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JUL 1 6 1953
ANALYTIC GEOMETRY AND
CALCULUS
BY
FREDERICK S. WpODS
AND
FREDERICK H. BAILEY
PROFESSORS OF MATHEMATICS IN THE
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
GINN AND COMPANY
BOSTON • NEW YORK • CHICAGO • LONDON
ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO
COPYRIGHT, 1917, BY FREDERICK 8. WOODS
AND FREDERICK H. BAILEY
ALL BIGHTS RESERVED
217.10
Ht fltltw«ttm >rt«
GINN AND COMPANY • PRO-
PRIETORS • BOSTON • U.SA.
\
9/7
PEEFACE
The present work is a revision and abridgment of the authors'
" Course in Mathematics for Students of Engineering and. Applied
Science." The condensation of a two-volume work into a single
volume has been made possible partly by the omission of some
topics, but more especially by a rearrangement of subject matter
and new methods of treatment.
Among the subjects omitted are determinants, much of the
general theory of equations, the general equation of 'the conic
sections, polars and diameters related to conies, center of curv-
ature, evolutes, certain special methods of integration, complex
numbers, and some types of differential equations. All these
subjects, while interesting and important, can well be postponed
to a later course, especially as their inclusion in the present
course would mean the crowding out, or less thorough han-
dling, of subjects which are more immediately important.
The rearrangement of material is seen especially in the bring-
ing together into the first part of the book of all methods for
the graphical representation of functions of one variable, both
algebraic and transcendental. This has the effect of devoting
the first part of the book to. analytic geometry of two dimen-
sions, the analytic geometry of three dimensions being treated
later when it is required for the study of functions of two
variables. The transition to the calculus is made early through
the discussion of slope and area (Chapter IX), the student being
thus introduced in the first year of his course to the concepts
of a derivative and a definite integral as the limit of a sum.
The new methods of handling the subject matter will be
recognized by the teacher in places too numerous to specify
here. The articles on empirical equations, the remainder in
Taylor's series, and approximate integration are new.
• • •
in
iv PEEFACE
It is believed that this book can be completely studied by
an average college class in a two years' course of 180 exer-
cises. Teachers who wish a slower pace, however, may omit
the last chapter on differential equations, or substitute it for
some of the work on multiple integrals.
The book contains 2000 problems for the student, many of
which are new. It is, of course, not expected that any student
will solve all of them, but the supply is ample enough to allow the
assignment of different problems for home work and classroom
exercises, and to allow different assignments in successive years.
F. S. WOODS
F. H. BAILEY
CONTENTS
Chapter I — CARTESIAN COORDINATES
Article Page
1. Direction of a straight line 1
2. Projection 2
3. Number scale 3
4. Coordinate axes 4
5. Distance 5
6. Slope 6
7. Point of division 8
8. Variable and function 9
9. Functional notation 12
Problems 13
Chapter II — GRAPHS OF ALGEBRAIC FUNCTIONS
10. Equation and graph 20
11. Intercepts 21
12. Symmetry and impossible values 23
13. Infinite values 25
14. Intersection of graphs 29
15. Real roots of an equation 34
Problems 36
Chapter III— CHANGE OF COORDINATE AXES
16. Introduction 39
17-18. Change of origin 39
19. Change of direction of axes 42
20. Oblique coordinates 43
21. Change from rectangular to oblique axes 44
22. Degree of the transformed equation 45
Problems 45
v
vi CONTENTS
Chapter IV — GRAPHS OF TRANSCENDENTAL FUNCTIONS
Article Page
23. Definition 49
24. Trigonometric functions 49
25. Inverse trigonometric functions 52
26. Exponential and logarithmic functions 52
27. The number e 53
Problems 55
Chapter V— THE STRAIGHT LINE
28. The point-slope equation 57
29. The slope-intercept equation 57
30. The two-point equation 58
31. The general equation of the first degree 58
32. Angles 59
33. Distance of a point from a straight line 63
Problems 64
Chapter VI — CERTAIN CURVES
34. Locus problems 69
35-37. The circle 70
38-40. The ellipse 74
41-43. The hyperbola . 77
44-45. The parabola 80
46. The conic 83
47. The witch 84
48. The cissoid 85
49. The strophoid 86
50. Use of the equation of a curve 88
51. Empirical equations 89
Problems 92
Chapter VII — PARAMETRIC REPRESENTATION
52. Definition 106
53. The circle 108
54. The ellipse 108
55. The cycloid . . % 109
56. The trochoid 110
CONTENTS vii
Article Page
57. The epicycloid Ill
58. The hypocycloid 112
59. The involute of the circle 112
Problems 113
Chapter VIII — POLAR COORDINATES
60. Coordinate system 118
61. The spirals 120
62. The straight line 121
63. The circle 122
64. The limacon 123
65. Relation between rectangular and polar coordinates . . . 124
66. The conic, the focus being the pole 125
67. Examples 126
Problems 127
Chapter IX — SLOPES AND AREAS
68. Limits 130
69. Theorems on limits 132
70. Slope of a curve 134
71. Increment 135
72. Continuity 136
73. Derivative 136
74. Differentiation of a polynomial 137
75: Sign of the derivative 138
76: Tangent line. 140
77. The differential 141
78. Area under a curve 143
79: Differential of area 146
80. The integral of a polynomial * ' 146
81. The definite integral 147
Problems 150
Chapter X — DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
r
82. Theorems, on derivatives 154
83. Derivative of w» 159
84. Formulas . 161
85. Higher derivatives 162
86. Differentiation of implicit functions 162
viii CONTENTS
Article Pag*
87. Tangents and normals 164
88. Sign of the second derivative 166
89. Maxima and minima 168
90. Limit of ratio of arc to chord 172
91. The differentials dx, dy, ds 174
92. Rate of change 175
93. Rectilinear motion 177
94. Motion in a curve 180
Problems 181
Chapter XI — DIFFERENTIATION OF TRANSCENDENTAL
FUNCTIONS
95. Limit of 55^ 192
h
96. Differentiation of trigonometric functions 193
97. Differentiation of inverse trigonometric functions . . . 196
i
98. Limit of (1 + A)* 199
99-100. Differentiation of exponential and logarithmic functions . 199
101. Applications 202
102. The derivatives in parametric representation 204
103. Direction of a curve in polar coordinates 205
104. Derivatives with respect to the arc in polar coordinates . 206
105. Curvature 207
106. Radius of curvature 208
107. Radius of curvature in parametric representation . . . 210
108. Radius of curvature in polar coordinates . . . . . . 211
Problems 212
Chapter XII — INTEGRATION
109. Introduction 222
110. Fundamental formulas 222
111. Integral of un 223
112. Integrals of trigonometric functions 226
113-114. Integrals leading to inverse trigonometric functions . . 230
115. Integrals of exponential functions 236
116. Collected formulas 236
117-118. Integration by substitution 238
119-120. Integration by parts 243
121. Integration by partial fractions 247
122. Reduction formulas 252
Problems 253
CONTENTS ix
Chapter XIII — APPLICATIONS OF INTEGRATION
Article Page
123-124. Element of a definite integral 260
125. Area of a plane curve in Cartesian coordinates .... 262
126. Infinite limits or integrand 264
127. The mean value of a function 265
128. Area of a plane curve in polar coordinates 267
129. Volume of a solid with parallel bases 268
130. Volume of a solid of revolution 270
131-132. Length of a plane curve 272
133. Area of a surface of revolution 274
134. Work 275
135. Pressure 275
136. Center of pressure 277
137. Center of gravity 278
138. Attraction 283
Problems 285
Chapter XIV — SPACE GEOMETRY
139. Functions of more than one variable 300
140. Rectangular coordinates 301
141. Graphical representation of a function of two variables . 301
142. Cylinders 303
143. Other surfaces 304
144. Surfaces of revolution 309
145. Projection 310
146. Components of a directed straight line 312
147. Distance between two points •. . . . 313
148. Direction cosines 314
149. Angle between two straight lines 315
150. Direction of the normal to a plane " 316
151. Equation of a plane through a given point perpendicular to
a given direction 316
152. Angle between two planes 316
153. Equations of a straight line 317
154. Straight line determined by two points 317
155. Straight line passing through a known point in a given
direction 318
156. Determination of the direction cosines of a straight line . 319
157. Distance of a point from a plane 320
158. Problems on the plane and the straight line 321
x CONTENTS
Article Page
15^ Space curves 322
160. Direction of space curve and element of arc 324
161. Tangent line and normal plane 326
Problems 326
Chapter XV — PARTIAL DIFFERENTIATION
162. Partial derivatives 335
163. Higher partial derivatives 338
164. Increment and differential of a function of two variables . 339
165. Extension to three or more variables 342
166. Directional derivative of a function of two variables . . 343
167. Total derivative of z with respect to a: 344
168. The tangent plane 345
169. Maxima and minima 348
170. Exact differentials 349
171. Line integrals . 353
172. Differentiation of composite functions 357
Problems 361
Chapter XVI — MULTIPLE INTEGRALS
173. Double integral with constant limits 369
174. Double integral with variable limits 371
175. Computation of a double integral 373
176. Double integral in polar coordinates 374
177. Area bounded by a plane curve 376
178. Moment of inertia of a plane area 377
179. Center of gravity of plane areas 379
180. Area of any surface 381
181. Triple integrals 385
182. Change of coordinates 388
183. Volume 389
184. Moment of inertia of a solid 390
185. Center of gravity of a solid 392
186. Attraction 393
Problems 394
- Chapter XVII — INFINITE SERIES
187. Convergence 405
188. The comparison test for convergence 406
189. The ratio test for convergence 407
CONTENTS xi
Article Page
190. Absolute convergence 409
191. The power series 410
192. Maclaurin's and Taylor's series 412
193. The remainder in Taylor's series 416
194. Relations between the exponential and the trigonometric
functions 418
195. Approximate integration 419
196. The theorem of the mean 422
197. The indeterminate form % 423
198. Other indeterminate forms 425
199. Fourier's series 427
Problems 431
Chapter XVIII — DIFFERENTIAL EQUATIONS
200. Definitions 438
201. The equation Mdx + Ndy = 0, when the variables can be
separated 441
202-203. The homogeneous equation Mdx + Ndy=0 442
204. The linear equation of the first order 444
205. Bernoulli's equation 446
206. The exact equation Mdx+Ndy = 0 447
207. The integrating factor 448
208. Certain equations of the second order 449
209. The linear equation with constant coefficients 454
210. The linear equation of the first order with constant
coefficients 455
211. The linear equation of the second order with constant
coefficients . • . . 456
212. The general linear equation with constant coefficients . . 460
213. Solution by undetermined coefficients 462
214. Systems of linear differential equations with constant
coefficients 465
215. Solution by series 466
Problems 471
Answers 481
Index 514
)
ANALYTIC GEOMETRY AND
CALCULUS
CHAPTER I
CARTESIAN COORDINATES
1. Direction of a straight line. Consider any straight line
connecting two points A and B. In elementary geometry, only
the position and the length of the line are considered, and
consequently it is immaterial whether the line be called AB or
BA ; but in the work to follow, it is often important to consider
the direction of the line as well. Accordingly, if the direction
of the line is considered as from , , ,
A to B it is called AB, but if ^ B °
the direction is considered from
B to A it is called BA. It will be seen later that the distinc-
tion between AB and BA is the same as that between +a
and - a in algebra.
Consider now two segments AB and BC on the same straight
line, the point B being the end of the first segment and the
beginning of the second. The segment AC is called the sum of
AB and BC and is expressed by the equation
AB + BC = AC. (1)
This is clearly true if the points are in the position of fig. 1, but
it is equally true when the points are in the position of fig. 2.
Here the line BC, being opposite • ^_
in direction to AB, cancels part of A c B
it, leaving AC. FlG- 2
If, in the last figure, the point C is moved toward A, the
sum AC becomes smaller, until finally, when C coincides with A,
we have AB + BA = 0, or BA = -AB. (2)
1
2 CARTESIAN COORDINATES
If the point C is at the left of A, as in fig. 3, we still have
AB + BC = AC, where AC=- CA by (2).
It is evident that this addition is analogous to algebraic
addition, and that this sum may be an arithmetical difference.
From (1) we may obtain by . t
transposition a formula for sub- C A B
traction; namely, Fig. 8
BC = AC-AB.
(3)
This is universally true, since (1) is universally true.
2. Projection. Let AB and MN (figs. 4, 5) be any two straight
lines in the same plane, the positive directions .of which are
respectively AB and MN. From A and B draw straight lines
perpendicular to MN, intersecting it at points A1 and B1 respec-
tively. Then ArBf is the projection of AB on MN, and is positive
if it has the direction MN (fig. 4), and negative if it has the
direction NM (fig. 5).
Denote the angle between MN and AB by <f>, and draw AC
parallel to MN. Then in both cases, by trigonometry,
AC = AB cos <f>.
But AC = AfBr, and therefore
A'B' = AB cos <f>.
Hence, to find the projection of one straight line upon a second,
multiply the length of the first by the cosine of the angle between
the positive directions of the two lines.
The projection of a broken line upon a straight line is defined
as the algebraic sum of the projections of its segments.
NUMBER SCALE 3
Let ABODE (fig. 6) be a broken line, MN a straight line
in the same plane, and AE the straight line joining the ends
of the broken line. B %
Draw AA', BB', CC\ D&, and A^rK^
ElS1 perpendicular to MN\ then , j \
A'B', B'C, CD*, D'E\ and A'E' M j, j, \ f
are the respective projections on \y^
MNoiAB,BC,CD,DE,axidAE. „ °a
7 7 ' Fig. 6
But, by § 1, A'Bf +BrCr + C'P +D'Ef = A'Ef.
Hence the projection of a broken line upon a straight line is
equal to the projection of the straight line joining the extremities
of the broken line.
3. Number scale. On any straight line assume a fixed point
O as the zero point, or origin, and lay off positive numbers in
one direction and negative numbers in the other. If the line
is horizontal, as in fig. 7, it is usual, but not necessary, to lay
off the positive numbers to the right of 0 and the negative
numbers to the left. The numbers which we can thus lay off
are of two kinds : the rational num~ ^ %,
bers, including the integers and the _| _j Js _j — ^~~[ — -J J ' I —
common fractions; and the irrational F -
numbers, which cannot be expressed
exactly as integers or common fractions, but which may be so
expressed approximately to any required degree of accuracy.
The rational and the irrational numbers together form the class
of real numbers.
Then any point M on the scale represents a real number,
namely, the number which measures the distance of M from 0 :
positive if M is to the right of 0, and negative if M is to the
left of 0. Conversely, any real number is represented by one
and only one real point on the scale.
Imaginary, or complex, numbers, which are of the form d+b V— 1,
cannot be laid off on the number scale.
The result of § 1 is particularly important when applied to
segments of the number scale. For if x is any number corre-
sponding to the point M, we may always place x = OM, since both
4 CARTESIAN COORDINATES
x and OM are positive when M ia at the right of 0, and both
x and OM are negative when Jf ia at the left of O. Now let Mt
and 3fa (fig. 8) be any two points, and let xy = OMl and xt = OMa.
Then
M,M = OM.- OM, = x-x,. j i i i ? i^ t ffii .
On the other band, FlQi 8
MtMx = OM1 - OMa = as, - *,= - 3f,3f^
It is clear that the segment MxMt is positive when Mt is at
the right of M , and negative when 3fa is at the left of JH^.
Hence (fo length and the sign of any segment of the number
scale is found by subtracting the value of the x corresponding to
the beginning of the segment from the value of the x corresponding
to the end of the segment.
4. Coordinate axes. Let OX and OY be two number scales at
right angles to each other, with their zero points coincident at O,
Let P be any point in the plane,
and through P draw straight lines
perpendicular to OX and 07 re-
spectively, intersecting them at M
and N. If now, as in § 3, we
place x = OM and y = ON, it is
clear that to any point P there
corresponds one and only one
pair of numbers x and y, and
to any pair of numbers corresponds one and only one point P.
If a point P is given, x and y may be found by drawing the
two perpendiculars MP and NP, as above, or by drawing only
one perpendicular as MP. Then
OM=x, MP=ON=y.
On the other hand, if x and y are given, the point P may be
located by finding the points M and N corresponding to the num-
bers x and y on the two number scales and drawing perpendiculara
to OX and OY respectively through M and N, These perpen-
diculars intersect at the required point P. Or, as is often more
DISTANCE
convenient, a point M corresponding to x may be located on its
number scale, and a perpendicular to OX may be drawn through
M, and on this perpendicular the value of y laid off. In fig. 9,
for example, M (corresponding to x) may be found on the scale
OX, and on the perpendicular to OX at Jf, MP may be laid off
equal to y. When the point is located in either of these ways it is
said to be plotted. It is evident that plotting is most conveniently
performed when the paper is ruled in squares, as in fig. 9.
These numbers x and y are called respectively the abscissa and
the ordinate of the point, and together they are called its coordi-
nates. It is to be noted that the abscissa and the ordinate, as
defined, are respectively equal to the distances from OY and OX
to the point, the direction as well as the magnitude of the distances
being taken into account. Instead of designating a point by writing
x = a and y = — b, it is customary to write P (a, — 6), the abscissa
always being written first in the parenthesis and separated from
the ordinate by a comma. OX and OY are called the axes of
coordinates, but are often referred to as the axes of x and y
respectively.
5. Distance. Let ^(xv y^) and P2(x2, y2) be two points, and
at first assume that PXP2 is parallel to one of the coordinate axes,
as OX (fig. 10). Then y2 = yx. Now MXM^
the projection of P^P2 on OX, is evidently
equal to %P2. But M1M%=x%—x1 (§3).
In like manner, if #2= x^ PXP2 is parallel
to OF, and
1
Pi
f
P,
1
1
I
1
i
i
i
i
Mx 0
M, "-
■?^=y,-yi-
(2)
Fig. 10
If x2 ¥= xx and y2 ¥= yv PXP2 is not parallel to either axis. Let
the points be situated as in fig. 11, and
through Px and P2 draw straight lines
parallel respectively to OX and OY. They
will meet at a point -B, the coordinates
of which are readily seen to be (#2, y^).
By (1) and (2),
PxR = x%- xv BP2 = y2- yx. Fig. 11
AC
CARTESIAN COORDINATES
But in the right triangle iJJSiJ,
whence, by substitution, we have
%%= VCs.-a^+Ov-fc)". (3)
It is to be noted that there is an ambiguity of algebraic sign,
on account of the radical sign. But since Iffi ^ parallel to
neither coordinate axis, the only two directions in the plane the
positive directions of which have been chosen, we are at liberty
to choose either direction of ifZJ as the positive direction, the
other becoming the negative.
It is also to be noted that formulas (1) and (2) are particular
cases of the more general formula (3).
Ex. Find the coordinates of a point equally distant from the three points
Px(l, 2), P2(- 1, - 2), and P8(2, - 5).
Let P (xf y) be the required point. Then
PXP = P2P and P2P = P9P.
But PXP = V(;r-l)2 + (y-2)2,
P2P = V(z + 1)2 4- (y 4- 2)a,
PZP = V(x - 2)2 4- (y 4- 5)2.
Therefore V(z - 1)2 4- (y - 2)2 = V(x 4- l)2 4- (y 4- 2)a,
V(x 4- 1)2 4- (y 4- 2)2 = V(x - 2)2 4- (y 4- 5)2;
whence, by solution, a: = § and y = — §. Therefore the required point is
(i - «•
6. Slope. Let %(xl9 y^) and J£(#a, y2) (figs. 12, 13) be two
points upon a straight line. If we imagine that a point moves
along the line from ij to P» the change in x caused by this
motion is measured in magnitude and sign by x2 — xx, and the
change in y is measured by y2— y . We define the slope of the
straight line as the ratio of the change in y to the change in x as
a point moves along the line, and shall denote it by the letter m.
We have then, by definition,
■.-*=*. (1)
X — x v J
2 1
SLOPE
A geometric interpretation of the slope is readily given. For
if we draw through ij a line parallel to OX, and through P2 a
line parallel to 07, and call R the point in which these two
lines intersect, then x2—x1=^PlR and y2— y 1=^^29 and hence
m =
R%
PXR
(2)
It is clear from the figures that the value of m is independent
of the two points ij and P2 and depends only on the given line.
We may therefore choose Px and P2 (as in figs. 12 and 13) so
that I(R is positive. There are then two essentially different
•
r
N5
R
1
i
i
i
i
i
_ -rr
0
JL
Fig. 12
Fig. 13
cases, according as the line runs up or down toward the right
hand. In the former case RP2 and m are positive (fig. 12) ; in
the latter case RP2 and m are negative (fig. 13). We may state
this as follows:
The slope of a straight line is positive when an increase in x
causes an increase in y, and is negative when an increase in x
causes a decrease in y.
When the line is parallel to OX, y2 = yv and consequently m = 0.
If the line is parallel to 07,a;2= x^ and therefore m = oo (§ 13).
Ez. Find a point distant 5 units from the point (1, — 2) and situated
so that the slope of the straight line joining it to (1, — 2) is j.
Let P(x, y) be the required point. Then
(a?-l)8+(y + 2)2=25,
y + 2^4
x-1 3*
Solving these two equations, we find two points, (4, 2) and (— 2, — 6).
8
CARTESIAN COORDINATES
7. Point of division. Let P(x, y) be a point on the straight
line determined by ^(x^ y^) and -§(#a, y2), so situated that
There are three cases to consider, according to the position of
the point P. If P is between the points Ij* and J?(fig. 14), the
Ml M Mt
Fig. 14
M j Ma M
Fig. 15
segments P^P and PXP2 have the same direction, and PXP < JfJJ ;
accordingly I is a positive number less than unity. If P is
beyond ij from J? (fig. 15), PXP and ij^ still have the same
direction, but PXP > Px^ ; therefore I is a positive number
greater than unity. Finally, if P is beyond ^ from P2 (fig. 16),
PXP and ^i^ have opposite directions,
and 2 is a negative number, its nu-
merical value ranging all the way from
0 to oo.
In the first case P is called a point
of internal division, and in the last
two cases it is called a point of ex-
ternal division.
In all three figures draw PYMV PM,
and I%M2 perpendicular to OX. In each figure OM— OMx-\-MxM\
and since ii-P = l(PJ%), MXM=^ l(MYM^) by geometry.
Therefore OM= OM1 + 1 (J^M^ ;
whence, by substitution,
x = zt+l(z%— xty (1)
By drawing lines perpendicular to OF we can prove, in the
same way, y = yx+Ky2~ y> (2)
Fig. 16
VAK1ABLE AND FUNCTION 9
In particular, if P bisects the line J^^ I = £, and these
formulas become , ,
X 2 y 2
Ex. 1. Find the coordinates of a point £ of the distance from Px(2, 3)
to P2(3, - 3).
If the required point is P(x, y),
z = 2 + S(3-2) = 2§,
y = 8 + |(-8-8)=J.
Ex. 2. Prove analytically that the straight line dividing two sides
of a triangle in the same ratio is parallel to the third side.
Let one side of the triangle coincide
with OX, one vertex being at 0. Then the
vertices of the triangle are O (0, 0), A (xv 0),
B (x2, y2) (fig. 17). Let CD divide the sides
OB and AB so that OC = I (OB) and
AD = l(AB).
If the coordinates of C are denoted by
(xv Vz) and tnose of D bv 0'4> V\)y tnen> hy
the above formulas, ^IG' ^
*8 = lxv Vz = lVv
xi = x1 + l(x2-xl), yA = lyr
Since y3 = y4, CD is parallel to OA.
8. Variable and function. A quantity which remains un-
changed throughout a given problem or discussion is called a
constant A quantity which changes its value in the course of
a problem or discussion is called a variaBle. If two quantities
are so related that when the value of one is given the value of
the other is determined, the second quantity is called a function
of the first. When the two quantities are variables, the first is
called the independent variable, and the function is sometimes
called the dependent variable. As a matter of fact, when two-
related quantities occur in a problem, it is usually a matter of
choice which is called the independent variable and which the
function. Thus, the area of a circle and its radius are two
related quantities, such that if one is given, the other is deter-
mined. We can say that the area is a function of the radius,
and likewise that the radius is a function of the area.
10
CARTESIAN COORDINATES
The relation between the independent variable and the function
can be graphically represented by the use of rectangular coordi-
nates. For if we represent the independent variable by x and the
corresponding value of the function by y, x and y will determine
a point in the plane, and a number of such points will outline a
curve indicating the correspondence of values of variable and
function. This curve is called the graph of the function.
Ex. 1. An important use of the graph of a function is in statistical
work. The following table gives the price of standard steel rails per ton
in ten successive years :
1695 (24.33 1900 f32.29
1896 28.00 1901 27.83
1897 18.76 1902 28.00
1898 17.62 1903 28.00
1899 28.12 1904 28.00
If we plot the years as abscissas (calling 1895 the first year, 1896 the
second year, etc.) and plot the price of rails as ordinates, making one unit
of ordinates correspond to
ten dollars, we shall locate
the points P„ Ps, . . ., P10 in
fig. 18. In order to study the
variation in price, we join
these points in succession by
straight lines. The resulting
broken line serves merely to
guide the eye from point to Fio. 18
point, and no point of it
except the vertices has any other meaning. It is to be noted that there
is no law connecting the price of rails with the year. Also the nature of
the function is such that it is defined only y
for isolated values of x.
Ex. S. As a second example we take the
law that the postage on each ounce or frac-
tion of an ounce of first-class mail matter is
two cents. The postage is then a known
function of the weight. Denoting each ounce
of weight by one unit of z, and each two
cents of postage by one unit of y, we have
a series of straight lines (fig. 19) parallel
to the axis of x, representing corresponding
values of weight and postage. Here the function is defined by United
States law for all positive values of x, but it cannot be expressed in
VARIABLE AMD FUNCTION 11
elementary mathematical symbols. A peculiarity of the graph is the series
of breaks. The lines are not connected, but all points of each line repre-
sent corresponding values of x and y.
Ex. 8. As a third example, differing in type from each of the preceding,
let ns take the following. While it is known that there is some physical
law connecting the pressure of saturated steam with its temperature, so
that to every temperature there is some
corresponding pressure, this law has not
yet been formulated mathematically.
Nevertheless, knowing some correspond- sso
ing values of temperature and pressure,
we can construct a curve that is of con-
siderable value. In the table below, the -^
temperatures are in degrees centigrade
and the pressures are in millimeters
of mercury.
150
11.9
Let 100 represent the zero point of
temperature, and let each unit of x repre- en
sent 5 degrees of temperature; also let
each unit of y represent 100 millimeters
of pressure of mercury, and locate the
points representing the corresponding
values of temperature and pressure given Fig. 20
in the above table. Through the points
thus located draw a smooth curve (rig. 20); that is, one which has no
sudden changes of direction. While only the eleven points located are
exact, all other points are approximately accurate, and the curve may be
used for approximate computation as follows : Assume any temperature,
and, laying it off as an abscissa, measure the corresponding ordinate of
the curve. While not exact, it will nevertheless give an approximate
value of the corresponding pressure. Similarly, a pressure may be assumed
12
CARTESIAN COORDINATES
and the corresponding temperature determined. It may be added that
the more closely together the tabulated values are taken, the better
the approximation from the curve ; but the curve can never be exact
at all points.
Ex. 4. As a final example, we will take the law of Boyle and Mariotte
for perfect gases ; namely, at a constant temperature the volume of a
definite quantity of gas is inversely proportional to its pressure. It follows
that if we represent the pressure by
x and the corresponding volume byy,
then j = - 1 where k is a constant
and x and y are positive variables.
A curve (fig. 21) in the first quad-
rant, the coordinates' of every point
of which satisfy this equation, repre-
sents the comparative changes in
pressure and volume, showing that
aa the pressure increases by a cer-
tain amount the volume is decreased
more or less, according to the amount
of pressure previously exerted.
This example differs from the y10_ 21
preceding in that the law of the
function is fully known and can be expressed in a mathematical
formula* Consequently we may find as many points on the curve as
we please, and may therefore construct the curve to any required
degree of accuracy.
9. Functional notation. When y is a function of a; it is cus-
tomary to express this by the notation
Then the particular value of the function obtained by giving x
a definite value a is written /(a). For example, if
/<»=*■+ 3 a*+l,
then /(2) = 2' + 3 ■ 2"+ 1 = 21,
/(0)=0'+3.0'+1 = 1,
/(-3)-(-3)'+3(-3)' + l-l,
/(«) = «■+ So'+l.
PROBLEMS 13
If more than one function occurs in a problem, one may be
expressed as f(x), another as F(x), another as </>(#), and so on.
It is also often convenient in practice to represent different
functions by the symbols fx(x)9 f2(x), /80O» etc
Similarly, a function of two or more variables may be ex-
pressed by the symbol f(x, y). Then /(a, 6) represents the
result of placing x = a and y = b in the function. Thus, if
f(x, y) = x2+3xy-4tfJ
f(a, 6)=a2+3a6-4 62.
PROBLEMS
1. Find the perimeter of the triangle the vertices of which are
(3, 4), (- 2, 4), (2, 2).
2. Find the perimeter of the quadrilateral the vertices of which
are (2, 1), (- 2, 8), (- 6, 5), (- 2, - 2).
3. Prove that the triangle the vertices of which are (— 3, — 2),
(1, 4), (— 5, 0) is isosceles.
4. Prove that the triangle the vertices of which are (—1, 1),
(1, S), (- V3, 2 + V3) is equilateral.
5. Prove that the quadrilateral of which the vertices are (1, 3),
(3, 6), (0, 5), (— 2, 2) is a parallelogram.
6. Prove that the triangle (1, 2), (3, 4), (— 1, 4) is a right
triangle.
7. Prove that the triangle the vertices of which are (2, 3),
(— 2, 5), (— 1, — 3) is a right triangle.
8. Prove that (8, 0), (0, - 6), (7, - 7), (1, 1) are points of a
circle the center of which is (4, — 3). What is its radius ?
9. Find a point equidistant from (0, 0), (1, 0), and (0, 2).
10. Find a point equidistant from the points (— 4, 3), (4, 2), and
(1, - !)•
11. Find a point equidistant from the points (1, 3), (0, 6), and
(- 4, !)•
12. Find the center of a circle passing through the points (0, 0),
(4, 2), and (6, 4).
13. Find a point on the axis of x which is equidistant from (0, 5)
and (4, 2).
14 CARTESIAN COORDINATES
14. Find the points which are 5 units distant from (1, 3) and 4
units distant from the axis of y.
15. A point is equally distant from the points (3, 5) and (— 2, 4),
and its distance from OF is twice its distance from OX. Find its
coordinates.
16. Find the slopes of the straight lines determined by the fol-
lowing pairs of points : (1, 2), (— 3, 1) ; (3, — 1), (— 5, — 1) ;
(2, 3), (2, - 5).
17. Find the lengths and the slopes of the sides of a triangle the
vertices of which are (3, 5), (— 3, 2), (5, 2).
18. A straight line is drawn through the point (5, 0), having the
slope of the straight line determined by the points (— 1, 2) and
(4, — 2). Where will the first straight line intersect the axis OY?
19. One straight line with slope — J passes through (2, 0), and a
second straight line with slope 1 passes through (— 2, 0). Where
do these two lines intersect?
20. The center of a circle with radius 5 is at the point (1, 1).
Find the ends of the diameter of which the slope is $.
21. Two straight lines are drawn from (2, 3) to the axis OX. If
their slopes are respectively J and — 2, prove that they are the
sides of a right triangle the hypotenuse of which is on OX.
22. Find the coordinates of a point P on the straight line deter-
mined by />>(- 2, 3) and P2(4, 6), where j£ = |.
23. A point of the straight line joining the points (3, — 1) and
(5, — 5) divides it into segments which are in the ratio 2 : 5. What
are its coordinates ?
24. On the straight line determined by the points ^(4, 6) and
P2(— 2, — 5) find the point three fourths of the distance from
pi to p*
25. Find the points of trisection of the line joining the points
pi(-3>-7) andPa(10, 2).
26. The middle point of a certain line is (—1, 2), and one end is
the point (2, 5). Find the coordinates of the other end.
27. To what point must the line drawn from (1, —1) to (4, 5) be
extended in the same direction, that its length may be trebled ?
PROBLEMS 15
28. One end of a line is at (2, — 3), and a point one fifth of the
distance to the other end is (1, — 2). Find the coordinates of the
other end.
29. Find the lengths of the medians of the triangle (3, 4),
(- 1, 1), (0, - 3).
30. The vertices of a triangle are -4(0, 0), B(— 2, 5), and C(4, 3).
Show that the slope of the straight line joining the middle points
of AB and BC is the same as \he slope of A C.
31. Find the slopes of the straight lines drawn from the origin
to the points of trisection of the straight line joining (— 2} 4)
and (4, 7).
32. Given the three points A(— 2, 4), 5(4, 2), and C(7, 1) upon
AD AB
a straight line. Find a fourth point D9 such that -— = — ■— •
DC BC
33. If P(x, y) is a point on the straight line determined by
P P I
pi(xv 2^i) and P2(xa> y*)> such that j?f = f' Prove that
2 2
34. Given four points Pv P^ P8, P4. Find the point halfway
between P1 and P# then the point one third of the distance from
this point to P# and finally the point one fourth of the distance
from this point to P4. Show that the order in which the points
are taken does not affect the result.
35. Prove analytically that the lines joining the middle points
of the opposite sides of a quadrilateral bisect each other.
36. Prove analytically that in any right triangle the straight line
drawn from the vertex of the right angle to the middle point of the
hypotenuse is equal to half the hypotenuse.
37. Prove analytically that the straight line drawn between two
sides of a triangle so as to cut off the same proportional parts, meas-
ured from their common vertex, is the same proportional part of the
third side.
38. OABC is a trapezoid of which the parallel sides OA and CB
are perpendicular to OC D is the middle point of AB. Prove
analytically that OD = CD.
16
CARTESIAN COORDINATES
39. Prove analytically that the diagonals of a parallelogram bisect
each other.
40. Prove analytically that if in any triangle a median is drawn
from the vertex to the base, the sum of the squares of the other two
sides is equal to twice the square of half the base plus twice the
square of the median.
41. Prove analytically that the line joining the middle points of
the nonparallel sides of a trapezoid is one half the sum of the
parallel sides.
42. Prove analytically that if two medians of a triangle are equal,
the triangle is isosceles.
43. Show that the sum of the squares on the four sides of any
plane quadrilateral is equal to the sum x)f the squares on the diagonals
together with four times the square on the line joining the middle
points of the diagonals.
44. The following table gives to the nearest million the num-
ber of tons of pig iron produced in the United States for the years
indicated. Represent the table by a graph.
1907
25,000,000
1911
23,000,000
1908
16,000,000
1912
29,000,000
1909
25,000,000
1913
31,000,000
1910
27,000,000
1914
23,000,000
45. The following table gives to the nearest thousand the number
of immigrants into the United States during the year 1914. Exhibit
this table in the form of a graph.
January .
45,000
July
60,000
February
47,000
August
38,000
March
93,000
September
29,000
April
120,000
October
30,000
May
108,000
November
26,000
June
72,000
December
21,000
PROBLEMS
17
46. The average yearly precipitation at the different meteoro-
logical stations of one of the states for the years indicated was as
follows. Represent the table by a graph.
1905
0.74 in.
1910
0.79 in.
1906
2.09 in.
1911
0.30 in.
1907
1.56 in.
1912
0.98 in.
1908
1.20 in.
1913
1.06 in.
1909
3.02 in.
1914
0.41 in.
47. The daily maximum temperatures at a certain town in the
United States for the first ten days of August, 1915, were respec-
tively 100°, 99°, 93°, 89°, 92°, 94°, 96°, 95°, 99°, 97°. Construct a
graph showing the variation in temperature. (Place 90° at the zero
point of the temperature scale.)
48. At a certain place the readings of the height of the barometer
taken at noon and midnight for a week were, in order, as follows :
29.4, 29.6, 29.8, 30.1, 30.4, 30.8, 30.9, 30.4, 29.8, 29.3, 29.4, 29.7,
29.6, 29.8. Construct a graph showing the changes in atmospheric
pressure.
49. On a certain Swiss railroad, the stations with their distance
in miles from the first station and their elevation in feet above
sea level are as follows. Represent graphically the profile of the
railroad.
Station
A
B
C
D
E
F
G
H
I
J
K
Distance
8 mi.
9.5mi.
12 mi.
13 mi.
15mi.
18mi.
20 mi.
22.5mi.
25 mi.
28 mi.
Elevation
1437'
1440'
1530'
1620'
1555'
1558'
1665'
2305'
216(K
3295'
I960'
50. In a test on the tensile strength of a steel rod, originally 10 in.
long, subjected to a varying load, the following readings were made,
the applied load being expressed in pounds per square inch of cross
section and the elongation being measured in inches. Illustrate the
test graphically.
Load
1000
5000
10,000
20,000
30,000
60,000
65,000
70,000
Elongation
0
.0015
.0031
.0070
.0108
.0212
.0230
.0248
18
CARTESIAN COORDINATES
61. A varying load, expressed in pounds per square inch of cross
section, was applied to the end of a concrete block originally 5 in. tall,
and the corresponding compression was measured in inches, with the
results expressed in the following table. Illustrate the test graphically.
Load
100
0
200
300
400
600
600
700
800
900
1000
Compression
.0004
.0010
.0021
.0036
.0064
.0078
.0107
.0130
.0178
62. A body is thrown vertically upward with a velocity of 100 ft.
per second. If v is the velocity at the time t, v = 100 — gt. Assuming
g = 32, construct the graph showing the relation between v and t.
63. The space s through which a body falls from rest in a time t
is given by the formula s = J gt2. Assuming g = 32, construct the
graph showing the relation between s and t
64. A body is thrown up from the earth's surface with an initial
velocity of 100 ft. per second. If s is the space traversed in the
time t, s = 100 1 — \ gt\ Assuming g = 32, construct the graph
showing the relation between s and t
66. Make a graph showing the relation between the side and the
area of a square.
66. Make a graph showing the relation between the radius and
the area of a circle.
E
67. Ohm's law for an electric current is C = — > where C is the
R
current, E the electromotive force, and R the resistance. Assuming
E to be constant, plot the curve showing the relation between the
resistance and the current.
68. Two particles of mass m1 and m2 at a distance d from each
other attract each other with a force F given by the equation
F= \P' Assuming m1 = 5 and m2=20, construct the graph
showing the relation between F and d.
69. If f(x) = a4 - 4x* + 6x - 1, find /(3), /(0), /(- 2).
60. If f(x) = a? - 3x* + 1, show that/(2) + 2/(0) =/(l).
61. If fix) = tf8- 3^+535- 6, find /(«),/(- a), f(a + h).
62. If f(x) = ^1^ , find /(3), /(0), /(- 1).
PROBLEMS 19
xk — 6 x2 4- 7
63. If f(x) = / , prove that/(- x) =/(*).
64. If f(x) ssa^ + Sx* — 9x, prove that/(— as) = —/(*).
66. If f(x) = xi+2ax-at, prove that /(a) +/(- a) = 0.
66. If /(*) = (x - i) (x> - £} , prove that /(a) =/@-
67. If /(*) = |:t|£» prove that /(a) . /(- a) = 1.
All „ ., N x44- 5 a;8 4- 5x + 1 ,, . .. . ^/IX
68. If/(x) = — ^ -j -1— > prove that f(x) =/(-)•
69. If /1(a;) = a;8+a8an(i/2(ir)= 2aa;, prove that/1(a)-a/2(a)=0.
70. If fx(x) = Va2-4 and /2(x) = Vs2 + 4, prove that
C/i(*)]* - C/2(*)]a = C/i(«)]a-
72. If f(x) = + , prove that /[/(«)] = as.
73. If /(*, y) = x* + y> - 5, find /(0, 0), /(l, 0), /(0, 1), /(l, 2)
74. If /(a;, y) = ^-^ > prove that /(a, 6) = -f(b, a).
x y
76. If f^x, y)=x + y and /„(«, y)=x — y, prove that
76. If /x(x, y) = ^ + 2 and /,(», y) = ^ + g, prove that
77. If fx(x, y)=x + 3y and f2(x, y) = 3 x + 9 y, prove that
CHAPTER II
GRAPHS OF ALGEBRAIC FUNCTIONS
10. Equation and graph. If f(x) is any function, and we place
y =/(*>
we may, as already noted, construct a curve which is the graph of
the function. The relation between this curve and the equation
y =f(x) is such that all points the coordinates of which satisfy the
equation lie on the curve; and conversely, if a point lies on the
curve, its coordinates satisfy the equation.
The curve is said to be represented by the equation, and
the equation is called the equation of the curve. The curve
is also called the locus of the equation. Its use is twofold:
on the one hand, we may study a function by means of the
appearance and the properties of the curve; and on the other
hand, we may study the geometric properties of a curve by
means of its equation. Both methods will be illustrated in
the following pages.
Similarly, any equation in x and y expressed by
represents a curve which is the locus of the equation. To
construct this curve we have to find enough points whose
coordinates satisfy the equation to outline the curve. This
may be done by assuming at pleasure values of #, substituting
these values in the equation, and solving for the corresponding
values of y. Before this computation is carried out, however,
it is wise to endeavor to obtain some idea of the shape of the
curve. The computation is then made more systematic, or in
some cases the curve may often be sketched free-hand with
sufficient accuracy.
20
INTERCEPTS
21
The following plan of work is accordingly suggested:
1. Find the points in which the curve intercepts the coordi-
nate axes.
2. Find if the curve has symmetry with respect to either of
the coordinate axes or to any other line.
3. Find if any values of one variable are impossible, since
they make the other variable imaginary.
4. Find the values of one variable which make the other
infinite.
Each of the above suggestions is illustrated in one of the
following articles:
11. Intercepts. The curve will have a point on the axis
of x when y = 0 and will have a point on the axis of y
when x = 0. Hence we may find the intercepts on one of the
coordinate axes by placing the other
coordinate equal to zero and solving the
resulting equation.
Ex. 1. y = .5(x + 2)(x + .o)(x-2).
If y = 0, x = — 2 or — .5 or 2, and there are
three points of the curve on the axis of x.
If x = 0, y =— 1, and there is one intercept,
on the axis of y.
If x < — 2, all three factors are negative ;
therefore y < 0, and the corresponding part of
the curve lies below the axis of z. If — 2 < x
< — .5, the first factor is positive and the other
two are negative ; therefore y > 0, and the
corresponding part of the curve lies above the
axis of x. If — .5 < x < 2, the first two factors
are positive and the third is negative ; therefore
y < 0, and the corresponding part of the curve lies below the axis of x.
Finally, if x > 2, all the factors are positive ; therefore y > 0, and the
corresponding part of the curve lies above the axis of x.
Assuming values of x and finding the corresponding values of y, we plot
the curve as represented in fig. 22.
Ex. 2. y = .5 (x + 2.5) (x - 1)2.
If y = 0, x = — 2.5 or 1, and there are two points of the curve on the
axis of x.
If x = 0, y = 1.25, and there is one intercept on the axis of y.
AC
Fig. 22
22
GRAPHS OF ALGEBRAIC FUNCTIONS
If x<— 2.5, the first factor is negative and the second factor is
positive ; therefore y < 0, and the corresponding part of the curve lies
below the axis of x. If — 2.5 < x < 1, both factors are positive ; therefore
y > 0, and the corresponding part of the curve
lies above the axis of x. Finally, if x > 1, we
have the same result as when — 2.5 < x < 1, and
the curve does not cross the axis of x at the
point x = 1 but is tangent to it.
Assuming values of x and finding the corre-
sponding values of y, we plot the curve as repre-
sented in fig. 23.
Since it will be shown in § 31 that an
equation of the first degree in x and y,
Az + By+C=Q,
Fig. 23
always represents a straight line, and since a straight line is
determined by two points, it is generally sufficient in plotting
an equation of the first degree to find the intercepts on the
two axes and draw a straight line through the two points thus
determined. The only exception is when the straight line passes
through the origin, in which case some point of the straight
line other than the origin must be found by trial.
Ex. 3. Plot the line 3x — 5 y + 12 = 0. Placing y = 0, we find x = — 4.
Placing x = 0, we find y = 2£ . We lay off OX = - 4, OK = 2§, and draw a
straight line through L and K (fig. 24).
Fig. 24
Fig. 25
Ex. 4. Plot the line 3 x - 5 y = 0. Here, if x = 0, y = 0. If we place
x = 1, we find y = £. The line is drawn through (0, 0) and (1, $ ) (fig. 25).
SYMMETRY
23
12. Symmetry and impossible values. A curve is symmetri-
cal with respect to the axis of x when to each value of x
in its equation correspond two values of yy equal in magni-
tude and opposite in sign. This occurs in the simplest manner
when y is equal to plus or minus the square root of a func-
tion of x. Any value of x which makes the function under
the radical sign positive gives two points of the curve equi-
distant from the #-axis. Values of x which make the function
under the radical sign negative make y imaginary and give
no points of the curve. These values of x we call impossible
values.
Similar remarks hold for symmetry with respect to the axis
of y. How symmetry with respect to other lines may sometimes
# be determined is shown by Ex. 5.
Ex. 1. y = ± V(x + 2) (x - 1) (x - 5).
If x = — 2, 1, or 5, y = 0, and the graph intersects the axis of x at
three points.
The lines x =— 2, x = 1, and x = 5 divide the plane (fig. 26) into four
sections.
If x < — 2, all three factors of the
product are negative; hence the radi-
cal is imaginary and there can be no
part of the graph in the correspond-
ing section of the plane. If — 2 < x <1,
the first factor is positive and the
other two are negative; hence the
radical is real and there is a part of
the graph in the corresponding section
of the plane. If Kx<5, the first
two factors are positive and the third
is negative; hence the radical is
imaginary and there can be no part
of the graph in the corresponding
section of the plane. Finally, if
x>5, all three factors are positive;
hence the radical is real and there
is a part of the graph in the corre-
sponding section of the plane.
Therefore the graph consists of two separate parts and is seen (fig. 26)
to consist of a closed loop and a branch of infinite length.
Fig. 26
24
GRAPHS OF ALGEBRAIC FUNCTIONS
Ex. 2. y = ±V(x + 2)(x-iy.
This will be written as
Fig. 27
y = ±(*-l)Vx + 2.
The line x = - 2 divides the plane (fig. 27)
into two sections.
Proceeding as in the previous example, we
find the radical to be real if x > — 2 and
imaginary if x < — 2. Therefore there is a
part of the graph to the right of the line
x = — 2, but there can be no part of the
graph to the left of that line unless x can
have a value that makes the coefficient of the
radical zero; and this coefficient is zero only
when x equals unity. Hence all of the graph lies to the right of the
line x = — 2, as shown in fig. 27.
To every value of x correspond two values of y which are in general
distinct but become equal when x = 1. Hence the curve crosses itself
when x = 1.
Comparing this example with Ex. 1, we see that by changing the
factor x — 5 to x — 1 we have joined the infinite branch and the loop,
making a continuous curve crossing itself at the point (1, 0).
Ex. 3. y = ±V(z + 2)2(x-l)
= ±(x + 2)Vx-l.
The line x = 1 divides the plane (fig. 28) into
two sections.
If x > 1, the radical is real and there is a
part of the graph in the corresponding sec-
tion of the plane. If x <1, the radical is
imaginary and there will be no points of the
graph except for such values of z as make
the coefficient of the radical zero. There is
but one such value, —2, and therefore there
is but one point of the graph, (—2, 0), to
the left of the line x = 1. The graph con-
sists, then, of the isolated point A and the
infinite branch (fig. 28).
Comparing this example also with Ex. 1,
we see that by changing the factor ar-5 to
x + 2 we have reduced the loop to a single
point, leaving the infinite branch as such.
Fig. 28
INFINITE VALUES
Ex. 4. y = ± V-(js + 4) (x + 2)«(i - 4)
= ±(x + 2)V,-(*+4)(z-4).
The lines a; = — 4 and x = 4 divide the plane
(fig. 29) into three sections.
If — 4 < x < 4, the radical is real and there is a
part of the graph in the corresponding portion of
the plane. If x<—4 or x > 4, the radical is imagi-
nary ; and since in the corresponding sections there
is no value of x which makes x + 2 zero, there can
be no part of the graph in those sections. It is
represented in fig. 29.
Ex. 5. 2x* + ya+Zx-4y-
Solving for y, we have
i = 0.
or, after the expression under the
radical sign has been factored,
, = a±v-»(i- j)(i + s>
The lines x = — 3 and x = § divide
the plane (fig. SO) into three sections,
and proceeding as before, we find that
the curve is entirely in the middle \---
section (that is, when — 3<i<§)
and that the line y = 2 is an axis of
symmetry,
If now we should solve for x in
terms of y, we should find another
aiis of symmetry, x =— J, and that
the curve is bounded by the lines
y = -1.2 and y = 5.2. Fig. 30
13. Infinite values. If the expression defining a function con-
tains fractions, the function is not defined for a value of x which
makes the denominator of any fraction zero. But if x = a is a
value which makes the denominator zero, but not the numerator,
and if a; is allowed to approach a as a limit, the value of the func-
tion increases indefinitely and is said to become infinite. The
graph of a function then runs up or down indefinitely, approach-
ing the line x = a indefinitely near but never reaching it.
26
GRAPHS OF ALGEBRAIC FUNCTIONS
This is expressed concisely by the formula
c
Other important formulas involving infinity are
GO
00 X c = GO,
00
— =00,
c
which may be explained in a similar manner. For example, to
obtain the meaning of — > we may write - and then allow x to
increase indefinitely. It is obvious that the quotient decreases in
numerical value and may be made as small as we please by tak-
ing x large enough. This is the meaning of the formula — = 0.
GO
Ex.1. y = -L-.
It is evident that y is real for all values of x ; also, if x < 2, y is nega-
tive, and if x>2, y is positive. Moreover, as x increases toward 2, y is
negative and becomes indefinitely great ; while as x decreases toward 2, y is
positive and becomes indefinitely great. We
can accordingly assign all values to x except 2.
The curve is represented in fig. 31.
It is seen that the nearer to 2 the value
assigned to x, the nearer the corresponding
point of the curve to the line x = 2. In fact,
we can make this distance as small as we
please by choosing an appropriate value for x.
At the same time the point recedes indefi-
nitely from OX along the curve.
Now, when a straight line has such a position
with respect to a curve that as the two are
indefinitely prolonged the distance between them
approaches zero as a limit, the straight line is
called an asymptote of the curve.
It follows from the above definition that the line x = 2 and also the
line y = 0 are asymptotes of this curve. In this example it is to be noted
that the asymptote x = 2 is determined by the value of x which makes
the function infinite.
It is clear that all equations of the type
1
y =
#— a
represent curves of the same general shape as that plotted in fig. 31.
Fig. 31
IKFINITE VALUES
:. 2. y = -
If x = - 2 or if x = 2, y
ia infinite; hence these two
values may not be assigned
to x, all other values, however,
being possible. The curve is
represented in fig. 32.
By a discussion similar
to that of Ex. 1 it may be
proved that the lines x = — 2
and x = 2, which correspond
to the values of x which make
the function infinite, and also
the line y = 0, are asymptotes*
of the curve.
This curve is a special case
of that represented by
and it is not difficult to see
will look for any number of terms.
:. 3. y =
-2)»
All values of x may be assumed except 2.
The curve is represented in fig. 33. It is
evident that the lines x = 2 and y — 0 are
asymptotes.
This curve is a special case of that repre-
sented by
1
which is itself a special case of
28
GKAPHS OF ALGEBRAIC FUNCTIONS
Ex. 4. f =
x-3
We solve for y, forming the equation y = ± -W • The line x = 3
If X ~~ o
(fig. 34) divides the plane into two sections, and it is evident that there
can be no part of the curve in
that section for which x < 3.
Moreover, this line x = 3 is an
asymptote, as in the preceding
examples. The curve, which is a
special case of that represented by
x — a
is represented in fig. 34. It is to
be noted that the axis of x also
is an asymptote.
Ex. 5. y =
** + !
Fig. 34
x
To plot this curve we write the equation in the equivalent form
y = x + -
x
(i)
It is evident that all values except 0 may be assigned to x, that value
being excluded as it makes y infinite. Let us also draw the line
y = *> (2)
a straight line passing through the origin
and bisecting the first and the third
quadrants.
Comparing equations (1) and (2), we
see that if any value xx is assigned to x,
the corresponding ordinates of (1) and (2)
are respectively xx H and xx and that
1 x\
they differ by — Moreover, the numerical
xi
value of this difference decreases as greater
numerical values are assigned to xv and
it can be made less than any assigned
quantity however small by taking xx Fig. 35
sufficiently great. It follows that the
line y = x is an asymptote of the curve. It is also evident that the line
x = 0, determined by the value of x which makes the function infinite,
is an asymptote. The curve is represented in fig. 35.
INTEKSECTION 29
14. Intersection of graphs. Let
/*>(*, y)=o ^ (i)
and /»(*,30=0 (2)
be the equations of two curves. It is evident that any point
common to the two curves will have coordinates satisfying both
(1) and (2), and that, conversely, any values of x and y which
satisfy both (1) and (2) are coordinates of a point common to
the two curves. Hence, to find the points of intersection of two
curves^ solve their equations simultaneously.
The simplest case which can occur is that where each
equation is of the first degree and hence (§ 31) represents
a straight line. In general there is a single solution, which
locates the single point of intersection of the two straight
lines. If no solution can be found, it is evident that the lines
are parallel.
Other important cases are the two following:
Case I. fx(x, y) = 0 and fn(x> y) = 0. Let
AC* y) = o, (i)
AC* v) = o, (2)
be a linear equation and an equation of the nth degree, where
n > 1. The degree of a curve is defined as equal to the degree
of its equation. Accordingly this problem is to find the points
of intersection of a straight line and a curve of the nth degree,
and the method of solution is as follows :
Solve (1) for either x or y and substitute the result in (2).
If, for example, we solve (1) for y, the result of substituting
this value in (2) will in general be an equation of the nth
degree in x, the real roots of which are the abscissas of the
required points of intersection. The ordinates of' the points of
intersection are now found by substituting in succession in (1)
the values of x which have been found.
If two roots x1 and x2 of the equation in x are equal, the cor-
responding ordinates are equal and the two points coincide.
30
GRAPHS OF ALGEBRAIC FUNCTIONS
We may regard this case as a limiting case when the position
of the curves is changed so as to make xx and x% approach each
other; tl*at is, so as to make the points of intersection of the
straight line and the curve approach each other along the
curve. Accordingly the straight line represented by equation
(1) is, by definition, tangent to the curve represented by equa-
tion (2). In general the tangent line simply touches the
curve, without cutting it, as in the case of the circle.
Ex. 1. Find the points of
intersection of
3a:-2y-4 = 0
and x2 — 4 y = 0.
a)
(2)
Solving (1) for y and sub-
stituting the result in (2), we
have x2— 6ar + 8 = 0, the roots
of which are 2 and 4. Substi-
tuting these values of x in (1),
we find the corresponding
values of y to be 1 and 4.
Therefore the points of inter-
section are (2, 1) and (4, 4)
(fig. 36).
Fig. 36
Ex. 2. Find the points of
intersection of
6a:-4y-9 = 0 (1)
and x2 - 4 y = 0. (2)
Solving (1) for y and sub-
stituting the result in (2), we
have r*-6a; + 9 = 0. The
roots of this equation are
equal, each being 3. Hence
the straight line is tangent
to the curve. Substituting 3
for x in (1), we find y = | ;
hence the point of tangency
is (3, |) (fig. 37).
Fig. 37
INTERSECTION
31
Ex. 3. Find the points of
intersection of
3ar-2y-5 = 0 (1)
and x2 - 4 y = 0. (2)
Proceeding as in the two
previous examples, we obtain
x2 — 6 x + 10 = 0, the roots of
which are 3 ± V— 1. Hence
the straight line does not
intersect the curve (fig. 38).
The corresponding values of
y are 2 ± § V— 1.
Ex. 4. Find the points of
intersection of
and
y = 2x (1)
y* = ar(*-3)2. (2)
Fig. 88
Substituting the value of y from (1) in (2),
we have
x[(o:-3)2-4a:] = 0,
a:[x2-10ar + 9] = 0.
or
Its roots are 0, 1, and 9. The
corresponding values of y are
found from (1) to be 0, 2,
and 18. Therefore the points
of intersection are (0, 0),
(1, 2), and (9, 18) (fig. 39).
itx. 5. Find the points of
intersection of
y = 3 x + 2 (1)
and y = x8, (2)
Substituting in (2), we have
x*-3x-2 = 0,
or (x - 2) (x + l)2 = 0.
Fig. 39
Fig. 40
Its roots are 2,-1,-1. The corresponding values
of y, found from (1), are 8, — 1, — 1. Therefore the
points of intersection are (2, 8) and (— 1, — 1), the
latter being a point of tangency (fig. 40).
32
GRAPHS OF ALGEBRAIC FUNCTIONS
Ex. 6. Find the points of intersection of
2a; + y-4 = 0
and y2 = x(x2-12).
G)
(2)
After substitution we have x8 — 4 x2 + 4 x — 16 = 0, or (x — 4) (x2 + 4) = 0,
the roots of which are 4 and ± 2 V— 1. The corresponding values of y,
found from (1), are — 4 and 4^4 V— 1. The
only real solution of equations (1) and (2)
being x •— 4 and y = — 4, the straight line
and the curve intersect in the single point
(4, - 4) (fig. 41).
Case II. fm(x, y)=0 and fn(x, y)=0.
Let
/.0*y) = o
(1)
be an equation of the with degree, and
/.(*y) = o
(2)
Fig. 41
be an equation of the nth degree, where
m and n are both greater than unity.
The method is the same as in the preceding case ; that is,
the elimination of either x or y, the solution of the resulting
equation, and the determination of the corresponding values
of the unknown quantity eliminated. The equation resulting
from the elimination is in general of degree wm, and the
number of simultaneous solutions of the original equations
is mn. If all these solutions are real and distinct, the corre-
sponding curves intersect at mn points. If, however, any of
these solutions are imaginary, or are alike if real, the correspond-
ing curves will intersect at a number of points less than mn.
Hence two curves of degrees m and n respectively can intersect at
mn points and no more.
Ex. 7. Find the points of intersection of
y2 - 2 x = 0
and x2 + y2 - 8 = 0.
(1)
(2)
Subtracting (1) from (2), we eliminate y, thereby obtaining the equa-
tion x2 + 2x — 8 = 0, the roots of which are — 4 and 2. Substituting 2
INTEKSECTION
33
and — 4 in either (1) or (2), we find the corresponding values of y to be
± 2 and ± 2 V— 2. The real solutions of the equations are accordingly
x = 2, y = ± 2, and the corresponding
curves intersect at the points (2, 2)
and (2, -2) (fig. 42).
From the figure it is also evident
that the value — 4 for x must make y
imaginary, as both curves lie entirely
to the right of the line x = — 4.
Ex. 8. Find the points of intersec-
tion of 2 Q A /I V
xz — 3 y = 0 (1)
and y2 - 3 x = 0. (2)
Substituting in (2) the value of y
from (1), we have x4— 27 x = 0. This
equation may be written
x {x - 3) (a;2 + 3 x + 9) = 0,
the roots of which are 0, 3, and — - • Substituting these values
Lt
of x in (1), we find the corresponding values of y to be 0, 3, and
Fig. 42
-3^3 V-3
Therefore the real solutions of these equations are x = 0,
0
y = 0 and x = 3, y = 3. If we had
substituted the values of x in (2),
we should have at first seemed to
find an additional real solution,
y = — 3 when x = 3. But — 3 for
y makes x imaginary in (1), as no
part of (1) is below the axis of x.
Geometrically, the line x = 3
intersects the curves (1) and (2)
in a common point and also
intersects (2) in another point.
Therefore the only real solutions
of these equations are the ones
noted above, and the correspond-
ing curves intersect at the two
points (0, 0) and (3, 3) (fig. 43).
We see, moreover, thai any results found must be tested by substitution in both
of the original equations.
The remaining two solutions of these equations, found by letting
-3±3V^3
x = » are imaginary.
Fig. 43
84
GRAPHS OF ALGEBRAIC FUNCTIONS
Ex. 9. Find the points of intersection of
2*2 + 3y2 = 35 . (1)
and xy = 6. (2)
Since these equations are homogeneous quadratic equations, we place
y=mx (3)
and substitute for y in both (1) and (2). The results are 2 Xs + 3 mV = 35
and mx* = 6, whence
r
x* =
35
2 + 3ma
(4)
and
Therefore
35
x2 = — •
m
2 + 3m2
6
m
(5)
(6)
Fig. 44
from which we find m = % or j.
If m = |, then, from (5), x = ± 2 ;
and from (3) the corresponding
values of y are ±3.
If ra = j, in like manner we
findar = ± |V5 and y = ± |V5.
Therefore the curves intersect at the four symmetrically situated
points (2, 3), (-2, -3), ($V6, | V5), (-|V6, -§V5) (fig. 44).
Sz. 10. Find the points
of intersection of
2 y2 = x - 2 (1)
and x2 - 4 y2 = 4. (2)
Eliminating y, we have
x2-2x = 0,
the roots of which are 0
and 2. When x = 0 we
find from either (1) or (2)
y = ± V— 1, and when x = 2 either (1) or (2) reduces to y1 = 0, whence
y = 0. Therefore these two curves are tangent at the point (2, 0) (fig. 45).
15. Real roots of an equation. It is evident that the real
roots of the equation f(x) = 0 determine points on the axis of
x at which the curve y=/(a?) crosses or touches that axis.
Moreover, if xx and x% (xt < rr2) are two values of x such that
Fig. 45
REAL ROOTS 35
f(xx) and /(a?2) are of opposite algebraic sign, the graph is on
one side of the axis when x = xv and on the other side when
x = xr Therefore it must have crossed the axis an odd number
of times between the points x = xx and x = x%. Of course it may-
have touched the axis at any number of intermediate points.
Now if /(#) has a factor of the form (x—d)\ the curve y =/(#)
crosses the axis of x at the point x = a when k is odd, and
touches the axis of x when k is even. In each case the equa-
tion f(x) = 0 is said to have k equal roots, x = a. Since then a
point of crossing corresponds to an odd number of equal root^
of an equation and a point of touching corresponds to an even
number of equal roots, it follows that the equation f(x) = 0
has an odd number of real roots between xx and *a, if f(xj
and /(#2) have opposite signs.
The above gives a ready means of locating the real roots of
an equation in the form f(x) = 0, for we have only to find two
values of x, as x and #2, for which f(x) has different signs. We
then know that the equation has an odd number of real roots
between these values, and the nearer together x and x^ the more
nearly dp we know the values of the intermediate roots. In locat-
ing the roots in this manner it is not necessary to construct the
corresponding graph, though it may be helpful.
Ex. Find a real root of the equation a^ + 2ar — 17 = 0, accurate to two
decimal places.
Denoting x* + 2 x — 17 ty f(x) and assigning successive integral values
to x, we find /(2) =—5 and/(3) = 10. Hence there is a real root of the
equation between 2 and 3.
We now assign values to x between 2 and 3, at intervals of one tenth,
as 2.1, 2.2, 2.3, etc., and we begin with the values nearer 2, since f(2) is
nearer zero than is/(3). Proceeding in this way we find/(2.3) =— .233
andy(2.4) = 1.624 ; hence the root is between 2.3 and 2.4.
Now, assigning values to x between 2.3 and 2.4 at intervals of one hun-
dredth, we find /(2.31) = - .054 and /(2.32) = .127 ; hence the root is
between 2.31 and 2.32.
To determine the last decimal place accurately, we let x = 2.315 and
find /(2.315) = .037. Hence the root is between 2.31 and 2.315 and is
2.31, accurate to two decimal places.
If y(2.315) had been negative, we should have known the root to be
between 2.315 and 2.32 and to be 2.32, accurate to two decimal places.
36 GRAPHS OF ALGEBRAIC FUNCTIONS
PROBLEMS
Plot the graphs of the following equations :
1. Sx + 4y-7=0. 18. y=(x+l)(x—4:)(x — 3)a.
2. 2x -5y + 6 = 0. 19. y =(x-l)(x + 3)(x* + 2).
3. x+7y = 0. 20. y=(x-l)\2x*+6x + 5).
4. 4a- 3 = 0. 21. ya = (aj-2)(x2-9).
5. 3y + 8 = 0. 22. if = (x + 3)(6x-x2-S).
. 6. y = 4xa+4*-3. 23. 9 / = (x + 2) (2 x - l)2.
7. y = 4:X2-2x + S. 24. 4 y2 = x8 + 4 a;2.
8. y = 6-x-x2. 25. 9 y2 = (x2 - 1) (4 x2 - 25).
9. y = -3xa + 4x. 26. 3^ = (l-ar^x3 -25).
10. y = (x + 2)(x-l)(x-3).. 27. 4 y2 = 9 a;4 - x6.
11. y=(x2-l)(2x + 9). 28. y2 = (2x + 3)(4xa- 9).
12. y = x8+4xa. 29. y» = (a; — 2)a(3 — 2x).
13. y = (x--3)(2x + l)2. 30. f = (2 + * - x*)(x + 2)a.
14. y = a8-8ar, + 16'x. 31. ^ = x^x - 5)a(2x - 3).
15. y = 2x8+3xa-14x. 32. 4Y=(x-l)a(4xa-4x-3).
16. y = x8-xa-4x + 4. 33. f = x (x + 2)\x + 3)a. .
17. y = «8-aaaj. 34. f = (2x - 3)(x2 + l).
35. y2 = (x-l)(2x--l)2(x2-j-3x + 3).
36. x2 + if — 4x + 6 y + 9 = 0.
37. ar1- 4y-f 4/= 0. 48. (y - x)a = 16 - a*
38. xa-ya-2x + 4y-4 = 0. 49. (x + y)2 = y2 0 + 1).
39. 9x2+36ya-96y+28 = 0. 50. x2-4x?/-5y2 + 9y4 = 0.
40. x8+ 3x2- f - x - 3 = 0. /x\l /y\l
41. y8=a-(x2-9). 61' W +W =L
42. ^ = ^ (x + 3).
43. (y + l)8=(x+l)(x2-9).
44. x2- y4(5 + y) = 0. 53. x* + y* = a*
45. aj»-y» + y»+2y = 0.
46. (y + 3)2 = x (x - 2)2.
47. (y-2)2=(x-2)2(x-5). 55. y,(a2-fx2)=xa(aa-x2).
-(!)*+(!)-
"• (I)'+(D-
PROBLEMS
37
56. aY + b*x4 = a?b*x2.
57. 16aY=b2x\a*-2ax).
58. xy = 16.
59. xy = — 16.
60. 2y — xy = 12.
61. (y + l)2= *
67. 4y = 2x + —
2a;
68. y = x + -£-
69. y-l= 3(^ — 2)4-
x-2
62. y =
x + 2
1 1
(x + l)2 (* + 4)s
a*
63'^ = ^-5a!-6-
o # —1
64. f =
65. xY + 25 = 9y*.
Q a; (a; 4- 1)
66. 2^= ^ H '•
9 x—1
70. y = x* + --
x
71. y = :•
y a; xs
72. ay = 4a2(2a — a;).
73. v2 = 7TJi
y 2a — a;
74. / = 5!£l±£>.
75. ^(a;2-fa2) = a2ar!.
76. y (x2 + a2) = a2 (a — #).
Find the points of intersection of the following pairs of loci :
77. 3a; — y — 2 = 0, 5x -3y + 2 = 0.
78. 6a; -24 y +19 = 0, 12a; + 3y + 4 = 0.
79. 2a; - y - 2 = 0, a;2 + y2 = 25.
80.t2a;-3y4-9 = 0, a;2 + y2 + 2x + 4y - 8 = 0.
81. 4a; + 5y -20 = 0, a;2 + f - 2a; - 3= 0.
82. 2y + 3a;-5 = 0, x2 - 2a; - 2y + 4 = 0.
83. 3a;-2?/ + 6 = 0, f + Ay + a; +7= 0.
84. « — 4^ + 1=0, 47/* + 4y — 4a; + 5 = 0.
85. 3a; + 2y-7=0, 5a;2 + 4ya = 21.
86. 7a;-2y + 4 = 0, 21a;2 - 4y2-12 = 0.
87. x - 2y = 0, a-Y + 36 = 25^.
88. 2a; -y -1=0, 42/2=(a;4 2)(2a;-l)2.
89. x 4- 2 y — 2 = 0, y + arfy = 1.
90. ar* + t/2 = 25, 16 x2 + 21 if = 576.
91. x2 + ^=12, arl-82/ + 8 = 0.
92. 4a;y=l, 2a;2 + 2^=1.
AC
38 GRAPHS OF ALGEBRAIC FUNCTIONS
93. 32y2-9x* = 0, 8^-9aj = 0.
94. 2if = S-x, tf = 7^
J ' J 2 — x
95. x2 - f = 0, x2 + f - ±y - 4 = 0.
96. 7z2 = 25-5y, 2/-2 = ^:T
97 • y~2 = x^i9 i6(y-2)=(*-i)*
Find the real roots, accurate to two decimal places, of the
following equations :
98. xs + 2x - 6 = 0. 101. x4 - 4sr8 + 4 = 0.
99. ;c8 + x+ll=0. 102. *8- 3a2 + 6* -11=0.
100. ;c4-ll* + 5 = 0. 103. a8 + 3a2 + 4a +7=0.
CHAPTER III
CHANGE OF COORDINATE AXES
16. Introduction. So far we have dealt with the coordinates
of any point in the plane on the supposition that the axes of coor-
dinates are fixed, and therefore to a given point corresponds one,
and only one, pair of coordinates, and, conversely, to any pair of
coordinates corresponds one, and only one, point. But it is some-
times advantageous to change the position of the axes, that is, to
make a transformation of coordinates, as it is called. In such a
case we need to know the relations between the coordinates of a
point with respect to one set of axes and the coordinates of the
same point with respect to a second set of axes.
The equations expressing these relations are called formulas
of transformation. It must be borne in mind that a trans-
formation of coordinates never alters the position of the point
in the plane, the coordinates alone being changed because of
the new standards of reference
adopted.
17. Change of origin. In this
case a new origin is chosen, but
the new axes are respectively
parallel to the original axes.
Let OX and 0 Y (fig. 46) be the -
original axes, and O'X' and O'Y'
the new axes intersecting at 0\ -
the coordinates of 0' with respect
to the original axes being x0 and y0>
Let P be any point in the plane,
its coordinates being x and y with respect to OX and 0 F, and
Xs and yf with respect to O'X' and Of F\ Draw PMM' parallel
to OF, intersecting OX and O'X' at M and M1 respectively.
30
0'
M
JV
Fig. 46
M'
X'
40 CHANGE OF COORDINATE AXES
Then OM=x, MP = y,
aM' = x', M'P = y\
NO'=xQy ON=yQ.
But OM = NM' = NO1 + O'M1,
and MP = MM1 + M'P = ON+ M'P.
Therefore x = x0 + a/, y = y0 + y ',
which are the required formulas of transformation.
Ex. 1. The coordinates of a certain point are (3, — 2). What will be
the coordinates of this same point with respect to a new set of axes
parallel respectively to the first set and intersecting at (1, — 1) with
respect to OX and OF?
Here x0 = 1, y0 = — 1, x = 3, and y = — 2. Therefore 3 = 1 + ar' and
— 2 = — 1 + %f, whence x' — 2 and tf = — 1.
Ex. 2. Transform the equation y2— 2y— 3a:— 5 = 0 to a new set of
axes parallel respectively to the original axes and intersecting at the
point (— 2, 1).
The formulas of transformation are x — — 2 + x', y = 1 + tf. Therefore
the equation becomes
(l + /)2-2(l + /)-3(-2 + ^)-5 = 0,
or y*2 - 3 xt = 0.
As no point of the curve has been moved in the plane by this transformation,
the curve has been changed in no way whatever. Its equation is different because
it is referred to new axes.
After the work of transformation has been completed the primes
may be dropped. Accordingly the equation of this example may be
written y2 — 3 x = 0, or y2 = 3 x, the new axes being now the only ones
considered.
18. One important use of transformation of coordinates is
the simplification of the equation of a curve. In Ex. 2 of
the last article, for example, the new equation y2 = Sx is
simpler than the original equation. It is obvious, however,
that the position of the new origin is of fundamental im-
portance in thus simplifying the equation, and we shall now
solve examples illustrating methods of determining the new
origin to advantage.
CHANGE OF ORIGIN 41
Ex. 1. Transform the equation y2 — 4y — x3 — Sx2 — 3ar + 3 = 0 to
new axes parallel respectively to the original axes, so choosing the
origin that there shall be no terms of the first degree in x and y in
the new equation.
The formulas of transformation are
x = x0 + x* and y = y0 + y/,
where suitable values of x0 and y0 are to be determined. The equa-
tion becomes
(y0 + 2O2- 4 (y0 + /)- (*o + *08- 8(*0 + *02- 3(*o + *') + 3 = o,
or, after expanding and collecting like terms,
y-* + (2y0- 4)/- x'8 -(3i0 + 3)z'2- (3 x* + 6x0 + 3)^
By the conditions of tl£ problem we are to choose x0 and y0 so that
2y0-4 = 0, 3x02 + 6x0+3 = 0,
two equations from which we find x0 = — 1 and y0 = 2.
Therefore (—1, 2) should be chosen as the new origin, and the new
equation is y'2 — x'z = 0, or y2 = x8 after the primes are dropped.
Ex. 2. Transform the equation
16 x2 + 25 y2 + 64 x - 150 y - 111 = 0
to new axes parallel respectively to the original axes, so choosing the
origin that there shall be no terms of the first degree in x and y in
the new equation.
We may solve this example by the method used in solving Ex. 1, but
since the equation is of the second degree, the following method is very
desirable.
Rewriting, we have
16 {x2 + 4 x) + 25 O2- 6 y) = 111 ;
whence 16 (x2 + 4 x + 4) + 25 (?/2 - 6 y + 9) = 400,
or 16 (x + 2)2 + 25 ( y - 3)2 = 4 00.
Placing now x = — 2 + xt, y = 3 + y',
we have as our new equation 16 x'2 + 25 y'2 = 400,
the new origin of coordinates being at the point (— 2, 3) with respect to
the original axes.
42
CHANGE OF COORDINATE AXES
19. Change of direction of axes.
Case I. Rotation of axes. Let OX and OY (fig. 47) be the
original axes, and OXr and OF be the new axes, making Z<j>
with OX and OY respectively. Then ZXOYr = 90° + <j> and
Z YOX' =90°-<f>.
Let P be any point in the plane,
its coordinates being x and y with
respect to OX and OF, and #'
and y' with respect to OX1 and
OY'. Then, by construction, OM= .r,
OJST=yy OM' = x', and J/'P = y'.
Draw OP.
The projection of OP on OX is fig. 47
OJf, and the projection of the *
broken line Oifcf'P on OX is OJf'cos<£ + Jf'Pcos(90° + <£), or
OM' cos <£ - Jf 'P sin <£.
Therefore OM == 0 JfcT ' cos <j> - iLT'P sin 0,
by § 2.
In like manner the projection of OP on OY is OiV, and the
projection of the broken line OM'P on OF is OM' cos (90°— <£>)
+ M'P cos <£.
Therefore ON = 0 Jf' sin <£ + Jf' P cos <£,
by §2.
Replacing OJtf", OiV, OJf ', J^P by their values, we have
x = xr cos <£ — y sin <£,
y = xf sin <j> + yf cos <£.
Ex. 1. Transform the equation a:?/ = 5 to new axes having the same
origin as the original axes and making an angle of 45° with them. ,
Here <£ = 45°, and the formulas of transformation are x =
*' + /
y=_vT
Substituting and simplifying, we have as the new equation x2 — y2 = 10.
Ex. 2. Transform the equation 34 x2 + 41 y2 — 24 xy = 100 to new axes
with the same origin as the original axes, so choosing the angle <f> that the
new equation shall have no term in xy.
_* -.V
z= — »
V2
OBLIQUE COORDINATES 43
The formulas of transformation are
x — x' cos <j> — if sin <j>,
y z= x' sin <f> + y' cos <£,
where <j> is to be determined.
Substituting' in the equation and collecting like terms, we have
(o4 cos2 <f> + 4 1 sin- <£ - 2 1 si n <f> cos $) x2
+ (;U siu-<£ + 41 cos"2 ^ + 24 sin <£ cos <t>)y2
+ (24 sin-<£ + 14 sin <£ cos <£ - 24 cos2<£).r# = 100.
By the conditions of the problem we are to choose <f> so that
24 sin2<j() + 14 sin <f> cos <f> — 24 cos2<£ = 0.
One value of <f> satisfying this equation is tan_1|. Accordingly we sub-
stitute sin <f> = § and cos <f> = -j, and the equation reduces to x2 + 2 y2 = 4.
Case II. Interchange of axes. If the axes of x and y are simply
interchanged, their directions are changed, and hence such a trans-
formation is of the type under consideration in this article. The
formulas for such a transformation are evidently x = y\ y = x1*
Case III. Rotation and interchange of axes. Finally, if the
axes are rotated through an angle <f> and then interchanged, the
formulas, being merely a combination of the two already found, are
x = y' cos <f> — x! sin <£, y — y! sin <f> + xr cos <f>.
A special case of some importance occurs when <f> = 270°. We
have then x = x\ y=— yf*
Cases II and III, it should be added, occur much less frequently
than Case I.
If both the origin and the direction of the axes are to be
changed, the processes may evidently be performed successively,
preferably in this order : (1) change of origin ; (2) change of
direction.
20. Oblique coordinates. Up to the present time we have
always constructed the coordinate axes at right angles to each
other. This is not necessary, however, and in some problems,
indeed, it is of advantage to make the axes intersect at some
other angle. Accordingly, in fig. 48, let OX and OY intersect
at some angle to other than 90°.
44
CHANGE OF COORDINATE AXES
We now define x for any point in the plane as the distance
from OF to the point, measured parallel to OX, and y as the
distance from OX to the point, measured parallel to OY. The
algebraic signs are determined accord-
ing to the rules adopted in §4.
It is immediately evident that the
rectangular coordinates are but a special
case of this new type of coordinates,
called oblique coordinates, since the new
definitions of x and y include those pre-
viously given. In fact, the term Carte-
sian, or rectilinear, coordinates includes
both the rectangular and the oblique.
Oblique coordinates are usually less convenient than rectangu-
lar ones and are very little used in tliis book. If necessary, the
formulas obtained by using rectangular coordinates can be trans-
formed into similar ones in oblique coordinates by the formulas
of the following article. When no angle is specified the angle
between the axes is understood to be a right angle.
21. Change from rectangular to oblique axes. Let OX and OY
(fig. 49) be the original axes, at right angles to each other, and
OX' and OY' the new
axes, making angles <f>
and <f>' respectively with
OX. Then a> = ^>'-<£.
Let P be any point in
, the plane, its rectangular
coordinates being x and
y, and its oblique coordi-
nates being x1 and y'.
Draw PM parallel to 0 F, PM' parallel to 0 F', M'N parallel
to OF, and RM'N' parallel to OX. Then Z.RM'P = <f>'.
But OM= ON + NM = ON + M'N' = OM' cos <£ + M'P cos <£',
MP=MN'+N'P=NM'+N'P=OM'sm<l> + M'P8m<l>'.
Therefore x = xf cos <£ + y' cos <£>',
y= xr sin <£ 4- y' sin <f>'.
PEOBLEMS 45
r_
Ex. Transform tjie equation — — 75 = 1 to the lines y = ± - x as axes.
or Ir a
Here let <f> = tan-1/ j> and <^/ = tan-1-. The formulas of trans-
formation become
x = a (^+/), y = * (-a^/).
a2+ ft2
Substituting and simplifying, we have as the new equation xy =
* h
Unless b = a, the axes are oblique and o> = 2 tan-1- •
a
22. Degree of the transformed equation. In reviewing this
chapter we see that the expressions for the original coordinates in
terms of the new are all of the first degree. Hence the result of
any transformation cannot be of higher degree than the original
equation. On the other hand, the result cannot be of lower
degree than the original equation ; for it is evident that if any
equation is transformed to new axes and then back to the original
axes, it must resume its original form exactly. Hence, ii the
degree had been lowered by the first transformation, it must be
increased to its original value by the second transformation.
But this is impossible, as we have just noted.
It follows that the degree of an equation is unchanged by
any single transformation of coordinates or by any number of
successive transformations. In particular, the proposition that
any equation of the first degree represents a straight line is
true for oblique, as for rectangular, coordinates.
PROBLEMS
1. What are the new coordinates of the points (3, 4), (— 3, 6),
and (4, — 7) if the origin is transferred to the point (2, — 3), the
new axes being parallel to the old ?
2. Transform the equation sc2 + 9 y1 — 4 .r + 18 y + 8 = 0 to new
axes parallel to the old axes and meeting at the point (2, — 1).
3. Transform the equation 2<r24-2^2 — 2x + 2y — 7=0 to new
axes parallel to the old axes and meeting at the point (J, — J).
4. Transform the equation x2 — y2 + 2x — 3 = 0 to new axes
parallel to the old axes and meeting at the point (— 1, 0).
46 CHANGE OF COORDINATE AXES
5. Transform the equation if — 3 y2 -f 3 x2 + 3 y + 12 x + 11 = 0
to new axes parallel to the old axes and meeting at the point (— 2, 1).
6. Transform the equation y2 -4/- (>// -}-.*> = 0 to new axes
parallel to the oM axes, so choosing the new origin that the new
equation shall contain only terms in //- and *>\
7. Transform the equation .>•- -j- 2.'* + 4 // — 3 = 0 to new axes
parallel to ilie ohl. so choosing mm* new origin that the new equation
shall contain onlv terms in .»•" and //. •
8. Transform the equal ion 2 .*•- — 4 if + 1 2 .>• + 1 (> // — 7 = 0 to
new axes parallel to the old, so choosing the origin that there shall
be no terms of the first degree in the new equation.
9. Transform the equation Ax'2 -f- 9//2 — 4.r +12 y + 4 = 0 to
new axes parallel to the old, so choosing the origin that there
shall be no terms of the first degree in the new equation.
10. Transform the equation xy — 3 // -+- 2 x — 1 2 = 0 to new axes
parallel to the old, so choosing the origin that there shall be no terms
of the first degree in the new equation.
11. Transform the equation 6xy — 10. >- -f- 3// — 19 = 0 to new
axes parallel to the old, so choosing the origin that there shall be
no terms of the first degree in the new equation.
12. Show that any equation of the form xy + ax + by -\- r = 0
can always be reduced to the form xy = k by choosing new axes
parallel to the old, and determine the value of k.
13. Show that the equation if -h ay -f bx + c = 0 (b ^ 0) can always
be reduced to the form f -h bx = 0 by choosing new axes parallel to
the given ones.
14 . Show that the equation ax2 + by2 + ex + dy + o = 0 (a -^0,b=^ 0)
can always be put in the form ax2 + by* = k by choosing new axes
parallel to the old, and determine the value of /»•.
15. What are the coordinates of the points (0, 2), (2, 0), (2, — 2)
if the axes are rotated through an angle of 60° ?
16. What are the coordinates of the points (1, 2), (2, 2), (2, — 1)
if the axes are rotated through an angle of 45°?
17. What are the coordinates of the points (1, 2), (— 1, — 2),
(1, —2) if the axes are rotated through an aeute angle tan"1-}?
18. Transform the equation 2 a?2 + 2f — Sxy — 7= 0 to a new
set of axes by rotating the original axes through an angle of 45°.
PROBLEMS 47
19. Transform the equation 4#2 4- 2 -wSxy + 2 y2 — 5 = 0 to a
new set of axes by rotating the original axes through a positive
angle of 30°.
20. Transform the equation 4 x2 — 12 xy + 9 ?/* — 14 = 0 to a new
set of axes making a positive angle tan_1§ with the original set.
21. Transform the equation 5 x2 -h 4 iry + 8 t/2 — 36 = 0 to a new
set of axes by rotating the original axes through a positive angle
tan-X- £).
22. Transform the equation 4 x2 -f- lo a"y — 4 ?/2 — 34 = 0 to a new
set of axes making a positive angle tan-1 J with the original axes.
23. Show that the equation x2 -\~ y2 = a2 will be unchanged in
form by transformation to any pair of rectangular axes if the origin
is unchanged.
24. Transform the equation x2 — if = 49 to new axes bisecting
the angles between the original axes.
25. Transform the equation 5x2 + 2xy + 5y2 — 12 = 0 to one
which has no xy-tevm, by rotating the axes through the proper
angle.
26. Transform the equation 6 x2 -f- 24 xy — y2 —150 = 0 to one
which has no xy-teim, by rotating the axes through the proper
angle.
27 . Transform the equation 16 x2 — 24 xy + 9 // — 30 x — 40 y = 0
to one which has no a^/-term, by rotating the axes through the proper
angle.
28. Transform the equation 41 x1 4- 24 xy + 34 y2 — 100 x — 50 y
— 100 = 0 to one which has no o-y-terin, by rotating the axes through
the proper angle.
29. Transform the equation ll;r2 — 6 V3.ry -h 5?/2 + (22 +
12 V3)z - (20 + 6 V3)y + 3 + 12 V3 = 0 to a new set of axes
making an angle of 60° with the original axes and intersecting at
the point (—1, 2) with respect to the original axes.
30. Transform the equation 4 x2 + 25 y2 = 100 from rectangular
axes to oblique axes witli the same origin and making the angles
tan-1 J and tan-1(— |) respectively with OX.
31. Transform the equation 9 x1 — 4 y2 = 36 from rectangular axes
to oblique axes with the same origin and making the angles tan-1 \
and tan-1(— J) respectively witli OX,
48 CHANGE OF COORDINATE AXES
32. Transform the equation 9 x* — 4 y1 = 36 from rectangular axes
to oblique axes with the same origin and making the angles tan_1|
and tan"1 3 respectively with OX.
33. Prove that the formulas for changing from a set of rectangular
axes to a set of oblique axes having the same origin and the same
axis of x are „ * , * .„
x = x -f~ y cos to,
y = y1 sin o>,
where o> is the angle between the oblique axes.
34. By rotating the axes through an angle of 45° and changing
the origin, prove that the equation x* -f- y* = a* can be transformed
into y1 = V2 ax, and sketch the curve.
35. The equation of the Folium of Descartes is xz -f y8 — 3 a#y = 0.
Rotate the axes through an angle of 45° and sketch the curve.
CHAPTER IV
GRAPHS OF TRANSCENDENTAL FUNCTIONS
23. Definition. Any function ofx which is not algebraic is called
transcendental. The elementary transcendental functions are the
trigonometric, the inverse trigonometric, the exponential, and the
logarithmic functions, the definitions and the simplest properties
of which are supposed to be known to the student. In this
chapter we shall discuss the graph's of these functions.
24. Trigonometric functions.
Ex. 1. y = sin x.
The values of y are found from a table of trigonometric functions.
In plotting it is desirable to express x in circular measure ; for example,
for the angle 180° we lay off
x = 7T = 3.1416. When a: is a 7
multiple of ir, y = 0 ; when
x is an odd multiple of -»
y = ± 1 ; for other values of x,
y is numerically less than 1. 'Fig. 50
The graph consists of an
indefinite number of congruent arches, alternately above and below the
axis of x, the width of each arch being ir and the height being 1 (fig. 50).
Fig. 51
The curve y = sin x may be constructed without the use of tables, by a
method illustrated in fig. 51.
Let Px be any point on the circumference of a circle of radius 1
with its center at C, and let AO be a diameter of the circle extended
49
50 UKAPHS OF TRANSCENDENTAL FUNCTIONS
indefinitely. With a pair of dividers lay off on AO produced a distance
ONx equal to the arc OJ\. This may be done by considering the arc OPx
as composed of a number of straight lines eacli of which differs inappre-
ciably from its arc. From X1 draw a line perpendicular to AO, and from
f\ draw a line parallel to AO, Let these lines intersect in Qv Then
yxQi = ^iPi = CJ\&hiOCI\. But (7^=1, and the circular measure
of OCPx is OI\ = OXv If, then, we take OX1 = x, XxQt = y, Qx is a
point of the curve y = sin x. By varying the position of the point Px
we may construct as many points of the curve as we wish. The figure
shows the construction of another point Q^.
Ex. 2. y = a sin bx.
When lis a multiple of -•> // = 0 ; when x is an odd multiple of -^-~<t
1 b 2 6
y =*■ ± a ; for all other values of x, y is numerically less than a. The
Fig. 52
curve is similar in its general shape to that of Ex. 1, but the width
IT
of each arch is now — » and its height is a. Fig. 52 shows the curve
b
when a = 3 and b = 2.
Ex. 3. y = a sin (bx + c).
Place x — — - + x', y = %f,
b
The equation then becomes %f = a sin bx*.
The graph is consequently the same as in Ex. 2, the effect of the term
+ c being merely to shift the origin.
Ex. 4. y = a cos bx.
This may be written
y = a sin f <
bx +
which is a curve of Ex. 3. Hence the graph of the cosine function differs
from that of the sine function only in its position.
TKIG0N0MET11IC FUNCTIONS
any integer. Hence
:. Between any
mtinuously from 0 to ± 1 and hack
the graph
consecutive two of these points y v:
to zero. It follows that as
x approaches 0, the corre-
sponding point on the
graph oscillates an infi-
nite number of times
back and forth between
the straight lines y = ± 1.
It is therefore physically
impossible to construct
the graph in the neigh-
borhood of the origin.
This is shown in fig. 54 Fig. 64
by the break in the curve.
The value of y can be calculated for any value of x, no matter how
12 .j h-
small. For example, if x =.,;/ = sin -■ - = .Bfi.')0. The value of y is
not defined for x — 0, and the function is discontinuous at that point.
Ex.7.
a multiple of
9 infinite,
When x is
it, y = 0 ; whf
multiple of -
in the sense of § 13. The
curve has therefore an un-
limited number of asymp-
totes perpendicular to OX,
namely,z = ±— , x = ±~-~- ,
. . ., which divide the plane
into an infinite number of
sections, in each of which is a distinct branch of the curve, a
Fiq. 55
52 GEAPHS OF TEANSCENDENTAL FUNCTIONS
25. Inverse trigonometric functions* The graphs of the inverse
trigonometric functions are evidently the same as those of the
direct functions but differently placed with reference to the
coordinate axes. It is to. be noticed particularly that to any
value of x corresponds an infinite number of values of y.
Ex. l.
x.
From this, x = sin y, and we may plot the graph
by assuming values of y and computing those of x
(fig. 56).
Fig. 66
Fig. 57
Ex. 2. y = tan-1 a:.
Then x = tan yt and the graph is as in fig. 57.
These curves show clearly that to any value of x corresponds an infinite
number of values of y.
26. Exponential and logarithmic functions. The equation
y = a*
defines y as a continuous function of x, called the exponential func-
tion, such that to any real value of x corresponds one and only
one real positive value of y. A proof of this statement depends
upon higher mathematics, but the student is already familiar with
the methods by which the value of y may be computed for simple
values of x. If x is an integer w, y is determined by raising a to
the nth power by multiplication. If a; is a positive fraction — > y is
the gth root of the pth power of a. If # is a positive irrational num-
ber, the approximate value of y may be obtained by expressing x
EXPONENTIAL AND LOGAKITHMIC FUNCTIONS 53
approximately as a rational number. If #=0, y = a° = l. Finally,
if x = — m, where m is any positive number, y = <r m = — .
a
Practically, however, the value of ax is most readily obtained
by means of the inverse function, the logarithm*, for if
y = «*»
then x = logay.
The quantity a is called the base of the system of logarithms
and may be any number except 1.
When a = 10, tables of logarithms are readily accessible. Sup-
pose a is not 10, and let I be such a number that
106=a;
y =a* = (106)*=:106a:.
fa = logio#'
that is,
Then we have
Hence
and
x = logi<>y = lQgi0y
l log10a'
Ex. 1. The graph of y = loga£)x is shown in fig. 58.
It is to be noticed that the curve has the negative portion of the y-axis
for an asymptote and has no points corresponding to negative values of x.
Ex. 2. The graph of y = (1.5)* is shown in fig. 59.
Fig. 58
Fig. 59
27. The number e. In the theory and the use of the expo-
nential and the logarithmic functions an important part is
played by a certain irrational number, commonly denoted by
the letter e. This number is denned by an infinite series, thus :
1111
e = 1 + T + [2 + [3 + l4 + ---
AC
54 GEAPHS OF TRANSCENDENTAL FUNCTIONS
It can be shown that this series converges ; that is, that the
greater the number of terms taken the more nearly does their
sum approach a certain number as a limit. Assuming this, we
may compute e to seven decimal places by taking the first
eleven terms. There results
e=2.7182818....
When y = e*, a: is called the natural, or Napierian, logarithm
of y. The use of Napierian logarithms in theoretical work gives
simpler formulas than would result from the use of the pommon
logarithm. Hence in theoretical discussions the expression log x
usually means the Napierian logarithm. On the other hand,
when the chief interest is in calculation of numerical values, as
in the solution of triangles, log a; usually means log10a^ In this
book we shall use log x for logex.
Tables of values of logea; and e* are found in many collections
of tables and may be used in finding the graphs. It is evident,
however, that the graphs
will not differ in general *"
shape from those in Exs. 1
and 2 of § 26.
In the following exam-
ples we give the graphs
of certain other functions
which involve e • and pre-
sent other points of interest.
Fig. 60
Ex. 1.
y = e-<
The curve (fig. 60) is symmetrical with respect
to OF and is always above OX. When x = 0, y = 1.
As x increases numerically, y decreases, approaching
zero. Hence OX is an asymptote.
X X
Ex- 2. y = Z(e* + e «).
This is the curve (fig. 61) made by a string
held at the ends and allowed to hang freely.
It is called the catenary.
Fig. 61
PKOBLEMS
55
Ex. 8. y = er** sin bx.
The values of y may be computed by multiplying the ordinates of
the curve y = e-** by the values of sin Jar for the corresponding
abscissas. Since the value of
sin bx oscillates between 1 and — 1,
the values of e-.°* sin bx cannot
exceed those of «-«*. Hence the
graph lies in the portion of the
plane between the curves y = e-**
and y = — e~ax. When a; is a multi-
ple of —> y is zero. The graph there-
b •
fore crosses the axis of x an infinite
number of times. Fig. 62 shows
the graph when a = 1, b = 2 tt.
i
Ex. 4. y = c*.
When a; approaches zero, being positive, y increases without limit.
When x approaches zero, being negative,
y approaches zero; for example, when
* = iota* y = «1000> and when x =- toW>
v = 0-iooo = . The function is therefore
9 glOOO
discontinuous for x = 0.
The line y = 1 is an asymptote (fig. 63), for
as x increases without limit, being positive or
negative, - approaches 0, and y approaches 1.
Fig. 62
Fig. 68
5. y =
10
l
1+e*
As x approaches zero positively, y ap-
proaches zero. As x approaches zero nega-
tively, y approaches 10. As x increases
indefinitely, y approaches 5.
The curve (fig. 64) is discontinuous when
a? = 0.
PROBLEMS
Plot the graphs of the following equations :
1. y = J sin 2x. 4. y = smlx — — h
5. y = 2sin3faj — j)
3. y
2. yssSsin^-
sin(*+f)
a— 1
6. y = Jsin— ^—
56 GRAPHS OF TRANSCENDENTAL FUNCTIONS
8. y = cos 3 a. \ */
9. y = 3 cos T 31. y = sin"1 •
4 *
0. y = 2 cos 3(z + 2). 32- V = cos"1^ + 2).
1. y = 2cos(2ic-l). 33# sin-i^5.
2. y = vers a. ^
o , • o 34. y = tan-Vac +1).
3. y = 2 + sin3a;. * 1
4. y = 2 - J cos x. 35. y = tan"1 t— -•
5. y = sin x + sin 3 x. 36. y = e1"*.
6. y = J sin a; — J sin 2 a. 37. y = xe"x.
. irx 1 . 38. y = a^e"*.
7. y = sin — -sui7nc. * x
39. y = see*.
8. y = J cos 2 x 4- J cos 3 x.
9. y =l+cosaj — Jcos3aj. 4(>. y
20. y = sin2aj. ^ t^-
y 41. y = e1"*.
21. y = sinsca. ^
1 42. y = e1"x.
22. y = x sin--
* a? 43. y = £(«*-*"*)•
. 1
a?
24. y = etna;. 45. y = ^ ^ ^_x
l-ac
ace x .
23. y = x2sini. 44' y=*(^ + «"">
e* — e~x
e° + e-
= e * cos a:.
25. y = Jtan2aj. 46 y
au
26. y = 2tan~- 47. y = e~txava.2x.
27. y = 4tan^. 48' ? = ^i^
28. y = sec x. 49. y = log sin x.
29. y = esc*. 50. y = log tana.
CHAPTER V
THE STRAIGHT LINE
28. The point-slope equation. If the slope of a straight line
and a point on the line are known, the equation of the line is
readily found. Let LK (fig. 65) be any straight line, If(xv yx)
a known point on it, and m its slope. Take P (#, y), any point
on the line. Then, by § 6,
^ ^1= m.
x — x.
If m is not infinite, we may clear
of fractions and obtain
y-y^mQc-xJ- (1)
Fig. 65
This is an equation which is
obviously satisfied by the coordi-
nates of any point on LK and by those of no other point.
Hence it is the equation of LK
If tjie line is parallel to OX, m = 0, and the equation of the
lineis jr-y, (2)
If the line is parallel to 0 F, m = oo, and the equation of the
line is ^\
X ^— *C|.
Ex. Find the equation of a straight line with the slope — $, passing
through the point (5, 7).
By substituting in the formula, we have
whence
2z + 3y-31 = 0.
29. The slope-intercept equation. The equation (1) of § 28
takes a special form when the point J% is taken at B (fig. 65),
57
58 THE STRAIGHT LINE
where LK cuts the axis of y. If OB = 6, the coordinates of B
are (0, 6). Then the equation of LK is
y — b = m(x — 0),
or, after a simple reduction,
y = mx -f- b. (1)
It is to be noticed that the equation of a line parallel to the
axis of y cannot be put in this form, since the line does not cut
0 F, but the equation of any other line can be given this f orm.
Conversely, any equation of the form (1), no matter what are
the values of m and 6, represents a straight line. For a straight
line can be drawn with any slope m and any intercept 5. The
equation of this line is then y = mx + 6, and this equation is sat-
isfied by no point not on the line.
30. The two-point equation. If a straight line is determined
by the two points ^(»x, yx) and J£ (#2, ya), then
m = 9
by § 6, and the equation of the line is by (1), § 28,
2 1
If y2 = yv the line is parallel to OX, and its equation is
y = yr • (2)
If x% = a^, the line is parallel to 07, and its equation is
x = xx. (3)
Ex. Find a straight line through (1, 2) and (— 3, 5).
By formula (1), . _ 9
y-2 = -^I(s--l),
or 3 a: + 4 y - 11 = 0.
31. The general equation of the first degree. The equation
Ax+By + C=0,
where Ay B, and C may be any numbers or zero, except that
A and B cannot be zero at the same time, is called the general
ANGLES
59
equation of the first degree. We shall prove: The general
equation of the first degree with real coefficients always represents
a straight line.
1. Suppose A 4s- 0 and B =£ 0. The equation may be written
A C
9 B B
This equation is of the form y = mx + b and therefore repre-
, sents a straight line, by § 29.
It follows that if the equation of a straight line is in the form
Ax+By + C=0y its slope may be found by solving the equation
for y and taking the coefficient of x.
2. Suppose A = 0, B ^= 0. The equation is then
By + C=0, or y = -|.
C
and represents a straight line parallel to OX at a distance
units from it.
3. Suppose A ^= 0, B = 0. The equation is then
Q
Ax + C=0, or x = — 7,
A
Q
and represents a straight line parallel to O Y at a distance
units from it.
Therefore the equation Ax+By + C=0 always represents a
straight line.
32. Angles. The slope of a straight line enables us to solve
many problems relating to angles, some of which we take up in
this article.
1. The angle between the axis of x
and a known line. Let a known line
cut the axis of x at the point L.
Then there are four angles formed.
To avoid ambiguity we shall agree
to select that one of the four which
is above the axis of x and to the
right of the line and to consider LX as the initial line of this
angle. We shall denote this angle by <f>. Then if we take any
Fig. 66
60
THE STRAIGHT LINE
point P on the terminal line of <£ and drop the perpendicular
MP, we have, in the two cases represented by figs. 66 and 67,
tan<£ =
MP
LM
MP
But is equal to the slope of the
line, by (2), § 6. Therefore
tan <f> = m.
If the straight line is parallel to
OF, <j> = 90° and tan<£ = oo. If the
line is parallel to OX, no angle <f>
is formed, but since m = 0, we may say tan <f> = 0 ; whence
<£ = 0° or 180°.
2. Parallel lines. If two lines are parallel, they make equal
angles with OX, and hence their slopes are equal. It follows that
two equations which differ only in the absolute term, such as
Fig. 67
and
Ax + Bff+Cx=0
represent two parallel lines. It is to be noticed that these two
equations have no common solution (§ 14).
Ex. 1. Find the equation of a straight line passing through (— 2, 3) and
parallel to 3 a; — 5 y + 6 = 0.
First method. The slope of the given line is J. Therefore the required
line is
y - 3 = g (x + 2), or 3 x - 5 y + 21 = 0.
Second method. AVe know that the required equation is of the form
3s-5y+C = 0,
where C is unknown. Since the line passes through (—2, 3),
3(-2)-5(3)+C=0,
whence C = 21. Therefore the required equation is
3 x - 5 y + 21 = 0.
ANGLES
61
3. Perpendicular lines. Let AB and CD (fig. 68) be two lines
intersecting at right angles. Through P draw PR parallel to
OX, and let RPD = ^ and RPB = <j>2. Then tan ^ = m± and
tan02 = 7w2, where rn1 and ?wa are the slopes of the lines. But,
*, = *,+ 90";
tan <j>2 = — cot <j>1 = —
1
by hypothesis,
whence
tan<£x
which is the same as
w2 = -
m.
That is, two straight lines are perpendicular when the slope of
one is minus the reciprocal of the slope of the other. This theorem
may be otherwise expressed
by saying that two lines are
perpendicular when the prod-
uct of their slopes is minus
unity.
It follows that two straight
lines whose equations are of
the type
•Ax + Btf + Ct = 0
and Bx — Ay + C2 = 0
are perpendicular. Fig. 68
Ex. 2. Find a straight line through (5, 3) perpendicular to7:r+9y+l = 0.
First method. The slope of the given line is — J. Therefore the slope
of the required line is ^. Therefore the required line is
y - 3 = $ (x - 5), or 9 x - 7 y - 24 = 0.
Second method. We know that the equation of the required line is of
the form 9z — 7 y + C = 0. Substituting (5, 3), we find C =— 24. Hence
the required line is 9 x — 7 y — 24 = 0.
Ex. 3. Find the equation of the perpendicular bisector of the line join-
ing (0, 5) and (5, — 11). The point midway between the given points is
(§» ~~ 3), by § 7. The slope of the line joining the given points is — V1,
by § 6. Hence the required line passes through (§, — 3), with the slope ^.
Its equation is
y + 3 = WO*- £)> or 10:r-S2y-121 = 0.
54 GRAPHS OF TRANSCENDENTAL FUNCTIONS
It can be shown that this series converges ; that is, that the
greater the number of terms taken the more nearly does their
sum approach a certain number as a limit. Assuming this, we
may compute e to seven decimal places by taking the first
eleven terms. There results
e = 2.7182818-.. .
When y = f, xia called the natural, or Napierian, logarithm
of y. The use of Napierian logarithms in theoretical work gives
simpler formulas than would result from the use of the common
logarithm. Hence in theoretical discussions the expression log x
usually means the Napierian logarithm. On the other band,
when the chief interest is in calculation of numerical values, as
in the solution of triangles, log x usually means logMa; In this
booh we shall use log x for logtx.
Tables of values of log„a: and e1 are found in many collections
of tables and may be used in finding the graphs. It is evident,
however, that the graphs
will not differ in general
shape from thoBe in Exs. 1
and 2 of § 26.
In the following exam-
ples we give the graphs
of certain other functions
which involve e • and pre-
sent other points of interest.
Ex. l. y=e-<
The curve (fig. 60) is symmetrical with respect
to OY and is always above OX. Wheu x = 0, y = 1.
As x increases numerically, y decreases, approaching
zero. Hence OX is an asymptote.
Ex. 3. y = -(<* + e ").
This is the curve (fig. 61) made by a string
held at the ends and allowed to hang freer/.
It is called the catenary.
^
the
Ex. S.
The values of y may
computed by multiplying the ordinates of
by the values of Bin bx for the corresponding
Since the value of
abscissas.
sin bx oscillates between 1 and — 1,
the values of e-,**sin6i cannot
exceed those of «-■*. Hence the
graph lies in the portion of the
plane between the curves y = e- ™
andy = — e-™. When x is a multi-
pleof — ,y is zero. The graph there-
fore crosses the axis of x an infinite
number of times. Fig. 62 shows
the graph when a = 1, h — 2 jr.
Ex. 4. y = e".
When x approaches zero, being positive, y
When x approaches zero, being negative,
y approaches zero; for example, when
* = t<fW V = el0M. and wnen x =~ toWi
y ^ e-vm — ., The function is therefore
discontinuous for x = 0.
The line y = 1 is an asymptote (fig. 63), for
as x increases without limit, being positive or
negative, - approaches 0, and y approaches 1.
Ex. 6. y= ;•
As x approaches zero positively, y ap-
proaches zero. As x approaches zero nega-
tively, y approaches 10. As x increases
indefinitely, g approaches 5.
The curve (fig. 64) is discontinuous when
Plot the grapha o
1. y = isin2
/
56 GEAPHS OF TEANSCENDENTAL FUNCTIONS
30. y = sec (as — — j
x-1
7. y = J sin (2 a; + 3).
8. y = cos 3 #.
9. y = 3cos| 81. y = sin- ^
0. y = 2 cos 3 (a +.2). 32- ? = Goa'\x + 2>
1. y = 2cos(2a-l). 33# y = sin-i Jn£.
2. y = vers x.
. « 34. y = tan_1(aj+l).
3. y = 2 + sm3aj. y 1
4. y = 2 - J cosaj. 35. y == tan"1 j^;'
5. y = since + sin 3 x. 36. y = e1"*.
6. y = £ sin # — J sin 2 cc. 37. y = ase"*.
. to 1 . 38. y = x*e~*.
^ 2 2 1
39, |# _. o^g0
8. y = \ cos 2 aj + \ cos 3 #. jl,
9. y =1+ cos a — \ cos 3a;. 40- y = xex.
20. y = sin2 a;. ., rrz
y 41. y = e1 -*.
21. y = sin«2. ^
1 42. y=zex~x.
22. y = ajsin-- ^ , . .
* a? 43. y = i(ex — e"*).
. 1
x
24. y = ctna;. 45« y =
23. y = ^sinl. 44' y = i(** + 0-
e* + e
— «
,— X
= e * cos a;.
25. y = Jtan2aj. 46 y
26. y = 2tan-« 47. y = e~8xsin 2a;.
„. y=s4tan*±l. 48. y = log^|.
28. y = secaj. 49. y = log sin as.
29. y = cscas. 50. y = log tan x.
CHAPTER V
THE STRAIGHT LINE
28. The point-slope equation. If the slope of a straight line
and a point on the line are known, the equation of the line is
readily found. Let LK (fig. 65) be any straight line, %(xx,y^)
a known point on it, and m its slope. Take P (#, y), any point
on the line. Then, by § 6,
If m is not infinite, we may clear
of fractions and obtain
y-yx = m(x-x^ (1)
Fig. 65
This is an equation which is
obviously satisfied by the coordi-
nates of any point on LK and by those of no other point.
Hence it is the equation of LK.
If tjie line is parallel to OX, m = 0, and the equation of the
*»*" y=y, (2)
If the line is parallel to OY,m = oo, and the equation of the
line is „ /«>.
•€/ — — •£..•
Ex. Find the equation of a straight line with the slope — §, passing
through the point (5, 7).
By substituting in the formula, we have
whence
y-7=-|(*-5);
2x + 3y-31 = 0.
29. The slope-intercept equation. The equation (1) of § 28
takes a special form when the point J? is taken at B (fig. 65),
67
58 THE STRAIGHT LINE
where LK cuts the axis of y. If OB = 6, the coordinates of B
are (0, 6). Then the equation of LK is
y — b = m(x — 0),
or, after a simple reduction,
y = mx + b. (1)
It is to be noticed that the equation of a line parallel to the
axis of y cannot be put in this form, since the line does not cut
OF, but the equation of any other line can be given this form.
Conversely, any equation of the form (1), no matter what are
the values of m and J, represents a straight line. For a straight
line can be drawn with any slope m and any intercept b. The
equation of this line is then y = mx + 6, and this equation is sat-
isfied by no point not on the line.
30. The two-point equation. If a straight line is determined
by the two points %(xv yx) and ij (#2, y2), then
m =
x — x
2 1
by § 6, and the equation of the line is by (1), § 28,
2 1
If y% = y^ the line is parallel to OX, and its equation is
y=yx- (2)
If x2 = x^ the line is parallel to 0 F, and its equation is
x = xx. (3)
Ex. Find a straight line through (1, 2) and (— 3, 5).
By formula (1), * _ 9
y-2 = -^-fI(*-i),
or 3 x + 4 y - 11 = 0.
31. The general equation of the first degree. The equation
Ax+By + C=0,
where A, B, and C may be any numbers or zero, except that
A and B cannot be zero at the same time, is called the general
ANGLES
59
equation of the first degree. We shall prove: The general
equation of the first degree with real coefficients always represents
a straight line.
1. Suppose A =£ 0 and B =£ 0. The equation may be written
A C
* . B B
This equation is of the form y = mx + b and therefore repre-
.sents a straight line, by § 29.
It follows that if the equation of a straight line is in the form
Ax+By + C=0, its slope may be found by solving the equation
for y and taking the coefficient of x.
2. Suppose A = 0, B •=£ 0. The equation is then
By + C=0, or y = -|»
C
and represents a straight line parallel to OX at a distance
units from it.
3. Suppose A =£ 0, B = 0. The equation is then
Q
^ + (7=0, or x = ,
A
Q
and represents a straight line parallel to O Y at a distance
units from it.
Therefore the equation Az+By + C=Q always represents a
straight line.
32. Angles. The slope of a straight line enables us to solve
many problems relating to angles, some of which we take up in
this article.
1. The angle between the axis of x
and a known line. Let a known line
cut the axis of x at the point L.
Then there are four angles formed.
To avoid ambiguity we shall agree
to select that one of the four which
is above the axis of x and to the
right of the line and to consider LX as the initial line of this
angle. We shall denote this angle by <f>. Then if we take any
Fig. 66
60
THE STRAIGHT LINE
point P on the terminal line of <j> and drop the perpendicular
MP> we have, in the two cases represented by figs. 66 and 67,
tan<£ =
MP
LM
MP
But is equal to the slope of the
LM
line, by (2), § 6. Therefore
tan <f> = m.
If the straight line is parallel to
OF, <£ = 90° and tan<£ = oo. If the
line is parallel to OX, no angle <f>
is formed, but since m = 0, we may say tan <j> = 0 ; whence
<j> = 0° or 180°.
2. Parallel lines. If two lines are parallel, they make equal
angles with OX, and hence their slopes are equal. It follows that
two equations which differ only in the absolute term, such as
Fig. 67
and
Ax + By+C^O
represent two parallel lines. It is to be noticed that these two
equations have no common solution (§ 14).
Ex. 1. Find the equation of a straight line passing through (— 2, 3) and
parallel to3x— 5 y + 6 = 0.
First method. The slope of the given line is J. Therefore the required
line is
y - 3 = g (x + 2), or 3 x - 5 y + 21 = 0.
Second method. AVe know that the required equation is of the form
3x-5y + C = 0,
where C is unknown. Since the line passes through (— 2, 3),
3(-2)-5(3) + C = 0,
whence C = 21. Therefore the required equation is
3 x - 5 y + 21 = 0.
ANGLES
61
3. Perpendicular lines. Let AB and CD (fig. 68) be two lines
intersecting at right angles. Through P draw PR parallel to
OX, and let BPD^^ and BPB = <j>2. Then tan ^ = 7^ and
tan <j>% = m%i where rnx and m2 are the slopes of the lines. But,
^ = ^+90°;
tan <j>2 = — cot <f>t = —
1
by hypothesis,
whence
tan<£x
which is the same as
w2 = -
m.
That is, two straight lines are perpendicular when the slope of
one is minus the reciprocal of the slope of the other. This theorem
may be otherwise expressed
by saying that two lines are
perpendicular when the prod-
uct of their slopes is minus
unity.
It follows that two straight
lines whose equations are of
the type
*Ax + By + Cx = Q
and Bz — Ay + C2 = 0
are perpendicular. Fig. 68
Ex. 2. Find a straight line through (5, 3) perpendicular to 7 a: + 9 y + 1 = 0.
First method. The slope of the given line is — J. Therefore the slope
of the required line is ^. Therefore the required line is
y - 3 = $ (x - 5), or 9 x - 7 y - 24 = 0.
Second method. We know that the equation of the required line is of
the form 9x-7y + C=0. Substituting (5, 3), we find C = - 24. Hence
the required line is 9 x — 7 y — 24 = 0.
Ex. 3. Find the equation of the perpendicular bisector of the line join-
ing (0, 5) and (5, — 11). The point midway between the given points is
(if ~~ 3), by § 7. The slope of the line joining the given points is — V1,
by § 6. Hence the required line passes through (§, — 3), with the slope -fy.
Its equation is
y + 3 = ^(3:- f), or 10:r-32y-121 = 0.
62
THE STEAIGHT LINE
<f>2 tan j>x
4. Angle between two lines. Let AB and CD (fig. 69) inter-
sect at the point P, making the angle BPDy which we shall
0 call /3. Draw the line PR parallel to OX, and place RPB = <f>1
and RPD = 6. Then 0 . .
hence tan 8 = tan (<b0— d>,) = - ^xZ 2l
vv2 yv 1 + tan*, tan*
Bnttan^^andtan*.
= t»2, where wi2 is the slope
of CD and m1 is the slope of
AB. Therefore
1 + Wl^
If <f>2 is always taken
greater than <j>iy tan y8 will
be positive or negative
according as ft is acute
or obtuse.
Fig. 69
rw.
Ex. 4. Find the acute angle between the two lines
2s-3y+5=0 and s + 2y+2 = 0.
Since the second line makes the larger angle with OX, we place
= — h mi = §•
Then, by substituting in the formula,
tanj8="^""^=-
1 7
7
Here /3 is an obtuse angle, and the supplementary acute angle is tan-1 J.
Ex. 5. Find the equation of a straight line through the point (— 2, 0),
making an angle tan-1 § with the line 3a;+4y+6 = 0.
Here tan fl is given as §, and one of the slopes m2 or mx is known to
be — j. Since it is unknown which of the slopes is — J, the problem
has two solutions :
(1) Place m2 =— |. Then, by substituting in the formula,
17
-= ., *, lf .whence
O 1— jf 7/lj
mx=-
6
or
The equation of the required line is then
y-0=- V-(* + 2),
17 a: + 6y + 34 = 0.
DISTANCE FROM A STRAIGHT LINE
63
or
(2)
Place wij =
— _ 3
Then
2_
3
whence m2
1
18
The
equation
of the
required
line is
then
y
-0 =
~I5(Z
+ 2),
s + 18y
+ 2 =
0.
33. Distance of a point from a straight line. Let LK (fig. 70)
be a given straight line with the equation
Ax+By + C=0,
and let ij(#l9 y^) be a given point. It is required to find the
length of the perpendicular PXR
drawn from ij to LK
Draw the ordinate JfiJ and
let it intersect the line LK in
the point Q. Then the abscissa
of Q is xxi and its ordinate may
be denoted by y2. Since Q is on
the line LK, we have
Axx+By2+C=0,
Fig. 70
whence
Then
y2 =
QPx=y-y* =
Axx+C
B
Axx+Byx + C
B
It is clear that this expression is a positive quantity when
(Xj y^) lies above the line LK and is a negative quantity when
(xi> #i) ^es below LK It is also evident from the triangle PXQR,
and from a like triangle in other cases, that the length of PXR is
numerically equal to QI^ cos <£. But tan <£ = — --> and hence
B
cos<f> =
B
±V^2 + i?2
We have, then, PXR = J*i + B!ti + G.
±^A2+B2
64 THE STEAIGHT LINE
We may, if we wish, always choose the + sign in the denomi-
nator. Then PR is positive for all points on one side of the line
Ax + By + C = 0 and negative for all points on the other side.
To determine which side of the line corresponds to the positive
sign, it is most convenient to test some one point, preferably
the origin.
Ex. Find the distance from the point (7, — 4) to the line 2 x + 3 y
+ 8 = 0.
By use of the formula,
pje_2(7)+3(r4)+8^ 10
1 Vl3 Vl3
Since the coordinates of the origin, similarly substituted, give a positive
sign to the result, the point (7, — 4) is on the same side of the line as the
origin. A plot verifies this.
PROBLEMS
1. Find the equation of the straight line passing through (1, — 3)
with the slope 2.
2. Find the equation of the straight line passing through (— 1,
— J) with the slope — 3.
3. Find the equation of the straight line passing through (5, — 1)
with its slope the same as that of the straight line determined by
(0, 3) and (2, 0).
4. Find the equation of the straight line passing through (2, — f )
with the slope zero.
5. Find the equation of the straight line passing through (J, f)
with an infinite slope.
6. Find the equation of the straight line of which the slope is 5
and the intercept on OF is — 4.
7. Find the equation of the straight line of which the slope is — 3
and the intercept on OF is \.
8. Find the equation of the straight line of which the slope is 0
and the intercept on OY is — |.
9. Find the equation of the straight line through the points
(-1, -4) and (0, 5).
10. Find the equation of the straight line through the points
(2, -i) and (-1, J).
PROBLEMS 65
11. Find the equation of the straight line through the points
(2, -1) and (2, 3).
12. What is the equation of a straight line the intercepts of which
on the axes of x and y are 3 and — 4 respectively ?
13. What is the equation of the straight line the intercepts of
which on the axes of x and y are — 5 and — 8 respectively ?
14. Derive the equation of the straight line the intercepts of
which on the axes of x and y are a and b respectively.
15. Find the equation of a straight line through ($, §) and the
point of intersection of the lines 3sc — 5y— 11 = 0 and 4:X+y — 7 = 0.
16. Find the equation of the straight line joining the point of inter-
section of the lines 2x — y — 1=0 and x — y + 7 = 0 and the point
of intersection of the lines x — 7 y — 1=0 and 2x — 5y +1=0.
17. Find the equation of the straight line passing through (2, — 3)
and making an angle of 120° with OX.
18. Find the equation of the straight line making -an angle of 30°
with OX and cutting off an intercept 3 on OK
19. A straight line making a zero intercept on OF makes an angle
of 45° with OX. Find its equation.
20. A straight line making a zero angle with OX cuts OF at a
point 3 units from the origin. Find its equation.
21. Find the equation of the straight line through (2, — 3) parallel
to the line 2 x + y = 7.
22. Find the equation of the straight line through (— J, — 2)
parallel to the line 3x — 2y + 2 = 0.
23 . Find the equation of the straight line passing through (—1,-1)
parallel to the straight line determined by (— 2, 6) and (2, 1).
24. In the triangle ^(-2, -1), 5(3, 1), C(— 1, 4) a straight
line is drawn bisecting the adjacent sides AB and BC. Prove by
computation that it is parallel to AC and half as long.
25. Find the equation of the straight line passing through the
point of intersection of as — Sy + 2 = 0 and 5x + 6y — 4 = 0 and
parallel to Ax + y +7=0.
26. Find the equation of the straight line parallel to the line
x + 3y — 5 = 0 and bisecting the straight line joining (— 2, — 3)
and (5, 5).
66 THE STRAIGHT LINE
27. Find the equation of the straight line through the origin
perpendicular to the line 3 sc + 4 y — 1=0.
28. Find the equation of the straight line through (2, — 3) per-
pendicular to the line 1 x — 4 y + 3 = 0.
29. Find the equation of the perpendicular bisector of the straight
line joining the points (— 5, — 1) and (— 3, 4).
30. A straight line is perpendicular to the line joining the points
(— 4, 6) and (4, — 1) at a point one third of the distance from the
first point to the second. What is its equation?
31. Find the equation of the straight line perpendicular to
2x — 3 y + 7 = 0 and bisecting that portion of it which is included
between the coordinate axes.
32. Find the equation of the straight line through the point of
intersection of 6x — 2 y + 8 = 0 and 4# — 6 y + 3 = 0 and per-
pendicular to 5x + 2y + 6 = 0.
33. Find the equation of the perpendicular bisector of the base of
an isosceles triangle having its vertices at the points (4, 3), (— 1, — 2),
and (3, - 4).
34. Find the acute angle between the lines x — y + 4 = 0 and
3x-y + 6 = 0.
35. Find the acute angle between the lines 2a — y + 8 = 0 and
2x + 5y-4 = 0.
36. Find the acute angle between the lines x + y — 5 = 0 and
4a + y-8 = 0.
37. Find the acute angle between the line 3 a; — 2y + 6 = 0 and
the line joining (4, — 5) and (— 3, 2).
38. Find the acute angle between the straight lines drawn
from the origin to the points of trisection of that part of the line
2x + 3y — 12 = 0 which is included between the coordinate axes.
39. Show that x — y + 3 = 0 bisects one of the angles between
the lines 4s - 3y + 11 = 0 and 3x - 4y +10 = 0.
40. Find the vertices and the angles of the triangle formed by
the lines 3x + 5y -14 = 0, 9a? - y + 22 = 0, and x — y - 2 = 0
41. Find the equations of the straight lines through the poinjb
(— 3, 0) making an angle tan-1 J with the line 3x — 5 y + 9 = 0.
42. Find the equations of the straight lines through (4, —3)
making an angle of 45° with the line 3 x + 4 y = 0.
PROBLEMS 67
43. Find the equations of the straight lines through the point
(— 1, — 1) making an angle tan-1 1 with the line Sx + 2y — 6 = 0.
44. Find the equations of the straight lines through the point
(2, 1) making an angle tan-1 2 with the line 2x — y + 4 = 0.
45. Find the equations of the straight lines through the point
(3, 1) making an angle tan"1 3 with the line x + Sy — 3 = 0.
46. Find the distance of (2, 1) from the line y = 3 x + 7.
47. Find the distance of (2, — f) from the line x + 2y — 4 = 0.
48. Find the distance of the point (b, — a) from the line
bx + ay = ab.
49. The equations of the sides of a triangle are respectively
Sx + 5y - 16 = 0, x - y = 0, and Sx + y + 4 = 0. Find the dis-
tance of each vertex from the opposite side.
50. The base of a triangle is the straight line joining the points
(— 3, 1) and (5, — 1). How far is the third vertex (6, 5) from the base ?
51. The vertex of a triangle is the point (5, 3), and the base is
the straight line joining (— 2, 2) and (3, — 4). Find the lengths of
the base and the altitude.
52. Find the equations of the medians of the triangle formed by
the lines 2a? — Sy + 11 = 0, Sx + y — 11 = 0, and x + 4y = 0.
53. Find the foot of the perpendicular drawn from the point
(— 1, 2) to the line Sx - 5y - 21 = 0.
54 . Find the distance between the two parallel lines 2x+Sy— 8=0
and 2x + 3y- 10 = 0.
55. Find the distance between the two parallel lines 3 x— 5 y+l=0
and Sx — 5y — 7 = 0.
56. A triangle has the vertices (2, 4), (3, — 1), and (— 5, 3). Find
the distance from the vertex (2, 4) to the point of intersection of
the median lines.
57. A straight line is drawn through (2, — 3) perpendicular to the
line Sx — 4 y + 6 = 0. How near does it pass to the point (6, 8) ?
58. Determine the value of m so that the line y = mx ■+• 3 shall
pass through the point of intersection of the lines y = 2 x + 1 and
y = x + 5.
59. A straight line passes through the point (— J, 4), and its
nearest distance to the origin is 2 units. What is its slope ?
68 THE STRAIGHT LIKE
60. One diagonal of a parallelogram joins the points (3, — 1) and
(— 3, — 3). One end of the other diagonal is (2, 3). Find its equation
and its length.
61. Perpendiculars are let fall from the point (9, 5) upon the
sides of the triangle the vertices of which are at the points (8, 8),
(0, 8), and (4, 0). Show that the feet of the three perpendiculars
lie on a straight line.
62. Find a point on the line 2x + 3y — 6 = 0 equidistant from
the points (4, 4) and (6, 1).
63. Find a point on the line &x — 3y + 15 = 0 the distance of
which from the axis of x equals § its distance from the axis of y.
64. A point is equally distant from (3, 2) and (— 3, 4), and the
slope of the straight line joining it to the origin is $. Where is the
point ?
65. A point is 8 units distant from the origin, and the slope of the
straight line joining it to the origin is — J. What are its coordinates ?
66. A point is 5 units distant from the point (1, — 2), and the
slope of the line joining it to (0, — 8) is £. Find the point.
67. Find the points on the straight line determined by (1, 1) and
(— 2, — 3) which are 15 units distant from either of the given points.
68. Prove analytically that the locus of points equally distant
from two points is the perpendicular bisector of the straight line
joining them.
69. Prove analytically that the medians of a triangle meet in a
point.
70. Prove analytically that the perpendiculars from the vertices
of a triangle to the opposite sides meet in a point.
71. Prove analytically that the straight lines joining the middle
points of the adjacent sides of any quadrilateral form a parallelogram.
72. Prove analytically that the perpendicular bisectors of the sides
of a triangle meet in a point.
73. Prove analytically that the perpendiculars from any two ver-
tices of a triangle to the median from the third vertex are equal.
74. Prove analytically that the straight lines drawn from a vertex
of a parallelogram to the middle points of the opposite sides trisect
a diagonal.
CHAPTER VI
CERTAIN CURVES
34. Locus problems. A curve is often denned as the locus of
a point which has a certain geometric property. It is then usually
possible to obtain the equation of the curve by expressing this
property by means of an equation involving the coordinates of
any point of the locus. This is illustrated in the following
examples :
Ez. 1. Find the locus of a point at a distance 3 from the straight line
4ar + 3y-6 = 0.
Let (a:, y) be any point of the locus. By § 33, the distance of (x, y) from
the given straight line is ± — • Hence, by the conditions of the
problem,
± i -3,
which reduces to 4x + 3y — 21 = 0, or 4a; + 3y + 9 = 0.
These are the equations of two straight lines parallel to the given line.
Ex. 2. Find the locus of a point at a distance 9 from the point
(- 5, - 3).
Let .(a;, y) be any point of the locus. Its distance from (— 5, — 3) is,
by § 5, V(ar + 5)a + (y + 3)2. Hence, by the conditions of the problem,
V(z + 5)2 + (y + 3)2 = 9,
which reduces to x2 + y2 + 10 x + 6 y — 47 = 0.
This is the equation of the required locus. The curve may be plotted
from the equation or may be drawn with compasses, as it is obviously
a circle. •
In the following articles we shall employ the methods just
illustrated, to obtain the equations of certain important curves.
An equation thus obtained may be used both for plotting the
curve and for examining its properties.
ac 69
70
CERTAIN CURVES
35. The circle. A circle is the locus of a point at a constant
distance from a fixed point The fixed point is the center of the
circle, and the constant distance is the radius.
Let (A, k) be the center C (fig. 71), and let r be the radius of
the circle. Then if P(x, y) is a point on the circle, x and y must
satisfy the equation
(x-hy+(y-ky=r*, (1)
by § 5.
Conversely, if x and y satisfy
the equation (1), the point
(x, y) is at a distance r from
(A, k) and therefore lies on the
circle.
Therefore (1) is the equation
of the circle.
Equation (1) expanded gives Fig. 71
x*+ y2- 2 hx - 2 hy + A2+ k*- r2 = 0 ;
and if this is multiplied by any quantity A, it becomes
Ax2+Ay*+ 2 Gx + 2 Fy + C = 0, (2)
aw n
where
h = -A
A A
Ex. The equation of a circle with the center (J, — J) and the radius J is
(*-i)2 + (y + 4)2 = f,
which reduces to 12^ + 12y2-12x + 8y-l = 0.
36. Conversely, the equation
• Ax*+Ay*+ 2Gx + 2Fy + C= 0,
where A ¥= 0, represents a circle if it represents any curve at all.
To prove this we will follow the method of Ex. 2, § 18, and
write the equation in the form
F\* G2+I*-AC
(*+f)a+4+f)a=
THE CIRCLE 71
There are then three possible cases:
1. G2+F2-AC>0. The equation is then of the type (1),
§ 35, where A = — -> k = — -, r2 = , and therefore
A A A
represents a circle with the center ( — -, ) and the radius
, — - — —- \ A A
\G*+F*-AC
N a
2. G2+F*- -4(7=0. The equation is then
(-IH
y+f)'-o,
which can be satisfied by real values of x and y only when
G F
x = — - and y = — -- Hence the equation represents the point
( j? — -J. This may be called a circle of zero radius, regarding
it as the limit of a circle as the radius approaches zero.
3. G2 + F2 — AC < 0. The equation can then be satisfied
by no real values of x and yy since the sum of two positive
quantities cannot be negative. Hence the equation represents
no curve.
:. 1. The equation 2sa + 2y2 + 2a; — 2y— 5 = 0 may be written
(*+i)* + (y-i)a = 3,
and represents a circle with the center (— i, i) and the radius v3. This
circle can now be drawn with compasses, the methods of Chapter II not
being required.
c 2. The equation a^ + y2 — 2a; + 4y + 5 = 0 may be written
(:r-l)2+O + 2)2 = 0,
and is satisfied only by the point (1, — 2).
Ex. 8. The equation aj' + y8 — 2ai + 4y + 7=0 may be written
(*-l)2 + (y + 2)2=-2,
and represents no curve.
72 CERTAIN CURVES
37. To find the equation of a circle which will satisfy given
conditions, it is necessary and sufficient to determine the three
quantities A, &, r, or the ratios of the four quantities -4, 6r, F, C.
Each condition imposed upon the circle leads usually to an equa-
tion involving these quantities. In order to determine the three
quantities it is necessary and in general sufficient to have three
equations. Hence, in general, three conditions are necessary and
sufficient to determine a circle.
It is not important to enumerate all possible conditions which
may be imposed upon a circle, but the following three may be
mentioned.
1. Let the condition be imposed upon the circle to pass
through the known point (x^ y^). Then (xx, y^) must satisfy
the equation of the circle ; therefore A, &, and r must satisfy the
condition
(*,_*)■ + (*-*)* = *
2. Let the condition be imposed upon the circle to be tangent
to the known straight line Ax + By + C = 0. Then the distance
from the center of the circle to this line must equal the radius ;
therefore, by § 33, A, &, and r must satisfy the condition
Ah+Bk + C
— =±r.
y/A2+B2
The sign will be ambiguous unless from other conditions of the
problem it is known on which side of the line the center lies.
3. Let it be required that the center of the circle should lie on
the line Ax + By + C = 0. Then A and k must satisfy the condition
Ah + Bk+C=0.
Ex. 1. Find the equation of the circle through the three points (2, — 2),
(7, 3), and (6, 0).
The quantities h, h, and r must satisfy the three conditions
(2 - hy + (- 2 - *)• = r*,
(7- A)1 + (3 -*)* = **
(6 - hf + (0 - k)2 = r».
THE CIECLE 73
Solving these, we have h = 2, k = 3, and r = 5. Therefore the required
equation is
(x - 2)a + (y - 3)2 = 25, •
or x2 + y2 - 4 a: - 6 y - 12 = 0.
■
Ex. 2. Find the equation of the circle which passes through the points
(2, — 3) and (—4, — 1) and has its center on the line 3 y + x — 18 = 0.
The quantities h, k, and r must satisfy the conditions
(2 - h)2 + (- 3 - k)2 = r3,
(-4-A)2 + (-l-l:)2 = ra,
. 3 k + A - 18 = 0.
Solving these equations, we find h = |, & = Jg^, r2 = ^f A. Therefore
the required equation is
(x - |)» + (y - ->,L)» = -4*.
or as + ys-3x-lly-40 = 0.
%
Ex. 3. Find the equation of a circle which is tangent to the lines
17a; + y- 35 = 0 and 13a; + Uy + 50 = 0,
and has its center on the line 88 x + 70 y + 15 =0.
The quantities h, k, and r must satisfy the conditions
17 A + £-35 ,
- = ± r,
V290
13 h + 11 k + 50
= ±r,
V290
88 A + 70 k + 15 = 0.
These equations have the two solutions
5 7 * ^290
a * k ^ ,3 3V290
and A = 5, £=— -^, r- — — —
Hence each of the two circles
3a? + 3y2 + 5a:-5y-20 = 0
and 40 x2 + 40 y2 - 400 x + 520 y + 2429 = 0
satisfies the conditions of the problem.
74
CEBTAIN CUEVES
38. The ellipse. An ellipse is the locus of a point the sum of
the distances of ythich from two fixed points is constant.
The two fixed points are called the foci. Let them be denoted
by F and Ff (fig. 72), and let the axis of # be taken through
them, and the origin halfway
between them. Then if P is
any point on the ellipse and
2 a represents the constant
sum of its distances from the
foci, we have
FlP+FP = 2a. (1)
From the triangle FrPF it
follows that
FfF< 2 a.
Hence there is a points on the axis of x and to the right
of F which satisfies the definition. We have, then,
F'A+FA = 2a,
or (FrO + OA) + QOA - OF)= 2 a;
whence OA = a.
OF
Let us now place
OA
= e, where e < 1.
The quantity e is called the eccentricity of the ellipse.
Then the points F and F1 are (± ae, 0). Computing the values
of F'P and FP by § 5 and substituting in (1), we have
V(x + aef+ y2+ V(z - ae)2+ /= 2 a.
(2)
By transposing the second radical to the right-hand side of the
equation, squaring, and reducing, we have
a — ex = V(a;--ae)2+y2 = FP.
(3)
Similarly, by transposing the first radical in (2), we have
a + ex =y/(x + ae)2+y* = F'P. (4)
THE ELLIPSE 75
Either (3) or (4) leads to the equation
(l_e2)^+/ = a2(l_e*), (5)
OT ^+ %,/ 2X=1' (6)
Since e < 1, the denominator of the second fraction is positive,
and we place
y as(l-0 = J2, (7)
thus obtaining "l+jj^l* (8)
We have now shown that any point which satisfies (1) has
coordinates which satisfy (8).
We may show, conversely, that any point whose coordinates satisfy (8)
is such as to satisfy (1). Let us assume (8) as given. We can then obtain
(6) and (5), and (5) may be put in each of the two forms
x2 + 2 aex + a2e2 + y2 = a2 + 2 aex + e2x2,
x2 — 2 aex + a2e2 + y2 = a2 — 2 a&r + A2,
the square roots of which are respectively
F'P = ± (a + ez),
i^P = ± (a - ear).
These lead to one of the four following equations :
F'P + FP = 2a,
F'P-FP = 2a,
-F'P + FP = 2a,
-F'P-FP = 2a.
Of these, the last one is impossible, since the sum of two negative num-
bers cannot be positive ; and the second and third are impossible, since the
difference between FP and F'P must be less than F'F, which is less than 2 a.
Hence any point which satisfies (8) satisfies (1), and therefore (8) is the
equation of the ellipse.
39. Placing y = 0 in (8), § 38, we find x = ± a. Placing x = 0,
we find y = ± b. Hence the ellipse intersects OX in two points,
A (a, 0) and Af(— a, 0), and intersects OF in two points, i?(0, 6)
and 2?'(0, — 6). The points A and Al are called the vertices of the
76
CEETAIN CUEVES
ellipse. The line AA\ which is equal to 2 a, is called the major
axis of the ellipse, and the line BBf, which is equal to 2 J, is
called the minor axis.
Solving (8) first for y and then for #, we have
y = ±-Va2— a?
a
and
x
=±?vF=y.
These equations show that the ellipse is symmetrical with
respect to both OX and OF, that x can have no value numeri-
cally greater than a, and that y can have no value numerically
greater than b. If we construct
the rectangle KLMN (fig. 73),
which has 0 for a center and
sides equal to 2 a and 2 b
respectively, the ellipse will
lie entirely within it; and if
the curve is constructed in one
quadrant, it can be found by
symmetry in all quadrants.
The form of the curve is shown
in figs. 72 and 73.
40. Any equation of the form (8), § 38, in which a > 6, repre-
sents an ellipse with the foci on OX. For if we place, as in § 38,
62= a2(l— 62), we find, for the eccentricity of the ellipse,
l
\f
B
L
r
~\
A'
^
'J
A '
1
V
B'
K
Fig. 73
e =
V^-J5
a
and may fix F and F\ which in § 38 were arbitrary in position,
by the relation OF = — OF1 = ae.
The foci may be found graphically by placing the point of a com-
pass on B and describing an arc with the radius a. This arc will
intersect A A1 in the foci ; for since OB= b and OF= Va2— b\ BF= a.
It may be noted that the nearer the foci are taken together, the
smaller is e and the more nearly b = a. Hence a circle may be
considered as an ellipse with coincident foci and equal axes*
THE HYPERBOLA 77
Similarly, an equation of the form (8), § 38, in which b > a,
represents an ellipse in which the foci lie on OF at a distance
V62— a2 from 0. In this case BBf= 2 b is the major axis and
AAf = 2 a is the minor axis.
Finally, any equation of the form
a' +~vr~-1
represents an ellipse with its center at the point (A, K) and its
axes parallel to OX and QY respectively; for if the axes are
shifted to a new origin at (A, &) by the formulas of § 17, this
equation assumes the form (8), § 38.
Ex. 1. Show that 4a? + 6y8 + 4x — 12 y — 1 = 0 is the equation of an
ellipse, and find its center, semiaxes, and eccentricity.
Following the method of Ex. 2, § 18, we may .write the equation in
2 J
Hence this curve is an ellipse with its center at (— £, 1) and its major
and minor axes equal respectively to 2 v2 and — - — Its eccentricity is —= •
3 V3
Ex. 2. Find the equation of an ellipse with the eccentricity J and its
foci at the points (— 1, 4), (7, 4).
Since the center is halfway between the foci, the center is the point
(3, 4). The major axis of the ellipse is parallel to OX, since it contains
the foci. Since each focus is at a distance ae from the center,
ae = 4.
But e = ^, therefore a = 12.
Then, from (7), § 38, 52 = a2 (1 - e2) = 128.
The equation of the ellipse is therefore
0-3)2 (y-4)» ■
144 128
which reduces to 8 s2 + 9 y2 - 48 x - 72 y - 936 = 0.
41. The hyperbola. An hyperbola is the locus of a point the
difference of the distances of which from two fixed points is constant.
78
CERTAIN CURVES
The two fixed points are called the foci Let them be F and
F* (fig. 74), and let FFr be taken as the axis of x, the origin
being halfway between F and F*. Then if P is any point on
the hyperbola and 2 a is the constant difference of its distances
from F and FrJ we have
either
or
F'P-FP=2a (1)
FP-F'P = 2a. (2)
Since in the triangle
F'PF the difference of the
two sides FP and F'P is
less than F'F, it follows
that F*F >2a.
There is therefore at least one point A between 0 and F
which satisfies the definition.
Fig. 74
Then
F'A-
-AF=2a,
or
whence
(FO + OA)-
(OF-
OA)=2ai
OA=a.
We may
therefore place
OF
e, where e > 1.
OA
The quantity e is called the eccentricity of the hyperbola.
Then the points F and Ff are (± ae, 0), and equations (1)
and (2) become
y/(x + aey+ y2- y/(x -aey+f=2a (3)
(4)
and
VO - aey+ y2- y/(x + aef+ y2 = 2 a.
By transposing one of the radicals to the right-hand side of
these equations, squaring, and reducing, we obtain from either
(3) or (4) (\-e^+f = a\l- e2), (5)
or
f _
^ I y 1
a2 a2(l-e2)
(6)
THE HYPEKBOLA
79
But since e > 1, a3 (1 — ea) is a negative quantity, and we
may write a3 (1 — e3) = — 6s, thus obtaining
a?
&=!•
m
an equation satisfied by the coordinates of any point which
satisfies (1).
Proceeding as in § 38, we may prove, conversely, that any point whose
coordinates satisfy (7) is such as to satisfy either (1) or (2), and hence is
a point of the hyperbola.
42. If we place y = 0 in (7), § 41, we have x= ± a. Hence
the curve intersects OX in two points, A and A', called the vertices.
If x = 0, y is imaginary. Hence the curve does not intersect 0 Y.
Solving (7), | 41, for y and x respectively, we have
b
and x-
These show that the curve is
symmetrical with respect to both
OX and OY; that x can have no
value numerically less than a, and
that y can have all values.
Moreover, the equation for y can
be written
6 \—£
As x increases, the term —^ decreases, approaching zero as a
limit Hence the more the hyperbola is prolonged, the nearer it
comes to the straight lines y = ± - x. Therefore the straight
lines y = ±-x are the atymptotes of the hyperbola. They are
the diagonals of the rectangle constructed as in fig. 75 and are
used conveniently as guides in drawing the curve. The line AA'
80 CERTAIN CURVES
is called the transverse axis, and the line BB' the conjugate axis,
of the hyperbola. The shape of the curve is shown in figs. 74
and 75.
43. Any equation of the form (7), § 41, where a and b are
any positive real values, represents an hyperbola with the
foci on OX. For if we place — #2 = a2(l— e2), we find for
the eccentricity of the hyperbola
Va*+ b2
e = ,
a
and may find the position of the foci from the equations
OF = -OF'=ae.
Similarly, any equation of the form
— — + £- = 1
a2^b2
represents an hyperbola with the foci on 0 Y.
If the two hyperbolas,
-2-C=l and -^ + 1 = 1
ar V a2 ¥
have the same values for a and J, each is said to be the conjugate
hyperbola to the other.
If b = a, the hyperbola is called an equilateral hyperbola, and its
equation is either x2— y2 = a2 or — a^-f y2 = a2.
Finally, it is evident that either
(x-hy (y-ky_t
a2 b2
Cx-h)2 , (y-k)2 .
or -^ — H-+ i2 =1
a2 b2
is the equation of an hyperbola with its center at the point (A, K).
44. The parabola. A parabola is the locus of a point equally
distant from a fixed point and a fixed straight line. The fixed
point is called the focus and the fixed straight line the directrix.
Let the line through the focus perpendicular to the directrix
THE PARABOLA
81
be taken as the axis of x, and let the origin be taken on this
line, halfway between the focus and the directrix. Let us denote '
the abscissa of the focus by p. In fig. 76 let F be the focus, RS
the directrix intersecting OX at D, and P
any point on the curve. Then F is (jp, 0),
£> is (— p, 0), and the equation of RS is
x — — p. Draw from P a line parallel to
OX, intersecting RS in N. If F is on the
right of RS, P must also lie on the right
of RS, and, by the definition,
SY
rt
<?
D
\F
FP = NP.
R
Eig. 76
If, on the other hand, F is on the left of RS, P is also on the
left of S& and „^ ^,^ „„
v . ^ "•"** FP PIP NP
In either case
FP =NP .
But, by § 5,
FP* = (%-£)*'+]?,
and
NP=x+p;
hence
(X-
-py+tf=(*+py,
which reduces to
y* = 4tpx.
(1)
Any point on the parabola then satisfies equation (1).
Conversely, it is easy to show that if a point satisfies (1), it
lies so that FP=±NP, and hence lies on the parabola.
Equation (1) shows that the curve is symmetrical with
respect to OX, that x must have the same sign as p, and that y
increases as x increases numerically. The position of the curve
is as shown in fig. 76 when p is positive. When p is negative, F
lies at the left of 0, and the curve extends toward the negative
end of the axis of x.
Similarly, the equation x2 = 4 py represents a parabola for
which the focus lies on the axis of y, and which extends
toward the positive or the negative end of the axis of y
according as p is positive or negative. In all cases 0 is
called the vertex of the parabola, and the line determined
by 0 and F is called its axis.
82
CERTAIN CURVES
A more general equation of the parabola is evidently
or (x — A)2= 4 p (y — A),
the vertex in either case being at the point (A, &). The work of
locating the parabola in the plane is illustrated in the following
example.
Ex. Show that y2 + y — 3 a: + 1 = 0 is a parabola, and locate it in
the plane.
The equation may be written
y2 + y = 3 x — 1,
or y2 + y + \ = 3 x - 1 + J,
which reduces to (y + J)2 = 3 (x — J).
Hence the vertex is at the point (J, — £); the equation of the axis is
y + J = 0, or 2 y + 1 = 0 ; the focus is at the point ( J + J, — J), or (1, — J);
and the equation of the directrix is a:— £=— J, or 2 a: + 1 = 0.
45, If .? (:cp yx) and i£(#2, y2) are two points on the parabola
tf = ±px (fig. 77), then
#? = 4^2;
2£_S -
whence
Vi
xn
a)
which may be written
#„
(2)
Fig. 77
if both numerator and denominator of the left-hand fraction
are multiplied by 4.
From the symmetry of the parabola, 2 yx = QXI{ and QJE^ = 2 ya;
and since xx= Oilfj and za = 0Jfa, (2) becomes
Q1P^ = OM1
(3)
That is, t he squares of any two chords of a parabola which are
perpendicular to the axis of the parabola are to each other as their
distances from the vertex of the parabola*
THE CONIC
83
The figure bounded by the parabola and a chord perpendic-
ular to the axis of the parabola, as QxOPx (fig. 77), is called a
parabolic segment The chord is called the base of the segment,
the vertex of the parabola is called the vertex hi the segment,
and the distance from the vertex to the base is called the
altitude of the segment.
46. The conic. A conic is the locus of a point the distance of which
from a fixed point is in a constant ratio to its distance from a fixed
straight line.
The fixed point is called the focus,
the fixed line the directrix, and the
constant ratio the eccentricity.
We shall take the directrix as the
axis of y (fig. 78), and a line through
the focus F as the axis of x, and
shall call the focus (c, 0), where c
represents OF and is positive or
negative according as F lies to the
right or the left of 0.
Let P be any point on the conic ;
connect P and F, and draw PN per-
pendicular to 0 Y. Then, by definition,
FP = ±e-NP, (1)
according as P is on the right or the left of OY. In both cases
FP2=e2.NP*.
But FP*=(z-cy+if, by § 5, and NP = x. Therefore for
any point on the conic
(x-cy+tf=e*x*. (2)
It is easy to show, conversely, that if the coordinates of P sat-
isfy (2), P satisfies (1). Hence (2) is the equation of the conic.
It is clear that the parabola is a special case of a conic, for the
definition of the latter becomes that of the former when e = 1.
It is also not difficult to show that the ellipse is a special case
of a conic, where the eccentricity is e of § 38 and < 1.
-
N
r
/
/
A
0
\ r
\
\
\
\
\
\
\
\
\
\
\
\
Fig. 78
84
CERTAIN CURVES
For if P (fig. 79) is a point on the ellipse ~2 + %
in § 38 that a b
F'P = a + ez;
= 1, we found
FP = a — ez,
or
FP = e(l-z\ F'P = e(- + z\
If now we take the point
D so that 0Z> = -> and D'
e
so that ODf = , and if we
e
draw the lines JDS and D'S'
perpendicular to OX, the
line N'PN perpendicular to
JDS, and the ordinate MP,
we have
a
Fig. 79
Therefore
--x = OD-OM=MD= PN,
e
- + z = D,0+OM=D,M=]VfP.
e
FP = e- PN, F'P = e . N'P.
a
The ellipse has, therefore, two directrices at the distances ± -
e
from the center. When the ellipse is a circle, e = 0 and the
directrices are at infinity.
In a similar manner we may show that the hyperbola is a
special case of a conic where e > 1.
47. The witch. Let OB A (fig. 80) be a circle, OA a diameter,
and LK the tangent to
the circle at A. From 0
draw any line intersect-
ing the circle at B and
LK at C. From B draw
a line parallel to LK and
from C a line perpendic-
ular to LK, and call the intersection of these two lines P.
The locus of P is a curve called the witch.
Fig. 80
THE CISSOID
85
To obtain its equation we will take the origin at 0 and the
line OA as the axis of y. We will call the length of the diameter
of the circle 2 a. Then, by continuing CP until it meets OX at
M and calling (x, y) the coordinates of P, we have
OM=z, MP = y, 0A=MC=2a.
MP OB OB . OC
In the triangle OMC,
MC OC
OC
(1)
Draw AB. Then OB A is a right angle, and consequently
OB-OC = OA*; also OCf= 0M2+ MC2.
Therefore
MP
¥-,:<*>
that is,
and finally,
MC OM+MC
2 a x*+±a2
8 a8
y =
^+4 a2
Solving (4) for a, we have
x=±2aJ*I
(3)
(4)
-y.
y
This shows that the curve is sym-
metrical with respect to OF, that y
cannot be negative nor greater than
2 a, and that y = 0 is an asymptote.
48. The cissoid. Let ODA (fig. 81)
be a circle with the diameter OA,
and let LK be the tangent to the
circle at A. Through 0 draw any
line intersecting the circle in D and
LK in K On OE lay off a distance
6>P, equal to DE. Then the locus of
P is a curve called the cissoid.
To find its equation we will take 0 as the origin of coordi-
nates and OA as the axis of #, and will call the diameter of the
circle 2 a. Join A and D and draw JfP perpendicular to OA.
AG
Fig. 81
86
CERTAIN CUKVES
Denoting angle MOP by 0, we have
OE=2asec0,
OD =2acos0;
whence DE = 0E-0D=2a (sec 0 - cos 0).
Therefore OP = 2 a (sec 0 - cos 0*).
Now x = 0M= OP cos 0
= 2a(l-cos20)
= 2asin20.
But
__MP __ y
sin#=-- — =
Substituting in (5), we have
OP Vz? +
It
_2atf
whence
f=
x
2 a — x
This equation is satisfied by the
coordinates of any point upon the
cissoid. It may be written
x
y=±*>fcr
x
From this it appears that the
curve is symmetrical with respect
to OX, that no value of x can be 2
greater than 2 a or less than 0,
and that the line x = 2 a is an
asymptote.
49, The strophoid. Let LK and
RS (fig. 82) be two straight lines
intersecting at right angles at 0,
and let A be a fixed point on LK.
Through A draw any straight line
intersecting BS in D, and lay off
on AD in either direction a distance DP equal to OD.
locus of P is a curve called the strophoid.
a)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
The
THE STKOPHOID 87
To find its equation take LK as the axis of x and ES as the
axis of y, and let OA = a. By the definition, the point P may
fall in any one of the four quadrants. If we take the positive
direction on AD as measured from A towards D, we have
OD = PD
when P is in the first quadrant,
OD=-PD
when P is in the second quadrant,
-OD = -PD
when P is in the third quadrant, and
-OD = PD
when P is in the fourth quadrant.
These four equations are equivalent to the single equation
OD =PD. (1)
Draw PM parallel to 0 Y and denote the angle MAP by 0.
Then OD = aton0 * (2)
and PD = x sec 0. (3)
Substituting in (1), we have
a2 tan2 0 = x2 sec2 0. (4)
But tan0 = ^ = -£--. (5)
MA a — x
Substituting in (4), we have
which may be reduced to y = ±x x (7)
ya + x
This shows that the curve is symmetrical with respect to OX,
that no value of x can be less than — a or greater than + a,
and that x = — a is an asymptote.
88
CERTAIN CURVES
50. Use of the equation of a curve. The use of the equation
of a curve in solving geometrical problems is illustrated in
the following problems:
Ex. 1. Prove that in the
ellipse the squares of the
ordinates of any two points
are to each other as the
products of the segments of
the major axis made by the
feet of these ordinates.
We are to prove that (fig. 83)
Mj>? _ A'MX • MXA
MoP
2* 2
* A' Mo • Mo A
Let Px be (xv yx) and let
P2 be (*2> %)• Then
Fig. 83
whence
b2
i2_
h2
«2 n'2 _r2 (n J. t \( n — t \
V2 a"-*a- (a + ^2)(«-g2)
But yx = MXPV a + xx = A'O + OMx = ii'Jfp a-xl = OA- OMx = MXA9
y2 = M2P2, a + x2 = -4'3/2, a — x2 = M2A. Hence the proposition is proved.
Ex. 2. If MXPX is the ordinate of a
point Px of the parabola y2 = 4joj:, and
a straight line drawn through the middle
point of MXPX parallel to the axis of x
cuts the curve at Q, prove that the inter-
cept of the line MXQ on the axis of y
equals %MXPV
Let Px (fig. 84) be (xv yx). Then
y2
a\ = — , from the equation of the
4/)
parabola.
By construction, the ordinate of Q
is ^- Since Q is on the parabola, its
abscissa is found by placing y = ^ in y2 = 4y?x. Then Q is ( — — > ^ ) ; and
/ y2 \ \ r r
M is (a?!, 0), which is the same as ( — *-> 0 J. Hence the equation of AftQ
is, by § 30, *? I
Spx + Zyiy-2y2 = 0.
The intercept of this line on 0 Y is § yx = § MXPV which was to be proved.
Fig. 84
EMPIRICAL EQUATIONS 8d
51. Empirical equations. We have met in § 8 examples of
related quantities for which pairs of corresponding values have
been found by experiment, but for which the functional relation
connecting the quantities is not known. In such a case it is often
desirable to find an equation which will represent this relation,
at least approximately. The method, in general, is to plot the
points as in § 8 and then fit a curve to them. At best this work
is approximate, the result depending largely on the judgment of
the worker, and in complicated cases it demands methods too
advanced for this book. We shall discuss a few simple exam-
ples, to illustrate merely the fundamental principles involved.
The simplest case is that in which the plotted points appear
to lie on a straight line, or nearly so. If the two related quan-
tities are x and y, the relation between them is expressed by the
equation y = mx + b, (1)
where m and b are to be determined to fit the data. In practice
the points are plotted, and it appears that a straight line may be
so drawn that the points either lie on it or are close to it and
about evenly distributed on both sides of it. The straight line
having been drawn, its equation may be found by means of two
points on it, which may be either two of the original data or any
two points of the graph. This method is illustrated in Ex. 1.
Closely connected with this case are two others. Suppose the
relation between the two quantities x and y is known or assumed
to be of a, fom y,rf (2)
or y = abx, (3)
where a, b, and n are to be determined to fit the given numerical
values of x and y. By taking the logarithms of both sides of
these equations, we have respectively
log y = n log x + log a (4)
and log y = (log 6) x + log a ; (5)
or, if we place log y = y\ log x = x\ log a = br, log b = m,
y' = nx' + V, (6)
y' = mx + b'. (7)
90
CERTAIN CURVES
We may now plot the points (a^, y'~) or (x, y1') and determine
the straight line on which they lie approximately. The equations
(6) and (7) having thus been found, the return to equations
(2) and (3) is easy. This method is illustrated in Ex. 2.
When the use of a straight line either directly or by aid of
logarithms fails, the attempt may be made to fit a parabola
y=a + hx + c2? (9)
to the points of the plot. Since three points are sufficient to de-
termine the constants of the equation, the parabola may be made
to pass through any three of the plotted points. This parabola
may then be tested to see if it passes reasonably near to the
other points. This method is illustrated in Ex. 3.
Other curves with equations of the form
y = a + bx + C3?+d3?-\ \-hf
may also be used. In this case the number of points through
which the curve may be exactly drawn is equal to the number
of arbitrary coefficients.
In all these cases it is often convenient to use different scales
for x and y, the proper allowance being made in the calculations.
This is illustrated in Ex. 2.
Ex. 1. Corresponding values of two related quantities i
given by the following table :
and y i
*
1
2
4
6
10
V
1.3
2.2
2.0
3.0
6.1
Find the empirical equation
connecting them.
We plot the points (x, y)
and draw the straight line as
shown in fig. 85. The straight
line is seen to pass through the
points (0, 1) and (2, 2). Its
equation is therefore
which is the required equation.
________^=£==
_^
~0~ '
EMPIRICAL EQUATIONS
Ex. 2. Corresponding values of pressure and volume taken from i
indicator card of an air compressor are as follows :
p
IS
21
26.5
33.5
44
62
V
.635
.556
.475
.397
.321
.243
Find the relation between them in the form pv" = c.
Writing the assumed relation in the form p =
logarithms of both sides of the equation, we have
Thee
y = logp, t = log ii, and b
irresponding values of x and y are
x = logo
- .1972
- .2549
-.8233
- .4012
-.4935
- .6144
j, = logp
1.2553
1.3222
1.4232
1.5250
1.6435
1.7824
We assume on the x-axis a scale twice as large as that on the y-asis, plot
the points (x, y), and draw the straight line as shown in fig. 86. The
construction should be made
on large-scale plotting paper.
The line is seen to pass through
the points (—.05, 1.075) and
(— .46, 1.6). Its equation is
therefore
y = - 1.28 s + 1.01.
Hence n = 1.28, log c = 1.01,
c = 10.2, and the required re-
lation between p and v is
\
N,
K
r_
El. >. Corresponding values of two related quantities
given by the following table :
V
1
2
3
4
5
1.37
.(18
.41
.54
1.05
Find the empirical equation connecting them.
92
CERTAIN CURVES
If we plot the points as in fig. 87, they suggest a parabola. Accordingly
we assume , » . o
y = a + bx + car
and determine a, b, and c, so that the curve will pass through the first,
third, and last points. The equations for a, b, and c are
1.37 = a + b + c,
.41 = a + 3 6 + 9 c,
1.05 = a + 5 b + 25 c ;
whence a = 2.45, 6 = — 1.28, and
c = .2. The required equation
is therefore
y = 2.45 - 1.28 x + .2 x2.
5 i I —^
Fig. 87
If we substitute for x in this equation the values 2 and 4, we find the
corresponding values of y to be .69 and .53. This shows that the curve passes
reasonably near to the points of the plot which were not used in computing
the coefficients.
PROBLEMS
1. Find the equations of the locus of a point the distance of which
from the axis of x equals five times its distance from the axis of y.
2. Find the equations of the locus of a point the distance of which
from the axis of re is 3 more than twice its distance from the axis of y.
3. Find the equation of the locus of a point the distances of which
from (2, — 1) and (— 3, 2) are equal.
4. Find the equations of the locus of a point equally distant
from the lines 3x — 5y — 15 = 0 and 5x — 3?/ + l = 0.
5. Find the equations of the bisectors of the angles between the
lines 9a;-f2?/-3 = 0and 7x-6y + 2 = 0.
6. Find the equations of the bisectors of the angles between the
lines 3a + 4?/— 7=0 and 12a: — 5y +1 = 0.
7. A point moves so that its distance from the axis of y equals
its distance from the point (5, 0). Find the equation of its locus.
8. Find the equation of the locus of a point the distance of
which from the axis of x is one half its distance from (0, 2).
9. A point moves so that the square of its distance from the
point (0, 3) equals the cube of its distance from the axis of y. Find
the equation of its locus.
PKOBLEMS 93
10. Find the equation of the locus of a point the distance of which
from the line x = 3 is equal to its distance from (4, — 2).
11. Find the equation of the locus of a point which moves so that
the slope of the straight line joining it to (a, a) is one greater than
the slope of the straight line joining it to the origin.
12. A point moves so that its distance from the origin is always
equal to the slope of the straight line joining it to the origin. Find
the equation of its locus.
13. Find the equation of the locus of a point the distance of which
from the line 3sc + 4y — 6 = 0is twice its distance from (2, 1).
14. Find the equation of the circle having the center (3, — 5) and
the radius 4.
15. Find the equation of the circle having the center (— f , $) and
the radius 2.
16. Find the points at which the axis of x intersects the circle
having as diameter the straight lin'e joining (1, 2) and (— 3, — 4).
17. Find the equation of the circle having as diameter that part
of the line 3se — 4y + 12 = 0 which is included between' the coor-
dinate axes.
18. Find the equation of the circle having as diameter the common
chord of the two circles
x2+y2+±x-±y-2 = 0 and x*+ y2- 2x + 2y - 14 = 0.
19. Find the equations of the circles of radius a which are tangent
to the axis of y at the origin.
20. Find the center and the radius of the circle
x* + ^+26aj + 16y-42 = 0.
21. Find the center and the radius of the circle
2x* + 2y* + 6x + Sy - 10 = 0.
22. Find the equation of a straight line passing through the cen-
ter of the circle cr2-|-?/a — 4aj + 2y — 5 = 0 and perpendicular to
the line x — 2 y + 1 = 0. How near the origin does the line pass ?
23. Prove that two circles are concentric if their equations differ
only in the absolute term.
94 CERTAIN CURVES
24. Show that the circles x2+ y* + 2 Gx + 2Fy + C = 0 and
a? + tf+2G'x + 2F'y + C'=0 are tangent to each other if
V(G-G')a+(F--F')2 = VG2 + JFT2-C±VG'3 + JF'a-C'.
25. Find the equation of the circle which passes through the
points (0, 2), (2, 0), and (0, 0).
26. Find the equation of the circle circumscribing the triangle
with the vertices (0, 1), (- 2, 0), and (0, - 1).
27. Find the equation of the circle circumscribing the isosceles
triangle of which the altitude is 5 and the base is the line joining
the points (—4, 0) and (4, 0).
28. Find the equation of the circle circumscribed about the tri-
angle the sides of which are x + 2y — 3 = 0, 3 a; — y — 2 = 0, and
2a;-3y-6 = 0.
29. Find the equation of the circle passing through the point
(— 3, 4) and concentric with the circle cc2 + 2/2 + 3x — 4y— 1=0.
30. A circle which is tangent to both coordinate axes passes
through (4, — 2). Find its equation.
31. Tl\e center of a circle which is tangent to the axes of x and
of y is on the line Sx — 5y +15 = 0. What is its equation ?
32. A circle of radius 5 passes through the points (4, — 2) and
(5, — 3). What is its equation ?
33. The center of a circle which passes through the points (— 2, 4)
and (— 1, 3) is on the line 2x — 3 y + 2 = 0. What is its equation ?
34. A circle which is tangent to OX passes through (—1, 2) and
(6, 9). What is its equation ?
35. The center of a circle which is tangent to the two parallel
lines a; — 2 = 0 and x — 6 = 0 is on the line y = 3 x — 6. What is
its equation ?
36. The center of a circle is on the line 2x + y + 3 = 0. The
circle passes through the point (3, 1) and is tangent to the line
4a; — 3y— 14 = 0. What is its equation ?
37. The center of a circle is on the line a; + 2y — 10 = 0 and the
circle is tangent to the two lines 2a; — 3y + 9 = 0 and 3a; — 2y + 1 = 0.
What is its equation ?
38. Given the ellipse 9 x2 + 25 y2 = 225, find its semiaxes,
eccentricity, and foci.
PROBLEMS 95
39. Given the ellipse 3 x2 + 4 y2 = 2, find its semiaxes, eccen-
tricity, and foci.
40. Find the vertices, eccentricity, and foci of the ellipse
41. Find the center, vertices, eccentricity, and foci of the ellipse
4:x2 + 9y2+16x-18y-ll=0.
42. Find the center, vertices, eccentricity, and foci of the ellipse
16a3 + 9y* -16z + 6y -139 = 0.
43. Find the equation of the ellipse when the origin is at the
left-hand vertex and the major axis lies along OX.
44. Find the equation of the ellipse when the origin is taken at
the lower extremity of the minor axis and the minor axis lies
along OY.
45. Determine the semiaxes a and b in the ellipse -5 + ^ = 1 so
that it shall pass through (2, 3) and (— 1, — 4).
46. Find the equation of an ellipse if its axes are 8 and 4, its
center is at (2, — 3), and its major axis is parallel to OX.
47. Find the equation of an ellipse if its axes are £ and £, its
center is at (1, —1), and its major axis is parallel to OY.
48. If the vertices of an ellipse are (±6, 0) and its foci are
(± 4=> 0)> nnd its equation.
49. Find the equation of an ellipse when the vertices are (± 4, 0)
and one focus is (2, 0).
60. Find the equation of an ellipse when the vertices are (0, 2)
and (0, — 4) and one focus is at the origin.
51. Find the equation of the ellipse the foci of which are (± 4, 0)
and the major axis of which is 10.
52. Find the equation of the ellipse the foci of which are (0, ± 3)
and the major axis of which is 12.
53. Find the equation of an ellipse when its center is at the
origin, one focus is at the point (— 4, 0); and the minor axis is
equal to 6.
54. Find the equation of the ellipse the foci of which are (1, ± 2)
and the major axis of which is 6.
56. Find the equation of an ellipse the eccentricity of which is §
and the foci of which are (0, ± 5).
96 CERTAIN CURVES
56. The center of an ellipse is at the origin and its major axis
lies on OX. If its major axis is 6 and its eccentricity is J, find its
equation.
57. The center of an ellipse is at (— 2, 3), and its major axis is
parallel to OF and 8 units in length. Its eccentricity is J. Find its
equation.
58. The center of an ellipse is at (1, 2), its eccentricity is J, and
the length of its major axis, which is parallel to OF, is 8. What is
the equation of the ellipse ?
59. Find the equation of an ellipse the eccentricity of which is \
and the ordinate at the focus is 4, the center being at the origin and
the major axis lying on OX.
60. Find the eccentricity and the equation of an ellipse if the
foci lie halfway between the center and the vertices, the center
being at the origin and the major axis lying on OX.
61. Find the equation and the eccentricity of an ellipse if the
ordinate at the focus is one third the minor axis, the center being
at the origin and the major axis lying on OX.
62. Find the eccentricity of an ellipse if the straight line connect-
ing the positive ends of the axes is parallel to the straight line joining
the center to the upper end of the ordinate at the left-hand focus.
x2 iF
63. Given the hyperbola — — y = 1, find its eccentricity, foci,
and asymptotes.
64. Given the hyperbola 4 x2 — 9^ = 36, find its eccentricity,
foci, and asymptotes.
65. Find the center, eccentricity, foci, and asymptotes of the
hyperbola 9x* - ±f - 36x - 24y - 36 = 0.
66. Find the center, eccentricity, foci, and asymptotes of the
hyperbola 2x2 - Sf + 4z + 12y + 4 = 0.
67. Find the equation of an hyperbola if its transverse axis is
V3, its conjugate axis Vf , its center at (1, — 2), and its transverse
axis parallel to OX.
68. Find the equation of an hyperbola if its transverse axis is
5, its conjugate axis 3, its center (— 2, 3), and its transverse axis
parallel to OY.
69. Find the equation of the hyperbola when the origin is at the
left-hand vertex, the transverse axis lying on OX.
PKOBLEMS 97
70. Find the equation of an hyperbola if the foci are (± 4, 0) and
the transverse axis is 6.
71. Find the equation of an hyperbola if the foci are (0, ± 3) and
the transverse axis is 4.
72. An hyperbola has its center at (1, 2) and its transverse axis
is parallel to OX. If its eccentricity is J and its transverse axis is
5, find its equation.
73. Find the equation of an hyperbola when the vertices are (7, 1)
and (— 1, 1) and the eccentricity is |.
74. Find the equation of an hyperbola the vertices of which are
halfway between the center and the foci, the center being at 0 and
the transverse axis lying on OX.
76. Find the equation of the hyperbola which has the lines
y = ± \ x f or its asymptotes and the points (± 2, 0) for its foci.
76. Find the equation of the hyperbola which has the asymptotes
y = ± J x and passes through the point (2, 1).
77. Find the equation of an equilateral hyperbola which passes
through (3, — 1) and has its axes on the coordinate axes.
78. Show that the eccentricity of an equilateral hyperbola is equal
to the ratio of a diagonal of a square to its side.
79. If the vertices of an hyperbola lie two thirds of the distances
from the center to the foci, find the angles between the transverse
axis and the asymptotes.
80. Express the angle between the asymptotes in terms of the
eccentricity of the hyperbola.
81. An ellipse and an hyperbola have the vertices of each at the
x* %P
foci of the other. If the equation of the ellipse is — + *- = 1, find
that of the hyperbola. Find the equations of the directrices of the
two curves.
x2 y2
82. Show that -5 —^ + i* 72 = *> where k is an arbitrary
a — fC O — K x2 y2
quantity, represents an ellipse confocal to —^ + ^ = 1 when h2<b2,
X2 1?
and represents an hyperbola confocal to -5 + jj = 1 when k2 > b2 but
< a3, a2 being greater than b2.
83. Find the vertex, axis, focus, and directrix of the parabola
y*+4y _6x + 7=0.
98 CERTAIN CURVES
84. Find the vertex, axis, focus, and directrix of the parabola
4^ + 4z + 3y-2 = 0.
85. Determine p so that the parabola y2 = 4jtcb shall pass through
the point (— 2, 4).
86. The vertex of a parabola is at the point (2, 3), and the parab-
ola passes through the origin of coordinates. Find its equation, its
axis being parallel to OX.
87. The vertex of a parabola is at the point (— 1£, 2), and the
parabola passes through the point (— 1, — 1). Find its equation, its
axis being parallel to OY.
88. Find the equation of the parabola when the origin is at the
focus and the axis of the parabola lies on OX.
89. Find the equation of the parabola when the axis of the curve
and its directrix are taken as the axes of x and y respectively.
90. The vertex of a parabola is (3, 2) and its focus is (5, 2). Find
its equation.
91. The vertex of a parabola is (— 1, 2) and its focus is (— 1, 0).
Find its equation.
92. Find the equation of the parabola of which the focus is
(2, — 1) and the directrix is the line y — 4 = 0.
93. The vertex of a parabola is at the point (— 2, — 5) and its
directrix is the line x — 3 = 0. Find its equation.
94. The vertex of a parabola is at (5, — 2) and its directrix is the
line y + 4 = 0. Find its equation.
95. The focus of a parabola is at the point (4, — 1) and its direc-
trix is the line y — x = 0. Construct the curve from its definition
and derive its equation. What is the equation of its axis ?
96. The altitude of a parabolic segment is 8 ft. and the length of
its base is 14 ft. A straight line drawn across the segment perpen-
dicular to its axis is 7 ft. long. How far is it from the vertex of
the segment ?
97. An arch in the form of a parabolic curve, the axis being verti-
cal, is 40 ft. across the bottom, and the highest point is 12 ft. above
the horizontal. What is the length of a beam placed horizontally
across the arch 3 ft. from the top ?
PROBLEMS 99
98. The cable of a suspension bridge hangs in the form of a
parabola. The roadway, which is horizontal and 300 ft. long, is sup-
ported by vertical wires attached to the cable, the longest wire being
90 ft. and the shortest being 20 ft. Find the length of a supporting
wire attached to the roadway 50 ft. from the middle.
99. Any section of a given parabolic mirror made by a plane pass-
ing through the axis of the mirror is a parabolic segment of which
the altitude is 8 in. and the length of the base is 12 in. Find the
perimeter of the section of the mirror made by a plane perpendicular
to its axis and 6 in. from its vertex.
100. Given the ellipse 4 x2 + 9 y2 = 36, find its foci and directrices.
101. Given the ellipse 5 x2 + 3 y2 = 1, find its foci and directrices.
102. Given the hyperbola 5 x2 — 10 if = 50, find its foci and
directrices.
103. Find the equation of an ellipse when the foci are (±3, 0)
and the directrices are x = ± 7.
104. Find the center, vertices, foci, and directrices of the ellipse
9a2 + 25 y2 + 30x + 40 y - 184 = 0.
105. Find the center, vertices, foci, and directrices of the hyper-
bola 5a?1 -- 4y* + 10z +16y - 31 = 0.
106. Find the equation of a circle through the vertex and the
ends of the double ordinate at the focus of the parabola y2 = Apx.
107. Find the equation of the circle through the vertex, the
focus, and the upper end of the ordinate at the focus of the parab-
ola y2 — 8 x = 0.
108. Find the equation of a circle which passes through the
vertex and the focus of the parabola t/2 = 8 x and has its center
on the line x — y + 2 = 0.
109. Find the equation of the locus of a point which moves so
that the slope of the straight line joining it to the focus of the
parabola x2 = 8 y is three times the eccentricity of the ellipse
16 a2 + 9f -144 = 0.
110. Find the equation of the cissoid when the origin is at the
center of the circle used in its definition, the direction of the axes
being as in § 48.
111. Find the equation of the cissoid when its asymptote is the
axis of y and its axis is the axis of x.
100 CERTAIN CURVES
112. Find the equation of the strophoid when the asymptote is
the axis of y, the axis of x being as in § 49.
113. Find the equation of the strophoid when the origin is at
A (fig. 82), the axes being parallel to those of § 49.
114. Show that the lines y = ± x intersect the strophoid at the
origin only, and find the equation of the curve referred to these
lines as axes.
115. Find the equation of the witch when LK (fig. 80) is the axis
of x and OA the axis of y.
116. Find the equation of the witch when the origin is taken at
the center of the circle used in constructing it, the axes being par-
allel to those of § 47.
117. Show that the locus of a point which moves so that the sum
of its distances from two fixed straight lines is constant is a straight
line.
118. Find the equations of the locus of a point equally distant
from two fixed straight lines.
119. A point moves so that its distances from two fixed points
are in a constant ratio k. Show that the locus is a circle except
when k = 1.
120. A point moves so that the sum of the squares of its dis-
tances from the sides of an equilateral triangle is constant. Show
that the locus is a circle and find its center.
121. A point moves so that the square of its distance from the
base of an isosceles triangle is equal to the product of its distances
from the other two sides. Show that the locus is a circle and an
hyperbola which pass through the vertices of the two base angles.
122. A point moves so that the sum of the squares of its dis-
tances from the four sides of a square is constant. Find its locus.
123. A point moves so that the sum of the squares of its dis-
tances from any number of fixed points is constant. Find its locus.
124. Find the locus of a point the square of the distance of
which from a fixed point is proportional to its distance from a
fixed straight line.
125. Find the locus of a point such that the lengths of the tan-
gents from it to two concentric circles are inversely as the radii of
the circles.
PROBLEMS 101
126. A point moves so that the length of the tangent from it to
a fixed circle is equal to its distance from a fixed point. Find its
locus.
127. Find the locus of a point the tangents from which to two
fixed circles are of equal length.
128. Straight lines are drawn through the points (— a, 0) and
(a, 0) so that the difference of the angles they make with the axis
of x is tan-1 - • Find the locus of their point of intersection.
a
129. The slope of a straight line passing through (a, 0) is twice
the slope of a straight line passing through (— a, 0). Find the locus
of the point of intersection of these lines.
130. A point moves so that the product of the slopes of the
straight lines joining it to A (— a, 0) and B (a, 0) is constant.
Prove that the locus is an ellipse or an hyperbola.
131. If, in the triangle ABC, tan A tan^JB = 2 and AB is fixed,
show that the locus of C is a parabola with its vertex at A and its
focus at B.
132. Given the base 2 b of a triangle and the sum s of the tan-
gents of the angles at the base. Find the locus of the vertex.
133. Find the locus of the center of a circle which is tangent to
a fixed circle and a fixed straight line-.
134. Prove that the locus of the center of a circle which passes
through a fixed point and is tangent to a fixed straight line is a
parabola.
135. A point moves so that its shortest distance from a fixed
circle is equal to its distance from a fixed diameter of that circle.
Find its locus.
136. If a straight line is drawn from the origin to any point Q of
the line y = a, and if a point P is taken on this line such that its
ordinate is equal to the abscissa of Q, find the locus of P.
137 . A OB and COD are two straight lines which bisect each other at
right angles. Find the locus of a point P such that PA • PB = PC • PD.
138. AB and CD are perpendicular diameters of a circle and M is
any point on the circle. Through M, AM and BM are drawn. AM
intersects CD in N, and from N a straight line is drawn parallel to
A By meeting BM in P. Find the locus of P.
AC
102 CERTAIN CURVES
139. Given a fixed straight line AB and a fixed point Q. From
any point R in AB a perpendicular to AB is drawn, equal in length
to RQ. Find the locus of the end of this perpendicular.
140. 0 is a fixed point and AB is a fixed straight line. A straight
line is drawn from 0, meeting AB at Q, and in OQ a point P is taken
so that OP-OQ = k\ Find the locus of P.
141. Let OA be the diameter of a fixed circle. From B, any point
on the circle, draw a straight line perpendicular to OA, meeting it
in D. Prolong the line DB to P, so that OD:DB=OA: DP. Find
the locus of P.
142. A perpendicular is drawn from the focus of an hyperbola to
an asymptote. Show that its foot is at distances a and b from the
center and the focus respectively.
143. Two straight lines are drawn through the vertex of a parab-
ola at right angles to each other and meeting the curve at P and Q.
Show that the line PQ cuts the axis of the parabola in a fixed point.
144. In the parabola ^ = ±px an equilateral triangle is so
inscribed that one vertex is at the origin. What is the length
of one of its sides?
145. Prove that in the ellipse half of the minor axis is a mean
proportional between ^4Pand FA'.
146. Show that in an equilateral hyperbola the distance of a point
from the center is a mean proportional between the focal distances of
the point.
147. If from any point P of an hyperbola PK is drawn parallel
to the transverse axis, cutting the asymptotes in Q and R, prove
PQ • PR = a2. If PK is drawn parallel to the conjugate axis, prove
PQ- PR=-b\
148. Prove that the product of the distances of any point of the
hyperbola from the asymptotes is constant.
149. Prove that in the hyperbola the squares of the ordinates of
any two points are to each other as the products of the segments of
the transverse axis made by the feet of these ordinates.
150. Straight lines are drawn through a point of an ellipse from
the two ends of the minor axis. Show that the product of their
intercepts on OX is constant.
PROBLEMS
103
151. P is any point of the parabola y2l=4^z, and PXQ, which
is perpendicular to OPv intersects the axis of the parabola in Q.
Prove that the projection of PXQ on the axis of the parabola is
always 4jp.
152. Show that the focal distance of any point on the hyperbola
is equal to the length of the straight line drawn through the point
parallel to an asymptote to meet the corresponding directrix.
153. Show that the following points lie approximately on a straight
line, and find its equation :
X
4
9
13
20
22
25
30
y
2.1
4.6
7
12
12.9
14.5
18.2
154. For a galvanometer the deflection Z>, measured in millimeters
on a proper scale, and the current /, measured in microamperes, are
determined in a series of readings as follows :
D
29.1
48.2
72.7
92.0
118.0
140.0
165.0
199.0
I
0.0493
0.0821
0.123
0.154
0.197
0.234
0.274
0.328
Find an empirical law connecting D and /.
155. For a copper-nickel thermocouple the relation between the
temperature t in degrees and the thermoelectric power in microvolts
is given by the following table :
t
0
50
100
150
200
p
24
25
26
26.9
27.5
Find an empirical law connecting t and p.
156. The safe loads in thousands of pounds for beams of the same
cross-section but of various lengths in feet are found as follows :
Length
10
11
12
13
14
15
Load
123.6
121.5
111.8
107.2
101.3
90.4
Find the empirical equation connecting the data.
104
CERTAIN CURVES
157. The relation between the pressure p and the volume v of a
gas is found experimentally as follows :
Pressure
20
23.5
31
42
59
78
Volume
0.619
0.540
0.442
0.358
0.277
0.219
Find an empirical equation connecting p and v in the form pvn = c.
158. The deflection a of a loaded beam with a constant load is
found for various lengths I as follows :
I
1000
900
800
700
600
a
7.14
5.22
3.64
2.42
1.60
Find an empirical equation connecting a and I in the form a = kF.
159. The relation between the length I (in mm.) and the time t
(in seconds) of a swinging pendulum is found as follows :
I
t
63.4
80.5
90.4
101.3
107.3
140.Q
0.806
0.892
0.960
1.010
1.038
1.198
Find an empirical equation connecting I and t in the form t = kF.
160. For a dynamometer the relation between the deflection $,
when the unit 6 = -r^r > and the current /, measured in amperes, is
as follows :
e
i
40
86
120
160
201
240
280
320
362
0.147
0.215
0.252
0.293
0.329
0.360
0.390
0.417
0.442
Find an empirical equation connecting / and 0 in the form / = k(F.
161. In a chemical experiment the relation between the concen-
tration y of undissociated hydrochloric acid is connected with the
concentration x of hydrogen ions as shown in the table :
X
y
1.68
1.22
0.784
0.426
0.092
0.047
0.0096
0.0049
0.00098
1.32
0.676
0.216
0.074
0.0085
0.00315
0.00036
0.00014
0.000018
Find an empirical law connecting the two quantities in the form y=kx*.
PROBLEMS
105
162. Show that the values of x and y as given in the following
table are connected by a relation of the form y = cax, and find c and a.
X
8
10
12
14
16
18
20
V
3.2
4.6
7.3
9.8
15.2
24.6
36.4
163. In a certain chemical reaction the concentration c of sodium
acetate produced at the end of the stated number of minutes t is
as follows :
t
1
2
3
4
5
c
0.00837
0.0070
0.00586
0.00492
0.00410
Assuming that the law is of the form c = ab*, find the equation
connecting the concentration with the time.
164. The molal heat capacity at constant temperature is for water
vapor at various temperatures as follows :
Temp.
10
100
500
700
1000
Cap.
8.8
8.6
8.4
8.6
9.1
Determine the law in the form C = a -f- bt + ct2.
165. Assuming Boyle's law, pv = c, determine c graphically from
the following pairs of observed values :
V
39.92
42.17
45.80
48 52
51.89
60.47
65.97
V
40.37
38.32
35.32
33.29
31.22
26.86
24.53
*
166. The distance p of an object from a lens and the distance p'
of its image are found by experiment as follows :
P
P'
320
240
180
140
120
100
80
60
21.35
21.80
22.50
23.20
23.80
24.60
26.20
29.00
Assuming the law - -\ — . = - > where / is the focal length of the lens,
P ¥ f
compute f graphically by plotting the reciprocals of p and p\
CHAPTER VII
PARAMETRIC REPRESENTATION
52. Definition. Consider the two equations
(i)
where ' fx(f) and f2(t) are two functions of an independent
variable t. If we assign to t any value in (1), we determine
x and y and may plot a point with these coordinates. In
this way a value of t determines a point in the plane. So
other values of t determine other points, which together deter-
mine a curve.
The two equations (1) then represent the curve. The vari-
able t is called a parameter, and the equations (1) are called
the parametric representation of the curve. It is sometimes
easy to eliminate t from the equations (1) and obtain thus
a Cartesian equation
of the curve, but this y t
elimination is not essen-
tial and is not always
desirable.
Ex. 1. x = t2, y = t.
Giving t in succession
the values — 3, — 2, — 1, 0,
1, 2, 3, we find the corre-
sponding points (9, — 3),
(4, -2), (1, -1), (0, 0),
(1, 1), (4, 2), (9, 3). These
points, if plotted, may be
connected by the curve of
fig. 88, and as many inter-
mediate points as desired may be found. In this case we may easily
eliminate t from the equations and obtain x = y2. The curve is a parabola.
106
O
1=0
t=-t
t=~l
Fig. 88
PAEAMETKIC REPRESENTATION
107
Ex. 2. x - f8 + 2 12, y = t*-t.
Giving t in succession the values — 2, — f , — 1, — £, 0, £, 1, we
find as corresponding points (0, — 6), (}, — *£•), (1, 0), (§, |), (0, 0),
(I. - I). (3. 0).
These points give the curve shown in
fig. 89. If more details as to the shape of
the loop are wanted, more values of t must
be assumed intermediate to those we have
used. Elimination of t in this example is
possible but hardly desirable.
Ex. 3. x = a cos8*, y = a sin8*.
If values of t are assumed at convenient
intervals between t = 0° and t = 360°, the
curve may be found to be as in fig. 90.
The elimination of t gives the equation
x$ + yi = a*. The curve is called the four-
cusped hypocycloid (§ 58).
As the examples show, the param-
eter t is in general simply an inde- FlG 8q
pendent variable to which values are
assigned at pleasure. In problems of mechanics, however, the pa-
rameter frequently represents time. In this case the curve of equa-
tions (1) represents the path
of a moving point, the position
of the point at any instant
being given by the equations.
Any of the above examples
may be interpreted in this
way. Other illustrations will
be found in the examples of
§§53 and 54.
In some cases, also, it is
possible to give a geometric
interpretation to the param-
eter t This is illustrated
by the curves which follow, where in each case the parameter is
a certain angle.
108
PARAMETRIC REPRESENTATION
53. The circle. Let P (#, y) (fig. 91) be any point on a circle
with its center at the origin 0 and its radius equal to a. Let <f>
be the angle made by OP and OX. Then, from the definition of
the sine and cosine,
x = a cos <f>,
y = a sin <f>y
are the parametric equations of the circle
with <f> as the arbitrary parameter.
Ex. A particle moves in a circle at a con-
stant rate k. Then, if s represents the arc
traversed in the time t,
s = kt and
, s kt
9 = ~ = —
a a
Fig. 91
Therefore the equations of the circle are
kt
x = a cos — ,
a
. kt
y = a sin —
9 a
54. The ellipse. In a circle with radius a let the abscissa of
every point Q (fig. 92) be left unchanged and its ordinate be
altered in a fixed ratio b : a, where b is any length whatever.
The point Q then takes such a position
as P, where in the figure b < a. The
parametric equations of the locus of P
are therefore, from § 53,
x = a cos <f>,
y = b sin <f>.
The elimination of <j> from these equa-
tions gives (-J + (^j=l, showing that
the locus of P is an ellipse.
<f> is called the eccentric angle of a point on the ellipse, and
the circle 7?+y2=^ a2 is called the auxiliary circle.
Ex. A particle Q moves at a constant rate along the auxiliary circle of
an ellipse ; required the motion of its accompanying point P.
kt
As in § 53, <£ = — • Hence the equations of the path are
Fig. 02
kt
x = a cos — >
a
. . kt
y = o sin — .
9 a
THE CYCLOID
109
55. The cycloid. If a circle rolls upon a straight line, each
point of the circumference describes a curve called a cycloid.
Let a circle of radius a roll upon the axis of x, and let C
(fig. 93) be its center at any time of its motion, N its point of
contact with OX, and P the point on its circumference which
Fig. 93
describes the cycloid. Take as the origin of coordinates, 0, the
point found by rolling the circle to the left until P meets OX.
Then OJST= arc PN.
Draw MP and CN, each perpendicular to OX, PR parallel to
OX, and connect C and P. Let
Then
angle JSFCP=(f>.
x = OM=ON-MJST
= arc NP-PR
= a<f> — a sin <f>.
y = MP = NC-RC
= a — a cos <f>.
Hence the parametric representation of the cycloid is
x = a($> — sin <£),
y = a (1 — cos <£).
By eliminating <f>, the equation of the cycloid may be written
a~y ±V2ay-y2,
x = a cos
-i
a
but this is less convenient than the parametric representation.
At each point where the cycloid meets OX a sharp vertex
called a cusp is formed. The distance between two consecutive
cusps is evidently 2ira.
110
PARAMETRIC REPRESENTATION
56. The trochoid. When a circle rolls upon a straight line,
any point upon a radius, or upon a radius produced, describes
a curve called a trochoid.
Let the circle roll upon the axis of x, and let C (figs. 94 and
95) be its center at any time, N its point of contact with the
Fig. 94
axis of #, P(x, y) the point which describes the trochoid, and
K the point in which the line CP meets the circle. Take as
the origin 0 the point found by rolling the circle toward the
left until K is on the axis of x. Then
ON=mcNK.
pi
1
C J
R J
1 >»?
\M
o) 2
f
Fig. 95
Draw PM and CN perpendicular to OX, and through P a line
parallel to OXy meeting CjV, or CN produced, in R. Let the
radius of the circle be a, CP be A, and angle NCP be <f>. Then
x =
y =
OM=ON-MN
arc NK- PR
a<f>-hsm<f>.
jrp = xc-RC
= a — h cos <f>.
THE EPICYCLOID
111
57. The epicycloid. When a circle rolls upon the outside of a
fixed circle, each point of the circumference of the rolling circle
describes a curve called an epicycloid.
Let 0 (fig. 96) be the center of the fixed circle, C the center
of the rolling circle, N its point of contact with the fixed circle,
and P(xy y) the point
which describes the
epicycloid. Determine
the point K by rolling
the circle C until P
meets the circumference
of 0. Then
arc JTiV= arc NP.
Take 0 as the origin
of coordinates and OK
as the axis of x. Draw
PM and CL perpendic-
ular to OXy PS parallel
to OXj meeting CL in
JR, and connect 0 and C.
Let the radius of the
rolling circle be a, that of the fixed circle 6, and denote the
angle OCP by 0, the angle KOC by <f>. Then
arc KN= b<f>, arc NP = ad ;
whence b<f> = ad.
We now have x = 0M= OL + LM
= OC cos KOC - CP cos SPC
= (a + 6) cos <f> — a cos (<£> + 0)
= ( a + 6 ) cos <f> — a cos p.
v ' a
y = MP = LC-BC
= OC sin KOC -CP sin SPC
= (a + J) sin <£ — a sin (<£ + 0)
= (a + 6)sin<f)-asin p.
v y a
Fig. 96
112
PARAMETRIC REPRESENTATION
The curve consists of a number of congruent arches, the first of
which corresponds to values of d between 0 and 2 7r, that is, to
2 air
values of <f> between 0 and — — . Similarly, the Mi arch corre-
sponds to values of <f> between — ^ — — - — and — - — . Hence
6 6
the curve is a closed curve when, and only when, for some value
2i left IT
of A, — - — is a multiple of 2 7r. If a and b are incommensurable,
this is impossible, but if T = ^ , where ~- is a rational fraction in
by. 9.
its lowest terms, the smallest value of k = q. The curve then con-
sists of q arches and winds p times around the fixed circle.
58. The hypocycloid. When a circle rolls upon the inside of
a fixed circle, each point of the rolling circle describes a curve
called the hypocycloid. If the axes and the notation are as in
the previous article, the equa-
tions of the hypocycloid are
x = (b — d) coscf> + « cos <f>,
Cv
y = (b — a) sin <f> — a sin <j>.
The proof is left to the student.
The curve is shown in fig. 97.
In the special case in which
the radius of the rolling circle
is one fourth that of the fixed
circle, we have b = 4 a. Then
Fig. 97
x = a (3 cos $ + cos 3 $) = 4 a coss<f> = b cos8<£,
y = a (3 sin <f> — sin 3 $) = 4 a sin8<£ = b sin8<£.
This is the four-cusped hypocycloid of Ex. 3, § 52.
59. The involute of the circle. If a string, kept taut, is
unwound from the circumference of a circle, its end describes
a curve called the involute of the circle. Let 0 (fig. 98) be
the center of the circle, a its radius, and A the point at which
PROBLEMS
113
the end of the string is on the circle. Take 0 as the origin of
coordinates and OA as the axis of x. Let P(#, y) be a point
on the involute, PK the line drawn from P tangent to the
circle at JT, and <f> the angle
XOK Then PK represents a
portion of the unwinding string,
and hence
KP = arc AK= a<f>.
Now it is clear that for all
positions of the point Ky OK
makes an angle <j> — — with OF.
Hence the projection of OK on
OX is always OK cos <f> = a cos <f>,
and its projection on OF is
OK cos ( <£ — — ) = a sin <j>. Also JTP always makes an angle <f> — -^
with OX and an angle 7r — <£ with OF. Hence the projection of
KP on OX is KP cos ( <£ — — J = a<£ sin <£, and its projection
on 0 F is KP cos (tt — <£) = — a<£ cos <£. The projection of OP on
OX is #, and on 0 F is y. Hence, by the law of projections, § 2,
x = a cos <f> + a<f> sin <£,
y = a sin <f>— atp cos <£.
PROBLEMS
Plot the graphs of the following parametric equations :
6 6*
Fig. 98
1. x = tf2, y = £ + 1.
4 4
3. a; =
, y =
a0
5. ar
±V4 + 9*2' * ±V4 + 9^
4. # = /, y =
2a
a2 + J2
2a
l+*2' y~~*(l+*2)'
o • 2i 2 a sin8<ft
6. # = 2 a sm26, ?/ = — -
™ J cos<£
7. x = e* sin £, y = e' cos £.
114 PARAMETRIC REPRESENTATION
8. x = a<l> + & sin <f>, y = a — a cos <£.
9. a; = — cos <£ — - cos 3 <£, y = — sin <£ — - sin 3 <£.
10. x = 2a cos <f> — a cos 2 <£, y = 2 a sin </> — a sin 2 <£.
11. A projectile moves so that the coordinates of its position
at any time t are given by the equations x = 60 1, y = 80t —16P.
Plot its path.
12. Find the parametric equations of the parabola y2 = &px when
the parameter is the slope of a straight line through the vertex.
x2 i?
13. Find the parametric equations of the ellipse -5 + jj =1 when
the parameter is the slope of a straight line through the center.
14. Find the parametric equations of the cissoid when the param-
eter is the slope of a straight line through the origin, the axes of
coordinates being as in fig. 81.
15. Find the parametric equations of the cissoid when the param-
eter is the angle A OP (fig. 81).
16. Find the parametric equations of the strophoid when the
parameter is the angle MAP (fig. 82).
17. When a circle rolls upon the outside of a fixed circle, a point
on the radius of the rolling circle at a distance h from its center
describes a curve called an epitrochoid. Find its equations.
18. When a circle rolls upon the inside of a fixed circle, a point
on the radius of the rolling circle at a distance h from its center
describes a curve called an hypotrochoid. Find its equations.
19. If a circle rolls on the inside of a fixed circle of twice its
radius, what is the form of the curve generated by a point of the
circumference of the rolling circle ?
20. AB is a given straight line perpendicular to OX at the point
C, where OC = a. Through 0 any straight line is drawn, meeting AB
at D. On OX a point M is taken, to the left of C, so that CM= CD.
Finally, through M a straight line is drawn perpendicular to OX,
intersecting OD at P. Find the parametric equations of the locus of
P, using the angle XOD as the parameter. Find also the Cartesian
equation, name the curve, and sketch the. graph.
PROBLEMS 115
21. A fixed circle of radius a with its center at 0 intersects OX
at A. The straight line BC is tangent to the circle at A. Through 0
any straight line is drawn, intersecting the circle at D and intersecting
BC at E. Through D a straight line is drawn parallel to OY, and
through E a straight line is drawn parallel to OX. These lines inter-
sect at P. Find the parametric equations of the locus of P in terms
of the angle XOD as parameter. Find also the Cartesian equation
and sketch the curve.
22. A circle of radius a has its center at 0, the origin of coordinates.
The tangent to the circle at any point A meets OX at M. Through M
a straight line is drawn parallel to OY, and through A a straight line
is drawn {Parallel to OX. These lines intersect at P. Find the par-
ametric equations of the locus of P, using the angle MOA as the
parameter. Find also the Cartesian equation and sketch the curve.
23. A circle of radius a has its center at the origin of coordinates
0. Through 0 any straight line is drawn, intersecting the circle at
A. The tangent to the circle at A intersects OY at B. Through B a
straight line is drawn parallel to OX, meeting OA produced at P.
Find the parametric equations of the locus of P in terms of the
angle X OA as parameter. Find also the Cartesian equation.
24. Let OA be the diameter of a fixed circle and LK the tangent
at A. From 0 draw any straight line intersecting the circle at B
and LK at C, and let P be the middle point of BC. Find the para-
metric equations of the locus of P, using the angle A OP as the
parameter, OA as the axis of y, and 0 as the origin. Find also the
Cartesian equation.
25. A circle of radius a has its center at the origin of coordinates
0, and the straight line AB is tangent to the circle at A (a, 0). From
0 any straight line is drawn, meeting AB at E and the circle at D.
On OE, OP is taken equal to BE. Find the parametric equations of
the locus of P in terms of the angle A OP as parameter.
26. The straight line AB is perpendicular to OX at A (a, 0). From
0 a straight line is drawn to any point C of AB. The straight line
drawn from C perpendicular to OC meets OX at M. The perpendic-
ular to OX at M meets OC produced at P. Find the parametric
equations of the locus of P in terms of the angle XOC as parameter.
Find also the Cartesian equation.
116 PARAMETRIC REPRESENTATION
27. OBCD is a rectangle with OB = a and BC'—c. Any line is
drawn through C, meeting OB in 2£, and the triangle EPO is con-
structed so that the angles CEP and EPO are right angles. Find
the parametric equations of the locus of P, using the angle BOP as
the parameter, OB as the axis of x, and 0 as the origin. Find also the
Cartesian equation of the locus.
28. A fixed circle has as diameter the straight line joining the
origin and the point ^4(0, 2a). Any point B of the circle is connected
with A and 0, and BM is drawn perpendicular to OX, meeting OX
at M. On MB, MP is laid off equal to BA. Find the parametric
equations of the locus of P in terms of the angle XOB as parameter.
Find also the Cartesian equation.
29. Let AB be a given straight line, 0 a given point a units from
AB, and k a given constant. On any straight line through 0, meet-
ing AB in My take P so that OM . AfP= &a. Find the parametric
equations of the locus of P, using 0 as the origin, the perpendicular
from 0 to AB as the axis of x, and the angle between OX and OP
as the parameter. Also find the Cartesian equation.
30. ABC is a given right triangle of which the sides AB and BC
about the right angle at B are always equal to a and b respectively.
The triangle moves in the plane XOYso that A is always on OF and
B is always on OX. P is the middle point of the hypotenuse AC.
Find the parametric equations of the locus of P, using the angle
XBC as the parameter.
31. Let 0 be the center of a circle with radius a, A a fixed point
on the circle, and B a moving point on the circle. If the tangent at
B meets the tangent at A in C, and P is the middle point of BC,
find the equations of the locus of P in parametric form, using the
angle A OB as the arbitrary parameter, OA as the axis of x, and 0
as the origin.
32. A fixed circle has as diameter the straight line joining the
origin of coordinates and the point A (2 a, 0), and LK is tangent to
the circle at A. From 0 any straight line is drawn, meeting the circle
at D and the tangent LK at E. On OE a point P is so taken that
PD=DE in both length and direction. Find the parametric equar
tions of the locus of P in terms of the angle AOE as parameter.
Find also the Cartesian equation.
PROBLEMS 117
33. A and B are two points on the axis of y at distances — a and
-f- a respectively from the origin. AH is any straight line through A,
meeting the axis of x at H. BK is the perpendicular from B on AH,
meeting it at K. Through K a straight line is drawn parallel to
the axis of x, and through H a straight line is drawn parallel to
the axis of y. These lines meet in P. Find the parametric equar
tions of the locus of P, using the angle BAK as the parameter.
Also find the Cartesian equation.
34. Q is the point on the auxiliary circle of the ellipse
x2 I/2
-, + $ = !> (•>»)
corresponding to the point P of the ellipse. The straight line
through P parallel to OQ meets OX at L and OF at M. Prove
PL = b, and PM= a.
35. If a projectile starts with an initial velocity v in an initial
direction which makes an angle a with the axis of x, taken horizontal,
its position at any time t is given by the parametric equations
x = vt cos a, y = vt sin a — \ gi?.
Find the Cartesian equation of the path of the projectile and its
nature and position.
36. From the equations of problem 35 determine when and where
the projectile strikes a point on the axis of x.
37. From the equations of problem 35 determine when, and for
what value of x, the projectile passes through a point which is at
a distance h below the horizontal.
38. From the equations of problem 35, what elevation must be
given to a gun that the projectile may pass through a point b units
distant from the muzzle of the gun and lying in the horizontal line
passing through the muzzle ?
39. From the equations of problem 35, what elevation must be
given to a gun to obtain a maximum range on a horizontal line
passing through the muzzle?
40. A gun stands on a cliff h units above the water. From the
equations of problem 35, what elevation must be given to the gun
that the projectile may strike a point in the water b units from the
base of the cliff ?
AC
CHAPTER VIII
POLAR COORDINATES
60. Coordinate system. So far we have determined the posi-
tion of a point in the plane by two distances, x and y. We
may, however, use a distance and a direction, as follows:
Let 0 (fig. 99), called the origin, or pole, be a fixed point, and
let OM, called the initial line, be a fixed line. Take P any point
in the plane and draw OP. Denote OP by r and the angle
MOP by 0. Then r and 0 are called the polar coordinates of
the point P (r, 0), and when given will completely determine P.
For example, the point (2, 15°) is plotted by laying off the
angle MOP = 15° and measuring OP = 2.
OP, or r, is called the radius vector, and 0
the vectorial angle, of P. These quantities may
be either positive or negative. A negative
value of 0 is laid off in the direction of the
motion of the hands of a clock, a positive
angle in the opposite direction. After the
angle 0 has been constructed, positive values of r are measured
from 0 along the terminal line of 0, and negative values of r
from 0 along the backward extension of the terminal line. It
follows that the same point may have more than one pair of
coordinates. Thus (2, 195°), (2, -165°), (-2, 15°), and
(— 2, — 345°) refer to the same point. In practice it is usually
convenient to restrict 0 to positive values.
Plotting in polar coordinates is facilitated by using paper ruled
as in figs. 100 and 101. The angle 0 is determined from the num-
bers at the ends of the straight lines, and the value of r is counted
off on the concentric circles, either towards or away from the
number which indicates 0, according as r is positive or negative.
When an equation is given in polar coordinates, the corre-
sponding curve may be plotted by giving to 0 convenient
118
1
^ w£~—
s«
-^M*
spC,
SST--/
Jrd
JjyQr\\j$^
u *
i/fm'
*joV"\
^>S»"
Fig. 100
POLAR COORDINATES 119
values, computing the corresponding values of r, plotting the
resulting points, and drawing a curve through tliem,
Ex. 1. r = acos6.
a is a constant which
may be given, any con-
venient value. We may
then nod from a table oi
natural cosines the value
of r which corresponds
to any value of $. By
plotting the points corre-
sponding to values of $
from 0° to 90°, we obtain
thearc,4.BC0(fig.lOO).
Values of 6 from 90° to
180° give the arc ODEA.
Values of 6 from 180"
to 270° give again the
arc ABCO, and those
from 270° to 360° give the arc ODEA. Values of 6 greater than 360° can
clearly give no points not already found. The curve is a circle (§ 63).
Ex. a. r = a sin 3 0.
As 6 increases from 0° to
80°, r increases from 0 to a ;
as 0 increases from 30° to 60",
r decreases from a to 0 ; the
point (r, d) traces out the
loop OA 0 (fig. 101). As 6
ocreases from 60° to 90°,
negative and decreases
tOto-a; Mi
i 90° to 120°,
i — a to 0; the point
V, 8) traces out the loop
OBO. As $ increases from
120° to 180°, the point (r, $)
traces out the loop OCO.
Larger values of $ give
points already found, since
(180° + &)■■
rin3(60°+ 0) = -
The three loops are congruent, because
his curve ia called a rose of three leaves.
120
POLAR COORDINATES
Fig. 102
Ex. 3. r2 = 2a2cos 2ft
Solving for r, we have r = ±a V2 cos 2 ft
Hence, corresponding to any values of 6 which make cos 2 6 positive, there
will be two values of r numerically equal and opposite in sign and two
corresponding points of the curve symmetrically situated with respect to
the pole. If values are assigned to 0 which make cos 2 6 negative, the cor-
responding values of r will be imaginary and there will be no points on
the curve.
Accordingly, as 6 increases
from 0° to 45°, r decreases nu-
merically from a to 0, and the
portions of the curve in the first
and the third quadrant are con-
structed ; as 0 increases from 45°
to 135°, cos 2 $ is negative, and
there is no portion of the curve between the lines 6 = 45° and 0 = 135°;
finally, as 0 increases from 135° to 180°, r increases numerically from 0
to a, and the portions of the curve in the second and the fourth quadrant
are constructed. The curve is now complete, as we should only repeat the
curve already found if we assigned further values to 0; it is called the
lemniscate (fig. 102).
61. The spirals. Polar coordinates are particularly well
adapted to represent certain curves called spirals, of which
the more important follow:
Ex.1. The spiral of Archimedes,
r = aft
In plotting, 6 is usually considered
in circular measure. When 0 = 0,
r = 0, and as $ increases, r increases,
so that the curve winds infinitely often
around the origin while receding from
it (fig. 103). In the figure the heavy
line represents the portion of the spiral corresponding to positive values
of 0, and the dotted line the portion corresponding to negative values of ft
Fig. 103
Ex. 2. The hyperbolic spiral,
r$ = a,
or
a
r = e
As $ increases indefinitely, r approaches zero. Hence the spiral winds
infinitely often around the origin, continually approaching it but never
THE STRAIGHT LINE
121
reaching it (fig. 104). As $ approaches zero, r increases without limit.
If P is a point on the spiral and NP is the perpendicular to the initial line,
sin#
NP = ram$ = a
$
Fig. 104
Hence, as $ approaches zero as
a limit, NP approaches a (§ 95).
Therefore the curve comes con-
stantly nearer to, but never reaches,
the line LK, parallel to OM at a
distance a units from it. This line is therefore an asymptote. In the
figure the dotted portion of the curve corresponds to negative values of 0.
Ex. 8. The logarithmic spiral.
When $ = 0, r = 1. As 6 increases, r
increases, and the curve winds around
the origin at increasing distances from
it (fig. 105). When 0 is negative and
increasing numerically without limit, r
approaches zero. Hence the curve winds
infinitely often around the origin, continu-
ally approaching it. The dotted line in the
figure corresponds to negative values of 0.
A property of this spiral is that it cuts
the radius vectors at a constant angle. The
student may prove this after reading § 103.
Fig. 106
We shall now give examples of the derivation of the polar
equation of a curve from the definition of the curve.
62. The straight line. Let LK (fig. 106) be a straight line
perpendicular to OD. Let the angle MOD be denoted by a,
and let OD =p ; then p is the normal
distance of LK from the pole. K
Let P(rj ff) be any point of LK
Then, by trigonometry,
or
OP cos D OP =OD,
rcos(0 — a)=jt>,
a)
which is the equation of the straight
line.
Fig. 106
122 POLAR COORDINATES
If a = 0 and p = a, we have the special equation
r cos 0 = a,
or r = a sec ft (2)
If the straight line passes through the origin, p = 0. The
equation of the line then becomes
, cos (0 — a) = 0,
or simply 0 = — -f- ay
which is of the form 0 = c. (3)
63. The circle. Let (7(6, a) be the center and a the radius of
a circle (fig. 107). Let P(r, 0) be any point of the circle, and
draw the straight lines 0(7, OP, and CP.
By trigonometry, we have
OP +OC -20P- OC cos POC = CP.
Noting that cos PO (7 = cos (0 - a), OP = r, 0(7= J, and
(7P = a, and substituting in the equation, we have the result
r2-2r6cos(0-a)+62 = aa (1)
as the polar equation of the circle. I s — ^-Jp/ e\
When the origin is at the center of
the circle, 6 = 0 and (1) becomes simply
r = a. (2) Fig. 107
When the origin is on the circle, b = a and (1) becomes
r — 2 a cos (0 — a) = 0 ;
which may be written r = aQ cos 0 + ax sin 0, (3)
where aQ and ax are the intercepts on the lines 0=0 and 0 = —
respectively.
When the origin is on the circle and the initial line is a
diameter, (3) becomes r = % cog 0 (4)
When the origin is on the circle and the initial line is tangent
to the circle, (3) becomes . ^ ,tv
v y r=a sinft (5)
THE LIMAQON
123
Fig. 108
64. The limacjon. Through any fixed point 0 (fig. 108) on the
circumference of a fixed circle draw any line cutting the circle
again at Z>, and lay off on this line a
constant length measured from D in
either direction. The locus of the
points P and Q thus found is a curve
called the limapon.
Take 0 as the pole, and the diameter
OA as the initial line, of a system of
polar coordinates, and call the diame-
ter of the circle a and the constant
length b. Then it is clear that the
entire locus can be found by caus-
ing OD to revolve through an angle of 360° and laying off
DP=b, always. in the direction of the terminal line of AOD.
Let P be (V, 0), where 0=AOD. Then r = OD+DP when
0 is in the first or the fourth quadrant, and r = — OD + DP
when 0 is in the second or the third
quadrant. But it appears from the
figure that OD = OA cos 0 when 0 is
in the first or the fourth quadrant,
and that OD = — OA cos 0 when 0 is
in the second or the third quadrant.
Hence, for any point on the lima§on,
r = a cos 0 + b.
/
*«>*«)
In studying the shape of the
curve there are three cases to be
distinguished:
e--co?u
Fig. 109
1. b >a. r is always positive; the curve appears as in fig. 108.
2. b < a. r is positive when cos 0 > > negative when
a
cos0< * and zero when cos0 = • The curve appears as
a a
in fig. 109.
3. b = (L The equation now becomes
q
r = a (cos 0+1)= 2 a cos2-.
124
POLAR COORDINATES
Here r is positive, except that when 0 = 180° r is zero. The
curve appears as in fig. 110 and is called the cardimd.
The cardioid is an epicycloid for
which the radius of the fixed circle
equals that of the rolling circle. The
proof of this is left to the student.
65. Relation between rectangular
and polar coordinates. Let the pole
0 and the initial line OM of a sys-
tem of polar coordinates be at the
same time the origin and the axis
of # of a system of rectangular
coordinates. Let P (fig. Ill) be
any point of the plane, (x, y) its
rectangular coordinates, and (r, 0) its polar coordinates.
Then, by the definition of the trigonometric functions,
a x
cos 0 = - >
r
Fig. 110
in0 = £.
r
sin
a)
Whence follows, on the one hand,
x = r cos 0,
y = r sin 6 ;
and, on the other hand,
r = y/z2+y2, sin0=--=i£
V**+y*
COS0 =
By means of (1) a transformation can be made from rectangular
to polar coordinates, and by means of (2) from polar to rectangular
coordinates.
Ex. 1. The equation of the cissoid (§ 48) is
y* =
2a — x
Substituting from (1) and making simple reductions, we have the polar
equation
r =
2osin20
COS0
THE STRAIGHT LINE
121
reaching it (fig. 104). As 6 approaches zero, r increases without limit.
If P is a point on the spiral and NP is the perpendicular to the initial line,
8in0
NP = r sin $ = a
e
Fig. 104
Hence, as 6 approaches zero as
a limit, NP approaches a (§ 95).
Therefore the curve comes con-
stantly nearer to, but never reaches,
the line LK, parallel to OM at a
distance a units from it. This line is therefore an asymptote. In the
figure the dotted portion of the curve corresponds to negative values of 0.
Ex. 8. The logarithmic spiral,
r = e<&.
When 6 = 0, r = 1. As 0 increases, r
increases, and the curve winds around
the origin at increasing distances from
it (fig. 105). When 0 is negative and
increasing numerically without limit, r
approaches zero. Hence the curve winds
infinitely often around the origin, continu-
ally approaching it. The dotted line in the
figure corresponds to negative values of 0.
A property of this spiral is that it cuts
the radius vectors at a constant angle. The
student may prove this after reading § 103.
Fig. 106
We shall now give examples of the derivation of the polar
equation of a curve from the definition of the curve.
62. The straight line. Let LK (fig. 106) be a straight line
perpendicular to OD. Let the angle MOD be denoted by a,
and let OD =p ; then p is the normal
distance of LK from the pole. K
Let P(r, ff) be any point of LK.
Then, by trigonometry,
or
OP cos D OP = OD,
r cos(0 — oc)=p,
a)
which is the equation of the straight
line.
M
Fig. 106
122 POLAR COORDINATES
If a = 0 and p = a, we have the special equation
r cos 0 = a,
or r = a sec 0. (2)
If the straight line passes through the origin, p = 0. The
equation of the line then becomes
, cos (0 — a) = 0,
or simply 5 = - + a,
which is of the form 0 = c. (3)
63. The circle. Let 0 (J, a) be the center and a the radius of
a circle (fig. 107). Let P(r, 0) be any point of the circle, and
draw the straight lines 0(7, OP, and CP.
By trigonometry, we have
OP + OC - 2 OP • 0(7 cos POC = (7P .
Noting that cos POO = cos (0 - a), OP = r, 00=5, and
OP = a, and substituting in the equation, we have the result
r2-2rbcos(0-a)+b2 = a2 (1)
as the polar equation of the circle. ( y^^^Jp(re)
When the origin is at the center of
the circle, 6 = 0 and (1) becomes simply
r = a. (2) Fig. 107
When the origin is on the circle, b = a and (1) becomes
r — 2 a cos (0 — a) = 0 ;
which may be written r — a0 cos 0 -f a± sin 0, (3)
IT
where a0 and a1 are the intercepts on the lines 0=0 and 0 = —
respectively.
When the origin is on the circle and the initial line is a
diameter, (3) becomes r = ^ cog 0 (4)
When the origin is on the circle and the initial line is tangent
to the circle, (3) becomes . n xfrx
v J r = a sin 0. (5)
THE LIMAQON
123
Fig. 108
64. The lima^on. Through any fixed point 0 (fig. 108) on the
circumference of a fixed circle draw any line cutting the circle
again at 2), and lay off on this line a
constant length measured from D in
either direction. The locus of the
points P and Q thus found is a curve
called the limapon.
Take 0 as the pole, and the diameter
OA as the initial line, of a system of
polar coordinates, and call the diame-
ter of the circle a and the constant
length 6. Then it is clear that the
entire locus can be found by caus-
ing OD to revolve through an angle of 360° and laying off
DP =6, always in the direction of the terminal line of AOD.
Let P be (V, 0), where 0 = AOD. Then r = OD+DP when
6 is in the first or the fourth quadrant, and r = — OD + DP
when 0 is in the second or the third
quadrant. But it appears from the
figure that OD = OA cos 0 when 0 is
in the first or the fourth quadrant,
and that OD = — OA cos 0 when 0 is
in the second or the third quadrant.
Hence, for any point on the lima§on,
r = a cos 0 + b.
/
*witi)
In studying the shape of the
curve there are three cases to be
distinguished:
Fig. 109
1. b >cl r is always positive; the curve appears as in fig. 108.
2. b<a. r is positive when cos0> > negative when
a
cos 0 < * and zero when cos 0 = • The curve appears as
a a
in fig. 109.
8. b = Ow The equation now becomes
0
r = a (cos 0 +1) = 2 a cos2- .
124
POLAR COORDINATES
Here r is positive, except that when 0 = 180° r is zero. The
curve appears as in fig. 110 and is called the cardioid.
The cardioid is an epicycloid for
which the radius of the fixed circle
equals that of the rolling circle. The
proof of this is left to the student.
65. Relation between rectangular
and polar coordinates. Let the pole
0 and the initial line OM of a sys-
tem of polar coordinates be at the
same time the origin and the axis
of a? of a system of rectangular
coordinates. Let P (fig. Ill) be
any point of the plane, (#, y) its
rectangular coordinates, and (n, 0) its polar coordinates.
Then, by the definition of the trigonometric functions,
Fig. 110
cos
0 =
X
'■ — *
r
sin
0 =
y
■ — .
r
(1)
Whence follows, on the one hand,
x = r cos 0,
y = r sin 0 ;
and, on the other hand,
r = ^z2+y2, sin0=—=J£
V^+
y
COS0 =
Vx*+f
By means of (1) a transformation can be made from rectangular
to polar coordinates, and by means of (2) from polar to rectangular
coordinates.
Ex. 1. The equation of the cissoid (§ 48) is
y2 =
x*
2a — x
Substituting from (1) and making simple reductions, we have the polar
equation
r =
2asin2fl
COS0
THE CONIC 125
Ex. 8. The polar equation of the lemniscate (Ex. 3, § 60) is
r2=2a2cos2 0.
Placing cos 20 = cos2 6— sin20 and substituting from (2), we have the
rectangular equation
(x2 + y2)2 = 2 a2(x2- y2).
66. The conic, the focus being the pole. From § 46, the equa-
tion of a conic when the axis of # is an axis of the conic and
the axis of y is a directrix is
We may transfer to new axes having, the focus of the conic as
the origin and the axis of the conic as the axis of x by placing
x = c + a/, y = yl ,
thus obtaining x,2+yf2 = e2(V + c)\
If we now take a system of polar coordinates having the focus
as the pole and the axis of the conic as the initial line, we have
xr=r cos 0, y'=r sin 0.
The equation then becomes
r*=e2(r cos 0 + cy,
which is equivalent to the two equations
ce ee
r = - -i r = —
1 — e cos 0 1+e cos 0
Either of these equations alone will give the entire conic.
To see this, place 0 = 0 in the second equation, obtaining
— ce
r =
1 1 + e cos 0X
Now place 0 = 7r + 0 in the first equation, obtaining r = — rx.
The points (rx, 0^) and (— rx, ir + 0^) are the same. Hence any
point which can be found from the second equation can be
found from the first.
Therefore r =
1 — e cos 0
is the required polar equation.
126
POLAR COORDINATES
67. Examples. Polar coordinates may be used with great
advantage in the solution of problems involving a number of
straight lines radiating from a given point, the given point
then being taken as the pole of the system of coordinates.
This use is illustrated in the following examples:
Ex. 1. Prove that if a secant is drawn
through the focus of a conic, the sum of the
reciprocals of the segments made by the
focus is constant.
Let PXP2 (fig. 112) be any secant through
the focus F, and let FP1 = rv FP2 = r2, and
the angle MFPX = $. Then the polar coordi-
nates of Px are (rv 0) and those of P2 are
(r2, $ + w). From the polar equation of the
conic, we have
r, =
r„ =
ce
1 — e cos 6
Fig. 112
ce
ce
Hence
i + i
1 — e cos (0 + ir)
2
ce
1 + e cos 0
Ex. 2. Find the locus of the middle points of a system of chords
of a circle all of which pass through a fixed point.
Take any circle with the center C
(fig. 113), and let 0 be any point in the
plane. If 0 is taken for the pole, and OC
for the initial line, of a system of polar
coordinates, the equation of the circle is
r2 - 2 rb cos $ + b2 - a2 = 0. (1)
Let PXP2 be any chord through 0, and
let OPx = rv OP2 = r2. Then i\ and r2 are
the two roots of equation (1) which corre-
spond to the same value of 0. Hence
rx + r2 = 2b cos 0.
Fig. 113
If Q is the middle point of PXP2 and we now place OQ = r, we have
But this is the polar equation of a circle through the points 0 and C.
PROBLEMS
127
PROBLEMS
Plot the following curves :
1. r = a sin 2 0.
2. r = acos30.
. 0
3. r = asm-*
e
4. r = a cos ^ •
*> • a^
5. r = asin8;r-
6. r2 = a2 sin 0.
7. r2 = a2sin30.
8. r2 = a2sin40.
9. r = a(l + sin0).
10. r = a(2 + sin0).
11. r = a(l-fcos2 0).
12. r = a(l — cos 20).
13. r = a(l-fcos3 0).
14. r = a(2 + cos20).
15. r = a(l + 2sin0).
16. r = a(l + 2cos20).
17. r = a(l + 2cos3 0).
3 0
18. r = 2 + sin — •
Zi
19. r2 = a2(l — cos0).
20. r2 = a2(l + 2cos2 0).
21. r = atan0.
22. r = a tan 20.
23. r = a tan-«
Z
24. r = a sec 2 0.
25. r = a sec-*
Z
26. r = a sec2-*
Z
27. r = a(l + sec0).
28. r = a(l + 2sec0).
29. r = a(2+ sec0).
30. r cos 0 = a cos 2 0.
a . a
31. r = 7 + -^*
cos 0 sin 0
32. r = l — 2 0.
2
33. r =
0-1
Plot each pair of the following curves in one diagram and find
their points of intersection :
34. rcos
35. rcos
^0 - jj = a, r cos^O + jj = a.
(0 — g) = a, r cosf 0 — g j = a.
36. r cos( 0 — — j = aV2, r = 2 a cos 0.
37. r2 = a2 sin 0, r2 = a2 sin 2 0.
38. r = a(l + sin2 0), r2 = 4a2sin20.
39. r2 = a2 sin 0, r2 = a2 sin 3 0.
128 POLAR COORDINATES
40. 0 is a fixed point and LK a fixed straight line. If any straigh
line through 0 intersects LK in Q, and a point P is taken on thi:
line so that OP • OQ = k2, find the locus of P.
41. A straight line OA of constant length a revolves about 0.
From A a perpendicular is drawn to a fixed straight line OM, inter-
secting it in B. From B a perpendicular is drawn to OA, intersecting
it in P. Find the locus of P, OM being taken as the initial line.
42. 0 is a fixed point of a circle of radius a, and OM is a fixed
straight line passing through the center of the circle. A straight
line is drawn from 0 to any point P1 of the circle, and from Px a
straight line- is drawn perpendicular to OM, meeting OM at Q. From
Q a straight line is drawn perpendicular to OPv meeting OPx at P.
Find the equation of the locus of P, taking 0 as the origin of coor-
dinates and OM as the initial line.
43. MN is a straight line perpendicular to the initial line at a
distance a from O. From 0 a straight line is drawn to any point B
of MN. From B a straight line is drawn perpendicular to OB, inter-
secting the initial line at C. From C a straight line is drawn per-
pendicular to BC, intersecting MN at D. Finally, from D a straight
line is drawn perpendicular to CD, intersecting OB at P. Find
the locus of P.
Transform the following equations to polar coordinates :
44. xy = 7. 46. x* + xhf — a2!/2 = 0.
45. x2 + if - Sax - Say = 0. 47. (x2 + y2)2 = a2^ - y2).
48. Find the polar equation of the strophoid when the pole is O
and the initial line is OA (fig. 82).
Transform the following equations to rectangular coordinates :
49. r cos (o — ^ j + r cos U + ^ j = 12.
50. r=asin0. 51. r = a tan 0. '
52. Find the Cartesian equation of the rose of four petals
r = asin20.
53. Find the Cartesian equation of the cardioid r = a(l — cos 0).
54. Find the Cartesian equation of the lima^on r = a cos 0 -f b. .
r
PROBLEMS 129
55. In a parabola prove that the length of a focal chord which
makes an angle of 30° with the axis of the curve is four times the
focal chord perpendicular to the axis.
56. A comet is moving in a parabolic orbit around the sun at the
focus of the parabola. When the comet is 100,000,000 miles from
i:he sun the radius vector makes an angle of 60° with the axis of the
orbit. What is the equation of the comet's orbit ? How near does
it come to the sun ?
57. A comet moving in a parabolic orbit around the sun is
observed at two points of its path, its focal distances being 5 and 15
million miles, and the angle between them being 90°. How near
does it come to the sun?
58. If a straight line drawn through the focus of an hyperbola
parallel to an asymptote meets the curve at P, prove that FP is one
fourth the chord through the focus perpendicular to the transverse
axis.
59. The focal radii of a parabola are extended beyond the curve
until their lengths are doubled. Find the locus of their extremities.
60. If Pt and P2 are the points of intersection of a straight line
drawn from any point O to a circle, prove that OPx • OP2 is constant.
61. If Pj and Pa are the points of intersection of a straight line
from any point O to a fixed circle, and Q is a point on the same
2 OP • OP
straight line such that OQ = ^ 2 > find the locus of Q.
8 OPt + OP2
62. Secant lines of a circle are drawn from the same point on the
circle, and on each secant a point is taken outside the circle at a
distance equal to the portion of the secant included in the circle.
Find the locus of these points.
63. From a point O a straight line is drawn intersecting a fixed
circle at P, and on this line a point Q is taken so that OP • OQ = k2.
Find the locus of Q.
64. Find the locus of the middle points of the focal chords of
a conic.
65. Find the locus of the middle points of the focal radii of
a conic.
66. If PXFP2 and QXFQ2 are ^w0 perpendicular focal chords of a
conic, prove that p p pp + QF,FQ ™ constant.
1
3
7
%* — ~z*
aj.= -7'
#_ = T*
1 2
2 4
8 8
CHAPTER IX
SLOPES AND AREAS
68. Limits. A variable is said to approach a constant as a
limit, when, under the law which governs the change of value of
the variable, the difference between the variable and the constant
becomes and remains less than any quantity which can be named,
no matter how small.
If the variable is independent, it may be made to approach a
limit by assigning to it arbitrarily a succession of values follow-
ing some known law. . Thus, if # is given in succession the values
2n-l
and so on indefinitely, it approaches 1 as a limit. For we
may make x differ from 1 by as little as we please by taking n
sufficiently great ; and for all L s T
larger values of n the differ- + J — T 4t x
ence between x and 1 is still
smaller. This may be made
evident graphically by marking off on a number scale the
successive values of x (fig. 114), when it will be seen that
•
1 ±-± rx 1 1 * J. 1
s 5 7 O s e 7 » 4
i HH 1 hH 1 T ■ T
Fig. 115 •
the difference between x and 1 soon becomes and remains
too minute to be represented.
Similarly, if we assign to x the succession of values
1111 r iwi 1
n + 1
x approaches 0 as a limit (fig. 115).
130
LIMITS
131
If the variable is not independent, but is a function of #, the
values which it assumes as it approaches a limit depend upon
the values arbitrarily assigned to z. For example, let y =/(i),
and let x be given a set of values,
x«
x
2'
x„
xA
X,
'»»
approaching a limit a. Let the corresponding values of y be
y* y,» y* y* -■» y»» ••••
Then, if there exists a number A such that the difference between
y and A becomes and remains less than any assigned quantity, y
is said to approach iasa limit as x approaches a in the manner
indicated. This may be seen
graphically in fig. 116, where
the values of x approaching a
are seen on the axis of abscissas,
and the values of y approach-
ing A are seen on the axis
of ordinates. The curve of the
function is continually nearer
to the line y = A.
In the most common cases the
limit of the function depends
only upon the limit a of the
independent variable and not
upon the particular succession of values that x assumes in
approaching a. This is clearly the case if the graph of the
function is as drawn in fig. 116.
Ex. 1. Consider the function
Fig. 116
y =
x2 + 3 x - 4
x-1
and let x approach 1 by passing through the succession of values
s = l.l, a; = 1.01, x = 1.001, 2 = 1.0001, •••.
Then y takes in succession the values
y = 5.1, y = 5.01, y = 5.001, y = 5.0001.
It appears as if y were approaching the limit 5. To verify this we place
x = 1 + h, where h is not zero. By substituting and dividing by h, we find
132 SLOPES AND AREAS
y = 5 + h. From this it appears that y can be made as near 5 as we
please by taking h sufficiently small, and that for smaller values of h,
y is still nearer 5. Hence 5 is the limit of y as a: approaches 1. More-
over, it appears that this limit is independent of the succession of values
which x assumes in approaching 1.
Ex. 2. Consider y — as x approaches zero.
1 — Vl — x
Give x in succession the values .1, .01, .001, .0001, • • • . Then y takes the
values 1.9487, 1.9950, 1.9995, 1.9999, • • •, suggesting the limit 2.
In fact, by multiplying both terms of by 1 + vl- x, we find
1 — Vl — x
y — 1 + Vl— x for all values of x except zero.
Hence it appears that y approaches 2asx approaches 0.
We shall use the symbol = to mean "approaches as a limit."
Then the expressions
Lim x = a
and x = a
have the same significance.
The expression Lim/(a;) = ^
x = a
is read "The limit of f(x), as x approaches a, is AS
69. Theorems on limits. In operations with limits the follow-
ing propositions are of importance :
1. The limit of the sum of a finite number of variables is equal
to the sum of the limits of the variables.
We will prove the theorem for three variables; the proof is
easily extended to any number of variables.
Let X, F, and Z be three variables, such that Lim X=A,
Lim Y=B, LimZ= C. From the definition of limit (§ 68) we
may write X=A+a, F = J5-f-6, Z=C+c, where a, 6, and c
are three quantities each of which becomes and remains
numerically less than any assigned quantity as the variables
approach their limits.
Adding, we have
.Y+ F+ Z = A +B + C+ a + b + e.
LIMITS 133
Now if € is any assigned quantity, however small, we may
make a, 6, and c each numerically less than - , so that a + b + c is
numerically less than €. Then the difference between X+ Y+ Z
and A + B + C becomes and remains less than e ; that is,
Lim (X+ F+ Z) = A + B + C= Lim X+ Lim F+ Lim Z.
2. The limit of the product of a finite number of variables is
equal to the product of the limits of the variables.
Consider first two variables X and F, such that Lim X=A and
Lim Y=B. As before, we have X= A+ a and Y=B+ b. Hence
XY= AB+bA+aB+ ab.
Now we may make a and b so small that bA, aB, and ab are
each less than - , where e is any assigned quantity, no matter how
small. Hence
Lim XY= AB = (Lim X) (Lim F).
Consider now three variables X, F, Z. Place XY= U. Then,
as just proved, Lim^ = (LimCr)(LimZ);
that is, Lim XYZ = (Lim XY) (Lim Z)
= (Lim X) (Lim F) (Lim Z).
Similarly, the theorem may be proved for any finite number
of variables.
3. The limit of a constant multiplied by a variable is equal to
the constant multiplied by the limit of the variable.
The proof is left for the student.
4. The limit of the quotient of two variables is equal to the
quotient of the limits of the variables, provided the limit of the
divisor is not zero.
Let X and Y be two variables, such that Lim X= A and
Lim F= B. Then, as before, X= A + a, Y=B + b.
TT X A+a , X A A + a A aB-bA
HeDCe F = *T*' md Y-B = B^b-B = ^bB>
AC
134
SLOPES AND AREAS
Now the fraction on the right of this equation may be made less
than any assigned quantity by taking a and b sufficiently SmalL
Hence
T. X A LimX
Lim — = — = «
Y B Lim Y
Fig. 117
The proof assumes that B is not zero.
70. Slope of a curve. By means of the conception of a limit
we may extend the definition of " slope," given in § 6 for a
straight line, so that it may be applied to any curve. Let
Px and Pz be any two points upon a curve
(fig. 117). If ij and B2 are connected by
a straight line, the slope of this line is
-± — ^. If B and B are close enough to-
x — x*
2 1
gether, the straight line I[Bi will differ only
a little from the arc of the curve, and its
slope may be taken as approximately the
slope of the curve at the point ij. Now
this approximation is closer, the nearer the point 2J is to ij.
Hence we are led naturally to the following definition:
The slope of a curve at a point Px (xv y^) is the limit approached
by the fraction — ^ where x2 and y2 are the coordinates of a
X2 ~~ Xl
second point B2 on the curve and where the limit is taken as ij
moves toward I[ along the curve.
Ex. 1. Consider the curve y = x2 and the point (5, 25) upon it, and let
*i = 5> Vi = 25-
We take in succession various values for x2 and y2f corresponding to
points on the curve which are nearer and nearer to (xv yx), and arrange
our results in a table as follows :
X9
V2
x2 — Xj
2>2-yi
x2 — xx
6
6.1
6.01
6.001
36
26.01
25.1001
25.010001
1
.1
.01
.001
n
1.01
.1001
.010001
11
10.1
10.01
10.001
INCREMENT 135
The arithmetical work suggests the limit 10. To verify this, place
x* = 5 + h. Then y2 = 25 + 10 h + h2. Consequently J _ * = 10 + h, and
as x2 approaches xv h approaches 0 and _ approaches 10. Hence the
x2 xx
slope of the curve y = x2 at the point (5, 25) is 10.
Ex. 2. Find the slope of the curve y = - at the point (3, J).
x
We have here xx = 3, yx = J.
1
We place x2 = 3 + h, y2 = :
3 + £
Then a:« — x* = A, y„ — y* = — ■ > and ^ — ^ = — - — — - .
As Pa approaches P! along the curve, A approaches 0, and the limit of
y9 — y. 1
— is — q ; hence the slope of the curve at the point (3, J) is — £.
x2 — Xj y
In a similar manner we may find the slope of any curve the
equation of which is not too complicated; but when the equa-
tion is complicated, there is need of a more powerful method
for finding the limit of — -• This method is furnished by
x*~~x\
the operation known as differentiation, the first principles of
which are explained in the following articles.
71. Increment. When a variable changes its value, the quan-
tity which is added to its first value to obtain its last value is
called its increment. Thus, if # changes from 5 to 5 J, its incre-
ment is J. If it changes from 5 to 4|, the increment is — \.
So, in general, if x changes from x1 to #2, the increment is
x2 — xx. It is customary to denote an increment by the symbol
A (Greek delta), so that
Ax = x2-xt, and x% = xx + te.
If y is a function of #, any increment added to x will cause
a corresponding increment of y. Thus, let y=f(x) and let x
change from xx to x%. Then y changes from yx to y2, where
Hence Ay = y2 - yx =f(xx + As) -/<»
136
SLOPES AND AREAS
x=a
■hrX
Fig. 118
72. Continuity. A function y is called a continuous function
of a variable x when the increment of y approaches zero as the
increment of x approaches zero.
It is clear that a continuous function cannot change its value
by a sudden jump, since we can make the change in the function
as small as we please by taking the increment of x sufficiently
small. As a consequence of this, if a continuous function has a
value A when x = a, and a value B
when x = b, it will assume any value
(7, lying between A and J5, for at
least one value of x between a and b
(fig. 118).
In particular, if f(a) is positive
and f(b>) is negative, f(x) = 0 for at
least one value of x between a and b.
When Ax and Ay approach zero together, it usually happens
that — - approaches a limit. In this case y is said to have a
Ax
derivative, defined in the next article.
73. Derivative. When y is a continuous function of x, the deriva-
tive of y with respect to x is the limit of the ratio of the increment
of y to the increment of x, as the increment of x approaches zero.
The derivative is expressed by the symbol -j-\ or, if y is
ax
expressed by /(#), the derivative may be expressed by /'(#).
Thus, if y =/(z),
§3L =/'(*) = Lim ^ = Lunf^ + ^-fto.
dX Ax = oA# . Ax = 0 Ax
The process of finding the derivative is called differentiation,
and we are said to differentiate y with respect to x. The process
involves, according to the definition, the following four steps :
1. The assumption of an increment of x.
2. The computation of the corresponding increment of y.
3. The division of the increment of y by the increment of x.
4. The determination of the limit approached by this quotient
as the increment of x approaches zero.
DIFFERENTIATION OF A POLYNOMIAL 137
Ex. 1. Find the derivative of y when y = x8.
(1) Assume Ax = h.
' (2) Compute Ay = (x + h)* - x* = 3 x2h + 3 zfc2 + A8.
(3) Find ^ = 3 a? + 3 xh + £2.
Ax
(4) The limit is evidently 3 x2. Hence ^ = 3 *2-
Ex. 8. Find the derivative of - .
1 x
(1) Place y = - and assume Ax = A.
(2) Compute Ay =
x + h x x2 + xh
(3) Find ^2 = - X
Ax x2 + xA -
(4) The limit is clearly - > and therefore -^ = .
It appears that the operations of finding the derivative of /(#)
are exactly those which are used in finding the slope of the curve
y =f(x). Hence the derivative is a function which gives the slope
of the curve at each point of it
74. Differentiation of a polynomial. The obtaining of a deriv-
ative by carrying out the operations of the last article is too
tedious for practical use. It is more convenient to use the
definition to obtain general formulas which may be used for
certain classes of functions. In this article we shall derive
all formulas necessary to differentiate a polynomial.
1. — ±z — -=naxn~1, where n is a positive integer and a any
ax
constant.
Let y = aa?.
(1) Assume Ax = h.
(2) Then Ay = a(x + K)n- ax*
= aimT-'h + *^[2 1V-2A2+ . . . + hn
(3) |2 = a/^-i+H^
(4) Taking the limit, we have-r^ = nax" _1.
ax
138
SLOPES AND AREAS
2. \ ^ = a, where a is a constant.
dx
This is a special case of the preceding formula, n being here
equal to 1. The student may prove it directly.
dc
3. — = 0, where c is a constant.
dx
Since c is a constant, Ac is always 0, no matter what the
A/? dc
value of x. Hence — = 0, and consequently the limit — = 0.
i ir ax
4. 2%e derivative of a polynomial is found by adding the
derivatives of the terms in order.
This is a special case of a more general theorem (3, § 82).
The proof of the special case before us may be easily givien
by the student or may be assumed temporarily.
Ex. Find the derivative of
f(x) = 6x5-3:r4 + 5:r8-7:r2+8a:-2.
Applying formulas 1, 2, or 3 to each term in order, 'we have
f'(x) = 30 s4 - 12a;8 + 15 x2 - Ux + 8.
75. Sign of the derivative. A function of a? is called an
increasing function when an increase in x causes an increase
in the function. A function of x is called
a decreasing function when an increase in x
causes a decrease in the function. The graph
of a function runs up toward the right hand
when the function is increasing and runs
down toward the right hand when the func-
tion is decreasing. Thus x2— x — 6 (fig. 119)
is decreasing when x < \ and increasing when
x>\.
The sign of the derivative enables us to
determine whether a function is increasing
or decreasing in accordance with the follow-
ing theorem:
When the derivative of a function is posi-
tive, the function is increasing; when the derivative is negative^ the
function is decreasing.
Fig. 119
SIGN OF THE DERIVATIVE
139
To prove this, consider y=f(x), and let us suppose that
-^ is positive. Then, since -=- is the limit of — , it follows that —
ax dx Ax Ax
is positive for sufficiently small values of Ax; that is, if Ax
is assumed positive, Ay is also positive, and the function is
increasing. Similarly, if -^ is negative, Ay and Ax have oppo-
site signs for sufficiently small values of Ax, and the function is
decreasing by definition.
Ex. 1. If y = x2 — x — 6, -f- = 2 x — 1, which is negative when x < £
ax
and positive when x> J. Hence the function is decreasing when x<\ and
increasing when x > J, as is shown in fig. 119.
Ex. 2. If y= \(x*-Sx2-9x + 27), ^-=%x*- *x- f = £(s+l)(:r-3).
\JLJU
Now -^ is positive when x< — 1, negative when -KK3, and positive
ax
when x > 3. Hence the function is increasing when x < — 1, decreasing
when a: is between — 1 and 3, and increas-
ing when x > 3 (fig. 120).
It remains to examine the cases in
dy
which -— = 0. Referring to the two ex-
dx
amples just given, we see that in each
the values of x which make the deriv-
ative zero separate those for which the
function is increasing from those for
which the function is decreasing. The
points on the graph which correspond
to these zero values of the derivative
can be described as turning points.
Likewise, whenever fr(x) is a continuous function of x, the
values of x for which it is positive are separated from those
for which it is negative by values of x for which it is zero
(§ 72). Now in most cases which occur in elementary work,
fr(x) ig a continuous function. Hence we may say,
The values of x for which a function changes from an increas-
ing to a decreasing function are, in general, values of x which
make the derivative equal to zero.
Fig. 120
uo
SLOPES AND AREAS
Fig. 121
The converse proposition is, however, not always true. A
value of x for which the derivative is zero is not necessarily a
value of x for which the function changes
from increasing to decreasing or from,
decreasing to increasing. For consider
£(a;8- 9^+27^-19).
Its derivative is x2— 6 x + 9 = (x — 3)2,
which is always positive. The function
is therefore always increasing. When
x = 3 the derivative is zero, and the
corresponding shape of the graph is
shown in fig. 121.
76. Tangent line. A tangent to a curve is the straight line
approached as a limit by a secant line as two points of intersection
of the secant and the curve are made to approach coincidence.
Let Px and i£ be two points on a curve. Then if a secant is
drawn through PY and P2 of a curve (fig. 122) and the point
P2 is made to move along the curve toward ij, which is kept
fixed in position, the secant will turn on ij as a pivot and will
approach as a limit the tangent PXT. The point ij is called the
point of contact of the tangent.
From the definition it follows that the slope
of the tangent is the same as the slope of the
curve at the point of contact; for the slope
of the tangent is evidently the limit of the
slope of the secant, and this limit is the slope
of the curve, by § 70.
The equation of the tangent is readily written by means of
§ 28 when the point of contact is known. Let (xv y^) be
-~ ) denote the value of -f- when
ax/i dx
Fig. 122
the point of contact, and let
x = xx and y = yx. Then (x^ y^) is a point on the tangent and
-^ J is its slope. Therefore its equation is
y-yx
(dy
x — x.
a)
THE DIFFERENTIAL
141
Ex. 1. Find the equation of the tangent to the curve y = x* at the
point (xv yx) on it.
Using formula (1), we have
y-yl = Sx12(x — x1).
But since (xv yx) is on the curve, we have yx = a:*. Therefore the equa-
tion can be written 0 2 n • •
y = 3 x{ x — 2 x°. t
Ex. 2. Find the equation of the tangent to
y = x2 + 3 x
at the point the abscissa of which is 2.
ax
If ,1 = 2, then * = 10 and (g) = 7. Fi(j m
Therefore the equation is
y — 10 = 7 (* — 2), or y = 7 a; — 4.
If PJ (fig. 123) is a tangent line and <£ the angle it makes
with OXy its slope equals tan <£, by § 32. Hence
tan<f> = -r^-
77. The differential. Let the function /(#) be represented by
the curve y =/(#), and let P and Q be two neighboring points
of the curve (fig. 124). Draw the
tangent PT and the lines PR and
RQ parallel to the axes, RQ and
PT intersecting at T. Then, from
the preceding work,
PR = Az,
RQ = Ay,
tan RPT=f(x).
RT= (tan RPT) PR =f(x) **-
Fig. 124
The quantity f!(x) Ax is called the differential of y and is
represented by the symbol dy. Accordingly
<fy =/'(*) A*-. (1)
142 SLOPES AND AREAS
This definition is true for all forms of the function/ (x) and is
accordingly true when y =/(#) = x. In this case fr(x) = 1, and
formula (1) gives dx = AiC, (2)
Substituting from (2) into (1), we have the final form
dy=f'(x)dx. (3)
To sum this up : The differential of the independent variable is
equal to the increment of the variable; the differential of the function
is equal to the differential of the independent variable multiplied by
the derivative of the function.
It is important to notice the difference between Ay and dy.
The figure shows that, in general, they are not equal, but that
they become more nearly equal as Ax approaches zero. Without
using the figure, we may proceed thus :
Since Lim -^ =/'(#),
Ax = 0 AX
where Lim e = 0 ; and hence
Ax = 0
Ay =fr(x) Ax + zAx = dy + eAx.
Ex. 1. Let y = x*.
We may increase x by an increment Ax equal to dx. Then
Ay = (x + dx)* - xs = 3 x2dx + 3 x (dx)2 + (dx)*.
On the other hand, by definition,
dy = 3 x2dx.
It appears that Ay and dy differ by the expression 3 x (dx)2 + (dx)*, which
is very small compared with dx.
Ex. 2. If a volume v of a perfect gas at a constant temperature is under
k
the pressure p, then v = - » where k is a constant. Now let the pressure be
P
increased by an amount A/? = dp. The actual change in the volume of the
gas is then the increment
A = * — ^ — *^P kdp I 1
p + dp p~ p(p + dp)~~ p2 I x + dp
The differential of v is, however, \ I
. __ kdp
AREA 143
It is to be emphasized that dx and dy are finite quantities,
subject to all the laws governing such quantities, and are not
to be thought of as exceedingly minute. Consequently both sides
of (3) may be divided by dx, with the result
dx
That is, the derivative is the quotient of two differentials.
This explains the notation already chosen for the derivative.
So, in general, the Hmit of the quotient of two increments i» equal
to the quotient of the corresponding differentials.
/'CO-
For let
y=f(x) and z=$(x).
Then
Ay=f'(x)Ax + €jAx,
Az=$(x)Ax + eaAx,
dy =f'(x) dx,
dz = <l>'(x) <&>
d
Az £'(» + <:/
Whence
LimA? LimW + e' ££* £.
Az <f>'(x) + ea 4>'(x) dz
78. Area under a curve. Let LK (fig. 125) be a curve with
equation y =/(#), and let OE = a and OB = b. It is required
to find the area bounded
by the curve LK, the axis
of x, and the ordinates at E
and B.
For convenience, we as-
sume in the first place that
a < b and that /(a;) is positive
for all values of x between a
and h. We will divide the
line EB into n equal parts
by placing Ax, and laying off the lengths EM^M^^
M1Mt=.--=M^iB = Ax. (In fig. 125, n = 9.)
r, il, MtMiMt H, M, B
Fiq. 125
144 SLOPES AND AREAS
Let OMx= xv OM2 = x2, . • ., OMn_1= xn_v Draw the ordinates
and BC. Draw also the lines DRV JJB,, %RSJ • • •, J^x-B* parallel
to OX. Then
/(a) A# = the area of the rectangle EDR%MV
f(x^)Ax = tke area of the rectangle My^R%Mv
/(#2)Aa; = the area of the rectangle M2I^RZMZ1
• ••••••••
/(a;n_1) Aa; = the area of the rectangle Mn_1^_1RnB.
The sum
f(a) Ax +/(*1) Ax +/(x2) A* + • • • +/(*.-i) Ax (1)
is then the sum of the areas of these rectangles and equal to
the area of the polygon EDRXI^R2 • • • Rn_1I^_1RnB. It is evident
that the limit of this sum as n is indefinitely increased is the
area bounded by ED, EB, BC, and the arc DC.
The sum (1) is expressed concisely by the notation
t = 0
where 2 (sigma), the Greek form of the letter S, stands for the
word " sum," and the whole expression indicates that the sum
is to be taken of all terms obtained from f(x{) A# by giving to i
in succession the values 0, 1, 2, 3, • • •, n — 1, where xQ = a.
The limit of this sum is expressed by the symbol
J f(^)dx,
where I is a modified form of S.
i=n-l
) j is a
Jf*b i = n-l
f(x) dx = Lim 5)/(a?<) ^x = the area EBCD.
It is evident that the result is not vitiated if ED or BC is
of length zero.
ATIEA
o find the area bounded by the curve y =
x, and the ordinate* x = 2 and x = 3 (fig. 126).
(1) We may divide the axis of x between x = 2 and x =
i - * 3~2 i
placing A* = -j^- = .1.
We make then the following calculation :
a = 2,
*, = 2.1,
ua
0882
xt = 2.2,
/(*S)A* =
0968
ra = 2.3,
/(Xjte =
1058
i4 = 2.4,
f(ti*x =
1152
*, = 2.5,
/(*£** =
1250
i, = 2.6,
f(zj*x =
1352
x, = 2.7,
/(«,)Ar =
1458
x, = 2.8,
/<*,)A* =
1508
i, = 2.9,
f(x,)Ax =
1082
1
2170
The first approximation to the ar
example, the value of the
sum (1) for n = 10. 1
(2) As a better approxi-
mation the student may
compute the sum for a = 20
and te = ^-^ = .05. The
ea is therefore 1.217, which i
A
,for
4
/
result is 1.2418.
(3) If we take n = 100
and A* = ^-^ = .01, the
^A
1
1
calculation is very tedious. 0
X
The result, however, is
1.26167. These successive
determinations appear to be approaching a limit. By subsequent
methods it will be shown that this
» 1A-
It is obvious that the direct calculation of the sum (1) is very
tedious, if not practically impossible, if the number of terms is
very large. Some other method must be found to determine the
limit of the sum as n increases indefinitely. This method is fur-
nished by the discussion in the following sections.
146
SLOPES AND AREAS
79. Differential of area. Let .any one of the rectangles of
fig. 125 be redrawn in fig. 127 and relettered, for convenience,
MNRP. Draw also QS and complete the rectangle MNQ&
Let A denote the variable area
EMPD. Then
MN=Ax, RQ=Ay,
MNQ£=AA,
MNRP = MP . MN= yAx,
MNQS = NQ- MN
= (y + Ay) Ax.
But, from the figure,
that is,
whence
MNRP < MNQP < MNQS;
yAx < AA < (y + Ay) As,
AA
y<-^<y+*y-
Now as Ax approaches zero as a limit,
AA
dA
approaches
Ax a j dz
y is unchanged, and y + Ay approaches y. Hence
lies between y and y + Ay, also approaches y; that is,
dA
which
dx
= y =/(*).
a)
In the differential notation we have
dA = f(x) dx. (2)
To find the area it is therefore necessary first to find a func-
tion whose derivative is f(x) and whose differential is f(x) dz.
80. The integral of a polynomial. The process by which a
function is found from its derivative or its differential is called
integration, and the result of the process is called the integral of
the derivative.
Integration is expressed by the symbol J ; thus,
jf(x)dx = F(x), (1)
THE DEFINITE INTEGRAL 147
where F(x) is a function of which the derivative is f(x). The
process may be carried out in the simpler cases by reversing the
rules for differentiation. Thus,
I 2xdx = a?+c, I Za?dx=zx*+cy
by the formulas of § 74.
In these results c may be any constant whatever, since — - = 0.
ax
In fact, any derivative has an infinite number of integrals dif-
fering by a constant. The most general form of formula (1) is
/•
f(x)dx=F(x) + C, (2)
where F(x) is any particular function whose derivative is f(x)
and C is any arbitrary constant, called the constant of integration.
To integrate a polynomial we need to know that its integral is
the sum of the integrals of its terms and that the integral of
each term is found either by the formula
/
. , az»+1
aaf'dx = + c
w+1
or by the formula / adx = ax + c.
These are simply the formulas of § 74 reversed.
Ex. f(x3+5z2+7z + 3)tfx = ^ + ^ + ^ + 3x4-0.
«/ 4 3 2
81. The definite integral. Return now to the problem of area.
From §79, dA=f(x)dx,
whence, by use of § 80, A = F(x) + C. (1)
This is the area of the figure EMPD (fig. 127), in which the
line MP can be drawn anywhere between ED and BC. But if the
line MP coincides with ED, A = 0 and x=j}. Substituting these
values in (1), we have Q = F , . + c
whence C = — F(a).
148 SLOPES AND AREAS
Formula (1) now becomes
A = F(x) - F(a).
The area A becomes the area EBCD when x = b. Then
area EBCD = F(b) - F(a).
This gives us our desired method of evaluating the limit of
the sum (1), § 78, and may be expressed by the formula
s:
f(x)dx = F(V)-F(a). (2)
The limit of the sum (1), § 78, which is denoted by J f(x)dx,
is called a definite integral, and the numbers a and b are called
the lower limit and the upper limit* respectively of the definite
integral.
This result gives the following rule for evaluating a definite
integral :
To find the value of I f(x)dx, evaluate \ f(x)dxy substitute
x = b and x = a successively, and subtract the latter result from
the former.
It is to be noticed that in evaluating I f(x)dx the constant
of integration is to be omitted, since — F(a) is that constant.
However, if the constant is added, it disappears in the sub-
traction, since
[F(b) + C]- IF (a) + C] = F(b) - F(a).
In practice it is convenient to express F(V) — F(a) by the
symbol [F(x)]ba, so that
x
f(x)dx = \_F(x)fa.
Ex. The example of § 78 may now be completely solved. The required
areais f8*2,/ -J*8!8-27 8 _19_1 4
h 5rf*-Ll5Jri5~15-15-llV
* The student should notice that the word f r limit " is here used in a sense
quite different from that in which it is used when a variable is said to approach
a limit (§ 68).
THE DEFINITE INTEGRAL
149
In the foregoing discussion we have assumed that f(x) is
always positive and that a< b. These restrictions may be
removed as follows:
\if(x) is negative for all values of x between a and b, where
a < b, the graphical representation is as in fig. 128. Here
f(a)Ax = — the area of the rectangle EMJtfi,
/(a>,) Ax = — the area of the rectangle Jfj JH^iJ, etc.,
bo that
£f(x)dx
- the area EBCD.
In case f(x) is sometimes positive and sometimes negative,
we have a combination of the foregoing results, as follows:
If a <b, the integral
X* °\ g JS ,v' 5 3f' ''f* M« ,lf' Jf*
f(x) dx represents
the algebraic sum of the
areas bounded by the
curve y =f(x), the axis
of x, and the ordinates
x = a and x = b, ike
areas above the axis of
x being positive and
those below negative.
If a > b, Ax is negative, since Ax = - — - • The only change
necessary in the above statement, however, is in the algebraic
signs, the areas above the axis of x being now negative and
those below positive. It is usual to arrange the work so that
Ax shall be positive.
It is obvious, however, that
J/<»,fc=-_£/<X)1&.
i areas involved,
jj(x)dx=£f(x)dx+jj(x)dx.
Also, from the areas involved,
150 SLOPES AND AREAS
PROBLEMS
Find approximately, by a numerical calculation, the slope of each
of the following curves at the point given :
1. y = a* at (2, 4). 6# y = lat(2,J).
2. y = x8 at (3, 9). x
3. y = z8 at (1, 1). 6.y-VZat(4»2).
4. y = z8 at (2, 8).
Find from the definition, without the use of formulas, the deriva-
tives of the following expressions :
7. 4s8. 9. x* — x. ^2
x2 12. V£.
Find by the formulas the derivatives of each of the following
polynomials :
13. 4x8-3aja + 2a;-l. 15. x* + 7s7 - 6z8 + 7x - 3.
14. a;4 + 7a;2-a; + 3. 16. %x* - $ x6 + %x* + a" - 1 x.
17. Prove that the derivative of ax9 + foe2 + ex + e is the sum of
the derivatives of its terms.
18. By expanding and differentiating show that the derivative
of (4s + 3)8isl2(4a; + 3)a.
19. By expanding and differentiating show that the derivative
of (x + a)n is n(x + a)n_1.
Find the values of x for which the following expressions are
respectively increasing and decreasing, and draw their graphs :
20. xa + 6aj-4. 23. x4-2arl + 7.
21. x'-3x* + 7. 24. 2z8- 15a* + 36s-270.
22. »4 + 4aj-6. 25. x8 - 3a8 - 9x + 27.
26. If a stone is thrown up from the surface of the earth with a
velocity of 100 ft. per second, the distance traversed in t seconds is
given by the equation s = 100 1 — 16 £2. Find when the stone moves
up and when down.
PROBLEMS 151
27. A particle is moving in a straight line in such a manner that
its distance x from a fixed point A of the straight line, at any time
t, is given by the equation x = t* — 9 12 + 24 1 + 100. When will the
particle be approaching A ?
28. A piece of wire of length 20 in. is bent into a rectangle one
aide of which is x. When will an increase in x cause an increase in
bhe area of the rectangle and when will it cause a decrease ?
29. In a given isosceles triangle of base 20 and altitude 10 a rec-
tangle of base x is inscribed. Find the effect upon the area of the
rectangle caused by increasing x,
30. A right circular cylinder with altitude 2x is inscribed in a
sphere of radius a. Find when an increase in the altitude of the
cylinder will cause an increase in its volume and when it will cause
el decrease.
31. A right circular cone of altitude x is inscribed in a sphere of
radius a. Find when an increase in the altitude of the cone will
cause an increase in its volume and when it will cause a decrease.
32. On the line Sx + y = 6 a point P is taken and the sum s of
the squares of its distances from (5, 1) and (7, 3) computed. Find
the effect on 5 caused by moving P on the line.
Find the turning points of the following curves and draw the
curves:
33. y = 2x* — 9x*. 35. y = \x*-2x* + \.
34. y=2sc8 + 3arl-12a;-18. 36. y = x* - 2xz + 4.
37 . Find the equation of the tangent to the curve y=4cca + 4a; — 3
at the point the abscissa of which is — 1.
38. Find the equation of the tangent to the curve y = x% -f- 4a^
at the point the abscissa of which is — 3.
39. Show that the equation of the tangent to the curve y = ax2 -f
2 bx + 6 at the point (xv yx) is y = 2 (axx + b)x — ax} -f- c.
40. Show that the equation of the tangent to the curve y = Xs +
ax + b at the point (xv y^) is y = (3 x? + «)« — 2 xf + b.
41. Find the area of the triangle included between the coordinate
axes and the tangent to the curve y = x* at the point (3, 27).
152 SLOPES AND AEEAS
42. Determine the point of intersection of the tangents to the
curve y = x* — 5x + 7 at the points the abscissas of which are
— 2 and 3 respectively.
43. Determine the point of intersection of the tangents to the
curve y = xz — 3 x -f 7 at the points the abscissas of which are
2 and 0 respectively.
44. Find the angle between the tangents to the curve y = x2 —
4 x + 1 at the points the abscissas of which are 1 and 3 respectively.
45. Find the angle between the tangents to the curve y = x* —
3 x2 + 4 x — 12 at the points the abscissas of which are — 1 and 1
respectively.
46. Find the equations of the tangents to the curve y = x8 + x2
that have the slope 8.
47. Find the equations of the tangents to the curve 2x* + 4 a* —
x — y = 0 that have the slope J.
48. Find the points on the curve y = 3 x* — 4 x2 at which it makes
an angle of 45° with OX.
49. Find the points on the curve y = x* — x* + 2x + 3 at which
the tangents are parallel to the line y = 3 x — 7.
50. How many tangents has the curve y = x* — 2x* + x — 2
which are parallel to the line 7x — 4 y + 28 = 0 ? Find their
equations.
51. Find approximately the area bounded by the straight line
y = 2 x + 3, the ordinates x = 1 and x = 2, and the axis of *, by
considering the area as the sum of rectangles the bases of which are
.2 in the first approximation and .1 in the second approximation.
Also find the area exactly by elementary geometry.
52. Find approximately the area between the axis of x and the
portion of the curve y = x — x2 which is above the axis of or, by
considering the area as the sum of rectangles the bases of which are
.2 in the first approximation and .1 in the second approximation.
53. Find approximately the area bounded by the curve y = -y
x
the ordinates x = 2 and x = 3, and the axis of x, by considering the
area as the sum of rectangles the bases of which are .2 in the first
approximation and .1 in the second approximation.
PROBLEMS 153
64. Find the area bounded by the curve y = V#, the ordinates
a? = 1 and x = 4, and the axis of x, by considering the area as the
sum of rectangles the bases of which are .5 in the first approxima- .
tion and .2 in the second approximation.
65. Find by integration the area described in Ex. 51.
66. Find by integration the area described in Ex. 52.
67. Find the area bounded by the curve y = x* — 2 x2 + 3 x — 1,
the ordinates x = 2 and x = 4, and the axis of x.
68. Find the area bounded by the axis of x and the portion of
the curve y = 9 — x2 above the axis of x.
69. Find the area between the axis of x and that part of the
curve y = 10 — llx — 6x2 which is above the axis of x.
60. Find the area between the axis oi.x and that part of the
curve y = x* — 3 x2 — 9 x + 27 which is above the axis of x.
61. Find the area bounded by the axis of x and the portion of
the curve y = x* + 3x* — 4 below the axis of x.
62. Find each of the two areas bounded by the curve y = 150 as —
25 x2 — x* and the axis of x.
63. Find the area bounded by the axis of x} the curve y=2x* +
3x* + 2, and the ordinates through the turning points of the curve.
64. Prove that the area of a parabolic segment is two thirds of
the product of its base and altitude.
66. Find the area between the parabola y = \ x2 and the straight
line Sx — 2y — 4 = 0.
66. Find the area of the crescent-shaped figure between the
curves y = x2 + 5 and y = 2 x2 + 1.
CHAPTER X
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
82. Theorems on derivatives. In order to extend the process
of differentiation to functions other than polynomials, we shall
need the following theorems:
1. The derivative of a /miction plus a constant is equal to the
derivative of the function.
Let u be a function of x which can be differentiated, let c be
a constant, and place , c
Then if a? is increased by an increment Ax, u is increased by
an increment Aw, and c is unchanged. Hence the value of y
becomes u + Au + c.
Whence Ay = (u + Au + c)—(u + c) — Au.
Therefore -^- = -— >
Ax Ax
and, taking the limit of each side of this equation, we have
dy _du
dx dx
Ex. 1. y = 4 Xs + 3.
2. The derivative of a constant times a function is equal to the
constant times the derivative of the function.
Let u be a function of x which can be differentiated, let c be
a constant, and place ,. _ „.
Give x an increment Ax, and let Au and Ay be the corre-
sponding increments of u and y. Then
Ay = c(u + Au)— cu = cAu.
164
THEOREMS ON DERIVATIVES 155
TT Ay Au
Hence —£ = <?-—,
Ax Ax
and, by theorem 3, § 69,
T . Ay T . Au
Lun —£■ = c Lim — - •
Ax Ax
Therefore -^ = <? — ->
by the definition of a derivative.
Ex.2. y = 5(x*+ 3:^ + 1).
^ = 5~(*8 + 3 a? + 1) = 5 (3 *2 + 6 x) = 15(xa + 2x).
ax dx
3. 2%« derivative of the sum of a finite number of functions
is equal to the sum of the derivatives of the functions.
Let u, v, and w be three functions of x which can be differen-
tiated, and let y = u + v + w.
Give x an increment Ax, and let the corresponding increments
of u, vy wy and y be Aw, Av, Aw, and Ay. Then
Ay = (w + Aw + v + Av + w + Aw) — (u + v + w)
= Au+Av+Aw;
, Ay Au Av , Aw
whence TiL = T- + T- + -i —
Ax Ax Ax Ax
Now let Ax approach zero. By theorem 1, § 69,
T . Ay T . AmiT. Av , T . Aw
Lun —2. — Lim (- Lim - — h Lim -— ;
Ax Ax Ax Ax
that is, by the definition of a derivative,
dy du . dv , rfw
dx dx dx dx
The proof is evidently applicable to any finite number of
functions.
Ex.3. y = a^-3x3 + 2x2-7x.
-^ = 4a:8-9x2 + 4x-7.
dx
156 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
4. The derivative of the product of a finite number of functions
is equal to the sum of the products obtained by multiplying the
derivative of each factor by all the other factors.
Let u and v be two functions of x which can be differentiated,
and let y __ UVu
Give x an increment Ax, and let the corresponding increments
of u, v, and y be Am, Av, and Ay.
Then Ay = (u + Au) (y + Av) — uv
= uAv + v Au + Au • Av
Ay Av , Au , Au A
and t*- = u — + v- — h — • Av.
Ax Ax Ax Ax
If, now, Ax approaches zero, we have, by § 69,
T . Ay T . Av , T . Au , T . Au T . A
Lim — - = w Lim - — f- v Lim - — \- Lim — - • Lim Av.
Ax Ax Ax Ax
But LimAt>=0,
and therefore -d^ = u — + v — •
aa; a.r az
Again, let y = uvw.
Regarding uv as one function and applying the result already
obtained, we have
dy dw , d(uv)
-£- = uv h w — - — -
dx dx dx
dw ,
= uv—- + w
dx
[dv du\
dx dx]
dw , dv , du
= uv — + uw — - + vw — •
a.c a# cte
The proof is clearly applicable to any finite number of factors.
Ex.4. y = (3x-5)(x2 + l)x8.
^ = (3x-5)(x2 + l)^ + (3x-5^^
dx dx dx dx
= (3 x - 5XZ2 + 1)(3 x2) + (3 x - 5)x»(2 x) + (x2 + l)x*(3)
= (18x* - 25 x2 + 12 x - 15)x2.
THEOREMS ON DERIVATIVES 157
5. The derivative of a fraction is equal to the denominator times
the derivative of the numerator minus the numerator times the deriva-
tive of the denominator, all divided by the square of the denominator.
u
Let y = — , where u and v are two functions of x which can be
v
differentiated. Let A#, Aw, At;, and Ay be as usual. Then
u + Au u v Au — uAv
Ay =
v + Av v v2+vAv
Au Av
Ay Ax Ax
and — = •
Ax tr+vAv
Now let Ax approach zero. By § 69,
T . Au T . Av
v Lim u Lim —
. Ay Ax Ax
Ax v*+v\j\mAv
du dv
v u —
. dy dx dx
whence -r- = ; •
dx tr
x*^l
x2 + l'
dy^(x* + l)(2x)-(x*-l)2x^ ±x
dx (r'+l)3 (x2+l)2
Ex. 5. y =
6. If y is a function of x, then x is a function of y, and the
derivative of x with respect to y is the reciprocal of the derivative
of y with respect to x.
Let Ax and Ay be corresponding increments of x and y. Then
Ay~ Ay'
Ax
Ax 1
whence Lim -— =
Ax
, dx 1
that is, — = — •
dy dy
dx
158 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
1. If y is a function of u and u is a function of x, then y is
a function of x, and the derivative of y with respect to x is equal
to the derivative of y with respect to u times the derivative of u
with respect to x.
An increment Ax determines an increment Aw, and this in turn
determines an increment Ay. Then evidently
Ay __ Ay Au
Ax Au Ax
whence Lim — - = Lim — - • Lim -— ;
Ax Au Ax
,, . . dy dy du
that is, -^ = -£.— .
ax au ax
Ex. 6. y = u* + 3 u + 1, where u = — .
x2
<ty_/o , ox/ 2\_ 2 + 3r* 2 4 + 6a^
=(2u + 3)(-|)=_i±
dx \ xzl x2 Xs x5
The same result is obtained by substituting in the expression for y the
value of u in terms of x and then differentiating.
This result has an important application to the differential. For
suppose we have y=/(ft)f u = <Kx> (l)
By substitution, we obtain
y=fW(x)-]=F(x), (2)
and the formula proved above gives us
*"(*) =/'0) • f(*> (3)
By use of § 77 we obtain from (1)
dy =fr (u) du, du = <f>f (x) dx, (4)
and from (2) we have dy = Ff(x) dx. (5)
It is important to know that the two values of dy in (4)
and (5) agree. In fact, by means of (3) and the second part
of (4), (5) becomes
dy =f'(u) 4>'(x) dx =/'(w) du.
Hence it is not necessary, in applying § 77 to find a differential,
to ask whether # is an independent variable or not.
DERIVATIVE OE U* 159
83. Derivative of if. If u is any function of x which can be
differentiated and n is any real constant, then
\ J =nun * — •
ax ax
To prove this formula we shall distinguish four cases:
1. When n is a positive integer.
rffiQ.rffiQ* (by 7, §82)
as du ax
= »«-lg. (Byl, §74)
2. When w is a positive rational fraction.
Let n = — > where jp and j are positive integers, and place
y = u9.
By raising both sides of this equation to the 5th power, we have
Here we have two functions of x which are equal for all
values of x. If we give x an increment Aa^ we have
A<y) = A«)>
Az A#
and therefore / = ^ ^ 5
whence 93/9~1-j=PuP~lmT-9
ax ax
since jp and q are positive integers. Substituting the value of y
and dividing, we have
dy __ p |-i c?w
Hence, in this case also,
d(unr) n xdu
(fa; cfo
160 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
3. When n is a negative rational number.
Let n = — m, where mis a positive number, and place
. 1
um
d(u*)
Then ^ = — ±- (by 5, §82)
dx u2M
^du
dx
rnum
u2m
(by 1 and 2)
« i du
ax
Hence, in this case also,
d(uH) m .du
\ / = nun~x — •
dx dx
4. When n is an irrational number.
The formula is true in this case also, but the proof will not
be given.
It appears that 1, § 74, is true for all real values of n.
Ex. 1. y = (x8 + 4 x2 - 5 x + 7)8.
^ = 3(x8 + 4X2 - 5x + 7)24-(x8 + 4x2 - 5s + 7)
ax ax
= 3 (3 x2 + 8 x - 5 ) (x8 + 4 x2 - 5 x + 7) 2.
Ex. 2. y - Vx~* + - = x* + x-«.
or
ax 3
= _2 3^
~zV-x *
Ex. 3. y = (* + l)Vx2 + l.
dx ax ax
= (x + 1)[J (x2 + I)" i • 2 x] + (x2 + l)i
(x2 + 1)*
_ 2 x2 + x + 1
Vx2TT
FORMULAS 161
4- y = vS = (^Tl)-
-*
dx 3Vr» + l/ dx\x* + V
3\ x )
(*» + 1)3
l-2x»
3ar*(x* + l)*
84. Formulas. The formulas proved in the previous articles are
a)
(2)
(3)
(4)
d(u + c)
du
dx
dx
d(cu)
dx
du
dx
)
d(u + v)
dx
du
dx
dv
dx
d(uv)
dx
dv
dx
du
+ V — i
dx
'©
du
dx
dv
dx
dx
v> '
dx
-nun~
, du
— »
dx
dx
1
dy'
' dy
dx
dy =
dx
du
dy
du
dx
dy_
dx
du
— ^^^_ •
dx
du
Formula (9) is a combination of (7) and (8).
(5)
(6)
(7)
(8)
(9)
162 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
85. Higher derivatives. If y=/(V), then -^ is in general a
ax
function of x and may be differentiated with respect to x. The
result is called the second derivative of y with respect to x
and is indicated by the symbol -T-(-jr)> which is commonly
j % d& \axj
abbreviated into -r-ir
dx*
Similarly, the derivative of the second derivative is called
the third derivative, and so on. The successive derivatives are
commonly indicated by the following notation :
y =zf(x), the original function ;
--£ =zf(x\ the first derivative ;
ax
ir( j ) = -t4 =/"(a0> the second derivative;
ax \ax/ axr
£(^ = S =/"'(*)» the third derivative;
dxXdx2/ da? K J
— £ =/(">(a;), the nth derivative.
ax>
It is noted in § 9 that f(a) denotes the value of /(#) when
x = a. Similarly, /'(#), /"(«), fln(&) are used to denote the
values of /'(#), /"(#), frff(pO respectively when x = a* It is
to be emphasized that the differentiation is to be carried out
before the substitution of the value of x.
*x. If/(*) = |r^,find/''(0>
J K } (x* + l)2
J W (x2 + l)8
Therefore /"(°) = 2-
86. Differentiation of implicit functions. Consider any equation
Of the form /O,y) = 0. (1)
By means of this equation, if a value of x is given, values of y are
determined. Hence (1) defines y as a function of x. When (1} is
IMPLICIT FUNCTIONS 163
solved for y, so that y is expressed in terms of x, y is an explicit
function. When (1) is not solved for y, y is an implicit function.
For example,
3 x2 - 4 xy + 5 f - 6 x + 7 y - 8 = 0,
which may be written
5y*+(7-4a;)y + (8^-6*-8)=0f
defines y as an implicit function of x.
If the equation is solved for y, giving
_ 7+ 4z±V209 + 643-44^
y = io '
y is expressed as an explicit function of x.
It is possible to find -^ from (1) without solving (1), for we
have in (1) a function of x which is always equal to zero. Hence
its derivative is zero. The derivative may be found by use of
tHe formulas of the previous articles, as shown in the following
examples :
Ex. 1. Given a2 + y2 = 5.
Then 'y+^-O,
dx
thatis, 2x + 2y^ = 0;
dx
, dy x
whence -£ = .
dx y
The derivative may also be found by solving the equation for y. Then
y = ± V5 - x\
dy _ — x _ _x
fc~ V5-x2~ y
Ex. 2. Given y8— xy — 1 = 0.
Then ^2-^M = 0.
dx dx
Hence 3«a^-x^-y = 0,
dx dx
and & = _!
tfx 3 y 2 — x
The second derivative may be found by differentiating the
result thus obtained.
164 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
Ex. 3. If x2 + y2 = 5, we have found -?• =
ax y
Therefore g = - £ft
or3 dx\yj
dy
y*
-•(-9
_ y2+ x2
5
Then
Ex. 4. If y8 — xy — 1 = 0, we have found -^ = — ■ f-
ax 3 y2 — x
<Py„ * } dx * dx
dx2 (3y2-x)2
(3y2-x)a
Cy2
(Sy2-*)2
= -2xy
(3y2-x)8'
87. Tangents and normals. It has been shown in § 76 that
the tangent to a curve y =/(#) at a point (x^ yx) is
where (-^) denotes the value of -^- at (ar, y.).
\dx/i ax
The normal to a curve at any point is the straight line perpen-
dicular to the tangent at that point
To find the equation of the normal we first find the slope
of the tangent and then use the method of § 32, 3.
TANGENTS AND NORMALS
165
Ex. 1. Find the equations of the tangent and the normal to the parabola
y2 = 3 x at the points for which x = 3.
When x = 3, y = ± 3.
By differentiation we have
2«!^
dy 3
or -f- = — -
ax 2#
Therefore the slope of the tangent at (3, 3) is J, and the slope of the
tangent at (3, — 3) is — £.
Hence at (3, 3) the equation of the tan-
gent is
y - 3 = J(x - 3), or x r- 2 y + 3 = 0,
and the equation of the normal is
y — 3 = - 2 (x - 3), or 2x + y-9 = 0;
and at (3, — 3) the equation of the tangent is
y + 3 = -£(x-3), or x + 2y + 3=0,
and the equation of the normal is
y + 3 = 2 (x - 3), or 2x-y-9 = 0.
Ex. 2. Prove that the normal to a pa- Fig. 129
rabola at any point makes equal angles
-with the axis of the parabola and the focal radius drawn to the point.
Let Pi(xv yx) be any point of the parabola y2 = 4px (fig. 129), and let
F(p, 0) be the focus. Then FPX is the focal radius of Pv and let PXN be
the normal to the parabola. To prove AFNPX = ZFPXN.
By differentiation we have 2y— = 4/?, whence the slope of the tan-
2» ^x v
gent at Px is — and the slope of the normal is — jp- • It follows that
Vi
2p
tan FNP.^p..
_ V\
therefore
yo Wl x j. x -
*1
-P
. Vi
Vx
tan T?P N
2p x j — p
vail ri ill
1-
2/A*
JL.)
l-JV
- Vi (xi
+/>)
2p(*i-p)-yi
2jo'
if we replace yf by 4 pxv since y* = 4/7XJ, Px being a point of the parabola.
Since tan FPtN = tan FNPV the angles are equal.
AG
166 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
The angle of intersection of two curves is the angle between
their respective tangents at the point of intersection. The
method of finding the angle of intersection is illustrated in
the following example:
Ex. 3. Find the angle of intersection of
the circle x2 + y2 = 8 and the parabola x2 = 2 y.
The points of intersection are Px (2, 2) and
P2 (— 2, 2) (fig. 130), and from the symmetry
of the diagram it is evident that the angles
of intersection at Pt and P2 are the same.
Differentiating the equation of the circle,
we have 2x + 2y-^ = 0, whence -^ = >
dx ax y
and differentiating the equation of the parab-
ola, we find -~ = x,
ax
Hence at Pt the slope of the tangent to the circle is — 1 and the slope
of the tangent to the parabola is 2.
Accordingly, if /J denotes the required angle of intersection,
tanft= ~~*~J* = S,
Fig. 130
or
1-2
)8=tan-13.
88. Sign of the second derivative. Since the second derivative
is the derivative of the first derivative, the sign of 3-^ shows
whether g „ a» *»*, . . dee^a^ ta— *
dx cPy
The significance of -7$ for the graph y =/(V) is obtained from
dy . <Py
the fact that -^ is equal to the slope ; hence -7^ is the deriva-
cPy
tive of the slope. Therefore, by § 75, if •— ^ is positive, the
cPy
slope is increasing ; if —4 is negative, the
slope is decreasing. We may have, accord-
ingly, the following four cases:
da?
1. ^>o, *3L
dx ' da?
>0.
Since both the ordinate and the slope are increasing, the
graph runs up toward the right with increasing slope (fig. 131).
SIGN OF THE SECOND DEEIVATIVE
167
2. 4y>o, $3L
dx ' dx2
<0.
The graph runs up toward the right
with decreasing slope (fig. 132).
3. ^<0, &
dx ' dx2
>0.
The graph runs down toward the
right. The slope, which is negative, is
increasing algebraically and hence is
decreasing numerically (fig. 133).
d?y
4. ^<0,
dx
da?
<0.
— X
Fig. 133
Fig. 134
The graph runs down toward the
right, and the slope is decreasing
algebraically (fig. 134).
The consideration of these cases leads
cPy
to the following conclusion: If --^ is posi-
tive^ the graph is concave upward ; if -z—
dx^
is negative, the graph is concave downward.
If a curve changes from concavity in one direction to con-
cavity in the other direction at any point, that point is called a
d2y
point of inflection. It follows that at such a point —^ changes
sign, either by becoming zero or by becoming infinite. These
two cases are illustrated in the following
examples :
Ex. 1. Examine the curve y *= ^ (a8 — 6 x2)
for points of inflection.
dy _ 1 v2 „
— — == — X — x%
dx 4
— - = - x — 1 = - (x — 2\
dx* 2 2^ ;
The curve (fig. 135) is concave downward
when x<2, is concave upward when x>2, and accordingly there is a
point of inflection when x = 2. The ordinate of this point is — 1 J.
168 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
Ex. 2. Examine the curve y = (x — 2)* for points of inflection.
dy _ 1 d?y _ 2
dx 3(a:-2)*
dx2
d*y_
9 (x - 2)t
It is evident that — - = oo if x = 2, and that
no finite value of x makes
<Py
&y
dx*
dx2
0. If x < 2,
, > 0 ; and if a: > 2, — ^ < 0. Hence the point
ax2 '/y*2
Fig. 136
for which x = 2 is a point of inflection, since
on the left of that point the curve is concave upward and on the right of
that point it is concave downward (fig. 136). The ordinate of this point is 0.
89. Maxima and minima. If f(x) changes from an increasing
function to a decreasing function (§ 75) when x increases through
Y
o
x=a
Fig. 137
Fig. 138
the value a and f(a) is finite, /(a) is called a maximum value of
f(x) (figs. 137, 138); and if f(x) changes from a decreasing
function to an increasing function when x increases through the
value a and f(a) is finite, f(a) is called a minimum value oif(x)
(figs. 139, 140).
Since the derivative of an increasing function is positive and
the derivative of a decreasing function is negative, it follows
MAXIMA AND MINIMA 169
that the derivative of the function must change sign at either
a maximum or a minimum value and hence must become either
zero or infinity. Accordingly we have two cases :
I. If -^ = 0 or oo whenx = a, and -~ >0 whenx<a, and-^-<0
dx dx ax
when x > a^f(a) is a maximum value of y=f(x*).
II. If ~ — 0 or oo when x = a, and -~- < 0 when x < a, and
, dx dx
-j- > 0 when x > a, f(a) is a minimum value of y = f(x).
If, however, -j- changes sign by becoming infinite and at the
same time y becomes infinite (fig. 33), the function is discon-
tinuous and there is no corresponding maximum or minimum
value.
In order to apply the above tests it is necessary to factor -d%
as shown in the following examples:
Ex. 1. Find the maximum and the minimum value of
f(x) = x5- 5 x4 + 5 x* + 10 x2- 20 x + 5.
We find /'(*) = 5 a:4- 20 x8+ 15 x2 + 20 x - 20
= 5(s2-l)(a;2-4a: + 4)
= 5(z + 1) (x - 1) (x - 2)2.
The roots of f\x) — 0 are — 1, 1, and 2. As x passes through — 1, f'(x)
changes from + to — . Hence x = — 1 gives/(ar) a maximum.value, namely 24.
As x passes through + l,f'(x) changes from — to +. Hence x = + 1 gives
f(x) a minimum value, namely — 4. As a: passes through 2, f'(x) does not
change sign. Hence x = 2 gives f(x) neither a maximum nor a minimum
value.
Ex. 2. A rectangular box is to be formed by cutting a square from
each corner of a rectangular piece of cardboard and bending the resulting
figure. The dimensions of the piece of cardboard being 20 by 30 in.,
required the largest box which can be made.
Let x be the side of the square cut out. Then if the cardboard is bent
along the dotted lines of fig. 141, the dimensions of the box are 30 — 2 x,
20 — 2 x, x. Let y be the volume of the box. Then
y = z(20-2:r)(30-2z)
= 600 x - 100 x2 + 4 x\
^ = 600-200ar + 12xa.
dx
170 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
Equating this to zero, we have
3 x2 - 50 x + 150 = 0.
25 ± 5 V7
x =
= 3.9 or 12.7.
Hence
^ = 12 (*- 3.9)0* -12.7).
ax
-p changes from + to — as x
ax
X
X
l
30— 2X
Fig. 141
passes through 3.9. Hence x = 3.9
gives the maximum value 1056 + for the capacity of the box. x = 12.7
gives a minimum value of y, but this has no meaning in the problem,
for which x must lie between 0 and 10.
Ex. 3. y = </(x-l)(x-2)2 = (x - l)*(x - 2)*.
dy 3 x — 4
** 3\/(x-l)2(x-2)
— = 0 when x = J, and changes from + to — as x passes through §.
dx i
Therefore x = $ gives a maximum value to the function, -j- = oo when
» ttX
a; = 1 or 2. When x = 1, -^ does not change
ax
sign. When x = 2, -p changes from — to
ax
+ . Then x = 2 gives a minimum value of
the function. Its graph is in fig. 142.
Referring to figs. 137 and 139, we
see that if -y- = 0 at a maximum value
of y, the curve is concave downward,
and that if -— = 0 at a minimum
dx
value of y, the curve is concave upward. Hence we may have
the following two cases :
dv d v
I. If -j- = 0 and -j^ < 0 when z = a, f(a) is a maximum
value of y =/(#).
dv d^v
II. If -j- = 0 and -t~ > 0 when # = a, f(a) is a minimum
value of y =/(#).
Fig. 142
MAXIMA AND MINIMA
171
It is evident that these tests can be used to advantage when
it may be difficult or impossible to factor -^» and that they
fail if
<£y_
also becomes zero.
Ex. 4. Light travels from a point A in one medium to a point B in
another, the two media being separated by a plane surface. If the velocity
in the first medium is v1 and in the second
v2, required the path in order that the time
of propagation from A to B shall be a
minimum.
It is evident that the path must lie in
the plane through A and B perpendicular
to the plane separating the two media,
and that the path will be a straight line
in each medium. We have, then, fig. 143,
where MN represents the intersection of
the plane of the motion and the plane
separating the two media, and ACB rep- Fig. 143
resents the path.
Let MA = a,NB = b, MN = c, and MC- x. Then A C = Va2 + x2 and
CB = V(c — x)2 + b2. The time of propagation from A to B is therefore
J Va2 + x2 , V(c - x)2 + b2
t = h — * i- ;
v.
v,
whence
and
dt
2
C — X
dx "xVa2 + x2 v2 V(c - x)2 + b2
cPt a2 b2
+
te* vx(a2 + x2)i v2 [(c - x)2 + ft2]*
cPt .
Since — - is always positive, the time is a minimum when
dx2
x
c — x
vx Va2 + x2 v2V(c-x)2 + b2
= 0.
(1)
This equation may be solved for x, but it is more instructive to proceed
as follows :
x MC . .
= sin <f>,
Vtf + x2 AC
CN
c — x
Then equation (1) is
V(c - x)2 + ft2 CB
sin <f> __ vx
= SUH/r.
sin^r
172 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
Now <j> is the angle made by A C with the normal at C and is called the
angle of incidence, and *// is the angle made by CB with the normal at C
and is called the angle of refraction. Hence the time of propagation is a
minimum when the sine of the angle of incidence is to the sine of the
angle of refraction as the velocity of the light in the first medium is to
the velocity in the second medium. This is, in fact, the law according to
which light is refracted.
In practical problems the question as to whether a value
of x for which the derivative is zero corresponds to a maxi-
mum or a minimum can often be determined by the nature of
the problem.
In Ex. 2 above, it is evident that there must be a maximum volume of
the box and that there can be no minimum value. Accordingly, when we
have found — = 0 if x = 3.9 or 12.7, since 12.7 is unreasonable in our
ax
problem we conclude, without further discussion, that x = 3.9 corresponds
to the maximum volume.
90. Limit of ratio of arc to chord. The student is familiar
with the determination of the length of the circumference of a
circle as the limit of the length
of the perimeter of an inscribed
regular polygon. So, in general,
if the length of an arc of any
curve is required, a broken line
connecting the ends of the arc is
constructed by drawing a series of
chords to the curve as in fig. 144. ^ ^ 144
Then the length of the curve is
defined as the limit of the sum of the lengths of these chords
as each approaches zero and as their number therefore in-
creases without limit. The manner in which this limit is
obtained is a question of the integral calculus and will not
be taken up here.
We may use the definition, however, to find the limit of the
ratio of the length of an arc of any curve to the length of its
chord as the length of the arc approaches zero as a limit ; that is,
as the ends of the arc approach each other along the curve.
Fig. 145
LIMIT OF RATIO OF ARC TO CHORD 173
Accordingly, let if and Pz (fig. 145) be any two points of a
curve, P^ the chord joining them, and I>T and J^T the tangents
to the curve at those points re-
spectively. We assume that the
arc J%J% lies entirely on one side
of the chord PXPZ and is concave
toward the chord. These condi-
tions can in general be met by
taking the points ij and P2 near
enough together. Then it follows
from the definition that
P1T + TP2>htcP1P2>P1P2;
whence i?T+^ arcjg
PXP2 P^
If TR is the perpendicular from T to iji£, and if the angles P2PXT
and PXP2T are denoted by a and j3 respectively, then PXT = I*R sec a,
and TP2=RP2 sec £ = (ijij -iJ.R) sec £.
Therefore PXT -f- TP2 = %R sec a + (PXP2 - iJJR) sec j3
= PXP2 sec /3 + PXR (sec a - sec £),
, I>T+TP2 P^P2 sec &+PxR (sec g - sec ff )
PP ~~ PP
= sec £ -f- -tj— (sec a — sec £).
*
Now, as ij and ij approach each other along the curve, a and fi
both approach zero as a limit, whence sec a and secyS approach
unity as a limit; and since -J— is always less than unity, it fol-
lows that the limit of — - is unity.
PP
arc PP * 2
Hence — * 2 lies between unity and a quantity approaching
x * arc PP
unity as a limit, and therefore the limit of * 2 is unity ; that is,
£Ae foVm£ of the ratio of an arc to its chord as the arc approaches
zero as a limit is unity.
174 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
91. The differentials dx, dy, ds. On any given curve let the
distance from some fixed initial point measured along the curve
to any point P be denoted by a, where 8 is positive if P lies in
one direction from the initial point
and negative if P lies in the oppo-
site direction. The choice of the
positive direction is purely arbi-
trary. We shall take as the posi-
tive direction of the tangent that
which shows the positive direction
of the curve and shall denote the
angle between the positive direc-
tion of OX and the positive direction
of the tangent by <j>.
Now for a fixed curve and a fixed initial point the position
of a point P is determined if 8 is given. Hence x and y, the
coordinates of P, are functions of 8 which in general are con-
tinuous and may be differentiated. We will now show that
dx . dy . .
_ = co8<fc ^ = smf
Let arc PQ = As (fig. 146), where P and Q are so chosen that
As is positive. Then PR = Ax and RQ = Ay, and
Ax PR chord PQ PR
Fig. 146
As arcPQ arcPQ chord PQ
chord PQ ___
= —r- • cos RPQ.
arc PQ
Ay RQ chord PQ RQ
As ~~ arc PQ ~~ arc PQ " chord PQ
chord PQ
arcPQ
sin RPQ.
Taking the limit, we have, since Lim =-r^ = 1
LimRPQ = <f>,
arcPQ
dx .
- = C08<fc
dy
-Z- = fll
ds
= sin<£.
BATE OF CHANGE
175
If the notation of differentials is used, equations (1) become
dx = ds • cos <£, dy = ds • sin <f> ;
whence, by squaring and adding, we obtain the important
equation 2 2 2
ds = dx + dy . (2)
This relation between the differentials of x, y, and s is
often represented by the triangle of fig. 147. This figure is
convenient as a device for memorizing formula (1), but it
should be borne in mind that RQ is not
rigorously equal to dy (§ 77), nor is PQ
rigorously equal to ds. In fact, R Q = Ay
and PQ = Asy but if this triangle is
regarded. as a plane right triangle, we
recall immediately the values of sin<£,
cos<£, and tan<£ which have been pre-
viously proved.
92. Rate of change. If y=f(x), a change of Ax units in x
causes a change of Ay units in y, and the quotient — - gives the
ratio of these changes. If this ratio is equal to m, Ay = mAx;
that is, the change in y is m times the change in x. Hence, if
m were independent of Ax, a change of one unit in x would
cause a change of m units in y, and — - would consequently
measure the change in y per unit of change in x. But m does
depend in general upon Ax, and hence does not give an unam-
biguous measure of the relative changes in x and y. To obtain
Ay
such a measure, it is convenient to take the limit of — - as Ax
Ax
approaches zero and to call this limit the rate of change of y
with respect to x. We have then
Fig. 147
dy _
dx
= rate of change of y with respect to x.
Ex. 1. Coefficient of expansion. Let a substance of volume v be at a
temperature t If the temperature is increased by At, the pressure remain-
ing constant, the volume is increased by Av. The change per unit of vol-
ume is then — > and the ratio of this change per unit of volume to the
v
176 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
change in the temperature is - — • The limit of this ratio is called the
coefficient of expansion ; that is, the coefficient of expansion equals — - •
In other words, the coefficient of expansion is the rate of change of a unit
of volume with respect to the temperature.
Ex. 2. Elasticity. Let a substance of volume v be under a pressure p.
If the pressure is increased by Ap, the volume is increased by — Av. The
change in volume per unit of volume is then • The ratio of this
V 1 A
change per unit of volume to the change in the pressure is — > and
v Ap
the limit of this is called the compressibility ; that is, the compressibility
is the rate of change of a unit of volume with respect to the pressure.
The reciprocal of the compressibility is called the elasticity, which is
therefore equal to — v -p •
dv
In many cases it is convenient to take time t as the inde-
pendent variable. Then — and -?- measure the rates at which
r dt dt
x and y respectively are changing with the time. If both x and
y can be expressed in terms of t, these rates may be found by
differentiating; but if y is expressed in terms of x and x is
expressed in terms of £, -j- may be found by the formula
dy __ dy dx
dt dx dt
which -is a special case of (8), § 84.
Ex. 3. A stone thrown into still water causes a series of concentric
ripples. If the radius of the outer ripple is increasing at the rate of 5 ft.
a second, how fast is the area of the disturbed water increasing when the
outer ripple has a radius of 12 ft.?
Let x be the radius of the outer ripple and A the area of the disturbed water.
Then A = irx2
and
By hypothesis,
Therefore
and when x = 12,
dA
— ■ = 120 ir, the required rate.
dt
dA
dt
= 2irx — *
dt
dx
Tt
= 0.
dA
dt
= 10
irx;
RECTILINEAR MOTION 177
This problem may also be solved by expressing A directly in terms of t.
By the conditions of the problem, x = 5 1,
and therefore A = 25 irP ;
whence — — = 50 irt
dt
dA
When x = 12, t = 2 $ and — - = 120 ir, as before.
dt
93. Rectilinear motion. An important application of the con-
cept of a derivative is found in the definition of the velocity of
a moving body. We shall confine ourselves in this article to
rectilinear motion, that is, to motion which takes place in a
straight line.
Let a body move along a straight line AL (fig. 148), and let
its distance from a fixed point A, at any time t, be denoted by 8.
Then, if the body is at the point P at . .
the time t, AP=s. * ? 3 £
The velocity of the body is then defined • FlG' 148
as the rate of change of the distance 8 with respect to the time t
More in detail, if t is increased by the increment At, let 8
As
be increased by the amount A8=PQ. Then — is the average
velocity of the body during the period A£. Since this average
velocity depends in general upon the magnitude of At, we take
the limit of — as At approaches zero, and call this limit the
velocity of the body at the point P.
Hence, if v denotes the velocity,
T . A« ds
v = Lim — = — •
At dt
If the velocity is constant and equal to k, the motion is said
to be uniform, and 8 = kt.
We note that if v > 0, an increase of time corresponds to an
increase of «, while if v < 0, an increase of time causes a decrease
of 8. Consequently, the velocity is positive when the body moves
in the direction in which * is measured, and negative if it moves
in the opposite direction.
178 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
Ex. 1. If a body is thrown up from the earth with an initial velocity
of 100 ft. per second, the space traversed, measured upward, is given by
the equation
s = 100 1 - 16 *2.
Then
v = * = 100 - 32 1.
dt
When t < 3 J, v>0; and when t>3 J, v<0. Hence the body rises for 8 J sec,
and then falls. The highest point reached is 100 (3 \) - 16 (3 J)2 = 156 J.
Ex. 2. A man standing on a wharf 20 ft. above the water pulls in
a rope attached to a boat at the uniform rate of 3 ft. per second.
Required the velocity with which the boat approaches the wharf.
Let A (fig. 149) be the position of the man
and C that of the boat. Let ^
^£ = A = 20, AC = s, and BC = x.
dx
We wish to find —
dt
h
1
\«
3 X
Fig. 149
Now x = Vs2 - 400 ;
,, - dx s ds
thprefnra — = .
dt v$2 - 400 dt
a
But, by hypothesis, s is decreasing at the rate of 3 ft. per second ; there-
ds
fore — - = — 3, and the required expression for the velocity of the boat is
dx
-3«
dt Vs2 - 400
To express this in terms of the time, we need to know the value of *
when t = 0. Suppose this to be s0 ; then
and
s
dx
~di
s0-St
- 3 s0 + 9 t
Vs2 - 400 - 6 .y + 9 t?
When the motion of the body moving in a straight line is not
uniform, the velocity at the end of ah interval of time is not
the same as at the beginning. Then, if v -f- Av denotes the velocity
of the body at Q (fig. 148), — is the average change of velocity
per unit of time during the period A£. The limit of this ratio is
called the acceleration; that is, the acceleration of a body moving
RECTILINEAR MOTION 179
in a straight line is the rate of change of the velocity with respect
to the time. Hence, if a denotes the acceleration,
__dv __ d Afo\
a~lh~di\di)~~
dt2
If a is constant, the motion is said to be uniformly accelerated,
and v = kt, where k is constant.
When a is positive, an increase of t corresponds to an
increase of v. This happens when the body moves with
increasing velocity in the direction in which 8 is measured
or with decreasing velocity in the direction opposite to that
in which 8 is measured.
When a is negative, an increase of t causes a decrease of v.
This happens when the body moves with decreasing velocity in
the direction in which 8 is measured or with increasing velocity
in the direction opposite to that in which 8 is measured.
The force which acts on a moving body is measured by the
product of the mass and the acceleration. Thus, if F is the force,
and m the mass of a body moving in a straight line,
„ dv d28
F=ma = m — = m—:'
dt dt2
From this it appears that a force is considered positive or nega-
tive according as the acceleration it produces is positive or nega-
tive. Hence a force is positive when it acts in the direction in
which 8 is measured and negative when it acts in the opposite
direction.
Ex. 3. Let
s=A + Bt+ \Ct*
Then
v=B + Ct,
a = C,
i
F=mC.
and
If *0 and v0 denote the values of s and v when t = 0, we have, from the
last equations, A _ ^
SQ —A, VQ— 15,
and the original equation may be written
S = sQ + v0t + J at2.
180 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
Fig. 160
94. Motion in a curve. When a body moves in a curve, the
discussion of velocity, acceleration, and force becomes more com-
plicated as the directions as well as the magnitudes of these quan-
tities need to be considered. We shall not discuss acceleration
and force, but will notice that the definition for the magnitude
of the velocity, or the speed, is the
same as before, namely,
ds
where s is distance measured on the
curved path, and that the' direction
of the velocity is that of the tangent
to the curve.
Also as the body moves along a curved
path through a distance PQ = As (fig. 150), x changes by an amount
PR = Ax and y changes by an amount RQ = Ay. We have then
Lim — = — = v = velocity of the body in its path,
At at
\nr. dx
Lim — = — = vx = component of velocity parallel to OX,
At at
Lim — - = -j- = vy = component of velocity parallel to OY.
Otherwise expressed, v represents the velocity of P, vx the
velocity of the projection of P upon OX, and vy the velocity
of the projection of P on OF.
Now, by (8), § 84, and by § 91,
dx dx
ds
= V COS <£>,
V =
x dt
ds
di
(1)
and
__dy _dy ds
v dt ds dt
= v sin <f>.
Squaring and adding, we have
v2 = vl + V*.
(2)
(3)
PROBLEMS
181
Ex. If a projectile starts with an initial velocity r0 in an initial direction
which makes an angle a with the axis of x taken as horizontal, its position
at any time t is given by the parametric equations
x = v0t cos a, y = v0t sin a — £ gfl.
Find its velocity in its path.
dx
It
We have
vx = — - = v0cosa,
»„ = -?■ = vn sin a— at.
v dt ° y
Hence
w = Vv02 — 2 ^rv0< sin a + g*P.
PROBLEMS
dy
Find -p- in each of the following cases :
ax
1. y = (2x + 3)(x* + 3x-l).
«.. y = (oja-|-4ar-3)(3x2-|-12x-|-12).
SC + « 13. y :
3. y =
x — a
xa-4
2aa + 4a-3
a1 + aj - 2
o o
8. 3/ = 3z*-f-2a^ 1""-;"
x* as*
, 9 2
9. y = #8 — 2# h-=*
* xx9
10. y= -n/x-— =•
6/~T 5/— 1 1
11. y=->/xA -V* +-i7=+-77=-
Va; Vr
12. y = (2z8 + 3xa+6)8.
AC
4. y =
5. y =
6. y =
7. y =
8. y =
9. y =
20. y-
21. y
(z8 - 1)4.
^4x8+6z2-5.
Vx4 -|- x2 - 2 x.
a8 + 8
3
xa + 4a + l
22. y
23. y:
24. y
25. y
-VJ + l
: (3 x - l)2(z - l)8.
(l-2a;2)V~3ic-|-l).
; (x - 1) Vza + 1.
2z-l
V*2 + l
=(aa-2z + 3)*(z8+l)*-
= Vl + x + Vl — x.
1
x + Va^
182 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
^ = NaHri
26. V = -J^-^. 29. y =
V^
x
2
3 — 1 "i/^724- rr2
y Vx2 - 1 30- y "" x
x 1
28. y = • 31. y = .
x -f- V a2 + x2 x — V a2 + x*
Find -p from each of the following equations :
32. x2 + 2xy + f=0. 34. xy = (a -f- y)\
33. a;6+5irV — 10ary4+y5=0. 35. (x + y)* + (« — y)* = a*.
Find -^ and j^ from each of the following equations :
CLX U/X
36. 3aj* + y* = l. 39. ar> + a-y + y* = 0.
37. x* + f = a\ 40. f = a(^ + y*).
38. x* + y^ = a^. 41. xif = x + y.
42. Find the equations of the tangent and the normal to the
curve 5 ax2 — 4 x2y = 4 y8 at the point (2 a, a).
43. Find the equations of the tangent and the normal to the
strophoid y = ± xy\ at the point ( — > -=-).
44. Find the point at which the tangent to the curve y = Xs at
(1, 1) intersects the curve again.
45. Find the equations of the tangent and the normal to the
ellipse Sx2 + by1 = 32 at a point the abscissa of which is equal to
its ordinate.
46. Find a point at which the tangent to the curve xy — 5x* — 4 = 0
has the slope 1.
47. Find the length of the portion of the normal to the parabola
f = 8 x at (2, 4) included between the axis and the directrix of
the parabola.
Find the equation of the tangent to each of the following curves
at the point (xv y^):
48. y2 = x6. 50. xi + y$ = ai.
49. Vx + Vy = Va. 51. x8 + if — 3 axy = 0.
PROBLEMS 183
52. Prove that the equation of the tangent to the parabola
y2 = 4 px at the point (xv y^) is yx y = 2 p (x + x^).
63. If the slope of a tangent to the parabola y2 = Apx is m, prove
that its equation is y = mx H
64. Prove that the equation of the tangent to the ellipse
-^ + yi = 1 at the point (xv y^) is -\- + *y^ = 1, and that the equa-
x2 iF
tion of the tangent to the hyperbola -5 — y^ = 1 at the point (a^, y^)
13 a2 b2 ~lm
65. Prove that the equations of the tangents with slope m to the
x2 y2 x2 y2
ellipse -^ + 75 = 1 and the hyperbola — — y^ = 1 are respectively
y = mx ± Va2m2 -|- &2 and y = ma; ± Va2ra2 — &2.
Draw each pair of the following curves in one diagram and #
determine the angles at which they intersect: .
56. x + y — 7 = 0, x2 — 4a - 3y + l = 0.
57. «2 + 2/l-16a; + 14 = 0, sc2 + y* - Sy + 6 = 0.
58. 2^-9a; = 0, 3z2 + 4?/ = 0.
59. x2 = 4ay, 2z2 + 22/2-5ax = 0.
a;8
60. y* = - , x2 + y*-2ax = 0.
L a — x
61. a2 + 3/* = 45, y2 = 12x.
62. x* + y*-12 x + 16 = 0, f =
xz
4: — X
Xs 8 a*
63. t? = a8, y2 = ^ • 65. ar* = 8 a2 — 4 ay, y =
64.^ = 4,2, = ^. 66. a*, = a2, 2^ = 2^-
67.^-3^ = a2,y = ^^.
68. ^ = 6(oj-3), 4^=(a?-3)a(a;-l).
69. Prove that the parabolas #* = 4 asc + 4 a2 and y2 = — 4 bx + 4 #2
are confocal and intersect at right angles.
184 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
70. Show that for an ellipse the segments of the normal between
the point of the curve at which the normal is drawn and the axes
are in the ratio a2 : b2.
71. Find the coordinates of a point on the ellipse -^ + ^ = 1
such that the tangent there is parallel to the line joining the
positive extremities of the major and the minor axes.
72. Find a point on the ellipse -5 + ^ = 1 such that the tangent
there is equally inclined to the two axes.
73. Prove that the portion of a tangent to an hyperbola included
by the asymptotes is bisected by the point of tangency.
74. If any number of hyperbolas have the same transverse axis,
show that tangents to the hyperbolas at points having the same
abscissa all pass through the same point on the transverse axis.
75. If a tangent to an hyperbola is intersected by the tangents at
the vertices in the points Q and R, show that the circle described on
QR as a diameter passes through the foci.
76. Prove that the ordinate of the point of intersection of two
tangents to a parabola is the arithmetical mean between the ordi-
nates of the points, of contact of the tangents.
77. If, on any parabola, P, Q, and R are three points the ordinates
of which are in geometrical progression, show that the tangents at
P and R meet on the ordinate of Q.
78. Show that the tangents at the extremities of the chord of a
parabola, which is perpendicular to the axis of the parabola at the
focus, are perpendicular to each other.
79. Prove that the tangents described in Ex. 78 intersect on the
directrix of the parabola.
80. Prove analytically that if the normals at all points of an
ellipse pass through the center, the ellipse is a circle.
81. Prove that any tangent to the parabola y2 = &px will meet
the directrix and the straight line drawn through the focus, per-
pendicular to the axis of the parabola, in two points equidistant
from the focus.
82. Find in terms of xx and p the length of the perpendicular
from the focus of the parabola y2 = Apx to the tangent at any
point (xv yx).
PROBLEMS 185
83. If from two given points on the axis of a parabola which
are equidistant from the focus perpendiculars are let fall on any
tangent, prove that the difference of their squares is constant.
84. Show that the product of the perpendiculars from the foci
of an ellipse upon any tangent equals the square of half the
minor axis.
85. Find the equation and the length of the perpendicular from
the center of the ellipse —2 + y2 = 1 to any tangent.
86. If two concentric equilateral hyperbolas are described, the
axes of one being the asymptotes of the other, show that they
intersect at right angles.
87. Prove that an ellipse and an hyperbola with the same foci
cut each other at right angles.
88. Prove that the normal to an ellipse at any point bisects the
angle between the focal radii drawn to the point.
89. Prove that the normal to an hyperbola at any point makes
equal angles with the focal radii drawn to the point.
Determine the values of x for which the following curves are
(1) concave upward ; (2) concave downward :
90. y = 4:z8-6a:a + 3. 91. y = xA - 12 #2 -f- 2.
Find the points of inflection of the following curves :
92. y = 2x* + 9x*-2x-5. 96# y= 1 +•- J—
93. y = 3xA-±x*-6x* + ±. X + 1 X~~1
94. y = (x + 6a)(x-a)K 97' «V = * + "V-
*
8«8
95. y =
x2 + 4ca?
»°-0Hf)-
Find the turning points and the points of inflection of each of
the following curves and then draw the curve :
99* p = (x — 2)\x + 2). 102. y = x4 - ±x* + 16.
100. y = x* - 3 a2 - 9sc - 5. 103. if = x(x* - 4).
101. y = x(x — l)8.
186 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
104. It is required to fence off a rectangular piece of ground
to contain a given area, one side to be bounded by a wall already
constructed. If the length of the side parallel to the wall is xy
will an increase in x cause an increase or a decrease in the total
amount of fencing ?
105. The hypotenuse of a right triangle is given. If one of the
sides is x, find the effect on the area caused by increasing x.
106. The stiffness of a rectangular beam varies as the product of
the breadth and the cube of the depth. From a circular cylindrical
log of radius a inches, a beam of breadth 2 as is cut. Find the effect
on the stiffness caused by increasing x.
107. A right cone is generated by revolving an isosceles triangle
of constant perimeter about its altitude. If x is the length of
one of the equal sides of the triangle, will an increase in x cause
an increase or a decrease in the volume of the cone ?
108. A gardener has a certain length of wire fencing with which
to fence three sides of a rectangular plot of land, the fourth side be-
ing made by a wall already constructed. Required the dimensions
of the plot which contains the maximum area.
109. A rectangular plot of land to contain 216 sq. rd. is to be
inclosed by a fence and divided into two equal parts by a fence
parallel to one of the sides. What must be the dimensions of the
rectangle that the least amount of fencing may be required ?
110. A gardener is to lay out a flower bed in the form of a sector
of a circle. If he has 20 ft. of wire with which to inclose it, what
radius will he take for the circle to have his garden as large as
possible ?
111. An open tank with a square base and vertical sides is to
have a capacity of 4000 cu. ft. Find the dimensions so that the
cost of lining it with lead may be a minimum.
112. A rectangular box with a, square base and open at the top is
to be made out of a given amount of material. If no allowance is
made for the thickness of the material or for waste in construction,
what are the dimensions of the largest box that can be made ?
113. Find a point on the line y = x such that the sum of the
squares of its distances from the points (— a, 0), (a, 0), and (0, b)
shall be a minimum.
PROBLEMS 187
114. A piece of wire 12 ft. in length is cut into six portions, two
f one length and four of another. Each of the two former portions
bent into the form of a square, and the corners of the two squares
fastened together by the remaining portions of wire, so that the
<3ompleted figure is a rectangular parallelepiped. Find the lengths
into which the wire must be divided so as to produce a figure of
^maximum volume.
115. The strength of a rectangular beam varies as the product of
its breadth and the square of its depth. Find the dimensions of the
strongest rectangular beam that can be cut from a circular cylindri-
C2al log of radius a inches.
116. What are the dimensions of the rectangular beam of great-
est volume that can be cut from a log a feet in diameter and b feet
Xong, assuming the log to be a circular cylinder ?
117. A log in the form of a frustum of a cone is 20 ft. long, the
diameters of the bases being 2 ft. and 1 ft. A beam with a square
«3ross section is cut from it so that the axis of the beam coincides
"with the axis of the log. Find the beam of greatest volume that can
\)Q so cut.
118. Find a point on the axis of x such that the sum of its dis-
tances from the two points (1, 2) and (4, 3) is a minimum.
119. Find the point on the circle x2 + y2 = a2 such that the sum
of the squares of its distances from the two points (2 a, 0) and (0, 2 a)
shall be the least possible.
120. A water tank to hold 300 cu. ft. is to be constructed in the
form of a right circular cylinder, the base of the cylinder being
horizontal. The tank is open at the top, and the material used for
the bottom costs twice as much per square foot as that used for
the lateral wall. What are the most economical proportions for the
tank?
»
121. A tent is to be constructed in the form of a regular quadran-
gular pyramid. Find the ratio of its height to a side of its base
when the air space inside the tent is as great as possible for a given
wall surface.
122. An isosceles triangle of constant perimeter is revolved about
its base to form a solid of revolution. What are the altitude and
the base of the triangle when the volume of the solid generated is
a maximum ?
188 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
- 123. Required the right circular cone of greatest volume which
can be inscribed in a given sphere.
124. The total surface of a regular triangular prism is to be &.
Find its altitude and the side of its base when its volume is as
great as possible.
125. The combined length and girth of a postal parcel is 60 in.
Find the maximum volume : (1) when the parcel is rectangular with
square cross section ; (2) when it is cylindrical.
126. A length I of wire is to be cut into two portions, which are
to be bent into the forms of a circle and a square respectively. Show
that the sum of the areas of these figures will be least when the
wire is cut in the ratio it : 4.
127. A piece of galvanized iron b feet long and a feet wide is to be
bent into a U-shaped water drain b feet long. If we assume that
the cross section of the drain is exactly represented by a rectangle on
top of a semicircle, what must be the dimensions of the rectangle
and the semicircle that the drain may have the greatest capacity:
(1) when the drain is closed on top ? (2) when it is open on top ?
128. A circular filter paper 10 in. in diameter is folded into a
right circular cone. Find the height of the cone when it has the
greatest volume.
129. It is required to construct from two equal circular plates of
radius a a buoy composed of two equal cones having a common base.
Find the radius of the base when the volume is the greatest.
130. Two towns A and B are situated respectively 10 mi. and
15 mi. back from a straight river from which they are to get their
water supply, both from the same pumping station. At what point
on the bank of the river should the station be placed that the least
amount of piping may be required, if the nearest points of the river
to A and B respectively are 20 mi. apart ?
131. A man on one side of a river, the banks of which are assumed
to be parallel straight lines 2 mi. apart, wishes to reach a point on
the opposite side of the river and 10 mi. further along the bank. If
he can row 3 mi. an hour and travel on land 5 mi. an hour, find the
route he should take to make the trip in the least time.
PROBLEMS 189
132. A power house stands upon one side of a river of width
b miles and a manufacturing plant stands upon the opposite side,
a miles downstream. Find the most economical way to construct
the connecting cable if it costs m dollars per mile on land and
n dollars per mile through water.
133. At a certain moment of time a vessel is observed at a point
A, sailing in the direction AB at the rate of 10 mi. per hour, and
another vessel is observed at C, sailing in the direction CA at the
rate of 20 mi. per hour. The angle between AB and A C is 60°, and
AC is 50 mi. When will the vessels be nearest to each other?
134. A vessel is sailing due north at the rate of 10 mi. per hour. «
Another vessel, 190 mi. north of the first, is sailing on a course
S. 60° E. at the rate of 15 mi. per hour. When will the distance
between them be the least?
135. Find the least ellipse which can be described about a given
rectangle, the area of an ellipse with semiaxes a and b being irab.
136. Find the isosceles triangle of greatest area which can be
cut from a semicircular board, assuming that the base of the triangle
is parallel to the diameter.
137. Find the isosceles triangle of greatest area which can be cut
from a parabolic segment, assuming that the vertex of the triangle
lies in the base of the segment.
138. The number of tons of coal consumed per hour by a certain
ship is 0.3 + 0.001 v8, where v miles is the speed per hour. Find the
amount of coal consumed on a voyage of 1000 miles and the most
economical speed at which to make the voyage.
139. The fuel consumed by a certain steamship in an hour is
proportional to the cube of the velocity which would be given to
the steamship in still water. If it is required to steam a certain
distance against a current flowing a miles an hour, find the most
economical rate.
140. The altitude of a variable cylinder is constantly equal to the
diameter of the base of the cylinder. If when the altitude is 8 ft. it
is increasing at the rate of 3 ft. an hour, how fast is the volume
increasing at the same instant ?
141. Find where the rate of change of the ordinate of the curve
y = se8 — 6 a;2 + 3 a; -|- 5 is equal to the rate of change of the slope
of the tangent.
190 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
142. The angle between the straight lines AB and BC is 60°, and
AB is 28 ft. long. A particle at A begins to move along AB toward B
at the rate of 4 ft. per second, and at the same time a particle at B
begins to move along BC toward C at the rate of 8 ft. per second. At
what rate are the two particles approaching each other after 1 sec. ?
143. A series of right sections is made in a right circular cone
of which the vertical angle is 90°. How fast will the areas of the
sections be increasing if the cutting plane recedes from the vertex
of the cone at the rate of 2 ft. per second ?
144. A roll of belt leather is unrolled on a horizontal surface at
the rate of 5 ft. of length per second. If the leather is \ in. thick, at
what rate is the radius of the roll decreasing when it is equal to
2 ft., if the roll is assumed to remain always a true circle ?
145. A trough is in the form of a right prism with its ends equi-
lateral triangles placed vertically. The length of the trough is 10 ft.
It contains water which leaks out at the rate of 1 cu. ft. per minute.
Find the rate, in inches per second, at which the level of the water
is sinking in the trough when the depth is 1 ft.
146. A solution is being poured into a conical filter at the rate of
3 cc. per second and is running out at the rate of 1 cc. per second.
The radius of the top of the filter is 10 cm., and the depth of the
filter is 30 cm. Find the rate at which the level of the solution is
rising in the filter when it is one third of the way to the top.
147. A peg in the form of a right circular cone of which the ver-
tical angle is 30° is being driven into the sand at the rate of 2 in.
per second, the axis of the cone being perpendicular to the surface
of the sand, which is a plane. How fast is the lateral surface of the
peg disappearing in the sand when the end of the peg is 10 in. below
the surface of the sand ?
148. A body is moving in a straight line according to the law
s = & — 9 tf2 + 15 £. Find its velocity and acceleration. When is
the body moving forward and when backward ?
149. A body is moving in a straight line according to the law
s = %t* — 2 £* + 4 tf3. Find its velocity and acceleration. When is its
velocity a maximum ? During what interval is it moving backward ?
150. The top of a ladder a units long slides down the side of a
vertical wall which rests on horizontal land. If the velocity of the
top is v0, what is the velocity of the bottom ?
PROBLEMS 191
151. Two parallel straight wires are a feet apart. A bead slides
along one of them at the rate of b feet per second. How fast is the
bead approaching a fixed point on the other wire ?
152. A boat with the anchor fast on the bottom at a depth of
30 ft. is drifting at the rate of 4 mi. an hour, the cable attached to
the anchor slipping over the end of the boat. At what rate is the
cable leaving the boat when 50 ft. of cable are out, assuming it
forms a straight line from the boat to the anchor ?
153. A lamp is 60 ft. above the ground. A stone is let drop from
a point on the same level as the lamp and 20 ft. away from it. Find
the speed of the shadow on the ground after 1 sec, assuming that
the distance traversed by a falling body in the time t is 16 P.
154. A particle moves in a plane so that its coordinates at any
2
time t are given by the equations x = 2 1, y = 2 • Find the
t + 1
Cartesian equation of its path, and its velocity in its path.
155. Two points, having always the same abscissa, move in such
a manner that each generates one of the curves y = xs — 12 #2 + 4 a;
and y = x* — 8 x2 — 8. When are the points moving with equal
speed in the direction of the axis of y?
156. A particle is moving along the curve y2 = 4 as, and when
x = 4 its ordinate is increasing at the rate* of 10 ft. per second. At
what rate is the abscissa then changing, and how fast is the particle
moving in the curve ? Where will the abscissa be changing ten
times as fast as the ordinate ?
157. A ball is swung in a circle at the end of a cord 5 ft. long, so
as to make 20 revolutions a minute. If the cord breaks, allowing the
ball to fly off at a tangent, at what rate will it be receding from the
center of its previous path T£^ sec. after the cord breaks, if no
allowance is made for any new force acting ?
168. The top of a ladder 32 ft. long rests against a vertical
wall and the foot is drawn along a horizontal plane at the rate of
4 ft. per second in a straight fine from the wall. Find the path of a
point one fourth of the distance from the foot of the ladder, and its
velocity in its path at any time t
CHAPTER XI
DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS
95. Limit of • In order to apply the methods of the dif-
h
ferential calculus to the trigonometric functions, it is necessary
• 7
to know the limit approached by — — as h approaches zero as
h
a limit, it being assumed that h is expressed in circular measure.
Let AOB (fig. 151) be the angle A,
r the radius of the arc AB described
from 0 as a center, a the length of AB,
p the length of the perpendicular BC
from B to OAy and t the length of the
tangent drawn from B to meet OA pro-
duced in D.
Revolve the figure on OA as an axis
until B takes the position B'. Then the chord BCB'= 2p, the
arc BAB1 = 2 a, and the tangent B'D = the tangent BD. Evidently
BD + DB' > BAB1 > BCB\
whence t> a>p.
Dividing through by r, we have
tap
r r r
that is, tan A > A > sin A.
Dividing by sin A, we have
1 A
^y^
P
a\\f
ys^W
C
A\d
r"^
i
«*
■ * /
«*
«*
i '/
"^
'/
^,
---y
B'
Fig.
161
cos A sin A
or, by inverting, cosA<— r— <1.
192
TRIGONOMETRIC FUNCTIONS 193
Now as A approaches zero, cos A approaches 1. Hence — z— »
A
which lies between cos A and 1, must also approach 1; that, is,
T . sin A H
Lim — — = 1.
a=o A
96. Differentiation of trigonometric functions. The formulas
for the differentiation of trigonometric functions are as follows,
where u represents any function of x which can be differentiated :
d . du _. N
— sinw = cosw — > (1)
dx dx
d . du .~N
— costt = — smw— ? (z)
dx dx
d , 2 du ,ON
— tan u = sec u — » (o)
— - ctn u = — esc w — - » (4)
aa dx
d du ,CN
— sec u = sec w tan u — » (5)
— esc w = — esc w ctn w — - • (o)
dx dx
1. By(8),§84, £sinM = AsinM.g.
To find — - sin w, we place v = sin u.
du
Then if u receives an increment Aw, y receives an increment Ay,
where . / Aw\ . Au
Ay = sin (w + Au) — sin w = 2 cos ( w + -jr J sin — *
the last reduction being made by the trigonometric formula
. , * a + b . a — b
sm a — sin b = 2 cos — - — sin — — •
Then we have A A
. Au . Au
sm --- sin —
Aw 0 / Aw\ 2 / Au\ 2
2
194 TRANSCENDENTAL FUNCTIONS
Let Au approach zero. By 2, § 69, .
sin —
Lim-7^ = Limcos(tt + -r-)Lim — - •
Au \ 2 / Au
But Lim-^ = -^,
Au du
Limcos(w + -^-)= cosw,
. Au
sin —
and Lim— — = 1. (By § 95)
Au
"2*
Hence — sin w = cos u,
au
d . du
and — - sni u = cos u — .
dx dx
2. To find — - cos u, we write
dx
cosw
Then — cos u = — sin( ^ — w )
oz a# \2 /
=cos(f -*)£(!-*) ^^
= _Co8^--Mj_
(fie
= — sin u — •
02/
rf
3. To find — tan u, we write
dx
sin w
tan u =
cosw
TRIGONOMETRIC FUNCTIONS 195
r™ d , d sin u
Then — tan u =
dx dx cos ?4
d . rf
cos w — sin t* — sin u — cos w
«^ dx m n ^ t
(by (5), § 84)
cos2w
(a m a >. (A/it
cos u + suru) —
cos2w
(by (1) and (2))
= sec u — - •
dx
4. To find — ctn w, we write
ax
cosw
ctn w = —
sm w
rr,, d , d cosw
lhen — ctn u = - — : —
dx dx sin u
d d .
sin u — cos w — cos u — sin w
(by (5), § 84)
sin2w
— sin2w — cos2w du
sin2w dx
du
dx
(by (1) and (2))
= — csc2w
3
5. To find — sec u, we write
dx
sec w = = (cos w) \
cos u
Then — sec u = — (cos u)~2-j- cos w (by (6), § 84)
_ sin u du ,
cos2w dx
du
= sec w tan u — - •
dx
■ /?
6. To find — esc w, we write
dx
esc w = — — = (sin u)~x.
sin u
\
196 TRANSCENDENTAL FUNCTIONS
Then — esc u = — (sin u)~ 2 — sin u (by (6), § 84)
ax ax
= — escwetnw — • (By (1))
Ex. 1. y = tan 2 x — tan2 a; = tan 2 x — (tan ar)2.
J? = sec2 2 a;—- (2 ar) — 2 (tanar) — - tana;
ax ax ax
= 2 sec2 2 a: — 2 tan a: sec2a\
Ex. 2. y = (2 sec4ar + 3 sec2 a:) sin x.
— = sin x 8 sec8z — (sec x) 4- 6 sec a: — (sec x) I + (2 sec4ar + 3 sec2ar) — (sin a:)
ax L ax dx J ax
= sin a: (8 sec4 a: tan x + 6 sec2x tan x) + (2 sec4 a: + 3 sec2 a;) cos x
= (1 — cos2 a:) (8 sec5a; + 6 sec8ar) + (2 sec8a; + 3 sec x)
= 8 sec6x — 3 sec x.
97. Differentiation of inverse trigonometric functions. The
formulas for the differentiation of the inverse trigonometric
functions are as follows:
1. — sin"1^ = = — - when sin_1w is in the first or the
V l — u fourth quadrant ;
— - when sin""1!* is in the second or
1 u the third quadrant.
j 1 fiti
2. — cos~1w = — when cos"~lw is in the first or the
V l — u second quadrant ;
= =— when cos""1^ is in the third or the
VI — u x fourth quadrant.
o d _x 1 du
3. -—tan 1u=- - — -•
dx 1 + u dx
a d , , 1 du
4. — -ctn"He = — -— -•
ax 1 -f u dx
5. — sec_1w = — - — = — -.— when sec_1w is in the first or the
dx uVu2-l dx third quadrailt .
— - when sec~lw is in the second or
u
u * the fourth quadrant.
INVERSE TKIGONOMETRIC FUNCTIONS 197
6. — csc~1w = — when csc""1^ is in the first or
dx Uy/u*-l dx the third quadrant;
-— when csc_1w is in the second or
u
' u ■*• the fourth quadrant.
The proofs of these formulas are as follows:
1. If y = sin""1^,
then sin y = w.
Hence, by § 96, cosy-^ = — >
dx ax
whence
dy _ 1 du
dx cos y dx
But cosy = Vl — u2 when y is in the first or the fourth quad-
rant, and cosy = — Vl — u2 when y is in the second or the third
quadrant.
2. If
y = COS"1!*,
then
cosy = u.
Hence
dy du
dx dx
whence
dy 1 du
dx sin y dx
But siny =Vl— u2 when y is in the first or the second
quadrant, and sin y = — Vl — u2 when y is in the third or the
fourth quadrant.
3. If
y = tan"1!*,
then
tan y = u.
Hence
„ dy du
sec y — = — »
dx dx
whence
dy 1 du
dx 1 + u2 dx
AC
I
198 TRANSCENDENTAL FUNCTIONS
4. If
y = ctn"1^,
then
ctn y = u.
Hence
„ dy du
-csc2y-f- = — >
dx ax
whence
dy 1 du
dx 1 + u2 dx
5. If
*
y = sec"1!*,
then
sec y = u.
Hence
dy du
secytany^-,
. dy 1 du
whence -f- = 7- ■
dx sec y tan y dx
But sec y = uy and tan y = Vw2 — 1 when y is in the first
or the third quadrant, and tan y = — Vw2 — 1 when y is in the
second or the fourth quadrant.
6. If y = csc"1^
then esc y = u.
TT , dy du
Hence — esc y ctn y-f- = — >
whence
rfy __ 1 rfw
<fa cscyctny<fa
But cscy = w, and ctn y = Vw2 — 1 when y is in the first
or the third quadrant, and ctn y = — Vw2 — 1 when y is in the
second or the fourth quadrant.
If the quadrant in which an angle lies is not material in a
problem, it will be assumed to be in the first quadrant. This
applies particularly to formal exercises in differentiation.
Ex. 1. y = sin-1 Vl — a?, where y is an acute angle.
*p Vi-(i-x2) <** VTT^
This result may also be obtained by placing sin-1 Vl — x* = cos-1*.
EXPONENTIAL AND LOGAKITHMIC FUNCTIONS 199
Ex. 2. y = sec-1 V4 x2 + 4 x + 2.
f-V4x2 + 4a: + 2
<** V4 x2 + 4 x + 2 V(4 ar2 + 4a: + 2)-l
4x4-2 1
(4a? + 4a: + 2)(2a; + l) 2a:2 + 2a: + l
l
98. Limit of (1 + A)*. In obtaining the formulas for the differ-
entiation of the exponential and the logarithmic functions it is
i
necessary to know the limit of (1 + A)* as h approaches zero, the
rigorous derivation of which requires methods which are too
advanced for this book. We must content ourselves, therefore,
with indicating somewhat roughly the general nature of the proof.
i
We begin by expanding (1 + A)* by the binomial theorem and
making certain simple transformations; thus,
(i+Ay=i+-A+^— h*+ ^ *•+...
^1^ [2 ^ [8
where B represents the sum of all terms involving A, Aa, h\ etc.
Now it may be shown by advanced methods that as h approaches
zero R also approaches zero, and at the same time
lH ur + r+...
1 H H
approaches e (§ 27). Hence
Lim (1 + hy = e.
99. Differentiation of exponential and logarithmic functions.
The formulas for the differentiation of the exponential and the
logarithmic functions are as follows, where, as usual, u represents
200 TRANSCENDENTAL FUNCTIONS
any function which can be differentiated with respect to z, log
means the Napierian logarithm, and a is any constant:
d , log„e du ,+ N
d , __ 1 dw .o\
da: w dr
£«- = a-logag, (3)
Txe=eTx (4)
The proofs of these formulas are as follows:
1. By (8)>m A ,<*.„ = J; !<*.».£.
To find — log0w, place y = log0w.
(XXL
Then if w is given an increment Aw, y receives an increment
y" W 6re Ay = log0(w + Aw) - log>
=^(i+^)
-*■*(
the last transformation being made by the f ormula p log M = log IP.
Then ^ = !loga(l + ^f.
Aw u \ u )
Now as Au approaches zero, the fraction — may be taken
as A of § 98. ^ M
Hence LimllH ) =e.
Therefore -^ = - log. 6
aw w
a# w as
EXPONENTIAL AND LOGARITHMIC FUNCTIONS 201
2. If y = log w, the base a of the previous formula is e ;
and since logee = l, we have
dy __ 1 du
dx u dx
3. If # = «",
ve have log y = log a* = u log a.
Hence, by formula (2),
1 dy , du
whence -~ = a* log a -7- •
dx dx
4. If y = e", the previous formula becomes
dy „du
dx dx
Ex. 1. y = \og(x*-±x + 5).
dy __ 2 a; — 4
cte a:2 — 4 a: + 5
Bz. 2. y = £-**•
Bz. 8. y = e-^cos fa.
-Jl = cos fa — (e-*") + e-** — (cos fa) = — ae-^cos fa — be"ax sin fa
flb dx dx
= — g- ^(a cos fa + 6 sin fa).
100. Sometimes the work of differentiating a function is
simplified by first taking the logarithm of the function.
*jl 1. Let y = J1'**.
\l + x*
Then logy = log ^i-j-^
= £ log(l -**)-£ log (1 + .T2).
202 TRANSCENDENTAL FUNCTIONS
Hence
and
1 dy _ x x
ydx 1 — x2 1 + a?
_ -2x
(l-a^Xl+a^)'
<fy _. - 2 xy
dx (1 - a;2)(l + x2)
-2ar /IEZ
(l-a^Xl + ar2) \l + a^
-2a:
(l + a*)Vl-a:4
This method is especially useful for functions of the form wr,
where u and v are both functions of x. Such functions occur
rarely in practice and cannot be differentiated by any of the
formulas so far given. By taking the logarithm of the function,
however, a form is obtained which may be differentiated.
Ex. 2. Let y = x^x.
Then log y = log (x****)
= sin x • log x.
Therefore - j = C8*11 x) " + cos x ' 1°8> x>
y dx x
and — = ar8inx_1 • sin x + x^* cos a: • log x.
dx
101. Applications. The applications of differentiation discussed
in the previous chapter are evidently applicable to problems in-
volving transcendental functions.
Ex. 1. Find the turning points and the points of inflection of the curve
(Ex. 5, § 24) . L , . o
v ' * ' y — sin x + \ sin 2 x.
We find — = cos x + cos 2 x = 2 cos2a: + cos x — 1
ax
= (2 cos x — 1) (cos x + 1).
Equating — to zero, we have cos x = — or cos x = — 1.
dx 2
If cos a; = -,a; = - + 2rwror — — + 2 n7r. As x passes through either of
2 3 o
these values, ~f- changes sign, and hence these values correspond to turning
dx o
points of the curve. In fact, x = — + 2 rwr gives maximum values of y = j V5,
and a: = — - + 2 nir gives minimum values of y = — t*^
APPLICATIONS
203
If cos a: = — 1, ar = Tr±2 nm As x passes through these values, -j- does
ax
not change sign. Hence these values do not correspond to turning points
of the curve.
To examine for points of inflection, we find
^ = -
dx*
sin x — 2 sin 2 x == — sin x (4 cos x + 1).
This is zero when a: = 0 + 2rwror7r + 2 iwr, or when x = cos~1(— ^).
As x passes through any of these values, — | changes sign. These values
correspond, therefore, to points of inflection.
Ex. 2. A particle of mass m moves in a straight line so that
s = k sin bt,
where t = time, s = space, and ft and & are constants.
ds
Then velocity = v = — = bk cos bt,
dt
d2s
acceleration = a = — - = — b2k sin 6* = — ft2*,
force = F= ma = — mft2*.
Let O be the position of the particle when t = 0, and let Oil = k and
05 = — k. Then it appears from the formulas for s and v that the particle
oscillates forward and backward between B and A. It describes the distance
IT 2 7T
Oil in the time — and moves from B to A and back to B in the time
2b b
The formula F= — ml?s shows that the particle is acted on by a force
directed toward 0 and proportional to the distance of the particle from 0.
The motion of the particle is called simple harmonic motion.
Ex. 8. A wall is to be braced by means of a beam which must pass over
a lower wall b units high and standing a units in front of the first wall.
Required the shortest beam which can be used.
Let AB = I (fig. 152) be the beam, and let C be the top of the lower
wall. Draw the line CD parallel to OB, and let EBC = 0.
Then
l = BC+ CA
= ECcsc$ + DC8ecO
= b esc 0 + a sec $.
— = — b esc 0 ctn $ + a sec $ tan 0
ad
a8in80 — ft cos8 0
Placing
dO
sin* 0 cos2 0
= 0, to find the minimum,
we have a sin80 = ft cos8 0, whence tan 0 = — - •
as
204 TRANSCENDENTAL FUNCTIONS
When 0 has a smaller value than this, a sin80 < b cos80 ; and when 6 has a
larger value, a sin8 $ > b cos8 $. Hence / is a minimum when tan 0 = — •
Then / = &csc0 + asec0
_b\gi + bi aVgi + bl
= (a* + &*)*.
102. The derivatives in parametric representation. When a
curve is defined by the equations
we have, by (9), § 84, ^ = ^' (1)
It
d\
If it is required to find — ^, we may proceed as follows:
(XX
d*y
_ d /dy\dt\dx)
dx\dx) dx
dx2
dt
Ex. For the cycloid
x = a (<£ — sin <£),
y = a(l-cos<£),
dy _ d<f> a sin <£ <£
dx dx a (1 — cos <£) 2
rf2y_ d
dd>\ 2/dx
dx2 d<f>
1 2*
— -cosecz-£
2 2
a(l — cos<£)
1
4 a sin4 ?
2
DIRECTION OF A CURVE
Formula (2) may be expanded as follows:
205
dtxdx/
d*y dx
l^~dt
d2xdy
dt?~di
/dxX
\dt)
a
Therefore
dx2
d*y dx
lfi~dt
d*x dy
d(?di
/dx
\dt
dxX
dt)
8
(3)
103. Direction of a curve in polar coordinates. The direction
of a curve expressed in polar coordinates is usually determined
by means of the angle between the
tangent and the radius vector. Let P
(r, 0) (fig. 153) be any point on the
curve, PT the tangent at P, and ^r
the angle made by PT and the radius
vector OP. Give 0 an increment A0 =
POQi expressed in circular measure,
thus fixing a second point Q(r + Ar,
0+A0) of the curve.
To determine Ar describe an arc of
a circle with center at O and radius OP, intersecting OQ at E.
Fig. 163
Then
and
OQ=r + Ar
RQ = Ar.
Draw also the chord PQ and the straight line PS perpendicular
to OQ and meeting it in S.
Then &P=rsinA0,
OS=roosA0,
SQ= OQ- OS= r+Ar-r cos A0
= Ar + 2 r sin -jr- •
206 TRANSCENDENTAL FUNCTIONS
As Ad approaches zero the chord PQ approaches the limiting
position PT and the angle EQP approaches yfr. But in the
triangle SPQ,
SP r sin Ad
tan SQP= — = -
Ar + 2 r sur —
sin Ad
r-*r
. Ad
. * sin -—
Ar . Ad 2
+ rsin
Ad ' 2 Ad
2
Hence, taking the limit, we have
T
tan-^ = — -• (1)
d0
If it is desired to find the angle MNP = <f>, it may be done by
the evident relation , , . /» ,^
<p = Y + V. (2)
104. Derivatives with respect to the arc in polar coordinates.
In the triangle PQS (fig. 153),
SP
sin SQP =
chord PQ
SP arc PQ
= r
arc PQ chord PQ
r sin Ad arcPQ
A* chord PQ
sin Ad Ad arcPQ
Ad A* chord PQ
A /J
As Ad approaches zero SQP approaches -^r, Lim = 1, and
Lim-^f|~ = l(§90).
chord PQ V3 y
Hence sin yjr = r — . (1)
as
CURVATURE
207
By dividing (1), just obtained, by (1) of the previous article,
dr
cos-^r =
d%
(2)
From (1) and (2) we obtain
rd0 = sin-^rrfs, dr = cos-^cfo ;
whence, by squaring and adding, we obtain
ds>=dr2+r2dd2.
(3)
The formulas of this and the foregoing article are correctly
represented by the triangle of fig. 154, which is a convenient
device for remembering the formulas. Here
the lines marked as differentials are really
increments, but as the size of the figure is re-
duced, they become more nearly differentials.
The correct formulas are obtained by using
the triangle as a straight-line figure. We have
ds = Vdr2+ r2d6\ tan yjr = T
Fig. 154
dv
cos sir = — - , sin sir =
ds
dr
rdd
ds
105. Curvature. If a point describes a curve, the change
of direction of its motion may be measured by the change of
the angle <f> (§ 91).
For example, in the
curve of fig. 155, if A% = «
and J%I% — As, and if <f>t and
<f>2 are the values of ^ for
the points J? and 1% respec-
tively, then <£2 — <f>x is the
total change of direction
of the curve between J\
and Z>. If ^-^ = A^,
expressed in circular meas-
ure, the ratio -^- is the average change of direction per linear
unit of the arc iJJ£. Regarding <f> as a function of 8 and
Fig. 155
208
TRANSCENDENTAL FUNCTIONS
Fig. 156
Hence -j- = - > and the circle is a curve of constant
ds a
taking the limit of — - as A* approaches zero as a limit, we
have -£» which is called the curvature of the curve at the
as
point 1J. Hence the curvature of a curve is the rate of change
of the direction of the curve with respect to the length of the
arc (§ 92).
If -£ is constant, the curva-
ds
ture is constant or uniform ;
otherwise the curvature is va-
riable. Applying this defini-
tion to the circle of fig. 156, of
which the center is C and the
radius is a, we have A<f> = PXCPV
and hence As = a A<f>. Therefore
A£_l
As a
curvature equal to the reciprocal of its radius.
106. Radius of curvature. The reciprocal of the curvature is
called the radius of curvature and will be denoted by p. Through
every point of a curve we may pass a circle with its radius
equal to p, which shall have the same tangent as the curve at
the point and shall lie on the same side of the tangent. Since
the curvature of a circle is uniform and equal to the reciprocal
of its radius, the curvatures of the curve and the circle are the
same, and the circle shows the curvature of the curve in a
manner similar to that in which the tangent shows the direction
of the curve. The circle is called the circle of curvature.
From the definition of curvature it follows that
__ ds
P~cUJ>*
If the equation of the curve is in rectangular coordinates,
ds
dx
It
by (9), § 84,
p =
RADIUS OF CURVATURE 209
To transform this expression further, we note that
d% = dx +dy ;
whence, dividing by dx and taking the square root, we have
dx N \dx)
Since <^ = tan-1(^V (by § 91)
<Py
d<f> dx2
dx~ \+(d]£\'
\dx)
Substituting, we have p = -^
dx1
In the above expression for p there is an apparent ambiguity of
sign, on account of the radical sign. If only the numerical value
of p is required, a negative sign may be disregarded.
sc2 v2
Ex. Find the radius of curvature of the ellipse — + fr = 1.
a2 o8
u dy IPx
Here -f = - —
dx ay
dx2 a2^8
Therefore (aY+ 6«**)»
a*b*
Another formula for py that is,
, Mf)T
a
dy
*nay be found by defining <f> as the angle between OF and the
tangent and interchanging x and y in the above derivation.
210
TKANSCENDENTAL FUNCTIONS
107. Radius of curvature in parametric representation. If x
and y are expressed in terms of any parameter t, the radius of
curvature may be found as follows :
But
d%
ds dt
P~ d<f>~d$
dt
(By (9), § 84)
ds 1/dxV (dys?
dt V\dt) \dt)'
(by (2), § 91)
and
*-
whence
~dt
dy
,dy ydt
tan-1-/ = tan"1 — ;
dx dx
dt
<Px
dt2
(dx\^l_(dy\
\dt/dt2 \dtj
(I)'
dx d2y dy d2x
di "dt2 ~~~dt "cfo*
\dt) \dt,
Therefore, by substitution,
nl
P =
[mm
dx d2y dy d2x
di' ~d?~"dt9~d^
Ex. Find the radius of curvature of the cycloid
x = a<f>— a sin <j>9
y = a — a cos <£.
Here the parameter t of the general formula is replaced by <£.
RADIUS OF CURVATURE
211
Therefore
dx
■— = a — a cos <p,
d<p
d2x . ,
— = «sm*;
~=fl81Il(/>,
Hence, by substitution, p =
= a cos <£.
[a2 (1 - cos 4>y + a2 sin2 <ft] *
a (1 — cos <j>) • a cos <f> — a sin ^ (a sin ^)
= 2*a(l-cos<£)£
= 2ia.(2sin2|)i
= 4 a sin
«« +
108. Radius of curvature in polar coordinates. For a curve
^pressed in polar coordinates the radius of curvature may be
fc>iand as follows:
di_d([
d(f> d<f>
Jo
(By (9), § 84)
From § 104,
and, from § 103,
Then
<f> = yfr + 6.
=1+W/_
7^ +
Substituting these values and simplifying, we have as the
required formula, __ , , > „ *
P =
Mm
r* + 2
m-
cPr
212 TRANSCENDENTAL FUNCTIONS
Ex. Find the radius of curvature of the cardioid r = a(l — cos ff).
fir (Pr
Here — -z — a sin 0 and —rz = a cos $.
dO d0*
_ [a2(l - cos fl)2 4 a2 sin8 0]*
9 a2(l-cos0)24 2a2sin20-a(l--co8 0)aco8d
= [2«*(l-cosfl)3*=2ta (1,C08(9a
a2(3-3cos0) 3 V ' '
or p = § (2 ar)i.
Therefore
PROBLEMS
dv z.
Find -p" in each of the following cases :
ax
1. y = Jsin42x. 2. y = J sin6 3 a; — }sin73x.
3- y = ""( T sm* ax — - sin9 ax I •
y a\4 6 /
4. y = \x — J sin 4 a;. • 5. y = \x — J sin (2 — 4a;).
6. y = cos* 3 as (-J*!- cos8 3 x — \ cos 3 a).
7. y = y\ ^cos 2 x (cos2 2 x — 7).
8. y = £ cos (2 x -f 1) [cos2 (2 a; 4- 1) - 3].
9. y = J tan8 a; 4- tan2 x 4- tan x.
2 x x
0. y = -tan82 — 2tan2 4-a.
1. y=- Jctn2(arl + a2).
3 Ax -X _ x
2. y = — - ctar- -f ctn8- — 3 ctn - — x.
o o «5 o
3. y = -sec6--
4. y = 25 ^s^1 ~ HSeCi + |[8eCl) '
5- V = J (csc bx — ctn fcc). 19. y = sin"1 2LZ_£
x4-2
6. y = sin"1 2x.
1 • i^ + 3
7. y = sin-1(2x-l). y 2 2V3
• ,*-2 _ ,2x-3
8. y = sin~1 — — • 21. y = cos-1 r
PEOBLEMS 213
22. y = cos-1 3a?j"1 - 29. y = ctn-1 vF — -•
. a;2 - a2 30. y = sec"1 3 x.
23. y = cos_1 — -•
24. y = tan-\ic — *). *
25. y = tan-1 V*2 + 2 x. QO .^ , 1\
* 32. y = sec Msc + - ) •
2R. */ = fan-1 -
a;" — a" ou. y = sec ^a;,
a2-!- a2" ,a; + 2
r,-2V 31. ^sec-1— ~
ox. ?/ = sec — —
24. y = tan-1(a;-2). U 2
26. y = tan"1
27. y = ctn~1^
y/tf-x2 33- y = csc"1(4aj2-|-4a:).
a2 9A .ViT^2
^- 34. y = csc"1
28. y = ctn-1-^-^ • 35. y = esc"1 ^ + -,) .
36. y = x sin"1 Vl - x2 - Vl - ar2.
a x
37. y = x2 tan-1 - + a2 ctn-1 - + ax.
x a
1/ . « x /7z -n . 3a2 . _xx — a
38. y = — -(a -|- 3 a) V2 asc — x2 + ~^- sin
a
39. y = ^(2^-4* + 3). ^ 1- Va2 + ar2 - a
7 46. y = — log •
40. y = logVa;2 + 4a; + 3. a *
= J_i 2s-3 47. y = log sin a;.
' y ~~ 12 g2a; + 3* 48. y = log(sec3cc + tan3cc).
42. y = -±= log3* -A l-sin|
2V3 3s + V3 49. y = log -.
i 1 1 + sin-
43. y = log =- 2
V3 — 4a;-j-a;2 0, , .
^ , 2 tan a? + 1
44. y = log(3a; + V9^T2). 50' ^ = log tana + 2 '
45« y = i log(a;8 -I- Vcc6 — a6). 51. y = log ctn a; — esc 2 a;.
52. y = log Va^-M +
a;2 -|-4
53. y = 3 Vcc4 - a4 + log (a;2 + Vaj4 - a4)2.
1 , 1 — cos 2 a? cos 2 a;
54- y=olog
AC
8 1 -|- cos 2 ar 4 sin2 2 a?
55. y = x [(log aa;)2 — 2 log oaj + 2].
56. y == log(ar* + Va;4 — l) — sec"1 a:2.
214 TRANSCENDENTAL FUNCTIONS
57. y = 2x tan"1 2x — log Vl + '4a2.
58. y = flog(2aj2 + l) + V2tan-1ccV2.
59. y = a? tan-1 oas log Vl + oV.
60. y = x sec-1 oa: log(oa? -|- Va2^2 — l).
61. y = e"= 69. y=iJ— (a2^-a-2^) + A
62. y = Je^+1. *
63. y = e**"1*. 70. y = 2* - a log (e* + 1).
ft 4 7/ — 1 P8in~ 1t ~ - e^ia sin t/wc — m cos ma; )
y"* ' 71. y = —* -jrr-i L'
65. y = -ia00B2x. a +m
66. y=$(e*x-e-s*)+3(e*-e-x).72' y = tan"1aaf.
* c(l-l-loga) ^
68. yaB_(aV-2« + 2). 74. yaa8in-^ + ^a,-
75. y = sec x ^ tan-1
76. y = log V2 - 2 e* + e2x + (e* - l)dtxr1(ele - 1).
77. y = tan-1Va7T2^-l0f(a; + 1)-
78. y = 31og(^2 + 4)+log>||^| + tan-1|.
i a + Va2 — a;2 /-= 5
79. y = a log — ! Vtf2 — a;2.
as
I ViC — "v ft \X
80. y = logAj-p -j= + tan-\—
81. y = sec-ia? — log(aj -h a -f Vaj2 + 2 oaA
a a; + a ov 7
82. y = 2 sin-1 V2 e* — e2x + V2 e* — e2* log (2 — *■).
84. y=(x)**.
87. y =(tana5)x*.
85. y = (x)*. 88. y = (a2 + a52)
86. y = (tan V^)8lnV^.
a
PROBLEMS 215
Find -j- in each of the following cases :
M x ,y , .„ 91. sm(x + 2y)+e2x+v = 0.
* 92. a* + y* = 0.
90. sin-1- + -vy--:z2 = 0.
y, * 93. #* — sec ccy — tan xy = 0.
Find ~j- and -r4 in each of the following cases :
ax axr
94. e? + ev = ec+v. 96. log (x2 + y*) — tan"1 ^ = 0.
a?
_ ^ .a; , . r-z 5 ~ 97. cos(a;-f-V)+cos(a;— y) = l.
95. tan-^+logVa^+^O. v ^vj^ \ if)
V 98. e*+v = yx.
99. Show that the portion of the tangent to the curve
a. a + Va2 — a2 .
y = 2 l0g / 2 o ~ Va2 - x2
^ a — V a2 — or
included between the point of contact and the axis of y is constant.
(From this property the curve is called the tractrix.)
100. Draw the curve y = e_aarcos bx, and prove that it is tangent
to the curve y = e""* wherever they have a point in common.
sin as
101. Draw the curve y = — — > and show that it is tangent to
1 x
the curve y = —^ wherever they have a point in common.
X
102. Find the angle of intersection of the curves y = sin as and
y = COS X.
103. Find the angle of intersection of the curves y = sin x and
y = sin (x + ^ j.
104. Find the angle of intersection of the curves y = sin x and
y = sin 2 as.
105. Find the point of inflection of the curve y = (x + 1) tan-1 a.
106. Find the points of inflection of the curve y = e~**.
i
107. Find the point of inflection of the curve y = el~x.
108. Draw the curve y = log tan2 x. Find a point of inflection and
the slope at that point.
216 TRANSCENDENTAL FUNCTIONS
109. Prove that the curve
y — % & — $ sin x + ^ sin 2 x
has an indefinite number of points of inflection, and that two of
them lie between the points for which x = 6 and x = 10 respectively.
Find the turning points and the points of inflection of the follpw-
ing curves, and draw the curves.
110. y = xe~< 113- y = sin^'
111. y = x*e-*. 114' y = 2sina + jsin2*.
112. y = xse~x. 115- xy = a2log~-
116. A tablet 10 ft. high is placed on a wall so that the bottoirm-
of the tablet is 8 ft. from the ground. How far from the wall should-
a person stand in order that he may see the tablet to the best ad van —
tage; that is, in order that the angle between the lines from th»
observer's standpoint to the top and the bottom of the tablet may
be the greatest?
117. One side of a triangle is 5 ft., arid the opposite angle is 40°-
Find the other angles of the triangle when its area is a maximum.
118. Above the center of a round table is a hanging lamp. What
must be the ratio of the height of the lamp above the table to the*
radius of the table in order that the edge of the table may be most
brilliantly lighted, given that the illumination varies inversely as
the square of the distance and directly as the cosine of the angle
of incidence ?
119. A weight P is dragged along the ground by a force F. If
the coefficient of friction is K, in what direction should the force be
applied to produce the best result ?
120. An open gutter is to be constructed of boards in such a way
that the bottom and the sides, measured on the inside, are to be each
5 in. wide, and both sides are to have the same slope. How wide
should the gutter be across the top in order that its capacity may be
as great as possible ?
121. A steel girder 27 ft. long is to be moved on rollers along a
passageway and into a corridor 8 ft. in width at right angles to the
passageway. If the horizontal width of the girder is neglected, how
wide must the passageway be in order that the girder may go around
the corner ?
PROBLEMS 217
122. Given that two sides and the included angle of a triangle
have at a certain moment the values 8 ft., 12 ft., and 30° respectively,
and that these quantities are changing at the rates of 4 ft., — 3 ft.,
and 12° per second respectively, what is the area of the triangle at
the given moment, and how fast is it changing ?
123. A particle of unit mass moves in a straight line so that
wt
8 = 6 — 5 sin2 — > where t is the time and s the distance from a
point 0, Find when the particle is moving forward and when
backward. Find also the greatest distance which the particle
reaches from 0, and the force which acts upon it.
124. A motion of a particle in a straight line is expressed by the
equation s = 5 — 2 co$H. Express the velocity and the acceleration
at any point in terms of s.
125. Two particles are moving in the same straight line, and
their distances from the fixed point O on the line at any time t are
respectively x = acoskt and x' = aeoslkt + — )> h and a being
constants. Find the. greatest distance between them.
126. If s = aeht + be~kt, show that the particle is acted on by a
repulsive force which is proportional to the distance from the point
from which s is measured.
127. If a particle moves so that
s = e~^ct(a sin ht -+- h cos ht),
find expressions for the velocity and the acceleration. Hence show
that the particle is acted on by two forces, one proportional to the
distance from the origin and the other proportional to the velocity.
Describe the motion of the particle.
128. A revolving light in a lighthouse £ mi. offshore makes one
revolution a minute. If the line of the shore is a straight line, how
fast is the ray of light moving along the shore when it passes the
point of the shore nearest to the lighthouse ?
129. A, the center of one circle, is on a second circle with center
at B. A moving straight line, AMN, intersecting the two circles at
M and N respectively, has constant angular velocity about A, Prove
that BN has constant angular velocity about B.
218 TRANSCENDENTAL FUNCTIONS
130. BC is a rod a feet long, connected with a piston rod at C, and
at B with a crank AB, b feet long, revolving about A. Find C's velocity
in terms of AB's angular velocity.
131. A body moves in a plane so that x = a cos t + b, y = a sin£ + £>
where £ denotes time and a, ft, and c are constants. Find the path of
the body, and show that its velocity is constant.
132. The parametric equations of the path of a moving point are,
in terms of the time t, x = a cos kt, y = b sin kt, where a, b, and k
are constants and a> b. Prove that the path is an ellipse. Find
the velocity of the point in its path. Find when the velocity is a
maximum and when a minimum.
133. A particle moves so that x = 2 cos t — cos 2 1, y = 2 sin t —
sin 2 £, where t is the time. Find its velocity in its path when t = — •
134. If a wheel rolls with constant angular velocity on a straight
line, required the velocity of any point on its circumference ; also of
any point on one of the spokes.
135. Prove that a point on the rim of the wheel of problem 134
is moving parallel to the straight line on which the wheel rolls, with
a velocity proportional to its distance from OX. •
136. Show that the highest point of a wheel rolling with constant
velocity on a road moves twice as fast as each of the two points in the
rim whose distance from the ground is half the radius of the wheel.
137. If a wheel rolls with constant angular velocity on the cir-
cumference of a fixed wheel, find the velocity of any point on its
circumference and on its spoke.
138. If a string is unwound from a circle with constant angular
velocity, find the velocity of the end in the path described.
139. A man walks along the diameter, 200 ft. in length, of a semi-
circular courtyard at a uniform rate of 5 ft. per second. How fast
will his shadow move along the wall when the rays of the sun are
at right angles to the diameter ?
140. How fast is the shadow in the preceding problem moving if
the sun's rays make an angle a with the diameter ?
141. A man walks across the diameter of a circular courtyard at
a uniform rate. A lamp, at one extremity of the diameter perpen-
dicular to the one on which he walks, throws his shadow on the walL
Required the velocity of the shadow along the wall.
PROBLEMS 219
*
142. A ladder b feet long leans against a side of a house. Its foot
is drawn away in the horizontal direction at the rate of a feet per
second. Find the path described by the center of the ladder and the
"velocity of the center in its path.
143. Find -j- and -r4 for the curve x = a(cos^ + ^sin^),
2/ = a (sin <l> — <l> cos ^).
144. Find -^ and -~ for the curve x = a cos8 <j>, y = a sin8 A.
oo5 aar
145. Find -^ and -r4 for the curve x = e* sin t, y = e* cos £.
aas aar
146. Prove that the logarithmic spiral r = e00 cuts all radius
vectors at a constant angle.
147. Prove that the angle between the normal and the radius
vector to any point of the lemniscate is twice the angle made by
the radius vector and the initial line.
148. Prove that the angle between the cardioid r = a(l — cos $)
and a radius vector is always half the angle between the radius
vector and the initial line.
149. If p is the perpendicular distance of a tangent from the pole,
prove that p =
^hW
150. If a straight, line drawn through the pole 0 perpendicular to
a radius vector OP meets the tangent in A and the normal in B,
show that OA = r*— and OB = — •
dr ad
151. Show that for any curve in polar coordinates the maximum
and the minimum values of r occur in general when the radius
vector is perpendicular to the tangent.
152. Sketch the curve r = 2 -f sin 3 0, and find the angle at which
it meets the circle r = 2.
Q
153. Sketch the curve ra= a2 sin — > and determine the angle at
which it intersects the initial line.
154. Sketch the curves r2 = a2 sin 2 0 and r2 = a2 cos 2 0, and
show that they intersect at right angles.
220 TRANSCENDENTAL FUNCTIONS
155. If a particle traverses the cardioid r = a(l — cos0) so that
$ makes uniformly two revolutions a second, find the rate at which
r changes, and the velocity of the particle in its path: (1) when
$ = ^ ; (2) when $ = tt.
156. Find the velocity of a point moving in a limacon
r = a cos $ + b
when $ changes uniformly.
157. When a point moves along the curve r = 4 sin8 — at a uniform
rate of 2 units per second, find the rates at which $ and r are
changing : (1) when 0 = - ; (2) when 0 = 7r.
158. Find the radius of curvature of the curve ar + y ■ = a*.
159. Find the radius of curvature of the catenary y = ^(e° + e °)
_ X X
a
160. Show that the catenary y = -(ea + e °) and the parabola
1 Z
y — a + ^— #2 have the same slope and the same curvature at their
common point.
161. Find the radius of curvature of the curve y2 = ^ (x — 2)8
at the point for which x = 3.
162. Find the radius of curvature of the cycloid
y = a cos~ 1 + V2aa; — x2
at a point for which x = - •
163. Find the radius of curvature of the curve y = e~2xsm3x
at the origin.
164. Find the least radius of curvature of the curve y = logos.
165. Find the points of greatest and of least curvature of the
sine curve y = sin x.
166. Show that the curvature of the parabola y = ax2 + bx + c
is a maximum at the vertex.
167. Show that the product of the radii of curvature of the curve
X
y = ae ° at the two points for which x = ± a is a2(e + e""1)8.
PROBLEMS 221
168. Find the radius of curvature of the four-cusped hypocycloid
x = a cOs8<£, y == a sin8^.
169. By use of the parametric equations of the ellipse find the
points where the radius of curvature is a maximum or a minimum,
and the values of these radii.
170. Find the radius of curvature of r = a (2 cos 0 — 1).
17 1 . Find the radius of curvature of the lemniscate r2 = 2 a2 cos 2 $.
172. Find the greatest and the least values of the radius of
curvature of the curve r = a sin8 - •
o
9
173. If the angle between the straight line drawn from the origin
perpendicular to any tangent to a curve and the radius vector to
the point of contact of the tangent is either a maximum or a mini-
r2
mum, prove that p = — , where p is the length of the perpendicular.
CHAPTER XII
INTEGRATION
109. Introduction. In § 80 the process of integration was
denned as the determination of a function when its derivative or
its differential is known, and was denoted by the symbol ( ;
that is, if f(x)dx = dF(z),
then f/O) dx = F(v)- C1)
The expression f(x) dx is said to be under the sign of inte-
gration, f(x) is called the integrand, and F(x) is called the
integral of f(x)dx; sometimes F(x) is called the indefinite
integral, to distinguish it from the definite integral denned
in § 81.
The determination of the indefinite integral is important in
a wide range of problems, and for that reason we shall now
deduce formulas of integration.
We ought to note first, however, that a more general form
of (1) is p
Jf(z)dz = F(x)+C, (2)
where C is the constant of integration (§ 80). In each of the
formulas we shall derive, C will be omitted, since it is inde-
pendent of the form of the integrand, but it must be added
in all the indefinite integrals determined by means of them.
110. Fundamental formulas. The two formulas
| cdu = c I du (1)
and / (du + dv + dw + • • •) = t du+ I dv + I dw + • • • (2)
are of fundamental importance, one or both of them being used
222
INTEGRAL OF U* 223
in the course of almost every integration. Stated in words
they are as follows:
(1) A constant factor may be changed from one side of the sign
of integration to the other.
(2) The integral of the mm of a finite number of functions is
the sum of the integrals of the separate functions.
To prove (1), we note that since cdu = d(cu)9 it follows that
/.*-/*(«)- «-./*u
In like manner, to prove (2), since
du + dv + dw+ ••• = rf(w + v + ^4- • • •)»
we have
| (du + dv + dw + • • •) = I d(u + v + w + • • •)
= u + v + w+ • • •
= I du+ I dv+ I dw + • • • .
The application of these formulas is illustrated in the follow-
ing articles.
111. Integral of n". Since for all values of m except m = 0,
d(y") = mum~1duy
or 4-)=wm"14
/u!*
um~1du = — •
m
Placing m = n + 1, we have
/
un+1
undu = z- (1)
n+1 v J
for all values of n except n = — 1.
In the case n = — 1, the expression under the sign of inte-
du
gration in (1) becomes — » which is recognized as dQogu).
u
Therefore / — = log u. (2)
J «
224 INTEGRATION
In applying these formulas the problem is to choose for u
some function of x which will bring the given integral, if pos-
sible, under one of the formulas. The form of the integrand
often suggests the function of x which should be chosen for u.
Ex. 1. Find the value of J (ax2 + bx -\ \- -^\dx.
Applying (2), § 110, and then (1), § 110, we have
f lax2 + bx + - 4- -jicfr = f ax2dx + f bxdx + I -dx + J —zdx
= a f x2dx + b I xdx + c J \- e jx-2dx.
The first, the second, and the fourth of these integrals may be evaluated
by formula (1) and the third by formula (2), where u = x, the results being
11 e
respectively - ax8, - bx2, > and c log x.
o & x
Therefore
1 1 ax2 + bx -\ H -7 W* = :r as8 + - bx2 + c log x V C.
J \ x x2/ 3 2 x
Ex. 2. Find the value of f(x2+ 2) xdx.
If the factors of the integrand are multiplied together, we have
f(x2+2)xdx=f(x*+2x)dx,
which may be evaluated by the same method as that used in Ex. 1, the
result being J x4 + x2 + C.
Or we may let x2 + 2 = u, whence 2 xdx = du, so that xdx = J du. Hence
f(x2+2)xdx=fiudu=ifudu
2 2
Instead of actually writing out the integral in terms of u, we may note
that xdx = %d(x2+ 2) and proceed as follows:
f(x2+ 2)xdx =f(z*+ 2)±d(x2 + 2)
= lf(x2+2)d(x2+2)
= t(*2+2)2+C.
Comparing the two values of the integral found by the two methods of
integration, we see that they differ only by the constant unity, which may
be made a part of the constant of integration.
INTEGRAL OF U* 225
[. 3. Find the value of ( (ax2 + 2 bx)z(ax + b) dx.
Let ax2 + 2 bx = u. Then (2 ax + 2 b) dx = du, so that (ax + b)dx = % du.
Hence f (ax* + 2 for)8 (ax + b)dx= f $u*du
= i (ax2 + 2 fcr)4 + C.
Or the last part of the work may be arranged as follows :
f (ax2 + 2 bx)*(ax + b)dx= C (ax2 + 2 &*)*}<* (oz» + 2 6x)
= i /"(a*2 + 2 6x)8rf (a*2 + 2 6x)
= J (a*2 + 2 6x)4 + C.
Ex. 4. Find the value of f i^E+ii^.
J ax2 + 2bx
As in Ex. 3, let ax2 + 2bx = u. Then (2ax + 2b)dx = du, so that
(ax + b)dx = I du. A , , rt ,
Hence f4(a» + t)<fe = r^du = 2rdu
J ax2 + *%bx J u J u
= 2 log u + C
= 21og(ax2+2 6x)+C
= log (ax2 + 2 bx)2 + C,
or
f4(ax + &)ax^ r 2 d (ax2 + 2 fcr)
J ax2 + 2 te ~ J ax2 + 2 6x
_ f d (ax2 + 2 for)
" J ax2 + 2bx
= 21og(ax2 + 2ix) + C
= log (ax2 + 2 6x)2 + C.
Ex. 5. Find the value of C (e™ + b)2e°xdx.
Let c"* + b = w. Then e^adx = aw.
Hence f (<?«* + b)2e°*dx = Cu2 —
2du
= — I u
a J
o a
o a
or
f (e«* + b^e^dx = f - (««* + i)2*/^"* + 6)
a J
= i-(e~ + br+c.
226 INTEGKATION
Ex. 6. Find the value of f * —*
J tan (ax + 0) + c
Let tan (ax + 0) + c = u. Then sec2 (ax + b)adx = du.
jj r sec2 (ax + b) dx _ rldu
J tan (ax + 0) + c J a u
_ 1 rdu
= *logu + C
= - log[tan(oj; + b) + c] + C,
or
/8ec2(a3: + 6)rfar _ /*1 c?[tan(qjr + b) +
tan (ax + 6) + c J a tan (or + o) +
_ 1 /* (/[tan (ax + 6) + c]
a •/ tan (or + 6) + c
1
= -log [tan(ar + b) + c] + C.
The student is advised to use more and more the second,
method illustrated in the preceding problems as he acquires*
facility in integration.
112. Integrals of trigonometric functions. By rewriting the
formulas (§ 96) for the differentiation of the trigonometric func-
tions we derive the formulas
/-
cos udu = sin u, (1)
8mudu=— cost*, (2)
sec2udu = tan w, (3)
esc2 udu = — ctn u, (4)
sec u t&n. udu = sec u, (5)
esc u ctn udu = — esc w. (6)
/
TRIGONOMETRIC FUNCTIONS 227
In addition to the above are the four following formulas:
taxi udu = log sec u, (7)
ctn udu = log sin w, (8)
/ 8ecudu = log(aecu + taxiu) = logtan( — + -]> (9)
u
esc udu = log (esc u — ctn u) = log tan — • (10)
sin u
To derive (7) we note that tan u = , and that — sin udu
js x on. cosw
= a(cosw). Then
/tanwc?w = — / — ^ ^
J co%u
= — log cos u
= log sec u.
T vi r j. j r cos udu ,
In like manner, I ctn Maw = l — : = log sin w.
Direct proofs of (9) and (10) will not be given here. At
present they may be verified by differentiation. For example,
(9) is evidently true since
d log (sec u + tan u) = sec udu.
The second form of the integral may be found by making a
(7T u\
1 ~*~ o)*
Formula (10) may be treated in the same manner.
Ex. 1. Find the value of I cos (ax2 + bx) (2 ax + b) dx.
Let ax2 + 6x = u. Then (2 ax + b) dx = rfti.
Therefore f cos (ax2 + &r) (2 ax + 6) ax = f cos (ax2 + 6x) d (ax2 + 5x)
= sin (ax2 + far) + C.
Ex. 2. Find the value of f sec (e*3* + 6) tan (f* + 6) e^xdx.
Let e**** + 6 = u. Then eax*2 axdx = aV
Therefore J sec (e*** + &) tan (e** +b)e°^xdx
= ~- f sec (e"* + b) tan (<***+ &)</(e"* + 6)
= JL sec (e"1 + b)+C.
2a v y
228 INTEGRATION
•
It is often possible to integrate a trigonometric expression by
means of formulas (1) and (2) of § 111. This may happen when
the integrand can be expressed in terms of one of the elementary
trigonometric functions, the expression being multiplied by the
differential of that function. For instance, the expression to be
integrated may consist of a function of sin x multiplied by cos xdz,
or of a function of cos x multiplied by (— sin xdx), etc.
Ex. 3. Find the value of f Vsina; coa*xdx.
Since d(ainx) = coaxdx, we will separate out the factor coaxdx and
express the rest of the integrand in terms of sin x.
Thus Vsina; cos8 xdx = Vsina; (1 — sin2 x) (cos x dx) .
Now place sin x = u, and we have
j Vsina; cos8 a; da; = fn* (1 — u2) du
= f (u2— u*)du
= %ui-$ ut+C
= ^sin*a;(7- 3 sin2*) + C.
Ex. 4. Find the value of f sec6 2 arete.
Since rf(tan 2x) = 2*sec22 xdx, we separate out the factor sec2 2 xdx and
try to express the rest of the integrand in terms of tan 2 x.
Thus sec6 2 x dx = sec4 2 x (sec2 2 x dx)
= (1 + tan2 2 x)2(sec22 xdx)
= (1 + 2 tan22 x + tan42 x) (sec22 xdx).
Now place tan 2 x = u, and we have
fsec62 xdx = i f(l + 2 m2 + u4) du
= £tan 2 x + Jtan82 x + ^ tan62 x + C.
Ex. 5. Find the value of j ten.6 xdx.
Placing tan6 a; = tan8 a: tan2a; = tan8a; (sec2x — 1),
we have f ta,n6xdx = f tan8 a; sec2 a; da; — ft&n*xdx
= J tan4 a; — j tan8 xdx.
TRIGONOMETRIC FUNCTIONS . 229
Again, placing tan8x = tanx (sec2 a; — 1),
we have / tan8xete = ftanxsec2xcte — / tanxete
= £ tan2x + log cos x + C
Hence, by substitution,
/ tan5 a: dx = \ tan4x — J tan2x — log cos x + C.
When the above method fails, the integral can often be
brought under one or more of the fundamental formulas by a
trigonometric transformation.
Sz. 6. Find the value of f cos2xcte.
Since cos2 x = J (1 + cos 2 x),
we have J cos2 arete = J I (1 + cos 2 x) dx
- $fdx + £ f cos 2 x cZ (2 x)
= £x + J sin 2 a; + C.
Sz. 7. Find the value of fsin2xcos4xcte.
Placing sin2 a: cos4 a: = (sin x cos x)2cos2x, .
we have sin2 a: cos4 a: = J sin2 2 x (1 + cos 2 a:).
Therefore jsia2x cos4 a: ete = £ / sin2 2 x ete + £ / sin2 2 x cos 2 a: ete.
Using the method of Ex. 6, we have
/sin2 2 x dx = £ f (1 — cos 4 a:) ete
= J x — \ sin 4 x.
Writing sin2 2 x cos 2 x ete = sin2 2 a: (cos 2 x dx)
and placing sin 2 x = u, we have
f sin2 2 x cos 2 a? ete = W u2e?u
= t"8
= £ sin8 2 x.
Combining these results, we have, finally,
f sin2x cos4xcte = ^ x + ^¥ sin8 2 x — ^ sin4 x + C.
AC
230 INTEGRATION
Ex. 8. Find the value of f Vl + cosxdx.
Since cos x = 2 cos2 - — 1,
Vl + cos x = V2 cos - •
Therefore \ Vl + cos x dx = j V2 cos - dx
= 2 V2sin|.
113. Integrals leading to inverse trigonometric functions. From
the formulas (§ 97) for the differentiation of the inverse trig-
onometric functions we derive the following corresponding
formulas of integration:
du . , ,
/
/;
Vl3^
du
T+v?
du
f
= sin"1t( or — cos~xw,
= tan_1wor — ctn_1w,
= sec-1w or — csc"1^.
wVw2 — 1
These formulas are much more serviceable, however, if u is
u
replaced by - (a > 0). Making this substitution and evident
reductions, we have as our required formulas
/
du , u ,+ N
sin-1-, (1)
Va2 - u* a
f:
du 1 _xu .0
= -tan *-, (2)
.2
f,
a* + u* a a
du 1 _xu /Qx
= -sec 1-« (o)
Uy/u2-a* « «
Only one of the possible values has been given for each
integral, as that single value is sufficient for all work.
Referring to 1, § 97, we see that sin"1 - must be taken in the
a
first or the fourth quadrant; if, however, it is necessary to
INVERSE TRIGONOMETRIC FUNCTIONS 231
have sin"1- in the second or the third quadrant, the minus
a
sign must be prefixed. In like manner, in (3), sec"1 - must be
a
taken in the first or the third quadrant or else its sign must
be changed.
/dx
_______
V9-4*2
Letting 2 x = u, we have du = 2 dx, and
r dx \ r d(2x) 1 . .2i , «
I , = - I v / — = - sin-1 -r- + C
J V9 - 4 a:2 2 J V9 - (2 x)2 2 3
— •
xVSx2-!
If we let v3 x = u, then */u = V3 <fc, and we may write
f dx _ r d(Vsx~) =l8ec_1V3ar | c
•>*V3**-4 J V3xV(V3^)2-4 2 2
V4 x - x2
Since V4ar-xa = V4 - (x - 2)2,
W« have f , dx = f , dx = f /(*-*)
J V4X-X2 J V4 - (a: - 2)2 J V4 - (x - 2)2
= 8in-i£r_2 + c.
Sz. 4. Find the value of f -— - — . .
«/2x2+3x + 5
To avoid fractions and radicals, we place
dx Sdx 0 4 dx
2ar2+3x + 5 16x2+24x + 40 (4x + 3)2+31
Therefore
/dx __ 2 r 4dx _ 2 /» r/(4s + 3)
2x3+3ar + 5~ J (4x + 3)2+ 31 ~ J(4x + 3)2+31
tan
-ii£±i + C.
V31 V31
The methods used in Exs. 3 and 4 are often of value in dealing with
functions involving ax2 + bx + c.
232 INTEGRATION
Ex. 5. Find the value of f (**+x)dx .
J 5 + 4a:4
Separating the integrand into two fractions, that is,
x8 x
5 + 4 x4 5 + 4**
and using (2), § 110, we have
/(a:8 + x) dx _ r x9 dx r xdx
5 + 4a:4 ~V 5 + 4a:4 J 5 +4x4'
t> x C x*dx 1 rlQxsdx 1 , ,- , . Ax
But /5T4^=i6/5T4^ = 16l0g(5 + 4x4>'
, /• xrfa: 1 f 4a:da; 1 , ,2 a:2
and I = - I = tan-1 •
J 5 + 4a:4 4 J 5+ (2a?)2 4V5 V5
Therefore
/(xz+ x)dx 1 , /K , A ,N , 1 , - 2 a:2 ~
5 + 4a:4 16 &v J 4V5 V6
Ex. 6. f ^-^L- = [tan-1*]^ = tan-* V3 - tan"1 (- 1).
J —1 1 + x*
There is here a certain ambiguity, since tan_1\/& and tan_1(— 1) have
each an infinite number of values. If, however, we remember that the graph
of tan-1x is composed of an infinite number of distinct parts, or branches, the
ambiguity is removed by taking the values of tan-1 VE and tan_1(— 1) from
Jr»6 dx
= tan-1& —
a 1 + a?
tan-1 a and select any value of tan-1 a, then if b = a, tan-1 b must be taken
equal to tan-1 a, since the value of the integral is then zero. As b varies
from equality with a to its final value, tan-1 b will vary from tan-1 a to the
nearest value of tan-1 b.
The simplest way to choose the proper values of tan-1 b and tan-1 a is
to take them both between — - and - . Then we have
2 2
/V? dx _ 7T __ /_ ir\ lir
-il + a?~3 \ 4/12'
a
Ex. 7. f2 ,dx =r«iTi-1-12 = sin-U - sin-10.
Jo Va2-*2 L a Jo
The ambiguity in the values of sin-1 \ and sin_10 is removed by notic-
x
ing that sin-1- must lie in the fourth or the first quadrant and that the
a
two values must be so chosen that one comes out of the other by continuous
change. The simplest way to accomplish this is to take both sin-1 \ and
IT TV
sin-10 between — - and -.
Then f 2 / « Q = ft""0 = ft*
Jo Va* — x2 6 «
LOGARITHMIC FUNCTIONS 233
114. Closely resembling formulas (1) and (2) of the last
article in the form of the integrand are the following formulas :
du
f
Vwa + a2
du
= log(M+V«* + aa), (1)
% = ]og(u+T&=<*), (2)
vr — a2
, C du 1 , w — a 1 , a — u XON
and 1-5 5 = tt— log or -r— log • (3)
J u2-a2 2 a Bu + a 2 a 6 a + u w
To derive (1) we place u = a tan <f>. Then du = a 8ec2<f>d<f>y and
vV+ a2 = a sec $. Therefore
/, = I sec <f>c?<f>
= log (sec <f> + tan £) (by (9), § 112)
= log(-
+vV+ a2>
= log (w 4-^7+a2) — log a.
But log a is a constant and may accordingly be omitted from
the formula of integration. If retained, it would affect the
constant of integration only.
To derive (2) we place u = a sec <f> and proceed as in the
derivation of (1).
Formula (3) is derived by means of the fact that the fraction
— - may be separated into two fractions, the denominators
w — a
of which are respectively u — a and u + a; that is,
u2 — a2 2 a\u — a u + a/
Then f_*L_ * r/JL-^A,
J u — a 2aj \u — a u + a J
— — { C du r du \
2a\J u — a J u + a)
= 2~a P0g (w *" a) ~ loS O + a)]
1 , u — a
= 7T- l0^
2a *u + a
234 INTEGRATION
The second form of (2) is derived by noting that
/du C ~ du , N
u-a J a-u BK J
The two results differ only by a constant, for
= -1 ;
a + u u + a
and hence log = log(— 1) 4- log ,
a + u u-\- a
and log(— 1) is a constant complex quantity which can be ex-
pressed in terms of V— 1.
/dx
— -«
V3x2 + 4x
To avoid fractions we multiply both numerator and denominator
by V3.
dx V3<fo ->/Edx
Then V3x2 + 4x " V9 x2 + 12 x "~ V(3x + 2)2-4
Letting 3 a: + 2 = u, we have du = 3 tfx, and
/<fa __ 1 /* 3tfx
V3xa + 4x ~ V3 ^ V(3x + 2)2=^4
= -i= log (3 x + 2 + V(3 x + 2)2 - 4 ) + C
V3
= -i=log (3 x + 2 + V9*2 + 12 s) + C.
V3
Ex
. 2. Find the value of j-
dx
2 x2 + x - 15
Multiplying the numerator and the denominator by 8, we have
dx ~ r 4rfx
J2x* + x-15 2J
(4 x + l)2 - (ll)2
_1 (4x-H)-ll
ll10g(4x + l) + ll+C*
This may be reduced to — - log + C, or —- log rrlog2 + C,
J 11 52x + 6 11 5x + 3 11 6
and the term — ^ log 2, being independent of xf may be omitted, as it will
only affect the value of the constant of integration.
Ex. 3. Find the value of f^-^- *-^- •
J 2 x2 + x - 15
If 2 x2 + x - 15 = u, du = (4 x + 1) dx.
Now 3 x + 4 may be written as | (4 x + 1) + *£■.
LOGARITHMIC FUNCTIONS 235
Therefore f <3* + 4>* = f M** + V> + ffl *
J2i2 + i-15 J 2x2 + ar-15
_3 /» (4a;-H)<fa 13 /* cfa
""4 ^ 2x2 + a;-15 4 J 2a:2 + a: -15*
The first integral is f log (2 a:2 + x - 15), by (2), § 111, and the last
13 2 x 5
1«gral is of the form solved in Ex. 2 and is — log — •
Hence the complete integral is
|log(2*2 + *-15) + glog^ + C.
Bx.4. Find the value of f(2x + b)dx.
J V3x2 + 4x
The value of this integral may be made to depend upon that of Ex. 1
the same way that the solution of Ex. 3 was made to depend upon the
lution of Ex. 2. For let 3 x2 + 4 x = u ; then du = (6 x + 4) dx.
Now 2 x + 5 = £(6 x + 4) + ty.
Therefore f <2* + 5>rf* = f [W» + *) + W»
J V3a?+4x J V3 a:2 + 4 a:
= g J (3 a;2 + 4x)-*[(6x + 4)tfx] + yj
dx
3J V3 x2 + 4 x
The first integral is § V3 x2 + 4 x, by (1), § 111, and the second inte-
gral is - log (3 x + 2 + V9 x2 + 12 x), by Ex. 1. Hence the complete
3 V3
integral is
| V3 a? + 4 x + -^7= log (3 * + 2 + V9 x2 + 12 x) + C.
3 3V3
Ex. 5. Find the value of ( sec xdx.
/, r dx r cos xdx
sec xdx = / = i - —
«/ cos a: t/ cosax
= _ /*rf(sinx) _ _ 1 1-sinx -
Jsin2x-1 2 gl + sinx
1 . 1 + sin x n
= 9logi ^T*+ C
1 1 — sin x
1. (1 + sin a:)2 , n
2 1 — sinzx
1 , /l + sin x\2 , ~
= log (sec x + tan x) + C.
236 INTEGRATION
/dx
1 + 2 cos x
As in Ex. 8, § 112, we place cos x = 2 cos2- — 1.
Then 1 + 2coss = 4cos2^ — 1,
dx r dx
and / - — = /
J 1 + 2 cos x J
4cos2|-l
2
x
Multiplying both numerator and denominator by sec2 - , we have
2
„„ . dx
2 / 2
sec2 - dx f sec2 - dx
4 — sec2 -
it
sec2 - dx
3 -tan2!
2
Now let tan ~ = *• Then sec2 -dx = 2 dz, and the integral assumes
the form
J3-*2 2V3 z +V3
tan^ + V3
/dz 1 2
= — — log -^— — + C.
l + 2cos* V3 tan?-V3
115. Integrals of exponential functions. The formulas
Ceudu = eu (1)
and fau du = - au (2)
are derived immediately from the corresponding formulas of
differentiation. The proof is left to the student.
116. Collected formulas.
un+1
undu = -> (1)
— = log u, (2)
u
FORMULAS 237
/ cos u du == sin u, (3)
/ sin u du = — cos w, (4)
I sec2u du = tanti, (5)
i cscaw rfw = — ctn w, (6)
I sec w tan udu = sec w, (7)
/ esc u ctn m^ = - esc w, (8)
/ tan m& = log sec % (9)
I ctn udu = log sin w, (10)
I secwc?w = log (sec u + tan w) = log tan (— + ~), (H)
I esc udu = log (esc u — ctn w) = log tan - , (12)
/cfa . _1u
V^7>=am -f (18)
r-^ = !tan-^, (14)
/^ , = - 8ec "> (15)
uwu2 — or a a
C-7== = log (« + V? + ^), (16)
r.7±==iog(M+v^^), (17)
/c?w 1 , u — a 1 , a — u _, ON
^^=2-al0S^Ta °r li^T+V (18)
fe"du = eu, (19)
faudu = - a". (20)
J log a
238 INTEGRATION
117. Integration by substitution. In order to evaluate a given
integral it is necessary to reduce it to one of the foregoing
standard forms. A very important method by which this may
be done is that of the substitution of a new variable. In fact,
the work thus far has been of this nature, in that by inspection
we have taken some function of x as u.
In many cases where the substitution is not so obvious as in
the previous examples, it is still possible by the proper choice
of a new variable to reduce the integral to a known form.
The choice of the new variable depends largely upon the skill
and the experience of the worker, and no rules can be given
to cover all cases. We shall, however, suggest a few substi-
tutions which it is desirable to try in the cases defined.
I. Integrand involving fractional powers of a + bx. The substi-
tution of a power of z for a + bx will rationalize the expression.
/x dx
-•
(1 + 2 x)i
Here we let 1 + 2 x = z8 ; then x = J (z8 — 1) and dx = § z*dz.
Therefore f — ^dx = % f (*7 - 2 2* + 2) dz
J (1 + 2*)* 8 J
= 3?ff *2(5 26 - 16 2» + 20) + C.
Replacing 2 by its value (1 + 2 xy and simplifying, we have
C_J^L_ 8 a + 2*)i(0 -12* + 20 *») + £.
J (i + 2xy S20
II. Integrand involving fractional powers of a-\- bx*. The sub-
stitution of some power of z for a + bx* may rationalize the
expression.
/Vx2 + a2
dx.
We may write the integral in the form
VxM- a2
x2
and place x2 + a2 = z2. Then xdx = zdz, and the integral becomes
p^e*)
'■^-/(•♦^-•♦i^**
SUBSTITUTION 239
Replacing z by its value in terms of x, we have
J cte = Vx2 + a2 + - log t + C.
Sz. 3. Find the value of fa* (1 + 2 xrfdx.
We may write the integral in the form
fx*(l + 2x*)l(x2dx),
d place 1 + 2 x9 = z2. Then a^cte = \zdz, and the new integral in z is
jy,(^-22)rf2=gvZ8^22_5)+c
Replacing z by its value, we have
fx6 (1 + 2 x*)Ux = ^ (1 + 2 a*)* (3 x8 - 1) + C.
Sz. 4.
Find the value of f(* + 2) ~ (* + 2) *<fa.
J (x + 2)i + 1
Here we assume a; + 2 = z4. Then a: = z4 — 2, and c/a; = 4 aft/z. On substi-
tution the integral becomes
4 C?JUtdz = ± f(z4-2z» + 2z2-2z + 2 ^rW
«/ z + 1 J \ z + 1/
' =4[iz6- iz4 + fz8-z2 + 2z-21og(z + l)] + C.
Replacing z by its value (x + 2)*, we have
f{^2)i-^x^2)i^ii_2Si^i
U (s + 2)i + l 5 i 8 i
+ S(x + 2)* - 8 log[(x + 2)* + 1] + C.
III. Integrand involving \V-V. Let a; = a sin 2.
Sz. 5. Find the value of fVa2 — x2dx.
Let ar = a sin z. Then dx — a cos z </z and V a2 — x2 = a cos z.
Therefore f-Va?-x*dx = a2 f cos2 zdz = £ a2 f (1 + cos 2 z)rfz
= £a2(z + £sin2z) + C.
x X I
But z = sin-1 - , and sin 2 z = 2 sin z cos z = 2 — Va2 — x2.
a or
Finally, by substitution, we have
fVa2 -x*dx = i (x Va2 - x2 + a2 sin"1 -\ + C.
240 INTEGRATION
IV. Integrand involving y/x2 + a2. Let x = a tan z.
/dx
(x2 + a2)*
Let x = a tan 2. Then dx = a sec2 2^2 and Var2 + a2 = a sec 2.
Therefore / = — 1 — — = — / cos zefe = — sin z + C.
(a:2 + a2)$ a sec 2 a • a
x x
But tan z = - > whence sin 2 = —-== > so that, by substitution,
a Va:2 + a2
*^(x2 + a2)i 'a2Vx2 + a2
If we try to find the value of TVa:2 + a2dx by the substitution x = a tan 2,
we meet the integral a2 f sec8 2^2, which is not readily found. Accordingly
for a better method see Ex. 6, § 119.
V. Integrand involving Vx2— a2. Let x = a sec z.
Ex. 7. Find the value of f xsVx2 — a2dx.
Let x = a sec 2. Then dx = a sec 2 tan 2rfz, and Va:2 — a2 = a tan 2.
Therefore I x*Vx2 — a2dx = a5 I tan2 2 sec42c/2
= a6 f (tan2 2 + tan4 2) sec22rf2
= a6(£ tan82 + £ tan5 2) + C.
a? "v ar2 -— o2
But sec 2 = - , whence tan 2 = > so that, by substitution, we have
a a
fx*Vx2-a2dx = TV V(x2-a2)3(2 a2 + 3 a:2) + C.
We might have written this integral in the form fx2Vx* — a2(xdx)
and let z2 — x2 — a2.
VI. Integrand of the form , =• Let
+ (Ax + B^vax* + DX + C
Ax + B = —
z
/dx
•
(2x + l)v5a^ + 8x + 3
Let2x + 1 =-• Then x = ^(- - 1) , dx =- —-dz, and V&x* + &x + S
2 2 \z I 2 2s
= _L V22 + 62 + 5.
22
SUBSTITUTION 241
Therefore
dx __ r dz __ __ /• dz
(2x + 1) V5s- + 8x + 3 J Vz2 + 6 z + 5 J V(a + 3)2 - 4
= - log(z + 3 + V*2 + 6 z + 5) + C.
But 2 = 9 and hence
2s + l
6 a; + 4 + 2 V5 a:2 + 8 x + 3
— log(*+ 3 +V*2+6z+ 5) = - log
2ar + l
1 2x + l
= log = - log 2.
3x + 2 + V5a:2 + 8 a: + 3
Therefore
/dx , 2 x + 1 ^
= log = + C,
(2 a: + 1) V5 x2 + 8 x + 3 3ar + 2+V5z2 + 8a; + 3
log 2 having been made a part of the constant of integration.
118. The evaluation of the definite integral I f(x)dx may
V*e performed in two. ways, if the value of the indefinite integral
is found by substitution.
One method is to find the indefinite integral as in the pre-
vious article and then substitute the limits.
Ex. 1. Find C^Ja2 - x2dx.
Jo
By Ex. 5, § 117,
fVa2-x2dx = Ux Va2 - x2 + a* sin-1-) + C.
Therefore C^/a2-x2dx - [^(x Va2 - x2 + a* sin-1-)]"
= -la Va2 — a2 + a2 sin-1-)
2\ a;
- i/o Va2-0 + a* sin-1-)
_ ira2
A better method is to replace the limits of / f(x)dx by the
corresponding values of the variable substituted. To see this, sup-
pose that in I f(x)dx the variable x is replaced by a function
of a new variable z, such that when x varies continuously from
242 INTEGRATION
a to 5, z varies continuously from zQ to zx. Let the work of
finding the indefinite integral be indicated as follows:
Cf(x) dx = C<f> 0) dz = * 0) = F(z),
where F(x) is obtained by replacing z in 4> (2) by its value in
terms of x. Then
But F(b)-F(a) = C f(x)dx,
and <I> (Zj) — <I> (z0) = ( l(f> (z) dz.
f(x)dx = I <f>(z)dz.
Applying this method to the example just solved, we have by Ex. 5, § 117,
' I Va2 — x2dx = a2 f cos2zdz
= \a2(z + i sin 2 2) + C,
where a: = a sin z. When a: = 0, z = 0, and when x = a, z = — , so that 2
varies from 0 to — as x varies from 0 to a.
2
Therefore f" Va2 - a?dx = a2 f 2 cos* zdz
Jo Jo
= [i«2(z+ism2*)J
ira2
In making the substitution care should be taken that to
each value of x between a and b corresponds one and only
one value of z between z0 and z^ and conversely. Failure to
do this may lead to error.
vr
Ex. 2. Consider f 2 cos <£ d<f>, which by direct integration is equal to 2.
«/_ir
Let us place cos <£ = x> whence <f> = cos-1 a: and d<^ = /" f where
Vl-33
the sign depends upon the quadrant in which <£ is found. We cannot,
INTEGRATION BY PARTS 243
eref ore, make this substitution in f 2 cos <f> d<f>, since <f> lies in two differ-
ent quadrants; but we may write -§
•n n it
f * cos <f>d<t>= f cos <£ d<l> + f 2 cos <£ d<l>,
<l in the first of the integrals on the right-hand side of this equation
dx ~~~ doc
I>l«*ce <f> = cos""1 a:, d<f> = — = , and in the second <£ = cos""1 x, d<f> =
en Vl - x2 Vl - x2
fhos +d<t>= f 1_££l= _ r °_£|l= = 2 fxdx^ = 2
J-Z JoVl-x2 JlVl-x2 Jo Vl - X2
2
119. Integration by parts. Another method of importance in
e reduction of a given integral to a known type is that of
^^^tegration by parts, the formula for which is derived from the
formula for the differential of a product,
d(uv) = udv -f vdu.
From this formula we derive directly that
uv= I udv + I vdu,
^iiich is usually written in the form
/ udv = uv — I vdu.
In the use of this formula the aim is evidently to make the
original integration depend upon the evaluation of a simpler
integral.
Ex. 1. Find the value of f x&dx.
If we let x = u and e*dx = dv, we have du = dx and t? = e*.
Substituting in our formula, we have
jxePdx = xe* — f e*dx
= xe*- e* + C
= (x-l)e* + C.
It is evident that in selecting the expression for dv it is desirable, if
possible, to choose an expression that is easily integrated.
244 INTEGRATION
Ex. 2. Find the value of f sin-1 xdx.
Here we may let sin-1 a: = u and dx = dv, whence du = and v = i
Substituting in our formula, we have VI — x
f 8m"1xdx = x sin""1* — I
xdx
VT^x*
= x sin-1* + Vl - x* + C,
the last integral being evaluated by (1), § 116.
Ex. 3. Find the value of f x cos2 xdx.
Since cos2x = J(l + cos 2x), we have
f x co8*xdx = — / (x + x cos 2x)dx = — + - f x cos 2 xdx.
Letting x = u and cos 2 xdx = dv, we have du = dx and v = J sin 2 x.
Therefore | x cos 2 xdx = -sin2x — - |sin2x</x
J 2 2 J
= - sin 2 x + - cos 2 x + C.
Therefore / x cos2xdx = — + -(- sin 2 x + - cos 2 x ) + C
J 4 2 \2 4 /
= J(2 x2 + 2 x sin 2 x + cos 2 x) + C.
Sometimes an integral may be evaluated by successive inte
gration by parts.
Ex. 4. Find the value of f x^rfx.
Here we will let xa = u and e*dx = dv. Then du = 2 xdx and v = <F.
«
Therefore f x%e*dx = xV* — 2 IxePdx.
The integral / xe^dx may be evaluated by integration by parts (see Ex. 1}
so that finally
fx^dx = iV - 2(x - l)e* +C = ^(x* - 2 x + 2) + C.
Ex. 5. Find the value of \elax8mbxdx.
Letting sin bx = u and ePxdx = dv, we have
/e"* sin &r dx = - e0* sin bx \ e** cos &r dx.
a a J
INTEGRATION BY PARTS 245
In the integral f e"* cos bxdx we let cos bx = u and e^dx = dv, and have
/e™0 cos bxdx — -€Fc cos bx + - / e!** sin toox.
a a «/
■Substituting this value above, we have
/e** sin bxdx = - e"* sin &r (- e"* cos &r + - / e^ sin fccax ).
a a \a a •/ /
Now bringing to the left-hand member of the equation all the terms
c^^^ataining the integral, we have
/ b2\ r 1 b
1 1 + — J / e"* sin bxdx = - e** sin bx e"* cos bx,
a a2
ence
/„«,... eax (a sin bx — b cos &r)
e"* sin 6x dx = — - l •
a2 + ft2
x^x
. 6. Find the value of [Vx2 + a2dx.
Placing Vx2 + a2 = w and a*x = aV, whence dw = — ^== and v = x,
have ^x + a
fVx2 + a2tfx = x Vx2 + a2 - f t_^£^=. (1)
J J V*2 + a2
Since x2 = (x2 + a2) — a2, the second integral of (1) may be written as
'(x2 + a2) rtx 2 r dx
r{x* + a-4) ax _ 2 /• ax
•J Vx2 + a2 ^ Vx2 + a
2
dx
which equals / Vx2 + a2tf*x — a2 ( - J
J J Vx2 + a2
*
Evaluating this last integral and substituting in (1), we have
JVz2 + a2dx = x Vx2 + a2 - f Vx2 + a2c?x + a2 log (x +Vx2 + a2),
whence f Vx2 + a2a*x = J [x Vx2 + a2 + a2 log (x + Vx2 + a2)].
120. If the value of the indefinite integral I f(x)dx is found
by integration by parts, the value of the definite integral
I /0*0 d& may be found by substituting the limits a and 6, in
the usual manner, in the indefinite integral.
AC
246 INTEGRATION
Ex. Find the value of C 2 x2 sin xdx.
Jo
To find the value of the indefinite integral, let x2 = u and sin xdx = dv.
Then / a^sin xdx = — x2 cos x + 2 1 x cos xdx.
In j x cos xdx, let a; = w and cos xdx = </#.
Then / x cos xdx — x sin x — / sin xrfx
= x sin x + cos a:.
Finally, we have
f x2 sin xdx =— x2cosx + 2xsinx + 2cosx + C.
Hence C 2 x2 sin xtfx = — x2 cos x + 2 x sin x + 2 cos a; I
2
10
= ir-2.
The better method, however, is as follows: ' 6
lif(x)dx is denoted by -wrfv, the definite integral I f(x)dx
udv, where it is understood that a and b
a
are the values of the independent variable. Then
I udv = [uv~\ — 1 vdu.
J a %J a
To prove this, note that it follows at once from the equation
S*b f*b f%b r*b
[uv] = J d(uv) = 1 (udv + vdu) = I udv -\- I vdu.
xJ a %Ja %J a %J a
Applying this method to the problem just solved, we have
ir
C 2 x2 sin xdx = I — x2 cos ar + 2 f^x cos xdx
ir
= 2 f 2" x cos a: rfx
Jo
ir
= I 2 x sin x — 2 fa" sin xdx
= IT + I 2 cos X
■0
= ir-2.
2cosx I
PAETIAL FRACTIONS 247
121. Integration by partial fractions. A rational fraction is a
fraction in which both the numerator and the denominator are
polynomials. If the degree of the numerator is equal to, or
greater than, the degree of the denominator, we may, by actual
division, separate the fraction into an integral expression and a
fraction in which the degree of the numerator is less than the
degree of the denominator.
For example, by actual division,
?=I6 =2*~1 + ?=ie (1)
It is evident, then, that we need to study the integration of
only those fractions in which the degree of the numerator is
less than the degree of the denominator.
If the denominator of such a fraction is of the first degree
or the second degree, the integration may be performed by
formulas (2), (14), (18), § 116, as in Ex. 3, § 114.
If the denominator is of higher degree than the second, we can
separate the fraction into partial fractions the sum of which will
equal the given fraction.
For example,
g»+8s?-4g + 20 1 1 x-1
x4-16 ~~~x-2 x+2*x*+±' ^
as the reader can easily verify.
The three fractions on the right-hand side of (2) are the
partial fractions of the fraction on the left-hand side of (2).
It is to be noted that their denominators are the rational
factors of the denominator of the fraction of the left-hand
side of (2).
Substituting, in (1), we have
2 ^- :r4+ z8+ 3 ar2- 36 # + 36
• —— ^-^— ^_ — _^-_^_^_^_^__^^^-^_ ^_ ^^
x4-16
248 INTEGRATION
„ , 2^-:z4+ a*+ Zx*-Z§x + 36 ,
Hence I -. — -r-z ax
x — lb
= a;a-a;+log(a;-2)-log(a;+2) + ilog(a!2+4)-itan
z
2 ov y 2 2
= *»-* + log <--^ 2^ 2'
The separation of a fraction into partial fractions, as in (2),
is evidently a great aid in integration. We shall illustrate this
process in the following examples:
x2 + llx + 14 ,
ax.
Ex. 1. Find the value of f rtv „ _ —
J (x + 3)(x2-
(x + 3) (x2 - 4)
The factors of the denominator are x + 3, x — 2, and x + 2. We assume
s2 + 11 a; + 14 = A B C a
(x + 3)(*2-4) x + 3 x-2 x + 2' ^)
where A, B, and Care constants to be determined.
Clearing (1) of fractions by multiplying by (x + 3) (x2 — 4), we have
x2 + 11 x + 14 = 4 (x - 2)(a? + 2) +B(x + 3)(x + 2) + C(z + 3)(x - 2), (2)
or ar2+llx + 14=(.4+B + C)x2+(5JB + C)2: + (-4.4+6JB-6C). (3)
Since A, B, and C are to be determined so that the right-hand member
of (3) shall be identical with the left-hand member, the coefficients of like
powers of x on the two sides of the equation must be equal.
Therefore, equating the coefficients of like powers of x in (3), we obtain
the equations A A- B-\- C = \
5 B + C = 11,
-4A+GB-BC = 14,
whence we find A — —2,B = % C = 1.
Substituting these values in (1), we have
x2 + lla; + 14 2 2 1
(a: + 3)(a^-4) x + 3 x-2 x + 2
r r» + ll:E+14 _ r2dx />2dx r_dx
and J(* + 3)(*2-4) -~Jj+^ + Jj^ + J^T2
= - 21og(x + 3) + 21og(x- 2) + log(x + 2) + C
8 (x + 3)2
PARTIAL FRACTIONS 249
Ex. 2. Find the value of f — x dx.
The real factors of xs — 1 are x — 1 and x2 + x + 1. Hence we assume
4 x2 + a: + 1 _ ^ Bx + C
Clearing of fractions, we have
4 a:2 + x + 1 = 4 (x2 + a; + 1) + (Bx + C) (a; - 1)
= (A + £)z2 + (4 - £ + C)x + (4 - C). (2)
Equating coefficients of like powers of x in (2), we obtain the equations
A + B = 4,
,4 - £ + C = 1,
^ - C = 1,
ence 4 = 2, 5 = 2, C = 1.
Hence ix* + x + X = * + 2* + 1
a:8 - 1 a: — 1 a:2 + a; + 1
J xs -1 f* Jj-1 J ar2 + a; + l
= 2 log(z - 1) + log(a;2 + x + 1) + C
= log[(a; - 1)2(*2 + x + 1)] + C.
2x2dx
Ex. 3. Find the value „_ .
./(* + 2)«<x-2)
Here we assume
off—
2*2 _ A B C
(z + 2)2(x-2) (x + 2)3 1 + 2 a:-2 w
Clearing of fractions, we have
2 x2 = A (x - 2) + B(x2 - 4) + C(x + 2)2
= (J5 + C)a:2 + (A + 4 C)ar + (- 2 ^ - 4 5 + 4 C). (2)
Equating the coefficients of like powers of x in (2), we obtain the
^uations J5 + C = 2,
A + 4 C = 0,
- 2 .4 - 4 B + 4 C = 0,
Whence jl =- 2, 5 = }, C = £.
Hence — = - — + — L + _i_
(a; + 2)2(ar-2) (a: + 2)2 x + 2 a: - 2
rfa:
2
and f 2 ^^ — — f ^dx 4- f %dx (* jd:
J (x + 2)2(a: - 2) " J (a; + 2)2 * J x + 2 J x -
2 + | log(x + 2) + l log(x - 2) + C
x + 2 2 ov ' 2
-|— + log V(ar + 2)8(a;-2) + C.
250 INTEGRATION
Sx* + Sx-G
Sz. 4. Find the value of / — — - — — - dx.
J (x + 1) (x8 + 1)
Now (x + 1) (x8 + 1) = (x + l)2(x2 — x + 1), and we assume
3a^ + 3x-6 A t B t Cx+D /1x
+ Z-TT + ^ 7TT- V1)
(a: + 1) (x8 + 1) (ar + 1)2 x + 1 x^x+l
Clearing (1) of fractions, we have
3 Xs + 3 x - 6 = A (x2 - x + 1) + B (x8 + 1) + (C x + 2)) (x + l)a
= (J5 + C)x* + (v4 + 2 C + D)x2 + (- 4 + C + 2Z>)x
+ 04 + J3 + Z)). (2)
Equating coefficients of like powers of x in (2), we obtain the equations
B + C = 3,
^ + 2C + D = 0,
-^ + C + 2D = 3,
,4 + £ + !)=- 6,
whence v4 =- 4, B = 0, C = 3, 2) =- 2.
Substituting these values in (1), we have
3 x8 + 3 x - 6 -4 3 x - 2
(x + 1) (x8 + 1) (x + l)2 x2 - x + 1
^3x8+3x-6^ = f _-4<*x f3x-2
J (x+lXx8*!) J(x + 1)2 Jx2-x + l
4 + |log(^2- * + l)---L=tan-i2ar *+ C,
* + l 2 ov y V3 V3
the last integral being evaluated as in Ex. 3, § 114.
We notice in the solution of the above examples the follow-
ing points :
1. The denominator is factored into linear or quadratic factor*,
or integral powers of such factors.
2. As many partial fractions are assumed as there are factors in
the denominator.
3. Corresponding to any single linear factor, as ax + b, one
A
fraction of the form is assumed, and corresponding to the
ax + b a
square of any linear factor, as (ax + J)2, two fractions 775
H are assumed, the numerator over the square of the factor
ax + 0
being of the same type as that over the first power of the factor.
PARTIAL FRACTIONS 251
4. Corresponding to any single quadratic factor, as ax2 -\-bx+c,
one fraction of the form — - — = is assumed.
axr + bx + c
5. The numerators assumed are determined and the integration
of the partial fractions is completed.
If (ax + by in 3 is replaced by (ax + i)n, and the correspond-
ing n fractions are assumed to be
* , B , . , , P
(ax + by (ax + by-1 ax + b*
and if ax2 + bx -f c in 4 is replaced by (ax2 + bx + c)w, the cor-
responding n fractions assumed being
Ax + B Cx + D Px+Q
(ax* + bx + cy (aaf + bx + cy-1 "" ax*+bx + c*
the above becomes a working rule for the integration of all
rational fractions in which the degree of the numerator is less
than the degree of the denominator; but the proof of the pos-
sibility of assuming the partial fractions in the form noted
above is omitted.
To make the work of this article complete we must discuss
the integral / - — -^ — dx, where n is any integer greater
j i cintr —j~ ox — f— c j
than unity.
Since d (ax2 4- bx + c) = (2 ax + J) dx, we may, as in Ex. 3,
A Ab
§114, let Ax + B = -£-(2ax + b) + B-=-, and obtain the
la 2 a
equation
r Ax + B A rd^af+bx+c) / Ab\ C dx
J(ax2+bx+cy X~~2aJ (ax*+bx+cy \ 2a)J (atf+bx+cj'
Proceeding as in Ex. 2, § 114, we may put the last integral in
the form / — - — — , which may be reduced to the integral I — — -
J (u2+a2y J -J u2+a2
by successive applications of the formula
/du 1 I" u .g _qn r du 1
(u2+a2y~2(n-l)d2l(u2+a2y-1 ^ U }J (u2 + a2yl\
This is a special case of (4), § 122.
252 INTEGRATION
122. Reduction formulas. The methods of integration derived
in this chapter are sufficient for the solution of most of the
problems which occur in practice. If the reader should meet
any integrals which cannot be evaluated by these methods, he
should refer to a table of integrals, in which the integrals have
been either completely evaluated or expressed in terms of simpler
integrals. Some of this latter type of integrals, known as reduc-
tion formulas, have been tabulated below for convenience.
farta + bafydx
tf"-n+1(a+6<)p+1 (m-w+l)a r n, , , _, xix
= — ^— - — rfr 7s /w / sT-Xa + bxrydz, (1)
(np + m+X)b (np + m+l)bj v J v J
Caf* (a + ba
J _ar+1(a
barydx
+ ba?y t npa
-"(q + <y npa />(a + fa.)P-^ (2)
np + m + 1 np + m+lj v J v J
Car(a + bzfydx
(m-fl)a (m-fl)a J v y w
CaTfa + barydx
w(jt?+l)a rc<j>+l)a J v 7 w
/ sinm#cosw2;d#
sin^+^cos""1^ , n — 1 /* . m . „ 7 xCN
= 1 I sinma; cosw~3:r ax, (o)
m + n m + n J
I sinma;cosn#£fo
sinm+12;cosn+1:r , m + n + 2 C • m * + * ^ ^
= ; — — i— \ sinma; cosn + 2:r<fe, (6)
n + 1 n + 1 J v 7
J sinma;cosw:r<fc
sinm"1a;cosn+1a; , m-1 f . „ 3 „ , ,-N
— 1 I smm " 2 a; cos" a:. dfa^ (7)
m + n m + nj
I sinmxcosnxdx
sinm+1#cosw+1a; m + n + 2 /• . fl
= 3 ^ — f sinm+aa; cosn.r dx. (8)
w+1 m + 1 J v J
PROBLEMS 253
These formulas do not always hold. For example, (1) and
(2) fail if np + m +1 = 0, (3) fails if m + 1 = 0, etc. In these
cases, however, it is not necessary to use these formulas, as the
integration may be performed by elementary methods.
There are also integrals which cannot be expressed in terms
/dx
V(l-aO(l-*W)
cannot be so expressed; and, in fact, this integral defines a
function of # of an entirely new kind.
PROBLEMS
Find the values of the following integrals :
' 1. f(±x* + 3xi + ±x-3)dx. 12. \*^eSxdx.
2. f(^-x2 + i-^W 13. C(2 + 3x)Ux.
/x2 + Vx~z + 3J _ C cosset
7= OX. 15. I — :
■y/x J a + bsmx
x2dx
4.
6. ffe + ffifc • 17. f ?+*>*> ,
J v? J ^/l+3x + x*
^2 + ?<fdx. 18. J -^±
8
cos 2x
e2xdx r dx
)[log(*-a)J
r e2xdx r
9J*" + 2 19'J(x-a
/dx riog^8 7
xlogx2 J X
J (x — cos x)2 t/ W + a; \a — ay
254 INTEGRATION
23
Cjte. 39. fco8*(2x-l)dx.
24. / . ; . 40. / (sin- + cos£)cte.
25. / cos8a;sinajefoj. 41. / tan8-cftc
26. fsin8(2a;+l)cos(2a;+l)G*B. 42* fsec*(3x + 2)dx.
(sec ewe + tan ax) sec ozda;. 43. | tan — sec8 -5- cfce.
28. 1 (esc » etna; 4- esc xfdx. 44. | sec4 2 a; Vtan~2#dx.
29. I cosa 2 x sin8 2 xdx: 45. | esc6 - dx.
30. I sin* 3 x cos* 3 scg&c. 46. I ctn* (a; + 2)dx.
o, C • s^ ,w T, 22x 42x.
31. I sin8-efec. 47. I ctn2 — esc4 — dx.
/cos* 4 x /*
, -dx. 48. I tan4aajefcc.
V sin8 4 x J
/cos2x _ ^ Tctn8
dx. 49. I
cos a? # J csc<
OS!/.
ace
34. I sinxsin2ajefec. 50. I tan8 -"CI sec 7:^.
35. I (tan ax -f- ctn axfdx. 51. I sin2(3a; +l)cfcc.
36. f(sec2a; + tan 2 ar)2dx. 52. f cos2 (2 — 3 x) dx.
/esc8 3 a; — ctn8 3 a _ lpo /\ . 0 0 xa,
— — — efo. 53. I (sin 2 a; — cos2aj)2c?x.
esc 3 x — ctn 3 x J '
38. I (tan2rt — ctn2-Wa;. 54. J sin2 3 a; cos2 3 xdx.
PROBLEMS 255
66. fcos*|<&. 71. f(l-coB4x)ldx,
66. ( ain'x caa'xdx. 72. 1 — im=.
J J V25-9za
67. J Vcosa* + 1 sin 2 astir. 73. I - ■
M f/»J^_c<»2*U ,4. f_*.
,/ \ sin ;c cos*/ Ji Viz* — 9
"J oo» . + ,,,■ . "•J4«« + 4I+10'
60. J.rn^.mtxdx. (.*») "•/jSTj^+r
61. | cosaxcos£x<&E. (a=£i) 77. I .■ - -■■ ■
«.J»n<i>.+3>oo,<2,-S),fc ■"■fy/T+fx_^-
63. fain 3 sin 2 a; sin 3 zda:. 79. ( —f==~-
J J v9a> — a?
/•»».* 80> r — <&_
J see 2 x tan2* J (a; + 1) Var1 + 2 1
... f-4-. 84. r^
to. r . a . so. i%=s=
-if
INTEGRATION
r <ix
J V9*'-2'
r dx
J *x>-2
r dx
J V4z».+ 6z'
f dx
J 2x>-6x'
f dx
J *" + 3* + l'
r dx
J V3z»-2x + 3
f dx
J 21-41-!-'
r dx
J V2*a + 4z-7
/"(5x-3)</x
J a? -I- 6* + 12'
r(» + 5)Ar
Ja?+x-6'
f*(2r+10)<fa
J 2ft' + 6z+l'
/•(6» + 20)ifa
J 6*- + 7x-3'
r (' + 1)**
J 2x- + 6x + <>
f (x-2).fc
J 3*> + 2x + 3
r (x + 2)dx
J V3 + 2x-x»'
T (6-2x),fa
J V8-4r-4aJ
119
PROBLEMS 257
dx
. fe*+1x*dx. 135. f-^
. fe6+cxa6+cxdx. 136. J -
122
123
126.
127.
131
+ 3)J
3 + 4a8
121. I ^— -^flte. 137. I x*->/tf~^-x2dx.
/rfaj ^ r dx
-o S- 138« I B
e?x-2 J (^-l)*
eta
Ce2x+lJ r
126. i,^±!+ifc i-4i. r^4^^.
3?
x6dx
i. I — , dx. 1*41. I -
r **** . 142. f
r_**_. 143. f-^4-
J (^ + 3)* J (a^ + 9)*
128. r_-£^=. 144. f(aJ - a;2)*^.
J V(4 - a8)8 J
129. f *** • 146. fx(z +!.)*<&;.
J V(9 - x*)& J
130. Cxt(4:+xi)idx. 146. J (^ ~ 4) da,.
. f *! 147. f f* .
J x V4 + 9*2 ./ *4 v 2 - x*
132. f^^dx 148. r^±Didx.
J X ' J X
1M r dx %Atk c xSdx
133. I . 149. I .
J «8V9x2-4 J V1+4X3
258 INTEGRATION
/x cLx f*
1- 166. / xsin23xdx.
152. | = • 167. I tf'amxdx.
J (aj-2)V2aja-4aj~l J
153. | , X = • 168. ( efcosSxdx.
J (2 z- 1)^16^ -12s + 3 J
154. f: .^ = • 169. f V?Tldaj,
J (2x-3)V4x2-12x + 5 J
155. I . = • 170. / sec8 xdx.
J (x + 2)Va2 + 2» + 2 J
log accede. 171. I — —* — ytdx-
157. I ccwlogxd!a. 172. I = - — ! — dx.
a8 — 4 a
x 4-4
1^ + 4)
I ccwlogxd!a. 172. /
158. / tan" 1axdx. 173. \ f «"*"*,. dx,
J J x(x
159. J log(x + Va;2 + a2)ete. 174. j * ~ ,<&;.
. / zsin3£cfa;. 175. |
160
3<fc
161. j sec-1 2 xdx. 176. \ j-TTTz^-
x*(x+l)
s*-4
162. fasec-^scda;. 177. f f + \ dx.
J J *(* +
j4
+ 2)
163.
I x2e*xdx. 178. J-
efte
i)(^+i)
164. f 0*008 2 xdx. 179. f7 ?SjX^ on
J J (a;-l)(xJ + 2a;-8)
166. f (log xydx. 180. f^z^^dx.
(2*+ IV
PROBLEMS
181. f *?-2x + 3
J <*-X)(x+l)(x-3)dx- f^*l
182 r 4*»_3 194' / ~**-
J (x-2)(a? + 2x + 5)d*- J C0S2
183. f^tL^ 195. f^^dx
J x (x - 4)* **■ J sin«4 x ax-
196. C-4^=-
Va^ + o1
dx
190.
J (1+Z8)*
198. / ^
IM.fatVx^Ttfdx.
201.
202. ' dx
203.
H
sin^a; cos8aj
dx
sin8a;
204. f dx
J si
206.
cos°a;
sura cos x
dx
259
CHAPTER XIII
APPLICATIONS OF INTEGRATION
123. Element of a definite integral. In §§ 78 and 81, by means
of the area under a curve, we have defined the definite integral
by the equation
f(x) dx = Lim 2)/(*0 ^ CO
I
n = 00 ;
» = 0
and have shown that this limit may be evaluated by the formula
Cf(x) dx = F(b) - F(a)y (2)
where dF(x)=f(x)dx.
Since any function f(x) may be graphically represented by
the curve y =/(#), formulas (1) and (2) are perfectly general.
We shall proceed to give certain applications. The general
method of handling any one of the various problems proposed
is to analyze it into the limit of the sum of an infinite number
of terms of the form f(x) dx. The expression f(x) dx^ as well
as the concrete object it represents, is called the element of
the sum.
124. In finding the element of integration, it is often not
possible to express the terms of the sum (1), § 123, exactly
as /(#,) A#, the more exact expression being [/(#<) + «,] Aa^
where the quantities et. are not fully determined but are
known to approach zero as a limit as Ax approaches zero.
It is consequently of the highest importance to show that
i = n-l
Lim V €{Ax = 0, so that
"=» / = 0
» = *-! i=*-l
Lim X U(x^ + *] A* = Lim 2 /O.) A* = ff(x) dx.
*-* / = 0 w=ao ,-=o */a
260
ELEMENT OF A DEFINITE INTEGRAL 261
For that purpose, let 7 be a positive quantity which is equal to
the largest numerical value of any et. in the sum. Then
- 7 ^ * ■ ^ 7
and — 27A2; ^ 2etAr = 27Arc
But 27A2; = 72 Ax = 7 (b — a)
and Lim 27A2: = 0 since 7 approaches zero as Ax approaches zero.
n = oo
Hence Lim Se.Arr = 0.
n=oo
Hence the quantities et. which may appear in expressing the sum
do not affect the value of the integral and may be omitted.
Quantities such as Ax and et., which approach zero as a limit,
are called infinitesimals. Terms such as f(x) Ax, which are formed
by multiplying Ax by a finite quantity, not zero, are called infini-
tesimals of the same order as Ax. Quantities such as e{Ax, which
are the products of two infinitesimals approaching zero together,
are called infinitesimals of higher order than either infinitesimal.
The theorem above proved may be restated in the follow-
ing way:
In forming the element of integration infinitesimals of higher
order than f(x)Ax may be disregarded.
Ex. Consider the area under a curve (§ 78). We have obtained it, by
means of rectangles, as t=n-i
LimV/(a0A*. (1)
n=fl° i=0
Suppose that in place of the rectangles we used the trapezoids formed
by drawing the chords DPV PxP2i etc. (fig. 125). The area of one such
trapezoid is /(^)Ar + ^^
But \Ay is a quantity which approaches zero as a limit when Ax
approaches zero, and may be denoted by €»•• Hence, if we used the trape-
zoids, we should have for the required area
Lim V [/(*,) + t<]Ar. (2)
n-» f=0
We see then directly that in this example
i=n-l i=n-l
Lim V [/(*;) + €{] Ax = Lim V f(x{) Ax.
AC
262
APPLICATIONS OF INTEGRATION
125. Area of a plane curve in Cartesian coordinates. This
problem was used to obtain the definition of a definite inte-
gral, with the result that the area bounded by the axis of
x, the straight lines x = a and x = b (a < J), and a portion
of the curve y=f(%) which lies above the axis of & is
given by the definite integral
r
ydx.
(X)
It has also been noted that either of the boundary lines z = a
or x = b may be replaced by a point in which the curve cuts OX.
Here the element of integration ydx represents the area of a
rectangle with the base dx and the altitude y.
Similarly, the area bounded by the axis of yy the straight lines
y = c and y = d(c < cZ), and a portion of the curve x =/(y)
lying to the right of the axis
of y is given by the integral
f
d
xdy,
(2)
where the element xdy repre-
sents a rectangle with base x
and altitude dy.
Areas bounded in other ways
than these are found by express-
ing the required • area as the
sum or the difference of areas of the above type, or by writing
a new form of the element as illustrated in Ex. 2.
Fig. 157
Ex. 1. Find the area of the ellipse — + fr = 1.
a"
ft2
It is evident from the symmetry of the curve (fig. 157) that one fourth
of the required area is bounded by the axis of y, the axis of x9 and the
curve. Hence, if A is the total area of the ellipse,
A = 4 faydx = 4: f-Vrf-xtdx
Jo Jo a
= — \x Va2 — x2 + a2 sin-1- = trab.
a L aj0
AREA
263
Ex. S. Find the area bounded by the axis of x, the parabola y" = i px,
and the straight line y + 2x — ip = 0 (fig. loS). The straight line and
the parabola intersect at the point C
(p, 2j>), and the straight line intersects
OX at S (2p, 0). The figure shows that
the required area is the sum of two
areas OCD and CBD. Hence, if A is
the required area,
A = f"Vipl:dx+f \ip-
2i)rfi
The area may also be found by con-
sidering it as the limit of the sum of
such rectangles as are shown in fig. 159.
The height of each of these rectangles
is Ay, and its length is ar2 — *,, where xs
the straight line and xt from that of the parabola. The values of y range
from y = 0 at the base of the figure
S = 2p at the point C. Hence
s taken from the equatio:
In the above examples we have
replaced y in I ydx by its value
f(x) taken from the equation of
the curve. More generally, if the
equation of the curve is in the parametric form, we replace both
x and y by their values in terms of the independent parameter.
This is a substitution of a new variable, as explained in § 118,
and the limits must be correspondingly changed.
Ex. 3. Let the equations of the ellipse be
Then the area A of Ex. 1 may be computed as follows :
A = i fydx = - 4 J°ab sin'^i/^ = iab J"7sina<H4> = »
264
APPLICATIONS OF INTEGRATION
126. Infinite limits or integrand. If the curve extends in-
definitely to the right hand, as in figs. 160-162, it is possible
to consider the area bounded by the curve, the axis of x, and a
fixed ordinate x = a, the figure being unbounded at the right
hand. Such an area is expressed by the integral
Lim f f(x) dx = Lim F(b) - F(a),
which may be written concisely as
f
f(x) dx = F(oo ) - F(a).
There is no certainty that this area is either finite or deter-
minate. Where it is so, the area bounded on the right by a
movable ordinate approaches a defi- T
nite limit as the ordinate recedes
indefinitely from the origin.
dx
Ex.1. r°°i^r=[2V^]00 = oo. (Fig. 160)
Ex-2Ts=Hr=i- (F*i6i>
Fig. 16
►0
•
z
\
"01
J
L
Fig. 161
Fig. 162
Ex. 3. I sin arete = [— cosx]°°
= indeterminate. (Fig. 162)
Similarly, the area may be unbounded
at the left hand, and the lower limit or
both limits of the definite integral may
be infinite.
In like manner let /(#) become infinite at the upper limit,
and the curve y =/(#) approach x = b as an asymptote. Then
the area bounded by the curve, the axis of x, an ordinate x = a,
and an ordinate near the asymptote x=b may approach a
definite value as the latter ordinate approaches the asymptote.
Such an area may be expressed by the integral
Lim f f(x) dx = Lim F(b - A) - F(a),
or, more concisely, I f(x)dx= F(V) — F(a).
68 Fio. 184
Failure to do this
MEAN VALUE
. (Fig. 164) Jl
Similarly, /(x) may become infi- jSi
irite at the lower limit or at both /Mm
limits. If it becomes infinite for y^H
any value c between the limits, liiia
the integral should be separated
into two integrals having c for the 10'
upper and the lower limit respectively,
may lead to error.
Ex. 6. Consider I — -
Since — becomes infinite when x — 0 (fig. 165),
separate the integral into two, thus ;
X+'rfi _ r" dx /"'(to _
i 1? ~ J-i~x* + Jo l^-"'
Had we carelessly applied the incorrect formula
we should have been led to the absurd result — 2.
127. The mean value of a function. In
curve DPC be the graph of the function /(%)■
J f(x) dx = area ADPCB.
Let m=AN and M=AH
be respectively the smallest
and the largest value assumed
by f(x) in the interval AB.
Coustruct the rectangle ABKH
with the base AB and the alti- " ""
tads AH =M. Its area is AB ■ AH=(b-a)M. Construct also
the rectangle ABLN with the base AB and the altitude AN= m.
Its area is AB- AN=(b-a)m.
■
H
S
D
N
K
G
T
L
/ \ /
' K/
u
i A J
266 APPLICATIONS OF INTEGRATION
Now it is evident that the area ABCD is greater than the
area ABLN and less than the area ABKH. That is,
(b-a)m< C f(x)dx<(b-d)M*
Consequently I /(#) dz = (b — a) fi,
where p is some quantity greater than m and less than My and
is represented on fig. 166 by AS. But since f(x) is a continuous
function, there is at least one value f between a and b such
that /(£:) = A&, and therefore
/(*) dx = (b- «)/(£). (1)
I
Graphically, this says that the area ABCD is equal to a rec-
tangle ABTS whose base is AB and whose altitude AS lies
between AN and AH.
From (1) we have
/(£> = ^ J[ /(*) <**, (2)
where f lies between a and J. The value
1 r6
- X /w
dx
is called the wean value of /(#) in the interval from a to J. This
is, in fact, an extension of the ordinary meaning of the average,
or mean, value of n measurements. For let y0, yv y2, • • •, yn_x
correspond to n values of #, which divide the interval from a
to b into n equal parts, each equal to Ax. Then the average of
these n values of y is
n
This fraction is equal to
Oo+ffi+y2+ « • - +y,-i)Aa? y0Aa?+y1Aa?+yaAa:+ ■ > ■ +y,.1Aa;
-^ ■^— — — — — — ^— ^ — ^— ^— — ^s ^— ^— — -^^— — — ^— — ^ — — — — — — — ^_ _ #
nAx b — a
*A slight modification is here necessary if /(x) = fc, a constant. Then
3f=m = fcand I f(x) dx = (b — a) k.
J a
AREA 267
As n is indefinitely increased, this expression approaches aa a
1 /** If"
limit - | ydx=- \ f(x)dx. Hence the mean value
t> — aja ° — aJ*
of a function may be considered as the average of an " infinite
number" of values of the function, taken at equal distances
between a and b.
Ex. 1. Find the mean velocity of a body falling from rest during the
The Telocity is gt, whert g is the acceleration due to gravity. Hence the
mean velocity ia — — -J gtdt = \g<v This ia half the final velocity.
Ex. 2. Find the mean velocity of a body falling from rest through a
The velocity is V2 gs. Hence the mean velocity is
This is two thirds the final velocity.
128. Area of a plane curve in polar coordinates. Let 0 (fig. 167)
be the pole, OM the initial line of a system of polar coordinates
(r, 0), OA and OB two fixed radius vectors for which 8 = a and
0 = /S respectively, and AB any curve for
which the equation is r =f(ff). Required
the area AOB.
The required area may be divided into
n smaller areas by dividing the
AOB=@ — a into n equal parts,
each of which equals = A0,
n
and drawing the lines 0Plt 0PS,
0Pt, -.., 0P„_V where AOPl =
PO%=I>OIZ= • - ■ = Pn_l0B=&6.
(In the figures =8.) The required 0— f"~~~~ Flo' 167
area ia the sum of the areas of these
elementary areas for all values of n. The areas of these small
figures may be found approximately by describing from 0 as a
center the circular arcs ASlt PB3, B.R& • • •, .?_,.#„. Let
OA = r., OPi=r„ OP~r„ ■ • ■, OPH_l = rn_v
268 APPLICATIONS OF INTEGRATION
Then, by geometry,
the area of the sector AOB1 = ^ro2A0,
the area of the sector IfOR2 = ^-r/Afl,
the area of the sector %_1ORn = ^rf^AO.
The sum of these areas, namely
t=0
is an approximation to the required area, and the limit of this
sum as n is indefinitely increased is the required area. Hence
the area A OB=i C r2d0.
-»/
The above result is unchanged if the point A coincides with
0, but in that case OA must be tangent to the curve. So also B
may coincide with 0.
Ex. Find the area of one loop of the curve r = a sin 3 6 (fig. 101, § 60).
As the loop is contained between the two tangents $ = 0 and 0 = — > the
required area A is given by the equation
2 Jo 2 Jo 2 12
129. Volume of a solid with parallel bases. Fig. 168 repre-
sents a solid with parallel bases. The straight line Off is drawn
perpendicular to the bases, cutting the lower base at A, where
h = a, and the upper base at B, where h = b. Let the line
AB be divided into n parts each equal to = AA, and let
n
planes be passed through each point of division parallel to the
bases of the solid. Let A0 be the area of the lower base of
the solid, Ax the area of the first section parallel to the base,
A J the area of the second section, and so on, An_x being the
area of the section next below the upper base. Then A Ah
represents the volume of a cylinder with base equal to A0 and
altitude equal to AA, AxAh represents the volume of a cylinder
standing on the next section as a base and extending to the
section next above, and so forth. It is clear that
A^k+A^Ak+A^hj + AK_lAh = 'j?A1Ah
is an approximation to the volume of the solid, and that the
limit of this sum as n indefinitely increases is the volume of
the solid. That is, the required
volume V is
-i:
AdL
To find the value of this integral
it is necessary to express A in terms
of h, or both A and h in terms of some
other independent variable. This is
a problem of geometry which must
be solved for each solid. It is clear
that the previous discussion is valid
if the upper base reduces to a point,
ie. if the solid simply touches a
plane parallel to its base. Similarly,
both bases may reduce to points.
Ex. 1. Two ellipses with equal major axes are placed with their equal
axes coinciding and their planes perpendicular. A variable ellipse moves
so that the ends of its
axes are on the two given
ellipses, the plane oi the
moving ellipse being per-
pendicular to those of the
given ellipses. Required
the volume of the solid
generated.
Let the given ellipses Fig. 1
be ABA'S" (Sg.lGd) with
semiaxes OA = a and OB = b, and A CA'C with s
OC = c, and let the common axis be OX. Let NMN'M' be one position
of the moving ellipse with the center P where OP = x. Then ii A is the
area of NMN'M% „ ™ ,
r" "*' (By Ex. 1, § 125)
= *PM-PN.
APPLICATIONS OF INTEGRATION
But from the ellipse AJiA'Ti'
Therefore PM ■ PN = ^ (a! - x*).
Consequently the required volume is
J — ^(as — x*)dx = - wabc.
The solid is called an ellipsoid (§ 143, Ex. 5).
Ex. 2. The axes of two equal right circular cylinders intersect at
right angles. Required the volume common to the cylinders.
Let OA and OB (fig. 170) be the axes of the v
cylinders, OY their common perpendicular at
their point of intersection 0, and a the radius
of the base of each cylinder. Then the figure
represents one eighth of the required volume V.
A plane passed perpendicular to OK at a dis-
tance ON = y from 0 intersects the solid in a
square, of which one side is
NP = ^TJP^-ON* = -vV-y*.
Therefore \V= ("NP3 dy = f" (a* - y*) dy = Sj
130. Volume of a solid of revolution. A solid of revolution is a
solid generated by the revolution of a plane figure about an axis
. in its plane. In such a solid a section made by a plane perpen-
dicular to the axis is a circle, or is bounded by two or more
concentric circles. Therefore the method of the previous article
can usually be applied to find the volume of the solid. No
new formulas are necessary. The following examples illustrate
the method.
Ex. 1. Find the volume of the solid generated by revolving about
OX the figure bounded by the parabola y* = 4px, the axis of x, and
the line x = a.
VOLUME
271
The area to be revolved is shaded in fig. 171. Let P(x9 y) be a point
on the parabola. Then any section of the solid through P perpendicular
to OX is a circle with radius MP = y. Hence in the formula of § 129 we
have A = try2 and dh = dx. Hence
the required volume V is
V= I iry2dx.
7
M
m
mm
A
0
M x=a
Fig. 171
But from the equation of the pa-
rabola y2 = 4px. Therefore
V=ip7rf xdx = 2pwa2.
Ex. 2. Find the volume generated by revolving around the line x =,a
the figure described in Ex. 1.
If P (fig. 172) is a point on the curve,
a section of the required solid through
P and perpendicular to AB is a circle
with radius PN =a — x. Hence in the
general formula of § 129 A = it (a — x)2
and dh = dy. When x = a, y = 2Vpa.
Hence the volume V is given by
2 Vpa
x-a
Fig. 172
V = f*^**w(a - xfdy = irf2y/pa(a2 -2ax + ^dy.
But from the equation of the parabola x = j— • Hence
V=7rf2^(a2-^ + ^-)d y = M
Jo \ 2p 16 W y 15
irp'a
Ex. 3. Find the volume of the ring solid generated by revolving a circle
of radius a about an axis in its plane b units from the center (b > a).
Take the axis of revolution as OF
(fig. 173) and a line through the center
as OX. Then the equation of the circle
is (x — b)2 + y2 = a2.
A line parallel to OX meets the circle
in two points, A where x = xx=b — Va2 — y2
and B where x = x2 = b + Va2 — y2. A sec-
tion of the required solid taken through
AB perpendicular to OF is bounded by
two concentric circles with radii xx
and x2 respectively. Hence in § 129
A = irx2 — wx2, and dh = dy. The summation extends from the point L
where y = — a to the point K where y = + a. Hence, for the volume V,
V=irf* Va - *i2) dlJ = 4 «*/ + °Va2 - y2dy = 2 >n*a%.
:
r
K
N
A4
%P
0
Fm
L
WE
Fig.
173
272
APPLICATIONS OF INTEGRATION
131. Length of a plane curve. To find the length of .any
curve AB (fig. 174), assume w — l points, if, P„ • • ., J^_v be-
tween A and B and connect each pair of consecutive points by
a straight line. The length of AB is
then defined as the limit of the sum
of the lengths of the n chords A^
i^ZJ, i^ZJ, • • ., I^^B as n is increased
without limit and the length of each
chord approaches zero as a limit. By
means of this definition we have
already shown (§§ 91 and 104) that
in Cartesian coordinates, and
Fig. 174
(i)
in polar coordinates.
Hence we have
and
(2)
(3)
(4)
To evaluate either (3) or (4) we must express one of the
variables involved in terms of the other, or both in terms of a
third. The limits of integration may then be determined.
It may be noticed that (4) can be obtained from (3). For
we have * • &
x = r cos 0, y = r sm u.
Then dx = cos Odr — r sin OdO,
dy = sin Odr + r cos 0d0,
and da? + djf = df* + f*dP.
Ex. 1. Find the length of the parabola y2 = 4px from the vertex to
the point (h, k).
From the equation of the parabola we find 2ydy = 4pdx. Hence
formula (3) becomes either
s
- jN^*
r
rh \x +
=k\—
dx
or
=foyF^^=fPfokV7TT?dy-
LENGTH 273
Either integral leads to the result
s = J-VF + 4 p2 + Plog * + V5 + 4;?2»
Ex. 2. Find the length of the epicycloid from cusp to cusp.
The equations of the epicycloid are (§ 57)
x = (a + b) cos <l> — a cos • <£,
a
« = (a + £>) sin <b — a sin <&.
a
Hence dx = — (a + b) sin <£ + (a + b) sin <£ d<j>,
dy = I (a + b) cos ^ — (a + &) cos <f> d<j).
Then cte = (a + V) -i/2 — 2 f sin <f> sin <£ + cos <f> cos <j>\ d<j>
= (a + b) <J2 - 2 cos- <f> d<j> = 2 (a + b) sin ^- <t> d<j>.
fc sin ^r- <l> d<j> = — (a + b).
132. The work of the previous article may be brought into
connection with § 124 as follows :
>Hl
Since *<**? + <*»? N VA*/A*
then T,;TnV(A^+(A^ = T.m>|l + (A9T.TnAg = 1
Vcfc2 + dtf
Hence V<f>'+ (**>'-,! + «,
1 , /<&x" ^
and V(Az)2+ (Ay)2 = Vrfar1 + dtf + ey/daf+dy*.
By § 124 the term eVd^+dtf will not affect the limit of
2V(Ax)* + (Ay)2.
274
APPLICATIONS OF INTEGRATION
Fig. 176
133. Area of a surface of revolution. A surface of revolution
is a surface generated by the revolution of a plane curve around
an axis in its plane (§ 130). Let the curve AB (fig. 175)
revolve about OH as an axis. To find the area of the surface
generated, assume n — 1 points, if, P2, Pz, • • .,
-?_r between A and B and connect each
pair of consecutive points by a straight
line. These lines are omitted in the figure
since they are so nearly coincident with the
arcs. The surface generated by AB is then
defined as the limit of the sum of the areas
of the surfaces generated by the n chords
AJ^y I(P2, i£J£, • • •, Pn_xB as n increases
without limit and the length of each chord
approaches zero as a limit.
Each chord generates the lateral surface of a frustum of a
right circular cone, the area of which may be found by
elementary geometry.
Draw the lines ANoy iji\^, iJ-ZV^ • • • perpendicular to O-ff, and
Place N0A = r0, Ntf = rv NtP2 = r„ • • • , K^P^ = rn.
Then the frustum of the cone generated by i?J?+1 has for the
radius of the upper base A^+1^+1, for the radius of the lower base
iV)i?, and for its slant height J?Pi+ v Its lateral area is therefore
equalt0 2v(M±^I±i1pP
A 7T ^ ^i^i + V
Therefore the lateral area of the frustum of the cone equals
This is an infinitesimal which differs from
27rr/fa
by an infinitesimal of higher order, and therefore the area
generated by AB is the limit of the sum of an infinite number
of these terms. Hence, if we represent the required area by #
we have r
S=2ir I rds.
WORK 275
To evaluate the integral it is necessary to express r and ds in
terms of the same variable and supply the limits of integration.
Ex. Find the area of the surface of revolution described in Ex. 1, § 130.
Here r = y and ds = Vdx2 + dy2, where x and y satisfy the equation
y2 = 4/wr. Consequently we may place r = 2 Vpx, and, as in Ex. 1, § 131,
<fc=-l/- — ~dx.
Then S = 4 irVp fa^/x + pdx = | irVp [(a + p$ — />*].
134. Work. By definition, the work done in moving a body
against a constant force is equal to the force multiplied by the
distance through which the body is moved. Suppose now that
a body is moved along OX (fig. 176) from A (x = a) to B (x = 6)
against a force which is not Q , , , , , , , , ^
constant but a function of a; A M^M%MzMtM<M< n
and expressed by /(#). Let the FlG* 176
line AB be divided into n equal intervals, each equal to Ax> by
the points M^M# Mz, . . ., Mn_v (In fig. 176, n = 7.)
Then the work done in moving the body from A to Ml would
be /(#) Ax if the force were constantly equal to f(a) through-
out the interval AMX. Consequently, if the interval is small,
f(a)Ax is approximately equal to the work done between A
and Mx. Similarly, the work done between Mx and M2 is approxi-
mately equal tof(x^) Ax, that between M2 and Ms approximately
equal to/(#2)A#, and so on. Hence the work done between A
and B is approximately equal to
f(a) Ax +/(*!> A* +/(*2) Ax + . . . +f(xu _,) Ax.
The larger the value of n, the better is this approximation.
Hence we have, if W represents the work done between A and B,
W= Lim 2) /(aQ A* = f /O) <**•
»=«> »=0 Ja
135. Pressure. Consider a plane surface of area A immersed
in a liquid at a uniform depth of h units below the surface.
The submerged surface supports a column of liquid of volume
hAy the weight of which is whAy where w (a constant for a
given liquid) is the weight of a unit volume of the liquid.
276 APPLICATIONS OF INTEGRATION
This weight is the total pressure on the immersed surface. The
pressure per unit of area is then wh, which is defined as the
pressure at a point h units below the surface. By the laws of
hydrostatics this pressure is exerted equally in all directions.
We may accordingly determine, in the following manner, the
pressure on plane surfaces which are perpendicular to the
surface of the liquid:
Let BRQ (fig. 177) be a plane surface so immersed that its
plane is perpendicular to the surface of the liquid and inter-
sects that surface in the line TS. Divide BRQ into strips by
drawing lines parallel to TS. Let
the depth of a line of the first
strip be A0, that of the second strip
be hv that of the third strip be A2,
and so on. Call the area of the
first strip (A^4)0, that of the second
strip (A^)x, that of the third strip *" FlG 177
(A^4)2, and so on. Then the pres-
sure on the first strip is approximately whQ(AA)0y that on the
second strip is approximately wh^LA)^ that on the third strip
is approximately wh2(AA)2j etc. Therefore the total pressure
on BRQ is approximately
w \_KCAAX + AX(AJ)X+ • • . + K^aax^ = w "V \(M)e
t = 0
This approximation is better the greater the number of strips,
since we have taken the whole strip as lying at the level of
the same line. Therefore the total pressure P is the limit of the
above sum as w = oo; that is,
P=w ChdA.
To evaluate the integral it is necessary to express h and A in
terms of the same variable and supply the limits. In finding dA
the strips may be taken as rectangles, as in finding the area.
Ex. A parabolic segment with base 2 b and altitude a is submerged so
that its base is in the surface of the liquid and its axis vertical.
Let RQC (fig. 178) be the parabolic segment and let CB be drawn
through the vertex of the segment perpendicular to TS. According to
CENTER OF PRESSURE
277
the data RQ = 2b, CB = a. Draw a horizontal strip LNNXLV with its
bottom line cutting CB at M. Let CM = x; then the depth A of the line
ZiV is a — x and the breadth JtOfj of the strip is dx.
Consequently dA = (Z iV) dx,
2
But, from § 45,
whence
and therefore
UT=CM.
RQ2 CB'
— 2 4 b2x
a
2bx*
dA =
a
i
dx.
Therefore, since x = 0 at C, and x = a at B, the total pressure P is
given by
P = w I -— (a — x)x* dx = -~wba*.
Jo a\ 15
136. Center of pressure. From mechanics we take the fol-
lowing principles:
1. The resultant of a set of parallel forces is equal to the
sum of the forces.
2. The moment of a force about a line at right angles to the
line of action of the force is defined as the product of the force
and the shortest distance between the two lines.
3. The moment about a line of the resultant of a number of
forces is equal »to the sum of the moments of the forces.
Now in the pressure problem of § 135, the pressure on each
one of the elementary strips is a force approximately equal to
whLA acting at right angles to the area. By the second principle
stated above, the moment of this force about TS is h(whAA),
and the limit of the sum of the moments of all the forces is
/ h(whd£) = w I h2dA.
By the first principle stated above, the resultant of the pres-
sures on all the rectangles is the total pressure P. If this acts
at a distance h below the surface of the liquid, we have, by the
third principle, _ '
hP = w J h2dA,
from which h can be found.
AG
278 APPLICATIONS OF INTEGRATION
The point at which P acts is called the center of pressure.
The formula above gives the depth of the center of pressure.
Ex. Find the depth of the center of pressure of the parabolic segment
of the example in § 135.
From the discussion just given,
Ph = w I — (a — X)Ax*dx = •
Jo aiK } 105
But P = A wba* (Ex-> § 135> Therefore h = \ a. By symmetry the
center of pressure lies in CBy and is therefore fully fixed;
137. Center of gravity. Consider n particles of masses ml9
™>2, ™8, •••> ™n> placed at the points I((xv yx), P2(x2, y2),
JS(*.: y.)» •••» Fn(?« y.) (fig. 179) re-
spectively. The weights of these particles
form a system of parallel forces equal to
mx9i m*9> m& • • • » ™»#> where # is the
acceleration due to gravity. The principles
of mechanics stated in § 136 are therefore
applicable. The resultant of these forces is
the total weight W of the n particles, where
i = n
JP= ^ + w^ + m8^ + . . . + m^^gVmi.
This resultant acts in a line which is determined by the con-
dition that the moment of W about any line through 0 is equal
to the sum of the moments of the n weights.
Suppose first the figure placed so that gravity acts parallel
to OYy and that the line of action of W cuts OX in a point
the abscissa of which is x. Then the moment of W about a
line through 0 perpendicular to the plane XOY is gx^m0 and
the moment of one of the n weights is gm^
Hence gx V mi = g V mpc
Similarly, if gravity acts parallel to OXy the line of action of
the resultant cuts OF in a point the ordinate of which is y,
where _^ —
9yZ,mi=0A/m&r
CENTER OF GRAVITY 279
These two lines of action intersect in the point G, the coordi-
nates of which are _-* ^
*=% — > y = % — C1)
Furthermore, if gravity acts in the XO Y plane, but not paral-
lel to either OX or 0 Y, the line of action of its resultant always
passes through G. This may be shown by resolving the weight of
each particle into two components parallel to OX and 0 Y respec-
tively, finding the resultant of each set of components in the
manner just shown, and then combining these two resultants.
If gravity acts in a direction not in the XO Y plane, it may still
be shown that its resultant acts through G, but the proof requires
a knowledge of space geometry not yet given in this course.
The point G is called the center of gravity of the n particles.
If it is desired to find the center of gravity of a physical body,
the solution of the problem is as follows : The body in question
is divided into n elementary portions such that the weight of
each may be considered as concentrated at a point within it.
If m is the total mass of the body, the mass of each element
may be represented by Am. Then if (x0 y^) are the coordinates
of the point at which the mass of the ith element is concentrated,
the center of gravity of the body is given by the equations
V#fAm _ ^.ViAm
a; = Lim^= > y = Lim^^ 1
I xdm I ydm
whence x = — — » y = — ; (2)
I dm I dm
To evaluate, the integrals must be expressed in terms of a
single variable and the limits supplied.
It is to be noticed that it is not necessary, nor indeed always
possible, to determine x$ y{ exactly, since, by § 124,
Lim ^? (x{ + et) Am = Lim V#. Am,
n sb oo i = \ n = oo t = 1
if et approaches zero as Am approaches zero.
280
APPLICATIONS OF INTEGRATION
Ex. 1. Find the center of gravity of a quarter circumference of the
circle x2 + y2 = a2, which lies in the first quadrant.
Let the quarter circumference be divided into elements of arc ds
(fig. 180) ; then, if p is the amount of mass per unit length,
dm = p ds.
The mass of each element may be considered concentrated at a point
(x, y) of the curve. Hence
jpxds fpyds
-> y =
X =
/<
)ds
!•
ds
If p is assumed constant, it may be removed from under the integral
signs and canceled. The denominator of each fraction is then equal
to s, a quarter circumference. To compute
the numerators, we have, from the equation
of the curve,
ds = Vcte2 + dy2 = - dx = dy,
where s is assumed as measured from A so
that dx is positive and dy negative.
Cxds = — J ady = a2,
Therefore
and
Hence
Cyds = | adx = a2,
/ds = — - f a quarter circumference.
2
Fig. 180
_ _ 2a
x-y-
7T
Ex. 2. Find the center of gravity of a quarter circumference of a circle
when the amount of matter in a unit of length is proportional to the length
of the arc measured from one extremity.
As in Ex. 1, dm = pds, but here p = ks, k being a constant. Then
dm = ksds.
The integration is best performed by use of the parametric equations of
the circle (§ 53). Then
x =
7T2
/sxds \2 as<l> cos d>d<b ,. ox
^_ Jo y y y _ (4tt— 8)a
/** n<i2<t>d<i>
Jo
ir
Csy ds C 2 aB<l> sin <j) d<j> R
y =
i
sds
IT
C2a2<f>d<t>
Jo
IT'
CENTER OF GRAVITY
281
Ex. 3. Find the center of gravity of the area bounded by the parabola
y2 = 4 px (fig. 181), the axis of x, and the ordinate through a point (A, k)
of the curve.
As in finding the area, let the area be divided into elementary rectangles
ydx, where (x, y) is a point on the curve. Then, if p is the amount of mass
per unit area, dm = pydx,
and this mass may be considered as concentrated at the middle point
(, A c ,
Then
its left-hand ordinate.
h
o
^x(pydx) X (l)(py^
JQpydx
/•"
dx
Fig. 181
If p is assumed constant, it may be removed from under the integral
signs, and canceled. Then, by aid of the equation of the curve, we compute
the integrals rh i ph „ x *
J xydx = 2p*j x*dx = tpH* = | h2k,
o
h
and
Therefore
J j y2dx = 2 p I xdx = ph2 = J hk2,
f ydx = 2pifhxidx = $ph* = § hk.
x= J A,
y
— 3 1.
ara
y*_
Ex. 4. Find the center of gravity of the segment of the ellipse -— + ^ = 1
C* (J
(fig. 182) cut off by the chord through the positive ends of the axes of
the curve. Divide the area into elements by y
lines parallel to OY. If we let y2 be the ordi-
nate of a point on the ellipse, and yx the
ordinate of a point on the chord, we have
as the element of area,
(y2 - Vd dx>
and hence dm = p (y2 — y2) dx,
where p, the amount of mass per unit of area, is assumed constant.
The mass of this element may be considered as concentrated at the
point (,, 2L+A).
Hence - — °
Fig. 182
f (y2-yi)xdx
x =
I
a yi + ft ,„ _
v =
(y2-yi)rf* ifo 0/2-y!)dx
f0 (y2-yi)^ f0 (y2-yi)dx
282 APPLICATIONS OF INTEGRATION *
From the equation of the ellipse, y2 = — Va2 — x2; from the equation of
b a
the chord, yx = - (a — x).
The denominator I # (y2 — yx) dx is equal to the area of the quadrant
of the ellipse minus that of a right triangle, i.e. is equal to — — •
- I x [Va2 — x2 — (a — x)] dx
x = ^^— —
Hence
2a
ab
b2
(H)
2^i C\(a2-x2)-(a-xy]dx
y =
ab
a - 1)
3(7T-2)
26
3 (tt - 2)
Ex. 5. Find the center of gravity of a spherical segment of one base
generated by revolving the area BDE (fig. 183) about OY, where OB = a,
and OE = c.
Let the volume be divided into elementary
cylinders as in § 130. Then the element of
volume is Ady = irx2dy, and hence
dm = pirx2dy,
where p is the density, assumed constant. The
mass of this element may be considered as con-
centrated at (0, y), the center of its base. Hence
the center of gravity of the entire volume is in
the line 0 Y, and its ordinate y is given by ^ , g^
fc y(p**dy) fc (a2y-y*)dv 3 (a + <o»
^— pa ~~ pa 4 2a + c
/ fmx2dy J (a2 — y2)dy
Ex. 6. Find the center of gravity of the surface of the spherical seg-
ment of Ex. 5.
Divide the surface into elementary bands as in § 133. Then
dm = 2 irpxds,
where p, the amount of mass per unit area, is assumed constant.
This mass may be considered concentrated at (0, y). Hence, using the
notation and the figure of Ex. 5, we have ds = — - > and therefore
x
_ ft'*** f'*d» a +
y =
J/% a pa
xds I dy
C v C
ATTRACTION
283
138. Attraction. Two particles of matter of masses m1 and
m2 respectively, separated by a distance r, attract each other
Tit Tit
with a force equal to k l 2? where k is a constant which de-
pends upon the units of force, distance, and mass. We shall
assume that the units are so chosen that k = 1.
Consider now n particles of masses m^ m%, mz, • • •, mn lying
in a plane at the points Pv P2, Pz, • • •, Pn (fig. 184). Let it
be required to find
their attraction upon
a particle of unit mass
situated at a point A
in their plane.
Let the distances APV
AJ^y • • • , Ag be denoted
by *v r2' • • •» r»- The
attractions of the indi-
vidual particles are
rw,
m
2
WL
Fig. 184
2
but these attractions cannot be added directly, since they are
not parallel forces. To find their resultant we will resolve
each into components along two perpendicular axes AX and AY
respectively. If we denote the angle XAPt by 0„ we have as
the sum of the components along AX,
2
n
and for the sum of the components along AY,
r=^sin^ + ^sin^ + ..v + ^sin^tt.
The resultant attraction is then
B=^/X2+ Y2
-iY
and acts in a direction which makes tan * — with AX.
284
APPLICATIONS OF INTEGRATION
Let it now be required to find the attraction of a material
body of mass m upon a particle of unit mass situated at a
point A. Let the body be divided into n elements, the mass of
each of which may be represented by Atw, and let i? be a point
at which the mass of one element may be considered as concen-
trated. Then the attraction of this element on the particle at A
> where ri = PiA^ and its component in the direction AX
is
Am
is — - cos 0p where 0i is the angle XAI?. The whole body, there-
fore, exerts upon the particle at A an attraction whose com-
ponent in the direction AX is
Similarly, the component in the direction AY is
X=Lim¥^^Am
dm.
dm.
Ex. Find the attraction of a uniform wire of length I and mass m
on a particle of unit mass situated in a straight line perpendicular to
the wire at one end, and at a distance
a from it.
Let the wire OL (fig. 185) be
placed in the axis of y with one
end at the origin, and let the par-
ticle of unit mass be at A on the
axis of x where A 0 = a. Divide OL
into n parts, OMv MXMV M2MV • • •,
Mn_x L, each equal to - = Ay. Then,
n
if p is the mass per unit of length of
the wire, the mass of each element
is Am = pAy. We shall consider the
mass of each element as concen-
trated at its first point, and shall in this way obtain an approximate
expression for the attraction due to the element, this approximation
being the better, the smaller Ay is made. The attraction of the element
MiMi+1 on A is then approximately
Fig. 185
AM? «2 + tf *x
PROBLEMS 285
"JL'he component of this attraction in the direction OX is
-B^L-cosoam, — e^-f
a + *t (a* + y«)i
«i the component in the direction OF is
-^L sin 0,0/,= PyAy ■
a* + *? (a* + tf)l
Then, if X is the total component of the attraction parallel to OX, and
the total component parallel to OF, we have
X = Lim V '-^4 = paf1-**—,
F=Lim=V1 P*** =pr' ydy
— iA (a2 + y*)i Jo (a2 + y2)*
To evaluate the integrals for X and Y, place y = a tan 0. Then, if
cc = tan-1- = OAL,
a P Ca /» /» P m
X = - I cos Odd = - sin a = — , sin a,
at/o a at
y = ^ f%in0c/0 = ^(l-cosa) = ^(l-cosa),
since Zp = m.
If J? is the magnitude of the resultant attraction and /3 the angle which
its line of action makes with OX,
R =VX* + Y*=^sinla,
al 2
a 4. *Y 4. ,1 — cos a 1
a = tan-1 •— • = tan-1 — : = - a.
X sin a 2
PROBLEMS
1. Find the, area of an arch of the curve y = sinx.
2. Find the area bounded by the portions of the curves y = J sin 2x
and y = sin x -f- J sin 2 x that extend between x = 0 and a = 7r.
3. Find the area of the three-sided figure bounded by the coordi-
nate axes and the curve x* + y * = a\
X
4. Find the area bounded by the catenary y = -(ea+e °), the
axis of xf and the lines x = ± h. 8
5. Find the area included between the witch y = Q , . — 5 and its
asymptote.
286 APPLICATIONS OF INTEGRATION
6. Find the area of one of the closed figures bounded by the
curves y2 = 16 x and y1 = sc8.
7. Find the area bounded by the curve y (x2 + 4) = 4 (2 — x), the
axis of xf and the axis of y.
8. Find the area bounded by the curve y2 = x (log x)\ the axis
of x, and the ordinates x = 1 and x = e,
9. Find the area bounded by the parabola y1 = 2 (x — 4) and the
line x = 3 y.
10. Find the area between the parabola x2 = 4 ay and the witch
8a8
y-"a2 + 4a2'
11. Find the area bounded by the parabola x2 — 9y = 0 and the
line x — 3 y + 6 = 0.
12. Find the area included between the parabolas y2 = ax and
x2 = by.
13. Find the area bounded by the curve xPy2 + a2b2 = ahp and its
asymptotes.
x2 v2
14. Find the area bounded by the hyperbola -5 — jz =1 and the
chord x = h.
15. Find the area bounded by the curve ^(x2 + a2) = a2^ and its
asymptotes.
16. Find the total area of the curve 81 y2 + 4sc4 = 36 x2.
17 . Find the area of the loop of the curve (y — l)2 = (x — 1)2(4 — x).
18. Find the area of the loop of the curve cy2= (as — a) (x — b)\
(a < b).
19. Find the area of the loop of the curve 16a8y2= bWfa — 2x).
x id "4"~ x\
20. Find the area of the loop of the strophoid y2 =* — * L •
a — x
21. Find the area of a loop of the curve y*(ci2 4- a2) = x2(a? — x2).
22. Find the total area of the curve ah/ = x*(2a — x).
23. Find the area of the loop of the curve (2x 4- yf = a2 (2 — x).
24. Find the area between the axis of x and one arch of the cycloid
x = a (<j> — sin <£), y = a (1 — cos <£).
25. Find the area inclosed by the four-cusped hypocycloid
x = a cos80, y = a sin80.
PROBLEMS .287
26. Find the entire area bounded by the curve x = a cos 0,
y = b sin80.
27. Find the mean value of the lengths of the perpendiculars
from a diameter of a semicircle to the circumference, assuming
the perpendiculars to be drawn at equal distances on the diameter.
28. Find the mean length of the perpendiculars drawn from the
circumference of a semicircle of radius a to its diameter, assuming
that the points taken are equidistant on the circumference.
29. Find the mean value of the ordinates of the curve y = sin x
between x = 0 and x = tt, assuming that the points taken are
equidistant on the axis of x.
30. A number n is divided into two parts in all possible ways.
Find the mean value of their product.
31. If the initial velocity of a projectile is v0, and the angle of
elevation varies from 0 to -=> find the mean value of the range,
using the result of problem 36, Chap. VII.
32. In a sphere of radius r a series of right circular cones is
inscribed, the bases of which are perpendicular to a given diameter
at equidistant points. Find the mean volume of these cones.
33. A particle describes a simple harmonic motion defined by the
equation « = asin&£. Show that the mean kinetic energy (-s~)
during a complete vibration is half the maximum kinetic energy if
the average is taken with respect to the time.
34. In the motion defined in problem 33, what will be the ratio
of the mean kinetic energy during a complete vibration to the
maximum kinetic energy, if the average is taken with respect to
the space traversed?
35. Find the area described in the first revolution by the radius
vector of the spiral of Archimedes r = ad.
36. Show that the area bounded by the hyperbolic spiral r$ = a
and two radius vectors is proportional to the difference of the
lengths of the radius vectors.
37. Find the total area of the lemniscate ^ = 2a2 cos 2 0.
38. Find the area of a loop of the curve r = a sin n$.
39. Find the area of a loop of the curve r* = a2 sin n$.
288 APPLICATIONS OF INTEGRATION
40. Find the area swept over by the radius vector of the curve
r = a tan $ as $ changes from 0 to j •
41. Find the total area of the cardioid r = a(l -f- cos &).
42. Find the area of the limaqon r = 2 cos $ -f- 3.
43. Find the area of the curved strip of the plane which has two
portions of the initial line for two boundaries and the arc of the
spiral r = aO between $ = 2 7r and $ = 6 it for the other boundary.
44. Find the area of the loop of the curve r3 = a2 cos 2 9 cos 0
which is bisected by the initial line.
45. Find the area of a loop of the curve r3 sin $ = a2 cos 2 6.
46. Find the area of the kite-shaped figure bounded by an arc of
a parabola and two straight lines from the focus making the angles
± a with the axis of the parabola.
47. Find the area bounded by the curves r = a cos 3 $ and r = a.
4
48. Find the area inclosed by the curves r = z 7 and
a 1 — cos 0
r =
1 4- cos $
49. Find the area cut off one loop of the lemniscate r8 = 2 a2 cos 2 9
by the circle r = a.
50. Find the area of the segment of the cardioid r = a(l + cos ff)
cut off by a straight line perpendicular to the initial line at a distance
J a from the vertex.
51. Find the area of the loop of the curve (7? -f- y3)8 = 4 dhhp.
(Transform to polar coordinates.)
52. Find the total area of the curve (a3 + y*)2 = 4 ah? + 4 #y.
(Transform to polar coordinates.)
53. Find the area of the loop of the Folium of Descartes,
#* 4- #* — 3 axy = 0, by the use of polar coordinates.
a3 1/3
54. On the double ordinate of the ellipse -5 4- 75 = 1 as base an
a* Ir
isosceles triangle is constructed with its altitude equal to the dis-
tance of the ordinate from the center of the ellipse and its plane
perpendicular to the plane of the ellipse. Find the volume gener-
ated as the triangle moves along the axis of the ellipse from
vertex to vertex.
PROBLEMS 289
55. Find the volume cut from a right circular cylinder of radius
a by a plane through the center of the base making an angle $ with .
the plane of the base.
56. Two parabolas have a common vertex and a common axis but
lie in perpendicular planes. An ellipse moves with its center on the
common axis, its plane perpendicular to the axis, and its vertices on
the parabolas. Find the volume generated when the ellipse has
moved to a distance h from the common vertex of the parabolas.
57. An equilateral triangle moves so that one side has one end in
OY and the other end in the circle x2 + y1 = a2, the plane of the
rectangle being perpendicular to OY. Required the volume of the
solid generated.
58. In a sphere of radius a find the volume of a segment of one
base and altitude h.
59. Find the volume of the solid generated by revolving about OY
the plane surface bounded by OY and the hypocycloid #» -f- y* = a*
60. Find the volume of the solid formed by revolving about the
line x = 3 the figure bounded by the parabola y1 = 8 x and the line
x = 2.
61. Find the volume of the solid formed by revolving about the
line y = — a the figure bounded by the curve y = sin x, the lines
x = 0 and x = — > and the line y = — a.
62. A right circular cone with vertical angle 2 a has its vertex at
the center of a sphere of radius a. Find the volume of the portion
of the sphere intercepted by the cone.
63. A variable equilateral triangle moves with its plane perpen-
dicular to the axis of y and the ends of its base respectively on the
parts of the curves y1 = 16 ax and y* = 4 ax above the axis of x.
Find the volume generated by the triangle as it moves a distance a
from the origin.
64. Find the volume of the solid formed by revolving about OX
x9
the plane figure bounded by the cissoid y1 = ~ > the line x = a,
and the axis of x.
66. A right circular cylinder of radius a is intersected by two
planes, the first of which is perpendicular to the axis of the cylinder,
and the second of which makes an angle 6 with the first. Find the
290 APPLICATIONS OF INTEGRATION
volume of the portion of the cylinder included between these two
.planes if their line of intersection is tangent to the circle cut from
the cylinder by the first plane.
66. On the double ordinate of the four-cusped hypocycloid
afi + y^ = a? as base an isosceles triangle is constructed with its
altitude equal to the ordinate and its plane perpendicular to the
plane of the hypocycloid. Find the volume generated by the
triangle as it moves from as = — a to x = a.
67. Find the volume of the solid formed by revolving about OY
8 a8
the plane figure bounded by the witch y = « , Aa and the line y = a.
68. Find the volume of the solid formed by revolving about the
line y = a the plane figure bounded by the line y = a and the witch
8 a8
69. Find the volume of the solid bounded by the surface formed
8 a8
by revolving the witch y = a 2 about its asymptote.
X ~f~ 4k Ob
70. Find the volume of the wedge-shaped solid cut from a right
circular cylinder of radius a and altitude h by two planes which
pass through a diameter of the upper base and are tangent to the
lower base.
71. Two circular cylinders with the same altitude h have the
upper base, of radius a, in common. Their other bases are tangent
at the point where the perpendicular from the center of the upper
base meets the plane of the lower bases. Find the volume common
to the two cylinders.
72. Find the volume of the ring solid formed by revolving the
ellipse v*""^ + £ = l around OY (d > a).
73. The cap of a stone post is a solid of which every horizontal
cross section is a square. The corners of all the squares lie in a
spherical surface of radius 8 in. with its center 4 in. above the plane
of the base. Find the volume of the cap.
74. Find the volume of the solid formed by revolving about the
line x = — 2 the plane figure bounded by that line, the parabola
y3 = 4 as, and the lines y = ±2.
PROBLEMS 291
75. Find the volume of the solid formed by revolving about the
line x = 2 the plane figure bounded by the curve y* = 4 (2 — x) and-
the axis of y.
76. A variable circle moves so that one point is always on OY, its
x2 if
center is always on the ellipse -5 + jg = 1, and its plane is always
perpendicular to OY. Required the volume of the solid generated.
77. Find the volume of the solid generated by revolving about
xs
the asymptote of the cissoid y2 = the plane area bounded by
the curve and the asymptote.
78. Find the volume of the solid formed by revolving about
OX the plane figure bounded by OX and an arch of the cycloid
x = a (<f> — sin <£), y = a (1 — cos <£).
79. Find the volume of the solid generated by revolving the
cardioid r = a (1 -f- cos 0) about the initial line.
80. A cylinder passes through two great circles of a sphere which
are at right angles to each other. Find the common volume.
81. Find the length of the semicubical parabola y2 = (a? — 2)8
from its point of intersection with the axis of x to the point (6, 8).
_ X X
d , -
82. Find the length of the catenary y = - (ea -f- e a) from x = 0
to x = h.
83. Find the total length of the four-cusped hypocycloid
8* -f* y* = a*.
84. Show that the length of the logarithmic spiral r = ea9 between
any two points is proportional to the difference of the radius vectors
of the points.
q
85. Find the complete length of the curve r = asin8--
86. Find the length of the curve y = a log -5 — -5 from the
CL ""~ • X>
... a
origin to x = 5 •
87. Find the length from cusp to cusp of the cycloid
x = a(<f> — sin <£), y = a (1 — cos <f>).
292 APPLICATIONS OF INTEGRATION
88. From equidistant points on an arch of the cycloid
x = a (4> — sin <f>), y = a(l — cos <£),
perpendiculars are drawn to the base of the arch. What is their
average length ?
89. From a spool of thread 2 in. in diameter three turns are
unwound. If the thread is held constantly tight, what is the length
of the path described by its end ?
e* 4- 1
90. Find the length of the curve y = log — — - from x = 1
to* = 2. e^"1
91. Find the mean distance of all points on the circumference
of a circle of radius a from a given point on the circumference.
92. Find the length of the spiral of Archimedes, r = a$, from
the pole to the end of the first revolution.
93. Find the length of the curve 8 a*y = x4 -f- 6 aV from the
origin to the point x = 2 a.
94. The parametric equations of a curve are
x = 50(1- cos 0) 4- 50(2 -0)sin0, y = 50 sin0 + 50(2 - 0)cos0.
Find the length of the curve between the points $ = 0 and 0=2.
95. Find the length of the cardioid r = a(l -f- cos $).
96. Find the mean length of the radius vectors drawn from the
pole to equidistant points of the cardioid r = -(1 -f- cos $).
q
97. Find the length of the curve r = a cos6- from the pole to
o
the point in which the curve intersects the initial line.
98. Find the length of the tractrix (§ 200)
a. a + Va2 — x2
V 210ga_V^T^ Va *
from x = h to x = a.
99. Find the area of a zone of height h on a sphere of radius a.
100. Find the area of the surface formed by revolving about OX
the hypocycloicj x = a cos8 0, y = a sin8 0.
101. Find the area of the surface formed by revolving about the
line x = a the portion of the hypocycloid x = a cos8 6, y = a sinty,
which is at the right of OY.
PROBLEMS 293
102. Find the area of the surface formed by revolving about the
a . - x
tangent at its lowest point the portion of the catenary y = -^(e^-\-e a)
between x = — h and x = h.
103. Find the area of the surface formed by revolving about the
initial line the cardioid r = a(l -f- cos 0).
104. Find the area of the surface formed by revolving an arch of
the cycloid x = a(4> — sin <f>), y = a(l — cos <£) about the tangent at
its highest point.
105. Find the area of the surface formed by revolving about OY
the tractrix (§ 200) y = % log a + f ,~~ ** - Va2 ^ x\
£ a — V a2 — x2
106. Find the area of the surface formed by revolving the lem-
niscate r2 = 2 a2 cos 2 $ about the initial line.
107. Find the area of the surface formed by revolving the lem-
niscate r2 = 2 a2 cos 2 $ about the line $ = 90°.
108. A positive charge m of electricity is fixed at O. The repul-
sion on a unit charge at a distance x from 0 is —^ Find the work
done in bringing a unit charge from infinity to a distance a from 0.
109. Assuming that the force required to stretch a wire from the
x
length a to the length a + x is proportional to - > and that a force
of 1 lb. stretches a wire of 36 in. in length to a length .04 in.
greater, find the work done in stretching the wire from 36 in.
to 39 in.
110. A body moves in a straight line according to the formula
x = erf8, where x is the distance traversed in the time t. If the re-
sistance of the air is proportional to the square of the velocity, find
the work done against the resistance of the air as the body moves
from x = 0 to x = a.
111. Assuming that below the surface of the earth the force of
the earth's attraction varies directly as the distance from the earth's
center, find the work done in moving a weight of m pounds from a
point a miles below the surface of the earth to the surface.
112. Assuming that above the surface of the earth the force of
the earth's attraction varies inversely as the square of the distance
AC
294 APPLICATIONS OF INTEGRATION
from the earth's center, find the work done in moving a weight of
m pounds from the surface of the earth to a distance a miles above
the surface.
113. A wire carrying an electric current of magnitude C is bent
into a circle of radius a. The force exerted by the current upon a
unit magnetic pole at a distance x from the center of the circle in
a straight line perpendicular to the plane of the circle is known to
be -• Find the work done in bringing a unit magnetic pole
(a2 + x*)*
from infinity to the center of the circle along the straight line just
mentioned.
114. A spherical bag of radius 5 in. contains gas at a pressure
equal to 15 lb. per square inch. Assuming that the pressure is in-
versely proportional to the volume occupied by the gas, find the
work required to compress the bag into a sphere of radius 4 in.
115. A piston is free to slide in a cylinder of cross section S.
The force acting on the piston is equal to pS, where p is the pres-
sure of the gas in the cylinder, and a pressure of 7.7 lb. per square
inch corresponds to a volume of 2.5 cu. in. Find the work done
as the volume of the cylinder changes from 2.5 cu. in. to 5 cu. in.,
(1) assuming pv = k, (2) assuming pv1A = Jc.
116. Find the total pressure on a vertical rectangle with base 8
and altitude 12, submerged so that its upper edge is parallel to the
surface of the liquid at a distance 5 from it.
117. Find the depth of the center of pressure of the rectangle
in the previous problem.
118. Find the total pressure on a triangle of base 10 and altitude 4,
submerged so that the base is horizontal, the altitude vertical, and
the vertex in the surface of the liquid.
119. Show that the center of pressure of the triangle of the pre-
vious problem lies in the median three fourths of the distance from
the vertex to the base.
120. Find the total pressure on a triangle with base 8 and altitude 6,
submerged so that the base is horizontal, the altitude vertical, and the
vertex, which is above the base, at a distance 3 from the surface of
the liquid.
PROBLEMS 295
.121. Find the depth of the center of pressure of the triangle of
the previous problem.
. 122. The centerboard of a yacht is in the form of a trapezoid in
which the two parallel sides are 3 and 5 ft. respectively in length,
and the side perpendicular to these two is 4 ft. in length. Assuming
that the last-named side is parallel to the surface of the water at a
depth of 1ft., and that the parallel sides are vertical, find the
pressure on the board.*
123. Find the moment of the force which tends to turn the center-
board of the previous problem about the line of intersection of the
plane of the board with the surface of the water.
124. A dam is in the form of a regular trapezoid with its two
horizontal sides 400 and 100 ft. respectively, the longer side being
Sbt the top and the height 20 ft. Assuming that the water is level
"with the top of the dam, find the total pressure.
125. Find the moment of the force which tends to overturn the
dam of the previous problem by turning it on its base line.
126. Find the total pressure on a semiellipse submerged with one
axis in the surface of the liquid and the other vertical.
127. Find the depth of the center of pressure of the ellipse of
tzhe previous problem.
128. The gasoline tank of an automobile is in the form of a
horizontal cylinder, the ends of which are plane ellipses 20 in.
tiigh and 10 in. broad. Assuming w as the weight of a cubic inch
of gasoline, find the pressure on one end when the gasoline is
~X5 in. deep.
129. A parabolic segment with base 15 and altitude 3 is sub-
merged so that its base is horizontal, its axis vertical, and its vertex
in the surface of the liquid. Find the total pressure.
130. Find the depth of the center of pressure of the parabolic
segment of the previous problem.
131. A circular water main has a diameter of 5 ft. One end is
closed by a bulkhead and the other is connected with a reservoir in
'which the surface of the water is 20 ft. above the center of the
bulkhead. Find the total pressure on the bulkhead.
* The weight of a cubic foot of water may be taken as 62 J lb. = ^ ton.
296 APPLICATIONS OF INTEGRATION
132. A pond of 10 ft depth is crossed by a roadway with vertical
sides. A culvert, whose cross section is in the form of a parabolic
segment with horizontal base on a level with the bottom of the pond,
runs under the road. Assuming that the base of the parabolic seg-
ment is 6 ft. and its altitude 4 ft., find the total pressure on the
bulkhead which temporarily closes the culvert.
133. Find the pressure on a board whose boundary consists
of a straight line and one arch of a sine curve, submerged so
that the board is vertical and the straight line is in the surface
of the water.
134. Find the center of gravity of the semicircumference of the
circle x2 + y2 = a2 which is above the axis of x.
135. Find the center of gravity of the arc of the four-cusped
hypocycloid x* -f- y* = a which is above the axis of x.
136. Find the center of gravity of a parabolic segment.
137. Find the center of gravity of the area of a quadrant of an
ellipse.
138. Find the center of gravity of a triangle.
139. Find the center of gravity of the area bounded by the
semicubical parabola ay2 = #8 and any double ordinate.
140. Find the center of gravity of the area bounded by the pa-
rabola 0^4-4^— 16 = 0 and the axis of x.
141. Find the center of gravity of half a spherical solid of con-
stant density.
142. Find the center of gravity of the solid formed by revolving
x2 %P
about OY the surface bounded by the hyperbola -r — ji = 1 and the
a
b2
lines y = 0 and y = b.
143. Find the center of gravity of a hemisphere.
144. Find the center of gravity of the surface of a right circular
cone.
146. Find the center of gravity of the area bounded by the curve
y = sin x and the axis of x between x = 0 and x = 7r.
146. Find the center of gravity of the area between the axes of
coordinates and the parabola «* -f- y* = a*.
PROBLEMS 297
147. Find the center of gravity of a uniform wire in the form of
the catenary y = - (e« -f- 0~") from x = 0 to x = a.
148. Find the center of gravity of the solid formed by revolving
about OX the surface bounded by the parabola y* = ±px, the axis
of a?, and the line x = a.
149. Find the center of gravity of the plane area bounded by the
two parabolas y2 = 20 x and x2 = 20 y.
150. Find the center of gravity of the plane area bounded by the
parabola y1 = 4 x, the axis of y, and the line y = 4.
151. Find the center of gravity of the solid formed by revolving
about OY the plane figure bounded by the parabola y2 = Apx, the
axis of y, and the line y =.k.
152. Find the center of gravity of the surface of a hemisphere
"when the density of each point in the surface varies as its perpen-
icular distance from the circular base of the hemisphere.
153. Find the center of gravity of that part of the plane surface
t>ounded by the four-cusped hypocycloid x = a cos80, y = a sin80,
"which is in the first quadrant.
154. Find the center of gravity of the plane area bounded by the
ellipse -5 + ^ = 1, the circle x* + y2= a2, and the axis of y.
b2
155. Find the center of gravity of the plane area common to the
X>arahola x2 — 8 y = 0 and the circle x2 + y* — 128 = 0.
156. Find the center of gravity of the plane surface bounded by
"the first arch of the cycloid
x = a(<£ — sin £), y = a(l— cos <£),
and the axis of x.
167. Find the center of gravity of the arc of the cycloid
x = a(4> — sin <£), y = a (1 — cos <f>),
between the first two cusps.
158. Find the center of gravity of the solid formed by rotating
about OX the parabolic segment bounded by y* = 4 x and x = h, if
the density at any point of the solid equals - •
x
298 APPLICATIONS OF INTEGRATION
169. Find the center of gravity of the plane surface bounded by
the two circles x2 + y2 = a2,x2 + jf — 2 ax = 0, and the axis of x.
160. Show that the center of gravity of a sector of a circle lies on
. a
' 2 8m2
the line bisecting the angle of the sector at a distance - a — from
o cc
2
the vertex, where a is the angle and a the radius of the sector.
161. Find the center of gravity of the solid generated by revolv-
ing about the line x = a the area bounded by that line, the axis of x,
and the parabola y1 = Apx.
162. Find the center of gravity of the plane area bounded by the
two parabolas x2 — 4p (y — b) = 0, x* — ±py = 0, the axis of y, and
the line x = a.
163. Find the center of gravity of the arc of the curve 9 ay2 —
x (x — 3 a)2 = 0 between the ordinates x = 0 and x = 3a.
164. The density at any point of a lamina in the form of a para-
bolic segment of height 8 ft. and base 6 ft. is directly proportional
to its distance from the base. Find the center of gravity.
165. Find the center of gravity of the portion of a spherical
surface bounded by two parallel planes at distances hx and h2
respectively from the center.
l&B. Find the center of gravity of the solid formed by revolving
about OY the plane area bounded by the parabola x2 = Apy and any
straight line through the vertex.
167. Find the center of gravity of the surface generated by the
revolution about the initial line of one of the loops of the lemniscate
/■*= 2a2 cos 20.
168. Prove that the total pressure on a plane surface perpendic-
ular to the surface of a liquid is equal to the pressure at the center
of gravity multiplied by the area of the surface.
169. Prove that the area generated by revolving a plane curve
about an axis in its plane is equal to the length of the curve multi-
plied by the circumference of the circle described by its center of
gravity.
170. Prove that the volume generated by revolving a plane figure
about an axis in its plane is equal to the area of the figure multiplied
by the circumference of the circle described by its center of gravity.
PROBLEMS 299
171. Find the attraction of a uniform straight wire of length 20
and mass M upon a particle. of unit mass situated in the line of
direction of the wire at a distance 3 from one end.
17 %> Find the attraction of a rod of mass M and length I, whose
density varies as the distance from one end, on a particle of unit
mass in its own line and distant a units from that end.
173. A particle of unit mass is situated at a perpendicular dis-
tance 5 from the center of a straight homogeneous wire of mass M
and length 12. Find the force of attraction of the wire.
174. Find the attraction due to a straight wire of length 2 I on a
particle of unit mass lying on the perpendicular at the middle point
of the wire and distant c units from the wire, the density of the wire
varying directly as the distance from its middle point.
175. Find the attraction of a homogeneous straight wire of neg-
ligible thickness and infinite length on a particle of unit mass at a
p>erpendicular distance c from the central point of the wire.
176. Find the attraction of a uniform wire of mass M bent into
sin arc of a circle with radius 5 and angle — upon a particle of unit
xnass at the center of the circle.
177. Find the attraction of a uniform circular ring of radius a
stnd mass M upon a particle of unit mass situated at a distance c
from the center of the ring in a straight line perpendicular to the
X>lane 0f the ring.
178. Find the attraction of a uniform circular disk of radius a
and mass M upon a particle of unit mass situated at a perpendicular
distance c from the center of the disk. (Divide the disk into con-
centric rings and use the result of problem 177.)
179. Find the attraction of a uniform right circular cylinder with
mass My radius of its base a, and length I upon a particle of unit
mass situated in the axis of the cylinder produced, at a distance c
from one end. (Divide the cylinder into parallel disks and use the
xesult of problem 178.)
180. Find the attraction of a uniform straight wire of length 5
and mass M upon a particle of unit mass situated at a perpendicular
distance 12 from the wire and so that lines drawn from the particle
7T
to the ends of the wire inclose an angle — •
CHAPTER XIV
SPACE GEOMETRY
139. Functions of more than one variable. A quantity z is
mid to be a function of two variables, x and y, if the values of z
are determined when the values of x and y are given. This rela-
tion is expressed by the symbols z =f(x, y), z = F(x, y), etc.
Similarly, u is a function of three variables, x, y, and s, if the
values of u are determined when the values of #, y, and z are
given. This relation is expressed by the symbols u =/(#, y, z),
u = F(x, y, z), etc.
Ex. 1. If r is the radius of the base of a circular cone, h its altitude,
and v its volume, v = £ irr*h, and v is a function of the two variables
r and h.
Ex. 2. If / denotes the centrifugal force of a mass m revolving with a
velocity v in a circle of radius r,/= — , and /is a function of the three
variables m, r, and r.
Ex. 3. Let v denote a volume of a perfect gas, t its absolute temperature,
and p its pressure. Then — = k, where k is a constant. This equation may
be written in three equivalent forms : p = k-, v = *-, t = \pv9 by which
v p k
each of the quantities p, v, and t is explicitly expressed as a function of
the other two.
A function of a single variable is defined explicitly by the
equation y =/(#), and implicitly by the equation F(x, y) = 0
(§ 86). In either case the relation between x and y is repre-
sented graphically by a plane curve. Similarly, a function of
two variables may be defined explicitly by the equation
z =/(#, y), or implicitly by the equation F(x, y, z) = 0. In
either case the graphical representation of the function of two
variables is the same, and may be made by introducing the
conception of space coordinates.
300
RECTANGULAR COORDINATES 801
140. Rectangular coordinates. To locate a point in space of
three dimensions, we may assume three number scales, OX, OY,
OZ (fig. 186), mutually perpendicular, and having their zero
points coincident at 0. They will determine three planes,
XOY, YOZ, ZOX, each of which is perpendicular to the other
two. The planes are called the coordinate planes, and the three
lines OX, OY, and OZ are called the axes of x, y, and z
respectively, or the coordinate axes, and the point 0 is called
the origin of coordinates.
Let P be any point in space, and through P pass planes
perpendicular respectively to OX, OY, and OZ, intersecting
them at the points L, M, and N respectively. Then if we
place x = OL, y = OM, and z = ON, it is evident that to any
point there corresponds one, and only
one, set of values of x, y, and z; and
that to any set of values of x, y, and
2 there corresponds one, and only one,
point. These values of x, y, and z are
called the coordinates of the point, which
is expressed as P(x, y, z).
From the definition of x it follows
that x is equal, in magnitude and direc-
tion, to the distance of the point from the
coordinate plane YOZ. Similar meanings are evident for y
and z. It follows that a point may be plotted in several dif-
ferent ways by constructing in succession any three nonparallel
edges of the parallelepiped (fig. 186) beginning at the origin
and ending at the point.
In case the axes are not mutually perpendicular, we have a system of
oblique coBrdinates. In this case the planes are passed through the point
parallel to the coordinate planes. Then x gives the distance and the direc-
tion from the plane YOZ to the point, measured parallel to OX, and similar
meanings are assigned to y and 2. It follows that rectangular coordinates
are a special case of oblique coordinates.
141. Graphical representation of a function of two variables.
Let/(a^ y) be any function of two variables, and place
h/
y
,/
0
L ,
/
y
u
J
302 SPACE GEOMETRY
Then the locus of all points the coordinates of which satisfy (1)
is the graphical representation of the function f(x, y). To con-
struct this locus we may assign values to x and y, as x = xx and
y = yi> and compute from (1) the corresponding values of z.
There will be, in general, distinct values of 3, and if (1) defines
an algebraic function, their number will be finite. The corre-
sponding points all lie on a line parallel to OZ and intersecting
XOY at the point Px(x^ yt), and these points alone of this line
are points of the locus, and the portions of the line between
them do not belong to the locus. As different values are
assigned to x and y, new lines parallel to OZ are drawn on
which there are, in general, isolated points of the locus. It
follows that the . locus has extension in only two dimensions,
i.e. has no thickness, and is, accordingly, a surface. Therefore
the graphical representation of a function of two variables is a
surface.*
If /(#, y) is indeterminate for particular values of x and y,
the corresponding line parallel to OZ lies entirely on the locus.
Since the equations z =f(x, y) and F(x, y, z) = 0 are equiv-
alent, and their graphical representations are the same, it follows
that the locus of any single equation in x, y, and z is a surface.
There are apparent exceptions- to the above theorem if we demand that
the surface shall have real existence. Thus, for example,
x2 + y2 + z2 = — 1
is satisfied by no real values of the coordinates. It is convenient in such
cases however, to speak of "imaginary surfaces."
Moreover, it may happen that the real coordinates which satisfy the
equation give points which lie upon a certain line, or are even isolated
points. For example, the equation
x2 + y2 = 0
is satisfied in real coordinates only by the points (0, 0, z) which lie upon
the axis of 2; while the equation
x2 + y2 + z2 = 0
* It is to be noted that this method of graphically representing a function
cannot be extended to functions of more than two variables, since we have but
three dimensions in space.
CYLINDERS 303
is satisfied, as far as real points go, only by (0, 0, 0). In such cases it
is still convenient to speak of a surface as represented by the equation,
and to consider the part which may be actually constructed as the real
part of that surface. The imaginary part is considered as made up of
the points corresponding to sets of complex values of x, y, and z which
satisfy the equation.
142. Cylinders. If a given equation is of the form F(x, y) = 0,
involving only two of the coordinates, it might appear to rep-
resent a curve lying in the plane of those coordinates. But if
we are dealing with space of three dimensions, such an inter-
pretation would be incorrect, in that it amounts to restricting
« to the value s = 0, whereas, in fact, the value of z corre-
sponding to any simultaneous values of x and y satisfying
the equation F(x, y)=0 may be anything whatever. Hence,,
corresponding to every point of the curve F(x, y) = 0 in the
plane Z07, there is an entire straight line, parallel to OZ,
on the surface F(x, y) = 0. Such a surface is a cylinder,
i/ts directrix being the plane curve F(x, y) = 0 in the plane
» = 0, and its elements being parallel to OZ, the axis of the
coordinate not present.
For example, x2 + y2 = a2 is the equation of a circular cylinder,
its elements being parallel to OZ, and its directrix being the circle
zt?+ y*= a2 in the plane XOY.
In like manner z2 = ky is the equation of a parabolic cylinder,
its elements being parallel to OX, and its directrix being the
parabola z2=Jcy in the. plane ZOY.
If only one coordinate is present in the equation, the locus is a
number of planes. For example, the equation x2— (a+ V)x+ab= 0
xnay be written in the form (x— a)(x— #) = 0, which represents
the two planes x — a = 0 and x — b = 0. Similarly, any equation
involving only one coordinate determines values of that coordinate
only and the locus is a number of planes.
Regarding a plane as a cylinder of which the directrix is a
straight line, we may say that any equation not containing all
the coordinates represents a cylinder.
If the axes are oblique, the elements of the cylinders are not
perpendicular to the plane of the directrix.
304 SPACE GEOMETRY
143. Other surfaces. The surface represented by any equa-
tion F(z, y, z) = 0 may be studied by means of sections made
by planes parallel to the coordinate planes. If, for example,
we place z = 0 in the equation of any surface, the resulting
equation in x and y is evidently the equation of the plane
curve cut from the surface by the plane XOY. Again, if we
place z = zv where z is some fixed finite value, the resulting
equation in x and y is the equation of the plane curve cut
from the surface by a plane parallel to the plane XOY and zx
units distant from it, and referred to new axes O'X1 and OfYf,
which are the intersections of the plane z = zx with the planes
XOZ and YOZ respectively; for by placing z — zx instead of
z = 0, we have virtually transferred the plane XOY, parallel to
itself, through the distance z.
In applying this method it is advisable to find first the three
plane sections made by the coordinate planes x = 0, y = 0, 2 = 0.
These alone will sometimes give a general idea of the appearance
of the surface, but it is usually desirable to study other plane sec-
tions on account of the additional information that may be derived.
The following surfaces have been chosen for illustration because
it is important that the student should be familiar with them.
Ex. 1. Ax + By + Cz + D = 0.
Z
Placing z = 0, we have (fig. 187) |
Ax + By + D = 0. (1) AV
/! \f2)
Hence the plane XOY cuts this surface / i . x
/o\/ t\ X.
in a straight line. Placing y = 0 and then (3y ,| — ~~Z^\ — x'
x = 0, we find the sections of this surface 1/^ — \£) \
made by the planes ZOX and YOZ to be t/J /O ^^^
respectively the straight lines L^^*^
Ax + Cz + D=0, (2) r/
and By + Cz + D - 0. (3)
Fig. 187
Placing z = zv we have Ax + By + Czx + D = 0, (4)
which is the equation of a straight line in the plane z = zv The line (4)
is parallel to the line (1), since they make the angle tan_1(— -— J with the
parallel lines O'X' and OX and lie in parallel planes. To find the point
SURFACES
305
where (4) intersects the plane XOZ, we place y = 0, and the result
Ax + Czx + D = 0 shows that this point is a point of the line (2). This
result is true for all values of zv Hence this surface is the locus of a
straight line which moves along a fixed straight line always remaining
parallel to a given initial position ; hence it is a plane.
Since the equation Ax + By + Cz -f D= 0 is the most general equa-
tion of the first degree in the three coordinates, we have proved that
the locus of every linear equation in rectangular space coordinates is a plane.
Ex. 2. z = ax* + by2, where a > 0,
6>0.
Placing z = 0, we have
ax2 + by* = 0, (1)
and hence the XOY plane cuts the
surface in a point (fig. 188). Placing
y = 0, we have
z = ax2, (2)
which is the equation of a parabola
with its vertex at 0 and its axis along
0Z. Placing x = 0, we have
z = by2, (3)
which is also the equation of a pa-
rabola with its vertex at 0 and its
axis along OZ.
Placing z = zv where zx > 0, we
may write the resulting equation in
the form
n h
(4)
Fio. 188
°s» + V=l,
which is the equation of an ellipse with semiaxes
\i^^t
As the
plane recedes from the origin, i.e. as zx increases, it is evident that the
ellipse increases in magnitude. It is also evident that the ends of the axes
of the ellipse lie on the parabolas (2) and (3).
If we place z = — zv the result may be written in the form
— x*-r — y = — 1>
and hence there is no part of this surface on the negative side of the
plane XOY.
The surface is called an elliptic paraboloid, and evidently may be gener-
ated by moving an ellipse of variable magnitude always parallel to the
plane XOY, the ends of its axes always lying respectively on the parabolas
z = ax* and z = by2.
SPACE GEOMETRY
which is the equation
hyperbola with its tn .
axis along OX and ita conjugate
which is the equation of an hyperbola with its transverse axis along OY
and its conjugate axis along OZ.
It we place 8= ± zv and write the resulting equation in the form
>-M) »hi)
:e that the section is an ellipse with s
1 + ZL and )
which accordingly increases in magnitude as the cutting plane recedes from
the origin, and that the surface is symmetrical with respect to the plane
XOY, the result being independent of the sign of z,.
Accordingly this surface, called the ttnparted hyperbaloid or the hyper-
boloid of one sheet, may be generated by an ellipse of variable magnitude
moving always parallel to the plai
XOY and with the ends of its i
always lying on the hyperbolas
SUEFACES
307
„ , x2 . y2 , z2 -
Ex. 5. — + ?r + — = 1.
a
2
ft2
,2
This surface (fig. 191) is the
ellipsoid.
a1
ft2
,2
Fig. 191
This surface (fig. 192) is the
biparted hyperboloid or the hyper-
boloid of two sheets.
The discussions of the last three surfaces are very similar to that of
the unparted hyperboloid, and for
that reason they have been left to
the student.
Ex. 7. z = ax2 — by2, where a > 0,
b>0.
Placing z = 0, we obtain the equa-
ti0n ax2 - by2 = 0, (1)
i.e. two straight lines intersecting
at the origin (fig. 193). Placing
y = 0, we have
(2)
z = ax\
Fig. 192
the equation of a parabola with its
vertex at 0 and its axis along the positive direction of OZ.
Placing x = 0, we have
z = - by*, (3)
the equation of a parabola
with its vertex at 0 and
its axis along the negative
direction of OZ.
Placing x = ± xv we
have „ _
z = ax{ — by*,
or 3^ = - ~(z-axf), (4)
a parabola with its axis
parallel to OZ and its
% vertex at a distance ax*
from the plane XOY. It is evident, moreover, that the surface is sym-
metrical with respect to the plane YOZ, and that the vertices of
Fig. 193
308
SPACE GEOMETRY
these parabolas, as different values are assigned to xv all lie on th
parabola z = ax2.
Hence this surface may be generated by the parabola z = — by* movin
always parallel to the plane YOZ, its vertex lying on the parabola z = ax2.
The surface is called the hyperbolic paraboloid.
The reason for the name given to this surface becomes more evident i
two more sections are made.
Placing z = zv where zx > 0, we have z1 = ax2 — by2, or
.ij
-x2 y2 = l,
(5).
P)
an hyperbola with its transverse axis parallel to OX, the ends of the
transverse axis lying on the parabola z = ax2.
If z = — zv we may write the equation in the form
-y2 x* = 1,
(6)
)
an hyperbola with its transverse axis parallel to OY, the ends of the
transverse axis lying on the parabola z = — by2.
Ex. 8. z = kxy, where
k>0.
This surface is a
special case of the hyper-
bolic paraboloid of Ex. 7,
in which b— a. The proof
of this statement is as
follows :
If b = a, the equation
of the surface of Ex. 7 is
z = a(x2-y2). (1)
Fig. 194
Revolve the planes
XOZ and YOZ about
the axis OZ, which is held fixed, through an angle of — 45° into new
positions X'OZ and Y'OZ. By § 19, the formulas of transformation are
z = /,
* + }f
x^ r>
V2
y
V2
Substituting these values in (1), and simplifying, we have
which is the equation given above with k = 2 a.
(2)
SURFACES 309
The discussion of the plane sections of the surface (fig. 194) made by
the planes parallel to the coordinate planes is left to the student.
If b j* a, we can make a similar transformation by using the formulas
of § 21, and the result will be zf = kx^, only the coordinates will not be
rectangular.
144. Surfaces of revolution. If the sections of a surface
made by planes parallel to one of the coordinate planes are
circles with their centers on the axis of coordinates which is
perpendicular to the cutting planes, the surface is a surface of
revolution (§ 133) with that coordinate axis as the axis of
revolution. This will always occur when the equation of the
surface is in the form F(y/a? + y2, z) = 0, which means that the
two coordinates x and y enter only in the combination Vx2 + y2 ;
for if we place z = z1m this equation to find the corresponding
section, and solve the resulting equation for a^ + y2, we have,
as a result, the equation of one or more circles, according to
the number of roots of the equation in x2 + y2.
Again, if we place x = 0, we have the equation F(y, z) = 0,
"which is the equation of the generating curve in the plane
1TOZ. Similarly, if we place y = 0, we have F(x, z) = 0, which
is the equation of the generating curve in the plane XOZ. It
should be noted that the coordinate which appears uniquely in
the equation shows which axis of coordinates is the axis of
revolution.
X V z
Ex. 1. Show that the unparted hyperboloid — — j^ + — = 1 is a surface
of revolution.
Writing this equation in the form
a2 ft2
'we see that it is a surface of revolution with 0 Y as the axis.
x v
Placing 2 = 0, we have — • — ^-. = 1, an hyperbola, as the generating
a1 £r
curve. The hyperbola was revolved about its conjugate axis.
Conversely, if we have any plane curve F(x, z) = 0 in the
plane XOZ, the equation of the surface formed by revolving it
about OZ as an axis is FQs/x2 + y2, z) = 0, which is formed by
simply replacing x in the equation of the curve by y/x2 + y2.
AC
SPACE GEOMETRY
Ez. S. Find the equation of the spher
1 + z- = oa about OX as an axis.
formed by revolving the circle*
Replacing z by Vy* + a', we have as the equation
phere,
•■+/+«'=.-.
Z
This equation may also be
found directly from a figure.
Let P[ (fig. 195) be any point
of the circle, and let P be any
PkV\
point of the sphere, on the circle /
described by Px. Since P, is a /
_____
|i 1 \
point of the circle, /,-'
0~L' + LP? = a*. (1) L
iA'TJ-
But lp,=lp='>Slm±+mF*- \
"'-•—.. /
'JL-^TI
Substituting this value of LP1 \
in (1), we have
\ Y
0~L'' + LMa + MP3 = a3,
or x1* + f + z1 = a\
as the equation of the sphere.
F10. 195
145. Projection. The projection of a point on a straight line
is denned as the point of intersection of the line and a plane
through the point perpendicular to the line. Hence, in fig. 186,
L, M, and N are the projections of the point P on the axes
of x, y, and z respectively. a
The projection of one straight
line of finite length upon a second
straight line is the part of the
second line included between the
projections of the ends of the first
line, its direction being from the
projection of the initial point of
the first line to the projection of
the terminal point of the first
line. In fig. 196, for example,
the projections of A and B on MN being A' and B' respectively,
the projection of AB on MN is A'B\ and the projection of BA
on MN is B'A'. If MN and AB denote the positive directions
respectively of these lines, it follows that A'B' is positive when
E^S
PROJECTION
311
it has the same direction as MN, and is negative when it has
the opposite direction to MN.
In particular, the projection on OX of the straight line F^
drawn from I[(xv yv z^) to ij(#2, y2, s2) is LXL2, where OLx = x
and OL2=x2. But L1L2=x2 — xx, by § 3. Hence the projection
of PXP2 on OX is x2 — xx; and, similarly, its projections on OF
and OZ are respectively y2 — yx and z2— z±m
If we define the angle between any two lines in space as the
angle between lines parallel to them and drawn from a common
point, then the projection of one straight line on a second is the
product of the length of the first line and the cosine of the angle
between the positive directions of the two lines. Then, if <f> is the
angle between AB and MN (fig. 196),
A'B' = AB cos <f>.
To prove this proposition, draw AlC parallel to AB and meet-
ing the plane ST at C. Then A'C = AB, and A' B' = A'C cos fa
by § 2, whence the truth
of the proposition is
evident.
Defining the projec-
tion of a broken line
upon a straight line as
the sum of the projec-
tions of its segments, we
may prove, as in § 2,
that the projections on
any straight line of a
Woken line and the straight line joining its ends are the same.
We will now show that the projection of any plane area upon
znother plane is the product of that area and the cosine of the
zngle between the planes.
Let X'OY (fig. 197) be any plane through OY making an
ingle <j> with the plane XOY. Let A'B1 be any area in X'OY
such that any straight line parallel to OX' intersects its
boundary in not more than two points, and let AB be its
projection on XOY.
Fig. 197
312 SPACE GEOMETRY
Then (§ 125) area AfBf = f \x'2 - x[) dy, (1)
the limits of integration being taken so as to include the
whole area.
In like manner, area AB = I (x2 — x^) dy, (2)
the limits of integration being taken so as to cover the whole area.
But the values of y are the same in both planes, since they are
measured parallel to the line of intersection of the two planes;
and hence the limits in (1) and (2) are the same. Since the x
coordinate is measured perpendicular to the line of intersection,
x2 = x'2 cos <£, x1 = x[ cos <f>, and (2) becomes
area AB = I (x'2 — #{) cos <f> dy
= cos <f> I (#2 — xi) dy
= (cos <j>) (area ArBf>).
146. Components of a directed straight line. Let PXP2 (fig. 198)
be a straight line, the direction of which is from ij to P2. Through
ij and ij pass planes parallel respectively
to the coordinate planes, thereby forming
on im as a diagonal a rectangular paral-
lelepiped with its edges parallel to the
coordinate axes. The lines PXQ, I^B, and
P^ considered with respect to both length
and direction, are called the components of
PXP2. It is evident that they are the projec- f Fig. 198
tions of P^ on OX, OY9 and OZ respectively.
Conversely, the components of a straight line will determine
its direction and length, but not its position ; for if the compo-
nents are given equal to a, 6, and e, we may lay off, from any
point ij, a straight line parallel to OX and equal to a in length,
a straight line parallel to OF and equal to b in length, and a
straight line parallel to OZ and equal to c in length. These
three lines determine the edges of a rectangular parallelepiped,
and hence determine the diagonal drawn from ij. That is, if J^Q
(fig. 198) is laid off equal to a, PXR equal to J, and I[S equal
DISTANCE 313
to e, the rectangular parallelepiped is determined, and hence the
diagonal I[P2 is determined in both length and direction.
It is evident that the direction of I[P2 will not be changed if
a, 6, and c are multiplied by the same number ; in other words,
the ratios of the components are the essential elements in fixing
the direction of the line. We shall accordingly speak of a
straight line as having the direction a: b: c.
On the other hand, the length of the line does depend upon
the values of a, 6, and c, for
PXP2 =V fPxQ2 +PXR2 +P1S* =Va2 + b2 + <?. (1)
147. Distance between two points. An important application
of (1), § 146, is in finding the distance between two points
ij^, y2, zj and P2(x2, y2, z2). Referring to fig. 198, we have,
by §145, a=P1Q=x2-x1, b=P1R=y2-y1, c = P1S = z2-zl;
whence ^ = V(*2 - Xiy + (y2 _ yi)» + (*2 - rf. (1)
Ex. 1. Find the length of the straight line joining the points (1, 2,
— 1) and (3, - 1, 3).
The required length is
V(3 - 1)2 + (- 1 - 2)2 + (3 + l)2 =V29.
Ex. 2. Find a point Vl4 units distant from each of the three points
CI, 0,3), (2, -1,1), (3,1,2).
Let P(x, y, z) be the required point.
Then (x - 1)2 + (y - 0)2 + (z - 3)2 = 14,
(x - 2)2 + (y + l)2 + (z - 1)2 = 14,
(x - 3)2 + O - l)2 + (z - 2)2 = 14.
Solving these three equations, we determine the two points (0, 2, 0) and
(4, - 2, 4).
Ex. 3. Find the equation of a sphere of radius r with its center at
**i(*v Vvzi>
If P (x, y, z) is any point of the sphere,
(x - xxy + (y - yxy + (z - zxy = r2.
Conversely, if P(x, y, z) is any point the coordinates of which satisfy
"this equation, P is at the distance r from Pv and hence is a point of the
sphere. Therefore this is the required equation of the sphere.
314
SPACE GEOMETRY
148. Direction cosines. If we denote by a, fi9 and 7 the angles
which a straight line makes with the positive directions of the
coordinate axes OX, O F, and OZ respectively, the cosines of
these angles, i.e. cos a, cos/3, cosy are called the direction cosirm
of the line.
If the line is drawn through the origin, as in fig. 199, it is
evident that the same straight line makes the angles a, /8, 7 or
7T — a, 7T— y8, 7T— 7 with the coordinate
axes, according to the direction in which
the line is drawn. Hence its direction
cosines are either cos a, cos/8, cos 7 or
— cos a, — cos& —cos 7. Hence the
straight line can have but one set of
direction cosines after its direction has
been choseni
The direction cosines may be deter-
mined directly from the components of the line; for, referring
to fig. 198, we see that
Fig. 199
cos* = 7hr cob£ = -£-, 0087 = ^-
Jrxjr2 jrxjr2 jtxj^
(i)
or
cosa =
cos 7 =
a
Vaf + ^ + ^
» cpsy8 =
Va2 + b* + r2
Va" + P + *>
(2)
Squaring and adding equations (2), we have
cos2<z + cos2j3 + cos27 = 1 ;
(3)
that is, the sum of the squares of the direction cosines of any
straight line is always equal to unity.
It follows that the direction cosines of any line are not
independent quantities.
Ex. Since the length of the line of Ex. 1, § 147, is V29, and its respec-
tive components are 2,-3, and 4, it follows that its direction cosines are
2 8_ _4_
V29 ' V29 ' V29
ANGLE 315
149. Angle between two straight lines. Given two straight
lines having the respective directions ax : bx : c and a2 : b2 : c .
If they are drawn from a common point P (#, y, z) (fig. 200),
let the segment of the first line extend to 7J and the segment of
the second line extend to i£, so that the coordinates of ij are
sc + av y + bx, and z -f c^ and the coordinates
of 1% are x + a2, y + J2, and z + c2. It follows
-fcliat the components of ij^ are a2— a , b2 — b^
and c — c .
"""■^ 2 1
Then if 0 is the angle between these two
lines, we have, by trigonometry,
e^PP* + PP2*-I&
cos.
2PP1-PP2
:2
But PP^ = a* + b? + <?*,
Pl£=al + bl + &
k2
Fig. 200
Ptf = («f - ax)2 + (bt - b^ + 02 - ^
"^rhence, by substitution in (1) and simplification,
on* fi = *A + *A + C*C* - . (2)
,2
'2
If cos a1<t cos ^ cos y1 are the direction cosines of PPV and cos «2,
<3os #2, cos 72 are the direction cosines of PP2, formula (2) may
Joe written, by (2), § 148,
cos 0 = cos a1 cos a2 + cos j3x cos y82 + cos yx cos y2. (3)
If the lines are perpendicular to each other, cos 6 = 0, and
(2) and (3) reduce respectively to
aA+*A+*A=0 (4)
and cos ax cos <*2 + cos fix cos £2 + cos yx cos y2 = 0. (5)
If the lines are parallel to each other,
cos ax = cos <x2, cos &x = cos £2, cos yx = cos y2,
whence it is easily shown that
.?! = *! = fi. (6)
a2 6.2 <?2
316 SPACE GEOMETRY
150. Direction of the normal to a plane. Let Px (xv yv zx) an
■^ (xv Vv zz) ^ any two points of the plane
Ax + By+Cz + D=0.
Substituting their coordinates in (1), we have
Axx + 3^+ Czx + D ^ 0
and Ax2 + By2+ Cz2 + 1) = 0.
Subtracting (2) from (3), we have
whence, by (4), § 149, the direction A:B:C is normal to th
direction x2— xxi y2 — yx\ z2— zx. But the latter direction is th
direction of any straight line of the plane. Hence the direc-
tion A: B: C is the direction of the normal to the plane Ax + B\
+ Cz + D = 0.
151. Equation of a plane through a given point perpendicular"^^
to a given direction. Let the plane pass through a given point
^(x1<t yv zx) perpendicular to a straight line having a given
direction A:B: C. Let P(#, y, z) be any point of the plane.
Then x — xx: y — y1: z — zx is the direction of I\P^ Le. is the
direction of any straight line through ij in the plane.
Since a perpendicular to a plane is perpendicular to every
line in the plane, it follows that
A(z-zJ + B(y-yJ+C(z-*J=0,
which is, accordingly, the required equation of the plane.
Since every plane may be determined in this way, and this
equation is a linear equation, it follows that every plane may
be represented by a linear equation.
Ex. Find the equation of a plane passing through the point (1, 2, 1)
and normal to the straight line having the direction 2 : 3 : — 1.
The equation is
2(x-l) + 3(y-2)-l(2-l) = 0,
or 2ar + 3y — z - 7 = 0.
152. Angle between two planes. Let the two planes be
A? + Bxy + Cxz + Dx = 0, (1)
4* + Bjt+Cf + D% = 0. (2)
STRAIGHT LINE 317
The angle between these planes is the same as the angle
Tbetween their respective normals, the directions of which are
^respectively the directions Ax : Bx : Cx and A2 : B2 : C2. Hence, if
^9 is the angle between the two planes, by (2), § 149,
coofl= A1A* + B1B%+C1Cl
y/A? + B* + C* ^A* + B* + Cx
2
2
The conditions for perpendicularity and parallelism of the
planes are respectively
and
A2 B2 C2
153. Equations of a straight line. In space of three dimen-
sions a single equation in general represents a surface ; hence, in
general, a curve cannot be represented by a single equation. A curve
may, however, be regarded as the line of intersection of two
surfaces. Then the coordinates of every point of the curve*
satisfy the equations of the surfaces simultaneously; and, con-
versely, any point the coordinates of which satisfy the equations
of thS surfaces simultaneously is in their curve of intersection.
Hence, in general, the locus of two simultaneous equations in x, y,
and z is a curve.
In particular, the locus of the two simultaneous linear equations
Axx + Bxy + Cxz+DX=Q,
A2x + B^ + C2z+D = 0,
is a straight line, since it is the line of intersection of the two
planes respectively represented by the two equations.
We will now find the equations of the straight line determined
by two points, and the equations of the straight line passing through
a known point in a given direction,
154. Straight line determined by two points. Let the given
points be I((xv yv ^) and J%(x2, y2, z2). Then the direction of
ijj£ is x2— Xi-y^—y^- z2— zv Let P(x, y, z) be any point of the
line. Then the direction of IfP is x — xx : y — y1 : z — zx.
318
SPACE GEOMETRY
*■-*!
Since I(P and 1(1% are parts of the same straight line, and
hence parallel, it follows that
x - xi = V - V\ _ z - gi
Here are but two independent equations in x, yy and z. This
result proves the converse of the statement above, that two
linear equations always represent a straight line ; for we have
any straight line represented by two linear equations.
It is to be noted that, if in the formation of these fractions
any denominator is zero, the corresponding component is zero,
and the line is perpendicular to the corresponding axis.
Ex. Find the equations of the straight line determined by the points
(1, 5, - 1) and (2, - 3, - 1).
x — 1_ y — 5 _ z + 1
2-l~-3-5~-l + r
Hence the two equations of the line are 2 + 1 = 0, since the line is par-
allel to the XOY plane and passes through a point for which z = — 1, and
8 x + y — 13 = 0, formed by equating the first two fractions.
155. Straight line passing through a known point in a given
direction. If the direction of the line is given as a:b:Cj the
equations of the line are evidently
x - xi = V ~ V\ = z - zi
a b c
(i)
Q
v-'-
Q
R
Fig. 201
for in the formula of § 154 we may place
If the direction of the line is given in
terms of its direction cosines, the derivation
of the equations is as follows:
Let Px(xv yv z^) (fig. 201) be a known point of the line, and
let Z, m, and n be its direction cosines. Let P(x, y, z) be any
point of the line. On I^P as a diagonal construct a parallele-
piped as in § 146. Then if we denote J^P by r, we have
%Q = lr, PXB = mr, %S= nr.
But I>Q=x-xv PlR=y-yv %S=z-zv
whence x — xx=lr, y — y^mr, z — z^nr.
STRAIGHT LIKE 319
^Eliminating r from these last three equations, we have
x-x^y-y^z-z^
I m n
■^rhich are but two independent linear equations.
156. Determination of the direction cosines of a straight line.
ZKi the equations of the straight line are in any one of the forms
of §§154 and 155, the determination of the direction cosines is
-"very easy, for the denominators of the fractions in those formu-
las are either the direction cosines of the line or else give the
components for the line, from which the direction cosines are
<juickly computed.
If the equations of the straight line, however, are in any
other form, as . ^ p , „ ^n n n.
Aj* + Bj + Cf+DM = 0, (2)
■
let its direction cosines be Z, tw, and n. Since the line lies in
both planes (1) and (2), it is perpendicular to the normal to
each. Therefore, by (4), § 149,
Axl -f Bxm + Cxn = 0,
A2l + B2m + C2n=0;
also I2 + m* + n2 = 1. (§ 148)
Here are three equations from which the values of /, m, and n
may be found.
Ex. Find the direction cosines of the straight line 2x + Sy + z — 4 = 0,
4* + y-z + 7=0.
The three equations for Z, m, and n are
2 I + 3 m + n = 0,
4Z + 771— n = 0,
P + m2 + n2 = 1,
the solutions of which are
2 3 *
1 = — ==» m = = » n =
V38 V38 V38
2 3 5
or I = = > m = — -==. j n = —
V38 V38 V38
320
SPACE GEOMETRY
Since cos (180° — <£) = — cos <£, it is evident that if the angles corre-
sponding to the first solution are C4, (&v yv the angles corresponding to
the second solution are 180° — av 180° — f}v 180° — yr Since these two
directions are each the negative of the other, it is sufficient to take either
solution and ignpre the other.
157. Distance of a point from a plane. Let it be required
to find the perpendicular distance from the point Px(x^ yl9 «x)
to the plane
Ax+By + Cz+D = 0. (1)
From % (fig. 202) draw the
required perpendicular PXN
and also a line parallel to
the axis of z and let it cut
the plane in E. Then for the
point B, x = x±J y = yv and z
is determined from the equa-
tion of the plane as
__ — Ax1 — By1 — D
~ C
z =
Fig. 202
Hence
£3 = *!-
— Ax1 — Byx — D
C
_Ax1+By1 + Czl+D
C
But PXN=RPX cos 7 where 7 is the angle RPXN which is equal
to the angle made by the normal to the plane with the line OZ.
Then, by § 150,
cos7=±
C
Hence
V^ + ^ + C2
PN=jLAxl+By1 + Cz1+D
1 \/A*+&+&
is the magnitude of the required distance, being positive for all
points on one side of the plane and negative for all points on
the other side. If we choose, we may take the sign of the
radical always positive, in which case we can determine for
which side of the plane the above result is positive by testing
for some one point, preferably the origin.
STRAIGHT LINE 321
Ex. Find the distance of the point (1, 2, 1) from the plane 2 x — 3 y
+ 6 2 + 14 = 0. The required distance is
2(l)-3(2)+6(l) + 14_2?
Furthermore the point is on the same side of the plane as the origin, for
if (0, 0, 0) had been substituted, the result would have been 2, i.e. of the
same sign as 2$.
158. Problems on the plane and the straight line. In this
article we shall solve some problems illustrating the use of
the equations of the plane and the straight line.
1. Plane through a given line and subject to one other condition.
Let the given line be
Axx + Bxy + (7^ + ^=0, (1)
A2x + Bj, + C2z+D=0. (2)
Multiplying the left-hand members of (1) and (2) by hx and k2
respectively, where kx and k2 are any two quantities independent
of #, y, and 3, and placing the sum of these products equal to
zero, we have the equation
Jc1(A1z+B1y+ C1z+DJ + k2(A2x+Bj,+ C2z + D2)=0. (3)
Equation (3) is the equation of a plane, since it is a linear equa-
tion, and furthermore it passes through the given straight line,
since the coordinates of every point of that line satisfy (3) by
virtue of (1) and (2). Hence (3) is the required plane, and
it may be made to satisfy another condition by determining the
values of kx and k2 appropriately.
Ex. 1. Find the equation of the plane determined by the point (0, 1, 0)
and the line 4s + 3y + 2z-4 = 0, 2 a: - 11 y - 4 z - 12 - 0.
The equation of the required plane may be written
^(4 s + 3y + 2z-4) + k2(2 x - 11 y - 4 z - 12) = 0. (1)
Since (0, 1, 0) is a point of this plane, its coordinates satisfy (1), and
hence i^ + 23 k2 = 0, or kx =- 23 kv
Substituting this value of kx in (1), and reducing, we have as the required
equation, 9z + 8y + 5z- 8 = 0.
322 SPACE GEOMETRY
Ex. 2. Find the equation of the plane passing through the line 4 x + 3 y
+ 2 z — 4 = 0, 2x — 11 y — 4 z — 12 = 0, and perpendicular to. the plane
2s + y-2z + l = 0.
The equation of the required plane may be written
^(4 z + 3y + 2z-4) + k2(2 x - lly - 4 z - 12) = 0, (1)
or (4ifc1 + 2ifc2)a: + (3^1-ll^2)y + (2^1-4^2)z + (-4^1-12^2) = 0.
Since this plane is to be perpendicular to the plane 2 a: + y — 2 z + 1 = 0,
2(4 kx + 2 fc2) + 1(3 kx - llfc2) - 2(2 *x - 4 *2) = 0,
whence k2 = — 7 ifcr
Substituting this value of £2 in (1), and reducing, we have as the required
equation, * - 8y- 3z- 8 = 0.
2. Plane determined by three points. If the equations of the
straight line determined by two of the points are derived, we
may then pass a plane through that line and the third point, as
in Ex. 1. The result is evidently the required plane.
Ex. 3. Find the equation of the plane determined by the three points
(1, 1, 1), (- 1, 1, 2), and (2, - 3, - 1).
The equations of the straight line determined by the first two points are
x — 1 _y — 1_2"~1
-1-1 1-1 2-1
which reduce toy — 1 = 0, # + 2z — 3 = 0.
The equation of the required plane is now written in the form
kx (y - 1 ) + k2 (x + 2 z - 3 ) = 0 .
Substituting (2, — 3, — 1) in this equation, we have
— 4 kx — 3 k2 = 0, or k2 = — J kv
Substituting this value of k2 in the equation of the plane, and simplify-
ing, we have as our required equation,
4z-3y + 8z-9 = 0.
159. Space curves. We saw in § 153 that, in general, the
locus of two simultaneous equations in x, y, and z is a curve —
CURVES 328
the curve of intersection of the surfaces represented by the
equations taken independently.
Let fx(x,y, *)=<>, /f(^y,*)=0, (1)
be the two equations of a space curve.
If we assign a value to one of the coordinates in equations
(1), as x for example, there are two equations from which to
determine the corresponding values of y and z, in general a
determinate problem. But if values are assigned to two of the
coordinates, as x and y, there are two equations from which to
determine a single unknown, zy a problem generally impossible.
Hence there is only one independent variable in the equations
of a curve.
In general, we may make x the independent variable and
place the equations in the form
y = <kO)> * = *,<»> (2)
by solving the original equations (1) of the curve for y and z
in terms of x. The new surfaces, y = <j>l(xyj z = <f>2(jx), deter-
mining the curve, are cylinders (§ 142), with elements parallel
to OZ and 0 Y respectively. The equation y = <j>x(x) interpreted
in the plane XO Y is the equation of the projection (§ 145) of
the curve on that plane. Similarly, the equation z = <j>%2(x),
interpreted in the plane ZOX, is the equation of the projection
of the curve on that plane.
Hence, to find the projection of the curve (1) on the XOY plane
we eliminate z from the two equations.
Similarly, to find the projection on the XOZ plane we eliminate
y, and to find the projection on the YOZ plane we eliminate x.
Finally, the three equations
are parametric equations of a curve. They may generally be
put in the form y = <f>1 (#), z = ^2 0O» ^y eliminating t from
the first and second equations, and from the first and third
equations.
324
SPACE GEOMETRY
Ex. The space curve called the helix is the path of a point which moves
around the surface of a right circular cylinder with a constant angular
velocity and at the same time moves parallel to the axis of the cylinder
with a constant linear velocity.'
Let the radius of the cylinder (fig. 203) be a, and let its axis coincide
with OZ. Let the constant angular velocity be o> and the constan
linear velocity be v. Then if 0 denotes the angle through which the
plane ZOP has swung from its initial position ZOX, the coordina
of any point P (x, y, z) of the helix are given by the equations
x = a cos 6,
y = a sin 0,
z = vt.
But 6 = (ot, and accordingly we may have as the
parametric equations of the helix,
x — a cos col,
y = a sin tot,
z = vt,
t being the variable parameter.
Or, since t = — > we may regard $ as the
Co
variable parameter, and the equations are
x = a cos 6,
y = a sin 0,
z = k$,
v
where k is the constant — •
ci)
Fig. 203
160. Direction of space curve and element of arc. Let PQc, yy z)
be any point of a curve, and Q (x + Arc, y + Ay, z + Az) be any
second point of the curve. Then the direction cosines of the
chord PQ are
Ax
Ay
Az
v Arc2 + Ay2 + Az2 v Arc2 + Ay2 + Az2 v Arc
+ Ay +Az
As the point Q approaches the point P along the curve,
Arc, Ay, and Az each approach zero as a limit, and the direction
cosines of the chord PQ approach the direction cosines of the
tangent to the curve at P as limits. To determine these limits,
denote by 8 the distance of the point P from some fixed point
CURVES 325
of the curve, 8 being measured along the curve. Then the
arc PQ = As; and A« = 0 as PQ approaches the tangent.
XT Ax Ax As
Now
y/Az*+Ay2 + Az2 As Va^+aT+A?'
Ax dx
whence Lim
T^2 ds
**^y/Ax2 + Ay2+Az
for Lim = 1.
v Ax*+Ay2+Az2
Proceeding in the same way with the other two ratios, we
dx dti dz
have — , -~, — as the direction cosines of the* curve at any
ds as as
point, since the directions of the tangent and the curve at any
point are the same.
whence ds = y/dx2 + dy1 + dz2, (1 )
a formula for the differential of the arc of any space curve.
It also follows from (1) that we may speak of the direction
of the curve as the direction dx: dy : dz.
Ex. 1. Find the direction of the helix
x = a cos 0, y — a sin 0, z — kO,
at the point for which 0 = 0.
Here dx = — a sin 0 dO, dy= a cos 0 dO, dz = k dd. Therefore, at the point
for which 0=0, the direction is the direction 0 : add : kdO, and the direction
rt a k
cosines are 0,
Va^ + k2 VV + fc2
Ex. 2. Find the length of an arc of the helix corresponding to an
increase of 2 tr in 0.
Using the values of dx, dy, and dz found in Ex. 1, we have
whence 8=1 Va2 + k*d0
= 2irV^2~+*2.
<fe = Va2+ P</0;
AC
326 SPACE GEOMETEY
161. Tangent line and normal plane. If %(xvyv z^) is thepoiix_ — t
of tangency, the equations of the tangent line are, by (1), § 15£ d,
x-x1^y-y1^z-zl^ r~^m\
dxx dyx dzx ^ '
where dxx, dyx, dzx are the respective values of dx, dy, dz at 1
point Py
The plane perpendicular to a tangent line at the point
tangency is called the normal plane to the curve.
By § 151, the equation of the normal plane to the curve at
dxx(x - xx) + dyx(y - yx) + dzx(z - zj = 0.
Ex. Find the equations of the tangent line and the equation of
normal plane to the helix
x = a cos 6, y — a sin 6, z — kO
at the point for which 0=0. Here xx = a, y± = 0, zx = 0, and dxx
dyt = add, dzx = kdO. Hence the equations of the tangent line are
x — a __y — Q _z — 0
0 ~ add ~ kdO '
which reduce to x — a = 0, ley — az = 0.
The equation of the normal plane is
0(x-a) + ad$(y - 0) + kdd(z - 0) = 0,
which reduces to ay + kz = 0.
PROBLEMS
1. Describe the surface y2 — 4y — 2# = 0.
2. Describe the surface y(z — 1)= 1.
3. Write down the equation of a right circular cylinder of radius
4. Show that the surface ax -f- by = cz2 is a cylinder, and descri
its directrix and generatrix.
5. Describe the surface x2 + y2 + z2 — 6 z = 0.
6. Describe the surface xyz — a8.
7. Describe the surface 9 a? + 4 s2 =12y — 24.
8. Describe the locus of the equation x — (z + 2)a = 0.
PROBLEMS 327
9. Show that the surface (ax.+ by)2 = cz is a cylinder, and
lescribe its directrix* and generatrix.
10. Describe the surface 36 x2 - 72 x + 9 y2 + 4 s2 == 0.
11. All sections of a given right cylinder made by planes parallel
*
X) the plane XOZ are ellipses of which the longest chord is 10 in.
ind the shortest is 8 in. What is the equation of the cylinder ?
12. Describe the surface x* + y* + z* = a*
13. Show that the surface z = a — Va^-f-y2 is a cone of revolution,
ind find its vertex and axis.
14. Find the equation of a 'prolate spheroid, i.e. the surface gen-
erated by revolving an ellipse about its major axis.
15. Find the equation of an oblate spheroid, i.e. the surface gen-
erated by revolving an ellipse about its minor axis.
16. Describe the locus of the equation x2 + 4 xy + 4 y3 — 4 z2 = 0.
8 a8
17. Describe the surface z = -5— — Q , . — 5-
r + t/2 + 4a2
18. Find the equation of the cone of revolution formed by revolving
the line z = 2 x about OX as an axis.
19. Describe the locus of the equation 2a3a — Sx — 2 = 0.
20. Find the equation of a parabolic cylinder the elements of which
are parallel to OX and the directrix of which is in the plane YOZ.
X2 V* z*
21. Describe the surface —z + 7^ + —. = 1.
a2 b2 c*
22. Describe the surface 3/* — (2 a — y) (z2 + x2) = 0.
23. Describe the surface x2 — y3 — 2# + 4y = 0.
24. Find the equation of the cone of revolution formed by revolv-
ing the line 3y = 2x -f-1 about the line y = 1 in the plane XOY as
an axis. What are the coordinates of the vertex of the cone ?
25. Show that the surface x2+ 2tf- 3z2+2x-12y + 12z + 7 = 0
is a cone with its vertex at the point (— 1, 3, 2). What are its cross
sections made by planes parallel to the plane XOY?
26. Describe the surface (x — a)x2 + (x + a) (y* + z2) = 0.
27. Find the equation of the ring surface formed by revolving the
ellipse - — Ia~ + ~£ = 1 (a > &) about OF as an axis.
328 SPACE GEOMETRY
28. Describe the surface 4«2 = ^(9 — sc2).
29. If P(x, y} z) is situated on the straight line drawn fronv
pi(xv Vv *i) t0 p*(x» Vv *a) so that pip = k(PiP*)> Prove that
x = xx + k(x2 -xj, y = yx + k(y2 - yx), z = zx + k(z2 - sj.
30. Prove that p(^y^, ^±J^, ^^) is the middle point of
the straight line joining P1(xv yv zj and P2(x2, y2) z2).
31. Find the equation of the sphere constructed on the straight
line joining (3, — 1, 3) and (5, 3, 5) as a diameter.
32. Find a point of the plane x + 3y + z = 0 equally distant
from the three points (1, 1, 1), (0, 2, 1), (2, 1, 2).
33. Find the points distant 5 from the points (— 2, — 2, 1),
(3, - 2, 6), (3, 3, 1).
34. Find the point of the plane x + 2y + 3z — 6 = 0 equally
distant from the points where the plane is pierced by the three
coordinate axes.
35. Find the equation of the sphere passing through the points
(- 1, 1, - 5), (- 2, 4, 3), (- 5, 0, - 2), (7, 1, - 1).
36. A point moves so that its distances from two fixed points are
in the ratio k. Prove that its locus is a sphere or a plane according
as k =£ 1 or k = 1.
37. Prove that the locus of points from which tangents of equal
length can be drawn to two given spheres is a plane perpendicular
to their line of centers.
38. A straight line makes the same angle with the three coordi-
nate axes. What is that angle ?
39. Prove that a straight line can make angles 60°, 45°, 60°
respectively with the coordinate axes.
40. Find the direction cosines of the straight line determined by
the points (1, 3, 5), (2, - 1, 4).
41. A straight line makes an angle of 30° with OX and equal
angles with O Y and OZ. What is its direction ?
42. Find the angle between the two straight lines joining the
origin to the points (1, 2, 1) and (3, — 1, 3).
PROBLEMS 329
43. Prove that the three points (5, 3, - 2), (4, 1, - 1), (2, - 3, 1)
lie on one straight line.
44. Through the point (1, — 3, 1) of the straight line having the
direction 1:2:3a straight line is drawn to the point (4, 2, 0). Find
the angle between the two lines.
45. Prove that Px(-1, 2, 1), P2(2, 3, 5), and P8(4, 5, 3) are the
vertices of a right triangle.
46. Find the equation of a plane passing through the point
(— 2, 3, — 4) parallel to the plane x — oy + 7 z — 11 = 0.
47. Find the equation of a plane passing through the point
(5, — 2, 7) equally inclined to the three coordinate axes.
48. Find the equation of a plane perpendicular to the straight
line joining the points (1, 3, 5) and (4, 3, 2) at its middle point.
49. Find the equation of a plane passing through the point
(1, 1, 2) perpendicular to the straight line determined by the points
(1, - 1, 1) and (3, 1, 3).
50. What is the angle between the planes 2x + y — 72 + 11 = 0,
5x-2y + 5z-12 = 0?
51. Find the angle between the planes 3x + 2y — 4 = 0, 2y + 3z
+ 13 = 0.
52. Find the equations of the straight line determined by the
.points (6, 2, —1) and (3, 4, — 4).
53. What are the equations of the straight line determined by
the points (2, 3, 5) and (1, - 1, 5) ?
54. Find the equations of a straight line passing through the
point (0, 3, 5) perpendicular to the plane x + 3y + 5z — 9 = 0.
55. A straight line is drawn through the point (4, 6, — 2) parallel
to the straight line drawn from the origin to the point (1, — 5, 3).
What are its equations ?
56. A straight line making angles 60°, 45°, and 60° respectively
with the axes of x, y, and z passes through the point (2, — 2, 2).
What are its equations ?
57. A straight line passes through the point (2, — 5, 2) parallel
to OF. What are its equations ?
58. Find the direction cosines of the line 4<c — 3^ — 4 = 0,
12* -3* -15 = 0.
330 SPACE GEOMETEY
59. Find the direction cosines of the line 3se + y — 1 z — 6 = O,
2x-Sy + 4«-7=0.
60. Find the equations of the straight line passing through*.
(1, 3, — 5) parallel to the line y = Sx — 14, 7x — 2z = 17.
61. Prove that the three planes x — 2 y + 1 = 0, 1 y — z — 4 = 0,
7se — 23 + 6=0 are the lateral faces of a triangular prism.
62. Find the angle between the line Sx — 2y — 4 = 0, y + 3 s?
+ 5 = 0 and the plane Sx + y — 2z + 31= 0.
63. Find the distance of the plane 2x + 3«+ll=0 from th^
origin.
64. Find the locus of points distant 3 from the plane x + y + z^
+ 3 = 0.
65. Find the locus of points equally distant from the planes-
s + 2y + 3* + 4 = 0, x-2y + Sz-5 = 0.
66. Find a point on the line Sx — 2y — 11 =0, 2x — y— z — 5 = 0
equally distant from the points (0, 1, 1) and (1, 2, 1).
67. Find the equation of the plane passing through the point
(2, —3,-2) perpendicular to the line 2x + y — 5z — 7=0, y + 2 s
-4 = 0.
68. Find the equation of a plane four units distant from the
origin and perpendicular to the straight line through the origin and
(1, - 5, 6).
69. A straight line is drawn from the origin to the plane 2x + y
+ 2 z — 5 = 0. It makes equal angles with the three coordinate axes.
Find its length.
70. Find the coordinates of a point on the straight line determined
by (- 1, 0, 1) and (1, 2, 3) and 3 units distant from (2, - 1, 1).
71. Find the foot of the perpendicular drawn from (3, — 2, 0) to
the plane 2x + y — 4 2 + 17 = 0.
72. Find the length of the projection of the straight line joining
the points (1, 2, 1) and (2, — 1, 2) upon the straight line determined
by the points (2, 1, 3) and (4, 4, 6).
*
73. Find the equation of the plane determined by the three points
(1, 3, - 2), (0, 2, -10), and (- 2, 4, - 6).
74. Find the direction of the normal to the plane determined by
the three points (1, 2, 3), (-1, - 2, - 3), (4, - 2, 4).
PROBLEMS 331
75. Find the point of intersection of the lines
fx + 2y-3 = 0 1 (3y + 5z + 15 = 01
\x + y-2z-9 = 0} l3a-2*-15 = 0J
76. Prove that the lines
fx-2y + 3 = 0 1 C7x + y-9 = 0 1
\2a-2y-* + 3 = 0J l5x-2*-3 = 0J
intersect at right angles.
77. Prove that the two lines
fa? + 2y -3 + 7 = 0 1 Ux-7y + 8z + 19 = 01
\2x-y + 2* + 11 = 0J \x-3y + 3* + 4 = 0 J
are coincident.
78. Prove that the two lines
f3a-2y-7=0\ fx-z = 0 1
\2y- 3* + 7 = 0J 13* -4^ + 3*- 8 = 0J
can determine a plane, and derive its equation.
79. Prove that the two lines
f2z + 3y + 4 = 0l (x + 2y + z + 2 = 0 1
\2y + 2 + 3 = 0 J l2x-y + 2*-9 = 0J
cannot determine a plane.
80. Prove that the two lines
rx _ 2^-10 = 01 r7z-3*-ll = 0 1
U2/_^ + 17=OJ \7x + 14^-* + 43 = 0J
can determine a plane, and derive its equation.
81. Find the equation of a plane passing through the line
x + y + 3z — 7=0, 3x + 2y — z = 0 and perpendicular to the
plane 2x + y — 2 s + 11 = 0.
82. Find the equation of a plane passing through the points
(— 2, 3, — 2), (2, — 1, 2) perpendicular to the straight line deter-
mined by the points (0, 0, 0), (1, 2, 1).
83. Find the equation of the plane determined by the point
(1, 5, — 2) and the straight line passing through the point (6, — 2, 4)
equally inclined to the coordinate axes.
84. Find the equation of the plane passing through the points
(0, 3, 2), (2, - 3, 4) perpendicular to the plane 6a + 3y-22 + 3 = 0.
332 SPACE GEOMETRY
85. Find the equation of a plane determined by the poirx
(2, 3, 2) and the straight line passing through (1, — 1, 1) in tlu
direction 1:2:3.
86. Find a point on the line 5x + 3y — 1=0, 3y — 5z — 11 =
equally distant from the planes 3x + 3y — 2 = 0, 4# + y + « + 4 = O-
87. Find the equations of the projection of the line x + y + zsz
— 2 = 0,x + 2y + z — 2 = 0 upon the plane 3x + y + 3z — 1 = O-
88 . Find the length of the projection of the straight line j oining th&
points (2, 3, 4), (0, —3, 1) upon the straight line 0 =*-- — = — — _
Ii 1 2i
89. Prove that the plane 5x + 3y — 4« — 35 = 0 is tangent tc*
the sphere (x + l)2 + (y - 2)2 + (z - 4)2 = 50.
90. Find the center of the circle cut from the sphere x*-\-if*
+ z2 = 49 by the plane 4 a; + 6y + 12s - 49 = 0.
91. Find the equation of a plane passing through the line
x + 3y + 3z + l = 0, 2/ + 2z + l = 0 and -parallel to the line
2x + y-z = 0y3x + 2z-7 = 0.
92. Find the center of a sphere of radius 7, passing through the
points (2, 4, — 4) and (3, —1,-4) and tangent to the plane 3 x — 6 y
+ 2* + 51=0.
93. What kind of line is represented by the equations x* + z2
-4y = 0, y-2 = 0?
94. What kind of line is represented by the equations x2 — 9 y
-36 = 0, x + 5 = 0?
95. What is the projection of the curve y2 + «2 — 6a; = 0, s2 = 4y
on the plane XOY?
96. What is the projection of the curve x2 + y2 = a2, y2 -f z2 = a2
on the plane XOZ ?
97. Find the projection of the curve £c2 + 3y2 — *2 = 0, a^ + y*
- 2 x = 0 on the plane XOZ,
98. Find the projection of the curve x2 + 2yi — «* = 1, 2x2 — y2
= 8 z on the plane FOZ.
99. Show that the curve x2+y2 = a2, y = « is an ellipse. (Rotate
the axes about OX through 45°.)
100. Find the projections of the skew cubic x = t,y = &, z = (?
on the coordinate planes.
PROBLEMS 333
101. Prove that the projections of the helix x = a cos 0,y = a sin 0,
z = kO on the planes XOZ and YOZ are sine curves, the width of
each arch of which is hir.
102. What is the projection of the curve x = e\ y = e~\ z = t V2
on the plane XOY?
103. Turn the plane XOZ about OZ as an axis through an angle
of 45°, and show that the projection of the curve x = el, y = e~ ',
z = t V2 on the new XOZ plane is a catenary.
104. Show that the curve x = #, y = 2 1> z = t is a plane section
of a parabolic cylinder.
105. Prove that the skew quartic x = t, y = tf8, z = t* is the inter-
section of an hyperbolic paraboloid and a cylinder of which the
directrix is the cubical parabola y = xs.
106. The vertical angle of a cone of revolution is 90°, its vertex
is at 0, and its axis coincides with OZ. A point, starting from the
vertex, moves in a spiral path along the surface of the cone so that
the measure of the distance it has traveled parallel to the axis of
the cone is equal to the circular measure of the angle through which
it has revolved about the axis of the cone. Prove that the equations
of its path, called the conical helix, are x = t cos t, y = t sin t, z = t.
107. Show that the helix makes a constant angle with the elements
of the cylinder on which it is drawn.
108. Find the angle between the conical helix x = t cos t,y = t sin t,
z = t and the axis of the cone, for the point t = 2.
109. Show that the angle between the conical helix x = t cos t,
y = t sin t, z = t and the element of the cone is tan"1 — ;=•
y V2
110. At what angle does the curve x = a (1 — cos 0), y = a sin 0,
z = a$ intersect the straight line passing through the origin and
making equal angles with the three coordinate axes ?
111. Find the length of the curve x = t2, y = 2 1, z = t from the
origin to the point for which t = 1.
112. Find the length of the curve x = e', y = e~ty z = t^/2
between the points for which t = 0 and t = 1.
113. Find the length of the curve x = t2, y = £ £8, z = 2 1 from
the origin to the point (9, 9, 6).
334 SPACE GEOMETRY
114. Find the length of the curve x = tcos2t, y = tsin2t,
3 z = 4 $ between the points for which t = 0 and t = 1.
115. Find the equations of the tangent line and the equation of
the normal plane to the curve x = t2, y = 2 1, z = t at the point foxr
which t = 1.
116. Find the equations of the tangent line and the equation of
the normal plane to the curve x = e\ y = e~*9 z = t V2 at the poinfc
for which t = 0.
117. Find the equations of the tangent line and the equation ofz
the normal plane to the curve x — 2 12 + 1, y = t — 1, z = 3 tz at th^
point where it crosses the plane XOZ.
118. Find the equations of the tangent line and the equation o^
the normal plane to the conical helix x = t cos t, y = t sin t, z = £r
at the point for which t = — •
119. Find the equations of the tangent line and the equation o£=
the normal plane to the skew quartic x = t, y = t*9 z = t* at thi
point for which t = 1.
CHAPTER XV
PARTIAL DIFFERENTIATION
162. Partial derivatives. Consider f(x, y), where x and y are
independent variables. We may, if we choose, allow x alone to
"vary, holding y temporarily constant. We thus reduce f(x, y) to
£i function of x alone, which may have a derivative, defined and
computed as for any function of one variable. This derivative
is called the 'partial derivative off(x, y) with respect to x, and is
denoted by the symbol ' • Thus, by definition,
ox
df& y) = Lim /(* + Ax, y) -/O, y)
dx ax=s=o Ax
Similarly, if x is held constant, /(#, y) becomes temporarily a
function of y, whose derivative is called the partial derivative of
8/fc y) = Lim /(*> y + Ay) -/fo y) (2)
fy av=o Ay
Graphically, if z =f(x^j/) is represented by a surface, the rela-
tion between z and x when y is held constant is represented by
the curve of intersection of the surface and the plane y = const.,
dz
and — is the slope of this curve. Also, the relation between z
ox
and y when x is constant is represented by the curve of inter-
dz
section of the surface and a plane x = const., and — is the slope
of this curve. &
Thus, in fig. 204, if PQSE represents a portion of the surface
z =/(#, y), PQ is the curve y = const., and PR is the curve
x = const. Let P be the point (#, y, 2), and LK = PKf = Ax,
LM= PMf = Ay.
336
336
PARTIAL DIFFERENTIATION
Then LP=f(r,y), KQ=f(x + Ax, y), MR=f(x,y + Ay
K'Q=f(z + Ax, y)-/(* y), M'R=f(x, y + Ay)-/<>, y),
/ = Lim — — = slope of P#,
— - = Lnn 7 = slope ol Pit.
ay iW F
Fig. 204
ct
Ex. 1. Consider a perfect gas obeying the law v = — • We may chang
the temperature while keeping the pressure unchanged. If At and At? are^
corresponding increments of t and r, then
and
cA*
Or we may change the pressure while keeping the temperature un-
changed. If Ap and At- are corresponding increments of p and r, then
and
Av =
or
Yp
ct
ct
ctAp
P + Ap p
Si
«2*
p* + pAp
So, in general, if we have a function of any number of variables
/(#, y, • • ., 2), we may have a partial derivative with respect to
each of the variables. These derivatives are expressed by the sym-
df cf cf
bols ^ , -jf- j • • •» j » or sometimes by/x(a:, y, • • •, 2), /y(a^ y, • • •, z),
DERIVATIVES
337
• -,fz(x, y, - • •, z). To compute these derivatives, we have to
apply the formulas for the derivative of a function of one variable,
regarding as constant all the variables except the one with respect
:o which we differentiate.
2- = 3a;2-6a;y,
Ex. 2. f=x*-Sx2y + y*,
dx
dl
dy
Ex. 3. f=sin(x2 + y2),
¥ = 2XCos(x2 + y2),
\JJb
-3 a:2
+ 3y2.
Ex.
4. / =
2£ =
ay
2 y cos (a:2
+
?2).
log Vx2 + y2
+ z2,
y_
X
dx
x2 + y2 + z2
y_
y
dy
x2 + #2 + z2
»/_
z
5,2 x2 + y2 + z3
Ex. 5. In differentiating in this way care must be taken to have the
functions expressed in terms of the independent variables. Let
x — r cos 0, y = r sin 6.
Then
dx
dr
= cos 6,
dx . /,
— = — r sin 0,
a0
^ = sin0,
-? = r cos 0,
a0
(1)
where r and 0 are the independent variables.
Also, since r = Va;2 + y2 and 0 = tan-1 - ,
dr
x
dx Va;2 + y2
5? = y_ =
da: x2 + #2
= cos 0,
sin0
x
dr
y
= sin0,
dy Vx2 + y2
dO _ x _ cos 0
dy x2 + y2 r
(2)
5r .
where x and y are the independent variables.
dx vi
It is to be emphasized that — in (1) is not the reciprocal of — in (2).
^ dr dx
vX
In (1) — means the limit of the ratio of the increment of x to an increment
v J dr
of r when $ is constant. Graphically (fig. 205), OP = r is increased by
dx PR
PQ = Ar, and PR = Ax is thus determined. Then — = Lim — — = cos 0.
dr PQ
338
PARTIAL. DIFFERENTIATION
dr
Also — in (2) means the limit of the ratio of the increment of r to that
dx v ■'
of x when y is constant. Graphically (fig. 206), OM = x is increased by MN =
PQ = Ax, and RQ = Ar is thus determined. Then — = Lim —^ = cos0.
~ ^ /» dx PQ
dx dr dx du
It happens here that — = — But -^ in (1) and — in (2) are neither equal
, dr dx du dx
nor reciprocal.
Fig. 206
In cases where ambiguity is likely to arise as to which variable is
constant in a partial derivative, the symbol for the derivative is sometimes
inclosed in a parenthesis and the constant variable is written as a subscript,
thus
\pr/B
163. Higher partial derivatives. The partial derivatives of
f(Xj y) are themselves functions of x and y which may have
partial derivatives, called the second partial derivatives oif(x, y).
They are !(§£), £(2), |/£\, * /£\. But it may ^ 8hown
cx\dx/ oy\dx/ cx\dy/ dy\dy/
that the order of differentiation with respect to x and y is imma-
terial when the functions and their derivative fulfill the ordinary
conditions as to continuity, so that the second partial derivatives
are three in number, expressed by the symbols
di?
XX >
dx\dy/ dy\dx)
d_/df
dy\dyj dy
dy
92/
dxdy
=/>
xy*
INCREMENT AND DIFFERENTIAL 339
Similarly, the third partial derivatives of f(x, y) are four in
number, namely,
j?/m = &f
dxKdx2) da?'
dy\da?) dx\dxdy) dx\dy) dx^dy
d /gy\ a / d*f\ d* /df\ d*f ^ .
dx\dy2) dy\dxdy) dy*\dx) dxdy2
_a_/ay\_ay
So, in general, - — ^- signifies the result of differentiating
j(x, y) p times with respect to x and q times with respect to
y, the order of differentiating being immaterial.
The extension to any number of variables is obvious.
164. Increment and differential of a function of two variables.
Consider z =f(x, y), and let x and y be given any increments
Ax and Ay.m Then z takes an increment Az, where
As =f(x + Ax,y + Ay*)- f(x, y). (1)
In fig. 204, N8 = zf =f(x + Ax, y + Ay}
and N'S = z' -z = Az. (2)
If x and # are independent variables, Ax and Ay are also
independent. Thus the position of S in fig. 204 depends upon
the choice of LK and LM, which can be taken at pleasure.
The function z is called a continuous function of x and y if Az
approaches zero as a limit when Ax and Ay approach zero as a
limit in any manner whatever.
Thus, in fig. 204, if z is a continuous function of x and y,
the point S will approach the point P as LK and LM approach
zero, no matter what curve the point N traces on the plane
XO Y or the point S on the surface.
We shall assume that z and its derivatives are continuous
functions.
340 PARTIAL DIFFERENTIATION
The expression for Az may be modified as follows:
The line N'S may be separated into two portions by drawing
from Q a line parallel to K'N1 meeting NS in N,r. Then
JST'S = N'N" + N"S = KfQ + N"S. (3)
The line K'Q is connected with the slope of PQ by the
relation R,g ^
Lim— — = slope of PQ = — ,
PK' ex
the limit being taken as PK1 = Ax approaches zero.
Hence — -, = he,,
PK1 dx *
where ex approaches zero as Ax approaches zero, so that
Also the line N" S is connected with the slope of QS by the
relation ^ff «
Lim -— -f = slope of QSy
the limit being taken as QNn = Ay approaches zero. But as
Ax = 0, the curve QS approaches the curve PP. Hence we are
justified in saying
J¥,fS dz
Lim — — = slope of PR = — ,
QN" F dy
the limit being taken as both Ax and Ay approach zero.
N"S_dz
QN" ~ dy
where e2 approaches zero as Ax and Ay approach zero, so that
fdz
Hence T^77 = j" + *2,
N»S = (™+^Ay, (5)
since QNn = PM, = Ay.
Substituting from (4) and (5) in (3) and then in (2), we
have s 0
vZ OZ
Az = — Ax + — Ay + exAx + e2Ay. (6)
DIFFERENTIAL 341
In a manner analogous to the procedure in the case of a
function of one variable (§ 77), we separate from the incre-
ment the terms eAz + eAy, call the remaining terms the
total differential of the function, and denote them by dz. The
differentials of the independent variables are taken equal to
the increments, as in § 77. Thus, we have by definition, when
z is a function of two independent variables x and y,
In (7) dx and dy may be given any values whatever. If,
in particular, we place either one equal to zero, we have the
partial differentials, indicated by d^z and dyz. Thus
* dx ' * dy u
A partial differential expresses approximately the change in
the function caused by a change in one of the independent
variables; the total differential expresses approximately the
change in the function caused by changes in all the inde-
pendent variables. It appears from (7) that the total differen-
tial is the sum of the partial differentials.
Ex. The period of a simple pendulum with small oscillations is
whence g = — .
Let I = 100 cm. with a possible error of J mm. in measuring and
T = 2 sec. with a possible error of ^ J^ sec. in measuring. Then dl = ± ^
andrfr=±Ii,j.
Also dg = t^dl-^dT,
and we obtain the largest possible error in g by taking dl and dT of oppo-
site signs, say dl = ^, dT = - T^.
Then dg = £- + tt2 = 1.05 tt2 = 10.36.
The ratio of error is
*1 = (1 - 2 *JL = .0005 + .01 = .0105 = 1.05%.
9 I T
AC
342 PARTIAL DIFFERENTIATION
165. Extension to three or more variables. The results of fcfcm^e
previous article may be extended to the cases of three or motre
independent variables by reasoning which is essentially that jix^st
employed, without the geometric interpretation, which is no -^y
impossible. For example, consider
Let x, y, z be given increments Ax, Ay, As, and let
**' =/0 + Ax, y + Ay, z + Az).
Then Aw = uf—u.
For convenience, introduce new functions
wi =/0 + Ax, y + Ay, z),
w2 =/(a; + Ax, y» *)•
Then Aw = w' — Wj+ wx— w2+ wa— w. * ^L^)
XT T . u„ — w dw , u9 — u du .
Now Lim -*- — = — , whence -*- — = — + €,,
ax=o Ax tfx Ax ox 1
T . w, — u2 duq , w, — u„ dwQ ,
Lun -i- — * = —2, whence -1- — ? = _» + €
Ay=o Ay ay Ay dy *
T . u' — u. du. , uf — u, du,
Lun — - — - = — - , whence — - — - = -^ +
6
A« = 0
A2 dz ' " " "" As fe ' '•'
so that wa— w = — - Ax + e,Ax,
2 dx 1
^U2 A , A
ui-s=g^Ay + eiAy»
-n j. ^*a J ^wi 1 du , du A , A
But — * and -^ approach — and — - as Ax and Ay appro:
dy cz dy dz
zero so that -^ = — + eA and — ^ = — + eK. Hence (2)
1 ... oy dy 4 dz dz 6 v y
be written v *
Aw = ^Ax + ^Ay + ^Az + eaAx + <Ay + ^As, CT^
where er2 = e2 + e4 and €3 = e8 + e5.
DIRECTIONAL DERIVATIVE 343
Then du is the part of this expression which does not contain
ev €3, or €3, with the increments of the independent variables
replaced as usual by their differentials. That is,
du = — dx + — dy + — dz. (4)
dx dy dz y
The extension to more variables is obvious.
166. Directional derivative of a function of two variables. The
result of § 164 may be used to find the slope of PS (fig. 204),
which is a curve cut out of the surface z=f(x, y) by any
plane through LP. Draw the lines PN1 and LN as shown in
the figure, and let £ N = pN, = ^
where r is the distance measured from some point on the line
LN produced. Denote by 0 the angle KLN=K'PNt, which
is equal to the angle made by the plane of PS with the plane
ZOX. Then __ A _ __ A
LK Ax ~ LM Ay . ^
= -— = cos 0, = — - = sm 0,
LN Ar LN Ar
N'S Az dz
and the slope of PS = Lim — — - = Lim —- = —-, the limit being
r PN* Ar dr 5
taken as S approaches P along PS.
From (6), § 164,
Az dz Ax . dz Ay . Ax , Ay
— = 1 £ + e 1- e — —
Ar dx Ar dy Ar x Ar 2 Ar
dz dz
= — cos 0 + 7T- sin 0 + e, cos 0 + e0 sin 0.
dx dy * 2
Taking the limit, we have
— = — cos 0 + — 8in0 = slope of PS.
dr dx dy
dz
Now — measures the rate of change of z in the direction
dz dx ' dz
LK, — the rate of change in the direction XJf, and — the rate
°y m dr, dz m
of change in the general direction LN The derivative — is
called the directional derivative in the direction of r.
344 PARTIAL DIFFERENTIATION
Ex. The temperature u at any point of a plate is given by the formula j&Xla
in the direction making an angle of 30° with OX,
u = - • Find at the point (2, 3) the rate of change of temperature «
We have — - ^2 + ^2» ay - ^ + ^2>
and at the point in question
du _ 4 du __ 6
to""!^' fy"~~169*
Hence the required rate of change is
^=--i-cos30o--|-sin30o
dr 169 169
=_2V3+3=_
169
167. Total derivative of z with respect to x. In fig. 204, leU*~
the point S approach the point P along any curve whatever on
the surface, and not along the curve PS, as in § 166. Then the
point N describes a curve on the plane XOY, the equation of "^^
which may be taken as y = <j> (#), and
-e
§=*'(*) c>
dx
is the slope of the curve described by N.
During this motion of the point S, z is a function of x ; since
it is in general a function of x and y, and y is a function of x.
Hence z has a derivative with respect to x and
dz __ T . Az
dx Ax
Dividing the expression for Az as given in (6), § 164, by
Ax, and taking the limit, we have
dz^dz_^dzdyt
dx dx dydx
This is the total derivative of z with respect to x.
TANGENT PLANE 345
This result (2) has an important application when the curve
along which S moves is on the plane XO Y. For then z = 0, a
constant, and, from (2),
ex eydx
"where -~ is the slope of this curve, as shown in (1). We express
this result in the following theorem:
1. The value of -£ may be found from the equation
(XX
by the formula ^ + 4- -r = °-
ex cy ax
Again, let z be denned as an implicit function of x and y by
the equation F(x,y,z)=Q.
If we hold y constant temporarily, the case reduces to the one
dz
discussed in theorem 1, with z in place of y and — in place
(I
of -j-* Similarly, if we hold x temporarily constant, we get
ax
theorem 1 with change of letters. Hence:
$2 3z
2. The values of — and — - may be found from the equation
ex cy
F(x, y, z) = 0
, , j, 7 dF dFdz A
by the formulas - — h — — = 0,
ex oz ex
dy dz dy
168. The tangent plane. In fig. 204, let P be given the fixed
coordinates (xv yl9 z^. The tangent line to PQ in the plane
y = yx is, by § 76, ,_
and the tangent line to PR in the plane x = xx is
*"v"(S)10r",ri> (2)
346
PARTIAL DIFFERENTIATION
Both of these lines lie in, and hence determine, the plane of
which the equation is
-z> = ^)1(x-^+(|)l^-^'
(3)
for this equation reduces to (1) when y = yv and reduces to
(2) when x = xY.
This plane is called the tangent plane to the surface at the point
We shall prove that the plane (3) contains all tangent lines to*
the surface z = /(#, y) which pass through P.
The line through the two points P and S has the directioi
Ax : Ay : Az. Its equations are therefore
x - xi _ V - Vx _ g - zi
Ax
or
Ay
Ay
Az
(*)■
As the point S approaches the point P, the line (5) ap-
proaches as a limit a tangent line at P, and the equations
of this tangent are
'-* = (!),<—*>
_ /dz dz dy\ f
(6)
An easy combination of these equations gives (3) as an equation
satisfied by any tangent line. Hence the theorem is proved.
If the equation of the surface is given in the form
F(x, y, z) = 0,
(7)
TANGENT PLANE 347
the equation of the tangent plane may be found without solving
for z. For, from theorem 2, § 167,
dF
dz dx
dx
dF*
dz
dF
dz
dy
■ •"—* ■
— S^BBM •
dy
dF
dz
Substituting these values in (3) and making a few simple
changes, we have as the equation of the tangent plane,
eaH)+®.('-*)+©.(--)- <*>
The straight line perpendicular to the tangent plane at the
point of contact is the normal to the surface. Its equations are
x-xx __y-y1__ g-gt
/dz\ /dz\ -1 w
W/i \dyJx
x — x% y — y% z — zy ,+ AN
or 1 = 2 — £i = 1. CIO)
[dH\ [dJ-\ (dJ-\
Wi \dyh \dzh
Ex. 1. Find the tangent plane and the normal line to the paraboloid
z = ax2 + by2.
Here — = 2 ax and — = 2 by. Hence the tangent plane is
dx dy
2axl(x- xx) + 2 hyx (y - yx) - (z - zx) = 0,
or 2 axxx + 2 %xy — 2 axf — 2 ft^2 — 2 + z1 = 0.
But since 2 ax^ + 2 by} = 2 zr this may be written
2 ^x + 2 %!# — 2 — zx = 0.
The normal is £J^£l = ^Zlt = i^_£l .
2 oxj 2 o^j — 1
348 PARTIAL DIFFERENTIATION
Ex. 2. Find the tangent plane and the normal line to the ellipsoid
_ dF 2x dF 2y dF 2z
Here — = — - > — = -£• > — = ~T '
dx a2 dy lr oz cr
Hence the tangent plane is
2-f (* - *x) + ^ O - yd + ^ (* - *i) = 0.
a2 Cr cz
3T ?/ 2
The normal line is - — il = - — 2l = f 1 .
£1 yi £l
a2 &2 c2
If a curve is denned as the intersection of two surfaces by
the equations
f(x, y, z) = 0, F(x, y, z) = 0,
its tangent line is evidently the intersection of the two tangent
planes to these surfaces. The equations of the tangent line are
therefore two equations of the form (8). The direction cosines
of the tangent line can be found by the method of § 156. The
normal plane may be found T)y the method of § 151.
169. Maxima and minima. In order that the function f(x, y)
shall have a maximum or a minimum value for x = x^ y = yv
it is necessary, but not sufficient, that the tangent plane to the
surface z =/(#, y) at the point (xx, yx, z^) should be parallel
to the plane XOY. This occurs when hM = 0, (jr)= °- These
are therefore necessary conditions for a maximum or a mini-
mum, and in case the existence of a maximum or a minimum
is known from the nature of the problem, it may be located
by solving these equations.
Ex. It is required to construct out of a given amount of material a cistern
in the form of a rectangular parallelepiped open at the top. Required the
dimensions in order that the capacity may be a maximum, if no allowance
is made for thickness of the material or waste in construction.
EXACT DIFFERENTIALS 349
Let x,yyzbe the length, the breadth, and the height respectively. Then
~fclie superficial area is xy + 2 xz + 2 yz, which may be placed equal to the
Lven amount of material, a. If v is the capacity of the cistern,
axy — x2y2
v = xyz = —2 2- .
Then 8y - (a — 2xy — x2)y2 dv = (a — Zxy — ffiz2
dx 2(x + y)2 ' fy 2(x + y)2
For the maximum these must be zero, and since it is not admissible to
ave x = 0, y = 0, we have to solve the equations
a — 2 ary — x2 = 0,
a - 2 a?y - y2 = 0,
hich have for the only positive solutions x = y = a/ - > whence 2 = - -%/ - •
onsequently, if there is a maximum capacity, it must be for these dimen-
sions. It is very evident that a maximum does exist; hence the problem
is solved.
More generally, if a function of three or more independent
^variables has a maximum or minimum when all the variables
change in any way, it must have a maximum or minimum when
^ach changes alone. Therefore, if f(x, y, z) has a maximum or
si minimum, it is necessary, by § 89, that
Jf-o. f=o, ^=0.
dx dy dz ,
170. Exact differentials. We have seen that if z =f(x, #), then
dz = — - dx + — dy. (1)
dx dy * v J
dz
"When the function /(#, y) is known, the partial derivatives — -
dz .
and — may be found, and the second member of (1) is of
dy
the form Mdx+Ndy, (2)
where M and N are functions, of x and y. In § 164, (1) was
called a total differential ; it will now be called an exact differ-
ential, to emphasize the fact that it may be exactly obtained
by differentiation.
350 PARTIAL DIFFERENTIATION
Now expressions of the form (2) arise in practice by other
methods than by differentiation, or they may be written down
at pleasure. For example, we may write arbitrarily the two
following expressions:
(4 z8- 2 xf) dx + (4 y2- 2 x*y) dy, (3)
(a? + xy) dx + yzdy. (4)
It is important, therefore, to know whether an expression
of the form (2) is always exact; that is, whether it is always
possible to find z =/(#, y) so that (2) is equivalent to (1).
In discussing this question we note first that if (2) is
equivalent to (1), we must have
dz dz
- = M, - = N, (5)
d2z dM dN
whence —- = — = — .
dxdy dy dx '
Hence, if Mdx + Ndy is an exact differential, it is necessary
that dM_d_N
dy" dx' ^
From this it appears that (4) is not an exact differential,
since — - = x and — = 0. On the other hand, (3) may possibly
dy dx
be exact, smce — — = — 4 xy and -— = — 4 xy.
dy * dx *
Let us now assume that the condition (6) is met, and try to
find z. We may integrate the first equation of (5) consider-
ing y as a constant. The constant of integration then possibly-
contains y and must be expressed as a function of y. Then
z= CMdx + <\>(yy (7)
Substituting this in the second equation of (5), we have
or ^(y) = ^-^- CMdx. (8)
EXACT DIFFERENTIALS 351
By hypothesis the first member of (8) does not contain x.
Hence the. second member of (8) must be free from x or the
^rork cannot go on. Now the condition that an expression shall
~be free from x is that its derivative with respect to x shall be
zero. Hence, from (8), we must have
*f_*fMdx=0. (9)
ox oxoyj
d2
ZBut
dxdy
The condition (9) is then simply (6), which is fulfilled by
Xiypothesis.
From (8), the value of <£(#) can now be found and substi-
tuted in (7). The value of z is thus found.
We have accordingly the following theorem, the converse of
"the one stated above.
If -— = — > the expression Mdx + Ndy is an exact differential
dy ox
<lz, and m—— TV— —
dx dy
The process of finding z is illustrated in Exs. 1 and 2. Ex. 3
shows how the process fails if it is wrongly applied to an
expression which is not an exact differential.
Ex. 1. (4 x8 - 2 xf)dx + (4 f - 2 x2y) dy.
Here = — 4 xy = Hence the expression is equal to dz, and
dy dx
g = 4*«-2*3,*, (1)
^ = 4y8-2*V (2)
Integrating (1) with respect to x, we have
z = x* - x*y2 + f(y). (3)
Substituting in (2), we have
-2x2y+/(y) = 4y8-2x2y;
whence f (y) = 4 yB,
and f(y) = t + C.
Substituting in (1), we have z = x4 — x*y2 + y4 + C.
352 PARTIAL DIFFERENTIATION
Ex. 2. ( : — / y )dx + dy.
\x x Vy2 - x2/ V^-x2
dM x dN
Here — = = — • The expression is therefore an exact
dy (^2 _ ^f dx
differential dz, and dz 1
cxm x x v y2 — x2
dz 1
dV Vy2 - x2
Integrating (2) with respect to y, we have
2+Ay) = ^;
X2
(1)
(2)
z = log (y + Vy2 - x2) + /(x). (3)
Substituting in (1), we have
+/(*) = r
Vy2 - x2 (y + Vy2 - x2) * xVy2-x2 '
whence /" (x) = 0,
and /(x) = C.
Substituting in (3), we have z = log (y + Vy2 — x2) + C.
Ex. 8. (x2 + xy)dx + y8^.
Here = x, -— = 0,
5y ox
and the expression is not exact. If one wrongly put
dz
— = x2 + xy, (1)
dx
jr = y*> (2)
and integrated (1) with respect to x, he would have
Substituting in (2), he would have
x2
whence f(y) = y8 — — •
Butt/V(y) should be a function of y alone, and the last equation is absurd.
Equations (1) and (2) are therefore false.
LINE INTEGRALS 353
171. Line integrals. The expression Mdx+Ndy occurs in
certain problems involving the limit of a sum as follows:
Let C (fig. 207) be any curve in the plane XOY connecting
the two points L and K, and let M and N be two functions of
x and y which are one-valued and continuous for all points
on C. Let C be divided into n segments by the points Pv ij,
-?>*••> Pn-v ^d let Ax be the pro-
jection of one of these segments on
OX and Ay its projection on OY.
That is, Ax = xi+1-x„ Ay = yi+1-y{,
where the values of Ax and Ay are
not necessarily the same for all
values of L Let the value of M for
Fig. 207
each of the n points X, i£, ij, • • •, °
JJ_! be multiplied by the correspond-
ing value of Axj and the value of N for the same point by
the corresponding value of Ay, and let the sum
t = H-l
be formed. '"°
X [M(.x» yd&v+N(xt, ft)4y]
The limit of this sum as n increases without limit and Ax
and Ay approach zero as a limit is denoted by
/
J(C
(Mdx + 2Tdy),
and is called a line integral along the curve C. The point K may
coincide with the point X, thus making C a closed curve.
Ex. 1. Work. Let us assume that at every point of the plane there acts
a force which varies from point to point in magnitude and direction. We
wish to find the work done on a particle moving from L to K along the
curve C. Let C be divided into segments, each of which is denoted by A*
and one of which is represented in fig. 208 by PQ. Let F be the force
acting at P, PR the direction in which it acts, PT the tangent to C at P,
and 6 the angle RPT. Then the component of F in the direction PT is
F cos $, and the work done on a particle moving from P to Q is F cos $ As,
except for infinitesimals of higher order. The work W done in moving
the particle along C is, therefore,
TT=LimV Fco8 0As= f FcosOds.
*4 J(C)
354
PARTIAL DIFFERENTIATION
Now let a be the angle between PR and OX, and <f> the angle between P~
and OX. Then 0 = <f> — a and cos 0 = cos <f> cos a + sin <f> sin a. Therefo:
' W= f (F cos ^ cos a + F sin <f> sin a) ds.
But Fcosa is the component of force parallel to OX and is usually
denoted by X. Also F sin a is the component of force parallel to 0
and is usually denoted by Y. Moreover,
cos <f>ds = dx and sin <f>ds = dy (§ 91).
Hence we have, finally,
W
= f (Xdx + Ydy).
0
Fig. 208
Ex. 2. Heat. Consider a substance in
a given state of pressure p, volume v, and
temperature t. Then p, v, t are connected
by a relation /( jo, v, t) = 0, so that any
two of them may be taken as independent
variables. We shall take t and v as the
independent variables and shall therefore work on the (t, v) plane.
Now if Q is the amount of heat in the substance and an amount dQ i
added, there result changes dv and dt in v and t respectively, and, excep
for infinitesimals of higher order,
dQ = Mdt + Ndv.
Hence the total amount of heat introduced into the substance by a
variation of its state indicated by the curve C is
Q= f (Mdt + Ndv).
Ex. 8. Area. Consider a closed curve C (fig. 209) tangent to the
straight lines x = a, x = b, y = d, and y = e, and of such shape that a
straight line parallel to either of the coordinate axes intersects it in
not more than two points. Let the
ordinate through any point M inter-
sect C in Px and P2, where MPX = yx
and MP2 = y2. Then, if A is the area
inclosed by the curve,
A = f y*dx ~ f Vidx
v a v a
Vidx- J yxdx
L
= - | ydx,
'(C)
Fig. 209
the last integral being taken around C in a direction opposite to the
motion of the hands of a clock.
T
ore
y
0Y
18
t
a
/
LINE INTEGRALS 355
Similarly, if the line NQ2 intersects C in Qx and Q2, where NQX = x1
and JVQ^ = xv we have
A = f x2dy - f xxdy
= C x2dy + C xxdy
= f xdy,
the last integral being taken also in the direction opposite to the motion
of the hands of a clock. By adding the two values of A, we have
2 A = f (— ydx + xdy).
If we apply this to find the area of an ellipse, we may take x = a cos <f>,
y = b sin <f> (§ 54). Then A = £ I abd<f> = irab.
Jo
If the equation of the curve C is known, the line integral
may be reduced to a definite integral in one variable. In
general, the value of the line integral depends upon the curve
C and not merely On the position of the points L and K. This
is illustrated in Ex. 4. If, however, Mdx + Ndy is an exact
differential dz, we shall have
Jf (Mdx + Ndy) = I dz = z1 — z(p
where zQ and zx are the values of z at the points L and K. This
result is, in general, independent of the curve C, though special
consideration may be necessary if z may take more than one
value at L or K.
The integral of an exact differential taken around a closed
path is, in general, zero; while the line integrals of other dif-
ferentials around a closed path are not zero.
(ydx — xdy).
"Let us first integrate along a straight line connecting O and Px (fig. 210).
The equation of the line is y = — x, and therefore along this line ydx— xdy ^=0,
xi
and hence the value of the integral is zero.
356
PARTIAL DIFFERENTIATION
Next, let us integrate along a parabola connecting 0 and Pv the equation
y?
of which is y2 = — x. Along this parabola
xi
JCydx — xdy) = — ^1_ I vxdx = —x*y*.
(o,o) v* J 2-y/7xJo 3 iyi
Next, let us integrate along a path consisting of the two straight
lines OMx and MXPV Along OMv y = 0 and dy = 0 ; and along MXPV
x = xx and <fo = 0. Hence the line integral
r*i
reduces to — J xxdy = — xxyv
Finally, let us integrate along a path consist-
ing of the straight lines ONx and NXPV Along
ONv x = 0 and dx = 0 ; and along NXPV y = yx
and dy = 0. Therefore the line integral reduces
to fXlyxdx = xxyv
Ex
Here
. 5. f
(*i,ifi) _ yfa + xdy
Fig. 210
(aro. Vo)
£2 + #S
-yd
X*
Therefore
x
to. vO
(*o» vd)
x+x9dy=d(\*^y)=d6.
+ y2 \ xl
ydx+xdz=ridd=ex-o0.
x2 + 3T
If the curve C does not pass around O, 0X will be the angle shown in
the figure (fig. 211). If, however, C is drawn around the origin, the final
value of 0 is 2 it + 0X, and the value of
the integral is 2tt + ^- 0o.
The value of this integral around a
closed curve is zero if the curve does not
inclose the origin, and is 2 ir if the curve
winds around the origin once in the posi-
tive direction.
Fio. 211
Ex. 6. Work. If X and Y are compo-
nents of force in a field of force, and
— = — i then the work done in moving
dy dx
a particle between two points is inde-
pendent of the path along which it is moved, and the work done on
a particle moving around a closed curve is zero. Also there exists a
function ^, called a force function, the derivatives of which with
respect to x and y give the components of force parallel to the axes
COMPOSITE FUNCTIONS 357
of x and y. Such a force as this is called a conservative. force. Examples
are the force of gravity and forces which are a function of the distance
from a fixed point and directed along straight lines passing through
that point.
If the components of force X and Fin a field of force are such that
— ^ — , then the work done on a particle moving between two points
dy dx
depends upon the path of the particle, the work done on a particle moving
around a closed path is not zero, and there exists no force function.
Such a force is called a nonconservative force.
Ex. 7. Heat. If a substance is brought, by a series of changes of tem-
perature, pressure, and volume, from an initial condition back to the same
condition, the amount of heat acquired or lost by the substance is the
mechanical equivalent of the work done, and is not in general zero. Hence
the line integral Q = f(Mdt + Ndv) around a closed curve is not zero,
and there exists no function whose partial derivatives are M and N. Id
fact, the heat Q is not a function of t and v, not being determined when
t and v are given.
172. Differentiation of composite functions. It is frequently
necessary to differentiate with respect to a variable a function
of a function of that variable. Several cases of this will now
be discussed.
1. Consider /(w), where u = <f>(x).
Then £.#*/»(«)* (1)
ax au ax ax
This has been proved in § 82.
2. Consider /(w), where u = (j>(x, y).
The proof of this formula is like that of (1), the only
difference being that
T . A/ df T . Au du
Lim — = — , Lim — = — •
Ax dx Ax dx
AC
♦58 PARTIAL DIFFERENTIATION
Ex. 1. If 2 = sin - , find — .
y dx
Place u = - . Then, by (2),
' V
dz x d (x\ 1 x
— = cos I - 1 = - COS - •
dx y dx \y/ y y
0. ., , dz x d /x\ x x
Similarly, — = cos - — ( - ) = - cos - •
dy y dy w y y
d* dz
Ex. 2. If z =f(x* + y2), show that x — — y — = 0.
dy ox
Place u = x2 + y2. Then
whence a: v — = 0.
dy ox
3. Consider /(w, v), where u = <\>(x), v = ty(x).
From (6), § 164, with a change of letters,
A/= / Au + /av + €1 Aw + €. At>.
If we divide by Ax, and take the limit, we have
df _ df du df dv ^_ ^ \
eta du dx dv dx
Ex. 8. Let 2 = tan - > where u = x2, r = log a:.
From (2), * = (sec* H) 1 M = * ^ « ,
e)2 / nii\ d /u\ u 9u
— = ( sec2 -) — (-)= sec8 -
dv \ 17 du \v/ v2 v
Hence, from (3), — = - (sec2 -J 2 x — —0 I sec2- ) -
dx v \ iv r2 \ vf x
(—)■
\log ar/
_ 2 a: log x — x
(log x)
2
sec
COMPOSITE FUNCTIONS 359
4. Consider f(u, v), where u = <j>(x, y), v = yfr (x, y).
Then df = dldJ£ + dl?l,
dx du dx dv dx
df__df_du df_dv_ ' ^
dy du dy dv dy
The proof is like that of (3).
Ex. 4. If z =/(* -y,y- *), prove ^ + ^1 = 0.
Place x — y = u, y — x = v. Then z =f(u9 v), and
dx du dx dv dx du dv
d± = dl ^ + dl <H = - f£ + dl .
dy du dy dv dy du dv
By addition the required result is obtained.
Ex. 5. Let it be required to change — and -^ from rectangular coor-
dx dy
dinates (x, y) to polar coordinates (r, 0), where x = r cos 0, y = r sin 0, and
f is a function of x and y.
dr dv
From Ex. 5, § 162, — = cos 0, — = sin 0,
dx dy
whence
d0 _
dx
sinO
9
r
d0_
dy
COS0
r
*1=
dx
:^C08tf-
dr
dd
sin#
9
r
dy
^sin0 4
dr
d0
cos0
r
5. If, in (4), we multiply the equations (4) by dx and dy, add
the results, and apply the definition of § 164, we have
df=d£du + l{-dv.
du dv
•
This result is easily generalized. Hence the form of the differ-
ential df is the same whether the variables used are independent
or not
6. Higher derivatives of a composite function may be found
by successive applications of the foregoing formulas.
360 PARTIAL DIFFERENTIATION
The extension of all the foregoing relations to cases involving
nore variables is obvious.
a polar coordinates, where V it
nfiefl
Ex. 6. Required
fonction of x aoc
lo express
I .-'
*v
From Ex. 5,
£f'_
Br
,.«-
Then, by (i).
3F_
8»
r
D« +
v _ srsr .
r» IrYlr
e6
!inj|| +
e«Ler
X&v a &v
nin« !
nr
«].
-[^cos0 i
■id
S
rvv
+ UrZ0
OS 6
B»F
Bfl*
ain
BV
*V »fl
-■-['
sin 0 gc
afl
c=T's
e+2^
^^v + ^~^v + ^
s<?=
py 2>v_d*r i b*f
Be* 3/ Brs r2 Btfa
= /, <z + a() + /a (j - al), show that -
-u,x-at = v. Then ^ = 1, ^ = a,
Sx du Sx rle Sx du du
Differentiatinj; these ei|imU"n-; ;i second time, we have
S?z __ d% Su -/y2 8p _ r/y, tf%
Pj-' rfu" Bj: ,/i>! Br (/u5 ife" '
Sf " dv> St " dv* SI " du1 " def '
PROBLEMS 361
PROBLEMS
Find | and £, when:
ox oy
1. «=—7JL=. 4. s = log(y + V^n^).
-y 5. « = sin — — •
2. z = tan-1-* x — y
x
3. 3 = 8in-i£Lzy. 6. * = *"* + log|.
7. If « = sin (x2 — 2 xy -f y3), prove tt- + tt- — 0.
8. From 2x* — 3y* + 6xy + 2z2 = 0, prove a;^ + ^ = ^
A TJ, ? . y dz , dz ^
9. If « = e* sin -> prove x -^- + y -r- = 0.
a; ox oy
1 $z %z
10. If » = y2 + tan (ye*), prove x27r- + y-£- = 2yk.
11. If ar = log (ay) + sin*1 - > prove sa; ^- + «y «- = 1.
-r*. r~5 J t V dz , dz
13. If z = Vy3 — or sin-1 -> prove a; ^- + y w~ = «.
14. Show that the sum of the partial derivatives of
u = (x — y)(y — z)(z — x) is zero.
16. If x = . + v, y = V-, find (g^and (g);
16. If a; = er sin 6 , y = er cos 0, find f g- J and I tt-) *
17. If x = e* sec v,y = e* tan v, find f g- J and ( ~- j •
18. If a = re», y = re"*, find (g-) and (^.
19. If « = (s2 + f) tan"1 ^. find ^-
362 PARTIAL DIFFERENTIATION
d*z
20. If z = & sin (x — y), find ~ % y
&z &z
21. If z = log(z2 + 2/*), prove ^ + ^ = 0.
22. If z = tan (y + ax) + (y — era)*, prove jr^ = a2 ^
&z &Z
Verify -r- — — — — , when :
oxoy oyox
2 . o J 25. 3 = log (a + Vy2 + «*).
23. z^xy2 + 2 ye*.
V^ + y2
■• * = >|
# — V 26. « =
24. * = aJ— ~ *
27. Calculate the numerical difference between Az and dz, wheX^
z = xs 4- if — 3 a-2y, x = 2, y = 3, &x = dx = .01, and Ay = dy = .001 -
28. The hypotenuse and one side of a right triangle are respec-
tively 5 in. and 4 in. If the hypotenuse is decreased by .01 in. and
the given side is increased by .01 in., find the total change made in-
the third side, the triangle being kept a right triangle. Find th^
error that would be made if the differential of the third side corre-
sponding to the above increments were taken for the change.
29. A right circular cylinder has an altitude 10 ft. and a radius
5 ft. Calculate the change in its volume caused by increasing the
altitude by .1 ft. and the radius by .01 ft. Calculate also the differ-
ential of volume corresponding to the same increments.
30. A triangle has two of its sides 8 in. and 10 in. respectively, and
the included angle is 30°. Calculate the change in the area caused
by increasing the length of each of the given sides by .01 in. and
the included angle by 1°. Calculate also the differential of area
corresponding to the same increments.
31. The distance between two points A and B on opposite sides
of a pond is determined by taking a third point C and measuring
AC = 80 ft., BC = 100 ft., and BCA = 60°. Find the greatest error
in the length of AB caused by possible errors of 6 in. in both AC
and BC, assuming that powers of the errors of measurement higher
than the first may be neglected.
32. The distance of an inaccessible object A from a point B
is found by measuring a base line BC = 100 ft. and the angles
PROBLEMS 363
CBA = a = 30° and BCA = 0 = 45°. Find the largest possible error
in the length of AB caused by errors of 1" in measuring a and ft,
assuming that powers of the errors of measurement higher than the
first may be neglected.
33. The density D of a body is determined by the formula
D = , 9 where w is the weight of the body in air and w1 the
weight in water. If w = 243,600 gr. and w' = 218,400 gr., what is
the largest possible error in D caused by an error of 5 gr. in w and
an error of 8 gr. in w\ assuming that powers of the errors of w and
w1 higher than the first may be neglected ?
34. If the electric potential V at any point of a plane is giyen by
the formula V= log Vsc2 -f- y2, find the rate of change of potential at
any point : (1) in a direction toward the origin ; (2) in a direction at
right angles to the direction toward the origin.
35. If the electric potential V at any point of the plane is given
V(a? — a)2 + y2
by the formula V= log p — ' => find the rate of change of
W(x + a)2 + y2
potential at the point (0, a) in the direction of the axis of y9 and at
the point (a, a) in the direction toward the point (— a, 0).
v
36. On the surface 2 = 2tan-1-, find the slope of the curve
through the point 1 1, 1, — ) whose plane makes an angle of 30° with
the plane XOZ.
37. On the paraboloid z = 2 x2 -f- 3 y2, what plane section perpen-
dicular to the plane XOY and through the point (2, 1, 11) will cut
out a curve with the slope zero at that point ?
38. In what direction from the point (xv y^) is the directional
derivative of the function z = kxy a maximum, and what is the
value of that maximum derivative?
39. Find a general expression for the directional derivative of
the function u = e~v sin x + - e~*v sin 3x at the point ( — > Oj. Find
also the maximum value of the directional derivative.
1 dz
40. If« = 4#2-f-2i/2 and y = -> find — •
X CuX
364 PAETIAL DIFFEKENTIATION
y
41. If a point moves on the surface z = k tan"1 - so that i
X dz
projection on the XOY plane is the circle x2 + y2 = a2, find —
cue
dy Z ■
42. Find -^ from the equation x = ce*.
ax
43. Find -^ from the equation 2 logee + I sin-1- ) = e.
dz dz
44. Find «- and ~- > when (a; + y)(y + z) (z + x) = c.
45. Find tt- and o~ » when a?8 + y2 + z2 — log ys2 = c.
3z 3z
46. Find y and ^-> when *8 + (a* + y2) * + x*y* = 0.
47. Find ^ and y , when (x2 + i/2 + a2)8 = 27 xyz.
48. Find the equations of the tangent plane and the normal li
to the ellipsoid x2 + 3 y2 + 2 z2 = 9 at the point (2, 1, 1).
49. Find the equations of the tangent plane and the normal lin
to the surface xy + yz -f- zx = 1 at the point (1, 0, 1).
50. Find the equations of the tangent plane and the normal lin
to the surface z = (ax + by)2 at the point (xv yv «x).
51. Find the tangent plane to the cone x2 + y2 — z2 = 0 and provi
that it passes through the vertex and contains an element of the cone
52. Show that the sum of the squares of the intercepts on tb
coordinate axes of any tangent plane to the surface x* + y* + «* = a
is constant.
53. Show that any tangent plane to the surface z = kxy cuts tb
surface in two straight lines.
54. Find the equations of the tangent line and the normal plan'
to the curve xyz = 1, y2 = x at the point (1, 1, 1).
55. Find the equations of the tangent line and the normal plana
to the curve x = sin z, y = cos z at the point ( 1, 0, -= )•
56. Find the equations of the tangent line and the normal plana
to the curve of intersection of the cylinders x2 + y2=25, y2-f-«*=25-
at the point (4, 3, — 4).
PEOBLEMS 365
57. Find the equations of the tangent line and the normal plane
^fco the curve of intersection of the ellipsoid 24 x2 -f- 16 y2 + 3 z2 = 288
sud the plane 2x + 8y + 5z = 0a,t the point (2, — 3, 4).
58. Find the angle at which the helix x2 + y2 = a2, z = k tan-1 -
intersects the sphere x2 -f- y2 + z2 = rL(r>a).
59. Find the angle at which the curve y2 — z2 = a2, x = b(y + z)
intersects the surface x2 -f- 2 zy == c2.
60. Find the minimum value of the function 2 = 4#2 — Sxy
+_9y* + 5x + 15y + 16.
61. An open rectangular cistern is to be constructed to hold
1000 cu. ft. Required the dimensions that the cost of lining should
be a minimum.
62. Divide the number a into three parts such that their product
shall be the greatest possible.
63. Find a point in a plane quadrilateral such that the sum of the
squares of its distances from the four vertices is a minimum.
64. Find the volume of the greatest rectangular parallelepiped
inscribed in an ellipsoid.
65. Find by calculus the point in the plane 2x + 3y — 62 + 5 = 0
which is nearest the origin.
66. Find the points on the surface 2 a2 + 4 1/2 — z2 — 6x + 5y
+ 18 = 0 which are nearest the origin.
67. Find the highest point on the curve of intersection of the
hyperboloid x2 + y2 — z2 = 1 and the plane x+y + 2z = 0.
68. Find the volume of the greatest rectangular parallelepiped
which can be inscribed in a right elliptic cone with altitude h
and semiaxes of the base a and b, assuming that two edges of the
parallelepiped are parallel to the axes of the base of the cone.
69. Through a given point (1, 1, 2) a plane is passed which with
the coordinate planes forms a tetrahedron of minimum volume. Find
the equation of the plane.
70. Find the point inside a plane triangle from which the sum of
the squares of the perpendiculars to the three sides is a minimum.
(Express the answer in terms of K, the area of the triangle ; a, b, c,
the lengths of the three sides ; and x} y, z, the three perpendiculars
on the sides.)
366 PARTIAL DIFFERENTIATION
Prove that the following differentials are exact and find thei
integrals :
71. (5x4 - 3x2y + 2xf)dx+(2xhj - x% + 5y*)dy.
wo 1 + 2x1/ + u*tf . 2jrt/ + l.
73. — •' J '' dx + '' T dy.
x2 + ?/ , 2 j* ,
74. _^_ ,,,__ dlh
X
75. dx-l-2e i, + -)dy.
y W y) '
xdx I y \
77. (cos2 a? — // sin x) dx + cos a* dy.
V It'
78. e2 sin (a* -f- y)dx + e'2 [sin (a; + y)— y cos (a? + y)~\dy.
J^(l,2)
[(*/ — x)dx + y dy~\,
(0,0)
(1) along a straight line,
(2) along a parabola with its axis on OX.
Jf»d, 0)
\_(x2 + y*)dx +xdy\
(0,1)
(1) along a straight line,
(2) along a circle with its center at 0.
[ifdx + (xy + y2) dy\
(0,0)
(1) along a parabola with its axis on OX,
(2) along a broken line consisting of a portion of the ar-axis
and a perpendicular to it.
82. Find the value of j^ (-^_ + ^,
(1) along the curve x = t, // = t2,
(2) along a broken line consisting of a portion of the a>axis
and a perpendicular to it.
83. Find the value of
(3,0)
PEOBLEMS 367
(«. *) _ y dx + x dy
x Var2 — y2
(1) along the curve x = 3 sec 0, y = 3 tan 0,
(2) along a broken line consisting of a portion of the aj-axis
and a perpendicular to it.
84. Find, by the method of Ex. 3, § 171, the area of the four-
cusped hypocycloid x = a cos3<£, y = a sin8<£.
85. Find, by the method of Ex. 3, § 171, the area between one
arch of a hypocycloid (§ 58) and the fixed circle.
86. Find, by the method of Ex. 3, § 171, the area between one
arch of an epicycloid (§ 57) and the fixed circle.
d2u
87. If u =/(#, y) and y = F(x), find -j-j •
(XX
88. If f(x, y) = 0, prove ^ = y- -JL* £
(X\ VZ VZ
-), prove xYx + 3,^ = 0.
90. If f(lx + my + nz, x2 + y2 + z2) = 0, prove
(ly — mx) -f- (ny — mz) j— -f- (Iz — nx) — = 0.
(1 \ 3z x2 oz
- + logy), prove - = 2y--^-
92. If /(*, y, .)- 0, show that (|X(|X(|X=-1-
93. Ii, = ^ + ^,Wo,e^ + 2Xy^- + y^ = 0.
94. If 2 = <f>(x + *y) + ^(^ — iy)> where i = V— 1, prove
+ S = o.
&c2 dy2
95. If V is a function of r only, where r = V#2 -f- y2, find the value
of -^-r + tit in terms of r and F.
oxr oy*
368 PAETIAL DIFFEKENTIATION
96. If x = eu, y = ev, and z is any function of x and yy find th<
value of a?r-T + w2ri + a!T- + V7" in terms of the derivatives of
ox2 - vtf ox a oy
with respect to u and v.
97. If x = u + v, y = > and V is any function of x and y3
Cb
.&V d2V qd2V
prove aA -j-g — -=-z = or
dx2 dy2 dudv
98. If x = eu cos v9 y = eu sin v, and V is any function of x and y^
find Q Q in terms of the derivatives of F with respect to a; and y.
cucv
99. If a; = eu cos vy y = eu sin v, and F is any function of x and y,
d*F , WV 9t&V , a2F\
z>0 I ^ — 0 pB __ p~ 0
100. If x = r ^ , y = r > and F is any function of
&V &V &V 1 8*V ldV
xtIAytWm — - — = — -- w + -—.
101. If x = e" sec m, y = e" tan m, and <£ is any function of x and y,
Pr0Ve C0S Mfe ~ d = Mw + V/ + V* + *) dxTy
102. If x + y = 2 e* cos <f>, x — y = 2ieP sin <£, and F is any func-
taon of x and y, prove _ + _ = 4 ay ^-
103. If x=f(u, v) and y = <f>(u, v) are two functions which
satisfy the equations ■£- = —} ^ = — -^ , and F is any function of
a2r , ^f /a2F , Pv\r/df*t /df\n
CHAPTER XVI
MULTIPLE INTEGRALS
173. Double Integral with constant limits. By definition,
f f(x) dx = Lim g /(*<) **,
and f(x) dx is the element of the integral. In the problems of
Chapter XIII it has been possible to form the element f(x)dx
immediately by elementary theorems of geometry and mechanics.
There are problems, however, in which it is advantageous to
determine the element itself as a definite integral.
For example, let us nod the volume bounded by the planes
z = 0, x = a, x = b(a<b~), y = c, y = d(c < &), and the surface
.=/(*y)(fig.212)
which lies entirely
on the positive side
of the XOY plane
for the volume to
be considered.
Divide the dis-
tance k — a on OX
into n equal parts
Ax, thus giving x
the series of values
a, xx = a + Ax, x3 = xx + Ax,
Through the points thus determined on OX pass planes par-
allel to YOZ, thus dividing the required volume into slices,
such as LM.
Divide the distance d — c on OY into m equal parts Ay, thus
giving y the series of values
c, y1 = c + Ay, ys = y1 + Ay, ....
U
5
J
'A
y
-/
•~
p
-/-
■
370 MULTIPLE INTEGRALS
Through the points thus determined on OY pass planes par —
allel to XOZ, thus subdividing the slices into volumes, such as
NQ, each of which stands on a base jVjE, AxAy in area.
If N has coordinates (xi9 %), NP =f(x., ^.), and the volume of
a prism with NE as a base and NP as altitude is f(xv t/j) AxAy—
If we hold x equal to x„ and give y the values e, yv y2, • • -
in succession, and take the limit of the sum as m = oo, we hav&
J = m — 1 *% d
Lim V f(z„ y,) AxAy= I f(x„ y) Ax dy (1>
m = 0° j = o t/c
as an approximate expression for the volume of the slice LM.
Using the definite integral (1) as an element, we now assign,
to x the values a, xv #2, • • • in succession and take the limit-
of the sum as n = oo. The result is
Lim 2) / f(x.,y)dyAx= I I I f(x,y)dy\dx, (2>
n = • i — o %J c J a \*Jc /
which is the required volume.
Removing the parentheses, we shall write (2) in the form
%) a %) c
d
f(x, y) dx dy, (3)
where the summation is made in the order of the differentials
from right to left, i.e. first with respect to y and then with
respect to x, and the limits are in the same order as the differ-
entials, i.e. the limits of y are c and d, and the limits of x are
a and b.*
Referring to fig. 212, we see that we could have made the sum-
mation first with respect to x, thereby finding an approximate
d& I f(x> V) &V1 *n which the order
o •/ c
of summation is first with respect to y and then with respect to x.
I /(3, V)dydx, which
a *> 'c
is merely (2) with the parentheses removed. In this form it is to be noted that
the limits and the differentials are in inverse orders, and that the order of
summation is the order of the differentials from left to right, i.e. first with
respect to y and then with respect to x. In this text this last form of writing
the double integral will not be used. In other books the context will indicate
the form of notation which the writer has chosen.
DOUBLE INTEGRALS
371
expression for the volume of a slice bounded by two planes par-
allel to the YOZ plane. The final summation would have been
with respect to y, the result being the volume expressed by (3).
If this order had been followed, the result would have appeared
in the form
jf (jf a* *>**)**-£*£*<* ****** <4>
Integrals (3) and (4) are called double definite integrate, the
limits in this case being constants. As any function f(x, y}
may be represented graphically by the surface z=f(x, y~), we
are led to the general definition of the double definite integral,
] fffafity*"*] \ f(x, y~)dxdy,
it, as m and n are botl
e sum
as equal to the limit, as
nitely, of the double sum
i and « are both increased indefi-
(5)
where A*, Ay, and (xa y}~) have the meanings already defined.
The integral is called the double integral of f(x, y) over the
area bounded by the lines x=a, x = b, y = c, y=d. This
definition is independent of the graphical interpretation, and
therefore any problem which leads to the limit of a sum (5)
involves a double integral.
174. Doable integral with variable
limits. We may now extend the idea
of a double integral as follows: In-
stead of taking the integral over a
rectangle, as in § 173, we may take
it over an area bounded by any
closed curve (fig. 213) such that a
straight line parallel to either OX or
OY intersects it in not more than
two points. Drawing straight lines
parallel to OY and straight lines parallel to OX, we form
rectangles of area AxAy, some of which are entirely within
"S
v%°
...
K "f
|_. ■'.'■. ]
"n h
Fig. 213
372 MULTIPLE INTEGRALS
the area bounded by the curve and others of which are only
partly within that area. Then
%%/(*, 2/)** Ay, (1)
where the summation includes all the rectangles which are wholly
or partly within the curve, represents approximately the volume
bounded by the plane XOY, the surface z =f(x, y), and the cyl-
inder standing on the curve as a base, since it is the sum of the
volumes of prisms, as in § 173. Now, letting the number of these
prisms increase indefinitely, while A2?=0 and Ay==0, it is evident
that (1) approaches a definite limit, the volume described above.
If we sum up first with respect to y, we add together terms
of (1) corresponding to a fixed value of x, such as xv Then
if MB is the line x = x{J the result is a sum corresponding
to the strip ABCD, and the limits of y for this strip are the
values of y corresponding to x = x{ in the equation of the curve.
That is, if for x = x{, the two values of y are MA=f1(x.*) and
MB =/2(#.), the limits of y are fx(x^) and f2(x^). As different
integral values are given to t, we have a series of terms corre-
sponding to strips of the type" ABCD, which, when the final
summation is made with respect to x, must cover the area
bounded by the curve. Hence, if the least value of x for the
curve is the constant a and the greatest value is the constant
6, the limit of (1) appears in the form
f(x,y)dxdy> (2)
where the subscript i is no longer needed.
On the other hand, if the first summation is made with,
respect to #, the result is a series of terms each of which corre-
sponds to a strip of the type AfB'CfDf, and the limits of x are
of the form <£x(y) and <£2(#), found by solving the equation of
the curve for x in terms of y. Finally, if the least value of y
for the curve is the constant c and the greatest value is the
constant d, the limit of (1) appears in the form
/0> y)dydx. (8)
ff
DOUBLE INTEGRALS
373
While the limits of integration in (2) and (3) are different,
5t is evident from the graphical representation that the integrals
are equivalent.
So far in this chapter f(x, y) has been assumed positive for
all the values of x and y considered, i.e. the surface z =/(#, y)
^was entirely on. the positive side of the plane XOY. If, however,
-/(^ V) *s negative for all the values of x and y considered, the
^reasoning is exactly as in the first case, but the value of the
integral is negative. Finally, if /(#, y) is sometimes positive and
sometimes negative, the result is an algebraic sum, as in § 81.
175. Computation of a double integral. The method of comput-
ing a double integral is evident from the meaning of the notation.
Ex.
.3 />2
:. 1. Find the value of / f xydxdy.
As this integral is written, it is equivalent to / [ / xy dy \ dx, the integral
in parentheses being computed first, on the hypothesis that y alone varies.
f02*ydy = [fj=**-
f 2xdx = [x2]* = 9.
Ex. 2. Find the value of the integral / / xydxdy over the first quadrant
of the circle x2 + y2 = a2.
If we sum up first with respect to y, we find a
series of terms corresponding to strips of the type
A BCD (fig. 214), and the limits of y are the ordinates
of the points like A and B. The ordinate of A is
evidently 0, and from the equation of the circle the
ordinate of B is v a2 — x2, where OA = x. Finally, to
cover the quadrant of the circle the limits of x are
0 and a. Hence the required integral is
0
i
i
O
A.D
Fig. 214
J xydxdy
,r\ Va>-x»
-urn ■*
-IT— -— 1°
"2L 2 4 Jo
= i<*
AC
374 MULTIPLE INTEGRALS
176. Double integral In polar coordinates. Let us assume that
we have a function f(r, #) expressed in polar coordinates and
an area bounded by a curve (fig. 215) which is also expressed in
polar coordinates. As in § 141, we may graphically express the
function by placing z =f(r, dy, where the values of z correspond-
ing to assigned values of r and 8 are laid off on perpendiculars
to the plane of r and 8 at the points determined by the given
values of r and 8. It follows that the graphical representation
of/(r, 8) is a surface.
Let us now try to find the volume bounded by this surface,
the plane of r and 8, and the cylinder standing on the curve
as a base. Proceeding in a manner
analogous to that in § 174, we divide
the area into elements, such as ABCD
(fig. 215), by drawing radius vectors
at distances A8 apart, and concentric
circles the radii of which increase by
Ar. The area of ABCD is the dif-
ference of the areas of the sectors
OBG and OAD. Hence, if OA = r,
area ABCD^^(r+AryA8-^A8=(r+i') ArA0, where e=JAr.
Then the volume of the element standing on the base ABCD.is
f(r, ff)rArA0+f(r, 8)eArA8,
and the total volume is the lunit of the double sum
2)2J[/(n 0)r*rM+f(r, 0)*ArAff\,
or, what is the same thing, the limit of the double sum
%%f(r, 8-)rArA8 = jjf(r, $)rdrd$. (1)
If the summation in (1) is made first with respect to r, the
result is a series of terms corresponding to strips such as A^t (7,2),,
and the limits of r are functions of 0 found from the equation
of the boundary curve. The summation with respect to 8 will
then add all these terms, and the limits of 6 taken so as to
cover the entire area will be constants, Le. the least and the
greatest value of 8 on the boundary curve.
DOUBLE INTEGRALS 375
If, on the other hand, the summation is made first with respect
to 0, the result is a series of terms corresponding to strips such
as AJS^Cfi^ and the limits of 0 are functions of r found from
the equation of the boundary curve. The summation with respect
to r will then add all these terms, and the limits of r will be the
least and the greatest value of r on the boundary curve.
Now f(r, 0) may be any function, and (1), which is inde-
pendent of the graphical representation, is called. the double
definite integral over the area considered. Furthermore, the area
of ABCD has been denoted in (1) by rdrdO, i.e. by the product
of AB and AD, for AB = dr and AD = rd$.
Kx. Find the integral of r1 over the circle r = 2 a cos 0.
If we sum up first with respect to r, the limits are 0 and 2 a cos 6,
found from the equation of the boundary curve, and the result is a
series of terms corresponding to sectors of the type A OB (fig. 216). To
sum up these terms so as to cover the circle,
the limito of & are — - and - - The result is
£X
-4ET
Fin. 213
We might have solved this problem as follows : Since the initial line
ia a diameter of the circle and the values of r2 at corresponding points of
"the two semicircles are the same, it is evident that the required integral
is twice the integral taken over the semicircle in the first quadrant.
By this method the result is
= %atj2coa*$d0
Such use of symmetry as was made in the second solution above is si
often of advantage that the student should always note when there i
symmetry, and arrange his work accordingly.
876 MULTIPLE INTEGRALS
177. Area bounded by a plane curve. Let us in (2), § 174,
denote /j(as) by y1 and /a(V) by ya, and omit f(x, y~). The
result is /•»/*»
/ J fody, (i)
which is evidently the area bounded by the curve in fig. 213.
But r„ -„ „„
J J dxdy = J (jtt-yjdx, (2)
where (ya — y^)dx is the area of the rectangle ABCD.
In the same way we may transform (3) of § 174 into
s:
(x^-xjdy,
(8)
which will represent the same area that is represented by (2),
(x^ — x^dy being the area of the rectangle A'B'CDt.
It is evident that, if the area r
bounded by a plane curve expressed
in rectangular coordinates is found
by double integration, the result of
the first integration is an integral of
the type given in § 125.
Ex. Find the area inclosed by the curve
ft-*-8)- = !-»•(% 217).
Solving the equation of the curve for y
in terms of x, we have
y = z + 3±Vi=x*.
Accordingly we let yl = x + 3 — V4 — x1
and yt — x + 3 + V4 — i!, whence y% — y, = 2 V4 — x\ and take for the
element of area a rectangle such as ABCD. Its area is 2 V4 — x*dx.
Since the curve is bounded by the lines x = — 2 and x — 2, — 2 and 2
are the limits of integration. Hence the area = f 2 V4 — x*dx = 4 ir.
In like manner the area bounded by any curve in polar
coordinates may be expressed by the double integral
JOT'
rdrdS,
(*>
MOMENT OF INERTIA 377
the element of area being that bounded by two radius vectors
the angles of which differ by A0, and by the arcs of two
circles the radii of which differ by Ar.
We may transform (4) into forms similar to (2) or (3), or
we may make the double integration, substituting both sets of
limits in each problem.
If the first integration of (4) is with respect to r, the result
before the substitution of the limits is \rldd, which is exactly
the expression used in computation by a single integration.
178. Moment of inertia of a plane area. The moment of inertia
of a 'particle about an axis is the product of its mass and the square
of its distance from the axis. The moment of inertia of a number
of particles about the same axis is the sum of the moments of
inertia of the particles about that axis. From this definition we
may derive the moment of inertia of a lamina of uniform thick-
ness &, and of density /o, about any axis as follows :
Divide the surface of the lamina into elements of area dA.
Then the mass of any element of the lamina is pkdA. Let R{
be the distance of any point of the ith element from the axis.
We may then take as the moment of inertia of the ith element
RfpkdA, the exact expression evidently being (i£? + ei>)pkdA
(§ 124). If the lamina is divided into n elements, ^?R?pkdA
t=i
is an approximate expression for the moment of inertia of the
lamina. Then, if I represents the moment of inertia of the
lamina, t.=n
I = Lim ]£ Rfpk dA= I R2pk dA, (1)
where the integration is to include the whole lamina.
If in (1) we let k=\ and />=1, the resulting equation is
-/
R*dA, (2)
where / is called the moment of inertia of the plane area which is
covered by the integration. When dA in (1) or (2) is replaced
by either dxdy or rdrd0, the double sign of integration must
be used.
378
MULTIPLE INTEGBALS
Ex. 1. Find the moment of inertia, about an axis perpendicular to the
plane at the origin, of the plane area (fig. 218) bounded by the parabola
y2 = 4 ax, the line y = 2 a, and the axis OY.
We divide the area into elements by straight lines parallel to OX and OY.
Then dA = dx dy, and R2 = x2 + y2, whence the expression for the moment of
inertia of any element is (x2 + y2) dx dy.
If the integration is made first with respect to x, the limits of that
integration are 0 and j— , since the operation is the summing of elements
of moment of inertia due to the elementary rectangles in any strip corre-
sponding to a fixed value of y ; the limit 0 is found from the axis of y,
v2
and the limit — is found from the equation
4a
of the parabola.
Finally, the limits of y must be taken so
as to include all the strips parallel to OX,
and hence must be 0 and 2 a.
2a „*!
Therefore / = C f*a(x2 + y2)dy dx
f2a(—
Jo \19
192 a8 4 a) y
178
105
a1
Fig. 218
Ex. 2. Find the moment of inertia about OF of the plane area bounded
by the parabola y2 = 4 ax, the line y = 2 a, and the axis 0 Y.
Since the above area is the same as that of Ex. 1, the limits of
integration will be the same as there determined, but the integrand will
be changed in that x2 + y2 is replaced by x2, for R = x.
Hence
2« „£-
1=1 J iax2dydx
1 r2a
= 192 a8 Jo ^dy
2 ,
= — a\
21
Ex. 3. Find the moment of inertia, about an axis perpendicular to the
plane at 0, of the plane area (fig. 219) bounded by one loop of the curve
r = a sin 2 0.
We shall take the loop in the first quadrant, since the moments of
inertia of all the loops about the chosen axis are the same by symmetry.
CEKTER OF GRAVITY 379
We divide the area into elements of area by drawing concentric circles
and radius vectors. Then dA = rdrdd (§ 176), and R* = r1, whence the
element of moment of inertia is rxdrd$.
If the first integration is made with respect to r, the result is the moment
of inertia of a atrip bounded by two successive radius vectors and a circular
arc ; hence the limits for r are 0 and a sin 2 &. Since the values of 0 for
the loop of the
evident that those values
6 in the final integration.
= **>*■
179. Center of gravity of plane areas. If the center of gravity
of any physical body can be expressed by two coordinates x and
y, we proved in § 137 that
I xdm j ydm
j dm j dm
where x and y are the coordinates of the point at which the
element of mass dm may be regarded as concentrated.
We may now place
dm=pdxdy, or dm = prdrdB,
where p is the mass per unit area, in which case the above
integrals become double integrals.
Ex. 1. Find the center of gravity of the segment of the ellipse — + ^ = 1
cut off by the chord through the positive ends of the ares of the curve.
This is Ex. 4, § 137, and the student should compare the two solutions.
The equation of the chord is bx + ay = ab.
Dividing the area into elements dxdy (fig. 220), we have dm = pdxdy,
but we may omit p since it is constant. Hence, to determine x and y, we
have to compute the two integrals fixdxdy and CCydxdy over the area
ACBD, and also find that area.
380
MULTIPLE INTEGRALS
The area is the area of a quadrant of the ellipse less the area of the
triangle formed by the coordinate axes and the chord, and accordingly is
\(irab) -%ab = \ab(Tr- 2).
For the integrals the limits of
integration with respect to y are
ab — bx , u /—g n
yx = — and y2 = - Va2- x\
a
b
a
yx being found from the equation
of the chord, and y2 being found
from the equation of the ellipse.
The limits for x are evidently
0 and a.
Fig. 220
|« xdxdy = / (-Wa*-z»- &* + —)<**
0 Jab-bx Jo \a a J
'ab — bx
a
= i ba2.
,b-Vtf~x^
f « ydxdy = — \ (- b2x2 + ab2x) dx
0 Jab-bx d2J0
a
Therefore
= i b2a.
2a
x =
y =
2b
3 (V - 2) ~ 3 (tt - 2)
Ex. 2. Find the center of gravity of the area bounded by the two circles
r = a cos 0, r = b cos 6. (b > a)
It is evident from the symmetry
of the area (fig. 221) that y = 0.
As p is constant, the denominator
of x is the difference of the areas of
the two circles, and is equal to
IT
©'-©'-i-<-< )•
Since x = r cos 0, and the ele-
ment of area is rdrdO, the numer-
ator of x becomes
Fig. 221
6 cos 0
f2 f ° r2 cos 0d0dr= J(68-a8) f* cos* Odd
J it J a cos 0 J tt
Therefore
x =
i7r(&8-a8).
b2 + aft + a2
2(6 + a)
. AREA. 881
180, Area of any surface. Let C (fig. 222) be any closed
carve on the surface f(x, y, z) = 0. Let its projection on the
plane XOY be C. We shall assume that the given surface is
such that the perpendicular to the plane XOY at any point
within the curve C meets the surface in but a single point.
In the plane XOY draw straight lines parallel to OX and OY, ■
forming rectangles of area AarAy, which lie wholly or partly in
the area bounded by C". Through these lines pass planes paral-
lel to OZ. These planes will intersect the surface in curves,
which intersect in points the
projections of which on the
plane XOY are the vertices of
the rectangles > for example,
M is the projection of P. At
every such point as P draw
the tangent plane to the sur-
face. From each tangent plane
there will be cut a parallelo-
gram * by the
drawn parallel
We shall now
define the area
of the surface
/fry, *)=«.„
bounded by the
curve C, as the limit of the sum of the areas of these parallelo-
grams cut from the tangent planes, as their number is made to
increase indefinitely, at the same time that Aar = 0 and Ay=0.
It may be proved that the limit is independent of the manner
in which the tangent planes are drawn, or of the way in which
the small areas are made to approach zero.
If AA denotes the area of one of these parallelograms in a
tangent plane, and 7 denotes the angle which the normal to the
tangent plane makes with OZ, then (§ 145)
Ar Ay =s AA cos 7, (1)
vn in the figure, since it coincides so nearly
382 MULTIPLE INTEGRALS ,
since the projection of AA on the plane XOY is AzAy. The
direction cosines of the normal are, by (9), § 168, proportional
dx dy
and hence
and
^+(£)W
-*+®+®-«
2^=S5^i+(£)V(g)W
According to the definition, to find A we must take the limit
of (3) as Ax = 0 and A# = 0; that is,
-JGU
1+
r+(g)"** w
the area in the
where the integration must be extended
plane XOY bounded by the curve C".
Ex. 1. Find the area of an octant of a sphere of radius a.
If the center of the sphere is taken as the origin of coordinates
the equation of the sphere is
«■+!■+*=* (i)
and the projection of the required
area on the plane XOY is the area
in the first quadrant bounded by
the circle
«• + f = <? (2)
and the axes OX and OY.
From (1), ~ =- -,
££=_?
Therefore A = f f**^- adxd* =l«, C,
J« Jo -vV - ** - f 2 Jo
AREA
383
Sx. 2. The center of a sphere of radius 2 a is on the surface of a right
circular cylinder of radius a. Find the area of the part of the cylinder
intercepted by the sphere.
Let the equation of the
sphere be
s2 + y2 + *2 = 4a2, (1)
"the center being at the
origin (fig. 224), and let
"the equation of the cylin-
der be
y2 + z2-2ay = 0, (2)
"the elements of the cylin-
der being parallel to OX,
To find the projection
of the required area on
the plane XOY it is y Fig. 224
necessary to find the
projection on that plane of the line of intersection of (1) and (2).
Hence (§ 159) we must eliminate z from (1) and (2). The result is
xA
+ 2 ay - 4 a2 = 0.
(3)
From (2),
^ = 0,
dx
to
dy
a-y
aF1W->P53-
a
ana
ha = adydx .
V2 ay — y*
^2ay-tf
Since the area on the positive side of the plane XOY is symmetrical
with respect to the plane YOZ, it is twice the area on the positive side of
the latter plane. Hence we may find this latter area and multiply by 2.
If the first integration is with respect to x, the lower limit is evidently
0 and the upper limit, found from (3), is V4 a2 — 2 ay. For the final
integration with respect to y the limits are 0 and 2 a, the latter being
found from (3).
Therefore A = 2 f*° f^^-£M^== 2aVTa f2"^= 8a*.
ay — y*
As an equal area is intercepted on the negative side of the plane XOY, the
above result must be multiplied by 2. Hence the total required area is 16 a2.
The evaluation of (4) may sometimes be simplified by trans-
forming to polar coordinates in the plane XOY.
884 MULTIPLE INTEGRALS
Ex. 3. Find the area of the sphere x1 + f + z1 = a* included i
cylinder having its elements parallel to OZ and one loop of the ci
r = a cos 2 $ (fig. 225) in the plane XOY as its directrix.
Proceeding s
1 Ex. 1, we find the integrand -
rming this integrand to polar coordinates, and multiplying by rdrrffl,
i have
irdrd$
dA =
vV-r3
It is evident that the required area
is twice the area cut out of the sphere
on one side of the plane XOY, and
that this latter area is twice the area
over the half of the loop of the curve
r = a cos 2 9 which is in the first
quadrant.
Hence we integrate first with respect to r from 0 to a cos 2 6, and then
integrate with respect to & from 0 to — . the limits of integration all being
Fig. 225
determined from the equatioi
i to obtain the required area.
2 $. This integral we multiply by
*2» ardBdr
Therefore A = i C* f '
= 4<r>J*(l-sin20)d0
= a»(w-2>
If the required area is projected on the plane YOZ, we have
where the integration extends over the projection of the area
on the plane YOZ; and if the required area is projected on the
plane XOZ, we have
-114
1+
J dxdz,
(6)
where the integration extends over the projection of the area
on the plane XOZ.
TRIPLE INTEGRALS
385
181. Triple integrals. 1. Rectangular co&rdinates. Let any
volume (fig. 226) be divided into rectangular parallelepipeds
of volume AxAyAz by planes parallel respectively to the coor-
dinate planes, some of the parallelepipeds extending outside
the volume in a manner similar to that in which the rectangles
in § 174 extend outside the area. Let (xit yp zk) be a point
of intersection of any three of these planes and form the sum
as in § 174. Then the limit of this sum as n, m, and p increase
indefinitely, while Az = 0, Ay = 0, Az = 0, so as to include all
points of the volume, is called the triple integral of f(x, y, z)
throughout the volume. It is
denoted by the symbol
fff'<-*>
z)dxdydz, (1)
the limits remaining to be substi-
tuted. If the summation is made
first with respect to z, x and
remaining constant, the result is
to extend the integration through-
out a column of cross section
AxAy ; if next x remains constant
and y varies, the integration is
extended so as to combine the columns into s
as x varies, the slices are combined so as to complete the
integration throughout the volume.
The volume of the parallelepiped with edges dx, dy, dz is the
element of volume dV, and hence
ss ; and, finally,
dV= dxdydz.
(2)
2. Cylindrical co&rdinates. If the x and the y of the rectan-
gular coordinates are replaced by polar coordinates r and 8 in
the plane XOY, and the z coordinate is retained with its original
386 MULTIPLE INTEGRALS
significance, the new coordinates r, 8, and z are called cylindri-
cal coordinates. The formulas connecting the two systems of
coordinates are evidently
x = r oos 8, y = r&md, z = z.
Turning to fig. 227, we see that z = z1
determines a plane parallel to the plane
XOY, that 8 = 01 determines a plane
MONP, passing through OZ and making
an angle 81 with the plane X0Zy and
that r = r, determines a right circular
cylinder with radius r1 and OZ as its
axis. These three surfaces intersect at v
the point P.
The element of volume in cylindrical coordinates (fig. 228)
is the volume bounded by two cylin-
ders of radii r and r + Ar, two planes
corresponding to z and z + Az, and
two planes corresponding to 8 and
8+A8. It is accordingly a cylinder
with its altitude equal to Az and the
area of its base approximately equal
to rA8Ar (§ 176). Hence, in cylin-
drical coordinates,
dV=rdrd8dz, (3)
and the triple definite integral in cylindrical coordinates is
ff(r, 8,z)rdrd8dz, (4)
JJJ*
the limits remaining to be substituted.
3. Polar coordinates. In fig. 229 the
cylindrical coordinates of P are 03f=r,
MP = z, and ZL0M=8. If instead of
placing 0M=r we place 0P = r, and
denote the angle NOP by 4>, we shall
have r, <f>, and 8 as the polar coordi-
nates of P. Then, since 0N= OP cos ^
TRIPLE INTEGRALS
387
and OM = OP sin <f>, the following equations evidently express
the connection between the rectangular and the polar coordi-
nates of P:
z = r cos <f>, x = r sin <f> cos 0, y = r sin £ sin 5.
The polar coordinates of a point determine three surfaces
"which intersect at the point. For 0 = 01 determines a plane
(fig. 230) through OZ, making
"the angle 0X with the plane XOZ\
*f> = £ determines a cone of revo-
lution, the axis and the vertical
angle of which are respectively
OZ and 2 ^ ; and r = r1 deter-
mines a sphere with its center
at O and radius rx.
The element of volume in
polar coordinates (fig. 231) is
the volume bounded by two
spheres of radii r and r + Ar,
two conical surfaces corresponding to <f> and <£ + A<£, and two
planes corresponding to 0 and 0 + A0. The volume of the
spherical pyramid O-ABCD is
equal to the area of its base
ABCD multiplied by one third
of its altitude r. To find the
area of ABCD we note first that
the area of the zone formed by
completing the arcs AD and BC
is equal to its altitude, r cos <f> —
r cos (<f> + A£), multiplied by 2 tit.
Also the area of ABCD is to the
area of the zone as the angle Ad
is to 2tt.
Fig. 230
Fig. 231
Hence area ABCD = rA0 [r cos <f> — r cos (<£ + A<£)]
and vol O-ABCD = l r*A0 [cos <f> - cos (<£ + A£)].
Similarly, vol 0-EFGH= J(r+Ar)8A0[cos<£-cos(<£+A<£)],
388 MULTIPLE INTEGRALS
Therefore vol ABCDEFGH = 1 [(r + Ar)8 - r8] A0 [cos <j>
— cos(£+A<£)].
£ [ (r + Ar) 8 - r8] = r2 Ar + r A^ 2 + *- A? 8
= (r2 + €1)Ar.
cos <f> — cos (<£ + A<£) = — A cos $
= (sin * + e2) A*. (§ 77)
Hence vol ABCDEFGH = (f sin <£ + €8) Ar A0 A<£,
which differs from r2 sin <f> Ar A0 A<f> by an infinitesimal of a
higher order.
Accordingly we let dV=r2 sui(f>drd<f)d0y (5)
and the triple integral in polar coordinates is
m
f(r, <£, 0) r2 sin <j> dr d$ d0, (6)
the limits remaining to be substituted.
It is to be noted that dV is equal to the product of the
three dimensions AB, AD, and AE, which are respectively rdfa
rsia<f>d0, and dr.
182. Change of coordinates. When a double integral is given
in the form I //(#, y)dxdy, where the limits are to be substi-
tuted so as to cover a given area, it may be easier to determine the
value of the integral if the rectangular coordinates are replaced
by polar coordinates. Then /(a?, y) becomes f(r cos 0, r sin 0),
i.e. a function of r and 0. As the other factor, dxdyy indicates
the element of area, we may replace dxdy by rdrd0. These two
elements of area are not equivalent, but the two integrals are
nevertheless equivalent, provided the limits of integration in each
system of coordinates are taken so as to cover the same area.
In like manner the three triple integrals
IIP
/(#, y, z)dxdydz,
f(r cos 0, r sin 0, z) rdrd0dzy
f(r sin <f> cos 0, r sin <f> sin 0, r cos $) r2 sin <f>drd<f>d0
VOLUME 389
are equivalent when the integration ia taken over the same volume
in all three and the limits are so taken in each as to include the
total volume to be considered.
183. Volume. In § 181 we found expressions for the element
of volume in rectangular, in cylindrical, and in polar coordinates.
The volume of a solid bounded by any surfaces will be the limit
of the sum of these elements as their number increases indefi-
nitely while their magnitudes approach the limit zero. It will
accordingly be expressed as a triple integral.
Ex. 1. Find the volume bounded by the ellipaoid —
r^-r
= 1.
From symmetry (fig. 232) it is evident that the required volume is
eight timea the volume in the first octant hounded hy the surface and the
coordinate planes.
In summing up the rectangu-
lar parallelepipeds dxdydz to
form a prism with edges paral-
lel to OZ, the limits for z are 0
the latter
being found from the equation
of the ellipsoid.
Summing up next with respect
to y, to obtain the volume of a
slice, we have 0 as the lower limit
of y, and &-\/l ;
by solving the equatii
the upper limit This latter limit i
^ + ^ = 1, found by letting z — 0 in
of the ellipsoid ; for it is in the plane z = 0 that the ellipsoid has the
greatest extension in the direction OY, corresponding to any value of x.
Finally, the limits for x are evidently 0 and a.
-n'^/.^
v.hdydz
=*<s:ir1~H-i-*<"»
=2'"cX"(1-i)'"
390 MULTIPLE INTEGRALS
It is to be noted that the first integration, when rectangular
coordinates are used, leads to an integral of the form
JJ(.z2-zDdxd^
where z% and zx are found from the equations of the boundary
surfaces. It follows that many volumes may be found as easily
by double as by triple integration.
In particular, if zx= 0, the volume is the one graphically repre-
senting the double integral (§ 174).
Ex. 2. Find the volume bounded by the surface z = ae-&+**> and the
plane 2 = 0.
To determine this volume it will be advantageous to use cylindrical
coordinates. Then the equation of the surface becomes z = ae~ *"*, and the
element of volume is (J 181) rdrdOdz, .
Integrating first with respect to z, we have as the limits of integration
0 and ae~^. If we integrate next with respect to r, the limits are 0 and oo,
for in the plane z = 0, r = go, and as z increases the value of r decreases
toward zero as a limit. For the final integration with respect to $ the
limits are 0 and 2ir.
Therefore
2 ir /» ao /» cut— r*
V= f f f rdOdrdz
Jo Jo Jo
I re-^d6dr
o Jo
J<*2n A
dO
0
= ira.
In the same way that the computation of the volume in Ex. 2
has been simplified by the use of cylindrical coordinates, the
computation of a volume may be simplified by a change to polar
coordinates ; and the student should always keep in mind the
possible advantage of such a change.
184. Moment of inertia of a solid. Following the method
of § 178 we divide the volume of the solid into n elements Av
and multiply each element by its density p. Then, if Rt is the
distance of any point of the ith element from the axis about
which the moment of inertia is to be taken, we may take RfpAv
MOMENT OF INERTIA 391
as the moment of inertia of that element. If I denotes the
moment of inertia of the solid, Vi^pAv is an approximate
i = i
expression for L Finally, if we let n = oo and at the same time
let each element of volume approach zero as a limit, we have
= Limit V^t.2/oAv= I R*pdv,
where Ry /o, and dv are to be expressed in terms of the same
variables and the proper limits of integration substituted. In
particular, if dv is replaced by any one of the three elements
of volume determined in § 181, the integral becomes a triple
integral.
Ex. Find the moment of inertia of a sphere of radius a about a diameter
if the density varies directly as the square of the distance from the diameter
about which the moment of inertia is to be taken.
We shall take the center of the sphere as the origin of coordinates, and
the diameter about which the moment is to be taken as the axis of z. The
problem will then be most easily solved by using cylindrical coordinates.
The equation of the sphere will be r2 + z2 = a2, and dv = rdrdOdz,
R = r, and p = fcr2, so that we have to find the value of the triple integral
kCCCr*dOdrdz.
Since the solid is symmetrical with respect to the plane z = 0, we
shall take 0 and Va2 — r2 as the limits of integration with respect to z,
the latter limit being found from the equation of the sphere, and double
the result.
Integrating next with respect to r, we have the limits 0 and a, thereby
finding the moment of a sector of the sphere. To include all the sectors,
we have to take 0 and 2 w as the limits of 0 in the last integration.
Therefore I = 21c I I I r^dOdrdz.
Jo Jo Jo
As a result of the first integration,
I=2kf2irfar6Va*-rid0dr.
Jo Jo
After the second integration,
I=j\hkaif2Wd0f
Jo
and, finally, / = ^y kira1.
392 MULTIPLE INTEGEALS
185. Center of gravity of a solid. The center of gravity of a
solid has three coordinates, xy y, 5, which are defined by the
equations
I xdm I ydm I zdm
x = ^- v y = - » s = - >
I dm I dm I dm
where dm is an element of mass of the solid and x, y, and z are
the coordinates of the point at which the element dm may be
regarded as concentrated. The derivation of these formulas is
the same as that in § 137.
When dm is expressed in terms of space coordinates, the
integrals become triple integrals, and the limits of integration
are to be substituted so as to include the whole solid.
The denominator of each of the preceding fractions is evi-
dently My the mass of the body.
Ex. Find the center of gravity of a body of uniform density, bounded
by one nappe of a right circular cone of vertical angle 2 a and a sphere of
radius a, the center of the sphere being at the vertex of the cone.
If the center of the sphere is taken as the origin of coordinates and the
axis of the cone as the axis of z, it is evident from the symmetry of the
solid that x = y = 0. To find z, we shall use polar coordinates, the equations
of the sphere and the cone being respectively r = a and <f> = a.
I II r cos (ji'r3 am <f>d$d<f>dr
Then z =
Iff r*8in€l>dOd€l>dr
The denominator is the volume of a spherical cone the base of which
is a zone of one base with altitude a (1 — cos a) ; therefore its volume
equals § ira* (1 — cos a).
f f f Is cos <f> sin (frdOdxfrdr = J a4 f f^cos^sin^ctftf^
= ia4(l-cos2a) f2nd6
Jo
= J7ra4(l — cos2 a).
Therefore 5 = | (1 + cos a) a.
■V"
*
ATTRACTION
Attraction. The formula
rcos 6 dm
/cos a dm
(S 188)
for the component of attraction in the direction OX is entirely
general. Similar formulas for the components in the direc-
tions OT and OZ may be deduced. The application of these
formulas requires us to express
R, cos. 6, and dm in terms of
the same variables, and to substi-
tute limits of integration so as
to include the whole of the
attracting mass. In general, the
integral, after the substitution of
the variables, will be a double
or triple integral.
Ex. Find the attraction due to a
homogeneous circular cylinder of
density p, of height A, and radius
of cross section a, on a particle in
the line of the axis of the cylinder
at a distance 6 units from one end
of the cylinder.
Take the particle at the origin of
coordinates (fig. 233), and the axis of
the cylinder as OZ. Using cylindrical
coordinates, we have dm = prdrd&dz
and RsVz' + r*.
From the symmetry of the figure the resultant
tion in the directions OX and OY are zero, and cc
resultant component in the direction OZ.
Therefore, letting A represent the component
we have A = Pf*w f fi*h-—H d6drdz,
where the limits of integration are evident from fig. 233.
A=rf'"f°( r -- T -\d»d,
Jo Jo VvVTr* V(ft + ft)* + rV
= pCi"(h +V6* + o*-V(6 + ft)* + a^dd
= 2 wp(h + Vi* + a1 - V(6 + ft)* + a*).
omponents of attrac.
$ = Z for the
the direction OZ,
394 MULTIPLE INTEGRALS
PROBLEMS
Find the values of the following integrals :
'•jfjfV* 4'ITlog^-
2 6. Express in two ways the mte-
w gral of /(x, y) over the smaller area
3. rrxsin(x + y)^y. bounded by the curves ^ + y = 2 a
Jo Jo and (x — a)2 + y2 = a2.
Find the values of the following integrals :
-»8co«4
r*am*0d0dr.
• 2ac<M0 ns /»8co«0
f I rdfldr. 8. I I
0 »/ o */ - | »/o
9. Find the integral of r over one loop of the curve r = a sin 2 0.
10. Find the integral of r over the area bounded by the initial
line and the curves r == a and r = a(l + cos 0).
11. Find the area bounded by the curves y = x2 and y = 2 — x2.
12. Find the area bounded by the hyperbola xy = 4 and the line
x + y — 5 = 0.
13. Find the area bounded by the confocal parabolas y2 = 4 ax
+ 4a2, tf=-4:bx + 4cb\
14. Find the area of the loop of the curve (x + y)2 = y2 (y + 1).
15. Find the area bounded by the curves x3 4- y2 = 25, 3 y2 = 16 x,
3x2 = 16y.
16. Find each of the areas bounded by the circle x2 -|- y2 = 5 a2
and the witch y = 9 , . — =•
9 x2 + 4 a2
17. Find the area bounded by the circles r = a cos 0, r = a sin 0.
18. Find the area cut off from a loop of the curve r— a cos 2 0
by the curve r = - •
19. Find the area cut off from the lemniscate r2 = 2 a2 cos 2 0 by
the straight line r cos 0 = -jr- a.
PROBLEMS 395
20. Find the area bounded by the limaQon r = 2 cos $ + 3 and
the circle r = 2 cos $.
21. Find the area which is outside the circle r = a and inside
the cardioid r = a(l + cos 0).
22. Find the area in the first quadrant bounded by the circle
r = 2 a sin 0 and the lemniscate r2 = 2 a2 cos 2 9.
23. Find the moment of inertia of the area bounded by the hyper-
bola xy = 4 and the line x + y — 5 = 0 about an axis perpendicular
to its plane at 0.
24. Find the moment of inertia of the area bounded by the curves
y = x2y y = 2 — x2 about an axis perpendicular to its plane at 0.
25.' Find the moment of inertia about OF of the area bounded by
OF and the parabola y1 = 1 — x.
26. Find the moment of inertia about an axis through 0 perpen-
dicular to the coordinate plane of that part of the first quadrant
included between the first two successive coils of the spiral r = e*9.
27. Find the moment of inertia of the entire area bounded by
the curve r*= a2 sin 3 0 about an axis perpendicular to its plane at
the pole.
28. Find the moment of inertia of the area of one loop of the lem-
niscate r2=2aa cos 2 0 about an axis perpendicular to its plane at
the pole.
29. Find the moment of inertia of the total area bounded by the
curve r2 = a2 sin $ about an axis in its plane perpendicular to the
initial line at the pole.
30. Find the moment of inertia about OX of the area bounded by
the parabolas y2 = 4 ax + 4 a2, y2 = — 4 bx + 4 b*.
31. Find the moment of inertia about OF of the area of the loop
of the curve y1 = x2(2 — x).
32. Find the moment of inertia of the area of the cardioid
r = a (1 4- cos 0) about an axis perpendicular to its plane at the pole.
33. Find the moment of inertia of the area of a circle of radius a
about an axis perpendicular to the plane of the circle at any point
on its circumference.
34. Find the moment of inertia about its base of the area of a
parabolic segment of height h and base 2 a.
396 MULTIPLE INTEGRALS
35. Find the moment of inertia about OX of the area bounded on
the left by an arc of the curve y1 = ax + a2 and on the right by an
arc of the curve x2 + f = «2-
36. Find the moment of inertia of the area of one loop of the
lemniscate r2 = 2 a2 cos 2 0 about an axis in its plane perpendicular
to the initial line at the pole.
37. Find the moment of inertia about the initial line as an axis
of the area of the cardioid r = a (cos 04-1) above the initial line.
38. Find the moment of inertia of the area bounded by a semi-
circle of radius a and the corresponding diameter, about the tangent
parallel to the diameter.
39. Find the moment of inertia of the area of a loop of the curve
r = a cos 2 0 about the axis of the loop as an axis.
40. Find the moment of inertia of the area of the circle r = a
which is not included in the curve r = a sin 2 9 about an axis
perpendicular to its plane at the pole.
41. Determine the center of gravity of the half of a parabolic
segment of altitude 9 in. and of base 12 in. formed by drawing a
straight line from the vertex of the segment to the middle point
of its base.
42. Find the center of gravity of a lamina in the form of a
parabolic segment of altitude 7 in. and of base 28 in. if the density
at any point of the lamina is directly proportional to its distance
from the axis of the lamina.
43. Find the center of gravity of the area of a loop of the curve
a4//2 = a2xk — a6.
44. Find the center of gravity of the area bounded by the parabola
#* 4- V = a an^ the circle x2 4- if = a2.
45. Find the center of gravity of the area bounded by the cardioid
r = a (cos 6 4- 1).
46. Find the center of gravity of the area bounded by the parabola
8 a8
x2 = 4 ay and the witch y = 8 a •
x —j- tc a
47. A plate is in the form of a sector of a circle of radius a, the
angle of the sector being 2 a. If the thickness varies directly as the
distance from the center, find its center of gravity.
PROBLEMS 397
48. Find the center of gravity of the area in the first quadrant
2 2 2
bounded by the curve x$ 4- y* = #3 and the line x + y = a.
49. The density at any point of a lamina in the form of a loop
of the curve r = a cos 2 9 is directly proportional to its distance
from the point of the loop. Determine its center of gravity.
50. Find the center of gravity of the area bounded by the limacon
r = 2 cos 9 + 3.
51. Find the center of gravity of the area bounded by the curve
Q
r = a sin - as 0 changes from 0 to 2 tt.
52. Find the center of gravity of the area bounded by the cissoid
xs
y1 = - and its asymptote.
Z a — x
53. Find the center of gravity of the area cut off from the lemnis-
V3
cate r2 = 2 a2 cos 2 9 by the straight line r cos 9 = —^- «•
Zi
X? 'Z/2
54. From a homogeneous elliptic plate, -5 + jz = 1, is cut a circular
plate of radius rlr < -) with center at I -> Oj. Find the center of
gravity of the part left.
5 5 . Find the area of the surface cut from the paraboloid y2-!- «2 = 4 ax
by the cylinder y1 = ax and the plane x = Sa.
56. Find the area of the surface of the cone x2 4- y2 — 4 z2 = 0
cut out by the cylinder x2 4- y2 — 4 % = 0.
57. Find the area of the surface cut from the cylinder x2-\-y2 = a2
by the cylinder y*-\-z2 = a2.
58. Find the area of the surface of a sphere of radius a inter-
cepted by a right circular cylinder of radius ^a, if an element of
.the cylinder passes through the center of the sphere.
59. Find the area of the sphere x2 -f y^ + z2 = a2 included in the
cylinder with elements parallel to OZ and having for its directrix in
the plane XOY a single loop of the curve r = a cos SO.
60. Find the area of the surface of the cylinder x2 4- if — 2 ax = 0
bounded by the plane A'OFand a right circular cone having its vertex
at 0, its axis along OZ, and its vertical angle equal to 90°.
398 MULTIPLE INTEGRALS
61. Find the area of the paraboloid x2 4- y1 = 2 az included in the
cylinder with elements parallel to OZ and having for its directrix in
the plane XOY one loop of the curve r2 = a2 sin 2 0.
62. Find the area of the surface z = xy included in the cylinder
63. Find the area of that part of the surface z = — 9 the projec-
tion of which on the plane XOY is bounded by the curve r2 = a2 cos 0.
64. Find the area of the surface of the cylinder sc2 4- y2 — 2 ax = 0
included in the cone x2 — y2 + 2 z2 = 0.
65. Find the area of the sphere x2 + y2 4- z2 = 4 a2 bounded by
the intersection of the sphere and the right cylinder the elements
of which are parallel to OZ and the directrix of which is the cardioid
r = a (cos 0 + 1) in the plane XOY.
66. Find the area of the surface of the sphere (x—ay+tf+sP—a*
included in one nappe of the cone x2 4- y* — «2 = 0.
Find the values of the following integrals :
1 f* p^^? dxdydz
67.
Iff.
nVJZ^ ^Vg'-aJ-y* yz dxdydz
I e*+y+zdxdydz.
Jo
f* f° f rzdddrdz
J--Ja»ineJo /a2 ^2 i g2\J
7o ii C°* Crdddrdz
Jl
rV
73. J I J r sin2 <£ cos <£ cos 0 ddd^dr.
Jo Jo J a sin 0
J^»»r s%2ir s+a CO* 0
I I I rsin8<£a>c?0dr.
0 »/0 »/0
PROBLEMS 399
76. Find the volume bounded by the surface x* + y^ + z^ = afr
and the coordinate planes.
76. Find the volume of a cylindrical column bounded by the
surfaces y = x2, x = y2, z = 0, z = 12 4- y — #2.
77. Find the volume bounded by the plane z = 0 and the
cylinders sc2 + y2 = a2, y2 = a2 — az.
78. Find the volume bounded by the surfaces r2 = &«, « = 0,
r ■= a cos 0.
79. Find the volume bounded by the sphere x2 + y* + z2 = 5 and
the paraboloid x2 + y2 = 4 z.
80. Find the volume bounded by the cylinder z2 = x + y and the
planes x = 0, y = 0, « = 4.
81. Find the volume of the paraboloid y2 + z2 = 2 a cut off by
the plane y = cc — 1.
82. Find the volume bounded by a sphere of radius a and a right
circular cone, the axis of the cone coinciding with a diameter of the
sphere, the vertex being at an end of the diameter and the vertical
angle of the cone being 90°.
83. Find the total volume bounded by the surface (x2 + y2 4- z2)*
= 27 a8xyz. (Change to polar coordinates.)
84. Find the volume bounded by the plane XOY, the cylinder
a^-l-y2 — 2ax =z 0, and the right circular cone having its vertex at
0, its axis coincident with OZ, and its vertical angle equal to 90°.
2/ 1i Z
85. Find the total volume bounded by the surface -5 + 75 + 1 = 1-
J aA oA c*
86. Find the volume bounded below by the paraboloid x2 + y* = az
and above by the sphere x2 -\- y2 -\- z2 — 2az = 0.
87. Find the volume bounded by the surface b2z2 = y* (a2 — x2)
and the planes y = 0, y = b.
88. Find the volume cut from a sphere of radius a by a right
circular cylinder of radius -> one element of the cylinder passing
Li
through the center of the sphere.
89. Find the total volume bounded by the surface (x2 + y2 + z2)2
= axyz.
400 MULTIPLE INTEGRALS
90. Find the volume in the first octant bounded by the surfaces
z = (x + y)2y x2 + y2 = a2.
91. Find the volume of the sphere x2 + y2 + z2 = a2 included in
a cylinder with elements parallel to OZ, and having for its directrix
in the plane XOY one loop of the curve r3 = a2 cos 2 $.
92. Find the volume bounded by the surfaces az = xy, x-\-y + z
= a, z = 0.
93. Find the total volume which is bounded by the surface x% + y*
+ s$ = a$.
94. Find the total volume which is bounded by the surface r2 + z2
= 2 ar cos 2 0.
95. Find the moment of inertia about its axis of a hollow right
circular cylinder of mass M, the inner radius and the outer radius
of which are respectively rx and ra: (1) if the cylinder is homoge-
neous; (2) if the density of any particle is proportional to its
distance from the axis of the cylinder.
96. A solid is bounded by the plane z = 0, the cone z = r (cylindri-
cal coordinates), and the cylinder having its elements parallel to OZ
and its directrix one loop of the lemniscate r2 = 2 a2 cos 2 $ in the
plane XOY. Find its moment of inertia about OZ if the density at
any point varies directly as its distance from OZ.
97. Find the moment of inertia of a homogeneous right circular
cone of density p, of which the height is h and the radius of the base
is a, about an axis perpendicular to the axis of the cone at its vertex.
98. A ring is cut from a homogeneous spherical shell of density p,
the inner radius and the outer radius of which are respectively 4 ft
and 5 ft., by two parallel planes on the same side of the center of
the shell and distant 1 ft. and 3 ft. respectively from the center.
Find the moment of inertia of this ring about its axis.
99. A mass M is in the form of a right circular cone of altitude
h and with a vertical angle of 120°. Find its moment of inertia
about its axis if the density of any particle is proportional to its
distance from the base of the cone.
100. The radius of the upper base and the radius of the lower
base of the frustum of a homogeneous right circular cone are respec-
tively ax and aa, and its mass is M. Find its moment of inertia about
its axis.
PROBLEMS 401
101. The density of any point of a solid sphere of mass M and
radius a is directly proportional to its distance from a diametral
plane. Find its moment of inertia about the diameter perpendicular
to the above diametral plane.
102. Given a right circular cylinder of mass M, height h, and
radius a, the density of any particle of which is k times its distance
from the lower base. Find the moment of inertia of this cylinder
about a diameter of its lower base.
103. Find the moment of inertia about OZ of that portion of the
surface of the hemisphere z = Va2 — x2 — y1 which lies within the
cylinder x2 + y2 = ax.
104. A homogeneous solid of density p is in the form of a hemi-
spherical shell, the inner radius and the outer radius of which are
respectively rx and r2. Find its moment of inertia about any diam-
eter of the base of the shell.
105. A homogeneous anchor ring of mass M is bounded by the
surface generated by revolving a circle of radius a about an axis in
its plane, distant b(b > a) from its center. Find the moment of
inertia of this anchor ring about its axis.
106. The density at any point of the hemisphere z = Va2 — x2 — y2
is k times its distance from the base of the hemisphere. Find the
moment of inertia about OZ of the portion of the hemisphere in-
cluded in the cylinder x2 + y2 = ax.
107. Through a homogeneous spherical shell of density p, of
which the inner radius and the outer radius are respectively ax and
aa, a circular hole of radius b(b< a^) is bored, the axis of the hole
coinciding with a diameter of the shell. Find the moment of inertia
of the ring thus formed about the axis of the hole.
108. Find the center of gravity of the portion of a uniform wire
in the form of the curve x = at2, y = %at*y z = %at*, between the
points for which t = 0 and t = 1.
109. Find the center of gravity of a uniform wire in the form of
the helix x = a cos 0, y = a sin 0, z = kd, between the points for
which 6 = 0 and $ = $v When will the center of gravity fall on the
axis of the helix ?
110. Find the center of gravity of a homogeneous solid bounded
by the coordinate planes and the surface #* + y* + z* = a*.
402 MULTIPLE INTEGRALS
111. Find the center of gravity of a homogeneous body in the
/j»2 ^2 g2
form of an octant of the ellipsoid —2 4- 75 + -5 = 1-
112. Find the center of gravity of the homogeneous solid bounded
by the surfaces z = 0, y = 0, y = b, b2z2 = j/*(a2 — x2).
113. Find the center of gravity of a homogeneous solid bounded
by the paraboloid a2x2 + b2y* = « and the plane z = c.
114. A ring is cut from a homogeneous spherical shell of density
p, the inner radius and the outer radius of which are respectively
4 ft. and 5 ft., by two parallel planes on the same side of the
center of the shell and distant 1 ft. and 3 ft. respectively from the
center. Find the center of gravity of this ring.
115. Find the center of gravity of a homogeneous solid bounded
by a spherical surface of radius b and two planes passing through
its center and including a dihedral angle 2 a.
116. Find the center of gravity of a hemisphere of radius a if
the density at any point varies directly as the distance of the point
from the base of the hemisphere.
117. Find the center of gravity of a homogeneous solid bounded
by the surfaces of a right circular cone and a hemisphere of radius
a which have the same base and the same vertex.
118. Find the center of gravity of an octant of a sphere of radius
a if the density at any point varies directly as its distance from the
center of the sphere.
119. Find the center of gravity of a right circular cone of altitude
a, the density of each circular slice of which varies directly as the
square of its distance from the vertex.
120. Find the center of gravity of a homogeneous solid bounded
by two concentric spherical surfaces of radii 4 ft. and 5 ft. respec-
tively and a plane through the common center of the two spherical
surfaces.
121. Find the center of gravity of a homogeneous solid in the
form of the frustum of a right circular cone, the height of which
is h and the radius of the upper base and the radius of the lower
base of which are respectively rx and ra.
PROBLEMS 403
122. A solid is bounded by a sphere of radius a and a right circu-
77"
lar cone, the vertical angle of which is — > the vertex of which is on
o
"the surface of the sphere, and the axis of which coincides with a
diameter of the sphere. Find its center of gravity if the density at
any point is k times its distance from the axis of the cone.
123. Find the attraction of a hemisphere of radius a on a particle
of unit mass at the center of its base if the density at any point of
the hemisphere varies directly as its distance from the base.
124. A homogeneous solid of density p is bounded by the plane
» = 3 and the surface z* = ar* 4- y2. Find the attraction of this solid
on a particle of unit mass at the origin of coordinates.
125. A portion of a right circular cylinder of radius a and uniform
density p is bounded by a spherical surface of radius b (b > a), the
center of which coincides with the center of the base of the cylinder.
Find the attraction of this portion of the cylinder on a particle of
unit mass at the middle point of its base.
126. A portion of a right circular cylinder of radius a is bounded
by a spherical surface of radius b (b > a), the center of which coin-
cides with the center of the base of the cylinder. Find the attraction
of this portion of the cylinder on a particle of unit mass at the
middle point of its base, the density of any particle of the cylinder
being proportional to its distance from the axis of the cylinder.
127. Show that the attraction of a segment of one base, cut from
a homogeneous sphere of radius a, on a particle of unit mass at its
vertex is 2 irhp 1 1 — - \l— )> where p is the density of the sphere
and h is the height of the segment.
128. A ring is cut from a homogeneous spherical shell of density
p, the inner radius and the outer radius of which are respectively
4 ft. and 5 ft., by two parallel planes on the same side of the center
of the shell and distant 1 ft. and 3 ft. respectively from the center.
Find the attraction of this ring on a particle of unit mass at the
center of the shell.
129. The density of a hemisphere of mass M and radius a varies
directly as the distance from the base. Find its attraction on a
particle of unit mass in the straight line perpendicular to the base
404 MULTIPLE INTEGRALS
at its center, and at a distance a from the base in the direction
away from the hemisphere.
130. A solid of mass M is bounded by a right circular cone of
vertical angle 90° and a spherical surface of radius 2 ft., the center
of the spherical surface being at the vertex of the cone. If the
density of any particle of the above solid is directly proportional to
its distance from the vertex of the cone, find the attraction of the
solid on a particle of unit mass at the vertex of the cone.
131. The vertex of a right circular cone of vertical angle 2 a is
at the center of a homogeneous spherical shell, the inner radius and
the outer radius of which are respectively ax and a2. Find the attrac-
tion of the portion of the shell outside the cone on a particle of unit
mass at the center of the shell, in terms of the attracting mass.
132. The density at any point of a given solid of mass M in the
form of a hollow right circular cylinder is directly proportional to
its distance from the axis of the cylinder. If the height of the
cylinder is 2 ft., and its inner radius and outer radius are respec-
tively 1 ft. and 2 ft., find its attraction on a particle of unit mass
situated on its axis 2 ft. below the base.
CHAPTER XVII
INFINITE SERIES
187. Convergence. The expression
01 + a2 + a8 + a4 + a6 + ---, (1)
where the number of the -terms is unlimited, is called an infinite
series. .
An infinite series is said to converge, or to be convergent, when the
sum of the first n terms approaches a limit as n increases without limit.
Thus, referring to (1), we may place
*2=ai+a2>
*8=ai+a2+a8'
Then, if Lim8n=^4,
n= ao
the series is said to converge to the limit A. The quantity A is
frequently called the sum of the series, although, strictly speaking,
it is the limit of the sum of the first n terms. The convergence
of (1) may be seen graphically by plotting sx, s2, «3, • • •, sn on the
number scale, as in § 3.
A series which is not convergent is called divergent This may
happen in two ways : either the sum of the first n terms increases
without limit as n increases without limit ; or sn may fail to approach
a limit, but without becoming indefinitely great.
Ex. 1. Consider the geometric series
a + ar + ar2 + ar* + • • •.
1 — r*
Here sn = a + ar + ar2 + • • • + ar11-1 = a • Now if r is numerically
1 — r
less than 1, r" approaches zero as a limit as n increases without limit ; and
ac 405
406 INFINITE SERIES
therefore Lim sn = If, however, r is numerically greater than 1,
r" increases without limit as n increases without limit ; and therefore sn
increases without limit. If r = 1, the series is
and therefore sn increases without limit with n. If r = — 1, the series is
a — a + a — a + ••',
and sn is alternately a and 0, and hence does not approach a limit.
Therefore, the geometric series converges to the limit when r is numeri-
1 — r
cally less than unity, and diverges when r is numerically equal to, or greater
than, unity.
Ex. 2. Consider the harmonic series
l + _ + _ + I + ± + i + _4.±-j....4.±+...
2 3 4 5 6 7 8 n '
consisting of the sum of the reciprocals of the positive integers. Now
and in this way the sum of the first n terms of the series may be seen to
be greater than any multiple of J for a sufficiently large n. Hence the
harmonic series diverges.
188. The comparison test for convergence. If each term of a
given series of positive numbers is less than, or equal toy the corre-
sponding term of a known convergent series, the given series
converges.
If each term of a given series is greater than, or equal to, the
corresponding term of a known divergent series of positive numbers,
the given series diverges.
Let 0J+ a2+ a8+ a4H (1)
be a given series in which each term is a positive number, and let
J1+J2+J8+J4+"- (2)
be a known convergent series such that ak=bk.
Then, if sn is the sum of the first n terms of (1), s^ the sum
of the first n terms of (2), and B the limit of s!n, it follows that
CONVERGENCE 407
since all terms of (1), and therefore of (2), are positive. Now as
n increases, sn increases but always remains less than B. Hence sH
approaches a limit, which is either less than, or equal to, B.
The first part of the theorem is now proved ; the second part
is too obvious to need formal proof.
In applying this test it is not necessary to begin with the
first term of either series, but with any convenient term. The
terms before those with which comparison begins form a poly-
nomial, the value of which is of course finite, and the remaining
terms form the infinite series the convergence of which is to be
determined.
Ex. 1. Consider
X "4* — + — + — + — + • • • + - —" —~— ~~ ~~~" "4* • • ••
1 2s 8» 4* (n-1)"-1
Each term after the third is less than the corresponding term of the con-
vergent geometric series
1 + i + -i + — + -— + • • • + — — + • • •.
2 22 28 24 2"-1
Therefore the first series converges.
Ex. 2. Consider
1 + --= + — + — + — - + ••• + — + •••.
V2 V3 V4 V5 V^
Each term after the first is greater than the corresponding term of the
divergent harmonic series
1 + 1 + 1 + I + -1+ .. . + _+....
2 3 4 5 n
Therefore the first series diverges.
189. The ratio test for convergence. If in a series of positive
numbers the ratio of the (n 4-l)ȣ term to the nth term approaches
a limit L as n increases without limit, then, if L < 1, the series
converges; if L>1, the series diverges; if L=l, the series may
either diverge or converge.
Let ai + a2 + a8"' r-0n + an+iH (1)
be a series of positive numbers, and let Lim-2L±1=Z. We have
three cases to consider. *aB0° n
408 INFINITE SERIES
1. L<1. Take r any number such that L<r<l. Then,
since the ratio -JLti approaches L as a limit, this ratio must
become and remain less than r for sufficiently large values of n.
Let the ratio be less than r for the with and all subsequent
terms. Then _ ^ n „
Now compare the series
with the series am + amr + amr* + aror8 + • • .. (3)
Each term of (2) except the first is less than the correspond-
ing term of (3), and (3) is a convergent series since it is a
geometric series with its ratio less than unity. Hence (2)
converges by the comparison test, and therefore (1) converges.
2. L>1. Since -JLtl approaches L as a limit as n increases
without limit, this ratio eventually becomes and remains greater
than unity. Suppose this happens for the rath and all subsequent
terms. Then _ x. _
am + 2 > am + l > am»
Each term of the series (2) is greater than the corresponding
term of the divergent series
Hence (2) and therefore (1) diverges.
3. Z=l. Neither of the preceding arguments is valid, and
examples show that in this case the series may either converge
or diverge.
In applying this test, the student will usually find -^^1 m
a
n
the form of a fraction involving w. To find the limit of this
CONVERGENCE
409
fraction as n increases without limit, it is often possible to
divide numerator and denominator by some power of n, so as
a
to be able to apply the theorem (§13) that Lim- = 0, or some
other known theorem of limits. n=w
Ex. 1. Consider
l + «+S5 + ^ + S7+-"- +
n
3 • 32 33 34
n
3n-l
The ratio of the
76 Tl "f* 1
The nth term is and the (n + l)st term is
3 n + 1 3"
+ l)st term to the nth term is — — , and
on 1
^1 1+i 1
T . n + 1 T . n 1
Lim — — = Lim t = - •
n = oo O Tl n = oo O O
Therefore the given series converges.
22 38 44 nn
Ex. 2. Consider 1 + — + — + — +... + --+....
11 \1 li \l
nn (n + l}**1
The nth term is — and the (n + l)st term is ^ '— — • The ratio of
\n ■ \n + 1
"tvhe (n + l)st term to the nth term is * - = ( ) > and
v J (n + l)n» \ n /
Lim (?±iy = Lim (l + iV = e. (§ 98)
n=oo \ n / »=oo \ n/
Therefore the given series diverges.
190. Absolute convergence. The absolute value of a real num-
ber is its arithmetical value independent of its algebraic sign.
Thus the absolute value of both + 2 and — 2 is 2. The abso-
lute value of a quantity a is often indicated by \a\. It is evi-
dent that the absolute value of the sum of n quantities is
less than, or equal to, the sum of the absolute values of
the quantities.
A series converges when the absolute values of its terms form a
convergent series, and is said to converge absolutely.
Let at + a2 + az + a4 -\ (1)
be a given series, and
a + a H (2)
a.
+ «. +
410 INFINITE SERIES
the series formed by replacing each term of (1) by its absolute
value. We assume that (2) converges, and wish to show the
convergence of (1). *
Form the auxiliary series
The terms of (3) are either zero or twice the corresponding
terms of (2). For «* = — |aj.| when ak is negative, and ^ = (^1
when ak is positive.
Now, by hypothesis, (2) converges, and hence the series
2|aJ + 2|aJ + 2|aJ+ 2|aJ + ... (4)
converges. But each term of (3) is either equal to or less
than the corresponding term of (4), and hence (3) converges by
the comparison test.
Now let sn be the sum of the first n terms of (1), srn the sum
of the first n terms of (2), and s" the sum of the first n terms
of (3). Then *=*"_*'
and, since «" and sfn approach limits, «n also approaches a limit.
Hence the series (1) converges.
We shall consider in this chapter only absolute convergence.
Hence the tests of §§ 188, 189 may be applied, since in testing
for absolute convergence all terms are considered positive.
191. The power series. A power series is defined by
aQ + aYx + a2a? + ajt? H h anx* -\ ,
where a0, a , #2, ag, • • • are numbers not involving x.
We shall prove the following theorem : If a power series con-
verges for x = x^ it converges absolutely for any value of x such
that \x\ < \xx\.
For convenience, let \x\ = X, \an\ = An, |#J = X^ By hypothe-
sis the series
a0 + a\X\ + Vl + azXl H 1" anX\ H (1)
converges, and we wish to show that
AQ + A1X+A2X* + AZX* + • • • +AnX*+ ... (2)
converges if X< Xx.
POWEK SERIES 411
Since (1) converges, all its terms are finite. Consequently
there must be numbers which are greater than the absolute
* value of any term of (1). Let M be one such number. Then
we have AnX* < M for all values of n.
Then
AnX" = AnX?(^J<M{
*
Each term of the series (2) is therefore less than the corre-
sponding term of the series
/Tr\8 /v\n
.... (3)
But (3) is a geometric series, which converges when X < X,
Hence, by the comparison test, (2) converges when X<X .
From the preceding discussion it follows that a power series
will converge for values of x lying between two numbers — E
and +-K, and diverge for all other values of x. In some cases
K may be infinity, that is, the series may converge for all values
of x. In other cases, less frequent, R may be zero, that is, the
series may converge only for x = 0.
In any case the values of x for which the series converges
are together called the region of convergence. If represented
on a number scale, the region of convergence is in general a
portion of the scale having the zero point as its middle point.
In some cases the region may extend to infinity or shrink to
a point. In practice the student will generally find it con-
venient to determine the region of convergence by applying
the ratio test, as shown in the examples.
Ex. 1. Consider
l + 2a: + 3a:2 + 4a:8+ ••• + nxn _1 + • • • .
The nth term is nx"-1, the (n + l)st term is (n + 1) x", and their ratio is
x. Lim x = x Lim (1 + - ) = x. Hence the series converges when
n nan n n=0o \ nj
|x|<l and diverges when \x\ >1. The region of convergence extends on
the number scale between — 1 and + 1.
412 INFINITE SERIES
Ex. 2. Consider l+± + £. + £ +
1 |2 |8 [n^l
The nth term is -, the (n + l)st term is — > and their ratio is -
n — 1 In
n
x
Lim - = 0 for any finite value of x. Hence the series converges for any
n = oo n
value of x and its region of convergence covers the entire number scale.
Ex. 3. Consider
l + a: + [2a:2 + |3a:8+ • • • + \n — la*"1 + •••.
The nth term is [n_— lxn_1, the (n + l)st term is [nx», and their ratio
is nx. This ratio increases without limit for all values of x except x = 0.
Therefore the series converges for no value of x except x = 0.
A power series defines a function of x for values of x within the
region of convergence, and we may write
f(x) = a0+ axx + «2^+ a^i h ana*H , (4)
it being understood that the value of f(x) is the limit of the
sum of the series on the right of the equation. The power
series has the important property, not possessed by all kinds
of series, of behaving very similarly to a polynomial. When a
function is expressed as a power series it may be integrated
or differentiated by integrating or differentiating the series term
by term. The new series will be valid for the same values of
the variable for which the original series is valid. If the method
is applied to a definite integral, the limits must be values for
which the series is valid.
Similarly, if two functions are each expressed by a power series,
their sum, difference, product, or quotient is the sum, the dif-
ference, the product, or the quotient of the series.
For proofs of these theorems the student is referred to ad-
vanced treatises.
192. Maclaurin's and Taylor's series. We have noted that any
convergent power series may define a function. Conversely,
it may be shown (see § 193) that any function which is con-
tinuous and has continuous derivatives may be expressed as a
power series. When a function is so expressed it is possible to
MACLAUEIN'S AND TAYLOE'S SEEIES 413
express the coefficients of the series in terms of the function
and its derivatives. For let
f(x) = aQ+ atx + ag?+ azx*+ ajt-\ \- an7?+ • • •.
By differentiating we have
f(x') = a1+2a2x + 3a82?+4a4a*+ - • • +nanafi-1+ . • .,
/"(#)= 2 a2+ 3 • 2 a8x + 4 . 3 a^-\ +n(n-V)ajf-*+ • • •,
f"(x) = 3 • 2a8+4 • 3 - 2a4aH +rc(w-l)(w-2)ans*-8 + . - .,
fn\x) = [n(n -1)0 - 2) • • • 8 • 2]aw+ • • •.
Placing a; = 0 in each of these equations, we find
«o=/(°)> «,=/(<>> «,= ||r'(0), «,= j|f "(0), • • -, W/^O).
Consequently we have
/(*)=/(<>)+/'(0> +-^^^*+ • • • +^W • • .. (1)
This is called MactaurirCs series.
Again, if in the right-hand side of
f(x) = aQ + axx + aj? + a%xz H + aua? -\
we place x=a + zr, and arrange according to powers of x', we have
f(x) = \ + bxa! + bjt!% + bf!*+ • • • + bj!*+ . . .,
or, by replacing x1 by its value x — a,
By differentiating this equation successively, and placing #= a
in the results, we readily find
Hence
/(3;)=/(a) + (a;-a)/(a)+^l^/'(a)+^lV''(«)+ . . .
+^^:r)(«)+---. (2)
This is Taylors series.
414 INFINITE SEEIES
Another convenient form of (2) is obtained by placing
x — a = hy whence x = a + h. We have then
/(a + h) =/(a) + hf(a) +^V'<>) +^/"<>) + • • •
+£/<»><>) + .... (3)
In
Maclaurin's series (1) enables us to expand a function into
a series in terms of ascending powers of x when we know the
value of the function and its derivatives for x = 0. By means
of the series the function may be computed for values of x for
which the series converges. Practically the computation is con-
venient for small values of x.
Taylor's series (2) enables us to expand a function in terms
of powers of x — a when the value of the function and its
derivatives are known for x = a. The function is said to be
expanded in the neighborhood of x = a, and the series can be
used to compute the value of the function for values of x
which are near a.
Ex. 1. Expand e* into a power series and compute its value when x = $.
Since /(*) = e*f f (x) = e*, f' (x) = e*, etc., /(0) = 1, f (0) = 1, /" (0) = 1,
etc. Hence, by Maclaurin's series,
x . x2 . x9 . x*
eat = 1 + l+[2+[3+il+--
This converges for all values of x. If we place x = J, we have e* = 1
+ i + T*g + T&7 + t At = 1-3956, correct to four decimal places. If x has
a larger value, more terms of the series must be taken in the computation,
so that the series, while valid, is inconvenient for large values of x.
Ex. 2. Expand (a + x)n into a power series in x. Here
f(x) = (a + x)% /(0) = a»,
f(x) = n(a + x)»-\ f(0) = nd*-\
f'(x) = n(n -l)(a + x)»-*, /"(°) = n(n -l)a»"2,
f"(x) = n(n -l)(n - 2)(a + *)— 8, f"(0) = n(n -l)(n - 2)a— ».
Hence, by Maclaurin's series,
(a + x)» = a» + na*-*x + n(" -1) a»-2:g2 + nO ~*)(n ~ 2) a*-8** + ••••
[2 . [3
MACLAUKIN'S AND TAYLOB'S SEEIES 415
This is the binomial theorem. If n is a positive integer, the expansion is
a polynomial of n + 1 terms, since j*n + X) (x) and all higher derivatives are
equal to 0. But if n is a negative integer or a fraction, the series converges
when x is numerically less than a.
Ex. 3. Find the value of sin 61°.
Let f(x) = sin xf then f (x) — cos xf f (x) = — sin x, etc., provided x is
expressed in circular measure. 61° expressed in circular measure is ^$ ir
— — + Since sin - and cos - are known to be respectively - V$
3 180 3 3 v J 2
1 IT
and — t it will be convenient to use Taylor's series with a = - • Formula
2 o
(3) gives • (if , A . if , , if h2 . ir hs ir
v / » sin ( - + A ) = sin - + A cos - — -sm- — - cos - + • • •
\3 / 3 3 [2 3 [3 3
IT
Placing h = and computing, we have
416 INFINITE SERIES
193. The remainder in Taylor's series. Let us write
/(*) =/(«) + (x - a)f (a) + ^j^-> («)+•••
+£^>«o+*"' a)
and attempt to determine R* For that purpose place
B=(x~ay+1p. (2)
ft+1 v J
In the right-hand member of equation (1), with R in the
form (2), replace a everywhere, except in P, by 2, and call
F(z) the difference between f(x) and this new expression.
That is, let F(z) be defined by the equation
p(?) =/(*) -/(«) -(*- *yr 00 - ^^f" 00 —
_QL^yr<^_^-^P, (3)
n ft + 1 *
[ft |ft +
where a: is considered constant.
Differentiate (3) with respect to s, still holding x constant.
All the terms obtained cancel, except the last two, and we have
f' (z) = - (*rg)n/(n+l) o) + (* 7 g)w p
= ^£}1[^-/("+1)(^]- (4)
Now when z = #, P(s) = 0, as is at once apparent from (3).
Also when z = a, i^(s) = 0, as appears from (3) with the aid
of (1). Hence F(z) must have a maximum or a minimum for
some (unknown) value of z between z = a and z = x. That is,
where f lies between a and x.*
* The theorem that if F(z) = 0forz = a and z = 6, Men F'(z) = 0/or some
value of z between z = a and z = b is called Rollers Theorem. It is geometrically
evident on drawing a graph. Of course F(z) and F'{z) must be continuous and
hence finite.
EEMAINDER IN TAYLOK'S SERIES 417
From (4), it follows that
and hence, from (2),
B=(X~^1f*+1>(g). (5)
This is the remainder in Taylor's Theorem. It measures the
difference between the value of the function f(x) and the sum
of the first n + 1 terms in (1).
It is evident that if R a^roaches zero as n is indefinitely
increased, the Taylor's series converges and represents the
function. We have, then, in this case, a proof of the possi-
bility of a series expansion for the function, which was assumed
in § 192.
Generally also it will be sufficient to test the convergence
of the series by one of the methods of §§ 188 and 189. For
usually if the series converges, it properly represents the func-w
bion. Examples can be given in which this is not true, but the
student will certainly not meet them in practice.
The remainder may be said to measure the error made in cal-
culating the value oif(x) by means of n + 1 terms of a Taylor's
or Maclaurin's series. It is therefore often important to know
something of the magnitude of R. Now R can usually not be
found exactly, since f is unknown, but it can sometimes be seen
that R cannot exceed some known value, and this is enough for
practice. This is illustrated in the examples.
Ex. 1. What error is made by calculating e* by 5 terms of Maclaurin's
series ? (See Ex. 1, § 192.)
When f(x) = eac,/(,l + 1>(a:) = eF. Hence, in Maclaurin's series for e*,
n+1
where ( lies between 0 and x.
In the present example n = 4 and x = ^.
Therefore R = ^ ef = — i— et
[5 29160
418 INFINITE SERIES
where £ lies between 0 and £. Since the largest value of £ gives the
largest value of e*f we may write
whence it appears that R < .00005.
The calculation of Ex. 1, § 192, is therefore correct to 4 decimal places.
Ex. 2. How many terms of Maclaurin's series must be taken to compute
e% correctly to 4 decimal places ?
As in Ex.1, R=£&L!lee
|n + l
#
where £ is between 0 and £. Hence
|n + l
and n + 1 must be so determined that
f^'t1 3* < .00005.
|n + l
fc This can be done only by trial. It results that n + 1 = 6. Then 6 terms
will be sufficient to assure the required accuracy, though from the nature
of the calculation fewer terms may do.
194. Relations between the exponential and the trigonometric
functions. By Maclaurin's series, we find
x a? . Xs . x
.4
^=1 + l+[2 + [l+[4 + --' CI)
* 3? ~'
• JU JU Jls *^
SmX=X--+---+.-; (2)
oosx-l-j| + jJ-j£+...f (3)
where the laws governing the terms are evident. It is possible
to show that in each case It approaches zero as the number of
the terms increases without limit, no matter what the value of x.
Hence the series converge and represent the functions for all
real values of x.
The series (1) may be used to define the meaning of e* when
a? is a pure imaginary quantity and the definitions of § 26 no
APPEOXIMATE INTEGBATION 419
longer have a meaning. We write as usual i = v^l and replace
x in (1) by iz. We obtain
^1 [2 [3 ^ [4 ^
Then, since i2 = — 1, i8 = — i, i* = + 1, etc.,
4*
-H4-H-H-)
But the two series here involved are equal to cos a? and sin x
respectively by (3) and (2). Hence we have
ete = cos x + i sin x. (4)
Similarly, e~ te = cos x — i sin x, (5)
and, from (4) and (5),
*\x ~ — ix
e — e
Shlx = — 2~T~ ' (6)
cosa?= • (7)
The results (4)-(7) are of great importance in some appli-
cations, notably to the simplification of certain results in the
solution of differential equations.
It may be proved from (1) that e*le** = e*l+x*. Then
e + iy = e Vy = f (cos y + i sin y), (8)
*>*-* = e*<r iy = e* (cos y - i sin y). (9)
195. Approximate integration. When it is not possible, or
convenient, to evaluate the integral
x
/(*) dx (1)
exactly, the function f(x) may be expanded into a power series
and the integral computed to any required degree of accuracy.
This procedure leads to the following three rules:
1. The prismoidal formula. Let us take the first four terms
of Taylor's series for f(x) in the neighborhood of x = a, writ-
ing them in the form
f(x) = A + B(x - a) + C(x -af + D(x- of. (2)
420 INFINITE SERIES
Substituting this in (1), we have
x
f(x)dx = A(b-a)+\B(b-aytJrlC(b-ay+\D(b-d)t
=^ [6 A+3B(b-a-)+2C(b-a,y+$l>(b-ay]. (3)
Now, from (2),
f(a) = A,
f(h) = A + B(b-a) + C(b- a)* + 2>(5 - a)*,
and f(^±^\ = A + \B(b-a) + \C(b-ay + \D(b-ay-,
from which it appears that equation (3) can be written in the
X6/(a:)^=^[/(a)+4/(^)+/(6)]* (4)
This is the prismoidal formula.
If the integral (1) is interpreted as an area, the result (4)
may be expressed as follows : The area bounded by the axis of x,
two ordinate*, and a curve may be found approximately by multi-
plying one sixth of the distance between the ordinates by the sum of
the first, the last, and four times the middle ordinate.
If the integral (1) arises in finding the volume V of a solid
with parallel bases, then formula (4) becomes
F=|(B + 4Jf+6), (5)
where h is the altitude of the solid, B the area of the lower
base, b the area of the upper base, and M the area of the section
midway between the bases.
Of course the prismoidal formula gives an exact result when
f(x) can be exactly represented in the form (2), where any of
the coefficients may be zero. The most important and frequent
cases in which (5) is exact are those in which f(x) is a quadratic
polynomial in x. In this way the student may show that the
formula applies to frustra of pyramids, prisms, wedges, cones,
cylinders, spheres, or solids of revolution in which the gener-
ating curve is a portion of a conic with one axis parallel to the
axis of revolution, and also to the complete solids just named.
APPROXIMATE INTEGBATIOH 421
The formula takes its name, however, from its applicability
to the solid called the prismoid, which we define as a solid hav-
ing for its two ends dissimilar plane polygons with the same
number of sides and the corresponding sides parallel, and for
its lateral faces trapezoids.
Furthermore, the formula is applicable to a more general
solid two of whose faces are plane polygons lying in parallel
planes and whose lateral faces are triangles with their vertices
in the vertices of these polygons.
Finally, if the number of sides of the polygons of the last
defined solid is allowed to increase without limit, the solid goes
over into a solid whose bases are plane curves in parallel planes
and whose curved surface is generated by a straight line which
touches each of the base curves. To such a solid the formula
also applies.
The formula is extensively used by engineers in computing
earthworks.
2. Simpson's rule. When f(x) is not exactly expressed by
(2), the prismoidal formula will in general give better results
the nearer J is to a. Hence we may obtain greater accuracy
by dividing the interval b — a into segments and applying the
prismoidal formula to each. Taking the interpretation of (1)
as an area, we divide the distance b — a into an even number
(2n) of segments, each equal to A#, and call the values of x
at the points of division a, xY, #2, #8, • • •, %2n-v ^ At eac^ point
of division we draw an ordinate of the curve, thus cutting the
required area into strips, and apply the prismoidal formula to
figures each of which is made up of two of these strips, so that
xi* x$ x& " ' *' x2n-i correspond to the middle ordinates of these
figures. Adding the results thus obtained, we have
Xb At
f(x) dz = ^ [/(a) + 4/(xx) + 2/0,) + 4/(z8) + 2/<X)
+ ••• + 4/(^.0 +/(6)]. (6)
This is Simpson's rule.
3. The trapezoidal rule. An area may also be computed
approximately as the sum of rectangles, as shown in § 78. It is
more exact, however, to replace the rectangles of fig. 125, § 78* by
AC
I
422 INFINITE SERIES
trapezoids. This amounts to replacing a small portion of the
curve y =/(#) by a straight line, which is equivalent to using
the first two terms of the series (2). If Ax and xv x$ x# • • •
are taken as in § 78, this method leads to the result
/(*) dx = ^ [/(a) + 2/0.) + 2/Oa) + 2/(2;,) + . . .
+ 2/(*.-0 +/(*)]• (7)
This is the trapezoidal rule. It is evident that it gives less
accurate results than those found by Simpson's rule.
Ex. Evaluate f (l + x*)%dx.
Jo
1. By the prismoidal formula.
/(0) = 1, /(§) = 5.859, /(3) = 31.623.
T8(l + a:2) * rfs = f [1 + 4 (5.859) + 31.623] = 28.030.
2. By Simpson's rule.
Take Ax = J. Then
/(0) = 1, /(£) = 1.398, /(l) = 2.828, /(f) = 5.859,
/(2) = 11.180, /(f) = 19.521, /(3) = 31.623.
T8(l + z2)^* = i [1 + 4 (1.398) + 2 (2.828) + 4 (5.859)
+ 2 (11.180) + 4 (19.521) + 31.623]
= 27.96.
3. By the trapezoidal rule.
Take Ax = £ and use the previous calculations.
T8(l + a;2)* cfe = J [1 + 2 (1.398) + 2 (2.828) + 2 (5.859)
+ 2 (11,180) + 2 (19.521) + 31.623]
= 28.55.
196. The theorem of the mean. If in the general form of
Taylor's series (1), § 193, with R in the form (5), § 193, we
take n = 1, we obtain
f(x) =/(«) + (x- a)f (£), (1)
or, placing x = a + A,
/(« + A) =/(«) + hf(£), (2)
where f is between a and a + A.
INDETERMINATE FORMS
423
This result either in the form (1) or the form (2) is called the
theorem of the mean, and has a very simple graphical interpre-
tation. For let LK (fig. 234) be the graph of y =f(x), and let
OA = a, OB=a + h. Then AB = h,f (a) = AD, f (a + A) = BE,
^^l^^ = the slope of the chord DE.
If now f is any value of x, /'(f) is the slope of the tangent
at the corresponding point of LK. Hence (2) asserts that there
is some point between D and
E for which the tangent is
parallel to the chord DE.
This is evidently true if /(#)
and f(x) are continuous.
Formula (1) may be used
to prove the proposition
which we have previously
used without proof ; namely,
If the derivative of a function
is always zero, the function is a constant. For let/' (x) be always
zero and let a be any value of x. Then, by (l),/(#) — /(«) = 0.
That is, the function is a constant.
From this it follows that two functions which have the same
derivative differ by a constant. For if f'(x) = fiQv)) then
^ [/(*) - <K*)] = 0 ; whence f(x) = 4>(x) + C.
Fig. 284
197. The indeterminate form - . Consider the fraction
0
(1)
and let a be a number such that /(a) =0 and <£(«) = 0. If
we place x = a in (1), we obtain the expression - , which is
literally meaningless.
It is customary, however, to define the value of the frac-
tion (1), when x = a, as the limit approached by the fraction
as -x approaches a.
In some cases this limit can be found by elementary methods.
424 INFINITE SERIES
a2 — x2
Ex. 1.
a — x
When x = a this becomes - • When i^owe may divide both terms
of the fraction by a — x, and have
q2 — x2
a — x
= a + x
for all values of x except x = a. This equation is true as x approaches a,
and hence
CL2 ~" x2
Lim = Lim (a + a?) = 2 a.
x = a O ~~ 3? x=a
„ „ 1-VT^2
Ex. 2.
x
When x = 0 this becomes -• When i^Owe have
1 - Vl - x2 1 - Vl - x2 1 + Vl - x2 x
* i + Vl-z2 1 + Vl-x2
1 «_ Vl — a:2 x
Hence Lim = Lim 7= = 0«
*=o a: x=o 1 + Vl — x2
The theorem of the mean may be used to obtain a general
method. For we have
f(x) _ f(a + ft) _ f(a) + hf(to
if, (x) <f> (a .+ K) if, (a) + W(£ 2) '
where f 1 and f f lie between a and a + A. By hypothesis,/(a) = 0,
if, (a) = 0. Therefore f or h ¥= 0
/Q) _ /(a + A) = /'(f)
As a; is made to approach a, A approaches zero, and £ and £
approach a. Hence
Lim/M=m. (2)
INDETEEMINATE FORMS 425
If, however, f'(a) = 0 and $'(«) = 0, the right-hand side of
(2) becomes -• In this case we take more terms of Taylor's
series and have ?2
/Q) = /(q + A) W" IT /"(ft)
whence Lim ^Q- = ,„^ c '
x±a <p(x) 9 («)
unless /"(a) = 0 and <£"(a) = 0. In the latter case we take still
more terms of Taylor's series, with a similar result.
Accordingly we have the rule :
To find the value of a fraction which takes the form — when
x = a, replace the numerator and the denominator each by its deriv-
ative and substitute x = a. If the new fraction is also - ? repeat
the process.
gX __ g — X
Ex. 8. To find the limit approached by — : when x = 0.
sin x
By the rule, Lim * ~ e~* = P* + e~*l = H = 2.
x=o sin a: L cos a; Jar=o 1
Ex. 4. To find the limit approached by : when x = 0.
x smz
If we apply the rule once, we have
T . e* — 2 cos x + e~x Ve? + 2 sin x — e-x~\ 0
Lim : = — : ■ =-•
x=o xsuix L smar + xcosx Ja:=0 0
We therefore apply the rule again, thus :
x . e* — 2 cos x + e~x re* + 2 cos x + erx~\ 4 ft
Lim : = — : = - = 2.
xao xsinx L 2 cos a; — x sin a; Jx=o 2
198. Other indeterminate forms. If /(a) = oo and <£(a) = oo,
the fraction . takes the meaningless form — when x = a.
$(x) ° °°
The value of the fraction is then defined as the limit approached
426 INFINITE SERIES
by the fraction as x approaches a as a limit. It may be proved—
that t he rule for finding the value of a fraction which becomes t—
holds also for a fraction which becomes ^.
The proof of this statement involves mathematical reasoning^
which is too advanced for this book and will not be given.
lo&r x
Ex. 1. To find the limit approached by ° (n > 0) as x becomes infinite
1
By the rule, Lim — — = Lim = Lim — = 0.
x=oo x" ar-ooiw^-1 x«oo nx*
There are other indeterminate forms indicated by the symbols.
0.QO, QO-QO, 0°, 00°, 1°°.
The form 0 -oo arises when, in a product /(#) «0(a;), w©
have /(a)= 0 and 0(a)= oo. The form oo — oo arises when, in.
f(x) — <f> (a:), we have f(a) = oo, <f> (a) = oo.
These forms are handled by expressing f(x) • <f>(x) or /(#) —
<£(#), as the case may be, in the form of a fraction which
0
becomes - or ~ when z = a. The rule of § 197 may then be
applied.
J£X* 25 • Xr€ •
When x = oo this becomes oo • 0. We have, however, xte-** = -3 > which.
becomes •£ when x = 00. Then
00
T. Xs T. 3x* T. Sx T. 3 ~
Lim — - = Lim — = Lim — — = Lim 3 = 0.
i=oo«^ x=*>2xer x=*>2er x=*>±xer
In the same manner Lim xPer** = 0 for any value of n.
IBOD
Ex. 8. sec x — tan x.
When x = - this is 00 — 00. We have, however,
. 1 — sinar
sec x — tan x = »
cos a:
0 7T
which becomes - when x = — • Then
0 2
T . , . x T . 1 — sin x T . — cos x ~
Lim (sec x — tan x) = Lim = Lim : — = 0.
- w cosar w — sina?
x±- x±— *i-
2 2 2
FOURIER'S SERIES 427
The forms 0°, oo°, 1" may arise for the function
[/0)]*(x) when x = a.
If we place u = [/(V)]*(:c),
we have log u = <f> (x) log f(x).
If Lim $ (x) log/(V) can be obtained by the previous methods,
x = a
the limit approached by u can be found.
i
Ex. 4. (1 — x)x.
When x = 0 this becomes 1". Place
i
u = (1 — x)x ;
then log u = — si -I .
x
Now UmW-'^fjlLI =-1.
Hence log u approaches the limit — 1 and u approaches the limit - •
e
199. Fourier's series. A series of the form
-£ + a>x cos x + a2 cos 2 x + • • • + an cos nx + • • •
+ 6X sin a; + b2 sin 2 a; + • • • + bn sin nx + • •' ., (1)
where the coefficients a0, ax, • • •, bv J2, • • • do not involve x> is
called a Fourier's series. Every term of (1) has the period*
2 7r, and hence (1) has that period. Accordingly any function
defined for all values of # by a Fourier's series of form (1)
must have the period 2 7r. But even if a function does not have
the period 2 7r, it is possible to find a Fourier's series which will
represent the function for all values of x between — ir and 7r,
provided that in the interval — it to ir the function is single-
valued, finite, and continuous except for finite discontinuities,!
*f(x) is called a periodic function, with period &, if f(x + k) =/(x).
t If xY is any value of x, such that f(xx — e) and f(xx + e) have different
limits as e approaches the limit zero, then f(x) is said to have a finite discon-
tinuity for the value x = xv Graphically, the curve y =/(x) approaches two
distinct points on the ordinate x = xv one point being approached as x increases
toward xv and the other being approached as x decreases toward xv
428 INFINITE SERIES
and provided there is not an infinite number of maxima or
minima in the neighborhood of any point.
We will now try to determine the formulas for the coeffi-
cients of a Fourier's series, which, for all values of x between
— 7r and 7r, shall represent a given function, f(x), which satisfies
the above conditions.
Let f(x) = -£ + ax cos x + a2 cos 2 x + • • • + an cos nx+ • • •
+ bx sin x + b2 sin 2 x + • • • + bn sin nx + • • •. (2)
To determine a0, multiply (2) by dx> and integrate from — ir
to 7r, term by term. The result is
f(x) dx = a 7r,
1 rff
whence ao==— / f(p)^ (8)
since all the terms on the right-hand side of the equation,
except the one involving aQ, vanish.
To determine the coefficient of the general cosine term, as
an, multiply (2) by cos nxdx, and integrate from — ir to 7r,
term by term. Since for all integral values of m and n
X
sin mx cos nx dx = 0,
— n
ir
cos mx cos n# cfc = 0, (m =£ w)
and J cos2 nxdx = Tr,
all the terms on the right-hand side of the equation, except
the one involving aM, vanish, and the result is
J f(x) cos nxdx = anir,
1 /**
whence an = — I f(x)co8nxdx. (4)
It is to be noted that (4) reduces to (3) when n = 0.
FOURIER'S SERIES
429
In like manner, to determine 6n, multiply (2) by sin nxdzy and
Integrate from — ir to 7r, term by term. The result is
.i£/W5i
sin nx dx.
(5)
For a proof of the validity of the above method of deriving the
formulas (3), (4), and (5), the reader is referred to advanced
treatises.
Ex. 1. Expand x in a Fourier's series, the development to hold for all
'values of x between — ir and ir.
By (3),
*>y (4),
and by (5),
l r*
aQ = — / xdx = 0,
7T«/-ir
l rw
an = — I x cos nxdx = 0,
irJ—v
on = — I x sin nxdx = cos rwr.
Hence only the sine terms appear in the series for x, the values of the
coefficients being determined by giving n in the expression for bn the values
1, 2, 3, • • • in succession. Therefore bx = 2, b2 = — §, 68 = §, • • •, and
* = 2(^-2i
sin 2 x sin 3 x
....)
c-5x,o)yx-j
The graph of the function x is the infinite straight line passing through
the origin and bisecting the angles of the first and the third quadrant.
The limit curve of the series coincides with this line for all values of x
between — ir and ir, but not for x = — ir and x = ir ; for every term of the
series vanishes when x = — ir
or x = 7r, and therefore the Y
graph of the series has the
points (± ir, 0) as isolated
points (fig. 235).
By taking xx as any value
of x between — ir and ir, and
giving k the values 1, 2, 3, • •*
in succession, we can represent all values of x by xx ± 2 kir. But the series
has the period 2 ir, and accordingly has the same value for xx ± 2 for as
for Xj. Hence the limit curve is a series of repetitions of the part between
x = — ir and x = ir, and the isolated points (±2 kir, 0).
It should be noted that the function defined by the series has finite discon-
tinuities, while the function from which the series is derived is continuous.
Fig. 235
430
INFINITE SERIES
It is not necessary that/(rr) should be defined by the same law
throughout the interval from — it to 7r. In this case the integrals
defining the coefficients break up into two or more integrals, as
shown in the following examples :
Ex. 2. Find the Fourier's series for /(#) for all values of x between — w
and «•, where /(x) = x + w if — ir<x<09 and /(a;) = tt — x if 0<x<ir.
Here a0 = - f (x + w) dx + f (tr — x) dx = tt ;
an = - 1 I (x + ir) cos nxdx + J (w — x) cos nxefa I
2 /i
= — 5 (1 — cos nir) ;
bn = — I / (a: + t) sin nxdx + j (w — x) sin nxdx I
= 0.
Therefore the required series is
w 4 /cos
2 »\1«
4 /cos x cos 3 x cos 5 x ,
H r^ r — ^ — +
32
52
■)■
The graph of f(x) for values of x be-
tween — tt and tt is the broken line ABC
(fig. 236). When x = 0 the series reduces to
Fig. 236
2+;rAP + 3-* + 5^+ '/^
for — — +
l2 32 52
TT2 *
. = — . When x = ± ir
the series reduces to 0. Hence the limit curve of the series coincides with
the broken line ABC at all points. From the periodicity of the series it
is seen, as in Ex. 1, that the limit curve is the broken line of fig. 236.
Ex. 3. Find the Fourier's series for/(x), for all values of x between — v
and w, where /(x) = 0 if — w <x <0, and /(a:) = w if 0 < x < ir.
a0 = — I / Odx+j trdx\= tt;
Here
1 /•*
a,, = — I ircosnxdx = 0;
ir Jo
1 /•*■ 1
6n = — I tt sin nxdx = — (1 — cos nir).
ir «/o n '
Therefore the required series is
+
ir 0 /sin x sin 3 a: sin 5 a:
•)•
*Byerly, Fourier's Series, p. 40.
PROBLEMS
431
The graph of the function for the values of x between — w and ir is the
axis of x from x = — ir to a: = 0, and the straight line AB (fig. 237), there
being a finite discontinuity
when x = 0.
The curves (1), (2), (3),
and (4) are the approxima-
tion curves corresponding re-
spectively to the equations
IT
' = 2'
IT
(1)
y = -+2sins, (2)
t-X
ir , n /sin x , sin 3 x\
y = 2-+2l_- + -3_)'
ir , n /sin x . sin 3 x , sin 5 x\
y=2+2n-+-3-+-5->
(3)
(4)
They may be readily constructed by the method used in § 24. It is
to be noted that all the curves pass through the point (0, -J, which is
midway between the points A and 0, which correspond to the finite
discontinuity, and that the successive curves approach perpendicularity
to the axis of x at that point.
PROBLEMS
1. Prove that the series
1+1+1+1+1+1+1+1+.. .,
^ 2° 2a 4a 4a 4° 4a 8a '
where there are two terms of the form — > four terms of the form
1 1 Z 1
-> eight terms of the form — > and 2* terms of the form
4a ° Sc
(k = 1, 2, 3, • • •), converges when a > 1.
(2*)"
2. By comparison with the series in problem 1 or with the
harmonic series (Ex. 2, § 187) prove that the series
^ 2° 3° T 4° ^ raa
converges when a > 1, and diverges when a s§ 1.
432 INFINITE SERIES
By comparison with a geometric or a harmonic series establish
the convergence or the divergence of the following series :
8. 1 + ,4 + fi + ,4 H h —XT + ' " "
[2 [3 |4 \n + 1
2 22 2* 2""1
33535.7' ' 3-5-7... (2n+l)
3 4 5 6 n + 2
2^3.2^4.3^5.4^ ^ (n 4- l)n
By comparison with the series of problem 2 establish the con-
vergence of the following series :
1,1.1, 1
6. ^ + 32 -h 52 + • • • + pn __ ^2 + • • -
Ill 1
1.3^5.7^9.11^ ^(4n-3)(4n-l)^
1.23.4^23.4.53.4.5.6
+ n(n + l)(n + 2)(n + 3) H '
1^5^ 10 ^ »■ 4- 1
By the ratio test establish the convergence or the divergence of
the following series:
2.1^28.3^2*.5^ ^22tt-1(2w-l)^
. 5 52 58 5""1
11. 1 + T + 777 + ,-77 + - • - + T + • • -
1 ' [2 [3 ' ' [n-1
2 22 28 2"
12. 1 1 1 H h
1.2^2.3^3.4^ n(n 4- 1)
5 ^ 3 • 52 ^ 4 . 58 ^ ^ (» 4-1)5- ^
12 3 n
3 32 38 ^3n
PROBLEMS
433
2a 32 4a
n
16. ! + _ + -+_ + ...+_ +
113 1 32
16. - H — -A — 4-
2^5 22 52 3a
^5»"1 wa ^
Find the region of convergence of each of the following series
17-*+f + f +
+
x
2n-l
2/1-1
+
.8
X X* X"
18- 22 + 42 + 62" +
+^„+
19.
a
+
x2
+
of
1.2 34 ' 56
(2»)«
+ ... +
xv
(2n-l)2n
+
1 x .x2
2 22 28
+
x
n-l
2»
+
IL« 1| + 1^ +
a;
2n-l
3 3 38 ' 5 36 ' "+^ ^ ^n-l'S8— 1 +
22 l-^+L^
1+1-2
a4-
1-3-5
1-2-3
jr6 +
1 - 3 - 5 - - - 2 7i — 3
Find the following expansions and verify the given region of
convergence :
— 1Y»-l
+ (-!>
23. sin* = x — .— + ,-r — •
|3 [6
24. cosa; = 1 - _ + + (- 1)— *
a2 . a;"
aj2n-
-l
2w-
-1
x2*-
-2
2n-
-2
+
(— oo<a;<oo).
+ •
aj"
25. log(l + »)= »--- + -- + (_!)— 1_ +
n
(— oo < a; < oo).
(-1<*<1).
26.1og^ = 2(, + f + |+... + £^ + ...)
27. tan"1® = a? - — 4- ■=• +(-l)"-1
a;
2n-l
2n-l
+
(-l<a;<l).
• • • •
434 INFINITE SERIES
Expand each of the following functions in a series of ascending
powers of x, obtaining four terms no one of which is zero :
28. ' =• 30. secx. 32. log(se + Vl+x2).
_ ^ v ' 31. e* sec a?. 33. log cos a?.
29. tana;. °
Find four terms of the expansion into a Taylor's series of each
of the following functions :
34. cos a, in the neighborhood of x = — •
35. logse, in the neighborhood of x = 5.
36. e*, in the neighborhood of x = 4.
37. tan"1^, in the neighborhood of a; =1.
38. Vl+a?, in the neighborhood of x = 2.
39. Compute sin 12° to four decimal places by Maclaurin's series.
40. Compute sin 46° to four decimal places by Taylor's series.
41. Compute cos 10° to four decimal places by Maclaurin's series.
42. Compute cos 32° to four decimal places by Taylor's series.
43. Using the result of problem 33, compute log cos 18° to four
decimal places.
44. Using the series in problem 25, compute log $ to five decimal
places.
45. Using the series in problem 26, compute log 2 to five decimal
places, and thence by aid of the result of problem 44 find log 3.
46. Using the series in problem 26, compute log J to five decimal
places, and thence by aid of the result of problem 45 find log 5.
47. Using the series in problem 26, compute log £ to four decimal
places, and thence by aid of the result of problem 45 find log 7.
(x* xb \ M N
x + -=- + -=-+ • • .Jwhere x = ———'
o o / M+N
49. Compute the value of ir to four decimal places from the
expansion of sin"1 a? (Ex.4, §192) and the relation sin"1 - = — •
.2 o
50. Compute the value of ir to four decimal places from the expan-
1 1 ir
sion of tan_1x (problem 27) and the relation tan"1 - 4- 2 tan"1 x = T'
PROBLEMS 435
51. By the binomial theorem find Vl7 to four decimal places.
52. By the binomial theorem find V26 to four decimal places.
53. Show that in the expansion of log (1 + x) (problem 25)
x*+1 |se*+1|
\R\< — rr when x>0, and \R\<-, — , ' ., , — ;— n when a;<0.
1 + x
54. Show that, in the expansion of log;: (problem 26),
2xn+2 1 — x 2\xH+2\
\R\<7 — . o\/i ^+2 wnen x>®> and \R\<-p — , 0\/i . \n+a
1 ' (n + 2)(1 — x)n+2 ' ' (n + 2) (1 + x)n+2
-when x < 0, where w is the exponent of x in the last term retained
in the expansion.
55. By integrating the expansion of 1 a to obtain the expan.
+ X \xn+2\
sion of tan"1 x, show that for the latter expansion \R I < ■ ^ > where
1 ' n 4- 2
n is the exponent of a in the last term retained in the expansion.
56. Show that, in the expansion of (1 + x)k,
1 ' n + 1 '
and \R\<k(JC7}l'[ ' S 7An+1l when x<0,
1 ' w + 1(1 + sc)n-*+1 ' '
if n - k + 1 > 0.
57. From the result of problem 53 estimate the error made
in commuting log 1.2 from three terms of the series. How many
terms of the series are sufficient to compute log 1.2 accurately to 6
decimal places ?
58. From the result of problem 53 how many terms of the
expansion of log (1 -f x) are sufficient to compute log .9 to 5
decimal places?
59. From the result of problem 54 how many terms of the expan-
1 + x
sion of log z are required to compute log J to 4 decimal places ?
60. Using the result of problem 55, find how many terms of the
expansion of tan_1x are sufficient to compute tan_1£ to four decimal
places. Also estimate the error made in computing tan"1 J from 5
terms of the series.
436 INFINITE SERIES
61. From the result of problem 56 find how many terms of the
binomial series are sufficient to compute V102 to four decimal places.
X10 dx
zr— — , approximately,
(1) by the prismoidal formula,
(2) by Simpson's rule, taking Aa; = 1,
(3) by the trapezoidal rule, taking Asc = 1.
/* dx
(1) by the prismoidal formula,
(2) by Simpson's rule, taking Ax = £,
(3) by the trapezoidal rule, taking Ax = £.
ir
64. Compute J log cos x dx, approximately,
(1) by the prismoidal formula,
(2) by Simpson's rule, taking Ax = — »
(3) by the trapezoidal rule, taking Ax = -— •
IT
i
IT
Find the limit approached by each of the following functions as
the variable approaches its given value :
-c 2cos2.sc— 1 . ir „„ x — sin"1^ . .
65. >s = -r- 71. n >x = 0.
ir 6 sin8se
72. >sc==0.
e8x__g-8x a; — tana;
66. : — „ > X = 0.
sm2x cot 5a;
a2x — b2x - cot a;
67. — ^ >a; = 0.
^a; 1— logo;2
('-f
74. J2— , a; = 0.
J „. 0*
**' 2X8in»ll'*-6" 75. logfr-T)^^
6*— e"x — 2x . ^ tan-
69. : >SC = 0. 2
a; — sin x
log sin ^
x ajw
76. ^,a; = oo(tt>0).
7T
70. > a;==7r. 77. (7r — 2-a;) tan a;, a; =. —
(a; — 7r) J
PROBLEMS 437
78. sin 3 x esc 5 x, x == 0.
79. e'^logbx, x = oo.
(1 \loga:
i
80. a-«-~ a = oo. 86. (^ + ^ a = oo.
1 1
81. 9X = 7T.
1
x — tt tanx * 87. x1-*, aj==l.
82. -i- - =-?- , x =1. 88. (cos «)— , x = 0.
X — 1 lOg # !
83. sc»lna!, x = 0. 89- i1 + sin *)*> * — °-
84. (sin ar)tanx, x = 0. 90. af, x = 0.
Expand each of the following functions into a Fourier's series for
values of x between — it and ir :
91. x*. 92. e«
93. f(x), where /(a) = — 7r if — 7r < a < 0, and /(a) = 7r if 0 < x < tt.
94. f(x), where/(oj) = — xif — 7r < a < 0, and/(ar) = 0 if 0 < x < ir.
95 . f(x), where /(x) = — 7r if — 7r < # < 0, and/(x) = x if 0 < x < ir.
96. f(x), where /(x) = 0 if — *ir<x<0, and/(x) = x2 if 0 < x < ir.
AC
CHAPTER XVIII
DIFFERENTIAL EQUATIONS
200. Definitions. A differential equation is an equation which
contains derivatives. Such an equation can be changed into
one which contains differentials, and hence its name, but this
change is usually not desirable unless the equation contains the
first derivative only.
A differential equation containing x, y, and derivatives of y
with respect to x, is said to be solved or integrated when a
relation between x and y, but not containing the derivatives,
has been found which, if substituted in the differential equa-
tion, reduces it to an identity.
The manner in which differential equations can occur in
practice and methods for their integration are illustrated in
the following examples: •= ns*,
Ex. 1. Required the curve the slope of which
at any point is twice the abscissa of the point.
By hypothes
Therefore
■* + C.
(1)
Any curve whose equation can be derived
from (1) by giving C a definite value satisfies
the condition of the problem (fig. 238). If it is
required that the curve should pass through
the point (2, 3), we have, from (1),
3=4+ C; whence C = -l,
and therefore the equation of the curve is
y = *»-l.
Fig. 238
But if it is required that the
have, from (1), i0 = 9 + C;
curve should pass through (— 3, 10)
whence C = 1,
and the equation is
».*> + !.
EXAMPLES
439
Ex. 2. Required a curve such that the length of the tangent from any
point to its intersection with OF is constant.
Let P (x, y) (fig. 239) be any point on the required curve. Then the
equation of the tangent at P is
where (X, Y) are the variable coordinates of a moving point of the tangent,
(x, y) the constant coordinates of a fixed point on the tangent (the point
of tangency), and -j- is derived from the,
dx
as yet unknown, equation of the curve.
The coordinates of jR, where the tangent
intersects 0 Y, are then X = 0, Y—y -x,
ax
and the length of PR is *Jr* + x2 (^V-
Representing by a the constant length
of the tangent, we have
a» +
«■(!)'-«•
or
dy __ Va2 — #s
dx x
(1)
Fig. 289
which is the differential equation of the required curve. Its solution
is clearly
y
-*/
Va2 - Xs
dx+ C
2 a±Va2-x2
+ C. (2)
t-u=o
The arbitrary constant C shows that
there is an infinite number of curves which
satisfy the conditions of the problem.
Assuming a fixed value for C, we see
from (1) and (2) that the curve is sym-
metrical with respect to OF, that x2 cannot
be greater than a2, that — = 0 and y = C
dx
when x = a, and that — becomes infinite
dx
as x approaches zero.
From these facts and the defining property the curve is easily sketched,
as shown in fig. 240. The curve is called the tractrix.
Fig. 240
440
DIFFERENTIAL EQUATIONS
Ex. 3. A uniform cable is suspended from two fixed points. Required
the curve in which it hangs.
Let A (fig. 241) be the lowest point, and P any point on the required
curve, and let PT be the tangent at P. Since the cable is in equilibrium,
we may consider the portion AP as a rigid body acted on by three forces, —
the tension t at P acting along PTf the tension h at A acting horizontally,
and the weight of AP acting vertically. Since the cable is uniform, the
weight of A P is ps, where s is the length of AP and p the weight of the
cable per unit of length. Equating the horizontal components of these
forces, we have M , ,
t cos 9 = A,
r
and equating the vertical components, we have
t sin <f> = ps.
From these two equations we have
tan <t> = £s,
h
or
dy
dx
Fig. 241
where - = a, a constant.
P
This equation contains three variables, x, y, and s, but by differentiating
with respect to x we have (§ 91)
dx* dx \1 + W'
a>
the differential equation of the required path.
dy
To solve (1), place -Z-=p. Then (1) becomes
dx
or
whence
dp _ dx,
a
Vl + »2
logO +Vl+/>2) = - + C.
a
(2)
Since A is the lowest point of the curve, we know that when x
p = 0. Hence, in (2), C = 0, and we have
= 0,
•T
or
whence, since P = -f- >
dx
p + Vl + />2 = e5,
p = h(<r-e~*);
y = -\e" + € a
= 5\ea + e *) + C.
VARIABLES SEPARABLE 441
The value of C depends upon the position of OX, since y = a + C when
x = 0. We can, if we wish, so take OX that OA = a. Then (7 = 0, and
we have, finally,
y = |(^ + e"«),
the equation of the catenary (fig. 61, § 27).
The order of a differential equation is equal to that of the
derivative of the highest order in it.
The simplest differential equation is that of the first order
and of the first degree in the derivative, the general form of
which is
dx
or Mdx + Ndy=Q, (1)
where M and N are functions of x and y, or constants.
In the following articles we shall consider some cases in
which this equation can be readily solved.
201. The equation M dx + Ndy = 0 when the variables can
be separated. If the equation (1), § 200, is in the form
/1(a0<fe+/1O)dy = 0,
it is said that the variables are separated. The solution is then
evidently
where c is an arbitrary constant.
The variables can be separated if M and N can each be
factored into two factors one of which is a function of x alone
and the other a function of y alone. The equation may then
be divided by the factor of M which contains y multiplied by
the factor of N which contains x.
Ex. 1. dy=f(x)dx.
From this follows y = \f(x) dx + c.
Any indefinite integral may be regarded as the solution of a differential
equation with separated variables.
442 DIFFERENTIAL EQUATIONS
Ex.2. Vl - y2dx + Vl - x2dy = 0.
This equation may be written
<** ■ dV =0;
whence, by integration, sin-1 a: + sin-1y = c. (1)
This solution can be put into another form, thus : Let sin-1 a: = <f> and
sin-1y = ^r. Equation (1) is then ^ + ty = c, whence sin(<£ + \fr) = sin c ;
that is, sin <f> cos \p + cos <f> sin ip = k, where ifc is a constant. But sin <f> — xy
sin \ff = y, cos ^ = Vl — x2, cos ^r = Vl — y2 ; hence we have
a: Vl - y2 + y Vl - r* = k. (2)
In (1) and (2) we have not two solutions, but two forms of the same
solution, of the differential equation. It is, in fact, an important theorem
that the differential equation Mdx + Ndy = 0 has only one solution involv-
ing an arbitrary constant. The student must be prepared, however, to meet
different forms of the same solution.
Ex. 3. (1 — a:2) -^ + xy = ax.
dx
This is readily written as
(1 — x2) dy + x (y — a) dx = 0,
dy , xdx A
or — — + z -a = 0;
y — a 1 — x*
whence, by integration,
log(y - a) - £ log(l - a:2) = c,
which is the same as log •; = c,
VT^~?
and this may be written y — a — k Vl — x2.
202. The homogeneous equation Mdx + Ndy = 0. A polynomial
in x and y is said to be homogeneous when the sum of the expo-
nents of those letters in each term is the same. Thus ax2 + bxy + cy*
is homogeneous of the second degree, ax* + hx*y -f cxy2 + ey8 is
homogeneous of the third degree. If, in such a polynomial, we
place y = vx, it becomes 3?f(v) where n is the degree of the
polynomial. Thus
ax2-}- bxy + cy*= x*(a + bv + crv2),
ax*+ bx*y+ cxy*+ ey*= x*(a + bv + cv2+ ev8).
HOMOGENEOUS EQUATION 443
This property enables us to extend the idea of homogeneity to
functions which are not polynomials. Representing by f(x, y)
a function of x and y, we shall say that/(#, y) is a homogeneous
function of x and y of the nth degree, if, when we place y = vx,
f(x, y*)=xnF(y). Thus Va^ + y2 is homogeneous of the first
degree, since y/z* -f y* = x V 1 + v\ and log - is homogeneous of
x
degree 0, since log — = log v = x° log v.
x
When M and N are homogeneous functions of the same degree
the equation Mdx + Ndy=Q
is said to be homogeneous and can be solved as follows :
Place y = vx. Then dy = vdx + xdv and the differential equa-
tion becomes ^^ dx + ^^ ^vdx + xdv^ = 0>
or [/x(tF) + */2<V)] dx + zf2<V) <ft> = 0. (1)
If fx(?) + yf2(y)^0, this can be written
<fo , f %(?)&> =0
* /i00+v/200
where .the variables are now separated and the equation may be
solved as in § 201.
If fjfv) + vf2(y) = 0, (1) becomes dv = 0 ; whence v = c and y=cx.
Ex. (x2-y2)dx + 2xydy = 0.
Place y = vx. There results
(1 — v2)dx + 2v (xdv + vdx) = 0,
dx 2 vdv __ .
or T + i+Ta
Integrating, we have log x + log (1 + v2) = c' ;
whence ar(l + u2) = c,
or a:2 + y2 = car.
203. The equation
(axx + \y + q) dar + (a^ + $2y + c2) rfy = 0 (I)
is not homogeneous, but it can usually be made so, as follows :
Place x = X* + h, y = / + k. (2)
Equation (1) becomes
(a^+ &y+ axh + \k + cx) dx'+ (a^H- 62/+ a2fc + b2k + c2) dyy= 0. (3)
444 DIFFERENTIAL EQUATIONS
If, now, we can determine h and k so that
axh + bxk + cx = 01 /£\
a2^+&2*+c,2 = °J
(3) becomes (a1x/ + b^) dx* + (a2^ + b2\f) d/ = 0,
which is homogeneous and can be solved as in § 202.
Now (4) cannot be solved if axb2 — a2bx = 0. In this case — = -* = kt
where k is some constant. Equation (1) is then of the form a* 1
(axx + bxy + cx) dx + [k{axx + \y) + c2] tfy = 0, (5)
so that, if we place axx + bxy = ar7, (5) becomes
• (x' + cjdx + (**' + c2) ^ "" a*dx = 0,
1
which is dx+ — — - — ^ dx' = 0,
and the variables are separated.
Hence (1) can always be solved.
204. The linear equation of the first order. The equation
£+/i(*)y=/,(*). (i)
where fx(x) and/2(a;) may reduce to constants but cannot con-
tain yy is called a linear equation of the first order.
An equation of the form M dx + Ndy = 0 may be put in
form (1) if, after transforming it to -*- -\ — = 0, — can be
dx N N
expressed as fx(x)y — /2(#) ; that is, as the difference of two
terms one of which is y multiplied by a function of x and the
other of which is a function of x only.
To solve (1) let .0.
v y y = uv, (2)
where u and v are unknown functions of a: to be determined later
in any way which may be advantageous. Then (1) becomes
dv du ~ , N - , x
or
•[£+*H+"i-''<*> ^
LINEAR EQUATION OF FIRST ORDER 445
Let us now determine u so that the coefficient of v in (3)
shall be zero. We have
or h/ (V) <fc = 0,
u l
of which the general solution is
logw+ lfx(x)dx = c
Since, however, all we need is a particular function which will
make the coefficient of v in (3) equal to zero, we may take <?= 0.
Then log u = — \fx(x) dx.
or u = eSfmdx. (4)
With this value of w, (3) becomes
or ^ ! -****/.(.),
whence v may be found by integration. Substituting the values
of u and v in (2), we have the solution of (1).
Ex. (1 — x2) -j- + xy = ax.
dx
Dividing the equation by 1 — x2, we have the linear equation
dy x _ ax -
Substituting uv for y, we have
Idu x \ dv ax 0
446 DIFFERENTIAL EQUATIONS
Placing the coefficient of v equal to zero, we have
^ + ^ = 0; • (4)
u 1 — X1
whence log u — J log (1 — x2) = 0,
so that u = Vl — x*.
Substituting this value of u in (2), we have
CLX
whence dv = dx (6)
(l-x2)i
and t? = m + c.
VT^2
Substituting these values of uv in the equation y = ui>, we have the
solution .
y = a + cvl — a:2.
This example is the same as Ex. 3, § 201, showing that the methods of
solving an equation are not always mutually exclusive.
If (1) is in the special form
fx-ay =/(*), . (5)
where a is any constant, its solution is
ax i Max
y = ceax + e
fe-m**. (6)
The proof is left to the student.
205. Bernouilli's equation. The equation
% +/1(*)y=/,.(*)y
may be solved by the same method that was used in solving
the linear equation.
EXACT EQUATION 44?
ax x
Placing y = uv, we have
„(*f-_«\+ttf£ = :cW. (1)
\dx xj ax
Placing = 0, we find u = x.
ax x
Substituting this value of u in (1), we have
£-*., (2)
1 a:6
whence — — - = — + c, (3)
3v8 6 v '
V 1 x8 c
and, finally, since i? = - > -s = "~ tt + -; •
x y3 2 Xs
206. The exact equation M dx + Ndy = 0. If the left-hand
member of the equation
Mdz + Ndy = Q (1)
is an exact differential, df(x, y) (§ 170), that is, if
dM_ dN
Jjj~te% ( }
(1) may be written df(x, y) = 0, (3)
the solution of which is evidently
/(re, y) = c. (4)
In this case (1) is called an exact differential equation.
The method of solving (1) is evidently to find /(re, y) as in
§ 170, and set it equal to a constant.
Ex. (4X8 + 10 xf - 3 y*)dx + (15 x*y* - 12 xtf + 5 yA)dy = 0.
Here — = 30 xy2 — 12 v8 = — , and the equation is therefore exact.
dy dx
Hence its solution is f(x, y) = c, where
^^ = 4x8 + 10xy8-3y* (1)
€/X
and ^^ = 15xV-12xy8 + 5^. (2)
448 DIFFERENTIAL EQUATIONS
Integrating (1) with respect to x, we have
/(*, V) = ** + 5 *Y - 3 xtf + F(y).
Substituting this value in (2), we have
15 *y - l2xf + *"&) = 15 *y- 12^ + 5y*;
whence *T/ (y) = 5 y4 and F(y) = y5.
Therefore f(x, y) = a?4 + 5 #y — 3 a:^4 + y6,
and the solution of the differential equation is
x4 + 5 x2y* — 3 xy4 + y5 = c.
207. The integrating factor. If the equation
Mdx + Ndy = 0 (1)
is not exact, i.e. it -r— =f -^— ,
it may be proved that theje exists an infinite number of func-
tions of x and y such that if (1) is multiplied by any one of
them it is made an exact equation. Such a function is called
an integrating factor.
No general method is known for finding integrating factors,
though the factors are known for certain cases, and lists can
be found in treatises on differential equations. Sometimes an
integrating factor can be found by inspection. In endeavoring
to do this the student should keep in mind certain common
differentials, such as
d(uv) = vdu + udv,
\v/ v
— udv
2
,. *u vdu — udv
rftan-1-= ,
v u + v
71 u vdu — udv
d log - = ,
V uv
d O2 + v2) = 2 (udu + v dvy
CERTAIN EQUATIONS OF SECOND ORDER 449
Ex. (x*-y2)dx + 2xydy = 0.
We may write this equation in the form
x2dx — y2dx + xd(y2) = 0.
The last two terms of the left-hand member of the equation form the
numerator of tf(— ).
Consequently we multiply the equation by — , and have
dx + *d(y*)-y*dx = 09
x*
,2
the solution of which is x + — = c,
x
or x2 + y2 = ex.
It is to be noted that it is not necessary to use the method of § 206 to
solve the equation, for when the integrating factor is found by inspection,
the solution is at once evident.
208. Certain equations of the second order. There are certain
equations of the second order, occurring frequently in practice,
which are readily integrated. These are of the four types:
We proceed* to discuss these four types in order:
By direct integration,
y= \\f(x)da?+cxx
+ <?..
This method is equally applicable to the equation — 2 =f(z).
(JLjC
450 DIFFERENTIAL EQUATIONS
Ex. 1. Differential equations of this type appear in the theory of the
bending of beams. Each of the forces which act on the beam, such as the
loads and the reactions at the supports, has a moment about any cross sec-
tion of the beam equal to the product of the force and the distance of
its point of application from the section. The sum of these moments
for all forces on one side of a given section is called the bending moment
at the section. On the other hand, it is shown in the theory of beams
EI
that the bending moment is equal to — - > where Ey the modulus of elas-
R
ticity of the material of the beam, and /, the moment of inertia of the
cross section about a horizontal line through its center, are constants, and
R is the radius of curvature of the curve into which the beam is bent.
Now, by § 106,
1 dx*
*>©?
where the axis of x is horizontal.' But in most cases arising in practice
-f- is very small, and if we expand — by the binomial theorem, thus :
dx R
R dx2l 2\dxJ J
O
Fio. 242
C B
we may neglect all terms except the first without sensible error. Hence
cPv
the bending moment is taken to be EI—^- This expression equated to
dx2
the bending moment as defined above gives the differential equation of the
shape of the beam.
We will apply this to find the
shape of a beam" uniformly loaded A
and supported at its ends. %
Let I be the distance between the
supports, and w the load per foot-run.
Take the origin of coordinates at the lowest point of the beam, which, by
symmetry, is at its middle point. Take a plane section C (fig. 242) at a
distance x from 0 and consider the forces at the right of C. These are
the load on CB and the reaction of the support at B. The load on CB is
(- — x\ acting at the center of gravity of CB, which is at tl^e distance
i-x «°(|_:e)
of — — from C. Hence the moment of the load is — i which is
taken negative, since the load acts downward. The support B supports
to
CERTAIN EQUATIONS OF SECOND ORDER 451
wl
half the load, equal to — • The moment of this reaction about C is therefore
t(R-
Hence we have
dx2 2 \2 / 2 \2 / 2 \4 /
The general solution of this equation is
EIy = 2[T ~ U)+ C>* + C*
But in the case of the beam, since, when x = 0, both y and — are 0, we
have cx = 0, c2 = 0.
Hence the required equation is
9 2 \ 8 12/
*2M«2>
The essential thing here is that the equation contains -f- and
d2v . . . **
—J, but does not contain y except implicitly in these derivatives.
\JyJs J fO J
Hence, if we place -£- = py we have -=~ = -J- » and the equation
■j dX .0L3/ (XX
becomes -/- =f(x, #), which is a differential equation of the
ax •
first order in which p and x are the variables. If we can find
p from this equation, we can then find y from -z-=p. This
method has been exemplified in Ex. 3, § 200.
s- 3 =4 !)•
The essential thing here is that the equation contains — and
-J^ but does not contain x. As before, we place -^=jp, but
now write -^ = ^=j3^==i?j' so that the equation becomes
, dsr ax ay ax ay
p -±- =/(#, jp), which is a differential equation of the first order
if
in which p and y are the variables. If we can find p from this
equation, we can find y from -f-=p.
ax
452 DIFFERENTIAL EQUATIONS
Ex. 2. Find the curve for which the radius of curvature at any point
is equal to the length of the portion of the normal between the point and
the axis of x. r /,/,A2-ii
The length of the radius of curvature is ± (§ 106). The
equation of the normal is (§ 87) — ^
dx &**
This intersects OX at the point (x + y — > 0 ) . The length of the normal
I ldv\2 V '
is therefore y \\1 + l-p) •
The conditions of the problem are satisfied by either of the differential
dx*
i
- ' ft' w
L1 + (fa) J L /rfy\i
—15 — = 'Y+fc#'
55 — = y\l1+(x)v (2)
Placing -r-=p and — 4 = /?— in (1), we have
dx dx2 dy
l+p*=py±-
whence — = J p ., •
y 1 +/>2
The solution of the last equation is
y = c1vT+pa;
Vm2 — c *
whence /> = — ^ L .
ci
Replacing p by — , we have * = dx.
dx Vy2 _ c. «
Transforming this equation to ,
and integrating, we have
FT
c ( *-Zh x~c*\
whence y = tM* ci +* °i /*
CERTAIN EQUATIONS OF SECOND ORDER 453
This is the equation of a catenary with its vertex at the point (c2, cx).
If we place -~ = P, and —4 = P -/- in (2), we have
ax ax2 ay
, dy —p dp
whence ~ = ., r •
y l + p2
The solution of this equation is y = — * ;
Vl+jt?a
Vc2-y*
whence p = —
y
Replacing p by -f- , we have
ax
A2 - y2
Integrating, we have — v c2 — y2 = x — c2,
or (x - c2)2 + y2 = c2.
This is the equation of a circle with its center on OX.
If we multiply both sides of this equation by 2-^dx, we have
or d
[®1-««»*
Integrating, we have ( -^ j = J 2/(y) dy + cx\
whence, by separating the variables, we have
dy
/
v2J/(y)<*y + 'i
= # + <?2.
Ex. 8. Consider the motion of a simple pendulum consisting of a parti-
cle P (fig. 243) of mass m suspended from a point C by a weightless string
of length I. Let the angle A CP = 0, where A C is the vertical, and let
AC
454
DIFFERENTIAL EQUATIONS
d*s
AP = s. By § 93 the force acting in the direction AP is equal torn—;
but the only force acting in this direction is the component of gravity.
The weight of the pendulum being mgy its component in the direction AP
is equal to — mg sin 0. Hence the differential equation of the motion is
d2s
m — - = —mg sin 0.
dt2 y
We shall treat this equation on the hypothe-
sis that the angle through which the pendulum
swings is so small that we may place sin 0 = 0,
a
without sensible error. Then, since 0 = - > the
equation becomes
d2s _ g
13 ~ Is'
ds
Multiplying by 2 — dt and integrating, we
, dt
have
where a2 is a new arbitrary constant. Separating the variables, we have
Vo2
whence
sin-^ = A/f
(< - Q>
where t0 is an arbitrary constant. From this, finally,
s = a sin a/^ (t — t0).
The physical meaning of the arbitrary constants can be given. For a
is the maximum value of s ; it is therefore the amplitude of the swing.
When t = t0, s = 0 ; hence t0 is the time at which the pendulum passes
through the vertical.
209. The linear equation with constant coefficients. The differ-
ential equation
dny dn~Yy
dx"
+a^+
+ «„-i^+«»y=/(a;>
(i)
where al9 a2, • . •, an_v an are constants, and where f(x) is a
function of x which may reduce to a constant or even be zero,
is called a linear differential equation with constant coefficient*.
LINEAR EQUATIONS 455
dti d u
To study (1) it is convenient to express -j- by By, -— -| by
iPy, • • •, -z-sL by Ity, and to rewrite (1) in the form
JTy + a^- V + • • • 4- a^Dy + a„y =/(*),
or, more compactly,
(D» + aJT-1 + • • • + a^D + «n)y =/(*). (2)
The expression in parentheses in (2) is called an operator,
and we are said to operate upon a quantity with it when we
carry out the indicated operations of differentiation, multiplica-
tion, and addition. Thus, if we operate on sin x with D8 — 2 D2
+ 3 D — 5, we have
(D8— 22)2 + 3D — 5)sin# = — cos# + 2 sin# + 3 cosrr — 5 sin a;
= 2 cos x — 3 sin x.
Also, the solution of (1) or (2) is expressed by the equation
1
IT + alIT-1+ • • • + an_xD + anm
y = t» ■ „ t»-i . ■ „ t, ■ „ /(*)» (3)
where the expression on the right hand of this equation is not
to be considered as a fraction but simply as a symbol to ex-
press the solution of (2). Thus, if (2) is the very simple
equation Dy=f(x), then (3) becomes
y = |/(*)= J/OO*** (4)
In this case — means integration with respect to x. What the
more complicated symbol (3) may mean, we are now to study.
210. The linear equation of the first order with constant
coefficients. The linear equation of the first order with con*
stant coefficients is
or, symbolically, (2>— a)y =/(s). (1)
456 DIFFERENTIAL EQUATIONS
The solution of this equation is given in § 204. Hence
y = 2^ f(x) = of + e°* fe-**f(x) dx. (2)
The solution (2) consists of two parts. The first part, ce"*, con-
tains an arbitrary constant, does not contain /(#)» ^d* ^ taken
alone, is not a solution of (1) unless f(x) is zero. The second
part, e0* I e~axf(x)dx, contains /(#), and, taken alone, is a solu-
tion of (1), since (1) is satisfied by (2) when c has any value,
including 0. Hence e"* j e'^ffx^dx is called a, particular inte-
gral of (1), and, in distinction from this, ce°* is called the
complementary function. The sum of the complementary func-
tion and the particular integral is the general solution (2). The
complementary function can be written down from the left-hand
member of equation (1), but the determination of the particular
integral requires integration.
Ex. 1. Solve ^ + 3 y = 5i».
ax
This equation may be written
(Z> + 3)y = 5x8.
Hence the complementary function is ce~Sx. The particular integral is
—i— (5 Xs) = 5 e~**fe**i*dx = § x8 - § x* + V * - ir-
Hence the general solution is y = ce~*x+ §x8 — §x2 + J^a? — J^.
Ex. 2. Solve — + y = sin x.
ax
The complementary function is ce~x. The particular integral is
— — - sin x = e~x f e* sin xdx = J (sin a: — cos x).
Therefore the general solution is y — ce~x + J (sin a; — coax).
211. The linear equation of the second order with constant
coefficients. The symbol (D — a)(D — b)y means that y is to
LINEAR EQUATIONS 457
be operated on with D — b and the result operated on with
D — a. Now (D — b)y = -j- — by, and hence
<"-)(i»-»)»-£(S-«r)-.(*-V
= (D* + pD + q)y, (1)
where p = — (a + J), j = db.
This result, obtained by considering the real meaning of the
operators, is the same as if the operators D — a and D — b had
been multiplied together, regarding D as an algebraic quantity.
Similarly, we find
(D-by(D-a')y = \_I>2-(a + b)D+ab]y=(D-aXZ>--t>)!/.
That is, the order in which the two operators D — a and D — b
are used does not affect the result.
Moreover, if (JP + pD+-q)y is given, it is possible to find a
and b so that (1) is satisfied. In fact, we have simply to factor
DP+pD + q, considering D as an algebraic quantity.
This gives a method of solving the linear equation of the
second order with constant coefficients. For such an equation
ha* the form d, .
or, what is the same thing,
(l?+PD + q)y=f(x), (2)
where p and q are constants ajid f(x) is a function of x which
may reduce to a constant or be zero.
Equation (2) may be written
(D - a) (D-i)y =/(*);
whence, by (2), § 210,
(D - h)y = f(x) = cfeax+ e** Ce-^fQc^dx.
JJ a j
458 DIFFERENTIAL EQUATIONS
Again applying (2), § 210, we have
y= n^h^C'eaX^eaX
fe-^fCx^dx^
= cjt* + $* fe~ bx Me** + e™ Ce~ ^/(x^dxj dx. (3)
There are now two cases to be distinguished:
I. If a =£ J, (3) becomes
y = <?/*+ c1eax+ ebx C(eia-b)x Ce'^f^dxj dx. (4)
II. If a = J, (3) becomes
y = (<?2 + Cia.) *«* + e«* CCe~ «/(*) da* (5)
In each case the solution consists of two parts. The one is
the complementary function ceP* + c2ebx or (<?2 + ex) e?*, involving
two arbitrary constants but not involving f(x). It can be
written down from the left-hand member of the equation, and
is, in fact, the solution of the equation (i> — a)(D — b)y = 0.
The other part of the general solution is the particular integral,
and involves f(x). Its computation by (4) or (5) necessitates
two integrations.
Formula (4) holds whether a and b are real or complex.
But when a and b are conjugate complex it is convenient to
modify the complementary function as follows: Let us place
a = m + in, b = m — in*
Then the complementary function is
c g(»» + in)x I c g(m - in)x
= e™ (c/™ + c2e~ inx) (§ 194)
= e1"* [^ (cos nx + i sin nx) + c2 (cos na; — i sin wa;)]
= emx(C1 cos na; + C2 sin tiz), (6)
where C^ = ^ + <?2, <72 = i^ — <?2). Since c1 and <?a are arbitrary
constants, so also are Cx and (72, and we obtain all real forms of
the complementary function by giving real values to Cx and C%.
LINEAR EQUATIONS 459
The form (6) may also be modified as follows: Whatever
be the values of Cx and C2 we may always find an angle a such
C C
that cos a = i ■. , sin a = 2 : . Then (6) becomes
y/Cf+Cl Vc?+CJ
Ice™* cos (nx - a), (7)
where a and h = y/C*+ C22 are new arbitrary constants. Or we may
— C C
find an angle j3 such that sin j3 = 1 = , cos j3 = 2 .
Then (6) becomes Vcf + G? Vc? + G?
for* sin (nz - /9). (8)
te.i.fa + gS + «, = «-.
cur a#
This equation may be written
(Z) + 2)(Z> + 3)y = e*
The complementary function is therefore c^-2* + c2e-8x. To find the
particular integral we proceed as follows :
1 r \
(2> + 3)y = — — -ex= e~2x I e^xdx = -ex.
K )y D + 2 J 3
y = — - — (-<A = e-*x f-e*xdx = —ex.
y D + 3\3 / J 3 12
Therefore the general solution is
y = cxer*x + c2€~Sx + T^e30.
This equation may be written (J) + 1)2# = a;.
Therefore the complementary function is (cj + c2x)e~x. To find the
particular integral we proceed as follows :
(D + 1) y = — — - x = e~x jxe30 dx = x — 1.
y = j- — - (x — 1) = e-*C(x — l)exdx = x — 2.
Therefore the general solution is
V ~ (ci + c^c)e~x + x ~ 2-
Ex. 3. Consider the motion of a particle of unit mass acted on by an
attracting force directed toward a center and proportional to the distance
of the particle from the center, the motion being resisted by a force pro-
portional to the velocity of the particle.
460 DIFFEEENTIAL EQUATIONS
If we take s as the distance of the particle from the center of force, the
ds
attracting force is — ks and the resisting force is — h —- > where k and h are
at
positive constants. Hence the equation of motion is
£ = _*_**
dt2 dt
or (Z>2 + hD + k)s = 0. (1)
The factors of the operator in (1) are
K-^)K+^)-
We have therefore to consider three cases :
I. h* - 4 k < 0. The solution of (1) is then
_ht
or
s = e ^(Cjcos *+C2sin t),
_*! / V4 k-h2 \
s=ae 2sinl * — /?).
The graph of s has the general shape of that shown in fig. 62, § 27. The
particle makes an infinite number of oscillations with decreasing ampli-
tudes, which approach zero as a limit as t becomes infinite.
II. h2 - 4 k > 0. The solution of (1) is then
s = c,e
(h VA»-4*\ (h Vft»-4fc\
= Cl6 ^ 2 / +c26 Va 2 ''
The particle makes no oscillations, but approaches rest as t becomes
infinite.
III. h2 - 4 k = 0. The solution of (1) is
s = (Cj + cj) e 2 .
The particle approaches rest as t becomes infinite.
212. The general linear equation with constant coefficients.
The methods of solving a linear equation of the second order
with constant coefficients are readily extended to an equation
of the nth order with constant coefficients. Such an equation is
or, symbolically written,
<jy + a^-1 +... + a^D + oy =/(*)• ' (2)
LINEAR EQUATIONS 461
The first step is to separate the operator in (2) into its linear
factors and to write (2) as
(2> - r,) (2> - r.) • • • (2> - rB) y =/(*), (3)
where rx, ra, • • • , rn are the roots of the algebraic equation
f» + a/1-1 H + a^r + att = 0.
It may be shown, as in § 211, that the left-hand members
of (2) and (3) are equivalent, and that the order of the factors
in (3) is immaterial.
The general solution of (1) consists now of two parts, the
complementary function and the particular integral.
The complementary function is written down from the factored
form of the left-hand side of (3), and is the solution of (1) in
the special case in which f(x) is zero. If rv r2, • ••,/•„ are all
distinct, the complementary function consists of the n terms
c^er<* + c2er** + • • • + cner*x, (4)
where ey <?2, • • •, cn are arbitrary constants.
If, however, D — r{ appears as a k-i old factor in (3), k of
the terms of (4) must be replaced by the terms
(cx + c2x + eg? + h ckz?-l)erf:.
Also, if two factors of (3) are conjugate complex numbers,
the corresponding terms of (4) may be replaced by terms in-
volving sines and cosines, as in (6), § 211.
The particular integral is found by evaluating
(P-r,)<.D-V...(p-rS<& (5)
This may be done by applying the operators
• • •
n ' n— 1
in succession from right to left. This leads to a multiple
integral of the form
er? fe<r*-rJ° fe<r*-r*>x . . . fe-r^f(x)daf. (6)
462 DIFFERENTIAL EQUATIONS
In evaluating (6) the constants of integration may be omitted,
since they are taken care of in the complementary function.
The general solution is the sum of the complementary function
and the particular integral.
213. Solution by undetermined coefficients. While theoreti-
cally the particular integral can always be found by evaluating
(6), § 212, practically the work may become very complicated,
and can be much simplified when the general form of the inte-
gral may be anticipated. The particular integral may then be
written with undetermined coefficients, and the coefficients deter-
mined by direct substitution in the differential equation. We give
below certain directions for such substitutions, which have been
obtained by studying formula (6) for different functions f(x).
We will denote the differential equation by
where -P(i^) is a polynomial in D, and will denote the particular
integral by I.
I. Iff(x) = aQa? + ajf~l + • • - + an_1x+ an, assume, in general,
I=Ai» + A1a»-1+... + An_1x+A%\
but if Dm is a factor of P(D), assume
I=ar(Aa»+A1afi-1+ . . . +An_lz + AJ.
II. If f(x) = ce**, assume, in general,
I=Aeax;
but if (D — a)m is a factor of -P(2>), assume
I=Axmeax.
III. If f(x) = c sin ax or c cos ax, assume, in general,
I=A sin ax + B cos ax ;
but if (2^+ a2)m is a factor of P(i>), assume
1= af1 (A sin ax + B sin ax).
IV. If f(x) = e^ffrQc), place y = e^z and divide out e**.
V. Iff(x) is the sum of a number of functions, take I as the mm
of the particular integrals corresponding to each of the functions.
LINEAR EQUATIONS 463
~«.3+*-«»-~
This may be written (D + 3) (D - 2) y = e*x.
The complementary function is c1e2x + c2e~8x. To find the particular
integral, we place I = A**9
and substitute in the equation. We obtain
!4Ae*x = (*x.
To satisfy the equation, we must have
14-4 =1, whence A = y1^.
Therefore the particular integral is
and the general solution is
y = Cle2x + c2e~*x + jyieix.
Ex/a. <ry + p> = Sin2x.
ax3 ax2
This may be written D2(Z) + 1) y = sin 2 a\
The complementary function is therefore c1 + c2a? + cse~x. To find the
particular integral, place
I = A sin 2x + B cos 2 a;,
and substitute in the equation. We obtain
(8£ - 44)sin 2x - (4£ + 8,4)cos 2 a; = sin 2a;.
To satisfy the equation, we must have
8B-4cA=l, 4£ + 84 = 0,
whence B = T^, A = — ^.
Therefore the particular integral is
/ = — 2^ sin 2 x + T*ff cos 2 x,
and the general solution is
y = ct + c2x + cse~x — ^ sin 2 x + y^ cos 2 x.
Ex.3. ^ + ^ = ^.
Substituting y = e*z, we have
This may be written (D + 1) (Z) + 2) 2 = x\ (2)
464 DIFFERENTIAL EQUATIONS
The complementary function is therefore c1e"x + c2e~2x. To find the
particular integral, place
/ = Ax* + Bx + C,
and substitute in (1). We obtain
2 Ax* + (6,1 + 2 B)x + (2 A + 3 B + 2C) = a;2.
Therefore 24=1, 64 + 2£ = 0, 24 + 3£ + 2C = 0,
whence A = J, i* = — §, and C = J.
Hence the particular integral is
and the general solution of (1) is
z = Cle-* + c2e~*x + %x* - $a: + J,
whence y = cx + c2e~x + £*(£ se2 — $ a: + J).
This may be written
(D - 1) (D + 2) (D - 2)y = e*x.
Since D — 2 is a factor of P(D), we place
J = 4are2*,
and substitute in the equation. We obtain
4cAe2x = e2x>
whence A = J and JT = J a;e2x.
The general solution is
y = c^ + c2e-*x + Cge2* + £a;e2x.
Ex. 5. —^ + y = sin x.
ax*
This may be written (D* + V)y = sin a:.
By HE, we write / = Ax sin x + Bx cos ar,
and substitute in the equation. There results
— 2 B sin x -\r2A cos a: = sin x.
1 x
Therefore B =— -, 4=0, and J = — - cosar.
The general solution is
a? a?
y = cx&* + c2e_ia: — - cos x = Ct cos a: + C2 sin x — - cos x.
LINEAR EQUATIONS 465
Ex.6. g_2f& + ifc = iM.. + «...
ax6 ax* ax
This may be written D(D — l)2 y = xe2x + e8x.
The complementary function is cx + (c2 + Cgz) ex. The particular integral
/ is the sum of I1 and J2, where Ix corresponds to the term xe2xy and /2
to 6s*. To find /1? place y = e2xz in the equation
(D» - 2 D2 + D)y = xe2x.
There results (D8 + 4D2 + 52) + 2)2 = ar.
Placing z = -4x + B, we find -4 = J, iJ = — f .
Therefore Ix = \ (2 x - 5) e2*
To find I2, we substitute y = AeSx in the equation
(D8-22>2 + D)y = e8a:.
We find ^2 = tV e8*
Hence I = J (2 a: - 5) e2x + tV <?8*.
The general solution of the equation is
y = Cl + (c2 + Cga;)^ + i(2x - 5)e2* + ^e8*.
214. Systems of linear differential equations with constant
coefficients. The operators of the previous articles may be em-
ployed in solving a system of two or more linear differential
equations with constant coefficients, when the equations involve
only one independent variable and a number of dependent vari-
ables equal to the number of the equations. The method by
which this may be done can best be explained by an example.
dt dt *
te+*K-2x-3y = e2<.
dt dt *
These equations may be written
(D - l)x + (D - 4)y = e", (1)
(D-2)x + (D-Z)y = e2<. (2)
We may now eliminate y from the equations in a manner analogous to
that used in solving two algebraic equations. We first operate on (1) with
Z> — 3, the coefficient of y in (2), and have
(B2 - 4 D + 3)ar + (D2 - ID + 12) y = 2 e", (3)
466 DIFFERENTIAL EQUATIONS
since (D - 3) e6t = 5 e6t - 3 e6t = 2 e6*. We then operate on (2) with D- 4
the coefficient of y in (1), and have
(Z)2 - 6D + 8)z + (i)2 - ID + 12)y = - 2 e2«, (4)
since (Z> — 4) e2t = — 2 e2t. By subtracting (4) from (3) we have
(2Z>-5)x = 2e« + 2e2<, ' (5)
the solution of which is x = c^a' + $ e6' — 2 e2'. (6)
Similarly, by operating on (1) with (D — 2) and on (2) with D —1, and
subtracting the result of the first operation from that of the second, we have
(2 D - 5)y = - 3 e*< + e2', (7)
the solution of which is y = c2e*' — $ e6' — e2'. (8)
The constants in (6) and (8) are, however, not independent, for the
values of x and y given in (6) and (8), if substituted in (1) and (2), must
reduce the latter equations to identities. Making these substitutions, we
have
whence it is evident that c2 = cv Therefore, replacing cx by c, we have
x = ce^ + I e6t - 2 e\
y = ce^' — § e6' — e2',
as the solutions of the given equations.
215. Solution by series. The solution of a differential equa-
tion can usually be expanded into a series. This is, in fact, an
important and powerful method of investigating the function
denned by the equation. We shall limit ourselves, however,
to showing by examples how the series may be obtained. The
method consists in assuming a series of the form
where m and the coefficients a0, a , a2, • • • are undetermined.
This series is then substituted in the differential equation, and
m and the coefficients are so determined that the equation is
identically satisfied.
SOLUTION BY SERIES 467
Ex. 1. ar^ + (a:-3)^-2y = 0.
ax2 ax
We assume a series of the form given above, and write the expression
for each term of the differential equation, placing like powers of x under
each other. We have then
cPy
x — ^=m(m— l)a0arm-1+(m+l)ma1a:mH h(m+r+l)(m+r)ar+1xm+rH ,
dy
x—-= matfcm+"* +(m+r)arxm+r+ •••,
ax
-3-^= — 3ma0xm-1-3(m+l)a1a:m — 3(m+r+l)ar+1arm+r ,
ax
— 2y= —2a0xm —2arxm+r .
Adding these results, we have an expression which must be identically
equal to zero, since the assumed series satisfies the differential equation.
Equating to zero the coefficient of xm~1> we have
m (m — 4) a0 = 0. (1)
Equating to zero the coefficient of xmf we have
(m + 1) (m - 3)ax + (m - 2)a0 = 0. (2)
Finally, equating to zero the coefficient of xm+r, we have the more general
relation ^ (m + r + l)(m + r - 3)ar+1 + (m + r- 2)ar = 0. (3)
We shall gain nothing by placing aQ = 0 in equation (1), since a0xm is
assumed as the first term of the series. Hence to satisfy (1) we must
have either __ n _ _ A
m = U or m = 4.
Taking the first of these possibilities, namely m = 0, we have, from (2),
ai = - § ao>
o
and from (3), ar+1 = - (r + 1)(r_3)<V (4)
This last formula (4) enables us to compute any coefficient, ar + lt when we
know the previous one, ar. Thus we find a2 = "~ i ai = i ao> a8 = 0> and
therefore all coefficients after az are equal to zero.
Hence we have as one solution of the differential equation the polynomial
yi = «o(1- §*+ bx*)- (5)
Returning now to the second of the two possibilities for the value of m,
we take m = 4. Then (2) becomes
5 ax + 2 a0 = 0,
and (3) becomes or + 1 = - (r + 5)(r + 1) <V (6)
468 DIFFERENTIAL EQUATIONS
Computing from this the coefficients of the first four terms of the series,
we have the solution
V*
= aJx*--xb + -^-x* - — x1 + ...V (7)
°\ 5 5- 6 5.6.7/ W
We have now in (5) and (7) two independent solutions of the differential
equation. A more general solution is
y = cxyx + c2y2,
and this may be shown to be the most general solution.
Ex. 2. Legendre's equation. (1 — x2) — \ — 2x-^ + n(n+l)y = 0.
ax* ax
Assuming the general form of the series, we have
d?v
— ^ = to(to — l)a0xm-2 + (m + l)7iia1a:m-1+ (to + 2)(to + l)a2xw + -..,
-*p-= -m(m-l)a0*» — -,
ax2
— 2x-^= — 27waftxm ,
ax
n(n + l)y= n(n + l)a0xm + •••.
Equating to zero the coefficients of xm~2, arm_1, and xm,we have
to (to — 1) a0 = 0, (1)
(TO + l)?wa1 = 0, (2)
(to + 2) (to + l)a2 - (to - n) (to + n + l)a0 = 0. (3)
To find a general law for the coefficients, we will find the term contain-
ing arm + r_2 in each of the above expansions, this term being chosen because
it contains ar in the first expansion. We have
d*v
_-£ = . .. + (m + r)(TO + r-l)arxm + r-a+ ...,
"^-T \= (TO + r-2)(TO + r- 3) fl^ J2a*»+'-2 - . . .,
-2x^= 2 (to + r - 2)ar_2x'"+r-2 ,
ax '
n(n +l)y = •• • + n(n + l)ar_2xm+r-a + ....
The coefficient of xm + r_2 equated to zero gives
(to + t)(to + r — l)ar — (to — n + r — 2) (to + n + r — l)ar_a = 0. (4)
SOLUTION BY SERIES 469
We may satisfy (1) either by placing m = 0 or by placing m = 1. We
shall take m = 0. Then from (2), ax is arbitrary ; from (3),
n(n + l)
(5)
and from (4), ar = - (»-r + 2)(i» + r-l)
r(r — 1)
By means of (6) we determine the solutiqn
y=a»(i-"(v1)j3+n(""2)(wi4+1)("+8):c4--)
+ ai(;c-("-l)(" + 2)j,+ (n-l)(»-3)(» + 2)(n + 4^_.^(7)
Since a0 and ax are arbitrary, we have in (7) the general solution
of the differential equation. In fact, the student will find that if
he takes the value m = 1 from (1), he will obtain again the second
series in (7).
Particular interest attaches to the cases in which one of the series in
(7) reduces to a polynomial. This evidently happens to the first series
when n is an even integer, and to the second series when n is an odd
integer. By giving to a0 or ax such numerical values in each case that the
polynomial is equal to unity when x is equal to unity, we obtain from
the series in (7) the polynomials
r> " 2 I '
3
2" 2
•* 8 — r\^ e\^y
P4 = ^^_2^^ + iJ:,
4 4-2 4-2 4-2
x6 — 2 XT + X.
4-2 4-2 4-2 '
each of which satisfies a Legendre's differential equation in which n has
the value indicated by the suffix of P. These polynomials are called
Legendre's coefficients.
AC
470 DIFFERENTIAL EQUATIONS
cPv dv
Ex. 3. BesseVs equation. x2 — -n + x~r + (x2— n2)y = 0.
ax2 ax
Assuming the series for y in the usual form, we have
x2—^ = m(m - l)a0a?» + (to + l)ma1xm+1 + (to + 2)(to + l)a2a^ + 2 + • • •,
dx2
dy
x-f-= ma^f?* + (to + l)a1xm+1 + (m + 2)a2xm + 2 + • ••,
— n2y = — n2a0arm — n2^"1-!-1 — n2a2xw + 2 — • • •,
a?y = * OfiX™*2 + •••.
Equating to zero the coefficient of each of the first three powers of x,
we have
(TO2-n2)a0 = 0, (1)
[(m + l)«- 11^ = 0, (2)
[(w + 2)2-n2]a2 + a0 = 0. (3)
To obtain the general law for the coefficients, we have
x2 ~- = . . . + (to + r) (to + r — 1) arxm+r + • • •,
dx2
x — = • • • + (to + r) arafn+r + • • •,'
dx
— n2y = ... — n2arxm+r — • • •,
x2^ = ••• + ar_2a;m+r + "" "•
Equating to zero the coefficient of x™ + r, we have
[(to + r)2 -n*]ar + ar_2 = 0. (4)
Equation (1) may be satisfied by to = ± n. We will take first m = n. Then
from (2), (3), and (4) we have
OOq Q>r — 2
> g2 = - o /o . ox> Or = -
2(2n + 2) r r(2n + r)
By use of these results we obtain the series
Vl = %X* V ~" 2(2n + 2) + 2.4-(2n + 2)(2n + 4) / ' (5)
Similarly, by placing to = — n, we obtain the series
PROBLEMS . 471
If, now, n is any number except an integer or zero, each of the series
(5) and (6) converges and the two series are distinct from each other.
Hence in this case the general solution of the differential equation is
y = Wi + c2Vr
If n = 0 the two series (5) and (6) are identical. If n is a positive
integer, series (6) is nfeaningless, since some of the coefficients become
infinite. If n is a negative integer, series (5) is meaningless, since some
of the coefficients become infinite. Hence, if n is zero or an integer, we
have in (5) and (6) only one particular solution of the differential equa-
tion, and another particular solution must be found before the general
solution is known. The manner in which this may be done cannot,
however, be taken up here.
The series (5) and (6) with special values assigned to a0 define new
transcendental functions of x, called BesseVs functions. They are important
in many applications to mathematical physics.
PROBLEMS
Solve the following equations :
1. x(l — y)dx + y(l — x)dy — 0.
2. sec2ydx + gob2 xdy = 0.
3. ytdx — (x2 + 2xy)dy = 0.
4. (x "y/x2 + y* — y*)dx + xydy = 0.
v v
5. [(# — y)e* + x]dx + xe*dy = 0.
6. (2x + 3y — 4)dx +(3x + y + l)dy = 0.
7. (x + y — 5)dx + (x + y — S)dy = 0.
8. (y — x2 — 1) dx + xdy = 0. 10. dx + (x — y) dy = 0.
9. xdy — (y + x*eZx)dx = 0. 11. (x + l)2y dx + (x + 1)8 dy = dx.
12. (y + xy*)dx — dy = 0.
13. (1 + xfdy - [(1 + x)y + rftf^dx = 0.
14. (2 x + ye^)dx + (cos y + xe?*)dy = 0.
15. (Sx2 + ^ + 2xtf -2^dx+(3tf + ^ + 2x2y + 2 jW=0.
\x-y x2 + 2/7 \y - x x2 + f) V ~~
472
DIFFERENTIAL EQUATIONS
18.
19.
20.
21.
22.
23.
24,
25
26.
27,
28.
29.
30.
31,
32.
33.
34.
35,
36.
37,
38.
39.
40.
41.
42,
43.
44.
45
46.
xdx + ydy=(x2 + y*)dx.
xdy — ydx = Vsc2 — y*dx.
X _x
ye vdx—(xe v + y*)dy = 0.
3 sin (x 4- y) dx 4- [3 sin (x + y) — 2 y cos (x + y)~\ dy
x cos2ydx — esc xdy = 0.
dx -\-(x tan y — sec y)dy =0.
(y — Var* + y^dx — xdy = 0.
(x2 - f)dx + 3xfdy = 0.
(2x — 5y + 5)dx+(4:x — y + l)dy = 0.
6-?i?)*+0?T?-;h-a
(2y - x2 + l)dx + (x2 - l)dy = 0.
a2 Vl-2/8 «fc - y1 Vl - a?% = 0.
2xydx — (4a8 + s/2)^ = 0.
scefo + yrfy + (a2 + if)* (ydx — a;dy) = 0.
e* (x2 + f + 2 x) dx + 2 ye30 dy = 0.
(3 xy + 2 exi)xdx — dy = 0.
(2 — scy) y cfcc + (2 4- #y) xdy = 0.
(5 i/2 — 6 a;y) c&e + (6 as2 — 8 scy + y2) dy = 0.
a>(l + x*)dy - [(1 -|- a»)y + xf\dx = 0.
a; dx + (y — Va^ + y2) rfy = 0.
[y cos 2 a; + 2 (sin 2 a;)*] efo + sin 2 a;<2y = 0.
= 0.
(*♦
a;
^
dx 4- ( 1 —
fc?)*-ft
Vo2 - y2/ \ Va?
(afy8 + y8)^ 4- («V ~ x*)dV = °-
(a;y — Xs) dx + rfy = 0.
(x — y tan"1 Mete + a; tan"1 ^ dy = 0.
(2a? - 3y 4- l)dx 4- (2a; - 3y + 2)dy = 0.
(ar 4- 2y)dx — (2a; — y)dy = 0.
(«" + f + y4)^ - 2a;y<fy = 0.
[ajy - a;8/ (1 + a?)]dx 4- (1 4- x*)dy = 0.
PKOBLEMS 473
-fa-p)
47. I xy* — -^ex\dx + x2ydy = 0.
48. (y + x2 - l)xdx - (x2 - l)dy = 0.
49. (2xtf -Sy)dx + (2x*y + x)dy = 0.
d2y 1 d*y /dy\2 ..
51.
dx2 x dx2
dx2 Vl-x2'
»»-g-(fWfj=<>-
dx2 (1 + a)2 ^ «*b
53. ^ = sec2 ax. 60. -7^ = - k2y.
dx2 dx2 *
54.^-^ = ^. 61.^ + ^-sin*.
cfcc4 ox aar ax
«-S=aFW- --3-*
65. Solve -r-j = - Vy, under the hypothesis that, when x = — 1
66. Solve 33 = 2 y8 -|- 6 y, under the hypothesis that, when x = 77 >
y=0and^ = 3.
67. Solve -t4 = tan8y + tan y, under the hypothesis that, when
w * * dy .
x = — ■ > y = 0 and -7*- = 1.
rf*y
68. Solve -r4 = tan y + tanfy, under the hypothesis that, when
. 7r _ dy .
x = 0, y = -r and -71 =1.
4 ox
474 DIFFERENTIAL EQUATIONS
Solve the following equations :
«»-S-t-6s"-6-T*- 3**-
oar dx '
73. g + 9y = 6e.«. 76. J|_3^ = 2-6*.
oar * ckc3 dx
76- § + 8^ + 15y = 2«-»*. 78. §+ 4^ + 5y = 20cos3x.
oar ax aar dx
79. 4^4-4-^-3y = 2sinx + cos2x.
oar ox
80. -pi-2^- + 3y = e-xco8x. 82. ^ + 9y = 2 sin 3x.
dx2 dx * dx* J
81- 3 + 4? + 4y = *W*. 83. §-?-2y = (x + sinxK.
dor dx dor dx v 7
84. ^-7^+10y = a;e2* + sina;.
dx* dx *
87. ^-4^ + 8y = 4a^-15cos3x.
88. ^ + y = x« + a;.
PROBLEMS 475
91' dx> da? *dx^*y ~^ + e +e-
92 ^_^ + 4^_4v = Cos2a;
9Z" dx* dx*+ dx *y-cos^-
93- !S+2§-^-2y-5siii2a:+12a*to.
dxr chr ax
97. ^ + 5^-36y.= 20<j2*cos3a;.
98. ^ + 2^ = 2x + 25e-*sin2a;.
99. -7^H-4-7^ = 4cos4a;.
101.- + y = e>, 105.--3j = ^,
„-_ eftc , dy . ^ <rv^ rf2^ t ndx t dy _ 0.
102- ^+i=sm<' 106- ^ + 2^+i = 2e*
103. ft=lx-y + 2, 107._g-aV=0,
|=2y_3,+l g + A-a
476 DIFFERENTIAL EQUATIONS
Solve the following equations by means of series :
m.(i + ^)g + ,|-wy = o.
114. Prove that any curve the slope of which at any point is
proportional to the abscissa of the point is a parabola.
115. Find a curve passing through (0, — 2) and such that its
slope at any point is equal to three more than the ordinate of
the point.
116. Find the curve the slope of which at any point is propor-
tional to the square of the ordinate of the point and which passes
through (1, 1).
117. Find the curve in which the slope of the tangent at any
point is n times the slope of the straight line joining the point to
the origin.
118. Find in polar coordinates the equation of a curve such
that the tangent of the angle between the radius vector and the
curve is equal to minus the reciprocal of the radius vector.
119. Find in polar coordinates the equation of a curve such that
the tangent of the angle between the radius vector and the curve is
equal to the square of the radius vector.
120. Find in polar coordinates the curve in which the angle be-
tween the radius vector and the tangent is n times the vectorial angle.
121. A point moves in a plane curve such that the tangent to the
curve at any point and the straight line from the same point to
the origin of coordinates make complementary angles with the axis
of x. What is the equation of the curve ?
PROBLEMS 477
122. Show that if the normal to a curve always passes through
a fixed point the curve is a circle.
123. Find the curve in which the perpendicular from the origin
upon the tangent is equal to the abscissa of the point of contact.
124. Find the curve in which the perpendicular upon the tangent
from the foot of the ordinate of the point of contact is a constant a.
125. Find the curve in which the length of the portion of the
normal between the curve and the axis of x is proportional to the
square of the ordinate.
126. Derive the equation of a curve such that the sum of the ordi-
nate at any point on it and the distance from the point to the axis of
x9 measured along the tangent, is always equal to a constant a.
127. Find the polar equation of a curve such that the perpendic-
ular from the pole upon any tangent is k times the radius vector of
the point of contact.
128. Find the curve in which the chain of a suspension bridge
hangs, assuming that the load on the chain is proportional to its
projection on a horizontal line.
129. Find the curve such that the area included between the
curve, the axis of x, & fixed ordinate, and a variable ordinate is
proportional to the difference between the fixed ordinate and the
variable ordinate.
130. Find the curve in which the area bounded by the curve, the
axis of x, a fixed ordinate, and a variable ordinate is proportional to
the length of the arc which is part of the boundary.
131. Find the curve in which the length of the arc from a fixed
point to any point P is proportional to the square root of the
abscissa of P.
132. Find the space traversed by a moving body in the time t if
its velocity is proportional to the distance traveled and if the body
travels 100 ft. in 10 sec. and 200 ft. in 15 sec.
133. In a chemical reaction the rate of change of concentration
of a substance is proportional to the concentration of the substance.
If the concentration is T^ when t = 0, and ^^ when t = 5, find the
law connecting the concentration and the time.
478 DIFFERENTIAL EQUATIONS
134. Assuming that the rate of change of atmospheric pressure
jo at a distance h above the surface of the earth is proportional to
the pressure, and that the pressure at sea level is 14.7 lb. per square
inch and at a distance of 1600 ft. above sea level is 13.8 lb. per
square inch, find the law connecting h and p.
135. The sum of $100 is put at interest at the rate of 5% per
annum, under the condition that the interest shall be compounded
at each instant of time. How much will it amount to in 50 yr. ?
136. If water is running out of an orifice near the bottom of a
cylindrical tank, the rate at which the level of the water is sinking
is proportional to the square root of the depth of water. If the level
of the water sinks halfway to the orifice in 20 min., how long will
it be before it sinks to the orifice ?
137. Find the deflection of a beam fixed at one end and weighted
at the other.
138. Find the deflection of a beam fixed at one end and uniformly
loaded.
139. Find the deflection of a beam loaded at its center and sup-
ported at its ends.
140. Find the curve whose radius of curvature is constant.
141. Find the curve in which the radius of curvature at any point
varies as the cube of the length of the normal between that point
and the axis of x.
142. A particle moves in a straight line from a distance a towards
a center of force which attracts with a magnitude equal to —• If the
particle was originally at rest, how long will it be before it reaches
the center ?
143. A particle moves in a straight line from a distance a towards
a center of force which attracts with a magnitude equal to fir~$. If
the particle was originally at rest, how long will it be before it
reaches the center?
144. A particle begins to move from a distance a towards a fixed
center of force which repels with a magnitude equal to fi times the
distance of the particle from the center. If its initial velocity is
V/xa2, show that the particle will continually approach, but never
reach, the center.
PROBLEMS 479
145. A particle moves along a straight line towards a center of
force which attracts directly as the distance from the center. If it
starts from a position of rest a units from the center, what velocity
will it have acquired when it has traversed half the distance to the
center?
146. A particle moves in a straight line from a distance a towards
a center of force which attracts with a magnitude equal to 77-^
2i IT
r denoting the distance of the particle from the center of force. If
the particle had an initial velocity of -j=> how long will it take to
traverse half the distance to the center ?
147. A body moves through a distance d under the action of a
constant force. Its initial velocity is vv and its final velocity is vr
Find the time required.
148. A particle moves from rest to a center of force which attracts
with a magnitude equal to ^ • Show that the average velocity on the
first half of its path is to the average velocity on the second half in
the ratio tt — 2 : tt + 2.
149. Assuming that gravity varies inversely as the square of the
distance from the center of the earth, find the velocity acquired by
a body falling from infinity to the surface of the earth.
150. Find the velocity acquired by a body sliding down a curve,
without friction, under the influence of gravity.
161. A bullet is fired horizontally into a sand bank in which the
retardation is equal to the square root of the velocity. When will it
come to rest if the velocity on entering is 100 ft. per second ?
152. A motor boat weighing 1000 lb. is moving in a straight line
with a velocity of 100 ft. per second when the motor is shut off. If
the resistance of the water is directly proportional to the velocity of
the boat, and is equal to 10 lb. when the velocity is 1 ft. per second,
how far will the boat move before its velocity is reduced to 25 ft.
per second ? How long will it be before this reduction of velocity
takes place?
153. A particle is projected vertically upward from the earth's
surface in a medium in which the resistance is k times the square
of the velocity. If vx is the velocity of projection and v2 is the
480 DIFFERENTIAL EQUATIONS
velocity with which the particle returns to its starting point, find
the value of va in terms of vv k, and the mass of the particle.
154. The force exerted by a stretched elastic string is directly
proportional to the difference between its stretched length and its
natural length. One end of an elastic string of inconsiderable mass
and of natural length 2 ft. is fastened at a point on the surface of a
smooth table. A particle of mass ^ lb. is attached to the other end
of the string and is drawn back till the string is stretched by an
amount 1 ft., and is then released. Find the time of a complete
oscillation of the particle if a force of J lb. is required to stretch
the string to double its natural length.
155. A particle of unit mass moving in a straight line is acted
on by an attracting force in its line of motion directed towards a
center and proportional to the distance of the particle from the
center, and also by a periodic force equal to a cos kt. Determine
its motion.
156. A particle of unit mass moving in a straight line is acted
on by three forces — an attracting force in its line of motion directed
towards a center and proportional to the distance of the particle
from the center, a resisting force proportional to the velocity of
the particle, and a periodic force equal to a cos kt. Determine the
motion of the particle.
157. Under what conditions will the motion of the particle in
problem 156 consist of oscillations the amplitudes of which become
very large as the time increases without limit?
ANSWEES
(The answers to some problems are intentionally omitted.)
Page 13 CHAPTER I
1. 5 + 3 V6 9. (J, 1). 11 (- 2 ^ *»)• 18- (- f , 0).
2. 10 + 2 V65. 10. (^, 2|f). 12. (- 3, 11).
Page 14
14. (4, 7), (4, - 1), (- 4, 3). 16. J, 0, oo.
"• a At A). (1#. ~ 5)' 17- 3 V5, £ ; 8, 0 ; Vl3, - f .
18. (0, 4). 22. (J, 4J). 25. (1J, -4), (6§, -1).
19. <- §, 1J). 23. (8*, - 2*), (4f , - 8f). 26. (- 4,-1).
20. (4, 6), (- 2, - 3). 24. (- £, - 2J). 27. (10, 17).
Page 15
28. (- 3, 2). 29. $ V5, £ V26, £ Vl49. 31. oo, 3. 32. (16, - 2).
Page 18
59. 62, - 1, - 13. 62. £ Vl5, 0, J VS.
Page 19
73. -5,-4,-4, 0.
Page 37 CHAPTER II
77. (2, 4). 83. (- 4, - 3), (- 3 J, -1J). 88. (J, 0), (2, 3).
78. (-£, §). 84. (6,1*). 89. (0, 1), (1, J).
79. (3, 4), (- If, - 4|). 85. (2, £), (1, 2). 90. (± 3, ± 4).
80. (- 3, 1). 86. (- 1, - 1£). 91. (± 2 V2, 2).
82. (-£±iV5,J^TfV6). 87. (±4, ±2), (±3, ±1}). 92. (± J, ± J).
Page 38
93. (0, 0), (2, ± l£). 96. (± 1, 3$). 99. - 2.07. 102. 2.41.
94. (1, ± 1). 97. (3, 2£), (- 1, 1£). 100. .46, 2.05. 103. - 2.62.
95. (±2V2, 2). 98. 1.46. 101. 1.12, 3.93.
„ -. CHAPTER III
Page 45
1. (1, 7), (- 6, 9), (2, - 4). 2. x2 + 9y2 = 5. 3. x2 + y2 = 4. 4. x2- y*= 4.
Page 46
7. x2 + 4y = 0. 9. 4x2 + 9y2=l.
8. 2x2- 4y2 = 9. 10. xy = 6.
15. (V3, l), (l, -V3), (1-V3, -1-V3).
16. (fcVS, *V2), (2V2, 0), (*V2, -f V2).
17. (2,1), (-2, -1), (-£, -2£).
18. x2 + 7y2 = 14.
417.10 481
5. y» + 3x2 =
= 0.
6. y2 = 4x.
11. Sxy = 7.
12. ab — c.
** c2 d2
14-4^ + 4-6-
- e.
482 ANSWEES
Page 47
19. 6x2 + y2 = 5. 24. 2xy ± 49 = 0. 28. 10x2+6y2-22x + 4y-20 = (X
20. 13 y2 - 14 = 0. 25. 3x2 + 2y2 = 6. 29. x2 + 7^ = 14.
21. 4x2 + 9y2 = 36. 26. 3x2 - 2y2 = 80. 30. 205x2 + 520y2 = 4264.
22. x2-y2 = 4. 27. y2 = 2x. 31. 4xy + 13 = 0.
Page 48
32. 24x2-15y2 = 200.
CHAPTER V
Page 65
14. bx + ay - ab = 0. 21. 2x + y — 1 = 0. 25. 12x + Sy — 2 = 0.
15. llx + 4y-18 = 0. 22. 6x-4y + l = 0. 26. 2x + 6y — 9 = 0.
16. 23x-14y + 26 = 0. 23. 5x + 4y + 9 = 0.
Page 66
27. 4x - 3 y = 0. 31. 36x + 24 y + 35 = 0. 35. tan-*12.
28. 4x + ly + 13 = 0. 32. 4x - lOy + 1 = 0. 36. tan-i J.
29. 4x + lOy + 1 = 0. 33. 2x — y - 6 = 0. 37. tan~i5.
30. 24 x - 21y + 109 = 0. 34. tan-1 J. 38. tan-*^.
40. (- 2, 4), tan-i \\ ; (- 3, - 5), tan-* \ ; (3, 1), tan-* 4.
41. 2x - 9y + 6 = 0, 7x - Qy + 21 = 0.
42. 7x + y-26 = 0, x— 7y-25 = 0.
Page 67
43. 8x + y + 9 = 0, 4x + 7y+ll=0. 51. V61, J^VST. 54. f^VlS.
44. y-l = 0, 4x + 3y-ll = 0. 52. 6x-2y = 0, 55. TVV34.
45. x- 3 = 0, 4x- 3 y -9 = 0. 4x + 6|/-ll = 0, 56. 2 V2.
' 49. fVlO, |fV34, 4V2. x-7y + ll = 0. 57.9$.
50. f^Vl7. 53. (2, -3). 58. f.
59. i^, -J.
Page 68
60. 5x-2y-4 = 0; 2\/29. 65. (±ffVl7, T A Vl7).
62. (24, TV).* 66. (4, -6),(2J, -6J).
63. (- 5, - 3 J), (- 2}, 1?). 67. (- 8, - 11), (10, 13);
64. (-11,-3). (- 11, - 16), (7, »).
Page 92 CHAPTER VI
3. 6x- 3^ + 4 = 0. 7. y2-10x + 25 = 0.
4. x+ 2/ + 8 = 0, 4x-42/-7 = 0. 8. x2- Sy2 - 4ty + 4 = 0.
5. 2x + 8^-5 = 0, 16x-4y-l = 0. 9. x2 + (y - 3)2 = ± x8.
6. 21x - 77 y + 96 = 0, 99x + 21y - 86 = 0.
Page 93
10. y2 + 4y - 2x + 11 = 0. 11. x2 = ay. 12. x* + x2y2 = ya.
13. 91x2 - 24xy + 84 y2 - 364x - 162y + 464 = 0.
14. x2 + |/2 - 6x + lOy + 18 = 0.
15. 6x2 + 52/2 + 8x-6y-16 = 0. 18. x2 + y* + 2x - 2y - 6 = 0
16. (- 1 ± 2 V3, 0). 19. x2 + y2 ± 2ox = 0.
17. x2 + y2 + 4x-3y = 0. 22. 2x + y-3 = 0; fV6.
ANSWERS
483
29. x2 + y2 + 3 x - 4y = 0.
80. x2 + V2 — 4x + 4y + 4 = 0,
x2 + y2 - 20x + 20y + 100 = a
Page 94
25. x2 + 2/2 — 2x — 2y = 0.
26. 2x2 + 2y2 + 3x-2 = 0.
27. 6x2 + 5y2±9y-80 = 0.
28. 7x2 + 72/2-19x + Hy-6 = 0.
31. 4x2 + 4y2 - 60x- 60yH- 225 = 0, 64 x2 + 64 y2 + 240x- 240y + 226 = 0.
32. x2 + y2 - 2x + 12 j' + 12 = 0, x2 + V2 - 16x - 2y + 40 = 0.
33. x2 + y2 + 26x + 16y - 32 = 0.
84. x2 + y2 - 6x - lOy + 9 = 0, x2 + y2 + 18x - 34 y + 81 = 0.
35. x2 + y2 - 8x - 12y + 48 = 0.
36. x2 + y2 + 4x — 2y — 20 = 0, x2 + y2 + 24x — 42 y — 40 = 0.
37. 13x2 + 13y2-156x-52y + 296 = 0, I3x2 + 13i/2- 52 x-104i/ + 269 = 0.
38. 6,3; J; (±4,0).
Page 95
39. *V5, iV2; J; (± jVS, 0).
40.(0, dziV2); JV2; (0, ± *).
41. (- 2, 1) ; (- 5, 1), (1, 1) ;
£V5; (_2±V5, l).
i^7; (fc-i±V7).
48. fr^x2 + a2y2 - 2a62x = 0.
44. ^x2 + a2y2 - 2 a% = 0.
45. |V386, jVl65.
46. x2 + 4 y2 - 4x + 24 y + 24 = 0.
47. 36x2 + 252/2-72x+50y+60 = 0.
48. 6x2 + 92/2 = 180.
49. 3x2 + 4y2 = 48.
50. 9x2 + &y2 + 16y - 64 = 0.
51. 9x2 + 25y2 = 225.
52. 4x2 + 32/2 = 108.
53. 9x2 + 25y2 = 226.
54. 13x2 + 9y2 - 26x - 104 = 0.
55. 36x2 + 20y2 = 1126.
62. JV2.
63. |V29; (±V29, 0) ;
2x± 6y = 0._
64. jVl3; (±Vl3,0);
2x±3y=0.
Page 96
56. 3x2+4i/2 = 27.
57. 9x2 + 82/2+36x-48i/-20 = 0.
58. 16x2 + 16y2 - 32x - QOy - 164 = 0.
59. 8x2 + 9y2 = 162.
60. £; 3x2 + 4y2 = 3a2.
61. 4x2 + 9y2 = 4a2 ; £ V5. _
65. (2, -3); jVl3; (2 ±Vl3, - 3) ; 3x - 2y -12 = 0, 3x + 2y = 0.
66. (-1,2); |VlO; (- 1, 2 ± J Vl05) ; y - 2 = ± J VS (x + 1).
67. 4x2 - 20y2 - 8x - SOy - 79 = 0.
68. 100x2 - 36 y2 + 400x + 216y + 301 = 0.
69. b*x2 - a2?/2 - 2 a^x = 0.
Page 97
70. 7x2-9y2 = 63.
71. 5y2-4x2 = 20.
72. 28 x2 - SQy2 - 56x + 144y - 291 = 0.
73. 9x2 - 16y2 - 64x + 32y - 79 = 0.
74. 3x2-y2 = 3a2. *
75. 153x2-4262/2 = 450.
81. 9x2-7y2 = 63; V7x± 16 = 0, 4x ± 7 = 0.
83. (£, - 2) ; y + 2 = 0; (2, - 2) ; x + 1 = 0.
76. 9x2-16y2 = 20.
77. x2 - y2 = 8.
79. ± tan-i^VB.
2-e2
80. cos-1
e2
^SWERS
484
90. Vs - *V . 8w _ 16
81,»+iov + ' 9 = 0.
96.X» + ^+V^
x+ v-»-
96. «**■
91. 20 ft-
0.
9». 27ifk.
PM**00
106. (-l<a)'; + ^ = 0,3*-^--
ta<2V\3 L^O-
Ka.*^, 2*_4„=0.
ioi. *a + va""2*'6«=:0'
109- SV7X- *« +
\\0. V* = - "
111. «"
116. V= ,„, .
u8. Two straigW lines-
l2a. Circle-
X2S. Circle-
184. Circle-
iHi. Concentric ^
115. V
^86. Straight line-
S. Straight toe-
188. Circle-
189. HyP61*01*'
Page 114 . 4P
12. x^-^'* *
IS. x
129. Straig^ toe.
132. Parabola.
133. Parabola.
140. Circle.
135. T*o parabola*
136. Parabola-
131. Bypert?la-
133. Parabola.
141. \tttcb.
144. BP^
CHA**** VI1
2ai* .
2asi
15. x^2asinH^==""~coi
2 at2
14. x = ^p
ANSWERS 485
16. x = ± a sin 0, y = a (1 3= sin 0) tan $.
17. x = (a + b) cos <f> — ft cos 0, y = (a + b) sin 0 — ft sin 0.
tt a
18. x = (6 — a) cos 0 4- ft cos 0, y = (b — a) sin <j> — ft sin 0.
19. Straight line.
20. x = a (1 — tan 0), y = a tan 0 (1 — tan 0) j x2 = a (x — y).
Page 115
21. x = a cos 0, y = a tan 0 ; x2 (a2 + y2) = a4.
22. x = a sec 0, y = a sin 0 ; x2 (a2 — y2) = a4.
28. x = a esc 0 ctn 0, y = a csc 0 ; y4 = a2 (x2 + y2).
24. x = a sin 0 (cos 0 + sec <p), y = a cos 0 (cos 0 + sec 0) ;
y(x24-y2) = a(x24-2y2).
25. x = a (1 — cos 0), y = a tan 0 (1 — cos 0).
26. x = a sec20, y = a sec20 tan0 ; x8 = a(x2 4- y2).
Page 116
Page 117
27. x = (a — c tan 0) sin20, 3/ = (a cos 0 — c sin 0) sin 0 ;
2/ (x2 + y2) = x (ay — ex) .
28. x = a sin 2 0, y = 2 a cos 0 ; y4 = 4 a2 (y2 — x2).
29. x = - (a2 + fc2 cos20j, y = -(a2 tan20 + fc2 sin0cos 0^ ;
a(x-a)(x2 + y2) = fc2x2.
30. x = J (a sin 0 + 6 cos 0), y = J (a cos 0 4- & sin 0).
«* o 0 CL / . 0\
31. x = a cos2 - 1 y = - J sin 0 4- tan - 1 .
2 2^ 2' 2a — x
32. x = 2 a cos 2 0, y = 2 a cos 2 0 tan 0 ; y2 = x2
2a4- x
33. x = a tan 0, y = a cos 2 <f> ; y (a2 4- x2) = a (a2 — x2).
35. 2 u2y cos2 a = v^ sin 2 a — ax2.
ft_ . 2 1> sin a v2 . _
36. t = : x = — sin 2 a.
37. x = - (t?2 sin a cos a + B cos a V2 aft 4- v2 sin2 a) :
9 '
t = - (v sin a 4- V2 aft 4- «2 sin2 a),
if
38. lsin-i^. 39.-.
2 v2 4
40. tan-i — (v2 + \V 4- 2t>2aft - a2&2).
CHAPTER VIII
Page 127
84. (aA0). „ (0,0) /±a^ A 39. (0,0), (± a, j),
35. f 1.035a, *V V X4 3' . \ , '
36. (8a, 0), (aV2, jV 38. (0,0), (± 2a, ^V ^ ^ 4" V ^' 4/
AC
486
ANSWERS
Page 128
40. Circle.
41. r = acos20.
42. r = 2acos80.
cos 29
43. r = a — -•
COS80
44. r2 sin 29 = 14.
45. r = 8 a(cos0 + sin0).
46. r = atan0.
Page 129
47. r^a2 cos 20.
48. r cos 6 =_a cos 2 0.
49. x = 4a/3.
50. x2 + y2 — ay = 0.
51. x2 (x2 + y2) = a?y*..
52. (x2 + y2f - 4 oW = 0.
53. (x2 + y2 + ox)2 - a2 (x2 + y*) = 0.
54. (x2 + y2 - ax)2 - 62(x2 + y2) = 0.
60000000 OCAAAAA/v .,
56. r = ; 25,000,000 miles.
1 — cos 6
57. 1.2 million miles, or 4.8 million miles.
59. Parabola.
61. Straight line.
CHAPTER IX
62. Circle.
63. Circle.
ce2 cos0
l-^cos2*
65. Confocal conic.
»■ -s-
12.
2Vx
Page 150
10. -~
x8
20. Increasing if x > — 3 ; decreasing if x < — 3.
21. Increasing if x < 0 or x > 2 ; decreasing if 0 < x < 2.
22. Increasing if x >— 1 ; decreasing if x <— 1.
23. Increasing if— l<x<0orx>l; decreasing ifx<— 1 or 0 < x <1.
24. Increasing if x < 2 or x > 3 ; decreasing if 2 < x < 3.
25. Increasing ifx<— lorx>3; decreasing if — 1 < x < 3.
26. Upward if t < 3 J ; downward if t > 3 J.
Page 151
27. 2 < t < 4.
28. Increase if x < 5 ; decrease if x > 5.
29. Increase if x <10 ; decrease if x >10.
a .„ a
30. Increase if x <
31. Increase if x <
vr
4a
decrease if x >
decrease if x >
Vs
4a
33. (0, 0), (3, - 27).
34. (- 2, 2), (1, - 26).
35. (- 2, - 3 J), (0, J),
(2, - 3'i).
36. (1J, 2fy).
37. 4x + y + 7 = 0.
38. 3x- 1/ + 18 = 0.
41. 64.
3 ' 3
32. s increases if x > § ; decreases if x < £ .
Page 152
«• <4§> 56§)- «. (1, - 1), (-i, -^).
43. (1£, 3). 49. (1, 6), (- J, 2^7).
44. tan-ij. 50. 7x-4y-17 = 0,
45. tan-i $. 189 x - 108 y - 209 = 0.
46. 8x - y + 12 = 0, 51. 5.8 ; 5.9 ; 6.
216x-27y- 176 = 0. 52. .100; .166.
47. x - 2 y + 9 = 0, 53. .423 ; .414.
27x-64y-7=0.
ANSWERS 487
Page 153
54. 4.411; 4.566. 57. 38f. 60. 108. 63. 2.6.
55. 6. 58. 36. 61. 6|. 65. £.
56. £. 59. 31 Jff. 62. 90,000; 677^. 66. 10f.
Page 181
CHAPTER X
1. 6x2+18x + 7. 4x(x + l)
X4*
2. 6(x + 2) (2x2 + 8x + 1). ^/(4x8 + 6x2 - 5)2
Q 2a 2x8 + x-l
o. — • 10.
(x-a)2 Vx* + x2-2x
- 16x %tt 21 x2
4. — 7777- 16. —
(x2 + 4)2 (x8 + 8)2
2x2(6-x) 6(x + 2)
(4-x)2 ' (x2 + 4x + l)2
6, 38(*+1> . 18. **
(3x2 + 6x+5)2 V(x2 + 1)*
x2 + 4x + l 19. 3(5x-3)(3x-l)(x-l)2.
(x2 + x + l)2' 20. 3(2x2-l)(4x8-10x2 + 2x + l).
8V xi *t xi/ ^+^
22. ^+2
(x2 + 1)*
9. (3x»-2)(l + |)
,A Vx + 1 „„ (6x*-7x»+6x2+3x-3)(x2-2x + 8)i
10. • 23. «
4xvx (x8 + l)J
5^x\ «/ 5\/?\ x/ 2VVl + x VT^x/
12. 18x(x + l)(2x8 + 3x2 + 6)2. 26 1 *
13. 12x2(x8-l)8. Vx2-1
Page 182
26. 1—=. 31. _*_±^!±Z. 36. -**; - *.
(l-x)Vx2-l a2Va^f^ y ^
27. - oo 2(x + y) 37 -^--i^8
(x + i)Vx2-i ' 2x + 3y2 y*' y*
2x2 + a2 2x oo 2y4-4x8y-x* i i
28. -. 33. _z * vt at
a2 Va2 + x2 a2 y* - 8 xy» + ** 38. - — ;
29.
34.-^+*.
xi 2x4
(a2-x2)* « + 22/ ,ft 2x + 2/.n
a2 „ Vx + y+Vg-y x + 2y
30. . 35. 9
x2 Va2 + x2 Vx + y — Vx — y
488
ANSWERS
40 3«3 . 2a(4o-8y)
' Sy2-2ay' (3y-2a)*
41 I-?2 . 2(y*-l)(4x + y)
' 2xy-l' (2xy-l)8
42. x-ly + 5a = 0;
7x + y — 15a = 0.
48. 31x + 82/ + 9a = 0;
8x-Sly + 42a = 0.
44. (- 2, - 8).
45. 3x + 6y ±16 = 0;
5x — Sy ±4= 0.
46. (± 1_, ± 9).
47. 8V2.
48. 2 |/Ty = 5 xx4x — 3 xx6.
49. xx~^x + yx~^y = a^".
50. xx~*x + y{~*y = a*m
51. xfx + yfy — ayxx — axxy = ax,yr
Page 183
56. tan-1 3, tan-if.
57. tan-1 2.
58. -, tan-i^.
tr
59. — , tan-1 7.
2
Page 184
60. -, tan-i2.
2
61. tan-iS.
62. 0.
63. 0, tan-1^.
64. tan-1 J.
65. 0, tan-1 J.
66. tan-1 3.
67. tan-1 J.
68. tan-iVi,
5V6
tan-1
24
\ 2 2 / V VaM^i2 Va2 + &V
Page 185
85. a?y1x — b2x1y = 0;
a2**2
V&*x* + a4y2
90. Upward if x > J ; downward if x < J.
91. Upward ifx< — V2 or x > V2 ; downward if — V2 < x < V2.
92. (-1J, 11^).
93. (1, - 3), (- J, 3^).
94. (0, 6al).
95. (±^V3,Ja).
96. (0, 0).
97. (0, 0).
99. (2, 0)t (- |f 9jf); (§, 4|$).
100. (- 1, 0), (3, - 32); (1, - 16).
101. (J, - rfft); a, 0), (J, - A).
102. (3,-11); (0, 16), (2, 0).
2 2^2
103. ±— , T
V3 a/3
•); (0,0),
(± 2, 0).
Page 186
104. Increase if x > V2a ; decrease if x < V2a (a = given area).
105. Increase if x < — - ; decrease if x > — — (h = hypotenuse).
V2 V2
106. Increase if x < - ; decrease if x > - .
2 2
107. Increase if x < jVp; decrease if x > ^p (p = perimeter).
ANSWERS
489
108. Length is twice breadth.
109. 12 rd., 18 rd.
110. 5 ft.
Page 187
114. 4 portions 1 ft. long ;
2 portions 4 ft. long.
115. Breadth = — V§,
3
depth = — VS.
116. Breadth = depth.
117. 13^ ft. long.
118. (2J, 0).
111. Side of base = 20ft., depths 10 ft.
112. Depth = one half side of base.
113
119.
(^•H
120. Height = twice radius of base.
121.
V2
122. Altitude = ? V2 :
P 4
base = - (p = perimeter).
4
Page 188
123. Altitude = j radius of sphere.
124. ;
3* 3f
126. 200 cu. in. ; 2547 cu. in.
127. Height of rectangle = radius
of semicircle ;
semicircle of radius — •
Page 189
132. a -
bm
Vn2
bn
miles on land ;
m*
miles in water.
Vn2 - m2
133. ljj hr.
134. 7hr.
135. Area of ellipse = - area of
rectangle.
ir
128. -A in.
V§
129. -VS.
3
130. 8 mi. from point on bank near-
est to A.
131. He travels 8 J mi. on land.
136. Altitude = J V2 radius of semi-
circle.
137. Altitude = § altitude of seg-
ment.
3a
138. Vl50 mi. per hour.
139. Velocity in still water —mi.
per hour.
140. 144 7r cu. ft. per hour.
141. (1,3), (5, -5).
Page 190
142. 2 V7 ft. per second.
143. 4 ir times distance from vertex.
144. .1 in. per second.
145. .02 in. per second.
146. .06 cm. per second.
147. 34.9 sq. in. per second.
148. Forward if t < 1 or t > 5 ;
backward if 1 < t < 5.
149. v max. when t = .85 ;
backward if 2 < t < 4.
150. - v0 (y = distance of top of lad-
der from ground, x — dis-
tance of bottom of ladder
from wall).
490
ANSWERS
Page 191
151. b cos <f> (<t> is the angle between
the wire on which the bead
slides and the straight line
drawn from the bead to the
fixed point).
152. 281 £ ft. per minute.
153. 150 ft. per second.
164 v- 8 .2V(t2 + l)' + 4<g
•" X2+4> (t2 + l)2
155. When x = £.
156. 20 ft. per second ;
10 V5 ft. per second ; (100, 20).
157. .22 ft. per second.
1M _„. 0 /144-2*2
158. Ellipse: 2-v n —
t^) \ 64 -«2
Page 212
CHAPTER XI
1. 2 sin8 2 x cos 2 x. 11. xctn(x2 + a2)csc2(x2 + a2).
2. 3 sin4 3 x cos8 3 x.
3. sin8 ax cos8 ax.
4. sin22x.
5. cos2(l-2x).
6. cos* 3 x sin8 3 x.
m sin8 2 x
7.
cost 2 x
8. sin8(2x + l).
9. sec2x(l + tanx)2.
10. tan4--
2
12. ctn6-.
3
13. sec6 -tan-.
2 2
14. seci -tan5-.
5 5
15. esc bx (esc bx — ctn bx) .
16. 2
17.
Vl-4x2
1
18.
19.
20.
21. -
V4x-x2
2
■ •
(x + 2) V2x
x
Vs-ex2-^
l
V3x — x2
Vx-x2
Page 213
22. -
23. -
24.
25.
26.
27.
28.
29.
V8-6x-9x2
2a
x2 + a2
1
x2— 4x+ 5
1
(x + 1) Vx2 + 2x
1
■ - ■■■■ •
Va2 - x2
2a2x
x4 + a4'
2
x2 + l
1
30.
31.
32.
xV9x2-l
2
(x + 2) Vx2 + 4x
x2-l
33. -
34.
(x2 + 1) Vx4 + X2 + 1
1
(x2 + x)V4x2 + 4x-l
35. -
1 + x2
4a2x
x4* a4
2 Vx - x2
36. sin-iVl-x2.
37. 2xtan~1-.
x
38.
39.
40.
41.
42.
43.
44.
45.
V2 ax — x2
4(x-l)
2x2-4x + 3
x + 2
x2 + 4x + 3
1
4x2-9
1
3x2-l
2 — x
3-4X + X2
3
V9xa + 2
X2
Vx6 — a6
ANSWERS 491
46. _». 50. A R6, ■ 53. «£+**.
xVa2 + x2 4 + 6sin2x Vx* - a*
47. ctnx. 51. 2csc2x(ctn2x — 1). 54. csc82x.
48. 3 sec 3 x. eo x8 55. (log ax)2.
52.
2 /x2-l
x \x2 + l
49. -sec?. ^ + 4)2 56
2
Page 214.
57. 2tan-i2x. 65. iacoo2ar.gin2x.loga. 4
3x + 2 ««• <*" + e-*)8. e2x+e-2**
68* r+2rf" 67' (f)b + c*- 75. 0.
59. tan- 1 ax. 68. x2e«*. 76. e*ctn-i(e*- 1).
60. sec-iax. 69* *ja + a > ' • 77 (» + l)log(x + l)
«l.Iri. 70. *±*. ' (x2 + 2x)l
x2 J 78 6x8 + 4x2-24x
^i 71.*~sinmx. ^~16
63. f^. 72. a^SSf . 79. - "^^
pain- 1x
64. *
1+ a2x x
78. aXl0«a . n V^
80.
2 v 1 — x2 V4ax — a2x ' x2 — a2 *
81. - — ?— - log(x + a + Vx2 + 2 ax),
(x + a)2 e v ■
82. V^^^ + (*-*«)log(»-1..
"v 2 ^ — c^sc
83. yx*(l + logx).
84. yx* - + logx + (logx)2]-
■ 85. 2/ea?/-+ logx).
86. — — (cos Vx • log tan Vx + sec Vx).
2Vx
Qrf y (x sec2x — 2 tan x • log tan x)
x8 tan x
88. —1—S% x tan-i - + a log (x2 + a2)].
x2 + a2L a J
Page 215
89 2/(l-x2-y2) 92# xylogy-y2
' x(l + x2 + y2) ' xylogx - x2
y (x — l)
90. -^ ^. y(ysecxy-logy)
2/ — X W3. :
o • / , o x / , o ^ x (1 — y sec xy)
fl 2sm(x + 2?/) — cos(x + 2y) v '
' 2cos(x + 2y)— sin(x + 2y)' 94. — &-x; e2*-*.
492 ANSWERS
Q_ x + y 2(x2 + V2) 104. tan-13, tan-U.
Wt J — •
x — y (x — y)»
2x + y. 10(x2 + y2)
105.
(••i)
■«• (* £• a
' x — 2 y ' (x — 2 y)9
_tanx tan2 2/ sec2 x + tan2 x sec2 y *W V^ VsT -\A
tan y' tan8y ,„ ^
as. ^ . »■(»-»«>. 107-(-2'p)-
x(y-x) x(y-x)8
" 102. tan-i2 V2. 108. -,0); 4.
103. tan-ij. ^4 '
Page 216 '
110. Turning points when x = ± \ V2 ;
points of inflection when x = 0 or ± £ V5.
111. Turning points when x = 0 or 2 ; points of inflection when x = 2 ± V2.
112. Turning point when x = 3 ; points of inflection when x = 0 or 3 ± Vs.
IT
113. Turning points when x = far or (2 k + 1) - ;
IT
points of inflection when x = (2 k + 1) - .
4
114. Turning points when x = cos-1 ;
2 2tt
points of inflection when x = far or 2 kir ± — •
115. Turning point when x = ae ; point of inflection when x = ae$.
116.12 ft. J_ 119. At an angle tan-1 IT 120. 10 in.
117.70°. U8* V2 with the ground. 121. 6 V6 ft.
Page 217
122. 24 sq. ft. ; 14.71 sq. ft. per second. 124. 2V(6- s)(s - 3) ; 4(4 - s).
123. Greatest distance = 6 ; 125. a.
force = — (7 — 2 s). 128. tr miles per minute.
Page 2l8
(ft2 sin 0 cos 9 \
b sin 0 H — ) times angular velocity of A B, where 0= (MB.
Va2 - ft2 sin2 0/
131. Circle.
132. k Va2 sin2 kt + b2 cos2 fc£ ; maximum at end of minor axis; minimum at
end of major axis.
133. 2V2.
134. 2 aw sin - ; w Va2 — 2 aA cos <f> + A2t where w is the constant angular
velocity.
, ft„ 2 aw (a + b) . 0 . (a + &) « /~s — ^— r ,, , ,Q
137. 1 — L__Z Sin - ; ^ — L—!— Va2 — 2 ah cos 0 + A2.
b 2 b
ANSWERS
493
188. aOio, where a is radius of circle, a& is distance through which point of
string in contact with circle has moved along circle, and w is the
constant angular velocity.
ft. per second, where x is distance of man from center
189.
140.
141.
VlOOOO - x2
of diameter.
600 sin a
VlOOOO - x2 sin2 a
center of diameter.
2 a2
ft. per second, where x is distance of man from
a2 + x2
diameter.
times man's rate, where x is distance of man from center of
Page 219
142. Circle;
ab
feet per
148. tan0;
2V&2-a2*2
second, where at is distance
of foot of ladder from side
of house.
sec8^
144. — tan0; — sec40csc0.
8a
145.
cost — sin t
a4
Page 220
155. 4 tto, 4 wa V2 ; 0, 8 ira.
156. w Va2 + 2o6cos0 + 62, where
<a = rate of 9.
cost + sin* '
152. tan-i §.
153. 0, - .
' 2
158. S(axy)i.
e*(cos£ + sinQ8
159.
VA
157. 2, a/3 ; §, 1.
a
161. 5J.
162. 2 a.
163. i VlO.
164. jV2.
7T
165. Greatest when x = (2 k + 1) - ; least when x = far.
Page 221
a;
168. 8 a sin 0 cos 0. 169. — ; — • 170.
b a
\* i*n a(5-4cos*)i
2a2
9-6cos0 3r * '
Page 253
22. 2 Va2 - x2.
Page 254
28. x-log(e* + l).
CHAPTER XII
24. log [log (x + Vx2 - a2)].
26. Jsin*(2x + 1).
27. — (tan ax + sec ax) .
a
28. - £ (ctn x + l)8.
29. ^(3cos62x-5cos82x).
30. sin* 3 x (J - fT sin2 3 x
+ -j^sin'Sz).
31. - cos8 2 cos - .
3 2 2
32. -
3 + sin24x
6 Vsin 4 x
33. 2 sin x — log (sec x + tan x) ,
494
ANSWERS
34. §sin8x.
35. - (tan ax — ctn ax).
a
36. tan 2 x + sec 2 x — x.
37. __ | ctn 3 x — £ esc 3 x — x.
(x x\
tan- + ctn-V
2 £/
39. £sin(2x - 1) - J sin8(2x - 1)
+ TVsin6(2x-l).
40. 8(sin^-cos|j+2(sin8|— cos8-V
41. - tan2 - + 3 log cos - •
2 3 3
42. Jtan(3x + 2) + £tan8(3x + 2).
J. 2 88iC
43. - sec8 — •
9 2
44. ^Vtan2x(7tan2x + 3tan82x).
Page 255
55. |x + Jsinx+ ^sin2x.
56. fax — 2^sin4x— ¥1gsin82x.
57. - § V(cos2x + l)8.
58. log (sec x + tanx).
59. cosx + sinx.
1 fsin (a — 6) x sin (a + b) x
60.
[sin (a — 6)x__
a — b
a + &
]
1 [~sin (a + b)x sin (a — 6)x"|
' 2|_ a + 6 a — 6 J
62. £xsin6— ^cos4x.
63. ^cosOx— T^cos4x — Jcos2x.
64. ^(tan2x + sec2x).
65. — log(l + cosx).
x
66. — 2 ctn x.
2
67. x + ctnx — cscx.
3 1
63. -x H (8 sin 2 ax + sin 4 ax).
8 32al '
69. -2V2COS-.
2
70. — — log (sec x + tan x) .
V2
V2 .
71. — (cos8 2 x — 3 cos 2 x).
x 8 x 4 x
45. — 4ctn — -ctn8- ctn6-.
4 3 4 5 4
46. logsin(x + 2) + £ctn2(x + 2)
— Jctn4(x + 2).
47. - 1 ctn8 — /s + 3 ctn2 — ) .
10 3 \ 3 /
48. — tan8 ax tan ox + x.
3a a
49. (esc ax + sin ax).
a_
50. -\ /sec -(sec2 — 7).
7\ 3\ 3 /
61- ix — T^sin(6x + 2).
52. Jx-1^sin(4-6x).
53. x + J cos 4 x.
54. Jx — ^sinl2x.
„_ 1 . . o X
72. -sin-1 —
3 5
V§, ,2x
tan-1— —
73. _Ll!tan-i-
6
74. -sec-1 —
Q Q
Vs"
»k *♦ i2x + l
75. -tan-1 —
6 3
76. 2
,4x + 5
— tan-1 •
V7 V7
. ,4x — 5
sin-1 — «
x — 2
77. sin-1 — — •
79. sin-1^!9.
80.
80. sec-1 (x + 1).
81. log Vx2+9 + I tan-1 ? .
3 3
82. 5log(3x2+2)--^=tan-1-^.
6 * ' V6 Ve
83. — Vl — x2 — 2sin-1x.
84. asin-lX--Va2-x2.
ism-1-
a
85. — tan-1
85. — tan - 1 (cos x).
86. Jlog(2x+V4x2+9).
ANSWERS 495
Page 256
87. £log(3x + V9x2-2). J_ 2x + 3-V5
88. ^glog3^^. ' ^ °g2x + 3 + V^
12 3x+^2 92. -i,log(3x-l+V9x2-6x+9)
89. Jlog(4x + 3 + 2 V4x2 + 6x). V5
90. llog?-^. 93. J-log*-^.
6 s x 10 Bx-3
94. J-log(2x + 2 + V4x2 + 8x - 14).
V2
95. ^log(x2 + 6x + 12)-6V3tan-i^tJ.
96. !log(x2 + x-6) + — log^l?.
ft* *i /o q , k . iv . 15V17. 4x + 5-Vl7
97. -log(2x2 + 5x + l) + log — .
2 34 4x + 5+Vl7
98. -log(6x2 + 7x - 3) + -log8g~ *.
2 8V '2 *2x + 3
99. ilog(2x2 + 6x + 9) - - tan-*2x + 3.
4 &v ' 6 3
100. -log(3x2 + 2x + 3)-^— ^tan-i8x + 1.
6 * 12 2V2
101. 3 sin-i ZZL*L _ V3 + 2 x - x2.
102. lV8-4x-4x2 + 7sin-i^-±i.
2 3
103. V4 x - xa + sin-i ?— - . 4 tan? + 3
2 2 2
104. tan-i (x + tan a). 110' ^= tan * -^
105. sec a tan-1 (e* sec a + tan a). ■. tan x
106. sin-i (e* cos a + sin a). m« — 7= tan_1 -7=- •
/ , — v ' 4V2 V2
107. og^+V^D + sec-x. a ten2;c_3_2^
108. -log(x2+Vx7Z^)--8ec-i-. na- ^^to^.g + aVS'
3tan?-l tan- + 2+V6
109. 1 log ? 113. -L log 2
3 3tan- + l v* tan? + 2-V5
2 2
Page 257
ion e6 + cara6+«B 123. log(e* — e-*).
c(l+loga)' 124. ^(x2 + 2x + 6)V2x-3.
121. 21og(^ + l)-x. 125. x + 2 + 4 Vx + 2 + 41og(Vx + 2 - l)
122. llog(e2*-2)-*. 126' *<*-«^^E+*-
4 *V ' 2 127. i (x» - 9) \^T8.
184.^(4*-l)(Sz + l)i
... z*-9s*-81a;-81 , „„,
V4-3« 2
1. V4~+8x*-2
2l0g z
Vxa-26-6sec->-.
5
V9x*-4 ,9 _,8x
ANSWERS
141. 2 log (2 3 + 741""
V4za-8
-»)
m x*-8*«-18
6V3 + 2a!»
1M.- + 18.
Vz* + 9
2(1 + 3)
g_l,„g,3 + 4,.).
144 ^z^a*-2z*)Va1-x*
im. A('«-6)(. + i)i
* + l
Vz* + 4 - 2
15 J
V3-V8-X2
-log-
148. J(x* + 4)Vx4 + l
TTT-i
s -vV + 1 + 1
149. ^i(21i!«-l)Vl + 4zS.
160. — u
a 2(a"+x>>
3(6 -x">*
■ _, 2x-3
(z-2)V6
164. iain
-2x
x + 2
VS V2z2 + 4x + 4-x
*(10g(HS-l).
[(» + l)10g*-l].
(m + 1)'
x tan-i ax - _L leg (i + a^). m ^ + ^
slogOz + Vx^ + a^-vV+a3.
161. a5Bec-12E-Jlog(2*+V4ia-l).
162. - sec-' 3 1 - 1 VBz*-l.
2 18
163. s>Te'^(9a*-6i + 2),
164. i[(2x2-l)Bm2H-2xcos2z].
165. x[(logx)2-21ogx + 2].
166. — - — sm0x- — cos6s.
4 12 72
167. ^(•■(Saina-eaM).
168. ^^(SsinSs + cosSx).
169. 4[xV^3T-log(z + V^ri)].
170. Jt^ictanx + logfsecx + tan*)].
loS;.,
ANSWERS 497
X 1 , X , „m y X2
17Q inir — I- -tan-1-- 177.log -
g Vx2~+4 2 2 &x + 2
, Vx2 + 1 ,„„ , Vx— 1 1,. .
174. x - tan-ix + log z — 178. log — - ten-ix.
« V x2 + 1 2
175. iog^-?. imhJ,v«-a;y+4>.
176-i^+1°g(X+1)- . 180.1^t^ + 3,og(2,+ l).
v ' 16x +8 4
Page 259
V(x + 1) (x - 3)8 1 1 , Vx8+2-V2
181. log V ^ Mft '— 193. — 7= + -^=log /— — -f'
6 x-2 3Vx*+2 6V2 Vx»+2+V2
182. log(x - 2) (x2 + 2 x + 5)i . x
1x4-1 ^ "^ 8m ~
+ itan"1-|- "* log 5-?sin|(3 + 8in*|Y
1-sin^
*r~x 4
1 1+ cos4x cos4x
101. X-Sx \\qg( X V 198*i610gl-co8 4x 88in24x*
x(x — 1) \x — 1/ ,r r-
v ' 196. j[xVx2 + a2
185. ing Vg-71 - a2log (x + Vx2 + a2)].
</x2 + x + l x
1 ,2x + l 197. •
-tan-1 — — . vT+x*
V3 V3
1M x2 , . Vx2-2x + 4 198. -^1 + g3.
186' 2+1°g— 7^
^2X1 1 1 199. lx(2x2 + a2) Vx2 + a2
187. log— =+-tan-ix-— •
Vx-1 2 * L --log(x + V^ + a2).
x 1 x ®
188. 1 tan-1 - • 4,
8(x2 + 4) 16 2 goo, ,^1 + x^
189. 17X<x + log Vx2 + 1 X
2 (x2 + 1) 201. Jx (x2 + 10) Vx2 + 4
+ i tan_1 *' +6 log (x + Vx^Ti).
19°' ^l0g^T^"-2VIT"X- 202.^^+?log(secx+tanx)
^1+x Vi6 2smxcos2x 2
191. x — 1+4 Vl — x coga. 1
' • / ./; \ 203. 1- - log (esc x — ctnx).
— 21og(Vl-x-x) *wo' 2sin2x 2 v
+ — log2 +1~ _. 204. log(secx + tanx) - cscx.
V6 2VT^+14.V5 sina;(2 4.3co82x)
192, sinx -hog1"8111*. ' 8cos4aj
2 cos2x 4 1+sinx + flog (sec x + tan x).
498 ANSWERS
CHAPTER XIII
Page 285
h _*
1. 2. 2. 2. 3. £a2. 4. a2(e«-e «). 5. 4 to*.
Page 286
7. »- StogS. "• -^V^^+aalog - ^ 21 i(,^2)a2.
8. §(ei + 2). 16. 4a2. . 227ra2
10. $(37r-2)a2. 17. ¥ V5. "•• »™
U. 13i- 8 /&3^ **• 3™2'
12. Ja6. 18- l6(6""a) \~7~' M' !*«*•
13. 2tto6. 19i ^
Page 287
26. f wad. 29. .6366. 32. Jwr8. tto2
27. jTra. 30. £n2. 34. §. ' In"
28.^. ».!* M-^ai- 89.1s.
7r 7T0 37. 2 a2. n
Page 288
40. J(4-7r)a2. 41. §7ra2. 42. 11 *r. 43. 16 iH^a2. 44. — V2 .
46. a2[V2-log(V2 + l)].
46. - ( 3 tan — + tan8 —J c2, where 2 c is the length of the chord through the
focus perpendicular to the axis of the parabola.
47. |7ra2. 49. £(3V3-7r)a2. 51. Jira2. 53. §a2.
48. 21£. 60. ^(Sw + 9Vs)a2. 52. 27r(a2+62). 54. §&a2.
Page 289
66. §a»tan«. 57. fa'Vs. «*<»*+ !) + *«.
66. 2 ***-/?,»., where „ irft2 .„ _. 4
Pl and p2 are the 68" T (3 ° " A)" 6a- § «* (1 - cos a).
respective param- 59, 32 vas. ag &^L_?.
eters of the pa- «o 38* ' ^^
rabolas. **"' 64. § 7ra*(81og2 - 2).
65. ira8tan0.
Page 290
66. T3<fra8. 69. 4 A*. 72. 2ir2aM.
67. 4 ?ra8 (2 log 2-1). 70. £ M2 (3 *r - 4). 73. 1152 cu. in.
68. 2 Tra8 (4 - ir) . 71. J Aa2. 74. 22^ *r.
ANSWERS 499
Page 291
75. tyirV^. 76. f ira%. 77. 2ir2a8. 78. o^a*. 79. § *ra».
80. §(2 + 7r)a8, where a is the radius of the sphere.
81. JV (10 VlO - 1). 83. 6 a. g6t « (log 9 _ 1}
82. |(e^-e~«). 85. fira. 87. 8a.
Page 292
88. %a. 89. 177.5 in. 90. log(e + e-i). 91. — .
92. -[27rV4ir2+l + log(27r + V47r2+l)].
93. 4aV2. 95. 8 a. &?• jf wa. 99. 2iraA.
94.100. 96. fa. 98. a log 5. JJ?' X ™!"
6 ft 101. J^-Tra2.
Page 293
_ « 2A 2ft A A
102. 2*raft + — (ea—e *— 4e° + 4e a).
108. ^ tto2. 106. 4 ^2 (2 _ V2) . 108 »
104. 4>*7ra2. o ,- a'
105. 2 *a*. 107.4^a2V2. 109. 9|ft.-lb.
110. ^f-kV&a1, where k is the constant ratio.
771
111. — (2 ar — a2), where r is the radius of the earth.
2rv "
www
112. , where r is the radius of the earth.
r + a
Page 294
113. 2ttO. 115. l.llft.-lb.; 116. 1056 w. 118. 53Ju>.
114. 139.5ft.-lb. 97ft.-lb. 117. 12^. 120. iq8w.
Page 295
121. 7£. 126. § wad2, where 2 a is 128. J- jp. (8 tt + 9 VS) w.
122. 3041 § lb. the length of the 129. 54 w,
123. 10,760ft.-lb. axis in the surface 139, 21.
124. 1250 tons. of the liquid. 131, 12.27 tons.
125. 10,416§ f t.-tons. 127. &wb.
Page 296
182. 4.2 tons. 136. On the axis of the segment, §
183. J ttw. of the distance from the ver-
184 /o ^ a\ tex to ***e ^ase*
\ 5 / 138. Intersection of the medians.
500
ANSWERS
189. (^, o),
where x = h is the
equation of the ordinate.
140. (0, |).
141. On the radius perpendicular to
the base of the hemisphere, £
of the distance from the base.
142. y = A*>.
Page 297
147. (**-
\e + l
148. x = — •
3
a(e*+4e2-l)
4e(e2-l)
149. (0, 9).
160. (fr, 3).
151. £ = £&.
152. On the axis, § of the distance
from the base to the vertex.
Page 298
159.
5a
;)•
2(4tt-3V3)>
161. On the axis of the solid, T5^ of
the distance from the base to
the highest point.
/a .6 , a2 \
. ( — » — I I •
\2 2 12 p)
(la aVs\
'\T*~~r/'
162
168
Page 299
M
171.
69
172. -^-(log — — )
I2 \ a I + a/
173.
Jf
sVel
174. — — [ = i» where M
I2 V Vc2+J2/
is the muss of the wire.
175. — , where p is the mass of a
c
unit length of the wire.
148. Middle point of the axis.
144. On the axis of the cone, § of
the distance from the vertex
to the base.
• (f • i>
266 a 256 a
/4 a
• ( —
\8jt
• (o,
)
315 ir 315tt
4(a + 6)
IT
3tt
1tc ,_ 352
155. ( 0,
5(2 + 3w)
156. (?ra, fa).
157. (ira, |a).
«»« - ^
158. x = -.
2
)
)
164. On the axis of the lamina, 3£ ft.
from the vertex.
165. On the diameter of the sphere
perpendicular to the planes,
and half way between them.
166. The middle point of the axis of
the solid.
167. x = ^(V2 + l).
176.
177.
25 ir'
Mc
(c2 + a2)*
178.^/1 J—J
a2 \c T/c* + a*/
23f
179. ^L (i+ ^/c*+a*- V(c+ i)2+a2).
a?l
ton *"
180. —
60
ANSWERS
501
CHAPTER XIV
Page 327
24. 9z2 + 9(y - l)2 = 4(x - l)2 ; (1, 1, 0).
Page 328
32. (-2,-1,5).
33. (3, - 2, 1), (- J, |, J£).
35. x2+y2+z2-2x+4y-2z = 43.
Page 329
, Vib
TXt
7
• m
46.
x — Sy + 7z + 39 = 0.
47.
« + y + z
-10 =
= 0.
48.
X — 2 + 1 :
= 0.
49.
X + # + Z
-4 =
0.
50.
7T
- ■ •
3
51.
cos_1T45.
Page
330
59.
17
26
11
V1086 V1086
V1086
60.
x — 1 _y
-3_
2 + 5
2 6
62. sin-i|^.
64. x+y+z+3± 3 V3 = 0.
65. 4y + 9 = 0 and 2x + 6z — 1 = 0.
66. (3, - 1, 2).
Page 331
75. (3, 0,-3).
78. x — z = 0.
80. 11 x — 2y- 5z — 25 = 0.
81. 19x + 16y + 27z-70 = 0.
Page 332
85. 5 x — y — 2 — 5 = 0.
86. (- 1, 2, - 1), (2, - 3, - 4).
87. 3x + y+ 3z-l = 0,
x-6y + z-2 = 0.
88. 5 J.
90. (1, |, 3).
91. 13x + lly-17z-15 = 0.
AC
An V2 2V2 V2
6 3 6
41. 2V3:V2:V2.
4
42. cos-1
VlU
53. z — 6 = 0, 4x — y — 5 = 0.
y_3 z-6
54. x =
55. x — 4 =
3 5
y-6 z + 2
-5
v + 2
56. x - 2 = ^_Lf = z-.2.
V2
*»• ft. A. H-
67. 7x — 4y + 2z - 22 = 0.
68. x — 6y + 6z ± 4 V62 = 0.
69. V3.
70. (0, 1, 2), (- §, J, |).
71. (1, - 3, 4).
72. ftV22.
73. 3x + 5y-z-20s=0.
74. 7:4:-5.
82. x + 2y + z — 2 = 0.
83. 13x-y-12z-32 = 0.
84. 3x + 8y + 21z- 66 = 0.
92. (0, 1, 2), (Aft?, W, - V*>-
95. y2 + 4y — 6x = 0.
96. x2 - z2 = 0.
97. 2x2 + z2-6x = 0.
98. 5y2-2z2 + 8z = 2.
100. y = x2 ; z = x* ; za = y8.
502
ANSWERS
Page 333
102. xy = 1.
110. cos-1
Ve
108. cos-1
Ve
111. J(12 +5 log 6).
112. e - e-i.
118. 16.
Page 334
114. 2.
115. ?^l = V^ = z-l; 2x + 2y + «-7 = 0.
2 2
116. x-l = ^i = -7^ x-y + V2« = 0.
*-l V2
117. ?L=i = y = ^^? ; 4x + y + 9* - 89 = 0.
4 9
118. 4x + 2iry-7r2 = 0, y-« = 0; irx - 2y - 2z + 2w = 0.
119. x-l = ^-^ = ^-^; x + 8y + 42!-8 = 0.
3 4
20. 2e*sin(x — y).
27. .0004.
28. -.0302; .0002.
29. 11.080 cu. ft. ; 10.996 cu. ft.
CHAPTER XV
Page 361
i ^8 X8
1 -? , 1 x _* , 1
6. — l-e v + -); — e * + -
\y x/ y2 y
" (x2 + y2)* ' (x2 + y2)4
2 * • x
15 l • X
x2 + y2 ' x2 + y2
"" 1+y' (1 + y)2
* X
16 V V
x V2 xy — y2 V2 xy — y2
' x2 + y2 ' x2 + y2
4
17 X V
Vy2-x2(y+Vy2-x2)
x2 - y2 ' x Vx2 - y2
1
1 1 /«
18. : -\ /-•
2y' 2 \y
Vy2 - x2
y2 *y
5. — -cos — ~ ;
(x — y)2 x — y
x2 xy
-cos — — • •
(x — y)2 x — y
19. »t«-i! a2aV
x x2 + y*
Page 362
•
80. .646 sq. in. ; .660 sq. in.
81. .49 ft.
82. .0218 ft.
ANSWERS 503
Page
363
33.
.0048.
34.
1
Vx2 + y2
;0,
85.
0; 0.
36.
1-V5
2
Page
364
41.
k
V
42.
x + y
X
48.
y Vx2 —
■v2
X
xsm-
iV.
37. 4x + 3y-ll = 0.
., xf + yj2 in direction tan- 1 -± .
39. --COS0- — sintf; 1.
2 2
3x22 . 2(1 -2y2)
' 2-222' y(222-2)'
2x2 + 3x2y8 2y2 + 3xV*
322 + x2 + y2; 3z2 + x2 + y2'
47 9yz~2x(x2 + y2 + 2!2)a
' 22(x2 + y2 + 2;2)2_9X2/»
x 9xz-2y(x2 + y2 + z2)2
^ (y + 2)(2x + y + 2) . 2z(x2 + y2 + z2)2-9xy*
(x + y)(z + y + 22)' 48. 2x + 3y+ 22-9 = 0;
(2 + x)(x + 2y + 2) x— 2 y — 1 _2 — 1
(x + y)(x + y+2z)' 2 ~ 3 "~ 2
49. x + 2y + 2-2 = 0;x-l = - = 2-l.
50. 2(ax1 + ty1)(os + ty) = 2 + 21; ZZih = VjzVl = ?_Il?i.
v it yV\ tv) t i, 2a(ax1 + 6y1) 26(ax1 + 6y1) -1
54. x + y + z — 3 = 0, x — 2y + 1 = 0; 2x + y — 32 = 0.
55. x — 1 = 0, 2y + 22- ?r=0; 2y-22 + ?r = 0.
56. 4x + 3y-25 = 0, 3y- 42 -25 = 0; 3x - 4y - 32- 12 = 0.
Page 365
57. 4x - 4y + 2 - 24 = 0, 65. (- J§, - J|, \%).
2x + 8y + 52 = 0; , j 6v/2v
14x + 9y- 202 + 79 = 0. 66. (1, --, ± — — j.
58. sin-ifc /^Z==» 67. (- 1, - 1, 1).
rVa2 + &2
59. * . 68.
Sabh
2' 27
60.6. 69. 2x + 2y + 2-6 = 0.
61. 10^2, 10^2, 5^2. 2aK
r. r. „ 70. X =
at% o, o, o>
3 3 8
63. Mean point of the vertices. y ~~
Sabc
©*• — -=. • z =
3V3 a2 + &2 + c2
a2
+ 62 + c2
2 6JBT
a2
+ 62 + c2
2cJT
504 ANSWERS
Page 366
71. x6 - x*y + xay2 + y5. 76. Vx2 + y* - y.
72. xy + logxy. 77. £(x + sin x cos x) + y cosx.
73. logx-—- — -. 78> _eTCos(x+y).
x8 79. 2J ; 2f .
74,3? + X' 80. I; i(4-ir).
_* 81. If; 1J.
75. _e y-logy. 82t i + log2; £(l+log2).
»
Page 367
83. tan-i|; sin-j. a2/ J2/ «£
84. fira2. ax2 5xay w &y* L WJ
85. !^ (3 & _ 2 a). + ^ F"(x).
86. — (36 + 2a). 95. — - +
b v ' dr* r dr
Page 368
96. — - + — -• 98. xy( — H+fc — y )
au» a»2 Vay2 ax2/ v '
a2^ ar . bv
y |-x —
dxdy dx dy
CHAPTER XVI
Page 394
1. 6 -log 2. 7. J*™2. 11. 2 J.
2. £ (2 - -7T). 8. 2$ . 12. j (is - 8 log 4).
8^-2- 9- ^ 13. |(a + 6)Vo6.
4. i(ll-161og2). ^± 14.1
5. j(5 + 6V3). 36l ' "*
15. 4 + V(sin-i|-sin-iJ). n< l2(x-2).
/ 2 \ 8
la /2-7r + 5sin-i_ )a2; „2
V Vs/ 18 " (3Vs + 2ir).
t 2 \ 48
67T — 2 — 5 sin-1 — = )a2. a2V§
V5/ 19. — -—•
(•
Page 395
20. 10 x. 24. 8£f. 29. t^™4.
21 a\s 4- ^ * *■ 8°- S?V a6f)VS'
21. — (8 + tt). 1024 V 2
f 26. _-(e8»ra_l)(eaira_l). 8 ' 316
22. 1(» + 3 - 3 V3). 10a 82. f $ira*.
6 **■ i™4' 83. §xa*.
23. 22}. 28. £7ra*. 34. t%oA8.
ANSWERS 505
Page 396
35. — (15x + 82). 88. -(16ir - 82). 41, W» *>"
120 24 42. (V, 0).
a* . _ _ a4
86. — (8 + 3 ir). 89. (45 ir - 128). ,<><>„ \
24v T ' 1920 V ' 48. 2ft0
37. f&Tra*. 40. T6^7ra*. \15* /
44.
/ 18a 18a \ / 3a(5tt- + 8)\
\6(8x-2)' 5(8ir-2)/' ' \ ' 5(6tt-4) f
its l^a t\ 47. On the axis of the sector,
x ' from the center of the circle.
Page 397
48. ( 804» , 304a V 68. /tt<8Tt2>, oY
\105(16-3ir) 105 (16- 3 ir)J \ 6V3 /
• (***• •>■ * (- ^- •)•
so. a*,o). ••*• y**-
VTT' ' 56. 4irV5.
51. (_H£, oV 57. 8 a2.
52.
58. 2a2(7r-2).
/5a A 2a2
60. 8 a2.
Page 398
61 tt%20 3^ 68' tV™4-
61. -(20-3^). 69. -K^-Se + lS).
62. £(20 -Sir). * _^?
68.^(20-3*). '3« 48'
9 V ' .71. a(2-V2).
64. 4a2[V2-log(l + V2)] 72. i(4 + ir)-log(VS + l).
65. 16 a2 [> - V2 - log (l + V2)]. a2
66. (ir-2)a2. 78, 40*
67. -(*r-2). 74. §*ra2.
8
Page 399
a8 80. 102$. 87. §ira*b.
' 90* 81. ^. .. 2a»
76. 4T|T. "' 4 W. -0-<Bir-4).
77. fira8. 82. *ra8. 8
78 8?rg4 88. | a8. 89. ~
'326' 84. Va8.
79. ^(5V5-4). 86' t1"*5-
3 ' 86. $*ra8.
5l
860
506
ANSWERS
Page 400
a*
90. — (w + 2).
o
,3
91. — (3tt + 20-16 V2).
V
a1
92. —(17 -12 log 4).
98. /jira8.
94. fira8.
"•fW + tf; *f
96. f&a6, where A; is the coefficient,
of variation.
97. ^V + 4**).
20 V '
98. 291 Trp.
99. jjitfA*.
mo. ^/^a
10 \« 8 - a*/
Page 401
101. JJfa2.
102. -(2A2 + a2).
2 a4
103. _(3tt-7).
9 '
104. 1^2 (rf-rft.
16 \ 2 1/
105. — (462 + 3a2).
4
106.
fara6
~48~
107. i^£ [(2 a 2 + 3 fi2) (a2 - tf2)*
-(2a12 + 362)(a2-62)2].
108. ( —
\15
152
525
a 11 a\
H 120/
109 /asin^i «(l--cos^i) *^i\.
' \ ex ex 2 / '
$x = 2 mir,
*** /3a 3a 3a\
U°- (»' 28* »)
Page 402
111
/3a 36 3c\
112. (0, — -, — -).
\ 3 9W
118
114. 2 = 2.
57r6sina
115.
16 a
from center of sphere.
116. On the axis of the hemisphere,
8a
15
base.
from the center of the
117. At the middle point of the axis
of the cone.
„„A /2a 2a 2a\
119. On the axis of the cone, § of
the distance from the vertex
to the base.
120. On the axis of the shell, 2JgJ
from the center of the spheri-
cal surfaces.
121. On the axis of the frustum,
h(r* + 2r1rt + Srf)
4(r12 + r1r2 + r2)
smaller base.
from the
ANSWERS
507
Page 403
122. On the axis of the cone,
a(407r + 27V3)
40 7T
vertex.
4Jf
from the
128.
3a2
124. 2ttP.
125. ^(26 -a).
o
Page 404
180.^(2 + V2).
O
kiroft
126. — — (36 — 2 a), where h is the
coefficient of the variation of
the density.
2irp
128.
129.
4Jf(7-4\/2)
15 a2
181 33f(l — cos a)
' 2 (a2 + axa2 + a2) "
132.
28
r4V2-5V6+Vl7+log *2(l + ^r I
L (1 + V2)*(l + Vl7)16J
Page 433
17. -Kx<l.
18. — 1<»<1.
CHAPTER XVII
19. — l<x<l.
20. — 2 < x < 2.
21. - 3 < x < 3.
1 1
<x < — -■
V2 V2
Page 434
28. 1 x2 + x4 x* +
2 2-4 2-4.6
29. x+ Jx8+ T25x6+ 3V5a;7+ *••
x2 5x* 61 x«
+
I?
+
80. 1 + L— +
I* 14
31. l + x + x2+§x8 + ....
1 x8 1 . 3 x6 1 • 8 • 5 x7 ,
32. x « • — h
2 3 2-4 5 2.4-6 7
83. -J*- A^-^rf-jH**-
84.
V2
35. log
-K)-
tl+tl
12
I*
5 (»-5) 1 (x-5)" ■ 1 ft ~ 5)"
"•"6 2' 5» 3 5*
«.+!=i_t=a:+i^E+....
38. V6[l+§(x-2)+5i(J(x-2)2-T43(x-2)84-...].
508 ANSWERS
89. .2079. 42. .8480. 45. .69315; 1.0986.
40. .7198. 48. -.0502. 46. .22314; 1.6094.
41. .9848. 44. .40546. 47. .8473; 1.946.
Page 435
51. 2.0305. 52. 2.9625. 57. \R\< .0004 ; 7. 60. | B | < .00005.
Page 436
62. 1.328; 1.300
1.308.
; 64. -.0962;
- .0991.
^~
0960 5 a» 1
' 67. log
a 70. —
&' 71. -
4-
74. 0.
75. 0
68. .14; .1325;
65. -2V§.
68. 0.
72. J.
76. 0
.1418.
66. 3.
69. 2.
73. J.
77. 2
Page 437
78. % .
79. 0.
80. 0.
81. 0.
82. - j{.
88. 1.
84. 1.
85. 1.
86. e.
87. i.
e
88. 1.
89. e,
90. 1
V3 . /COSX
9L 3 4( I-
cos 2 x cos 3
22 32
X
■^ . . . 1 .
92te--e— _
ate0* — e-m) /
cosx cos2x f
cos3x
A
2<wr ir
-e-a*/smx 2sin2x 3sin3x \
^ \12 + a2 ~~ 22 + a* 3s + a2 /'
«« ,/sinx , sin3x . sin6x , \
9S.4(— + — + — +.••).
__ ir 2 /cosx , cos3x . cos6x . \ /sinx sin2x sin3x \
"■ *-A-w + ^r+-^+ ■■)-(— —!T+—s •)
__ Sir 2 /cosx , cos3x . cos6x . \
95. ( + h •..)
4 tt\ l2 32 S2 /
/3sinx sin2x 3 sin 3 x sin4x \
+ \— 2~ + ^ 4~+"T
-- 7T2 _ /cosx cos2x . cos3x • \ . ir/'ra 4\ . IT2 . rt
96. — — 2( — — + — •••) + -! I =)sin« 8in2x
6 \ l2 22 82 / tL\1 W 2
/ir2 4\ . . tr2 . . , 1
\T " 3»7 - -sm4x + • • -J •
CHAPTER XVIII
Page 471
1. x + 2/ + log(x-l)(y-l) = c. 7. x2 + 2xy + y2-10x-6y+c = 0
2. 4 tan x + 2 y + sin 2 y = c. xa c
3. y(s + y) = ex. 8,y=3" + i + 1-
4. Vx2 + y2 + xl0gX = CX. e8x
y 9. y = — (3xa-x) + cx.
5. x(e* + l) = c. 9V '
6. y2 + 6xy + 2x2 + 2y-8x = c. 10. x = y-l + ce-y.
11. y =
12. y =
ANSWERS 509
14. x2 + d*y + sin y = c.
I + l (X + l)1 11 ?/2
«/ + ' i6. rf + ^-i-i+xV + ^c.
XV X4
c — (x — l)** 16. y = cx.
13. y = ^ (* + *>. 17. log(x - y) + tan-1? = c.
V2(c-x8) y
Page 472
18. log(x2 + y2) = 2x + c. 84. log^ - — = c.
19. sin-i^ = logx + c. x *y
x „ , V2 , 2x
a: « 85. log — ^— - + — = C.
20. e-* + ^ = c. *-8* *
2 x
86. y=-
21. y2+ logcos^x + y) = c. * * log yx + x2 + c
22. sinx-xcosx-tany = c. 8T z2 = c2 + 2ey.
23. x = siny + ccosy. 88. y Vsta2x" = cos 2 x + c.
24. x2 = c2— 2cy. «„ /-= s
25.x2+y3=cT 39.x + y+V^ = c.
26. (y + 2x-l)2 = c(y-x-l). 40. x + y- — + -^ = c.
27. y = cx. ^x ^
28. y(x-l) = (x + l)[a; + c tt< y = xa _ 2 + ce*"2".
-log(x + l)2]. y y X2
oa • i A ?. /; « 42. - tan-1 - + log
29. sm-1x-xVl-x2 + yVl-y2 x x Vx2 + y1
-sin-iy = c. 43> 5x + 5y-log(10x-16y + 7)=c.
= c.
30. y*=c(y2+2x2).
44. log Vx2 + y2 - 2 tan-1- = c.
31. tan-1? = = c. x x
y Vx2 + y2 4fi* x + tan-1 — = c.
32. (xt + y»)c«=c. ^
33. y = (x2 + c) e*8. 46. y =
Page
473
47.
x2y2 + 2e* =
y = x2 - 1 +
c.
48.
cVx2-l.
49.
2xy + log-^-:
X8
= c.
50.
y = x(logx +
ct) + c2.
Vl+ x2 Vc- x2 + log (1+ x2)
58. y + c1logy = x + c2.
59. y = e2 + et sin x — x — J sin 2 x.
60. y = cx sin k (x — c2).
61. y = c2 + c1e-x— ^(cosx+sinx).
62. y = — (e^*-**)* e-*i<*-c»>).
/ 2 c,
51. y = x sin-1 x + VI — x2 + cxx + c2.
52. y = cxx + c2 - log(l + x). 68. y = -± (e*<*-<V + «-*<*- V).
1 3x
53. y = cax + c2 - log cos ax. 64. 6y + c8 = c2x8
54. y = e*(x - 1) + c^2 + c2. 65. 1296y = (x + 1)*.
55. y = c1logx-- + c2. W- y = V3tan Vs(x- 1).
56. 4y + c2 = c1x2--logx. 67. siny = x--.
57. y = cx sec Cj (x — c2) . 68. x = log ( V2 sin y).
510 ANSWERS
Page 474
«« «~ . o~ . x* . -. 71. y = c.er*+ c„e-6x + *Ve2ar.
69. y = c1e8x+ c2e-2x + — hx — 1. * * 2 *T
70
74
75
76
77
78
79
80
81
82
88
84
85
86
87
88
89
90
91
92
93
94
95
96
y = Cle*+ c2e-2x 2 -72- V = (cix + c2)c"2x + (* - 2)2.
— £(6sin2x + 2cos2x). 73. y = c1cos3x + c2sin8x + Je8x.
|/ xV3 , . xV3\ , , , 0 ,
y = e2 1 cx cos 1- c2 sin 1 + x8 + 3x2.
y = (x + c1)e~8x + c2e~6x.
y = cx + c2e8x + x2.
y = Cje~ 2x + c2e2x (sin 2 x + 2 cos 2 x).
y = e~2x(c1cosx + c2sinx) + J(3sin3x — cos3x).
x 8x
y = cxe 2 + c2e 2 + ^5 (8 cosx — 14 sin x) — ? J3 (8 sin 2 x + 19 cos 2 x) .
y = ^(qcosx V2 + c2sinx V2) A (5 cosx — 4sinx).
41
y = (cxx + c2)e-2x + ]Z (6 - 20x + 25x2).
020
y = Cj sin 3 x + ( c2 ) cos 3 x.
gx X6X Gx
y = c,e2x + c„e-x (3 sinx + cos x).
1 2 4 2 10
(x x2\ 1
c, ) + c9e6x H (9 sinx + 7 cosx).
1 9 6/ 2 130 v '
y = e~x(c1cosxV3 + c2sinxV§) + £(2x — 1+ e2x).
y = c1e8x+ c2e~x (4 sin 2 x + 7 cos 2 x).
4 65
y = e2x(c1cos2x + c2 sin2x) + J (1 + 4x + 4x2) + ^ (3cos3x + 36sin3x).
y = c1e_x + ^1 c2 cos J- c8 sin I — 6 + x + x8.
y = cle?c+ c2e-x + c8e-8x— £(20— 6x + 9x2)+ J(cos3x — sinSx).
y = Cle-2x + e2x(c2 + c8x - ^ + ^ + i (1 + x + x2).
Page 475
(x\ / x\ e8x
ci - 3) + e2x (c2 + -j + c3e-2x + — .
y = ^e* + (c2 - ^jcos 2x + f c8 - ^jsin 2x.
1 e2x
y = cr€F + c2e~x +c8e-2x + - (cos 2x — sin 2x) + — (12x — 19).
4 1.Z
y = e~x(c1 + CjjX + c8x2) H (2 sinx — 11 cosx).
y = (Cj + c2x) cos x -f (c8 + c4x) sin x — 4 + x + x2.
y = e23^ + CgX + jj + e-2x(c8 + c4x).
ANSWERS 511
o%x
97. y = Cje2*^- c2e~2x+ c8cos3x + c4sin3x (sin3x + 3 cos 3 x).
98. y = c1+c2z + cz cosxv2 + c4 sin x V 2 H (13 sin 2 x + 16 cos 2 x).
6 17
99. y = ct + CgX + CgX2 + c4e~4a? + T J^(cos 4x — sin 4x).
100. x = (c^ + c2) <* + e2«, y = (cxt + c2) e< + c8 + c2«.
101. x = ctsini + c2cost -f J(e« + e~'), y = c2sin£ — ca cosi + J(e< + c~').
102. x = ct + c2e* -f c8e-', y = c4 — c2e< -f c8e-< — cost.
103. x = cae< + c2e6* — 1, y = 3cxe« — c2e6< _ 2.
104. x = c1e-2« + c2e2«+ $e8< + ^ (7 sin 3 1 + 3 cos 3 Q,
c 3 1
y = c,e-2* _ _2e2* _ _C8« + _(sin3« - 6 cos 3*).
3 5 13
105. x = cx + c2e2< + c8e-8'-T£¥(49* - 33 «2 + 6t%
y = 2ct - —M + ^e2< - c.e-8« + i_ (n « _ st* - 6*8).
1 54 3 8 54 v '
106. x = cx + c2e-2« + c8e8< + £(3*2 - *),
y = Cl -^-SCgeS'H- e2«- J(3*2+2*).
107. x = 6^/2/^008 — — + c2sin— -) + e V2[c8cos-— + c4sin — J,
\ V2 V2/ \ V2 V2/
y = e V2 ( c2 cos — = — c, sin — = J + e V2 ( c8 sin — - — c4 cos — = ) .
\ V2 V2/ V V2 ' V2/
age 476
*A . 3 2x , 3-4 22x2 , ,8.4.5...(r + 2) ^ \
108. y = c,x8 / 1 + + h • • • + v ' h • • • I
\7 |1 7 • 8 12 7 . 8 • 9 • • • (r + 6) \r J
+ f?(15 + 18x + 9x2 + 2x8).
x8
109. y = ^(35 - 42x + 21 x2 - 4x8) + £|(3- 14x + 21 x2).
tin CMA x . »A 4 x , 4-5 x2 4.5.6 x8
4.5.6...(r + 3) g; \
v ' 6 • 7 . 8 • • • (r + 6) (r /
m.y = Cl(i + ^ + ^i)^ + n<"-4><"-16)ai«+...
n(n-4)(n-16)---[n-(2r-2)"] \
[2r +V7
*\ |S [6 [I
(n-l)(n-9)(n-25)---[n-(2r-l)'] \
|2r4-l /'
512
ANSWERS
(X2 X4
1 + — + +
2-5 2.4-6.9
x2r
[2 • 4 • 6 • • • 2r] [6 • 9 • 13 • • .(4r + 1)]
1 / x2 x*
X2r
+ [2 • 4 • 6 . . . 2r] [3 • 7 • 11 • • «(4r - 1)]
113.
(x4
1
22|;
+
x°
22[3 24|5
+ (- l)r
xAr
22rj2r + l
■)
)•
+
Ml~^
x»\ 22|5
x!
+ ^77 + (-1)1
X*r
22|2 ' 2* [4 ' * ' 22r|2r
115. x=log(y + 3). 118. r(0 + c) = 1.
116. k(x - l)y - y + 1 = 0, where 119. r2 = 2(0 + c).
& is the constant ratio. 130. i* = c sin n0.
117. y = cx». 121. xa— y2 = c.
Page 477
123. x2 + y2 = ex.
a
ar+ c
x+c
124. y = -(c a + e ° ).
125. y = — (e*<*- c> + e-«*-«>),
where k is the constant ratio.
Va2 — 2 ay — a
126. x + c = ± a log
Va2 — 2 ay + a
±2Va2-2ay.
127. r = ce* * *.
128. y = ax2 + 6.
Page 478
134. p = 14.7 e— 0000**.
135. % 1218.
136. 68min.
x — a
129. y — ce k , where x = a is the
fixed ordinate, and k is the
constant ratio.
x — c
x — e
130. y = -(c * + e * ), where *
is the constant ratio.
131. y = c± jVJk*x-4x2
, *2 . .2Vx , . .
± -sin-1— — , where k is
4 &
the constant ratio.
132. « = 25 (2) * .
133. c=.01e-«8«.
140. (x - cx)2 + (y - c2)2 = c*.
,2
141. c1y2--l(x + c2)2 = l.
142. a*
143.
2a$
ANSWERS 513
Page 479
*** <* /7TT v , . iL _. A 149. About 7 nu. per second.
145. - V3 Jfc, where ac is the constant „ „_ /r— r
2 150. V2^.
ratio. 151. 20 sec.
146. £ (4 - V2) Vo». 152. 234§ ft. ; 4.3 sec. (g = 32.)
2(2
147. — ■ 153
vx + v2
*mg + kv{
Page 480
154. 1.8 sec. (g = 32.)
155. s = c, cos At + c„ sin to + — cos Art, where K* is the constant ratio ;
at
« = cx cos At + c2 sin At H sin Art, if A = A;.
A fC
156. s = e~ 2 (cxe 2+036 2 )
. a(A2 — A;2) cosArt + aA^sinArt ... _.
(A2 - fc2)2 + (£)2
»/ tV4A2-Z2 ' . iV4A2-P\
s = e 2 I cx cos 1- c, sin J
, a(h* — A;2) cos Art + aArt sin Art ... _ .
+ (A2l*2)2+W2 ""<"'
a = (Cl+c2Qe-^+a^~fea>COsA;e+2a^sinfct,if^2A.
V1 2' (A2-fc2)2 + (2Afc)2
(A2 and 2 are the constant ratios.)
157. h = k,l very small.
INDEX
(The numbers refer to the pages)
Abscissa, 5
Acceleration, 178
Angle, 69, 166, 206, 316, 316
eccentric, 108
vectorial, 118
Arc, differential of, 174, 206, 324
limit of ratio to chord, 172
Archimedes, spiral of, 120
Area, of any surface, 381
as double integral, 376
as line integral, 364
of plane curve, 143, 262, 267
of surface of revolution, 274
Asymptote, 26, 79
Attraction, 283, 393
Axis of conic, 76, 80, 81
Bernouilli's equation, 446
Bessel's equation, 470
Bessel's functions, 471
Bisection of straight line, 9
Cardioid, 124
Catenary, 54, 441
Center of gravity, 278, 379, 392
Center of pressure, 277
Change of coordinates, 39, 124, 386,
387, 388
Circle, 70, 108, 122
auxiliary, 108
involute of, 112
Cissoid, 85, 124
Comparison test, 406
Components, of a straight line, 312
of velocity, 180
Concavity of plane curve, 167
Cone, 306
Conic, 83, 125
Constant of integration, 147, 222
Continuity, 136, 339
Convergence, 405, 409
Coordinates, Cartesian, 1, 4, 301
cylindrical, 385
oblique, 43, 301
polar, 118, 386
Curvature, 207
Curves, in space, 322
intersection of, 29
Cusp, 109
Cycloid, 109
Cylinder, 303
Degree of plane curve, 29
Derivative, 136
. directional, 343
higher, 162, 338
in parametric representation, 204
partial, 335
sign of, 138, 166
total, 344
Differential, 141, 339, 369
of arc, 174, 206, 324
of area, 146
exact, 349
total, 341
Differentiation, 136
of algebraic functions, 154
of composite functions, 367
of implicit functions, 162
partial, 335
of polynomial, 137
successive, 162, 338
of transcendental functions, 192
Direction in polar coordinates, 206
514
INDEX
515
Direction cosines, 314, 319, 324
Directrix of conic, 80, 83
Discontinuity, finite, 427
Distance, of point from plane, 320
of pointtf rom straight line, 63
between two points, 5, 313
Divergence, 405
Division of straight line, 8
e, the number, 63
Eccentricity of conic, 74, 78, 83
Element of definite integral, 260
Ellipse, 74, 108
area of, 262
Ellipsoid, 307
volume of, 270
Epicycloid, 111
Epitrochoid, 114
Equations, differential, 438
empirical, 89
parametric, 106
roots of, 34
Factor, integrating, 448
Focus of conic, 74, 78, 80, 83
Force, 179
Forms, indeterminate, 423, 425
Fourier's series, 427
Fractions, partial, 247
Function, 9, 300
BessePs, 471
complementary, 456, 458, 461
composite, 357
implicit, 163, 300
notation, 12
periodic, 427
transcendental, 49
Graph, 10, 20, 49, 301
Helix, 324
Hyperbola, 77
Hyperboloid, 306, 307
Hypocycloid, 112
four-cusped, 107
Hypotrochoid, 114
Increment, 135, 339
Infinitesimal, 261
Infinity, 25
Inflection, point of, 167 *
Integral, 146, 222
constant of, 147, 222
definite, 147, 260
line, 353
multiple, 369
particular, 456, 458, 461
triple, 385
Integrand, 222, 264
Integration, 222
approximate, 419
collected formulas of, 236
by partial fractions, 247
by parts, 243
of a polynomial, 146
by reduction formulas, 252
by substitution, 238
Intercepts, 21
Intersections, 29
Involute of circle, 112
Legendre's coefficients, 469
Legendre's equation, 468
Lemniscate, 120, 125
Length of a curve, 272, 325
Limagon, 123
Limit, 130
of ratio of arc to chord, 172
of 5^, 192
h
l
of (1+ A)*, 199
theorems on, 132
Limits of definite integral, 148, 264
Line, straight, 57, 121, 317
Line integral, 353
Locus, 20, 69
Logarithm, Napierian, 64
Maclaurin's series, 412
Maxima and minima, 168, 348
Mean, theorem of, 422
Moment of inertia, 377, 390
516
INDEX
Motion, in a curve, 180
rectilinear, 177
simple harmonic, 203
Normal, to curve, 164, 826
to plane, 816
to surface, 347
Number scale, 3
Operator, 465
Order of differential equation, 441
Ordinate, 5
Origin, 3, 118
Parabola, 80
segment of, 88
Paraboloid, 305, 308
Parameter, 106
Parts, integration by, 248
Plane, 305, 316
Point, of division, 8
turning, 189
Pole, 118
Polynomial, derivative of, 137
integral of, 146
Pressure, 275
center of, 277
Prismoid, 421
Prismoidal formula, 410
Projectile, path of, 181
Projection, 2, 310, 323
Radius of curvature, 208
Radius vector, 118
Rate of change, 175
Ratio test, 407
Reduction formulas, 252
Region of convergence, 411
Remainder in Taylor's series, 416
Revolution, surface of, 270, 274, 809
Rolle's theorem, 416
Roots of an equation, 34
Rose of three leaves, 119
Segment, parabolic, 83
Series, 405
Fourier's, 427
geometric, 406
harmonic, 406 •
Maclaurin's, 412
power, 410
Taylor's, 412
Simpson's rule, 421
Slope, 6, 134
Space geometry, 800
Sphere, 310, 313
Spirals, 120
Strophoid, 86
Substitution, integration by, 288
Surfaces, 304
of revolution, 270, 274, 809
Symmetry, 23
Tangent, to plane curve, 80, 140, 164
to space curve, 826
to surface, 345
Taylor's series, 412
Tractrix, 439
Transformation of coordinates, 89,
124, 386, 387, 388
Trapezoidal rule, 421 .
Trochoid, 110
Value, absolute, 409
infinite, 25
mean, 265
Variable, 9
Vector, radius, 118
Velocity, 177, 180
Vertex of conic, 75, 79, 81
Volume, of any solid, 889
element of, 385, 386, 887
of solid with parallel bases, 268
of solid of revolution, 270
Witch, 84
Work, 275, 353, 866
LANE MEDICAL LIBRARY
This book should be returned on or before
the date last stamped below.
FEB*- i«B
FEB 2 5 1965
MAY 2 4196b
N
\
N*
389
A205 Woods. £.u.
W89 Analytic geometry and
1917 calculus.
NAME
DATE DUE
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