is a new constant. Transposing y and squaring, 
we get 



Hence, finally, 



y' 



k" = [cie^j - 2 cie*2/ + y^. 



k^ -I 



y = 2' -^27.' 



k. 



61. Exact Differential Equations. — An equation 

du — 0, 

obtained by equating to zero the total differential of a func- 
tion u of X and y, is called an exact differential equation. 
The solution of such an equation is 

u = c. 

The condition that M dx -\- N dy be an exact differential 
is (Diff. Cal., Art. 100) 

dy dx 
This equation, therefore, expresses the condition that 

M dx + N dy = 
be an exact differential equation. 



(61) 



134 Differential Equations Chap. 10 

An exact equation can often be solved by inspection. To 
find u it is merely necessary to obtain a function whose total 
differential is M dx -\- N dy. 

If this cannot be found by inspection, it can be determined 
from the fact that 

du = M dx -{-N dy 
and so 

— = M. 
dx 

By integrating with y constant, we therefore get 
w = / Mdx-\-j{y). 

Since y is constant in the integration, the constant of inte- 
gration may be a function of y. This function can be found 
by equating the total differential of u to M dx -\-'N dy. 
Since df (y) gives terms containing y only, / (y) can usually 
be found hy integrating the terms in N dy that do not contain x. 
In exceptional cases this may not give the correct result. 
The answer should, therefore, be tested by differentiation. 

Example 1. (2x — y) dx + (iy — x) dy = 0. 

The equation is equivalent to 

2xdx + 4:ydy— (ydx+xdy) = d {x^ -\- 2 y"^ — xy) =0. 
It is therefore exact and its solution is 
x^ -\- 2 y^ — xy = c. 



Ex.2. (\iiy-2x)dx + {^-2'^dy=Q. 



In this case 



-r- =-^{\ny -2x) =-) 
dy dy^ "^ ' y 

dx dx\y V y' 



Art 62 Integrating Factors 135 

These derivatives being equal, the equation is exact. Its 
solution is 

x]ny— 3^ — y' = c. 

The part x In y — x^ is obtained by integrating (In 2/ — 2 x) dx 
with y constant. The term — y^ is the integral oi — 2 y dy, 

which is the only term in f ^yjdy that does not contain x. 

62. Integrating Factors. — If an equation of the form 
M dx + N df/ = is not exact it can be made exact by 
multiplj-ing b}' a proper factor. Such a multipher is called 
an iiUegrating factor. 

For example, the equation 

xdy — ydx = 

is not exact. But if it is multiplied by -j* it takes the form 



xdy— ydx 



-0-^ 



which is exact. It also becomes exact when multiplied by 

-5 or — . The functions -;»-;? — are all integrating factors 

y- xy x^ y^ xy 

oi xdy — y dx = 0. 

While an equation of the form M dx -\- N dy = always 
has integrating factors, there is no general method of finding 
them. 

Example 1. y (I -\- xy) dx — x dy = 0. 

This equation can be written 

y dx— xdy -\- xy^dx= 0. 
Dividing by y^, 

ydx — X dii 



y' 



+ X dx = 0. 



Both terms of this equation are exact differentials. The 
solution is 

^ 1 1 2 

y 2 



136 Differential Equations Chap. 10 

Ex. 2. (i/2 + 2 xy) dx + {2x^ + 3 xy) dy = 0. 
Tliis is equivalent to 

y'^ dx + Z xy dy -\- 2 xy dx ■\- 2 x^ dy = 0. 
Multiplying by y, it becomes 
y^dx + 3 xy^ dy + 2 xy^ dx + 2x^ydy = d {xy^ + x V) = 0. 

Hence 

xy^ + x^y^ = c. 

63. Linear Equations. — A differential equation of the 
form 

% + Py = Q> (63a) 

where P and Q are functions of x or constants, is called 
liJiear. The linear equation is one of the first degree in one 
of the variables (y in this case) and its derivative. Any 
functions of the other variable can occur. 

If the linear equation is written in the form (63a), 

fPdx 

e 

is an integrating factor; for when multiplied by this factor 
the equation becomes 

The left side is the derivative of 

Hence 

y/^'^^ J/^'^Qdx + c (63b) 

is the solution. 



Example 1. ^ + ^ ^ = ^* 



i 



Art. 64 Equations Reducible to Linear Form 137 

In this case 



P'^-f 



-dx = 2hix = hix^. 



Henc3 

fpdx lnx» „ 

The integrating factor is, therefore, x^. Multiplying by x* 
and changing to differentials, the equation becomes 

j^ dy -\- 2 xy dx = a^ dx. 
The integral is 

x^y = I x« + c. 
Ex. 2. a+y')dx-ixy-\-y+y')dy = 0. 
This is an equation of the first degree in x and dx. Divid- 
ing by (1 + t/2) dy^ it becomes 

dx y _ 

di'TTy^'^-y- 
P is here a function of y and 

fpdy 1 



vTT? 

Multiplying by the integrating factor, the equation becomes 
dx xydy _ ydy 

whence 

= vr+7 + 



r-i-c 



and 

X = 1+ 2/2 -I- c Vl +?/2. 

64. Equations Reducible to Linear Form. — An equation 
of the form 

^4-Pi/ = Q2/», (64) 



138 Differential Equations Chap. 10 

where P and Q are functions of x, can be made linear by a 
change of variable. Dividing by ?/", it becomes 

If we take I 

as a new variable, the equation takes the form 
1 du 
1 — ndx 
which is linear. 

Examvle. f^ + lv =%^ 



+ Pu = Q, 



Division by y^ gives 

-3 _ 

dx ^ x^ x^ 



.dy . 2 „ 1 



Let 
Then 



u = y~^. 



dx dx' 

whence 

-3 ^ — —1 — 

^ dx~ 2dx' 
Substituting these values, we get 

2dx'^x'^ x'' 
and so 

dw_4 ^ _2 
dx X x^ 

This is a linear equation with solution 

3x2 



w = t:;:2 + ex*' 



or, since u = y~^, 



y2-3x2 + ^^' 



Art. 65 HoMOGEXEOTJS Equations 139 

65. Homogeneous Equations. — A function / (x, y) is 
said to be a homogeneous function of the nth degree if 

fitx,ty) =t-f{x,y). 

Thus Vx^ -f- y^ is a homogeneous function of the first 
degree; for 

Vx¥T^ = t Vx2 + y2. ( 7\ 

It is easily seen that a polynomial whos^ terms are all of 
tnfe nth degree is a homogeneous function of the nth degree. 
The differential equation 

Mdx+Ndy = 

is called homogeneous if M and N are homogeneous functions 
of the same degree. To solve a homogeneous equation 
substitute 

y = vx. 

The new equation will be separable. 
dy 



Example 1. x-^ — y = Vx^ + y^. 

This is a homogeneous equation of the first degree. Sub- 
stituting y = vx, it becomes 



whence 



(v+^j-)- vx = Vx^+v^ 



.| = v/IT7. 



This is a separable equation with solution 
X = civ + Vl-\-v^). 

Replacing t; by -, transposing, squaring, etc., the equation 

becomes 

x2 - 2 q/ = c*. 



"140 Differential Equations Chap. 10 

dy 
dx 



Ex.2. ym\2x%-y = (i. 



Solving for j-, we get 

dy^ _ — X ± Vx^ + y^ 
dx~ y 

or 

ydy+xdx= ± Vx^ + y^ dx. 

This is a homogeneous equation of t he first degree. It is 
much easier, however, to divide by Vx^ + y^ and integrate 
at once. The result is 

xd x-\-ydy ^_^^^^ 
v x^ + y^ 
whence 

VxH-^ = cztx 
and 

2/2 = c2 ± 2 ex. 

Since c may be either positive or negative, the answer can be 
written 

y2 = c2 + 2 crc. 

66. Change of Variable. — We have solved the homo- 
geneous equation by taking as new variable 

X 

It may be possible to reduce any equation to a simpler form 
by taking some function w of a; and y as a new variable or by 
taking two functions u and v as new variables. Such func- 
tions are often suggested by the equation. In other cases 
they may be indicated by the problem in the solution of 
which the equation occurs. 

Example, (x ~ VY^ =" ^** 



Art. 66 Change of Variable 14 J 

Let X— y = u. Then 

dy du 
dx dx 

and the differential equation becomes 
whence 

2 2 2<^" 

u^ — a^ = w* . 
dx 

The variables are separable. The solution is 

X =u +7^ In—— he 

2 u -\- a 



or 



, a, X — y — a , 

= x — 7/+?: In ^— he, 

2 a; — 1/ + o 



a x — y — a 

2/ = - In ^— h e. 

^ 2 X — 2/ + a 



EXERCISES 
Solve the following differential equations: 
L x'dy -jfdx = 0. 

2. tan X sin* y dx + cos* x cot ydy = 0. 

3. (xy* + x) dx + (i/ - x*y) dy = 0. 

4. (xy* + x) dx + (x*j/ -y)dy = 0. 

6. (3x* + 2xy-y*)dx+(x»-2xy-3t/*)di/ = 0. 

6. ;c^-t/ = y». 12. (2x2/»-y)dx + xrfy = 0. 

7. xdx + ydy = a(x» + y»)dy. 13. (i_a^)^+2xi/ = (l-x*)». 

, 14. tan x— — y = a. 

#v dy fc_ dx 

9. -f- — ay = er'. 

10. Jf?-2xv = 3. W.x|-3,+xV.O. 

dx 

11. x*g-2xy = 3t,. 16. % + y = xy». 



142 Differential Equations Chap. 10 

17. (x2 - 1)3 dy + (x^ + 3 xy VjnTi) dx = 0. 

18. X dx -{- {x + y) dy = 0. 

19. (a;2 + i/2) dx-2xydy = 0. 

20. 2/da; + (a: + ?/)dy = 0. 

21. (x3 _ 3 x2y) dx + (2/3 - a;') dy = 0. 

22. 2/e'' dx = {f + 2 xe^) dy. 

23. I xye^ + j/^j dx - x^e^ dy = 0. 

24. {x + y - 1) dx + (2x + 2 y - 3) dy = 0. 

25. 3y2^-y3 = a;. 






30. The differential equation for the charge 5 of a condenser having 
a capacity C connected in series with a circuit of resistance R is 

^dl + c-^' 

where E is the electromotive force. Find v as a function of i if ^ is 
constant and g = when t = 0. 

31. The differential equation for the current induced by an electro- 
motive force E sin at in a circuit having the inductance L and resist- 
ance R is 

di 
L-r, + Ri = ^ sin at. 
dt 

Solve for i and determine the constants so that i = I when t = 0. 

Let PT be the tangent and PN the normal to a plane curve at 
P (x, y) (Fig. 66a). Determine the curve or curves in each of the 
following cases' 

32. The subtangent TM = 3 and the curve passes through (2, 2). 

33. The subnormal MN = a and the curve passes through (0, 0). 

34. The intercept OT of the tangent on the a>axis is one-half the 
abscissa OM. 

35. The length PT of the tangent is a constant a. 

36. The length PN of the normal is a constant a. 

37. The perpendicular from M to PT is a constant o. 



Art. 67 Certain Equations of the Second Ordeb 



143 



Using polar coordinates (Fig. 666), find the curve or curves in each 
of the following cases: 




Fig. 66a. 



Fig. 666. 



38. The curve passes through (1, 0) and makes with OP a constant 
angle lA = ^• 

39. The angles i^ and d are equal. 

40. The distance from O to the tangent is a constant a. 

41. The projection of OP on the tangent at P is a constant a. 

42. Find the curve passing through the origin in which the area 
bounded by the curve, x-axis, a fixed, and a variable ordinate is pro- 
portional to that ordinate. 

43. Find the curve in which the length of arc is proportional to the 
angle between the tangents at its end. 

44. Find the curve in which the length of arc is proportional to the 
difference of the abscissas at its ends. 

46. Find the curve in which the length of any arc is proportional to 
the angle it subtends at a fixed point. 

46. Find the curve in which the length of arc is prop>ortional to the 
difference of the distances of its ends from a fixed point. 

47. Oxj'gen flows through one tube into a liter flask filled with air 
while the mixture of oxygen and air escapes through another. If the 
action is so slow that the mixture in the flask may be considered uniform, 
what percentage of oxygen will the flask contain after 10 liters of gas 
have passed through? (Assume that air contains 21 per cent by volume 
of oxygen.) 

67. Certain Equations of the Second Order. — There 
are two forms of the second order differential equation that 



144 Differential Equations Chap. 10 

occur in mechanical problems so frequently that they de- 
serve special attention. These are 






The peculiarity of these equations is that one of the vari- 
ables (y in the first, x in the second) does not appear directly 
in the equation. They are both reduced to equations of the 
first order by the substitution 

dy 
This substitution reduces the first equation to the form 

This is a first order equation whose solution has the form 

p = F{x, ci), 

dy 
or, smce p = -f-^ 

This is again an equation of the first order. Its solution is 
the result required. 

In case of an equation of the second type, write the second 
derivative in the form 

^ _ dp _dp dy _ dp 
dx^ dx dy dx dy 

The differential equation then becomes 



p| = /(y,p). 



Art. 67 Certain Equations of the Second Ordeb 145 

Solve this for p and proceed as before. 

Example 1. (1 + x^) g + 1 + [^J = 0. 

Substituting p for --^, we get 

(l+x2)g + 14-p^ = 0. 
This is a separable equation with solution 

Ci — X 

whence 

dy = T^- dx. 

The integral of this is 

By a change of constants this becomes 

2/ = ex + (1+ c2) In (c - x) + c'. 

Substituting 



dy ^ _ _ ^P 

we get 



dx ^' dx2 ^dj/' 



The solution of this is 

y2p2 = t/2 _|_ c^_ 

dv 
Replacing p by -^ and solving again, we get 

i/2 + c, = (a; + C2)2. 



146 Differential Equations Chap. 10 

Ex. 3. Under the action of gravitation the acceleration of 

k 
a falhng body is -^ , where k is constant and r the distance 

from the center of the earth. Find the time required for the 
body to fall to the earth from a distance equal to that of the 
moon. 

Let Ti be the radius of the earth (about 4000 miles), r2 the 
distance from the center of the earth to the moon (about 
240,000 miles) and g the acceleration of gravity at the surface 
of the earth (about 32 feet per second). At the surface of 
the earth r = ri and 

k 
a = — 2= — g. 

The negative sign is used because the acceleration is toward 
the origin (r = 0). Hence k = — gr-^ and the general value 
of the acceleration is 

ar r 

where v is the velocity. The solution of this equation is 

r 
When r = r2, v = 0. Consequently, 



C=-24 



and 



The time of falling is therefore 



=XV: 



^ dr = 116 hours. 



2 gri' {1-2 — r) 

This result is obtained by using the numerical values of n 
and Ti and reducing g to miles per hour. 



Art. 68 Constant Coefficiknts 147 

68. Linear Differential Equations with Constant Coeffi- 
cients. — A differential equation of the form 

d'^y , d"~^y , d"~^y , , r/ \ /^o \ 

where ai, ch, . . . an are constants, is called a linear differen- 
tial equation with constant coefficients. For practical appUca- 
tions this is the most important type of differential equation. 
In discussing these equations we shall find it convenient 

to represent the operation — by D. Then 

^ = Dt, ^ = D^ etc 
dx ^' da^ ^^, etc. 

Equation (68a) can be written 

(D- + aiD"-i + a^D"-'- + • • • + a,)y = / (x). (68b) 

This signifies that if the operation 

Z)» + aiZ)"-i + aiD"-^ + • • • + a„ (68c) 

is performed on y, the result will be /(x). The operation 
consists in differentiating y, n times, n — 1 times, n — 2 
times, etc., multiplying the results by 1, Oi, as, etc., and 
adding. 

With the differential equation is associated an algebraic 
equation 

r" + air"-i + a^r''-'^ + • • • + On = 0. 

If the roots of this auxiliary equation are ri, r^, . . , r„, the 
polynomial (68c) can be factored in the form 

(D - n) (D - r^) . . . (D - r,). (68d) 

If we operate on y with D — a, we get 

{D-a)y = ^-ay. 



148 Differential Equations Chap. 10 

If we operate on this with D — 6, we get 

^D - b) ' (D - a)y = {D - b)(^^- ay^ 

The same result is obtained by operating on y with 

iD-a){D-h) =D^- (a-i-b)D+ ab. 

Similarly, if we operate in succession with the factors of 
(68d), we get the same result that we should get by operating- 
directly with the product (68c). 

69. Equation with Right Hand Member Zero. — To solve 
the equation 

(D» + aiZ)"-i + (hD^-^ + • • • +an)y = (69a) 

factor the symbolic operator and so reduce the equation to 
the form 

(D - ri) (D - ra) . . . (D - Vn) y = 0. 

The value y = Cie^^^ is a solution; for 

(D— n) Cie'"'^ = Cirie""!^ — riCie'"'^ = 

and the equation can be written 

(D-r^) • • • (Z)-r„) . {D-n)y=iD-r,) ■ • • (D-r„).0 = 0. 

Similarly, y = C2e'"'^ y = c^e^^, etc., are solutions. Finally 

y = cie^>^+ 026"'^ + • • • + c„e'""^ (69b) 

is a solution; for the result of operating on y is the sum of 
the results of operating on Cie"^^, c^d^^, etc., each of which 
is zero. 

If the roots Vi, 7*2, . . . , /•„ are all different, (69b) contains 
n constants and so is the complete solution of (69a). If, 
however, two roots ri and r^ are equal 

CiC*' + C2e''»^ = (ci + Co) e*"" 



Art. 69 Equatiox vrira Right Hand Member Zero 149 

contains only one abitrary constant Ci + C2 and (69b) con- 
tains only n — 1 arbitrary' constants. In this case, however, 
xe"^^ is also a solution; for 

{D — n) xe^^ = Tixe^^ -\- e'*' — nxe^^ = e'** 
and so 

(D - ri)2 xe*-!' = (D - rO e^ = 0. 

If then two roots r^ and rt are equal, the part of the solu« 
tion corresponding to these roots is 

(Ci + CiX)e^. 

More generally, if m roots n, r2, . . . r^ are equal, the part 
of the solution corresponding to them is 

(ci + CoX + C3X2 + • • . + c^x'^^)e^^. (69c) 

If the coefl5cients Oi, Oo, . . . On, are real, imaginary roots 
occur in pairs 

ri = a + /3 V- 1, r2 = a - /3 V^T. 

The terms Cie*"'', C2e''»^ are imaginary' but they can be replaced 
by two other terms that are real. Using these values of r^ 
and r2, we have 

{D-T,){D-r^) = {I)-aY + ^. 

By performing the differentiations it can easily be verified 
that 

[{D - a)2 + /32] . e«^ sin ^x = 0, 
[(D - a)2 + /32] . e« cos /3x = 0. 

Therefore 

ef^ [ci sin /3x + C2 cos |3x] (69d) 

is a solution. This function, in which a and /3 are real, can, 
therefore, be used as the part of the solution corresponding^ 
to two imaginary' roots r = a ± /3 V— 1. 
To solve the differential equation 

(D- + aiZ)»-i + aj)--^ + • • • + a„) y = 0, 



160 Differential Equations Chao. 10 

let ri, r2, . . . , r„ be the roots of the auxiliary eqvxition 
r" + air"-i + a^r""-^ + • • • + an = 0. 

If these roots are all real and different, the solution of the 
^qvMion is 

y = cie^>^ + c^e''^ + • • • + c„e'"''^ 

// m of the roots ri, rz, . . . , rm are equal, the corresponding 
part of the solution is 

(ci + CiX + C3x'^ + ' ' ■ + CmX"^-^) e'-'^ 

The part of the solution corresponding to two imaginary roots 
r = ado^ V— 1 is 

e°^ [ci sin ^x + Cz cos ^x], 

This is equivalent to 

{D^-D-2)y = 0. 
The roots of the auxiUary equation 

r2 - r - 2 = 
a,re — 1 and 2. Hence the solution is 
y = Cie~' + de^', 

Ex.2. f^ + *|-5^ + 3s, = 0. ■ 
dx^ dx^ dx ^ 

The roots of the auxihary equation 

are 1,1, — 3. The part of the solution corresponding to the 
two roots equal to 1 is 

(ci + c^x) e". 
Hence 

2/ = (ci + Gjx) e" + Cie~^'. 



Art. 70 Right Hand Member a Function of x 151 

Ex.Z. (D2 + 2D + 2)i/ = 0. 

The roots of the auxiliary equation are 

- izb v^ni. 

Therefore a = — 1, /3 = 1 in (69d) and 

y = e~' [ci sin a: + Co cos x]. 

70. Equation with Right Hand Member a Function of x. — 

Let y = uhe the general solution of the equation 

(I>» + aiD"-i + a2D"-2 + • • • +a„) y = 

and let y = t; be any solution of the equation 

(D" + axZ)»-i + a^D--' + - - - an) y = f (x). (70) 

Then 

y = u + v 

is a solution of (70) ; for the operation 

D" + ai2)»-i + azD'^^ + • • • + a» 

when performed on u gives zero and when performed on v 
gives / (x). Furthermore, u -\- v contains n arbitrary' con- 
stants. Hence it is the general solution of (70). 

The part u is called the complementary function, v the 
'particular integral. To solve an equation of the form (70), 
first solve the equation with right hand member zero and 
then add to the result any solution of (70). 

A particular integral can often be found bj' inspection. 
If not, the general form of the integral can usually be deter- 
mined by the following rules: 

1. If fix) = ax" + flix"-! + • • • + a„, assume 

y = Ax" + Aix"-i + • • • + A,. 

But, if occurs m times as a root in the auxiliary equation, 
assume 

1/ = x™ [Ax" + Aix"-i + • • • + A«]. 

2. If / (x) = ce", assume 

y = Ae^. 



152 Differential Equations Chap. 10 

But, if a occurs m times as a root of the auxiliary equation, 
assume 

y = Ax"'e'". 

3. If f (x) = a cos ^x -\-b sin /Sx, assume 

y = A cos ^x -\- B sin /Sx. 

But, if cos fix and sin fix occur in the complementary function, 

assume 

y = x[A cos fix -^ B sin fix]. 

4. If / (x) = ae°^ cos fix + 6e"^ sin fix, assume 

y = Ae"^ cos /3a; + /3e°^ sin ySa;. 

But, if e"^ cos jSx and e"^ sin /3rr occur in the complementary 
function, assume 

y = xe"^ [A cos fix -\- B sin j8a;]. 

If / (x) contains terms of different types, take for y the 
sum of the corresponding expressions. Substitute the value 
of y in the differential equation and determine the constants 
so that the equation is satisfied. 

Example 1. -^^ -\- 4 y = 2 x + S. 

A particular solution is evidently 
y = i{2x-{-3). 
Hence the complete solution is 

y = ci cos 2 X + C2 sin 2 a; + J (2 x + 3). 
Ex.2. (D2 + 3 Z) + 2) 1/ = 2 + e\ 
Substituting y = A -\- Be'', we get 

2A -^QBe' = 2-{-e'. 



Hence 
and 


2A = 2, QB = 1 

y = 1 + i e^ + cie-' + C2e-2*. 


Ex.3. 


d'y . <i'y _ ^2 



Art. 71 Simultaneous Equations 153 

The roots of the auxihary equation are 0, 0, — 1. Since 
is twice a root, we assume 

y = x'{A3^ + Bx-{-C)=Ax*-\-Bx^-\- Cx'. 

Substituting this value, 

12 Ax2 + (24 A + 6 5) z + 6 B + 2 C = x2. 

Consequently, 

12.4 = 1, 24A+6B = 0, 6B + 2C = 0, 

whence 

A=^\, B=-l C = l. 

The solution is 

y = j\x^ - Ix^ +x~ -\-Ci +C2PC -{■ c^-'. 

71. Simultaneous Equations. — We consider only linear 
equations with constant coefficients containing one independ- 
ent variable and as many dependent variables as equations. 
All but one of the dependent variables can be eliminated by 
a process analogous to that used in soh-ing linear algebraic 
equations. The one remaining dependent variable is the 
solution of a linear equation. Its value can be found and 
the other functions can then be determined by substituting 
this value in the previous equations. 

Example. -z--\-2x — 3y = t, 

Using D f or -r., these equations can be written 

{D-\-2)x-3y = t, 
{D-\-2)y-3x = ^'. 

To ehminate y, multipU' the first equation by Z) + 2 and the 
second by 3. The result is 

(D + 2)2x-3(Z) + 2)i/= 1+2^, 

3(D+2)i/-9x = 3€2'. 



154 Differential Equations Chap. 10 

Adding, we get 

[(D 4-2)2 -9]a: = l+2« + 3e2'. 
The solution of this equation is 

x= -M-H + ye^'+Cie' + Cae-^'. 
Substituting this value in the first equation, we find 
y = i (^ + 2) X - i « = - f < - if + ^ e^ ' + cie' - C2e-". 

EXERCISES 

Solve the following equations: 

2. (, + i)g_(. + 2)|+x + 2 = 0. 



d<2 ~ S^" rfx3 ■"dx^'^'^dx 



12. g+j,=0. 25. g-a=!, = e". 

14. ^=V. 27. g -,-..- >.. 

«-S-^| + ^''='>- 28.g-4|+3v = e...„.. 



Art. 71 SufCTLTANEOTrs Equations 155 



29. ^-9y = e»«co8x. 31. ^ + 4y = co82x. 



ax 


-y 


= e-*. 


dy 


= X 


-y + 2 


dx 
dt 


+ 2/ 


= cos<. 



33. f+. = e., 

34. g=x-2y + l, 
„- . dx dy , n ■ s 

3^- ■*di-i + ^^ = ^"^'' 

37. Solve the equation 

^+f^V = i 

do^^\dx] 
and determine the constants so that y = and -^ = \ when x = 0. 

38. Solve j^ = 3 v^ under the hypothesis that y = 1 and 3^ = 2 

when X = 0. 

39. When a body sinks slowly in a Uquid, its acceleration and 
velocity approximately satisfy the equation 

a = g — ko, 

g and A; being constants. Find the distance passed over as a function 
of the time if the body starts from rest. 

40. The acceleration and velocity of a body falling in the air approxi- 
mately satisfy the equation a = g — kv^, g and A; being constants. 
Find the distance traversed as a fimction of the time if the body falls 
from rest. 

41. A weight supported by a spiral spring is lifted a distance b and 
let faU. Its acceleration is given by the equation a = — fc*s, fc being 
constant and s the displacement from the position of equiUbriima. 
Find s in terms of the time t. 

42. Find the velocity with which a meteor strikes the earth, assimi- 
ing that it starts from rest at an indefinitely great distance and moves 
toward the earth with an acceleration inversely proportional to the 
square of its distance from the center. 

43. A body falling in a hole through the center of the earth would 
have an acceleration toward the center proportional to its distance from 
the center. If the body starts from rest at the surface, find the time 
reqxiired to fall through. 



156 Differential Equations Chap. 10 

44. A chain 5 feet long starts with one foot of its length hanging 
over the edge of a smooth table. The acceleration of the chain will be 
proportional to the amount over the edge. Find the time required to 
slide off. 

45. A chain hangs over a smooth peg, 8 feet of its length being on 
one side and 10 on the other. Its acceleration will be proportional to 
the difference in length of the two sides. Find the time required to 
slide off. 



SUPPLEMENTARY EXERCISES 



xdx 



J a + bi^' 
J{a + hx)*dx. 



a -\-hx 



dx. 



J p + qx 
4. fxV2 -3i*dr. 

5 f (x + a)dx 
J Vx» + 2 ax + 6 



7. J*(i-l)(x»-2x)'dr. 



/x- 1 dx 

X*' 



CHAPTER II 

18. 
19. 
20. 

21. 
22. 

23. 



8. r. 

J SI 



dx 



0. 
1. 



sin ax 
■in ox 



dx. 



COS* ax 
cos2x 
sin2x 
sec^xtanxdx 



/cos 2.., 
\ r-^T- dx. 
1 — su 

/sej 
a + bsec*x 
/ dx 
secx 
cot xdx 



/i 

/ 

/ 

/ 



sinx 
sin ax 



cos ax 
dx 



dx. 



cos X — sin X 
dx 

sec X — tan x 
dx 

sin' ax COS* ax 



cosxdx 
cos X + sin X 

dx 
Vtan' X + 2 
dx 



X Vx» - 1 



dx. 



24. 
25. 
26. 
27. 
28. 
29. 
30. 
31. 
32. 
33. 
34. 



cot X dx 
Vl + sin' X 
dx 



/ 
/ 

/ 
/ 

/ dx 
sin* X — cos* 

/ 

^ (2x-l) V4i*-4x 
fxe-^dx. 

/• f^dx 
J 6 + ce"' 

j sec* xe**" * d!x. 
jah^dx. 

e - 

j tanxlncoexdx. 

/ 
/ 



e' + l 
dx_ 
e" 



dx 

a*x* + 2a6x + 6** 
xdx 

Vx2 _ 2x + 3 
(2x + 3)dx 
(i - 1) Vx* -2x' 



167 



158 Supplementary Exercises 

35. r — f'^ • 56. fx3 Va2 - x^ dx. 

•^ X V 2 X + 3 J 



•^ x2 Vax2 + 6* -^ ^^2 - x^ 

37. j {a? - x')'^ dx. 58. /^^r^^ 



38 



49. 



x^dx 

IP 
dx 



1 f ^^ 69. f ^^• 

J X Vax^ + 6x -^ («' - ^')* 

39. JV3-2x-x^dx. 60. /^^^- 

40. r(a3-x?)3dx. 61. Je^ + in*dx. 

41. fcosSxsm^xdx. 62. J e'^sinftxtte. 

42. r(l+cosx)^dx. 63. J^ d^- 

43. rtan2xsec6 2xda;. 64. Tx In (ex + &) da;. 

44. jTcot^xdx. 65. Jll^^^da: 



45. fi ^r — r- 66. Cxcoi-^xdx. 

J tan X + cot X J 

46. /(secx+tanx)^dx, 67. j^^^—0^^-^, 
tan X — 1 ,_ „„ r x" dx 



47 

._ rcos^xdx ^^ r 

48. I —^-, 69. I 



tan X + 1 ■ "°- J (r> + 1) (x* - 2) • 

cos'xdx „„ r xdx 



x^ + 1 

/sin 2 X cos^ X dx. 70. C - — ^--r- dx. 

J (x — 1)* 

50. JVI + cos2 X sin 2 X dx. 71, fx^cosixdx. 

51. C dx ^ f 2x^ + 3x 

•^ X Va^x + 62 '^- J (x-l)(x-2)(x4-3) 

^2 f ^^ 70 f (3x-5)dx 

■ ^ x^ V^3-2 "^- J X (x + 3)2 * 

-^ (x - 1) Vx + 2 J ^-^ 

64. Jx(ax + 6)«dx. 75. /jj^^* 

65. (^^M.dx. 76. f ^^^ , . 
J Vax + 6 •' (1 -2x2)» 



StIPPLEMENTART EXERCISES 159 

x2 - 3j + 2 



dx. 



77. faec^xtaxi^xdx. 84. f- , 

J J Vj^ _ 4 X + 3 

78. I sin Sxcos 4xdx. a- CKrLUJl dx. 

J y a + X 

79. j'sin2xcos2idx. 86. J(sinx - co8x)»(ir. 

80. fsinxsinSxdx. 87 f ^^ — 



. rco8 2xco6 3xdx. 88 r log (x + Vx^ - l) 

J Vx2- 1 

. J (cotx + cscx)^dx. 89 Jgec^^dx. 

/(3x — l)c?x /. , 

-===• 90. /(x«+««)idx. 



cir. 



83. 



CHAPTER IV 

91. Find the area bounded by the x-axis and the parabola y = x* 
-4x + 5. 

92. Find the area bounded by the curves y = 3?, y^ = x. 

93. Find the area bounded by the parabola y* = 2 x and the witch 
1 

y- + 1 

94. Find the area within a loop of the curve j/* = x= — x*. 

95. Find the area of one of the sectors boimded by the hyperbola 
X* — y* = 3 and the lines x = =fc 2y. 

96. Find the area bounded by the parabolas y' = 2 ax + a-, ^ -\- 
2ax = 0. 

97. Find the area within the loop of the curve x = ; — ■. r, y = 

1 + wi 
3aw^ 



98. Find the area bounded by the parabola x = a cos 2<t>, y = 
o sin (^ and the line x = —a. 

99. Find the area inclosed by the curve x = a cos^ <l>, y = b sin* 4>. 

100. Find the area bounded by the curve x = asin^, y=a cos* 0. 

101. Find the area of one loop of the curve r = a cos nd. 

102. Find the area of a loop of the cxu^'e r = a (1 — 2 cos 0). 

103. Find the area between the curves r = a (cos -{- 2), r = a. 

104. Find the total area inclosed by the cxu-ve r — a sin \ 0. 

105. Find the area of the part of one loop of the curve 1^ = 0^ sin 3 & 
outside the curve r^ = a sin 0. 



100 Supplementary Exercises 

106. By changing to polar coordinates find the area within one loop 
of the curve (x^ + ifY = a?xy. 

107. By changing to polar coordinates find the area of one of the 
regions between the circle x"^ -\- y^ = la? and hyperbola x^ — y* = a^. 

108. Find the area of one of the regions bounded by d = sin r and the 
line e = \. 

109. Find the volume generated by revolving an ellipse about the 
tangent at one of its vertices. 

110. Find the volume generated by revolving about the y-axis the 
area bounded by the curve y^ = 3? and the line x = 4. 

111. Find the volume generated by rotating about the y-axis the area 
between the x-axis and one arch of the cycloid x = o (<^ — sin ^), 
y = o (1 — cos<^). 

112. Find the volume generated by rotating the area of the preced- 
ing problem about the tangent at the highest point of the cycloid. 

113. Find the volume generated by revolving about the x-axia the 
part of the ellipse ^ — xy -\- y^ = I'va. the first quadrant. 

114. Find the volume generated by revolving about 9 —-^ the area 

enclosed by the curve r" = (^ sin B. 

115. The ends of an ellipse move along the parabolas z* = ox, 
J/'' = ax and its plane is perpendicular to the x-axis. Find the volume 
swept out between x = and x = c. 

116. The ends of a helical spring lie in parallel planes at distance h 
apart and the area of a cross section of the spring perpendicular to its 
axis is A. Find the volume of the spring. 

117. The axes of two right circular cylinders of equal radius intersect 
at an angle a. Find the common volume. 

118. A rectangle moves from a fixed point, one side varying as the 
<iistance from the point, and the other as the square of this distance. 
At the distance of 10 feet the rectangle becomes a square of side 4 ft. 
What is the volume then generated? 

119. A cylindrical bucket filled with oil is tipped until half the bot- 
tom is exposed; if the radius is 4 inches and the altitude 12 inches find 
the amount of oil poured out. 

120. Two equal ellipses with semi-axes 5 and 6 inches have the same 
major axis and lie in perpendicular planes. A square moves with its 
•center in the common axis and its diagonals chords of the ellipses. 
Find the volume generated. 

121. Find the volume bounded by the paraboloid 12 3 = 3 x'^ -f y* 
And the plane 3 = 4. 



StIPPLEMENTARY ElXERCISES 161 

CHAPTER V 

122. Find the length of the arc of the curve 

y = i X Vx2 - 1 — ^ In (x + Vi* - l) between i = 1 and x = 3. 

123. Find the arc of the curve 9 !/- = (2 x — 1)' cut off by the line 
X = 5. 

124. Find the perimeter of the loop of the curve 

9^ = i2y-l)(y-2r-. 

125. Find the length of the curve x=P+t, y = t^— t below the 
a>axis. 

126. Find the length of an arch of the curve 

z = aV3(2</. — sin2</>), y = ^ (1 — co8 3<^). 

127. Find the length of one quadrant of the curve 

X = a cos* 0, y = b sin* ^. 

128. Find the circumference of the circle 

r = 2 sin » + 3 cos fl. 

129. Find the perimeter of one loop of the curve 



(I) 



130. Find the area of the surface generated by revolving the arc of 
the curve 9 y* = (2 x — 1)* between x = and x = 2 about the y-axis. 

131. Find the area of the surface generated by revolving one arch of 
the cycloid x = a {<f> — sin <f>), y = a (1 — cos (/>) about the tangent at 
its highest point. 

132. Find the area of the surface generated by rotating the curve 
t" = a^ sin 2 about the x-axis. 

133. Find the area generated by revolving the loop of the curve 
9 X* = (2 J/ - 1) (i/ - 2)2 about the x-axis. 

131. Find the volume generated by revolving the area within the 
ciu^e j/2 = X* (1 — x^) about the y-axis. 

135. The vertical angle of a cone is 90°, its vertex is on a sphere of 
radius a, and its axis is tangent to the sphere. Find the area of the cone 
within the sphere. 

136. A cylinder with radius b intersects and is tangent to a sphere of 
radius a, greater than b. Find the area of the surface of the cylinder 
within the sphere. 

137. A plane passes through the center of the base of a right circular 
cone and Ls parallel to an element of the cone. Find the areas of the 
two parts into which it cuts the lateral surface. 



162 Supplementary Exercises 

CHAPTER VI 

138. Find the pressure on a square of side 4 feet if one diagonal ia 
vertical and has its upper end in the surface. 

139. Find the pressure on a segment of a parabola of base 2 b and 
altitude h, if the vertex is at the surface and the axis of the parabola is 
vertical. 

140. Find the pressure on the parabolic segment of the preceding 
problem if the vertex is submerged and the base of the segment is in the 
surface. 

141. Find the pressure on the ends of a cylindrical tank 4 feet in 
diameter, if the axis is horizontal and the tank is filled with water under 
a pressure of 10 lbs. per square inch at the top of the tank. 

142. A barrel 3 ft. in diameter is filled with equal parts of water and 
oil. If the axis is horizontal and the weight of oil half that of water,. 
find the pressure on one end. 

143. Find the moment of the pressure in Ex. 138 about the other 
diagonal of the square. 

144. Weights of 1, 2, and 3 pounds are placed at the points (0, 0),. 
(2, 1), (4, — 3). Find their center of gravity. 

145. A trapezoid is formed by connecting one vertex of a rectangle 
to the middle point of the opposite side. Find its center of gravity. 

146. Find the center of gravity of a sector of a circle with radius a 
and central angle 2 a. 

147. Find the center of gravity of the area within a loop of the curve 
yi = x^ — x^. 

148. Find the center of gravity of the area boimded by the curve 

y^ = ^ and its asymptote x = 2a. 

149. Find the center of gravity of the area within one loop of the 
curve r^ = a^ sin 6. 

150. Find the center of gravity of the area of the curve x = a sin' </», 
y = b sin' 4> above the x-axis. 

151. Find the center of gravity of the arc of the curve 9 y^ = 
(2x — l)3cutoff by thelinex = 5. 

152. Find the center of gravity of the arc that forms the loop of the 

curve 

9 2/2 = (2 X - 1) (x - 2)2. 

153. Find the center of gravity of the arc of the curve x = P + 1, 
y = t? — t below the x-axis. 

154. Show that the center of gravity of a pyramid of constant 
density is on the line joining the vertex to the center of gravity of the 
base, J of the way from the vertex to the base. 



StrPPLEMENTART EXERCISES 163 

155. Find the center of gravity of the surface of a right circular cone. 

156. Show that the distance from the base to the center of gravity 
of the surface of an obhque cone is j of the altitude. Is it on the line 
joining the vertex to the center of the base? 

157. Find the center of gravity of the soUd generated by rotating 
about the line x = i, the area above the 3>-axis bounded by the parab- 
ola y* = 4 X and the Une x = 4. 

158. The arc of the curve x* + y* = a* above the x-axis is rotated 
about the j/-axis. Find the center of gravity of the volume and that 
of the area generated. 

159. Assuming that the specific gravity of sea water at depth h in 
miles is 

find the center of gravity of a section of the water with vertical sides 
five miles deep. 

160. By using Pappus's theorems, find the center of gravity of the 
arc of a semicircle. 

161. The eUip>se 

is rotated about a tangent inchned 45° to its axis. Find the volume 
generated. 

162. The volume of the ellipsoid 

^+^+- =1 
is ^vabc. Use this to find the center of gravity of a quadrant of the 
elhpse -5 + ^ = 1. 

163. Find the volume generated by revolving one loop of the curve 

r = a sin d about the initial line. 

164. A semicircle of radius a rotates about its bounding diameter 
whUe the diameter sUdes along the line in which it hes. Find the 
volume generated in one revolution. 

165. The plane of a moving square is perpendicular to that of a fixed 
circle. One comer of the square is kept fixed at a point of the circle 
while the opposite comer moves around the circle. Find the volume 
generated. 

166. Find the moment of inertia about the x-axis of the area bounded 
by the x-axis and the curve y = 4 — x*. 

167. Show that the moment of inertia of a plane area about an axis 
perpendicular to its plane at the origin is equal to the siun of its moments 
of inertia about the coordinate axes. Use this to find the moment of 



164 Supplementary Exercises 

inertia of the ellipse ~5 + rj = 1 about the axis perpendicular to its 
plane at its center. 

168. Find the moment of inertia of the surface of a right circular cone 
about its axis. 

169. The area bounded by the x-axis and the parabola ^^ = 4 ax — x- 
is revolved about the x-axis. Find the moment of inertia about the 
X-axis of the volume thus generated. 

170. From a right circular cylinder a right cone with the same base 
and altitude is cut. Find the moment of inertia of the remaining 
volume about the axis of the cylinder. 

171. A torus is generated by rotating a circle of radius a about an 
axis in its plane at distance h, greater than a, from the center. Find 
the momeat of inertia of the volume of the torus about its axis. 

172. Find the moment of inertia of the area of the tonos about its axis. 

173. The kinetic energy of a moving mass is 



j 5 y- dm, 



where v is the velocity of the element of mass dm. Show that the 
kinetic energy of a homogeneous cylinder of mass M and radius a 
rotating with angular velocity w about its axis is \ Mw'^a'^. 

174. Show that the kinetic energy of a uniform sphere of mass M and 
radius a rotating with angular velocity co about a diameter is ^ Mu^a'. 

175. When a gas expands without receiving or giving out heat, its 
pressure and volume are connected by the equation 

pv^ = k 
where y and k are constant. Find the work done in expanding from 
the volume vi to the volume V2 . 

176. The work done by an electric current of i amperes and E volts 
is iE joules per second. If 

E — Eo cos (Jit, i = h cos {wt + a), 
where Eo, h, w are constants, find the work done in one cycle. 

177. When water is pumped from one vessel into another at a higher 
level, show that the work in foot poimds required is equal to the product 
of the total weight of water in pounds and the distance in feet its center 
of gravity is raised. 

CHAPTER VII 

178. Find the volume of an ellipsoid by using the prismoidal formula. 

179. A wedge is cut from a right circular cylinder by a plane which 
passes through the center of the base and makes with the base an angle a. 
Find the volume of the wedge by the prismoidal formula. 



StrPPLEMENTABY ExERCISES 16& 

180. Find approximately the volume of a barrel 30 inches long if its 
diameter at the ends is 20 inches and at the middle 24 inches. 

181. The width of an irregular piece of land was measured at inter- 
vals of 10 yards, the measurements being 52, 56, 67, 49, 45, 53, and 62 
yards. Find its area approximately by using Simpson's rule. 

Find the values of the following int^rals approximately by Simpson's 
rule: 



i2. f Vj-5 + 1 dx. 
53. t -ilnxdx. 

Jl X* 




+ J* 



186. Find approximately the length of an arch of the curve j/ = sin x. 

187. Find approximately the area bounded by the a!>-axis, the curve 



y = , and the ordinates x = 0, x = r. 

X 



CHAPTER VIII 

E.xpress the following quantities as double integrals and determine 
the limits: 

188. Area bounded by the parabola y = x- — 2 x + 3 and the line 
y = 2x. 

189. Area bounded by the circle x- + y^ = 2 a- and the curve 

y- =:r: — :• 



190. Moment of inertia about the x-axis of the area within the circles 

x*-fy* = 5, x' + y' — 2x — 'iy = 0. 

191. Moment of inertia of the area within the loop of the curve 
y* = x^ — x* about the axis perpendicular to its plane at the origin. 

192. Volume bounded by the xy-plane the paraboloid z = x- -{- if 
and the cylinder x* + y^ = 4. 

193. Volmne boimded by the xy-plane the paraboloid z = x- + y^ 
and the plane z — 2 x -\- 2y. 

194. Center of gravity of the soUd bounded by the xz-plane, the 
cylinder x- -\- z- = o*, and the plane x •\- y -{- z = \ a. 

195. Volume generated by rotating about the x-axis one of the areas 
bounded by the circle x^ -\- y- = b a^ and the parabola y^ = A ax. 

In each of the following cases determine the region over which the 
integral is taken, interchange dx and dy, determine the new limi ts, and 
so find the value of the integral: 



166 Supplementary Exercises 

, ix+y)dydx. 199. | I Vx^ + xy dydz. 

Express the following quantities as double integrals using polar 
■coordinates: 

200. Area within the cardioid r = a (1 + cos d) and outside the 
circle r = | a. 

201. Center of gravity of the area within the circle r = a and 
outside the circle r = 2 a sin 9. 

202. Moment of inertia of the area cut from the parabola 

2a 

T = 

1 — COS 9 
by the line y = x, about the x-axis. 

203. Volume within the cylinder r = 2 a sin B and the sphere 

x^ + 2/2 + 2= = 4 a-. 

204. Moment of inertia of a sphere about a tangent line. 

205. Volume bounded by the paraboloid z = x- -{■ y^ and the plane 
z = 2x + 2y. 

206. Find the area cut from the cone x^ + y^ = z"^ by the plane 
x = 2z-Z. 

207. Find the area cut from the plane by the cone in Ex. 206. 

208. Find the area of the surface z^ + (x + yY = a^ in the first 
octant. 

209. Determine the area of the surface z^ = 2 x cut out by the planes 
3/ = 0, 2/ = X, X = 1. 

CHAPTER IX 

Express the following quantities as triple integrals: 

210. Volume of an octant of a sphere of radius o. 

211. Moment of inertia of the volume in the first octant bounded by 

the plane - + ^ + - = 1 about the x-axis. 
a c 

212. Center of gravity of the region in the first octant boimded by 
the paraboloid z = xy and the cylinder x"^ ■]- y^ = o^. 

213. Moment of inertia about the z-axis of the volume boimded by 
the paraboloid z = x^ + y"^ and the plane z = 2 x + 3. 

214. Volume bounded by the cone x"^ = y^ + 2z'^ and the plane 
3z + j/ = 6. 

Express the following quantities as triple integrals in rectangular, 
cylindrical, and spherical coordinates, and evaluate one of the integrals; 



SuPPLEBfENTAKY ExERCISES 167 

215. Moment of inertia of a right circular cylinder about a line 
tangent to its base. 

216. Moment of inertia of a segment cut from a sphere by a plane, 
about a diameter parallel to that plane. 

217. Center of gravity of a right circular cone whose density varies 
as the distance from the center of the base. 

218. Volume bounded by the xy-plane, the cylinder 3? -\- y- = 2 ax 
and the cone z^ = x* + j/*. 

219. Find the attraction of a uniform wire of length I and mass M 
on a particle of unit mass at distance c from the wire in the perpen- 
dicular at one end. 

220. Find the attraction of a right circular cylinder on a particle at 
the middle of its base. 

221. Show that the attraction of a homogeneous shell bounded by 
two concentric spherical surfaces on a particle in the enclosed space 
is zero. 

CHAPTER X 

Solve the following differential equations: 

222. ydx + {x-xy)dy =0. 

223. sin X sin y dx + cos x cos ydy = 0. 

224. (2x2/-y2+6x«)dx + (3y*+x*-2xy)di/ = 0, 

225. x^+y = xh/. 

226. X :r^ + y = cot X. 

ax 

227. xdy - \y + e') dx = 0. 

228. (1 + x*) dy + (xy + x) dx = 0. 

229. xdx -\- ydy = x dy — y dx. 

230. (sin X -\- y) dy -\- (y cos x — x*) dx = 0. 

231. y (e* + 2) dx + (e* + 2x) dy = 0. 

232. (xy* - x) dx + (y + xy) dy = 0. 

233. (l+x2)^+xy = 2y. 

234. xdy — ydx = Vi* + y2 dx. 

235. (x - y) dx + X dy = 0. 

236. xdy -ydx = x^3^-ry^dx. 

237. c*^-^' dy + (1 + e*) dx = 0. 

238. (2x + 3y- l)dx + (4x + 6y-5)dy =0. 

239. (3y« + 3xy + x»)dx = (x* + 2xy)dy. 

240. (1+ x») dj^ + (xy - x») dx = 0. 



168 Supplementary Exercises 

241. {x'y + t/*) dx - (x3 + 2 xy^) dy = 0. 

243. 2^ +y + x2/» = 0. 

ax 

244. ydx = (y^ — x) dy. 

245. y -1^ + t/2 cot X = cos x. 

246. (x2 - y2) (dx + dy) = (x« + r/^) (dy - dx). 
247 x-'^ 4-^ = 1 



253. f2-!^.e". 
dx^ dx 

258. j^ + aV = sin ox. 

260. ^-^-2y = e-*sin2x. 
dx'^ dx 

261. j^ + 9 y = 2 cos 3 X - 3 cos 2a;. 

262. g + 6^+5y = (e- + l)^ 



Supplementary EIxercises 169 

263. ^ - y = xe''. 
dx- " 

265. ^+2x = sm/, ^-2y = co6<. 
at at 

267. According to Newton's Law, the rate at which a substance cools 
in air is proportional to the difference of the temperature of the sub- 
stance and the temperature of air. If the temperature of air is 20° C. 
and the substance cook from 100° to 60° in 20 minutes, when will its 
temperature become 30°? 

268. A particle moves in a straight line from a distance a towards a 
point with an acceleration which at distance r from the point is k f^i. 
If the particle starts from rest, how long will be the time before it 
reaches the point? 

269. A substance is undergoing transformation into another at a 
rate proportional to the amount of the substance remaining untrans- 
formed. If that amount is 34.2 when t = 1 hour and 11.6 when t = 3 
hours, determine the amount at the start, t = 0, and find how many 
hours will elapse before only one per cent will remain. 

270. Determine the shape of a reflector so that all the rays of light 
coming from a fixed point will be reflected in the same direction. 

271. Find the curve in which a chain hangs when its ends are sup- 
ported at two points and it is allowed to hang under its own weight. 
(See the example solved in Art. 57.) 

272. By Hooke's Law the amount an elastic string of natural length 
I stretches under a force F is JdF, k being constant. If the string is held 
vertical and allowed to elongate under its own weight w, show that the 
elongation is 5 kwl. 

273. Assuming that the resistance of the air produces a negative 
acceleration equal to k times the square of the velocity, show that a 
projectile fired upward with a vdocity vi will return to its starting point 
with the velocity 



^-V' 



g + kvr' 



g being the acceleration of gravity. 

274. Assuming that the density of sea water imder a pressure of p 
pounds per square inch is 

p = 1 + 0.000003 p. 



170 SUPPLEMKNTART EXEKCISES 

show that the surface of an ocean 5 miles deep is about 465 feet lower 
than it would be if water were incompressible. (A cubic foot of sea 
water weighs about 64 pounds.) 

275. Show that when a liquid rotating with constant velocity is in 
equilibrium, its surface is a paraboloid of revolution. 

276. Find the path described by a particle moving in a plane, if its 
acceleration is directed toward a fixed point and is proportional to the 
distance from the point. 



ANSWERS TO EXERCISES 

Page 6 

2. |x3 + l+C. 

3. |x' + 2x* + C. 

4. i v^(2I-3)+C.- 
~6. a' X - 2 ox' + f a* x^ - I x' + C. 

8. 2x + 31nx + C. 

9. ^^ + 4y + 41nj/ + C. 

10. xiCAx^-^x^-e) +C. 

11. In (x + 1) + C. 23. - i (a* - P)» + C. 

12. - ■:r^ + C. 24. X + 3 In (x - 2) + C. 

13. V2X + 1+C. 25. 5 In (2x^ + 1) 



+ C. 



x + 1 

27+1 

14. i In (x^ + 2) + C. . " 1 

15. Vx^ - 1 + C. ~ 4 (2 X* + 1) 

^^- - 46(a+6xy +^- 26. ^ (l - 1]' + C. 

17. - 1 (o^ - x*)' + C. 1 

18. iln(a3 + x3)+C. 27. "„ („_i) (^n_^o)»-i + C 

20 ln(x2 + ax + 6)+C. 28. ^'^ -^^ ^"^ +(7. 



21. 2 Vx« + ax + 6 4- C. 
1 



29. ix3-|ln(x3 + 2)+C. 



^ in (1 - a^) + C. 30. i.i:8_|a;5^ia^^C'^ 

Pages 12, 13 

1. 138. 2. |^ + 30<. 

3. A = - ^ ff^ + 100 < + 60. It reaches the highest point when 
t = 3.1 sec, h = 215.3 ft. 
4- Tis^ffT sec. 

171 



172 Answers to Exercises 

5. X = (^ — t + I, y = t — ^ t^ + 2. These are parametric equa- 
tions of the path. The rectangular equation is x^ + 4 xy + 4 y^ _ 
12x-22t/ + 31 =0. 



6. About 53 mUes. 10. (-i -^) 

11. 6y = 

12. -12 



7. x = jV3fi,y = ^fi + Vot. 11. 6i/ = x3-3x2 + 3x + 13. 



4 ■ "-.^ 4' 

8. J/ = 2 X - -^ x2 - f . 

9. y = e*. 14. About 4 per cent. 



15. X = xoe**, where x is the number at time t, xo the number at time 
t = 0, and k is constant. 

17. 17 minutes. 19. 11.6 years. 

18. 11.4 minutes. 



Pages 18, 19 



1. 

2. 

3. 

4. 


- (1 cos 2 x+l sin 2 x) + C. 
5 • f2x-3\ 


6. 
7. 
8. 
9. 


\ sin2 e + C. 
tan X + C. 

- ^ cot 2 X + C, 

— CSC X + C 


- i cos (nt + a) +C. 
n 


5. 


-4csc^ + C. 


10. 


J sec* X + C. 



11. 2 (esc I - cot 1^ +C. 

12. i sin (x« - 1) + C. 

13. I (tan 3 X + sec 3 x) + C. 

14. tan X + X — 2 In (sec x + tan x) + C. 

15. hi (1 + sin x) + C. 

16. e + cos2 + C. 

17. sin X + hi (esc x — cot x) + C. 

18. ism'x + C. 

19. ltan*x + C. 

20. Itan^x + C. 

21. -icos«x + C. 

22. Hn(l +2tanx) + C. 

23. -Jhi(l -sin2x) + C. 

24. i In (1 + tan ax) + C 
a 

_. 1 . . x V2 . ^ 

25. — = sin-' — -=7- + C. 
V2 V3 









Answers to Exebcisbs 173 

27. isec-i^ + C! 

28. itan-i2y + C. 

29. -^ In (x V7 + V7 1* + l) +C. 

v7 

30. |sec-'Y+C. ^^ 

1 , 2J + V3 , ^ 

31. 7= m 7= + ^- 

4 V3 2 X - V 3 

32. In (2z + V4x^-3) + C. 

33. -3 VT^T^ - 2sm-i | + C. 

34. 2 V:?T4 + 3ln (x + Vi« + 4) + C. 

35. 5ln(4x2-5)+-^ln|^-=-^ + C. 
8 ' V5 2 X + V 5 

36. J V3x»-9 - -|= In (x + V?"=^) + C. 

V3 

37. ein-.(^-^)+C. 46. -i.->^ + e. 

38. -V2-sin«x + C'- 47. ^ (e*« - e"*-*) + 2 x + C 

39. tan-i (sin x)+C. 48. i In (1 + e^*) + C. 

40. sec-i (tan x) + C. ^^ 49. In (e* + e"') + C. 

41. In (sec x + v'sec' x + 1). _1 

Art n / _ "0. e "T C. 

42. 2 VT^-^e + C. 51^ t^-, (^) ^ c. 
._ 1, 2 + lnx , ^ 1 — e* 

43- 4^2 ^n^ + ^- 52. In f:^ + C. 

44. - ^ Vcos^x-8in«x + C. 53 1 ^^_, ^^^ ^ ^ 

1 _ X* a 

45. 2 sin ' ^ + C. 54 tan-i (e*) + C. 

Page 20 
^l*-i^ + 3.„ -l._, 2x-l,- 

3. -^ In (3 X + 2 + V9x* + 12x + 6) + C. 
V3 

. 1 . _, (2x- 1) V5 . ^ . 1 _, (x-3)V6,_ 

^- VB'^" 3 +^- 5- Vl""" 3 + ^- 



174 Answers to Exercises 

6. 1 in^ + C. 
— a X + 

7. -ln(4x^-4a:-2)+ — In^^-^-p-^ + C. 

8. I V3x2 -6x + 1 + 4= In [3 (a; - 1) + VQx^ - 18x+3] + C. 

3 Va 

9. 1 In (3 x2 4- 2 X + 2) + -^ tan-i ^^^ + C. 
6 ^ 3 V 5 V 5 

10. 4= sec-i ^^ .t^ + ^ln (2x + 1 + V4x'^ + 4x-l) + C. 
V2 V2 2 

11. -— =J=+C. 

Vx2 - 2 X + 3 

12. Vx2 - X - 2 + I In (2 X - 1 + V4x2 -4x-8) + C. 

13. JL,„(iil±lii^Uc. 

Vl7 V4e* + 3 + Vl7/ 

Page 26 

1. — cos X + 3 cos' X + C. 

2. sin X — I sin^ x + ^ sin^ x + C. 

3. sin X — f cos' x + f sin' x — cos x ■{• C. 
4.-3 cos' X + 5 cos^ X + C. 

5. I sin^ 1 X - f sin'' ^ X + I sin' I X + C. 

6. -Jy sin« 3 (9 - ^ sinS 3 + C. 

7. - f cos'0 + cos0 + C. 

8. sin X + 5 sin^ x + C. 

9. cos X + In (esc x — cot x) + C. 

10. cos^ d — I cos* - In cos + C 

11. tan X + i tan' x + C. 

12. - (cot y + i cot' 2/ + I cots y _|. ^ cot^ y + i cot» y) + C. 

13. tan X — X + C'. 

14. 2 tan - sec - tf + C. 

15. |sec'|x + C. 

16. t'j sec^ 2 X - i sec^ 2 x + i sec' 2x + C. 

17. — I csc^ X — In sin X + C. 

18. I sec* X — f sec* x + | sec'' x + In cos x + C. 

19. - i cot« X - i cot' x + C. 

20. ^tan^x + lntanx + C. 

21. |-^ sin (2 ax) + C. 

22. 1 + j^sin(2ox) + C. 



Answers to Exercises 17a 



23. t^ X - sV sin 4 a; - tV sin' 2 X + C. 

24. xV^- V2sin2x + ism»x + C. 

25. ^ X — I sin 2 X + ^\ sin 4 z + -^^ sin' 2 x + C. 

26. tan x + sec x + C*. 

27. tan ^ X + C. 

28. 2 f sin ^ - cos I j + C. 

29. I Vx2 -oT- - |- In (x + ^3? - a") + C. 

30. I Vx2 + d^ + |. In (x + Vx* + a?) + C. 

31. I Vx2 + a^ - ^' In (x + Vx* + a?) + C. 

32. -^ +C. 

a^ V x^ — a- 

33. -In - ^ +C. 

« O + V o2 — X* 



34. -:^I^Zi + C. . 

ax 

35. . ^ +C. 
Vo* -x* 

36. i (x2 + a2)« - I (x^! + a^)* + C. 

37. -^^7^ + C. 

a-x 



38. 



X - 2 



\/x2-4x + 5 + iln(x-2 + V'x*-4x+5) + C. 



2 -^ ^-^ ' " ' 2 



39. ^^ - ^^ V 2 - -.x - 4x^ + ^ sin-^ ^^ + ^ + C 
32 ^•^^64 3 ^^' 

Page 30 

1. ^ + 4x-2In(x- 1) + 121n(x-2) +C. 

2. 31nx -ln(x + 1) 4-C. 
3 j^(x_-i^(x+2)+c. 

4. l + lnx- jgln(2x-l)-^ln(2x + l)+C. 

5. f In (X + 3) - i In (X + 1) - f In (X + 5) + C. 

6. ^ln(2x- 1) -31n(2x-3) +|ln(2x-5) + C. 

I . X -\ h In h C. 

J> X 



376 Answers to Exercises 

8. lln(x + l)+|ln(x-l)-2^^j+C. 



»--Ur4^+'"l^)+- 



10. x-81n(x + l) 3(^ + 1)3 +g- 

13- 2(4^ + ^- 

14. x + ^ln^^-Kan-iz + C. 

4 X + 1 2 

„ 1, J + 1 1 , ,2x-l . - 

15. 5 In , ^: H — tan-i 7= h C, 

3 Vx2 - X + 1 Vs Va 

16. i In (x3 + 1) + C. 

17 1 . 1, X- 1 2V3^^ , 2x-l . ^ 

6(x + l) 4 x + l 9 \/3 

19. --^+lln— "-=^ + 1 tan-x^ + C. 

a^ - 8 6 Vx2 + 2 X + 4 2 V3 V3 

20. 3 (x + D* + In [(x 4- D* - 1] - ^^ tan-i ^ (x + 1 ) +1 ^ ^^^ 

v3 

21. -1 A+1 i_l i_:,A+llnl±£5+c. 



1-x^ 



5 

22. 3|,(ax + 6)i»-|A(a, + ,)i + C. 

23. 2 Vx + 2 - In (x + 3) - 2 tan-i V^T2 + C. 

24. 4x* + 21n (x* - l) + In (x* + l) - 2tan-» x* + C. 

25. Hx + l)* + i(x-l)« + C. 



Page 34 

1 X 

1. 2[ cos 2 X + 2 sin 2 X + C 3, x sin"! x + Vl - r* + C. 

2. |lnx-| + C. 4. ^-|^tan-ix-|x + C. 
6. X In (x + Va» + 3?) - Vo« + x« + C. 



Answebs to Exercises 177 

6. 2 VF^^ In X - 4 Vj - 1 + 4 taii-i Vx - 1 + C. 

7. Inxln(lnx) -Inx + C. 

8. ^ x»8ec-ix - I V^^n _ 1 In (x 4. V^^TT) + C. 
600 

9. X - (1 + e-') In (1 +6*) + C. 

10. (x» - 2 X + 2) e* + C. 

11. - (x» + 3x» + 6x + 6)e^ + C. 

12. ^^sm2x-^^^^^^^^^cos2x + C. 

13. ^ Vx'-a^ - ^ In (x + V^i:^») + C. 

14. I Va* + x* +|ln (x + V^*T^) + C. 

15. ^(28in3x-3coe3x) +C. 

lo 

16. -^ (cob X + sin x) + C'. 

17. -?(sin2x + 2cos2x) +C. 

o 

18. H9ec0tan0 + ln(sec» + tan0)] + C. 

19. 5 cos X — T^jj cos 5 X + C. 



1- f 

2. 2.829. 



1. 1 - i V3. 



2. 


3' 


3. 


-20. 


4. 


2. 


5. 


2 

3- 


6. 


1.807. 


7. 


0.2877 


8. 


0. 


9. 


\a. 


10. 


2. 


11. 


00. 



Page 38 




3. 


-0.630. 


Pages 45, 46 




12. 


2* 


13. 


1 

2fc» 


14. 


0.5493. 


15. 


r + 1. 
4^2 


16. 


1.786. 


17. 


0.4055. 


18. 


0.2877. 


19. 


f (l-ln2) 



178 Answers to Exercises 



Pages 49, 50 

1. 11. 13. 27r + |, 67r-f 

2. |V3(4-V2). 14. 4V3 4-Y'r. 



4. i 

5. 9.248 

6. irab. 

7. ia?. 

8. 51. 

9. A. 

11. ^aK 

12. f. 



2. >2. 
4 



3. fa^Vs. 

4. ^ (e"- - 1). 



5 ^■ 
^- 2 



17. 

18. 
19. 
20. 
21. 
22. 


37ra2. 

lira". 

Tr{¥ + 2ab). 

X V3. 

3ira\ 

f 7ro6. 


Page 52 

9. 


2a2(l+|V2). 


10. 


167r3a2. 


11. 


^'{i+H- 


13. 


f(.-i). 


14. 


(lO» + 9V3)g. 


15. 


m'. 


Pages 56, 66 




5. 


^^'+4-4V 



6. Itt. 

7. iira^. 

8. |a2. 



3. iU- 

. irO' 

4- -6"' 6. 1x2. 

7. f xa'(l — cos* a), where a is the radius of the sphere and 2 a 
the vertical angle of the cone 

8. \^xa». 13. fxa'. 

9. Mxa». 14. Jji^x^a^. 
10- Vto^. 15^ 8xV3. 

11. 5ir^aK ^gz 

12. ^ira\ 16. "16 V2. 



Page 69 

2. f a' tan a. 

3. la". ^• 

4. ixo62. 9. ta^A. 

5. ^ira%. j^ 4 ^ 3 

6. .4A. • 3 v'2 '^^ ' 



l-'O+i) 



Answers to Exebcises 179 

Page 63 



2. ^(lOVlO-l). 6. 6a 



S. ln(2 + V3). ^- "(^-ij* 

4. f + ^ln2. 8. 8a. 

5. 2.003. 9. 2x2a. 

Page 64 

3. ^^^^ (e - 1). 5. 2a [V2 + In (l + ^l 

. ±a 6. i^a. 

• Vs' 7. 8a.~ 

8. ixa. 

Pages 66, 67 

\ V a^ - 6* a / 

4. V^a^. 7. V'o*- 

''■2^ 6== + ^; 9. 8x[V2+ln(l + v^)]. 

6. V»-a*. 10. 4xa». 

Page 69 

1. fT»o». 6. fxa'a -cos a). 

2- 2a«. 7. 2a/i(x-2). 

^- 1^°*- _ 8. §xaVA» + 4a^ 

4. ixaKa + 26V3). g ia«A(9x-16). 

5. |xoK2 V2 - l). 

Pages 71, 72 

1. 45,000 lbs. 3. i«*A». 

2. 33,750 lbs. 4. |u'6/j«. 

5. § ipofc*, where a is the semi-axis in the surface and 6 the vertical 
semi-axLs. 

6. 300,000 IP. 7. 40 xw. 

Pages 78-80 

1. i pd'b, where p is the pressure per unit area, a the width, and 6 
the height of the door. 

2. ^wa^b. 4. The intersection of the 

3. ^wbh* {4:C + 3h). medians. 



180 Answers to Exercises 



5. 


(!a,0). 




_ 4a _ 


6. 


x= 77—, y 




Stt' " 


7. 


(a a\ 


8. 


/ 256a\ 
P 3157rj 




_ IT 


9. 


^=8' 


10. 


(^ a, ^ a) 


11. 


(5 16a' 



46 



12. y = ia. 

13. X = Itt V2a. 

14. At distance — from the 

TT 

bounding diameter. 

1^- y- 4e(e^-l) 

16. (^a, fa). 

17. (0.399, 1.520). 

18. y = ia. 

19. On the axis i of the distance from the base to the vertex. 

20. At distance f a from the plane face of the hemisphere, where a 
is the radius, 

21. (iO). 

22. Its distance from the plane face is ^ of the radius. 

23. On the axis at distance f a (1 + cos a) from he vertex, a being; 
the radius and a the angle of the sector. 

24. (|a,0). 

25. The dista'nce of the center of gravity from the base of the cylinder 
is ^j ira tan a. 

26. At the middle of the radius perpendicular to the plane face. 

28. x.:«^^. 
15 v^ - 5 

Pages 82, 83 

2. 2 x^a^b. 6. J xa» [3 hi (1 + V2) - V2). 

3. ^(12V3-1). 7. §7ra3(3a + 2sina). 

4. 7r(36x + V6)V6. 8- '^'«'- 

5. "/Tra^. 

10. § TTo' (4 sin a — sin' a) tan a, where a is the radius of the cylinder 
and 2 a the vertical angle of the cone. 

Pages 84, 85 

1. ^ a%, where b is the edge about which the rectangle is revolved- 

2. T^ bh^, where b is the base and h the altitude. 

3. i b¥, where b is the base and h the altitude. 

4. ^a*. 

6. ^^a*. 
6. ^ ira*. 



Answers to Exercises 181 



7. i Ma^h, where h is the altitude. 

8. iMa\ 

9. — — : — , where p is the density. 

LO 

10. I 3/a*, where M is the mass and a the radius. 
12. I TO* 



Pages 83, 83 

k (b - a)» 



2a 

2. au; ft. lbs., where a is the radius of the earth in feet. 

„ , Vt — h , a a 

3. c In J H • 

Vi — Vt Vi 

4. 25,133 ft. lbs. 

5. ^ TTfiPa, where a is the radius of the shaft. 

6. 2~T ^ - > where A is the altitude of the cylinder. 

7. -T— ( ^ ~ T ) > where a and b are the inner and outer radii. 

kh 

8. —7 , where a and b are the radii of the ends and h the altitude 



t>. 


irab' 


9. 


2xi 


r 


2. 


8.5. 


7. 


0.785392. 


8. 


1.26. 


9. 


4.38. 


10. 


21.48. 



1. In If. 

2. h^a\ 

3. ^. 

5. -1. 

„ TO* 

6- -^- 



10. 2i. 
c 




Pages 95, 96 




11. 4.27. 




12. 0.9045. 


13. aX - 


3L3 5L5"*" * * * 


14. 1.91. 




Pages 102, 103 




7. f. 


15. 4. 


8. ^a\ 


16. |a*. 


9. X. 


17. 16 hi 2 --V 


10. 13§. 


18. fa*. 


11. 3x. 


19. |o*. 


12. X. 


20. (i -f). 


13. h 

14. ia*. 


21. (Va, -2o) 



182 Answers to Exercises 

Pages 107, 108 

, va* . ica^ 

2^ I^. 6. J aM2 a - sin 2 a). 

^ 7. i^aaMeTr-S). 

8. On the bisector at the distance — x from the center. 

o a 

9. ^xo*. 11. aU4ir -If). 
10. \T,a\ 12. 3xa^ 

15. X5 ilfa^, M being the mass and a the radius of the base. 

16. ia3(3^_4). 18, ^^Trpa\ 

i7. l-a3. jg (8V2-9)4^. 

lUo 

Page 111 

1. 3 Vii. 

2. There are two areas between the planes each equal to 2 mofi. 

3. Two areas are determined each equal to ira^ V2. 

4. ixa2 V3. 7. ^iva' (3 Vs - l). 

5. 4. ' 8. a? (tt - 2). 

ۥ 8 a'- 9. 8anan-iiV2. 

Page lie 

1. \. 4. wahc. 

2. A. 6. ^^a^A. 

3. Its distance from the base 7. ^. 

is -/j TTO. 

Page 121 

1- ^s^"-- 4. I^ (2/i2 + 3a«>. 

2. f A, where A is the altitude. * 60 

3. IT. 5. fTra'. 

6. On the axis of the cone at the distance f o (I + cos a) from the 
vertex. 

7. If the two planes are = ± ?, the spherical coordinates of the 

9 T 

■center of gravity are r = T^a, 0=0, <^=k' 

10 ^ 

8. Httos. 



Answers to Exercises 183 



Page 125 
kM 2 kMc ri 1 ~ | 



2. 2M. 

xa- 



4. 2 xArpA (1 — cos a), where p is the density, A the altitude, and 
2 a the vertical angle of the cone. 

6. The components along the edge through the comer are each 
equal to 



2Mk fx , , 2 + >/2l 



Pages 130, 131 

dx^ dx " d^ d3? dx 

Q. X dy — y (x + 1) dx = 0. — 4 y = 0. 

7. ^ + j/ = 0. 9. ydx =xdy. 

Pages 141-143 

10. u=cx* • 

2. tan' X — cot* y = c. x 

3. 1^ + 1 = c (x2 - 1). -? 
.,,,,, 11. u = cx*e *. 

4. x-(^ + i» - ys = c. 19 L, _ ^ . „, 

5. x^ + x-^/-x^/*-y3=c. jg ^ = (i _ ^) (;, + c). 

6. y^ =cx^{y^ + 1). 14 ^ ^ csinx - a. 

7. x' + ^ = ce*"*. 15. 7 1» = y (x' + c). 

8. xy = c (y - 1). 1 , 1 , .^ 

16. — = X + - + «?»* 

■^ ^h-a^' 17. x< + 4y(x»-l)*=c 

18. hi (x» + xy + y*) + A tan'i ^-^^ = c. 

V3 xVs 

19. X* - y* = ex. 25. ys = ce* - X - 1. 

20. y* + 2xy=c. 26. e» = |e* + ce"*. 

21. x*-4x»y + y* = c. „ c x» 

27. y =--— . 

22. 4 = c - e-". "^ ^'^ 

y 28. y = ^ X* + c, or y = og». 

f 29. y* = 2 ex + c*. 

23. e" + In X = c. / , ^ 

24. x + 2y + ln(x + y-2) =c. 30. q = Ec\\ - f Re] , 
31. i =/e~l' + ^i-q^^^ /i!sina/-Lafcosa/-e"iM . 



184 Answers to Exercises 

32. 2/3 = 8 6=^-2. 34. y = cifi. 

33. 2/2 = 2 ax. 

35. X = o In ^ + Vo" - 1/2 + c. 

a + Va2 - j/2 

36. y2 + (x - c)2 = a2. 

37. 2/ = |e^ + |3e--. 

38. r = e". 

39. r = c sin 0. 

40. r = a sec (0 + c). 

41. o0 = Vr2 - a2 - o sec"! - + c. 

a 

42. y = e*. 

43. A circle. 

44. A straight line. 

45. A circle with the fixed point on its circumference or at its center. 

46. The logarithmic spirals r = ce^o. 

47. 0.999964. 

Pages 164^156 

1. y = Ci In X — I a:2 + C2. 

2. y — X + Cixe^ + C2. 

3. y = cie'^ + C2e-°*. 

4. y = Ci sin ax -\- d cos ax. 



6. s = ^ In (cie°*< - g-""") + cj. 
a^ 

^- J/ = l^'-2^1nx + c. 

8. y = 5^ [e<=i*+<'2 + e-i'^'^t)]. 

9. y = ci + c-ifi^'. 

10. 2/ = Cie*^ + Cae"*. 

11. y = (ci + cix) e'^. 

12. J/ = Ci cos X + C2 sin x. 

13. y = Ci + Cje"* + cse'*. 

14. y = cic^ + C2e~* + Cs cos x + C4 sin x. 

15. y = e* [ci cos (x V2) 4- Ci sin (x V2)]. 

1ft -i«r xVs , . xVsl 

16. 2/ = e »* ci cos — 2 f- Ci sm — g— • 



Answers to EIxercises 186 

17. y = ae' + cie-' + Cje"^ -{- Cxer' ^. 

18. y = (ci + cjx + Car*) e*. 

19. !/ = X + 3 + Ci cos X + cj sin X. 

20. y = cic^i + C2€-^* -he'. 

21. y = cie^^ + c^e-^"^ - 5 -c' - tt^ - xk- • ^ 

22. J/ = ce* — J (sin x + cos x). .^ ' ^ 



23. 2/ = Ci + ce*' - i X* + X. y /K/^ \ 

24. y = Cxe-' + (^-^* + ix-A + ^e''. ^ J (^Z ^ 

25. y = cie" + cje"" + #- e". */ '^ 

r a;V3 xVsl 1 *^^ 

26. y = e»* Ci cos — ^ h cj sin , I — y^ (2 sin 2 x + 3 cos 2 x). 

[X v's . X V^~| 

Cx cos — X H cj sin — ^ I — x* + x* — 6. 

28. y = cie* + ce»'- ^e»*sinx. 

29. y = Ci(?' + Cj€~" + jV ^' (6 sin X — cos x). 

30. y = Ci + Cjx + cjx^ + 046"^ + x^y (4 cos 4 X — sin 4 x). 

31. y = Ci cos 2 X + (ci + J x) sin 2 X. 

32. y = e-' (ci + cx + ^x*) + ie*. 

33. X = Ci cos <H-Cj8inf + |(e* — e~*)» 
y = Ci sin < — cj cos < + 3 (e* — e~'). 

34. y — Cx cos < + Cj sin < — 1, 

X = (ci + C2) cos ^ + (cj — Ci) sin < — 3. 

35. X = CiC* + C3€~", 

y = Cie~' + 3 Cje~" + cos I. 

36. X = Cic' + C26~* + cj cos < 4- C4 sin f, 
y = Cie' + CjC"' — C3 cos < — C4 sin t. 

37. y = X. 

38. 2 y* = X + 2. 

39. « = f<+|(e-*'-l). 

^•«=jk^l 2 j- 

41. s = 6 cos {kt). 

42. About 7 miles per second. 

43. About 42? minutes. 

44. < = V-ln (5 + ^^). 

45. < = -^ In (9 + 4 V5). 



TABLE OF INTEGRALS 

u" du = — -— , if n is not — 1. 

n + 1 

2. r^ = lnu. 
J u 

r du \ u 

3. I -;;— — = - tan 1 - • 

J V? -\- a^ a a 

. r du \ , u — a 

4- I —^ -o = ^7-1^^ — \ 

J w — c^ 2 a u -\r a 

5. { e^ du = c". 



6. Ca'^du 



In a 



Integrals of Trigonometric Functions 

7. I sin udu = — costt. 

8. I sin^ udu = lu — lsm2u = I (u — sinu cos m). 

9. 1 sin* w dw = I If — J sin 2 M + ^^j sin 4 u. 

10. j sin^ udu = f'^u — ls\n2u + :^g sin' 2 u + ?? sin 4 u, 

11. J costtdw = sin u. 

12. J cos^ « du = § M + -J- sin 2 M = 5 (w + sin u cos u). 

13. j cos* M (iu = I u + i sin 2 M + -g^j sin 4 u. 

14. j cos^ udu — ^^u + \sm2u — ^g sin' 2 u + ^'j sin 4 u. 

15. I tan udu = —In costt. 

16. j cot udu = In sin u. 



186 



17. 
18. 
19. 
20 
21. 
22. 

23. 
24. 

25. 
26. 
27. 
28. 
29. 
30. 
31. 

32. 
33. 



Table of Integrals 

Jsec udu = In (sec u + tan u) = In tan ( o "I" I )* 

I sec* u du = tan u. 

j.'sec' udu = ^ sec M tan t* + i In (sec t* + tan u). 

esc u du = In (esc u — cot «) = In tan -x* 
I CSC* udu = — cotu. 
jcsc^ udu = — ^cscucotu + |ln (esc u — cot u). 



187 



Integrals containing Va* — u* 
JV^^^rV* du = I Va«-tt» + ^ sin-i ^• 

fu^ V^i^r;r« du = 3 (2 u* - a») Vo* - u* + ^* sin"! -• 

/Va- - u- J ^/-z ;. , I o - "^q ' - «* 
du = Va* — u* + a In 
u u 

/u' du u /— , a* . , u 

.- -, = - :5 ^«* - "' + 2 «^"" n ' 
va- — u- -^ z a 

r du _ 1 , a - V^s^T^ 

J u v'a* - u- " 



In 



u 



/ du _ V g- — 

fid' - u*)' du = '^ (5 a* - 2 u*) V^^^TII^ + ^ sin-» H 
«/ o 5 a 



du 



dh 

(n- — 



(.0- — u*)* a* Va* — u* 



Integrals containing Vu* — a* 

J* ViiT^^i du = I Vu« - o* - I" In (u + Vu^-a*). 

J*u* Vt^^T^ du = I (2 u»-a*) Vi^TT^ ~ ^ ^ ^^+ Vu»-rf). 



188 Table of Integrals 

34 r Vu^ - g^ ^„ ^ ViF^r^^ - a sec"! -• 
J u a 



35. 



J Vl/2 — rt2 



Vyi - o2 



36. ^-^l^^ = HV^73T2 + ^^in(t* + V^ir^^. 
-' V M* — a- '^ ^ 



37. r_^^_. = 1 sec- H. 
•^ w V «« _ a2 a a 

dw V ^2 _ q2 



S-, 



39. J(m« - a2)*du =|(2u2-5a2) V«2 _ ^2 4.§|*i„ (u+ Vu^-a^). 
40 • '^^ 



Integrals containing V^* + a* 

41. J V^^Srp^ '^^ = I Viii"4r^2 + ^ In (u + Vu^+a^). 

42. Ji/2 Vm2 + a^ dtt = I (2 m2 4- a2) Vm2 + «« _ |'ln (u + V^iH^). 

43. r^^:Z±^d. = V^I-F^ + aln^^^^+^^«. 
J u u 

44. f— JL= = ln(«+V^lHr^2). 

*' V U^ -|- ^2 ^ ^ 

46. f /^ ^l,^V^^ + a^-a 
•^ w Vm2 -[- o2 a 

du 



Vm2 -[- o2 a w 



48. J(m* + 0^)3 du = |(2 u2+5 a^) v^To^ +^'ln U+ ^vH^^)' 



49. f ^^ = ^ . 



Table of Integrals 189 

Other Integrals 



60. l\/?^±|dx 



^ 



J \ ax + b 

= - V{ax + b)(j)x + q) 

- ^7^ 1n(Vp(qx + 6) + Va(px + g)) 
a Vap 

= ^ V (ax + b) (px + g) - ^-^^ tan-i :^ -«Pjg^ ^\ 
° aV—ap oVpx + 5 

. et r n* • 1. J e" (a sin 6x — b cos bx) 
^51. Je"sm6xdx^ ^-^ 1. 

-62. fe-cosbxdx = ^"^ (^ ^in bx + a cos 6x) . 



190 



Natural Logarithms 



0-609 



N 





1 


2 


3 


4 


6 


6 


7 


8 


9 







0.0000 


0.6931 


1.0986 


1.3863 


1.6094 


1.7918 


1.9459 


2.0794 


2.1972 


1 


2.3026 


2.3979 


2.4849 


2.5649 


2.6391 


2.7081 


2.7726 


2.8332 


2.8904 


2.9444 


2 


9957 


3.0445 


3.0910 


3.1355 


3.1781 


3.2189 


3.2581 


3.2958 


3.3322 


3.3673 


3 


3.4012 


4340 


4657 


4965 


5264 


5553 


5835 


6109 


6376 


6636 


4 


6889 


7136 


7377 


7612 


7842 


8067 


8286 


8501 


8712 


8918 


5 


9120 


9318 


9512 


9703 


9890 


4.0073 


4.0254 


4.0431 


4.0604 


4.0775 


6 


4.0943 


4.1109 


4.1271 


4.1431 


4.1589 


1744 


1897 


2047 


2195 


2341 


7 


2485 


2627 


2767 


2905 


3041 


3175 


3307 


3438 


3567 


3694 


8 


3820 


3944 


4067 


4188 


4308 


4427 


4543 


4659 


4773 


4886 


9 


4998 


5109 


5218 


5326 


5433 


5539 


5643 


5747 


5850 


5951 


10 


6052 


6151 


6250 


6347 


6444 


6540 


6634 


6728 


6821 


6913 


11 


7005 


7095 


7185 


7274 


7362 


7449 


7536 


7622 


7707 


7791 


12 


7875 


7958 


8040 


8122 


8203 


8283 


8363 


8442 


8520 


8598 


13 


8675 


8752 


8828 


8903 


8978 


9053 


9127 


9200 


9273 


9345 


14 


9416 


9488 


9558 


9628 


9698 


9767 


9836 


9904 


9972 


5 0039 


15 


5.0106 


5.0173 


5.0239 


5.0304 


5.0370 


5.0434 


5.0499 


5.0562 


5.0626 


0689 


16 


0752 


0814 


0876 


0938 


0999 


1059 


1120 


1180 


1240 


1299 


17 


1358 


1417 


1475 


1533 


1.591 


1648 


1705 


1761 


1818 


1874 


18 


1930 


1985 


2040 


2095 


2149 


2204 


2257 


2311 


2364 


2417 


19 


2470 


2523 


2575 


2627 


2679 


2730 


2781 


2832 


2883 


2933 


20 


2983 


3033 


3083 


3132 


3181 


3230 


3279 


3327 


3375 


3423 


21 


3171 


3519 


3566 


3613 


3660 


3706 


3753 


3799 


3845 


3891 


22 


3936 


3982 


4027 


4072 


4116 


4101 


4205 


4250 


4293 


4337 


23 


4381 


4424 


4467 


4510 


4553 


4596 


4638 


4681 


4723 


4765 


24 


4806 


4848 


4889 


4931 


4972 


5013 


5053 


5094 


5134 


5175 


25 


5215 


5255 


5294 


5334 


5373 


5413 


5452 


5491 


5530 


5568 


26 


5607 


5645 


5683 


5722 


5759 


5797 


5835 


5872 


5910 


5947 1' 


27 


5984 


6021 


6058 


6095 


6131 


6168 


6204 


6240 


6276 


6312 


28 


6348 


6384 


6419 


6454 


6490 


6525 


6560 


6595 


6630 


6664 


29 


6699 


6733 


6708 


6802 


6836 


6870 


6904 


6937 


6971 


7004 


30 


7038 


7071 


7104 


7137 


7170 


7203 


7236 


7268 


7301 


7333 


31 


7366 


7398 


7430 


7462 


7494 


7526 


7557 


7589 


7621 


7652 


32 


7683 


7714 


7746 


7777 


7807 


7838 


7869 


7900 


7930 


7961 


33 


7991 


8021 


8051 


8081 


8111 


8141 


8171 


8201 


8230 


8260 


34 


8289 


8319 


8348 


8377 


8406 


8435 


8464 


8493 


8522 


8551 


35 


8579 


8608 


8636 


8665 


8693 


8721 


8749 


8777 


8805 


8833 


36 


8861 


8889 


8916 


8944 


8972 


8999 


9026 


9054 


9081 


9108 


37 


9135 


9162 


9189 


9216 


9243 


9269 


9296 


9322 


9349 


9375 


38 


9402 


9428 


9454 


9480 


9506 


9532 


9558 


9584 


9610 


9636 


39 


9661 


9687 


9713 


9738 


9764 


9789 


9814 


9839 


9865 


9890 


40 


9915 


9940 


9965 


9989 


6.0014 


6.0039 


6.0064 


6.0088 


6.0113 


6.0137 


41 


6,0162 


6.0186 


6.0210 


6.0234 


0259 


0283 


0307 


0331 


0355 


0379 


42 


0403 


0426 


0450 


0474 


0497 


0521 


0544 


0568 


0591 


0615 


43 


0638 


0661 


0684 


0707 


0730 


0753 


0776 


0799 


0822 


0845 


44 


0868 


0890 


0913 


0936 


0958 


0981 


1003 


1026 


1048 


1070 


45 


1092 


1115 


1137 


1159 


1181 


1203 


1225 


1247 


1269 


1291 


46 


1312 


1334 


1366 


1377 


1399 


1420 


1442 


1463 


1485 


1506 


47 


1527 


1549 


1570 


1591 


1612 


1633 


1654 


1675 


1696 


1717 


48 


1738 


1759 


1779 


1800 


1821 


1841 


1862 


1883 


1903 


1924 


49 


1944 


1964 


1985 


2005 


2025 


2046 


2066 


2086 


2106 


2126 


60 


2146 


2166 


2186 


2206 


2226 


2246 


2265 


2285 


2305 


2324 


N 





1 


2 


3 


4 


6 


6 


7 


8 


9 



Natural Logarithms 



191 



60O-10O9 



N 





1 


2 


3 


4 


5 


6 


7 


8 


9 


50 


6.2146 


6.2166 


6.2186 


6.2206 


6.2226 


6 2246 


6.2265 


6 2285 


6.23a5 


6 2324 


51 


2344 


2364 


2383 


2403 


2422 


2442 


2461 


2480 


2500 


2519 


52 


2538 


2558 


2577 


2596 


2615 


2634 


2653 


2672 


2691 


2710 


53 


2729 


2748 


2766 


2785 


2804 


2823 


2841 


2860 


2879 


2897 


54 


2916 


2934 


2953 


2971 


2989 


3008 


3026 


3044 


3063 


3081 


55 


3099 


3117 


3135 


3154 


3172 


3190 


3208 


3226 


3244 


3261 


56 


3279 


3297 


3315 


3333 


3351 


3368 


3386 


3404 


3421 


3439 


57 


3456 


3474 


3491 


3509 


3526 


3544 


3561 


3578 


3596 


3613 


58 


3630 


3648 


3665 


3682 


3699 


3716 


3733 


3750 


3767 


3784 


59 


3801 


3818 


3835 


3852 


3869 


3886 


3902 


3919 


3936 


3953. 


60 


3969 


3986 


4003 


4019 


4036 


4052 


4069 


4085 


4102 


4118 


61 


4135 


4151 


4167 


4184 


4200 


4216 


4232 


4249 


4265 


4281 


62 


4297 


4313 


4329 


4345 


4362 


4378 


4394 


4409 


4425 


4441 


63 


4457 


4473 


4489 


4505 


4520 


4536 


4552 


4568 


4583 


4599 


64 


4615 


4630 


4616 


4661 


4677 


4693 


4708 


4723 


4739 


4754 


65 


4770 


4785 


4800 


4816 


4831 


4»46 


4862 


4877 


4892 


4907 


66 


4922 


4938 


4953 


4968 


4983 


4998 


5013 


5028 


5043 


5053 


67 


5073 


5088 


5103 


5117 


5132 


5147 


5162 


5177 


5191 


5206 


68 


5221 


5236 


52.50 


5265 


5280 


5294 


5309 


5323 


5338 


5352 


69 


5367 


5381 


5396 


5410 


5425 


5439 


M53 


5468 


5482 


5497 


70 


5511 


5525 


5539 


5554 


5568 


5582 


5596 


5610 


5624 


5639 


71 


5653 


5667 


5681 


5695 


5709 


5723 


5737 


5751 


5765 


5779 


72 


5793 


5806 


5820 


5834 


5*48 


5862 


5876 


5889 


5903 


5917 


73 


5930 


5944 


5958 


5971 


5985 


5999 


6012 


6026 


6039 


6053 


74 


6067 


6080 


6093 


6107 


6120 


6134 


6147 


6161 


6174 


6187 


75 


6201 


6214 


6227 


6241 


6254 


6267 


6280 


6294 


6307 


6320 


76 


6333 


6346 


6359 


6373 


6386 


6399 


6412 


6425 


fr438 


6451 


77 


6464 


6477 


6490 


6503 


6516 


6529 


6542 


6554 


6567 


6580 


78 


6503 


6006 


6619 


6631 


6644 


66.57 


6670 


6682 


6695 


6708 


79 


6720 


6733 


6746 


6758 


6771 


6783 


6796 


6809 


6821 


6834 


80 


6846 


6859 


6871 


6884 


6896 


6908 


6921 


6933 


6946 


6958 


81 


6970 


6983 


6995 


7007 


7020 


7a32 


7044 


7056 


7069 


7081 


82 


7093 


7105 


7117 


7130 


7142 


7154 


7166 


7178 


7190 


7202 


83 


7214 


7226 


7238 


7250 


7262 


7274 


72S6 


7298 


7310 


7322 


84 


7334 


7346 


7358 


7370 


7382 


7393 


7405 


7417 


7429 


7441 


85 


7452 


7464 


7476 


7488 


7499 


7511 


7523 


7534 


7546 


7558 


86 


7569 


75S1 


7593 


7604 


7616 


7627 


7639 


7650 


7662 


7673 


87 


7685 


7696 


7708 


7719 


7731 


7742 


7754 


7765 


7776 


7788 


88 


7799 


7811 


7822 


7833 


7845 


7856 


7867 


7878 


7890 


7901 


89 


7912 


7923 


7935 


7946 


7957 


7968 


7979 


7991 


8002 


8013 


90 


8024 


8035 


8046 


8057 


8068 


8079 


8090 


8101 


8112 


8123 


91 


8134 


8145 


8156 


8167 


8178 


8189 


8200 


8211 


8222 


8233 


92 


8244 


8255 


8265 


8276 


8287 


8298 


8309 


8320 


8330 


8341 


93 


8352 


8363 


8373 


8384 


8395 


^405 


8416 


8427 


8437 


8448 


94 


8159 


8469 


8180 


8491 


8501 


8512 


8522 


8533 


8544 


8554 


95 


8565 


8575 


8586 


8596 


8607 


8617 


8628 


8638 


8648 


8659 


96 


8669 


8680 


8690 


8701 


8711 


8721 


8732 


8742 


8752 


8763 


97 


8773 


8783 


8794 


8804 


8814 


8824 


8835 


8845 


8855 


8865 


98 


8876 


8886 


8896 


8906 


8916 


8926 


8937 


8947 


8957 


8967 


99 


8977 


8987 


8997 


9007 


9017 


9027 


9037 


9048 


9058 


9068 


100 


9078 


9088 


9098 


9108 


9117 


9127 


9137 


9147 


9157 


9167 


N 





1 


2 


3 


4 


6 


6 


7 


8 


9 



i 



INDEX 



The numbers refer to the pages. 



Approximate methods, 90-96. 
Area, by double integration, 97. 

bounded by a plane curve, 47-52. 

derivative of, 39. 

of a surface of revolution, 65. 

of any surface, 108. 
Attraction, 121. 

Center of gravity, 73-80. 
Change of variable, 44. 
Constants of integration, 1, 128. 
Curves with a given slope, 7. 
Cylindrical coordinates, 116. 

Definite integrals, 36. 

properties of, 41. 
Differential equations, 126-156. 

exact, 133. 

homogeneoas, 139. 

linear, 136, 147. 

of the second order, 143. 

reducible to linear, 137. 

simultaneous, 153. 

solutions of, 128. 

with variables separable, 132. 

Exact differential equations, 133. 

Formulas of integration, 14. 

Homogeneous differential equa- 
tions, 139. 

Infinite limits, 42. 

Infinite values of the inteerand, 43. 



Integral, definite, 36. 
definition of, 1. 
double, 97. 
indefinite, 36. 
triple, 112. 
Integrals, containing ax- + 6x -f- 
c, 19. 

p 
containing (ax + &) « , 29. 
of rational fractions, 26-29. 
of trigonometric functions, 21— 

23. 
relation of definite and indefinite, 
40. 
Integrating factors, 135 
Integration, 1. 

by substitution, 15-19. 

constant of, 1. 

formulas of, 14. 

geometrical representation of, 

36. 
in series, 94. 
of rational fractions, 23-29. 

Length of a curve, 61, 63. 
Limits of integration, 36, 42. 
Linear differential equations, 136, 
147. 

Mechanical and physical applica> 

tions, 70-89. 
Moment, 72. 

of inertia, 83. 
Motion of a particle, 5. 



193 



194 



Index 



Order of a differential equation, 
126. 

Pappus's theorems, 80. 
Physical and mechanical applica- 
tions, 70-89. 
Pressure, 70. 
Prismoidal formula, 90. 
Polar coordinates, 50, 63, 103. 

national fractions, integration of, 

26-29. 
Reduction formulas, 33. 

Separation of the variables, 9, 132. 
Simpson's rule, 93. 



Spherical coordinates, 119. 
Summation, 35. 

double, 100. 

triple, 113. 

Trigonometric functions, integrals 

of, 21-23. 
Trigonometric substitutions, 23. 

Variables, separation of, 9, 132. 
Volume, by double integration, 99. 

of a solid of revolution, 52. 

of a soUd with given area of sec- 
tion, 56. 

Work, 85. 







^<;q'^~V 



WNOING SECT. AUG 2 6 1971 



QA 
P5 



Phillips, Henry Bayard 
Differential calculus 



Applier' 
Physic 



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