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Differential calculus
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WORKS OF
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PUBLISHED BY
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PH.D.
Inc.
Analytic Qeometry.
Tii+197 pages.
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Differential
DIFFERENTIAL CALCULUS
BY
H. B. PHILLIPS, Ph.D.
AssisioTii Professor of
Mathematics in
the
Massachusetts
Institute of Technology
FIRST EDITION
FIRST THOUSAND
NEW YORK
JOHN WILEY & SONS,
London:
CHAPMAN
Inc.
& HALL, Limited
1916
Copyright, 1916,
BY
H. B.
PHILLIPS
StanDope
F
pcex
H.GILSON COMPANY
BOSTON, U.S.A.
PREFACE
on
have continued the
Geometry, wherein a few central methods are expounded and appUed to a la,rge variety
of examples to the end that the student may learn principles,
and gain power. In this way the differential calculus makes
only a brief text suitable for a term's work and leaves for the
integral calculus, which in many respects is far more important, a greater proportion of time than is ordinarily devoted
In
this text
plan adopted for
to
differential calculus I
my Analytic
it.
As material
for review and to provide problems for which
answers are not given, a supplementary list, containing about
haK as many exercises as occur in the text, is placed at the
end of the book.
I wish to acknowledge my indebtedness to Professor H. W.
Tyler and Professor E. B. Wilson for advice and criticism
and to Dr. Joseph Lipka for valuable assistance in preparing
the manuscript and revising the proof.
H. B. PHILLIPS.
Boston, Mass., August,
1916.'
CONTENTS
Chapter
I.
Pages
1- 9
Introduction
Derivative and Dipfbrbntial
10- 18
III.
Differentiation of Algebraic Functions
IV.
Rates
19-31
32-38
II.
v.
VI.
VII.
Maxima and Minima
39- 48
Differentiation of Transcendental Functions.
49- 62
Geometrical Applications
63- 84
VIII. Velocity
IX. Rolle's
and Acceleration in a Curved Path
Theorem and Indeterminate Forms
X, Series and Approximations
XI. Partial Differentiation
.
....
85- 93
94-100
101-112
113-139
Supplementary Exercises
Answers
154-160
Index
161-162
140-153
DIFFERENTIAL CALCULUS
CHAPTER
I
INTRODUCTION
—A
Definition of Function.
1.
quantity y is called a
function of a quantity x if values of y are determined by values
of X.
.
Thus, if 2/ = 1 — x^, 2/
determines a value of y.
function of
its
is
a function of x; for a value of x
Similarly, the area of a circle is
a
radius; for, the radius being given, the area
is
determined.
It
not necessary that only one value of the function
is
correspond to a value of the variable.
Thus,
determined.
if
3?
then y
is
a function of
Several values
may be
x and y satisfy the equation
—
To each
x.
=
2 xy ]- y^
X,
value of x correspond two
values of y found by solving the equation for y.
quantity u is called a function of several variables if u is
determined when values are assigned to those variables.
A
Thus,
if
2
=
x^
+ y^,
then
values being given to x and
Similarly, the
volume
2 is
y,
of a cone
a function of x and y;
a value of
is
z is
a function of
for,
determined.
its
altitude
and radius of base; for the radius and altitude being assigned,
the volume is determined.
of Functions.
2. Kinds
An expression containing
variables is called an explicit function of those variables.
—
Thus
Vx +
2/
is
an expUcit function
y
y
is
an
= Vx-j-
explicit function of x.
1
of
1,
x and
y.
Similarly,
if
DIFFERENTIAL CALCULUS
2
A
is
called
an implicit function.
x^
is
I.
quantity determined by an equation not solved for that
quantity
y
Chap.
—
an implicit function
2 xy
=
X,
Also x
is
-\-
of x.
Thus,
y^
if
an implicit function
of y.
Explicit
and
do not denote properties of the func-
implicit
tion but of the
way
expressed.
it is
rendered explicit by solving.
tion
imphcit function
is
equivalent to
is
x±
=
y
in
An
For example, the above equa-
Vx,
which y appears as an
explicit function of x.
A
one representable by an algebraic
rational function
is
expression containing no fractional powers of variable quantities.
For example,
a;
V5 +
x^
is
a rational function of
An
+
3
2x
x.
irrational function is
one represented by an algebraic
expression which cannot be reduced to rational form.
Vx + y is an irrational function of x and y.
A function is called algebraic if it can be represented
algebraic expression or
is
Thus
by an
the solution of an algebraic equa-
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.
tion.
—
3. Independent and Dependent Variables.
In most
problems there occur a number of variable quantities connected 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 variables;
Which
those
determined are called
dependent
variables.
and which as dependent is usually a matter of convenience. The number of
independent variables is, however, determined by the equations.
variables are taken as independent
Chap.
INTRODUCTION
I.
For example,
3
in plotting the curve
y
=
31?
+ X,
values are assigned to x and values of y are calculated.
independent variable is x and the dependent variable y.
The
We
might assign values to y and calculate values of x but that
would be much more difficult.
4.
— A particular function of x
Notation.
is
often repre-
sented by the notation / (x), which should be read, function
For 'example,
of x, or / of X, not / times x.
/
means that/
(x) is
= Vx^+
(x)
a symbol for Vx^
=
2/
means that y
function of
If it is
is
some
/
+
1.
Similarly,
(aj)
definite (though perhaps
unknown)
x.
necessary to consider several functions in the same
discussion, they are distinguished
by the use
f"
1
(x), g (x)
by
subscripts or accents or
Thus,
of different letters.
/i (x), /a (x), /' (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 writing
the variables. For example,
commas between
V
expresses that
«;
is
The /
in the
is
h)
a function of r and h and
V
expresses that «
=f{r,
=
f{a,
a function of
symbol
b, c)
a, b, c.
of a function should be considered as
representing an operation to be performed on the variable or
variables.
Thus,
if
f{x)
= Vx^
+
1,
/represents the operation of squaring the variable, adding 1,
and extracting the square root of the result. If x is replaced
.
DIFFERENTIAL CALCULUS
4
by any other quantity, the same operation
on that quantity. For example,
/
= V22+
(2)
f{y+l) =
Similarly,
then
If
/
(x, y)
(1, 2)
{x, y, z)
-3,
f (2,
1)
0^
=
3
to be performed
= Vs.
1
Vf + 2y + 2.
+1)2+1 =
= x^
= P
= x^
= 2^
+ xy + 1 2 - 22 = -1.
+ y^ +
+ (-3)2 + 1 = 14.
2/2,
.
z^,
EXERCISES
Given x + y
a express !/ as an explicit function of x.
Given logio (x) = sin y, express x as an explicit function
Also express 2/ as an explicit function of x.
3. If / (x) = x^ - 3 X + 2, show that/ (1) = / (2) = 0.
4. If F (x) = X* + 2 x^ + 3, show that F (-a) = F (a).
1.
,
2.
+ Xi,
+
5.
If
F
(x)
=
6.
If
.*
(x)
= Vx2 -
7.
If^(x)=^3,find^(i). Alsofind^.
8.
If /i (x)
9.
If/
(x,
I.
if
/
/
then
V(2/
is
Chap.
X
find
1,
F
(x
find
J/)
10.
Given /
11.
On how many
=
F
Also find 2
(2 x).
= 2% /2 (x) = x^
= X - i, show
Also find
l).
find /i
!/2 (j/)].
that/
(2, 1)
(x)
+
1.
(x).
.^
Also find f,
= 2/ (1,
of y.
2)
=
[/i (^)].
1.
+
x^
xj/, find / {y, x)
independent variables does the volume of a right
circular cylinder depend?
12. Three numbers x,y, z satisfy two equations
(x,
j/)
X2
X
How many
5.
Limit.
of these
—
+
+
2,2
y
+
32
=
5,
+z =1.
numbers can be taken as independent variables?
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 constant
is
The
said to be the limit of the variable.
use of limits
is
well illustrated
by the incommensurable
Chap.
INTRODUCTION
I.
and the determination
cases of geometry
circle or
5
of the area of
a
the volmne of a cone or sphere.
—
Limit of a Function.
As a variable approaches a
a function of that variable may approach a limit. Thus,
as X approaches 1, x^
1 approaches 2.
We shall express that a variable x approaches a limit a by
6.
limit
+
the notation
x
The symbol — thus means
Let /
(x)
expressed
=
a.
" approaches as a limit."
A
approach the limit
as x approaches a; this
is
by
lim/(x)
=
A,
x=a
which should be read, " the limit of /
is A."
Example 1. Find the value of
lim (x
x=l V
As x approaches
or
as x approaches
a,
-)•
+ xl
the quantity
1,
(x),
x+ X- approaches 1 +t1
Hence
2.
lim
Ex.
2.
Find the value of
sin 6
,.
hm
e=o 1
As
d
+ cos 9
approaches zero, the function given approaches
Hence
sin e
hm —
,.
6=5=0
7.
1
+ cos 6
=
0.
Properties of Limits.
— In finding the limits of func-
made
of certain simple properties that
tions frequent use
is
follow almost immediately from the definition.
DIFFERENTIAL CALCULUS
6
The
1.
limit of the
equal to the
sum
sum
a
of
finite
I.
number of functions
is
of their limits.
Suppose, for example, X, Y,
proaching the limits A, B,
is
Chap.
C
Z
are three functions ap-
X+Y+Z
Then
respectively.
A + B + C. Consequently,
{X + Y + Z) = A + B + C = \imX + limY + \im Z.
approaching
lim
The
2.
limit of the product of
a
finite
number
of functions
is equal to the product of their limits.
3.
X, Y,
for example,
If,
then
Z
approach A,
B,C
respectively,
XYZ approaches ABC, that
lim XYZ = ABC = lira X lim Y lim Z.
is,
If 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,
X
Then ^
zero.
A
approaches „
,.
X
,
B
that
A
and suppose
B
is
not
is,
limX
A is not zero, will be infinite. Then
X
A
X
y cannot approach „ as a limit; for, however large y. may
If
B
is
zero
and
-?,
become, the difference of -^ and infinity
8,
The Form
value,
jr
,
— When
x
is
will
not become small.
by a
replaced
a function sometimes takes the form ^
•
particular
Although
this
symbol does nob represent a definite value, the function may
have a definite limit. This is usually made evident by writing the function in a different form.
Example
1.
Find the value of
,,
iim
x=l
x^-1
r-
X-l
1
Chap.
INTRODUCTION
I.
When
X
is
replaced by
1,
7
the function takes the form
1-1
1-1
0'
Since, however,
X
—
the function approaches 1
=X+l,
T
1
+1
^=
lim—
x=l
Ex.
2.
Find the value
,.
lim
x—
X
=
2.
1
of
(\/r+^ -
1)
X
a:=0
When
Therefore
or 2.
the given function becomes
1-1 ^0
"O"
Multiplying numerator and denominator by
Vl+x-
^
~
X
As X approaches
X
X {Vl
Vl + +
a;
1
+x+
~ Vl
1)
+x+1
the last expression approaches
0,
1,
|.
Hence
^^(vr+^-i)^i
X
x=o
Infinitesimal.
9.
—A
2
variable
approaching zero as a
an infinitesimal.
Let a and fi be two infinitesimals.
limit is called
If
lim^
is finite
and not
the same order.
|8.
If
than
a.
zero,
If
a and ^ are
the hmit
is
the ratio ^ approaches
said to be infinitesimals of
zero,
a
is
infinity,
of higher order
/3
Roughly speaking, the higher the
the infinitesimal.
is
than
of higher order
order, the smaller
8
DIFFERENTIAL CALCULUS
>
For example,
let
x approach
The
zero.
Chap.
I.
quantities
X, 0?, 3?, 3^, etc.
are infinitesimals arranged in ascending order.
higher order than
lim -;
=
—
lim x^
of lower order than
is
x* is of
0.
1=0
x=oX''
Similarly, 3?
Thus
x^; for
3^,
since
X
3^
approaches infinity when x approaches zero.
As x approaches ^
same
,
cos x and cot x are infinitesimals of the
order; for
lim
T,
which
is finite
cos x
—
— = lim sin
,.
,.
-
cot X
and not
.
cc
=
1,
1=0
zero.
EXERCISES
Find the values
.
1.
of the following limits:
hm a;2-2x ^+
,.
„
2-
4.
«
5.
,.
hm
By
a'
—
3a;
X
—
+2
sin
hm
s=osin 2 9
the use of a table of natural sines find the value of
Imi
a:=0
9.
Iim^i5_£.
9=0 tan 9
...
6.
r-'
\
,.
8.
a;"
^
,.
x=\
7.
hm
x=0
J
3.
Vl - s^— - Vl +
...
3
X- 5
sin 9 + cos 9
>^"^.sin29 + cos2e«=
;=D
/
sinx
^
Define as a Umit the area within a closed curve.
Define as a limit the volume within a closed surface.
10.
Define V2.
11.
On
the segment
angles reaching from
PQ
P
(Fig. 9a) construct
to Q.
a
As the number
series of equilateral tri-
of triangles is increased,
Chap.
INTRODUCTION
I.
their bases
PQ.
Does
approaching zero, the polygonal line
its length approach that of PQ?
A \/
9
PABC,
etc.,
approaches
\/
Fig. 9a.
12.
as
Inscribe a series of cylinders in a cone
shown
in Fig. 9b.
As the number
of cyl-
inders increases indefinitely, their altitudes
approaching zero, does the sum of the volumes of the cylinders approach that of the
cone? Does the sum of the lateral areas of
the cylinders approach the lateral area of the
Fig. 9b.
cone?
13.
Show
that
when x approaches
zero,
tan - does not approach a
limit.
14.
is
As X approaches
1,
which
of the infinitesimals 1
—
x and
Vl —
x
of higher order?
15. As the radius of a sphere approaches zero, show that its volume
an infinitesimal of higher order than the area of its surface and of the
same order as the volume of the circumscribing cylinder.
is
CHAPTER
II
DERIVATIVE AND DIFFERENTIAL
10.
Increment.
— When
a variable changes value, the
algebraic increase (new value
crement
and
is
minus
old)
called
is
its in-
by the symbol A written
represented
before
the variable.
Thus,
if
X changes from 2 to
Ax = 4
If
4, its
-
2
=
increment
is
2.
X changes from 2 to —1,
-1-2 =
Ax =
When
the increment
is
-3.
positive there
is
an increase
in
value,' when negative a decrease.
Let y he
a,
function of
x.
When x
receives
an increment
Ay will be
Ax, an increment
determined.
The increments
of
X and y thus correspond.
To
illustrate this graphically
X and y be the rectangular
coordinates of a point P. An
equation
let
y =fix)
to
some other
position
Q on
represents a curve.
When
changes, the point
P changes
the curve.
The increments
x
of
X and y are
Ay = RQ.
Ax = PR,
—
(10)
Continuous Function.
A function is called continuous if the increment of the function approaches zero as
the increment of the variable approaches zero.
11.
10
Chap.
'
II.
In Fig.
DERIVATIVE AND DIFFERENTIAL
2/
is
approaches zero,
Q
10,
a continuous function of x;
approaches
P and
so
11
for,
as
Ay approaches
Ax
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
Y
DIFFERENTIAL CALCULUS
12
The
slope of the tangent at
Chap.
II.
P is called the slope of the curve
at P.
Example. Find the
slope of the parabola
y
= x^
P
Q
at the point
(1, 1).
Let the coordinates of
be X, y.
Those of
are a;
Ay.
Ax, y
+
+
Since
P and Q
are both
on the curve,
y =x^
and
y
+ Ay=-{x + Axy =
+ 2 Ax + (Aa;)^.
x2
we
Subtracting these equations,
Ay =
Dividing by
a;
get
2xAx+
{Axy.
Aa;,
^^2a=+A.T.
Aa;
As
Aa; approaches zero, this
approaches
Slope at
This
is
then
X.
Let w be a function of
Derivative.
Ax
slope at
is
is
It is represented
=
lim
Ai=o
represented
Ay
Ax
^
Ax
called
by the
by /
(13a)
(x)
=
derivative with
(x), its
often represented by/' (x).
f
is
-r^
is,
D^y
a function
to x.
If
x.
approaches zero, that limit
the derivative of y with respect
notation Dxy, that
respect to x
The
2-1=2.
approaches a limit as
If
2
the slope at the point with abscissa x.
(1, 1) is
13.
P=
Thus
= Z)^(x).
lim^^
iiX
^=0
(13b)
Chap.
DERIVATIVE AND DIFFERENTIAL
II.
13
In Art. 12 we found that this Umit represents the slope of
is, in fact, a function
the curve y = fix). The derivative
of X whose value is the slope of
the curve at the point with abscissa X.
The derivative, being the limit
Aw
of
-T-^
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
its limit
Dxy are
positive.
If
Fig. 13.
they have opposite signs D^y
Therefore Dxy is positive when x and y increase
and decrease together and negative when one increases as the
is
negative.
other decreases.
2.
Example. y = x? — 3x
Let x receive an increment Ax. The new value of x is
Ay. Since these satisfy
X
Ax. The new value oiy isy
+
+
+
the equation,
y
+ Ay={x + Axy -3{x + Ax)+ 2.
Subtracting the equation
y
= oi^-3x +
Ay = 3x^Ax
+ 3x (Ax)^ +
2,
we get
(Ax)'
-
3 Ax.
Dividing by Ax,
Ay
= 3x^
Ax
As
fyx
+ 3xAx + {Axy - 3.
approaches zero this approaches the limit
Dxy
=
3
a;2
-
3.
DIFFERENTIAL CALCULUS
14
Chap. U.
The graph is shown in Fig. 13. At A (where x = 1) y =
= 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 D^y is positive and x and y inand Dxy
crease
is
and decrease
Between
.together.
A
and
B
the slope
negative and y decreases as x increases.
EXERCISES
Given y = Vx, find the increment of y when x changes from
2 to X = 1.9. Show that the increments approximately satisfy
1.
I
=
the equation
A^
Ax
1_.
2
VS
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 approximately half the first.
.3.
Find its slope by
3. The equation of a certain line iay = 2 x
+
Aw
calculating the limit of -p4.
Construct the parabola y = x'' — 2 x.
is 2 (x — 1).
Find
the point with abscissa "x
what point
is
Show
its
that
slope at
its
slope at
(4, 8).
At
the slope equal to 2?
Construct the curve represented by the equation y = x* — 2 x'.
its slope at the point with abscissa x is 4 x (x^ — 1).
At what
points are the tangents parallel to the x-axis? Indicate where the slope
5.
Show
is
that
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.
Chap.
As
DERIVATIVE AND DIFFERENTIAL
II.
A.X approaches zero, -i— approaches
15
Dxy and
so
«
ap-
proaches zero.
The increment
of y
is
+ eAa;.
Ay = DxV Ax
The
part
D^y Ax
(14)
called the principal part of Ay.
is
As Ax approaches
amount eAx.
eAx becomes an
so
It differs
zero, e
from Ay by an
approaches zero, and
It is an
than Ax. If then the principal
used as an approximation for Ay, the error will be
indefinitely small fraction of Ax.
infinitesimal of higher order
part
is
only a small fraction of
When
Example.
mate value
be
1-
change in y
for the
In exercise
9,
Ax when Ax
page
Hence the
The exact increment
15.
and
is
2%
small.
-
principal part of
-7(.1)
4
=
Ay
- was found
is
-0.0250.
Ay with an
error less than 0.002
of Ax.
Differentials.
^Let
x be the independent variable
y be a function of x.
called the differential of y and
The
let
dy
=
is
principal part of
denoted by dy; that
D^^y Ax.
Ay
ii
y
=
x,
D^y =
dx
is,
=
1,
is
is,
(15a)
This equation defines the differential of any function y of
In particular,
that
to
is
principal part represents
which
=
14, the derivative of
-\Ax=
x^
The
is sufficiently
x changes from 2 to 2.1 find an approxi-
x.
and so
Ax,
(15b)
the differential of the independent variable is equal to
DIFFERENTIAL CALCULUS
16
its
Chap.
II.
increment and the differential of any function y is equal to
and the increment of the independent
the product of its derivative
variable.
Combining 15a and
15b,
is,
the quotient
dy
-^
get
dy
= Dxy dx,
(15c)
X
=
(1^^)
whence
that
we
"^"V'
is equal to the derivative
ofy vnth
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-
PRT (Fig. 15) with
hypotenuse PT extending
along the tangent at P.
On this diagram, Aa; and Ay
are the increments
angle
Ax = PR,
occurring
from
P to
Ay = RQ,
the
change
The
differen-
in
Q.
tials are
dx
= PR,
A
point
differential
dy
x
+
= RT.
the
moving when it
the tangent PT. The
curve
passes through
dy
describing
P
is
in the direction of
then the amount y would increase when x
Ax if the direction of motion did not change.
is
changes ixi
In general the direction of motion does change and so the
If the inactual increase Ay = RQ is different from dy.
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
continuous functions of an independent variable
dx
=
Dtx
At,
dy
=
Dty At.
x and y are
For then
t.
Chap.
The
II.
DERIVATIVE AND DIFFERENTIAL
identity
Ay _ Ay
Ax
Ai~Ax"Ai
gives in the limit
Dty
= D^y
•
DfX.
17
DIFFERENTIAL CALCULUS
18
The
error in the area
dA=d
which
is
2%
(a;2)
Chap.
II.
approximately
is
=2xdx=
±0.02
x^
= ±0.02 A,
of the area.
EXERCISES
1.
=
Let n be a positive integer and y
A?/
=
(a;
Show
by using the binomial theorem.
Expand
a;".
+ Ax)" — x"
that
= nx-K
^
ax
What
the principal part of Ay?
Using the results of Ex. 1, find an approximate value for the increment of a^ when x changes from 1.1 to 1.2. Express the error as a
is
2.
percentage of Ax.
3.
If
A
the area of a circle of radius
is
r,
show that
dA
-j-
is
equal to the
circumference.
4.
If
the radius of a circle
is
measured and
using the result, show that an error of
radius will lead to an error of about
5.
If V is
Let
V
be the volume
Show
that
if
r is
if
If
that the
is
constant
constant
dv
-7- is
-jr is
If
an
-y- is
equal
h.
equal to the lateral area.
= / (x) and for all variations ia x,
graph oi y = f (x) is a straight line.
2/
the
show that
r,
equal to the area of the base of cylinder
8. If y is the independent variable and x
showing dx, dy. Ax, and Ay.
9.
of the
in the area.
a cylinder with radius r and altitude
of
dv
h
area calculated by
measurement
its surface.
6.
7.
its
in the
the volume of a sphere with radius
to the area of
and
2%
1%
2/-axis is vertical,
zontally from the origin with
the
dx
=
=f
a;-axis horizontal,
a velocity
of 50
ft.
Ax, dy
(y),
=
Ay,
show
make a diagram
a body thrown horiper second will in
t
seconds reach the point
X
=
50t,
y
= -16 tK
path at that point.
10. A line turning about a fixed point P intersects the x-axis at A
and the y-swa at B. If Ki and K2 are the areas of the triangles OPA and
Find the slope
OPB, show
of its
that
dKi^PA^
dKi
PB^'
CHAPTER
III
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
The
16.
called
is
method
process of finding derivatives and differentials
Instead
differentiation.
applying the direct
of
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-jior the
the derivative with respect to
To
d {u
+ v)
J- (m
+ w) =
=
Thus
x.
differential of (m
derivative of {u
operation of taking
+
w),
+ v) with respect to x.
we proceed as in
everywhere replaced
obtain the derivative with respect to x
finding the differential except that d
""^dTx
17.
Formulas.
single variable x,
I.
II.
III.
IV.
V.
VI.
— Let
and
c,
u,
n
v,
w
be continuous functions of a
constants.*
= 0.
d {u-\- v) = du + dv.
d {cu) = c du.
d (tfv) = u dv -\- V du.
V du — ti dv
fu\
3
d[-) =
dc
d
(it")
=
nu"'-'-
du.
assumed that the functions u,
continuous functions,
* It is
exist
is
u =f{x),
19
v,
w
have derivatives.
There
DIFFERENTIAL CALCULUS
20
Chap.
III.
—
Proof of I.
The differential of a constant is zero.
a variable x takes an increment Ax, a constant does
18.
When
not vary.
dc
-^ =
0.
Proof of n.
19.
Ac
0,
=
-r—
and
0,
in the limit
Clearing of fractions,
= dx-0 =
dc
number
=
Consequently, Ac
— The
0.
differential of the
sum
of functions is equal to the
sum
a
of
finite
of their differentials.
Let
y
When
to
«;
+
=u+
V.
X takes an increment Ax, u will change to w
Ay. Consequently
Ay, and y to j/
+ Am,
v
+
y
+ Ay = u + Au + v-j- Av.
Subtraction of the two equations gives
Ay = Au
+ Av,
whence
Ay _ Aw
~ Ax
Aa;
.
.
As Ax approaches
,
respectively.
Ay
Ax
Au Av
dy du
^^ approach ^, ^,
Ay
,
^, ^,
zero,
dv
^
Therefore
dy
dx
_du
dv
dx
dx'
and so
=
dy
By
the same method
d{u±v±w dc
•
du
+ dv.
we can prove
•
•
)
=
du±dv±dw±
•
•
•
.
such that
Au
Ax
does not approach a limit as
Ax approaches
no derivative DxU and therefore no
du
=
zero.
differential
DxU
dx.
Such a function has
Chap.
ALGEBRAIC FUNCTIONS
III.
20.
Proof of in.
function
is
— The
differential of
21
a constant times a
equal to the constant times the differential of the
function.
= cu.
Ay = c(u +
Ay = c~"Am,
Ay _ Am
Let
y
Then
y
+
and so
A
A
1
As Ax approaches
Ax
Ay
""
,
Am)
Ax
Am
zero, -r-.and c-ir-
dy
du
approach j- and c-j-,
,
Therefore
dy
dx
_ du
dy
=
dx'
whence
c du.
Fractions with a constant denominator should be differentiated
by
Thus
this formula.
'^©='^(^)=^"21.
.
Proof of IV.
functions is equal
— The
to the first
differential of the product of two
times the differential of the second
plus the second times the differential of the
Let
Then
y
+
y =
Ay =
=
first.
uv.
(u
+ Am)
+ Av)
+ (m + Am) Av.
(v
uv -\-v
Am
Am +
(m
Subtraction gives
Ay =
V
+ Am) Av,
whence
Aw
Am
,
,
,
.
,
Aw
m is a continuous function, Am approaches zero as
approaches zero. Therefore, in the limit,
Since
dy
dx
_ du
do
dx
dx
Ax
DIFFERENTIAL CALCULUS
22
Chap.
III.
and so
dy
= V du + u dv.
way we can show that
d (uvw) = uv dw -\- uw dv
In the same
22.
Proof of V. — The
-\-vw du.
a fraction
differential of
the denominator times the differential of the
the
is
equal
to
numerator minus
numerator times the differential of the denominator,
all-
divided by the square of the denominator.
Let
y
=u
Then
and
+ Am
w
,
Ay V
-\-
u-
Av
V
V
Am — m
V {v
At;
+ Av)
Dividing by Ax,
.
At/
-r^
Ax
Am
= V-:Ax
V (v
At;
M-7-
Ax
+ Av)
t; is a continuous function of
approaches zero. Therefore
Since
du
dx
_
dy
dx
x,
Av approaches
zero as
Ax
dv
dx
v^
whence
,
dv23.
Proof of VI.
— The
V
du
— udv
^
differential of
a variable raised
to
a
constant -power is equal to the product of the exponent, the variable
raised to a power one
and
less,
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.
page 61.
For the case
of irrational exponent, see Ex. 25,
Chap.
ALGEBRAIC FUNCTIONS
III.
Let n be a positive integer and y
(1)
=
23
Then
u".
and
Ly =
nu"-^
Am + ^ ^\~
^^
m"-^ (Am)^
+
•
•
.
Dividing by Am,
n
Ah
-^ =nM"-iH
As Am approaches
—
—
(n
,
,
1)
,
,
,
+
,
•
•
.
approaches
zero, this
=
-^
du
„
^M"-2 (Am)
^^"2
nM"-!.
Consequently,
=
dy
du.
nu"~''-
V
Let n be a positive fraction
(2)
=
2/«
and y
-
—
u''= u".
Then
M".
Since p and g are both positive integers, we can differentiate
both sides of this equation by the formula just proved.
Therefore
qy^~^ dy
=
du.
pu'^~'^
p
Solving for dy and substituting m' for
dy
=
VU^
qu
«
P--p
=
M"
=
M"""
Therefore,
we
=
nu"~^ du.
get
— m. Then
=
M"
m is positive, we can find d (m")
above.
,
du
n
q
*
W
"
dy
V ^
= -u
Let n be a negative number
(3)
Since
du
y,
by the formulas proved
by V,
—
= u"'d(l)
=
^ „^„ld{u'^)- = —7rmr~Hu
^
—
„,, = nu"-^,,du.
—mu-'^-^du
DIFFERENTIAL CALCULUS
24
.
n
Therefore, whether
an
is
Chap.
III.
integer or fraction, positive or
negative,
d
If
(m")
=
the numerator of a fraction
=
d -j
(
=
y
1.
is
constant, this formula can be
Thus
used instead of V.
Example
nw"""' du.
d
icu~^) ==
—cu~^ du.
43^.
Using formulas III and VI,
=
dy
Ex.2, y
4d
= Vl +
=4
(x^)
~
+
Vx
3
•
a;2
dx
=
12
a;2
da;.
3.
This can be written
y
=
x^
+
2+3.
x
Consequently, by II and VI,
dy
dx
_d (xi)
d (x-i)
dx
1
d
—idx
—sdx
1
(3)
dx
dx
,
= 2^^d^-2^^di +
Ex.
3.
y
2VZ 2V'^
= (x + a) (x^ - b^).
Using IV, with
=
=
Ex.4.. y
n
(x
3
u = x + a,
v
=
x^
— V,
+ a)(2x-0) + (3?- ¥)
+ 2 ax -
a;2
= $^-
62.
(1
+ 0)
Chap.
ALGEBRAIC FUNCTIONS
III.
Using V, with u
,
'^'^
(x"
~
= x^ + 1, v = x^ — 1,
-l)d (x' + 1) - (x^ +l)d
{?? -
^ {x^-\)2xdx_
+
i
-
1)
+ l)2xdx
-1)2'
(a;2
Ex. 5. y = Va;2 - 1.
Using VI, with u = x^
=
jx"
{x^
ly
Axdx
~
'~
25
—
1,
(a;2
-
1)-^ (2
—
2/^
=
a;)
=
Vx2-
1
Ex.
6.
We
can consider y a function of x determined by the equa-
cc^
1.
Then
tion.
d
that
a;2/
ix')
+ d (xy) - d (y2) = d (1) = 0,
is,
+ X dy + y dx ~ 2 y dy = 0,
(2x + y)dx+{x-2y)dy = 0.
2 X dx
Consequently,
d^
dx
Ex.7.
a;
=
<
+ -,
y
^
2x-\-y
2y —
= t
x
-.
In this case
dx
=
dt
— ^,
dy
=
dt
+ ^-
Consequently,
%^
dx
^ ^ l±l
t^
1
t^-l
f
Ex.
X
=
8.
0.2.
Find an approximate value of
2/
=
/I
(
_ x\i
.
1
when
:
DIFFERENTIAL CALCULUS
26
When
=
X
0,
=
y
=
a;
this
2dx
3(l-a;)^(l+a;)*'
becomes
= -f da;.
dy
If
in
III.
Also
1.
^~
When
Chap.
we assume that dy is approximately equal to Ay, the change
y when x changes from to 0.2 is approximately
dy
The
required value
= -f
is
(0.2)
=
-0.13.
then
y
=
l-
=
0.13
.87.
EXERCISES
In the following exercises show that the differentials and derivatives
have the values given
1.
y
=
+
3x*
^-
y
4.
y
5.
J/
6-
2/
x^
+
1
'
5
(,x
=
a;
+ 2a){x- ay,
(2a;- 1) (3a; + 2),
=
2X
\
(ix (x
+3
I)''
(x
(4x-5J^
2x
-
3 x'
-
2
3 (y?
-
^=
ISa;^
+ 2a; - 2.
_
-
-2a;cfa;
(i
^"-
a") dx.
d9
+ if
1
Ve^-l-e
0+V^2Tri= ^/W^^l
^
ds« va-
Vo^
a:
=
1)'
x'
x) dx.
5
{l-t)di
/
+ x^ -
dy
dx
dy
-22dx
/
dx Vx
11 {x"
Zx^-2
(a;2
-2x _
-
^
dy
''y
^?:pi-
'•^"4x-5'''^
d
di
1
=
_
+ 5,
3,
-
g3
~
-6x'
ix'
3
J
-
x^
y
-
a2-2s2
s>
= y^, _
a?dy
^j
'
'
'
Chap.
18.
ALGEBRAIC FUNCTIONS
III.
27
= (x+l)(2-3a;)2(2x-3P,
2/
^ = (24 + 13a; -36a;2) (21-1
19
dy
a;"
=
*
"^
20.
=
.
^^V.-^.
3
+
(x
vr+T')"+^
+
,«
«
I
1
(x
.x2
+
2/2
=
a2,
23.
x3
+
,3
=
3 ax..
25.
2 x^
+4
X2/
=
2/
2/'"
2/2
d^
dx
3 X,
?/
when
1
V
_ 4x — 3)/ —
~ 3X — 8
-1
3
2/
0.
da
"
dx
1
- - n
Vl+x'+Vr+a*"
I
+ x'"2/" = x^™,
2
mydx =nx dy.
+3
J
d2/
29.
x=i-V<2-l, y=t+^e-l,
31.
Given y
=
-
= 2 (x + VrTx'2)"dx.
_ X
ydx-xdy =
I
X
1
+
| P^^=
a;'
27.
Va:2
=
+1 = h
1
n~
26.
3
jfi
K
d^
dx
22.
24.
_i_ 1
+ vrr^o"-'
'
1
d?/
-
3)=.
*
"^
t
dx
'
ij
-
(a-|-6x")«+'
ar*
'
(2x
maa:"'-'
da;
+ ia;")""
(a
_
~
3a;)
=
^
+ ydx =
xdy
0.
X
.^
^
„
,
an approximate value
find
for
4.2.
32.
Find an approximate value
of
i/5
T X^
X
when X
=
X+
1
+
1
-|-
X
.3.
= x^ find d2/ and Ay when x changes from 3 to 3.1. Is
dy a satisfactory approximation for Ay? Express the difference as a
33.
Given y
percentage of Ay.
34. Find the slope of the curve
2/
at the point x
=
=X
(x5
+
31)^
1.
35. Find the points on the parabola y^
incUned at an angle of 45° to the x-axis.
= iax
where the tangent
is
.
DIFFERENTIAL CALCULUS
28
36.
Chap.
III.
Given y = (a + x) Va — x, for what values of x does y increaae
and for what values does y decrease' as x increases?
Find the points P {x, y) on the curve
as X increases
37.
=
y
+,1-
x
where the tangent is perpendicular to the
38. Find the angle at which the circle
x^
+
P
to the origin,
<
2x-Zy
=
y'^
line joining
intersects the a;-axis at the origin.
39.
A line through
2/-axis at (0, y).
40.
If
—
a;2
a;
the point
Find
+2
=
(1, 2)
outs the x-axis at
{x,
and the
0)
-—
0,
why
:^
ax
is
(x2
the equation
-
a;
+
=
2)
not satisfied?
41.
The
distances x, x' of a point
and
its
image from a lens are con-
nected bv the equation
-=
+
x^x'
-
-
r
/ 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
.
MIT
approximately, x and x' being the distances of the object and
from the
24.
Higher Derivatives.
function of x.
is
its
image
lens.
— The
Its derivative
derivative ^^
with respect to
called the second derivative of
is,
first
-^
dx^
dx \dx)
/%Y
Similarly,
d^^d^ l^\
dx \dxy'
^ = 1W
etc
a
written ^-^
y with respect to
^=
db?
x,
is
x.
That
Chap.
The
/'
(a=),
ALGEBRAIC FUNCTIONS
III.
29
derivatives of / (x) with respect to x are often written
/"
Thus,
(.x),f"' (x), etc.
g=
| = /'(x).
Exafnple
=
y
1.
ii
y
/"(x),
=
f
(x),
g = r(x),etc.
x^.
Differentiation with respect to x gives
All higher derivatives are zero.
x^
JEx. 2.
+
+
xt/
=
2/'
^
1.
,
,
\
'^^
Differentiating with respect to x,
+ ,+x|+2,|=0,
2x
whence
dy
dx
The second
derivative
^
+
+2
2x
j/
X
J/
is
^= __ddx /2xJ-_^\
+2
U
dx2
dv
Replacing -~ by
Y
its
6
(Py
(x^
(x
last expression is
curve
x^
+ xy
derivative
we
-\-
y^
(x
+ 2 y)'
value in terms of x and y and reducing,
dx2
The
2//
3x^-Sm
^_2dx__^
+ x?/ +
+ 2 yy
6
2/^)
(x
+2
2/)'
obtained by using the equation of the
=
1.
By
differentiating
this
could find the third derivative, etc.
second
DIFFERENTIAL CALCULUS
30
Change
25.
second derivative by -r^
tient obtained
by dividing a second
The value
(dxy.
=
d^ =
If,
however,
=
^
df
30
^«
we
=
x
x^,
2 (dxY
=
is
is
=
differential
however depend on the
Then
t'.
2 (3
vari-
differentiated.
<2
dty
=
d^y
-r^
18
=
1^
differentiate with respect td
2 and so
{dty.
t,
since y
=
t^,
and
=
d?y
which
have represented the
d (dy)
of d^y will
able with respect to which y
Thus, suppose y
III.
This can be regarded as the quo-
.
d^y
by
— We
Variable.
of
Chap.
30
t"
{dty,
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
dj^ti
derivatives like
-^ and
dr
fractions because d^y does not
d^v
-r^
ax'
must not be combined
like
have the same value in the two
cases.
If
we have
derivatives with respect to
t
and wish to
derivatives with respect to x, they can be found
find
by using the
identical relation
du
d
dx
_ du
dt
dt^
_
dx
dt
,„
dx
Tt
For example,
dx\dtj
dt\di) dx
dx
V
Chap.
ALGEBRAIC FUNCTIONS
III.
=
Given x
Example.
— -1
t
y
,
=
1
t
+ -,
31
d'y
find -r|
In this case
dy
dx
f_
^
,,
dl
+
,
^f -
dt
-.
f
+
1
l'
Consequently,
d^y
_
d ff
dx \f
- 1\_
+
1/
it
dt
4t
_
_
1
{('
EXERCISES
Find -r and -r4 in each
ax
ax'
1.
y
2.
y
3.
y
4.
!/^
9.
If
=
X
—
5.
r1
= Va2 - x2.
= (x - 1)3 (x +
= 4 X.
a and
of the following exercises:
2)<.
b are constant
dx'
10.
and y
dx
=
x^
6.
x''
7.
X2/
8.
x^
ax^
+
2/"
=
a?.
-2y^ = 1.
= a +
+ t/» = ai
+ bx,
?/.
show that
4f
CHAPTER
IV
RATES
26.
—
Rate of Change.
the change in a quantity z
If
proportional to the time in which
at a constant rate.
Az
If
is
it
occurs, z
is
is
said to change
the change occurring in an in-
terval of time Ai, the rate of change of z
is
Az
Af
If the rate of
constant
if
change of
z is
the interval Ai
not constant,
it will
very short.
Then
is
be nearly
Az
-r- is
ap-
proximately the rate of change, the approximation becoming
The exact rate of
greater as the increments become less.
change at the time
t
is
consequently defined as
0^'
^fr-%'
that
is,
the rate of change of
any quantity
respect to the time.
If
if
the quantity
is
increasing, its rate of
decreasing, the rate
27.
negative.
is
Velocity Along a Straight Line.
move along a
is its derivative with
<
straight
hne
(Fig. 27).
s
change
— Let a
Let s
is
positive;
particle
= OP
P
be con-
As
P
Fig. 27.
on one side of 0, negative on the other. If
the particle moves with constant velocity the distance As in
sidered positive
the time Ai,
its
velocity
is
As
At'
If the velocity
is
very short.
is
not constant,
Therefore
As
-rr is
it will
be nearly so when A<
approximately the velocity, the
32
RATES
Chap. IV.
33
approximation becoming greater as At becomes
velocity at the time t is therefore defined as
less.
„=liin^ = g.
At
dt
The
(27)
^
A(=o
'
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
the velocity
s is increasing,
decreasing, the velocity
Example.
A body
starting
=
s
feet in
The
t
Find
seconds.
t,
At the end
=
positive;
from
16
rest falls approxinaately
i2
t is
= 32
^
dt
of 10 seconds
«
ft./sec*
it is
320
ft./sec.
Acceleration Along a Straight Line.
tion of a particle
moving along a
the rate of change of
=
That
is
defined as
is
5=g-
This equation shows that Uv
if
— The accelera-
straight line
its velocity.
«
in the time dt
s is
dv
V •=
28.
when
velocity at the end of 10 seconds.
its
any time
velocity at
is
negative.
is
is
the
(^'
amount
v
would increase
the acceleration remained constant.
is positive when the velocity is increasing,
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
The
acceleration
negative
h
Find
rising,
*
= 100t-lQ
f.
and acceleration. Also find when it is
when falling, and when it reaches the highest point.
its
velocity
The notation
ft./sec.
means
feet per second.
Similarly, ft./sec.^,
used for acceleration, means feet per second per second.
DIFFERENTIAL CALCULUS
34
The
The
velocity
-^ =
and acceleration are
v
=
j^ (100 - 32
a
=
^=
be
It will
3|.
highest point
-32
be rising while
ball will
Chap. IV.
when
ft./sec.2.
v is positive,
falling after
=
<
ft./sec,
i
=
3|.
that
It will
3|.
until
is,
t
=
be at the
—
Angular Velocity and Acceleration.
Consider a
body rotating about a fixed axis. Let d be the angle turned
through at time t. The angular veloc^P
ity is defined as the rate of change of
29.
that
9,
is,
angular velocity
Fig
of
The angular
change of angular velocity, that
,
..
,
angular acceleration
Example
=
-^
at
•
=
a
acceleration is the rate
=
do}
-tt
A wheel
1.
is
=
(P9
ttt
dt^
•
turning 100 revolutions per minute
Find its angular velocity.
turned
through in one minute
angle
The
its axis.
CO
Ex.
to
is,
dt
about
=
'^9
2.
constant
A
=
100
•
2
=
TT
will
be
200 T radians/min.
wheel, starting from rest under the action of a
moment
about
(or tv/ist)
its axis,
will
turn in
t
seconds through an angle
8
k being constant.
at time
By
Find
its
=
kt\
angular velocity and acceleration
t.
definition
0)
=
a=
jT
-jT
at
=
2
ftf
rad./sec,
= 2k rad./sec*.
RATES
Chap. IV.
— In matny cases the rates of change
Related Rates.
30.
of certain variables are
be calculated.
This
35
is
known and the rates of others are to
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.
Exam,ple
and
its
The
1.
radius of a cylinder
is
increasing 2
ft. /sec.
Find the rate of change
altitude decreasing 3 ft./sec.
of its volume.
Let r be the radius and h the altitude.
=
V
The
By
irr'-ll.
derivative with respect to
t is
,dh,^
dv
Then
dr
,
hypothesis
dr
_
dh
_
di~^'
_
dt~
~^-
—
TT^-
Hence
-r-
=
4
irrh
3
dt
This
h.
is
If r
when the
the rate of increase
=
10
ft.
and
=
/i
-77=
6
radius
is
r
and altitude
ft«,
— 60
TT
cu. ft./sec.
dt
Ex. 2. A ship B sailing south at 16 miles per hour is northwest 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
point where their paths cross.
ships
is
The
then
s
This distance
is
= Vx^
+
2/2.
changing at the rate
dx
ds
^
dy
^dt^^di
Va;2
+
9
A
and
B
from the
distance between the
DIFFERENTIAL CALCULUS
36
By
;t7
Chap. IV.
hypothesis,
=
-|
10,
=
-16,
Vx'
"
=
,
dt
at
+ y^
=
,
VX2
+
cos45°
—^•
=
V2
2/2
Therefore
10
ds
dt
The
- 16
=
V2
-3\/2mi./hr.
negative sign shows that s
decreasing, that
is
the
is,
ships are approaching.
EXERCISES
From
1.
the roof of a house 50
upward with a speed
t
of
100
ft.
above the
ft.
thrown
above the ground
street a ball is
Its height
per second.
seconds later will be
A
Find
its velocity
+
50
100
-
<
and acceleration when
What
tinue to rise?
=
is
t
16
=
fi.
2.
How
A body moves in a straight line according to
s = it* - il' + 16P.
2.
long does
the highest point reached?
it
con-
p4 /
the law
During what interval is the velocity
moving backward?
3. If V is the velocity and a tht acceleration of a particle jnoving
along the a;-axis, show that
Find
and
velocity
its
decreasing?
When
acceleration.
is it
adx =
4.
If
a particle moves along a line with the velocity
=
t)2
where g
is
2
gs,
constant and s the distance from a fixed point in the
that the acceleration
6.
V dv.
When
is
line,
show
constant.
a particle moves with constant speed around a circle with
its shadow on the a^-axis moves with velocity v
center at the origin,
satisfying the equation
2,2
n and
r
being constant.
proportional to
its
A
is
6.
wheel
angular velocity?
it
drive a belt?
4-
„2j;2
Show
=
^2^2^
that the acceleration of the
shadow
is
distance from the origin.
turning 500 revolutions per minute.
If
the wheel
is
4
ft.
in diameter, with
What
is
its
what speed does
RATES
Chap. IV.
37
.
A
rotating wheel is brought to rest by a brake. Assuming the
between brake and wheel to be constant, the angle turned
through in a time t will be
7.
friction
e
a, b, c
When
=
a
+ bt -
cfi,
being constants. Find the angular velocity and acceleration.
wiU the wheel come to rest?
8. A wheel revolves according to the law w = 30 <
<', where m is
the speed in radians per minute and t the time since the wheel started.
+
second wheel turns according to the law 9 = Ifi, where t is the time
and 9 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 v is the velocity and a
the acceleration of its center, a the angular velocity and a the angular
acceleration about its axis, show that
A
in seconds
V
10.
The depth
increasing 1
=
a
Tec,
=
ra.
water in a cylindrical tank, 6 feet in diameter, ia
foot per minute.
Find the rate at which the water is flowof
ing in.
A
stone dropped into a pond sends out a series of concentric
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?
11.
ripples.
12.
At a
of its base 3
its
certain instant the altitude of a cone
ft.
If
base decreasing
the altitude
1 ft./sec,
is
is
7
increasing 2 ft./sec.
how
fast is the
and the radius
and the radius of
ft.
volume increasing or de-
creasing?
13. The top of a ladder 20 feet long slides 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.
The cross section of a trough 10 ft. long is an equilateral triangle.
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 ground. WTien 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.
Assuming
that the water is 1000 ft. deep, the cable is attached to the bottom and
stretches in a straight line to the stern 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 surface forms a cone
with constant angle ^ at the vertex, depending on the material. If the
14.
If
DIFFERENTIAL CALCULUS
38
sand
is
poured at the rate of
creasing
when
it
c cu. ft./sec,
at
what
rate
Chap. IV.
is
the radius in-
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.
19.
An
elevated car running 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
face car 8 mUes, at
is
16 miles per hour
what rate are the
and that
of the sur-
cars separating 10 seconds after
they meet?
of the sun make an angle of 30 degrees with the horidrops from a height of 64 feet. How fast is its shadow
moving just before the ball hits the ground?
20.
zontal.
The rays
A ball
CHAPTER V
MAXIMA AND MINIMA
31. -A function of x
is
said to
have a maximum
a,t
x
=
a,
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
if
sufficiently near a.
If
we
represent the function
=
maximum
/ Wi
bottom of a wave.
y
If
is
^
the derivative
plot the curve
minimum
at the
continuous, as in Fig. 31a, the tangent
wave and
maxima and minima
horizontal at the highest and lowest points of a
the slope
of
is
by y and
occurs at the top, a
is
zero.
a function /
(a;)
Hence
we
in determining
first
look for values of x such that
lfix)=nx) =
If
a
is
a root of
minimum,
this equation,
or neither.
T
/
(a)
o.
may
be a maximum, a
DIFFERENTIAL CALCULUS
40
=
atx
a,
iff
(x) is positive for values of
for values a
tive
If
the slope
is
Chap. V.
xa little less and negor
greater than a.
little
negative on the
left
and positive on the right,
and the ordinate is a
as at B, the curve rises on both sides
minimum.
That
is,
negative for values of
/
has a
(x)
xa
minimum
and
little less
=
x
at
a,
iff
positive for values
(x) is
a
little
greater than a.
If
the slope has the same sign on both sides, as at C, the
curve
rises
neither a
falls on the other and the ordinate
nor a minimum. That is, / (x) has
a minimum at x = a if f (x) /los the
on one side and
maximum
maximum nor
neither a
is
same sign on both sides of a.
Example 1. The sum of two numbers
mum
Find the maxi-
is 5.
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
is
=
X
ib
maximum.
a
—
x)
The
=
5 X
— x^
derivative
is
f = 5-2x.
ax
This
Fig. 31b.
ti
is
zero
when x =
the derivative
is
f
than f the derivative
a;
a;
=
=
J the graph then has the shape
f the function has
its
f (5
Ex.
its
2.
.
If
positive.
is
shown
x
is
If
x
than
less
is
negative.
greater
Near
At
in Fig. 31b.
greatest value
-
f)
= ¥•
Find the shape of the pint cup which requires for
construction the least
amount
of tin.
Let the radius of base be r and the depth
tin used is
A = TTT^ 2 irrA.
h.
The area
+
Let V be the number of cubic inches in a pint.
v
=
irr^h.
Then
of
MAXIMA AND MINIMA
V
Chap.
Consequently,
41
=
h
irr'
4 = Trr^
and
t and
Since
—=2 ^~
—
2
is
zero
if tit'
=
v.
"
(
J.2
^
If there is
a
dr
it
2v
v are constants,
dA
This
+
^2
V
y
maximum
or
minimum
must then occur when
= V!;
has any other value,
for, if r
dA
-p
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
This must then be the value
there must be a least amount.
of
A
when
that r
=
v
=
Also
in^.
The cup
h.
v
=
We
irr%.
therefore conclude
requiring the
thus has a depth equal to
least tin
the radius of
base.
its
The strength of a rectangular beam is proportional to
the product of its width by the
Ex.
3.
square
of
strongest
from a
Find the
depth.
its
beam
that can be cut
24 inches in
circular log
diameter.
In Fig. 31c
is
shown a
and beam.
beam. Then
of the log
of the
Let x be the breadth and y the depth
a;2
The
strength
of.
the
+
beam
S=
Fig. 31c.
section
kxy^
2/2
=
(24)2.
is
=
kx (242
_
a.2)^
.
DIFFERENTIAL CALCULUS
42
k being constant.
The
dS
S
derivative of
=
dx
k (24^
x = ±8 Vs.
cannot be negative.
-
3
Since x
beam,
Hence
the only solution.
there
maximum
2/
=
Ex.
Since no other value can
8
minimum, a; = 8 v3 must be
beam. The corresponding depth
or a
the width of the strongest
is
the breadth of the
Since the log cannot be infinitely strong,
must be a strongest beam.
give either a
is
= 8^3
a;
is
is
x2).
If this is zero,
it
Chap. V.
Ve.
Find the dimensions of the largest right circular
4.
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 ABCj
DE^AB
EC
that
h
y
r
—
X
V
Fig. 31d.
its
=
derivative to zero,
2 rx
-
3
y
r
The volume
Equating
BC'
is,
=
-
of the cylinder
irx'^y
we
a;^
=
—
{rx^
-
(r
x).
is
—
x^)
get
=
0.
Hence x =
or a; = | r.
The value a; = obviously does
maximum. Since there is a largest cylinder, its
radius must then be a; = | r. By substitution its altitude is
then found to be 2/ = ^h.
32. Method of Finding Maxima and Minima.
The
method used in solving these problems involves the following
not give the
—
steps:
MAXIMA AND MINIMA
Chap. V.
Decide what
(1)
it
be
to be a
maximum
or
minimum.
Let
y.
Express y in terms of a single variable. Let it be x.
be convenient to express y temporarily in terms of
(2)
It
is
43
may
several variable quantities.
If
our present methods, there
nate all but one of these.
will
the problem can be solved
Calculate -p and find for
(3)
by
be relations enough to elimi-
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.
determine the signs of
If not,
a
-j^
when
a;
is
a
and
and apply the
greater than the values in question
little
little less
criteria given in Art. 31.
EXERCISES
Find the
1.
2
2.
6
Show
maximum and minimum
_
a;2
+
x\
6.
a?
9.
Show
11.
x*
-2x^
+
The sum
Q.
^^
that x
4.
•
Va? - x'
maxima
+-X
has a
-
7.
6
8.
X Va^
a;*
15 a^
+
or minima:
+
10 x\
a;".
maximum and a minimum and
that the
than the minimum.
is less
of the square
and the
reciprocal of a
number
is
a mini-
Find the number.
Show
square.
12.
3.
7.
x\
+ 4 X.
maximum
mum.
+
-
a;
that the following functions have no
5.
10.
5x
12
values of the following functions:
that the largest rectangle with a given perimeter
is
a
<
Show that
the largest rectangle that can be inscribed in a given
a square.
Find the altitude of the largest cylinder that can be inscribed in
circle is
13.
a sphere of radius
a.
A rectangular box with square base and
open at the top is to be
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?
14.
made out
of
DIFFERENTIAL CALCULUS
44
15.
A
cylindrical tin
Show
capacity.
Chap. V.
can closed at both ends is to have a given
amount of tin used wiE be a minimum when
that the
the height equals the diameter.
16. What are the most economical proportions for an open cylindrical
water tank if the cost of the sides per square foot is two-thirds the cost
of the bottom per square foot?
17. The top, bottom, and lateral surface of a closed tin can are to be
cut from rectangles of tin, the scraps being a total loss. Find the most
economical proportions for a can of given capacity.
18. Find the volume of the largest right cone that can be generated
by revolving a right triangle of hypotenuse 2 ft. about one of its sides.
19. Four successive measurements of a distance gave ai, 02, Oa, 04 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?
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
20.
If the
maximum
capacity.
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 which is twice as intense as the other, at
which the illumination is a minimum.
22.
23.
wide.
The
sides of a trough of triangular section are planks 12 inches
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
A beam of square section is to be cut
ft. and 1 ft.
of its ends being 2
from the
26.
A
log.
Find
length
if
the perimeter is 30 ft.,
circle.
If
amount
of Ught
27.
its
window has the form
may
beam is a maximum.
a rectangle surmounted by a semifind the dimensions so that the greatest
the volume of the
of
be admitted.
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
The. whole forms a model of a right cylinder. Calculate the
lengths into which the wire must be divided to produce the cylinder of
greatest volume.
pieces.
MAXIMA AND MINIMA
Chap. V.
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.
If 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.
Find the length
31.
Une which
of the shortest
will divide
an equi-
lateral triangle into parts of equal area.
A
32.
angle
is
an oval curve.
triangle is inscribed in
a
maximum, show
If
the area of the
tri-
graphically that the tangents at the vertices
of the triangle are parallel to the opposite sides.
33.
A
and C
are points on the
Ught passes from
A
same
side of a plane mirror.
A
ray of
C by way of a point B on the mirror. Show that
path ABC will be a minimum when the lines AB,
the length of the
to
CB make
equal angles with the perpendicular to the mirror.
Let the velocity of light in air be Vi and in water Vi. The path of
a ray of light 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 fli and 0i 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 wiU be least if B is so
34.
placed that
sin $1
sin
35.
The
62
_
vi
Vi
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 proporand is $20 per hour for a speed of 10 knots,
tional to the cube of her speed
and if the other expenses amount to $100 per hour,
ical
speed in
33.
atill
Other Types
given in Art. 31
find the
most econom-
water.
of
Maxima and Minima.
is sufficient
to determine
—
The method
maxima and minima
the function and
its derivative are one-valued and continuIn Figs. 33a and 33b are shown some types of maxima
and minima that do not satisfy these conditions.
At B and C, Fig. 33a, the tangent is vertical and the deif
ous.
rivative infinite.
At
D the slope on the left is different from
DIFFERENTIAL CALCULUS
46
that on the right.
and
E
The
derivative
is
Chap. V.
At
discontinuous.
A
This happens in problems where
values beyond a certain range are impossible. According to
the curve ends.
Fig. 33a.
our definition, y has
and E.
If
maxima
more than one value
at A, B,
D
and minima
at
C
of the function corresponds to a
single value of the variable, points like
B, Pig. 33b,
A
may
and
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 deFig. 33b.
the function ceases to
rivative
exist,
is
discontinuous,
or two values of the function
coincide.
Exatnyle
Find the
1.
on the curve
maximum and minimum
In this case,
dy
dx
No
finite
ordinates
= 7?.
y = x^ and
y^
2 _i
3
value of x makes the derivative zero, but
a;
=
MAXIMA AND MINIMA
Chap. V.
makes
Since y
it infinite.
minimum
is
never negative, the value
47
is
a
(Fig. 33c).
Y
Ex. 2. A man on one side of a river J 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.
i_
2
Fig. 33e.
Fig. 33d.
Let
tion.
(Fig. 33d) be the starting point and B the destinaSuppose he rows to C, x miles down the river. The
A
+
time of rowing will be ^ Va;^
J and the time of walking
J (2 — a;). The total time is then
t
=
i
ViH=l + H2 -
Equating the derivative to
'6
which reduces to 5
a;^
Vx^
+
f
zero,
+i
=
0.
we
x).
get
4
This has no real solution.
:
DIFFERENTIAL CALCULUS
48
The
Chap. V.
—
a;) is the time of walking only
above B. If C is below B, the time is i (r — 2).
The complete value for t is then
if
C
trouble
that J (2
is
is
t
the sign being
+
=
i
v^n ±
if
x
<
2 and
the equation connecting x and
i (2
—
t
if a;
is
- x),
> 2. The
shown
graph of
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
maximum and minimum values
Find the
1.
x'
=
+
2/'
=
x'{x
a^.
2/3
of y
on the following curves
= X* = t' +
1.
X
i^, y = t' - tK
6. Find the rectangle of least area having a given perimeter.
6. Find the point on the parabola y^ — ix 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 required 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
2.
y^
-
3.
4.
1).
+
resistance of J
ohm
is
C =
60 xE
20
s2
How many
current?
cells
+
should be placed in each row to give the
maximum
CHAPTER
VI
DIFFERENTIATION OF TRANSCENDENTAL
FUNCTIONS
Formulas for Differentiating Trigonometric Func-
34.
tions.
— Let u be the circular measure of an angle.
= cos u du.
— — sin u du.
= sec^ u du.
= — csc^ u du.
= sec u tan u du.
= — CSC u cot M du.
d sin u
d cos v,
d tan u
d cot u
d sec u
d CSC u
VII.
VIII.
IX.
X.
XL
XII.
The negative
sign occurs in the differentials of all co-
functions.
35.
The Sine
of natural
of a Small Angle.
sines will
— Inspection of a table
show that
the sine of a small angle
is
very
nearly equal to the circular meas-
Thus
ure of the angle.
sin 1°
180
We
=
0.017452,
=
0.017453.
should then expect that
sin 6
lim-^=
O
,.
1.
(35)
Fig. 35.
9=0
To show
this graphically, let d
PM perpendicular to OA.
is
defined
The
= AOP
circular
by the equation
e
=
arc
arc
AP
OP
rad.
49
(Fig. 35).
Draw
measure of the angle
DIFFERENTIAL CALCULUS
50
Also sin d
=
jyp
Hence
.
e
6
MP _ chord QP
_
~
sing
As
Chap. VI.
arc
AP ~
arc
QP
approaches zero, the ratio of the arc to the chord ap'
proaches
—r-
Therefore the limit of
1 (Art. 53).
Proof of VII, the Differential of the Sine.
36,
y
=
sinu.
+ Ay
=
sin (u
is 1.
— Let
Then
y
-\-
and so
Ay =
shown
It is
;
If
Am)
—
sin u.
in trigonometry that
sin A -
sin
A = u+
then
+
sin (w
Am)
~
5 =
2 cos I (A
B=
u,
Ay = 2cos
{u
Am,
+ B) sin |
(A
-
B).
+ i Am) sin § Au,
whence
Am
^ = 2cos(m +
„
/
,
1
.
,
i Am)
As Am approaches
Am
=
^^
sin A
,
cos (m
,
.
,
+ |Am)
,
sin i
^
Am
^^
.
zero
Am _ sin 8
Am ~ ~e~
sin §
I
approaches
1
and cos (m
+ § Am)
approaches cos u.
There-
fore
dy
-rr
au
=
cos M.
Consequently,
dy
37.
Proof of
Vm,
=
cos
u du.
the Differential of the Cosine.
trigonometry
cosM
=
sin(^
—
MJ-
— By
.
'
TRANSCENDENTAL FUNCTIONS
Chap. VI.
51
Using the formula just proved,
dcosu = d
38.
sin
(^
— wj =
— u]dl- — u]=
cosf^
—sin u du.
— Differentiating both
Proof of IX, X, XI, and XII.
sides of the equation
tan u
and using the formulas
sin u and cos u,
d tan u
=
cos
Md
sin
5
=
By
u
sec^
wd
cos
u =
cos^
differentials of
udu +
„^
U
sin^
u du
. „.
COS''
u
du.
differentiating both sides of the equations
cot
u =
COS
—.
sin
M
u
1
u =
sec
,
cos
and using the formulas
we
sin
u
COSM
sin
proved for the
just
—
M
COS^
=
u
CSC
,
1.
y
=
1
=
sin
for the differentials of sin
obtain the differentials of cot u, sec
Example
u
sin^
(a;^
u and
M
u and cos
u,
esc u.
+ 3).
Since
sin2
we use the formula
dy=-2 sin
Ex.
-r
+ 3) =
for u^
(a;^
and
[sin (x^
+ 3)]^,
so get
+ 3)d sin (x^ + 3)
+ 3) cos {x^ + 3) d {x^ +
+ 3) cos {x^ + 3) dx.
= 2 sin (a;2
= 4 X sin (a;2
= sec2x tan 2 x.
3)
sec 2
+ tan 2 -5dx
= sec 2 sec^ 2 (2) + tan 2 x sec 2 tan 2 x (2)
= 2 sec 2 (sec^ 2 + tan^ 2 x)
=
dx
y
2.
(a;2
sec 2
a;
T- tan
ax
2x
a;
a;
a;
a;
a;
a;
a;
EXERCISES
In the following exercises show that the derivatives and differentials
have the values given:
1.
2/
=
2 sin 3
a;
+ 3 cos 2
a;,
-p
=
Q (cos 3 x
—
sin 2 x).
,
,
'
•
•
DIFFERENTIAL CALCULUS
52
=
Chap. VI.
y
=
sin'
3.
2/
=
2 cos a; sin 2 a;— sin a; cos 2 s, dy
=Z cos x cos 2 x dx.
,
V
==^
1 —
—
^
2.
4»
-
dy
ix
cos
dy
—
—
,
sin i
dx
dy
a;
6.
2/
=
tan 2
e
6.
y
=
cot'^csc'^,
7.
X
=
a cos
+ sec 2
a;
,,a;
,a;
=
y
t,
asin^
/,
—
X
—x
cos \
Z sin' J
a;
=
2 sec 2 x (sec 2
d?/
^
=
.a;
.xt
,x
-cot2Csc'2(csc'2
-^
= —3 sin
-3^
a;,
^ cos ^ dx.
sin
t
cos
a;
+ tan 2
a;).
,,x\
+ cot'-j,
i.
da;
8.
a;
=
a
—
(</>
«
sin 0),
+
=
a
—
(1
cos
—
=
3^
ax
</>),
cot ^•
i
d?v
-rK
ax'
1
=
9.
.-c
=
10.
2/
=|cot'i— icot'x+cotx+x, dy = —cot'xdx.
cos
i
sin
i
y
<,
=
sint
t
cos
t,
.
12.
13
V
'•
^.
y
i tfirf s
y
=
_
15.
y
=
cosx/1 ..
=— ^sm'x
I
1
_
—
18.
y
II
=
y
:
j
—
sin
— tan x
X
+ tan x
(cotx
=A
dy
-rdx
dy
dx
X
sec X
sec
17.
J,'
dy
,5. \,5
+,5.,
YoSin'x + gSinx + T^x,
+—sin X
-^r-^
1
^„
+ f tan' x+ tan x,
I
14.
.,
'
c
= sec' x dx.
^ sec' 9 — f sec' 9 +^ sec' 9, du = tan" 9 sec' 9 d9.
-J x dx.
J
= x\/cos'x
^ sm X, dy = x sm'
—q— —cos ^ \,1.,
+ Q sin' X +,2.
=
M =
11.
r-COS''
t
— 3tanx)
cos
Vcotx,
+ B sin
(?ix)
-,-
"^
(rax),
=
=
-Bj
sm' x dx.
2 cos x
,ri
(1
_
~
,
dy
rz— sin x)'
-
—
2 sec x (tan x
=
sec x)
+ tan x
sec x
5—
2cot^x
A
where
and
B
are constant,
show that
19.
Find a constant
A
such that
g+
20.
Find constants
satisfies
A
and
j/
=A
sin 2
52/
= 3sin2x.
B
such that y
x
satisfies
= A
sin 6
the equation
x
+ 5 cos 6 x
the equation
S+4x+«^=^^'"^21.
TT
Find the slope
of the curve
2/
=
2 sin
a;
+
3 cos
a;
at the point
TRANSCENDENTAL FUNCTIONS
Chap. VI.
53
Find the points on the curve y = x + sin2x where the tangent
y = 2 x + 3.
If
23. A weight supported by a spring hangs at rest at the origin.
the weight is lifted a distance A and let f aU, its height at any subsequent
time t will be
y = A cos {2imt),
22.
is
parallel to the line
n being
constant.
origin.
Where
24.
A
is
Find
its
and
acceleration as
Where
is
it
passes the
the acceleration
revolving light 5 miles from a straight shore makes one com-
plete revolution per minute.
beam
velocity
the velocity greatest?
when
Find the velocity along the shore of the
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 respectively, and
that these quantities are changing at the rates of 3 ft., —2 ft., and 10
of light
it
how fast is the area of the triangle changing?
a crank and AB a connecting rod attached to a piston B
moving along a line through 0. If OA revolves about
with angular
prove that the velocity of B is MC, where C is the point in
velocity.
which the Une BA cuts the line through O perpendicular to OB.
28. An alley 8 ft. wide runs perpendicular to a street 27 ft. wide.
What is the longest beam that can be moved horizontally along the
degrees per second,
27.
OA
is
fc),
street into the alley?
29.
A needle rests with one end in a smooth hemispherical bowl.
needle will sink to a position in which the center
is
The
as low as possible.
If the length of the needle equals the diameter of the bowl,
what
will
be
the position of equilibrium?
30.
A rope with a ring at one end is looped over two pegs in the same
and held taut by a weight fastened to the free end. If
the rope slips freely, the weight will descend as far as possible. Find
horizontal line
the angle formed at the bottom 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 be made by bending into shape a strip of copper so
that the cross section shall be
is a,
an arc
of a circle.
find the radius of the cross section
when
If
the width of the strip
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.
in
If
the bicycle
rolls straight
ahead,
what position will the outer ends of the two spokes be closest together?
DIFFERENTIAL CALCULUS
54
Trigonometric
Inverse
39.
X
sin~i
is
— The
used to represent the angle whose sine
=
y
x
sin~* X,
cot""' X, sec~' x,
=
sin
is x.
symbol
Thus
y
Similar definitions apply to cos~^ x
are equivalent equations.
tan~^ x,
Functions.
Chap. VI.
and
csc~' x.
by mulan indefinite number of
angles are represented by the same symbol sin"' x. The
algebraic sign of the derivative depends on the angle differentiated.
In the formulas given below it is assumed that
sin~' u and csc~' u are angles in the first or fourth quadrant,
COS"' u and sec~' u angles in the first or second quadrant.
Since supplementary angles and those differing
tiples of
2
have the sam'e
7r
If angles in other
must be
sign
are valid in
40.
all
—
quadrants are differentiated, the opposite
The formulas for tan-'w and cot~'M
used.
quadrants.
Formulas
Functions.
for Differentiating Inverse Trigonometric
du
Xm. d sin-'
vT w
du
XIV.
Vl-u^
du-
XV.
l
XVI.
XVII.
dcot-'tt
d
sec-'
+ u"'
du
=
1+1?'
u =
xvni.
41.
sine,
du
u Vw^ _
i'
du
u Vm^ _
Proof of the Formulas.
1
— Let
y
=
sin-i u.
siny
=
u.
cos y dy
=
du,
Then
Differentiation gives
TRANSCENDENTAL FUNCTIONS
Chap. VI.
whence
dy
'^
=
55
'
cosy
But
cos
=
2/
y is an angle
Hence
If
± Vl — sin^
2/
± Vl — w'.
=
in the first or fourth quadrant, cos y is positive.
cos y
= Vl —
u^
and so
,
dy
The
=
du
Vl -m2
other formulas are proved in a similar way.
EXERCISES
1.
^
=
sin-i
-
{Sx
3dx
1),
V6x -
9a;2
V2 ax - x2
3.
2/
=
tan-'|^,
4.
2/
=
cof
5.
6.
2/
=
dv_ -2
-(i-rj.
sec~^
dx
(4 X
dy
dx
2/=^csc-"
2
1.
8.
o
9.
y
X
2/
=
tan
+a
csc~' (seo0),
=
-1
sin 1-
11.
^
12.
=
2/
=
2/
_,
= itan^
„
13.2/
1)
Vx
1
V2 + 2x-4x2
_
a
x'
~
a?
-\-
~^'
^2/
- x2
d^;
sec-i
I
,
V 1 — x^
,
v^^:^^ + 1 sin-i^,
3
4sinx
,
+ 5COSX
.
= sec-^,^,
(a2
^
=
dx
^
y
HO
+
Hfi
Va?
10.
dy
dx
dx
a
=
+l
d^/
1,
1,3
4x-l
—
x^
dx
V4x +
X
X
6dx
dx
-
3;2)
Vo^
^
.
Vl — x^
g ^ v^^r^s.
rf,.
,
_
5»
ckc
4
1
=
5
+ 3COSX
g=__^.
-
2 x^
DIFFERENTIAL CALCULUS
56
14.
y
"
15.
2/
=
+ V^TT^,
asin-ia
= 2(3x
=
<k
dx
Chap. VI.
J^^^ a + X
+ l)* + 4cot-i^^-^-±^.
L
dj/
ic
16.
= 51*tan-1
6
•
2/
17.
3^
=
"
2/
+
_,x
cos
+ 2a;2
2
'
-
g'
=
tan-'
20.
y
=
x sin-' x
21.
2/
=
x2 sec-'
Let
22.
x2
.
+
2/2
=
,
I
Let
-
+
2
Show
A
(3a;
+ 1)^'
1_
dt/
j
1)8+4
j-
=
dx
17a;2+4
+
a:
_
(3
:
=•
1
V3 - 2x ,
a;)
a;2
1
dx~ajVx2 -
o^'
^
^ -^Vx^+4x-4
1
^
.
^ = 2xsec-i|-
Vx^":^,
be the arc from the x-axis to the point
s
a*.
(x,
y)
on the
=
a?,
circle
that
= _«,
*=«,
ay
y
ds2=rfx.+d^..
x
be the area bounded by the circle x^
through (x, y). Show that
-\- y'^
the
y-a,jaa
vertical line
A =
The end
24.
4a;*
^=
+ Vl - x^
*
23.
a'
^
ax
and the
03*'^"
+ Vx2 + 4 X - 4
y
dy
dx
^
_jX
]_
2
19.
+
.
^~ 2oV
X
(3a;
2
2x
Tocos';;
3 - a;
V3
l
s
2
_ Vx'
dx
of
xy
+ a^ tan"' -
,
dA =
2
2/
dx.
a string wound on a pulley moves with velocity v
Find the angular
along a line perpendicular to the axis of the pulley.
velocity with which the pulley turns.
A tablet 8 ft. high is placed on a wall with its base 20 ft. above the
25.
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?
level of
—
Exponential and Logarithmic Functions.
If a is a
and u a variable, a" is called an exponential
function. If m is a fraction, it is understood that a" is the
42.
positive constant
positive root.
li
That
y
=
a",
then u
is
called the logarithm of
y to base
a.
is,
2/
=
a",
u =
logay
(42a)
TRANSCENDENTAL FUNCTIONS
Chap. VI.
are
by
definition equivalent
57
Elimination of u
equations.
gives the important identity
aio8.!'=2/.
(42b)
This expresses symboUcally that the logarithm
is the power
which the base must be raised to equal the number.
43. The Curves y = O'.
Let a be greater than 1.
The graph of
to
—
=
y
O"
has the general appearance of Fig. 43
increment A, the increment in y
If
X receives a small
is
Aj/
=
—
a^+*
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 joining
A
and
(0, 1)
P
g''-
(x,
y) is
1
X
As Pi moves toward A the slope
of APi increases.
As P2 moves toward
decreases.
A
the slope of
AP2
Furthermore the slopes of AP^ and APi approach
equality; for
1
-k
=-e-^')'
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
,.
a^
—
lim
1=0
X
1
=
m.
(43)
DIFFERENTIAL CALCULUS
58
This
is
Definition of e.
44.
number
e
a" at the point
where
— We shall now show that there
is
it
a
such that
(44)
^
1=0
In
=
the slope of the curve y
crosses the y-axis.
Chap. VI.
fact, let
=
e
m
where
is
^ — —1I =
limbO
a"
the slope found in Art. 43.
X
,.
Inn
bO
a™
a"
—
1
—1 = —
X
,.
a"*
a"
Then
—
lim
1
—1
=
m m
1.
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 .4 is 1. If a > e the slope is greater than 1. If
a < e, the slope is less than 1.
Fig. 44.
We
shall find later that
e
=
2.7183
Logarithms to base e are called natural
In this book we shall represent natural log-
approximately.
logarithms.
arithms by the abbreviation
logarithm of u.
In.
Thus In u means the natural
TRANSCENDENTAL FUNCTIONS
Chap. VI.
45.
tions.
Differentials of Exponential
—
=
=
XIX.
<«e"
XX.
d!a"
XXI.
dhiu =
59
and Logarithmic Func-
e" du.
a" In a du.
—u
log„e du
V VTT d^ !.,„
XXII.
lOga'U' = —^
ttl
46.
Proof of XIX, the Differential of e».
y
=
— Let
e".
Then
y
+
Ay =
e"+^",
Ay =
6"+^"
whence
—
e"
=
e"
=
e" du.
(e'^"
—
1)
and
Am
Am.
As Am approaches
zero,
by
(44),
gAu
_
1
Am
approaches
1.
Consequently,
-^
47.
=
dy
e",
Proof of XX, the Differential of a".
a"
_
=
e"'"".
a)
=
e"
^
gives
— The identity
glna
Consequently,
da"
48.
=
e"
Proof of
Let
arithm.
—
'"""
d (m In
" " In a dw
XXI and XXn,
=
=
u.
dy
=
dM.
Inu.
Differentiating,
e^
a" In a dit.
the Differential of a Log-
e"
y
Then
=
DIFFERENTIAL CALCULUS
60
Chap. VI.
Therefore
du
J
The
derivative of logo
Example
J
dy
Ex.
y
1.
X
d
^—
sec' x
sec^
=
y
2.
=
=
=
dy
]n
u
is
du
found in a similar way.
(sec'' a;).
2 sec X (sec x tan x dx)
—
^^
sec^
x
=
„ ^
,
2 tan x dx.
2t='°"'^
2'»°"'»'
d (tan-i x)
In 2
=
2""'"'Mn2cia;
+ x^
1
EXERCISES
-
2.
y
=
dy
'-
1
g = 2 a*»°^^ln a
a*^2^,
x-l
5.
2/
=
I"
6.
2/
=
a^2«,
7.
2/
=
In (3
y
=
9.
2/
=
In (x
10.
y
=
\n (sec ax
8.
V+
e-^'
+ re%
=
+ n^ln
^
dx
^ = a^x»-i(a + a;lna).
»ia;"-i
a;2
+ 5X +
ln sec^ x,
+
—
+ tan ax),
12.
3/
=
In sin X
+ i cos" x,
1
X
2/
=
.
cosx
2sin2x
1
=
"2
tan
x.
1
a?),
+ ^.),
o *° t^'^ 2
2
3i2
dx
dy
ln(a^
,
dz
-y-
Vx2
+5
+ 5a; + i
6l
1),
=
13.
n.
a-;
11.2/
,n
2 x.
-e"
=
e*
2/
4.
sec^^
x-l
-?-
ox
=
a sec
ax.
g,
a-lna
u?/
-;^
COS X
ax
dy
dx
=
—.
sin
X
1
sin^x
+ 5^1n.
.
TRANSCENDENTAL FUNCTIONS
Chap. VI.
^
.
J
"•
1
_^
,
,.1,
= -m
15.
16.
17.
18.
19.
20.
1_
4*^3:2-4
^
2/
"
X
-
a
+
Va^
—
x2
dy
dx
dy
—
dx
8
^
a; (a;^
—
-
4)2'
1
^j
Va^
X''
2/=ln(V^T3 + Vx + 2)+V(x + 3)(x + 2),
2/
2/
=
In
(VxTa +
V^),
y
=
ef^ (sin
2/
=
i tan*
ax
X—
—
cos ax),
J tan*
!^^
x
i(.+
2
Vx2
+ ax
^
ax
=
tan-i--
j^
=
2 ae<" sin ax.
— In cos x,
i}
J = V|-^-
^
=
"^
=xtan-i--^ln(x'i+a^),
a z
21.x = aln,2. =
22.
a;2-4'
61
a
-3^
—
tan^ x.
i=i(-J)-
DIFFERENTIAL CALCULUS
62
Sl.^From equation
(44)
show that
=
e
lim (1
x=o
32.
If the space described
acceleration
33.
is
+ »)»•
by a point
is s
=
ae'
Assumfng the
and k being
+
fee"*,
show that the
equal to the space passed over.
resistance encountered
by a body sinking
to be proportional to the velocity, the distance
g
Chap. VI.
constants.
Show
in water
descends in a time
t
is
that the velocity v and acceleration a
satisfy the equation
=
a
it
Also show that for large values of
g
t
—
kv.
the velocity
is
approximately con-
stant.
34.
Assuming the resistance of air proportional to the square
a body starting from rest will fall a distance
of the
velocity,
_
g
fc2
in a time
tion
t.
Show
."
™\,
eM
+ e-^ N
2
/
that the velocity and acceleration satisfy the equa-
a
=
kV
g--.
Also show that the velocity approaches a constant value.
CHAPTER
VII
GEOMETRICAL APPLICATIONS
— Let
Tangent Line and Normal.
a given curve at Pi{xuyi). It
geometry that a line
through (xi, yi) with slope
49.
of
rui
is
represented
is
mi be the slope
shown in analytic
by the
equation
y
—
yi
= mi(x -
xi).
This equation then represents the tangent at
where the slope of
(xi, 2/1)
the curve
The
is
mi.
PiN
line
perpen-
dicular to the tangent at
its
point of contact
Fig. 49.
is
caUed the normal to the curve at Pi.
tangent
is
is
mi, the slope of a perpendicular line
the equation of the normal at
Example
1.
to the ellipse
The
At
and
so
(xi, yi).
+
any point
the slope
is
of the curve
dy
dx
then
^ _
mi
The equation
is
Find the equations of the tangent and normal
2y^ = 9 at the point (1,2).
x^
slope at
(1, 2)
Since the slope of the
of the tangent
2/
-
2
is
X
2y
=
-|.
is
= -i
63
(a;
-
1),
DIFFERENTIAL CALCULUS
64
and the equation. of the normal
2/
Ex.
2.
2
=
4
is
(a;
!)•
Find the equation of the tangent to x^
at the point
The
-
—
y^
=
a^
{xi, yi).
Xi
slope at
(xi,
j/i)
is
The equation
2/1
is
Chap. VII.
then
2/
-
2/1
= - (a; -
a;i)
of the tangent
GEOMETRICAL APPLICATIONS
Chap. VII.
Find the angles determined by the Une y
Example.
and the parabola y
=
parabola
(1, 1)
and
any point
x
we
find that the fine
at
intersect
The slope
The slope at
(0, 0).
of the line is 1.
=
x^.
Solving the equations simultaneously,
and
65
of the parabola
y
\
is
= 2x.
^
ax
At
(1,
the slope of the
1)
parabola
is
then 2 and the
Fig. 50b.
angle from the line to the
parabola
is
then given by
tan
1
|3i
1
+2
3'
whence
l3i
At
(0, 0)
from the
=
tan-ii
=
18°26'.
the slope of the parabola
fine to
the parabola
tan
j32
=
is
is
and so the angle
given by the equation
0-1 =
1+0
-1,
whence
ft
The
:
-45'
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.
2."
3.
4.
5.
6.
The circle x^ + y^ ='5 at (-1, 2).
The hyperbola xy = i&t (1, 4).
The parabola y' = ax a.t x = a.
The exponential curve y = aVa,tx =
The
The
sine curve
x^
ellipse -j
2/
=
V^
+ fe
3 sin
=
1,
a;
at
at
a;
=
(xi, yi).
^
0.
DIFFERENTIAL CALCULUS
66
8.
The hyperbola x' + xy The semicubical parabola
9.
Find the equation
7.
x
at the point
<t>
—
=
a
sin
Show
<t>i'
- 2 x, a,t (2,0).
= x', at — 8, 4).
y"
y'
(
normal to the cycloid
of the
—
(<t>
Chap. VII.
y
<j)),
that
it
=
a
—
{1
cos
<^)
passes through the point where the
rolling circle touches the z-axis.
Find the angles at which the following pairs
=
4
10.
3/*
11.
x^
12.
x''
16.
Show
x'
a;,
=
iy.
+ y' = 9, x^ + y^— 6 X = 9.
+ y^ + 2x = 7, = 4 x.
all
y
14.
y
of curves intersect:
=
=
=
sin x,
logw
y
=
cos x.
=
y
x,
hix.
e^), 2/ = 2
i (e^
values of the constants a and 6 the curves
x^
—
y^
=
y
15.
2/2
that for
IZ.
xy
o'',
+
e*.
= V
intersect at right angles.
Show
17.
that the curves
y
=
e°*,
^
=
+ c)
e"^ sin (bx
are tangent at each point of intersection.
Show
18.
that the part of the tangent to the hyperbola xy == a' inis bisected at the point of tan-
tercepted between the coordinate axes
gency.
19.
Show
20.
Let the normal to the parabola
that the projection of
The
focus
F of
the parabola
that the tangent at any point
with
21.
FP and
The
the line through
foci of
are the points F' (
y^
= ax
at
P cut the x-axis at
N.
PN on the x-axis has a constant length.
j/^
=
ax
the point (i
is
P of the parabola makes
P parallel to the axis.
a, 0).
Show
equal angles
the ellipse
- Va^ - 6^,
P of the
tangent at any point
o)
and F (Va^ - b\ O)
Show that the
makes equal angles with FP and
.
ellipse
F'P.
22.
Let
P
I
be any point on the catenary
2/
23.
_^x\
"), Af the
M
P on the x-axis, and N the projection of on the tangent
Show that MN is constant in length.
Show that the portion of the tangent to the tractrix
projection of
at P.
X
= sU^ + e
"
2
\a- vV - xV
intercepted between the y-axis
length.
and the point
of
tangency
is
constant in
GEOMETRICAL APPLICATIONS
Chap. VII.
Show
24.
Une joining
67
that the angle between the tangent at any point
P to the
same at
origin is the
In
+ "y' =
Vx'
all
P
and the
points of the curve
k tan-'-X
25. A point at a constant distance along the normal from a given
curve generates a curve which is called parallel to the first. Fiad 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
Direction of Curvature.
51.
P
if
the part of the curve near
P
cave
y
(t>,
upward
at a point
=
b sin
<t).
— A curve
is
said to be con-
above the tangent at PIt is concave downward at Q
if the part near Q hes below
lies
the tangent at Q.
At "points where -p^ispositive,
Fig. 51.
upward; where
the curve is concave
-7-^ is
negative, the
curve is concave downward.
For
d^y
^
dx'
If
then -T^
is
positive,
by
d^/dy\
dx \dx/
Art. 13, -- the slope, increases as x
,
and decreases as x decreases. The curve therefore
above the tangent on both sides of the point. If,
increases
rises
however,
d^
-7-| is
negative, the slope decreases as x increases
and increases as x
decreases,
tangent.
so the curve falls below the
— A point like A
52. Point of Inflection.
one side of which the curve
concave downward, is
assumed that there is a,
flection.
and
is
called
(Fig. 52a),
on
concave upward, on the other
a point of
inflection.
It
is
definite tangent at the point of in-
A point like B is not
called
a point of
inflection.
DIFFERENTIAL CALCULUS
68
The second
Chap. VII.
on one side of a point of
Ordinary functions
change sign only by passing through zero or infinity. Hence
inflection
derivative
is
positive
and negative on the
to fimd points of inflection
other.
we find where t^
is
zero or infinite.
Fig. 52a.
If
the second derivative changes sign at such a point,
point of inflection.
sign
on both
If the
sides, it is
it is a
second derivative has the same
not a point of inflection.
Fig. 52b.
Example
1.
Fig. 52c.
Examine the curve By = x*
and points of inflection.
tion of curvature
The second
derivative
.
dx'
is
= 4
(^^
-
1).
—
6x^ ior
direc-
GEOMETRICAL APPLICATIONS
Chap. VII.
This
zero at
is
a;
=±
cave upward on the
It is positive
1.
and the curve conon the right of
= —1 and
x
left of
69
= +1.
It is negative and the curve concave downward
between x = —1 and x=+l. The second derivative
changes sign at A ( — 1, — #) and B (+1, — ^), which are
a;
therefore points of inflection (Fig. 52b).
Examine the curve y =
cc*
In this case the second derivative
is
Ex.
This
2.
is
of x.
is
for points of inflection.
zero when x is zero but is positive for all other values
The second derivative does not change sign and there
consequently no point of inflection (Fig. 52c).
Ex.
If
3.
> 0,
a;
—
sin
x
>
= smx-
x
+ —^-
show that
a;
^*
Let
y
We
show that y
are to
-p-
dx
When X
-p
a;
—
-rr
da;
is
1
x
Differentiation gives
0.
+ 2!
>
0,
* If
from
1
-r%= —sin
,
pT,
dx^
is less
then positive when
Since -^
dx
a;
= when x = 0, y
which was to be proved.
n
is
to
ra.
any
positive integer
ra!
> 0,
is
+
is
zero
a;.
is
positive.
when x
is
and so y increases with
therefore positive
when
represents the pi:oduct of the integers
Thus
3!
a;
than x and so -r^
increases with x.
Since y
X.
a;
cos
positive, sin
is
Therefore
zero,
=
>
= l-2-3 =
6.
DIFFERENTIAL CALCULUS
70
Chap. VII.
EXERCISES
Examine the
following curves for direction of curvature
of inflection:
1.
y
2.
2/-
3.
y
=a^-3x + 3.
= 2x' -3x2 -6z +
= x* - 4 x3 + 6 1^ + 12 1.
4.
2/3
=
l.
X
-
1.
and poiats
GEOMETRICAL APPLICATIONS
Chap. VII.
The
On
AB,
projection of a chord, such as
uct of
its
by the
length
the arc
is
71
equal to the prod-
makes with PQ.
Let a be the
PQ makes with the
cosine of the angle
it
AB is a tangent RS parallel to AB.
largest angle that
any tangent on the arc
Fig. 53.
chord PQ. The angle between RS and PQ is not greater
than a. Consequently, the angle between AB and PQ is not
Therefore
greater than a.
proj.
AB = AB cos a.
Similarly,
proj.
proj.
PA = PA cos a,
BQ = BQ cos a.
Adding these equations, we get
PQ = {PA + AB
It
line
is
+ BQ)
cos a.
evident that this result can be extended to a broken
with any number of
sides.
As the number
of sides in-
creases indefinitely, the expression in parenthesis approaches
the length of the arc PQ.
Therefore
PQ =
that
arc
PQ
cos a,
is,
chord
PQ
curve is continuous, the angle a apapproaches P- Hence cos a approaches
If the slope of the
proaches zero as
land
Q
chord
PQ
,
hm
57P =1.
Q=p arc PQ
DIFFERENTIAL CALCULUS
72
Since the chord
is
be greater than
1.
always
less
than the
arc,
Chap. VII.
the limit cannot
Therefore, finally.
,.
lim
PQ
jqTT^ =
PQ
chord
Q=p arc
— Let
,
1.
(53)
be the distance measured
to a variable point P
Then s is a function of the coordinates of P. Let be the
angle from the positive direction of the x-axis to the tangent
Differential of Arc.
54.
s
along a curve from a fixed point
A
<t>
PT drawn
in
the direction of increasing
s.
FiQ. 54a.
If
X, y,
Fig. 54b.
P
moves to a neighboring position Q, the increments
and s are
Aa;
From
= PR,
the figure
Ay = RQ,
it is
rr>T>r,\
(RPQ)
=
sin
(RPO)
-
RPQ
As
1.
arc
PQ.
As
_ = Ax
-—,
Ao;
^ - ^ Ai
approaches
_
PQ ~
approaches
=
seen that
cos
As Q approaches P,
As
in
arc
and
PQ
PQ
chord
The above equations then
cos^
=
dx
ds'
.
•
,
give in the Hmit
dy
as
(54a)
'
GEOMETRICAL APPLICATIONS
Chap. VII.
73
These equations express that dx and dy are the
sides of a
right triangle with hypotenuse ds extending along the tangent
(Fig. 54b).
be read
All the equations connecting dx, dy, ds,
One
off this triangle.
=
ds^
55.
its
|3
Curvature.
chord, the
—
If
amount
+ dy\
can
is
(54b)
everywhere concave toward
bent can be measured by the angle
it is
its
_
j3
is
dx''
an arc
between the tangents at
arc
and
of particular importance
is
ends.
The
-
_
~ As
<j}'
PP' ~
ratio
.
Atj}
4>
As
the average bending per unit length along PP'.
The
limit as P' approaches P,
lim
A<t>
As=o
is
called the curvature at P.
bends more sharply,
less
_d4
.
As
ds'
It is greater
where
it is
Fig. 55a.
In case of a
straight.
Fig. 55b.
circle (Fig.
<t>
where the curve
more nearly
=
55b)
B
+ 2'
d<t>
_
s
Consequently,
ds
dd
1
ado
a
=
ad.
DIFFERENTIAL CALCULUS
74
that
is,
the curvature of
reciprocal of
56.
its
Radius
any curve
radius of curvature of
curvature, that
to the
— We have just seen that the
the reciprocal of
is
and equal
circle is constant
of Curvature.
radius of a circle
its
a
radius.
Chap. VII.
is
curvature.
its
The
defined as the reciprocal of
is,
radius of curvature
It is the radius of the circle
=
p
=
ds
(56a)
yr-
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
= tan-|.
Consequently,
^^dx
1
Also
ds
dx^
ldy\
= Vdx^ + dy^.
Substituting these values for ds and
d<t>,
we
get
(56b)
"
cPy
dx^
If
the radical in the numerator
the same sign as
-r-^
,
that
is,
is
will
the radius will be positive
have
when
is concave upward.
If merely the numerical value
wanted, the sign can be omitted.
By a similar proof we could show that
the curve
is
taken positive, p
dy'
GEOMETRICAL APPLICATIONS
Chap. VII.
Example
y^
= Ax
75
Find the radius of curvature of the parabola
1.
at the point
At the point
dy
dx
(4,
(4, 4).
4) the derivatives
2-1
y
2'
Therefore
U + &)1
\dx/
_
have the values
DIFFERENTIAL CALCULUS
76
same slope
at P.
-r-
Since they have the
Chap. VII.
same radius
of
curvature, the second derivatives will also be equal at P.
T
Fig. 57.
The
-rr
ax
circle of
curvature
is
thus the circle through
and to have the same values
P such that
for the circle as for the curve
ax'
at P.
EXERCISES
1.
is s
=
The
x''
+
length pf arc measured from a fixed point on a certain curve
X.
Find the slope of the curve at x = 2.
cos s,y = sin s, represent a curve on which s is the length
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
2.
Can X =
of arc
point indicated:
3
^-
1+^-1
+ ~
4.
x^
62
o^
at
^'
+ xy + y'
=
(0, b).
3,
at
(1, 1).
Find an expression for the radius
5.
r
=
e'.
6.
r
=
a
at9=2(1
of ciuvature at
+ cos 6),
any point
at S
of
=
0.
each of
the following curves:
7.
y
=
l[^ +
8.
X
=
In sec y.
11.
Show
infinite.
e ^)-
9.
X
=
ly^
10.
r
=
a
—
ilny.
sec^ \ 9.
that the radius of curvature at a point of inflection
is
GEOMETRICAL APPLICATIONS
Chap. VII.
12.
A
77
point on the circumference of a circle rolling along the x-axia
generates the cycloid
= a
X
—
{<t>
y
sm<l>),
—a
—
(1
cos
<j>),
a being the radius of the rolling circle and the angle through which it
has turned. Show that the radius of the circle of curvature is bisected
by the point where the rolling circle touches the x-axis.
ij>
13.
A
string held taut
unwound from a
is
The end
fixed circle.
of
the string generates a curve with parametric equations
X
=
a cos
6
aB sin
-{-
= asmB —
y
8,
aS cos
8,
a being the radius of the circle and d the angle subtended at the center
by the arc unwound. Show that the center of curvature corresponding
to any point of this path
the point where the string
is
is
tangent to the
circle.
14.
Show that
cycloid x^
the radius of curvature at any point
+ y^ =
to the tangent at
o'
of the hjrpo-
(x, y).
cos X
1
58.
(x, y)
three times the perpendicular from the origin
is
Limit of
It
is
shown
in
trigonometry
that
—
1
cos
a;
=
2 sin" ^
Consequently,
1
—
2.x
= sm
2 sm^ ^
COS X
=
X
X
sm
I
5
2
X
2 /
\
.
-
z
X
smAs X approaches
approaches
zero,
,.
hm
1
—
1=0
59.
Derivatives
of
angle from the outward
cosa;
Therefore
0.
X
—
Arc in Polar Coordinates.
The
drawn radius to the tangent drawn
in the direction of increasing s
letter ^.
= 0-1 =
1.
is
usually represented
by the
DIFFERENTIAL CALCULUS
78
Let
r,
those of
B
be the polar coordinates of P, and r
Q
Draw QR
(Fig. 59a).
As = arc PQ.
let
Chap. VII.
-\-
Lr,d
perpendicular to
M
and
Then
sin Ag A9 _As
= RQ _ (r+ Ar) sin ^e _..
-^^+^^) AO
As' PQPQ~
PQ
(r
At-)
r
PR
cos
A9
_ +
(,RPQ) =
.
.
sin (/2FQ)
cos
\-
PR
'
PQ~
=
Ar
cos (Ae)
p^
- r (1 —
cos
Afl)
PQ
/•
^°'(^^)ii-^-
(1
-
—
cos Ad)
A0
PQ
Fig. 59b.
Fig. 59a.
As Ad approaches
As
Ag
As
zero,
sin
lim(i2FQ)=^, lim-
A&
A6I
= 1, hm
The above equations then
sm^ =
,
,.
l-cosA0
^g
As
,
^0, limp^=l.
give in the limit,
rdd
.
dr
COSl^
ds
(59a)
ds
These equations show that dr and rcW are the sides of a
right triangle with hypotenuse ds
and base angle
^.
this triangle all the equations connecting dr, dd, ds,
can be obtained.
The most important
tan^ =
^.
d^ =
dr^
From
and
\l/
of these are
+ r^ dS^.
(59b)
GEOMETRICAL APPLICATIONS
Chap. VII.
The
Example.
In this case, dr
=
logarithmic spiral r
=
ae^-
and so
aeP dd
tan
79
= —;— =
yL
1.
dr
The angle ^ is
The equation
dr
,
''''l'
shows that
-j-
is
and equal to 45
therefore constant
=
ds
also constant
degrees.
1
=
V=2
and so
and
r
s increase
propor-
tionally.
EXERCISES
Find the angle i^ at the point indicated on each of the following curves:
=
1.
The
spiral r
=
2.
The
circle r
= asinflate =
3.
The
straight line r
4.
The
ellipse r (2
6.
6.
ad, a,t 6
-
=
^^
o
-:
4
=
a sec '9, at 9
cos 6)
=k,a.td
^•
= |-
= | ir.
The lemniscate
2 a? cos 2 9, at
Show that the curves r = oe», r = ae~« are perpendicular
r''
=
fl
at each
of their points of intersection.
FLad the angles at which the curves
7.
r
=
a cos
(1
—
cos 6) where the tan-
r
B,
=
a sin 2
9
intersect.
Find the points on the cardioid
8.
gent
is
9.
r
=
a
parallel to the initial line.
Let
P
(r,
8)
be a point on the hyperbola
that the triangle formed
r-^
sin 2 9
by the radius OP, the tangent
=
c.
at P,
Show
and the
a;-axis is isosceles.
10.
Find the slope
of the curve r
=
e^* at
the point where
Angle between Two Directed Lines in Space.
directed line is one along which a positive direction
60.
A
9=2"
assigned.
This direction
is
usually indicated
—
is
by an arrow.
DIFFERENTIAL CALCULUS
80
Chap. VII.
An
angle between two directed lines is one along the sides
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 defined as that between intersecting lines respectively parallel
of
to the given lines.
M
i^&iC-
Fig. 61.
Fig. 60.
61.
try* that the angles a,
the line
cos a
0^2
=
— It
is shown in analytic geomey between the coordinate axes and
P1P2 (directed from Pi to P2) satisfy the equations
Direction Cosines.
— Xl
P1P2
fi,
-yi
=
cos /3
^""''
'
PlP2
22
cos
"""
'
7
'
— Zi
P1P2
These cosines are called the direction cosines of the
They
line.
satisfy the identity
cos^
If
(61a)
a
+ cos^ +
/3
cos^
7 =
1.
(61b)
the direction cosines of two lines are cos ai, cos /3i, cos 71
a^, cos j32, cos 72, the angle 6 between the lines is given
and cos
by the equation
cos 6
=
cos ai cos
In particular,
if
a.2
+ cos
182
cos ^2
+ cos 71 cos 72.
the lines are perpendicular, the angle B
(61c)
is
90
degrees and
=
cos ai cos a2
* Cf.
H. B.
+ cos
|8i
cos
1S2
+ cos 7i cos 72.
Phillips, Analytic Geometry, Art. 64, et seq.
(61d)
GEOMETRICAL APPLICATIONS
Chap. VII.
81
Direction of the Tangent Line to a Curve.
62.
— The
a point P of a curve is defined as the limiting
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 j8,
cos 7 the direction cosines of the tangent
tangent
line at
position
PT
drawn
X
in the direction of increasing
If
X,
+
Aaj,
y,
y
+
s.
A?/, s
+ Lz, those of
^'°'
P,
are the coordinates of
z
^'^^'
Q, the direction cosines of
PQare
Ay
PQ'
Lx
PQ'
Az
PQ'
As Q approaches P, these approach the
the tangent at Pcos
a
=
,.
lim
Q=P
On
the curve,
,.
lim
direction cosines of
Hence.
x, y, z
Ax dx
— -r
TAs
ds
Ax
7^7;
PQ
Ax
= am -r—
,.
As
are functions of
s.
As
7^7,
•
PQ
Hence
As
arc
hm r-^ = hm -j 7 =
PQ
chord
,
,.
,.
^
1.*
Therefore
cos
a =
-7-
(62a)
•
Similarly,
d^
^y
a
cos^=^,
These equations show that
if
along the tangent, dx,dy, dz are
nate axes (Fig. 62b).
*
The proof that
a distance ds
its
is
measured
projections on the coordi-
Since the square on the diagonal of a
the limit of arc/chord
is
1
was given
for the case of plane curves with continuous slope.
can be given for any curve, plane or space, that
direction.
fan \
(62a)
cosT=^^.
A
is
ia Art. 53
similar proof
continuous in
DIFFERENTIAL CALCULUS
82
rectangular parallelepiped
is
equal to the
dx^
+ dif + dz\
Chap. VII.
sum of
the squares
of its three edges,
ds2
=
(62b)
Fig. 62b.
Example.
Find the direction cosines of the tangent to the
parabola
X
=
at the point where
At
t
=
=
t
=
bt,
z
=
\ cf
2.
2 the differentials are
dx
ds
y
at,
=
=
a
dy
dt,
=
b
dt,
± Vdx^ + dy^ + dz^
=
dz
=
\ctdt
=
c dt,
± Va^ + ¥ + 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
ds
ds
dy
Va^
+ ¥ + c^'
ds
Va^
+ + c"'
fe^
c
ds~ Vcf+W+~^
—
Equations of the Tangent Line.
It is shown in
analytic geometry that the equations of a straight line
63.
GEOMETRICAL APPLICATIONS
Chap. VII.
through a point Pi (xi,
tional to A, B,C are
yi, Z\)
83
with direction cosines propor-
x-^i _ y-yi _ z-zi
The
direction cosines of the tangent
to dx, dy, dz.
we
then
If
,„
.
hne are proportional
C hy numbers pro-
replace A, B,
portional to the values of dx, dy, dz at Pi, (63) will represent
the tangent line at Pi.
Example
Find the equations of the tangent to the curve
1.
X
=
t,
t
=
at the point where
y
=
dy
:
The equations
=
dz
:
dt
2.
4
3,
x^
1,
.2tdt -.Sf
—
1
_
y
—
_
1
=
1,
=
dt
1
2/1
:2
=
1,
Zi
=
1.
-.3.
g
—
1
~~3
2
1
Ex.
a;i
i^
of the tangent line are then
X
=
=
1.
The point of tangency is i =
At this point the differentials are
dx
z
t^,
Find the angle between the curve Sx + 2y — 2z
and the line joining the origin to
j/2 = 2 2^
+
(1,2,2).
The curve and
y and
satisfj''
S dx
+ 2 dy - 2 dz = 0,
At the point
3dx
Along the curve
line intersect at (1, 2, 2).
can be considered functions of
the equations
z
The
x.
8xdx + 2y
dy
of intersection these equations
+ 2dy -2dz = 0,
Solving for dx and dy in terms of
dx
=
izdz.
become
8dx + 4:dy = 8dz.
dz, we get
dy
2 dz,
^
differentials
= —2 dz.
Consequently,
ds
= Vdx^ + */ + dz^ =
S dz
and
dx
'=°^"
2
= d-s=3'
„
'°''^
—2
dy
=
d^
=
-3-'
'^°'^
=
dz
ds
1
=
3-
DIFFERENTIAL CALCULUS
84
The line joining the origin and
equal to
The angle
d
between the
2
3-
31
31
line
2) has direction cosines
(1, 2,
12
Chap. VII.
and curve
satisfies
the equation
2-4+2
= 0.
^
cos 6 =
.
The
line
and curve
intersect at right angles.
EXERCISES
Find the equations
hues to the following curves at the
of the tangent
points indicated:
1.
X
=
sec
2.
X
=
e',
3.
X
=
e'
4.
On the
a;
2/
sin
=
y
/,
=
tan
e-',
y
(,
=
=
z
t,
z
=
e'
cos
P,
at
=
z
t,
=
«
TT
=
a,tt
at,
-•
1.
at
kt,
=
i
^•
circle
= ocos9,
+
= acos(e
^
o'^)i
= acos(9
z
+ ^7rJ
show that As is proportional to dS.
6. Find the angle at which the heUx
X
—
a cos
e,
=
y
aeai.6,
cuts the generators of the cylinder x'
6.
I
3?
+
z
=
a^
on which
kO
it lies.
Find the angle at which the conical heUx
=
t
cos
t,
y
cuts the generators of the cone x'
7.
+ y^ =
=
tsint,
+ y^
=
Find the angle between the two
y^ + ^ = liby the planes x — y +
z^
z
circles
z
=
t
on which
=
it lies.
cut from the sphere
and x
+y—z =
2.
CHAPTER
VIII
VELOCITY AND ACCELERATION IN A CURVED
PATH
64.
Speed
a curve,
its
of
a
speed
Particle.
is
— When a particle moves along
the rate of change of distance along the
path.
Let a particle P move along the curve AB, Fig. 64. Let s
The speed of the parbe the arc from a fixed point A to P
ticle is then
ds
(64)
'"=dt-
Fig. 65a.
Fig. 64.
65.
Velocity of a Particle.
at the point
— The
velocity of a particle
P in its path is defined as the vector* PT tangent
to the path at P,
drawn
in the direction of
equal to the speed at P.
To
motion with length
specify the velocity
we must
then give the speed and direction of inotion.
A vector
a quantity having length and direction. The direction
by an arrow. Two vectors are called equal when
they extend along the same line or along parallel hnes and have the
*
is
is
usually indicated
same length and
direction.
85
DIFFERENTIAL CALCULUS
86
The
Chap. VIII.
can be considered as moving instantaneously
particle
The
in the direction of the tangent.
velocity indicates in
magnitude and direction the distance
unit of time
if
it
would move
in
a
the speed and direc-
tion of motion did not change.
Example.
A wheel 4 ft. in diameter rotates at the rate of 500
Find the
revolutions per minute.
speed and velocity of a point on its
rim.
Let
center of the wheel
OA
to a
and
Then
moving point P.
s
The speed
of
OA be a fixed line through the
the distance along the wheel from
s
=
2dit.
P is
= 2^=2 (500) 2 X = 2000 x ft./min.
^
at
at
Its velocity is
at
2000 t ft./min. in
The speeds
P.
Their velocities
66.
If
is,
its
the direction of the tangent
on the rim are the same.
differ in direction.
Components
velocity in a plane
that
of all points
of Velocity in a Plane.
it is
customary to give
— To specify a
its
components,
projections on the coordinate axes.
PT is the
velocity at
P
(Fig. 66), the
FQ = Prcos</.= ^cos<^ =
and the ^/-component
^m
a;-component
is
^^ = ^,
is
r>m
,
The components are thus
rfs
.
,
the rates of
ds dv
change of the coordinates.
Since
pya = PQ2
-I-
dv
QT\
VELOCITY AND ACCELERATION
Chap. VIII.
the speed
is
87
expressed in terms of the components
by the
equation
(dsY
[dtj
^ (dxV
(dyf
[dt) '^[dtj
Fig. 67.
FiQ. 66.
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
PQ = PT cos a =
Since
PT? = PQ^
-77
T-
dt ds
+ QR^ + RT^
=
-77
the speed and compo-
nents are connected by the equation
(dsV
\dt)
^ /dx\
(dyV
.
dt
(dzV
\dt) '^[dtj '^[dtj
DIFFERENTIAL CALCULUS
88
Chap. VIII.
—
In this book we shall indicate a vector
68. Notation.
with given components by placing the components in brackets.
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 a;-component
a,
y-component
b,
and 2-component
c will
be represented by the symbol [a, b, 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
=
a;
and a
t
«
Vs
vertical distance
=
y
in
500
Find
seconds.
500
<-
velocity
its
16.1
«2
and speed at the end
of 10
seconds.
The components
of velocity are
^ = 500 V3,
= 500 - 32.2
§
at
at
At the end
of 10 seconds the velocity
7=
and the speed
Ex.
into
a
2.
A
fixed
then
[500 V3, 178]
is
=
j^
is
t.
V
(500
V3)' + (178)2 = 384
ft./sec.
point on the thread of a screw which
nut describes a helix with equations
X
—
r cos
d,
y
=
rsind,
z
=
is
turned
k9,
being the angle through which the screw has turned, r the
radius,
and k the pitch
speed of the point.
of the screw.
Find the velocity and
VELOCITY AND ACCELERATION
Chap. VIII.
The components
dx
.
of velocity are
„dd
dy
„dd
dz
dt
dt
dt
dt
at
d8
Since
-37
the angular velocity
is
V=
its
speed
ds
—=
which
is
[—na
&>
sin 6,
rco
dd
,
dt
with which the screw
moving point
rotating, the velocity of the
and
89
cos
is
is
9, kia]
is
y/rW sin2 d
+
r^oj^
cos^ 6
+
fcW
=
co
Vr^
+ k^,
constant.
X^
m,vy
^^
Fig. 69b.
Fig. 69a.
—
Composition of Velocities.
By the sum of two
and V2 is meant the velocity Vi
F2 whose
components are obtained by adding corresponding compo69.
+
velocities F]
nents of Vi and V2.
Similarly, the difference V2
—
Vi
is
the
whose components are obtained by subtracting the
components of Vi from the corresponding ones of Vi.
velocity
Thus,
Fi
+
Fa
if
=
[ai
Fi
=
[ai, 61],
+
02,
h + h],
Vi = [02, 62],
F2 — Fi = [02—
ai, 62
— &i].
If Vi and F2 extend from the same point (Fig. 69a),
F2 is one diagonal of the parallelogram with Fi and
Fi
F2 as adjacent sides and F2 — Fi is the other. In this case
Fg — Vi extends from the end of Vi to the end of F2.
+
DIFFERENTIAL CALCULUS
90
By
Chap. VIII.
mV of a vector by a number is meant a
m times as long as V and extending in the same direcm is positive but the opposite direction m is negative.
evident from Fig. 69b that the components of mV are
the product
vector
tion
It
if
is
if
m times those of V.
1
V
The quotient — can be considered as a product —
Acceleration.
— The
ing along a curved path
is
acceleration of a particle
the rate of change of
A=
lim
AV
and
-jrr is
ing
the
mov-
velocity
dV
'
.
In this equation
by
its
.
M=0 At
AV
Its
V by m.
components are obtained by dividing those of
70.
V.
dt
AF is a vector
obtained
by
components
divid-
of
AF
A*.
Let the particle
move from
P where the velocity
the point
is V to an adjacent point P'
AF.
where the velocity is 7
dx dy
to
The components of velocity will change from
+
It' dt
da;
di'^
.dx
di'
dy, .dy
dt'^
dt
Consequently,
Subtraction and division
by At
give
dx
A
^^'b%^t]
As At approaches
At
At
zero, the last equation
,
dV
^1
AV
[d'x
d^yl
'
At
approaches
,,„
,
VELOCITY AND ACCELERATION
Chap. VIII.
91
In the same way the acceleration of a particle moving in
is found to be
space
(70b)
Equations 70a and 70b express that the components of the
a moving particle are the second derivatives of
acceleration of
its
coordinates with respect to
the time.
A
Example^
particle
moves with a constant
speed V around a circle of
Find its velocity
r.
and acceleration at each
radius
point of the path.
Let
e
= AOP.
The
The
co-
P are
ordinates of
velocity of
a;
=
P
is
r cos
0,
.del
^dd
'df
Since
s
=
ds
rd,
-y.
=
=
v
r
-j-
.
The
^""
dtj
velocity can therefore be
dt
dt
written
V=
Since v
is
[— «;sin5,
.dd
r
Replacing 37 by
-
,
[—cos
9,
is
.
.del
this reduces to
sm0 = -[ — cosS, — sin9].
COS0,
Now
«;cose].
constant, the acceleration
—sin
9]
is
a vector of unit length directed
DIFFERENTIAL CALCULUS
92
along
PO toward the
Chap. VIII.
Hence the acceleration of P is
and has a magnitude
center.
directed toward the center of the circle
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 shdes with its ends in the x- and j/-axes.
If the
end in the s-axis moves with constant speed v, find the velocity and speed
of the middle point of the rod.
3. A wheel of radius o rotates about its center with angular speed u
while the center moves along the x-axis with velocity v. Find the velocity
and speed of a point on the perimeter of the wheel.
4. Two particles Pi (xi, yi) and Pi (xa, 2/2) move in such a way that
= 1+ 2
% = 3 + 2<2,
xi
Find the two
6.
Pi
=
2
-3
«2,
-4<'.
2/2=
and show that they are always parallel.
xi, yi, zi) and Pi (xa, y^, Zj) move in such a way
velocities
Two particles
2/1
<,
(
that
xi
X2
=
=
a cos
o sin
e, yi
9, 2/2
= a cos (5 + 4
= o sin (9 + J
= o cos
= a sin
jt), Si
tt), 32
{0
(9
+%
+f
ir),
ir).
and show that they are always at right angles.
He wishes to cross
6. A man can row 3 miles per hour and walk 4.
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
Fiud the two
velocities
take to reach his destination in the least time.
Neglecting the resistance of the air a projectile fired with velocity
b, c] moves in t seconds to a position
shall
7.
[a,
X
Find
8.
=
its speed, velocity,
A
particle
and
=
bt,
z
=
ct
—
\
gf'.
acceleration.
moves along the parabola
Show that
yr is constant.
y
at,
its
acceleration
3?
is
=
ay in such a way that
constant.
at
9.
When
a wheel
rolls
along a straight
line,
a point on
its
circum-
ference describes a cycloid with parametric equations
X
=
a
(<^
—
sin 0),
y
=
a
{1
—
cos
<l>),
the angle through which it has>
a being the radius of the wheel and
rotated. Find the speed, velocity, and acceleration of the moving point.
<t)
VELOCITY AND ACCELERATION
Chap. VIII.
10.
Find the acceleration
of
93
a particle moving with constant speed v
along the cardioid r = o (1 — cos 9).
11. If a string is held taut while it
is
unwound from a
fixed circle, its
end describes the curve
X
—
a cos 9 -{ a 8 Bin
d,
y
=a
sin d
—a
ff
cos
9,
subtended at the center by the arc unwound. Show
that the end moves at each instant with the same velocity it would have
9 being the angle
if
the straight part of the string rotated with angular velocity -w about
the point where
A piece
it
meets the fixed
circle.
mechanism consists of a rod rotating in a plane with
constant angular velocity w about one end and a ring sUding along the
12.
of
rod with constant speed
v.
(1)
If
when
of rotation, find its position, velocity,
time.
if
t
=
the ring
is
at the center
and acceleration as functions of the
Find the velocity and acceleration immediately
(2)
after
/
=
U,
at that instant the rod ceases to rotate but the ring continues sliding
with unchanged speed along the rod. (3) Find the velocity and acceleration immediately after t = ii ii at that instant the ring ceases sliding
but the rod continues rotating. (4) How are the three velocities re-
How
lated?
are the three accelerations related?
B and lie in a plane. A is
w about A, and BC rotates with
angular speed 2 a 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 = ii 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 horizontal line, while an insect crawls along the rim with speed oua.
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 aun, the rotation of the hoop about its center with angular speed wi, and the crawl
Find the three
of the insect advancing its radius with angular speed £02.
velocities and accelerations which result if at the time t = h two of these
13.
fixed,
Two
AB
rods
AB,
BC
are hinged at
rotates with angular speed
actions cease, the third continuing unchanged.
velocity
tions?
and
How
are the actual
acceleration related to these partial velocities
and
accelera-
CHAPTER IX
ROLLE'S
Rolle's
71.
least
off
THEOREM AND INDETERMINATE
FORMS
Theorem.
one real root of
(x)
=
f
— If
(x)
=
/' ix) is
continuous, there is at
between each pair of real roots
0.
To show this consider
the curve
y =f(x).
Let / (x) be zero at
X = a and x = b.
Between a and b there
must be one
Fig. 71a.
distance from the
horizontal
points
a;-axis.
or
more
maximum
\t such a point the tangent
=f
(x)
=
is
0.
That this theorem may not hold if /' {x)
shown in Figs. 71b and 71c. In both
Fig. 71b.
cro.sses
at
and so
dy
dx
is
P
discontinuous
cases the curve
Fig. 71c.
the x-axis at a and b but there
point where the slope
is
is zero.
94
is
no intermediate
THEOREM
ROLLE'S
Chap. IX.
Show
Example.
95
that the equation
a;S
+ 3a;-6 =
cannot have more than one real root.
Let
+ 3x-&.
f{x)=x'
Then
+S
=3x'
fix)
= 3(x2+l).
Since /' (x) does not vanish for any real value of
=0
/ (x)
cannot have more than one real root; for if there were two
between them.
there would be a root of /' (x) =
The expressions
72. Indeterminate Forms.
x,
—
^,
-,
O-oo,
00-00,
00
are called indeterminate forms.
No
r,
0°,
00°
definite values
can be
assigned to them.
If
when X
form, there
=
a
a,
function / (x) assumes an indeterminate
may however
be a definite limit
lim/(x).
In such cases this limit
function at x
=
is
usually taken as the value of the
a.
For example, when x
=
the function
2x0
~0"
X
It is evident,
however, that
lim
x=0
—X =
lim
(2)
=
2.
This example shows that an indeterminate form can often be
made definite by an algebraic change of form.
73.
for
The Forms
-r
U
and
—
00
.
— We shall now show that,
a particular value of the variable a fraction
the form
t;
or
—
,
f (x)
.
,
if
assumes
numerator and denominator can be replaced
DIFFERENTIAL CALCULUS
96
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 b.
= 0, F (a) = 0, and F (b) is not zero, there is a number
Jff
Xi between a and b such thai
W
f(b)_f(x^)
Fib)
To show
this let
|^ = R.
t
/
(73a)
F'{x,)
Then
(p)
- RF (6) =
(6)
0.
Consider the function
/
This
(a)
there
is
F
- RF
(a;).
when x = h. Since / (a) = 0,
By Rolle's Theorem
0, it also vanishes when x = a.
then a value Xi between a and b such that
function
=
ix)
vanishes
/' (xi)
-
RF'
(xi)
=
0.
Consequently,
jf^rM
fib)
F{b)
"-
F'ixO'
which was to be proved.
2.
and
Let/'
F
For,
if
(x)
=
(a)
we
and F'
(x)
be continuous near
by
x, (73a)
replace b
fix)
Fix)
xi
being between a and
proaches
(a)
=
x.
becomes
^f'ixQ
F'ix,)'
Since xi approaches a as x ap-
a,
iS F
3.
Iff
a.
0, then
ix)
'^a F'
In the neighborhood of x
(Xi)
=
Ta
F'
(X)
a, let /' (x)
and F'
(x)
be
'
THEOREM
ROLLE'S
Chap. IX.
continuous at
approach infinity as x approaches
,•
^^a
To show
this let c
— / (c)
Since / (x)
by Theorem
where
Xi is
If J
a.
and
F
(x)
fix)
/(^)
,
F
x=a F' (X)
(X)
{x)
a,
be near a and on the same side as
F {x) — F (c) are zero when x =
and
x.
c,
1,
IM
F'
=
points except x
all
97
He)
1
= /W-/(c) =
F
(xi)
(x)
- F (c)
between x and
F
(x)
F
{c)/F (x) approach zero.
/W
F_(^
F(x)
As x approaches
c.
increase indefinitely.
/M
i^ (x)
The
The
a,
/
and
(x)
quantities / (c)// (x) -and
right side of this equation
therefore approaches
x=a
Since Xi
a the
is
between
left side of
c
and
a,
F
by taking
the equation can be
™ F'
Since the
two
(x)
c sufficiently
made
near to
to approach
(x)
sides are always equal,
we
therefore conclude
that
= lim^^limm
F
.r. F'
(x)
Example
1.
(x)
Find the value approached by
approaches zero.
Since the numerator and denominator are zero when x
we can apply Theorem 2 and so get
,.
lim
Ex.
2.
sinx
cosx
= lim—
—=
,.
Find the value of Um-^
xi^ (x
—
r^-
x)2
1.
•
as x
=
0,
DIFFERENTIAL CALCULUS
98
When
zero.
=
X
Chap. IX.
the numerator and denominator are both
IT
Hence
l
,.
Iml 7
i=,
Since this
is
+ cosx
- TV =
(—sin
,.
Iim
x=T
xy
(tt
-^^r-j
—2
(tt
we apply
indeterminate
a;)
— X)r =
-•
the method a second
time and so obtain
sin
,.
X=T 2
The value
Ex.
required
a;
1
a;
—2
x=„
therefore
is
cos
,.
— X)
(TT
2,
\.
Find the value approached by —
tan X
3.
as x ap-
proaches ^-
When
this fraction
approach
Iim tan 3
TT
a:=2
When
X
TT
-^
X approaches
is
a;
—.
Therefore,
oo.
,.
=
lim
tan X
replaced
the numerator and denominator of
3sec^3a;
5
sec^ x
by ^ the
by Theorem
—
= hm 3cos^a;
5-;r—
,.
cos^ 3
last expression
3,
•
x
takes the form
7:
•
Therefore
,.
hm
—
a;
cos^ 3
a;
3 cos^
TTTT-
6 cos x sin x
,.
= hm
6 cos 3 x sin 3 x
eos^
lim
X
— sin^ x
- sin2 3 x)
1
3 (cos2 3 X
74.
The Forms
•
00
and
the
t< a fraction
e expression to
it
00
will
—
00.
— By
3
transforming
take the form
tt
or
—
00
For example,
xlnx
has
the
THEOREM
ROLLE'S
Chap. IX.
form
when
oo
•
=
a;
99
can, however,
It
0.
be
written
,
a;
In
a;
=
In
a;
-j—
i
X
which has the form
The
—
00
expression
— tan x
sec X
has the form
oo
—
IT
when x =
oo
It can,
^-
however, be
written
sec x
—
tan x
cos X
which becomes ^ when x
75.
The Forms
sin
1
=
—
sin
cos x
x
,
— The logarithm of the given
From
oo.
•
1
=
-•
=
0°, 1"=^, oo°.
function has the form
a;
cos x
the limit of the log-
arithm the limit of the function can be determined.
1
Example.
Find the limit of
+ xY
(1
as x approaches
zero.
1
Let
y
= {1+ xf-
Then
,
In 2/
When
X
is
+
1
I
N
a;)
=
+ x)
In (1
'-
zero this last expression becomes ^
,.
bm
1=0
The
1
/I
=-ln(l
limit of
ln(l +
—
X
^^
lay being
1,
a;)
'-
,.
= hm
1
=
1 -|- a;
the hmit of y
is e.
1.
.
Therefore
-
-
DIFFERENTIAL CALCULUS
100
Chap. IX.
EXERCISES
1.
Show by RoUe's Theorem that
X* - 4x -
the equation
1
=
cannot have more than two real roots.
Determine the values of the following limits:
~ 1r—1
-1
^'
„
T
2. Lim-n;
•
ii xi"
x"
3.
Lim
.^^-1'
4.
Lim
6.
7.
Lim
„
T
•
X
Lim
=0
10.
1-2
.,
-
11.
T
Lun
Lim tan x
21.
Lim
(x
+ a) Inf 1 +2).
22.
Lim
(x
-
"
24.
1
+
COS
cos X
X
(2 sin
—
sin
x
—
X
1)
t:^ logioCsmx -sina)
13.
Li^
=^^-
^(s-i^ij-
26.
Lim
27.
Limftanx-^
2g_
j^j^
_
X'
Sec'-^
-2tan.»
(cot
x
—
In x).
^^^-^r-l
smx-sm'^xj
.-t L
2
6sinx-6x+x3
x=0
1
Limx^a^.
x=o
-.,
2)2
-
3) cot (ttx).
X
sin^x
(a;
cos 3 x.
1=3
a;=3
"• ,l™lo!r,„rtanx-tan«V
i=5=a logio (tan X — tan a)
Lup
20.
1=
to
14.
Lim X cot x.
a;
Lim,
hi)
19.
In (x
In cos
a;
=^=0
„,
- 2)
„
—6
,
—
—X
Limi-t*2iL2x.
-^|sec
x^cosx
cosx — 1
x=3
9.
18.
-
-
,
8.
sec 3 X
^= «
x=a X — o,
tan X — X
Lim
1=5=0 X — sm X
j:=?=0
•
Lim
smo;
Lim^
Lim
T
^ sec X
^".""^^
^0
6.
<iT
17.
^_
x=0
29.
Lim
(sin
x)*™^
+ cos 4
1
2
1
15.
Limi5^.
16.
Lim
30.
x=0
x=0 cot X
a:=a)
1
Inx
•
a;
Lim (1+ ax^.
31.
Lim
a;=QO
(x™-
—
a"") 'i *.
CHAPTER X
AND APPROXIMATIONS
SERIES
76.
Mean
Value Theorem.
tinuous from X
= atox =
m^
such that
To show
and /
(6)
b
the arc
=
{x)
and
a value
/' (x) are con-
Xi between
/' (-0.
AB
—a
let
=
y =/(x).
a and x = b,
Qf ^j^Qj.^
gjQ
Pi be a point at
a and b
(76)
this consider the curve
are the ordinates at x
/(&)-/(«) ^
On
— If f
h, there is
Since f (a)
j^g
maximum
distance from
Fig. 76.
the chord.
and so its
The tangent at Pi
slope/'
(xi) will
will
be parallel to the chord
Therefore
equal that of the chord.
i»l^) = ;,(,,,
which was to be proved.
Replacing 6 by x and solving for / (x) equation
,
fix)
=f{a)
+
101
(x-a)f'{xi),
(76)
becomes
DIFFERENTIAL CALCULUS
102
Xi
being between o and
This
x.
we
general theorem which
— 7/ /
Taylor's Theorem.
77.
a special case of a more
is
now
shall
Chap. X.
prove.
and
(x)
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)
To prove
+ ix- a)f' (a) + i^^^/" (a)
this let
<t>{x)=f(x)-f{a)-(x-a)f'(a)
_
jx
- ay
_
.^s
^''^
^
2!
.
.
_
.
-
(X
a)"-i
(n-
1)!
^
*
^
It is easily seen that
<t>
(a)
=
0,
=
0' (a)
0,
.
When
X
=
=
(a)
(t)"
.
0,
r-'
.
(a)
=
r
0,
(x)
= /"
(x).
a the function
4>{x)
—
(x
therefore assumes the
value
3i
form
a)"
By
^
zz.
Art. 73 there
then a
between a and x such that
_
0(a;)
—
(x
4,'(zi)
n
a)"
{zi
—
This new expression becomes ^ when
sequently a value
Zi
between
Zi
a)"~^
=
z,
and a (and
a.
There
_
<^^(zi)
n
(zi
—
a)"-i
(n
—
1) (Z2
—
a)""^
continuation of this argument gives finally
(t>(x)
(x
—
a)"
^
<P^(Zn)
n\
^
con-
between x and
ct>" (z,)
71
is
so
such that
A
is
/" (Zn)
^
n\
a)
,
Zn
being between x and
Equating
/
AND APPROXIMATIONS
SERIES
Chap. X.
we
(x),
If Xx
a.
=
Zn
103
we then have
this to the original value of
^
(x)
and solving
for
get
/W=/(a) +
(x-a)/'(a)
+ ^>(«)+---+^"/"(^0.
which was to be proved.
Example. Prove
In X
where
=
xi is
When
X
(x
-
(x
between
=
1
-
1)
^
2
1
and
ly
{x
"^
,
,
3
ly
{x
-
ly
^ ^
4x1*
x.
the values of In x and
/(x)=ln(x),
-
its
derivatives are
DIFFERENTIAL. CALCULUS
104
called the remainder.
is
value of the function
/W
-/(o)
+
If this is small,
Chap. X.
an approximate
is
(l-a)f(o)
the error in the approximation being equal to the remainder.
To compute/
of/
(a;)
/(«).
close
by this formula, we must know the values
We must then assign a value to a such
(a), /' (o), etc.
knovm.
/' («)i ete., are
that
Furthermore, a should be as
as possible to the value x at which / (x) is wanted. For,
a, the fewer terms {x
a)^, (x
a)', etc.,
—
the smaller x
—
—
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
known
are
is
45°.
Therefore
we
let
a
/(a;)=tana;,
-7
•
4
^(l)'"
f'{x)=eec^x,
^'
•^'(i)=2.
(x)
=
2 sec2 x tan
/'" (x)
=
2 sec* x
/"
=
+
4:
x,
f"
sec^
(^ =
4,
x tan^ x.
Using these values in Taylor's formula, we get
and
tan46°
=
l
+ 2(^) + 2(^)^=1.0355
approximately.
Since xi
does not differ
much from
/'" (46°)
is
=
between 45° and 46°, /'"
8
+8=
16.
(x{)
The
AND APPROXIMATIONS
SERIES
Chap. X.
error in the
above approximation
f(lioJ<3|o)-3<4oko
It
is
thus very nearly
is
=
105
0-««««25.
'
therefore correct to 4 decimals.
Ex.
Find the value
2.
of e to four decimals.
The only value of x for which e^ and its derivatives
known is a; = 0. We therefore let a be zero.
/
{x)
/(O)
By
=
=
=
=
/' (x)
e-,
/' (0)
1,
f'{x)=e-,
e,
/"
1,
(0)
=
/" {x)
,
1,
,
/" (xi)
are
= e^,
= e\
Taylor's Theorem,
x^
,
Letting x
=
=
1
6
1,
this
x^
,
if
,
a;"e^i
becomes
11
+ 1 + 21 + 3! +
In particular,
a;"-^
,
,
n =
1
•
•
•
+(;r^i)!
e'^
+ ^r
2,
e
=
2
+
i
e^'.
+
between
and 1, e is then between 2^ and 2
therefore
between
To get a better ape,
and
2^ and 4.
J
proximation let n = 9. Then
Since xi
is
e
=
l
+ l + ^ + ^+--- +3^=2.7183
approximately, the error being
9!^
The value
2.7183
is
9-,
^
9!
<
-00002.
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.
ln(T%).
3.
cot 43°.
V.
Ve.
4.
tan 58°.
9.
Given In
tan"' (^j).
1.6094, find In 17.
8.
3
=
1.0986, In 5
=
DIFFERENTIAL CALCULUS
106
— As
Taylor's and Maclaurin's Series.
79.
indefinitely, the
Chap. X.
n
increases
remainder in Taylor's formula
n
In that case
often approaches zero.
/(x)=lim[/(a) + (a;-a)/'(a)+
This
is
•
•
+ ''"^ /""Ha)]-
•
usually written
f{x)=f
+
(a)
{x-
a)f' (a)
+ ^^—^ f" (a)
(^
,
- a)l^„,
(^)
3!
the dots at the end signifying the limit of the
number of terms is indefinitely
sum is called an infinite series.
.
.
.
^
sum
Such an
increased.
This one
+
as the
infinite
called Taylor's
is
Series.
In particular,
if
a
=
0,
Taylor's Series becomes
f{x)=fiO)+xf'{0)+^f"{0)+f^f"iO)+
This
is
Show
that cos x
x''
represented
is
x^
x^
cosx=l-2| + 4;-gj+
The
series
=
that
0,
•
.
called Maclaurin's Series.
Example.
a
•
,
given contains powers of
is,
/
when
(x)
f'{x)
f"ix)
/'" {x)
/""
(x)
by the
series
,
•
•
•
.
This happens when
x.
Taylor's Series reduces to Maclaurin's.
= cos X,
= -smx,
= -cosx,
= sm x,
= cos X,
f
(0)
=
1,
/'(0)=0,
/"(0) = -l,
f"
(0)
f""
(0)
=
=
0,
1.
These values give
cosa;=l-2]
+ 4j-
•
±-,/"(xi).
Chap. X.
AND APPROXIMATIONS
SERIES
107
The nth. derivative of cos x is =tcos x or ±sin x, depending
on whether n is even or odd. Since sin x and cos x are never
greater than 1, /„ (xi) is not greater than 1. Furthermore
/viTC
I*
f
'>•
'y*
n\^l'2'3'
can be
made
'
'
'n
you please by taking n sufficiently
Hence the remainder approaches zero and so
large.
as small as
= l-2; + 4|-g;+
cosa;
•
•
•
.
which was to be proved.
EXERCISES
1.
1
„
/
1
7r\
1
ttV
/
1
,
irV
(
,
3.
2^
(^^ + Mi2)!+.
+
= l+xln2
ryl
X
T'G
/r5
'1-3
..
T'
4s*
4x5, 8
,.e-cosx = ,,
l+x-^-^--gj+ -^+
2a;3
a;'
...
5.
+ x)" =
a"
+
—
6.
(o
7.
V-x
8.
9.
=
2
+
+ ^'^gT^
na"-'x
'*''"^^'
i-^+i^j^
+
•
•
ln3+^-^^=^' + ^^=3?l!
ln(x + 5)=ln6+^^-i|=^+M!
80.
Convergence and Divergence
the
,
if
|x|*<
,iflx-3|<l.
,if|x-l|<l.
of Series.
sum
— An
of the first
a
*
out
The symbol
its
-\-
ar
|x| is
algebraic sign.
in-
n terms
approaches a limit as n increases indefinitely. If this
does not approach a limit, the series is said to diverge.
The series for sin x and
The geometrical series
|o|.
,if|x-4|<l.
lnx =
finite series is said to converge if
•
sum
cos x converge for all values of x.
ai^ \- ar^ -{
-{- ar'^ -\-
•
•
used to represent the numerical value of x withThus, — 3 = 3 =3.
|
|
|
1
when r is numerically
n terms is
converges
the
Chap.
X
For the sum
of
DIFFERENTIAL CALCULUS
108
first
iS„
If r is
=
+ ar + ar^ +
a
•
•
numerically less than
than
1.
+ ar"~i
=
less
•
a
1
-.
\
—
—
r"
r
approaches zero and
1, r"
/S„
approaches
1
as
is
n increases
The series
—
r
indefinitely.
1-1+1-1+1-1+
•••
sum oscillates between and
The geometrical series
2
+ + 4 + 8+16+ •••
divergent, for the
not approach a
1
and does
limit.
1
diverges because the sum increases indefinitely and so does
not approach a limit.
The convergence of a
81. Tests for Convergence.
series can often be determined from the problem in which it
—
Thus the
occurs.
series
_ E!
e! _ ^
2!"'"4!
_i_
_l_
e!"*"
converges because the
sum
of
n terms approaches cos x
as
n
increases indefinitely.
The terms near the beginning
of a series (if they are all
have no influence on the convergence or divergence of
the series. This is determined by terms indefinitely far out
finite)
in the series.
82.
General Test.
Ul
+
to converge it is
ll2
— For the series
+ U3+
beyond m„ approach zero as
For,
if
+ Un+
•
•
necessary and sufficient that the, sum of terms
n
increases indefinitely.
the series converges, the
proach a Umit and so the
approach zero.
sum
sum
of terms
of n terms must apbeyond the nth must
Chap. X.
AND APPROXIMATIONS
SERIES
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 sum will approach
and the
zero
84.
series will converge.
Ratio Test.
approaches a limit r
— //
asn
is convergent if r is
'
increases indefinitely, the series
•
numerically
numerically greater than
of consecutive terms
Un
+ U3+
U1+U2
-^
the ratio
'
+ Un + Un+1 +
•
•
•
than 1 and divergent
less
if r is
1.
by taking n sufficiently large the ratio
made as nearly r as we please.
be a fixed number between r and 1. We can
Since the limit
is r,
of consecutive terms can be
<
If r
1, let
n
take n so large that the ratio of consecutive terms
is less
than
series are therefore less
than
Then
ri.
Un+l
<
nUn, U„+2
<
<
riUn+1
Beyond Un the terms of the given
etC.
T-^Un,
those of the geometrical progression
+
Un
r-iUn
which converges since
?'i
+ r^n +
is
'
•
•
numerically
than
less
1.
Con-
sequently the given series converges.
If,
however,
r is greater
ultimately increase.
and
their
than
1,
the terms of the series must
The terms do not then approach
sum cannot approach a
Example.
limit.
Find for what values of x the
a;
+ 2a;^ +
3a^
+ 4a^+
•
series
•
converges.
The
ratio of consecutive
Mn+i
^
(w
+
terms
is
1) a:"+^
=
(-S
•
zero
—
DIFFERENTIAL CALCULUS
110
The
Chap. X.
limit of this ratio is
=
r
lim
H
1
(
The
converge
series will
Power
85.
Series.
if
x
—A
is
x
]
nj
n=oo \
=
x.
less than
powers of {x
numerically
series of
1.
—
a) of
the form
P
{x)
=
+ ai
ao
(a;
—
o)
+ 02 (x — a)^ + as (x — a)' +
•
•
,
where a, ao, ai, Oa, 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
In
will
fact, if
ao
+
be
less
—
<
a|
the series converges
ai (6
—
a)
than a
+
02 (&
a\.
when x =
— ay +
maximum
|o„ (b
—
|&
03 (&
M^ that
< M.
value
-
a)"|
b,
—
each term of
o)'
+
•
•
•
is,
Consequently,
The terms
Oo
+
of the series
Oi (x
—
a)
+ 02 (x —
a)2
+
03 (x
—
o)'
+
•
•
•
are then respectively less than those of the geometrical series
in
which the ratio
is
\b-a\
If
then
\x
—
a\
<
\b
—
a\,
the progression and consequently
the given series will converge.
// a power series diverges when x
values of
x further from a than
|x
^
o|
=
&, it
will diverge for all
b is, that is, such that
>
|6
—
a\.
Chap. X.
AND APPROXIMATIONS
SERIES
111
could not converge beyond b, since by the proof just
would then converge at b.
This theorem shows in certain cases why a Taylor's Series
For
given
is
it
it
not convergent.
Take, for example, the
^2
series
^4
/wo
+ x)=x-^ + ^-^+
ln(l
•
•
•
.
+
As X approaches —1,
In (1
x) approaches infinity.
Since
a convergent series cannot have an infinite value, we should
when x = —1.
must then
from a = 0.
The series in fact converges between a; = — 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 poljoiomials.
In
expect the series to diverge
diverge
when
at a distance greater than
is
a;
It
1
—
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
sin
a;
tana;=
cos
=
~
^
x^
a:'
^ 120 ~
6
z
a;
'
'
'
,
_ ^ "^ ^ _
1
a;^
3
I
24
2
EXERCISES
1.
Show
that
and that the
2.
By
series
converges
expanding cos 2
sin''
X
Prove that the
=
1
—
cos
2
series
x,
when
|a;|
<
1.
show that
2x
_
a;2
„„
a;*
2 21-2'
2!
4!
converges for
2
a;^
=a;+ir +-t?-+
-j
all
,
+
„, a?
"'
'
values of x.
616!
15j
•
•
.
DIFFERENTIAL CALCULUS
112
3.
Show
that
(l+e^)2
= l+2e* + e=* =
4
+ 4a; + 3a^+ix'+fx*+---
and that the series converges for all values
4. Given / (x) = sin"' x, show that
/' (X)
Expand
this
=
1— =
VI —
(1
of x.
-
x^)-\.
x^
by the binomial theorem and determine /"
Hence show that
differentiating the result.
,
1 a^
,
1 .3 a?
,
3 5x'
1
and that the series converges when |a;| < 1.
6. By a method similar to that used in Ex.
-lO
*i^3
U.n-^x
and that the
6.
Chap. X.
=
4,
show that
^7
x-^+j-j+---
series converges
when
\x\
<
1.
Prove
cosx
For what values
of
2
24
x do you think the
series
converges?
{x), etc.,
by
CHAPTER XI
PARTIAL DIFFERENTIATION
87.
u
is
—
A quantity
Two or More Variables.
a function of two independent variables x and y,
Functions of
called
u =f{x,
if
u
is
y),
determined when arbitrary values (or values arbitrary
within certain limits) are assigned to x and y.
For example,
M =
is
a function of x and
+
y.
Vl If
a;2
-
2/2
u is to be real, x and y must be
so
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 independent variables. An illustration of a function of variables 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 belonging 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 inThus, if
crease when all the variables change.
chosen that x^
y^ is not greater
u =f(x,
y),
u+ Au =f(x + ^x,y + Ay)
and
so
Au=fix +
Ax,y
A function is called continuous if
zero
when
all
its
+ Ay) - f {x, y).
increment approaches
the increments of the variables approach zero.
113
DIFFERENTIAL CALCULUS
114
88.
Chap. XI.
— Let
Partial Derivatives.
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
g
Similarly,
stant,
we
if
we
or
fAx,y).
differentiate with respect to y with
x con-
get the partial derivative with respect to y denoted
by
^
For example,
or
fy(x,y).
if
u =
x^
+ xy — y^,
then
du
^
_=2x
+
du
„
-^-x-2y.
'
,
Likewise,
if
m
is
j/,
a function
any number
of
of independent
variables, the partial derivative with respect to one of
is
them
obtained by differentiating with the others constant.
89.
Higher Derivatives.
are functions of the
functions partially,
we
— The
first
By
variables.
partial derivatives
differentiating
For example, the derivatives of
— with respect to x and y
are
d
ldu\
tdu\
dhi
^dhi
\dx)
dx \dx/
~
/du\
^
d
d^
tdu\
ldu\
dy \dx)
\dxj
dx^ '
dhi
_ dhb
~ dy
dydx
dx
Similarly,
d_
d^u
dx dy
dx \dy/
It
these
get higher partial derivatives.
/du\
^ 6^
dy \dy/
dy^
d^
'
can be shown that
dhc
dxdy
_
dhi
dydx'
PARTIAL DIFFERENTIATION
Chap. XI.
if
both derivatives are continuous, that
is,
115
-partial derivatives
are independent of the order in which the differentiations are
performed*
Example,
u =
— = 2 xy +
af|
= ^(^^
90.
+
x^y
xy^.
—=
y",
+
x^
2 xy,
+ 2^^)=2- + 2,,g=A(,. + 2.,)=2..
— It often happens that some
Dependent Variables.
F6r example,
of the variables, are functions of others.
u =
+
x^
y^
let
f
-\-
let 3 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
and
—
= 2X
dx
-\-
2 z
—
dx
Similarly, the partial derivative with respect.to y with x con-
stant
is
+ 2zp.
P=2y
dy
dy
"
If,
however, we consider
z constant,
the partial derivatives
are
du
-^
dx
The value of a partial
=
2x,
du
—
= 2
dy
2/.
derivative thus depends
on what quantities
are kept constant during the differentiation.
The
quantities kept constant are sometimes indicated
subscripts.
*
Thus, in the above example
For a proof see Wilson, Advanced
Calculiis,^ 50.
by
DIFFERENTIAL CALCULUS
116
It will usually be clear
variables
u
is
Chap. XI.
from the context what independent
considered a function
Then
of,
-r— will
ox
repre-
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
From
the triangle
a
P
=
Differentiating with
da
=
sin
it is
c sin
A
c,
find
(—
j
•
seen that
A.
constant,
we
get
A,
dc
which
is
the value required.
—
Geometrical Representation.
Let z = f (x, y) be
the equation of a surface. The points with constant ycoordinate form the curve AB (Fig. 91a) in which the plane
y = constant intersects the surface. In this plane z is the
Consequently,
vertical and x the horizontal coordinate.
91.
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
3
At the lowest
=
a;2
+ 2/2-2x-4j/ + 6.
point, the curves
= 2a;-2 =
f
dx
AB and CD
(Fig.
Hence
have horizontal tangents.
0,
=
p-=2y-i
dy
0.
91b) will
Chap.
XL
PARTIAL DIFFERENTIATION
Consequently, x
=
1,
y
=
These values substituted
2.
the equation of the surface give 2
is
then
(1, 2, 1).
That
this
is
117
=
1.
The point
really the lowest point
in
required
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.
2.
4.
x^
u
z
x
=
+
+y
(.x+a}(.y
u =\n
du
i
{x^
-\-xy
,
du
""Tx-^^Vy-
+ b),
-\- y'^),
x —ox
=
+y—
dy
2.
18
PARTIAL DIFFERENTIATION
Chap. XI.
By
}{x
mean value theorem,
the
+ ^x,y-\- Ay)
xi lying between x
/
{x,
+
y
+ Ay) + Axf^
+
Similarly
+ Ay.
Aw = Ace/, (xi, y
As Ax and Ay approach
If fx (x, y)
y.
/x
y
(xi,
Ayfy
y
+ Ay),
(.x,.yi),
Ayfy
(x, yi).
approaches x and
zero, Xi
2/)
+
«i
= fy (X,
2/)
+
€2
^+
—+
= du
=
ap-
€2,
and e2 approaching zero as Ax and Ay approach
These values substituted in (92b) give
+~
yi
«i'
€i
Au = -^ Ax
(92b)
y) are continuous,
{x,
(x,
fa:
(xi,
Using these values in (92a),
+ Ay)+
and fy
+ Ay) =
fy (X, yi)
Ax.
=f {x, y) +
Ay)
yi being between y and y
we get
proaches
Art. 76,
=f{x,y
and x
119
Ay
+
eiAx
+
e2
Ay.
zero.
(92c)
The quantity
du
~
Ax + — Ay
dx
dy
du
.
.
,
It differs from Am by an
As Ax and Ay approach zero, ei and e^
approach zero and so this difference becomes an indefinitely
small fraction of the larger of the increments Aa; and Ay.
We express this by saying the principal part differs from Am
by an infinitesimal of higller order than Aa; and Ay (Art. 9).
When Aa; and Ay are sufficiently small this principal part then
is
called the principal part of Am.
amount
ei
Ax
+
62
Ay.
gives a satisfactory approximation for Am.
Analogous results can be obtained for any number of independent variables. For example, if there are three independent variables x, y, z, the principal part of Am is
du
du
—
-^Ax+—Ay
Az.
+ du
dx
dy
dz
In each case,
if
.
,
.
,
.
the partial derivatives are continuous, the
DIFFERENTIAL CALCULUS
120
Chap. XI.
from Am by an amount which becomes
comparison with the largest of the increments 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 diameter decreases from 2 ft. to 23 in.
= irr%, the exact change is
Since the volume is
principal part differs
indefinitely small in
?;
At;
The
=
TT
(1
-
1^)2 (6
+
tV)
-
P
T
6
.
principal part of this increment
=
-0.413 x
cu.
ft.
^
is
|Ar + |A/. = 2../.(-l) + ..^(l)=-0.417.cuft.
^
Total Differential.
93.
m
If
pendent variables x and y, the
principal part of Aw, that is,
'- =
is
a function of two inde-
total differential of
u
is
the
(93^)
fx^'+fy^y-
This definition applies to any function of x and
u = x and u = y give
dx = Ax,
dy = Ay,
y.
The
particular values
that
the differentials of the
is,
independent variables are equal
increments.
to their
Combining
(93a)
and
<^w
We shall show later
if
(93b)
(93b),
=
^
we
«^a;
get
+ ^ dy.
(93c)
(Art. 97) that this equation is valid
even
X and y are not the independent variables.
The
quantities
J
dxU
are called partial
that the
—
= 5m
,
dx,
differentials.
total differential of
J
dyU
—= du
.,
dy
Equation (93c) expresses
a function
is
equal
to the
sum
of the
partial differentials obtained by letting the variables change one
at
a time.
PARTIAL DIFFERENTIATION
Chap. XI.
121
Similar results can be obtained for functions of
For instance,
of variables.
pendent variables
= Tdx
=
+—
dy
Aa;
u =
particular values
dx
any number
a function of three inde-
is
x, y, z,
dw
The
w
if
Ax,
x,
dy
At/
+—
dz
u = y,u =
=
A2.
z give
dz
A.y,
=
A2.
The previous equation can then be written
=
du
dx
-:—
dx
+^—
dy
dy
+
^r dz
(93d)
dz
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
and
u =
By
+
x^y
xy'^.
equation (93 c)
du
,
du =
,
-j-
i2xy
+
dx
=
,
dx
-:r-
du
T— dy
dy
y^) dx
-,
+
(a;^
+ 2 xy) 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 x, y, z. The volume is then
V
The
If
error in
v,
=
xyz.
due to small errors
Aa;, Aj/,
Az
in x, y,
z, is
Aw.
the increments are sufficiently small, this will be approxi-
mately
dv
Dividing
by
v,
we
= yzdx
+ xz dy + xy dz.
get
dv
_
yz dx
+ xz dy
=
Now — expresses the
dx
-\-
xy dz
xyz
V
dx
—
H
xyz
dy
,
dz
•
error dx as a fraction or percentage of x.
DIFFERENTIAL CALCULUS
122
The equation
just obtained expresses that the percentage
volume
error in the
equal to the
is
errors in the edges.
If,
not more than one per cent, the
more than three per cent.
of the percentage
error in the
was assumed that
functions of a single variable.
It
volume
is
not
— In proving the formu-
Calculation of Di£ferentials.
las of differentiation it
sum
for example, the error in each edge
is
94.
Chap. XI,
is
u,
etc.,
v,
were
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
du +-T-
(uv) dv
= vdu
-\-
u dv,
dv
which is the formula IV of Art. 17.
Example, u = y^ + ze".
Differentiating term by term, we get
—
du
We
y^"
dx
+ ^ dy -{
ze'"
dy
+ e" dz.
obtain the same result by using (93d)
;
for that formula
gives
du
=
95.
— dx + — dy + — dz = y€!''dx+
Partial Derivatives
(d''
+ ze") dy + e" dz.
as Ratios of Differentials.
—
The equation
dxU
shows that the
=
-— dx
dx
du
partial derivative -r- is the ratio of
dx
two
dif-
and dx. Now d^u is the value of du when the
same quantities are kept constant that are constant in the
ferentials dxU
calculation of
-r--
dx
Therefore, the f
partial derivative v- is the
'
Q^
value
which
to
same
du
"^dx-
the
123
PARTIAL DIFFERENTIATION
Chap. XI.
reduces
-r-
dx
are determined with
when du and dx
quantities constant that are constant in the calculation
.
Given u
Example.
Differentiating the
we
=
x^
+ y^ +
z',
=
v
two equations with
xyz, find
v
and
^
f
j
•
z constant,
get
du
= 2xdx + 2y dy,
=
yz dx
+
xz dy.
Eliminating dy,
du
=2xdx-2'^
x
Under the given conditions the
du
dx
^
2
=
dx
2 f
\
^'
ratio of
(x^
-
~
X
^'
) dx.
/
du to dx
is
then
y")
x
Since v and z were kept constant, this ratio represents
that
(
—
j
;
is,
(-
\dx/v
EXERCISES
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
1.
its
One
side of
principal part.
2.
A
3.
and
and 6 in. deep,
Find approximately the volume
closed box, 12 in. long, 8 in. wide,
material J inch thick.
used.
Two
A =
sides
45°.
and the included angle
By
a2
4.
and
A
=
62
of
a when
20, c
=
30,
6 increases 1 unit, c decreases
a simple pendulum
is
r = 2xv^Find the error in
of
+ c2 - 2 6c cos A,
increases 1 degree.
The period
=
made
using the formula
find approximately the change in
i unit,
of a triangle are 6
is
of material
T due to small errors in
I
and
g.
)
Chap. XI.
DIFFERENTIAL CALCULUS
124
If
6.
(/
computed by the formula,
is
s
=
J gfi,
and
find the error in g due to small errors in s
The area
6.
a triangle
of
is
K
t.
determined by the formula
=
i ab sin C.
K
due to small errors in a,b,C.
Find the error in
Find the total differentials of the following functions:
7.
xf^.
"
y
xysm(x
8.
-+^z + -X
9.
+ y).
tan"' -
10.
X
+ tan"' -•
y
The pressure, volume, and temperature of a perfect gas are connected by the equation pv = kt, k being constant. Find dp in terms of
11.
dv
and
dt.
y are rectangular and
show that
12.
If X,
point,
xdy - ydx =r^de,
13.
lix =
14.
llu = xy
16.
11
+
+ yz + zx,
x^
=
tix
+
y
-\- z^
+
(^
—1 yz,
z',
find
da
cos
96.
Let u
s
and
a;
+
cos
f
dff^.
•
find
(
—
•
J
—
b, c
and opposite angles A, B, C
that
db
A
j
+ r"
dr'^
same
\OZJu,x
Show
inscribed in a fixed circle.
=
find
««,
= uy
vx
tfi,
+ dy^
dx^
A variable triangle with sides a,
16.
is
yz
=v?
u-v,
6 polar coordinates of the
r,
_
dc
,
B
cos
_
C
Derivative of a Function of Several Variables.
=
t.
f {x,y)
When
Ax and y
t
+
Am =
—
x and y be functions of two variables
changes to i
Ai, a; and y will change to
and
let
+
The
Aj/.
—
Aa;
resulting increment in
+—
Aj/
+
€1
Ax
+
€2
u
will be'
Ay.
Consequently,
^
Am _duLx du
bM
Ax
Ay
~ dx Ai'^ dy At'^ ^^Ai'^ ^~Ai
'Kt
K
As Ai approaches
zero,
Ax and Ly
will
approach zero and ^o
'
ei
and
£2
PARTIAL DIFFERENTIATION
•
Chap. XI.
approach
will
_
~
It
If
a;
or
is
2/
Taking the
zero.
du
125
limit of
both
sides,
du dx
du dy
dx dt^ dy dt'
,
,
^^^>
dx
a function of
t
only, the partial derivative
-r-.
at
du
or T7
replaced
is
dx
by a
total derivative
dt
and y are functions
of
i,
w
-77
or -^
dt
at
a function of
is
du
t
If
.
both x
with total deriva-
tive
du
dt
Likewise,
depend on
if
w
is
dt
As
before,
is
dx
du dy
dt'^ dy dt
(q(\w\
^
^
a function of three variables
that
x, y, z,
t,
du
rivative
_du
~ dx
if
_ du
~ dx
a variable
replaced
any number
The term
dx
'di
dz
^ '^ du Tt'
,
*
'dz
a function of
is
by a
du dy
^ dy
t
,
^^
only, its partial de-
Similar results hold for
total one.
of variables.
du dx
dx
is
dt
u with respect to t, leaving all
u except x constant. Equations (96a) and
that if u is a function of several variable quanti-
the result of differentiating
the variables in
(96c) express
ties,
— can
be obtained by differentiating with respect to
only one of those quantities were variable at a time
t
as if
and adding
the results.
Example
1.
Given y
The function
=
du
x", find
-^-
x^ can be considered a function of
ables, the lower x
and the upper
x.
If
two
the upper x
is
vari-
held
constant and the lower allowed to vary, the derivative (as in
case of
oj") is
X
•
cc^~i
=
a?.
DIFFERENTIAL CALCULUS
126
If
the lower x
Chap. XI.
held constant while the upper varies, the
is
derivative (as in case of
a?') is
iU'lnx.
The
actual derivative of y
-^
ax
Ex.
~
Given u
2.
=
=
x^
f {x,y,
sum
then the
is
+
z),
a?'
In
a;.
y and z being functions of
a;,
du
^-•
ax
,
find
By
equation (96c) the result
_
du
dx
du
dx
is
du dz
dz dx
du dy
dy dx
.
In this equation there are two derivatives of u with respect
If y and z are replaced by their values in terms of x, u
to X.
will
is
J-
If
.
2/
and
function of x.
Ex.
3.
Find the
It
The
be a function of x only.
is
z are replaced
derivative of that function
by constants, u
Its derivative is -r-
Given u
will
be a secopd
•
dx
=
f {x,y,
z),
partial derivative of
z being a function of
u with
x and
y.
respect to x.
understood that y is to be constant in this partial
Equation (96c) then gives
differentiation.
du
dx
_ du
dx
du dz
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
A way to avoid the confusion is to use the
derivatives.
PARTIAL DIFFERENTIATION
Chap. XI.
letter/ instead of
u on the right side
127
It then
of the equation.
becomes
dx
dx
It is
understood that /
^
dx
and that
the derivative obtained with
is
Change
we have
a definite function of
{x, y, z) is
but X constant.
97.
dz dx
of Variable.
—
w
If
is
all
x, y, z
the variables
a function of x and y
said that the equation
,
du
==
5m
—- dx
,
dx
du
+ ^dy
,
I
ay
and y are the independent variables or not.
and t be the independent variables and x
and y functions of them. Then, by definition,
is
true whether x
To show
this let s
= du
-r-
,
du
Since m is a function
by equation (96a),
du
ds
_ du
~ dx
dx
ds
,
and
of x
du dy
dy ds
j/"
du
+ -rr
Of
,
as
dS
,,
dt.
which are functions of
du
Tt
'
_
^
du dx
dx
s
and
t,
du dy
dy 'di\
'di
Consequently,
/du dx
,
du dy\
,
,
,
/du dx
,
du dy\
,.
'^''=[di-d-s+d^-d-sr+[d^^+dimr
du /dx
=
,
d-x[d-s'^
,
dx ,\
.
du /dy
which was to be proved.
A similar proof can be given
variables.
98.
,
,
dy
,
A
du
,
du
,
,
+ -dt'^V + d^[ds^'+mV = d-x'^'' + d-y'^y'
Implicit Functions.
—
If
in
case of three or
more
two or more variables are
connected by an equation, a differential relation can be obtained
by equating the
the equation.
total differentials of the
two
sides of
DIFFERENTIAL CALCULUS
128
Example
1.
/
(x,
y)
=
Chap. XI.
0.
In this case
d-f{x,y)=^dx
Consequently,
+ ^dy
= d-0 =
0.\
PARTIAL DIFFERENTIATION
Chap. XI.
rivative of this function
dz
The
for the ratio
is
129
found by eliminating dy and solving
result
is
dx'
99.
dfidf2
dfidh
dz
dy dx
dx dy
dx
dfidf2
dfidf,
dz dy
dy dz
Directional Derivative.
— Let u
P
point
in the aitz-plane,
(x, y)
move away from
u has a
=f
At each
If we
PQ, x and y will
{x, y).
definite value.
P in any definite direction
Fig. 99.
The
be functions of the distance moved.
with respect to s is
du dx
dx ds
du
ds
This
is
when
•r-
ds
is
if
du
dx
u
,
,
du
.
u
,
dy
in the direction
PQ.
The
— and — are special values of — which
PQ
Similarly,
du
du dy
dy ds
called the derivative of
partial derivatives
result
,
derivative of
is drawn in the
m = / (x, y, z),
direction of
OX or
OY.
du dz
du
dudy
du dx
„
du
= ^COSa + du
= T-j-T-COS|8 + -r-COS7
+ T--f
+ T-jds as
dx
dy ds
dy
dx ds
dz
,
,
,
u with respect to s as we move along a
with direction cosines cos a, cos |3, cos y. The partial
the rate of change of
line
,
DIFFERENTIAL CALCULUS
130
derivatives of
u are the values
to
which
-r-
Chap. XI.
reduces
when
s is
ds
measured in the direction of a coordinate
Example. Find the derivative of a;^ +
<t>
=
45° at the point
The
result
^(x^
+
axis.
y^ in
the direction
(1, 2).
is
yi)
=2x^+2y^ = 2xcos<t> + 2ysm<l>
=
2~
+ 4~= 3V2.
V2
V2
—
100. Exact Differentials.
If P and Q are functions
two independent variables x and y,
Pdx
may
and
or
y.
may
of
+ Qdy
not be the total differential of a function w of
the total differential of such a function,
If it is
P dx + Q dy =
Since dx and dy
du
=
-T-
ax
dx
dy.
+ -ray
a;
PARTIAL DIFFERENTIATION
Chap. XI.
Similarly,
if
+ Qdy +
Pdx
is
131
the differential of a function u of
Rdz
x, y, z,
dPdQ
dQ_dR
dRdP
^-'di'
'di-^'
'di'dz'
,.„„,.
^^""^^
and conversely.
Example 1. Show that
(x2
is
+
2 xy) dx
+
(x^
+
2/2)
dy
an exact differential.
In this case
The two
partial derivatives being equal, the expression is
exact.
Ex.
U
2.
In thermodynamics
being the
shown that
it is
dU = TdS -pdv,
internal energy, T the absolute
temperature,
S
the entropy, p the pressure, and v the volume of a homogeneous substance. Any two of these five quantities can be
assigned independently
Show
and the others are then determined.
that
The
result to
/an ^
/dv\
\dp/s
\dSjj,
be proved expresses that
TdS + vdp
is
an exact
differential.
That such
is
the case
is
shown by
T dS by its value dU + pdv. We thus get
TdS + vdp = dU + pdv + vdp = d(U + pv).
replacing
EXERCISES
1.
If
M =/
(x, y),
2.
If
M =/
(x, y, z),
y
==
(t>
(x),
find
^•
z =<t> (x), find
f
^j
—
DIFFERENTIAL CALCULUS
132
=/
3.
If
u
4.
If
« =/
and
(a;,
y, z), z
=
y
{x, y),
=
(x, r),
<l>
=
r
=^
<{'
(x),
~
find
find
(x, s),
,
[J^j
\J^j^'
\ax),
5.
If/ fe
6.
If
F
y,z)=Q,z = F
y, z)
(a;,
=
7.
If
w =
8.
If
u =/
If z
9.
If
u =f
-,
+ ay),
X
(x, y),
_
dx~
s
== T
cos
—+y— =
=y + bt,
=
show that t"
"*"
Urj
Bz^
show that x
= a;+<rf,
r
(r, s),
= / (x
=
z
a;/ (z),
find^-
{X, y),
show that
0,
Sx ay
dy dz
10.
y
(x, y),
<t>
Chap. XI.
B,
y
=
u.
show that -rr = a
ot
«
-r
ax
\-b -r--
dy
^"
r sin 9,
show that
\ax) '^\ay)
\r Be]
moving in a plane is
determined by the coordinates h, k of the moving origin and the angle
between the moving x-axis and a fixed one. A variable point P has coordinates x', y' with respect to the moving axes and x, y with respect to
11.
The
position of a pair of rectangular axes
Then
the fixed ones.
X
=f {x',
Find the velocity
y', h, k,
of P.
= F
y
<t>),
Show
that
it is
the
(x', y', h, k, 4>).
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,
9 of
a point in a plane.
Show that they
are identical with the derivatives with respect to s given in Arts. 54
and
59.
13.
point
14.
is
Find the derivative
At a
y=
15.
of
x''
—
y''
'va.
the direction
ij>
=
30° at the
(3, 4).
-
.
distance r in space the potential due to
Find
an
electric charge e
its directional derivative,
r
Show that the
the curve 3?
—
y^
=
derivative of xy along the normal at
c^ ia zero.
any point
of
PARTIAL DIFFERENTIATION
Chap. XI.
= f {x,y), show
Given u
16.
133
that
and sj are measured along perpendicular directions.
Determine which of the following expressions are exact
17. y dx — X dy.
18. (2x + y)dx + {x -2y) dy.
19. exdx-\- eydy + (x + y) ezdz.
20. yz dx — xzdy + y'^dz.
21. Under the conditions of Ex. 2, page 131, show that
if Si
\dT)p
\av]T
{dTJv
Kdph'
In case of a perfect gas, pw
22.
=
Using
hT.
differentials:
this
and the equation
-p dv,
dU = TdS
show that
=
1^
dp
Since
U
that J7
always a function of p and T,
a function of T only.
is
is
Direction of the
101.
0.
Normal
this last
equation expresses
at a Poiat of a Surface.
—
Let the equation of a surface be
F
(x, y, z)
=
0.
Differentiation gives
dF,
-T-dx
dx
Let PN
through P
be
(x,
the
line
y, z)
with
cosines
direction
.dF,
,dF, =
-r-dz
+^-dy +
dy
_
0.
(101a)
dz
propor-
tional to
dF^SF^dF
dx' dy
If
P
'
dz
moves along a curve
on the
surface, the direc-
tion cosines of its tangent
PT
are proportional to
dx
:
dy
:
dz.
Fig. 101.
Equation (101a) expresses that PN and PT are perpendicular to each other (Art. 61).
Consequently PN is perpendicu-
DIFFERENTIAL CALCULUS
134
lar to all
saying
Chap. XI.
This is expressed by
normal to the surface at P. We conclude
the tangent lines through P.
PN is the
that the normal
to the
surface
F
=
y,z)
{x,
at
P
{x, y, z)
has
direction cosines proportional to
f:f:fdx dy dz
102. Equations of the
Normal
(101)
^
^
—
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
at Pi
_ y-yi _
z
^4~~~B~~
(asi,
-
yi, z^.'
..„„,
z,
^^^^>
C
—
Equation of the Tangent Plane at Pi (xi, yi, «i).
All the tangent lines at Pi on the surface are perpendicular
103.
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.
in analytical geometry that ii A, B, C are prothe direction cosines of the normal to a plane
passing through (xi, yi, Zi), the equation of the plane is
It is
shown
portional
,to
A(x-xO+B(y- yi) +
*
C
(z
-
z^)
= 0*
See Phillips, Analytic Geometry, Art. 68.
(103)
PARTIAL DIFFERENTIATION
Chap. XI.
135
A, B, C are proportional to the direction cosines of the
normal to a surface at Pi, this is then the equation of the
It
tangent plane at Pi.
Example. Find the equations of the normal line and tangent plane at the point
+
a;2
The equation given
+
a;2
The
2
is
2
—1,
(1,
+
2/2
3
2) of the ellipsoid
=
z2
3
a;
+
12.
equivalent to
+
1/2
-
^2
3
direction cosines of its
3x
-
=
12
0.
normal are proportional to the
partial derivatives
2x At the point
(1,
3 -.iy :6z.
—1,' 2), these are proportional to
A :B :C = -1 -4
The equations of the normal are
:
:
'
X
1
The equation
_
~
y
1
+
4
1
1)
:
4
:
-12.
2
~ - 12
4
+
(?/
=
+1_g-
of the tangent plane
-
X
I
12
-
is
12
(2
-
2)
=
0.
EXERCISES
Find the equations
of the
normal and tangent plane to each
of the
following surfaces at the point indicated:
+ y^ + = 9, at (1, 2,
+ xy + y' = 7, at (2,
= x' + y', at (3, 4, 5).
1.
Sphere, t'
2.
Cylinder, x^
3.
Cone, z'
Hyperbolic paraboloid, sj/ = 3 z — 4, at (5, 1, 3).
Elliptic paraboloid, x = 22/^
3 z^, at (5, 1, 1).
Find the locus of points on the cylinder
{y-z)' = i
{x
zy
4.
5.
6.
z''
+
7.
3).
is
+
parallel to the iiz-plane.
that the normal at any point P (x, y, z) of the surface
4 X makes equal angles with the x-axis and the line joining
Show
22
=
PandA
8.
-3,
+
where the normal
j^2 _]_
2).
(1,0,0).
that the normal to the spheroid
Show
9
at
P
A'
(0,
{x, y, z)
-4,
0)
"^25
determines equal angles with' the lines joining
and
A
(0, 4, 0).
P
with
DIFFERENTIAL CALCULUS
136
Maxima and Minima
A maximum value
Chap. XI.
Functions of Several
a function m is a value
greater than any given by neighboring values of the variables.
In passing from a maximum to a neighboring value, the func104.
Variables.
—
tion decreases, that
is
Am <
A minimum
of
of
value
is
a value
(104a)
0.
than any given by neighIn passing from a minimum
less
boring values of the variables.
to a neighboring value
Aw >
If
(104b)
0.
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.
rilj
If -r- is continuous,
ox
by
Art. 31,
g=o.
(104c)
Therefore, if the first partial derivatives of
u
with respect
to the
independent variables are continuous, those derivatives must be
zero
when u
When
is
zero.
is
a
maximum
or
minimum,.
the partial derivatives are zero, the total differential
For example,
if
x and y are the independent vari-
ables,
du
dx +^ dy =
=^
ax
ay
Therefore, if the
total differential of
dx
+
dy
first partial derivatives are
u
is zero
when u
is either
=
0.
(104d)
continuous, the
a
maximum
or a
minimum.
To find the maximum and minimum values of a function,
we equate its differential or the partial derivatives with respect to the independent variables to zero and solve the
resulting equations.
minimum, or
neither.
from
a maximum,
It is usually possible to decide
the problem whether a value thus found
is
PARTIAL DIFFERENTIATION
Chap. XI.
Example
1.
137
Show that the maximum rectangular parallele-
piped with a given area of surface
Let
X, y, z
is a cube.
be the edges of the parallelopiped.
A the area of its
A =
V = xyz,
volume and
Two
If
V is
the
surface
2 xy
-\-
2 xz
-\-
2
of the variables x, y, z are independent.
yz.
Let them be
Then
X, y.
_ A — 2xy
^~2{x + y)'
Therefore
y ^xy{A-2 xy)
+ y)
2{x
dV^f FA
dz
-2x^
2L
{x
-4 xyl
+ yy
J-"'
dV ^^ VA -2y^ -Axy
dy~ 2l
The values x =
A -
2x^
y
0,
{x
=
+ yy
'\
" "
J
cannot give maxima.
- 4xy =
A -
0,
2y^
Hence
- Axy =
0.
Solving these equations simultaneously with
A =
we
+
2 xy
2 xz
+
2
yz,
get
x^y
We know
there
only one solution
Ex.
2.
is
it
a
=
z
=V-g-
maximum.
Since the equations give
must be the maximum.
Find the point
x
+
in the plane
2y
+ iz=
14
nearest to the origin.
The distance from any point
origin
{x, y, z)
is
D =
Vx^
+ y^ +
z^.
of the plane to the
DIFFERENTIAL CALCULUS
138
If this is
a minimum
,
j^
+ ydy + zdz ^
+ + 2^
^ xdx
Va;2
that
Chap. XI.
^
2/2
is,
xdx
From
+ ydy + zdz = 0.
the equation of the plane
dx
+ 2dy +
we get
Zdz =
(104e)
(104f)
0.
The only equation connecting x, y, z is that of the plane.
Consequently, dx, dy, dz can have any values satisfying this
If x, y, z are so
last equation.
chosen that
D is a minimum
by all of these values. If two linear
equations have the same solutions, one is a multiple of the
Corresponding coefficients are proportional. The
other.
(104e)
must be
satisfied
coefficients of dx, dy, dz in (104e) are x, y,
are
z.
Those
in (104f)
Hence
1, 2, 3.
_z
_y
~2~3'
X
1
Solving these simultaneously with the equation of the plane,
we get X = 1, y = 2, z = 3. There is a minimum.
we get only one solution, it is the minimum.
Since
EXERCISES
(
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
1.
minimum.
3.
When an
resistance
R
electric current of strength
the heat produced
are connected
by three wires
is
/ flows through a wire of
proportional to I^R.
Two
terminals
of resistances Ri, Ri, Ra 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
4.
A
Ii,
h,
h in the three wires will satisfy the equations
particle attracted
toward each of three'points A, B,
force proportional to the distance will be in equilibrium
C
with a
when the sum
PARTLAi DIFFERENTIATION
Chap. XI.
of the squares of the distances
from the points
139
is least.
Find the
posi-
tion of equilibrium.
5.
Show
that the triangle of greatest area with a given perimeter
is
equilateral.
6.
Two
starting at
adjacent sides of a room are plane mirrors. A ray of light
P strikes one of the mirrors at Q, is reflected to a point B on
the second mirror, and is there reflected
same horizontal pline find the positions
PQBS may
7.
A
to S.
of
If
P
Q and B
and S
are in the
so that the path
be as short as possible.
table has four legs attached to the top at the
Ai, Ai of a square.
A weight
comers Ai, Ai,
W placed upon the table at a point of the
diagonal AiAs, two-thirds of the way from Ai to As, will cause the legs
to shorten the amoxmts Si, Si, Sj, St, while the weight itself sinks a dis-
The
increase in potential energy due to the contraction of a
where k is constant and s the contraction. The decrease in
potential energy due to the sinking of the weight is Wh. The whole
system will settle to a position such that the potential energy is a minimum. Assuming that the top of the table remains plane, find the
tance
h.
leg is
fcs^,
ratios of
Si, Si, Ss, St,
SUPPLEMENTARY EXERCISES
CHAPTER
Find the
III
difierentials of the following functions:
2
^_
+ 1) (2 + 7)'
(2a; + 5)»
(a + 2)'
+ 4)'
+ l)na: + 3)«'
(2
7
b^ax?+h
,
2
Vaa:'
+ hx
„
.
5-
(ax
+6 _
+ 6i + c
12.
13.
14.
15.
16.
(2x''-l)
_
+ 6)'H^ ~ fa(ai+&)'H-i
+ 2) aUn + 1)
,„,
'
^0.
a;(a;"
+
,5=1.
»
7i)
in each of the following cases:
+ 3j/2 = 6x-42/ + 18.
+ Zx'y = y\
2 =
+ 22/'.
= 2o2(x'-2/'').
(x2 +
^
x = + r^, "
t-V y = 2t- {t-iy
2a;='-4xj/
x'
32/''
2/2)2
«
X
=
/:;--—
Vl+
02 + o2)*
'
y
^
t2
18.
=
X =
19.
x"
20.
The volume
17.
V^HH;
a^
aHn
Find -y
11.
(a;
2o3;
Vax'
a;
(a;
6a;
^
a;
X
<
z' + 2z,
+ 2^ = aS
2/
2
yz
=
Vl - <2
= < «2 + a2)*.
= + 2^.
= V.
2/2
elasticity of
according to Boyle's law, pv
=
a &uid iae
constant,
= —v-^.
show that
When a
e
If
=
a gas expands
p.
gas expands without receiving or giving out heat, the
pressure, volume, and temperature satisfy the equations
21.
B, n, and
C being
pv
= BT,
constants.
Find
pv"
=
C,
^ and ^-
140
^
1
1
SUPPLEMENTARY EXERCISES
22.
yr
is
If « is
141
the volume of a spherical segment of altitude
show that
h,
equal to the area of the circle forming the plane face of the segment.
23.
a polynomial equation
If
W=
/
has two roots equal to
/
r,
(x) is
has
=
(x)
/
where /i
(x)
a polynomial in
— r)^ as a factor,
- ryf, (x),
{x
(x
Hence show that
x.
W
/'
=
that
r is
is,
a root of
0,
the derivative of/ (x).
Show by the method of Ex. 23 that each of the following equations
has a double root and find it:
where/'
(x) is
24. x3
-
25.
- x^ -
x=
3x2
-3=
5X
4x'-8x'i-3x
4x4 - 12x3+
26.
27.
a;2
Find '^ and
28.
+ 4 = 0.
y
=
j^
^
- x\
c
ax
by
X = 2
3(,
^^
33.
x
34.
If, =
x^findgandg.
35.
Given
x^
—
=
y'
1,
is
If
~
d!/2
Compare
38.
If
=
v are functions of x,
= d%
^-«
,
,
(x)
= 7:ri'
+
*
\,dx/
.d'u
dv
+ 4^-^^
,
constant.
show that
du
d^
+ «^.
+6^.^,+4^.^
,
-dhi
cPv
this with the binomial expansion for
/
y
= 4-5<.
a positive integer, show that
M and
d^
^(w)
+
2/
verify that
d"
-r^ x"
dx"
37.
——T
=
a2.
dx^
«
+
'
xj/
If
+ = 0.
+
32.
31.
30.
36.
=0.
in each of the following eases.
3.2
x Va?
=
0.
+ 9 =0.
+ i2x + 4
=
(x
—
rYfi
(x),
where /i
(x) is
/'W=/"«=0.
.
(4t
.
dh)
,
+ w)*.
a polsmomial, show that
DIFFERENTIAL CALCULUS
142
CHAPTER
A particle moves
39.
along a straight line the distance
particle
When
Two
velocity
its
t
When
acceleration.
is
is
the
the velocity
decreasing?
trains start
from
different points
track in the same direction.
in
and
When backward? When
moving forward?
increasing?
40.
Find
seconds.
t
+ 3&t + l
= it>-21P
s
feet in
IV
and move along the same
the train in front moves a distance 6 1'
hours and the rear one 12 P, how fast will they be approaching or
If
At the end
separating at the end of one hour?
wUl they be
41.
If s
of
When
two hours?
closest together?
=
Vt, show that the acceleration
negative and propor-
is
tional to the cube of the velocity.
The
42.
Find
its
moving along a
velocity of a particle
acceleration
k
=
-,
s
when
where k
/
V
= 2P -3t.
=
2.
straight line
is
constant, find the acceleration.
43.
li
44.
Two
by a
is
is
What
What
is
tf'
is
wheels, diameters 3 and 5 ft., are connected
the ratio of their angular velocities and which
belt.
greater?
the ratio of their angular accelerations?
Find the angular velocity of the earth about its axis assuming
that there are 365j days in a year.
46. A wheel roUs.down an inclined plane, its center moving the
45.
distance s
= 5P
wheel about
47.
in
An amount
that the acceleration of the
constant.
money
of
Show
seconds.
t
its axis is
interest is immediately
is
drawing interest at 6 per cent.
added to the
principal,
what
is
If
the
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?
If the
49. A kite is 300 ft. high and there are 300 ft. (rf cord out.
kite moves horizontally at the rate of 5 nules an hour directly away
from the person fls^ng it, how fast is the cord being paid out?
50. A particle moves along the parabola
100 y
in such a
way
that
its
the velocity and acceleration of
51.
10
ft.
How
The
side of
is
16a;2
its
projection on the
an equilateral triangle
per minute and
large
=
abscissa changes at the rate of lO ft./ sec.
its
the triangle?
is
Find
j/-axis.
increasing at the rate a!
area at the rate of 100 sq.
ft.
per minute.
SUPPLEMENTARY EXERCISES
143
CHAPTER V
52.
The
velocity of waves of length X in deep water
is
proportional to
a
X
\^a'^X
when o
X
=
is
a constant.
Show
that the velocity
ia
a minimum when
a.
53.
The sum of the surfaces
sum of the volumes is
that the
of
a sphere and cube
least
is
when the diameter
Show
given.
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
How
material each 4 inches wide, the middle one being horizontal.
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 will 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 6, 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
sphere of radius
of the smallest cone that can contain
a
a.
60. To reduce the friction of a liquid against the walls of a channel,
the channel should be so designed that the area of wetted surface 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 channel, the
making an angle B with the vertical.
62. Find the least area of canvas that can be used to make a
tent of 1000 cu.
63.
ft.
banks
conical
capacity.
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; assuming 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.
2
DIFFERENTIAL CALCULUS
144
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
minimiiTn
66.
.
A vertical telegraph pole at a bend in the line is to be supported
from tipping over by a stay 40
How
stake in the ground.
long fastened to the pole and to a
from the pole should the stake be
ft.
far
make the tension in the stay as small as possible?
The lower comer of a leaf of a book is folded over so
driven to
67.
as just to
the width of the page is 6 inches,
find the width of the part folded over when the length of the crease is
reach the inner edge of the page.
If
a minimum.
68. If the cost of fuel for running a train is proportional to the
square of the speed and $10 per hour 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 functions:
X
sin
70.
76. sec^
X
sm'-secs"
+ cos 9
78.
tan:
sin 9
__
.
1
1
72.
— cos 9
73. sin
74.
2
CSC
tan 2x —
X
1
ax cos ax.
^.^0
cot K —
75.
— tatf x.
77.
sm „
71.
x
--..?>
X
'
2
cot
—
2 tan
80.
.,
81.
;;•
A
X
x
— tan' x
- 7 setf 9.
„
sec I esc I — 2 cot x.
1
5 sec' 9
.
x.
Differentiate both sides of each of the following equations
that the resulting derivatives are equal.
X
82.
sec'
83.
sin
84.
sin 3
85.
sin (x
86.
87.
+ csc'a; = sec' x esc' x.
2X
a;
=
=
2 sin
a;
3 sins
cos x.
—
4 sin' X.
+ o) = sin s cos a + cos x sin a.
sec'x = 1 + tan'x.
ainx + maa == 2mn^ (x + a) cos i (x — a).
and show
^
SUPPLEMENTARY EXERCISES
88.
cos a
— cos s =
Find —- and
=
=
X =
X =
X =
=
2 sin 5
X
a
cos^
e,
y
90.
X
a
cos*
e,
2/
92.
93.
94.
a;
—
ta.ne
y
e,
sec2e,
y
sec
2/
e,
esc 8
—
+ o) sin |(s — a).
in each of the following cases:
-3-^
89.
91.
(a;
145
cot
8,
y
=
=
=
=
=
=
a sin* fl.
a sitf 8.
cos
9.
tan*
9.
tan 8.
csc8
+ cot 0.
Differentiate the following functions:
95.
sin-i
102.
96.
cos-'
103.
97.
tan-i
2x
2
98.
:7|tan-
99.
cos-i
,
+l
V3
Vx^
+1
a
csc-i
-
+ Va* -
a;«-
cot-^a;-
l+a:^
104.
Vl —
105.
sec->^i4 + siii"'^^-
sin-i
X
a;
—
106.
sm
10',-
1
2 cos-i X
108.
Vx*
1
.r
- Vi-
1
r—+a
a;
+1
cos a;
+ 2 VT
X*-
V5
100.
CSC-'
2a;
101.
-1'
—
a*
e^.
110.
V^.
111.
(Vi)^
118.
119.
120.
+ ihi(a2 + x2)e-*' cos (a + 6/).
hi(a + Va' + x*).
(x + yin(x + y-x-L
itan-ii
112.
5thxt.
113.
7^.
114.
a^lna;.
115.
In sin" x.
116.
hihi.-!;.
123.
tan-i i (e^
U7.
.{1=^.
124.
hi(Vx
1
126.
a sec"
sec-i
109.
125.
—
121.
122.
ln^^ + °
+ 2x + 5)+|tan-i-^^.
secixtanjx + ln (secix +tan Jx).
(x
+
l)hi(x*
+ ^^^zJ.
vx + a ~ V X — a
+
+ ^^1^2).
e-"^).
DIFFERENTIAL CALCULUS
146
127.
|(.+i)-M
+ -j-In(a;2 + l)
xsec->^|a;
+
+
iln (HX^
128.
-z^[l^'
l)-|a;
')-
+ tan-'|a;.
CHAPTER
VII
Find the equations of the tangent and normal to each of the following curves at the point indicated:
130.
= 2x + y,a.t{l,2).
x^-yi = 5, at (3, 2).
131.
x'
129.
y^
132.. I*
+
+
=
2/
a^{x
136.
2/*
= x + 3y,Sit{-l,l).
= 2, at (1, 1).
hix, at
133.
134.
135.
j/^
(1, 0).
+ y) = a" {x - y),
X
=
2 cos
r
=
a
9,
+
(1
=
2/
3 sin
at
9,
=
cose), at 9
(0, 0).
at 9
=
2
2"
Find the angles at which the following pairs
n
137.
x^->ry''
138.
y'-
139.
x'
140.
2/2
=
Sx, y^ (2
-
x)
=
of curves intersect:
x'.
= 2ax + a?, x^ = 2hy + ¥.
= Aay, (x^ + 4a2) = Sa^.
= 6x, x^ + y' = 16.
= He' + €-"), y = i.
= sin X, = sin 2 X.
i/
141.
y
142.
2/-
143.
Show that
2/
all
the curves obtained by giving different values to
in the equation
are tangent at
144.
Show
(a, 6).
that for
x'
all
-
3
values of a and b the curves
xi/'
=
a,
3 x^ -2/3
=
6,
intersect at right angles.
Examine each
of the following curves for direction of curvature
points of inflection:
1
i.ir
146.
147.
y
X
=
=
—X
tan x.
6y' -22/3.
and
SUPPLEMENTARY EXERCISES
= 2«-i,
=
+ i-
148.
x
149.
Clausius's equation connecting the pressure, volume, and tem-
perature of a gas
2/
2«
is
RT
a, b, c
c
v-a
^
B,
being constants.
T(.v
T
If
is
of a point, this equation represents
T
147
+ by'
constant and
an isothermal.
p, v
the coordinates
Find the value
of
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' (o), but /" (a) is not equal to F" (o), show that the curves
are tangent and do not cross at x = a. Apply to the curves y = 7?
and 2/ = s' at X = 0.
151. If two curves y = f {x), y = F (x) intersect at x = a, and
f (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
curves y = x^ and y = x^ + (x — ly at x = 1.
for
Find the radius
on each
of curvature
of the following curves at the
point indicated:
152.
Parabola
153.
x^
Ellipse -;
a-'
154.
155.
=
ax at
i/^
=
+ r:
0'
Hyperbola
2/
=
y^
-
^'
In CSC x, at
(
s
vertex.
at its vertices.
1
|'
its
=
1
at x
= Va^
+ ¥.
•
I
,
x = Jsin2/ — iln (sec y + tan y), at any point (x, y).
X = a cos' 6, y = a sin' 6, at any point.
158. Find the center of curvature of j/ = In (x — 2) at (3, 0).
156.
157.
Find the angle
at the point indicated
</'
on each
of the following
curves:
=
159.
r
=
2",
160.
r
=
a
+ 6 cos
161.
r (1
—
cos 0)
162.
»•
=
at e
0.
9,
=
at e
=
2
=
k, a.t e
a sin 2 9, at 8
=
=•
•
-J
4
Find the angles at which the following pairs
r = a (1 — cos9).
163. r (1 — cose) = a,
a
164.
r
=
asec2-,
a
r
—
bcsc^--
of curves intersect:
DIFFERENTIAL CALCULUS
148
=
=
165. r
r
166.
o cos 9,
r
asec9,
r
= o cos 2 9.
= 2awi0.
Find the equations of the tangent Unes to the following curves at
the points indicated:
167.
X
=
2 «,
2/
=
sin
t,
168.
X
169.
x^
170.
z
2
=
-
y
=
,
z
=
at
fi,
cos t, z
=
<
=
sec
2.
a,tt
t,
1/2
-1).
a,t
A
171.
0.
2/
CHAPTER
VIII
point describes a circle with constant speed.
projection on a fixed diameter
its
=
+ 2^ = 6, X + + 2 = 2, at (1, 2,
= x' + y^, !? = 2x - 2y,
(1, -1, 2).
+
Show
that
moves with a speed proportional
to the distance of the point from that diameter.
The motion
172.
Show
that
its
x
= lVa^~p
2/
=
I
speed
Find the speed,
by the equations
of a point (x, y) is given
is
Vo^
+ ^Bbi-^-,
+ + ^ In + v^T+TO.
«2
constant.
velocity,
and acceleration
in each of the following
cases:
X
174.
X
=
=
=
=
173.
177.
+ 3<,
a cos
2/
{uit
=
4 -9«.
2/ = a sin
+ a),
(w(
+ a).
= b + pt, z = c + yt.
= e' cos z = kt.
e' sin
The motion of a point P (x, y) is determined by
X = o cos (nl
a), y = b sin (nt + a).
X
176. X
175.
2
a-\- ad,
i,
y
y
t,
the equations
-\-
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 parabola
2/2
179.
2/2
=
=
2
A
2 ex.
180.
(x, y)
ex.
Show
particle
Find
that the acceleration
moves with
is in
parallel to the x-axis.
[a,
o]
along the parabola
its velocity.
Show that the
and
is
acceleration
vector
I
j-^
'
jz
I
extends along the normal at
magnitude equal to the curvature at
(x, y).
—
SUPPLEMENTARY EXERCISES
149
CHAPTER IX
Show
181.
that the function
x^
-1
= — 1 and x =
vanishes at x
1, but that its derivative does not vanish
between these values. Is this an exception to RoUe's theorem?
182. Show that the equation
i^
-
=
5a;H-4
has only two distinct real roots.
Show
183.
that
x^sin-
=
Lim —
x = o sins
:
0,
but that this value cannot be found by the methods of Art.
Show
184.
,
.
Lim
1=0
Why
1
—
cos
a;
=
cosx
.
0.
cannot this result be obtained by the methods of Art. 73?
Find the values
185.
,„^
186.
of the following limits:
^11^.
Lim
lifll— cos2x
T-
Lim
!
189.
187.
V3i - Vl2 - X
Lima;"e-=^.
191.
Lim -^^
a;===osm2a;
1
+ esc (s —
r
xcotx
1
192.
1)
1
The
a
cot
Lim ^sec x) x'
=
(ttx)
area of a regular polygon of n sides inscribed in a circle
is
no'
sm-
n
cos
—
n
that this approaches the area of the circle
indefinitely.
194.
-
Lim'^^^-^)
of radius
Show
/
Vl9 -Ss
tan-;f
2
Lim
=
193.
^
190.
1
188.
- '-2^1
-
= 3 2a:-3
1
Lim f i
= oV^
i
TTX
Show
that the curve
x^
is
Explain.
73.
that
+
2/'
=
3
xj/
tangent to both coordinate axes at the origin.
when n
increases
DIFFERENTIAL CALCULUS
150
CHAPTER X
Determine the values of the following functions correct to four
decimals:
195.
cos 62°.
198.
v^O.
196.
sin 33°.
199.
tan-' (iV)-
197.
hi (1.2).
200.
esc (31°).
201.
Calculate
ir
by expanding
=
I
202.
203.
=
Given hi 5
Prove that
tan"' x
tan-i
and using the formula
(1).
1.6094, calculate hi 24.
D=
Vfh
is an approximate formula for the distance of the horizon,
distance in miles and h the altitude of the observer in feet.
X
Prove the following expansions indicating
which they converge:
if
D being the
possible the values of
for
204.
+ x') = In 10 + 4 (x - 3) - A (x - 3)2 +
ln(e^ +
+ 5+ ••
hi(l +sinx) =x - ix2+Ji= -t^eX<+
e^secx = l+x + x2 + |xs + Jx<+
In (x + VT+^') =x-i| + if|J+--..
hi (1
0=|-g
yr>2
205.
206.
207.
208.
•
•
209. In
e»--
=
l+x+|-,-^
«*'°"
=
l+^+|+^' + ^+---
212.
In
x
Determine the values
rt»2
01R
216.
+ 7x*
-^ +
=
211.
215.
X*
+
x
In tan
214.
•
•
.
•
•
.
.
5^ = 2g + 3^ + ^ + ...].
210.
213.
•
'y6
-J^
1^
-5-
of
x
for
•
•
•
.
.
which the following
series converge:
-M
1+x + l+f + j +
•
•
•
.
(^V(^V^^+....
+
1 +2x + 3x2 + 4x'+
+ 2 (x + 2)' (x + 2)»
»
2 + -Y-2-+^73-+-3TT-+'
(x-l)
•
•
•
•
.
a:
I
•
'
•
SUPPLEMENTARY EXERCISES
151
CHAPTER XI
In each of the following exercises show that the partial derivatives
satisfy the equation given:
«,n
2n.u =
218.
z
=
x«
du
du
du
x~
+ z~=y--
+ yV,
, ,
,
xy
-
,
+
2 a;V
^
2/*,
M
+ 2/)lnxz,
^ +^E "
du\
du
^3m
^^_--j=^-.
219.^ =
(x
220.. =
(.+ i)tan-.(.-J),
221.
/
,
u = xy
°-
i+.^g=.^
z
223.
,*
=
hi
+ t),
(a:^
Shi
Shi^ _L
dydz
dxdy
{x
dx^'
dz'
+y
= z+e^,
du
du
u =
z{x^
-
't
dx
If
a;
dz
_
dz
y dy
if),
—hx = (^-y'){y%+-'i)
y"£+-'£
du
227.
_
y'di~''dx
""to
226.
du dz
dx dy
du dz
du
dx ~ dy dx
dy
u
z iS'a function of
- zf,
du
225.
2/^-,.
^ = ^.
Prove the following relations assuming that
224.
dz
+ ^^ = °-
^^
= ^ + 5,
y —z
u =
-r—
+ -,
^'^ + 2..z—^ + z^^ =
222.
'Z
,
9m
-Tdy
=^
(e*
+ e"'),
=
o',
show that
2/
= | (e* - e''),
Xdrje
228. If xyz
/dy\ /dj\
\dxjy
/dA ^
\dx)i\dylx\dz]y
show that
x and
y:
—
DIFFERENTIAL CALCULUS
152
In each of the following exercises find As and its principal part,
assuming that x and y are the independent variables. When Aa; and
A^ approach zero, show that the difference of Az and its principal part
is an infinitesimal of higher order than Ax and Ay.
229.
z
230.
2
=
=
231.
^
= ,-^-
xy.
a;2
232.
-y^
Find the total
-
z
+ 2x.
Vx''
+ y^.
the following functions:
differentials of
+ bxV + cy*.
+22).
hi (a? +
233. ax^
234.
2/2
235.
a'tan-'^ -^Han-'-"
X
y
236.
yze'
237.
If
u = x"f (z),
238.
li
u — f{r,s),
+ zxe" + xye'.
=
z
r
show that
^,
=
x
aw
+ y,
s
=
x
-
aw
aw _
dxdy~
239.
If
M = / (r,
0,
s,
»•
=
du
dx
240.
a
If
P
is
s
du
,
\-y
"
t.
show that
dr'
= ^,
du
r
dy
,
a;-:;
origin to
-,
— y,
z
r-
dz
«
=
=
-,
show that
.
0.
the angle between the a>axis and the line OP from the
find the derivatives of a in the directions parallel
(x, y, z),
to the coordinate axes.
241.
Show
that
(cot
is
an exact
y
—
y sec x tan x) dx
—
{x esc'
y
+ sec x) dy
differential.
Find the equations
of the
normal and tangent plane to each of the
following surfaces at the point indicated:
242. x^
+ 2y' -z'' = 16,
+ 32/ -42 = 4,
243.
22;
244.
z'
245.
y
246.
Show
surface
is
= 8x2/, at (2,
= z^-x'' + l,
1,
at
-1).
at
(3, 2,
at
(1, 2, 1).
-4).
(3, 1,
-3).
that the largest rectangular paraUelopiped with a given
a cube.
SUPPLEMENTARY EXERCISES
153
An open rectangular
box is to be constructed of a given amount
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 largest body that can be constructed from
247.
of material.
a given amount of material.
249. Find the volume of the largest rectangular parallelepiped that
can be inscribed in the eUipsoid.
a2
250.
Show
"^
62
"•"
&
that the triangle of greatest area inscribed in a given
circle is equilateral.
251. Find the point so situated that the sum of its distances from
the vertices of an acute angled triangle is a minimum.
252. At the point (a;, y, z) of space find the direction along which a
given function F (x, y, z) has the largest directional derivative.
ANSWERS TO EXERCISES
Page 8
1.
-i
4.
-1.
2.
V2.
5.
1.
3.
-1.
6.
|.
3.
2.
5.
The tangents are
The
(1, —1).
Page 14
and on the
parallel to the x-axis at
slope
right of
10.
Negative.
11.
Positive in 1st
31.
When
is
(1,
and 4th quadrants, negative
32.
(0, 0),
and
and (0,0)
=
in
2nd and
3rd.
27, 28
da;.
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.
When a; = 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
x
A,
y
^ and dy
exact value
is
0.072
*
0.754.
35.
34.
18.
36.
Increases
37.x =
38.
=
— 1, —1),
( — 1, —1)
—1).
Pages
=
(
positive between
when x
<
±^.
2
(a,
when x
s. decreases
a).
>
-x-
2
^^--(^r^riy^
tan-i|.
Page 31
2
,
3.
4
(x-iy
{x-iy(.x
+
_J:
(x-i)»
2y(7x +
2
2),
(a;
-l)(x
y
y'
X
x_
2y'
'
v^;rz^2'
_4.
2
-
a'
7
l^Zjy
„
y*
x-r
_2_.
{.x-iy
rii
Sx^y
_J_
At
154
^a'-x'^^
+ 2)2(42x2 + 24 a;-
12).
ANSWERS TO EXERCISES
12
tPa;
+ 3 0''
<i2/'
_
(ft/
1Q
di'
<(2
155
12
_
<(2-3 0''
Pages 36-38
1. e
«
2.
-
=
100
h
=
206.25.
=
«'
-
32
12
i^
between
is
w =
7.
i
«,
a
= -
+ 32
=
«,
1.691
32.
Rises
6
2
cf,
13.
Highest point
SJ.
8.
a = —2 c. Wheel
comes to
rest
when
15.
12ift./sec.,
16.
4
,
c
Vs
<
=
•^
7|ft./sec.
mi./hr.
tan fl
,
,
"• -l^"/^*^"-
V-3:l.
18.
i V3 in./see.
Neither approaching nor separating.
19.
25.8 ft./sec.
14.
=
= 3P - 24t + 32. Velocity decreasing
Moving backward when t
t — 6.309.
a
9 ir cu. ft./min.
11. 144 JT sq. ft./sec.
12. Decreasing 8 ir cu. ft./sec.
10.
t
and
negative or between 4 and
—
until
20.
64
Vs ft./sec.
Pages 43-45
1.
3.
4.
Minimum 3|.
Maximum at x = 0, minima
Minimum when x =0.
Minimum — 10, maximum 22.
= — 1 and x = +1.
2.
at
a;
13.
-a
Vs.
^2.
10.
J
14.
Length of base equals twice the depth of the box.
Radius of base equals two-thirds of the altitude.
4
16.
17.
Altitude equals - times diameter of base.
TT
IGirVs
20.
Girth equals twice length.
27
21.
Radius equals 2
i (ai
22.
The
23.
12 V2.
24.
[5^-|-6¥.
Radius of semicircle equals height of rectangle.
26.
27.
Ve
inches.
+ 02 + 03 + 04).
19.
distance from the more intense source
tance from the other source.
25.
194
ft.
4 pieces 6 inches long and 2 pieces 2 ft. long.
of the sector is two radians.
28.
The angle
29.
At the end
of 4 hours.
,
is
'^2 times the
dis-
DIFFERENTIAL CALCULUS
156
30.
He
o
31.
35.
should land 4.71 miles from his destination.
V2
—s—
2
<*
.
'
being the length of
side,
2imi./hr.
13.6 knots.
36.
Page 48
1.
2.
3.
4.
10.
Maximum = a, minimum = — n.
Maximum = 0, minimum = — (jV)
Minimum = — 1.
Minimum = 0, maximum = fy.
Either 4 or
5.
Pages
52, 63
23.
A = -i„ B = -A.
Vs - |.
Velocity = — 2imA, acceleration =
24.
26.
20.
21.
^
•
miles per minute.
'JT
± g, n bemg any mteger.
0.
f+f^rVs.
13 Vl3.
,.
„
,
} radians per hour.
29. The needle will be inclined to the horizontal at an angle of about
28.
„.
25.
32° 30'
30.
120°.
31.
120°.
33.
If
„„
a
^^-
V
the spokes are extended outward, they will form the sides of an
isosceles triangle.
Page 66
24.
a =
25.
by the
4V35.
27.
X
30.
a;
-
cos
c^,
r being the radius of pulley
string
and
line along
which
Page 61
= nir + cot~' 2, n being any integer.
< -3, x>2, or -2 < z < 1.
Pages
-x
+ 4a
1.
2y
2.
3/
3.
2 2/ =F
a;
66,
= 5, y + 2x = G.
= 8, iy — x = \b.
= ±a, ^ ± 2 2 = ±3 a.
66
its
and
the angle formed
end moves.
ANSWERS TO EXERCISES
4.
2/
5.
y
6.
=
a (x In 6
—
2
2
+ 1),
7.
a
a;
9.
2/
ay]nb = a?]nb.
-\-
^-i+i^Ki)=«-
^(--i)'
^ + ^' =
8.
x
1.
- a^ViX =
fc'aJiy
+ = 2, - = 2.
+ 3 = 4, j/-3a; = 28.
=
tan ^ 0i.
+ tan
a;
2/
2/
2/
a:
10.
90°,
11-
4^°-
(6^
- tf)
12.
90°.
13.
tan-i2V2.
1^.
_ In 10 — 1
tan ij^^^-j-^-
15.
tan-' 3
CKt>i
h<t>i
157
tan-'f.
ajij/i.
Vs.
Page 70
1.
Point of inflection
point,
2.
Point of inflection
point,
3.
Concave upward on the
(0, 3).
downward on the
— f).
(J,
downward on
The curve
is
the
right of this
left.
Concave upward on the
right of thia
left.
everywhere concave upward.
There
is
no point of
inflection.
4.
Point of inflection
ward on the
5.
Concave on the
(1, 0).
Point of inflection
I
downward on the
—2,
6.
7.
Points of inflection
— ^)-
a;
=
the points of inflection,
8.
X
<
or
a;
±—V2
upward
(0, 0),
3
^
4.
3V2.
5.
e^ v'2'
8.
fo.
right,
Concave downward between
outside.
(±3, ±1).
Concave upward when
> 3.
Pages
±2V6.
Concave upward on the
;=.
Point of inflection at the origin.
the origin.
1.
down-
left.
Points of inflection at
-3 <
left of this point,
right.
Concave upward on the
left of
DIFFERENTIAL CALCULUS
1158
Page 79
There are two angles 4/ depending on the direction in which s is
measured along the curve. In the following answers only one
of these angles is given.
' --(I)2
3
^-
r
t
1.
7.
4
8.
Z.
•-*
0°, 90°,andtaii-i3V3.
fl= ±ijr.
10.
3.
3
Page 84
I
-V2 _ yv-l
—1
^"T
^-
*^"'
6.
tan
4__
k
1.
X
—e _
^
2/«
^
g
—
V2
1
2
e
'—=•
69° 29'.
7.
irfc
'.-e^
2
y
3.
Pages
1.
The angular speed
is
^
where
,
^
92, 93
a;
is
the abscissa of the
moving
point.
2.
If xi is
the abscissa of the end in the a>axis and yi the ordinate of
j/-axis, the velocity of the middle point is
the end in the
'£].
2yi_
[*".
the upper signs being used
right, the
3.
The
lower signs
velocity is
[v
where
6 is
moving
—
7.
point.
Speed
the end in the avaxis
to the
left.
moves
The speed
to the
is
—
^
2j/i
aa cos fl],
<ia sin 0,
The speed
is
+ aW — 2
The boat should be pointed
Velocity
if
moves
the angle from the a>axis to the radius through the
Vt^
6.
if it
30°
ooi!)
sin 0.
up the
river.
= [a, h, c — gt], Acceleration =
= Vo" + 62 + (e - gt)^.
[0, 0,
—g].
ANSWERS TO EXERCISES
Velocity
Speed
Acceleration
10
12.
3
r
L
X
1)'
4a
=
«J
=
=
=
[oa (1
ao>
X
=
coaf
St!"
6in|fl'
4ajSinJeJ
cos
ait,
y
sum
=
q
The
vt sin aj<.
velocity
but the acceleration
-\-b cos
2 at,
j/
(^,
aa'cos<^].
<^,
sin 1 9
a cos ait
the
(^
[a«^ sin
tial velocities,
13.
— cos<^), ousin^],
Vz — 2 cos = 2 aw sin J
159
=
a sin ui
of the partial velocities
is
is
the
sum
of the par-
The
velocity is
not.
+ 6 sin 2 at.
and the
acceleration the
sum
of the partial accelerations.
14.
X
=
(Mit
the
—
sum
+
a sin (ui
uj) <,
y = a cos (coj + aj) t. The velocity is
of the partial velocities and the acceleration the sum
of the partial accelerations.
Page 100
2.
A.
DIFFERENTIAL CALCULUS
160
Pages 123, 124
1.
Increment
4.
-ip=n{-j
=
—0.151, principal part
——)
=
T may
negative, the percentage error in
5.
The percentage
may be
Since dl and dg
percentage errors in
and
I
error in g
—0.154.
either positive or
be | the sum of the
g.
may be
as great as that in s plus twice
that in T.
13.
-'^+^.
2''-"^
15.
zx
14.
dx
dx
dx
dy dx
dx
Bydx
131, 132
"'
dz\dx
_
d'x
1.
2.
dy dx
WW"
dy
dz dy
£^1 =
^
3
3.
—5
.X^—=5
4.
.
5.
6.
= 1:^2,
z
—
_
(a;
5
„
^15"
z
—
x
3x
•
3
Z3-'
— 5y— IZ—
= '^-^'
g—
—^
X
3V3-4.
14,
--3(a:co8Q:+2/oos/3+zcos7).
Page 135
—4
4^ =
y — 1
2/
dx
g
Normal, y + ix = H, z = 3.
Tangent plane, a; — 4 2/ — 14
„x— —3~=
dz
13.
dx dy
^
dx
\dxjy
dydx'
df_aF_aldF
dz
2uv
+ y' + xy-z'^).
-{x'
Pages
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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
surfaces in the ratio 2
4.
The
:
in the ratio 3
3.
center of gravity of the triangle
ABC,
:
2 and lateral
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 increment of a fvmction, 14, 15,
118-120.
Arc, differential of, 72.
Continuous function, 10, 113.
Convergence of infinite series, 107111.
Differentiation, of transcendental
functions, 49-62.
partial, 113-139.
Directional derivative, 129.
Direction cosines, 80, 81.
Direction of curvature, 67.
Divergence of infinite series, 107111.
Exact
differentials, 130, 131.
Exponential functions, 56-62.
Function, continuous, 10, 113
discontinuous, 10.
Curvature, 73.
explicit, 1.
center and circle of, 75.
direction of, 67.
radius of, 74.
Curve, length of, 70.
slope of, 11.
implicit, 2, 127, 128.
irrational, 2.
of one variable, 1.
of several variables, 113.
rational, 2.
Functions, algebraic, 2, 19-31.
Dependent
exponential, 56-62.
inverse trigonometric, 54-56.
logarithmic, 56-62.
transcendental, 2, 49-62.
trigonometric, 49-53.
Functional notation, 3.
variables, 2, 115.
Derivative, 12.
directional, 129.
higher, 28, 29, 114.
of a function of several variables, 124-127.
partial, 114.
Differential, of arc, 72.
of a constant, 20.
of a fraction, 22.
of an nth power, 22.
of a product, 21.
of a sum, 20.
total, 120, 121.
Differentials, 15.
exact, 130, 131.
of algebraic functions, 19-31.
of transcendental functions, 4962.
partial, 120.
Differentiation, of algebraic functions, 19-31
Geometrical applications, 63-84.
Implicit functions,
2, 127.
Increment, 10.
of a function, 14, 15, 118, 119.
Independent variable, 2.
Indeterminate forms, 95-100.
Infinitesimal, 7.
Infinite series, 106-112.
convergence and divergence
107-111.
Maclaurin's, 106.
Taylor's, 106.
Inflection, 67.
161
of,
INDEX
162
Rate
of change, 32.
Rates, 32-38.
Length
of a curve, 70.
Limit, of a function, 5.
related, 35.
sinS
of—
,49.
,
.„
Limits, 4-9.
properties
Related rates, 35.
Rolle's theorem, 94.
of, 5, 6.
Series, 106-112.
Loganthms,
56, 58.
natural, 58.
convergence and divergence
107-111.
of,
Maclaurin's, 106.
Maclaurin's
power, 110, 111.
series, 106.
Maxima and minima,
exceptional
types, 45, 46.
method of finding, 42, 43.
one variable, 39-48.
several variables, 136-138.
Mean value theorem, 101.
Power
series, 110, 111.
operations with. 111.
134.
series, 106.
Total differential, 120, 121.
of,
Partial, differentiation, 113-139.
differential, 120.
Plane, tangent, 134.
Point of inflection, 67, 68.
Polar coordinates, 77-79.
Tangent plane,
Tangent, to a plane curve, 63.
to a space curve, 81-83.
Taylor's, theorem, 102.
Natm-al logarithm, 58.
Normal, to a plane curve, 63.
to a surface, 133, 134.
Partial derivative, 114.
geometrical representation
116, 117.
Taylor's, 106.
Sine of a small angle, 49.
Slope of a curve, 11.
Speed, 85.
Variables, change of, 30, 127.
dependent, 2, 115.
independent,
2.
Vector, 85.
notation, 88.
Velocities,
Velocity,
composition
components
of, 89, 90.
of, 86, 87.
along a curve, 85-89.
along a strai^t line, 32.
angular, 34.