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DIFFERENTIAL AND INTEGRAL
CALCULUS
WITH EXAMPLES AND APPLICATIONS
BY
GEORGE
OSBORNE,
A.
WALKER PROFESSOR OF MATHEMATICS
IN
S.B.
THE MASSACHUSETTS
INSTITUTE OF TECHNOLOGY
REVISED EDITION
BOSTON,
D.
C.
HEATH &
U.S.A.
CO.,
1906
PUBLISHERS
ao2>
a
A
UBRARY of CONGRESS
Two
Copies Received
JAN 111907
A
Cooyriarht Entry
Kioto
L.
It).
iSS
o^ XXc„
COPY
No.
B.
Copyright,
1891
By GEORGE
A.
and
1906,
OSBORNE
PREFACE
In the original work,
tile
author endeavored to prepare a text-
book on the Calculus, based on the method of
limits, that
should
be within the capacity of students of average mathematical ability
and yet contain
all
that
is
essential to a
working knowledge of the
subject.
In the revision of the book the same object has been kept in view.
Most
the
of
text
been rewritten, the
has
demonstrations have
been carefully revised, and, for the most part, new examples have
been substituted for the
of subjects in a
There has been some rearrangement
old.
more natural
order.
In the Differential Calculus, illustrations of the " derivative"
aave been introduced in Chapter
"ion will
be found,
also,
II.,
and applications of
among the examples
differentia-
in the chapter
imme-
diately following.
Chapter VII.. on Series,
is
entirely new.
In the Integral Calculus,
immediately after the integration of standard forms, Chapter XXI.
has been added, containing simple applications of integration.
In both the Differential and Integral Calculus, examples
illustrat-
ing applications to Mechanics and Physics will be found, especially
in Chapter X. of the Differential Calculus, on
and
in
ter has
Maxima and Minima,
Chapter XXXII. of the Integral Calculus.
been prepared by
my
The
latter chap-
colleague, Assistant Professor N.
It.
George, Jr.
The author
also acknowledges his special obligation to his col-
leagues, Professor H.
W. Tyler and
important suggestions and criticisms.
Professor F.
S.
Woods,
for
CONTENTS
DIFFERENTIAL CALCULUS
CHAPTER
I
Functions
PAGES
AF.T*.
I.
-7, 9.
8.
Variables and Constants
Definition
and
1
Classification of
Examples
Notation of Functions.
CHAPTER
Limit.
10.
Definition of Limit
11.
Notation of Limit
12.
Special Limits (arcs
13-15.
16.
17-21.
22.
Increment,
1-5
Functions
Increment.
5-7
II
Derivative
8
8
and chords, the base
e)
8-10
.
Expression for Derivative
Derivative.
Illustration of Derivative.
.
.
.11,
Examples
Three Meanings of Derivative
Continuous Functions. Discontinuous Functions.
CHAPTER
12
13-15
10-21
Examples
.
22-25
III
Differentiation
Algebraic Functions.
Examples
....
Logarithmic and Exponential Functions.
Examples
Trigonometric Functions. Examples
Inverse Trigonometric Functions. Examples
.
Relations between Certain Derivatives.
Examples
26-39
39-45
45-51
51-57
57-60
CHAPTER IV
Successive Differentiation
57, 58.
Definition and Notation
60.
The nth
60.
Leibnitz's Theorem.
Derivative.
Examples
Examples
61
63-65
65-67
CONTENTS
VI
CHAPTER V
Differentials.
Infinitesimals
PAGES
61-63.
Definitions of Differential
64.
Formulae for Differentials.
65.
Infinitesimals
68-70
Examples
71-73
.
73,74
CHAPTER VI
Implicit Functions
66.
Examples
Differentiation of Implicit Functions.
CHAPTER
Series.
75-77
VII
Power
Series
Convergent and Divergent Series.
Positive and Negative
Terms. Absolute and Conditional Convergence
69-71. Tests for Convergency.
Examples
Convergence of Power Series. Examples
72, 73. Power Series.
67,68.
.
CHAPTER
.
78, 79
.
85-87
79-85
VIII
Expansion of Functions
74-78.
Maclaurin's Theorem.
Examples
88-93
Huyghens's Approximate Length of Arc
80,81. Computation by Series, by Logarithms
82. Computation of -w
83-87. Taylor's Theorem.
Examples
93
94-96
79.
96,97
97-100
.
89.
90-93.
94.
Rolle's
Theorem
Remainder
101
.
Mean Value Theorem
....
101-104
.
105
CHAPTER IX
Indeterminate Forms
95.
96, 97.
98-100.
Value
106
of Fraction as Limit
Evaluation of
106-110
Examples
Evaluation of g, 0- oo, oo- ex),
0°,
1",
oo°.
Examples
110-113
CONTESTS
vn
CHAPTER X
Maxima and Minima of Functions of One Independent
Variable
PAGES
AUTS.
101.
102-104.
Definition of
Maximum and Minimum
Conditions for
Maxima and Minima,
105.
When
100.
Maxima and Minima by
Values
Examples
.
.
.
.
.
.
114
114-119
Examples
119-121
d.r
Taylor's Theorem.
Problems
121-129
CHAPTER XI
Partial Differentiation
Two
More Independent Variables
Examples
....
107.
Functions of
108.
Partial Differentiation.
109.
Geometrical Illustration
110.
Equation of Tangent Planes.
or
130. 131
131. 132
133
Angle with Coordinate Planes
Examples
Ill, 112.
114-116.
Order of Differentia
Examples
tion.
113.
133-136
Partial Derivatives of Higher Orders.
Total Derivative.
Differentiation
136-139
Examples
Total Differential.
of Implicit Functions.
140-144
Taylor's Theorem
Examples
144-147
CHAPTER
XII
Change of the Variables
117.
118.
119.
120,121.
Change Independent Variable x
Change Dependent Variable
Change Independent Variable z
in
Derivatives
148, 149
to y
149
Examples
Derivatives from Rectangular
Transformation of Partial
Polar Coordinates
to
z.
.
.
150-152
to
152-154
CHAPTER
XIII
Maxima and Minima of Functions of Two or More
Variables
122,123.
124.
Definition.
Conditions for
Maxima and Minima
Functions of Three Independent Variables
155, 156
156-161
.
CHAPTER XIV
Curves for Reference
120-127.
Cirsoid.
"Witch.
Folium
of Descartes
.
162, 163
CONTENTS
Vlll
PAGES
ARTS.
Cubical Pa-
Parabola referred to Tangents.
Semicubical Parabola
128-130.
Catenary.
131-134.
Epicycloid.
135-145.
Spiral of Archimedes.
HyperCircle.
Polar Coordinates.
Logarithmic Spiral. Parabola. Cardioid.
bolic Spiral.
rabola.
Hypocycloid.
(-)*+{?)* =1
Equilateral Hyperbola.
r
Lemniscate.
164, 165
a*2/
-
2
=« 2^-«6
166,167
Four-leaved Rose.
= asm*-
167-172
o
CHAPTER XV
Direction or Curves.
Tangents and Normals
Subtangent. Subnormal. Intercepts of Tangent
Angle of Intersection of Two Curves. Examples
148. Equations of Tangent and Normal.
Examples
149-151. Asymptotes.
Examples
Polar Coordinates. Polar Subtangent
152, 153. Direction of Curve.
and Subnormal
154. Angle of Intersection, Polar Coordinates.
Examples
155, 156. Derivative of an Arc
146.
.
.
147.
.
.
.
.
.
.
.
173
174-176
176-179
179-182
182, 183
183-186
186-188
CHAPTER XVI
Direction op Curvature.
157.
158.
Points of Inflexion
Concave Upwards or Downwards
Point of Inflexion. Examples
189
190-192
CHAPTER XVII
Curvature.
Radius of Curvature.
Involute
157-161.
Curvature, Uniform, Variable
162-164.
Circle of Curvature.
193, 194
Radius of Curvature, Rectangular Co-
ordinates, Polar Coordinates.
165.
Evolute and Involute
168-170.
Properties of Involute aud Evolute.
173.
Order of Contact
Osculating Curves
.
.
.
195-200
200
201, 202
CHAPTER
Order of Contact.
.....
Examples
Coordinates of Centre of Curvature
166, 167.
171,172.
Evolute and
Examples
.
.
.
202-205
XVIII
Osculating Circle
206-208
208,209
CONTENTS
IX
PAGES
Order of Contact at Exceptional Points
175. To find the Coordinate of Centre, and Radius, of the Osculating Circle at Any Point of the Curve
174.
209
.
....
170.
Osculating Circle at
Maximum
Minimum
or
209-211
Ex-
Points.
amples
211-213
CHAPTER XIX
Envelopes
177.
Series of Curves
214
Envelope
178, 179.
Definition of Envelope.
180-182.
Equation of Envelope
Evolute of a Curve is the Envelope of
183,
is
Tangent
214,215
215-217
.
.
Examples
Normals.
its
217-221
INTEGRAL CALCULUS
CHAPTER XX
Integration of Standard Forms
184, 185.
Definition of Integration.
186-190.
Fundamental Integrals.
Elementary Principles
.
223-225
225-240
.
Examples
Derivation of Formulae.
CHAPTER XXI
Simple Applications of Integration.
Integration
191,192.
195.
Area
Examples
Derivative of Area.
Illustrations.
of Curve.
Constant of
Examples
241-244
.
244-248
CHAPTER XXII
Integration of Rational Fractions
194, 195.
Formulae for Integration of Rational Functions
Operations
196.
Partial Fractions
197.
Examples
Case II. Examples
Case III. Examples
Case IV.
Examples
198.
199.
200.
Case
p reliminary
249
250
250-253
.
I.
254-256
256-259
»
.
260-262
CONTENTS
CHAPTER
XXIII
Integration of Irrational Functions
PAGES
202.
Integration by Rationalization
263
p
203, 204.
Integrals containing (ax
206, 207.
Integrals containing
208.
+
&)«,
+
(ax
V± x + q z +
2
Examples
Examples
b) s
b.
263-266
.
266-268
Integrable Cases
CHAPTER XXIV
Trigonometric Forms readily Integrable
209-211.
Trigonometric Function and
212, 213.
Integration of tan n x dx, cot" x dx, sec n xdx, cosec n xdx
its
Examples
Differential.
.
.
270-272
273, 271
214.
Integration of tanm x sec n x dx, cot m x co&ec n xdx.
274-276
215.
Integration of
276-278
Examples
sin™ x cos w x dx by Multiple Angles. Examples
CHAPTER XXV
Integration by Parts.
by
216.
Integration
217.
Integration of
218-222.
Parts.
e ax sin
279-282
dx.
Examples
.
.
Reduction Formulae for Binomial Algebraic Integrals.
vation of Formulae.
223, 224.
Reduction Formulae
Examples
nx dx, e ax cos nx
283,284
Deri-
Examples
284-291
Examples
Trigonometric Reduction Formulae.
.
.
.
291-294
CHAPTER XXVI
Integration by Substitution
p
226.
Integrals of f(x 2 )xdx, containing (a
227.
Integrals containing Vcfi
Substitution.
228-232.
y/x 2
Examples
bx2 )i.
± a2
295, 296
by Trigonometric
Examples
Integration of Trigonometric
tion.
233.
-f
296-299
Forms by Algebraic
Substitu-
Examples
299-304
Miscellaneous Substitutions.
Examples
304, 305
.
CHAPTER XXVII
Integration as a Summation.
234.
235-237.
Integral the Limit of a
Area
of Curve.
Integral
Definite Integrals
Sum
Definite Integral.
306
Evolution of Definite
,
306-309
CONTEXTS
XI
PAGES
ART?.
of Definite Integral.
288, 239.
"Definition
240-242.
Sign of Definite Integral.
243-245.
Change
Constant of Integration.
Examples
310-314
Infinite Limits.
Infinite
Values
314-317
of/(.r)
Definite Integral as a
of Limits.
Sum
.
.
.
317-319
CHAPTER XXVIII
Application- of Integration to Plane Curves.
Application to Certain Volumes
Areas of Curves, Rectangular Coordinates. Examples
Areas of Curves. Polar Coordinates. Examples
249. Lengths of Curves. Rectangular Coordinates.
Examples
250. Lengths of Curves, Polar Coordinates.
Examples
246. 247.
24S.
251.
252, 253.
.
Volumes
of Revolution.
Volumes by Area
.
.
.
.
Examples
Examples
Examples
of Section.
320-324
325-327
327-330
330-332
333-335
Derivative of Area of Surface of Revolution.
faces of Revolution.
254.
.
.
Areas of Sur-
....
336-339
340-342
CHAPTER XXIX
Successive Integration
255-25'
Definite Double Integral.
Examples
Variable Limits.
Triple Integrals.
343-345
.
CHAPTER XXX
Applications of Double Integration
258-262.
Moment
Double Integration, Rectangular CoPlane Area as a Double
Examples
of Inertia.
ordinates.
Integral.
Variable Limits.
Double Integration, Polar Coordinates. Moment of Inertia
Variable Limits. Examples
266. Volumes and Surfaces of Revolution, Polar Coordinates
Examples
346-350
26:5-265.
350-353
353, 354
CHAPTER XXXI
Surface, Volume, and
267.
To
Moment of Inertia of Ant Solid
Area of Any Surface, whose Equation
between Three Rectangular Coordinates, x, y,
find the
amples
is
given
z.
Ex355-360
CONTENTS
Xll
To
268.
find the
Volume
whose Equation
ordinates,
Moment
269.
jc,
is
y, z.
of Inertia of
of
Any
Solid
bounded by a Surface,
given between Three Rectangular Co-
Examples
Any
361-363
Examples
Solid.
.
.
.
363, 364
CHAPTER XXXII
Centre of Gravity. Pressure of Eldids.
Eorce of Attraction
Examples
Examples
Pressure of Liquids. Examples
Centre of Pressure. Examples
Attraction at a Point. Examples
365-369
Centre of Gravity.
270,271.
Theorems
272, 273.
274.
275.
276.
of Pappus.
.
369, 370
.
370-373
373-375
375-377
CHAPTER XXXIII
Integrals for Reference
277.
Index
.
.
....
378-385
386-388
DIFFERENTIAL CALCULUS
CHAPTER
I
FUNCTIONS
1.
Variables and Constants.
unlimited number of values
is
A
quantity which
may assume an
called a variable.
A
quantity whose value is unchanged is called a
For example, in the equation of the circle
x 2 +y°
=a
constant.
2
,
For as the point whose
is a constant.
moves along the curve, the values of x and y
x and y are variables, but a
coordinates are
x, y,
are continually changing, while the value of the radius a remains
unchanged.
Constants are usually denoted by the
a, b, C, a,
(3,
first letters
of the alphabet,
y, etc.
Variables are usually denoted by the last letters of the alphabet,
*, y, z,
2.
<t>,
«A,
etc
-
Function.
When
one variable quantity so depends upon an-
other that the value of the latter determines that of the former, the
former
is
said to be a function of the latter.
For example, the area of a square is a function of its side the
volume of a srjhere is a function of its radius the sine, cosine, and
;
;
tangent are functions of the angle
x2
are functions of
x.
,
log
(V9
;
the expressions
+ 1), V*(* + l),
DIFFERENTIAL CALCULUS
2
A quantity may be a function of two or more variables. For
example, the area of a rectangle is a function of two adjacent sides;
either side of a right triangle is a function of the two other sides
the volume of a rectangular parallelopiped is a function of its three
dimensions.
The expressions
x 2 + xy
and
are functions of x
+y
g(x2 + y2), a x+ ^,
2
,
\
y.
The expressions
+ yz + zx, ^^~~f
xy
are functions of
3.
and
x, y,
log(x 2
+ y~z),
z.
Dependent and Independent Variables.
If y
is
a function of
x,
as in the equations
y
x
=x
2
,
y
= tan
4
y
a?,
=e
x
-f-
and y the dependent variable.
a function of x, x may be also regarded
called the independent variable,
is
It is evident that
as a function of
y,
variables reversed.
when y is
and the positions of dependent and independent
Thus, from the preceding equations,
x=Vy,
x
= ±tan-
1
y,
x
= log
e
(y-l).
In equations involving more than two variables, as
z
+ x — y = 0,
iv
+ wz + zx + y = 0,
one must be regarded as the dependent variable, and the others as
independent variables.
Algebraic and Transcendental Functions.
An algebraic function
one that involves only a finite number of the operations of addi-
4.
is
tion, subtraction, multiplication, division, involution
with constant exponents.*
and evolution
All other functions are called transcen-
Included in this class are exponential, logarithmic,
dental functions.
and inverse trigonometric, functions.
Note.
The term "hyperbolic functions" is applied to certain
forms of exponential functions. See page 00.
trigonometric or circular,
—
*
A more general
tion to the variable
definition of Algebraic Function
is
is,
a function whose rela-
expressed by an algebraic equation.
FUNCTIONS
A
Functions.
Rational
5.
integral powers of
x,
is
polynomial involving only positive
an
called
integral function of x\
as,
for
2 + x - 4 .r + 3 x\
example,
A. rational fraction is a fraction
whose numerator and denominator
are integral functions of the variable
a,r3
x*
A
3
example,
as, for
;
+ 2q;-l
+ x*-2x'
rational function of x is an algebraic function involving no frac-
powers of x or of any function of x.
of such a function is the sum of an integral
function and a rational fraction as, for example,
tional
The most general form
;
2
6.
Explicit
,r
rf-
3x -2x
2
x-1+
and Implicit
x-
+l
When
Functions.
one quantity
pressed directly in terms of another, the former
explicit
is
ex-
function of the latter.
For example, y
y
is
an explicit function of x in the equations
= a? + 2x,
y
"When the relation between y and x
= Va + 1.
2
is
given by an equation con-
taining these quantities, but not solved with reference to
to be
is
said to be an
an
y,
y
is
said
implicit function of x, as in the equations
axy
-f
bx 4- cy
Sometimes, as in the
+d=
first of
y
3
-\-
log y
these equations,
= x.
we can
solve the
y, and thus change the function from
Thus we find from this equation,
equation with reference to
implicit to explicit.
_bx±d
m
ax
7.
Single-valued
and
Many-valued
y
for every value of x, there
Expressing x
in
+
terms of
= X- - 2
Functions.
In
y,
the
.r,
one and only one value of
is
x
c
we have
= 1 ± Vy + 1-
y.
equation
DIFFERENTIAL CALCULUS
4
Here each value
case,
y
is
In the
An
of y determines two values of
a single-valued function of x.
x
latter case,
is
a two-valued function of
w-valued function of a variable x
values corresponding to each value of
The
8.
Notation
and the
is
of
like, are
x.
x,
we may
has an unlimited num-
x.
Functions.
The symbols
used to denote functions of
a function of x,"
?/.
a function that has n
is
inverse trigonometric function, tan -1
ber of values for each value of
In the former
x.
F(x),f(x), <£(#), \f/(x),
Thus instead of " y
x.
write
y=f(x), or y
= <f>(x).
A functional symbol occurring more than once in the same problem or discussion is understood to denote the same function or
Thus
operation, although applied to different quantities.
•
f(x)
then
f(y)=y*
if
= x* + 5,
+ &,
/( ft )
(1)
= a + 5,
2
+ l) = (a + l) + 5 = a* + 2a + 6,
/(1) = 6.
/(2) = 2 + 5 = 9,
2
/(a
2
In
all
these expressions
fined
by
(1)
;
that
/(
denotes the same operation as de-
)
the operation of squaring the quantity and
is,
adding 5 to the result.
Functions of two or more variables are expressed by commas be-
tween the
Thus if
then
variables.
/ (x,
2
then
xy
- f,
= a + 3 ab - b\
f(b, a) = b + 3 ba - a
jf(3, 2) = 3 + 3-3-2 - 2 = 23.
f(a, 0) = a
4>(x, y, z) = x + yz-y + 2,
l,-l) = 3 + l(-l)-l + 2 = 27;
<f>(3,
*(a, 0, 0) = a - b + 2
<K0, 0, 0) = 2.
f(a, b)
2
2
2
2
2
If
=x +3
y)
.
2
s
2
3
3
2
;
2
.
FUNCTIONS
9.
y
and
if
by
If y is a given function of x, represented
Inverse Function.
from
5
this relation
+(*),
we express x
'
then each of the functions
=
(i)
<f>
=
and
in terms of
y,
so that
(2)
•/<(.'/),
\p
said to be the inverse of the
is
other.
.
For example,
= * = <£<»>
x =VS = ^O/)V
if
then
Here
\p,
the cube root function,
the
is
inverse
of
cf>,
the cube
function.
= a* = <Kx),
x = log y = xp(y).
V
If
then
a
Here
the logarithmic function,
«/r,
function.
x
y =-,
Again, suppose
is
the inverse of
<f>,
the exponential
n
-4- 2i
— x-=<f>(x)
v — 2
x =
j = \p(y)
(3)
r
From
this
we derive
'
(4)
Here xp as defined by (4) is the inverse of
as defined by (3).
The notation ^> _1 is often employed for the inverse function of
<f>
Thus,
if
y
if
=
=
x=f~
x
<£(#),
y=f(?),
The student
is
<£
0/).
1
(y).
already familiar with this notation for the inverse
trigonometric functions.
If
1.
y
Given
<j>.
_1
=
x
sin x,
= sin -1
EXAM PLES
2x* — 2 xy + y = a
2
?/.
2
;
change y from an implicit to an explicit function.
y
= x ± Va — x
2
2
.
DIFFERENTIAL CALCULUS
6
Given
2.
sin
(a;
ft
— y)~ m
sin
?/
change y from an implicit to an explicit function,
Ans.
^3.
Given
that
+
=/(*)
ft)
/ (a +
1)
+ (6
-6
2
a;
F(x)
s
fc
.
;
Show
/ (2a?) -/(- 2 ag =
}
<£
Show
,
2
2
= ^t^l:
If
2
a;
2
Griven.f(x)
that
/(0).
1),
=6
+ l)h + (6x- 3)/i + 2
-/(*)
i»
;
s
show that
%K^.
/(-
= (x -1)
i^(a? + 1) - F(x — 1) = 8 X
H. Given
h.
m + cos x
= 2^-3^+^+2
/(a?)
Show
—
1
/(l), /(2), /(|),
find
fix
= tan'
y
(0)
=
c»,
<£
find/(0),
(a -\-b)
/(*)+/(-«).
[/ (a:)]
=
.
- [/ (- a?)]
2
2
.
cj>(a)cj> (6).
that the same relation holds for the function
= cos 4- V"^ sin
(a) ^(6).
if/(a + b) =
$(0)
giving
(9,
if,
"
If
7
>«-sfey
-
show that the inverse function
8.
If
<f>(x)
=
X
ax—
,
is
If
find the inverse function of
/(«)
show that /-1 (a;)
10.
Show
/ (a? +
ft,
<j>.
b
=
c.
= log.(«+'V?=l),
= a* + *y
•
/ (a^) = ax* + 2 toy + q/
If
same form.
c
Compare the two functions when
V9.
of the
2
;
find /(l, 2),
/ (y, - a?).
that
y
+
ft)
=/(«,
2/.)
+ 2(aa? + by)h + 2(bx + cy)
ft
+f(h,
ft).
FUNCTIONS
\m-\-n
11.
Given
«£(m,n)
\m
where m,
<f>
12.
Given
(m, w
»,
|w
are positive integers;
+ 1) +
<£ (m
/ fe
show that f{y +
+ 1,
& »)
z,z
show that
») = <K m + 1? + !)•
?i
z
«,
?/,
2,
X,
y
y»
*>
X
+ x, x + y) = 2/
(a;,
y, 0).
CHAPTER
LIMIT.
10.
When
Limit.
II
INCREMENT. DERIVATIVE
the successive values of a variable x approach
a, in such a way that the difference
becomes and remains as small as we please, the value a is
nearer and nearer a fixed value
x
—a
called the limit of the variable
The student
meaning
x.
supposed to be already somewhat familiar with the
is
which the following
of this term, of
may
illustrations
be
mentioned.
The limit of the value of the recurring decimal
number of decimal places is indefinitely increased, is
The limit of the sum of the series 1 + + + -J-H
-J-
of terms
The
is
indefinitely increased,
limit of the fraction
x
The
is
.3333
—
-J-
...,
as the
^.
,
number
as the
is 2.
~a
—a
,
as x approaches a, is 3
circle is the limit of a regular polygon, as the
a2
number
.
of sides
indefinitely increased.
The
limit of the fraction
—
—
,
as 6 approaches zero,
is 1,
provided
6
6
is
expressed in circular measure.
11.
Notation
"Lim^"
of
Limit.
"The
denotes
Special Limits.
2
(2 x
2
2
There are two important limits required
in the following chapter.
(a)
Lim 0=o
,
will be used
a, of."
—a == 2.
xr — ax
- hx + h ) =2x>.
x2
Lim A=0
Some
x approaches
Lim z=a —
For example,
12.
The following notation
limit, as
6 being in circular measure.
;
LIMIT
Let the angle
ADA' =
2
From geometry, ABA'
that
2 a sin
is,
<
that
2a0
is,
sin
<2 a tan
> cos
and
sin
a be the radius of the arc
let
^-
2 aO,
Also from geometry,
0.
v
(i)
ACA' < ADA
sin
fl
0,
^1)
and
>e,
(2)
is
(2),
mediate in value between
As
approaches zero, cos
Hence
;
cos
0.
Hence by
ACA!.
< AC A'
u
1
inter-
and cos
0.
approaches
1.
Lini„ =0
sin 9
..
The student
and
sin
6,
will do well to compare the corresponding values of
taken from the tables, for angles of 5°, 1°, and 10'.
Angle
sin 6
5°
—
= .0872605
36
.0871557
= .0174533
.0174524
=.0029089
.0029089
1°
-5180
10'
'
1080
(b)
Lim 2=ae
M + —IV
z )'
]
Before deriving this limit
the value of the. ex pression for increasing values of
(1
(1
Thus,
z.
+ 1) = 2.25
+ 1) = 2.48832
2
5
+ to) = 2.59374
= 2.70481
(1.01)
= 2.71692
(1.001)
= 2.71815
(1.0001)
= 2.71827
(1.00001)
= 2.71&18
(1.000001)
(1
us compute
let
•
10
100
1000
10000
100000
1000000
.
DIFFERENTIAL CALCULUS
10
The required
last
limit will be found to agree to five decimals with the
number, 2.71828.
By
the Binomial Theorem,
which may be written
iy =
+
1\ /
/
1
i
zj
When
^
+ i + izl
+v
[2
|3
,
z increases,
the fractions -, z
T
have
This quantity
is
approach
zero,
and we
e,
so that
= 1+7 + %+£ + £ + '—1
The value
etc.,
z
usually denoted by
e
2
~w~~z/
|2
[3
of e can be easily calculated to
|4
any desired number of
decimals by computing the values of the successive terms of this
series.
For seven decimal pi ices the calculation
is
as follows:
1.
2) 1.
3)
.5
4)
.166666667
5)
.041666667
6)
.008333333
7)
.001388889
8)
•000198413
9)
.000024802
10)
.000002756
11)
.000000276
.000000025
e=
This quantity
e is
2.7182818--.
the base of the Napierian logarithms.
* For a rigorous derivation of this limit, the student
tensive treatises on the Differential Calculus.
is
referred to
more ex-
DERIVATIVE
11
An
increment of a variable quantity is any addidenoted by the symbol A written before this
quantity. Thus Ax denotes an increment of x, Ay, an increment of y.
For example, if we have given
Increments.
13.
tion to its value,
and
is
y
= 10, then
and assume x
of y
is
if
we
=A
increase the value of x by
increased from 100 to 141, that
In other words,
if
is,
by
2,
the value
41.
we assume the increment
of x to be
A& = 2, we
increment of y to be A?/ = 44.
If an increment is negative, there is a decrease in value.
shall find the
For example, calling x = 10 as before, in y = x
Ax
if
y
and the same
initial
value of
us calculate the values of
A.'-.
We
Ax =
It
zero.
=x
2
,
= 10,
Ay corresponding
to different values of
then Ay
=
and -^-
=
Ax
3.
09.
23.
2.
44.
22.
1.
21.
21.
20.1
0.1
2.01
0.01
0.2001
20.01
0.001
0.020001
20.001
20 h
h
The
equation,
thus find results as in the following table.
If
ments
= - 36.
x,
z
let
,
Ay
then
With the same
Derivative.
14.
= - 2,
2
+
20+
1C-
ft
third column gives the value of the ratio between the increof
./
and of
y.
appears from the table that, as
Ay
also diminishes
Ax diminishes and approaches
and approaches
zero.
DIFFERENTIAL CALCULUS
12
The
ratio
—^
proaches 20 as
its limit.
This limit of
and
is
^=
dx
20.
diminishes, but instead of approaching zero, ap-
—^ is
called the derivative of y with respect to x,
denoted by -^.
**
In
this case,
It will be noticed that the value
tion y
=x
2
,
Without
and partly on the
-^ from y = x2
dx
Increase x by Ax.
the derivative
20 depends partly on the func-
initial value
restricting ourselves to
tain
when x = 10,
10 assigned to
any one
initial value,
x.
we may
ob-
.
Then the new value
of y will be
= (x + Ax)
Ay = y — y = (x + Ax) — x = 2xAx + (Ax)
2
y'
;
therefore,
Dividing by Ax,
The
limit of this,
Ay =
2x+
Ax
2
.
^
when Ax approaches
—=2
Hence,
2
2
1
zero, is 2x.
x.
dx
The
derivative of a function
may
then be defined as
the limiting
value of the ratio of the increment of the function to the increment of the
variable, as the latter increment approaches zero.
It is to be noticed that -2 is not here defined as & fraction, but as
dx
a single symbol denoting the limit of the fraction
will find as he advances that
—
has
many
A ^.
—
The student
of the properties of
ordinary fraction.
The
derivative
is
sometimes called the
differential coefficient.
15. General Expression for Derivative.
In general,
y -/(»)
Increase x by Ax, and
we have
the
new
value of y }
y'=f(x-\-Ax).
let
an
:
.
DERIVATIVE
Ay =
13
- y =/(* + A ^) — /(*)i
A?/
= /(* + A.r)-/(.r)
y'
A#
A.y
A#
The process of
(f.i'
Geometrical Illustration.
finding the
derivative
from y = x-, may be illustrated by a squared
Let x be the length of the side OP, and y the area of the square
on OP.
That is. y is the number of square units
corresponding to the linear unit of x.
When
area
the side
is
Ay
increased by PP', the
increased by the space between the
is
y
X
squares.
That
Ay_
is,
Ay=2xAx+(Ax)* =^=2x+Ax,
Ax
dy
ax
From
16.
T
A?/
•
p
the definition of the derivative
process for obtaining
Ax
X
2x.
Ax
we have
P'
the following
it
(a) Increase x by Ax, and by substituting x -f- Ax for x, determine y + Ay, the new value of y.
(b) Find Ay by subtracting the initial value of y from the new
value.
(c)
(d)
limit
Divide by Ax, giving
—
Ax
Determine the limit of _
is
•
_?,
as
Ax approaches
zero.
—
the derivative
dx
Apply
1
.
y
this process to the following examples.
=2
3
.r
— 6 x + 5.
Increasing x by Ax,
EXAMPLES
we have
y + Ay = 2(x + Axf-6(x + Ax) + 5;
therefore,
Ay = 2 (x + Ax) — 6 (x + Ax) + 5 — 2 x' + 6x—5
= (6 x - 6) Ax + 6 x(Ax) + 2(A x)
3
2
2
3
.
This
DIFFERENTIAL CALCULUS
14
Dividing by Ax,
-^ = 6x -6 + 6xAx + 2(Ax)
v
}
2
2
.
Ax
*
= Lim A ^/ = 6x
Ax
dx
2.
y
y
=
x
+1
x + Ax
+ Ax + 1
x + Ax
Ay =
x + Ax + 1
+ Ay =
-6.
2
x
Ax
x
z
+1
(x-f Ax-|-l)(x
+ l)
Ay =
1
Ax (x + Ax-fl)(x + l)'
^
= _J:
^-Lira
^°Ax (x + 1)
dx~
2
2/= Vx.
3.
y
-\-
Ay
= Vx -f Ax,
A?/
= Vx + Ax — Vx,
Ay _
Ax
The
limit
Vx
of
-f-
Ax —
Ax
this
V#
_
takes the
rationalizing the numerator,
indeterminate form
Az=0
^4.
4. 2/
y=x
-2x + 3x-4,
4
aj -
5.
Ax
2
?/=(x-a) 3
,
But by
J
we have
Ay _
Ax
Ax Ax(Vx + Ax+Vx)
dx
-.
1
Vx+Ax-fVx
2V
^
= 4x -4x +
dx
3
^/ =
dx
3(x-a) 2
.
3.
DERIVATIVE
6.
y=(« + 2)(3-2«)
11
g = -4*-l.
J
%
mo;
=
—X
it
x
15
>
\n-xf
dx
= 2.V-T.
»*
+
»
mn
da
9
"<*
cty
= .r + a
+ *>*
2
10.
.-/
dx
V
,=
*12.
^/=V^T2
dx _
14.
15.
16.
y
17.
dy
dx
"We shall
_
dt~
that the derivative of the area of a
is its
that the derivative of the
now
'circle,
'
X
-Vcr—x2
1
2$
with respect
circumference.
respect to its radius,
tive.
Zx%
2
dy _
dx
=±
Show
=
dx
2
Show
+l
dy _
1
dx 2Vz-r-2
= x\
to its radius,
+ a)
dt
2/=V« -^,
s
«
~
(*-!)"
8
13.
(«j
is
volume of a sphere,
with,
the surface of the sphere.
give some illustrations of the
meaning of the deriva-
DIFFERENTIAL CALCULUS
16
17. Direction of a Plane Curve.
This
is
one of the simplest and
most useful interpretations of the derivative.
Let P be a point in a curve determined by
and PT the tangent at P.
= y.
Let OM=x,
its
equation y
= f(x),
MP
If
Ax
we
give x the increment
= MN,
y will have the
Draw PQ.
Now
will
Then
Ax diminish and
Ay will also
if
approach
approach
in-
= PQ.
crement Ay
zero,
point
the
zero,
move along
PQ
towards P, and
Q
curve
the
will
approach in direction PT as
(1), we have
its
limit.
Taking the limit of each member of
TPR = Lim Ax
tan
That
the derivative
is,
—
,
at
^
=^
Ax dx
_
any point of a curve,
is
the trigono
dx
metric tangent of the
clination to
tangent
OX
line
in-
of the
that
at
point.
This
quantity
The
de-
is
noted by the term
slope.
slope of a straight
line is the tangent of its
inclination
of
to
the
axis
X
The
slope
of
a plane
curve at any point
is
slope of its tangent
the
at
that point.
Thus, -^, at any point of a curve,
point.
is
the slope of the curve at that
—
DERIVATIVE
For example, consider the parabola x2
The
At
slope
of the curve
1
y
•
2p
= 0, the slope = 0, the direction being
x = 2 p, the slope = 1, corresponding to an
At L, where
Beyond
L
all
horizontal.
inclination
X
the slope increases towards
towards the limit
For
—=—
dx
= -x—
2
= ±py,
where x
0,
of 45° to the axis of
is
is
17
oo,
the inclination increasing
90°.
points on the left of
x
Y",
is
negative, and hence the slope
negative, the corresponding inclinations to the axis of
X
being
negative.
Velocity in
Terms
over the distance
OP =
18.
of a Variable
s in
A
denoting Time.
t
the time
t,
s
body moves
being a function of
t;
it is
required to express the velocity at the point P.
Let As denote
the
PP
distance
in the interval At.
it
£
ir
would be equal
P and
-
±-,
P, and
is
were uniform during this interval,
If the velocity
to
As
For a variable velocity,
make
—
traversed
—
As
is
the average or
more nearly equal
mean
velocity between
to the velocity at
P the
less
we
At.
That
is,
the velocity at
— -r.'
P — Lim Afe0 -r.
At
dt
If v denote this velocity, v
Thus,
—
is
—
= ds
•
the rate of increase of
s.
dt
Similarly,
-vr
and
-rr
are the rates of increase of x
and y respectively.
DIFFERENTIAL CALCULUS
18
19.
The
Acceleration.
rate of increase of the velocity v is called
acceleration.
If
we denote
this
by
a,
we have by
a
—
the preceding article,
—
dv
dt
For example, suppose a body moves so that
s
Then the
v
velocity,
= t\
= ds
_
3?
dt'
dv
and the acceleration,
6t.
dt'
Rates
20.
For further
two following problems
of Increase of Variables.
derivative, consider the
illustrations of the
Problem 1.
A man
walks across the street
from A to B at a uniform
rate of 5 feet per second.
A lamp
at
L
throws his
shadow upon the wall
MN. AB is 36 feet, and
BL
4
feet.
How
fast is
the shadow moving when
he
16 feet from
is
When
26 feet
?
A?
When
30 feet?
Let P and Q be simultaneous positions of man and shadow.
Then
When
That
is,
y _ BL
x~PB
4
36
he walks from
—x
P
Let
AP = x, AQ = y.
4x
V
(1)
36 —a;
to P', the
shadow moves from Q
to Q'.
when Ax=PP', \y=QQ'.
Let At be the interval of time corresponding
Then we may
write
Ay =
Ax
to
Ax and
Ay.
At^
'
Ax
At
(2)
DERIVATIVE
If
now we suppose
Ay and A.i* will
and we have for the limits of the two
At to diminish indefinitely,
also diminish indefinitely,
members
19
of ^2),
f?
dy
— dt
dx
dx
rate of increase of
?/
rate of increase of
x
Art. 18.
dt
velocity of
That
shadow
at
is,
velocity of
Finding the derivative of
(1),
any point
Hence,
shadow
at
any point
See Ex.
144
-
=
=
=
Problem
The
2.
—
-
;
when x
;
20 feet per second, when
of a ladder 20 feet long rests against a wall.
moved away from the wall
12 feet
from the wall ? AVhen 16
feet from the wall ?
Suppose PQ to be one
position of the ladder.
Let
AP=
x,
AQ = y.
Then
y
= 16
= 26
x = 30.
when x
7.2 feet per second,
is
= V400 - x
2
.
man)
feet per second
the top moving,
is
the foot
of
X)-
rate of 2 feet per second.
How fast
(velocity
2
(5 feet per second)
foot of the ladder is
when
— x}
1.8 feet per second,
The top
Art. 16.
x)
720
(oo
8,
144
Q=
(36
(36
dy
dx
we have
144
dy =
dx
(36 - xj
velocity of
Q_
man
(3)
at a
uniform
DIFFERENTIAL CALCULUS
20
When
That
Pto
the foot moves from
is, if
Ax
In the same
= PP',
way
Ay
=
as in
P', the top
moves from Q
to
Q\
QQ'.
Problem
1,
Ay
Ay
Ax
_ At
~ Ax
At~
dy
And from
dy
this,
_dt
dx~ dx'
dt
that
Q_
velocity of top at
is,
velocity of foot
From
_
dy
(3),
dy
dx
—x
See Ex. 14, Art. 16.
<^~V400-;/
Hence,
velocity of top at any point
Q
=
= (velocity of foot)
V400-X
2
2x
V400-^
feet per second.
2
The negative sign is explained by noticing from the figure that y
when x increases. Hence the rates of increase of x and y
decreases
have different signs.
When
x
—
12,
velocity of top
=—
11 feet per second.
When
x
= 16,
velocity of top
=—
2| feet per second.
From
these problems
it
the increments of *y and x,
'
of these variables.
appears that, while
—
dv
dx
is
Aw
-^
is
the ratio between
the ratio betiveen the rates of
J increase
DERIVATIVE
Increasing and
21.
junction of x
is
Decreasing Functions.
if
If the derivative
of a
when x increases; and
function decreases ivhen x increases.
positive, the junction increases
the derivative is negative, the
For
21
—
the derivative
,
which
if
the ratio between the rates of
is
dx
increase of the variables (see conclusion of Art. 20),
is positive, it
follows that these rates must have the same sign
is,
when x
But
increases,
-^
ax
if
must have
negative, the rates
increases,
different signs
;
that
is,
and increases when x decreases.
also evident geometrically
is
y increases
and decreases when x decreases.
when x
y decreases
This
is
;
that
by regarding -^
as the slope of
a curve.
A to B,
As we pass from
C,
y increases as x increases, but from
B to
y decreases as x in-
creases.
A
Between
slope
—
B
and
positive
is
;
the
be-
tween B and C, negative.
In the former case y is
said to be an increasing
function
;
the
in
latter
case, a decreasing function.
For example, consider
= x3 from which we find -^ =
^x
rh
the function y
Since
—
dx
is
,
positive for all values of
x,
3x*.
the function y
=x
3
is
an
increasing function.
If
we take y
1
—
we
,
find
—
1
dx
x
Here we have a decreasing function with a negative derivative.
Another illustration is Ex. 1, Art. 16,
y
When
When
x
is
x
is
=2
3
ar
- 6 x + 5,
^
=6
ax
(a?
- 1).
numerically less than 1, y is a decreasing function.
numerically greater than 1, y is an increasing function.
DIFFERENTIAL CALCULUS
22
22.
A
Continuous Function.
y=f(x), is said to be
when y = f (afc) is a definite
AxQ approaches zero, Ax being
function,
continuous for a certain value x
,
of x,
quantity, and A?/ approaches zero as
positive or negative.
The
latter condition is
sometimes expressed, "when an infinitely
small increment in x produces an infinitely small increment in y."
The most common
case of discontinuity of the elementary functions
(algebraic, exponential, logarithmic, trigonometric
nometric, functions)
is
when
the function
and inverse
trigo-
is infinite.
Y
^
a
"-——
A
°
-
\
For example, consider the function y
for all values of x except x = a.
When
taking x
x
= a, y =
when x>a, y
There
oo,
that
is,
sufficiently near a.
is
=
1
which
is
continuous
y can be made as great as we please by
Also when x<a, y is negative, and
is positive.
a break in the curve
when x
to be discontinuous for the value x
= a.
= a,
and the function
is
said
•
.
DERIVATIVE
The function y
(x-ay
is
23
likewise discontinuous
This function being positive for
all
values of
x,
when x = a.
the -two branches
of the curve are above the axis of x.
Likewise the functions, tan
In general,
if/(.r)
= oo,
y=f(x) corresponding
x,
sec
x,
when x = a,
to x
= a, and
are discontinuous
there
when x = —
Z
a break in the curve
is
both the curve and the function
are then discontinuous.
I
X
Another form of discontinuity
is
seen
2
in the function y = —
when a = 0.
limits, according
Lim z = +
-^—
=1.
see that
when x =
curve jumps from
from
B to
The function
ous for x
y=2
to
the
y=l,
A.
is
discontinu-
= 0.
It is to be noticed that the
definition
of
the
Lim
i
r+i
2'+l
is
,
as x approaches zero
through positive or negative values.
that
2
2*+l
Here y approaches two
We
4-
derivative
= 2.
DIFFERENTIAL CALCULUS
24
implies the continuity of the function.
limit, unless
Ay approaches
The converse
zero
For —^ cannot approach a
Ax
when Ax approaches
zero.
There are continuous functions which
have no derivative, but they are never met with in ordinary
not true.
is
practice.
EXAMPLES
The equation
1.
Find
(a)
of a curve is y
its
Find
where
points
curve
parallel
is
+ 2.
inclination to the axis of
when x =
(b)
x>
x,
when x = 0, and
0°
Ans.
1.
and
135°.
the
the
to
the axis of X.
x=0
Ans.
and x=2.
Find the
where
the
(c)
points
slope
unity.
is
Ans.
= (1±V2).
£C
(d)
Find
the
point where the direction
is
the same as
that at x
>
2.
= 3.
Ans. x
In Problem
1,
when
man ?
Art. 20,
be the same as that of the
When
= — l.
will the velocity of the
Ans.
one quarter, and when nine times, that of the
Ans.
^3.
A
When
AP = 12
circular metal plate, radius r inches,
the radius being expanded
m
At what
rate is the
conditions of Ex. 3 ?
is
inches per second.
the area expanded ?
4.
When
ft.
man ?
ft.,
and 32
ft.
expanded by heat,
At what rate is
Ans. 2 irrm
volume
shadow
AP= 24
sq. in.
of a sphere increasing
Ans. 4 Trrra cu.
per
sec.
under the
in.
per sec.
DERIVATIVE
5.
The radius
of a spherical soap bubble is increasing uniformly
at the rate of yL inch per second.
volume
is
increasing
when
Find the rate
the diameter
Ans.
6.
In Exs.
"
Is
7.
»+l
5, 7,
is
at
which the
3 inches.
^ = 2.827
cu. in.
per
sec.
Art. 16, is y an increasing or a decreasing function ?
an increasing or a decreasing
o function of x ?
In the Example
ing function of
8.
25
x,
and
1,
for
above, for what values of x is y an increaswhat values a decreasing function ?
Find where the rate of change of the ordinate of the curve
+ o, is equal to the rate of change of the slope of
s
y = X — 6.i~ + 3.r
the curve. Ans.
x= 5
or 1.
—
3
9.
When
is
the fraction
—-x
x +a
2
-
2
increasing at the same rate as
Ans.
When
x2
a??
=a
2
.
10. If a body fall freely from rest in a vacuum, the distance
through which it falls is approximately s = 16 t 2 where s is in feet,
and t in seconds. Find the velocity and acceleration. What is the
velocity after 1 second ?
After 4 seconds ? After 10 seconds ?
,
Ans. 32, 128, and 320
ft.
per
sec.
CHAPTER
III
DIFFERENTIATION
23.
The process
of finding the derivative of a given function is
called differentiation.
trate the
meaning
The examples
in the preceding chapter illus-
method of
any but the
of the derivative, but the elementary
differentiation there used
becomes very laborious
for
simplest functions.
Differentiation
is
more readily performed by means of certain
general rules or formulae expressing the derivatives of the standard
functions.
In these formulae u and v will denote variable quantities, funcand c and n constant quantities.
tions of x
;
It is frequently convenient to write the derivative of a quantity u,
—
u instead of —
dx
dx
the symbol
Thus
24.
A^ ~r
dx
v)
^
t
—
dx
denoting " derivative of ."
k e derivative of (u
-f-
v\
may be written
Formulae for Differentiation of Algebraic Functions.
i.
^=1.
dx
II.
TTT
—
=
dx
0.
—
dx
4-
(
.
}
——
dx
26
—
dx
-l-
—
dx
(u-\- v),
•
DIFFERENTIATION
du
d
—
(«r) = v
dx
dx
TTT
f
IV.
d
dx
V
N
^ __
,
vi
h
—
?<
dx
du
dx
u
dx
dx
.
v2
dx\vj
VII.
—
dv
.
V
7
A/ ^—
27
(vr)=nun ~ 1 —.
dx
dx
These formulae express the following general rules of
ation
differenti-
:
I.
TJie derivative
II.
TJie derivative
of a variable with respect
of a constant is zero.
III.
TJie derivative
of the
sum of two
of
product of two variables
to itself is unity.
variables is the
sum of
their
derivatives.
IV.
TJie derivative
the products
tJie
of
tJie
eacJi variable
by
tJie
is tJie
sum of
derivative of the other.
TJie derivative of tJie product of a constant and a variable is
V.
product of the constant and.tJie derivative oftJie variable.
VI.
TJie derivative of a fraction is tJie derivative of tJie numerator
multiplied
by
tJie
nator multiplied by
denominator minus
tJie
tJie derivative of tJie denominumerator, this difference being divided by the
square of'the denominator.
VII.
TJie derivative of any power of a variable
exponent,
tJie
power
exponent diminisJied by
witJi
is tJie
product of the
&nd
the derivative
1,
of the variable.
25.
Proof of
a derivative.
26.
I.
For, since
Proof of II.
vary.
Hence Ac
This follows immediately from the definition of
=
and
A
\w
—
=
Ax
1, its limit
constant
—
=
Ax
j
is
1.
a quantity whose value does not
therefore
•
dx
—
=
dx
its
limit
—
=
dx
0.
DIFFERENTIAL CALCULUS
28
Let y
Proof of III.
27.
= u-\-v,
and suppose that when x reAu and Av,
ceives the increment Ax, u and v receive the increments
respectively.
Then the new value
y
therefore
Divide by Ax
of y,
+ Ay = u + Au + v + Av,
Ay = Au-{- Av.
then
;
Ay _Au
Ax Ax
Now
suppose
Ax
Av
Ax
and approach
to diminish
and we have for
zero,
the limits of these fractions,
dy _du
dx
dx
If in this
we
substitute for
u
y,
—
d /
(u
dx
dv
dx'
-{-v,
we have
+ v) = du
N
.
28.
and
We
,
s
,
,
du
dx
.
dv
dx
should then have
,
= uv;
y + Ay = (u + Au) (y + Av),
Ay = (u + Au) (y + Av) — uv = vAu + (u
Divide by
Ax
dw
,
dx
-f-
Au) Av.
;
^/ = „^ + M + Au) ^.
Ax
Ax (
Aaj
then
Now
u + Ait
,i
that
.
dx
Let y
Proof of IV.
then
dv
1
same proof would apply to any number of
It is evident that the
terms connected by plus or minus signs.
d ,
dx
,
dx
suppose Ax to approach
is u,
dy
-£
= v du
dx
dx
—d (uv)=v du
dx
/
is,
zero, and, noticing that the limit of
we have
dx
k
,
\-u
,
\-u
—
dv
dx
dv
—
dx
;
\
a
.
DIFFERENTIATION
Formula IV. may be extended
Thus we have
Product of Several Factors.
29.
29
to
the product of three or more factors.
—
d ,
v
d ,
d ,
dw
x
as
(uv -w)=w
(uv)J + UV
( uviv)
y
K
K
)
}
dx
dx
dx
dx
—
—
•.
= w[ v du
,
,
\-u
= vw du
—
dv\
.
-{-uv
]
,
\-uw
dx
dw
dx
dx
dx
dv
,
[-uv
dx
dw
—
dx
from the preceding that the derivative of the product
two or three factors may be obtained by multiplying the derivative of each factor by all the others and adding the results.
This rule applies to the product of any number of factors. To
prove this, w& assume
It appears
of
—d
(
I
Wh
•••
Then ^-(
u n \j
l(
\
u
-2
=u
2
u3
un
•••
—
du,
-f
u u z uA
...
x
— w w n+1 = un+l -—( UjU
B
2
un
••
•
—
du„
-
h u x u 2 ~-u,
uA + u&v
•••
un
J
=u u
2
+
?,
Thus
7i-|-l
lW2
3
• • •
...
un+1
— + Uju u
du*
duo
,
i
3
ax
4
• • •
u n+l —? H
ax
.
p.
du n
dx
.
H
w^
• • .
du„^
dx
w n _!W n+1
du
—
dx
n
Wn _n±i.
ax
it
appears that
factors,
and
if
is
the rule applies to n factors,
it
holds also for
consequently applicable to any number of
factors.
Tlie derivative
of the product of any number of factors
is the
sum of
the products obtained by multijjlying the derivative oj each factor by all
the other factors.
DIFFERENTIAL CALCULUS
30
30.
This
Proof of V.
we may
derive
it
is
a special case of IV.,
dc
—
being
dx
zero.
But
independently thus
= cu,
y
y
+ Ay = c(u + Aw),
Ay = cAw,
Aw
Aw
—
= —
Ax
Ax
c
dy
-^
dx
31.
= c du
—
y
x
c
du
dx
=u
+ Aw
w
+ Av
v+Av
p
'
v
Aw
w
Av
Aw_ ^ x
^x
Ax~ (y + Av)v
and
Now
,
dx
Ay = «±* tt - ? = »* M ~ "f »
therefore
v
or
~Lety
Proof of VI.
Then
— (cw) =
d,
,
dx
Ax to approach
we have
suppose
+ Av is
v,
dy
dx
zero,
_
V
and noticing that the limit
dw
dx
~~
u
v2
Or we may derive VI. from IV. thus
Since
y
= -,
v
therefore
yv
= u.
dv
dx
of
,
.
DIFFERENTIATION
t>
By
dy
ttt
v-f +
dx
IV.,
.
dv
du
=—
y—
dx
dx
dy __ du
dx
dx
v
_
dy
therefore
u dv,
v dx
du
dx
dx
Proof of VII.
32.
y
and
that
2
= un
+ Ay = (u + Aw) n
Ay = (u + A?*)' —
,
,
1
Putting
Ay
—
dv
dx
v
y
then
u
1
suppose n to be a positive integer.
First,
Let
31
=
u"
.
u + A?<, we have
= (V — u) (u" ~ + u ~ u + ~3 w H
Ay = Au (u'*- + it" - u + u*^V
w*= (u*- +u' u + m'- ^ + a- —
u' for
—
un
l
l
2
hl
«4
1
is,
2
1
let
1
n
2
2
8
1
)
...
then, u being the limit of
terms within the parenthesis becomes u n
-/
ax
= nun ~
l
~l
;
—
u',
each of the n
therefore
•
dx
of Art. 29,
,
— (O =u — +u
n ~ lC
dx
du
dx
-1
n
dx
+
...
to
dx
Second, suppose n to be a positive fraction,-?.
Let
y
then
therefore
if
= u",
= up
;
—
Of) = — (u
dx
dx
p
KJ
J
),
Ax
Or it may be proved by regarding this as a special case
where u l7 u 2 ••• and u n are each equal to u.
Then
~l
),
'
;
un
1
N
Ax diminish
+
2
,B
f-
^
Ax
Now
un
K
).
}
n terms
DIFFERENTIAL CALCULUS
32
But we have already shown VII.
a positive integer
equation.
;
hence we
may
to be true
apply
when
the exponent
member
to each
it
This gives
qyq ijL
=pvP
i
.
dx
dx
4=«!^*
~
therefore
dx
q yq
l
dx
2
Substituting for
y,
uq gives
,
dy
dx
which shows VII.
Third, suppose
by
is
u
_p
P
q~ l
q
du
dx'
q
to
—
m.
= u~ m =: —m
u
—=
dx
su 2m
dy
/- =
dx
Hence, VII.
du
q p _pdx
n to be negative and equal
y
T7T
VI.,
l
to be true in this case also.
Let
i
_p up ~
—
dx
u 2m
—
= - mic m - ,du
1
dx
true in this case also.
EXAMPLES
Differentiate the following functions
1.
y
= x\
*
= !(*•).
dx dx
If
we apply
VII., substituting
u = x and w
— (z )=4ar —
=4ar>.
dx
4
dx
Hence,
K
}
J
^
= 4^.
dx
by
J
= 4,
I.
we have
is
of this
a
DIFFERENTIATION
2.
y
= 3» + 4aj
4
8
.
*? = J*(3a;*+4a?)=
by
III.,
making
33
u
= 3x
4
and
u
= 4#
— (3
+-^(4
x*)
8
),
3
.
byV.
|(3^)=3|(A
= 3-4^ = 12^.
= 4 — (or) = 4
Similarly,
—
Hence,
^= 12a
(4
x3 )
3
.
3 x2
12«a = 12 (a?
+
= 12 a?.
+
2
a?
).
e2as
3.
y = ** + 2.
dx
dx
|(,f)
dx
= |,i
by
|(2)=0,
Hence,
4.
;
by VII.
II.
^
= 1**
dx
2
y=3Vx-A + | +a
dx
dx
.
y
J
^
3 -I
= 2*
-
K
2 (-_1
--*- + _1 _
2x*
dx
3
'
x4
J
dx
K
'
dx
DIFFERENTIAL CALCULUS
34
5.
y
= %£
ar + 3
dy_
d
/
+ 3\
x
dx\x 2
dx
-{-3
Applying VI., making
u=x
+
(-
dfx+3\
<toV<s + 3y
2
2
3 and v
= x + 3,
2
we have
+ 3)|(, + 3)-(, + 3)|(^ + 3)
(x + 3f
2
3-6a;-^
_ f + 3-(x + 3)2x =
~~
(z + 3)
(> + 3)
2
dy
=
dx
Hence,
9
6.
S-6x-a*
(x + 3)
m
= (x + 2)t.
6
2/
we apply
dx
making
VII.,
^
=x +2
2
and w
= 2-,we
3(»2
Hence,
dx
y
have
o
3
7.
2
2
2
dx
If
2
2
3(x2
+ 2)3
= (x + l)Vx*-x.
2
^=A[(aj»
+ l)(aJ»-aOH
+ 2)i
,
DIFFERENTIATION
If
we apply
IV.,
m
=
35
making
as?
+1
and v
= (x — x)-,
3
we have
£[(rf + i)(#-.)»]
=
dx
+ 1) |- (X - »)* + (*" - K)i|s
(x2
^=
=
i(.T 2
(j
2
+ 1)(3 x - 1) (ar
+ 1) (3
8
10.
y
= (x + 2a)(x-a)
,
s,= (»*-ai)«
'
13.
+ ar -5,
?
,
Example 11
~
v= 2x 1
(x-if
J
y
= z(a*+5)*,
=7
-»)*2a
-2 -1
2{x - a>)*
4
2
a;
a;
z
^ = 3 (10 a
-4z
9
^=3 (a -a
2
2
,11..
Differentiate
a;)
(s3
3
= 3x
-2a;6
-*)-£ +
- 1) + 4 afo (.r - xft
2
x-
^
10
3
2
8.
12
+ 1).
+ l)=2x.
2
11.
2
dx
2
dx
Hence
2
(a-
<fy
S=
4
(a; 3
—
5
2
).
q3)3
3ic f
also after expanding.
dy =
dx
g=5
2x
(x-iy
(ar
+ l)(x + 5)i
3
2
-far )-
DIFFERENTIAL CALCULUS
14.
y
'
Va - x
2
2
(^-a )^
—x
la
16
dy
dx
x
members
Differentiate both
18.
/'^-±A Y
2
a=
21
^^
y
y
* (*
(
2
= ^ + 2a + a »2
^
2
~ 3)
+ 2)
f=
25.
2
+ a )(* + a )T
2
2
2
2
3
n-i
aJ_a
(?
dx
'
+ l) (3z-8)
V2 aa; -
(nt
dt
= *(*» + n)->
</=
.
<fy_6(3* +4Q(2; 2 -3) 2
'
(2a-3a) 9
N
<,
4
2
4
(z
(2a-3a?) 10
2
-
oj
24) (a;
+ 1)
2
(3
dy_ n(xn + l)
^d2/
2
+ 2f
+ 2^
^ = 3 (13 a
24.
2
.
3
2
—x
2
+ <*)*&
2
(?
2,= (z
4
^_ = 1+
2
,
20.
23.
2 xy/ax
)
2
*
2
of the identical equations, Exs. 17-19.
2
2
(x 2
'
-xf
+ ax -h a )(x — ax + a = x* -f aV + a
17.
19.
(a
2
dy = 3aWx*-a 2
dx
2
15
a2
dy
dx
x
=
*»
(ajn
+ n) l
a'
(2
aa - x2)%
a;
- 8)
3
(x
+ 2)
5
.
x
1
1
DIFFERENTIATION
*"•
*
~~
'
~4
37
4
^.4
Q x
3
,7^
x
rf
cty
28.
g=
d*
4
*=(*^2)JJ7i;
\
29.
30.
t,
(x3
_
— «?)»
if
c7y
l'
y
V
= ^jX-
6 x2 4- 6 *
31.
+
(.r
=
(x
+ Vx
2
+ 8az +
n
4- l) ("
15«
For -what values of x
Arts.
A
2
)3
)
- 9a )
_
3 cu.) 5(4 rf + 8 ax + 15 a )*
(rf
4(2
3
3
2
2
te
34.
9
or
Vx + 1 - *),
cZ.V
ing function of x
2
4
(4« + l) f
*b
y
+x + 1) Vx + x +
12
c7?/_
33.
)V-«') f
x-'-l
*
2/== ( iC2_3ax)^(4^
3_h a3
(a*
dy_
(4*+l) f
32.
2 a 3x 2
cfy_
ax
~X-\-l
t
+ 1)1
_
d*
\^+x + l'
3
(ti
is
= (,r-l)(z+Vz + l)».
3 x4
2
—8x
3
an increasing or a decreas-
?
Increasing,
when x
>2
;
decreasing,
when x
<
2
form of an inverted circular cone of semiwith water at the uniform rate of
one cubic foot per minute. At what rate is the surface of the water
rising when the depth is 6 inches ? when 1 foot ? when 2 feet ?
35.
vessel in the
vertical angle 30°, is being filled
Arts.
.76 in.
;
.19 in.
;
.05 in., per sec.
DIFFERENTIAL CALCULUS
38
36.
The
side of an equilateral triangle
of 10 feet per minute,
How
second.
37.
vessel,
hour'.
and the area
is
increasing at the rate
10 square feet per
Ans. Side = 69.28 ft.
at the rate of
large is the triangle ?
A
Another
vessel is sailing due north 20 miles per hour.
40 miles north of the first, is sailing due east 15 miles per
At what rate are they approaching each other after one
After 2 hours ? Ans. Approaching 7 mi. per hr. separating
hour ?
15 mi. per
When
;
hr.
will they cease to
approach each other, and what
is
then
their distance apart ?
After 1 hr. 16 min. 48
Ans.
38.
A train
starts at
being represented by
From Worcester,
f.
= 24
Distance
noon from Boston, moving west,
=9
s
sec.
its
mi.
motion
forty miles west of
Boston, another train starts at the same time, moving in the same
motion represented by s' = 2 f. The quantities s, s',
and t in hours. When will the trains be nearest together, and what is then their distance apart ?
Ans. 3 p.m., and 13 mi.
direction, its
are in miles,
When
will the accelerations be equal ?
Ans.
=
If a point moves so that s
y7, show that the acceleration
negative and proportional to the cube of the velocity. How is
39.
is
1 hr. 30 min., p.m.
the sign of the acceleration interpreted ?
40.
Given
41.
A
= - + bt
s
body
starts
the coordinates of
its
from the
2
;
find the velocity
origin,
and
acceleration.
and moves so that in
t
seconds
position are
o
Find the
rates of increase of
Also find the velocity in
ds
its
x and
path,
vdHty
vn
,
,
y.
which
is
Ans.
5f + o.
DIFFERENTIATION
42.
axis of
Two
bodies move, one on the axis of
and
y,
At what
1 minute ?
in
t
x
5=
39
and the other on the
aj,
minutes their distances from the origin are
2
2
f
-
6
t
and y = 6 1 — 9
feet,
feet.
rate are they approaching each other or separating, after
After 3 minutes
Ans.
?
Approaching 2
ft.
per min.
\Yhen will they be nearest together
separating 6
;
per min.
ft.
Ans. After 1 min. 30
?
sec.
M
are the middle points of BC
43. In the triangle ABC, L and
and CA respectively. A man walks along the median AL at a uniform rate. A lamp at B casts his shadow on the side AC. Show
2
3 2 4 2 and
that the velocities of the shadow at A, M, C, are as 2
3
3
3
3 4
that the accelerations at these points are as 2
:
:
:
Suggestion.
line parallel to
33. Formulae
— P being
:
;
.
any position of the man, draw from
L
a
BP.
for
and Exponential
Logarithmic
Differentiation of
Functions.
du
VIII.
-^-log u «
= log e^.
a
dx
TV
IX.
d
du
dx
— log w=—u
i
—
a
dx
u
*e"
XI.
= log a-au —
e
=e
v
d
-
dx
•
dx
—
dx
dx
YTi
XII.
•
e
dx
X.
u
"='//
.du
dx
,
-,
h log e
?^
•
uv
dv
—
dx
•
DIFFERENTIAL CALCULUS
40
34.
Lety = loga u,
Proof of VIII.
then
y
+ Ay = \og
a
(u
+ Au),
Ay = \og a (u + Aw) - log a u = lo go
^±^
= log/l + ^=^log„(l+^f
".
u J
\_
u
\
u J
Dividing by Ax,
^logA+^f^
u J
u
Ax
If
Ax approach
Au
zero,
\
likewise approaches zero.
Now Lim^^l + ^Y^Lim^Jl + l
For,
if
—
=
Au
we put
i+
(
and as An approaches
But
in Art. 12
fr=(i+
zero, z
Hence,
we
infinity.
we have found
Lim AM=0 [ 1
if
t:
approaches
Lim^M +-j
therefore
z,
—
-\
=e;
- r" =
e.
take the limit of each
dy
dx
i
member
du
dx
u
of (1),
CD
.
.
DIFFERENTIATION
This
35. Proof of IX.
41
a special case of VIII.,
is
when a = e.
In this case
log a e
Note.
— Logarithms
Hereafter,
e
to base e are
when no base
understood; that
= log e = l.
is specified,
called Napierian logarithms.
Napierian logarithms are to be
is,
log u denotes log e
36.
u.
Proof of X.
Let
= au
y
.
Taking the logarithm of each member, we have
log y
dy
dx
= u log a;
—
— = loga du
dx
therefore by
J IX.,7
,
y
Multiplying by y
= au
,
we have
dy
—
= loga
dx
t
37.
Proof of XI.
38.
Proof of XII.
This
is
•
au
du
—
dx
a special case of X., where a
Let y = u v
—
e.
.
Taking the logarithm of each member, we have
\ogy=v\ogu;
therefore
dy
dx
by
J IX.,
—=
'
y
Multiplying by y
=u
v
,
du
dx
do
+ logw—
dx
u
.
n
we have
—
dv
dy —
r-idu
vuv l
hlog^-^ v
dx
dx
dx
.
-?-
The method
each member,
of proving X.
may
This exercise
i
and XII. by taking the logarithm of
be applied to IV., VI., and
is left
to the student.
VII
DIFFERENTIAL CALCULUS
42
EXAMPLES
(See note, Art. 35.)
1.
y
= log
2.
y
= xn log (a# + 6),
3-
4.
y
2/
(2^
+ 3 r%
10
(3
ox
dy
dx
ax+b
dy _ _
dx
xlogx
= log
dy _ 6 (a; + 1)
dx 2 a? 2 + 3 x
x
+ 2),
riy
da
1
+ log x
(x log x) 2
= 3 log e = 1.3029
3x + 2 3a+2
10
5.
#
= log ax — b
aa? + 6
dy _ 2 ab
dx a2x'2 — b 2
I.
y
= lo{ 3^ + 1
«+3
dy
dx
'
dy
eft
8.
y
=ae
9.
y
= log(a* + 6*),
10.
y
+nlog(ax+b)
3a + 10a;-f 3
2
_
2
(f-
- 1)
^4-^-j_^
|2=(l + loga)a*e".
x x
,
dy _ a x log a +
~
dx
ax +
dy
=8
dx
= (««-_ 1)*,
e
2x
(e
2x
&*
b
log b
x
- l)
s
.
Differentiate Ex. 10 also after expanding.
11.
12.
y
V
= 6* +
x—3
y=(3x-l)
<fy
»,
cte
2 3x
e
-2
,
dy
dx
= 24a--10^
&
(x-S) 2
3 (9 x2
-
l)e 3x -\
v
DIFFERENTIATION
13.
y
=x
5
5x
~JL
,
=x
4
43
o T (o
+x
log
5).
da;
14.
= log
y
d#
dx
1
log x -
log x
members
Differentiate both
+e
15.
(x
16.
(a*
17.
log
18.
.i
..
20.
4
.r
x
2
(e *
+ 4 .rV + 6 xV" +
+
2a
e'
)
= log
x
(e
~a
3 a x e 2x
+e
a"x
)
4 a* 3*
+e
+ x + a.
.
x
/
i
(Vx~HTa
,
+
dy
^
.
Vx),
log x
dx
dy _
da
22.
y
= log
^±i-^
+ +
dy _
ax
1
= x [(log x) - 2
2
1
log x
+ 2],
23.
y
oa
24.
r = log(VJ+7-V—,fc
26
,,
i
/
V4r>- 1
1
xVx + 1
4
= (log X)
UX
"'•'
/
S
,r
'.-"'
- 1 + *'"'
+ 1 + <r"
•''!
,
dx
ax
2
2-
— ^
-
+
(x
a)*
+ (x - a)*
..'
.
= log(V* + 3 + V* + 2)+V(* + 3)(z +
= W-
+
2Vx*
= log (2a-+V4x*-l),
a.,-
1
<ty_
y
25.
4 *.
- e*.
21.
i
a;)
log x
los:
= log
x
"
=a
x
y
y=
2
x (log
of the identical equations, Exs. 15-18.
- e f = a3 - 3 cr*e +
Jog a
Q
x
_ 1 + log X
2),
f
g=^/|±|
= 2 „(,'-?-<)
c-"' + 1 + e"-"
DIFFERENTIAL CALCULUS
44
87.
lo« x
y = log
1
+
®L =
,
log x
,
x log
dv
30
+ log x)
(1
a;
.
e°*
= 3 log ( Vx + 3 -3) + log ( Vx + 3 + 1),
2
2
2/
1
dx
4
fly __
a;
^"~^-2V^T3
30.
y
= log
(a
+ V2ax-a ) +
2
Va 4- V2 — a
a;
0>
dx
+ l4-V^ + 3a + l
2
a;2
v _,i og
31
;
j
x
1
2(x
+ V2 ax - a )
2
dy =
x2 -!
dx x^/x^ + Stf + l
The following may be derived by XII. or by differentiating after
taking the logarithm of each member of the given equation.
^| = nx™
33.
y
= xnx
34.
y
= (ax>) x
35.
2/
= x**
36.
2/
= (logx)*,
37.
y
= x<
38
,»(-£_£
\x
.
,
,
2
*
^
+ aJ
1
2
)
+ log x).
[2
+ log (ax )].
2
=
+ 21ogx).
^
ax
= ° x)/-i— + log log x V
^
ax
Vlogx
y
* (w +
V <^
ax°*
,
10
(ax
^=
ax
(1
2+1
(l
(log
te
*
dx
rf
1} (log
,,)
io,
f-fL,W-i-+il€g^-A.
x + a)
a
\x-\-aJ \x + a
.
DIFFERENTIATION
The method
expression
of differentiating after taking the logarithm of the
may
This
tions.
45
is
often be applied with advantage to algebraic funcsometimes called logarithmic differentiation.
In this way differentiate Exs. 21-26, pp. 36, 37.
I
39.
Find the slope of the catenary y = ~(e a
What
is
When
X?
x=
at
),
0.
inclined 45°
is
Ans. x = a log e (1
+ V2).
does log 10 # increase at the same rate as #?
When
Ans.
When
a
the abscissa of the point where the curve
to the axis of
40.
X
e
-f-
at one third the rate?
x
= log
When
Ans.
10
e
= .4343.
x = 1.3029.
Verify these results from logarithm tables.
41.
show that the acceleration
42.
by a point
If the space described
If a point
is
is
given by
s
= ae + be~
c
moves so that in
seconds
t
s
= 10
log
feet,
and acceleration at the end of 1 second. At the end
= — 2 ft., and — .5 ft. per sec.
Acceleration = .4, and .025.
- 2) - 9 ^~ 36 x + 32
~ )
\
x > 3; decreasing when x < 3.
y = log (x
an increasing or a decreasing function ?
For what values of
a?
is
Ans. Increasing when
39.
+4
Ans. Velocity
of 16 seconds.
43.
,
equal to the space passed over.
£
find the velocity
f
Formulae
for
Differentiation
Trigonometric Functions.
of
the following formulae the angle u
3
is
circular measure.
XIII
d
du
—
sin w = cos u—dx
dx
XIV
—cos u =
XV.
dx
- sin u—
tanw = sec
—
dx
dx
2
In
supposed to be expressed in
M—
dx
.
DIFFERENTIAL CALCULUS
46
du
d
cot u = — cosec w —
—
dx
dx
o
2
,
XVI
see u=
—
dx
XVII.
u tan %
sec
•
—
dx
—
— cosec u = — cosec u cot dx
XVIII.
?t
-
dx
—-versw=sm?t—
XIX.
ax
ax
40.
Proof of XIII.
then
= sin u,
y-\-Ay = sin
+ Am);
A y = sin (w A u) — sin
Let y
(it
therefore
?t.
-f-
But from Trigonometry,
sin
If
we
A - sin B = 2 sin* (A - B) cos
substitute
^4
= u + A w,
and
\
(A
B = m,
—-)sin—
Av = 2cos(mH
we have
+ B).
-
9
Aw
Hence,
Now when Ax
as
Au
is
=
H
Ax
u
cos
+ *g)
2
V
approaches zero,
Aw
likewise approaches zero, and
in circular measure,
•
sin
Au
2
Lim AM=0 -^-
= 1.
T
TT
l^
An Ax
Hence,
dy
—
= cos a du
dx
dx
See Art. 12.
—
DIFFERENTIATION
41. Proof of XIV.
for u,
may
This
47
be derived by substituting in XIII.
£-?/.
Then
d
cos?*
dx
—
or
Proof of XV.
42.
= cos(|-„)|(| r «)
|-(l-«)
—
du\
dxj
f
= sin
?/[
V
=-sm« d»
dx
= ^-^,
Since tan u
cos u
cos
„T
.
bv
tf
~y~
\ 1.,
d*
tan "
— sm
tf
?/
.
v
— sm
d
—
cos u
?/
(to
ete
=
5
cos-
>/
dv
dn
b—
t sm- u —
,
cos 2
.
.
COS"
du
—
o
COS"
?'
w.
sec- "
CfeB
ing
This 'may be derived from XV. by substitut-
Proof of XVI.
43.
——
v for
?/.
44. Proof of XVII.
Since sec u
=
—
cos u
d
,
„,
f?
3~ sec
bv
J A L,
ax
7
u
.
=
=
5
cos-
= sec
sm
cos u
eta
=
cos-
?<
c?u
dx
u
—
du
.
v tan
w
>/
dx
45.
Proof of XVIII.
stituting
^
— u for
be derived from XVII. by sub-
u.
46. Proof of XIX.
relation
may
This
This
vers
is
readily obtained from
u= 1 — cos
1
u.
XIV. by
the
DIFFERENTIAL CALCULUS
48
EXAMPLES
y
u
=3
2.
y
= log cos
3.
?/
= log
1.
sin
— 2 cos
3x cos 2x
Sx sin
= 5 cos 3x cos
2x, -^
2x.
dx
2
x-\-2x tan
a?
—
2
a?
-^-= 2x tan 2
,
—= m
(sec ra# -f tan mx),
#.
sec wa?.
C*3J
?/
u
= log5
(a
v
y
= cos
a log sec
i
>i
4.
c
5.
6.
/
sm
•
2
+ 6*
i
cc
//)
i
y=(m — 1)
(8
sec m+1
dy
tan #
—
= 2—(a—b)
r
dx
a tan x +
\
cos 2 a),
;
1
2
— a) + a sin
\
dv
•
ot,
sin
—
=
-^
dv
,
b
cos (0
— a)
.
#— (ra + l)sec m_1 a;,^==(m — l)secm-1 a;tan
2
3
a;.
cia;
7.
2/
8.
r
9.
= log tan
= log&
(
aa?
— ^"j,
tan 6 (sec
v
[sec
L
=— 2a sec 2ax.
-^
+ tan
2
0)y J
]
,
'
— = (sec 6 + tan ^
dB
2
-
tan
= cosecm ax cosec" bx,
—^ = — cosec™ ax cosec" bx (ma cot ax + n& cot
2/
bx).
dx
10.
= 2x
w
2
sin
2x
+ 2x cos 2# — sin 2x,
-&
==
4a2 cos
2a\
da;
11.
=2
?/
tan3
a;
sec
a;
+ tan
a;
sec x
— log
+ tan #),
-^ = 8 tan x sec
(sec
x
2
3
a;.
dx
»- sin ?-f cos x
'
-
12
-
13.
y
dy
a;
'
= e '(sin 2a; - 5 cos 2a;),
3
V
2 sin x
Tx
^ =13e
ax
to
(sin2x
- cos2z).
«
DIFFERENTIATION
14.
y
+
cos (x
15.
,
16.
17.
y
= sin
y
= log
2/
3
t
=
dy
dx
cos x
= log
a)
sin
x
4- vers
sin
x
— vers
!
(sin 2
-^
dx
cos 4 3x,
4.r
= 12
x
sin a
+ a)
cos x cos (x
sin 2 4a; cos 3 3a; cos 7x.
dy
-^
,
= secx.
dx
a;
—=
1
a;)
49
,
2/
(log sin
+ 2a;
2x
cot
2a;).
da;
18.
19.
20.
= (tan
?/
y
a;)
= (sin x)
7/
— = y (cos
8inz
,
log c0 * x
-^
,
dx
7/
= (tan -3 cot
a;
sin^fl
22.
y
23.
24.
y
y
a;)
sin
a;
Vtana;,
— «)
a;
+ sec
- = 2 sec
3
x.
da;
dy
d*
= 3sec
dy
_
4
a;
2 tan* x
sine*
d0~"cos«-cos0
+ a)'
= a log (a sin + 6 cos
a;)
+ 6a,
^ = _^_±&!_,
da;
a tan
da;
1-|- sin 4a;
dy
d*
4-5 sW
a;
-f b
=2-
sin
(
+ i)
tan25.
a;
a;
= log
sini^
log tan
-2
y = log
2tan?-l
a?).
= y (cot x log cos — tan x log sin x).
= tana;seca; + log J^L±4^,
—
*
1
21.
a;
dx
3
DIFFERENTIAL CALCULUS
50
oa
«D.
= a sin + b vers x
a sm x — b vers x
2 ab vers x
dy
—
- —
if
;
])
,
dx
—
;
(a
sm x — b
In each of the following pairs of equations derive by
two equations from the other:
vers x)~.
differentia-
tion each of the
27
.
= 2 sin x cos x,
— sin
cos 2 x = cos
sin 2
a;
2
28.
sin 2
2
a?
ic.
2 tan x
a;
1 -f tan 2 aj'
—
1
cos 2
tan
2
aj
a*
tan 2 #
29.
= 3 sin — 4 sin
—
4 cos — 3 cos #.
cos 3 x
sin 3
3
a;
a?
3
30.
sin 4
cos 4
31.
32.
onds,
a?
a:,
a;
= 4 sin cos — 4 cos
= 1 — 8 sin x cos x.
3
a;
a:
(m + n) x =
33.
a;
sin 3 #,
2
mx cos + cos mx sin wa;,
cos (m -\-n)x= cos mx cos wa; — sin mx sin wa\
sin
sin
rase
made in ir
when = 0°,
If 6 vary uniformly, so that one revolution is
show that the
rates of
increase of
45°, 60°, 90°, are respectively 2,
of tan
a;
2
V3, ^/%
when
= 0°,
1. 0,
0,
30°, 45°, 60°, 90°,
sec-
30°,
per second.
show that the
If 6 is increasing uniformly,
0,
sin
rates of increase
are in harmonical progres-
sion.
34.
For what values of
0,
less
than 90°,
is
sin 6 -f cos
an increas-
ing or a decreasing function ?
Find
its rate
of change
when
=
15°.
The crank and connecting rod
Ans.
vr
and 10
35.
feet respectively, and the crank revolves uniformly, making two
At what rate is the piston moving, when
revolutions per second.
of a steam engine are 3
DIFFERENTIATION
makes with the
the crank
line of
51
motion of the piston
0°,.
45°, 90°,
135°, 180° :
If a,
and
the triangle,
are the three sides of
b, x,
opposite
$ the
angle
b,
=a
x
+ V6 — a2
cos
Ans.
A
OP
PQ
0,
sin2
6.
32.38, 37.70, 20.90, 0,
ft.
per
sec.
O with angular velocity <o, and a
hinged to it at P, whilst Q is constrained to
Prove that the velocity of Q is w. OP,
move in a fixed groove
where R is the point in which the line QP (produced if necessary)
36.
crank
connecting rod
revolves about
is
OX
meets a perpendicular to
OX drawn through
The inverse trigonometric
Inverse Trigonometric Functions.
47.
functions are many-valued functions; that
x.
there are an infinite
For example, sin
But
-1
the angle
if
number
is,
where n
any given value of
for
of values of sin
-= ±-± 2mr,
is
O.
-1
is
#,
any
tan
-1
#,
&c.
integer.
restricted to values not greater numerically
than a right angle, sin -1 x will have only one value for a given value
of
x.
cosec
the
Then
-1
x,
first
sin
tan
-1
-1
x,
-
= -,
2
6'
and
cot
sin
_1
_1
.T,
=—
6
2y
V
as taken between
(
]
•
We
—-
thus regard sin
and
-, that
-1
is,
x,
in
or fourth quadrants.
But cos _1 .r,
-1
and vers -1 a;, must be taken between and v,
that is, in the first and second quadrants, which include all values of
the cosine, secant, and versine.
These restrictions are assumed in the following formulae of differsec
x,
entiation.
48. Formulae for Differentiation of Inverse Trigonometric Functions.
flu
XX.
— sin
VI - u2
dx
clu
XXI.
i^cos-S<
dx
=
x
i
DIFFERENTIAL CALCULUS
52
du
dx
XXII.
~l + w
dx
XXIII.
d
—
cot
dx
XXIV.
sec
—
dx
i
,
2
du
dx
u
l+u
2
du
therefore
XX.
2
—
d
vers
dx
i
du
dx
##
u-Vu 2
1 jl
-
1
du
dx
=
•
V2w
y— sin-1 %;
sin y = u.
Let
'
By XIIL,
«V« -1
—
XXVI.
Proof of
dx
u
d
cosec
dx
XXV.
49.
-1
cos
du
dx
2// =
du
—
dx
;
du
therefore
But
dy _ dx
dx ~ cos y
cos y
_
= ± Vl- sin
2
2/
If the angle y is restricted to the first
(Art. 47), cos y is positive.
Hence
and
cosy
dy
dx
= Vl_
du
dx
VI
<
= ± VT
and fourth quadrants
;
.
;
DIFFERENTIATION
Proof of XXI.
50.
Let y
cos y
therefore
— u.
du
dx
du
dx
dy_
dx~
therefore
But
sin y
If the
= cos _1 w;
dy
dx
A, TT ^
angle
?/
is
53
smy
= Vl — cos' y = Vl — uK
2
and second quadrants
restricted to the first
(Art. 47), sin y is positive.
Hence
VI -
sin y - =
51. Proof of XXII.
VI -u
Let y
—
dy
dx
sec 2 y
But
dy
dx
therefore
cot
-1
Proof of XXIII.
u = tan -1 -
= u.
o
dy
du
sec^-^ =
dx dx
therefore
52.
2
= tan -1 u
tan y
therefore
By XV,
,
du
dx
dy_
dx~
and
u2
This
may
_
du
dx
sec 2 ?/
= 1 -h tai
du
dx
_
1
+u
2
be derived like
XXIL,
or
from
DIFFERENTIAL CALCULUS
54
53. Proof
sec
u=
-1
cos
XXIV.
of
-1
This
may
XXI.
be obtained from
Since
-,
u
=—
-11
cosec
XXV.
Proof of
54.
=
w
d /1\
dx \uj
d
_,1
cos l - =
dx
u
d
_!
sec * u
dx
mav
du
du
./
i
I
This
1
u2 dx
dx
«V% -1
2
i
XX.
be obtained from
Since
- ll
sin ->
•
u
d
d
_!
,
n -1
d 1
f )
dx
1
—
:
1
w
du
du
dx
u 2 dx
,-\/ir
^-i +-i
55.
XXVI.
Proof of
vers _1 w
This
may
XXL
be obtained from
—
Since
= cos -1 (1 — w),
d
vers
-1
m
= — cos
x
dx
dx
(1
/-.
_
du
v
— u)
Vl-(l-w)
V2
2
EXAM PLES
!
1.
•
3.
4.
y
^
2/
y
u
?/
= 4.-1
tan
x
= sec
= sin
x
x
-
———
5®
1
-
dy
dx
——
^.V
7
(8#
2
5.r- — 2a? -h 1
da;
—
= vers -1
o
dt/
9
^V4x -9
2
dy
4
V(aj
_
),
da;
5.
3/
= tan
x
>
1
_
dx
— 8#
3
_
%_
- 5)(2 -
4
VI a
x2
1
X)
1
1
,
1
DIFFERENTIATION
6.
7.
y
?/
= tan
*
dy
d6
(3 tail 0),
= sec -1
55
dy_
sec 2
9
.
10.
11.
+
r
//
= cosec -1
= tan
*-
x
(-
'
cot
6x
y
=
cos
13.
y
J
=
o
14.
>/
-1
Vvers
cot-
^ +—^a
16.
y
Sill
= sin _i
•
y
A
= sin
2a
e
tan-
dy
*c
_
-^
= 0.
*
*=
6
da;
j/
dy
y
=
sin
-1
y
= cot
'
1
.,-
4-a 2)<>2
+&
-
-
,
2
)
a
_
+a
2
a;
1.
ty
k -6
2
1
_
•
+
-a; 2
(
•#_.1
a?)
-*
x 4- 1)
2
dx
- cut"
1
(a;
-
e
x
dy
1)
da;
2
'
&a
dy
-
a7.
(a'-ftV
2
,
(sec x 4- tan
e' 4- e
19.
4- sec
dx
da*
18.
+ 2b - 1
SB
f
= tan -1
-2«x
1
(a;
da;
17.
e
a V8ar
da;
— COS
V2
;
2ax_|_
,
X
+ l'
da;
fa;
15.
2
*»
— = — 4 Vl
a
=
>
sc,
-b
tan- ?
+
+1
Vsec 2
dy
dx
5
12.
2
dy _
dx
1
y=cot -i *"+':'
er* — e
//
cos 2
0,
9
-1
y = vers
-4
o
dO
8.
3
_
2
a;
4-
4- e~
x
DIFFERENTIAL CALCULUS
56
20.
y=:tan- l4
+ 5tanfl;
21.
2/
= cos-
22.
2/
=^
2
members
^±i = cos-
23.
2cos-
24.
3 vers- 1 x
25.
sin
26.
tan -1 mse +. tan -1
ol-
27.
-1
,2
*
a6(l
OQ
29
'
^
].
01
= 2lQg
__
_
z
-\-x
^
/a;
,
+ tanx)
^
fa
\b
^
.
a;
—x
2
•
\
^
_
J
2
aj
Vl — a ).
1
-2a; + 5 14 _!^-5
+tan
17^
^ + 2, + 5
4
2
2
= tan -1 ^—
1 — mnxr
"'tan*-* =
tan-
s
dy
&
.^(a^-a
—^—- )^
i—
4
3 V3 a a
^
y
2
32.
What
/
2
2
V4 a - or
2
a?
value must be assigned to a so that the curve
y
may
2\ x -a
dy =
-£
x
.
a-
54^F+3'
3
.
2/=sm
b
12x 2 -20
=
dy
2
Q1
31.
sec" 1 -•
a;
a.
2
+-—=2tan
_i
\\ —+^2
+3
=[
sin 2a;
of the identical equations, Exs. 23-28.
1
nse
=2
da;
= vers" [x(2x- 3)
a;
»
28.
^/
+ sin -1 a = sin -1 (a y 1 — x
x
vers
1
+4
% = V6av-a^
=l,
--2V^
2
Differentiate both
5
da?
^^-2 x /^
1
sec- 1
1
*
?
3
=
log e (x
be parallel to the axis of
— 7 a)
-f-
tan -1 a«,
X at the point x = 1 ?
^4ns.
J or
— i.
DIFFERENTIATION
57
A man
walks across the diameter, 200 feet, of a circular
courtyard at a uniform rate of 5 feet per second. A lamp at one
extremity of a diameter perpendicular to the first casts his shadow
upon the circular wall. Required the velocity of the shadow along
the wall, when he is at the centre when 20 feet from centre when
33.
;
50 feet
;
when 75
feet
when
;
;
at circumference.
10, 9 T8g, 8,
Ans.
56. Relations between Certain Derivatives.
6-f,
5
ft.
per
sec.
It is necessary to notice
the relations between certain derivatives obtained by differentiating
with respect to different quantities.
To
may
have
express
—
dx
in
terms of
'
—
If y
•
dy
be regarded as a function of
-^,
and from the
latter,
dx
is
From
y.
—
a given function of
x,
then x
the former relation,
we
These derivatives are connected
dy
by a simple
relation.
It is evident that
—£ = -—,
Ax A#
*9
however small the values of Ax and Ay. As these quantities approach zero, we have for the limits of the members of this equation,
w
ft=i
dx
That
is,
(i)
dx
dy
—
the relation between -2 and
-,.
-
dx
..
ordinary tractions.
is
the same as
if
they were
dy
a
For example, suppose
w
x=-^~.
y
Differentiating with respect to
y,
By( i),
•
we have
dx
dy
(2)
+1
a
G/
+ 1)
2
!=_M!=-!,
by(2)
.
DIFFERENTIAL CALCULUS
58
This
is
ence to
y,
the same result as that obtained by solving (2) with refergiving
= - 1,
y
and differentiating
x
this with respect to
—
x.
To express -^ in terms of -^ and
that is, to find the derivative
dz
dx
dx
If y is a given function of z, and z a
of a function of a function.
given function of x, it follows that y is a function of x. This relation
may often be obtained by eliminating z between the two given
:
equations, but -^ can be found without such elimination.
dx
By
differentiating the
two given equations, we
find
—and—, and
dz
dx
from these derivatives,— may
J be obtained by
J the relation
dx
dy_dy^dz^
dz dx
/on
dx
For
it is
evident that
—^=—^
Ax
however small Ax, Ay, and
of this equation
if
we
obtain
—
Az Ax
Az.
By
(3).
That
the derivatives were ordinary
For example, suppose
y
=*
taking the limits of the members
is, the relation is the same as
fractions.
5
,
\
,
2
Differentiating these equations, the
the second with respect to
x,
dz
By
(3),
%L = bz\- 2x)
dx
x
4
K)
z^tf-x .}
first
with respect to
we have
dx
= - 10x(a - x )\
2
2
by
(4).
z,
and
DIFFERENTIATION
The same
(4),
result
59
might have been obtained by eliminating
z
between
giving
and differentiating
The
relation (1)
substituting y
= x.
with respect to x.
be obtained as a special case of
this
may
_
dx dz _ dx
dx
dz dx
Another form of
(3)
by
This gives
..
(3) is
dy
dx
dy
(5)
dz~dz'
dx
which
is
of frequent use.
EXAMPLES
In Exs. 1-4, find
—
and thence
i
,-
—
ay
by
2.
3.
x
x
dy _ _ 2
dx
+ logy'
V2-X
2
2z-l'
~
dz
and
dy
=
2^/tf
dx
—
,
dx
:;./;-
_
xy
f
—x
2z
= olog Vy + q hV^,
In Exs. 5-8 find
2
y
cos y
logy
dx
Va
9
Vl + sin
dy._(l-flog;V ) 2
=
"
a?
(1).
dy _ (by — k) 2 __ bh — ak
dx
bh — ak
(bx — a) 2
~^
— k'
= vl +siny,
1
4.
— by
dx
dy
2'
and thence
-j-
dx
by
J
/
2
(3).
v '
'
dx
2x
+ cu = e«-e
a
(a>+2) 2
2
DIFFERENTIAL CALCULUS
60
6.
y = log£±±,
dy e-e~*_
= x
x
dx
e + e~
= «,
,
z
7.
y=ze z
+e
2z
= \og(x-x
z
,
^/
2
),
=4
a)
3
-6
+ l.
ar>
dec
8.
?/
= log
!
62
—
z
,
= sec x -f tan #,
-fa
%
2 ab
da?
Differentiate
9.
(a?
+ 2)
2
Let y = (x 2 + 2) 2 and
,
^=
2 ==
2
.
4aj(aj
dy
w
(5)
2
— = 3a;
+ 2),
(7 7/
-j*-
2
.
dx
= ±x(x + 2) = ±(x + 2)
2
3x
dz
2
3x
2
—6 + —6
Find the derivative of
10.
x
dz
cte
BJ
(a -f 6 2) cos
n fi
nrl
It is required to
find
3
2
2
3
sb .
with respect to
ar
—5
a2
-\-
;
;
x
a
'
with respect
to c
a
+x
Sfc + l + £\
Ans.
a;
Find the derivative of
11.
sin 3a;
with respect4o sin
Find the derivative
12.
of taii
-1
/
x.
3 (4 cos 2 x
Ans.
2
— 3).
+ x).
^/^ with respect to log (1
Ans.
2-Vx
Find the derivative of log
13.
—
:
a sin x
—
j
2
2
a sin x
14.
—b
2
cos x
cos x
a
<f>
— cos
5cf>,
y
—5
sin<£
—b
+b
ab (a 2 tan x
2
Given x = 5 cos
with respect to
— sin
5<£;
2
find
2
cot x)
—
.
ciaj
u.-i?t.s.
= tan 3
—
dx
4>.
CHAPTER IV
SUCCESSIVE DIFFERENTIATION
57.
Definition.
As we have
seen, the derivative is the result of
This derivative being generally
we thus obtain
called the second derivative; the result of three successive
differentiating a given function of
also a function of x,
what
is
differentiations
For example,
is
may
x.
be again differentiated, and
the third derivative; and so on.
if
y
= %\
dx
-1^=12^,
dxdx
dx dx dx
58.
Notation.
denoted by
J
That
The second
derivative of
—^«
dx2
is,
dry
_
dx2
Similarly,
d dy
dx dx
2
_ d d y
d dy
dx dx dx
dhj __ d
dx3
d 4y
_
4
dx
dry
dx n
_
dx dx2
d d d dy
dx dx dx dx
d d" _1 .y.
dx dxn ~
l
01
_
d d 3y
dx dx3
DIFFERENTIAL CALCULUS
62
Thus,
if
=
y
A
dy_
4a3
=
,
dx
= 12a;',
dx 2
fry
==
24x.
dxs
The successive
third,
.
.
.
derivatives are sometimes called the
If the original function of x
rivatives are often denoted
/'(•).
59.
tive of
first,
second,
differential coefficients.
is
denoted by f(x),
/'"(•).
/?(«)»
its
successive de-
by
The rtth Derivative. It
some functions.
•••
/*(?).
/-(*)
possible to express the nth deriva-
is
For example,
(a)
From y = e ax we have
,
dy
— ae ax
(b)
From y =
fry
'
dec
ax
+b
dx2
_ae
2
ax
= (ax + b)~\
g = (-l)a(aa +
6)-*,
fry
. . .
dxn
g=
(-1)(
3)a 8 (aa
2JL = (_ l)"|na-(aa!
+ 6)"Bj1 -
11
'
—
y
.
we have
g = (- 1)(- 2)(dx
= a"e ax
- 2)a\ax + &)-»,
+ 6)~ = (- 1)^3 a
4
—+
(ax
-
n+l
6)
3
(ax
+ 6)"
4
,
SUCCESSIVE DIFFERENTIATION
(c)
From y =
sin ax,
=a
f
-^-
cos ax
63
we have
=a
*"
sin
(
GKE-f-
da;
*fc
=a
2
dx
&=
cos
faa
crcos f«e
+ 1) = a
+ —\= a
—i = a n- sin /ckb H
(/"?/
2
.
,
sin few
3
+ ^\
sin few
+^\
?nr\
].
f
efo"
1.
2
y = Za*- 5
rt
+ 20
EXAMPLES
-or' + 2^
^=
ar»
a^M
= {x -^,
2
!/
- + 1).
120(0?
ax
Q
= 20(x
2
a?
-l)(x 2 -4:)i
ctx~
3.
y
= xr + x~ m
,
-{ — m(m - l)(m — 2)aJm "3 — ra(m + l)(m + 2)ar
J
5.
6.
//
7
da*
L
da5
_24
X
'
=x
=
4
2
a*
log
log
a-*,
(a*
[3
8.
9.
v
.,
/•
= 4^_2)e*+(a,'-l)e
2x
.
d 2y_
(«
=
4 x(e x
dx2
= ^_3^ + — -3)ea
= log sec 0,
J9
cfy_ 2(x 2 - -3a; + 3)
— 1),
dx"'
7.
|6
d 2x
,
:4ts e 2t
-I) 3
+ *)•
-
df
dfr
-
6 sec 4
0-4 sec
2
(9-
ro
-3
.
DIFFERENTIAL CALCULUS
64
10.
^=
y=ze-x (llsin2x-\-2eos2x),
dx3
n
11.
12.
= tan _,
.
?/
e— sin 2 x.
125
A2 = 24a<l-a;
—
1
2
)
>->
^
a?,
'
A
dx
^tan-f^,
J
2
da;
a;
)
2(^ + ^-6).
g,
'
+
(1
2 4
3
(
e*
+ e-*)3
13.
2/
= logV^ + a- + tan-'-,
+
^y_2(x — aYx
(^ + y
14.
y
= (e + e-"*) sin a6,
^ + 4a
,—
,_
?/
=
16.
2/
= e- %
18.
19.
a;e
x
(sma;
— cos a;)-}- 3 e
w
x
2/
^+
= arYsin log + cos log x),
x 2 —\
6
dar
.
?^2/
+n
2
^=
2
0.
o.
— 3 x-^- + 5 v = 0.
^=
*,
= 0.
(tana;-l)
2
+ coswa?
=a
t/
—-^ = 4 a;e* cos x.
cos a?,
^+
a;
4
du
tan
sinwg
! ax + a 2)
2
a
a»
15.
l7
d3
a;
da;
&w (loga) na6a! .
(XX
20.
2/
9
= log(3a;+2),y
,
,
22.
2/
—
^ = (—1Y- + t==.
I
0*
^
da;
d"2/
n
i-
3n \n
(3a;
2)
1
M
(-l) w " 1 1.3.5...(2n-3)
= sin5a;sin2a;, ^=y~3w cos/3a;+
^V
?n
cosf7a;+^Yj
•
:
SUCCESSIVE DIFFERENTIATION
The following
65
fractions should be separated into partial fractions
before differentiating.
1
23
i/
fin..
2
oa
24.
n
=-1
x -1
3w
=
*
n
\n
+1
2
d"y
—
£ = (— l)
,
,
K
da?
13
2x--
26.
— l)
daT
3a?-4
2.r-' + 3.r-2
_
5
(
1
. n|
n
J
LH
2X
+ 1)"*
2"
2
|~
-\_( x
+ 2y
+1
—
(2a>-l)»+1
3" +1
2 " +1
^1 = (- ±\n\ f
+x+l
(x
L(aj-1)"
+z
2.r'-x-l
-1
On+1
V
da"
27.
x2
y=
(x
28.
60.
jf
3f
^/
_ (- l)"
dry
+ 2)
2
'
n
dx
= /ax + 1 \
\ax - 1/
2
L L3(x-l) n+1
)
+
2)'
4 ( - i)" 01 " |*
dx"
(ax
This
is
(
a?
+ 1)
- n + 1)
4[w (x
(x
d" y
Leibnitz's Theorem.
+1
3(2
l
+2
a:c
+ n)
— iy
a formula for the
?ith
derivative
the product of two factors in terms of the successive derivatives
those factors.
A
special case of Leibnitz's
d
Tx
For convenience
,
du
s
{ttV)
=Hx
let us
vi
Uj
Theorem, when n
= 1,
is
Formula
v+u dv
m\
(1)
<r
x
use the following abridged notation
—
= dv
dx
= du
—
dx
?
v2
d?v
=—
o>
u2
"
—
dx
= cPu
2
,
=d
'"
vn
•••
un =
dxr
,
IV.,
n
v
dx n
—u
dn
dx"
DIFFERENTIAL CALCULUS
66
Then
(1)
becomes
d
(uv)=u 1 v+uv 1
(2)
Differentiating (2),
—d (uv) = u v +
2
-
2
ihVi -h VqVi
+ uv = u v + 2 u v
2
2
l
1
uv 2
4-
,
dx~
—d
3
(uv)
dx>
= u v + U V + 2 iu>v + 2u V + 0^2 4- uv
= u v 4- 3 w ^i 4- Sufis 4~ wy3
2
s
2
x
3
.
2
s
We
x
1
law of the terms applies, however far we
shall find that this
continue the differentiation, the coefficients being those of the Bino-
mial Theorem
—
(uv)
so that
;
= UnV + Wn-lVl + -^T^
Z
Un-2» 2
1-
-\
£
(XX
UU^O^ + UV n
This may be proved by induction, by showing that,
dn+l
(uv).
This exercise is
it is also true for
J
J
nK(uv),
dx n+iy
d
—
dx
if
(3)
.
true for
n
left for
the
student.
In the ordinary notation
(3)
becomes
n
n
d f N d u
v
}
nK(uv)=
dxn
dx
dn
~l
udv
+ n dxn ~
,
l
dx
dud n ~
.
l
v
dx dx"*
nn(n-l)d
ud*v
—
2
.
h
[2
,
h
dxn -*da?
dnv
dxn
,
1
EXAMPLES
1.
Given 9y
From
—X
s
sin 2 x
;
— (uv)
d*
-
(3),
find
by Leibnitz's Theorem
== u#)
—^.
dx4
+ 4 u^ + 6 u v + 4 w^ 4
2
2
ttv4 .
Cta?
^
v
v4
=
a;
3
,
i<!
=3
2
a?
,
= sin 2 %\ — 2 cos 2 x,
= 16 sin 2
a?,
a;.
u2 = 6
v2
a?,
Wg
= 6,
= — 4 sin 2 x,
u4 = 0.
v3
— — 8 cos 2
a?
5
•••
—
.
•
SUCCESSIVE DIFFERENTIATION
<P
^ = dx
!&/
,
.r sin 2 x)
= 0.
sin 2
+ 4-6-
x
4
-J-
4 3 x3 (— 8
•
- 16
2.
Given
y
[(.i-
cos 2 «)
- 9 x)
3
= *«*•
+
3
a?
x
sin 2
2 cos 2
16 sin 2
-f 0-6
.r
67
(- 4 sin 2 »)
a;
aj
+ (3 - 6 a )
2
cos 2 »].
25
find
d.r"
= e"
v = x,
Here
u
x
Substituting in
,
(3),
^=—
3.
,
Mj
= oe",
vl
=l
v.2
)
= 0,
(e**x)
^^
=a
n ax
e
dA y
-ilo
J=
/o
,31
2/
e
= 0,
ax
= ane
un
,
ax
x
+ m"" ^ =a
1
1
tl
-1
e
ax
(ax
+ w).
,
= sin x log cos
in
a,*,
4
c?
V
8(»-l)*
€
J,l
X
°Z X
,46,8
48(a;4-l)(2«2
—- = sin
a;
6\
+ x-^^-x*)
+ 2x + l)
[log cos x
— 2 tan
2
#(3 tan 2 x
dar
7.
y=
^=
x*a*,
ax (loga)"- 2 [(ajloga
+
ft)
\
8.
9v
.
•••.
= ( . + i )V^ri; ^3(5^-14,4-13).
K
6.
v3
l
we have
d7?
a
1
u,^ = a n ~
•••
=
!
(a-,+ 1)
3
>
—-=( — 1)
dx n
K
}
*
ft-
-
(x
!
+ l) n+
!
*
—
2
-ft].
-*-
5)1.
CHAPTER V
INFINITESIMALS
DIFFERENTIALS.
61.
The
derivative -^ has been defined, not as a fraction having a
ax
numerator and denominator, but as a single symbol representing the
limiting value of
—
,
as
Ax approaches
In other words, the
zero.
derivative has not been defined as a ratio, but as the limit of a ratio.
We
have seen (Art. 56) that derivatives have certain properties
and there are some advantages in treating them as such,
of fractions,
thus regarding -^ as the ratio between dy and dx.
dx
Various definitions have been given for dx and dy, but however
defined, they are called differentials of x and y respectively.
The
symbol d before any quantity is read " differential of."
62.
The
Definition
of
One
Differential.
definition
is
the following:
any variable quantity is an infinitely small increment in that quantity.
That is, dx is an infinitely small Ax, and
dy an infinitely small Ay.
By the direct process (Art. 16) of finding the derivative of an
differential of
algebraic function,
Ay
generally expressed in a series of ascending
is
powers of Ax, beginning with the
For example,
and
if
y
=x
Ay
first.
y+
3
,
= 3x
2
Ay=(x + Ax)
3
,
Ax + 3x(Axy+(Ax)\
...
(1)
In finding the derivative we have
^=
in which, as
Ax approaches
as its limit, the second
3x?
+ 3xAx + (Ax)
zero, the
second
2
,
member approaches
and third terms approaching the limit
3x?
zero.
DIFFERENTIALS
If
we
let A.r
zero, but there
approach zero in equation
is
nevertheless a
marked
69
(1),
every term approaches
distinction between them, in
that the second and third terms, containing powers of
the
first,
Ax higher than
diminish more rapidly than that term.
Thus we have
Ay=3x 2 Ax
and the closeness of the approximation increases
approximately,
as
Ax approaches
zero.
From
this point of view, regarding dx
increments,
we may
and dy as
infinitely small
write
dy
= 3x
2
dx,
not in the sense that both sides ultimately vanish, but in the sense
that the ratio of the two sides approaches unity.
dy=3x
2
Thus
dx,
and
^- = 3x 2
dx
,
two modes of expressing the same relation.
According to the first,
An infinitely small increment of y is 3x2 times
are
nitely small increment
the corresponding infi-
of x.
According to the second,
Tlie limit
of the ratio of
increment approaches zero,
the
increment of y
to that
of
x,
as the latter
is 3xr.
Just as we sometimes say
"An infinitely small arc is equal to its chord," instead of
"The limit of the ratio between an arc and its chord, as these
quantities approach zero,
So in general,
if
is
unity."
y =f(x),
Lim Aae=0 -^=/'(aj),
Ax
that
^/ =/'(*) +
Ax
is,
where
c
e,
approaches zero as Ax approaches zero.
DIFFERENTIAL CALCULUS
70
Ay
Hence
= f'(x)Ax-\-eAx,
and as the term eAx diminishes more rapidly than the term f'(x)Ax
we have
Ay = / '(a?) Ax approxim ately
}
dy=f'(x)dx.
or
Corresponding to every equation involving differentials, there is
another equation involving derivatives expressing the same relation,
and the former may be used as a convenient substitute
for the
more
rigorous statement of the latter.
Thus the use
of differentials is not indispensable, but convenient.
always be kept in mind that their ratio only is important,
the derivative being the real subject of mathematical reasoning.
It should
63. Another Definition
of
are sometimes defined as any
derivative
Differentials.
differentials dy, dx,
ratio equals the
dy
Y
dx
Let us see what this
tion
The
two quantities whose
defini-
means geometrically.
If
we regard
the derivative
o/
as the slope of a curve,
dy
tan
?^<k\ TR
RPT.
dx
By
dx
this definition of differen-
dx may be any distance
PR taken as the increment of
x, and dy is then RT, the corresponding increment of the orditials,
y
X
nate of the tangent line at P.
That the two definitions are
consistent
will
appear,
if
we suppose
PR
diminished.
to
be
indefinitely
The smaller we take PR, the more nearly
is
RT equal to unity, or
^RQ
RT equal
to
RQ.
in other words, the
PR is
more nearly
is
supposed to be infinitely small, this definition of
entials becomes that of the preceding article.
If
differ-
DIFFERENTIALS
The second may be
tions,
but the
said to be the
more rigorous of the two
71
defini-
has the advantage of being more symmetrical, and
first
better adapted to the various applications of the calculus to mechanics
and physics.
64.
may
The formulae for differentiation
by omitting dx in each
Formulae for Differentials.
be expressed in the form of differentials
member.
To each
of the formulae for a derivative, corresponds a formula
for a differential.
Thus we have
II.
III.
= 0.
d(u + v) = du + dv.
dc
—vdu
IV.
d(uc)
VI.
jfu\ _ vdu —
VII.
d(u")
= nu
=
-f udv.
n~l
udv
du.
—
(l
"
IX.
d7 log u
XI.
= e 'du.
d sinw = cosu du.
d cos u = — sin u du.
d tan u = sec udu.
d cot u= — cosec u du.
d sec u = sec u tan u du.
d cosec u = — cosec u cot u du.
1
u.
XIII.
XIV.
XV.
XVI.
XVII.
XVIII.
XX.
XXII.
,
de"
2
2
•
O7
7
,i
_i
Sill
*.
tan
-1
U
u
—
fl
•
Vl-u
(l
=
d sec
XXVI.
il
7
vers
i
=
u
-,
u
2
"
-•
1
XXIV.
"
=
+U*
du
"VuY^l
du
V2u
•
DIFFERENTIAL CALCULUS
72
Differentiation
by the
by the new formulae
old, differing
substantially the
is
only in using the symbol d instead of
same
—
as
.
(XX
For example,
to,
ay
let
y
=
.
X2
-\-
- (x
_ rif^±l\
~
a
2
O
+3)d(x + 3)-(x + 3)d(x2 + 3)
[x^T3j~
(z
2
+ 3)
2
~ (x + 3) dx - (x + 3) 2 xdx
=
2
(x 2
_
(a;
2
If
we wish
by
2
+ 3 - 2 -6 apcfa = (3 - 6 - a
(x + 3)
+ 3)
2
a;
2
to divide
+ 3)
2
a;
2
2
) cfa
2
(a;
to express the result as the derivative,
we have only
dx, giving
dy_ 3 — 6x — x
dx~ (x -\-3)
2
2
2
EXAMPLES
Differentiate the following
functions, using differentials in
process
1.
y=(a?-l)(2-3a>)(2a> +
2
x
3.
4.
= (t-l)(t-2)
(* + l)(* + 2)'
y=
r
-vVTT Va7^2,
y
dx= 6(f-2)dt2)«"
(t+l)«(*
+
Wx'-\- 1
rdx~Z=
Var — 2
dy =
^ = (2 +cosm^)sin^
= 5J2l-*,
S
4
cos 3
5.
dy= (-IS x + 2 x + ll)dx.
2
3),
= e*(x -6x + 2£x-4:0), dy=
3
2
e
2
(^ + Adx.
the
INFINITESIMALS
7.
r
= sin 6
w
= tan
log tan
4.r
*
4
<^
4
,
8
x2
,
dy
4
cfa
+x
^,
.
2
= SVI — 9 x* dx.
3 tan 2
d^ =
3
tan
4
6>
may
infinitesimal
dO
spoken of
be defined as a variable whose limit
several infinitesimals that approach
If there are
neously, one of them,
a,
and called the principal
Then a
(9
— tair^+1'
In Art. 62 we have
65. Order of Infinitesimals.
f6.
small or infinitesimal increments.
An
+ sec 6 dO.
cos 6 log tan 6 dO
sc
+ 3.r Vl - 9
= tan-Han
=
o?/=
5
-,
-
sin- 1 3a
9.
dr
9,
73
may
zero
infinitely
is zero.
simulta-
be taken as the standard of comparison
infinitesimal.
a3, a* are said to be infinitesimals of the second, third,
2
,
with respect to
?ith orders,
a.
In general the order of an infinitesimal is defined as follows: An
infinitesimal (3 is said to be of the nth order with respect to a when
Lim a=0 ^ =
When n=
When
n
1, (3 is
= 2,
(3 is
A',
a finite quantity, not zero.
.
.
.
(1)
of the first order with respect to a.
of the second order with respect to
a.
From the definition it may be shown that the limit of the ratio of
an infinitesimal to one of the same order is finite, and to one of a
lower order, zero.
Equation
If
we
write
(1),
Art. 62, illustrates infinitesimals of different orders.
it
dy
= 3x
2
dx + 3x (dx) 2
+
(dx)8,
and regard dx as the principal infinitesimal, the terms of the second
member are infinitesimals of the first, second, and third orders, with
respect to dx.
i),
are Bin a;
if we regard x as the principal
and \tvsx, with respect to x?
infinitesimal, of
what orders
DIFFERENTIAL CALCULUS
74
By
Hence by
to
x.
T
Lim
x=0
.
Limx=o ?HL£!
Art. 12,
(1) sin
x
is
=l
T
cosx
1
.
a finite quantity.
an infinitesimal of the
—— = Lim ———— ——
versa;
?
x=0
x-
2
sin'
*
siii
2
or
sc
== T
first
order with respect
1
-
Lim x=0
l
/sinscV
+ cosa?y
x J
1
I
a finite quality.
Hence by
respect to
Show
(1) versa? is
that tan 6
respect to
an infinitesimal of the second order with
x.
6.
—
sin 6
is
an infinitesimal of the third order with
CHAPTER VI
IMPLICIT FUNCTIONS
(See also Art. 114.)
66.
explicit
In the preceding chapters differentiation has been applied to
The same
functions of a variable.
rules or formulae of differ-
—*,
—4,
dx2 dx5
entiation are sufficient for deriving -^,
° dx
'
•••,
'
'
when yy
is
an
implicit function of x
that is, when the relation between y and x is
expressed by an equation containing these variables, but not solved
;
with respect to y.
For example, suppose the relation between y and x to be given by
the equation
9
cry
9,799
2
b xr —
2
+
Differentiating with respect to
97 9
orb'.
x,
dx
(
2ahJ hl + 2bix = o,
dx
2
b x^
dx
a 2y
Having thus obtained the
first
derivative,
we may by another
differentiation find the second derivative.
cC-ylfJ
dx-
d.ca 2
y~
- Hm»3K
dx
<i^f
7-3
b
2
fy
-x&S
dx)
V
d2y 2
DIFFERENTIAL CALCULUS
76
Substituting
now
dry
dx
-^
dx
for
its
+bx
b\a2f
_
2
a
2
The
entials
3
2
b*
)
y
ay
we may
differentiatin g again,
d3y _
dx8
value,
obtain
6\
'
a 4y 5
first differentiation
may
be conveniently performed by
differ-
Thus we should have from the
instead of derivatives.
equation
a 2y 2
-f b
2
x2 =a2 b 2
,
2a ydy + 2b xdx = 0,
2
2
dn
-^ =
dx
giving
In deriving
differentials.
—
^, —*,
dx dx
"hi*
—
,
as before.
a 2y
...,
derivatives should be used rather than
EXAMPLES
Find the following derivatives.
1.
(x- a) 2 + (y-b) 2 =
2
,
x—a
dv _
dx
2.
c
y
x=y\og(xy),
— b'
d 2y
4.
2
+x
2
'
dB
d 3y_
dx*
3c 2 (x-a)
(y-by
<frtanfl
Iogcos0-Ocotcf>
1,
<fy_ _axjj-hy
hx + by'
dx
xy
gj = logBin* +
(coB^=(8in^,
ry
v
7
ax2 +2 hxy-\-by 2 =
''_
(y-by
dy_xy- y
dx
3.
_
dx 2
d 2y _ lr — ab
dx2
(hx + by) 3
d2x _
'
dy 2
h 2 —ab
(ax
+ hyy
IMPLICIT FUNCTIONS
5.
aa?
6.
tan
.
„
+ 2hx!r + bf =
.
.
,
tand>
r
o
/=x
}
—_x6
rtd>
:
i
2
d =
_x
d>
.
sin2
ch/
s
/
2)/
—x
d2
T*+ cos 2 0)z
<f>
(x
)/
-
25
— y)
2
+ y + 6y(3y-3x + 2)=c, Jj=y=|e.
(3x
9.
,W« + 2r + l =
0,
(|J
diJ
e-=crtr
c?£
=
+ 2,eot.|=^
?/
"~ logft
a;
— log
«+
hilog (,-+,-) = 2o tan -^, g=^-J
/
ii.
2 sin 2 Afcos
2
^v
sin2 0'c?0 2
'd0
\
/
«.
sin 2
= w,—" —
8.
10.
d
dx
77
o
,
oN
.
i
v
r?y
d2y
?/
y
,
,
6'
^y = %-iog«)
2
(to
2x?
(a;
+ 2w
2
J=-^r#-
— log b)
2
j
'
CHAPTER
POWER SERIES
SERIES.
67.
VII
Convergent and Divergent Series.
%+w
2
-f
u8 +
The
+un + un+1
•••
series
.
,
...
.
.
.
(1)
composed of an indefinite number of terms following each other
according to some law, is said to be convergent when the sum of the
terms approaches a
this
n terms of
(1),
Lim
When
the series
a
convergent
when
number
of terms is indefi-
does not approach a finite
Sn
is, if
denote the
definite finite quantity.
when
series,
+ ar +ar
2
-f-
ar 3
-f-
r is numerically less
than unity, and divergent
Sn = a + ar + ar* +
•••
~
+ ar"" = a (} — **)
1
1
When
When
When
* \r
|
|
r
|
=1,
r
.
r
|
<
1,
Lim^ Sn = ^_i_
1 —r
|
>
1,
Liin n=00
the series
is
Sn = oo.
also divergent.
Series of Positive and Negative Terms.
Convergence.
Absolute and Conditional
In the case of series composed of both positive and
negative terms, a distinction
and
of the
r is numerically greater than unity.
For
68.
sum
when
convergent,
this condition is not satisfied, the series is divergent.
Thus the geometrical
is
is
Sn = some
)(=00
sum
That
limit, the series is divergent.
first
the
finite limit, as
But when
nitely increased.
is
made between
absolute convergence
conditional convergence.
* r
|
|
denotes the numerical value of
78
r.
SERIES
79
Before defining these terms, the following theorem should be
noticed:
A series
formed by
whose terms have different signs
tailing the absolute values
convergent if the series
is
of the terms of the given series
is
con vergent.
"Without giving a rigorous proof of the theorem,
we may regard
the given series as the difference between two series formed of the
positive ami negative terms respectively.
The theorem is then equivalent to this
sum of two series is convergent,
:
If the
their difference is also
convergent.
A
series is said to be absolutely convergent,
when
terms is convergent.
series whose terms have different signs
absolute values of
A
the series of the
its
Such a
without being absolutely convergent.
may
be convergent
series is said to
be
conditionally convergent.
For example
1
:
converges to the limit log e
but
it is
2,
not absolutely convergent, since
2
is
3
4
divergent (see Art. 70).
Series (1)
is
accordingly conditionally convergent.
l_i + I_I +
But
is
(1)
h
1
32
22
...
42
absolutely convergent (see Art. 70).
69.
Tests for Convergence.
The following
are
some
of the
most
useful tests.
In every convergent series the nth term must approach zero as a
limit, as n is indefinitely increased.
That
is
is,
the series
convergent, only
For
when
+ Mg+
Lim =x u — 0.
Sn = #__, + u n
v^
u2
-f-
#,
n
.
•••
+ un
-\
DIFFERENTIAL CALCULUS
80
sum
If the
of the series has a definite limit,
Lim n =
Lim n=00
Hence
For a decreasing
series
negative, this condition
^
is
1
+ log
e
2,
(1)
series
2
_3
4
1
2
3
does not satisfy
4
3
_5
4*"
Lim M=00 u n
(1), since
according as the number of terms
series
not sufficient.
an
v
— - + = + -••*
2
it
.
=
two
of this series oscillates between
series is called
For a
1
But the decreasing
divergent, as
The sum
Mn
whose terms are alternately positive and
'
convergent.
Sn = Lim n=0O Sn
is sufficient.*
For example,
is
a0
is
= 1.
and
Such a
limits, loge 2
even or odd.
oscillating series.
whose terms have the same
For example, the harmonic
sign, the condition (1) is
series
^2^3 4^
is
divergent (see Art. 70).
70.
We may
Comparison Test.
series of positive
terms
is
often determine whether a given
convergent or divergent, by comparing its
terms with those of another series known to be convergent or divergent.
In this way the harmonic
1
may
series
+ -2 + 3- + -++ -6 + -7 + 8- + —
4
5
be shown to be divergent, by comparing
1+ I +
HHH
+
* The proof of this
+
is
it
•
.
•
.
w
(1)
with
+-
omitted.
.
(2)
SERIES
Each term
term of
(2)
may
of (1) is equal to, or greater than, the corresponding
Hence
(2).
81
if (2) is
divergent, (1)
But
also divergent.
is
be written
T 2 T 4816
=1+1+1+1+1+
The sum
therefore
of this series is unlimited
;
...
hence
(2) is divergent,
and
(1).
Consider
now
the more general series
h+h + h + h+If
If
p = 1,
p < 1,
the series (3) becomes
every term of
responding term of
(3)
which
is
divergent.
after the first is greater
Hence
(1).
(1),
(3)
(3) is
than the
> 1, compare
! + !+!+L + L + !+!+L + ...+JL +
15"
1»
If p
2'
3'
1'
o»
6'
Every term of
But
(5).
(4) is
(5)
equal
may
or less than,
to,
the
4:
a geometrical series whose
ratio,
2
—
,
is less
whenp^l,
the series (3)
when p >1,
the series (3)
with which others
-
•
®
corresponding
...
'
than unity.
Hence by Art. 67, (5) is convergent and consequently
Thus it has been shown that
series (3) together
:
(4)
v i
S
'
The
.
be written
L + ^+l+^
+
9P
\P
P
p
series,
...
8'
7'
™th h + h + h+h + h + h + h+h +, " + h + "
term of
cor-
divergent in this case also.
is
is
(4).
divergent;
convergent.
with the geometrical series are standard
often be compared.
may
DIFFERENTIAL CALCULUS
82
This depends upon the ratio of any
In the series
Cauchy's Ratio Test.
71.
term to the preceding term.
u2
+u
3
h un
H
+ un+1 +
•
(1)
•
this ratio is
Let us
first
consider,
series
a
Here the
We
ratio
from
+ ar + ar
-^ =
r,
this point of view, the geometrical
and
is
|r|<
,
,
That
is,
(2) is
ratio
now
Wi
(1) is
is
(2)
„
1,
series is convergent or divergent,
-
i
or |?-|> 1.
,
,
convergent or divergent according as
<
If
....
+
the same for any two adjacent terms.
have seen (Art. 67) that this
according as
arn 4- ar n+1
\-
-\
any
1,
>1.
or
other than the geometrical series, the
series
The
not constant, but a function of n.
then
series is
convergent or divergent, according as
<
Liim
We
Wi >1.
or Lim, J=
(3)
will first suppose (1) to be a series of positive terms.
=P
Lim,
Let
Suppose p
—
1,
<
approach
1.
its
By
.
taking n sufficiently
limit p as nearly as
we
large
we can make
please.
n
There must be some value m, of
-^- <
Hence
r,
u m+1 < ujr3
n,
such that when n
^
m,
a proper fraction.
u m+2 < u m+lr
< u mr
u m + u m+l + m ot+2 +-- <u m + u m r+
2
,
etc.
uy+
••-.
.
•
W
2
SERIES
But
since
< 1,
the second
member
of
convergent, and therefore the
series, is
is
r
83
Consequently
convergent.
> 1.
Suppose p
By
v.
Since r
> 1,
m
r,
is
the geometrical
member
convergent.
similar reasoning,
-^ >
Hence
(1) is
which
(4),
first
when
n
> m,
an improper fraction.
^ > u m r,
u m+2
> u m+1r > u f+ etc.
n
the second member, and therefore the
must be divergent.
Thus the theorem
first
member,
proved for a series of positive terms.
have different signs, it is evident from Art. 68
will be absolutely convergent if
is
If the terms of (1)
that the series
u.n+l
l
Lim,
<1.
It is also true that for different signs, (1) will be divergent if
,n+l
>1.
Lira,
The proof
of this latter statement is omitted.
If
U n+1
Lira,
_
-i
may be either convergent or divergent. There are other
such cases, but they will not be considered here.
the series
tests for
EXAMPLES
1.
Is the following series convergent?
1-2
Applying
(3),
Art. 71,
2-
2
«2»
3.2 s
we have
n
^*±*
un
2(w
+ l)
1
2 (n
As
Its limit is
+ 1)
2
than unity, the given series
log, 2, as will appear later.
this is less
is
convergent.
DIFFERENTIAL CALCULUS
84
Determine which of the following
series are convergent,
and which
divergent.
2
'
2
+ \3 + + \5 +
10
4
-
+ 10
l+ +
l
1
5
l-f?
.
1+
10 4
10 3
H
+
By
+
1
2
7
II
2
By
'*
1
9
'
2
3
5
1
+ Vl
L§
(1),
Art. 69.
Compare with
(3),
Art. 70.
Compare with
(3),
Art. 70.
(1),
Art.
Ji=
9
7
11
i
,
n2 + l
4—i-_ +
1
+ V2
+
1
1
+ V3
11.
log?-log| + log|-log| +
12.
sec^-see^+sec^-sec^ +
13.
sin 2
^+
2
By
4
10
2
1
Art. 70.
+* *
1+I+A.+
_r 5"r
10.
(1),
I
i_?+?-^+A_
3
(3), Art. 71.
± ^
+l
2
2
2
L?
8
Art. 71.
1
1
I
1
6.
By (3),
g
sin 2
- + sin 2 ^ + sin2 ^
3
4
5
By
+
..
69.1
POWER
14.
15.
12
+1
+1
1
+1
1
2+1
2
+1
'
'
OS
+1
+l
3
+1
4+1
+l
i
.
.19
4
,
-t
42
85
'
+1
+ l~ "2 +l~t S + l~ "4 + l +
2+l + 3±l 4 ± l + 5 ± l
+
+
t
2
l»
16.
+l
32
3
,
22
SERIES
2^-1
f
2
5 S_ ±
43_1
33_1
s
Answers
Exs.
2, 5, 6, 9, 11,
Exs.
3, 4, 7, 8, 10, 12, 15,
Exs.
8, 12, oscillating.
72. Power Series.
A
13, 14, 16, convergent.
terms containing the positive
series of
tegral powers of a variable
divergent.
+ a x + a # + a$?
2
a
-\
2
x
,
called a power series in x.
The quantities
supposed to be independent of x.
is
For example,
in-
arranged in ascending order, as
x,
+ 2a; + 3a,- + 4^
2
l
3
H
a,-,,
a l5 a2
,
are
,
1_ t+ I*_ t +
12
are
power
series in
li
1$
x and y respectively.
73. Convergence of Power Series.
A
power
series
is
generally
convergent for certain values of the variable and divergent for
others.
If
we apply
the ratio test,
a
we have
{)
(3),
+a x + a#?-\
for the ratio
\-a nx
x
power
n
-\
= Lim,
anx
an x
= |a?|Liin
series
(1)
,
between two terms
Un±X
Lim.
Art. 71, to the
BS
«n-l
DIFFERENTIAL CALCULUS
86
The
series (1) is convergent or divergent according as
x Li
|
that
is,
|
<
m^
1,
|
>i;
Lim,
a;
|
according as
|<c|<Lim n=
The
or
case
or
= Lim w=Q
\x\
+ 2x + 3x + 4lX
2
=
Here
n
an
Hence
(2) is
We may
t-naT-1
3
-\
Lim,,^,
71
+ (n + 1) a? +—.
+1
(2)
= 1.
convergent or divergent, according as
|aj|
say that (2)
—1
terval from
series
,
+
7t=
requires further examination.
For example, consider the
1
|#|>Lim
+
to
1
<
1
or
\x\
> 1.
when
— 1 < x < 1,
and the
is
convergent
is
called the interval of convergence.
in-
EXAMPLES
Determine the values of the variable for which the following
+x+x +x
2
1.
l
2.
x
x-J^-j-_^_
+
,
1-2
2-3
/v*—
3.
A
4
+---
-^ +
3-4
rW*
»+'-+- + - + —.
,2^3
X
-
/y»3
s
Xs
5
4
7
X
x
-3 + 5~7 + -'
,
.
POWER SERIES
5-
- + 1—-T3 ~ + 1n
.1
.r
+
3
.r*
.i-
•
m3
7.-
7.
V-
)•"*
}•'
2
i
[6
a-
12
•
3
,
7
T> .r
-+
,
,.
•
/>v*
1-— + — -=-+•'.
3
Q
,
,
^
87
:
««x
6
.
•
sc
a*'
12
II
.
Exs. 1-5, convergent
when
Exs. 6-8, convergent for
all
—1 < <
a?
values of
1
jc.
CHAPTER
VIII
EXPANSION OF FUNCTIONS
When
74.
by any process a given function
of
a variable
expressed as a power series in that variable, the function
is
is
said to
be expanded into such series.
Thus by ordinary
1
By
division
+x
= l_ + ^_ x +
3
a;
...
(1)
the Binomial Theorem
+ a) = a + 4 a x+ 6 a¥ + 4 ax + x*.
(l-£ )- = l + 2x + 3» + 4ar +...
4
4
(x
C
2
The methods employed
2
method
3
(2)
in these expansions are applicable only to
We are now
functions of a certain kind.
general
3
3
of expansion, of
about to consider a more
which the foregoing are only
special
cases.
when a function is expanded into a
an unlimited number of terms, as (1) and (2), the
expansion is valid only for values of x that make the series conFor such values, the limit of the sum of the series is the
vergent.
given function, to which we can approximate as closely as we please
by taking a sufficient number of terms.
The general method of expansion is known as Taylor's Theorem
It should be noticed that
power
series of
and as Maclaurin's TJieorem.
These two theorems are so connected that either may be regarded
as
involving
the
other.
We
shall
Theorem.
88
first
consider
Maclaurin's
'
:
EXPANSION OF FUNCTIONS
This is a theorem by which a function
expanded into a power series in x. It may be
Maclaurin's Theorem.
75.
of x
89
may
be
expressed as follows
/(*) =/(0)
in
which f(x)
f"(x),
•••,
its
is
the given function to be expanded, and/' (x),f
•
••,
Derivation
as the notation implies, denote the values of
when * = °-
'
Maclaurin's
of
Theorem.
If
power
the following manner:
possibility of the expansion of f(x) into a
determine the series in
Assume
where A, B,
(x),
successive derivatives.
f(O),f(0),f'(0),
/(*), / (*)> /' (*)> ;
76.
+/(Q)j+/' (0)|+/"(0)|+ -,
/(x^i + m'+CV + DxH^i
C,
•••
we assume
series in x,
•••,
.
-
.
the
we may
.
(1)
are supposed to be constant coefficients.
Differentiating successively, and using the notation just defined,
we have
f\x)
= B + 2Cx + 3Dx + 4:Ex +
f"(x)
= 2C + 2.3Dx + 3.±Ex
f"'(x)
2
i
= 2.3D + 2-3-4;Ex +
..-
....
2 ...
...
Now
since equation (1),
true for all values of
x,
and consequently
•••
(2), (3),
when x = 0.
we have
from
(1),
f(0)=A,
A=f(0),
from
(2),
f(0)=B,
B=f(0),
from
(3),
/'(0)
= 2C,
(4)
(5)
they will be true
zero for x in these equations,
(3)
.
r(x) = 2'Z-±E+...
(2)
C
=^'
are supposed
Substituting
DIFFERENTIAL CALCULUS
90
from
(4),
from
(5),
/»'(0)
/
iv
(0)
=2- 3D,
D = OS,
= 2-34^
E = G^-,
li
Substituting these values of A, B, C,
•
•
we have
in (1),
•
...
/(*)=/(0)+/ (0)j+/"(0) | + /"'(0)^+--,
(6)
|
77.
it
As an example
in the application of Maclaurin's
be required to expand log (1
fix)
f(x)
-f x) into
= log (1 + x),
= -±- =
/(0)
+ x)-\
(l
/'(0)
f'(x)=-(l + x)-*,
Substituting in
(6),
4
Art. 76,
,
=0 + 1.^-1- 2
/v?2
= 1.
/"(0)
= - [3.
/-
=14.
(0)
we have
x2
log(l+x)
= log 1 = 0.
/'"(0)=2.
,
= - [3 (1 + *)/*.(*) = 14(1+.*)^,
let
x.
/"(0)=-l.
/'»(*)= 2 (1+z)- 3
/* (x)
Theorem,
a power series in
/yiO
,
2 xs
[3*
4
,
+^--^J
/yi4
5
Li*
+ ^g----
/yi5
iog(i-M)=*-f +!-!+!-•••.
78.
If
in the application of Maclaurin's
Theorem
to a
given
function, any of the quantities, f(0), /'(0), /"(0), ••• are infinite,
this function does not admit of expansion in the proposed power
series in x.
In
x
some of its derivatives are discontinuous for
and the conditions for Maclaurin's Theorem are not satisfied
this case f(x) or
= 0,
(see Art. 94).
The
functions logic, cotcc, x2} illustrate this case.
:
EXPANSION OF FUNCTIONS
91
EXAM PLES
Expand the following functions
Theorem
3
1.
e
x4
t
= 1 + x + -r- + '—
x
4- '— H
|4
|3
:
\2
power
into
•
series
by Maclaurin's
Convergent for
all
values of
x.
Convergent for
all
values of
x.
Convergent for
all
values of x.
2*^
2
3
4.
X'
= x — — + '—X° — ~
H
sin x
cos a;
(a
[3
[5
,2
,4
j2
|4
•
[7
6
=1 ——+-
—+
[6
+ jc)« = a" + va»-\v -f "^~^ a"~V
+ "O*- 1 )!"- 2
)
^-3^3
+
...
< a.
Convergent when
\x\
Convergent when
|a;|< 1.
[3
5.
log u (l
+ = log ae ^-|+|-j+a;)
(1—*) =
6.
log
7.
tan- 1 ;/— x
2
— — + — — ^--\
O
Here
Q
.
/(a?)
.
4
3
— x- —————
O
Convergent when x
.
|
Convergent when
.
= tan -1
/" (a?)
= -2a? + 4ar - 6a? +
1
-o'+^-aM3
.r
'
\x\
< 1.
\x\
<1.
a;,
=^-1-^ =
1
< 1.
4
/'(*)
,
|
,
1.3
8
a?
...,
5
1
•
3
•
5
x7
.
Convergent when
DIFFERENTIAL CALCULUS
92
Here
f(x)
— sin -1 #,
VI — XT
Expanding by the Binomial Theorem,
= 1 +aar>+ bx + cx +
1 3
a = -, 5 = —, c^
2*
2-4'
/'
where
6
'
•
••,
'
5
,
2.4.6'
/"
9.
4
(a;)
sin (x
(x)
»
= 2 ax + 4&c + 6ca + —
3
+ a) = sin a +
— — sin a — — cos a +
cos a
a;
5
Convergent for
10.
log(l
3
11.
e*
sin x
= +
cc
a3
12.
e*
cos #
=1+
a;
2
x
.
4
o
x
|
x
|
< 1,
5
H
.
.
1
3
o
Convergent when
Convergent for
all
values of
x.
Convergent for
all
values of
x.
6
tana = a + ^ + ? ?+"-.
r
3
lo
2
=l+ +
| |f+-.
14.
S ec*
15.
logsec*
= -+-+-+
....
Defining the hyperbolic sine, cosine, and tangent by
sinh x
—
= ——
2
16.
sinh
x.
+ x + x*) = x + ^-?f + ^ + ^
^
13.
values of
all
,
cosh x
=—
,
2
'
«=* + - + -+.-.
[3
|5
'
tanh x
=
,
e-H-e-'
show that
EXPANSION OF FUNCTIONS
x"
= 1 + ar
;- + ;-+. •
,
17.
cosh 3
18.
tanh x = x
3
5
x
2 r
——
4- 7715
3
Show by means
19.
93
of the expansions of Exs. 1, 2, 3, that
,xiA-l
e
-x
v/- 1
cos
4-
as
V— 1 sin
_ cos _ -y/ _ 1
a-
sin
x,
^
#
These are important relations.
79. Huyghens's Approximate Length of a Circular Arc.
ACB, a
may be shown that
If s denote the length of the arc
chord of half the
s
——
= 8 b -- a
arc, it
chord, and b the
its
.,
approximately.
.
,
o
Let
be the half angle
<£
Then
s
a=2r
=2
sm<£
r<j>,
AOC.
and by Ex.
= 2W
*!
<f>
+
2,
Art
72,
£
[5
[3
5
in| = 2r^-^4--*
3
2
Combining so
b-a =
V2
25
2 [3
'
as to eliminate
1^
3
<£
,
2rfs^ + ^=3s(l-^.
480
If
s is
If s is
an arc of
an arc of
&
,T
30°,
G0°,
<*>
= —,
and the error
12'
= |,
and the error
<
-
102000
< ^q
DIFFERENTIAL CALCULUS
94
80.
Computation by
Compute by Ex.
Series.
1, p. 91,
Ve
to 5 decimal places.
Ans. 1.64872.
Compute
^\/e
Compute by Ex.
7T
Ans. 1.1051709181.
to 10 decimal places.
2, p. 91,
sin 1° to 8 decimal places.
= 3.14159265.
Compute
Ans. 0.01745241.
to 4 decimal places the cosine of the angle
whose arc
81.
log (1
Calculation
+ x),
is
Ans. 0.5403.
equal to the radius.
Logarithms.
of
By means
the expansion of
of
Art. 77, the Napierian logarithms of
numbers may be
computed.
Let us find the logarithms in the following
log 9
=0.6931,
=1.0986,
=1.3862,
=1.6094,
=1.7917,
=1.9459,
=2.0793,
=2.1972,
log 10
= 2.3025.
log 2
log 3
log 4
log 5
log 6
log 7
log 8
table.
It is only necessary to calculate directly the logarithms of the
numbers 2, 3, 5, 7, as the others can be expressed
We have from Art. 77,
lo g
This series
is
2
prime
in terms of these.
= log(l + l)=l-i + |-i+....
convergent, but converges so slowly that 100 terms
would give only two decimal places correctly. But we may obtain
a series converging much more rapidly by taking
log2
= log
?
~3
= log
(1
+
|)-log (1-|).
EXPAXSIOX OF FUNCTIONS
For log
1
+x
log (1
1-x
=X
.v-
x3
,
2
+ X) a*
(-a
,
3
- x)
log (1
+
95
4
ar
_ ar _
.t
2~
3
4
4
_
.
}
Convergent when
log2 = 2/^ + -i- + -i- + -i-+...\
Thus
Four terms of
this series give
losr
2
The computation may be arranged
J-
!a;|<l.
~ .333333
= .6931.
as follows:
i =.333333
3
1
—
.037037
1
3"'
_ .004115
1
= .012346
3-3 3
33
1
=.000823
5-3
5
1
.000457
.
= .000065
37
1
3
1
.000051
9
,J
r
•
= .000006
3"
.34657
2
.69314
The numbers
cessively
by
in the first
Any number may
may
column may be obtained by dividing
9.
be put in the form
be found like log
2.
1
+x
T
1
an(j
\ Q(T
3
+
suc-
DIFFERENTIAL CALCULUS
But having log 2,
easier to
it is
1
log -
compute
1
+
= log
and then
log3
= log| + log2.
Let the student make this computation
i
Find log 5 from
.
5
log -
=!log
i+i
'4-
1In a similar way find log 7 from log 5.
Having obtained the logarithms
of 2,
3, 5, 7,
rithms in the table at the beginning of this
To
obtain the
common
logarithm, that
find the other loga-
article.
is,
logarithm 10
,
it is only-
necessary to multiply the Napierian logarithm by .4343, the modulus
of the common system.
Find thus the common logarithms of the numbers in the foregoing
first, of 2, 3, 5, 7, and from these the others.
tables,
—
82. Computation of
From Ex.
ir.
I=
a slowly converging
To obtain a
taa->l
7, p. 91,
= l-! +
by letting x =
H
+ ..,
series.
series converging
tan" 1 1
more rapidly, we may use
= tan"
1
-
1
from which
4
2
3 2s
^3
3-3 3
•
+ tan +5
'
1
3'
1
•
1
25
5-3 5
7
•
27
7-3 7
+
+
1,
we have
:
:
EXPANSION OF FUNCTIONS
By
taking 9 terms of the
first series
97
and 5 of the second, the
student will find
= 0.463647
4
and
= 3.14159
7T
Other forms of tan -1 1
more rapidly,
may
...
+
0.321751...
••..
be used, giving series converging even
as
1=2
tan" 1
tan
-1
1
tan" 1 -
= 4 tan -1
+ tan"
By
-
•
tan-i
5
mal
1
239
these formulae the computation has been carried to 200 deci-
places.
This is a theorem for expanding a function
two quantities into a power series in one of these
83. Taylors Theorem.
sum
of the
of
quantities.
As the Binomial Theorem expands (x + h) n into a power series
It
in h, so Taylors Theorem expands f(x + h) into such a series.
may
be expressed as follows
/(x
84.
+ ft)=/(x)+/'(x)A+/"(x)|+/"'(»)|+-.
The proof
of Taylor's
Theorem depends upon the following
principle
If
we
differentiate
stant, the result is the
to h, regarding
That
is,
For, let
f(x + h) with respect to x, regarding h consame as if we differentiate it with respect
x constant.
-ff(x
ax
+ h) =-£
f(x + h).
all
z
= x + h,
(1)
DIFFERENTIAL CALCULUS
98
then by
(3), Art. 56,
dx
dx
Jy
dh
But from
dh
dh
'
J
dh
Derivation of Taylor's Theorem.
of the expansion of
series
f(x
where A, B,
C,
+ h)
into a
we assume
If
power
• • •
= A + Bh + Ch +
2
we may
Assume
article.
j.,
,
dA
7N
dx,
.
dB
— f(x + h)=B + 2 Ch
Art. 84, the first
members
7
s
-f
x,
(1)
3Dh 2 +
of these
h.
then with respect to
dC.o
dx
.
dx
deter-
m+
are supposed to be functions of x but not of
Differentiating (1), first with respect to
d
dx
the possibility
series in h,
by the aid of the preceding
f(x+h)
By
K J
dz
—=1
and
1.
dx
mine the
}
Af(
f (x '+ h)
JK x +
^ h)= — JKJ
therefore
85.
K
J
dJi
=
—
dx
(1),
v
;'
dx
dz
,
dD JS
h,
,
dx
....
two equations are equal
to-
each other, therefore
§A + QRn + Mltf +... = B + 2Ch + 3Dh +
2
dx
dx
....
dx
Equating the coefficients of like powers of h according to the
Undetermined Coefficients, we have
principle of
^=
B = ^.
B,
dx
dx
dB_2f
l
p_
dx
dC=3D,
dx
j)
2
1 d A
2 dx2
=~
[3
~
c
dx 3
EXPANSION OF FUNCTIONS
The
99
= 0,
A may be found from (1) by putting h
supposed true for all values of h.
coefficient
equation
is
Then
A =/(«).
Hence
£=^ =/'(»)•
Substituting these expressions for A, B, C,
•••
in (1),
as that
we have
3
/(.r
+ /o=/«>o+/'(^A+rw|Vr'(^,4
[2
Maclaurin's Theorem
86.
by substituting x
= 0. We
+---.
.
.
(2)
[3
may be obtained from Taylor's Theorem
then have
/CO =/(0) +./(0)» +/"(0)jf +/"'(0)jf + •This
87.
is
Maclaurin's Theorem expressed in terms of h instead of
As an example
in the application of Taylor's
+ h) into a power
f(x + h) = sin (x + h)
/ (x) = sin x,
f'(x) = cos x,
f"(x) = — sin x,
/"'(a*) = — cos x,
= sin x.
/
be required to expand sin (x
Theorem,
series in h.
;
hence
iv
(a;)
Substituting these expressions in
sin (x
+
/ij
=
(2),
Art. 85,
we
find
4
sin x -f h cos
x
/r
h
h*
——
——
sin
cos x + — sm
a:
re.
a:
-f
x.
let it
DIFFERENTIAL CALCULUS
100
EXAM PLES
Derive the following expansions by Taylor's Theorem
h2
+ h) = cos x — h sin x — 7F> C0S x +
1.
cos (x
3.
(x
+
/i)
4.
(x
+
7i)
5.
log
7
w
= x + 7a;
7i+
= af + nxn~
x
6
7
1
+ h) = log x
(a;
7
tan
(a;
h2
a?
7Tl>
2 ar
Compute from Ex.
8.
Compute from Ex.
fjx + h) +/(»,-
.
As
2
a;
A4
+ 3^Js~T^4
+
ar
4 ar
4- 7r sec
2
a;
'
tan x
2
7.
10
+
+
h3
h
-\
a;
4
9
(n— 1)
n
+ h) = tan + h sec
+ ^(3sec aj-2sec x) + ....
6.
:
•••.
+
h
h3
75 s i n ^
1,
cos 62°
6,
tan 44°
h)
=f(x)
= 0.4695.
= 0.9657,
tan 46°
= 1.0355.
+ | r(x)+ V /iv(a;) + _
/(«+*) -ffr-Q ^hfim)
+
g /"'(g) + g/» + -;
a special case of Ex. 10, derive
ii og.«±» - * + j!. + i. + ..,
2
# — 7i
a;
oar 5X5
11. /(2a.)
= /(a )+x/^) +
/
/{«)
12.
1
;
+ aJ
1
+
2
^
/'
,
W + |r'W +
/'(*)
f"
+
(1
05
+ *)
-...
(x)
2
1
x3
(1
13.
If 2/= /(#),
show that
y
da;
^cto2
[2
^da:3
[3
+ *)
f'"(x)
3
[3
+
EXrAXSIOX OF FUNCTIONS
101
88. In the preceding derivations of Taylor's and Maclaurin's
Theorems, the possibility of the expansion in the proposed form has
been assumed. In the remainder of this chapter we shall show how
Taylor's Theorem may be derived without such assumption.
=a
If a given
Theorem.
Rolle's
89.
x
and when x=b, and
well as
its
derivative
is
<£'(.r);
function
<£(.r)
zero
is
when
continuous between those values, as
then
must be zero for some value
x between a and b.
<f>\x)
of
Let the function be represented
Let
curve y = <f>(x).
by the
6 A = a,
OB = b.
Then
accord-
=
when
ing to the hypothesis,, y
x
= a,
= b.
and when x
Since the curve
is
O/A
B\
continuous
between A and B. there must be
some point P between them, where the tangent
and consequently <f>'(x) = 0.
90.
x=
b,
is
parallel to
If/(#) is continuous from x = a
Mean Value Theorem.
must be some value x x of x, for which
there
b-a
This
may
J
K
J
be stated geometrically
thus:
The
OX,
1
difference of the ordinates of
A
two points of a continuous curve,
divided by the corresponding difference of abscissas of these points, equals
the slope of the curve at some inter-
P
A(
/Y
mediate point.
In the figure
let the
represent y=f(x).
curve
M
PRQ
K
B
to
DIFFERENTIAL CALCULUS
102
Let
OA = a, OB = b.
Then
PM
b-a
At some point
of the curve, as R,
PQ.
be drawn parallel to
Call
between
OK=x
1
P and
Q, a tangent can
Then the
.
R is/'(a^), which equals tan QPM.
HS~f(a) =f(Xl
where
slope of the
tangent at
Hence
—
AB = b — a = h,
a
),
< x, < b.
.
(1)
CL
If
we
let
+ h)=f(a)+hf(a +
f(a
= a + h,
(1)
important, in that
it
may
may
be written
0<<£<1.
where
cf>h),
The following method
91. Another Proof.
is
b
.
(2)
of deriving (2), Art. 84,
be extended to higher derivations of
f(x), as appears in Arts. 92, 93.
Let
R be
defined by
f(a
That
is,
let
+ h)-f(a)-hR =
/( a
R denote
+ ?0
(1)
-/(«)
.
h
Consider a function of x whose expression
x substituted for
That
h.
Call this function
is,
<£(#)
Differentiating,
It is evident
<j>'
from
=f(a + x) -f(a)
Hence by
-xR
= f (a + x) - R
<j>(x) = 0, when x = h,
<j}(x) = 0, when x = 0.
(2) that
Eolle's Theorem, Art. 89,
and
the same as (1) with
<j>'(x)
x, 6h,
f'(a -f Bh)
Substituting this value of
f(a
we have from
-R =
R in
(3)
by
== 0, for
h.
Calling this value of
(2)
(x)
also
between
is
4>(x).
(3)
0.
(1),
+ h)=f(a) + hf'(a + $h).
(1)
;
some value of x
EXPANSION OF FUNCTIONS
92.
103
We may extend the
Extension of Mean Value Theorem.
method
and
of the preceding article so as to include the second derivative,
obtain
+ h) =/(a) + hf (a) +|V" (* + Oh).
f(a
Hence
From
=f (a + —f
4>"(x)=f"(a + x)-B
(a)
x)
<f>'(x)
(2) it is
evident that
also
and
}
Also
<f>'
Hence
<f>"
and
tween
= 0,
(x) = 0,
(x)
h.
<f>(x)
= 0,
when x =
for
when
Substituting this value of
value,
= Oh,
be noticed that
=a
E
Define
It
to
some value, x1}
x2 between
,
and x1} that
is,
be-
(3),
in (1),
we have
assumed that /(a;), f'(x), and /"(a;) are
— a + h.
This
assumed that fix) and
a
for
+ 0h)-E = O.
it is
to x
Taylor's Theorem.
—
(1);
= 0.
= 0,
we have from
may now
preceding method so as to include the
x
by
+ h) =f(a) + hf (a) + ~f'{a + Oh).
continuous from x
It is
<f>'(x)
a;
h,
0.
some
f'(a
It is to
when x =
h.
Writing x2
f(a
(2)
(3)
Rolle's Theorem, Art. 89,
x between
from
(1)
— xE,
<ft(x)=0,
Hence by
93.
6
$(x)=f(a + x)-f(a)-xf(a)-^R
Let
of
~M =
f(a + h)-f(a)-hf(a)
Define i? by
x
be derived by extending the
?*th derivative.
its first
n derivatives are continuous
= a + h.
by
f(a + h)-f(a)-kf(a)-gf»(a)
[2
^/»-i(a)-f"i^0.
\n-l
\n
(1)
DIFFERENTIAL CALCULUS
104
Let
J^/^»(
= fr(a + x)-f'(a)-xf»(a)
if(x)
/^(a)-^
*
+(x)=f(a+z)-f($-xfXa)-£f"(a)
)-
tt
$«-* (»)
=/w "
1
+
(a
a;)
-/"- 1 (a)
*-(oj)=/
As
l
(a
= 0,
= 0,
= 0,
<f>"(x) = 0,
Hence
But
- a>2J,
+ aj)-iJ
in the preceding articles,
cj)(x)
-gL j»
|
[
when x =
(2)
evident that
it is
and also when x=0.
h,
<f>'(x)
when x = x where
<f>'(x)
when # =0
1,
0<x
x
<h.
hence
;
when x = x2 where
,
<
x2
< xv
<
xn
< xn_u
Continuing this reasoning, we find
n
<fr
that
is,
(x)
= 0,
where xn
Hence from
(2)
is
<£"
when x = x n where
,
(Oh)
=f
n
Substituting this value of
f(a
and
between
(a
h.
+ 6h) - R = 0.
R in
(1),
+ h)=f(a)+hf («)+! /"(«) +
Since a
is ara/
quantity,
we may
...
we have
+ |r^/-'(a)+^/"(a +
write x in place of
a,
fl»).
giving
/( + A) =/() + V(«) +|/» + - + J3I /- o*)
1
|
+ .-/•(* + »).
.....
(3)
EXPANSION OF FUNCTIONS
94.
Remainder.
The
105
term of this equation
last
I*
is
called the remainder after n terms in Taylor's Theorem.
limit of this remainder
is zero,
Theorem gives a convergent
When
the
as n is indefinitely increased, Taylor's
series.
We
have already seen (Art. 86) that Maclaurin's Theorem is a
special case of Taylor's Theorem, so that corresponding to (3) of the
preceding
/(*)
write Maclaurin's
Theorem
= /(0) +.r/'(0) + |/"(0) + .- + -^_/..-'(0) +|/»(te),
f(x) and
from
we may
article,
its first
to
n derivatives being assumed continuous for values
sb.
Thus the remainder
after n terms in Maclaurin's
Theorem
is
™»
When
=
Lim n=I=e
0,
(1)
Maclaurin's Theorem gives a convergent series.
Applying
(1) to
f(x)
=e
x
we have
,
= 0,
Lim,
which
is
evidently satisfied for
The same
llf(x)
is
true for
= log (1 4- *),
f(x)
(1)
all
values of
= sin x,
x.
and f(x)
= cos x.
becomes
Lhn_ \* (-1)-|!L=11
(i + Bxy
J
U±
=--[ =Ffe)-]=»!
This
is satisfied
It is to
when
|aj|
<
1.
be noticed that the preceding test for convergence
is
of no
practical use, unless the nth. derivative of f(x) can be expressed.
CHAPTER IX
INDETERMINATE FORMS
Value
95.
"#
any assigned value
•
This
When
,
is
The value
of Fraction as Limit.
i
ot x, as x
=a,
•
is
^v y
<K«)
a definite quantity, unless
of the fraction ^;
'
•
or
<j>(a)
\{/(a) is
zero or infinity.
we may, by regarding the fraction
define its value when x equals a, as its
as a
this is the case,
continuous variable,
when x approaches
That
limit
a.
the value of
is,
for
^>
^'
'
,
Lim x _ a ^
when x
or
,
what
= a,
is
is
defined to be
the same thing,
±(a+M.
+ h)
Lim h
-°t(a
There
is
no
difficulty in
determining this limit immediately, when
the numerator only, or the denominator only,
when one
We
x,
is
will
zero
now
and the other
zero or infinity
is
;
or
infinity.
consider the cases where, for some assigned value of
the numerator and denominator are both zero or both infinity.
The
fraction
is
then said to be indeterminate.
96. Evaluation
of
the Indeterminate
Form -
formation of the given fraction will determine
— —— — =
Thus,
But
if
j-
x2
we reduce
LinL,_,
* _1
-
-
z
value.
its
l.
l
the fraction to
2
when x =
,
Frequently a trans-
•
its
lowest terms,
+.t-2 = TLim^i x +—2 = -3
.
[
Vhl
x*-l
106
2
•
we have
—
INDETERMINATE FORMS
a
T
—
—2 =
— -, when x = 2.
z-1 -1
r
Again,
By
rationalizing the denominator,
a;
.
-2
T
(a;
.
- 2) (V» -1 + 1)
g—'2
Vai-1-1
= Lim^Va-l +
As another
1)
= 2.
illustration,
008
cos
4
7
cos 6
Lillian (cos 6
+
(9
4
2
(9
.
:
:
fl
(9
sin 0)
=
cos4
Form
-f-
sin-=
4
-y^.
T/>>-
variable,
We
value of this
the given
new
fraction, for the assigned value of the
now show how
this rule
' —-,
*W °
Suppose the fraction ^
= o.
By
of the given numerator
denominator for a new denomi-
limiting value of the given fraction.
is /la-
will
is
derived.
when x
=
a-,
that
Art. 95 the required value of the fraction
.
By
as h approaches zero.
h)
-
the
Mean Value Theorem,
(2),
Art. 90
= (a) + h4'(a + Oh),
^{a^h) = ^{a)-r h^(a-r-e h)
(ft
(a
+h)
<t>
i
where
6
and
method:
yen- fraction, taking the derivative
<•
new numerator, and of
<<
nator.
-
= ?.
whenfl
Differential Calculus furnishes the following general
97.
for
= °,
2
4
The
2g
0- sin
cos 6- sin
—7 — Lim e _zr
;r~— w
~4 cos^-siu^
— sin
cos 2
"But Liui ep ~
_zr
=
107
X
are proper fractions.
t
is, <f>(a)
is
=0, and
the limit of
DIFFERENTIAL CALCULUS
108
But
since
= 0, we have
4 h) = h<f>'(a 4 Oh) = (a +
^(a 4
^'(a 4 dji)
4 0^)
cj>(a)
—
and
if/(a)
f
<t>(a
which
If
that
is
Lim
^^±£ = £M
the theorem expressed by the rule.
<j>'(a)
is,
*
t
i}/'(a
/i)
Hence
flft)
<fr
=
and
ij/'
(a)
= 0,
it
follows likewise that
the process expressed by the rule must be repeated, and as
often as
may
be necessary to obtain a result which
is
not indeter-
minate.
For example,
let
us find the limiting value of the fraction in
Art. 96.
x2
$(x)
y,>
/
-l
^
=
= 2'
-,
2x
xp\x)
0'
when
05
= 1.
q
Thus the required limiting value
is
-
•
Z
For another example, let us find the limiting value,
e*
1
i//
_2
e
— COS
_|_
(a?)
<£'(<e)
-x
£C
1
— cos
= -,
_ e — e~* _
x
ij/'(x)
sin
a;
xj/"(x)
cos
a?
when
a;
=0.
a?
when x = 0.
0'
when x = 0.
Thus the required
limiting value
is 2.
when x = 0,
INDETERMIXATE FORMS
109
EXAMPLES
Find the limiting values of the following fractions for the assigned
values of the variable.
1.
*0g ~ 1 ;r~ 2
x-
%
—2X
+ s-3)
log(3*»
~
a
,
.r— tan
_1
.T
108(^-4 8 + 6)
log cos (#
6
whenz = 2.
Ans.
n
whenx = l.
Ans.
7.
when x = 0.
Ans.
log 6 a.
when x = 0.
-4?is.
— 2.
— 2)
w-log(x + l)
>
when* = 2.
^n S
wh enx = 0.
4«&
.
XT
vers
-2.
g.
Jj
fl
sin 2
8.
+ \.
L
\ogx
3.
5
,
= 0.
Ans.
whenm = «.
Ans.
when
0.
'
sinm "- sin
sin (?n
"",
— n)«
sin^ + |)-l
when0 = ?«
^4??s.
when a =6.
^w.
cos no.
-
log sin 2
6
10.
5^=±
b
a —b"
„
log >g
a
,
-log.6
—b
^-2ar' + 2x-l
when
a
=
6.
6
^+J2S-
1— log 6
Ans.
b log b
.
DIFFERENTIAL CALCULUS
110
13>
,
A
14.
-
c
15.
faums-ntans
wh en
nsmx — sin7ix
—
— n tan x
n sin x — sin nx
tan nx
wnen?i =
-,
;
;
,
(a;
eto
98.
+ e"
•
,
— x cos x
A
Ans.
m —m
4
Lmu=
o
The method
°°.
00
is
the
form -
in Art. 97 for the
been shown in that
article that
T^-ZL_1
if/
0,
—
s
when x = A0.
,
x
Evaluation of the Indeterminate Form
<£(«)=
xsec 2 ^— tana;
— af
same as that given
if
A
Ans.
^
l.
2.
-
log sec 2 x
It has
Ans.
sin x
m sin x — sin mx
x
(x — 1) e +
+ 1) e~
16.
= 0.
z
(a
+
= T-±-l
(1)
,
i//(a)
li)
and ^(a)=0.
may be shown that (1) is true also, if <f>(a) = oo and \p (a) = oo
For the proof of this the student is referred to more extensive
treatises on the Differential Calculus.
It
.
°x
For example, find the limiting value of
,
when x = 0.
cot x
&£)=M^ = »
cot
\f/(x)
x
when z = 0.
oo
1
sin 2 a;
—=
7?H=
— cosec^a?
y(x)
6'(a?)
d>"(x)
,.) (
\\>"(x)
a;
=
2 sin a; cos
=
a;
a;
= - = A0,
1
Thus the required limiting value
A
whenx-0.
,
a'
when
-,
= A0.
is 0.
—^The form — can in most cases be avoided by transformation into —
oo
Note.
a;
1.
GO
J
.
INDETERMINATE FORMS
111
99. Evaluation of the Indeterminate Forms
x
Transform the expression into a fraction, which
•
the form - or
—
cc
,
— cc
will
.
assume either
•
x
For example, find the value of
—
(?r
2 x) tan
This takes the form
But
•
when x = ~+
x,
oc
(tt-2x)J taiix=
7r
v
^
~ 2x =®,
cot
0'
a;
~2
=
,
-cosec 2 a-
i//(.r)
Thus the required limiting value
when« = J.
2
= 2j
when
*«*
2
is 2.
For another example, find the limiting value of
1
1
log X
X
—1
-But
1
log
1
x
sc
^
'«
=
when x = 1.
x—x
This takes the form
-p,-,4.
,
= b — 1— log° = — 1) log x
a;
—1
__£_=4 when
1-1+fcg.
°
1
„
*(*)
=-
=
12 + 1
a?
-,
when x
.
(a;
2
.
'
when
x
a;
Thus the required limiting value is--
=l
c
a-
= 1.
=
.,
1.
DIFFERENTIAL CALCULUS
112
100. Evaluation
Take the logarithm
form - or —
•
The
Exponential Indeterminate Forms,
of the
of the given expression,
which
0°,
will
l
00
,
limiting value of this logarithm will determine the
oo
given function.
For example, find the limiting value of
This takes the form 0°.
Let
then
y
log y
=x
x*,
when x
= 0.
x
;
= x log x = —^—
= GO
—
-\
,
when x
= 0.
x
1
—
iLLA
= — x = 0,
when x = 0.
_1_
ij/'(x)
x2
Thus the limiting value of log y
is 0.
Hence the limiting value
e°
of y is
=
1.
For another example find the limiting value of
i
(1
ax)* t
-f-
This takes the form of l
when x = 0.
00
.
i
Let
y
log y
= (l + ax)x,
= lpg (! + ax = %
)
when * = 0.
a;
y
f
v
oo°.
have the
y
^' (a)
=—
=
!
a,
when x = 0.
1
The limiting value of log y being a, the limiting value of y
is e
a
,
::
INDETERMINATE FORMS
113
EXAMPLES
Find the limiting values of the following expressions for the
assigned values of the variable.
x
—3
when #
a; = 00
00.
3.
x tan v
4.
'°g tan "*,
log tan &#
2T
—
sec x,
j:
*
»
Ans.
a;
when
a'
=
sec 3 x
2
fx
7T
+ x\-\
:
2
V
Ans.
3.
-
1.
0.
when x = 0.
'
Ans.
^±ns.
6*
a;
sec o x
when x ~2*
Ans.
when x
= 1.
Ans.
when
6
~2*
when
when
5
3'
3
J
tan
8.
(cosec0)
9.
(tan0) cos<
>
/ 2 sec 2 6
(9
- Il\V
10
=
^.
~2"
7T
when
=
2x tan
2x?
„
== 0.
=
2
2 X tan
6.
a;
or
2.
5
when
log(lH-aj),
1.
Ans.
vz
0: "2"
Ans.
1.
0: _
"4"
Ans.
when x = 0.
Ans.
= e.
Ans.
tan2 '
c"\i
7T
7i'
7r
4
e
3-
abc.
i
12.
(log a>>--,
13.
(a
«*
/ cos
when
a;
i
\
z
+ z)x,
ax
4- cos
2
when x =
0.
Ans.
when a; = = 0.
Ans.
=
bx\
J
xZ
ae.
1
CHAPTER X
MAXIMA AND MINIMA OF FUNCTIONS OF ONE
INDEPENDENT VARIABLE
101. Definition. A maximum value of a function is a value greater
than those immediately preceding or immediately following.
A minimum value of a function is a value less than those immediately preceding or immediately following.
If the function is represented
represents a
minimum
102.
maximum
by the curve y=f(x), then
value of y or of /(»), and
QN
PM
represents a
value.
Maximum
Conditions for a
It is evident that at
both
or a
P and
Q
Minimum.
the tangent
is
parallel to
OX,
and therefore we have for both maxima and minima,
clx
Moreover, as we move along
the curve from left to right,
at
P
the slope changes from
positive to negative
;
but at Q,
from negative to positive.
In other words,
At
P the
slope decreases as
...
x increases.
At Q the
x increases
By
when
.
(a)
slope increases as
Art. 21
(b)
we have
the case (a)
d
(slope)
dx
114
<
0.
•
(1)
MAXIMA AND MINIMA OF FUNCTIONS
*
But
Iv
dx
and
(1)
(slope)J
*
L15
%
= ^f^\
=
dx\dxj lW
—\ <
becomes
0.
efar
^=
Hence when
and
0,
dx
there
By
is
a
maximum
value of
similar reasoning
0,
dx2
.
(2)
y.
we have
the case
$y >
—
\
when
(6),
0.
dx2
S=
Hence when
and
0,
is
a mi id mum value of
For example,
let
^>
w
0,
(3)
dar
r/.v
there
^<
?/.
us find the
the function
maximum and minimum
values of
«
Put
|_2o; + 3^ + l.
2
2/==
^ = a*-4a?^&
Then
(4)
da;
fH*- 4
Putting
(
4)
= 0,
s?
•
-
•
•
•
- 4 + 3 = 0,
a;
whence
Substituting those values of
when a =1,
a;
in (5),
we
= _2<0;
find
<5 >
DIFFERENTIAL CALCULUS
116
Hence by
(2)
and
(3),
when x = 1,
y has a
maximum
value
when x = 3,
y has a
minimum
value.
From
the given function
that the
maximum
we
value of y
and the minimum value of
103.
^=
0,
In exceptional cases
makes
^
find
y,
it
== 0, so that
is
= 2J,
y = 1.
y
may happen that
a value of x given
neither (2) nor (3), Art. 102,
by
is satisfied.
dxdx
This would be the case for a
point of inflection
B
158) whose tangent
to
OX.
is
neither a
(see Art.
parallel
is
Here the ordinate
maximum
RL
nor a
minimum.
But there may be a maximum
or
minimum
when —4=0.
2
value of
This
is
dx
considered
in
Art.
following article
is
y,
even
more fully
J
106.
The
also appli-
cable to such cases.
104. Second Method of determining Maxima and Minima.
Maxima
—
and minima may be determined from the first derivative
alone.
dx
d2
—
without using
%.
dx2
We have seen in Art. 102 that when y is a maximum, as at
P, the slope, that
is,
—
,
changes from
ax
mum,
as at Q, -& changes
dx
from
—
to
-f-
+
pass along the curve from left to right.)
.
to
—
;
and when y
(It is
is
a mini-
understood that
we
.
MAXIMA AND MINIMA OF FUNCTIONS
By examining
the form of
—
which should be expressed
,
form,
we may determine whether
—
+
it
changes from
any assigned value of x.
Let us apply this method to the example
to
,
^ = X - 4a? + 3 =
-
s
dx
,
or
from
in Art. 102,
1)(*
-
3).
— can change
sign only when x = 1 or x =
°
°
3.
J
dx
By
—
-f to
in factor
for
c
Here
117
supposing x to change from a value slightly
greater than
we
1,
find that (x
— 1)
less, to
—
changes from
to
one slightly
+
;
but since
the factor (x — 3) is then negative, it follows that — changes from
dx
+ to — when x = 1, and denotes a maximum. In the same way, we
find that — changes from — to +, when x = 3, and denotes a mini,
dx
mum.
Again, consider the function y
=
(x
— 4) (# + 2)
4
5
.
Differentiating and writing the result in factor form,
c
Il
dx
= 3 (3 x -
2)(x
"When x =-.
3
since
(a;
As
this
.
changes from
—
When
=
-^ does woi change
dx
4
4,
2
+
+ to —
-2 changes from
dx
x
to
dx
= — 2,
— 4)
=—
3
"When x
•
sign,
cannot be negative.
Hence we conclude that y
when x
-*-
- 4:)\x + 2)
;
but neither a
is
a
minimum when #
=
2
-;
a
maximun
o
maximum
nor
method does not require —%,
2
minimum when x
it
is
= 4.
preferable to thai
dx
of Art. 102,
when the second
differentiation of y involves
much work
DIFFERENTIAL CALCULUS
118
EXAMPLES
1.
Find the maximum value of
Find the
32
—x
a;
maximum and minimum values
4
Ans.
.
of the following functions.
2.
2X
— 3 x — 12 x + 12.
Ans.
x
3.
2 x3
— 11
^4ns.
a?
2
= -,
gives a
x
— 3,
gives a
4.
s
«3
2
+9
(a
2
ic
+ 12 x + 10.
— #)
3
= —l, gives a maximum 19.
x = 2, gives a minimum — 8.
x
^tfts.
.
#
= 3-—a
maximum
_._
gives a
,
1
".-.
.
.
gives a
minimum
=2
•
= gives a
,
1
-\
V3
2 (3
7.
Show
8.
^_
a
+ 2) — 3 #
2
6.
a?
that
4
-2
maximum
^-^—^t
x—
^
~
= 2,
aj
gives a
maximum
has no
maximum
80.
nor minimum.
+ _^_ )W herea > 6.
~x
a
x
a2
=
a
x
Show
+b
,
a2
=
a
,
a
gives a
that the greatest value of
.
minimum
(a-b)1-2
^
a
—no*^—
n
/
1
Show
—— '—
maximum ^(a4-bY
.
—b
that the greatest value of
gives a
y*
a;
10.
minimum
1
Ans.
9.
a
3V3
-4ns.
.
.
°
gives
,
9 a3
3V3"
•
=
'
16
V3
x
9 a3
:
4
—-,
= 3a
= 2o
a>
1.
„
maximum
(x-l)(x-2)(x-3).
A
-4ws.
13-J-f-
minimum
4
5.
48.
cos 2
is
—
ne
+-
sin
is
9
8
-.
,
;
.
MAXIMA AND MINIMA OF FUNCTIONS
Show
11.
maximum and minimum
that the
sin 2 $
+
sin 2
f
,
—
Find the maximum value of
13.
Find the
maximum value of
values of
- and
are
)
2
/
12.
taken in the
6
3
a sin x
tan
-1
x
+
b cos x.
— tan -1 ~,
quadrant.
first
119
2
Ans.
Va + b
2
2
.
the angles being
.
,
_j3
4'
Show that the least value
14.
that of
15.
16.
ere**
y
J
=
^
a
a
+ &V",
~x
-2x
)
a tan
2
2
-f b cot
2
is
the same as
to 2 ab.
u4?is.
.
A minimum when x = -
4
= (.r-l)%r + 2)
Ans. A maximum
3
2/
.
neither
17.
of
and equal
2
7/=0-2)
J./<s.
5
(2z
when a =
when
+ l)
a;
;
a
minimum when x = 1
= — 2.
4
A maximum
.
when x
=—-
;
a
minimum when x =
2
neither
105. Case where -^
when x = 2.
= oo
dx
sign by passing through
Hence
—
lo
.
It is to be noticed that
-^
dx
may change
infinity instead of zero.
^=oo,
if
dx
for a finite value of
# this value should be examined, as well as
;
those given by
''"
dx
=0.
DIFFERENTIAL CALCULUS
120
For example, suppose
y
= a — b(x — c)
dy_
Then
dx
25
3(x-cy
hence we have
dy
dx
oo
2
3
.
Y
,
when x = c.
It is evideDt that
when x
dx
—
changes from
4- to
maximum
y, which is a.
shows the maximum
The
indicating a
value of
figure
ordinate
,
PM,
corresponding
to
'
a
cusp at P.
On
the other hand, suppose
dy
Then
dx
But
dy
dx
y
b
= a — b(x — c)"
3".
= oo
as -^ does not change sign
mum nor minimum.
Y
when x = c.
3(«-c) 8
when x = c,
The corresponding curve
is
there
shown
is
no maxi-
in the figure.
MAXIMA AND MINIMA OF FUNCTIONS
121
EXAMPLES
Find the maximum and minimum values of the two following
functions:
y=(x + l)i{*-oy-
1.
Ans.
a
2.
y
= 5;
x = — 1.
A minimum when
minimum when
x
a
maximum when
a;
=
;
= (2x-ay(x-a)*'
Ans.
106.
A maximum
Conditions for
when x =
—
;
a
minimum when x =
a.
Maxima and Minima by Taylor's Theorem.
to be a maximum when x = a.
Then, by
Suppose the function /(.r)
the definition in Art. 101,
/(o)>/(a + ft),
f(a) >f(a
and also
where h
is
any small but
— h),
finite quantity.
of a for x in Taylor's Theorem,
Now, by the
substitution
we have
S
/(a + A)-/(a)=
Rf(a) + ^/"(a)
/"'(a) + -.
+i
[»'
/(a-;0-/(a) = -7i/'(a) + f/"(a)-|V'» + -By
and
the hypothesis
also
+
/(a
— h)-f(a) < 0.
Hence the second members
*
The rigorous form
context.
of Art. 93
of both (1)
may
•
(2)
— /(a) < 0,
/(a
/*)
(1)*
•
and
(2)
must be negative.
be used here without any change in the
DIFFERENTIAL CALCULUS
122
By
taking h sufficiently small, the
cally greater than the
sum
Thus the sign of the
term.
As these have
entire second
first
term can be made numeri-
of all the others, involving h 2 , h3
member
different signs in (1)
,
etc.
will be that of the first
and
(2),
the second
mem-
bers cannot both be negative unless
/(a) = 0.
Equations
(1)
and
f(a
(2)
then become
+ A) -/(a.) = | /"(a) + |/"'(a) +
...»
/(a-A)-/(a)=|/"(a)-|/''(a)+....
The term containing h 2 now determines the sign of the second
members. That these may be negative, we must have
f"(a)<6.
/'(a) =
If then
f(a)
is
a
and
/"(a)<0,
maximum.
Similarly,
it
may
be shown that
/'(a)
if
=
and
/"(a)>0,
=
and
/"(a) =
f(a) will be a minimum.
/'(a)
If
similar reasoning will
show that
for a
0,
maximum we must
also have
iv
f'"(a)=0
and
f (a)<0;
/"'(a)=0
and
f
and for a minimum
The
conditions
may
lv
(a)>0.
be generalized as follows
Suppose that
f(a) = 0,
and that fn+1 (a)
Then if n
If
n
is
is
is
/"(a)
= 0,
f"'(a)
= 0,
...
f(a) = 0,
not zero.
even, /(a)
is
odd, f(a) will be a
maximum nor a minimum.
maximum or a minimum, according as
neither a
/M+1 (a)<0
or
>0.
MAXIMA AND MINIMA OF FUNCTIONS
PROBLEMS
123
MAXIMA AND MINIMA
IN
Divide 10 into two such parts that the product of the square
and the cube of the other may be the greatest possible.
1.
of one
Let x and 10
mum.
— x be the parts. Then
= x (10 — x) we find
2
Letting u
:>.i'(4-.t)(10-.t) 2
from which we find that u
required parts are 4 and 6.
A
— x)
3
is
to be a maxi-
,
du
dx
2.
x 2 (10
3
maximum when
a
is
= 0,
square piece of pasteboard whose side
Find the side of
square cut out at each corner.
x
= 4.
is
Hence the
a has a small
this square that the
remainder may form a box of maximum contents.
Let x = the side of the small square. Then the contents of the
box will be (a — 2 x) 2 x. Eepresenting this by u, we find that u is a
maximum when x=
which
-,
the required answer.
is
3. Find the greatest right cylinder that can be inscribed in a
given right cone.
AD = a, DC=b.
Let
Let x
= DQ,
the radius of the base of the cylinder, and y
= PQ,
its altitude.
From the
we find
ADC, PQC,
similar triangles
b-x
The volume
.r).
y=i«
y
b
of the cylinder is
= 7r-X
2
7r:r?/
(6
— x).
a
This will be a
is
a
maximum when u^bxP—x3
maximum.
This
is
found
to be
when
x
=|
b,
the radius of the base of the
required cylinder.
From
this,
y
= -,
o
the altitude of the cylinder.
DIFFERENTIAL CALCULUS
124
Determine the right cylinder of the greatest convex surface
4.
that can be inscribed in a given sphere.
Let r — OP, the radius of the sphere x = OR, the radius of base
and y = PR, one half its altitude.
;
of cylinder
From
;
the right triangle
OPR
x2 -jg y2 =r2
we have
.
i
The convex
2
surface of the cylinder
ttx
•
2y
= 4 7rx Vr — x
2
is
2
.
We may put u equal to this expression,
and determine the value of x that gives a
maximum value of u. But the work may
be shortened by the following considerations
:
4
irx
Vr — x
2
when
x V?*2
and
x-y/r
when
its
square
2
i^x
Hence we may put u =
maximum, when x
2
—#
2
2
—x
—x
2
2
r^x
is
a
maximum,
is
a
maximum
a
maximum,
is
4
a
is
—x
maximum.*
from which we find u
k
,
is
a
=
V2
From
this
y
= —-,
giving for the altitude of the cylinder,
V2
2?/
Another Method.
The convex
= rV2.
The equations
surface
= 4 irxy,
2
x
may
u
+y =
2
be used without substituting in
* Since
we
= xy,
(1)
i*,
(2)
(1)
the value of y from
are only concerned with the positive root of
(2).
Vr2 — x 2
.
.
MAXIMA AND MINIMA OF FUNCTIONS
and differentiating
Substituting in
%
^L = y
dx
Differentiating (1),
(2).
x
+ X dx
dx
—
=y
dx
y
=*
y
y
Since x and y are positive quantities,
x=y,
—
dx
changes from
Combining
-f to
x
= y,
x
= —=
V2
—
,
giving a
with* (2),
,
y
= —V2
(3)
*= _5
c
+ y ^=0,
dx
(3),
125
it
is
when
evident that
maximum
value of
u.
we have
,
as before.
In some problems this method has some advantages over the
first.
5. Divide 48 into two parts, such that the sum of the square cd
one and the cube of the other may be a minimum.
Ans. 42f, 5-|.
6.
Divide 20 into two parts, such that the sum of four times the
and nine times the reciprocal of the other may be
reciprocal of one
a
minimum.
7.
A
Ans.
rectangular sheet of tin 15 inches long and 8 inches wide
has a square cut out at each corner.
so that the remainder
may form
Find the side of
a box of
maximum
this square
contents.
Ans.
8.
8, 12.
How
1J
in.
from the wall of a house must a man, whose eye is
window 5 feet high, whose
9 feet from the ground, may subtend the greatest angle ?
far
5 feet from the ground, stand, so that a
sill is
Ans.
9.
is
A
wall 27 feet high
is
8 feet from the side of a house.
6
ft.
What
the length of the shortest ladder from the ground over the wall
to the house ?
Ans.
13
V 13= 46.87
ft.
DIFFERENTIAL CALCULUS
126
10.
A
person being in a boat 5 miles from the nearest point of the
beach, wishes to reach in the shortest time a place 5 miles from that
point along the shore; supposing he can run 6 miles an hour, but.
row only
at the rate of
4 miles an hour, required the place he must
land.
Ans. 929.1 yards from the place to be reached.
11.
Find the maximum rectangle that can be inscribed in an
whose semiaxes are a and b.
Ans. The sides are aV2 and &V2; the area, 2ab.
ellipse
12.
A
rectangular box, open at the top, with a square base,
What must
be constructed to contain 500 cubic inches.
dimensions to require the least mate'rial
Ans.
13.
A
is
to
its
?
Altitude, 5 in
cylindrical tin tomato can
a given capacity.
is
be
;
side of base, 10 in.
made which
to be
Find what should be the
shall
have
ratio of the height to
the diameter of the base that the smallest amount of tin shall be
Ans.
required.
14.
drical
What
Height
= diameter.
are the most economical proportions for an open cylin-
water tank,
if
the cost of the sides per square foot
is
two
thirds the cost of the bottom per square foot ?
Ans. Height
=f
diameter.
15. (a) Find the altitude of the rectangle of greatest area that
can be inscribed in a circle whose radius is r.
Ans. rV2; a square.
(b)
Find the altitude of the right cylinder
can be inscribed in a sphere whose radius
of greatest
volume that
Ans.
is r.
—^«
V3
16.
(a)
Find the altitude of the
inscribed in a circle of radius
(b)
r.
isosceles triangle of greatest area
Ans.
Find the altitude of the right cone
in a sphere of radius
r,
—
;
equilateral triangle.
of greatest
volume inscribed
Ans.
—
•
MAXIMA AND MINIMA OF FUNCTIONS
17.
Find the altitude of the
(a)
isosceles triangle of least area
circumscribed about a circle of radius
r.
Arts.
3r; equilateral triangle.
Find the altitude of the right cone of
(b)
scribed about a sphere of radius
A
18.
least
volume circumAns. 4
r.
maximum volume
right cone of
127
is
r.
inscribed in a given right
cone, the vertex of one being at the center of the base of the other.
Show
that the altitude of the inscribed cone
is
one third the altitude
of the other.
19.
Find the point of the
line,
of the squares of its distances
2x + y = 16,
from
(4,
such that the
sum
— 3) may
be a
and
5)
(6,
minimum.
20.
-
a
Find the perpendicular distance from the origin
-f -
= 1,
b
'
i
21.
Ans.
A
bv
& the minimum distance.
J rinding
what
is
2
Another
hour on a
will they be nearest together, and
When
is
sailing 15 miles per
A
In 7 hrs.
Distance 15 Vo7
= 113.25
mi.
vessel is anchored 3 miles off the shore.
5 miles farther along the shore, another vessel
from the shore. A boat from the first vessel
on the shore and then proceed to the other
is
Opposite a point
anchored 9 miles
to land a passenger
is
What
vessel.
The
velocity of
the
is
Ans. 13 mi.
shortest course of the boat ?
23.
Va + &*
their least distance apart ?
Ans.
22.
'
due north 10 miles per hour.
vessel is sailing
vessel 190 miles north of the first
course East 30° South.
to the line
Ans.
'
(7, 2).
waves of length X in deep water
is
propor-
7
tional to
\
*
the velocity
24.
a
+ -,
is
where a is a certain linear magnitude.
X
a minimum when A = a.
Assuming that the current
Show
C=
in a voltaic cell is
v
being the electromotive force, r the internal, and
resistance; and that the
is
a
maximum when
R=
power given out
r.
is
R
P=RC
-j-
R
that
,
E
the external,
2
\
show that
P
DIFFERENTIAL CALCULUS
128
From a
25.
given circular sheet of metal, to cut out a sector, so
that the remainder
may form
Angle
Arts.
maximum
a conical vessel of
of sector
=
(
_
1
capacity.
^-)2 ir = 66°
14'.
3,
Find the height of a
26.
on a wall so as best to illuminate
assuming that the illumiinversely as the square of the distance from the light, and
light
a point on the floor a feet from the wall
nation
is
;
directly as the sine of the inclination of the rays to the floor.
Ans.
V2
At what point on the
27.
must a
line joining the centres of
amount
light be placed, to illuminate the largest
two spheres
of spherical
surface ?
Ans.
The
centres being A,
AP :PB =a?:b
2
2
point;
28.
(a)
The strength
B
;
the radii,
a, b
;
and
P the required
s
.
of a rectangular
beam
varies as the breadth
and the square of the depth. Find the dimensions of the strongest
beam that can be cut from a cylindrical log whose diameter is 2 a.
Ans. Breadth
The
beam
=
—
V3
Depth = 2a A /?.
.
\3
and
Find the dimensions of the stiffest beam
that can be cut from the log.
Ans. Breadth = a. Depth = a VS.
(6)
stiffness of a rectangular
varies as the breadth
the cube of the depth.
29.
The work
of propelling a steamer through the water varies
as the cube of her speed.
Find the most economical speed against
a current running 4 miles per hour.
The
A?is.
6 mi. per hr.
consumed in propelling a steamer through
is $ 25 per hour when
the speed is 10 miles per hour. The other expenses are $ 100 per
hour. Find the most economical speed.
30.
cost of fuel
the water varies as the cube of her speed, and
Ans. -^2000
31.
is to
A
minimum.
mi. per hr.
hanging 2 feet from one end of a lever
Supposing the
per fpot, find its length that the force may be
Ans. 20 ft.
weight of 1000
lbs.
be raised by an upward force at the other end.
lever to weigh 10 lbs.
a
= 12.6
MAXIMA AND MINIMA OF FUNCTIONS
32.
(a)
The lower corner
of a leaf,
whose width
over so as just to reach the inner edge of the page.
of the part folded over,
when the length
of the crease
is
129
a, is
folded
Find the width
is
a minimum.
Ans.
(b)
J a.
In the preceding example, find when the area of the triangle
Ans. When the width folded is -| a.
is a minimum.
folded over
33. A steel girder 25 feet long is moved on rollers along a passageway 12.8 feet wide, and into a corridor at right angles to the
passageway. ISTeglecting the horizontal width of the girder, how
wide must the corridor be, in order that the girder may go around
Ans.
the corner ?
34.
Find the altitude of the
5.4
ft.
least isosceles triangle that can
be
circumscribed about an ellipse whose semiaxes are a and
of the triangle being parallel to the
35.
b,
A
tangent
is
drawn
major
to the ellipse
axis.
its
length
is
a -f-
b.
the base
Ans. 3
b.
whose semiaxes are a and
is a minimum.
Show
such that the part intercepted by the axes
that
b,
CHAPTER XI
PARTIAL DIFFERENTIATION
107. Functions of Two or More Independent Variables. In the preceding chapters differentiation has been applied only to functions of
one independent variable.
than one variable.
We
shall
now
consider functions of more
u=f(x,y)
Let
be a function of the two independent variables x and
y.
Since x and
we may suppose x to
vary while y remains constant, or y to vary while x remains conWe
stant or we may suppose x and y to vary simultaneously.
must distinguish between the changes in u resulting from these dif-
y are independent of each other,
;
ferent suppositions.
Let
only,
A x u denote the increment in u resulting from a change
and A y u the increment in u from a change in y only.
Let Aw, called the
increment of
u,
total
be the
in-
N
r
crement when x and y both
C
Suppose u the area of a
rectangle whose sides are x
and y.
Then
u = xy.
from
OA
to
y
u
OA', while y remains conu is increased by the rectangle
stant,
That
r
& 3u
Ay
change.
If x changes
is,
in x
AM.
Ax u = area AM.
130
u
A^u
x
Ax
'
:
PARTIAL DIFFERENTIATION
OB
y changes from
If
increased by the rectangle
That
If x
of
131
to OB', while x remains constant,
u
is
BX.
\u = area BN.
is,
and y both change together, we have
A« = area
u,
for the total increment
AM+ area 5A"+ area MN*
108. Partial Differentiation. This supposes only one of the independent variables to vary at the same time, so that the differentiation is performed by the same rules that have been applied to
functions of a single variable.
we
If
we
constant,
If
we
obtain
—
dx
differentiate,
du
obtain —
= f(x,
u
differentiate
y),
supposing x to vary, y remaining
•
supposing y to vary, x remaining constant,
we
•
dy
The
derivatives,
— —
dy
,
,
thus derived, are called partial deriva-
dx
tives,
and a special notatioD,
For example,
u
if
—
—
ox
dy
,
is
= x + 2 x?y — y3
-f
— = 2x
— 3y
2
used for them.
3
—
=3x
dx
2
,
4 xy,
2
,
,
the ^derivative of u.
the ^/-derivative of
u.
In general, whatever the number of independent variables, the
by supposing only one to vary at a
partial derivatives are obtained
time.
EXAMPLES
Derive by partial differentiation the following results
du
xy
1.
•'•
2.
.
du
V.
2
+y
= (ax + 2
2
bxy
+ off,
(to
+„)*_(«.+«„)*
DIFFERENTIAL CALCULUS
132
o
-
5.
«
6.
r
= AVx
/ T +#>
m
%
2
vt
-
t
.
ii
x
e
=
l
x
12.
+
2
3
c?/ ),
0g
,
1
3-
du
du
+2/
oy
ox
+2 du = A0.
du
,
dz
.
,
w
2 2)i
y
sin
vw
-
V^y
1
(^
+ r+^-)
a5logaj-^+ylogy-^ =
da;
2
0.
d?/
a?,
2+
(tan x
)
l
gY+ gl]f+ glY,
J,
+
.
a;
a?/
^
=logy aj4-log x y,
= log
—=
dy
= °+ *¥+
*£
oy
ox
x~ + y
2/
2/
u= e z sin y + e
u
y
(
^)
13.
dy
\-
ofl?
2/
^
1 w
^ + ^ = + 2/- iN
£^l + 2tan-^,
+
dx
du
'
2
du
,
a-
,
y
4-
.
dr
—
—
= x dr
dx
y
+e
du
.
a
xy
..
2
foci/
.
,
(^
11.
2
x
= £+-z + -,
a;
10
w
y
2
?/
*=
M
?/
3
e
.
,
= log (x + ace +
z
8
du
\
.
4.
7
\/
\/
/
= e2x + e2 + 2 e +
*
(fr)*
"
*
sin
(*+*)
+ tan ^ + tan z),
—
o du
sm 2x
+
'
da;
,
o
sin 2w
du
.
f-
a?/
sm o2z
.
du
—
=2.
dz
PARTIAL DIFFERENTIATION
Geometrical Illustration of Partial Derivatives.
109.
133
Let
z
= f(x, y)
be the equation of the surface
APCIL
The ordinate
PN
is
thus
given for every point A" in the
plane
Let
XT.
APB
and
CPD
tions of the surface
be
sec-
by planes
to XZ and
through P. parallel
YZ. respectively.
If x and y both vary,
moves
to
P
some other position
on the surface.
y
remaining conmoves on the curve of intersection
If x vary, y
stant,
P
Hence
—
dx
is
If y vary, x remaining constant,
Hence
—
APB
the slope of the curve
at P.
P moves
the slope of the curve
is
CPD
APB.
on the curve CPD.
at P.
110. Equation of Tangent Plane.
Angles with Coordinate Planes.
In the figure of the preceding article, let P be the point (a?', y', z')
PT, the tangent to APB in the plane APNM-, and PT, the tangent
;
to
CPD
in the plane
It is evident
CPNL.
from the preceding
article that the equations of
PT
are
,
6V
-,(»-«'),
dx
and of
= y',
y
(1)*
PT,
*"*'* #<*-&>
.
ax'
^dy'
denote the values of &-,
dx
X
$?
dy
,
=X
'-
respectively, for (x', y' z').
(2)
DIFFERENTIAL CALCULUS
134
The plane tangent
PT and PT'.
z
For
to the surface at
- z,==d
6
(3) is of the first
x, y, z,
and
the tangent lines
- x )+
'
W'
{y
~ y,)
(3)
degree with respect to the current variables
by
is satisfied
(x
and
(1),
also
by
(2).
P
are those of a line through
of the normal through
perpendicular to (3). Its equations are
TJie equations
(x', y', z')
P contains
Its equation is
—
—
d^_
dz^_
dx'
dy'
x'y
-jn= ?-^-
x
The angles made by
i/'
the tangent
...
,
=
K
-(z-z')
'
y
(4)
v J
plane with the coordinate planes are
equal to the inclinations of the normal to the coordinate axes.
By
analytic geometry of three dimensions, the direction cosines of
the line perpendicular to (3) are proportional to
dj_
dz[
dx' '
dy'
_1
?
Hence, if a, {3,
OZ, respectively,
y,
are the inclinations of the normal to
cos a
cos 2 a
Also
From
(5)
and
(6)
/3
dz;
w
dx
dy'
+ cos
we
_ cos _ cos y
+ cos
2
/3
find,
OX, OY,
2
y
=1
dropping the accents,
dz
dx
(5)
(6)
PARTIAL DIFFERENTIATION
135
-(DM!)
For the inclination of the tangent plane
XY, we have from
to
-"- i+ (D* + (IJ
(7),
<
8»
The term slope used in geometry of two dimensions may thus be
extended to three dimensions, as the tangent of the angle made by
In this sense,
the tangent plane with the plane
XY
=v(IJ + (SJEXAMPLES
Find the equations of the tangent plane and normal,
sphere x 2 — y 2 + z- = a2 at (V, y,' z').
to the
1.
,
dz
_
dx
„
Hence
Substituting in
z
dz
z'
By
—=
x'
dx
z'
dz'
(3),
x
"''-
xx'
y
(x-x')
t
y
z
—
=—
Oy
dz'
-,
- z' = -
_
y'
-
-;
•
•
z'
-£
(y
- y'),
+ yy' + zz' — x' + y'
2
2
-j- z'
2
=a
2
,
vlws.
DIFFERENTIAL CALCULUS
136
From
we
(4)
find for the normal
(y-y')- = ^-^
^-^-,=
x
y
— — 1 = ^ — 1 = - — 1,
y'
x'
2.
- = 2- = -.
x
z'
Find the equations of tangent plane and normal
3x
2
-y + 2z = 0,
2
2
at
y\
(x',
Find the equation of the tangent plane
z= 3 x2 + 2
2
,
x
at the point
Find the
5.
of tangent plane
2
and normal
to the ellipsoid,
2
3y = 4;Z + 2.
slope of this tangent plane.
Ans.
Find the equation of the tangent plane
x2 + y 2 + z 2 -2x + 2y =
Ans.
111.
2z'
6x+8y — 2 = 11.
+ 2y + 3z =20,
x = 3, y = 2, z being positive.
Ans. 3x + ±y + 3z = 20; x = z + 2,
2
icaj'-f-
-§.
to the sphere,
l,
at
yy'
+ zz' — x — x' + y + y' =zl.
(x', y', z').
Partial Derivatives of Higher Orders.
By
successive differ-
entiation, the independent variables varying only one at a time,
may
'
to the elliptic parabo-
at the point (1, 2, 11).
Find the equations
—y'
a;'
Ans.
4.
to the cone,
z').
3
loid,
Ans.
z'
^-^ = y-^- = z ~ z
3xx'-yy'+2zz' = 0,
Ans.
3.
y'
we
obtain
d 2u
dx
If
we
differentiate
to y,
we
obtain
—
2'
d 2u
dy 2
dhc
dhc
dx*
dy
'
u with respect
I -^ \ which
dy\dxj
is
4>
to x, then this result with respect
written
—-—
-.
dydx
PARTIAL DIFFERENTIATION
Similarly,
J
—
is
the result of three successive differentiations,
dyd.r
two with respect
that this result
and one with respect to y. It will now be shown
independent of the order of these differentiations.
to x,
is
n
*\
In other words, the operations
— and —
dx
.
That
137
are commutative.
dy
is,
dydx
112. Given
dydx 2
dxdy
(1)
«fe)"Sfe>
Supposing x to change in
(1),
y being constant,
Au = f(x +
Ax
supposing y to change in
AVAtA = f( x +
dx*dy
u=f(x,y),
to prove that
Now
dxdydx
As, y
As, y) -f(x, y)
(2)
^
"
Ax
x being constant,
(2),
+ Ay) -f(x, y + Ay) -/(a? + As, y) +f(x,
gfl
Ax
Ay
Ay\Ax)
Reversing the above order, we find
A»
Ay
= f(x, y + Ay) -f(x, y)
Ay
A (*^\ = f( x + Ax y + A
>
y)
-f( x + Aa:
Ax\Ay)
Hence
—
= f'(x+0>Ax),
As
y)
— f(x
>
y
+ A ^) +
f fa y)
.
Ax Ay
±/A«\
A/Aii\
Ay\AxJ
Ax\Ay)
The mean value theorem,
form
>
and
(2),
Art. 90,
(3)J
may be
where u =f(x).
expressed in the
0< 6< 1.
DIFFERENTIAL CALCULUS
138
In the present
ty{lx)
Similarly,
and
By
=
where u =f(x,
case,
iy
fx {X
+ &1
AX
'
V)
+ ^' AX
=fyx(X
'
V
+ $rAy)
'
x ,y + B^Ay),
^=f(
Ay
= f* y<KX + 6i
Ax\^r)
(3)
'
y) f
/^(aj+^.Ax, y
Taking the limits as
'
Ax> V
+ ° '^'
+ Or Ay) =f
Aaj,
xy (x
approach
A?/,
3
+ 6fAx, y +
8 .Ay).
and assuming the
zero,
functions involved to be continuous,
fyx(%,y)=fxy(%,y)mi
That
,
•
is,
d fdu\
-d fdu\
— =-—
ay\axj
2
2
ay ax
ax ay
d u
d u
_—_
= _-—
or
dx\&yj
This principle, that the order of differentiation
be extended to any number of differentiations.
Thus
d 3u
dydx
d
=
2
2
=
fdu\
d
f
d 2u \
dx\dydxj
It is evident that the
same
is
fdu\
d
f
d 3u
dx 2 dy
true of functions of three or
variables.
dx
may
dxdydx
d 2u \
dx\dxdyj
immaterial,
&u
=
dxdy\dxj
dydx\dx)
=
d2
is
ay
more
PARTIAL DIFFERENTIATION
139
EXAMPLES
=
A erlfy
dx dy
1.
u
——
in Lxs. l-o.
dy dx
= ax +
ay + bx
l) /
!
w = fBylog?.
2.
,
Derive the following results
4.
s.
6.
u
7.
= a.x* -f 6 Saf^ +
8
dx2 dy 2
,
,
,
n
g_^_«g.
n
*d u
dhi
ar—
-f 2a^-—- +
2
>
d 2q
a
= log
u
11.
i^ztan
(e
*
1
u
dar
*
io
13.
u
d 2ii
2
2
"
.
:
2
999,00').
999
- y-z-e+ z-.cre- + afyV,
u
= sin
(jj
+ z) sin (z +
2
1 d q
.
a;)
dy z
A
r2 d#-
5?'
d 2u
t-=2
dy
d 2u
+ dz
,
*
1%,
U
IV
?
^
14.
2
2/
3 «.
+ dy + dz- = o.
—
+
dx
/".>,
^
= log (ar
-f y + z ),
i
r
-,
?/
10
12.
1 dq
,
2
= 2e^+t4V
dxdydz
+ e" + <*)
—
odhi
.
oxdy
-^ = ££
d>
10.
c,
,
dar
„\
,
d*u
find
y
,.
y) e*-».
dx dy 2 dx
dx dy dx dy
find
= ,= tan-' 2 - r tan- 5,
/
+
(a;
g+g=°-
2
1/
8.
d a
,
x
= - + log-,
a*
=
4
d u
4
c?/
,=(3. + ,)3 + sin(2,-,),
tt
?*
:
4
.=iog(*» +!o,
-
3.
y
^
1
2
x2 -\-y 2 -\-z 2
= e-0,9,0
+ er + e
V
*
2
.
sin (x -f y),
2cos(2a;
dxdydz
+ 2?/ + 2z).
DIFFERENTIAL CALCULUS
140
113.
Total
This change
is
In Art. 107
Total Differential.
Derivative.
referred to the change in u
when x and y vary
Thus the
called the total increment of u.
mentof
we have
simultaneously.
total incre-
u=f(x,y)
= f(x + Ax, y + Ay)~ f{x, y).
Au
is
The terms
total derivative
and
example,
u
let
and suppose x and y
total differential are also used.
= x y-3x y
s
dt
2
(1)
,
to be functions of a variable
Differentiating with respect to
dt
2
y)
K
dt
= tfty. + 3^2*
dt
For
t.
t,
J)
K
dx
dt
_ 6x2J
dy
_6
dt
dx
2
y
dt
= (Zx y-$xtf)f +(x*-(Sx y) dX
2
....
2
(2)
t
But from
(1)
we
find
—
= 3 x y — 6 xy
ox
Cj7/
So that
(2)
may
2
— = x — 6x
fi?f
2
z
,
dy
2
y.
be written
_du dx du dy
dt~dxdtdydt'
du
If
we had used
,q\
r
differentials in differentiating (1)
^ '
we should have
obtained
du
= — dx H
dx
dn
—
in
(2)
and
(3) is called
the
dy
w
dy
J
total derivative,
(4)
and du
in (4) the total
aii
differential, of u.
We
x and
proceed to show that (3) and (4) are true for any function of
y.
PARTIAL DIFFERENTIATION
Noticing that
and A x u, A
u,
An
is
the total increment of
the partial increments,
L/
let
141
u,
when x and y vary
= f(x, y),
x and y
u'=r\x + Ax,y),
u"=f(x + Ax,y + Ay).
u
separately,
being functions of
t.
= u' —
A u' = w" —
Ait = u" — m.
Axm
Then
it,
m',
y
Am
Hence
r
x
Am
and
At
Taking the
approach
= A m + A,.M
Axu A.U A„w'
_
~~
clu
Am
Ay At
Aa A£
limits of each
zero,
,
member,
as At,
= dudx
du dy
dxdt
dy dt'
dt
"
and consequently Ax, Ay,
(5)
K
}
since the limit of u' is u.
This
may
be written in the differential form
du
=—
+—
dx
dy
dx
In the same way,
if
dt
and
ri„
u =f(x,
~
y, z),
dx dt
dy
—
that
is,
are functions of
x, y, z
^
dz dt'
dt
dy
t,
(8)
dz
write in (8)
du
ox
giving
where
(6)
= ^rl< + ^dy + ^dz
ox
We may
w
dy
J
,
tfaj
= d,
_
dy = d„w,
dy
cht
7
tt
du
,
= dxu + dvu
the total differential of m
is
du
—
,
the
4-
-
,
cZz
= d u,
1
z
dz
dz u,
sum
of its partial differentials.
DIFFERENTIAL CALCULUS
142
=du+d
This principle, as expressed by du
x
by the figure of Art. 107, from which
u,
may be illustrated
+ A u + area MN,
Aw == A w + A u Ax Ay.
Aw
that
=A
y
we have
is,
x
u
y
x
-+-
u
As Ax and Ay approach zero, the last term diminishes more
and we may write
rap-
idly than the others,
Aw
the
closeness
approach
of
x
+A
u
u,
y
approximately,
K
the approximation increasing
we suppose
t
Ax and Ay
== x,
u=f(x, y),
then
(5)
as
zero.
If in (5)
and
=A
du^du
becomes
Similarly, if in (7),
dx
ox
u =f(x,
y, z),
t
=.x
+
3udy_
By dx
;
(9)
}
y and
du
—=—
+ du dv
dy dx
dx
du
dx
,
whence
y being a function of x
.
z
being functions of x
du dz
,
-h
/iAN
(10)
'
-=
dx
dz
;
EXAMPLES
Find the
1.
total derivative of
u=f(x,y,z),
where
x
w by
=
(5) or (7) in the three following
2
t
y
}
=
z—-.
1?,
——2 —
dx
4- ^
t
t
w
= log (x ~ y
2
2
),
where
x
= a cos
t,
= a sin
y
2
— —- —
dy
dt
2.
t
2
dz'
t.
^ = - 2 tan 2
dt
3-
w = tan -1
-,
y
where
x
= 2t,
r
y
.r
=1—
:
2
t
.
—=
dt
—
l
+ t\
t.
:
:
.
:
PARTIAL DIFFERENTIATION
Apply
4.
u
two following
(10) to the
=/(*,
143
y, z),
where
y
= x — x,
2
z
= x? — x
2
.
h = ^+(2x-l)^ + (3x*-2x)^.
(
dx
o
.
u
= tan
-»
where
,
v
dx
y
= d— ar,
K
dy
2
=1—o
)
dz
flr,
1
da?
Find the
6.
u
total differential
= ax + 2 6«y + cy
2
by
(6) or (8) in
2
du
,
9
u
=
l
sin
i( x + y)
sin
i
rr
v.
du
— y)'
2
du
»
y
= sinydx-smxdy
cos — cos
a)
= ax + ty + cz + 2/#2 + 2
2
.
2
gr«aj
it
= af'
11.
?/
= tan
+ 2 fta#,
du
2
= a?*-1 (yz cZx + zx log
a:tair?/tairz,
du
a;
c?y
+ xy log
a;
dz)
= 4uf-^— + -^_ + __^_\
\sin2a;
sin2y
sin 2zy
t in (5) and
(7) denotes the time, we have the
between the rates of increase of the variables.
If the variable
For
illustration consider the following
12.
One
cond.
;
another side
The included angle
is
re-
example
side of a plane triangle is 8 feet long,
inches per second
At what
m
?/
=2 (ax + hy+gz) dx + 2(hx + by+fz) dy + 2(gx +Jy + cz)dz.
10.
lation
'
= 2 (ax + &?/) da; + 2 (6aj + c#) dy.
j
(a?
2
the following
v
8
+x
and increasing 4
5 feet, and decreasing 2 inches
is 00°,
and increasing
rate is the area of the triangle increasing?
2° per second.
DIFFERENTIAL CALCULUS
144
A=
The area
dA =
—
dt
,db
c
- sin^l
•
2
from which
-be sin A,
,
V-
b
-
sm^lA
•
2
dt
dc
.
be
cos
—
AA dA
2
dt
dt
= ^sin^.i + 5sin^.-i + ^cos^-^
32
2
= .4934
= 70.05
sq. ft.
2
6
90
sq. in. per sec.
13. One side of a rectangle is 10 inches long, and increasing uniformly 2 inches per second. The other side is 15 inches long, and
At what rate is the area
decreasing uniformly 1 inch per second.
increasing ?
Ans. 20 sq. in. per sec.
At what rate after the lapse of 2 seconds ?
Ans. 12 sq. in. per sec.
14.
The
altitude of a circular cone
is
100 inches, and decreasing
10 inches per second, and the radius of the base
increasing 5
At what
inches per second.
is
is
50 inches and
the volume in-
Ans. 15.15 cu.
creasing ?
15.
rate
In Ex. 12, at what rate
is
Ans. 8.63
of
per sec.
the side opposite the given angle
increasing ?
114. Differentiation
ft.
an Implicit Function.
may
derivative of an implicit function
in.
per sec.
(See Art. 66.)
The
be expressed in terms of
partial derivatives.
The equation connecting y and
one member,
may
by transposing
x,
all
the terms to
be represented by
+ (*,y)=o
From
=
u
Let
(9),
Art. 113,
we have
du
dx
cj>(x, y).
for the total derivative of u,
_ du
da dy
dx
dy dx
(i)
PARTIAL DIFFERENTIATION
But by
(1)
x and y must have such values that u
a constant
is,
;
and therefore
its total
dx
may
be zero, that
derivative— must be
dx
du dy _
=
By dx
du
Hence
145
zero.
0,
du
dy = _-_
_ dx
^
and
dx
For example,
find
—
(2)
du
dy
+ ^y
3
x?y 2
from
•om
dx
Let
u
J?
o.r
By
- 3 a^ + 2 a-r,
dy
(2)
=
cfa
115.
Extension
of
If
-+-
—a
xry3
f?
dy
2^ +
first
Taylor's
3a^?/
2
.
"
h,
y
*y.
derivatives in the examples of Art. 67
Theorem
to
Functions of
we apply Taylor's Theorem
regarding x as the only variable, we have
f(x+
3
3ay + 2y 2
2z 2 + 3a*/'
f(x+7i,y+k),
to
.
5
= 2 &y +
3.i-y+ 2ay» =
In the same way find the
pendent Variables.
= xPy-
= --a5
+ k) =f(x, y+k) + h~f(x,
y
+
k)
Two
Inde-
DIFFERENTIAL CALCULUS
146
Now
expanding f(x, y
f(x, y
+ k) =f(x,
Substituting this in
+
i
/i2
may
This
£* /(
^
V
h—
if
dx
and
regarding y as the only variable,
+ k —d f(x,
y)
=f(x, y)+h±f(x,
^
+2J
f(x
h&
—
dyj
k~
)
y)
is to
y f(x,
2
y)
+
....
>
y)
+ h^-f(x, y)
*>+"£/<*« + .».
(2)
be expanded by the Binomial Theore m, as
were the two terms of the binomial, and the
}
result-
y).
Taylor's Theorem applied to Functions of
dependent Variables.
f(x
d2
+ fhfx + kjj-\ f(x, y)
ing terms applied separately tof(x
article
k
+-
dy
dx
116.
2
y)
be expressed in the symbolic form thus
f(x + h,y + Jc) =f(x,
where (h
7c),
(1),
+ h,y + k)
fix
+
we
By
Any Number
of In-
a method similar to that of the preceding
shall find
+ h,y + k,z^l)=f(x,y,z) + (lij- + kj- + l£^f(x,y
i
+ |(*s +
This expansion
may
*S
+,
s)
y
z)
^^
f>
+: -
be extended to any number of variables.
-
PARTIAL DIFFERENTIATION
147
EXAM PLES
Expand
1.
T
Let
.
log
-,
N
u =/(.i\ y)
(a?
+ h) log (y + A).
= log x log y,
i
i
x
—
= —a^y — = -£-,
ay
dx
los:
d>/
losr
d?<
,
a;
d 2 ??
_
32 m
log?/
?/
2
_ 1
d'
u
_
log
?
d.r
By
.
4-
L
A*
i
+ -loga-,
a;
2.
log
(2), Art. 115,
-logy
2/
(a
+ A)
(y
(»+*) log
_
*&
A2
,
.
log?/ H
(y
AA
xy
re
2
dy~
scy*
y
+ A) = log a logy
A_2
,
5
i loga5+—
2y-
+ A) = xY + 3 A.r y + 2 fc^y
2
3
dyda;
ar
2
2
+ 3 h xy + 6 AAa; y + A~V +
2
3.
sin
[(a:
2
2
• •
••
+ A) (y + A)] = sin (xy) + Ay cos (xy) -f A» cos (#?/)
7 2
2
- ^-sin (xy) + 7ik [cos (xy)-xy sin (a-y)] -
—
V, 9 ,2
'
sin(a;y)
+
•••.
CHAPTER
XII
CHANGE OF THE VARIABLES IN DERIVATIVES
117. To express
This
is
dy d2y d3y
dx dx* dx3
••in terms
'
of
(1),
Art. 56,
By
(3),
Art. 56,
dy
= ±_
dx
dx
dy
to y.
(1)
dry_dL dy_d_dy
dx2
'
'
changing the independent variable from x
By
dx d 2x d3x
dy dy 2 dy3
dxdx
dy
dydx dx
d 2x
From
(1),
d dy _ d 1
dy dx ~ dy dx ~
dy
dy 2
(ax*
[dy,
d?x
d2y _
dx2 ~
dy2
fdx\ 3
.
Similarly,
W
dPj[_dL d^_dL
dx3
dx dx 2
(2),
d d 2y
dy dx 2
d^y
dy
dy dx 2 dx
2
From
(2)
fd x\
\dy 2)
2
_ dx (Px
dydy 3
dx^ 4
dy
fdrx^ 2
d3y
dx3
_ dx d x
3
dydy 3
\dy 2)
dx^ 5
148
(3)
CHANGE OF THE VARIABLES
It is
IN DERIVATIVES
149
sometimes necessary in the derivatives,
dy
dx
to introduce a
new
d*y
effy
dar
dx?
variable z in place of x or
y, z
being a given
function of the variable removed.
There are two
118
where y
By
J
To express 4, g, g,
-,
dx dxr dx6
First.
is
cases, according as z replaces y or x.
a given function of
dydh__ d 2y/dz\ 2
dz\dxj
,
<fy d?z
dz dx2
find
—
—4,
^,
(XX
•-.,
o^ydz_dh,dy
may
dz dx3
be expressed in terms of
For example, suppose
and
y
=
no change of the
x.
z*.
= 3z ^.
2
dx
dx2
=
^
dx
3
z
x.
CIX
It is to be noticed that in this case there is
*V
dx
d?z_
'
dxdx2
dz 2
dependent variable, which remains
Then
••,
'
dz dx2
d3y _ cPy/dzY,
dx3
dz3 \dx)
Similarly,
**
dx dxr dxr
z.
d 2y _ d fdy\dz
dx2 dx\dz Jdx
we
*
*
= **.
dx
dz dx
(3),
v
" Art. 56,'
Similarly,
in terms of
\dx)
6
dx1
18**^ + 3*^.
f*Y+
\dxj
dxdx
dx
2
3
in-
DIFFERENTIAL CALCULUS
150
119.
To express^, **, *M
dx dx2 dx3
Second.
9
'
in
terms
of
-2
dz
This
is
—a
_Jl
dz 6
dz 2
where
•
a;
is
,
...,
'
'
a given function of
changing the independent variable from x
to
z.
z.
dy
By
(3),
dy _ dy dz _
dx dz dx
Art. 56,
dz
dx
dz
d fdy\
d*y
=
dx2
dL fay \dz _dz\dx)
dx
dz \dx )dx~
y
dz
2
dx d y
dy d x
dz dz 2
dz dz 2
may
Similarly, higher derivatives
is
generally easier to
2
be expressed.
work out each case by
For example, suppose
x
=
z
In practice
it
itself.
3
.
dy _ dy dz
dx dz dx
But
Hence
dx
_
Z
dz~
dz
2
'
_ z~
2
~dx~~3'
ySdy_l z -2 <ty
dx
d 2y
dx2
3
_
d_
(1)
dz
fdy\ _
dx\dx)
d_ fdy\
dz
dz\dxjdx
From(l),
*(&\u.±(f**!-i r*®\.
v
" dz\dx) S\ dz 2
dz)
Hence
'&*(,-««& _2r*#\
dx2
9\
dz 2
dz)
w
(2)
:
::
CHANGE OF THE VARIABLES
h */*3t\
2
(2),
v
151
S-#(§V£.
dz \darj dx
dar
Similarly,
From
IN DERIVATIVES
dz\dx J
= */V^- 6z^^+ 10z~^
9\
dz 2
dz"
dz
SU
A(r*S*- 6r^+10 *+$L\
dx
27
Hence
3
c?2
V
dz 2
3
dz
1
EXAMPLES
Change the independent variable from x to y in the two following
equations
1.
3
*£* - *W=
(Wyiry
dx\dxj
4»s.
0.
3
\
J\dx~J
c?oj
\
dx
«fff-i1
^~
+
,
to z in the
2(i+ yy%Y
TTFW'
\dy
Jdif
two following equations
„_ tan/
y
iK
^-2^Y=eos**.
2
da;
v
Change the independent variable from x
'cfo3
Veto:/
dxdx 2
to z in
the following
equations
'/.'
,J
das'
./'
1
(for
'A''
+ a^rfx
(1
+z)
2 2
0.
dy 2
(**\U** + ay*
\dy-J
3
&=
Jdx dxs
Ans.
Change the variable from y
*5s +
dy
dsccfcc
C?Z
ciz
J
J
DIFFERENTIAL CALCULUS
152
(2aj-l) 3
7.
g+
(2a
J
-l)^ = 2
2» = l + e'.
2/,
^
s
= 0.
4^-12^-f-9^-2/
dr
dr
dz
.
+ 9^ + 3^ +
^^ + 6^
dx
8.
da;
4
3
dar
2/
= ^^,
a;
=
e*.
da;
^
d?
+
+2/
-*'
d**
120. Transformation of Partial Derivatives from Rectangular to Polar
Coordinates.
u — f(x, y),
Given
to express
—
-
and
dx
—
dy
—
dr
in terms of
—
dO
and
,
where
x, y,
are rec-
and r, 6, polar coordinates.
have from (5), Art. 113, regarding u as a function of r and
tangular,
We
du _ du dr
dx
dr dx
,
du_dudr
dy~drdy
The values
tions
between
These are
But
of
a;,
—
dx
y,
,
—
-,
ay
and
a;
,^
dudO
,
are
now
.
^x
U
<j$dy
— —
dy
,
du dO
d6 dx
0,
to be found
from the
rela-
ox
r, 0.
= rcos0,
y
= rsmO
—^, —^ and
in the partial derivatives
,
dy
da;
garded as functions of x and
These are, from (3),
(3)
—
—
dx dy
,
y.
^ = a^ +
2
2/ ,
d
= tan-
1
^.
x
,
r
and 6 are
re-
CHAXGE OF THE VARIABLES IN DERIVATIVES
Differentiating,
dx
86
we
find
r
8y
r
_
_ sin 6
y
x2
dx
153
+ y-
r
Substituting in (1) and
(2),
86
_
_
x
x2
8y
'
+y
cosfl
2
r
we have
8u
—
= cos0-8u
dx
dr
sin 6
»
r
8u
—
86
//IN
(4)
,
8u
8u
8u
—
= sm0 —
+ cos —
dr
r
86
*
•
6
.
/trx
(5)
8y
121.
Transformation
of
—
from Rectangular
A
dxr
to Polar Co-
8y 2
*\
By substituting in (4), Art. 120)
ordinates.
8rv
L
_
8_(8u\_
dx*~dx\dxj
Differentiating
(4),
(8v\_
8r\8x)~
86\8x)
q 8
for u,
dr
2
sin
r2
8r86
fl
dr
r
r,
da
,o\
K)
8$
Art. 120, with respect to
drd6
,*.
^'
~8~6\8x)
sinfl 8 u
r
we have
(8u\
sin 6 8
Art. 120, with respect to
2
(4),
(8u\_
8i\dx)
2
q8 h
8
Differentiating
C°S
—
dx
862
6,
r
86
W
DIFFERENTIAL CALCULUS
154
Substituting (2) and (3) in
—=
dhi
2
dx
2
ad u
cos 2 6
dr2
2 sin
cos
we have
(1),
dhi
sin 2
,
drdd
r
Similarly by using
_
•
2
dy 2
a dhi
2
u
,
(5),
2 sin
dr2
sin 2
du
r
dr
1
r
2
2 sin
d 2u
d
1
d$
2
du
cos
...
Art. 120, instead of (4),
d2u
cos
drdO
r
cos 2
d 2u
r2
dO 2
2 sin
we
find
cos 2 6 du
r
cos
dr
du
.(5)
d$
Adding
(4)
and
(5^
=*>*
^ 2u
c%
dx
2
.
obtain
d 2u
dhi
dy
2
_ d5
2
u
dr
2
.
1 du
r dr
.
1 d 2u
r
2
d&
:
CHAPTER
XIII
MAXIMA AND MINIMA OF FUNCTIONS OF TWO OR MORE
INDEPENDENT VARIABLES
122.
A
Definition.
said to have a
is
sufficiently small
function of two independent variables, f(x, y)
value when x
a, y
b; when, for all
}
=
maximum
numerical values of h and
=
k,
+ h,b + k),
(a)
5)</(a + *»& + &).
(6)
f(ti,b)>f(a
and a minimum value, when
/(a,
123.
Conditions for
Maxima
or
Minima.
u=f(x,y),
If
we
find that a necessary condition for both (a)
—
=
ox
This
may
0,
and
— = 0,
when x =
dy
(&)
must hold when
Jc
maximum
f(a,b)>f(a + h,b),
for a
y
(6) is
that
= b.
be shown as follows
Conditions (a) and
and
a,
and
minimum
f(a,
6)</(a +
for sufficiently small values of
//.
155
ft,
&),
= 0,
and we have
for a
DIFFERENTIAL CALCULUS
156
We
By
thus have for consideration a function of only one variable.
Art. 106,
we must have
— f(x,
for both
b)
= 0,
maximum and minimum,
when x = a,
dx
x
jrf(
ox
that is
Similarly,
by
'
^ = ®>
—
letting h
=a
wnen x
in (a) and
>
(6),
V =b>
we may
derive
~\
f(%, V) =
T
dy
These conditions for a
not sufficient.
As
0,
when
= a,
a;
y
= b.
maximum
in the case of
or minimum are necessary but
maxima and minima of functions
of one variable, there are additional conditions involving derivatives
These we shall give without proof, as their
beyond the scope of this book.
The conditions for a maximum or minimum value of u = f(x, y)
of higher orders.
rigorous derivation
is
are as follows
For either a maximum or minimum,
=
£
ox
0,
and
2
For a maximum,
-^ < 0,
and
For a minimum,
—
>
ox
and
S<*
-
2
0,
2
< 2>
—^ <
(3)
—>
(4)
w
dy 2
Functions of Three Independent Variables.
maximum or minimum value of u =f(x,
For either a maximum or minimum,
for a
£?!f
dx
^—
—o
'
dy
^u
'
—
dz
(^L\< d u d u
2
and
(1)
6S-Y<SS
\dy oxJ
dx dy
also
124
|5_0s
dy
\dx dy)
2
2
dx dy 2
y, z)
The
conditions
are as follows
MAXIMA AND MINIMA OF FUNCTIONS
For a
for a
2
d u
maximum
-<0, and
A<0;
and
A>0;
dx2
minimum,
|^>0,
d2u
dh
dx dy
dx dz
d2u
dx
where A
2*
d2u
=
d 2u
d 2u
2'
dy dz
d 2u
d2u
d2u
dx dz
dy dz
dz1
dx dy
dy
EXAMPLES
Find the
1.
maximum
value of
u = 3 axy
Here
»
.
= 3ay-3x —=3ax-Sy
^
dx
dy
2
2
,
d2U
n
d2U
a
=_6flj
^j
dx
Also
Applying
— xP — y3
2
(1),
>
Art. 123,
x
dy
d2U
= - a6 y>
=Sa"
r~^dx
dy
we have
ay — x2 =
whence
-z-i2
.
0,
= 0, y = 0;
and
or x
ax
— y = 0;
2
= a, y = a.
The values x = 0, y = 0, give
t-,
dx-
= 0,
= —— =
—
dy
,
0,
3ct,
dx dy
which do not satisfy (2), Art. 123.
Hence they do not give a maximum or minimum.
157
DIFFERENTIAL CALCULUS
158
The values x — a, y = a,
— = — n6
give
2
-•
dx
u —a
—
= ba,
by
d2
d u
-
a,
2
d2u
-
2
= 6o a,
dx By
which satisfy both (2) and (3), Art. 123.
Hence they give a maximum value of u, which
2.
Find the maximum value of
c
and as xyz
is
fr
.
1
(1)
c-
2
a2
b
1
maximum when
numerically a
a3
xyz, subject to the condition
t + vl + t
a2
is
x^yh 2
a
is
maximum,
we put
.
A
2
2« 2 \
2a.-
50a52/
From
(2),
—
=
dx
and
—
=
dy
0,
we
find, as
the only values satisfying
Art. 123,
x=
^
——
V3
,
y
= ——
which give
V3
'
9
'
2
fy
9
'
da%
9
.,;
MAXIMA AND MINIMA OF FUNCTIONS
As
a
these values satisfy (2) and
Art. 123,
(3),
it
159
follows that xyz
is
maximum when
,
_
_
a
b
The maximum value
c
~
V3
a 3
V3
—
of xyz is
3v3
3.
Find the values of
+ y + + x — 2 z — xy
2
ar
a
that render
x, y, z
z
2
minimum.
#=— 2 y=—-,1
A
Ans.
-,
3
4.
2
= 1.
Ans.
—
i
3
Find the maximum value of
— x)(a — y){x + y — a).
(a
jut
5.
Find the minimum value of
+
x2
6.
Find the values
of x
sin
a
maximum
or
xy
+ y — ax— by.
Ans. - (ab
o
2
—a —b
2
2
)
and y that render
x
+ sin y + cos (x + y)
minimum.
A
.
A
Ans.
minimum, when x==y =
.
.
-,
—
3tt
25
a
maximum, when
x=y = -.
6
7.
Find the maximum value of
— + 6w +
(cu;
v
y
„
^+
8.
„
-
2
z/
c)y
2
A
.
+i
+ &- 4.
7
2
,
2
<r.
Find the maximum value of afyV, subject to the condition
2x + 3y-t-4z = a
9.
9
^Ins. or
Find the minimum value of - + - + -, subject
a
b
c
Xyz
= abc.
Ans.
to the condition
Ans.
3.
DIFFERENTIAL CALCULUS
160
Divide a into three parts
10.
may
such, that their
continued product
be the greatest possible.
Let the parts be
x, y,
and a
— x — y.
u = xy (a — x — y), to be a maximum.
Then
—
= ay-2xy-y =
dx
2
—
= ax-x*-2xy =
oy
0,
0.
These equations give x = y = ^.
o
Hence a
Note.
a
is
divided into equal parts.
— When, from the nature of the problem,
maximum
or
minimum,
it is
it is
evident that there
is
often unnecessary to consider the second
derivatives.
11.
Divide a into three parts,
x, y, z,
such that xm ynzp
may
be a
maximum.
Ans. *
m
12.
the
= Z = *-=^
n p m + n+p
.
Divide 30 into four parts such that the continued product of
the square of the second, the cube of the third, and the
first,
fourth power of the fourth,
may
be a
maximum.
Ans.
13.
when
Given the volume a of a rectangular parallel opiped
the surface
is
An
;
find
a minimum.
When
Ans.
14.
3, 6, 9, 12.
3
open vessel
is
to be
the parallelopiped
is
a cube.
constructed in the form of a rec-
tangular parallelopiped, capable of containing 108 cubic inches of
water.
What must
be
its
dimensions to require the least material iu
construction ?
Ans.
Length and width, 6
in.
;
height, 3 in.
Find the coordinates of a point, the sum of the squares of
whose distances from three given points,
15.
0*i> 2/i,)>
(x2
2/2),
(a*
2/s),
MAXIMA AND MINIMA OF FUNCTIONS
is
^
a minimum.
|<*,
+ *, + **),
161
§<* + *+*>,
the centre of gravity of the triangle joining the given points.
16.
If x, y, z are the perpendiculars
a, b, c of
a triangle of area A, find the
from any point
minimum
P on the sides
+y +z
value of x2
.
4
Ans.
a2
17.
2
2
A
2
+ &2 + c
2
Find the volume of the greatest rectangular parallelopiped
that can be inscribed in the ellipsoid,
+ S-1
+S
^
a
c
b
2
18.
The
electric
2
3V3
time constant of a cylindrical
u
=
mx^z
ax + by + cz
«*
Ans.
2
coil of
wire
is
,
where x is the mean radius, y is the difference between the internal
and external radii, z is the axial length, and m, a, b, c are known constants. The volume of the coil is nxyz = g. Find the values of x, y, z
which make u a minimum if the volume of the coil is fixed also the
;
minimum
value of
u.
Ans.
ax=by=cz=Jj^\
?i
'
u = ™j[jEL
3 Vabctf
CHAPTER XIV
CURVES FOR REFERENCE
We
give in this chapter representations and descriptions of some
of the curves used as examples in the following chapters.
RECTANGULAR COORDINATES
Y
125.
The
Cissoid,
f
This curve
the circle
may
ORA
any oblique
2a —
line
be constructed from
(radius a)
by drawing
OM, and making
PM= OR,
The equation above may be
obtained from this construction.
line
AM parallel to T
is
easily
The
an asymp-
tote.
The
polar equation of the cissoid
r=2a sin 6 tan 0.
162
is
CURVES FOR REFERENCE
The Witch
126.
of Agnesi,
y
=
163
8 a3
a2 4-4 a 2
This curve may be constructed from the circle OBA (radius, a) by
drawing any abscissa ME, and extending it to P determined by ORN,
by the construction shown in the figure.
The equation above may be derived from this construction. The
axis of
is an asymptote.
X
The Folium
127.
b
s
3
-f-
V
of Descartes,
— 3 axy = 0.
The point A, the vertex
the loop,
fSa
S_a\
\2>
The equation
tote
of
is
MN
2/
of the
asymp-
a
x + y + a = 0.
The polar equation
folium
a o
is
is
3 a tan 6 sec 6
l
+ tan
8
of
the
DIFFERENTIAL CALCULUS
164
128.
This
The Catenary,
is
2/
= 7jV + e
«)•
the curve of a cord or chain suspended freely between two
points.
129.
The Parabola,
l
Latus Rectum, x 2
+
JL
y^
referred to Tangents at the Extremities of the
1
=a
2
.
OL=OL' = a.
Y
The
OFM,
line
LL'
is
the latus rectum
the axis of the parabola ;
;
its
middle point F, the focus
OF, the vertex.
A the middle point of
CURVES FOR REFERENCE
165
n
y = .v where one coordinate is proportional
nth
power
of
the
other,
is sometimes called the parabola of
to the
The curve
130.
an
~l
,
the nth degree.
If n
= 3,
we have the
Cubical Parabola, a 2y
=x
Y
If
n=
-,
we have the Semicubical Parabola,
a 2 y= x 2 , ay 2 = x3
.
5
.
DIFFERENTIAL CALCULUS
166
The Two-arched Epicycloid.
131.
= —-cos — -cos 3
x
4>
—-sm<£--sm3<£.
y=
The Hypocycloid
132.
2.
a;
3
This
scribed
+y
is
2.
2
3
a3
-
the curve de-
by a point
P
in the circumference of
the circle
PR, as it rolls
within the
circumfer-
ence of the fixed circle
ABA', whose
is
radius a
four times that of
the former.
The equation above
may
be
</>,
given in the
form
#=acos 3 <£, 2/=asin 3 <£.
of
Four Cusps sometimes
called the Astroid,
CURVES FOR REFERENCE
"/+ (*)*=* *'
The equation
is
the same as that
of the ellipse with
exponent of
second term
changed from 2
the
the
tof.
134.
The Curve,
<ry-
— on? —
a:'
POLAR COORDINATES
135. The
The
a)
circle
is
OPA
tangent to the
OX at
= a sin
Circle, r
the origin 0.
$.
(diameter,
initial
line
167
DIFFERENTIAL CALCULUS
168
136.
The Spiral
"
In
of Archimedes, r
= aO.
this
curve
r
is
proportional
to
Lay-
6.
ing off
r
=
OA,
when
= 2tt,
then
OP = \OA, OP = iOA, OP = \OA, OP = iOA,
0B = 2 0A, OC = SOA.
2
x
The dotted portion corresponds
to negative values of
137. The Hyperbolic or Reciprocal Spiral, rO
In this curve r
varies inversely as
0.
The
MN
line
an asymptote,
which the curve
is
approaches,
as
approaches
zero.
Since
when
r=0
= oo
,
only
it fol-
lows that an
finite
number
revolutions
6
in-
of
are
necessary to reach the origin.
5
z
a.
6.
CURVES FOR REFERENCE
The Logarithmic
138.
Spiral
r
169
=e
Starting from A,
0=0 and r=l,
where
r increases
but
with 6
:
we suppose
if
negative,
/•
decreases
as 6 numerically in-
Since
creases.
only
it
that
an
number
of
follows
infinite
revolu-
retrograde
tions
r=0
when 0=—x:,
from
A
is
re-
quired to reach the
origin 0.
A property of this spiral is that the radii vectores
make
139. The Parabola, Origin at Focus, r(l
The
initial line
the parabola;
focus
OP, OPj,
OP
a constant angle with the curve.
;
OX
the
is
— cos 0) = 2 a.
the axis of
origin
is
the
LL', the latus rectum.
140. The Parabola, Origin at Vertex (see preceding figure),
/•
The
initial line is
sin $ tan 6
the axis
AX;
= 4a.
the origin
is
the vertex A.
2,
DIFFERENTIAL CALCULUS
170
141. The Cardioid, r
This
= a(l — cos 6).
the curve described
is
by a point
P
in the circum-
ference of a circle
ameter, a) as
it
PA
(di-
upon
OA.
rolls
an equal fixed circle
Or it may be constructed
by drawing through 0, any
line
OB
the circle
in
OA,
OR to Q and
BQ=RQ'=OA.
and producing
making
The given equation
Q',
fol-
lows directly from this construction.
142. The Equilateral Hyperbola,
The
origin
is
the
centre, of the
hyperbola,
and the initial line
is
OX
the trans-
verse axis.
If
Or
is
taken as the
initial
line,
the equation
of the hyper-
bola
2
is
r sin 2
6=
2
.
2
r cos 2
2
CURVES FOR REFERENCE
The Lemniscate
143.
r=a
This
It
is a
may
OA
referred to
2
171
(see preceding figure),
cos 2
0.
curve of two loops like the figure eight.
be defined in connection with the equilateral hyperbola, as
the locus of P, the foot of a perpendicular from
on PQ, any
tangent to the hyperbola.
The loops
are limited
by the asymptotes of the hyperbola, making
TOX =T'OX= 45°. OA = a.
The lemniscate has the following property:
If
two
points,
F and
F', called the foci, be taken
OF=
that
on the
axis,
such
OF'
V2
then the product of the distances P'F, P'F', of any point of the
curve from these fixed points, is constant, and equal to the square
of OF.
If
J"
is
taken as the
initial line,
r2
144. The Four-leaved Rose,
=a
2
the equation of the lemniscate
sin 2
r=a sin
6.
0.
is
DIFFERENTIAL CALCULUS
172
145.
The Curve,
r
= a sin
3
o
:
CHAPTER XV
DIRECTION OF CURVES.
We have
plane curve
146.
the
seen in Art. 17 that the derivative at any point of a
is
the slope of the curve at that point.
some further applications
sider
ordinate
PX
is
will
now
con-
— Let
PT
called the
and
subnormal,
.
MN
corre-
sponding to the point
P.
To
find expressions
for these quantities
Let
angle
<f>
clination
gent to
By
donote
PTX,
of
the
the
the
in-
tan-
OX.
tan
Art. 17,
Subtangent
PTX = tan = ^-.
<f>
= TM= PM cot PTM= y cot
<£
dx
=^ =y—
dx
Subnormal
= MN= PM tan MPX= ytsmcf> = y dy
dx
Intercept of tangent on
Intercept of tangent on
But as OT'
is
be
the normal, to a curve at the point P, whose
PM.
==
y
is
MT
subtangent,
the
We
of differentiation to curves.
Subtangent, Subnormal, Intercepts of Tangent.
tangent, and
Then
TANGENTS AND NORMALS
negative,
Intercept of tangent on
OX= OT = OM— TM=x — y dx
dy
OY= OT' PS - PM= x tan —
<f>
we have
Y= y — x tan = y — x
cf>
dx
y.
DIFFERENTIAL CALCULUS
174
147. Angle
of Intersection of
Two
Suppose the two curves
Curves.
intersect at P.
Let
PT
PT
and
be the
tangents at P.
PTX=<j>, PT'X=cf>',
and
let
I be
the
angle
TPT between the tangents.
Then
I=<f>'-<]> and
tan<fr
tan/=
f
- tan<
ft-.
(i)
v
l+tan<£'tan<£
From
the
'
equations of
the given curves find the
coordinates of the point of
intersection
T
P; then using
these equations separately, find by tan
<f>
^
= dy
the values of tan
dx
and tan
for the point P.
Substituting in (1) gives tan
For example find the angle at which the circle
<j>'
a*
+ 3^ 13,
.
.
(2)
.
intersects the parabola
2y 2 = 9x.
The
.
.
intersection
found to be
(3)
.
P of
(2)
and
(3) is
(2, 3).
Differentiating (2),
^
= -- = -?
dx
y
3
From
forP, tan<£
(3),
Substituting in (1),
dy
dx
tan/
=
9
2
3*
=?forP,
4cy
17
1=70° 33',
tan
<f>'
=
/.
<£
—
TANGENTS AND NORMALS
DIRECTION OF CURVES.
175
EXAMPLES
1.
Find the direction at the origin of the curve,
(a 4
What must
parallel to
2.
— b )y = x (x - a) - b x.
i
4
4
Ans. 45° with OX.
be the relation between a and
OX at the
b,
so that it
=2 a?
point x
•
may be
3a = b
2
Arts.
2
.
Find the poiuts of contact of the two tangents to the curve,
= 2 X + 9 x - 12 + 2,
s
6y
2
aj
parallel to the tangent at the origin to
the curve,
3.
y
+ ay = 2 ax.
= 4tax,
Ans.
Subtangents,
1
= -1
-V
f 1,
(
— 4, 11)-
Show
that the
parabola (Art. 129),
Show
#
2
2a?, -;
subnormals, 2
a,
—
in the cissoid (Art. 125),
at the point (a, a).
,
Ans.
-,
2
a.
Li
iii
+ =
sum
4:<iy.
X
Li (X
6.
and x2 =
Find the subtangent and subnormal
y
5.
Ans.
Find the subtangents and subnormals in the parabolas,
7f
4.
2
of the intercepts of the tangent to the
y
2
a2
,
is
equal to a.
that the area of the triangle intercepted from the co-
ordinate axes by the tangent to the hyperbola,
2 xy
7.
Show that
2
2
8.
—a
2
3
,
=a
2
,
is
equal to a2
.
the part of the tangent to the hypocycloid (Art. 132),
intercepted between the coordinate axes,
At what angle do the
parabolas, y 2
An s. At
= ax
(0, 0), 90°;
is
equal to
a,
and x2 = 8 ay intersect?
at another point, tan
3
-1 -•
DIFFERENTIAL CALCULUS
176
9.
At what angle does the
3y=7x
10.
3
— 1,
Show
at their
x2 + y2
circle,
common
point
= 5 x, intersect the' curve,
Ans.
(1, 2) ?
45°.
that the ellipse and hyperbola,
7
'
+
2
'
2~
3
'
intersect at right angles.
11.
Find the angle of intersection
x* + y2-. x + 3y + 2 = 0,
12.
Show
x2
of the circles,
+ if-2y = 9.
Ans.
tan" 1
\
that the parabola and ellipse,
y
2
= ax,
2 x2
+y =6
2
2
,
intersect at right angles.
13.
Show
that the parabolas,
2
y
intersect at
14.
an angle
= 2 ax + a
2
,
and #2 = 2
%+6
Find the angle of intersection of the parabola,
= -8
^bis.
15.
y
,
of 45°.
x2 = 4 ay, and the witch (Art. 126), y
2
2
tan- 1 3=71°34'.
Find the angle of intersection between the parabola,
and its evolute, 27 a?/ 2 = 4 (x — 2 a) 3 (See Fig., Art. 167.)
= 4 a#,
.
-4ns.
tan
-1
V2
Equations of the Tangent and Normal. Having given the
148.
equation of a curve y=f(x), let it be required to find the equation
of a straight line tangent to it at a given point.
.
TANGENTS AND NORMALS
DIRECTION OF CURVES.
Then
Let (V. y 1) he the given point of contact.
straight line through this point is
y-y'
177
the equation of a
= m{x-x'),
(1)
which x and y are the variable coordinates of any point in the
and m, the tangent of its inclination to the axis of X.
But since the line is to be tangent to the given curve, we must have,
by Art. 17,
in
straight line
;
m = tan
(f>
dx'
(
-^ being derived
and applied
from the equation of the given curve y =f(x),
to the point of contact
If we denote
ii
m\
equation (1),
by
this
—
(»', ?/').
we
,
have,
substituting
dx'
m = -^-
in
dx'
y-y'=%(*-*),
(2)
for the equation of the required tangent.
Since the normal
tangent,
we have
is
a line through
(a?',
y')
perpendicular to the
for its equation
r-f—^C— 0— g<—
>
(3)
dx'
For example, find the equations of the tangent and normal to the
x2 + if = a 2 at the point (x', y').
Here, by differentiating ar + ?/ 2 = cr, we find
circle
,
—
= — -, from which -&- = — —
tw
dx'
y
Substituting in
(2),
we have
y-y' =
x ~ x%
--X
y
x'
as the equation of the required tangent.
y'
DIFFERENTIAL CALCULUS
178
It
may be
simplified as follows
yy'-y' 2 = -xx'
xx' -\-yy'=x'
The equation
of the
2
+ x'
2
,
+ y' = a
normal to the
2
2
.
found from
circle is
(3) to
be
y-y'=-,(x-x'),
X
which reduces
to
x'y
= y'x.
EXAMPLES
Find the equations of the tangent and normal to each of the three
following curves at the point
1
.
The
parabola,
y
2
= 4 ax.
Ans. yy'
2.
The
ellipse,
=2 a{x + x'),
The
2
— y')+y'(x — x ) = 0.
r
2
^ + WL =
equilateral hyperbola,
+
yx'
Find the equation of the tangent
3x2 -4:xy + 2y 2 + 2x = 2.
2
1, b x'(y
2xy = a 2
Ans. xy'
4.
2a(y
= 1.
+^
^
b
a
Ans.
3.
(x'} y')
- y')= a y'(x 2
a?').
.
=a
2
,
y'(y
— y') = x'(x — x').
at the point (x',y') to the
ellipse,
Ans. 3 xx'
5.
Find the equations
x5 — asy 2
to the curve,
+ 2yy'—2 (x'y + y'x) + x + x' = 2.
of tangent
and normal at the point (V,
y')
.
Ans.
=
^-^
x
y
3,
2xx'
+ 5yy' = 2x' + 5y'
2
2
.
:
DIRECTION OF CURVES.
TANGENTS AND NORMALS
In the cissoid (Art. 125), y 2
6.
3
=
ar
2a — x
find the equations of the
,
tangent and normal at the points whose abscissa
Ans. At
(a,
In the witch (Art. 126), y
— a),
=
—So
Ans. x
Find the equation of the tangent
8.
+ xy = a
Xs y
2
-f-
= 3 a.
x
2y — x — 3a.
3
4r + x-
find
,
the equations of
the tangent and normal at the point whose abscissa
curve,
2y
y
t
7.
is a.
= 2 x — a,
y + 2x = a,
(a, a),
At
179
2
is
a.
= 2 x — 3 a.
+ 2y = ka,
y
at the point
(x', y')
to the
3
.
xy\2 x' + y')
Ans.
+ yx'(2 y' + x') = 3 a
3
.
Find the equations of tangent and normal to the three following
curves
x'+f =3
9.
axy (Avt. 127), at
+ y = 2e
x-y
10.
:c
11.
fjY + (llX= 2,
\aj
,
Ans.
at (1, 1).
at (a, b).
\bj
Ans.
/^, 2^).
Ans. a
x+y=3a, x=y.
3y = x-{-2, 3^ + y = 4.
+ = 2,
U.
Find the equations of the two tangents to the
12.
2
x
+ y — 3y =
2
14:,
=a -b
ax -by
2
2
.
b
circle,
7y = 4:X + l.
= 4# + 43, ly = 4 x — 22.
parallel to the line,
Ans. 7
2/
Find the equations of the two normals to the hyperbola,
13.
4 x2
- 9 y + 36 = 0,
2
parallel to the line, 2 y
^[?is.
Asymptotes.*
149.
"When the tangent
+ 5 x = 0.
8y
+ 20 =±65.
a;
to a curve approaches a
limiting position, as the distance of the point of contact from the
origin
is
*
indefinitely increased, this limiting position
The
limits of this
work allow only a
is
called an
brief notice of this subject.
DIFFERENTIAL CALCULUS
180
In other words, an asymptote
asymptote.
within a
at
an
We
finite distance of
is
a tangent which passes
the origin, although
its
point of contact
is
infinite distance.
have found in Art. 146, for the intercepts of the tangent on the
coordinate axes,
Intercept on
OX=x — y —
OY= y — x -^-.
c
Intercept on
,
dy
dx
If either of these intercepts is finite for x
= oo,
or y
= oo,
the cor-
responding tangent will be an asymptote.
The equation
intercepts, or
of this asymptote
may
from one intercept and
be obtained from
its
two
the limiting value of -^.
dx
Let us investigate the conic sections with reference to asymptotes.
(1)
The
parabola, y 2
Intercept on
^ = ^.
y
= 4 ax,
dx
dx
—x
OX — x — y—
dy
t
x.
,
Intercept on
When x = oo, y=ao,
^
Tr =
OT
y
— a=—
dy
— x—
=y
ii~
^—
2
2 ax
'
dx
x,
= ^y
2
y
and both intercepts are also
infinite.
Hence the parabola has no asymptote.
(2)
= l,
The hyperbola, .^-«j
2
2
a
Intercept on
b
OX = —
,
&=.$£.
a 2y
dx
OY =
Intercept on
.
x
y
These intercepts are both zero when x = oo, and there is an
To find its equation, it is
asymptote passing through the origin.
necessary to find the limiting value of -^,
dy
dx
_bx_
2
bx
a 2y
a V^2 _-«2
-± -b
when x
1
\/i-
a2
X2
when x =oo.
Hence
dx
a'
=
9
oo.
:
TANGENTS AND NORMALS
DIRECTION OF CURVES.
181
There are then two asymptotes, whose equations are
=±
,
y
The
(3)
ellipse,
no
having
b
-x.
a
infinite
branches,
can
have no
asymptote.
Asymptotes Parallel
150.
equation of the curve, x
y
=
a
to the Coordinate Axes.
gives a finite value of
the equation of an asymptote parallel to
is
And when
to
= oo
y
= oo
gives x
=
a,
then x
= a is
When,
y, as y
in the
= a,
then
OX.
an asymptote parallel
OY.
Asymptotes by Expansion.
151.
may
Frequently an asymptote
be determined by solving the equation of the curve for x or
y,
and
expanding the second member.
For example, to find the asymptotes of the hyperbola
=±
y
As x
y
=± —
V - ^=
a
increases
± *?(l - f)i=
a\
arj
indefinitely,
the
±*(l*-.
2x
a\
curve
2
approaches the
lines
the asymptotes.
,
EXAMPLES
Investigate the following curves with reference to asymptotes
1.
?
= 32
3
x3
= 6x —
2.
y
3.
The
Asymptote, ?/=
-
3 a»
,
2
Asymptote,
5
or
.
cissoid (Art. 125) y 2
=
^
a;
Asymptote,
z.
+ y = 2.
a;
= 2a.
DIFFERENTIAL CALCULUS
182
4.
5.
6.
Asymptote, x + y = 0.
+ =a
— 2 a)?/ = x — a
Asymptotes, # = 2 a, x + a = ±
Asymptote, x + + a = 0.
X -f = 3 cm/ (Art. 127).
= vx in the given equation and in the expressions
(Substitute
3
a?
3
3
?/
.
2
(a?
3
3
?/.
.
s
3
?/
2/
?/
for the intercepts.)
152. Direction
Polar Coordinates.
of Curve.
In this case the angle
OPT between the tangent
and the radius vector may
be most readily obtained.
Denote this angle by
Let r, 0, be the coordinates of P; r+Ar,0+A0,
if/.
the
coordinates
PR
Draw
to
of
Q.
perpendicular
OQ.
Then tanPQP
= |^ =
RQ
r
+ A> — rcos A0
Ar
+ 2rsin
2
^
A0
A0
sin
.
sin
—
Ar
-f r
A0
-
Now
angle
let
PQR
A0 approach
approaches
Hence
tan
zero
its
if/
=
;
•
sin
2
A0
2
the point
limit
—A0
—
—
A6»
Q
approaches P, and the
if/.
Lim A
=o tan
PQR =
(1)
d0
The
inclination
<f>
of the tangent to
<f>
OX
= t+0
may be found by
(2)
.
TANGENTS AND NORMALS
DIRECTION OF CURVES.
183
153. Polar Subtangent and Subnormal.
If through O,
NT
pendicular to OP,
polar subtangent, and
subnormal,
be drawn per-
OT
is
called the
ON
the polar
corresponding
the
to
point P.
OT=OP tan OPT;
that
is,
2
Polar subtangent
r
= r tan = —
\b
a>
dO
ON= OP cot PNO;
Polar subnormal
that
is,
—
= r cot Y = dr
xb
•
dO
154. Angle
OPT=xj/
OPT' =
,
Then the angle
and
tan
I-
By
this
PT and PT\
x\,'.
of intersection,
tan ^-tani/r
1
section
Suppose the two curves intersect at P,
of Intersection.
and have the tangents
+ tan
i//
tan ^
a)
formula the angle of interbe found in polar coordi-
may
same way as by (1), Art. 147, in rectangular coordinates.
For example, find the angle of intersection between the curves
nates, in the
= asin20,
r = a cos 20.
r
and
From
(2)
and
(3)
we have
for the intersection
tan20 =
l.
(2)
(3)
DIFFERENTIAL CALCULUS
184
From
tan
(2),
if/'
= - tan 2 = -,
From
tan
(3),
if/
= — - cot 2 = — -,
Z
Substituting in
The curves
for the intersection.
Z
-i
for the intersection.
Z
tan I — o
(1),
are that in Art. 144,
and the same curve revolved 45
c
about the origin.
EXAMPLES
1.
In the
circle (Art. 135), r
— a sin 6,
find
if/
and
Ans.
2.
In the logarithmic spiral (Art. 138),
r
<£.
= 6,
if/
= e ad
and $
show that
,
= 20.
is
\J/
constant.
3.
In the spiral of Archimedes (Art. 136),
tan \j/=0; thence find the values of
when
^,
r
=2
= ad,
and 4
-k
show that
-k.
Ans. 80° 57' and 85°
Also show that the polar subnormal
4.
The equation
2
at its center is r
5.
27'.
constant.
of the lemniscate (Art. 143) referred to a tangent
=a
Arvs.
is
if/
2
sin 2
Find
0.
= 2 6;
if/,
= 30;
<£
In the cardioid (Art. 141), r
<f>,
and the polar subtangent.
subtangent
= a tan 2
= a (1 — cos 0),
find
<£,
Vsin 2 0.
^,
and the
polar subtangent.
Ans.
6.
<f>
=
—
z
;
if/
=
z
;
subtangent
= 2 a tan - sin —
2
z
z
Find the area of the circumscribed square of the preceding
formed by tangents inclined 45° to the axis.
—
27
cardioid,
AnS
'
2
f^ + V3)a
2
.
TANGENTS AND NORMALS
DIRECTION OF CURVES.
7.
In the folium of Descartes (Art. 127),
185
r= 3atan ^ sec ^
+
= tan - 2 tail 6
4
show that
8.
tan
<£
2tan 3 0-l
Find the area of the square circumscribed about the loop of
the folium of the preceding example.
Ans. 2 a/2 a2
9.
Show
that the spiral of Archimedes (Art. 136), r
reciprocal spiral (Art. 137), rO
10.
Show
= a,
.
= ad, and the
intersect at right angles.
that the cardioids (Art. 141),
r
= a (1 — cos 0),
r
= b (1 + sin 6),
intersect at an angle of 45°.
11.
Show
that the parabolas (Art. 139),
r =:
m sec- -,
2'
r
= n cosec- 2'
intersect at right angles.
12.
r
Find the angle of the intersection between the
and the curve (Art. 144), r = a sin 2 6.
At
Ans.
13.
r
circle (Art. 135),
= a sin 0,
origin 0°
;
at
two other
tan -1 3 V3
points,
Find the angle of intersection between the
and the cissoid (Art. 125), r = 2 a sin
= 2 a cos 6,
circle (Art. 135),
tan
0.
Ans.
14.
At
v%
hat angle does the straight line, r cos 6
circle (Art. 135), r
Show
/-cos 26 = b
= 5 a sin 6 ?
= 79° 6'.
tan -1
= 2a, intersect the
^
tan
that the equilateral hyperbolas (Art. 142), r2 sin 2 6
15.
2
,
intersect at right angles.
2.
_,
3
4
=a
2
,
DIFFERENTIAL CALCULUS
186
Find the angle of intersection between the
16.
= a sin + b cos 0,
r
r
circles
= acos$-\-b sin 0.
Ans. tan"
2ab
17.
r2
=a
2
Find the angle of intersection between the lemniscate (Art.143),
2
0, and the equilateral hyperbola (Art. 142), f^sin 2 6 — b
sin 2
.
Ans.
Derivative of an Arc.
155.
Rectangular Coordinates.
Let
2 sin
s
—
-1
denote
the length of the arc of the curve measured from any fixed point of
Then
arc
We have
Now
AP,
sec
Ax
suppose
it.
As = arc PQ.
QPR
PQ
PR
to approach zero,
and consequently the point
Q
to approach P.
Then
Lim
QPR= sec TPR= sec
PQ = PQ arcPQ
PR arcPQ* PR
sec
<£.
'
since
Lim _p_e_ =1
arcPQ
,
Lim^=l4m^§
P#
PR
T
Hence
.
As
c?s
Ax
dx
sec<£
=
ds
dx'
therefore
|= VH^-V^g)'
(1)
,
TANGENTS AND NORMALS
DIRECTION OF CURVES.
187
It is evident also that
SlU(i
= -,
COS
6=
It
may
(2)
.
.
ds
ds
be noticed that these
relations
(1)
and
(2)
are cor-
by a right
whose hypothennse is
dx and dy, and angle
rectly represented
triangle,
ds, sides
at the base
<j>.
ds= ^(dxf+(dyy,
Here
or
V
dx
156. Derivative
Art. 152,
we
sec \b
r
of
have, as
an Arc.
[dxj
Polar Coordinates.
A0 approaches
From
the figure of
zero,
= Lim sec FOR
^ = Lim —-^ = Lim
ar °
RQ
RQ
^ = Lim
—
RQ
J.
As
As
A0
*' + »'-»¥
A,,.^ sm A.
A0
AF T
As
.
ds
.
sec ^
Y
T
= Lim
•
As
RQ
ds
=—=—
d9
dr
(i)
dr
dO
Hence
-^
r
= Vl+tan
<l»_'l*<]r_
V' + <|J,
2
i/>
.
(2)
/'*'
.
(3)
Vctf
DIFFERENTIAL CALCULUS
188
It
may
(1), (2), and (3), are corby a right triangle, whose hypothenuse is ds, sides
and angle between dr and ds, \p.
be noticed that these relations
rectly represented
dr and rdd,
Here
and thence
ds
=
V (drf +
(r
ddf,
^ = Ji + W™Y,
or
*
=V
r2
I I)-
CHAPTER XVI
DIRECTION OF CURVATURE.
A
Concave Upwards or Downwards.
157.
cave upwards at a point P,
P
POINTS OF INFLEXION
it lies
when
in the
curve
is
said to be con-
immediate neighborhood of
wholly above the tangent at P, as in the first figure below.
it is said to be concave downwards, when in the immediate
Similarly,
neighborhood of
P
it lies
wholly below the tangent at P, as in the
second figure below.
It will
now
when
be shown that
the equation of the curve
is
in
rectangular coordinates, the curve is concave upwards or downwards,
d2 v
according as —- is positive or negative.
ax~
Suppose
-^ >
0,
that
oar
of the slope
Then by
is
is,
— > 0; in other words, the derivative
—
dx\dxj
(
)
positive.
Art. 21 the slope increases as x increases.
This case
is
illustrated in the first figure above,
evidently increases as we pass from
P
Y
to
P
2.
where the slope
The curve
is
then con-
upwards.
But
if --»
<
0, it
follows that the slope decreases as x increases.
189
DIFFERENTIAL CALCULUS
190
We
then have the case of the second figure, where the slope dewe pass from 1 to 2 The curve is then concave down-
P
creases as
P
.
wards.
A
158.
Point of Inflexion
a point
is
P
where
—^2
changes sign,
dx
the curve being concave upwards on one side of this point, and concave downwards on the other.
This can occur, provided -^ and
dx
are continuous, only
dx2
when
^=
(1)
dx2
-^ and —\ are infinite, we
dx
dx2
may have a point of inflexion
But
if
J=
when
X*
oc
dx2
It is evident that the tangent at a point of inflexion crosses the
curve at that point.
For example, find the point of inflexion of the curve
2y = 2-8x + 6x2 -x3
d 2y_
3(2-a>).
dx2
Here
Putting this equal to zero,
flexion,
.
x
= 2.
If
'
Hence the curve
wards on the
we have
x<2, ^ >0; and
dx
is
o
2
'
for the required point of inif
a>>2,^<0.
concave upwards on the
right, of the point of inflexion.
dx2
left,
and concave down-
:
DIRECTION OF CURVATURE.
POINTS OF INFLEXION
191
EXAM PLES
Find
five
1.
the points of inflexion
and the direction of curvature
of the
following curves
y=(r-iy.
x= ± —-;
Ans.
concave downwards between these points, con-
V3
cave upwards elsewhere.
2.
y
^Lns.
= x* - 16 x" + 42 x - 28 x.
= 1 and x= 7; concave
2
05
downwards between these
points,
concave upwards elsewhere.
3.
= x (x — a) + a x.
a 4y
A
4
a
x=^;
r>
Ans.
concave downwards on the left of this point, con-
o
cave upwards on the right.
4.
The witch
Ans.
± —*-,
(
(Art. 126), y
—
)
concave
:
downwards
between these points,
concave upwards outside of them.
5.
The
Ans.
curve, y
[—3a,
a?
a?
+ 3 a*
-j,
(0,0),
(3
a,
—
);
concave upwards on the
of first point, downwards between first and second,
upwards between second and third, and downwards on the
left
right of third point.
DIFFERENTIAL CALCULUS
192
Find the points of inflexion of the following curves:
= „a4:Xx a
x> + ±
m
4t
6
-
7.
y
y=
8.i/
9.
10
.
11.
y
'
x=
Ans.
«*.
T
(x — a)
Ans.
2
= (#
2
-f
a?)
e
-x
Ans.
.
= e-~-e-K
gy
and
Ans.
6
=
(Art. 134).
Ans.
2 V3.
x = -2a.
and
a;
= 3.
= ^2S^zMH.
a—b
^
+ g)U.
aV = a¥ - »
x
x
±
„. ± -.
x=±% ^21 - 3 V33.
6
CHAPTER XVII
CURVATURE. RADIUS OF CURVATURE. EVOLUTE AND
INVOLUTE
159.
If a point
Curvature.
moves
in a straight line, the direc-
motion is the same at every point of its course, but if its
path is a curved line, there is a continual change of direction as it
moves along the curve. This change of direction is called curvature.
We have seen in the preceding chapter that the sign of the second
derivative shows which way the curve bends.
We shall now find
that the first and second derivatives give an exact measure of the
tion of
its
curvature.
The
direction at
any point being the same as that of the tangent
may be measured by comparing the
at that point, the curvature
linear motion of the point with the simultaneous angular motion of
the tangent.
160- Uniform Curvature.
The curvature
is
uniform when, as the
point moves over equal arcs, the tangent turns through equal angles.
The only curve
uniform curvature is the circle. Here the measthe ratio between the angle described by the tangent and the arc described by the point of contact.
In other words,
it is the angle described by the tangent while the point describes a unit
of
ure of curvature
is
of arc.
Suppose the point Pto move in the circle AQ.
Let s denote its distance
from some initial position
the angle PTXmade by the tangent
with OX.
AP
<j>
A
}
and
PT
Then
and
<f>
As
as the point
by the angle
moves from
P
to Q, s is increased
by
PQ = As,
QRK= A<£.
the point describes the arc As, the tangent turns through the
angle A(£.
DIFFERENTIAL CALCULUS
194
The
form,
'
curvature, being uni-
is
then equal
to
^
—
As
•
If we draw the radii CP,
CQ, and let r denote the
radius, then
angle
PCQ = QBK= A</>.
But
arc
that
PQ = CP (angle PCQ)
;
is,
As = rA<f>,
—^ = —
As
r
Hence
the curvature of a circle is the reciprocal of its radius.
For example, suppose the radius of a circle to be 50 feet.
Then
its
curvature
1
=
~ 50'
A<j>
is
As
where A<f> is in circular measure, and As in feet.
In other words, for every foot of arc, the change of direction
— in circular measure = 1°
is
8' 45".
50
161. Variable Curvature.
curvature varies as
arc As,
—*
As
is
the
For
curves except the circle the
all
we move along
the curve.
mean curvature throughout
ture at the beginning of this arc
we take As.
Hence the curvature
is
In moving over the
the arc.
more nearly equal
shorter
at
any point of a curve
-L<ini As=o
——
-
-7-
As
as
is
equal to
The
to
curva-
— *,
the
RADIUS OF CURVATURE
CURVATURE.
Circle of Curvature.
162.
having
its
A circle
concavity turned in
195
tangent to a curve at any point,
the same
direction,
and having the
same curvature as that of the curve at that point, is called the circle
of curvature ; its radius, the radius of curvature; and its centre, the
centre of curvature.
The
figure
ellipse.
It
is
C is
shows the
circle of curvature
the centre of curvature, and
MPN for the point P of the
CP the
radius of curvature.
to be noticed
that the circle of curv-
B
n>\
ature crosses the curve
i!
VL
This can be
easily proved.
At P the circle and
ellipse have the same
at
P.
curvature, but as
go
towards
P
2
,
/
p!v
AJ \
we
_j^C^
the
M
curvature of the ellipse
increases, while that of
N
the circle continues the same.
Hence on the right of P the circle is outside of the
Moving from P to P2 the curvature of the ellipse
,
therefore on the left of
So in general
contact.
P
the circle
ellipse.
decreases, and
the circle is inside of the ellipse.
of curvature crosses the curve at the point of
DIFFERENTIAL CALCULUS
196
The only exceptions to this rule are at points of maximum and
minimum curvature, as the vertices A and B of the ellipse.
As we move from A along the curve in either direction, the curvature of the ellipse decreases
hence the
;
circle of curvature at
entirely within the ellipse.
Similarly it appears that the circle of curvature at
without the
163.
B
A lies
lies entirely
ellipse.
Radius
The curvature
of Curvature.
ture being that of the given curve,
denote the radius of curvature by
is
of the circle of curva-
-^
equal to
(Art. 161).
If
we
then by Art. 160,
p,
m
ds
^
p=
(1)
ds
To
obtain p in terms of x and
y,
we may
write
(1),
p
= — = —dcf>
•
d<f>
dx
From
(1) Art. 155,
Also,
= Jl + f§Y
^
dx
\ ^ \dx)
tand>
= ^,
<f>
= tan-
dx
Differentiating,
dcf>
^>
=
dx
1
^\
\dxj
2
dx'
^—-
(2)
1+ (^y
\dxj
Hence
P=
Mm
^
d 2y
dx2
3
2
^
.
RADIUS OF CURVATURE
CURVATURE.
always to be considered positive
It is to be uoticed that p is
the sign of
|~1
+ f^t
197
is
taken the same as that of
;
that
—
dor
By
interchanging x and
we have
y,
Kf)t
dtf
which
sometimes the more convenient expression.
is
As an example,
parabola ay 2
=x
3
find the radius of curvature of the semicubical
(Art. 130).
Substituting in
^ = _J
*_**,
Differentiating,
(3),
we
find
a£(4 a
P==
164. Radius
Art. 163,
of
p = —q,
+9x)i
Wa
Curvature in Polar Coordinates.
let
us express p in terms of
ds_
We mayJ
write
d*
do
d<f>
d±
p
H
dO
From
(3),
Art. 156,
ds
= J?
I
From
(2),
o
,
fdr^
\
0.6
Art. 152,
*
*'
06
dO
r
and
Eesuming
6.
(1),
DIFFERENTIAL CALCULUS
198
From
(1),
Art. 152,
T
tan \b-=—,
T
dr r
il/
= tan _1
dr
dO
ld$)
'dr\
_ rd r
2
d&
d&__ \dd)
Differentiating,
T
f
d$J
.2,2^—Y-r
\d0)__
dcf>
Substituting,
dO
—
dO 2
dr
dO
.2,[dr
Hence
P
\dO.
=
(1)
^y
d0
2
EXAMPLES
Find the radius of curvature of the following curves
1.
y
= (x-l)
2
(x-2), at
(1, 0)
and
:
=\
Ana. p
(2, 0).
-L
and
Y2
2
= log x,
2.
?/
3.
The
4.
The
when
a;
=f
Ans. p
cubical parabola (Art. 130),
parabola, y 2
2
ct
y
The
3
.
Ans. p
—
T
4
.
= 4ax.
Find the point of the parabola where p
5
=#
= 2||.
equilateral hyperbola,
= 54 a.
2xy = a 2
Ans. x
= 8a.
2 -4-
.
Ans. p
= ±(x—
y2 )
2
•
2
:
CURVATURE.
6.
The
t.
ellipse,
+ 1 = 1.
Am.
P
=
199
W+
**)*
.
a 464
b-
ar
What
RADIUS OF CURVATURE
are the values of p at the extremities of the axes ?
2
Ans.
- and
a
7.
Show
£L .
b
that the radius of curvature of the curve,
+ y + 10 x — 4 y + 20 =
2
x*
is
constant,
and equal
to 3.
Find the radius of curvature of the following curves
8.
sH-log(l-x*)
9.
sin y
=e
= 0.
z
XT
The catenary
11.
The hypocycloid
12.
The curve aY = aV— a?
(Art. 128), y
(Art. 132),
a?«
^%s. p
+ y* = a*.
^^-
p
(Art. 133), at the points
Ans. p
cycloid, x
= a(0 — sm0),
y
=-
14.
Show
15.
r
Show
= a sin 6 416.
The
.
= £.
= 3 (aa;y)*0)
and
and p
= a.
(0,
= a(l — cos0).
Ans. p
(Art. 138), r
= e~x
°
= *(e° + e~"«).
(a, 0).
The
<}+%
Ans. p
.
10.
13.
=
Ans. P
= 4asin -'
that the radius of curvature of the logarithmic spiral
= e ae
,
is
proportional to
= r Vl +
p
r.
a1
.
that the radius of curvature of the curve,
b cos
0, is
spiral of
constant.
Archimedes (Art. 136),
p
r
= - Va + 6
2
= aO.
Ans. p
(r^-Mry
r
l
+2a
2
2
.
DIFFERENTIAL CALCULUS
200
17.
The
cardioid (Art. 141), r
18.
The
curve, r
19
The parabola
.
= a sin
165.
- cos 6).
(1
— a sec
(Art, 143), r
2
2
2
-
Ans. p
•
cos2
CP
curvature.
= 2a sec
^4ws. o
s
=
3r
Let
Coordinates of the Centre of Curvature.
of
= - ar.
2 -'
Ans. p=-c&sin
H
4
3
=a
AB, and C
ordinates of P, any point of the curve
centre
Ans. p 2
(Art. 145).
-
(Art. 139), r
The lemniscate
20.
3
=a
x,
y be the co-
the corresponding
is
then the radius of curvature,
and
normal to the curve.
is
Draw
also the tangent
CP =
Then
angle
Let
a,
;
PCR = PTX=
/?,
<f>.
be the coordinates
OM- RP,
LC=MP+RC;
OL =
of C.
that
p
PT.
is,
=x — p sin
P=y+P
a
<£,
cos'<f>.
To
and
express « and
(1)
in terms of x
/?
and
y,
we
have, by
(2), Art. 155,
(1), (2), Art. 163,
dy
ds dy
p sin</>
dcf>
_
ds
dy
d<j>
_dy dx _ cto
dx
d<fi
1
+
dys
dx
d 2y
dx2
pCOS
dx
dx
cji
\dxj
d 2y~
dx 2
Hence
dy
dx
1
+
~¥y
dx
2
£=2/ +
d*y
dx
2
(2)
CURVATURE.
RADIUS OF CURVATURE
Every point
166. Evolute and Involute.
corresponding centre of curvature,
their
respective
centres
Thus,
of a curve
P
201
AB
P>, P,, etc.,
x,
has a
have for
of
C C C etc, The
UK, which is the locus
curvature
curve
1}
2,
3,
of the centres of curvature, is
AB.
called the evolute of
To
express the inverse relation,
AB
is
the
called
involute
HK.
of
167.
To
find
the Equation
of the Evolute of a
By
(2),
Given Curve.
Art. 165, a and
/?,
the
coordinates of any point of the required evolute,
in terms of
x and
y,
may
be expressed
the coordinates of any point of the given curve.
These two equations, together with that of the given curve, furnish
three equations between a, /3, x, and y, from which, if x and y are
eliminated, we obtain a relation between a and /3, which is the
equation of the required evolute.
For example, find the equation of the evolute of the parabola
y-
Here
— a?x
-J-
dx
Substituting in
(2),
'
Art. 165,
a
Eliminating
x,
?,
— 4 ax.
-^ =
'
2
we have
= 3 x + 2 a,
we have
crx
dx2
j3
=
for the equation of the evolute,
«/3
?
= i(«-2«)
3
.
This curve is the semicubical parabola (Art. 130).
The figure
its form and position.
is the focus of the given parabola.
shows
F
OC=2a = 2 0F.
DIFFERENTIAL CALCULUS
202
As another example,
let
us
Y
find the equation of the e volute
of the ellipse,
9
2
-+2- =1.
a2
dy
dx
= _b
2
x
d 2y
ay
dx2
2
2
b
=
__&*_ #
a 2y
(Art. 66)
Substituting in
Art. 165,
(2),
To eliminate a; and y between these equations and that
of the ellipse,
x3
a3
a2
we
find
M-
s
aa
y
- b*
b
s
a2
b2
'
,V_ (q«)* + (&£)* __
u
(a - b y
x2
"
2
2
giving, for the equation of the evolute,
(aa)$+(60)*=(tt2 ^& 2)"*.
The evolute is EF'EFE. E is
P; Ffor JB; ^' for ^ i^ for B'.
centre of curvature for
A\
C
for
7
f
'
;
In the figure
F and _F' are outside the
ity is decreased, so that
168.
a
ellipse,
< b V2, these points
Properties of the Involute and Evolute.
equations, (1), Art. 165,
= x — p sin
£ = y -h p cos
a
<f>,
<f>.
but
fall
if
the eccentric-
within the ellipse.
Let us return to the
*
CURVATURE.
RADIUS OF CURVATURE
203
Differentiating with
respect to
sin
s,
da
_ dx
dp
ds
ds
ds
(p
— p cos —
a)
<f>
ds
ds
ds
ds
coscp-psincp^-.
Substituting in
P
= ^-
(2)
(1),
and cos 4,=
r ^,
ds
d<f>
two terms
(Art. 155),
cancel
each
other,
giving
Similarly in (2),
p=—
and sin
d<p
Dividing
But JS
(4)
i
s
by
(3),
dy
<f>
ds
d^
dp
ds
ds
3£=
(Art. 155), giving
(4)
L_
da
tan
(5)
<£
the slope of the tangent to the evolute at any point
d,
(see fig., Art. 166), and tan
at the corresponding point
</>
the slope of the tangent to the involute
Since by (5) one is minus the recip-
Pv
rocal of the other, these tangents are perpendicular
to each other.
In other words, a tangent to the evolute at any point
C\
normal to the involute at Pv
is
C^,
the
DIFFERENTIAL CALCULUS
204
Again, from (3) and
169.
daV
dsj
where
s
(4),
+ M?Y
\dsj
Art. 168,
fdp\*
\dsj
0T
Hence,
Hence,
s'
since, if a derivative is
nor decrease, but
It follows
is
from
ds\
ds
ds
±p=a
always
3
3,
170.
(1)
zero, the function
can neither increase
constant.
(1) that
As'
= ± Ap.
the difference between any two radii of curvature
PC
equal to the corresponding included arc of the evolute
CC
is,
is
J
constant,
A 0' ± p) = 0,
That
\dsj'
denotes the length of the arc of the evolute m&asured from
r
a fixed point.
P (7
(d*y = (dp
fdp\
\ds J
'
From
the involute
X,
X
X
Z.
the two properties of Arts. 168 and 169, it follows that
may be described by the end of a string unwound
AB
from the evolute
UK.
From
the word evolute
this property
is
derived.
It will
be noticed that a curve has only one evolute, but an infinite
number of involutes, as may be seen by varying the length of the
string which is unwound.
EXAMPLES
1.
Find the coordinates of the centre of curvature of the cubical
parabola (Art. 130), ahj
= x\
a<
lg x<
2
6ax
2.
^_9^
2 a4
Find the coordinates of centre of curvature of the semicubical
parabola (Art. 130), ay 2
= x\
Ans.
.
= --!£
^(* + |)^i
(
RADIUS OF CURVATURE
CURVATURE.
3.
205
Find the coordinates of the centre of curvature of the catenary
(Art. 128), y
=|
-»
e
+ <T°).
^ins.
Ans.
a
—
= xx-^^/
2_ a
y
a
4.
Show
relation a
5.
that in the parabola (Art. 129), x?
+ —3
/3
(as
p == 2y.
2
,
+ y).
Find the coordinates
of the centre of curvature,
Ans.
X
a=a +3x*y*
(«
}
s
and the equa-
-j-y 1
(3
=d
(y+ a?)3
« + j8=
g
a-
Thence derive the equation
,
s
.
= y + 3 xf y
s
,
+ j8)* + (« - P)*=2a\.
Given the equation of the equilateral hyperbola 2 xy
show that
j
+ y* = a we have the
tion of the evolute, of the hypoeycloid (Art. 132),
6.
2
«-P=
(y
—a
2
,
~^\
ar
of the evolute,
(«+/^-(a-/?)3 = 2al
7.
Find the equation
f = _ -— x
(X
.
of the evolute of the cissoid (Art. 125),
Ans.
4096 asa
+ 1152 a £ + 27 ^ = 0.
2
2
CHAPTER
XVIII
ORDER OF CONTACT. OSCULATING CIRCLE
Let us consider two curves whose equa-
171. .Order of Contact.
tions are
y=
and
<f>(x)
= if/(x).
y
If for a definite value a, of x, the value of y is the
curves, that
the
same
for both
is, if
common
have 'a
curves
point P.
If,
moreover, for x
value of
—
also
dx
is
= a,
the
the same for
,
both curves, that
<f}(a)
=
and
if/(a)
is, if
<£'(a)
the curves have a
The curves
common
tangent at P.
are then said to have a contact of the^rs^ order.
If besides, for
curves, that
= ^'(a),
x=
a,
the values of
—^
are
the
same
for both
x
is, if
<£(a) =ip(a),
<£'(a)
= i//(«),
and
<f>"(a)
=
if/"
(a),
the curves have contact of the second order.
In general, the conditions for a contact of the nth order at the
point x
*(a)
and
= a,
= ^(a),
n+1
<£
are
<£'(a)
= f(a),
*"(a)
(a)9^w+1 <>)206
=
f
(a),
••,
<£»(a)
=
^(a),
ORDER OF CONTACT.
OSCULATING CIRCLE
207
= a,
In other words, for x
dny
dx
must
daf
dx*'
have the same values, respectively, taken from the equations
all
of both curves
and
:
must have
different values.
dx"
When
172.
of
Contact
the Order of Contact
but
;
when
Even, the Curves cross at the Point
is
the Order is Odd, they do not cross.
Let us
dis-
tinguish the ordinates of the two curves by
T=€f»(x),
Y—y
= if/(x).
Y
figure, the curves
do not cross at P; but
P and negative on the other,
side of
y
refers to the full curve, and y to the clotted curve.
has the same sign on both sides of P, as in the first
In the figures
If
and
OM=a,
Let
MM =
1
Y—y is
positive on one
h.
Y-y==4>(a + h) -
AQi=
Then
if
the curves do cross at P.
if/(a
+ K).
R^rT
>
F
p^
^^
_
.
or
^y^
.
//
//
rs
/
/
'
/
F
y%~
s
//
Qs
QyY
/
/
/
/
// p
7
J
M9
M
Mj
2
M.
X
M
Mi
Expanding by Taylors Theorem,
a
\o
^(a)-^'(a)-|V"(a)-|V"(«)
13
(1)
DIFFERENTIAL CALCULUS
208
Suppose the contact of the
<£
(a)
= ^ (a),
^=|
For
order
first
<j>'(a)
— ^'(a)
*»(a)-f'(a)
;
then
and
,
becomes
(1)
+ |T*'"(«)-f "(«)] + •-
sufficiently small values of h the sign of the lowest
power
de-
PQ
will
termines that of the second member, and hence the sign of
remain unchanged when — li is substituted for h, giving
Y
PQ
the
(2)
•
2
2,
X
as in
first figure.
Thus when the contact
is of
the
first order,
the curves do not cross
at the point of contact.
Again, suppose the contact of the second order
cf>"(a)
™=i V
Now
= «//"(a),
"'(«)
and
- *"'(«)]
PiQi will change sign with
(2)
+ jj
J\,
;
then
becomes
[+"(«)
so that
- r («)] +
PQ
2
2
and
PQ
X
•
X
will
have
different signs, as in the second figure.
Thus when the contact
is
of the second order, the curves cross at
the point of contact.
By similar reasoning
may be of service
the general proposition
points,
when they have
consecutive
order
;
common
as having
n
established.
two curves as having two consecutive common
to think of
ciple,
is
to the student, in connection with this prin-
It
contact of the
first
order; as having three
when they have contact of the second
consecutive common points, when they have
points,
+1
contact of the nth order.
An
odd number of common points implies the crossing of the
where there is an even number of common points, the
curves, but
curves do not cross.
173.
and
Osculating Curves.
its first
Contact of the nth order requires that y
definite value of x, have
n derivatives should, for some
the same values for both curves.
This implies n
+1
conditions.
ORDER OF COXTACT.
OSCULATING CIRCLE
=
209
+
The equation of the straight line, y ax b, having only two
Hence
arbitrary constants, can satisfy only two of these conditions.
a straight line can have contact of the first order with a given curve,
and cannot, in general, have contact of a higher order.
The equation of the circle x2 + y 2 -\- ax + by + c = 0, having three
arbitrary constants, can satisfy three of the conditions.
may have
circle
Such a
Hence the
contact of the second order with a given curve.
circle is called the osculating circle.
whose equation contains four constants,
and the general conic, whose
equation contains five constants, may have contact of the fourth
These are called the osculating parabola
order with a given curve.
and the osculating conic.
Similarly, the parabola,
may have
174.
contact of the third order
;
Although the tangent
Order of Contact at Exceptional Points.
has generally contact of the
first order, it
may
at exceptional points
of a curve have a contact of a higher order.
For example, since the tangent at a point of inflexion crosses the
it follows from Art 172, that the order of contact must be
curve,
Hence
even.
at a point of inflexion
the tangent has contact of at
least the second order.
The
osculating circle, which has generally contact of the second
order, has a higher order of contact at points of
mum
maximum
evident from the symmetry of the ellipse with reference to
tices,
It is
its ver-
that no circle tangent at these points would cross the curve at
the point of contact.
odd,
or mini-
curvature, as, for example, the vertices of an ellipse.
Hence, by Art. 172, the order of contact
is
— at least the third.
175.
To Find the Coordinates
lating Circle at
Any
of the Centre,
and Radius,
of the Oscu-
Point of a Given Curve.
Let the equation of the given curve be
y =/(»)•
The
general equation of a circle with centre (a, b) and radius
(x-a?+(ij-by-=^
r, is
(1)
DIFFERENTIAL CALCULUS
210
Differentiating twice successively,
we have
(2)
HW
From
(3),
y
+(y - b)d
A=°-
(3)
-b=
(4)
dx 2
From
(2),
x— a =
dy
1
dx
+ f^V"
dx
1
(5)
c^y
dx
Substituting (4) and (5) in
1
2
(1),
iff
dx
(6)
dx )
2
Hence
a
— x—
dy
dx
x\_
\dx
dx
and
,
h
d*y
Wt
C£#
2
2
>
= y+
.
\dx)
^f-'
dx
fTs
()
2
(8)
ORDER OF CONTACT.
In these expressions,
x, y,
—
,
OSCULATING CIRCLE
—
,
211
refer to (1), the equation of the
but since the osculating circle bv definition has contact of the
second order with the given curve, these quantities will have the
same values if derived from the equation of this curve y=f(x), and
circle
;
applied to the point of contact.
By comparing
Arts. 163, 165,
and
and
with the expressions for
a,
ft,
evident that the osculating circle
is
the same as
(7)
it is
(8)
p,
in
the circle of curvature.
At a Point
176.
of
Maximum
If
we regard equation
(8) in
the given curve y =f(x),
minimum
value of
Minimum
or
the preceding article as referring to
we have
as a condition for a
maximum
r,
^
=
dx
We
Curvature, the Osculating
Third Order.
Circle has Contact of the
thus obtain from
(8),
M2'
dx\da?)
3
tlL
dx*
from which
0.
=
dx*
dyfdy
* x[
']
f\
.
im(<&\*
\dx)
Again,
if
we regard
(8) as referring to
the osculating circle
(x-af+(y-by=r\
we
shall also
since r
is
—
=
dx
have
constant for
all
0,
points on the circle.
or
DIFFERENTIAL CALCULUS
212
Thus we
obtain, both for the curve
have, at
follows that
circle,
the same ex-
J,
and since -^ and —\2 in the second member of
dx
dx
the point of contact, the same values for both curves, it
pression (1) for
(1)
and the
d3 v
t
—
dx3
has likewise the same value.
Hence the contact
is
dx3
of the third order.
EXAMPLES
Find the order of contact
1.
=
y
and
x?,
of the
y
two curves,
= 3 x — 3 x + 1.
2
By combining the two equations,
common to both curves.
the point x
=
1,
y
= 1,
is
found
to be
Differentiating the two given equations,
y
=x
3
dx>-
^=
da?
when
but
—4
dx3
^
2
6aj--3,
dx
bX
d2y_
=
dx2
6,
6,
d3y_
dx?~
0.
'
dy
=
^
dx
=
x
= 1,
3,
in both curves
x
= 1 ^4 = 6,
in both curves
,
dx2
has different values in the two curves.
Hence the contact
2.
= 3 x — 3 x + 1,
=
'
dx
When
y
,
is
of the second order.
Find the order of contact- of the parabola, 4?/= a?2 and the
= x — 1.
Ans. First order.
straight line, y
,
ORDER OF CONTACT.
OSCULATING CIRCLE
Find the order of contact of
3.
9y = a*-3x* + 27 and
)
9//
+ 3.r = 2S.
Second order.
Ans.
4.
Find the order of contact of the curves
y
at the
5.
= log(x — 1),
common
point
- 6a; + 2y + 8 = 0,
— 2 y = 3.
What must
may have
Second order.
^l»s.
(2, 0).
4,y
= x~ — 4,
Ans.
be the value of
y
a, in
and the
Third order.
order that the parabola,
= x + l + a(x— l)
2
,
contact of the second order with the hyperbola,
xy
7.
and x2
Find the order of contact of the parabola,
circle, or -f y-
6.-
213
— 3x — 1?
Ans.
a——
1.
Find the order of contact of the parabola,
(x-2a) 2 +(y-2ay=2xy,and the hyperbola,
xy
=a
2
.
^4?is.
Third order.
CHAPTER XIX
ENVELOPES
177.
When,
Series of Curves.
values are assigned to one of
in the equation of a curve, different
its
constants, the resulting equations
represent a series of curves, differing in position, but all of the same
kind or family.
For example,
if
we
give different values to a in the equation of
=
the parabola y 2
£ ax, we obtain a series of parabolas,
common vertex and axis, but different focal distances.
Again, take the equation of the circle (x
giving different values to
centres are on the line y
we have
a,
=
— a)
2
having a
all
— b) = c
2
2
-f (y
.
By-
a series of equal circles whose
b.
The quantity a which remains constant for any one curve of the
series, but varies as we pass from one curve to another, is called the
parameter of the
series.
Sometimes two parameters are supposed to vary simultaneously,
so as to satisfy a given relation between them.
Thus, in the equation of the circle (x — a) 2 + (y — b) 2 = c 2 we may
suppose a and b to vary, subject to the condition,
,
tf
We
+ b = tf
2
then have a series of equal
circles,
whose centres are on
another circle described about the origin with radius
178.
The
fc.
two curves
two curves approach
we suppose the parameter to vary by infinitesi-
Definition of Envelope.
intersection of any
of a series will approach a certain limit, as the
coincidence.
Now,
if
mal increments, the locus
curves
is
of the ultimate intersections of consecutive
called the envelope of the series.
214
ENVELOPES
The Envelope
179.
of a Series of
Curves
215
is
Tangent to Every Curve
of
the Series.
P
Suppose L. My
intersection of
N
to be
Q
auy three curves of the
series.
M with the preceding curve L, and Q
its
P
is
the
intersection
with the following curve N.
As the curves approach coincidence,
P and Q will ultimately be
two consecutive points of the envelope and of the curve M. Hence
the envelope touches M.
Similarly, it may be shown that the envelope touches any other
curve of the series.
180. To
find the
Equation
cf the
Envelope
Before considering the general problem
of a
Given Series
let
us take the following
special example.
Required the envelope of the
by
series
of straight lines represented
m
= ax -f —
.
y
a
a being the variable parameter.
Let the equations of any two of
these lines be
and
y
= ax + m
y
= (a +
From
(1)
equations,
tion of the
and
(1)
(2)
we can
two
+h
a
as
find
lines.
(2)
simultaneous
the inters
Subtracting (1) from
(2),
of Curves.
DIFFERENTIAL CALCULUS
216
hm
= to-
+ A)
a(a
—
™
=£
or
From
(3)
and
(1),
m
if
(2a
.
we suppose h
to
.
.
.
.
(4 )
of the intersection.
approach zero in
(4),
we have
for the ulti-
of consecutive lines
m
m
2
a
ar
By
+ h)m
,
which are the coordinates
Now
(3)
we have
x=
mate intersection
,
eliminating a between these equations
2
y
we have
= 4 ma?,
which, being independent of
a, is
the equation of the locus of the in-
tersection of any two consecutive lines, that
is,
the equation of the
required envelope.
The
figure
shows the straight
lines,
and the envelope, which
is
a
parabola.
181. We will now give the general solution.
Let the given equation be
f(x,y,a)
= 0,
which, by varying the parameter
To
find the intersection of
a,
represents the series of curves.
any two curves of the
series,
we com-
bine
f(x,y,a)
and
f{x, y,a
= Q,
+ h) =
(1)
(2)
,
ENVELOPES
From
(1)
and
217
we have
(2),
f{x,
a + h)-f(x,
?/,
y, a)
_n
and
(1)
it
is
and
^x
(;
'
A
evident that the intersection
may
be found by combining
instead of (1) and (2).
the two curves approach coincidence, h
(3),
When
and we have, by Art.
|-/(.r,2/,«)
Thus equations
(1)
approaches zero,
15, for the limit of equation (3),
and
(4)
=
(4)
determine the intersection of two con-
By
eliminating a between (1) and (4) we shall
obtain the equation of the locus of these ultimate intersections,
secutive
curves.
which
the equation of the envelope.
is
182. Applying this method to the preceding example,
= ax + m
—
.
y
we
differentiate
a
with reference to
a,
and obtain for
(4) Art. 181,
= x --a2
Eliminating a between these equations gives the equation of the
envelope,
y
183. The Evolute
This
is
2
=
4:
of a
mx,
r
as found'in Art. 180.
Given Curve
is
the Envelope of its Normals.
indicated by the figure of Art. 166, and the proposition
may
be proved by the method of Art 181,
The general equation of the normal
Art. 148,
x -x! +
as follows
:
at the point (V, y') is
&(S -y = 0,
by
l
)
(1)
DIFFERENTIAL CALCULUS
218
in
which the variable parameter
is
x',
dn'
the quantities
y',
-~, being
dx'
functions of
From
Differentiating (1) with reference to
x'.
and
(1)
(2)
we
we have
the intersection of consecutive
for
find
x',
normals,
\dx'
y=y +
,
,
dx' 2
dy'
HSf\
dx'
x = x' —
d 2y'
dx' 2
As
these expressions are identical with the coordinates of the
centre of curvature in Art. 165,
it
follows that the envelope of the
normals coincides with the evolute.
EXAMPLES
1.
y
Find the envelope of the
— 2 mx
-f-
m
m
4
,
series of straight lines represented
by
being the variable parameter.
Differentiating the given equation with reference to m,
= 2^ + 4m
Eliminating
Find the envelope
y
2
.
m between the two equations, we have for the envelope,
16?/ 4-27x = 0.
3
2.
3
= a (x — a),
4
of the series of parabolas
a being the variable parameter.
ky2 = x2
Ans.
.
3. Find the envelope of a series of circles whose centres are on
the axis of X, and radii proportional to (m times) their distance
from the
origin.
Ans. y 2
=m
2
(x2
+y
2
).
m
\
ENVELOPES
219
Find the evolute of the parabola y 2 = 4ax according to Art.
l^o. taking the equation of the normal in the form
4.
y
5.
= ni (x — 2 a) — a
Find the evolute of the
normal in the form
s
Ans. 21aif
.
— + ^=1,
ellipse
= 4 (x — 2 a)
3
.
taking the equation
of the
by
where
<£
is
= ax tan —
<j>
(a
2
—b)
2
<j>,
the eccentric angle.
Ans.
6.
sin
Find the envelope of the straight
(ax)*
+ (by)* = (a — b
2
lines represented
2
)
by
+ y sin 3 = a(cos 2 0)*,
x cos 3
6 being the variable parameter.
Ans.
(x
7. Find the envelope of the
and whose area is constant.
The equation
2
+y =a
2
2
2
)
— y ),
(x2
2
series of ellipses,
the lemniscate.
whose axes coincide
of the ellipses is
£,+£=i
a-
a and
b
a)
cr
being variable parameters, subject to the condition
ab
calling the constant area
-n-k
=k
2
,
.
Substituting in (1) the value of b from
(2),
+ &-1,
K
ak
4
in
which a
is
reference to
the only variable parameter.
a,
we have
(2)
2
Differentiating (3) with
-^+^ =
a8
Eliminating a between (3) and
4 xhf
Jr
(4),
=
k\
(3)
we have
(4)
DIFFERENTIAL CALCULUS
220
Differentiate
Second Solution.
regarding both a and b as
(1),
variable.
x?da
a3
j
b
Differentiating (2) also,
b
From
(5)
and
(6),
da
2
y db
=^
/
+ a db =
(6)
.
we have
9
9
2
2
a
(7)
and
b
W
(7)
(1),
t=t=\
a2
Substituting (8) in
b
2
Find the envelope of the
(8)
=
&4
.
whose diameters are the double
circles
2
ordinates of the parabola
9.
W
2
(2),
4 x 2y 2
8.
y
= 4 ax.
Find the envelope of the straight
Ans. y 2
-
lines
a
when
10.
an
+ bn = kn
Find the envelope of the
ellipses
a
+b=
= 4 a (a + x).
+ ^ = 1,
b
.
_^
JL_
_
—
X2
a2
when
n
we have
£=£
From
5
3
V
2
\-¥b-
—l
9
2
-4ns.
fc.
#3
+
2
3
2/
11. Find the envelope of the circles passing through the
2
whose centres are on the parabola
y = 4 ax.
Ans.
(x
+
_n_
2a)y 2
+
=&
2
3
.
origin,
x?
= 0.
ENVELOPES
12.
an
Find the envelope of
circles described
221
on the central radii of
ellipse as diameters, the equation of the ellipse being
a~
o-
13. Pind the envelope of the ellipses whose axes coincide, and
such that the distance between the extremities of the major and
minor axes
is
constant and equal to
Ans.
A
k.
square whose sides are (x
±
2
y)
=
k2
.
INTEGRAL CALCULUS
:>**c
CHAPTER XX
STANDARD FORMS
INTEGRATION.
184. Definition
entiation
is
The operation inverse
of Integration.
By
called integration.
to differ-
differentiation Ave find the dif-
and by integration we find the function
ferential of a given function,
corresponding to a given differential.
This function
is
called the
integral of the differential.
For instance;
2xdx
since
x2
therefore
The symbol
following
I
the differential of
is
x
2
,
the integral of 2xdx.
is
used to denote the integral of the expression
is
it.
Thus the foregoing
relations
d{x2 )
=
would be written,
2xdx,
p2xdx
=
x\
same thing, whether we consider this integral
whose differential is 2xdx, or the function whose
It is evidently the
as the function
derivative
As
is
2x.
regards notation, however,
|
2zdx
=
2
:c
,
it is
customary to write
and not
223
j
2x
=
xr.
/d
INTEGRAL CALCULUS
224
In other words,
is
Thus the general
the inverse of
definition of
differential is <^{x)dx
—
dx
that function whose
<1>(x)dx is
I
the symbol
;
and not of
d,'
denoting " the function whose
|
same way that the inverse symbol, tan -1
denotes "the angle whose tangent is."
differential is," in the
Integration
sists in
known
,
not like differentiation a direct operation, but con-
is
given expression as the differential of a
recognizing the
function, or in reducing
it
form where such recognition
to a
is possible.
185.
Elementary Principles.
(a) It is evident that
I
we may write
= x* + 2,
2 x dx
or
I
=x
2 x dx
as well as
x2
-f-
2,
J
where
C denotes
;
x2
as well as of
2 x dx
In general
2
2
J
since the differential of
= x — 5,
2 x dx
=x +
—5
2 x dx.
is
2
(7,
an arbitrary constant called the constant of integra-
tion.
Every integral
+
(6)
it
in
its
most general form includes
this
0.
d(u ±v ± w)
Since
= du ±
dv
±
=
±
i
div,
follows that
I
(du
±
dv
±
dw)
I
du
dv
±
I
div.
term,
.
STANDARD FORMS
INTEGRATION.
That
we
is,
integrate a polynomial
225
by integrating the separate
terms, ami retaining the signs.
(c)
it
d(au)
Since
follows that
That
is,
symbol
|
aclu
may
a constant factor
= adu,
=a
dw.
j
be transferred from one side of the
without affecting the integral.
to the other,
J
Fundamental Integrals.
186.
differentiation, to integrate
Since integration
is
the inverse of
any given function we must reduce
it
to
one or more of the differentials of the elementary functions, expressed by the fundamental formulas of the Differential Calculus.
Corresponding to these formulae we
may
write a
which may be regarded as fundamental, and
should,
if possible,
in this chapter
be ultimately reduced.
II.
III.
frcfe-Jf-L.
J
n-f-1
Jy = logu.
Ca*du=
J
,—
log a
u
»• /•e"du = e
.
[
r
V.
.
VI.
We
list
of integrals,
which
all integrals
shall then consider
such examples as are integrable by these formulae,
some simple transformation.
either directly, or after
I.
to
I
j
cos u du
I
sin u
du
= sin u.
— — cos u.
INTEGRAL CALCULUS
226
VII.
)
sec 2
VIII.
I
cosec 2 w
IX.
|
sec
X.
u
= tan u.
die
du
= — cot w.
u tan udu
= sec w.
cosec u cot icdu
= — cosec w.
J
= log sec u.
XI.
j
tan
XII.
I
cot
udu = log
XIII.
j
sec
u du
XIV.
it
du
sin u.
= log (sec w + tan w) = log tan
cosec u du
7T
-
f
= log (cosec u — cot w) = log tan
I
<S
^t-xt
XV.
S XVI.
vvt
\X
,
XVII.
XVII.
XVIII.
1
C du = -tan
-1 u
-,
a
J vr + a a
XIX.
XX.
or
=
2
i
-
=
f—
I
Vcr — u2
= sin
*
—
,
1
,_i1
cot
a
u—a
C -^ du = —1-log-—
—
u-\- a
J vr — a? 2 a
or
,
or
u
-a
1
a
= —log,
2 a
= — cos
a
l
a
+u
a
du_
r_J?^=
log (u+v^tf)
•/
ATT AT
i
,
j
VV ± a
C
WU
I
J
y
2
M
Vw _ a
2
—-sec _1 -,
2
J-
'<*.
a
a
"
f V2 aw — w =
r/
•^
2
vers
'
a
or
=
cosec
a
a
tfc
+2
,
-
:
STANDARD FORMS
INTEGRATION.
INTEGRALS BY
187.
To
Proof of
derive
I.
and
I.
AND
227
II.
II.
I.,
d(u n+l ) = (?i
since
-f-
n
1) u du,
therefore
«
n+1
= f(n + l)u n du = (n + 1)
/
Formula
u
n
du,
by
(c),
Art. 185.
u n+l
=
du
Cu n
•
n
+1
II. follows directly
from
,j
log u
d
i
/
= du
u
It is to be noticed that
For this value
it
v
/•
Formula
II.
I.
applies to all values of n except
n = —1.
gives
l
du
u°
=—
=oo
„
provides for this failing case of
I.
EXAMPLES
Integrate the following expressions
1.
CaHkx.
If we apply L,
we have
j
A
(a),
<hr
=
'—
+
Art. 185.
C,
calling u
= x,
and
n=4;
then du —dx.
adding the constant of integration
Then
C, according to
INTEGRAL CALCULUS
228
C(x 2
2.
+ l)hdx.
we apply
If
We
I.,
=x
2
u
calling
-j-
and n
1,
=-
;
then cfa*=2 #d#.
must then introduce a factor 2 before the
quently
its reciprocal
f
(<e*
- on the left of
2
(a;
2
2
and conse-
|
+ l)*a> da; = i f(» + 1)*2 x dx,
_l
icda?,
by
Art. 185.
(c),
+ l)f _0g_+l}j
3
3
2
r (a; --a )cfa = l %3a - 3a )(fo
-3a x
J x*-3a x 3 J
2
2
2
2
3
2
/
3
'
2
a;
= |log (a - 3 aV) = log (or - 3 aV)* +
3
5
C.
o
By
introducing the factor
4.
we make
3,
f(2x»-3x«+12x -3)dx = — 3
O
%s
5.
f(»f_l- +
c/
\
^f
2_2V
5
ct-
f(aj 2
3
5
8
-2)Vcto=- -^- +
*/
7.
f(a^-2) 3a?da; =
«y
10
4
^~ 2
8
—
4
)
-|-0.
differ-
II.
?jl
+ 3x*-3x +
C.
4
3»t_
a;y
}
6.
the numerator the
and then apply
ential of the denominator,
o
^_
1 _
2
2x*
-2* +
4
<7.
_
STANDARD FORMS
INTEGRATION.
CI
}\
\
3
9ctiv$
7
9oU
,
x-
,
229
~
2a
10
(7 *F
4 /«
v *a
11.
3
Y.
^
4a#
12 a£
13a*
7a*
^ + i^+
fcil ^ = ^^ +
jy»i + aJ-lY(to =
,
5^
6» + 12**-3ar* +
S
12.
J
J
=
14.
JV +
15.
j%.x~
+ 6)Wi\
17.
C(aa?
+ b)*a?dx.
21.
5
x2 da;
Ju-i-l/"^'.
(
a?3
a
4
7
xi
1)
32,-log y+
2
3
7/
11,4a*', „
,
6
+ 1 ) + C.
,
n
16.
C(ax
18.
f(oaf + ft^a?"- 1 da?.
22.
-\-b) dx.
J(a + logOy-
0.
INTEGRAL CALCULUS
230
23.
25
-
f.
x
-\-xe
$\J^zf + *£-
27.
f(e +
28.
I
2
*
f(sinm
31.
f(sec
10
<£>
-1
r
A rational
w
-f cos 0)
+ tan 0)
("(sin
sin
34
sin 20)(e 2
e
+
2x
da.
— log
10
/
(
sin5
29.
I
sec 5 a tan
cos
x)
dx
x
(9
^6>.
cos 20)d0.
x dx.
cos 6 dO.
sin
sec <9d0.
+ cos <£) w (sin
2
<f>
- cos
2
a £fa
C(a x
33.
<£)d<£.
+
&)
3
a x da.
da
35
'
fraction,
j (a
26.
tan 6 x sec 2 x dx.
30.
32.
24.
-dx.
e
whose denominator
J(l + x~)
is
tan -1 a
of the first degree,
may
be integrated directly or after being reduced to a mixed quantity.
36
-
37
33
•
/iSri log(4 *Jfffi^ = ^-^
2
4
J
2x-l
3>
+c
-
1(
3
2
2
4
V
y
STANDARD FORMS
INTEGRATION.
on
39.
+
J
.
f
(- r
«y
42.
.17
(
A
log
n
C.
+ a)\ + r\
,
,
(&as
b~
—+
1
« 2 -«
2
3
+a
8
log (»
+ flV <& = f + 2 as - fcc + (a f
'
- a) +
&)'9
G.
log (x
+ 6) +
<7.
—
-}-
s
+ ^
J x—a
f
1
,
b
—a
x
2
.
a
C^LtJ^ civ = - 4-
40
41
J
,
dte
;
bx
ax
6 —a
=—
+
s
Cost -f b
231
<*b
= - + 2 oas + 7 a * + 8 a
2
2
3
log (x
3
INTEGRALS BY
III.
AND
- a) +
C.
IV.
188. Proof of III. and IV. These are evidently obtained directly
from the corresponding formulae of differentiation.
EXAMPLES
.
f(^ + a* + 3b-*ydx=^.+ i^
5 log a
2.
-l^-f
2 log b
f<>« + e- ax fdx = -\ e-^ + 3e ax -3 e~ ax Cl\
a\_
o
3
*J
.
&e
,
r
4.
I
I
j
,
1
,
/2a 2
4x
7a
9a 3
,
18a 6
fa
J
L
%&*
cto =
4
—
6 e3
h
2
C
+ 0.
INTEGRAL CALCULUS
232
5.
C(e x+a
- eax+h ) dx.
6.
C(e 5inx cosx
8.
fte
— acos2x sm2x)dx.
i
C(e
7.
x
\ + -\dx.
J
e
tan
*
J
2*
sec
6>
V
J
-e nan0) sec 0d<9.
sec
;
log(«6 2 )
log (a m b p )
v
J
;
21og6
2 log a
log (a&)
INTEGRALS BY V.— XIV.
—
—
X. are obtained
XIV. It is evident that V.
of V.
from the corresponding formulas of differentiation.
To derive XI. and XII.,
189. Proof
directly
tan
udu = —
/, udu= J
,
cot
= — log cos u —
|
J
cos
— udu = log sin
(* cos
I
log sec
u.
u
i
:
u.
sinw
To derive XIII. and XIV.,
/'
-j
_ r sec u (tan a 4- sec u) du _ /~ sec u tan u du + sec-u du
J
J
sec u + tan u
sec 4- tan-%
it
= log (sec u 4- tan u)
/«„„«„
cosec
udu=
-
,7
.
Tcosec w (—
cot
i
I
J
cosec
z*
4- cosec w)
u— cotu
= log (cosec u — cot w).
l
aw
—
•
STANDARD FORMS
INTEGRATION.
By
Trigonometry,
cosec u
——sm
substitute in this,
we have
sec u
2 sirr -
cos u
— cot u = 1
we
If
^+u
a
1.
o
•
•
I
.
u
it
—
_
•
2,
2 sin -cos-
for u,
+ tan w = tan
Cf sm 3o x + cos o- x —
= tan u-
2
=
:
Thns we obtain the second forms
,
233
sb\
sin -
-
(
+-
of XIII.
7
ax
and XIV.
—-
cos 3
—
aj
.
—x
sin 5
-\
+ 2cos^+C.
J
o
3.
\
sm —-*m
1-
—
—n!—
cos
cto
J
= — m cos ^ + n sm —^— +
m
n
1
CI 4-—sin mx dx = —
(tan mx
J COS' 0KC
,,
7
:
|
,
-\-
mx)x
sec
,
AT
4- C.
??i
4.
(sec 5 x
I
J
c>
8.
9.
f*
a?)
sec 5 x dx.
(sec 2
4-
tan 2
^——y = cos x — 2 log (1 4- cos #) 4- C.
—vers
sm./;
y
C^pdx.
10.
c
r
I
J
da;
I
J
J
5.
smdj
\cos-6
•/
n
— tan 5
seo»rf»
a sin
£ + & cos
=1
<£
a
,
,,,
tan
4>
'.fl?™!***
J
sin-x
+ b) + a
6) dO.
C.
INTEGRAL CALCULUS
234
— cot# + l)
12.
I
(tan x
13.
I
(sec 2x-\- tan 2
a?
2
= tan + cot — 3x — 2 log sin 2 x + C.
dsc
—
cc
a;
cot 2
a?)
2
c?#
= tan 2x
— log tan — 4 x
I
(sec
<f>
+ cosec — I)
cot 2
a;
= + tan — cot <£
2
<£
4>
dcf>
<f>
2x — -
sec
-j- (7.
ce
14.
-\-
°s
+ 2Iog*1 + csin
+ f+a
<£
The following may be
integrated after trigonometric transforma-
tion.
sin2 a?
=
Csin*xdx
15.
\.
a
lb.
1
17.
in
19.
I
Ccos^
J
.
(
J
= -»
7
2
|
a;
*-™^+a
4
cfe
—
,
sin 2 x
1-
vers 2a; dx.
18.
f.sin mOa sin nOn dOn =
J
.
oi
21.
C sm m0a cos w0a d0
ja =
-
•
t> a?
—— + C.
=
x
——
16
sin(m
+ %)0
J
^
—+
2 (m -f w)
,
n
C.
— ?i)0
sin(m + w)d
— + -ttt
—r~ + ^"•
—
2(m-j-w)
2{m n)
cos
J
,
,
(m
— w)0
—
.
1
—
sin 1 x
14
—
!
2(m— w)
sin 3 x
j
o £(&«:=
cos 2
6
+
(m w)0
f—-)
2 (m+n)
cos
'-
J
|
x cos 2 x dx
\
j
Tcos K
2
'-
sin(m
7/1
|
J
sin
(m — n)0
)
2 (m — n)
-
C cos m0a cos w0a c?0 =
J
on
22.
j
J
sin
-,
I
-
^
C.
4
20.
nr\
.
-\
2
,
\-
n
C.
'
n
-f C\
STANDARD FORMS
INTEGRATION.
fsin (3 x + 2)
23.
nA
24.
25.
27.
Tsm
J
I
sm 3o
.->
sin 2 x
.r
C^MdO.
J
.r
7
=
dx
—24—6
cos
<
cos 4
a;
fl
J
—^— = tan0-sec0 +
-dx —
rfg
— cosec
vers x
VI + sm x
30
31.
dx
fVveraada?.
= C
I
J
f-^-.
J vers
— log vers + C.
a;
,
a;
+ cos
To
— XX.
Proof of XV.
derive XV.,
r^^ = i f_W
a
c_±!_ = i
J + a a .7
2
2
it
i
i
?L
1
cr
To
(-]
\aj
derive XVII.,
du
I
•J
•=
Va —
2
<•-
— SID-1-.
I
^
}_'_!!
\
a2
C.
= J- log tan (^ + ^+
87
V2
V2
INTEGRALS BY XV. — XX.
190.
^
.x*
^
sin
,
-f C.
— — 2„ VI — sin +
a; cfce
f-
32.
J
cos
V 1 — sin
J
sc
8
= 2si n g + 2gin3fl + C.
28.
O.
—cot x
a?
cos 2
o
sin
Jf 1 + sm
a;
- C os( ^+^ + C.
10
rg?°i-
26.
sin
29
**(* + *)
+ 3) dx =
cos (4 •
235
a
= i taa -i«
a
a
C.
INTEGRAL CALCULUS
236
To
"
derive XIX.,
K
/du
1
2
uJ u ly?_
1
2
a^a
To
derive
— 1 SeC _ ,u
'
a
a
2
XX.,
du
/du
C
a
J \u
^/2au — u 2
\2
* a
tan-1 ^
a
Since
it is
_
I
3;:;:
wVw -a
du
a
r
evident that
d tan-1 -
=
7r
vers"
u2
2
or
-cot-
2
1
^,
a
= d ( — cot -1 -
)
•
Hence either expression may be used as the integral in XV.
In the same way we obtain the second forms of XVII. and XIX.
The formulae XVI. and XVIII. are inserted in the list of integrals,
because their forms are similar to XV. and XVII., respectively, with
different signs.
To derive XVI.,
u2
—a
2
2 a\u — a
u
+ aj
hence
/
du
u2
—a
2
1 Cf du
2aJ \u — a
h
[log (w
du \
u
+ a)
- a) ~ log (M + a)] =
£ hi
log
:
STANDARD FORMS
INTEGRATION.
237
Or we may integrate thus
r (hi _ \_ r r -du _
J ir~ a 2 a J \a — u
2
=
To
A
[log («
(7?/
+u
a
- u) - log (a + «)] =
A
log
j a
a
a
+U
derive XVIII.
-\vF±a- =
assume
u2
Then
a
z,
±a = z
2
new
variable.
2
,
2udu = 2zdz;
therefore
du
dz
z
u
du
f^ =
(
Hence
1 1S '
J v^±^'
+ dz
u+z
C d " + (fe = log (u + z)
= log (w + v%2 ± a2)
;
-
EXAM PLES
J4.r + 9
3.
4.
5.
I
—====== = -sm
J VT-25^
P
f
•^
dj;
J4a; 2 -9
3
6
5
*
\-
V5 a^ + 1
C.
2
= I log (5 + V25 ar - 4) +
a;
r? 'r
=
—
Vo
12
log (ars/5
C.
+ Voa?+l) + 0.
°2a; + 3
INTEGRAL CALCULUS
238
J3-12x
2
&
12
2a?-l^
dy
/
12
C
dx
^V9ar — 4
/:V ma? —
15.
16.
2
vers -lx
a?
J&VaV-16
2
a?
=-sec-
1
4
1
l
2
17.
18
a?
da;
P
= sin"
1
J a?V4— (log a?)
20.
3
a;
da?
2
.
h C.
g2 + g
—x
-8
7
log
l0g(^ +9)
8
~2
Jf V9-a?
J V3~a^T2
2
Vx + C.
2
-/SS^ = i
J4a?2 -5
C dx
J 3^-5*
^
'
m
dx
_,
^ = -±1 vers
f _====r
vers"
2
V7 — 4
a?
'
2
2
^
f
q
-3°
= lsec-^+G
=
dec
»
2
J V2 - 3 a
'
2
•^
dw
'Jm<;
^ V3* -2
13.
C
g
3*
f+
^i
"1 +a
tatl
2x+V5
4V5
= - 3 V9 - x -2sm- ^+C.
2
l
3
a
STANDARD FORMS
INTEGRATION.
21.
22.
23.
24.
25.
x
_± ___
JfV .r + 4
1
dx
J V3.r-9
J
Jf1 +
rf
»
y2
<#>
si °
1'
f e_!l±^!
same
/j
2
.1-
formulae
or
J
**
f
r7x
x
+
may
(x
+ J_ tan -i
2a;-4) + (7.
be
applied
_£_
to
+c
integrals
by completing the square
+ 3/ + 4
_-
= f
J
a 9
= Jl.
a7+ 2
2
"
involving
with,
Thus,
x.
Jf v/8 + 4 a; — 4 .r
-'a
2x
2
Ja* + 6a?+13
30.
(e
/
= -^Iog(3cos2s+V9cos
— x + ax + b,
terms containing
29.
-
— 4 sin- 2 i
= 1 log
tfa.
2
V
2£J_^1
Vo cos
+
C.
6
fl
S
- ax
V3 + V3 ^ - 9) +
=CO gV«*JUft
+
V3
The
log (a
V2
^
g
/
27.
«6
2
A/Sp.ns
cos- fl4-a.siTi
4 sin-2
f
*^
s
3
(7.
= -Ltaii-'gg4 + C.
cos-
r
—
V3
= 5 V3 X - 9 -
+ 6- cos-
cr siu-
•/
26.
5x
C
= ^ ^T4 -f 3 log (x + V.^+4) +
c?.r
239
V3
W
2
2
«**
(2
(3
a?
jc
- l)
2
^sin-'g^
+ C.
3
2
- 2 + V9ar>- 12 + 6) + C.
a;
the
INTEGRAL CALCULUS
240
31.
32.
x2
2 ar
/'—
or
33.
*5
f
J
—3 +5
dx
— ax + 9
O/i
35.-
,
(a;
when a = 4
+ 6)
2
eos2ecW
;
when a = 6
1
=
^ ^= =
===
r—
I
V31
V31
A
/si
V31
-*?
Jf-(a> + a) +
2
34.
= -4,tan-i^4?+C.
a;
a?
Va + &
2
1
-
sm -1
•
1
2
4
2
JB
——1
37.
8.
1
b'
C.
+n
.
sin2e
= ±log
Jfsin 20+msin20 2m & m + sin2
+a
- =sm-i 2x a b +C.
dx
JfV(a-a)(&-
when a =
tan-^^±^±J + a
Vet +
2
36.
;
a-&
a-)
2
a
&)V3 +a
*L
=
tan-i(^ + +
—
Jf(x + af-(x
a) — (a? + by
/q
b
6V
a
_
(„
a
m*
V3(a-6)
(a?
3
3
2
CHAPTER XXI
CONSTANT OF IN-
SIMPLE APPLICATIONS OF INTEGRATION.
TEGRATION
Before continuing
the integration of functions,
we
will consider
the relation of integration to the determination of the area bounded
by a given curve, and show how the constant
of integration
may
be
determined.
191. Derivative
curve
P
(„
OPv
and
let x,
an Area.
of
Suppose a point
y, be the coordi-
Let y=f(x) be the equation of a given
move along the curve starting from
to
nates of any position P.
At the same time the ordiof the moving point
starts
from the position
nate
P^lf,,,
and
generates
When
sweeps over or
a
certain
area.
moved
PqMqMP.
the point has
to P, this area is
Denote this area by A.
A is a function of x, and it
will now be proved that its derivative with respect to x
A.? = MN.
AA = PMNQ.
AA > y Ax, and AA <
AA
and
is
equal to
y.
Give to x the increment
Then
-r->y>
Ax
Hence
rl_A
= Lim
—
<
Ax
(y
+ Ay) Ax,
y + Ay.
AA
.'/•
dx
P
In case the curve descends from
to Q, the
will be reversed, but the result will be the same.
241
above inequalities
INTEGRAL CALCULUS
242
Let it be required to find the area PqMqMxP^
192. Area of Curve.
between the curve, the axis of X, and the two ordinates P
Q and
M
Pallet
0M = a, and 0M =
From
X
Q
b.
—
dA =
the preceding article
y.
dx
A=
Hence
Let
ydx=
j
i
f(x)dx.
Cf(x)dx=F(x)
•
0}
A = F(x) + G
then
To determine C, we have
when x = a.
is, A —
(1)
the condition that
A begins
when x = a
;
.that
C=- F(a).
(1), A = F(x) - F (a) = P M MP
= F(a) + C,
Hence
Substituting in
It is to be noticed that
corresponding to the
If
now we
let
x
C is
determined by the
initial ordinate
=b
(2)
initial
value a of
x,
PoM
Q.
we have
in (2),
A = F(b) - F(a) = PqMoM^
For example,
let
A=
Then
To determine
the given curve be the parabola y 2
G,
A=
Cydx= fxidx=
when x
o
—+
= a.
3
G.
= x.
...
(3)
.
SIMPLE APPLICATIONS OF INTEGRATION
Substituting in
248
(3),
A=Z^-^ = P<>M MR
(4)
Q
To
find
PJUfP..
let
s=
b in (4).
"
3
3
EXAM PLES
In the curve of Ex.
1.
P
OJIP,
1
1
2
2
Find the area included between the equilateral hyperbola
the axis of X. and two ordinates x = a, x = 2 a.
Ans. cr los: V2.
= cr.
2xy
3.
Find the area included between the witch of Agnesi (Art. 126),
Ans. ircr.
and I", and the ordinate x = 2 a.
the axes of
4.
X
Find the area included between the catenary (Art.
the axis of X, and the ordinates x
= a, x = 2a.
—
An
2
s.
5.
Find the area
6.
Find the area included
(Art. 129),
7.
".
-
= ~'
P M M P = ^-
Also
2.
show that
1, p. 24,
=
of one arch of y
and the axes of
= sin
2
(ft
— e + e _1 — e~
x.
between the parabola x*
X and
Y.
128),
Ans.
2
)
2.
+ y- = a*
Ans.
<r
Find the area included between the semicubical parabola
A rt. 130 the axis of 1", and two abscissas, y = 8 a, gl =* 27 a.
633
A
•
•
INTEGRAL CALCULUS
244
Conversely, instead of determining the area from the integral,
may
from the area, when
For example
metrically from the figure.
find the integral
we
can be obtained geo-
it
:
Va~ — x
2
Find
8.
|
about 0, radius
dx,
by means of the curve y =
Va — x
2
2
,
circle
a.
A = BOMP = OMP+ BOP
= -xy-\
y
2
<}>=
2 *
-x-y
2
a~
— xr
a
2
If the initial ordinate, instead of
had been some other
have had
Hence
A=
9.
Find
10.
Find
2
ydx—
j
(3
J
Ans.
1
i
a?
Va —
2
2
j
+ 2) eta,
V2 ax — x
2
OB,
we should
ordinate,
A = - Va - x + - sin- - + C,
2
-
sin
-\
sc
where
cZas
C
= Va —
*|
by means
CV2ax-x* dx=
J
independent of
2
2
by means of the
dx,
is
a,*
+ ^- sin -1 - +
line y
V2aa-a
-^
2
2
-f
(7.
= 3x + 2.
of the curve y
X
x.
= V2 ax — x'
sin"
£
2
2
1
^^
+
a
.
C.
193. Other Illustrations. In order to further illustrate the determination of the constant of integration, we will work three examples,
involving geometrical or physical properties.
Ex.
1.
Determine the equation of a curve through the point
which the slope of the tangent is equal to the
at every point of
rocal of twice the ordinate of the point of contact.
(4, 3),
recip-
SIMPLE APPLICATIONS OF INTEGRATION
By
dy_
the hypothesis
from which
1
dx
2y
2ydy
= dx.
Integrating,
y-
245
=x+ a
(1)
2=
£-8
This equation represents a series of parabolas whose axes coincide
with the axis of x.
If
now we impose
the additional condition that the curve must
pass through the point
fe),
(4, 3), its
giving
9
The equation
=4+
C,
0=5.
of the particular curve
2
y
A
coordinates must satisfy equation
is
therefore
= x -h 5.
body starting from
'
with a given initial velocity ^
moves with a constant acceleration g. Find the space passed over
in any time.
Ex.
2.
rest,
,
,
dv
In Art. 19, acceleration
dt
Here
rj
}
dv==gdt.
dt
Integrating,
From
v
= gt+C.
the conditions of the example, v
v
= 0+C,
=v
C=Vq.
when
t
=
;
therefore
INTEGRAL CALCULUS
246
Hence
v
= gt-\- v
Since
v
—
= ds
.
(Art. 18),
= gtdt-\- % dt.
ds
tit
Integrating,
From
= - gt + v$ + C.
2
s
the conditions of the example,
O=0, and
s
= - #£ + v
2
is
£
=
s
when
=
t
;
therefore
the complete solution.
Ex. 3. A body is projected at an angle a with the horizon, and
Eind the equation of its path.
with a velocity v
Represent the horizontal and vertical components of the velocity
by vx and v y respectively. Then, since gravity is the only force acting on the body, we have
.
^=
and
0,
dt
Integrating,
«L=C,
When t = 0,
vx
Hence
vx
at
dx
-r-
is,
=v
=v
=
i<
vy
cos
a,
vy
cos
a,
vy
cos a,
-^-
dt
=v
x
When t — 0,
x and
Hence
x
£
= -gt+0.
= v sin a.
= -gt + v sina-,
l)
= — gt + v
sin
a.
(*5
Integrating,
Eliminating
*
dt
=v
t
cos a
?/,
£
+ C,
2/
and therefore
cos
a,
and
=—-
C and
y
2
gtf
+v
C, are
= — -gf-\- v
£
sin a
+
C".
zero.
t
sin a.
between these equations, we have as the equation of
the path of the projectile,
?/=#tan a
qxr
2 v? cos 2 a
This evidently represents a parabola whose axis
axis of Y.
is
parallel to the
SIMPLE APPLICATIONS OF INTEGRATION
247
EXAMPLES
Find the equation of the curve whose subnormal (Art. 146)
has the constant value 4, and which passes through the point (1, 4).
4.
= 8x + 8.
Ans. y 2
is
5. Find the equation of the curve whose subtangent (Art. 146)
twice the abscissa of the point of contact, and which passes through
the point
6.
The
Ans
(2, 1).
slope of the tangent to a curve at
the curve passes through the point
(3, 2).
any point
Find
its
is
7.
8.
2
Find the equation
is
of the curve
whose polar subnormal
(2, 0).
Ans. r
Find the equation of a curve through the point
A
balloon
Ans.
(Art.
(3,
is
—
j,
is
= 2 ew
in
.
which
half the
r = 6(1 - cos 6).
ascending with a velocity of 20 miles an hour.
stone dropped from the balloon reaches the ground in 6 seconds.
Find the height of the balloon when the stone
is
dropped,
Ans.
11.
of
.
3 times the length of the corresponding radius vector, and
the angle between the radius vector and the tangent
10.
(Art.
Ans. r = 2e s
(2, 0).
vectorial angle.
A
and
3 times the length of the corresponding radius vector, and
which passes through the point
9.
.
+ 9?/ = 72.
Find the equation of the curve whose polar subtangent
is
which passes through the point
153)
,
2
equation.
Ans. 4ar
153)
= 2y
x
,
If a particle
X and Y
moves
400
so that its velocities parallel to the axes
are ky and. kx respectively, prove that its path is
equilateral hyperbola.
ft.
an
INTEGRAL CALCULUS
248
A
from the origin of coordinates, and in t seconds
is 6 1, and its velocity parallel
2
Find (a) the distances traversed parallel
to the axis of Y is 3 1 — 3.
(b) the distance traversed along the path
to each axis in t seconds
12.
its
body
starts
velocity parallel to the axis of
X
;
(c)
the equation of the path.
Ans.
(a)
(b)
= 3f; = f-3t.
s =
+3
=
xfr-$y.
27tf
x
tj
t
(c)
3
t.
13. If a body, projected from the top of a tower at an angle of
45° above the horizontal plane, falls in 5 seconds at a distance from
the bottom of the tower equal to
tower
14.
(gi
its
height
;
find the height of the
= 32).
When
Ans. 200
the brakes are put on a train,
stant retardation.
its
velocity suffers a con-
If the brakes will bring to a
utes a certain train running 30 miles an hour,
should the brakes be applied,
if
the train
is
ft.
dead stop in 2 minfar from a station
how
to stop at the station ?
Ans. Half a mile.
—
CHAPTER XXII
INTEGRATION OF RATIONAL FRACTIONS
194. Formulae for Integration
of
On examin-
Rational Functions.
ing the fundamental integrals in Art. 186,
it
will be seen that only
four apply to the integration of rational algebraic functions,
XV.. and XVI. and of these only I.,
since XVI. depends directly upon II.
;
It will be
shown
II.,
in this chapter that
by these three formulae any
The
rational function can be integrated.
I., II.,
and XV. are independent,
polynomial has been explained in Chapter
integration of a rational
XX.
We
will
now
con-
sider the integration of rational fractions.
195.
equal
Preliminary Operation.
to,
numerator
If the degree of the
is
greater than, that of the denominator, the fraction
or
should be reduced to a mixed quantity, by dividing the numerator
by the denominator.
For example,
a*-2o* =
x'
+l
4
2a5 -3tf
+l—
—'
The degree
X*
+ X~
of the
1
2.t
2
a^
Qx
z
— oo
+1
'
+l
.
_2ar3 + 3a 2
-\
X*
+
+l
•
XT
numerator of the new fraction will be
less
than
that of the denominator.
The entire part of the mixed quantity is readily integrable, and
thus the integration of any rational fraction is made to depend upon
the integration of one whose numerator is of a lower degree than the
denominator.
240
INTEGRAL CALCULUS
250
196.
A
Partial Fractions.
composing
into
it
by dewhose denominators are the
rational fraction is integrated
partial fractions,
The complete
factors of the original denominator.
We
Partial Fractions belongs to Algebra.
discussion of
shall only consider here
the form of these partial fractions and the processes of determining
them.
Factors of the Denominator.
It is
shown by the Theory
of
tions that a polynomial of the nth. degree, with respect to x,
resolved into n factors of the
(x
—a
x)
(x
first
Equa-
may
be
degree,
— a ) (x — a )
2
3
•
••(x
— an
).
These factors are real or imaginary, but the imaginary factors
occur in pairs, of the form
x
whose product
— a + &V— 1,
is (x
— a) + b
2
2
,
and x — a
— b V — 1,
a real factor of the second degree.
any polynomial may be resolved into real factors
or second degree, and only such factors will be considered
It follows that
of the first
in the denominators of fractions.
There are four cases to be considered.
Where the denominator contains factors of the first degree
only, each of which occurs but once.
Second. Where the denominator contains factors of the first
First.
degree only, some of which are repeated.
Where
the denominator contains factors of the second
which occurs but once.
Fourth. Where the denominator contains factors of the second
degree, some of which are repeated.
Third.
degree, each of
197.
Case
I.
Factors of the Denominator
all
of the
First
Degree,
and none repeated.
The given fraction may be decomposed
shown by the following example,
/
x2
+6x—8
Xs — 4 x
,
m
into partial fractions, as
—
.
INTEGRATION OF RATIONAL FRACTIONS
sume
.r
where
^1.
+
.r
5.
C
-8=
are
x2
AB
+6 -8
.v
unknown
251
C
§
;
^
constants.
Clearing (1) of fractions,
x*
...
+ 6x-8 = Ax(x + 2)+Bx(x-2)+C(x-2)(x + 2)
(2)
=(A + B+C)x + 2(A-B)x-4:a
i
Equating the
Coefficients,
two memUndetermined
of like powers of x in the
coefficients
bers of the equation, according
method
to the
of
we have
A + B+C=l, 2(A-B) = 6, _4C=-8.
4 = 1, B = -2,
whence
—+ 6a>-8 =
and
x3 —
x
4:X
r x* + 6x-S dx =
1()g
2,2
1
aj*
Hence
C=2.
(--?
—2
(flj
x-i-2
^ 2) _ 2 log
x
(flJ
+ 2) + 2 log
2 ).
= log^^8
(a
+ 2)*
The following
is a shorter method of finding A, B, C:
Suppose the denominator of the given fraction to contain the
tor x — a, not repeated.
Then the fraction may be expressed as
fix)
(x
Hence
— a)
<f>
(a?)
= A
x— a
tfx)
|
<f>(x)
&* = A + (x - a) ^M
This being an identical equation
is
true for'all values of
x.
fac-
INTEGRAL CALCULUS
252
we put x = a, we have A = ^f^, since by hypothesis <£ (x) does
^
when x = a.
Thus we have the following rule
To find A, the numerator of the partial fraction
put x = a
If
w
not vanish
'
:
x
in the given fraction, omitting the factor x
For example, having written equation
=2
ing x
x-2.
—a
(1),
——^
,
itself.
we
l
in the given fraction
—a
,
find
A by
substitut-
omitting the factor
(x-2)(x + 2)x
This gives
4
+ 12-8^
4(2)
To
find B, substitute x
= — 2,
omitting the factor x-\-2.
4-12-8 = _ 2
-4(-2)
To
find C, substitute x
= 0,
omitting the factor
x.
EXAMPLES
of integration C will be omitted in the examples in
and the following chapters on the integration of func-'
The constant
this chapter,
tions.
J
2
x-
— 3#-j-2
3
(x*+x+l)dx = l
1
f
*Ja?-4a? + + 6 12
a;
3
#—1
2
r(zo+iydW = i loy , (2^
(x
*
+ l)(x-3r
(o;-2) 28
+ i)(2^-i)»
INTEGRATION OF RATIONAL FRACTIONS
h
+ 5)
l)(.i-+3)(.c
°
8
(.c
+ l)(x + 5)
_1,
"
J
(2as-l)(3ir-l)(3*-.2)
as + ta
/
6
J
(CKB
r
y
— 0)(OSE —
Cv
+ aYclr
r
.,
'
12.
2
ft
)
(6\r
C2
x
— a2)
i
2 a6
fa + i)dx
_ l
r
J (a? - 19/ - i (x + 8f 360
C
(2.i--l) a
0-
b-a
=
«y
J (oV —
GO—
= 1 lQ
J (x+d)(x + b)
9
(8a»-l)»(8«»-8)
18
(fa= _^L^lo g (to
CI)
_ „) +—*_,log (ax-6).
CIO
— (T
+ <Q (s- «y
oV
log
fr+io'fr-i)- 1
° (ax
(,-
°
&v
a-b
'
(a;
— b)(bx
-j-
a)
+ 5)%,---y
+ 3) - 1)
8
(a;
"
5
>^m:L_
J1j;>-17a-
!
+
4.»;
= jLlog^|
-1
253
:^:,:.<.,-
+
l)(2a.-l)']
+ ilog*.
;
INTEGRAL CALCULUS
251
198. Case
Factors of the denominator
II.
all of
the
first degree,
and
some repeated.
Here the method
of decomposition of Case
a3
+l
(x-iy
J x(x
If
we
a?
since the
member
+l =A
— I) X
3
tion,
x
B
—1
common denominator
we should
write
D
C
x
—1
—1
x
of the fractions in the second
— 1), their sum
denominator x(x — l)
of this equation is x(x
given fraction with the
requires modifica-
7
ax.
follow the method of the preceding case,
x(x
But
I.
we have
Suppose, for example,
tion.
cannot be equal to the
To meet
3
.
this objec-
we assume
x?
+l
_A
x(x-Xf
B
,
(x-iy
x
D
G
(x-iy
x-1
Clearing of fractions,
X
s
+ 1 = A(x -l) + Bx +
3
Cx(x
- 1) + Dx(x - 1)
2
= (A + D)x* + (- 3 ^ + <7- 2 D)x + (3 A+ B-C + D)x-A.
2
A + D = l,
(1)
= SA+C-2D = 0,
(2)
Hence
3A + B-C+D = 0,
-4=1.
WhenceTherefore
A= -1,
= 2,
(7=1,
D = 2.
+ *) = - - +
(x-iy
x
1)
(f -
x(x
23
8
(x
- 1)
2
x-1
1
INTEGRATION OF RATIONAL FRACTIONS
+*
Hence
f—
Jfjc(a;
1)°8
1
-logs-
dte==
(.x-
i_
— 1)-
x
—1+
255
21«g>(»--l)
(x-1)*2
=The numerators
given for Case
I.,
.r
;
,
+ log*
.
~rr2
A and 5 may be determined by the short method
and then C and D may be found by (1) and (2).
EXAMPLES
»'
2
— ar
ar
2
x
2x'-
a;
r
a*dx
aftfc;
2-3
1
.It
= 2-3a;
lQ
°
(a
J + l)(aJ l) 4(a? 1)- 8
+1
a?
a;
3
a;
3.
*^=_1
h
_
—
— +
-44
f
J
r
J
)-'
2
ar
x
4)2
4
x\ ar
(19
-fa
—
7.
f
'
'* =
J
4(2
a;
-3)
ar-
a2 — 3
aas
4)
216
+
„ 2
a;
°g
4.i-
-3
+l
3 x
+2
2
.
a?
t
a;
,
4
+J_l 0g8 ^2.
*
".':/
-
3
j
18(9
,
*
-4
1
2
^
g
=
f
J(9ar-4/-'
° ar9
4
x-S2)dx
+ l;(2a;-3)
(4;/-
4
—a
W—**l - -J*
+ 8 log (, _ 1).
INTEGRAL CALCULUS
256
8
.
9
"
10
= x - ia ~ 6 >' - 3 (a ~ ^ + 3(a - b) log&\(x t
+ b).
2(x + b)
x +b
2x ~ 1
xdx
x-2 + V3
C
=
+
6(a _4x + l)
J(aj»_4aj + 1)»
6 V3
C(*±*\*<te
J\x+b)
K
2
+ a + &)
J (a+a) (x + 6)
±_(_V_ + J*\
3
(*
2
2
(a-6) 2 V« + a
"
3 a& 2
.
199. Case III.
j
— W\_2-W
2
f
J
—b
x
3
/
-,
N
.
Denominator containing
a3
.
+ b)
—3a6
Factors
2
of
t
the
,
,
,
N
Second
Degree, but none repeated.
The form
of
decomposition
'5cc
from the following
appear
will
example,
+ 12 dx.
,
f-x(x2 + 4)
w
We assume
5x
Bx+C
+ 12 = A
^+ 4) - + +y 4
/-,
.
,
a?(ar
N
(1)
,
a;-
a;
and in general for every partial fraction in this case, whose denomiis of the second degree, we must assume a numerator of the
form Bx + O.
nator
Clearing (1) of fractions,
+ 12 = (A + B)x +Cx + 4,A.
A + B = 0, 0=5, 4 A = 12.
.4 = 3, B = -3, 0=5;
2
5 x
Whence
therefore
5
a;
x(x 2
+ 12
+ 4)
3,-3^+5
03
x2
+4
f-3«+s fe= _ 3
a2 + 4
r^. +6 r
J
x'
+i
Ja? + 4
-|log(^ + 4) + ?tan-^.
1
'
:
INTEGRATION OF RATIONAL FRACTIONS
°
Hence
Take
'*'•/"
f
Vn dx
*
= 3 log
+
.
r,
tan_1 %
for another example,
r
J
This fraction
(a
(2x2 -3x-3)dx
(x-l)(.r--2.i'
is
+ 5)'
decomposed as follows
2a?-3x-3
— l)(jr — 2# + 5)
:
3.v-2
1
a?
—1
x2 — 2x + 5
dx
C
J x*-2x + 5
r *3x-2)dx = r (3x-3)dx
J g*-2x + 5 J x*-2x + 5
= |log(^-2 + 5) + |tana;
(*-2s+5)t
r <2*>-3x-S)dx =1
J _ l) (^ _ 2 2 + 5)
°
(.-,
The
257
a?
-
1
^i
-i£
+
^2 tan =
2
l
l,
integration of any fraction with a quadratic denominator like
~~
the *
preceding.
°'
(
\
'
—V—
,
J x*-2x + 5'
may
J
Having written the denominator
r
.
in the
form
as follows
(a;
+a +
2
)
r (g — pa)dx
+ g)dx __ C p(x + a)dx
+ o.f + b ~J {x + ay + W J (x + a) + b
px
(
(x
shown
be
b
2
,
we have
.
2
2
=£
J
log [(x
+ a)^ + &*] + *=»
b
2
tan- 2±5
b
.
INTEGRAL CALCULUS
258
EXAMPLES
+3
+ 3a?
32 x>
r32_
"
J
4
9
J
x4 +ar
x
•
'
a;
1,
A
4^ + 3
-2
6
+ rf
x3
1
J^I=3 + I
When
,
i
l0g
2
V3
+ 2V2 tan J
V2
(^-l)(^ + 2)
(®+iy
^l l0 &.
1
- + 3V2tan-
:
2rf
+ 3V3tan-
(
'4^-30 + 1, t _6£-l +
* ""2^"
r rfdx
„
4
8
7
(2s»-aQ(to
/»
'
ar
3
1
(a-V2).
a?
x-1,1,tan _i *•
^Tl + 2
the given fraction and the denominators of the partial
fractions contain only even powers of x, they
functions of x2 and
,
we may assume A, B,
may
be regarded as
C, etc., as the
numerators
of the partial fraction.
may
In the following example, the partial fractions
x-
+
a2
x2
be assumed as
B
+ b-
*
5
x dx
C
=x+
J (x + cr)
(x- + &-)
a )(x
or — b~ V
2
2
3
ft
tan -1
a3 tan
2
(per
s
- x )do
3x
- = - tan"
= - tan-1 2 x — tan -1i^l^-i
2J
2+ 2
6
+ l)(a? + 4) 6 V
2
(1
1
2
(4a>
]
Zr>
J (a¥+5 )(5¥ + a )
2
2
a6
'
a&(l
-a)
2
,
J
9
a;(ar
-
6
a?
+ 13)
26
°8
8
jc
13
_i
x
-3
2
a,-
1
INTEGRATION OF RATIONAL FRACTIONS
(3x>-2x-20)dx = l lo/ (2^-6a? + 6y
r
9
,
+
10.
11.
12.
ly- 1/
AlAJlna
=*
*
I
s* + 3^-4-1
Jf^±g^
—-— =
J ^ +
4
1Q
13.
6
r
|-
- 2 tan-
V3
V3
y.+
^—
log
—
-
Jtf-^ + ^-l
2
+ w^2+l
=—
1
6(* +
;
(2
^
a?
- 3).
;
V3
?
iv
1
2
1
.r-
:
dx
tan- 1 ^-
= li og& ^ + a + l + vg tail -x+1 2
4
4V2
l
—
V3
,-i2y±l
_u_l,t.
a
-j
tan
-A3
V+
losj
f
Jt?-1
I
259
4
- a?
-tan
H
]
I-™
2V2
ay-1
2
*x+l
V
,1^
r-+Tlog
l)
W3.
1
3
,
.
2
tan
_!2aj-l
1
'3
200. Case IV.
V3
Denominator containing Factors of the Second Desome of which are repeated.
This case is related to Case III., as Case II. to Case I., and requires
gree,
a similar modification of the partial fractions.
For
illustration take
s
We
2*+*+*^
O-s
+ l)
2
assume
2s 3 + s 2
+ 3 _ Ax + B
(a^-f-l)
*
3
(a?+l)*
+3=
-4*-2,
/*
Oc8
Cx +
x*
+ A* +
= 2, C=2,
D
+
M+
£==1.
C)a>
+ B + Z>.
INTEGRAL CALCULUS
260
lt±_t±l = -2* + 2 + l*±A
(a^ + 1)
+l) ^^ + l
Therefore
2
/»
-2a + 2
J
(x'
/»
J
+ iy
2
a:
To
2
2
(a;
2
2a?da?
(^ +
"*"
+l
vf
+ i)
(o^
2
'
J <V+1)
we use
integrate the last fraction,
dx
2_r 2
i)
2
the following formula of
reduction,
J
CISC
(x2
U/
J-
+« ) w ~ 2 (n - 1) a
2
2
[_(o5
8
.
+a
2
1
)'
"1
/(y
^*
q\
i
}
J
'
/die
— —
depend upon
integral
—tt~~~^—
j
made
is
(ar-f-cr)'
-By successive
'
-!
by making
the given
which
2
x~ -\-d
itan- -.
a
a
1
* This formula
£
*
j^ r
[
dx
may be
+a
2
|_(x
_,
d
2
= A[x(x
+
1
2
)
MJ
*
2
%a 2 f
+
a 2 )"
*5
J(x2
+a ) w+i
2 (w
v
(1
-
+a 2 )" w -
2
nx2 (a?
a 2 )-"- 1
+
=
(x2
n
—
f
J (x 2
1
.
after multiplying
(?X
J (x 2
to,
1)J a 2
(x 2
to
= (l-2n)f
y
v
2
Substituting for
=
+ a 2 )-" - 2 [(x 2 + a2 ) - a 2 ] (x 2 + a*)-"- 2 to)(x2 + a 2 )~ n + 2 na 2 (x 2 + a2)-""1
(x2
members
Integrating both
a 2 )-"]
:
dx
=
=
(x 2
derived as follows
x_^
+a ) w
2
a2) n
n_
x)y
dx,
+2na 2
Jf
,
(x 2
f_dx
**,
+
a2 ) n + l
.
§
J (x 2 +a 2 ) M
we have
1,
^
-f
,
v2
+
by
a 2 )"
=
?
(x2
+ a2 ) "-
1
+
(2 to
v
-
3)
it
1
_
|
J
&.
2
applications
depend ultimately upon
to
CltC
+ a )-^
(a2
(—-J**J (x 2
+
—
.
a 2 )"- 1
•
is
;
INTEGRATION OF RATIONAL FRACTIONS
n
Substituting in the formula
J ur + 1)-
2
C—
A
4. a
[_.r
= 2 and a = 1,
2
° r 4- 9
partial fraction
h
+ l)^2
2(.r
1
1
the form
of
we have
- tan -1 sc
s2 +
+1 J
^x + ^
+
|> a) + 6 ]»'
becomes P
z,
been explained.
For example,
^9a
^
";
C^ +
ft
x-3 =
if
£>
j
z,
2
the integration of which has already
,
2
}'
=J
2
2
J
2
(*
__*,
f?2
5
4<V 4-3V
By
by
y substituting
,
2
=
2G1
+ 16 f
J (^ + 3y
i
the formula of reduction,
+ 3)
3
12L(z
12(^ 2
2
+ 3)
+ 3)
2
12(z 2 4-3) 2
2
+ J
+ 3) J
2
2
(*
Jz + 3j
4 6[_z 2 + 3
2
— V31
+ 24(z
—
2
H
4-3)
.
tan
7
2
(a
Jf-
(*«
+ 2)tfa
6a?4-12) 8
tan -i
,
,
2
2
'
e
z
V3
24
162-15
2z
/5z
5z + 16
dz = tt——
+ 3(z
4-'
(z
12(z
4-3)
+ 3)
(2
2
_i*
"
2
3V3
160,-63
12 (a2
— 6a?4-12)
2
+ 3(s*-6a 3)
+ 12)
(a?
3
2
3V3
tan
Un -i
a?-3
V3
INTEGRAL CALCULUS
262
EXAMPLES
4^ + 3
(4a;2
+ 3)
3
—+3)
+ 5x~-2^_
,_4
= x*,.„
^r^
t
"'
d%
8(4a;2
2
.,2*
1
= tan"
16 V3
V3
For the following example, see note preceding Ex.
r 36s»(s? + l) + 25a,*
J (4a + 9) (9a; + 4)
a
4
'
2
2
x*
2(4a?+9)(9aj 8
156
r ^ + Sx-21
J (^ + 4x + 9)
— 6x
2
2
2(3
'
+ 7)
+4«4+ 9)
^
^
i
C
J
(
(a^
+
a;
xS
-2
)
_|_i)(aj2
=
clx
+ a,_h 2)
+
2
^
+ l)
[( X
+ 2) -
2
a;
3
+l
2
da;
4
(a?
+ l)
2
4
]
=
~"
V5
+4
+ + 2)
a^
7(ar*
tan-
7V7
(a^
+ 4)
3(q;
,
2
6
Case III.
+x
*
2
2
5,
°a; 2
a;
2£±1 _ _2_
V7
V3
2x±t
-a; + l
V3
+ 36a; + 29
+ 3) (2 a + 6 x + 5)
Stan" (2 + 3).
12a;2
2
2 (2 a;
1
a;
V3
—
-
CHAPTER
XXIII
INTEGRATION OF IRRATIONAL FUNCTIONS
We have shown in the preceding chapter that the integral
any rational function can be expressed in terms of algebraic, logarithmic, and inverse-trigonometric functions.
201.
of
We
shall
202.
radicals
now
consider the integration of irrational functions.
may
Some
them
by Rationalization.
Integration
be integrated, by reducing
by a change of
limited number
variable.
This
of cases.
This process
integrals
involving
to rational integrals
possible, however, in only a very
is
is
sometimes called
integra-
tion by rationalization.
p
203.
containing
Integrals
rationalized
(ax
+ b)
by the substitution ax
For example, take
(
—
Such an integral may be
q
.
-{-b
=z
q.
x 2clx
(2x + sy
Assume
Then
2x
+3=z
i
x= —-—
,
dx =
—
:rax
/ (2x +
(1 «. _i_ 3)*
=l(t-¥
,
<
z
J
SJ
)=¥(7-¥ +9)=if2 f2a;+3)S(8a;2 - 18a:+81 >
+9s
203
INTEGRAL CALCULUS
264
Another example
,J
is
Assume
x
C_^x_ =
Then
J Vx + i
=z
'
_
Va? + i
i
J
2
dx
,
•
= 2zdz.
= 2 C,U^
f2^
J \
+i
J
z
+ 1 _J_\ dz
»
2
+v
= 2|~- - - + ^ - log (2 + 1)1 = ^ - x + 2 a* - log (a>* + 1)
204.
n
is
the least
common
(aaj
+ &)%•••
multiple of
••,
q, s,
.
In this case
where
the denominators of the
by the substitution ax
rationalized
is
+ 6)%
(ax
Integrals containing
the integral
2
.
+ b = zn
,
fractional exponents.
Take, for example,
j
'
Assume
(
x
(x
- 2)* +
x
—2=z
_
2 )i
=z
(x
- 2f
,
dx = 6z 5 dz,
,
(x-2)i = z\
6
3
2
= 6p- -2-}-log(z + l) = 3(aj-2)*-6(aj-2)* + 61og[(a;-2)* + l].
EXAMPLES
L
f
•^
L dx = 2Vx-2 + V2tan- J^=J-
^+
xVx —
1
*
2
INTEGRATION OF IRRATIONAL FUNCTIONS
4.
5
'
(V*
+*
2
—1
80
f.r Vrt7+6
10.
11.
3
-6?/ 2
+ 2 ^ - 4 .^ + 4 log (s* + 1).
-6?/-l
24(4y + l)*
2
=
A 2
M>
—1+1
= 2 (q Y + 6
-
da;
J
9.
^
=
civ
^-
J W+V
8.
+-1
f !?cbf = 4?/
J (4y + l)*
f
6
l
+ 2 log (V2w-1 + 1).
a
)
(15 a *».«
105 a3
_ 12 a&a + g &2\
.
*
= 2(3a; + l)*-4tan-^+i)L
2
Jf (3* + l)* + 4(3a; + l)*
fA
-'
+ Xo~ 1 dte = 2 Va^fl - log
(as
+ 3) - 2 V2 tan" J^±^.
1
3)l
fa ~ 2 fe
f
= g(2s-3)Mlog (2 *- + 3
8
^ (2z-3)* + 6a;-9 4
a
(2 -3)*
)
V3
4
12
.
265
r
J
V'2:/:
+ l+Vo;-l
= 2V2^Tl - 2V^1 + 2 V3 (tan-iyjxEl _ tan" J?^±3
1
= 2V2^+1 - 2V^1 + Vsfcos^
1
^^
- oc»-il=*Y
x+2
x
+ 2J
INTEGRAL CALCULUS
266
=2xMlog(2**-l)+!log(** + 2)
;—^-5
f
9
9
J (2aj*-l)(a>*+2)
13.
16 V2,
—
— tan
9
-1,
-
J
(a?
«*
—
;
V2
3
+ l)* + l
4
--log(l + V^+l).
205.
Roots
of
Polynomials of Higher Degrees.
tion of irrational integrals
the
first
Here
we now
— In the rationaliza-
pass from roots of binomials of
degree to roots of polynomials of higher degrees.
rationalization
is
limited to the square root of an expression
of the second degree.
206.
ized
Integrals containing Vtf2
+ ax -f b.
may
This
be rational-
by the substitution
yx + ax + b = z — x.
2
For example, consider
dx
I
x -y/x
If,
2
—x+2
following the method of the preceding articles,
^x -x + 2 = z, x?-x + 2 = z
2
the expression for
x,
and consequently that for
will involve radicals.
This difficulty
Vx — x-\- 2 = z — x,
2
is
we assume
2
,
dx, in
— x-\-2 = z — 2zx,
2
cancelling x 2 in both members.
z
2
-2
2z-l'
V^^x~T~2 = z-x =
= 2(z -z + 2)dz
2
dx
-z +
2z-l
(2z-iy
l
z2
terms of
avoided by assuming
.
^
z,
-
INTEGRATION OF IRRATIONAL FUNCTIONS
267
Hence,
-z + 2)c7z
2(z 2
(2z-l)
-I)
(2*
dx
J
rw
-_ x +2
«-*/rf_*j.9
.
_o
2
z--^
s*-2
2
I
<-/
z
x-Vx
,
z-V2
2z- -1
2z-l
= Va,- — x + 2
2
c&
2
1
2
2
,
-1
2z-l
2z
Substituting
r<2dz
g
— + 2^ Jz
z -z
y-2
-2~A>°
Jz--Z
V2 z + V2
Z
-
2
2
1
—x+2
Var
,
V2
V— x
207. Integrals containing
-f- a?,
- x + 2 -f x - V2
— + 2 + + V2
2
Var'
2
a#
-f
a;
+ b.
a?
This
may
be rational-
ized by the substitution
V—x + ax+ b = V(« —
2
where «
—x
and
/?
+
a;
values of
—z? + ax-{-
V— + a#
unless
real,
or
ar*
-f-
= (/? + #)z,
b.
6 is
imaginary for
x.
—
dx
Take, for example,
Assume
+ ») = (« — #)z
are the factors of
These factors will be
all
a;) (/?
J
(
a;V2
+x—x
2
V2 + x — x = V(2 — x) (1 + x) = (2 — aj)z.
2
l
+ z = (2-a0z
V2 + a?
2
a;
2
,
a
?
= 2 f-" 1 dx= 6zdz
z + l'
(z + l)
,
2
= (2 - z)z =
2
2
3z
j^
•
Therefore,
r
J
da;
W2 + X-.X
2
= r 2dz _ 1 log gV2-l
J2z -1 V2 zV2-fl
2
2
—
a;
1
dx
**
V2 + 2a;-V2-a;
= -J-log
^f a^
aj-ar
^/2 + x-x
V2 + 2a; + V2-*
V2
,
22
INTEGRAL CALCULUS
268
EXAMPLES
/
f
2.
a;V'ar
_ x + Va; + 4 x — 4
__•,
_
tan
2
dx
x
+4x—4
Va^ + 4a; cfa =
/
J
+V +4
—T)
(2 aa? _
2
2\
a 2 V2
2
a;
—
asc
;
^
+ 2 + Va;
2
-f
4
a;).
2
a;
2(n
—^±A)^=_ =
2
2
Va;2
3a?
ft
+ 4a
= z— + a
2
2
-f
,
dx
62
2
4a
Integrable
Cases.
2(»-l)
tl-x
3
n
9
— The
2
rt
tan"
V3
—
—+2
,
4
+x
1
l
_ia;
x
r
— log 2 H
= x + Va^ + a
where
+ l)
^
COS
208.
(a;
a;
x— a
da;
.
a;
/;(3-a;)V3-2a;-a;
/
+ log
2
a;
cos
COS
5o,
log
9
a;
_i
yS
ij3-3c
* 3+
*
2
a;
3-x
—a
z + 3a'
3
2!
2
.
preceding
articles
include
those
forms of irrational integrals that can be rationalized. In general,
integrals containing fractional powers of polynomials above the first
degree
except the square root of polynomials of the second degree
cannot be rationalized, and cannot be integrated in terms of the
elementary functions, that is, cannot be expressed in terms of alge-
—
—
braic, exponential, logarithmic, trigonometric, or anti-trigonometric
functions.
INTEGRATION OF IRRATIONAL FUNCTIONS
269
Every integral may be regarded as defining a certain function.
been shown in Art. 192 that if f(x) is any continuous func-
It has
tion of
Xj
I
f(x)dx
is
a function of
x,
which may be geometrically
represented by an area bounded by the curve y =f(x) but this
cannot always be expressed in terms of the elementary functions.
;
CHAPTER XXIV
TRIGONOMETRIC FORMS READILY INTEGRABLE
any power
It is to be noticed that
209.
tion
may
be integrated by Formula
I.,
of a trigonometric func-
when accompanied by
its
differential.
Thus,
/.sin" x cos xdx = sin w+1 x
_
n
/,tanw_ x sec
+1
1
I
,
w+1
2
= tan —x C cot"n x eosec
l
J
n+1
>
7
x dx
2
I
.
sec n x sec x tan x
cos n+1 x
^ =
C cos'» x sin x ax
J
7
,
x dx
,
7i+l
cot w+1 x
=
-
Ti
71
+1
'
secw+1 x
dx
71+1
J
'
M+1
/
Having
in
mind these
cosec w
a?
cosec x cot
a;
da?
= — cosec
+1
#
71
integrals, the student should readily under-
stand the transformations in the following articles.
210. To
&vnn
find
xdxov
I
eos n xdx.
When
n
is
an odd
posi-
J
tive integer,
sin5
J
we may
xdx=
integrate as in the following examples
sin 4 x sin x
I
dx
=
(1
— cos
2
2
a,*)
sin
x dx
J
//h
(1
— Io cos^ x + cos
2
,
4
\
ax =
7
•
x) sin
a?
— cos
a?
,2
COS 3 X
H
3
270
COS 5 X
—
o
•
TRIGONOMETRIC FORMS READILY INTEGRABLE
Another example
|
cos G 2 x dx
271
is
=
cos 2 2
i
cos 2 x dx
as
=-
— sin
(1
2
2 x) cos 2 # 2 dx
J
= -1/ sin 2 x
•
sin 3
2aA
s
)
A
211. To
sin m x cos' x dx.
When
1
find
I
positive integer, this
]•
m
either
or
n
an odd
is
form may be integrated in the same manner as
For example,
in the preceding article.
sin 4
.r
cos 5 x dx
=
sin 4 x cos 4 x cos x
I
dx
=
J
/,
•
(sin
4
x
— 2o sm + sm
sin3 x cos - x
•
i
aj
Another example
I
6
b
•
dx
=
sin 4 x (1
j
sin x
—
2 sin 7 x
o
7
5
8
\
a;)
7
cos x dx ==
— sin
2
,
is
I
cos T as sin 2 x sin #
c7ru
=
cos*
|
as
—
(1
cos 2 #) sin #da;
5
9
EXAMPLES
1
o
2.
o
Csm'
-
.
3.
i
x dx
I
J
C COS
J
n
C sin
J
-
I
J
= — cos x -f cos
7
.r r/.v;
I
= sm X
J
COS 5 a;
3 cos
25
x
4 sin 3 x
a;
-,
•''
7
- das
2
T
.
23
V
^
9
a;
5
.
,
6 sin 5 a?
4 sin 7
3
.
.
sin 9 a;
7
—
252
11
a;
a?
h
5
= — 2o cos - + 4- cos o«2
cos
as
cos 7
.
,
(-
3
5
snr a;
h
J
1
2
x) cos x dx
55
#
-
•
13
15
INTEGRAL CALCULUS
272
s
5.
6
.
sin 5 2
cos 3 2
= sin
6
+ cos
6
3
(cos 3
I
<£
+
sin3
<£)
(cos 2
= sin
cos
rsm £_£
ydy = _
-JL
2
?/
C <x**te = fa m + _2
J sin x
sm
9
4
/~ cos 3
^Q
Vsin
•^
11.
I
a;
(sin™
dx
3
a;
dx
4 cos 5
_ _ 2 sin
2
x
sin 2
4>
.
<f>)
_ 6 cos
7
a;
7
2
</>
sin <£
+ -2 (sin
.
iq
13.
I
/*
I
J
(sin 2
•
,
sin 4
cos 3
a;
— cos w
+ cos 2
«
sm3
•
a?
<j>
+ cos
5
<f>)
.
6
3Vsina?
a?
a?
5
-,
a;
sin3 x)
dx
a?
7
a;
cto
a?)
cos 3 x dx
= 4 sin
5
5
x
= - (sin
5
8 sin 7 a?
..
7
+ cos TO+3
m+3
sin™+ 3 x
m+1
.
"3"
3 cos 2 y
^.j-iogsec?/.
J
_ sin m+1 x + cos m+1
12
cos 9 x
d<f>
+ cos
1
3 sin3 x
a;
-f-
a?
5
_-*h
4
o
cos y
cos
<£
I
./
—
<£
3 cos 4 ?/
6
7
o
8.
16
x) sin3 x cos 2 x
x
_ cos
~3~
7.
sin 8 2 6
2
12
(sin6 x
I
<9d<9
x — cos 5 a;)
+ sin x cos
a;
2
a;.
.
\
TRIGONOMETRIC FORMS READILY IXTEGRABLE
212. To
( tan"
find
x dx, or
( cot"
x dx.
These forms can be readily integrated when n
tan"
.i-
=
dx
J
tan"
x (sec 2 x
is
any integer.
— 1) dx
J
=
|
tan"
tan
Thus
-2
273
tan" a; da;
.i'
sec
x
made
is
-2
2
ado;—
tan'
I
1-2
x dx
tan" -#«#.
depend upon
to
mately, by successive reductions, upon
tan
tan n2 a*
j
J
a;
da;
or
I
cZa;,
and
ulti-
dx.
J
For example,
tan5 x dx
I
=
tan 3 x (sec 2 x
J
= tan
4
a
/*,
o
tan-*
I
J
4
tan3 as dx
I
=
I
tan
—
= tan
Hence
I
2
a;
a;
(sec
log sec
C
./*
I
J
5
—
5
= — cot
O
a;
.
log sec
2
,
cot2 a;(cosec2 aj
<N
— l)dx=s
,
-^^cot5 a;
,
cot3 a;
.
h
o
C (cosecnnat%eAtm
x—
I
%J
1
1)
,
dx
feot 4 x dx
cot 3
a;
,
\-
\-
5
1
x.
Z
is
.
9
x.
f-
4
4
cot 5
x—l)dx
,
Ccotfxdx= Ccot x(cosec x-l)dx =
=
2
x dx.
=
tan5 x dx
*s
Another example
— 1) da;
3
—
Do
= - cot
5
a;
,
H
cot8 a;
/•
|
i9
cot 2
a;
dx
«/
cot x
- x.
INTEGRAL CALCULUS
274
To
213.
find
j
positive integer,
j
sec 6
xdx=
sec" x
dx
we may
(teLn
2
or
When n
cosec" x dx.
|
x+T) 2 sec 2 xdx=
+ 2 tan + 1) sec
(tan 4 a?
2
—
2 tan 3 x
5
= tan —x H
.
cosec 4 x dx
=
x+1)
da;
,
.
htana;.
cosec 2 x dx
x
— — co
— cot x.
J
To
214.
(cot 2
a;
o
o
j
2
a;
J
is
an even
integrate as follows
J
Another example
is
find
I
tanm x sec n xdx or
When n
cot™ x cosec" x dx.
J
is
an even positive integer, these forms may be integrated in the same
For example,
as in the preceding article.
manner
j
(tan8 x
When m
is
j
=
tan6 x sec 4 x dx
+ tan
6
I
tan 6 x (tan 2 x
x) sec 2 x
tan 5 x sec 3 x dx =
=
j
j
I
j
a?
integrate as follows
— l)
2
x
(sec 6
x—2 sec
a;
sec 2 x sec
4
a;
aj
tan x dx
+sec 2 x)
sec3
2 sec5 x
cfcc
sec
a;
tan
a;
dx
a;
3
5
is
cot3 $e
# cosec 5 x dx
(cosec6
may
dx
—
1
(sec 2
7
Another example
sec 2 x
tan4 x sec 2 x sec x tan a?
sec 7
=
=
an odd positive integer, we
=
j
dx
+ 1)
— cosec
4
=
(
j
cot
c
2
x) cosec
x cosec 4 x cosec
a?
cot x dx
=
a;
cot a? dx
cosec'
a;
cosec x
TRIGONOMETRIC FORMS READILY INTEGRABLE
275
EXAMPLES
tan s x
I
^_t^ + tan^_
da.
2.
3.
fse c"ya'
i/
*/
4.
5.
= -|
Jcotf 5 dx
=^ +
= 1-
(tan 9 x
4tau ^
a;)
f(sec 3
+
tan 3
2
<£)
.
5
.
<£
Q
o.
Q
9.
10-
/"tan 7
I
J
tan
se+l
'—
a-
rsec 5
;/*
I
J
sec x
Tsec 6 x
I
J
I
*/
4tan
3
+1
,
dx
tan
=—
5
sec
+
-7
tan
3
a;
nnn5D
sec
,v
tan—3 -*<f>
.
<£)
-\
5
+ cot 3
A
+ 1-
(tan 5 x
— sec
2sec 3
= tan
aJ
£
<f>
-*-
3
.
.
.
<f>
.
2 sec 3 a;
Hsecar+a;.
,
3
4
,
dx
=
tan
—
2
x
.
2
cot a? 4- 3 log tan x.
-,
,
2
.,.
cosec-<9cot
J
5
x).
—
2 tan
+ „.
,
h tan x -f log cos x.
.
4
a;
7
tan 4 a;
a;
a;
a;
2/.
o
3
,
7
cte
|.
^ + tan
00 V 3 x
- (tan 7 #+sec 7 a;)
o
+ tan
+ tan
sin 2
dx
2
+
^
- log
^1* +
5
7.
1
f
cty
- (tanB
= 2/.
+
»
2
7
<£
_
o
5
4
sec # tan x
9
- cot
S^ +
+
1
— sec
|
7
y
6.
cot*
= - (*£?? +
— tan x)
x
(sec
| +|
9
Jcosec* 3 x dx
I
catf
taua;
3
5
i
#0
=
r,
2
f tan-
5
- sin
2
0)
-+-
log (sin 6 tan
0).
y
.
<fi.
INTEGRAL CALCULUS
276
H-
Vsec 3 x tan x (Vsec5 x— Vtan 5 x) dx
J
9
7
=-
(sec
12.
m x* tan 5 x
— tanm_1
(tan 2
"
a?
y
^— sec* x )
-f-
9
3
^ (tan*
3
a;
+ sec"
2
#).
sec 6 x) dx
J
_ sec m+4 x — tanOT+4
m+4
A
term
even,
I
sec m
2(sec m+2 o? + tan m+2 a;)
a;
a;
m+2
m
sec x cosec M
may
x dx
=
cosec a
by substituting
be integrated,
when
— tan
m
TO
a;
m+n
is
-.
tan x
13.
14.
cosec 3 x dx
fsec 5
J
j
= *^^ + 3tan ^ - ^^ +
4
(sec4
a;
— cosec
2
x) 2
215. To find
tion of this form,
211.
j
7
x
,
3 tan5 a;
.
tan3 x
H
1
sin™ x cos n x dx
when
3 log
5 tan x.
dx
—
too
tan
^
=—
2
2
either
The following method
m
is
—
— cot
,
3 tan x
3
x
,
\-
cot x.
o
The
by Multiple Angles.
integra-
or n is odd, has been given in Art.
applicable
when
m
and n are any
positive integers.
By
trigonometric transformation sin" x cos n
1
x,
when
m
and n are
positive integers, can be expressed in a series of terms of the
first|
and cosines of multiples of x.
If we use the method of Art. 211 for integrating terms with one
odd exponent occurring during the process, the following formulae
degree, involving sines
for the double angle will be sufficient for the transformation of the|
terms with even exponents
:
.
TRIGONOMETRIC FORMS READILY 1NTEGRABLE
sm
2
x*
= - (1 — cos 2
cos 2 x
= - (1 -f cos 2 #).
Csm
•
Hence
sin 4
I
aj
cos 2 a
s
= (sin X cos xf sin
1
tt
=- sm A x,
sin 2 x
For example, required
sin 4 x cos
x cos x
I
sb
9
4
.i'
cos- x
#
efo.
= - sin
.
cos 2 x-\
1
(1
2
.t(1
— cos 2
— cos 4 x).
sin
sm 3 22a;
=
7
ax
2
8
_«_
o __
-
sin2 2
2
a?),
a:
x
,.
sin 4
oj
h
16
48
«/
64
EXAMPLES
i
rsnr x ax = 1- /3— x
•
1
I
2.
fcos
3.
I
4
4
sin 2
x
.'.-
dr.
•
—
,
x-\
si n
4x
= - f^ + sin 2 x + sin 4 x
cos 2
.'-
dx=*
1
fsin«aj(ia?=
J
sm 2o
7
1
-
f
7>
16^
.,-
f
x
—
sin 4
a
- 4 sin 2x + sm
*
3
2
,7 ?
4;A
+ -sin 4
a?)
277
DIFFERENTIAL CALCULUS
278
5.
6.
i?
7.
8.
fcos 6 xdx
J
16 \
fsin
4
x cos 4 x dx
-9-,
/~
I
J
I
J
= — f5ff + 4sin2%- sm32a; +-siii4A
6
cos b x snr #
aa,*
)
= -i/3 a - sin 4 a + !^£\
1
==
^
sin8 x
4
3
,
/,
8
\ox-\- -snr 2
3
,
.
— sm 4a;
.
a;
A
128\
dx
=—
f
16\ 8
4 sin 2
a?
4- - sin3 2
3
sin 8 aA
)•
8
a
+ - sin 4 x
8
+
sin8aA
64 J'
/
4
CHAPTER XXV
INTEGRATION BY PARTS.
From
216. Integration by Parts.
d(uv)
Hence
I
the differential of a product
= udv + v du,
uv =
we have
REDUCTION FORMULAE
i
u dv
udv=-uv
—
+
I
v du.
Ivdu
This formula expresses a method of integration, which
(1)
called
is
integration by x>^rts.
For example,
let
us apply
it
j
u = log a
Let
whence
'
du
j,
= dx
to
x log x dx
then
dv = = xdx;
and
v-
X
Substituting in (1)
,
Jbg;x
X?
2*
we have
•
x dx
= log5 x — -/?• dx
2
X
-
=-
-
X2
loga>- 4*
279
•
(2)
INTEGRAL CALCULUS
280
Integration by parts may be regarded as a process, which begins
by integrating as if a certain factor were constant.
Thus
in (2), if in
we
dx we treat
•
2
x
—
log x
as if
it
were a con-
From this we must subtract a
Z
integral formed as indicated by the following connecting lines.
stant factor,
new
logx
|
.
obtain log x
•
•
= logX'--J *x—dx
2
logx>xdx
x
This method of remembering the process
Another example
= cos x,
Assuming u
I
As
the
is
x cos
j
if
we take u
= x,
I
x cos x dx.
cos x
•
——
|
— (— sin x dx).
contains a higher power of x than the original
by
integral, nothing is gained
But
be found useful.
we have
xdx —
new integral
may
we
this application of the process.
find
x cos x dx = x sin x —
I
sinscefa;
= x sin x + cos
a?.
EXAMPLES
4
1.
a?
J
2.
logx dx = y flogcc
fa (e a* +
J
x (sin
3.
J
e~ ax ) dx
x
—-
= - (e — eax
a
— cos 3x)dx=(
a
*)
— i (e ax + e~
a2
(--) sin
3x
a:c
).
— (- + -) cos
x.
..
REDUCTION FORMULAE
INTEGRATION BY PARTS.
4.
5.
6
I
J
x log -
2
V
+ sm -
dx
= — losr
fa;(e»»-l) 2 Qte=ete (-
Clog (ax
2
2
2)
w)-
+ 6) dx = fx -f -
)
log
8
10>
11
12
a?
?
14.
T"2
+
(oaj
)
•
2
+2
— x.
ft)
fsec
4
log sin
<£
rlog(.,
J
<£ d<f>
<£
1
(?
ftan- - dx
a
]
= x tan" - - - log (x + a
2
1
= - tan"
3
sin
f (3
-1
2
x-
- dx
a
1
x
)-
-- + - log (x + 1).
6
6
2
= x sin -1 - + Va — x
- 1) sin-
2
2
a
pc2 tan" 1 aj da
I
= f 1^^ + tan ^ log sin - ^L* - 2&
2
J
J
a;
+ 2) ^_log(. + 21 +
*±1.
*x+2
x+1
+ l)
(.i'
a;
;
c/
13.
e
+ 4 sin -
+ - 3 log( + 3)-g + |l)log(a + 3)cfa =
+
^
J(
a?2
7.
2 x cos 2
4
281
2
2
.
a
1
x dx
=
(x-
- x) sin"
1
a;
-
(*
~^
.
INTEGRAL CALCULUS
282
n
K
15.
C a?snrxaa; = x /cos
•
3
,1
3
x
\
— tan
x (sec 6 x
16.
6
a;)
dx
= x tan
3
J
(^ +
„
/ 1o S
18.
flog (a
+
l)
(to
Var
2
sin 3
.
cosicJH
I
x
2 sin x
.
1
a; -f-
—HL®
—
= ,_gl±l l08(e + 1)
_j_ i
g sec Xt
.
a
+ a ) cte =
2
a;
log (a
+
-Vx2
+ a log
.
+a)
2
(a;
-f
Va^2
+ a )—
2
a?.
In each of the following examples integration by parts must be
applied successively.
19.
20.
21
CaPe-^dx =
-^!(4ar + 6a; +
2
5
JV - xfdx = £ - ^(4
Car- 1 (log a;) 2 da;
J
x
6a;
+ 3^
- 1) + ^(2x> - 2 x + 1) -
^
= - [(log xf - ?i°S« + -1
n
n 2J
n
|_
N
22.
rx3 sin2a;^ = ('?^-l )sin2aj-^--— )cos2x.
23.
Tartan- 1 a;) 2 dx
x log
(a;
= ^^(tan"
+ a) log
(a;
1
— a) dx =
x) 2
-
a;
—-—
-
tan" 1 a;
log
(aj
+ 1 log (x + 1)
2
.
+ a) log (x — a)
2
Q^!log(x + a)-^-^\og(x-a) + x
^
INTEGRATION BY PARTS.
217. To
e™ sin nx
find
J
dx, and
ax
=e
e
We
ax
•
sin nx
required integrals,
the
e
j
ax
(a2
+ iv)
e
ax
sin
sin
e
ax
cos nxdx.
C e"ax cos nx dx.
j
n
nx
e
I
by n 2 and
az
(2)
,
M sin nx dx = e™ (a
cos nx dx.
•
—
-,
(1)
.
.
(2)
/o\
.
sin
nx
— w cos nx)
;
/0 \
(3)
'-
2
Eliminating
,
J
I
.
by a 2 and adding, gives
sin no; — n cos ?i#)
e^ia
r e"smtia5CW5=
ax
*
J
a +n
v
hence
.
two equations containing the two
nx dx and
(1)
I
= sinnic,
(2) are
by multiplying
latter,
nx dx.
cos
,
f--
=€
7
dx
and
see that (1)
ax
ax
sm nxdx =
Integrating the same, with u
/„
e
283
J
Integrating by parts, with u
e
REDUCTION FORMULAE
2
Substituting this in (1) and transposing, gives
—
C e ajax cos nx dx = e" (a?i sin
a
-
'
7
*
I
»J
a
—
C e^ax cos nx dx = e" (w
I
hence
"
7
I
J
*
+a
+n
?ia;
2
cos
!
2
—
as -f a cos
a2 + ?i 2
sin
na;)
z-
2
wa;)
!
L
w
, AS
(4)
EXAMPLES
The student
is advised to apply the process of Art. 217 to Exs.
For the remaining examples he may substitute the values of
a and n in (3) and (4).
1-4.
'
I
'
/e
3*
sin 5 x
cfcr
= — (3 sin 5 — 5 cos 5
cos 5 x
dx
= — (5 sin 5
aj
34
as
-f
3 cos 5
a;),
x).
/e~
INTEGRAL CALCULUS
284
e~
2.
2x
\
e~ 2x cos
e
ax
sin aa;
da;
a;
dx
=
2x
— (2 sin x
=
sin x dx
—
(sin
x
— 2 cos
a?).
2a
hos^dx = ~^ Y2sin?
fe
5.
S
— 2 + cos 2 a;,
f*sin
x
sin 2
a;
-f-
13
(e
2x
+ sin 2
(e*
a?)
fe 2* cos 2 3 x dx
8.
I
x
a;e
sin 2
2x
cos
a;
s
(sin
1
—
.
7.
5 cos 2 a;
e
+ cos x)dx —
H
/
a?),
= — K(sin ax — cos ax).J
4.
e
cos
-f-
e* /
.
(sin 2
a?
dx =
sin
a;
+ cos
5 sin 5 x 4- co s 5
2x
by which the
13
be
made
to
a?
— 4 sin x — 3 cos
]
a;].
These
integral,
j
may
.
a?).
218. Reduction Formulae for Binomial Algebraic Integrals.
are formulae
2 cos 3 a;
N
a;)
a;
xdx = — [5x (sin a; + 2 cos a;)
g
— 2o cos 2o
a;)
3
a;
a?
+ 2 cos
5
= — + — (3 sin 6 + cos 6
sin 3
x
x m (a
+ bxn) p dx,
depend upon a similar integral, with either m or p
There are four such formulae, as follows:
numerically diminished.
INTEGRATION BY PARTS.
+ m + l)b
(i\p
Cx m (a
-f
(np
REDUCTION FORMULAE
K
+ m + l)bJ
}
285
K
'
J
bx n ) p dx
= ar»(a + bary
*pa
UTarfr + toy^ifr
.
(B)
x^ja + bx")^ _ (»j) + « + m + l)5 C m+n
_
b yd
*")«*'
~
J * &+
(ro + l)a
m l)a
(c
(C )
+
np
m-\-l
np
+m+
.
.
.
+ bx ) p dx
n
Cx"\a
1
,
+
(
r.r
m
(a
+ ^)
p
^
^^^
=_
n(p + l)a
¥ + M+w + l ^
n(jp
+ l)a J
^
(
K
J
lfe
(
v
J
Formulae (A) and (5) are used when the exponent to be reduced,
or p, is positive, (^L) changing m into ?ti — w, aud (B) changing p
intop — 1.
m
m
Formulae (C) and (D) are used when the exponent to be reduced,
or p. is negative, ((7) changing m into ?/i + ?i, and (D) changing p
into
p
+ 1.
If, in
the application of one of these formulae to a particular case,
any denominator becomes
For this reason,
formula
zero, the
Formula (C)
fails,
when np
when
Formula (D)
fails,
when
Formulae (A) and (B)
In these exceptionable
case's
fail,
is
then inapplicable.
+ m + 1 = 0.
m + 1 = 0.
p 1 = 0.
-f-
the required integral can be obtained
without the use of reduction formulae.
INTEGRAL CALCULUS
286
219. Derivation
Let us put for brevity
Formula (A).
of
X = a + bx
n
,
dX = nbxn ~
Cxm Xp dx = Cxn -n+l
Then
Integrating by parts with u
Cxm
J
X
p
= xm
dx
=
m-n+1
se
~ n+lXP+1
7ib(p
+ l)
,
l
dx.
X dX
p
nb
we have
_ m 7 + 1 Cxm ~ n X>+
nb(p + l)J
yi
1
dx.
.
.
(1)
K J
Comparing the integrals in (1), we see that not only is m diminished
by n, but p is increased by 1.
In order that p may remain unchanged, further transformation is
necessary.
By
substituting
the last integral
X ^ = (af bx X
n
p
may
p
)
,
be separated into two.
Cxm ~ n Xp+1 dx = a Cxm - n Xp dx-\-b Cxm Xp dx.
Substituting this in (1) and freeing from fractions,
nb(p
+
1)
Cxm Xp dx = xm
~ n+1
— (m — n + l)fa
Transposing the last integral to the
(iip
+ m + 1) b
Cx m
X
p
dx
Xp+1
Cxm ~ n Xp dx
first
+ bfx X
(A).
dx\
number,
= xm ~ n+1 Xp+1 — (m — n + l)a
which immediately gives
p
Cxm ~ n
X dx, (2)
p
+
x
INTEGRATION BY PARTS.
u
220.
Derivation
=X
we have
p
,
(
J
Formula
of
%-X* dx =
REDUCTION FORMULA
287
by parts
with
Integrating
(B).
X*^—-m+1
I
—
Jm
pX^bnx"- dx
1
1
= ^^_m>brxm+nXP - ldx ^
m + 1 m + lJ
Comparing the
m
but that
To avoid
is
we
integrals,
increased by
see that not only is
p
.
.
.
(1)J
decreased by
1,
n.
the change in m, substitute in the last integral of (1)
bx n
=X-a.
Also freeing from fractions,
+ 1)
(m
Cxm X p dx
Transposing to the
(np
+ m + 1)
= x m+l Xp - npf
X
Cx m
p
which immediately gives
221.
(m
dx
+ 1) a
l
dx).
Cx m Xp~1 dx,
...
(2)
(B).
((7).
by transposing the two
by m.
dx-a Cx m Xp ~
the last integral but one,
= xm+l Xp + npa
Derivation of Formula
Art. 219,
m—n
member
first
Cx m Xp
may be obtained from (2),
and replacing throughout,
This
integrals,
This gives
Cx m Xp dx =
from which we obtain
222. Derivation
of
afl+1 X^p+1
— (np + m + n + 1)
Cx m+n
X dx,
p
(C).
Formula (D).
Art. 220, by transposing the
two
This
may
integrals,
be obtained from
and replacing
p— 1
(2),
by p.
This gives
cte
=-
from which we obtain
(Z>).
n(p
+ l)a
f afX*
m+1
Xp+1 + (np + n + m + 1)
Cx m Xp+1 dx,
INTEGRAL CALCULUS
288
EXAMPLES
f ^da; = Cx\a -x )~^dx.
J Va —
^
2
Here
2
2
2
sc
Apply
(A),
making
m = 2,
= — -,
n = 2, p
a
fx\a 2 - x 2y* dx = < *-f)*
—^
«/
=
2
2.
fV<* 2
+a
2
cte
(jB),
2
_ J^ f( a _ xyi dx
— zu
2
— x )* 4- — sin-1 -.
2
2
Z
2
sc
).
making
w
= 2, p = -,
z
JV + a0**« = f (*'+af)* + f/
a
*( a2
/die
V# — a
^
2
,
6
= 1.
2
+
x2y
+ a?)i + 1 log(z + V^+^).
V#— a2
=a
da?
(a
2
a
2
'
Z
m = 0,
ar
&= — 1.
,
= | Vtf + a^ + ^logfc + V« +
»/
Apply
(a
V
a = a2
ax
"
.
1
^a
,0?
a
-
.
INTEGRATION BY PARTS.
Apply
(C),
n
= 2, p = — -, a= — a
JV^ - tff**- ^~
2
,.
r
dz
l
Apply
(D),
rf >*
aW
1
-S6C
.
_,»
?i
3
= 2, p = — -,
a
=—a
.see
a 2(^_
7.
-x
2
dx
VflS?
f(a
2
—a
2
8.
2
2
2
2
2 x2 )
8
<fe
6
,
= 1.
_iiC
1
-
a3
a
1 1'
a
^
- a )* <fo = -(5 a -
(V-a?)$
*J
2
^
»/
a 2)i
= -^a -x + - sin"
2
2
QT*J
1
J
= 1.
a*
CI
C^Ja 2
b
,
making
«/
J
2
+ ^/^(^ - *>-»*
2o8
1
—
m = — 1,
5.
289
making
m = — 3,
*j
REDUCTION FORMULAE
V^^ + ^
2
= *(2 x - 5 a )V?=7+^
2
8
sin_1 -•
a
8
2
8
4
log
(a?
+ VaF+d).
INTEGRAL CALCULUS
290
9.
10.
2
fx
-x
^/a 2
2
fxWx^+ti
2
dx
= ^(2 x - a ) Va - x + - sin"
2
2
dx = ^(2x2
+a
2
o
»/
)
2
2
^/xT~+a~2 -
1
-•
^ log (a + V^T^"
2
8
).
C?iC
2
(a
12.
a2 Va 2 -aJ*
-a )t
2
Derive the formula of reduction used in Case
IV
of Rational
Fractions.
OjX
/
J
1Q
is.
(x 2
+a
f
)
X
J.
2(n-l) a
da;
2
P
*
v
|"
—x
2
2
(x3
+a
5
2
s'cfa
?
a
and apply (^) twice
2
.
= ^* V2 ax - x + f sin" 5=
2
2
a
3
1
= —-— V2 a# — # + — vers -1
2
2
U'37
I
V
sec
J V2 a — x
x\/2ax—a?dx
or
Q\
__ + 3^ ^^
aiC
az - x 2 dx
"~
l
1
2
V2 a# — x
f
^ V2
/(~\
.
+ a )*-
dx
/*
*^
asc
2
—a
^^
x V2
[_(x
3 x2
^jx_
•^
17.
=
r_
=_
^ V 2 aa? — or
Write
*g
n
2
2
a
n~1
2
)
_
REDUCTION FORMULA
INTEGRATION BY PARTS.
V2 ax —
Write
J
n Ex.
2
a*
d.r
=
2
o «-
«y
f*
1ft
19.
— a)
2
I
3
2
\
,
vers
-1,3!
-.
a
a
a 2
r? ' 1
-
a.i-
"- 1
—
~-
'
1
'
7
a-
V2 aa —
2
(2
it-
m — l)a
»»
.r
/»
a;™- 1 eta
^ V2 aa; - or
w»
*
5
**
21.
f xm
*^
V2 ax — x
2
V 2 ax— x + m — 1
m - l)ax m (2 m - l)a J
2
|
.
x m V2
aa;
-x
2
xm ~^2 ax
f i'-
aa:
—x
2
i
- ar
2
dx
m+2
m+2
t
cfa
/•
,
(2
23
-
*
a;
f
J
+ ax - 1 * V o^r^ + g
o
2
V2 ooj — x2 dx = V2
/«
ax — x- + a vers -i#.
'"'
22
and substitute
rife,
...
»/
20
(x
o.
fx V2^T? fe = _
18.
Va —
I
291
•/
dx
- x )*
(2 m - 3)aa;"
(2 aa;
2
223 Trigonometric Reduction
m-3
(2 ra
r ^2ax-x
- 3)aJ
Formulae.
— The
1
a;"
"1
2
dx
.
methods explained
q Arts. 211, 214 are applicable only in certain cases.
By means
I
of the following formulae,
sin m x cos n x dx,
tan" x sec n x dx,
1
and
J
lay be obtained for all integral values of
eduction.
J
m
cot m x cosec n x dx,
and
n>
by successive
292
INTEGRAL CALCULUS
sin" x cos" x dx = Jf
slnW-1 ^ CQS " +1 ^
Psin"
a;
cos" x
= smm
dx
*/
/
sin TO
*2?^l1 fsin—
+ ra
+ nJ
m+n
^ x cosW "
1
x
2
aj
+ -^— f sin™ a cosw m + n*/
^-
m+n
cos w x dx. (1)
2
a da;.
l
1
sin"
a?
sinm+1 x cos w+1 x
—JC sm
m-\-n-\-2
+1
,
J.
,
cos w x dx
x cos
a?
1
m
= sina;cos
.
M_1
a?
,
h
1
cotm a; cosec w
a?
—— 1
n
T cos wn _
J
.
(3)
.
.
a;
da;
sa\
(4)
/KN
(5)
-.
xdx
2
/a
(o)
,
,
x dx
m~1
m+n—
,T
U
("tan"
1
"2
x see" a? da;.
(7)
da?
m — lC
n „ ,7™
m x cosec
„,>„«„*
x dx.
cot±m-2
m+n
secn_2 x tan
n—1
a;
/q>
z
I
m+ —
sec n
2
I
= tanW sec " x
m+n—l
cot m_1 a; cosec" x
_ £2
7i
1
/-, =
m x cos n+J
n+2
^
x dx.
I
~1
tan™ a? sec' x dx
-
I
m — 1 C sin™m _
m J
n
7i
I
.
71+1
sin m_1
sm m xdx =
/_
.
'
71
_
.
cos w x dx
=
Jf
(2)
x cos n x dx
JS»J«3»
ii-., M .,fc
+ Si!±!f
m+1
m-\-l J
I
.
.
n
— 2 C w2
„_
sec
— 1*/
7i
,
2
I
1
a;
da;
.
\6 J
/
(9
>
1
REDUCTION FORMULAE
INTEGRATION BY PARTS.
cosec x dx
1
=
—1
n
Ctmi"xdx =
Ccot n xdx
224.
tan "~ lx
-
= - Qotn
I
sin-
-
,v
by parts with
sin- x cos" x dx = Jf
-2
x cos"^ 2 as dx
n—
cosec n ~- xdx.
U
I
.
(10)y
.
.
Cteti*-*xdx
Ceot n
u
xdx
(12)
— To
derive
we
(1),
= sin m_1 x.
sm ""
1
n
=
-2
(11)
Preceding Formulae.
Derivation of the
integrate
1
293
x cos " +1 x
+
sin"
1-2
a;
+ Azil f sin—
n-\-lJ
7
cos" x
—
dx
J
2
x cos n+2 x dx.
sin w x cos n
x cfa.
J
Substituting this in the preceding equation, and freeing from
fractions,
(m
we have
+ »)
sin m
I
a;
cos'
1
a;
cto
= — sin"m_1 » cos" +1 # + (m — 1)
sinTO_2 a? cos"
L
which gives
To
derive
a?
cfa,
J
(1).
(2),
integrate
by parts with u
= cos
71
"1
x,
and proceed
as
in the derivation of (1).
Formula (3) may be derived from
and replacing m — 2 by m.
Formula (4) may be derived from
and replacing n — 2 by n.
To
derive (5),
The derivation
make
ra
=
in (1);
of (7), (8), (9),
(1)
by transposing the
integrals,
(2)
by transposing the
integrals,
and
and (10)
to derive (6),
is left
have already derived (11) and (12) in Art. 212.
make
to the student.
m=
We
INTEGRAL CALCULUS
294
EXAMPLES
n
n
2.
o
3.
C
5
•
6
cos x /sin x
7
Ccosec xdx =
5
7
/•
J
,
7
sec x dx
—f
sin x
=
,
f-
-
4
3 \
„
2sin 2 xJ
5
,
2cos oA3cos #
\
.
5x
,
3i
X
+ 8 logB tan-2
.
,
t:
— + -5
N
.
|
4
2
5
.
.
—
1
»
.
—
1
xf -—
\sm*x
cos
I
J
I
5
.
12 cos- a?
8,
5
log (sec x + tan x).
+—
A
4.
K
5.
sina/
Ccos * #cta
a =—
— cos
8 \
J
8
/*
I
J
•
7.
J
1
-r-y-*8=
3cosa?
5
4
3
\
35a;
+—
cos#+tt—
16
128
y
—
,35 cos 3 # ,35
#+
sin 3
24
,
—- +
16
sinaA
ic
.
a;
-
]
12
^ 3
——4 cos
-p-
snr x
+,7-cos
6
2
f cos*x
a
cos # /sin 5 x
^ f
^
=
sm4 x cos o2 x dx
%
c
6
7
I
8
3coscc
a;
o
sm x
A
4
•
3,
x
,
+ 8g log tan-2
,
8 sir2 a;
J
J-^^-^^+^-lsin;
sin 4
a;
cos 3 x
cos 2 # V 3 sin 3 x
3 sin a;
2
+ -5 log (sec x + tan #).
8.
Atan
I
J
4
4
o
sc
,
sec 3 a cte
=
/tan 3 a;
\
6
tanaA
-—
8
/
sec 3 aH
,
sec
a?
tan #
•
16
+ — log (sec x + tan x).
9.
rcot^cosec5
J
^^- 60566 ^^ 008602 ^^-^^^ 60^ 0086
2
V
3
-ilogtan|
12
8
•
CHAPTER XXVI
INTEGRATION BY SUBSTITUTION
The
225.
XXIII,
new
substitution of a
variable has been used in Chapter
for the rationalization of certain irrational integrals.
We
some other cases where, by a change
given integral may be made to depend upon a new
shall consider in this chapter
of variable, a
variable of simpler form.
We
some substitutions applicable to integrals
and afterward those applicable to integrals
shall first consider
of algebraic functions,
of trigonometric functions.
226.
Integrals of form
I
f(a?) xdx, containing (a
the most obvious substitutions,
By
is
this,
any integral of the form
Integrals containing (a
By
for
+
example
C
is
is
One
of
z.
ifix^xdx
p
2
6.x* )?
-
-
J
the substitution
This
applicable,
2
- \f{z)dz.
changed into
Take
when
+ bzr)q.
x =
x'dx
are often of this form.
x3 dx
VI -x 2
x2
=
z,
_
1
r
of the form of Art. 203,
l-z = v:\
295
zdz
and
is
rationalized
by putting
INTEGRAL CALCULUS
296
The two
substitutions in succession are equivalent to the single
Applying
2
2
2
.
this to the given integral,
x2
^ dx
Jf vi - «
— x = id
1
substitution
=1—w
xdx
2
,
= — w dw.
=- r a-"')wdw = _ r(X _ w r)dw
J
J
w
= -^_| ) = -|(3-^) = -^^(^ + 2)
3
EXAMPLES
J V2s*+1
2.
3.
far* (a
2
- x )^dx = -£-(6 a - aV - 5 a )(a - x )*.
4
2
dx
C
30
JaVa + a
2
=
2
1
log
2a
4
^' + ^-^ ^-Llog
Va +
2
a2 + a
2a
1
.
r
°g?
vx + 1 — l
2
•^
5.
f
^
a^
2
2
aj
t
Var
2
+a +a
2
= |r^±i)! + (^ + ,i)i + iog(^+i-i)'
^L
xdx
+ 2V3 -
«*
(yV + a + a)
a
4
2
2
^
* log
OJ
2
4
V3~=^ + 1) + j log ( V3^¥ -3).
2
(
4
Va — ar
V&*2 ± a 2
by a Trigonometric Substitution. Frequently the shortest method of
treating such integrals is to change the variable as follows
227. Integration
of
Expressions containing
2
2
or
,
INTEGRATION BY SUBSTITUTION
For a
a*
— .r
,
let
as
= a sin
or x
= a cos 0.
o-,
let
x
= a tail
or
a;
= a cot
let
#
= a sec
or
a*
= a cosec 0.
For
vr +
For
v .r — a
2
,
Jdv-
297
0.
.
(a2
= a sin 0,
Let x
cr
/
Take
= a tan
dx
2
— or = a- — a
2
— a?)% ^
_
+a
2
)*
dO,
= <r cos- 0.
smi
1
r
C
tan
cJO
a 2 J cos 2 6
a3 cos 3
for another exai~~
nple
Let x
/ x^/x
= a cos
do __
_ a, cos eOde
c7x
(a*
cfcc
2
a,-
j
J
(9
a2
a 2 Vd2
—
x^Jx2
+
0.
r a sec
^ a tan
2
/" sec
__ 1
dO
a sec
•
a J tan
d
n_l C
- cot 0) = 1 log Vx +a -a
2
2
- log (cosec
a
Again, find
Let x
J
a;
=
I
a sec
x
a
^-« \te
^
'
0.
J
J
a sec
= a f (sec - l)d0 = a (tan 0-0)
2
Vx — a — a sec -1 2
dO
a J sin
2
a
•
—x
2
INTEGRAL CALCULUS
298
EXAMPLES
1.
n
Va — x
2
I
J
2
dx
= - Va — x
2
sin
-j
-1
-
•
a
2
C dx
_ _ Va^ + a
J Va?Ta
2
2
^
2
a;
3.
—
,2
2
2
f dx = \og(x4-VxT^~a1\
J ^Jx — a2
2
4
5.
dx
C
I
J
—
(2a;
=
+ a )Va -^
8.
9.
1_
2
log (x
.
+ Var -h a
2
)-
rr.
x4
3a2xs
C_dx__ =z
~ (2x -l)Vx + l
2
2
7
2
2
^— dx — — ——-L
ctr
J
2
J x'Vx^+l
3x*
fv/^±i da? = f
r
^
io.
f
11.
f
*
+1
da;
——
=\/^^x + 1
*?
=J5±i.
g*
(a3
+ l)V^ -l
2
*<* g
^V3 + 2x-aj
==
2
= V^I + log (x + V^l).
_V 8 + 2cc-a; + sm-^
2
2
'
INTEGRATION BY SUBSTITUTION
V3 4- 2 x - x* = V4— (x— l)
For
r
J (j* + 2x + 3)*
"+1
2 Va 2 + 2 a;
<f.r
12.
0.
+3
2
let
+ 1 = V2 tan 0.
a;
r v
x—a
—
C
J (2ax-x*)% aW2ax-x*
*'
13
UC
.
ja
J \i
V
—x
I
— x cfo = wax—x/
= C—a^=r
+ «,
_
da;
,
J V2^"T~'~
•
sin
_i2.T-ffl
l
a
2
(3a-\-x) V2
'
'
s
I
^ Vaaj-ai
2
*
a 2 ^x
(*
15
—
3a 2
Sm_! «; a
2
a
—x
2
aa;
.
~
2
228. Substitutions for the Integration
A
x — 1 = 2 sin
let
,
= V(<e+ 1) + %
Var>+2 x + 3
For
2
299
"
Trigonometric Functions.
of
trigonometric function can often be integrated by transforming
it,
by a change of variable, into an algebraic function. For this purpose
two methods of substitution may be used, as shown in the two following articles.
229. Substitution,
x=z,
sin
C
Consider, for example,
cos x
=
sin x cos x
sin
x = z,
then x
= sin -1
z,
dx
=
z.
dx
J 1 — sin x + cos
Let
tan x =
or
z,
2
x
dz
VI -z
J
sin x cos x
1
dx
— sin x + cos
2
x
zdz
C
=
J (2 + z)(l-z)
= - ? log
o
(2
C
_
+ 2) - I
o
J
2
3
z
1
Vl — z
_ C
dz
— z + l—z
r dz
J 2 +2
log (1
2
2
^/\
_2
1
r_dz_
3
J 1-2
- 2) = - |
o
log [(2
zdz
J
2
+ sin
a;)
2
—z—z
2
(l
2
- sin «)].
INTEGRAL CALCULUS
300
EXAMPLES
1.
C
Ja +bft3in
2
2
2
3.
,
2
x
=-1—
r^-^taii-^tan^l
a
\a
a -b
— —
i
8
a?
g
v
,
j
6
2
= lio
f^^^
4
sin
)]
[_
r dx
=x
J 1 + tanx 2
J
2
2
v
cos
in
;
l-sm^ J_ log l + V2sinx
+
1 + sinsc
1 — V2sina;
4V2
>
Let sin a; =
»/
sinaj
/• sin
K
5.
dx
°
f
4.
sin
+
— cos x
+ 2 cos
a?
dx
I
log
5
7
=
dx
3
sin
1
= x-\
x
.
,
-\
—+
(1
a;
i
cos x)'-.
oj
1
1
•
log (sin x
5
,
i
log
= loj
^3
tan x
tan x
= z.
T
Let
cos
Let
tan x =
,
(l+2cosa;) 2
5
tan x
7.
n
a;
/'tan
'tan 3 x
a;-,,
6.
3
x
I
J
+
= -1
sin2x
a;
2.
+l02 cos x)
\
z.
— V -.
+ ^J§
Show, by transforming into algebraic functions, that only one of
the following integrals can be expressed in terms of the elementary
functions.
(See Art. 208.)
where
/Vtana?d!a;= Cxi!*?,
2
J
J Vsina?daj= J
I
I
1
z
= tana;.
+z
—
Vl-z
=
2
|
—
J Vz-z
,
3
where
z
— sin x.
INTEGRATION BY SUBSTITUTION
230.
=
The Rational Substitution, tan -
sin x, cos x, tan x, and
By
z.
301
this substitution,
dx are expressed rationally in terms of
2 tan 2
siu x
z.
For
22
-
1
+ tan--
1-tan 2 *
2
COS X-
1+tan
2
2
1-Z
~1+Z
2
2
2 tan 9
tana;
i-*
1-tan *
2
2
2
From
= tan-
-
1
dx
2,
= ^-
c
9.
any trigonometric function of
It follows that the integral of
may
not containing radicals,
of a rational function of
z,
of elementary functions of
231. To
preceding
find
and can therefore be expressed
Applying
J s the substitution of the
™
|
tan -
=
in terms
x.
C dx
J a + bsmx
article,
x,
be made to depend upon the integral
z,
/i
z riz
2dz
i
&
+*
2
2bz
,
~r z
1
+2
«
'
2
2dz
r
J a (l + *) + 2ki
2fttfe
/
a 2z 2
+ 2abz + a*
_
2
/"
J
(az
r/rfc
+ b) + a - b
2
2
2
INTEGRAL CALCULUS
302
If
a>
by
numerically,
x
r_dx__ =
J
a
+ 6 sin
< 6,
If a
/
+
6 sin
t
Va -6
2
a;
^_
az
Y
+b =
Va -6
2
2
2
^^
2
Va -6
2
2a&
_ r
cc
J
aZ
(i
2
+ 6 — V& — a
2
2
find
:
atan| +
&
+ V& -a
2
2
dx
Jfa
-j-
b COS
/
£C
2^2
6(1— z 2 ) ~~J ( a —
2
/"
b) z
2
+ a -f &
dz_
&''
+
a—b
> b,
numerically,
dx
Jfa + b COS
b
t
2
-6-V6
a2 + &+ V & 2_ a;
a^
2 dz
a
+
2
log
V& - a
If
2
Va -6
2
log
y 6 2_ a 2
+ &)2_(52_ a2)
a tan -
232. To
a tan -
numerically,
da?
a
2
£C
a-6^a+6
_2
tan-'f
Va+6
J«L^ tan
;
INTEGRATION BY SUBSTITUTION
If a
<
b,
303
numerically,
dz
2
r az
2_
—
b
aJ «a _ b + a
b — a
dx
r
=
J a + b cos x
gv6-g-V& + a
sV&-a+V& + a
1
1
V& - a
2
2
.r
— a tan h + V b + a
Vfr
V& — a
2
log
2
V 6 — a tan #« — V & + a
/
/
EXAMPLES
Integrate the following functions by means of the rational substitution.
1.
2
"
3.
= W^2tan*
3cosa?
2
J,-^2
V
*
= 1 j
f
J l + 2sin2a 2V3
,
+ 2-V3
tana; + 2 + V3
tanaj
(?
&
3tan^
in^ + 2
*
=i
Jf 5 sin x + 12 cos a 13 log&
I
,
2 tan
x
^-3
&tan£-a + Va +
2
J a sin* + 6 cob*
5.
f
J
sin
./,*
*vers
-f-
V^+6
5
tan
= logas
^
°g
6 taD
x
2
H- tan
2
*_a_
&
V^+6
2
5
INTEGRAL CALCULUS
304
tan--l
6.
'•
Jf 3 —
*»
sm x + 2 cos a?
1=
T
a;
(1
= ^log
— cos x
/dx
—
2
3tan| + l
^
Jo—
+ sm
I
= tan-_J?
.
5
.
+ sina + cosaj)
2
tan^ + 2
= -1.x
tan
2
233. Miscellaneous Substitutions.
reciprocal substitution, x
1+tan
—
»
i
log
A1 + tan
f
V
Various substitutions applicable
by experience.
to certain cases will be suggested
The
1
2
= ~, may
be mentioned as simplify-
z
many
ing
integrals.
EXAMPLES
Apply
L
J
trie
reciprocal substitution x
3aV
x4
r___dx__
2
'
J
_ _ Va?-|-a
fvV + a
r V2 ax x
J
3
2
2
a;
"
^=
(2
ax
.
- x )?
2
g
**
=ii og
r
J j;VV ± x a
« + Va ±
r ^-f)* ^
J
x
4
^
Sax3
2
2
5
2
a^»
s
4.
=-
4
3JX-Xill,
)
3
,4
8tf
x1
-
to Exs. 1-6.
INTEGRATION BY SUBSTITUTION
r
J ±\
7.
f
*b
2
8x
+2x-1
=
"***
305
x-1
= snr
3x
§_ + _6_ +
5
i
g(a
.
+ 2).
Let
8
.
^f
r ix+a)(x+brdx=
J
n
-\-
+(a _
b)
Z
a;
+ 2=
(^i
+
1
91
Let x-\-b
J
x (a
+ &a?)
3
+ &a>)
a3 _2(a
2
a
+ bx
- dx =
tan
11
.
(a;
+ a)
—^
f-
sin
a;
— sin a log tan
*
"T
Let
•
= - - «- + log Ve & + e* + 1
+ 6a; =
a;
—-dx = V(a -
Substitute b
+x=z
a)(&
2.
1 ^.,2^ + 1
—
tan
a?
Jf\^ b +
4-x
a;z.
+a=
V3
V3
Let
12.
— z.
x
Let a
10.
2.
e
x
=
z.
+b
+ a) + (a + 6) sin" J*
1
a
2
,
and the integral takes the form of Ex.
5,
Art. 222.
13.
= V(a; + a)(x + b) + (a- 6)log(Va; + a + Va; + 6).
Substitute x
Art. 222.
+6=z
2
,
and the integral takes the form of Ex.
2,
CHAPTER XXVII
INTEGRATION AS A SUMMATION.
An
Integral the Limit of a Sum.
234.
DEFINITE INTEGRAL
integral
may
be regarded
and defined as the limit of a sum of a series of terms, and it is in
this form that integration is most readily applied to practical
problems.
Area
235.
of curve the limit of a
quired to find the area
PABQ
sum
of rectangles.
Let
it
be
the axis of X, and the ordi-
nate s
Let
AP and
y=x?
BQ.
be the equa-
tion of the given curve.
Let OA = a, and OB=b.
Suppose AB divided into
n equal parts (in the figure,
n = 6), and let Ax denote
one of the equal parts,
A\A2
AA
1}
•••
,
Then
At A
AB = b — a = n Ax.
lf
Ac,,
•••,
draw the
••, and complete the rectangles
2
2
PA P A
= x?,
QB= b?.
PA = a\ P A =(a + Ax% P A = (a + 2 Ax)?,
Area of rectangle PA — PA x AA = a? Ax.
Area of rectangle P±A = P^A x A A = (a + Ax)? Ax.
Area of rectangle P A = P A x A A = (a + 2 Ax)? Ax.
ordinates
From
AP AP
X
X,
,
the equation of the curve y
X
1
X
2
Area
of rectangle
3
2
PB=PA
5
•••,
2
2
X
2
5
re-
included between the given curve OS,
5
X
X
2
2
2
3
x
A B = (b — Ax) ? Ax.
306
X,
Y
2,
INTEGRATION AS A SUMMATION.
The sum
of all the n rectangles is
1
+ Aac)* A.r -f (a + 2 A.r)*- As H
a - A.r -f (a
^
which may be represented by >
5
.r-'
+ A.r, a + 2 A.r,
evident that the area PABQ
a
a,
•••,
rectangles, as n increases,
That
and
is
series,
the limit of the
V x*
A.r
The
.r-
x°-
From
Ax.
the preceding article
= a* A.r + (a 4- Aa)* Ax'+ (a -f 2 Ax) * Ax H
That
is,
by
of the
a
h (6
limits of this sum, as A.r approaches zero, is
dx.
sum
A.r decreases.
'
Definite Integral.
x taking in succes-
— A.r.
b
PABQ = Lim Ax=0 >
Area
is,
236.
— As)* Aa?,
A.r,
each term of the
A.r represents
sion the values
It is
h (&
i
a
l
where x-
307
— Ax) i A.r.
denoted by
definition,
J
«2
|
dx
= Lim A:e=0 > & Ax.
^
a
h
X
a
1
1
x-
is calleji
f?.r
It is to
bol
The
I
.
be noticed that a
237.
new
definition is thus given to the
sym-
anti-differential.
between these two definitions will be shown in the
fol-
article.
Evaluation
of the Definite Integral
by finding a function whose derivative
dx\ 3
By
of a^da?.
6,
which has been previously defined as an
relation
lowing
the definite integral, from a to
I
x% dx.
This
is a?*
J
the definition of derivative, Art. 15,
—
da\
f
,
]
3 J
= Lim
^
- (a
-f
Ax) 2
= s*.
.
A,;
is effected
INTEGRAL CALCULUS
308
= X* +
T
Ax
where
e is
a quantity that vanishes with Ax.
2
2x*
^—
3
-(x4-Ax) 2
Hence
o
o
i
=x Ax + eAx.
2
Substituting in this equation successively for
a,
Ax,
a-\-
a
| (a + Ax)*
o
^ (a
o
(a
|
o
26f
.
C,
o
+ 2 Ax,
^=
•••,&
a* Ace
o
x,
— Ax,
+
<a
Ax,
+ 2Aa)* — - (a + Ax)^ = (a + Ax)? Ax + e
2
o
Ax,
+ 3 Ax) * - ? (a 4 2 Ax)* = (a + 2 Ax)* Ax + €
3
o
_ o|
Ax,
_ Ax) = (6 - Ax)?Ax + en Ax.
3
(ft
Adding and cancelling terms
in the first
members,
2
?A _ 1^1 = a i Ax + (a + Ax)*Ax 4- (a 4- 2 Ax)? Ax H
+
e1
Ax + e 2 Ax 4
e3
Ax
4
•••
-f eM
1-
(6
- Ax)?Ax
Ax
= ^^
V^Ax + Y^Ax
a
Comparing with the
seen, represents the
the
sum
(1)
a
figure, Art. 235,
sum
Xx
of the rectangles,
2
Ax, as
and 2j
we have
e
a
Ax
before
represents
of the triangular-shaped areas between these rectangles
and
the curve.
The
latter
sum approaches
the limit zero, as
Ax approaches
zero.
INTEGRATION AS A SUMMATION
For
if e A is
.
the greatest of the quantities
V
1
that
As
>,
is,
et
c
Xv
<
e,T
e
&x
<
e /t(^
^^ a
a
c1} u, •••£„, it
809
follows that
Ax,
—a
)-
vanishes with Ax,
Lim
Taking the
^^
eAaj
= 0.
limit of (1),
a?s
Thus the value
of
da;
= ^iL-— = Area PABQ.
x*
I
dac is
found from the integral
2
S xldx = 4>
by substituting for
a;,
b
and a
in succession, thus giving
3
3
The process may be expressed
(Mo.-**
Ja
This
is
6
I
2b%
3
2
J
3
called integrating between limits, the initial value a of the
variable being the lower limit, and the final value b the upper limit.
In contradistinction
fx
is
l
dx
called the indefinite integral of
= 2-^+C,
xMx.
INTEGRAL CALCULUS
310
238. General Definition of Definite Integral. In general if f(x) is a
given function of x which is continuous from a to b, inclusive, the
definite integral
/ (x)dx = Lim Aa = o
/(a)Aa>
.
J
+ f(a + Aa) Ax + /(a +
+ - +/(&-Ax)Ax
If
j
f(x) dx
= F(x),
2 Ax)Ax
.
the indefinite integral,
l
f(x)dx = F(b)-F(a)
--.(I)
may
be illustrated by the area bounded by a curve as in Art.
to be the equation of the curve OS.
The proof of Art. 237 may be similarly generalized by substitut-
This
235,
by supposing y—f(x)
ing /(» for
a?*,
and F(x) for
~-
Geometrically the definite integral
I
f(x)dx denotes the area
swept over by the ordinate of a point of the curve y =f(x), as x
from b to a.
It is to be noticed that in Art. 192, by a somewhat different course
of reasoning, we have arrived at the same result,
varies
Area
239. Constant
constant
G
PABQ = F(b) - F(a).
of Integration.
in the
indefinite
It is to
integral
be noticed that the arbitrary
disappears in the definite
integral.
Thus,
if in
evaluating
x* dx,
J
/
wefind
Or
if
we
take for the indefinite integral
3
'
jyUx = *f + C-(?f+c)== 236*
Cf (x) dx = F (x) +
Cf(x)dx = F(b)
2
J
g
C,
+ G- IF (a) + C]=F.(b)-F(a).
INTEGRATION AS A SUMMATION
EXAMPLES
.0
1.
Compute
When Ax =
X/ Ax =
2 -^Ax for different values of Ax.
.2,
(l
2
2
+ L2 + 1A + D?
2
2
+ r8 )(.2) = 2.01.
When Ax = .1,
2^x Ax = (Is + 13* + V? +
2
•
•
-
+ L9 )(.1) = 2.18.
2
When
A.?
= .05,
Ax
Lirn Ax=0 5j a~
= 2 26
Jj'asPAa
=
x2
(
-
c?x
-
=—
2.33.
Curve OS, y=x*. 0A=1, 0B=2.
Area PABQ = 2.33 square units.
2.
Compute
—x
V 'Ax
*"*1
When Ax =
4
Ax
(\
for different values of Ax.
Y |R
1,
1
s.T-e+i+i^- 1 ".
When Ax = .5,
When Ax
o
,
^A Ax
1XX _
AT"
Ax
^1
X
1.593.
1.426.
311
INTEGRAL CALCULUS
312
— = Jl —X =
Lim Aa;=0 >
i
0A = 1, 0B = ±.
= ->
Curve RQ, y
= log 4 - log 1 = log 4 = 1.386.
log x
**l X
PABQ = 1.386
Area
square
units.
3.
Y x Ax, when Ax = 1
Compute
;
when Ax = .5
when Ax = .2.
;
3
^Ins.
Find Lim A3;= o 2y ^ Acc
4.
Computed
Ax =.05.
Find Lim Ax=0
5.
Compute
Am
V
when Ax =
log 10 xAx,
V
^^
log 10 xAx.
tan<£ A<£,
when Ax =
l;
Ans.
'
>
.5;
.316;
Ans. log e
A<£
.328;
r
3
(ic
2
-4) 2x<!x =
/* 9
125
Va
Jo
e
2*
-2/
+l
2
_9
J Vx + 144
2
5
r
2
x dx
'
6
-;«
.785.
when
3
2
Ji x2
4
dx
-x + l
.340.
V2 = .346.
4
7.
=
oU
.4ns.
180
<f>
when
3.121; 3.150; 3.161.
ou
tan
-
when A<£ =3°=^; whenA<£ = -^-; when
IT
V
20
Ans. 13 log 10 13 — 3 log 10 e — 10=3.177.
= -^-.
Find Lim AcE=0
.l;
-
.833; .810; .798.
Ans. £
4
10
Compute
when Ax =
Ax =.2;
2
Find Lim Aa;=0
A<£
V
when
+x
Y* -**L.
^o 1 + x
Ax =.3.
6.
^ns
-,
ol
18; 19; 19.6.
-
*
3V3
:
INTEGRATION AS A SUMMATION
13.
15.
C
flv
jTsin^
<
-
=
313
rf
ff .
M = ^_|.
16
8
r vers22e ^ = ^_7V3.
_
4
«/o
is
,;
cos 20
"f.
cos
18.
Jo
(aj
— cos 2 a dO
1A
= 1 + - cos a.
— cos a
-,
6
1
+ 2)
=
3
f
Jo
2.
—
1
19.
6 log
&
4
l
+ 2x + 2x + 2x + x*
2
tt
,
s
= -log& (2 e).
;
^
4
7T
20.
.i*-
sin
.r
dx
2a
21.
a log (x
f
%J a
(5)
and
(6),
= - log 4.
tt
4
Art. 223,
sin"
J
;/:
we
dx
find
= ~
?/
n
o
~2
a;
cfo,
7T
n
I
sin n
(
Jo
7T
Jo
4
A
•/o
By
7r
+ a) cfc = ^log (3 a) - 1.
f'tan- 1 -dx
22.
= — 2.
cos x ax
=
I
n
Jo
from which derive the following results
cos n-_ x ax
;
16
INTEGRAL CALCULUS
814
n
If
24.
even,
is
E
E
xdx = C\os> n xdx =
J"\in
o
n
1 S b '"( n
'
Jo
n
If
25.
E
2
J"
sin"
I
C
x dx =
I
/ (x) dx, we
a series
Jf
2
cos M # d#
(x)
dx
is
sum
^
•
of
x— a
negative
•
£•
• • •
n
In considering the definite
have supposed a <
to
terms,
x
6,
= b,
is
inte-
and
/ (x)
V
/(a?) A#,
being the sum
and
consequently
negative,
to be positive be-
negative.
If /(a;) changes sign
algebraic
= 2 »4. 6 •••(71 — D
3 5 7
Sign of Definite Integral.
tween the limits a and b.
Iif(x) is negative from
of
)
2-4-6.--H
c/o
o
gral
~V
odd,
is
E
240.
'
between x = a and x=b,
of a positive
and a negative quantity.
£
— 1 = area ^4 0J5.
Y
For exa mple,
I
cos x dx
cos x dx ==
— 2 = area
JBCZ).
I
/(#)
cto
is
the
J
:
INTEGRATION AS A SUMMATION
315
3ir
2
cos x dx
i
f
The change
= —1 = 1 — 2.
dx=0 = l — 2 + 1.
"""
cos x
of sign resulting
from a
<
b is
considered in Art. 243.
Infinite Limits.
In the definition of a definite integral the
have been, assumed to be finite. When one of the limits is
infinite, the expression may be thus defined
241.
limits
:
J
For example, consider Ex.
o
s
12, following Art. 239.
,
A ^,,tan _ —b\ = 2tt a
s
b
dx
,
fix) dx.
\
= TLi m 5=» r Sa dx = TLnn
ar + 4cr
Jo or + 4 a
r~8a
Jo
= Lim 6=w
fix) dx
.
.
I
o
Keferring to Art. 126,
o
we
o
2
,
,
x
4
\
6=ao
2
.
2a
find the geometrical interpretation of
this result.
The area included between the
curve, the axes of
variable ordinate, approaches the limit
ordinate
2-n-a
2
,
X
and Y, and a
as the distance of the
indefinitely increased.
is
J^dx
—
we
,
find
x
6
dx
—
= Lim^^ log
J" x
b
— oo
.
i
00
Jr
1
meaning.
242.
is
Infinite
assumed
/(#)
is
Values of f(x).
to be a continuous function
continuous for
is infinite,
In the definition of
all
from x
to
values from a to b except x
the definite integral
fix) dx
may
be defined thus
= Lim A=0
I
/ (x) dx.
has no
25
f(x)dx, f(x)
I
=a
dx
—
x
= a,
= b.
where
If
it
INTEGRAL CALCULUS
316
= oo, f(x) being
If f(b)
continuous for other values of
Xbf(x) dx = Lim^
For example, consider Ex.
9,
x,
S*b-h
f(x) dx.
I
following Art. 239,
dy
r Va -/
2
2
Here
—
= oo
,
Va — y
2
,
when y = a.
2
Hence
J* ^/a 2
J
—y
2
-y
-Va2
l
2
_
Another example
is
6J
IT
7T
Ex. 13, following Art. 239.
r
Here
a
V
7T
—
-
*
y/(x-2)(3-x)
= oo,
when x = 2, and
also
when # = 3.
V(a>-2)(3-a;)
Hence
j
—
= Lim
J2 yj(x_2)(3-x)
Lim^o sin -1 (2 x —
'"*=
5)
ft=0
j
—
j2 + h V(s
Lim^Csin- (1-2 ft)- sin- (-l + 2 ft)]
1
1
= sin -1 l — sin _1 (— 1)= w.
If /(a?) is infinite for
some value
c
between a and
6,
and
is
con-
tinuous for other values, the definite integral should be separated
into two.
Cf(x)dx = Cf(x)dx+ Cf(x)dx.
These new
definite integrals
may
See Art. 243.
be treated as already explained.
INTEGRATION AS A SUMMATION
317
EXAMPLES
r
3
dx
-
=l0 g(2+V3).
C
4.
^-— = -log2.
243. Change of Limits. The sign of a definite integral
by the transposition of the limits,
a
varies from a to
a,
the sign of
will be changed, if the limits a
A
may
definite integral
by the
and
Ax
also
is
from the
and
—
definition.
opposite to that where x
signs of all the terms of
f(x)dx =
integrals
f(fi) dx.
(1), Art. 238,
Hence the
b.
changed
h
f f(?) dx = -£
This is evident from
For if x varies from b to
is
]x f(x) Ax
b are transposed.
I
f(x) dx.
be separated into two or more definite
relation,
f 7(a)
dx =
f /(*) dx + f /(a)
(to.
This follows directly from the definition.
244.
variable
of Limits for a Change of Variable.
When a new
used in obtaining the indefinite integral, we may avoid
Change
is
retaining to the original variable, by changing the limits to corre-
spond with the
new
variable.
For example, to evaluate
p_* _
L
Jo 1
-f-
V#
i
assume
y/x
=
z.
INTEGKAL CALCULUS
318
Now when
x
1+Vz
=
4, z
—2
1
+z
and when
;
f—^p = C
Hence
2zdz
dx
Then we have
^
=4-
x
= 0,
« 2 [* -
2 log
l
og
2!
(1
= 0.
+ *)]
3.
EXAMPLES
1.
CxVx~T2dx = ^.
J2
2.
f(
a;
-2)" !K c«a; =
*/2
f s ^
J o(x -\-iy
2
3
8
2 ax
6-
7
8
- 3m + 5
+ 3^ + 2
8
—x
2
+ ^g).
dx =
7rcr
r v3 + 2^-^ &=V3 _T
c/2
I
(a;
— l)
V(x —
2
Let
x—2
Let
a2 +
Let
x
Let
a
a)(b
Let
#
= a cos
Let
x
= a sin
—x)dx = ^
(b
— a)
x^a -x dx^(^ + ^\a\
2
Let
z
= asin0
=
.
2.
.
l=z
3
.
— a = a sin 6.
=sl+ 2sinft
3
j^V-^) f ^ = 3^
32
\
a-f2 = z3
n2
3 (3
2
5.
Let
15
2
V64
48
2
.
2
<£
3
0.
+ 6 sin
2
<£.
INTEGRATION AS A SUMMATION
245.
it is
Definite Integral as a
Sum.
319
In the application of integration
often convenient, in forming the definite integral from the data
/(as)
dx as the sum of an
infinite
f{x)dx being called an element
of infinitely small terms,
number
of the
required definite integral.
From
this point of view,
ff(x) dx =f(a) dx + f(a.
This
may
+ dx) dx+f(a + 2dx)dx+
•••
+/(&)
dx.
be regarded as an abbreviation of the definition of a
definite integral given in Art. 238.
CHAPTER XXVIII
APPLICATION OF INTEGRATION TO PLANE CURVES.
APPLICATION TO CERTAIN VOLUMES
246. Areas of Curves. Rectangular Coordinates. We have alreadyused this problem as an illustration of a definite integral. We will
now
consider
it
more generally, and derive the formula
for the area
in rectangular coordinates.
To
247.
find the area
between a given curve, the axis
AP and BQ
moving from AP
given ordinates
ordinate
that
;
is,
of
X, and two
by the
to find the area generated
to
BQ.
Let OA = a, OB = b.
Let x and y be the coordinates of any point P2 of the
curve then
;
x
+ Ax,
y
+ Ay,
will be the coordinates of
The area
P
3.
the rectangle
of
P A A is
P A x A A = y Ax*
2
2
3
2
2
The sum
2
Z
of all the rectangles
PAA U P A^A P A A
X
2,
2
2
3,
•••,
maybe
represented by 2\ a y &v*
as
The required area PQBA is the limit
A# is indefinitely diminished. That
=
*
By
J
of the
of the rectangles,
y dx,
Art. 245, one readily sees that this rectangle
320
sum
is
is
an element of area.
APPLICATION OF INTEGRATION TO PLANE CURVES
the lower limit a
limit b
= OB,
=
OA, being the
the final value of
initial
value of
x,
and the upper
x.
Similarly the area between the curve, the axis of
given abscissas,
GP and
IIQ,
321
Y,
and two
is
=J
vdij,
the limits of integration being the initial and final values of
g=OG,
and
h=
y,
OH.
EXAMPLES
Find the area between the parabola y 2 = kax and the axis of
X, from the origin to the ordinate at the point (h, k).
1.
h
A = f ydx= f 2a^dx
Here
4a^i*
4a*7i-
Since k 2 =lah, k = 2
aw,
which
gives
A = hi2ah* = -hk = -OMPN*
2.
Find the area of the
ellipse
* + t = l.
a2 ^b 2
Area
|
BOA
y dx = -
Va-
I
— x- (he
aJo
o
bV x /-=
'= -\
-V« -ar + a-sin _iX~\
2
2
a\_2
t,
a
.
.
2
aj
Tr«b
* In finding areas, after the element of area and the limits of integration are
chosen, the problem becomes purely mechanical.
INTEGRAL CALCULUS
322
The
entire area
=
-n-ab.
Or we may integrate by
letting x
= a sin
<£.
a
f Va*-a?dx = a* C\m
Then
2
C
4>d<t>=^
(1
-cos 2 <l>)d<f>=
—
Areai^^.^^
4
a 4
3.
Find the area included between the parabola x 2 = 4 ay, and the
witch y
8<r
=
#2
+4d
^Ins.
2
Having found the point
of intersection P, (2 a, a),
(2,r-*V.
we proceed
as
follows
Area
AOP= AOMP- OMP*
= C
Jo
2a
cc
+ 4a
Area between two curves
4.
r 2a a?dx =
$a?dx
2
2
=
(
2v
2
4a
Jo
—
2 a2
3
)a 2
.
Find the area of the parabola
(y-5)* = 8(2-x),
on the right of the axis of Y.
Length
of element of area
is
Arts.
104.
the y of the witch minus the y of the parabola.
x
APPLICATION OF INTEGRATION TO PLANE CURVES
5.
Show
x2 — y 2
that the area of a sector of the equilateral hyperbola
= a*
X and a diameter
included between the axis of
the point (xf y) of the curve,
6.
323
Find the
— log
is
i'
through
•
entire area within the curve (Art. 133)
= 1.
f-Y + f£\
Ans.
- wab.
4
7.
Find the
Let x = a
ft + y$= a\
8.
9.
sin3
within
hypocycloid
the
132)
(Art.
3
^?is.
<£.
8
Find the
and the
area
entire
line x
entire area
= 2 a,
Find the area
between the cissoid (Art. 125) y 2 =
asymptote.
its
2a —
3
Arts.
of one loop of the curve (Art. 133) a*y 2
?ra
=ax —
4
2
Ans.
2
.
6
jc
,
TO
??.
8
Also from x
=-
to
a;
= a.
^3
Find the area
10.
(*r)^
of
the
evolute
of
the
+ (6y)*= (a - b )K
11.
ellipse
2
What
and
is
its
2
the ratio between a and
6,
when the
167)
?ffe 8 \b
*
a
areas of the
evolute are equal ?
6
12.
(Art.
ellipse
Ans.
8 ;4
V3
Find the area included between the parabolas
2
y
= ox
and
x*
= by.
Ans.
—
•
INTEGRAL CALCULUS
324
13.
Find the area included between the parabola
2
y
14.
=2x
and the
2
circle
y
=4x—x
2
Ans. 0.475.
.
Find the area included between the parabola
2
y
==
and
4 ax
= 4 (x — 2 a)
evolute (Art. 167) 27 ay 2
its
3
.
Ans 352V?„2
.
.
15
Parametric Equations. Instead of a single equation between x and y
for the equation of a curve, the relation between x and y may be expressed by means of a third variable. Thus the equations
x = a sin
represent a circle; for
x
Equations
if
we
+y = a
2
2
<f>
(sin
y
from
<£
+ cos
2
</>
(1)
<j>,
2
<£)
we have
(1)
=a
2
.
parametric equations of the
circle,
and
called the parameter.
is
A=
The formula
= a cos
eliminate
2
(1) are called the
the third variable
y
<£,
I
y
clx is
= a cos
For a quadrant of the
applied to (1) by substituting
dx
<}>,
= a cos
</>
d<j>.
circle
n
A=
15.
ydx=
a 2 cos 2
I
cf>
d$ =
7T(T
T
Find the area of one arch of the cycloid
x = a (0
16.
I
— sin 6),
y
= a (1 —
The parametric equations
cos
Ans. 3
6).
of the trochoid, described
at distance & from the centre of a circle, radius
a,
which
ira
2
.
by a point
upon a
rolls
straight line, are
x
Find the area
= a6 — b sin 0,
y
= a — b cos
0.
of one arch of the trochoid above the tangent at the
lowest points of the curve.
Ans.
it
(2
a
+ b) b,
when
b
<a
or b
> a.
APPLICATION OF INTEGRATION TO PLANE CURVES
to find the area generated
is,
To
Polar Coordinates.
Areas of Curves.
248.
PQ
included between a Given Curve
Two
find the
POQ
Area
Given Radii Vectores
;
that
by the
OP
radius vector turning from
and
325
to
OQ.
Let
Let
POX=a, QOX =
r
and
any point
P of the curve,
r + Ar,
+ A0,
The area
P2 OR, is'
of
"
"
The sum
P
3.
the circular sector
- OPo x P.Mo = -r -r A0
"
then
2
will be the coordinates of
2
f3.
be the coordinates of
= Ir
2
POP, PxOR^
of the sectors
resented by
as
A<9.
2
2
s
P OR
2
2,
•
••,
may
be rep-
-i
The required area POQ is the
A0 approaches zero. That is,
limit of the
sum
of the sectors,
-IX'"*
the initial value
final
value of
6,
of
= POX,
a
6,
being the lower limit, and the
j3= QOX, the upper
limit.
EXAMPLES
1.
Find the area
of one loop of the curve (Art. 144)
r
= a sin 2
2
A = lC
,*dO
-V/Q
= l fa
J»/o
2
sin 2 2
-IT-
f\l - cos 4 0)d0
f
4
*/0
-
'
0d$ =
0.
INTEGRAL CALCULUS
326
The
which
2.
7rd
entire area of the four loops
is
2
half the area of the circumscribed circle.
Find the
entire area of the circle (Art. 135)
r
= a sin
0.
—
Ans. —
4
In the two following curves find the area described by the radius
vector in
moving from
=
=-.
to
3.
r
= sec0 + tan0.
4.
r
= a(l-tan
5.
Find the entire area of the cardioid
Ans. 1
+ V2 - ^
8
Ans.
fc-*\<A
2
0).
(Art. 141) r
= a(l — cos 0).
"*
Ans.
,
or six times the area of the generating circle.
Also find the area from
=-
=
to
6.
Ans. (Sir
v
— 2) —
;
8.
Find the area described by the radius vector in the parabola
r
= a sec
2
=
-, from
Also find the area from
=5
to $
Find the
= -.
Ans.
= —^.
to
^4ns.
3
3
7.
—
4
4
—
—
9V3*
2
entire area of the lemniscate (Art. 143) r
=a
2
cos 2
^.fts.
8.
Show
.
that the area bounded by any two radii vectores of the
reciprocal spiral (Art. 137) r0
between the lengths of these
9.
0.
a2
=a
is
proportional to the difference
radii.
In the spiral of Archimedes (Art. 136), r =
a$, find the area
described by the radius vector in one entire revolution from
Ans.
^
= 0.
2
2
Also find the area of the strip added by the nth revolution.
Ans. 8 (w —
1)
3
ir
a2
.
APPLICATION OF INTEGRATION TO PLANE CURVES
Find the area of the part of the
10.
=a
/
sin
+ b cos 6,
circle
r
=
from
to 6
= -.
Ans. «•(«'+*$
8
Find the area common to the two
11.
2
2
2
r
=
3 a tan
Show
13.
r=
that the line
sec
+ tan
1
3 a2
Ans.
+ tanfl'
divides the
x + y=2a),
J
J
(
V
area of the loop of the preceding example in the ratio
2:1.
Find the entire area within the curve (Art. 145) r
14.
(Art. 127)
3
2asec0
l
> 6.
where a
Find the area of the loop of the Folium of Descartes
12.
2
r
2
^iws.
+ «»
circles
= a cos + b sin 0.
+ 2 tan-_, b\ a + b a b
= a sin + b cos 0,
327
= a sin
a
3
-,
no
o
part being counted twice.
249. Lengths
Length
of the
of
Ans. (10tt
Rectangular
Curves.
PQ between Two
OB = b.
Arc
Coordinates.
P
Given Points
+ 9V3)
§-.
To
the
find
and Q.
Let OA = a,
Denoting the required length of
arc by s, we have from (1), Art. 155,
ds
=nMS
dx.
\dx
a
P.
Hence
wi + aw«.
and between the given limits
s>m
dx,
•
•
(1)
B
A
the limits being the initial and final values of
x.
X
INTEGRAL CALCULUS
328
We may
also use the formula
-&+&«
the limits being the initial and final values of
g=OG,
and
«
y,
h=OH.
EXAMPLES
1.
Find the length of the arc of the parabola y 2 =
from the
4=ax,
vertex to the extremity of the latus rectum.
4 = 2!.
Here
therefore $
This
may
=j[Yl + ~Y
<& =
f Y£±*V
<fo,.
be integrated by Ex. 13, p. 305, making 6
rfa±x\l
j^Va +
»y
dx
= 0.
= ^ ax + x + a log V"^+^ + Vas)
2
(
^ = o j-^2 + log £ + ^2)] =2.29558
Or we may use the formula
(2),
*=jhr
_
2
?/
4a
=
dx
cfa/
M| V^HU + ^
2
= a[V2 + log(l+
V2)]
_
y
2a
log
+ V2/ + 4a )T
2
(2/
2
a.
..
APPLICATION OF INTEGRATION TO PLANE CURVES
2.
Find the
(Art. 130) ay
3.
2
length of
=x
3
,
=-
to x
= 5a.
—
Ans.
Find the entire length of the arc of the hypocycloid (Art. 132)
2.
2.
2
3
3
x*+y = a
4.
semicubical parabola
the arc of the
from x
329
'
Ans. 6
a.
Find the length of the arc of the catenary (Art. 128)
= a-( e + e —-),
^
-
y
from x
5.
=
a
a
-
Ans. ^{e a
to the point (x, y).
= log sec x,
=
from x
Find the length
6 aw
*
+ V3).
of the curve
= x* + 3,
from x
'
=1
to
a:
= 2.
—
12
17
^4ns.
Find the perimeter of the loop of the curve
9 a y-
8.
°).
= -.
to x
Ans. log (2
7.
-x
Find the length of the arc of the curve
y
6.
—e
=(x-2 a) (x - 5 a)
Ans. 4 V3 a.
2
.
Find the length of that part of the evolute of the parabola
= 4 (x — 2 a) 3 included within the parabola y2 = 4 ax.
(Art. 167) 27 ay2
Ans.
9.
y
10.
4(3V3-l)a.
Find the length of the curve
= log —=—
eZ
+l
,
from #
=1
to
x=
2.
Aiis. log(e
Find the length of one quadrant of the curve
(
-
j
+
(
-
+ e _1
)
).
= 1.
2
Ans.
11.
The parametric equations of a curve are x =
Find the length of arc from
=
to 6
=
jj.
40
e 9 sin 0,
-4n*.
^±^±&_
a+
y
=e
e
.
b
cos
0.
V2 (e*- 1).
INTEGRAL CALCULUS
330
The parametric equations
12.
fixed circle being a,
of the epicycloid, the radius of the
and that of the
rolling circle ~, are
z
x = - (3 cos
y
<f}
4>
— cos 3 $),
— - (3 sin — sin 3
<£
being the angle of the fixed
circle,
<£),
over which the small circle has
rolled.
Find the
Lengths
250.
of Curves.
Let
We
Ans. 12
To
Polar Coordinates.
PQ between Two Given
POX = a, QOX^fi.
Arc
of the
ds
entire length of the curve.
Points
P and
find the
a.
Length
Q.
have from (3), Art. 156,
=
p + (1)72
rei ore
2
s
fdrVJA
\dB) _
».
•
•
(1)
the limits being the limiting values
of
q'
0.
Or we have
therefore
ds
1
Off
+r
dr
2
s
(2),
Art, 156,
dr
dr,
-f-r
(2)
dr
the limits being the limiting values of
r.
That
is,
OP— a, OQ =
b.
EXAMPLES
1.
(Art.
Find the length of the arc of the
136), r = aO, from the origin to the end
Here
—=
d6
a,
and we have by
spiral of
Archimedes
of the first revolution.
(1),
APPLICATION OF INTEGRATION TO PLANE CURVES
s= C~\a
2
6
2
+ a )?d0 = a
2
C*"
331
2
(1+0 )"'dO
=0 [^l±i: + |i og( , + vr+^)J
+ 47T +hog (2 + Vl + 47T
2
TtVI
Or
formula
^ve ma}' use the
s
r
dO
1
a
ar
a
= Jf»2jra
(2)
+r h
Vl
\
a
2
=1
dr
2
p V?"^
2
"
2
2.
7rV47r 2
dr
a «/o
= * [~| V^+d + flog (r + Vt M^)1
=a
)"].
'06
Jf \i
a
2
7T
2
:
+ l+^log(27r +
Find the entire length of the
vW +
l)l.
circle (Art. 135) r
= 2 a sin 0.
Ans. 2
3.
Find the length of the arc of the
r
4.
= a sin +
& cos
0,
from
=
ira.
circle
to
(r, 0).
^Lns.
Va + b
2
2
0.
Find the length of the logarithmic spiral (Art. 138) r = e ae from
(r1} 0j) to (r2
2 ), using the formula (2), and the equation
the point
,
O^Ml.
a
Ans.
Va2
+ 1 (r - ri
2
a
).
INTEGRAL CALCULUS
332
5.
Find the
entire length of the cardioid (Art. 141)
r
=a (1 — cos 0).
Ans. 8
Also show that the arc of the upper half of the curve
bisected by
is
6.
Solve Ex. 5 by using formula (2) and the equation 6
7.
Find the arc of the reciprocal
$
=£
to 6
12
8.
9.
Find the
=
to 6
Also show that the arc
10.
= a,
from
(~ + log-V
Ans.
V15
(Art. 139) r
= |.
= vers -1 -.
— a sec
^4ns. (sec
2
- from
*Lz)a.
^ + log tan
entire length of the arc of the curve (Art. 145)
r
Hence OA,
= ?.
r6
4
Find the arc of the parabola
$
spiral (Art. 137)
a.
^.
= a sin 1
3
AB is
one third of
?£«
OABG.
AB BG are in arithmetical progression.
Find the
}
entire length of the curve r
=
a sin n -, n being a posi-
tive integer.
See for integration Exs. 24, 25, p. 314.
Ans.
2.4-6...
2
1.3.5...(?i-l)
1.3-5-w
2.4-6... (n-1)
a,
7ra,
when n
when n
is
is
even
odd.
APPLICATION OF INTEGRATION TO PLANE CURVES
251. Volumes
of Surfaces of Revolution.
by revolving about OX the Plane Area
Let OA = a, OB = b.
Let x and y be the coordinates
of any point P 2 of the given
ated
To
find the
Volume
333
gener-
APQB.
curve.
It is evident that the rectan-
gle P,A.2 A~ will generate a right
cylinder,
whose volume
is
TTlf A.i\
The sum
may
of all these cylinders
be represented by
Ai
A2
A3
A4
B
sum
of the cylinders, as
Similarly the volume generated by revolving
PGHQ about OF is
The required volume is the
Ax approaches zero. That is,
limit of the
V = ir\
y
where
OG = g,
and
OH =
x2
cbj,
It.
EXAMPLES
1.
Find the volume generated by revolving the
9
a2
about
its
major
— =ttI
axis,
2
y dx
=
7r
OX.
+
2
'
Z>
This
is
called the prolate spheroid.
— (a — a?)dx = —2
I
ellipse
9
F=%a6
2
.
crx
arY
27ra^ 2
INTEGRAL CALCULUS
334
2.
minor
Find the volume generated by revolving the
axis,
This
OY.
is
ellipse about its
called the oblate spheroid.
|_j^=^j>-^=
y=*
2
7ra?b
a 2 b.
7T
3
3.
volume from
to x = 4 a.
4.
= kax is revolved about OX, show that the
x=2a is one third the volume from x = 2 a
If the parabola y 2
a?
=
to
Find the volume generated by revolving the segment LOL' of
the parabola about the latus rectum LL'.
Here
-?= tt
C\pN)
2
C
y-
"-"(
dy
=
V
r,
ir
C\a - x)
2
Y
dy
t
P/
16
N
7rCT
2a
Ans.
32
a
irCT.
5.
about
Find the volume generated by revolving
OX
6.
about
Ans.
—
3
-n-a .
x*+y*
L'
35
Find the entire volume generated by revolving about
hypocycloid (Art. 132)
7.
v
one loop of the curve (Art. 134)
aAy 2 — a2 xA — xe
X
F
15
= a*.
Ans.
OX the
32
7r(r.
105
Find the volumes generated by revolving about OX, and
OY,
the curve (Art. 133)
Ans.
V = ^irab
x
oo
b.
V = ^a
o
2
2
.
y
APPLICATION OF INTEGRATION TO PLANE CURVES
The part
8.
dinate axes,
is
of the line
- + | = 1, intercepted between the
a
o
revolved about the line x
= 2 a.
Ans. -vefb,,
The segment
of the parabola, x2
revolved about OX.
A
10.
its
Show
circle is
is
8a '
=
+4a
2
is
(0,
2 a) to y
= a.
-.
OF the witch
Ans. 4 (log 4-1) 7ra3
[-
]
+
(
half,
-|-
3oJ
OX
Find the volume generated by revolving about
—
ABA',
^-ffW
4
included between the ellipse
.
-) =1, about the tangent at B.
Ans.
13.
is
equal to that of a
Find the volume generated by revolving the upper
of the curve (Art. 133)
81
equal to the chord.
from
„,
or
—
Ans.
Find the volume generated by revolving about
(Art. 126), y
above OX,
revolved about a diameter parallel to
that the volume generated
sphere whose diameter
12.
— 3 x + 2 y = 0,
Find the volume generated.
segment of a
chord.
11.
coor-
Find the included
volume.
9.
335
the area
^ = 1, and the parabola 2 ay = 3 b x.
2
2
Ans.
—
irab
2
.
48
X
14.
A
segment of the catenary (Art. 128), y
chord through the points x
at the vertex.
= ± a log 2,
is
tum
a
X
-\-e
a
),
by a
revolved about the tangent
Find the volume generated.
Ans.
15.
= ~{e
3nog2-^J7ra
Find the volume generated by revolving about the latus
of the ellipse
rectum.
— + ^ = 1,
the
segment cut
Ans. 2ir{ab
2
off
3
.
rec-
by the latus
-—- ab^/d 2
2
b sin"
1
-\
INTEGRAL CALCULUS
336
252. Derivative of Area of Surface
Revolution.
of
In order
to
obtain the formula for the surface generated by the revolution of a
given arc, it is necessary to find the derivative of this surface with
respect to the arc.
Let
S
denote the surface gen-
erated by revolving about
the arc
s,
Using
OX
AP.
for
abbreviation
expression "Surf
(
the
)" to denote
"the surface generated by re( ) about OX" we have
volving
S = Surf (s), AS =
This
AS
Now
may
Surf (As).
be written
Surf (As)
Surf (Chord
PQ)
PQ)
Surf (Chord
the surface generated by the chord
of the frustum of a right cone,
which
is
Surf (Chord
PQ)
=2
As,
is
the convex surface
the product of the slant
the bases.
(?E±M\ chord PQ
tt
=2
^±1^ Chord PQ
=
(2y + Ay) Chord PQ.
7r
Substituting this for the last factor in
by
PQ
midway between
height by the circumference of a section
Hence
(1)
(1),
and dividing both sides
we have
Ag =
As
Surf (A,)
Surf (Chord
PQ)
, (2
Taking the limit of each member,
that
Lim As=0
^
+A)
9>
^
9
as
Chord PQ
As
As approaches
Surf (As)
Surf (Chord
PQ)
1,
zero, noticing
APPLICATION OF INTEGRATION TO PLANE CURVES
Lim A
and
„ =>
ChOTdP(? =
ds
Y is the
if
we ha*
l,
As
f = Lim
Similarly
337
AS
irf
Try.
As
axis of revolution,
dS
2ttx.
ds
of Surfaces of Revolution.
To find the Area
by revolving about OX the Arc PQ.
the preceding article we have
253.
Areas
of the Sur-
face generated
By
dS =
27ry
ds
S=
hence
y,
I 2-rry ds.
To express this in terms of x and
we have from (1), Art. 155,
ds
MSI
dx,
which gives
*—£{HWJ
>n*
If
Y is
B
dx.
X
(1)
.
the axis of revolution,
Sy = 2 7T Cxds = 2 7T
Or we may use
and instead of
(1)
and instead of
(2)
rls
=
\
1
C x\l +
+
(
-^
)
dyYf dx.
dx)
(2)
dy,
we have
Sv = 2ir Cx[l + (—Y~]
dy.
(20
INTEGRAL CALCULUS
338
EXAMPLES
1.
Find the area of the surface generated by revolving about
the hypocycloid (Art. 132)
Here
(y
OX
+ y* = a*.
o$
^ = - (a* -x^x~\
= J - a,i)4,
ctx
2.
Using
±Sx = 2ir£(a}
(1)
£C
3
1
:
)
a3
+
—x
2._,1
~\~
s
J
x*
2
•.{
/Y3
x*y—dx = 2 7ra
/^ a
j
2
—#
(a 3
3
|
2
3
3
) * a;
-1
cto
i«3
6
Or we may use
Show
2
2
1
+
a3
from
3.
3
2 2/"
1
3
f*
I
2
F
W
a
1
= 6-n-a
y^dy
2
J
that the area of the surface generated by revolving the
from
a?
=
to x
= 3 a,
is
one eighth of
Find the area of the surface generated by revolving about
the loop of the curve 9 ay2
4.
— (a — 2/*)
— y:2_1 dy = 2 ?ra
= 4 ax, about OX,
x — 3atox = 15a.
parabola y 2
that
12
#* =
-^ =
(1')
& = 2 -X'
2.
7ra
= x(3a— xf.
Ans. 3
OX
tto
2
.
Find the surface generated by revolving about OX, the arc
a2 xy = $4 + 3 a4 from x = a to x = 2 a.
*„
of the curve 6
,
Ans.
-—tto
2
.
16
5.
The
about
6.
4y
Y.
arc of the preceding curve from x
What
is
the surface generated
?
=a
to
x
= 3 a, revolves
+ log 3)?ra
Find the surface generated by revolving about
from x = 1 to x = 4.
= x — 2 log x,
2
2
Ans. (20
.
OF the curve
Ans. 24
tt,
APPLICATION OF INTEGRATION TO PLANE CURVES
7.
ellipse
8.
Find the entire surface generated by revolving about OX the
3.r + 4?/- = 3cr.
Ans (8 + _*\rCp
VsJ
\2
OY the
Find the entire surface generated by revolving about
preceding
9.
339
Am
ellipse.
(4
Find the surface generated by revolving about
of the curve S
3 iog 3)i£
+
OX
— x*.
aV
J = a-x
the loop
2
ira?
A
Ans.
4
10.
An
arc.
revolves about
11.
to
The
x=2a,
subtending an angle 2
its
chord.
a,
Find the surface generated.
Ans. 4 7ra 2 (sina
arc of the catenary (Art. 121) y
revolves about
whose radius
of a circle
OT.
f = -±le«
+e
ci
— -\
«
is a,
— a cos a).
=a
from x
),
Find the surface generated.
Ans. (e> + 2e-l -3e- z)ira2
.
12.
The parametric equations
x
=e
e
sin
6,
of a curve are
y=
ee
cos
0.
.
Find the surface generated by revolving the arc from
0=-,
about
2
13.
OX
6
=
to
Ans. i*(e»-2).
5
Find the surface generated by revolving about
the preceding example.
AnSm
OT the
^ 7r
(2c
arc of
77
1
1).
o
14.
The parametric equations
fixed circle being a,
are
x
=—
of the epicycloid, the radius of the
and that of the
cos
<f>
cos
3$,
rolling circle
y=
—
sin
<£
- (Ex.
12, p. 330)
— - sm 3
<£.
Find the entire surface generated by revolving the curve about OX.
Ans.
15.
- 7r 2 a 2
.
Find the surface generated by revolving one arch of the pre-
ceding curve about OY.
Ans.
drra 2
.
INTEGRAL CALCULUS
340
254. Volume by Area of Section. The volume of a solid may be
found by a single integration, when the area of a section can be expressed in terms of its perpendicular distance from
a fixed point.
Let us denote this
tance by
x,
dis-
and the area
of the section, supposed to
be a function of
x,
by X.
The volume included
between two sections separated by the distance dx
Xdx,
will ultimately be
and we have for the volume
of the solid
V=
the limits being the
initial,
Cxdx,
and
final,
values of
x.
EXAM PLES
1. Find the volume of a pyramid or cone having any base.
Let A be the area of the base, and h the altitude.
Let x denote the perpendicular distance from the vertex of a section parallel to the base
Calling the area of this section X, we
have, by solid geometry,
X = tf
A
h2
X = Ax
'
2
2
'
Ji
Hence,
Xdx = -
I
)
/iVO
x 2 dx
Ah
Ah
s
'
2
3
h 3
2. Find the volume of a right conoid
with circular base, the radius of base being
a, and altitude h.
OA = BO=2 a,
The
section
BTQ,
BO=CA =
h.
perpendicular to OA,
is
an
isosceles triangle.
APPLICATION OF INTEGRATION TO PLANE CURVES
=
Let x
341
OP; then
X = area PTQ = PTxPQ = h V2 ax - x\
f* Xdx = h «/oC**^/2ax- x
V=
Hence,
1
dx
This
3.
Find the volume of the ellipsoid
0~
(V
Let
us
area
the
find
a
of
....
C"
7
section
c
C'B'D' perpendicular
OX,
the
at
^^^fT
/
dis-
tance from the origin
031= x.
This section
is
an
^y
whose semiare MB' and
ellipse
MC
To
z
=
find
—
C
>£i
\
\
*~7
a
x
axes
—
one half the cylinder of the same base and altitude.
is
* + £ + $=!•
to
=
2
«/o
i
\
"1sk\
yn
—*
/
y/
MB', let
aud we
in (1),
7i
have
y
= MB' = -^d - XT.
2
a
To
find J/C", let y
=
in (1),
z
and we have
= MC = -->/a
2
-x*.
a
The area
of the ellipse (Ex.
—
Hence,
and
2, p.
V= 2
PX
«A>
da;
321)
is
(MB ) (MC).
1
it
(«--.i~),
= " ^ P (a - s ) da; = Kabc.
a- J
3
"
2
2
a
INTEGRAL CALCULUS
342
A
4.
rectangle moves from a fixed point, one side varying as the
distance from the point, and the other as the square of this distance.
At the distance
What
5.
The axes
radius
of 2 feet the rectangle
the vohime then generated
is
of
?
becomes a square of 3
Ans. 41 cubic
feet.
feet.
two right circular cylinders having equal bases,
Find the volume common to
16 a3
intersect at right angles.
a,
the two.
,
Ans.
3
6.
A
torus
is
generated by a
circle,
radius
b,
revolving about an
axis in its plane, a being the distance of the centre of the circle
Find the volume by means
the axis.
Ans.
the axis.
A
from
of sections perpendicular to
2
2
7r
a2 b.
16 inches long, and a plane section containing a
an ellipse 8 inches broad. Find the volume of
the ball, assuming that the leather is so stiff that every plane crosssection is a square.
Ans. 341^ cu. in.
7.
seam
football
is
of the cover is
8. Given a right cylinder, altitude h, and radius of base a.
Through a diameter of the upper base two planes are passed, touching the lower base on opposite sides. Find the volume included
^
between the planes.
9.
Two
for their
/
_ |\ %
_
cylinders of equal altitude h have a circle of radius a,
common upper
each other.
base.
Their lower bases are tangent to
Find the volume common
to the
two
cylinders.
Ans.
i^.
CHAPTER XXIX
SUCCESSIVE INTEGRATION
255. Definite Double Integral.
—A
double integral
is
the integral
of an integral.
Thus, x and y being independent variables, the definite double
„a „c
integral,
f(x,y)dxdy
I
}
indicates the following operations:
Treating x as a constant, integrate f(x. y) with respect to y
between the limits d and c then integrate the result with respect
to sb between the limits b and a.*
For example,
;
J
&(P
- y)dx dy =
J
x-l
by
1j
dx =
Xotice that the order of the two integrations
is
J
x~~dx
indicated in the
given definite integral by the order of the differentials dxdy, taken
being nsed in the same
|
Jo
order.
It should be said, however, that the order of the integrations
denoted differently by different writers.
256. Variable Limits.
— The limits of the
first
is
integration, instead
of being constants, are often functions of the variable of the second
integration.
*
Using parentheses, this might be repmsfntert by
343
I
[( f(z,y)dy\<l.r.
INTEGRAL CALCULUS
344
For example,
p
JP
2
xy dy dx = jT°g
j dy = | jf
"(3 jf
+2
J"
V-
a*y)dy
=
^
As another example,
J J^^(x-\-y)dxdy=J f^ + |-)
When
the limits are
may
integrations
Where
x2 (b
— y)dxdy —
*S0
I
x 2 (b
I
*^
is,
— y)dy dx.
*/a
changed only by new limits adapted to the new order.
Triple Integrals.
cessive integrations.
III
— A similar notation
is
x 2y 2zdxdydz
=
——x y dxdy
2
I
I
Evaluate the following definite integrals
n
—
2
6
®V 0»
- V) dx dy = a b
C Cr sin 6 dr dB =
2
used for three suc-
Thus
EXAMPLES
2.
That
the definite integral has variable limits, the order of integra-
tions can be
257.
Art. 248, the order of the
all constants, as in
be reversed without affecting the result.
I
i
a
^"^dx
o
a3
2
(a
- b).
~ 63
(cos
/3
- cos a).
2
SUCCESSIVE INTEGRATION
Jb Jo
r
=
dr d0
—
5.
r'sin^?^)-^.
I
Jo Jo
3
nlO
y
Vxy — y
2
r^ear =
fT'
c/o c/acose
10.
dy dx
=6a
3
.
2
«/0 «A)
9-
•
--±
2
8.
345
1
L- ±\^.
I5y 10
^
J^jTsin (2 +-«)<»** = 1.
fTfsirf **«** = £+5.
»/o
»/o
Jo
J^
2
6
2?
+^
2
8
&
a
B
12
-
13.
14.
X XX ^ +
(
j
'
C
J" f
J
+*
sin (xyz)
j
%/o
15
y2
2)
dx dy dz
dx dydz
v
f uvw du dvdw=
Ju-%
S*f*
VeX+y+zdxd
y dz
=
=f
—
18
=-
t
(a2
+ 52 + c2)
-
CHAPTER XXX
APPLICATIONS OF DOUBLE INTEGRATION
258.
Moment
of Inertia.
If r1} r 2
?*
,
a given line of n particles of masses
m^ + m r + ?%r +
2
2
3
••,
rn are the distances
• • •
2)
1}
2
2
•
3,
m m m
-f
m
n rn
2
3,
=
•
•,
mn
the
,
X (w?*
from
sum
2
defined in treatises on mechanics as the moment of
system about the given line.
is
)
inertia of the
The moment of inertia of a continuous solid about a given line is
sum of the products obtained by multiplying the mass of each
infinitesimal portion of the solid by the square of its distance from
the
the given line.
The summation is then effected by integration, and we have
moment of inertia of a body of mass M,
for
the
-/- dM.
Moment
259.
of Inertia of a
The moment
Plane Area.
of a given plane area about a given point
sum
of the products obtained,
may
of
inertia
be defined as the
by multiplying the area of each
infini-
tesimal portion by the square of its distance from 0.
This
may
be regarded as the
moment
of inertia of a thin plane
sheet of uniform thickness and density, about a line through
per-
pendicular to the plane, the mass of a square unit of the sheet being
taken as unity.
We
shall illustrate double integration
inertia of certain areas.
346
by finding the moment of
APPLICATIONS OF DOUBLE INTEGRATION
OACB
of inertia of the rectangle
Let OA
Suppose
the
into
elements
by
rectangle
M'
rectangular
parallel
lines
which are
to
be
N'
t
i
Let
to the coordinate axes.
y,
find the
about 0.
= a. OB = b.
divided
sb,
To
Rectangular Coordinates.
Double Integration.
260.
moment
847
[Q
——
re-
i.
T
garded as independent vari-
i
ables, be the coordinates of
M
any point of intersection as
P, and x + dx, y + dy the
coordinates of Q.
Then the area
N
of the element
PQ
dxdy.
is
Moment of inertia of PQ = OP' dx dy = (x2 + y ) dx dy.
The moment of inertia of the entire rectangle OACB is the sum
2
•
all
the terms obtained from (x 2
and y from
to
+ y )dx dy,
2
by varying x from
b.
we suppose x
to be constant, while y varies from
to
have the terms that constitute a vertical strip MNN'M'.
If
of
to a,
b,
we
shall
Hence
Moment
of inertia of
MNN'M' = dx
I
(x2
dx[(.x2 ii
V
Having thus found the moment
+y
2
)
dy
+ V3yo
of a vertical strip,
M+
we may sum
these strips by supposing x in this result to vary from
is,
the
moment
of inertia of
1=
re
,2
_ a?b + ab
+ 'A dx —
,
The preceding operations
C
all
That
7
(pf
3
,
?)
are those represented
b
C
to a.
OACB,
integral,
T=
^yiae.
+ f)dxdy.
by the double
INTEGRAL CALCULUS
348
If
sum
we
first collect all
the elements in a horizontal strip, and then
these horizontal strips,
1=
I
(x 2
f
261. Variable Limits.
OAO about 0.
Let OA = a, AC=b.
equation of 00 is
we have
To
+ ab
+ y ) dy dx =
2
find the
moment
3
of inertia of the right
triangle
V
3
This
The
Y
= —b x.
a
differs
from the pre-
ceding problem
only in the
limits of the first integration.
to
In collecting the elements in a vertical strip MN, y varies from
MN. But MNis no longer a constant as in Art. 260, but varies with
OM.
according to the equation of OC, y
= -x.
a
Hence the
limits of y are
and
b
-x.
a
In collecting all the vertical strips by the second integration, x
to a, as in Art. 260.
varies from
Thus we have for the moment of inertia of OAO,
*/0
By
we
+ y*)dxdy = ab(?- +
supposing the triangle composed of horizontal strips as
shall find
Y
'=
I-
12
I
I
(xP
+ y^dydx
T
G + 5>
HK,
APPLICATIONS OF DOUBLE INTEGRATION
Plane Area as a Double Integral.
262.
factor
(ar^
+y
2
),
we
260 we omit the
If in Art.
shall have, instead of the
349
moment
of inertia, the
area of the given surface.
That
Area =
is,
)
}
=
dx dy
dy dx,
I
J
the limits being determined as before.
EXAMPLES
1.
Find the moment of inertia about the origin of the right
tri-
angle formed by the coordinate axes and the line joining the points
b(a-*)
»),(0,6).
2.
Find the moment of inertia about the origin of the
r +>/- = «-.
Ans.
i
£ j^-
x
\x*
circle
+ y^dxdy = ^.
3. Find by a double integration the area between a straight line
and a parabola, each of which joins the origin and the point (a, 6),
the axis of
X being the axis of the parabola.
^
h\l x
Ans.
I
I
dx dy
=
I
I
dy dx
=
—
4.
Find the moment of
5.
Find by a double integration the area included between the
+ y- = 10 ay, the line 3 x -j- y = 10 a, and the axis of T.
inertia about the origin of the preceding
circle xr
10
>~a
/M0a
/»Vl0ay-y2
dy dx
Ans.
=
/-Sa
/*10a-3z
I
I
,
+
I
dx dy
•»
q-y
3
dy dx
= g—-y3 +5
.
sin
x
gy
-
INTEGRAL CALCULUS
350
6.
Find the moment of
tween the
—
f-*_
ellipse
ab-
inertia about the origin of the area be-
=l
and the
+ " = 1.
line -
a
b
(jL-A-\ asb + ab 3J
Arts. .[-
\16
).
12/
7. Find the moment of inertia about the origin of the area between the parabola ay = 2 (x 2 — a2), the circle x2 + y 2 = a 2 and the
,
axis of
Y
.35
8.
Find by a double integration the area included between the
= 3x, and y 2 = 12 (60 — x).
Ans. 960.
parabolas y 2
9.
Find the moment of
parabola y2
parabola.
= 4 ax,
x
= 4=a
inertia of the area included between the
and the axis of X, about the focus of the
2336
A
}
.
Ans.
.
a\
35
10.
lines y
Find the moment
= 2x,
of the first
x
two
f
lines.
Ans.
263. Double Integration.
the quadrant of a circle
between the
and the axis of X, about the intersection
125 a4
of inertia of the area included
+ 2y = 5a
To
Polar Coordinates.
AOB, whose
radius
is
find the area of
a
In rectangular coordinates, Art.
260, the lines of division consist of
two systems, for one of which x
constant, and for the other, 2/
is
b
is
l
constant.
So in polar coordinates, we have
one system of straight lines through
is
the origin, for each of which
constant, and another system of
circles
about the origin as centre, for
each of which r is constant.
Let r, 0, which are to be regarded
\M'
M
L
/
;
"'
l/0c^lxA"\'"\
N
N' A
as independent variables, be the coordinates of any point of intersec-
APPLICATIONS OF DOUBLE INTEGRATION
tion as P,
of
PQ
is
and
r
+ d)\
coordinates of Q.
4- dO, the
ultimately
Then the
351
area
PBxPQ=rdO'dr.
we first integrate, regarding constant while r varies from
we collect all the elements in any sector MOM'.
The second integration sums all the sectors, by varying 6 from
If
to a.
*>l
Hence
Area
BOA = f
C
~
r
dO dr
=
—
we
reverse the order of integration, integrating first with respect
and afterwards with respect to r, we collect all the elements in
a circular strip XLL'X', and sum all these strips. This is written
If
to
6,
BOA =
Area
264. If the
the
moment
BOA
moment
XT'
of inertia about
of inertia of
dr dO.
required, we have for
Hence, the moment of
is
PQ, r-rdOdr.
is
1=
C C r\Wdr=
265. Variable Limits.
of the semicircle
OBA
To
2
f f
find
r*drdO
=
by a double integration the area
with radius
OC=a,
the pole being on the
circumference.
/*
The polar equation
If we
= 2 a cos 0.
of the circle
integrate
is
first
with respect to r, then with respect
6, we shall have
to
Area
OB
-IT
dOdr =
-<i-
Here, in collecting the elements in a radial strip OM, r varies
to OM.
But 03/ varies with 0, according to the equation
of the circle r = 2 a cos 6.
Hence the limits are and 2 a cos 6.
from
INTEGRAL CALCULUS
352
In collecting
varies
By
about
from
these radial strips for the second integration, $
all
to -
•
supposing the area composed of concentric circular strips
as
LK, we
find
X
A
Area
'(£)
/»c
OBA X2a
r dr d6
= Tror
EXAMPLES
1.
Find the moment of
cluded between the two
inertia about the origin of the area in-
= a sin
no
circles, r
and
sing
Find the moment of
2.
inertia
between the parabola (Art. 139), r
= b sin 0, where a >
b.
O
32
about the origin of the area
= asec
0X
r
sin G
2
-, its latus rectum,
and
4
'
i^-
Ans.
.
35
3.
Find the moment
lemniscate (Art. 143)
4.
of inertia about its centre of the area of the
r2
=a
2
cos 2
0.
Ans.
irCC
Find by double integration the entire area of the cardioid
Sva'
r = a(l-cos0).
(Art. 141)
Ans.
5. Find the moment of inertia about the origin of the area of the
preceding cardioid.
35 wa4
A
Ans.
-_-
APPLICATIONS OF DOUBLE INTEGRATION
6.
Find the moment of inertia about
Df the four-leaved rose (Art.
K
144)J r
its
353
centre of the entire arc
= a sin 2 0.
A
Ans.
3
wci*
16
7.
Find by
a double
area of one loop of the
integration the
lemniscate (Art. 143) outside the circle
2r = a
2
2
.
tt\ a
/ /«
(V3T
i
Ans.
5)
preceding
8. Find the moment of inertia of the area of the
example about the centre of the lemniscate.
/V3 ir\ a 4
*
RS
+ 3)l6
'
V~2~
266. Volumes and Surfaces
of Revolution.
Polar Coordinates.
If
263 we suppose a revolution about OX, the
volume generated by the infinitesimal area PQ is the product of
this area by the circumference through which it revolves, that is,
in the figure of Art.
r
r sin
•
Hence
r
dO
dr.
for the entire
volume
V=2tt f frsmedBdr,
the limits being determined as in Art. 263.
If the revolution
is
about OY,
V= 2 f Cr cos OdO dr.
7T
The area
of the surface generated about
OX is
(Art. 253)
U
EXAMPLES
1.
=a
2.
Find the volume generated by revolving the cardioid (Art. 141)
(1
— cos 6)
Show
about OX.
that the entire
leaved rose (Art 144) r
circumscribed sphere.
.
Ans.
8
-
,
irflr,
L
.
^
.,
,
,
twice the inscribed sphere.
volume generated by revolving the
= a sin 2 6 about OX is —
of the
four-
volume of the
INTEGRAL CALCULUS
354
3.
Find the volume generated by revolving one loop of the
= a sin 2 about the axis of the loop.
leaved rose r
Ans. fs
4.
V2 - 9
=a
2
cos 2
about
(Art,
vk
s
-^—
Find the volume generated by revolving the lemniscate about
OX
^4ns.
L
L log(V2 +1 )-|;
Find area of surface generated, by revolving the cardioid
r=a(l
7.
105
^a
Y.
-4ns.
6.
V
Find the volume generated by revolving the lemniscate
2
143) r
5.
four-
-- cos 0)
about OX.
,
32 -n-a 2
Find the moment of inertia of a sphere (radius a) about a
8 7ra5 m
m being the mass of a unit of volume.
Ans.
diameter,
15
CHAPTER XXXI
SURFACE, VOLUME. AND
267.
To
find
MOMENT OF INERTIA OF ANY
the Area of
Any
Surface,
between Three Rectangular Coordinates, x3
Let this equation be
whose Equation
is
SOLID
given
y, z.
„,
z=f(x,y).
Suppose the given surface to be divided into elements by two
XZ and YZ. These planes
will also divide the plane
into elementary rectangles, one of
which is PQ. the projection upon the plane
of the corresponding
element of the surface P'Q'.
series of planes, parallel respectively to
XY
XY
Let
of
(>'.
x,
y,
z
be the coordinates of P' and x
+ dx,
y
+ dy, z + dz,
INTEGRAL CALCULUS
356
Since
PQ
is
by the cosine
AT. This angle
with the plane
plane
is
Area
see
from the
PQ = Area P'Q'
(8),
dy
dx
the given surface
If
S
z
sec y
y,
•
sec
y.
dz
fdz
dx J
\dy
partial derivatives, taken
=f(x,
from the equation
of
y).
Area P'Q' = [l
Hence
cos
PQ = dx dy.
Art. 110,
— and — are
where
= Area PQ
•
figure that
Area
Also from
equal to that
made by the tangent plane
Denoting this angle by y,
Area P'Q'
We
is
of the inclination of P'Q' to the
evidently that
XY
at P'
PQ
the projection of P'Q', the area of
of P'Q' multiplied
+ (^Y + (^
dx dy.
denote the required surface,
S
©V:
dz
-ff[ i+
dxdy,
(i)
\dy
the limits of the integration depending upon the projection, on the
plane
XY,
of the surface required.
For example, suppose the surface
surface of a sphere whose equation is
X2
dz
Here
+
—
dx
A2
1
+
W
+z =a
2
2
2/
X
—
dz
>
x
,
l
to be one eighth of the
2
.
__
dy
z
-1
ABC
+f
z
a2
a2
—x —y
2
2
SURFACE. VOLUME, AXD
Substituting in
(1),
MOMENT OF INERTIA
357
we have
dx dy
J J Va —
2
This
to be integrated over the region
is
the required surface on the plane
The equation
boundary
of the
s»
Integrating
a strip
limits
+
we
y,
collect all the elements in
zero to
ABC,
Another example
centre
x,
that
we sum
2
between the
all
the strips, to
to a.
dx dy
Va — x* — y
is,
7r(r
l
the following
is
a
of
sphere, whose radius
a, is
Jo Jo
ML,
x varying from
\/a'2—x2
5
Hence
the projection of
a".
Integrating afterwards with respect to
obtain the required surface
OB A,
XY.
with respect to
first
ir
AB is
M'X'KL. y varying from
and v<r — .r\
The
s
is
on the surface of a
right circular cylinder,
the
base
radius
a
is
surface
of
of
whose
Find the
the
sphere
by
the
the
equa-
intercepted
cylinder.
Take
for
tions of the sphere
and
cylinder,
and
xr
CPAQ
+ y = ax.
2
is
one fourth
the required surface.
Since this surface
is
a part of the sphere, the
INTEGRAL CALCULUS
358
—
—
ox
dy
partial derivatives
must be taken from x2
,
-f y
2
-f z
2
giving, as in the preceding example,
J
2
n^2 — x
nt&
— y2
-\a
*S
-\i
to be integrated over the region
the plane
ORA,
the projection of
CPAQ
on
17.
The equation
of the curve
±S =
Hence
ORA is
x2 + y2
f f^
#=(2ir-4)a2
= ax.
<*****
_
frA
.
Let us now find the surface of the cylinder intercepted by the
sphere, one fourth of
Since this
is
x
+y
2
=ax, we
The formula
CPARO.
is
must be taken from
derivatives in (1)
2
which
a part of the cylinder x2
—
=
ox
find
is,
(1)
—
=
oy
oo,
+y =
2
ax',
this equation.
the partial
But from
co-
then, inapplicable in this case.
from the figure that the surface CPARO cannot
projection on the plane XY, since this projection
It is also evident
be found from
is
the curve
The
its
ORA.
difficulty is
removed by projecting on the plane XZ, and
using, instead of (1),
x2 + y 2 =
We now find from
a—
dy
dx~
Substituting in
(2),
ax,
2x
y
dy
dz
and simplifying,
1 o
C C adx dz
2
^Jax—x2
= 0.
SURFACE, VOLUME, AND
MOMENT OF INERTIA
This must be integrated over the region CP'AO,
projection on
To
XZ of
CP'A
359
being the
CPA.
find the equation of
x2 + y 2
CP'A, we eliminate y from
+z =a
2
2
2
=a
~li
i
giving,
z
2
and x 2
\
2
—ax.
v/ qt:
S
Hence
+ y = ax,
dx dz
v ax — ar
EXAMPLES
The axes of two equal right circular
of base, intersect at right
1.
cylinders, a being the
radius
angles
:
find the surface of one inter-
cepted by the other.
Take
the
for
equations
the
of
cylinders,
x2
-\-z
2
=a
2
,
and x2
-{-7j
2
=a
Ans.
2
.
8 a2
.
Find the area of the part of the
2.
plane
a
in
the
cepted
first
o
c
octant,
inter-
by the
coordinate
^V^c-' +
c^ + a^.
planes.
An*-
3.
Find the area of the
surface
x2
2
-f-
y
the
of
=
a2
,
be-
z=mx
and
tween the plane
the plane
IT.
cylinder
included
Ans ± ma?
.
.
ccvJ
INTEGRAL CALCULUS
360
Find the area
4.
2
y
-\-z
= 4 ax,
2
parabolic cylinder
x=3
the plane
of the surface of the paraboloid of revolution
intercepted by the
y
2
=
a.
Ans.
and
ax,
—^—
^q
*
In the preceding example,
5.
area of
find the
the surface of
the cylinder intercepted by the
paraboloid of revolution and the
given plane.
(13VT3-1)
Ans.
V3
Find the area of that part of the surface
6.
2
z -f (x cos
which
is
The
situated in the first octant.
surface
right
circular
inder,
whose
is
the
x cos
= a?
a + y sin a) 2
line
a+y
is
a
cyl-
axis
= 0,
z
sin
a=0,
and radius of base a.
a2
—
a
Ans.
.
sin a cos a
7.
a
A
diameter of
whose
sphere
radius is a
axis
of
a
is
the
right
prism with a square
base, 2 b being the
side of the square.
Find the surface of the sphere intercepted by
the prism.
Ans.
8a[2bsmV
1
—
—
Va -6
2
— a sin -1 —
2
-J.
a2 -b 2 J
MOMENT OF INERTIA
SURFACE, VOLUME, AND
To
268.
Volume
of
Any
Solid bounded
by a Surface, whose
given between Three Rectangular Coordinates, x, y, z.
solid may be supposed to be divided, by planes parallel to the
Equation
The
find the
361
is
coordinate planes, into elementary rectangular parallelopipeds. The
volume of one of these parallelopipeds is dx dy dz, and the volume of
the entire solid
is
rr r
v= jjj
dxdy^
the limits of the integration depending upon the equation of the
bounding surface.
For example, let us find the volume of one
whose equation is
,
2
2
a-
b
2
eighth, of the ellipsoid
<?
c
^H
vif
m'
4
/ 4.
<
P
y
/'k
K
'X
^
PQ
is
represents one of the elementary parallelopipeds whose volume
dx dy
If
we
column
dz.
integrate with respect to
MN\
z
z,
we
varying from zero to
from
to
z
collect all the elements in the
MM*\
that
= c x h_''"_ 'L
V
a1
lr
is,
INTEGRAL CALCULUS
362
Integrating next with respect to
the slice
KLN'H,
y,
we
collect all the
KL
y varying from zero to
Vx
This value of y
which
is
that
columns in
is,
2
1
a
;
.
2
taken from the equation of the curve
ALB,
is
^4-^ = 1
a ^b
2
Finally,
we
2
'
integrate with respect to
the entire solid
ABC.
Here x
to collect all the slices in
x,
varies from zero to
OA
that
;
is,
from
to a.
The y and x
integrations are said to be over the region
at
a2
I
For the entire ellipsoid
b2
dxdydz =
AOB.
abc
V= 4^?.
EXAMPLES
Find the volume of one of the wedges cut from the cylinder
+ y2 = a2 by the plane z = x tan a and the plane XY. (See Figure,
1.
x2
Ex.
3,
Art. 267.)
Ans.
2.
2)1
|
dxdydz =
Find the volume of the tetrahedron bounded by the coordinate
planes and by the plane
x
'
y
z
abc
3.
,
u
A
abc
o
Find the volume included between the paraboloid of revolution
=
— ax, and the plane x=3a.
2
2
4 ax, the parabolic cylinder y 2
y -f z
(See Figure, Ex. 4, Art. 267.)
Ans. (6
tt
+ 9 V3)a
3
.
AND MOMENT OF INERTIA
SURFACE, VOLUME,
4.
ar
-j-
363
Find the volume contained between the paraboloid of revolution
2
= «.:. the cylinder x2 -f y- = 2 ax, and the plane XT.
1/
2
5.
Find the volume of the cylinder
paraboloid of revolution y 2
6.
2
The
x'
+y
2
= ax, intercepted
+ z = 2 ax.
2
centre of a sphere (radius a)
circular cylinder, the radius of
by the
2\
fir
is
whose base
on the surface of a right
is
-.
the part of the cylinder intercepted by the sphere.
Find the volume of
(See second Figure,
^•267.)
.
2/ _4\.,
3
3V
7.
Find the volume in the
&
8.
Find the
entire
bounded by the surface
first octant,
tt+Pt-l.
Am-
*
volume within the surface
aj*
269. Moment
3
+ y* + «*»<!#.
of Inertia of
Any
Solid.
Ans.
This
may
^f.
be expressed by a
triple integral.
Thus, the
of volume,
moment
of inertia about
OX,
m
being the mass of a unit
is
I=mJJf(y + z )dxdydz,
2
2
with similar formulae for the moments of inertia about the axes
OY, OZ.
EXAMPLES
1.
Find the moment of inertia about OX of the rectangular paralbounded by the planes x = a, y = b, z = c, and the co-
lelepiped
ordinate planes.
.
I
,, 2
.
»<
mabc
INTEGRAL CALCULUS
364
2. Find the moment of
bounded by the plane
OZ
inertia about
a
b
c
Ans. (a 2
and by the coordinate planes.
3.
Find the moment
cylinder x2 -f y 2
=a
2
of the tetrahedron
of inertia about
OX
z
=h
and
oj
(
= — h.
z
fa
A
Ans.
irmarh —
Find the moment of
.
of the portion of the
included between the planes
2
4.
——
2
-f b )
2
h
+ 2-=.
inertia of the preceding cylinder about OZ.
Ans. TrmaVi.
5.
Find the moment
of inertia of a sphere (radius a) about a
diameter.
6.
Find the moment of
A
inertia about
OZ of
- + - +- - 1
2
6
2
2
'
the ellipsoid
^
s
-
—15—
8 -n-ma5
CHAPTER XXXII
PRESSURE OF FLUIDS.
CENTRE OF GRAVITY.
FORCE OF ATTRACTION
CENTRE OF GRAVITY
The
270. Definition.
body
centre of gravity of a
is
a point so
situated that the force of gravity acting on the body produces no
tendency to rotate about an axis passing through the point.
271.
Coordinates
gravity. C, of
given body,
PQ' the
direction
of
and
MN
of
Centre
any body, take
Gravity.
of
P
as
To
find
the centre of
any infinitesimal part of the
line
gravity,
of
any horizontal
through C.
axis passing
Let
BD
be the
perpendicular
common
between
MN and PQ. Take the
axis of X parallel to BD
and represent by x and
OL and OL', the x
coordinates of P and C re-
x,
spectively.
tance
The
Then
the dis-
BD = L'L = x — x.
force exerted
by gravity on
P
is
proportional to and there-
may be measured by its mass. Denoting its mass by dm, the
moment of this force about MN would be (x — x)dm; and if dm is
fore
an infinitesimal in one, two, or three dimensions, the tendency of
the whole body to rotate about
MN
365
is
equal to
(as
J
x)dm.
INTEGRAL CALCULUS
366
Since this must equal zero,
(x— x)dm
I
= 0,
xdm
J
and
a)
/
dm.
Similar formulae
—-The
Note.
may
be derived for y and
mass of a
represent the density by
p,
unit's
volume
is
z.
called density.
the differential mass or
dm
is
If
we
equal to p
multiplied by the differential of the arc, area, or volume.
Find the centre of gravity of a quarter of the arc of a
1.
Let the equation of the circle be x2 + y 2 = a 2
Ex.
circle.
.
Here dm
= p ds.
Substituting in
Pxds
x
Art. 271,
(1),
we have
aC^xia'-x )
2
^dx
=
i
9 a
wa
2
From
the symmetry of the figure y
= 2a
Find the centre of gravity of the surface bounded by a
parabola, its axis, and one of its ordinates.
Ex.
2.
Let the equation of the parabola be y 2 = ±px, B being (9p, 6p).
Here dm = p dx dy, and substituting in formula (1), Art. 271.
x
x=J
o
Jo
x dx dy
i
Jo
Jo
VI p
I
x 9-dx
Jo
V4 pJo
a
x 2 dx
_ 27
CENTRE OF GRAVITY.
PRESSURE OF FLUIDS
307
Similarly,
*/
=
rx**
r(g *-£
%
9p
4
3.
Find the centre of gravity of a circular disk of radius
whose density varies directly as the distance from the centre, and\
from which a circle described
upon a radius as a diameter
has been cut.
Let the equation of the
large circle be r = a
and
Ex.
a,
;
the equation
circle be r
The
of
= — a cos
disk
is
small
the
0.
s}*mmetrical
with respect to OX, hence
»=o.
Here
dm
= pr dd dr = kt
2
dd
dr,
(if p = Kr).
Also x =
OM= r cos
0.
r
Jo
3
cos
?*3
dd dr
«yo
%Jtt_
%J —
I
cos
dO dr
r2 cos
dd dr
cos 9
Therefore x
2
|
T
x
r2 cos d dd dr
%/0
c/o
+11J
«/7r
f f cos
dd
4-
f * (cos
$
—acos 9
-
cos 5 0) dB 1
=
[P*+£
5(3tt-2)
= 0.1016
(1
+ cos 0)
3
c76>
]
a.
INTEGRAL CALCULUS
368
Ex.
4.
Find the centre of gravity of a cone of revolution, the
radius of the base being 2 and the altitude
The equation of OB is y = ±x.
Here dm = piry 2 dx, and substituting
|
xy-
6.
in (1), Art. 271,
Y
dx
3
.Jo
X
2
y dx
^^
The cone is symmetrical with
respect to OX, hence y = 0.
Note.
— On
i
\
*
.1
o^-
X
cl
comparing the
1
formulae for the centre of gravity of arc, area,
and volume,
/xds
4V
—
j
f
we
is
*Aj
—
xdA
I
J
*As
y
;
—
far
f dA
ds
»
notice that, in each case, the element of the numerator integral
x times the element of the denominator integral.
5.
Find the centre of gravity of the arc of the hypocycloid
2
2
_
_ 2
Ans. x = y — -a.
125) x* + y* =a f in the first quadrant.
2.
(Art.
o
6.
Find the centre of gravity of the arc of the cycloid
x
7.
= a vers"-iV± — V2 ay — y
1
2
.
Ans.
x
=
-n-a,
y
Find the centre of gravity of a straight rod of length
= -a.
a,
the
density of which varies as the third power of the distance of each
point from the end.
Ans.
x
= -a.
5
8.
Find the centre of gravity
when the
of the surface
of a hemisphere
density at each point of the surface varies as
dicular distance from the base of the hemisphere.
*
its
perpen-
-
_ 2_a
,
CEXTRE OF GRAVITY.
Find the centre of gravity
9.
PRESSURE OF FLUIDS
of a semiellipse.
Ans.
369
= —-.
z
3
Find the centre of gravity of the area between the
10.
y-
=—
2 a
its
asvinptote.
= 8 x,
bounded by the paraband the axis of X.
of gravity of the area
the line y -f- x
— 6 = 0,
= 2.48; y=lA.
Find the centre of gravity of one loop of the curve
Ans. x
A
13.
—
Ans
,
=
a sin 2
_
;
Find the centre of gravity of the upper half
= «(1-«*<D-
r
128 a
105
'
=
y
J
lOOTT
7r'
of the cardioid
aj
_
y
14.
Find the centre of gravity of one loop
° 0s2(''
of
$.
128 a
= _5 a
.
6
rW
= - a.
Ans. x
Ans. a
12.
cissoid
3
Find the centre
11.
ola y-
—x
and
7T
16 a
'
= ——=.57a.
rrr
the lemniscate
W *a-55a
—
— ——
xX
^cxllb.
Ans
LI
.OO
LI.
8
15.
Find the centre
of gravity of a hemisphere.
3
Ans. x = -a.
8
16.
Find the centre of gravity of a hemispheroid.
Ans. '%
= 3- a.
8
17.
A
is scooped out of a cylinder of the same
Find the distance of the centre of gravity of the
cone of height h
height and base.
remainder from the vertex.
,
3
8
272.
Theorems
of
Pappus.
Theorem I. If a plane area be revolved about an axis in its plane
and not crossing the area, the volume of the solid generated is equal
to the product of the area and the length of the path described by
the centre of gravity of the area.
INTEGRAL CALCULUS
370
Theorem II. If the arc of a curve be revolved about an axis in
plane and not Crossing the arc, the area of the surface generated is
equal to the product of the length of the arc and the path described
by the centre of gravity of the arc.
its
273.
and let
have
Proof of the Theorems.
it
XY
Let the area be in the plane
Then by (1), Art. 271, we
revolve about the axis of X.
y
{
(ydxdy
I
I
=
dxcly
yj j dx dy=j \y dx dy.
Or
Then
2
Try
(*2
f Cdx dy = C Try dx dy
But the right-hand member of equation
by revolving the area through the angle 2
(1) is
2
it,
path described by the centre of gravity, and
(1)
the volume described
tig is
j
the length of the
dxdy
I
is
the plane
area.
The first theorem is thus seen to be
proved true in a similar manner.
true,
and the second can be
EXAMPLES
1.
Find the volume and surface generated by revolving a recc units from the centre
tangle with dimensions a and b about an axis
of the rectangle.
2.
Am
2
^^ and 4 ^ +
Eind the volume and surface generated by revolving an
lateral triangle each side
from the centre of the
a units in length, about an axis
triangle.
2
equi-
c units
r=
7rac
'
Am.
&)c
^6
2
and 6
fl-ac.
CENTRE OF GRAVITY.
3.
PRESSURE OF FLUIDS
371
Find the volume and surface generated by revolving a
an axis c units from the centre of the circle.
circle
of radius a about
Ans. 2 irarb and 4
4.
-r^ab
Find the volume generated by revolving an ellipse, semiaxes
b, about an axis c units from the centre of the ellipse.
a and
Ans. 2
2
7r
abc.
PRESSURE OF LIQUIDS
274. The pressure of a liquid on any given horizontal surface is
equal to the weight of a column of the liquid whose base is the
given surface and whose height
is
equal to the distance of this sur-
face below the surface of the liquid.
The pressure on any
method of determining
Ex.
Suppose
1.
rectangular board
it
and the
by the following examples.
vertical surface varies as the depth,
it is
illustrated
required
is
OABC,
find
to
the edge
OC
the
pressure
on
the
being at the surface of the
water.
BC= a, and AB —
Let
Suppose
the
C
b.
rectangle
into horizontal strips one of
is
which
X
HK.
OH=x,
Let
the strip
a
the
a
then the width of
is dx.
H
on this strip
were uniform throughout and the
same as it is at the top of the strip,
the pressure on the strip would be
wbx dx, where iv is the weight of
If
K
pressure
cubic unit of the
water.
b
is
evidently the integral of this
expression.
is,
Entire pressure
B
And
the entire pressure on the board
That
Y
divided
-r- bx dx =
a-biv
INTEGRAL CALCULUS
372
2. Find the pressure on that part of the board in Examwhich is below tile diagonal.
In this case the area of
is y dx, and the entire pressure on the
Ex.
ple
1,
UK
triangular board
is
f
wyx
dx.
b
But
hence entire pressure
bw C a
bwa?
=—
x~dx =
o 7
I
a Jo
Ex.
One
3.
face of a
box immersed in water
a square, the diagonals being 8 feet in length.
square
is
vertical.
is
in the
The
form of
centre of the
6 feet below the surface of the water, and one diagonal
is
Eind the pressure
on the square face.
Let
be the surface
of the water.
Taking the
SW
~
YV
axes as in the figure, the
equations of
are
AB
and
BO
y=4-f x, and ?/=4— x,
respectively.
Then,
entire
if
P represents the
pressure
on
the
board,
P=2iv( C
(6-{-x)dydx
= 512 tv = 15872
lbs.
Ex. 4. Eind the pressure on a sphere 6 feet in diameter, immersed in water, the centre of the sphere being 10 feet below the
surface of the water.
CENTRE OF GRAVITY.
SWbe
Let
PRESSURE OF FLUIDS
373
W
the surface of
the water.
Take the axes as in the
and let the elemen-
figure,
tary
surface
The
area of
be
a
zone.
a
zone at a
from the cen-y ds.
The pressure on the zone
distance
x
tre of the sphere is 2
is
2
-"vmIO
Then,
if
+
P
a?)efc.
represents
the entire pressure on the
sphere.
L
m
But
y
= V9 —
y (10
+ x)ds.
x 2 and ds
,
=-
civ.
y
Hence
P=far«fl
= 360
5.
A
C
ttv:
(10
=
+ x)dx,
22320tt lbs.
rectangular flood gate whose upper edge
the water,
is
is
in the surface of
divided into three parts by two lines from the middle
of lower edge to the extremities of upper edge.
Show
that the parts
sustain equal pressures.
6. A rectangular flood gate 10 feet broad and 6 feet deep has its
upper edge in the surface of the water. How far must it be sunk to
double the pressure ?
Ans. 3 ft.
7.
A
board in the form of a parabolic segment by a chord perpenis immersed in water.
The vertex is at the sur-
dicular to th< axis
j
and the axis vertical.
Find the pressure in tons.
far-p
It is
20 feet deep and 12 feet broad.
Ans.
59.52.
INTEGRAL CALCULUS
374
How
8.
pressure
far
must the board in Ex. 5 be sunk
Ans. 12
Suppose the position of the parabolic board
9.
the chord being in the surface
How
10.
to double the
?
far
;
what
is
in
Ex. 5 reversed,
the pressure ?
Ans. 39.38 tons.
must the board in Ex. 7 be sunk
to double the
Ans. 8
pressure ?
A trough 2
11.
elliptical ends.
and 2
feet deep
12.
ft.
feet broad at the top has semi-
If it is full of water, find the pressure
on one end.
Ans. 165^
closed
ft.
lbs.
One end of an unfinished water main 2 feet in diameter is
by a temporary bulkhead and the water is let in from the
Find the pressure on the bulkhead if
below the surface of the water in the reservoir.
reservoir.
its
centre
Ans.
is
30 feet
18607r lbs.
A water tank is in the form of a hemisphere 24 feet in diamesurmounted by a cylinder of the same diameter and 10 feet high.
Find the total pressure on the surface of the tank when the tank is
13.
ter
A
14.
is filled
is 12 inches and base a
with equal parts of water and
to be half as
heavy as the water, show that
whose depth
cylindrical vessel,
20 inches diameter,
circle of
oil.
Ans. 148. 8?r tons.
within 2 feet of the top.
filled to
Assuming the
oil
the pressure on the base equals the lateral pressure.
275.
Centre of Pressure.
cal surface varies as
Since the pressure of a liquid on a verti-
the depth, there exists a horizontal line about
which the statical moment of the entire pressure on the surface is
Such a line passes through the centre of pressure and the
zero.
abscissa of this point may be found by the method used in the following example.
Ex.
1.
Find the abscissa
of the centre
=
of
liquid pressure on
X
and
a vertical surface bounded by the curve y
f(x), the axis of
Given that the origin is at a distance
the two ordinates yQ and y v
W
=
CENTRE OF GRAVITY.
PRESSURE OF FLUIDS
h below the surface of the liquid, the axis of
X vertical,
375
and the
-weight of a cubic unit of liquid is w.
Let P^JPJIQ be the surface bounded by the curve y = f(x), the
X, and the two ordinates ?/ = QP and y x —
Divide
V
the surface into horizontal strips of width dx, one of which is HK.
= x. Let
Let
pass through the centre of liquid pressure,
HP
axis of
MN
OH
031—
and
x.
Then the pressure on the
HK
+
x) dx,
and the moment of
this
strip
wy
is
pressure about
(h
MN
is
|%
+ x)(x —I-)dx.
wy(h
moment
Therefore, the
of the entire pressure
the
—
integral
of
this
Y
is
ex!
pression between the abscissas of
is,
But
P
between
and
P
x
and
that
l}
must equal
this
Q
x1
Pn
X
\x
.
zero,
therefore
M
£
wy
(Jt
+ x) (x - x) dx
0.
\
M
i«»
l
H
IS.
R
I
J
1
1
xy (h
Or
-f
a;)
dx
x=^
Jy(h + x)dac
.
X
x
2. Find the centre of pressure of the water on the parabolic
board given in Ex. 7, Art. 274.
Ans. 14f in. below vertex.
3.
Find the centre of pressure of the water on the bulkhead given
ft.
Ans. x = T
^
in Ex. 12, Art. 274.
4.
A
rectangular flood gate a feet deep and b feet broad, with
upper edge at the surface,
How
far
down must
turn about
it ?
is
its
to be braced along a horizontal line.
the brace be put that the gate
may
not tend to
Ans.
|
a
ft.
INTEGRAL CALCULUS
376
One end
5.
of a cylindrical aqueduct 6 feet in diameter
half full of water
by a
How
brace.
is
far below the centre of the
What
brace be put ?
pressure must
it
A water
is
bulkhead should the
be able to withstand ?
Ans. x
6.
which
closed by a water-tight bulkhead held in place
= TF
9
7T ft.
;
P = 1116 lbs.
pipe passes through a masonry dam, enters a reservoir,
by a
The diameter
which is hinged at the
and the depth of its centre
below the water level in the reservoir is 12 feet. Find the pressure
on the valve, and the distance of the centre of pressure below the
hinge.
Ans. P = 1674 tt lbs. and ff ft.
and
is
top.
closed
cast-iron circular valve
of the valve is
3
feet,
7. Water is flowing along a ditch of rectangular section 4 feet
deep and 1 foot wide. The water is stopped by a board fitting the
ditch and held vertical by two bars crossing the ditch horizontally,
one at the bottom and the other one foot from the bottom of the
ditch.
How
high must the water rise to force a passage by upsetAns. To within 1 ft. of top of ditch.
ting the board ?
ATTRACTION AT A POINT
276. A particle of mass m is situated at a perpendicular distance
and
from one end of a thin, straight, homogeneous wire of mass
length I. Required to find the attraction on the particle due to the
M
c
wire.
X
Y
and
be the
Let
be the particle and AB the wire. Let
and ^respectively.
components of the attraction along the axes of
Divide AB into elements of length dy
and let PQ be one of these elements.
X
The mass
of
— dy,
PQ is M
since
M
—
is
L
l
the mass of a unit's length.
If the
mass
PQ
were concendue
according to Newton's Law
of
trated at P, the attraction at
to
PQ
is,
of Attraction,
KmJ(f(%
l(c
2
+ f)'
CENTRE OF GRAVITY.
and the components along
PRESSURE OF FLUIDS
377
OX and OF are
«"»* d
and
J* cos 6,
*™Md
l sin
fl,
respectively.
and sin
Substituting for cos
^ = KinMc C
X
l
=
d\j
•-
I
JoQS + y*)?
I
TT-_
their values
KinM C
^°
I
2
(c'
(c
= KinM sm
.
-_
cW^+T
2
+ yyrf
y
2
~|
c
V?+T
Vc + J
2
L
cl
a
6.
Cl
_ KinM f^
y dy
1
—KinM
——
we have
2
_ KinM (1 — cos
Denoting by
•K=
It the total attraction of the
VI +
2
Y = —£- V2(l - cos 0)
2
2
KmM sm-0.
1„
.
cl
The
line of attraction evidently
tangent
is
2
makes with
OA
an angle whose
J^l-cos^lg
X
The
wire on the particle,
sin
2
resultant attraction, therefore, bisects the angle
—
6.
Xote.
If we take as our unit of force the force of attraction between two unit masses concentrated at points which are at unit's
distance apart, k becomes unity.
INTEGRAL CALCULUS
378
EXAMPLES
Find the attraction perpendicular to the wire in Example 1
1.
when the
particle
is
at a distance - above 0.
o
Ans.
21
SiM
c
[_V9
c-
I
.
+ <U
2
V9c-
+ >]
Find the attraction of a thin, straight, homogeneous wire
upon a particle or mass m which is situated
I and mass
a distance c from one end of the wire and in its line of direction.
2.
length
M
Ans.
of a
m
at
KinM
c(c
3. Find the attraction
upon a particle of mass
of
+ l).
homogeneous circular disk of radius a
and at a distance c from the
in its axis
disk.
Ans.
2 Kirmpl 1
—
Vc + a J
2
where p
is
the density of the disk.
2
Find the attraction due to a homogeneous right circular cylinder
I and radius a upon a mass m in the axis produced of the
cylinder and distant c from one end.
4.
of length 2
Ans.
2
TTKinp [2
?+Va + c - Va + (c + 2
2
2
2
2
J) ].
CHAPTER XXXIII
INTEGRALS FOR REFERENCE
277.
We
give for reference a list of some of the integrals of the
preceding chapters.
/
2.
o
3.
I
J
x"
dx
r n+l
=
ra-fl
—X = log
05.
C dx— =-tan
I-—
J xr -r cr a
-f- a."
-
4.
a
r — dx = 1 log # — a
J x —a 2a
x-\-a
—
I
2
2
,
EXPONENTIAL INTEGRALS
5.
ax
Ca x dx =
log a
6.
j
e
x
dx
=e
x
.
TRIGONOMETRIC INTEGRALS
7.
8.
dx =
I
j
sin x
I
cos £ dx
— cos x.
= sin x.
379
INTEGRAL CALCULUS
380
9.
I
tan x dx
= log sec x.
10.
I
cot x dx
= log sin x.
11.
a
sec x
= log (sec x + tan x)
dx
= logtan(|+|).
12.
I
cosec x dx
= log (cosec x — cot x)
= log tan -
dx = tan
13.
j
2
sec x
14.
j
cosec 2
15.
I
sec x tan x
16.
j
cosec
17.
j
sin 2 xd£
18.
cc
a?
dx
=-
2
a?
I
cot
a?.
= sec x.
cot x dx
= — cosec #.
--sin2a\
4
2
/<cos
x.
—
dx =
•
dx = - -f -
sin 2 #.
4
2
INTEGRALS CONTAINING Vcr-a; (CHAP. XXV. AND ART.
2
-
n
19.
C
I
— dx
J Va —
2
20.
— sin _!#-•
.
2
sc
a
_^L_ = _ 2 V^^t? + - sin2
Jf Va -^ ^ 2
2
1
a
•
227)
.
.
INTEGRALS FOR REFERENCE
21.
C
dx
I
^
22
r
24.
25.
26.
J
— or
.
r
1
a a
'
\
a
cr
__
_a
x
=
a
+
a2
-.r
a
— x*
(r>r
_ x'
2 a-.r
((-
ar
'
2
^^
(a
da?
2 a 3 °° a
= | Vtf^a? + - sin-
1
-
+ ya -^
2
2
2
2
ar
8
f
—
8
=
f (a - :r) I (to = 2b (5 a - 2 x ) Va^=^ +
2
2
2
INTEGRALS CONTAINING
29.
^f a
+
A
_
Jdaj
x\
31
f
J
.- -
da;
2
+
Va?2
a 2 (CHAP.
—
sin" 1 ?
a
8
XXV. AND ART.
= log(a?4-VS*"+^).
a2
jrVj?
('
c/
**
.r
a
cr-x2
*/
28.
1
(Art. 227.)
«2 a
(a*- a?2)*
2
-
f.r\'^7 da? = % (2 - a )Va -r + - sin"
J
•/
27.
= -1,log
da?
2
a*
•^
23
a
sca
2
381
=^A^T^-giog(.x-4-A^T^).
J
2
,,-
— = 1 log
a
a
= 1- — v
.x*
,
2
a -f Var' -f
= _ Vx + a
2
a?2
s
t
1
a
2
ct
2
a?
4-
a2
—
a
227)
INTEGRAL CALCULUS
382
J
33.
34
.
35.
or
Vx
5
2aV
a2
4-
3
2 a
a
fV^T^tto = | V^T^ + f log (a + -y/iF+a*).
Jo
fa^Va?
+a
2
/dec
(x
36.
2
f(x
2
2
2
= x (2x + a^Va^ + a - - log
dx
2
2
8
+a )*
aV^ + a
+a
= ^ (2 x + 5 a ) V^+"^ +
)
+ VV + a
1 da>
2
2
8
~
8
log (»+ Va?+a*).
INTEGRALS CONTAINING Va - a (CHAP. XXV. AND ART.
2
2
38.
oa
39.
Vec2
f
/
—a
= log(a? + V^a»).
2
^_g_, = ? V* 2 - a 2 + £ log (« + V?=^)
dx
-———= -1 sec -1 x-
{*
,
I
•^
40
da;
f
J
xV# —
2
a2
2
-\/x
2
—a
2
2
^==== =
dec
x^x ~a
2
a
a
Vx — a
c?»
/:x
).
2
%/
37.
2
%
__
2
2
(jc
2
2
'
ax
V# — or
2aV
2
.
1
1
3
2a
sec
_i
a
a
•
227)
•
.
INTEGRALS FOR REFERENCE
42.
43.
(\
faj
=|a
.r-crV.r
2
\
2
8
dg
f
•*
45.
a--
—
2
2
2
^ -^
2
)^7.r
2
47.
—
I
•^
= vers
2
-a*).
a.r
-x
2
-•
a
= — V 2 a.? — a^+
a vers
'
V2oa?— x
2
V2
a*£
48
!
sr
—
+ Vtf
4rr
^).
)V^^ + ^log(a;+V^
8
INTEGRALS CONTAINING V2
I
log («
b
2
= ^(2^-5a
8
J V2 cub
).
•
«Vr — a
»y
46.
2
*
=
cr)^
f^r-a
+ Vx - a
log (x
Z^T? <fe = 5 (2 or - a ) V^
•J
44.
- «- - |
.r
383
«.i*
—x
a
2
V2 aa — x'
1
49
50.
51.
(
J
v 2 ax —
2
a;
tfte
—
=-
-•
2
V2 ax — x + ^- vers -1 -a
2
2
•
2*
VU^rf + £2 vers-*.a
f«v?^=7* = - 3a + <f6
'
J
Jf>^^-^^=
u;
v
aax-^ + avers-
1
*.
a
INTEGRAL CALCULUS
384
W2ax — x
.
53.
s
x
/»
dx
2
_ (2 ax — x
dx _
3
f
3 ax
)
2
—a
x
a 2 V2 ax — x2
(2 ax — x-y
xdx
54.
2
3
x
f (2ax — x y
aV2ax —
2
INTEGRALS CONTAINING ±ax +bx + c
2
C_J*
55
'
J
aar
+
6a;
or
56.
57.
I58.
+
f
J
I
^
V aar +
,
9
,
2
=
,
~-\og
V6 — 4ac
,
-y/ax2 -\-bx
= log
a
c
+ cdx =
62
,
2
+
- 6- V6 -4ac
2aa; + 6 + V6 — 4 etc
2
2aa;
,
2
=+ = —V
6a;
V4ac-6
2
1
=
2^ + 6
tan-i
V4ac-6
c
(2 aa;
2
+ 6 + 2vWaa; + 6a; + c).
2
.
—4a
— 4 ac log
Vax +
2
6a'
+ 6 + 2VaV aa; + 6a; +
2
(2 aa;
c).
8 a*
59.
1
60.
^
f—
I
•^
j
V — aa^ + 6a; + c
V—
aa;
2
-\-bx
2—aa;
—6
4a
=
1
= sin
_ix
—2
aaj
——
6
V6 +
Va
2
-i
ac
+ cdx
V — ax *——,bx— —
;
/
2
-{-
-\-
c
,
-{
6
2
+
— 4 ac sin _,
.
*
8a
|
2
—6
z^=.
aa;
V6 + 4ac
2
INTEGRALS FOR REFERENCE
385
OTHER INTEGRALS
61.
CJ^t^cU
J \6 4-a+
a {a
62.
+
.i-)
(b
+ x) + (a - 6) log ( Va + + V& + x).
fJjL^cfa = V(a - a)(& + 0) + (a +
a?
1
6) sin"
J^|-
INDEX
Acceleration
Angles, between two curves
with coordinate planes
Arc, derivative of
Area, any surface
derivative of
of curve
242, 306, 320,
surface of revolution
.
18
.
.
.
.
.
Asymptotes
Attraction at a point
Cauchy's test for convergence
Centre, of curvature
Change
of gravity
of pressure
of variable
.
.
.
Curves, direction of
16, 174, 182
for reference, higher plane
162
length of
327, 330
174, 183
133
186, 187
355
241
325, 349
337, 353
179
375
82
200
365
373
as a
convergence
Computation, by logarithms
of n
Constant, definition of
test for
.
.
...
double
sign of
Differential coefficient
Differentials, definition of
.
...
.
.
.
.
.
.
.
.
.
.
.
.
lae
order of
partial
successive
.
.
.
....
....
51
39
137
130
61
trigonometric formu45
lae
Derivative, definition
general expression of
.
.
.
.
307
319
310
343
314
12
68, 70
71
26, 29
26
.
inverse trigonometric formulae
logarithmic and exponential formu-
....
...
.
....
....
definition of
1
.
.
formulae
Differentiation, algebraic formulae
209
80
94
96
derivative of
26
notation of
1
of integration
224, 315
Contact, order of
206, 207, 209
Convergence, absolute and conditional
78
interval of
86
tests for
79
Curvature, centre of
200
circle of
195, 200
direction of
189
radius of
193, 196, 197
uniform
193
variable
194
Curves, angle of intersection of
197
area of
242, 306, 320, 325, 349
continuous and discontinuous
22
sum
definition of
195, 200
osculating
Comparison
208
Definite integrals
.57, 58, 148, 263,
....
.
osculating
299, 304, 317
Circle, of curvature
.
.
.
.
.
illustrations of
meanings
of an arc
386
.
...
.
.
....
.
.
241,336
of area
of function of a func58
tion
130
partial
partial, of higher order
136, 137
relation
.
.
of
11
12
13
16, 17, 18
186, 187
.
between
dv
—
ax
and *L
total
......
57
140
INDEX
TACK
320
319
Element, of area
of an integral
Envelopes, definition of
.
.
214, 215
.
equation of
of normals
Equation, of envelopes
21S
217
215
ofevolute
of
201
133
133
324
201
217
201
202, 204
normal
of tangent
parametric
Evolute
an envelope
equation of
properties of
...
.
387
PAGE
P
Integration, containing (ax+b)?
.
.
263
.
264
containing
r
p
(ax+b)v, (ax
+ b)*
containing
V± x + ax + b ...
2
266
307
definition of
223
double
343, 347, 349, 350
evaluation of
307
for reference
378
fundamental
225
indefinite
309
of see" x dx, cosec" x dx
274
of sin" x dx, cos' x dx
270
of sin m x cos n x dx
271, 276, 291, 293
of tan" x dx, cot" x dx
273
of tan" x sec" x dx,
cot" x cot" x dx 274, 291, 293
of e^ sin nx dx,
e ax cos nx dx
283
of x m (a
bx n )Pdx
284
2
of f(x )xdx
295
of rational fractions
249
proofs of formulae
227
successive
343
triple
344, 361, 363
Intercepts of tangent
173
Involute
201
properties of
202, 204
definite
.
....
....
.
Function, algebraic
continuous
2
22, 315
definition
1
discontinuous
expansion of
22
88
implicit) differentiation of
75, 144
increasing and decreasing
inverse trigonometric
logarithmic
of two or more variables
transcendental
trigonometric
.
21
51
39
15G
2
45
.
.
....
.....
1
.
.
+
.
.
Gravity, centre of
365
Higher plane curves
Huyghens's approximate length of
162
arc
.
.
.
.
.
.
.
...
...
.
93
Leibnitz's theorem
65
11 Length of curves
93, 327, 330
309 Limit, change of
317
Indeterminant forms
106
definition of
8
Inertia, moment of
340, 363
infinite
315
Infinite, limits
315
Napierian base
10
variables
315
notation of
8
Infinitesimals, order of
73
relation of arc to chord
9
Inflexion
190
variable
343, 348, 351
Integration and derivative of integral 270 Liquids, pressure of
370
as a summation
306 Logarithmic functions
2, 39
between limits
309 Logarithms, computation by
94
by algebraic substitution 299
Napierian
41
by parts
279
substitution
218-22*;
by
constant of .
224, 310 Maclaurin's theorem
89
containing
Maxima and minima
114, 155
Va*— x*,
296 Moment of inertia
346, 363
....
Increment
Indefinite integral
.
....
....
.
.
.
.
....
.
....
.
.
.
.
.
INDEX
388
Napierian base
.
.
„
Normals
,
Order, of contact
....
PAGE
8
133
.......
of differentiation
of integration
Osculating curves
order of contact of
Osculating circle, coordinates of centre
.
.
209
209
radius of
Pappus, theorems of
Parameter
Parametric equations
Power
series
Rates
Reduction formulae
Remainder, Taylor's theorem
324
85
373
370
70
173
133
97, 103, 145
65
89
84, 86
369
83
97, 103, 145
152, 153
2, 45
intercept of
Tangent planes
Taylor's theorem
Theorem, Leibnitz's
....
Maclaurin's
mean value
.
.
.
Pappus's
Rolle's
.
...
...
Taylor's
Transformation
Trigonometric functions
18
284, 291
...
computation by
convergence of power
convergent and divergent
of positive and negative
terms
...
power
Slope of a curve
of a line
of a plane
Subtangent
Subnormal
Tangent
369
.....
.
.
214, 324
Pressure, centre of
of liquids
Series,
206
137
343
208
209
PAGE
Surfaces, area of any
355
Surfaces of revolution, areas of
337, 353
derivative of
area of
336
volumes of . 333
.
.
.
105
94
85
78
78
85
16
16
133
173, 183
173. 183
Uniform curvature
Unit of force
.
.
.
„
.
„
.
.
.
.
.
...
.
.
193
376
*
Variable, change of
57, 58, 148,
.
263, 299,304, 317
curvature
......
194
definition of
1
dependent
independent
2
2
notation of
1
17, 18
Velocity
Volumes, any solid
by area
of section
.
surfaces of revolution
.
361
340
333, 353
.