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ELEMENTARY TEXTBOOK
ON THE
CALCULUS
BY
VIRGIL SNYDER, Ph.D.
AND
JOHN IRWIN HUTCHINSON, Ph.D
Of Cornell University
NEW YORK •:• CINCINNATI •:• CHICAGO
AMERICAN BOOK COMPANY
THE MODERN MATHEMATICAL SERIES.
Lucien Augustus Wait,
{Senior Professor of Mathematics in Cornell University ,)
General Editor.
This series includes the following works :
BRIEF ANALYTIC GEOMETRY. By J. H. Tanner and Joseph Allen.
ELEMENTARY ANALYTIC GEOMETRY. By J. H. Tanner and Joseph
Allen.
DIFFERENTIAL CALCULUS. By James McMahon and Virgil Snyder.
INTEGRAL CALCULUS. By D. A. Murray.
DIFFERENTIAL AND INTEGRAL CALCULUS. By Virgil Snyder and
J. I. Hutchinson.
ELEMENTARY TEXTBOOK ON THE CALCULUS. By Virgil Snyder
and J. I. Hutchinson.
HIGH SCHOOL ALGEBRA. By J. H. Tanner.
ELEMENTARY ALGEBRA. By J. H. Tanner.
ELEMENTARY GEOMETRY. By James McMahon.
COPYRIGttl', 1912, BY
AMERICAN BOOK COMPANY
EL. CALCULUS.
W. P. 1
Q fV3 03
53/
PREFACE
The present volume is the outgrowth of the requirements
for students in engineering and science in Cornell University,
for whom a somewhat brief but adequate introduction to the
Calculus is prescribed.
The guiding principle in the selection and presentation of
the topics in the following pages has been the ever increasing
pressure on the present-day curriculum, especially in applied
science, to limit the study of mathematics to a minimum of
time and to the topics that are deemed of most immediate use
to the professional course for which it is preparatory.
To what extent it is wise and justifiable to yield to this
pressure it is not our purpose to discuss. But the constantly
accumulating details in every pure and applied science makes
this attitude a very natural one towards mathematics, as well
as towards several other subjects which are subsidiary to the
main object of the given course.
This desire to curtail mathematical training is strikingly
evidenced by the numerous recent books treating of Calculus
for engineers, for chemists, or for various other professional
students. Such books have no doubt served a useful purpose
in various ways. But we are of the opinion that, in spite of
the unquestioned advantages of learning a new method by
means of its application to a specific field, a student would
ordinarily acquire too vague and inaccurate a command of the
fundamental ideas of the Calculus by this one-sided presenta-
tion. While a suitable illustration may clear up the difficulties
3
262792
4/ ( • ;/f ;'l :*•/;• \ * ., 'breface
of an abstract theory, too constant a dwelling among applica-
tions alone, especially from one point of view, is quite as likely
to prevent the learner from grasping the real significance of
a vital principle.
In recognition of the demand just referred to, we have made
special effort to present the Calculus in as simple and direct
a form as possible consistent with accuracy and thoroughness.
Among the different features of our treatment, we may single
out the following for notice.
The derivative is presented rigorously as a limit. This does
not seem to be a difficult idea for the student to grasp, espe-
cially when introduced by its geometrical interpretation as
the slope of the line tangent to the graph of the given func-
tion. For the student has already become familiar with this
notion in Analytic Geometry, and will easily see that the
analytic method is virtually equivalent to a particular case of
the process of differentiation employed in the Calculus.
In order to stimulate the student's interest, easy applications
of the Differential Calculus to maxima and minima, tangents
and normals, inflexions, asymptotes, and curve tracing have
been introduced as soon as the formal processes of differentia-
tion have been developed. These are followed by a discussion
of functions of two or more independent variables, before the
more difficult subject of infinite series is introduced.
In the chapter on expansion, no previous knowledge of series
is assumed, but conditions for convergence are discussed, and
the criteria for determining the interval of convergence of those
series that are usually met with in practice are derived.
A chapter on the evaluation of indeterminate forms and
three chapters on geometric applications furnish ample illus-
PREFACE 5
tration of the uses of infinite series in a wide range of
problems.
By reason of its significance in applications, it does not seem
advisable to omit the important principle of rates. Arising
out of the familiar notion of velocity, it affords an early glimpse
into applications of the Calculus to Mechanics and Physics.
We do not propose to make the Calculus a treatise on Mechanics,
as seems to be the tendency with some writers; but a final
chapter on applications to such topics of Mechanics as are easy
to comprehend at this stage is thought advisable and sufficient.
Especially in treating of center of gravity, the formulas have
been derived in detail, first for n particles, and then, by a limit-
ing process, for a continuous mass. This was considered the
more desirable, as textbooks in applied mathematics frequently
lack in rigor in discussing the transition from discrete particles
to a continuous mass. Besides, the derivation of these formu-
las affords a very good application of the idea of the definite
integral as the limit of a sum. This idea has been freely and
consistently used in the derivation of all applied formulas in
the Integral Calculus. However, as the formula for the length
of arc in polar coordinates is especially difficult of derivation
by this method, we have deduced it from the corresponding
formula for rectangular coordinates by a transformation of the
variable of integration.
In-order to make the number of new ideas as few as possible,
the notions of infinitesimals and orders of infinitesimals have
been postponed to the last article on Duhamel's principle. This
principle seems to flow naturally and easily from the need of
completing the proof of the formulas for center of gravity.
The teacher may omit this article, but its presence should at
6 PREFACE
least serve the important end of calling the attention of the
student to the fact that there is something yet to be done in
order to make the derivations complete.
Some teachers will undoubtedly prefer to do a minimum
amount of work in formal integration and use integral tables in
the chapters on the applications. For such the first chapter of
the Integral Calculus might suffice for drill in pure integration.
The problems in this chapter are numerous, and, for the most
part, quite easy, and should furnish the student a ready insight
into the essential principles of integration.
The characteristic features of the books on the Calculus
previously published in this series have been retained. The
extensive use of these books by others, and a searching yearly
test in our own classroom experience convince us that any far-
reaching change could not be undertaken without endangering
the merits of the book. The changes that have been made are
either in the nature of a slight rearrangement, or of the addi-
tion of new illustrative material, particularly in the applications.
We wish to acknowledge our indebtedness to our colleagues,
who have added many helpful suggestions ; to Professor I. P.
Church, of the College of Civil Engineering, for a number of
very useful problems in applications of integration (See Exs.
14-18, pp. 318-320, and Exs. 6-7, pp. 323-324), and particu-
larly to Professor James McMahon, who has carefully read all
the manuscript, assisted throughout in the proof reading, and
made many improvements in the text.
CONTENTS
DIFFERENTIAL CALCULUS
CHAPTER I
Fundamental Principles
ARTICLE PAGE
1. Elementary definitions 15
2. Illustration : Slope of a tangent to a curve . . . .16
3. Fundamental theorems concerning limits 17
4. Continuity of functions 19
5. Comparison of simultaneous increments of two related variables 20
6. Definition of a derivative 21
7. Process of differentiation 22
8. Differentiation of a function of a function .... 23
CHAPTER II
Differentiation of the Elementary Forms
9.
Differentiation of the product of a constant and a variable
25
10.
Differentiation of a sum .
26
11.
Differentiation of a product
27
12.
Differentiation of a quotient
28
13.
Differentiation of a commensurable power of a function .
29
14.
Differentiation of implicit functions
33
15.
Elementary transcendental functions ....
34
16.
Differentiation of log a x and log a u
34
17.
Differentiation of the simple exponential function
36
18.
Differentiation of an incommensurable power .
37
19.
Limit of ^— as 6 approaches
38
20.
Differentiation of sin u
39
21.
Differentiation of cos u
40
22.
Differentiation of tan u
40
23.
Differentiation of sin -1 u
42
24.
Table of fundamental forms
44
CONTENTS
CHAPTER III
Successive Differentiation
ARTICLE PAGE
25. Definition of the nth derivative 47
26. Expression for the nth derivative in certain cases ... 49
CHAPTER IV
Maxima and Minima
27. Increasing and decreasing functions 51
28. Test for determining intervals of increasing and decreasing . 51
29. Turning values of a function 53
30. Critical values of the variable 55
31. Method of determining whether <f>'(x) changes its sign in pass-
ing through zero or infinity ....... 55
32. Second method of determining whether <p'(x) changes its sign
in passing through zero ... ..... 57
33. The maxima and minima of any continuous function occur
alternately .59
34. Simplifications that do not alter critical values .... 59
35. Geometric problems in maxima and minima .... 60
CHAPTER V
Rates and Differentials
36. Rates. Time as independent variable 68
37. Abbreviated notation for rates 72
38. Differentials often substituted for rates 74
39. Theorem of mean value 74
CHAPTER VI
Differential of an Area, Arc, Volume, and
Surface of Revolution
40. Differential of an area 78
41. Differential of an arc 79
42. Trigonometric meaning of — , — .80
dx dy
43. Differential of the volume of a surface of revolution ... 81
CONTENTS
ARTICLE
44. Differential of a surface of revolution
45. Differential of arc in polar coordinates
46. Differential of area in polar coordinates
81
82
83
CHAPTER VII
Applications to Curve Tracing
47. Equation of tangent and normal ....
48. Length of tangent, normal, subtangent, and subnormal
49. Concavity upward and downward
60. Algebraic test for positive and negative bending
51. Concavity and convexity toward the axis .
52. Hyperbolic and parabolic branches .
53. Definition of a rectilinear asymptote .
85
85
89
90
94
95
96
Determination of Asymptotes
54. Method of limiting intercepts 96
55. Method of inspection. Infinite ordinates, asymptotes parallel
to axes 97
56. Method of substitution. Oblique asymptotes .... 99
57. Number of asymptotes 102
Polar Coordinates
58. Meaning of p — 104
dp
59. Relation between ^ and p— 105
dx dp
60. Length of tangent, normal, polar subtangent, and polar sub-
normal 105
CHAPTER VIII
Differentiation of Function
61. Definition of continuity
62. Partial differentiation
63. Total differential
64. Total derivative . •
65. Differentiation of implicit functions
66. Geometric interpretation .
67. Successive partial differentiation
68. Order of differentiation indifferent
s of Two Variables
109
110
112
115
116
118
122
122
10 CONTENTS
CHAPTER IX
Change of Variable
article page
69. Interchange of dependent and independent variables . . 124
70. Change of the dependent variable 125
71. Change of the independent variable 126
72. Simultaneous changes of dependent and of independent variables 126
CHAPTER X
Expansion of Functions
73. Convergence and divergence of series 132
74. General test for convergence , 133
75. Interval of convergence 138
70. Remainder after n terms , . 140
77. Maclaurin's expansion of a function in a power series . .141
78. Taylor's series 148
79. Rolle's theorem 150
80. Form of remainder in Maclaurin's series 150
81. Another expression for the remainder 153
CHAPTER XI
t
Indeterminate Forms
82. Definition of an indeterminate form » 157
83. Indeterminate forms may have determinate values . . . 158
84. Evaluation by development 160
85. Evaluation by differentiation ....... 161
86. Evaluation of the indeterminate form g- 165
CHAPTER XII
Contact and Curvature
87. Order of contact 167
88. Number of conditions implied by contact 168
89. Contact of odd and of even order 169
90. Circle of curvature 172
91 . Length of radius of curvature ; coordinates of center of curvature 1 72
92. Limiting intersection of normals 174
93. Direction of radius of curvature . . . ' • • 175
CONTENTS
11
ARTICLE PAGE
94. Total curvature of a given arc ; average curvature . . . 176
95. Measure of curvature at a given point 177
. 96. Curvature of an arc of a circle 178
97. Curvature of osculating circle 178
98. Direct derivation of the expressions for k and B in polar co-
ordinates 180
EvOLUTES AND INVOLUTES
99. Definition of an evolute 182
100. Properties of the evolute 184
CHAPTER XIII
Singular Points
101. Definition of a singular point
102. Determination of singular points of algebraic curves
103. Multiple points
104. Cusps .........
105. Conjugate points
191
191
193
194
197
CHAPTER XIV
Envelopes
106. Family of curves
107. Envelope of a family of curves
108. The envelope touches every curve of the family
109. Envelope of normals of a given curve
110. Two parameters, one equation of condition
200
201
202
203
204
12 CONTENTS
INTEGRAL CALCULUS
CHAPTER I
General Principled oe Integration
ARTICLE PAGE
111. The fundamental problem 209
112. Integration by inspection 210
113. The fundamental formulas of integration .... 211
114. Certain general principles 212
115. Integration by parts 216
116. Integration by substitution 219
117. Additional standard forms 222
118. Integrals of the forms C(^x + B)dx and C ( Ax + B)dx # »M
J ax 2 + bx + c J ^/ ax -2 + bx + c
119. Integrals of the forms f — and
J (Ax + B) Vax 2 + bx + c
dx
5
(Ax + B) 2 Vax 2 + bx + c
121. Decomposition of rational fractions
122. Case I. Factors of the first degree, none repeated
123. Case II. Factors of the first degree, some repeated
124. Case II J. Occurrence of quadratic factors, none repeated
125. Case IV. Occurrence of quadratic factors, some repeated
126. General theorem
226
CHAPTER II
120. Reduction Formulas 229
CHAPTER III
Integration of Rational Fractions
238
240
241
243
245
247
CHAPTER IV
Integration by Rationalization
127. Integration of functions containing the irrationality y/ax + b 248
128. Integration of expressions containing Vax 2 + bx + c . . 249
129. The substitution V± t 1 ± k 2 -z 253
CONTENTS
13
CHAPTER V
Integration of Trigonometric Functions
ARTICLE
130. Integration by substitution
181. Integration of ( sec 2n a; dx, \ csc 2 »x dx
132. Integration of I sec TO x tan' 2n+1 x dx, j csc m x coV^+h; dx
133. Integration of i tan"x dx, \ cot n x dx
134. Integration of ( sin'"x cos n x dx
135. Integration of f — , f-
J a + b cos nx J a
k
dx
b sin nx
dx
+ b sin nx -f c cos nx
136. Integration of I e ax sin wxcJx, je^cos nxdx
PAGE
255
255
257
258
260
264
266
CHAPTER VI
Integration as a Summation. Areas
137. Areas 268
138. Expression of area as a definite integral 270
139. Generalization of the area formula 273
140. Certain properties of definite integrals 274
141. Maclaurin's formula 276
142. Remarks on the area formula - 277
143. Precautions to be observed in evaluating definite integrals . 283
144. Calculation of area when x and y are expressible in terms of a
third variable 289
145. Areas in polar coordinates 291
146. Approximate integration. The trapezoidal rule . . . 292
147. Simpson's rule 294
148. The limit of error in approximate integration .... 295
14
CONTENTS
CHAPTER VII
Geometrical Applications
ARTICLE PAGE
149. Volumes by single integration 298
150. Volume of solid of revolution 302
151. Lengths of curves. Rectangular coordinates .... 306
152. Lengths of curves. Polar coordinates 309
153. Measurement of arcs by the aid of parametric representation . 310
154. Area of surface of revolution 312
155. Various geometrical problems leading to integration . 315
CHAPTER VIII
Successive Integration
156. Functions of a single variable .
157. Integration of functions of several variables
158. Integration of a total differential
159. Multiple integrals . . .
160. Definite multiple integrals
161. Plane areas by double integration .
162. Volumes
321
324
326
328
329
330
334
CHAPTER IX
Some Applications of Integral Calculus to
Problems of Mechanics
163.
Liquid pressure on a plane vertical wall .
. . . 338
164.
Center of gravity
. 340
165.
Moment of inertia .
. 346
166.
Duhamel's theorem .
. 348
Trigonometric Formulas
. 352
Logarithmic Formulas
. 353
DIFFERENTIAL CALCULUS
^XKc
CHAPTER I
FUNDAMENTAL PRINCIPLES
1. Elementary definitions. A constant number is one that
retains the same value throughout an investigation in which it
occurs. A variable number is one that changes from one value to
another during an investigation. If the variation of a number
can be assigned at will, the variable is called independent; if
the value of one number is determined when that of another is
known, the former is called a dependent variable. The depend-
ent variable is called also a function of the independent variable.
E.g., 3 x 2 , 4vx — 1, cos x, are all functions of x.
Functions of one variable x will be denoted by the symbols
/(#), <f>(x), • ••, which are read as "/of x" " <f> of x" etc. ; simi-
larly, functions of two variables, x, y, will be denoted by such
expressions as
f(?,y),F(x,y), ••••
When a variable approaches a constant in such a way that
the difference between the variable and the constant may be-
come and remain smaller than any fixed number, previously
assigned, the constant is called the limit of the variable.
15
16
DIFFERENTIAL CALCULUS
2. Illustration : Slope of a tangent to a curve. To obtain the
slope of the tangent to a curve at a point P upon it, first take
the slope of the line joining P = (x ly y x ) to another point (x 2 , y 2 )
upon the curve, then determine the limiting value of the slope
m
as the second point approaches to coincidence with the first,
always remaining on the curve.
Ex. 1. Determine the slope of the tangent to the curve
2-FM + fc
at the point (2, 4) upon it.
Here, x\ = 2, y\ = 4. Let x 2 = 2 + h,
yi = 4 + k, where h, k are so related that the
point (x 2 , y*) lies on the curve.
Thus 4 + k = (2 + h)\
or h = 4 A + A 2 - (1)
The
slope
m = y* -
x 2 -
-Xi
becomes
4+ Tc
2 + h
-4
_ 2
k
Fig. 1
which from (1) may be written in the form
k
= 4 + h.
(2)
The ratio k : h measures the slope of the line joining (x h y x ) to
(ar 2 , ys) • When the second point approaches the first as a limiting
position, the first member of equation (2) assumes the indeterminate
form -, but the second member approaches the definite limit 4. When
the two points approach coincidence, a definite slope 4 is obtained,
which is that of the tangent to the curve y = x 2 at the point (2, 4).
It may happen that h, k appear in both members of the equation
which defines the slope, as in the next example.
FUNDAMENTAL PRINCIPLES
17
Fig. 2
Ex. 2. If x 2 + y 1 — « 2 ? find the slope of the tangent at the point
Oi> yd-
Since
Xl * + i/! 2 = a 2 , (asi + ny+ (t/1 + ky = a\
hence 2 hx 1 + A 2 + 2 /fr/i + fc 2 = 0,
from which - = — - — — —
h 2 ?/i + k
k
To obtain the limit of -, put h, k
h
each equal to zevo in tlie second member.
lim * = _*!.
h±o h ?/i
This step is more fully justified in the next article.
This result agrees with that obtained by elementary geome-
try. The slope of the radius to the circle a 2 + y 2 = a 2 through
the point (x lf y x ) is — , and the slope of the tangent is the nega-
tive reciprocal of that of the radius to the point of tangency,
since the two lines are perpendicular.
3. Fundamental theorems concerning limits. The following
theorems are useful in the processes of the Calculus.
Theorem 1. If a variable a approaches zero as a limit, then
lea will also approach zero, k being any finite constant.
That is, if a = 0,
then Jca = 0.
For, let c be any assigned number. By hypothesis, a can be-
come less than -, hence ka can become less than c, the arbi-
k ■ '
* For convenience, the symbol = will be used to indicate that a variable
approaches a constant as a limit; thus the symbolic form x = a is to be read
" the variable x approaches the constant a as a limit."
el. calc — 2
18 DIFFERENTIAL CALCULUS
trary, assigned number, hence ka approaches zero as a limit.
(Definition of a limit.)
Theorem 2. Given any finite number n of variables
a, (3, y, •••, each of which approaches zero as a limit, then their
sum will approach zero as a limit. For the sum of the n
variables does not at any stage numerically exceed n times the
largest of them, which by Theorem 1 approaches zero.
Theorem 3. If each of two variables approaches zero as a
limit, their product will approach zero as a limit. More gen-
erally, if one variable approaches zero as a limit, then its
product with any other variable having a finite limit will have
the limit zero, by Theorem 1.
Theorem 4. If the sum of a finite number of variables is
variable, then the limit of their sum is equal to the sum of
their limits ; i.e.,
lim (x + y + • • •) = lim x + lim y +
For, if x = a, y = b, • • •,
then x = a + a, y = b -\- (3, •-•,
wherein a = 0, fi = 0, •• • ; (Def . of limit)
hence x + y+ ••• = (o+ &+ •••) + (« + fi+ •••)>
but a + p+->- =0, (Th. 2)
hence, from the definition of a limit,
lim (x + y + •••) = a-\-b-\- ••• = lim x -f- lim y + •••.
Theorem 5. If the product of a finite number of variables
is .variable, then the limit of their product is equal to the
product of their limits.
For, let x = a + a, y = b+($,
wherein a = 0, (3 = 0,
so that lim x = a, lim y = b.
FUNDAMENTAL PRINCIPLES 19
Form the product
xy = (a + a)(b + fi) = ab + «6 -f /3a + «0.
Then lim sc?/ = lim (ab + ab + (5a + a(5)
= ab + lim ab + lim 0a + lim a/3 (Th. 2)
= ab. (Th. 1)
Hence lim xy = lim a; • lim y.
In the case of a product of three variables x, y, z, we have
lim xyz = lim xy • lim z (Th. 5)
= lim x lim y lim 3,
and so on, for any finite number of variables.
Theorem 6. If the quotient of two variables as, y is vari-
able, then the limit of their quotient is equal to the quotient
of their limits, provided these limits are not both infinite or
not both zero.
(Th. 5)
y lim y
4. Continuity of functions. When an independent variable x,
in passing from a to b, passes through every intermediate
value, it is called a continuous variable. A function f(x) of an
independent variable x is said to be continuous at any value x l}
or in the vicinity of x Xi when f(x^) is real, finite, and determi-
nate, and such that in whatever way x approaches a^,
From the definition of a limit it follows that corresponding
to a small increment of the variable, the increment of the
For, since
X
x = -y,
y
lim x = lim - lim y y
y
and hence
,. x lim x
lim - =
20
DIFFERENTIAL CALCULUS
function is also small, and that corresponding to any number
c, previously assigned, another number 8 can be determined,
such that when h remains numerically less than 8, the differ-
is numerically less than c.
2/,+e
a>.+5
Fig. 3
Thus, the function of Fig. 3 is continuous between the values
x x and x Y -f- 8, while the functions of Fig. 4 and Fig. 5 are dis-
continuous. In the former of these two the function becomes
infinite at x = c, while in the latter the difference between the
value of the function at c + h and c — h does not approach
zero with h, but approaches the finite value AB as h ap-
proaches zero.
When a function is continuous for every value of x between
a and b, it is said to be continuous within the interval from a
to b.
5. Comparison of simultaneous increments of two related vari-
ables. The illustrations of Art. 2 suggest the following general
procedure for comparing the changes of two related variables.
Starting from any fixed pair of values x 1} y^ represented graph-
ically by the abscissa and ordinate of a chosen point P on a
given curve whose equation is given, we change the values of
FUNDAMENTAL PRINCIPLES
21
x and y by the addition of small amounts h and k respectively,
so chosen that the new values x L + h and y l + k shall be the
coordinates of a point P 2 on the
curve. The amount h added to x x
is called the increment of x and is
entirely arbitrary. Likewise, k is
called the increment of y ; it is not
arbitrary but depends upon the
value of h ; its value can be calcu-
lated when the equation of the curve
is given, as is shown by equation (1). These increments are
not necessarily positive. In the case of continuous functions, h
may always be taken positive. The sign of k will then depend
upon the function under consideration. The slope of the line
PjP 2 is then - and the slope of the tangent line at Pj is the
limit of - as h and consequently k approach zero.
The determination of the limit of the ratio of k to h as h and
k approach zero is the fundamental problem of the Differential
Calculus. The process is systematized in the following ar-
ticles. While the related variables are here represented by
ordinate and abscissa of a curve, they may be any two related
magnitudes, such as space and time, or volume and pressure of
a gas, etc.
6. Definition of a derivative. If to a variable a small incre-
ment is given, and if the corresponding increment of a contin-
uous function of the variable is determined, then the limit of
the ratio of the increment of the function to the increment of
the variable, when the latter increment approaches the limit
zero, is called the derivative of the function as to the variable.
22 DIFFERENTIAL CALCULUS
k
That is, the derivative is the limit of - as h approaches zero,
or
liin (k
For the purpose of obtaining a derivative in a given case it is
convenient to express the process in terms of the following steps:
1. Give a small increment to the variable.
2. Compute the resulting increment of the function.
3. Divide the increment of the function by the increment of
the variable.
4. Obtain the limit of this quotient as the increment of the
variable approaches zero.
7. Process of differentiation. In the preceding illustrations,
the fixed values of x and of y have been written with sub-
scripts to show that only the increments h, k vary during the
algebraic process of finding the derivative, also to emphasize
the fact that the limit of the ratio of the simultaneous incre-
ments h, k depends upon the particular values which the
variables x, y have, when they are supposed to take these in-
crements. With this understanding the subscripts will hence-
forth be omitted. Moreover, the increments h, k will, for
greater distinctness, be denoted by the symbols Ax, Ay, read
" increment of x," " increment of y."
If the four steps of Art. 6 are applied to the function
y = <£ (x), the results become
y + £fy=<f>(x + \x),
Ay = <j>(x + Ax) — <f>(x) = A<f> (x),
Ay _<f>(x + Ax) — <j> (x) _ A<£ (x)
Ax Ax Ax
. lira - A J = Km { +(* + **)- »(*) | =Um Aj>^
A# Ax Ax
' FUNDAMENTAL PRINCIPLES 23
The operation here indicated is for brevity denoted by the
symbol — , and the resulting derivative function by <f>'(x); thus
dx
dy _d<f>(x) _ lim f <f>(x + Ax)-<j>(x) '
dx dx Aa; =
Ax
= +-(*).
The new symbol -^ is not (at the present stage at least) to
ax
be looked upon as a quotient of two numbers dy, dx, but rather
as a single symbol used for the sake of brevity in place of
the expression " derivative of y with regard to x."
The process of performing this indicated operation is called
the differentiation of <f> (x) with regard to x.
EXERCISES
Find the derivatives of the following functions with regard to x.
5. I.
X 3
6. x n , n being a positive integer.
7-2
7.
1.
x 2 - 2x-, 2x\ 3; x.
2.
3x*-4:x + 3.
3.
1
4*'
4.
**-2 + i.
X 2
9. y = Vx.
10. y = x~$.
8.
ar+1
x
f 1
[Put # 2 = x, and apply the rules.]
8. Differentiation of a function of a function. Suppose that y,
instead of being given directly as a function of x, is expressed
as a function of another variable u, which is itself expressed
as a function of x. Let it be required to find the derivative
of y with regard to the independent variable x.
Let y =f(u), in which u is a function of x. When x changes
to the value au + Aaj, let u and y, under the given relations,
24 DIFFERENTIAL CALCULUS
change to the values u + Aw, y + A?/. Then
A?/ _ Aw Aw ,
Aa; — Aw Ax
hence, equating limits (Th. 5, Art. 3),
dy _dy da _ df(u) du
dx ~~ du dx~ du dx
This result may be stated as follows :
The derivative of a function of u with regard to x is equal to
the product of the derivative of the function with regard to w, and
the derivative of u with regard to x.
EXERCISES
1. Given v = 3u 2 -l, M = 3x 2 + 4; find ^-
dy du
du dx
dx du dx K '
2. Given ?/ =3m 2 -4u+ 5,« = 2x 3 -5; find ^ •
3. Given y = -,w = 5a; 2 -2x + 4; find ^ •
1 -r 3 3 _ , rfv
3 m 2 3 a: 3 aar
CHAPTER II
DIFFERENTIATION OF THE ELEMENTARY FORMS
dv
In recent articles, the meaning of the symbol -f was ex-
ctx
plained and illustrated ; and a method of expressing its value,
as a function of x, was exemplified, in cases in which y was a
simple algebraic function of x, by direct use of the definition.
This method is not always the most convenient one in the dif-
ferentiation of more complicated functions.
The present chapter will be devoted to the establishment of
some general rules of differentiation which will, in many cases,
save the trouble of going back to the definition.
The next five articles treat of the differentiation of algebraic
functions and of algebraic combinations of other differentiable
functions.
9. Differentiation of the product of a constant and a variable.
Let
y = cu,
Then
y + Ay = c(u + Au),
A?/ = c(m + Au) — cu = cAu,
Ay Au
Ax Ax'
therefore
dy du
dx~~ dx
Thus
d(cu) _ du
dx dx
25
(1^
26 DIFFERENTIAL CALCULUS
The derivative of the product of a coyistant and a variable is
equal to the constant multiplied by the derivative of the variable.
10. Differentiation of a sum.
Let 2/ == M_[_<y_ W ;_j_ ...
in which u. v, w, ••> are functions of x.
Then y + Ay = u + Au + v + Av — w — Aw + • • •,
Ay = Au + A?; — Aiv + • • •,
Ay _ Au ,Av_ Aw
Ax Ax Ax Ax '
dy _du dv dw
dx dx dx dx
Hence -f(u + v- w+ . ..)=f^+^-^+ .. (2)
doc doc doc doc
The derivative of the sum of a finite number of fractions is
equal to the sum of their derivatives.
Cor. If y = u + c, c being a constant, then
y + Ay = u + Au + c,
hence Ay = Au,
and dy = du
dx dx
This last equation asserts that all functions which differ
from each other only by an additive constant have the same
derivative.
Geometrically, the addition of a constant has the effect of
moving the curve y = u(x) parallel to the y-axis ; this opera-
tion will obviously not change the slope at points that have
the same x.
-c /rtN dy du , dc
From (2), -f- = — + — ;
dx dx dx
DIFFERENTIATION OF THE ELEMENTARY FORMS 27
but from the fourth equation above,
dy _du %
dx dx'
dc
hence, it follows that — = 0.
dx
The derivative of a constant is zero.
If the number of functions is infinite, Theorem 4 of Art. 3 may not
apply; that is, the limit of the sum may not be equal to the sum of
the limits, and hence the derivative of the sum may not be equal to
the sum of the derivatives. Thus the derivative of an infinite series
cannot always be found by differentiating it term by term.
11. Differentiation of a product.
Let y = uv, wherein u, v are both functions of x.
Then ^=( U + * U W + * V )- UV = u ^ + v^ + ^ . to,.
Ax Ax Ax Ax Ax
Now let Aa; approach zero, using Art. 3, Theorems 4, 5, and
noting that if — has a finite limit, then the limit of Avf—)
Ax \AxJ
is zero. '
The result may be written in the form
d(uv) = u dv + v du (3)
doc dx doc
The derivative of the product of two functions is equal to the
sum of the products of the first factor by the derivative of the sec-
ond, and the second factor by the derivative of the first.
This rule for differentiating a product of two functions may
be stated thus : Differentiate the product, treating the first
factor as constant, then treating the second factor as constant,
and add the two results.
28 DIFFERENTIAL CALCULUS
Cor. To find the derivative of the product of three functions
uvw.
Let y = uvw.
By (3), *y = w ±(uv)+uv^
dx dx dx
= w(u
dv du\ dw
dx dx ) dx
The result may be written in the form
d(uvw) =uv dw + vw du + wu dv (4
doc dx dm dx
By induction the following rule is at once derived :
The derivative of the product of any finite number of factors is
equal to the sum of the products obtained by midtiplying the de-
rivative of each factor by cdl the other factors.
12. Differentiation of a quotient.
Let y = - , u, v both being functions of x.
Then
u -f Au u Au Av
! V U
A?/ v -\- Av v _ Ax Ax
Ax ~~ Ax v(y + Av)
Passing to the limit, we obtain the result
v du- u dv
d (u\- dx dx (5)
dx\v J v 1
TJie derivative of a fraction, the quotient of two functions, is
equal to the denominator multiplied by the derivative of the nu-
merator minus the numerator multiplied by the derivative of the
denominator, divided by the square of the denominator.
DIFFERENTIATION OF THE ELEMENTARY FORMS 29
13. Differentiation of a commensurable power of a function.
Let y = u n , in which it is a function of x. Then there are
three cases to consider :
1. n a positive integer.
2. n a negative integer.
3. n a commensurable fraction.
1. n a positive integer.
This is a particular case of (4), the factors u, v, w, ••• all
being equal. Thus
dy n _ x du
dx dx
2. n a negative integer.
Let n = — m, in which m is a positive integer.
Then y = u n = u~ m = — ,
«* d l = ^- d l by (5), and Case (1)
hence
dx
u 2m
dx
— mu~ m ~
idu,
dx'
dy _
dx
n-l dtt
wit" — •
dx
3. 7i a commensurable fraction.
Let n=*-, where p, g are both integers, which may be either
• q
positive
or nega
tive.
p
Then
y = u n = u 9 ;
hence
if = fir,
and
i.e.
dec cte
30 DIFFERENTIAL CALCULUS
Solving for the required derivative, we obtain
dx V dx J
hence -*- U n = nu n - 1 4". (6)
dx, doc
The derivative of any commensurable power of a function is
equal to the exponent of the power multiplied by the power with
its exponent diminished by unity, multiplied by the derivative of
the function.
It should be noticed that Vu = it 2 ,
u
hence ±^=±*« > ±(1\ = = 1*».
dx 2^/ u dx dx\u) u 2 dx
These theorems will be found sufficient for the differentia-
tion of any function that involves only the operations of addi-
tion, subtraction, multiplication, division, and involution in
which the exponent is an integer or commensurable fraction.
The following examples will serve to illustrate the theo-
rems, and will show the combined application of the general
forms (1) to (6).
ILLUSTRATIVE EXAMPLES
t 3 x 1 - 2 ~ , dy
1. y = ; find -*-•
J x + 1 dx
(x + l)-f(3 *»- 2) - (3x3 - 2) L (x-r 1)
d / = ~ 7—TO ^ ^^
dx (x + l) 2
± (3^-2)=^ (3^)--f(2) (by 2)
dx dx dx
= 6x. (by 1,6)
f(si-l)=£? = l. (by 2)
ax dx
DIFFERENTIATION OF THE ELEMENTARY FORMS 31
Substitute these results in the expression for -f- . Then
dx
dy = (x + 1)6 x - (3 x 2 - 2) = 3 x* + 6 x + 2
. <fc (*+l) 2 (x + 1) 2
2. u = (3s 2 + 2)Vl + 5s 2 ; find ^?.
^ = (3 s 2 + 2) - VI + 5 s 2 + Vl + 5s 2 • — (3 s 2 + 2) . (by 3)
ds ds ds
— vTToT 2 = — (1 + 5 s 2 )*
</* ds .
= l(l+5s 2 )-^(l + 5s 2 ) (byG)
2 as
5s
Vl + 5 s 2
-^(3s 2 + 2>=6s. (by 6)
ds
Substitute these values in the expression for — Then
ds
du = 5j jS* + 2) , 6gVrTT72 = 45*»+.16»
rf* Vl + 5s 2 Vl~+5s 2
3 v= vr+^ + vr^g ;find $.
Vi + x 2 - vT^ 2 d*
First, as a quotient,
(VT+T 2 - VI - x *)^(VTTx* + VI^ 2 )
</a;
(Vl + z 2 -Vl-x 2 ) 2
(Vi + x 2 + vi - x 2 ) — ( vrr^ 2 - vi - x 2 )
, ^ , (by 5)
(Vl + z 2 - Vl-z 2 ) 2
— ( vr+T 2 + vr^) = -^ vrr^ 2 + — vn^ 2 . (by 2)
dx dx dx
•-f VIT^ = # (1 + * 2 )* = 1 (1 + * 2 ) _i -7- (1 + * 2 )- (by 6)
dx dx 2 dx
lL(l+x*)=2x. (by 2 and 6)
dx
32 DIFFERENTIAL CALCULUS
Similarly for the other terms. Combining the results, we have
dx x 3 { VI - X*J
Ex. 3 may also be worked by first rationalizing the denominator.
EXERCISES
Find the ^-derivatives of the following functions :
1.
y = x xo .
2.
y — x~ s .
3.
y = cVx.
4.
2 Vi
'' Vx 3
5.
y = v / ^'~ 5 -
6.
y = (x + a) n .
7.
y — x n -\- a n .
8.
X
y = .
14.
y = (2 a\ + x2) V a i + ^i.
15
y-i * r
|l + Vl-**J
16.
1.7-
:r n -f 1
v — — - —
S x n - 1
18
1 1
(a + x) m (b + x) w
19.
3x 3 + 2
v = — -
Va 2 - x* X ( x s + !)|
9 y== «jL?. 20. y = 3(x 2 + l)t(4x 2 -3).
x 2 + 2
21. y = 3 m" - 7.
10. 2/=(* + l)vW2. 22> y = 4w 3_ 6w2+1 o M _ 3 .
11. Va + a 23. y =(1-3 u 2 + 6u 4 )(l + u 2 ) 3 .
v « + V * 24. 7/ = us.
i2. y = JT±J 25 - $ =■ M * + 3 * w2 + * 4 -
1 ~ X " 26. y = "" .
13.,= *==. (« + *>•
x + Vl — x 2 27. y = t* 2 j? 8 io.
28. Given (a + a:) 5 = a 5 + 5 a 4 x + 10 « 3 -r 2 + 10 a 2 x 3 + 5 a* 4 + x* ;
find (a + x) A by differentiation.
29. Show that the slope of the tangent to the curve y = x 3 is never
negative. Show where the slope increases or decreases.
DIFFERENTIATION OF THE ELEMENTARY FORMS 33
30. Given b 2 x 2 + a 2 y 2 = d 2 b 2 , find d JL : (1) by differentiating as to
dx
x ; (2) by differentiating as to y ; (3) by solving for y and differen-
tiating as to x. Compare the results of the three methods.
31. Show that form (1), p. 25, is a special case of (3), p. 27.
32. At what point of the curve y 2 = ax 3 is the slope ? — 1 ? + 1 ?
33. Trace the curve y = x 3 + 3x 2 + x — 1.
34. y - - 3 ^ 2 + 7 and u = 5 x 2 - 1 ; find (J l .
V7 u 2 + 5 tf*
35. At what angle do the curves y 2 = \2x and y 2 + x 2 + ;r — G3 =
intersect ?
14. Differentiation of implicit functions. When a functional
relation between x and y cannot be readily solved for y, the
preceding rules may be applied directly to the implicit function.
The derivative will usually contain both x and ?/. Thus the
derivative of an algebraic function, defined by equating a poly-
nomial in x and y to zero, may be obtained by the process illus-
trated in the following examples :
Ex. 1. Given the function y of x, defined by the equation
x 5 4- y r ° — 5 xy + 1 = 0,
find^.
dx
Since
dx
- 5 xy + 1) =
0,
hence
5x 4 + 5</^-5
. dx
y -
r d
- 5 x ,- =
dx
o,
G>:
f2,
3)
Solving for -*. , we obtain
dx
<]y _
dx
•'■ 4
x -
-7/
Ex. 2.
xy 2 -\- x 2 y = 1. Find
dy
dx
Ex. 3.
*+ y + (x-y) 2 +(2
x —
3 2/) 3 = 0.
Find
EL. CALC. — 3
34 DIFFERENTIAL CALCULUS
15. Elementary transcendental functions. The following func-
tions are called transcendental functions :
Simple exponential functions, consisting of a constant
number raised to a power whose exponent is variable, as
4 X , a* ;
the logarithmic functions, as log a x, log 6 u ;
the incommensurable powers of a variable, as x^ 2 , u n ;
the trigonometric functions, as sin u, cos u ;
the inverse trigonometric functions, as sin -1 u, tan -1 x.
There are still other transcendental functions, but they will
not be considered in this book.
The next four articles treat of the logarithmic, the two ex-
ponential functions, and the incommensurable power.
16. Differentiation of log a od and log a u.
Let
y = log a x.
Phen
y-\-Ay = log a (x + Ax),
Ay log a (x + Ax) — log a x
Ax Ax
Lio g /*_±M
iX \ X J
Ax
For convenience writing h for Ax, and rearranging, we obtain
Ay
Ax
_l.«Wl+»)
x h \ xj
H-;»»K'+;)i-
-)
DIFFERENTIATION OF THE ELEMENTARY FORMS 35
SB
f h\*
To evaluate the expression ( 1 + - ) when h = 0, expand it
by the binomial theorem, supposing - to be a positive inte-
ger m.
The expansion may be written
\ mj m 1-2 m 2 1-2-3 m 3
which can be put in the form
\ mj 1 2 ^1 2 3
1 2
Now as m becomes very large, the terms — , — , ••• become
in m
very small and m increases without limit as h approaches zero.
As m = cc the series approaches the limit
1 + 1+— + — + — + •••,
2 ! 3 ! 4 !
which will be discussed later.
The numerical value of this limit can be readily calculated
to any desired approximation. This number is an important
constant, which is denoted by the letter e, and is equal to
2.7182818-..; thus
lim A + IV = e = 2.7182818-...*
* This method of obtaining e is rather too brief to be rigorous ; it assumes
that — is a positive integer, but that is equivalent to restricting Ax to ap-
Ax
proach zero in a particular way. It also applies the theorems of limits to the
sum and product of an infinite number of terms. The proof is completed on
p. 315 of McMahon and Snyder's " Differential Calculus,"
36 DIFFERENTIAL CALCULUS
The number e is known as the natural or Naperian base;
and logarithms to this base are called natural or Naperian log-
arithms. Natural logarithms will be written without a sub-
script, as logcc; for other bases a subscript, as in \og a x, will
generally be used to designate the base. The logarithm of e to
any base a is called the modulus of the system whose base is a.
X
li / 7A*
If the value , ™ f 1 + - ) = e is substituted in the expres-
sion for _?, the result is
dx
dy 1 ,
/ = - • log, e.
ax x
More generally, by Art. 8,
d , logffle du f ms
s log„ M = - s -^. (7)
In the particular case in which a = e,
d . _ 1 du /ox
doc ~~ u dx
The derivative of the logarithm of a function is the product of
the derivative of the function and the modulus of the system of
logarithms, divided by the function.
17. Differentiation of the simple exponential function.
Let / y = a u .
Then log y = u log a.
Differentiating both members of this identity as to x, we obtain
1 dy , du ,, oX
dy , du
therefore -j- <*> u = l«g a • a ' 1 " ^ ■ ^
DIFFERENTIATION OF THE ELEMENTARY FORMS 37
In the particular case in which a = e,
d du , + n x
as •"=•"• a*- (10)
The derivative of an exponential function with a constant base
»
is equal to the product of the function, the natural logarithm of
the base, and the derivative of the exponent.
18. Differentiation of an incommensurable power.
Let y = u n ,
in which n is an incommensurable constant. Then
log y = n log u,
1 dy _ n du
ydx u dx'
dy_ y du
dx u dx'
d _ __, du
dx dx
This has the same form as (6), so that the qualifying word
" commensurable " of Art. 13 can now be omitted.
EXERCISES
Find the x derivatives of the following functions
1.
y-
= log(x+ a).
2.
y-
= log (ax + b).
3.
y =
= log (4 x 2 - 7 x
4-2)
4.
y =
= logj + *.
1 — X
5.
2/ =
1 — X 2
6.
y =
= x log X.
7.
y =
- x n log X.
8.
y =
- x n log x m . N
9. y = log Vl — X 2 .
10. y = Vx — log ( y/x + 1) .
11. y = \og a (3x 2 -V2 + x).
12. y = log 10 (x 2 + 7 x).
13. y = log x a.
14. y = e xa .
154 y = e 4 *+ 5 .
l
16. y=-e l+x .
38
DIFFERENTIAL CALCULUS
17. y = — - — .
18. y = e x (l -z«).
-X . — X
19. y
23.
24.
log
Vq -f Vx
e* + e~ x
20. y = log (e* - c-*).
21. y = log (x + e x ).
22. v = x n a*.
Va
1
logx'
25. 3/= (log*)*.
26. y = log (log a).
27. y
28. y = a l °s
x log - .
x
The following functions can be easily differentiated by first taking
the logarithms of both members of the equations.
31. ¥ =*&+*!.
32. y = x\a -f 3 x)\a - 2 x) 2 .
33. y = V (*+ g )'.
Vx — a
29.
(*-*)■
(a;-2)*(a;-3)*
30. y = xVl - x(l + a:).
19. Limit of
sin
as 9 approaches 0. Before proceeding to
determine the derivatives of the trigonometric functions it is
necessary to prove the following lemma :
lim sin
= 1.
=
With as a center and OA = r
as radius, describe the circular
arc AB. Let the tangent at A
-4 meet OB produced in D; draw
Fig. 7 BC perpendicular to OA, cutting
OA in C. Let the angle OAB = in radian measure,
then arc AB = r#,
(75 < arc AB < .4Z), by geometry
i.e. r sin < r6 < r tan 0,
. sin 6 < < tan 0.
DIFFERENTIATION OF THE ELEMENTARY FORMS 39
By dividing each member of these inequalities by sin 0,
1 <^-<sec0;
sin
but sec = 1, when =
sin
, lim ., A lim sin -,
hence > 0=o^ = 1 > and »=o-vr = 1
20. Differentiation of sin ?/.
Let y = sin u.
,p. Ay _ sin (u 4- A?/) — sin w Ait
A./ 1 Aw A/
To evaluate the expression
sin (u + Aw) — sin w,
we make use of the formulas for the sine of the sum and the
sine of the difference of two angles. Since
sin (a + b) = sin a cos b 4- cos a sin b,
sin (a — b) = sin a cos 6 — cos a sin 6,
hence, by subtracting the second equation from the first,
sin (a + b) — sin (a — b) = 2 cos a sin b.
This equation is true for all values of a and of b. In particu-
lar, then, putting
' ' F 5 a + 6 = w + Aw,
and a — b = u,
that is, a = u + — , and 6 = — ,
we obtain
sin (w + Aw) — sin w = 2 cos (u-\ — - ) sin — .
40
DIFFERENTIAL CALCULUS
The expression for — J. may now be written in the form
Ax
Aw
sm —
A?/ / Au\ 2 A?*
— ^ = cos [u + — — - ,
Ax V 2 J Au Ax
hence
dy ..
-f- = cos u • li m
dx Aw =
sm
2
Aw'
Aw
2~
c?w
dx~'
hence, by Art. 19, ^- sin w = cos u —
rfx doc
(ID
77>e derivative of the sine of a function is equal to the product
of the cosine of the function and the derivative of the function.
21. Differentiation of cos u.
Let ?/ = cos u — sinf - — u \
bl
dx dx
du
-w)-^( -u\
dx\2
d
— i
doc
u = — sinu
doc
(12)
The derivative of the cosine of a function is equal to minus the
product of the sine of the function and the derivative of the function.
22. Differentiation of tan u.
Let
Then
y = tan u
sin?*
cos?*
d . d
cos u • — sin u — sin u • — cos u
dy dx
dx
- (bv 5)
dx ~~ cos- u
• o du . • Q . du
cos J u h sm- U —
dx dx
du
dx
(by 11, 12)
cos 2 u
cos 2 u
that is,
^tant* = sec 2 t*^
doc doc
(13)
DIFFERENTIATION OF THE ELEMENTARY FORMS 41
The derivative of the tangent of a function is equal to the
product of the square of the secant of the function and the deriva-
tive of the function.
Since the remaining elementary trigonometric functions
can be expressed as rational functions of those already con-
sidered, their derivatives can be obtained by means of the
preceding rules. The results are
4-wtu= -csc 2 t*^. (14)
dx dx
~ sec u = sec u tan u — • (15)
dx dx
lL esc u = - esc w cot w — (16)
dx dx
EXERCISES
Find the ^-derivatives of the following functions :
1. y — sin 7 x.
2. y — cos 5 x
3. y = sin x 2 .
4. y = sin 2 x cos x.
5. y = sin 8 x.
6. y = sin 5 x 2 .
7. y = sin 2 7 x.
8. y — \ tan 3 x — tan x.
9. y = sin 3 x cos x.
10. ?/ = tan x + sec x.
11. y = sin 2 (1-2 x 2 ) 2 .
12. # = tan (3 -5 a; 2 ) 2 .
13. y — tan 2 a: — log (sec 2 a;).
14. y =log tan ($# + Jir).
■ 15. y = logsinVx. 29. ?/ — tan (x -f y) .
16.
* i
y = tan a*.
17.
y = sin narsin n a;.
18.
y = sin (w + b) cos (w — J).
19.
sin m nx
y = .
cos n mx
20.
y = x + log cos ( x - j J.
21.
?/ = sin (sin t/).
22.
?/ = sin 2 ^.
23.
y = sin e x • log a\
24.
?/ = Vsin x 2 .
25.
y = esc 2 4 x.
26.
y — sec(4 x — 3) 2 .
27.
r/ = cot x 2 -f sec Var.
28.
y ~ sin a:?/.
42 DIFFERENTIAL CALCULUS
30. Find — (cos u) directly from the definition of the derivative.
dx
Also — (tan u).
dx
31. Find — (cos u) from the relation sin 2 u + cos 2 u = 1.
dx
23. Differentiation of sin- 1 u-
Let y = sin~ 1 u.
Then sin y = u,
and, by differentiating both members of this identity,
dy ,du
cos y — = — ;
d# dx
hence
dy _ 1 fZw 1 d?t.
da; cos ?/ d.r _j_ Vl— sin 2 y dx '
d • _i 1 dw
i.e. — - sm -1 it = ±
d# VI _ tf d^
The ambiguity of sign accords with the fact that sin -1 u is a
many- valued function of u, since, for any value of u between
— 1 and 1, there is a»series of angles whose sine is u : and, when u
receives an increase, some of these angles increase and some
decrease; hence, for some of them, sm — - is positive, and
du
for some negative. It will be seen that, when sin -1 u lies in
the first or fourth quarter, it increases with u, and, when in
the second or third, it decreases as u increases. Hence, for the
angles of the first and fourth quarters,
jL Sin- 1 !* = --f- COS" 1 !*:= + ■— 1— gg. (17)
€loc doc VI — «*2 dor
In the other quarters the minus sign is to be used before
the radical.
DIFFERENTIATION OF THE ELEMENTARY FORMS 43
The derivatives of the other inverse trigonometric functions
can be easily obtained by the method employed in the present
article. The most important of the remaining ones are tan -1 u,
sec -1 u ;
4- tan" 1 u = - 4- cot 1 u = -L-^.
doc ddc l + u 2 doc
x sin -1 a;.
2. y = cos -1 Vl — x' 2 . 17 # y
3. y = sin-i (3 x - 1). 18. y = e**" 1 *.
4. y = sin- 1 (3x -4 x 3 ). 1Q _., 1
1 — x 2
5. y = sm-i- — -•
1 + r
6. y = vsin -1 x.
1. y — tan -1 ?*.
8. y = cos -1 log x.
9. y = sin -1 (tan a-).
10. y = sec -1 —
VI -x 2
11. # = CSC -1 -•
a:
12. y^tan^f 1
V Va: 2 - 1
13. y = tan- 1 -*
14. y = sin" 1 Vsin x.
15. y = tan- 1 V^^
1 -f cos
(18)
-^-sec ~ 1 u = --^-csc- 1 m = !__*•. (19)
das dx u^u 2 —l (lQC
EXERCISES
Find the ar-derivatives of each of the following functions :
1. y = sin -1 2 x 2 . 16. y = tan a; • tan -1 x.
2 a; 2 - 1
20.
y = sec" 1 *L±i.
^ a; 2 - 1
21.
y-tan- lV ^+ V «.
1 — V«ar
22.
y = cos x
23.
y = tan -1 (n tan a).
24.
y = cos -1 (cos 2 a-).
25.
y = cos -1 (2 cos a:).
26.
y = tan -1 (Vl + x 2 -
X).
27.
y = 2 tan" 1 aL— '
M + a:
28.
y = tan -1 — |- tan
iV3
.^b-x
a:V3
44 DIFFERENTIAL CALCULUS
24. Table of fundamental forms.
d(cu) = c du . (1)
dx dx
Jt {u + V -. w)= du+dv_dw. 2
dx dx dx dx
*&»1 = u^ + v^ (3)
dx dx dx
d -(uvw) = uv m + vw m + wu <M. (4)
dx dx dx dx
v du_ u dv
d u _ dx dx
dx v »,2
(5)
^a w =loga-a u ^ (9)
da? da?
A e « =e ud ^ (10)
<ia? da?
-f-sinw =costi~> (11)
<?a? da?
^-cosi* =-smu dw . (12)
c?a? da?
^tanw =sec*tt^. (13)
da? da?
-£- cot ft = - esc 2 u d ^ (14)
da? da?
DIFFERENTIATION OF THE ELEMENTARY FORMS 45
4- sec u = sec u tan u ^- (15)
doc doc
^cscu = -cscMcotw~ ( 16 )
doc doc
JL sin l u=-4- COS" 1 M = 1 *L (17)
«a? «a? Vi _ ^2 «a?
4- tan" 1 t* = - # cot 1 w = —1— *!. (18)
dx doc i + w 2 da?
4- sec 1 u=-~- esc 1 t* = -J ^* (19)
doc doc u y/u* -i d &
EXERCISES ON CHAPTER II
Find the ^-derivatives of the following functions :
1. y = 3 Z 2 + 5 a,- 3 - 7.
o 3 5 1
3. y = (x + 5) Vx - 3.
4. ?/ = xVa' 2 — x 2 .
5. ?/ = x log sinx.
a
10.
y = log-- .
a*
11.
1-x 2
y =
Vl + x 2
12.
y = e x cos x.
13.
, = 00^(1).
14.
. , 4 sin x
?/ = tan -1
3 + 5 cos x
15
v — ( x 4- rtH.an- 1 -!/-
6 - y=-V«2_ x 2.
( r 2 ?
> 7. y= c - ec-
x
8. J ,= ta„2z,z = t.n-i(2»-l). ^
. 16. w = cot x — — —
2 9. y = e Vu , u = log sin x. a;
17. ?/ = tan 4 x — 2 tan 2 x +.log(sec* x).
18. y = £l2«£+log(l- a; ).
19. y = C0S-i 3 + 5c0S:C .
5 + 3 cos x
46 DIFFERENTIAL CALCULUS
20. y.= log (l±£^-|tan-i*.
21. # = log (x + Vx 2 - a 2 ) + sec" 1 --
a
22. y = e u , u — log x. 25. x 2 ?/ 2 + x 8 + y 3 = 0.
23. y = log s 2 + e", s = sec x. 26. x 3 -f x = y + y 8 .
24. x s + y s — 3 axy = 0. 27. xy 2 + x 2 # — x-\-y.
28. y = sin (2 u - 7), u = log x 2 .
29. By means of differentiation eliminate the constant p from the
equation y =px 2 .
30. At what points is the tangent to the curve y = cos x parallel to
the x-axis ?
31. Show that the x-derivative of tan -1 \ ~~ cos x is not a func-
* 1 + cos X
tion of x.
x 2 ?/ 2
32. Find at what points of the ellipse h *— — 1 the tangents cut
_, . a 2 Z> 2
off equal intercepts on the axes.
33. Find the points at which the slope of the curve y = tan x is
twice that of the line y = x.
34. Find the angle which the curves y = sin x and y = cos x make
with each other at their point of intersection.
CHAPTER III
SUCCESSIVE DIFFERENTIATION
25. Definition of nth derivative. When a given function
y~<f>(x) is differentiated with regard to x by the rules of
Chapter I, then the result d y
is a new function of x which may itself be differentiated by the
same rules. Thus, 7 , , . ,
dx\dxj dx
The left-hand member is usually abbreviated to _J?, and the
dx 2
right-hand member to <f>"(x) ; that is,
Differentiating again and using a similar notation, we obtain
±(cPy\_d*i_
dx\dx 2 J dx 3
and so on for any number of differentiations. Thus the sym-
d 2 y
bol -r^ expresses that y is to be differentiated with regard to x,
and that the resulting derivative is then to be differentiated.
d s y . . d
Similarly, -^ indicates the performance of the operation —
three times, j-( j~(;p ))• ^ n general, the symbol -~ means
that y is to be differentiated n times in succession with regard
to x.
47
48 DIFFERENTIAL CALCULUS
Ex. 1. If y = x 4 + sin 2 x,
-£ = 4 x 8 + 2 cos 2 x,
ufx
g = 12x 2 -4sin2x,
?^ = 24x-8cos2x,
«x 3
^ = 24 + 16sin2x. ■
dx 4
If an implicit equation between x and y is given and the
derivatives of y with regard to x are required, it is not neces-
sary to solve the equation for either variable before perform-
ing the differentiation.
Ex. 2. Given x 4 + y 4 + 4 a 2 xy = ; find v^ .
dx' 2
^-(x* + y*+±a 2 xy)=0,
4 x 8 + 4 ?/ 3 ^ + 4 a 2 * ^ + 4 a 2 # = 0.
The last equation is now to be solved for -j-,
dy x s + a 2 y
dx y s + a 2 x
Differentiating again, we obtain
(i)
d 2 y _ d fx 8 + a 2 y\
dx 2 ~ dx\y 8 + a 2 x]
w
(y 8 + a 2 x) — (x 3 + a 2 y) - (x 8 + a*y) JL (y 8 + a 2 x)
dx ' dx
(y T T~a 2 7y 2
(y 8 + a 2 x) ('dx 2 + a 2 '- 1 !') - (x 8 + a 2 ?/) fsv 2 ^ + a")
V '/x/ " V ax /
(*/ 3 + d 2 x) 2
SUCCESSIVE DIFFERENTIATION 49
The value of JL from (1) is now to be substituted in the last
dx
equation, and the resulting expression simplified. The final form
may be written :
dhj 2 a*xy - 1 a 2 x s y s - a 4 O 4 + y 4 ) - 3 x 2 y 2 O 4 + # 4 )
dx 2 ~ (y 8 + a 2 x) 8
In like manner higher derivatives may be found.
26. Expression for the nth derivative in certain cases. For cer-
tain functions, a general expression for the nth derivative can
be readily obtained in terms of n.
where n is any positive integer. If y = e ax , ~- = a n e ax .
Ex. 2.
If
y '= sin x,
dy 1 tt\
^|=cos:r = sin^+^j,
g=cos(x + |) = sin(x+^),
6/ n v . / n7r\
If
y~-
- sin ax.
d n y 1 mr\
d^ = aSln [ ax+ 2-)-
EXERCISES ON CHAPTER III
1. y=3x*+5x2+3x-9; find^. 5. ij = tana:; find ^l.
dx 3 J dx 8
2. y = 2x 2 + Zx + 5; find v|. 6. y = e*logz; find ^X
3 ' ^ = * ; fmd S' 7."y = *«loga:j find g.
4. y = ^ _I 2 ; find g. 8. y = sec**; find g.
EL. CALC. — 4
50 DIFFERENTIAL CALCULUS
9. y = logsina:; find g. 18. 3^^-p find J*.
10. y = sin x cos a; ; find -r\> 19. y z= cos ma: ; find ~^-
U. , = ^L ; flnd % 20. y = , X . ; find £».
12. y = ** logs'; find ^. 21. y = log (a + *)"; find J. ,
13. y = sin,; find g. 22. y 2 = 2^; find g-
14. y = log (* + *-) ; find g • 23. J + £ = 1 ; find g.
15. y = (^- 3^ + 3)^; find g. 24. a* + y 3 = 3axy; find g- t
16. 3/ = a; 4 log a;; find g. 25. «■+» = zy; find g.
17. y = e ax I find g- 26. y = 1 + xe* ; find ^-
d u (in
27. y = «*sina;; prove ^J,-2-^+2y = 0.
28. y = aa;sina;; prove x 2 ~ - 2a; -£ + 2 + 2)y = 0.
d 2 y
29. y = ax n + 1 + bx~ n \ prove a; 2 ^J> = n(n + l)y.
30. y =(sin-^)2; prove (1 - z 2 ) ^ 2 - X-+ = 2.
»■» = ££ P^g- 1 -*
_ , d n y
32. y = x n_1 loga:; find -=-£•
33. „ = *.; prove ^ = 2^-^ + 9«-.
34. y = cos 2 a:; find -^.
CHAPTER IV
MAXIMA AND MINIMA
27. Increasing and decreasing functions. A function is said
to be increasing if it increases as the variable increases and
decreases as the variable decreases. A function is said to be
decreasing if it decreases as the variable increases and increases
as the variable decreases. When the graph of the function is
known it will indicate whether the function is increasing or
decreasing for an assigned value of x ; conversely, a knowledge
of the fact whether a function is increasing or decreasing is of
great assistance in drawing the graph. Usually a function is
increasing for certain values of x and decreasing for others.
28. Test for determining intervals of increasing and decreasing.
Let y — 4>(x) be a continuous function having a derivative for
all values of x from a to b. By the above definition y is in-
creasing or decreasing at a point x l , according as
k = Qfa + h) - <£(zi)
has or has not the same sign as h, where h is a sufficiently
small number. Hence <f>(x) is an increasing or a decreasing
function at the value x x according as
is positive or negative.
51
52
DIFFERENTIAL CALCULUS
Thus, the function y = <f>(x) is increasing, if <f>'(x) is positive ;
if 4>'(x) is negative, the function is decreasing.
In order that a function shall change from an increasing
function to a decreasing function or vice versa, it is necessary
and sufficient that its derivative shall change sign. If the
derivative is continuous, this can happen only when the deriva-
tive passes through the value zero. The derivative may also
change sign when it becomes infinite, and, notwithstanding
this discontinuity of the derivative, the original function may
still be continuous. In the graph of the function this requires
that at such a point the tangent to the locus shall be parallel to
the y-axis. The process will be illustrated by a few examples.
Ex. Find the intervals in which the function
<f>(x) = 2 a; 8 - 9a; 2 + 12a; - 6
is increasing or decreasing. The derivative is
<f>'(x) = 6a; 2 - 18 a; + 12 = Q(x - \){x - 2);
hence, as x passes from -co to 1, the derived function <f>'(x) is posi-
tive and <f>(x) increases from <£( — oo)
to <£(1), i>e. from <f>=—ccto<f>= — 1;
as x passes from 1 to 2, <f>'(x) is nega-
tive, and <f>(x) decreases from <£(1) to
<£(2), i.e. from — 1 to - 2; and as x
passes from 2 to + co, <£'(*) is posi-
tive, and <f>(x) increases from <£(2) to
<£(ao), i.e. from — 2 to + oo . The
locus of the equation y = <£(a;) is shown
in Fig. 8. At points where <f>'(x) = 0,
the function <f>(x) is neither increas-
ing nor decreasing. At such points
the tangent is parallel to the axis of x. Thus in this illustration, at
x = 1, x = 2, the tangent is parallel to the x-axis.
Fig. 8
MAXIMA AND MINIMA 53
EXERCISES
1. Find the intervals of increasing and decreasing for the function
<f>(x) = x s + 2x 2 + x - 4.
Here <£'(*) = 3x 2 +4.r+l = (3z+ l)(x + 1).
The function increases from x = — coto x = — 1 ; decreases from
a: = — ltoar=-— |; increases from a: = — \ to a: = go .
2. Find the intervals of increasing and decreasing for the function
y = X s - 2 x 2 + x - 4,
and show where the curve is parallel to the a>axis.
3. At how many points can the slope of the tangent to the curve
# = 2x 3 -3x 2 + l
be 1 ? - 1 ? Find the points.
4. Compute the angle at which the following curves intersect ;
y = 3 x 2 - 1, y = 2 x 2 + 3.
29. Turning values of a function. It follows that the values
of x at which <f> (x) ceases to increase and begins to decrease
are those at which <j>'(x) changes sign from positive to nega-
tive ; and that the values of x at which <f> (x) ceases to decrease
and begins to increase are those at which <f>'(x) changes its
sign from negative to positive. In the former case, <£(#) is
said to pass through a maximum, in the latter, a minimum,
value.
Ex. 1. Find the turning values of the function
<£0) = 2 x s - 3 x 2 - 12 x + 4,
and exhibit the mode of variation of the function by sketching the
curve , , x
V = <K X )-
Here <f>'(x)=6x 2 - Qx- 12 = 6(z + 1)0 - 2),
54
DIFFERENTIAL CALCULUS
Fig. 9
9, - 16, - 5, 36, oo.
hence <j>'(x) is negative when x lies between - 1 and
+ 2, and positive for all other values of x. Thus <f)(x)
increases from x = - oo to x = - 1 ; decreases from
{ x =— 1 to a: = 2; and increases from x = 2 to x = oo.
Hence </>(- 1) is a maximum value of <f>(x), and
<f>(2) a minimum.
The general form of the curve y = <f>(x) (Fig. 9)
may be inferred from the last statement, and from
the following simultaneous values of x and y :
x = - oo, - 2, - 1, 0, 1, 2, 3, 4, oo.
y = -co, 0, 11, 4,
Ex. 2. Exhibit the variation of the
function <f>(x) = (x—l)~$+ 2,
especially its turning values.
Since <f>'(x) = - -,
6 (x - 1)*
hence <f>'(x) changes sign at x = l,
being negative when x < 1, infinite
if x = 1, and positive if z> 1. Thus
<f>(l) = 2 is a minimum turning value
of tf>(x). The graph of the function
is as shown in Fig. 10, with a vertical tangent at the point (1, 2).
Ex. 3. Examine for maxima and minima the function
Fig. 10
<f>(x) = (x
1
iy + 1,
Here <f>'(x) =
3 0-l)§
Fie. 11
hence cf>'(x) never changes sign, but is always
positive. There is accordingly no turning-
value. The curve y =' <f>(x) has a verti-
cal tangent at the point (1, 1), since -^
dx
is infinite when x = l. (Fig. 11.)
MAXIMA AND MINIMA 55
30. Critical values of the variable. It has been shown that
the necessary and sufficient condition for a turning value of
<f>(x) is that <f>'(x) shall change its sign. Now a function
can change its sign only when it passes through zero, as in
Ex. 1 (Art. 29), or when its reciprocal passes through zero,
as in Ex. 2. In the latter case it is usual to say that the
function passes through infinity. It is not true, conversely,
that a function always changes its sign in passing through
zero or infinity, e.g. x 2 and x~ 2 .
Nevertheless all the values of x, at which <j>'(x) passes
through zero or infinity, are called critical values of x, be-
cause they are to be further examined to determine whether
<f>'(x) actually changes sign as x passes through each such
value ; and whether, in consequence, <f> (x) passes through a
turning value.
For instance, in Ex. 1, the derivative <f)'(x) vanishes when
x = — 1, and when x = 2, and it does not become infinite for
any finite value of x. Thus the critical values are — 1, 2,
both of which give turning values to <f> (x). Again, in
Exs. 2, 3, the critical value is x = 1, since it makes <f>'(x)
infinite ; it gives a turning value to <£ (x) in Ex. 2, but not
in Ex. 3.
31. Method of determining whether <f>'(&) changes its sign in
passing through zero or infinity. Let a be a critical value of x;
in other words, let <£'(a) be either zero or infinite, and let h
be a very small positive number, so that a — h and a + h are
two numbers very close to a, and on opposite sides of it. In
order to determine whether <f>'(x) changes sign as x increases
through the value a, it is necessary only to compare the signs
of <f>'(a -f- h) and <f>'(a — h). If it is possible to take h so
56
DIFFERENTIAL CALCULUS
small that cf>'(a —h) is positive and <f>'(a + h) negative, then
cf>'(x) changes sign as x passes through the value a, and
<f>(x) passes through a maximum value <f>(a). Similarly, if
<j>\a — h) is negative and <f>'(a + h) positive, then <f>(x) passes
through a minimum value <£ (a).
If <f>'(a — h) and <j>'(a + h) have the same sign, however
small h may be, then <f> (a) is not a turning value of <£ (x).
Ex. Find the turning values of the function
4>0) = 0-i) 2 + i) 8 .
Here <f>'(x) = 2(x- 1)0 + 1) 8 + S(x - l) 2 (x + l) 2
= (a:-l)(x+ 1) 2 (5 x- 1).
Hence <j>'(x) becomes zero at x = — 1, \, and 1 ; it does not become
infinite for any finite value of x.
Thus, the critical values are — 1, |, 1.
Fia. 12
When x has any value less than — 1, the three factors of <f>'(x)
take the signs — + — , hence <f>'(x) is +, and when x has a value
MAXIMA AND MINIMA 57
between — 1 and .J they become — + — , and <f>'(x) is still + ; hence
<f>(— 1) = is not a turning value of <f>(x).
When x has any value between \ and 1, the signs are — + + and
<f>'(x) is — ; hence <f>($) is a maximum.
Finally, if x has any value greater than 1, the signs are + + +;
hence <f>'(x) changes sign from — to + as x increases through 1, and
<£(1) = is a minimum value of <j>(x).
The general march of the function may be exhibited graphically
by tracing the curve y = <{>(x) (Fig. 12), using the foregoing results
and observing the following simultaneous values of x and y :
x^-co, - 2, - 1, 0, \, 1, 2, go.
y = - oo, - 9, 0, 1, 1 • 1 ••-, 0, 27, oo.
32. Second method of determining whether <!>'(») changes sign in
passing through zero. The following method may be employed
when the function and its derivatives are continuous in the
vicinity of the critical value x = a.
Suppose, when x increases through the value a, that <j>'( x )
changes sign from positive through zero to negative. Its
change from positive to zero is a decrease, and so is the change
from zero to negative ; thus <f>'(x) is a decreasing function at
x = a, and hence its derivative <£%e) is negative at x= a.
On the other hand, if <f>'( x ) changes sign from negative
through zero to positive, it is an increasing function and <f>"(x)
is positive at x = a ; hence :
TJie function cf>(x) has a maximum value <f>(a), when <f>'(a) =
and <f>"(a) is negative; $(x) has a minimum value <f>(a), when
cf>'(a) = and <f>"(a) is positive.
It may happen, however, that <f>"(a) is also zero.
. To determine in this case whether <£(#) has a turning value,
it is necessary to proceed to the higher derivatives. If <f>(x) is
4
58 DIFFERENTIAL CALCULUS
a maximum, <j>"(x) is negative just before vanishing, and
negative just after, for the reason given above ; but the change
from negative to zero is an increase, and the change from zero
to negative is a decrease ; thus <f>"(x) changes from increasing
to decreasing as x passes through a. Hence (f>'"(x) changes
sign from positive through zero to negative, and it follows, as
before, that its derivative <f> IV (x) is negative.
Thus <f>(a) is a maximum value of <f>(x) if <t>'(a) = 0, <f>"(a) = 0,
<f>'"(a) = 0, <j> JV (a) negative. Similarly, <f>(a) is a minimum
value of <f>(x) if *'(«) = 0, <f>"(a) = 0, <£'"(a) = 0, and <£ lv (a)
positive.
If it happens that <f> lv (a) = 0, it is necessary to proceed to
still higher derivatives to test for turning values. The result
may then be generalized as follows :
The function <£(#) has a maximum (or minimum) value at
x = a if one or more of the derivatives <£'(a), <f>"(a), <f>'"(a) vanish
and if the first one that does not vanish is of even order, and
negative (or positive).
Ex. Find the critical values in the example of Art. 31 by the
second method.
<f>"(x) = (x+iy(5x-l)+2(x-l)(x+l)(ox-l) + 5(x-l)(x + l)' i
= 4(5a: 8 + 3a: 2 -3x-l),
<£"(1) = 16, hence <£(1) i s a minimum value of <f>(x)',
<f>"(— 1) = 0, hence it is necessary to find <£'"( — 1) ;
<£'"(V) = 12(5 a: 2 + 2 a; - 1),
<£'"( — i.) — 24, hence <£(— 1) is neither a maximum nor a minimum
value of <j>(x).
Again, <£"(£) = 5(J - l)(i 4- l) 2 is negative, hence <£(£) is a maxi-
mum value of <£(V).
MAXIMA AND MINIMA 59
33. The maxima and minima of any continuous function occur
alternately. It has been seen that the maximum and minimum
values of a rational polynomial occur alternately when the
variable is continually increased, or diminished.
This principle is true also in the case of every continuous
function of a single variable. For, let <£(a), <f>(b) be two
maximum values of <f>(x), iu which a is supposed less than
b. Then, when x = a + h, the function is decreasing ; when
x = b — h, the function is increasing, h being taken sufficiently
small and positive. But in passing from a decreasing to an
increasing state, a continuous function must, at some inter-
mediate value of x, change from decreasing to increasing, that
is, must pass through a minimum. Hence, between two maxima
there must be at least one minimum.
It can be similarly proved that between two minima there
must be at least one maximum.
34. Simplifications that do not alter critical values. The work
of finding the critical values of the variable, in the case of any
given function, may often be simplified by means of the follow-
ing self-evident principles.
1. When c is independent of x, any value of x that gives a
turning value to c<jy(x), gives a turning value to <f>(x) also; and
conversely. These two turning values are of the same or
opposite kind according as c is positive or negative.
2. Any value of x that gives a turning value to c + <f>(x) gives
a turning value of the same kind to <f>(x) also ; and conversely.
3. When n is independent of x, any value of x that gives a
turning value to [<£(aj)] B gives a turning value to <f>(x) also;
and conversely. These turning values are of the same or
opposite kind according as ?i[<£(a,*)] n_1 is positive or negative.
60 DIFFERENTIAL CALCULUS
EXERCISES
Find the critical values of x in the following functions, determine
the nature of the function at each, and obtain the graph of the function.
1. M = X(x 2 — 1).
2. u = 2 x 3 - 15 x 2 + 36 x - 4.
3. u = (x - l) 3 (x - 2) 2 .
4. u = sin x + cos x.
5. u =Q-*) 3 .
a - 2x
7.
u = 5 + 12 x - 3 x 1 - 2 x 8 .
8.
logx
X
9.
m = sin 2 x cos 3 x.
10.
^ _ x 2 - X + 1
X 2 + X - 1
11.
w _(x + 3)(x+l)
(x - l)(x-2)
6. w = x(x + l) 2 - 5.
12. Show that a quadratic integral function always has one maxi-
mum, or one minimum, but never both.
13. Show that a cubic integral function has in general both a
maximum and a minimum value, but may have neither.
14. Show that the function (x — b)* has neither a maximum nor
a minimum value.
35. Geometric problems in maxima and minima. The theory
of the turning values of a function has important applications
in solving problems concerning geometric maxima or minima,
i.e. the determination of the largest or the smallest value a
magnitude may have while satisfying certain stated geometric
conditions.
The first step is to express the magnitude in question
algebraically. Jf the resulting expression contains more than
one variable, the stated conditions will furnish enough relations
between these variables, so that all the others may be expressed
in terms of one. The expression to be maximized or minimized,
being thus made a function of a single variable, can be treated
by the preceding rules.
MAXIMA AND MINIMA 61
Ex. 1. Find the largest rectangle whose perimeter is 100. Let x,
y denote the dimensions of any of the rectangles whose perimeter is
100. The expression to be maximized is the area
u = xy, (1)
in which the variables x, y are subject to the stated condition
2x + 2r/ = 100,
i.e. y =50 -x; (2)
hence the function to be maximized, expressed in terms of the single
variable x, is u _ <£(V) = ^50 -x)= 50 x - x 2 . (3)
The critical value of x is found from the equation
<P(x) = 50-2x =
to be x = 25. When x increases through this value, <f>'(x) changes
sign from positive to negative, and hence <f>(x) is a maximum when
x = 25. Equation (2) shows that the corresponding value of y is 25.
Hence the maximum rectangle whose perimeter is 100 is the square
whose side is 25.
Ex. 2. If, from a square piece of tin whose side is a, a square be
cut out at each corner, find the side of the latter square in order that
the remainder may form a box of maximum
capacity, with open top.
Let x be a side of each square cut out.
Then the bottom of the box will be a square
whose side is a — 2 x, and the depth of the box
will be x. Hence the volume is
v = x(a - 2 x) 2 ,
which is to be made a maximum by varying x.
Here — = (a-2i) 2 ^i(a-2z)
dx
= (a-2x)(a -6.i).
This derivative vanishes when x — -, and when x — - • It will be
2 6
found, by applying the usual test, that x = ^ gives v the minimum
62
DIFFERENTIAL CALCULUS
value zero, and that x = - gives it the maximum value " a • Hence
6 27
the side of the square to be cut out is one sixth the side of the given
square.
Ex. 3. Find the area of the greatest rectangle that can be inscribed
in a given ellipse.
An inscribed rec-
tangle will evidently be
symmetric with regard
to the principal axes
of the ellipse.
Let a, b denote the
lengths of the semi-
axes OA, OB (Fig. 14) ;
let 2 x, 2 y be the dimen-
Fig. 14
sions of an inscribed rectangle.
Then the area is
u = 4 x?j, (1)
in which the variables x, y may be regarded as the coordinates of the
vertex P, and are therefore subject to the equation of the ellipse
2 b l
(2)
It is geometrically evident that there is some position of P for
which the inscribed rectangle is a maximum.
The elimination of y from (1), by means of (2), gives the function
of x to be maximized, 4 j }
u = — x
a
V^
(3)
By Art. 34, the critical values of x are not altered if this function
is divided by the constant — , and then squared. Hence, the values
a
of x which render u a maximum, give also a maximum value to the
function +(x) = x*(a* - &) = aW - x*.
Here <f>'( x ) = 2 a 2 x - 4 x z = 2 x(a* - 2 x°-),
cf>"(x) =2a 2 -12r 2 ;
MAXIMA AND MINIMA
63
hence, by the usual tests, the critical values
± ■ — render cb(x),
V2
and therefore the area w, a maximum. The corresponding values of
y are given by (2), and the vertex P may be at any of the four points
denoted by
, a , b
±— :, y = ±— ,
V2 V2
giving in each case the same maximum inscribed rectangle, whose
dimensions are a V2, b V2, and whose area is 2 ab, or half that of the
circumscribed rectangle.
Ex. 4. Find the greatest cylinder that can be cut from a given
right cone, whose height is k, and the radius of whose base is a.
Let the cone be generated by
the revolution of the triangle 0^45
(Fig. 15), and the inscribed cylin-
der be generated by the revolution
of the rectangle A P.
Let A = h, AB =a, and let the
coordinates of P be (x, ?j) . Then
the function to be maximized is
Try' 2 (h r- x) subject to the relation ^ = -
x h
This expression becomes
v = ^.x*a-
/i 2 v
Fig. 15
The critical value of x is f h, and V
x).
ijraVi
27
EXERCISES ON CHAPTER IV
1. What is the width of the rectangle of maximum area that can
be inscribed in a given right segment of a parabola?
..
2. Divide 10 into two parts such that the sum of their squares is
a minimum.
3. Find the number that exceeds its square by the greatest pos-
sible quantity.
64 DIFFERENTIAL CALCULUS
>^4. What number added to its reciprocal gives the least possible
sum?
5. Given the slant height of a right cone; find its* altitude when
the volume is a maximum.
6. A rectangular piece of pasteboard 30 in. long and 14 in. wide
has a square cut out at each corner. Find the side of this square so
that the remainder may form a box of maximum contents.
7. Find the altitude of the right cylinder of greatest volume in-
scribed in a sphere of radius r.
8. Determine the greatest rectangle that can be inscribed in a
given triangle whose base is 2 b, and whose altitude is 2 a.
9. A rectangular court is to be built so as to contain a given area
c 2 , and a wall already constructed is available for one of its sides.
Find its dimensions so that the expense incurred in building the walls
for the other sides may be the least possible.
10. The volume of a cylinder of revolution being constant, find
the relation between its altitude and the radius of its base when the
entire surface is a minimum.
11. Assuming that the stiffness of a beam of rectangular cross
section varies directly as the breadth and as the cube of the depth, what
must be the breadth of the stiffest beam that can be cut from a
log 16 in. in diameter?
12. A man who can row 4 mi. per hour, and can walk 5 mi. per
hour, is in a boat 3 mi. from the nearest point on a straight beach,
and wishes to reach in the shortest time a place on the shore 5 mi.
from this point. Where must he land ?
13. If the cost per hour for the fuel required to run a given
steamer is proportional to the cube of her speed and is $20 an hour
for a speed of 10 knots, and if other expenses amount to $135 an hour,
find the most economical rate at which to run her over a course s.
MAXIMA AND MINIMA 65
14. If the cost per hour of running a boat in still water is propor-
tional to the cube of the velocity, find the most economical rate at which
to run the steamer upstream against a current of a miles per hour.
15. A Norman window consists of a rectangle surmounted by a
semicircle. If the perimeter of the window is given, what must be its
proportions in order to admit as much light as possible?
16. Find the most economical proportions for a cylindrical dipper
which is to hold a pint.
17. The gate in front of a man's house is 20 yd. from the car
track. If the man walks at the rate of 4 mi. an hour and the car on
which he is coming home is running at the rate of 12 mi. an hour,
where ought he to get off in order to reach home as early as possible?
18. How much water should be poured into a cylindrical tin dip-
per in order to bring the center of gravity as low down as possible ?
[Omit until after reading Art. 164.]
19. A statue 10 ft. high stands on a pedestal that is 50 ft. high.
How far ought a man whose eyes are 5 ft. above the ground to stand
from the pedestal in order that the statue may subtend the greatest
possible angle?
20. The sum of the surfaces of a sphere and a cube is given. How
do their dimensions compare when the sum of their volumes is a
minimum ?
21. An electric light is to be placed directly over the center of a
circular plot of grass 100 ft. in diameter. Assuming that the inten-
sity of light varies directly as the sine of the angle under which it
strikes an illuminated surface and inversely as the square of its dis-
tance from the surface, how high should the light be hung in order
that the most light possible shall fall on a walk along the circumfer-
ence of the plot ?
22. Find the relation between length of circular arc and radius, in
order that the area of a circular sector of a given perimeter shall be a
maximum.
el. calc. — 5
66 DIFFERENTIAL CALCULUS
23. On the line joining the centers of two mutually external
spheres of radii r, R, find the distance of the point from the center of
the first sphere from which the maximum of spherical surface is visible.
24. The radius of a circular piece of paper is r. Find the arc of
the sector which must be cut from it so that the remaining sector
may form the convex surface of a cone of maximum volume.
25. Describe a circle with its center on a given circle so that the
length of the arc intercepted within the given circle shall be a maxi-
mum.
26. Through a given point within an angle draw a straight line
which shall cut off a minimum triangle.
,27. What is the length of the axis, and the area, of the maximum
parabola which can be cut from a given right circular cone, given
that the area of the parabola is equal to two thirds of the product of
its base and altitude? A parabola is cut from the cone by a plane
parallel to an element.
28. Through the point (a, b) a line is drawn such that the part
intercepted between the rectangular coordinate axes is a minimum.
Find its length.
29. The lower corner of a leaf, whose edge is a, is folded over so
as just to reach the inner edge of the page. Find the width of the
part folded over when the length of the crease is a minimum.
30. What is the length of the shortest line that can be drawn tan-
gent to the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 and having its ends on the co-
ordinate axes ?
31. Given a point on the axis of the parabola y 2 = 2 px at a dis-
tance a from the vertex. Find the abscissa of the point of the curve
nearest to it.
32. A wall 6 ft. high is parallel to the front of a house and 8 ft.
from it. Find the length of the shortest ladder that will reach the
house if one end rests on the ground outside the wall.
MAXIMA AND MINIMA 67
33. It is required to construct from two circular iron plates of
radius a a buoy, composed of two equal cones having a common base,
which shall have the greatest possible volume. Find the radius of
the base.
34. A weight W is to be raised by means of a lever with force F
at one end and the point of support at the other. If the weight is
suspended from a point at a distance a from the point of support, and
the weight of the beam is w pounds per linear foot, what should be
the length of the lever in order that the force required to lift the
weight shall be a minimum?
35. A load is hauled up an inclined plane by a horizontal force ; it
is required to find the inclination of the plane so that the mechanical
efficiency may be greatest, assuming that the efficiency 77 is defined by
the formula tan Q
' tan(0+4>)'
where <f> is the angle of friction; i.e. tan <f> — fi, the coefficient of fric-
tion between the load and the plane.
36. If the plane is of cast iron and the load is steel, and if the
coefficient of friction between these substances is fi = 0.347, at what
angle is the efficiency of the inclined plane a maximum?
37. Prove that a conical tent of given capacity will require the
least amount of canvas when the height is v2 times the radius of
the base.
38. If given currents c and c' produce deflections a and a' in a
tangent galvanometer, so that tan «/tan «' = c/c', show that a — a' is
a maximum when a + a' = -.
CHAPTER V
RATES AND DIFFERENTIALS
36. Rates. Time as independent variable. Suppose a particle
P is moving in any path, straight or curved, and let s be the
number of space units passed over in t seconds. Then s may
be taken as the dependent variable, and t as the independent
variable. The motion of P is said to be uniform when equal
spaces are passed over in equal times. The number of space
units passed over in one second is called the velocity of P,
The velocity v is thus connected with the space s and the time
t by the formula s
v — -.
t
The motion of P is said to be non-uniform when equal spaces
are not passed over in equal times. If s is the number of space
units passed over in t seconds, then the average velocity during
these t seconds is denned as -. If during the time A£ the num-
ber of space units As are described, then the average velocity
As
during the time A£ is — . The actual velocity of P at any in-
stant of time t is the limit which the average velocity
approaches as At is made to approach zero as a limit.
rr, lim As ds
Thus v = A , . n — = —
A * = °A* dt
is the actual velocity of P at the time denoted by t. It is
evidently the number of space units that would be passed over
68
RATES AND DIFFERENTIALS 69
in the next second if the velocity remained uniform from the
time t to the time t -f 1.
It may be observed that if the more general term, " rate
of change,'' is substituted for the word " velocity," the above
statements will apply to any quantity that varies with the
time, whether it be length, volume, strength of current, or any
other function of the time. For instance, let the quantity of
an electric current be C at the time t, and C -\- AC at the time
t + At. Then the average rate of change of current in the in-
terval At is — ; this is the average increase in current-units
At
per second. And the actual rate of change at the instant de-
noted by t is k rt jn
J lim AC dC
te±° At~ dt'
This is the number of current-units that would be gained in
the next second if the rate of gain were uniform from the time
t to the time t + 1. Since, by Art. 8,
dy_dy. dx
dx dt dt
hence -^ measures the ratio of the rates of change of y and
dx
oi x.
It follows that the result of differentiating
2/=/(*) (1)
may be written in either of the forms
*-m (2)
*-/«!■ (3)
70 DIFFERENTIAL CALCULUS
The latter form is often convenient, and may also be obtained
directly from (1) by differentiating both sides with regard to
t. It may be read: the rate of change of y is f(x) times the
rate of change of x.
Ketnrning to the illustration of a moving point P, let its
dx
coordinates at time t be x and y. Then measures the rate
dt
of change of the ^-coordinate.
Since velocity has been defined as the rate at which a point
is moving, the rate —may be called the velocity which the
dt
point P has in the direction of the a>axis, or, more briefly, the
^-component of the velocity of P.
It was shown on p. 68 that the actual velocity at any instant
t is equal to the space that woiild be passed over in a unit of
time, provided the velocity were
*lrt\B / c^r&- uniform during that unit. Ac-
cordingly, the ^-component of
dt velocity — may be represented
by the distance PA (Fig. 16)
which P would pass over in the
direction of the a>axis during a
unit of time if the velocity remained uniform.
Similarly -# is the ^/-component of the velocity of P, and
may be represented by the distance PB.
The velocity — of P along the curve can be represented by
Civ
the distance PC, measured on the tangent line to the curve at
P. It is evident from the parallelogram of velocities that PC
is the diagonal of the rectangle PA, PB.
^.dx
Fig. 16
RATES AND DIFFERENTIALS 71
Since PC 2 = PA 2 + PB 2 , it follows that
S)'=(f)" + (*)' «
Ex. 1. If a point describes the straight Hue 3 x + 4y = 5, and if x
increases h units per second, find the rates of increase of y and of s.
Since y = £ - £ x,
hence -7-— — 7-r"
rf< 4 eft
When — = A,
dt
it follows that ^ = - f A, ^ = VW^TJ^i - 5 A<
eft </£
Ex. Si. A point describes the parabola # 2 = 12 r in such a way that
when x = 3 the abscissa is increasing at the rate of 2 ft. per second ;
at what rate is y then increasing? Find also the rate of increase of s.
Since
y*=12*,
then
"(It '" dt
dy Qdx 6. dx ,
dt y dt Vl2x dt '
hence when x = 3 and — = 2, it follows that ^ = ± 2.
A ^- (D 2 = (fT + B) ' •-- |=^ft. per second.
Ex. 3. A person is walking toward the foot of a tower on a hori-
zontal plane at the rate of 5 mi. per hour. At what rate is he ap-
proaching the top, which is 60 ft. high, when he is 80 ft. from the
bottom ?
72 DIFFERENTIAL CALCULUS
Let x be the distance from the foot of the tower at time t, and y
the distance from the top at the same time. Then
x 2 + 60 2 = if,
and *** = «&
dt -dt
When x is 80 ft., y is 100 ft.; hence if — is 5 mi. per hour,
dy ■ A • , dt
-^ is 4 mi. per hour.
dt k
37. Abbreviated notation for rates. When, as in the above
examples, a time derivative is a factor of each member of an
equation, it is usually convenient to write, instead of the
symbols — , -^, the abbreviations dx and dy, for the rates of
J dt dt J '
change of the variables x and y. Thus the result of differen-
tiatillg y=m (l)
may be written in either of the forms
*-/«, (2)
dt ' } dt' w
dy=f'(x)dx. (4)
It is to be observed that the last form is not to be regarded
as derived from equation (2) by separation of the symbols, dy,
dx\ for the derivative -M- has been defined as the result of
dx
performing upon y an indicated operation represented by the
symbol — , and thus the dy and dx of the symbol -M- have
dx dx
been given no separate meaning. The dy and dx of equation
(4) stand for the rates, or time derivatives, ~ and •-— occur-
dt - dt
RATES AND DIFFERENTIALS 73
ring in (3), while the latter equation is itself obtained from
(1) by differentiation with regard to t, by Art. 8.
In case the dependence of y upon x is not indicated by a
functional operation/, equations (3), (4) take the form
dy
dy dx
dt
dx dt '
dy =
-fax.
dx
In the abbreviated notation, equation (4) of the last article
is written (ds) 2 — (dx) 2 -f (dy) 2 or ds 2 = dx 2 + dy 2 .
Ex. 1. A point describing the parabola y 2 — 2 px is moving at the
time t with a velocity of v ft. per second. Find the rate of increase
of the coordinates x and y at the same instant.
Differentiating the given equation with regard to t, we obtain
ydy = pdx.
But dx, dy also satisfy the relation
dx 2 + dy 2 = v 2 ;
hence, by solving these simultaneous equations, we obtain
dx = — ' v, dy = ^ v, in feet per second.
v V + p 2 Vy* 2 + p 2
Ex. 2. A vertical wheel of radius 10 ft. is making 5 revolutions per
second about a fixed axis. Find the horizontal and vertical velocities
of a point on the circumference situated 30° from the horizontal.
Since x = 10 cos 6, y = 10 sin 6,
then dx = -10 sin OdO, dy = 10 cos Odd.
But dO = 10 7T = 31.416 radians per second,
hence dx = — 314.16 sin $ = — 157.08 ft. per second,
and dy = 314.16 cos = 272.06 ft. per second.
Ex. 3. Trace the changes in the horizontal and vertical velocity
in a complete revolution.
74
DIFFERENTIAL CALCULUS
38. Differentials often substituted for rates. The symbols dx,
dy have been denned above as the rates of change of x and y
per second.
Sometimes, however, they may conveniently be allowed to
stand for any two numbers, large or small, that are propor-
tional to- these rates ; the equations, being homogeneous in
them, will not be affected. It is usual in such cases to speak
of the numbers dx and dy by the more general name of differ-
entials ; they may then be either the rates themselves, or any
two numbers in the same ratio.
This will be especially convenient in problems in which the
time variable is not explicitly mentioned.
39. Theorem of mean value. Let f(x) be a continuous func-
tion of x which has a derivative. It can then be represented
by the ordinates of a curve whose
equation is y —f(x).
In Pig. 17, let
x = ON, x + h = OR,
f(x) = NH, f(x + h) = RK.
Then f(x + li)-f(x) = MK, and
f(x+iC\-f(xS MK
Fig. 17
= ^±± = tan MHK.
h HM
But at some point 8 between H and A" the tangent to the
curve is parallel to the secant IIK. Since the abscissa of S is
greater than x and less than x + h it may be represented by
x-\-0h, in which 6 is a positive number less than unity. The
slope of the tangent at S is then expressed by/'(.x* + 6h), hence
f(x + h) -f(x) = / {x + m)t
from which
f(x + h)=f(x)+hf'(x+Oh).
RATES AND DIFFERENTIALS 75
The theorem expressed by this formula is known as the
theorem of mean value.
If in this equation we put
f(x + K) — f (a?) = dy, h = dx,
in which h is an arbitrary increment, then the relation between
the increment of the variable and the actual increment of the
function will be expressed by the equation
dy=f'(x + 6dx)dx,
whereas if dy, dx are regarded as differentials (dy not an
actual but a virtual increment), then the relation becomes
dy=f\x)dx.
This more clearly illustrates that the differential dy is de-
fined as the change that would take place in the function y,
corresponding to the actual change dx in the independent vari-
able x } provided the rate of change remained constant.
EXERCISES
1. When x increases from 45° to 45° 15', find the increase of
logio sin x, assuming that the ratio of the rates of change of the func-
tion and the variable remains constant throughout the short interval.
Here dy = logio e • cot xdx = .4343 cot xdx = .4343 dx.
Let dx— .004163 (the number of radians in 15').
Then dy = .001895,
which is the approximate increment of logio sin x.
But log 10 sin 45° = - \ log 2 = - .150515,
therefore log 10 sin 45° 15' = - .148620.
2. Show that log, ft .r increases more slowly than x, when
x > log 10 e, that is, x > 0.4343.
76 DIFFERENTIAL CALCULUS
3. A man is walking at the rate of 5 mi. per hour towards the
foot of a tower 60 ft. high standing on a horizontal plane. At what
rate is the angle of elevation of the top changing when he is 80 ft. from
the foot of the tower ?
4. An arc light is hung 12 ft. directly above a straight horizontal
walk on which a man 5 ft. in height is walking. How fast is the man's
shadow lengthening when he is walking aw T ay from the light at the
rate of 168 ft. per minute?
5. At what point on the ellipse 16 x 2 + 9 y 2 — 400 does y decrease
at the same rate that x increases?
6. A vessel is sailing northwest at the rate of 10 mi. per hour.
At what rate is she making north latitude?
7. In the parabola y 2 = 12 x, find the point at which the ordinate
and abscissa are increasing equally.
8. At what part of the first quadrant does the angle increase twice
as fast as its sine ?
9. Find the rate of change in the area of a square when the side
b is increasing at a ft. per second.
10. In the function y = 2 x z + 6, what is the value of x at the point
where y increases 24 times as fast as x ?
11. A circular plate of metal expands by heat so that its diam-
eter increases uniformly at the rate of 2 in. per second. At what rate
is the surface increasing when the diameter is 5 in.?
12. What is the value of x at the point at which x 8 — 5 x' 2 + 17 x and
x s — 3 x change at the same rate ?
13. Find the points at which the rate of change of the ordinate
y = x 3 - 6 x 2 + 3 x + 5 is equal to the rate of change of the slope of the
tangent to the curve.
14. The relation between s, the space through which a body falls,
rates and differentials 77
and t, the time of falling, is s = 16 t 2 . Show that the velocity is equal
to 32 t.
The rate of change of velocity is called acceleration and is denoted
by a.
TT dv d 2 s
Hence a = — = -— •
dt dt 2
Show that the acceleration of the falling body is a constant.
15. A body moves according to the law s — cos (nt + e). Show that
its acceleration is proportional to the space through which it has
moved.
16. If a body is projected upwards in a vacuum with an initial
velocity v , to what height will it rise, and what will be the time
of ascent?
17. A body is projected upwards with a velocity of a ft. per second.
After what time will it return?
18. If A is the area of a circle of radius x, show that the circum-
dA
ference is — . Interpret this fact geometrically.
19. A point describing the circle x 2 + y 2 = 25 passes through (3,4)
with a velocity of 20 ft. per second. Find its component velocities
parallel to the axes.
20. Let a point P move with uniform velocity on a circle of radius
a with center ; let AB be any diameter, and Q the orthogonal projec-
tion of P on AB. Find an expression for the velocity of Q in terms
of the angular velocity of P, and show how this velocity varies during
a revolution of P. The motion of the point Q along AB is called
harmonic.
21. A point P moves along the curve y = x z at the rate of 3 ft. per
second. At what rate is the angle <£, which the tangent to the curve
makes with the x-axis, increasing when P is passing through the
point (1, 1)?
V
CHAPTER VI
DIFFERENTIAL OF AN AREA, ARC, VOLUME, AND
SURFACE OF REVOLUTION
40. Differential of an area. If the coordinates of P are (x, y)
and those of Q (x 4- Ax, y-\- Ay), then
MN= PR = Ax, and PS = RQ=Ay.
If the area OAPM is denoted by A,
then A is evidently some function
of the abscissae; also if area OAQN
is denoted by A 4- AA then the
area MNQP is AA ; it is the incre-
ment taken by the function A, when
x takes the increment Ax. But MNQP lies between the
rectangles MR, MQ ; hence
i/Aa; < A.4 < (y + Ay)Aa;,
and
A. 4
y<—-<y + ^y-
Ax
Therefore, when Ax, Ay, AA all approach zero,
lim
A A dA
Ax
(lx
= ?/
Hence, if the ordinate and the area are expressed each as a
function of the abscissa, the derivative of the area function
with regard to the abscissa is equal to the ordinate function.
78
AREA, ARC, VOLUME, AND SURFACE
79
In the notation of differentials we may say : The differential
of the area between a curve and the axis of x is measured by the
product of the ordinate and the differential of x.
dA = ydx.
Ex. If the area included between a curve, the axis of x, and the
ordinate whose abscissa is a:, is given by the equation
A = x s ,
find the equation of the curve.
Here
=M = 3z2.
dx
41. Differential of an arc. A segment of a straight line is
measured by applying the unit of measure successively to the
segment to be measured. In the case of a curve this is gen-
erally impossible. We define the length of a given curve
between two points upon it as the limit of the sum of the
chords joining points on the curve when the lengths of these
chords approach the limit zero. We shall then assume that the
ratio of the arc to the chord approaches the limit 1 when the
length of the chord approaches the limit zero. [Compare § 19.]
Let PQ be two points on the curve (Fig. 19) ; let x, y be the
coordinates of P ; x + Ax, y + Ay Y
those of Q ; s the length of the arc
AP; s -+- As that of the arc AQ.
Draw the ordinates MP, NQ ; and
draw PR parallel to MN. Then
PR = Ax, RQ = Ay ; arc PQ = As. -
Hence chord PQ = V(A xf+ (Ay) 2 ,
M N
Fig. 19
PQ
Ax
Therefore ' =
Ai
Ax
As
1>Q
AXy
PQ = As
Ax PQ
A?/Y
v i+ is-
80 DIFFERENTIAL CALCULUS
Taking the limit of both members as Ax approaches zero
lim As
and putting ™ Q -~ = 1, we obtain
%-Mif
(1)
Si ,u„„l y , |-V' + (*/.
(2)
Moreover, from Art. 36,
(ds\*/dx\* (dy\*
[dtj \dtj \dtj'
(3)
or in the differential notation,
ds 2 = dx 2 -f dy 2 .
W
42. Trigonometric meaning of — , — .
doc dy
cs- A.r A.r PQ »r>/^
Since — = • — -3S = cos PPQ
As PQ As
As'
it follows by taking the limit that
dx ,
— = cos <£,
ds
wherein <f>, being the limit of the angle RPQ, is the angle
which the tangent at the point (x, y) makes with the o-axis.
Similarly, -* = sin </> ; whence — = sec <£, — = esc <£.
ds dx dy
By using the idea of a rate or
differential, all these relations may
be conveniently exhibited by Fig.
20.
These results may also be de-
rived from equations (1), (2) of
Art. 41, by putting -'^tan d>.
Fig. 20 J F dx
*■ ■ .»
QL^dx
AREA, ARC, VOLUME, AND SURFACE
81
43. Differential of the volume of a solid of revolution. Let
the curve APQ (Fig. 21) revolve about the x-axis, and thus
generate a surface of revo-
lution ; let V be the volume in-
cluded between this surface,
the plane generated by the
fixed ordinate at A, and the
plane generated by any ordinate
MP.
Let A V be the volume gener-
ated by the area PMNQ. Then AV lies between the vol-
umes of the cylinders generated by the rectangles PMNR
and SMNQ) that is,
Try 9 Ax < A V < 7T (y -f Ay) 2 Ax.
Y
s
A
/
R
X
M A
J
Fig. 21
Dividing by Ax and taking limits, we obtain
dV
dx
Try 2 , dV = 7ry 2 dx.
44. Differential of a surface of revolution. Let S be the area
of the surface generated by the arc AP (Fig. 22), and AS that
generated by the arc PQ, whose length is As.
Draw PQ', QP parallel to OX
and equal in length to the arc PQ.
Then it may be assumed as an
axiom that the area generated by
PQ lies between the areas gen-
iL erated by PQ' and P'Q; i.e.
2 iry As < AS < 2 tv (y + Ay) As.
I
M N
Fig. 22
EL. CALC. —
DIFFERENTIAL CALCULUS
Dividing by As and passing to the limit,
f = 2^ (1)
dx ds dx \ \dx)
45. Differential of arc in polar coordinates. Let p, 6 be the
coordinates of P (Fig. 23) ; p ■+ Ap, 6 + A6 those of $ ; s the
length of the arc KP ; As that of the
L arcPQ; draw PM perpendicular to
OQ. Then
PM = p sin A0,
p
MQ = OQ-OM=p + A P -p cos A<9
= p (1 — cos A0) + A/a
Fig. 23 = 2 P sin2 i A # + V
Hence PQ 2 = (p sin A0) 2 + (2 p sin 2 1 A0 + Ap) 2 ,
/PQV = ^sinA^V + ( p sin J A0 . S Ai^ + V)\
Replacing the first member by f — - %■ • -^ ] , passing to the
\ .AN -Af7 /
limit when A0 = 0, and putting lira— ^ = 1, lira — — = 1,
As A9
lira sin * A6 = 1, we obtain
i^0
^V = , 2 + ^Y.
that is, ^ = V^ +/
d<9 ' Vd0 y
AREA, ARC, VOLUME, AND SURFACE
83
In the rate or differential notation this formula may be
conveniently written ^ = rf * + 2 ^
46. Differential of area in polar coordinates. Let A be the
area of OKP (Fig. 24) measured from a fixed radius vector
OK to any other radius vec-
tor OP ; let A^4 be the area of
OPQ. Draw arcs PM, QN,
with as a center. Then the
area POQ lies between the
areas of the sectors OPM and
ONQ; i.e. Fig. 24
1 p 2 A0 < A.-l < J (p + Ap) 2 A0.
Dividing by A0 and passing to the limit, when A0 = 0, we
obtain
dA
d$
= ip 2 -
Hence, in the differential notation we may write the formula
dA = \ phlO.
EXERCISES ON CHAPTER VI
1. In the parabola y 2 = 4 ax, find — , — , — - ,
dx dx dx dx
2. Find — and — for the circle x 2 + y 2 = a 2 .
dx dy
ds
3. Find — for the curve e v cos x = 1.
dx
4. Find the x-derivative of the volume of the cone generated by
revolving the line y = ax about the axis of x.
5. Find the it-derivative of the volume of the ellipsoid of revolu-
tion, formed bv revolving — + •*- = 1 about its major axis.
a 2 b 2
84 DIFFERENTIAL CALCULUS
ds
6. In the curve p = a 9 find —
du
7. Given p = a (1 + cos 6) ; find ^
8. In p 2 cos 2 0, find ^.
9. The parabolic arc y 2 = 9 x measured from the vertex to a variable
point P = (x, y) is revolving about the a>axis. If P moves along the
curve at the rate of 2 in. per second, what is the rate of increase
of the surface of revolution when P is passing through the point
(4, 6)? What is the rate of increase of the volume of revolution?
10. The radius vector to the cardioid p = 2 (1 — cos $) is rotating
about the origin with an angular velocity of 18° per second. Find
the rate at which the extremity P of the radius vector is moving along
the curve, taking the inch as unit of length. At what points of the
curve will P be moving fastest? slowest? Find the velocities at
these points.
CHAPTER VII
APPLICATIONS TO CURVE TRACING
47. Equation of tangent and normal. The function y=f(x)
may be represented by a plane ciu-ve. It will now be shown
how to obtain several of the properties of this curve by means
of the principles already established. The tangent line at a
point (x-t, y{) on the curve passes through the point and has
the slope — , the symbol meaning that the coordinates x x , y x
are substituted in the first derivative after the differentiation
has been performed. Its equation may be written in the form
y- 2/1 = ^' (*-*.)• (1)
The normal to the curve at the point (x u y Y ) is the straight
line through this point, perpendicular to the tangent. Since
the slope of the normal is the negative reciprocal of that of
the tangent, its equation may be written in the form
*-*i + p(*- *)-0. (2)
48. Length of tangent, normal, subtangent, subnormal. The
segments of the tangent and normal intercepted between the
point of tangency and the axis OX are called, respectively,
the tangent length and the normal length, and their projections
on OX are called the subtangent and the subnormal.
85
DIFFERENTIAL CALCULUS
C
Fig. 25 a
Fig. 25 6
Thus, in Fig. 25, a, b let the tangent and normal to the curve PC
at P meet the axis OX in T and N, and let i/P be the ordi-
nate of P. Then TP is the tangent length,
PNthe normal length,
TM the subtangent,
MN the subnormal.
These will be denoted, respectively, by t, n, r, v.
Let the angle XTP be denoted by cf>, and write tan <f> = -^*. "
Then
hence
d^x
2/i
1 +
(for,
y
H§,
fdjh
\dx x
The subtangent is measured from the intersection of the
tangent to the foot of the ordinate ; it is therefore positive
when the foot of the ordinate is to the right of the intersec-
tion of tangent. The subnormal is measured from the foot
of the ordinate to the intersection of normal, and is positive
when the normal cuts OX to the right of the foot of the ordi-
APPLICATIONS TO CURVE TRACING 87
nate. Both are therefore positive or negative, according as
<£ is acute or obtuse.
The expressions for t, v may be obtained also by finding
from equations (1), (2), Art. 47, the intercepts made by the
tangent and normal on the axis OX. The intercept of the
tangent subtracted from x l gives t, and x x subtracted from
the intercept of the normal gives v.
Ex. Find the intercepts made upon the axes by the tangent at the
point (x v y x ) on the curve Va: + Vy = Va, and show that their sum
is constant.
Differentiating the equation of the curve, we obtain
2 V* 2V~ydx
Hence the equation of the tangent is
x i
The x intercept is x l + y/x l y v and the y intercept is y x 4- y/x l y v
hence their sum is , ,— ,— N0
If a series of lines is drawn such that the sum of the intercepts of
each is the same constant, account being taken of the signs, the form
of the parabola to which they are all tangent can be readily seen.
EXERCISES
1. Find the equations of the tangent and the normal to the ellipse
X 2 V 2
(- V- — 1 at the point (x v ?/,). Compare the process with that era-
a' 2 b 2 i
ployed irr-analytic geometry to obtain the same results.
2. Find the equation of the tangent to the curve
x\x +y)= a 2 (x -y)
at the origin.
88 DIFFERENTIAL CALCULUS
3. Find the equations of the tangent and normal at tho point
(1, 3) on the curve y' 2 = 9 x 3 .
4. Find the equations of the tangent and normal to each of the
following curves at the point indicated :
(a) y = — , at the point for which x = 2 a.
4 a 2 + x 2
(jg) y 2 = 2 x 2 — x 3 , at the points for which x = 1.
(y) y 2 ~ ipx, at the point (p, 2p).
5. Find the value of the subtangent of y 2 = 3 x 2 — 12 at x = 4.
Compare the process with that given in analytic geometry.
6. Find the length of the tangent to the curve y 2 = 2 x at x = 8.
7. Find the points at which the tangent is parallel to the axis
of x, and at which it is perpendicular to that axis for each of the fol-
lowing curves : (ft) a,? + 2 hxy + bf = 1.
(/J) , =?^iL\
ax
(■y) y 3 = x 2 (2 a — x).
8. Find the condition that the conies
ax 2 4- by 2 = 1, a'x 2 -f b'y 2 — 1
shall cut at right angles.
9. Find the angle at which x 2 = y 2 + 5 intersects 8 x 2 -f 18 y 2 = 144.
Compare with Ex. 8.
10. Show that in the equilateral hyperbola 2 xy — a 2 the area of
the triangle formed by a variable tangent and the coordinate axes is
constant and equal to a 2 .
11. At what angle does y 2 = 8 x intersect 4 x 2 + 2 y 2 =48 ?
12. Determine the subnormal to the curve y n — aP-^x.
13. Find the values of x for which the tangent to the curve
7/3 = ( x - a ) 2 (x - c)
is parallel to the axis of x.
APPLICATIONS TO CURVE TRACING 89
14. Show that the subtangent of the hyperbola xy = a 2 is equal to
the abscissa of the point of tangency, but opposite in sign.
15. Prove that the parabola y 2 = 4 ax has a constant subnormal.
16. Show analytically that in the curve x 2 + y 2 = a 2 the length of
the normal is constant.
17. Show that in the tractrix, the length of the tangent is con-
stant, the equation of the tractrix being
x = Vc*-Tj* + £ log c-V c 2 -y\
- c + vV 2 — y 2
X
18. Show that the exponential curve y = ae c has a constant sub-
tangent.
19. Find the point on the parabola y 2 = 4 px at which the angle
between the tangent and the line joining the point to the vertex shall
be a maximum.
49. Concavity upward and downward. A curve is said to be
concave downward in the vicinity of a point P when, for a
finite distance on each side of P, the curve is situated below
the tangent drawn at that point, as in the arcs AD, FTL It
is concave upward when the curve lies above the tangent, as
in the arcs DF, HK.
90 DIFFERENTIAL CALCULUS
By drawing successive tangents to the curve, as in the fig-
ure, we easily see that if the point of contact advances to the
right, the tangent swings in the positive direction of rotation
when the concavity is upward, and in the negative direction
when the concavity is downward. Hence upward concavity
may be called a positive bending of the curve, and downward
concavity, a negative bending.
A point at which the direction of bending changes con-
tinuously from positive to negative, or vice versa, as at F or
at D, is called a point of inflexion, and the tangent at such a
point is called a stationary tangent.
The points of the curve that are situated just before and just
after the point of inflexion are thus on opposite sides of the
stationary tangent, and hence the tangent crosses the curve, as
at D, F, H.
50. Algebraic test for positive and negative bending. Let the
inclination of the tangent line, measured from the positive end
of the ^axis toward the forward end of the tangent, be denoted
by <£. Then <£ is an increasing or decreasing function of the
abscissa according as the bending is positive or negative ; for
instance, in the arc AD, the angle <f> diminishes from 4- —
Z
through zero to — - ; in the arc DF, cfy increases from — -
4 4
through zero to - ; in the arc FIT, <£ decreases from -f - through
o o
zero to — y ; and in the arc HK, <f> increases from — y through
zero to + -.
At a point of inflexion $ has evidently a turning value which
is a maximum or a minimum, according as the concavity changes
from upward to downward, or conversely.
APPLICATIONS TO CURVE TRACING 91
Thus in Fig. 26, cf> is a maximum at F, and a minimum at D
and at II.
Instead of recording the variation of the angle <f>, it is gen-
erally convenient to consider the variation of the slope, tan <£,
which is easily expressed as a function of x by the equation
tan <£ = -^.
dx
Since tan <f> is always an increasing function of <f>, it follows
that the slope function -^ is an increasing or a decreasing
dx
function of x, according as the concavity is upward or down-
ward, and hence that its x-derivative is positive or negative.
Thus the bending of the curve is in the positive or negative
d-y
direction of rotation, according as the function — ^ is positive
dx-
or negative.
At a point of inflexion the slope — is a maximum or a
dx
d 2 v
minimum, and therefore its derivative — - changes sign from
dx 2
positive to negative or from negative to positive. This latter
condition is evidently both necessary and sufficient in order that
the point (x, y) may be a point of inflexion on the given curve.
Hence, the coordinates of the points of inflexion on the curve
may be found by solving the equations
/"O) = o, /"(*)=«,,
and then testing whether f"(x) changes its sign as x passes
through the critical values thus obtained. To any critical
value a that satisfies the test corresponds the point of inflexion
92
DIFFERENTIAL CALCULUS
Ex. 1. For the curve y = (x 2 — l) 2
find the points of inflexion, and show the mode of variation of the
slope and of the ordinate.
Here
^/ = 4a-(> 2 -l),
dx
(Py_
dx 2
= 4(3:r 2 -l),
1
hence the critical values for inflexions are x = ± — -. It will be seen
i vs
that as x increases through = , the second derivative changes sign
V3
from positive to negative, hence there is an inflexion at which
the concavity changes from upward to downward. Similarly, at
x = ^ — =3 the concavitv changes from downward to upward. The
V3
following numerical table will help to show the mode of variation of
the ordinate and of the slope, and the direction of bending.
As x increases from — oo to
= the bending is positive, and
Va
the slope continually increases from
— go through zero to a maximum
I Q
value — — , which is the slope of
3V3
the stationary tangent drawn at
the point f , - ).
V V3 9/
As x continues to increase from
L_ to H = , the bending is neg-
V3 V3
Q
ative, and the slope decreases from + — - through zero to a minimum
oV3
o
value _, which is the slope of the stationary tangent at
3V<3
X
y
dy
dx
d*y
dx 2
— 00
+ co
— GO
+
_ 2
+ 9
-24
+
-1
+
V3
-1
8
3V3
1
_
V3
H
8
3V3
1
+
+ CO
+ 00
+ GO
+
^■tl
APPLICATIONS TO CURVE TRACING
93
Finally, as x increases from -\ to + oo, the bending is positive
V3
and the slope increases from the
o
value through zero to + oo.
3V3
The values x = — 1, 0, +1, at
which the slope passes through zero,
correspond to turning values of the
ordinate.
Ex. 2. Examine for inflexions
the curve
x + 4 = (y - 2) 3 .
V
J
Fig. 27
In this case
y = 2+(x + 4)\
dy 1
dx
d 2 y
(x + 4)"
dx 2
+ 4)
dy
Fig. 28
Hence, at the point ( - 4, 2), ^
dx
and — ¥ are infinite. When x< - 4,
dx' 2
d 2 u d 2 v
—± is positive, and when x > — 4, — ^ is negative.
dx 2 dx 2
Thus there is a point of inflexion at (— 4, 2), at which the slope
is infinite, and the bending changes from the positive to the negative
direction.
Ex. 3. Consider the curve
dy _
dx
4 a: 8 ,
y = ar
d 2 y
dx 2
12 x 2 .
d 2 y
At (0, 0), ^ is zero, but the
dx 2
curve has no inflexion, for — &
dx 2
never changes sign (Fig. 29).
Fig. 29
94 DIFFERENTIAL CALCULUS
51. Concavity and convexity toward the axis. A curve is
said to be convex or concave toward a line, in the vicinity of
a given point on the curve, according as the tangent at the
point does or does not lie between the curve and the line, for
a finite distance on each side of the point of contact.
Fig. 30 a Fig. 30 6
First, let the curve be convex toward the #-axis, as in the left-
hand figure. Then if y is positive, the bending is positive
cl 2 v
and — " is positive : but if y is negative, the bending is nega-
dx 2
tive and —^ is negative. Hence in either case the product
dx 2
sin
dx 2
y—^- is positive.
Next, let the curve be concave toward the a>axis, as in the
right-hand figure. Then if y is positive, the bending is nega-
tive and — ^- is negative ; but if y is negative, the bending is
dx 2
positive and — -^ is positive. Thus in either case the product
dx 2
d 2 v
y — -- is negative. Hence :
dx 2
In the vicinity of a given point (x, y) the curve is convex or
concave to the x-axis. according as the product y —^ is positive or
dx 2
APPLICATIONS TO CURVE TRACING 95
EXERCISES
1. Examine the curve y — 2 — 3(x — 2) 5 for points of inflexion.
2. Show that the curve a 2 y = x(a 2 — x 2 ) has a point of inflexion
at the origin.
3. Find the points of inflexion on the curve y
x 2 + 4 a 2
■m
4. In the curve ay = x«, prove that the origin is a point of in-
flexion if m and n are positive odd integers.
5. Show that the curve y = c sin - has an infinite number of
a
points of inflexion lying on a straight line.
6. Show that the curve y(x 2 -f- a 2 ) = x has three points of inflexion
lying on a straight line ; find the equation of the line.
7. If y 2 =f(x s ) is the equation of a curve, prove that the abscissas
of its points of inflexion satisfy the equation
[/'(*)] 2 = 2/(z) •/''(*).
8. Draw the part of the curve a' 2 y = ~ — ax 2 -f 2 a 3 near its point
o
of inflexion, and find the equation of the stationary tangent.
9. Show that the curve y = x 2n has no points of inflexion, n being
any positive integer. Sketch the curve.
10. Show that the curve (1 + x 2 )y =1 — x has three points of in-
flexion, and that they lie in a straight line.
52. Hyperbolic and parabolic branches. When a curve has a
branch extending to infinity, the tangents drawn at successive
points of this branch may tend to coincide with a definite fixed
line, as in the familiar case of the hyperbola. On the other
hand, the successive tangents may move farther and farther out
of the field, as in the case of the parabola. These two kinds
of infinite branches may be called hyperbolic and parabolic.
The character of each of the infinite branches of a curve can
always be determined when the equation of the curve is known.
96 DIFFERENTIAL CALCULUS
53. Definition of a rectilinear asymptote. If the tangents at
successive points of a curve approach a fixed straight line as
a limiting position when the point of contact moves farther
and farther along any infinite branch of the given curve, then
the fixed line is called an asymptote of the curve.
This definition may be stated more briefly but less precisely
as follows : An asymptote to a curve is a tangent whose point
of contact is at infinity, but which is not itself entirely at
infinity.
DETERMINATION OF ASYMPTOTES
54. Method of limiting intercepts. The equation of the tan-
gent at any point (x 1} y x ) being
y-yi = ~(^-^i),
ClXi
the intercepts made by this line on the coordinate axes are
dx l
Xq — x x — y x —
dy x J
(1)
Suppose the curve has a branch on which x == oo and y = oo .
Then from (1) the limits can be found to which the intercepts
x , y approach as the coordinates x lf y x of the point of contact
tend to become infinite. If these limits are denoted by a, b,
the equation of the corresponding asymptote is
a b
Except in special cases this method is usually too compli-'
cated to be of practical use in determining the equations
of the asymptotes of a given curve. There are two other
methods, which together will always suffice to determine the
APPLICATIONS TO CURVE TRACING 97
asymptotes of curves whose equations involve only algebraic
functions. These may be called the methods of inspection
and of substitution.
55. Method of inspection. Infinite ordinates, asymptotes parallel
to axes. When an algebraic equation in two coordinates x and
y is rationalized, cleared of fractions, and arranged according
to powers of one of the coordinates, say y, it takes the form
a^+(6aj + c)r- 1 + (^ + ^+/)r" 2 + - +«»-# + «» = <>,
in which u n is a polynomial of the degree n in terms of the
other coordinate x, and u n . l is of degree n — 1.
When any value is given to x, the equation determines n
values for y.
Let it be required to find for what value of x the correspond-
ing ordinate y has an infinite value.
For this purpose the following theorem from algebra will
be recalled :
Given an algebraic equation of degree n,
ay n + (3y n - 1 + yy n - 2 + ••• =0;
if « = 0,one root y becomes infinite; if a = and /? = 0, two
roots y become infinite; and in general if the coefficients of
each of the k highest powers of y vanish, the equation will
have k infinite roots.
Suppose at first that the term in y n is present ; in other
words, that the coefficient a is not zero. Then, when any
finite value is given to x, all of the n values of y are finite,
and there are accordingly no infinite ordinates for finite
values of the abscissa.
Next suppose that a is zero, and b, c, not zero. In this
case one value of y is infinite for every finite value of x, and
el. calc. — 7
98 DIFFERENTIAL CALCULUS
hence the curve passes through the point at infinity on the
y axis.
There is one particular value of x, namely, x = ^—. for
b
which an additional root of the equation in y becomes infinite.
For, when x has this value, the coefficient bx -f- c of the high-
est power of y remaining in the equation vanishes.
Geometrically, every line parallel to the y axis has one
point of intersection with the curve at infinity, but the line
bx-\-c = has two points of intersection with the curve at
infinity. A line having two coincident points of intersection
with a curve is a tangent to the curve ; and when the coinci-
dent points are at infinity, but the line itself not altogether at
infinity, the tangent is an asymptote. Hence, an ordinate that
becomes infinite for a definite value of x is an asymptote.
Again, if not only a, but also b and c are zero, there are
two values of x that make y infinite ; namely, those values
of x that make dx 2 + ex +f= 0. The equations of the
infinite ordinates are found by factoring this last equation;
and so on.
Similarly, by arranging the equation of the curve according
to powers of x, we can easily find what values of y give an
infinite value to x.
Ex. 1. In the curve
2 x s + z 2 y + zy* = x 2 — y 2 — 5,
find the equation of the infinite ordinate, and determine the finite
point in which this line meets the curve.
This is a cubic equation in which the coefficient of y 8 is zero.
Arranged in powers of y it is
y 2( x + 1) + yx 2 + (2 x 8 t- z 2 + 5) =0.
APPLICATIONS TO CURVE TRACING
99
When x = — 1, the equation for y becomes
• y 2 + y + 2 = 0,
the two roots of which are y = oo , y = — 2 ; hence the equation of
the infinite ordinate is x + 1 = 0. The infinite ordinate meets the
curve again in the finite point (— 1, — 2).
Since the tern^in x 8 is present, there are no infinite values of x
for finite values of y.
Ex. 2. Show that the lines x = a, and y = are asymptotes to the
curve a 2 x = y(x — a) 2 (Fig. 31).
Fig. 31
Ex. 3. Find the asymptotes of the curve x 2 (y — a) + xy 2 = a 3 .
56. Method of substitution. Oblique asymptotes. The as-
ymptotes that are not parallel to either axis can be found by
the method of substitution, which is applicable to all algebraic
curves, and is of especial value when the equation is given in
the implicit form
/(*»y) = o. (i)
Consider the straight line
y = mx + b,
(2)
100 DIFFERENTIAL CALCULUS
and let it be required to determine m and b so that this line
shall be an asymptote to the curve /(a*, y) = 0.
Since an asymptote is the limiting position of a line that
meets the curve in two points that tend to coincide at infinity,
then, by making (1) and (2) simultaneous, the resulting equa-
f(x, mx + b) = 0,
is to have two of its roots infinite. This requires that the
coefficients of the two highest powers of x shall vanish.
These coefficients, equated to zero, furnish two equations
from which the required values of m and b can be determined.
These values, substituted in (2), will give the equation of an
asymptote.
Ex. 4. Find the asymptotes to the curve y 3 = x 2 (2 a — x).
In the first place, there are evidently no asymptotes parallel to
either of the coordinate axes. To determine the oblique asymptotes,
make the equation of the curve simultaneous with y = mx + b, and
eliminate y. Then
(mx + b) 3 = x 2 (2a-x),
or, arranged in powers of x,
(1 + m 3 )x 3 + (3 m 2 b - 2 a)x 2 + 3 b 2 mx + b s = 0.
Let m 3 + 1 = and 3 m 2 b - 2 a = 0.
Then m = — 1, /> = =—;
o
hence y = - x + —
o
is the equation of an asymptote.
The third intersection of this line with the given curve is found
2 a
from the equation 3 mb 2 x -f b 8 = 0, whence x = —•
APPLICATIONS TO CURVE'. TRACING:
161
Fig. 32
This is the only oblique asymptote, as the other roots of the equation
for m are imaginary.
Ex. 5. Find the asymptotes to the curve y(a 2 + x 2 ) = a 2 (a - x).
Y
Fig. 33
Here the line y = is a horizontal asymptote by Art. 55. To find
the oblique asymptotes, put y = mx + b.
Then (mx + b) (a 2 4- x 2 ) = a 2 (a - x),
i.e. mx* + bx 2 + (ma 2 + a 2 )x + (a 2 b - a 8 ) = ;
hence m = 0, b = 0, for an asymptote.
Thus the only asymptote is the line y = already found.
1#2 DIFFERENTIAL CALCULUS
57. Number of asymptotes. The illustrations of the last
article show that if all the terms are present in the general
equation of an nth degree curve, then the equation for deter-
mining m is of the nth degree and there are accordingly n
values of ra, real or imaginary. The equation for finding b is
usually of the first degree, but for certain curves one or more
values of m may cause the coefficients of x n and x n ~ x both to
vanish, irrespective of b. In such cases any line whose equa-
tion is of the form y = m y x + c will have two points at infinity
on the curve independent of c ; but by equating the coefficient
of x n ~ 2 to zero, two values of b can be found such that the re-
sulting lines have three points at infinity in common with the
curve. These two lines are parallel ; and it will be seen that
in each case in which this happens the equation defining m
has a double root, so that the total number of asymptotes is
not increased. Hence the total number of asymptotes, real
and imaginary, is in general equal to the degree of the equation
of the curve.
This number must be reduced whenever a curve has a para-
bolic branch, since in this case a value of m which makes the
coefficient of x n vanish does not correspond to any finite value
of b.
Ex. 6. Find the asymptotes of the curve (x — y) 3 = 2 x. The
equation in m is (m — l) 3 = 0. The coefficient of z 3 vanishes identi-
cally when m = 1 ; that of x is 3(?n — l)b 2 - 2 which cannot be made
to vanish for any finite value of b when m = 1* The curve has no
asymptotes.
Ex. 7. Find the asymptotes of the curve
0r-l)( 2-^
* ' x - 3
and trace the curve. (Fig. 34.)
APPLICATIONS TO CURVE TRACING
103
Fig. U
EXERCISES
Find the asymptotes of each of the following curves:
1. y(a 2 - x 2 ) = b(2z + c).
9 x 2 -2ax
3. x 2 y 2 = a\x 2 - y 2 ).
b*
4. y = a +
5. y z = x 2 (a — x).
6. y\x - l) = x 2 .
15. x 8 + 2 x 2 y - xy 2 - 2 y s + 4 y 2 + 2 xy + y = 1
7. (* + a)y«=(y+6)*«.
8. x 2 y 2 = a: 3 -f # + y.
9. xy 2 + x 2 y = a 3 .
10. y(x 2 + 3a 2 )=a*
11. x 3 - 3 axy + y 8 = 0.
12. j; 3 + y 3 = a 3 -
13. x 4 - x 2 y 2 + a 2 x 2 + b* = 0.
14. x 4 - y 4 - a 2 *^.
104
DIFFERENTIAL CALCULUS
POLAR COORDINATES
58. When the equation of a curve is expressed in polar
coordinates, the vectorial angle is usually regarded as the
independent variable. To determine the direction of the
curve at any point, it is most convenient to make use of the
angle between the tangent and the radius vector to the point
of tangency.
Let P, Q be two points on the
curve (Fig. 35). Join P, Q with
the pole 0, and drop a perpendic-
ular PM from P on OQ. Let p,
6 be the coordinates of P ; p + Ap,
+ A6 those of Q. Then the angle
POQ = A0', PM=p sin A0; and
Fig. 35 MQ = OQ- OM=p + Ap-p cos A0.
Hence
tan MQP
p sin A6
p -f Ap — p cos A0
When Q moves to coincidence with P, the angle MQP
approaches as a limit the angle between the radius vector
and the tangent line at the point P. This angle will be
designated by if/.
Thus
tan if/
lim *
A0 = O
p sin A0
p + Ap — p cos A0
But p (1 - cos A0) = 2 P sin 2 i A0,
hence tan i/> = A0 l ™
p sin A0
1 A/i sin i A0 . Ap
sm i A 0--r-hr- + ^
±A0
APPLICATIONS TO CURVE TRACING
105
Since A l lr ? ft sm - == 1, the preceding equation reduces to
i±0
tan if/ =
dp
dB
dO
dp
(3)
Ex. 1. A point describes a circle of radius p.
Prove that at any instant the arc velocity is p times
the angle velocity,
ds
dt
' dt '
dt'
Ex. 2. When a point describes a given
curve, prove that at any instant the velocity
— has a radius component l -&- and a com-
dt dt
ponent perpendicular to the radius vector
p — , and hence that
r dt
Fig. 37
cos \p
ds
i dO , . r<
, sin \p = p — , tan \p = p -
ds
This furnishes a dynamical proof of equation (3).
If
59. Relation between -£- and p V
doc dp
the initial line is taken as the axis of x,
the tangent line at P makes an angle $
with this line.
Hence -f \p = <£ ;
dd\ .. _Udy y
i.e., + tan"
p — i = tan"
d Pj
dx
Fig. 38
60. Length of tangent, normal, polar subtangent, and polar sub-
normal. The portions of the tangent and normal intercepted
between the point of tangency Pand the line through the pole
perpendicular to the radius vector OP, are called the polar
106
DIFFERENTIAL CALCULUS
tangent length and the polar normal length; their projections on
this perpendicular are called the polar subtangent and polar
subnormal.
Fig. 39 a
Fig. 39 b
Thus, let the tangent and normal at P (Figs. 39 a, b) meet the
perpendicular to OP in the points jVand M. Then
PN is the polar tangent length,
PM is the polar normal length,
ON is the polar subtangent,
OM is the polar subnormal.
They are all seen to be independent of the direction of the
initial line. The lengths of these lines will now be determined.
Since PN=OP- sec OPN= p sec ^
flO
4
•\|j +1
hence polar tangent length = p
dp
(16
d P
\P
' + D
v
'<$)'■
Again, ON= OP tan OPN= p tan if/ = p
hence
polar subtangent = p' 2 —
dp
2 d$
dp'
APPLICATIONS TO CURVE TRACING 107
PM = OP • esc OPN= p esc ^ = yj/ + (&\ ,
hence polar normal length = -v/p 2 -h ( — ^* ) •
OM= OP cot OPN = *£,
dO
hence polar subnormal = -£ •
The signs of the polar tangent length and polar normal
length are ambiguous on account of the radical. The direc-
d6
tion of the subtangent is determined by the sign of p 2 — •
09. dp
When — is positive, the distance ON should be measured to
dp
the right, and when negative, to the left of an observer placed
at and looking along OP; for when increases with
d6
p, — is positive (Art. 28), and \p is an acute angle (as in
dp
d6
Fig. 39 b) ; when 6 decreases as p increases, _ is negative,
and if/ is obtuse (Fig. 39 a).
EXERCISES
1. In the curve p = a sin 0, find \J/.
2. In the spiral of Archimedes p = aO, show that tan xp = 6 and
find the polar subtangent, polar normal, and polar subnormal. Trace
the curve.
3. Find for the curve p 2 = a 2 cos 2 the values of all the expres-
sions treated in this article.
4. Show that in the curve p6 = a the polar subtangent is of con-
stant length. Trace the curve.
5. In the curve p = a(l — cos 0), find if/ and the polar subtangent.
108 DIFFERENTIAL CALCULUS
6. Show that in the curve p = b • e 9cota the tangent makes a con-
stant angle a with the radius vector. For this reason, this curve is
called the equiangular spiral.
7. Find the angle of intersection of the curves
p = a{l+ cos 0), p = h(l — cos 6).
a
8. In the parabola p = a sec 2 -, show that cp + $ = tt.
EXERCISES ON CHAPTER VII
Trace the following curves. Find asymptotes, intervals of in-
creasing and decreasing ordinate and direction of bending, as well as
intercepts on the axes.
1. y = x s + 2 x 2 - 7 x + 1. 5. y 2 = x s .
2. y 2 = x 3 + 2 x 2 -7 x + 1. 6. ay 2 = x 3 - hx 2 .
3. y = 2 -l) 2 . 7. z 4 -7/ 4 = 2x.
4. x 3 + ?/ 3 = 1.
In the following curves find \p, determine whether p can become
infinite, and obtain the (angular) intervals of increasing and decreas-
ing p.
8. p = a cos 2 6. 10. p = «(1 - cos 0).
a
9. p = a sin 3 0. "■• p = «sec 2 -.
CHAPTER VIII
DIFFERENTIATION OF FUNCTIONS OF TWO VARIABLES
Thus far only functions of a single variable have been con-
sidered. The present chapter will be devoted to the study of
functions of two independent variables x, y. They will be
represented by the symbol
2 =/(?, y)-
If the simultaneous values of the three variables x, ?/, z are
represented as the rectangular coordinates of a point in space,
the locus of all such points is a surface having the equation
61. Definition of continuity. A function z of x and y,
z=f(x, ?/), is said to be continuous in the vicinity of any point
(a, b) when /(a, b) is real, finite, and determinate, and such
h Y Tof(a + h,b + k)=f(a,b),
k =
however h and k approach zero.
When a pair of values a, b exists at which any one of these
properties does not hold, the function is said to be discontinu-
ous at the point (a, b).
E.g., let z= x -+JL.
x - y
When x = 0, then z — — 1 for every value of y ; when y = then
2=4-1 for every value of x. In general, if y — mx,
- - 1 + m
1 — m
109
110 DIFFERENTIAL CALCULUS
and z may be made to have any value whatever at (0, 0) by giving an
appropriate value to m.
Geometrically speaking, when the point (x, y) moves up to (G, 0),
the limiting value of the ordinate z depends upon the direction of
approach.
62. Partial differentiation. If in the function
* =/to y)
a fixed value y x is given to y, then
is a function of x only, and the rate of change in z caused by a
change in x is expressed by
dz = — dx, (1)
dx
dz
in which — is obtained on the supposition that y is constant.
dx
To indicate this fact without the qualifying verbal state-
ment, equation (1) will be written in the form
d^Ax. (2)
OX
dz
The symbol — represents the result obtained by differentiat-
ed
ing z with regard to x, the variable y being treated as a con-
stant; it is called the partial derivative of z with regard to x.
From the definition of differentiation, Art. 6, the partial
derivative is the result of the indicated operation
dz = lim f(x + Ax, y)-f(x,y)
dx Ax ~° Ax
Similarly, the symbol — represents the result obtained by
differentiating z with regard to y, the variable x being treated
FUNCTIONS OF TWO VARIABLES 111
as a constant; it is called the partial derivative of z with
regard to y.
The partial derivative of z with regard to y is accordingly the
result of the indicated operation
dz = lim f(x, y + &y)-f(x,y)
dy A y = ° Ay
dz
d x z = — die is called the partial x-differential of z, and
dx
dz
cLz = — dy is called the partial y-differential of z.
by
EXERCISES
1. Given u = x* + 3 x*y* - 7 xy s , prove that x ~-hy^=iu.
ox By
2. Given u = tan -1 ^, show that x— + y — = 0.
x dx dy
3. u = log (e* + e») ; find ^ + &
dx dy
4. n = sin ary; find ^ + ^.
dx dy
5. u = log (x + V^T^) ; find x~ + y|^
6. m = log (tan x + tan y + tan 2) ; show that
sin 2x^ + sin 2 y^ + sin 2 *<?H = 2.
dx dy dz
7. II = log (a; + y) ; show that f? + ^ = 1.
5a: dy e w
8. ti = -3L; show that *^ + «£!?=«.
*+3/ dx *dy
9. u = (y-«)( 2 -x)(x-y); show that ^ + f^ + ^ = 0.
dx dy dz
10 - » = ^ show that g+g = (» + »-D«.
11. u = log (x 3 + y 8 + z 8 - 3 xyz) ; show that
du , du .Qu 3
dx dy dz x + y-\- z
112 DIFFERENTIAL CALCULUS
63. Total differential. If both x and y are allowed to vary
in the function z=f(x, y), the first question that naturally
arises is with regard to the meaning of the differential of z.
Let z x = f(x x ,%),
and z x + Az = f(x l -\- Ax, y x -\- Ay)
be two values of the function corresponding to the two pairs
of values of the variables a^, y x and x x + Ax, y x + Ay.
The difference
Az = f(x x + Ax, y x + Ay) - f(x x , y x )
may be regarded as composed of two parts, the first part being
the increment which z takes when x changes from x x to x x + Ax,
while y remains constant (yz=y x ), and the second part being
the additional increment which z takes when y changes from
2/i to y x -f- Ay, while x remains constant (x = x x + Ax). The
increment Az may then be written
Az = f{x x + Ax, ^ + Ay) - ffa + Ax, ^)
4-/(«i + Ax, 2/0 -f(x x , y x )
- /fa + Aa? > 2/1 + A -y) ^ /fa + A:g > yQ a?/
Ay
+ f(x x + Ax,y x )-f(x x ,y l ) ^ x
Ax
From the theorem of mean value, Art. 39, the last equation
may be written
Az = A/fa + 6 Ax, y x ) Ax + A/fa + Ax, y, + $ x Ay) Ay. (3)
dx dy
It represents the actual increment Az which the dependent
variable z takes when the independent variables x and y take
the increments Ax and Ay.
FUNCTIONS OF TWO VARIABLES
113
To illustrate, let z = f(x, y) be the equation of a surface (Fig. 40).
Let Ai = (xi, y\), A 2 = (xi + Ax, y\), A%= (xi + Ax, y\ + Ay), so that
AiPi=f(x h y x ), AiPi~f{xi + Ax, y x ), A z Pi=f{x l + Ax, y x + Ay),
Q 2 P 2 =/(2i + Ax, yi) -f(x h yi) = Aiz,
Q z Pz=f(xi + Ax, yi + Ay) -f(xi + Ax, y{) = A 2 z,
RzP&=f{xi + Ax, y x + Ay) -/(a*, yi) = A x z + A 2 z = Az.
As the moving point P passes from P to
P-2 along the plane curve P\P 2 , the ordinate
takes the increment
T t
n
i?
f-' Q 3
'
Q*
/
i?a
A
'
V
A 2
A 3
Fig. 40
where the derivative is taken at the inter-
mediate point x = x\ + 6 Ax, y — y\ (Art. 39).
Similarly, as P passes from P 2 to Pz along /
the plane curve P 2 Ps, the ordinate takes the
further increment
where the derivative is taken at the intermediate point y = y\ 4- diAy,
x = xi -f Ax.
The sum of these two partial increments gives the total increment Az.
In the preceding equation (3) let Ax, Ay, Az be replaced by
c • dx, € • dy, e • dz respectively, in which dx, dy are entirely
arbitrary. After removing the common factor e, let e approach
zero. The result is
(4)
dx dy
The differential dz denned by this equation is called the total
differential of z. It is not an actual increment of z, but the
increment which z would take if its change continued uniform
while x changed from x x to x x + dx and y changed from y x to
Vi + dy.
EL. CALC — 8
114 DIFFERENTIAL CALCULUS
In other words, dz is the rate of change of the variable 2
when the independent variables x and y change simultane-
ously at the rates of dx, dy respectively.
Equation (4) may be written in the form
dz = ^dx + -^dy = d x z + d y z,
dx dy
from which the following theorem can be stated ; the total dif-
ferent icd of a function of two variables is equal to the sum of iU
partial differentials taken with regard to the separate variables,
or the total rate of change of z is equal to the sum of its par-
tial rates.
The same method can be applied directly to functions of
three or more variables. Thus, if u is a function of the vari-
ables x, y, z, u=^(x,y J z) )
then du = ~dx-\-^- dy + — * dz.
ox dy dz
Ex.1. Given z = axy' 2 -\-bx 2 y + ex 3 + cy,
then dz = (ay 2 + 2 bxy + 3 cx 2 )dx + (2 axy + bx' 2 + c)dy.
Ex. 2. Given u = tan -1 --, show that du — ^ ~ " —
x x' 1 + y' 2
Ex. 3. Assuming the characteristic equation of a perfect gas,
vp = Rt, in which v is volume, p pressure, t ahsolute temperature, and
11 a constant, express each of the differentials dv, dp, dt, in terms of
the other two.
Ex. 4. A particle moves on the spherical surface x 2 + y 2 + z 2 = a 2
in a vertical meridian plane inclined at an angle of 60° to the em-
plane. If the a; -component of its velocity is ^- feet per second when
x — - , find the y-component and the ^-component velocities.
Since z = Va 2 — x 2 — y 2 ,
then dz = ^ ■ y - V
Va' 2 — x 2 — y' 2 vV- — x 2 — y' 2
FUNCTIONS OF TWO VARIABLES 115
But since dx = — , and the equation of the given meridian plane is
V = x tan 60°, hence dy — VS dx = a — , and y = ^—— • Therefore
u j 10 4
dz = - — - & = - — in feet per second.
2V3 2 15
Ex. 5. A triangle has a base of 10 units and an altitude of 6 units.
The base is made to increase at the rate of 2 units and the altitude
to decrease at the rate of \ unit. At what rate does the area change?
Ex. 6. A point on the hvperboloid x 2 — &- = 1 in the position
4 5
x = 2, y — 2 moves so that x increases at the rate of 2 units per sec-
ond, while y decreases at the rate of 3 units per second. Find the
rate of change of z.
Ex. 7. If the area of a rectangle A = xy is incorrectly measured
owing to a small error dx, dy in the length of each side, how close
is dA = xdy + ydx to the actual error in the area?
64. Total derivative. If in the relation z=f(x, y), the vari-
ables x, y are not independent, but both are functions of
another variable s, the process of the preceding article can
still be applied. The variable z is now a function of s, and
its derivative as to s may be expressed in the form
dz _ dz dx dz dy
ds dx ds dy ds
In particular, if y is not independent, but is a function of
x, then s may be chosen as x itself, and the preceding equation
becomes '
dz _ dz dz dy
dx dx dy dx
If the functional relation between x and y is given,
y = <*>0)>
116 DIFFERENTIAL CALCULUS
dz
then the same result will be obtained, whether — is deter-
dx
mined by the present method, or y is first eliminated from the
relation
*=f(®, y),
and the resulting equation is differentiated as to x. The
method of this article frequently shortens the process.
It is here well to note the difference between — and — •
dx dx
The former is the partial derivative of the functional expres-
sion for z with regard to x, on the supposition that y is con-
stant. The latter is the total derivative of z with regard to
x, when account is taken of the fact that y is itself a func-
tion of x.
In the former case the differentiation with regard to x is
merely explicit ; in the latter it is both explicit and implicit.
dz
Ex. 1. Given z = Vx 2 -f y 2 , y = log x; find
dx
dz _ x y dy
dx y/ x 2 _f. yi \Ac 2 + y 2 dx
d]L-_
dx
_1
x
dz
X 2
+ y
dx
x yJx
2 + yl
hence
Ex. 2. If z = tan-i ^- and 4 x 2 + y 2 = 1, show that ^ = — .
2x dx y
65. Differentiation of implicit functions. If, in the relation
z =f(x, y), z is assumed to be constant, then
dz = :
FUNCTIONS OF TWO VARIABLES 117
hence ¥-dx + d fdy = 0, (1)
dx By
from which ^ = -37- ( 2 )
dy_
dx
df
dx
dy
In all such cases either variable is an implicit function of
the other, and thus the last equation furnishes a rule for
finding the derivative of an implicit function.
Ex. 1. Given x 3 + y 3 + 3 axy = c, find -^.
dx
Since (3 x 2 + 3 ay) + (3 y 2 + 3 ax) ^ = 0, ^ = - x * + ay .
dx dx y 2 + ax
Ex. 2. /(a* + ty) = c ; J£ = a/'(ax + by) ; |£ = &/'(«* + by) ;
<7y _ _ a
</x~ &'
Ex. 3. If ax 2 + 2 fary 4- % 2 + 2 gx + 2fy + c = 0, find ^.
</x
Ex. 4. Given x 4 - y4 = c, find^.
</x
Ex. 5. If x increases at the rate of 2 inches per second as it passes
through the value x = 3 inches, at what rate must y change when
y = 1 inch in order that the function 2 xy 2 — 3 x 2 y shall remain
constant ?
If
u = 2 xy 2 - 3 x 2 y,
then
du n i a du a
— = 2y 2 -Qxy, ^-=± X y-
dx dy
du dy
dy _ dx _ 2y 2 -6xy _dt
dx ~ du ~ 4 xy — 3 x 2 ~~ </x
dy dt
Since x = 3, y — 1, -y = 2, hence, -y = - 2 T 2 5 inches per second.
118 DIFFERENTIAL CALCULUS
Ex.
6.
u — v 2 -f vy, v = log s, y = e 8 . Find —
Ex.
7.
u = sin" 1 (r - a), r = 3 7, s = 4 * 3 . Fin
Ex.
8.
e » _ c * + a^ = o. Find ^.
Ex.
9.
sin (*#) - e** - xhj = 0. Find ^.
du
dt
It is to be noticed that the result of differentiating any implicit
function of x, y by the method of the present article will agree with
the result of differentiation according to the rules of Chapter II.
66. Geometric interpretation. Geometrically, the equation
z =f(x, y) represents a surface. The equation y = y x defines
a plane parallel to the ^-coordinate plane. The two equations
treated simultaneously therefore define the plane section made
on the surface z = f(x, y) by the plane y = y x . The derivative
dz
— - defines the slope of the tangent line to this curve at the
ax x
point (x l} y u z x ).
Similarly, the plane x = x l cuts the surface in a section
parallel to the yz-coordinate plane. The slope of the tangent
dz
line to this second curve is defined by — -. The equations of
these two lines are
V = 2/i, z - z 1 = — i (x - x x ),
dx x
dz
x = x x , z-z x = -± (y - ?/,).
They have the point (x x , y x , z x ) in common ; hence the two lines
will define a plane. The equation of any plane through the
first line will be of the form
dXi
+ *(//-*/,)= o,
FUNCTIONS OF TWO VARIABLES 119
and similarly, the equation of any plane through the second
line will be of the form
oyi
+ k'(x-x 1 )=0.
These two equations will be identical when
dz, , dz-,
K = ~ ^H~> k = — — -j
tyi Mi
hence the equation of the plane containing both lines is
2 - *i = -p- (x - a^H ^-{y - vi).
dx x dy x
It is called the tangent plane to the surface z = f(x, y) at the
point («!, y v z x ).
From the equation
dM = %Ldx+pLdg, (3)
ax l dy 1
it is seen that if x, y receive the arbitrary increments dx, dy,
then the increment dz is defined by the sums of the products
of these increments by the corresponding partial derivatives.
Thus, if dx = x — X], dy = y — y l9 dz = z — z l9 it is seen that
the point (x, y, z) always lies in the tangent plane to the sur-
face z =f(x, y), however the increments dx and dy approach
zero.
Moreover, the equations of the line joining (x lf y l9 z x ) to
Xj -f A#, ?/! + Ay, z y + Az on the surface will be of the form
x - :> \ __ y - y l _ z - z x
Ax A?/ Az
120 DIFFERENTIAL CALCULUS
Now as Ax, Ay approach zero, the point always remaining
on the surface, the line becomes a tangent in the limit, and its
equations are
x~x 1 = y-y 1 = z-z x (i)
dx dy dz ' W
wherein dx, dy depend upon the direction of approach, and dz
is defined by (3).
But a tangent line to the surface is also tangent to any plane
section passing through the line, and the line (4) is seen to lie
in the tangent plane, hence :
TJie tangent lines to all the plane sections of the surface
z =f(x, y) passing through the point (x x , y x , Zj) lie in the tangent
plane at that point.
The line through (x Y , y ly z ± ) perpendicular to the tangent
plane dz dz
t-^^-^+dy 1 ^-^
is called the normal to the surface at the point (x^ y x , z^). Its
equations are
X - Xl y-y x . z-z x
dz x dz x — 1
dx x dy x
If the equation of the surface is given in the implicit form
F(x, y, z) = 0, then since
dF, , dF , , dF, ft
-—dx-\ rt?y -| dz = 0,
ox dy ' dz
the equation of the tangent plane becomes, if F(a; 1 , y lf z^) = F 1?
dxt dt/t dz l
FUNCTIONS OF TWO VARIABLES 121
and those of the normal are
3-3* 1 = y—V\ _ g-g i
bFy dF\ BF X '
dx x dy 1 dzj
EXERCISES
1. Show that the plane z = touches the surface z = xy at (0, 0, 0).
2. Find the equation of the tangent plane to the paraboloid
z = 2 x 2 + 4 y 2 at the point (2, 1, 12).
3. Find the equations of the normal to the hyperboloid
x 2_4y 2 +2z a =6at (2,2,3).
4. Show that the normal at any point (x h yi, z{) on the sphere
x 2 + y 2 + z 2 = 16 will pass through the center.
5. Find the equation of-the tangent plane at any point (x\, y h z\)
of the surface x 1 + y* + z 5 = a* and show that the sum of the squares
of the intercepts which it makes on the coordinate axes is constant.
6. Show that the volume of the tetrahedron cut from the coor-
dinate planes by any tangent plane to the surface xyz = a 3 is constant.
7. The sphere x 2 + y 2 + z 2 = 14 and the ellipsoid 3 x 2 +2 y 2 + z 2 = 20
pass through the point (—1,-2,-3). Determine the angle at
which their tangent planes at this point intersect.
8. How far distant from the origin is the tangent plane to the
ellipsoid x 2 + 3 y 2 + 2 z 2 = 9 at the point (2, -1, 1) ?
9. Find the equation of the tangent plane and of the normal to
the cone z 2 = 2 x 2 + y 2 at (x ly yi, z\) on the surface. Show that the
plane will always pass through the vertex of the cone.
10. Find the equations of the tangent line to the circle
x 2 + y 2 + z 2 = 25,
x + z = 5,
at the point (2, 2V3, 3).
122 DIFFERENTIAL CALCULUS
67. Successive partial differentiation. The expressions
dz dz
— , — which were denned in Art. 62 are functions of both
ox oy
x and y.
dz
If — is differentiated partially as to x, the result is written
ox
d_fdz\ == dh m
dx\dxj dx 2
This expression is called the second partial derivative of
z as to x.
Similarly, the results of the operations indicated by
JL{<>?\ JL(te\ JL(te\
dy\dxf dx\dyf dy\dyj
dh dh d 2 z
are written -— -, — — , -— respectively.
oy ox ox oy oy z
Beginning with the left, we call these expressions the
second partial derivative of z as to x and y, the second partial
derivative of z as to y and x, and the second partial derivative
of z as to y.
68. Order of differentiation indifferent.
Theorem. The successive partial derivatives
d 2 z dh
By dx dx dy
are equal for any values of x and y in the vicinity of which
z and its first and second partial x- and ?/-derivatives are
continuous.
The truth of this theorem will be assumed. Tt should be
verified for special cases as in the following examples.
FUNCTIONS OF TWO VARIABLES 123
Cor. It follows directly that under corresponding conditions
the order of differentiation in the higher partial derivatives is
indifferent.
dh d s z . ffhs
E.g.,
dx dy dx dx 2 dy dy dx 2
EXERCISES
1. Verify that -^_ = J£}L. % w hen u = xh/\
dx dy dy dx
2. Verify that -_^H_ = J^, when u = xhj + xy 9 .
dx dy 2 dy' 2 dx
3. Verify that -^L = -^-, when u = y log (1 + xy).
dx dy dy dx
4. In Ex. 3 are there any exceptional values of x, y for which the
relation is not true?
5. Given u = (x 2 + y 2 )l, verify the formula
* 2 ^ + 2*^ + ^ = 0.
x 2 dx dy dy 2
6. Given u = (x s -f t/ s ) I, show that the expression in the left
member of the differential equation in Ex. 5 is equal to — .
7. Given u = (x* + y* + z *yh ■ prove that B ~ + ^ + = 0.
8. Given u = sec (y+ax) + tan (?/ - ax) ; prove that — = a^' u -
dx 2 dy 2
9. Given M=sinx cosy; verify that d u = ^ = ^ u .
J dy 2 dx 2 dxdydxdy dx 2 dy 2
d 2 u d 2 u
10. Given «=(4a&- c 2 )-| ; prove that :rr = r - ^!*
11. If «=*&L, show that o;^ + „-i^-=2^.
* + # dx* J dxdy dx
12. Given u = log (a: 2 + y 2 ), prove 5 ^ + — = 0.
a< 2 ■ ay
13. If u = (x n + y n y, show that the equation of Ex. 5 is satisfied.
14. Given u = (x* + y* + z 2 + w 2 )" 1 , prove ^ + <*% + ^f + S^M. =
d* 2 9^ a* 2 Qw*
CHAPTER IX
CHANGE OF VARIABLE
69. Interchange of dependent and independent variables. If
y is a continuous function of x, defined by the equation
f(x, y) = 0, the symbol — represents the derivative of y with
regard to x, when one exists. If x is regarded as a function of
y, defined by the same equation, the symbol — represents the
dy
derivative of x with regard to y, when one exists. It is re-
quired to find the relation between -^ and —
dx dy
Let x, y change from the initial values x ly y x to the values
x 1 + Ax, y x + Ay, subject to the relation f(x, y) = 0.
Then, since A
Ay _ 1
Ax Ax
Ay
it follows, by taking the limit, that
dx dx
dy
(1)
Hence, if y and x are connected by a functional relation, the
derivative of y with regard to x is the reciprocal of the derivative
of x with regard to y.
This process is known as changing the independent variable
from x to y. The corresponding relations for the -higher de-
124
CHANGE OF VARIABLE
125
rivatives are less simple. They are obtained in the following
manner :
d y dx d x
To express — 4 in terms of — , — - differentiate (1) as to x ;
dx 2 dy dy 2
d 2 y _ d [ 1
d
[ 1
dy _ d
\ 1 1
dx 2 dx 1 dx
dy
dx
dx dy
dx
idy
dy\
idyl
dx
dy
But
hence
In a similar manner,
— l 1 }-
dy dx 1
idyl
d 2 x
dy 2 ,
/dx\ 2
[dyj
d?x
d 2 y
dx 2
dy 2
fdx\ 3 '
\dy)
ner,
d 3 y
d 3 x dx o fd 2 x\i
d?dy WJ
dx 3
fdx\ 5
\dy)
(2)
(3)
70. Change of the dependent variable. If y is a function of z,
let it be required to express — — , •••in terms of — , — -, •••.
dx dx- dx dx-'
Suppose y = <f>(z). Then
dy ^dycte = ^u\dz_
dx dz dx dx
dx 2 dxV KJ dx
dz d .i^\,,i/^d 2 z
126 DIFFERENTIAL CALCULUS
But ^-<l>'(z) = -cp'(z)- = <p"(z) — ■
dz* K ; dz v ; <to v v "eta
S-^S'+^S- (4)
The higher ^derivatives of y can be similarly expressed in
terms of ^-derivatives of z.
71. Change of the independent variable. Let y be a function
of x, and let both x and y be functions of a new variable I. It
is required to express — in terms of -3L. and — in terms of
1 , cto eft' dx 2
-■- and — a-
eft eft 2
By Art. 8, e%
% = d* ? (1)
die do?
hence
d 2 ?/ C?£C d 2 ^ e%
^y_ ~dfdi~~d^di (9)
fdx\*
dt
In practical examples it is usually better to work by the
methods here illustrated than to use the resulting formulas.
72. Simultaneous changes of dependent and of independent vari-
ables. Suppose, for example, that an equation involving x, y,
— , . . . is given, and it is required to transform the equation
into polar coordinates by means of the formulas x = p cos 0,
y = p sin 6. Since the variables x and y are connected by some
equation (y being a function of x), we may regard x, y, p as
CHANGE OF VARIABLE
functions of 6. E.g., consider the function
12'
11 =
HtJf
dry
dx 2
rom Art. 71,
dy
dy _d0 f
dx dx
dO
dx d-y dy drx
d 2 y dO ' d6- dO ' d0\
dx-
/dxX 3
tie.
By substituting these values in the expression for E, it becomes
R
dx\ 2 , (dy
dO \d0
dx d~y dy ^ d 2 x
dO ' ~d&- ~ dd ' d¥
This is in terms of a new independent variable 0. We have
now to express these ^-derivatives of x and y in terms of p
and 6.
From the relations x = p cos 0, y = p sin we have
g = - p cos*-2sin*| + cos*g,
dhj
d$-
' = - p sin + 2 cos =c 4- sin
dp
d$
de 2
128 DIFFERENTIAL CALCULUS
Upon substituting these values in the last expression for B,
we obtain
B =
W]
f + 2 («!>)*- p *E
(16
EXERCISES
1. Change the independent variable from x to z in the equation
x 2 — \ + x — '- + y = 0, when x = e*.
dx 2 dx a
dx dz
dx 2 dz 2 dz
Hence x 2 ^- + x^- + y = becomes ^ + y = 0.
2. Interchange the function and the variable in the equation
dx 2 \dxj
3. Interchange x and y in the equation
</x 2
4. Change the independent variable from ar to y in the equation
JdY\ 2 _dy d*«_d 2 y (djY = o
V^ 2 / dx dx* dx 2 \dx)
5. Change the dependent variable from y to z in the equation
!^= 1 +2(1 +.V)fM', when y = tans.
</x2 1 + y* \dxl *
CHANGE OF VARIABLE 129
6. Change the independent variable from x to y in the equation
x 2 h x \- u = 0, when y = log x.
dx 2 dx
7. If y is a function of x, and x a function of the time t, express
the ^/-acceleration in terms of the ar-acceleration, and the x-velocity.
Since dy = dydx
dt dx dt
hence
d 2 y _ dy (Px .dx d fdy\
dt 2 ~ dx dt 2 dt <it\dx)'
c ]l( ( 111\ = ^_((J][\flx = d 2 y(lx
it\dxl ~ dxKdxj dt dx 2 dt'
But
dt
d 2 y _ dy d 2 x d 2 y ldx\ 2
hence ^-s^ + sUi'
In the abbreviated notation for /-derivatives,
dx dx 2
8. Change the independent variable from x to u in the equation
dry 2 x dy y
dx 2 1 + x 2 dx (1 + x 2 ) 2
] y + y = 0, when x = tan u.
9. Change the independent variable from x to t in the equation
(1 _ X 2) A _ x ( Il = o, when x = cos t.
dx 2 d-x
10. Show that the equation
d 2 y , dy . A
dx 2 dx J
remains unchanged in form by the substitution x = .
EL. CALC 9
130 DIFFERENTIAL CALCULUS
11. Interchange the variable and the function in the equation
dx 2 \dxl y \dx)
12. Change the dependent variable from y to z in the equation
|j*+ a - y)% + y 2 = °' when y = z ' 2 -
Change the independent variable from x to t in the equations :
13. CI — x 2 ) — — x— + y = 0, given a: = cos t.
K clx 2 dx y ' h
14. x 3 — - + 3 x 2 \- x 1- v = 0, given x — e l .
dx 3 dx' 1 dx
15. x 4^ +fl 2^ = 0, x = -.
dx 2 t
dy
X dx~ V
16. Transform by assuming x = p cos 6, y = p sin 0.
17. Given z = 7 + * 2 , ?/ = 3 + * 2 - 3 l\ Find — .
CHAPTER X
EXPANSION OF FUNCTIONS
It is sometimes necessary to expand a given function in a
series of powers of the independent variable. For instance,
in order to compute and tabulate the successive numerical
values of sin x for different values of x, it is convenient to
have sin x developed in a series of powers of x with coeffi-
cients independent of x.
Simple cases of such development have been met with in
algebra. For example, by the binomial theorem,
(a + x) n = a n + na n ~ l x + n l - 1 ) a — v + • . . ; (1)
J. • A
and again, by ordinary division,
j-L-=l + s + 3» + aj »+.... (2)
It is to be observed, however, that the series is a proper
representative of the function only for values of x within a
certain interval. For instance, the identity in (1) holds
only for values of x between — a and -f a when n is not a
positive integer ; and the identity in (2) holds only for values
of x between — 1 and -f 1. In each of these examples, if a
finite value outside of the stated limits is given to x, the sum
of an infinite number of terms of the series will be infinite,
while the function in the first member will be finite.
131
132 DIFFERENTIAL CALCULUS
73. Convergence and divergence of series.* An infinite series
is said to be convergent or divergent according as the sum of the
first n terms of the series does or does not approach a finite
limit when n is increased without limit.
Those values of x for which a series of powers of x is con-
vergent constitute the interval of convergence of the series.
For example, the sum of the first n terms of the geometric
series , , 2 , s ,
a -f- ax 4- ax z + ax? -\
ail — x n )
is s n = -1 >-.
1 — x
First let x be numerically less than unity. Then when n is
taken sufficiently large, the term x n approaches zero ;
hence Hm s n = -^--
W = co l_ x
Next let x be numerically greater than unity. Then x n be-
comes infinite when n is infinite ; hence, in this case
lim
,,-od *» = «>•
Thus the given series is convergent or divergent according
as x is numerically less or greater than unity. The condition
for convergence may then be written .
-l<a<i,
and the interval of convergence is between — 1 and -|- 1.
Similarly the geometric series
l-3z + 9z 2 -27^+..., >
* For an elementary, yet comprehensive and rigorous, treatment of this
subject, see Professor Osgood's " Introduction to Infinite Series" (Harvard
University Press, 1897).
EXPANSION OF FUNCTIONS 133
whose common ratio is — 3 x, is convergent or divergent accord-
ing as 3 a; is numerically less or greater than unity.
The condition for convergence is — 1 < 3 x < 1, and hence
the interval of convergence is between — ^ and -f- ^.
74. General test for convergence.
Let £ = ?<! + u 2 + u 3 -\ hM n + M B+1 H
be a series of positive terms having the property that w+1 < r
(r a fixed proper fraction) for all values of n that exceed a def-
inite integer k that can be assigned. We wish to prove that
under these conditions S is convergent. This is called the ratio
test for convergence.
According to hypothesis we have the inequalities
^±l<r, ^*±*<r, ^±- 3 <r, etc.
U k U k+\ U k+2
By multiplying the first two equalities together we obtain
^^ < r 2 ; then, multiplying this result by the third of the
u k
given inequalities we deduce further -^ ±? <r 3 ; and so on. These
results may be written in the form
%+i < ru k , u k+2 < r*u k , w 4+3 < rhi k , .-• , u k+p < r'u k .
Hence we have the inequality
S < u x -f u 2 H (- u k -f ru k -f r 2 ^ + rX -\
But the series in the right member, which may be denoted by
S'j can be put in the form
S' = ^ + W2 -f — + m*_i + u h (1 -f r + r 2 + r 3 + •••)
= w x + w 2 H f- ?^_! + -^*- •
1— r
134 DIFFERENTIAL CALCULUS
The terms u x , u 2 , ••• , u k being assumed finite, it follows that
S' is finite and hence S, which is less than S', also is finite.
Since S is formed by the successive addition of positive terms,
it follows that the series S converges towards a definite finite
limit.
If the series S contains an infinite number of negative, as
well as of positive, terms, it converges whenever the series
formed by the positive, or absolute, values of its terms eon-
verges. The series is then said to be absolutely convergent.
In order to prove the preceding theorem, we obeerve that
the positive terms of S taken alone form a converging series,
whose limit will be denoted by P, and the negative terms taken
alone will form a converging series whose limit will be denoted
by — J¥. t Let S m denote the sum of the first m terms of S and
suppose that these consist of p positive terms whose sum is
denoted by P p and of n negative terms whose sum is — N n .
Then we have S m = P p — N n . Now when m becomes infinite,
p and n also become infinite, and hence
a lim - e . . Km p _ lira jy __ p _ *r
m i oo m p = oo p n = oo »
Therefore, S is convergent.
When a series is convergent, but the series formed with
the absolute values of its terms is not convergent, the given
series is said to be conditionally convergent*
The absolute value of a real number u is its numerical value
taken positively, and is written | u |.
If a series consists of terms that are alternately positive
* The appropriateness of this terminology is due to the fact that the terms
of an absolutely convergent series can be rearranged in any way, "without
altering the limit of the sum of the series ; and that this is not true of a con-
ditionally convergent series. For a simple proof, see Osgood, pp. 43, 44.
EXPANSION OF FUNCTIONS 135
and negative, and if, after any definite term of the series, each,
succeeding term is numerically less than the preceding one,
then the series is convergent.
For, suppose that beginning with the term n k , the series is
&' = V k ~ V k + l + %+2 — U k+3 + U k+i ~ '" >
in which u k , v k+l , etc. represent positive numbers and n k+l < u k ,
u k+2 < Uk+u ••' j n m+i < l 'mi f° r every value of m greater than 7c.
By grouping the terms in pairs, (u k — ii k+l ), (u k+2 —u k+ ^), ••• , each
of which is positive, it is seen that S' has a positive value,
which .may be finite or infinite.
But S' may also be written in the form
& = u k - [(%+i - *<*+ 2 ) + (m*+8 - u k+i) + •••],
wherein the terms in brackets are all positive, hence S' has a
value less than u k . It therefore converges towards a definite
finite limit.
It now follows that the approximate value of S' obtained by
algebraically adding u k , u k+l , ••• , u m differs from the true value
of the series by a number less than u m . This fact can be
shown in precisely the same way as that by which S' has just
been shown to have a value less than u k .
Ex. 1. Is the series 1 1 \- ... + (— l)' 1-1 - + ••• con-
2 3 4 n
vergent i
Since the terms are alternately positive and negative and their
numerical values are always decreasing, it follows at once from the
preceding paragraph that this series is convergent. It will be found
later that its value is log 2.
Ex. 2. Prove the convergence of the series met with in Art. 16,
a+ i. + l + i + ...+J- + ...-.
•2! 31 4! T n\
136 DIFFERENTIAL CALCULUS
In this case u n = 1 , w n+1 = 1— . Hence ^±1 = — 1— . This
n ! (w + 1) ! m„ n + 1
ratio is less than \ for all values of n greater than 2, and the ratio
condition for convergence is satisfied.
Ex. 3. Prove the divergence of the harmonic series
The ratio u n+ i : u n becomes greater than r when n is sufficiently
large. By grouping the terms it may be written in the form
1 + *+(* + *) + (* + * + * + »+-,
the succeeding groups having 2 3 , 2 4 , ••• , 2 n , ••• consecutive terms re-
spectively. The sum of the terms in any group is greater than \.
For, in the nth group the last term — has the least value, and as there
are 2 n_1 terms in the group their sum is greater than 2 n_1 — = -•
As there is an infinity of such groups, their sum is infinite.
Ex. 4. The series
£=1 + 1 + 1-+ ... +1+...
is convergent for p > 1.
Let the terms of S be grouped in the following manner :
the nth group beginning with and containing 2 n_1 terms.
The nth group is accordingly less than its first term multiplied by
the number of terms in the group, that is, < 2 n_1 • — = — — .
& r ' ' (2" _1 ) p (2 n— 1 ) p_1
Hence we deduce the inequality
the right member of which is a geometric series having — — as the
common ratio. It is therefore convergent, and hence S is convergent,
EXPANSION OF FUNCTIONS 137
if < 1. This inequality is satisfied for every value of ^greater
than unity. Moreover, it was shown in Ex. 3 that ior p = 1 the series
5 is divergent. When p < 1, S is divergent. For in that case
— ->-, n is any positive integer (except 1), and therefore the
n p n
terms of S are greater than the corresponding terms of the harmonic
series.
Hence :
The necessary and sufficient condition that the series 1 -\ — + - \- •••
may converge is p > 1.
Ex. 5. Show that the- series 1 = — \-
12 2-3 3.4+ n(n+l)
is convergent.
This may fc>e proved by comparison with the series in Ex. 4 for the
particular case p = 2.
1 .-, 1 .1 1 .1 1 .1
Since
l-2 < ' 2-3 < 2 2 ' 3-4 < 3 2 ' '" ' n(n + 1) < n*' "'
it follows that the value of the given series is less than that of
which is known to be convergent on account of the theorem deduced in
the preceding example.
Ex. 6. Examine for convergence the series whose nth term is — — ■
n 2 +l
r n = * ^ - 1
Hi nt. n 2 + 1 1 ^ n + 1
n -\
n
Ex. 7. Examine for convergence the series
l_ 2 f i _ . , (~l) w ~^
2 5 10 n 2 + 1
Ex. 8. Determine whether the series whose nth term is
n 2 + 1
convergent or not; the series whose general term is — — .
n 3 + l
138 DIFFERENTIAL CALCULUS
75. Interval of convergence. If the terms u 1} u 2 , ••• of a given
series are functions of a variable x, then the series will usually
converge for .some values of x and diverge for all others. In
such a case the problem is to determine the interval of conver-
gence, that is, the range of values of x for which the series is
convergent The following examples will illustrate the method
of procedure.
Ex. 1. Determine the interval of convergence of the series
1 + x -f 2 x 2 + 3 x s + ••• + nx n + ... .
In this case u n = (n — l)^"" -1 and u n+1 = nx n ..
Hence, ^ = "*" = -JL- x.
u n (n-l)x n - 1 n-1
According to the ratio condition for convergence, it is necessary
that this ratio shall be numerically less than 1 for all values of n
exceeding a fixed number k. As n increases, the fraction
71—1
approaches unity. Hence if [ x \ has any fixed value less than 1, the
given series is absolutely convergent. The interval of convergence
is defined by the inequalities — 1 < x < 1.
It is evident from the preceding example that the ratio con-
dition for the absolute convergence of a series ma}^ be written
lim
l ±n±l
< 1, (3)
which is especially convenient for application.
Ex. 2. Find the interval of convergence of the series
1 + 2 • 2 x + 3 • 4 x 2 + 4 • 8 x* + 5 • 16 x* +.- .
EXPANSION OF FUNCTIONS 139
Here the nth term u n is n 2 n_1 x n_1 , and the (n -f l)th term u n+1 is
(n + \)2 n x n ;
i ^n+i (n 4- l)2"z* (n + 1) 2 x
lience -J!±±=*-— — ^ = i — — i
w n w2 n - 1 x" » n
therefore when n = co, -^ = 2 x.
It follows by (3) that the series is absolutely convergent when
— 1 <2#<1, and that the interval of convergence is between — \
and + \. The series is evidently not convergent when x has either
of the extreme values.
Ex. 3. Find the interval of convergence of the series
x
1-3 3
x 3 x 5 x" > (_l)n-i^2 M -i ^
• 3 3 5-3 5 7-3 7 "" (2n-l)3 2 "- 1
Here
t/ H 4.i I _ 2n - 1 3 2 ""* a; 2 "* 1 = 2n-l z 2 .
w ft 2 n + 1 ' b 2 "* 1 ' x 2 "-i 2 n + 1 ' 3 2 '
v Inn
hence
n = co
3 2 '
a: 2
thus the series is absolutely convergent when — < 1, i.e., when
b
— 3 < x < 3, and the interval of convergence is from — 3 to + 3.
The extreme values of x, in the present case, render the series con-
ditionally convergent.
Ex. 4. Determine the interval of convergence of the series
2 ! 4 ! 6 ! . ' (2 n - 2) !
Since even powers of x are positive, the terms of this series are
alternately positive and negative. The term w n+1 is derived from u n
by multiplying it by . For all values of n such that
(2 n — 1)2 n
this fraction is less than 1, we shall have the condition [ u n+1 | < | n n \
and the series is convergent on account of the property of series with
alternately positive and negative terms.
140 DIFFERENTIAL CALCULUS
Ex. 5. Prove the convergence of the series
Inthiscase | u n |=z 2 - f 1 +-+••• + — Y Notice that h . m I u J
\9 4 2"- 1 / n=<x> ' n[
is not zero. The series is nevertheless convergent, but not absolutely
convergent.
Ex. 6. Determine the interval of convergence for the series
16 n
Ex. 7. Determine the interval of convergence for the series
-A_ + 2 3 + ... + _n _ + ....
x-1 (x-1) 2 (ar-l)* (ar-l) w
Ex. 8. Find the interval of convergence for the binomial series
-, , , a(a — 1) 9 , a(a — Y)(a — 2) „ ,
1 4-rtxH — i — *- x 2 H — ^ ^ '- x 3 +
in which a is any constant.
Ex. 9. Show that the series
has the same interval of convergence as that of Ex. 3 ; but that the
extreme values of x render the series absolutely convergent.
76. Remainder after n terms. The last article treated of
the interval of convergence of a given series without reference
to the question whether or not it was the development of any
known function. On the- other hand, the series that present
themselves in this chapter are the developments of given func-
tions, and the first question that arises is concerning those values
of x for which the function is equivalent to its development.
EXPANSION OF FUNCTIONS 141
When a series 1ms such a generating function, the difference
between the value of the function and the sum of the first n terms
of its development is called the remainder after n terms. Accord-
ingly, if f(x) is the function, S n (x) the sum of the first n terms
of the series, and li n (x) the remainder obtained by subtracting
S„(x) from fix), then
in which S n (x), R n (x) are functions of n as well as of x.
If JToO R -^) = > tllei1 Hi" «-(*) =/(*) i
thus the limit of the series S n (x) is the generating function
when the limit of the remainder is zero. Frequently this is
a sufficient test for the convergence of a series.
If a series is expressed in integral powers of x — a, the pre-
ceding conditions are to be modified by substituting x — a for
x ; in other respects each criterion is to be applied as before.
77. Maclaurin's expansion of a function in a power-series.*
It will now be shown that all the developments of functions
in power-series given in algebra and trigonometry are but
special cases of one general formula of expansion.
It is proposed to find a formula for the expansion, in
ascending positive integral powers of x — a, of any assigned
function which, with its successive derivatives, is continuous
in the vicinity of the value x = a.
The preliminary investigation will proceed on the hypothe-
sis that the assigned function f(x) has such a development,
♦Named after Colin Maclaurin (1608-1746), who published it in his
" Treatise on Fluxions " (1742) ; but he distinctly says it was known by
James Stirling (1692-1770), who also published it in his " Methodus Differ-
entialis "(1730), and by Taylor (see Art. 78).
142 DIFFERENTIAL CALCULUS
and that the latter can be treated as identical with the former
for all values of x within a certain interval of equivalence that
includes the value x = a. From this hypothesis the coeffi-
cients of the different powers of x — a will be determined. It
will then remain to test the validity of the result by finding
the conditions that must be fulfilled, in order that the series
so obtained may be a proper representation of the generating
function.
Let the assumed identity be
fix) = A+B(x-a) + C(x- a) 2 + D(x- a) 3
+ E(x-ay+-, (1)
in which A, B, C, ••• are undetermined coefficients, indepen-
dent of x.
Successive differentiation with regard to x supplies the
following additional identities, on the hypothesis that the
derivative of each series can be obtained by differentiating it
term by term, and that it has some interval of equivalence
with its corresponding function :
f(x) = B + 2C(x-a)+ 3D(x-a) 2 + ±E(x-af+ •••
f"(x)= 2(7 +3-2D(x-a) + 4 . 3E(x-af+ •-.
f"(x)= 3.2Z> +4.3.2^(»-o)+-.-
If, now, the special value a is given to x, the following
equations will be obtained:
f(a)=A,f(a) = B, /"(a) = 2 C,f"(a) = 3-2D, -.
Hence,
A =/(«), B=/(a), C= f -^f, D= f -^, ■-
EXPANSION OF FUNCTIONS
143
The coefficients in (1) are now determined, and the required
development is
/(a5)=/(a) + /(a)(a5-a)+^^(aj-a)«+£^(a5-a)»
(2)
2!
This is known as Maclaurin's series, and the theorem ex-
pressed in the formula is called Maclaurin's theorem.
Ex. 1. Expand log x in powers of x — a, a being positive
1
x
Here /(*) = log x,f(x) = £,/"(*)
1 1.0
V'"(*) =:L T
Hence, /(a) = log a,f'(a) = If" (a) = - I, /"'(a) = ^
1-2
/«(a)
and, by (2), the required development is
(-l)-i( w -l)|
log x = log a + 1 (* - a) - J- O - a) 2 + -L (a; - a) J
a 2 a 2 o a 3
The condition for the convergence of this series is
lira I (x — a)^ 1 . (x - a)>
(n + l)a n + 1
x — a
na n
<1,
<1,
\x - a \<a,
0<x<2a.
• This series may be called the development of log x in the vicinity of
x — a. Its development in the vicinity of x = 1 has the simpler form
log X = X - 1 - 1(X - I)* + \(X - l) 3 - ..,
which holds for values of x between and 2.
144 DIFFERENTIAL CALCULUS
In using this series for the computation of a table of logarithms
we may put for a any number whose logarithm is already known, and
for x any number near a in magnitude. It is a great advantage to
keep x — a so small that the power-series in x — a may be not merely
convergent, but may converge to its limit so rapidly that all powers of
x — a above the fourth or fifth may be neglected without affecting the
desired degree of accuracy.
E.g., being given log 10 = 2.302585, suppose it is required to com-
pute log 11, log 12, •••, log 20. Put a = 10, and x = 11. Then
logi^iogio+^-Kr^HKi^^K^+Ki^-Ki^HKiV) 7 ---
The numerical work may be tabulated in the following form :
+ 2.30258509
.10000000 - .00500000
.00033333 .00002500
.00000200 .00000017
.00000001 - .00502517
2.40292043
- .00502517
2.39789526
Hence log 11 = 2.397895-.,
correct to six places of decimals. To make sure of the sixth figure it
is well to carry the work to seven or eight figures. The remaining
terms of the series after KrV) 7 cannot affect this result, because their
sum is less than an infinite decreasing geometric progression whose
first term is IGV) 8 and whose ratio is fo. From the formula
1 - r
i
it follows that the remainder is less than
72 • 10 9
To calculate log 12, log 13, ••• we could now keep a = 10 and let
x = 12, 13, ••• successively, but in order to secure rapid convergence it
is better to change the value of a, choosing for a the nearest number
EXPANSION OF FUNCTIONS 145
whose logarithm has been found. Thus, in computing log 12 we can
use either of the two series
log 12 = log io + a - KA) 2 + K&Y - -.
log 12 = log li + A - KA) 2 + Kt 1 !) 3 - - ;
but it will be found that five terms of the second furnish as close an
approximation as nine terms of the first. The practical advantage
of the step-by-step process will depend on how many of the intermedi-
ate values we actually require. If we are given log 10 and wish to
compute log 15, it may be easier to compute the latter directly with-
out determining the intermediate values.
Ex. 2. Develop f(x) = x s — 2 x 1 + 5 x — 7 in powers of x — 1 and
use the result to compute/(1.02),/(1.01),/(.99),/(.98).
Ex. 3. Develop 3 y 2 — 14 y + 7 in powers of y — 2.
Ex. 4. Expand sin x in powers of x — a and use the result to com-
pute sin 31°.
Let a = 30°, x = 31°. In the formula
sin x = sin a + cos a(x - a)-^^(x - a) 2 - ^^ (x - a) 3 •»,
the difference x — a becomes 1° or .001745 radians, and the coefficients
of its various powers are all known ; since sin a = .5, cos a = .866025 the
work is now reduced to numerical calculation in which three terms
are sufficient to obtain the result correct to six places of decimals. In
general, to calculate sin x or cos x, take for a the nearest value for
which sin a, cos a are known.
The expansion of a function f(x) in a series of ascending
powers of x can be obtained at once from formula (2) by giv-
ing a the particular value zero. The series then becomes
A*)=A0) + /-(0)x + q|i^ + ... + ^^ + .... (3)
EL. CALC. — 10
146 DIFFERENTIAL CALCULUS
Ex. 5. Expand sin x in powers of x, and find the interval of con-
vergence of the series.
Efeve /0) = sin x, /(0) = 0,
f(x) =cosx, /'(()) =1,
f"(x) = - sin a;, /"(0) = 0,
/'"(*) = - cos*, /"'(())= -1
f"(x) = sinx, /iv(0)=0,
/v(ar) = cos x, / v (0) = 1,
Hence, by (3),
+ 1 • x + • x 2 - i. x- 3 + • x 4 +
o !
thus the required development is
sin a: = a; x s H x 5 x 7 + ••• + -^ — 2!! — a: 2 "- 1 + ....
3! 5! 7! (2w-l)!
To find the interval of convergence of the series, use the method
of Art. 74. The ratio of u n+1 to u n is
u n +\_ a> +1 Z 2 "" 1 x 2
« n (2n + l)! (2n-l)l (2n + l)2n
This ratio approaches the limit zero, when n becomes infinite, how-
ever large may be the fixed value assigned to x. This limit being less
than unity, the series is convergent for any finite value of x, and
hence the interval of convergence is from — oo to -f oo.
The preceding series may be used to compute the numerical value
of sin x for any given value of x. It is rapidly convergent when x is
small. Take, for example, x = .5 radians. Then
sin(.5)_.o-— +2>3>4>5 2.3.^.5.6.7+ ->
EXPANSION OF FUNCTIONS 147
= .5000000
- .0208333
+ .0002604
- .0000015
+ .0000000
sin (.5) = .4791256 ••
Show that the ratio of u 5 to u A is 2 \-g ; and hence that the error in
stopping at u 4 is numerically less than u t [^i? + Ciis)^ ••*]» ^ na ^
is, < ?fa u 4 .
When x is not small, it is better to use the more general series in
powers of x — a.
Ex. 6. Show that the development of cos x is
cosx = l - J L + ±--±- + ... +i ii ± + ....
2! 4! 6! (2n-2)! '
and that the interval of convergence is from — oo to + oo.
Ex. 7. Develop the exponential functions a x , e x .
Here
/(*) = «*, /'(*)=a*loga, /"(*) = a*(log «) 2 , .», /*■>(*) =a-(loga)»;
hence /(0)=l; /'(0)=log a,/"(0) = (loga)2, ...,/W(0) = (log a)»,
and a-H (log a)* + ^ a )V + ... + ( lo g q )V + ....
•2! n !
As a special case, put a = e.
Then log a = log e = 1,
and ex^i + ^ + rL + rL-f ... +£!+ ....
2.! 3! n!
These series are convergent for every finite value of x.
Ex. 8. Compute 10* when x = .1.
Ex. 9. Compute 10* when r = 2.01.
148 DIFFERENTIAL CALCULUS
Ex. 10. Defining the hyperbolic cosine and the hyperbolic sine
by the equations
cosh x = \(e x + e x ), sinh x = \{e x — e~ x ),
prove — cosh x = sinh x, — sinh x = cosh x,
dx dx
cosh 0=1, sinh 0=0; and hence expand cosh x and sinh x in powers
of x. Verify that cosh x + sinh x = e x , and cosh x — sinh x = e~ x .
Compute cosh 2 and sinh 2 to four decimal places. Show that the
error made in stopping the series at any term is much less than the
last term used.
78. Taylor's series. If a function of the sum of two num-
bers a and x is given, f(a + x), it is frequently desir-
able to expand the function in powers of one of them,
say x.
In the function /(a + x), a is to be regarded as constant, so
that, considered as a function of x, it may be expanded by
formula (3) of the preceding article. In that formula, the
constant term in the expansion is the value which the func-
tion has when x is made equal to zero, hence the first term
in the expansion of f(a 4- x) may be written /(a). In the
same manner the coefficients of the successive powers of x
are the corresponding derivatives of f(a + x) as to x, in which
x is put equal to zero after the differentiation has been per-
formed. The expansion may therefore be written
f(a + x) = f(a) + f(a)x +£^r-x 2 + ••• + f< [ a) oc n + ••••
9. ' m T
This series, from the name of its discoverer, is known as
Taylor } s series, and the theorem expressed by the formula is
known as Taylor's theorem.
EXPANSION OF FUNCTIONS 149
Ex. 1. Expand sin (a + x) in powers of x.
Here f(a + x) — sin (a + x),
hence f(a) = sin a,
and f'( a ) — cos a,
Hence sin (a + x) = sin a + cos a • x - s i^x2 _ cos_a 3
v ' 2! 3!
Ex. 2. Compute sin 61°, by putting a = 60°.
EXERCISES
1. Expand tan x in powers of x. Obtain three terms.
2. Compare the expansion of tan x with the quotient derived by
dividing the series for sin x by that for cos x.
3. Find a limit for the error which occurs in replacing cos x by
the first three terms of its expansion in powers of x when x = \ of a
radian.
4. Prove that log (x + Vl + x 2 ) = x - -±— + -^- •
2-32-4
5. Provelogc'os^-^-^- 1 -^ 6 -?!^!....
8 2 4! 6! 8!
6. Compute sin 1° cdrrect to six places of decimals.
7. Expand Vl — x 2 in powers of x, and compare with the expan-
sion by the binomial theorem.
8. Expand cos x in powers of x — -.
9. Expand e x+h in powers of h.
10. Arrange 3 x 3 — 5 x 2 + 8 x — 5 in powers of x — 2.
11. Expand log (x + K) in powers of h.
12. Arrange x 4 — 1 in powers of x + 1.
X"~N w ! Cl n ~ r X r
13. Prove the binomial theorem (a + x) n = a n + ••• = x, - 7~, \T-
r=0 v J
Find the form of the series when n is not an integer, and determine
the interval of convergence.
150 DIFFERENTIAL CALCULUS
14. Find V12G = V125 + 1 = 5 vl + T ^ 3 to three places of deci-
mals by the binomial theorem.
15. Find S^Lm
16. Calculate log 31.
17. What is the greatest value of m that will permit the approxi-
mation (1 + m) 4 = 1 + 4 m with an error not exceeding .001 ?
18. Expand - in powers of x — 1 and find the interval of con-
x
vergence.
79. Rolle's theorem. By Art. 76 a series can be the correct
representation of its generating function only when the re-
mainder after n terms can be made as small as desired by
taking n large enough. Before obtaining the form of this
remainder it is necessary to introduce the following lemma.
Rolle's theorem. If f(x) and its first derivative are continu-
ous for all values of x between a and b, and if f(a), f(b) both
vanish, then f'(x) will vanish for some value of x between a
and b. *
The proof follows immediately from the theorem of mean
value (Art. 39). See Figure 41.
80. Form of remainder in Maclaurin's series. Let the re-
mainder after n terms be de-
noted by B n (x, a), which is
a function of x and a as well
as of n. Since each of the
succeeding terms is divisible
by (x — a) n , R n may be con-
Fig. 41 veniently written in the form
R n (x, a) = ^ — pi- <f> (x, a).
EXPANSION OF FUNCTIONS 151
The problem is now to determine <£(#, a) so that the
relation
f(x)=f(a)+f(a)(x-a)+^(x- af+ .'..
+ 7 \f, (x - a) H J + yv ' ' (z - ci) n (1)
(n — 1)1 n\
may be an algebraic identity, in which the right-hand mem-
ber contains only the first n terms of the series, with the
remainder after n terms. Thus, by transposing,
f(x) _/(«)_/(«)(» - a) - f J£p(x-ay - -
_/->) (a! _ B w_ »fe«) (a _ o) .= o. (2)
(h-1)! V ; n! ^ ; w
Let a new function, F(z), be defined as follows:
F(z) =f(x) -/(*) -f\z){x - z) -£& (x - zf - -.
-O^**-'^ 1 -*^*-*- (3)
This function F(z) vanishes when z—x, as is seen by
inspection, and it also vanishes when z = a, since it then
becomes identical with the left-hand member of (2) ; hence,
by Rolle's theorem, its derivative F'(z) vanishes for some
value of z between x and a, say z v But
F\z) = -/(«) -|-/(2) -/»(»)(* - »)+/"(*)(» -«)-...
(n-l)! V ' ^(n-l)! 1 - '
152 DIFFERENTIAL CALCULUS
These terms cancel each other in pairs except the last two ;
hence _ x
Since F'(z) vanishes when z = z 1} it follows that
4>(x, a) =/<»>&). (4)
In this expression z x lies between x and a, and may thus be
represented by
J Zl = a + 0(x- a),
where 6 is a positive proper fraction. Hence from (4)
4>(x, a)=f»[a + d(x-a)],
and R n {x, a) = / w l> + fl(*-«) 1 {x _ o) - .•
The complete form of the expansion of f(x) is then
/(a>) =/(«) + /»(«) fas- a) +^f^ (oc-a) 2 + -
+f"^ (*-a)»-* + / (W We(g--«)) («-«)". (5)
(n — 1); w!
in which n is any positive integer. The series may be carried
to any desired number of terms by increasing n, and the last
term in (5) gives the remainder (or error) after the first n
terms of the series. The symbol / (n) (« 4- (x — a)) indicates
that f(x) is to be differentiated n times with regard to x, and
that x is then to be replaced by a + 0(x — a).
* This form of the remainder was found by Lagrange (1736-1813), who pub-
lished it in the Memoires de l'Academie des Sciences a Berlin, 1772.
EXPANSION OF FUNCTIONS 153
81. Another expression for the remainder. Instead of putting
RJx, a) in the form ( x _ a \»
r-^fo a),
n\
it is sometimes convenient to write it
R n (x, a) = (x — a) ^(#, a).
Proceeding as before, we find the expression for F'(z),
(w-1)!
In order for this to vanish when z = z lf it is necessary that
(n-l)l
in which z t = a + 0(x — a), x — z l = (x — a)(l — 0).
Hence *fe a) = / ( ">(a + <?(*- a)) (1 _ e) .- 1(a . _ a) »-. ;
(n-1)!
and !}„(*, «)=(!- g)-' /'"y^- a ) >(s - a)".*
An example of the use of this form of remainder is fur-
nished by the series for log x in powers of x— a, when x—a
is negative, and also in the expansion of (a -+- #) m -
Ex. 1. Find the interval of equivalence for the development of
log x in powers of x — a, when a is a positive number.
Here, from Art. 77, Ex. 1,
X n
hence /<»>(a f 6(x -a)) = (-l)n-i (" >- 1) »
(a + 0(x — a))'*
and, by Art. 80,
^ fl) = ^--'c*-*)- = (-D- 1 r *-« r.
V ' n(a+0{x-a)) n n la + 6(x-a)\
* This form of the remainder was found by Cauchy (1789-1857), and first
published in his "Le<,:ons sur le calcul infinitesimal," 1826.
154 DIFFERENTIAL CALCULUS
First let x — a be positive. Then when it lies between and a, it
is numerically less than a + 0(x - a), since is a positive proper
fraction ; hence when n = oo
x ~ a T = 0, and RJx, a) = 0.
Again, when x — a is negative and numerically less than a, the
second form of the remainder must be employed. As before
hence R n (x, a) = (1 - 0)«-i . f" 1 ^"^
= (l_0)»-i. - («-')'
[a-(9(a-.r)]»
t (a — x) — 0{a — x) l "- 1 a — x
a -0(a - x) J ' a - ${a - x)'
The factor within the brackets is numerically less than 1, hence
the (n — l)th power can be made less than any given number, by tak-
ing n large enough. This is true for all values of x between and a.
Therefore, log x and its development in powers of x — a are equiva-
lent within the interval of convergence of the series, that is, for all
values of x between and 2 a.
_i
Ex. 2. Show that the development of x ? in positive powers of
x — a holds for all values of x that make the series convergent ; that
is, when, a; lies between and 2 a.
If the function is expanded in powers of x, the complete
form will be
ffljj. , /'-"(0)^-i
/(*)=/(0)+/'(0)*+-^^+ ... +^~y;*"
+ /^M*. (1)
EXPANSION OF FUNCTIONS 155
for the first form of remainder, and
/(*)=/(<>) +/'(0)* + /^-'+ ... +M^-i
£W(1 _ gy-l . x , ( 2 )
for the second form of remainder.
Similarly, the complete form of Taylor's series (Art. 78)
becomes
0|
(n-l)l
/»>(q + flaQ ,
for the first form of remainder, and
/(a + x) =f(a) +f(a)x + -^z 2 + • •• +
/•"«„ 2 , , /'-"(a) „„-,
<> |
(»-l)
(4)
for the second form of remainder.
These forms are of no value for numerical computation
unless f (n) (x) can be determined, but may sometimes be used
to advantage to obtain a maximum error, corresponding to
small values of n. It should be observed that when n = 1,
the theorem of mean value results. (Art. 39.)
Ex. 3. Obtain the limit of error in retaining but two non-vanish-
ing terms in the expansion of log (x + Vl + x' 1 ) when x = \.
log (x + viTi" 2 ) = x- -*. + P- ?/ - n - ?/ ; r
wherein y = 6x. LV J J ■
24
156 DIFFERENTIAL CALCULUS
The next step is to obtain the largest and the smallest value which
the expression in brackets assumes for values of y within the interval
to \. For this purpose, consider the function
„- 3.v(3-2.y*)
(1 + y*)! ^
We find that — is positive for all values of y between y = and y = \;
dy
hence u has its largest value when y = J, and the corresponding value
of the last term is .000284.
Ex. 4. How many terms should be used in the expansion of e x in
powers of x to insure a result correct to four places of decimals when
x = |?
Ex. 5. In the expansion of logio (1 + x) in powers of x how many
terms should be used in order to obtain the value of logio (1 • 8) cor-
rect to 5 decimals ?
CHAPTER XI
INDETERMINATE FORMS
82. Hitherto the values of a given function f(x), corresponding
to assigned values of the variable x, have been obtained by direct
substitution. The function may, however, involve the variable
in such a way that for certain values of the latter the correspond-
ing values of the function cannot be found by mere substitution.
For example, the function
sin x
for the value x = 0, assumes the form -, and the correspond-
ing value of the function is thus not directly determined. In
such a case the expression for the function is said to assume
an indeterminate form for the assigned value of the variable.
The example just given illustrates the indeterminateness of
most frequent occurrence; namely, that in which the given
function is the quotient of two other functions that vanish for
the same value of the variable.
Thus if f(x) = *M,
and if, when x takes the special value a, the functions <f>(x)
and \p(x) both vanish, then
J K ' *(a)
is indeterminate in form, and cannot be rendered determinate
without further transformation.
157
158
DIFFERENTIAL CALCULUS
83. Indeterminate forms may have determinate values. A
case has already been noticed (Art. 2) in which an expression
that assumes the form - for a certain value of its variable takes
a definite value, dependent upon the law of variation of the
function in the vicinity of the assigned value of the variable.
As another example, consider the function
x 2 — a 2
y =
x— a
If this relation between x and y is written in the forms
y(x — a)=x 2 — a 2 , (x — a)(y — x — a) = 0,
it will be seen that it can be represented graphically, as in Fig.
42, by the pair of lines
x — a = 0,
y — x — a = 0.
Hence when x has the value of a, there
is an indefinite number of corresponding
points on the locus, all situated on the
line x = a; and accordingly for this
value of x the function y may have any
value whatever, and is therefore indeterminate.
When x has any value different from a, the corresponding
value of y is determined from the equation y = x-\-a. Now,
of the infinite number of different values of y corresponding
to x = a, there is one particular value AP which is continuous
with the series of values taken by y when x takes successive
values in the vicinity of x = a. This may be called the de-
terminate value of y when x — a. It is obtained by putting
x = a in the equation y = x + a, and is therefore y = 2 a.
Fig. 42
INDETERMINATE FORMS 159
This result may be stated without reference to a locus as
follows. When x = a, the function
x— a
is indeterminate, and has an infinite number of different
values ; but among these values there is one determinate value
which is continuous with the series of values taken by the
function as x increases through the value a ; this determinate
or singular value may then be defined by
liui xr— a 2 ^
x=a x — a
In evaluating this limit the common factor x— a may be re-
moved from numerator and denominator, since this factor is
not zero while x is different from a; hence the determinate
value of the function is
lim a? + « _o„
Ex. 1. Find the determinate value, when x = 1, of the function
x 3 + 2 x 2 -3x
3 x b _ 3 x a _ x + i '
which, at the limit, takes the form —
This expression may be written in the form
(x* + Zx)(z- 1)
(3s 3 - 1)0- 1)'
which reduces to x + ' x • When x = 1, this becomes - = 2.
3 x 2 - 1 2
Ex. 2. Evaluate the expression
x 8 + ax 2 + a 2 x 4- a 3
, x 8 4- b 2 x + az 2 + a& 2
when x = — a.
160 DIFFERENTIAL CALCULUS
Ex. 3. Determine the value of
x 8 - 7 x 2 + 3 x + 14
x 3 + 3 x 2 - 17 x + 14
when x = 2.
Ex. 4. Evaluate - — — — when x = 0.
(Multiply both numerator and denominator by a -f Va 2 — x 2 .)
84. Evaluation by development. In some cases the common
vanishing factor can be best removed after expansion in series.
Ex. 1. Consider the function mentioned in Art. 82,
e* - e~*
sin x
When numerator and denominator are developed in powers of x,
the expression becomes
21 31^ V 21 31 /
*-£+...
3!
2, + ^+... 2+f+...
x 3 , - X 2 .
— ST+- 1 -« + -
which has the determinate value 2, when x takes the value zero.
Ex. 2. As another example, evaluate, when x= 0, the function
x — sin- 1 a:
sin 8 x
By development it becomes
« +
X J +
Removing the common factor, and then putting x = 0, we obtain
INDETERMINATE FORMS 161
In these two examples the assigned value of x, for which the
indeterininateness occurs, is zero, and the developments are made
in powers of x. If the assigned value of x is some other number,
as a, then the development should be made in powers of x — a.
Ex. 3. Evaluate, when x = -, the function
2
COS X
1 — sin x
IT
By putting x — - = h, x = - + h, the expression becomes
eo.(=+») . -»+£-... _l+£
\ 2 / — sm 7i 6 6
. . hr , ,\ 1 -cos A A 2 A 4 , /i A 8 ,
1 — sin — \- h) — h ••• h •••
\2 / 2 24 2 24^
which becomes infinite when 7* = ; that is, when x = -•
2
TJ lim cos x
Hence . m - = ± oo,
x=fl-siuo: '
according as h approaches zero from the negative or positive side.
85. Evaluation by Differentiation. Let the given function
be of the form ^~, and suppose that /(a) = 0, cf>(a) = 0. It
is required to find ^ ^-L-i .
4 x = a <f>(x)
We assume that f(x), <j>(x) can be developed in the vicinity
of x = a, by expanding them in powers of x — a. Then
m /(«)+/'(«)(* - «)+-^ ( f (»- «) 2 + -
*(*) <l>(a) + 4>'(a)(x - a) + *!M (x - uy+ -
/•(«)( a( _ a ) + ^2l( ( ,_«)i + ...
^«)(*-a) + ^(z-a) 2 +...
EL. CALC. 11
162 DIFFERENTIAL CALCULUS
By dividing by x — a and then letting x = a, we obtain
lim f(x)_f(a)
«-« + («) +'(o)
By hypothesis the functions f'(a), <£'(a) will both be finite.
If /'(a) = 0, <£'(a) =£ 0, then £i& = 0.
<j>(a)
lf/'(a)*0, *'(a) =0, then ^ = oo.
<£(a)
If /'(a) and <£'( a ) are both zero, the limiting value of J-^-l
is to be obtained by carrying Taylor's development one term
farther, removing the common factor (x — a) 2 , and then letting
f"(a)
x approach a. The result is J —~ •
Similarly, if /(a), /'(a), /"(a); </>(a), <^)'(a), <^>" (a) all vanish,
it is proved in the same manner that
lim f(x) = f'"(a)
x = a '<j>(x) <£'"(«)'
and so on, until a result is obtained that is not indeterminate
in form.
Hence the rule :
To evaluate an expression of the form -, differentiate numer-
ator and denominator separately ; substitute the critical value of
x in their derivatives, and equate the quotient of the derivatives
to the indeterminate form.
Ex. 1. Evaluate * ~ T ^ when = 0.
Put f(0) = 1 - cos 6, <f>(0) = e*.
Then f(0) = sin 0, <f>'(0)=2$ t
and /'(0) - 0, <£'(0) = 0.
INDETERMINATE FORMS 163
Thus, the function \ * J is also indeterminate at = 0. It is there-
9 (0)
f"({y\
fore necessary to obtain </ ,,„ ' •
Accordingly, f"{6) = cos 0, p"(0) = 2,
/'(0)=1, 9"(0)=2,
hence lim 1 — cos _ 1 J
= 02 - 2 *
Ex.2. Find x= ,
lim i* + e~* + 2 cos a: — 4 lim e x — e~ x — 2 sin a:
x =
4 a: 3
lim e* + e
* — 2
COS X
x=0
12 a: 2
lim e x — e
-* + 2
sin x
x =
24 x
lim e z + e
-* + 2
cosx
x = 24
1
6'
Ex. 3. Find lim a; - siD3rcosa: .
x = x 8
Ex.4. Find M" « , -2»'i**+»«-*.
* = 1 a: 4 - 2 a: 3 + 2 a: - 1
In this example, show that z — 1 is a factor of both numerator and
denominator.
^ _ „. , lim 3 tan x - 3 a: - a: 8
Ex. 5. Find „ . n
In applying this process to particular problems, the work
can often be shortened by evaluating a non-vanishing factor
in either numerator or denominator before performing the
differentiation.
164 DIFFERENTIAL CALCULUS
Ex. 6. Find ";■> (»-*)' tan* .
X == " X
The given expression may be written
lim . tanx _ lim lim tans
= 16-1 = 16.
In general, if f(x) = *P(%)x( x )> an( ^ ^ «M°0 = 0> x( a )^0,
(h (a) = 0, then ,. ., N ,,, *
llm /M = v r^^M.
* = «*(*) XU ^>'(a)
For
lim iKx)x(x) = lim , v . lim </<») a) . ^!M
x = a $(x) x = a^ K > x = a^{x) * ; <f>'(a)
lim sin x cos 2 x
Ex. 7. Find
x = $ (2 x - tt) 2
Ex.8. Find lim (.-3) 2 log(2 x } >
x = 1 sin(x - 1)
EXERCISES
Evaluate the following expressions :
1. 1 - CO8ar whenx = 0. 7. e * + e ~' ~ 2 when x = 0.
2. e * . e when a; = 0. 8 tan a;- sin x cos a: when ^_ a
3.
* 3 - 1 U 1
when x = 1.
x- 1
4.
when x = 0.
&* - 1
5.
sin ax , „
when x = 0.
sin frx
l
9. when x = 0.
tan -1 a;
10. g*B"i*-*-* when a- = 0.
x 2 + a;log (1-a:)
>. (1 4- arV» — 1 , n -.-. tan a; — sin a; , n
6. V x ^ x ' 1 when x = 0. 11. - Q when x = 0.
x X s
There are other indeterminate forms than -• They are
£,00-00,0°, r, 00°.
INDETERMINATE FORMS
165
86. Evaluation of the indeterminate form ® .
GO
fix)
Let the function J \ ' become 5°- when x = a. It is re-
<j>(x) <*>
quired to find lim &&•
This function can be written
<K*) _1_'
which takes the form - when x = a, and can therefore be
evaluated by the preceding rule.
When x = a,
— vv,
lim /(») _ lim
x = a <£(#) £ = a
1 4,\x)
*(*)_ lim [*(*)]*
1 ic = a f'(x)
lim
~" " x — a
-/W T+'(*).
(1)
If both members are divided by im -£-^J when this limit
is not nor co, the equation becomes
1= lim f(x) +\x)
*±<*<f>(x)f'(x)'
therefore
lim
x = a
d,{x)
= /'(<0
+'(«)
(2)
This is exactly the same result as was obtained for. the form
; hence the procedure for evaluating the indeterminate forms
-, g, is the same in both cases.
^
166 DIFFERENTIAL CALCULUS
When the true value of ■->--' is or go , equation (1) is satis-
fied, independent of the value of ,~ ; but (2) still gives the
correct value. For, suppose ™ ^ w = 0. Consider the
function
which has the form °°- when a; = a, and has the determinate
GO 7
value c, which is not zero. Hence by (2)
lim fix) + c<f> (x) = /' (a) + c<f>' (a) = /' (a) c
* = a <f>(x) 4>'(a) <j>' (a)
Therefore, by subtracting c,
lim /(a?) = /(a) ,
• £ = «<£(&) </>'(a)
If "F-fiS. =00, then "™ ^ = 0, which can be treated
x ~ a <f>(x) x - a f(x)
as the previous case.
The forms • go and go-go can usually be evaluated by
putting them in one or the other of the forms already dis-
cussed. In the case of the others, in which the indetermi-
nateness appears in the exponent, the logarithm of the
function can be reduced to one of the preceding forms.
EXERCISES
Evaluate the following expressions:
1. lo g si » 2 * when x = 0.
log sin x
2. !2££ ^
4.
tan x
tan 5 x
when
x
_ IT
5.
sec 3 a:
when
X
_ IT
cot x sec 5 x 2
6. x s[nx when x = 0.
7. (cos ax) csc2cx when x = 0.
^n 6. x B[QX when x = 0.
3. — when x = go.
IT)
CHAPTER XII
CONTACT AND CURVATURE
87. Order of contact. The points of intersection of the two
are found by making the two equations simultaneous ; that is,
by finding those values of x for which
Suppose x = a is one v?lue that satisfies this equation.
Then the point x = a, y = A(i) = if; (a) is common to the curves.
If, moreover, the two curves have the same tangent at this
point, they are said to touch each other, or to have contact
with each other. The values of y and of — are thus the
J dx
same for both curves at the point in question, which requires
that *(«)=</<(«),
<j,\a)=4,\a).
d 2 v
If, in addition, the value of — " is the same for each curve
dx~
at the point, then
and the curves are said to have a contact of the second order
with each other, provided cf>'"(a) =£ i//"(a).
If <f>(a) =\f/(a), and all the derivatives up to the nth order
inclusive are equal to each other, that is, <f>'(a) = if/' (a),
<}>"(a) = i(,"(a), .-., cf> (n \a) =<A (n) (a), but <f> (n+1) (a) =£ f n+1) (a),
the curves are said to have contact of the ?ith order.
167
168 DIFFERENTIAL CALCULUS
88. Number of conditions implied by contact. or j equation
of the curve y = <f>(x) is given, and it is requir ,. to determine
the equation of a second curve y = \\j (x) that shall have contact
of any given order with y = <f>(x) at a specified point, then,
from the definition given in the preceding article for contact
of the nth order, n -f 1 conditions must be imposed upon the
coefficients in \p(x). The required curve must therefore con-
tain at least n -f- 1 arbitrary constants in >rder to fulfill the
required conditions.
A straight line has two arbitrary constants, which can be
determined by two conditions ; accordingly a straight line can
be drawn which touches a given curve at any specified point.
For if the equation of a line is written y = mx -f b, then
dy dr.; p.
dx dx
hence, through any arbitrary point x = a on a given curve
y=cf>(x), a line can be drawn which has contact of the first
order with the curve, but which has not in general contact of
the second order; for the two conditions for first-order con-
tact are ma + 6 = <£(«),
m = <£'(«),
which are just sufficient to determine m and b.
In general no line can be drawn having contact of an order
higher than the first with a given curve at a given point ; but
there are certain special points at which this can be done.
For example, the additional condition for second-order contact
is = <f>"(a), which is satisfied when the point x = a is a
point of inflexion on the given curve y=<j>(x). (Art. 49.)
Thus the tangent at a point of inflexion on a curve has contact
of the second order with the curve.
CONTACT AND CURVATURE 169
The equation of a circle has three independent constants.
It is therefore possible to determine a circle having contact of
the second order with a given curve at any assigned point.
The equation of a parabola has four constants, hence a
parabola can be found which has contact of the third order
with the given curve at a given point.
The general equation of a central conic has five independent
constants, hence a conic can be found which has contact of the
fourth order with a given curve at any specified point.
As in the case of the tangent line, special points may be
found for which these curves have contact of higher order.
89. Contact of odd and of even order.
Theorem. At a point where two curves have contact of an
odd order they do not cross each other; but they do cross
where they have contact of an even order.
For, let the curves y = <f>(x), y=\f/(x) have contact of the ?ith
order at the point whose abscissa is a ; and let y x , y 2 be the
ordinates of these curves at the point whose abscissa is a + h.
Then y 1 = cf>(a + h), y 2 =z ij,(a-\-h),
and by Taylor's theorem
Vl = *(a) + *'(a) • h + *^) • ft 2 + • • •
+ *^.». + *^« (a)+ ....
n ! (?i -f 1) !
fc .^(o)+^(a).*+*^i.V
+ •
n ! (n + 1) !
170 DIFFERENTIAL CALCULUS
Since by hypothesis the two curves have contact of the nth
order at the point whose abscissa is a, hence
4(a) =$(a), <j>'(a)=ip' a), ..'., <y(a) = ^(a),
and yi -y^-J^i r+1(a)+ ... _ r+ i (a) _ ... ];
but this expression, when h is sufficiently diminished, has the
same sign as .,,, r „,.> , , , v _
Hence, if n is odd, y l — y 2 does not change sign when h is
changed into — h, and thus the two curves do not cross each
other at the common point. On the other hand, if n is even,
ll\ — V* changes sign with 7i ; and therefore when the contact
is of even order the curves cross each other at their common
point.
Geometrically, we may say that two curves having contact
of the nth order pass through n -\- 1 common points which
approach coincidence at the point of contact. For let y = <f>(x),
y = xf/(x) touch each other at x = a. This means that they have
two coincident points in common at (a, <f>(a)), and the condi-
tions to be satisfied are
<f>(a)=^(a), *'(a) = *'(a).
If the curves also have a point in common for x = a -f- h, then
<f>(a + h)=ip(a + h).
Expanding by Taylor's series and making use of the preced-
ing conditions, we may cancel the common factor Jr. If now
this condition is still satisfied when h approaches zero, so that
the third point of intersection approaches the position of the
two coincident ones, then we must have the further condition
CONTACT AND CURVATURE 171
<£"(a)= \J/"(a). Thus, three coincident points of- intersection
imply contact of the second order. By repeating this argu-
ment the above theorem results.
For example, the tangent line usually lies entirely on one
side of the curve, but at a point of inflexion the tangent crosses
the curve.
Again, the circle of second-order contact crosses the curve
except at the special points noted later, in which the circle
has contact of the third order.
EXERCISES
1. Find the order of contact of the curves
4 y = x 2 and y = x — 1.
2. Find the order of contact of the curves
x = y s and x — 2 y + 1 = 0.
3. Find the order of contact of the curves
4 y = x 2 — 4 and x 2 — 2 y — 3 — y 2 .
4. Determine the parabola having its axis parallel to the #-axis,
which has the closest possible contact with the curve a 2 y — x 3 at the
point («, a). (The equation of a parabola having its axis parallel
to the y-axis is of the form
y = Ax 2 + Bx + C.)
5. Determine a straight line which has contact of the second order
with the curve y = x 8 - 3x 2 - 9 x + 9.
6. Find the order of contact of
y = log(x — 1) and x 2 — 6x + 2y + 8 =
at the point (2, 0).
7. What must be the value of a in order that the curves
y=x+l-\-a(x— l) 2 and xy = 3 x — 1
may have contact of the second order?
172 DIFFERENTIAL CALCULUS
8. Determine the parabola which has its axis parallel to the y-axis
and has contact of the second order with the hyperbola xy = 1 at the
point (1, 1).
9. Determine the point and order of contact of the curves
(a) y=x*, y = 6 x 2 - 9 x + 4 ;
(6) y = x s , y = -Q x 2 -12x-8.
10. Determine the parabola which has its axis parallel to the y-axis,
passes through the point (0, 3), and has contact of the first order
with the curve y = 2 x 2 at the point (1, 2). Similarly for a parabola
having its axis parallel to the x-axis.
11. Show that the curve y = sin x has contact of the sixth order
with the curve x z , x 5
v = x
* 6 120
at the origin. Show that y = sin x, y = sinh x, have contact of the
second order at the origin. Draw these curves.
12. Find the order of contact of the curves y = cos x, y = cosh x
at the point (0, 1). Sketch the curves.
13. Find the order of contact at the origin of the curves
, ■ , , _ sinh x
y = tan x, y = tanh x = .
cosh x
90. Circle of curvature. The circle that has contact of the
closest order with a given curve at a specified point is called
the osculating circle or circle of curvature of the curve at the
given point. The radius of this circle is called the radius of
curvature, and its center is called the center of curvature at the
assigned point.
91. Length of radius of curvature; coordinates of center of
curvature. Let the equation of a circle be
(X-ay + (Y-pf = R\ (1)
in which R is the radius, and a, ft are the coordinates of the
center, the current coordinates being denoted by X, Yto dis-
CONTACT AND CURVATURE
173
tinguish them from the coordinates of a point on the given
curve.
It is required to determine R, a, /?, so that this circle may
have contact of the second order with the given curve at the
point (x, y).
From (1), by successive differentiation, we deduce
(X-o) + (F-/8)g=0,
\dX K H 'dX 2
(2)
If the circle (1) has contact of the second order at the point
(x, y) with the given curve, then when X = x it is necessary
that
dY = c]y d 2 Y = d 2 y
dX dx dX 2 dx 2 '
(3)
Substituting these expressions in (2), we obtain
dy
(x- a) +(3,-^)^ = 0,
whence
y-P-
H%
dx 2
dy
dx
x~ a= —
L + (D"
<Py
dx*
(4)
(5)
and finally, by substitution in (1),
R
.MIT)
dx 2
(6)
174 DIFFERENTIAL CALCULUS
Ex. 1. For the curve y = sin x, show that a = x-\- cot x(l + cos 2 x),
s
j3 = — 2 cos a: esc x, it = — (1 + cos' 2 x) 2 esc x. Find the numerical
values of a and B when a: = 0, — , — , — , and locate the corresponding
r 6 3 2
points («, /3) on a drawing. Sketch roughly the path of this point
as x varies. Write the equation of the osculating circle for the point
x = - , and for x = -. Draw these circles.
3 2
Ex. 2. For the curve y = x s , find a, /?, R in terms of x. Compute
their numerical values at x — 1, .7, .5, .3, 0. Show that i? is a mini-
mum when x — — — = .39 •••, and that the value of R is .57 ••••
92. Limiting intersection of normals. Let P= (# 1? y-^) and
P' = (a? 2 , y 2 ) be Wo points on a given curve f(x, y) = 0. The
equations of the normals at these points are
( a ._ a . 2 ) + ( 2/ _ 2/2 )^? = 0<
d# 2
If («', /?') is the point of intersection of these two lines, then
dx x
(^_ iK2 ) + ( /3 '_ 2/2 )^ == 0.
Now consider the function if/(x) of a: denned by the equations
f(x) = (x- a') + (y- /?') £, f(x, y) = 0.
Since i/a(^ 1 ) = and ^(# 2 ) = 0, hence by Rolle's theorem
(Art. 79) it follows that
t'(x) = 0,
CONTACT AND CURVATURE 175
in which x is defined by the inequalities
U/j <^, X < C X 2 >
Hence a', ft' may be determined by the simultaneous
equations ^(^) = 0, f(5) = 0.
When P' = (x 2 , y 2 ) approaches coincidence with the point
P=(x x , y x ), then if/' (x) = i// (a;,), and therefore from (4), the
point («', /8') becomes the center of curvature, hence :
The center of curvature at a point Pon a curve is the limiting
position of the point of intersection of the normal at P with the
normal at the point P', when P' approaches the position of P.
93. Direction of radius of curvature. Since, at any point P
on the given curve, the value of -* is the same for the curve
dx
and the osculating circle for that point, it follows that they
have the same tangent and normal at P, and hence that the
radius of curvature coincides with the normal. Again, since
the value of — is the same for both curves at P, it follows
dx 2
from Art. 50, that they have the same direction of bending
at that point, and hence that the center of curvature lies on
the concave side of the given curve (Fig. 43).
It follows from this fact and Art. 87 that the osculating
circle is the limiting position of a circle passing through three
points on the curve when these points move into coincidence.
The radius of curvature is usually regarded as positive or
negative according as the bending of the curve is positive or
d 2 v
negative (Art. 49), that is, according as the value of — -~ is
ctx~
positive or negative ; hence, in the expression for R, the radi-
cal in the numerator is always to be given the positive sign.
176
DIFFERENTIAL CALCULUS
The sign of R changes as the point P passes through a point
of inflexion on the given curve (Fig. 44). It is evident from
the figure that in this case R passes through an infinite value ;
Fig. 43 Fig. 44
for the circle through the points N, P, Q approaches coinci-
dence with the inflexional tangent when N and Q approach
coincidence with P, and the center of this circle at the same
time passes to infinity.
94. Total curvature of a given arc ; average curvature. The
total curvature of an arc PQ (Fig. 45) in which the bending
is in one direction, is the angle through
which the tangent swings as the point
of contact moves from the initial point
P to the terminal point Q ; or, in other
words, it is the angle between the tan-
gents at P and Q, measured from the
former to the latter. Thus the total
curvature of a given arc is positive or negative according as
the bending is in the positive or negative direction of rotation.
The total curvature of an arc divided by the length of the
arc is called the average curvature of the arc. If the length of
Fig. 45
CONTACT AND CURVATURE 177
the arc PQ is As centimeters, and if its total curvature is A<£
radians, then its average curvature is — — radians per centimeter.
As
95. Measure of curvature at a given point. The measure of
the curvature of a given curve at a given point P is the limit
which the average curvature of the arc PQ approaches when
the point Q approaches coincidence with P.
Since the average curvature of the arc PQ is — — , the
As
measure of the curvature at the point P is
lim A<£ _ dcf>
As=0 As d
and may be regarded as the rate of deflection of the arc from
the tangent estimated per unit of length ; or, as the ratio of
the angular velocity of the tangent to the linear velocity of the
point of contact.
To express k in terms of x, y, and the derivatives of y, we
observe that
whence
and
tan <£ =
dx '
* =
dx
d<f>_
ds
ds \ dxj
dx\ dxj ds
<l 2 y
dx 2 1
\dxj dx
cPy
defy
dx 2
K ~ds =
Mtw
therefore k = ^ = "^ 7 9 • [Art. 41.
EL. CALC. — 12
178 DIFFERENTIAL CALCULUS,
96. Curvature of an arc of a circle. In the case of a circular
arc the normals are radii ; *
hence As = r.A0, ^ = -, (1)
As r
and therefore k = -.
r
It follows that the average curvature of all arcs of the same
circle is constant and equal to - radians per unit of length.
r
For example, in a circle of 2 feet radius the total curvature
of an arc of 3 feet is § = 1.5 radians, and the average curva-
ture is .5 radian per foot.
It also follows from (1) that in different circles, arcs of the
same length have a total curvature inversely proportional to
their radii.
Thus on a circumference of 1 meter radius, an arc of 5 decimeters
has a total curvature of .5 radian, and an average curvature of .1
radian per decimeter ; but on a circumference of half a meter radius,
the same length of arc has a total curvature of 1 radian and an
average curvature of .2 radian per decimeter.
97. Curvature of osculating circle. A curve and its osculat-
ing circle at P have the same measure of curvature at that
point.
For, let k, k' be their respective measures of curvature at
the point of contact (x, y). Then from Art. 95,
&y
dx 2
\>+m\
CONTACT AND CURVATURE 179
But this is the reciprocal of the expression for the radius of
curvature (Eq. (6), p. 173) ; hence
1
K = •
R
That is : the measure of curvature k at a point P is the recipro-
cal of the radius of curvature R for that point. Since a curve
and its osculating circle have the same radius of curvature
(Art. 90) at their point of contact, it follows from this result
that the measure of curvature is also the same for both; *=*'.
It is on account of this property that the osculating circle
is called the circle of curvature. This is sometimes used as
the denning property of the circle of curvature. The radius
of curvature at P would then be denned as the radius of the
circle whose measure of curvature is the same as that of the
given curve at the point P. Its value, as found from Art. 95
and Art. 96, accords with that given in Art. 91.
EXERCISES
1. Find the radius of curvature of the curve y 2 = 4 ax at the origin.
2. Find the radius of curvature of the curve y s + x z + a(x 2 + y 2 )
= cfiy at the origin.
3. Find the radius of curvature of the curve ahj — bx z + cx 2 y at
the origin.
Find the center and the radius of curvature for each of the following
curves at the point (x, y) and their numerical values at the special
point indicated. Find where the curvature is greatest and least on
each curve.
4. Rectangular hyperbola x$ = m 2 at (m, m).
5. Hyperbola — - -^ = 1 at (a, 0).
a 2 b 2
180
DIFFERENTIAL CALCULUS
6. General parabola a n ~ x y = x n at (a, a).
7. Parabola Vx + Vy = Va at (a, 0).
1 2 2
8. Hypocycloid x 3 -f y s = of 3 " at the point at which x = y.
9. Cissoid y"
a,t x = a.
10. Catenary y—-( e a _j_ e <*) at £ = 0.
11. In what points of the curve y = x 3 is the curvature greatest ?
98. Direct derivation of the expressions for k and R in polar
coordinates. Using the notation of Art. 58, we have
hence
But
_d$_dB_
~ ds~ ds
dO
d O
ds
dO
('+2)
■'%
tan^ = p — , ^=tan _1
dp
P_
dp
dO
(1)
[Art. 44.
therefore, by differentiating as to and reducing, we obtain
'dp^ 2
dij/ _ \d6
d0~
P 2 +
(IT '
CONTACT AND CURVATURE 181
which, substituted in (1), gives
HUT
Since k = — , it follows that
H
R =
Mm
f j
*d0 2
+ 2
(U
This result should be compared with that of Art. 72.
When u = - is taken as dependent variable, the expres-
P
sion for k assumes the simpler form
K =
»r
Since at a point of inflexion k vanishes and changes sign,
hence the condition for a point of inflexion, expressed in polar
72
coordinates, is that u -\ shall vanish and change sign.
cW
EXERCISES
Find the radius of curvature for each of the following curves :
1. p = a . 3. p = 2 a cos $ - a. 5. p 2 cos 2 = a 2 .
2. p 2 = a 2 cos 2 6. 4. p cos 2 J $ = a. 6. p = 2 a(l - cos 0) .
7. P = a.
182 DIFFERENTIAL CALCULUS
EVOLUTES AND INVOLUTES
99. Definition of an evolute. When the point P moves along
the given curve, the center of curvature C describes another
curve which is called the evolute of the first.
Let f(x, y) = be the equation of the given curve. Then
the equation of the locus described by the point C is found by
eliminating x and y from the three equations
x— a
dy
dx
i+(&
\dx
dx 2
*-p- — it '
dx 2
and thus obtaining a relation between a, /3, the coordinates of
the center of curvature.
No general process of elimination can be given ; the method
to be adopted depends upon the form of the given equation
/(*,y) = o.
Even when the elimination cannot be performed, the evolute
can be traced from point to point by computing successive
values of (a, /3) corresponding to successive values of (x f y).
Ex. 1. Find the evolute of the parabola ?/ 2 = 4 px.
Since «=2pW, ^ = »H <?t> = _ M*-! ,
dx dx 2 2
hence x — a = — p^x~^(l + px~ *)2 p~?x% — — 2 (x + p),
and y - p = (1 + px~ x )2 p"~W = 2(p~%x% + pW) ;
therefore a = 2p + 3 a:, ft = — 2 p*x*.
CONTACT AND CURVATURE
183
Fig. 46
The result of eliminating x between the last two equations is
i.e., 4(«- 2 P y = 27 p/3 2 ,
which is tho equation of the evolute of the parabola, a, {3 being the
current coordinates.
Use the expressions for a and /3 to compute their values, and to
locate the points («, /3) when x — 0,
P
Ex. 2. Find the evolute of the ellipse
-2 ?<2
Here
dx 2
ir-
x ir
a 2 b 2
(1)
y_ .dy = o 'hi- -
b 2 X
b 2 dx
dx
dy
dx -ft 2 / , b 2 x 2 \ -b 2 , 2 „ , , 2 „. -ft*
r a # \ a 2 y / a 4 #° a 2 // 3
184 DIFFERENTIAL CALCULUS
whence
y
H aW \ b* a>Y J \ ¥ by-
Therefore - (3 = ^ ~ b ' V - (2)
b*
Similarly, a = a —^x^. (3)
On eliminating x, y between (1), (2), (3), the equation of the
locus described by (a, (3) is
(aa)f + (bp)i = (a 2 - 6 2 )l (Fig. 51)
Use (2), (3) to locate various values of («, /3), and trace the evolute.
Take a = 2b; a =—.
3
100. Properties of the evolute. The evolute has two im-
portant properties that will now be established.
I. The normal to the curve is tangent to the evolute. The
relations connecting the coordinates (a, ft) of the center of
curvature with the coordinates (x, y) of the corresponding
point on the curve are, by Art. 91,
x- a + (y-/J)^ = 0, (1)
1+ (!)' + (*-«g=°- < 2 >
By differentiating (1) as to x, considering a, (3, y as functions
of*,weobtain + ^_^_<L«_ M = . (3)
\dxj K HJ dx 2 dx dx dx w
Subtracting (3) from (2), we obtain
da + dgdy = ^ ^
dx dx dx
, dB dx
whence — = — — •
da dy
CONTACT AND CURVATURE 185
But c -l=- is the slope of the tangent to the evolute at (a, /?),
da
and is the slope of the normal to the given curve at
dy
(x, y). Hence these lines have the
same slope ; but they pass through the
same point (a, f3), therefore they are
coincident.
II. TJie difference between two radii '
of curvature of the given curve, which
touch the evolute at the points C lt C 2 Fig. 47
(Fig. 47), is equal to the arc C X C 2 of the evolute.
Since R is the distance between the points (x, y), (a, (3),
hence (aj - a) 2 + (y - /3) 2 = Br. (5)
When the point (x, y) moves along the given curve, the
point («, /?) moves along the evolute, and thus a, /?, R, y, are
all functions of x.
Differentiation of (5) as to x gives
(-^-SVMs-i?)-*^ (6)
hence, subtracting (6) from (1), we obtain
( X - a) p +(y -f }) f = -l t ™. (7)
dx dx dx
Again, from (1) and (4),
da d(3
dx dx
x — a y — (3
(8)
186 DIFFERENTIAL CALCULUS
Hence, each of these fractions is equal to
4
da\ 2 ,fdg\ 2 do-
dx) \dxj dx
V(a-a) 2 + (^=^ _± ^'
in which <r is the arc of the evolute. (Compare Art. 41.)
Next, multiplying numerator and denominator of the first
member of (8) by x — a, and those of the second member by
y — /?, and combining new numerators and denominators, we
find that each of the fractions in (8) is equal to
(x-af + iy-py ■
which equals —
R 0B
dx
R 2
by (7) and (5).
By combining
with
(9), we obtain
d<r_ dR
dx dx'
that is,
^(cr±R)=0.
dx
Therefore <r ± R = constant, (10)
wherein o- is measured from a fixed point A on the evolute.
Now, let (?!, C 2 be the centers of curvature for the points
P ly P 2 on the given curve ; let P X C X = 7^, P 2 C 2 = R 2 ; and let
the arcs AC X , AC 2 be denoted by p- lt <r 2 . Then
<t 1 ±R 1 = <t 2 ± R 2 , by (10) ;
that is a- 1 — (T 2 =± (R 2 — R } ) ;
hence, arc C X C 2 — R 2 — R v (11)
CONTACT AND CURVATURE
187
Fig. 48
Thus, in Fig. 48,
P 2 C 2 +CA = P S C 3 , etc.
Hence, if a thread is wrapped
around the eve-lute, and then is un-
wound, the free end of it can be ^ N
made to trace out the original curve.
From this property the locus of the
centers of curvature of a given
curve is called the evolute of that curve, and the latter is called
the involute of the former.
When the string is unwound, each point of it describes a
different involute; hence, any curve has an infinite number
of involutes, but only one. evolute.
Any two of these involutes intercept a constant distance
on their common normal, and are called parallel curves on
account of this property.
Ex. Find the length of that part of the evolute of the parab-
ola which lies inside the curve.
From Fig. 46 the required length is twice the difference between
the tangents C 3 P 3 and P Q C , both of which are normals to the
parabola.
To find the coordinates of the point P 3 , write the equation of the
tangent to the evolute at C 3 , and find the other point at which it
intersects the parabola.
The coordinates of C 3 , the point of intersection of the two curves,
are (8p, 4joV2), and the equation of the tangent at C 3 is
2x -V2y - Sp = 0.
This tangent intersects the parabola at the point (2p, — 2 V2 />),
which is Po.
188 DIFFERENTIAL CALCULUS
The value of the radius of curvature is ~{ x ~*~P) f hence P C =2p,
Vp
P3C3 = GV3p, hence the arc C C S is 2p(3V3 - 1), and the required
length of the evolute is therefore 4jt>(3V3 — 1).
EXERCISES
Find the coordinates of the center of curvature for each of the
following curves :
1. x 2 + y 2 = a 2 . 3. y 3 ^ a 2 x.
2. x = a log a + ^ a ' 2 ~ y2 - Va 2 - y 2 . 4. y=-^( e a + e «).
Find the equations of the evolutes of the following curves :
5. xy = a 2 . 6. a 2 y 2 - b 2 x 2 = - a 2 b 2 . 7. x% + y* = ah
8. Show that the curvature of an ellipse is a minimum at the end
of the minor axis, and that the osculating circle at this point has con-
tact of the third order with the curve.
Fia. 49
This circle of curvature must be entirely outside the ellipse
(Fig. 49). For, consider two points Pi, P 2 , one on each side of B,
the end of the minor axis. At these points the curvature is greater
CONTACT AND CURVATURE
189
than at B, hence these points must be farther from the tangent at B
than the circle of curvature, which has everywhere the same curva-
ture as at B.
9. Similarly, show that the curvature at A, the end of the major
axis, is a maximum, and that the circle of curvature at A lies entirely
within the ellipse (Fig. 49).
10. Show how to sketch the circle of curvature for points between
A and B. The circle of curvature for points between A and B has
three coincident points in common with the ellipse (Art. 93), hence
the circle crosses the curve (Art. 89). Let K, P, L be three points
on the arc, such that K is nearest A and L nearest B. The center
Fig. 50
of curvature for P lies on the normal to P, and on the concave side
of the curve. The circle crosses at P, lying outside of the ellipse at
K (on the side towards .4), and inside the ellipse at L; for the bend-
ing of the ellipse increases from B to P and from P to K, while the
bending (curvature) of the osculating circle remains constant
(Fig. 50).
11. Two centers of curvature lie on every normal. Prove geo-
metrically that the normals to the curve are tangent to the e volute.
190 DIFFERENTIAL CALCULUS
12. Show that the entire length of the e volute of the ellipse is
4 (- — — ). [From equation (11) above, take B v R 2 as the radii of
\b a J
curvature at the extremities of the major and minor axes.]
13. If E is the center
of curvature at the vertex
A (Fig. 51), prove that
CE = ae 1 , in which e is
the eccentricity of the
ellipse; and hence that
CD, CA, CF, CE form
a geometric series whose
common ratio is e. Show
also that DA, AF, FE
form a similar series.
14. If H is the center
of curvature for B, show
that the point 77 is with-
out or within the ellipse,
according as a > or
< bV2, or according as e 2 > or < J. Sketch the evolute when b = -— -•
o
15. Show by inspection of the figure that four real normals can
be drawn to the ellipse from any point within the evolute.
16. Find the parametric equations of the evolute of the cycloid
x = a(d — sin $), y = a (I — cos 6).
CHAPTER XIII
SINGULAR POINTS
101. Definition of a singular point. If the equation f(x, y) =
is represented by a curve, the derivative -^, when it has a
cix
determinate value, measures the slope of the tangent at the
point (x, y). There may be certain points on the curve, how-
ever, at which the expression for the derivative assumes an
illusory or indeterminate form ; and, in consequence, the slope
of the tangent at such a point cannot be directly determined
by the method of Art. 5. Such values of x, y are called sin-
gular values, and the corresponding points on the curve are
called singular points.
102. Determination of singular points of algebraic curves.
When the equation of the curve is rationalized and cleared of
fractions, let it take the form /(a?, y = 0.
This gives, by differentiation with regard to x, as in Art. 65,
df + dfdy^Q
dx dy dx
3f
du dr
whence Tx = ~W' (1)
By
In order that — may become illusory, it is therefore neces-
dx
sarytohave # = 0, ^ = 0. (2)
dx dy
191
192 DIFFERENTIAL CALCULUS
Thus, to determine whether a given curve f(x, y) = has
singular points, put J- and f- each equal to zero and solve
dx dy
these equations for x and y.
If any pair of values of x and y, so found, satisfy the equa-
tion f(x, y) = 0, the point determined by them is a singular
point on the curve.
To determine the appearance of the curve in the vicinity
of a singular point (x lf yx), evaluate the indeterminate form
dy _ dx
dx = ~aj = ^
by finding the limit approached continuously by the slope of
the tangent when x = x x , y = y v "**
dy dx\dx y
Hence
dx d fdf
dx
dx 2 dx dydx
d 2 f , d*fdy
dx dy dy 2 dx
[Arts. 64, 85.
at the point (x ly y{).
This equation cleared of fractions gives, to determine the
slope at (ajj, ?/i)> the quadratic
dy 2 [dxj dxdy\dx) dx 2 V }
This quadratic equation has in general two roots. The
only exceptions occur when simultaneously, at the point in
SINGULAR POINTS 193
in which case -^ is still indeterminate in form, and must be
dx
evaluated as before. The result of the next evaluation is a
cubic in — , which gives three values to the slope, unless all
dx
the third partial derivatives vanish simultaneously at the
singular point.
The geometric interpretation of the two roots of equation
(3) will now be given, and similar principles will apply when
the quadratic is replaced by an equation of higher degree.
The two roots of (3) are real and distinct, real and coinci-
dent, or imaginary, according as
H f a 2 /y d 2 f d*f
\dx dyj dx 2 By 2
is positive, zero, or negative. These three cases will be con-
sidered separately.
103. Multiple points. First let 77 be positive. Then at the
point (x, y) for which -J- = 0, -J- = 0, there are two values of
dx dy
the slope, and hence two distinct singular tangents. It fol-
lows from this that the curve goes through the point in two
directions, or, in other words, two branches of the curve cross
at this point. Such a point is called a real double point of
the curve, or simply a node. The conditions, then, to be satis-
fied at a node (x lf y{) are
ox v oy L
and H(x u y x ) > 0.
Ex. Examine for singular points the curve
3 x 2 _ xy _ o f + x s -8y* = 0.
EL. CALC. — 13
194 DIFFERENTIAL CALCULUS
Here $f=6x-y+3x*, & = - x - 4 y - 21 v*.
dx dy 9 9
The values x = 0, y = will satisfy these three equations, hence
(0, 0) is a singular point.
Since
|^=6+ Qx = 6 at (0,0),
Fig. 52
hence the equation determining the slope is, from (3),
-*a)"-«e)+-*
of which the roots are 1 and — f . It follows that (0, 0) is a double
point at which the tangents have the slopes 1, — f.
Find the equation of the real asymptote, and the coordinates of the
finite point in which it meets the curve.
104. Cusps. Next let H=0. The two tangents are then
coincident, and there are two cases to consider. If the curve
recedes from the tangent in both directions from the point of
tangency, the singular point is called a tacnode. Two distinct
branches of the curve touch each other at this point. (See
Fig. 53.)
SINGULAR POINTS
195
If both branches of the curve recede from the tangent in
only one direction from the point of tangency, the point is
called a cusp.
Here again there are two cases to be distinguished. If the
branches recede from the point on opposite sides of the double
tangent, the cusp is said to be of the first kind ; if they recede (
on the same side, it is called a cusp of the second kind.
The method of investigation will be illustrated by a few
examples.
. Ex. 1. f(x, y) = aY - <* 2 x A + * 6 = 0.
^ = -4a2z« + 6a? s ; & = 2 a*y.
dx By
The point (0, 0) will satisfy /(x, y) = 0, $£■ = 0, |^= 0; hence it
dx dy
is a singular point. Proceeding to the second derivatives, we obtain
§y = - 12 a 2 * 2 + 30 x* = at (0, 0),
dx 2
BY =
dxd/
a 2 / =
o,
The two values of -* are there-
fore coincident, and each equal to Fig. 53
zero. From the form of the equation, the curve is evidently sym-
metrical with regard to both axes ; hence the point (0, 0) is a tacnode.
No part of the curve can be at a greater distance from the y-axis
than ± a, at which points -^ is infinite. The maximum value of y
corresponds tox = ± aVf. Between x = 0, x = aVj there is a point
of inflexion (Fig. 53).
196 DIFFERENTIAL CALCULUS
Sketch the curves obtained by giving larger and larger values to the
parameter a.
Ex.2. f(x,y) = y*-x* = 0;
dx~ 6X ' d y- 2y '
Hence the point (0, 0) is a singular point.
Further, ^ = - 6a: = at (0, 0) ;
d 2 f =0 . dY =2
dxdy ' dy 2
Therefore the two roots of the quadratic equation defining -^ are
both equal to zero. So far, this case is exactly like the last one, but
here no part of the curve lies to the left of the axis y. On the right
side, the curve is symmetric with regard to the x-axis. As x increases,
y increases ; there are no maxima nor minima, and no inflexions
(Fig. 54). .
p]x. 3. f(x, y) - x 4 - 2 ax' 2 y - axy 2 -f ahf- = 0.
The point (0, 0) is a singular point, and the roots of the quadratic
defining -j- are both equal to zero, hence the origin is a cusp, and the
cuspidal tangent is the a; -axis.
To show the form of the curve near the cusp, solve the equation
for ?/. Then ._
a — x\ y a I
First suppose that a is positive.
When x is negative, y is imaginary ; when x = 0, y = ; when x is
positive, but less than a, y has two positive values, therefore two
branches are above the a>axis. When x = a, one branch becomes in-
finite, having the asymptote x = a ; the other branch has the ordinate
\ a. The origin is therefore a cusp of the second kind (Fig. 55).
Next suppose that a is negative. When x is positive, y is imagi-
rary; when x is negative, y is real. The same reasoning as before
SINGULAR POINTS
197
shows that there is a cusp of the second kind in the second quarter,
with the #-axis as a cuspidal tangent.
Examine the transition case in which a = 0.
Fig. 54
Fig. 55
105. Conjugate points. Lastly, let H be negative. In this
case there are no real tangents ; hence no points in the im-
mediate vicinity of the given point satisfy the equation of the
curve.
Such an isolated point is called a conjugate point.
Ex. 1. /(*, y) = ay 2 - x* + bx 2 = 0.
Here (0, 0) is a singular point of the
locus, and at this point we find
dy =
dx
both roots being imaginary if a and b
have the same sign.
To show the form of the curve, solve
the given equation for y.
Then
y=±x\
x — b
Fig. 5G
and hence, if a and b are positive, there are no real points on the
curve between x = and x - b. Thus O is an isolated point (Fig. 56).
Examine the cases in which a or b is negative.
198 DIFFERENTIAL CALCULUS
These are the only singular ities that algebraic curves can
have, although complicated combinations of them may appear.
In each of the foregoing examples, the singular point was
(0, 0) : but for any other point, the same reasoning will apply.
Ex. 2. f(x, y) = x 2 + 3 y* - 13 y 2 - 4 x + 17 y - 3 = 0,
3£=2.r-4, 3/"= 9 y 2 - 26 y + 17.
dx dy
At the point (2, 1), /(2, 1) = 0, &=0, ^=0; hence (2, 1) is
, . 4. dx dy
a singular point.
Also^ = 2; JE£-=0; |2f= 18y - 26, = - 8 at (2, 1).
dx 2 da; cty d// 2
Hence — = ± - ; and thus the equations of the two tangents at the
dx 2
node (2, 1) are y - 1 = \(x - 2), y - 1 = - | (a: - 2).
When JT is negative, the singular point is necessarily a con-
jugate point, but the converse is not always true. A singular
point may be a conjugate point when H = 0. [Compare
Ex. 4 below.]
EXERCISES ON CHAPTER XIII
Examine each of the following curves for multiple points and find
the equations of the tangents at each such point ; also find the
asymptotes and sketch the curve :
1. a 2 x 2 = b 2 y 2 + *V-
2 a — x
3. x3 + y\ = «f ; or, in rational form, (x 2 +y 2 — a 2 ) 3 + 27 a 2 x 2 y 2 = 0.
4. y 2 (x 2 - a 2 ) = x 4 .
5. y— a+ x + bx 2 + cx% ; or, in rational form,
( y - a -x- bx 2 ) 2 - c 2 s« = ( ) .
SINGULAR POINTS 199
When a curve has two parallel asymptotes it is said to have a node
at infinity in the direction of the parallel asymptotes. Apply to
Ex. 6.
6. (x*-y*y—4y* + y = 0.
7. x*-2 ay* - 3 a 2 y 2 - 2 d 2 x 2 + a 4 = 0.
J. y 2 _ x ( x + a )2. a > o . a < o.
9. x s — 3 axy + y z = 0. Find the asymptote and sketch the curve.
10. y' 2 = x* + x 5 .
11. Show that the curve y = x log a: has a terminating point at the
origin. Find the minimum value of y and sketch the curve.
12. y — x 2 log x.
CHAPTER XIV
ENVELOPES
106. Family of curves. The equation of a curve,
/(»,y) = o,
usually involves, besides the variables x and ?/, certain coeffi-
cients that serve to fix the size, shape, and position of the
curve. The coefficients are called constants with reference
to the variables x and y, but it has been seen in previous
chapters that they may take different values in different
problems, while the form of the equation is preserved. Let
a be one of these "constants." Then if a is given a series
of numerical values, and if the locus of the equation, corre-
sponding to each special value of a is traced, a series of curves
is obtained, all having the same general character, but differ-
ing somewhat from each other in size, shape, or position. A
system of curves so obtained is called a family of curves.
For example, if h, k are fixed, and }> is arbitrary, the equa-
tion (y—k) 2 = 2p(x — h) represents a family of parabolas,
each curve of which has the same vertex (h, k), and the same
axis y = k, but a different latus rectum. Again, if k is the
arbitrary constant, this equation represents a family of parab-
olas having parallel axes, the same latus rectum, and having
their vertices on the same line x = h.
The presence of an arbitrary constant a in the equation of
a curve is indicated in functional notation by writing the
200
ENVELOPES 201
equation in the form ,/(#, y, a) = 0. The quantity a, which
is constant for the same curve but different for different
curves, is called the parameter of the family. The equations
of two neighboring curves are then written
f(x, y, a) = 0, />, y, a + h) = 0,
in which h is a small increment of a. These curves can be
brought as near to coincidence as desired by diminishing h.
107. Envelope of a family of curves. A point of intersection
of two neighboring curves of the family tends toward a limit-
ing position as the curves approach coincidence. The locus of
such limiting points of intersection is called the envelope of
the family.
Let f(x, y, a) = 0, f(x, y, a + h) = (1)
be two curves of the family. By the theorem of mean value
'(Art. 39)
f(x, y, a + h) = f(x, y, a) + h d /- (as, y, a + 6k), (2)
da
which, on account of equation (1), reduces to
d £{x,y,« + 6h)=().
Hence, it follows that in the limit, when h = 0,
J-(x, y, «)=0
da
is the equation of a curve passing through the limiting points
of intersection of the curve /(a?, y } a) = with its consecutive
curve. This determines for any assigned value of a definite
limiting points of intersection on the corresponding member of
202 DIFFERENTIAL CALCULUS
the family. The locus of all such points is then to be obtained
by eliminating the parameter a from the equations
f(x, y, a) = 0, -f (x, y, a) = 0.
da
The resulting equation in x and y represents the fixed enve-
lope of the family.
108. The envelope touches every curve of the family.
I. Geometrical proof. Let A, B, C (Fig. 57) be three consec-
utive curves of the family ; let A, B intersect in P, and B, C inter-
sect in Q. When P, Q approach coincidence, PQ will be the
direction of the tangent to the envelope at P; but since P, Q
are two points on B that approach coincidence, hence PQ is
also the direction of the tangent to B at P, and accordingly B
and the envelope have a common tangent at P. Similarly for
every curve of the family.
II. More rigorous analytical proof. Let — f(x, ?/, a) =
da
be solved for a, in the form a = <£(#, y). Then the equation
of the envelope is
f(x, y, <f>(x, y)) = 0.
Equating the total avderivative to zero, we obtain
dx dy dx d<f>\dx By dxj
ENVELOPES 203
but -J- = -J- = 0, hence the slope of the tangent to the enve-
d<f> da
lope at the point (x, y) is given by
df + dfdy =
dx dy dx
But this equation defines the direction of the tangent to the
curve /(a?, y, a) — at the same point, and therefore a limit-
ing point of intersection on any member of the family is a
point of contact of this curve with the envelope.
Ex. Find the envelope of the family of lines
Differentiate (1) as to m,
y = mx+£-, (1)
• m
obtained by varying m.
=*-■£. (2)
To eliminate m multiply (2) by m and square ; square (1) and sub-
tract the first from the second. The envelope is found to be the
parabola « A
r y 2 = Ipx.
Draw the lines (1) corresponding to
m = 1, 2, 3, 4, oo ; m - - 1, - 2, - 3, - 4.
109. Envelope of normals of a given curve. The evolute
(Art. 99) was defined as the locus of the centers of curvature.
The center of curvature was shown to be the point of intersec-
tion of consecutive normals (Art. 92), whence by Art. 107 the
envelope of the normals is the evolute.
Ex. Find the envelope of the normals to the parabola y 2 = ±px.
The equation of the normal at (x v y\) is
204 DIFFERENTIAL CALCULUS
or, eliminating x\ by means of the equation yf = ipx v we obtain
J Ul 8^ 2 p K }
The envelope of this line, when y x takes all values, is required.
Differentiate as to ?/,, n „
_ i - ^JL\ _ JL
8p* 2p'
On substituting this value for y l in (1), the result,
27 pif = 4(> - 2 jo) 3 ,
is the equation of the required evolute. Show that this semi-cubical
parabola has a cusp at (2p, 0). Trace the curve.
110. Two parameters, one equation of condition. In many
cases a family of curves may have two parameters which are
connected by an equation. For instance, the equation of the
normal to a given curve contains two parameters x Yi y x which
are connected by the equation of the curve. In such cases
one parameter may be eliminated by means of the given rela-
tion, and the other treated as before.
When the elimination is difficult to perform, both equations
may be differentiated as to' one of the parameters, a, regard-
ing the other parameter /? as a function of a. This gives four
equations from which a, 8, and — may be eliminated, the
da
resulting equation being that of the desired envelope.
Ex. 1. Find the envelope of the line
a b
the sum of its intercepts remaining constant.
ENVELOPES ■ 205
The two equations are - -\- -J- — 1,
a b
a + b = c.
Differentiate both equations as to a ;
— x _y_db _ „
• a 2 H 2 da~ '
1+*
da
Eliminate
Then — = *-. which reduces to
a 2 b 2
db i
da
x y x + y
a b a h
- — - ; whence a = Vex, b = Vey.
a b a + b c
Therefore Vx + Vy = Vc
is the equation of the desired envelope. [Compare Ex. p. 87.]
This equation when rationalized is
(x- yy-2c(x + y) + c 2 = 0.
By turning the coordinate axes through 45°, show that this repre-
sents a parabola whose axis bisects the angle between the original
axes. Show that the curve touches both these axes. Draw different
lines of the family, corresponding to a = 4, b = 4 ; a = 5, b = 3 ; a = 6,
Z> = 2 ; a - 7, b = 1 ; a = 8, 6 = 0; etc.
Ex. 2. Find the envelope of the family of coaxial ellipses having
a constant area.
Here x - + f- = \-
a* l b 2
a& = £ 2 .
For symmetry, regard a and 6 as functions of a single parameter /.
206 DIFFERENTIAL CALCULUS
Then by differentiation as to t,
&<]±j r t_db _
a s dt b* dt.~ '
hence
-.da . db A
b h a — =
dt dt
& = T
i 2 6 2 2
a = ±xV'2, b =±yV2,
and the envelope is the pair of rectangular hyperbolas xy = ± \ k' 2 .
Y
Fig. 58
Note. A family of curves may have no envelope; i.e., consecutive
curves may not intersect; e.g., the family of concentric circles x 2 +y 2
= r 2 , obtained by giving r all possible values.
ENVELOPES 207
If every curve of a family has a node, and the node has
different positions for different curves of the family, the enve-
lope will be composed of two (or more) curves, one of which
is the locus of the node.
Ex. Find the envelope of the system
in which X is a varying parameter.
Here 3L = — 2(y — A.) = 0; by combining with /= to eliminate
dX
X, we obtain x 2 = 0, x — 1 = 0, x + 1 = 0.
From Art. 103 it is seen that the point
x = 0, y = X
is a node on /; moreover, the various curves of the family are ob-
tained by moving any one of them parallel to the y-axis. The lines
x — 1=0, £ + 1 = form the proper envelope, and x = is the locus
of the node.
EXERCISES ON CHAPTER XIV
Find the envelope of each of the following families of curves ;
draw to scale various members of the family, and verify that the en-
velope has been correctly found.
1. The family of straight lines x cos a + y sin a = p, when « is a
parameter.
2. A straight line of fixed length a moving with its extremities
in two rectangular axes.
3. Ellipses described with common centers and axes, and having
the sum of the semi-axes equal to c.
4. The straight lines having the product of their intercepts on
the coordinate axes equal to k 2 .
5. The lines y - ft = m (x -«)+ rvT+m*, m being a variable
parameter.
208 DIFFERENTIAL CALCULUS
6. A circle moving with its center on a parabola whose equation
is y 2 = 4 ax, and passing through the vertex of the parabola.
7. A perpendicular to any normal to the parabola y 2 = 4 ax,
drawn through the intersection of the normal with the x-axis.
8. The family of circles whose diameters are double ordinates of
the ellipse b 2 x' 2 + a 2 y 2 = a 2 // 2 .
9. The circles which pass through the origin and have their
centers on the hyperbola x 2 — y 2 = c 2 .
10. The family of straight lines y = 2 mx + m A , m being the vari-
able parameter.
11. The ellipses whose axes coincide, and such that the distance
between the extremities of the major and minor axes is constant and
equal to k.
12. From a fixed point on the circumference of a circle chords are
drawn, and on these as diameters circles are described.
13. With the point (xi, //i) on a given ellipse as center, an ellipse
is described having its axes equal and parallel to those of the given
ellipse. Let (xi, y{) describe the given ellipse.
14. Show that if the corner of a rectangular piece of paper is
folded down so that the sum of the edges left unfolded is constant,
the crease will envelop a parabola.
15. In the " nodal family " (y - 2 a) 2 =(x - «) 2 + 8 x s - y 3 , show
that the usual process gives for envelope a composite locus, made up
of the "node-locus" (a line) and the envelope proper (an ellipse).
16. The family of curves (y — x 2 ) + a (x — y 2 ) = 0.
INTEGRAL CALCULUS
:>>*<
CHAPTER I
GENERAL PRINCIPLES OF INTEGRATION
111. The fundamental problem. The fundamental problem
of the Differential Calculus, as explained in the preceding
pages, is this :
Given a function f(x) of an independent variable x, to deter-
mine its derivative f'(x).
It is now proposed to consider the inverse problem, viz. :
Given any function f'(x), to determine the function f(x) hav-
ing f'(x) for its derivative.
The solution of this inverse problem is one of the objects
of the Integral Calculus.
The given function f(x) is called the integrand, the func-
tion f(x) which is to be found is called the integral, and the
process gone through in order to obtain the unknown function
f{x) is called integration.
The operation and result of differentiation are symbolized
by the formula c j
Tx f{x) = m ' (1)
or, written in the notation of differentials,
df(x)=f\x)dx. (2)
el. c/xc. — 14 209
210 INTEGRAL CALCULUS
The operation of integration is indicated by prefixing the
symbol j to the function, or differential, whose integral it is
required to find. It is called the integral sign, or the sign of
integration. Accordingly, the formula of integration is written
thus :
/(*)=//'(*)
dx.
Following long established usage, the differential, rather
than the derivative, of the unknown function /(a?) is written
under the sign of integration. One of the advantages of so
doing is that the variable, with respect to which the integration
is performed, is explicitly mentioned. This is, of course, not
necessary when only one variable is involved, but it is essential
when several variables enter into the integrand, or when a
change of variable is made during the process of integration.
112. Integration by inspection. The most obvious aid to
integration is a knowledge of the rules and results of differen-
tiation. It frequently happens that the required function /(a?)
can be determined at once by recollecting the result of some
previous differentiation.
For example, suppose it is required to find
/
cos x dx.
It will be recalled that cos x dx is the differential of sin x, and
thus the proposed integration is immediately effected ; that is,
/
cos xdx = sin x.
Again, suppose it is required to integrate
x n dx,
/•
GENERAL PRINCIPLES OF INTEGRATION 211
in which n is any constant (except — 1). This problem sug-
gests the formula for differentiating a variable affected by a
constant exponent [(6), p. 44]. The formula referred to may
be written . n+1
d(- )= x n dx,
\n + y
/» zy.n + 1
and hence we conclude,
s, f x n dx
?i + l
An exception to this result occurs when n has the value — 1.
For in that case we deduce from (8), p. 44, the formula of
integration
I x L dx = I — = log x.
The method used in the above illustrations may be designated
as integration by inspection. This is, in fact, the only practical
method available. The object of the various devices suggested
in the subsequent pages is to transform the given integrand
or to separate it into simpler elements in such a way that the
method of inspection can Ipe applied.
113. The fundamental formulas of integration. When the
formulas of differentiation, pp. 44-45, are borne in mind, the
method of inspection referred to in the preceding article leads
at once to the following fundamental integrals. Upon these,
sooner or later, every integration must be made to depend.
clu =
n + l
n+ 1
I. f u tl
n.Jf = ,. g „.
III. Ca^du=~ lU -
J log a
212 INTEGRAL CALCULUS
IV. Ce u du = e u .
V. I cos u du = sin u,
VI. I sin u du = — cos u.
VII. I sec 2 udu = tan w.
VIII. I esc 2 «* rfi* = — cot u,
IX. J sec u tan ^ d«* = sec u.
X. I esc u cot ?^ f/«j = — esc u.
XI. f rfM =sin-iM. or-cos-t^.
XII. f ^ M , = tan-i w, or - cot-i t*.
J 1 + w 2
114. Certain general principles. In applying the above for-
mulas of integration certain principles which follow from the
rules of differentiation should be made use of.
(a) Tlie integral of the sum of a finite number of functions is
equal to the sum of the integrals of the functions taken separately.
This follows from Art. 10.
For example,
I dx = I x dx — I — ■ = - — lo
J x *s J x "J
sx.
GENERAL PRINCIPLES OF INTEGRATION 213
(b) A constant factor may be removed from one side of the
sign of integration to the other.
For, since d(cu) = cdu,
it follows that I cdu—cu-c I du.
To illustrate, let it be required to integrate
x 2 dx.
/<
The numerical factor 5 is first placed outside the sign of
integration, after which formula I is applied. Accordingly,
n ^3
I 5 x 2 dx = 5 I x 2 dx = -—.
Again, suppose the integral
/,
x dx
is to be found. We notice that if the numerator had an addi-
tional factor 2, it would be the exact differential of the
denominator, and formula II would be applicable. All that is
required, then, in order to reduce the given integral to a known
form, is to multiply inside the sign of integration by 2 and
outside by \. This gives
Cxdx 1 C2xdx lAI(x 2 + l) li /o liN
In this connection it must not be forgotten that:
An expression containing the variable of integration cannot be
moved from one side of the sign of integration to the other.
(c) An arbitrary constant may be added to the result of
integration.
214 INTEGRAL CALCULUS
For, the derivative of a constant is zero and hence
du = d(u + c),
from which follows
d(u + c) = u + c.
f au =f'
This constant is called the constant of integration.
From the preceding remark it follows that the result of
integration is not unique, but that any number of functions
(differing from each other, however, only by an additive con-
stant) can be found, each of which has the same given expres-
sion as its derivative. [Compare Art. 10, Cor.]
Thus, any one of the functions x 2 — l, x* + l, tf + a 2 ,
(x—a)(x-\-a) may serve as a solution of the problem of inte-
grating f 2 x dx.
It often happens that different methods of integration lead
to different results. All such differences, however, can occur
only in the constant terms.
For example,
f 3 (x + 1) 2 dx = 3 |f(aj + l) 2 d{x + 1) = (x + l) 3
= aj 8 + 3a 2 -f3a;+l.
Integration of the terms separately gives
Csx 2 dx-\- C$xdx+ ^3^ = ^ + 3^4-3^,
a result that agrees with the preceding except in the constant
term.
Again, from formula XII,
dx
s
tan -1 x, or — cot -1 x.
ar' + l
GENERAL PRINCIPLES OF INTEGRATION 215
It does not follow from this that tan _1 & is equal to — cot -1 #.
But they can differ at most by an additive constant. In fact,
it is known from trigonometry that
— cot -1 x = tan -1 x -f for + j ,
_
in which k is any integer.
In a similar manner the different results in formula XI can
be explained.
EXERCISES
Integrate the following :
1. (Vxdx. 11. f CSc2j: ^.
J J cot X
[Hint. For the purpose of in-
tegration this may be written 12. f - sin
(x% dx.~]
J ~ dx
fw*. 13. f * L/ *
J J x log x\_ t/ log x
+ cos a;
r
2. \ x a dx.
3
log x L «' log a:
f ,/.* 14 C 5x 2 dx
' J 3-' ' J X * + l '
V x
4 Cmt*JJz m 15. Jtana:^[ = -| -^ s ^ ].
5. jV -**)•**. 16 ^otxdx.
6 C5x*-^x + l dx 17 C eaxdx ,
7. (x(x 2 + a 2 )' 2 dx. H8. (e**xdx.
8. J(aar+6)»rfar. * 19. f («-!-&) »»+n*a*a:.
20. I cos 2 xdx.
9 ' J^i- 20- J'
, n C(n -x)dx r .
u " J Ti 7 * 21. i sin nz rt.r.
^ 2 ax - x 1 J
216 INTEGRAL CALCULUS
, 22. fco**xdx[= P+ c ™ 2x dx~].
23- \sln 2 xdx. i24. i sin(m -f n)x dx. 25. Kxsmx^dx.
26. ( co$?xdx\ = i (1 — sin 2 a:)cos xdx . 27. ( sin 3 a;G?a;.
28. rtan2xr/xT= f(sec 2 a; - l)r/af|. 29. f tan 2 x sec 2 ztf*
30. \ esc 2 (ax -\- b)dx. 31. ( Vcot a: • esc 2 x dx.
32 f ^ r f — r sec 2 .rr/.r ~j
•/ sin x cos a; L J tan a: J
33. \ sec 3 a: tan a; dx.
34. r tnTirr/r . 35. r
J sec a- ./
r/s
Va 2 - x 2
[Hint. Divide numerator and denominator by a and then write
in the form
J ^w
36.
J
dx
Vl - 4 x 2
37.
J a 2 + m 5
38. f-^-
J a 2 a- 2 +
fta
39. f ** r = f ^- 2 > i.
J x-2 _ 4 3 + 5 L J (a: - 2) 2 + 1 J
115. Integration by parts. If u and v are functions of x, the
rule for differentiating a product gives
d(uv) = v du -J- u dv,
whence, by integrating and transposing terms, we have
I u dv — uv — | v du.
GENERAL PRINCIPLES OF INTEGRATION 217
This formula affords a most valuable method of integration,
known as integration by parts. By its use a given integral is
made to depend on another integral, which in many cases is
of a simpler form and more readily integrable than the
original one.
Ex. 1. J \ogxdx.
Assume u = log x, dv = dx.
Then du = — , v = x.
x
By substituting in the formula for integration by parts, we obtain,
i log xdx — x log x — \ dx
= x log x — x = xQog x — 1 )
= x(\og x — log e) = x log - .
e
Ex. 2. (xe x dx.
Assume u = x, dv = e* dx.
Then du = dx, v = e x ,
and ( xe x dx = xe x — ( e x dx = e?(x — 1).
Suppose that a different choice had been made for u and dv in the
present problem, say u = ^ rfy = ^
From this would follow
r 2
du — e x dx. v = — ,
2
and j xe x dx = | x 2 e x — I —e x dx.
C x 2
It will be observed that the new integral j —e*dx is less simple in
form than the original one ; hence the present choice of u and dv
is not a fortunate one.
No general rule can be laid down for the selection of u and dv.
Several trials may be necessary before a suitable one can be found.
218 INTEGRAL CALCULUS
It is to be remarked, however, that dv should be so chosen that its
integral may be as simple as possible, while u should be so chosen
that in differentiating it a material simplification is brought about.
Thus in Ex. 1, by taking u — log x, the transcendental function is
made to disappear by differentiation. In Ex. 2, the presence of either
x or e x prevents direct integration. The first factor x can be removed
by differentiation, and thus the choice u = x is naturally suggested.
Ex. 3.
( x 2 a x dx.
From the preceding remark it is evident that the only choice which
will simplify the integral is
u = x 2 , dv = a x dx.
a
Hence du = 2 xdx, v = - ,
log: a
and
(x 2 a*dx = -^- — (xa x dx.
J lotr a log: a J
Apply the same method to the new integral, assuming
u = x, di' = a x dx,
whence du = dx, v — ,
log a
and f xa x dx = ^- - — !— f a x dx
■J log a log a J
x(t ar
log a (log a)' 2
By substituting in the preceding formula, we have
J log a L log a (logo) 2 J
GENERAL PRINCIPLES OF INTEGRATION 219
EXERCISES
1.
i sin~ i xdx.
7.
i x cot -1 x dx.
2.
f «* tan" 1 (€*)</*.
8.
\ x sin 3 x dx.
3.
1 a: 2 cos xdx.
9.
\ e* cos x dx.
4.
\x n \ogxdx.
10.
\ e x sin xdx.
5.
1 a* 2 tan -1 xdx.
11.
\ cos a: cos 2 a: /7a*.
6.
f sec a: tan x log cos a:
dx.
12.
I x sec 2 a- </a\
116. Integration by substitution. It is often necessary to
simplify a given differential f'(x)clx by the introduction of a
new variable before integration can be effected. Except for
certain special classes of differentials (see, for example, Arts.
127-129) no general rule can be laid down for the guidance of
the student in the use of this method, but some aid may be
derived from the hints contained in the problems which follow.
Ex. 1. | xdx .
Va 2 - x 2
Introduce a new variable z by means of the substitution a-— x 2 = z.
Differentiate and divide by - 2, whence xdx— — — . Accordingly,
The details required in carrying out this substitution are so simple
that they can be omitted and the solution of the problem will then
take the following form :
f XdX = (V - **)"* xdx = - I ( (« 2 - x*y*( - 2 xdx)
J Vf/ 2 - a- 2 J • - J
220 INTEGRAL CALCULUS
In this series of steps the last integral is obtained by multiplying
inside the sign of integration by — 2 and outside by — I, the object
being to make the second factor the differential of a 2 — x 2 . Think-
ing of the latter as a new variable, the integrand contains this
variable affected by an exponent (- i) and multiplied by the differ-
ential of the variable, in which case formula I can be applied.
Ex. 2. r !°£* dx .
J x
Assume log x = z..
Then -- = dz,
x
and fl^?^ = f^ = l 2 = -Q^.
J x J 2 2
Here again it is not necessary to write out the details of the sub-
stitution, as it is easy to think of log a; as a new independent variable
and to perform the integration with respect to it. It is then readily
seen that the expression to be integrated consists of the variable
dx
log # multiplied by its differential — : , and that the integration is
x
accordingly reduced to an immediate application of the first formula
of integration. Thus
k2
ij\ogx.d(\ogx) = Off*)'
Ex. 3. j
Ex. 3. \ e ian *
dx
gives
1 + x 2
Think of tan -1 x as a new variable and apply formula IV. This
r e tan-'*_^ _ fgtan-'x, /(tan-la-) = ^tan^x.
J 1 + x 2 J
Ex. 4. C*™- 1 ***.
dx
j-
Ilegard sin -1 a: as a new variable and - as the differential
Vl — x 2
of that variable. Apply formula I.
GENERAL PRINCIPLES OF INTEGRATION 221
Ex. 5. j*0 2 + 2 x + 3)0 + l)dx.
Multiply and divide by 2. The integral then takes the form
If (a? + 2 x + 3) • (2 x + 2)rfar.
Observing that (2 x -\- 2)dx is the differential of x 2 + 2 x + 3, and
using the latter expression as a new variable, we see that formula
I is directly applicable, leading to the result
Ex. 6. flog cos (x 2 + 1) sin (a; 2 + 1) • xdx.
Make the substitution x" 2 + 1 = z.
The given integral takes the form
- I log cos z sin z dz.
Make a second change of variable,
cos z — y.
Then sin zdz = — dy.
The transformed integral is
-lijlogydy,
to which the result of Ex. 1, p. 217, can be at once applied.
It will be observed that two substitutions which naturally suggest
themselves from the form of the integrand are made in succession.
The two together are obviously equivalent to the one transformation,
cos(x 2 + 1) = y.
Ex.7, f du ■ Ex.8, f du .
J y/cfi _ u * J u- + a 2
-[Hint. Substitute u = as.]
222
INTEGRAL CALCULUS
Ex. 9. f
dx
/„.9
— • Ex. 10.
^
dx
V2 ax — x*
Hint. Substitute x = - • [Hint. Substitute x = z + a.]
Ex. 11. | esc m du.
Multiply and divide the integrand by esc u — cot u. It will then
be seen that the integral has the form i —
J z
Another method would be to use the trigonometric formula
n • u u
sin u = 2 sin -cos -,
2 2 \
2 V2/ Cdt
=1—1
tan^ J l
2
Ex. 12.
\ sec u du.
Put u = z and use Ex. 11.
2
Solve the problem also by means of substitutions similar to those
used in the preceding example.
sin 3 x
Ex.13. (x*y/a*-x*dx. Ex.15. f£°l£
J J sin-
Ex.14, f * a ' 7ar • Ex.16, f
J(a;-1) 3 J
dx
cos 2 a; -f 2 sin 2 x
Put tan x — z.
Ex. 17. Prove that I — — can be integrated by a substitution,
J (a + bx) n
when m is a positive integer.
117. Additional standard forms. The integrals in Exs. 7, 8,
11, 12 of the preceding article, and in Exs. 15, 16 of Art. 114,
are of such frequent occurrence that it is desirable to collect
GENERAL PRINCIPLES OF INTEGRATION 223
the results of integration into an additional list of standard
forms. Two other very useful formulas are also included, the
derivation of which we now give.
du
Integration of f du
Make the substitution
u + Vw 2 4- « = z.
From this equation, we obtain, by differentiation,
(l + U } du = dz ;
\ Vw 2 + a J
du
that is, ( VV + a + u) — — = = dz,
Vw* + a
whence,
du dz dz
Vm 2 + a Vw 2 + a-\-u z
This gives, on integrating,
>_* ,r*_ log ,
J Vm 2 +« ^ z
= \og(u -f Vw 2 + a).
Integration of f f u o •
J w 2 — a 2
The fraction — may be written as the sum of two
u 2 — a"
simpler fractions,
_l_ = j_[^ l_i
u- — a 2 2a\_u —a u + a J
224 INTEGRAL CALCULUS
whose denominators are the factors of u 2 — a 2 . Hence,
r du _ 1 rr_du_ du 1
J u 2 — a 2 2 a J [_u — a u -f a J
li / x .1 1 , u— a
J a it 4- ct
= 2^ log (w - a)- log (w 4- a)
XIII. du =S in-^
Va 2
/;
XIV. f **" - log (u 4 VttHo).
a
u — a
u + a
XVII. I tan udu = - log cos u - log sec u.
XVIII. J cot udu = log sin u.
XIX. J sec w du = log (sec ™ + tan u) = log tan (tt + jY
XX. J esc m r Jw = log (esc u — cot w) = log tan „ •
118. Integrals of the forms
/ •(Ax + B)dx and /» (^g; + B)dx
ax 2 + bx + c J ■ y /ax*+ bx + c
Such integrals occur so frequently that they deserve special
mention. The integration is facilitated by the substitution
of a new variable t which reduces the affected quadratic
ax 2 4- bx 4- c to a pure quadratic of the form mt 2 + n. The
mode of procedure will be readily understood from the follow-
ing illustrative problems.
.
GENERAL PRINCIPLES OF INTEGRATION 225
x , , C xdx
I,x. 1. \ — -.
J 2 x 2 + 2 x + 3
The first step is to complete the square of the x terms in the
denominator. After the factor 2 has been placed outside the integral
sign, the quadratic expression may be written
(x* + x + 1) + a - \) = (x + \y + 1 .
Now substitute a new variable t in place of x+\. Since x = t — \ and
dx = dt, we obtain for the new form of the given integral
1 r(t-\)dt = \ r'ltdt If dt
**+*
1L
= ilog(* 2 +-^ — tan
2V5 V5
Ex.2, f (2^-1)^
J VI +2^-3^'
Divide out V3 from the denominator; since the coefficient of x 2 is
negative, put the x terms in parentheses preceded by the negative sign
and complete the square. The integral then becomes
\)dx.
V'SJ v'| - (x - i)*
Now make the substitution x — \= t. Since dx = <//, the integral
reduces to
V ( -^=-^(M 4 (-- -^-
V.jJ V| - ** ViiJ \9 / 3V3J V-f - t' 2
2
'3A9 ; 3 V3 V 2 /
2 /r~2 I 1 • J3x-1\
-v/- -f - :r — a; 2 sin -1 -
Vo - ^ 3 3 V3 V 2 ;
- -a/1 + 2 a: - 3 x 2 - — siu-*f 3 * ~ * V
3^ 3V3 V 2 J
EL. CALC — 15
226 INTEGRAL CALCULUS
It is seen from the two preceding examples that the met)
here used contains two essential steps :
(1) Completing the square of the x terms in ax 2 -\- bx + c ;
(2) Substituting a new variable for the part in parentheses.
If the numerator of the new integral contains two terms,
separate into two integrals and integrate each one separately.
EXERCISES
1 C dx 8 C (' 2x ~ 3 ) rix
J3x 2 -2x + 5' 9. )\ }~ X dx.
dx
J 8 + 4 x - 4 x 2
4
r dx
^ V30 x - 9 x i - 24
5.
C x dx
* Vx* + 2 x + 2
6.
J VI + 2 x - x 2
7
C (4x + 5)dx
* V8 - 4 x - 4 x 2
119. Integrals of the forms
+
[Rationalize the numerator.]
1Q r (3x + 2)rfz ,
11.
]•#
12 f (2 x _+_l),/a .
J V-2^- 3z- 1
13 r (x-Z)dx
' J V-3s 2 -2a; + l
f- ** and f
dx
(Ax + ByWax* + bx + c
Integrals of these types can be reduced to forms given in
the preceding article by means of the reciprocal substitution
GENERAL PRINCIPLES OF INTEGRATION 227
I x EXERCISES
C dx 7 f_. dx
* xVx* + a 2 ^ (x + 2) V- a: 2 - 10x-7
8. f dx .
J ar 2 Va 2 — x 2
^
6
xVx 2 + a 2
rfx
xVa 2 — x 2
dx
x Vo x- — 4 x +
1
rte
(x + l)Vx 2 + 2
X + 3
tfs
(x + l)Vx 2 + x
+ 1
tfx
^ x 2 v x 2 - a 2
10. f_
rfj;
5. f- dx ' JxWx* + a*
r ^/x ii r - f i x
* (\ - x^ \f'2 x 2 - 4 r. + 1 ^ f2 x -
(1 - x) v/2 x 1 - 4 x + 1 J (2 x- - 1) V4 x' 2 - 3
EXERCISES ON CHAPTER I
1. ( e** e x dx. 7. \x(a' 2 - x 2 )* dx.
8. f *•« ■
9. f •" ■
J y/ X + 1 + Vx - 1
10. i cos 8 x dx.
11. ( sec 3 x dx.
12- 1 f r sin e* rfx.
2.
f 5 x z dx
J 4 + x 8
3
f (2 + 3x 2 ) f /x
J 6 x 3 + 12 x + 5
4.
f 1 + *rf*.
J Vx
5.
f * .
J ^8 - 2 «
6
r //x
228 INTEGRAL CALCULUS
dx
13
C sin x dx 23 f_
J a cos x + b * 6xVl - log a;
C dx
' 1 , «/. C e x dx
J Vl - e 2 * 24. \ -•
14
. [Put e— = «.]
15
r s<fo . 25. j*
e x + e-
cos dO
y/l _ ^4 vl + cos' 2 - sin ^
16.
S , dX ■ 26. f-
dx
V'4 a: 4 + 8 a; 2 J a: (log a:) 2 + x
17 l e T^- 27. f( seC * Vd,.
e e J \a - b tan x/
18. \% 4 tan -1 a; da;.
28. J.
(a: — a) dx
C x~ dx J Va*-a 2 (x — a) 2 —(x-a) 4
J a x
2,j_
™ f ^0 r fl-sin^./n
20 - )m^L = 3^s^ rfe ]- "«-v8^ +2 «+i
._ f tan (9^ r
Ja + fttan'0" 31. J sin a: log tan x dx.
32 r ^ r— f s * n x ^ x
J 1 + cot a: L J sin a: 4- cos x
— 1 f ( sm x + cos x) — (cos x — sin a:) ,
2 J sin a; + cos x
= 1 rA.cosar-sinarX^l
2 J V sin a; -f cos a;/ J
[Another method would be to multiply numerator and denominator
by sin a:(cos x — sin x) and express in terms of the double angle.]
CHAPTER II
REDUCTION FORMULAS
120. In Arts. 118, 119 the integration of certain simple ex-
pressions containing an irrationality of the form ^/atf+bx+c
was explained. As was shown in Art. 118, the radical can
be reduced to the form V ± x 2 ± a? by a change of variable.
It remains to show how the integration can be performed in
in such cases as, for example,
x n dx
j z w V ± x 2 ± a 2 dx, J -
V ± a? ± ci-
ri being any integer.
For this purpose it is convenient to consider a more general
type of integral of which the preceding are special cases, viz.,
Car (a + bx n ) p dx, (1)
in which m, n, p are any numbers whatever, integral or frac-
tional, positive or-negative.
It is t$> be remarked in the first place that n can, without
loss of generality, be regarded as positive. For, if n were
negative, say n = — n', the integrand could be written
x mf a + IV = x m f axn 'j' 5 V = x m -* n '(b + ax*'y
This expression, which is of the same type as x m (a + bx n ) p , is
such that the exponent of x inside the parentheses is positive.
229
230 INTEGRAL CALCULUS
It will now be proved that an integral of the type (1) can in
general be reduced to one of the four integrals
(a) A j x m ~ n (a + bx n ) p dx, (b Ai x m+n (a + bx n ) p dx,
(c) A I x m (a + bx n ) p -*dx, (d) A ( x m (a + bx n ]*+ l dx f
plus an algebraic term of the form
Bx K (a + bx n y.
Here A, B, \, fx are certain constants which will be deter-
mined presently.
Observe that in each of the four cases the integral to which
(1) is reduced is of the same type as (1), but that certain
changes have taken place in the exponents, viz.,
the exponent m of the monomial factor is increased or dimin-
ished by n,
or, the exponent p of the binomial is increased or dimin-
ished by unity.
The values of A. and fi are determined by the following rule :
Compare the exponents of the monomial factors in the given
integral and in the integral to which it is to be reduced. Select
the less of the two members and increase it by unity. The result
is the value of\. In like manner, compare the exponents of the
*
binomial factors in the two integrals, select the less, and increase
it by unity. This gives /x.
Thus, if it is desired to reduce the given integral to
A j x m ~ n (a 4- bx n ) p dx,
first write down the formula
j x m (a + bx n ) p dx — A \ x m ~ n (a 4- bx n ) p dx + Bx x (a 4- &»■)*.
REDUCTION FORMULAS 231
The exponents of the monomial factors in the two integrals
are m and m — n respectively, of which m — n is the less.
This, increased by unity, gives the value of X; that is,
X = m — n + 1.
Again, the exponent of the binomial factor in each integral
is the same, namely p, so that there is no choice as to which of
the two is the less. Increase this number p by unity to obtain
the value of /x. Hence /x = p -f 1.
The above formula may now be written
J x m (a -(- bx n ) p dx
= A Cx m -' l (a + bx n ) p dx + Bx m - n+1 (a + bx n ) p+ \ (2)
In order to determine the values of the unknown constants
A and B, simplify the equation by differentiating both mem-
bers. After being divided by x m ~ n (a + bx n ) p the resulting
equation is reduced to
x n = A + Ba(m — n + 1) + Bb(m + np + l)a,- n .
By equating coefficients of like powers of x in both members,
we find the values of A and B to be
A _ a(m - m + 1) B= 1
b(m + np -h 1) ' b(m + np + l)'
When these values are substituted in formula (2), it becomes
J x m (a +bx n ) p dx
ci{ m — n + 1)
+ np+l)J v 7 6(m+np + l) LJ
Notice that the existence of formula (2) has been proved
by showing that values can be found for A and B which make
the two members of this equation identical.
232 INTEGRAL CALCULUS
There is one case, however, in which this reduction is
impossible, viz., when
m + np + 1 = 0,
for in that case A and B become infinite. [See Ex. 4, p. 235.]
In a similar manner the three following formulas may be
derived :
J x m (a + bx n ) p dx
a(m + l) J k -r j a (m + l)
I £ m (a + bx n ) p dx
: l I x m (a + bx n ) p x dx H *■—-*- f- • TC]
m + np + 1 •/ m + ?ip + 1
I x m (a-\-bx n ) p dx
: — ! — i — — I x m (a + bx n ) p +hlx i — ■ - TDl
an(j> + l) J awQj + 1)
The cases in which the above reductions are impossible are,
For formulas [A] and [C], when m + np + 1 = 0;
for formula [B] , when m + 1 = ;
for formula [D] , when _p + 1 = 0.
.. (x s Va
Ex.1. \ x 8 Va* - x* <lx.
If the monomial factor were x instead of x 3 , the integration could
easily be effected by using formula I. Since in the present case
m — 3, n = 2, formula [A], which diminishes in by n, will reduce the
above integral to one that can be directly integrated.
REDUCTION FORMULAS 233
Instead of substituting in [A], as might readily be done, it is best
to apply to particular problems the same mode of procedure that was
used in deriving the general formula. There are two advantages in
this. First, it makes the student independent of the formulas, and
second, when several reductions have to be made in the same problem,
the work is generally shorter. [See Ex. 4.]
Accordingly assume
f x\d 2 - xrf dx = A ( x(a 2 - x 2 y dx + Bx 2 (a 2 - x 2 )\
the values of X and ft having been determined by the previously given
rule.
Differentiate, and divide the resulting equation by x(a 2 — x 2 y.
This gives .
6 x 2 = A + B(2a 2 -5x 2 ),
from which, on equating coefficients of like powers of x,
o 5
hence,
(x 3 ^d 2 -x 2 dx = —( (a 2 ~ x 2 ) * xdx-\ x\a 2 - z 2 )*
= - ^(2 a 2 + 3x 2 ) (a*-xrf.
:. 2. (Vx 2
Ex. 2. \ Vx 2 -2x -6 dx.
By following the suggestions of Art. 118, this integral can be re-
duced to the form
C Vz 2 - 4 dz,
in which z = x — 1.
Assume
jV - 4)Klz = A§(z 2 - 4)~^ dz + Bz(z 2 - 4)i
In determining A notice that m = in both integrals, so that
X = + 1 = 1. Also, fji = - £ + ! = £.
234 INTEGRAL CALCULUS
Ex.3. ( V2 ax - x l dx.
The mode of procedure of Ex. 2 may be followed. Another method
can also be used, as follows.
On writing in the form
fa* (2 a -xy dx,
and observing that the integration of
jV* (2 <z -aO"~*tf*= J
V2 ax - x 2
can be performed (see Ex. 10, p. 222), it will be seen that integration
may be effected in the present case by reducing each of the exponents
m and p by unity. This is possible since n = 1 and m can accordingly
be diminished by 1. Hence assume
(V (2 a- xy dx = A ' jV * (2 a- x)* dx + B'x? (2 a- x)% .
The exponent of the binomial in the new integral may be reduced
in turn by assuming
f aT* (2 a- a?)* dx = A" ( x~? (2 a- x)~^ dx + B" x?(2a-x)K
When this expression is substituted for the integral in the second
member of the preceding equation, the result takes the form
( V2 ax - x 2 dx = A ( dx + Bx*(2 a - a:)* + Cx?(2 a - xf,
J . a/2 ax - x 1
in which A, B, C are written for brevity in the place of A' A", A'B",
B' respectively. The values of A, B, C are calculated in the usual
manner by differentiating, simplifying, and equating coefficients of
like powers of x.
The method just given requires two reductions, and hence is less
suitable than that employed in Ex. 2, which requires but one reduction.
The rule for determining the values of A. and /x may now be
advantageously abbreviated. Let m,p be the exponents of the
REDUCTION FORMULAS 235
two factors in the given integral, and m', p' the corresponding
exponents in the new integral. Of these two pairs, m, p and
m', p', one of the numbers in the one pair is less than the cor-
responding number in the other pair. This fact will be ex-
pressed briefly by saying that the one pair is less than the
other pair. With this understanding the preceding rule may
be expressed as follows ;
Select the less of the two pairs of exponents m, p and m', p'.
Increase each number in the pair selected by unity. Tliis gives
the pair of exponents A., /jl.
Ex.4, f * dx .
(x 2 + a 2 )*
Assume successively
( x\x 2 + a 2 )~~ 2 dx = A' (x\x 2 + a 2 )~2 dx + B'x\x 2 + a 2 )"^,
(x 4 (x 2 + a 2 )~* dx = A" (x 2 (x 2 + a 2 )~* dx + B"x*(x* + a 2 )*,
(x 2 (x 2 + a 2 )"* dx = A'"1j(x 2 + a 2 ) - ^ tte + i?'".r(.r 2 + a 2 )i
These equations may be combined into the single formula
( x\x 2 + a*)~*dx = A fj (x 2 + a 2 )~^dx + Bx(x 2 + a 2 y
+ Cx*(x 2 + a 2 )* + Dx*(x 2 + a 2 )""i
The values of the coefficients are found to be
A=-$a*, B=-, C = -—. D = —-
2 a 2 a 2
Hence
( x\x 2 + a*)~* dx = ** +%<** * _ 3 a -2 log ( X + VX 2 + a 2).
J 2Vx 2 •+- a' 2
236 INTEGRAL CALCULUS
In this example three reductions were necessary; first, a reduction
of type [X)], second, and third, a reduction of type [-4]. Can these
reductions be taken in any order?
The different possible arrangements of the order in which these
three reductions might succeed each other are
(1) UL Ul [2>] ; (2) 01], [Z>], [A] ; (3) [Z>], Ul [A],
of which number (3) was chosen in the solution of the problem. Of
the other two arrangements, (2) can be used, but (1) cannot. For,
after first applying \_A.~] (which would be done in either case), the new
integral is „ _ 3
\ x 2 (a 2 + x 2 ) J dx.
If \_A~\ were now applied it would be necessary to assume
f x\a 2 + x 2 )~i dx = A ( (a 2 + x 2 )~i + Bx(a 2 + x 2 )~\.
This equation, when differentiated and simplified, becomes
x 2 = A + Ba 2 ,
a relation which it is clearly impossible to reduce to an identity by
equating coefficients of like powers of x, since there is no x 2 term in
the right member to correspond with the one in the left member. It
will be observed that this is the exceptional case mentioned on page
232, in which m + np + 1 = 0.
EXERCISES
1. f (a 2 - x 2 y 2 dx. 5. ( Va 2 -
2. f— ^ 6. f_
J(> 2 + 4)2 J x *
x 2 dx.
dx
3. f ** • 7 C—**
(x 2 +a)
. C x 2 dx /» 3
REDUCTION FORMULAS 237
9. (Vx^T^dx. 12. ( dx •
10. f a;v2ox - x*dx. 13 f ^f
J ' J (y+ + i x + .3)3
r y/'^nx # 2 , /•
11. J -^— dx. i4. j VI - 2 x - x 2 dx.
15. Show that
f dx _ 1 f a? (9 _o\ C dx _~|
J 2 + c)» ~ 2 c(> - 1) L (a; 2 + c)»-i x K " U ' } J (x 2 + c)«-U '
16. f_^ 19. f
17 "^
sin0<70
(1 + e sin 2 0)t
[Substitute cos = z.]
C xdx
J (* 2 +7) 2 ' 20. jWa
ia J : — Hh^" 21. f (« 2 - x 2 )'2
2 - s 2 rfa:.
dx.
CHAPTER III
INTEGRATION OF RATIONAL FRACTIONS
121. Decomposition of rational fractions. The object of the
present chapter is to show how to integrate fractions of the
form <f>(x)
wherein <f>(x) and if/(x) are polynomials in x.
The desired result is accomplished by the method of sepa-
rating the given fraction into a sum of terms of a simpler
kind, and integrating term by term.
If the degree of the numerator is equal to or greater than
the degree of the denominator, the indicated division can be
carried out until a remainder is obtained which is of lower
degree than the denominator. Hence the fraction can be re-
duced to the form
iM =aX n + bx n-l + ... +./M,
xP{x) x];(x)
in which the degree of f(x) is less than that of if/(x).
As to the remainder fraction ^ > ' . it is to be remarked in
the first place that the methods of the preceding articles are
sufficient to effect the integration of such simple fractions as
A A' _. Mx + n M'x+N ' . P^±_Q _ (1)
x—a(x — a)' 2 ' ' x ,2 ±a 2 ' (# 2 ±a 2 ) 2 ' ' x 2 +mx+ri
Now the sum of several such fractions is a fraction of the
kind under consideration, viz., one whose numerator is of
238
INTEGRATION OF RATIONAL FRACTIONS 239
lower degree than its denominator. The question naturally
arises as to whether the converse is possible, that is : Can
f(x)
every fraction ^-^ be separated into a sum of fractions of as
simple types as those given in (1)?
The answer is, yes.
Since the sum of several fractions has for its denominator
the least common multiple of the several denominators, it fol-
lows that if -7^- can be separated into a sum of simpler frac-
tions, the denominators of these fractions must be divisors of
$(x). Now it is known from Algebra that every polynomial
xf/(x) having real coefficients (and only those having real coeffi-
cients are to be considered in what follows) is the product
of factors of either the first or the second degree, the coefficients
of each factor being real.
This fact naturally leads to the discussion of four different
cases.
I. When if/(x) can be separated into real factors of the
first degree, no two alike.
E.g., xf,(x) = (x-a) (x - b) (x - c).
II. When the real factors are all of the first degree, some
of which are repeated.
E.g., if; (x) = (x-a)(x- b) 2 (x - cf.
III. When some of the factors are necessarily of the sec-
ond degree, but no two such are alike.
E.g., + (x) = (x 2 + a 2 ) (x 2 + x + 1) (x - 6) (x - c) 2 .
IV. When second degree factors occur, some of which are
repeated.
E.g., t(x) = (x 2 + a 2 ) 2 (x 2 -x + l)(x-b). ■
240 INTEGRAL CALCULUS
122. Case I. Factors of the first degree, none repeated.
When ij/(x) is of the form
ijz (xj = {x — a) (x — b)(x — c) • • • (x — n),
f(x) A . B G . N
assume J -\J- = f- H h • • • H >
j/a(.t) a; — a x — b x — c £ — ft
in which .4, 5, C, •••, N are constants whose values are to be
determined by the condition that the sum of the terms in the
right-hand member shall be identical with the left-hand
member.
Ex. C*-** + *dz.
J x\ - 3 x + 2
Dividing numerator by denominator, we obtain
x 3 — 3 x" 2 -f a: a:
.r-
a; + 2 x 2 - 3 ar + 2
Assume — — ■ = — - — - -f
(x-l)(a:-2) x-1 x - 2
By clearing of fractions, we have
(1) x = A(x -2) +B(x- 1).
In order that the two members of this equation may be identical
it is necessary that the coefficients of like powers of x be the same in
each.
Hence 1 = ^+5, = - 2 A - B,
from which A = - 1, B = 2.
Accordingly the given integral becomes
$( x+ ^--^dz = ^+]og(x-l)-21og(x-2)
Alog *-* .
2 & (a; - 2) 2
INTEGRATION OF RATIONAL FRACTIONS 241
A shorter method of calculating the coefficients can be used.
Since equation (1) is an identity, it is true for all values of x. By
giving x the value 1 the equation reduces to 1 = A(— 1), or A = — 1.
Again, assume x = 2. Whence 2 = B.
EXERCISES
1 r dx . r (x 2 - ab) dx
J x 2 - a 2 ' " J (x - a) (x - b) '
2. f 1 - 3 * ^. 5. r__^f
J x s - x J x 2 - 4 x + 1
3 r (x*-V2)dx 6 f_(x 2 - l)dx
J x 2 + 4 x + 3 " ' J (x*
? f x 2 -2cx+flc- q& + &
J (x — a)(x — &)(x — c)
8. fx 2 (x + a)- 1 ^*^) -1 ^-
9 r (3s + l)tfar 12 f__^£_
J2x 2 + 3x-2' " J x 2 + 7 a; + 12
1Q f (x 2 + qft)<7x 13 f tf*
Jx(x -a)(x+ &)' ' J a 2 x 2 - i 2 '
4)(4x 2 - 1)
Jx.
11 f (' + *)*« . 14. f_!
J 2 x - x 2 - x 3 J 1
sec 2 x e/x
tan* 2 x
[Put tan x = t.~\
123. Case II. Factors of the first degree, some repeated.
Fx r (ox 2 -3x+l)^x
J x(x - l) 3
Assume
(1) 5 x 2 - 3 x + 1 ^ ,1 [ B , C , D
x(x-l) 3 x x-1 (x-1) 2 (x-1) 3
To justify this assumption, observe that:
(a) In adding the fractions in the right-hand member, the least
common multiple of the denominators will be x(x — l) 3 , which is
identical with the denominator in the left-hand member.
el. calc — 16
242 INTEGRAL CALCULUS
(6) Further, the expressions x, x — 1, (x — l) 2 , (a; — l) 3 are the
only ones which can be assumed as denominators of the partial
fractions, since these are the only divisors of x (x — l) 3 consisting of
powers of a prime factor.
(c) When equation (1) is cleared of fractions, and the coefficients
of like powers of x in both members are equated, four equations are
obtained, exactly the right number from which to determine the four
unknown constants A, B, C, D.
Instead of the method just indicated in (e) for calculating the
coefficients, a more rapid process would be as follows.
By clearing of fractions, the identity (1) may be written
5 x 2 - 3 x + 1= A(x - l) 3 + Bx(x - l) 2 + Cx(x - 1) + Dx.
Putting x = 1 gives at once 3 = D.
Substitute for D the value just found, and transpose the corre-
sponding term. This gives
5a;2_ 6z + 1 = A(x-1)* + Bx(x- 1)-+ Cx (x - 1).
It can be seen by inspection that the right-hand member of the
result is divisible by x — 1. As this relation is an identity, it follows
that the left-hand member is also divisible by x — 1. When this
factor is removed from both members, the equation reduces to
5 x - 1 = A (x - l) 2 + Bx (x - 1) f Cx.
Now put x = 1. Then C = 4.
Substitute the value found for C, transpose, and divide by x — 1.
The result is 1 = A (x - 1) + Bx.
By giving x the values and 1 in succession, we find that
A =-1, 5=1.
Accordingly, we have
r (5s»-3ar+l)tf* = (7 1 + _L_ + 4 » ) dx
J X(X-1Y J\ XX- (x-1)* (X-
= log ;
x(x-iy J\ x x-\ o-i) 2 (x-iy
x - l Sx - 5
2(x - l) 2 '
INTEGRATION OF RATIONAL FRACTIONS 243
EXERCISES
C dx 4 f (V2r+l)</g
" y{x-m* + v ' J^ (x+ V2)2
Jx 3 (x-1) J 38(3 + 1)3
T xr/x 6 C 2(.r 3 + ft 2 *)*/*
J (x 2 - a 2 ) 2 ' ' J a: 4 - 2 a 2 x 2 f a 4 '
7 f V2^
f —
J (2 +
(2 + V2 - V2 x) 8
8 r ax* + n' 2 x 2 + (a + \)x+a dx
J x 2 (a + x)
9 r(.r 3 - \)dx ±1 C (x 2 -Ux+2G)dx
J x 3 +3x 2 '* 'J
10. f (ax 2 + fa 8 )- 1 rfx.
[Substitute .r — 3 = 2.]
12. r x ' 2<lx
, T .r 2 r/x
J (x-aY
(-
[Substitute a; — a = 2.]
124. Case III. Occurrence of quadratic factors, none repeated.
Fx ± C (4x 2 + ox + ±)dx
' J (x' 2 + l)(x 2 + 2x + 2)'
Assume
.jn 4 x 2 + 5 a: + 4 _ J x + i? Cx + />
O 2 + l)(x 2 + 2 x + 2) x 2 + 1 x 2 + 2 3 + 2
Then
(2) 4x 2 + 5x+4=(^x + £)(z 2 + 2a; + 2) + (Ca; + Z))(x 2 + 1).
By equating coefficients of like powers of x
= A + C, 5 = 2 ,4 + 2 5 + C,
4 = 2/1+5 + A 4 = 2/3 + Z),
from which 4 = 1, 5 = 2, C - - 1, Z> = 0.
Hence the given integral becomes
C(x + '2)r/x C X'lx „. , 1/ , ix, 11 * 2 +l
1 —5—1 J ~ — ; ; = 2 tan- 1 ^ + tan" 1 ^ + l)+£log- — -^ — -.
J x 2 +\ Jx 2 +2x + 2 x 2 + 2.r + 2
244 INTEGRAL CALCULUS
To make clear the reasons for the assumption which was made con-
cerning the form of equation (1), observe that since the factors of the
denominator in the left member are x 2 + 1 and x 2 + 2 x -f 2, these
must necessarily be the denominators in the right member. Also,
since the numerator of the given fraction is of lower degree than its
denominator, the numerator of each partial fraction must be of lower
degree than its denominator. As the latter is of the second degree in
each case, the most general form for a numerator fulfilling this re-
quirement (i.e., to be of lower degree than its denominator) is an ex-
pression of the first degree such as Ax + B, or Cx + D.
Notice, besides, that in equating the coefficients of like powers of x
in opposite members of equation (2), four equations are. obtained
which exactly suffice to determine the four unknown coefficients
A,B, C,D.
dx
Ex.2. (
J (x 2 +
2 +l)0 2 +2)
We can assume in this case - —
(«+!)(« + 2)
(.r 2 + l)(a; 2 + 2) x 2 + 1 x 2 +
;ion x 2 = t, the
, to which Case I is applicable.
For if we make the substitution x 2 = t, the given fraction becomes
1
EXERCISES
± C 4dx 5 C (ix-6)dx
J x 8 + 4 x J x 4 + 2 x' 2
„ C xdx 6 C x dx
J (x + \){x 2 +1)' J x* + x 2 + 1*
3 f fix 7 C xdx
Jx 3 + a 8 ' ' J (x - a)\x 2 + a 2 )'
4 C (a 2 -b 2 )dx 8 C (x* + 2x + 2)dx
J (x 2 + a 2 )(x 2 + b 2 )' ' J (x - l)(x 2 + 2x + 2)
9. f 2d *
J (x-lX^ + 1)
INTEGRATION OF RATIONAL FRACTIONS 245
125. Case IV. Occurrence of quadratic factors, some repeated.
This case bears the same relation to Case III that Case II
bears to Case I, and an exactly analogous mode of procedure is
to be followed.
Ex. f 2 * 5 -* 4 * 8 * 8 + *,/*.
J O 2 + 2) 3
Assume
2x 5 -x 4 + 8x* + i = Ax + B Cx + D Ex + F
(x' 2 + 2)* " x 2 + 2 (x a + 2) a (^ + 2) 3 '
Whence, by clearing of fractions,
2x*-x* + 8xS+4=(Ax+B)(x 2 + 2) 2 +(Cx + D)(x 2 + 2)+Ex + F. (1)
Instead of equating coefficients of like powers of x, as might be
done, we may calculate the values of A, B y C, ••• by the following
briefer method.
Substitute for x 2 the value — 2, or, what is the same thing, let
x = V— 2. This causes all the terms of the right member to drop
out except the last two, and equation (1) reduces to
• - 8V^2 = EV^2 + F.
By equating real and imaginary terms in both members, we obtain
- 8 = E, = F.
Substitute the values found for E and Fin (1), and transpose the
corresponding terms. Both members will then contain the factor
x 2 -f 2. On striking this out the equation reduces to
2x* -x 2 + ±x + 2 = (Ax + B)(x 2 + 2) + Cx + D.
Proceed as before by putting x 2 = — 2. Whence
4 = CV^2 + D,
and therefore = C, 4 = D.
246 INTEGRAL CALCULUS
Substitute these values, transpose, and divide by x 2 -f 2. This gives
2x -1 -Ax + B,
whence A = 2, i? = - 1.
The given integral accordingly reduces to
J x 2 + 2 J (x- 2 + 2) 2 ^ (** + 2) s
The first term becomes
J x 2 + 2 ^ z 2 + 2 yo v 2
The second, integrated by the method of reduction (Chap. II),
8" ives x 1 . . x
tan -1
& + 2 V2 V2
Finally, by using formula I the last term is integrated immediately.
Hence
f 2 x b - x A + 8 x s + 4 , , / o , \ , a:
J (x l -4- 2 V ir- 2 -4- 2
(z 2 + 2) 3 ° v x 2 + 2 (x- 2 + 2)2
EXERCISES
+ 1/ •> a: 2 (ar 2 +l) s
J V^ 2 + 1/ J x 2 (x 2 + l) 2
2. f^ + ^' + ^fe. 5. f^ + ^'^rft
J (x 2 -ha 2 ) 2 J (x 2 + a 2 ) 2
3 C 2xdx 6 C x 6 dx
J (1+ x)(l+x 2 ) 2 ' * J(l + x 2 ) 8 "
[Ex. 6 can also be integrated, and more easily, by means of the
substitution 1 + x 2 = £.]
The principles used in the preceding cases in the assump-
tion of the partial fractions may be summed up as follows :
Each of the denominators of the partial fractions contains one
and only one of the prime fac'ors of the given denominator.
INTEGRATION OF RATIONAL FRACTIONS 247
When a prime factor occurs to the nth power in the denominator
of the given fraction, all of its different powers from the Jirst to
the nth must be used as denominators of the partial fractions.
TJie numerator of each of the assumed fractions is of degree
one lower than the degree of the prime factor ichose power occurs
in the corresponding denominator.
126. General theorem. Since every rational fraction can be
integrated by first separating it, if necessary, into simpler frac-
tions in accordance with some one of the cases considered
above, the important conclusion is at once deducible :
Hie integral of every rational algebraic fraction is expressible
in terms of algebraic, logarithmic, and inverse-trigonometric
functions.
CHAPTER IV
INTEGRATION BY RATIONALIZATION
At the end of the preceding chapter it was remarked that
every rational algebraic function can be integrated. The
question as to the possibility of integrating irrational func-
tions has next to be considered. This has already been
touched upon in Chapter II, where a certain type of irrational
functions was treated by the method of reduction.
In the present chapter it is proposed to consider the sim-
plest cases of irrational functions, viz., those containing
■y/ax -f- b and yW 2 -f bx -4- c, and to show how, by a process
of rationalization, every such function can be integrated.
127. Integration of functions containing the irrationality
•y/anc + b. When the integrand contains -y/ax + b, that is,
the nth root of an expression of the first degree in x, but no
other irrationality, it can be reduced to a rational form by
means of the substitution
■y/aoc -\-b = z.
Ex. 1.
Ki
dx
x + 3 -
- 1
Assume
V2 x + 13 = z,
that is,
2 x + 3 = z 2 .
Then
dx = z <h,
and
f
dx _ C zdz
z+ log(z-l)
V2 x + 3 - 1 sz-i
= V2 .rT~3 + log (V2T+3 - 1).
248
INTEGRATION BY RATIONALIZATION 249
j-l + «* r **-
1 + X* — X 1 — VI
x z + a:
It would appear at first sight that this integrand contains several
irrationalities, viz., Vx, Vx, Vx. It is readily seen, however, that
they are all powers of Vx, and hence the substitution Vx = z will
rationalize the expression to be integrated.
EXERCISES
4.
5.
f
dx
K
Vx + 1
S;
fix
/x+ Vx
f
dx
dx
(x - l)Vx
dx
(x — a — b 2 ) Vx — a
5-
2 Vx - 1 -f x x 7 + x
i l+^l dx .
When two irrationalities of the form Vaa 4- b, -Vex -f d
occur in the integrand, the first radical can be made to dis-
appear by the substitution
Vax -\-b = z.
The second radical then reduces to
V
C a (z*-b) + d,
and the method of the next article can be applied.
128. Integration of expressions containing Vaoc 2 +bx + c.
Every expression containing Va.T 2 + bx + c, but no other
irrationality, can be rationalized by a proper substitution.
Two cases are distinguished.
(a) When ax 2 -f bx + c has real factors. We may then write
the quadratic expression in the factored form
ax 2 -f bx + c = a (x — a) (x — /?), (1)
250 INTEGRAL CALCULUS
in which a and ft are real. Introduce a new variable t by
means of the formula
VaX 2 + bx + c = t(x — a) . (A)
Square both members of this equation and replace the left
member by means of (1). This gives
a(x-a)(x-(3) = t 2 (x-a) 2 .
On canceling x — a and solving for x we obtain as the equa-
tion of transformation .9
x= «tr-aP m (2)
Hence x (and therefore dx) is rationally expressible in terms
of t, while the radical reduces to
[ at 2 -a(S 1
_ t 2 -a
at(a — /?)
f-a
(3)
which is also rational in t. The substitution of these expres-
sions in the proposed integrand gives a rational fraction which
may be treated by the methods of the preceding chapter.
(b) When a, the coefficient of x 2 , is positive.
Make the substitution
Vase 2 + bx + c = Va- x + t. (B)
By squaring both members and solving for x we obtain
b - 2Vat
while the radical is expressible in the form
■Vat 2 — bt-\-^fac /K x
-p > \»)
2Vat-b
INTEGRATION BY RATIONALIZATION 251
and hence the integrand becomes rational when expressed in
terms of t.
The only case that is not included in (a) or (6) is that in
which the factors of ax 2 + bx + c are imaginary and the coeffi-
cient a is negative ; the radical is then imaginary for all values
of x. Although the integral can be obtained (in an imaginary
form) by either of the preceding substitutions, this case does
not arise in practical applications of the calculus and will not
be considered further.
Ex. 1. f dx
J x + vV-2 + 2 x - 1
Formula (B) gives
Vx 2 + 2x-l = x + t,
whence, by solving for x, we obtain
,_ ' 2 +l
2(1 -
_ /2 i 9 j J. J
and accordingly dx = — — — - dt,
h y 2(1 - 2
^ + Dx-U - fi + 2< + 1
2(1 -
When these expressions are substituted in the above integral it
reduces to
r (- t * + 2t+l)dt
2(1 + 2
The work of integrating may be facilitated by means of the trans-
formation 1 -f t = 2. The result, in terms of x, is
h(x - VV 2 + 2x - 1) +
1 - x + Vx 2 + 2 x - 1
+ 2 log (1 - x + Vx 2 + 2 x - 1 ).
252 INTEGRAL CALCULUS
Ex.2, r Vl +*</*_.
^ (1 - *) VT - X
By rationalizing either numerator or denominator we obtain
Vl — x 2 as the radical part of the integrand.
Formula (A) gives Vl — x' 2 = 7(1 — x),
whence JI±* = *, (1)
'1 — x
or J- ± ^ = <», (2)
1 — x
and hence, by differentiation,
2dx
(\- x y
= 2tdt. (3)
Add 1 to both members of (2) and combine the two terms of the left
member. The result is
(4)
o
1 -X
= t 2 +l.
1)
ividing (3) by (4),
we
have
rfx
_2tdt
Now multiply (1) and (5) together and integrate. We obtain
f /l + x dx r 2 t 2 dt
J^l -x ' 1 -x~ J t' 2 + 1
= 2 jrr^_ 2tan -iJr±Z
'1 — a: '1 — a:
-J
EXERCISES
(1 -x)(l -Vl -x' 2 )
dx
r dx
J V2 x 2 - 3 x + 1 [V2x a -3x-+ 1 + V2(a; - 1)]
INTEGRATION BY RATIONALIZATION 253
We can rationalize also by means of a trigonometric substi-
tution. First reduce ax 2 -\-bx + c to the form ±t 2 ±k 2 , as in
Art. 118, and then make one of the following transformations:
In k 2 — t 2 put t = k sin 0,
in t 2 — k 2 put t = k sec 0,
in t 2 + k 2 put t = k tan 0.
Since V — t 2 — 1& is imaginary, we shall exclude this case from
consideration.
The resulting trigonometric functions can then be integrated
by methods to be explained in the next chapter.
129. There is one case in which a different transformation
leads more rapidly to the desired result. If, after reducing
the terms under the radical sign to one of the simple forms
mentioned in the preceding paragraph, the integrand can be
expressed as the product of t dt and a function containing only
even powers of t, then we may substitute
y/ ± t 2 ±k 2 = Z.
For this gives t 2 = ± (z 2 ± k 2 )
and tdt = ±zdz,
and hence the integral takes a rational form in z.
EXERCISES ON CHAPTER IV
1 f (— ^ s + 4rx)dx
[Notice that Art. 129 is applicable.]
2. f t(x-a)%-\-\dx 3 C y/J+ldx
Jo(x-a)$-(x-a)i' V7+1 + 2
254 INTEGRAL CALCULUS
4. f dx 8. f ,lx
J x + Vx-1 J * + Vx 2 - 1
5- j-^4- 9. fi+^tf*
*^ (« + *)* J 1 + %x
6. f(2- 3 *-*)<**. 10 . f_
x 3 dx
x — 3 x 6 + 5 x* v a' 2 — x 2
7 f dx -,, ( y/a*-x*dx
[Use trigonometric substitutions in the following exercises.]
</x
12
13.
14.
*" x 2 "" (x 2 + a 2 )s
f ^ . 17. f rf * .
J (x' 2 +a 2 )^ J (a?-x*)%
CHAPTER V
INTEGRATION OF TRIGONOMETRIC FUNCTIONS
130. In regard to the integration of trigonometric functions,
it is to be remarked in the first place that every rational trigo-
nometric function can be rationally expressed in terms of sine
and cosine.
It is accordingly evident that such functions can be inte-
grated by means of the substitution
sin x — z.
After the substitution has been effected, the integrand may
involve the irrationality
Vl — 2T [ = COS X ].
This can be removed by rationalization, as explained in the
preceding chapter, or the metnod of reduction may be employed.
The substitution cos x — z will serve equally well.
It is usually easier, however, to integrate the trigonometric
forms without any such previous transformation to algebraic
functions. The following articles treat of the cases of most
frequent occurrence.
131 . Csec 2n x dx, f csc 2w x dx.
In this case n is supposed to be a positive integer.
If sec 2n # dx is written in the form
sec 2n ~-.r • setfxdx = (1 + tan 2 x) n ~ l d (tan a?),
255
256 . INTEGRAL CALCULUS
the first integral becomes
f (tan 2 cc + l) n - 1 d(tan x).
If (tan 2 a? + 1)" -1 is expanded by the binomial formula and
integrated term by term, the required result is readily
obtained.
In like manner,
I csc 2n x dx = f csc 2 "~ 2 # «csc 2 a; dx
= — f(cot 2 ^ + l) n ~V/(cot x).
This last form can be integrated, as in the preceding case,
by expanding the binomial in the integrand.
The same method will evidently apply to integrals of the
form
I tan m .T sec 2n # dx, I cot TO # csc 2h x dx,
in which m is any number.
EXERCISES
1 f ( Jx 5 f (1 -cosxydx
J cos 4 z * sin 4 .*:
r 6 C tjx
2. J CSC 4 X dx. ' J sin 4 x . CQS 4 X ( CQS 4 X _ S j u V)4
3. \sec e xdx. 7. i - [ = 1 tan~ 8 .r sec 4 xdx'] .
J J sin 3 x cos x J
4 C ^ x q r cos 2 ar dx
J ain^x cos 8 * J siii°x
INTEGRATION OF TRIGONOMETRIC FUNCTIONS 257
132. fsec w x tan 2n+1 sc dx, fcsc m ^ cot 2n+1 sc da.
In these integrands n is a positive integer, or zero, so that
2)i + l is any positive odd integer, while m is unrestricted.
The first integral may be written in the form
I sec m-1 aj tan 2n a; • sec x tan x dx
= I sec m_1 a;(sec 2 # — l)'V?(sec x),
which can be integrated after expanding (sec 2 .c — l) n by the
binomial formula.
Similarly,
I csc TO # cot 2n+1 # dx = I csc m_1 # cot 2n # • esc x cot x dx
= — I csc m_1 x(csc 2 a; — l) n c?(csc x).
EXERCISES
1. \ sec 2 * tan 3 x dx. 5. I tan 5 * dx.
2. f csc 8 * cot 5 * dx. 6. f sm3 - r dx [ = f sec*" 8 * tan 8 * tfx] .
•J J cos"* J
f sec a* , r
Jcot 5 a* 7/ J tan a:
dx.
4. l : in x cot 3 * (/*. 8 i cot x dx.
EL. CALC. 17
258 INTEGRAL CALCULUS
133. CtsiVL n acdx, Ccot n ocdx.
The first integral can be treated thus :
I tsin n x dx = M tan n ~ 2 • tan 2 # dx
= f tan n_2 ^(sec 2 ic — l)dx
^""^ 'Un-'xdx.
When n is a positive integer, the work of integration may-
be rapidly carried out by writing t for brevity in place of tan x
and then putting t n dx in a different form by means of the
following process. First, divide t n by £ 2 + 1; the quotient is a
polynomial of the form t n ~ 2 — t n ~~ A + t n ~ G — •••, while the re-
mainder R is either ± 1 or ± t according as n is even or odd.
Then, since the dividend equals the product of divisor and
quotient plus the remainder, we have
«•=(«•-" - r- 4 + r~ 6 )(t 2 + 1) + R.
But since (tan 2 # + l)dx = sec 2 x dx = d(tan x) = dt,
we have
Ctsm n x dx = C(t n ~ 2 - r~* + * n ~ 6 )^ + f ^ da; -
For example,
f tan 8 # dx = C{f -t* + t 2 -- l)dt + Cdx,
and J tan 7 x dx = \ (f — f + t) dt — (tan x dx.
INTEGRATION OF TRIGONOMETRIC FUNCTIONS 259
The integral J cot n # dx can be treated in a similar manner,
in case n is a positive integer.
For any value of n we have
j cot n # dx = I cot n-2 # aot 2 xdx
= I cot n_2 #(csc 2 # — l)dx
= _ COt- 1 .T _ C cot n-2 xdXt
n-1 J
Since tan x and cot x are reciprocals of each other, the above
method is sufficient to integrate any integral power of tan x or
cot x.
Another method of procedure would be to make the substi-
tution tan x = z, whence
j tan n xdx= j — -
(h
If the exponent n is a fraction, say n = -, the last integral
can be rationalized by the substitution z = u q .
It is evident from this that any rational power of tangent
or cotangent can be integrated.
EXERCISES
1. j cot*xdx. 3. ( (tan x - cotx) 3 dx.
2. \ tau s axdx. 4. ( (tan n x + ta,n n - 2 x)dx.
5. j tan 8 x dx.
260 INTEGRAL CALCULUS
When n is a positive integer show that
^ d. 9n i tan 2n_1 x tan 2 " -3 a; , , , n „ 1A x
6. 1 tan 2 " x dx = h ••• 4- (— l) n-1 (tan x — x).
J 2n- 1 2n-3 V J K '
r> Ct. 9„a.i j tan 2n x tan 2 " -2 a:
7. \ tan 2 " +1 xdx =
+ ••• + (— l) n_1 (J tan 2 a; + log cos a;).
134. fsiu m x cos** x dx.
(a) Either m or n a positive odd integer.
If one of the exponents, for example m, is a positive odd
integer, the given integral may be written
j sin" 1-1 x cos n x sin xdx = — I (1 — cos 2 x) 2 cos n x d (cos x).
Since m is odd, m — 1 is even, and therefore — - — is a
Z
positive integer. Hence the binomial can be expanded into
a finite number of terms, and thus the integration can be
easily completed.
E
x. ( sin 5 xVcos xdx.
According to the method just indicated this integral can be re-
duced to
— ( sin 4 xVcos x d(cos x) = — \ (1 — cos 2 a;) 2 (cos a;) 2 c?(cos x)
= — | cos 2 X + \ COS * x — T 2 T cos *~ X.
EXERCISES
1. (sin* xdx. 3. (-
2. j*sin3a:cos 4 a:rfx. 4> J cos 2 x j/^
_ C sin 8 x dx
J VI - cos a;
'cos 5 X
sin x
dx.
' sin f
' x dx
INTEGRATION OF TRIGONOMETRIC FUNCTIONS 261
(b) m + n an even negative integer.
In this case the integral may be put in the form
J^H^. cos TO+n x dx = f tan m a; sec- (m+n) x dx,
COS m X J
which can be integrated by Art. 131, since the exponent
— (in -+- n) of sec x is an even positive integer.
fcx. \ dx.
COS^ X
The integration is effected in the following steps :
C y/s\\\xdx Ct- h 4 /
J — ^= = \ tan 2 x sec 4 x dx
Vcos x cos 4 x J
= \ tan^ x(tan 2 x + 1) d (tan x)
= 2 tan^2(! + \ tan 2 a;).
EXERCISES
dx
1. (^dx 4. f_
J sin 4 ar ./ sm 4 x cos- 6 x
2. r^_. 5. f
J sin 6 a: *
Vsin 3 x cos 5 x
3. f ^*. 6. p""^**,.
J sin 8 x J cos n+2 x
(c) Multiple angles.
When m and ?i are both even positive integers, integration
may be effected by the use of multiple angles. The trigono-
metric formulas used for this purpose are
1 — cos 2 x
sin- x =
cos-a;
sin x cos x
2
1 4- cos 2 x
2
sin 2 #
262 INTEGRAL CALCULUS
Ex. ( sin 2 x cos 4 x dx.
\ sin 2 x cos 4 x dx = I (sin a: cos x) 2 cos 2 x dx
sin 2 2x1 + cos 2 x
-J
= - f sin 2 2 x rfx + — f sin 2 2 x cos 2 x rf(2 x)
111 -
=§J- - dx+
1 f 1 — cos 4 x , ,1 sin 3 2 x
dx -\
2 16 a
= j & x - e¥ sin 4 x + 3 X 8 - sin 8 2 x.
EXERCISES
1. ( cos 2 x sin 2 x </x. 3. j sin 4 x cos 4 x tfx.
2. ( sin 2 x cos 6 x g?x. 4. I (sin 4 x — cos 4 x) 4 </x.
5 C^lldx= f C 1 -cos 2 x) 2 ^ = f(sec 2 x-2 + cos 2 xyx.
J cos 2 X J COS 2 x J
(d) Reduction formulas. Integrate J sin m # cos n x dx by parts,
taking u = cos* _1 a*, cfa = sin m # cos x dx,
sin m+1 a*
whence du = — (n — 1) cos n -a; sin x dx, v = ,
m+- 1
and therefore
frill** cos'a! dx = sin " + ' a;cos "'" 1 - c + '^4 f sin"* 2 cos"" 2 * dft
c/ m + 1 7/1 + 1./
In the last term replace sin 2 a- by 1 — cos 2 # and separate the
integral into the two terms
I sin m # cos n_2 x dx — I sin m a; cos n # dx.
INTEGRATION OF TRIGONOMETRIC FUNCTIONS 263
Transpose the second integral and unite with the similar
integral in the left member. After dividing the resulting
equation by we obtain the formula of reduction
m + 1
/„ j sin m+1 # cos M_1 x . n—1 C - m n-i i
sm m # cos w # ax = 1 I sm m # cos n z x dx
ra + ?i m + n J
by means of which the exponent of the cosine factor may be
diminished or increased by 2 according as the integral in the
left member or that in the right member is taken as the given
integral.
In like manner a reduction formula may be deduced which
decreases or increases the exponent of the sine factor by 2.
The details are left to the student as an exercise. The
result is
/- m n i sin w_1 x cos" +1 x . m—1 C - m-2 n ^
sm m x cos n x ax = 1 I sm m l x cos n # ax.
m 4- n m 4- nJ
The two preceding formulas, when solved for the integrals
in the right members, and m (or n) increased by 2, become
/• m n i sin m+1 # cos n+1 . / c . m-\-n + 2 C . m n+2 j
sm m aJcos n a;aa;= ! ! — I sm m xcos n +'xax,
n + 1 n + l J
Csin m x cos n xdx = smW+la? cos n+1 a; + m + n + 2 C sin m + 2 X G0S n xdx ,
J m + 1 m-j-1 J
Whenever the values of m and n are such that one of the
three preceding cases, (a), (6), (c), is applicable, the integration
can generally be performed more quickly by one of those
methods.
264 INTEGRAL CALCULUS
EXERCISES
1. (sin*xdx. 2. (99&*dx.
J J sin 2 x
[In Ex. 2 after one reduction, diminishing the exponent of cos x by
2, Art. 133 may be applied.]
3 r sin 4 :r , . f dx _ C co^xdx
J cos x J sin 3 2 a; J sin 4 a;
135.
/ eta r doc r
a+bnosna J a + bsmnw J a +
bcosncc' J a + bsinnw' J a + b sin noc+c cos nx
These forms can be integrated by expressing them in terms of
the half angle and then in terms of tan - -.
Ex.l. J
dx
5 + 4 cos x
By making use of the trigonometric relations
cos 2 - + sin 2 - = l,
2 2
cos x = cos 2 - — sin 2 - ,
2 2
the denominator may be written in the form
5^cos 2 |+sin 2 |)+ 4(cos 2 |-sin 2 -V
which becomes sin 2 - + 9 cos 2 - on collecting the terms ; whence
2 2
dx
1
sin 2 ! +9
,,x
cos 2 -
2
INTEGRATION OF TRIGONOMETRIC FUNCTIONS 265
Now divide numerator and denominator by cos 2 - and bear in mind
1 x
that = sec 2 -. This gives
„ t 2
cos 2 -
i'(S)
J
- = ? tan- 1 (i tan-
tan 2 - +9 3 Vd 2
Ex.2, J-
t/x
sin 3 a: + 1
Express the denominator in the form
4 sin — cos — + ( sin 2 — - + cos 2 — - J .
3 a:
Then, after dividing both terms of the fraction by cos 2 — , the given
integral becomes r^ 3 x
sec 2 — dx
2
J
tan 2 ^+4 tan— + 1
Now make the substitution tan— = t and apply Art. 118.
It will be observed from these two problems that the aim is
to put the denominator in the form of a homogeneous quadratic
expression in sine and cosine functions. Then, when both terms
of the fraction are divided by the square of the cosine, the
denominator becomes quadratic in the tangent function while
the numerator can be expressed as the differential of the tangent.
EXERCISES
1. ( ^ . 5. f ^ .
J 5 + 3 cos 2 x J (a sin x + b cos x) 2
2. (—«* . 6. f <"
J 5 — 3 sin x J a 2 sin 2 x + b 2 cos 2 x
dx „ C dx
t. ( — dx . 7. f—
J 1 - 2 sin 2 x J 1 +
t. f * 8. f—
J a sin x -f b cos x J 1 +
cos 2 a:
ax
sin x + 2 cos a:
266 INTEGRAL CALCULUS
136. I e ax sin nx dx, I e ax cos nx dx.
Integrate | e ax sin nx da; by parts, assuming
u = sin nx, and cfa = e ax dx.
This gives ^. *
/e oz sin nx c?x = - e ax sin nx | e ox cos nx dx. (1)
Integrate the same expression again, assuming this time
u = e ax , dv = sin nx dx.
Then
/e ax sin nxdx = e ax cos nx + - ( e ax cos nx dx. (2)
n nJ
Multiply (1) by - and (2) by - and add. The integrals in
the right members are eliminated, and the result is
J„ . 7 e ax (a sin nx — n cos nx)
e ax sin nx dx = — * ■ L •
a 2 + n 2
By subtracting (1) from (2), the formula
/, e ax (n sin nx 4- a cos nx)
e ax cos nx dx = — * — L
a~ + iv
is obtained.
EXERCISES ON CHAPTER V
1. Derive the reduction formula
C, ec n x dx = tanssec*-** + n-2 r ecn _ 2jc d ^
J n — 1 n — \J
[Integrate by parts, taking u = sec M ~ 2 :r, dv - sec 2 a: dx.~]
2. Derive
f csc»x & = - cot * csc ;~ 2 * + ^? f ce-'* <**
J W — 1 n — JV
J sin x cos a; J cos 5 x
INTEGRATION OF TRIGONOMETRIC FUNCTIONS 267
5. f **, 9. ( S ™^dx.
J cos x sin 2 x J e x
10. \ e 2x sin 2 x dx.
6 - y^-^dx. 10 . f
J cos 3 x J
7- I • 11. i <? x sin 2 a: sin a: efa.
J (1 -x)Vl - a.- 2 J
[Put a; = cos $] . [Hint. 2 sin 2 a: sin a: = cos x — cos 3 a\]
f - a:
8. i e 2 cos~dx.
12. Show that
f sin ax sin fta: tfa; = sin ( a " & ^ - sin + *>*.
J 2(a - ft) 2(a + ft)
Use the trigonometric formula
sin a sin ft = i [cos(cc — j3)— cos(ct + /?)].
13. Show that
r • 7 7 cos (a — ft)ar cosfa + b)x
\ sin aa: cos bx dx = * *- * — ^ ' •
J 2(a - ft) 2 (a + ft)
14. Show that
f cos a* cos bxdx = sin (g- ft) a; sin(q + ft)*
J 2(a-ft) 2(a + ft)
15. i sin n x cos 3 a; da:. 19. I— — ■ - :~~;
J J sin x cos 3 x — sin 3 x cos a:
16. j"^L_. 20. f^^-.
^ '.A 4? ^ sin 4 .r oos 4
sin^a: cos-* a:
sin* a- cos* x
17. j*(tan a: + cot a-) 6 dx. 21. j* 4 ^ <fo.
18. ( «« f a sin a: + ft co s a:
■J (1 + cos a:) 3 zz ' J a Bin x + o cos ~ dx -
ft COS X
[Hint. Assume
a sin x + 6 cos a: = .4 (a sin x + ft cos x) -f- i?(« cos x — ft sin a:)
and determine J. and Z? by equating like terms. Treat Ex. 23 in
like manner.]
23. ( aeX + he ~ X dx. 24. f sin (* + q > <fa.
^ «e* + fte~ x J sin (a; + ft)
CHAPTER VI
INTEGRATION AS A SUMMATION. AREAS
137. Areas. The problem of calculating the area bounded
by given straight or curved lines can be solved by means of
the Integral Calculus provided that the equations of the boun-
dary curves are known and satisfy certain restrictions.
Suppose it is required to determine the area limited by a
continuous arc of a curve whose equation, in rectangular coor-
dinates, is written
in the form
y =/(»), (i)
by the two ordinates
x — a and x = b, and
by the it*-axis; that
is, the area APQB
(Fig. 59).
We proceed as
follows. It is as-
sumed in the first
place, for the sake of simplicity, that/(#) is always increasing
(or always decreasing) between x = a and x = b, so that a vari-
able point on the arc PQ is continually rising (or falling) as
its abscissa x increases. Suppose, further, that every ordinate
between x = a and x = b cuts the arc PQ in but one point. Let
2G8
Fig. 59
INTEGRATION AS A SUMMATION. AREAS 269
the interval A to B (Fig. 59) be divided into n equal intervals
AA V A l A 2 '", A n _ 1 B, each of length Ax, so that
interval AB= b — a = n • Ax.
At each of the points of division A, A lf A 2 , • ••, B erect ordi-
nate^ and suppose that these meet the curve in the points
P, P x , P 2 , • ••, Q. Through the latter points draw lines PPi,
P 1 R 2 ) PzR& ' ' ' Pn-i Rn parallel to the x-axis.
A series of rectangles PA X , P X A>, • • • is thus formed, each of
which lies entirely within the given area. These will be re-
ferred to as the interior rectangles. By producing the lines
already drawn, a series of rectangles SA 1} SiA 2J ••• is formed
which will be called the exterior rectangles. It is clear that
the given area will always be greater than the sum of the in-
terior rectangles and always less than the sum of. the exterior,
or, expressed in a formula, "^
PA X + iVl 2 + ••• +P n - 1 B< Area APQB < SA 1 + S 1 A i + •••
■+S.-1-B. (2)
The difference between the sum of the exterior and the sum
of the interior rectangles is
SP 1 + ^fi 2 +- + £„_! R n = rectangle S n _ x T = TQ • Ax. (3)
As we suppose the curve to be continuous between P and Q,
the line TQ is of finite length.
If the number n of equal parts into which AB is divided is
increased, the first sum in (2) increases in value and the
second sum in (2) decreases. Moreover, as their difference
TQ • Ax, given in (3), approaches the limit zero, it follows
that the limit of the sum of the exterior rectangles is equal to the
limit of the sum of the interior rectangles when n = cc, that is,
when Ax = 0.
270 INTEGRAL CALCULUS
Since the required area always has a value intermediate
between the two sums, it follows that the area is equal to the
limit of either sum. So that, for example, we have
^^ = A^O t PA > + P * A * + - + P n~l B l ( 4 )
The second member of this equation may be expressed in
terms of the function f(x) which appears in the equation (1)
of the given curve. For,
area PA± = AP • Ax = f(a)Ax,
since AP is the ordinate y when x = a.
Similarly,
area PiA 2 = A X P X • Ax = f(a + Ax) • Ax,
area P 2 A 3 = A 2 P 2 • Ax = f(a + 2 Ax) • Ax,
area P n - Y B = A n _iP n _i • Ax =f(a + n — 1 Ax) • Ax.
If these expressions are substituted in (4), it takes the form
area = A^o[/( a )+/( a + A ' T ) + /( a + 2Aa; )+ -
+ f(a + n-lAx)-]Ax. (5)
As it now stands, the formula just derived is of little prac-
tical value for computing areas. This is due to the fact that
there is no general method for calculating the sum of the n
terms given in brackets in the second member of (5).
Fortunately, the value of the limit of this sum when n = cc
and Ax = can be calculated by integration as we shall now
proceed to show.
138. Expression of area as a definite integral. Denote the
function arising from the integration of f(x) by F(x), that is,
INTEGRATION AS A SUMMATION. AREAS 271
let F(x) = ff(x) dx,
By definition of the derivative of F(x) we have
lim F(x + Ax)-F(x) _ f( .
A
The quotient F ( x + Ax ) F ( x ) may be written in the form
f(x) -J- <f>, in which <f> approaches zero at the same time as Ax,
otherwise the limit of the quotient when Ax = could not be
f(x). From this relation follows, on multiplying by Ax,
F (x + Ax) - F (x) =f(x) • Ax + <£ • Ax. (6)
Next, in equation (6) substitute for x the successive values
a, a + Ax, a + 2 Ax, • • •, a + (n — 1) Ax.
We thus deduce the following series of n equations, in which
<£i> 02? ••* are used to denote the different values which <£ may
take: F(a + Ax)-F(a)=f(a) -Ax+fa-Ax,
F(a + 2 Ax) — F(a + Ax)=f(a + As) • Ax + <£ 2 • Ax,
F(a + 3 Aaj)-2P(a + 2 Ax)=f(a + 2 Ax) • Ax -J- <£ 3 ■ Ax,
F(a + n — l • Ax)-F(a + n-2 • Ax)=f(a + n- 2 ■ Ax)Ax
+ <£ n -i • Ax,
F(a + nAx)—F(a + n — 1 • Ax) = /(a -fn- 1 • Ax)Ax
4- <£* • Ax.
Let these n equations be added ; then all but two of the
terms in the left member of the sum cancel each other and the
272 INTEGRAL CALCULUS
result may be written
F(b)-F(a) = [f(a)+f(a+Ax) + ... +/( a + n-l • Ax)]&x
+ W>i + 02 + -+0 B ]Aa>,
in which b is written for a-\-n A#, since ?i A# = 6 — a.
Now let Aa; approach zero. The expression
(<fc + <fe-h ... +0 B )Aa>
vanishes at the limit. For, let <£ denote the positive value of
the numerically largest term of the set <f> u <f> 2 , • ••, cf> n ; then we
have evidently
|(0i + <f>2 H +0») Aa;|^ (^-|-<I> ...(w terms)) A# = n<£ • Aa;
= wAa; • <£ = (6 — a) • 3>.
Hence, from the fact that ^ m 3> = and that b — a is finite, it
Ax=0 '
follows from Art. 3 that
Hm(<k + 4> 2 + .••0 n )Aa J = O;
and therefore 2^(6) - F(a) = J^J[/(a) +/(a + Aa>) + ...
+f(a+n - 1 • Aa;)]Aa;. (7)
Now the right member of this equation is exactly the ex-
pression previously derived for the area APQB; hence,
area APQB = F(b) - F(a). (8)
To compute the value of the right member of (8), first obtain
F(x) by integrating f(x) dx. Having determined F(x), substi-
tute the values b and a which x takes at the extremities of the
arc bounding the given area and then subtract the second from
the first. This result may conveniently be represented by the
symbol
ff( x ) dx >
INTEGRATION AS A SUMMATION. AREAS
273
which indicates both the integration to be performed and the
substitution of the two limiting values a and b for x. It is
called the definite integral of the function f(x) between the limits
a and b.
We thus obtain, as a final formula for area,
area APQB = f / (») 'dx. (9)
139. Generalization of the area formula. Instead of taking the
limit of the sum of the interior (or exterior) rectangles, a more
Y
Fig. 60
general procedure would be to take a series of intermediate
rectangles. Let x 1 be any value of x between a and a + Ax, x 2
any value between a -{-Ax and a -+- 2 Ax, etc. Then / (x r ) Ax
would be the area of a rectangle KLA X A (Fig. 60) intermediate
between PA X and SA X ; that is,
PA.Kf^AxKSA,.
Likewise P X A 2 <f(x 2 )Ax < S^A 2 , etc.
EL. CAI.C
18
274 INTEGRAL CALCULUS
Hence,
sum of interior rectangles < [/(a?i) +/(^) + •••]^ aJ
< sum of exterior rectangles,
and therefore (cf . Fig. 59),
area APQB = ^o [/(&) + f(x 2 ) + ... + /«)] Ax. (10)
This result combined with (9) gives for the definite integral
the more general formula :
fj(x)dx=£l [f{x 1 )+f(x 2 )+ -f(x„)-]Ax. (11)
140. Certain properties of definite integrals. From the defini-
tion of the definite integral | f(x)dx as the limit of a par-
ticular sum, certain important properties may be deduced.
(a) Interchanging the limits a and b merely changes the sign
of the definite integral.
For, if x starts at the upper limit b and diminishes by the
addition of successive negative increments (— Ax), a change
of sign will occur in formula (7), giving
F(a)-F{b)=£f{x)dx.
Hence,
f a f(x) dx = - Cf{x) dx.
(b) If c is a number between a and b, then
f b f(x) dx = Cf(x) dx + Cf(x) dx.
INTEGRATION AS A SUMMATION. AREAS
275
(c) Tlie Mean Value Theorem.
The area APQB (Fig. 61), which represents the numerical
value of the definite integral may be expressed as follows.
Let an ordinate MN be drawn
in such a position that
area PSN = area NRQ.
If £ denotes the value of x cor-
responding to the point N, then
MN = f($), and
area APQB = rectangle ASRB
= MN-AB=f{Z){b-a).
Hence,
Cft*)** = f(gft> -a), (12)
%J a
in which £ is some value of x between a and b. This result
is known as the Mean Value Theorem (compare Art. 39),
dx
is called the mean ordinate
f/(*)<
and the ordinate /(£) =^ a -
b — a
between x = a and x = b. This is also called the mean value
of the function /(a?) between these limits.
The theorem may be expressed in words as follows :
The value of the definite integral
£f{x)dx
is equal to the product of the difference between the limits by the
value of the function f(x) corresponding to a certain value x = f-
between the limits of integration.
(d) It is frequently desirable to make a change of variable
in the definite integral in order to facilitate the work of inte-
gration. It is obvious, from the nature of the definite integral,
276 INTEGRAL CALCULUS
that the limits of integration must be changed so that in the
new integral the limits shall be the values of the new variable
corresponding to those of the old variable.
Ex. Evaluate \ v a 2 — x 2 dx.
j>~ 2
Make the change of variable x — a sin 0, whence dx = a cos dO,
and therefore
n
f ° y/d'-x* dx = a 2 ( * cos 2 d6.
Here the limits for the new integral are determined by inspection
of the equation connecting x and 0, namely, sin 6 = -. It is seen that,
a
as x varies from to a, sin 6 varies from to 1. This corresponds to
a variation of 6 between the limits and -. The indefinite integral
is, by Art. 134 (c),
The substitution of the limits gives the value — •
4
141. Maclaurin's formula. As an application of the mean
value theorem (Art. 140 (c)), we derive Maclaurin's formula
with the remainder term.
Let s and t be independent variables. Suppose J[s — 1\ to-
gether with its first n derivatives with respect to t, to be
continuous within the interval to t v Then we have by inte-
gration
£f\s-t)dt = -f{s-t)
=f(s)-f(s-t l ).
On the other hand if we integrate by parts, taking u =f'( s — 0>
dv = dt, we obtain
£ l f'(s - t)dt =f'(s - t) • tT + jTV"(« -t)-tdt
*/o
INTEGRATION AS A SUMMATION. AREAS 277
Integrate the last term by parts, taking u =/"{s — t), do = t dt.
By successive applications of this process we deduce the
formula
/(*)_/(* - tl )=f'(s - (,)«. +/'(•-■ ttfl +f'"(s - trfl + -
+ ~^-r-. Cf\s-t)t^dt.
(n — 1) \Jo
(n-l)
By the mean value theorem we have
f V°(* - t n ~ l dt =f n) (s - et^idhf- 1 • t x ,
Jo
in which 6 is a positive fraction and Qt x is the same as £ of
(12). Inserting this in the preceding equation and substituting
s = x, t-L — x — a (hence s — t x = a) we obtain as a final form
f(x)=f(a) +f' { a)(x-a) +-^(x - af + ...
+ (iS)l /< " )(a; _ 6{X - a ^ X - °>"-
If we replace by 1 — 0', the remainder term takes the form
given on p. 153, with 6' written in the place of 0.
142. Remarks on the area formula, (a) It is noticed that
the formula
Cf(x)dx= *™ [/(a) + /(a + Ax) + - +/(o + ^=1 . Az)]A*
indicates two steps, — a summation, and a process of passing
to a limit. The differential /(x) dx which appears under the
integral sign may be regarded as representing the general
term/(x) Ax of the series to be summed, while the process of
taking the limit of this sum is indicated by replacing Ax with
the differential dx and prefixing the sign of integration.
278 INTEGRAL CALCULUS
The general term f(x) Ax represents the area of an arbitrary
rectangle (of the set of interior rectangles) whose altitude is
the ordinate corresponding to an arbitrary x and whose width
is Ax. This is called an element of area. The definite integral
may then be thought of as indicating the limit of the sum of
all contiguous elements of area between x — a and x = b.
This notion of summation (followed by passing to the limit
Ax = 0) is a very useful one in applying the calculus to prob-
lems of geometry, mechanics, and physics. In each case an
application of this notion consists in finding the general ex-
pression for an element of the given magnitude (element of
area, element of mass, element of moment of inertia, etc.) and
then indicating the two steps of summation and taking the
limit by changing Ax to dx and prefixing the symbol * of the
definite integral. It must not be forgotten that in every case
it is necessary to prove that the limit of the sum gives pre-
cisely the desired result, f This we have already done in case
of the area formula.
(6) The element of area/(x) • Ax is positive when the cor-
responding rectangle is above the x-axis, since in that case f(x)
is positive, while Ax is positive if b>a. Accordingly, the
formula \ f(x) dx gives a positive value for an area above the
x-axis provided we take b>a.
Similar considerations show that the same formula gives a
negative value for an area below the x-axis.
(c) If the curve y=f(x) crosses the x-axis between the two
points A, B, then the area consists of a positive part APC,
* This symbol originated historically from the initial of the word sum.
f In some cases the limit of the sum is used as a definition of the magnitude
in question, as, for example, in the definition of the length of arc. (Art. 151.)
INTEGRATION AS A SUMMATION. AREAS
279
B
Fig. (52
Q
represented by the integral J f(x) dx, and a negative part CBQ
represented by the integral I f(x) dx. The sum of these two
integrals, which (by Art. 140 b) is Y
equal to I f(x) dx, would accord-
ingly give the algebraic sum of the O
positive and the negative area,
((f) Some of the restrictions
placed upon the function f(x) in Art. 137 can be removed. In
the first place, suppose that f(x) is not always increasing (or
decreasing) as x increases from
a to b. Let ordinates be drawn
at the maximum and minimum
points of the given arc PQ (Fig.
63). These divide the required
area into several parts A', A",
A'" for each of which the ordinates satisfy the original condi-
tion of Art. 137, hence we conclude that
area = A' + A" + A"' =Cf{x) dx -f C*f(x) dx + Cf(x) dx
= Cf(x) dx, by Art. 140 (6).
A discussion of the methods to be employed in case f(x) be-
comes discontinuous, or is not singly valued in the assigned
interval, is postponed to Art. 143.
(e) Since f(x)=y, formula (9) may be written more briefly
area
APQB= Cydx.
(13)
280
INTEGRAL CALCULUS
X
B
y=b r
/
/ q
{'
V=a' L
p
X
Fig. 04
(/) By exactly the same process used
in deriving (9), or (13), it may be shown
that the Me&A'PQB' (Fig. 64) bounded by
the curve PQ, the ?/-axis, and the two lines
y = a', y = b' is given by the formula
area A'PQB' = | x dy,
(g) If it is required to find the area bounded by several
arcs such as PQ, QR, ES, etc. (Fig. 65), we may calculate by
formula (9) the simple areas
APQB, BQRC, etc., and by
proper additions and sub-
tractions obtain the desired
area. Thus the area in Fig.
65 would be expressed by
Xb /*c r*c S*d
f Y {x) dx + J & f 2 (x) dx - | f 3 (x) dx -J f 4 (x) dx.
1. Find the area bounded
by the curve y = log x, the
ar-axis, and the ordinates
— o <r —
* The symbol
la
- Area APQB (Fig. 66) =
/*8 -i8*
I \ogxdx = x(logx — 1)
= 3(log3-l)-2(log2-l) (
Fig. 66 =log-^-l.
indicates that the values 3 and 2 are to be substituted for
x in the expression which precedes the symbol and the second result sub-
tracted from the first.
INTEGRATION AS A SUMMATION. AREAS 281
2. Find the area bounded by the arc of the parabola y 2 = 4px
measured from the vertex to the point whose abscissa is a, the x-axis
and the ordinate x = a.
From the result show that the area of the parabola cut off by a
line perpendicular to the axis of the curve is two thirds the area of
the rectangle circumscribing this segment.
Does this result hold good for all parabolas?
3. Find the area between the x-axis and one semi-undulation of
the curve y = sin x.
4. Find the area bounded by the semicubical parabola y 2 = 25 x 3
and the line x = 3.
5. Find the area bounded by the curve y 2 = 4(x-f 5) 3 and the^-axis.
6. Find the area bounded by the cubical parabola y = x 3 , the
y-axis, and the line y = 1.
7. Find the area bounded by the curve x + y 3 = 2 and the coordi-
nate axes.
8. Find the area bounded by the parabola y = 2 x 2 and the line
y = 2x.
9. Find the area bounded by the parabola y = x 2 and the two lines
y — x and y =2 x.
10. Find by integration the area of the circle x 2 + y 2 = r 2 .
11. Find the area between the curve y = x(x — l)(x — 3) and the
x-axis.
12. Find the area bounded by the coordinate axes, the witch
8a 3 .
y = — — - — -, and the ordinate x = x v By increasing x\ without limit,
x 2 -f 4 a 2
ftnd the area between the curve and the x-axis.
13. Find the area of the ellipse — + y — = 1.
a 2 b 2
(* 14. Find the area included between the hyperbola xy = 36 and the
- line x -f y = 15.
282 INTEGRAL CALCULUS
15. Find the area bounded by the logarithmic curve y = a x , the
x-axis, and the two ordinates x = x v x = x 2 . Show that the result is
proportional to the difference between the ordinates.
16. Find the area between the curve y = (x 2 — l)(x 2 — 2) and the
x-axis.
17. Find the area cut off from the parabola (x — l) 2 = y — 1 by
the line y = x.
18. Find the area of the oval in the curve y 2 =(x — a)(x — b) 2 ,
given a < b.
19. Prove that the area of the curve a 2 y 2 = x 3 (2 a — x) is equal to
that of a circle of radius a. Draw figures of the two curves (center
of the circle at the point (a, 0)) and compare.
20. Find the area of the loop of the curve y 2 = x 4 4- x B .
21. Given the curve of damped vibrations y = e~ x sin x. Show
that the areas contained between successive semi-undulations of the
curve, and the positive x-axis form a geometrical series of alternately
positive and negative terms.
Find the sum of this infinite series and verify that the same result
may be obtained by integrating between the limits and go.
Find the total area included between the positive x-axis and the
curve (changing the negative areas to positive).
22. Find the area bounded by the hyperbola xy = a 2 , the x-axis,
and the two ordinates x = a, x — na.
From the result obtained, prove that the area contained between
an infinite branch of the curve and its asymptote is infinite.
23. Find the area contained between the curves y s = x and x 8 = y.
24. Take the segment of the equilateral hyperbola xy = k 2 , be-
tween two points P and Q. Show that the area between this arc and
the x-axis is the same as that between the same arc and the y-nxis.
25. Find the area bounded by the parabola Vx + Vy = Va and
the coordinate axes.
INTEGRATION AS A SUMMATION. AREAS 283
26. Find the area between the curve y 2 (y 2 — 2) = x — 1 and the
coordinate axes.
27. Find the area common to the two ellipses
a 2 b* ' b* a 2
28. Find the area enclosed by the curves y = sin x, y — cos x be-
tween two consecutive intersections.
29. Find the mean ordinate of the curve y = tan x between the
limits x = and x = - (see p. 275).
4
30. Find the mean value of the function sin x between the limits
and -; also of the function e^sinar.
31. Find the area of the loop of the curve
o a — x
V 2 = x 2 .
* a + x
143. Precautions to be observed in evaluating definite integrals.
The method given above for determining plane areas in rec-
tangular coordinates involves two essential steps :
(1) To find the integral of the given function f(x) ;
(2) To substitute for x the two limiting values a and b, and
subtract the first result from the second.
Erroneous conclusions may be reached, however, by an in-
cautious application of this process. The case requiring par-
ticular attention is that in which f(x) becomes infinite for
some value of x between a and b, or at a or b. When that
happens, a special investigation must be made. The method
of procedure will be brought out in the following examples.
Ex. 1. Find the area bounded by the curve y(x - l) 2 =c, the
coordinate axes, and the ordinate x = 2.
284
INTEGRAL CALCULUS
A direct application of the formula gives
n cdx
^]:
2 c,
]6
is a sign of substitution, indicating that the
values b, a are to be inserted for x in the expression immediately
preceding the sign, and the second result subtracted from the first.
This result is incorrect. A glance at the equation of the curve
shows that f(x)\ = - becomes infinite for x = 1. It is
L (x — 1)' 2 J
Fig. 67
accordingly necessary to find the area OCPA (Fig. 67) bounded by an
ordinate AP corresponding to a value x — x', which is less than 1.
For this part of the area/(x) is finite and positive, and formula (9)
can be immediately applied, with the result
area
OCPA = (* cdx , = c - T = c — -c. 0<x'<l.
J Q (x _l } 2 O-l)J X'-I
If now x' is made to increase and approach 1 as a limit, the value
of the expression for the area will increase without limit.
A like result is obtained for the area included between the ordi-
nates x = 1 and x = 2. Hence the required area is infinite.
Ex. 2. Find the area limited by the curve y 8 (x 2 — a 2 ) 2 = Sx 3 , the
coordinate axes, and the ordinate x = 3 a.
INTEGRATION AS A SUMMATION. AREAS 285
Since /(x) = — becomes infinite for x = a, it is necessary
L O 2 - a 2 )t J
in the first place to consider the area OPA (Fig. 68) and determine
Fig. 68
what limit it approaches as AP approaches coincidence with the
ordinate x = a. Accordingly
area OPA = £* 2xdx ^ = 3(x 2 - a 2 )$T
(x 2 — a 2 )f -*
whence
= 3(>' 2 -a 2 )* + 3a*,
[area OP.4] = 3ai
lim
x 1 = a
In the same manner, the area A'P'QB has the value
3a 2 xdx P 2 l
6 a^ — 3(x 2 — a 2 ) s,
£
0<x'<a
a < x' < 3 a.
(x 2 - a 2 )*
As x' diminishes towards a, the area increases to the limiting value
2
6 aL Hence, by adding the two results, the required area is found
tobe 8al + 6a*=9ai
The same result is found by a direct application of (9), viz. :
J-_2^ = 3(x2 _ a2)r |" = 9nS)
•" (x 2 -a 2 )f Jo
so that in this case an immediate use of the area formula gives the
correct result.
286
INTEGRAL CALCULUS
Some of the details in such problems as the two preceding
may be omitted. It is unnecessary first to put x — x', a value
] ess than the critical one, and, after integration and substitu-
tion of limits, to let x' approach the critical value as a limit.
For this is clearly equivalent to taking the critical value at
once as the upper limit for the portion of the area to the left
of the infinite ordinate (or as the lower limit for the area to
the right of this ordinate).
Thus, in case of an infinite ordinate, the rule of procedure
becomes :
Calculate separately, by formula (9), the two portions of area on
each side of the infinite ordinate and add the two results. If one
of these portions is infinite, it is not necessary to calculate the
other ; the required area is infinite.
The formula (9) for area has been deduced under the as-
sumption that the limits a and b are finite. It may happen,
however, that the curve y =f(x) approaches the x-axis as an
asymptote. It .might then be required to determine the strip
of area extending to infinity between the curve and its asymp-
tote. The method of procedure for such a case will be ex-
plained in the following example.
Ex. 3. Find the area bounded by the curve y(x 2 + 1) = 1 and
the ar-axis.
INTEGRATION AS A SUMMATION. AREAS 287
This curve being symmetrical with respect to the y-axis, it is
sufficient to calculate the area in the first quadrant. As our formula
of integration does not take account of the case b = co, we integrate
from to x' and in the result cause x' to increase without limit.
This limit will be defined to mean the area between the arc in the positive
quadrant, its asymptote, and the y-axis. It is evident that these steps in
the evaluation amount to a direct application of the area formula,
using the limits and co . The half area is, accordingly,
V = tan- ] a; = tan -1 co — tan- 1 0.
Jo 1 + x 2 Jo
We are here confronted with the difficulty that the anti-tangent is
a many-valued function and there is a question as to which of its
values should be chosen. It is necessary in such a case to go back
and examine the limiting process just explained. The area OPQN is
equal to tan -1 a;' — tan-*0. If x' approaches zero, this expression
should approach zero; and as x' increases continuously the area also
increases continuously. Accordingly, whatever value we choose for
tan _1 0, the limit of tan- 1 a:' should be the value obtained by a continu-
ous increase in this function as x' increases without limit. The sim-
plest value for tan- 1 is 0. If tan- x' increases continuously from 0,
it reaches the limit — when x' becomes infinite. Hence
2
. lim (tan-V-tan^O) = £•
x' = co v 2
If we choose tan _1 = n7r, 7i any integer, then , . tan -1 a;' = nT+ — »
J x' = x 2
and the difference gives * , as before.
Ex. 4. Find the area bounded by the curve y(x 2 + a 2 ) 2 = x and
the positive x-axis.
Ex. 5. Find the area bounded by the curve y — tan -1 x, the coordi-
nate axes, and the line x — 1.
In this problem we have to deal with a many-valued function of x.
In fact, to each value of x corresponds an infinite number of values of
288
INTEGRAL CALCULUS
tan -1 a:. The problem, accordingly, has an indefiniteness, which must
be removed by making some additional assumption.
The curve y = tan -1 x consists of an infinite number of branches,
corresponding ordinates of which differ by integer multiples of tt.
Each branch is continuous for all finite
values of x (see Fig. 70). It is evidently
necessary to select one of these branches
for the boundary of the proposed area,
and discard all the others. Suppose, for
example, the branch A B is selected. The
ordinate to this branch has the value
it when x is zero, and increases con-
_ 5 _
tinuously to it -\ — = - — as x increases
4 4
continuously to 1. Hence the required
area is
Y
y^"
A
~B
C
X
— ^
X
-1
Fig. 70
| tan -1 x dx = \ x tan -1 x— \ log (a; 2 -f 1)
5* -1 log 2.
4 2 s
EXERCISES
1. Find the area bounded by the curve y 2 (x — 1)= 1, the asymp-
tote x = 1, and the line x = 2.
2. Find the area bounded by the curve y 3 (x — l) 4 = 1 and its
asymptota, the x-axis.
3. Find the area bounded by the curve of Ex. 2, the x-axis, and
the ordinate x = 2.
4. Find the area inclosed by the curve x 2 y 2 = a 2 (y 2 — x' 2 ) and its
asymptote.
5. Find the area bounded by the curve a 2 x — y(x — a), the rr-axis,
and the asymptote x = a.
INTEGRATION AS A SUMMATION. AREAS 289
x 3
6. Find the area between the cissoid y 1
2a
and its asymp-
tote x = 2 a.
7. Find the area between the curve y 2 (l — x 2 ) = 1 and its
asymptotes.
144. Calculation of area when x and y are expressible in terms
of a third variable. When the rectangular coordinates of any
point of the boundary arc of the required area are given as
functions of a third variable 6, we may substitute in I ydx
the expressions for y and dx in terms of 6 and integrate be-
tween the corresponding new limits for in accordance with
Art. 140(d).
Area of the cycloid. This curve is traced by a point P in
the circumference of a circle of radius r as the circle rolls on
a straight line, without sliding.
Y
O M
Fig. 71
Let the point Pbe in contact with the given line at when
the circle begins to roll. Suppose that an arbitrary arc PQ
has rolled over the segment OQ. Let (x, y) denote the rec-
tangular coordinates of P, and let 6 represent, in radian meas-
ure, the angle at the center C subtending PQ ; then,
OQ = arc PQ = rO.
EL. CALC. — 19
290 INTEGRAL CALCULUS
Dropping a perpendicular PR on the line CQ, we have
PR = r sin 6, RC = r cos 0.
Accordingly,
x=OM=OQ-MQ = rO-r sin 6 = r (0 - sin 0),
y = MP= QC- RC= r - r cos = r(l - cos 0).
These are called the two parametric equations of the cycloid,
being a varying parameter. One complete arch of the cy-
cloid is generated as 6 varies from to 2 7r, that is, as x varies
from to 2 ttt. The maximum ordinate for this arc occurs at
x — -n-r, and the arc is symmetrical with respect to this ordinate.
The area inclosed by the arc OP A and the cc-axis is
ydx= \ r(l - cos 0) • r(l - cos 0)d6 = 3 ttt 3 .
The area is three times that of the rolling circle.
EXERCISES
1. Find the area of the ellipse when x and y are expressed in
terms of the eccentric angle, x = a cos <£, y = b sin <f>.
What is the meaning of the negative sign in the result?
2. 2 2
2. Find the area of the hypocycloid x 3 + y* = a 3 by expressing x
and y in the form x = a cos 3 0, y = a sin 3 6.
3. Find the area of the loop of the folium of Descartes
x s + y s — 3 xy = 0.
This area may be calculated either by expressing x and y in the
6*+ 1 3 + 1
obtained by putting y = Ox and solving for a: and ?/, or by transform-
ing to polar coordinates and using the polar formula for area, Art. 145.
INTEGRATION AS A SUMMATION. AREAS
291
4. Find the area within the curve y 2 = (1 — a; 2 ) 8 by assuming
x = cos 0, y = sin 3 0.
5. Find the area of (aa:)*+ (by)* = (a 2 - b 2 )*, the evolute of the
ellipse. (See Fig. 51, p. 190.) Express x and y in the form,
ax = (a 2 - b 2 ) sin 8 0, by = (a 2 - b 2 ) cos 3 0.
145. Areas in polar coordinates. Let PQ be an arc of a curve
whose equation is given in polar coordinates (p, 0). It is re-
quired to find the area bounded
by this curve and the two as-
signed radii QP and OQ.
Let A and B be any two
points of the curve with coordi-
nates (p, 0) and (p + Ap, + A<9)
respectively. Through ^4 draw
an arc AC of a circle with radius
p and center 0. The element of
area OAC is a sector of a circle
of angle A0. The arc AC is, FlG> 72
therefore, p A0 and the sectorial area is | p 2 A0. The limit of
the sum of all such elements contained between OP and OQ is
!fV«
(14)
That this is the actual area sought remains to be proved by
showing that the sum of the elements of area has the required
area for its limit. This may be done by steps exactly analo-
gous to those used in Art. 137, which would consist in
proving that the sum of all interior sectors, such as OAC, has
the same limit as the sum of all exterior sectors, such as ODB.
The details are left to the student as an exercise.
292 INTEGRAL CALCULUS
EXERCISES
1. Find the area of the three loops of the curve p = a sin 3 0.
From the symmetry of the figure it is seen that one sixth of the
total area is described as varies from to — . Hence the area is
6 j* Q 6 ± a 2 sin 2 S$ d$= % a 2 J * (1 - cos 6 0)dO
r_a?
This is one fourth the area of the circumscribing circle.
2. Find the area of the lemniscate p 2 = a 2 cos 2 0.
3. Find the area of the circle p = 2 r cos 0.
4. Find the area of the cardioid p = r(l — cos $).
5. Find the area of the circle p = 10 sin 0.
6. Find the area bounded by the hyperbolic spiral p$ = c . and
radii drawn to two arbitrary points (p v #,) and (p 2 , 2 ). Show that
the area is proportional to the difference between the radii.
7. Find the area of the four loops of the curve p = a sin 2 0.
8. Find the area of the loop in the spiral of Archimedes p = aO
generated between the limits — - and + - for 6.
9. Find the area bounded by the lituus p 2 6 = k and two arbitrary
radii, making angles 0\ and 6 2 with the polar axis.
10. Find the area of one loop of the curve p 2 = a 2 cos nO.
11. The radius vector of the logarithmic spiral p = e~ e starts at
the angle = and rotates positively about the origin an infinite
number of times. Determine the area swept over by the radius
vector.
12. Find the area of the curve p 4 = sin 2 cos 0.
13. Find the area within the curve p = cos 2 0.
14. Find the area of the innermost loop of the double spiral p = 2 .
146. Approximate integration. The trapezoidal rule. As
shown in Art. 138, the numerical value of the definite integral
INTEGRATION AS A SUMMATION. AREAS
293
f y dx is the same as that of the area bounded by the curve
y z=f(x), the z-axis, and the two ordinates x = a, x = b.
When a, b, and the coefficients in f(x) are numerically given,
the approximate value of this area, and therefore of the defi-
nite integral, can be found by adding the n terms of the series
[/(«) + f( a + Aa + ' * * + f( a + n — 1 • Ax)] Ax. The close-
ness of the approximation improves with increasing values of
n. A much more rapid method of approximation is now to be
considered.
Instead of forming rectangles,
as in Fig. 59, p. 268, draw the
chords PP X , P X P 2 , • • ., P n _!.Q, thus
making trapezoidal elements of
area, APP X A X , A x P x P 2 £ 2i etc.
Denote the ordinates at A, A ly A 2 ,
5—p,
A A x A z
Fia. 73
A^B'
A n _ ly Bby y , y lf y 2 , >-
?/„_!, y n respectively. Also for brevity write Ax = h.
the areas of the several trapezoids are
APP X A X = 1^ + y x )h,
A X P X P 2 A 2 = l( yi + y 2 )h,
Then
A>-iPn-iQB = i(y n _ 1 + y n )h.
Hence, by adding, we obtain for the approximate value of
the definite integral the expression
*[*¥* + * + » + ••• + !,„-,].
This is known as the trapezoidal formula for the approximate
value of J ydx and this method of computing its numerical
value is called the trapezoidal rule.
294 INTEGRAL CALCULUS
147. Simpson's rule. With three ordinates. Instead of draw-
ing the chords PP 1} P X P 2 pass a parabola, having its axis ver-
tical, through the three points P, P x , P 2 and determine the
area of the double strip bounded by the two ordinates y , y 2 ,
the #-axis, and the parabolic arc.
The equation of the parabola is of the form
y = k -+- Ix -f- mx 2 .
For convenience take the origin at the foot of the middle
ordinate y x . Then the abscissas of the three ordinates may
be represented by — h, 0, + h, and the area under the para-
bolic arc is given by the formula
x
\k + Ix + mx 2 )dx = | (6 k + 2 mh 2 ).
This result can be expressed in a simple form in terms of the
three ordinates 2/ , y l} y 2 . For,
y =k — lh-\- mh 2 ,
Vi = k,
y 2 = k + Ih + mh 2 ;
therefore, y •+- y 2 = 2 k + 2 ra/i 2 ,
hence, 6 ft + 2 m/i 2 = y + 4 ?/j + ?/ 2 ,
and, accordingly,
parabolic area APP X P 2 A 2 — - (y -f- 4 y x -f ?/ 2 ). (IS)
o
This is Simpson's parabolic formula for three ordinates.
With n ordinates. In like manner the area bounded by the
two ordinates y 2 , y 4 and a parabolic arc through P 2 , P 3 , P 4 is
| (2/2 + 4 2/3 + 2/*), (16)
INTEGRATION AS A SUMMATION. AREAS 295
and so on. If the number of ordinates y , y u •••, y n is odd, we
obtain, by adding together the expressions (15), (16), etc.
J[2/o + * Z/i + 2 2/2 + 4 2/3 +2 1/4 + • + 2 2/ n -2 + * V n -l + 2/J-
o
This is Simpson's formula for the approximate value of I ydx.
148. The limit of error in approximate integration. The ap-
proximate value obtained for I f(x)dx by means of Simpson's
s
formula differs from the true value by an amount which does
not exceed*
_ { Q — a ) j l \£)rr
j
180
in which f IV (£) is the value of the fourth derivative of f(x)
when x is given a certain value £ between a and b. The limit
of error for the trapezoidal rule is *
(b-a)f"(£W
12
Since £ is not definitely known, in applying the above
formulas to find the limit of error it is necessary to choose £
so that / IV (£) or /"(£) has its greatest value in the interval
from a to b. The result so obtained may be considerably
larger than would be given by the formula if £ were actually
known. In some cases the result will be so large as to give no
useful information in regard to the closeness of our approxi-
mation. In other cases it will be small enough to indicate
that the required degree of approximation has been attained.
For example, suppose it is required to evaluate
'login a
dx.
* See Markoff, "Differenzenrechrmnij," § 14, pp. 57, 59.
296 INTEGRAL CALCULUS
Since f(x) = x~ l \og l0 x, we obtain by successive differentiation
/^(a;)=a;- 6 (241og 10 a;-50*iW"), M =log 10 e== 0.4343, very nearly.
As we cannot readily determine by inspection the largest nu-
merical value of f lv (x) in the interval 20 <^ x <J 30, we obtain
the next derivative . , s . /rt _. ,_„«-..
/v(a?) = ar 6 (274 Jf- 120 log 10 x).
The first factor x~ 6 is positive. The second factor takes a nega-
tive value for x ^20 and hence / v (<r) is negative in the given
interval. Therefore, / IV (x) is a decreasing function for all the
values of x under consideration. But f lv (x) is positive for
x = 30, and accordingly its greatest numerical value occurs for
x = 20, which is / Iv (20) = 0.000003.
The limit of error for Simpson's formula is, therefore,
_ 10(0.000003) y = _ (a0OO0OO2)y .
If we use 3 ordinates, then h=5 and the error does not exceed
— 0.0001 + ; that is, the error is less than two units in the fourth
place of decimals.
EXERCISES
In the following problems use Simpson's formula whenever an odd
number of ordinates is given. Determine the limit of error and,
when possible by direct integration, the exact error. Also evaluate
by using the trapezoidal rule, and compare the degree of accuracy
attained by the two different methods.
1. Evaluate ( x 2 dx by the trapezoidal rule, using 5 ordinates ;
9 ordinates.
In the case of 9 ordinates, n = 8 and h = — ^ — = - , y = 0,
3/i = G) 2 >!/2 = l,y 3 = (§)V--, </8 = 42.
2. Prove that Simpson's rule gives the exact value of f x 2 dx y
( h x s dx, C\ ax* + /?.r 2 + yx + 8) dx.
INTEGRATION AS A SUMMATION. AREAS 297
3. Evaluate ( cos xdx, using 3 ordinates ; 5 ordi nates ; 7 ordi-
nates; 9 ordinates. Notice the variation of error with increasing
values of n.
4. Evaluate ( Vxdx, using 5 ordinates.
5. Evaluate ( Vl + X s dx, using 4 ordinates ; 7 ordinates.
6. Evaluate f cos x dx, using 7 ordinates.
Jo
C 12
7. Evaluate \ log 10 x dx, using unit intervals.
r to ,/ r
8. Evaluate \ '— , using 7 ordinates.
Jio loo- r
9. Evaluate ( Vl — x i dx, using 6 ordinates.
Jo °
10. Evaluate ( e~ x idx, using 11 ordinates.
This integral (with any upper limit) is called the Probability Inte-
gral since it plays an important role in the theory of probabilities.
IT
11. Evaluate f Vl — 3 sin' 2 xdx, using ? ordinates.
r 10
12. Evaluate \ x 2 dx by the trapezoidal rule, using 11 ordinates.
13. Calculate the value of 7rfrom the formula — = ( ( x , using
1 J» 1 + x' 2
11 ordinates.
Determine the error by comparison with the known value of 7r.
r l
14. Evaluate i Vcos6d$, taking at intervals 15°, 10°, 0°.
This, like Ex. 11, is an Elliptic Integral and cannot be integrated
by any formula given in the present volume. It occurs in the prob-
lem of calculating friction in journals. (See " Engineering Mathe-
matics " by Prof. V. Kakapetoff, Part I, p. 16. Wiley, 1912.)
15. Evaluate ( °%w x dx, using 3 ordinates.
J20 x &
CHAPTER VII
GEOMETRICAL APPLICATIONS
149. Volumes by single integration. The volumes of various
solids may easily be calculated by a summation process exactly
similar to that used in computing areas. The following prob-
lems will make the mode of procedure clear.
Ex. 1. A woodman fells a tree 2 ft. in diameter, cutting halfway
through on each side. The lower face of each cut is horizontal and
the upper face makes an angle of
60° with the lower. How much
wood does he cut out?
The portion cut out on one side
forms a solid bounded by a cylindri-
cal surface whose equation may
be taken in the form x 2 +
1,
and by two planes whose intersec-
tion may be chosen for the y-axis.
Imagine this wedge-shaped solid
divided into thin plates by means
of planes parallel to the a;2-plane
and at equal distances Ay. The
volume of an arbitrary plate PQRP'Q'R' is approximately equal to
the area of the triangular face multiplied by the thickness Ay.
Area PQR = \ RP ■ PQ= \xz =^x i
GEOMETRICAL APPLICATIONS 299
since - = tan 60° = V3. The element of volume is therefore
x
V3 x 2 A
Since the figure is symmetrical with respect to the arz-plane, it is suffi-
cient to calculate the volume between the limits and 1 for y and
double the result.
The limit of th3 sum of all elements of volume in the first octant
is
^j/^ = ^j-; ( i- y s )dy= -L.
That this limit is the volume to be determined may be seen on
observing that the element of volume falls short of the total amount
contained in the plate PQRP'Q'R' by the prismatic piece PNP'QMQ'.
The sum of all these neglected portions, in the first octant, is less
than the volume of the maximum plate (having the a;z-plane for base),
and hence approaches zero as Ay diminishes.
Therefore the total volume of wood cut out is cu. ft.
V3
Ex. 2. Calculate the volume in Ex. 1, by dividing the solid of
Fig. 74 with equidistant planes parallel to the yz-plane.
Ex. 3. Find the volume of the ellipsoid
Imagine the solid divided into a number of thin plates by means
of planes perpendicular to the x-axis and at equal distances Ax. Re-
gard the volume of each plate as approximately that of an elliptic
cylinder of altitude Ax, whose base is the section of the ellipsoid by
one of the cutting planes. If the equation of this plane is x = A,
the equation of the elliptic base of the plate is (in y, z coordinates)
300 INTEGRAL CALCULUS
** z* _ A 2
+ -=1-
r^ 2 c 2 a 2
X 2
Dividing by 1 — — , we obtain
a 2
The semiaxes of the ellipse are
= 1.
Since the area of the ellipse is the product of the semiaxes multi-
plied by 7r (Ex. 13, p. 281), it follows that the area of the elliptic base
is irbcy 1 -J. On replacing A. by x, the element of volume may
be written
The sum of all such elements for values of x varying by equal
increments Ax between and a differs from the volume of the half
ellipsoid by a series of ring-shaped portions, the total sum of which
is less than the volume of the maximum plate of the figure. It
readily follows from this that the total volume of the ellipsoid is
C a I x 2 \ 4
2 \ irbc 1 ) dx = - rrabc.
Jo V a 2 / 3
Ex. 4. Solve Ex. 3 by taking the cutting planes parallel to the xz-
plane and at equal distances At/.
Ex. 5. Solve Ex. 3 by taking the cutting planes parallel to the
x^-plane.
Ex. 6. Find the volume of the portion of the elliptic paraboloid
v 2 z 2
'— H — = x cut off by the plane x — 1.
a 2 b 2
GEOMETRICAL APPLICATIONS
301
Ex. 7: Find the volume of the elliptic cone ^ -\ — = (x — l) 2 raeas-
, a 2 b-
ured from the ^2-plane as base to the vertex (1, 0, 0).
Ex. 8. Find the volume of a pyramid of altitude h and of base
area A.
[Hint. Take the base on the ar^-plane, the altitude coinciding
with the 2-axis. Cut the solid into thin plates by planes parallel to
the base.]
e=i.
b 2
On the major axis a plane
D
Fig. 75
Ex. 9. Given an ellipse —
a 2
rectangle A BCD is con-
structed perpendicular to
the plane of the ellipse.
Through any point P of
the line CD a plane is
constructed perpendicu-
lar to CD. The two
points ii and 5 in which
the latter plane meets the
ellipse are joined to P
by straight lines. The
totality of all lines so determined forms a ruled surface called a conoid.
Given A C — p, find the volume of the above conoid.
Ex. 10. A rectangle moves from a fixed point P parallel to itself,
one side varying as the distance from P, and the other as the square
of this distance. At the distance of 2 ft., the rectangle becomes a
square of 3 ft. on each side. What is the volume generated?
Ex. 11. The center of a square moves along a diameter of a given
circle of radius a, the plane of the square being perpendicular to that
of the circle, and its magnitude varying in such a way that two oppo-
site vertices move on the circumference of the circle. Find the vol-
ume of the solid generated.
302
INTEGRAL CALCULUS
Ex. 12. A right circular cone having an angle 2 at the vertex has
its vertex on the surface of a sphere of radius a and its axis passing
through the center of the sphere. Find the volume of the portion of
the sphere which is exterior to the cone.
Ex. 13. Find the volume of the paraboloid — + ^— = z cut off by the
plane z = c.
Ex. 14. A banister cap is bounded by two equal cylinders of revo-
lution of radius r whose axes intersect at right angles in the plane of
the base of the cap. Find the volume of the cap.
150. Volume of solid of revolution. Let the plane area,
bounded by an arc PQ of a given curve (referred to rectangular
axes) and the ordinates
p n-\ ^-^L at the extremities P and
Q, be revolved about the
x-axis. It is required to
find the volume of the
solid so generated.
Let the figure APQB
be divided into n strips
of width Ax by means
of the ordinates A l P tJ
A 2 P 2 ,.-, A-iPn-v In
revolving about the
x-axis, the rectangle APR X A X generates a cylinder of altitude
Ax, the area of whose base is tt • AP . Hence
2
volume of cylinder = -k • AP • Ax.
The volume of this cylinder is less than that generated by
the strip APP X A X by the amount contained in the ring gen-
erated by the triangular piece PR X P V Imagine this ring
Fig. 76
GEOMETRICAL APPLICATIONS
303
pushed in the direction of the x-axis until it occupies the posi-
tion of the ring generated by CDE. If every other neglected
portion (such as is generated by P^P^) is treated in like
manner, it is evident that the sum is less than the volume
generated by the strip A^P^QB, and hence has zero for
limit as Ax approaches zero. Therefore the sum of the n cylin-
ders generated by the interior rectangles of the plane, viz.
tt(.IF + A^ 2 + -+A n _ 1 P n J)Ax,
has for limit the volume required. But the limit of this sum
is the definite integral I wy 2 dx, and hence
volume = 7r I y 2 dx.
The volume generated by revolution about the y-axis is found
by a like process to be expressed by the definite integral
rj x?dy,
in which a' and b' are the values of y at the extremities of the
given arc.
When the axis of revolution does not coincide with either of
the coordinate axes, a similar procedure will usually give at
once the element of volume, y Q y
Examples 1-3 will illustrate.
Ex. Find the volume of revo-
lution of the segment of the
parabola y 2 = x cut off by the
line y = x, the axis of revolu-
tion being the given line.
Let OQ be the axis, and P
any point of the parabolic arc. °\ Fig. 77
304 INTEGRAL CALCULUS
If v denotes the perpendicular distance PR from P to OQ and u the
length of the line OR, then the element of volume is
7T?> 2 Am.
The formula of analytic geometry for the distance from a point
to a line gives
_ y — x _ Vx — x
V ~~ V2~~ V2 '
in which (x, y) are the coordinates of P. The second form for v is
obtained by substituting for y the expression given by the equation
of the parabola.
Since Am is measured on a line making an angle of 45° with the
.r-axis, it follows that Am = V'2 ■ Ax.
Hence the required volume is
1 [Vx- xV
i
7T
V2 dx
V V2 / 30 V2
EXERCISES
1. A quadrant of a circle revolves about its chord. Find the
volume of the spindle so generated.
[Hint. Take the equation of the circle in the form x 2 + y 2 = r 2
and the equation of the chord x + y = r.]
2. Find the volume of revolution of the segment of the circle
x 2 + y 2 — r 2 cut off by the line x = a, this line being the axis of
revolution.
3. Find the volume of the truncated cone obtained by revolving
about the ?/-axis the segment of the line 3 # + # = 5 between the
points (2, - 1) and (1, 2).
4. Find the volume generated by the revolution of the cissoid
x 3
= about the x-axis from the origin to the point (x v y^).
2a — x
What is the limit of this volume as xi approaches 2 a?
GEOMETRICAL APPLICATIONS 305
5. Find the volume obtained by revolving the entire cissoid about
its asymptote, the line x = 2 a.
[Hint. The element of volume is tt(2 a — x)' 2 Ay. For the pur-
pose of integration express x and y in terms of a third variable t by
means of the equations
z = 2asin 2 *, 2/ = 2a— 1
' * cost J
6. Find the volume of the oblate spheroid obtained by revolving
2 b*
the ellipse — -f 2_ = 1 about its minor axis.
7. Find the volume of the sphere obtained by revolving the
circle x' 2 + (y — k) 2 — r 2 about the y-axis.
8. The arc of the hyperbola xy = k 2 , extending from the vertex
to infinity is revolved about its asymptote. Find the volume
generated.
What is the volume generated by revolving the same arc about the
other asymptote?
9. Find the entire volume obtained by rotating the hypocycloid
xi 4- y* = a* about either axis.
10. Find the volume obtained by the revolution of that part of
the parabola Vx -f Vy = Va intercepted by the coordinate axes about
one of those axes.
11. Find the volume generated by the revolution of the witch
8 a*
' 2 +4« 2
about the a: -axis.
12. Find the volume generated by the revolution of the witch
about the #-axis, taking the portion of the curve from the vertex
(x = 0) to the point (xi, y\).
What is the limit of this volume as the point (x v y { ) moves toward
infinity?
EL. CALC. 20
306
INTEGRAL CALCULUS
13. Find the volume obtained by revolving a complete arch of
the cycloid x = a(0 — sin 6), y = a(l — cos 6) about the a>axis.
Volume = Tri ifdx = ira 3 l (1 - cos 0) 3 dO.
14. Find the volume obtained by revolving the cardioid
p = a(l — cos 6) about the polar axis.
Assume x = p cos 0, y = p sin B.
Then dx = d(p cos 6) = d[a{\ — cos 0)cos &]
= a sin 0( - 1 + 2 cos 6) dO.
Hence
volume = it f ifdx = - 7m 3 f "sin 3 0(1 - cos 0) 2 (1 - 2 cos 0) dd.
151. Lengths of curves. Rectangular coordinates. Let it be
required to determine the length of a continuous arc PQ of a
curve whose equation is written in rectangular coordinates
It is first necessary to define what is meant by the length
of a curve. For this pur-
pose, suppose a series of
points Pj, P 2 , '•', P„_i taken
on the arc PQ (Fig. 78), and
imagine the lengths of the
chords PP X , P X P 2 , ---to have
been determined. The limit
of the sum of these chords as
the length of each chord ap-
proaches zero will be taken,
in accordance with accepted
usage, as the definition of the length of the arc PQ ; * that is,
arc PQ = Lt (chord PP, + chord P,P 2 + •• • + chord P n _ x Q). (1)
* That this limit is always the same no matter how the points Pi are chosen,
as long as the curve lias a continuously turning tangent, and the distances
Fig. 78
GEOMETRICAL APPLICATIONS 307
This definition is immediately convertible into a formula
suitable for direct application.
For, let the points P x , P.,, • •• be so chosen that
PB 1 = P 1 R 2 = .. = Ax,
the lines PBi, etc., being drawn parallel to the ic-axis.
Denote by Ay the increment R x Px of y. Then the length
of the chord PP X is
V(A*)» + (Ayy = yjl+ ( '^Y A* = X /l + ( |^) V (2)
\AxJ * \&yj
Now — ^ is the slope of PP V It is, therefore, equal to the
l\X
slope of that tangent to the arc PP X which is parallel to the
chord. If (x 1} i/i) denote the coordinates of the point of con-
tact of this tangent line, then we have
Ay _ dy l
Ax dx x
Hence the length of chord PP X may be expressed in the form
/(x^Ax, in which
'w-V 1 *® 1 - (3 >
Similarly
P,P 2 =f(x 2 ) Ax, P 2 P S =f(x 3 ) Ax, • • • ,
in which x 2 is the abscissa of a certain point on the arc P X P 2 ,
and so for x s , •••. When these expressions are substituted in
(1), it becomes
arC PQ = Ax™ ^"W +/&) + • ' ' +/(*-)] **•
Pi-lPi are all made to tend towards zero, admits of rigorous proof. The
proof is, however, unsuitable for an elementary textbook. (See Rouche et
Comberousse, " Traite de geome'trie," Part I, p. 189, Paris, 1891).
308 INTEGRAL CALCULUS
But, by (11), p. 274, this limit is J f(x)dx. -Substituting for
f(x) from (3), we obtain the formula
-"e-rv+ffi)'** w
day
in which a and b are the abscissas of P and (^respectively.
Taking for PP X the second form in (2), namely,
vwiw
we deduce in like manner
-^-xv+sy*
\d>j;
in which a' and b f are the ordinates of P and Q.
EXERCISES
1. Find the length of arc of the parabola y 2 = 4 px measured from
the vertex to one extremity of the latus rectum.
In this case ^=-W",
dx y x
hence length of arc = f P \il + £ dx = (" x+p dx.
J0 " x > Vx*+px
2. Find the length of arc of the semicubical parabola ay 2 — x z
from the origin to the point whose abscissa is -•
3. Find the length of arc of the curve y = log cos x, measured
from the origin to the point whose abscissa is — •
2 2 2
4. Find the entire length of the hypocycloid x* + y^ = a 5 .
GEOMETRICAL APPLICATIONS 309
6. Find the length of arc of the catenary y = - (e a + e a ) from
the point (0, a) to the point whose abscissa is a.
6 ' 2x
x 3 1
7. Find the length of arc of the curve y = — + — - between the
limits x = 1 and x = 2.
8. Find the length of the logarithmic curve y = log x from a: = 1
to x = V3.
9. Find the length of arc of the evolute of the ellipse
(ax)s + (byfs = (a*-b 2 )l
10. Find the length of arc of the curve y = a log (a 2 — x 2 ) from
x = to x = - .
152. Lengths of curves. Polar coordinates. The polar
formulas for length of arc may be derived from those of the
previous article by transformation from rectangular to polar
coordinates.
Since x = p cos 0, y = p sin 6, we obtain by differentiating
with respect to 6
dx = (dp cos 0-p sin dXie, dy = ( d ? sin 6 + p cos $) cl0,
hence
V i +© , *-va^v-v@y+7*
Therefore the length of arc is
arcPe = £^ + (|y<W, (5)
the limits of integration being the values of 6 at P and Q.
310 INTEGRAL CALCULUS
If p instead of is taken as the independent variable, we
deduce in like manner
arcPQ=jf;Vl + (^)V
the limits being the values of p at P and Q.
EXERCISES
1. Find the length of arc of the logarithmic spiral p = e a between
the two points (p v 0j) and (p 2 , 2 ), and show that it is proportional
to the difference of the two radii p\ and p 2 .
2. Find the length of arc of the circle p = 2 a sin 0.
3. Find the entire length of the cardioid p = a(\ — cos 6).
Q
4. Find the length of the parabola p = a sec 2 - between the points
(Pv #]) and (p 2 , (9 2 ).
5. Find the length of the spiral of Archimedes p = aO between
two arbitrary points.
6. Find the length of arc of the spiral p = 0' 2 measured from
= to 6 = 7t.
7. Find the entire length of the curve p = cos' 2 0.
A
8. Find the entire length of the curve p = a sin 3 r -
o
3. Find the length of arc of the cissoid p = 2 a tan sin 6 between
the limits and — •
4
[Hint. For the purpose of integration, express the integrand in
terms of sec as the independent variable.]
153. Measurement of arcs by the aid of parametric representa-
tion. Suppose the rectangular coordinates of a point on a
given curve are expressed in terms of a third variable t. Then,
GEOMETRICAL APPLICATIONS 311
since in rectangular coordinates — =-%/( — J +( — ) (Art. 41),
we have
in which s = arc PQ, and t lt t 2 are the values of t corresponding
to the points P and Q. In like manner, if the polar coordi-
dates (p, 0) are expressed in terms of t, the formula for length
of arc is
-JWaW'Sr'
dt
since
iwijH'f)'
EXERCISES
1. Find the length of a complete arch of the cycloid
x = a(t — sin t), y = a(l — cos /).
2. Find the length of the epicycloid
x = a(m cos t — cos mt), y = a(m sin t — sin m/)
from t = 0tot= '
m — 1
3. Find the length of arc of the hypocycloid x* -\- y^ = a? by ex-
pressing x and y in the form x — a sin 3 t, y — a cos 3 t.
4. Find the length of the involute of the circle
x = a(cos t + t sin <), y = a(sin t - t cos f)
from < = to t = t v
5. Find the length of arc of the curve x* — yt = a% from (a, 0)
to (x v yj by assuming x = a sec 3 t, y = a tan 3 /.
312 INTEGRAL CALCULUS
6. Find the length of arc of the curve x = e* sin t, y = e l cos t from
t = to t = t\.
7. Find the length of arc of the curve x = a + t 2 , y = b 4 t 3 , meas-
ured from the point t = to the point t = t v
154. Area of surface of revolution. Let AQ be a continuous
arc of a curve whose equation is expressed in rectangular coordi-
nates x and y. It is required to
determine a formula for the area
of the surface generated by revolv-
ing the arc AQ about the a>axis.
It has been shown in Art. 44,
if p. 81, that if S denotes the area
of the surface generated by the
FlG - 79 rotation of AP (P being a variable
point with coordinates (x, y)), then A# satisfies the conditions
of inequality
2 7t y As < AS < 2 7r(y + Ay) As. (6)
Let the arc AQ be divided into n equal parts of length A.s.
For each segment of arc there will be a set of conditions such
as (6), the values of y, Ay, AS being in general different for
the different segments. Let the n sets of inequalities thus
obtained be added. In what follows, the symbol ^, will be
used as an abbreviation of the expression, "The sum of the
n terms of the form." Since 2j A>S = S (in which S now
denotes the entire surface generated by arc AQ), we have
2 7r ^yAs<S<2Tr^(y + Ay)As. (7)
Now let As (and hence Ay) approach zero. The first mem-
ber of (7) becomes 2-k | yds, which changes to
GEOMETRICAL APPLICATIONS 313
on making x, or y, the independent variable. The limit of the
last member of (7) may be written
lim
As.
™o X t y As + Ay As ~i = fy ds + lim S A ^ As -
The last term is zero. For, let 8 represent the maximum
value of Ay in any of the terms of 2 Ay As. Then follows
]T Ay As ^ 8 ]P As = 8 • arc AQ,
and since 8 approaches zero, we conclude that lim Va^As = 0.
Hence lim ^?y As = lim ^P (y -f Ay) As,
and therefore
In like manner the area of the surface obtained by revolving
arc AQ about the y-axis is
EXERCISES
1. Find the surface of the catenoid obtained by revolving the
X _x
catenary y = ~ (e a +e a ) about the y-axis, from x — to x = a.
, XX
Since -^- — \{e a -e a ),
ax
it follows that
1 + (dyy_(e*+e~°)\
+ \dx) ' 4
314 INTEGRAL CALCULUS
hence, by using the first formula of (8), the required surface has the
area
[e a + e a )dx.
r jo *(<
2. Find the surface obtained by revolving about the y-axis the
quarter of the circle x 2 -\ y 2 + 2x + 2y + l = contained between
the points where it touches the coordinate axes.
3. Find the surface generated by revolving the parabola y 2 = 4px
about the £-axis from the origin to the point (p, 2 p).
4. Find the surface generated by the revolution about the ?/-axis of
the same arc as in Ex. 3.
5. Find the surface generated by the revolution of the ellipse
a 2 + b 2 ~ '
(a) about its major axis (the prolate spheroid) ;
(b) about its minor axis (the oblate spheroid).
6. Find the surface generated by the revolution of the cardioid
p = a(l + cos 0) about the polar axis.
Regarding the figure as referred in the first place to rectangular
axes such that x — p cos 0, y = p sin we have
surface = 2 tt jj y ds = 2 tt f "" p sin B^p 2 +(^Y d 0>
ds =^p 2 + (& YdO by Art. 45.
7. Find the surface of the cone obtained by revolving that por-
tion of the line - + £ = 1 which is intercepted by the coordinate axes,
a b
(«) about the x-axis ; ((S) about the y-axis.
8. Find the surface of the sphere obtained by revolving the circle
p = 2 a cos about the polar axis. [Cf. Ex. 6.]
GEOMETRICAL APPLICATIONS
315
9. Find the surface generated by the revolution of a complete
arch of the cycloid x — a(0 — sin 0), y = a(\ — cos 0) about the z-axis.
10. Find the surface of the ring generated by revolving the
circle x 2 + (y — k) 2 = a 2 , k>a, about the x-axis. Also find the vol-
ume of this ring.
11. Find the surface generated by the rotation of the involute
of the circle
x = a (cos t + t sin /), y = a (sin t — t cos t)
about the x-axis from t — to t = t v
155. Various geometrical problems leading to integration.
Ex. 1. A string AB of length a has a weight attached at B. The
other extremity A moves along a straight line OX, drawing the weight
in a rough horizontal plane XOY. The path traced by the point B
is called the tractrix. What is its equation ?
Let OF be the initial position of the string and AB any intermedi-
ate position. Since at every instant the force is exerted on the weight
316
INTEGRAL CALCULUS
B in the direction of the string BA, the motion of the point must be
in the same direction ; that is, the direction of the tractrix at B is
the same as that of the line BA and hence BA is tangent to the curve.
The expression for the tangent length is (Art. 48, p. 86)
dy
dx
=wir
+ l = a.
dx
Solving for • — , we obtain
dy
dx
V a -
Integrating with respect to y gives
Vo2
-J-
y
dy = Va* - f - a log « + Va 2 - y* + c
y
The constant of integration is determined by the assumption that
(0, a) is the starting point of the curve. Substituting these coordi-
nates in the above equation, we find C = 0.
Ex. 2. The equiangular spi-
ral is a curve so constructed
that the angle between the ra-
dius vector to any point and
the tangent at the same point
is constant. Find its equation.
Ex. 3. Determine the curve
having the property that the
line drawn from the foot of
any ordinate of the curve per-
pendicular to the correspond-
ing tangent is of constant
length a.
If the angle which the
Fig. 81
GEOMETRICAL APPLICATIONS 317
tangent makes with the a>axis is denoted by <f>, it is at once evident
(Fig. 81) that
a . 1 1
- = cos </>
VI + tan** A /l+(^) !
From this follows
x - i g ( ?y + Vf - a 2 ) + C.
When the tangent is parallel to the x-axis, the ordinate itself is the
perpendicular a. If this ordinate is chosen for the y-axis, the point
(0, a) is a point of the curve, and hence
C = - log a.
The equation can accordingly be written
v + ^v ~ a ' 2 = c «. (i)
a
From this follows, by taking the reciprocal of both members,
y + Vyt- a 2
whence, on rationalizing the denominator,
y - Vy2 _ q a _ ^
a
o
Adding (1) and (2) and dividing by -, we obtain
(2)
y = |(c« + « «),
which is the equation of the catenary.
Ex. 4. Find the equation of the curve for which the polar subnor-
mal is proportional to (is a times) the sine of the vectorial angle.
318 INTEGRAL CALCULUS
Ex. 5. Find the equation, in rectangular coordinates, of the curve
having the property that the subnormal for any point of the curve
is proportional to the abscissa.
Ex. 6. Find the equation in polar coordinates of the curve for
which the angle between the radius vector and the tangent is n times
the vectorial angle. What is the curve when n = 1 ? When n = | ?
Ex. 7. Find the rectangular equation of the curve for which the
slope of the tangent varies as the ordinate of the point of contact.
Ex. 8. Find the equation of the curve for which the polar sub-
tangent is proportional to the length of the radius vector.
Ex. 9. Find the volume generated by the revolution of the trac-
trix (see Ex. 1) about the positive x-axis.
Ex. 10. Find the area of the surface of the revolution described
in Ex. 9.
Ex. 11. Find the length of the tractrix from the cusp (the point
(0, a)) to the point (an, yi).
Ex. 12. Derive the following formulas for the length of arc s of a
twisted curve, in space of three dimensions, limited by the points
(#i, yi, zi), (#2, yi-, Z2), the coordinates being rectangular:
-i:v 1+ (gr + (i) a -£v 1+ (i)v(i)>
Ex. 13. Using the formula of Ex. 12, find the length of the helix
x = a cos t, y = a sin t, z = bt,
in which a and b are constants, and t is a variable parameter.
Ex. 14. A plate of steel is \ inch thick and has the form of a right
segment of a parabola. It weighs 490 lb. per cubic foot. Find the
total weight of a plate 30 in. broad and 16 in. long.
Take the equation of the parabola in the form y 2 = ±px. Since
y= 15 when x — 16, we may find the value of p by substituting these
GEOMETRICAL APPLICATIONS
319
coordinates in the assumed equation, namely, 4/>
the parabolic plate is therefore
*tf. The area of
/*16 1
2 \ l £ x^ dx sq. in.
The volume and hence the weight are now
readily obtainable.
Ex. 15. A plate of wrought iron of heavi-
ness 480 lb. per cubic foot is \ in. thick and
is bounded by three straight edges at right
angles to each other, as shown in the figure,
while the curved boundary is a hyperbola Fig. 82
with the equation (x + 5) y = 40, the base of the figure being on the
ar-axis. Calculate the weight.
Ex. 16. A metal plate, in the form of an
equilateral triangle, is \ in. thick and has an
altitude of 4 in. Any very narrow vertical
strip, as AB, of length 2y and width Ax, is
of nearly uniform density- The density varies
from one strip to another in such a way that Fig. 83
the weight y per cubic inch is determined by the condition
= 0.26 1 +
100 \
9 + xV'
Find the weight of the plate.
[Hint. Calculate the weight of the strip AB, then take the limit
of the sum of all such strips con- T
tained in the figure.] r —
Ex. 17. A trapezoidal plate ABCD
is | in. thick. The weight y per cubic
inch is constant along any vertical
line, but varies with x according to
the law
y = 0.05 x 1 oz. per cubic inch.
320 INTEGRAL CALCULUS
The first strip DA is 4 in. from the origin. What altitude h must
be adopted for the trapezoid in order that the total weight of the
plate may be just three ounces?
Ex. 18. The frustum of a paraboloid of revolution has vertical par-
allel bases five inches apart. The equation of the meridian curve, with
the inch as the linear unit, is y = Vx. The heaviness y is constant
over a vertical plane section, but varies with x according to the law
y = 0.06 VlOO — x 2 lb. per cubic inch. Find the total weight from
x = 4 to x = 9.
CHAPTER VIII
SUCCESSIVE INTEGRATION
156. Functions of a single variable. Thus far we have con-
sidered the problem of finding the function y of x when -&■
dx
only is given. It is now proposed to find y when its nth.
d"v
derivative — & is given.
dx n
The mode of procedure is evident. First find the function
— '" which has — - for its derivative. Then, by integrating
dx n ~ l dx n
d n ~ 2 v
the result, determine -„, and so on until after n successive
dx n ~ 2
integrations the required result is found. As an arbitrary
constant should be added after each integration in order to
obtain the most general solution, the function y will contain
n arbitrary constants.
Ex.1. Given ^ = I,findy.
dx 3 x 3
Integration of — with respect to x gives
dx 1 2 & T l '
A second integration gives,
dx 2 x
and finally y = \ log x + \ Cix 2 + C 2 x + C 8 .
EL. CALC— 21 321
322 INTEGRAL CALCULUS
The triple integration required in this example will be symbolized by
which will be called the triple integral of — with respect to x.
x s
Ex. 2. Determine the curves having the property that the radius
of curvature at any point P is proportional to the cube of the secant
of the angle which the tangent at P makes with a fixed line.
If a system of rectangular axes is chosen with the given line for
x-axis, it follows from equation (6), p. 173, and from Art. 42, that
\dxl J i r 1 ,(dyyii
~y =«L + UJ J '
dx 2
in which a is an arbitrary constant. This equation reduces to
£* = «,
dx*
from which follows
y = jjf[a(dxy = «[f + Cix + e 2 ],
Ci and Cz being constants of integration. Hence the required curves
are the parabolas having axes parallel to the ?/-axis.
The existence of the two arbitrary constants Ci, C 2 in the preceding-
equation makes it possible to impose further conditions. Suppose,
for example, it be required to determine the curve having the prop-
erty already specified, and having besides a maximum (or a minimum)
point at (1, 0).
Since at such a point ( -2L = 0, it follows that
dx
= «(1+ Ci),
whence C\ = — 1.
SUCCESSIVE INTEGRATION 323
Also, by substituting (1, 0) in the equation of the curve,
= a{\ - 1 + CV),
Accordingly the required curve is
y=l(x-iy.
Ex. 3. Find the equation (in rectangular coordinates) of the
curves having the property that the radius of curvature is equal to
the cube of the tangent length.
[Hint. Take y as the independent variable.]
Ex. 4. A particle moves along a path in a plane such that the
slope of the line tangent at the moving point changes at a rate pro-
portional to the reciprocal of the abscissa of that point. Find the
equation of the curve.
Ex. 5. A particle starting at rest from a point P moves under the
action of a force such that the acceleratiop (cf. Ex. 14, p. 77) at each
instant of time is proportional to (is k times) the square root of the
time. How far will the particle move in the time /?
Ex. 6. In connection with a certain curve referred to rectangular
axes, we know in advance that it passes through a point A on the
y-axis at a distance 1.12 in. above the origin. It also passes through
a point B of the first quadrant which is at a distance of 12 in. from
the y-axis, and the slope of the tangent to the curve at this point is
0.09. At each point P of the curve the second derivative of y satis-
fies the relation
— { = 0.0012 x.
It is required to find the general expression (in terms of x) of the
ordinate and the slope of the tangent line for any point P of the
curve. In particular, find the ordinate and slope when x = 20 in.
324 INTEGRAL CALCULUS
Ex. 7. For a certain curve ADN situated in the first quadrant we
have given
1000^ = 1.5 -0.276 x.
dx 2
The point A has the coordinates (0, 0.04) and the abscissa of D is
10. At the point B of the curve, whose abscissa is 5, the slope of the
tangent line is 0.002.
A second curve DC is tangent to the first at the point D, and for
each point of it we know that
1000 ^ = 0.2 x -0.115.
dx 2
Find the equations of both curves.
157. Integration of functions of several variables. When
functions of two or more variables are under consideration,
the process of differentiation can in general be performed
with respect to any one of the variables, while the others
are treated as constant during the differentiation. A repeti-
tion of this process gives rise to the notion of successive
partial differentiation with respect to one or several of the
variables involved in the given function. [Cf. Arts. 62, 67.]
The reverse process readily suggests itself, and presents
the problem : Given a partial (first, or higher) derivative of a
function of several variables with respect to one or more of these
variables, to find the original function.
This problem is solved by means of the ordinary processes
of integration, but the added constant of integration has a
new meaning. This can be made clear by an example.
Suppose u is an unknown function of x and y such that
dx J
SUCCESSIVE INTEGRATION 325
Integrate this with respect to x alone, treating y at the
same time as though it were constant. This gives
u = x 2 -f- 2 xy -f- <f>,
in which <f> is an added constant of integration. But since
y is regarded as constant during this integration, there is
nothing to prevent <£ from depending on it. This depend-
ence may be indicated by writing <f>(y) in the place of <f>.
Hence the most general function having 2x + 2y for its
partial derivative with respect to x is
u = x 2 + 2xy+<f>(y),
in which <f>(y) is an entirely arbitrary function of y.
Again, suppose
d 2 u 2 2
= x z y .
dxdy
Integrating first with respect to y, x being treated as though
it were constant during this integration, we find
where i{/(x) is an arbitrary function of x, and is to be regarded
as an added constant for the integration with respect to y.
Integrate the result with respect to x, treating y as constant-
Then
u = i «V + *0) + $>(?/).
Here ®(y), the constant of integration with respect to x,
is an arbitrary function of y, while
*(x) = J ij/(x)dx.
Since if/(x) is an arbitrary function of x, so also is <l'(x).
326 INTEGRAL CALCULUS
158. Integration of a total differential. The total differential
of a function u depending on two variables has been defined
(Art. 63) by the formula
, du 7 . du -
du = — dx-\ ay.
dx dy
The question now presents itself: Given a differential ex-
pression of the form
Pdx+Qdy, (1)
wherein P and Q are functions of x and y, does there exist
a function u of the same variables having (1) for its total
differential f
It is easy to see that in general such a function does riot
exist. For, in order that (1) may be a total differential of a
function u, it is evidently necessary that P and Q have the
forms
P=* Q = p. (2)
dx ay
What relation, then, must exist between P and Q in order
that the conditions (2) may be satisfied? This is easily
found as follows. Differentiate the first equation of 2 with
respect to y, and the second with respect to x. This gives
dP = d 2 u dQ = d 2 u
dy dy dx' dx dxdy
from which follows (Art. 68)
8P = 8Q (3
By dx
This is the relation sought.
The next step is to find the function u by integration. It
is easier to make this process clear by an illustration.
SUCCESSIVE INTEGRATION 327
Given (2 x + 2y + 2)dx +(2y + 2 x + 2)dy,
find the function w having this as its total differential.
Since P=2x + 2y + 2, Q=2y + 2x + 2,
it is found by differentiation that
^=2 and 98-2,
dy ox
hence the necessary relation (3) is satisfied.
From (2) it follows that
p = 2x + 2y + 2.
ox
Integrating this with respect to x alone gives
u = x 2 + 2xy + 2x + <£(?/). (4)
It now remains to determine the function <j> (y) so that
d f[=Q]=2y + 2x + 2. (5)
Differentiating (4) with respect to y alone gives
1^ = 2*+^),
dy
where <t>'(y) denotes the derivative of <f>(y) with respect to y.
The comparison of this result with (5) gives
2y + 2x+2 = 2x + <l>\y) 1
or <f>'(y) = 2y + 2, (6)
whence, by integrating with respect to y,
4>(y)=y 2 + 2y + C,
in which O is an arbitrary constant with respect to both x
and y.
Hence u = x 2 + 2 xy + 2 x + if + 2 y -f C.
328 INTEGRAL CALCULUS
EXERCISES
Determine in each of the following cases the function u having the
given expression for its total differential :
1. y dx + x dy.
2. sin x cos y dx + cos x sin y dy.
3. y dx — x dy.
4 ydx-xdy
xy
5. (3 x 2 - 3 ay)dx + (3 y 2 - 3 ax) dy.
g y dx x dy
x' 1 + y' 1 y' 2 + x' 2
7. (2 x 2 + 2xy + 5) dx + (x 2 -\- y 2 - y) dy.
8. O 4 + i/ 4 + x 2 - y 2 ) dx + (1 y s x - 2 xy + y - y 2 + 2) dy. <~
159. Multiple integrals. The integration of — — was con-
dx ay
sidered in Art. 157. If F(x, y) is written for the given func-
tion, the required integration will be represented by the symbol
= || F(x, y).dx dy,
and the function sought will be called the double integral of
F(x, y) with respect to x and y.
Likewise I F{x, y, z) dx dy dz
will be called the triple integral of F(x, y, z). It represents
d 3 u
the function u whose third partial derivative — is the
dx dy oz
given function F(x, y, z). It will be understood in what fol-
lows that the order of integration is from left to right, that is,
SUCCESSIVE INTEGRATION 329
we integrate first with respect to the left-hand variable x, then
with respect to y, and lastly with respect to z.
Such integrals (double, triple, etc.) will be referred to in
general as multiple integrals.
160. Definite multiple integrals. The idea of a multiple
integral may be further extended so as to include the notion
of a definite multiple integral in which limits of integration
may be assigned to each variable.
Thus the integral I I x 2 y s dy dx will mean that x 2 )/ is to
be integrated first with respect to y between the limits and 2.
This gives
►2
x-y* dy =4 x 2 .
j:
The result so obtained is to be integrated, with respect to x
between the limits a and b, which leads to
x
h 4:X 2 dx = ±(b 3 -a 3 )
as the value of the given definite double integral.
In general the expression
x'j;
Fix, y)dydx
will be used as the symbol of a definite double integral. It
will be understood that the integral signs with their attached
limits are always to be read from right to left, so that in the
above integral the limits for y are b and b', while those for x
are a and a'.
Since x is treated as constant in the integration with re-
spect to y, the limits for y may be functions of x. Consider,
330 INTEGRAL CALCULUS
for example, the integral J J xydydx. The first integra-
tion (with respect to y) gives
Jxy dy = x
, x 2 x\ X s — X 2
By integrating this result with respect to x between limits
and 1 the given integral is found to have the value — ^.
EXERCISES
Evaluate the following definite integrals :
1. f * ( 2 xcos (xy)dydx. 5 - j/jo ' p 2 sin dp dO.
2. j o j o x 2 ^ <Zx. 6. J o j y Vx*, - y^ dx dy.
3 r« f l °ev dxdy 7 f 2 f « C xs/z xdy dxdz
' Ji Jo y ' Ji J<> Jo a; 2 + # 2 *
4
'dzdydx
. J j^sec 2 ^),/,^. 8. J ( J x J o - +y +
161. Plane areas by double integration. The area bounded
by a plane curve (or by several curves) can be readily ex-
pressed in the form of a definite double integral. An illus-
trative example will explain the method.
Ex. 1. Find by double integration the area of the circle
( x _ fl )2 + (y - b) 2 = r 2 .
Imagine the given area divided into rectangles by a series of lines
parallel to the y-axis at equal distances Ax, and a series of lines
parallel to the x-axis at equal distances A?/.
The area of one of these rectangles is Ay • Ax. This is called the
element of area. The sum of all the rectangles interior to the circle
SUCCESSIVE INTEGRATION
331
will be less than the area required by the amount contained in the
small subdivisions which border the circumference of the circle.
All these neglected portions are contained within a ring bounded
by the given circle and a circle concentric with it, whose radius
is less than r by the
length of diagonal of
an element of area,
that is, of radius
r - V(A;r) 2 + (Ay)K
In other words, the
amount neglected is
less than the area of a
circular ring whose
width is
V(Ax) 2 + (A*/) 2
and which therefore
approaches zero simul- ! r " 5
taneously with Ax and Ay. Hence the area of the circle is the limit
of the sum of all the elements of area included within it.
To find the value of the limit of this sum it is convenient
first to add together all the elements contained between two con-
secutive parallels. Let P X P 2 be one of these parallels having the
direction of the ar-axis. Then y remains constant while x varies
from a — vV 2 - (y — b) 2 (the value of the abscissa at P\) to
a + vV 2 —(y — b)' 1 (the value at P 2 ). The limit, as Ax approaches
zero, of the sum of rectangles in the strip from PiP 2 is evidently
A#[limit of sum (Ax + Ax +••■)] = Ay I
a+vV-Cy-fi) 2
;-\/r2-(y-6) 2
dx. (1)
Now find the limit of the sum of all such strips contained within
the circle. This requires the determination of the limit of the sum
of terms such as (1) for the different values of y corresponding to
the different strips. Since y begins at the lowest point A with the
332 INTEGRAL CALCULUS
value b — r, and increases to b + r, the value reached at B, the final
expression for the area is
rb+r i- /*a+v'r 2 -(2/-&)2 ~1 fb+r r a +\Zr^-(y~b^
\ \ . . dx \dy = \ \ , dx dy.
J b-r LJa-VrZ-(y-b)* J Jh-r ^a-Vr 2 -(y-b) 2
Integrating first with respect to x gives
e-w*=s=ir i>^p^ = 2V r ,_ (y _ 6)2 ,
J a -vV2-(y-&)2 Ja-\/»*-(y-6)' w '
This result is then integrated with respect to y, giving
f >+r 2Vr 2 - (y - b) 2 dy = (y-b)Vr 2 - (y - by + r 2
sm ^ —irr
r
J»-
If the summation had begun by adding the rectangles in a strip
parallel to the y-axis, and then adding all of these strips, the expres-
sion for the area would take the form
) ) , dydx.
It is seen from the last result that the order of integration in a
double integral can be changed if the limits of integration are properly
modified at the same time.
Ex. 2. Find the area which is included between the two parabolas
y 2 = 9 x and y 2 = 72 - 9 x.
Ex. 3. Find the area between y 2 = 5a; and y = x.
Ex. 4. Find by double integration the area of the segment of the
circle x 2 + y 2 = 16 cut off by the line x + y = 4.
Ex. 5. Find the area between the two curves
y 8 = x and y = x z .
Ex. 6. Find the area between the two curves
y 2 = x s and y 2 = x.
Ex. 7. Find by double integration the area
_J2 of one loop of the polar curve p = a sin 2 0.
Fi<;. W Imagine the area divided into small ele-
SUCCESSIVE INTEGRATION 333
merits by means of concentric circles whose radii vary by equal
increments Ap and by means of radii drawn from the origin, the
angle between two consecutive radii being A0. (See Fig. 86.)
The area of an arbitrary element may be expressed as the differ-
ence of two circular sectors with a common angle A0 and with radii
p + Ap and p respectively. That is,
element of area = \(p + Ap) 2 A0 - \ p 2 A0
= pA0Ap + iA0(Ap) 2 .
The sum of all the complete elements within the loop may then be
represented by the formula
^pA0Ap + i^A0(Ap) 2 .
Reasoning precisely as in Ex. 1, we find the limit of the first sum
to be
r\ /-a sin 2
The second sum may be written \ Ap2 A0 Ap, hence its limit is
n
I • lira Ap • lim ^A(9 Ap= i * ° * J ~ Jo d P dB = °'
Following the analogy of Ex. 1, we can easily see that all the
neglected incomplete elements of area lie within a narrow band along
the boundary of the given area, the width of which band approaches
0. Their sum therefore approaches zero in passing to the limit.
• It follows from the preceding discussion that the general formula
for area in polar coordinates is
§§pdpdO,
the limits of integration being determined by the boundary of the
given area.
Ex. 8. Find by double integration the area of the cardioid
p = a(l — cos 0).
334 INTEGRAL CALCULUS
Ex. 9. Find the area of the lemniscate p 2 = a 2 cos 2 0.
Ex. 10. Express by double integrals the three areas between the
cardioid (Ex. 8) and the circle p = a.
Ex. 11. Find by double integration the area of the triangle whose
vertices have the rectangular coordinates (5, 2), (—3, 6), (7, 6).
Ex. 12. Find the area common to the two circles
x 1 - 8 x + y 2 - 8 y + 28 = 0,
x 2 - 8a:+ y 1 - ±y + 16 =.0.
162. Volumes. The volume bounded by one or more surfaces
can be expressed as a triple integral when the equations of the
bounding surfaces are given.
Let it be required to find the volume bounded by the surface
ABC (Fig. 87) whose equation is z—f(x, y), and by the three
coordinate planes.
Imagine the figure divided into small equal rectangular
parallelopipeds by means of three series of planes, the first
series parallel to the ?/z-plane at equal distances Ax, the second
parallel to the ccz-plane at equal distances Ay, and the third
parallel to the a?2/-plane at equal distances Az. The volume
of such a rectangular solid is Ax Ay Az ; it is called the element
of volume. The limit of the sum of all such elements con-
tained in OABC is the volume required, provided that the
bounding surface ABC is continuous. For the sum of the
neglected incomplete elements, which border the surface, is
less than the volume of a shell whose outside boundary is
the given surface and whose thickness is V(A#) 2 -f-(Ay) 2 -f-(Az) 2 ,
the diagonal of the element of volume. Hence the error ap-
proaches zero as the three increments diminish.
To effect this summation, add first all the elements in a
SUCCESSIVE INTEGRATION
33;
vertical column. This corresponds to integrating with respect
to z (x and y remaining constant) from zero to f(x, y). , Then
add all such vertical columns contained between two consecu-
tive planes parallel to the ?/z-plaiie (x remaining constant),
which corresponds to an integration with respect to y from
y = to the value attained on the boundary of the curve AB.
Fig 87.
This value of y is found by solving the equation f(x, y) = 0.
Finally, add all such plates for values of x varying from zero
to its value at A. The result is expressed by the integral
XX X dzd ' Jdx >
336 INTEGRAL CALCULUS
in which <f>(x) is the result of solving the equation f(x, y) =
for y, and a is the ^-coordinate of A.
Ex. 1. Find the volume of the sphere of radius a.
The equation of the sphere is
x 2 + ?/ 2 + z % _ a2?
or z = Va 2 — x 2 — y'K
Since the coordinate planes divide the volume into eight equal
portions, it is sufficient to find the volume in the first octant and
multiply the result by 8.
The volume being divided into equal rectangular solids as described
above, the integration with respect to z is equivalent to finding the
limit of the sum of all the elements contained in any vertical column.
The limits of the integration with respect to z are the values of z
corresponding to the bottom and the top of such a column, namely,
z — 0, and z = Va 2 — x 2 — y 2 , since the point at the top is on the sur-
face of the sphere.
The limits of integration with respect to y are found to be y =
(the value at the x-axis), and y — Va 2 — x 2 (the value of y at the cir-
cumference of the circle a 2 — x 2 — y 2 = 0, in which the sphere is cut
by the a^-plane).
Finally, the limiting values for x are zero and a, the latter being
the distance from the origin to the point in which the sphere inter-
sects the x-axis. Hence
V[= volume of sphere] = 8 j^ £ V " 2 -* 2 ^ V " 2 -**-** dz dy dx.
Integration with respect to z gives
V = 8 i \ Va 2 - x 2 - y 2 dy dx;
then with respect to y and x,
V = 8 j> [| V^^V, + ^^-^-J^
SUCCESSIVE INTEGRATION 337
Ex. 2. Find the volume of one of the wedges cut from the cylinder
x 2 + y 2 = a 2 by the planes 2 = and z = mx.
Ex. 3. Find the volume common to two right circular cylinders
of the same radius a whose axes intersect at right angles.
Ex. 4. Find the volume of the cylinder (x — 1) 2 + (y — l) 2 = 1
limited by the plane z = 0, and the hyperbolic paraboloid z = xy.
Ex. 5. Find the volume of the ellipsoid
a 2 b 2 c 2
Ex. 6. Find the volume of that portion of the elliptic paraboloid
z=l-*-£
a 2 b 2
which is cut off by the plane z = 0.
Ex. 7. Find by triple integration the volume of the tetrahedron
formed by the three coordinate planes and the plane x + 2 y + 3z — 1.
Ex. 8. Find the volume of the elliptic paraboloid 2 y 2 -f 3 z' 1 — 6 x
cut off by the plane x = 2.
el. calc. — 22
Jf-
CHAPTER IX
SOME APPLICATIONS OF INTEGRAL CALCULUS TO
PROBLEMS OF MECHANICS
163. Liquid pressure on a plane vertical wall. The pressure
exerted by the liquid upon any point of a plane vertical wall
is proportional to the depth
Surface B
7 of that point below the sur-
f / face of the fluid. To calculate
\ i y
the pressure upon the entire
wall we divide it into nar-
row horizontal strips of equal
areas Ai. Denote the breadth
FlG - 88 ' of the Jcth strip PQ (Fig. 88),
counting from the top, by h k . The pressure exerted on the
ftth strip is equivalent to the weight of a column of fluid
standing on a base of the same area AA and having an
altitude intermediate between the least depth x and the
greatest depth x + h k of points on the given strip. This
altitude may be represented by x 4- k h k in which k has a
value between and 1. If w denotes the weight of a cubic
unit of the fluid, the pressure on PQ is w(x + k h k ) Ai.
Summing the pressures for all the strips of the wall, we
obtain for the total pressure
^w(x + 6 k h k ) AA.
338
SOME APPLICATIONS TO MECHANICS 339
In order to evaluate this sum we take its limit as &A
approaches zero. This gives, by separating into two terms
and observing that w is constant,
Km y x &A + w lim V #A A A
w
A.4=b0
The second term reduces to zero. For,
]£0A A.4 = AA^e k h k < ^A • H (since 6 k < 1),
in which H denotes the total altitude of the wall ; as A^l =
the right member of this inequality approaches zero. Hence
pressure = iv I x dA.
In order to evaluate the integral, it is most convenient to
make x the variable of integration. Denote by y the width of
the wall at the depth x. Then &A = y k Ax in which y k is a
certain value of y between y and 2/+A?/. (Compare Art. 40.)
Dividing by A£ and passing to the limit we obtain, since
lim y k = y,
dA = dx
dt ~ V dt '
or in the differential notation, dA — y dx. The substitution of
this in the above integral gives
pressure
= w j xy dx,
the limits of integration being the values of x at the top and
the bottom of the given wall or surface.
If the liquid is water and the unit of length is a foot, then
w = 62^ lb.
340 INTEGRAL CALCULUS
EXERCISES
1. Find the pressure on the end of a rectangular tank full of water
that is 10 ft. long, 8 ft. wide, and 5 ft. deep.
2. A watermain 6 ft. in diameter is half full of water. Find the
pressure on the gate that closes the main.
3. A vertical masonry dam in the form of a trapezoid is 200 ft.
long at the surface of the water, 150 ft. long at the bottom, and 60
ft. high. What pressure must it withstand ?
4. A vertical cross section of a trough is a parabola with vertex
downwards, the latus rectum lying in the surface and being 4 ft.
long. Find the pressure on the end of the trough when it is full of
water.
5. One end of an unfinished watermain, 4 ft. in diameter, is closed
by a temporary bulkhead and the water is let in from the reservoir.
Find the pressure on the bulkhead if its center is 40 ft. below the
surface of the water in the reservoir.
164. Center of gravity. (1) For a system of n particles. Let
P 1? P 2 be two particles of matter of masses (or weights) m x and
p m 2 , respectively, and let x lf x 2 be their
O " f *~~ distances from a chosen point O on the
Fig. 89 straight line through them. There
exists a point P such that the segments P X P and PP 2 are in-
versely proportional to the masses of the two points, that is,
PiP = %_ ^
PP 2 %'
Let x represent the distance OP. Then formula (1), expressed
in terms of the abscissas of the points, is
x — x l _m 2
SOME APPLICATIONS TO MECHANICS 341
whence, by solving for x,
m l x 1 -\-m 2 x 2 /ON
m! + m 2
The point P is called the center of gravity, or, the center of
mass, of the system formed by the two points P ly P 2 . If we
imagine the line P X P 2 to consist of a rigid, weightless rod
with the two given particles fastened at its extremities, and if
we suppose this object to rest on the point P as a base, it will
remain in equilibrium, without any tendency in either of the
end points to move downward under the force of gravity.
In other words, the system of two particles is equivalent, as
far as the action of gravity is concerned, to a single particle,
of mass m x -+- m 2 , placed at the point P.
Let P 3 be a third point of mass m 3 situated on the same line
with P x and P 2 . Then the abscissa x of the center of gravity
of the system of three points may be found by calculating the
center of gravity of the pair P 3 and P (the center of gravity
for P lf P 2 ), the mass of P being taken as m 1 -f ra 2 , the sum of
the masses of P x and P 2 . This gives
Ox + m 2 ) 1 l ; — =2 + m 3 x 3
w&i + m 2 m^ 4- m 2 x 2 -f- m 3 X3 m
(m x -+- WI2) + m s wh + m 2 + n h
x =
In like manner the center of gravity for any number n of
particles situated on a straight line is given by the formula
- _ m^ + m 2 x 2 -{ + m n x n ^ ,^
m 1 + m 2 + ••• +m n
If the n particles are not on a straight line but are situated
in the same plane at the points (a^, y^), (x 2 , y 2 ), • ••, (x n1 y n ),
342 INTEGRAL CALCULUS
then the center of gravity of the system has its abscissa given
by (3) and its ordinate y is
y = WW + m 2 y 2 + h m^ ri
m^mz-l \-m n
If the n particles are not situated in one plane, there will
be a third and similar formula for z.
(2) For a continuous solid. Imagine the solid divided up
into small elements, precisely as in determining its volume, by
means of three series of planes parallel to the coordinate
planes and at distances Ax, Ay, Az. If we regard any par-
ticular element as being very nearly of uniform density, then
the mass of an arbitrary element is approximately p Ax Ay Az,
in which p is the weight of a cubic unit of homogeneous mat-
ter having the same density as the given element. This num-
ber p is usually called the density. For a finite number of
elements the ^-coordinate of the center of gravity is determined
approximately by (3) in the form
(pigq + P1F2 + — + P A) Aa; Ay Az
0>i + p 2 + >~ + p n )AxAyAz
in which x ly x 2 , •• are the abscissas for the different elements
and p lt p 2 , ••• are their densities. The abscissa of the center of
gravity of the given continuous solid is obtained by making
Ax, Ay, Az approach zero as a limit.* This gives
J I I pxdxdydz
I I Ipdxdydz
* A proof of this statement will be found in Art. 166.
SOME APPLICATIONS TO MECHANICS 343
the limits of integration being determined just as in calculat-
ing the volume of the solid. If the solid is homogeneous,
p is constant and cancels out of numerator and denominator.
Otherwise, it is a function of x, y, z.
In precisely the same manner the values of y and z are
obtained. The coordinates of the center of gravity are thus
found to be
x
M
m j j j 9 x dx dy dz ' y = mSSS p y dx dy dz '
'^SSS pzdxdydZi
in which p is the density at the point (x, y, z) and M is the
total mass of the given solid, that is,
M= J J (pdxdydz.
The coordinates of the center of gravity of a plane area are
found in like manner to be
x = — j Jpxdxdy, T/ = ~JJpydxdy, M=Jjpdxdy.
EXERCISES
In the following problems p is understood to be constant unless
otherwise specified. The abbreviation C. G. will be used for " center
of gravity."
1. Find the C. G. of the tetrahedron whose faces are the three
coordinate planes and the plane x + 2y + 3z = 6.
2. Find the C. G. of the volume bounded by the coordinate planes
the plane x -f y = 1, and the surface z = xy.
344 INTEGRAL CALCULUS
3. Find the C. G. of the volume bounded by the hyperboloid
- — ^ — — = 1 and the plane x = k, k > a.
a* b 2 c 2 r
4. Find the C. G. of the semiellipsoid on the positive side of the
xy-plane, the equation of the ellipsoidal surface being — -f ^- -\ — = 1.
a 2 b 2 c 2
5. Find the C. G. of a thin hemispherical shell of thickness h
bounded by two concentric hemispheres of radii a and a + h.
6. A hemispherical iron bowl of uniform thickness a is filled with
water. If the density of iron is seven times that of water, find the
C. G., supposing the radius of the interior of the bowl to be r.
[Hint. Find the C. G. of the iron bowl by means of Ex. 5. Find
the C. G. of the hemisphere of water and combine the centers of grav-
ity of the iron and the water by means of (2).]
7. Show that the C. G. of a triangular plate one inch thick is one
half inch below the intersection of the medians of the upper face.
8. Find the C. G. of a T-iron one inch thick, the vertical bar being
a inches wide and b inches high, and the horizontal bar a' inches wide
and V inches long.
9. Find the C. G. of a sector of a circle of radius a and angle 0.
10. Find the C. G. of the segment of the circle x 2 + y* = r 2 cut off
by the line x = a, < a < r.
11. Find the C. G. of the quadrant of an ellipse.
12. Find the C. G. of the segment of an ellipse cut off by the
chord joining the extremities of the major and minor axes.
13. Find the C. G. of the area bounded by the parabola
Vx + y/y = Va
and the line x + y = a.
14. Prove that the volume of a solid of revolution is equal to the
product of the generating area by the length of path described by its
center of gravity.
SOME APPLICATIONS TO MECHANICS 345
15. Find the C. G. of an octant of an ellipsoidal mass.
16. Find the C. G. of the preceding mass when the density varies
directly as the distance from the plane x = 0.
17. Find the C. G. of an octant of a sphere. From this result find
the C. G of an octant of a spherical shell of thickness h and inner
radius a.
18. Find the C. G. of an octant of a sphere if the density varies
directly as the distance from the center of the sphere.
[Hint. Divide up into thin concentric shells of equal thickness h,
the density of a particular shell being regarded as constant. Let A.
denote the radius of an arbitrary shell, X the distance of its C. G. from
the origin, and m its mass. Calculate X in terms of A by means of
Ex. 17, measuring it on a line equally inclined to the x, y, z axes.
Then use the different values of X in place of xi, X2, ••• , formula (3),
and pass to the limit.
19. Find the C. G. of a right circular cone of altitude h and base-
radius r.
This problem can be solved by single integration if we suppose the
solid divided up into thin plates of equal thickness by means of
planes parallel to the base. Then find the approximate expression
for the C. G. of any plate, apply (3), and pass to the limit.
20. Find the C. G. of the portion of the elliptical cone
— a + f- = (z — l) 2 between the vertex (0, 0, 1) and the zv-plane.
a 2 b 2
21. A cone of vertical angle 2 has its vertex on the surface of a
sphere, its axis passing through the center of the sphere.
(a) Find the C. G. of the mass outside the cone and inside the
sphere.
(b) Find the C. G. of the mass inside the sphere and inside the
cone.
346 INTEGRAL CALCULUS
165. Moment of Inertia. The moment of inertia of a small
particle of matter of mass m about an axis is defined as the
product of the mass by the square of the distance of the
particle from the axis. It measures the resistance of the par-
ticle to rotation about the axis.
To find the moment of inertia of a homogeneous solid body,
imagine it divided up into small rectangular blocks (or ele-
ments) of dimensions Asc, Ay, Az. Then the moment of
inertia of a single element about the x'-axis is approximately
p(y 2 + z~)& x ty Az,
in which p is the density, that is, it is the weight of a cubic
unit of the given solid. Summing up these elements over the
whole body and taking the limit of the sum, we find the
moment of inertia to be *
///<
p(y 2 + z 2 )dxdydz f (4)
the triple integral being extended over the entire solid, just
as was done in finding its volume.
If the solid is not homogeneous, then p is variable. Its
value at a specified point P of the given body is equal to the
weight of a homogeneous cubic unit of matter having the
same density throughout as the particle of matter at the point
P. It is a function of x, y, z which is to be determined by the
conditions of the given problem.
Similarly, the moment of inertia of a plane area about the
a>axis is defined as the limit of the sum of terms formed by
multiplying each element of area by the square of its distance
from the axis. This gives the formula
y 2 dx dy.
//•
* See the next article for a completion of the proof.
SOME APPLICATIONS TO MECHANICS 347
EXERCISES
In the following problems M.I. is used for brevity to denote
'•• moment of inertia." Unless the contrary is stated, the body is
homogeneous and of density p.
1. Find the M. I. of a rectangular parallelopiped of dimensions a,
b, c about an edge a.
Take three edges a, b, c meeting in a common point as the x, y, z
axes, respectively. Then by formula (4) the M. I. is
"!oTJ> + -~ 2) *"- yrf *-
2. Find the M. I. of a circular cylinder of radius a and altitude h
about its axis.
3. Find the M.I. of the cylinder of Ex. 2 about a line perpendicu-
lar to, and bisecting, the axis.
4. Find the M. I. of a circular cone of altitude a and base-radius
r about its axis.
Hint. If the axis of the cone is taken for the x-axis and its vertex
at the origin, the equation of the conical surface is
x 2 _ y 2 + z 2 "I
a 2 r 2 J
5. Find the M. I. of an elliptical right cylinder about its longi-
tudinal axis, the axes of the elliptical bases being 2 a, 2 b and the
altitude h.
6. Find the M.I. of the preceding solid about the minor axis of
an elliptical base.
7. Find the M.I. of the same body about a line bisecting the
longitudinal axis and parallel to the major axes of the elliptical
348 INTEGRAL CALCULUS
8. Find the M. I. of a sphere about a diameter. Hence find the
M. I. of a spherical shell of uniform thickness h about a diameter, as-
suming that the M. I. of a solid consisting of two parts is the sum of
the moments of the separate parts.
9. Find the M.I. of a spherical solid of radius r about a diameter
if the density varies directly as the nth power of the distance from
the center.
[Hint. Imagine the sphere divided into concentric shells of equal
thickness AA and denote by X the interior radius of any shell. Using
the preceding problem, write down the element of M. L, that is, the
M.I. of the shell of radius X and thickness AA.. Take the limit of
the sum of all such elements as AA. = 0. The required M. I. is thus
obtained by a single integration.]
10. Find the M. I. of a cube of edge a about its diagonal.
[Hint. Take three faces of the cube as coordinate planes. Obtain
an expression for the square of the distance from any point (#, y, z)
to the diagonal of the cube that passes through the origin. This,
multiplied by AxAyAz, will be the element of M.I. Then take the
limit of the sum.]
11. Find the M. I. of a cylindrical shell, of length a, about its
axis, the radius of the inner surface being r and that of the outer
surface being R.
12. Find the M. I. of a rectangle of sides a, b about the side b.
13. Find the M. I. of a triangle of base b and altitude h about an
axis through a vertex parallel to the opposite side.
14. Find the M. I. of a circle of radius a about a diameter.
166. Duhamel's Theorem. In order to complete the proof
of the formulas for center of gravity and moment of inertia,
we make use of the following theorem which is of very general
use in applications of the Integral Calculus.
SOME APPLICATIONS TO MECHANICS 349
Duhamel's Theorem. Let a ly ol>, •••, a n be positive variables,
each of which approaches zero as n increases without limit, and
suppose that the sum ai + c^-j- ••• + « n approaches a finite limit
as n = oc . Let /3i, (3 2 , •••, ft be variables having the same prop-
erty as the a's and such that ™ ^ = 1 for k=l, 2, • •-, n.
Then "*
^(ft + ft* - +« = n 1 ^ oo (« 1 + « 2 + ...+« B ).
Since lim & = 1, we may write & in the form l + e k in
« = «>«* « fc
which e k approaches zero as n = oc. Hence,
ft = a k + e t %,
and therefore 2 ft = S "* + S **<**•
Let e denote the positive value of the numerically greatest
term of the series e y , c 2 , •••, e„. Then we have the inequalities
— €a 1 '^€ l a l < 4-««u
— ca 2 5^ e 2 a 2 ^ 4- ca 2 ,
and by adding we obtain
— «(«! + «o 4- • • • 4- «„) ^ 2 e *«* ^ + «(«i 4- « 2 4- h <*»)•
Now let n increase without limit. Since by hypothesis e =
and («!4-«2+ ••• 4- a n ) has a finite limit, it follows that the
first and last members of the preceding inequalities vanish at
350 INTEGRAL CALCULUS
the limit and therefore .
Hence Jf^A- J?.S* #
As an application of the above theorem, consider the sum
occurring in the approximate formula for center of gravity,
nam6ly ' <«% + *%+•••+ PA) AF
in which AF= A# A?/ Az.
Let p k , x k be the minimum, and p k ", x k " the maximum values
of p, x in the ftth element of volume. For brevity write
Pk 'x k '*V=a k , Pk "x k ''±V=(3 k .
Then we have a k <^ p k x k AV^ /3 k
hence, by taking the sum,
6 o "x "
But £-* = ^* * which approaches 1 as n increases since p k ', x k
approach equality with p k ", x k ". Hence
lim X a * = lim ]£ A = lira ^ p a AF.
In obtaining this result no restriction is placed on x k and p k
* A variable which has zero as a limit is often called an infinitesi-
mal. Hence ai, a2, •••, a n are infinitesimals. If we write Sj. = e^, then
„ V^L — = lim e k = 0. When two infinitesimals, 5 and a, are so related that
the ratio of 5 to a has the limit zero, then 5 is said to be infinitesimal with
respect to a, or it is called an infinitesimal of a higher order than a.
Since, by Duhamel's Theorem, lim ^(a*. -f 5*) = lim 2 a *i tn * 8 theorem is
equivalent to saying that the limit of a sum of infinitesimals is not affected
by dropping from each term an infinitesimal of a higher order.
ANSWERS
DIFFERENTIAL CALCULUS
Page 23. Art. 7
2x-2; 2; 0; 1. 6. wx"- 1 .
2. 6x-4.
3.
1
4 a*
4.
4x 3 -
5.
3
X*'
7 x 2 + 2 a;
(x + l) 2 '
« 1-x 2
(X* + 1)2
9. _L.
2Vx
io. -fx-i
Page 24. Art. 8
(6«-4)6*. 3. -i(10x-2). 4. (e—A) (rf-2)
Page
32.
Art.
13
1.
10X9.
g
2 - 6 x - x 2
2.
- 8 x-9.
(a: 2 + 2) 2
3.
4.
c
2Vx
1
Vx3
_ 5 4/ —
4 VX"
1
^9.
10.
11.
3x-f 5
Vx + 2
Va(Vx,— y/a)
5.
2 y/x ( y/x -fa) ( Va + Vx) 3
6.
n(x + a)
nx* 1-1 .
,n-l.
12.
1
7.
Vl-a; 2 (l-x)
8.
a 2
3
13.
1
(a 2 - x 2 )s
2 x(l - x 2 ) + vT^x 2
355
356
ANSWERS
14.
15.
16.
17.
18.
19.
20.
21.
4 oft + 3 aft
4A/x^ a ! + a i
2 x 3 - 4 a
(l-x 2 )^(l+z 2 )*
— 2 nx"- 1
(x n -l) 2 '
_ m (ft + x)+n(g + x)
(a + x)™+! • (b + x)«+!
x 2 (x 3 + 1)T
56x 3 (x 2 + l)i
6
dx
12(m 2 - t* + l) — .
ax
I. 60 M 5(1 4-^2)2^?.
ax
24. w + x^.
ox
du
25. (2m + 6xw) — + 3w 2 + 4x 3 .
dx
dx nu n
(a+z) n (a + se) n + 1
27- 2 wx 3 w ^ + w 2 x 3 — + 3 uWw.
dx dx
30.
6 2 x
&x
Va 2
32. (0,0), (i,--il
\9o 27 a/
\9a' 27a/'
34 (21m 3 - 19m)10x
(7 w 2 + 5)*
35. At right angles at (3, ± 6).
y 2 + 2 xy
2xy + x 2 '
x + a
a
ax-}- b
8x-
■7
4x 2 -7
x + 2
2
1-X 2 ~;
4x
Page 33. Art. 14
3 6(2x-3y) 2 +2(x-y)+l
9(2 x - 3 y)*+ 2(x - y) - 1 '
Page 37. Art. 18
7. ?ix n_1 log x + x M_1 .
8. nx"- 1 log x m + mx n_1 .
x
1 -x*
log x + 1
9.
10.
x 2 -l
1
2 (\/x+l)
11. log e-
12. log 10 e
12xV2 + x — 1
2 V2+x (3 x 2 - V2+x)
2x + 7
*x 2 + 7x*
xlogx
14.
ae ax .
15.
4 e 4x + 5 .
16.
i
(i+x) 2 "
17.
e x
(1 + e*) 2
18.
2/ - 3 x 2 e*
19.
1 - y 2 .
ANSWERS
25.
21ogx
26
1
xlogx
27.
- (log x + 1)
28.
a logxlPg«.
X
357
e? + e~ 3
-(x-l)^(7x 2 +30x-97)
12(x-2)^(x-3)^
2 + x - 5 x 2
e 1
2V1-X
21 1 + e * 31 1 + 3 x 2 - 2 x*
' X+e *' * (l-x 2 )f
. 5 x*(a + 3 x) 2 (a - 2 x)
(a 2 + 2 ax - 12 x 2 ).
22. wx"- 1 ^ + x n a x log a.
23. V«
Vx(a — x)
24. 1 33. («-2«)vY
x(logx) 2 ' • (x-a)*
Page 41. Art. 22
1.
7 cos 7 x.
12.
- 20 x (3 - 5 x 2 ) sec 2 (3- 5 x 2
2.
— 5 sin 5 x.
13.
2 tan x sec 2 x — 2 tan x.
3.
2 x cos x 2 .
14.
secx.
4.
2 cos 2 x cos x — sin 2 :
t sin x.
15.
cot Vx
5.
3 sin 2 x cos x.
2Vx
1 1
6.
10 x 09s 5 x 2 .
16.
log a • a x • sec 2 (a x ).
X 2
7.
14 sin 7 x cos 7 x.
17.
w sin w_1 x sin (n + l)x.
8.
sec 2 x (tan 2 x— 1).
18.
cos2 W ^.
9.
3 sin 2 x cos 2 x — sin 4 x.
ax
10.
sec x (tan x + sec x).
19.
mn sin m_1 nx • cos(w — n)x
COS n+l mx
11.
-16x(l-2x 2 )sin(l
cos(l-2x 2 ) 2 .
-2x 2 ) 2
20.
2
1 4- tan x
358 ANSWERS
21. cos (sin «) cos u^ 26 ' ~ 8 csc2 4 * cot 4 *■
dx
J. 2 ae 8 * sin e 8 * • cos e 8 *.
26. 8 (4x - 3) sec (4x - 3)2
tan (4x - 3) 2 .
27. -2a?oatfrf+? eoV5tanVS -
cose^logx+^i-. 2Vx
28.
y cos xy
xcosx 2 1-xcosxy
VsTnx 2 29. - csc 2 (x + y).
Page 43. Art. 23
4 * 6. 1 11. *
\/l-4x 4 2Vsin- 1 xVl — x 2 Vl — x 2
1 7. 1 • 12. ^*-
Vl -.x 2 e * + erx * Vx 2 - 1
3 3 8. - 1 13. -ZJL.
V 6x-9x 2 xVl-(logx) 2 l+.x 2
o n SPC 2 X .
4. * . 9 /t t • 14. |Vl + cscx.
^1 _ x z VI - tan 2 x
5. _^_. 10. — J=- 15. h
l + x 2 vT^T 2
16. sec 2 x.tan-ix + -^i£L. 17. sin-ix+-
1 + x 2
18.
gtan 1 x 21, . _
1 + x 2 " 2 Vx (x + 1)
2 22.
19. — - e x + e~ x
VI - x*
23
r
25.
L-x 2
2 sin x
26.
27.
Vl — 4 cos 2
- 1
X
2(1 + x 2 )
-1
,. — 2 cos -2 x + w 2 sin 2 x V 1 — x 2
' " x 2 + 1 ' 24. 2. 28. 0.
Page 45. Exercises on Chapter II
1. 6x+ 15 x 2 . 4 a 2 - 2 x 2
2 - 6 15 V^^x 2
x 3 x 4 5. log sin x + x cot x.
3 x - 1 « - a 8
3.
2Vx-3 *Va 2 -x 2
7,
8.
19.
1 J a + a
x 'x — a
ANSWERS 359
x V xj 1— iC 4
2 x 2 - 2 x 4- 1
2(x-x 2 ) 2 ' ""' x
_ e^ 22. 1.
9> = ' cot «• «« „
2 Vu 23. 2 tan « + e 8ec x • sec x tan x.
JO. i-loga. 24. 2^ST.
x 5 a* - y 2
11 -(3x4- a*) 25 2xy 2 + 3x 2
3x 2 + l
(i + x 2 )i '
,12. e*(cos x - sin x). 26. -
3 r + 1
13 - 1^3=^- 27. y+a^-i.
2 xy — x 2
^Vx^n: 27. £
14. J 9R 4 cos (21ocrx 2 -7)
5 + 3cosx 28 k — ^ l '
15. tan-i J*. 29. x^ = 2y.
y a dx y
16. 1 30. x = rnr.
(- 1 + ^) 32. x, y are determined from
17. 4 tan* x. a 2 y _ ± b 2 x and equation
18 logx of curve.
' (1-x) 2 '
33. x = kit ± -.
4 4
5 + 3 cos x 34. tan- 1 2 V2.
Pages 49, 5Q Exercises on Chapter III
1.
72 x.
8.
8 tan x sec 2 x (3 sec 2 x -
-1)
2.
0.
9.
2 cot x esc 2 x.
3.
3!
X 4 '
5!
X6 '
6 sec 4 x - 4 sec 2 x.
10.
11.
12.
16 sin x cos x.
24
4.
5.
(1-x)*
48
X
6.
7.
e* logx + Ai*—^!.
x x 2
2 log x + 3.
13.
14.
sinx.
_ 8(e* - e-*)
360
ANSWERS
15.
16.
17.
18.
19.
21.
8 z 2 e2*.
_4J
x 2 '
a n e ax .
(-l)"w!
(x -!)"+!'
\lmx + ra^Y
(- !)"• (m+ n-1)!
(w— 1) !(a + sc) w +"
m ( - l)n-i . ( W _ 1) ;
(a + x) n
3jp a
2/ 6 '
ft 4
24 - 2 a3 *y
(y 2 -ax) 3 '
25. -yr(*-i) 2 + Q/-n 2 i
a:2( y _i)3
e 2 *.
34. 2"- 1 cos['2a;+ — V
3-
-?/
(2-
-2/) 3
(n-
-1)!
2.
and 1
3.
r , max. ; . — , min.
V3 V3
Page 53. Art. 28
Inc. from -co to i ; dec. from 1 to 1; inc. from 1 to + oo ; |
Two. + 1 at x = \ ± V^ ; - 1 at x = J ± VJ. 4. ± tan-* ^.
Page 60. Art 34
6. — 1, max. ; — £, min.
7. — 2, min. ; 1, max.
2, max. ; 3, min. 8. e, max.
2, min. ; f, max. 9. 2n7r, min. ; also tan-i ± V|
(2 w+J)w, max. ; (2ra + £)7r, for angles in 2d and 3d
min. for all integral values quarter. (2 n + 1) 71- ,
of w. tan -1 ± V|, 1st and 4th
quarter, max.
10. 2, min. ; — 1, max.
11. x = 3m + 4 m 2 + 44w + 4=0.
2(u-l)'
-, mm.
4
Pages 63-67. Exercises on Chapter IV
Two thirds the length of the segment.
The parts are equal.
h 5. -*=• 7. 2r
V3
6. 3 inches.
V3
8. Area is ~
2
ANSWERS 361
9. The side parallel to the wall is double each of the others.
10. The altitude is equal to the diameter of the base.
11. 8 inches.
12. One mile from stopping point.
13.— Most economical per hour at 15 knots.
14. fa.
15. The altitude of the rectangle is equal to the radius.
16. The altitude is equal to the radius of the base.
20
17. — - yards from the nearest point.
> 4 2 - 12 2
19. 15 V2 feet.
20. The diameter of the sphere equals the edge of the cube.
21. ilfeet.
V2
22. Circular arc is double the radius.
23. — , T> being the distance between the centers of the spheres.
24. Arc = 2 *r(l — Vf) .
25. Angle at center of variable circle denned by d — cot 6.
26. The line should be bisected at the given point.
27. The altitude is f the slant height of the cone.
28. (a* + &*>*. 31 - x = a 'P- 34. *=J2jD?feet
29. | a. 32. 20 ft. X »
35. tan 6 = sec 0— tan </>•
30. a + 6. 33. aV\. 36. 6 = 35° 20'.
Pages 76, 77. Art. 39
3. About 3° 58' per second. 5. (3, -\ e ).
4. 120 feet per minute. 6. At 5a/2 miles per hour.
7. (3,6). 9. 2ab. 11. 5 w.
8. At 60°. 10. i 2. 12. 2.
362 ANSWERS
13.
1 and 5.
16.
S ~64'
'~32
17.
a
16'
Page
19. ± 16, =F 12 feet per second.
20. sin • d<t>.
36
21. — :rrL radians per second.
V37*
Page 83. Exercises on Chapter VI
1. -J^+JE, 2V^x, 4tt\/^+ ax, 4wax. 2. 2, 2.
* x y x
3. secx. 5. ,-^(a 2 -a: 2 ). 7. V2 ap.
a z
4. Tra 2 ^ 6 p Vl + (loga) 2 . p
9. 30 7t, 72 ir. 10. f sin d, 90°, 270° ; 2, - 2.
Pages 87-89. Art. 48
I Xix . yiy _ 1 4. (a) x + 2 */ = 4 a,
a 2 6 2 ' y-2x + 3a = 0.
y yi Mb/ y = T2x±3.
2. y = x. (7) */ = x+p, x + y-3i>=0.
3. 2y = 9x -3, 9y + 2x = 29. 5. 3. 6. 4^17.
7. (a) Parallel at points of intersection with ax + hy = 0.
Perpendicular at points of intersection with hx + &?/ = 0.
(j8) Parallel at f—^-z, 8 *y 2 > ) ; perpendicular at X = 0.
(7) Parallel at (—, fL^iiA . perpendicular at (0, 0); (2 a, 0).
8. _ = , i.e. they must be confocal.
a b a' b'
IT
12.
2 2 nx
13. 2c + a . 19. (2p, ±2pV2).
o
Page 95. Art. 51
1. An inflexion at x = y = 2.
, /2o 3a\ /-2o 3o\
ANSWERS
363
8. Point of inflexion at (a, $a), tangent is x + y=-£. Bending
o
changes from negative to positive.
10. (- 1 ,l),(2 ± V3,^^g).
Page 103. Art. 57
1. y = 0, x = a, x = — a.
2. x = 0, x = 2 a, 2/ = a, y = — a.
3. y = a, y=—a ; two imaginary.
4. y = « ; x = c twice.
5. y = — x + - ; two imaginary.
o
6. x = 1 ; one parabolic branch.
8. a* = twice ; one parabolic
branch.
9. x - 0, y - 0, x + y = 0.
10. y = # ; two imaginary.
11. x + y + a=0; two imaginary.
12. y -f sr, = ; two imaginary.
13. x = twice ; x = y, x = — y.
14. y=x, y=— x; two imaginary.
7.- x = — a, y = — 6, y=x + b-a. 15. a; + 2y-0,x+y=l,a--y = — 1.
Page 107. Art. 60
1. yff = 6.
2. Polar subtangent = — , Polar normal = y/a?+p\ Polar subnormal = a.
3. yff = - + 2 0, Subtangent = - p cot 2 0, Tangent = q2p .
2 Va 4 -V
Subnormal =- a ' 2sin2 * , Normal = £
P P
5. ^,2asin2^.tan^.
2 2 2
7. They have a common tangent at the pole ; elsewhere, -.
3. 1.
Page 111. Art. 62
4. (ar + y) cos xy.
5. 1.
Page 115. Art 63
5. \ square units. 6. 5Vlo!
Differs by dxdy.
364 ANSWERS
Page 117 Art. 65
3 _ ax + hy + g . ? 3
hx + by +f ' VlT^'
4. x ~. 8. *=1.
y 3 &> + x
6 . 2l + V y+U. 9. yCco8(gy)-e^-2or1 >
s s x [x + e 2 * — cos {xy)~\
Page 121. Art. 66
2. 8 a; + 8 y - z - 12 = 0. 5. — + JL + JL = at •
£Ci* ^ z\^
3. »=J s= t^=«=» 7 . 008 -i_i<L. 8 . JL.
1 - 4 3 V119 V17
9. 2* 1 * + „»-« 1 . = 0; ^fl = r=H= s t^l.
2 5Ci «/i — *i
10. 2x + 2V3y + 3s = 25, x + s = 5.
Pages 128-130. Art. 72
2. **-2y^ = 0. 11. ^ + ^+y=0.
dy' 2 dy dy 2 dy
Q
r h(f)T
d 2 x
dy 2
4.
fd % x cPx\ dx_Q
\dy s dy 2 ) dy
5.
£* = cos's + 2(-T
dx 2 \dxJ
6.
dy*
8.
^L + y = 0.
du 2
9.
SUO.
12. 2z
d 2 z , n f dz
+ 2(^\
dx 2 \dxj
+ (l-z 2 )2z — +z* = 0.
dx
13. ff+«/=0.
dt 2
14. ^ + ^ = 0.
d£ 3
d« 2 < d«
16.
dl> 17. -6.
Page 137. Art. 74
6. Divergent. 7 Convergent. 8. Convergent in both cases.
ANSWERS 365
Page 140. Art. 75
6. -1<x<1. 7. |s|>l. 8. -a<x<a.
Page 145. Art. 77
2. ' /(s) = (x-l) 3 +(x-l)2 + 4(a:-l)-3.
/(1.02) = - 2.919592, /(1.01) = - 2.959899.
/(.99) = - 3.038901, /( .98) = - 3.079608.
3. 3(y-3) 2 +4(y-3)-8. 4. sin 31° = .51508.
Page 149. Art. 78
1. x+ ??+2-x*+B. 3. .000002.
3 15
6. .017452. 7. 1 — — — — + ^.
2 8
y/S l/„ tt\ V3
2 2
9. ? + e?K + —h* + B.
10. 15 + 24 (a: - 2) + 13 (x - 2) 2 + 3 (a - 2) 8 .
U.log« + *- ". + -»-».+*.
8 . a; 2 a; 2 3 X s 4 a: 4
12. - 4 (a; + 1) + 6 (x + l) 2 - 4 (x + 1)3 + (x + l) 4 .
14. 5.013. 16. 3.433987.
15. 11.0087. 17. .0127.-..
18. 1 - (* - 1) + (x - l) 2 - (x - \f + B. to 2.
Pages 159, 160. Art. 83
2.
2 a 2
a 2 + 6 2
3.
13
7 '
4. JL.
2a
Pages 163,
164.
Art. 85
3.
f
5.
f
8. -4,
4.
4.
7.
*■
366 ANSWERS
Page 164. Art. 85
1. 0.
6
9. 1.
2- 2. 6. I. 10. _?,
n 3
3- 3. 7. 1. 11. I
]oga
log b '
4. !^?. * 1.
Page 166. Art. 86
1- 1- 3. 0. 5. -$.
2. 0. 4. 5. 6. 1.
«• e 2c2.
Page 171. Art. 89
1. First. 6. Second.
2. They do not touch. 7. a = — 1.
3. Third. 8. y = 2 a; 2 - 5 x + 4.
4. 3i(x-a) = a(2/-a). 9 a. (— 2, - 8), First.
5. y + 12x = 10. 96. (- 2, -8), Second.
10. y=-x 2 + 2x + 3, $ x =-3y 2 + Uy-36.
2.
First.
13. Second.
2 a.
a .
2*
Page 179. Art. 97
6 (* 2 + rcV)2
n(n — l)xy
i.
2.
9 aVx(8a-3x)*
3(2 a - xy
3.
4.
00 .
(z 2 + ?/ 2 )*
2 m 2
7 2(z + 2/)^
Va
10. «!.
a
5.
(e 2 x 2 - a 2 )^
a&
8. 3(aa*/)*-
Page 181. Art. 98
3 a(5-
9-
11. ( * _j_y
V3V2 54V2"/
1.
2.
pVl + (log a) 2 -
- 4 cos 0)*
- 6 cos
3,
ANSWERS 367
2p* 6 4x/ ^
3
Va
5. -if.
a 2
1. a = 0, /3 = 0.
7 q(l + gg)*
Page 188. Art. 100
2. a _alog-— - , p--. b (« + i8 )!-(«-/S)i = (4a)t.
. a=x-^(e a -e a ),p = 2y.
. (a + )3)f-(a-/3)§ = (4a;
. , 1K . , . 6. (aa)*- (6/3)'- = ( a 2 + 6 2 )t.
3 ce = q + 15 y 8 = ay ~ ° y
6 a-V 2 a* 7. (a + B)% + (a - /3)t = 2 ai
16. a = a(0' - sin 6"), p = a(l - cos #'), 0' = - tt.
Pages 198, 199. Exercises on Chapter XIII
1. (0, 0) ; ax±by = 0. 6. Two nodes at infinity ; the
asymptotes arex = v-|-l, x+y=±\.
2. (O,0);cuspoffirstkind,y=0. m , n x , ' / _
V ' . *" 7. (0, -a); (+a, 0); (-«,0:)
3. Four cusps of first kind ; the tangents are, respectively,
(0, ± a), (± a, 0) ; y = 0,a; = 0. V% + a) = ±V2a;
2(.z+a) = ±\/3y;
4. (0, 0) ; conjugate point with 2 (x -a) = ±VSy.
real coincident tangents, y = 0. / Ax
° ' " 8. (— a, 0) ; conjugate points.
5. (0, a); y = a + *J cusp of 9 (0, 0) ; » = 0, y =0.
second kind. 10. (0, 0) ; is a tacnode ; y = 0.
12. Terminating point at (0, 0) .
Pages 207, 208. Exercises on Chapter XIV
1. x 2 + y l = p 2 . 7. y 2 = 4 a(2 a - x).
2. xf + i/f = ai »• & 2 x 2 +(a 2 fW=6 2 (a 2 + & 2 ).
3. x$ + yl = cf. 9 - (* 2 + ?/ 2 ) 2 = 4 c2 (^ 2 " y' 2 )'
4. 4xy = k*. 10 - 16y 3 + 27^ = 0.
5. (x - a) 2 + (y - py = r 2 . 11 - y±s±fc = 0.
6. y 2 (x + 2a) 4-a* = 0. 12. (x 2 +y 2 -ay) 2 =a 2 (x 2 +0/+a) 2 ).
13. 6 2 x 2 + a 2 y 2 = 4 a 2 6 2 .
INTEGRAL CALCULUS
Pages 215-216. Art. 114
1.
„ 3
21.
— cos nx.
2.
x a+l
n
a+ 1
22.
x . sin 2 x
2 4
3.
4.
2 to X «H
am ~\L ,.
23.
24.
x sin 2 x
2 4
cos (to 4 n) x
5.
ax - \ aH* + | a *05* - £ x 2 .
m+n
6.
5kJx-f -J* JL.
e 2x 2 3x 8
25.
26.
— | COS X 2 .
sin x — ^ sin 3 x.
7.
H^ 2 + « 2 ) 3 -
27.
— cos x + I COS 3 X.
8.
(atx + &) n+1
28.
tan x — x.
JoT+1)
29.
| tan 3 x.
9.
10.
log;(x 4 a).
| log (2 ax - x 2 ).
30.
— cot (ax 4 6).
a
11.
— log cot X.
31.
- f (cot x)l
12.
— log (1 + cos a).
32.
log tan x.
13.
log (logx).
33.
\ sec 3 x.
14.
flog(x 3 4l).
34.
— cos X.
15.
16.
— log COS X.
log sin x.
35.
sin-i^.
a
17.
- e ax .
36.
i sin- 1 2a;.
18.
a
37.
a a
19.
(a 4- b) m + nx
w log (a 4- 6)
38.
A tan-i^-
ab b
20.
| sin 2 x.
39.
tan- 1 (x-2).
Page
219.
Art. 115
1.
x sin^x + Vl — x 2 .
3. _X* Sill x -f- Z X COS X -
2.
e x tan -1 e x — \ log (1 4- e
*).
4. J2±i(log* *
n + 1 V n + 1
ANSWERS 369
5. A[2z 3 tan-ix-z2 + log(l+*2)]. Q £e*(sin x + cos x).
6. sec x [log cos x + 1J. 10. | e* (sin x - cos x) .
7. J [(*» +1)001-1* + *;]. 11. icosxsin2x-icos2xsinx.
8. Ksin3*-3*cos3*]. 12. x tan x + log cos x.
Pages 220-222. Art. 116
4. Ksin-i*)2. i, , ♦ «
11. log tan-.
6. Jcos(*2 + l)[l-logcos(*2 + i)]. 2
12. log tan g + |).
13. -§(a 3 -x*)i
7.
sin-i «
a
8.
I tan-i «
a a
9.
±cos-i«.
a *
10.
sin-i^-^
a
14. log (x - 1) -
15. 1
2
x-1 2(x-l)
2 sin 2 *
16. -^tan-^v^tan*).
V2
Page 226. Art. 118
1. -i-logV^Cg+l)-!. 3 . _± lo|f *=*.
2V2 V2(x+1)+1 12 °x+l
2. -i_ tan-ii^l. 4. £sin-i(3* - 5).
Vl4 V14 '
5. Vx 2 + 2 x + 2 - log(x + 1 + Vx 2 + 2 * + 2).
6. - V-*2 + 2 x + 1 + sin-i^ni.
V2
7. .-V / 8-4*-4*2 + 3 sin -i2*_+l -
2 3
8 ^V3x^ + x-'2-~ log(aj + | + Vx 2 + i x - f ).
9- ^Vl+*-2*2 + -JL_ sin -i 43-l t
2 4V2 3
10. ^ log (x- 5) -| log (x-1).
11. Vax - x 2 + ? sin-i 2-^J? .
2 a
el. calc. — 24
370 ANSWERS
12. - V- 2 x 2 - 3 x - 1 -
sin" 1 (4x + 3).
13. -lVi-2*-3^--^-sin-i 3a; + 1
3
2 V2
10
3V3
i log («_±^)
Page 227. Art. 119
2
3. -lo
4.
l-2x + V5^-4a;+ 1
/ a+V« 2 -a 2 \
1 , /V2 + Vx 2 + 2z + 3\
~VS lQg ( "¥+1 J
lo<
/ 1 — a; + 2 Vz 2 + x+J \
6. sin-
V2 (x - 1)
7 . _l ]o(f /l-x+V-x*-10x-r
9.
a 2 z
10
5C + 2
Vx 2 + a' 2
— Va 2 — x 2
11. -JLrin-i 3 ~ 2 *_
V2 V3(2 x - 1)
Pages 227-228. Exercises on Chapter I
3. i log (6 x^ + 12 a; + 5)
2. 5 tan-i-.
8 2
4. *£(! + **). 5. -f(3-2, )t 6. ^
7. _| (d «_rf)
10. I sin 8 x.
12. — cos e*.
V, 2 + 1
Ks + l) 1 - *(*-!)*.
11. 1 log tan ^^ + -1
3 s V 2 4J
13. log (a cos se + b).
a
14. - log (e~ x + Ve~ 2x - 1) .
15. I sin-i x 2 .
2V2 L * J 2 & e* + l
18. i a* tan-i x - ^ * 4 + rV * 2 - tV log (* 2 + 1) ■
ANSWERS 3T1
19. -f2 + 2xloga + (x\ogaf}^ 2Q tiind _ secd .
a* (log a) 3
.21. -cot?. 22. logCgcoB'g + ftBin'^
2 2(6 -a)
23. -JVl-logx. 24. ilog(e'^ + l). 25. sin -i / 2 sin ^ + 1 V
26. tan- 1 (logic). 27
6(a — 6 tana;)
2'
28. irin-ir ^-^+^ 1
L V5 a 2 J
V 3x 2 + 2x+l + log p + l+V3*» + 2s+r
30. - — log (a; + Vx 2 - a 2 ) +
Vx 2 — a 2
2 x 2 2 a 2 x
31. — cos x log tan x + log tan - . 32. \x — \ log (sin x + cosx).
Pages 236-237. Exercises on Chapter II
1. *(s a -2 _ 2 x 2 ) Va 2 - x 2 + — sin-i? .
2. ir_JL_ + ltan-i2l. 5. - Va 2 - x 2 + ^sin-i?-
8|_x 2 + 4 2 2j 2 2 a
3. 2a; - 1 -f-i-tan-i^l. 6. - ^ZEZ.
3(x 2 -x + l) 3V3 V3
4. _?v / a 2 -x 2 +— sin-i-. 7. + ' 2x _
8. ~ (2 x 2 + 5 a 2 ) Vx 2 + «" + — log (x + Vx 2 + a 2 ) .
9. - Vx 2 + a + ^log(x -f Vx 2 + a).
10. -(2 x 2 - ax - 3 a 2 ) V2 ax - x 2 + ^ sin-i — - •
6 2 a
n (2ax-x 2 )? 12> _Vl
3 ax 3 2 x*<
13 3fa + 2)«-5(g + 2) 3 t x + 1
8(x 2 + 4 x + 3) 2 16 ° x + 3
372 ANSWERS
14. l( x + l)Vl-2x-x 2 + sin-i «±i
2 V2
V^EI-lrin-ll. 18-
17.
SILL • / - r-
2 x 2 2 x aVa + bx 2
-1 19. - cos *
2(x 2 + 7)' (1 + e) Vl + esin 2
20. -(2x 2 - a 2 )Va 2 - x 2 + ^siiT^-
8 8 a
21. — (33 a 4 - 26 a 2 x 2 + 8 x*) Va 2 - x 2 + ^- sin-i - •
48 16 a
Page 241. Art. 122
1. JLlog^. 3. ^-4,+^logi^I 3
2c x + a 2 T 2 ° x + 1
2. w ( x + 1 ) 2 ■ 4- x + log(x-a)«(x-&)».
x(x — 1)
5. 2 _±^§log(x-2-V3)-^^log(x-2+V3).
2V3 2V3
6 li 0(y (^-l)^-2) . 7 . i og («-«)(« - 6 > .
20 & (2x+l)(x + 2) x-c
8. x + ^— [a 2 log(x + a)-& 2 log(x + 6)].
& — a
9. log[(x + 2)V2x-l]. 12. ^-7x + 641og(x + 4)
10 . log (x- a )(, + 6 J, -271og(x + 3).
2 a& ax + &
t1 1. x 6 14 I i og ! + ^ .
u - 3 l0g (2-T^xr^^ 2 log i-.
Page 243. Art. 123
L "2(x-l) + I l0g ^l* 3 " 2(a 2 - x 2 ) '
2 - ^ + log ^ 4 ' v^Tvii*
5 . ^_2x + ^^ + logx(x4-l) 2 .
2 x 2 + x
ANSWERS 373
6. log(x 2 -a 2 )--^ a2
x*-a* 4(V2 + l-a;)2
8. ax-- + log
x x + a
9. x +-L-l[28\og(x + S)-\ogxl
ox y
10. Aiog^±_^_±. ll. a;__A^_51og(x-3).
a 2 x ax x — S
Page 244. Art. 124
1. log *— . 2. llog aj, + 1 a + -tan-ig.
°V^T4 4 *> + l) 2 ^2
3. — [log(x + a)--log(x*-ax+a 2 ) + V3 tan-* 2 * ~_ a ~| .
3a 2 L 2 aV3 J
4. - 1 tan-* - + - tan-i ? . 7. — tan-i*
a aft b 2 a (x — a) 2 a 2 a
6. __Ltan-i-^5— . 9. log x ~ 1 -tan-* a.
V3 2x 2 + l V^+T
Page 24b. Art. 125
1. tan-i* + _^-. 2. Atan-i«U ?7 2< *
x 2 + 1 2 a a 2 (x 2 + a 2 )
3. llog * 2 + 1 4- z - 1 .
4 & (x + l) 2 2(z 2 +l)
4 -g + Slog^^- 3 + 2 * -Stan-ix.
x x 2(.r 2 + l)
5. + log (x 2 + a 2 ) - J_ tan -i? #
2 (x 2 + a 2 ) 5 V J 2 a a
6. — - +ilog(x 2 + l).
x*+l 4(x 2 + l) 2 2 &K '
374 ANSWERS
Page 249. Art. 127
1. log Vx + 1 ~i . 2. 2Vx-3?/x + Qfyx-6log(fyx + l).
y/x + 1 + 1
3. 2 log ( Vx - 1 + 1) +
Vx - 1 + 1
4. 2tan-iVx-2. 5. llog ^ a ^ - &
& Vx — a -f 6
6. 14 (xT? - i -xt + J xt? - | x7 + i XT?).
Page 252. Art. 128
L »l-xj x-l + Vl-a: 2
2 . _ 21og [-V2 + V^].
Pages 253-254. Exercises on Chapter IV
1. 2 V3 tan-i V^V^ ~ V ^ ~ 1#
. 3 l 3 V2(x-a)*-l
2. o(x-a)^ -log *> '- .
* 4V2 V2(x-a)^ + l
3. x-4Vx+ 1 + 81og(Vx + l +2).
2 2Vx^T + l
4. log(x+Vx-l) -tan" 1
V3 V3
5. M2x-3a)(a + x)i ? 1 1q „ V^IT^r + sVS
6. 6 log (** - 3xU 5). " 2 ^« 2 ° V * 2 - ^ " * V *
8. i[x 2 -xVx 2 - 1 +log(x+ Vx 2 -1)].
9. f x^ - f x^ + f x* + 2 x^ - 3 x* - 6 x^ + 3 log (x* + 1) + 6 tan" 1 !*;*.
10. -$(2a* + s 2 )V^V. v gr^
,, . a, „ Va 2 — x' 2 — a a 2 x
11. Va 2 -x 2 + o l0 S
2 Va 2 -x 2 4-a
Va 2 - x' 2 • _, x
14.
12. - — ^_sin-i-. a 2 Vx 2 + a 2
ANSWERS 375
15 x(2x 2 + 3a 2 ) _ lg 1^ Va*+z*+x Va 2 +
3a 4 (x 2 + « 2 )^ 2 ° Va^+x^-x x
17.
«Va 2 - x 2
Page 256. Art. 131
5. f esc 3 x — cot x — | cot 3 X.
6. - 64 [cot 4 x + -|- cot 3 4 a] .
3. tan x + | tan 3 a; + A tan 5 x. 1
3 7. — + log tan x.
4. — 128 [cot 2x + cot 3 2 a; 2 tan- a;
+ |cot 5 2x+ 1 cot 7 2x]. 8. - \ cot 3 x - i cot 5 x.
Page 257. Art. 132
1.
£ sec 4 £ — \ sec 2 x.
5.
\ sec 4 x — sec 2 x +
2.
— 4. esc 7 x + | esc 6 x
— | esc 3 a:.
6.
sec" -1 x sec n-3 x
n — 1 w — 3
3.
- [ -sec 5 ax — sec 3 ax + sec ax ) .
a\5 3 /
7.
log sec x.
4.
— (sinx + cscx).
8.
— log CSC X.
Pages 259-260. Art. 133
1. _i C ot 3 x + cotx + x. tan-ix
4.
2. — tan 2 ax log sec ax. w
2a a °
3. i (tan 2 x + cot 2 x) 5 - i tan? K - I tan5 « + i tan3 «
+ 4 log (sin x cos x) . — t ' an x + x -
Page 260. Art. 134 (a)
1. — COS X + | COS 3 X. _4 2 8
_ . , 4. | cos 3 x + 3 cos 3 x — | cos' x.
2. — |cos 5 x + \ cos' x. 8
3. log sin x - sin 2 x + J sin 4 x. 5. |(1 - cosx) z - |(1 - cosx)*.
Page 261. Art. 134 (6)
1. — $cot 3 x.
2. -cotx-fcot 3 x-icot 5 x. 5 - !Vtanx(tanx-3cotx).
3. _ cot 5 x (i + \ cot 2 x). 6. tann " ly + taD " +lx -
w — 1 w + 1
4. — I cot 3 x — 2 cot x + tan x.
376 ANSWERS
Page 262. Art. 134 (c)
1. | x — ■£% sin 4 x.
2- xis (^ # + t s i n3 2 £ — sin 4 x — \ sin 8 x) .
3. T |s(3x — sin 4 a; + fsin8x).
4. |(3 x + sin 4 a + | sin 8x). 5. tan x + £ sin 2 x — § x.
Page 264. Art. 134 (d)
1. |(x — sin x cos x) . 2. ^ cot x(cos 2 x — 3)— §x.
3. — | sin 3 x — sin x + log tan (- + -].
4. — \ cot 2 x esc 2 x + J log tan x.
5. | x + 2 cot x + \ sin x cos x — £ cot 3 x.
Page 265. Art. 135
1. Jtan-i(^tanx). 2. - tan-i f~2 tan (*■ - -\~|.
1 , tanx-2-V3
3.
2 V3 tan x - 2 + V3
x
b tan - - o + vV + b 2
4. — — log .
v a 2 + 6 2 6 tan ^ - a - Va 2 + 5 2
-1 7. J-tan-if^Y
V2 W2 /
a (a tan x + &)
6. i-tan-if^i^y 8 ' 2
a& \ b J
l i 2
log
tan - - 3
2
Pages 266-267. Exercises on Chapter V
3. 2 Vtan x. . 4. J tan 4 x + \ tan 2 x.
5. - cscx + log tan I- • + - j.
6. - tan 2 x sin x + § fsin x - log tan (* + *\~] . 7. ^ + x • •
ANSWERS 377
?, x 11. I e x (sin x + cos a: — f sin3x
8. 62^sin- + cos-j. _; C os3x).
9. — $ e _I (sin 2 a; + 2 cos 2 x). 15 sin^x _ sin"+ 3 x
10. i e 2x( 2 - sin 2 x - cos 2 x). w + X n + 3
16. §tan*x — 2Vcotx.
17. - 32 cot 2 x(l + 1 cot/ 2 2 x + $ cot 4 2 x).
18. J- tan | x(l + | tan 2 \ x + \ tan 4 § x).
19. log tan 2 x. 20. - 8 [cot 2 x + ^ cot 3 2 x] .
21 1 / Va 2 -x 2 \«
22 ^^ + ^6^^^, + ^^
-• i(S + |)- + i(5-|) ta «^ + ^>-
24. x cos (« — &) + sin (a — b) log sin (x + 6).
Pages 281-283. Art. 142
2. fp*A 3. 2. 4. 36 V3. 5. 40V5.
6. f. 7. fv^. 8. i. 9. £. 10. Trr 2 . 11. f|.
12. 4 a 2 tan- 1 ^i- ; 4 Tra 2 . 13. *•«&. 14. if* - 72 log 2.
2a' 2
15.
*.-<* 16 24-8^ 17 f lg< ^ fl) |
log a 5
20.
3 2 OJ * . ^ -|- I
To3 ' ' 2' 2(6'- 1) "
22. a 2 logra. 23. 1. 25. -.
& 6
26.
A. 27. 4a&tan- 1 5. 28.
T5 a
2V2. 29. I°£i. 30. * . IzlT"
7T 7T 7T
31. ^(4-tt).
2 V ;
Pages 287-289. Art. 143
Ex. 4. 2-. 1. 4. 2. oo. 3. 3. 4. 4 a 2 .
2
5. oo. 6. 3tt« 2 . 7. 2tt.
378 ANSWERS
Pages 290-291. Art. 144
1. vab. 2. lira 2 . 3. f.
4 lZ. 5 3y(g2-68)a |
4 8a6
Page 292. Art. 145
2.
a\
7.
2
n.
h
3.
4.
Ti-r 2 .
3 7rr 2 #
8.
« 2 7r3
24
12.
!•
5.
2
25 7T.
9.
-log
2 *
ft
;)•
13.
3tt
8 '
6.
c /
2 (P1 "
-/>2).
10.
n
14.
7T 5
5 '
Pages 296-297. Art. 148
In the following answers the values are given for Simpson's formula
only, unless the trapezoidal formula is called for in the problem.
1. 22 ; 21.5. 3. 0.500014 ; 0.500002 ; 0.5000014 ; 0.5000011.
4. 5.2523. 5. 37.8555; 36.5261. 6. 0.9996. 7. 8.0047.
8. 39.6465. 9. 0.7593. 10. 0.7468. 11. 0.4443.
12. 335. 13. 3.006. 14. 1.1873; 1.1830; 11931.
15. 0.5633.
6. \irab.
7. \irab.
8. \AK.
Pages 300-302.
Art.
149
a irabp
9. _.
10. 4£ cu. ft.
12.
| wa 8 cos 4 6.
13.
1 trabc*.
11. fa 3 .
14.
fr 8 .
Pages 304-306. Art. 150
1. jrL(io_3 7r). 2. 27rrl(2r 2 +a 2 )Vr 2 -^-ar 2 sin-i^2^1.
6V2 L3 « J
3. 7tt. 4. -7T
3
|«i« + ax{* + 4 eflxt + 8 a 3 ] og (~=^\ ]
ANSWERS 379
5. 2 7r 2 a 3 . 6 . i*^. 7. firr 3 . 8. x/fc 3 ; oo.
o
9. ^^- 10. Sf. 11. 4 «■*««.
105 15
12. 7rr8a 3 log^-4a 2 (2a-?/i)]; oo. 13. 5*- 2 a 3 . 14. ^~-
5. 7rr8a 3 log^-4a 2 (2a-yi)l; oo. 13. 57r 2 a 3 . 14.
Pages 308-309. Art. 151
1. p[V2 4-log(l+ V'2)]. 6la 3 logVS
'216' • 5 •
4. 6 a. 5. 2irr. 6. - (e - e" 1 ). 7. }f-
8. 2-V2 + log 1 i^. 9 . 4 (« 3 - &3 > . 10. alogS-^a.
V3 aft & *
Page 310. Art. 152
1. (p 2 -pi)Va 2 + 1. 2. 2ira. 3. 8 a.
4. a [tan* sec I + log tan (1+^~|* 2 .
5. ^[flV^ + l + log (5 + V^ + l)] \
c (7r 2 + 4)t 8 w A , 2 ,_
•• — f- -3 7. 4+ — log(V 3+2).
9. 2a[~V5-2-v / 31og V5+V5 .T
L \/2(2 + V3)J
8 . 3ira a o-f,/5 o ,/5u„ V3 + V5 1
2
Page 311 Art. 153
1. 8a. 2. JJWL. 3. 6 a . 4. la*! 2 .
ra — 1
7. ^[(4 + 9ti«)*-8].
Pages 313-315. Art. 154
1. Tra^l-^V 3. f^Cv/S-l).
2.^-2).'' 4 ' f[3V2-log(l + V2)].
380 ANSWERS
5. (a) 2irb(b + - — cos- 1 -
V Va 2 -& 2 a
(6 ) 2 ^ + -Jgg- log r « + vtf'-y -|.
Va 2 - b 2 La - Va 2 - 6 2 J
«• ¥« 2 - 7. («) tiWoH^; 9 ¥»*•
8. 4*-d* (/S) 7ra\/« 2 + & 2 . 10. 4 7r 2 aA:; 2w 2 a 2 k.
11. 2 7ra 2 (3 sin f i — 3 ii cos <i — t\ 2 sin £i).
Pages 316-320. Art. 155
2.
p = e a .
4. p = — a cos + C.
5.
y*= ax 2 + 6.
6. p n = c sin n0. Straight line. Cardioid.
7.
y = e ax + c .
8. p:
-h).
_ ce a 9 ™1 10. 2 7ra 2 . 11. a log -2.
3 */i
13.
Va 2 + b 2 (t 2
14. 22.7 lb. 15. 0.9627 lb. 16. 4.4312.
17.
h = 5.28 in.
18.
h = 43.17 lb.
Pages 323-324. Art. 156
3. xy = ay 2 + by — \. 4. y = kx (log x - 1) + ax + b. 5. T 4 3 &«£.
6, y = 0.0002 x 3 + 0.0036z + 1.12, slope = 0.0006 a 2 + 0.0036 ; a =20,
y = 2.792, slope = 0.2436.
7. 1000 y= -0.046 x 3 + 0.75 x 2 - 2 05 x + 40; 1000?/= ^ a 8 -0.0575 x 2
-9.7x+117.91§.
Page 328. Art. 158
1. xy+C. 3. Impossible. 5. x 3 + y 3 — 3 axy + C.
2. - cos a cos y + C. 4. log-+C. 6. tan-^+C.
y y
7. | a 3 + x 2 ?/ + 5 x + £ ?/ 3 - \y 2 + <7.
8. £ x 5 + xy 4 + $ x 3 - xy 2 + | y 2 - \ y* + 2 ?/ + (7.
Page 330. Art. 160
1. — =■ 3. J. 5. | a". 7 - «•
2\/2 Z 3 2
2. 4. 4. 1. 6. 6& 3 . 8. J log 2.
ANSWERS 381
Pages 332-334. Art. 161
6. &. c Srfl;
■Kd
2
2. 64. 4. 4 7T - 8.
3. 4£. 5. 1. '8 9. a 2 .
/•y /» a /"a /*7r /*ir /*o(l — cos 0)
10. 2l"f pdpdtf;2j I _ Vi p .pfZ^p;2l | prfpdfl.
JO Ja(l-co8 0) JO Jcos {}-*) JnJa
a 2
1L 20. 12. §JH— 2V3.
o
Pages 336-337. Art. 162
3 3 * 2 ^
Page 340. Art. 163
1. 62251b. 2. 1120.51b. 3. 9337.5 tons. 4. 66.41b. 5. 15.645 tons.
Pages 343-345. Art. 164
i. (t,M).
2 (fcfA). 4 . fo,o,^- c V
3. x = 8 (*+ a > g ,y = g = 0. V 8;
4(& + 2a)'^
- 3|"(a + ft) 4 a 4 "! from cen ter of sphere.
' 8|_(a + ft) 3 -a 3 J
3r7(r + «) 4 -6^-| from herical
8L7(r + a) 3 -6^J
center.
g af + 2 fl W + a^ above tne base .
2(ab + a'b')
o
9. from the vertex.
4 a sin T
~3
2(r 2 — a' 2 ^
10. from center of circle.
3 (V 2 cos- 1 --aVr 2 - a 2 )
» (& £)• 1S (¥• t)-
10 / 2fl 2 6 \ « /3a 36 3c\
12 [S(^Y) ' 3(7325)- 15 ' U ' 8 ' 8 j
382 . ANSWERS
16 (*u l 6 -^ 16 c\
V15' 157r' 15 tt/
17 (<L* 8a 3«\. - - = ^ = 3r (« + ft)*-g* -i
' \ 8 ' 8 ' 8 /' 8L(a + ^) 3 -a 3 J
18. x = y = z = — -
5
19. (0, 0, — j, the base of the cone being in xy-pl&ne. 20. (0, 0, \)
21. (a) a cos 2 ; (6) a ( 1 H — ] both measured from the vertex,
\ 1 + cos 2 d )
Pages 347-348. Art. 165
1. «^(&2 + c 2). 6. !!^(3« 2 + 4^ 2 ).
o 12
2. ^ ira^h. ^ Tra&fr /g &2 i #2)
3. 7 ^(3 a 2 + /l 2) <
12 8. *£; ^[(r + fc).-,*].
4 ttoH 15 15
10 ' g 8 7IT"+ 5
5. \ Trabh(a 2 + tf) . ' 3(n + 5)'
10. ia 5 . 11. ^ira(2?*-r*). 12. ^a 3 6. 13. \bh*. 14. iira*
INDEX
(The numbers refer to pages)
Absolute value, 134.
Absolutely convergent,
134.
Acceleration, 77.
Actual velocity, 68.
Approximate integra-
tion, 292.
Arc, length of, 306.
Area, by double integra-
tion, 330.
derivative of, 79.
formula for, 273, 280.
in polar coordinates,
291.
in rectangular coordi-
nates, 273.
Asymptotes, 96.
Average curvature, 176.
Bending, direction of,
90.
Cardioid , area of, 292, 333.
Catenary, 180, 316.
length of arc, 309.
volume of revolution,
312.
Catenoid, 313.
Cauchy's form of remain-
der, 153.
Center of curvature, 172.
Center of gravity, 340.
Change of variable, 124.
Circle, area by double
integration, 331.
of curvature, 172.
Cissoid, 180.
area of, 289.
Component velocity, 70.
Concave, 89.
toward axis, 94.
Conditionally conver-
gent, 134.
Conditions for contact,
168.
Conjugate point, 197.
Conoid, 301.
Constant, 15.
factor, 25, 213.
of integration, 214.
Contact, 167.
of odd and even order,
169.
Continuity, 19, 109.
Continuous function, 19.
Convergence, 132.
Convex, 94.
to the axis, 94.
Critical values, 55.
Cubical parabola, 281.
Cusp, 194.
Cycloid, length of, 311.
surface of revolution,
315.
Decreasing function, 51.
Definite integral, 270.
geometric meaning of,
273.
multiple integral, 329.
Dependent variable, 15.
Derivative, 21, 22.
of arc, 79.
of area, 78.
of surface, 81.
of volume, 81.
Determinate value, 158.
Development, 131, 160.
Differentials, 74, 210.
integration of, 326.
total, 112.
Differentiation, 23.
of elementary forms,
44, 45.
Direction of curvature,
175.
Discontinuous function,
20.
Divergent series, 132.
Duhamel's theorem, 348.
383
Ellipse, area of, 281.
evolute of, 190, 309.
parametric form, 290.
Ellipsoid, volume, 299.
Envelope, 200.
Epicycloid, length of, 311.
Equiangular spiral, 316.
Evaluation, 160, 165.
Evolute, 182.
of ellipse, 190, 183, 291.
of parabola, 183.
Expansion of functions,
131.
Exterior rectangles, 269.
Family of curves, 200.
Formula for integration
by parts, 216.
Formulas of differentia-
tion, 44, 45.
of integration, 211, 224.
of reduction, 229, 262.
Function, 15.
Helix, 318.
Hyperbolic branches, 95.
spiral, area of, 292.
Hypocycloid, area of, 290.
length of arc of, 308,
311.
volume of revolution
of, 305.
Implicit function, 33.
Impossibility of reduc-
tion, 232.
Increasing function, 51.
Increment, 21.
Independent variable, 15.
Indeterminate form, 157.
Infinite, 20.
Infinite limits of integra-
tion, 287.
ordinates, 97.
Infinitesimal, 350.
384
INDEX
Integral, 209.
definite, 270.
double, 330.
multiple, 329.
of sum, 212.
triple, 328, 334.
Integration, 209.
by inspection, 211.
by parts, 216.
by rationalization, 248.
by substitution, 219.
formulas of, 211, 224.
of rational fractions,
238.
of total differential,
326.
successive, 321.
summation, 268.
Interior rectangles, 269.
Interval of convergence,
138.
Involute, 187.
of circle, 311.
Lagrange's form of re-
mainder, 152.
Lemniscate, area of,
334.
Length of arc, 306.
of e volute, 185.
of space curve, 318.
polar coordinates, 309.
rectangular coordi-
nates, 306.
Limit, 15.
change of, in definite
integral, 276.
Limits, infinite, for defi-
nite integral, 287.
Liquid pressure, 338.
Logarithm, derivative
of, 34.
Logarithmic curve, 280.
spiral, length of arc,
310.
Maclaurin's series, 141,
276.
Maximum, 53.
Mean value theorem, 74,
275.
Measure of curvature,
177.
Minimum, 53.
Moment of inertia, 346.
Multiple points, 193.
Natural logarithms, 36.
Normal, 85.
Notation for rates, 72.
Oblique asymptotes, 99.
Order of contact, 167.
of differentiation, 122.
of infinitesimal, 350.
Osculating circle, 172.
Osgood, 132.
Parabola, 108, 89, 171.
semi-cubical, 308.
Parabolic branches, 95.
Paraboloid, 314.
Parallel curves, 187.
Parameter, 201.
Partial derivative, 110.
Point of inflexion, 90.
Polar coordinates, 104.
subnormal, 106.
subtangent, 106.
Problem of differential
calculus, 21.
of integral calculus,
209.
Radius of curvature,
172.
Rates, 68.
Rational fractions, inte-
gration of, 238.
Rationalization, 248, 249.
Rectangles, exterior and
interior, 269.
Reduction, cases of im-
possibility of, 236.
formulas, 229, 262.
Remainder, 150.
Rolle's theorem, 150.
Simpson's rule, 294.
Singular point, 191.
Slope, 16.
Solid of revolution, 81.
Sphere, volume' by triple
integration, 336,.
Spheroid, oblate, 305, 314.
prolate, 314.
Spiral, of Archimedes,
107.
equiangular, 108 s 316,
310.
hyperbolic, 292.
logarithmic, 292.
Standard forms, 211, 224.
Stationary tangent, 90.
Steps in differentiation,
22.
Stirling, 141.
Subnormal, 86.
Subtangent, 86.
Summation, 268.
Surface of revolution, 81.
area of, 312.
Tacnode, 194.
Tangent, 85.
Tangent plane, 119.
Taylor, 141.
Taylor's series, 148.
Tests for convergence,
133.
Total curvature, 176.
differential, 112.
Tractrix, 315.
length of, 318.
surface of revolution
of, 318.
volume of revolution
of, 318.
Transcendental func-
tions, 34.
Trapezoidal rule, 292.
Trigonometric functions,
integration of, 255.
Variable, 15.
Volume of solid* of revo-
lution, 302.
Volumes by triple inte-
gration, 334.
Witch, area of, 286.
volume of revolution
of, 305.
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