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WORKS OF H. B. PHILLIPS, PH.D.
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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 3238
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 94100
X. Series and Approximations 101112
XI. Partl\l Differentiation 113139
Supplementary Exercises 140153
Answers 154160
Index 161162
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), fz (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
11
110*
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
11 ^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; ifl^
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 = 12= 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.
Exampie. 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 21 = 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 = x33xt2,
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 xaxis? 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 yaxis is vertical, the xaxis 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 xaxis at A
and the yaxis 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 = ui Auiv + 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 yT Mr
r^ = Ax Ax
Ax — , 1 — 7T"*
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
Pi 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) (2x0) + (x262) (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
Vx21
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 + ix2y)dy = 0.
Consequently,
dy ^ 2x + y
dx 2y — X
Ex.7. x = t{j, y = t.
In this case
dx = dtrp, dy = dt{^*
Consequently,
dy ^ ^ t^ ^ fhl
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 ^ITig
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)(23i)»(2x3)',
^ = (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. 2x23xy + 4t/» = 3x,
ax
dy _ 4x  3y  3
dx ~ 3x 8y
25.  + ?^ = 1, ydxxdy = 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=<Vi2l, y = <+V<2l, 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 xaxis.
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 = 2x32/
intersects the xaxis at the origin.
39. A line through the point (1,2) cuts the za^ 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^
Exampie 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. 'KC/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
^ = 520:.
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 twothirds 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 onevalued 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 • fcj5»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 ofS 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
sini 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'(3xl),
2. y=coe(l?).
5. y = 8ec» V4 1 + 1,
ft 1 . 3 .
V 7. y = tani'
dy =
dy =
Zdx
v6x  9i»
rfx
V2axx*
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 + 2x4x»
dx~ 3? +a*'
d9 ^•
^^ ^
<** (o»  x») Va»  2 1»*
dy 1
11. y = §V^^^ + ^'Bin», ^ =
< o i I 4 sin I
12. y = tani 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 coti ^^^±^,
rfy^ i
dx (3x + l)* + 4(3x4l)**
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 3x dx (3_a;)V32xx»
1ft _ ^x'  g* 1_ _jX dy 1
"• ^~ 2a^x« 2a»^'' o' dx ~ ^a v^TTT^j*
19
, ,x + Vi2 + 4x4 '^I/.
I. y = tani ^ 1 ^Z 
2 <te xVx2+4x4
20. y = X sini x + Vl  x^, ^ = , ^
dx2 Vl x'
21. y = x2 seci I  2 Vx^^Tl^ ^ = 2xseci
Z uX i2
22. Let 8 he the arc from the xaxis 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 yaxis
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.
2taniiin2da;
EXERCISES
I. 2/ = ea;,
\/ 2. y = a*^'^^*^
zl
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
6x45
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 tani % 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=^(21nx3).
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 iaxis.
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
yyi=  — (^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(xl).
63
Fig. 49.
64
DIFFERENTIAL CALCULUS
Chap. VII.
and the equation of the normal is
y2 = A(xl).
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
yyi = (.^ ^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 counterclockwise 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 =
21
whence
1
3'
1+2
/3i = tan4 = 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 hjperbola 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 xaxis.
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 xaxis at A'^.
Show that the projection of PN on the xaxis 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 xaxis, 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 tani •
^ 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! = l23 = 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.
QP 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 xaxis 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 hjpotenuse 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 tangents 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 xaxis
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 xaxis.
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 , ,. 1cosA^
\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
Xaxis 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 = 5B ' (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 = ^^°^chord = ^
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
xxi _ y yi _ zzi ,^„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\2dy2dz = 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
24 + 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 xcomponent 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 xcomponent a, ^/component h, and 2component 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], VzVt= [0201,626,].
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 yaxes. If the
end in the xaxis 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 xaxis 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 xaxis.
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 xaxis 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
x0 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^6sinx6x + 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 — : rr— 
.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(xa)/'(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){xa)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) + (xa)r(ay
+ (^V'(a)+...+^V"(xO.
which was to be proved.
Example. Prove
hix=(xl) 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(xl)i(xl)2 + ?(xl)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) + ixa)ria)
(x  ay „ , . (x  g)"^ . , , .
+ ^j / (a) + • • • + (^jJTiyr /" («)'
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(j5) + 2(j5)'= 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 = 00^25.
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 TVi + — 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+ +3^27183
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) + {xa)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=l2 + 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 = l2! + 4!~6!"^ ' ' * '
which was to be proved.
EXERCISES
1. smx = x3J + ^yj +  • • .
  , ^ 2x» 4x« 4x6 ,8x^,
6. (a + x)" = a» + no»»x + " %7 ^^ """^ + • • • , if xl*< o.
8. lnx = ln3+^^^^' + ^^^ .iflx3I<l.
9. ln(x + 5)ln6+^^^^'+^^' ,ifxl<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
11 + 11 + 11+.. .
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\
\ba\
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
8mx=x + ^ + .^+^.^.gy+...
and that the series converges when x < L
6. By a method similar to that used in Ex. 4, show that
tani x = x5+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
=2x42 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
^=2x2 = 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^xy) = u\x~yy
^ ' 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 = 2xdx2^ 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. tani  + tani •
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 xyplane, 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 af^ 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~ + 44== 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 xaxis 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 + (x2y) 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 _ yyr _ 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{xXx)^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 22
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 xyplane.
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 xaxis 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, twothirds 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)'H2 6(ax+b)"+i „_,
^ o2 (n + 2) a^ (n + 1) ' 10 a; (x" + n) » '
Find T in each of the following cases:
11. 2x24x2/43^2 = 6x4y + 18.
12. x^ + 3 x«7/ = 2/3.
13. x = Sy^ + 2yK
14. (x2 + 2/2)2 = 2a2(x22/«).
15. X = 1 + ^, y = 2t ^
i1' ^ " (<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 = —vj.
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 = 45/.
''•'(?+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 + cscx = 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. BinV/^ ^^2. ac8C^ + V^^^.
96. cos (^). 103. pq.cot
97. tan
tan
(^)
104. Vl — X sini X — "n/x
,. ,x + 1 , . .z — 1
 ^ , , lOo. seci r+sin * — : •
2 2x + l x1 x + 1
VS"^ ^^ 106. sm^ ;+^^^^ 
_ + a cos X
99. cos^^^py jQ. 1 osx + ^vT:^
V5 ^
100. CSC ' 2xl ' 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)xi.
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)+tani^^.
126. sec§xtanx + ln(secx + tanx).
146 DIFFERENTIAL CALCULUS
127. x seci I /^x + ]  In (x2 + 1) .
128. I In f I x2 + 1 )  I a; + tani 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 i1,1).
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 = 62/22i/».
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
^ va 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 %ini 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/ = 49<.
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 xaxis.
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
xi1
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 = olcos2x 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. tani (t^)
197. hi (1.2). 200. esc (31°).
201. Calculate tt by expanding tan~* x and using the formula
\ = tani (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)=fg + 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 + J3 + ^ + ...].
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 +2x43x2 + 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(gg)..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* tani ^  y^ tani •
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 xaxis 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 xaxis 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. tani.
Page 31
1 2 4 9 ^ «'
{xiy (xl)» • V^rr^j* (a2x2)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' xr (xl)3
8.
^ y^ x* 3xij/S
X 1
6
2y' 42/»
154
ANSWERS TO EXERCISES 155
13.
^ 12 gj _ 12
dl» ~ U2 + 30'' rfy* U230*'
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 4345
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 twothirds 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. 2mi./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. 2yx = 5, y + 2x = 0.
2. 7/ + 4x = 8, 4yx = 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, 2y = 2. 12. 90°.
8. x + 3i/ = 4, y3x = 28. 13. tan»2V2.
9. y \xX&a.\4>x = a<iniaiih<h _, In 10  1
10. 90°, tanif. ^^' ^^ In 10 + 1'
11 45°. 15. tani3V3.
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 xaxis and yi the ordinate ct'
the end in the yaxis, the velocity of the middle point is
the upper signs being used if the end in the xaxis 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 xaxis 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. 3V34.
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(z2)=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: + 52/32l =0.
a;5 7/lzl cicn
5. ^rr'"~^ = ~6~' a;42/6z + 5=0.
6. x + z = yz = 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,
118120.
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, 124127.
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, 1931.
of transcendental functions, 49
62.
partial, 120.
Differentiation, of algebraic func
tions, 1931
Differentiation, of transcendoital
functions, 4962.
partial, 113139.
Directional derivative, 129.
Direction cosines, 80, 81.
Direction of curvatiu^, 67.
Divergence of infinite series, 107
111.
Exact differentials, 130, 131.
Exponential fimctions, 5662.
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, 1931.
exfKinential, 5662.
inverse trigonometric, 5456.
logarithmic, 5662.
transcendental, 2, 4962.
trigonometric, 4953.
Ftmctional notation, 3.
Geometrical apphcations, 6384.
ImpUcit functions, 2, 127.
Increment, 10.
of a function, 14, 15, 118, 119.
Independent variable, 2.
Indeterminate forms, 95100.
Infinitesimal, 7.
Infinite series, 106112.
convergence and divergence of,
107111.
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, 49.
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, 3948.
several variables, 136138.
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, 113139.
differential, 120.
Plane, tangent, 134.
Point of inflection, 67, 68.
Polar coordinates, 7779.
Power series, 110, 111.
operations with, 111.
Rate of change, 32.
Rates, 3238.
related, 35.
Related rates, 35.
RoUe's theorem, 94.
Series, 106112.
convergence and divergence of,
107111.
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, 8183.
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, 8589.
along a straight line, 32.
angular, 34.
\
V
r
WORKS OF H. B. PHILLIPS, PH.D.
PUBLISHED BY
JOHN WILEY & SONS, Inc.
Differential Equations. Second Edition, Rewritten.
vi+116 pages.
$1.50 net.
Analytic Geometry.
vii + 197 pages.
$175 net.
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V + 162 pages.
$1.50 net.
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5 by 7}4. Illustrated. Cloth,
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Differential and Integral Calculus.
In one volume. $2.50 net.
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.
525
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 twopage
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 113 ^
I. Integration
II. Formulas and Methods of Integration 1434
III. Definite Integrals 3546
IV. Simple Areas and Volumes 4/59
V. Other Geometrical Applications 6069
VI. Mechanical and Phtsical Appucations 7089
VII. Approximate Methods 9096
VIII. Double Integration 97111
IX. Triple Integration 112125
X. Differential Equations 126lo6
Supplementary Exercises 157170
Answers 171185
Table of Integrals 186189
Table of Natural Logarithms 190191
Index.
193194
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 + Ax22x + 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
4x42 ^ , 4
2x 1 ' 2xl
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)y2) ^^^ • //,;_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
= g0 + 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 xaxis, and the vertical line as
yaxis. 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 3inch 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 xaxis 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
r74i= = 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 V23x»
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 V2Zx^ J Va2  u^ ^J Va2  u^
= VSsini  + C = V3 sini^^ + 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^4C = ^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. ' ^ ^4x"
/• 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
secxdx , _ r^'dx
40.
/ secxdx ^ ^ rj
tanx Vtan^x l' * ^ 1 + e*'
Vlcosd" 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 V2dxx^ J
Ex.3. ^ (2^1)^^
, = 2sini f7^ + C.
/;
V4 x2 + 4 a: + 2
Since the numerator contains the first power of x, we
resolve the integral into two parts,
dx
r (2xl)dx ^1 r (Sx\4:)dx ^ C
J V4rr2 44 3:42 4 J V4rr2 44 3:42 J ^
V4x2f4a:+2 4«^ V 4x^+4 xh2 ^ 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 / . „ , . rx
>/4x2+4x + 2 2
 In (2x f lf V4x2 + 4x + 2) + C.
EXERCISES
1 r__— ^£__ 7 r (2 X f 5) dx
' Jx2f6x413" J4x«— 4x— 2*
J V9 1 4 r J. tJ
(2 X  1) dx
^ •^V 2H4x4x i' ^Vax^ex + i
3^ C dx ^ J 3 x2 + 2 X h 2
>•/
X dx
j2x
(2x
rf:c "" J (2x 41)^4x^1 4 x1
VS x2 + 4 X h 2 f (2 X 1 3) dx.
/;
VI+6X5X' 11. r .f'^r^V'^,. 
J (x^— 2x 4 3)1
•^ (x 3) V2x2 12x415 ^2. J y
6
dx.
dx ^^ r e'dx
• J (io)(xf 6)" ^^' J
(io)(x46) c/ 2e*^43e*— 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 = ^sini+^Va2x2 + 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"S0 = ^/^^^'^^^
= iT^ (^ + sin cos e) + C.
Since
a; = otan0, = tan^, sin cos = ti — :.■
' a a^ \ x^
Hence
/ dx _ 11" _i 5 , «a; "I ^
(a;2 + a2)22a3L 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^ 2xS
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'2xZ ' ' x22a;+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) x3"^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
= xl +
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\2BC)x2A
are identical. That is
A+B + C = 3, A\2BC = 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+^^ + ^^^
= x2x 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:. Cf^A'^.dx.
(X2+ 1)2'
Art 14 I^'TEGRAI;s 29
Assume
x' + 1 ^A BxhC 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 , + ^ tani 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
(2x3)^+l
30
Formulas and Methods of Integration
Chap. 2
To rationalize both (2 a; — 3)* and (2 a; — 3)», let
2xZ = z\ Then
J (2a; 3^ + 1 J 2' + 1 J\ 2'+!
dz
= 3(y  f + 1  2 + tani2) + 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
(2xl)(2x3)(2x5)'
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 Vo2x2 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. Jsinixdx. 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+()2Axh
(1)2 Ax + (1)2 Ax
= «/Ax = ^3.i = 1.97.
Ex. 2. Find approximately the area bounded by the
Xaxis, 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 icaxis.
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
Xaxis 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 Xaxis, 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
J2t
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 xaxis, 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
taniO = 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 Chai. 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 {xl)^dx.
. (^ xdx  ra\a2( £ _£\
^4 Vj2 + 144 Jo >• ^
^ sin' do. 10. J^ ;^
46 Definite Integrals Chap. 3
J 13. f ek'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 ^^. —; rTf 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 Xaxis 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 yaxis 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 + yy^)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 \/l6x2da: = ^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 xaxis,
and the ordinates x = 1, x = 3.
.^. 2. Find the area bounded by the parabola y = 3 x, the yaxis, 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
Xaxis.
.y 6. Find the area bounded by y = In x, the xaxis, 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 Xaxis and the ordinates at its ends, is equal to the area bounded by
the same arc, the yaxis 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 xaxis 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
yaxis.
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 tiuns 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
Xaxis.
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 irydx. (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 xaxis.
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^ = bx=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 xaxis.
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 =
xa'
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 zaxis the area
X)unded by the iaxis 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 xaxis the
area bounded by the catenary y = x ye" j e"y , the xaxis 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 yaxis, 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 xaxis.
12. Find the volume generated by revolving the curve
X = o cos' 0, 2/ = o sin' <>
about the yaxis.
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 xaxis.
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 xaxis at the distance x
from the center is an ellipse
^ J ?! = 1 _ ^
521^2 A o2*
58 Simple Areas and Volumes Chap. 4
The semiaxes 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 cjlindrical 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
cjlinders.
10. .\ circle moves with its center on the 2axis and its plane parallel
to a fixed plane inclined at 45° to the 2axis. 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
Onefourth 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 xaxis.
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 yaxis the part of the curve y = \ — x*
above the xaxis.
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
yaxis the area bounded by the parabola x^ = y — 1, the
Xaxis, and the ordinates a: = ±1.
Resolve the area into slices by ordinates at distances Aar
apart. When revolved about the yaxis, 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 onefourth 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^ = Jasmddd.
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 xaxis. 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 jhV2oh^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 semiellipse 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
onefourth 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 xaxis 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 Xaxis 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 yaxis (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 Xaxis 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 xaxis 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;yplane 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 xyplane
! 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^ijiyfydy =
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 xaxis bounded
by the curve x* + y = a*.
9. Find the center of gravity of the area bounded by the xaxis 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 xaxis 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 xaxis. 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 Xaxis. 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 Xaxis. 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 = 27ra2 7r6 = 4:irab.
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 crosssection 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 crosssection of the band is a semiellipse, 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 xaxis 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 xaxis.
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=(ar).
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 xaxis.
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 xaxis. Find its moment of inertia about the xaxis.
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
keiCz
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 crosssection 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.
Exampie 1. Find the
area bounded by the
Xaxis, 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 = eS 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 xaxis.
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 crosssections 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 xaxis 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.^ + 4i + ^
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 xaxis.
11. Volume of the solid generated by revolving about the xaxis 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 rcaxis. The area is
y2
/2a f2ay 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 xyplane.
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 xyplane.
z
Fig. 47a.
Fig. 476.
Fig. 476 shows onefourth 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
J2\^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
Xaxis. 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(yi4)
11 dydx
«/2 t/ J (1^24)
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 ekr^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 xaxis.
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 xyplane.
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 xaxis.
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 V2aii2 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 onefourth of the required volume. Take
a system of polar coordinates in the xyplane. 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 zrdddr 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 yaxis of the area within
the cucle {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 xyplane.
17. Find the volume bounded by the xyplane, 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 2axis 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 xaxis
or waxis 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^zplane. 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 yzplane. 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 zyplane bounded
by the planes j/ = 1, x = 2.
6. Find the area of the surface of the cylinder x^ + y^ = 2ax
between the lyplane 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 oneeighth 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 xcoordinate 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 2axis. The distance of any
point (x, y, z) from the 2axis 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 xyplane. 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 2axis 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 xaxis,
Fig. 53c. The square of the distance from the element PQ
to the Xaxis 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 xyplane, 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 2axis
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— (^Hi*) and the
xyplane.
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 xyplane.
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 yends 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 yaxis. The volume of such a
ring is
2 irx dx dy
and its mass is
M 2M
dm = — 57 '2Kxdxdy = —zjxdx 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'\(chh)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 xaxis 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
aJff_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')dx2xydy = 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. ycix + ' ^ 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 _ A2) =^ + 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. (\iiy2x)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
multipljing 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')dxixy\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 +
ric
and
X = 1+ 2/2 I c Vl +?/2.
64. Equations Reducible to Linear Form. — An equation
of the form
^4Pi/ = 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~^,
y23x2 + ^^'
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) =tf{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* + 2xyy*)dx+(x»2xy3t/*)di/ = 0.
6. ;c^t/ = y». 12. (2x2/»y)dx + xrfy = 0.
7. xdx + ydy = a(x» + y»)dy. 13. (i_a^)^+2xi/ = (lx*)».
, 14. tan x— — y = a.
#v dy fc_ dx
9. f — ay = er'.
10. Jf?2xv = 3. W.x3,+xV.O.
dx
11. x*g2xy = 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) dx2xydy = 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
Lr, + 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 onehalf 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, xaxis, 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 + 14p^ = 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' {12 — 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
{Da)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
iDa){Dh) =D^ (aib)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
(Dr^) • • • (Z)r„) . {Dn)y=iDr,) ■ • • (Dr„).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
{DT,){Dr^) = {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^D2)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' + C2e2*.
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 sohing 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)x3y = t,
{D\2)y3x = ^'.
To ehminate y, multipU' the first equation by Z) + 2 and the
second by 3. The result is
(D + 2)2x3(Z) + 2)i/= 1+2^,
3(D+2)i/9x = 3€2'.
154 Differential Equations Chap. 10
Adding, we get
[(D 42)2 9]a: = l+2« + 3e2'.
The solution of this equation is
x= MH + 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. ga=!, = e".
14. ^=V. 27. g ,.. >..
«S^ + ^''='> 28.g4+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=x2y + l,
„ . dx dy , n ■ s
3^ ■*dii + ^^ = ^"^''
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*(il)(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*
/
^ (2xl) 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. JV32xx^dx. 60. /^^^
40. r(a3x?)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 (xl)(x2)(x43)
^2 f ^^ 70 f (3x5)dx
■ ^ x^ V^32 "^ 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 xaxis 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 cxuve 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 yaxis the
area bounded by the curve y^ = 3? and the line x = 4.
111. Find the volume generated by rotating about the yaxis the area
between the xaxis 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 xaxia 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 xaxis. 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 semiaxes 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^ = i2yl)(y2r.
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 yaxis.
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 xaxis.
133. Find the area generated by revolving the loop of the curve
9 X* = (2 J/  1) (i/  2)2 about the xaxis.
131. Find the volume generated by revolving the area within the
ciu^e j/2 = X* (1 — x^) about the yaxis.
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 xaxis.
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 xaxis.
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 xaxis 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 xaxis of the area bounded
by the xaxis 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 xaxis and the parabola ^^ = 4 ax — x
is revolved about the xaxis. Find the moment of inertia about the
Xaxis 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 Vj5 + 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 xaxis 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 xyplane the paraboloid z = x { if
and the cylinder x* + y^ = 4.
193. Volmne boimded by the xyplane the paraboloid z = x + y^
and the plane z — 2 x \ 2y.
194. Center of gravity of the soUd bounded by the xzplane, the
cylinder x \ z = o*, and the plane x •\ y { z = \ a.
195. Volume generated by rotating about the xaxis 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 xaxis.
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 = 2zZ.
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 xaxis.
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 zaxis 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 xyplane, 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 + {xxy)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 + 6y5)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^(2I3)+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. ix3ln(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^ _
12x22t/ + 31 =0.
6. About 53 mUes. 10. (i ^)
11. 6y =
12. 12
7. x = jV3fi,y = ^fi + Vot. 11. 6i/ = x33x2 + 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 • f2x3\
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. iseci^ + C!
28. itani2y + 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^  2smi  + C.
34. 2 V:?T4 + 3ln (x + Vi« + 4) + C.
35. 5ln(4x25)+^ln^=^ + C.
8 ' V5 2 X + V 5
36. J V3x»9  = In (x + V?"=^) + C.
V3
37. ein.(^^)+C. 46. i.>^ + e.
38. V2sin«x + C' 47. ^ (e*«  e"**) + 2 x + C
39. tani (sin x)+C. 48. i In (1 + e^*) + C.
40. seci (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^x8in«x + C. 53 1 ^^_, ^^^ ^ ^
1 _ X* a
45. 2 sin ' ^ + C. 54 tani (e*) + C.
Page 20
^l*i^ + 3.„ l._, 2xl,
3. ^ In (3 X + 2 + V9x* + 12x + 6) + C.
V3
. 1 . _, (2x 1) V5 . ^ . 1 _, (x3)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) + ^ tani ^^^ + C.
6 ^ 3 V 5 V 5
10. 4= seci ^^ .t^ + ^ln (2x + 1 + V4x'^ + 4xl) + C.
V2 V2 2
11. — =J=+C.
Vx2  2 X + 3
12. Vx2  X  2 + I In (2 X  1 + V4x2 4x8) + 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.
ax
38.
X  2
\/x24x + 5 + iln(x2 + V'x*4x+5) + C.
2 ^ ^^ ' " ' 2
39. ^^  ^^ V 2  .x  4x^ + ^ sin^ ^^ + ^ + C
32 ^•^^64 3 ^^'
Page 30
1. ^ + 4x2In(x 1) + 121n(x2) +C.
2. 31nx ln(x + 1) 4C.
3 j^(x_i^(x+2)+c.
4. l + lnx jgln(2xl)^ln(2x + l)+C.
5. f In (X + 3)  i In (X + 1)  f In (X + 5) + C.
6. ^ln(2x 1) 31n(2x3) +ln(2x5) + C.
I . X \ h In h C.
J> X
376 Answers to Exercises
8. lln(x + l)+ln(xl)2^^j+C.
»Ur4^+'"l^)+
10. x81n(x + l) 3(^ + 1)3 +g
13 2(4^ + ^
14. x + ^ln^^Kaniz + C.
4 X + 1 2
„ 1, J + 1 1 , ,2xl . 
15. 5 In , ^: H — tani 7= h C,
3 Vx2  X + 1 Vs Va
16. i In (x3 + 1) + C.
17 1 . 1, X 1 2V3^^ , 2xl . ^
6(x + l) 4 x + l 9 \/3
19. ^+lln— "=^ + 1 tanx^ + C.
a^  8 6 Vx2 + 2 X + 4 2 V3 V3
20. 3 (x + D* + In [(x 4 D*  1]  ^^ tani ^ (x + 1 ) +1 ^ ^^^
v3
21. 1 A+1 i_l i_:,A+llnl±£5+c.
1x^
5
22. 3,(ax + 6)i»A(a, + ,)i + C.
23. 2 Vx + 2  In (x + 3)  2 tani V^T2 + C.
24. 4x* + 21n (x*  l) + In (x* + l)  2tan» x* + C.
25. Hx + l)* + i(xl)« + 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. ^^tanixx + C.
6. X In (x + Va» + 3?)  Vo« + x« + C.
Answebs to Exercises 177
6. 2 VF^^ In X  4 Vj  1 + 4 taiii Vx  1 + C.
7. Inxln(lnx) Inx + C.
8. ^ x»8ecix  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. ^(28in3x3coe3x) +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 (lln2)
178 Answers to Exercises
Pages 49, 50
1. 11. 13. 27r + , 67rf
2. V3(4V2). 14. 4V3 4Y'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.
^^'+44V
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. ^(lOVlOl). 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(x2).
^ 1^°* _ 8. §xaVA» + 4a^
4. ixaKa + 26V3). g ia«A(9x16).
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 semiaxis in the surface and 6 the vertical
semiaxLs.
6. 300,000 IP. 7. 40 xw.
Pages 7880
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. ^(12V31). 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^aaMeTrS).
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. la3. jg (8V29)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).
€• 8 a' 9. 8ananiiV2.
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 141143
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 +  + «?»*
■^ ^ha^' 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 + y2) =c. 30. q = Ec\\  f Re] ,
31. i =/e~l' + ^iq^^^ /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 + ei'^'^t)].
9. y = ci + cifi^'.
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' + (^^* + ixA + ^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 <HCj8inf + (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. < = Vln (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 = ^71^^ — \
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» + ^ sini ^•
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 =(2u25a2) 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^IF^ + 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)  ^^^ tani :^ «Pjg^ ^\
° aV—ap oVpx + 5
. et r n* • 1. J e" (a sin 6x — b cos bx)
^51. Je"sm6xdx^ ^^ 1.
62. fecosbxdx = ^"^ (^ ^in bx + a cos 6x) .
190
Natural Logarithms
0609
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
60O10O9
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, 9096.
Area, by double integration, 97.
bounded by a plane curve, 4752.
derivative of, 39.
of a surface of revolution, 65.
of any surface, 108.
Attraction, 121.
Center of gravity, 7380.
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, 126156.
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, 2629.
of trigonometric functions, 21—
23.
relation of definite and indefinite,
40.
Integrating factors, 135
Integration, 1.
by substitution, 1519.
constant of, 1.
formulas of, 14.
geometrical representation of,
36.
in series, 94.
of rational fractions, 2329.
Length of a curve, 61, 63.
Limits of integration, 36, 42.
Linear differential equations, 136,
147.
Mechanical and physical applica>
tions, 7089.
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, 7089.
Pressure, 70.
Prismoidal formula, 90.
Polar coordinates, 50, 63, 103.
national fractions, integration of,
2629.
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, 2123.
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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