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foT L\v,coU *»«,
?vU ft.**, Ck.(if.
yiw. 2.4, Watr.
MATHEMATICAL TEXTS
FOR COLLEGES
EDITED BY
PERCEY F. SMITH, Ph.D.
PROFESSOR OF MATHEMATICS IN THE SHEFFIELD
SCIENTIFIC SCHOOL OF TALE UNIVERSITY
1 Jl
w ^15H
Silt Isaac Newton
ELEMENTS OF THE DIFFERENTIAL
AND INTEGRAL CALCULUS
(BE VISED EDITION)
BY
WILLIAM ANTHONY GRANVILLE, Ph.D.. LLD.
FORMERLY PRESIDENT OT GETTYSBURG COLLEGE
c # »««»*. i » *>
« * %
WITH THE EDITORIAL COOPERATION OF
PERCEY F. SMITH, Ph.D.
PROFESSOR OF MATHEMATICS IK THE SHEFFIELD SCIENTIFIC SCHOOL
YALE UNIVERSITY
GINN AND COMPANY
BOSTON • NEW YORK • CHICAGO • LONDON
ATLANTA • DALLA8 • COLUMBU8 • SAN FRANCISCO
5/1
eia
1 r* r *
50
COPYRIGHT, 1904, 1911, BY
WILLIAM ANTHONY GRANVILLE AND P
PRINTED IN THE UNITED STATES OF AMERICA
628.6
Ce^^>
<?£
CINN AND COMPANY • PRO
PRIETORS • BOSTON • U.SJL
PREFACE
That teachers and students of the Calculus have shown such a gen*
erous appreciation of Granville's w Elements of the Differential and
Integral Calculus " has been very gratifying to the author. In the last
few years considerable progress has been made in the teaching of the
elements of the Calculus, and in this revised edition of Granville's
w Calculus" the latest and best methods are exhibited, — methods that
have stood the test of actual classroom work. Those features of the
first edition which contributed so much to its usefulness and popu
larity have been retained. The introductory matter has been cut down
somewhat in order to get down to the real business of the Calculus
sooner. As this is designed essentially for a drill book, the pedagogic
principle that each result should be made intuitionally as well as
analytically evident to the student has been kept constantly in mind.
The object is not to teach the student to rely on his intuition, but, in
some cases, to use this faculty in advance of analytical investigation.
Graphical illustration has been drawn on very liberally.
This Calculus is based on the method of limits and is divided into
two main parts, — Differential Calculus and Integral Calculus. As
special features, attention may be called to the effort to make per
fectly clear the nature and extent of each new theorem, the large
number of carefully graded exercises, and the summarizing into
working rules of the methods of solving problems. In the Integral
Calculus the notion of integration over a plane area has been much
enlarged upon, and integration as the limit of a summation is con
stantly emphasized. The existence of the limit e has been assumed
and its approximate value calculated from its graph. A large num
ber of new examples have been added, both with and without
answers. At the end of almost every chapter will be found a col
lection of miscellaneous examples. Among the new topics added are
approximate integration, trapezoidal rule, parabolic rule, orthogonal
vi PREFACE
trajectories, centers of area and volume, pressure of liquids, work
done, etc. Simple practical problems have been added throughout;
problems that illustrate the theory and at the same time are of
interest to the student. These problems do not presuppose an ex
tended knowledge in any particular branch of science, but are based
on knowledge that all students of the Calculus are supposed to have
in common.
The author has tried to write a textbook that is thoroughly modern
and teachable, and the capacity and needs of the student pursuing a
first course in the Calculus have been kept constantly in mind. The
book contains more material than is necessary for the usual course of
one hundred lessons given in our colleges and engineering schools ;
but this gives teachers an opportunity to choose such subjects as best
suit the needs of their classes. It is believed that the volume con
tains all topics from which a selection naturally would be made in
preparing students either for elementary work in applied science or
for more advanced work in pure mathematics.
WILLIAM A. GRANVILLE
Gettysburg College
Gettysburg, Pa.
CONTENTS
DIFFERENTIAL CALCULUS
CHAPTER I
COLLECTION OF FORMULAS
SECTION PAGE
1. Formulas from Algebra, Trigonometry, and Analytic Geometry 1
2. Greek alphabet 3
3. Rules for signs in the four quadrants 3
4. Natural values of the trigonometric functions 4
5. Tables of logarithms 5
CHAPTER II
VARIABLES AND FUNCTIONS
6. Variables and constants 6
7. Interval of a variable 6
8. Continuous variation 6
9. Functions \ 7
10. Independent and dependent variables 7
11. Notation of functions 8
12. Values of the independent variable for which a function is defined 8
CHAPTER III
THEORY OF LIMITS
13. Limit of a variable 11
14. Division by zero excluded 12
15. Infinitesimals 13
16. The concept of infinity (oo) . 13
17. Limiting value of a function 14
18. Continuous and discontinuous functions 14
19. Continuity and discontinuity of functions illustrated by their graphs . 16
20. Fundamental theorems on limits 18
21. Special limiting values 20
22. The limit of — as x = 21
x
23. The number e 22
24. Expressions assuming the form ^ .23
vii
viii CONTENTS
CHAPTER IV
DIFFERENTIATION
KCTIOX MOB
25. In trod notion 25
26. Increments 25
27. Comparison of increments ......... 26
28. Derivative of u function of one variable 27
29. Symbols for derivatives 28
80. Differentiablu functions 29
31. Kenei'iil ruli' for differentiation 29
82. Applications of the derivative to Geometry SI
CHAPTER V
RULES FOR DIFFERENTIATING STANDARD ELEMENTARY FORMS
33. Importance of General Rule 34
34. Differentiation of a constant 36
35. Differentiation of a variable with respect to itself 37
36. Differentiation of a sum 37
37. Differentiation of the product of a constant and a function . . .37
38. Differentiation of the product of two functions 38
39. Differentiation of the product of any finite number of functions . . 38
40. Differentiation of a function with a constant exponent . . . .39
41. Differentiation of a quotient 40
42. Differentiation of a function of a function 44
43. Differentiation of inverse functions 45
44. Differentiation of a logarithm 46
45. Differentiation of the aim pie exponential function. . . . 48
48. Differentiation of the general exponential function . . . .49
47. Logarithmic differentiation 50
48. Differentiation of sin o 54
49. Differentiation of cos v 55
50. Differentiation of tan v 56
61. Differentiation of cote 56
52. Differentiation of sec v . 56
53. Differentiation of esc o 57
54. Differentiation of verso 57
55. Differentiation of arc sine 61
66. Differentiation of arc coar 62
67. Differentiation of arc tan v 62
58. Differentiation of arc cote 63
69. Differentiation of arc sec d 63
60. Differentiation of arc esc e 64
CONTENTS ix
motion PAG*
61. Differentiation of arc vers v 65
62. Implicit functions . . 60
63. Differentiation of implicit functions 60
CHAPTER VI
SIMPLE APPLICATIONS OF THE DERIVATIVE
64. Direction of a curve 73
65. Equations of tangent and normal, lengths of subtangent and subnormal.
Rectangular coordinates 76
66. Parametric equations of a curve 70
67. Angle between the radius vector drawn to a point on a curve and the
tangent to the curve at that point 83
68. Lengths of polar subtangent and polar subnormal .... 86
69. Solution of equations having multiple roots 88
70. Applications of the derivative in mechanics. Velocity .... 90
71. Component velocities .......... 91
72. Acceleration 92
73. Component accelerations 93
CHAPTER VII
SUCCESSIVE DIFFERENTIATION
74. Definition of successive derivatives 97
75. Notation 97
76. The nth derivative 98
77. Leibnitz's formula for the nth derivative of a product .... 98
78. Successive differentiation of implicit functions 100
CHAPTER VIII
MAXIMA AND MINIMA. POINTS OF INFLECTION. CURVE TRACING
79. Introduction <#v 103
80. Increasing and decreasing functions 106
81. Tests for determining when a function is increasing and when de
creasing 108
82. Maximum and minimum values of a function 109
83. First method for examining a function for maximum and minimum
values Ill
84. Second method for examining a function for maximum and minimum
values 112
85. Definition of points of inflection and rule for finding points of in
flection 125
86. Curve tracing 128
CONTENTS
CHAPTER IX
DIFFERENTIALS
8EOTIOK PAGE
87. Introduction 131
88. Definitions 131
89. Infinitesimals 132
90. Derivative of the arc in rectangular coordinates .... 134
91. Derivative of the arc in polar coordinates 135
92. Formulas for finding the differentials of functions .... 137
93. Successive differentials ... 139
CHAPTER X
RATES
94. The derivative considered as the ratio of two rates • „ . 141
CHAPTER XI
CHANGE OF VARIABLE
95. Interchange of dependent and independent variables . . 148
96. Change of the dependent variable ....... 149
97. Change of the independent variable 150
98. Simultaneous change of both independent and dependent variables . 152
CHAPTER XII
CURVATURE. RADIUS OF CURVATURE
99. Curvature 155
100. Curvature of a circle 155
101. Curvature at a point 156
102. Formulas for curvature . 156
103. Radius of curvature 159
104. Circle of curvature 161
CHAPTER XIII
THEOREM OF MEAN VALUE. INDETERMINATE FORMS
105. Rolle's Theorem 164
106. The Theorem of Mean Value 165
107. The Extended Theorem of Mean Value 166
CONTENTS xi
OCTIOY PAGE
108. Maxima and minima treated analytically 167
109. Indeterminate forms 170
110. Evaluation of a function taking on an indeterminate form 170
111. Evaluation of the indeterminate form  171
112. Evaluation of the indeterminate form §£ 174
113. Evaluation of the indeterminate form • oo 174
114. Evaluation of the indeterminate form oo — oo . . . .175
115. Evaluation of the indeterminate forms 0°, 1", oo° . . .176
CHAPTER XIV
CIRCLE OF CURVATURE. CENTER OF CURVATURE
116. Circle of curvature. Center of curvature 178
117. Second method for finding center of curvature 180
118. Center of curvature the limiting position of the intersection of nor
mals at neighboring points 181
119. Evolutes . 182
120. Properties of the e volute . .186
121. Involutes and their mechanical construction • 187
CHAPTER XV
PARTIAL DIFFERENTIATION
122. Continuous functions of two or more independent variables
128. Partial derivatives
124. Partial derivatives interpreted geometrically
125. Total derivatives ....
126. Total differentials ....
127. Differentiation of implicit functions .
128. Successive partial derivatives
129. Order of differentiation immaterial .
190
191
192
194
197
198
202
203
CHAPTER XVI
ENVELOPES
180. Family of curves. Variable parameter . . 205
131. Envelope of a family of curves depending on one parameter . . 205
132. The evolute of a given curve considered as the envelope of its normals 208
133. Two parameters connected by one equation of condition .  . 209
xii CONTENTS
CHAPTER XVII
SERIES
SECTION PAGE
134. Introduction 212
135. Infinite series 213
136. Existence of a limit 215
137. Fundamental test for convergence 216
138. Comparison test for convergence 217
139. Cauchy's ratio test for convergence 218
140. Alternating series 220
141. Absolute convergence 220
142. Power series 223
CHAPTER XVIII
EXPANSION OF FUNCTIONS
143. Introduction 227
144. Taylor's Theorem and Taylor's Series 228
145. Maclaurin's Theorem and Maclaurin's Series 230
146. Computation by series 234
147. Approximate formulas derived from series. Interpolation . . . 237
148. Taylor's Theorem for functions of two or more variables . . . 240
149. Maxima and minima of functions of two independent variables . 243
CHAPTER XIX
ASYMPTOTES. SINGULAR POINTS
150. Rectilinear asymptotes 249
151. Asymptotes found by method of limiting intercepts .... 240
152. Method of determining asymptotes to algebraic curves 250
153. Asymptotes in polar coordinates 254
154. Singular points 255
155. Determination of the tangent to an algebraic curve at a given point
by inspection 255
156. Nodes 258
157. Cusps 259
158. Conjugate or isolated points 260
159. Transcendental singularities 260
CHAPTER XX
APPLICATIONS TO GEOMETRY OF SPACE
160. Tangent line and normal plane to a skew curve whose equations are
given in parametric form 262
161. Tangent plane to a surface 264
CONTENTS xiii
SBOTIOV PAOB
162. Normal line to a surface 266
163. Another form of the equations of the tangent line to a skew curve 268
164. Another form of the equation of the normal plane to a skew curve . 269
CHAPTER XXI
CURVES FOR REFERENCE
INTEGRAL CALCULUS
CHAPTER XXII
INTEGRATION. RULES FOR INTEGRATING STANDARD ELEMENTARY
FORMS
165. Integration 279
166. Constant of integration. Indefinite integral 281
167. Rules for integrating standard elementary forms .... 282
168. Trigonometric differentials . . . . . . . . 298
169. Integration of expressions containing Va a — x* or Vx a ± a a by a trigo
nometric substitution . 804
CHAPTER XXIII
CONSTANT OF INTEGRATION
170. Determination of the constant of integration by means of initial con
ditions 307
171. Geometrical signification of the constant of integration . . 807
172. Physical signification of the constant of integration .... 809
CHAPTER XXIV
THE DEFINITE INTEGRAL
173. Differential of an area 314
174. The definite integral 314
175. Calculation of a definite integral 316
176. Calculation of areas 318
177. Geometrical representation of an integral 319
178. Mean value of <£(x) 320
179. Interchange of limits 320
180. Decomposition of the interval 320
181. The definite integral a function of its limits 321
182. Infinite limits 321
183. When y = <f>(x) is discontinuous 322
XIV
CONTENTS
CHAPTER XXV
INTEGRATION OF RATIONAL FRACTIONS
SECTION PACK
184. Introduction 325
185. Case I 325
186. Case H 327
187. Case HI 329
188. Case IV 331
CHAPTER XXVI
INTEGRATION BY SUBSTITUTION OF A NEW VARIABLE.
RATIONALIZATION
189. Introduction 335
190. Differentials containing fractional powers of x only .... 335
191. Differentials containing fractional powers of a + bx only . . . 336
192. Change in limits corresponding to change in variable . . . 336
193. Differentials containing no radical except Va + bx + x 3 . . . 338
194. Differentials containing no radical except Va + bx — x a . . . 338
195. Binomial differentials . . 340
196. Conditions of integrability of binomial differentials .... 341
197. Transformation of trigonometric differentials 343
198. Miscellaneous substitutions 345
CHAPTER XXVII
INTEGRATION BY PARTS. REDUCTION FORMULAS
199. Formula for integration by parts 347
200. Reduction formulas for binomial differentials 350
201. Reduction formulas for trigonometric differentials .... 356
202. To find f «** sin nxdz and fe * cos nxdx 359
CHAPTER XXVIII
INTEGRATION A PROCESS OF SUMMATION
203. Introduction 361
204. The fundamental theorem of Integral Calculus 361
205. Analytical proof of the Fundamental Theorem 364
206. Areas of plane curves. Rectangular coordinates .... 365
207. Area when curve is given in parametric form 368
208. Areas of plane curves. Polar coordinates 370
209. Length of a curve 372
210. Lengths of plane curves. Rectangular coordinates .... 373
211. Lengths of plane curves. Polar coordinates ... .375
CONTENTS xv
SECTION PAGE
212. Volumes of solids of revolution 377
213. Areas of surfaces of revolution 381
214. Miscellaneous applications 385
CHAPTER XXIX
SUCCESSIVE AND PARTIAL INTEGRATION
215. Successive integration 893
216. Partial integration 395
217. Definite double integral. Geometric interpretation . . 396
218. Value of a definite double integral over a region .... 400
219. Plane area as a definite double integral. Rectangular coordinates 402
220. Plane area as a definite double integral. Polar coordinates . 406
221. Moment of area 408
222. Center of area 408
223. Moment of inertia. Plane areas 410
224. Polar moment of inertia. Rectangular coordinates .... 410
225. Polar moment of inertia. Polar coordinates 411
226. General method for finding the areas of surfaces .... 413
227. Volumes found by triple integration 417
CHAPTER XXX
ORDINARY DIFFERENTIAL EQUATIONS
228. Differential equations. Order and degree 421
229. Solutions of differential equations 422
230. Verifications of solutions 423
231. Differential equations of the first order and of the first degree . . 424
232. Differential equations of the nth order and of the first degree . . 432
CHAPTER XXXI
INTEGRAPH. APPROXIMATE INTEGRATION. TABLE OF INTEGRALS
233. Mechanical integration 443
234. Integral curves 443
235. The integraph 445
236. Polar planimeter 446
237. Area swept over by a line 446
238. Approximate integration 448
239. Trapezoidal rule 448
240. Simpson's rule (parabolic rule) 449
241. Integrals for reference 451
INDEX 461
DIFFERENTIAL CALCULUS
CHAPTER I
COLLECTION OF FORMULAS
1. Formulas for reference. For the convenience of the student we
give the following list of elementary formulas from Algebra, Geome
try, Trigonometry, and Analytic Geometry.
1. Binomial Theorem (n being a positive integer) :
n(nl)(n2)...(nr + 2) tr _,„ ft ,_ 1
Ir — 1
2. n I = [n = 1 • 2 • 8 • 4 • • • (n — 1) n.
3. In the quadratic equation ax 2 + fox + c = 0,
when ft 2 — 4 ac > 0, the roots are real and unequal ;
when 6* — 4 ac = 0, the roots are real and equal ;
when 6* — 4 ac < 0, the roots are imaginary.
4. When a quadratic equation is reduced to the form x* + px + q = 0,
j> = sum of roots with sign changed, and q = product of roots.
6. In an arithmetical series,
i = a + (nl)d; s =  (a + =  [ 2 « + ( n ~ l)fl
6. In a geometrical series,
, ^., W — a a (r« — 1)
/ = at 41  1 ; * = = — i ' .
r1 r1
7. log ab = log a + log 6. 10. log Va =  log a. 13. log  = — log a.
n a
8. log  = log a — log b. 11. log 1 = 0. 14. Circumference of circle=2wr.
o
9. log a* = n log a. 12. log a a = 1. 15. Area of circle = m*.
• In formulas 1425, r denotes radius, a altitude, B area of base, and s slant height.
1
DIFFERENTIAL CALCULUS
16. Volume of prism = Ba.
17. Volume of pyramid = \ Ba.
18. Volume of right circular cylinder = vr*a.
19. Lateral surface of right circular cylinder = 2 vra.
20. Total surface of right circular cylinder = 2 wr (r + a).
21. Volume of right circular cone = \m*a.
22. Lateral surface of right circular cone = *rs.
23. Total surface of right circular cone = wr (r + a).
24. Volume of sphere = Jxr 1 .
25. Surface of sphere = 4wr*.
26. sin x = ; cos x = ; tan x =
cscx secx cotx
__ A sinx . cosx
27. tanx = ; cotx = 
cosx sinx
28. sin 2 x + cos 2 x = 1 ; 1 + tan 2 x = se^x ; 1 + cot 2 x = csc'x.
~ . /* \ 31. sin (x + y) = sinx cosy + cosx sin y.
29. sinx = cos ( — xl; \ »/ » ^
. 32. sin (x — y) = sin x cosy — cosx sin y.
cosx = sin (^ — x);
\2 / 33. cos(x ± y) = cosx cosy ¥ sin x sin y.
tanx = cot(x). tanx + tany
\2 / 34. tan(x + y) =  — .
1 — tan x tan y
30. sin (*■ — x) = sin x ;
cos(ir — x) =— cosx; •• * # v tanx — tany
v ' ' 36. tan (x — y) = — .
tan (ir — x) = — tan x. 1 + tanxtany
2 tanx
36. 8in2x = 2 sinx cosx; cos2x = cos 2 x — sin*x; tan2x =
1 — tan 2 x
„. n .xx « X 9 X * 2 tan J x
37. sin x = 2 sin  cos ; cos x = cos 2 sin 2  ; tan x = J — .
2 2 2 2 1— tan 2 Jx
38. cos 2 x = J + { cos 2 x ; sin 2 x = \ — J cos 2 x.
X X
39. 1 + cos x = 2 cos 2  ; 1 — cos x = 2 sin 2  .
2 2
. . X /l — COS X X /l + COS X A X /l —
40. 8m  = ± ^—— ; c08  = ±y J—^—, tan = ±^—
— cosx
cosx
41. sin x + sin y = 2 sin } (x + y) cos J (x — y).
42. sin x — sin y = 2 cos J (x + y) sin J (x — y).
43. cos x + cos y = 2 cos 1 (x + y) cos J (x — y).
44. cos x — cos y = — 2 sin J (x + y) sin J (x — y).
45. = = ; Law of Sines.
sin A sini? sinC
46. a 2 = 6 s + c 2 — 2 6c cos A ; Law of Cosines.
47. d = V(x x — x 2 ) 2 + (y x — y 2 ) 2 ; distance between points (x l , y t ) and (x 2 , y 8 ).
48. d = AXl + By *+ £ . distance from line 4x + By + C = to (x,, y t ).
iV^Tu 2
COLLECTION OF FORMULAS
3
48. x = ****» , y = V} * V * ; coordinates of middle point.
50. x = x + x 7 , y = y + if; transforming to new origin (a^, y^.
51. x = jf cos0 — if sintf, y = X 7 sin0 + y' oos0; transforming to new axes making
the angle $ with old.
52. x = p cosl, y = /> sintf; transforming from rectangular to polar coordinates.
53. p = Vx a +y a , = arc tan  ; transforming from polar to rectangular coordinates.
x
54. Different forms of equation of a straight line :
(a) y ~ y \ = y *~ y \ , twopoint form ;
X "™ X. Xa ~ — • X«
X V
(b)  f £ = 1, intercept form ;
(c) y — y l = m(x — x 1 ), slopepoint form ;
(d) y = mx + 6, slopeintercept form ;
(e) x cos a + y sin a = p, normal form ;
(f) .4x + By + C = 0, general form.
55. tanl = — i 2 , angle between two lines whose slopes are wij and m,.
m x = mj when lines are parallel,
and Hi} = when lines are perpendicular.
56. (x — a)* + (y — £)* = r 8 , equation of circle with center (a, /9) and radius r.
2. Greek alphabet.
Let
tere
Names
Letters
Names
Letters
Names
A
a
Alpha
I i
Iota
p p
Rho
B
/9
Beta
K K
Kappa
2 0* 5
Sigma
r
7
Gamma
A X
Lambda
T T
Tau
A
S
Delta
M fi.
Mu
T v
Upsilon
E
6
Epsilon
N *
Nu
4> 4>
Phi
Z
f
Zeta
B f
Xi
x X
Chi
H
*7
Eta
O o
Omicron
¥ ^
Psi
e
Theta
n ir
Pi
il »
Omega
3. Rules for signs of the trigonometric functions.
Quadrant 1 Sin
Cos
Tan
Cot
Sec
C«c
First ....
+
+
+
+
+
+
Second ....
+
—
—
—
—
+
Third ....
—
+
+
—
—
Fourth ....
—
+
—
—
+
—
DIFFERENTIAL CALCULUS
4. Natural values of the trigonometric functions.
Angle in
Radiant
Angle in
Degrees
Sin
Cos
Tan
Cot
See
Om
0°
1
00
1
00
2
90°
1
00
00
1
w
180°
1
00
1
CO
3tt
2
270°
1
CO
00
1
2tt
360°
1
CO
1
CO
Angle in
Radians
Angle in
Degrees
Sin
Cos
Tan
Cot
Sec
Csc
0°
1
00
1
CO
ir
6
30°
1
2
V3
2
3
V3
2V8
3
2
ir
4
45°
V2
2
V2
2
1
1
V2
V2
ir
3
60°
2
1
2
V3
V3
3
2
2V5
3
2
90°
i ! o
i
CO
CO
1
Angle in
Radians
Angle in
Degrees
Sin
Cos
Tan
Cot !
1
.0000
0°
.0000
1.0000
.0000
00
90°
1.5708
.0175
1°
.0175
.9998
.0175
57.290
89°
1.5533
.0349
2°
.0349
.9994
.0349
28.636
88°
1.5359
.0524
3°
.0523
.9986
.0524
19.081
87°
1.5184
.0698
4°
.0698
.9976
.0699
14.300
86°
1.5010
.0873
5°
.0872
.9962
.0875
11.430
85°
1.4835
.1745
10°
.1736
.9848
.1763
5.671
80°
1.8963
.2618
15°
.2588
.9659
.2679
3.732
75°
1.3090
.3491
20°
.3420
.9397
.3640
2.747
70°
1.2217
.4363
25°
.4226
.9063
.4663
2.145
65°
1.1346
.5236
30°
.5000
.8660
.5774
1.732
60°
1.0472
.6109
36°
.5736
.8192
.7002
1.428
55°
.9599
.6981
40°
.6428
.7600
.8391
1.192
50°
.8727
.7854
45°
.7071
.7071
1.0000
1.000
45°
.7854
Cos
Sin
Cot
Tan
Angle in
Degrees
Angle in
Radians
COLLECTION OF FORMULAS t
5. Logarithms of numbers and trigonometric functions.
Table of Mantissas of the Common Logarithms of Numbers
1
2
3
4
6
6
7
8
l.KI
0414
0792
1139
1461
1781
2041
2304
2563
2788
3010
3222
3424
3617
3802
3976
4160
4314
4472
4624
a
4771
4014
6051
6185
Ml 5
5441
5563
5682
6798
6911
i
6021
6128
6232
6336
11435
6632
6628
6721
6812
0602
B
6H0O
707fl
7160
7243
7324
7404
7482
7550
7634
7709
6
77B2
7853
7024
7903
KI*!J
8120
8196
8261
8326
8388
7
8451
3513
B573
8633
f <;■.'•■
8761
8808
8866
8621
8676
1
90S1
was
P1S8
0101
6243
0294
9346
8365
9446
9404
10
ii
6642
9500
0638
9685
6731
6777
9828
0868
6612
0656
(XHXi
0043
DOSfl
oi as
01 TO
esia
nasi
02H4
0684
U374
0414
0453
0492
0631
0566
0607
0646
0682
0719
0765
a
0702
0828
0804
0896
0634
0669
1004
1038
1072
1106
13
1130
1173
1206
1236
1271
1303
1336
1367
1369
1430
14
1461
1462
1623
1563
1584
1614
1644
1073
1703
1732
u
1761
1700
1618
1847
1876
1903
1931
1966
1987
2014
16
2041
2068
2005
2122
2148
2176
2201
2227
2263
2276
17
2804
2330
2365
2380
2406
2430
2455
2480
2604
2628
18
2653
2577
2601
2625
2648
2672
2665
2716
2742
2765
19
2788
2810
3838
2866
2878
2600
2623
3946
2967
2989
Table of Logarithms of the Trigonometric Functions
Angle In
Angle In
Degree!
togiln
logeo.
logun
log cot
.0000
0°
0.000
90°
1.6708
.0175
1°
8.2410
0.6000
8.2419
1.7681
80°
1.5638
.0340
2°
8.5428
9.9997
8.6431
1.4569
88"
1.5366
.0624
a 6
8.7188
9.0094
8.7104
1.2806
87"
1.5184
MM
4"
8.8436
6.6969
8.8446
1.1664
86°
1.5010
.0878
6°
8.6403
6.8683
8.9420
1.0580
86°
1.4835
.1745
10°
9.2397
6.6634
9.2463
0.7687
80°
1.3963
.2618
15"
9.4130
8.6840
9.4281
0.6719
76°
1.8090
.3461
20°
9.5341
9.0780
6.6611
0.4380
70°
1.2217
.4363
25°
9.6266
9.9573
0.6687
0.3313
66°
1.1345
.5236
30°
9.6890
9.6375
8.7614
0.2386
00°
1.0472
.6109
86°
0.7686
9.8184
6.8462
0.1548
66°
0.8509
.6981
40°
9.8081
9.8843
8.9238
0.0762
60°
0.8727
.7854
46°
0.8495
9.8465
0.0000
0.0000
46°
0.7854
lofoo.
...
*..
*«.
Angle Id
Angle In
CHAPTER II
VARIABLES AND FOKCTIONS
6. Variables and constants. A variable is a quantity to which an
unlimited number of values can \ie assigned. Variables are denoted
by the later letters of the alphabet. Thus, in the equation of a
straight line,
a b
x and y may be considered as the variable coordinates of a point
moving along the line.
A quantity whose value remains unchanged is called a constant.
Numerical or absolute constants retain the same values in all prol>
lems, as 2, 5, v7, w, etc
Arbitrary constants, or parameters, are constants to which any one
of an unlimited set of numerical values may be assigned, and they
are supposed to have these assigned values throughout the inves
tigation. They are usually denoted by the earlier letters of the
alphabet. Thus, for every pair of values arbitrarily assigned to a
and b, the equation
represents some particular straight line.
7. Interval of a variable. Very often we confine ourselves to a
portion only of the number Bystem. For example, we may restrict
our variable so that it shall take on only such values as lie between
a and b, where a and b may be included, or either or both excluded.
We shall employ the symbol [a, i], a being less than b, to represent
the numbers a, b, and all the numbers between them, unless otherwise
stated. This symbol [a, 6] is read the interval from a to b.
8. Continuous variation, A variable x is said to vary continuously
tlirough an interval [a, 6], when j starts with the value a and increases
until it takes on the value b in such a manner as to assume the value
VARIABLES AND FUNCTIONS 7
of every number between a and b in the order of their magnitudes.
This may be illustrated geometrically as follows:
The origin being at 0, lay off on the straight line the points A and
B corresponding to the numbers a and b. Also let the point P corre
spond to a particular value of the variable x. Evidently the interval
[a, 6] is represented by the segment AB. Now as j varies continuously
from a to 6 inclusive, i.e. through the interval [a, J], the point P gen
es the segment AB.
9. Functions. When two variable! are so related that the value of the
firtt variable depends on the value of the »econd variable, then the first
variable is said to be a funrtion of the secoml variable.
Nearly all scientific problems deal with quantities and relations
of this sort, and in the experiences of everyday life we are con
tinually meeting conditions illustrating the dependence of one quan
tity on another. For instance, the weight a man is able to lift
depends on his strength, other things being equal. Similarly, the
distance a boy can run may be considered as depending on the
time. Or, we may say that the area of a square ia a function of
the length of a side, and the volume of a sphere is a function of
its diameter.
10. Independent and dependent variables. The second variable, to
which values may be assigned at pleasure within limits depending on
the particular problem, is called the independent variable, or argument;
and the first variable, whose value is determined as soon as the value
of the independent variable is fixed, is ealled the dependent variable.
Or function.
Frequently, when we are considering two related variables, it is in
our power to fix upon whichever we please as the independent variable ;
but having once made the choice, no change of independent variable
is allowed without certain precautions and transformations.
One quantity (the dependent variable) may be a function of two
or more other quantities (the independent variables, or arguments).
For example, the cost of cloth is a function of both the quality and
quantity; the area of a triangle is a function of the base and altitude;
the volume of a rectangular parallelepiped is a function of its three
dimensions.
8 DIFFERENTIAL CALCULUS
11. Notation of functions. The symbol /(i) is used to denote a
function of x, and is read / of x. In order to distinguish between
different functions, the prefixed letter is changed, as f(x), <j>(x),
/'(», etc
During any investigation the same functional symbol always indi
cates the same law of dependence of the function upon the variable.
In the simpler cases this law takes the form of a series of analytical
operations upon that variable. Hence, in such a case, the same func
tional symbol will indicate the same operations or series of operations,
even though applied to different quantities. Thus, if
/(*) x'Sx + U,
then /(y) = y9y + 14.
Al.o /(«) «* 9 « + K
/(6 + l)(S + l)'9(J + ll + 14}'7J + 6,
/(0)0'90+14 = 14,
/(3) = 3'98 414 4,
/(7)7'97 + 14 = 0, etc.
Similarly, ^(x, y) denoteB a function of x and y, and is read <t> of
x tind y.
If +(«,») .in (» + ,),
then ^ (a, 6) = sin (a + 6),
and *(.0j=sui = l.
Again, if F(x, y, z) = 2z+3y — 12 z,
then F(m,~ m, m)=2m3m12m = 13m,
and F(3, 2, 1)= 23 + 3 212 1 = 0.
Evidently this system of notation may be extended indefinitely.
13. Values of the independent variable for which a function is defined.
Consider the functions
x* — 2 x + 5, sin a, arc tan x
of the independent variable x. Denoting the dependent variable in
each ease by y, we may write
y = x 1 — 2 a: +5, y =* sin x, y = arc tan x.
VARIABLES AND FUNCTIONS ft
In each case y (the value of the function) is known, or, as we
say* defined, for all values of x. This is not by any means true of all
functions, as the following examples illustrating the more common
Exceptions will show.
Here the value of y (Le. the function) is defined for all values of x
except x = 6. When x = b the divisor becomes zero and the value of y
cannot be computed from (1).* Any value might be assigned to the
function for this value of the argument.
(2) y=V£
In this case the function is defined only for positive values of x.
Negative values of x give imaginary values for y, and these must be
excluded here, where we are confining ourselves to real numbers only.
(3)y = log.s. a>0
Here y is defined only for positive values of x. For negative values
of x this function does not exist (see § 19).
(4) y = arc sing, y = arccosz.
Since sines and cosines cannot become greater than + 1 nor less
than — 1, it follows that the above functions are defined for all values
of x ranging from — 1 to + 1 inclusive, but for no other values.
1. GiYen f(x) = z»  10x f + 81 x — 80 ; show that
/(0)=80, /(y) = V»10^ + 81y80,
/(2) = 0, /(a) = a»  10a 1 + 81a  80,
/(8) = /(6), f(yz) = yV  10 i/*z* + 81 yx  80,
/(I) >/(_ 8), f( z _ 2) = z*  16x* + 88x  140,
/(I) =6/(0).
». If f(x) = x*  8x + 2, flnd/(0), /(l), /( 1), /( J), /(1J).
3. If /(x) = x«  10x f + 81x  80, and* (x) = x*  65x«  210x  216, show that
/(2) = *(2), /(8) = *(8), /(6) = *(4), /(0) + *(0) + 246 = 0.
4. If F(x) = 2", find F(0), F( 8), F(J), F( 1).
ft. GiYen F(x) = x(x — 1) (x + 0) (x — i) (x + J) ; show that
F(0) = F(l) = F( 0) = FQ) = F( i) = 0.
• See § 14, p. 12.
10 DIFFERENTIAL CALCULUS
6. If /(m,) = m \" 1 , ghow that
n "h + 1
1 +/(m 1 )/(m 2 ) 1 + m^n,
7. If 0(x) = a* show that 4>{y)4> (z) =4>(y + *)'.
1 — x
8. Given 0(x) = log ; show that
*<*) + ,(,) = , (i±£).
9. If /(^) = cos ^, show that
/(*) =/( *) = /(»  *) = /(» + *)•
10. If F(6) = tan B, show that
F<2*)= "«
1  [*<*)]■
11. Given ^ (x) = x 2 * + x* m + 1 ; show that
f(l) = 8, *(0) = 1, *(a) = *(a).
12. If /(x) = ?^=y , find /(V2). 4iw.  .0201
CHAPTER HI
THEORY OF LIMITS
13. Limit of a variable. If a variable v takes on successively a series
of values that approach nearer and nearer to a constant value I in such
a manner that  v — 1\* becomes and remains less than any assigned arbi
trarily small positive quantity, then v is said to approach the limit 2, or
to converge to the limit I. Symbolically this is written
limit v = /, or, v = /.
The following familiar examples illustrate what is meant :
(1) As the number of sides of a regular inscribed polygon is indefi
nitely increased, the limit of the area of the polygon is the area of the
circle. In this case the variable is always less than its limit.
(2) Similarly, the limit of the area of the circumscribed polygon is
also the area of the circle, but now the variable is always greater than
its limit.
(3) Consider the series
The sutn of any even number (2 n) of the first terms of this series is
2 ' 4 8 2 8 " 1 2 f — 1
J— i
Similarly, the sum of any odd number (2 n+1) of the first terms of
the series is ill 11
° f  +l 2 4 8 2*— 1 • 2 1 *
1 i
o»»+i o 1
(<?) fl.+i _ i _ 1 =3 + 3^;' By6,p.l
* To be read the numerical value of the difference between v and I.
11
12 DIFFERENTIAL CALCULUS
Writing (£) and (C) in the forms
o i
we have
and
limit /2
1
5,.
limit 1
limit 1
Hence, by definition of the limit of a variable, it is seen that both
S an and <S alil are variables approaching  as a limit as the number of
terms increases without limit.
Summing up the first two, three, four, etc., terms of (4), the sums
are found by (B) and (C) to be alternately less and greater than ij,
illustrating the case when the variable, in this case the sum of the terms
of (A), is alternately less and greater than its limit
In the examples shown the variable never reaches its limit. This is
not by any means always the case, for from the definition of the limit
of a variot'le it is clear that the essence of the definition is simply that
the numerical value of the difference between the variable and its limit
shall ultimately become and remain less than any positive number we
may choose, however small.
(4) As an example illustrating the fact that the variable may reach
its limit, consider the following. Let a series of regular polygons
be inscribed in a circle, the number of sides increasing indefinitely.
Choosing any one of these, construct the circumscribed polygon
whose sides touch the circle at the vertices of the inscribed polygon.
Let />. and % be the perimeters of the inscribed and circumscribed
polygons of n sides, and C the circumference of the circle, and sup
pose the values of a variable x to be as follows:
P.v
G Z+s
etc.
Then, evidently,
and the limit is reached by the variable, every third value of the variable
Iteing C.
14. Division by zero excluded. — is indeterminate. For the quotient
of two numbers is that number which multiplied by the divisor will
give the dividend. But any number whatever multiplied by zero gives
THEORY OF LIMITS 18
aero, and the quotient is indeterminate ; that is, any number whatever
may be considered as the quotient, a result which is of no value.
 has no meaning, a being different from zero, for there exists no
number such that if it be multiplied by zero, the product will equal a.
Therefore division by zero is not an admissible operation.
Care should be taken not to divide by zero inadvertently. The following fallacy
is an illustration.
Assume that a = b.
Then evidently ab = cfl.
Subtracting 6», ab  6* = a*  6*.
Factoring, 6(a — b) = (a + b) (a — 6).
Dividing by a — 6, b = a + 6.
But a = 6,
therefore 6 = 25,
or, 1 = 2.
The result is absurd, and is caused by the fact that we divided by a — 6 = 0.
15. Infinitesimals. A variable v whose limit is zero is called an
infinitesimal.* This is written
limit v = 0, or, v == 0,
and means that the successive numerical values of v ultimately become
and remain less than any positive number however small. Such a
variable is said to become indefinitely small or to ultimately vanish.
If limit v =s Z, then limit (y — 7) = ;
that is, the difference between a variable and its limit is an infinitesimal.
Conversely, if the difference between a variable and a constant is an
infinitesimal, then the variable approaches the constant as a limit.
16. The concept of infinity (oo). If a variable v ultimately becomes
and remains greater than any assigned positive number however large,
we say v increases without limit, and write
limit v = f oo , or, v = + oc .
If a variable v ultimately becomes and remains algebraically less
than any assigned negative number, we say v decreases without limit,
and write r ..
limit r= — oo, or, v = — oo.
• Hence a constant, no matter how small it may be, is not an infinitesimal.
14 DIFFERENTIAL CALCULUS
If a variable v ultimately becomes and remains in numerical value
greater than any assigned positive number however large, we say v,
tn numerical value, increases without limit, or v becomes infinitely great,*
and write limit* = 00, or, v = oo.
Infinity (00) is not a number; it simply serves to characterize a
particular mode of variation of a variable by virtue of which it
increases or decreases without limit
17. Limiting value of a function. Given a function /(x).
If the independent variable x takes on any series of values such that
limit z = a,
and at the same time the dependent variable f(x) takes on a series of
corresponding values such that
limit f(x) = A,
then as a single statement this is written
and is read the limit off(x), as x approaches the limit a in any manner,
is A.
18. Continuous and discontinuous functions. A function f(x) is said
to be continuous for x = a if the limiting value of the function when x
approaches the limit a in any manner is the value assigned to the
function for x = a. In symbols, if
££/(*) =/oo.
then f(x) is continuous for x = a.
The function is said to be discontinuous for x = a if this condition
is not satisfied. For example, if
the function is discontinuous for x = a.
The attention of the student is now called to the following cases
which occur frequently.
•On account of the notation used and for the sake of uniformity, the expression
v  + 00 is sometimes read v approaches the limit plus infinity. Similarly, v =  oc is read
v approaches the limit minus infinity, and v = oc is read t\ in numerical value, approaches
the limit infinity.
While the above notation is convenient to use in this connection, the student must not
forget that infinity is not a limit in the sense in which we defined a limit on p. 11, fox
infinity is not a number at all.
limit
x
THEORY OF LIMITS 15
Case I. As an example illustrating a simple case of a function con
tinuous for a particular value of the variable, consider the function
For x = 1, f(x) =f(l) = 3. Moreover, if x approaches the limit 1
in any manner, the function f(x) approaches 3 as a limit. Hence the
function is continuous for x = l.
Case II. The definition of a continuous function assumes that
the function is already defined for x = a. If this is not the case, how
ever, it is sometimes possible to assign such a value to the function for
x=a that the condition of continuity shall be satisfied. The following
theorem covers these cases.
Theorem. Iff(x) is not defined for x = a, and if
**/(x)=*.
then f(x) will be continuous for x = a, if B is assumed as the value of
f(x) for x=a. Thus the function
^4
x 2
is not defined for x = 2 (since then there would be division by zero).
But for every other value of rr,
and "=2(*+ 2 ) = 4 ;
therefore limi * ^=i = 4.
x = 2 x—2
Although the function is not defined for rr= 2, if we arbitrarily assign
it the value 4 for x = 2, it then becomes continuous for this value.
A function f(x) is said to be continuous in an interval when it is
continuous for all values of x in this interval.*
* In this book we shall deal only with functions which are in general continuous, that is,
continuous for all values of z, with the possible exception of certain isolated values, our
results in general being understood as valid only for such values of x for which the function
in question is actually continuous. Unless special attention is called thereto, we shall as a
role pay no attention to the possibilities of such exceptional values of x for which the function
is discontinuous. The definition of a continuous function f(r) is sometimes roughly (but
imperfectly) summed up in the statement that a small change in x shall produce a small
change inf(x). We shall not consider functions having an infinite number of oscillations
in a limited region.
16
DIFFERENTIAL CALCULUS
19. Continuity and discontinuity of functions illustrated by their
graphs.
(1) Consider the function a?, and let
If we assume values for x and calculate the corresponding values
of y, we can plot a series of points. Drawing a smooth line freehand
through these pointy a good representation of the gen
eral behavior of the function may be obtained. This
picture or image of the function is called its graph.
It is evidently the locus of all points satisfying
equation (^4).
Such a series or assemblage of points is also called
a curve. Evidently we may assume values of x so near
together as to bring the values of y (and therefore the points of the
curve) as near together as we please. In other words, there are no
breaks in the curve, and the function x 2 is continuous for all values of x.
(2) The graph of the continuous
function sin a: is plotted by draw
ing the locus of
y = sin x.
It is seen that no break in the curve occurs anywhere.
(3) The continuous function e* is of very frequent occurrence in
the Calculus. If we plot its graph from
y = **, (e = 2.718 • • •)
we get a smooth curve as shown. From this it is
clearly seen that,
(a)whenrr = 0, ^K=0 = 1;
(b) when a;>0, y (=<?*) is positive and increases
as we pass towards the right from the origin ;
(c) when x < 0, y (= e*) is still positive and decreases as we pass
towards the left from the origin. Y
(4) The function log e :r is closely related to the
last one discussed. In fact, if we plot its graph 77
from y = log.*,
it will be seen that its graph has the same rela
tion to OX and OY as the graph of tf has to Y and OX
THEORY OF LIMITS
17
Here we see the following facte pictured :
(a) For x = 1, log,a: = log. 1 = 0.
(b) For x > 1, log,a; is positive and increases as x increases.
(c) For 1 >x> 0, log,! is negative and increases in numerical value
as x diminishes, that is, ^'jjj'jj log x s= — « .
(d) For ia0, log,x is not defined ; hence the entire graph lies to
the right of OF.
(5) Consider the function i and set
1
If the graph of this function be plotted, it
will be seen that as x approaches the value
zero from the left (negatively), the points of
the curve ultimately drop down an infinitely great distance, and as x
approaches the value zero from the right, the curve extends upward
infinitely far.
The curve then does not form a continuous branch from one sicfe
to the other of the axis of r, showing graphically that the function
is discontinuous for x = 0, but continuous for all other values of x.
(6) From the graph of
1x*
it is seen that the function
2x
1x*
is discontinuous for the two values x = ± 1, but continuous for all
other values of x.
(7) The graph of
jfmtanx
shows that the function tanx is dis
continuous for infinitely many values
of the independent variable x, namely,
x =* 5* where n denotes any odd positive or negative integer.
(8) The function
18
DIFFERENTIAL CALCULUS
has infinitely many values for a given value of x, the graph of equation
y = arc tan x
consisting of infinitely many branches. If, however, we confine our
selves to any single branch, the function is continuous. For instance,
if we say that y shall be the arc of smallest numeri
cal value whose tangent is r, that is, y shall take
on only values between — ~ ami ^. then we are
limited to the branch passing through the origin,
and the condition for continuity is satisfied.
(9) Similarly, i
K J J »m tun '
5 X
Confining ourselves to c
is found to be a manyvalued function.
branch of the graph of . l
° r y = arc tan  ,
we see that as x approaches zero from the left, y approaches the
limit — ^i and as x approaches zero from the right, y approaches the
limit + ^ Hence the function is discon
tinuous when x = 0. Its value for x =
can be assigned at pleasure.
Functions exist which are discontinuous
for every value of the independent vari
able within a certain range. In the ordinary applications of the Cal
culus, however, we deal with functions which are discontinuous (if
at all) only for certain isolated values of the independent variable ;
such functions are therefore in general continuous, and are the only
ones considered in this book,
20. Fundamental theorems on limits. In problems involving limits
the use of one or more of the following theorems is usually implied.
It is assumed that the limit of each variable exists and is finite.
um of a finite number of vari
of tlie limits of the several
Theorem I. The limit of the alye,
ablet it equal to the like algebra i
variable t.
Theorem II. The limit of the product of a finite number of variables
is equal to the product if the limit* of the newral variables.
Theorem III. The limit of the quotient of two variables is equal to the
quotient of the limits of the separate variables, provided the limit of the
denominator it not zero.
THEORY OF LIMITS
19
Before proving these theorems it is necessary to establish the fol
lowing properties of infinitesimals.
(1) The sum of a finite number of infinitesimal* is an infinitesimal.
To prove this we must show that the numerical value of this sum can
be made less than any small positive quantity (as e) that may be
assigned (§ 15). That this Is possible is evident, for, the limit of each
infinitesimal being zero, each one can be made numerically less than
 (m being the number of infinitesimals), and therefore their sum can
be made numerically less than e.
(2) The product of a constant c and an infinitesimal is an infinitesimal.
For the numerical value of the product can always be made less than
any small positive quantity (as e) by making the numerical value of
the infinitesimal less than .
(3 ) The product of any finite number of infinitesimals is an infinitesimal.
For the numerical value of the product may be made lees than any
small positive quantity that can be assigned. If the given product
contains n factors, then since each infinitesimal may be assumed less
than the nth root of e, the product can be made less than t itself.
(4) If v is a variable which approaches a limit I different from zero,
then the quotient of an infinitesimal by v is also an infinitesimal. For if
limit v = I, and k is any number numerically less than I, then, by defini
tion of a limit, v will ultimately become and remain numerically greater
than k. Hence the quotient , where e is an infinitesimal, will ulti
mately become and remain numerically less than > and is therefore
by (2) an infinitesimal.
Proof of Theorem I. Let v v t» f , »,, • be the variables, and l t , l t , l t , • 
their respective limits. We may then write
where * t , e a , t a , • • are infinitesimals (i.e. variables having zero for a
unit). Adding
(A) (»,+ »,+ *. + 0(*,+ VK+ ••) = («,+ «.+ «,+ ■■■>
20 DIFFERENTIAL CALCULUS
Since the righthand member is an infinitesimal by (1), p. 19, we
have, from the converse theorem on p. 18,
• •>
limit (1^+ v t + v 8 + . . .) = ^h J a + Z g +
or, limit (Vj + v a + v 8 H ) = limit 1^+ limit v a + limit v g H ,
which was to be proved.
Proof of Theorem II Let r, and v t be the variables, l x and Z f their
respective limits, and e 1 and € 2 infinitesimals; then
and v f '«+««•
Multiplying, v x v % = (l x + € t ) (7 2 + e a >
'A+'A+'A+V*
or,
Since the righthand member is an infinitesimal by (1) and (2), p. 19,
we have, as before,
limit (»a) = ^ 2 = limit v x . limit v a ,
which was to be proved.
Proof of Theorem UL Using the same notation as before,
"l Jl+ 'l Jl { (h+ *1 l x\
v t *,+ €, lj\l t +e t Ij
or,
^^.^ V I I € — I €
Here again the righthand member is an infinitesimal by (4), p. 19,
if Z # ¥= 0; hence , v , ,. ..
limitft) = ^^,
\v t / l t limit v t
which was to be proved.
It is evident that if any of the variables be replaced by constants,
our reasoning still holds, and the above theorems are true.
21. Special limiting values. The following examples are of special
importance in the study of the Calculus. In the following examples
a > and e *f* 0.
THEORY OF LIMITS
21
Written in the farm of limits.
(i)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
limit € __ _.
z = 0«>'
limit ^ „_ .
X = 00 '
Abbreviated form often used.
c
= 00.
oo;
limit f _
x = co (?
limit £ = A.
x = co a: »
!L"!V ^ = + oo, when a<l
X = — 00
X
"^ a* = 0, when a < 1
= + 00 '
""IIL «* = Of when a > i
a* =s f oo, when a > 1
im ^log a :r= + oo, when a < 1
(10) x ^Mog a * = ao, when a <1
(11) ^ log a :r =  oo, when a > 1
X =— 00
limit
x = + oo
limit
x
C • 00 = 00.
00 — ~N
— = 00.
00
a""" =f oo.
a +  = 0.
a
+ • _
= f oo.
lo g« = + 00.
lo g« (+ CO) =  00.
l°g«° = — oo
l°g«(+ oo) = + oo.
(12) x J™Mog.* = +ao, when a>l
The expressions in the second column are not to be considered as
expressing numerical equalities (oo not being a number); they are
merely symbolical equations implying the relations indicated in the
first column, and should be so understood.
22. Show that Umit A !^5 = l.*
x = x
Let be the center of a circle whose radius is unity.
Let arc AM = arc AM'=x, and let MT and M'T be tangents drawn
to the circle at M and M\ From Geometry,
MPM' < MAM' < MTM';
or 2 sin x < 2 x < 2 tan x.
Dividing through by 2 sin a, we get
.. x 1
sm x cos x
• If we refer to the table on p. 4, it will be seen that for all angles less than 10° the angle
in radian 8 and the sine of the angle are equal to three decimal places. If larger tables are
consulted, fiveplace, say, it will be seen that for all angles less than 2.2° the sine of the angle
and the angle itself are equal to four decimal places. From this we may well suspect that
limit sins
X=0 z
= 1.
22
DIFFERENTIAL CALCULUS
If now x approaches the limit zero,
limit x
x = sin x
limif 1
must lie between the constant 1 and A » which is also 1.
x = U cos x
Therefore limi * A = 1, or, limi * 5™£ = 1. Th. Ill, p. 18
# = U sin a: a: = U x
It is interesting to note the behavior of this function from its graph,
the locus of equation m
y =— •
Although the function is not defined for x = 0, yet it is not discon
tinuous when x = if we define
sin
~0~
= 1.
Case II, p. 15
23. The number e. One of the most important limits in the Cal
culus is
limi t (i + x y = 2.71828 •  • = e.
x= v y
To prove rigorously that such a limit e exists, is beyond the scope
of this book. For the present we shall content ourselves by plotting
the locus of the equation i
1
and show graphically that, as x = 0, the function (1 + #)*(= y)
X
V
X
y
10
1.2710
5
1.4310
2
1.7320
1
2.0000
.5
2.2600
.5
4.0000
.1
2.5937
.1
2.8680
.01
2.7048
.01
2.7320
.001
2.7169
.001
2.7195
\ Yi
e
1
ye
A
l
yi
1
O
X
takes on values in the near neighborhood of 2.718 •••, and therefore
e = 2.718 • • • approximately.
THEORY OF LIMITS 23
As x = from the left, y decreases and approaches e as a limit. As
x = from the right, y increases and also approaches e as a limit.
As x = oo, y approaches the limit 1 ; and as x = — 1 from the right,
y increases without limit.
In Chap. XVIII, Ex. 15, p. 233, we will show how to calculate the
value of e to any number of decimal places.
Natural logarithm* are those which have the number e for base.
These logarithms play a very important role in mathematics. When
the base is not indicated explicitly, the base e is always understood
in what follows in this book. Thus log t v is written simply log v.
Natural logarithms possess the following characteristic property:
If x = in any way whatever,
limit _*± ^ = limitlog(l + :r)i = log*=l.
x
24. Expressions assuming the form § • As oo is not a number, the
expression oo * oo is indeterminate. To evaluate a fraction assuming
this form, the numerator and denominator being algebraic functions,
we shall find useful the following
Rule. Divide both numerator and denominator by the highest power of
the variable occurring in either. Then substitute the value of the variable.
^ * t. i . limit 2x» — 3x* + 4
Illustrative Example 1. Evaluate m _  5 r—r •
x = co6x — x 2 — 7 x 3
i* "f 2 x* 3 x 2 1 A
Solution. Substituting directly, we get 1,nm — — = £, which isindeter
e ° x = co 6xx 2  7x» °°
minate. Hence, following the above rule, we divide both numerator and denominator
byz*. Then
2? + l
limit 2x«3x 2 + 4 = limit x x* = _ 2 A ^
x = oo6x — x 2 — 7x* z = 0O J>._l_ 7 7
x 2 x
EXAMPLES
Prove the following :
limit /* + l\ =1
x = co\ x /
^ limit /* + lX limit/ 1\
x = m\ x ) x = x>\ T c)
= J' mit (1) + J imit () Th. I, p. 18
= 1 + = 1.
24 DIFFERENTIAL CALCULUS
2 limit ( z* + 2x \ ^ _ 1
p. limit / z 2 + 2z \ _ limit [ *
rroo1  x = ooV63xV z==00 lj>_ a
[Dividing both numerator mud denominator by *•.]
limit L 2\
• = ^  Th. Ill, p. 18
limit /1X , limit /2\
limit /6\ _ limit ,«v
x = go \ x s/ x = 00 * '
= l+0 = 1
""03"" 3*
Th. I, p. 18
3 limit x«2x + 5 _l limit a J ±
„ limit 3x3 + 6x 2 2 . limit 2x* + 3x*
4. /x = • A*« __A = 00.
limit ^+1 _ 6 15 . "niit 5x*^2x =
5 x=2 a . + 3 ° 3 = « x
6. I imi £ (3ax 2 2Zix+6* 2 ) = 3ax 2 . 16. , limit * = 1.
7. i imit (ax 2 + bx + c) = 00. i 7 Hmit n(n + l) = x
Z ~ °° ' n = • (n + 2) (n + 8)
8 limit (x^2 fc x» = 1 „_,
* = x(x + A:) 18 ' « = 1 737 = 8 *
o limit *±1 =1. 1Q limit (x + ft)'  x«
»• x = oo 3x 2 + 2a ;_i 3 19 /i = ft = nx 1.
10 limit 3 + 2x _ HmitT //3j . M 8in/n
10  x = co x2 _ 5x " u ' 20. ft = ^cos((9 + *)— J = cos*
11. ] imi i C ° 8(Q  a) =  tan a. 21. limit **** =  * .
2C08(2a — a) * = c <>4— 3x 2 3
12 limit a3;2 + to + C =  22 limit 1 "" C08 ^ = 1
* * = «>ax 2 + ex+/ d' ' * = ° tf 2 2*
23. ' _ =— co, if x is increasing as it approaches the value a.
24. _ im / a =+ co, if x is decreasing as it approaches the value a.
CHAPTER IV
DIFFERENTIATION
25. Introduction. We shall now proceed to investigate the man
ner in which a function changes in value as the independent variable
changes. The fundamental problem of the Differential Calculus is to
establish a measure of this change in the function with mathematical
precision. It was while investigating problems of this sort, dealing
with continuously varying quantities, that Newton * was led to the
discovery of the fundamental principles of the Calculus, the most
scientific and powerful tool of the modern mathematician.
26. Increments. The increment of a variable in changing from one
numerical value to another is the difference found by subtracting the
first value from the second. An increment of x is denoted by the
symbol A#, read delta x.
The student is warned against reading this symbol delta times x,
it having no such meaning. Evidently this increment may be either
positive or negative * according as the variable in changing is increas
ing or decreasing in value. Similarly,
Ay denotes an increment of y,
A<f> denotes an increment of <£,
Af(x) denotes an increment of /(#), etc.
If in y =f(x) the independent variable x takes on an increment Axy
then Ay is always understood to denote the corresponding increment
of the function f(x) (or dependent variable y).
The increment Ay is always assumed to be reckoned from a definite
initial value of y corresponding to the arbitrarily fixed initial value of x
from which the increment Ax is reckoned. For instance, consider the
function y = ^r 2 .
* Sir Isaac Newton (16421727), an Englishman, was a man of the most extraordinary
genius. He developed the science of the Calculus under the name of Fluxions. Although
Newton had discovered and made use of the new science as early as 1670, his first published
work in which it occurs is dated 1687, having the title PhiloAophiae Naturalis Principia
Mathematical. This was Newton's principal work. Laplace said of it, " It will always remain
preeminent above all other productions of the human mind." See frontispiece.
t Some writers call a negative increment a decrement,
25
26
DIFFERENTIAL CALCULUS
Assuming x = 10 for the initial value of x fixes y = 100 as the initial
value of y.
Suppose x increases to rr = 12, that is, Ax = 2;
then y increases to y = 144, and Ay = 44.
Suppose x decreases to x = 9, that is, A# =— 1 ;
then y decreases to y = 81, and Ay = — 19.
It may happen that as x increases, y decreases, or the reverse ; in
either case Ax and Ay will have opposite signs.
It is also clear (as illustrated in the above example) that if y =f(xy
is a continuous function and Ax is decreasing in numerical value, then
Ay also decreases in numerical value.
27. Comparison of increments. Consider the function
00 y = **.
Assuming a fixed initial value for re, let x take on an increment Ax*
Then y will take on a corresponding increment Ay, and we have
y + Ay = (x + Ax)\
or, y 4 Ay = 2^4 2 x • A# 4 (Aa;)'.
Subtracting (.4), y =3?
(#) Ay= 2 a; . As + (Aa;)*
we get the increment Ay in terms of a: and Ax.
To find the ratio of the increments, divide (2?) by As, giving
Ay_
Ax
= 2x4 Ax.
If the initial value of a: is 4, it is evident that
limit Ay _ ft
Ax = 0Ax"
Let us carefully note the behavior of the ratio of the increments of
x and y as the increment of x diminishes.
Initial
New
Increment
Initial
New
Increment
Ay
value of x
value of x
Ax
value of y
value of y
Ay
AX
4
5.0
1.0
16
25.
9.
9.
4
4.8
0.8
16
23.04
7.04
8.8
4
4.6
0.6
16
21.16
5.16
8.6
4
4.4
0.4
16
19.36
3.36
8.4
1
4
4.2
0.2
16
17.64
1.64
,' 3.2
4
4.1
0.1
16
16.81
0.81
8.1
4
4.01
0.01
16
16.0801
0.0801
8.01
DIFFERENTIATION 27
It is apparent that as Ax decreases, Ay also diminishes, but their
ratio takes on the successive values 9, 8.8, 8.6, 8.4, 8.2, 8.1, 8.01 ;
illustrating the fact that — can be brought as near to 8 in value as
Ax
we please by making Ax small enough. Therefore
limit ^ = 8 *
Ax = Ax
28. Derivative of a function of one variable. The fundamental
definition of the Differential Calculus is :
The derivative * of a function is the limit of the ratio of the increment
of the function to the increment of the independent variable, when the lat
ter increment varies and approaches the limit zero.
When the limit of this ratio exists, the function is said to be differ
entiable, or to possess a derivative.
The above definition may be given in a more compact form symbol
ically as follows : Given the function
00 y =/(*)«
and consider x to have a fixed value.
Let x take on an increment Ax ; then the function y takes on an
increment Ay, the new value of the function being
(£) y + Ay=f(x + Ax).
To find the increment of the function, subtract (4) from (2?), giving
(C) Ay=f(x + Ax)f(x).
Dividing by the increment of the variable, Ax, we get
( m *l = /(s + AaQ/(aQ
K J Ax Ax
The limit of this ratio when Ax approaches the limit zero is, from our
definition, the derivative and is denoted by the symbol —• Therefore
\AfJu
rF , dy_ limit /(x + Ax)/(x)
w di~Ax=0 ■&
defines the derivative of y \prf\x)] with respect to x.
•The student should guard against the common error of concluding that because the
numerator and denominator of a fraction are each approaching zero as a limit, the limit of
the value of the fraction (or ratio) is zero. The limit of the ratio may take on any numerical
value. In the above example the limit is 8.
f Also called the differential coefficient or the derived function.
28 DIFFERENTIAL CALCULUS
From (D) we also get
v J * <ty = limit Ay
dx Ax = 0a*'
The process of finding the derivative of a function is called differ
entiation.
It should be carefully noted that the derivative is the limit of the
ratio, not the ratio of the limits. The latter ratio would assume the
form » which is indeterminate (§ 14, p. 12),
29. Symbols for derivatives. Since Ay and A# are always finite and
have definite values, the expression
Ay
Ax
is really a fraction. The symbol
dy^
dx
however, is to be regarded not as a fraction but as the limiting value of
a fraction. In many cases it will be seen that this symbol does possess
fractional properties, and later on we shall show how meanings may
be attached to dy and dx, but for the present the symbol ^ is to be
considered as a whole.
Since the derivative of a function of x is in general also a function
of x, the symbol f f (x) * s ^ so use d to denote the derivative of f(x).
Hence, if y =/(#),
we may write ~ =f'Cx),
ax
which is read the derivative of y with respect to x equals f prime of x
The symbol d
dx
when considered by itself is called the differentiating operator, and
indicates that any function written after it is to be differentiated with
respect to x. Thus
~ or — y indicates the derivative of y with respect to x ;
dx dx
— /(#) indicates the derivative off(x s ) with respect to x;
— (22*+ 5) indicates the derivative of 2^45 with respect to x*
d
y' is an abbreviated form of JL.
DIFFERENTIATION 29
A
The symbol D x is used by some writers instead of — • If then
ax
y =/(*),
we may write the identities
*'=  = £* = £**) =*,/(*) =A*>
30. Differentiable functions. From the Theory of Limits it is clear
that if the derivative of a function exists for a certain value of the
independent variable, the function itself must be continuous for that
value of the variable.
The converse, however, is not always true, functions having been
discovered that are continuous and yet possess no derivative. But
such functions do not occur often in applied mathematics, and in this
booh only differentiable functions are considered, that is, functions that
possess a derivative for all values of the independent variable save at
most for isolated values.
31. General rule for differentiation. From the definition of a deriv
ative it is seen that the process of differentiating a function y =/(a?)
consists in taking the following distinct steps :
General Rule for Differentiation*
First Step. In the function replace x by x + As, giving a new value
of the function, y + Ay*
Second Step. Subtract the given value of the function from the new
value in order to find Ay (the increment of the function).
Third Step. Divide the remainder Ay (the increment of the function)
by Ax (the increment of the independent variable).
Fourth Step. Find the limit of this quotient, when Ax (the increment
of the independent variable) varies and approaches the limit zero. This
is the derivative required.
The student should become thoroughly familiar with this rule by
applying the process to a large number of examples. Three such
examples will now be worked out in detail.
Illustrative Example 1. Differentiate 3z* + 5.
8ohitta. Applying the successive steps in the General Rule, we get, after placing
y = 3z» + 5,
Fir* step. y + Ay = 3(x + Ax)* + 5
= 3x* + 6x . Ax + 3(Ax)» + 5.
* Also called the Fourstep Rule.
30 DIFFERENTIAL CALCULUS
Second step. y + Ay = 3x 2 + 6x • Ax + 3(Ax) 2 + 6
y = 3x 2 +6
Ay = 6xAx + 3(Ax) 2 .
Tfeirdstep. — = 6x + 8Ax.
Ax
dy
Fourth step. 2. = 6x. Ans.
dx
We may also write this
i?(3x 2 + 6) = 6x.
ax
Illustrative Example 2. Differentiate x 8 — 2x + 7.
8olntk>n. Place y = x»2x + 7.
First step. y + Ay = (x + Ax)» — 2(x + Ax) + 7
= x 3 + 3x 2 • Ax + 3x • (Ax) 2 + (Ax)»  2x 2 . Ax + 7.
Second step. y + Ay = x 8 + 3x 2 • Ax + 3x • (Ax)* + (Ax)» — 2x — 2 . Ax + 7
y = x* — 2x + 7^
Ay= 3x 2 Ax + 3x.(Ax)» + (Ax)» 2 Ax.
Third step. — = 3x 2 + 3x • Ax + (Ax) 2  2.
Fourth step. ^ = 3x 2 — 2. ^ns.
dx
Or, ^ (x»  2x + 7) = 8x 2  2.
ox
(•
Illustrative Example 3. Differentiate —  •
x 2
c
Solution. Place y = — •
First step. y + Ay =
Second step. y + Ay =
y =
(x + Ax) 2
c
(x + Ax) 2
c
X 2
_ c c __ — c Ax(2x + Ax)
(x + Ax) 2 x 2 ~ x 2 (x + Ax) 2
TO . , . Ay 2x + Ax
Third step. — = — c
Ax x 2 (x + Ax) 2
Fourttstep. *=« 2z
dx x 2 (x) 2
= • Ans.
x"
n d/c\ 2c
DIFFERENTIATION
31
Use the General Rule, p. 29, in differentiating the following functions :
1. y = 3x 2 . An*. ^ = 6x.
dx
7. y = x». ^*«. ^ = 3x 2 .
dx
2. y = x 2 + 2. $^ = 2x.
dx
8. y = 2x 2 3.
dx = 4x '
3. y = 54x. ^=4.
dx
9. y = l2x».
^=6x 2 .
dx
n^ ^ <** as
4. 3 = 2** — 4.  = 4f.
dt
10. p=a$*.
% = 2aB.
d*
K 1 dy 1
5. y = . _= — .
x dx x s
11. y = 4
X 2
dx" x«'
x + 2 dy__ 2
*>. y — ■■ • . — •
x dx x s
12. y= , 3 •
X 2 — 1
dy 6x
dx (x 2  l) 2
13. y = 7x 2 + x. 18.
y =
ftx 8 — ex.
23. y = Jx 2 + 2x.
14. 8 = at*2bt. 19.
p =
3^2^.
24. z = 4x — 3x 2 .
16. r = 8* + 3t*. 20.
y =
Jx 2 Jx.
26. p = 3^ + ^.
16. y = ~ 21.
X s
y =
x 2 5
X
26. y = aX + 6 .
X 2
17 fl  fl 99
p =
1 + rf
<~ * g + 2
"" 21 + 3
X
28. y = x 2 3x + 6.
^7i«. ^ = 21
3.
29. 3 = 20 + 6* 8.
« / = 4( + 6.
30. p = 60«20+6.
p' = 160
'2.
31. y = ax* + 6x + c.
y'ss 2 ax
+ 6.
32. Applications of the derivative to Geometry. We shall now
consider a theorem which is fundamental in all applications of the
Differential Calculus to Geometry. Let
be the equation of a curve AB.
Now differentiate (4) by the General Rule
and interpret each step geometrically.
y + Ay =f(x + Ax)
yfAy =/(a:f A*)
First Step.
Second Step.
= MP = NR
Thibd Step.
Ay =/<> + As) f{x) = i*0.
Ay ^ f(x + Ax) /O) _RQ _RQ
Ax Ax
= tan JRP# = tan<£
= slope of secant line PQ.
JfAT PR
32 DIFFERENTIAL CALCULUS
Fourth Step.
i of the derivative at P.
limit A# _ limit f(_x + Ag ) /(»
A* = Ax Aj = Aj
(*)
■2'
But when we let Aj = 0, the poiut <> will move along the curve and
approach nearer and nearer to P, the secant will turn ahout P and
approach the tangent as a limiting position, and we have also
limit Au limit <_ . .
a o r 1 — * a ta11 * = tan t
Q y A* = A* Ax — T
(£7) = slope of the tangent at P.
Hence from (B~) and (C),
= slope of the tangent line PT. Therefore
Theorem. The value »f the de r iv ati v e at any paint of a curve it eipial
to the tlope of the line drawn tangent to the curve at that point.
It was this tangent problem that led Leibnitz * to the discovery of
the Differential Calculus.
Illustrative Exam flu I. Find the slopes at Hits tangents to the parabola y = x 1
at tbe vertex, mid at the point where x = \.
Solution. Differentiating by General Rule, p. 29, we get
(A) — = 2z = slope of tangent line at any point on curve.
To find slope of tangent at vertex, substitute x = in {A),
giving — = 0.
Therefore the tangent at vertex baa the slope zero ; that is, it is
parallel to the axis of x and in this case coincides with it.
To find slope of tangent at the point P, where x = \, substitute
in (X), giving dy
dx~ ''
that 1b, the tangent at the poiut P makes an angle of 45° with tbe
Gottfried Wilbelm Leibnitz (16461716) was a native of Leipzig. His remarkable abili
ties were shown by original Investigations in several branchesol learning. He was first to pub
lish his discoveries in Calculus in a short essay appearing In the periodical Acta fimditonim
at Leipzig in 1684. It is known, however, that manuscripts on Fluxions written by Newton
were already in existence, and from these some claim Leibnitz got tbe new ideas. The decision
of modem times seems to be that both Newton and Leibnitz invented tbe Cali'nlus independ
ently of each other. The notation used today was introduced by Leibnitz. See fi
DIFFERENTIATION 83
1. y = x»  4,
where x = 2.
2. y = 6 — Sx*
where x = 1.
3. y = x«,
where x = — 1.
4. y = ,
X
where x = — 2.
6. y = x — x*,
where x = 0.
6. y = — L,
x — 1
where x = 3.
1 >
7. y = x*.
2 '
where x = 4.
Find by differentiation the slopes of the tangents to the following curves at the
points indicated. Verify each result by drawing the curve and its tangent.
Ans. 4.
6.
8.
_1
2*
1.
_1
i'
4.
8. y = x» — 2x + 3, where x = 1. 0.
9. y = 9 — x a , where x = — 3. 6.
10. Find the slope of the tangent to the curve y = 2x* — 6x + 5, (a) at the point
where x = 1; (b) at the point where x = 0. Ans. (a) ; (b) — 6.
11. (a) Find the slopes of the tangents to the two curves y = 8x a — 1 and y = 2 x* + 3
at their points of intersection, (b) At what angle do they intersect ?
Ans. (a) ± 12, ± 8; (b) arc tan ^y.
12. The curves on a railway track are often made parabolic in form. Suppose that
a track has the form of the parabola y = x* (last figure, p. 82), the directions OX and
OY being east and north respectively, and the unit of measurement 1 mile. If the train
is going east when passing through 0, in what direction will it be going
(a) when \ mi. east of OY ? Ans. Northeast.
(b) when } mi. west of OF? Southeast.
Vs
(c) when — mi. east of OY ? N. 30° E.
2
(d) when ^ mi. north of OX? E. 30° S., or E. 30° N.
13 A streetcar track has the form of the cubical parabola y = x 8 . Assume the
same directions and unit as in the last example. If a car is going west when passing
through O, in what direction will it be going
(a) when — mi. east ot OY? Ans. Southwest.
V8
(b) when — = mi. west of OF? Southwest.
V8
(c) when J mi. north of OX ? S. 27° 6& W.
(d) when 2 mi. south of OX ?
(e) when equidistant from OX and OY?
CHAPTER V
RULES FOR DIFFERENTIATING STANDARD ELEMENTARY FORMS
33. Importance of General Rule. The General Rule for differentia
tion, given in the last chapter, p. 29, is fundamental, being found
directly from the definition of a derivative, and it is very important
that the student should be thoroughly familiar with it. However, the
process of applying the rule to examples in general has been found
too tedious or difficult ; consequently special rules have been derived
from the General Mule for differentiating certain standard forms of
frequent occurrence in order to facilitate the work.
It has been found convenient to express these special rules by
means of formulas, a list of which follows. The student should not
only memorize each formula when deduced, but should be able to
state the corresponding rule in words.
In these formulas w, t>, and w denote variable quantities which are
functions of x y and are differentiate.
Formulas for Differentiation
d , . du do dm
dx K J dx dx dx
d / . do
IT dk W = C dx
d , N do du
y — (uv)=u \v — 
dx dx dx
d , _ N _ . do
Yi . (V 1 ) = m/ 1 ' 1 —
Yla ^.(x«)=nx" 1 .
dx
34
RULES FOR DIFFERENTIATING 86
TO
vna
nn
du
dp
v
u —
dx\v)
v*
du
dx\c)
dx
c
do
d .
dx
1( 1 °8««')
= log„«
• iHB •
dx
A;
V
d .
dr
(log.)
V
nna
d do
H — ((f) =0*108 a —
dx K J * dx
d do
d * du  do
— (if) =w° 1 — + loguvr—.
dx K J dx * dx
d do
II — (sini?) =cosi? — •
d!r v y dx
m — (cos?) = — sint> — .
— ftantA = sec a t> — .
un — (tani>)
dx dx
d .do
IIY — (cot?) = — csc*v — .
dx K J dx
d do
dx K J dx
d do
IYi — (cscv) = — csctrcoti? — «
d!r ' dx
IVU — (vers i/) = sin i> — 
dr J dx
dv
d dx
IYIII — (arc sin i/) =
dx Vl — i^
M DIFFERENTIAL CALCULUS
dv
d dx
IH — (arc cos v) = —
dx Vii^
dv
d dx
II — (arc tan v) =
if dr
IH — (arccottr) = —
dv
d dx
Iffl — (arc sec i>) =
dx irvV — i
db
Hin —(arc esc v) = —
A;
<f , dx
IHY — (arcversi>) =
dx V20 — i^
{ft/ {ft/ {fp
HY  = — — , y being a function of p.
dx dv dx
dy 1
XIYI — = — , y being a function of x.
dx dx
m
dy
34. Differentiation of a constant. A function that is known to have
the same value for every value of the independent variable is constant,
and we may denote it by y = &
As x takes on an increment Ax, the function does not change in
value, that is, Ay = 0, and a
Ax
But ?"**(¥)%*•
dc „
I .'. — = 0.
dx
The derivative of a constant is zero.
BULES FOB DIFFERENTIATING 87
35. Differentiation of a variable with respect to itself.
Let y = x.
Following the General Rule, p. 29, we have
First Step. y+Ay^x + Ax.
Second Step. Ay = Ax.
Third Step. t^**!
Ax
Fourth Step.
ax
dx
The derivative of a variable with respect to itself is unity.
36. Differentiation of a sum.
Let y = u f v — w.
By the General Rule,
First Step. y f Ay = u + Au + v + Av — w— Aw.
Second Step. Ay = At* + Av — Aw.
ma Ay Au t Av Aw
Third Step. ^=: — \ — •
At At At At
_ a dy du , dv dw
Fourth Step. & = — f — — •
dx dx dx dx
[Applying Th. I, p. 18.]
d , N du , du dw
Similarly, for the algebraic sum of any finite number of functions.
The derivative of the algebraic sum of a finite number of functions is
equal to the same algebraic sum of their derivatives.
37. Differentiation of the product of a constant and a function.
Let y = cv.
By the General Rule,
First Step. y + Ay = c (y f Av) = cv + cAv.
Second Step. Ay = c • Av.
38 DIFFERENTIAL CALCULUS
r« Ay Av
Third Step. —* = <?—•
Ax Ax
Foubth Step. j =c 1"
dx dx
[Applying Th. II, p. 18.]
d do
The derivative of the product of a constant and a function is equal to
the product of the constant and the derivative of the function.
38. Differentiation of the product of two functions.
Let y = uv.
By the General Rule,
First Step. y + Ay = (u + Aw) (v + Av)
Multiplying out this becomes
y + Ay = Mt> + tt'At; + ti' Aw f Au • Av.
Second Step. Ay = u • Av f v • Aw f Au • Av.
™ « Av Av , Aw , A Av
Third Step. — £• = w  — h v  — h Aw — •
At A y AT At
~ c?V rfv rfw
Fourth Step. ^ = w — + v — •
az dx ax
[Applying Th. II, p. 18, since when Ax *©, Au *0, and (au —J *0. 1
d , . do da
dx K J dx dx
The derivative of the product of two functions is equal to the first
function times the derivative of the second, plus the second function
times the derivative of the first
39. Differentiation of the product of any finite number of functions.
Now in dividing both sides of T by uv, this formula assumes the
form , , ,
d , N du dv
— Cuv) — —
dx __dx dx
uv u v
RULES FOR DIFFERENTIATING 39
If then we have the product of n functions
y = Vs a "*»'
we may write
d , N dv. d N
— i — ? — (V v • • • v 1
_ dr dx dx K * 4 nJ
rf^ rft> 2 rfv 8 <fo w
__ (£r dx dx dx
Multiplying both sides by vj) % • • • v„, we get
dx y * * nJ v * 8 " y cfo; v l 8 " y cfo;
Tfo derivative of the product of a finite number of functions is equal
to the sum of all the products that can be formed by multiplying the
derivative of each function by all the other functions.
40. Differentiation of a function with a constant exponent. If the
n factors in the above result are each equal to v, we get
dx K } dx
= n— .
tf v
d , _ v __, dv
dx K J dx
When v = x this becomes
We have so far proven YI only for the case when n is a positive
integer. In § 46, however, it will be shown that this formula holds true
for any value of n, and we shall make use of this general result now.
The derivative of a function with a constant exponent is equal to the
product of the exponent^ the function with the exponent diminished by
unity, and the derivative of tlie function.
40 DIFFERENTIAL CALCULUS
41. Differentiation of a quotient.
Let
u
V
v*Q.
By the General Rule,
First Step.
, u+Au
v + Ay
Second Step.
u + Am ti
A y = =
yH Av v
v An — uAv
v(y fAv)
Third Step.
Am Av
. v u —
Ay Ax Az
Ax v (y + Av)
du dv
Fourth Step.
v u —
dy dx dx
dx v 2
[Applying Theorems II and III, p. 18.]
du dv
v u —
d /u\ dx dx
The derivative of a fraction is equal to the denominator times the
derivative of the numerator, minus the numerator times the derivative
of the denominator, all divided by the square of the denominator.
When the denominator is constant, set v = c in TO, giving
du
Vila
dx\c)~ c
[8lnce^^.0.1
L dx dx i
We may also get VII a from IY as follows :
du
d /u\ __ 1 du _ dx
dx\c) c dx c
The derivative of the quotient of a function by a constant is equal to
the derivative of the function divided by the constant.
All explicit algebraic functions of one independent variable may be
differentiated by following the rules we have deduced so far.
&ULES FOB DIFFERENTIATING 41
Differentiate the following :
1. y = x».
Mutton. ^ = ^(x») = 3x*. Ana. By Via
ax ax
[n3.]
2. y = ax 4 — foe 2 .
Solution. ^ = A( ax 4_ te 2) = i.( ax 4)i.(te«) by III
di dx ax ax
= 4ax 8 2foc. An*. By Via
3. y = x» + 6.
= f x» . An*. By VI a and I
3x* 7x , q T/r
4. y = — — = + 8 Vx*.
v^x 2 #?
Solution. ^ = ^(3xV)  l(7x~i) + ^(8x*) by III
dx dx dx dx
= y x* + Jx"* + Y x ~* 4n*. " By IV and Via
5. y = (x*  8)«.
Solution. ^ = 6(x 2 3)*^(x 2 8) by VI
ax ax
[r  x*  3 and n  5.]
= 6(x 2 3) 4 2x=10x(x 2 8) 4 . An*.
We might have expanded this function by the Binomial Theorem and then applied
III, etc., but the above process is to be preferred.
6. y = Va 2  x 2 .
Solution. $L = A (a 2  x 2 )* = 1 (a 2  x 2 )~* ~ (a 2  x 2 ) by VI
dx dx 2 ax
[r  a*  x* t and n — J.]
= !( a 2x 2 )*(2x) = * Ana.
2 Va 2  x 2
7. y = (8x 2 + 2)Vl + 5x 2 .
Solution, p = (3x 2 + 2)^(1 + 6x 2 )* + (1 + 5x 2 )* ^(8x 2 + 2) by V
ax ax ax
[u3x* + 2, and * (l + Sx*) 1 ".]
= (8x 2 + 2)1(1 + 5x 2 )* A(l + Sx 2 ) + (1 + 6x 2 )*6x by VI, etc.
2 dx
= (8x 2 f 2) (1 + bz*f*bx + 6x(l + Sx 2 )*
5x(3x 2 + 2) n n T? 45x»+16x .
Vl + 5x 2 Vl + 5x 2
• When learning to differentiate, the student should have oral drill in differentiating
rimple functions.
42 DIFFERENTIAL CALCULUS
8. y =
Va«  x»
(a 3  * 2 )*^(a 2 + x*)  (a 2 + x 2 )^ (a 2  x 2 )*
—  I" ~ —* *™
_ 2x(a 2  x») + z(a* 4 x 2 )
(a»x»)i
[Multiplying both numerator and denominator by (a* a*)'.]
3a*XX» .
= An*.
(a 2  x 2 )i
9. y = 5x* 4 8x«  6. ^ = 20x* + 6x.
ax
10. y = 3cx 2 8dx + 6e. ^ = 6cx8d.
dx
11. y = x»+». ^ = (a + &)x«+»i.
ax
12. y = x» 4 nx + n. ^ = nx* 1 + n.
dx
13. /(x) = 1 x»  1 x* 4 5. /Xx) = 2x 2  3x.
14. /(x) = (a 4 6)x* 4 ex 4 d. Ax) = 2(a + &)x + c.
15. A/ a + te + ex*) = 6 4 2cx. 21. ^(2x» + 6) = dx 1 .
dx ax
16. — (6jr3/46) = 6mjr I 3. 22. ^(SL*  2t«) = 15* 4  4*.
dy ax
17. — (2x»43x«)=4x«»x*. 23. £(a0* 4 W) = 4a*» + b.
dx off
18. A(3* 4 a)=12*«l. 24. A(5 2a*) = 3a*.
cfo aa
19. — (4x* 4 x*) = 2x _ * + 2x. 26. ^(9*J + **) = 15t*  «*.
dx dt
20. — (y 2 4y"*) = 2y«42y"i. 26. ^(2x 12  x 2 ) = 24x" 9x»
dy dx
TTt. r = c0«4d0*4e*. r y = Zc0* + 2d* + e.
28. y = 6xJ 4 4xi 4 2x'. /= 21x4 + 10xi + 8xL
29. ysVSx" + *£ + !• •= —?= + £=  ^
x 2V3x 3*x« **
_ _ a 4 te 4 ex* a
80. y = ^=0.
x z*
31. y = ^ — ^ y / =fx*5x^42xi + Jxt.
x*
ao x*x — x*4a 2X? 4x42xi3a
o«. y = • y  •
z \ 2x*
33. y = (2x» 4 x 2  »)*. /= 6 r (3x 4 D (2z» 4 x 2  6) f .
RULES FOR DIFFERENTIATING 43
34. f{x) = (a + bz*)l f(x) = ^? (a + te»)* .
36. /(x) = (l + 4x«)(l + 2x*). /"(*) = 4x(l + 3x + 10x«).
36. /(x) = (a + x) Va"^. /* (x) _ a ~Z}*L •
2va — x
37. /(x) = (a + x)*(& + x)«. /"(x) = (a + *)*(6 + *)" f^ + j51
La + x o + xj
38. » = 1 * B
z" dx x«+ l
39. y = x(«« + x») VSH^i. * «< + «'x«4x«
<k Vo»  x»
40. Differentiate the following functions :
(a) £(2x»  4x + 6). (e) (6 + of)*. (i) £(xt  at).
(b) i(aP + «•  9). (0 £{x*  a*)L (j) (5 + 2t)l.
( c) 1 (8*1 _ 2*1 + 6*). (g) A (4 _ #*). (k)  VaOVi.
(d) £ (2 x» + x)*. (h) i VT+9P. (I) 5 (2 ** 4 2 x»).
., 2x« dy 86*x«4z«
41. y = —
42. y =
f
43. « =
6»  x* dx (6 s  x») 2
a — x dy __ 2a
a + x" dx""(a + x) f "
fr ds SP + fr
44. /w =<i±^!. rw= (<+2)( : x t 4)
46. /(*) = — 1=. /*(*) = a
60.
y =
a
X*.
61.
y =
= («*
xl)l.
VaW* ( a _ ftp)l
46. F(r) = v/i 4 ^ . J"/,) _ *
YK ' \lvt rw (ly)*»
xvT+x* x'tl + s 1 )*
49. y=V2px. /=.
6*x
a*y
■*
44 DIFFERENTIAL CALCULUS
52. r=V^ + cV?. ^VZ + Sc*
63. u = — 4 — u'= +
cd d c
V ^ lr (gi)l
55. Differentiate the following functions:
(a)
x*
d / q«  s« \ */_^!_\ d
(b) 1 f^V (e) * ( a **). (h) A _i±£!_
dx\vT
Wlx/ w dx x W (tt\l*«
42. Differentiation of a function of a function. It sometimes happens
that y, instead of being defined directly as a function of x, is given as
a function of another variable v, which is defined as a function of x.
In that case y is a function of x through v and is called a function of
a function.
2v
For example, if y = zr~ — 5'
and v = 1 — x 2 ,
then y is a function of a function. By eliminating v we may express y
directly as a function of x, but in general this is not the best plan
when we wish to find ~
dx
If y =/(l , ) and v = <f> (x), then y is a function of a; through v. Hence,
when we let x take on an increment Ax, v will take on an increment At>
and y will also take on a corresponding increment Ay. Keeping this
in mind, let us apply the General Rule simultaneously to the two
functions y =/(l0 and v=(f> (a:) .
First Step, y + Ay =f(y + Av) t> + Ai> = <£ (x + Ax)
Second STEP.y f Ay =/(v f Av) v f A t> = <f> (x f Ax)
y =/0) *> =<ftO)
Ay =/0 + Av) — /O), Av = <£ (a; + Ax) — <f> (x)
Third Step. *l = /(HAti)/(ti) t Av = 0(g+As)0(*) .
Av Av Ax Ax
RULES FOR DIFFERENTIATING 45
The lefthand members show one form of the ratio of the increment
of each function to the increment of the corresponding variable, and
the righthand members exhibit the same ratios in another form. Before
pissing to the limit let us form a product of these two ratios, choos
ing the lefthand forms for this purpose.
This gives — * . which equals — "■
fi Ai> Ax * Ax
Write this &&■£.
Ax Ac Ax
Fourth Step. Passing to the limit,
dy dv
dv dx
w %!■%■ »*»>»
This may also be written
(*)
// y =/('') and v = ^> (x), the derivative of v with respect to x equals
\t product of the derivative of y with respect to v and the derivative of v
rith respect to x.
43. Differentiation of inverse functions. Let y be given as a function
of i by means of the relation « =f(x\
It is usually possible in the case of functions considered in this book
to solve this equation for x, giving
»+o>:
t is, to consider y as the independent and x as the dependent
iable. In that case /(x)and<J>(y)
said to be inverse functions. When we wish to distinguish between
two it is customary to call the first one given the direct function
mi the second one the inverse function. Thus, in the examples which
follow, if the second members in the first column arc taken as the
direct functions, then the corresponding members in tiie second column
will be respectively their inverse functions,
yx»+i, x=±Vyl.
y = a*, x = log.y.
y = sin x, x = arc sin y.
Let us now differentiate the inverse functions
y=/(x) and x = <f>(y)
mltaneously by the General Rule.
46 DIFFERENTIAL CALCULUS
First Step. y+Ay=/(a;f Aa;) a;+Aa;=:^(yf Ay)
Second Step. yf Ay=/(a;+Aa;) a:f Aa; = <£(yf Ay)
y =/(*) a; =<Ky)
Ay =/<> 4 Aa:)/(a;)' As = <f> (y + Ay) <f> (y>
Third Step. Ay^ /(ar+AaQ/(aQ Aa: = <ft(y+Ay)<Ky)
Ax Aa; Ay Ay
Taking the product of the lefthand forms of these ratios, we get
Ay Aa; =1
Aa; Ay
or, Ay 1
Ar Ax
Fourth Step. Passing to the limit,
or,
*'00
Tfte derivative of the inverse function is equal to the reciprocal of the
derivative of the direct function.
44. Differentiation of a logarithm.
Let y = log a v.*
Differentiating by the General Rule, p. 29, considering v as the
independent variable, we have
First Step. y f Ay = log a (y f Ai>).
Second Step. Ay = log a (y f Av) — log a v t
[By 8, p. 1.]
* The student mast not forget that this function is defined only for positive values of the
base a and the variable v.
t If we take the third and fourth steps without transforming the righthand member,
there results:
Third step. Ay_log n (« + A P )log a t>
Av Av
Fourth step. —   . which is indeterminate. Hence the limiting value of the righthand
do
member in the third step cannot be found by direct substitution, and the above transform
mation is necessary.
RULES FOR DIFFERENTIATING 47
Av
[Dividing the logarithm by r and at the same time multiplying the exponent of the]
parenthesis by v changes the form of the expression hat not its value (see 9, p. 1). J
Fourth Step. ^ = locr e.
dv v Ba
V
[whenA»*0,^*0. Therefore ^^(l + ^^'e, from p. 22, placing x=^l
Hence
Since v is a function of x and it is required to differentiate log a t>
with respect to x, we must use formula (A), § 42, for differentiating
a function of a function, namely,
dy _ dy dv
dx dv dx
dv
Substituting value of — from (^4), we get
ty _ i 1 dv
dx v dx
dv
d dx
Tin /. — (log a ») = log a e
dx v
When a = c, log a e = log e e = 1, and VIII becomes
A;
Till a — (log v) = — .
dx v
The derivative of the logarithm of a function is equal to the product
of the modulus * of the system of logarithms and the derivative of the
function, divided by the function.
* The logarithm of e to any base a ( logae) is called the modulus of the system whose
base is a. In Algebra it is shown that we may find the logarithm of a number JV to any
base a by means of the formula
loga JV loga e • loge JV p^
The modulus of the common or Briggs system with base 10 is
log !0 e«. 434294 •••.
48 DIFFERENTIAL CALCULUS
45. Differentiation of the simple exponential function.
Let y = a 9 , a >
Taking the logarithm of both sides to the base e, we get
logy = v logo,
log a
logy.
log a
Differentiate with respect to y by formula Villa,
dv 1 1.
ty log a y'
and from (C), § 43, relating to inverse functions, we get
or,
 = loga.y,
0*) Ah 10 *"'""'
Since v is a function of £ and it is required to differentiate a 9 with
respect to x, we must use formula (A), § 42, for differentiating a
function of a function, namely,
dy dy dv
<£r dv dx
Substituting the value of — from (4), we get
dy , „ dv
3^ = logaa r — •
flte CLE
d dv
H /. — (a») = log a • a?
dx dx
When a = e, log a = log e = 1, and IX becomes
d dv
Ha — (€•) = €*—.
The derivative of a constant with a variable exponent is equal to (he
product of the natural logarithm of the constant, the constant with the
variable exponent, and the derivative of the exponent.
RULES FOR DIFFERENTIATING 49
46. Differentiation of the general exponential function.
Let y = u p .*
Taking the logarithm of both sides to the base e,
!og.y = v Jog."*
or, y = e vh * ".
Differentiating by formula II a,
=^.£ (v iog„)
=*e Ph * m
(v du , , dv\ , „
u /v du , , dv
)
• d , ^ ^ mdu * dv
dx y J dx * dx
The derivative of a function with a variable exponent is equal to the
mm of the two results obtained by first differentiating by VI, regarding
the exponent as constant; and again differentiating by II, regarding the
function as constant
Let ti = n, any constant ; then I reduces to
d du
But this is the form differentiated in § 40; therefore TI holds true
for any value of n.
Illustrative Example 1. Differentiate y = log(x* + <*)•
Solution. ay_ax_ by Villa
dx z* + a '
[v  x» + a.]
= — Ana.
z* + a
Illustrative Example 2. Differentiate y = log Vl — z*.
Solution. ay = ax tyr Villa
** (1  x 2 )*
i(is»r*(2x)
(1  x 2 )*
Ans,
by VI
x 2 l
* it can here assume only positive values.
60 DIFFERENTIAL CALCULUS
Illustrative Example 8. Differentiate y — a***.
Solution. ? = log a • a 8 ** — (3x») by IX
ax dx
= Oxloga'tt 3 **. 4n«.
Illustrative Example 4. Differentiate y = te *** 1 .
Solution. ^ = &iL (<*• + **) b yIV
dx dx
 fc^**^ (c 2 + x*) by IX tf
dx
= 2foxc cf +**. Ans.
Illustrative Example 5. Differentiate y = x*  .
Solution. ^ = e***' 1 — (x) + «*" logx ^ (e*) by X
dx ax ax
= e*x e ' 1 + x«" logx • e*
= e*xW + logx). 4ns.
47. Logarithmic differentiation. Instead of applying VIII and Yin a
at once in differentiating logarithmic functions, we may sometimes
simplify the work by first making use of one of the formulas 710
on p. 1. Thus above Illustrative Example 2 may be solved as follows:
Illustrative Example 1. Differentiate y = logVl — x 2 .
Solution. By using 10, p. 1, we may write this in a form free from radicals as
follows :
y = ilog(lx*).
— (1  **)
Then d l = \«± — by Villa
dx 2 1x 2
1 — 2x x
• Ans.
2 1x 2 x 2 l
.
1 ~ ~ X
Solution. Simplifying by means of 10 and 8, p. 1,
y = il>g(l+x 2 )log(lx 2 )].
dy = l
dx 2
f(l + x 2 ) £(lx 2 )
dx dx
1 + x 2 1x 2
x , x 2x
+ 1 = . • Ans.
by VIII a, etc
1 + x* 1x 2 1x 4
RULES FOR DIFFERENTIATING 51
In differentiating an exponential function, especially a variable
with a variable exponent, the best plan is first to take the logarithm
of the function and then differentiate. Thus Illustrative Example 5,
p. 50, is solved more elegantly as follows:
Illustrative Example 8. Differentiate y = X**.
Solution. Taking the logarithm of both sides,
log y = c* logx. By 0, p. 1
Now differentiate both sides with respect to z.
dy
— = e*^ (log*) + logxA (c*) by VIII and V
y ax dx
= «*. + logxe*,
! = «..„(! + tog.)
g + logx)
= e*x*[ + logx) . Ans.
Illustrative Example 4. Differentiate y = (4x 2 — 7)* +Vat»i.
Solution. Taking the logarithm of both sides,
log y = (2 + Vx*  5)log (4 x 2  7).
Differentiating both sides with respect to x,
ldy_ /o . ,/I5 ^ 8x
ydx 4x 2 7 Vx 2 6
^ = x(4x^7)^V^P(^^^ + ^(^7)1, ^
dx [ 4x»7 Vx^^l J
In the case of a function consisting of a number of factors it is some
times convenient to take the logarithm before differentiating. Thus,
^ * ^.« . /(x  l) (x — 2)
Illustrative Example 5. Differentiate y =\ zr* '
\(x3)(x4)
Solution. Taking the logarithm of both sides,
log y =  [log (x  1) + log (x  2)  log (x  8)  log (x  4)].
Differentiating both sides with respect to x,
1 dy = ir 1 1 1 11
y dx 2Lxl x — 2 x3 x4j
2x 2 10x + ll
(xl)(x2)(x3)(x4)
dy 2x 2 10xll .
or, — = — — • Ans.
to ( X  i)i < x _ 2) J (x  3)i (x  4)i
52 DIFFERENTIAL CALCULUS
Differentiate the following :
1. y = log(x + a).
2. y = log (ax + &)•
1 — x 1
4. y = log(x f +x).
5. y = log(x» — 2x + 6).
6. y = log«(2x+x«). w ~ ow _ o..^
2X + X*
7. y = x logx. y 7 = logx + 1.
8./(x) = logx». /'(*) = •
X
9. /(z) = log» x. y(x) = ^^ .
X
Hint, log 8 x = Gog x) s . Use first VI, t?  log i, n  3 ; ard then VTU a.
«/v .*, v i a + x ... . 2a
10. /(x) = log X . /'(x) = — — j .
a — x a* — x"
11. /(x) = log (x + VTTx*). /'(x) = —=L
dy _
dx~
1
x + a
dy_
a
ax
ax + 6
dy _
dx~~
4x
1x*
/ =
2x + l
X 1 + X
• =
8x«
2
x»2x
+ 5
»/ =
2 + 8x«
: lOff « . —
Vl + x*
12. —e°* = ae«*. " 17. — «*+*■ = 2xe*+«\
dx dx
13. Ae4x,6 = 4c**+«. 18. ^a**9 = \aW\oga.
dx dtf
14. —a** = Sa**\oga. 19. — &»"= 2slog&.&»\
dx ds
15. !log(32t*)=i^— 20. lae^ = ?Li.
dt 6X ' 2**3 dv 2 Vt>
16. A log L±l = — ?— . 21. ^a*=loga.a*.e«.
dy 1 — y 1 — y 2 dx
22. y = 7*" + 2*. y / =21og7.(x + l)7« , +«».
23. y = ***. y's^xlogcc*—".
c* dy 1
24. y = log
1 + e* dx 1 + e*
25. ^[e»(lx«)] = c»(l2xx«).
dx
26. A (^— ^) = 2c * . 27. 1 (x 2 *~) = x««(ax + 2).
RULES FOR DIFFERENTIATING
58
28. y = ?(e^e «).
29. y =
e*— er x
30. y = x«a*.
31. y = x*.
l
32. y = x*.
33. y = x»«f *.
34. f(y) = logy.c».
36./ W = !^.
36. /(x) = log(logx).
37. JF» = log*(logx).
38. 4>(x) = \og{\og*x).
39. *(y)
*v£i
40. f(x) = log
Vx 2 + 1  x
Vx 2 + 1 + X
Hint. First rationalize the denominator.
l
41. y = x 5 **.
42. y = e*.
43. y = —
x*
44
—er
45. ii7 = ««*.
«•«©•
47. y = x*".
48. y = x**.
49. i/ = a Vaf **.
dx 2 V ;
4
dx (e*+e*) 2 "
/ = a«x"  1 (n+ x log a).
y / = x^(logx + l).
v=
x*(l — logx)
X 1
1^=
logx 2 * 1 "** 1 .
/'(») =
eulogy + 1).
/'(») =
1 — 8 log a
a*
/'<*) =
1
xlogx
F'(x) =
4 log 8 (logx)
xlogx
*'(*) =
4
xlogx
r<v)=
1
/'(*) =
2
Vl + x*
dy _
:0.
dx
: e x (l+ logx)x
S ©K'>
do \ v )
= (£}\\oga\ogtl).
dz /aV
H
dy
— = x a: "+" 1 (nlogx + 1).
dx
— = x**x*nogx + log 2 x+J
dy _ xy log a
** (a*x 2 )*
54 DIFFERENTIAL CALCULUS
60. Differentiate the following functions :
j j ja
(a) ^logx. (f) —<F\ogz. (k) — log(a*+6«).
(b ) A(eax_ i)4. » £**• © £ lo «io<* 8 + *»>
. . d . 8x + l ,, v d 1 d 2 + x*
(c) — log . (h) . (m)  — .
v ' dx 8 x + 8 w dxxlogx v ' dx e**
(d) llogi^L. (i) £logx*vT+"x». (n) A (x» + a*) <* + «".
ax vi + x <** «*
»£«* 0) I© X  «>)(x. + 4)x.
51. y =
dx
(x + 1)* dy (x + l)(5x* + 14x + 6)
(x + 2) 8 (x + 8)* dx (x + 2)*(x + 8)«
Hint. Take logarithm of both sides before differentiating in this and the following
examples. . .
KO («!)* dy (xl)«(7x* + 80x97)
(x2)*(x3)l ** 12 (x  2)1 (x  8) V
eo Pi /* . v d V 2 + x — 5x 2
53. y = xvl — x(l + x). ^ = — ' — .
** 2Vlx
_ x(l + x 2 ) dy_ l + 3x 2 2x 4
~~ Vlx* ' d» "" (i  x «)i
66. ,«H. + I4.(.1*.  = 6x«(a + 3x)M«2x)(a. + 2ax12x^
48. Differentiation of sin v.
Let y = sin v.
By General Rule, p. 29, considering v as the independent variable,
we have
First Step. y + Ay = sin (v f At;).
Second Step. Ay = sin (y ■+■ Av) — sin v*
/ , Av\ . Av t
= zcoslv + — 1 "Sin—  T
* If we take the third and fourth steps without transforming the righthand member,
there results : A . . . .
Third step. & = «* D (» + At»smt, •
d Av
Fourth step. r~, which is indeterminate (see footnote, p. 46).
dv '* '
tLet A—v + Av Av + Av
and B—v B—v
Adding, A + B2v + Av Subtracting, ABAv
Therefore l(A + B)v+ — . l(AB)—.
2 2 2 V ' 2
Substituting these values of A, B> \ {A + B) ,\ {A  B) in terms of v and Av in the formula
from Trigonometry (42, p. 2),
sin A  sin B  2 cos \{A + B) sin H4  B) ,
we get sin (0 + Ar)  sin v ■ 2 cos ( t; + — ) sin — •
EULES FOR DIFFERENTIATING 55
Third Step.
Fourth Step.
^Uby^p.*, .nd j^"(" + T>
«»<* il I !!oTrl 1 ^522,p.21, and ^>o.(«; + ^)..o(»t>.
Since v is a function of x and it is required to differentiate sin v
with respect to x, we must use formula (A), § 42, for differentiating
a function of a function, namely,
dy dy dv
dx dv dx
Substituting value f from Fourth Step, we get
dv
dy dv
dx dx
d do
n .'. — (sini>) = cosi>— 
dx dx
The statement of the corresponding rules will now be left to the
student.
48. Differentiation of cos v.
Let y = cos *.
By 29, p. 2, this may be written
Differentiating by formula II,
dy /ir
cos
2 )dx\2 )
(l)(D
dv
= 8inti — •
dx
[Since ooe f ^  v j  sin v, by 29, p. 3. 1
d d*
HI .\ — (cost;) = — sin* — <
56 DIFFERENTIAL CALCULUS
50. Differentiation <rf tan v.
Let y = tan v.
By 27, p. 2, this may be written
sinv
y =
cost;
Differentiating by formula VII,
cos v — (sin v) — sin v — (cos v)
dy dx dx
dx cos 2 v
cos v —  + sin a —
dx dx
C08 2 V
dt>
dx
Q
dv
= sec'ti
> — ••
cos 2 v
dx
•'• — (tanp)
= sec 1 !)
dp
) •
dx
im
51. Differentiation of cot v.
Let y = cot v.
By 26, p. 2, this may be written
1
tan v
Differentiating by formula VII,
cfcr tan 2 v
a dv
sec v —
= — esc v
tan 2 i; (£r
HV .*. — (C0tl>) = — CSC 1 !?— «
52. Differentiation of sec v.
Let y = sec v.
By 26, p. 2, this may be written
1
y =
COS V
RULES FOR DIFFERENTIATING 57
Differentiating by
formula TO,
dy =
dx
d , ^
(cost,)
COS 2 ?
=
dv
sinv —
dx
C08 2 V
1 sinv
dv
cos V cos V
dx
sec v tan v
dv
^— ^ •
dx
xv .\ — (sect?) = sec? tan? —
53. Differentiation of esc v.
Let y = esc v.
By 26, p. 2, this may be written
1
y = •
sint;
Differentiating by formula VII,
d , . N
— (sin v)
dy dx
dx sin a v
dv
cos fl
ax
sin 2 i;
rfv
= — escveotv— «
ok
if do
tvi .•. — (csctO = — csctrcotp — .
54. Differentiation of vers v.
Let y = vers v.
By Trigonometry this may be written
y = 1 — cos v.
58 DIFFERENTIAL CALCULUS
Differentiating,
dy dv
^• = sini;— •
dx dx
d dv
ivii .. — (versiO = sin?— •
* fU dx y J dx
In the derivation of our formulas so far it has been necessary to
apply the General Rule, p. 29 (i.e. the four steps), only for the
following :
ttt & , . . du dv dw K . . .
Ill  r (u + titc) =  r +  — • Algebraic sum.
dx dx dx dx
V — (uv) = M + t;7 Product.
dx dx dx
du dv
VH ±(*\ ^~"^ Quotient
cto \ vl tr
VIII — (log a v) = log a e — • Logarithm.
ax v
XI — (sin v) = cos t; 7 • Sine.
dx dx
XXV ^ = ^.*!.
da; dv dx
Function of a function.
XXVI — = — Inverse functions.
dx dx
dy
Not only do all the other formulas we have deduced depend
on these, but all we shall deduce hereafter depend on them as
well. Hence it follows that the derivation of the fundamental
formulas for differentiation involves the calculation of only two
limits of any difficulty, viz.,
limit sinj, = 1 by §22, p. 21
= v
and "™o (1 + v y  e. By § 23. p. 22
RULES FOB DIFFERENTIATING 59
Differentiate the following :
1. y = sin ax*.
dy . d . ^
^ = cosax*(ax*) by XI
= 2 ax cos cue*.
2. y = tan Vl — x.
dy
■£ = sec*Vlx — (1x)* by XIII
= sec 8 Vlx • i (1  x)~*( 1)
sec*Vl — x
2Vlx
3. y = co8*x.
This may also be written
y = (cosx)*.
/ = 8 (cosx)* ^ (cosx) by VI
ax ax
[v«OM«udn«3.]
= 8cos 1 x(— sinx) by XII
= — 3 sin x cos* x.
4. y = sinnxsin*x.
dy d d
 i = 8innx(sinx)" + sin«x — (sinnx) by V
dx dx dx
[«— sinnx and t> sin»x.]
= sin nx • n (sin x)"  1  (sin x) + sin*x cos nx —  (nx) by VI and XI
dx dx
= nsinnx sin" 1 x cosx + nsin*xcosnx
= n sin" *x (sinnx cosx + cos nx sinx)
= nsin» 1 xsin(n + l)x.
 dy
5. y = secax. An$. — = a sec ax tan ax.
dx
6. y = tan(ax + 6). ^ = a sec 1 (ax + 6).
dx
ds
7. • = cos 3 ax. — = — 8 a sin 3 ax.
dx
8. • = cot(2<* + 8). !?! = _4*csc*(2<* + 8).
d£
9. /(y) = sin2ycosy. f(y) = 2cos2ycosy — sin 2 y sin j/
10. F(x) = cot«5x. F / (x)=10cot6xcsc*6x.
11. F(0) = tan00. F'(0) = tan*0.
12. /(^) = ^sin0+ cos0. /'(*) = 0cos*.
IS. f{t) = sin 1 1 cost. /'(*) = sin** (8 cos 1 1 sin 1 *).
14. r = acos2l.  =— 2asin20.
dd
60 DIFFERENTIAL CALCULUS
15.  r sin 1 x = sin2s. 23. — cos^ = ^sin?.
dx dt t P t
16. T coe 8 x 2 =6a^coB a x 2 8inx 8 . 24. sin  = — cos—
dx du %P u (P
d P P P d
17. — csc = — (esc— cot — . 26. — e iiIIZ = e rinx cosx.
dt 2 2 2 dx
,« d i r a sin 2 8 „ d . .. x cos(logx)
18. — avcos2s= 26. — sin(logx) = •
<& Vcos2« dx x
19. 4.a(lcos0) = asin0. 27. j tan (log x) = 8ec (logg) .
d0 dx x
20. — (logcosx)=— tanx. 28. —  a sin 8  = a sin*  cos 
dx dB 8 3 8
21. — (logtanx) = . 29. — sin (cos a) = — sin a cos (cos a)
dx 8in2x da
AA d „ . . « d tan x — 1
22. — (log sm 2 x) = 2 cot x. 30. = sin x + cosx.
dx
31. ylog Jl±^.
\ 1 — sin x
32. y = logtan0 + !)
dx
secx
dy
dx"
1
C08X
dy
1
dx cosx
33. f(x) = sin (x + a) cos (x — a). /' (x) = cos 2 x.
34. y = a** "*. y' = na Un,Mr sec 3 nx log a.
35. y = eoo^sinx. y* = a 00 ** (cosx — sin 1 x).
36. y = e* log sin x. y^e^cotx + log sin x).
37. Differentiate the following functions :
A A A
(a) — sin6x 2 (f) — csc(logx). (k) — <*»*«>".
dx dx dt
(b) — cos (a — fee), (g) — sin 8 2 x. (1) — sin  cos 8  •
w dx v ' V6/ dx v (ft 8 8
(C) dx^ n T ' < h > di °°* (l0gX) ' (m) d* ^ * *
(d) —cot Vox. (i) — tan* Vl  x 2 . (n) ^Vl + cos 2 *,
dx dx d>
j j j
(e) — sec e**. (j) f log (sin 2 ax). (o) f log Vl2sin 2 *.
dx dx as
38. — (x*e***) = x**e lA **(n + xcoRx).
ax
39. — (c^cosmxjrr e«* (a cos mxm sin mx).
40./W = 1 + CO8 ^. /'(*)= 2sin *
1— CO80 (1— costf) 2
^i //a ^(asin*— cos*) _,, . . .
41. /(0) = — * T ZL. /' (<f>) = e°* sin <f>.
a 2 + 1
42. f(8) = (* cot «) 2 . /'(a) = 2 8 cot s (cot « — * esc 2 s).
RULES FOR DIFFERENTIATING 61
48. retail 1 * tan0 + 0. ^ = tan*0.
d0
44. y = x**»*.
^. = x rinx/ j. logXCOSXj.
46. i/ = (sin x)*. y* = (sin x)* [log sin x + x cot x] .
46. y c (sinx)*«*. 1^= (sin x)* *^* sec* x log sin x)
47. Prove —  cos » = — sint> — t using the General Rule.
dx dx
48. Prove — cot v = — esc 2 v — by replacing cot © by •
dx ax sine
55. Diffcrentiatioii of arc sin v.
Let y = arcsinv;*
then i; = sin y.
Differentiating with respect to y by XI,
dv
— = cosy;
dy
therefore % = By (C), p. 46
dv cosy
But since v is a function of x, this may be substituted in
dy^dy dv^ 4g
dx dv dz
dy 1 dv
giving * = —
oar cosy aa?
rfv
[
y/\~fdx
OM y — Vl  sin* y ■ Vl  t* t the positire sign of the radical being taken,"!
since cosy is positire for all values of y between ^ and ^ inclusire. I
if dr
ivm .*. — (arcsini>) =
dx Vi 1^
* It should be remembered that this function is defined only for values of v
between —1 and +1 inclusive and that y (the function) is manyvalued, there
being infinitely many arcs whose sines all equal v. Thus, in the figure (the
locus of y arc sin »), when v = OM, y = MP\, JfPj, MP 9 , • . • , M (fo, MQt* —
In the above discussion, in order to make the function singlevalued, only
values of y between   and  inclusive (points on arc QOP) are considered ;
It £t
that is, the arc of smallest numerical value whose sine is v.
62
DIFFERENTIAL CALCULUS
56. Differentiation of arc cos v.
Let y = arc cos v ;
then v = cos y.
Differentiating with respect to y by HI,
dv
dy
dy = 1
. *
= — siny;
therefore
By (C), p. 46
dv amy
But since v is a function of z, this may be substituted in the
00, P 45
1 dv
formula
dy dy dv
giving
dy = __
dx siny dx
1 dv
XIX
Vl — v 2 <te
[sin y mm Vl cos* y «Vi «*, the plus sign of the radical being taken,!
since sin y is positive for all values of y between and » inclusive. J
dv
d dx
.\ — (arccosi>) = *
57. Differentiation of arc tan v.
Let
then
y = arc tan v ; *
v = tan y.
Differentiating with respect to y by XIV,
therefore r =
rfy
sec 2 y ;
*l.
1
dv
sec 2 y
By (C), p. 46
* This function is defined onlv for values of v between 1 and +1 inclu
sive, and is manyvalued. In the figure (the locus of j/ = arc cost?) , when
v mmOM, y~MPi, MP 2 ,  • •, MQy MQz,
In order to make the function singlevalued, only values of y between
and it inclusive are considered ; that is,
the smallest positive arc whose cosine is v.
Hence we confine ourselves to arc QP of
the graph.
t This function is defined for all values of ?.\ and is many
valued, as is clearly shown by its graph. In order to make it
single valued, only values of y between • aud — are con
It A
sidered ; that is, the arc of smallest numerical value whose
tangent is v (branch AOB).
RULES FOR DIFFERENTIATING
63
But since v is a function of x, this may be substituted in the formula
giving
IX
dy __ dy dv
dx dv dx
dy __ 1 dv
dx sec*y dx
1 dv
^ — — — ^— •— — •
l + v 2 dx
[9ec*y  1 + tan'y » 1+ 1?*.]
dv
d dx
— (arc tan i?) = .
dx K J l + i*
(A), p. 45
58. Differentiation of arc cot v.*
Following the method of the last section, we get
dv
m
d dx
— Care cot v) =
59. Differentiation of arc sec v.
Let y = arc sec v ;
then v = sec y.
.t
* This function is defined for all values of t\ and is manyvalued, as is seen from its
graph (Fig. a). In order to make it singlevalued, only values of y between and v are
considered ; that is, the smallest positive arc whose cotangent is v. Hence we confine our
•elves to branch AB.
Fig. b
t This function is defined for all values of v except those lying between 1 and +1, and is
seen to be manyvalued. To make the function singlevalued, y is taken as the arc of smallest
lomerical value whose secant is v. This means that if v is positive, we confine ourselves to
points on arc AB (Fig. b), y taking on values between and  (0 may be included) ; and if v is
negative, we confine ourselves to points on arc DC, y taking on values between  w and  —
( r may be included).
64
DIFFERENTIAL CALCULUS
Differentiating with respect to y by IV,
dv
— = secy tany;
dy
4. I
dv sec y tan y
therefore
By (C), p. 46
But since v is a function of x, this may be substituted in the formula
(4), p. 45
giving
UII
dy __ dy dv
dx dv dz
dy
dx
dv
sec y tan y dz
1 dv
v y/v*l<te
sec y  », and tan y — Vsec*yl<s Vr*l, the plus sign of the
radical being taken, since tan y is positive for all values of y
between and — and between — t and » including and — t.
dv
.\ — (arcseci;) = —
dx uVi^ 1
60. Differentiation of arc esc v.*
Let
then
y = arc esc v;
t; = esc y.
Differentiating with respect to y by IVI and following the method
of the last section, we get
dv
dx
XIHI — (arc esc i>) = — .
dx irvV_i
* This function is defined for all values of v except those
lying between  1 and + 1, and is seen to be manyvalued. To
make the function singlevalued, y is taken as the arc of small
est numerical value whose cosecant is v. This means that if v is
positive, we confine ourselves to points on the arc AB (Fig. a), y
taking on values between and  [maybe included); and
if v is negative, we confine ourselves to points on the arc CD, y
taking on values between  r and  — (  — may be included ) .
RULES FOR DIFFERENTIATING
65
61. Differentiation of arc vers v.
Let y = arc vers v ;
then v = vers y.
Differentiating with respect to y by IYII,
dv
dy
. *
= siny;
therefore
By (C), p. 46
dv siny
But since v is a function of x, this may be substituted in the formula
(4), p. 45
dy __ dy dv
dz dv dx
giving
dy 1 dv
dx siny <£r
1 dv
y/2 V — V* dx
F«ln y Vl _ cos* y Vi  (l  versy)*= V2t>  v\ the plus sign of the radical!
L being taken, since sin y is positive for all values of y between and r inclusive. J
inv
dv
d dx
.. — (arc vers v) =
Differentiate the following :
1. y = arc tan ax 1 .
Solution.
2. y = arc sin (8 x — 4 X s ).
Solution.
d_
dx
dx 1 + (ax 7 )*
_ 2ax
(3x4x«)
<k~ Vl(3x4x«) 8
S12x*
bj
byXVIU
Vl9x 2 + 24x 4 16x« Vlx*
* Defined only for values of v between and 2 inclusive, and is many
valued. To make the function continuous, y is taken as the smallest positive
are whose versed sine is v ; that is, y lies between and r inclusive. Hence
we confine ourselves to arc OP of the graph (Fig. a).
Fio. a
66 DIFFERENTIAL CALCULUS
ai 1 + 1
3 y = arc sec
x a 1
d (z* + 1\
dx \x 2  1/
s 2 + l // x 2 + i y
e»l\v* a l/
Solution. ^ = ^ y " by XXII
** x 2 + 1 //x 2 + l\ 2
(z*l)2x(x 2 H)2x
(x*  1)* 2
x 2 + l 2x x 2 + l
x a _ x ' x % _ x
^ d , x • 1 d /: 1
4. — arc sin  = — — = . 9. — arctanvl— x =
dx a Va 2  x 2 <& 2Vlx(2x)
d — 2x d 3 9
5. — arc cot (x 2 — 5) = . 10. —  arc cosec — =
dx * ' l + (x 2 6) 2 dx 2x V94x 2
 d 4 2x 2 ,, d 2x 2 2
6. — arc tan  = • 11. — arc vers
dx 1x 2 1 + x 2 dx 1 + x 2 l + x ;
_ d 1 2 ,o d * xa
7. — arc cosec — =  . 12. —arc tan 
2
dx 2x 2 — 1 Vl — x 2 d* a a 2 + x 2
8. — arc vers 2 x 2 = . 13. — arc sin
dx Vl^tf dx V2 Vl2xx 2
14. /(x) = x Va 2  x 2 + a 2 arc sin. /' (x) = 2 Va 2  x 2 .
15. /(x) = Vtf^tf + aarcsin?. /'(x) = (j^j) •
16. x = r arc vers  — V2 ry — y 2 .
dx
17. = arc sin (8 r — 1).
r <*V V2ry — y 2
d* 3
18. tf> = arc tan
19. a = arc sec
<fr V6r9r*
r + a d0__ 1
1ar' dr ~"l + r*"
1 ds 1
vr^s dt vr^?
d x
20. — (x arc sin x) = arc sin x +
dx VT^x 2
21. —  (tan $ arc tan 6) = sec 2 arc tan I + —
d0 v ' 1 + 2
d i
22. — [log (arc cos Q] = —
* arc cost Vlt 2
23. f(y) = arc coe(logy). /'(y) = 1
yVl(logy) 1
24. f(6) = arc sin Vain 0. /'(0) = } Vl + csc0.
RULES FOR DIFFERENTIATING 67
26./(*) = arctan J^ZSSHl. f'(4>) = \
\1 + CO8 2
A « 4 dp c M « tMI «
<** l + tf»
27. u = arc tan 
2 do e*+r*
c«e« dfl 2
28. * = arc cos
e*+e« eft e«+e«
29. y = *»**. y = x™W arC8inx + *<** \
v * vt^x*/
30. y = e**arctanx. if = e*\  + x*arctanx(l+logx) .
31. y = arc sin (sin x). y 7 = 1.
^ 4sinx , 4
32. y = arc tan . y 7 =
3 + 5C08X 5 + Scosx
a . i_ /x — a ./ 2fl « 8
33. y = arccot + log^/?_Zi!. y' =
Vx — a
xTi
* \x + a x*  a*
* 1 ., *'
34. y = logl J — arc tan x.
1x*
«e fk « _i ^ x arc sin x
35. y = vl— x'arcsinx — x. y^ m
VT^x 5
38. Differentiate the following functions :
(a) — arcsin2x*. (f) — Parcsin^. (k) — arc sin Vl  y 2 .
dx at o dy
(b) arctana'x. (g) .&****<". (I) —arc tan (log 3 az).
ax at az
tt *m it it
(c) — arc sec  . (h) — tan <f? • arc tan ^i. (m) — (a 2 + a*) arc sec  .
dx a cup ds 2
/JV d ,. v d , $ . . d 2 or
(d) — xarccosx. (i) —arc sin a . (n) — arc cot
' dx d0 da 3
(e) — x 8 arc cot az. (j) j arc ten Vl + 0*. (o)  Vl — I 2 arc sin t.
Formulas (A), p. 45, for differentiating a function of a function, and (C), p. 46,
for differentiating inverse functions, have been added to the list of formulas at the
beginning of this chapter as XXV and XXVI Tespectively.
In the next eight examples, first find j and — by differentiation and then
substitute the results in
** = *L,to byXXV
d • dx dv dx
to find ^ •
dx
* As was pointed out on p. 44, it might be possible to eliminate v between the two given
expressions so as to find y directly as a function of x, but in most cases the above method
is to be preferred.
68 DIFFERENTIAL CALCULUS
In general oar results should be expressed explicitly in terms of the independent
dy dz . dd>
variable : that is. — in terms of x, — in terms of y, — ^ in terms of 0, etc.
' dz dy dB
37. y = 2« 2 4, » = 3x 2 4l.
? = 4 r ; — = 6x ; substituting in XXV,
dv dz
^ = 4r.6x = 24x(3x« + l).
dz
38. y = tan2«, v = arc tan (2 x — 1).
^ = 2sec 2 2u; ^ = _— L ; substituting in XXV,
dv dz 2x a 2x + l
dy _ 2sec 2 2p _ tan 2 2i?4l _ 2x 2 — 2x41
dx~~2x 2 2x4l~ 2x 2 2x4l~ 2(xx 2 ) 2
[2x — 1 1
Since v arc tan (2 x 1), tant?«2xl, tan2t>= — — .
2* — 2x* J
39. y = 3c a 4u + 6, « = 2x»6. ^ = 72 x«  204 x 2 .
dx
._ 2v x dy 4
40. y =  » v =
3u2 2xl dx (x2) 2
dy
41. y = log (a 2 — v 2 ), v = a sin x. ^ = — 2 tan x.
42. i/ = arc tan (a + «),© = e*.
dx
dy _ e*
dz "" 1 + (a + e*) 2
43. r = e*' +;<*,« = log (***)• ^ = 4*» 6* 2 4 1.
dt
dx
In the following examples first find — by differentiation and then substitute in
dy
dy
dv = ±
dz dz
by XXVI
to find ^. *
dx
44. x = y Vlf y.
45. x = Vl + cosy.
4fi * — ^
14 logy
„ i a + ^ a * 
47. as = a lOflr
 v »
dy 2Vl + y 2x
dx 2 + 3y 2y48y 2
dy 2Vl + cosy 2
dx siny V2 — x 2
dy_ (14 lo gy) 8
^^ ' ■■■in «
dx logy
dy __ y Va 2 — y 2
y dx a 2
r dz M y
49. Show that the geometrical significance of XXVI is that the tangent makes
complementary angles with the two coordinate axes.
RULES FOR DIFFERENTIATING 69
62. Implicit functions. When a relation between x and y is given
by means of an equation not solved for y, then y is called an implicit
function of x. For example, the equation
s*4y =
defines y as an implicit function of x. Evidently x is also defined by
means of this equation as an implicit function of y. Similarly,
s* + y a + z 2 a* =
defines any one of the three variables as an implicit function of the
other two.
It is sometimes possible to solve the equation defining an implicit
function for one of the variables and thus change it into an explicit
function. For instance, the above two implicit functions may be solved
for y, giving j
y=S 4
and y = ±Va a z*2 a ;
the first showing y as an explicit function of z, and the second as an
explicit function of x and z. In a given case, however, such a solution
may be either impossible or too complicated for convenient use.
The two implicit functions used in this article for illustration may
be respectively denoted by fr x y) =
and F(xy y, z) = 0.
83. Differentiation of implicit functions. When y is defined as an
implicit function of x by means of an equation in the form
00 /(*, y) = o,
it was explained in the last section how it might be inconvenient to
solve for y in terms of x ; that is, to find y as an explicit function of x
so that the formulas we have deduced in this chapter may be applied
directly. Such, for instance, would be the case for the equation
(J5) az*+2z*yy 1 x10 = 0.
We then follow the rule :
Differentiate, regarding y as a function of x, and put the result equal
to zero.* That is,
(C) £f(x, y) = 0.
* This process will be justified in §127. Only corresponding values of z and y which
satisfy the given equation may be substituted in the derivative.
70 DIFFERENTIAL CALCULUS
Let us apply this rule in finding ^ from (2?).
^(aa*+2 2fyy T s10)=0; by (C)
ax
i<»+£<w»£<w>i<M»».
«r dx
(2z*7xj/ , )& = y'Gax t 6sfy;
rfx 2^72^ '
The student should observe that in general the result will contain
both x and y.
EXAMPLES
Differentiate the following by the above rule :
1. y 8 = ipx.
2. x 8 + y* = r 8 .
3. Wx 8 + aV = <**&*.
4. y 8 3y + 2ax = 0.
5. x* + V* = a*.
6. x* + y* = a*.
8. y» — 2xy + 6* = 0.
9. x^ + y 8 — Saxy = 0.
10. x* = y*.
11. p* = a*co6 20.
12. /> 8 coe0= a J sin30.
13. cos (ut?) = co.
14. 0= cos (0 + 0).
dy_
dx
2p
y
dy_
dx
X
y
dy =
dx
a*y
dy __
2a
dx
Sfly 8 )
dy __
dx~"
%!■
dy __
dx~
*
«y_
dx
36*xy*
a 8
dy _
. y
dx~
y — x
^ =
ay — x*
dx
y 8 — ax
dy _
y 8 — xylogy
dx"
X s — xylogx
d$
a 8 sin 20
P
*>_
3a 8 cos30 + p 8 sin0
d$
2pcos0
du_
c + u sin (ut?)
do"
— v sin (ut>)
<W_
sin(0 + *)
d>~~
l + sin(0 + *)
RULES FOR DIFFERENTIATING 71
dy
15. Find = from the following equations :
dx
(a)x f = ay. (f) xy + y* + 4x = 0. (k) tanx + v* = 0.
(b) x* + 4y« = 16. (gJyx^i^S. (1) cosy + 3x a = 0.
(c) Wx*  aV = <**&*. (h) x*  2x« = y». (m) x cot y + y = 0.
(d)y 1 = x 8 + a. (i)xV + 4y = 0. (n)y* = logx.
(e)x*y« = 16. (j)y* = sin2x. (o) e** + 2 y» = 0.
16. A race track has the form of the circle x* + y* = 2500. The directions OX and
OY are east and north respectively, and the unit is 1 rod. If a runner starts east at
the extreme north point, in what direction will he be going
(a) when 26 V2 rods east of OY? Ans. Southeast or southwest.
(b) when 25 V2 rods north of OX? Southeast or northeast.
(c) when 80 rods west of OY? E. 86° 52 / 12" N. or W. 86° 62 / 12" N.
(d) when 40 rods south of OX ?
(e) when 10 rods east of OY?
17. An automobile course is elliptic in form, the major axis being 6 miles long and
running east and west, while the minor axis is 2 miles long. If a car starts north at
the extreme east point of the course, in what direction will the car be going
(a) when 2 miles west of the starting point ?
(b) when \ mile north of the starting point?
MISCELLANEOUS EXAMPLES
Differentiate the following functions :
1. arc sin Vl — 4 x a . Ans.
2
Vl4x*
2. xe*». ef^(2x a +l).
8. logsin. i^i'
 a a.
4. arc cos*
V yVy»a«
5.
6.
a*
Va* — x 8 (a 2 — x*)$
x logx
1 + logx (1 + logx) 2
7. logsec(l2x). 2tan(l2x).
8. z*&**. xc a »»(28x).
9. 1" g J 1 ~ cmt • csc*.
\1 + cost
10. arcsinVj(l — cosx). \.
2a 2
11. arc tan
12. (2x
(l68*)Vs 2 l
7 + 4x »/ 2
3(l + x) \l + x
x»arcsinx , (x 2 + 2)Vlx 2 . x 2 arcsinx.
W  8 + 9
72 DIFFERENTIAL CALCULUS
? + lo*sec 2 ?. 31. (log tan8 x«)«.
82 23t* + 4t*+0
14, tan 8  + log sec s
8 8
15. arc tan \ (<P* + e~ **). t
16 /Sy*. 33 (l + x)(l2x)(2 + x)
17. x««»*. 34. arc tan (log 3 x).
, Q (x + 2)i (x»  1)* 36. "J'p  ax)«.
* f 36. logV(a 2 te 8 )*
19. e««tt 8 *>. r
20. arctanVTT?. **' l0g \pTTi'
21. ** 38. e«««« 2 «.
C08 2
22. e*"«*.
27.
39.
/(23x)«
\ l + 4x
23. logsin 2 **. ^— ;
24. e«* log sin ax. 40 ' COBX —
26. sin s 0co8*. 41# e* log sin x.
26. — . 42. arc sin
2V(6cx»)"» Vl + x*
m + x e marcUnx 43* arc tan a*.
1 + m 2 Vl + x* 44. a*»*««.
28. tan 2 x — log sec 2 x. 46. cot 8 (log ax).
^ 31o g (2co8x + 38inx) + 2x 46 (1 _ B ~±
13
30. arc cot  + log \f^^ • 47. log ^, 1 ""*' .
CHAPTER VI
SIMPLE APPLICATIONS OF THE DERIVATIVE
64. Direction of a curve. It was shown in § 32, p. 31, that if
V =/0)
is the equation of a curve (see figure), then
du
^=rtanr = slope of line tangent to the curve at any point P.
dx
The direction of a curve at any point is defined to be the same as
the direction of the line tangent to the curve at that point. From
this it follows at once that
du ~
f = tanr = slope of the curve at any point P.
dx
At a particular point whose coordinates are known we write
[— = slope of the curve (or tangent) at point (x x , y x ).
At points such as Z>, F, H, where the curve (or tangent) is parallel
to the axis of X, .
r = 0° ; therefore — = 0.
dx
At points such as A, B, G, where the curve (or tangent) is per
pendicular to the axis of X,
du
r = 90° ; therefore — = co.
dx
IS
74 DIFFERENTIAL CALCULUS
At points such as E, where the curve is rising*
du
r = an acute angle ; therefore — = a positive number.
dx
The curve (or tangent) has a positive slope to the left of B,
between D and F, and to the right of Q.
At points such as C, where the curve is falling,*
du
t = an obtuse angle ; therefore — = a negative number.
dx
The curve (or tangent) has a negative slope between B and
Z>, and between F and G.
Ii.iith.mivk Ex ihi'i.i; 1. Given the curve y = — — z*
(see figure) .
(a) Find r when x = I.
(b) Find t when x = 3.
(c) Find the points where the curve ;. parallel to OX.
(d) Find the points where t = 45°.
(e> Find the points where the curve is parallel to the I:
2l8i/ = 6(liiie^B).
(a) t
(b) t
Differed tinting,   = x* — 2z = slope at any point.
r =1—1 = 1  2 = 1 ; therefore r = 186°. Am
Ldx±,i
\ ''■* I
= — = 3; therefore t
n3. Am
(c) t = 0°, tan t = j = ; therefore z x — 2x = 0. Solving this equation, we And
that i = or 2, giving points C and D whore the curve (or tangent) is parallel to O.J.
(d) t = 45", lurxSaij therefore 3»2a = l. Solving, we get a: = 1±V2,
giving two points where the slope of the curve (or tangent) Is unity.
(e) Slope of line = j; therefore i a — 2i= }. Solving, we get x = 1 ±Vj, givin
points E and F where curve (or tangent) is parallel to line AB.
Since a curve at any point haw the same direction as its tangent s
that point, the angle between two curves at a common point will t
the angle between their tangents at that point.
Example 2. Find the angle of
x* + y*4x = l,
z* + y*Sy = 9.
• When moving from left to right
of the circles
SIMPLE APPLICATIONS OF THE DERIVATIVE 75
Solution. Silking simultaneously, we find tbe points of intersection to be (8, 2) and
I, 21.
5 =, ? * f rom (J). By 5 63, p, 60
 from (fl).
its 1 — w"
I bs slope of tangent I
By  63, p. 69
M) at (3,8).
fel.
3 = slope of tangent to (B) at (3, 2).
lines whose slopes
The formula for finding the angle between
Substituting, tan S = —  *—  = ) ; therefore # = 45°. An*.
This is also the angle of intersection at the point (1, — 2).
EXAMPLES
The corresponding figure sliuuld be drawn in each of the following examples:
1. Find the slope of y = at the origin. Aim. 1 = tan r.
2. What angle does the tangent to the curve ify* = o*(x + y) at the origin make
with tbe axis of X? Ana. t = 136°.
3. What is the di reel ion In which the point generating the graph of y — 3z* — x
lends to move at the Instant when x = 1 ? An*. Parallel to a line whose slope is 6.
4. Show that — (or slope) is
for a straight line.
= z"3z*9i + 5 is parallel to the a
5. Find the points where the curve i
of X. ^718. x = 3, z=_l.
6. At what point on y* = 2 1* is the slope equal to 3 1 An*. (2, 4).
7. At what points on the circle z> + y 1 = r* is the slope of the tangent line equal
r
An*. 
8. Where will a point moving on the parabola y = z* — 7 1 + 3 be moving paral
lel to the line y = a jr + ■; " An*. (6,  3).
9. Find the points where a particle moving on the circle z 1 + y* = 160 moves per
pendicular to the line hi + 12y = 60. An*. (± 12, T 5).
10. Show that ail the curves of the system y = log kz have the same slope ; i.e. the
slope is independent of k.
11. Tbe path of the projectile from a mortar cannon lies on the parabola y =
2ii': the unit is I mile, OX being horizontal aud OY vertical, and the origin
being the point of projection. Find the direction of motion of the projectile
(a) at instant of projection ;
(b) when it strikes a vertical cliff 1 j miles distant.
(c) Where will the path make an inclination of 46° with the horizontal ?
(d) Where will the projectile travel horizontally ?
An*, (a) arc tan 2 ; (b) 136= ; (c) (J, ) ; (d) (1. !)■
DIFFERENTIAL CALCULUS
76
12. If the cannonin the preceding example was situated on a hillside of inclination
46", at what angle would a shot tired up strike the hillside ? Ana. 45°.
13. At what angles does a road following the line 3 ^ — 2x — 8 = intersect a rail
way track following the parabola y 1 = %x. Ana. arc tan j, and arc tan j.
14. Find the angle of Intersection between the parabola y' = Ox and the circle
** + y» = 16. Ana. arctanSVS.
15. Show that the hyperbola I* — y* = 5 and the ellipse 1 — = 1 intersect at
right angles.
16. Show that the circle x + y* = Sax and the ciseoid ;/'' i
(a) are perpendicular at the origin ,
(b) intersect at an angle uf 45° at two other points.
17. Find the angle of intersection of the parabola x* = iay and the witch
V=£—  — j' Ana. arctanS = 71°3S\B.
18. Show that the tangents to the folium of Descartes X s + y* = Saxy at the point*
where it meets the parabola y s = ax are parallel to the axis of 1".
19. At how many points will a particle moving on the curve y = x* — 2x* 4
be moving parallel to the axis of X ? What are the points ?
Ana. Two ; at (1,  4) and (J,  '{>.<).
20. Find the angle at which the parabolas y — 3z a — 1 and y — 2z E + 3 intersect.
An.. «««.
21. Find the relation between the coefficients of the conies «j** + 6,?* = 1 and
of* + b 3 y*  1 when they intersect at right angles, I 11 1
65. Equations of tangent and normal, lengths of subtangent and
subnormal. Rectangular coordinates. The equation of a straight
line passing through the point (x# y t ) and having the slope m is
y — y=m(z — x~). 54, (c), p, 3
jn
a fi If this line is tangent to the curve AB at the
^ s^i^ P oint F ( x * ?!>' th en from § 64, p. 73,
~ r *»* m = tanT= i. =£■
Hence at point of contact JJOc,, #,) the equation of the tangent
line TP l is
(i) yy, = ^(x Xl ).
or,
* By this notation is me ant that we should first find — , then in the result substitute i t
for x and Vi for V The student is warned against interpreting tin symbol
that has no meaning whatever, siueu x, and y 1 I
SIMPLE APPLICATIONS OF THE DERIVATIVE 77
The normal being perpendicular to tangent, its slope is
=p By 55, p. 3
And since it also passes through the point of contact J?^, y,), we
MVe for the equation of the normal ij JV
(*>
Vtfi
That portion of the tangent which is intercepted between the point
of contact and OX is called the length of the tangent (= 27J), and its
rejection on the axis of X is called the length of the subtangent ( = TM).
ailarly, we have the length of the normal (= FN) and the length of
t subnormal (= MN).
In the triangle 27' A/, tan r = %£ ; therefore
tanr * dy.
In the triangle MPjN, tan t
(3,
TM*
TM
= length of subtangent.
MN
MP'
 ; therefore
(4)
MN 1 = MP, ta.nr=
= length of subnormal.
The length of tangent (= 27?) and the length of normal (= I>X~)
lay then be found directly from the figure, each being the hypotenuse
f a right triangle having the two legs known. Thus
TP=^TM % +MP
+W
+ 1 = length of tangent.
%X= J Ml? + UN' = ^(j,,)' + (y, &j
(6) =y 1 »i +lM±\ = length of normal.
The student is advised to get the lengths of the tangent and of
the normal directly from the figure rather than by using (5) and (6).
When the length of subtangent or subnormal at a point on a curve
is determined, the tangent and normal may be easily constructed.
78
DIFFERENTIAL CALCULUS
EXAMPLES
1. Find the equations of tangent and normal, lengths of subtangent, subnora
tangent, and normal at the point (a, a) on the cissoid j* =
dy _ 3 ox* — i'
dz~ y{3aK)*'
^1 = — I = = 2 = slope of tangent.
Substituting in (1) gives
y — 2 1 — a, equation of tangent.
Substituting in (2) gives
2 y + i = 3 (i, equation of norm?.] .
Substituting in (3) gives
TW =  = length of subtangenl.
length of subnormal.
 V5 — length of tangent,
t 1 = a VZ = length of normal.
Also PT=V{TM)' + {MP)*
ami subnormal
= ••£'—■
and P.V r_ V^VJV)* + (.»/•]= = \ / 4
2. Find equations of tangent and normal to the ellipses* + 2y* — 2xy — x = at
the points where x = 1. Ana. At (I, 0), 2 y = x  1, y + 2x = 2.
At(l,l),2j/=i + l,V + 2je =
3. Find equations nf tangent and normal, lengths of subUingenl and subr
at the point (*,, y,) on the circle x 5 + y 1 = r 3 ."
Ana. i,i + y t y = r 3 , x v y 
4. Show that the subtangent to the parabola y* = 4 px is bisected at the vertex,
and that the subnormal is constant and equal to 2p.
6. Find the equation of the tangent at (i,, y,) to the ellipse — + — = 1.
witch y at the point
n*. s + 2p = 4a,y = 2z3a.
(e^ + e  ") the lengths of sub
7. Prove that at any point o
norma! and normal are  (e a — «
■} and — respectively.
8. Find equations of tangent and normal, lengths of subtangent and subnormal, to
each of the following curves at the points indicated i
(a) v = xl«t (j,]). (e)y = 0 Z 'at(3,0).
(b) y* = 4 x at (9,  6). (f) x 1 = 8 y where x =  6.
(c) x* + 5y 3 H where y = l. (g) x* xy + 2s0 = 0, (3,2).
(d) x* + i/»=26at(_8,4). (h) 2a>if* = 14at(3,2).
SIMPLE APPLICATIONS OF THE DERIVATIVE
9. Prove tliiit tin length of subtanecnt to y — a* is constant and equal to . .
log a
10. Get the equation of tangent to the parabola y = 20x which makes an angle
of 45° with the axis of X. Ana. y = x + 6.
Hint. First find point of contact b
11. Find equations of tangents t
toe 2a; + Zy = fl.
12. Find equations of tangenW
pendicular to the line2y + 61 = 10.
13. Show that m the equilateral hyperbola 2xy
med by a tangent mid the (."Ordinate axes is eonsta
14. Find equations of tangents and normals to the curve y 5 = 2l' — a
points where as = 1. Ana. At (1, \),2y = x + 1, y + 2i = !
At(l,l),2y=s_l,yS
16. Show that the sum of the intercepts of the tangent to the parabola
.1 + .U.I
method of Illustrative Example 1, (d), p. T4.
the circle SB* + y* = 52 which are parallel to t
^its. 2x + Sy± 28 =
o the hyperbola 4z» 0y> + 3fl  which a
a 5 the area of the triangle
and equal to a 1 .
n the coordinate axes in constant and equal t>
IS. Find the equation of tangent to the cu
" ! (* + J) = a * f 1  V) « ^e origin
17. Show that for the bypocycloid z* + yi = a' that portion of the tangent
icluded between the coordinate axes is constant and equal to o.
18. Show that the curve y = ae° has a constant subtangent.
66. Parametric equations of a curve. Let the equation of a curve be
(A) F(x, j0=0.
If x is given as a function of a third variable, t soy, called a,param
tr, then by virtue of (.4) y is also a function of t, and the same func
tional relation (A) between x and y may generally be expressed by
leans of equations in the form
m f.AO.
ly<K0;
each value of ( giving a value of x and a value of y. Equations (5)
are called parametric equations of the curve. If we eliminate t between
equations (£), it is evident that the relation (.X)
must result. For example, take equation of circle
2 r' + ^ = r a , ory = \?Sx'.
Let x = r cos t ; then
y = /sin t, and we have
m {;::::;;
b parametric equations of the circle in the figure, ( being the parameter.
80 DIFFERENTIAL CALCULUS
If we eliminate t between equations (C) by squaring and add
ing the results, we have
x* + f = ^(cos* t + sin* t) = r 2 ,
the rectangular equation of the circle. It is evident that if t varies
from to 2 7t, the point P (x> y) will describe a complete circumference.
In § 71 we shall discuss the motion of a point P, which motion
is defined by equations such as
{.
We call these the parametric equations of the path, the time t being
the parameter. Thus in Ex. 2, p. 93, we see that
{
x = v Q cos a • £,
2
y = — \gv + v &ma*t
are really the parametric equations of the trajectory of a projectile,
the time t being the parameter. The elimination of t gives the rectan
gular equation of the trajectory
y = 3tang f
2 v; cos' a
Since from (2?) y is given as* a function of t, and t as a function of
*> wehave dy dy dt
£ = £.— by H Y
byuvi
It
that is,
(P)
dx
dt
dx
dy
1
' dt '
dy
dx'
~dl
dy
dt
_*'(0.
dx~
= dx
dt
AO
Hence, if the parametric equations of a curve are given, we can find
equations of tangent and normal, lengths of subtangent and subnor
mal at a given point on the curve, by first finding the value of * at
ax
that point from (D) and then substituting in formulas (1), (2), (3),
(4) of the last section.
SIMPLE APPLICATIONS OF THE DERIVATIVE
81
Illustrative Example 1. Find equations of tangent and normal, lengths of
sabtangent and subnormal to the ellipse
at the point where <f> = —
4
fx = acos0,
\y = 6sin0,*
dx
Solution. The parameter being 0, — = — a sin 0,
dtp
— = 6 cos*.
cUp
Substituting in (D), — =  = slope at any point.
dx asin0
Substituting =  in the given equations (E), we get
4
contact. And
Wi' Vi)
as the point of
JL becomes f± = , when * = .
dx dx x a 4
or.
or,
Substituting in (1), p. 76, y L =   ( x M,
V2 <*\ V2/
6x + ay = V2 a&, equation of tangent.
Substituting in (2), p. 77, y  ^ = ? /x  4=\
V2 6 \ V2/
V2 (ax — by) = a 8 — 6*, equation of normal.
Substituting in (3) and (4), p. 77,
(*) =
6*
= = length of subnormal.
aV2
— = length of subtangent.
V2
• As in the figure draw the major and minor auxiliary circles of the ellipse. Through
two points B and C on the same radius draw lines parallel to the axes of coordinates.
These lines will intersect in a point P (z, y) on the y
ellipse, because
and
or,
x  OA  OB cos  a cos <t>
y^P OD OCsin06sln0,
  cos 4 and £  sin 0.
Now squaring and adding, we get
p + jcos** + sin**l,
the rectangular equation of the ellipse. <p is sometimes
called the eccentric angle of the ellipse at the point P.
82 DIFFERENTIAL CALCULUS
Illuhthatiyk EiiitrLR 2. Given equation of the cycloid* in parametric form
\j( = a(lcoB0),
9 being tbe variable parameter; find lengths, of subtangent, subnormal, tangent,
and normal at the point where tf = — .
Solution.
Substituting in (D), p. 80,
sintf , , .
= slope at any point.
Since = , tbe point of contact is ( a, a), and?* = 1.
2 \2 / da:,
Substituting in (3), <4), (5), (8) of the last section, we get
length of subtangent = a, length of subnormal = a,
length of tangent — aV2, length of normal = oV2. Aim.
EXAMPLES
Find equations of tangent and normal, lengths of subtangent and subnormal to
each of the following curves at tbe point Indicated :
Tangent Normal But*. Sub*.
1. x = P,2y = t; Isl. 14^ + 1=0, 8* + 2y_9 = 0, 2, J.
2. z = t,y = P; ( = 2. 12ii/16 = 0, s + 12y98 = 0, , 06.
S. x = P,y = l*; ( = 1. 323yl = 0, 2* + 3? 6 = 0, J, 1.
4. a: = 2e<, i/ = e'; ( = 0. z + 2y_4 = 0, 2iyS = 0, 2, J.
Sy + 4z
= 0, 4y2*l =
h
* The path described by a point on the circumference of a circle which rolls without
sliding on a fixed straight line is called the cycloid. Let the radius of the rolling circle be a, P
tbe generating point, and M the point of contact with the fixed line OX, which la called the
base. If arc PSf equala OM in length, then P will touch at O if the circle Is rolled to the left.
We hare, denoting angle PCM by 8,
zt>W".VWaDaain9<i(0sln0),
y*pyMCACaaeoB8a(\v*f),
the parametric equations of the cycloid, the angle 8 through which the rolling circle turns
being the parameter. 0D2ra Is called the base of one arch of the cycloid, and tbe point F
la called the vertex. Eliminating 8, we get the rectangular equation
, I^A.ji^nzi
SIMPLE APPLICATIONS OF THE DERIVATIVE
83
= cost, y = sin2t; t = — .
8
= 3e«, y = 2#; t«=0.
ss sin t, y = 2 cost ; t = — .
= 4 cost, y = 8 sin t ; t = — .
= log(t + 2), y = t; t=2.
In the following curves find lengths of (a) subtangent, (b) subnormal, (c) tangent,
(d) normal, at any point :
fx = a (cos t + t sin t),
\ y = a (sin t — t cost).
e. x = i — i, y = i»; t = ».
11. X
7. s = 8t,y = 6t — t»; t = 0.
12. x
8. z = P,y = t; t = 2.
13. x
9. x = £, y = P; t=— 1.
14. x
10. x = 2 t,y = 3^; t=l.
15. x
16. The curve
( x = 4 a cos* t,
17. Thehypocycloid (astroid) <
{ y = 4 a sin* t.
4ns f (a) ycott, (b) y tant, (c) £,
sint
z = 4 a cos* t,
(d)
cost
.4n*. (a) — y cot t, (b) — y tan t, (c) ^ ,
sint
(d)
cost
18. The circle
19. The cardioid
rx = rcost,
\y =s rsint.
rx = a (2 cost — cos2t),
\ y = a (2 sin t — sii
sin 2 1).
20. The folium
. 3t
* = i + *'
8t»
y =
l+t»
21. The hyperbolic spiral
x = cost,
t '
y =  ein t.
t
67. Angle between the radius vector drawn to a point on a curve
and the tangent to the curve at that point. Let the equation of
the curve in polar coordinates be p =/(0).
Let P be any fixed point (/>, 0) on the curve.
If By which we assume as the independent vari
able, takes on an increment A0, then p will
take on a corresponding increment A/?. Denote
by Q the point (/> + Ap, + A0). Draw PR perpendicular to OQ.
Then OQ = p + Ap, PR = p sin A0, and OR = p cos Aft Also,
tanP&R = ^ =
P^
/? sin A0
P>Q OQ—OR p + Ap — /jcosA0
84 DIFFERENTIAL CALCULUS
Denote by ^r the angle between the radius vector OP and the
tangent PT. If we now let A0 approach the limit zero, then
(a) the point Q will approach indefinitely near P;
(b) the secant PQ will approach the tangent PT as a limiting posi
tion; and
(c) the angle PQB will approach yfr as a limit
Hence
tan*= ^ p 8in Ag
T A0 = O p + A/j/jcosA0
2$/ _ limit psinAfl
A0 = O m
2psin 2 — + Ap
T X *
I Sinoe from 39, p. 2, ppco8A0p (1cos A0)2psin» — • I
psin Afl
limit A0
A0 "'"A*
[Dirldlng both numerator and denominator by At.]
sin A^
P
limit r Afl
A0 = O ~a5
. A0 2 Ao
^T^T + A*
2
c . limit /A/A dp M , limit / . A0\ n _,„ limit / sin Afl\ .
Smce A^=o(A?j = i ^ AO=o( 8m Tr°' al8 ° A^=o(A7j=
. A0
sin —
and ^j™^—— =1 by § 22, p. 21, we have
From the triangle OPT we get
SIMPLE APPLICATIONS OF THE DERIVATIVE 85
Having found t, we may then find tan t, the slope of the tangent
to the curve at P. Or since, from (£),
4r 4. ^i L i\ tan0 + tan*^r
tan r = tan (0 + ^) = = —  — T >
v 1 — tanPtan^r
we may calculate tan yjr from (A) and substitute in the formula
(C) slope of tangent = tanr= i _ tangtan ^ .
Illustrative Example 1. Find V and r in the cardioid p = a(l — coa0). Also
find the slope at 9 =  .
6
Solution. £ = a sin 0. Substituting in (A) gives
^ 2 a sin 2 
p a(l — CO80) 2 * « ««. « „ «
tan » = ^ = «.in* % , 6 * =Un 2 By89,p. 2 ,and87,p.8
£ 2 a sin cos 
d(9 2 2
9 9 89
Since tan ^ = tan  , ^ = . <4tw. *Substituting in (B), t = +  = 4n*.
tanT = tan = l. Awt.
4
To find the angle of intersection <f> of two curves C and C 1 whose
equations are given in polar coordinates, we may proceed as follows :
angle TFT 1 — angle OPT'— angle OPT,
or, <f> = ^r' — *^r. Hence
(D) tanrf = tan^tanf
w r 1 + tan^'tan^
where tan ^r' and tan yfr are calculated by
(<A) from the two curves and evaluated
for the point of intersection.
Illustrative Example 2. Find the angle of
intersection of the curves p = a sin 20, p = a cos 2 9.
Solution. Solving the two equations simultaneously, we get at the point of inter
section t8Ln2 9 = 1, 29 = 46°, 9 = 22i°.
From the first curve, using (A),
tan^' = J tan 29 = J, for 9 = 22°.
From the second curve,
tan^=Jcot20 =  J, for 9 = 22°.
Substituting in (D), , . i
tan = ^ — I = . .«. <p = arc tan . Ana.
86 DIFFERENTIAL CALCULUS
68. Lengths of polar subtangent and polar subnormal. Draw a line
NT through the origin perpendicular to the radius vector of the
X point P on the curve. If PT is the tangent and PN
the normal to the curve at P, then
OT= length of polar nubtangent,
and ON ~ length of polar subnormal
of the curve at P.
OT
In the triangle OPT, tan J/ = — — Therefore
P
,d9
(7) OT = ptaiiiir = p t — = length of polar subtangent.*
dp
lu the triangle OPN, tan^r = ?■ Therefore
(8) OiV= —  — = £ = length of polar subnormal.
The length of the pular tangent (= PT) and the length of the polar
normal (= PN) may be found from the figure, each being the hypot
enuse of a right triangle.
Illustrative Example 3. Find lengths o( polar aubtangent anil subnormal to the
lemnisciue p' = a* cos 20.
Solution. Differentiating the equation of the curve aa an implicit function with
respect to 0,
.ft.
n 20,
d P
Substituting in (T) and (8), we get
length of polar subtangent = — ° — ,
length of polar subnormal =
If we wish to express the results in terms of 8, find p i n term s of 8 from the given
equation and substitute. Thus, in the above, p = ±aVeo820; therefore length of
polar subtangent = ± a cot 20 Vcos20.
* When 9 increases with p, — Is positive and * Is an acute angle, as In the above figure.
Then the suh tan gent 07" Is positive and is measured ti> (lie right of an observer placed at Oand
SIMPLE APPLICATIONS OF THE DERIVATIVE 87
Z. In tip circle p = r sin0, find V and r in terms of 0. Ans. ^ = 0, r = 20.
2. In the parabola p = a sec*  » show that t + ^ = x.
3. In the curve pi* = a* cos 20, show that 2 ^ = x + 40.
4. Show that V is constant in the logarithmic spiral p = e*. Since the tangent
makes a constant angle with the radius vector, this curve is also called the equi
angular spiral.
6. Given the curve p = a sin 8  , prove that r = 4 ^.
3
6. Show that tan ^ = in the spiral of Archimedes p = off. Find values of V
when = 2 x and 4 x. 4ns. ^ = 80° 67' and 86° 27'.
7. Find the angle between the straight line p cos0 = 2 a and the circle p —
5 a sin 0. 4n*. arc tan .
8. Show that the parabolas p = a sec*  and p = b esc*  intersect at right angles.
9. Find the angle of intersection of p = a sin and p = a sin 20.
iln«. At origin 0° ; at two other points arc tan 8 VS.
10. Find the slopes of the following curves at the points designated :
(a) p = a(l — 0080). = — Ans. — 1.
(b) p=asec*0. p = 2a. 8.
(c) p = a sin 40. origin. 0, 1, co, — 1.
(d) p* = a* sin 40. origin. 0, 1, oo, — 1.
(e) p = a sin 80. origin. 0, V§, — V&
(f) p = a cos 80. origin.
(g) p = a cos 20. origin.
(h) p=asin20. ^ = t*
4
(i) p=asin80. * = £•
6
(J) p = a$. = .
(k)p0=a. = .
(1) p = €*. 0=0.
a
11. Prove that the spiral of Archimedes p = ad, and the reciprocal spiral p =  »
intersect at right angles.
12. Find the angle between the parabola p = a sec* and the straight line
psin0=2a. Ans. 46°.
13. Show that the two cardioids p = a (I + cos0) and p = a(l — cos0) cut each
other perpendicularly.
14. Find lengths of subtangent, subnormal, tangent, and normal of the spiral of
Archimedes p = ad. A . . p* . p /=— — z
r iliw. 8iibt. = *— , tan. = * v a 2 + p*,
a a
subn. = a, nor. = Va* + fi*.
The student should note the fact that the subnormal is constant.
DIFFERENTIAL CA
15. Get lengths of subtaugent, subnormal, tangent, and normal in the logarithmic
ir:il
P
A.
'2:
. P t
logo'
^ploga, n
" 1  = P V 1 + io^'
or. = p Vl + log' a.
When a  e,
we notice that iubt. =
subn..
and tan.
= nor.
16.
Find the
angles between the cu
rvesp
= a(l +
uxS),p =
6(1 cwtf).
An*. Oand
17.
Show that the reciprocal spiral
'=£
has a constant subtangent.
18. Show that the equilateral hyperbolas p ! sin 26 = a', p* cos 26
right angles.
69. Solution of equations having multiple roots. Any root which
occurs more than once in an equation is called a multiple root.
Thus 3, 3, 3, 2 are the roots of
(A) x<7x' + $x t +27x54 = 0;
hence 3 is a multiple root occurring three times.
Evidently (A) may a]so be written in the form
(i3)'(»+2)=0.
Let/(ar) denote an integral rational function of x having a multiple
root a, and suppose it occurs m times. Then we may write
W /O) (* *)*(*),
where 0(V) is the product of the factors corresponding to all the roots
of /(a) differing from a. Differentiating (5),
' CO /'(*) (» «)» [(*  «) f <*) + too w).
Therefore /'(a:) contains the factor (x — n) repeated m — 1 times
and no more; that is, the highest common factor (H.C.F.) of f(x)
and/'(x) has m — 1 roots equal to a.
In case f(x) has a second multiple root occurring r times, it is
evident that the H.C.F. would also contain the factor (at — fly 1 , and
so on for any number of different multiple roots, each occurring once
more in f(x) than in the H.C.F.
We may then state a rule for finding the multiple roots of an equation
f(x) = as follows :
First Step. Findf(x).
Second Step. Find the H.C.F. off(x) andf(*).
Third Step. Find the roots of the H O.F. Each different root of the
H.C.F. will occur once more inf(x) than it doet in the H.C.F.
SIMPLE APPLICATIONS OF THE DERIVATIVE
89
If it tarns out that the H.C.F. does not involve z\ then f(x) has
no multiple roots and the above process is of no assistance in the
solution of the equation, but it may be of interest to know that the
equation has no equal, i.e. multiple, roots.
Illustrative Example 1. Solve the equation x 8 — 8x 2 + 18x — 6 = 0.
Solution. Place /(x) = x« — 8x 2 + 18x — 6.
First step. /'(x) = 8x 2  16x + 13.
Second step. H.C.F. = x — 1.
Third step. x — 1 = 0. .\ x = 1.
Since 1 occurs once as a root in the H.C.F., it will occur twice in the given equa
tion ; that is, (x — 1)* will occur there as a factor. Dividing z* — 8x 2 + 13 x — 6 by
(x — 1)* gives the only remaining factor (x — 6), yielding the root 6. The roots of
our equation are then 1, 1, 6. Drawing the graph of the function, we see that at
the double root z = 1 the graph touches OX but does not cross it.*
Solve the first ten equations by the method of this section :
Ans. 2, 2, 3.
1,1,1,8.
8, 3, 3, — 2.
3, 8, 3, — 4.
1,1,4,4.
8, 3,  1, 4.
2, 2,1 ± V5.
1,1,1,2,2.
2, 2, 2,  8,  8.
1,1, 1,8 ±VH.
1. x»7x« + 16x12 = 0.
2. x 4 — 6x f — 8x — 8 = 0.
3. x*7x» + 9x 2 + 27x64 = 0.
A. x 4  6x*  9x 2 + 81x  108 = 0.
5. x 4 + Ox 8 + x 2  24x + 16 = 0.
6. x*  9x* + 23x 2  3x  36 = 0.
7. x 4 6x» + 10x 2 8 = 0.
8. x* — x 4 — 6x» + x 2 + 8x + 4 = 0.
9. x«  15x» + 10x 2 + 60x 72 = 0.
10. x*  Sx 4  6x* + 13x 2 + 24x + 10 = 0.
Show that the following four equations have no multiple (equal) roots :
11. x 8 + 9x 2 + 2x48 = 0.
12. x 4  15x*  lOx + 24 = 0.
13. x 4  8x*  6x* + 14x + 12 = 0.
14. x» — <?• = 0.
15. Show that the condition that the equation
x* + 8gx + r =
shall have a double root is 4 g 8 + r 2 = 0.
16. Show that the condition that the equation
x* + 3px 2 + r =
shall have a double root is r (4p* + r) = 0.
* 8ince the first derivative vanishes for every multiple root, it
follows that the axis of X is tangent to the graph at all points corre
sponding to multiple roots. If a multiple root occurs an even number
of times, the graph will not cross the axis of X at such a point (see
figure) ; if it occurs an odd number of times, the graph will cross.
DIFFERENTIAL CALCULUS
70. Applications of the derivative in mechanics. Velocity. Recti
linear motion. Consider the motion of a point P on the straight line
AB. Let « be the distance meas
3 — ~p~^p % ure( i f rom some fixed point as A
to any position of P, and let t
be the corresponding elapsed time. To each value of t corresponds
a position of P and therefore a distance (or space) *. Hence $ will
be a function of t, and we may write
«=/(<)■
i an increment At ; then a takes on an increment
— = the average velocity
of P during the time interval At. If P moves with uniform motion,
the above ratio will have the same value for every interval of time
and is the velocity at any instant.
For the general case of any kind of motion, uniform or not, we
define the velocity (time rate of change of g) at any instant as the
As
limit of the ratio — as A< approaches the limit zero; that is,
limit A*
At = At '
TJie velocity is the derivative of the dittance (= space'} with
to the time.
To show that this agrees with the conception we already have of
velocity, let us find the velocity of a falling body at the end of two
.scroll lis.
By experiment it has been found that a body falling freely from rest
in a vacuum near the earth's surface follows approximately the law
« = 16.lt',
where s = space fallen in feet, ( = time in seconds. Apply the Gen
eral Rule, p. 29, to (J?).
'* Aa being the space or distant.'? pMMd over in the lime At.
SIMPLE APPLICATIONS OF THE DERIVATIVE 91
First Step. i + A* = 16.1(t + Af)*=16.1t , + 82.2£ At + 1&1(A£)*.
Second Step. As = 32.2 1 At +16.1 (At) a J
Third Step. — = 32.2 ( +16.1 At = average velocity throughout
the time interval At.
Placing f = 2,
A*
((7) — = 64.4 + 16.1 At = average velocity throughout the
time interval At after two seconds of falling.
Our notion of velocity tells us at once that (C) does not give us
the actual velocity at the end of two seconds ; for even if we take A(
very small, say jjg or j^Yi of a second, (C) still gives only the
average velocity during the corresponding small interval of time. But
what we do mean by the velocity at the end of two seconds is the
limit of the average velocity when At diminishes towards zero; that is,
the velocity at the end of two seconds is from (C), 64.4 ft. per second.
Thus even the everyday notion of velocity which we get from experi
ence involves the idea of a limit, or in our notation
limit /AA
At
: 64.4 ft per second.
The above example illustrates well the notion of a limiting value.
The student should be impressed with the idea that a limiting value
is a definite, fixed value, not something that is oniy approximated.
Observe that it does not make any difference how small 16.1 At may
be taken ; it is only the limiting value of
64.4 + 16.1 At,
when At diminishes towards zero, that is of importance, and that
value is exactly 64.4.
71. Component velocities. Curvilinear motion. The coordinates x
and y of a point P moving on a locus in the JTI'plane are also
functions of the time, and the motion may be denned by means of
two equations, *}(& y = «Q.«
These are the parametric equations of the path (see § 66, p. 79).
octangular coordinates may bo found by eliminating f
n
DIFFERENTIAL CALCULUS
The horizontal component v x of v * is the velocity along OX of the
projection M of P, and is therefore the time rate of ehange of x.
Hence, from (9), p. 90, when « is replaced by a, we get
Y
N
v
J^C
sf
A*?)
1 M X
(10)
dt
In the same way we get the vertical i
ponent, or time rate of change of y,
<»>
dt
Representing the velocity and its components by vectors, we have
t once from the figure
(12
■iW®
giving the magnitude of the velocity at any instant.
If t be the angle which the direction of the velocity makes with
the axis of X, we have from the figure, using (9), (10), (11),
dy dx dy
dt
dt
72. Acceleration. Rectilinear motion. In general, v will lie a function
of r, and we may write v = &(t'\
Now let ( take on an increment At, then v takes on an increment
Aw, and
— = the average acceleration of P during the time interval A/.
We define the acceleration a at any instant as the limit of the ratio
— as At approaches the limit zero ; that iB,
limit /A«\
or, v '
..*
The acceleration is the derivative of the velocity with respect to the time.
• Tha direction of v is along tho tangent to tbe path.
SIMPLE APPLICATIONS OP THE DERIVATIVE
98
73. Component accelerations. Curvilinear motion. In treatises on
Mechanics it is shown that in curvilinear motion the acceleration is
not, like the velocity, directed along the tangent, but toward the
concave side of the path of motion. It may be resolved into a tan
gential component, «,, and a normal component, a., where
dv v*
<"«> a, = — ; a.=~.
* ' ' (ft ' R
(.H is the radius of curvature. See § 103.)
The acceleration may also be resolved into components parallel to
the axes of the path of motion. Following the same plan used in § 71
for finding component velocities, we define the component acceleration*
parallel to OX and OY,
dv
a t = — *■ Also,
which gives the magnitude of the acceleration at any instant.
EXAMPLES
1. By experiment it has been found that a body falling freely from rest in a vac
uiim near the earth's surface follows approximately the laws = 16.lt 9 , where s = space
(height) in feet, ( = time in seconds. Find the velocity and acceleration fa) at any
instant ; (b) at end of the first second ; (c) at end of the fifth second.
Solution. id) s = 16.1t».
(a) Differentiating,
Differentiating again,
= 32.2.
ir, from ((»),
■, from (14),
32.2tft. per sec.
32.2 ft. per (sec.)*,
in other words, the
n (B) and (C) ;
n (B) and (C) ;
which tells us that the acceleration of a falling bod; is ci
Telocity increases 32.2 ft. per sec. every second it keeps on falling.
(b) To End o and a at tlie end of the first second, substitute ( 
e = 32.2 ft. per sec, a = 32.2 ft. per (sec.) 5 .
(c) To find d and a at the end of the fifth second, substitute t =
b = 161 ft. per sec, a = 32.2 ft. per (sec)*.
2. Neglecting the resistance of the air, the equations of motion for a projectile are
i = B[COB#.t, 11 = 6, sin*. (16.lt*;
where o, = initial velocity, <p = angle of projection with hori
zon, t = time of flight in seconds, z and y being measured in
feet. Find the velocity, acceleration, component velocities,
and component accelerations (a) at any instant ; (b) at the end
of the first second, having given o, = 100 ft. per sec, <p = 30° ; (c) find directioi
motion at the end of the first second.
64 DIFFERENTIAL CALCULUS
Solution. From (10) and (11),
(a) .  Ox =' "i cos^ ; v v = v l Bin<f> — 82.2 U
Also, from (12),' « = Vcf ~ 64.4 fy sin <p + 1086.8 t« .
From (16) and (16), a* = ; a, = — 82.2 ; a = — 82.2.
(b) Substituting t = 1, u t = 100, ^ = 80° in these results, we get
v x = 86.6 ft. per sec. a x = 0.
©y = 17.8 ft. per sec. a w = — 82.2 ft. per (sec.)*.
© = 88.4 ft. per sec. a = — 32.2 ft. per (sec.) 2 .
v 17 8
(c) r = arc tan * = arc tan — '— = 11° 86'.6 = angle of direction
of motion with the horizontal. Vx
3. Given the following equations of rectilinear motion. Find the distance, velocity,
and acceleration at the instant indicated :
(a) a = t 8 + 2f»; t = 2. , ; %  Ana. a = 16, « = 20, a = 16.
(b) a = (* + 2*; t = 8. a = 15, v = 8, a=2.
(c) a = 3 — 4*; * = 4. a = 13, * = — 4, a = 0. ,
(d) x = 2Jf 2 ; 4 = 1. x = 1, c = 0, a = — 2.
(e) y = 2 *  i* ; t = 0. y = 0, * = 2, a = 0.
(f) A = 20t.+ 16* 2 ; 4 = 10. A = 1800, u = 840, a=82.
(g) a = 2sin*; ( = ^. a = V2, u = V2, a =  VS.
,.. iri. a waV8 nr*a
(h) y = acos— , * = 1. y = , * = _,* = _..
(i) a = 2c 8 '; 4 = 0. a = 2, v = 6, a= 18.
(j) * = 2t*8*; ( = 2.
C (k) x = 4 + l«; t = 3.
(1) y = 5cos2t; * = ^«
irf 6
(m) a = 6sinY» t = 2.
(n) x = oer %t \ t = 1.
• *(o) * = * + 6P; « = V
" ^)a = 101og4;t = l.
4 + *
;• » ;. '•■.: .
4. If a projectile be given an initial velocity of 200 ft. per sec. in a direction
Inclined 45° with the horizontal, find
(a) the velocity and direction of motion at the end of the third and sixth seconds ;
(b) the component velocities at the same instants.
<n~ Conditions are the. same as for Ex. 2.
ilji*. (a) When* = 8, v = 148.3 ft. per sec., t= 17° 86%
when t = 6, c = 150.6 ft. per sec., t = 160° 63' ;
(b) when t = 3, v x = 141.4 ft. per sec, v 9 = 44.8 ft. per sec.
when t = 6, v x = 141.4 ft. per sec, ty = — 51.8 ft. per sec.
fc. The height (= a) in. feet reached Jn t seconds by a body projected vertically
HHWWris with, ft velocity k of t> v ft* per sec. is given by the formula
a = ^ — 16.1 **. <
y
SIMPLE APPLICATIONS 0* THE DERIVATIVE
95
Find (a) velocity and acceleration at any instant ; and, if n,=: 300 ft. per sec, find
velocity and acceleration (b) at end o( 2 seconds; (c) at end of 15 seconds. Resists
ance of air is neglected. Am. (a) = o, — 32.2 1, a = — 32.3 ;
(b) a = 335.0 ft. per sec. upwards,
a = 82.2 ft. per (Bee.) 1 downwards;
(o) ii = 163 ft. per sec. downwards,
a = 33.3 ft. per (sec.) downwards.
6. A cannon ball la fired vertically upwards with a muzzle velocity of 641 ft. per
aec. Find {a) its velocity at the end of 10 seconds ; (b) for how lung it will continue
to rise. Conditions same as for Ex. 5. Ant. (a) 323 ft. per sec. upwards;
(b) 20 seconds.
7. A train left a station and in i hours was at a distance (space) of
s = (" + 2P + 8J
miles from the starting point. Find its acceleration (a) at the end of 1 hours ; (b) at
the end of 2 hours. Am. (a) a = 6 I + i ;
(b) a = 16 miles per (hour)*.
8. In t hours a train bad readied a point at the distance of j (* — 4 I s + 16 P miles
from the starting point, (a) Find its velocity and acceleration, (b) When will the
train stop to change the direction of its motion ? (c) Describe the motion during the
first 10 hours. Am. (a) o = P12P + 32t, a= 3P 241 + 32;
(b) at end of fourth and eighth hours;
(c) forward first 4 hours, backward the next
4 hours, forward again after 6 hours.
9. The space in feet described in ( seconds by a point is expressed by the formula
s = 48t~16t'.
Find the velocity and acceleration at the end of Ij seconds.
Ana. v = 0, a = S2ft. per (sec./*.
10. Find the acceleration, having given
(a) D = i a + 2i; t = 3. Am. b = 8.
(b) B = Stt>; 1 = 2. a = 9.
(c> o = 4sin^; t=~. n=V5.
(d) t
>s3(; t
11. At the end of ( seconds a body has a velocity of 3i* + 2< ft. per sec; find its
acceleration (a) in general ; (b) at the end of 4 seconds.
A?u. (a) a= Q( + 2 ft. per (sec.)'; (b) a = 28ft. per (see.)*.
12. The vertical component of velocity of a point at the end of I seconds is
r r = 3 P — 2 1 + 8 ft, per sec.
Find the vertical component of acceleration (a) at any Instant; (b) at the end of 2
seconds. Ana. (a) tr r =8(2; (b) 10 ft. per (see.)'.
13. If a point moves in a fixed path so that
..VI,
sfaow that the acceleration is negative and proportional to the cube of the velocity.
96 DIFFERENTIAL CALCULUS
14. If the space described is given by
a = ae* + be« t
show that the acceleration is always equal in magnitude to the space pa wind over.
16. If a point referred to rectangular coordinates moves so that
z = acoet + 6, and y = a sin t + c,
show that its velocity has a constant magnitude.
16. If the path of a moving point is the sine carve
y = 6 sin at,
show (a) that the xcomponent of the velocity is constant ; (b) that the acceleration
of the point at any instant is proportional to its distance from the axis of JT.
17. Given the following equations of curvilinear motion, find at the given instant
«*, IV, v ; <**» <*r» a *t position of point (coordinates) ; direction of motion. Also find
the equation of the path in rectangular coordinates.
(a) x = f», y = f; t = 2. (g) x = 2sint,y = 8co8t; * = a\
{;
(b) z = t,y = P; t = l.
(h) x = sint,y = co82t; t = 
(c) x = <», yfi; t = 8. 4
(d) x = 2t,y = <» + 8; t = 0. (i) x = 2t, y = 8*; t = 0.
(e) x = l?,y = 2t; t = 2. (j) * = H y = log*; t = 1.
(f) x = asint,y = aco8t; t = ?^. (k) x = t, y = 12**; t = 8.
CHAPTER VII
SUCCESSIVE DIFFERENTIATION
74. Definition of successive derivatives. We have seen that the
derivative of a function of a: is in general also a function of x. This
new function may also be differentiable, in which case the derivative
of the first derivative is called the second derivative of the original
function. Similarly, the derivative of the second derivative is called
the third derivative; and so on to the nth derivative. Thus, if
y=3s 4 ,
! = 12 * 8 '
ax
dx\dxj
dx [dx \dx)\
75. Notation. The symbols for the successive derivatives are
usually abbreviated as follows:
dxxdxl da?
d_ X£ /^Al = £ /ays.
dx [dx \dx) J dx \dx*/
teftg.
dx\da? 1 ) da?
If y =/(x), the successive derivatives are also denoted by
/'(*), /"(*), /"'(*), f\x), • • •, f\xy,
y 7 , y", y"', y", •••, *";
m ' ic f ^ J? /Cr) ' l? /(x) ' J? /(a °' "•' d% f< &'
07
98 DIFFERENTIAL CALCULUS
76. The nth derivative. For certain functions a general expression
involving n may be found for the nth derivative. The usual plan is to
find a number of the first successive derivatives, as many as may be
necessary to discover their law of formation, and then by induction
write down the nth derivative.
(jny
Illustrative Example 1, Given y = e ax 1 find — ^.
dx*
dy
Solution. f = a***,
dz
** = «»««
• • •
d*y
.\ — — = aw*. Am.
dx"
d*y
Illustrative Example 2. Given y = logx, find ~*
ax*
dy 1
Solution. r = >
dx x
dx*~ x«'
d*y _l2
dx»"~x»~'
d*y __ 12 3
dx 4 " x 4 '
dv .llLzJL
d*i/
Illustrative Example 3. Given y = sin x, find
J.7U.
dx"
Solution. f = cos x
dx
dty _ a ~
dx 2
= gin(x + )
d*y
dx»
= = sin ( x + — ) = cos ( x +  ) = sin [ x + — ) t
ax \ 2/ \ 2/ \ 2 /
d ^ / , 2ir \ / . 2ir \ ; / , 3 *\
= — sinlx + — ) = cosix + — ) = sinlx + — )
dx \ X 2/ V 2/ \ 2/
= sin(x + ^)
— = 8in(x + — )• An*.
dx»
77. Leibnitz's Formula for the nth derivative of a product. This
formula expresses the nth derivative of the product of two variables
in terms of the variables themselves and their successive derivatives.
SUCCESSIVE DIFFERENTIATION 99
If u and v are functions of x, we have, from Y,
d . n_^ • & v
dx dx dx
Differentiating again with respect to x,
d? , N d*u , dudv , du dv , d?v d*u , ~dudv . rf*t;
— 5 CUV) = ; V + + + U j =  V + 2 + M •
efcr efcr d& {& (4r dx da? dor dx dx dx 9
Similarly,
da? dx* dx 2 dx da? dx dx da? dx da? da?
_d?u Q^?udv o^f^ i <^V
da? da? dx dx da? da?
However far this process may be continued, it will be seen that the
numerical coefficients follow the same law as those of the Binomial
Theorem, and the indices of the derivatives correspond to the expo
nents of the Binomial Theorem.* Reasoning then by mathematical
induction from the with to the (m + l)th derivative of the product,
we can prove Leibnitz's Formula
,«■* ' , x d*u d?~ x udv n(n—l)d**ud*v
(17) — (tip) = — v + n — + —^ ' —  h
dx* K J dx* dx» l dx [i dx"* dx?
du d*~ x v d*v
+ n 1 u — •
dx dx*' 1 dx*
Illustrative Example 1. Given y = e*logx, find — ^ by Leibnitz's Formula.
Solution. Let a = 6*,
and v = logx;
du
then — = €*,
dx
dv 1
dx x
cPu
— ffC
dx* '
d*v _ 1
dx*~ X*'
d*u
d*i>_ 2
dx*~x*'
Substituting in (17), we get
_ = e*logx + 
8
~x !
x J x 8 /
* To make this correspondence complete, u and t; are considered as —^ and — •
100 DIFFERENTIAL CALCULUS
Illustrative Example 2. Given y = x , c ax , find —  by Leibnitz's Formula.
ox*
8olotk>n. Let u = x s ,
and « = e  * ;
then — = 2x,
dx
& = <"*>
<Ptt_
&9 ^^
—  = a"^.
dx»
Substituting in (17), we get
*r oliriiiiftr t Q«Hi1«i«c I m /•• 1\ <i
im — fliMX <•■_•
= z f <r»e°» + 2na« 1 x€*«* + n(n — l)a»*e°* = a"»e«[x f a f + 2nax + n(n — 1)].
dx*
78. Successive differentiation of implicit functions. To illustrate the
process we shall find ^ from the equation of the hyperbola
bWaY=a*b\
Differentiating with respect to x, as in § 63, p. 69,
ax
or,
Differentiating again, remembering that y is a function of x,
d*y y dx
dx*~ a*f
Substituting for %• its value from (A),
ax
d*y 9 W bXbVa*}/')
dx* ay ay
But from the given equation, 6V— ay = a*b*.
dy V_
'' dx*~ a y'
SUCCESSIVE DIFFERENTIATION 101
Differentiate the following :
1. y = 4x«6x* + 4x + 7. ^ = 12(2* 1).
car
2. /(x) = ?.. /*(*) =
4. y = xMogx.
18. y = log(l + x).
19. y = cos ax.
20. y = x* 1 logx.
1x w (1x)*
3. /(y) = y«. /*(y) = [6.
d*y_6
dx 4 x
• # j
6. y = ±. n(n + l)c
.x" x»+*
6. y = (x— 3)e** + 4xe* + z. y" = 4e*[(x2)e* + x + 2].
7. y = ?(J + e~«). y"=±.(2 + e ) = % i .
* a a Or
8. /(x) = ax* + te + e. /'"(x) = 0.
9. /(x) = log (x + 1). f*(x) =  6
(x + 1)*
10. /(x) = log(^ + r«). /'"(x) = ?fc^3
dV
11. r = 8ina0, — = a*Bina0 = afir.
dfir
12. r = tan0. — = 6 sec 4 — 4860*0.
13. r = logsin0. r"'= 2cot0csc*0.
14. /(*) = e*coet. / lT (0 = 4c«cost =  4/(Q.
15. /(*) = Vsec20. /"(*) = 8 [/(*)]« /(*).
16. p = te* + a*)arctan? <f * P  ***
a dg» (a* + g*)«
17. y = a*. ^ = (loga)«a x .
dx»
(jny n — 1
dx» v ' (l + x)«
—  = a* cos (ax + — ).
ax» \ 2/
(j»y nl
dx» x
[n a positive integer.]
21.y = l^. ^ = 2(l)» ^
1 + * dX« x ' (l + 2)"+»
O
Hixt. Reduce fraction to form  1 + before differentiating.
22. If y = 0*8inx, prove that 4 — 2— + 2y = 0.
dx* dx
28. If y = a cos (log x) + 6 sin (log x), prove that x* ~ + x~ + y = 0.
dx dx
102 DIFFERENTIAL CALCULUS
Use Leibnitz's Formula in tfcfi ijext four examples :
24. y = *»a*. .. •.*"* —£ = a*(loga)»*[(xloga + n j« _ n i
» * CEX*
• % * " ^^
26. y*j&. ;. ^ = tf r (x + n) .
;26>/$ = 6* sin x. /(«)(x) = ( V2> c* sin (x + — ) .
•\ •!•• * \ 4/
•••;* •"' **" /( ^ = C08a ^ C08W  /^W = (a "t 6) " co8r(a + b)6 + — ]
28. Show that the formulas for acceleration, (14), (15), p. 02, may be written
, (a &)"
+  008
2
d 2 * (Px (Py
> Ctx — » (Xm = —
dP x dP " dP
29. y« = 4ax.
30. Px 2 + oV = «W.
31. x 2 + y 2 = r«.
32. y 2 + y = x 2 .
33. ax 2 + 2 Axy + &y* = L
34. y 2 — 2xy = a 2 . _ ,
dx 2 (yx) 8, dx» (yx)*
oc ^ * (P0 tan 2 — tan 2
35. sec <p cos = c. — = — .
d> 2 tan 2 *
d> 2 0«
37. Find the second derivative in the following :
(a) log(u + v) = u — «. (e) y 8 + x 8 — 8oxy = 0.
(b) c" + u = e» + v. (f)j^ 2mxy + x 2  a = 0.
(c) * = 1 + (*. (g) y = sin(x + y).
(d) e« + st — c = 0. (h) e?+* = xy.
(Py __ 4 a 2
dx 2 ~"""y r '
(Py _ &* <Py __
dx 2 ~ a 2 y» ' dx 2 ~"
3&*x
aV
cPy __ r 2
dx 2 y»*
(Py 24 x
dx 2 (l + 2y)»
(Py A 2 aft
dx 2 (Ax + by) 9
cPy_ a 2 (Py
8(Px
CHAPTER Vni
KAXIMA AND MINIMA. POINTS OF INFLECTION. CURVE TRACING
79. Introduction. A great many practical problems occur where
we have to deal with functions of such a nature that they have a
greatest (maximum) value or a least (minimum) value," and it is
vtry important to know what particular value of the variable gives
such a value of the function. For instance, suppose that it is required
to find the dimensions of the rectangle of greatest area that can be
inscribed in a circle of radius 5 inches. Consider the circle in the
following figure:
Inscribe any rectangle, as BD.
Let CD = i ; then DE = VlOO — x 3 , and the area of the rectangle is
evidently _____
(1) ___VlO0 — °.
at a rectangle of maximum area must exist may be seen as follows :
'jet the base CD (— z) increase to 10 inches (the diameter); then
t,Je altitude _D J ? = Vl0O — x* will decrease to
zero and the area will become zero. Now let
the base decrease to zero ; then the altitude
w "iH increase to 10 inches and the area will
a g a iB become zero. It is therefore intuitionally
t'vulent that there exists a greatest rectangle.
"J a careful study of the figure we might sus
pect that when the rectangle becomes a square
its ftfpa W ould be the greatest, but this would at best be mere guess
work, A better way would evidently he to plot the graph of the
function (1) and note its behavior. To aid us in drawing the graph
o> (1), we observe that
(ft) from the nature of the problem it is evident that x and A must
both be positive ; and
(b) the values of x range from zero to 10 inclusive.
DIFFERENTIAL CALCULUS
Now construct a table of values and draw the graph.
What do we learn from the graph t
*
*
1
0.0
2
19.0
3
28.0
4
30.0
1
43.0
48.0
7
49.7
8
48.0
9
S9.0
10
0.0
so fi ^T*x '
40 q/ \
r/\ \
I A\
"/I
6 /!
(\ I JAf I VI  ,HJ
I 4 " 1 7 8 9]
(a) If carefully drawn, we may find quite accurately the area of
the rectangle corresponding to any value of x by measuring the length
of the corresponding ordinate. Thus,
when x = OM= 3 inches,
then A = MP = 28.6 square inches ;
and when x = ON= 4^ inches,
then A =NQ — about 39.8 sq. in. (found by measurement).
(b) There is one horizontal tangent (RS). The ordinate TH from
its point of contact T is greater than any other ordinate. Hence this
discovery: One of the inscribed rectangles has evidently a greater area
than any of the others. In other words, we may infer from this that
the function defined by (1) has a maximum value. We cannot find
this value (= HT~) exactly by measurement, but it is very easy to
find, using Calculus methods. We observed that at T the tangent was
horizontal ; hence the slope will be zero at that point (Illustrative
Example 1, p. 74). To find the abscissa of T we then find the first
derivative of (1), place it equal to zero, and solve for x. Thus
a)
dA
viooi"
= W100i"
100 2*»
MAXIMA AND MINIMA
105
Solving,
Substituting back, we get DE = VlOO— 2* = 5 V2.
Hence the rectangle of maximum area inscribed in the circle is a
square of area
A = CD x DE= 5 V2 x 5 V2 = 50 square inches. The length of
HT is therefore 50.
Take another example. A wooden box is to be built to contain
108 cu. ft It is to have an open top and a square base. What must
be its dimensions in order that the amount of material required shall
be a minimum ; that is, what dimensions will make the cost the least ?
Let x = length of side of square base in feet,
and y = height of box.
Since the volume of the box is given, how 1
ever, y may be found in terms of x. Thus
volume = x*y = 108; ,'.y = — y
We may now express the number (= M~) of square feet of lumber
required as a function of x as follows :
area of base = x* sq. ft,
and area of four sides = 4xy = sq. ft Hence
(2) M=S + ™
y%
X
V
1
433
2
320
3
163
4
124
5
111
6
108
7
111
8
118
D
120
10
143
is a formula giving the number of square feet required in any such box
having a capacity of 108 cu. ft Draw a graph of (2).
106
DIFFERENTIAL CALCULUS
i the graph f
What do we learn fro
(a) If carefully drawn, we may measure the ordinate correspond
ing to any length (=3") of the side of the square base and so deter
mine the number of square feet of lumber required.
(b) There is one horizontal tangent (J?S). The ordinate from its
point of contact T is less than any other ordinate. Hence this dis
covery : One of the boxes evidently takes less lumber than any of the
others. In other words, we may infer that the function defined by
(2) has a minimum value. Let us find this point on the graph ex
actly, using our Calculus. Differentiating (2) to get the slope at any
point, we have dM 482
dx x*
At the lowest point T the slope will be zero.
that is, when x = 6 the least amount of lumber will be needed.
Substituting in (2), we see that this is
3f=108 sq.ft.
The fact that a least value of M exists is also shown by the follow
ing reasoning. Let the base increase from a very small square to a
very large one. In the former case the height must be very great and
therefore the amount of lumber required will be large. In the latter
case, while the height is small, the base will take a great deal of
lumber. Hence M varies from a large value, grows less, then
increases again to another large value. It follows, then, that the
graph must have a "lowest" point corresponding to the dimensions
which require the least amount of lumber, and therefore would involve
the least coet.
We will now proceed to the treatment in detail of the subject of
maxima and minima.
80. Increasing and decreasing functions.* A function is said to be
increasing when it increases as the variable increases and decreases as
the variable decreases. A function is said to be decreasing when it
decreases as the variable increases and increases as the variable
MAXIMA AND MINIMA
107
The graph of a function indicates plainly whether it is increasing
or decreasing. For instance, consider the function a* whose graph
(Fig. a) is the locus of the equation
y = a*, a > 1
As we move along the curve from left to right the curve is rising ;
that is, as a; increases the function (= y) always increases. Therefore a*
is an increasing function for all values of x.
Y
Ay
A*
X
Fig. a
Fig. b
On the other hand, consider the function (a — a;) 8 whose graph
(Fig. 6) is the locus of the equation
y=(axy.
Now as we move along the curve from left to right the curve is
falling ; that is, as # increases, the function (=y) always decreases.
Hence (a — a;) 8 is a decreasing function for all
values of x.
That a function may be sometimes increas
ing and sometimes decreasing is shown by the
graph (Fig. (?) of
As we move along the curve from left to right p I0# c
the curve rises until we reach the point 4, then
it falls from A to B, and to the right of B it is always rising. Hence
(a) from z = — cotoz = lthe function is increasing ;
(b) from x = 1 to x = 2 the function is decreasing ;
(c) from x=2tox=: + co the function is increasing.
108 DIFFERENTIAL CALCULUS
The student should study the curve carefully in order to note the
behavior o£ the function when x = 1 and x = 2. Evidently A and B
are turning points. At A the function ceases to increase and com
mences to decrease ; at B, the reverse is true. At A and B the t
gent (or curve) is evidently parallel to the axis of A", and therefore
the slope is zero.
81. Tests for determining when a function is increasing and when
decreasing. It is evident from Fig. c that at a point, as C, where a
function ., .
is increasing, the tangent in general makes an acute angle with the
axis of Xj hence
slope = tan t = ~= /'(*) = a positivi
Similarly, at a point, as D, where a function is decreasing, the tan
gent in general makes an obtuse angle with the axis of X; therefore
slope = fan t = ~£ = f'(x) = a "negative numlet.*
In order, then, that the function shall change from an increasing to
a decreasing function, or vice versa, it is a necessary and sufficient
condition that the first derivative shall change sign. But this can only
happen for a continuous derivative by passing through the value zero.
Thus in Fig. e, p. 107, as we pass along the curve the derivative
(= slope) changes sign at A and B where it has the value zero. In
general, then, we have at turning points
m g=/'(,) = o.
The derivative is continuous in nearly all our important applica
tions, but it is interesting to note the case when the derivative
(= slope) changes sign by passing through ao.f This would evidently
•Conversely, for any given value of x,
iff'{x)—*, «*»/(*) U inenating;
f/'(r)  , thfn/d) it demoting.
When /'(*)  0, we cannot decide without further investigation whether/(x) is increas
ing or decreasing.
I By this Is meant that ita reciprocal passes through the value »
hetber/(x) is increM
MAXIMA AND MINIMA
109
happen at the points B, E, G in the following figure, where the
tangents (and curve) are perpendicular to the axis of X. At such
exceptional turning points
a.
kdx
what amounts to the same thing,
=/(*)
/'CO
 = o.
. Maximum and minimum values of a function. A maximum
value of a function is one that is greater than any values immediately
preceding or following.
A minimum value of a function is one that is le»» than any values
immediately preceding or following.
For example, in Fig. c, p. 107, it is clear that the function has a
maximum value MA (= y = 2) when x = l, and a minimum value NB
(=y = l) wheni = 2.
The student should observe that a maximum value is not neces
sarily the greatest possible value of a function nor a minimum value
the least. For in Fig. c it is seen that the function (= y) has values
to the right of B that are greater than the maximum MA, and values
to the left of A that are less than the minimum NB.
A function may have several maximum and minimum values.
Suppose that the above figure represents the graph of a function
At B, D, G, I, K the function is a maximum, and at C, E, H, J a
minimum. That some particular minimum value of a function may
be greater than some particular maximum value is shown in the figure,
tie minimum values at C and H being greater than the maximum
value at K.
110
DIFFERENTIAL CALCULUS
At the ordinary turning points C, D, H, I, J, K the tangent (or
curve) ia parallel to OX; therefore
_<ty_
slope i
dx
=/'(*)= 0.
At the exceptional turning points B, E, G the tangent (or curve) is
perpendicular to OX, giving
**fA»>—
One of these two conditions is then necessary in order that the
function shall have a maximum or a minimum value. But such a con
dition is not sufficient; for at Ft\\e slope is zero and at A it is infinite,
and yet the function has neither a maximum nor a minimum value at
either point. It is necessary for us to know, in addition, how the
function behaves in the neighborhood of each point. Thus at the
points of maximum value, B, D, G, I, K, the function changes from an
increasing to a decreasing function, and at the points of minimum value,
C, E, H, J, the function changes from a decreasing to an increasing func
tion. It therefore follows from § 81 that at maximum points
.
slope = ~ =/'(*) must change from + to
points
slope
= f'(x)must change fro
when we move along the curve from left to right.
At such points as A and F where the Blope is z>
which are neitlitir maximum nor minimum points,
) or infinite, but
slope = j 
fix) does not change sign.
We may then state the conditions in general for maximum and
minimum values of/(a) for certain values of the variable as follows:
(19) /(*) is a maximum if/'(.r) = 0, and /'(x) changes from +
to,
(20) f(x) is a minimum if /'(x) = 0, and /'(x) changes from —
to + .
The values of the variable at the turning points of a function are
called critical values; thus x = \ and x =2 are the critical values of
MAXIMA AND MINIMA
111
variable for the fuuction whose graph is shown in Fig. c, p. 107.
The critical values at turning points where the tangent is parallel to
QJ are evidently found by placing the first derivative equal to zero
and solving for real values of x, just as under § 64, p. 73."
To determine the sign of the first derivative at points near a par
ticular turning point, substitute in it, first, a value of the variable just
a little less than the corresponding critical value, and then one a
little greater.* If the first gives + (as at L, Fig. d, p. 109) and the
second — (as at JQ, then the function (= y) has a maximum value in
that interval (as at /).
If the first gives — (as at P) and the second + (as at IT), then the
function (= y) has a minimum value in that interval (as at C).
If the sign is the same in both cases (as at Q and R), then the
function (= y) has neither a maximum nor a minimum value in that
interval (as at F).\
We shall now summarize our results into a compact working rale.
63. First method for examining a function for maximum and mini
mom values. Working rule.
First Step. Find the first derivative of the function.
Second Step. Set the first derivative equal to zeroh and solve the
rendting equation for real roots in order to find the critical values of the
variable.
Thied Step. Write the derivative in factor form ; if it is algebraic,
write it in linear form.
Fockth Step. Considering one critical value at a time, test the first
derivative, first for a value a trifle less and then for a value a trifle greater
than the critical value. If the sign of the derivative is first + and then — ,
the function has a maximum value for that particular critical value of the
variable; but if the reverse is trite, then it has a minimum value. If the
sign does not change, the function has neither.
•Similarly, if we wish to examine* function at exceptional turning points where the tan
pel it perpendicular to OX, we set tbe reciprocal of the first derivative equal to zero and
■oto to find critical values.
Iln this connection the term "little less," or " trifle less," means any value between the
Mat mailer root (critical value) and the one under consideration; and the term "tittle
ptater," or "trifle greater," means any value between the root under consideration and
aeaen larger one.
[A omlUr discussion will evidently hold for tbe exceptional turning points IS, E, and A
«B*rtiTe!y.
t When the first derivative becomes infinite for a certain value of the independent vari
able, ihen the function should be examined for audi a critical value of tbe variable, for it
Bay jive ■"»■■— or minimum values, as at B, E, or A (Fig. d, p. 10B). See footnote on
112 DIFFERENTIAL CALCULUS
In the problem worked out on p. 104 we showed by means of the
graph of the function
that the rectangle of maximum area inscribed in a circle of radius
5 inches contained 50 square inches. This may now be proved ana
lytically as follows by applying the above rule.
Solution.
First Hep.
Second alep.
= 5V2,
which ia the critical value. Only the positive sign of the radical is taken, since, from
the nature of the problem, the negative sign has no meaning.
Tliird step.
^ w »(SVS*)(bVS + «)
V(iOB>(10 + e)
FourUtatep. Whe»x<sV5,
V(+K+)
Wh«H>5Va,
V{+)(+)
co the sign of the first derivative
changes from + to — at x = 5 v'a, the function
baa a maximum value
/(6Va) = 5v^6V2 = 60. Am.
84. Second method for examining a function for maximum and mini
mum values. From (19), p. 110, it is clear that in the vicinity of a
maximum value of/(:r), in passing along the graph from left to right,
f'(x) change* from + to to — .
Hence /'(V) is a decreasing function, and by § 81
we know that its derivative, i.e. the second deriv
X~ ative [=/"(*)] of the function itself, is negative
or zero.
Similarly, we have, from (20), p. 110, that in the vicinity of a
minimum value of/(x)
f(x) changes from  to to +.
Hence f'(z) is an increasing function and by § 81 it follows that
/"(x) is positive or zero.
MAXIMA AND MINIMA
113
The student should observe thnt/"(V) is positive not only at mini
mum points (as iit X) but also at points such as P. For, as a point
passes through F in moving from loft to right,
tan T = ~=f(x') is an increasinq function.
dx
\.t such a point the curve is said to be concave
wards.
Similarly, f'(x') is negative not only at maximum points (as at B)
ut also at points such as Q. For, as a point passes through Q,
slope ■
dx
=f'(x~) is a decreasing function.
A.t such a point the curve is said to be concave downwards.*
We may then state the sufficient conditions for maximum and mini
mum values otf(x) for certain values of the variable as follows:
(21) f(x) is a maximum if/'<jr> = and/"(x) = a negative number.
(22) f(x) is a minimum iff'(x) = and/"(.r) = a positive number.
Following is the corresponding working rule.
Fibst Step. Find the first derivative of the function.
Second Step. Set the first derivative equal to ir.ro and solve the result
ing equation for real roots in order to find the critical values of the variable.
Tiiird Step. Find the second derivative.
Fourth Step. Substitute each critical value for the variable in the
necond derivative. If the result is negative, then the function is a maximum
for that critical value; if the result is positive, the function is a minimum.
When f"(x) = 0, or does not exist, the above process fails, although
there may even then he a maximum or a minimum ; in that case the
first method given in the last section still holds, being fundamental.
Usually this second method docs apply, and when the process of find
ing the second derivative is not too long or tedious, it is generally the
■hottest method.
Let us now apply the above rule to test analytically the function
M = a.* + ^
X
found in the example worked out ou p. 105.
DIFFERENTIAL CALCULUS
Solution.
/**+=■
First step.
mum.
Second atep.
—9%
x = 6, critical value.
Third step.
m**9
Fourth step.
/"(6) = +. Hence
The work of finding maximum and minimum values may frequently
be simplified by the aid of the following principles, which follow at
once from our discussion of the subject.
(a) The maximum and minimum values of a continuous function must
occur alternately.
(b) Wlten c is a positive constant, c f(x) is a maximum or a minimum
for such values of x, and such only, as makef(x') a maximum or a minimum.
Hence, in determining the critical values of x and testing for max
ima and minima, any constant factor may be omitted.
When c is negative, cf(x~) is a maximum whenf(x) is a minimum,
and conversely.
(c) // . U a cm.tant, /(>) — ( +/(>)
have maximum and minimum values for the same values of x.
Hence a constant term may be omitted when finding critical values
of x and testing.
In general we must first construct, from the conditions given in
the problem, the function whose maximum and minimum values are
required, as was done in the two examples worked out on pp. 103—
106. This is sometimes a problem of considerable difficulty. No rule
applicable in all cases can be given for constructing the function, but
in a large number of problems we may be guided by the following
General directions.
(a) Express the function whose maximum or minimum is involved in
the problem.
(b) If the resulting expression contains more than one variable, the
conditions of the pruhlem will furnish enough relations between the varia
bles so that all may be expressed in terms of a single one.
MAXIMA AND MINIMA
115
(c) To the resulting function of a single variable, apply one of our two
ride* for finding maximum and minimum values.
(d) in practical problems it is usually easy to tell which critical value
will give a maximum and which a minimum value, so it U not always
necessary to apply the fourth step of our rides.
(e) Draw the graph of the function (p. 104) in order to check the work.
1. It Is desired to make an opentop box of greatest possible volume from a square
piece of tin whose Bide is a, by cutting equal squares out of the corners and then fold
ing up the tin to form the aides. What should be the length of a side of the squares
i = side of small square a depth of box ;
x. — side of square forming bottom of box.
2x) = a*Sax + 12 z'
gives critical values x
which Is the function to be made a ma:
Applying rule.
First step. ~ = (a— Sse)« ix(a
Seeond step. Solving a ! — 8« + 12 x 1 =
It is evident from the figure that x =  ithim gi»e w minimum, for then all the tin
would be cut away, leaving no material out of which to make a box. By the usual
a 2n»
test, x =  is found to give a maximum volume Hence ihe side of the square to
be cut out is one sixth of the side of the given square.
The drawing of the graph of the function in this and the following problems is
left to the student.
2. Assuming; that the strength of a beam with rectangular cross section varies
directly as the breadth and as the square of the depth, what are the dimensions of
the strongest beam that can be sawed out of a round log whose diameter is d?
Solution. If a = breadth and y s depth, then the beam will have
sngth when the function xy" is a maximum. From the
— x 3 ; hence we should test the function
f{x) = z{d*x*).
First step. f(x) =  2x* + <P  x 1 = <P 8z 5 .
"Vs"
Second step. <PSi» = 0. ..
Therefore, if the beam is cut so that
depth = V of diameter of log,
and breadth = Vj of diameter of log,
the beam will bare maximum strength.
critical value which gives a maximum.
DIFFERENTIAL CALCULUS
6. Find the altitude of the cylinder
& given right
Hence the function
x::h:h~y. .:
e tented in
Altitude = \h, \^y=3
6. Divide a Into two pans such that their product is a maximum.
7. Divide 10 into two such parts that the si
the other mar be a minimum.
8. Find the number that exceeds its square by the greatest possible quantity.
Am. \.
9. What number added to its reciprocal gives the least possible sum ? Anx, 1.
10. Assuming that the stiffness <>i a In 'am "f i^cliiii^nlur cross .miiion varies directly
as the breadth and the cube of the depth, what must be the breadth of the stiffest beam
that can be cut from a lug 10 inches in diameter ? Anx. Breadth = 8 inches.
11. A water tank Is to be constructed with a square base and open top, and is to
hold 64 cubic yards. If the cost of the Hides is SI a square yard, and of the bottom
S2 a square yard, what are the dimeiisiuiis when the cost is a minimum? What is
the minimum cost f An». Side of base c= 4 yd., height = 4 yd., cost £96.
12. A rectangular tract of land is to be bought for the purpose of laying out a
quartermile track with straightaway sides and semicircular ends. In addition a
strip 36yard« wide along ear.li straightaway is to be bought for grand stands, training
quarters, etc. If the land cost* $200 an acre, what will be the maximum cost of
•Jie land required t Amt, MH
MAXIMA AND MINIMA
117
13. A torpedo bunt is anchored miles from the nearest point of a beach, and it is
desired to send a messenger in the shortest possible time lua military camp si dialed
15 miles from that point along the shore. If he can walk u miles an hour but row only
4 miles an hour, required the place he must land. Ana. 3 miles from the camp.
14. A gas holder is a rylindrieal vessel i/losid at lilt top and open at the bottom,
where it sinks into the water. What should be its proportions for a given volume to
require the least material (this would also give least weight)?
Ann. Diameter = double the height.
16. What should be the dimensions and weight of a gas holder of 8,000,000 cubic
l..t capacity, built in the most economical manner out of sheet iron ^ of an inch
thick and weighing 2J lb. per sq. ft. ?
Ana. Height = 137 ft., diameter = 273 ft., weight = 220 tons.
16. A sheet of paper is to contain 18 m). in. of printed matter. The margins at the top
and bottom are to l«. 2 i nlus each and at the sides 1 inch each. Determine the dimen
sions of the sheet which will require the least amount of paper, jins. 6 in. by 10 in.
17. A paper box manufacturer has in stoek a quantity of si raw hoard 30 inches by
14 inches, ( mt. i jf thi.j niiitii'ial In wi>1nsi.i>iii:iki' ii]. i'ni.ii ln.ixi's by flitting equal squares
out of each corner and then folding up to form the sides. Find the side of the square
that should be cut out in order to give the boxes maximum volume. A in. 3 inches.
18. A roofer wishes to make an open gutter of maximum » i
capacity whose bottom and sides are each 4 inches wide and \ , /
whose sides have the same slope. What should be the width \ /
across the top ? Ann. 8 Inches. \ i /
19. Assuming thai tl nervy expended in driving a steamboat through the water
YiirieeaB the cube of her velocity, find her most economical rate per hour when steam
ing against a current running c miles per hour.
Hist. Let c must eeiiuoniical speed;
op*= energy cxpendeil each lumr, n being a constant depending njion the partic
ular conditions,
i  c = act mil distance advanced per hour.
— Is the energy expended pur mil. i.f ili,tauce advanced, and i! is therefore the
whose minimum Is wanted. '
 :■■■
20. Prove that a conical lent of a given capacity will require the l< a.t amount of
when the height is v'2 times the radius of the base. Show that when the canvas
is laid out flat it will be a circle with a sector of 152° W cut out. A belt tent 10 ft.
high should then have a base of diameter 14 ft. and would require 272 sq. ft, of canvas.
21. A cylindrical steam boiler is to be constructed having a capacity of 1000 cu. ft.
The material for the side costs S2 a square foot, and for the ends S3 a square foot.
Find ratlins when the cost is the least. 10 ,
Ant. — — ft.
22. In the corner of a field bounded by two perpendicular roads a spring is situated
6 rods from one road and 8 rods from the other. How should a straight road be run
by this spring and across the corner so as to cut off as little of the field ag possible ?
Ana. 12 and It) rods from corner.
What croold bo the length of the shortest road that, could be run across?
Ans. (a' + 8S)!roda
118
DII'KKKENTIAl, CALCULUS
23. Show that a square is the rectangle of maximum perimeter that can be inscribed
in a given circle.
24. Two poles of height a And ft feet are stamliiii; upright and are e feet apart. Find
the point on the lim joinim; their buses such I hat the sum of the squares of the distances
from this point to the tops of the poles is a minimum. Ans. Midway between the poles.
When will the sum of Ihese distances he a minimum 7
25. A conical tank with open top is to be built to contain V cubic feet. Determine
the shape if the material used is a minimum.
26. An isosceles triangle has a base 12 in. long and altitude 10 in. Find the rec
tangle of maximum area that can be inscribed in it, one side of the rectangle coincid
ing with the base of the triangle.
27. Divide the number 4 into two such parts that the sum of the cube of one part
and three times the square of the ether shall have a maximum value.
28. Divide the number a into two parts such that the product of one part by the
fourth power of the other pan shall be a maximum.
29. A can buoy in the form of a double cone is to be made from two equal circular
iron plates of radius r. Find the radius of the base of the cone when the buoy has the
greatest displacement (maximum volume). Ana. rV\.
30. Into a full conical wineglass of depth a and generating angle a there is care
fully dropped a sphere of such size as to cause the greatest overflow. Show that the
radius of the sphere is a sin a
31. Awall27ft.highis8ft.fromahouse. Find the length of theshorteatiadderthat
will reach the house if one end rests on the ground outside of the wall. Jm. 13 Vis.
32. A vessel is anchored 3 miles offshore, and opposite a point 5 miles further
along the shore another vessel is anchored II miles fmni the shore. A l«iat from the
first vessel is to land a passenger on the shore and then proceed to the other vessel.
What is the shortest course of the boat? Am. 13 miles.
33. A steel girder 25 ft. long is moved on rollers along a passageway 12.8 ft. wide
and into n corridor at right angles to the passageway. Neglecting the width of the
girder, how wide must the corridor be ? Ans. 5.4 ft.
34. A miner wishes to dig a tunnel from a point A ton point B 300 feet below
and GOO feet to the east of A. Below the level of /I it is bed rock and above A is soft
earth. If the cost of tunneling through earth is 31 mid through rock 83 per linear foot,
find the minimum cost of a tunnel. Arts. 51348.53.
35. A carpenter has 108 si. ft. of lumber with which to build a box with a square
base and open top. Find I he dimensions of the largest possible box he can make.
36. Find the right triangle of
length ft as hypotenuse.
that can he constructed
length of both legs.
of
V2
37. What is the isosceles triangle of maximum area that can he inscribed in a
given circle 7 Ana. An equilateral triangle.
38. Find the altitude of the maximum rectangle that can be inscribed in a right
triangle with base b and altitude A. , ft
° Ana. Attitude = ■
MAXIMA AND MINIMA
119
39. Find the dimensions of the rectangle of maximum area that can be inscribed
nthe ellipse Px* + oV = o*6*. Ana. aV2and&V2; area = 2afi.
40. Find the altitude of the right cylinder of maximum volume that can be inscribed
a sphere of radius r. , ,, , , „ , 2r
Aia. Altitude of cylinder = —  •
Va
41. Find the altitude of the right cylinder of maximum convex (curved) surface
it can be inscribed in a given sphere. Ans. Altitude of cylinder = r V2.
42. What are the dimensions uf the right hexagonal prism of minimum surface
'hose volume Is 80 cubic feet ? A?it. Altitude = 2 Vs ; side of hexagon = 2.
43. Find the altitude of the right cone of minimum volume circumscribed about a
en sphere. Ans. Altitude = 4r, and volume — 2 x vol. of sphere.
44. A right cone of maximum volume is inscribed in a given right cone, the vertex
of the inside cone being at the center of the base of the given cone. Show that the
altitude of the inside cone is one third the altitude of the given cone.
45. Given a point on the axis of the parabola j/ 1 = 2px at a distance a from the
; find the abscissa of the point of the curie nearest to it. Ans. x — a — p.
46. What is the length of the shortest line that can be drawn tangent to the ellipse
+ a 1 )/ 8 = a?tF and meeting the coordinate axes ? Ans. a + 6.
47. A Norman window consists of a rectangle surmounted by a semicircle. Given
the perimeter, required the height and breadth of the window when the quantity of
light admitted is a maximum. Ans. Radius of circle = height of rectangle.
48. A tapestry 7 feet in height is hung on a wall so that its lower edge is feet
an observer's eye. At what distance from the wall should he stand in order to
favorable view ? Ans. 12 feet.
Hint. The vertical angle subtended by the tapestry in the eye of the observer must bs
;:
light a
above
obtain
i which shall have a
49. What are the most economical proportions of a tin c
given capacity, making allowance for waste ?
Ans. Height = — — x diameter of base.
Hint. There Is no waste in cutting out tin tor the side of (he can,
but lor lop and bottom a hexagon of tin circumscribing the circular
liews required is used up.
Note 1. If no allowance is made for waste, (lien lnidil — diameter.
Note 2. We know that the shape of a bee cell is hexagonal, giving a certain
cipacity for honey with the greatest possible economy of wax.
50. An open cylindrical trough is constructed by bending a given sheet of tin of
breadth 2 a. Find the radius of the cylinder of which the trough forms a part when
&i capacity of the trough is a maximum.
Ans. Rad. = — ; i.e. It must be bent in the form of a semicircle.
51. A weight Wis to be raised by means of a lever with the force fat one end and
th[ point of support at the other. If tiio weight is suspended from a point at a distance
_ a from the point of support, and the weight of the beam is u; pounds
' per linear foot, what should lie the length of the lever in order that
the force required to lift it shall be a minimum ? it — —
Ana. x — \ feet.
120
DIFFERENTIAL CALCULUS
J
52. An electric arc light is to be plaoed directly over tlie center of a circular plot
of grass 100 feet in diameter. Assuming that the intensity uf light varies directly !
the sine of the angle under which it strikes an illuminated surface, and inversely
as the square of its distance from the surface, lew high should the light be hung
In order that the best possible tight shall fall on a walk along the
circumference at the plot? . 50 ,
Vb
53. The lower corner of a leaf, whose width is a, is folded over bo
as just to reach the inner edge uf the page, (a) Find the width of the
part folded over when tin length of the crease is a minimum, (b) Find
the width when the urea folded over is a minimum. Ana. (a) la; (b) Ja.
54. A rectangular stockade is to be built which must have a certain area. If a
stone wall already constructed in available for one of the sides, find the dimensions
which would make the cost of construction the least.
Atus. Side parallel to wall = twice the 1cilltt.1i of each end,
55. A cow is tethered by a perfectly smooth rope, a
slip ii'. uise in tin' rope be ins thrown over a large square
post. If the cow pulls the rope taut in the direction
shown in the figure, at what angle will the rope leave
the post ? Ann. 30°.
56. When the resistance of air i* taken into account, the inclination of a pendulum
to the vertical may be given by the formula
S = «<r«cos(n( + <).
Show that the greatest elongations occur at equal intervals — of time.
57. It is required to measure a certain unknown magnitude x with precision.
Suppose that n equally careful observations of the magnitude are made, giving the
resulla «i, « a , «., a*
The errors of these observations are evidently
»«,, zOj, zQ„ ■■', i0>,
some of which arc positive and smite negative.
It has been agreed that the most probable value of x is such that it renders the
sum of the squares of the errors, namely
(x  a,)* + (i.  off + (x  a,)' + + {*  a,) 5 ,
a minimum. Show that this gives the arithmetical mean of the observations as the
most probable value of z.
68. The bend i itc moniciil at II of a beam of length I, uniformly B
loaded, is given by the formula
where w = load per unit length. Show that (tic maximum bending moment is :
center of the beam.
•fc=r
.the
59. If the total waste per mile h
in electric conductor is
W = cV f  , [c = constant]
■where e = current in amperes, r = resistance in ohms per mile, and ( = a constant
depending on the interest on the investment and the depreciation of the plant, what
is the relation between c, r, and ( when the waste is a minimum ? Ana. cr = (.
MAXIMA AND MINIMA 121
60. A submarine telegraph cable consists of a core of copper wires with a covering
made of nonconducting material. If x denote the ratio of the radius of the core to the
thickness of the covering, it is known that the speed of signaling varies as
xMogi.
x
1
Show that the greatest speed is attained when x = ~~p m
Ve
61. Assuming that the power given out by a voltaic cell is given by the formula
P= EiR
(r + R) 2
where E = constant electromotive force, r = constant internal resistance, R = exter
nal resistance, prove that P is a maximum when r = R.
62. The force exerted by a circular electric current of radius a on a small magnet
whose axis coincides with the axis of the circle varies as
(a 2 + x 2 )*
where x = distance of magnet from plane of circle. Prove that the force is a maxi
mum when x =  .
2
63. We have two sources of heat at A and B with intensities a and b respectively.
The total intensity of heat at a distance of x from A is given by the formula
I = ^ + _»
x 2 (d  x) 2
v ' A P B
Show that the temperature at P will be the lowest when i;.«i J. a *
dx _ Vb
that is, the distances BP and AP have the same ratio as the cube roots of the corre
sponding heat intensities. The distance of P from A is
x =
a* + 6*
64. The range OX of a projectile in a vacuum is given by the formula
v* sin 2 4> .
R =
9
where v x = initial velocity, g = acceleration due to grav
* ity, <p = angle of projection with the horizontal. Find the
angle of projection which gives the greatest range for a given initial velocity.
Ans. <t> = 46°.
66. The total time of flight of the projectile in the last problem is given by the
formula
;r _ 2t? 1 sin^
~ 9
At what angle should it be projected in order to make the time of flight a maximum ?
Ans. = 90°.
DIFFERENTIA
CTJLTT8
i inclined plane AB is give
Neglecting friction,
quickest descent ?
, what must be the value of # t
l) 1 (x + l) 1 for maximi
67. Examine the function f.
Use the first method, p. Ill,
Solution. fix) = (z  1)' (x + 1)'.
First step, f(x) = 2(x  l)(z + 1)* + S(:
Secondsiep. (z l)(z + l)»(6z 1) = 0,
Z = 1, — 1, \, which are critical values.
Third step. /'(z) = 5(3  1) (* + l)>(z _ j).
Fourth step. Examine flint fur critical value x — I (C in
flgure).
make the
i. # = 46°.
nt and minimi
!)■ (* + !)■ =><*l)C» + l)»(B*
1.11 (= ordinate
When z < l,/'(z) = 6() ( + )•( + ) =.
When b>1,j"(x) = «(+)(+)»<+)=+•
Therefore, wlienx=l the function haa a minimum value/(l) = 0,(= ordinate of C).
Examine now for the critical value x =  (B in flgure).
Wnen i c<i,/'(r) = 5()(+)»()=+.
Whe n *>J,/( I ) = 6()( + )'{ + )= —
Therefore, when z = J the function has a maximum value /( j) =
IB).
Examine lastly for the critical value x = — 1 ( A In figure).
Wben*<l,/'(s) = 5()()'<)=+.
When x >  1, /'(i) = 5 (_)(+)»()= + .
Therefore, when x — — 1 the function has neither a maximum nor i
68. Examine the function a — b(x — c)* for n
Solution. f(x) = a  6(as — e)*.
m s»
but for
Since x — e is a critical value for which f'{x) =
which f(x) is not infinite, let us test the function for
and minimum values when 1st
When*<c,/'(i)= +.
When x>c, /'(*)= .
Hence, when z = e = O.tf the function has a maximum value /(c) = a = JUT
Examine the following functions for maximum and minimum values:
69. (z — 3)* (z — 2). Ana. x = J, gives max. = J
z — 3, gives min. = 0.
7ft (zl)'(xS)". m = \, gives max. = .1
1 = 2, gives min. =
x = 1, gives neither.
MAXIMA AND MINIMA 123
71. (x — 4)*(x + 2) 4 . An*, z = — 2, gives max.;
x = ], gives min.;
x = 4, gives neither.
72. (x2)*(2x + l)«. *=J» gives max.;
x = J, gives min.;
x = 2, gives neither.
73. (x + l)*(x — 6) f . x = J, gives max.;
x = — 1 and 5, give min.
74. (2x — a)*(x — a)*. x = — » gives max.;
3
z = o, gives min.;
x =  » gives neither.
75. x (x — l) f (x + 1) 8 . x = J, gives max. ;
x = 1 and — J, give min.;
x = — 1, gives neither.
Q
76. x (a + x) f (a — x) 8 . x = — a and > give max.;
3
x = 1 gives min.;
x = a, gives neither.
77. 6 + c(x — a)*. x = a, gives min. = 6.
78. a — b(x — c)*. No max. or min.
x* — 7 x + 6 , .
79. _J_ — x = 4, gives max.;
x = 16, gives min.
a
4
81. a ~~t~ x = J, gives min.
80. — — — . x = » gives min.
X
10
<«
X) 8
a —
2x
1
x + x 8
1 +
X — X 8
x 8 
8x + 2
* x« + 8x + 2
x = V2, gives min. = 12 V5 — 17 ;
x = — V5, gives max. = — 12 V2 — 17;
x = — 1, — 2, give neither.
_ (x — a)(5 — x) 2a6 , (a6) 8
83. ^ — ^ ^. x = — — » gives max. = v ; •
X s a + 6 4a&
~. a 8 P a 2 , .
84. — + x = t gives min.;
x a— x a— o
x = » gives max.
a + b
124
DIFFERENTIAL CALCULUS
85. Examine x 8 — 8x 2 — 9x + 5 for maxima and minima. Use the second method,
p. 113.
Solution. f(x) = x*  3x 2  9x + 5.
First step. f'(z) = 3 x 2  6 z  0.
Second step. 3x 2  6x  9 = 0;
hence the critical values are x = — 1 and 3.
Third step. /"(x) = 6 x  6.
Fourth step. /"(1)=12.
.*. /(— 1) = 10 = (ordinate of A) = maximum value.
/"(3) = + 12. .. /(3) = — 22 (ordinate of B) = minimum value.
86. Examine sin 2 x cosx for maximum and minimum values.
Solution. f(x) = sin 2 x cosx.
First step, /'(x) = 2 sin x cos 2 x — sin 8 x. Y
Second step. 2 sin x cos 2 x — sin 8 x = ;
hence the critical values are x = nir
and z = nir ± arc tan V2 = nir ± a.
Third step. f"{x) = cosx (2 cos 2 x — 7 sin 2 x).
Fourth step. /"(0) = + • •'• /(°) = = minimum value at O.
/"(it) = — . .*. /(ir) = = maximum value at C
/"(a) = — . .. f(a) = maximum value at A.
f"{ir — a) = + . •*• /(«* — <*) = minimum value at B, etc.
Examine the following functions for maximum and minimum values.
87. 3x*9x 2 27x + 30.
88. 2x*  21 x 2 + 36x  20.
89. ^2x 2 + 3x + l.
90. 2x 8 15x 2 + 36x + 10.
91. x 8 9x 2 + 15x3.
92. x 8 — 3x 2 + 6x + 10.
93. x 5 — 5x 4 + 5x 8 + l.
94. 3x 6 125x 8 + 21fl0x.
95. 2x 8 3x 2 12x + 4.
96. x(x3) 8 .
97. x*2x 2 + 10.
Ans. x = — 1, gives max. = 45 ;
x = 3, gives min. = — 51.
x = 1, gives max. = — 8 ;
x = 6, gives min. = — 128.
x = 1, gives max. = J ;
x = 3, gives min. = 1.
x = 2, gives max. = 88 ;
x = 3, gives min. = 87.
x = 1, gives max. = 4 ;
x = 6, gives min. = — 28.
No max. or min.
x = 1, gives max. = 2 •,
x = 3, gives min. = — 26 ;
x = 0, gives neither.
x = — 4 and 3, give max.;
x = — 3 and 4, give min.
98. x*  4.
99. x»  8.
100. 4  x«.
MAXIMA AND MINIMA 125
o
101. sinx(l + cobx). An*. x = 2nx + , give max. =  V5 ;
3 4
x = 2nir— , give min. =— V8;
8 4
x = twt, give neither.
102.  . x = e, gives min. = c ;
log*
111.
XCOSX.
112.
sinx + cos2x.
113.
2 tanx — tan*x.
114.
sinx
1 + tanx
115.
X
1 + xtanx
x = 1, gives neither.
103. log cos x. x — n7T, gives max.
104. ae** + 6c 1 *. x =  log \/, gives min. = 2 Va6.
105. x*. x =  » gives min.
i
106. x*. x = e, gives max.
107. cosx + sinx. x = » gives max. =V2;
x = — , gives min. = — V5.
108. sin2x — x. x = , gives max.;
6
x = , gives min.
109. x + tan x. No max. or min.
_ A
110. sin'x cosx. x = nv +  , gives max. = — Vs ;
8 16
x = imt , gives min. = Vs :
8 16
x = wr, gives neither,
x = cotx, gives max.
x = arc sin J, gives max.;
z = ~ , gives min.
IT
x =  , gives max.
4
IT
x = — , gives max.
4
x = cosx, gives max,;
x =— cosx, gives min.
85. Points of inflection. Definition. Points of inflection separate arcs
concave upwards from arcs concave downwards.* Thus, if a curve
y =/(x) changes (as at 2?) from concave upwards (as at A) to con
cave downwards (as at C), or the reverse, then such a point as B is
called a point of inflection.
* Points of inflection may also be defined as points where
d*?/ (Py
(a) — = and — ^ changes sign,
dx* dx 2
(Px cPx
or (b) —   and — changes sign.
<ly dy*
126 DIFFERENTIAL CALCULUS
From the discussion of § 84 it follows at once that at A, f"(x) = +,
and at C t f"(x) = ~. In order to change sign it must pass through
the value zero ; • hence we have
(23)
at points of inflection, /"(*) = 0.
Solving the equation resulting from (23) gives the abscissas of the
points of inflection To determine the direction of curving or direc
tion of bending in the vicinity of a point of in
flection, test/"(x) for values of x, first a trifle
lesB and then a trifle greater than the abscissa
at that point.
If/"(*) changes sign, we have a point of in
flection, and the signs obtained determine if the curve is concave
upwards or concave downwards in the neighborhood of each point
of inflection.
The student should observe that near a point where the curve is
concave upwards (as at A') the curve lies above the tangent, and at
a point where the curve is concave downwards (as at C) the curve
lies below the tangent. At a point of inflection (as at B~) the tangent
evidently crosses the curve.
Following is a rule for finding points of inflection of the curve whose
equation is v =f(x). This rule includes also directions for examining
the direction of curvature of the curve in the neighborhood of each
point of inflection.
First Step. Findf"(x)
Second Step. Setf"(x) = 0, and tolve the resulting equation for real
root*.
Third Step. Write f"(x) in factor form.
Fourth Step. Teat f'(x') for values of z,fir»t a trifle lest and then a
trifle greater than each root found in the second atep. Iff"(x) changes
sign, we have a point of inflection.
Whenf"(x') = +, the curve is concave upward* y^. t
Whenf"(x) = ~ , tke curve is concave doienwarda *"C^«.
* It is assumed that/'(z) and/"(i) are continuous. The solution of Ex. 2, p. 1ST, shows
how to discuss a case where /'(*) and f"(x\ are both infinite. Evidently salient point* (see
p. 268) are excluded, since at such points /'(i] Is discontinuous.
t This may be easily remembered If we say that a vessel shaped like the curve where
It is concave upwards will hold (+) water, and where it Is concave downwards will spill
H water.
MAXIMA AND MINIMA
Examine tbe following curve* for points of Inflection and direction of bending.
1. t/ = Ss*4* , + l.
Solution.
Firal aUp.
Second step.
Third ttep.
Fourth stop.
/(x) = 8x*4a i + l.
/"{x)=SBx* 24 a.
36 x 1  24/ 
d and a = 0, critical values.
/"(*) = M*<*  «■
Whenx<0,/"(.r.) = 4 ; and when x >0, f'(x)=— ,
.. carve is concave upwards to the left and concave downwards to tbe right of x =
(A inflgure). When x< (,/"(*) = ; and when >§,/"(*) = + .
■. curve is concave downwards to the left and concave upwards to the right of
x = J (Bin figure).
The curve Is evidently concave upwards everywhere to the left of A, concave down
wards between .4 (0, 1) and B(},jl), and concave upwards everywhere to the right of B.
2. („2>' = (z4).
Solution. y = 2 + (x — 4)1.
Firal sltjj. ? =  (as  4)"',
Second step. When x = 4, both first and second derivatives a
Tftird stop. When as
^<te"
;+ ; but when x > 4,
.'/" ;
We may therefore conclude that the tangent at (4, 2) is perpendicular to the axis
of X, that to the lefv of (4, 2) the curve is concave upwards, and to the right of (4, 2)
it is concave downwards. Therefore (4, 2) must be considered a point of inflection.
3. y = *'. Ana. Concave upwards everywhere.
Concave downwards everywhere.
Concave downwards to the left and concave up
wards to the right of (0, 0).
Concave downwards to the left and concave up
wards to the right of (1, — 2).
Concave downwards to the left and concave up
wards to the right of (6, a).
Concave downwards to the left and concave up
wards to the right of (a, —J.
Concave upwards everywhere.
Concave upwards to tbe left of * = 2, concave
downwards between x = 2 and x = 4, concave
upwards to the right of x = 4.
Points of inflection are x = iw, n being any Integer.
4
V
= 52*
«•,
5
V
sft
6
v=
zx*8x*
9a
7.
V
= ■> + <*
6)'.
8. a'? = —  ax* + 2 o 1
DIFFERENTIAL CALCULUS
13. ystani. Aiu. Points of inflection are s = nr, n being any integer.
13. Show that no conic section can have a point of inflection.
14. Show that the graphs of t J and log s have no points of inflection.
86. Curve tracing. The elementary method of tracing (or plotting)
a curve whose equation is given in rectangular coordinates, and one
with which the student is already familiar, is to solve its equation for
y (or x), assume arbitrary values of x (or y), calculate the correspond
ing values of y (or x), plot the respective points, and draw a smooth
curve through them, the result being an approximation to the required
curve. This process is laborious at best, and in case the equation of
the curve is of a degree higher than the second, the solved form of
such an equation may be unsuitable for the purpose of computation,
or else it may fail altogether, since it is not always possible to solve
the equation for y or x.
The general form of a curve is usually all that is desired, and the
Calculus furnishes us with powerful tuethods for determining the
shape of a curve with very little computation.
The first derivative give3 us the slope of the curve at any point ;
the second derivative determines the intervals within which the curve
is concave upward or concave downward, anil the points of inflection
separate these intervals; the maximum points are the high points and
the minimum points are the low points on the curve. As a guide in
his work the student may follow the
Rule for tracing curves. Rectangular coordinates.
First Step. Find the first den'mtice ; place it erpml to zero; golfing
gives the abscissas if waxiitntm end minimum points.
Second Step. Find the second derivative ; place it equal to zero ; solv
ing gives the abscissas of the points of inflection.
Third Stkp. Calculate the corresponding ordinate* of the points whose
abscissas were found in the first two steps. Calculate as many more points
as may be necessary to give a good idea of the shape of the curve. Fill out
a table such as is ikotPH in the example worked out.
Fourth Step. Plot the points determined and sketch in the curve to
correspond with the results shown in the table.
If the calculated values of the ordinates are large, it is best to
reduce the scale on the Faxis so that the general behavior of the
curve will be shown within the limits of the paper used. Coordinate
plotting paper should be employed.
MAXIMA AND MINIMA
Trace the following curves, making use of the above rule. Also find the equator
of the tangent and normal at each point of inflection.
1. y = j*9x* + 24x7.
Solution. Use the above rule.
First ttep.
3z a 
1^ = 3*' 18*
8z + 24 = 0,
z = 2,*.
y"=6x18,
6a 18 = 0,
*
v
r
V"
R E »ARH«
D.RKCT.Ol.OrCORVR
2
3
4
6
7
18
11
9
29
+
+
+
+
pt. of infl.
S concave down
L concave up
Fourth ttep. Plotting the points and sketching in the curve, we get the figure shown.
To find the equations of the tangent and normal to the curve at the point of inflec
tion F,(3, 11), use formulas (I), (2), pp. 76, 77. This gives 3x + y  20 for the tangent
and S y — x = 80 for the normal.
2. y = a" — 6a: 1 — 38* + 5.
Ana. Max. ( 2, 45); rain. (0,  211); pt. of lnfl.
(2,88)1 tan. V + 48* 13 = 0; nor.
48? 1 + 3986 = 0.
3. y = x*2x* + 10.
Am. Man. (0, 10) ; min. (± 1, 9) ; pt. of lnfl.
4. y = {x*Sx* + 2.
Ant. Max. (0, 2); min. (± V3, — j); pt. of infl.
(± 1.  t>
Ani. Mai. (1,8); min. (1,3); pt. of infl. (0,0),
6. y = 12zx».
7. 4y + *"_Sx>+4 = 0.
Ant. Max. (2, 16) ; min. ( 2,  10) ; pt. of infl.
(0, 0).
180
DIFFERENTIAL CALCULUS
8. y = x? — Sz* — 9z + 9.
9. 2y + X 8 — 9x + 6 = 0.
10. y = x*  6x*  15x + 2.
11. y(l + x 8 ) = x.
12. y = •
y x* + 4a«
13. y = e*\
, 4 + x
14. y = — —
x*
16. y = (x + l)*(x  5)«.
16. y = ^.
x 8
17. y = x»3x 8 24x.
18. y = 18 + 36x  3x*  2x».
19. y = x— 2cosx.
20. y = 8x — x 8 .
21. y = x 8  9x* + 15x — a
22. x*y = 4 + x.
23. 4y = x* — 6x f + 5.
25. y = sinxH —
2
26. „ = *+!
X
27. y = 5x2x 2 Jx i .
28.„ = !±^\
2x
29. y = x — 2 sin x.
30. y = logcosx.
31. y = log(l + x«).
CHAPTER IX
DIFFERENTIALS
87. Introduction. Thus far we have represented the derivative of
y =/(x) by the notation jy
We have taken special pains to impress on the student that the
symbol <j y
dx
was to be considered not as an ordinary fraction with dy as numerator
and dx as denominator, but as a single symbol denoting the limit of
the quotient Av
Ax
as Ax approaches the limit zero.
Problems do occur, however, where it is very convenient to be able
to give a meaning to dx and dy separately, and it is especially useful
in applications of the Integral Calculus. How this may be done is
explained in what follows.
88. Definitions. If f r (x) w the derivative of f(x) for a particular
value of z, and Ax is an arbitrarily chosen increment of x, then the differ
ential off(x) y denoted by the symbol df(x^ is defined by the equation
00 df(x)=f(x)Ax.
If now f(x) = x, then f'(x) = 1> *&& (^) reduces to
dx = Ax,
showing that when x is the independent variable, the differential of
x(=z dx) is identical with Ax. Hence, if y =/0&), (4) may in general
be written in the form
(B) dy=f'(x)dx*
• On account of the position which the derivative f(x) here occupies, it is sometimes
called the differential coefficient.
The student should observe the important fact that, since dx may be given any arbi
trary value whatever, dx is independent of x. Hence, dy is a /Unction of two independent
variables x and dx.
131
132
DIFFERENTIAL CALCULUS
The differential of a fund ion equals its derivative multiplied by tfo
differential of the independent variable.
Let us illustrate what this means geometri
cally.
Let/'(V) be the derivative of y =f(x) at P.
Take dx  PQ, then
QT
/'(*) *" = tan T p Q =
PQ
PQ = QT.
Therefore dy, or df(x'), is the increment (= QT~) of the ordinate of
the tangent corresponding to dx*
This gives the following interpretation of the derivative as a fraction.
If an arbitrarily chosen increment of lite independent variable x for
a point P(x, y) on the curve y=f(x) be denoted by dr, then in the
derivative
 «/(«)= fa. r,
dy denotes the corresponding increment of the ordinate drawn to the
89. Infinitesimals. In the Differential Calculus we are usually con
cerned with the derivative, that is, with the ratio of the differentials
dy and dx. In some applications it is also useful to consider iir as
an infinitesimal (see § 15, p. 13), that is, as a variable whose values
remain numerically small, and which, at some Btage of the investiga
tion, approaches the limit zero. Then by (B), p. 181, and (2), p. 19,
dy is also an infinitesimal.
In problems where several infinitesimals enter we often make use
of the following
Theorem. In problems involving the limit of the ratio of two infinites
imals, either infinitesimal may he. replaced by tin infinitesimal so related
to it that the limit of their ratio is unity.
Proof. Let a, {3, a', ft' be infinitesimals so related that
(<0
• The student should note especially that Hie differential (= ilii) and the increment (. Ajf)
<d the function MUMWWUng tu tbe sBuia value of ilx ( ii) are not iu geoerel equal. For,
la the figure, dyQT, nut dy=QF'.
DIFFERENTIALS
138
I
We have
and
a a! a B 1 . , . ,,
£ = ^^ ldentlcally '
limit zz = limit r. • limit ; • limit r
a'
a'
&
= limit — • 1 • 1.
p
CD)
limit — = limit ^
P P
Th. II, p. 18
By(0
Q. S. D.
Now let us apply this theorem to the two following important limits.
For the independent variable a, we know from the previous section
that Ax and dx are identical.
A.X
Hence their ratio is unity, and also limit — = 1. That is, by the
above theorem,
dx
(JET) In the limit of the ratio of Ax and a second infinitesimal. Ax
may be replaced by dx.
On the contrary it was shown that, for the dependent variable y, Ay
and dy are in general unequal. But we shall now show, however, that
in this case also \y
limit =2 = 1.
dy
Since . _ n —■ =f'(x), we way write
Aa?=0 Ax
£./«+«,
where € is an infinitesimal which approaches zero when Ax = 0.
Clearing of fractions, remembering that Ax = dxy
Ay =/'(:r) dx + e • Ax,
or Ay = dy + € Ax.
Dividing both sides by Ay,
OT by by
(B), p. 131
Ay
by
. limit dy _..
' ' Ax = Ay '
and hence A _ ft p = 1. That is, by the above theorem,
(F) In the limit of the ratio of Ay and a second infinitesimal. Ay may
be replaced by dy.
184 DIFFERENTIAL CALCULUS
L Derivative of the arc in rectangular coordinates. Let • be the
length of the arc AP measured from a fixed point A on the carve.
. qs Denote the increment of $ (= arc PQ) by A*.
vy\±i The definition of the length of arc depends on
aJ] the assumption that, as Q approaches P,
J ^/clKmlPgX
— ^ \ ucPQ )
If we now apply the theorem on p. 132 to this, we get
(6r) In the limit of the ratio of chord PQ and a $econd infautenmal,
chord PQ may be replaced by arc P@(= A*).
From the above figure
(H) (chord PQ)*= (Ar) f + (Ay) f .
Dividing through by (Ax) 2 , we get
<o e^)=i+(£)
Now let Q approach P as a limiting position ; then Ax i and we
have
(£)' i+ (£)"
Similarly, if we divide (IT) by (Ay)* and pass to the limit, we get
(25)
! = >((1F
Also, from the above figure,
a Ax . a Ay
cos0 = : — r^:» sin0=s— — ,* •
chord P© chord PQ
Now as Q approaches P as a limiting position = t, and we get
dx A/
(26) cosr = — , sinr =— .
ds ds
f Since from (O) limit — ^ limit— — , and limit — ^ » limit ^^.1
• Defined in f 209.
DIFFEBENTIALS
185
' Using the notation of differentials, formulas (24) and (25) may be
written
[•♦GOT
(27)
dx.
(28)
*[(lM*
Substituting the value of ds from (27) in (26),
(29)
cosr =
<ftfVl*
sinr =
nm
<frY"l*
h(i)l
An easy way to remember the relations (24)(26) between the
differentials dx, dy y ds is to note that they are
correctly represented by a right triangle whose
hypotenuse is ds, whose sides are dx and dy,
and whose angle at the base is r. Then
Y
/
X
V
rt?«
'0
X
<fo = V(dr)*+(dy)«,
and, dividing by dx or dy, gives (24) or (25) respectively. Also, from
the figure,
dx . dy.
cosr = — t sinr = — J
ds ds
the same relations given by (26).
91. Derivative of the arc in polar coordinates. In the derivation
which follows we shall employ the same figure and the same notation
used on pp. 83, 84.
From the right triangle PRQ
(chord PQY
= (PBy+(RQy
= (/> sin A0 y + (/> + A/>  p cos A0 ) 2 .
Dividing throughout by (A0 )*, we get
(
chord PQ\*
A0
hmH% + >
1  cos A0V
A0
)'
136 DIFFERENTIAL CALCULUS
Passing to the limit as A0 diminishes towards zero, we get *
(sy'K*)
In the notation of differentials this becomes
( 3i) t^+ffif*.
These relations between p and the differentials ds, dp, and dO
% are correctly represented by a right triangle
jCj^Li, whose hypotenuse is d% and whose sides are
dp and pd0. Then
ds = ^{[pd0y+(dpy t
x and dividing by d0 gives (30).
Denoting by ty the angle between dp and d$, we get at once
dO
tan* = ,>,
which is the same as (4), p. 84.
Illustrative Example 1. Find the differential of the arc of the circle x* + y* = r 8 .
Solution. Differentiating, — =
dx y
To find ds in terms of x we substitute in (27), giving
.[. + 5r*F^r*i5r
Vr 8 **
To find cb in terms of y we substitute in (28), giving
—[•♦g*— F^'T— Er
Vr 8 ^
Illustrative Example 2. Find the differential of the arc of the cardioid p =
a (1 — cos 0) in terms of 0.
Solution. Differentiating, £ = a sin 0.
Substituting in (31), gives
d«=[a a (lco8^) 8 +a«8in^]*(W = a[2~2cos^]*(W = ar48in 2 l d0 = 2asincW.
SSS^l By J 22. p. 21
DIFFERENTIALS 187
EXAMPLES
Find the differential of arc in each of the following curves :
1. ]/* = 4x.
4 718
* d * = \ x dx *
2. y = ax 2 .
d«=Vl + 4a 2 x 2 dx.
3. y = x*.
da =Vl + 9x* ax.
4. y* = 35 s .
d»= J V4 + 9ydy.
6. x* + y* = a'.
d * = \^ dy 
6. 6*x* + aV = a 2 6«.
a 2
d* = x^P —  ax.
\ a 2  x 2
7. e*co8X = 1.
d« = secxdx.
8. p = a cos 0.
d* = ad0.
9. p 2 = a 2 cos20.
d* = aVsec20d0.
10. p = ae*«**.
d« = pcscadO.
11. p = a*.
d« = a*Vl + log 2 ad0
12. paB.
d* = Va 2 + p 2 d>.
13. (a) x 2  y 2 = a 2 .
(h)
x* + y* = a*.
(b) x 2 = 4 ay.
(0
y 2 = ax 8 .
(c) y = C + c~ x .
a)
y = logx.
(d) xy = a.
«
4x = y*.
(e) y = log sec x.
(f) p = 2atan0sin0.
(1)
p = a sec 2 .
r 2
(g) p = asec«g.
(m) p = l + sin0.
(n) p$ = a.
92. Formulas for finding the differentials of functions. Since the
differential of a function is its derivative multiplied by the differen
tial of the independent variable, it follows at once that the formulas
for finding differentials are the same as those for finding derivatives
given in § 33, pp. 3436, if we multiply each one by dx.
This gives us
I d(<?) = 0.
II d(x) = dx.
III d(u + v — w) = du + dv — dw.
IV d(cv) = cdv.
V d(uv) = u dv + v du.
VI d(v")=nv"" 1 dv.
i
188 DIFFERENTIAL CALCULUS
Via d&^ruf^dx.
— udv
VII
r /w\ du
VIII <*(log.t, ) = log.«
Vila ill
\cl e
dv
IX ^( a *) = a " l°g a ^ v «
IX a d(e 9 ) = e v dv.
X df(w w ) = vM , '~ 1 rfti + logw«w , 'dfv. '
XI df (sin v) as cos v <fo.
XII d(cos t>) = — sin v dv.
XIII df(tan v) = sec*vdto, etc.
XVIII df(arc sin v) = > etc
vl — v*
The term "differentiation" also includes the operation of finding
differentials.
In finding differentials the easiest way is to find the derivative as
usual, and then multiply the result by dx.
Illustrative Example 1. Find the differential of
x + 8
v =
x 2 + 3
Solution. ^ d (j^\=V + W + ** + WV + 1>
V* 2 + 3/ (x* + 8)»
_ (x a + 8)dx  (x + 3)2xdx _ (3  6x — z*)dx .
(x 2 + 8)» " (x« + 8) 2 * **■
Illustrative Example 2. Find dy from
&*x 8  aV = <**&*.
Solution. 2 6*xdx  2 atyfy = 0.
.\dy = —dx. An*,
dry
Illustrative Example 3. Find dp from
p a = a f cos20.
Solution. 2pdp =— a 2 sin20 • 2d0.
. a 2 sin 20 .„
P
Illustrative Example 4. Find d [arc sin (St — 4 £*)] .
Solution. d[a JC flin(St4t»)] = — ^*~ 4 * ,) = Sdt . ^rw.
Vl(8i4i«)» Vlt*
DIFFERENTIALS 139
93. Successive differentials. As the differential of a function is in
general also a function of the independent variable, we may deal with
its differential. Consider the function
V =/(*)•
d(dy) is called the second differential of y (or of the function) and
is denoted by the symbol <py.
Similarly, the third differential of y, d [rf(dy)]» is written
and so on, to the nth differential of y,
dry.
Since dz, the differential of the independent variable, is independ
ent of x (see footnote, p. 131), it must be treated as a constant when
differentiating with respect to x. Bearing this in mind, we get very
simple relations between successive differentials and successive deriva
tives. For dy=f(x)dx,
and d>y=f\xXdxy y
since dx is regarded as a constant.
Also, d?y=f'\x)(dxy,
and in general &y =f**\z)(dxy.
Dividing both sides of each expression by the power of dx occur
ring on the right, we get our ordinary derivative notation
3rw. §/"'«• •• % /»'«•
Powers of an infinitesimal are called infinitesimals of a higher order.
More generally, if for the infinitesimals a and /3,
limit £ = 0,
a
then fi is said to be an infinitesimal of a higher order than a.
Illustrative Example 1. Find the third differential of
y = x» — 2x s + 3x — 5.
8olotkm. dy = (6z i 6x* + S)dx,
d*y = (20x*l2x)(dx)*,
d»y=:(60x*12)(dx)». An*.
Note. This is evidently the third derivative of the function multiplied by the cube
of the differential of the independent variable. Dividing through by (dx) 8 , we get the
third derivative «
?_? = e0x 2 12.
dx*
140 DIFFERENTIAL CALCULUS
Differentiate the following, using differentials:
1. y = ax 8 — 6x* + ex + d. An*, dy = (8 ax* — 2 to + c)dx.
2. y = 2x48xt + 6xi + 6. <ty = (6xl 2x~i 6x*)dx.
3. y = (a*  x*) 6 . dy =  10x(a 2  x*) 4 dx.
4. y = Vl + x f . dy =; — ds.
Vl + x*
5 v = dy = dx.
* (1 + rf) (l + x*)» +1
/ „ 3x*dx
6. y = logVT3^. dy = ___.
7. y = (e* + e*) 8 . dy = 2(<** e~**)dx.
8. y = e* logx. dy = e*Mogx + Adz.
c« + c« Vrf+ev
. 1 + sin 
10. p = tan* + sec*. * = ^^ <*»
11. r = $tan»0+tan0. d^sec^dtf.
12./(x) = (logx).. /^.Mo**
X
t 8 St*dt
13. *(t) = —  — . <t>'(€)d£ =
(1  **)! (1  t*)l
1 cf 1 SB + w,4].SK*
15. d [arc tan log y] = — ? — .
L ^ J y[l + (logy) a ]
16. dr arc vers^  V2ry  y*l = ^
L r J V2ryy*
_2sin«0 2 8 2 J sin 8 *
CHAPTER X
RATES
94. The derivative considered as the ratio of two rates. Let
be the equation of a curve generated by a moving point P. Its coordi
nates x and y may then be considered as functions of the time, as
explained in § 71, p. 91. Differentiating r
with respect to t, by UV f we have
(32)
dt K J dt
'At any instant the time rate of change °
of y (or the function) equals its derivative multiplied by the time rate of
change of the independent variable.
Or, write (32) in the form
dy
(33)
dx dx
dt
The derivative measures the ratio of the time rate of change of y to
that of x.
Hat
— being the time rate of change of length of arc, we have from
dt
(12), p. 92,
(84)
IWdHf
which is the relation indicated by the above figure.
As a guide in solving rate problems use the following rule :
First Step. Draw a figure illustrating the problem. Denote by oc, y, s,
etc., the quantities which vary with the time.
Second Step. Obtain a relation between the variables involved which
will hold true at any instant.
141
142
DIFFERENTIAL CALCULUS
Third Step. Differentiate with respect to the time.
Fourth Step. Make a list of the given and required quantitiet.
Fifth Step. Substitute the known quantities in the result found bg
differentiating (third step), and solve for the unknown.
EXAMPLES
1. A man Is walking at the rate of 5 mhes per hour towards the foot of a tower
60 ft. high. At what rule Is he approaching the top when he is 80 It. from the foot
of the tower ?
Solution. Apply the above rule.
First step. Draw the figure. Let x = distance of the man from the foot and y = hi*
distance from the top of the tower at any Instant.
Second step. Since we have a right triangle.
Third step. Differentiating, i
2*^ = 2*^. or,
yet
(Bate of change of y) = lj {rate of change of x).
Fourth step
1 = 80,
— = 6 miles
dt
= 6x6281
y Js>
= 100
+ 8600
dt
Fifth step.
Substituting back in (A
dt
100
4 miles
x 5280 ft. per
jer hour. Ant
2. A point moves on the parabola fl y = x 1 in eucti a way that when x = 6, the
abscissa is Increasing at the rate of 2 ft. per second. At what rates are the ordinate
tad length of arc increasing at the same Instant ?
Solution. First step. Plot the parabola.
Second step.
Third step.
dv,
dt '
This means that at any pain
(Bate of change nf ordinate) :
(«l
z dz
on the parabola
j1 (rate of change of ab*ci#\a).
BATES 143
dx
Fourth step. — =2 ft. per second.
dt ^
x = e. ^ = ?
dt
x* ds
6 dt
Fifth step. Substituting back in (£),
— =  x 2 = 4 ft. per second. Ana.
dt 8
Substituting in (34), p. 141,
^ = V(2)« + (4)* = 2 V6 ft. per second. Ana.
dt
From the first result we note that at the point P (6, 6) the ordinate changes twice
as rapidly as the abscissa. ,
If we consider the point V (— 6, 6) instead, the result is ~ = — 4 ft. per second, the
dt
minus sign indicating that the ordinate is decreasing as the abscissa increases.
3. A circular plate of metal expands by heat so that its radius increases uniformly
at the rate of .01 inch per second. At what rate is the surface increasing when the
radius is two inches ?
.Solution. Let x = radius and y = area of plate. Then
y = wz*.
Thai is, at any instant the area of the plate is increasing in
square inches 2irx times as fast as the radius is increasing
in linear inches.
Substituting in (C),
— = 2w x 2 x .01 = .04ir8q. in. per sec. Ana.
dt
4. An arc light is hung 12 ft. directly above a straight horizontal walk on which
a boy 6 ft. in height is walking. How fast is the boy's shadow lengthening when he
is walking away from the light at the rate of 108 ft. per minute ?
Solution. Let x = distance of boy from a point directly
under light X, and y = length of boy's shadow. From the
figure,
y : y + x : : 5 : 12,
or y = f x.
Differentiating, dy = 6dz
I.e. the shadow is lengthening $ as fast as the boy is walking, or 120 ft. per minute.
5. In a parabola y* = 12 x, if x increases uniformly at the rate of 2 in. per second,
at what rate is y increasing when s = 8 in. ? Ana. 2 in. per see.
144
DIFFERENTIAL CALCULUS
8. At what point on the parabola of the last example do the abscissa and ordinate
Increase at the same rate ? Ins. (3, 6>.
7. In the function y = 2x' + 6, what is the value of x at the point where y
increases 24 times as fast as z ? An*, x = ± 2.
8. The ordinate of a point describing the curve x* + j/ 5 = 2;"i is decreasing nt the
rate of 1} in. per second. How rapidly is the abscissa changing when the ordinate is
4 inches? , dx .
Ann. — = 2 in. persec.
9. Find the values of z at the points where the rate of change of
**12s* + 45j:
lszero.
13
An*. x = 3 and 5.
10. At what points on the ellipse loz s
rate that x increases ?
11. Where !d the first quadrant, on £
fast as the ordinate ?
A point geoentM Baeb "f the following
increasing in each case :
'+V»!
= 400 does y
 r 3 , does the
. Find the rai
decrease at the same
arc increase twice as
Ana. At OOP.
e at which the arc is
12. y*=2x; ^ = 2, x = 2.
a™, f ( = Vh.
13. XV = 6 ; ^ = 2, y = 3.
dt 3
14. x' + 4v s = 20[ ^=— 1, V\.
15. y = x*;~ = 3,x=3.
16. y t =x'>; d J = *, V = 8.
17. The side of an equilateral diiiiule is 24 inches long, and is Increasing at
rate of 3 inches per hour. How fust is tin* area increasing ?
Am, 36 V5 sii. in. per hour.
13. Find the rate of change of the area of a square when the side b is increasing
at the rate of a units per second. Am. 2aisq. units per sec.
19. (a) The volume of a spherical soap bubble increases how many times as fast as
the radius? (b) When Its radius is 4 in. and increasing at the rate of I in. per second,
how fast is the volume increasing? Ana. (a) im" times as fast ;
(b) 32ireu. in. per sec.
How fast is the surface increasing hi the last case ?
20. One end of a ladder 50 ft. long is leaning against a perpendicular wall stand
ing on a horizontal plane. Supposing the foot of the ladder to be pulled away from the
wall at the rate of 3 ft. per minute ; (a) how fast is the top of the ladder descending
when the foot is 14 ft. from the wall ? (b) when will the top and bottom of the ladder
move at the same rate ? (c) when is the top of the ladder descending at the rate of
4 ft. per minute ? .1 im. (a) \ ft. per tnin. ;
(b) when 25 V2 ft. from wall ;
(c) when 40 ft. from wall.
21. A barge whoso deck is 12 ft. below the level of a dock is drawn up to it by
means of a cable attached in a ring in the floor of the dock, the cable being hauled in
by a windlass on deck at the rate of 8 ft. per minute. How fast is the barge moving
towards the dock when 16 ft. away ? Ana. 10 ft. per minute.
RATES
145
22. An elevated car is 40 ft. immediately above a surface car, their tracks inter
secting at right angles. K the speed of the elevated car in 18 miles per hour and of
the surface car 8 miles per hour, at what Rite are the earn separating 5 minutes after
tbeymeet? An», 17.0 miles per hour.
23. One ship was sailing south at the rate of miles per hour ; another east at the
r»te of 8 miles per hour. At 4 p.h. the second crossed the track of the first where the
was two hours before ; (a) how was the distance between the ships changing at
i.l (b) howatSp.M.? (c) when was the distance between them not changing?
Atut, (a) Diminishing a.B miles per hour;
(b) incrta.il ig H.~;i miles per hour ;
(c>8 : 17*.K.
4. Assuming the volume of the wood in a tree to be proportional to the cube of
its diameter, and that the latter increases uniformly year by year when growing,
show that the rate of growth when the diameter is S ft. is ao times as great as when
the diameter is It inches.
25. A railroad train is running 16 miles
he track having the form of the parabola
i ho
) ft. long,
and situated as shown in the figure,
the shadow & of the locomotive L is
it reaches the end of the wall.
If the sun is just rising in the east, find how fast
uoviug along the wall of the station at the instant
dx _ y &y
dt _ 800di'
Substituting this value of — in
Now
(!H^£Hi)'
— = 16 miles per hour
= 22 ft. per sec.
V = 400 and ~ = 1
Substituting back in (D), we get
= 13} ft. per second. Ant,
26. An express train and a balloon start from the san
The former travels 50 miles an hour and the latter rises ai
How fast are they separating ?
point at the same instant.
he rate of 10 miles an hour.
Am. 51 miles an hour.
DIFFERENTIAL CALCULUS
28. The rays of the aim make aa angle of 30" with the horizon. A ball is thrown
vertically upward to a height of 84 ft. How fast ia the shadow of the ball moving
along the ground just before it strikes the ground ? Ana. 110.8 ft. per se
29. A ship is anchored In 38 ft. of water. The cabie passes over a sheave on t
bow 6 ft. above the surface of the water. If the cable la taken in at the rate of 1 ft.
a second, how fast is the ship noting when there are 30 ft. of cable out ?
Ana. Jj ft. per sec.
30. A man is hoisting a chest to a window 60 ft. up by meanB of a block and tackle.
If he pulls in the rope at the rate of 10 ft. a minute while walking away from the
building at the rate of 6 ft. a minute, how fast is the chest rising at the end of the
second minute ? Ana. 10.98 ft. per mill.
31. Water Hows from a faucet Into a hemispherical basin of diameter 14 inches
at the rate of 2 en. in. per second. How fast is the water rising (a) when the wa
Is halfway to the top ? (b) just as it runs over ? (The volume of a spherical segment
= J nT 3 h + \ it/i', where h = altitude of segment.)
32. Sand i;, Iniiis; ]inii](.il nil thr uminul from the orifice of an elevated pipe, and
forms a pile which has always the shape of a riglil circular rone whose height Is equal
to the radius of the base. If sand is falling at the rate of cu. ft. per sec, how fi
is the height of the pile increasing when the height is 5 ft. ?
33. An aeroplane is 628 ft. directly above an automobile and starts cast at t
rate of 20 miles an hour at the same instant the automobile starts east at the rate of
40 miles an hour. How fast are they separating ?
34. A revolving IL'lit M.mliiigcuit a bundle of parallel rays is at a distance of \ a
mile from the shore and makes 1 revolution a minute. Find how fast the light is
traveling along the straight beach when at a distance of 1 mile from the nearest point
of the shore. Ann. 16.7 miles per min.
36. A kite is 160 ft. high and 200 ft. of Btrlng are out. If the kite starts drifting
away horizontally at the rate of 4 miles an hour, how fast is the string being paid out
at the start? Ana. 2.04 miles an hour.
36. A solution is poured into a conical filter of base radius cm. and height 24 cm.
at the rate of 2 cu. cm. a second, and filters out at the rate of 1 cu. cm. a second.
How fast is the level of the solution rising when (a) one third of the way up? (b) at
the top? Ana. (a) .079 cm. per sec. ;
(b) .000 cm. per sec.
37. A horse runs 10 miles per hour on a circular track in the center of which Is an
arc light. How fast wi!l his shadow move along a straight board fence (tangent to the
track at the starting point) when he has completed one eighth of the circuit P
Ana. 20 miles per hour.
38. The edges of a cube are 24 inches and are increasing at the rate of .02 in. per
minute. At what rate is (a) the volume increasing? (b) the area increasing?
39. The edges of a regular tetrahedron are 10 inches and are increasing at the rate
of .3 in. per hour. At what rate is (a) the volume increasing? (b) the area increasing?
40. An electric light bangs 40 ft. from a stone wall. A man is walking 12 ft. per
second on a straight path 10 ft. from the light and perpendicular to the wall. How fast
is the man's shadow moving when he la 30 ft. from the wall ? Ant. 48 ft. per sec.
BATES
147
41. The approach to a drawbridge has a gate whose two arms rotate about the
same axis as shown in the figure. The arm over the driveway is 4 yards long and
the arm over the f ootwalk is
8 yards long. Both arms ro fr I ©
tate at the rate of 5 radians
per minute. At what rate is
the distance between the ex
tremities of the arms chang
ing when they make an angle
of 46° with the horizontal ? Ana. 24 yd. per min.
42. A conical funnel of radius 8 inches and of the same depth is filled with a solu
tion which filters at the rate of 1 cu. in. per minute. How fast is the surface falling
when it is 1 inch from the top of the funnel ?
Ana. 7— in. per min.
4ir
48. An angle is increasing at a constant rate. Show that the tangent and sine are
increasing at the same rate when the angle is zero, and that the tangent increases
eight times as fast as the sine when the angle is 60°.
CHAPTER XI
CHANGE OF VARIABLE
95. Interchange of dependent and Independent variables. It is some
times desirable to transform an expression involving derivatives of y
with respect to x into an equivalent expression involving instead deriv
atives of x with respect to y. Our examples will show that in many
cases such a change transforms the given expression into a much
simpler one. Or perhaps x is given as an explicit function of y in a
problem, and it is found more convenient to use a formula involving
— » rs» etc., than one involving ~> ^p etc. We shall now proceed
dy df 6 dx dx* r
to find the formulas necessary for making such transformations.
Given y =/(a;), then from IXVI we have
dy i dx , A
(35) .SS f Ty**
dy
giving j in terms of — • Also, by XXV,
dx dy
or
dx 2 dx \dx) dy \dx) dx
(A ) £e„i./!\&.
^ J dx* dy\dx\dx
w
d*x
But i./J\ = _^; and ^ = ifrom (35).
dy\dx\ (dx? dx dx
\dy] \dy) dy
Substituting these in (A), we get
(36) tt *
' (%1
148
CHANGE OF VABIABLE 149
giving t§ in terms of — and —5 Similarly,
dor dy djf
er
and so on for higher derivatives. This transformation is called changing
the independent variable from xtoy.
Illustrative Example 1. Change the independent variable from x to y in the
equation AfVf _ *V &V _ ** /*(V =
\dx 2 / dxdx 8 dx a W
Solution. Substituting from (35), (36), (37),
<Px \* / cPx dx a (dhtf)
dy 8 dy \dy*}
81 *
W// W W/ / \ \dy)l\dyj
Reducing, we get
dy 8 dy*
a much simpler equation.
96. Change of the dependent variable. Let
and suppose at the same time y is a function of 2, say
(29 y =*(*)•
We may then express ~» ^p etc., in terms of — » 33* etc., as
 „ dx dsr dx ax
follows.
In general, z is a function of y by (E), p. 45 ; and since y is a func
tion of x by (4), it is evident that z is a function of x. Hence by
HV we have
K } dx dzdx * K } ix
150 DIFFERENTIAL CALCULUS
Similarly for higher derivatives. This transformation is called
changing the dependent variable from y to z, the independent variable
remaining x throughout. We will now illustrate this process by
means of an example.
Illustrative Example 1. Having given the equation
v ' dx* T l + y« W
change the dependent variable from y to t by means of the relation
(F) y = tan z.
Solution. From (F),
dy 9 dz (Py , d*z , _ « A /ck\*
^ = sec 2 2» — 4 = sec**—  + 2 8ec*ztanz ( — ) .
dz dz dz 2 dz* \dz/
Substituting in (E),
^ d 1 * . o t * /<k\* t . 2(1 + tan*)/ t dz\*
sec , «— + 2 8ec f «tan«(— ) =1+ ; — irM 8 ® *rJ ,
dx f \dx/ 1 + tan 2 z \ dz]
d*z /dz\*
and reducing, we get •— — 2 (— ) = cos 1 *. Ana.
dz* \dz/
97. Change of the Independent variable. Let y be a function of x,
and at the same time let x (and hence also y) be a function of a new
variable t It is required to express
dv d*ti
&' *5' etc "
in terms of new derivatives having t as the independent variable.
ByHT dy = dydx
dt dz dt
dx dx
dt
00
Ak0 S=:(§)=l(j ! )# =
oar aa; \ax/ dt \dx/ dx
But differentiating (A) with respect to £,
ldy\ dx d*y dy d*x
dx
"dt
dt \<fc/ dt I dx
w
A dt 9 dt d#
/dx\*
\dt)
Therefore
(*)
CHANGE OF VARIABLE 151
dxd*y dyd*x
d*y dt dt* dt dt* _
(f)'
and so on for higher derivatives. This transformation is called changing
the independent variable from x to L It is usually better to work out
examples by the methods illustrated above rather than by using the
formulas deduced.
Illustratitb Example 1. Change the independent variable from z to t in the
equation.
by means of the relation
Solution*
w
x =
>#.
dz_
dt"
. e*; therefore
dt __
dx"
:««.
dy_
dx
dy dt,
dtdx'
therefore
dy_
dx
dy
dt
Also
AUO da;* dx\<B/ dt dx dl\dt) dx dt dz
Substituting in the last result from (E),
(W) d*»  ' dt> dt € •
Substituting (D), (F), (G) in (C),
«"(<"£f<')+('f) + '= 0;
cPy
and reducing, we get ■—■ + y = 0. An*.
of 1
Since the formulas deduced in the Differential Calculus generally
involve derivatives of y with respect to x, such formulas as (A) and
(B) are especially useful when the parametric equations of a curve
are given. Such examples were given on pp. 82, 83, and many others
will be employed in what follows.
152
DIFFERENTIAL CALCULUS
98. Simultaneous change of both Independent and dependent variables.
It is often desirable to change both variables simultaneously. An im
portant case is that arising in the transformation from rectangular to
polar coordinates. Since
x = pcosd and y=p sind,
the equation
/(**)=
becomes by substitution an equation between p and 0, defining p as a
function of 0. Hence p, x, y are all functions of 0.
Illustrative Example 1. Transform the formula for the radius of curvature
(42), p. 159, p ^ tll
R =
[(1)1
d*y
dx 2
into polar coordinates.
Solution. Since in (A) and (B), pp. 150, 151, t is any variable on which x and y
depend, we may in this case let t = 0, giving
dy
W
(C)
* = 2£, and
dx dx
dd
dx d?y dy cPz
d*y <WdP~d0d0*
dx 2 ~~ /dx\ 8
/dxV
Substituting (B) and (C) in (A), we get
1 fix d?y dy dfa
d0dP~Md$*
dx\ 8
R =
m
R =
o
[fiff* (I)']'
dx d?y dy dfa
(i
or
d0d$* dBdff*
But since x — p cos0 and y = p sin0, we have
dx . A , A dp dy A , . A dp
Substituting these in (D) and reducing,
MS)']
aii
fl =
.. ilna.
CHANGE OF VARIABLE 158
EXAMPLES
Change the independent variable from x to y in the four following equations :
dry <Px
dx 2 dtf
dx 2 \dx/ dy 2 dy
Change the dependent variable from y to z in the following equations :
<M. + ,>(§»,) + (!M. + .)ig..>.
2 2 + 22.
. , , ^d*z ds d 2 * , 9 , _
a *V i^2(l + y)/dy\ 2 , . d*z a /dz\* .
6. t4 = 1 + t F(^) V = tanz. Ana. — — 2( — l = cos 2 *.
dx 2 1 + y 2 \dx/ dx 2 \dx/
^ M . *2x£ + 8z«* + x» = 0.
dx 8 dx 2 dx
Change the independent variable in the following eight equations :
 (Py x dy . y . . . cPy
9 < 1  x *)S x i= ' X=C08 * 0=°
„_ ,„ «,» d^ du , _ cPu .
10. (ly 2 )^r~y T  + a 2 u = 0, y = sinx. ^ + a 2 u = 0.
dy* dy dx 2
.d*i? ^ ePtf do d*v
dx* dx 2 dx dt*
d*y , 2x dy , y A A . d*y
—  f J h  = 0, x = tan 6. — 
dx 2 ^l + x 2 <fa> (1 + x 2 ) 2 ' d* 2
d 2 !* du „ .
14. H «*—  + sec 2 a = 0, s = arc tan J.
** "* (Pu du
4n*. (l + ^)^ + (2t + uarctant) — + 1 = 0.
at 2 dt
U>.*p i + a* V = 0, x = l An,. g + ?^ + a*» = a
dx 2 z d* 2 z dx
154 DIFFERENTIAL CALCULUS
In the following seven examples the equations are given in parametric form.
Find — and 4 in each case:
dx dx*
16. * = 7 + <*, y = 3 + t*3t«. Ana. *? = l6t*, ^ = 6.
dx dx*
17. x = cotl, y = sin«*. Ana. ^ =  3sin*lcost, —^ = 3sin 6 l(4 — Ssin 1 *).
dx dx*
18. x = a(coet + taint), y = a(8int — t coat). Ana. ^ = tan t, f^ =
dx dx* of cos 8 *
19. * =  1 y = 
1 + t " 1+t
20. x = 2t, y = 2 — t 1 .
21. * = 1 — 0, y = (•.
22. x = a cosl, y = b sin t.
dy
23. Transform — by assuming x = p cos'l, y = p sin0.
v^W
Ana.
24. Let /(x, y) = be the equation of a curve. Find an expression for its
slope (M in terms of polar coordinates. pcostf + sintf^
\dx/ A dy dB
Ana. — =
48 pstorf + corfg
CHAPTER XII
CURVATURE. RADIUS OF CURVATURE
99. Curvature. The shape of a curve depends very largely upon
the rate at which the direction of the tangent changes as the point of
contact describes the curve. This rate of change of direction is called
curvature and is denoted by K. We now proceed to find its analytical
expression, first for the simple case of the circle, and then for curves
in general.
100. Curvature of a circle. Consider a circle of radius R. Let
t = angle that the tangent at P makes with OX y and
t + At = angle made by the tangent at a neighboring point P f .
Then we say
At = total curvature of arc PP*.
If the point P with its tangent be
supposed to move along the curve to
f*, the total curvature (= At) would
measure the total change in direction,
or rotation, of the tangent; or, what
is the same thing, the total change in
direction of the arc itself. Denoting by * the length of the arc of
the curve measured from some fixed point (as A) to JP, and by As
the length of the arc PP*, then the ratio
At
A*
measures the average change in direction per unit length of arc*
Since, from the figure,
or
A«=2?'At,
At = 1
A* R
• Thus, if At  — radians ( 30°) , and As  3 centimeters, then —  — radians per centi
6 Aa lo
totter *iop per centimeter  average rate of change of direction.
166
156 DIFFERENTIAL CALCULUS
it is evident that this ratio is constant everywhere on the circle. This
ratio is, by definition, the curvature of the circle, and we have
The curvature of a circle equals the reciprocal of its radius.
101. Curvature at a point. Consider any curve. As in the 1
At = total curvature of tlw arc PP',
and ~  = average cvri'ature of the arc PP 1 ,
More important, however, than the notion of the average curvature
of an arc is that of wr w l iw at a point. This is obtained as follows.
Imagine P' to approach P along the curve ; then the limiting value of
the average curvature / = — J as i" ap
proaches P along the curve is defined as
the curvature at P, that is,
Ourvat ure at a point = A limit „ {**) = *I .
r A* = (H A»/ ds
(89) .. K=
 = curvature.
Since the angle At is measured in radians and the length of arc As
in units of length, it follows that the unit of curvature at a point is
one radian per unit of length,
102. Formulas for curvature. It is evident that if, in the last sec
tion, instead of measuring the angles which the tangents made
with OX, we had denoted by r and t + At the angles made by the
tangenta with any arbitrarily fixed line, the different steps would
in no wise have been changed, and consequently the results are
entirely independent of the system of coordinates used. However,
since the equations of the curves we shall consider are all given
iu either rectangular or polar coordinates, it is necessary to deduce
formulas for K in terms of both. We have
taiiT = 
§ 32, p. 31
CURVATURE
157
Differentiating with respect to x, using
d*y
dx
00
da?
W
dx [ \dx) J
From (24), p. 134
Dividing (A) by (2?) gives
(fa 8
But
dr
dx dr
— = — =JT. Hence
ds as
(40)
dr
r=
dy^V
nm
If the equation of the curve be given in polar coordinates, K may
be found as follows :
From CB), p 84,
CO)
But
t = 6 + yjr. Differentiating,
dd ^ d0
tanilr = — •
r dp
dd
.'. ^r = arc tan
From <>4), p. 84
dp
d0
Differentiating by
w
with respect to and reducing,
\de) p de*
it
de
' + ®
158 DIFFERENTIAL CALCULUS
Substituting (D) in ((7), we get
<*> 9 TTT^' Also
Dividing (^) by (^) gives
But —.= 2= if. Hence
a* a*
d0
(41)
Illustrative Example 1. Find the curvature of the parabola y 2 = 4pz at the
upper end of the latus rectum.
snin^. dy _2p d*y _ 2p dy _ 4p*
Solution. — = — Tl = i" :r = T '
dx y dx* y* dx y*
Substituting in (40), K =  4p *
giving the curvature at any point. At the upper end of the latus rectum (p, 2p)
(4p 2 + 4 p»)* 16 V2p* 4 V2p
Illustrative Example 2. Find the curvature of the logarithmic spiral p = e**
at any point.
Solution. £ = a«<* = ap ; § = a*e<* = a*p.
dd r d0* r
Substituting in (41), K = — . 4n*.
pVl + a*
* While in our work it is generally only the numerical value of K that is of Importance,
yet we can give a geometric meaning to its sign. Throughout our work we have taken the
positive sign of the radical %j 1 + ( ~ ) • Therefore K will be positive or negative at the same
time as— ff » that is (§ 86, p. 126), according as the curve is concave upwards or concave
dx*
downwards.
CURVATURE
159
In laying out the curves on a, railroad it will not do, on account of
the high speed of trains, to pass abruptly from a straight stretch of
track to a circular curve. In order to make the change of direction
gradual, engineers make use of transition curves to connect the straight
part of a track with a circular curve. Arcs of cubical parabolas are
generally employed as transition curves.
I u.rTi:\Ti vt Example 3. Tbe transition curve on a railway track baa the shape
of an arc of the cubical parabola y = j i*. At what rate Is a car on this track changing
its direction (1 mi. a unit of length) when It is passing through (a) the point (3, 9) ?
(b) the point (2, J) ? (c) the point (1, \) t
Solution.
Substituting in
(a) At (8, 6),
0») At (2, ),
<<:> At (1, 1),
 = 2x
U7)i
a _ l
" (2)* ~ V2
ns per mile = 8° 18' per mile. Ant.
radians per mile = 40" 30' per mile.
103. Radius of curvature. By analogy with the circle (see (38),
p. 156), the radius of curvature of a curve at a point is denned as the
reciprocal of the curvature of the curve at that point. Denoting the
nn lius of curvature by R, we have
» 1 .
K
or, substituting the values of A" from (40) mid (41),
[i+istfr
(12) « =
;§)]'
['+(S)T
• Hence the radius of ci
. according as the curve is concave upwards or concave downwards.
1 1n S 96, p. 152, (43) U derived from (42) by transforming from rectangular i
coordinates.
160 DIFFERENTIAL CALCULUS
Illustrative Example 1. Find the radios of curvature at any point of the cate
a  — 
naryy = (ei + e *)•
Substituting in (42),
_ L 1 + \ 2 /J \~2 / a(<* + e"^) tf .
R= = = = — . An*.
* * * _* 4 a
2a 2a
If the equation of the curve is given in parametric form, find the
first and second derivatives of y with respect to x from (&) and (2?),
pp. 150, 151, namely: d
w £ 
dt
dx d*tf dy d*x
d*v di~df~~di~d?
da?
[dt)
and then substitute the results in (42).*
Illustrative Example 2. Find the radius of curvature of the cycloid
x = a (t — sin t),
y = a(l — cost).
Solution. — = a (1 — cos t), — = asint;
at at
cPz . , d*y ,
—  = asini, —£ = acost.
Substituting in (G) and (£T), and then in (42), p. 159, we get
dy _ sin t <Py _ a (1 — cos Q a cos f — a sin t a sin t _ 1
dx"~"lcost' dx*~" a»(lcosQ» ~" a(lcost) 2
sin* \«1
[■ cay ]
B= l \; — "»»/ J = .2aV22co8t. .An*.
a(l — cost)*
Til
, rgr»arr
* Substituting (G) and (IT) in (42) gives   , J# A J .
cfz a"y ay cr*ai
CURVATURE
161
104. Circle of curvature. Consider any point P on the curve C.
The tangent drawn to the, curve at P has the same slope as the curve
itself at P (§ 64, p. 73). In an analogous man
ner we may construct for each point of the curve
a circle whose curvature is the same as the cur
vature of the curve itself at that point To do j
this, proceed as follows. Draw the normal to the
curve at P on the concave side of the curve. Lay
off on this normal the distance PC = radius of
curvature (= K) at P. With C as a center draw the circle passing
through P. The curvature of this circle is then
which also equals the curvature of the curve itself at P. The circle
so constructed is called the circle of curvature for the point P on
the curve.
In general, the circle of curvature of a curve at a point will
cross the curve at that point. This is illustrated in the above
figure.
Just as the tangent at P shows the direction of the curve at P, so
the circle of curvature at P aids us very materially in forming a geo
metric concept of the curvature of the curve at P, the rate of change
of direction of the curve and of the circle being the same at P.
In a subsequent section (§ 116) the circle of curvature will be
defined as the limiting position of a secant circle, a definition analo
gous to that of the tangent given in
§ 32, p. 31.
Illustrative Example 4. Find the radios
of curvature at the point (3, 4) on the equilat
eral hyperbola zy = 12, and draw the corre
sponding circle of curvature.
dy y <Py 2y
Solution. T = —
For (3, 4),
dz
dx"~ 8'
dz*
d*y
dz*
8
^ i
9
24
The circle of curvature crosses the curve at two points.
162 DIFFERENTIAL CALCULUS
1. Find the radius of curvature for each of the following curves, at the point indi
cated ; draw the curve and the corresponding circle of curvature :
(a) Wx« + aV = aV, (a, 0). Ana. R = .
a
(b) Wx* + aV = «*&*, (0, b). B = — .
b
(c) y = x* 4x« 18x*, (0, 0). R = £.
(d) lfy« = 4x*  «*, (2, 0). B = 2.
(e) y = x», (x t , Vl ). B = (L^lL.
(f) y 2 = x»,(4, 8). B = J(40)*.
(g) y« = 8x,(f,3). «=7H
B=*
(i) x 2 = 4ay, (0,0). B = 2a.
(j) (y  x 2 ) 2 = x*, (0, 0). fi=}.
(k) b*x*  aV = a 2 * 2 , <x n *). B = <P*f+£*{) .
(1) e>* = sin y, (x lt y^. (p) 9 y = x 2 , x = 3.
(m) y = sinx,(,l). (q) 4y 2 = x«, x = 4.
(n) y = cosx, f ^, V2V (r) x 2  y 2 = a 2 , y = 0.
(o) y = logx, x = e. (s) x 2 + 2y 2 = 9, (1,  2).
2. Determine the radius of curvature of the curve a*y = 6x 2 + cx*y at the origin.
Ana. B = — .
26
3. Show that the radius of curvature of the witch y 2 = — ' "*" ' at the vertex is  .
x 2
4. Find the radius of curvature of the curve y = log sec x at the point (x v y x ).
An*. R = secx 1 .
5. Find K at any point on the parabola x* + y * = a*. Ana. K =
a*
2(x + y)i
6. Find B at any point on the hypocycloid xi + yi = ai. 4 ns. B = 8(axy)>.
7. Find B at any point on the cycloid x = r arc vers — — V2 ry — y*.
r Ana. B = 2\/5ry.
Find the radius of curvature of the following curves at any point :
8. The circle p = a sin 9. Ana. B = .
(0* + a*\t
9. The spiral of Archimedes p = 00. B = ^V 3 — * .
r p 2 + 2a»
10. The cardioid p = a(l — cos 0). B = f Vifcap.
11. Thelemniscatep 2 =a 2 cos20. B = ~.
12. The parabola p = a sec 2  • B = 2 a sec 8 J .
13. The curve p = a sin* J R = } a sin 2 J.
CURVATURE
14. The trisectrii p = 2acoe<? — i
15. The equilateral h ypcrbola p' t
_ „(;,_ 4, ■.,.,#)?
JSg + e')*
18. The hyp.jtyct.iii
21, Find the radius of
indicated ; draw the curve and the corresponding circle c
for each of the following curves at the point
(a) .
(d)i = l
; sin I, y = cos 21; t =s —
to*
(h)a
= C + 2t;
22. An automobile rate track has the form of the ellipse 2? + IB j/* = 16, the unit
being one mile. At what rate is a car on this (rack changing its direction
(a) when passing through one end of the major axis ?
(b) when passing through one end of the minor axis?
(c) when two miles from the minor axis ?
(d) when equidistant from the minor and major axes ?
Ant. (a) 4 radians per mile ; (b) T ^ radian per mile.
23. On leaving her dock a steamship moves on an arc of the semicublcal parabola
4 y* = af. If the shore line coincides with the axis of y, and the unit of length is one
mile, bow fast is the ship changing its direction when one mile from the shore ?
Ant. ?_\ t radians per mile.
24. A battleship 400 ft. long has changed its direction 30° while moving through
a distance equal to its own length. What is the radius of the circle in which it is
moving? Ana. 704 ft.
26. At what rate is a bicycle rider on a circular track of half a mile diameter
changing his direction ? Ana. 4 rad. per mile = 43' per rod.
26. The origin being directly above the starting point, an aeroplane follows
approximately the spiral p = 6, the unit of length being cine mile. How rapidly is the
aeroplane turning at the instant it. has circled the starting point once?
27. A railway track has curves of approximately the form of arcs from the follow
ing curves. At what rate will an engine change its direction when passing through
it of length) :
(d) i/ = e*,a: = 0?
the points indicated (1 mi. = ur
(!)? = *■, (2, 8)?
(b) y = x\ (3, (t) ?
(c) ft*  y* = 8, (3, 1) ?
(f) ptf = 4, <? = !?*
CHAPTER XIII
THEOREM OF MEAN VALOE. OTDETERIOICATE FORMS
105. Rolle's Theorem. Let y =/(*) be a continuous singlevalued
function of x, vanishing for z = a and x = b, and suppose that f(x)
changes continuously when
x varies from a to b. The
function will then be rep
resented graphically by a
continuous curve as in the
figure. Geometric intuition
shows us at once that for
at least one value of x be
tween a and b the tangent is parallel to the axis of X (as at P);
that is, the slope is zero. This illustrates Rolle's Theorem :
If f(z) vanishes when x = a and x = b, andf(x) and f'(x) are con
tinvous for alt values of x from x—a to x = b, then f'(x) will be zero
for at least one value of x between a and b.
This theorem is obviously true, because as x increases from a to 6,
f(x) cannot always increase or always decrease as x increases, since
/(a) = and /(6) = 0. Hence for at least one value of x between a
and b, f(x~) must cease to increase and begin to decrease, or else cease
to decrease and begin to increase ; and for that particular value of x
the first derivative must be zero (§ 81, p. 108).
That Rolle's Theorem does not apply when f(x) or /'(*) are discontinuous is illus
trated as follows :
Fig. a shows the graph
of a function which is
discontinuous (= no) for
/ ■. ii value lying be
tween a and b. Fig. 6
shows a continuous func
tion whose first derivative
is discontinuous (= ■•)
for such an intermedials Flo. a
value x = c. In either cose it is seen that at no point
and i = b does the tangent (or curve) become parallel
o T ' / 3
THEOREM OF MEAN VALUE
165
106. The Theorem of Mean Value.* Consider the quantity Q defined
by the equation
(*) /(*) /(«)  (P  «) Q = o.
Let F(x) be a function formed by replacing b by x in the lefthand
member of (J5) ; that is,
(<7) F(x) =/(*) /(a)  (x  a) ©.
From (J?), P(6) = 0, and from (C), P(a) = ;
therefore, by Rolle's Theorem (p. 164) F'(x) must be zero for at least
one value of x between a and 6, say for x x . But by differentiating ((7)
WG g6t F'(*) =/'(*)  C
Therefore, since F'( x i) ~ ®» *^ en a ^° /'( x i) — G = 0»
and Q =/'(*,).
Substituting this value of Q in (.4), we get the Theorem of Mean
Value,
(**> /(&)/(«) =fixdf a< Xl <b
where in general all we know about x x is that it lies between a and b.
The Theorem of Mean Value interpreted Geometrically. Let the curve
in the figure be the locus of
Take OC = a and OD = b; then
/(a) = CA and /(6) = DB, giving
^# = 6 a and EB=f(V)f(a).
Therefore the slope of the chord
AB'w
(2>) tani^ = gg = ^?^> 
AE
b — a
There is at least one point on the curve between A and B (as P)
where the tangent (or curve) is parallel to the chord AB. If the
abscissa of P is x x , the slope at P is
(if) tan * =/'(*,) = tan i^B.
* Also called the Law? 0/ tAe Jfean.
166 DIFFERENTIAL CALCULUS
Equating (D) and (E), we get
ba J K lJ '
which is the Theorem of Mean Value.
The student should draw curves (as the one on p. 164) to show
that there may be more than one such point in the interval; and
curves to illustrate, on the other hand, that the theorem may not be
true if f(r) becomes discontinuous for any value of x between a and
b (Fig. a, p. 164), or if /'(x) becomes discontinuous (Fig. b, p. 164).
Clearing (44) of fractions, we may also write the theorem in the form
(45) /(ft) =/(fl) + (6  a)f(x x y
Let b = a + Au ; then b — a m An, and since x t is a number lying
between a and b, we may write
x x =a + 0Aa,
where 6 is a positive proper fraction. Substituting in (48), we get
another form of the Theorem of Mean Value.
(46) /(o + Ac)/(a) = Aqr*'(fl + ffAfl). O<0<1
107. The Extended Theorem of Mean Value.* Following the method
of the last section, let Jt be defined by the equation
(A) f(l) f(a) (b a)f'(a) \(b *)*X = 0.
Let J*"(x) be a function formed by replacing b by x in the lefthand
member of (J) ; that is,
(B) r($ =/<*) /(a)  (x  a)/'(a)  $(« «)' Jt
From (.4), F(b) = ; and from (B), F(a) = ;
therefore, by Rolle's Theorem (p. 164), at least one value of x between
a and h, say x % , will cause J*"(x) to vanish. Hence, since
F\x) /'<» /'(a)  (.  a) R, we get
*"(*,) A*,) fOO Ol<0* °
Since J^'Ot,) = and F'(a) = 0, it is evident that J*"(x) also satisfies
the conditions of Rolle's Theorem, so that its derivative, namely J""'(x),
must vanish for at least one value of x between a and x , say at , and
therefore x a also lies between a and i. But
F"(x~) =/"(x)  R ; therefore F%x t ) =/"(*,)  R = 0,
and J?=/"(x a ).
•Also called the Ettemled Lata of the Mean.
THEOREM OF MEAN VALUE 167
Substituting this result in (A), we get
(CT) f(V) =/(a) + (b  «)/'(«) + i (b  ay/»(xj a<x t <b
In the same manner, if we define S by means of the equation
/<$> /(«)  (J>  «)/'(«)  re (*  «)*/"(«)  4 (*«)•* 0.
we can derive the equation
(D) /(J) =/(a) + (b  a)f(a) + ^(b «)'/"(«)
+ 1 (P  «) , /'"(* i ), «<*,<i
where x, lies between a and 6.
By continuing this process we get the general result,
+ ^>'(a) + "+^^/ 1 >(«)
+ ^T^WO. a < x, < 6
where x 1 lies between a and b. (^) is called the Extended Theorem of
Mean Value.
108. Maxima and minima treated analytically. By making use of
the results of the last two sections we can now give a general discussion
of maxima and minima of functions of a single independent variable.
Given the function f(x). Let A be a positive number as small as
we please; then the definitions given in § 82, p. 109, may be stated
as follows :
If, for all values of x different from a in the interval [a — h> a + A],
(A) f(pO ~"/( a ) ^ a negative number,
then f(x) is said to be a maximum when x = a.
If, on the other hand,
(jB) f(p0 ""/( a ) = a positive number,
then f(x) is said to be a minimum when x = a.
Consider the following cases :
I. Let f '(a) *=0.
From (45), p. 166, replacing b by x and transposing /(a),
(C) /(*) /(«) = (*  <0/'(*x> a<x x <x
DIFFERENTIAL CALCULUS
Since f'(a') =£ 0, and/'(r) is assumed as continuous, A may be chosen
so small that /'(> j will have the same sign as/'u') for all values of x
in the interval [a — A, a + A]. Therefore /'(x^) has the same sign as
/'(a) (Chap. III). But x— a changes sign according as x is less or
greater than a. Therefore, from (C), the difference
will also change sign, and, by (A) and (B), /(a) will be neither a
maximum nor a minimum. This result agrees with the discussion in
§ 82, where it was shown that for all value* of x for which f(x~) is a
maximum or a minimum, the first derivative f (x) must vanish,
II. Letf'{a)=0, awi/»*0.
From (C), p. 187, replacing 6 by z and transposing /(a),
OT
/(*)/(«) =
TT
VO,).
Since f"(&) ^ 0, and/"(:r) is assumed as continuous, we may choose
our interval [a — A, a + A] so small that./*"^) will have the same sign
as f"(a) (Chap. III). Also (x — a) 1 does not change sign. Therefore
the second member of (D~) will not change sign, and the difference
will have the same sign for all values of x in the interval [i — A,
a + A], and, moreover, this sign will be the same as the sign of/"((i).
It therefore follows from our definitions (A) and (B) that
(.ff) /(a) w a maximum iff^a) = andf"(a) = a negative number;
(/") /(«) it a minimum iff'(a) = andf"(a) = a positive number.
These conditions are the same as (21) and (22), p. 113.
III. Let /'(a) =/"(«) =0, and f'"(a) * 0.
From (Z>), p. 167, replacing b by x and transposing /(a),
1
CO)
/C«>/C«)rgC*«)V w Cai>
As before, /'"(£,) will have the same sign as f'"(a). But (2 — a)*
changes its sign from — to + as x increases tlirough a. Therefore
the difference /<*>/C*3
must change sign, and /(a) is neither a maximum nor a minimum.
THEOREM OF MEAN VALUE
169
IV. i e (/'(a)=/"(a) = '=/ < "~ ,> («)=0. and f'\a) ¥= 0.
By continuing the process as illustrated in I, II, and III, it is seen
that if the first derivative of f(x) which does not vanish for x = a is
of even order (=n), then
(47) /(a) is a maximum if/(">(c) = a negative number;
(48) f(a) is a minimum if / ( "'(a)= a positive number.*
If the first derivative of f(x) which does not vanish for x = a is of
odd order, then /(a) will be neither a maximum nor • minimum.
i Example 1. Examine z" — 9z' + 24z — 7 for maximum and mini
...T(2) = 0,and/"<4)=0.
and minimum
Solution. /(i) = x>  9x« + 24x
f(x) = Zx*lBz + n.
Solving Sx 1 — 18z + 24 =
gives the critical values x = 2 and z = 4.
Differentiating again, /"(z) = 6z — 18.
Since/" (2) =  0, we know from (47) tliat/(2) =
Since/" (4) = + 6, we know from (48) that /(4) = if is a minimum.
e Example 2. Examine i J + 2 c
h /(z) = e' + 2co8Z + e '
f'(x) = e r 2isin* e 1
0, for X m
/"(*) = « 2 coax + e—
= 0, for x =
/'"(z) = e* + 2sinxr*
= 0, for x =
/»(z) = e* + 2co«x + e»
= 4, for x =
i from (48), /(0) = 4 is a minimum.
EXAMPLES
Examine the following functions for maximum and minimum values,
method of the last section :
1. Zx* — 4a« + l. Ant. z = l gives
2. z» Bi , + 12z + 48.
3. (zl)'(z + l)>.
* = 2 giv,
K = 1 giv.
z = i giv,
i. = 0;
neither,
neither.
4. Investigate x* — 5z* + flz* — l,atz = l and x = 3.
5. Investigate z* — 3z* + 3z + 7, at z = 1,
6. Show that if the first derivative of /(z) which does not vanish for z
odd order (= n), then f{z) is an increasing or decreasing function when z =
ingas/<»(a) is positive or negative.
= a is of
:, accord
A* in % 82, a critical vail
solving the resulting rqiintiu
t x — is the only root of the equation *'
found hyplncingthflflrstderivntivepqusl tc
DIFFERENTIAL CALCtTLUS
109. Indeterminate forms. When, for a particular value of the
independent variable, a function takes on one of the forms
, £, ()•*>, «_», 0*, «', 1*,
00
it is said to be indeterminate, and the function is not defined for that
value of the independent variable by tbe given analytical expression.
For example, suppose we have
where for some value of the variable, as x = a,
/OOo, ro)~a
For this value of x our function is not denned and we may there
fore assign to it any value we please. It is evident from what has
gone before (Case II, p. 15) that it is desirable to assign to the
function a value that will make it continuous when x = a whenever
it is possible to do so.
110. Evaluation of a function taking on an indeterminate form. If
when x = a the function /(x) assumes an indeterminate form, then
it taken a* the value off(x) for x = a.
The assumption of this limiting value makes /(x) continuous for
x = a. This agrees with the theorem under Case II, p. 15, and also
with our practice in Chapter III, where several functions assuming the
indeterminate form  were evaluated. Thus, for x = 2 the function
x*4
t the form , but
limit x*—i _ .
x=1x2~
Hence 4 is taken as the value of the function for x = 2. Let us
now illustrate graphically the fact that if we assume 4 as the value
of the function for x = 2, then the function is continuous for ;
x*4
V = ^T2
This equation may also be written in the form
r, O2)(yx2)0.
Let
* The calculation of this limiting value U called •
ting the indetem
INDETERMINATE FORMS 1T1
Placing each factor separately equal to zero, we have
x = 2, and y = x + 2.
In plotting, the loci of these equations are found to be the two
lines AB and CD respectively. Since there are infinitely many points
on the line AB having the abscissa 2, it is clear that when x=2
(= 0.1/), the value of y (or the function) may be taken as any num
ber whatever; but when x is different from 2, it is seen from the
graph of the function that the correspond
ing value of y (or the function) is always
found from
V = x + 2,
the equation of the line CD. Also, on CD,
when x— 2, we get
which we saw was also the limiting value of y (or the function)
for x=2\ and it is evident from geometrical considerations that if
we assume 4 as the value of the function for x = 2, then the function
is continuous for x = 2.
Similarly, several of the examples given in Chapter III illustrate
how the limiting values of many functions assuming indeterminate
forms may be found by employing suitable algebraic or trigonometric
transformations, and how in general these limiting values make the
corresponding functions continuous at the points in question. The
most general methods, however, for evaluating indeterminate forms
depend on differentiation.
111. Evaluation of the indeterminate form ^ Given a function of
the form iS* such that f(a) = and
F(jx)
F(a~) = ; that is, the function takes on
the indeterminate form  when a is sub
stituted for x. It is then required to find
limit f(x)
x = af(x)'
Draw the graphs of the functions /(^) and F(x). Since, by
hypothesis, /(«) = and F (a) = 0, these graphs intersect at (n, 0).
172
DIFFERENTIAL CALCULUS
Applying the Theorem of Mean Value to each of these functions
(replacing h by a:), we get
/«=/(>) + <>•■)/'<>,). a<x t <x
F<x) = F(a) + (xa)F>(xJ <•<*,<*
Since /(<*) = and J"(a) = 0, we get, after canceling out (x — a),
/(«) /'fa)
'w no
Now let x = a ; then x = a, x .= a, and
(49)
. limit /(■*) _ /'( fl )
J"(a)^0
the indeterminate form * and the theorem can be applied anew to
the ratio
giving us
a F(x) F'(a)
Rule for evaluating the indeterminate form  Differentiate the
numerator for a new numerator and the denominator for a new denom
inator* The value of thi* new fraction for the a»»igned value ' of tlie
variable will be the limiting value of the original fraction.
In ease it bo happens that
f'(a) = and F'(a) = 0,
that is, the first derivatives also vanish for x = a, then we still have
limit /O) = /»0Q
x = aF(x) F"(a)
When also /"(a) = and F"(a) — 0, we get in the same manner
limit /Q) f"(a)
x^a F (z) F'%a)'
and so on.
It may be necessary to repeat this process several times.
* The student is warned against the very careless but common mistake of differtutl sting
the whole expression as a fraction by VII.
t If 5», the substitution x reduces the problem lt> the evaluation of the limit for
I ° ,/l\ 1 ,/l\
Thua llmlt/W limit W*' limit W limit /'<*)
Therefore the rule holds in this case also.
INDETERMINATE FORMS 178
Illustrative Example 1. Evaluate 'LL1. ~ — when x = 1.
F(x) x*  x f  x + 1
«w**~ fO) x»8x + 2 V 18 + 2 ,_ . ,
Solution. ^J : z=— =  —  —  —  = . ..Indeterminate.
F(l) x»x f x + lJ XlB i 111 + 1
f'(\) 8x*8 1 88 . . . . .
^ = 3x»2xl L = ir^T = 0' ..indeterminate.
m = J*fL_l = * = ?. A*.
F"(l) 6X21,,! 62 2
limit e*— e* — 2x
Illustrative Example 2. Evaluate
x = * _
x — sinx
0.1 «. /(0) e*e*2xl 110 ... , A
Solution. ^^ = = =  . /. indeterminate.
F(0) xainx _L» 00
/'(0) e«+e*21 1 + 120 . . . t ,
i^rdr = — z = —z — 7~ = x • •'• indeterminate.
F'(0) 1cosx JLo 11
/"(0) **«*! 110 . . . . .
^W = sinTI.o = 0 = 0 ••• indetermin » te 
*""(0) cos x l.o 1
Evaluate the following by differentiation : *
. limit *' ~ 16 ,,„, 8 limit ^  arc sin 6 , 1
1 * = 4x« + x20' 9' 9 ' ^ = 0— gin^ At "' ~6*
2 limit x^1 1. limit Binx sin »
x = l x i n 10. x = ^___ cos*.
o limit logg !
°* x = 1 x^TT 11# limit e» + sinyl
^ limit e*c~*
2.
y = log(l + y)
x = sinx " 12 limit tan^ + sec^1 j
e limit tanx — x A * = °tan0 — sec0 + l"
* Ox — sinx „ , «^ o* ^ ,
limit sec»02tan0 1
a limit log sinx m _1 13. ____. .
x = ^(*r2x) f 8 *
7 limit s &» } a !4. l imi l——^^——. +oo.
x = — x g V f = fl o* 2a»* + 2« 8 **
limit r««r««V + «» lg limit, (a** 1 )' 6e * #
r = a r*a« * = 2 (x 4)e« + e»x
ta limit g'fx 2 limit sin 2x OA limit log cos (x1)
x = l X s — 1 x = x x = l _ . wx
* 1 — sin —
2
„ Umit s* + 8 1Q limit s sinx 9 limit tanx sinx
i7# x=2 x » + 32 # X x = xl *x = sinix "
* After differentiating, the student should in every case reduce the resulting expression
to its simplest possible form before substituting the value of the variable.
174 DIFFERENTIAL CALCULUS
112. Evaluation of the indeterminate form g. In order to find
limit f(x)
x = aF(x)
when J?L/(*)«o and ^^(x) = oo,
that is, when for x = a the function
M
F(x)
assumes the indeterminate form
00
00'
we follow the same rule as that given on p. 172 for evaluating the
indeterminate form ~ • Hence
Rule for evaluating the indeterminate form g. Differentiate the
numerator far a new numerator and the denominator for a new denomi
nator. Hie value of this new fraction for the assigned value of the vari
able will be the limiting value of the original fraction.
A rigorous proof of this rule is beyond the scope of this book and
is left for more advanced treatises.
lOff X
Illustrative Example 1. Evaluate — ^— for x = 0.
C8CX
flotation, {j^ = ^M = — • ... indeterminate.
.F(0) C8CXJx,o °°
F'{0) — csc x cot x ] xwm
m= __2Bto £ co 5 *_ = _0 = Q Am
F"(0) coex — xsinxJr.o 1
sin'xl _0
XCOSXJoO
indeterminate.
'(0)
113. Evaluation of the indeterminate form Ooo. If a function
f(x) • $(x) takes on the indeterminate form • oo for x = a, we write
the given function
so as to cause it to take on one of the forms  or ^, thus bringing it
under § 111 or § 112.
INDETERMINATE FORMS 175
Illustrative Example 1. Evaluate 8ec8xcos5xforx=.
2
8olntton. sec 8 x cos 6x] w = oo • 0. .. indeterminate.
*"•
Substituting for sec8x, the function becomes = ^M"
cosSx cosSx F(x)
\2/ cos5x~, w , . . . A
' — indeterminate.
1 =•
/w\ cosSxX.' 0*
J \2/ dB6z_J>l __6
If)
— sin8x
• Ana.
114. Evaluation of the indeterminate form soso . It is possible in
general to transform the expression into a fraction which will assume
either the form — or ££•
°°
Illustrative Example 1. Evaluate secx — tanx for x =  .
2
Solution, secx — tanx]  = oo — oo. .*. indeterminate.
C08Z COBX COSX F(x)
f \V 1rinxl 110 . . . . .
— = = — — =  • .: indeterminate.
F /*\ COBX JL_
V2/ = cosxI ^
■'©
— sinxj,.' — 1
EXAMPLB8
Evaluate the following expressions by differentiation : *
, limit (&±b , a limit log sin 2 x a
1. _ _ — z • AIW. • O. A — . .4.71*. 1.
* = ®cx s + d c « = 0log8inx
9 limit cotx 7 limit tan*
2 '* = 0io^x" "*• 7 '* = tan3*
3. llmit !SK. 0.
log(*f)
x = oo x« 8. Umi l  —• 0.
^ = E tan*
* limit * f A 2
4 x = co^ ° 9 limit log*.
* = C otx
K limit ^ limit
5 '* = «i^z" °° 10. ^x log Bin x. 0.
* In solving the remaining examples in this chapter it may be of assistance to the student
to refer to § 24, pp. 23, 24, where many special forms not indeterminate are evaluated.
176 DIFFERENTIAL CALCULUS
11.
i a x cot «x. An*.
x = u
1
12.
limit X.
y = co go/
0.
13.
limit
*(*• — 2 x) tan x.
z "" 2
2.
14.
j imit xsin?.
X = 00 x
a.
15.
JJ^xMogx. [wpotitlve.]
0.
16.
17.
limit
a__E (1 — tan0)sec20.
* = a v r/ 2a
1.
4a*
IT
18. limitl" 2 L_].^._l
x=lLaji_i xlj 2
19 limitl" 1 s 1
" x = ljogx logxj
1.
20. limi * [sec*tan*]. 0.
*=2
21 limit [2 1_1
22. wmi j r*  ji.
1
2
1
2
* = 0L4« 2«(e" + l)J 8
115. Evaluation of the indeterminate forms 0*, 1*, oo*. Given a func
tion of the form *, NA ,_v
•
In order that the function shall take on one of the above three
forms, we must have for a certain value of x
/<V) = 0, <K*) = 0, giving 0°;
or, f(x) =1, <f> (x) = oo, giving 1* ;
or, f(x) = oo, <f>(x) = 0, giving oo°.
Let y=/(*)* (x) ;
taking the logarithm of both sides,
In any of the above cases the logarithm of y (the function) will
take on the indeterminate form
oo.
Evaluating this by the process illustrated in § 113 gives the limit
of the logarithm of the function. This being equal to the logarithm
of the limit of the function, the limit of the function is known.*
Illustrative Example 1. Evaluate x? when x = 0.
8alutfcm. This function assumes the indeterminate form 0° for x = 0.
Let y = x* ;
then logy = xlogx = 0 — co, whenx = 0.
By 1 113, p. 174, logy = !2£i = Zl2, when x = 0.
1 »
X
* Thus, if limit log«y  a, then y  e«.
INDETERMINATE FORMS
177
By § 112, p. 174, logy = ^ = x = 0,
Since y = x«, this gives log,x* = ; i.e., x* = 1. Ans.
when x = 0.
Illustrative Example 2. Evaluate (1 + x)* when x = 0.
Solution. This function assumes the indeterminate form 1* for 2 = 0.
Let
then
By 1 118, p. 174,
By  111, p. 171,
y = (l + x)i;
logy =  log(l + x) = oo . 0,
z
. log(l + x)
1
logy =
1 + s
1
1 + x
= 1,
when x = 0.
when x — 0.
when x = 0.
8ince y = (1 + x)*, this gives logg(l + x)* = 1 ; i.e. (1 + x)* = e. Ans.
Illustrative Example 8. Evaluate (cotx) 14 " 1 * for x = 0.
Solution. This function assumes the indeterminate form oo° for x = 0.
Let y = (cotz) ai **;
then logy = sin x log cotx = • oo,
_ log cotx __ oo
cscx ~~ oo'
— C8C*X
By 1 118, p. 174,
logy
when x = 0.
when x = 0.
By 1 112, p. 174,
logy =
cotx
sinx A
= — r = '
when x = 0.
— cscx cotx cos s x
Since y = (cotx) 1111 *, this gives log, (cot x) ,lax = 0; i.e. (cotx)* 111 * = 1. Ans.
Evaluate the following expressions by differentiation :
X = 1
« limit /lY"*
3. lhl ^ (sin *)*»•.
y = co\ ^y)
5. ^q(1 + tin *)*••.
Ans. •
e
1.
1.
e°.
e.
e*.
7 "= (' + *>'"•
4n*. 6 s .
s ] l = o («*xys=.
1
e'
•• ,"S <» + «)••
c".
*j3Hfr*
1
11. ^'J; («*»»*)*.
c»—.
12. S (cotx)*.
1.
i3.. UmIt /2?\n.
1
CHAPTER XIV
IV.^iW
CIRCLE OF CURVATURE. CENTER OF CURVATURE
116. Circle of curvature.* Center of curvature. If a circle be drawn
through three points ijj, Jf, ^ on a plane curve, and if 7J and P t be
made to approach Jl along the curve as a limiting position, then the
circle will in general approach in magni
tude and position a limiting circle called
the circle of curvature of the curve at the
point ij. The center of this circle is
called the center of curvature.
Let the equation of the curve be
(1) y =/<» [
and let x a , x lt x t be the abscissas of the
points J£, ij, ij respectively, (a', £') the coordinates of the center,
and R' the radius of the circle passing through the three points.
Then the equation of the circle is
O «<)' + (if #)".B";
and since the coordinates of the points JJ, P lr P % must satisfy this equa
tion, we have , , . , . ol .^ „. .
(2) J (*,  ay + (y, fSyR°= 0,
ta«y+Gv*rjp*
Now consider the function of x defined by
F(x ) = (x<xy+(j,fryR';
hi which y has been replaced hy f(x) from (1).
Then from equations (2) we get
*•(*.)= o. *0O
y«)o.
circle. The circle c
CIRCLE AND CENTER OF CURVATURE
179
Hence, by Rolle's Theorem (p. 164), F'(x) must vanish for at least
two values of x, one lying between x Q and x v say X*, and the other
lying between x x and x t , say x" ; that is,
F f (*0 = 0, J^'Or") = 0.
Again, for the same reason, F"(x) must vanish for some value of
x between x 1 and x*\ say x % ; hence
F»(x^ = 0.
Therefore the elements a', £', R 1 of the circle passing through the
points JJ, JJ, JJ must satisfy the three equations
2P(a; o ) = 0, *"<y) = 0, F"(xJ = 0.
Now let the points P x and JJ approach ij as a limiting position ; then
Xf x t j a/, x*\ x % will all approach x as a limit, and the elements a, £, It
of the osculating circle are therefore determined by the three equations
f(x ) = o, *"(*.) = <>, *•"(*„)=<>;
or, dropping the subscripts, which is the same thing,
00 (xaf+Q, ($)*=&,
(z — a) + (y — £) ^ = 0, differentiating (A).
dx
1 + f $Y+ <y  P) Pi = 0, differentiating (5).
Wa;/
(fa 5
Solving (j5) and (C) for x — a and y — & we get ( tJ[ =£ V
0*>)
y/3=
dx>
HI
hence the coordinates of the center of curvature are
(*)
a = x—
lh(l)'] g (I)'
d*y
d*y
d?
(SH
180
DIFFERENTIAL CALCULUS
Substituting the values of x — a and y — fi from (D) in (A), and
solving for 2?, we get
R = ±
MS)T
fy
dx*
which is identical with (42), p. 159. Hence
Theorem. The radius of the circle of curvature equals the radius of
curvature.
117. Second method for finding center of curvature. Here we shall
make use of the definition of circle of
curvature given on p. 161. Draw a
figure showing the tangent line, circle
of curvature, radius of curvature, and
center of curvature (ff, /3) corresponding
to the point P(x, y) on the curve. Then
a=OA = ODAD=ODBP=x~BP,
0=AC=AB+BC=DP+BC=y+BC.
But BP = It sin t, BC = It cos t. Hence
oo
a = z — It sin t, fi = y+It cos t.
From (29), p. 135, and (42), p. 159,
dy
sinT = —
dx
Hm
COST =
HDT
It =
KD7
dx*
Substituting these back in (A), we get
i+
(50)
a = x —
d*y
; p=u+
(IT
d*y
d?
From (23), p. 126, we know that at a point of inflection (as Q in
the next figure) «
CIRCLE AND CENTER OF CURVATURE
181
Therefore, by (40), p. 157, the curvature K=0; and from (42),
p. 159, and (50), p. 180, we see that in general a, ft
without limit as the second derivative approaches
zero. That is, if we suppose P with its tangent
to move aloug the curve to P', at the point of
inflection Q the curvature is zero, the rotation of
the tangent is momentarily arrested, and as the
direction of rotation changes, the center of cur
vature moves out indefinitely and the radius of
curvature becomes infinite.
Ii.i.i sriiiTiiK Example 1. Find the coiirdi nates of the
center of curvature of the parabola y 1 = 4 px corresponding
a) to any point on the curve ; (u)
any point o:
fb) (2 p, 0) is the center of curvature corresponding to the vertex (0, 0).
118. Center of curvature the Limiting position of the intersection of
normals at neighboring points. Let the equation of a curve be
w »/(*>
The equations of the normals to the curve at two neighboring
points J£ and P are * Crajsi
0,—r> + (y,r)
ax.
If the normals intersect at C'(a\ £'), ^("VVa)
the coordinates of this point must satisfy both equations, giving
w
kv
«') + (!.
»?
o.
jo,
■O+0/,
"«t
= 0.
From (2), p. T7
Vatid Y being the var
able coordinates
82 DIFFERENTIAL CALCULUS
Now consider the function of x defined by
+ (*) =00 +(,«!
in which y has been replaced by f(x) from (A).
Then equations (B) show that
But then, by Rolle's Theorem (p. 164), </>'(*) must vanish for Borne
value of x between *, and a*,, say x 1 . Therefore a' and ff are deter
mined by the two equations
*(*.)= 0, +'(0=0.
If now P, approaches P> as a limiting position, then x' approaches x c ,
& v ' m s <KO = °> *'(* ) = °;
and C'(a', ff~) will approach as a limiting position the center of cur
vature C(a, #) corresponding to JjJ on the curve. For if we drop the
subscripts and write the last two equations in the form
dx
m^
tn
dx*
= 0,
it is evident that solving for a' and fi' will give the same results as
solving (J?) and (C), p. 179, for a and (3. Hence
Theorem. The center of curvature C corresponding to a point P on a
curve is the limitinij position of the intersection of the normal to thr >i,r' , >'
at P with a neiyhhorhi;; normal.
119. Evolutes. The locus of the centers of curvature of a given
curve is called the evolute of that curve.
Consider the circle of curvature corre
sponding to a point J* on a curve. If
P moves along the given curve, we may
suppose the corresponding circle of curva
ture to roll along the curve with it, its
radius varying so as to be always equal to
the radius of curvature of the curve at the
point P. The curve CC, described by the
center of the circle is the evolute of PP V
It is instructive to make an approximate construction of the evolute
of a curve by estimating (from the shape of the curve) the lengths
CIRCLE AKD CENTER OF CURVATURE 188
of the radii of curvature at different points on the curve anil then
drawing them in and drawing the locus of the centers of curvature.
Formula (£), p. 179, gives the coordinates of any point (rr, ff) on
the evolute expressed in terms of the coordinates of the cor respond ing
point (x, y) of the given curve. But y is a function of x ; therefore
M
M Ux
(Il
dx*
dx*
give us at once the parametric equations of the evolute in terms of the
parameter x.
To nnd the ordinary rectangular equation of the evolute we elimi
nate x between the two expressions. No general process of elimination
can be given that will apply in all cases, the method to be adopted
depending on the form of the given equation. In a large number of
cases, however, the student can find the rectangular equation of the
evolute by taking the following steps:
General directions for finding the equation of the evolute in rectangular
coordinates.
Fikst Step. Find a and & from (50), p. 180.
Second Step. Solve the two resulting equations for x and y in terms
of a and 0.
Tfimi) Step. Substitute these values of x and y in the given equation.
This gives a relation between the variables a ami which is the equation
of the evolute.
Illustrative Example 1. Find the equation of the evolute of the parabola y*= ipz.
Solution.
ay 2p <fty_ 4p"
ax y it' y*
First step.
Second step.
Third step
(4 P »ffll = 4 P (^P);
27
 *P1F
Remembering that a denotes (lie abscissa and p the
ordinate of a rectangular system of coSrdinates, we see
that the evolute of the parabola A OB is the semi cubical parabola DQ'E; the c
ol curvature for 0, P, P,, P, being at C, C, C,, C, respectively.
184
of the evolute of the ellipse
Third mep. (na)' + (63)' = (a 5  ft 5 ) 1 , the equa
tion of the evolute EHE'H' of the ellipse ABA'B'. E, E', £
curvature corresponding to the points A, A', B, B\ on the cur
spond to the points P, P", P".
i, and C, C, C" corre
When the equations of the c
i are given in parametric form, we
C^O
dy
dt_ dV
~'dz' <k*
dx d 1 ;/
and then substitute the results in formulas (50), p. 180. This gives
the parametric equations of the evolute in terms of the same parameter
that occurs in the given equations.
Example 3. The parametric equations of a curve are
Find the cquat:
find the radius of
c!e of curvature.
Substituting ii
IB)
i of the evolute in parametric form, plot the curve and the evolute,
t the point where I = 1, and draw the corresponding c:r
above formulas (A) and then in (60), p. 180,
_ 1  P  B I* i (' + 3 t
CIRCLE AND CENTER OF CURVATURE
185
the parametric equations of the evolute. Assuming values of the parameter t, we cal
culate x, y ; a, /? from (E) and (C) ; and tabulate the results as follows :
Now plot the curve and its evolute.
The point (i, 0) is common to the given curve
and its evolute. The given curve (semicubical
parabola) lies entirely to the right and the evo
lute entirely to the left of x = £.
The circle of curvature at A (, J), where
t = 1, will have its center at A' (— }, j) on
the evolute and radius = AA\ To verify our
work find radius of curvature at A. From
(42), p. 150, we get
2
t
X
V
a
3
i
f
2
i
t
¥
V
t
H
A
H
8
1
i
i
1
i
i
i
1
1
2
i
i
i
f
it
A
tt
8
2
i
i
V
V
3
i
!
This should equal the distance
** = V(l + 1)» + (J  {)* = VI.
Illustrative Example 4. Find the parametric equations of the evolute of the
cycloid,
rx = a(tsint),
ly = a(lcost).
Solution. As in Illustrative Example 2, p. 100, we get
dy
sint cPy
dx 1 — cost dz* a (1 — cost) 1
Substituting these results in formulas (60), p. 180, we get
Ja = a(t + sin t),
\fi =— a(l — cost), ^liw.
«
186
DIFFERENTIAL CALCULUS
Notk. If we eliminate t between equations (D), there results the rectangular equa
tion of the evolute 00" Q* referred to the lies 0" a and O"0. The coordinates of Owith
respect to these axes are (— ira, — Bn).
Let us transform equations (D) to the
new set of axes OX and OT. Then
Substituting in (D) and reducing, the
equations of the evolute become
Since (£) and (C) a
The erotutc of a cycloid
given cycloid.
identical in form, we have :
itself a cycloid whose generating circle equals that of the
120. Properties of the evolute. Fn
(^) a = x — R sin t.
m (J), p. 180,
$ = y + B cos t.
Let us choose as independent variable the lengths of the arc on the
given curve ; then x, y, Ji, t, «, /9 are functions of s. Differentiating
(J) with respect to s gives
da dz _ dr .
— = — — R cos t — — sin 1
d» d» ds
it it i
iS
it'
is
iy
cost, ^ = 8inT, from (36), p. 134; and 
(38) and (33), p. 156.
Substituting in (#) and (C), we obtain
ia
ill
(Pi
(*i
Dividing (2?) by (.D) gives
m f= cotT= 
— flCOST — — 8
iR
1 , iS dS
hmST =C08T 
R at a*
1_
CIRCLE AND CENTEE OF CURVATURE 187
Jut '— = tan t' = slope of tangent to the evolute at C, and
£ = tan t = slope of tangent to the given curve at the corre
dx
sponding point P(x, y~).
Substituting the last two results in (F), we get
tan t' = .
taiiT
Since the slope of one tangent is the negative reciprocal of the
slope of the other, they are perpendicular. But a line perpendicular
to the tangent at P is a normal to the curve. Hence
A normal to tlie given curve is a tangent to its evolute.
Again, squaring equations (/>) and (E) and adding, we get
(SMfHf)"
But if s 1 = length of arc of the evolute, the lefthand member of
(ff) is precisely the square of — (from (34), p. 141, where t = s,
r = (f,x = a,y = ff). Hence (<?) asserts that
/ds'\* /dR\* d»' dR
\s) (,■*> ***$•
That is, the radius of curvature of the given curve increase* or decreases
at fast as the arc of the evolute increases. In our figure this means that
P^ PC =a.rc CO t
The length of an arc of the evolute is equal to the difference between
the radii of curvature of the given curve which are tangent to this arc at
its extremities.
Thus in Illustrative Example 4, p. 186, we observe that if we fold
£*/*(= 4 «) over to the left on the evolute, P" will reach to 0', and
we have:
The length of one arc of the cycloid (as OO'Q'') is eight times the length
of the radius of the generating circle.
121. Involutes and their mechanical construction. Let a flexible
ruler be bent in the form of the curve C t C f the evolute of the curve
iJ/J, and suppose a string of length R t , with one end fastened at C t , to
188
DIFFERENTIAL CALCULUS
be wrapped around the ruler (or curve). It is clear from the results
of the last section that when the string is unwound and kept taut,
the free end will describe the curve
%%. Hence the name evolute.
The curve T^iJ is said to be an invo
lute of O.Cy Obviously any point on
the string will describe an involute,
bo that a given curve has an infinite
number of involutes but only one
evolute.
The involutes JfJJ, %'%', %"!;" are
called parallel curves since the distance
between any two of them measured
along their common normals is con
stant.
The student should observe how the parabola and ellipse on pp. 183,
184 may be constructed in this way from their evolutes.
Find the coordinates of the center of
each of the following curves. Draw the
circle of curvature.
1. The hyperbola 5 — jj = 1.
_ (a* + W/S
(a' + 6«)y* .
2. The hypocycloid if +j/t 
. Find the coordinates of the center of
evolute (aa)l  (68)1 = (a* + 6»)1.
Ins. a = x + Zxtyl, P = V + 3xiyil
evolute (a + 8)1 + (a  8)1 = 2 at.
re of the cubical parabola y* = a*x.
4 M rt + W a^fly*
ea'j/ ' P 2o« '
A. Show that in the parabola it + yt = at wb have the relation a + S = 8(j: + y).
6. Oiven the equation of the equilateral hyperbola 2iy = a 5 ; show that
From this derive the equation of the evolute {a + 8)1 — [a — 8)1 = £ at.
Find the parametric equations of the evoluWs of the following c
of the parameter £. Draw the curve and its evolute, and draw at least one circle
of curvature.
»•[, fa = acairt + 8acOB(sin*(,
6. The hypocycloid
CIRCLE AND CENTER OF CURVATURE
189
8. The curve
9. The curve
10. The curve
11. The curve
12. The curve '
13. The curve
14. The curve
15. The curve
16. The curve
x = a(cost + tsinQ,
y = a (sin J — tcmt).
x = 3t,
y = P  6.
{
{
X
X
y
X
6P,
2t.
P2.
4*.
8 + A
2t.
{
{
x = 2i,
y = P.
x = 2l,
8
faj = 2i
H
■*"•• {1=
= a cost,'
a Ant.
17. x=4P, y = 2t.
18. x = 2l, y = 160.
19. x = I, y = sin t.
_, y = 8l.
20. x
21. x = t», y = i*.
,=•*!.
a = 48P,
/J = 2(«.
a=2(«,
/J=8t».
* = (»,
/J = 11 + 80.
a = 78P,
/J = 2(«.
4lt«
— 4~ '
12 + 5f*
6t
4t» + 12l
— 8
2t» + «*
ar =
=
a =
= 
a =
=
2
12t* + 9
27 + 4t«
6t
22.
x = t, y = <•.
23.
x = sint, y = 8 cost.
24.
x = 1 — cost, y = t — sin t.
25.
z = cos 4 *, y = sin 4 t.
26.
x = asect, y = 6tant.
CHAPTER XV
PARTIAL DIFFERENTIATION
122. Continuous functions of two or more independent variables.
A function /(z, y) of two independent variables x and y is denned
as continuous for the values (a, i) of (j, y) when
limit
«««/<*, »)=/(«, 6>
no matter in what way x and y approach their respective limits <
and b. This definition is sometimes roughly summed up in the state
ment that a very small change in one or both of the independent variables
shall produce a very small change in the value of the function.*
We may illustrate this geometrically by considering the surface
represented by the equation _ f, „.
Consider a fixed point P on the surface where x = a and y = b.
Denote by Ax and Ay the increments of the independent variables
and y, and by Az the corresponding increment of the dependent
variable z, the coordinates of P' being
(x + A*, y + Ay, z + Az).
At P the value of the function is
2=/(a, b) = MP.
If the function iB continuous at P, then, however
Ax and Ay may approach the limit zero, Az will
also approach the limit zero. That is, M'P' will approach coincidence
with MP, the point P' approaching the point P on the surface from
any direction whatever.
A similar definition holds for a continuous function of more than
two independent variables.
In what follows, only values of the independent variables are
considered for which a function is continuous.
it again reads over !
PAETIAL DIFFERENTIATION 191
123. Partial derivatives. Since x and y are independent in
x may be supposed to vary while y remains constant, or the reverse.
The derivative of z with respect to x when x varies and y remains
constant* is called the partial derivative of z with respect to x\ and is
dz
denoted by the symbol — • We may then write
ox
(A . 3? = limit r /(* + A^y)/(^y) 1
K ' dx A*=0 L A* J*
Similarly, when x remains constant* and y varies, the partial
derivative of z with respect to y is
fS \ 0£ = limit [" /(>> y + Ay) /(a, y) 1
^ J dy Ay = 0_ Ay J'
^ is also written —f(x> y), or / .
ex dx ex
Similarly, — is also written — fix. v\ or — •
In order to avoid confusion the round dt has been generally
adopted to indicate partial differentiation. Other notations, however,
which are in use are
(£) * (^) ; rj & ti'M* y) ; fj & y) ' f & *> ; D *f* D > f; z * z *
Our notation may be extended to a function of any number of
independent variables. Thus, if
u = F(x, y, z) f
then we have the three partial derivatives
du du du m dF dF dF
dx dy dz 9 ' dx dy dz
Illustrative Example 1. Find the partial derivatives of z = ax 2 + 2bxy + cy*
Pa
Solution. — = 2 ax + 2 by* treating y as a constant,
dx
— = 2 te + 2 cy, treating x as a constant.
dy
• The constant values are substituted in the function before differentiating.
t Introduced by Jacobi (18011851).
i
DIFFERENTIAL CALCULUS
[ve Kv.imh.f. 2. Find the partial derivatives of 11 = sin (ai
Solution. — = a coh {ax + by + cz) , treating y and z as constants,
— =■ 6 cos {ox + by + ex), treating x and z m e>
— bo cos (az + tr/ + cz), treating y and imci
Again turning to the function
z =/0. y).
we have, by (A~), p. 191, defined— as the limit of the ratio of the
increment of the function (y being constant) to the increment of x, as
the increment of x approaches the limit zero. Similarly, (B), p. 191,
has defined — It is evident, however, that if we look upon these
partial derivatives from the point of view of § 94, p. 141, then
may be considered as the ratio of the time rates of change of z and
x when y is constant, and g z
dy
as the ratio of the time rates of change of z and y when x is constant.
124. Partial derivatives interpreted geometrically. Let the equa
tion of the surface shown in the figure be
* =/(* y)
Pass a plane EFGH through the
point P (where x = a and y = 6) on
the surface parallel to the JfOZplane.
Since the equation of this plane is
the equation of the section JPE cut
out of the surface is
i the axis of X. In this
if we consider EF as the axis of Z and EH i
dx
^tanMTP
= dope of section JK at P.
PARTIAL DIFFERENTIATION 193
Similarly, if we pass the plane BCD through P parallel to the
FO^plane, its equation is x _. „
and for the section DPI, — means the same as — • Hence
dy dy
dz dz
— = — = — tanMT'P = slope of section DI at P.
dy dy r
g2 jJt gl
Illustrative Example 1. Given the ellipsoid ( — H — = 1 ; find the slope
^ 24 12 8
of the section of the ellipsoid made (a) by the plane y = 1 at the point where x = 4
and z is positive ; (b) by the plane x = 2 at the point where y = S and z is positive.
Solution. Considering y as constant,
2x , 2z dz A dz x
= 0, or — = .
24 6 8x dx 4z
2 y 2 z dz dz y
When x is constant, — Z + — = 0, or — = — f •
n 12 6 dy dy 2z
(a) When y = 1 and x = 4, z = %/ . .. — = — %/ . Ana.
\2 dx \8
(b) When x = 2 and y = 8, z = — =. .. — = v2. 4na.
V ' V2 *V 2
EXAMPLES
du
3.
1. u = x , + 8x«yy I . 4na. — = 3x f + 6xy;
dx
^ = 8x»8y«.
2. u = 4x f + Bxy + Cy 1 + Dx + Ey + F. — = 2Ax + By + D ;
dx
— = Bx + 2 Cy + E.
dy
«,«*••. du 2anxu
3. u = (ax« + 5y« + «*)». t~= , , , 2 ,  S
dx ax* + 6y 2 + cz*
du 2bnyu
dy ax* + &y 2 + cz 1
. . x du 1
4. u = arc sin 
V & Vy*x*
du x
dy y Vy 2 — x a
« dU 
6. u = x*. —iiiyxy 1 ;
dx
— = x*logx.
dy
6. u = aa^y«z + bxy*z* + cy* + dxz*. — = 8ax*y 2 z + by*z* + dz 8 ;
dx
a,.
— = 20x^3 + 8&xy*z* + 6cy«;
Sy
— = ax»y* + 4tey»z« + Sdxz*.
dz
194 DIFFERENTIAL CALCULUS
7. u = xV — 2xy* + 8xV : show that x — + w — = 5u.
dx by
8. w = — — ; show that x — \y— = u.
z + y bx by
9. u = (y — z) (z — x) (x — y) ; show that 1 \ — = 0.
dx by dz
10. w = log (0° + ei) : show that 1 = 1.
bx by
11. u = ; show that = (x + y — l)w.
eF + # bx by v '
12. u = xty* ; show that x \y — = (x + y + log w) u.
dx dy
13. u = log(x» + y* + z*  8xyz) ; show that — + — + — = 8
dx by bz z + y + i
14. u = e^siny + ersinx ; show that
(^V+ (^V= «** + «** + 2e*+*siii(x + y).
15. u = log (tan x + tan y + tan z) ; show that
bu bu bu
sin 2 x H sin 2 y H sin 2 z — = 2.
dx by bz
16. Let y be the altitude of a right circular cone and x the radius of its base.
Show (a) that if the base remains constant, the volume changes  vx* times as fast as
the altitude ; (b) that if the altitude remains constant, the volume changes } vxy times
as f ast as the radius of the base.
x s y a
17. A point moves on the elliptic paraboloid z = — h — and also in a plane par
allel to the XOZplane. When x = 8 ft. and is increasing at the rate of 9 ft. per
second, find (a) the time rate of change of z ; (b) the magnitude of the velocity of
the point ; (c) the direction of its motion.
An*, (a) v a = 6 ft. per sec. ; (b) v = 8 Vl§ ft. per sec. ;
(c) r = arc tan f , the angle made with the XOFplane.
18. If, on the surface of Ex. 17, the point moves in a plane parallel to the plane
YOZ, find, when y = 2 and increases at the rate of 5 ft. per sec., (a) the time rate
of change of z ; (b) the magnitude of the velocity of the point ; (c) the direction of
its motion. Ana. (a) 5 ft. per sec. ; (b) 5 V2 ft. per sec. ;
(c) t = , the angle made with the plane XOY.
4
125. Total derivatives. We have already considered the differ
entiation of a function of one function of a single independent
variable. Thus, if
y=/(v) and v = <f>(z),
it was shown that
dy _ dy dv
dx dv dx
PAETIAL DIFFERENTIATION 195
We shall next consider a function of two variables, both of which
depend on a single independent variable. Consider the function
where x and y are functions of a third variable t
Let t take on the increment A£, and let Ax, Ay, Aw be the corre
sponding increments of £, y, u respectively. Then the quantity
Aw =/(a; + Aa^ y + Ay) /(a, y)
is called the totoZ increment of w.
Adding and subtracting /(a^ y + Ay) in the second member,
(A) At*=[/<>+As, y+Ay)/(s, y+Ay)] +[/(*, y+Ay)/(*, y)].
Applying the Theorem of Mean Value (46), p. 166, to each of
the two differences on the righthand side of (^), we get, for the
first difference,
(5)/(a ? + A^y + Ay)/(^y + Ay)=/,'(a: + ^.A^y + Ay)Aa;
fa« x, Aa= Ax, and since x varies while y + Ay remains]
[ constant, we get the partial derivative with respect to x.J
For the second difference we get
(C) /(*> V + Ay) /O, y) =/,'(*, y + 9 % • Ay) Ay.
fa— y, Aa« Ay, and since y raries while x remains oon1
[ stan t, we get the partial derivative with respect to y.J
Substituting (B) and (C) in (A) gives
(2>) Aw=/;(a; + ^.A^y + Ay)Aa:+/;(icy + ^.Ay)Ay,
where X and a are positive proper fractions. Dividing (2>) by At,
Now let At approach zero as a limit, then
[Since Ax and Ay converge to sero with At, we get "
aSJo/^ + V A*. V + **) /.'(a?, y), and JJf^/ f '<*. y + », Ay) //(*, y),
/■'(*» y) and// (x, y) being assumed continuous. J
Replacing /(:e, y) by u in (P), we get the total derivative
du __dudx dudy
196 DIFFERENTIAL CALCULUS
In the same way, if
a =/<>, », *),
and x, y, z are all functions of (, we get
du du dx Oudy audi
di = dxdi + dydi + dldt'
and so on for any number of variables."
In (51) we may suppose t~x; then y is a function of x, and u is
really a function of the one variable a, giving
du du Su dy
dx ~ dx dydx
In the same way, from (52) we have
du du du dy du dz
dx~dx + dy'dx + dz'dx'
(52)
(53)
(M>
du
, du.
The student should observe that 
meanings. The partial derivative — is formed on the supposition that
dx
the particular variable x alone varies, while
du_ limit /Am\
dx Az=0\Ax/'
where Am is the total increment of u caused by changes in all the vari
ables, these increments being due to the change Ax in the independent
variable. In contradistinction to partial derivatives, — i — are called
dt dx
fatal derivatives with respect to t and x respectively. f
•This is really only a special case of a general theorem which may he staled as follows:
If u Is a furiiiii.Ti of I lb' iurli'».'iii!<>i)! variables x, y, t,  , each of these in turn being a
function of the independent variables r, i, t, . . ., then {with certain assumptions a
continuity) ^ Su0 g u8l
It 6t
t It should be observed that — has a perfectly definite value for any point (i,y), while —
depends not only on the point (si, >/), but also on the particular direction chosen to reach tl
PARTIAL DIFFERENTIATION 197
Illustrative Examfle 1. Given u = sin  , x = c*, y = P ; find — .
y at
«. , ^ du 1 x du x x, dx . dy a .
Solution. — =  cos  , — =  cos  ; —  = e*. f = 2 <.
dx y y dy y* y dt dt
du t* €*
Substituting in (61), — = (t  2) ^ cos  • Ana.
dt ft 2
Illustrative Example 2. Given u = e ax (y — z), y = a sinx, z = cosx ; find — .
dx
du v du du dy dz
Solution. — = aeF* (y — z). — = c", — = — e«* ; = = a cosx, — = — sinx.
ax v ' dy dz dx ' dx
Substituting in (54),
du
— = ae ttX (y — z) + a* 1 * cosx + c^sinx = e ax (a* + l)sinx. 4n*.
dx
Note. In examples like the above, u could, by substitution, be found explicitly in
terms of the independent variable and then differentiated directly, but generally this
process would be longer and in many cases could not be used at all.
Formulas (51) and (52) are very useful in all applications involv
ing time rates of change of functions of two or more variables. The
process is practically the same as that outlined in the rule given on
p. 141, except that, instead of differentiating with respect to t (Third
Step), we find the partial derivatives and substitute in (51) or (52).
Let us illustrate by an example.
Illustrative Example 3. The altitude of a circular cone is 100 inches, and
decreasing at the rate of 10 inches per second ; and the radius of the base is 60 inches,
and increasing at the rate of 5 inches per second. At what rate is the volume changing ?
Solution. Let x — radius of base, y = altitude ; then u = ^ irz*y =
volume, — = incy, — = irx a . Substitute in (51), /
dx 8 dy 3 /
du 2 dx 1 9 dy / V\ \
dt 8 ctt 8 ctt / ! \
dx du A" — v ' "*ae~' s \
But x = 60, y = 100, ^ = 5, ^ =  10. e — 2 ~3
dt dt ^^ ^"^
.\ — = ir • 5000 • 6 ir . 2600 • 10 = 16.15 cu. ft. per sec, increase. Ana.
dt 8 8 *~ t
126. Total differentials. Multiplying (51) and (52) through by dt,
we get
f«, jl du . du.
(55) du = — dx + — dy,
dx dy
.** *. dU. dU. dU. m
(56) du = — dx\ — dy\ — dz;
K ' dx T dy ^ dz
and so on.* Equations (55) and (56) define the quantity du, which
is called a total differential o/wora complete differential,
, du , du , du ,
and —dx, —dy, —dz
dx dy dz
• A geometric interpretation of this result will be given on p. 266.
198
DIFFERENTIAL CALCULUS
are called partial differential*. These partial differentials are some
times denoted by dji, d f u, d t u, so that (5d) is also written
du = d x u + du + d,u.
Illusthative Example 1
Solution.
Substituting in (56),
Bx
x x + V* im x* + V s
_ xdy — ydx .
& + V* ""'
Illustrative Euiru 2. The base and altitude of a rectangle are 6 and 4 Inches
respectively. At a, certain instant they are increasing continuously at the rate of '1
itichea and 1 inch per second respectively. At what rate is the area of the rectangle
increasing at that instant?
Solution. Let x = base, y = altitude ; [hen u = xy = area, — = y, — = x.
dx &y
Substituting in (61),
... du dx dy
But
■i In,
dx
dy_
:=(e + 5)aq.i
= 13 sq. in. per sec. Aw.
Notb. Considering du aa an Infinitesimal increment of area due to the infinitesimal
increments dx and dy, du is evidently the sum of two thin strips added on W the two
sides. For, in du = ydx + xdy (multiplying {A) by dl),
ydx — area of vertical strip, and
xdy = area of horizontal strip.
But the total increment A« due to the incremented* and
dy Is evidently ^ u = y dx + xdy + dxdy.
Hence the small rectangle in the upper righthand corner
(= dxdy) is evidently the difference between An and du.
This Bgon illustrates the fact that the total increment and the total differential of a
function of several variables arc not in general equal.
127. Differentiation of implicit functions. The equation
GO /(*y)0
defines either x or y as an implicit function of the other,
sents any equation containing x and
transposed to the first member. Let
TO •/<* g):
Al du ehi du dy
then — = — H *
dx tix dy dx
* We aasume that a small change in the value of x causes only a small
It repre
when all its terms have been
(S3), p. 196
all change in the
PARTIAL DIFFERENTIATION 199
du,
But from (4), f(xj y) = 0. .\ u = and —  = ; that is,
dx
(<r> ^+^4y = o.
Solving for p»* we get ^
(57) * — * £*0
dr flu ay
a formula for differentiating implicit functions. This formula in
the form (C) is equivalent to the process employed in § 62, p. 69,
for differentiating implicit functions, and all the examples on p. 70
may be solved by using formula (57). Since
(i>) /(*. y) = o
for all admissible values of x and y, we may say that (57) gives the
relative time rates of change of x and y which keep f(x> y) from changing
at aH Geometrically this means that the point (x, y) must move on
the curve whose equation is (2>), and (57) determines the direction
of its motion at any instant. Since
*> =/(*> y),
we may write (57) in the form $f
dx df dy
dy
dy
Illustrative Example 1. Given x*y* + siny = 0, find p.
dx
8olntfon. Let/(x, y) = x*y* + sin y.
2£=2xy\ ^=4xV + cosy. .. f rom (57 a), ^ =  A **** ^ns.
dx ay * ^ * v "dx 4xV + cosy
Illustrative Example 2. 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 2xy 2 — 8x*y shall remain constant ?
Solution. Let /(x, y) = 2xy a — &z*y ; then
^=2y*6xy, ^=4xy3x a .
ax * "' dy
Substituting in (67 a), dy
fr = _2y'6xy £ = _2y'6xy B y (88), p. 141
dx 4xy8x* dx 4xy8x* J v " F
dt
But x = 8, y = 1, — = 2. .\ — = — 2A ft. per second. An*.
at dt
du du
• It is assumed that — and — exist.
dx dy
200
DIFFERENTIAL CALCULUS
Let P be the point (x, y, z) on the surface given by the equation
Off) u = F(x,y, s) = 0,
and let PC and AP be sections made by planes through P parallel to
the rOZand XOZplanes respectively. Along the curve AP, y is
constant; therefore, from (2£), z is an im z
plicit function of £ alone, and we have,
from (57 a),
dF
(58)
dz
dx~
dx
"dF
~dz
giving the slope at P of the curve AP, § 122, p. 190.
dz . . dz
— is used instead of — in the first member, since z was originally,
dx dx J
from (2£), an implicit function of x and y ; but (58) is deduced on the
hypothesis that y remains constant.
Similarly, the slope at P of the curve PC is
dF
dz ~dy
dy~~dF'
Tz
(59)
EXAMPLES
Find the total derivatives, using (51), (52), or (53), in the following six examples:
1. u = z a + y 8 + zy,z = sinx,y = e x . Ana. — = 3e** + extern 2 + cos 2) + sin2x.
dx
2. u = arc tan (xy), y = c*.
3. u = log (a* — p 2 ), p = a sin 0.
4. u = u 1 + vy, v = log «, y = e*.
5. u = arc sin (r — «), r = 8 £, a = 4 1 8 .
<4n*.
du _ e*(l + x)
dx 1 + xM x
— =— 2tan0.
d0
du _Zv + y t
— — "J"
(U
du 8
* Vlt*
— rre^sinx.
dx
c* (y — z)
6. u = — ^ ', y = asinx, z = cosx.
a a + 1
Using (55) or (56), find the total differentials in the next eight examples :
7. u = &y*x +cx* + gy* + ex. 4?w. du = (6y a + 2cx + e)dx + (2&y» + 8gy*)dy
8. u = log x*.
du =  dx + log xdy.
x
PARTIAL DIFFERENTIATION
Ana. du = y^" \og j/ cob xdx + 
 2 (adt — <f
=*
2U1
'•'{I/id* + zzlogxdy + xy logxdz).
14. u = tfti)'«tan 5 eiAn'^. du = 4u( — — +  
15. Assuming the characteristic equation of a perfect gas to be
where t> = volume, p = pressure, ( = absolute temperature, and R a constant, what ia
the relation between the differentials dt, dp, dt ? 4ns. vdp + pdD = Rdt.
16. Using the result in the last example as applied to air, suppose that in a given
cage we have found by actual experiment that
t = 300° C, p = 2000 lb. per aq. ft., o = 14.4 cubic feet.
Find the change in p, assuming it to be uniform, when ( changes to 301°C., and B
to 14.5 cubic feet, ft = 98. Am. 7.22 1b. per
ft.
17. One side of a triangle is 8 fl. long, and increasing 4 inches per second ; another
Hide is 5 ft., and decreasing 2 inches per second. The included angle is 00", and
increasing 2° per second. At what rate ia the area of the triangle changing ?
An». Increasing 71.05 s<. in. per sec.
18. At what rateia the side opposite the given angle in the last example increasing ?
Am. 4.1>
19. One aide of a rectangle is 10 in. and increasing 2 in. per sec. The other side
is 15 in. and decreasing 1 in. per sec. At what rate is the area changing at the end of
two seconds? Ann. Increasing 12 sq. in. per sec.
20. The three edges of a rectangular parallelepiped are 3, 4, 5 inches, and are each
increasing at the rate of .02 in. per min. At what rate ia the volume changing ?
21. A boy starts Hying a kilo. If it moves horizontally at the rate of 2 ft. a sec.
and rtseB at the rate of 5 ft. a aec, bow fast Is the string being paid out ?
22. A man standing on a dock is drawing in the painter of a boat at the rate of 2
ft. a sec. Ilis hands are 6 ft. above the 1k>w of the boat. How fast Is the boat moving
when it is 8 ft. from the dock ? An*, j ft. a sec.
23. The volume and the radius of a cylindrical boiler are expanding at the rate
of 1 ca. ft. and .001 ft. per min. respectively. How fast, is the length of the boiler
changing when the boiler contains tiO cu. ft. and has a radius of 2 ft. ?
Ana. .078 ft. a min.
24. Water is running out of an opening in the vertex of a conical filtering glass,
8 inches high and 6 inches across the top, at the rate of .005 on. in. per hour. How
fast is the surface of the water falling when the depth of the water is 4 inches ?
202
DIFFERENTIAL CALCULUS
26. A covered water tank is made of sheet iron in the form of an inverted cone
of altitude 8 ft. surmounted by a cylinder of altitude 5 ft. The diameter is 6 ft. If
the sun's heat is increasing the diameter at the rate of .002 ft. per inin., the altitude
of the cylinder at the rate of .003 ft. per inin., and the altitude of the cone at the rate
of .0026 ft. per minute, at what rate ie (a) the volume increasing ; (b) the total area
increasing ?
In the remaining examples find—, using formula (67 a):
ix
26. {x* + y*y
37. *»«■ +
28. sinflj,)
.<*(*
V*) =
JU. ^z
2(x* + y')
C* — z*y =
_ y[coa(xy)e*r—2x]
" z[x + e<*  co*{xy)] '
128. Successive partial derivative;;. Consider the function
/(*. y) ;
— and —
8x dy
then, in general,
are functions of both x and y, and may be differentiated again with
respect to either independent variable, giving successive partial deriva
tives. Regarding x alone as varying, we denote the results by
#u SV tfu
ftr 1 * dx*' dx*'
or, when y alone varies,
&u d\ a*w
ifu
the notation being similar to that employed for functions of a single
variable.
If we differentiate u with respect to x, regarding y as constant, and
then this result with respect to y, regarding x as constant, we obtain
dy\dxj
rhich we denote by
dydx'
Similarly, if we differentiate twice with respect to x and then once
with respect to y, the result is denoted by the symbol
PARTIAL DIFFERENTIATION
203
189. Order of differentiation immaterial. Consider the function
f(x, y). Changing x into x + Ax and keeping y constant, we get from
s Theorem of Mean Value, (46), p. 166,
(A) f(x +Ax, 30/0* y) = Ax ./j (x + 6  Ax, y). 0<8<\
If we now change y to y+Ay and keep a; and Ax constant, the
total increment of the lefthand member of (A~) is
1(B) [/( a: +Ax,y+Ay)/(x,y+Ay)][/( a: + Ax,y)/(x,y)].
The total increment of the righthand member of (^) found by the
Theorem of Mean Value, (46), p. 166, is
(C) Ax/J (x +  Ax, y + Ay)  Ax/j (x + B Ax, y) < <?,< 1
= AyAx/£(x+.9   Ax, y + tV Ay). 0<^<1
[a,. A* A*. „„.l *.«. i. vtrl* «hl1» x«nd it Mnuinl
[eonetutt. we get the pirtiel derivative wilt respect toy. J
Since the increments (£) and (C) must be equal,
(0) [/(x+Az, »+<>»)/(* jr+Ajr)][/(»+A«i »>/(•!»)]
= AyAx/£ (a: + 0, . Aat, y + 6 % Ay).
In the same manner, if we take the increments in the reverse order,
(E) t/(*+A*Jl + 4j()/(* + 4i,jr)][/(>4»+Ajt)/(*Jl)]
= AxbyfZ (i + 1, A*, y + 0, • Ay),
flj and d t also lying between zero and unity.
Tile lefthand members of (Z)) and (r?) being identical, we have
»(') /i (* + «, • Ai, y + 8, • Ay) /J (x + 9, • Ar, y + », Ay).
T,
Taking the limit of both sides as A,c and Ay approach zero as limits,
b have
(G) .C (**)=.£(* 30.
since these functions are assumed continuous. Placing
«=/(*, y).
( G) may be written
(60)
Sydx Bxdy
That is, (A* operation* of differentiating with rctpect to x and with
•d to y are commutative.
204 DIFFERENTIAL CALCULUS
This may be easily extended to higher derivatives. For instance,
since (58) is true,
dx*dy~ dx\dxdy)~ dxdydx dxdy\dx) dydx\dx) dydx**
Similarly for functions of three or more variables.
Illustrative Example 1. Given u = x*y — 8 xV ; verify
dydx dzdy
Solution. — = 8x*y  6xy», ^— = 8x«  18xy«,
dx dydx
— = x«  9x*y» ?— = 3x*  18xy* ; hence verified.
dy ' axdy
1. u = cos (x + y) ; verify
2. u = ^ ; verify
y 2 — x 2
3. u = ylog(l + xy); verify
4. u = arc tan; verify
8
5. u = sin (0*0) ; verify
6. u = 6#ty 2 z + 3e»x*z* + 2e*xfy — xyz : show that " " = 12(e*y + e*s + rtc).
dx*dydz
d*u
7. u = e*** : show that = (1 + 8xyz + xhfl&Su.
dxdydz
~ xhi* , . 5*u a*tt ^0u
8. u = — — : show that x —  + y = 2 — .
z + y dx 1 dzdy dx
9. u = (x* + y*)i; show that 8x^ + &y?^ + — = 0.
v ' dxay ay« ay
? ? * a«u ? ? f
10. u = j^ a e 2 + «*x a e! a + x*y 2 e a : show that — — — —  = & + ^ + ^.
dx*0y a 0z*
11. u = (x 2 + y* + **)*; show that^J + ^ + ^ = 0.
v ; ax* ay* dz*
dydx
dJ%
d*u
d*u
dydx
ax^y
8*u
a*u
dydx
axay
&u
a»u
dr*ds
a*ar*
&u
a»u
m<p
dtpdB
how th:
d*u
At
CHAPTER XVI
ENVELOPES
130. Family of curves. Variable parameter. The equation of a
curve generally involves, besides the variables x and y, certain con
.ts upon which the size, shape, and position of that particular
 e depend. For example, the locus of the equation
W (*«)■ + »■=>*
is a circle whose center lies on the axis of X at a distance of a from
the origin, its size depending on the radius r. Suppose a to take on
a series of values ; then we shall have a
corresponding series of circles differing
in their distances from the origin, as
shown in the figure.
Any system of curves formed in this
way is called a family of curves, and the
quantity a, which is constant for any one curve, but changes in pass
ing from one curve to another, is called a variable parameter.
As will appear later on, problems occur which involve two or more
parameters. The above series of circles is said to be e. family depending
on one parameter. To indicate that a enters as a variable parameter it
8 usual to insert it in the functional symbol, thus:
f(x,y,a)=0.
131. Envelope of a family of curves depending on one parameter.
The carves of a family may be tangent to the same curve or groups
of curves, as in the above figure. In that case the name envelope of
the family is applied to the curve or group of curves. We shall now
explain a method for finding the equation of the envelope of a family
of curves. Suppose that the curve whose parametric equations are
W « = *(«), y = >K<0
touches (i.e. has a common tangent with) each curve of the family
(*) /O,y,a) = o,
805
206
DIFFERENTIAL CALCULUS
the parameter a being the same in both cases. The slope of (.4) at
any point is
and the slope of (B) at any point is
<*>>
(57 a), p. 199
Hence if the curves (^) and (B) are tangent, the slopes (C) and
(Z>) will be equal (for the same value of a), giving
f(") = /to*") or
fCO. /JO**)
(J) /[(*, y, <r) *'(«) +/•(* y, a) *'(*) = 0.
By hypothesis (.4) and (B) are tangent for every value of a ; hence
for all values of a the point (x, y) given by (v4) must lie on a curve
of the family (B). If we then substitute the values of x and y from
(A) in (B), the result will hold true for all values of a ; that is,
(F) /[*<*), +(«),*] = 0.
The total derivative of (7') with respect to « must therefore vanish,
and we get
C«) /;<>. J. *) +'« +/;(*• ft O +'(«) +/K* ft «>" *.
where z = <£ (a), y = ^ (a).
Comparing (£) and (G) gives
Therefore the equations of the envelope satisfy the two equations
(B) and (H), namely,
(7) /(*, y, «) = and /[(* y, «) = ;
that is, the parametric equations of the envelope may be found by
solving the two equations (7) for x and y in terms of the parameter a.
General directions for finding the envelope.
First Step. Differentiate with respect to the variable parameter, con
sidering all other quantities involved in the given equation as constant*.
Second Stkp. Solve the result and the given equation of the family of
curves for x and y in terms of the. parameter. These solutions will be the
parametric equations of the envelope.
ib,
ENVELOPES
207 1
Note. In cate the rectangular equation of the envelope t» required we
may eitker eliminate the parameter from the parametric equation* of the
envelope, or else eliminate the parameter from the given equation (B~) of
the family and the partial derivative (.ff).
Illcstbativb Example 1. Find the envelope of the family of straight lines
zcaaa + ysiaa — p, a being the variable parameter.
Solution. (A) I con a + ya\na = p.
Fir* Hep. Differentiating (A) with respect to a,
<fl) xsina + ycoaa = 0.
Second step. Multiplying {A) by coatz and (B) by sin a and subtracting, we get
Similarly, eliminating x between (A) and (B)
y = p sin a.
e equation! of the envelope are therefore
a being the parameter. Squaring equations (C*) ami add
ing, we get 2* + **=^,
the rectangular equation of the envelope, which la a circle.
s 2. Find the envelope of a linf
o fixed rectangular axes.
XjUMFI
extremities move along t"
Solution. Let AB = a in length, and let
{A) scosa + vsiuarp =
be its equation. Now as AB moves always touching the two axes, both t
will vary. But p may be found in terms of a. For AO = AB
p = JOsina = asinacosit. Substituting In (A),
(B) i cos a 4 y sin a  a sin a cos or = 0,
where a U the variable parameter. Differentiating (it)
witb respect to a,
(Q — x sin a + y cos a + a sin 1 a
Solving (B) and (C) for x and y in
W
I the parametric equations of the envelope, a. hypocycloid.
The corresponding rectangular equation is found from equatioi
la follows i , ,
yi = ai coo* a.
I Adding, jA + yr = a!,
I rectangular equation of the hypocycloid.
Iu.ustbative Example 1. Find the rectangular equation of the envelope of the
traight lino y = m& + — , where the slope m is the variable parameter.
«k
.Substitute in the given e.iuatin
* = ±\l*±\yP = ±*'fa
and squaring, j/ ! = 4jjj;, a parabola, is the equation of the envelope. The family of
straight lines formed by varying the slope m is shown in the figure, each line being
tangent to the envelope, for we know from Analytic Geometry that y — mi + — ia
the tangent to the parabola y 1 — 4 pi expressed in terms of its own slope m.
132. The evolute of a given curve considered as the envelope of its
normals. Since the normals to a curve are all tangent to the evolute,
§ 118, p. 181, it ia evident that the evolute of a
curve may also he defined as the envelope of its
normals; that is, as the locus of the ultimate
intersections of neighboring normals. It is also
interesting to notice that if we find the para
metric equations of the envelope by the method
of the previous section, we get the coordinates
x and y of the center of curvature ; so that we
have here a second method for finding the a
dinates of the center of curvature. If we then elk
parameter, we have a relation between x and y which is the rectan
gular equation of the evolute (envelope of the normals).
Illustrative Example 1. Find the evolute of the parabola j> = ipx considered
as the envelope of its normals.
Solution. The equation of the normal at any point (x\ if) is
ate the variable
from (2), p. 77, As we are considering the normals all along the curve, both x' and y"
will vary. Eliminating!' by meansofy' < = 4pz', we get the equation of the normal to be
Considering y' as tba variable parameter, we wish t
family of normals. Differentiating (A) with respect to y'
mil solving for x,
(B) * = >*1±M.
*P
Substituting this value of x in {A) and solving for y,
4p*
(?) and (IV) are then tbe coordinates of the center of curvature of the parabola.
Taken together, (ii) and (C) are the parametric equations of tbe evolute In terms of
the parameter y'. Eliminating y' between (B) and (C) gives
27py* = *(x2p)»,
the rectangular equation of the evolute of the parabola. This is the same result we
obtained in Illustrative Example 1, p. 183, by the first method.
133. Two parameters connected by one equation of condition. Many
problems occur where it ia convenient to use two parameters con
nected by an equation of condition. For instance, the example given
in the last section involves the two parameters a/ and y' whicli are
connected by the equation of the curve. In this case we eliminated
£, leaving only the one parameter y'.
However, when the elimination is difficult to perform, both the
given equation and the equation of condition between the two param
eters may be differentiated with respect to one of the parameters,
regarding either parameter as a function of the other. By studying
the solution of the following problem the process will be made clear.
coincide and whose a
Solution. (A) ^ + j£= 1
ia the equation of tbe ellipse where a and
b are the variable parameters connected by
tbe equation
(B) rab = k,
rob being the area of an ellipse whose semi
aies are a and b. Differentiating [A) and
(II), regarding a and b as variables and x and
V as constants, we have, using differentials,
x*da y s db
mi
= 0, from (4),
bda + adb = 0, from (0).
DIFFERENTIAL CALCULUS
Transposing one term In each U> the second member and dividing, <
* = £
Therefore, from {A), — =  and — = .
giving a = ± x Vi »nd 6 = ± y V5.
Substituting these values in (B), we get the envelope
a pair of conjugate rectangular hyperbolas (see last figure).
EXAMPLES
1. Find the envelope of the family ol straight lines y = ims + n
variable parameter. An*. x=8m», y =— Sm"; or 18 y* + 27z« = 0.«
2. Find the envelope of the family of parabolas y s = a (x — a), a being the
variable parameter. Ana. z  2 a, y = ± a ; or jr = ± \ z.
3. Find the envelope of the family of circles z* + {]/ — £)* = r*, p being the
variable parameter. Ant. X = ± r.
4. Find the equat ion of the curve having as tangents the family of straight lines
y = mx ± Va'nt' + b>, the slope m being the variable parameter.
Ant. The ellipse oV + oV = aV,
6. Find the envelope of the family of circles whose diameters are double ordi
nate of the parabola y' = 4px. 4ns. The parabola y 1 = 4 p {p + x).
6. Find the envelope of the family of circles whose diameters are double ordl.
nates of the ellipse Vz* + b¥ = o*ft*. . _. ,„ i 1 y* ,
*^ T * 4n». The ellipse 1 — = 1.
7. A circle moves with its center on the parabola y* = 4 az, and its circumference
passes through the vertex of the parabola. Find the equation of the envelope of the
circles. Ant. The cissoid jr* (x + 2 a) + *• = 0.
8. Find the carve whose tangents are y — tx ± y/aP + W + c, the slope I being
supposed to vary. Ant. Hay' + toy + cz 1 ) — iac — 6".
9. Find the evotute of the ellipse JrV + aV o 1 **. taking the equaUon of nor
mal in the form Jy = <wtan#(a'b')sin*,
the eccentric angle $ being the parameter.
a i _ fii &i _ a * «
Ant. z = cos 1 *, y =   sin" * ; or (az)* + ((*)* = (a*  6 1 )*.
10. Find the evolute of the hypocyclold A + y* = a', the equation of whose
normal is „cost zsinT = «cob2t,
t being the parameter. Am. (x + y)i + (x — y)I = 2 a*.
* When two answers are given, the first is in parametric form and the second in rec
ENVELOPES 211
11. Find the envelope of the circles which pan through the origin and have
their centers on the hyperbola x s — y 8 = c 1 .
Ana. The lemniscate (x* + y 9 )* = 4e* (* f — *■).
12. Find the envelope of a line such that the sum of its intercepts on the axes
equals c. Ana. The parabola x T + y T = c T .
13. Find the equation of the envelope of the system of circles x 1 + y* — 2 (a + 2) x
+ o 1 = 0, where a is the parameter. Draw a figure illustrating the problem.
Ans. y* = Ax.
14. Find the envelope of the family of ellipses 6*x* + a*y* = aty*, when the sum
of its semiaxes equals c. Ana. The hypocycloid x* + y* = c*.
15. Find the envelope of the ellipses whose axes coincide, and such that the dis
tance between the extremities of the major and minor axes is constant and equal to I.
Ana. A square whose sides are (x ± y)* = P.
16. Projectiles are fired from a gun with an initial velocity v . Supposing the gun
can be given any elevation and is kept always in the same vertical plane, what is the
envelope of all possible trajectories, the resistance
of the air being neglected ?
Hint. The equation of any trajectory is
yztan<r— f
2 00 cos* <r
a being the variable parameter. a
Ana. The parabola y = ^ — ¥—*•
17. Find the equation of the envelope of each of the following family of curves,
t being the parameter ; draw the family and the envelope :
(a) (x 1 + y* = 1  f". (i) (x t)« + y* = 4*.
0>) x* + (y0*=2*. (j) x* + (y9* = 40.
(c) (x9« + y« = lt»l. (k) (x9« + (y9« = f».
(d) x« + (yt)« = Jt». (1) (*0 f + (r + 9 f = *.
(e) y = tx + **. (m) y = Px + t.
(f) x = 2*y + i*. (n) y = t(x2€).
(g)y = (x + l. (o)x = £ + f.
t t
(h) y» = t(x + 2t). (p) (x 9* + 4y« = t.
I
CHAPTER XVH
SERIES
134. Introduction. A series is a succession of separate numbers
which is formed according to some rule or law. Each number is
called a term of the series. Thus
I) M) 4) O) * * *) **
is a series whose law of formation is that each term after the first is
found by multiplying the preceding term by 2 ; hence we may write
down as many more terms of the series as we please, and any particu
lar term of the series may be found by substituting the number of that
term in the series for n in the expression 2"~\ which is called the
general or nth term of the series.
In the following six series :
(a) Discover by inspection a law of formation ;
(b) write down several terms more in each ;
(c) find the nth or general term.
Series nth term
1. 1, 3, 9, 27, ... 8»i.
2. — a, + a*, — a 8 , + a 4 , • • • . (— a)".
3. 1, 4, 9, 16, ... n*.
A X* X* X 4 x»
4x ' T T T "" n'
5. 4, 2, +1, J, .... 4(})i.
Sy 5y» 7y» 2n + l
2 6 10 n f + l
Write down the first four terms of each series whose nth or general term is given
below :
nth term Series
7. n*x». x, 4x f , Ox 1 , 16 x*.
x"
S.
1+Vn
n + 2
9 *n» + l
212
X
X*
2
.8
X*
2'
3
2'
1+V2
4 5
9' 28*
1 +
6
66*
V§'
l+Vi
SERIES 213
n 12 8 4
10, » I' V §' 16'
ii (logo)*!* logax log 8 a.x 2 log 8 a.x 8 log 4 a«a5 4
\n 1 2 6 24
io (l) 1 * 2 " 2 1 x* x* z«
j2nl 1 [3 [ [7
135. Infinite series. Consider the series of n terms
,A\ 1 1 1 1 1
<A) . 1» g' i' 8' • ¥^ ;
and let S n denote the sum of the series. Then
(B) £=1 + 1 + 1 + 1.] + _i —
v J "248 2"" 1
Evidently S n is a function of n, for
when n = l, & = 1 =1,
1
when w = 2, S s =l{ = 1£,
when n = 3, S 8 =l + 1 + 1 = 1,
whenn=*4, Jr 4 l +  + j +  If,
when n = n, 3=1 +  +  +  + ... +_ =2^.*
Mark off points on a straight line whose distances from a fixed
point correspond to these different sums. It is seen that the point
' k HHK
corresponding to any sum bisects the distance between the preceding
point and 2. Hence it appears geometrically that when n increases
without limit limit S n = 2.
We also see that this is so from arithmetical considerations, for
limit a = limit (^ __ _JL_\ = 2 t
w = oo " n = oo\ 2"" 1 /
I Since when n increases without limit — — approaches aero as a limit. I
* Found by 6, p. 1, for the sum of a geometric series.
t Such a result is sometimes, for the sake of brevity, called the sum of the series ; but
the student must not forget that 2 is not the sum but the limit of the sum, as the number of
terms increases without limit.
214 DIFFERENTIAL CALCULUS
We have so far discussed only a particular series (J) when
number of terms increases without limit Let us now consider
general problem, using the series
(C) «,. «, «, «.. .
whose terms may be either positive or negative. Denoting by S t the
Bum of the first n terms, we have
S.= u i +u J +u J + . .. + «„,
and S m is a function of n. If we now let the number of terms (= n]
increase without limit, one of two things may happen : either
Case I. S„ approaches a limit, say «, indicated by
Ca.se II. S„ approaches no limit.
In either case (C) is called an infinite serins. In Case I the infinite
series is said to be convergent and to converge to the valve u, (
the value «, or to have the sum «. The infinite geometric series <
cussed at the beginning of this section is an example of a convergeni
series, and it converges to the value 2. In fact, the simplest example
of a convergent series is the infinite geometric series
a, ar, at*, ar*, ar*, •■,
where r is numerically less than unity. The sum of the first n t
of this series is, by 6, p. 1,
a(lr) _ a «r_
l r l_ r l_ r '
If we now suppose n to increase without limit, the first fraction
the righthand side remains unchanged, while the second approaches
zero as a limit. Hence
& =
limit
" ~1
a perfectly definite number in any given case.
In Case II the infinite series is said to be noncvnvergent,* Series
under this head may be divided into two classes.
First Class. IHvergent series, in whieh the sum of n terms increae
indefinitely in numerical value as n increases without limit ; take f
example the series
* Same writers use divergent u equivalent to nonconeergent.
SERIES
215
As n increases without limit, S, increases without limit and there
fore the series is divergent.
Sxcond Class. Oscillating series, of which
is an example. Here <S„ is zero or unity according as n is even or odd,
and although S t does not become infinite as n increases without limit,
it does not tend to a limit, but oscillates. It is evident that if all the
terms of a series have the same sign, the series cannot oscillate.
Since the sum of a converging series is a perfectly definite number,
while such a thing as the sum of a nonconvergent series does not ex
ist, it follows at once that it is absolutely essential in any given prob
lem involving infinite series to determine whether or not the series is
convergent. This is often a problem of great difficulty, and we shall
consider only the simplest cases.
136. Existence of a limit. When a series is given we cannot in
general, as in the case of a geometric series, actually find the number
which is the limit of S h . But although we may not know how to
compute the numerical value of that limit, it is of prime importance
I to know that a limit does exist, for otherwise the series may be non
convergent. When examining a series to determine whether or not it
is convergent, the following theorems, which we state without proofs,
are found to be of fundamental importance.*
Theorem I. If S m is a variable that always increases as n increases,
but always remains less than some definite fixed number A, then as n
increases urithout limit, S n will approach a definite limit which is not
greater than A.
Theorem II. If S, is a variable that always decreases as n increases,
but always remains greater than some definite fixed number B, then as n
increases without limit, S^ will approach a definite limit tr/u'rh is not less
than B.
Theorem III. The necessary and sufficient condition that S, shall
approach some definite fixed number as a limit as n increases without
limit is that ,..,
„ = „(£.+, 3.)=°
Jor all values of the integer p.
See Osgood'* Introtluct'ton to Infinite Seria, pp. 4, 14, 6i
210 DIFFERENTIAL CALCULUS
137. Fundamental test for convergence. Summing up first n i
then n + p terms of a series, we have
OO S,= «, + ",+ ",+ •■■ + «,
(#) "S. + ,= «,+ ",+ ",+ ■■• + «.+ «. + ! + "• + *W
Subtracting (.4) from (B),
(C) S, +p 5,= «, + 1 + « 11 + a ++ Mll+p .
From Theorem III we know that the necessary anil sufficient eondi
tion that the series shall be convergent is that
M u ^(5. t/ ,5,)=0
for every value of p. But this is the same as the lefthand mer
ber of (C) ; therefore from the righthand member the conditi
may also be written
0°) J = » C", +1 + H . + »+ • • + *.♦,) = o
Since (/>) is tnie for every value of p, then, letting p = 1, a necessary
condition for convergence is that
„"(».. .) = <»;
or, what amounts to the same thing,
Henee, if the general (or nth) term of a series does not apprc
zero as n approaches infinity, we know at once that the series is r,
convergent and we need proceed no further. However, (£') is n
sitfficit'nt condition ; that is, even if the nth term does approach zett
we cannot state positively that the series is convergent ; for, eonsidt
the harmonic series 111 1
Here
^0O = n = «() = 0;
that is, condition (J?) is fulfilled. Yet we may show that the harmoni
series is not convergent by the following comparison :
GO i + l + [l + i] + [t + t + t + H+[l + '"A] + "
C) l + iKl + HKl + l + l + U+U. +■■•+&]+•
SERIES 21T
We notice that every term of ((r) is equal to or less than the cor
responding term of (2^), so that the sum of any number of the first
terms of (J?) will be greater than the sum of the corresponding terms
of (£). But since the sum of the terms grouped in each bracket in
(6r) equals £, the sum of (6?) may be made as large as we please by
taking terms enough. The sum (6r) increases indefinitely as the num
ber of terms increases without limit; hence (6r), and therefore also
(jP), is divergent.
We shall now proceed to deduce special tests which, as a rule, are
easier to apply than the above theorems.
138. Comparison test for convergence. In many cases, an example of
which was given in the last section, it is easy to determine whether or
not a given series is convergent by comparing it term by term with
another series whose character is known. Let
be a series of positive terms which it is desired to test for convergence. If
a series of positive terms already known to be convergent, namely,
can be found whose terms are never less than the corresponding terms in
the series (A) to be tested, then (^4) is a convergent series and its sum
does not exceed that of (E).
Proof. Let *„ = w 1 +w 2 +w g H \u n ,
and S n = a x + a 2 + a g H \a n ;
and suppose that J*™ 1 ^ S m = A.
Then, since £»<^ *&& *»—£»»
it follows that s H < A. Hence, by Theorem I, p. 215, s n approaches a
limit ; therefore the series (4) is convergent and the limit of its sum
is not greater than A.
Illustrative Example 1. Test the series
(C) i + i + I + .L + _L + ....
Solution. Each term after the first is less than the corresponding term of the geo
metric series
(D) 1 + 1 + I + 1 + 1 + ...
which is known to be convergent; hence (C) is also convergent.
218
DIFFERENTIAL CALCULUS
Following a line of reasoning similar to that applied to (^) and
(5), it is evident that, if
(E} « 1 + m 1 +w,H
it a teriet of pontine termt to be tested, which are never lett than
corresponding termt of the teriet of positive termt, namely,
(/") *!+&,+ &, + •.
hwum, to be divergent, then (£) is a divergent teriet.
Illustrative Example 2. Test the scries
VS V5 VI
Thiss.
s is divergent, since its terms are greater than the corresponding
terms of the harmonic series
i+l+j+j...,
which is known (pp. 216, BIT) to be divergent.
Illustrative KiurLS 3. Test the following series (called the p scries) for d
ferent values of p i ill
(O) '+S+S+5+V
Solution. Grouping the terms, we have, when p > 1,
1' , 1
*si<^. + «
1
2p 8» Sp 2p 2p 2p
I + I + ± + ±<± + ±
iP 6* 8" IP *p iP
, Construct the series
if ir 4c \2*V
m
•♦*£.♦ fcy*fcy'
When p > 1, series (/if) is .i geometric serum with the common ratio less than unity,
and is therefore convergent. But the sum of {(,') is leas than the sum of (/f), as shown
by the above imiuatitii:«; therefore (G) is also convergent.
When p = 1, series (G) becomes the harmonic series which we saw was divergent,
and neither of the above tests apply.
When p < 1, the terms of series (G) will, after the first, be greater than the a
spending terms of the harmonic series; hence (G) is divergent.
139. Cauchy's ratio test for convergence. Let
00 «,+«,+*,+•■
be a series of positive terms to be tested.
219
Divide any general term by the one that immediately precedes it ;
l form the test ratio ^ •
limit m. +i
As n increases without limit, let
" = />■
I. When p<\. By the definition of a limit (§ 13, p. 11) we can
choose n so large, say n = m, that when n 3B M the ratio s^ shall
differ from p by as little as we please, and therefore be less than a
proper fraction r. Hence
I«„ + i<'V; "„,,< v m¥l r<nj*i »„ ta <«./;
ind so on. Therefore, after the term «„, each term of the series (A)
is less than the corresponding term of the geometrical series
(B) , v .+ « i /+ H y + ....
But since r < 1, the series (B), and therefore also the series (A),
is convergent.*
II. When p > 1 (or p = oo). Following the same line of reasoning
as in I, the series (.I) may be shown to be divergent.
III. When p = l, the series may be either convergent or divergent;
that is, there is no test For, consider the p series, namely,
1
+ ...
The test ratio is 
(» + !)'
md limit /w^A limit A L_Y(iyl(«o>
Hence /> = 1, no matter what value p may have. But on p. 218 we
showed that t ^ , .i_ • j
when p > 1, the series converges, and
when;* s 1, the e
s diverges.
Thus it appears that p can equal unity both for convergent and for
divergent series, and the ratio test for convergence fails. There are other
tests to apply in cases like this, but the scope of our book does not
admit of their consideration.
* When examining a Berlea (or convergence we are at liberty to disregard an; finite
tnber of terms; the rejection of aucli terms would affect the valut but not
oi the limit.
!20 DIFFERENTIAL CALCULUS
Our results may then be stated as follows :
Given the series of positive terms
find the limit „ = „( „) = ''•
I. When p < 1,* the series is convergent.
II. When p > 1, the series is divergent
Ill. When p — 1, there is no test
140. Alternating series. This is the name given to a series whose
terms are alternately positive and negative. Such series occur fre
quently in practice and are of considerable importance.
If U,— "a + w,— w,H
is an alternating aeries whose terms never increase in numerical value,
and if ^^ «„ m 0,
then the series is convergent.
Proof, The sum of 2 n (an even number) terms may be written in
the two forms
00 fil.(« t O+(« l t0+C' B .«0+* K«i.i*0. or
(B) &,„= «,(«,«,)(«, w 6 ) «,„.
Since each difference is positive (if it is not zero, and the assump
tion °' ^ w h = excludes equality of the terms of the series), series (A)
shows that <£,„ is positive and increases with «, while series (.#) shows
that S. ln is always less thaiiM,; therefore, by Theorem I, p. 215, 5 a .must
approach a limit less than u t when n increases, and the scries is convergent.
Illustrative Example 4. Test the alternating series 1 ( 1 .
Solution. Since each term i« lens in immrrirn] value tliitu tin pitceiling one, and
limit , , _ limit /
le series is convergent.
141. Absolute convergence. A i
= 0,
said to be absolutely or
unconditionally convergent when the series formed from it by making
all its terms positive is convergent Other convergent series are said
• It is not enough that i<» + i/Un becomes and remains less than nutty tor nil values of n.
but this test requires that the limit of u„ + 1/" shall be leas than unity. For instance, in the
ease of the harmonic scries this ratio la always less than unity ami yet the series diverges aa
we have seen. The limit, however, is not less than unity but equals unity.
t The terms of the new aeriea are the numerical (absolute) values of the terms of the
given series.
SERIES 221
to be not absolutely convergent or conditionally convergent, To this
latter class belong some convergent alternating series. For example,
the series 1111
is absolutely convergent, since the series (C), p. 217, namely,
r 2 a 3' 4 4 5 6
is convergent. The series
2^3 4^5
is conditionally convergent, since the harmonic series
1 + + +  +  + 
is divergent.
A series with terms of different signs is convergent if the series deduced
from it by making all the signs positive is convergent.
The proof of this theorem is omitted.
Assuming that the ratio test on p. 219 holds without placing any
restriction on the signs of the terms of a series, we may summarize
our results in the following
General directions for testing the series
tt x+ w a+ w »+ w 4^ — y w »+ w »+i ^ — •
When it is an alternating series whose terms never increase in numer
ical value, and if limit .. _ n
then the series is convergent.
In any series in which the above conditions are not satisfied, we deter
mine the form of u n and u n+l and calculate the limit
limit M. + A
n==Q0 \ K )
I. TFA^np< 1, the series is absolutely convergent
II. When \p\> 1, the series is divergent
III. When p = l, there is no test, and we should compare the series
with some series which we know to be convergent, as
a + ar + ar*+ ar*\ ; r < 1, (geometric series)
1 + '& + ]T> + 4J> + ''' ; p>1% (^series)
222 DIFFERENTIAL CALCULUS
or compare the given series with some series which is known to be
divergent, as ., 1 1 1 ,. . N
1l._ +  + t I • (harmonic series)
2 8 4
1 + & + & + '& + '" ; ! ><1  (^series)
Illustrative Example 1. Test the series
1111
1 +
E + [? + [I + tI + ""
Solution. Here u» = , u m + i= r .
In — 1' w
. limit/u» + i\ limit/ » \_ limit / l»nl \_ limit/l\ 0/ _ .*
• • n = ooVUrV " n = oo I ~ I " n = *>\W ) ~ n = * W " "
and by* I, p. 221, the series is convergent.
II 12 [8
Illustrative Example 2. Test the series  + — + tt= + • • ••
10 10* 10*
In ln + 1
8olatioii. Here u„ = — , u„ + i = — .
. limit/u» + i\ limit /llL±i v 10»\ limit/n + 1\ . .
••n==4^n=«^isrri x ^n=4ir;" oo <^
and by II, p. 221, the series is divergent.
Illustrative Example 8. Test the series
(C) — + — + _L + ....
v ; 12 84 66
1 1
8otation. Here u„ =  tttt 9 w»+i =
(2nl)2n T (2n + l)(2n + 2)
. limit / u» + i\ _ limit [ (2nl)2n "I = oo
" n = fl0 \ u* / n = oo[(2n + l)(2n + 2)J oo"
This being an indeterminate form, we evaluate it, using the rule on p. 174.
Differentiating, limit (*±1*) = 2 .
^ n = oo\8n + 6/ oo
Differentiating again, ^^ (?) = 1 ( = p).
This gives no test (III, p. 221). But if we compare series (C) with (0), p. 218,
making p = 2, namely,
(D) l + 2_ + A+J_ + ... f
v ' 2 s 8* 4 1
we see that (C) must be convergent, since its terms are less than the corresponding
terms of (D), which was proved convergent.
SERIES
228
Show that the following ten aeries are convergent t
1 f« + 5 + i + —
2.1 + 1+1 + 1 + ..
22*2*2*
3. — + — + — +
12 84 56
M 1 18 1.86
4 tl + rr + zrzrz +
6. 1 + * +i + _L +
2V2 8V5 4Vi
7.12+II + I...
j? T ^ 7* 0*
ft 1 11,11 11
Q # — ^ — • — . 4 — • — . — — » • •— 4 <
2 2 2*^8 2* 4 2*
5.
8 8*6 8*6*9
*jl X m J 4.
B + Il + l! + '
9.
1
log 2
log8 +
1
log 4
10.
*♦
2 + 1 +
8*^4*^
• • ••
Show that the following four series are divergent :
n. 1 + 1 + 1 + ....
12.1 + 1+1 + 1+1 + 1+1 +
T l + 2» l + 8» 1 + 4» T
18 i+i+l+
14  1 +5+o+ 1 7+
• • •,
142. Power series. A series of ascending integral powers of a vari
able, say x, of the form
where the coefficients a , a 1% a 2 , • • • are independent of x, is called a
fewer series in x. Such series are of prime importance in the further
study of the Calculus.
In special cases a power series in x may converge for all values of x,
but in general it will converge for some values of x and be divergent for
other values of x. We shall examine (A) only for the case when the
coefficients are such that r ., / \
limit l a *+\ \ = L
n = Q0 \<W '
where L is a definite number. In (<4)
limit K +1 \ = limit ( a n +i** +1 \ = limit /a.+A m
x
Lx.
Referring to tests I, II, III, on p. 221, we have in this case p = Lx,
and hence the series (A) is
I. Absolutely convergent when  Lx  < 1, or  x \ <
II. Divergent when  Lx \ > 1, or  x  >
III. No test when 1X2*1 = 1, or 1^1 =
1
L
224
DIFFERENTIAL CALCULUS
We may then write down the following
General directions for finding the interval of convergence of the
power series,
(A) %+ a x x + aj?+ a % a?l .
First Step. Write down the series formed by coefficients, namely,
a +a 1 +a a +a g +... + a ll +a II+1 + ...
Second Step. Calculate the limit
limit / a »+A _
L.
Third Step. Then the power series (A) is
I. Absolutely convergent for all values of x lying between
and +
1
L
1
L
II. Divergent for all values ofx less than
III. No test when x = ±
1
L
or greater than +
1
L
; but then we substitute these two values of
x in the power series (A) and apply to them the general directions on p. 221.
Note. When Z = 0, ± y=±oo and the power series is absolutely
convergent for all values of x.
Illustrative Example 1. Find the interval of convergence for the series
(B)
X a . X 8 X 4 .
2 » T fca 4a T
2 a T s» 4» T
Solution. First step. The series formed by the coefficients is
(C)
Second 4«. 1,mit (±±l) = limlt [ *_1 = £.
Differentiating, ^ (_ _!i_) =
Differentiating again, i^«("" )=""" * (= X >
oo
00
T/itrd step.
1
1
= 1.
By I the series is absolutely convergent when x lies between — 1 and 4 1.
By II the series is divergent when x is less than — 1 or greater than + 1.
By III there is no test when x = ± 1.
SERIES 225
Substituting z = 1 in (B), we get
11 + .UI + ...
2* 8* 4*
which ia an alternating series that converges.
Substituting z = — 1 in (£), we get
2* 8* 4* "'
which is convergent by comparison with the p series (p > 1).
The series in the above example is said to have [—1, 1] as the interval of conver
gence. This may be written  1 5 x § 1, or indicated graphically as follows :
1
For what values of the variable are the following series Graphical representation* of
convergent ? intenralB o£ 7™**™ *
16. 1 + x + x* + x* + ... Ans. — 1<x<1.
t
+1
*4
16. z r — — — + •••• Ans. — 1< x ^ 1.
2 3 4 i +i
17. z + x 4 + x 9 + x ie + • • Ana. — 1< x < 1. q  q
+1
■*
18. x + ^= + ^= + • • •. Ans.  1 S x < 1. I c
x* x 8
_ + _ T .... _. _. __. — J _^
19. 1 + X + I— + I — I . Ans. All values of x. Z*.
[2 "'[S  r ""
i—*
0* . 0* P t ,_ All , ^ « I +00
20. l — i — Hi i — J • • • . An*. All values of $. .^
[§ [i d
I
21. ©> — nr+iT — &■ + "•• ^»*. All valuesof o>. "j° 
H + U"Iz
00 sin a sin 3 a sin 6a
JSA. — — — —  — •+*
I+co
1* 8* 6* o
4 n*. All values of a.
• End points that are not included in the interval of convergence have circles drawn
about them.
226 DIFFERENTIAL CALCULUS
Hot. Neither the sine dot eeeine eta emeed 1
tally.
^l + xlog a + ^ + ^ + ....
An*. AD value* of s.
+  = + s r + * «4»« »>!•
• • • ,
28. 10x + 100z J + 1000x« + ....
29. l + x + [2x J + [3x , + ...
* End poinU that are not included In the Interval of convergence nave circles drawn
about them.
CHAPTER XVIH
EXPAHSIOK OF FUNCTIONS
143. Introduction. The student is already familiar with some
methods of expanding certain functions into series. Thus, by the
Binomial Theorem,
(A ) (a + x) A = a 4 + 4 a % x + 6 a V + 4 ax* + x*>
giving a finite power series from which the exact value of (a + x) A
for any value of x may be calculated. Also by actual division,
(B) q^— ^l+s + ^+^+.. + af^ + ZV^V'
1 — x \1 — xf
we get an equivalent series, all of whose coefficients except that
of s* are constants, n being a positive integer.
Suppose we wish to calculate the value of this function when
x= .5, not by substituting directly in
1
i *
1 — a?
but by substituting x = .5 in the equivalent series
((7) (l + x + xt+J+'' + tf^ + fj—Jx*.
Assuming n = 8, (C) gives for x = .5
(2>) r?— =; 1.9921875 + .0078125.
v 1 — £
If we then assume the value of the function to be the sum bf
the first eight terms of series ((7), the error we make is .0078125.
However, in case we need the value of the function correct to two
decimal places only, the number 1.99 is as close an approximation
to the true value as we care for, since the error is less than .01.
It is evident that if a greater degree of accuracy is desired, all we
need to do is to use more terms of the power series
(E) l + z + x*+x*+..
227
DIFFERENTIAL CALCULUS
Since, however, we see at c
there is no necessity for the above discussion, except for purposes
of illustration. As a matter of fact the process of computing the
value of a function from an equivalent series into which it has
been expanded is of the greatest practical importance, the values
of the elementary transcendental functions such as the sine, cosine,
logarithm, etc., being computed most simply in this way.
So far we have learned how to expand only a few special forms into
series ; we shall now consider a method of expansion applicable to an
extensive and important class of functions and called Taylor's Theorem.
144. Taylor's Theorem * and Taylor's Series. Replacing b by x
in (£), p. 167, the extended theorem of the mean takes on the form
(61)/(x)=/(a)
/"'(«)
■/""CO.
where as, lies between a and x. (61), which is one of the most fi
reaching theorems in the Calculus, is called Taylor's Theorem. We
see that it expresses /(») as the sum of a finite series in (x — a).
(x — a)"
The last term in (61), namely  — , — — /""0O' ' 8 sometimes called
the remainder in Taylor's Theorem after n terms. If this remain(
converges toward zero as the number of terms increases without limit,
then the righthand side of (61) becomes an infinite power series
called Taylor's Series.* In that case we may write (61) in the form
ar
Ilea
der
nit,
I) /(*)=/(«)
■A")
(*<•}'
/"«M
!■/"'(«)+■
and we say that the function has been expanded into a Taylor's Series.
For all values of x for which the remainder approaches zero as n
increases without limit, this series converges and its sum gives the
exact value of f(x), because the difference (= the remainder) between
the function and the sum of n terms of the series approaches tl
limit zero (§ 15, p. 13).
:
EXPANSION OF FUNCTIONS 229
If the series converges for values of x for which the remainder
does not approach zero as n increases without limit, then the limit
of the sum of the series is not equal to the function f(x)>
The infinite series (62) represents the function for those values of sc,
and those only, for which the remainder approaches zero as the num
ber of terms increases without limit
It is usually easier to determine the interval of convergence of
the series than that for which the remainder approaches zero ; but in
simple cases the two intervals are identical.
When the values of a function and its successive derivatives are
known for some value of the variable, as x = a, then (62) is used
for finding the value of the function for values of x near a, and (62)
is also called the expansion off(x) in the vicinity of x = a.
Illustrative Example 1. Expand logx in powers of (x — 1).
Solution. f(x) = log x, /(l) = ;
.T(x) = l, /Xl) = l;
/(x)=i, r(i)=i;
r / (x) = , /'"(I) = 2.
Substituting in (62), logx = x — 1 — J (x — 1) 2 + J(x — l) 8 . Ana.
This converges for values of x between and 2 and is the expansion of logx in
the vicinity of x = 1, the remainder converging to zero.
When a function of the sum of two numbers a and x is given,
say/(a + :z), it is frequently desirable to expand the function into
a power series in one of them, say x. For this purpose we use another
form of Taylor's Series, got by replacing x by a + x in (62), namely,
(63) f(a + x) = f(a) + * f(a) + tfXd) + if" (a) + ■...
Illustrative Example 1. Expand sin (a + x) in powers of x.
Solution. Here /(a + x) = sin (a + x).
Hence, placing x = 0,
f{a) = sin a,
/'(a) = cos a,
/"(a) =  sin a,
/'"(a) =  cos a,
Substituting in (61),
X X s X s
sin (a + x) = sina + cos a — ,— sin a — r— cos a +'•••. An*.
v ' 1 [2 Li
DIFFERENTIAL CALCULUS
1. Expand e* Id powers of z — 2. Ana. e* = «• +
2. Expand z* — 2 z* + 5x — 7 in powers of z — 1.
dm. 3 + 4(al) + (il)' + (*l)>.
3. Expands? 1  14k 4 Tinpowereof y3. 4m. — 8 + 4(v 3) + 3(y 3)'.
4. Expand 6*' + 7* + 3 in Powersoft 2. Ant. 87 + 27 <a — 2)+ 5(j 2)'.
6. Expand 4x» 17** + 11* + 2 in powers of x  4.
6. Expand 6y* + 6y« 17 y* 4 18 ji 20 in powers of y + 4.
7. Expand f In powers of z + 1.
8. Expand Bin z in powers of x — a,
9. Expand cosz in powers otx — a.
10. Expand cos(a + z) in powers of x.
11. Expand log (z + A) in powers of x.
A 2A> 3 ft*
h Asec ! z + A'sec 1 ! tani + .
12. Expand tan (z )■ A) in powers of A.
Aim. tan (z + A) =
13. Expand the following in powers of h.
(a) <z + A)=z + nz"'A + '^l^z'
«"■' («♦»♦£♦!)•
145. Madaurin's Theorem and Madaurin's Series. A particular c
of Taylor's Theorem is found by placing a — in (61), p. 228, giving
(64, /(x) =/(0) + * A°) + 1/"(") + Tj/"'(0) + 
V
where Xj lies between and jr. (64) is called Maclauritfs Theorem.
The righthand member is evidently a series in a; in the same sense
that (61), p. 228, is a series in x  a.
Placing a = in (62), p. 228, we get Maclaurin'i Series,*
>[r'w + [i'
V
* In these examples we assume that the functions can be developed Into a power «
t Named after Colin Maclaurln (16081746), being first published in his Treatut a
/Vinton*, Edinburgh, 1743. The series is really due to Stirling (16921770).
cert
in <
ordi
nlif
by
8U1
_
EXPANSION OF FUNCTIONS 281
special case of Taylor's Series that is very useful. The statements
made concerning the remainder and the convergence of Taylors Series
apply with equal force to Maclaurin's Series, the latter being merely
a special case of the former.
The student should not fail to note the importance of such an
expansion as (65). In all practical computations results correct to a
certain number of decimal places are sought, and since the process
question replaces a function perhaps difficult to calculate by an
Unary polynomial with constant coefficients, it is very useful in sim
plifying such computations. Of course we must use terms enough to
give the desired degree of accuracy.
In the case of an alternating series (§ 139, p. 218) the error made
by stopping at any term is numerically less than that term, since the
m of the series after that term is numerically less than that term.
Illustrative Example 1. Expand cosz into an infinite power series and determine
(or what values of z it converges.
Solution. DiBorentiiiting first and then placing x — 0, we get
/W»«««i /(0) = i,
/(z)=sln», /'(0) = 0,
/"(x)=cosz, /"(0)=l,
/"'(i) = sin as, /'"(0) = 0,
/*(s) = cosz, /*(0> = 1,
/•(z)=slnz, /*(<>) = 0,
/"*(!) =Coez, /^(0)=l,
Substituting in (66),
(A)
..=..
x 1 x* x>
Comparing with Ex. 20, p.
In the same way for sin z.
225, we i
lee that the series
«■ * *
"5 + 5~(I +
converges for all values of z.
iMch converges for nil values of x (E
t. 21,
p. 226)."
• Since here /"'(j) 
sin U * ~\ and fe>(T. x )  sin (x x *
?)
we have, by puhnt ii nl in^
a the last term of 'Mi. [
.231,
remainder
•B'
..(?)'
< z, < z
BMli.^^™.
..,„
•Mr,
and from Ex
, IB,
p. 236, ^^0 for all
■alues of z. Hence
«Mr)
k 2. Using the scries (I!) found in the last eiamr.
late sin 1 correct to four decimal places.
Solution. Here x = 1 radian ; that iR, the angle is expressed in circular
Therefore, substituting x = 1 in (B) of the last example,
•"" = '(H[5 + "
Summing up the positive and negative terms separately,
rl = 0.16667...
1 = 1.00000.
 o.«
o.oooio..
Hence
which is c<
sinl = 1.00833  0.KH188 = 0.84147. ■•,
> live decimal places, since the error made must be li
i.e. less than .000003. Obviously the value of sin 1 may be calculated to any desired
degree of accuracy by simply including :i sulli. i.'in nuriilrr of additional tern
1. « = ! + * +
Verify the following expansions of functions into power series by Maclaurln't
Series and determine for what values of the variable they are convergent:
X> X* I* x»
, T'log'n sHog»a
fl* t*l" tV
"1" + "H""TT + "'
1M **z» fi«
f TF"TF + T""
6. iog(i +I) = *■■■" •*•■"
Convergent for all values of i
Convergent for all values of x.
Convergent for all values of x.
Convergent for all values of x,
k being any constant.
Convergent for all values of z,
k being any constant.
Convergent if l<igl.
Convergent if — 1 3 S < 1,
Convergent if ISiSL
Convergent If — 1 3i = 1.
Convergent for all values of r
Convergent for all values of 4.
Convergent for all values of S.
EXPANSION OF FUNCTIONS 233
13. Find three terms of the expansion in each of the following functions :
(a) tanx. (b) secx. (c) e*»*. (d) cos2x. (e) arc cos x. (f) a*.
14. Show that logx cannot be expanded bj Maclaurin's Theorem.
Compute the values of the following functions by substituting directly in the equiv
alent power series, taking terms enough until the results agree with those given below.
16. e = 2.7182...
Solution. Let x = 1 in series of Ex. 1 ; then
e=1+1+ d + ii + [i + [i + 
First term =1.00000
Second term = 1.00000
Third term = 0.60000
Fourth term =0.16667.. (Dividing third term by 3.)
Fifth term = 0.04167 • • • (Dividing fourth term by 4.)
Sixth term =0.00883". (Dividing fifth term by 5.)
Seventh term = 0.00180 * t < (Dividing sixth term by 6.)
Eighth term = 0.00019  • • , etc. (Dividing seventh term by 7.)
Adding, e = 2.71826 •• Ana.
16. arc tan (£) = 0.1078 • • • ; use series in Ex. 0.
17. cos 1 = 0.5403 • • • ; use series in Ex. 2.
18. cos 10°= 0.0848* * * ; use series in Ex. 2.
2jP Q& X^
19. sin. 1 = .0998..; use series x— . — li • — )•••.
13 16 [7
20. arc sin 1 = 1.6708 • • • ; use series in Ex. 8.
21. sin ^ = 0.7071 . . . ; use series (£), p. 281.
4
28. <« = l+2 + 2 + ^ + ... = 7.8891.
22. sin .6 = 0.4794 • • • ; use series (B), p. 281.
2* 2*
(2 + [f
24  v * =1+ ^ + 2 4 + ^I + " =1  6487 
In more advanced treatises it is shown that, for values of x within
the interval of convergence, the sum of a power series is differentiate
and that its derivative is obtained by differentiating the series term
by term as in an ordinary sum. Thus from (2?), p. 231,
of of x 1
Bin* = «j + jg + ....
Differentiating both sides, we get
1X> X 3/
— i — 4i i — + ••••
HE H
DIFFERENTIA! CALCULUS
which is the series of Ex. 2, p. 232. This illustrates how we i
obtain a new power aeries from a given power aerie* by differentiation.
Differentiating the power series of Ex. 6, p. 232, we obtain
1
1 + *
= 1
r + z*x* + x
In the same way from Ex. 8, p.
1
1 jijj. 1  8 tJ _ ls 
r'+
146. Computation by series. I. Alternating aerie*. Exs. 1524 of
the last exercise illustrate to what use series may be put for pur
poses of computation. Obviously it is very important to know the
percentage of error in a result, since the computation must necessarily
stop at some term in the series, the sum of the subsequent terms
being thereby neglected. The abaolute error made is of course equal to
the limit of the sum of all the neglected terms. In some series this
error is difficult to find, but in the case of alternating series it has
been shown in § 140, p. 220, that the sum is less than the first of
these terms. Hence the absolute error made is legs than the first term
neglected. Fortunately a large proportion of the series used for Com
putation purposes are alternating series, and therefore this easy method
for finding the upper limit of the absolute error and the percentage of
error is available. Let us illustrate by means of an example.
Illustkative Example 1. Determine the greatest possible error and percentage
of error made in computing the numerical value of the sine of one radian from the
due series, & & x i
(a) when all terms beyond the second are. neglected [
(b) when all terms beyond the third are neglected.
Solution. Let x n
E + [jE + 
(a) Using only the first two terms,
Binl = l =  = .8SS8,
tbo absolute error is less than j; l.e.< — ( = .0083), and the percentage o
lew (ban 1 per cent." L°
EXPANSION OP FUNCTIONS
(b) Using only tbe first three terms,
FJ' 5040
is less than ^, of 1 per cent."
Moreover, the exact nalue of sin 1 lies between
nating series S, ia alternately greater and less than '
i + T4o = .841«Kt,
( .000108), and the percentage of e
EXAMPLES
Determine the greatest possible error and percentage of error made in computing
the numerical value of each of the following functions from its corresponding series
(a) when all terms beyond the second are neglected ;
(b) when all terms beyond the third are neglected.
I..
al.
7,
2. sin 2. 5. e~*.
l«.. 6..i„.
II. The computation of ir by eerie*.
From Ex. 8, p. 232, we have
, 1z* , 1
arc tan 2.
sin 16°.
arc Bin x = x + 
l8g* l85a*
245 + 2 46 7"*
Since this series converges for values of x between
we may let x = J, giving
1 and +1,
r=U! I/IY+L 8 . iflY
6 2 T 2 3\2/ 24 h\V
or w = 3.1415 ■•■.
Evidently we might have used the series of Ex. 9, p. 232, instead.
Both of these series converge rather slowly, but there are other series,
found by more elaborate methods, by means of which the correct value
of it to a large number of decimal places may be easily calculated.
III. The computation of logarithm* by aeries.
Series play a very important role in making the necessary calcula
tions for the construction of logarithmic tables.
From Ex. '
. 2:42,
i have
log(l + x) = :r
= 1 in (J),
DIFFERENTIAL CALCULUS
converges for x = 1, and we can find log 2 by placiii
giving
But this series is not well adapted to numerical computation, because
it converges so slowly that it would be necessary to take 2000 terms
in order to get the value of log 2 correct to three decimal places. A
rapidly converging series for computing logarithms will now be
deduced.
By the theory of logaritlims,
i * + *
(*)
 = iog(i + *)!<*(!*).
By*
Substituting in (5) the equivalent Beries for log(l+a)
log (1 — *) found in Exs. 6 and 7 on p. 232, we get •
m Iog i±£^ +  + ^ + ...],
which is convergent when x is numerically less than unity. Let
1 + x M . MN
s = — i whence x = *
\x N M + N
and we see that x will always be numerically less than unity for all
positive values of M and N. Substituting from (Z>) into (C), we get
W
(2?)
rJWlogiV
n l MN
~"[m+N~
\(M
■i\M+
M+N J
a series which is convergent for all positive values of M and jV; i
it is always possible to choose M and N so as to make it couvei
rapidly.
Placing M= 2 and N=\ hi (\ff), we get
%!*[i+i?+i?+f? + ]  6M14nB  !
EXPANSION OF FUNCTIONS 237
Placing M = 3 and JST= 2 in (#), we get
log 3 = log 2 + 2r~ + ~.i + ii + .. 1 = 1.09861229...
It is only necessary to compute the logarithms of prime numbers in
this way, the logarithms of composite numbers being then found by
using theorems 710, p. 1. Thus
log 8 = log 2 $ = 3 log 2 = 2.07944154 • . .,
log 6 = log 3+ log 2 = 1.79175947...
All the above are Napierian or natural logarithms, i.e. the base is
e = 2.7182818. If we wish to find Briggis or common logarithms, where
the base 10 is employed, all we need to do is to change the base by
means of the formula j
locr n = —&* — •
^ 10 log, 10
Thus iog tft 2 = Jggj = °' 693 '' = 0.301..,
log. 10 2.302..
In the actual computation of a table of logarithms only a few of
the tabulated values are calculated from series, all the rest being
found by employing theorems in the theory of logarithms and various
ingenious devices designed for the purpose of saving work.
Calculate by the methods of this article the following logarithms :
1. log, 6 =1.6004.*.. 3. log, 24 = 3.1781....
2. log, 10 = 2.3025*... 4. log l0 5 = 0.6090 • •.
147. Approximate formulas derived from series. Interpolation. In
the two preceding sections we evaluated a function from its equivalent
power series by substituting the given value of a: in a certain number
of the first terms of that series, the number of terms taken depending
on the degree of accuracy required. It is of great practical importance
to note that this really means that we are considering the function as
approximately equal to an ordinary polynomial with constant coefficients.
For example, consider the series
•
238 DIFFERENTIAL CALCULUS
This is an alternating series for both positive and negative vali
of x. Hence the error made if we assume sin x to be approximately
equal to the sum of the first n terms is numerically less than the
(n + l)th term (§139, p. 218). For example, assume
(B) sin x = x,
and let us find for what values of x this is correct to three places
decimals. To do this, set *i
(<7) U<001.
This gives x numerically less than v.00b'(=.1817); i.e. (B) is cor
rect to three decimal places when x lies between +10.4° and —10.4°.
The error made in neglecting all terms in (A) after the one in
x*~ x is given by the remainder (see (64), p. 230)
hence we can find for what values of x a polynomial represents the
functions to any desired degree of accuracy by writing the inequality
(j£) \R\< limit tf
and solving for x, provided we know the maximum value of f*(i
Thus if we wish to find for what values of x the formula
7?
Cry
amx=x — —
ine
dity
is correct to two decimal places (i.e. error <. 01), knowing that
/ ! *'(z,)  ^ 1, we have, from (73) and (Z? ),
120 '
:«i;
rzs
Therefore ;
 ~ gives the correct value of sin x to two decimi
places if *Sl; ie if x lies between + 57° and —57°. This agrees
with the discussion of (A) as an alternating series.
Since in a great many practical problems accuracy to two or three
decimal places only is required, the usefulness of such approximate
formulas as (B) and (.F) is apparent.
Again, if we expand sin x by Taylor's Series, (62),
powers of x — a, we get
sin x = sin a + cos a (x  a)  2512 (* _ a ) f + • .
.
EXPANSION OF FUNCTIONS 239
Hence for all values of x in the neighborhood of some fixed value a
we have the approximate formula
(G) sin x = sin a + cos a(x — a).
Transposing sin a and dividing by x — a, we get
sin x — sin a
= cos a.
a; — a
Since cos a is constant, this means that
The change in the value of the sine is proportional to the change in the
angle for values of the angle near a.
For example, let a = 30° = .5236 radians, and suppose it is required
to calculate the sines of 31° and 32° by the approximate formula (G).
Then
sin 31° = sin 30° + cos 30° (.01745)*
= . 5000 + . 8660 x. 01745
= .5000 + .0151
» .5151.
Similarly, sin 32 p = sin 30° + cos 30° (.03490) = .5302.
In general, then, by Taylor's Series, we have the approximate
formula
(E) /(*) =/(a) +f(a) (x a).
If the constant /'(a) =£ 0, this formula asserts that the ratio of the
increments of function and variable for all values of the latter differing
little from the fixed value a is constant.
Care must, however, be observed in applying (H); for while
the absolute error made in using it in a given case may be
small, the percentage of error may be so large that the results are
Worthless.
Using one more term in Taylor's Series we get the approximate
formula
(0 /(*) =/(a) + /(«)(xa) +^(x«)».
* x  a = 1° = .01745 radian.
240
DIFFERENTIAL CALCULUS
The values of sin 31" and sin 32° calculated on p. 239 from (ff)
are correct to only three decimal places. If greater accuracy thi
this is desired, we may use (/), which gives, for /(*■)= sin a:,
IT
(»«)■•
(</) sin x = sin a + cos a (x — a)
Let a = 30° = .5236 radian.
Then sin 31° = sin 30°+ cos 30°(.01745)  ^~
= .50000 + .01511 .00008
= .51503.
ain 32° = sin 30° + cos 30° (.03490)  ^~
.50000 +.03022 .00030
=.52992.
These results are correct to four decimal places.
 (.01745)*
 (.03490)'
EXAMPLES
1. Using furmnla (H) fur interpolation by first differences, calculate (.hi; following
f unctioua !
(c) ain 85.1°, taking a =
(d) cot 70.3°, taking a =
n by BBCoud difference*, iijilculate the following
(c) (
:15.2° taking a = 15°.
1 09°, taking a = 70=.
" fs " * ~ fs + [& rej *>* ctiTel y< * nd °
(a) cub HI", taking a =
(b) tan 46°, taking a =
2. lining formula (I) for iivt«rpolatlor
functions :
(a) sin 11°, taking a = 10°.
(b) cos 86°, taking a = 86°.
3. Draw the graphs of the f unctions J
pare them with the graph of sin a.
148. Taylor's Theorem for functions of two or more variables. The
scope of this book will allow only an elementary treatment of the
expansion of functions involving; more than one variable by Taylor's
Theorem. The expressions fur the remainder are complicated and
will not be written down.
Having given the function
w /<* s0.
it is required to expand the function
(») f(x + l h y + k)
in powers of h and k.
Consider the function
i
EXPANSION OF FUNCTIONS 241
Evidently (J5) is the value of (C) when t = 1. Considering (C7) as
a function of t, we may write
(2>) /(x + fe,y + fe) = /(0,
which may then be expanded in powers of t by Maclaurin's Theorem,
(64), p. 230, giving
(20 F(f) = 2P(0) + tf"(0) + £ *"'(0) + £ *""(0) + ....
Let us now express the successive derivatives of F(f) with respect
to t in terms of the partial derivatives of F(f) with respect to x
and y. Let
(JO a = x+ht, /3 = y+kt;
then by (51), p. 195,
But from (**),
and since F(t) is a function of x and y through a and y8,
dF^dFda , dF_dFdfi %
dx~~ dadx dy~ d/3 dy 9
or, since from (^), — = 1 and — = 1,
dF dF , a^ &F
(/) — = — and — = — •
w & 2a dy d/3
Substituting in (G) from (J) and (#)»
Replacing JP(£) by 2?"(e) in (<7), we get
w dx dy \ da* dxdyj \ dxdy dy* j
In the same way the third derivative is
and so on for higher derivatives.
212 DIFFERENTIAL CALCULUS
When I 0, we have from (2>), (C), (J), (X), (X),
■f (»)/(*. y). "> 'CO ■» replaced by/O y),
J"(0)V
j+2«r
aj ay"
lW.
ay
*(0) =
and so on.
Substituting these results in (.ff), we get
r/
+*¥,
a
<*»!? 8y"
(«) /(*+«, y+«)=/(*,iD + <(jtg+*Q
To get f(x + A, y + £). replace I by 1 h (66), giving Taylor' t Theorem
for a function of two independent variables,
,67) /( J r + M H *) = /(x,J0 + »g+*^
gV m" "x*y if)
which ia the required expansion in powers of A and k. Evidently (67)
is also adapted to the expansion of f(x + h, y 4 k) in powers of x and y
by simply interchanging z with A and y with k. Thus
(670)
nh 0k
II .Iff . a'/ .a*/'
Similarly, for three variables we shall find
+ (*
aj* 4 ar* x
dxdy
6zdx
and so on for any number of variables.
OydzJ
EXPANSION OF FUNCTIONS
243
EXAMPLES
Given f(x, y) = Ax' + Bxy + Cy', expand f(x + ft, y + k) in powers of ft and k.
¥=2Ax + By, ¥=Bx + 2Cy;
£ V
2A,
ixBy
mp
= 2C.
The third and higher partial derivatives are all zero. Substituting In (87),
/(z + ft, y + k) = Ax' + Bxy + Cy* + {2Ax + By)h + (Bx + 2Cy)k
+ Ah* + Bhk + Ck*. Ant.
•■■'' __
Zzcy tyZz Zzdx
The third and higher partial derivatives are all zero. Substituting in (B8),
/(x + 1, y + m, i + n) = Ax' + By + Ci' + 2 Axl + 2 Bym + 2 Cza
+ AP + Bm' + Cn*. Ant.
3. Given /(z, y) = Vx tan y, expand /(z + ft, y 4 1) in powera of A and k.
4. Given/(i, y, t) s Ax' + By' + Cz> + Dxy + Eyz + Fzx, expand f{z + h,y + k,
z + I) in powers of A, k, I.
149. Maxima and minima of functions of two independent variables.
The function f(x, y) ia said to be a maximum at x = a, y = b when
f(a, by is greater than/(£, y) for all values of x and y in the neigh
borhood of a and 6. Similarly, /(a, 6) is said to be a minimum at
x^a, y = b when /(a, fi) is less than f(x, y~) for all values of x and
y in the neighborhood of a and 6.
These definitions may bo stated in analytical form as follows:
If, for all values of A and k numerically less than some small posi
tive quantity,
(^) f(a + h, b + A) —/(a, 6) = a negative number, then /(a, 6) is a
maximum value of/(j:, y). If
(£) /(a } A, 6 + ft) — /(«, 6) = a positive number, then /(«, J) is a
minimum value oi/(x, y).
These statements may be interpreted geometrically as follows: a
point P on the surface _ , *
DIFFERENTIAL CALCULUS
is a maximum point when it is " higher " than all other points on the
surface in its neighborhood, the coordinate plane XO Y being assumed
horizontal. Similarly, P* is a minimum point on the surface when it is
" lower " than all other points on the surface in its neighborhood. It
is therefore evident that all vertical planes through P cut the surface in
curves (as APB or DPE
in the figure), each of
which has a maximum
ordinate z (= MP') at P.
In the same manner all
vertical planes through
P 1 cut the surface in
curves (as BP'C or
FP'G), each of which
has a minimum ordinate
z(=JVP') at P 7 . Also,
any contour (as HIJK) cut out of the surface by a horizontal plane
in the immediate neighborhood of P must be a small closed curve.
Similarly, we have the contour LSfiT ne&r the minimum point J".
It was shown in §§ 81, 82, pp. 108, 109, that a neeetsary condition
that a function of one variable should have a maximum or a minimum
for a given value of the variable was that its first derivative should
be zero for the given value of the variable. Similarly, for a function
f(x, y) of two independent variables, a necessary condition that /(a, 6)
should be a maximum or a minimum (i.e. a turning value) is that for
x = a, y = b,
(<0
dx
= 0,
*y
Proof, Evidently (A) and (S) must hold when k = ; that is,
/(« + MW(M)
is always negative or always positive for all values of h sufficiently
small numerically. By §§ 81, 82, a necessary condition for 1
= a, or, what amounts to the sai
thing, t/(^ y) shall vanish for x = a, y = b. Similarly, (A~) and (B
must hold when h = 0, giving as a second necessary condition t
— /(if, y) shall vanish for x = a, y = b.
EXPANSION OF FUNCTIONS
245
In order to determine sufficient conditions that /(a, £) shall be a
maximum or a minimum, it is necessary to proceed to higher deriva
tives. To derive sufficient conditions for all cases is beyond the scope
of this book.* The following discussion, however, will suffice for all
the problems given here.
Expanding /(a + A, 5 + A) by Taylor's Theorem, (67), p. 242, re
placing x by a and y by b, we get
+m
ey
: + 2 *±
*L .
dx* ~"~8z8y ' " dy'J
where the partial derivatives are evaluated for x = a, y = b, and Ji
denotes the sum of all the terms not written down. All such terms
are of a degree higher than the second in h and k.
.U
V
Since Sd _ and %■
ex Sy
"g/O. J ).
0, from (C), p. 244, we get, after transpos
R.
If /(a, £) is a turning value, the expression on the lefthand side of
(J?) must retain the same sign for all values of h and k sufficiently small
in numerical value, — the negative sign for a maximum value (see (J),
p. 243) and the positive sign for a minimum value (see (5), p. 243) ;
U. f(a, 6) will be a maximum or a minimum according as the right
hand side of (ff) is negative or positive. Now JE is of a degree higher
than the second in h and k. Hence as h and h diminish in numerical
value, it seems plausible to conclude that the numerical value of £ will
eventually become and remain lets than the numerical value of the MM
of the three terms of the second degree written down on the righthand
tide of (J?) J Then the sign of the righthand side (and therefore also
of the lefthand side) will be the same as the sign of the expression
* * Bz 1 dxdy dy*
But from Algebra we know that the quadratic expression
hU+ZhkC + PS
always lias the same sign as A (or i?) when AB — C* > 0.
• See Court tf Analyst. Vol. I, by C. Jordan.
t FmboIim shown limit liisL'diiciiiiiiou dues do till Hajii liuli). See tile urtkde on " Maxima,
and Minim* of FDnctionn of Several Variable*." by Professor James Pierpoot Id the Bulletin
if tat American MatheiiijtLcjl Sicitty, Vol. IV.
246
DIFFERENTIAL CALCULUS
Applying this to (F), A
(lim fore
(3)
ay
, C = ^4< and
(F~), and therefore also tlie lefthand member of (£'), has the
.Sf, 
sign as
to*\
 when
>0.
Hence the following rule for finding maximum and minimum values of
function /(x, ff).
First Step, AWci; (//!■ *im.\tliauM>u# vitiations
SfH Co.
Second Step. Calculate for these values of x and y the value of
Third Step. The function will have a
maximum ;f A > and ^(or^J<0;
neither a maximum nor a minimum if A < 0,
The question is undecided (/'A = 0.*
The student should notice that this rule does not necessarily give
all maximum and minimum values. For a pair of values of x and y
determined by the First Step may cause A to vanish, and may lead to a
maximum or a minimum or neither. Further investigation is therefore
necessary for such values. The rule is, however, sufficient for sol'
many important examples.
The question of maxima and minima of functions of three c
independent variables must be left to more advanced treatises.
1. Examine the function Zuxy — x* — v' for
if A > and
>o ;
and minimum values.
Solution.
First ttep. 
Solving these t
} simultaneous equations, we get
EXPANSION OF FUNCTIONS 247
aeeottd tlep. — =— 6z. ^ = 8 a, ^=— 6v:
ax 2 ay* Vaxay/
2TOrtZ step. When x = and y = 0, A = — 9a 2 , and there can be neither a maxi
mum nor a minimum at (0, 0). ^ f
When x = a and y = a, A = + 27 a* ; and since — ^ = — 6 a, we have the conditions
0X 2
for a maTimnm value of the function fulfilled at (a, a). Substituting x = a, y = a in
the given function, we get its maTimnm value equal to a*.
Illustrative Example 2. Divide a into three parts such that their product shall
be a maximum.
Solution. Let x = first part, y = second part ; then a — (x + y) = a — x— y = third
part, and the function to be examined is
/(x, y) = xy(axy).
First step. — = ay — 2xy — y 2 = 0, — = ax — 2xy — x f = 0.
dx dy
a a *
Solving simultaneously, we get as one pair of values x =  , y =  •
Second**?. ^L = ^2y, ^ = a2x2y, ^ = 2x;
ax 1 ' axay ay 1 '
A = 4xy — (a — 2x — 2y) 2 .
7%wi step. When x =  and y =  , A = — ; and since — ^ = , it is seen that
3 8 8 ex 1 8
our product is a maximum when x =  , y = . Therefore the third part is also  , and
a 8
the maximum value of the product is — •
F 27
1. Find the minimum value of x 2 + xy + y 2 — oz — by. Ana. J (ab — a 2 — o 2 ).
Sir
2. Show that sin x + sin y + cos (x + y) tea minimum when x = y = — , and a
v *
maximum when x = y = — .
3. Show that z&+**** has neither a maximum nor a minimum.
4. Show that the maximum value of ' qx ^ ' is a 2 + ft 2 + c 2 .
x 2 + y 2 + 1
5. Find the greatest rectangular parallelepiped that can be inscribed in an ellipsoid
That is, find the maximum value of 8xyz(= volume) subject to the condition
x 2 y 2 z 2 8«&c
a 2 ft 2 c 2 8v3
Hint. Let uzyz, and substitute the value of z from the equation of the ellipsoid. This
*~ *w(iss).
where u is a function of only two variables.
•z0, y0 are not considered, since from the nature of the problem we would then
have a minimum.
248 DIFFERENTIAL CALCULUS
6. Show that the surface of a rectangular parallelepiped of giTen Yolnme la least
when the solid is a cube.
7. Examine z* + y 4 — sc f + xy — y* for maTimnm and minimnm value*.
An*. Maximum when x = 0, jr = ;
minimum whenz = jr = ± J, and when «=— y = ±jV5.
8. Show that the most economical dimensions for a rectangular tank to hold a
giTen Yolume are a square base and a depth equal to one half the side of the base.
9. The electric time constant of a cylindrical coil of wire is
mxyz
u =
ax + by + cz
where z is the mean radius, y is the difference between the internal and external
radii, z is the axial length, and m, a, 6, c are known constants. The volume of the
coil is nzyz = g. Find the values of x, y, z which make u a maximum if the volume of
the coil is fixed. . a !abca
Ana. ax = 6y = a=\;—^*
FT
t n i
CHAPTER XIX
ASYMPTOTES. SINGULAR POINTS
150. Rectilinear asymptotes. An asymptote to a curve is the limit
ing position* of a tangent whose point of contact moves off to an
infinite distance from the origin. f
Thus, in the hyperbola, the asymptote
is the limiting position of the tangent
'T as the point of contact P moves off
to the right to an infinite distance. In
the ease of algebraic curves the following
definition is useful : an asymptote is the
limiting position of a secant as two points
of intersection of the secant with a branch
of the curve move off in the same direction along that branch to an
infinite distance. For example, the asymptote AB is the limiting posi
tion of the secant PQ as P and Q move upwards to an infinite distance,
151. Asymptotes found by method Of limiting intercepts. The equa
of the tangent to a curve at (st, y : ) is, by (1), p. 76,
yVi
dx.
(**,)•
First placing y = and solving for x, and then placing x = and
olving for y, and denoting the intercepts by x, and y t respectively,
e f^t dx,
x i = x 1 — y t j 1 = intercept on OX;
ft— 3ft
.ih.
intercept on OY.
Since an asymptote must pass within a finite distance of the origin,
one or both of these intercepts must approach finite values as limits
when the point of contact (x lt y^) moves off to an infinite distance. If
limi t (x^) = a and limit (y,) = 6,
A line that approaches a filed straight line as a limiting position cannot be wholly at
Infinity ; hence it follow* that an asymptote mint pass within a finite distance of the origin.
It 1b evident that a curve which ha* no infinite branch can bare no real asymptote.
tOr, leas precisely, an asymptote to a curve Is sometimes defined as a tangent whose
Joint of contact is at an infinite distance.
2oO DIFFERENTIAL CALCULUS
then the equation of the asymptote is found by substituting the lin
ing values a and 6 in the equation
If only one of these limits exists, but
then we have one intercept and the slope given, so that the equatioi
of the asymptote is
y = mx + ft, or x = $ + a.
Illustrative Example 1. Find the asymptotes to the hyperbola ^ = 1.
Also I, = — and j/ ( =
Therefore the asymptotes pass through th
equations are .
y0 = ±(x0x),
hence these intercepts are zero when x~y
igin (see figure on p. M9) ud
their
This method is frequently too complicated to be of practical use.
The most convenient method of determining the asymptotes to alge
braic curves is given in the next section.
152. Method of determining asymptotes to algebraic curves. Given
the algebraic equation in two variables,
(4> f<& JO = o
If this equation when cleared of fractions and radicals is of degree «,
then it may be arranged according to descending powers of one of the
variables, say y, in the form
(5) ay+(te + c)y' + (./*'+ f*+.O/ ,  I t■ = 0. , '
For a given value of r this equation determines in general n
values of y.
• For nse in this section the attention of the student in called lei the following theorem
from Algebra : Given an algebraic eqnatirm of dogine »,
When A approaches rrv«, one rent (value of ;/) approaches
When A and R approach zero, two roots approach
When A, H, and C approach zero, three root? it>pr<i»i'h
oachea »
tch « , etc
ASYMPTOTES
251
Case I. To determine the asymptotes to the curve (B) which are
parallel to the coordinate axes. Let us first investigate for asymptotes
parallel to Y. The equation of any such asymptote is of the form
and it must have two points of intersection with (#) having infinite
ordinates.
First. Suppose a is not zero in (2J), that is, the term in j" is
present. Then for any finite value of x, (B) gives n values of y, all
finite. Hence all such lines as (C) will intersect (£) in points having
finite ordinates, and there are no asyinptutes parallel to OY.
Second. Next suppose a = 0, but b and e are not zero. Then we
know from Algebra that one root (=y) of (B) is infinite for every
finite value of x\ that is, any arbitrary line (C) intersects (#) at only
one point having an infinite ordinate. If now, in addition,
bx + c = 0, or
— I'
then the first two terms in (£) will drop out, and hence two of its
roots are infinite. That is, (/>) and (B) intersect in two points having
infinite ordinates, and therefore (i>) i* the equation of an asymptote to
(B) which is parallel to Y.
Third. If a = b = c = 0, there are two values of x that make y in
(B~) infinite, namely, those satisfying the equation
(£) dx 3 + ex+f=Q.
Solving (i?) for x, we get two asymptotes parallel to OY, and so on
in general.
In the same way, by arranging f(x, y') according to descending
powers of x, we may find the asymptotes parallel to OX. Hence the
following rule for finding the asymptotes parallel to the coordinate axis :
First Step. Equate to zero the coefficient of the highest power of x in
e equation. This gives all asymptotes parallel to OX
Second Step. Equate to zero the coefficient of the highest power of y
« the equation. This gives all asymptotes parallel to Y.
Note. Of course if one or both of these coefficients do not involve
x (or y), they cannot be zero, and there will be no corresponding
mymptote.
252 DIFFERENTIAL CALCULUS
I ll(t it bat! v* Eiufle 1. Find the asymptotes of the curve
Solution. Arranging the terms according to powers of x,
yz* — (2 ay + a 1 ) x + a*y = 0.
Equating to zero the coefficient of the high
est power of x, we get y — as the asymptote
parallel to OX. In fact, the asymptote coin
cides with the axis of x. Arranging the terms
according to the powers of y,
(z — a)*ya*x = 0.
Placing the coefficient of y equal to zero,
we get z = a twice, showing that AB is a
double asymptote parallel to OT. If this cur"
a by the method explained below,
y = 0ai
examined for asymptotes oblique to
>e seen ttiat there are none. Hence
the only asymptotes of the given curve.
Case II. To determine asymptotes oblique to the coordinate axes.
Given the algebraic equation
(.F) /(**>=».
Consider the straight line
(G) y = mx+t.
It is required to determine m and k so that the line ('..') shall be
an asymptote to the curve (J 1 ).
Since an asymptote is the limiting position of a secant as two points
of intersection on the same branch of the curve move off to an infinite
distance, if we eliminate y between (F) and (G~), the resulting equa
tion in x, namely,
(#) f(x,mx + k)=0,
must have two infinite roots. But this requires that the coefficients
of the two highest powers of x shall vanish. Equating these coeffi
cients to zero, we get two equations from which the required values
of m and k may he determined. Substituting these values in (G)
gives the equation of an asymptote. Hence the following rule for
finding asymptotes oblique to the coordinate axes :
Fibst Step. Replace y by mx + k in the given equation and expand.
Second Step. Arrange the term* according to descending powers ofx.
Third Step. Equate to zero the coefficient* of tlie two highest poweri *
of x, and solve for m and k,
* If the term involving r. 1 Is missing, or if tha value of m obtained by placing the first
Coefficient equal to zero causes the second coefficient to vanish, llien by placing the coeffi
cients of X* and a** equal to zero we obtain two equations from which the values of m
and k may be found. In this case we shall, in general, obtain two k'% for each m. that Is.
pairs of parallel oblique asymptotes. Similarly, If the term in V* is also missing, each
value of m furnishes three parallel oblique asymptotes, and so on.
ASYMPTOTES
258
Foubth Step. Substitute these values of m and k in
y = mx + h
This gives the required asymptotes.
Illustrative Example 2. Examine y 8 = 2 ax* — x 8 for asymptotes.
Solution. Since none of the terms involve both
x and y, it is evident that there are no asymptotes
parallel to the coordinate axes. To find the oblique
asymptotes, eliminate y between the given equation
and y — mx + fc. This gives
(mx + *) 8 = 2ax 8 — x»;
and arranging the terms in powers of x,
(1 + m*)z* + (Sro 8 *  2a)x* + &&mz + *» = 0.
Placing the first two coefficients equal to zero,
l + m 8 = and 3m 1 *— 2a = 0.
2a 2a
Solving, we get m = — 1, fc = — . Substituting in y = mx + fc, we have y = — x + —
3 3
the equation of asymptote <42?.
EXAMPLES
Examine the first eight curves for asymptotes by the method of { 151, and the
remaining ones by the method of { 152 :
1. y = e*. Ans.y = 0. 2. y = e«*. Ana. y = 0.
3. y = logx.
4. y
=H)
.An*, x = 0.
y = e, x = — 1.
5. y = tanx.
l
6. y = eFl.
7. y , = 6x f + x».
f
8. Show that the parabola has no asymptotes.
9. y 8 = a 8  x 8 .
x*
10. The cissoid y 8 =
2rx
11. tfa = y*x + «*.
12. y«(x« + l) = x»(x f l).
13. y«(x2a) = x«a«.
14. xV = o l (* l + y l ).
15. y(x*S6x + 2&*) = x*8ax* + a 8 .
16  y==C + (x^6)i
17. The folium x 8 + y 8 — 8axy = 0.
18. The witch x*y = 4a* (2a — y).
19. xy f + x*y = a 8 .
n«
n being any odd integer, x =
x = 0, y = 0.
y = x + 2.
y + x = 0.
x = 2r.
x = a.
y= ±x.
x = 2a, y = ±(x + a).
x = ± a, y = ± a.
x = 6, x = 26 y + 3a = x + 36.
y = c, x = 6.
y + x + a = 0.
y = 0.
x = 0, y = 0, x + y = 0.
20. x 8 + 2x t yxy 8 2y 8 + 4y 8 + 2xy + y = l. x + 2y = 0, x + y = 1, xy = l
254
DIFFERENTIAL CALCULUS
153. Asymptotes in polar coordinates. Let f(p, $)= be the e
tion of the curve PQ having the asymptote CD. As the asymptote
must pass within a finite distance (as OE)
of the origin, and the point of contact is
at an infinite distance, it is evident that
the radius vector OF drawn to the point
of contact is parallel to the asymptote,
and the subtangent OE is perpendicular
to it. Or, more precisely, the distance of the asymptote from the
origin is the limiting value of the polar subtangent as the point C
contact moves off an infinite distance.
To determine the asymptotes to a polar curve, proceed as follows :
First Step. Find from the equation of t/ie curve the valves of uhi
make p = ».* These values of 8 'jive the directions of the asymptotes.
Second Step. Find the limit of the polar subtangent
,d8
¥
by <Th p
as 8 approaches each such value, remembering that p approaches oo at
same time.
Third Step. If the limiting value of the polar mbtangent is finite, there
is a corresponding asymptote at that (Ustauce from the origin and parallel
to the radius vector drawn to the point of contact. When this limit is pos
itive the asymptote is to the right of the origin, and. irhen negative, to the
left, looking in the direction of the infinite radius vector.
1. Examine the hyperbolic spiral p =  t or asymptotes.
*,
Solution. When B =
subtangent = p 1
dp (P a.
a, which Is finite.
It happens in this case that the subtangent U the same for all values of 0. The
curve has therefore an asymptote BC parallel to the initial line OA and at a dis
unite a above ft.
SINGULAR POINTS 255
Examine the following curves for asymptotes :
2. /ocoe0=acofl20.
Arts. There is an asymptote perpendicular to the initial line at a distance a to
the left of the origin.
3. p = a tan 6.
Ans. There are two asymptotes perpendicular to the initial line and at a dis
tance a from the origin, on either side of it.
4. The lituus p$± = a. Ans. The initial line.
5. p = a sec 2 $.
Ana. There are four asymptotes at the same distance  from the origin, and
inclined 45° to the initial line. 2
6. (p — a) sin $ = b.
Ans. There is an asymptote parallel to the initial line at the distance b above it.
7. p = a(sec20 + tan20).
Ana. Two asymptotes parallel to — — , at distance a on each side of origin.
4
8. Show that the initial line is an asymptote to two branches of the curve
p*sin0:= a* coe 20.
9. Parabola p = . Ans. There is no asymptote.
r 1— CO80
154. Singular points. Given a curve whose equation is
Any point on the curve for which
^ = and ^ =
dz dy
is called a singular point of the curve. All other points are called
ordinary points of the curve. Since by (57 a), p. 199, we have
dy __ dx
dx~ df
'dy
it is evident that at a singular point the direction of the curve (or
tangent) is indeterminate, for the slope takes the form • In the next
section it will be shown how tangents at such points may be found.
155. Determination of the tangent to an algebraic curve at a given
point by inspection. If we transform the given equation to a new set
of parallel coordinate axes having as origin the point in question on
the curve, we know that the new equation will have no constant term.
Hence it may be written in the form
00 /(£> y) = ax + by + (cz* + dxy + ey 2 )
+ (fa* + gx*y + hzy* + iy') + ... = 0.
256
DIFFERENTIAL CALCULUS
The equation of a tangent to the curve at the given point (ri
the origin) will be
(*> * =(1?)* By(l),p.
\dx/'
n
Let y w mx be the equation of a line through the origin and
a second point P on the locus of (J). If then P approaches O along
the curve, we have, from (fl),
(C)
limit m =
<2x
Let be an ordinary point Then, by § 155, a and b do not both
vanish, since at (0, 0), from (J), p. 255,
£_,, g_8.
to 3y
Replace # in (J) by mr, divide out the factor z, and let x approach
zero as a limit. Then (J) will become *
a + bm = 0.
Hence we have, from (£) and (C),
or + by = 0,
the equation of the tangent. The lefthand member is seen to consist
of the terms of the first degree hi (.().
When is not an ordinary point we have a = b = 0. Assume that
e, d, e do not all vanish. Then, proceeding as before (except that we
divide out the factor X s ), we find, after letting x approach the limit
zero, that (.4) becomes c , j m , eBI * = o
or, from (C),
m c+d m +e (f)U.
Substituting from (.fi), we see that
(#) ex* + dxy + ey* =
is the equation of the pair of tangents at the origin. The lefthand
member is seen to consist of the terms of the second degree in (.4).
Such a singular point of the curve is called a double point from the
fact that there are two tangents to the curve at that point.
* After dividing by x an algebraic equation in m remains whose coefficients are functions
of x. If now x approaches! zero an a limit, the theorem holds that one root of this equation
In m will approach the limit u+ ft.
SINGULAR POINTS
257
Since at (0, 0), from (.4),
ay
a = 2c,
ay
=d,
*f
= 2e,
(*)
dxdy \dx/ By* \dx/
%x* ' dxdy "' dtf
it is evident that (2>) may be written in the form
In the same manner, if
a = b = c = d = e = Q,
there is a IrcpZe jpotVtf at the origin, the equation of the three tangents
being fJ+gxty + hxtf+iif^Q,
and so on in general.
If we wish to investigate the appearance of a curve at a given point,
it is of fundamental importance to solve the tangent problem for that
point. The above results indicate that this can be done by simple
inspection after we have transformed the origin to that point.
Hence we have the following rule for finding the tangents at a given point.
Fibst Step. Transform the origin to the point in question.
Second Step. Arrange the terms of the resulting equation according to
ascending powers of x and y.
Thikd Step. Set the group of terms of lowest degree equal to zero.
This gives the equation of tJie tangents at the point (origin).
Illustrative Example 1. Find the equation of the *■
tangent to the ellipse
5x 2 + 5y*+ 2xy — 12x — 12y =
at the origin.
Solution. Placing the terms of lowest (first) degree
equal to zero, we get
12x12y = 0,
or x + y = 0,
which is then the equation of the tangent PT at the origin.
Illustrative Example 2. Examine the curve
8x* — xy — 2y f + x 8 — 8y* = for tangents at the
origin.
Solution. Placing the terms of lowest (second)
degree equal to zero,
8x 2 xy2y 2 = 0,
or (x — y)(3x + 2y) = 0,
z — y = being the equation of the tangent AB, and 8x + 2y = the equation ©f
the tangent CD. The origin is, then, a double point of the curve.
DIFFERENTIAL CALCULUS
Since the roots of the quadratic equation (!•'). p. 257, namely,
dy\dx) dx%y\dx) ^ dx* '
may be real and unequal, real and equal, or imaginary, there are
three cases of double points to be considered, according as
(0)
is positive, zero, or negative (see 3, p. 1).
156. Nodes. (*£(*">*.
\dxdy) dx* By*
In this case there are two real and unequal values of the slope
jt\ found from (.F), so tliat we have two distinct real tangents
(»
to the curve at the singular point in question. This means that
the curve passes through the point in two different directions, or,
in other words, two branches of the curve cross at this point. Such
a singular point we call a real double point of the curve, or a node.
Hence the conditions to be satisfied at a node tn
'* J ' dx ' dy ' \0xdy/ dx* dy*
ILLUSTI
ative Exam
ple 1. Examine tbe lemniscate y 1 = i*
— z' for ainguli
ir point*.
Solution.
Here
W
, If) = v* — ** + ** = °
Also
as
dy
The point (0, 0) is o
, singulai
point, since its coordinates satisfy the above three
equations.
We have a
S (0, 0)
dziy ' 8p* " f
w
and the origin la a double point (node) through which
two branches of the curve pass in different directions. By placing the terms of tl
lowest (second) degree equal to zero we get
y* — & = 0, or y = z and y =s— x,
o tangenU AH and CD at the singular point or node (0, 0).
the equations of the t
SINGULAR POINTS
157. Cusps.
\dxdyj
dx'dy'
In this case there are two real and equal values of the slope
found from (F); hence there are two coincident tangents. This
means that the two branches of the curve which pass through the
point are tangent. When the curve recedes from the tangent in both
directions from the point of tangency, the singular point is called a
point of osculation; if it recedes from the point of tangency in one
direction only, it is called a cusp. There are two kinds of cusps.
First kind. When the two branches lie on opposite sides of the
common tangent
Second kind. When the two branches lie on the same side of the
common tangent."
The following examples illustrate how we may determine the nature
of singular points coming under this head.
i*y* — a'x* — z* for singular points.
o*x* + a* = 0,
Solution. Here /(z, y) =
^ = _4aV+6x' = 0, ^ = 2a* V = Q,
Sx By
igiilar point, since it satisfies the above three equations. Also, i
and (0, 0) is a
(0, 0) we have
B*/_
3V =
ay_
\dzdyJ ex* By'
since the curve is symmetrical with respect to OT, the
origin is a point of osculation. Placing the terms of lowest
(second) degree equal to zero, we get y* = 0, showing that the
coincide with OX.
Illobthativi Example 2. Examine y
BolntioD. Here f{',V) = V* x * =
* for singular points.
Sl s = 0,
V.
2j, = 0,
■bowing that (0, 0) is a singular point. Also, at (0, 0) we have
ftt" BxBy By* \Bxiy) Bz* By*
This is not a point of osculation, however, for if we solve the given equation for y
we get y = ± Vz",
• Meaning in the neighborhood of the singular point.
260 DIFFEBENTIAL CALCULUS
which shows that the curve extends to the rt«;ht only of OT, for negative value* of *
make y imaginary. The origin is therefore a cusp, ami since the branches lie on oppo
site sides of the common tangent, it is a cusp of the first kind. Placing the terms of
lowest (second) degree equal to zero, we get y a = 0, showing that the
tangents coincide with OX.
Illustrative Example 3. Examine (y — x 1 )* = x t for
singular points.
Solution. Proceeding as in the last example, wo find a
cusp at (0, 0), the common tangents to the two branches
coinciding with OX. Solving for y,
y = z'>± xi.
If we let x take on any value between and 1, y takes \
on two different positive values, showing that in the vicinity of the origin both
branches lie above (lie common tangciH. Hence the .insular point (0,0) is a cusp of
the second kind.
158. Conjugate or isolated points. lSCl — *% f£ <t.
\Sxdyl ax* dy*
In this case the values of the slope found are imaginary. Hence
there are no real tangents ; tho singular point is the real intersection of
imaginary branches of the curve, and the coordinates of
no other real point in the immediate vicinity satisfy the
equation of the curve. Such an isolated point is called a
• ■•"ijni/ate point.
Illustrative Example I. Examine the curve y* = x'—z t for singular 
points.
Solution. Here (0, 0) is found to be a singular point of the curve at
which — =±V— 1. Hence the origin is a conjugate point. Solving the
equation for y, y = ±e t/z — 1.
This shows clearly that the nrijnn i* sin isolated point nf the curve, foi
of x between and 1 give real values of y.
159. Transcendental singularities. A curve whose equation involve*
trans cenden till functions is called a transcendental curve. Such a curve
may have an end point at which it terminates abruptly, caused by i
discontinuity in the function : or a salient point at which two branches of
the curve terminate without having a common tan
gent, caused by a discontinuity in the derivative.
Illustrative Example 1. Show that y = xlogx has an
end point at the origin.
Solution, i cannot be negative, since negative numbers
have no logarithms ; hence the curve extends only U
right of OT. When x = 0, y = 0. There being only
value of y for each positive value of x, the curve consists of a single branch terminating
at the origin, which is therefore an end point.
SINGULAR POINTS
261
Illustrative Example 2. Show that y =
i 1 + e*
dy 1 e*
Solution. Here — = 1 .
dx 1 l
Y has a salient point at the origin.
1 + e* x(l + e*) 2
If x is positive and approaches zero as a limit, we
have ultimately
y = and — = 0.
dx
If x is negative and approaches zero as a limit, we get ultimately
y = and — = 1.
dx
Hence at the origin two branches meet, one having OX as its tangent and the other,
AB, making an angle of 45° with OX.
EXAMPLES
1. Show that y 2 = 2 x 2 + x 8 has a node at the origin, the slopes of the tangents
being ±V2.
2. Show that the origin is a node of y 2 (a 2 + x 2 ) = x 2 (a 2 — x 2 ), and that the tan
gents bisect the angles between the axes.
3. Prove that (a, 0) is a node of y 2 = x (x — a) 2 , and that the slopes of the tangents
are ±Va.
4. Prove that a 8 y 2 — 2 abz 2 y — x 5 = has a point of osculation at the origin.
5. Show that the curve y 2 = x 5 + x 4 has a point of osculation at the origin.
6. Show that the cissoid y 2 —
has a cusp of the first kind at the origin.
2a — x
7. Show that y 8 = 2 ax 2 — X s has a cusp of the first kind at the origin.
8. In the curve (y — x 2 ) 2 = x" show that the origin is a cusp of the first or second
kind according as n is < or > 4.
9. Prove that the curve x 4 — 2 ax*y — axy* + oV = has a cusp of the second
kind at the origin.
10. Show that the origin is a conjugate point on the curve y 2 (x 2 — a 2 ) = x*
11. Show that the curve y 2 = x (a + x) 2 has a conjugate point at (— a, 0).
12. Show that the origin is a conjugate point on the curve ay 2 — x 8 + 6x 2 = when
a and b have the same sign, and a node when they have opposite signs.
13. Show that the curve x 4 + 2 axfy — ay z = has a triple point at the origin, and
that the slopes of the tangents are 0, + V2, and — V2.
14. Show that the points of intersection of the curve ( J +() =1 with the axes
are cusps of the first kind. ^ a ' ^ b '
15. Show that no curve of the second or third degree in x and y can have a cusp
of the second kind.
16. Show that y = e * has an end point at the origin.
17. Show that y — x arc tan  has a salient point at the origin, the slopes of the
mm X
tangents being ± — •
160. Tangent line and normal plane to a skew curve whose equations
are given in parametric form. The student is already familiar with the
parametric representation of a plane curve. In order to extend this
notion to curves in space, let the coordinates of any point P (x, y, 2)
on a skew curve be given as functions
of some fourth variable which we shall
denote by t, thus,
'2£)&*«i
(j) «=+c0. y=+C0. »=xC0
The elimination of the parameter t
between these equations two by two
will give us the equations of the pro
jecting cylinders of the curve on the 3
coordinate planes.
Let the point P(x, y, z) correspond to the value t of the param
eter, and the point P'(x + Ax, y + Ay, 2 + A2) correspond to the value
t+At; Ac, Ay, Az being the increments of x, y, 2 due to the incre
ment A( as found from equations (.4). From Analytic Geometry of
three dimensions we know that the direction cosines of the secant
(diagonal) PP 1 are proportional to
Ax, Ay, A.:;
or, dividing through by At and denoting the direction angles of t
secant by a', ft', y',
, ~ , Aa Ay Az
cos a : cos £ : cos t' : : — : r"r—
Now let i* approach P along the curve. Then At, and therefoi
also Ar, Ay, Az, will approach zero as a limit, the secant PP 1 1
approach the tangent line to the curve at P as a limiting positioi
and we shall have
a dx dy dz
cos a : cos p : cos 7 : : — : ~f : — ,
APPLICATIONS TO GEOMETBY OF SPACE
263
where a, /S, y are the direction angles of the tangent (or curve) at P.
Hence the equation* of the tangent line to the curve
*=*(*). y=*(9, *=x(9
at ike point (x, y, z) are given by
X—x T—y Z—x
(69)
dx
dt
dt
IT'
dt
and the equation of the normal plane, i.e. the plane patting through (se, y, z)
perpendicular to the tangent, it
(TO)
(Xx) + (F y) + (Zx) = 0,
X, Y, Z being the variable coordinates.
Illustrative Example 1. Find the equations of the tangent and the equation of
the normal plane to the helix * {6 being the parameter)
'x = acosl,
' y = a sin 0,
,z = b$,
(a) at any point; (b) when = 2 x.
Sototkm.
=a*n0= V ,  = ««** = *, =b.
Substituting in (60) and (70), we get at (x, y, z)
X—x T—y Z—z
y
, tangent line;
and — y (X — x) + x(Y— y) + b(Z — z) = 0, normal
plane.
When $—2x, the point of contact is (a, 0, 2 for), J;
^▼ing xa F0 Z26ir
OT.
a b
X=a, 6r=aZ — 2a5r,
the equations of the tangent line ; and
aY+bZ2Vhc = 0,
the equation of the normal plane.
• The helix may be defined as a curve traced on a right circular cylinder so as to cut all
the elements at the same angle.
Take OZ as the axis of the cylinder, and the point of starting in OX at P . Let a  radios
of base of cylinder and $ angle of rotation. By definition,
fff PN  — *(const.), or *<0*.
8N arcPo^ a$
Let aifc6; then zb0. Also y JTZVa sin0, zOlfaoosA
264 DIFFERENTIAL CALCULUS
EXAMPLES
Find the equations of the tangent line and the equation of the normal plane to
each of the following skew curves at the point indicated :
1. x = 2l,y = t»,* = 4t*; t = 1. Ana. ^ = ^ = i=i;
x + y + 8z — 35 = 0.
2. x = £l,y = t + l,* = P; t=r2. Ana. ?LZ1? = ?^li? = L^? ;
4x + y + 12z — 111 = 0.
3. x = Pl l y = t + t a l z = 4*»3f + l; t = 1. 4?w. ? = ^? = ?^;
3 3 9
x + y + 3s — 8 = 0.
.... . . v . 4z — ir V%y — 1 V2z — 1.
4. x = f, y = sin t , 2 = cos t ; t = T . 4?w. —  — =  = ;
4 4 1—1
4x + 2 V2y — 2 V2* — *■ = 0.
5. x = a*, y = M*, 2 = ct 8 ; t = 1.
6. x = t, y = 1P, 2 = 30 + 41; t=2.
7. x = i, y = e*, * = e'; t = 0.
8. x = a sin f, y = b cost, z = t ; t = — .
6
9. Find the direction cosines of the tangent to the curve x = 0, y = t 8 , 2 = f 4 at
point x = 1.
161. Tangent plane to a surface. A straight line is said to be tan
gent to a surface at a point P if it is the limiting position of a secant
through P and a neighboring point P' on the surface, when P f is
made to approach P along the surface. We now proceed to establish
a theorem of fundamental importance.
Theorem. All tangent lines to a surface at a given point * lie in
general in a plane called the tangent plane at that point.
Proof. Let
00 F(x> y, z) =
be the equation of the given surface, and let P(x, y, z) be the given
point on the surface. If now P f be made to approach P along a curve
C lying on the surface and passing through P and P', then evidently
the secant PP f approaches the position of a tangent to the curve C
at P. Now let the equations of the curve C be
9 The point in question is assumed to be an ordinary (nonsingular) point of the surface,
i.e. — » — • — are not all zero at the point.
dx dy dt
APPLICATIONS TO GEOMETRY OF SPACE 265
Then the equation (A) must be satisfied identically by these values,
and since the total differential of (4) when z, y, z are defined by
(2?) must vanish, we have
^~v dFdx dF du dF dz ^ ^ ^ . + *~
(£) T^ + T^ + lT3:=  By (52), p. 196
v J dx dt dy dt dz dt J v y> *
This equation shows that the tangent line to (7, whose direction
cosines are proportional to
dx cty dz
(ft eft <ft
is perpendicular * to a line whose direction cosines are determined by
the ratios dF dF dF
dx* dy * dz *
and since (7 is any curve on the surface through P, it follows at once,
if we replace the point P(x^ y, z) by I{(x v y x , « x ), that all tangent
lines to the surface at P x lie in the plane *
&*! 0Jf x ^1
which is then the formula for finding the equation of a plane tangent at
(x , y , 3j) to a surface whose equation is given in the form
F(x, y, z) = 0.
In case the equation of the surface is given in the form z =/(#, y), let
(D) F(z, y, *)=/(*, y)* = 0.
Then ££ = §£=*, ££ = §£ = !*, ^ = 1
ftc &c 3a: #y dy dy dz
• From Solid Analytic Geometry we know that if two lines haying the direction cosines
Coaa lt cos ft, cos 71 and cos tf 2 , cos/3 s , cos y% are perpendicular, then
cos <Xi cos a 2 + cos ft cos ft + cos 71 cos 72 ™ 0.
oFt vF\ cF\
t The direction cosines of the normal to the plane (71) are proportional to g— * » g— 5 » •=— * •
Hence from Analytic Geometry we see that (C) is the condition that the tangents whose
direction cosines are cos a, cos ft cos 7 are perpendicular to the normal ; i.e. the tangents
must lie in the plane.
% In agreement with our former practice,
dF\ dT\ BF\ d£t dz\
bx x ' byi' dzi ' fa x ' by 1
denote the values of the partial derivatives at the point (z lt y lf z x ).
266 DIFFERENTIAL CALCULUS
If we evaluate these at (j^, y t , z^ and substitute in (71), we j
CD §C* r .) + ^W»,)^2,) = o,
which is then the formula for finding the equation of a plane tangent a
(z$ y,, *,) to a mrfaee whote equation it given in the form z =f(x, }f).
In 5 128, p. 187, we found (65) the total differential of a (unction u (or t) of x
(B) dx = — dx + — dy.
We hate now a means of interpreting this result geometrically. For the t
piano to the surface z = /(z, y) at (x, ji, *) is, from (72),
X, r, Z denoting the variable coSrdinates at any point
X = x + dx and J" = y + dy
in (F), there results
»*— 5* + S*'
Comparing (E) and (».'), we get
(ff ) d* = Z  i. Hence
Theorem. 7"fte (oCai diflerenital
c/a/unri«m/(z, V) corresponding
lo (Ae increments dz and dy o/uii/.t
i/ie corresponding increment of (Ae
zco6riinale of the tangent plane to
the lur/ace z ~f(x, y).
Thus, in the figure, PP" is the
plane tangent to surface PQ St
P(z, y, l).
Let j4B = i
e substitu
P'(J.r.g)
s^f
X
4
Ly^
j'lX.V,*,;
/
/A dx
/* X
/&
and CD = dy ;
i = DP'DE = EP'.
162. Normal line to a surface. The normal line to a surface at e
given point is the line passing through the point perpendicular to t
tangent plane to the surface at that point.
The direction cosines of any line perpendicular to the t
plane (71) are proportional to
*■
dF,
dF,
APPLICATIONS TO GEOMETEY OF 8PACE 267
are the equations of the normal line * to the surface F(z, y, z)=Q at
( x i' Vv *5
Similarly, from (72),
(74> xXi ^ VVx ^ z*x
dz t dz x l
are the equations of the normal line * to the surface z =/(s, y) at (x 9 y v z % y
1. Find the equation of the tangent plane and the equations of the normal line
to the sphere z» + y 1 + *» = 14 at the point (1, 2, 8).
Solution. LetJ^z, y, z) = z» + y*+z* 14;
then g=2z,^=2y,?=2*;* l = 1,^ = 2,^ = 8.
,3 = 2,^ = 4,^ = 6.
d*i &V\ fa\
Substituting in (71), 2(zl) + 4(y 2) + Q(zS) = 0, z + 2y + 8s=14, the
tangent plane.
Substituting in (78), Xj1 = y2 = 5^ ^
giving % = Zx and 2 z = 8 y, equations of the normal line.
2. Find the equation of the tangent plane and the equations of the normal line to
the ellipsoid 4x« + 9y* + 864 s = 86 at point of contact where z = 2, y = 1, and * is
positive. 4ns. Tangent plane, 8(x  2) + 9(y  1) + 6 Vll (z — J Vll) = 0;
normal Une,^ = ^ = ^=^.
3. Find the equation of the tangent plane to the elliptic parabola z = 2x f + 4y*
at the point (2, 1, 12). Ans. 8z + 8y — s = 12.
4. Find the equations of the normal line to the hyperboloid of one sheet
** — 4y* + 2z* = 6 at (2, 2, 8). Ana. y + 4x = 10, Sx  * = 8.
5. Find the equation of the tangent plane to the hyperboloid of two sheets
** V 1 ** 1 a / v a *& ViV *i* 1
5 " e? " * = * " (x » *• '•>• ^ * * " * = *'
6. Find the equation of the tangent plane at the point (z^ y v s 1 ) on the surface
ax* + 6y* + c*» + d = 0. 4ns. aa^x + ty x V + c^s + d = 0.
7. Show that the equation of the plane tangent to the sphere
x a + y 1 + * t + 2Xx + 21fy + 23rs + D =
at the point (z^ y lt z x ) is
«i* + y x y + ^ + £(z + x x ) + If (y + y x ) + iV(« + * x ) + D « 0.
9 See second footnote, p. 266.
208
DIFFERENTIAL CALCULUS
8. Find the equation of tlie tangent plane at any point of the surface
i' + V* + *' = a*,
and show that the sum of tin' squires <>f die iiil.oircpls on tin' axeri in;iik> liy the tangent
plane is constant.
9. Prove that the tetrahedron formed hy the coordinate planes and any tangent
plane to the surface xyz — a a is of. constant volume.
10. Find the equation of the tangent piano and the equations of the normal line to
the following surfaces at the points indicated:
(a) 2z' + 4), , i = 0; (2, 1, 12). <d) Sx* + j/«  2e = ; x = 1, y = 1.
(I)) z» + 4tf 5 ** = I6; (1,2, 1). (e) i'j' + 2i + ** = 16; I = 2, y = I.
(c) z s + j,'M' = ll; (3, I, 1). (f) l s + 3^ + 2l' = 9; Jf = l,* = l.
163. Another form of the equations of the tangent line to a e
curve. If the curve in questiun be the curve of intersection AB
of the two surfaces F(x, y, z) = and
G(x, y, z) = 0, the tangent line FT at
F(x^ y t , Zj) is the intersection of the
tangent planes CD and CE at that point,
for it is also tangent to both surfaces and
hence must lie in both tangent planes.
The equations of the two tangent planes
at P are, from (71),
\SF.
<H
BF.
U
BG L . bg 1
Taken simultaneously, the equations (75) are the et/ttatiofis of tht
tangent line FT to the tkew curve AB. Equations (75) in more com
pact form are
'— x. y — ff, z — z l
(76,
t>F, 3G, dF, 3G, dF, dG i BF 1 BG t BF l BG l BF X aG l
By x 3z l Bz x By x Bz x Bx L Bx t 3z x dx l 3y t 8y x Bx±
flF, BF [
By x Bz x
BG i BG }
By L Bz,
BF 1 BF 1
Bz t Bx,
e>G, BG %
Bz x dX x
ftp, 9F t
BXt By x
BG t BG 1
Bx x By x
using the notation of determinants.
APPLICATIONS TO GEOMETRY OF SPACE 269
164. Another form of the equation of the normal plane to a skew
carve. The normal plane to a skew curve at a given point has already
been defined as the plane passing through that point perpendicular to
the tangent line to the curve at that point Thus, in the above figure,
PHI is the normal plane to the curve AB at i*. Since this plane is
perpendicular to (77), we have at once
(iiO = o,
dF , dF t
aG 1 oG 1
(**d+
dF 1 dF t
dG L aC,
Grft) +
aF t dF x
9G ± aG 1
dx x dy x
the equation of the normal plane to a skew curve.
1. Find the equations of the tangent line and the equation of the normal plane at
(r, r, r V£) to the curve of intersection of the sphere and cylinder whose equations
we respectively z* + y l + *■ = 4 r*, ** + y* = 2 re
solution. Let F = 4* + y* + z* — 4r* and Q = & + y*2re.
3 = 2r, "i = .r, 3 = 1,/!,;
Substituting i
(77),
= «rVS .
_Va o i '
or, y = r,z+ V2* = 8r,
the equations of the tangent FT at P to the
curve of Intersection.
Substituting in (76), we get the equation
of the normal plane,
 V5 {«  r) + (y  r) + (z  r V5> = 0,
or, YiX — z = 0.
2. Find the equations of the tangent line to the circle
X* + y* + *• = 25,
z + * = 6,
at the point (2, 2 VS, 8). 4ns. 3z + 2 VJJ? + Si = 25, x + z =
8. Find the equation of the normal plane to the curve
x 1 + v* + ■* = *",
x*re + y* = 0,
I* (*it Vn «i). <*"*• 2y,s,z(2z 1 i')* 1 jfri< 1 I =
270 DIFFERENTIAL CALCULUS
4. Find the equations of the tangent line and the normal plane to the ourre
2x 1 + 8y* + * f = 9, s^Sx^ + y 1
at (1,  1, 2).
5. Find the direction of the curve
xyz = l, i/* = z
at the point (1, 1, 1).
6. What 1b the direction of the tangent to the curve
y = x\ z* = 1 — y
84(0,0,1)?
7. The equations of a helix (spiral) are
x 1 + y 1 = t^ l
y = xtan.
c
Show that at the point (X, y v zj the equations of the tangent line are
c(**i) + Vi(**i) = °i
c(VVi)*i (**i) = 0;
and the equation of the normal plane is
y x x — ztf — c(z — Sj) = 0.
x f v* z*
8. A skew curve is formed by the intersection of the cone — + ^r —  = and
a* W cr
the sphere x a + y a + s* = r*. Show that at the point (x 1? y lf z t ) the equations of the
tangent line to the curve are
c»(a«6«)x l (xx ! )=a a (6 1 +c«)* 1 (s« l )»
c*(a*~ b t )y l (yy l ) = + 6*(c« + a*)*^*  * t ) ;
and the equation of the normal plane is
<*■(&* + cfyyftx — 6*(c f + a^z^y — ^(a*— ft^x^* = 0.
CHAPTER XXI
CURVES FOR REFERENCE
For the convenience of the student a number of the more common
lnres employed in the text are collected here.
Cubical Parabola
Srmicubical Parabola
o
y=*aaf.
jfzsaa?.
The Witch of Agnksi
The C188OID of Diocles
«V = 4o s (2ay).
y*(2a *) = *».
271
272
DIFFERENTIAL CALCULUS
The Lemniscate of Bernoulli The Conchoid of Nicomedes
ae
/D*=a*cos 2 0.
Y
O
p=a esc + b.
Cycloid, Ordinary Case
x = a arc vers
Z^2aytf.
a (0 — sin 0),
a(l — CO8 0).
Cycloid, Vertex at Origin
£ =
a arc vers+v2
a
a (0 + sin 0),
a(l — cos#).
«yy*
Catenary
s
y=
2(^+« a >
Parabola
<r*+y*=a*.
CURVES FOR REFERENCE
273
Hypocycloid of Four Cusps Evolute of Ellipse
{
x = a cos 8 0,
y = a sin 1 0.
(as)* + (ty)*=(a'i*)*
Cardioid
Folium of Descartes
at+tf+OZ
1 ^^»
aVz* + y*.
a^+y , 3aay = 0.
/> = a(l
cos 0).
Sine Curve
Cosine Curve
Y
T
.i ,.
"hi
>. 1
r
i ^/
*
T
f\
1*
s
>^ li
yx
•
y=
sin z.
y = coss.
274
DIFFERENTIAL CALCULUS
LlMACON
p = b — a cos .
/=**
a + x
a — x
Spiral of Archimedes
p = a0.
Logarithmic or Equiangular
Spiral
p = 4* , or
log p = ad.
Hyperbolic or Reciprocal
Spiral
LlTUUg
p0 = a.
p*0 = a*.
CURVES FOR REFERENCE
275
Parabolic Spiral
Logarithmic Curve
(pa)'=4ac0.
y = logs.
Exponential Curve
Probability Curve
y = «*.
y = a"  *.
Secant Curve
Tangent Curve
y = 8eca^
y = tang.
DIFFERENTIA! CALCULUS
ThreeLeaved Rose
p = a cos 3 ft
FourLeaved Robe
FourLeaved Robe
TwoLeaved Rose Lemnisgate
EiqhtLeaved Rose
^ a* sin 2ft
CURVES FOR REFERENCE
277
Cubve with End Point
at Origin
Curve with Salient Point
at Origin
y = x\ogz.
y(l + <) = s.
Curve with Conjugate (Isolated)
Point at the Origin
Curve with Cusp of Second
Kind at Origin
j^s 8 * 2 .
(y^y=^.
p = a sec  •
Equilateral Hyperbola
O
xy = a.
USTTEGRAL CALCULUS
CHAPTER XXn
integration, rules for integrating standard
elementary forms
165. Integration. The student is already familiar with the mutu
ally inverse operations of addition and subtraction, multiplication
and division, involution and evolution. In the examples which fol
low, the second members of one column are respectively the inverse
of the second members of the other column :
y = sin Xy x = arc sin y.
From the Differential Calculus we have learned how to calculate the
derivative/'^) of a given function f(x), an operation indicated by
or, if we are using differentials, by
df(x)=f(x)dx.
The problems of the Integral Calculus depend on the inverse operation,
namely :
To find a function f(x) whose derivative
(A) /(*)♦(*)
is given.
Or, since it is customary to use differentials in the Integral Calculus,
we may write
(#) df (x) =/'<>) dx = 4> (x) dx,
and state the problem as follows :
Having given the differential of a function, to find the function itself.
279
The function /(*) thus found is called an integral * of the given
differential expression, tho process of finding it is called integration,
and the operation is indicated by writing the integral sign^ I in front
of the given differential expression ; thus
m
J>C.)*« /<»),
read an integral off'(x)dx equals f(x). The differential dx indicates
that x is the variable of integration. For example,
(a) If f(x) = 3*, then f'(x) dx = 3 x*dx, and
/■
%o*ax=x*.
(b) If f(x) = sin x, then f'(x) dx = coh xdx, and
( cos ir<fe = sin x
/«
(c) If /(x) = arc tana, then f'(x)dx = r  — ^* an ^
i ; ■= — arc tan r.
' 1 + a^
A
Let ua now emphasize what is apparent from the preceding expla
nations, namely, that
JMfferentiation and integration are inverse operations.
Differentiating (C) gives
CD) dff<t)dx=f(x)dx.
Substituting the value of f'(r)dx[=df(x)] from (B) in ((7), we get
c*) _/V(*)/<»
Therefore, considered as symbols of operation, = and J •••dx are
inverse to each other; or, if we are using differentials, d and I are
inverse to each other.
• Called antriJrjrerend'rt/ liy some writers.
t Hlstorieally thin sign ia a distorted S, the initio! letter of the word num. Instead of
defining integration n* tin irivcrw nl rlifTi>r>'^tiiilii>u, we may nVtine it as a proves*, of si
matloD, a very important notion which wo will consider In Chapter XXVIII.
J Some authors writo tills D~ f'{z) when they wish to emphasize the fact that it ii
inverse operation.
INTEGRATION 281
When d is followed by / they annul each other, as in (i>), but
when / is followed by d, as in (^), that will not in general be the
case unless we ignore the constant of integration. The reason for this
will appear at once from the definition of the constant of integration
given in the next section.
166. Constant of integration. Indefinite integral. From the pre
ceding section it follows that
since d (V) = 3 x*dx, we have J 8 x*dx = x* ;
since d(z*+ 2) = 3 x*dz, we have f 3 x*dz = x*+ 2;
since d (a" — 7) = 3 z*dx, we have J %x i dx = x t — 7.
it.since d(x* + C) = 3z*dz,
where C is any arbitrary constant, we have
f3z'dx = x'+C.
I*
A constant C arising in this way is called a constant of integration.'
Since we can give C as many values as we please, it follows that if
a given differential expression baa one integral, it has infinitely many
differing only by constants. Hence
i>
and since C is unknown and indefinite, the expression
/« + <>
is called the indefinite integral off'(x)dx.
It is evident that if <£(■*■) is a function the derivative of which is
/(i), then <j>(x)+C, where C is any constant whatever, is likewise
a function the derivative of which is f(x). Hence the
Theorem. If two functions differ by a constant, they have the same
derivative.
It is, however, not obvious that if <t>(f) is a function the derivative
of which k/(x), then all functions having the same derivative /(a;)
are of the form
+W + G
where C is any constant. In other words, there remains to be proved the
* Coaitant here means that it is independent of the variable of integration.
282
INTEGRAL CALCULUS
Converse theorem. If two function* have the same derivative, tht
difference is a constant.
Proof. Let </>(V) and ^(V) be two functions having the commoi
derivative /(«). Place
F(x) = ^(x)^(xy i then
(A) n*>SC+C»>*<*)3A*)/<»)« Byhypothesis
But from the Theorem of Mean Value (46), p. 166, we have
F(x + Az) F(x) = tW'(* + 8 Ar). < t
.'. F(z + bx)  F(x) = 0,
[Bines by {J) the il.Tivi.iiv,. ..I F(x) is lero for nil values of *.]
and F(x + Ax) = F(x).
This means that the function
does not change in value at all when Z takes on the increment i
i.e. <f>(V) and ^"fV) differ only by a constant.
In any given case the value of C can be found when we know the
value of the integral for some value of the variable, and this will be
illustrated by numerous examples in the next chapter. For the pres
ent we shall content ourselves with first learning how to find the
indefinite integrals of given differential expressions. In what fol
lows we shall assume that every continuous function lias in indefinite
integral, a statement the rigorous proof of which is beyond the scope
of this book. For all elementary functions, however, the truth of
the statement will appear in the chapters which follow.
In all cases of indefinite integration the test to bo applied in veri
fying the results is that the differential if the integral must be equal
to the given differential expression.
167. Rules for integrating standard elementary forms. The Dif
ferential Calculus furnished us with a General Rule for differentiation
(p. 29). The Integral Calculus gives us no corresponding general
rule that can be readily applied in practice for performing the inverse
operation of integration.* Each case requires special treatment and
we arrive at the integral of a given differential expression through
• Even though the Integra! of a Riven flifferiTitiiil I'* invasion may lie known In r\ist. yet
It may not be possible for ua actually to find it in terms of know 11 f timis, because there am
functions other than the elementary functions whose derivatives are elementary functions.
INTEGRATION 283
our previous knowledge of the known results of differentiation. That
is, we must be able to answer the question, What function, when dif
ferentiated* will yield the given differential expression?
Integration then is essentially a tentative process, and to expedite
the work, tables of known integrals are formed called standard forms.
To effect any integration we compare the given differential expression
with these forms, and if it is found to lie identical with one of them,
the integral is known. If it is not identical with one of them, we
strive to reduce it to one of the standard forms by various methods,
many of which employ artifices which can be suggested by practice
only. Accordingly a large portion of our treatise on the Integral Cal
culus will be devoted to the explanation of methods for integrating
those functions which frequently appear in the process of solving
practical problems.
From any result of differentiation may always be derived a formula
for integration.
The following two rules are useful in reducing differential expres
sions to standard forms :
(a) The integral of any algebraic sum of differential expressions equals
i* same algebraic sum of tlie integrals of these expressions taken separately.
Proof, Differentiating the expression
i du+ j dv— i die,
i, v, w being functions of a single variable, we get
du + dv — dw. By III, p. 34
(l) .. I (du + dv— dm) = j du+ I do— I dw.
(b) A constant factor may be written either before or after the integral
Proof. Differentiating the expression
•/*
adv. By IT, p. 34
.'. j adv = a j dv.
On account of their importance we shall write the above two rules
is formulas at the head of the following list of
284 INTEGRAL CALCULUS
Standard Elementary Forms
(1)
I (du + dv — dw) = I du + I dv — J dw.
(2) J ado = a I dv.
(3) fdx = x + C.
t*fo =  + C. n=?tl
n + l
/do
_ = logi>+C
= log v + logc = log cv.
[Placing C»logc]
/a
crdv =  + C.
logo
(7) ( e>dv = e>+C.
J sin v dv = — cos v + C.
I cos i> <fo = sin v + C.
I sec*vdv = taixv+C.
I csc f vdv = — cot i> + C.
I 8eci?tani?<fc = seci? + C.
J CSC V COt VdV = — C8CV+C.
[14) I tan vdv = logsecv + C.
[15) I cot v dv = log sin v + C.
[16) / secvd^ = log(secv + tanv) + C.
[17) / esc p <fc = log (csc v — cot i?) + C.
<*)
W
[10)
[11)
[12)
[13)
INTEGRATION 285
/dv i v
z — 3 = arctan + C.
(19)
(20) / , = arc sin + C,
a
(21)
r7=^==iog(»+v^=fcfl«)+c.
t = arc vers
(22) f , = = arc vers^ + C.
(28)
f * 1 i» _
I — 7== = arcsec + C.
Proof of (3). Since <*(* + <?) = <&, n, p. 84
we get I dx=zz+C.
Proof of (4). Since , +i .
rf( Y + c, )= ^*» "f P 84
tftfo = + C.
This holds true for all values of n except n = — 1. For, when
n = — 1, (4) gives
/
which has no meaning.
The case when n = — 1 comes under (5).
Proof of (5). Since ,
d(logv + CT) = — , YIHa, p. 35
we get / — = log v +C.
The results we get from (5) may be put in more compact form if
we denote the constant of integration by log c. Thus
/
— = log V + log C a= log CV.
Formula (5) states that if the expression under the integral sign is a
fraction whose numerator is the differential of the denominator, then the
integral is the natural logarithm of the denominator.
286 INTEGRAL CALCULUS
EXAMPLES*
For formulas (l)(6).
Verify the following integrations :
/x e + 1 x 7
x*dx =  —  + C = ~ + C, by (4), where v = x and n = 6.
2. f Vxdx = f xi dx = y + C = x* + C, l>y (4)
where t? = x and n = J.
where o = x and n = — 8.
ax*
4. J\ix*dx = afx*dx = jp + C By (2) and (4)
B./x«dx = f + C. 12. r^ox = ^ + C.
* * 4
6. fxl dx = y + C. is. Cs~*d8 = 2 Va + C.
7. forf(tt = — + C. 14.^80^0^ = 0^ + 0.
fox 1 , „ 15. f 6m**«dz = ^!^ + C.
J 8x 2 8x ,
9 r 20x = 5xt + c 16 J^2 + C 
J ox4 2 a
lM 1
10
. fsydy = *!? + C. "• /(nx)  ax = (nx)*+ C.
11. fV2pxdx = ?x V2px + C. 18. Cy—iidy = — + C.
19. f(2x»  6x« 8x + 4)dx = f 2x«dx  f 6x*dx  f 8xdx + f 4dx by (1)
= 2fx*dx  6 f x*ax  8 fxdx + 4 fax by (2)
Note. Although each separate integration requires an arbitrary constant, we write down
only a single constant denoting their algebraic sum.
20. jY^ — + 8c\/x 5 )dx = f2ax"iax ftar'dx + fscxidx by (1)
= 2a fx'idx — bfx~*dx + Scfxidz by (2)
= 2a~ &.— + 8c — + C by (4)
i  1 *
= 4a Vx +  + cxt + C.
x 6
* When learning to integrate, the student should have oral drill in integrating simple
functions.
INTEGRATION 287
21. f(2x»3x« + 12x«8)dx = ^^ + 3x*.8x + C.
J 6 7
22. f(<& 4= + l\dx = ?*8xi L + C.
23. f(af  xfydx = a*x + afx* ?aW  ~ + C.
J 7 6 8
Hint. First expand.
26. /( Va  Vt)»<B = ah  2 arf + *^  *£ + C.
(x*  2) Vdx = =  ^f + 2 x«  2 x* + C.
10 4
27./(a« + « B *)l«to = ^±^)? + C.
Hnrr. This may be brought to form (4). For let t> =» a a + We* and ni; then dv  2 o%edx.
If we now insert the constant factor 2 6 s before xdx, and its reciprocal rrj before the integral
sign (so as not to change the value of the expression) , the expression may be integrated, using
(4), namely,
J n + 1
Thus, f(a*+b»z*)*zdx^ f (a*+Me«)*2Mecfa;i f ( a a + &**«)*ci(aa + M**)
"2M 1 + C 3P~ + C 
Notb. The student is warned against transferring any function of the variable from one
side of the integral sign to the other, since that would change the value of the integral.
28. fVa*  x*xdx = f (a«  x*)lzdz = J(* a  x*)* + C.
29. f(Saz* + 4ax»)t(2ax + 46x*)dx= T (8ax* + 4te»)i + C.
Hnrr. Use (4), making v 3a* 8 + 4 6a* eft;  (6 ax + 12 fcc*) dx and n
30. r6(6ax a + 8W)J(2ax + 4te^<fe = ^(6ax* + 8te»)* + C.
J 16
J (a* + x 8 )* »
Hnrr. Write this f (a* + x*)~*&dz and apply (4).
32. f *?_ =2VT^z + C.
J vr^x
*• /*■*(!? + 0** = sj (ya +p8)i + c#
34. f (1 + c*)*e»tfx = f(l + «*)* + C.
35. fain* x cos x<Zx = /"(sin x)» cos x dx = i?!^ + C = ?!^ + C.
Hnrr. Use (4), making v sin z, dvcoszdz, and n2.
288 INTEQEAL CALCULUS
36. fcof*xsinxdx=^^ + C.
J 6
/I
sin 8 ax cos axdz = — sin 4 ax + C
4a
38. fcos*8xsin8xdx= — — cos*8x + C.
J 15
39. f *** =Vfl«gg4C.
J Va*x a
J (&*)• (6t) 5
41. f vTTx'xdx = f (1 + x a )* + C.
^ /» ttf > " 1 dtt = (a***!**) 1 ™ , c
' J (a + bu*) m an(l — m)
r 2a8d8 a
Solution.
/to— /iw *•
This resembles (6). For let t> = d 2 + e*x* ; then dv = 2 ePxdx. If we introduce the
factor 2 e* after the integral sign, and — — before it, we have not changed the value
of the expression, but the numerator is now seen to be the differential of the denom
inator. Therefore
Jw + &t>' 2«V6»+<*r» 2<?J W + At* 2e* BV ' ' w
48 7i^6 =log(10x,+16)l+c 
Hint. First divide the numerator by the denominator.
57.
INTEGRATION 289
/X*"" 1 — 1 1
<te = log(x«— nx) + C.
x*— nx n
56. f(loga)«— = 1 (log a)« + C.
J a 4
56. ril±idr = £ + r + 21og(rl) + C.
•/ r — 1 2
•/ a + 6cosx b
69. f : J^M = llog(l + 8tan^) + C.
J l + 8tan0 8 ox '
J er — 1 l
61. r?Jlldr = log(<r + l) a r + C.
•/ c + 1
02. Integrate the following and verify your results by differentiation :
(a)/(4«*?)<to.
Solutioa. fUz*  \dz = ifzVx  2 J— = ^  2 log* + C.
Ferjfleation. <*^ 21ogx + c\ = fc • 8** 2 • \<te = ^4x» ?W
<b)JV(fc. Qi)f *>+•**. (n) f '"'££ • (t)/sin«^cos^<fe.
W /tV5fc. <!)/<**<*. (o)/5?*i «/££
(d) /*;*.. o)/^ („) / <"*—*** wjvrite*.
Proofs of (6) and (7). These follow at once from the corresponding
formulas for differentiation, H and Ha, p. 35.
290
INTEGRAL CALCULUS
For formulas (8) and (7).
Verify the following integrations:
1. fbcfi*dz = ^ + c.
J 2 log a
8olntk»&. fbcfl*dz = bfcfl*dz.
By (2)
This resembles (6). Let r2x ; then <*t>2<ix. If we then insert the factor 2 before dx and the
factor £ before the integral sign, we have
2. f8e«dz = 8e* + C.
e»dx = ne" + C.
4. f e^'cosxdx = e  "* + C.
log i
7. fer*dz =— e* + C.
8. fe«dx = — + C.
J a
aaarfs = _!! + C.
By(<5)
2
10
2 log a
&V1
/yaw— *
21oe8
11. Jatfti* = , a T + C.
1 + loga
4. f(a~ 6*)<fe = J£  £L. + C.
•^ nloga mlogo
/* * * *
(*» + e~")dz = ate 5 — e~ a ) + C.
6. /V + er')*dy = ♦(«■» e**) + 2y + C.
7 . r(^^ax = ^l^.2x + C.
«/ a*o* log a — 1<
— q«ft*
log a— log 6
c* x a** 86**
9 /( ea<+e ~ a< ) 8<a = "P
+ 8e°<8e a<  ^
8 8
20. Integrate the following and verify your results by differentiation :
(a) fe*«d«. (e) fe**dx. (i) fsc^dx. (m) fafxdz.
o«*to. (f) J2ftdt. Q)Je*dx. (n)Je»dV.
(c) fc°*dz. (g) C&HFdx. (k) faer~*dx. (o) C(#*)*dz.
(d)/— • (h)/ 0)/^ (p)/j
fc»7
INTEGRATION 291
(q) f (£■*»♦ cos *d*. (s) fe*"»hfoi 0d0. (u) fefi**eec?tdt.
(r) f («* + e "*«)*(fe. (t) C0**xdx. (v) fa***— •
Proofs of (8)(13). These follow at once from the corresponding
formulas for differentiation, H, etc., p. 35.
sin vdv
cost*
— sin vdv
Proof of (14). f tan vdv = f
— r
J COS V
/ df(wv)
cost?
= — log cos v + (7 by (5)
= log sec t>+C.
[Since  log ooar log — — — logl + logieo v— logMotr.l
■00 V J
_   , € _ N Z 1 , , rcos vdto /*rf(sin v)
Proof of (15). I cot vdfv = I — : = I — ^ *
v ' J J smv J smv
= log sin v + C. By (5)
„ e  X<1AN . secv+tanv
Proof of (16). Since sec v = sec v
v ' secv+tanv
sec v tan t? + sec a v
sec v + tan v
/, rsec v tan v + sec a t> ,
sec vdv= I  dv
J sec v + tan v
/ d (sec t? + tan v)
sec v + tan v
= log (sec v + tan v) + C. By (5)
n— *_c/«i.x c cscv — cotv
Proof of (17). Since esc v = esc v —
v C8CV — cotv
— cscvcottt+cscSt
C8C V — COt V
/, Z 1 — csc vcot t; + csc*0 ,
esc vdv = I av
j csc v — cot v
/ rf(cscv — cot v)
csc v — cot V
e= log (csc v — cot v) + C. By (5)
292
INTEGRAL CALCULUS
For formulas (8)(17).
Verify the following integrations:
.. C . « j cos2oz . „
1. I sin 2 axdx = + C.
J 2a
Solution. This resembles (8). For let v = 2 ax ; then dt> = 2 adz. If we now insert
the factor 2 a before ax and the factor — before the integral sign, we get
2a
/sin 2 axdx = — (sin 2 ax • 2 adz
2aJ
1 /• 1
= — I sin 2 ax • d(2 ax) = cos 2 az + C.
2aJ v ' 2 a
By (8)
cos 2 ax
2a
+ C.
2. /cos mxdx = — sin mx + C. 7. f csc ay cot aydy = csc ay + C.
J m J a
3. f tanftzdz = log sec te + C. 8. f csc*8zdz =— JcotSz + C.
/I /* z z
sec axax=s log (sec oz+ tan ax)+C. 9. I cotdx = 21ogsin + C.
a «/ 2 2
6. f csc  dz = a log (esc cot ) + C. 10. f sec^z 8 • z*dz = J tan z 8 + C.
6. fsec8itan8fctt = isec8i + C. 11. f— — = cotz + C.
12. f —  = tan« + C.
«/ cos 2 *
3. f (tan* + cot0)*d0 = tan* cot* + C.
4. /(sec a — tan a)*da = 2 (tan a — sec a) — or + C.
6. f(tan 2a — 1)% = £ tan 2s + log cos 2a + C.
6. fl cos — sin 80
7. M sin ax + sin
. 1
cos ax— a cos  + C.
a
k f/cos — sin80jcW = 88in^ + ^co880 + C.
. /(sin ax + sin) ax =
/It
kcos(a + by)dy =  sin(a + by) + C.
9. Tcosec'z 8 • z*dz =— J cotz 8 + C.
20. fcos(logz) — = sin (log z) + C.
J x
21. f = — cotz+ C8CZ + C = tan + C.
J 1 +
cosz
2
Hint. Multiply both numerator and denominator by 1  cos z and reduce before inte
grating.
22.  = tanz — secz + C
J 1 + sinx
INTEGRATION 298
S3. Integrate the following and verify the results by differentiation :
W /*.?**. 0>)f~ <o)/(tan4.cotj)d..
(b) /cot e* • «te. (i) /tan  <fe. W /<«*» !)' <**•
/> « a r fa) f(aectl)*dt.
(c) / sec  tan ^ «W. (j) J csc*(ate)(fa. ^
0) f(lcscy)«<»y.
<d)/csc£cot^. (k)/ * J
sin*40 ,_* f <&
dy
(e) fcos(& + ax)d2>. 0) f^TT M f (fa
(f) fsec* 2 axdz. (m) f(aee20 cac?\d0. ^ r 2adt
fe) /c^W (D) / (tan * + "° +**+ (V) / 6 m
C08X
COB 8^
Proof of (18). Since
dP^tai + c)^^^ byH,p.36
we get I z 5 =  arc tan  + ft*
J t^+a a a a
Proof of (19). Since ^—, = J (1 LV
= ^{log(va)log(t; + a)} + C by (5)
* log^ + ft
•Also <f/arccot + C^ — z^and f— ^1— larccot + C". Hence
\a a / t^ + o* J w 2 + a a a a
f^larctan% Clarccot? + C'.
J tp + a a a a a a
Since arc tan  + arc cot    , we see that one result may be easily transformed into the other.
<* a 2 v v
The same kind of discussion may be given for (20) involving arc sin  and arc cos  , and for
a a
(28) involving arc sec  and arc esc  .
a a
t By breaking the fraction up into partial fractions (see Case I, p. 325).
294 INTEGRAL CALCULUS
Proof of (20). Since
a v 
d ( arc sin  + C )==— =4— = t=^L=> by XVIII, p. 35
^ a } ^og 5 ^^
* f dv . v . _
we get I , = arc sin  + C.
Proof of (21). Assume v = atanz, where z is a new variable;
differentiating, dv = a sec* 2J2. Hence, by substitution,
/ <to _ C a sec* g(to _ /* sec* g(to
Vv*+~a*J Va* tan* 2 + a* "J Vtan 2 « + 1
= / sec akfe = log (sec 2 + tan z) + C by (16)
= log(tan z +Vtan*« + 1) + e. By 28, p. 2
But tan z = ; hence,
a
JV£W^
+ c
= log + c
a
= log(w +\V+ a*)— log a + &
Placing (7 =— log a + <?, we get
In the same manner, by assuming v = a sec z, dv = a sec 2 tan zcfe
we get , ^ ,
P dv ra sec 2 tan zoz /* ,
I = I = 1 sec 2k*z
J Vt^— a* J Va* sec* « — a* J
= log (sec « + tan 2) + by (16)
an log(sec z +Vsec*« — l) + e by 28, p. 2
Proofs of (22) and (23). These follow at once from the corre
sponding formulas for differentiation, XXII and IHY, p. 36.
INTEGRATION
295
A large number of the fractional forms to be integrated have a
single term in the numerator, while the denominator is a quadratic
expression with or without a square root sign over it The following
outline will assist the student in choosing the right formula.
Numerator op First
Degree
Numerator op Zero Degree
No radical
in denominator
/£ = !og„ + C
r do 1 v
I —  =  arc tan  + C, or,
J v* + a* a a
f dv = Mog'' + C
J«*a* 2a % + a
Radical
in denominator
/ rrdo = + C
J n + 1
f = arc sin  + C, or,
J Va 2 «* a
f J__log(i> + V«*±a») + C
J V«* ± a*
Students should be drilled in integrating the simple forms orally
and to tell by inspection what formulas may be applied in inte
grating examples chosen at random.
For formulas (18)(28).
Verify the following integrations :
, C dx 1 . 2x t ,
1. I —  —  = arctan —  + C.
J 4** +9 6 S
Solution. This resembles (18). For, let «* = 4 x 2 and a 3 = 9 ; then v = 2 x, dv = 2 dx,
and a = 8. Hence if we multiply the numerator by 2 and divide in front of the
integral sign by 2, we get
2<2x \ r d(2x)
/dx _ 1 r 2dx _ 1 r
4x 2 +9 " 2 J (2x) 2 +(8) 2 ~ 2 J
(2x) 2 +(8) 2
=  arc tan — + C.
6 8
By (18)
. C dx 1 . 8x2 , n
•/■
6
■/
dx
3
dx 1 . 8x lP
— =  arc sin \C.
Vl69x 2 8 4
Vx 2 +9
6dx 6
= log(x+ Vx 2 +9) + C.
 /* odx 6 A x , _
7. / — —  = arctan + C.
J x 2 + 9 8 8
A r dx . x . n
4. I , = arc sin  + C
J VlT^x 2 8
Wx
/» MX
ax — c
2ac ax + c
log — + C.
x» + Vo"
/dx , / rz — «\ ~ A r7z*dz 7 . x' + VS ^
296 INTEGRAL CALCULUS
10. f = >rc sinx 2 + C. 13. f m  = arc vera  + C
J Vl^x* * J Vdxx 2 »
,_ /• dx 1 2x , „ , /• edt e . W + a
11. I = arc aec — 4» C 14. I — — r^r = log — — —
J xV4x 2 9 ^ 8 Jc 2 ^ 2a6^«a
12 . r^T = ^arctan^+C. 16. /^ = arc sin * + C.
Jx* + c* 2C 2 e 2 J vT^
16. /—= = —= arc sin <%/• + C.
•/ VSSs 2 V6 ™
17. f _^_ = 4= 1 °g(^ p + Vgt?I " 6 ) + cr 
J VaxPb Vo
, Q /• cob ada 1 /sina\ t „
18. I — — — =  arc tan ( 1 + C.
J a 2 + «ln* a a \ a /
19. f — — ====: = arc8ln(logx) + C.
J xVllog^x
20. f — J?= = log(ex + Vo 2 + <*x 2 ) + C.
21. C—JL= = i log (ty + V6V  a*) + a
J V&V  a 2 *
22. I — — = arc sin —  — + C.
J Va 2 (u + 6) a «
_ /• ads a * — c ~
23. I — — =  arc tan —  — + C.
J ( Z _ c ) 2 + ft 2 6 6
24. I  — =  arc tan J C.
Jx 2 +2x + 6 2 2
Hint. By completing the square in the denominator, this expression may be brought to
a form similar to that of Ex. 17. Thus,
Here »  x + 1 and a = 2.
26. I — = arc sin + C.
J V2 + x  x 2 3
Hint. Bring this to the form of Ex. Id by completing the square. Thus,
/dx r dx /• dx r dx . 2sl ^^
/ / / / , = / 7=arcsinr+a Bj(20
V2+sa:* J V2(* 2 a;) •/ V2(a 2 a+J)+J J Vf(xJ) 2 8
Here t; a; J and a.
26. f_^_ = :4 = arctan^± 1 + C.
•J 1 + x + x 2 Vs V5
o /• dx 1 /• dx 1 Sx1 , n
*<• I ^—3 — « : = « I = r = — ;= ar c tan — —  + C.
•/8x 2 2x + 4 8Jx 2 jx + J VTI Vll
INTEGRATION
29T
/• dx _ 1 r dx _ 1 r dx
" J V28x4x*~~2./ Vjfxx*~2J Vj(x»+ Jx + ^+tf*
29
dr
1 . 8x + 8 , n
=  arc gin — —  + C.
* Vii
= arcsin(2x8) + C.
. r7
J v8x — x a — 2
31. r__^_ = JLiog^±izL^ + c.
82. f * = log(t + l + V<»+* + lWo.
J Vl + 1 + V 2 /
83. f **  = arctan(2«l) + C.
84. f _^L_ = log(s+o + V2a* + s») + a
J V2M + S*
86. f —
„ /• 8x*dx 1 <n . , „
36. /——== =  arc vera 18 x* + C.
J Vx»9x« 3
3T. r^±^ = 5arctan? + log(a»+x») + C.
J a* + x* a a 2
= — arc sec — + C.
oft
Hint. A fraction with more than one term in the numerator may be broken np into the
ram of two or more fractions having the several terms of the original numerator as numer
tors, all the denominators being the same as the denominator of the original fraction. Thus,
the last example may be written
/ (b + ez)dz r bdx C exdz . r dx r zdz
a* + x 2 "J o 2 + x 2 J a* + x»" J a a + ** e J a* + x*
each term being integrated separately.
6 . xVsVI ,,
log —   + C.
2V5 xVS + V2
8a2
40. f *? i ds = 3 V9^~?~ 2 arc sin j[ + C.
41. r4==^ = Vx«T4 + 8iog(x + Vx«Ti) + o.
J Vx* + 4
42. f^M = lVsW^^±\o g (tVs+Vs^r9) + o.
J VaFo 8 vs
!98 INTEGRAL CALCULUS
43. Integrate the following expressions and verify your remilts by differentiation
W J S  12 z»
J VB4coh»S
nin SdS
J V26x'4
(j) f M " .
J i/3$«2
g)f * ,
J xV9x»16
■/6i»+l
V4(logi)»"
J Vl+9.
■* V8 + 4j
(r) r ■** .
J Va3fl»
0/. «
J V7e4t» + 6
, , /• <"*dr
■* Vl9«>
(w) f t 2 '* 3 )** .
wf M *
J V««— 4t + S
168. Trigonometric differentials. We shall now consider some trigo
nometric differentials of frequent occurrence which may be readily
integrated by being transformed into standard forms by means of
simple trigonometric reductions.
nple I. To find j sin" x cos" xdx.
When either m or n is a positive odd integer, no matter what
the other may be, this integration may be performed by means of
formula (4), „ ^+1
I v*dv =  .
For the integral is reducible to the form
/ (terms involving only cos r) sin xdx,
when sin x has the odd exponent, and to the form
I (terms involving only sin x) cos xdx,
when cos x Itas the odd exponent We shall illustrate this by
of examples.
+ J"<siu*)«<
INTEGRATION
I LLC STEATITE EXAMPLE 1. Find I sill' J (MS 1 JlZj.
Solution, j aiD'a; cos'xdx = fsin'zcos'lcos.wiz
= faln'xfl  sin'i)' cosxdz
= J* (sin's  2rin«a + Bln»z)co8;ette
= /(sin*)* cos wis 2j(8in*)« coaxd
_dn»i 2 sin's , do's , „
—5 — + ~ +c 
Here t> = aim, do = cosidr, and n = 2, 4, and 6 respectively.
> Example 2. Find fco&xtU.
Solution. j cob^/iIz = iuuti'i ciw/dz = f (1 — ain'jjw
EXAMPLES
wx + C. 6. /'sin'aScOBUfliW^^ + C'.
hC. 6. fcotf26aln26d0=
J sin**
S. J co8*Bwn'«tr = — Jcos's6 + }cos'* + C.
10. fvafix&i = cosx + ?cos*s  ^^ + C.
11. fcos'sd3 = BinxBin*z + ^^rC.
12. fsill' O^COS 1 *Q> = ^sinVfl, ^sillV^ + C.
* This was Integrated by the power formula taking ti — 1, i> slni, rftJ = co9 zrtz. To illus
trate how an answer may take on different tonus when more than one method ot integration
is possible, let us take n— 1, v cos z, dv sin xdx, and again integrate by the power
formula. Then
sin x cos xdx
  I (cos x) ( sin
x dx) =
a result which differs from the first one Id the arbitrary ci
coa'a „, 1  sln*a „, 1 sto'i .
2 2 ~22
Hence, comparing the two answera, 0—  + 0*i
300 INTEGRAL CALCULUS
13. f flini 9 cos* «dS = \ sit J 8  fl sin V S + A «inV* + C.
14. r^i^j^/l
•' Vcosy \
U + COB*!/) +C.
Vaini
16. Integrate the following expressions and prove jour results bj differentiation :
(b) fsln'ndcos^midi.
(a.) fs\n*2$d0. (t) fcosfax sin oi
(b) fca*\&. (g) Jsin^cos^dz. (1) Jsin'nWi.
(c) rsin2jco9 2ziii. (h) f coa?& zslu&zdz. (i n ) f siii*.c cosnti.
(d) fsiu'tco&tdt. (i) f 'sin« ftaccwtsd*. (n) fcos*y smydy.
(e)fcos*8inile. (j) f C0H '£ *w\** (o) fcoe»(a + M)dt
Example II. To jbd jtm'xdx, or fcofxdx.
These forma can be readily integrated, when n U an integer, on
somewhat the same plan as the previous examples.
Illustrative Example 1. Find I tiLii* xdx.
Solution. fta.n«xdj = ftan a *(sec*a  l)4r by 28, p. 2
= f tan'xsec'xtlz — ftan s zdx
= f(lJu>*)>d(tan;B) f(«ec*xl)<fc
x + a + C.
tan'/
Example III. To find j »e<fxdx, or j cs<Txdx.
These can be easily integrated when n is a positive even integer,
s follows:
Illustrative Example 2. Find j sec'wiz.
faec , tdi= Atan'i + lJ'sec'ajdi by 28,
= r(tana)*eec s nte + 2 MtanxJ'sec'xdx + ^see^xdx
When n is an odd positive integer greater than unity, the best plan is to redm
ine or cosine and then use reduction formulae on p. 303.
INTEGRATION
801
Example IV. To find j tan m xted t xdx, or j cot" x cttf xdx.
When n is a positive even integer we proceed as in Example
in.
Illustrative Example 3. Find f\a.n*xiec i xdx.
Solabcm. f Un»z sec* ids = f tan*z(tan's +■ l)sec ! zds by
28, p. 2
= f (tan;t) , MC , »fcc+ f tan' ;c sec 1 iris
BrM
tan's tan's , „
Here e = tan s, dt — sec' wis, etc.
When m is odd we may proceed as in the following example.
Illditbative Example 4. Find I tan's gee* ids.
Solution. j tan'zsec'sds = f tan's sec'z Bees tan sds
= Msec' 2 — l^sec's sees tan xdz bj 28, p. 2
b f(sec'z — 2sec*s + sec'sjsecs tansds
see's 2 aec*s Bee's „
By (*)
Here b = sees, do = sec* tan nil, etc.
EXAMPLES
I. fuux'zto a ^ + logcoaz + C. 3. fco\*xdx * ^  logait
ix + 0.
2. f t*n«2sds = ^^  s + C. 4. /cot'sds =  cots  s + 0.
5. ("ootids =cot'? + 3 cot? + s + C.
6. J* cot' ode =  J cot' a + i cot" a + log sin a + C.
7. /"tan"^di/ = tan*?2tan"^ + 41og8ec^ + C.
„ /• , , tan T z Stan's .... . n
8. 1 sec'sds — —  +  —  — + tan's + tans + C.
9. fcsc'zds =— cots— f cot's— lcot*s + C.
10. j"t*n* + *Ki**at = H^* + ^* + c.
11. /"taD«esec t SdS = J*ec'tflsec'ff + C.
._ f , , , cot's cot'j. „
12.  cot's esc* j&r = rC
J 6 8
... r 1 j „ Stands 2tan"z „ _.
13. J tan' x Bee' ids = — —  + —  — + C.
302 INTEGRAL CALCULUS
14./t an . I ,^^=2 W cl 1 ,(^'^ + i) + C.
J tan« a 3
16. f(tan a * + lan«*)i( = iU»n'r + C.
17. f(t»n( + cot ()' & = t (tan a l  eot'i) + log tan'i + C.
18. Integrate the following expressions and prove your results ly differentiation
(a) /tan' 2 tdl. (g) feec 1 S tan' fttf.
(»>/^7*
(b) j"cot* i to. (h) j" escV cot># <*#.
""/£=r fc
(c) ("tan'owla:. (ft f adx .
(»)/*■*■
(d) Jcof ?di. (j) rtan'taee'&a.
(p) 1 IMC* xdx.
< e) /J£ (kj/cot^W
(tj) f tanz sec'idi.
W Jcot*4tf W Jcot*S
w /«»»•«*■
Example V. To find j sin M x cotT xdx by means of multiple anglei.
When either m or w is a positive odd integer, the shortest method
is that shown in Example I, p. 298. When m and n are both positive
even integers, the given differential expression may lie transformed by
suitable trigonometric substitutions into an expression involving sines
and cosines of multiple angles, and then integrated. For this purpose
we employ the following formulae :
Bin u cob u = J sin 2 u.
86, p. 2
sin'w = J — £ cos 2 m,
88, p. 2
COB 1 !* = J + J cos 2 u.
39, p. 2
Illcbtkative Example 1. Find fona'xdx,
Solution. fcoB*xdx= f{i +  cos 2 x) At
86,9.1
.j/» + I/«.«*=j+
!ln2z + C.
Illustrative Example 2. Find fsin'xcofl'zdx.
Solution rBln'icos'mlt = i fsin'2niE
M.p.S
= if UiCOBi^dX
88, p. 2
= — ain4i + C.
8 32
INTEGRATION 308
Illustrative Example 8. Find fain 4 xcoe* zdx.
Solution. J sin 4 * co& xdz = J (sin x cos x) a gin* zdx
= y*isin s 2x(iicos2x)dx 86, p. 2 ; 88, p. 2
t =i fain* 2 xdx — i fain* 2 x cos 2 xdx
= i f(i co«4x)dxi r8ln s 2xcoe2xdx
_ x sin4x rin > 2x ^
""16 64 48
Example VI. To find J rinmxcoinxdx, J sinmxsinnxdxjor I cosmz
eo% nxdxy when m^n.
By 41, p. 2, sin mx cos nx = £ sin (wi + n) a? + J sin (wi — n) a^
.*. I sin mx cos rmir = J I sin (m + n)xdx + \ J sin (m — n)xdx
_ coa(m + ri)x cos(ro — n)a ^
"" 2(m + n) "" 2(ron) *
' Similarly, we find
, sin (m + n)a; , sinCm — tOa: , ^
sin wis sin nxdx = ^ f H — ^ ^ — h C,
2(m + n) 20w) *
, sinCwi + n^a; . sin(wi — n)x , ~
cos mx cos raor = — ^ ^— + 7TT r— + &
2(m + n) 2(win)
1. fcos*xdx = ? + sin2x + C.
J 2 4
o C • 4 j 8ac sin2x , sin4x 
3^co«*xdx = ^ + ^ + 5lal£ + C.
4. r 8 in«xdx==lf6x48ln2x + ^^ + 7 Bln4aJ )+ C7 
* 16 \ 8 4/
J 16 \ 8 4/
a fi 4 « _j sin 1 2a . a sin 4a , ,
6. ( sin 4 a cos 2 ada = H + (7.
J 48 16 64
7." Jsin 4 « cos*&tt = L (st  sin4t + ^^)+ C.
8. fco««xsin*«i* = — ^6x + sin»2x sin4x^lf > \ + C.
J 128 \ 8 8 /
INTEGRAL CALCULUS
11. /cos4»co«7ailt =  — — + ^^ + C.
J 22 6
169. Integration of expressions containing Va 1 — x* or Vx' ±a* by
a trigonometric substitution. In many eases the shortest method of
integrating such expressions is to change the variable as follows:
When v a' — x s occurs, let x = a sin Z.
When v a s + x* occurs, let x = a tan z.
When V** — a a occurs, let x=a tea X.*
ILLUSTRATIVE ExiKI'LE 1. Find J
Solution. Let x = a sin i ; then it/ = i
»zdr, and
f—
_ /• acosrde _ racoi
^1 "J (a«a'sln«r)l "^ ?*
i follows:
Since ilni.: . dm • right triangle with x u the oppoalte
leg to the acnte angle , anil o u the hvpotenme. Tuen
Ibe ad]»oent leg will be V«"^ and la
ocoti, and t— Ocac I respectively.
INTEGRATION
305
1. / dx = Vx 2 — a 2 — a arc sec  + C.
J x a
2
3
4
6
6
+ C
fVa*x*dx = ? Va 2  x 2 + arc sin? + C.
J 2 2a
/Vx 2 + a 2 . . / . /r — =\ Vx 2 + a 2 t ~
— dx = log(x + Vx 2 + a 2 ) + C.
X" X
/» x 2 dx 1 . x Vl  x 2
I =  arc sin x
J VT^* 2 2 2
/» dx = (2x 2 l)Vx 2 + l j c
/» dx __ Vx 2 + a a . g
J x 2 Vx 2 +a 2 ~ «"*
Vx 2 a 2 ,, _(x 2 a 2 )>
■■/■
dx = * *+ C.
x* SaV ^
MISCELLANEOUS EXAMPLES
dx
sln 4 x
(x 2 + 1V&
2
«/ x +
3 /» (ax + 6)dx
"J Vx 2 ia 2
4. Jtan'^cW.
^ Vl6x 2
"^ V2 + 2xx 2
68*
'/vT3
<tt.
Vl  a 2 ! 2
8 . r_** —
J 2*30 + 1
9
4xdx
l4x*
10. f(tan8xl) 2 dx.
11. f tan»0 sec«0d0.
12./
/• d*
sin 4  dx.
2
13
008*0
14
dt
Sco e OdO
sin0
/ at
/» Scos
16 J6^
16 r(^+y fc
J Va*
<fc
Vl + 8** 2
18. fcos'dx.
dx
19  fT
J x 2 +
2x + l
20
x*dx
/x°a
x~3
x8
2i. r
22. r
d3
sin 2^
dt
23
cos8t
6dx
/oax
24./
3
dy
vV6y + 10
26./^.
CX J
xdx
"'fl?T*r m
27
/dx
(a + 6
(a + te)»
l + sec 2
tanl
dx
dtf.
x 4 Vx 2  1
30. f(a — Sx 2 )* 2xdx.
r 2 x 2 dx
31. / ^Zf —
J (a»  x*)*
32. ffe±£*.
^ vx
J X
34. f e«*x 2 dx.
36. fJ^±dx.
J x 2 + 4 m 2
J 9x 2 n 2
37. Tco^ax sin axdx.
38. foot 4 3 aydy.
39. f sin 2 6xdx.
306 INTEGRAL CALCULUS
40. The following functions have been obtained by differentiating certain func
tions. Find the functions and verify your results by differentiation.
(a) 6x* + 8in2x.
8olntion. In this example (6 x* + sin 2 x) dx is the differential expression to be inte
/6x* 1
(6x* + sin2x)dx =  cos2x+C. An*.
4 2
d /fix* 1
Verffteatton. ^pf£cos2x + C)= 6x« + sin2x.
dx\ 4 2
(b) 6x*  6x. mx + n (l2y)«
0) v?=w m (r) ~VT'
(c)
2x*— 8x —
4.
(d)
cos*ax + sin
a
(e)
Va + 6x.
(0
ax + 6
te + a
(K)
X
6 + 2x
3 + 2x
x*+l
m
l8x
ax — m
1
W VfW (8) ^+4xl
M+c (t) sec*^.
, x 56. ( u ) ,„ .
W j3^ V4x*+2*
(n) * (') <*" «"')*•
.v 6 ." 2 * (w)x»(l + x«)l.
(o) sin mx cos mx. ,
(P) co^^j*. W x ^^=
4^3? (q> tan». w ^VHTxi.
CHAPTER XXIII
CONSTANT OF INTEGRATION
170. Determination of the constant of integration by means of initial
conditions. As was pointed out on p. 281, the constant of integration
may be found in any given case when we know the value of the
integral for some value of the variable. In fact, it is necessary, in
order to be able to determine the constant of integration, to have
some data given in addition to the differential expression to be
integrated. Let ua illustrate this by means of an example.
Illustrative Example 1. Find ft function whose first derivative is 3/ 5 — 2e + 6,
and which shall have the value 12 when sl,
Solution. (3* 1 — 2z + 5)<fx is the differential expression to be integrated. Thus
f{3x*—2x + b)dx = z?—& + 6z + C,
where C is the constant of integration. From the conditions of our problem this
result must equal 12 when x, = 1 ; that Is,
12 = 1  1 + 6 + C, or C = 7.
Hence «*— i>+ 5z+ 7 is the required function.
171. Geometrical signification of the constant of integration,
shall illustrate this by means of examples.
Illustrative Example 1. Determine the equation of
the curve at every point of which the tangent has the
Solution. Since the slope of the tangent to a curve at
any point is ^, we have, by hypothesis,
* = 2*,
We
dx
or, dy = 2idz.
Integrating, y = ijxdx
W V = ** + c 
where C is the constant of integration. Now If we gi'
3, 0,  S, (A) yields the equatii
C a series of values, say
whose loci are parabolas with axes coinciding with the axis of y and having 6, 0, 
respectively as Intercepts on the axis of Y.
308 INTEGRAL CALCULUS
All of the parabolas {A) (there are an infinite number of them) have the same
value of = ; that is, they have the same direction (or slope) for the same value of x
It will also be noticed that the difference in the lengths of their ordinates remains
the same for all values of x. Hence all the parabolas can be obtained by moving any
one of them vertically up or down, the value of C in this rase not w fleeting the slope
of the curve.
If in the above example we impose the additional condition that the curve shall
pass through the point (1, 4), then the coord inal.es of this point must satisfy (A), giving
A = 1 + C, or C = 8.
Hence the particular curve required is the parabola y = j* + 3.
Illubthative Example 2. Determine the equation of a curve such that the slope
of the tangent to the curve at any point is the negative ratio of the abscissa to the
ordinate.
Solution. The condition of the problem is expressed
by the equation $„ x
r, separating the variables,
V$y=
Integratlng, — =
xdx.
or, x' + y l = 2G.
This we see represents a series of concentric circles with their centers at the origin.
If, in addition, we impose the condition that the curve must pans through the point
(8,4), then B + lfl=2C.
Hence the particular curve required is the circle x 1 + y s = 25.
The orthogonal trajectories of a system of curves are another sys
tem of curves each of which cuts all the curves of the first system
at right angles. Hence the slope of the tangent to a curve of the
new system at a point will be the negative reciprocal of the slope of
the tangent to that curve of the given system which passes through
that point. Let us illustrate by an example,
iLLnsTHATivB Example 8. Find the equation of the orthogonal trajectories of the
system of circles in Illustrative Example 2.
Solution. For the orthogonal system we will then
have dy y
fa~ x
or, separating the variables,
Integrating, logo = togz + logc = logo
Hence the orthogonal trajectories of the system of circlos i* + j/ a = C Is the system
of straight lines which pass through the origin, as shown in the figure.
CONSTANT OF INTEGRATION 309
172. Physical signification of the constant of integration. The fol
lowing examples will illustrate what is meant
Illustrative Example 1. Find the laws governing the motion of a point which
moves in a straight line with constant acceleration.
Solution. Since the acceleration I = — from (14), p. 92 I is constant, say /, we
have L * J
or, dv=fdt. Integrating,
W v=/l + C.
To determine C, suppose that the initial velocity be « ; that is, let
v = v when 1 = 0.
These values substituted in (A) give
»o = ° + Gt 0P i c = V
Hence (A) becomes
(B) v=/l + v .
Since v = — [(9), p. 90], we get from (B)
at
or, ds =ftdt + Vffit. Integrating,
(C) s = Ifl* + v + c.
To determine C, suppose that the initial space (= distance) be * ; that is, let
8 = 8 when t = 0.
These values substituted in (C) give
« = + 0+C, or, C = « .
Hence (C) becomes
By substituting the values /= g, v = 0, « = 0, « = A in (B) and (D), we get the
lawB of motion of a body falling from rest in a vacuum, namely,
(Ba) * = gt, and
(Da) h = lfffi.
Eliminating t between (B a) and (D a) gives
v = V2gh.
Illustrative Example 2. Discuss the motion of a projectile having an initial
velocity v inclined at an angle a with the horizontal, the resistance of the air being
neglected. y\
Solution. Assume the .XTplane as the plane of mo
tion, OX as horizontal, and OF as vertical, and let the
projectile be thrown from the origin.
Suppose the projectile to be acted upon by gravity
alone. Then the acceleration in the horizontal direc ° v ^ coaa
tion will be zero and in the vertical direction — g. Hence from (16), p. 98,
dv x _ , dv v
 = 0, and '=„.
810 INTEGRAL CALCULUS
Integrating, v x = G v and v y =— gt + C t .
But t cos a = initial velocity in the horizontal direction,
and t> sin a = initial velocity in the vertical direction.
Hence C t — v cos a, and C % = t? sin a, giving
(£) 9 x = 9 cosa, and v, =— pf + 8 sina. v
But from (10) and (11), p. 92, t>, = — , and v v = — f therefore (E) gi
dt dt
— = « cos a, and i =— gt + » sin a,
— = tj fl cos a, and —
or, dx = c cob aett, and dy —— gidt + v sin adt.
Integrating, we get
(F) x = « cos a • t + C 8 , and y = — j gP + » sin a • t + C 4 .
To determine C 8 and C 4 , we observe that when
f = 0, 2 = and y = 0.
Substituting these values in (F) gives
C, = 0, and 4 = 0.
Hence
(G) x = v Q co8at, and
(ff) y = — lgt* + « sina»*.
Eliminating t between (O) and (ff ), we obtain
(I) y = 2tana •*
2 v * cos 1 a
which is the equation of the trajectory, and shows that the projectile will move in a
parabola.
BXAMPLB8
1. The following expressions have been obtained by differentiating certain functions.
Find the function in each case for the given values of the variable and the function ;
Derivative of
function
(h) bx* + ca + 4.
(i) ^+^
(j) COt — CSC* 0.
(k) Z#tt.
Value of
variable
Corresponding
value of function
2.
0.
6.
20.
2.
0.
2*
2.
1.
0.
0.
6.
a.
6.
10.
4.
a.
w
— •
2
a
8.
7
i'
Answers
2*
(a) 28. 2. 9. 82 + 18.
(0)8 + 262*. 6. 20. 804 + 82+— —
^ M*» 2 8
(c)y»6*y. 2. 0. £_?j: + [2&* 4.
(d) sin a + cos a. —• 2. sin a — cos a + 1.
(6) 7~2^T *' °' log(2tt*).
(f) sec*0 + tan 9. 0. 6. tan + log sec $ + &
1 v 1 A 2
„ — z m ° z— arctan .
2 a +a a 2a a a 4a
CONSTANT OF INTEGRATION 811
2. Find the equation of the system of curves such that the slope of the tangent
at any point is :
x a
(a) x. Ana. Parabolas, y = — + 0.
(b) 2x — 2. Parabolas, y = x a —2x + C.
Parabolas, * = x + 0.
JB
t/^ x 8
Semicubical parabolas, — = — + C.
Semicubical parabolas, = = — fC.
*^ ^82
Cubical parabolas, y = x 8 + C
x 8 6
Cubical parabolas, y = — + x a + C.
8 2
Cubical parabolas, ^ = x + C.
8
Equilateral hyperbolas, y a — x a = C.
Equilateral hyperbolas, xy = G.
Hyperbolas, aV — IW = C.
Ellipses, 6*y a + a*x f = C.
x 2 
logy = — + C, or y = ce*.
logy = x + C, or y = ce*.
Straight lines, y = mx + C.
w , , Circles, x a +y a +2x2y + C = 0.
1 — y
3. Find the equations of those curves of the systems found in Ex. 2 (a), (c), (d),
(i), (j), (m), which pass through the point (2, — 1). x ' 4
Ana. (a) x* — 2y — 6 = ; (m) y =— e * ; etc.
4. Find the equations of those curves of the systems found in Ex. 2 (b), (e), (g), (h),
(o), (p), which pass through the origin. An*, (b) y = x a — 2x ; (o) y = mx ; etc.
5. Find the equations of the orthogonal trajectories of the following systems of
curves found in Ex. 2 :
x a
(a) y = — + C, Ex. 2 (a). Ana. y = — logx + 0.
(b) £ = x + C, Ex. 2 (c). logy = x + C.
JB
(c) 7 = 7 + c ' Ex ' 2 < d > logy = i + c.
(d) y a  x a = C, Ex. 2 (I). xy = C.
(e) xy = C, Ex. 2 (j). y a x a = C.
(f) y^op, Ex. 2 (n). ^ =  x + C
(g) y = mx + C, Ex. 2 (o). my + x = C.
(h) x* + y* + 2 x  2 y + C = 0, Ex. 2 (p). y  1 = c (x + 1).
(0)
1
y*
(d)
x a
1/
(•)
X
(0
8x a .
(g)
x a +6x.
1
(h)
S 5 "
<n
X
w
y'
(J)
y
X
6*x
00
<■*¥.
#*%
a*x
0)
■^ •
\ /
6*y
(m) xy.
w
y.
(0)
m.
fnl
1 + x
312
INTEGRAL
CALCULUS
6. Find the equation of the curve whose subnormal is constant and equal to 2 a.
Hibt. From (4), p.
7, subnormal *= # —
A its. y — 4 (u + C. a parabola.
7. Find the curve
whose subtangent
s constant and eqnal to a (see (3), p. 77).
Ann. alogy = x + C.
8. Find the curve
whose subnormal equals the abscissa of the point of contact.
A
ns. y 1 ~ x 1 = 2 C, an equilateral hyperbola.
9. Find the curv
whose normal is
constant ( = K), assuming that y = B when
1 = 0.
Ana. i* + y 1 — fi 1 , a circle.
Hint. From (6), p. 77, length of normal 
v \ l *{tef or < & =*<k , ¥TW
10. Find the curve
whose subtangent
quale three times the abscissa of the point
of contact.
Ana. x = ey*.
11. Show that the
curve whose polar
subtangent (see (7), p. 88) is constant ia the
reciprocal spiral.
12. Show that the
curve whose polar
subnormal (see (1), p. 88) is constant is the
spiral of Archimedes.
13. Find the curv
in which the pol
r subnormal is proportional to the length
of the radius vector.
Ana. p = <x*.
14. Find the curve
in which the polar
subnormal is proportional to the sine of the
vectorial angle.
Ana. p = e a costf.
15. Find the curvt
in which the pol
r subtangent is proportional to the length
of the radius vector.
ins. p = ce"*.
16. Determine the
curve in which the
polar subtangent and the polar subnormal
are in a constant ratio
Ans. p c ce**.
17. Find the equal
on of the curve in
which the angle between the radius vector
and the tangent is one
half the vectorial
uigle. .ins. p = c(lcos0).
18. Determine the
curves in which the subtangetit is n times the subnormal ; and
find the particular cur
:• v'liidi passes, ttmragb (2, 8).
Ana. Vny = x + C; Vii(y 3) = xS,
19. Determine the
curves iu which the length of the subnormal is proportional to
the square of the ordinate.
Am. y = ee**.
20. Find the curvet
in which the angl
between the radius vector and the tangent
at an; point is n times
the vectorial angle
Ans. p" = c sin n6.
Assuming that >: =
„ when ( = 0, tind the relation between v and (, knowing that the
acceleration is i
21. Zero.
An,. B = e .
22. Constant = k.
.*+«
23. a + bt.
.* + « + ».
Assuming that s =
) when I = 0, find the relation between a and I, knowing that the
velocity la :
24. Constant (= n
Ant. s = t>„t.
25. m + ni.
.—+£.
26. 84'2(8P.
. = sin>_!
(c)(* +
2tl.
(d) J
(«)**
4.
«s
CONSTANT OF INTEGRATION 313
27. The velocity of a body starting from rest is 6 P feet per second after t seconds,
(a) How far will it be from the point of starting in 8 seconds ? (b) In what time will
it pass oyer a distance of 860 feet measured from the starting point ?
Ans. (a) 45 ft. ; (b) 6 seconds.
28. Assuming that 8 = 2 when t = 1, find the relation between s and i, knowing that
the velocity Is :
(a) 8. Ans. « = 8t — 1.
(b) 2t8. « = t*8t + 4.
a = log* + 2.
« = t*4t + 6.
29. Assuming that v = 8 when t = 2, find the relation between v and i, knowing
that the acceleration is :
(a) 2. Ans. t> = 2il.
(b)8£ + l. » = f8 + t7.
(c)P2l. v = £fi + s.
(d)J + t. t = logi + ^ + l.
80. A train starting from a station has, after t hours, a speed of t 8 — 21 t a + 80*
miles per hour. Find (a) its distance from the station ; (b) during what interval the
train was moving backwards ; (c) when the train repassed the station ; (d) the dis
tance the train had traveled when it passed the station the last time.
Ana. (a) 1 1 4 — 7<» + 40 & miles ; (b) from 6th to 16th hour ;
(c) in 8 and 20 hours ; (d) 4658} miles.
31. A body starts from the origin and in t seconds its velocity in the X direction
is 12 1 and in the T direction 4 1 2 — 9. Find (a) the distances traversed parallel to each
axis ; (b) the equation of the path. r
Ans. (a)s = 6P, y = *«0t; (b) r = ( g »9j^.
32. The equation giving the strength of the current i for the time t after the source
of E.M.F. Is removed is (B and L being constants)
Find i, assuming that I = current when t = 0. Ans. i = Ie L '
33. Find the current of discharge i from a condenser of capacity C in a circuit of
resistance £, assuming the initial current to be I , having given the relation (C and R
being constants) di dt —
34. If a particle moves so that its velocities parallel to the axes of X and Y are
ky and kx respectively, prove that its path is an equilateral hyperbola.
36. A body starts from the origin of coordinates, and in t seconds its velocity parallel
to the axis of X is 6 f , and its velocity parallel to the axis of Y is 8 P — 8. Find (a) the
distance traversed parallel to each axis in t seconds ; (b) the equation of the path.
Ans. (a) x = 8t«, y = C8<; (b) 27 y* = x (x  0) 1 .
CHAPTER XXTV
THE DEFINITE INTEGRAL
173. Differential of an area. Consider the continuous function 4>(x),
be the equation of the curve AB. Let CD be a fixed and MP a
variable ordinate, and let u be the measure of the area CMPD*
When x takes on a sufficiently small increment Ax, u takes on an
increment Am (= area MNQP). Completing the rectangles MNHP
and MNQS, we see that
area MNRP < area MNQP < area MNQS,
or, MPAx< Au< NQAx;
and, dividing by Ax,
MP < =£ < JV10.1
Now let At approach zero as a limit ; then since MP remains fj
and NQ approaches MP as a limit (since y is a continuous functi
or, using differentials,
dm
du = ydx.
Theorem. The differential of the area bounded by any curve, the e
of X, and two ordinate* is equal to the product of the ordinate i
imitating the area and the differential of the corresponding absciita.
174. The definite integral. It follows from the theorem in the )
section that if AB is the locus of
then du = ydx, or
(A) du = 4>(x)dx,
• We rany suppose (his urea to be generated by a variable ordinate starting out from Ct>
and moving to tin right ; hence u will he a fasettoci at  r which rinlnhw when x a,
t In this figure MP is less than NQ; if MP happens to be greater than SQ. simply
reverse the Inequality signs.
THE DEFINITE INTEGEAL 315
where du is the differential of the area between the curve, the axis
of x, and any two ordinates. Integrating (Ay, we £
Since i <b(x)dx exists (it is here repre
sented geometrically as an area), denote
it by/(*) + C.
(£) .:u=f(x) + C.
We may determine C, as in Chapter XXIII, if we know the value
of v for some value of x. If we agree to reckon the area from the
axis of y, ie. when
(C) x = a, « = area OCDG,
and when x = b, u = area OEFG, etc,
it follows that if
(D) x = 0, then u = 0.
Substituting (D) in (S), we get
0=/(0)+C, or, G=/(0>
Hence from (B) we obtain
(i) «=/(*) /(»>.
giving the area from the axis of y to any ordinate (as MP).
To find the area between the ordinates CD and EF, substitute
the values (C) in (S), giving
(J) area OCDG =/(«) /(0),
(G) area Off^G =/(6) /(0).
Subtracting (J? ) from (G),
(ff) area CEFD=f(b)f(a~)*
Theorem. TAe difference of the values of f ydx for x = a and x = b
gives the area bounded by the curve whose ordinate is y, the axis of X,
and the ordinates corresponding to x — a and x = b.
This difference is represented by the symbol *
(/) jrfx, or, jTVoOd*
* The student should observe that under the present hypothesis fix) will be a ringl*
Valued function which changes cimtimtotuly from/(a) to/(6) a* x changes from a to 6.
1 This Dotation la due to Joseph Fourier (1 7881830).
316 INTEGRAL CALCULUS
and is read "the integral from a to 6 of ydx" The operation is
called integration between limits, a being the lower and o the upper
limit*
Since (/) always has a definite value, it is called a definite iiUet/rtil.
Fop, if
then fjb (i) 4r =[/(!)+ c]'
or jT*C»)*»/(»)/C»).
the constant o/  integration having disappeared.
TPe may accordingly define the symbol
I tf>(x)dx or I ytfc
a» (*/.<? numerical measure of the area bounded bp the curve y =
the axis of X, and tJie ordinate* of the curve at z = a, x = b. This
definition presupposes that these lines bound an area, i.e. the curve does
not rise or fall to infinity, and both a aiui b are finite.
We have shown that the numerical value of the definite integral
is always f(b~)~ /(a), but we shall see in Illustrative Example 2, p. 324,
that/(£)— /(a) may be a number when the definite integral has no
meaning.
175. Calculation of a definite integral. The process may be sum
marized as follows :
Fikst Step. Find the indefinite integral of the given differential ex
pression.
Second Step. Substitute in this indefinite integral first the upper
limit and then the lower limit for the variable, and subtract the last
result from the first.
It is not necessary to bring in the constant of integration, since
it always disappears in subtracting.
" Tlio word limit In this connection means merely the value of the variable at one end of
its range (end value), and should oat be confused with the meaning of the word in t
Theory of Limits.
1 * (*) U continuous and singlevalued throughout the Interval [a, b].
THE DEFINITE INTEGRAL
317
Illustrative Example 1. Find ]x*dz.
Solution. fx*dz = T?T= ?s = 21  *"«•
Ji LSji 8 8
Illustrative Example 2. Find f sinxdx.
Solution
. r*slnxdx = rco8xl ,r =r(l)lrll = 2. A
— •
J— dx Tl x~l a 1 1
— = I  arc tan  I =  arc tan 1 arc tan
o a* + z* \_a a J o a a
=  = — . An*.
4a 4a
ns.
1. r 8 6x«dx = 88.
2. r'(a«xx«)dx = ^.
Jo 4
X*
4. r*i.
Ji x
5. f ^x*  2x + 2)(x  l)dx = f .
J * V82x
7.   = log8.
Jox + 1 8
■'• V28x* 4V8
9. f 8 «^ = =^126.
J * 2^x*4
10 . /%*l^j
Jo y» — y + 1 8v8
ii r» m lo S 2
u 'Ji i + * = "
13. f 4 sec*0(W = f
Jo
14. I — =dx = 4r.
Jo Vx
15. f i (fV f iA^)<B = *V55.
r rax _*i
f o Vr*x a ~^
17. r 2 ^ = 8r.
Jo V2r— v
19. 2af (2 + 2cos0)ld0 = 8a.
Jo
20. J * sin 1 a: cos 8 ada = ^.
21. f\txnada = 0.
4
 /l + V2\
22 J%8ec*d* = log( — —y
12. r a sin0d> = l.
23
o l + 8in 2 0~~ 4
318
OTTEGBAL CALCULUS
176. Calculation of areas. On p. 316 it was shown that the area
between a curve, the axis of X, and the ordinates x = a and x =
given by the formula
Area = / ydx,
where the value of y in terms of x is substituted
from the equation of the given curve.
Illustrative Example 1. Find tbe area bounded by
the parabola y — x 1 , the axia of X, and the ordinates x = 2
Solution. Substituting in the formula
Area ABDC^f'z'dx = l^V
1. Find the area bounded by (lie parabola y — x 1 , the axis of X. and the ordinal*
1 = 8. Ant. P.
2. Find the area above the axis of X, under the parabola y 1 = ix, and included
between the ordinates x = 4 and x = 9. An*. 25}.
8. Find the area bounded by the equilateral hyperbola xy = a 2 , the axis of X, and
the ordinatesz = ti and x = 2a. Ana. a 2 log 2.
4. Find the area between the parabola y = 4 — x* and the axis of X. An*. 10].
5. Find the area intercepted between the coordinate axes and the parabola
zl + yl=oi. Ana.~
6. Find the area by integration of the triangle bounded by the line y = 5i, the
axiB of X, and the ordinate x = 2. Verify your result by finding the area as one
the product of the base and altitude.
7. Find the area by integration of the triangle bounded by the line y = 2x
the axis of X, and the ordinate x = 4. Verify your result as in the last example.
8. Find the area by integration of the trapezoid bounded by the line x — y + 4 =
the axis of X, and the ordinates x = — 2 and x = 4. Verify your result hy finding i
area as one half tbe product of the sum of the parallel sides and the altitude.
9. Find the area by integration of the trapezoid bounded by the line x + 2y
6 = 0, the axis of X, and the ordinates x = and x = 3. Verify your result
the lust example.
10. Find the area by integration of the rectangle bounded by the line y = 6, the
axis of X, and tbe ordinates x = 2 and z = 6. Verify your result geometrically.
11. Find by integration tbe area bounded by the lines x = 0, x = 9, y = 0, y = T.
Verify your result geometrically.
12. Find the area bounded by the semicubical parabola y" = **, the axis of X, and
the line a = 4. An*. J ^JoM.
6 '
the
iilf
i
THE DEFINITE INTEGRAL 319
13. Find the area bounded by the cubical parabola y = x 8 , the axis of JT, and the
ordinate x = 4. Ans. 64.
14. Find in each of the following cases the area bounded by the given curve, the
axis of X, and the given ordinates :
(i) y = — x*. x = — 8, x = 8. Ans. 86.
(b) y = —^— . x = 0, x = 8. log V66.
1 + x*
(c) y = sinx. x = 0, x = . 1.
(d) y = x* + 3x a + 2x. x = 0, x = 8. 56J.
(e) y = x 2 + x + 1. x = 2, x = 8. 9J.
(f) y = x* + 4x* + 2x« + 8. x = 1, x = 2. 28^.
(g)y a =4x. x=l, x = 0. t 6
(h)xy = *«. x = a,x = 6. tflog.
(i)y = 2x + 8. x = 0, x = 4. " a
(j) y 2 = 4x + 16. x = 2, x = 0.
(k) y = x a + 4x. x=4, x=2.
(1) y = cosx. x = 0, x = .
4
(m) xy = 12. x = 1, x = 4.
15. Find the area included between the parabolas y 3 = 4x and x s = 4y. Ans. 6j.
16. Find the total area included between the cubical parabola y = x 8 and the
line y = 2x. Ans. 2.
17. Prove that the area bounded by a parabola and one of its double ordinates
equals two thirds of the circumscribing rectangle having the double ordinate as
one side.
18. Find the area included between the parabolas y 3 = 4 + x and y 3 = 4 — x.
x x
19. Find the area between the curve y = and the line y =  •
l + x a y 4
Ans. log 4—}.
20. Find by integration the area of the triangle bounded by the lines
x + 3y — 8 = 0, 6x — y — 16 = 0, x — y + l = 0. Ans. 8.
177. Geometrical representation of an integral. In the last section
we represented the definite integral as an area. This does not neces
sarily mean that every integral is an area, for the physical interpre
tation of the result depends on the nature of the quantities represented
by the abscissa and the ordinate. Thus, if x and y are considered as
simply the coordinates of a point and nothing more, then the integral
is indeed an area. But suppose the ordinate represents the speed of
a moving point, and the corresponding abscissa the time at which the
point has that speed; then the graph is the speed curve of the motion,
and the area under it and between any two ordinates will represent
the distance passed through in the corresponding interval of time.
That is, the number which denotes the area equals the number which
denotes the distance (or value of the integral).
820 INTEGRAL CALCULUS
Similarly, a definite integral standing for volume, surface, mass,
force, etc., may be represented geometrically by an area On p. 366
the algebraic sign of an area ia interpreted.
178. Mean value of <f>(x). This is denned as follows:
f *(x)dx
Mean value
from x = a to x = b
Since from the figure
J $(x)dz = vre*APQB,
this definition means that if we construct on the base AB(= b — a) a
rectangle (as ALMS') whose area equals the area of APQB, then
. laeaALMB ABCR , lV , „„
T
«_
I.
,
J *—""
1
/ ,
r
»
^
"
"
16
179. Interchange of limits.
Since jf"«Kz)<fe=/(S)/0),
and jy^d, =/(«) /(J) =  [/(J) /(a)],
we have / $(x)dx = — f $(x)dx.
Theorem. Interchanging the limits is equivalent to changing the sign
of the definite integral.
180. Decomposition of the Interval of integration of the definite
integral.
Since
and
<t>( X )dx /Or,) /(a>
<K*)<&=/(6) /<>,),
we get, by addition,
But
jf"V<»o>+J"«S0
> <k=/(6) /(«)•
«\ (i) *:_/(*) /(«);
therefore, by comparing the last two expressions, we obtain
£+(*)*"■ f*w* + fi'w*
THE DEFINITE INTEGRAL
821
Interpreting this theorem geometrically, as in § 174, p. 815, we
see that the integral on the lefthand side represents the whole
area CEFD, the first integral on the right
hand side the area CMPD, and the second
integral on the righthand side the area ,
MEFP. The truth of the theorem is there
fore obvious.
Even if x % does not lie in the interval
between a and b, the truth of the theorem
is apparent when the sign as well as the magnitude of the areas is
taken into account Evidently the definite integral may he decom
posed into any number of separate definite integrals in this way.
181. The definite Integral a function of its limits.
From
f\(x)dx=f(b)f(a)
we see that the definite integral is a function of its limits. Thus
I <f>(z)dz has precisely the same value as I <p(x)dz.
Theorem. A definite integral it a function ofitt limit*.
182. Infinite limits. So far the limits of the integral have been
assumed as finite. Even in elementary work, however, it is some
times desirable to remove this restriction and to consider integrals
with infinite limits. This is possible in certain cases by making use
of the following definitions.
When the upper limit is infinite,
and when the lower limit is infinite,
_£*oo*.!S?.jr*G0*t
provided the limits exist
IlXCiTXiTlYE EZAMPLI 1. Find ( — .
/*" dx _ limit f"dx _ limit [ 11*
OBWt * B  J, j5i6 = + gaJ, I i& = + »L xji
Now as the ordinate <,* moves indefinitely to the right.
Is always finite, and
which is also finite. In such cases we cull ike result tlie area bounded by the
ordinate OP, and OX, although strictly speaking this area is not completely
E Example 3. Find J
C*"dx_ limit r*dx_
J, 7~6 = + «Ji 7 =
he curve, the
ly bounded.
The limit of log 6 as b increases without limit does n.it. exist ; hence the integral
has in this case no meaning.
183. When y = <pt x j is discontinuous. Let us now consider cases
when the function to be integrated is discontinuous for isolated
values of the variable lying within the limits of integration.
Consider first the ease where the function to be integrated is con
tinuous for all values of x between the limits a and b except x = a.
If a < 6 and e is positive, we use the definition
w
jf+c»)i.=;^_£+(»)^
and when $(*) is continuous except at ;
provided the limits are definite quantities.
THE DEFINITE INTEGRAL
328
Illustrative Example 1. Find
1
Solution. Here
Va*  x*
ax
J* a ax
• Va*x f '
becomes infinite for x = a. Therefore, by (B),
arc sin 1 =  . J.n*.
2
r * == u«it / *» _ limit r «1«
«/o Va a — x f e ~ u ^o Va a — x f e  u L <*Jo
»[— (»3]
J^dx
X 1
Solution. Here — becomes infinite for x = 0. Therefore, by (J.),
r l dx _ limit f l dte _ limit /l \
In this case there is no limit and therefore the integral does not exist.
If c lies between a and 6, and <f> (z) is continuous except at x= c, then,
€ and €* being positive numbers, the integral between a and b is defined by
provided each separate limit is a definite quantity.
•*« 2xdx
Illustrative Example 1
. Find f
r
(x«a*)*
Solution. Here the function to be integrated becomes infinite for x = a, i.e. for a
va!ae of x between the limits of integration and &a. Hence the above definition
(C) must be employed. Thus
2xdx _ limit /••• 2xdx limit r*° 2xdx
(x*a«)*~~ < = Jo (x«  a")* </ = °J«+«' (x i« a i)f
= e 1! ^[8^(a«)»a» + 8af] + J^ 1 * [8 ^8^  8 ^(a + •)*  a 2 ]
= 8a* + 6a* = 9a*. 4n*.
To interpret this geometrically, let us plot
the graph, i.e. the locus, of
2x
V = r»
(x f  a*)*
and note that x = a is an asymptote.
^«™ r ~* 2xdx
area OPE = I
= 8v^(ac) f a f +8aT.
Now as PJB moves to the right toward the asymptote, i.e. as c approaches zero,
8^( f) 2 a 2 + 8a*
is always finite, and .. SA
limit [3 y (a _ f)i _ a , + 3 a f j = 8 a f
324
INTEGRAL CALCULUS
area S'Q.
=/:;..
= 3v'W3J / <<i + O s 
whicli is also finite. As in Illustrative Example 1, p. 323, 3 «' is called the ;
bounded by OP, the asymptote, and OX. Similarly,
2idx
•' (a*  o')*
is always finite as QE' moves to the left toward the asymptote, and as t' approaches
zero, the result 6a' is also finite. Hence 6n> Is called the area between QR, the
asymptote, the ordinate I = 3 a, and OX. Adding these results,'we get 0o», which
is then called the area to the right of OY between the curve, the ordinate
and OX.
iLLUBTSATtTB EXAMPLE 2. Find I
Solution. This function also becomes infinite between the limits of integration.
Hence, by (C),
/■*" dx _ limit /"■' dx , limit
Jo £T^i = . = »J (TT^i + «' =
_ limit T
_ limit n \\, limit/
In this case the limits do
pra I has no meaning.
If we plot the graph of this functh
limits, the condition of thinps appears very much the
same as in the last example. It turns out, however, that the shaded portion cannot
be properly spoken of as an area, and the integral sign has no meaning
That it is important to note whether or not the given function becomes infinite
within the limits of integration will appear at once if we apply our integratii
formula without any investigation. Thus
I exist ami the inte
Jo (*a)» L xaU
a result which is absurd it
n of the above discussions.
CHAPTER XXV
INTEGRATION OF RATIONAL FRACTIONS
184. Introduction. A rational fraction is a fraction the numerator
and denominator of which are integral rational functions.* If the
degree of the numerator is equal to or greater than that of the
denominator, the fraction may be reduced to a mixed quantity by
dividing the numerator by the denominator. For example,
= ar + x — 6 +
x*+2x + l a* + 2x + l
The last term is a fraction reduced to its lowest terms, having
the degree of the numerator less than that of the denominator. < It
readily appears that the other terms are at once integrable, and hence
we need consider only the fraction.
In order to integrate a differential expression involving such a
fraction, it is often necessary to resolve it into simpler partial frac
tions, Le. to replace it by the algebraic sum of fractions of forms such
that we can complete the integration. That this is always possible
when the denominator can be broken up into its real prime factors
is shown in Algebra. *
185. Case I. When the factors of the denominators are all of the first
degree and none repeated.
To each nonrepeated linear factor, such as re — a, there corre
sponds a partial fraction of the form
A
^— — — .
x — a
Such a partial fraction may be integrated at once as follows :
/ Adx . C dx
x—a J x—a
= A\og(xd)+C.
• That is, the variable is not affected with fractional or negative exponents.
t See Chap. XIX in Hawkes's "Advanced Algebra/' Ginn and Company, Boston.
825
326 INTEGRAL CALCULUS
r (2 x + 8) dx
Illustrative Example 1. Find  ^ « ' •
J x* + x*2x
Solution. The factors of the denominator being x f x — 1, x + 2, we assume*
(4) »» + « = 4+* + C
x(x— l)(x + 2) x x — 1 x + 2
where J., B, C are constants to be determined.
Clearing (A) of fractions, we get
(B) 2x + 8 = A(x  1) (x + 2) + B(x + 2)x + C(x  l)x,
2x + 8 = (^L + B + C)x* + (A + 2BC)x 2 A,
Since this equation is an identity, we equate the coefficients of the like powers
of x in the two members according to the method of Undetermined Coefficients,
and obtain three simultaneous equations
(C)
f A + B + C = 0,
^1 + 2BC = 2,
 2 A = 8.
Solving equations (C), we get
A=~l B = f, C=i.
Substituting these values in (A),
2x + 3 __jL+ 6 1
x(xl)(x + 2) 2x 8(xl) 6(x + 2)
/ 2x + 8 , __ 8 rdx 6 /» dx 1 r dx
x(xl)(x + 2) ~ 2Jx~8«/xl~~6«/x + 2
=  1 logx + } log(x  1)  i log(x + 2) + logc
. c(xl)* .
= log— i £— . jlns.
x*(x + 2)*
A shorter method of finding the values of A, B, and C from (B) is the following
Let factorx = 0; then 3=— 2 A, or .4 = —}.
Let factor x— 1 = 0, orx = l; then 6 = 3 2?, or B = J.
Let factor x + 2 = 0, or x = — 2 ; then — 1 = 6C, or C = — J.
A useful exercise is to integrate without determining the constants
A y By Cy etc For instance, in the above example,
Js(zl)(a:+2)~~J * +J zl+J z+2
= A\ogz + B\og(zl)+C\og(z + Z).
* In the process of decomposing the fractional part of the given differential neither the
integral sign nor dx enters.
INTEGRATION OF RATIONAL FRACTIONS 827
J (xl)(x2) ^ x1
/• xdr l (z + 8)«
J (z + l)(z + 8)(x+6) 8 ^(x + 5)«(x + l)
, fd1)* ^tog^ + O*.
./ z* + 6x + 8 (» + 2)1
B 7 ( ^z» 1 6f =1< ^^< x  2), <^ 3),+a
«/x f + 8x + 2 ^ x + 1
_ rx* + z*8 . x 8 , x f , . f . x*(x2) 8 f „
7. I — I —  — dx = — h — Mx + log —  1 + C.
J x»4x 8 2 ^ (x + 2) 8
J (x*l)(x + 2) 2 ^6 g (x + l) 8 ^ 8 lu 8^f^i^
/ (a6)ydy = log fr  *)« + c .
J y»  (a + b) y + ab * (y  6) »
ia r (*+**)<» = iog<''>('+g> + c.
11. /^ — ', = — plog  + — log t= + C.
•/^S^ + C 2V2 z + V2 2V8 z+V5
"•f^Tii" i^i
is. f $ *— : = l0gH.
Jo l + 3x + 2x f 6 6
14. / ^ — ' = log
Js x 8 — 4x
200
81 '
186. Case II. When the factors of the denominator are all of the first
degree and some repeated.
To every wf old linear factor, such as (re — a)*, there corresponds
the n partial fractions
(re — a)* (re— a)"" 1 re — a
The last one is integrated as in Case I. The rest are all integrated
by means of the power formula. Thus
r Adx = r^aynte ^ rT + C.
J (*«)" J (1  n) (re  a) 1
828 INTEGRAL CALCULUS
/x 8 + 1
dx.
x(x — l) 8
{Solution. Since x — 1 occurs three times as a factor, we assume
x» + l A , B , C D
x(xl) 8 x (x1) 8 (xl) a x1
Clearing of fractions,
x 8 + 1 = A (x  l) 8 + Bx + Cx(x  1) + Dx(x  l) a .
x 8 + 1 = (A + DJx 8 + ( 3 A + C  2D)x a + (8 A + B  C + D)x  A.
Equating the coefficients of like powers of x, we get the simultaneous equations
A + D = 1,
3A + C2D = 0,
8A + BC + D = 0,
.1 = 1.
Solving, A = — 1, B = 2, C = 1, D = 2, and
x« + l _ 1, 2 , 1 , 2
x(xl) 8 x (x1) 8 (xl) a x1
... f g8 + 1 , dx=logx ?— L_ + 21og(xl) + C
Jx(x1) 8 6 (xl) a x1 6V '
(x  1)<
+ i og <^ + c.
*' / (x !)»(* 2) = x^l + 1< *x^I + °'
z*dx 4
2 / (x + 2)»(x + l) = ^ + 1 ° g(X + 1) + a
Jx 8 — 4x 2 + 4x x — 2 x a
./•x a + l. 1 2., * v ^
4. / . 1x> <*g=7 7S  + log(xl) + C.
«/ (x — l) 8 (x — l) a X — 1
R Wx'x^ljdx x a 1 1. x1 , n
6 J x^x 8 = 2 +g + 2^ + x fl0g nr + C 
W3x + 2)dx = 4x + 3 _x>_
J x(x + l) 8 2(x + l) aT *(x + l) a ^
/» x a dx _ 5x + 12 / x + 4 \«
'J (x + 2) a (x + 4) a x a + 6x + 8 + ° g \x + 2 > / +
8 7 ^ +6 ^% y+ 4 =irf2 +l0 ^ +1 > +c '
INTEGRATION OF RATIONAL FRACTIONS 329
r dt t 1 t + V5
*J(t*2)« 4(^2) + 8 V2° g t .V2 '
13. r_^_ = lag2i.
Ji x + 2z« + x» 8 5^5
187. Case III. TTAm tfo denominator contains factors of the second
degree but none repeated.
To every nonrepeated quadratic factor, such as x*+pz + j, there
corresponds a partial fraction of the form
Az + B
This may be integrated as follows :
r (Ax + B)dz ^ I \ 2 2/
J z*+pz + q J z*+pz + q
Adding and subtracting — in the numerator. I
z*+pz + q J z*+pz + q
_A r (2z+p)dz /2BAp\ r dx
*j *+,*+, ys)j (.+$+(,{)
[Completing the square in the denominator of the second integral.]
z V4j — p % V4j — p %
Since a?+pz + q = has imaginary roots, we know from 3, p. 1,
that 4 qp*>0.
880 INTEGRAL CALCULUS
/4(2x
— .
x* + 4x
4 A Bx + C
Solution. Assume — = — h— g—  — .
x(x* + 4) x x* + 4
Clearing of fractions, 4 = A (z* + 4) + x(Bx + C) = (^L + B)x* + Ox + 4^4.
Equating the coefficients of like powers of x, we get
^+B = 0, = 0, 4.4=4.
This gives A = 1, B =— 1, C = 0, so that — =  —
6 x(x* + 4) x x* + 4
x(x* + 4)~«/ x~ Jx«"+4
1 ex
= logx —  log(x* + 4) + logc = log m Aw.
3 Vx a + 4
3 x . ^.
— arc tan — = + C
V2 V2
«T «*« 1, x* + 4 . 2 A x „
2. I = — log — — h  arc tan  + 0.
J(x + l)(x* + 4) 10 8 (x+l)«^5 2 +
/• (2x*8x3)dx , (x*2x + 6)i ,1 x l
3. I — i — — —  — — ' — — = log ^  — '— +  arc tan — — + O.
J (xl)(x 2 2x + 6) ° x — 1 2 2
A c x2dx 1 , 1 + x 1 . , _
4. I — r = Tlog arctanx + C.
J 1 — x* 4 1 — x 2
/dx 1 x* 1
= ; log arc tan x + O.
(x* + l)(x* + x) 4 8 (x + l)*(x a + l) 2 ^
„ /• (X s — 0)dx . x f + 4 ,8 A x
0 I t — z~i — a = l°g . +  arc tan
Jx* + 6x« + 8 VxHHa 2 2
_ r (6x f l)dx , x a 2x + 5 , 6 . xl 2 A x
7  V + 3)(x»2x + 6) = l08 ^T8 + 2 arCtan — ^ arCto °vl + a
8. f * = JtogJL±i)L + * arctanii^ + C.
/• ««<fe i . /«  1\ . V5 . % , _
/• 4<tt 1 , P + <V2 + 1 , / , ivl,,,
'/^ = ^ l0 ^^tV2,.i +Viarctan T^ + C 
H. r^ = li og ^+JLti_ + 1 arctan^ + 1 + c.
12 f 8 2zdz _i /s
'Jo (1 + x») (3 + x 8 )"" ° g \2*
fi(2x» + x + 8)dx = «
Jo (x + l)(x» + l) * T 4
9
10
INTEGRATION OF RATIONAL FRACTIONS 331
188. Case IV. When the denominator contains factors of the second
degree some of which are repeated.
To every »fold quadratic factor, such as (x*+px + j)", there cor
respond the n partial fractions
C A) Ax + B % + f Cx+D % _ i + ...+ Lx+M
(z*+pz + qy (x*+pz + qy 1 x*+px + q
To derive a formula for integrating the first one we proceed as
f oUows :
Ax+B ,
ax =
J (J+px + qy J (a?+px + qy
[Addtog »d ^.btr^UB, f ta th. numerator.]
J (x*+pz+qy + J 1
f+B^dx
(a>+px+qy «/ (x'+px+qy
dx
The first one of these may be integrated by (4) p. 284; hence
c*) f
Ax + B dx= A
(a»+px + qy 2(lii)(2?+pz+q)
«i
/ 2BAp \ P dx
\ 2 )J (x t +px +
qy
Let us now differentiate the function •—. v ,
Thus (J+px+qyi
A
x + £ \ t 2(nl)( a; +) t
z rl = r^ : : r= ; » Or
dx\(3?+px + qy l l Qf+pz + qy 1 (pf+px + qj
Co „ !_)( <»»> + 2( ° 1) ('^) , fc
V J \Q*+px + qy 1 W+px + qy 1 ^ Qt+px + qy '
*+£
[sinoe^+px + g(x + ) , +(g^),and(x + )(af«+pa ? + «)(7^).]
332 INTEGRAL CALCULUS
Integrating both sides of (C),
x + 2
2 C dx
or, solving for the last integral,
x 4^
(2>) J ( aP +i«+«> i "2(iii)(,^ +1 «+,r 1
2n3 /• cfc
Substituting this result in the second member of (5), we get *
(Ax + B)dx _ A(p*  4q) + (2B  Ap)(2x + p)
(x*+px+qy 2(n^(^qp*)( k x* + px + qy 1
w /i
(2 B Ap~) (2 n  8) /* dr
+ (n_l)(4«^) J (x* + px + qy*
J.
It is seen that our integral has been made to depend on the inte
gration of a rational fraction of the same type in which, however, the
quadratic factor occurs only n — 1 times. By applying the formula
(E) n — 1 times successively it is evident that odr integral may
be made ultimately to depend on
dx
and this may be integrated by completing the square, as shown on
p. 296.
In the same manner all but the last fraction of (A) may be inte
grated. But this last fraction, namely,
Lz + M
x*+px + q
may be integrated by the method already given under the previous
case (p. 329).
• 4 q _pa>o f since « a +pz + q  has imaginary roots.
INTEGRATION OF RATIONAL FRACTIONS 333
Illustrative Example 1. Find f  — ^—r — •
J (x 2 + 2) a
Solution. Since x a + 2 occurs twice as a factor, we assume
z* + x* + 2 _ Az + B Cx + D
(x a + 2)* ~ (x s + 2) a x a + 2 '
Clearing of fractions, we get
x 8 + x a + 2 = Ax + B + (Cx + D) (x a + 2).
x 8 + x* + 2 = Cx* + Dx a + (4 + 2 C)x + B + 2D.
Equating the coefficients of like powers of x,
= 1, D = l, A + 2C = 0, B + 2D = 2.
This gives 4 =2, B = 0, C = l, D = l..
Hence * + *V°,.'* , + 4 ±1 »^
(x a + 2)* (x a + 2) 2 x a +2
/ (x* + x f 4 2)dx __ __ r 2xdx /• xdx r dx
(x a + 2) a " J (x s + 2) a J x a + 2 J x* + 2
11 x 1
+ = arc tan —  + log(x a + 2) + C.
x a + 2 V2 V2 2
Illustrative Example 2. Find / —  dx.
J (x a + l) a
Solution. Since x a + 1 occurs twice as a factor, we assume
2x» + x + 8 _ ,Ax + B Cx + D
(x a + l) a " (x a + l) a x a + 1 "
Clearing of fractions,
2x* + x + 3 = Ax + B + (Cx + D) (x a + 1).
Equating the coefficients of like powers of x and solving, we get
4=l, 2*=3, C = 2, D = 0.
Hence f 2 * + * + 3 ds = fz^±» &+ f *■*
J (x a + l) a J (x a +l) a J x a + 1
= 108(^ + 1)+/^^.
Now apply formula (E), p. 832, to the remaining integral. Here
A=— 1, B = 3, p = 0, g = l, n = 2.
Substituting, we get
/ — x + 8 , l + 3x 3 r dx l + 3x 3
, „ dx = T +  lr— = T — + arctanx.
(x a + l) a 2(x a + l) 2Jx a + l 2(x a + l) 2
Therefore
/•2x» + x + 3. . ,,,., l + 3x ,3 . ,_
I  ^z—dx — log(x a + 1) + —  h  arc tanx + (7.
J (x a + l) a 6V ' 2(x a + l) 2
INTEGRAL CALCULUS
J («' + !)'
<«■ + !)• 2(." + i) a
Hog(i" + 2)1 ==«
°(l + l)' 2(i + 1)
/e
6(4j* + 8)* , 8 VS
/• g' + x' + a' + s
J (a* + 2)* (*« + «)«
/• <4g'Bs)<fa
J (a 1)' (* + !)■"
/■ (3z + 2)dj! _
J (f 1  3
3(ii
sx + s) avs
V5
+ c.
Since a rational function may always be reduced to the quotient
of two integral rational functions, i.e. to a rational fraction, it follows
from the preceding sections in this chapter that any rational function
whose denominator can be broken up into real quadratic and linear
factors may be expressed as the algebraic sum of integral rational
functions and partial fractions. The terms of this sum have forms all
of which we have shown how to integrate. Hence the
Theorem. The integral of every rational function whose denominator
can be broken vp into real quadratic and linear factors may be found,
and is expressible in terms of algebraic, logarithmic, and inversetrigono
metric functions ; that is, in terms of the elementary functions.
CHAPTER XXVI
IHTEGRATION BY SUBSTITUTION OF A NEW VARIABLE.
RATIONALIZATION
189. Introduction. In the last chapter it was shown that all rational
functions whose denominators can be broken up into real quadratic
and linear factors may be integrated. Of algebraic functions which
are not rational, that is, such as contain radicals, only a small number,
relatively speaking, can be integrated in terms of elementary functions.
By substituting a new variable, however, these functions can in some
cases be transformed into equivalent functions that are either in the
list of standard forms (pp. 284, 285) or else are rational. The method
of integrating a function that is not rational by substituting for the
old variable such a function of a new variable that the result is
a rational function is sometimes called integration by rationalization.
This is a very important artifice in integration and we will now take
up some of the more important cases corning under this bead.
190. Differentials containing fractional powers of x only.
Such an expression can be transformed into a rational form by meant
of the substitution a: = z"
where n is the least common denominator of the fractional exponents of x.
For x, dx, and each radical can then be expressed rationally in terms of z.
Illuitsatitk Kiiin.t 1
**
I*
Solution. Since 12 is the L.C.M. of the denominators of the fractional exponents,
<fe = lSx"(b t l* = z", 3* = ** !* = *'■
= $»» _ (z» + C = »i*  Ji* + C.
[Su but! tuting back the rd» of « In term, uf z, namely, ii 1 ^,]
The general form of the irrational expression here treated is then
B(af)fa,
lere R denotes a rational function of x*.
336 INTEGRAL CALCULUS
191. Differentials containing fractional powers of a + bx only.
Such an expression can be transformed info a rational form by mean*
of the substitution a + 6a = z",
wliere n is the hast common Jew* mi nut or <■/ the fractional exponents of
the expression a + bx.
For r, dx, and each radical can then be expressed rationally in
terms of z.
Illustrative Example 1. Find / —
J (l
Solution. Assume
l + x = i*\
2 zdi, (1 + x)i = r", and (1 + x)i = »
J (i + /)! + (i + X )i "J *" + * " J<« + 1
e + C =
n<l + *)'
when we substitute back the value of z in terms of x.
The general integral treated here has then the form
R [x, {a + bxy] dx,
where £ denotes a rational function.
193. Change in limits corresponding to change in variable. When
tegntting liv (.lie substitution of a new variable it is sometimes rather
troublesome to translate the result back into the original variable.
When integrating between limits, however, we may avoid the process
of restoring the original variable by changing the limits to correspond
with the new variable." This process will now be illustrated by
an example.
Illustrative Example 1. Calculate f .
Jo l + i»
Solution. Assume x — z*.
Then dx — i z s ih, x* = r*, z*" = z. Also to change the limits we observe that
when x = 0, z = 0,
and when x = 16, x — 2.
Jo l + ^i J. ! + «■ J» \ 1 + W
Jo Jo Ja 1 + z' L 3
l>>
* The relation between the old and the new variable should ho suoh thai to each value
of one within the limits of Integration there is always one. and only one. finite value of the
other. When one is given as a manyvalued function of the oilier, care must be taken to
choose the right values.
INTEGRATION BY RATIONALIZATION 337
/• xidx 4 i 4. i r*ztdx w 4
J^T = 3 x *3 l0g(x,+1)+c  8 /.i+i s= ii
J 6x i 3\9 13 / ^o (2 + z )VT+^ 2
5. f5±i cfe = 4 + ^T + 21ogx 241og(xA + 1) + C.
X* +Z* X* XtT
dx 8 a . ~, x* — 1
6. f = x*+21og + 4arctanx* + C.
J x f _ x * 3 x * + j
7 r^^ = «l 8 r^ + ^ + i* + 4xHl6xi + 321og(xi2)l + 0.
J 2 v x — v x 2 L o 2 8 J
8. I _=4 — 2 log 8. 10. I * * =
Jo i + Vx Jl V2 + 4y 2
9 p(x2)fdx =8 + 3Vg ir n ," xdx = g/ n »\
J » (x2)* + 3 2 Jo (2x + 3)t 8 \ ^3/
•V + y** 1/ 2^3 4^ 6 _T
l r^ = iog ( g+1 ^ 1 + c.
•x(x + l)* (x + l)* + l
4 /» xdx ^ 2(2afto) ^ c
""' (a + 6x)* ^Va + te
/» x*dx __ 6x*+ 6x + l ^
^ (4x + l)* 12(4x + l)'
6. f y ^a + ydy = A( 4 ?  3a) (a + y)t + C.
7.* f fjti +1 ax==x + l + 4Vx+l + 41og(VxTTl) + C.
J Vx + 11
8. f %= = (* + 1)* 3(x + 1)* + 31og(l + Vx~+T) + C.
J 1 + vx + l 2
9. f z "t_L_ dx = 2Vx2 + V2 arc tan • x p^ + C.
J xVx — 2 * 2
— ^ — = 6.31. 21. f " =8.886.
1 2rf + <* J * 1 Vx
22.
• f T^ 1 = 3{(« + 1)* + 2(x + 1)* + 21og[(x + 1)*  1]}+ C.
J (x + l)*(x + l)*
INTEGRAL CALCULUS
193. Differentials containing no radical except Va+ bx + x*.*
Stick an expression can be transformed into a rational form by rr>
of tke substitution .
va + bx + x 3 = z — z.
For, squaring and solving for x.
b + 2z'
■*/a + bx + z'(=zx') =
(4 + 2«)'
s* + bz + a
b + 2z
Hence *, dx, and Va +■ bx + x* are rational when expressed in terms
of z.
i Example 1. Find f
** Vi + i + 1«
Solution. Assume Vl + z + z'
Squaring and solving for x,
Vl + Z + Z'(:
a* + t + l
2* + l '
2(z' + t + l)tU
/ a^ ti t i)aj
* 3 + z + 1 J2z+1 "*Ll"1
2z + l
~ log[(2i + 1 + 2 Vl + z + x»)e],
when vre substitute back the value of z in terms of x.
194. Differentials containing no radical except Va + bx — jt*.*
iSmcA on expression can be transformed into a rational form by means
of the substitution
Va + te^[W(x«)03*)J«(*a)«;or = 09*)z],
where x — a and fi — x are real * factors ofa + ke — x*.
•If tlir radical Is of the form Vn+pz + tfi', o>0, it may be written V»«5 + £* + »*,
and therefore cornea under tbe above head, where a— . fi"
t If the radical is of the form Vn + pxjz 1 , g>0, it may be written Vq \~ + E x  x*.
and therefore comes under the above head, where a — ■ ft = 
1 1
I If the factors of a + bx  * a are imaginary, Va + 6z  x i is Imaginary for all values of x.
For if one of the factors is x  m + in, the other must be  (i  in  in), and therefore
b + azx*.{xm*in)ixmin) — [(i  m )i + „»],
which Is negative for all values of x. We shall consider only those cases where the factors
INTEGRATION BY RATIONALIZATION 839
For if Va + bx — x* = V(aj — a) (£ — x) = (x — a) 2, by squaring,
cancelling out (x — a), and solving for x, we get
*— ?+r ; then * V+v '
and Va + fcr — s*[= (a; — «)g] = ^V~ j* * »
Hence ic, die, and Va + &r — a? are rational when expressed in
terms of 2.
/dx
V2 + x  x a
Solution. Since 2 + xx a = (x + l)(2x),
we assume V(x + 1) (2 — x) = (x + 1)«.
2 — * a
Squaring and solving for x, x = — —  •
Hence dx = f——, and V2 + x  x a [= (x + 1)*] = — £.
/. f — =2/* = 2arctan* + C
J V2 + XX 8 J* + l g3j
= — 2 arc tan*/  + C,
\x+ 1
when we substitute back the value of z in terms of x.
EXAMPLES
/GX 1 . V*" — X + 2 + X — V* ^
. = ^tog==^=^ p + C.
x Vx*  x + 2 V2 Vx*x + 2 + x+V2
2. f — "*  =r2arctanfg4Vx a 42xl)4.C.
J xVx» + 2xl
tt /• dx 1 . V2 + 2xV2x , 
J x. V2 4 * — »« V 2
x V2 + x — x» V2 V2 + 2x + V2 — x
dx A x+Vx a + 4x — 4
= arc tan —
Vx a + 4x4 2
4. f "* =ar C tan " rv "" r "*" »C.
J 2
rVx+_4x dx= 8_ + log ( x + 2 + Vx« + 4 x) + C.
J & x+Vx a + 4.x
/xdx 8 + 6x „
— = T  + C.
(2 + Sx — 2x f )i 26V2 + 3X — 2x»
7.r * = *— +0.
(2 ax — x*)l a* V2 ax — X s
8. r (2» + ^ = log(g + 1+ V2xT^) ±=+0.
J & x+V2x + x 2
^ r dx . x — 1 + Vx* + x + 1 ,
9. / — w =log — T * Z_ 4.C.
J x Vx a + x + 1 x + 1 + Vs* + x + 1
10 . f * = _J aroten JgHg + c.
•>xV6x6x* V3 \8(x2)
340
INTEGRAL CALCULUS
The general integral treated in the last two sections has then t
form S(x, ^a + bx + cx") dx,
where B denotes a rational function.
Combining the results of this chapter with the theorem on p. 33
we can then state the following
Theorem. Every rational function of x and the square root of a poly
nomial of degree not higher than the second can be integrated and the
remit expressed in terms of the elementary functions.*
195. Binomial differentials. A differentia! of the form
xia + bafydx,
where a and J are any constants and the exponents m, n, p are ratior.
numbers, is called a binomial differential.
Let x = z" ; then dx = af~ l dz,
and af(« + bz*)'dx = t&" + — '(a + bf*)'d&.
If an integer a be chosen such that ma and na are also integers,
we see that the given differential is equivalent to another of the s
form where m and n have been replaced by integers. Also
3f(a + bx") p dx bs x" , +* P (ax % + bydz
transforms the given differential into another of the same form where
the exponent n of x has been replaced by — n. Therefore, no matte
what the algebraic sign of n may be, in one of the two differential
the exponent of x inside the parentheses will surely be positive.
When p is an integer the binomial may be expanded and the dU
ferential integrated termwise. In what follows p is regarded i
fraction ; hence we replace it by , where r and s are integers. 1
We may then make the following statement :
Every binomial differential may be reduced to the form
where m, r, r, .
3f(a + bx"ydx,
5 integers and n is positive.
•As before, however, it in assumed that in each
function can be broken up into real quadratic and lim
t It la always possible to choose a so that ma and
the L.C.M. of the denominators of m and n.
I The case where p is an integer h not excluded, but appears
INTEGRATION BY RATIONALIZATION 341
196. Conditions of intestability of the binomial differential
(4) jr*»(a + bx^'dx.
Case I. Assume a + &r"= z*.
1 r
Then (a + &r") ; = z, and (a + &r")* = 2T;
1 M
also x = ( —  — ), and rc""=[ —  — J;
n
w + l_ 1
hence ^ x = jZ^ 1 ( — I — ) ^
Substituting in (ii), we get
The second member of this expression is rational when
m+1
n
is an integer or zero.
Case II. Assume a + fa? = zV.
Then a?= , and a + fa?=z a Qf=
r r r
Hence (a + kfy=<f(* 6) 'V;
1 1 m m
also « = a"(z'6) ", 2^= a ,, (z , 6)" ,, ;
« 1 li
and <£r = a n z— l (z*— V) n dz.
n
Substituting in (ii), we get
2T(a + fa*ydx = a n •(^~ J ) ' V+— x <fe.
AM _1_ 1 4*
The second member of this expression is rational when h 
n *
is an integer or zero.
Hence the binomial differential
r
2T(a + fa?ydx
can be integrated by rationalization in the following cases : *
•Assuming as before that the denominator of the resulting rational function can be
broken up into real quadratic and linear factors.
342 INTEGRAL CALCULUS
Case I. When = an integer or zero, by assuming
n
a + bz?=&.
m "4" 1 r
Case II. When h  = an integer or zero, by assuming
n s
a + bx?=ssf2?.
r x*dx r , I2a + h* tn
Solution. m = 8, n = 2, r = — 8, 8 = 2; and here — — = 2, an integer. Hence
this comes under Case I and we assume n
a + 6x* = z* ; whence x = ( — ^— ) , dx = , and (o + &c*)i = *•.
\ 6 / 6l(**a)i
/» s'dx _ /» / g* — a\ t gdg 1^
^ (o + 6x«)l""^ \ 6 / &i(*«a)i *"
= i/(l  a**)<fe = i (z + cur*) + C
12a + te J
^Va + te*
J x*VTTj? ««■
+ c.
Solution, m = — 4, n = 2, = ; and here — — —  — = — 2 V an integer. Hence
« 2 n s
this comes under Case II and we assume
l + ****o>», z = {1 + xt) ;
x
z*
whence x* = — * 1 + x* = » Vl + x a =
*«_1 *«_i (*«l)i
11 zdz
also x = , x* = — rJ and dz= —
(*»l)i <**l> f (z«l)*
_L_._j_ J
= z ^ +c = (2x«_lHl+^ + a
8 3x»
3. f^l + ^ldx^'W + ^ + C.
J 15
4./*= * + C.
J (1 + *•)! Vl + s*
INTEGRATION BY RATIONALIZATION 348
J Vl + x* « 'o Va»x* 16
?(<** + v 8 )* va a + v 8 + «
13. f*(l + 2P)l<tt = (l + 3*)i — — i+C.
J 70
14. f U (l + u)ldu = AO + m)*(6u 2) + C.
jWJ** *— +a
16. J> (1 + *)*<# = A (1 + *) V  * (1 + *)* + A (1 + *>* + O.
dx 3x t + 2a
se»)i 2a 1 z(a + x t )4
197. Transformation of trigonometric differentials.
From Trigonometry
(A) sina:=2sincos» 37, p. 2
it. r — ??— 2E+i£_+c.
(5) cos a: = cos* jr — sin* • 37, p. 2
. . 1 1 *"?
But Bm o := ^
esc
f ^cot'f + 1 Jl + tan'f
. a; 1 1
and cos=
sec Jl + tan«
If we now assume
tan  = 2, or, £ = 2 arc tan 2,
. . X Z X 1
we get sm  = , cos  =
2"VTT7' 2  VTT7
844 INTEGRAL CALCULUS
Substituting in (A) and (2?),
2z 13*
sin x =  r , cos a; =
2(fe
Also by differentiating x = 2 arc tan a we have dx =
1 + * 2
Since sin x, cos x, and <£r are here expressed rationally in terms of
z, it follows that
A trigonometric differential involving sin x and cos x rationally only
can be transformed by means of the substitution
x
tan = z,
or, what is the same thing, by the substitutions
2s 1s* , 2dz
into another differential expression which is rational in z.
It is evident that if a trigonometric differential involves tan x, cot x,
sec X, esc x rationally only, it will be included in the above theorem,
since these four functions can be expressed rationally in terms of
sin x, or cos x, or both. It follows, therefore, that any rational trigono
metric differential can be integrated.*
EXAMPLES
, r (l + sinx)dx 1 A „x , . x , 1. . *.„
1. / x  =  tan 2  + tan  +  log tan  + C.
J sinx(l + cosx) 4 2 2 2 6 2
(1 + cosx)
Solution. Since this differential is rational in sinx and cosx, we make the above
substitutions at once, giving
\ i + Wi
2z / 1
l + z*\ 1 +
z*
/ 1 + _l£\ «L
\ 1 + zVl + z 5
j
C (l + z 2 + 2z)dz 1 Ci . « .
=i(f +2z+l0KZ ) +c
= itan» + tan + Ilog(tan) + C.
♦See footnote, p. 341.
INTEGRATION BY RATIONALIZATION 345
dx ^ r dx 2
2. p_£? 1. 8. f— ^— =
Jol + smx J 1 — sinx
3. f *_*_ = 1.
•/«• 1 — coex
a
dx
1 — tan
2
+ 0.
* r dx i.
' Jo 1 + cos a coex sin a
5 f w *y  * r dx 1 / x\
••/•3 + 2co 8y V6* 10 J638^^ = 2 arCtan ( 2tan 2) + C 
6 fa da  w r dx 1
'Jo2 + cosa 3V8 n  f * — :  = arctan(8tanx) + 0.
ovo J 6 — 4cos2x 3
7. f*:: = 1Vs. .« r (ft 2 / / t\
•/ 3ir l + co8x 12. / = — arctan(v8tan)+0.
T J2cos( V8 \ 2/
o f <to 1 /6tanx + 4\ 
13. I = arctan( — 1+0.
J6 + 4sin2x 3 \ 3 /
14. C^. = 2arctan(tan5)tan? + = xtan5 + 0.
J 1 + cosx \ 2/ 2 2
15. Derive by the method of this article formulas (10) and (17), p. 284.
16.
•/1 + sinx 1
2 2
L r^^ = ^L_^ + 2arctan(tan)+0 = ^L_^ + x + C.
Jl + sinx 1 + tan x \ 2/ 1 + tan?
198. Miscellaneous substitutions. So far the substitutions considered
have rationalized the given differential expression. In a great number
of cases, however, integrations may be effected by means of substitu
tions which do not rationalize the given differential, but no general
rule can be given, and the experience gained in working out a large
number of problems must be our guide.
A very useful substitution is
1 , dz
# = » dz = r>
z r
called the reciprocal substitution. Let us use this substitution in the
next example.
— n — *&•
x*
1 dz
8olation. Making the substitution x =  » dx =  , we get
z z*
846 INTEGRAL CALCULUS
1. i = loff h C Assume x* = *.
Jx(a« + x«) 8a»^a» + x« T ^uwim**
2. i — — — dx = log (x — 2) h C. Assume x — 2 = z,
J (x2)* ^ v ' (x2)*
ft /• x*dx 18x* + 27x + 11. . l1Vlfl
dx x . ^, . I
Assume x = — •
Assume x =  •
z
4. /■_*=_■, + (;.
m r dx l . ex
5. I — =log .
J xva 2 + x» a a + Vo» + x*
^ /• dx , ex . l
6. I —  = log  . Assumes = .
J xVl + x + x* 2 + x + 2Vl + x + x a *
7 r Vl + lpgs ^ _ 2 (l + logx) j + c Assume l + logs = z.
X o
/^^dx 4 i
= — (3 e» — 4) (e* + 1)* + C. Assume e» + 1 = *.
(e« + l)i 21
** /• dx 1 x 1 .
9 /= £— = = 7 + Tlog(e*2) + C. A8sumee» = *.
Jc»*2e* 2e» 4 4 ox '
A r *dx 1, * 1 1 T 2x , t T , _
J (l + x»)t 2 (x» + l)*x V8 V8[(x« + l)i J
Assume x* =
Assume x =  •
z
J i x*  *
f? (sin0+CO60)d0 _log8 An«.rf«^ —a
•Jo 3 + sln20 "T" Assume sin* coe* = *
J* 1 dx IT
= arc tan e —  . Assume e* = z.
o er + c* 4
Jf dx
— =^= = it. Assume x = a sin'z.
K r l0f5 c*Vc*l
J ^. , Q dx = 4 — ir. Assume e»  1 = z*.
6. f V2t + P<ft = V§  J log(2 + V3). Assume * + 1 = z.
Jo
JwlogS . 4 — IT
ve*— ldx = —  — Assume e* — 1 = z.
o 2
/•« + ^ (x a + l)dx , a 1
. I — 7===== = log 8. Assume x = *.
xVx* + 7x* + l
CHAPTER XXVn
INTEGRATION BT PARTS. REDUCTION FORMULAS
199. Formula for integration by parts. If u and v are functions of
a single independent variable, we have, from the formula for the dif
ferentiation of a product (Y, p. 34),
d(uv) = udv + vd%
or, transposing, udv = d(uv) — vdu.
Integrating this, we get the inverse formula,
(A) I udu=uv— j vdUy
called the formula for integration by parts. This formula makes the inte
gration of udv, which we may not be able to integrate directly, depend
on the integration of dv and vdu, which may be in such form as to be
readily integrable. This method of integration by parts is one of the
most useful in the Integral Calculus.
To apply this formula in any given case the given differential must
be separated into two factors, namely, u and dv. No general directions
can be given for choosing these factors, except that
(a) dx is always a part of dv;
(b) it must be possible to integrate dv; and
(c) when the expression to be integrated is the product of two func
tions, it is usually best to choose the most complicated looking one that it
is possible to integrate as part of dv.
The following examples will show in detail how the formula is
applied:
Illustrative Example 1. Find jxcoBxdx.
SolntJon. Let u = x and dv = coaxdx;
then du — dx and t) = f cosxdx = sinx.
Substituting in (A), .
° x ' u dv u v v du
f x cos xdx = x sin x — f sin x dx
= xsinx + C06X + C
847
348 INTEGRAL CALCULUS
Illustrative Example 2. Find j xlogxdx.
8d8tfan. Let u = logx and dv = xdx;
then du = — and v = fxdx = — .
x J 2
Substituting in (A),
/xlog«fe = logx./.
= 2 iogx T + a
Illustrative Example 8. Find JztP*dz.
Solntkm. Let u = e*** and dt? = xdx ;
then du = efi* • adz and v = ( xdx = — «
J 2
Substituting in (A),
/jp2 /* X 3
xe**dx = *»* I — c a *adx
2 */ 2
x , e a * a
2
— ? f x'c^dx.
Bat x*e?*dx is not as simple to integrate as xe"^, which fact indicates that we did
not choose our factors suitably. Instead,
let u = x and dv = e ax dx;
r c"*
then du = dx and v = I e°*dx = — •
J a
Substituting in (A),
xeP*dx = x I — dx
a J a
= Eff_!f + c = !!!( x _i) + c.
a a 3 a \ a/
It may be necessary to apply the formula for integration by parts
more than once, as in the following example :
Illustrative Example 4. Find fx 2 e ax dx.
Solution. Let u = x s and dv = e ax dx ;
r c*
then du = 2xdx and v = l e°*dx = — .
J a
Substitating in (A),
fx^e^dx = x» . —  f — . 2xdx
J a J a
z*e°* 2
« t!/
xefdx.
INTEGRATION BY PARTS 349
The integral in the last term may be found by applying formula (A) again, which
81768 r *"( 1\
J xe^dx = — I* j.
Substituting this result in (2J), we get
z*e**dx = —(x  ) + C = — (x« + — ) + C.
a a* \ a/ a \ a a 2 /
Among the most important applications of the method of integration
by parts is the integration of
(a) differentials involving product*,
(b) differentials involving logarithms,
(c) differentials involving inverse circular functions.
1. fz* log zdx = ir( lo 8 x  zj + °'
2. fa sin ada =— a cos a + sin or + C.
3. Tare sin xdx = z arc sin x + Vl — x a + C.
Hint. Let u  arc sin z and dt>  dx, etc.
4. Clog zdx = x(logx — 1) + C.
5. Tare tanxdx = x arc tan x — log(l + x 2 )» + C
/x" + 1 / 1 \
xMogxdx = rflogx rl + C. n?4 1.
n + 1 \ n + 1/
7. fx arc tanxdx = arc tan x jC
J 2 2
8 Tare cotydy = y arc cot y + + log(l + y 2 ) + C.
9. fxaf*dx = a*^  — — + C. a>0and;*l.
•/ Uog <* log 2 aj
10. ffiaHU = a' ["? ?! + 41 + C a>0and*l.
J Lioga log 2 a log 8 a J
11. f cosllogsinldl = sin B (log sin  1) + C.
12. Cz*e*a& s e*(x* 2x + 2) + C.
13. fxsinxcosxdx = ^sin2x — £xcos2x + (7.
14. Cz*er*dx = e*{2 — 2x  x 2 ) + C.
16. fare tan Vxdx = x arc tan Vx — Vx + arc tan Vz + C.
INTEGRAL CALCULUS
200. Reduction formulas for binomial differentials. It was shown in
§ 195, p. 840, that any binomial differential may be reduced to th»
where p is a rational number, m and n are integers, and n is positive—
Also in § 196, p. 341, we learned how to integrate such a differential
expression in certain cases.
REDUCTION FORMULAS
351
In general we can integrate such an expression by parts, using (A),
p. 347, if it can be integrated at all. To apply the method of integra
tion by part* to every example, however, is rather a long and tedious
process. When the binomial differential cannot be integrated readily
by any of the methods shown so far, it is customary to employ reduc
tion formula* deduced by the method of integration by parts. By
means of these reduction formulas the given differential is expressed
aa the sum of two terms, one of which is not affected by the sign of
integration, and the other is an integral of the same form as the origi
nal expression, but one which is easier to integrate. The following
are the four principal reduction formulas :
wf.
x m (a + bx*ydx=
l (a + bx n y ,t
mf.
x m (a+bx n ydx
(np + m+t)b
(mn+ l)a
x m \a + bx n ydx.
*^rJ'^ i '>'^
(C) [*"(a + bxy>dx
(D) Cx m (a + bx' r )*dx =
x m * 1 (a + bx w ) 1 '
(m+ij
(np + n +
4n + m + i~)b r
While it is not desirable for the student to memorize these formulas,
lie should know what each one will do and when each one fails. Thus :
I
Formula (4) diminishes m by n. (A) fails when np + m+l = 0.
Formula (B) diminishes p by 1. (£) fails when np + m + 1 = 0.
Formula (C) increases m by n. (C) fails when m+l= 0.
Formula (D) increases p by 1. (i>) fails when p+l—0.
352 INTEGRAL CALCULUS
I. To derive formula (JL). The formula for integration by parts is
(A) I udv = uv— J vdu. (A), p. 847
We may apply this formula in the integration of
fz m (a + bx*ydx
by placing m = x" , ~" +1 * and <fo = (a + &r") p :E ,,  1 dir;
then du = (m — n + V)z?*~*dx and v = ^— 4r—
Substituting in (4),
But J2r—(a + fa*y +l dz= Jar—(a + bz n y(a + baf i )dx
= a J2r n (^a + bx*ydz
+ b fz m (a + bz*ydx.
Substituting this in (2?), we get
J ^ ' nJQ>+l)
_ W i«+l />
Transposing the last term to the first member, combining, and solv
ing for I x"(a + bx'ydx, we obtain
v ' 6(n£ + m + l)
a(m — ii + l") /* ,
• In order to integrate dv by (4) it is necessary that x outside the parenthesis shall ha'
the exponent n  1. Subtracting n  1 from m leaves m  n + 1 for the exponent of z in u.
REDUCTION FORMULAS 353
It is seen by formula (A) that the integration of x m (a + bz*ydx is
made to depend upon the integration of another differential of the
same form in which m is replaced by m — n. By repeated applications
of formula (A), m may be diminished by any multiple of ru
When np + m + 1 = 0, formula (A) evidently fails (the denominator
vanishing). But in that case
m + 1 A
n
hence we can apply the method of § 196, p. 341, and the formula is
not needed.
II. To derive formula (B). Separating the factors, we may write
(C) \oT(a + bz*ydz = \x m (a + bx*y\a + bx**)dx
= a fz M (a + bz n y 1 dx
+ b ix m+n ^a + bx*y l dx.
Now let us apply formula (A) to the last term of (C) by substi
tuting in the formula m + n f or tw, and p — 1 for p. This gives
J npjm+1 np+m+lj
Substituting this in ((7), and combining like terms, we get
^np + m + lj K J
Each application of formula (JET) diminishes p by unity. Formula
(B) fails for the same case as (A).
III. 2b derive formula (C). Solving formula (A) for
/■
iP+i
x m ' u (a + bx n ydx y
and substituting m + n f or tw, we get
354 INTEGRAL CALCULUS
Therefore each time we apply (C), m is replaced by m + n. When
m + 1 = 0, formula (C) fails, but then the differential expression can
be integrated by the method of § 196, p. 341, and the formula is not
needed.
/■
iP+i
IV. To derive formula (D). Solving formula (2?) for
x m (a + bx*y l dx,
and substituting p + 1 f or p, we get
Each application of (D) increases p by unity. Evidently (D) fails
when p + 1 = 0, but then p = — l and the expression is rational.
l. f ;?^L =l(s» + 2)(lg»)* + C.
J Vix* 8
Solution. Here m = 3, n = 2, p = — 1, a = 1, b=— 1.
We apply reduction formula (A) in this case because the integration of the differen
tial would then depend on the integration of f x (1 — 1 2 )~ » dx, which comes under (4),
p. 284. Hence, substituting in (A), we obtain
fs'azr**^'"'*'* 1 *'*"** 1  1 < 3  2 + 1 > rx..(ix*)*dx
=  *x*(l  x 8 )* + ifz(l  a;*)***
=  *x*(l  x 8 )*  (1  x*)*+ C
= *(x* + 2)(lx*)* + C.
 , (a*x*)* \4 8 / T 8 a
Hint. Apply (4) twice.
3. C(a* + x*)*dx =  Va* + x* + — log(x + Va 2 + x*) + C.
Hint. Here m0,n2,p«},a*a J ,6 =1. Apply (B) once. '
A r dx (x«l)* 1 ^ _
4. I == =  — '— +  arc sec x + C
•'x'Vx*! 2x» 2
Hint. Apply (C) once.
1 x 2 dx _ x
Va«  x* " 2^~ " 2 a
/x 2 dx x / = « a 8 . x ^.
*"* = f Va 2  x« + — arcsin  + C.
^/„a _ ~a 2 2
REDUCTION FORMULAS 355
6. f , h,i9~t}^„t + ~a + n
J Vl — x a \6 16 16/
8. fx 2 Va 2 x 2 dx = f(2x 2 a 2 )Va 2 x 2 + arcrin+C.
•/ 8 8 a
Hdtt. Apply (A) and then (B).
Hnrr. Apply (2>) once.
0. f * = . V ^ r ^ + J_io g xrr.
J x« Va 2  x* *«W 2a« a + Va*x«
/• x»dx x 2 + 2a 2 , _
1. i = + o.
J (a* + x*)l (a* + x 2 )i
if ** = ( 8 ° 2  ***>* + a
J (a 2 x 2 )* 8a*(a*x*)*
3. f (x 2 + a«)idx = ?(2x» + 6a 2 )Vx 2 + a 2 + ^ i og ( z + Vx 2 + a 2 )+C.
•7 8 8
.4. fx 2 (x 2 + a*)*dx = f (2x* + a 2 ) Vx 2 + a*  — log(x + Vx 2 + a*) +C.
•/ 8 8
te /• x 2 dx x + 8a,_ «vi . 3a 8 x , ^
16. I m = (2 ax — x 2 ) T \ arc vera  + C.
J V2oxx* 2 ' 2 a
Hiirr. f X = fa;* (2 a  s)~ * dx. Apply (<4) twice.
•J v2axx 2 •/
 /• a*x Va 2 — X s ^
16. I = — — +C.
J x 2 (a*x 2 )* ax
,_ /• l^dV 2i/ f + 6r(v + 8r) /r r 6 . y „
17. I y y = r rur Vif ru J V2ry y 2 + r»arcversg + C.
J V2ryy 2 2 r
= (2<rt f 2 )* + a arc vers + C.
V2ait» «
19. 1 = _i arc tan  + C.
J (a 2 + « 2 ) 8 4a 2 (a 2 + s 2 ) 2 8a 4 (a 2 + 8 s ) 8a* a
20. r^ = =A(3r* + 4r» + 8)Vr=^ + C.
J VI — r* 46
2i. r ** . 26. r*v?+?*. 29. r * T<fa .
J (a»  x»)» J J (a + W) %
«./—*—: */#=• 30. f**L..
^ *«(l + x s )i ^ V«« + 9 J V8^a»
23. f^=. 27 f^EIE*. 31 f ^EZ^.
^ VlX* ' J X J y*
24. f^=. 28 r O^)^ 32 f * , '
356
INTEGRAL CALCULUS
201. Reduction formulas for trigonometric differentials. The method
of the last section, which makes the given integral depend on another
integral of the same form, is called successive reduction.
We shall now apply the same method to trigonometric differentials
by deriving and illustrating the use of the following trigonometric
reduction formulas:
oo
00
(G)
OD
/'
J'
xcos"xdx =
sin'' 1
m + n
8in"xcos n " 1 jrir,.
sin B, jrcos n ardr= —
+
x cos" xdx =
m1 r. _
m + nj
n + 1
m+n+2 r
n+1 J
m + 1
m + n +2
sin" , jrcos n+ *jriir.
+ — : / sm DI+ *xcos"x4r.
Here the student should note that
Formula (£) diminishes n by 2.
Formula (F) diminishes m by 2.
Formula (G) increases n by 2.
Formula (H~) increases m by 2.
(E~) fails when m + n = 0.
(F) fails when m + n = 0.
(C) /<n7« wA«i b + 1 = 0.
(ff) fails when m + 1 = 0.
To derive these we apply, as before, the formula for integration
by parts, namely,
I vdv = uv~ I vdu. (A), p. 347
09
Let
then
rf« =s — (m — l)eos""
nd dt> = sin* x cos wfo ;
, , sin"
i sin xdx, and u =
REDUCTION FORMULAS 367
Substituting in (4), we get
si>\ C • m » j . sin m+1 a?cos* 1 aj
CB) I sm m xcoa n xdx = + 
J m + 1
m+lj
In the same way, if we
let ttsssin"" 1 ^ and dv = co8 n zainzdx>
we obtain
•/*n C • m » j sin"" 1 xcos* +1 a?
J J n + 1
' sin" , ~ , a?cos* +i iwiR
But I sin" ,+ *a:cos*~*:rd!r = / sin m a;(l — cob*z) co&*~*zdx
= / sin"xco8*~ a a^a?— I sin m a?cos"a^a^
Substituting this in (2?), combining like terms, and solving for
/ sin m a? co8*xdxy we get
,jg. r ._.,._ 8in" +1 xco8" 1 x
sin m xcos n xdr =
m + n
/sin m
H / sin m xcos"~*xi£r.
m
Making a similar substitution in ((7), we get
8in w " 1 xcos n+1 x
<n f
sin m xcos n xdr= —
m + n
H / sin m " a xcos"xdr.
Solving formula (J?) for the integral on the righthand side, and
increasing n by 2, we get
sin w+1 xcos n+1 x
w /
sin"xcos" jrdr = —
n + 1
m + n + 2 ^,^+1^
n + 1 J
358 INTEGRAL CALCULUS
In the same way we get, from formula (F),
sin m+1 xc08 n+1 x
w /
sin M xco8"jafr =
m + l
H — — — / sin^+'xcos'jcir.
m + l J
Formulas (J5) and (F) fail when m + n = 0, formula ((?) when
w + l=0, and formula (IT) when w + l = 0. But in such cases we
may integrate by methods which have been previously explained.
It is clear that when m and n are integers, the integral
sin m 2; cos":rdi
/
may be made to depend, by using one of the above reduction
formulas, upon one of the following integrals:
 dx y  sin xdx, I cos xdx, I sin x cos xdx, I — — = I esc xdx,
= / sec xdx, I : — , / tan xdx, \ cot xdx,
cosx J J cos a; sin 2: J J
all of which we have learned how to integrate.
r . « a , 'sinxco^x sin x cos 8 z 1 . , _
1. I sin a x coe^xax = ^ \ (sin x cosx + x) + C.
J 6 24 16 v '
Solution. First applying formula (F), we get
(A) f sin a x cos 4 xdx = 1  f co^xdx.
[Here m = 2, w = 4.]
Applying formula (E) to the integral in the second member of (A), we get
/T1V r a j sinxcos'x , S r # ,
(B) I cos 4 xdx = 1  I cos 1 xdx.
[Here m = 0, n4.]
Applying formula (E) to the second member of (E) gives
/_tv T « j sin x cosx , x
(C) Jco&xdz =  + .
Now substitute the result (C) in (£), and then this result in (<4). This gives the
answer as above.
_ r . . « . cosx /sin'x sin'x sinx\ x „
2. [sin«xcc« a xek; = ^^(^^=^^^) + — + C.
J 2 \ 3 12 8/16
3. f — = tan x — 2 cot x —  cot 8 x + C.
J sin 4 xco8 2 x 3
REDUCTION FORMULAS 359
4. / — = (3co6?x) — + C.
J sin*x 2 v ' 2
/I i
sec'xdx = secz tanx + log (sec x + tanx) + C.
/I 1
C8c*xdx=— cscxcotx + log(c8cx — cotx) + C
2 2
_ /• cos 4 ada cosa 8. A a . _
7. I = r cosa log tan +C
J sin'a 2sin*a 2 8 2
o f • « ^ cosa/sin'a , 6 . . ,6. \ , 6a , _
8.JWada = — (_ + 8in«a + smaj + — + C.
9. fc8c«fl(tt=^/ r — + 3 Ulogtang + C.
J 4 \sin*0^2sin*0,r 8 K M * U 2 ^
10. fsec 7 *(fy = ^£( — ^ + 7^ 5 + «) + ^log(sec0 + tan0) + C.
J 2cod*0\3cos*0 12cos a 8/ 16 ^ v '
11. J COf^fctt = — ( COS 7 * +  COS 6 * + — COS 8 * + — COSM + — + C.
Jsln 4 yco8 8 y co6*y\8sin 8 y Ssiny 2/2
— Q  QIC
13.  f gin e xco8*xdx = 16. f a co8 8 ada =
Jo 612 Jo 266
14. f Wxdx = ?*E. 17. fWxdx = *£.
Jo 266 Jo 8
16. f %!*•« = —. 18. pcos**tt = ^.
Jo 16 Jo 16
202. To find / e^sinmrdr and / e^cosmrdr.
Integrate e"* sin nxdx by parts,
letting u = ^ ax J and tfv = sin nxdx ;
then rfw = aefdz, and v = •
n
Substituting in formula (A), p. 347, namely,
I wdv = uv — I vrfw,
we get
(4) I ^ sin nxa!z = h l e^ cos nxdx.
J n nj
Integrate e?* sin 7mfc again by parts,
letting u = sin nx, and dv = e^cfe ;
eT*
then <£u = n cos nxdx, and v = — •
a
860 INTEGRAL CALCULUS
Substituting in (A), p. 347, we get
(2?) I fBmnxdx = I fcosnxdx.
Eliminating f fcoenxdx between (A) and (2?), we have
(o*+ ft*) I 4"* sin nxdx =eT* (a sin nx — n cos na?),
or
Similarly, we may obtain
J (*+n*
In working out the examples which follow, the student is advised
not to use the above results as formulas, but to follow the method by
which they were obtained.
1. fe*rin»fe = (8inx — cosx) + (7.
/(CC
e* cob xdz = — (einx + cosx) + C\
3. re 3 *cos3xdx = ^(3 8in8x + 2co83z) + C.
J 13
A r&nxdx sinx + cosx , _
4. I = — — +U.
J & 2e*
_ /* COS 2 XdZ 1 /o • o o o v . si
6. I = (2 sin 2s — 3cos2x) + C.
e./ e »Bin»xd»=;(i 2gin2x 6 +C082z ) + c.
/ x / z x\
^cosdx = e 2 (sin + cos J + O.
~ C ii vj e*" sin aa . ~
9. J e a *(sin(ur+ co6Ckr)da = J C.
e* x (sln2z — cofl2x)dx = — (sin2x— 6 cos 2 x) + O.
18
J mm 1 /•• 8
e*siii«dx = . I e 8 *cos2xdz = — .
o 2 Jo 18
CHAPTER XXVIII
INTEGRATION A PROCESS OF SUMMATION
203. Introduction. Thus far we have defined integration as the
inverse of differentiation. In a great many of the applications of the
Integral Calculus, however, it is preferable to define integration as
a process of summation. In fact, the Integral Calculus was invented in
the attempt to calculate the area bounded by curves, by supposing
the given area to be divided into an " infinite number of infinitesimal
parts called elements, the sum of all these elements being the area
required." Historically, the integral sign is merely the long S, used
by early writers to indicate "sum."
This new definition, as amplified in the next section, is of fun
damental importance, and it is essential that the student should
thoroughly understand what is meant in order to be able to apply
the Integral Calculus to practical problems.
204. The fundamental theorem of Integral Calculus. If $<.>■') is the
derivative of f(x), then it has been shown in § 174, p. 315, that the
value of the definite integral
(J) _£♦»<*» /<?)/«
gives the area bounded by the curve
y = tjj (x~), the .Yaxis, and the ordin&tes
erected at x = a and x = ft.
Now let us make the following con
struction in connection with this area.
Divide the interval from x = a to x = ft into any number n of equal
subintervals, erect ordinates at these points of division, and complete
rectangles by drawing horizontal lines through the extremities of the
ordinates, as in the figure. It is clear that the sum of the areas of
these n rectangles (the shaded area) is an approximate value for the
area in question. It is further evident that the limit of the sum of
the areas of these rectangles when their number n is indefinitely
increased, will equal the area under the curve.
tT
J
362
Let us now carry through the following more general construction.
Divide the interval into n suhintervals, not necessarily equal, and e
ordinates at the points of division. Choose a point within each sub
division in any manner* erect ordinates
at these points, and through their ex
tremities draw horizontal lines to form
rectangles, as in the figure. Then, as
before, the sum of the areas of these
n rectangles (the shaded area) equals
approximately the area under the curve ;
and the limit of thi* mm as n increases
without limit, and each subinterval ap
proaches zero as a limit, is precisely the area under the curve. The?
considerations show that the definite integral (A) may be regarded a
the limit of a gum. Let us now formulate this result.
(1) Denote the lengths of the successive suhintervals by
.i ' ' . . Aa  ,, A. '■ . , Ar .
(2) Denote the abscissas of the points chosen in the suhintervals by
x,, x,,
Then the ordinates of the cui
these points are
H'.t, ♦<*,>. +(*.). . +(0
(8) The areas of the successive rec
tangles are obviously
ttn?
^OgAz,, oS^Ar,, $(xJAff ( , ■■, tj>(x„)Ax„.
(4) The area under the curve is therefore equal to
But from (A) the area under the curve = j <f> (_ r) d r.
Therefore our discussion gives
(3) j V (x) dx = Jj£ L <>,) Az, 4 (*0 A*,+ • ++(aO A(J:
namely, when the point U
INTEGRATION A PROCESS OF SUMMATION 368
This equation has been derived by making use of the notion of area.
Intuition has aided us in establishing the result. Let us now regard (2?)
simply as a theorem in analysis, which may then be stated as follows :
Fundamental Theorem op the Integral Calculus
Let <f> (x) be continuous for the interval x = atox = b. Let this interval
be divided into n subintervals whose lengths are Ax^ A# 2 , • • •, Ax n , and
points be chosen, one in each subinterval, their abscissas being x v x %y • • •, x %
respectively. Consider the sum
(C) *(z 1 )As 1 + *( i r a )A* 2 + ••■+*(OA*.=2*(aOAa%
Then the limiting value of this sum when n increases without limit, and
each subinterval approaches zero as a limit, equals the value of the definite
integral /»*
Equation (2?) may be abbreviated as follows :
dx.
(I>) f V O) dx = ^ j> to) **
The importance of this theorem results from the fact that we are
able to calculate by integration a magnitude which is the limit of a sum
of the form (CT).
It may be remarked that each term in the sum (C) is a differen
tial expression, since the lengths Arp , A# s , • • •, Ax n approach zero as
a limit. Each term is also called an element of the magnitude to be
calculated.
The following rule will be of service in applying this theorem to
practical problems.
Fundamental Theorem. Rule
Fibst Step. Divide the required magnitude into similar parts such
that it is clear that the desired result will be found by taking the limit of
a sum of such parts.
Second Step. Find expressions for the magnitudes of these parts such
that their sum will be of the form (C).
Third Step. Having chosen the proper limits x = a and x=b, we
apply the Fundamental Theorem
and integrate.
364 INTEGRAL CALCULUS
305 Analytical proof of the Fundamental Theorem. As
tbe interval from i = a lo x = b into any number
n of sub intervals, not necessarily equal, and de
note the abscissae of these points of division by
6,, 6j, • • , 6»_i, and tbe lengths of tbe snbiuter
vals by Az,, Az 5 , • , Ax.. Sow, however, we let
x'i, xi, • , *i denote abscissas, one in each inter
val, determined by tbe Theorem of Mean Value
(44), p. 1(J5, erect ordinates at these points, and
through their extremities draw horizontal lines
to form rectangles, as in tbe figure. Note that
here <p(i) tabes tbe place of $'{z). Applying
(44) to the first interval (a = a, b = &,, and z\ lies between
/(h,)/(a)
the loot section, divide
/&) /«*=♦ W) ah
/(&,) —/(&,) m #(j:j) Ac,, for the second interval,
/(&,>— /(*>,) = v> fit Ar „ for the third interval,
/( b > A&ni) = *{*;) A*., forthentb interval.
#(*0 Ac, + 0(ii)Ar, +  + $(*£) Ae».
area of the first rectangle,
area of the second rectangle, etc.
Hence the sum on the righthand side of (E) equals the sum of tbe areas of the
rectangles. But from (.4), p. 881, the lefthand side of (L") equals the area between
the curve y = *(z), the axis of T, and the ordinates at x = a and x = b. Then the at
Adding these, i
ieget
(E)
/(&)/(<*)
But
(XDAt,
*(£*")  AC,
IF)
£♦<■<)*
Pljllills this .ir
a. Aud while the corresponding
(0)
%*C)to
[Where x, le any abscissa of tbe nublntenal A,.]
(formed as in last section) does not also give the area, nevertheless we may show that
the two sums (F) and (K) approach equality when n increases without limit and each
subinterval approaches zero as a limit. For the difference <P{*i) — *(*i) does a
ceed in numerical value the i title re n<.i.< of the greatest and smallest ordinates il
And furthermore it is always possible* to make all these differences less in numerical
value than any assignable positive number t, however small, by continuing the process
of subdivision far enough, i.o. by choosing n sufficiently large. Hence for such a choice
of n the difference of the sums (F) ;iml ('■'} is less in miuurical value than e(6 — a),
•That ii
le is shown in advanced works on ths dilinlm.
INTEGRATION A PROCESS OF SUMMATION 36f
:. less than any assignable positive quantity, however small. Accordingly as n in
s without limit, the sums (F) and (G) approach equality, and since (F) i
ways equal to the area, the fundamental result follows that
£«*><& ="rLi> < "<>'"*
any manner whatever, and Zi Is any
] which the interval [n, b] is subdivided i;
in the corresponding subinterval.
206. Areas of plane curves. Rectangular coordinates. As already
xplained, the area between a curve, the axis of X, and the ordinates
x = a and x =■ b is given by the formula
i=I ydx,
u
w
i value of y in terms of x being substi
uted from the equation of the curve.
Equation (Ay is readily memorized by
«rving that ydx represents the area of
i rectangle (as CK) of base dx and altitude y. It is convenient to
t of the required area ABQP as the limit of the sum of all such
;ctangles (strips) between the ordinates AP and BQ.
Let us now apply the Fundamental Theorem, p. 368, to the calcu
lation of the area bounded by the curve x = $ (,'/)' (.AB m figure), the
s of Y, and the horizontal lines y = c and
= d.
First Step. Construct the n rectangles
s in the figure. The required area is clearly
s limit of the sum of the areas of these
rectangles as their number increases with
out limit and the altitude of each one ap
proaches zero as a limit
Second Step. Denote the altitudes by
Ay ( , Ay 3 , etc. Take the point in each inter
val at the upper extremity and denote their ordinates by y t , y % , etc.
Then the bases are <f> (yj), Q/ t ~), etc., and the sum of the areas of
the rectangles is
p(y l )Ay i + ^0^)Ay !i +...+^0.)Ay B =2;aS(y ( )Ay r
Thihd Step. Applying the Fundamental Theorem gives
866 INTEGRAL CALCULUS
Hence the area between a curve, the axis of Y, and the horizonts
lines y = c and y — d is given by the formula
c»)
1 =/•"*■
the value of x in terms of y being sub
stituted from the equation of the curve.
Formula (B) is remembered as indicating
the l imi t of the sum of all horizontal strips
(rectangles) within the required area, x and dy being the base i
altitude of any strip.
Illustrative Example I. Find the area iudiiriiil lnlivion the semicublcal paral
ola y = x a and the line x = i.
Solatlon. Let us first find the urea OMP, half of the required area OPP". For the
upper branch of the curve y = w, and summing up all the strips between the limits
z = and x = 4, we get, by substituting in (A),
area OMP =f ydx = f i'iLe = ^ = 12J.
ne inch, t
Hence area OPF = 2 I
If the unit of length l
squaro inches.
Note. For the lower branch y =— z' ; hence
ut
areaO.W7*= f'(* ! )<k =
Tfti* area lies below the axis ofx and has a negative sign becausi
the ordinate are negative.
In finding the area OjtfP above, the result was positive because the ordinates v
positive, the. area lying above the axis of x.
The above result, 25, was the total area regardless of sign. As ne shall i
trate in the next example, it is important to note the sign of the area wben t
curve crosses the axis of X within the limits of integration.
Ii
arch of the sine curve y =
Solution. Placing y =
z = 0, r,
Substituting in (A), p. !
f, 2. Find the area of
aCMB= f ydx=f*Anxdx = 2.
areaBCD = f ydx=f "sin»L:=2,
a OABCD = C ydx = f 'sinzdx = 0.
This last result takes Into account the signs of the two separate a
the whole. The total area, regardless of these signs equals 4.
INTEGRATION A PROCESS OF SUMMATION
367
Illustrative Example 3. Find the area included between the parabola x 2 = 4 ay
and the witch
8a«
V x* + 4a*
Solution. To determine the limits of integra
tion we solve the equations simultaneously to find
where the curves intersect. The coordinates of
A are found to be (— 2 a, a), and of C (2 a, a).
It is seen from the figure that
But
area DECBA
and
area DECOA
Hence
vre&AOCB
area A OCB = area DECBA — area DECOA .
a « 8a 8 dx
""7; 21TO*
x 8 + 4 a 8
ta r 2
J ,J,a X"
— dx =
4a
2ira a  — = 2a»(irf). 4n*.
3
4a*
8
Another method is to consider the strip PS as an element of the area. If y 7 is the
ordinate corresponding to the witch, and y" to the parabola, the differential expression
for the area of the strip PS equals (/ — y")dx. Substituting the values of y' and y"
in terms of x from the given equations, we get
area AOCB = 2 x area OCB
2a
= 2f WV")**
•/o
Jo \x 8 + 4a 2 4o/
= 2a«(irf).
x 2 . y %
Illustrative Example 4. Find the area of the ellipse — + ~ = 1
Solution. To find the area of the quadrant OAB, the limits are x = 0, x = a ; and
y = Va 2 x 2 .
a
Hence, substituting in (A), p. 366,
area OAB =  f °(<* 2  xrfdx
= r^ (a 2_ x . ) i + ^ arc8in ?r
1.2a v ' 2 aj
irad
4
Therefore the entire area of the ellipse equals
Tab.
868 INTEGRAL CALCULUS
207. Area when equation of the curve is given in parametric fort
Let the equation of the curve be given in the parametric form
We then have # = <£(()■ and dx=f '(€)&,
which substituted* in (A), p. 365, gives
0*)
J I,
where
i = f when a: = a, and £ = < s when a
We may employ this formula (A) when finding the area under a
curve given in parametric form. Or we may find y and dx from t
parametric equations of the curve in terms of ( and dt and tin
substitute the results directly in (.A), p. 365.
Thus in finding the
Lave been simpler to u
wliere the eccentric i
rea of the ellipse in Illustrative Example 4, p. 307, it wot
ise the parametric equations of the ellipse
x = acos#, y = bsiu^i,
igle $ is the parameter (§ 68, p. 81).
y = bain $, and dx =— usin^iff .
When x = 0, # = 
Substituting these in (.1), above, i
and when x 
'get
$ = 0.
area OA B = f'ydx = f'ab sin* ■pdf, = — .
Hence the entire area equals woo. j4n».
EXAMPLES
1. Find the area bounded by the line y = bx, the axis of J, and the ordinal*
e = 2. Am. 10.
2. Find the area bounded by the parabola y 1 = 4x, the axis of T, and the lines
y = 4 and y = 8. .4n*.
3. Find the area of the circle x* + y' = r*. Am. i
4. Find the area bounded by y 1 = Ox and y = Sx. ,4iu.
5. Find the area bounded by the coordinate axes and tl
6. Find the area bounded by the curve p^logx, the axis of y, and the li
j/ = 0andy = 8. ' Ant. e>
7. Find the entire area of the curve i* + v*=o*. dna. 4»i
INTEGRATION A PROCESS OF SUMMATION 369
9. Find the area between the curve y = logx, the axis of X, and the ordinates
x = 1 and x = a. Ana. a (log a — 1) + 1.
10. Find the entire area of the curve
11. Find the entire area of the curve d*y % = x , (2az). Ana. ira s .
12. Find the area bounded by the curves
x (y — e*) = sinx, and 2 xy = 2 sinx + x 8 ,
the axis of F, and the ordinate x = 1. Ana. f (eF — J x 8 ) dx = e —  = 1.66 + • • •.
8 a 8
13. Find the area between the witch y —
and the axis of X, its asymptote.
x 8 + 4 a 8 a a 9
• T .4n*. 4ira 8 .
x 8
14. Find the area between the cissoid y % = and its asymptote, the line
x = 2a. 2a x An% 3iro ,
16. Find the area bounded by y = x 8 , y = 8, and the axis of T. Ana. 12.
16. Find the area included between the two parabolas y 2 = 2px and x 8 = 2py.
Ana. *'
3
17. Find the area included between the parabola y 2 = 2x and the circle y % = 4 x — x 2 ,
and lying outside of the parabola. Ana. 0.476.
18. Find the area bounded by y = x 8 , y = x, y = 2 x. <4n*. J.
19. Find an expression for the area bounded by the equilateral hyperbola
x 8 — y 8 = a 8 , the axis of X, and a line drawn from the origin to any point (x, y).
. a 8 . x + y
Ana. —log — —.
2 a
20. Find by integration the area of the triangle bounded by the axis of Y and
the lines 2x + y + 8 = and y =— 4. uln*. 4.
21. Find the area of the circle
(x = rcos0,
\y = r8in0,
6 being the parameter. £na. xr 2 .
22. Find the area of the ellipse
x = a cos 0,
y = 6 sin 0,
where the eccentric angle is the parameter. Ana. irab.
23. Find the area of the cardioid
x = a (2 cost — C082Q,
y = a (2 sin t— sin 2 Q. <4n*. j ira 8 .
24. Find the area of one arch of the cycloid
rx = a(0sin0),
$ being the parameter. LV — V J»
Hint. Since x varies from to 2 ra, $ varies from to 2 r .
Ana. 3 ira 2 ; that is, three times the area of the generating circle.
370
INTEtiKAI, CALCULUS
26. The locus of A in the figure, p^ 82, ia called the "companion to the cycloid."
Its equation* are x = a$,
y = a(lC08tf). F"
Find the area of one arch. Ans. 2m
26. Find the area of the bypocycloid
f"x = acos*0,
$ being the parameter.
27. Find the a
z» + j/* = 3 axy.
Hint. Let
Sun*
t of the loop of the folium of
Ths limits foe t are auJ «.
28. Find by integration the a
( a .)( 3 ,z)* = #,V~0.
(b) (I  v*f = jl s , I = 0.
(c) o*y «»<** tt*), i/ = l
(d)x(i + y 3 ) = i, z = o.
(e) v = *(!«*), y = 0.
(g) V* = a*(2x + 1). Areaofloop. rfl,.
(h) y s = 2 s (2x + l). Areaofloop. ft.
b bounded by the following loci :
•In*. ,' . (i) » = x + 4, v = 2x + 4, y =
;'[•• 0) » = **"+ 5, tf = 0, 1 = 0, i =
J a'. <k> V = 2z», * = 0. V = 2, » = 4.
r. (I) ** = V + 6, 1/ = 0.
J. W»'4 + I = 0,x =
A (n) s* = »*l,tf = 0,x = i t x=l.
(0)^ = 4, y =
(p) * = 10., y = j, j, =
208. Areas of plane curves. Polar coordinates. Let it be required
to find the area bounded by a curve and two of its radii vectors. For
this purpose we employ polar coordinates. Assume the equation of
the curve to be __ frg\
and let OP and OD be the two radii.
Denote by a and jS the angles which the
radii make with the polar axis. Apply
the Fundamental Theorem, p. 363.
First Step. The required area is
clearly the limit of the sum of circular
sectors constructed as in the figure.
Second Step. Let the angles of the successive sectors be A(?j, AS e
etc., and their radii p^ p s , etc. Then the sum of the areas of the s
i ptA$ t + \ p^e t +   • + i p ;M n ^ J) £ ft'Atf,
*(>?«,.
■a uf n ..irinljtr at:
rtiectorJp,. />,£»,
INTEGRATION A PROCESS OF SUMMATION
Third Step. Applying the Fundamental Theorem,
limit i
^2ift'A»,£V<»
Hence the area swept over by the radius vector of the curve in mov
ing from the position 07^ to the position OD is given by the formula
(4) area = { I p*d0,
the value of p in terms of 8 being substituted from the equation of
the curve.
Illvstkativk Example 1. Find the entire area of the lemniscnte p" = a'cos20.
Solution. Since the figure is symmetrical with respect to both OX and OY, the
whole area = i times the area of OAB.
Since p = when 8 = — . we see that if 6 varies
from to — , the radios vector OP sweeps over the
area OAB. Hence, fa nb6ti tilting i
entire area = 4 x area OA B =
= 2o» J*'ct
1 r*
■if.""
i of a square constructed o
EXAMPLES
1. Find the area swept over in one revolution by the radius vector of the spiral
of Archimedes, p = a$, starting with 9 = 0. How much additional area is swept over
in the second revolution ? .4 ir*a J
e loop of the m
e p = a sin 2 8 equals c
; 8*»a".
wo"
An. —
e half the
of
4. Find the entire area of the cardioid p
Ana. ; that is, s
6. Find the area of the circle p = a cosfl.
: tt (lC««).
: times the area of the generating circle.
(i sin 3 8 equals one fourth of the
7. Prove that the area generated by tho radius vector of the spiral p = «• equals
one fourth of the area of the square described on the radius vector.
8. Find the area of that part of tho parabola p = a sec 3  which is intercepted
between the curve and the latus rectum. , 8 a 1
9. Show that the area bounded by aay two radii vectors of the hyperbolic spiral
pff = a is proportional to the difference between the lengths of these radii.
INTEGKAL CALCULUS
10. Find the area of the ellipse p' ! 
11. Find the entire area of the cun
12. Find the area of one loop of the curve p'i
13. Find the area below OX within the curve
14. Find the area bounded by p* = a»ain4S
IB. Find the area bounded by the fallowing
Ant. wab.
(10» + 27V3)~
An». a*.
nnd the given radii vecloi
; fl + 6 1 cos a fl
,(ain20 + cos20).
anin'
(*)/> =
(b)p =
n e, e = o, 3 =
',« = *« =
(<!)/« =
(e) p a si
(0 p = a
■0 +
 +
16. Find the area inclosed by iach nf tin: fiillnwiiin r
(a) p I = 4sln20, (d) p = l + 2co8?.
<b} p = acosStf. (e) p = 3 + cos 0.
(c) p = 8aln4tf. (f) p = 2  sin 0.
0, = 0. = 
8 , = 0,0 = *
2 ' 4
fl + ocosff, 6 = 0, 6 =
(g)p' = a'(l
<h) p = u(l + «l
0)p = acoo5*
209. Length of a curve. By the length of a straight line we com
monly mean the number of times we can superpose upon it another
straight line employed as a unit of length,
as when the carpenter measures the length
of a Iward by making endtoend applica
tions of his foot rule.
Since it is impossible to make a straight a
line coincide with an arc of a curve, we cannot measure curves in the
same manner as we measure straight lines. We proceed then as follows :
Divide the curve (as AB~) into any number of parts in any manner
whatever (as at C, D, E~) and connect the adjacent points of division,
forming chords (as AC, CD, DE, SB).
The length of the curve is defined as the limit of the run of the chords
as the number of points of division increases without limit in suck a wa$
that at the same time each chord separately approaches zero as a limit.
Since this limit will also be the measure of the length of some straight
line, the finding of the length of a curve is also called " the rectification
of the curve."
The student has already made use of this definition for the 1
of a curve in his Geometry. Thus the circumference of a circle i
defined as the limit of the perimeter of the inscribed (or circumscribed)
regular polygon when the number of sides increases without limit.
INTEGRATION A PROCESS OF SUMMATION
373
The method of the next section for finding the length of a plane
■e is based on the above definition, and the student should note
carefully how it is applied.
210. Lengths of plane curves. Rectangular coordinates. We shall
now proceed to express, in analytical form, the definition of the last sec
tion, making use of the Fundamental Theorem,
Given the curve _ ,...
and the points .f (a, <•), Q(b, d) on itj to find
the length of the arc P"Q.
First Step. Take any number n of points on
the curve between P' and Q and draw the chords
joining the adjacent points, as in the figure. The required length of
arc P'Q is evidently the limit of the sum of the lengths of such chords.
Second Step. Consider any one of these chords, P'P" for example,
I let the coordinates of P' and P" be
**(/, y0 and P"(z?+Az>, y' + Ay').
Then, as in § 90, p. 134,
P'P"
P'P"
m
x/.
; I ihi.iinc Inilde the radical by .i.i>= nn.l multiplying nnUlde ly Ax\]
But from the Theorem of Mean Value, (44), p. 165 (if Ay' is denoted
iy/(i) /(a) and A^ by 6  a), we get
r, being the abscissa of a point P on the curve between P 1 and P" at
which the tangent is parallel to the chord.
Substituting, P'P"** [1 +/'(*, ) a ]W = length of first chord.
Similarly, P"P'" = [1 +./"'(^) a ]*Ax" = length of second chord,
/<*>.
^<2.<j/+Ay
P ( "0 = [l+/'(^,) , ]*ic' nl = length of nth chord.
The length of the inscribed broken line joining P' and Q (sum of
the chords) is then the sum of these expressions, namely,
[1 +/'( 3  ) ) 5 ]iA 3 /+ [1 +/'(^)']lA 3 /'+  + [1 +f(xyfi&aP>
374 INTEGRAL CALCULUS
Third Step. Applying the Fundamental Theorem,
Umit j p + / (a . ) .]l 4 yo_jf"[l +/'<>y]iir.
Hence, denoting the length of arc F'Q by *, we nave the formula for
the length of the arc n*
s=J [l+/'(V) a ]Uz,or
where ~ must ho found in terras of x from the equation of the
ax
given curve.
Sometimes it is more convenient to use y as the independent variable.
To derive a formula to cover this case, we know from (35), p. 148, that
— = — ; hence dx — — dy.
ax ax ay
Substituting this value of dx in (A), and noting that the corre
sponding y limits are c and (/, we get * the formula for the length e
(fi) ^ 'JLlUM*
where — in terms of y must be found from the equation of the
given curve.
imple I . Find the length of the circle x* + y 2 = r 8 .
Solution. Differentiating,
Substituting in {A),
[circle iu or Jar to get evcrytliiug In terrai of i.J
Hence the total length equals 2 jtt. Am.
3
■<^m%'H\(%)^m''mh
INTEGRATION A PROCESS OF SUMMATION 375
1. Find the length of the arc of the semicubical parabola ay 3 = x* from the origin
to the ordinate z = 6 a. . 886 a
Arts. •
27
2. Find the entire length of the hypocycloid zl + yl = a^. An*. 6 a.
8. Rectify the catenary y =  (e* + e~ a ) from z = to the point (x, y).
2 a  
Ans.  (e» — e a ).
4. Find the length of one complete arch of the cycloid *
z = r arc vers — V2 ry — y*. .4n*. 8r.
r
Hnrr. Use (B). Here—  
{;
dy V2 ry  y a
5. Find the length of the arc of the parabola y a = 2 pa from the vertex to one
extremity of the latus rectum. © V5 o • /^
Ans. ^ + log(l+V2).
6. Rectify the curve 9ay a = x(x — 3a) 2 from x = to x = 8a. An*. 2a Vs.
7. Find the length in one quadrant of the curve lj + ( j =1. a% , ^ , ^
a + &
c* + 1
8. Find the length between x = a and x = 6 of the curve = •
6*— 1
e 86 — 1
4ns. log — 1 a — 6.
9. The equations of the involute of a circle are e 8 " — 1
x = a(cos0 + Isinl),
y = a(sin0 — OcoeO).
Find the length of the arc from = to 6 = 9 V Arts. \a8*.
10. Find the length of arc of curve \ ~~ _ A from = to =  •
6 \y = e»cos0 2 »
An*. V2(e*1).
11. Find the lengths of arcs in the following curves :
(a) y = log^ ; x = 1, x = 2. (d) y = logx ; x =1, z = 4.
(b) y = log(l  x*) ; x = 0, x = . (e) y = logsecx ; x = 0, x = .
x 2 1 ^ *r *•
(c) y = T logx; x = l, x = 2. (f) y = logcscx; * = . x = 
% z 6 2
211. Lengths of plane curves. Polar coordinates. Formulas (A) and
(2?) of the last section for finding the lengths of curves whose equa
tions are given in rectangular coordinates involved the differential
" M KS)T*  d [(!H*
In each case, if we introduce the differential of the independent vari
able inside the radical, they reduce to the form
376
INTEGRAL CALCULUS
Let us now transform this expression into polar coordinates by means
of the substitutions x = pcos8, y = pB,m6.
Then dx =  p sin 0d8 + cos ffdp,
dy = p cos 8d8 + sin 8dp,
and we have
[dx ! + d ! / ! \ i = [( p sin Ode + cos 8d P y + (p cos Bd6 + sin 8dp\
If the equation of the curve is
then a f =f(6)de = f g ae.
Substituting this in the above differentia! expression, we get
If then a and yS are the limits of the independent variable 6 corre
sponding to the limits in (A) and (B~), p. 874,
we get the formula for the length of the arc,
where p and
•R'+Vtt*
terms of 8 must be
tuted from the equation of the given curve.
In case it is more convenient to use p as the independent variable,
and the equation is in the form
then d8 = 4>'(P) 'ty = T" <fa
dp
Substituting this in [//V# I + dp*y
gives \ P \dp ) +1 \ dp '
Hence if p l and p t are the corresponding limits of the independent
variable p, we get the formula for the length of the arc,
where — in terms of p must be substituted from the equation of the
dp
INTEGRATION A PROCESS OF SUMMATION
Exai
s L Find the perimeter of the cardioid p = a(l + cosfl).
Solution. Here^=— agintf.
let 6 vary from OIot, the point P will generate
>f the curve. Substituting in (.4), p. 878,
= a f*(2 + 2cos<!)*d6 , = 2a f*cos.M = 4a.
I. Find the length of the spiral of Archimedes, p = aS, from the origin to the end
of the first revolution. Aju _ „, Vl+ 4jt> + ? log(2 a + Vl + 4ir*).
'
2. Rectify the spiral p — e°* from the origin to the point (p, 8). Ana. '■
Hnrr. U»a <B).
3. Find the length of the curvep = a sec' from =
Jns. [V2 + log(^ + l)]a.
4. Find the circumference of the circle p = 2rsin#.
5. Find the length of the hyperbolic spiral pff = i
from (p„ 6,) to (p„ ^J. , ^ ■•.
An,. V^Tp?^T7i + a\og^ "j
faV + V^ + rf)
6. Show that the entire length of the
.IB, BC are in arithmetical progression.
7. Find the length of arc of the cissoid p = 2
212. Volumes of solids of revolu
tion. Let V denote the volume of r'
the solid generated by revolving
the plane surface ABCJ) about the
axis of X, the equation of the plane
curve DC being
y/c»>
First Step. Construct rectangles
within the plane area ABCD as in
the figure. When this area is re
volved about the axis of X, each
rectangle generates a cylinder of
revolution. The required volume is clearly equal to the limit of
the sum of the volumes of these cylinders.
378
INTEGRAL CALCULUS
Second Step. Denote the bases of tlie rectangles by Aa^, Ax t , etc
and the corresponding altitudes by y f y % , etc. Then the volume <
the cylinder generated by the rectangle AEFt> will be ir^Ax^
the sura of the volumes of all such cylinders is
«yJA3^4 w^Aa^H 1 *y^9a^"*VirftAt&f
Tried Step. Applying the Fundamental Theorem (using ]
Hence the volume generated by revolving, about the axis of .
the area bounded by the curve, the axis of X, and the ordinatea x = a
and x = b is given by the formula
{A) V x =n f t/'dx,
where the value of */ in terms of a must be substituted from t
equation of the given curve.
This formula is easily remembered if we consider a slice or disk r
the solid between two planes perpendicular to the axis of revolutioi
as an clement of the volume, and regard it as a cylinder of iiifiiiitesini
altitude dx and with a base of area 7ry a , and hence of volume Trifdjr,
Similarly, when OY is the axis of revolution we use the formula
(B) V„=ir f x*dy,
where the value of x in terms of y must be substituted from I
equation of the given curve.
iLLtiaTtumvE Examflb 1, Find the volume generated by revolving the ellipa
^ + ~ = 1 about the axis of X.
Solution. Since y 3 = — (n 1 — I s ), and the re
quired volume is twice the volume generated
bj (MB, ira get. substituting in {A),
r T Vie = ir f"  z (a 5  z') ttl
.. V x :
To verily this result, let b = a. Then F,
, the volume of a sphere, i
Is only a special case of the ellipsoid. When the ellipse is revolved about its a
a*is, the solid generated is called a prolate spheroid ; when about its minor axis, a
obi ate spheroid.
INTEGRATION A PROCESS OF SUMMATION 379
EXAMPLES
1. Find the volume of the sphere generated by revolving the circle z* + y 2 = f*
about a diameter. Ans.  wr*.
2. Find by integration the volume of the right cone generated by revolving the
triangle whose vertices are (0, 0), (a, 0), (a, 6) about OX. Also find the volume gen
erated by revolving this triangle about OF. Verify your results geometrically.
3. Find the volume of the torus (ring) generated by revolving the circle
x* + (y  6) 2 = a 2 about OX. Ans. 2 **a*b.
4. Find by integration the volume of the right cylinder generated by revolving
the area bounded by x = 0, y = 0, z = 6, y = 4 (a) about OX; (b) about OT. Verify
your results geometrically.
. 5. Find by integration the volume of the truncated cone generated by revolving
the area bounded by y = 6 — z, y = 0, z = 0, z = 4 about OX. Verify geometrically.
6. Find the volume of the paraboloid of revolution generated by revolving the arc
of the parabola y 2 = 4 ax between the origin and the point (x v j/j) about its axis.
Ans. 2 wax* = — L_ ! ; i.e. one half of the volume of the circumscribing cylinder.
7. Find the volume generated by revolving the arc in Ex. 6 about the axis of Y.
tryf l
Ans. —  =  irxfy* : i.e. one fifth of the cylinder of altitude y, and radius of base z, .
80a 2 6
8. Find by integration the volume of the cone generated by revolving about OX
that part of the line 4z — 5y + 3 = which is intercepted between the coordinate axes.
Ans. — .
9 Find the volume generated by revolving about OX the curve 100
(z— 4a)i/* = az(z — So)
7TQ?
between the limits x = and z = 3 a. Ans. (15 — 16 log 2).
82 va*
10. Find the volume generated by revolving about OX the areas bounded by the
following loci :
(a) The hypocycloid z* + y\ = a*. Ans.
__ i x. 1 irofi
(b) The parabola z* + y* = a», z = 0, y = 0. — — .
15
IT 2
2
(d) The parabola y 2 = 4 z, z = 4. 32 *\
(e) y = xer, x = 1, y = 0. J (* "~ a )*
(c) One arch of y = sin z.
(f) y 2 = 9z, y = 3z.
8o«
3ir
2
(g) The witch y =  — — » y = 0. 4ir 2 a 8 .
z" + 4 a*
(h) y«(4 + z 2 ) = 1, y = 0, z = 0, z = ao. (1) y (1 + z 2 ) = z, y = 0, z = 0, z = 8.
(i) y = z*, y = 0, z = 1. (m) y(x  2) 2 = 1, y  0, z = 3, z = 4.
0) y*(6  z) = z 2 , y = 0, z = 0, z = 4. (n) y 2 = (z + 2) 8 , y = 0, z = 1, z =
(k) 4y 2 = z*, z = 4. (o) (z  l)y = 2, y = 0, z = 2, z = 5.
INTEGRAL CALCULUS
II. Find the volume generated by revolving the ai
is bounded by the following loci i
About OX About OT
(b) y = z>, i = 2, y = 0.
(c) B / = x», V = 0, x = a
®+&
M
(f) y l = Ox, y = 0,x = 9.
(g) y* = 4  x, x = 0.
(h) y* = * + 9, x ~ 0.
(i) *» si + V ,y = 0.
 of
86
(J) i» = 16  y, y = 0.
(It) z« + 9if' = Sfl
(1) v = 2*,v = 0, j: = 3.
(iu> y = z + 2, V = 0, * = 0, a
12. Find the volume generated by revolving one arch of the cycloid
about OX, its base. r
Himt. Substituted*— =^J=i and limits b = 0, » = 2r,in (J), p. 378. ^ M fi]r :
14. Find the volume of the solid generated by revolving the cissoid y 1 = 
about its asymptote i = 2 a. .
IB. Given the slope of tangent to the tractrix ^ = — find I
generated by revolving it about OX. "* Vn 1 — y> ^
16. Show that the volume of a conical cap of height a cut from the solid gi
by revolving the rectangular hyperbola z* — y~ — a 1 about OX equals the volume of a
sphere of radius a.
17. Using the parametric equations of the hypocycloid
rz = acos>0,
find the volume of the solid generated by revolving it about OX. A\
18. Find the volume generated by revolving one arcli of the cycloid
rx = a(flsinfl),
about its base OX Li' — «t ), ^ m fi ^
Show that if the arch be revolved about 01", tbe TOllUM gc unrated is 8jHu*.
19. Show that the volume of the egg generated by revolving the curve
*V + (*  a) {*  b) = 0, [a< b)
about OX is *(a + b) log 2 (6 a) \.
20. Find the volume generated by revolving the curve ** — u'i 1 + aty* = i
INTEGRATION A PROCESS OF SUMMATION 38
213. Areas of surfaces of revolution. A surface of revolution :
generated by revolving the arc CD of the curve
»A»
about the axis of X.
It is desired to measure this sur
face by making use of the Funda
mental Theorem.
First Step. As before, divide
the interval AB into subintervals
Aatj, Azg etc., and erect ordinates
at the points of division. Draw the
chords CE, EF, etc., of the curve.
When the curve is revolved, each
chord generates the lateral surface of a frustum of a cone of revolu
tion. The required surface of revolution is denned as the limit of the
sum of the lateral surfaces of these frustums.
Second Step. For the sake of clearness let us draw the first frus
tum on a larger scale. Let M be the middle point of the chord CE.
Then
{A) lateral area = 2 irNM CE*
In order to apply the Fundamental Theorem it
is necessary to express this product as a function
of the abscissa of some point in the interval Ai r
As in § 210, p. 373, we get, using the Theorem of
Mean Value, the length of chord
(#) cE=[i+f(x 1 y\ i &x I ,
where z, is the abscissa of the point ^(j;,, y,) on
the arc CE, where the tangent is parallel to the
chord CE. Let the horizontal line through M intersect QF, at R, and
(note RPj by e,.* Then
Substituting (£) and (C) in (A),
eget
2 ""(#!— £ ])[l+/'C c i) 5 ] A ^,= lateral area of first frustum.
* The lateral area of the frustum of a rone of revolt]
middle section multiplied l,y ttii_ slant height.
' The student will observe that as Ai! approaches z>
he circumference of
also approaches the
182 INTEGRAL CALCULUS
Similarly,
2 7r(i/ a — < a )[l +/'(Xy] A*,= lateral area of second frustum,
w (jfi ~ O [1 +/'( x ») a ]*^ 3 '» = l ate ral ^ea of last f rastum.
V2ir(i/ i  e ( )[l +/'(ay) i ] Aij = sum of lateral areas of frustums.
This may be written
(fl) j 2 «*P +/0,)']*<H  j,, [1 +/'(*,)']*Ar,.
Thikd Stkp. Applying the Fundamental Theorem to the first sum
(using the limits OA = a and OB = 6), we get
The limit of the second sum of (J3) forn = so is zero." Hence the area
of the surface of revolution generated by revolving the arc CD about
OX is given by the formula
(g) S ,= 2 ^V[. + ()']U
where ii and ^ in terms of x must be substituted from the equation
nf the revolved curve, and S denotes the required area. Or we may
write the formula in the form
5=2*
remembering that
ds = (da* + tfy 1 )* = [l + (J J 1 dr. (27). p. 135
This formula is easily remembered if we consider a narrow band of
the surface included between two planes perpendicular to the axis of
revolution as the element of area, and regard it as the convex surface
it \ yd»,
&a«5jci+/'c»wk*.
INTEGRATION A PROCESS OF SUMMATION
883
of a frustum of a cone of revolution of infinitesimal slant height d&>
and with a middle section whose circumference equals 2 wy, hence of
area 2wyds.
Similarly, when OY is the axis of revolution we use the formula
where the value of x and — in terms of y must be substituted from
dy
the equation of the given curve.
Illustrative Example 1. Find the area of the surface of revolution generated
by revolving the hypocycloid xi + y* = ai about the axis of X.
dy
A
Solution. Here ^ = — — , y = (at — xl)l.
& x i
Substituting in (E), p. 382, noting that the arc BA generates only one half of the
surface, we get yi
= 2 ,/>f*t)i(^
= 2*alf (alxi)txldz
6»a f
.*. S x —
6
12ira»
1. Find the area of the surface of the sphere generated by revolving the circle
as* + y* = r* about a diameter. Ans. 4«r*.
2. Find the area of the surface generated by revolving the parabola y* = 4ax
about OX, from the origin to the point where x = 3 a. ^^ —<*(&
3
3. Find by integration the area of the surface of the cone generated by r evolving
about OX the line joining the origin to the point (a, b). Ans. ir6Va s + 6*.
4. Find by integration the area of the surface of the cone generated by revolving
the line y = 2 x from x = to x = 2 (a) about OX ; (b) about OY. Verify your results
geometrically.
6. Find by integration the lateral area of the cylinder generated by revolving the
line x =s 4 about OY from y = to y = 6, and verify your result geometrically.
6. Find by integration the lateral area of the frustum of a cone of revolution
generated by revolving the line 2 y = z — 4 about OX from x = to x = 8, and verify
your results geometrically.
384
INTEGRAL CALCULUS
7. Parnbolic mirrors and refleuUms have the shape of a paraboloid of revolution.
Find lie area of the reflecting surface of such a mirror 2 feet deep and 8 feet
wide. Ana. Y*.
This equals the ares of n circle 7 feet in diameter.
8. Find the surface of the torus (ring) generated by revolting the circle
** + (V  &)' = Q* about OX. Ana. 4w*a&.
Hint. Using the positive value of Vo' J  J a gives the outside surface, and the negative
nhw
< inside surface.
Find the surface generated by ri
128 xa*
evolving an arch of the cycloid
about iW base. 3
10. Find the area of the surface of revolution generated by revolving each of the
following curves about OX :
(a) y = i», from x = to x = 2. Am. —[(146)1  1].
(b) y = e 1 , from x = to i = ». jt[V2 + log(l + Vz)].
(c) Theloopof Bay» = z(3a:r)». 3ira'.
(d) 6 a*xy = x* + 3 a«, from i=atoi = Sa, fj ira 1 .
(e) Theloopof 8 aV = a'i* 1*. —
(f) y* + 4 z = 2 log y, from y = 1 to y = 2. V ""•
(g) y = e», from i =— ootox = 0. «[Vz + log(l + Vz)]
( I) The cardioid [ Z = ° f* c ™" "
Kl \y = a(2si»
(j) i/ + 2 j = 4, from a = ti
(k) 3y— 2z = 6, fromz =
fl) y = z", from z = to x =
(m) z* + *y« = ie.
(o) 8s' + y« = 88.
(0) y" = 8 x, from x = to z :
11. Find the area of the surface of revolution generated by revolving
following curves about OY :
(a) z + 2 y = 8, from y = to y = 3.
(b) 3z + 2 y = 12, from y = to y = i,
<c) x* = 4y, from y = to y = 3.
(d) x* + 10)/* = 18.
(e) 4i* + y> = 100.
(f) 8x = y* t from y = to y = 1.
(g) i = y>, from y = to y = 8. ,4m. ^ [(780)11]
(h) 6a*zy = z* 8a', from a = a to j: = 3a. (20 + logS)*v'
(1) 4y = z*— 21ogx, from it = 1 to z = 4. 84 it.
cos 2 6),
sin 2d).
each of the
0) 2y =
n*
INTEGRATION A PROCESS OF SUMMATION
385
12. Find the area of the surface of revolution generated by revolving each of the
following curves :
(a) The ellipse 2L + L = i.
Hint, e  eccentricity of ellipse
a
(b) The catenary y =  (e" + e a ),
from x = to x = a.
(c) x 4 +8=6scy,fromz=l tox=2.
fx = e*sin0,
( d H y = «*cob0, from B = to  .
46otd OX
2 wo* 1 arc sin c.
4.60U* OF
f , wo*. 1 + e
2wo f + — log— ^~
« 1 — e
ira
^(ei+4c*).
4
2V^ir
(«*  2).
+
Vs)
2wa»(lei).
»W + log2).
^(2^+1).
wa a
(4 + 81og3)
wa a
(e) 3x s + 4y s = 3a s .
(f) x + y = 4, from x = to x = 4.
(g) y = 2x + 4, from y = 4 to y = 8.
(h) x s +2y«=16.
214. Miscellaneous applications. In § 212 it was shown how to
calculate the volume of a solid of revolution by means of a single
integration. Evidently we may con
sider a solid of revolution as gen
erated by a moving circle of varying
radius whose center lies on the axis
of revolution and whose plane is per
pendicular to it. Thus in the figure
the circle ACBD, whose plane is per
pendicular to OX, may be supposed
to generate the solid of revolution — EGFH, while its center moves
from to N, the radius MC(= y) varying continuously with OM (= x)
in a manner determined by the equa
tion of the plane curve that is being
revolved.
We will now show how this idea
may be extended to the calculation
of volumes that are not solids of
revolution when it is possible to ex
press the area of parallel plane sec
tions of the solid as a function of their distances from a fixed point.
Suppose we divide the solid shown in our figure into n slices by
sections perpendicular to OX and take the origin as our fixed point.
386
INTEl.HAL CALCULUS
Let FDE be one face of such a slice. Construct a right prism upon
FDE as a base, the second base lying in the other face of the slice.
Since, by hypothesis, the area of FDE is a function of ON, or a
let /(x) = area of FDE— area, of base of prism, and let Az =
tude of prism. „
Hence f(x) &x = volume of prism, and 2)/(.c,) Aa, = sum of voluir
of all such prisms. It is evident that the required volume is the limit
of this sum ; hence, by the Fundamental Theorem,
and we have the formula
w
rf/m*
where f(x) is the area of a section of the solid perpendicular to OX ex
pressed in terms of its distance (= x) from the origin, the jlunits being
chosen so as to extend over the entire region R occupied by the solid.
Evidently the solid O — ABC may be considered as being generated
by the continuously varying plane section DEF as ON(=x) varies
from zero to OM. The following examples will further illustrate this
principle.
i Calculate the volume of the ellipsoid
by means of a single integration.
Solution. Consider a section of the ellipsoid perpendicular to OX, aa ABCD, with
. i :i i. ..  and c'. The equation of the ellipse EEJQ Id the TOyplane Is
Solving this for k(=o') ii
x[= OM) gives
Similarly, from the equation of the
ellipse KFG1 in the XOZplatie we get
Substituting in (A),
a of the ellipse (section) A BCD is
INTEGRATION A PROCESS OF SUMMATION 387
We may then think of the ellipsoirl as being generated by a variable
ellipse ABCD moving from G to E, its center always on OX and its
plane perpendicular to OX.
e 2. Find the volume of ft right conoid with circular base,
the radius of base being r and altitude a.
Solution. Placing the conoid ns shown in the figure, consider a section PQR per
pendicular to OX. This section is an isosceles triangle; and since
(found by solving x 1 + y % = 2 rx, the equation of the circle
OR A Q, for v) and M p m 0)
av / 2rai a =/(x).
Substituting in {A), p. 888,
This is one half the volume of the cylinder of the «
P C
J...A9
"Z'TsSt* 171
e base and altitude.
Surface of fluid
We will now take up the study of fluid pressure and learn how to
calculate the pressure of a fluid on a
vertical wall.
Let ABCD represent part of the area
of the vertical surface of one wall of a
reservoir. It is desired to determine the
total fluid pressure on 'this area. Draw
the axes as in the figure, the Faxis lying
in the surface of the fluid. Divide AB
into n subintervals and construct hori
zontal rectangles within the area. Then
the area of one rectangle (as EP) is //A/.
If this rectangle was horizontal at the depth x, the fluid pressure on
it would be
Wxybx,
[The proMnrnof a
where W= the weight of a unit volume of the fluid. Since fluid pres
sure is the same in all directions, it follows that Wxy&x will be ap
proximately the pressure on the rectangle EP in its vertical position.
Hence the sum ,
"% Wxg£x,
888 INTEGRAL CALCULUS
represents approximately the pressure on all the rectangles. The pres
sure on the area ABCD is evidently the limit of this sum. Hence, by the
Fundamental Theorem,
limit '
IJfyrtyMfitov**
Hence the fluid pressure on a vertical submerged surface bounded
by a curve, the axis of X, and the two horizontal lines x = a and x = b
is given by the formula h
(B) fluid pressure = W I yxdx,
where the value of y in terms of x must be substituted from the equa
tion of the given curve.
We shall assume 62 lb. (= W) as the weight of a cubic foot of water.
Illustrative Exahpie 3. A circular water main
8 ft. in diameter is half full of water. Find the pressure
on the gate that closes the main.
Solution. The equation of the circle in 7? + y* = 0.
Hence y =V9 — a*,
W=62,
and the limits are from x 
(B>, we get the pressure o
to be
to i = 8. Substituting in
the right of the axis of X
12 f V9  i» • idx = [ V (0  «*>*]£ = 658.
Hence the total pressure = 2 x 558 = 1116 lb. Am.
Let us now consider the problem of finding the work done in emp
tying reservoirs of the form of solids of revolution with their axes
vertical. It is convenient to assume
the axis of A" of the revolved curve
as vertical, and the axis of Y on a
level witb the top of the reservoir.
Consider a reservoir such as the
one shown ; we wish to calculate the
work done in emptying it of a fluid
from the depth a to the depth b.
Divide AB into n subintervals, pass
planes perpendicular to the axis of
revolution through these points of
division, and construct cylinders of
revolution, as in § 212, p. 377. The volume of any such cylinder
will be vi/'Ax and its weight WirtfAx, where IT = weight of a cubic
INTEGRATION A PROCESS OF SUMMATION 389
unit of the fluid. The work done in lifting this cylinder of the fluid
out of the reservoir (through the height a:) will be
WirifxAx.
[Work done in lifting equals the weight multiplied by the vertical height.]
The work done in lifting all such cylinders to the top is the sum
»
2} Wiryfx^Xr
The work done in emptying that part of the reservoir will evidently
be the limit of this sum. Hence, by the Fundamental Theorem,
Therefore the work done in emptying a reservoir in the form of a
solid of revolution from the depth a to the depth b is given by the
formula
(C) work = Wn J ifxdx,
where the value of y in terms of x must be substituted from the
equation of the revolved curve.
ILLUSTRATIVE EXAMPLES
1. Calculate the work done in pumping out the water filling a hemispherical reser
voir 10 feet deep.
Solution. The equation of the circle is x* + y* =100.
Hence y a = 100 — x 2 ,
TT=62,
and the limits are from x = to x = 10.
Substituting in (C), we get
work = 62 » f 10 (100  x 2 ) xdx = 156,000 » ft. lb.  x
2. A trough 2 ft. deep and 2 ft. broad at the top has semielliptical ends. If it is
full of water, find the pressure on one end. Ans. 1661 lb*
3. A floodgate 8 ft. square has its top just even with the surface of the water. Find
the pressure on each of the two portions into which the square is divided by one of
its diagonals. Ans. 6290* lb., 10,681* lb.
4. Find the pressure on one face of a submerged vertical equilateral triangle of
side 4 ft., one side lying in the surface of the water. Ans. 496 lb.
5. A horizontal cylindrical oil tank is half full of oil. The diameter of each end is
4 ft. Find the pressure on one end if the oil weighs 50 lb. per cubic foot.
Ans. 206* lb.
INTEGRAL CALCULUS
6. Find the work done in pumping out a seinielliptical reservoir filled wit
The tup is a circle of diameter (I ft. and the depth is 5 ft. Ana. 3487} a ft. ill.
7. Find the pressure o
8. Find the pressure o
the surface of (In reservoir in Example 1
the surface of the reservoir in Example fl.
8, A conical reservoir 12 ft. deep is filled with a liquid weighing 80 lb. per cubic
foot. The top of the reservoir is a circle 8 ft. in diameter. Find the energy expended
in pumping it out. Ana. 15,860* ft. lb.
10. The cross section of a trough is a parabola with vertex downward, the latui
rectum lying in the surface and being 4 feet long. Find the pressure on one end of
the trough when it is full of a liquid weighing 82 t lb. per cubic foot. Ant.
<
Hist. Pressure  2 lit I y(lt> + z)il8, and da dz.
v An*. 223201 lb.
12. A board in the form of a parabolic segment by a chord perpendicular t
axis is immersed In water. The vertex is at the surface and the axis is vertical.
20 feet deep and 12 feet broad. Find the pressure in tone. Ana. 68.52.
13. How far must the board in Example VI be mint to double the pressure ?
Am. 12 feet.
14. A water tant is in the form of a hemisphere 24 feet in diameter, surmounted
by a cylinder of (he same diameter and 10 feet high. Find the work done in pumping
it out when tilled within 2 feet of the top.
15. 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 opposite vertices move on the circumfen
circle. Find the volume of the solid generated. Ann. {a 1
16. A woodman fella a tree 2 ft. in diameter, cutting halfway through on each side
The lower face of each cut is horizontal, ami the upper fact: makes an angle of
with the lower. How much wood does he cut out 1
Ana. _ w
Va
17. A variable equilateral triangle moves with its plane perpendicular to the x
and the ends of its base on the points on the curves y* = 10a* and y 5 = :
tlvely above the jraxis. Find the volume generated by the triangle as it moves from
the origin to the points » lnw abscissa is a. ^/§
18. A rectangle moves from a fixed poii
tance from this point, and the other equal
volume generated while the rectangle movi
t, one side being always equal to the dis
o the square of this distance. What is the
i a distance of 2 ft.? Ana. 4 cu. ft.
19. On the double ordinates of the ellipse — + *j = 1, isosceles triangles of verti
cal angle 90" are described in planes perpendicular to that of the ellipse. Find the
volume of the solid generated by supposing such a variable triangle moving from one
extremity to the other of the major axis of the ellipse.
INTEGRATION A PROCESS OF SUMMATION 391
20. Determine the amount of attraction exerted by a thin, straight, homogeneous
rod of uniform thickness, of length 2, and of mass Jf, upon a material point P of mass
in situated at a distance of a from one end of the rod in its line of direction.
Solution.* Suppose the rod to be divided into equal infinitesimal portions (ele
ments) of length dx.
— = mass of a unit length of rod ;
M
hence — dx = mass of any element.
Newton's Law for measuring the attraction between any two masses is
 . _ At product of masses
force of attraction = —  —   — ;
(distance between them)*
therefore the force of attraction between the particle at P and an element of the rod is
—mdz m 4 g ...^*
I
i
(x + a)*
which is then an element of the force of attraction required. The total attraction between
the particle at P and the rod being the limit of the sum of all such elements between
x = and x = J, we have ^, ,,
— mdx
I __Mm r l dx _ Mm
(x + a)* ~~ ~T~Jo (x + a)*~ a(a + I)'
force of attraction = I m = — — l = H An*.
I (x + a)« I Jo(x + a)» a(a + l)
21. Determine the amount of attraction in the last example if P lies in the per
pendicular bisector of the rod at the distance a from it. A 2 mM
aV4a 2 +l*
22. A vessel in the form of a right circular cone is filled with water. If h is its
height and r the radius of base, what time will it require to empty itself through
an orifice of area a at the vertex ?
Solution. Neglecting all hurtful resistances, it is known that the velocity of dis
charge through an orifice is that acquired by' a body falling freely from a height
equal to the depth of the water. If then x denote depth of water,
v = V2gx.
Denote by dQ the volume of water discharged in time dt,
and by dx the corresponding fall of surface. The volume of
water discharged through the orifice in a unit of time is
aV2gx,
being meas ured as a right cylinder of area of base a and alti
tude v(=V2gx). Therefore in time <tt,
(A) dQ = aV2gxdt.
Denoting by S the area of surface of water when the depth is x, we have, from
Geometry, s x t irr % x t
— = — > or. 5 = •
HT* h* ' tf
* The two following examples indicate commonly employed " short methods," the detailed
exposition followed in the preceding sections being omitted. The student should however
supply this.
892 INTEGRAL CALCULUS
But the volume of water discharged in time dt may also be considered as the vol
ume of cylinder AB of area of base 8 and altitude dz; hence
(B) d Q = Sdx = ^^.
Equating (A) and (B) and solving for dt,
.. m*z*dx
at =
Therefore t= l = —=. Am.
JO ^Mv/q^v R/.>k/oT.
aA 2 V20Z
* ?n*x a ds _ 2wr*Vh
aA 2 V20Z~6aV20
23. A perfect gas in a cylinder expands against a piston head from the volume v Q
to the volume v v the temperature remaining constant. Find the work done.
Solution. Let c = area of cross section of cylinder.
If dv = increment of volume,
then — = distance piston head moves while volume takes on the increment do.
By Boyle's Law, pv = k{= const.).
.\ p =  = pressure on piston head.
k dv
Hence element of work done =  • — (= pressure x dist.).
... , . r»ikdv k r*idv k. c,
•\ total work done = I — =  I — =  log 1 .
CHAPTER XXIX
SUCCESSIVE AND PARTIAL INTEGRATION
215. Successive integration. Corresponding to successive differenti
ation in the Differential Calculus we have the inverse process of
successive integration in the Integral Calculus. We shall illustrate
by means of examples the details of this process, and show how
problems arise where it is necessary to apply it
d*v
Illustrative Example 1. Given — ^ = 62. to find y.
dx*
Solution. We may write this
= 6x,
dx
or. d(^)=6zdx.
m
Integrating, :r? = 6xdx,
dx 1 J
or, ^zzSzi + Ci.
ax"
This may also be written .
dx
or, d(^ = (Sz* + cJdz.
Integrating again, ^ = J (8 z* + c t ) dx, or,
(A) j = * 8 + Ci* + <V
Again dy = (x 8 + c 1 x + c 8 ) dx, and integrating,
(B) V = rf^ + c 2 x + c 8 . ilTW.
4 A
The result (4) is also written in the form
^ = / / 6 xdxdx (or = J J 6 aair 2 ),
and is called a double integral, while (2?) is written in the form
y = J I J 6 xdxdxdx (or = 1/16 xdx*) y
803
394
INTEGRAL CALCULUS
and is called a triple integral. In general, a multiple integral require
two or more successive integrations. As before, if there are no lie
assigned, as in the above example, the integral is indefinite ; if there are
limits assigned for each successive integration, the integral is definite.
Illustrative Example 2. Find the equation of a curve for every point of wbi
the second derivative of the ordinate with respect to the absciss* equals 4.
Solution. Here
= 4. Integrating as in Illustrative Example I,
<»>
y = 2x* + c 1 x + t.
This is the equation of a parabola with its axis parallel to OT and ext
upward. By giving the arbitrary constants of integration c ( and c, all possible
we obtain all such parabolas.
In order to determine c, and e„ two more conditions are necessary. Suppose »
say (a) that at the point where z = 2 the slope of the tangent to the parabola is a
and (b) that the parabola passes through the point (2, — 1).
(a) Substituting a and ~ = in (C)
gives = 8 + r,.
Hence r, = — B,
and (D) becomes y = 2x* — 8z + c ? .
(b) The coordinates of (2, — 1) must satisfy this equation ; therefore
 1 = 8  16 + e„ or, c, = + 7
Therefore the equation of the particular parabola which satisfies all three c
ditiona is y _ 2x *  81 + 7.
1. Given ^ = a
9. fjfvon ?J? — n
dx*
4. Given ^sinfl, find p.
6. Given — = SP , finds.
dP (*
6. Given tPp = sin •p coa 5 «tf# a , find p.
7. Determine the equations of all curves having 7.1
Hiht. — "O.tronx (40), p. 157, since ff0.
" = ao +
+V + *.
y = loga + ?& + v + c,.
p=cos$ + 'j— + f t 9+e l .
' dz"
Ana. y = c,i + c,, a doubly infinite system of straight lines.
SUCCESSIVE AND PARTIAL INTEGRATION 395
8. The acceleration of a moving point is constant and equal to/; find the distance
(space) traversed.
d*« ffi
Hnrr. — /. Ana. a — ^ + c x t + c.
dt* 2
9. Show in Ex. 8 that e x stands for the initial velocity and c, for the initial
distance.
10. Find the equation of the curve at each point of which the second derivative of
the ordinate with respect to the abscissa is four times the abscissa, and which passes
through the origin and the point (2, 4). Arts. Sy = 2x (x 2 — 1).
11. Given !L = x cos x, find y. Ana. y = x cosx — 4 sinx + 1 — r— * + CjX + c r
dxr 6 2
 ~, dfy . . n , . 7 cosx cos^x , c.x 2 ,
12. Given —  m = sin*x, find y. Ana. y = J * — h cx + c t .
dx 1 f 9 27 2 ^ *
216. Partial integration. Corresponding to partial differentiation
in the Differential Calculus we have the inverse process of partial
integration in the Integral Calculus. As may be inferred from the
connection, partial integration means that, having given a differ
ential expression involving two or more independent variables, we
integrate it, considering first a single one only as varying and all the
rest constant. Then we integrate the result, considering another
one as varying and the others constant, and so on. Such integrals
are called double, triple, etc., according to the number of variables,
and are called multiple integrals.*
Thus the expression ~~
w=J j/0> y^dydx
indicates that we wish to find a function u of x and y such that
In the solution of this problem the only new feature is that the
constant of integration has a new form. We shall illustrate this by
means of examples. Thus suppose we wish to find u, having given
ox
Integrating this with respect to x, considering y as constant, we
u = x*+ xy + 3 x + <f>,
* The integrals of the same name in the last section are special cases of these, namely,
when we integrate with respect to the same variable throughout.
396
INTEGRAL CALCULUS
where tf> denotes the constant of integration. But since y was re
garded as constant during this integration, it may happen that tf>
involves y in some way ; in fact, <f> will in general be a function of y.
We shall then indicate this dependence of <f> on y by replacing <f> by
the symbol <f> (jy). Hence the most general form of u ii
u = x' + xy + 3x + <j>(y) l
where #(#) denotes an arbitrary function of y.
As another problem let us find
w
This means that we wish to find it, having given
My "
Integrating first with respect to y, regarding x as constant, we j
,1,
«jyW»Vyfc
8*
■*>+% + *<».
where ^(^) is an arbitrary function of z and is to be regarded i
the constant of integration.
Now* integrating this result with respect to x, regarding y as con
stant, we have _i u „ j
where <& (y) is the constant of integration, and
217. Definite double integral. Geometric interpretation. Let/(z,y)
be a continuous and smglevalued function of x and y. Geometrically,
is the equation of a surface, as KL. Take some area .5 in the XY
plane and construct upon S as a base the right cylinder whos
elements are accordingly parallel to OZ. Let this cylinder inters
KL in the area S', and now let us find the volume V of the i
bounded by S, S\ and the cylindrical surface. We proceed as follows:
At equal distances apart (= Ax) in the area S draw a set of 1
parallel to OY, and then a second set parallel to OX at equal distance
apart (= Ay). Through these lines pass planes parallel to YOZ a
SUCCESSIVE AND PARTIAL INTEGRATION 397
XOZ respectively. Then within the areas S and S' we have a net
work of lines, as in the figure, that in S being composed of rectangles,
each of area Ax Ay. This construction divides the cylinder into a
number of vertical columns, such as MNPQ, whose upper and lower
bases are corresponding
portions of the networks
in S' and S respectively.
As the upper bases of
these columns are curvi
linear, we of course can
not calculate the volume
of the columns directly.
Let us replace these col
umns by prisms whose
upper bases are found
thus : each column
is cut through by
a plane parallel to XT passed through that vertex of the upper base
for which x and y have the least numerical values. Thus the column
MNPQ is replaced by the right prism MNPM, the upper base being
in a plane through P parallel to the X OYplane.
If the coordinates of P are (x, y, z), then MP = z =f(x, y), and
therefore
(B) volume of MNPR=f(x, y)AyAz.
Calculating the volume of each of the other prisms formed in the
same way by replacing x and y in (B) by corresponding values, and
adding the results, we obtain a volume V approximately equal to
V; that is,
(C) K'=22;/C*y)AyAz;
where the double summation sign XX indicates that there are two
variable* in the quantity to be summed up.
If now in the figure we increase the number of divisions of the
network in S indefinitely by letting Ax and Ay diminish indefinitely,
and calculate in each case the double sum (0), then obviously V will
approach V as a limit, and hence we have the fundamental result
IP)
= Ajf = o^^/(x, y)AyAa%
398 INTEGRAL CALCULUS
The required volume may also be found as follows : Consider s
one of the successive slices into which the solid is divided by the
planes parallel to YZ; for example, the slice whose faces are
FIGH and TL.JK. The thickness of this slice is Ax. Now the
values of z along the curve HI are found by writing x= OD in
the equation z=f(x, y); that is, along HI
Hence the area FIGR
\£/ (oa *•'*'
[ual
is,
b of
The volume of the slice under discussion is approximately equal
to that of a prism whose base is FIGH and altitude Ax\ that is,
equal to f>a
Ax  area FIGH = Ax / /( OD, y)dy.
The required volume of the whole solid is evidently the limit t
the sum of all prisms constructed in like manner, as x(= 0D~) varies
from OA to OB ; that is, oB Dg
(#) V=f dxj f(r,y)dy.
Similarly, it may he shown that
m
The integrals (£) and (.r 7 ) are also written in the more
form r OB /"" r 0F r EU
 / f(x,y)dydx and / I f(x,y)dxdy.
Joa Jdf Joe Jew
In (B) the limits DF and DG are functions of x, since they are
found by solving the equation of the boundary curve of the base of
the solid for y.
Similarly, in (F) the limits E W and EU are functions of y. Now
comparing (Z>), (£ ), and {F) gives the result
limit _. . /"*■ /■">
(ff) r=to = * j XXf(x,y)AyAx=j J f(x, y)dydx
Jqc Jew
« compact
=J j '/(* y^dxdy.
where v^ and v t are, in general, functions of y, and «j and w 3 functions
of x, the second integral sign applying to the first differential and
being calculated first.
SUCCESSIVE AND PARTIAL INTEGRATION
Oar result may be stated ii
The definite double integral
the following form :
/ / f(x,y)dydx
may he interpreted at that portion of the volume of a truncated right
cylinder which it included between the plane XOY and the surface
the hate of the cylinder being the area bounded by the curvet
Similarly for the second integral.
It is instructive to look upon the above process of finding the vol
ume of the solid as follows :
Consider a column of infinitesimal base dydx and altitude z as an
element of the volume. Summing up all such elements from y = DF
to y = DG, x in the meanwhile being constant (say = OD~), gives the
volume of an indefinitely thin slice having FQHI as one face. The
volume of the whole solid is then found by summing up all such
slices from x = OA to x = OB.
In partial integration involving two variables the order of integra
tion denotes that the limits on the inside integral sign correspond to
the variable whose differential is written inside, the differentials of the
variables and their corresponding limits on the integral signs being
written in the reverse order.
Illuwkitivk Example 1. Find the value of the definite doable Integral
Bulntlon. f f ~ (i + y)dydx
= J." [f^*** + *> *] *
Interpreting this result geometrically, it means that we have found the volume of
the solid of cylindrical shape standing on OAB as base and bounded at the top by the
surface (plane) z = x + y.
400
INTEGRAL CALCULUS
AX)}
from y lia'iti
> from x li
The attention of the student is now particularly called to the manner in which tl
limiU do bound the base OAII, which [.'unvsponds lo the area S in the figure.
Our solid here stands on a base In the .1" l"plaue bounded by
y = (line OB )
y = Va a — x" (quadrant of circle A B
z = (line OA)\
218. Value of a definite double integral over a region S. In the 1:
section we represented the definite double integral as o volume. This
does not necessarily mean that every definite double integral is a vol
ume, for the physical interpretation uf the result depends on the nature
of the quantities represented by i\ y, z. Thus, if x, y, z are simply con
sidered as the coordinates of a point in spate, and nothing more, then
the result is indeed a volume. In order to give the definite double
integral in question an interpretation not necessarily involving the
geometrical concept of volume, we observe at once that the variable z
does not occur explicitly in the integral, and therefore we may confine
ourselves to the .XTplane. In fact, let us consider simply a region
S in the AT'plane, and a given function
f(x, y~). Then, drawing a network as be
fore, calculate the value of
f(x, y)AyAx
for each point* of the network, and i
up, finding in this way
and finally pass to the limit as Ax anil Ay approach zero. This op(
tion we call integrating the function f(x, y) over the. region S, and it it
denoted by the symbol rr
JJJXx,y)dydx.
S
If S is bounded by the curves x=a l ,z=a t ,y=u 1 ,y = n a , then, by (G'
JJfd* y)dydx=pjyrx, y^dydx.
* Mure generally, divide tl
OFIntoAjd, Ay t , , iif m . E
tily a cornoij choose a point z,
iterral on OX Into •iiibintorvals A.'i, A*,, •■
r the network, and In each rectangle AiiAy* (not n
. Then it is clear i,,fuitionaI!'j that
ff/fr, ridMl*m = «]£5)/(Si, *)A*AjA
SUCCESSIVE AND PAETIAL INTEGRATION 401
We may state our result as follows :
Theorem. To integrate a given function f(x, y) over a given region S
in the XOYplane means to calculate the value of
limit i\
ax = oy\/(i, y)AyA*,
Ay = ^ ^
as explained above, and the result is equal to the definite double integral
I I f(x, y) dydx, or, I I f(x, y) dxdy,
the limits being chosen so that the entire region S is covered. This process
is indicated briefly by rr
JJf(fr V) fyd*
S
In what f ollow8 we shall show how the area of the region itself and
its moment of inertia may be calculated in this way.
Before attempting to apply partial integration to practical problems
it is best that the student should acquire by practice some facility in
evaluating definite multiple integrals.
J nib pa 7a s 6*
I (a — y) z*dydz = .
6 Jo 6
Jnib pa r* b r V s "l a r ib a* 7a*6*
, j <t («V)x*d„d X =j i \a V *\z*dz=) h _*.<&. = _.
pa p y/cp — a* 2 <Z*
Illustrative Example 2. Verify I / xdydx — .
vo'r
J%a py/a*a* paT "Wo 1 a 1
I xdydx = I I XV I dx
JJaTZ* J » L J_V?T^
o/; a . v*=*,u =[_?(««  *fl=  #.
In partial integration involving three variables the order of inte
gration is denoted in the same way as for two variables ; that is, the
order of the limits on the integral signs, reading from the inside to
the left, is the same as the order of the corresponding variables whose
differentials are read from the inside to the right.
J»s p% pi 35
l l xy i dzdydx = — •
s Ji J* 2
Solution, f f f zy*dzdydz = f f \ f xy*dz \dydx =f f Ixtfz] dydx
= %{* CxyHydz = 8 f T f 'xytylcte
402
INTEGRAL CALCULUS
Verify the following :
o Jo
1. J o J xy(xy)dycfe = — (a6).
6 /.?
10 /o7. v5<i ^=l at 
2. f prdfidr =
76*
24
. .xl
2 ./J 4>
•• /»*F
3. f f xydxdy =
JO Jj«
11 g*
24 '
i2. rv x r F xv«wi^=~
Jo Jo Jo 90
Jo Jo Jo 4
J*v /«a(l + eM#) 4/rt
f >sin0drfW= ^
o Jo
13
o Jo Jo 90
>a /*y/c? — ax dldz
Jo Jo
VOX — X 1
yl
14. J^ J o «(x + y)(W* = ^.
= 4a*
>r /taeosf
o Jo 3
3
.2a /»x
16 a*
>6 /*10(
iff V5^dadt = 66».
,»a^
8. f f a (w + 2v)dwdv =
J* 1 /••*  1
/ e»duxto = .
o Jo
143 a»
30
2
16. f f (x* + y»)dydx =
l7 ,f« = ? log 2.
Jo J*«x* + y* 2 ^
"■/. ? £./**=(H)r.
 r/;
a 8 — 6»
r* sin tf^dr = — r — (cos/3 — cos a).
3
1 r x /.* + *
J i px /txtjr
1 1 e"+'+*d£dydx =
o Jo Jo
e«3 3e* .
1 T + c '
2i. r a r 6 rv+y*+« i )d«dy da; =^( ai + 6, + ci )
Jo Jo Jo 3
22.
Jf Vzy — y*dxdy = 66*.
o J*
' J\ Jo Jo x* + y* ~~ 2 *
219. Plane area as a definite double integral. Rectangular coordinates.
As a simple application of the theorem of the last section (p. 401), we
shall now determine the area of the region S itself in the XOFplane
by double integration.*
• Some of the examples that will be given in this and the following articles may be solved
by means of a single integration by methods already explained. The only reason in such
cases for using successive integration is to familiarize the student with a new method for
solution which is sometimes the only one possible.
SUCCESSIVE AND PAKTIAL INTEGRATION
403
As before, draw lines parallel to OX and OY at distances Ax and
Ay respectively. Now take any one of the rectangles formed in
this way, then
element of area = area of rectangle PQ = Ay • A&,
the coordinates of P being (x, y).
Denoting by A the entire area of region S, we have, using the
notion of a double summation,
limit
We calculate this by the theorem on
p. 401, setting /(:r, y) = l, and get
0») a
OA J CD
dt/dxj
Y
jJrr
•
At
fr
/i
f 1
•
•
!
1
1
l
^^^^^
i
i
i
A c
J& ^
where CD and CE are, in general, functions of x, and OA and 02? are
constants giving the extreme values of x, all four of these quantities
being determined from the equations of the curve or curves which
bound the region S.
It is instructive to interpret this double integral geometrically by
referring to our figure. When we integrate first with respect to y,
keeping x (= OC) constant, we are summing up all the elements in a
vertical strip (as DF). Then integrating the result with respect to x
means that we are summing up all such vertical strips included in
the region, and this obviously gives the
entire area of the region S.
Or, if we change the order of inte
gration, we have
I dxdy,
OK J JIG
where HG and HI are, in general, func
tions of y, and OK and OL are constants
giving the extreme values of y, all four of these quantities being
determined from the equations of the curve or curves which bound
the region S. Geometrically, this means that we now commence by
summing up all the elements in a horizontal strip (as £«/}, and
then find the entire area by summing up all such strips within the
region.
INTEGRAL CALCULUS
Correspontling to the two orders of summation (integration), the
following notation and figures are sometimes used :
(*>)
A^jjdydx, AJftety,
Referring to the result stated on p. 401, we may say:
The area of any region is the value of tfte, double integral of the function
f(x, »/) = I taken over that region.
Or, also, from § 217, p. 396,
The area equals numerically the volume of a right cylinder of unit
height erected on the base S.
Illcstbativk Example 1. Calculate the area of the circle
integration.
Solution. Summing up first (lit' elements in a verticil] strip, iv
A=f C dyfo.
From the equation of the boundnry curve (circle)
;* + j/ 1 = r 5 by double
have from (B), p. 403,
When the region whose area we wish to find is symmetrica] with
respect to one or both of the coordinate axes, it sometimes saves
us labor to calculate the area of only a part at first In the above
example we may choose our limits so as to cover only one quadrant
of the circle, aud then multiply the result by 4. Thus
'S:r*«Hr
*dz=
SUCCESSIVE AND PARTIAL INTEGRATION
405
Illustrative Example 2. Calculate that portion of the area which lies above OX
bounded by the semicubical parabola y 2 = x* and the straight line y = x.
Solution. Summing up first the elements in a horizontal strip, we have from (C),
p. 408,
OD /*AC
\ dxdy.
o Jab
ab
From the equation of the line, AB = y, and from the
equation of the curve, AC = y», solving each one for x. To
determine OD, solve the two equations simultaneously to
find the point of intersection E, This gives the point (1, 1);
hence OD = 1. Therefore
'n ,, **jr* , »*ri ! a
= i — 1 = A AnB 
1. Find by double integration the area between the straight line and a parabola
with its axis along OX, each of which joins the origin and a 6 _/* ^
the point (a, 6). Ans. f f
•/0 Jhx
'to
a
dydx = —
2. Find by double integration the area between the two parabolas 8y* = 26 x and
5x 2 = 9y. Ana. 5.
3. Required the area in the first quadrant which lies between the parabola y 2 = ax
and the circle y 2 = 2 ax — x 2 .
4na.
yq* 2 a*
4. Solve Problems 2 and 8 by first summing up all the elements in a horizontal
strip, and then summing up all such strips.
2a*
Am. Ex. 2, J^ J v *dxdy = 5. Ex. 8, f  f a
vcP
26
dxdy = __ —
5. Find by double integration the areas bounded by the following loci :
(a) x* + y* = a*, x + y = a.
(b) y 2 = 9 + x, y 2 = 9  3x.
(c) y = sin x, y = cosx, x = 0.
(d)y = ^^i' 2, ' = a!,X = 
(e) x* + y» = a», x + y = a.
(f) y 2 = x + 4, y 2 = 42x.
(g) y 2 = 4a 8  x 2 , y 2 = 4a 2  4ax.
(h) x 2 + y 2 = 25, 27 y 2 = 16x».
(i) 4y 2 =x«, y = x.
(j) y 2 = ax, y 2 = 
2a — x
(k) x 2  y 2 = 14, x 2 + y 2 = 36.
An*. — .
3
48.
V21.
a 2 (»l).
2
32
400
INTEGRAL CALCULUS
220. Plane area as a definite double Integral. Polar coordinates.
Suppose the equations of the curve or curves which bound the
region S are given in polar coordinates.
Then the region may be divided into
checks bounded by radial lines drawn
from the origin, and concentric circles
drawn with centers at the origin. Let
jP5 = A/j and angle POR = A8. Then
arc PR = pA8, and the area of the shaded
check, considered as a rectangle, is p\8 • Ap. The sum of the areas
of all such checks in the region will be
Since the required area is e
the formula
idently the limit of this sum, we have
.ff*..
Here, again, the summation (integration) may be effected in I
ways.
When we integrate first with respect to 8, keeping p constant, it
means that we sum up all the elements (checks) in a segment of a
circular ring (as ABCD), and next integrating with respect to p, that
we sum up all such rings within the entire region. Our limits t
appear as follows:
(B) A= f {"" pdddp,
the angles XOA and XOB being, in general, functions of p, and OB
and OF constants giving the extreme values of p.
Suppose we now reverse the order of integration. Integrating first
with respect to p, keeping 8 constant, means that we Bum up all the
elements (checks) in a wedgeshaped
strip (as GKLH). Then integrating
with respect to 8, we sum up all such
strips within the region S. Here
(C) A
■err
OH and OG being, in general, functions of 8, and the angles .Y0.7aud
XOl being constants giving the extreme values of 8.
SUCCESSIVE AND PARTIAL INTEGRATION 407
Corresponding to the two orders of summation (integration), the
following notation and figures may be conveniently employed :
. CfpdpdO, A = CCpdOdp.
These are easily remembered if we think of the elements (checks) as
being rectangles with dimensions pd8 and dp, and hence of area pdBdp.
Illustrative Example 1. Find the area of the circle p = 2rcos0 by double
integration.
. Slimming up all the elements In a sector
(as OB), the limits are and 2 r cob 6 ; and summing
up all such sectors, the limits are and — for the
semicircle OXB. Substituting in (D), 2
i /.'/.""*'*•
1. In the above example find the area by Integrating first with respect to 9.
2. Find by double integration the entire areas in Examples 116, pp. 368, 860.
3. Find by double integration the area of that part of the parabola p  a sec* 
intercepted between the curve and Its latus rectum. • »
<"■ '/.'JT <**x
4. Find by double integration the area between the two circles p = acw9,p=bcoe6,
b>a; integrating first with respect to p. ,■" ,>»wt T
Ant. 21 ' J pdpd0= (V — a*).
5. Solve the last problem by first Integrating with respect to 6.
6. Find by double integration the area bounded by the following loci :
(a) p = 6 si n 8, p . VI sin 6. Ant. 27 w.
(b)pcos0 = 4,p = 8. ^IwVs.
(c)p = «aec»!.p = 2a. 2<*r S g.
<d)p = a<l + costf), p = 2ocosA ^~
(e) p Bin = 6, p = 10.
(f) p = 8cos0, p cob = 2.
(g) p = 2cosfl, p = 8cos*.
408
INTEGRAL CALCULUS
221. Moment of area. Consider an element of the area of the region
S, as PQ, the coordinates of P being (z, y). Multiplying the area
of this element (=AyAx) by the distance
of P from the Faxia (=#), we get the
product
(A) x&y&z,
which is called the moment of the element
PQ with respect to the Faxis. Form a
similar product for every element within
the region and add all such products by a double summation. Then
the limit of this sum, namely,
(B) A* = 0^5):rAyA:e= j j xdydx,
defines the moment of area of the region S with respect to the Yaxis.
Denoting this moment by Jtfj,, we get
(C)
JOT 1 **'
for
the limits of integration being determined in the same way
finding the area.
In the same manner, if we denote the moment of area with respect
to the A'axis by M^ we get
(J>) M x = ffydydx,*
the limits being the same as for (C).
222. Center of area. This is defined as the point (x, y) given by
the formulas
<:*)
c)
K
II"
II*
II ydydx
From (E~), area x = M t and arcai/=J/ I .
Hence, if we suppose the area of a region to he concentrated at (i, y),
the moments of area with respect to the coordinate axes remain iinchanyi'l.
SUCCESSIVE AND PAETIAL INTEGRATION
409
The center of area of a thin homogeneous plate or lamina is the
same as its center of mass (or center of gravity).*
If a coordinate axis is an axis of symmetry of the area, it is evident
that the corresponding coordinate of the center of area will be zero.
In polar coordinates x = p cos d, y = p sin 0, and element of area
= pApAd replaces AyAar. Hence formulas (JP) become
(?)
CCp cos Odpdd CCp sin ddpdO
x= „„ > y= „„
Am
IIk*'
the limits being the same throughout and determined (as before) in
the same way as for finding the area.
Illustrative Example 1. Find the center of the area
bounded by y* = 4 x, z = 4, y = 0, and lying above OX.
8olntion. By (C), p. 408, M, = f f * zdydx = if i.
By (D), p. 408,
f ydydx = 16.
t/0
/• 4 /» 2 at
Area = f f dydx = y.
Substituting in (E), p. 408,
x = if a ^ y = i£, and y = 16 ~ y = . 4ns.
EXAMPLES
1. Find the centers of the areas bounded by the following loci :
(a) The quadrant of a circle.
(b) The quadrant of an ellipse.
(c) y = sin x, y = 0, from x = to x = ir.
(d) A quadrant of x^ + yl = ».
(e) y 2 = 4 ox, x = A.
(f) y = 2x, y = 0, x = 3.
(g) y 2 = 8x, y = 0, y + x = 6.
(h) (2a  x)y 2 = x 8 , x = 2a.
(i) y 2 (a*  x a ) = a 4 , x = 0, x = a.
(j) x* + i/* = aix = 0,y = 0.
(k) Cycloid x = a (0 — sin 0), y = a(l — cos 0).
^Liw. x = ■— = y.
Sir
_ 4a _ 46
8ir 3»
_ ir _ w
2 8
_ 256a _
x = = y.
815 ir y
x =  A, y = 0.
5
x = 2 = y.
x = 2.48, y = 1.4.
_ 5a _ A
 2a _ A
x = — » y = 0.
ir ¥
a
x = v = y.
_ 5 _ 5a
x = air, V = ~q'
* If the plate is supported loosely on a horizontal axis through its center of gravity, there
will be no tendency to rotate, whatever the position of the plate may be.
2. Find the centers of the a:
(a) One loop of p* = n* coe 2 8.
(b) One loop of p  a sin 2 6.
INTEGRAL CALCULUS
by the following c
(c) Cardinal p = a(l + cosfl).
(d) p = 6sm0, p = 12sin0.
tlf"
"»,f
(e)pco,« = 4,p»8.
(g)p = 2c
»«,P
F
r
i
223. Moment of inertia of plane areas. Consider an element of t
area of the region S, as PQ, the coordinates of P being (x, y). Mul
tiplying the area of this element (= AyAz)
by the square of the distance (= x~) of P
from the Yaxis, we get the product
(A} x'AyAx,
which is called the moment of inertia * of
the element PQ with respect to the Faxis.
Form a similar product for every element
within the region and add all such products by a double summation.
Then the limit of this sum, namely
(B) to  j 5 £ x*Ayte =jjx*dydx,
define* the moment of inertia of the area of S with respect to the Ya.
Denoting this moment by I t , we get
(C) l,=Jj^dydx,
the limits of integration being determined in the same way as fot
finding the area.
In the same manner, if we denote the moment of inertia of the >■
with respect, to the Xaxis by I lt we get
(B) I^fjtfdydx,
the limits being the same as for (C).
224. Polar moment of inertia. Rectangular coordinates. Considei
an element of the area of region S, as P Q. I f the coordinates of P a
(£, y), the distance of P from is VJf + y 3 . Multiplying the area c
" Because the element of area Is multiplied by the tquttr' at il« distance from the I"*
it Is sometimes called the second moment, to conform with the definition of m
15221. p.«W).
SUCCESSIVE AND PARTIAL INTEGRATION
411
lenient (= Ai/Ax) by the square of the distance of P from the origin,
we have the produet (W+ y)AyAx,
which is called the polar moment of inertia of the element PQ with
spect to the origin. The value of the double sum
(E) £=§SSc*'+ jO by** Ea j[jr<* i+ y 1 ) <&*"
'.ties thf polar moment of inertia of the area
ithin the region S with respect to the origin.
Denoting this moment of inertia by 7 ,
eget
(F) I t =jj(x* + y*)dydx*
le limits of integration being determined in the same way e
iiding the area.
From (F*),
I t =JJ(S+ ?) dydx =JJx'dydx +jjfdydx.
By comparison with (C) and (D) we get
(C) J. = /,+ /„
and bence the
Theorem. Tfie polar moment of inertia of a plane area with respect to
any point equal* the sum of its moments of inertia with respect to any
perpendicular axes through that point
25. Polar moment of inertia. Polar coordinates. Since the element
area is nowpApAt?, and x* + y' = /)*, we get, by substitution in (A'),
(H)
) limits of integration being the same
i for finding the area.
Since the element of area (= A(/Aj
= p&pA0) is essentially positive and x 3 , °
', p* are always positive, it follows that moment of inertia is never
zero, hut always a positive number. Moments of inertia arise fre
quently in engineering problems, the principal appUcation being to
the calculation of the energy of a rotating body.
We may then say that I„ in the value of the double integral of the fiuictioii /(i, i/i
+ <j over the area.
;=JfjO*.
INTEGRAL CALCULUS
1. Find I a over the ares bounded by the lines x = a, y = 0, y = x
Solution. These lines Muml a tviai^lt 1 0A!<. Summing up all the elements in 3
vertical strip (as PQ), the [/limits are zero and 1 (fouiid
from the equation of the line "/i'. Sin tuning up all such
strips within the 1
and a { 0£). Hence, by (F),
;), the fliuiiis a
= ////V+l'V!/<k = '>*(^ + ^V Am.
suppose the triangle to be composed of horizontal
S(d*>
ir„o
I. and the coOrdi
I'i'tariglu li'.iiml'.'ii by the lines x 
" ftW ""■ ins. ff<? + V)*V<!x = *
3. Find I over the right triangle formed by the coordinate axes and the line join
ing the points (a, 0), (0, b).
•■/x
{x* + y3)dydx =
4. Find T x for the region within the circle x 1 + y" =
6. Find tg for the ellipse ~ + ^ = 1,
;:
6. Find J„ over the region between the straight line and a parabola v
along O.Y, each of which joins the origin and the point (a, b).
"■■fX^'+^'W+l)
7. Find I over the region bounded by the parabola y' 1 = 4 ax, the line x + y — So
= 0, and OX . /•» **sffi «.«« ,>!■■ 814o*
^ns. J J (J: a + y 2 ) rfy<Le + J J (x + y') dydz. = ^ ,
°xirv +*>*<* = 3 ^
8. Find J„ over the region bounded by the
circle p s 2rcos6.
Solution. Summing up the elements in the tri
angularshaped strip OP, the pliniits are zero and
2r cosfl (found from the equation of the circle).
Summing up all such strips, the 0limits are
and  . Hence, by (II),
Summing up first the elements in a circular strip (as QR), we have
10. Find I c
t the entire area of the cardioid p =
SUCCESSIVE AND PARTIAL INTEGRATION 413
0. Find /„ over the area bounded by the parabola p = a sec 1  . its latus recti.m,
and the initial line OX. *
11. Find 7 forthe lemniscate p J = a 5 cos2 0. jlns. —
12. Find 1, and J r for area bounded by y* = 4 ax, y = 0, x = z t .
16 * 7
13. Find the moment of inertia of the area of a right triangle with respect to the
vertex of the right angle, a and 6 are the lengths of the perpendicular Bides.
An. ^(a' + W).
14. Find I t for tbe area bounded by y' = 4 ta, x + y — 8 a, y — 0. An*. I r = —a*.
IS. Find the
■ids through its
of inertia of a rectangle whose sides are 2 a, 26, about an
parallel to the side 2 6 ; to the side 2 a. . a*6 a&*
AnB  T ; T"
An* — *a*
16. Find J, for a* + yt = o*.
17. Find I„ over the area of one loop of p = acos2ff.
226. General method for finding the areas of surfaces. The method
given in § 213 for finding the area of a surface applied only to
surfaces of revolution. We
shall now give a more gen
eral method. Let
be the equation of the sur
face KL in the figure, and
suppose it is required to cal
culate the area of the region
S' lying on the surface.
Denote by S the region on
the .TOFplane, which is the
orthogonal projection of 5*
on that plane. Now pass planes parallel to YOZ and XOZ at com
mon distances Ax and Ay respectively. As in § 217, these planes
form truncated prisma (as PB~) bounded at the top by a portion
(as PQ~) of the given surface whose projection on the TOFplane
414
INTEGRAL CALCULUS
is a rectangle of area AjA# (as AB), which rectangle also forms t
lower base of the prism, the coordinates of P being (x, y, z).
Now consider the plane tangent to the surface A'L at P. Evidentl
the same rectangle AB is the projection on the AWplane of I
portion of the tangent plane (PS) which is intercepted by
prism PB. Assuming y as the angle the tangent plane makes \
the XOFplane, we have
area AB — area PR cos y.
AyAz =
a PR  cos 7.
1
W W
)T
8z\'"l l
KsMdi
area PR = \l +
$*$!«*
which we take as the element of area of the region S'. We then def
the area of the region S' as
baWfrffi**
the summation extending over the region S, as in § 21 7. Denot ing h
A the area of the region S', we have
the limits of integration depending on the projection on the XOT^plant
of the region whose area we with to calculate. Thus for (B) we choose
our limits from the boundary curve or curves of the region S in the
A'OFplane precisely as we have been doing in the previous for
sections.
SUCCESSIVE AND PARTIAL INTEGRATION 415
If it is more convenient to project the required area on the XOZ*
plane, use the formula
S
where the limits are found from the boundary of the region S> which
is now the projection of the required area on the XOZplane.
Similarly, we may use
<*> 'JOrhtsMsfi**
s
the limits being found by projecting the required area on the YOZ
plane.
In some problems it is required to find the area of a portion of one
surface intercepted by a second surface. In such cases the partial
derivatives required for substitution in the formula should be found
from the equation of the surface whose partial area is wanted.
Since the limits are found by projecting the required area on one
of the coordinate planes, it should be remembered that
To find the projection of the area required on the XOYplane, elimi
nate z between the equations of the surfaces whose intersections form the
boundary of the area.
Similarly i we eliminate y to find the projection on the XOZplane> and
x to find it on the YOZplane.
This area of a surface gives a further illustration of integration of
a function over a given area. Thus in (i?), p. 414, we integrate the
KSHDT
over the projection on the XO Fplane of the required curvilinear
surface.
Illustrative Example 1. Find the area of the surface of sphere z* + y* + z a = r*
by doable integration.
{fetation. Let ABC in the figure be one eighth of the surface of the sphere. Here
dz _ x dz _ y
dz~~ z* by" z'
y*
410
INTEGRA! CALCULUS
The projection of the area required on the A'Ol'pIani' is AOB, a regioi
br* = 0, (OB); » = 0, (OA); t* + y> = r; (BA).
Integrating Aral with respect to y, we sum up alt the elements along n strip (as
DEFG) which is projected on the .TOFplane in a
strip also (aa MJfFO ); that is, our yUmita are zero
and MF (= Vr* — x*). Then integrating with respect
to x sums up all such strips composing the surface
ABC; that is, our ilimits are zero and OA ( = r).
Substituting in (B), we get
r.£
Illustrative Example 2. The center of a sphere of radius r is on the surface of
a right cylinder, the radius of whose base is  Find the surface of the cylinder
Intercepted by the sphere.
Solution. Taking the origin at the center of the sphere, an element of the cylinder
for the zaxis, and a diameter of a right section of the cylinder for the zaxis, the
equation of the sphere is z 5 f y* + t 3 = r 3 , and of
the cylinder I s + y' = rx. ODA PB is evidently one
fourth of the cylindrical surface required. Since
this area projects into the semicircular arc ODA on
the XOrplane, there ia no region 5 from which to
determine our limits in this plane ; hence we will
project our area on, say, the JTOZplane. Then
the region S over which we Integrate is OACB,
which is bounded by * = 0, (OA) ; x = 0, {OB) ;
i' + re = r*, (ACB) ; the last equation being found
by eliminating y between the equations of the two
surfaces. Integrating first with respect to z means
that we sum up all the elements in a vertical strip '■
(as PD), the zlimits being zero and Vr*  rx.
Then on integrating with respect to * we sum up all such strips, the xlimlts b
Since the required surface lies on the cylinder, the partial derivative* required for
formula (C), p. 415, must be found from the equation of the cylinder.
*ss
»y
= 0,
Substituting In (C), p. 416,
Substituting the value of y in tei
AliJ^f^VSS
1(^)7*
is of x from the equation of the cylinder,
2 rf' l^^dx = 2 r f'J dx = 4r a .
SUCCESSIVE AND PARTIAL INTEGRATION
417
EXAMPLES
1. In the preceding example find the surface of the sphere intercepted by tbe
Jo Jo Vr* **!/<
:. The axes of two equal right circular cylinders, r being the radius of their
iteraect at right angles. Find the surface of one intercepted by the other.
. Take x 3 + z> = r* and **' + y 3  r* as equations of cylinders.
"/;/.
= 8r<.
. Find by integration the area of that portion of the surface of the sphere
I 1 + z' = 100 which lies between the parallel planes x = — 8 and x = 6.
4. Find the surface of the cylinder z 3 + y 3 =
dtheXOrplane.
. Find the surface of the cylinder z' + {z cos a + y sin a)* = r* which is situated
in the positive cumpartment of coordinates.
Hint. The axis uf this cylinder is the line z0, zvona + y sin a  0; and the radius of
6. Find the area of that part of tbe pi:
e coordinate planes.
= 1 which is intercepted by
.ins. * i/b>c" + c*a* + a'
Find the area of the surface of the paraboloid y 3 + z 3 = 4 a* intercepted by the
. v™ f 
a of the surface of the cylinder inter
im. (l8Vl3l)i
cylinder y* = a* and the plai
8. In the preceding usample find the a
d by the paraboloid and plane.
_v vi'
. Find the area of that portion of the surface of the cylinder y* + z^ = at
bounded by a curve whose projection on the Tl'plane is*« + y* = a*. Ant. V*
10. Find the area of that portion of the sphere x* + y 3 + z 1 = 2 ay cut out by one
nappe of the cone *' + z 3 = y*. Am. 2ra 3 .
227. Volumes found by triple integration. In many cases the vol
ume of a solid bounded by surfaces whose equations are given may
be calculated by means of tliree successive integrations, the process
being merely an extension of the methods employed in the preceding
sections of this chapter.
Suppose the solid in question be divided by pianos parallel to the
coordinate planes into rectangular parallelepipeds having the dimen
sions A«, At/, Ar. The volume of one of these parallelepipeds is
&z Aw A J",
and we choose it as the element of volume.
Now sum up all such elements within the region R bounded by
the given surfaces by first summing up all the elements in a column
INTEGRAL CALCULUS
parallel to one of the coordinate axes; then sum up all such columns
in a slice parallel to one of the coordinate planes containing that axis,
and finally sum up all such slices within the region in question. The
volume V of the solid will then be the limit of this triple sum as Az,
Ay, Ax each approaches zero as a limit That is,
limit
Az = Q R
the summations being extended over the entire region R bounded by
the given surfaces. Or, what amounts to the same thing,
R
the limits of integration depending on the equations of the boundi
surfaces.
Thus, by extension of the principle of § 218, p. 401, we speak of
volume as the result of integrating the function f(x, y, z) = 1 through
out a given region. More generally, many problems require the integra
tion of a variable function of r, y, and z throughout a given region,
this being expressed by the notation
R
/(*> y< z ) dzdydx,
which is, of course, the limit of a triple sum analogous to the double
sums we have already discussed. The method of evaluating this triple
integral is precisely analogous to that already explained for double
integrals in § 218, p. 401.
Jllusthative Example 1. Find the volume of that portion of the ellipsoid
* + ^ + !* = l
a* 6* c»
which lies In On drat octant.
Solution. Let — A BC be that portion of the
ellipsoid whose volume is reemired, the equations
of the bounding surfaces being
<» £+£+£•.<»«>.
(2) z = 0, {OAB),
(S) j, = 0, (0.4(7),
(i) 1 = 0, (OflC). 1
PQis an element, being one of the rectangular parallelepipeds witb dimensions Az,
Ay, Ax into which the planes parallel to the coordinate planes have divided the regioa.
SUCCESSIVE AND PARTIAL INTEGRATION 419
Integrating first with respect to z, we sura up all suc h elements In a column
(as BS>, the (limits being zero [from (2)] and TB = e Jl  *  j [from (1) by
solving for *],
Integrating next with respect to ^, we sum up all such cohimns in a slice (as
DEMNGF), the ylimita being zero [from (8)] ami HO = b + ) \  ~ [from equation
of the curve AOB, namely — + ~ = 1, by solving for y].
Lastly, integrating with respect to x, we sum np all such slices within the entire
region O  ABC, the x limits being zero [from (4)] and OA = a.
Hence Y= J j "f " "'dzdydz
web {•", . ... wabe
= 4rfJo ( *' ~6~'
Therefore the volume of the entire ellipsoid is —
Illustrative Example 2. Find the volume of the solid contained between the
paraboloid of
revolution x 3 + y 3 = az,
the cylinder x 3 + y 3 = 2 ax,
and the plane t = 0.
Solatkm. The zlimits are zero and N
by solving equation of paraboloid for z).
. The ^limits are zero and MN(=
solving equation of cylinder for y).
The zlimits are zero and OA (=2a).
The above limits are for the solid ONAB, one half of the solid whose volume U
required. „ tr ,
Hence — = f f C dzdydx = ——.
Therefore V = — —
1. Find the volume of the sphere x 1 + y' + z' ! = r* by triple Integration.
M. 
2. Find the volume of one of the wedges cut from the cylinder a a \ y 3 = r* by
the planes z = and z = mi. »r Wr*» /•■» 2r*m
jIim. 2J J I dzdydx = — —
3. Find the volume of a right elliptic cylinder whose axis coincides with the
Xaxis and whose altitude = 2 a, the equation of the base being cV + &*** = ** c *
dzdydx = 2 wabe.
»/.xr
&•&+&
4. Find the entire volume bounded by the surface
coordinate planes.
5. Find the entire volume bounded by the surface xt + yl + *' = a*.
A ~
6. Find by triple integration the volume of the solid bounded by the planes
x = a, y = b, ; = mi and the coordinate planes XOY and XOZ. Ana. i mba*.
7. The center of a sphere of radius r is on the surface of a right circular cylinder
the radius of whose basis is   Find the volume of the portion of the cylinder inter
cepted by the sphere. Ana. j I* — J) r*.
8. Find the volume bounded by the hyperbolic paraboloid ea = xy, the XOT
plane, and the planes x = a„ x = o„ y = b v y = b r (a*  a, 1 ) (6*  o»)
9. Find the volume common to the two cylinders x* + y 1 — r 3 and x* + x* = r*.
Ju
lflr»
10. Find the volume of the tetrahedron bounded by the coordinate planes and the
plane? +  + ? = l. Ana. hite.
11. Find the volume bounded by the paraboloid i 5 + y 1 — z = 1 and the XFplane.
ns. .
12. Find the voiume common to the paraboloid y* + z 1 d 4 ax and the cylinder
z 1 + y' = 2ax. Ana. 2ro? + ya'.
13. Find the volume included between the paraboloid y 1 + z 2 =
cylinder y 1 = ox, and the plane x = 8a. An.
14. Find the entire volume within the surface I" + y* + z' =t a
15. Compute the volui
the two parabolas x = y 1 ,
: of a cylindrical column standing
= x 2 as base and cut off by the surface z = 12 + y
16. Find the volume bounded by the surfaces y 3 =
= z + i.
17. Find the volume bounded by 2
■* + 2 1/ 5 , x + y = 1, and the coordinate
r, the parabolic
;air + oV3)o*.
,rea common to
• u + r *
* + !,«=«,
the coordinate
plan
18. Given a right circular cylinder of altitude a and radius of base r. Through a
diameter of the upper base pass two planes which touch the lower base on opposite
sides, Find the volume of the cylinder included between the two plar
Am. (Tl)ar 1 .
CHAPTER XXX
ORDINARY DIFFERENTIAL EQUATIONS*
228. Differential equations. Order and degree. A differential equa
tion is an equation involving derivatives or differentials. Differential
equations have been frequently employed in this book, the following
being examples :
< 2 >( 8 «!+ 2 )(SH^+>)t
(3) tan* & = p. (4>,p.84
(4) ft = 12 (2 x  1). Ex. 1, p. 101
ax
(5) dy = ^dx. Ex. 2, p. 138
ay
(6) dp =  a * sm20 dd. Ex. 3, p. 138
(7) d*y = (20 x> 12 x)dx*. Ex. 1, p. 139
(8) x ^ + y^=5u. Ex.7,p.l94
dx dy
(10) !f = (1 + 3 xyz + * y*») u. Ex. 7, p. 204
cxcycz
In fact, all of Chapter XI in the Differential Calculus and all
of Chapter XXIII in the Integral Calculus treats of differential
equations.
An ordinary differential equation involves only one independent
variable. The first seven of the above examples are ordinary differ
ential equations.
* A few types only of differential equations are treated in this chapter, namely, such as
the student is likely to encounter in elementary work in Mechanics and Physics.
421
422
INTEGRAL CALCULUS
A partial differential equation involves more than one independent
variable, as (8), (9), (10).
In this chapter we shall deal with ordinary differential equations
only.
The order of a differential equation ia that of the highest derivative
(or differential) in it. Thus (3), (5), (6), (8) are of the first order;
(1), (4), (7) are of the second order; and (2), (10) are of the thin,
order.
The degree of a differential equation which is algebraic in I
derivatives (or differentials) is tiie power of the highest derivative
(or differential) in it when the equation is free from radicals and
fractions. Thus all the above are examples of differential equations
of Ha first degree except (2), which is of the second degree.
229. Solutions of differential equations. Constants of integration. A
solution or integral of a differential equation is a relation between the
variables involved by which the equation is identically satisfied. Thus
GO , = <•,«,*■
is a solution of the differential equation
w 3+r*
For, differentiating (^),
rt = — P , sin x.
ax* '
(C)
Now, if we substitute (4) and (C) in (B), we get
— c 1 sin x + Cj sin x = 0,
showing that (A) satisfies (/i) identically. Here <?, is an arbitrary
constant. In the same manner
(0) y = c a cosj;
may be shown to be a solution of (B) for any value of c^. The relation
(i?) y = e, sin x + c a cos x
is a still more general solution of (B). In fact, by giving particular
values to c 1 and c. t it is seen that the solution (2?) includes the s
tions (ii) and (2>).
The arbitrary constants c, and c % appearing in these solutions a
called constants of integration. A solution such as (#), which cor
tains a number of arbitrary essential constants equal to the order c
ORDINAEY DIFFERENTIAL EQUATIONS 428
the equation (in this case two), is called the general solution or the
complete integral.* Solutions obtained therefrom by giving particular
values to the constants are called particular solutions or particular
integrals.
The solution of a differential equation is considered as having been
effected when it has been reduced to an expression involving integrals,
whether the actual integrations can be effected or not.
230. Verification of the solutions of differential equations. Before
taking up the problem of solving differential equations it is best to
further familiarize the student with what is meant by the solution of
a differential equation by verifying a number of given solutions.
Illustratite Example 1. Show that
(1) y = CjXcoslogx + c 2 X8inlogx]+ xlogx
is a solution of the differential equation
(2) *«g*! + 2„ = *log*.
Solution. Differentiating (1), we get
(3) ^ = (c, — c«) sin logx + (c 2 + c,) cos logx + logx + 1.
ax
^v <&V . .sin logx . . .cos logx 1
Substituting (1), (8), (4) in (2), we find that the equation is identically satisfied.
EXAMPLES
Verify the following solutions of the corresponding differential equations :
Differential equations Solutions
'•'(D' +2 *!— •
y* = 2cx + c 1 .
y*cx« + ^ = 0.
1 + c
M d 8 ]/ 3 (JPy „ e m
dx* dx * x ' ri (k  1) 2
It is shown in works on Differential Equations that the general solution has n arbitrary
constants when the differential equation is of the nth order.
Differential equal iota
INTEGRAL CALCULUS
Solution*
*V
+ <•,*» + v .
■,;,i,r'
10.(1^)
11 *v±*i
y = 6in/l + ee"*»*.
231. Differential equations of the first order and of the first degree.
Such an equation may be brought into the form Mdx + Ndy=* 0, in
which M and N are functions of x and y. Differentia! equations
coming under this head may be divided into the following types:
Type I. Variables separable. When the terms of a differential
equation can be so arranged that it takes on the form
(A) f(x)dx+F(y)dy = 0,
where f(x) is a function of x alone and F (y) is a function of y
alone, the process is called separation of the variable*, and the solu
tion is obtained by direct integration. Thus integrating (J), we
get the general solution
w
ff(x)dx+fr<j,)dy e .
where c is an arbitrary constant.
Equations which are not given in the simple form (A) may
be brought into that form by means of the following rule for
the variables.
First Step. Clear of fraction!, and if the equation involve* deriva
tives, multiply through by the differential of the independent variable.
Second Step. Collect all the terms containing the same differential
into a single term. If, then, the equation takes on the form
XYdx+X'Y'dy = Q,
where X, X' are functions of x alone, and Y, Y' are functions of y alone,
it may be brought to the form (J) by dividing through by X'Y.
Third Step. Integrate each part separately, as in (B).
ORDINARY DIFFERENTIAL EQUATIONS 426
Illustrative Example 1. Solve the equation
dy 1 + y*
dx (l + x s )xy
Solution. First step. (1 + x*)xydy = (1 + y*)dx.
Second step. (1 + y a )dx  x(l + x*)ydy = 0.
To separate the variables we now divide by x (1 + x 2 ) (1 + ?*), giving
dx ydy
x(l + x*) 1 + y*
dx r ydy
= 0.
Instep, f « fJgL^p,
f*f^ fJ*LC, p.829
log*  llog(l + x«)  i log(l + y«) = C,
log(l + x 8 ) (1 + y a ) = 2 logs  2C.
This result may be written in more compact form if we replace — 2 C by log c, i.e.
we simply give a new form to the arbitrary constant. Our solution then becomes
log(l + a;*) (1 + y a ) = logx 8 + logc,
log(l + x*)(l + V*) = logcx 2 ,
(1 + x*) (1 + V s ) = ex 9 . Ana.
Illustrative Example 2. Solve the equation
Solution. First step. axdy + 2 aydx = xydy.
Second step. 2aydx + z(a — y)dy = 0.
To separate the variables we divide by xy,
2adx {av)dy _
z V
Third step. 2af^+ af^fdy = C,
2 a logx + a logy — y = C,
alogx«y = C + y,
logeX a y =  + £.
a a
By passing from logarithms to exponentials this result may be written in the form
^+?
x*y = e° a ,
£ ?
or, x*y = c° • e«.
c
Denoting the constant e° by c, we get our solution in the form
9
x*y = ce*. Ana.
426 INTEGRAL CALCULUS
EXAMPLES
Differential equations Solutions
1. ydx — xdy = 0. y = ex.
2. (l + y)dx(lx)dy = 0. (1 + y)(lx) = c.
3. (1 + x)ydx + (1 — y)xdy = 0. logxy + * — y = c.
x — a
4. (x 2  a*)dy  ydx = 0. y* a = c
x + a
6. (x*yx*)^ + y« + xy«=0. £±y + log? = c .
ax xy x
l
6. tAto + (c — a)du = 0. © — a = ce*.
_ du 1 + u* t> + c
7. — = — — . u = — — .
do 1 + c 2 1 — co
8. (l + a^eftAtorzO. 2f* arc tans = c.
9. dp + p tan 0d0 = 0. o = c cos0.
10. sin B cos 0dl — cos B sin 0d> = 0. cos^ = c cos0.
11. sec s ^tan0d^ + sec s ^tan^d> = O. tan0tan0 = c.
12. sec s ^tan0d> + 8ec , ^tan^d^ = O. sin 2 + sin f = c.
13. xydz — (a + x) (b + V)dy = 0. x — y = c + log(a + xj^y 6 .
14. (1 + x 8 ) dy — Vl — y*dx = 0. arc sin y — arc tan x = c.
16. Vl  z*dy + Vl  y 2 dx = 0. y Vlx 2 + xVl y« = c.
16. 8e*tanydx + (le*)sec 2 ydy = 0. tany = c(l e«) 8 .
17. 2x*ydy = (l + x*)dx. y * = .l + x + c.
x
18. (x  y*x)dx + (y  x*y)dy = 0. x* + y* = x*y* + c.
y
19. (x*y + x)dy + (xy 3 — y)dx = 0. xy + log = c.
x
Type II. Homogeneous equations. The differential equation
Mdx+Ndy = Q
is said to be homogeneous when M and N are homogeneous functions
of x and y of the same degree.* Such differential equations may be
solved by making the substitution
y = vx.
This will give a differential equation in v and x in which the vari
ables are separable, and hence we may follow the rule on p. 424.
* A function of z and y is said to be homogeneous in the variables if the result of replacing
x and y by Xx and Xy (X being arbitrary) reduces to the original function multiplied by some
power of X. This power of X is called the degree of the original function.
ORDINAEY DIFFERENTIAL EQUATIONS 427
Illustrative Example 1. Solve the equation
A , *dy dy
dz dz
Solution. y 1 dz + (x 2 — zy)dy = 0.
Since this is a homogeneous differential equation, we transform it by means of the
substitution y = vz. Hence dy = vdz + zdv,
and our equation becomes
xWdx + (x 2  vz 9 ) (vdz + zdv) = 0,
zhrtz + z*(l v)dv = 0.
To separate the variables divide by vz*. This gives
dz . (1 — v)dv
/?+/*■
/* =
logx + logfl — t> =
•c,
logeta =
= c +
»,
«x =
e C +<
» = «£.«»,
«x =
rce".
But « =  • Hence the solution is
z
y =
ce*.
4iw.
EXAMPLES
Differential equations
5o2uttons
i« (x + y)<*x + xdy = o.
x 2 + 2xy = c.
2. (x + y)dx + (y  x)dy = 0.
log(x 2 + y 2 )*— arc tan = c
3. zdy — ydx = Vac 2 + y 2 dx.
l + 2cyc 2 x 2 = 0.
4. (8y+10x)dx + (5y + 7x)dy =
0.
(x + y)«(2x + y)» = c.
5. zj/*dy = (a5» + y*) dx.
y» = 3 X 3 log CX.
6. (x*  2y«)dx + 2xydy = 0.
y 2 = — x 2 log ex.
7. (x 2 + y 2 )dx = 2xydy.
y* =r X 2 + CX.
r
8. (2VSs)dt + td« = 0.
ur't = c.
g
9. (t8)dt + td8 = 0.
te« = c.
„_ y dy y
10. xco8~ = y cos — x.
x ax x
da?
xe * = c.
11. x cos (ydx + xcty) = V sin  (xdfl
r — ydx).
y
xycoa = e.
X
Type III. Linear equations. A differential equation is said to be
linear if the equation is of the first degree in the dependent variable
(usually y) and its derivatives (or differentials). The linear differen
tial equation of the first order is of the form
where P, Q are functions of x alone, or constants.
428 INTEGRAL CALCULUS
To integrate (4), let
where z is a new variable and u is a function of a; to be determined.
Differentiating (2?),
dx dx dx
Substituting (C) and (2?) in (4), we get
dz du ^ _
u— + z — +Puz = Q y or,
dx dx
Now let us determine, if possible, the function u such that the term
in z shall drop out. This means that the coefficient of z must vanish ;
***** **> du n A
as
Then — = Pd^
it
and log,w = — / Pdx + C, giving
(JZT) u = c 1 e'S pdx .
Equation (D) then becomes
dz ^
To find z from the last equation, substitute in it the value of u
from (i£) and integrate. This gives
c x dz = Qef Pd *dx>
(JF) ^ = CQeS Pdx dx + C.
The solution of (4) is then found by substituting the values of
u and z from (2£) and (P) in (J?). This gives
(<?) y = eS Pdx ( CQeS Pdx dx + CJ.
ORDINABY DIFFERENTIAL EQUATIONS 429
The proof of the correctness of (6) is immediately established by
substitution in (4). In solving examples coming under this head
the student is advised to find the solution by following the method
illustrated above, rather than by using (G) as a formula.
Illustrative Example 1. Solve the equation
» !,¥i =<*>'•
Solution. This is evidently in the linear form (A), where
P=?andQ = (x + l)i
X + 1
Let y = uz; then ? = u — + z — . Substituting in the given equation (1), we get
ax ax ax .
dz , du 2uz <v 5
Now to determine u we place the coefficient of z equal to zero. This gives
du 2u
= 0,
dx 1 + *
du_ 2dx
"u"~l + x'
log,u = 21og(l + x), t
(8) it = e**a+*P = (l + *)*•*
Equation (2) now becomes, since the term in z drops out,
u = (* + !)«•
Replacing u by its value from (3),
l = (x + l)i,
dz = (x + l)idx,
(4) ,.!fc±2» +0 .
Substituting (4) and (3) in y = uz, we get the solution
= ^±^ + C(x + l)« ^«.
V
9 3
• Since log* u  log. e^f fl + «)•  log (1 + x)« • loge e  log (1 + at)*, it follows that u  (1 + z)*.
For the sake of simplicity we have assumed the particular value zero for the constant of
integration.
430 INTEGRAL CALCULUS
Diferential equation* Sohtiioiu
*• % ~ vh = <* + ^ 2y = (x + 1)« + e(x + 1)V
ax x + 1
dy qy _ x + 1 „. _ _ , x 1
dx x x 1 — a a
3. x(l  x*)dy + (2x 2  l)ydx = ax*dx. y = ax + cx Vl  x*.
6. — coet + asint = 1. a = sin t + c cos t.
dt
6.  + 8C08t = i8in2t. 8 = sint— 1 + eer***.
dt
7. ?  y = «»•. y = x»(e* + c).
dx x
8. ^ +  y = — • x»y = ox + c.
dx x x*
~ dy 1
9. _l + y = — . e*y = z + c.
dx e*
10. g + l^y = t y = x»(l + c^).
Type IV. Equations reducible to the linear form. Some equations
that are not linear can be reduced to the linear form by means of a
suitable transformation. One type of such equations is
where P, Q are functions of x alone, or constants. Equation (4) may
be reduced to the linear form (A), Type III, by means of the substitution
2 = y"" + 1 . Such a reduction, however, is not necessary if we employ
the same method for finding the solution as that given under Type
III, p. 427. Let us illustrate this by means of an example.
Illustrative Example 1. Solve the equation
(1) %.+l = alogx.*.
ax x
Solution. This is evidently in the form (A), where
P= , Q = alogx, n = 2.
x
x . ., dy dz m du
Let y = uz ; then ^ = u— + *— .
dx dx ax
Substituting in (1), we get
dz , du t uz ,  
u— + S—H = a logx • uv,
dx dx x
(2) tt  + (£ + i)' = a, °8 xUV 
ORDINARY DIFFERENTIAL EQUATIONS 431
Now to determine u we place the coefficient of z equal to zero. This gives
du
dx
i =0 '
du_
u
_dx
X
logu =
logx
= log,
2
tt =
1
X
(«)
Since the term in z drops out, equation (2) now becomes
u— = alogxu 2 ? 2 ,
dx
—  = a logx >uz 2 .
dx
Replacing u by its value from (3),
dz . z 2
— = alogx«— t
dx x
dz . dx
 = alogx— ,
z 2 x
l^ q(logx) 2
2 2 '
(4) z= 2
a(logx) 2 + 2C
Substituting (4) and (3) in y = uz y we get the solution
1 2
y=
x o(logx) 2 + 2C
or, xy [a (logx) 2 + 2 C] + 2 = 0. Arts.
Differential equations Solutions
1. % + xy = xV. y 2 = x 2 + 1 + ce<
dx
2. (1  x 2 )^  xy = oxy 2 . y = (cVlx 2  a)*.
dx
*» «. A dv . .. • x + 11
3. 8y 2 ^ay» = x + l. y* = ce°*  5 .
dx a a 8
A/ £ £
4. ^(xV + »y)=l. x[(2y 2 )e 2 + c] = e 2 .
dx
6. (y logx — l)ydx = xdy. y = (ex + logx + 1) *.
dy , „ . . tanx + secx
6. y — cosx — = y a co8x(l — sinx). y =
dx sin x + c
41/
dx
7. x^ + y = y 2 logx. y 1 = logx + 1 + ex.
432
INTEGRAL CALCULUS
232. Differential equations of the nth order and of the first degree.
Under this head we will consider four types which are of importance
in elementary work. They are special cases of linear differential eqi
lions, which we defined on p. 427.
Type I. The linear differential equation
w
i+A=
■+p n y=
in which the coefficients p t , p t , ■■, p, are constants.
The substitution of e" for y in the first member gives
0" + PS " ' + P/" ' + ' ' + Pn> «"'■
This expression vanishes for all values of r which satisfy
equation
(S) T»+p 1 r— , +p 1 r 1 +  +p,= ;
and therefore for each of these values of r, e" is a solution of (.A).
Equation (B) is called the auxiliary equation of (4). We obser\
that the coefficients are the same in both, the exponents in (B) c
responding to the order of the derivatives in (4), and y in (ii) beic
replaced by 1. Let the roots of the auxiliary equation (B) be j
■■, r n ; then
(C) e r ' x , ?*, ■■., e'
are solutions of (A). Moreover, if each one of the solutions (C) I
multiplied by an arbitrary constant, the products
(.D) c/i 1 , c/f, ■■■, cjt'
are also found to be solutions.* And the sum of the solutions (7)),
namely,
(£) y = v r '* + V r "* + ' • + V""«
may, by substitution, be shown to be a solution of (A). Solution (jP)
contains n arbitrary constants and is the general solution (if the r
are all different), while (C) are particular solutions.
CASE I. When the auxiliary equation has imaginary roots. Since
imaginary roots occur in pairs, let one pair of such roots be
r l = a + bi, r 3 =a — bL i = V— 1
* Substituting <j? 1 " for y in (A), the lefthand member becomes
W + Piri"" 1 *Pt r i*~* + + p.) c,e*i'.
But this vanishes since r, is » root of (B); hence Cie">" is * solution of <J). Similsrlj
ORDINARY DIFFERENTIAL EQUATIONS 438
The corresponding solution is
y = c e<° +w >*+ c J a ">*
= er x (c 1 e* x +c 2 e**)
= ^{^(cos bx + i sin bx) + c a (cos bx — f sin bx)\*
= e°*\ (c x + c^) cos bx + i(c x — c^) sin bx\,
or, y = « flX (4 cos&r+^sin&r),
where A and 2? are arbitrary constants.
Case II. When the auxiliary equation has multiple roots. Consider
the linear differential equation of the third order
where p v p^ p % are constants. The corresponding auxiliary equation is
(G) **+Pi'*+Pf+P § = : °
If r x is a root of (G), we have shown that e r * x is a solution of (i^).
We will now show that if r x is a double root of (G), then xe r * x is also
a solution of (F). Replacing y in the lefthand member of (F) by
a^**, we get
(IT) vftf+ptf+pft+pj + e^( 3 r* + Zpfr+pJ.
But since r x is a double root of (G),
ri+Pfi+Pfi+P*** '
and 3 r* + 2p 1 r 1 +p t = 0. By § 69, p. 88
* Replacing x by ibz in Example 1, p. 232, gives
a. i^fc & ** 2 iMas« Ma« IM*»
e ^ aal + t6x __ [ _._^_ + ^__ + _ I ___... f or ,
and replacing x by  ibz gives
*. i j. b*x* ^ib*x* ^bh* ibh*
Bat, replacing * by foe in (4), (B), p. 231, we get
(3) cosdajl^ + T^.
(4) sin6«6a ! ^f + ^?....
Hence (1) and (2) become
•**coafce + isinte, e** cos toe  i sin &«.
INTEGRAL CALCULUS
Hence (//) vanishes, and vfr is a solution of (/'). Correspom
to the double root r, we then have the two solutions
More generally, if ^ is a multiple root of the auxiliary equatioi
p. 432, occurring s times, then we may at once write down 8 distl
solutions of the differential equation (4), p. 482, namely.
In case a + In and a — bi are each multiple roots of the auxiliary
equation, occurring s times, it follows that we may write down 2*
distinct solutions of the differential equation, namely,
c^cosbx, c J a , (!°*cos bx, Cjt?e°* eos bx, •■, c,x'~ , e"ccmbx;
cje"sui6j, rfjZtf* sin bx, <£rV"sini:r, ■■■, *Ja: ,_ Vain fee;
Our results may now be summed up in the following rule for solri
differential equations of the type
d"y
1 + ft
dx"
\ + +P„y = o,
where p it p x , ■■, p M are constant*.
First Step. Write down the corresponding auxiliary equation
r"+jy— '+;>/'*+•••+;>„= 0.
Skcokd Step. Solve completely the auxiliary equation.
Third Step. From the roots of the auxiliary equation write down t
corresponding particular solutions of the differential equation as follow*:
Auxiliary Equation Diffekk.ntial Eijuatiom
(a) Each distinct real 1
root r
V gives a particular solution e r < 1 .
(b) Each distinct pair
of imaginary roots a ± bt
(c) A multiple, root o.
ring s tunes
giren
l give*
{two particular solutions c" cosbx,
e"sinbx.
* particular solutions obtained I
multiplying the particular solution
(a) or (b) by 1, x, x*, ••,
Fourth Step. Multiply each of the n* independent solutions by a
arbitrary constant and add the results. This gives the complete t«!uti»
* A check on the lu'rurai'v of the work is tumid in the (act that the first three steps m
give n independent solutions.
ORDINARY DIFFERENTIAL EQUATIONS 486
dry dPy
Illustrative Example 1. Solve r\ — 8— ^ + 4 y = 0.
car car
Solution. Follow above rule.
First step, r 8 — 8r* + 4 = 0, auxiliary equation.
Second step. Solving, the roots are — 1, 2, 2.
Third step, (a) The root — 1 gives the solution er*.
(b) The double root 2 gives the two solutions 4 s *, ze**.
Fourth step. General solution is
y = c x er* + c 2 e* x + CjXe 8 *. Ans.
Illustrative Example 2. Solve ^4^ + 10??12^ + 6y = 0.
car car car dz
Solution. Follow above rule.
First step, r* — 4 r* + 10 r* — 12r + 6 = 0, auxiliary equation.
Second step. Solving, the roots are 1, 1, 1 ± 2 i.
Third step, (b) The pair of imaginary roots 1 ± 2 i gives the two solutions e* cos 2 x,
e*sin2x(a = l, 6 = 2).
(c) The double root 1 gives the two solutions e*, ze*.
Fourth step. General solution is
y = Cje* + CjXe* + c 8 e* cos 2 x + c^sin 2 x,
or, y = (q + V + c 8 cos 2x + c 4 sin 2x) e*. <4n*.
EXAMPLES
Differential equations General solutions
1. —  = 9y. y = c l e Sx + c a e**.
(Pv
2. — ^ + y = 0. y = Cj sin x + c, coex.
3. ^£ + 12y = 7^. y = c^* + c a e«*.
ax* ax
5 0" 4 i = O  »* + <*■»+*•»«.
6. ^+ 2—^ — 8y = 0. y = Cj«*^+c,e*Vi+c,8in2x + c 4 cos2a
(ZX^ uX
_ dr8 (Ps OS  ...•*.
8. ^?i2^? + 27p = 0. p = c 1 e»« + c 2 e»« + c 8 e«vi + c 4 e«^».
9. — 6— + 18u = 0. u = (c 1 sin2t> + c,co82tj)c«».
dvr dv l
10. ^ + 2n J ~ + n*y = 0. y = ( c i + CjXjcosnx + (c 8 + c 4 x)sinnx.
dx 4 ax 3
drs / *Vs iV?
11. — = S. B = C^ + C 2 ^C,8in— + C.COS —
436 INTEGRAL CALCULUS
Differential equations General solutions
d?n ds
12. :Lf_7^ + 6« = 0. s = c.e** + cjb* + (&*'.
dtr dt
14. ^ + 8^10y = 0. y = c 1 tf«*+c 1 e«*.
ax* ax
16. f? + 2^ + 10y = 0. y = «*(<?, coeSx + crinSx).
(ZX (ZX
16. 2^ 8^ + 2^ + 27 = 0. y = Cl c** + e^ccosx + c.rinx)
ax 8 ax 3 ax
Type II. The linear differential equation
(/) &+*dP^ + *dF* + mmm+PJ, = J '
where X is a function ofx alone, or constant, andp v p 2 , • • ,p n are constants*
When X= 0, (J) reduces to (4), Type I, p. 432,
The complete solution of («7) is called the complementary function
of (J).
Let w be the complete solution of ^eT*), Le. the complementary
function of (J), and v any particular solution of (7). Then
d n v , rf— x v , rf—H* , ^
, <f*w d*~ l u d n ~*u ~
Adding, we get
_( w + v ) + ^ i __( w + 1 ,) + ^ 2 __( w + t ;) + ... +A ( w + ^ = ^
showing that u + v is a solution* of (J).
To find a particular solution v is a problem of considerable diffi
culty except in special cases. For the problems given in this book
we may use the following role for solving differential equations of Type n.
First Step. Replace the righthand member of the given equation (J)
by zero and solve by the rule on p. 434. This gives as a solution the
complementary function of (/), namely,
y = u.
9 In works on differential equations it is shown that u + v la the complete sohttton.
ORDINARY DIFFERENTIAL EQUATIONS 437
Second Step. Differentiate successively the given equation (/) and
obtain, either directly or by elimination, a differential equation of a
higher order of Type I.
Third Step. Solving this new equation by the rule on p. 434, we get
its complete solution y = u + v
where the part u is the complementary function of (J) already found
in the first step,* and v is the sum of the additional terms found.
Fourth Step. To find the values of the constants of integration in
the particular solution v, substitute
y = v
and its derivatives in the given equation (/). In the resulting identity
equate the coefficients of like terms, solve for the constants of integration,
substitute their values back in y _. u __ v
giving the complete solution of (/).
This method will now be illustrated by means of examples.
Note. The solution of the auxiliary equation of the new derived differential equa
tion is facilitated by observing that the lefthand member of that equation is exactly
divisible by the lefthand member of the auxiliary equation used in finding the com
plementary function.
Illustrative Example 1. Solve
<*> g+I 2 *— •
Solution. First step. Replacing the righthand member by zero,
Applying the rule on p. 434, we get as the complete solution of (X)
(if) y = c,0* + c 2 e* x = u.
8econd step. Differentiating (K) gives
Multiplying (K) by 2 and adding the result to (N), we get
a differential equation of Type I.
Third step. Solving by the rule on p. 434, we get the complete solution of (0) to be
y = c x eF + c % er* x + c^zer 2x ,
or, from {M ), y = u + cjmt** = u + v.
• From the method of derivation it is obvious that every solution of the original equation
must also be a solution of the derived equation.
488 INTEGRAL CALCULUS
Fourth Hep. We now determine c, so that e^ee~ tx shall be a particular solution i
ot(K) dy 
Substituting y = CjXe 2 ', ~ = c.e 2 *(l 2x), 4 = &e**(4x — 4) in (JT), we get
ax ax 8
.: —Zc % = a, or, c, = — i a.
Hence a particular solution of (K) is
c=— ioxe 2 *,
and the complete solution is
y = u + c = Cje* + c 2 e 2x — i axe 1 *.
Illustrative Example 2. Solve
(P) —£ + n«y = cosox.
Solution. First step. Solving
we get the complementary function
(R) y = c x sin nx + c, cos nx = tc
Second step. Differentiating (P) twice, we get
(S) g + n .g = _ a . C08(MS .
Multiplying (P) by a 2 and adding the result to (&) gives
(T) g + (n * + a ^g + «^ = o.
27iird rtep. The complete solution of (T) is
y = c x sin nx + c s cosnx + c 8 sin ax + c 4 cos ax,
c
or, y = M + c 8 sin ax + c 4 cos ax = u + t?.
Fourth step. Let us now determine c 8 and c 4 so that c 8 sin ox + c 4 cos ax shall be a
particular solution v of (P).
Substituting
y = c 8 sin ax + c 4 cos ax, p = c 8 a cos ax — c 4 a sin ax, —^ = — c 9 a* sin ax — c 4 a* cos ax
in (P), we get , n _ v , _ _ v .
(n 2 c 4 — a 2 c 4 ) cos ax + (n 2 c 8 — a 2 ^) sin ax = cos ax.
Equating the coefficients of like terms in this identity, we get
n 2 c 4 a 2 c 4 = l and n\ — a 2 ^ = 0,
or, c 4 = n a _ fl2 and c 8 = °
Hence a particular solution of (P) is
cosox
and the complete solution is
cosox
yssu + vssCj sin nx + c a cosnx + 5— 3*
ORDINARY DIFFERENTIAL EQUATIONS
439
EXAMPLES
Differential equation*
LgTg + l.,...
y = c x e** + CjC** +
Complete solutions
12x + 7
144
* ^" 2 ^ +2 0"" 2 ^ +v=:a " y = c i 8inx + c » cosx + ( c 8 + c 4 aj )^+ fl 
3. !L?_a*« = t + l.
(tt*
I ft*.*
6. —  + a*s = coeaz.
dx*
10. ft_4^ + 8y = *.
U * *?'£+ •» = •"■
tt'gs? + «, = «.
(XX s ax
14. ^ + 4« = *sin*«.
Type III.
P=(c 1 + C ft 4 + ^«' + <V
y = Cje"* + e 2 er °* + c, sin ax + c 4 cosax —  .
a 4
« = <^ sin ax + e s cosax +
* = (c l + cj)e« +
x sin ax
2a
(al)«
y = e*(c 1 x + c,) + 6«*.
y = Cje* + c % er* — 6x — 2.
1 4
y = c 1 e"+c 2 e««+x + 
y = Cjf* + c 2 e»* +
y = c 1 eF+c s ^*+
e**
n 2  6n + 6
xe"* (2n— 8)e"»
n s 3n + 2 (n»8n + 2)*
4
= (c l i) 8 i„2t + (c s l)
cos2t + 
8
=X,
where X is a function of x alone, or constant.
To solve this type of differential equations we have the following
rule from Chapter XXIX, p. 893 :
Integrate n times successively. Each integration will introduce one
arbitrary constant.
440 INTEGRAL CALCULUS
Illustrative Example 1. Solve — ? = xe*.
dz*
Solution. Integrating the first time, ~ = f xe^dx,
° r ' g = xe*6f» + C r Bj(A), V .U7
Integrating the second time,
JL = xe« 2e* + C,x + CL.
dz
Integrating the third time,
y = fze*dz — fzeFdx +fc i xdz +fcjz
= a**  8 e* + ^ + C,x + C 8 ,
or, y = xe* — 3 c* + CjX a + c,x + c,. «4.tis.
Type IV. £ = Y f
where F is a function of y alone.
The rule for integrating this type is as follows :
First Step. Multiply the lefthand member by the factor
ax
and the rightshand member by the equivalent factor
2dy,
and integrate. The integral of the lefthand member will be *
Second Step. Extract the square root of both members, separate the
variables, and integrate againJ
(JPy
Illustrative Example I. Solve — ^ + cPy = 0.
dx a
cPv
Solution. Here — \ = — a*y, and hence is of Type IV.
ax 2
First step. Multiplying the lefthand member by 2 p dx and the righthand membei
by 2 dy, we get
dx dx*
Integrating, & \ =  oV + O v
\dxl dx dx 2
t Each integration introduces an arbitrary constant.
ORDINARY DIFFERENTIAL EQUATIONS
441
Second step.
dy
dx
= y/C x aW,
taking the positive sign of the radical. Separating the variables, we get
dy
Integrating,
or.
This is the same as
or.
. = = dz.
Vc^aty
 arc sin — ^ = x + C v
a Vcl
ay ~
arc sin —Z— = ax + aC a .
Vc,
ay
Vc[
= sin (ax + aC 2 )
= sin ax cos aC a + cos ax sin aC v
= <^ sin ax + c a cos ax. Ans.
81, p. 2
y = * cos aC 9 • sin ax H * sin aC« cos ax
a * a f
EXAMPLES
Deferential equations
1. _!?? = x 2 — 2 cos*.
2 'aV = 2 
3. ^ = sin 2 *.
d£P
4. — =/sin?it
at a
dx»
9. **=*.
d* 2 y«
10. — =
eft 2
11. ^4 = x 2 sinx.
#y
ax 2
d 2 *
Sofitfion*
y =  + 2 sinx + c^x 2 + c,x + c,.
u = v 2 log V + c^ + c 2 u + c t .
cos 8 ^ . 7cos0
p=
27
+ ^0* + c,0 + c r
/
8 =— ^sinni + cd + C.
n
« = $0i 2 + 0^ + 63.
[mx m +
y =
+ c 1 x" 1 +...+c»_iX + cw
[ro + n
ax = log (y + Vy 2 + c,) + c„ or,
3* = 2a*(«* 2c l )(s* + Cl )T + c 2 .
(Cji + c,) 2 + a = c^y 2 .
/— , Vtfe"* + 11
t V2 n = ^ log 1 c 2 .
VqV* + 1 + 1
y = (^ + CjX + (6 — x 2 ) sinx — 4x cosx.
ds
12. — = —  • Find t, having given that — = and s = a, when t = 0.
<n 2 4 s dt
J.TW. t = v/ — 1 (arc vers *•) — V« — i 2 *.
442 INTEGRAL CALCULUS
MISCELLANEOUS EXAMPLES
Solve the following differential equation* :
<te« dx* dx* dx
2 (P >'_ n dy
Z d*? 9 I + V = ^'
3. ^ + ^ = 8x*y*. ln d» l2x ,
4 8$f+ 2 "
dx x + 1 y 9
u.„,=^ + ).
6. <4y + 8x) + y = 2x. 12. xV**^ + V*)<fo = 0.
*»S + »S».a 18 .g8g + 4 V = 0.
7.^6^ + 6,, = ** 14.^ + ytenx = l.
ax* ax dx
CHAPTER XXXI
INTEGRAPH. APPROXIMATE INTEGRATION. TABLE OF INTEGRALS
233. Mechanical integration. We have seen that the determination
of the area bounded by a curve C whose equation is
y =/(*)
and the evaluation of the definite integral
jf(x)dx
are equivalent problems.
Hitherto we have regarded the relation between the variables x and
y as given by analytical formulas and have applied analytic methods
in obtaining the integrals required. If, however, the relation between
the variables is given, not analytically, but, as frequently is the case
in physical investigations, graphically, i.e. by a curve,* the analytic
method is inapplicable unless the exact or approximate equation of the
curve can be obtained. It is, however, possible to determine the area
bounded by a curve, whether we know its equation or not, by means
of mechanical devices. We shall consider the construction, theory,
and use of two such devices, namely, the Integraph, invented by
AbdankAbakanowicz,* and the Polar Planimeter. Before proceed
ing with the discussion of the Integraph it is necessary to take up
the study of integral curves.
234. Integral curves. If F(x) and f(x) are two functions so
related that
then the curve
(2?) y=F(x)
is called an integral curve of the curve
(C) y =/(*).*
* For instance, the record made by a registering thermometer, a steamengine indicator,
or by certain testing machines.
t See Lea InUgraphes ; la courbe integrate et ses applications, by AbdankAbakanowicz,
Paris, 1889.
t This curve is sometimes called the original curve.
443
444
INTEGRAL CALCULUS
The name integral curve is due to the fact that from (C) it is set
that the same relation between the functions may be expressed i
follows :
(Z>) jf ><» dx = *<*> J(O) =
Let us draw an original curve and a corresponding integral carte in such a way
aa easily to compare their corresponding points.
y = F(s)
v =m
r
yi\
'
IT
,
1/ i.v
li] r
/^
C
1
/ ._
\L.
t' '■
Hi
,V^
\i
1
/r
To find an expression for the shaded portion (OM/T*') of the area undet
curve we substitute in {A), p. 886, giving
area Vlfp = /""/(i) dr.
But from (D) this becomes
area trJTf = j" ''/(/)c£r = [F^)],.^ = *•(*,) = JfP.«
Theorem. For the same abscissa i, the number gluing the length of the ordinate of the
integral curve {JS) is the same as the number thai gives the area between the original a
the axes, and the ordinate correiponding to this abscissa.
The student should also observe that
(a) For the same abscissa i; the number giving the slope of the integral run
the name as the number giving the length of the corresponding ordinate of the original
curve [from (C)]. Hence (C) is sometimes called the curve of slopes of (B). In the
figure wo see that at points 0, R, T, V, where the integral curve is parallel to OX,
the corresponding points O', R', I", V on the original curve have zero ordinates, and
corresponding to the point II' the original curve is discontinuous.
• When x, = OVf. the positive area O'iFRP' is represented by the maximum ordinal*
XR. To the right of If the area is below the axis of .V and therefore negative ; consequently
the ordinates of the Integral curve, which represent the algebraic sum of the areas in "
will decrease in passing from If to T.
The most general integral curve is ot the form
Id which case the difference of the ordinates for z0 and zX\ gives the area under Ike
original curve. In the integral curve drawn i'F(O) = 0. i.e. the general Integral curve is
obtained if this integral curve be displaced the distance C parallel loOY.
DiTEGEAPH
(b) Corresponding to points of
maximum or minimum ordinates
nfiection Q, 8, (7 o
o the original curv
i tfc*
integral
curve we
MT
For example, since
1)
\
r
It follows that
V
a?
\
p,A
(f
is an integral curve ol
the parabola
V
a*
= 8"
s.
//'''
.'c
/ ..'/
X
Since from (F)
area OM l P l
■r
3 B
/
— SB— H
and from (E)
JfjPi
_ J! 1
/
it is seen that — indicates the number of linear units in the ordinate Jf.Pi, and also
the number of units of area in the shaded area OJf,* 1 ,.
Also, since from {E)
and from (F)
x,p i . 1
it Is seen that the w
slope of the tangent
Evidently the origin is a point of inflection of the integral
minimum ordinate on the original curve.
number — indicates the length of ordinate ^f,P, and the
I a point with
235. The integraph. Tlie theory of this instrument is exceedingly
simple and depends on the relation between the given curve and
a corresponding inte
gral curve.
The instrument is con
structed as follows; A rec
tangular carriage C moves
on rollers over the plane
in a direction parallel to
the axis of X of the
V =/<*>■
Two sides of the carriage
are parallel to the axis
of X; the other two, of
course, are perpendicular
to it. Along one of these
perpendicular sides moves
a small carriage C t bear
ing the tracing point T, and along the other a Bmall carriage C 2 bearing a frame F
which can revolve about an axis perpendicular to the surface, and which carries
the sharpedged disk h, to the plane of which it Is perpendicular. A stud S, is fixed
446 INTEGRAL CALCULUS
in the carriage C x so as to be at the same distance from the axis of X as is the trac
ing point T. A second stud 8 t is set in a crossbar of the main carriage C so as to be
on the axis of X. A split ruler R joins these two studs and slides upon them. A
cro8shead H slides upon this ruler and is joined to the frame F by a parallelogram.
The essential part of the instrument consists of the sharpedged disk D, which
moves under pressure over a smooth plane surface (paper). This disk will not slide,
and hence as it rolls must always move along a path the tangent to which at every
point is the trace of the plane of the disk. If now this disk is caused to move, it is
evident from the figure that the construction of the machine insures that the plane of
the disk D shall be parallel to the ruler R. But if a is the distance between the ordi
nates through the studs 8 V S„ and r is the angle made by R (and therefore also plane
of disk) with the axis of X, we have
(A) tanr = £;
and if y* = F(xT)
is the curve traced by the point of contact of the disk, we have
Comparing (A) and (B), ^ = , or,
ax a
(C) V = \ fvdx = 1 ff(x) dz = F(xr).t
That is (dropping the primes), the curve
V = F(x)
is an integral curve of the curve
w v = \m
The factor  evidently fixes merely the scale to which the integral curve is drawn,
and does not affect its form.
A pencil or pen is attached to the carriage C 2 in order to draw the curve y = F(z).
Displacing the disk D before tracing the original curve is equivalent to changing the
constant of integration.
236. Polar planimeter. This is an instrument for measuring areas
mechanically. Before describing the machine we shall take up the
theory on which it is based.
237. Calculation of the area swept over by a moving line of con
stant length.
Consider the bre&ABQ&A'PA swept over by the line AB of constant length /.
Let PQ and Fty be consecutive positions of the line, dd = angle POP's* change in
• Since x  if + df, where d width of machine, and therefore —  — • —  — .
dx* dx dx' dz
t It is assumed that the instrument is so constructed that the abscissas of any two corre
sponding points of the two curves differ only by a constant; hence x is a function of z*.
INTEGRAPH 447
direction of I'Q, and dt = circular arc described about by the middle point It of the
tine. Using differentials, we have
area. of 0<j(?= } 0y a d5,"
area of OPP'= i OP*d6.
.: weaof PQffP'=i 6q'd6 * OP'dS
= i(0«+OP)(0(3OP)<«
= 0RPQd6
= l0Rd8 = UU.
Bumming up all each elements,
{A) vnaAB^APA =Ji
where s = displacement of the center of the line in a direction always perpendicular
to the line.' To find s, let the line be replaced by a rod having a small wheel at the
center R, the rod being the axis of the wheel. Now as the rod is moved horizontally
over the surface (paper), the wheel will, In general, both slide and rotate. Evidently
s = distance it rolls
= circumference of wheel x number of revolutions.
(B) .. s = 2*r»,
where r == radius of wheel, and n = number of revolutions.
Substituting (fl) in (A), we get
(0) area swept over k 2rrln.
So far we have tacitly assumed that the areas
same direction. It Is easy to see, however,
that the results hold true without any such
restriction, provided areas are taken as posi
tive or negative according as they are swept
over towards the side of the line on which da
is taken positive, or the reverse. Choose signs
as indicated in the figure. If the line AB
returns finally to its original position, A and
B having described closed curves, it is evi JS
dent that the formula above will give (taking
Account of signs) the excess of the area in
closed by the path of A over that inclosed by
a»e path of B.
Por positive area = ABQB'A'PA = ABDBA'PA + closed curve BQRDB,
negative area = B'A'CABDB' = ABD^A'PA + closed mneAPA'CA.
finding the difference, we have
net area = dosed curve BQITDB — closed curve A PA'CA.
• Area of circular sector } radius * arc } OQ OQd8~i OQ 3 d8.
t It should be observed that s will not be the length of the path described by the cei
B unless A A and RB' are the arcs of circles with the center at 0.
448
INTEGRAL CALCULUS
Now if the area of one of these closed curves (as APA'CA) is zero, that Is, A keeps
to the same path both going and returning, the area meept over by the tine iotU 07110! the
area of the closed curoe BQ&DB.
A simple and widely iised typo of polar planimeter was Invented by Amsler, of
SchafThausen, in 1864, This consists essentially of two bars 0.4 and A B, freely jointed
at A, OA rotating about a fixed point and AB being the axis of awheel situated at
its center K, and having a tracing point at B.
Now if the tracing point completely describes
the closed curve, A will oscillate to and fro
along an arc of a circle (as CD), describing a
contour of zero area,. Hence the area swept
over by the bar AB exactly equals the area of
the closed curve, and is given by the formula (
{D) area of closed curve = 2 ttfln,
where I = length of bar AB,
r = radius of wheel,
n = number of revolutions indicated on the wheel after the tracing p
has made one complete circuit of the ci
238. Approximate integration. Since the value of a definite integral
is a measure of the area under a curve, it follows that the accurate
measurement of such an area will give the exact value of a definite
integral, and an approximate measurement of this area will give e
approximate value of the integral. We will now explain two apprc
imate rules for measuring areas.
239. Trapezoidal rule. Instead of inscribing rectangles within the
area, as was done in § 204, p. 361, it is evident that we shall get a
much closer approximation to the
area by inscribing trapezoids. Thus
divide the interval from x = a to
x = b into n equal parts and de
note each part by Ax. Then, the
area of a trapezoid being one half
the product of the sura of the
parallel sides multiplied by the
altitude, we get
4 (So + &i) ^ = area °^ fi 1 " 8 ' trapezoid,
i C?i + &) & x — area °^ second trapezoid,
^(y,_i — y„)Ax = area of nth trapezoid.
Adding, we get
4(y + 2^4 2y f H ( 2 #._! + #„) Ax = area of trapezoids.
'
igral
irate
inite
2 an
irox
Y>
/
/
<■

«"
"
/
/
v t
K
K
O
APPROXIMATE INTEGRATION
449
Hence trapezoidal rule is
04) area = (lff i + ff 1 + ff, + ...+y,_ 1 +ff„)Ajr.
It is clear that the greater the number of intervals (Le. the smaller
Ax is) the closer will the sum of the areas of the trapezoids approach
the area under the curve.
Illustjutiyk Examplb 1. Calculate f x'dz by the trapezoidal nils, dividing
x = 1 tc * = 12 into eleven intervals. '
131
Solution. Here —
y = x a . Substituting the abscissas
nates y = 1, 4, 9, • • , 144. Hence, from (.4),
area = < T j 4 + + 16 + 26 + 3*5 + 40 h 64 + 91
= Ai. The area in question Is under the curve
1, 2, 8, • • • , 12 Id this equation, we get the ordi
me third of 1%,
100 + 121 + i. 144). 1 = 677,.
676} . Hence, in this example, the trapezoidal
By Integration f aftlz
rule is in error by less than
240. Simpson's rule (parabolic role). Instead of drawing straight
lines (chords) between the points of a curve and forming trapezoids,
we can get a still closer approximation to the area by connecting the
points with arcs of parabolas and
summing up the areas under these
arcs. A parabola with a vertical
axis may be passed through any
three points on a curve, and a series
of such arcs will fit the curve more
closely than the broken line of
chords. We now divide the inter
val from x=a=OM il to x=b=OM,
into an even, number (= ri) of parts,
each equal to Ax. Through each successive set of three points ij, if,
■^i* J*t ft JJi etc., are drawn arcs of parabolas with vertical axe?
From the figure
area of parabolic strip Jf ( J > J , i , Jf  = area of trapezoid 3^7*7*3^
+ area of parabolic segment JJiJi*.
But the area of the trapezoid3f ^3f a = £ (y„ + y^) 2 Ax
and the area of the parabolic segmenti*J*i >
= two thirds of the circumscribing parallelogram I^P^Tl
Y>
J>
^
;
1
V
<4
%
u
&
,
, .
t, A
r,
j
r. i
450 INTEGRAL CALCULUS
Hence area of first parabolic strip M t) P^P i M i
= (yo+y 1 )^+(2y 1 y y il )Aa:
="3(yo+4y 1 +y 1 ).
Ace
Similarly, second strip = — (y 2 + 4 y 8 + y 4 ),
third strip = — (y 4 + 4 y 6 + y 6 ),
• • • • •
nth strip = g 0—*+ * y„_ ! + &,)•
Adding, we get
as the sum of these areas. Hence Simpeon's rale is (n being even)
Ax
(5) area = — (y, + 4y 1 + 2y 1 + 4% + 2y 4 + ...+y ll ).
8
As in the case of the trapezoidal rule, the greater the number of
parts into which M M n is divided, the closer will the result be to the
area under the curve.
x*dx by Simpson's rule, taking ten intervals,
o
Solution. Here = = 1 = Ax. The area in question is under the curve
n 10
y = x*. Substituting the abscissas x = 0, 1, 2, • • • , 10 in y = x 8 , we get the ordinates
y = 0, 1, 8, 27, • . ., 1000. Hence, from (B),
area = i (0 + 4 + 16 + 108 + 128 + 500 + 432 + 1872 + 1024 + 2016 + 1000) = 2500.
By integration, J x*dx = I y I = 2500, so that in this example Simpson's rule
happens to give an exact result.
1. Calculate the integral in Illustrative Example 1 (above) by the trapezoidal rule,
taking ten intervals. Ans. 2525.
•*dx
2. Calculate I — by both rules when n = 12.
3. Evaluate f x*dx by both rules when n = 10.
Ans. Trap. 1.6182; Simp. 1.6096.
Ana. Trap. 8690; Simp. 8660.
4. Calculate J log 10 xdx by both rules when n = 10.
1 Ans. Trap. 6.0656; Simp. 6.0896.
X* dx
by both rules when n = 6.
1 + x Ans. Trap. 1.0885; Simp. 1.0906.
TABLE OF INTEGRALS 451
& Calculate f sin xdx by both roles for tendegree intervals.
x*<to by both rules for n = 12.
8. Find the error in the evaluation of I x*dx by Simpson's rule when n = 10.
9. Evaluate f e*dx by Simpson's rule when n = 10.
241. Integrals for reference. Following is a table of integrals for
reference. In going over the subject of Integral Calculus for the first
time, the student is advised to use this table sparingly, if at all. As
soon as the derivation of these integrals is thoroughly understood, the
table may be properly used for saving time and labor in the solution
of practical problems.
SOME ELEMENTARY FORMS
1. C(du ± dv ± dw ± • • • ) = fdu ± jdv ± jdw ± • • • .
2. fadv = aCdv. 4. fx*dx =  —  + C,n^1.
3. fdf(x) = ff(x) dx =/(«)+ C. 6. J — = log x + C.
Forms containing Integral Powers of a + bx
6. f^ = ilog(a + ta) + C.
J a + bx b
7. f{a + bx)*dx = (< * + te) " * * + C, n *  1.
J 0(71 + 1)
8. f F(x, a + bx)dx. Try one of the substitutions, z = a + to, zz = a + to.
9 r>^^ = I[a + axalog(a + te)] + C.
J a + ox o*
/r*dx 1
^= = ±[k(a + te)«2a(a + te) + a 2 log(a + bx)] + C.
a + ox tr
1. I = log V C.
J x (a + bx) a x
a C &* 1 b . a + bx . „
2.  — = H— ;log h C.
J x 2 (a + bx) ax a 2 x
J(a + te) 2 o^L a + tej
•/ (a + to) 2 6" L a + ox J
/■ (to 1 1 « a + to «
5. f — —  =  i^og +C
•/ x (a + to) 2 a (a + to) a 2 x
6 r «te = 1[ 1 o 1 c
" J (a + to) 8 ^L a + to 2(a + to) 2 J
452 INTEGRAL CALCULUS
Forms containing a* + x*, a* — **, a + &e", a + &r»
17. r_^_ = Itani5 + Cj r_^_ = tan>x+C.
J a* + as» a a Jl + x"
, a /• dx 1 . a + x , ~ /• dx 1 . as — a , ~
•/a 8 — x a 2a a — x J z % — ar 2 a x + a
19. f_^_ = J_ tan 1*^ +C.
J a 9 — oV 2 oo a — to
21. fx»(a + ox»y>dx
6(np + m + l) 6(np + m+l)J
22. f *»(a + to.)*d* = g " +1 (a + te»)> + _^ f^ + <„.)„,&.
J np + m + 1 tip + m + 1 •/
23. f— «
J x»(a +
dx
(a + ox»)i>
(m — n + up — 1)6 /* dx
____ l (m — n + np — 1)0 /* ax
"" (m — l)ax J » 1 (a + 6x»)i >  1 (m — l)a J x"»»(a + ox*)**
24. f fe
J x"»(a + ox*)J>
_ 1 mn + np1 r dx
~"an(p — l)x* 1 (a + 6x«)i» 1 an(p — 1) J x"»(a + ax*)* 1 *
g /» (a + ox»)J>dx _ (a + ox*)** 1 6(m — n — np — 1) /» (a + ox»)»dx
'J £* a(m — ljx* 1 a(m — 1) J x*»
/» (a + ox»)J>dx _ (a + b&)p anp r (a + &x*)i > ~ 1 dx
"J x* (np — m + ljx* 1 np — m + lJ z m
^ /» x"dx x*** 1 a(ro — n + 1) /» x»»dx
' J (a + ox*)»~~o(m — np + l)(a + ox*)* 1 6(m — np + l)J (a + ox")*
M /» x»dx __ x 1 "* 1 ro + n — np + 1 r x»*dx
' J (a + ox*)* ~~ an(p — l)(a + ax*)* 1 an(p — 1) J (a + ox*)**'
^ J (a 2 + x 2 )» = 2(n  l)a 2 L(a 8 + x 2 )— i + < 2w " 8 )J ( a a + x 2)» lj '
30. f * = * [ ? + (2n8)f * 1.
J (a + ox 2 )" 2(n  l)aL(a + ox 2 )"* J (« + ox 2 )"*J
31. f —   =  f — , where z = x 2 .
J(a + ox 2 )" 2J(a + te)»
32 f ****  Z± i * C **
" J (a + ox 2 )* 26(n l)(a + ox 2 )** 26(n 1)J (a + te 2 )***
33. f— % = ±l 0g ^+C.
J x (a + ox") an a + ox*
J x 2 (a + ox 2 )» "" a J x 2 (a + ox 2 )** ~~ a J (a + ox 2 )*'
TABLE OF INTEGRALS 453
x*dz x a r dx
gg r x*az _x ar a
' J a + 6x f 6 6 J a + ox*
w> /^T*^ = 2^ log rfl*' + c '
te*
dx
38. f * ± b .f*
J x 2 (a + te a ) ax ava +
39 f to = g . _L f d
'J (a + ftx 2 ) 2 2a(a + &c 2 ) 2aJ a + te 2
Forms containing Va + te
•/ loo*
«. fs*V7Ttad» = 2 < 8g ' 12 ^ + 16 ^ v < ;r ^+o,
J 1066*
42. f / fg_= 2(2< — te >V7Tte + C
43. f^'^^+^vs+s+c.
aa C &* 1 i Va + ox — Va .  . 
44. I — = ~7:log , F +C,fora>0.
^ x V a + ox v a v a + 6x + V a
46. f_i = = 2 tani A H^+C,fora<0.
J xVa + bx V— a * — a
46 f dfc _ — V« + te &_ /* dx
'x 2 VaTte~~ « 2a 'xVa + te"
_ rVa + bxdx ft / — — r . r dx
47. I = 2va + ax + al — ,
J * J xVa + bx
Forms containing Vx* + a*
48. f (x 2 + a)ldx = 5 Vx 2 + a + ^log(x + Vx 2 + a 2 ) + (7.
49. f (x 2 + a 2 )dx = ?(2x + 6a 2 ) Vx 2 + a 8 + ^log(x + Vx 2 + a 2 ) + O.
•/ 8 8
50
d i . *xL x(x 2 + a 2 ) 2 , na 2 /• , **.
. J (x 2 + a 2 ) 2 dx = * — —  ' + __ / (x 2 + a 2 ) 2 dx.
•* n + l n + l«/
» » + »"
61. fx(x 2 + a 2 ) 2 dx = ( *' + g8) a + C.
J N ' n + 2
62. f x 2 (x 2 + drfdx = ? (2x 2 + a 2 ) Vx 2 + a 2  ^ log(x + Vx 2 + a 2 ) + C.
•J 8 8
63.
dx
64
. f  = log(x + Vx 2 +a 2 ) + C.
J (x 2 + a 2 )*
^ (x 2 +a 2 )* a*Vx 2 + a*
454 INTEGRAL CALCULUS
J (x 2 + a 2 )*
J (x 2 +a 2 )i 2 2
57. f_i^=^ = + log(x + Vx*T^ 2 ) + C.
J (x 2 + a*)l Vx 2 +a 2
68. f fe = *log £= + 67.
J x(x 2 + a 2 )* * o + vx J + c J
69. f— ^— =^I^+a.
J x 2 (x 2 + a 2 )* "
/dx Vx 2 +a 2 1 , a + Vx^Hha* ^
x»(x 2 + a 2 )i 2oaxf 2a8 *
„ f(x 2 + a 2 )*dx / , , , , a + Va 2 + x 2 „
61. I  —  — = va s + x* — a log h C.
J x x
~* /*(x 2 +a 2 )idx Vx 2 + a 2 , , / , /=: — s\ , ~
62. I ^ — ^—^ = — + log (x + Vx 2 + a 2 ) + C.
J x 2 x
Forms containing Vx 2 — a 2
63. f (x 2  a 2 )idx = ? Vx 2  a 2  ^log(x + Vx 2  a 2 ) + C
64. f (x 2  a 2 )*dx = ^(2x 2  5a 2 )Vx 2 a 2 + ?L l g(x + Vx 2  a 2 ) + C.
•/ 8 8
nH
x(x 2  a*)*dx = S H + C.
v ' n + 2
67. fx 2 (x 2  a^ldx = ?(2x 2  a 2 ) Vx 2  a 2  ^log(x + Vx 2  a 2 ) + C.
•/ 8 8
— = log(x + Vx 2  a 2 ) + C.
(x 2  a 2 )*
/dx x ^.
(x 2  a 2 )* a 2 Vx 2  a 2
70. f_2*E — = Vx 2 a 2 +C.
J (x 2  a 2 )*
71. f **** = Va^^ + ^iog( g +Vg3tf) + c.
J (x 2 a 2 )4 2 2
72. f_^==^ = + log(x + Vx^ 2 ) + C.
J (x 2 a 2 )* Vx 2 a 2
73. f *! = l6eci?+C; f— ^= = «*!« + 0.
J x(x 2 a 2 )i a ° J xVx 2 l
«./
TABLE OF INTEGRALS 455
dx
Vx 2 
a 2
4 (7
a*z
■f v/.
Vx 2 
fl9  X
2a 2 x s
i 2a»
Vx 2 
a 2 — ace
vx 2 "
«\i.
x 2 (x 2 a«)i
„ /• dx Vx 2 — a 2 ,1 .x , ~
J x»(x 2 a»)i 2a '*' 2a * a
™ r(x a a 2 )ldx Vx 2 a 2 , , / , /= $\ , n
77. /   1 = V log (x + vx 2 — a 2 ) + C.
•/ x 2 x
Forms containing Va 2 — x 2
78. f(a 2  x 2 )idx =  Va 2  x 2 + — sin*  + C.
•7 2 2 a
79. f(a 2 x 2 )idx = (5a 2 2x 2 )Va 2 x 2 + ^8ini5+C.
•/ 8 8 a
80. f(a 2  z*)*dz = »(^«V + *L />  ^" \fc.
J v ' n + 1 n + lJ v '
•±3
x (a 2  x 2 ) 2 dx =  £ ±J + c.
7 n + 2
82. fx 2 (a 2  x 2 )idx = (2x 2  a 2 ) Va 2 x 2 + rini? + C.
J 8 8 a
83. f fc =dni?; f— ^ = = rinix.
^ (a s a;*)l a ^ Vlx*
84, r_^_ = « +c .
(a 2  x 2 )i a 2 Va 2  x 2
xdx
(a 2  x 2 )*
86. f — 5 a ^__ = Va 2 x 2 +C.
J (a 2 
Q . /» x 2 ^ x /r 5 , a 2 ,x , „
86. I = va 2 — x 2  — sin 1  + C
J (a 2 x 2 )* 2 2 a
^ /• x 2 dx x . ,x , ^
87. / = ffln*+<7.
J (a 2  x 2 )* Va 2  x 2 a
_ /• x"«dx x 1 " 1 /= ; . (m — l)a 2 /• x* 2 
88. / = vo j z 2 + i ' — I dx.
J (a 2 x 2 )* m m J (a 2 x 2 )i
««. /* dx 1 , x . .
89. / r = log . +C.
J x(a 2 x 2 )t a o + Va«a!«
90. f ** V^s* 
^x 2 (a 2 x 2 )i a2a;
01 T ** Va 2  x 2 , 1 . x . n
91. I = — —  log + O.
J x 2 (a 2 x 2 )i 2a ^ 2a ^ a + y/a*x*
456 INTEGRAL CALCULUS
^ f(a a * a )K A! 5 i a + Va 2 x 2 n
92. / ^ '—dx = va 2 — x 2 — alog + C.
J x x
M f(a*s 2 )i. Va 2  x 2 ,x , _
83. I — ^— <fsB = sin 1  + C7.
J x 2 x a
Forms containing V2ax — x 2 , V2ax + X s
a
94. f V2ax  x 2 dx = ^^V2ax  x 2 + i. vers* + C.
J 2 2 a
/dx . x /* dx » ^»
 = vers *  ; I —== = vers * x + C
V2 ax  x 2 a J V2xx 2
AA T /£ 5^ x« 1 (2ax — x*)t , (2m + l)or , /r ; _
96. I x* v 2 ax — x 2 dx = * — +  z — — I x m ~ l V2ax — x*dx.
J m+2 m+2 J
/» dx __ _ V2 ax — x 2 m — 1 /• dx
" •/ ~m ^/o /»* _ /»2 "~ (2m — l^ax" 1 (2m— Da*/ X" 1  1
98
x»V2axx 2 (2ml)ax* (2ml)a*/ x" ,  1 V2axx 2
x«dx x"   1 V2ax — x 2 (2m — l)a /• x^idx
/x«ax __ x m ~ l wax — z* (zw i)a r
V2 ax  x 2 . m m J V2axx 2
/» V2axx 2 " _ (2axx 2 )t m8 r V2axx 2 "
'J x m "" (2m — 3)ax"« (2m3)a t ' X 1 "*
^ /• /r 5 . 3 a 2 + ax — 2 x 2 /r 5 a 8 t x
00. Ixv2ax — x 2 dx = v2axrx 2 + — vers 1 .
J 6 2 a
/vi /* <& V2 ax — x 2 , „
01. / — = + C.
J xV2axx 2 **
02. f r _^_ = V2axx 2 + avers 1 C.
J V2axx 2 °
„ /• x 2 dx x + 3 a rz 5 , 3 9 , x . „
03. I , =  V 2 ax — x 2 +  a 2 vers 1  + C.
J V2axx 2 2 2 a
04. f Z " dx = V2ax  x 2 + avers 1  + C.
J x a
M /• V2 ax — x 2 2 V2 ax  x 2 ,x , _
05. I  dx = vers 1  + C.
J x 2 x a
„ /• V2axx 2 , (2 ax  x 2 )i , „
OT.f *» = * a +C.
^ (2 ax — x 2 )« a 2 V2ax — x 2
08. r ^ = _; — 4.ff.
J (2ax — x 2 )§ a V2 ax — x 2
09. f ^(x, V2 ax  x 2 ) dx = f ^(z + a, Va 2  z 2 )(fe, where z = x — a.
____ = log(x + a + V2 ax + x 2 ) + C.
V2ax + x 2
11. f ^(s, V2ax + x 2 )dx =Cf(z  a, Vz 2 a 2 )ds, where z = x + a.
TABLE OF INTEGRALS 457
Forms containing a + bx ±cx*
112. f ^ = 2 tani 2cz + * _ .  . when 6» < 4 ac.
J a + te + ex 2 V4oc6» V4ac&*
110 /• dx l . 2cx + 6 — V& 2 — 4ac „ . .. ,
113. / = log  . + C, when 6* > 4 ac.
Ja + te + cx 2 V6«4ac 2cx + 6 + V& 2  4ac
H4 ./—£—, = _L=log^S±!^_ 6 + 0.
./a + 6xcx 2 V6«+4ac V&* + 4ac 2cx + 6
— == ^ = __ = JL log (2 ex + 6 + 2 Vc VaTteT^) + 0.
Va + te + ex 2 Vc
116. fVa + bz+cz*dz
= 2C * * & Va + frc + ex 2  y ~ ^ aC log(2cx + 6 + 2 VcVa+ te + ex 2 ) + C.
4c 8c*
... r dx l.i 2cx — b , _
117. I . _  = 8ini = +C.
•'Va + 6xcx 2 Vc V6 2 +4ac
118. rVa + tecx 2 dx = :^^Va + te. gC 2 + 6a+4qC 6ini ^ZL+C
J 4c 8ci V& 2 +4ac
,,. r xdx Va + 6x + ex 2 6 . /rt . « / / t s\ «
119. /— ===== — Z log(2cx + b + 2 Vc Va + te + cx 2 )+ C.
J Va + te+cx 2 c 2ci
m / xdx ^Vg + tec^J^., 2cx6 ^ r
J Va + te  ex 2 c 2c* V6 2 +4ac
Other Algebraic Forms
21  fVT^ <** = V(a + x) (6 + x) + (a  6) log ( Va + x + Vft + x) + C.
•/ * o + x
22  fV/r 115 ** = V(ax)(6 + x) + (a + 6) sin ufc+i + C.
•/ \6 + x \a + 6
23  f\IP^ dx =  V(a + x) (6  x)  (a + 6) sin ^^^ + <?•
•/ \ o — x \a<f o
24. r^/lil^ax= Vlx 2 + slnix + C.
26. f / * = 2 sin u£Ei + C.
J V(xa)03x) \0a
Exponential and Trigonometric Forms
26. fa*dx = ^— + C. 129. f sin xdx = — cosx + C.
J log a •/
27. fc^dx = e* + C. 130. f cosxdx = sinx + C.
e ax dz = — + C. 131. ftanxdx = logsecx= — log coax + C.
32. f cot xdx = log sin x + C.
33. (sec xdx = f = log (sec x + tan x) = log tan (^ + ?)+ C.
•/ •/ cosx \4 2/
458 INTEGRAL CALCULUS
//* if x x
coeec xdz= I — = log(cosecx — cotx) = log tan + C
J sinx 2
186. /«*■**, = tan* + C. 188. /coeec*cot*<fe = coeec,+ C.
136. f coeec*xdx = cot* + C. 139  J«n*xdx =   jsin2x + C.
co8*xdx =  + jsin2x + 0.
.„ r . . . sin»ixcosx n — 1 f . . « 
141. I sin"xdx = 1 I 8in»«sBdx.
J n n J
ijo C ~ a coep^xsfnx , n — 1 /• — , .
142. lcos»xdx = 1 I cos»*xdx.
•/ n n J
148. f*L= 1 cog£_ + n2/ <te
•/ sin"* n — 1 an" 1 * n— 1«/ sin» a x
I*. r_*L=_i — *i£_+«zJr
•/ cos»x n — lcos* 1 * n — !•/
cos" a x
cos m  1 xsin» + 1 x . m— 1
— 145. I cos m xsin"xd£ = 1 I coeP^xsinaxdx.
•/ m + n m + n«/
i^/» C _ . .  j sin» 1 xcoflP + 1 x , n — 1 /• _ . . ,
146.  coflPxsin^xdxss 1 / cos" , X6in , H a xax.
•/ m + n m + nJ
147. f £ = J 1 + m + " 2 f *
J sm»»xcos»x n — 1 sin^^xcos* 1 * n— 1 J sin m xcos" a x
148. r  =   + ro+n  2 r *
•/ sin m xcos*x m— 1 sin m  1 xcofi^ 1 x m— 1 J sin" ,  a xco6*x
t  a f coflPxdx co8P*+ 1 x m — n + 2 fco&*xdz
rco#*xaz __ co8P*+ 1 x m — n + 2/*
' •/ mn»x (n— l)sin ,,  1 x n — 1 J
sin» 2 x
150
/cofiFxdx^ co«?" 1 x m — 1 r co#**xdx
sin»x ~" (m — n) sin* 1 x m—nJ mn n z
co6" + *x
/coe^"'" 1 x
sinxcos"xdx = h C
n + 1
/gin* + 1 x
8in*xco8xcZx = \ C.
n + 1
163. ftanxctx = ten,, " lg  ft&n*^xdx + C.
/cot?* — ^ X (*
cott»xdx = J cott» a xdx + C.
... /* ,  sin(m + n)x , sin(m — n)x , ^
165.  sin mx an nxdx =  —  — '— H  — + C.
J 2(ro + n) 2(ron)
1M /• , sin(m + n)x , sin(m — n)x , _
166. f cos mx cos nxdx = —  —  — — H 7 — + C.
./ 2(m + n) 2(mn)
, tw /• . . cos(m + n)x cos(m— n)x . „
167. I em mx cos nxdx =  — —  — + C
./ 2(m + n) 2(m — w)
168. f — = , 2 tan i( A /iE tan?) + 0, when a > b.
Ja + fccosx Va a — 6* \\a + 6 2/
TABLE OF INTEGRALS 459
V6— a tan  + V6 + a
/dx 1 2
= = log + C, when a < b.
a + 6co8X VV=* vf^ tan? V&Ta 
a tan + b
/dx 2 2
— — = m tani — m + C, when a > 6.
a + 6sinx Va 2 ^ Va 2 — 6*
atan^ + feV^a 2
. f — — = , log — — — + C, when a < 6.
Ja + omnx V^=^ otan ? + 6 + V6sT^i
161
162. f  = ltani/^^Uc.
./ c^co^x + ^sh^x o6 \ a /
„„„ /• . , e * (a sin nx — n cos nx) , _ /• . . e*(8inx— cosx) . „
163. I e°*sin nxax = —   + C; I c flf sinxdx = — ^ ' + G
J a* + n* J 2
./>.. /*— j ^(nsinnx + acosnx) , „ /• . e*(8inx + co8x) . ^
164.  e^cosnxdx = —  — — ' + G : / ePcosxdz = — i  ^ + C.
J a 2 + n 2 'J 2
166. fxe"*dx = ^ (ax  1) + G.
J a 2
166. fx*e°*a"x = fx^ie^dx.
J a a J
167. fa"«Wx = ^V^ ^— fa***z*idz.
J m log a m log a J
1AA r a*dz __ a* log a r cpdx
J x* ~"~ (m — ljx* 1 mlJx* 1 '
„_ /• , e ox cos» 1 x(aco8X + nsinx) , n(n — 1) r tt ,
169. le«*cos*xax = 5l _l +— s ^ I 6 oar cos» 2 xdx.
J a 2 + n 2 a 2 + n 2 J
x" cos axdx = — —(ax sin ax + mcosax) *— ^ —  J X" 1 ~ 2 cos axdx.
Logarithmic Forms
171. flogxax = xlogx — x + G.
/dx 1
= log(logx) + logx + — log^ + . . • .
logx £ A
173 /^ =log(logx)+c 
...v* T i ^ e^logx 1 / , e K ,
175. J e^logxax = 5 J — dx.
x* , log"xax = log"x  x* , log" 1 «ix.
° m + 1 B m + W
iyy rawfe^^ x m + 1 m + 1 /*_xjjdx_
'J log"x~~ (n — l)log* 1 x n — lJlog" 1 x
INDEX
(The numbers refer to pages.)
Absolute convergence, 220
Acceleration, 02
Approximate formulas, 237
Archimedes, spiral of, 274
Area, moment of, 408 ; center of, 406
Areas of plane curves, polar coordinates,
870, 406 ; rectangular coordinates, 866,
402
Areas of surfaces, 881, 413
Asymptotes, 249
Auxiliary equation, 434
Bending, 118
Binomial differentials, 840
Binomial Theorem, 1, 99
Cardioid, 273
Catenary, 272
Cauchy's ratio test, 218
Center, of area, 408 ; of gravity, 409
Change of variables, 148
Circle of curvature, 178
Cissoid, 271
Computation by series of e, 283 ; of loga
rithms, 235 ; of *, 235
Concave up, 126 ; down, 126
Conchoid of Nicomedes, 272
Conditional convergence, 221
Cone, 2
Conjugate points, 260
Constant, 8; absolute, 8; arbitrary, 8;
numerical, 8 ; of integration, 807
Continuity of functions, 14
Convergency, 214
Coordinates of center of curvature, 178
Cosine curve, 237
Critical values, 110
Cubical parabola, 271
Curvature, center of, 178 ; circle of, 178 ;
definition, 155 ; radius of, 155
Curve tracing, 128
Curves in space, 262
Cusp, 259
Cycloid, 82, 272
Cylinder, 2
Decreasing function, 106
Definite integration, 814
Degree of differential equation, 426
Derivative, definition, 27
Derivative of arc, 134
Differential coefficient, 27
Differential equations, 421
Differential of an area, 314
Differentials, 141
Differentiation, 29; of constant, 36; of
exponentials, 48 ; of function of a func
tion, 44 ; of implicit function, 69 ; of
inverse circular functions, 61 ; of in
verse function, 45; of logarithm, 46,
60 ; of power, 89 ; of product, 88 ; of
quotient, 40 ; of sum, 87 ; of trigono
metrical functions, 64
Double point, 266
Envelopes, 206
Equiangular spiral, 274
Evolute of a curve, 182
Expansion of functions, 227
Exponential curve, 275
Family of curves, 206
Fluid pressure, 388
Fluxions, 25
Folium of Descartes, 278
Formulas for reference, 1
Function, continuity of, 14; definition,
7 ; graph of, 16 ; implicit, 69 ; increas
ing, decreasing, 106 ; inverse, 46 ; many
valued, 17 ; of a function, 44
461
462
INDEX
Gravity, center of, 400
Order of differentia! equations, 426
Greek alphabet, 8
Ordinary point, 265
Osculation, 260
Helix, 263
OHgood, 215
Homogeneous differential equation
486
Hyperbolic spiral, 274
Hypocycloid, 2T3
Parabola, 277 ; cubic, 271 ; semicubical,
271 ; spiral, 276
Increasing functions, 106
Parabolic rule, 4411
Increments, 25
Parameter, 6, 206
Indeterminate forma, 170
Pa nin if trie equations, 79
Infinitesimal, 13, 132
Partial derivative*!, 191 ; Integration, 893
Infinity, 13
Pierpont, 245
Inflection, 125
Planimeter, polar, 446
Integral curves, 446 ; definition, 31
; In
Points, conjugate, 200 j end, 260 ; isolated,
definite, 281
260 ; of inflection, 125 ; salient, 260 ;
Integraph, 443
singular, 256 ; turning, 108
Integration, by rational fractions, 325 ;
Polar planimeter, 446
by parts, 347 ; by rationalization,
335;
Pressure, fluid, 388
by transformation, 343 ; definition
270;
Probability curve, 275
mechanical, 443
Interpolation, 237
Quadratic equation, 1
Involute, 187
Radius of curvature, 160
Laplace, 25
Rates, 141
Leibnitz, 32 ; formula, 00
Rational fractions, 825
Lemniscate, 272
Reciprocal spiral, 274
Length of curves, 375
Reduction formulas, 360
Limacon, 274
Hollo's Theorem, 164
Limit, Interchange of, 320 ; of a variable,
11 ; of integration, 816 ; theory of
11
Secant curve 275
Linear differential equation, 427
Lituus, 274
Logarithmic, curve, 275 ; spiral, 274
Logarithms, Hriggs's, 237 ; common,
Napierian, 237
Seiuicubical parabola, 271
237;
Series, alternating, 220 ; arithmetical,
1; convergent, 214; divergent, 214;
geometrical, 1 ; infinite, 213 ; noncon
vergent, 214; oscillating, 215; power.
Maclaurin's Theorem and Series, 230
223
Maxima and minima, 103
Signs of trigonometric functions, 3
Mean value, extended theorem of,
166;
Simpson's rule, 449
theorem of, 165
Sine curve, 273
Mechanical integration, 443
Singular points, 265
Moment of area, 408
Slope of curve, 73
Moment of inertia, 410
Solution of differential equations, 422
Multiple roots, 60
Sphere, 2
Stirling, 230
Natural logarithms, 4
Stroplioid, 274
Newton, 26
Subnormal, 77
Node, 258
Subtangent, 77
Normal, 7fl
Successive differentiation, 97
Normal line, 266
Successive integration, 898
Normal plane, 262
Surface, area of, 381, 413
INDEX
468
Tangent, to plane curves, 76 ; to space
curves, 262
Tangent curve, 275
Tangent line to surface, 264
Tangent plane t 264
Taylor's Series, Theorem, 228
Test, comparison, 217
Total differentiation, 194
Trajectory, orthogonal, 808
Trapezoidal rule, 448
Triple integration, 417
Variable, definition, 6; dependent, 7;
independent, 7
Velocity, 90
Volumes of solids, 877, 417
Witch of Agnesi, 271
Work, 889
•7
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MAR  1 1967
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