THE LIBRARY
OF
THE UNIVERSITY
OF CALIFORNIA
LOS ANGELES
\j
The RALPH D. REED LIBRARY
DKPARTMKNT OF GEOLOGY
UNIVER.S1TY of CALIFORNIA
L08 ANGELES, CAUF.
Digitized by the Internet Archive
in 2007 with funding from
IVIicrosoft Corporation
http://www.archive.org/details/courseinmathemat01woodiala
A COURSE IN
MATHEMATICS
FOR STUDENTS OF ENGINEERING AND
APPLIED SCIENCE
BY
FEEDERICK S. WOODS
AND
FREDERICK H. BAILEY
Professors of Mathematics in the Massachusetts
Institute of Technology
Volume I
ALGEBRAIC EQUATIONS
FUNCTIONS OF ONE VARIABLE, ANALYTIC GEOMETRY
DIFFERENTIAL CALCULUS
GINN AND COMPANY
BOSTON • NKW YORK • CHICAGO • LONDON
ATLANTA • DAT.T.A8 • COLUMBU8 • SAN KKANCISCO
Copyright, 1907, by
Frederick S. Woods and Frederick H. Bailby
ALL RIGHTS RESERVED
620.12
Cbe gtbenttum l^rtet
GINN AND COMPANY • PRO-
PRIETORS • BOSTON • U.S.A.
Geology
Library
PKEFACE
This book is the first volume of a course in mathematics
designed to present in a consecutive and homogeneous manner
an amount of material generally given in distinct courses under
the various names of algebra, analytic geometry, differential
and integral calculus, and differential equations. The entire
course covers the work usually required of a student in his first
two years in an engineering school, the first volume containing
the work of the first year. In arranging the material, however,
the traditional division of mathematics into distinct subjects is
disregarded, and the principles of each subject are introduced as
needed and the subjects developed together. The objects are to
give the student a better grasp of mathematics as a whole, and
6f the interdependence of its various parts, and to accustom him
to use, in later applications, the method best adapted to the
problem in hand. At the same time a decided advantage is
gained in the introduction of the principles of analytic geometry
and calculus earlier than is usual. In this way these subjects
are studied longer than is otherwise possible, thus leading to
greater familiarity with their methods and greater freedom and
skill in their application.
In carrying out this plan in detail the subject-matter of this
volume is arranged as follows:
1. An introductory chapter on elimination, including the use
of determinants. This chapter may be postponed or omitted, if a
teacher prefers, without seriously affecting the subsequent work.
2. Graphical representation. Here the student learns the use
of a system of coordinates and the definition and plotting of a
function.
3. The study of the algebraic polynomial. This includes the
analytic geometry of the straight line, the more important
t GS'
iv PREFACE
theorems of the theory of equations, and the definition of a
derivative. Simple applications of the calculus to problems
involving tangents, maxima and minima, etc., are given. In this
way a student obtains an introduction to the piinciples of the
calculus, free from the difficulties of algebraic computation.
4. The study of the algebraic function in general. The knowl-
edge of analytic geometry and calculus is here much extended
by new applications of the principles already learned. Simple
applications of integration are also introduced. The study of the
conies forms part of the work in this place, but other curves are
also used and care is taken to avoid giving the impression that
analytic geometry deals only with conic sections ; in fact, the
chapters which deal especially with the conies may be omitted
without affecting the subsequent work.
5. The study of the elementary transcendental functions. It
has been thought best to assume the knowledge of elementary
trigonometry, since that subject is often presented for admission
to college, — a tendency which should be encouraged. Tlie
chapter discusses the graphs, the differentiation of transcendental
functions, and the solution of transcendental equations.
6. The work closes with chapters on the parametric represen-
tation of curves, polar coordinates, and curvature. In the first of
these chapters the solution of locus problems, which, from some
standpoints, is the most important part of analytic geometry,
finds its natural place ; for this problem involves, in general,
the expression of the coordinates of a point on a locus in terms
of an arbitrary parameter, and possibly the elimination of the
parameter.
As compared with the usual first course in analytic geometry,
there will be found in this volume fewer of the properties of the
conic sections, 'except as they appear in problems set for the
student. On the other hand, a greater variety of curves are
given, and it is believed that greater emphasis is placed on the
essential principles. All work in three dimensions is postponed
to the second year, and is to be taken up in the second volume
in connection with functions of two or more variables, partial
differentiation, and double and triple integration.
PREFACE V
This volume contains the matter usually given in a first course
in differential calculus, with the exception of differentials, series,
indeterminate forms, partial differentiation, envelopes, and some
advanced applications to curves. These subjects will find their
appropriate place in the further development of the course in
the second volume. Integration has been sparingly used as the
inverse operation of differentiation, and without employing the
integral sign. Simple applications to areas and velocities are given.
To do more would require the expenditure of too much time on
the operation of integration, and the introduction of too many
new ideas into one year's work. The integral, as a limit of a
sum, with its many applications, will form an important part of
the second year's work.
In the preparation of the text the needs of a student who
desires to use mathematics as a tool in engineering and scientific
work have been primarily considered, but it is believed that the
course is also adapted to the student who studies mathematics
for its own sake. Abstract discussions are avoided and frequent
applications and illustrations are given. Illustrations, however,
which are beyond the range of a first-year student's knowledge
of physical science are omitted. The proofs are made as rigorous
as the maturity of the student will admit. It is to be remembered
in this connection that the earlier chapters are to be studied by
students w^ho have just entered college.
In the preparation of the book the authors have had the advice
and criticism of the mathematical department of the Massachu-
setts Institute of Technology. In particular, they are indebted
to the head of the department. Professor H. W. Tyler, at whose
invitation the book has been written, and whose suggestions have
been most valuable.
Massachusetts Institute of Technology
September, 1907
CONTENTS
CHAPTER I — ELIMINATION
Article Page
1, 2. Determinant notation 1
3. Properties of determinants . 6
4. Solution of n linear equations containing n unknown quantities,
when the determinant of the coefficients of the unknown
quantities is not zero 12
5. Systems of n linear equations containing more than n unknown
quantities 15
6. Systems of n linear equations containing n unknown quantities,
when the determinant of the coefficients of the unknown
quantities is zero 17
7. Systems of linear equations in which the number of the equa-
tions is greater than that of the unknown quantities ... 18
8. Linear homogeneous equations 21
9. Eliminants 23
Problems 25
CHAPTER II — GRAPHICAL REPRESENTATION
10. Real number 28
11. Zero and infinity 29
12. Complex numbers 31
13. Addition of segments of a straight line 32
14-15. Projection 34
16. Coordinate axes 35
17. Distance between two points 36
18-19. CoUinear points 38
20. Variable and function 40
21. Classes of functions 43
22. Functional notation 44
Problems 45
vii
viii CONTENTS
CHAPTER III — THE POLYNOMIAL OF THE FIRST DEGREE
Article Page
23. Graphical representation 50
24-26. The general equation of the first degree 52
27. Slope 54
28. Angles 55
29. Problems on straight lines 58
30-31. Intersection of straight lines 61
32. Distance of a point from a straight line 63
33. Normal equation of a straight line 64
Problems 65
CHAPTER IV — THE POLYNOMIAL OF THE ^Yth DEGREE
34-36. Graph of the polynomial of the second degree 70
37. Discriminant of the quadratic equation 73
38. Graph of the polynomial of the nth degree 74
39. Solution of equations by factoring 77
40-41. Factors and roots 78
42-43. Number of roots of an equation 80
44-45. Conjugate complex roots 82
46. Graphs of products of real linear and quadratic factors ... 83
47. Location of roots 86
48. Descartes' rule of signs 87
49-51. Rational roots .89
52. Irrational roots 92
Problems 94
CHAPTER V — THE DERIVATIVE OF A POLYNOMIAL
53. Limits 97
54. Slope of a curve 99
55. Increment 100
56. Continuity 101
57. Derivative .102
58. Formulas of difFerentiation 103
59. Tangent line 104
60. Sign of the derivative 106
61. Maxima and minima 108
62. The second derivative 110
63. Newton's method of solving numerical equations 114
64. Multiple roots of an equation 116
Problems 118
CONTENTS ix
CHAPTER VI — CERTAIN ALGEBRAIC FUNCTIONS AND
THEIR GRAPHS
Article Page
65-66. Square roots of polynomials 121
67. Functions defined by equations of the second degree in ?/ . . 127
68. Functions involving fractions . • 128
69. Special irrational functions 131
Problems 133
CHAPTER Vll — CERTAIN CURVES AND THEIR EQUATIONS
70-72. The circle 134
73-75. The ellipse 139
76-78. The hyperbola 142
79-80. The parabola 146
81. The conic 148
82. The witch 149
83. The cissoid 151
84. The strophoid 152
85. Examples 154
Problems 155
»
CHAPTER Vni — INTERSECTION OF CURVES
86. General principle 161
87-89. /i(x,y) = 0 and /2(x,.?/) = 0 161
90. fi(x,y)='0 iiTidf„(x,y) = 0 10(3
91. /„(z, ?/) = 0and/,(x, .y)=0 168
92-93. If^(x,y) + kf„(x,y)=0 171
Problems 175
CHAPTER IX — DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
94. Theorems on limits 178
95. Theorems on derivatives 179
96. Formulas 184
97. Derivative of u" 185
98. Higher derivatives . 187
99. Differentiation of implicit algebraic functions 188
100. Tangents 190
101. Normals • 191
102. Maxima and minima 192
103. Point of inflection 194
X COKTENTS.
Akticle Page
104. Limit of ratio of arc to chord 195
105. The derivatives — and -^ . ; 196
(Is d,i
106. Velocity 198
107. Components of velocity 200
108. Acceleration and force 202
109. Other illustrations of the derivative 203
110. Integration 205
Problems 209
CHAPTER X — CHANGE OF COORDINATE AXES
111. Introduction 217
112-114. Change of origin without change of direction of axes . . 217
115. Change of direction of axes without change of origin . . 221
116. Oblique coordinates 223
117. Change from rectangular to oblique axes without change of
origin 224
118. Degree of the transformed equation 225
Problems 225
CHAPTER XI — THE GENERAL EQUATION OF THE SECOND DEGREE
119. Introduction 229
120. Removal of the xy-tevm 229
121. The equation Ax^ + By'^ + 2 Gx + 2 Fy + C = 0 . . . .231
122. The limiting cases 234
123. The determinant AB - H^ 235
124. The discriminant of the general equation 236
125. Classification of curves of the second degree 237
126-127. Center of a conic 238
128. Directions for handling numerical equations 240
129. Equation of a conic through five points 241
130. Oblique coordinates 244
Problems 244
CHAPTER XII — TANGENT, POLAR, AND DIAMETER FOR CURVES
OF THE SECOND DEGREE
131. Equation of a tangent 246
132. Definition and equation of a polar 247
133. Fundamental theorem on polars 247
CONTENTS xi
Article Page
134. Chord of contact 248
135. Construction of a polar 249
136. The harmonic property of polars 249
137. Reciprocal polars 251
138-140. Definition and equation of a diameter 252
141. Diameter of a parabola 254
142. Parabola referred to a diameter and a tangent as axes . . 255
143. Diameters of an ellipse and an hyperbola 256
144. Conjugate diameters 258
145. Ellipse and hyperbola referred to conjugate diameters as axes 259
146-147. Properties of conjugate diameters 260
Problems 262
CHAPTER XIII — ELEMENTARY TRANSCENDENTAL FUNCTIONS
148. Definition 266
149. Graphs of trigonometric functions 266
150. Graphs of inverse trigonometric functions 269
151. Limits of and '- — 270
h h
152. Differentiation of trigonometric functions 272
153. Differentiation of inverse trigonometric functions . . . .276
154. The exponential and the logarithmic functions 279
155. The number e 280
156. Limits of (1 + A)* and ^1^ 283
157-159. Differentiation of exponential and logarithmic functions . 284
160. Hyperbolic functions 288
161. Inverse hyperbolic functions 291
162. Transcendental equations 293
Problems 296
CHAPTER XIV — PARAMETRIC REPRESENTATION OF CURVES
163. Definition 302
164. The straight line 302
165. The circle 303
166. The ellipse 303
167. The cycloid 305
168. The trochoid 306
169. The epicycloid 307
170. The hypocycloid 309
xii CONTENTS
Article Page
171. Epitrochoid and hypotrochoid , 309
172. The involute of the circle 311
173. Time as the arbitrary parameter 313
174. The derivatives 315
175-176. Applications to locus problems 316
Problems 323
CHAPTER XV — POLAR COORDINATES
177. The coordinate system 329
178. The spirals 331
179. The conchoid 334
180. The limacon 336
181. The ovals of Cassini 338
182. Relation between rectangular and polar coordinates . . . 341
183. The straight line 342
184. The circle 342
185. The conic, the focus being the pole 343
186. Examples 344
187. Direction of a curve 345
188. Derivatives with respect to the arc 347
189. Area 348
Problems 349
CHAPTER XVI — CURVATURE
190. Definition of curvature 353
191-192. Radius of curvature 354
193. Coordinates of center of curvature 356
194. Evolute and involute 357
195. Properties of evolute and involute . 359
196. Radius of curvature in parametric representation .... 360
197. Radius of curvature in polar coordinates 361
Problems 362
Answers 365
Index 381
A COURSE m MATHEMATICS
CHAPTEE I
ELIMINATION
1. Determinaiit notation. Elimination is the process of obtain-
ing from a certain number of equations containing two or more
unknown quantities one or more equations which do not contain
all of these quantities. The quantities removed are said to have
been eliminated. The solution of equations is essentially the elim-
ination of all but one of the unknown quantities. The process of
elimination leads to the formation of certain expressions in the
coefficients, for which a special name and a corresponding notation
have been invented. In this chapter we shall consider equations of
the first degree, or linear equations. These are equations in which
no term contains more than one unknown quantity, and that in
the first degree.
Ex.1. aix + 6i2/ + ci = 0,
a2X + 62^ + C2 = 0.
To eliminate y, multiply the first equation by 62, the second by — 61, and
add. To eliminate x, multiply the first equation by — a^, the second by ai,
and add. There results
(ai62 - a26t) a; + (C162 - C261) = 0, /o\
(aib2 — a^bi) y + (aiC2 — OiCi) = 0.
Unless 0162 — (tibi = 0, equations (2) give at once the solution of (1).
If 0162 — a^bi = 0, the method used to eliminate y also eliminates x, and the
equations need further discussion, to be given in § 6.
1
ELIMINATION
Ex. 2.
ai« + hy + ciz + di = 0,
a^x + b^y + CiZ + d2 = 0,
Osx + bsV + C3Z + dz = 0.
(1)
To eliminate y and 2, multiply the first equation by {baCs — ftgCa), the second
by - {bid - bsci), the third by (61C2 - 62C1), and add. There results
[ai(&2C8 - bsCn) -(h{biC3 - bgCi) + 03(6100 - ftgCi)]*
+ [di (62C8 - bsCi) - d^ (61C3 - 63C1) + da (61C2 - 62C1)] = 0,
or (oibjCs + ajftsCi + OsftiCa — aihc^ — Oa^iCa — 036201) x
+ {dibuPs + d^ci + ds&iCa - ^16300 - ^06103 - dzb^Ci) = 0. (2)
To eliminate z and «, multiply the first equation by — (0203 — a^c^), the
second by (OiCg — agCi), the third by — (ciCj — 0201), and add. There results
{ai62C8 + a263f-"i + 036102 — 016302 — O261C8 — 036201) y
+ (aid2C8 + a^fiaCi + asdiC2 — axd^c^ — a^diCs — 03^201) = 0. (3)
To eliminate x and y, multiply the first equation by (0263 — 0362), the second
by — (O163 — 0361), the thii-d by (0162 — O261), and add. There results
(O1&2C8 + O263C1 + 036102 — O163C2 — 026103 — O362O1) z
+ (aibjds + 0263(^1 + 0361^2 — Oi63d2 — 0261^3 — 0362^1) = 0. (4)
Equations (2), (3), and (4) give the solution of (1), unless
Oi&aCs + OibzCi + 036102 — 016302 — O261C3 — O362C1 = 0.
The exceptional case will be considered in § 6.
The binomials which occur in the solution of Ex. 1 are called
determinants of the second order. The symbol
is used to denote the determinant afi^— afi^. Then equations (2)
of Ex. 1 may be written
«i ^1
» +
= 0,
«i ^1
'y +
a, c,
2
1 "1
= 0.
DETERMINANT NOTATION 3
The polynomials which occur in the solution of Ex. 2 are called
determinants of the third order. The symbol
«1
\
Ci
^2
h
^2
«3
h
Cs
is used to denote the determinant
«AC8+ aA(^l+ «8*1^2— «^A^2— «2^lC8— CCsK^^l-
The results of Ex. 2 may then be written
a.
\
Ci
a^
h
^2
« +
a.
K
Cs
C?j ftj Cj
d^ &2 Cg
= 0,
C?3 63 Cg
«i
h Ci
«2
h ^2
2/ +
«8
i^3 Cs
«!
d.
«1
«2
d.
^2
^8
d.
Cz
= 0,
«1
h
H
a.
h
C2
2 +
«8
h
^8
«1
\
d.
^2
K
d.
«8
h
ds
= 0.
By the work of Ex. 2,
a^ &j Cj
^2 ^2 «2
= a.
^2 «2
^3 ^3
^ ^1
-fa.
5„ c„
which may be taken as the definition of a determinant of the
third order.
Similarly a determinant of the fourth order is indicated by the
symbol
a, b, c,
d.
^2 \ «2
d.
(^S K ^8
d.
a, h, c.
d.
and is defined as equal to
&2 C2 d^
h ^3 ^3
?>. c, d.
I ^1 ^1 d^
-a\\ C3 <^3
+ a.
^
c, d^
?>2
^2 C?2
-«4
h
c, c?.
^
Cl
<
&2
^^2
d.
^'3
^3
d.
4 ELIMINATION
If now each of these determinauts of the third order is expressed
in terms of determinants of the second order, we shall have finally
the determinant of the fourth order expressed as an algebraic
polynomial of twenty-four terms.
2. In general a determinant of the nth order is an algebraic
polynomial involving n^ quantities, called elements. The symbol
of the determinant is obtained by writmg the elements in a square
of n rows and n columns. If in such a symbol a row and a col-
umn are omitted, there is left the symbol of a determinant of the
next lower order. This new determinant is said to be a minor of
the original determinant, and is said to correspond to the element
which stands at the intersection of the omitted row and column.
We shall now give as definition :
A determinant is equal to the algebraic sum of the products
obtained by multiplying each element of the first column by its
corresponding minor, the signs of the products being alternately
plv^ and minus.
By repeated application of the same definition to the minors
obtained, we eventually make the value of the determinant depend
upon determinants of the second order, and thus obtain the poly-
nomial indicated by the original symbol.
Students who desire a more general definition and discussion of
determinants are referred to treatises on the subject. We shall
derive here, as simply as possible, only those properties which are
of use in solving equations. Before doing so, however, we need to
show that the word "column" may be changed to "row" in the
above definition, thus : A determinant is also equal to the sum of
the products obtained by multiplying each element of the first row
by the corresponding minor, the signs of the products being alter-
nately plus and minus.
For a determinant of the third order the student may verify
that '.
= a.
-&.
-^-Cx
DETERMIKANT NOTATION 6
The theorem thus shown to he true for a determinant of the
third order may be proved for one of the fourth order as follows :
= a.
= «1
C2 ^2
C3 d,
c, d^
-«2
h
c, d,
C3 d,^
c, d]
+ ^3
\ d,
2 ^2
'4 ^4
-a^
b^ c^ d,
K c, d^
h C3 d^
(by definition)
K C2 d^
K C3 d^
b, c, d^
-ah ""^ ^^
-Ci
63 £?3
^4 d.
+ d.
\ «3 1
^4 ^4 J
H ' c, d.
-^1
K d,
b, d.
+ d^
&2 C, 1
^4 <^j
-Ci
^2 ^' +d
\ C2I
^3 ^3 J
(as already proved)
\ c^ d^
h C3 ^3
b, c, d^
H ' G, d.
-«3
^2 d-2
c, d.
+ a.
^2 ^2 "1
C3 d^ f
r ^-i d„
. ^'^V^b\d\
-«3
K d„
\d.
K d. 1
-.{
' ^3
"C3
C4
-^3
\ «2
K C4
+ ^4
h ^2
^3 ^3
}
(by a rearrangement)
= a.
h C2 d^
^2 ^2 ^2
^2 ^2 ^2
«2 ^2 ^2
h C3 ^3
-h
^3 C3 ^3
+ c,
«3 *3 ^3
-d.
<^3 h h
64 C4 d.
^4 C4 ^4
^4 &4 d^
^4 &4 C4
(by definition)
In a similar manner the theorem may be proved successively
for determinants of the fifth, the sixth, and, eventually, any order.
g ELIMINATION
3. Properties of determinants.
1. A determinant is unchanged in value if the rows and the
columns are interchanged in such a manner that the first row
becomes the first column, the second row the second column, and
so on.
The student may verify that
«1 «2
«1
\
«i
as
h
^2
=
*8
K
^3
a.
a„
c, c„
This proves the theorem for determinants of the second and
the third orders. To prove it for one of the fourth order, proceed
as follows :
a^ 6j Cj d^
^2 h ^2 ^2
C'S h ^i ^3
^4 ^4 ^4 ^4
\ c, d^
*1 Ci <^l
6, C, d.
^1 «i ^1
= «!
h Cs ds
-«2
^3 ^'S ^8
+ ^3
h C2 ^2
-a^
^2 ^2 ^2
K C4 ^4
\ C, rf.
\ c, <^,
h Cs ^3
«! ^2 «8 «4
h b h b
h h K
^ ^'3 ^'4
^1 ^2 ^4
^1 ^2 *3
t'l ''2 "s "4
C, C^ C3 c,
d^ d^ d^ d^
= «i
^2 C3 C4
-^2
C, C3 c.
+ ^3
C, C^ C,
-a.
<^1 ^2 ^3
^2 ^3 <
^1 < ^4
^1 C?2 ^4
^1 ^2 ^3
The expressions on the right of these equations are equal, and
hence the determinants of the fourth order are equal. In the same
manner the theorem may be proved for determinants of higher
order.
It follows from this theorem that any property which is true of
the rows is also true of the columns, and vice versa. The following
theorems are stated for both rows and columns, but are proved for
the rows only.
PROPERTIES OF DETERMINANTS
2. If two consecutive rows {or columns) of a determinant are
interchanged, the sign of the determinant is changed.
The student may verify that
a^ b„
=
= —
a„
5
«1
\
«i
K h ^2
«i
h
^1
a^
^
^i
a^
\
^'2
=
«1 ^ ^1
>
«2
h
^2
= —
%
&3
C3
^3
h
^3
a
3 ^3
^3
^3
h
^^3
«2
K
^2
The theorem is then proved for determinants of the second and
the third orders. To prove it for a determinant of the fourth order,
consider
\ c,
d.
h ^2
^3 ^3
d,
d.
and
I, c.
d.
h C3
By definition,
«j h^ Cj d
a, &2 ^2 do
% &3 ^^3 ^3
«^ l^ c^ d^
a^ h^ Cj c?j
«, &3 Cg 6?3
^2 ^2 ^2 ^2
«4 ^4 C4 f?^
&2 C2 ^2
^1 ^1 ^1
^1 ^1 ^1
\ Cj d^
= «1
^3 C3 f^3
-S
^3 C3 ^5
+ «3
^2 «2 <^2
— a^ b^ Cg d^
&4 C4 ^4
^4 C4 f?4
I, c, d^
h C3 ^3
K C3 ^3
\ ^x dx
^1 ^1 ^1
&^ Cj c?^
= «i
^2 ^2 ^2
-^3
&2 Cg d^
+ «2
63 C3 (^3
-«4
&3 C3 C?3
^'4 C4 «^4
K H d\
^4 C^ ^4
^2 ^2 ^2
Comparing these two expressions, it will be noticed that the
minors which multiply a^ and a^ (the elements of the unchanged
rows) differ in the two expressioils by the interchange of two con-
secutive rows, and that the minors which multiply a^ and a^ (the
elements of the interchanged rows) are the same in the two expres-
sions but are preceded by opposite signs. It is evident on reflection
that these laws always hold ; and hence, if the theorem is true
for determinants of any order, it is true for determinants of the
8
ELIMINATION
next higher order. The theorem is known to be true for determi-
nants of the third order ; hence it is universally true.
3. A determinant is equal to the algebraic sum of the products
obtained by multiplying each element of any row {or column) by its
corresponding minor, the sign of each product being ]}lus or minus
according as the sum of the number of the row and the number of
the column in which the element stands is even or odd.
For the hth. row may be made the first row of a new determi-
nant by ^ — 1 interchanges of two consecutive rows. By theorem 2,
if k is odd, the new determinant is equal to the original one ; and
if k is even, the new determinant is equal to minus the original one.
The new determinant may now be expressed by definition as the
algebraic sum of the elements of its first row multiplied by their
mmors, which are the same as those of the ^th row of the original
determinant. Hence the original determinant is equal to the alge-
braic sum of the elements of its kth. row multiplied by their minors,
the products being alternately plus and minus when h is odd, and
alternately minus and plus when k is even. From this the law of
signs as given in the theorem at once follows.
Ex.
a\
bi
Cl
di
(h
b.
C2
d.
as
b.
C8
ds
a*
64
Ci
di
ai
61
Cl
di
02
ft2
C2
d.2
a*
h
Ci
d4
as
63
ca
da
61 Cl di
04 62 C2 di
1 63 C8 da
+ 64
«x
61
Cl
di
Oi
64
Ci
d*
02
62
C2
do
as
63
C3
ds
ai Cl di
^2 C2 d^
as C3 da
-Ci
at
64
C4
di
ai
61
Cl
di
02
62
C2
d2
as
63
C3
ds
a\ 61 di
02 62 d2
03 63 d3
+ di
ai 61 Cl
tta 62 C2
as 63 C3
When a determinant is thus expressed it is said to be expanded
according to the elements of the kth row. We shall call the
coefficient of an element the quantity which multiplies it in
the expansion.
Then the coefficient of an element is plus or minus the corre-
sponding minor according as the nu7nber of the row added to the
number of the column is eveii or odd.
PROPERTIES OF DETERMINANTS
9
The coefficient of a^ shall be denoted by A^, that of h^ by B^,
and so on. Then
= b,B, + b,B,+ b,B,
= c^C^ + CgCg + C3C3
= a^A^+\B^+c^C^
= a^A^+\B^+c^C^.
4. If any two rows {or columns) of a determinant are inter-
changed, the sign of the determinant is changed.
For suppose the determinant is expanded as in theorem 3, and
that two rows other than that used in the expansion be inter-
changed. A similar interchange takes place in the minors of the
expansion. Hence, if the theorem is true for each of the minors,
it is true for the determinant. In other words, if the theorem is
true for determinants of any order, it is true for those of the next
higher order. But the theorem is certainly true for determinants
of the second order. Hence it is always true.
5. If two rows (or cohtmns) of a determinant are the same, the
determinant is equal to zero.
Let a determinant with two rows the same be expanded accord-
ing to the elements of some other row. Each minor of the expan-
sion has two rows the same. Hence, if the theorem is true for
determinants of any order, it is true for determinants of the next
higher order. But the theorem is certainly true for determinants
of the second order, for
afi^-
^1^1 ~ ^•
Hence it is universally true.
10
ELIMINATION
6. Tlie sum of the products obtained by multiplying the elements
of any row {or column) by the coefficients of the corresponding ele-
ments of some other row {or column) is zero.
Consider, for example,
= a,A, + b,B^ +c^C,+ d,D^
If we replace a^, b^, c^, d^, on the right-hand side of this equation
by a^, b^, c^, d^, the same substitution must be made on the left-
hand side. Then we have
= a^A^ + b^B^ + c^C^ + dj)^.
«1
\
H
d
«2
K
^2
d,
«S
\
^3
d,
a.
K
^4
d^
«i
\
<^X
d.
«4
h
^4
<
«3
h
«3
d.
a.
K
'^4
<
But the determinant is zero, by theorem 5 ; therefore
a^A^+b,B^+ cf^+ d^D^ = Q.
It is evident that the proof is general and establishes the
theorem.
7. If each element of any row {or column) is multiplied by the
same quantity, the determinant is multiplied by the same quantity.
This follows at once from theorem 3. For example,
8 h
kc^
d.
kc^
d.
kc^
ds
kc^
d.
= kc^ C^ + kc^ C^ -f- kc^ C3 + kc^ (7^
= kic^c,+ c,c^^+ c,c,+ c^c;\
«1
K
^1
d.
«2
b.
c.
d.
«3
h
^3
d.
«4
h
^4
d.
PROPERTIES OF DETERMINANTS
11
8. If each of the elements of any row {or column) is increased
hy the same multiple of the corresponding element of any other
row {or column), the value of the determinant is unchanged.
We wish to show, for example, that
(1)
r
«1 \
h ^1
«2 h ^1 ^2
^3 ^3 Cg d^
*4 ^4 ^4 ^4
a^ b^ + kd^
c, d^
a^ \+M^
^2 ^2
«3 &3+K
^3 dz
a^ h^+kd^
c, d.
(2)
Let the coefficients of the elements in the second column of
(1) be B^, B^, B^, B^. It is evident that these are also the coeffi-
cients of the elements of the second column of (2). Hence (2) is
{\ + kd;) B^ + {b^ + kd^) B^ + (&3 + Mg) B^ + (&, + kd^ B^,
which equals
b^B^ + b^B^ + b^B^ + b,B, + k {d^B^ + c?2^2 + d^B^ + d^B^).
The coefficient of k in this equation is zero, by theorem 6, and
the remaining terms equal the determinant (1). Hence (2) = (1).
It is evident that the proof is general. The following are special
cases : If A; = 1, the elements of one row or column are added to
the corresponding elements of another row or column ; if ^^ = — 1,
the elements of one row or column are subtracted from those of
another row or column.
This theorem is often used in simplifying determinants.
Ex. 1. Consider
1-2 1 2
3-5 3 5
- 1 2 3-4
3-5 2 5
(1)
If the elements of the second cohimn are added to those of the fourth column
this becomes
1-2 1 0
3-5 3 0
-1 2 3-2
8-5 2 0
(2)
12
ELIMINATION
If twice the elements of the first column are added to those of the second
column, (2) becomes
10 1 0
3 13 0 (3)
-10 3-2- ^ '
3 12 0
If the elements of the first column are subtracted from those of the third
column, (3) becomes
10 0 0
3 10 0
-10 4-2
3 1-1 0
(4)
Expressing (4) as the sum of the product of the elements of the first row
and their coefficients, it becomes
and this is equal to
1 0
0 4 -
1-1
14-21
- 1 0
= -2.
Ex, 2. Consider
X
y 1
Xi
yi 1
X2
2/2 1
By successive subtraction of the elements of one row from those of another
we have
X y I
xi yi 1
=
«2 2/2 1
x-Xi 2/ - 2/1 0
Xi ^1 1
^2 2/2 1
X -Xi 2/ - 2/1 0
SCl -X2 yi - 2/2 0
X2 2/2 1
\x Xi 2/ 2/1 ^ijg |f^g{ transformation being made by
X1-X2 2/1-2/2 ', „ *
theorem 3.
4. Solution of n linear equations containing n unknown quan-
tities, when the determinant of the coefficients of the unknown
quantities is not zero. We are now prepared to show that the
method used in § 1 to solve equations with two or three unknown
quantities can be so generalized as to apply to any system of
equations of the first degree in which the number of equations is
equal to the number of the unknown quantities. For convenience
we will take the case of four equations, but the student will readily
see that the method is perfectly general.
n EQUATIONS WITH n UNKNOWNS
Cousider the equations
a^x + bji/ + c^z + djW + e^ = 0,
a^x + b^y + c^ + d^w + e^ = 0,
a^x + h^y + C32 + d^w + e^ = 0,
a^a; + &42/ + c^z + f^^w + e^ = 0.
13
(1)
(2)
(3)
(4)
«1
^
^1
^1
a^
K
^2
^2
«8
h
C3
^3
^4
K
^'4
d.
Let the determinant of the coefficients of the unknown quan-
tities X, y, z, w be denoted by D, so that
D =
and let A^ denote the coefficient of a^, B^ the coefficient of &^, and
so on. We assume i> ^ 0.
If now we multiply (1) by A^, (2) by ^„ (3) by A^, (4) by A^,
and add the results, we have, by theorems 3 and 6, § 3,
Dx + c^A^ + ^2^2 + ^3^3 + ^4^4 = 0- (5)
Similarly, by using B^, B^, B^, B^ as multipliers, we have
By + e,B^ + e^B^ + ^3^3 + e^B^ = 0 ; (6)
by using C^, C^, C^, C^ as multipliers, we have
Bz + e^C,+ e,C,+ e^C,+ e^C, = 0; (7)
and by using D^, D^, D^, D^ as multipliers, we have
Dw + e^D^ + ej)^ + e3Z>3 + ej)^ = 0. (8)
Now it is clear that any values of x, y, z, w which satisfy (1),
(2), (3), (4) satisfy also (5), (6), (7), (8). Conversely, any values
which satisfy (5), (6), (7), (8) satisfy also (1), (2), (3), (4). For if
we multiply (5) by a^, (6) by h^, (7) by c^ (8) by d^, and add, we
obtain (1). Similarly (2), (3), (4) can be obtained from (5), (6),
(7), (8). Hence (1), (2), (3), (4) and (5), (6), (7), (8) are equivalent
equations.
14
ELIMINATION
Now e^A, + e^A^ + e^A^ + e^A^ =
e,B^^e^B^-^e,B^ + e,B,=
e,C,+ e,C,+ e^C,+ e,C,=
«,
^
<^1
d.
^.
K
^2
d.
««
h
^8
<
«4
\
^4
d.
«1
H
'^l
d.
^2
^2
Cg
d.
«8
«3
H
d.
a.
«4
H
d.
tj 5j e^ d^
«3 ^3 ^8 ^8
*4 ^4 ^4 ^4
and
6,i),+ e,l>2+e3A+«4^4 =
^1 ^1 ^1 ^1
^2 62 c^ e^
^3 ^3 ^3 ^3
^4 &4 C4 «^
Hence the solution of (5), (6), (7), and (8) is
X —
«1
\
\
^.
««
K
«2
«!,
««
h
<^3
^8
«4
h
<^,
<
a,
\
^1
d.
«2
K
«2
d.
«8
■K
^3
<
^4
h
C4
^4
2/ =
«1
«1
«1
^X
«2
«2
^2
d.
«3
«3
^8
d.
^4
«4
«4
d.
«1
^
^1
d.
^2
^2
^2
d„
a^
&3
C3
d.
^4
&4
^4
d.
a.
\
«i
^X
«2
K
«2
«^2
«8
h
«3
«^8
«4
K
«4
<
«t
h
^'l
d.
«2
K
^2
d.
««
K
<^3
d.
«4
h
^4
<
«i
^
Ci
«1
^2
^0
^2
^2
«3
^3
«3
«3
«4
K
C4
«4
«1
K
Cl
^1
«2
K
«2
«^2
«3
h
«8
«58
«4
h
C4
d.
and this is the solution, and the only solution, of (1), (2), (3), (4).
n EQUATIONS WITH MORE THAN n UNKNOWNS 15
Hence we may state the following important theorem :
Any system of n linear equations containing n unknown quan-
tities has one and only one solution when the determinant formed
by the coejfficients of the unknown quantities is not zero.
This solution may be written down at once, for each unknown
quantity is equal to mmus a fraction, of which the denominator is
the determinant of the coefficients and the numerator is a similar
determinant formed by replacing the coefficients of that unknown
quantity by the absolute terms.
Ex. 1.
Ex. 2.
3x + 52/-4 = 0,
2a; -32/ + 7 = 0.
-4
5 3
-4
7
-3 23 2
7
29
3
6 " 19' ^^ 3
6
~ 19
2
-3 2
2x-3y+ z-l = 0,
4x + 5y-22 + 2 = 0,
x-2 7/ + 32-3 = 0.
-3
1-3 1
2 5-2
3-2 3
2-3 1
4 5-2
1-2 3
= 0,
y = -
2
-3
-1
4
5
2
1
-2
-3
2
-3
1
4
5
-2
1
-2
3
2-1 1
4 2-2
1-3 3
-3 1
5-2
-2 3
= 1.
= 0,
5. Systems of n linear equations containing more than n un-
known quantities. When in a set of linear equations the number
of equations is less than the number of unknown quantities, the
equations have usually an infinite number of solutions, but may
have none. The general method of procedure in solving them is to
pick out a number of the unknown quantities equal to the number
of the equations and having the determinant of their coefficients
IQ ELIMINATION
not zero. These are solved by the method of § 4. We then have
these unknown quantities expressed in terms of the others.
Ex.1. 2x + Sy+ 2 + 4 = 0,
x-2y + 3z + 2 = 0.
I> = \' ?l=-7.
If we choose x and y for the unknown quantities, we have
12 31
|l -2|
Then, solving as in § 4, we have
I z + 4 31
3z + 2 -2 11
a; = — 1 — - = z — ^,
2 3 7 '
|l -2|
12 z + 4|
1 3z + 2 5
« = — —. — ^ = -z :
" 2 3 7
|l -2|
and since z may be given any value whatever, the equations have an infinite
number of solutions.
Ex.2. 2x + 3y+ z + 4 = 0,
2x + 3y + 2z + 3 = 0.
If we choose to solve for x and y, we have
12 3
2 3
I) = \l f| = 0.
But if we choose to solve for y and z, we have
n— ^ — ^
-^-13 2|-'^-
The solutions are y = — ^x — ^,
z = l.
It is possible that no selection of the unknown quantities will
lead to a determmaut of the coefficients which is not zero. In this
case the equations may have no solution. The discussion is too
complex for this book, but the student will probably have no diffi-
culty with the cases likely to occur in practice.
n EQUATIONS WITH n UNKNOWNS 17
Ex.3. 2x + Sy + z + 4 = 0,
2x + 3y + z + S = 0.
The determinant of any pair of unknown equations is zero. By subtracting
the second equation from the first we have 1 = 0, showing the equations to be
contradictory.
6. Systems of n linear equations containing n unknown quan-
tities, when the determinant of the coefficients of the unknown
quantities is zero. Consider again equations (1), (2), (3), (4) of
§ 4, but with the assumption that Z> = 0. We may proceed exactly
as in § 4, but equations (5), (6), (7), (8) do not now contain the
unknown quantities. In fact, these equations are, in general, con-
tradictory, and consequently equations (1), (2), (3), (4) have, in
general, no solution,
Ex. 1. X - 2/ + z + 3 = 0,
2x-{-y + Sz + 1 = 0,
« + 2y + 2z + 4 = 0.
Here B
1-11
2 13
1 2 2
0.
Eliminating y and z by the method of § 4, we have 0 x — 24 = 0, which is
absurd. Hence the equations have no solution.
It is, of course, possible that when D = 0 each of the other
determinants in (5), (6), (7), (8) Ib also zero. Each of these equa-
tions is then simply 0 = 0, and gives no direct information about
the solutions of (1), (2), (3), (4). As a matter of fact, in this case,
(1), (2), (3), (4) have, in general, an infinite number of solutions,
but may, under special conditions, have no solutions.
The general discussion is too complex to be given here. We
shall simply state the following theorem :
A set of linear equations containing n unknown quantities has,
in general, no solution when the determinant of the coefficients of
the unknoivn quMntities is zero, but may, under certain conditions,
have an infinite number of solutions.
13 ELIMINATION
In practice, one of the n equations may be temporarily set aside,
and the other w — 1 equations, which contain 11 unknown quan-
tities, may be examined by the method of § 5. If these equations
can be solved, the solution can be tested in the equation which
has been set aside.
Ex. 2. 2x- 32/+ 2-1 = 0,
X- 22/ + 3z + 4 = 0,
7a;-ll?/ + 6z+ 1 = 0.
If the method of § 4 is used, the result is 0 = 0. Solving the first two equa-
tions for X and y, we have
ic = 7 z + 14,
y = 5 z + 9,
and these results are found on trial to satisfy the last of the given equations.
Since z may have any value, the equations have an infinite number of solutions.
7. Systems of linear equations in which the number of the
equations is greater than that of the unknown quantities. If
there are more equations of the first degree than there are unknown
quantities, there will be, in general, no values of the imknown
quantities which satisfy all equations. There may be such values,
however, when certain relations exist among the coefficients of
the equations. To obtain these relations we may pick out a num-
ber of equations equal to the number of tlie unknown quantities
and solve them. If the solution is substituted in the remainmg
equations, there will result certain expressions in the coefficients
which must be zero if the equations are to be satisfied.
The most important case is that in which there are n-\-l equa-
tions containing n unknown quantities. For example, consider
a^x + l^y -f- Cj2 + d^ = Q,
a^x + hju + c^ + rf., = 0,
a^x -h &^ -h c„z + <3 = 0,
a, A' -\-'^i^y + c^z + (l^ = 0.
n + 1 EQUATIONS WITH n UNKNOWNS
19
The solution of the first three equations, if
^0, is (§4)
x =
d,
^
<^X
d.
(>.
^2
ds
K
^3
«i
\
'^l
«2
K
«2
«3
^3
^3
^
^'i
^1
*2
^2
«!,
*3
^3
d.
«1
K
^1
a^
^2
.«2
«3
^3
«3
«1
d.
^1
«2
d.
<^2
«3
ds
C3
«1
\
«1
«2
b.
^2
«3
^
C3
«1
«1
^.
«2
^2
d.
«3
^3
d.
«1
^
h
«2
*2
Cg
«3
^
«3
«1
^
^,
«2
62
^2
«3
*3
«53
«1
^'l
«1
«2
K
^2
«3
*3
^3
Substituting these values in the first member of the last equa-
tion, we have
— a.
h ^1
+ K
\ d^
". d^
— c.
a^ \ d^
^2 h ^2
+ d.
«3 ^3 ^3
b, c,
\ ^2
«2 K ^2
which, by theorem 3, § 3, is the same as
h
«i
^^
K
^2
^^2
\
C3
d.
b.
^4
d.
^2 «2
20
ELIMINATION
a,
\
h
rf
«2
K
^2
d.
^8
h
^3
d.
«4
h
^4
d
Hence, in order that the last equation may be satisfied, we must
have
= 0.
Extending this to any number of variables, we have the theorem :
In order that a system of n + 1 linear equations containing n
unknovm quantities shall have a solution, it is necessary that the
determinant formed from the coefficients of the unknown quantities
and the absolute terms shall he zero.
Ex. 1.
x+ 2/+ z-2 = 0,
2x+ y- z + 3 = 0,
x-22/-3« + 4 = 0,
5x-3y-4« + l = 0.
Here
111-2
2 1-1 3
1 _2 -3 4
6-3-4 1
= 0,
showing that if the first three equations have a solution it will satisfy the fourth
equation. In fact, the solution is x = 1, y = — 2, z = 3.
It should be noted that the converse of the theorem stated is
not necessarily true. All that has been proved is that if n of the
equations have a solution, that solution satisfies the {n + l)st equa-
tion when the determinant is zero. But the determinant may be
zero when the equations are contradictory.
Ex.2.
2x-3?/+ 2 + 1 = 0,
2x-3y + 52 + 2 = 0,
2x-3i/-6z-3 = 0,
2x-3?y + 2z-8 = 0.
Here
2-311
2-362
2 -3 -6 -3
2-3 2-8
= 0,
but any three of the equations may be seen to be contradictory by the method
of §8.
LINEAR HOMOGENEOUS EQUATIONS 21
8. Linear homogeneous equations. An equation is homoge-
neous with respect to the unknown quantities when the sum of
the exponents of the unknown quantities is the same in each term.
In particular an equation of the first degree is homogeneous when
each of the terms contains one of the unknown quantities; for
example,
where x^, x^, x^, x^ are the unknown quantities.
Tliis equation is, of course, satisfied by placing x^ = 0, x^ = 0,
x^ = 0, x^ = 0, but in practice tliis solution is generally unimportant.
In such equations, in fact, it is usually the ratios of the unknown
quantities wliich are important ; for if each unknown quantity is
multiplied by the same number, the equation is unaltered. In fact,
if we place
X,
X
1 "^2 "^3
the homogeneous equation just written becomes the non-homogene-
ous equation
a^ -H a^ + a^ + a^ = 0.
In this manner a set of homogeneous equations containing n
unknown quantities may be reduced to a set of non-homogeneous
equations containmg ii — \ unknown quantities by dividing each
equation by one of the unknown quantities. The methods of the
previous articles may then be used. But tliis method of proced-
ure is open to the objection that the unknown quantity by which
the equations are divided may possibly be zero when the division
is invalid. It is better, therefore, to handle the homogeneous equa-
tions as they stand, slightly modifying the methods used for non-
homogeneous equations in a manner which will be clear from the
examples.
Ex. 1, a^xx + a'^X'!, + 03X3 + ^4X4 = 0,
61X1 + 62X2 + 63X3 + &4X4 = 0, (1)
C]X\ + C2X2 + CgXs + C4X4 = 0.
'>')
ELIMINATION
We will handle this by the method of § 4, in that we temporarily look upon
Xi, Xa, Xa as the unknown quantities. We have, in the first place,
= 0,
61
02
62
63 Xi +
O4X4 fl2 O3
64X4 62 &3
Cl
C2
C3
C4X4 C2 Cs
61
02
62
03
63
X2 +
Oj 04X4 03
6i 64X4 63
Cl
C2
C3
Cl C4X4 C3
ai
61
O2
62
03
63
X8 +
Oi C2 O4X4
61 62 64X4
Cl
Ca
cs
Cl C2 C4X4
= 0,
= 0,
which may be written as
Oi
02
03
61
fe2
&3
Xi +
Cl
C2
C3
Oi
02
as
h
62
63
X2-
Cl
C2
C3 1
Ox
O2
O3
61
62
63
X3 +
Cl
C2
Cs
O2
as
04
62
hs
64
C2
Cs
C4
fli
61
Oa
&3
O4
64
Cl
Cs
C4
Oi
02
62
O4
hi
Cl
C2
Ci
X4 = 0,
X4 = 0,
X4 = 0.
(2)
(3)
(4)
From these follow :
Xi : X2 : Xs : X4 =
02
O3
04
62
63
64
: —
C2
Cs
C4
Oi
03
04
1
61
63
64
• 1
Cl
Cs
C4
1
Oi
02
O4
&1
62
h
: —
Cl
C2
C4
ai
02
as
61
62
63
Cl
C2
Cs
(5)
The result (5) holds even when one or more, but not all, of the determinants
involved are equal to zero. Then the corresponding unknown quantities are
equal to zero. For example, if
Oi
03
04
61
h
64
= 0,
Cl
Cs
C4
Ol
02
04
61
62
h
Cl
C2
C4
and the other determinants in (5) are not zero, (3) and (4) show that X2 = 0 and
Xs = 0, while (2) shows that the ratio of Xi and X4 are correctly given by (5).
If all the determinants in (5) are zero, the values of the unknowns are not
thereby determined. In this case, two of the equations (1) should be solved for
two of the unknown quantities in terms of the others, and the results tested for
the last equations.
ELIM^ANTS 23
It should be noted that contradictory equations cannot occur. The student
should compare the contradictory equations
2x-3y + 4 = 0,
2x-3y -2=0,
with the homogeneous equations
2xi — 3iC2+4 Xs = 0,
2 »i — 3 a^ — 2 ajs = 0.
By subtracting one equation from the other we have
6x3 = 0,
whence X3 = 0 and Xi : X2 = 3 : 2.
Ex. 2. The four equations
aiXi + aix<2. + 03X3 + 04X4 = 0,
hyxx + 623:2 + 63X3 + 643:4 = 0,
CiXi + C2X2 + C3X3 + C4X4 = 0, .
d\Xx + ^23:2 + ^3X3 + ^4X4 = 0,
have, of course, the common solutious, xi = 0, X2 = 0, X3 = 0, X4 = 0. In order
that they may also be satisfied by the same ratios of the unknown quantities, it
is necessary that
a\ a<i, as 04
61 62 63 64
C\ Ci C3 C4
d\ d^ ds (^4
= 0.
The proof is as in § 7. The condition is also sufiBcient, for the proof of § 7
shows that if three of the equations have a solution, that will also be a solution
of the fourth equation ; and, as just noted, three homogeneous equations always
have a solution.
9. Eliminants. The result of eliminating all the unknown
quantities from two or more equations is an equation the left-
hand member of wliich is called the eliminant, or resultant, of
the given equations. The following cases are important:
1. w + 1 non-homogeneous linear equations with n unknown
quantities. To eliminate the unknown quantities, we may solve
n of the equations and substitute the solutions in the remaining
equation. The work and the result are as in § 7 ; that is,
The eliminant of n + 1 non-homogeneous equations with n utv-
known quantities is equal to the determinant of the coefficients and
the absolute terms.
24 ELIMINATION
2. n homogeneous linear equations with n unknown quantities.
To eliminate the unknown quantities, we may solve n-1 equa-
tions for their ratios and substitute the results in the remaining
equation. The work and the result are as in § 8 ; that is,
The eliminant of n homogeneous equatwns with n unknown quan-
tities is equal to the determinant of the coefficients.
3. Two equations containing one unknown quantity. Let it be
required to eliminate x between the equations
a,ar^+M + ^i = ^' (^)
a,.r^+6,a^ + C3 = 0. (2)
If we multiply each equation by x, we have
a^n? + h^a?' + c^x = 0, (3)
and a^x^ + h^a? + c„x = Q. (4)
These four equations may now be considered as Hnear in the
three unknown quantities a?, yp-, and x. Elimination gives, by 1,
= 0. (5)
It is clear that if equations (1) and (2) have a common solution,
equation (5) must be true. Conversely, it may be shown that if (5)
is true, (1) and (2) must have a common solution ; but this proof is
too long to be given here.
The method used in the above problem may be used for
equations of any degree and is known as Sylvester's method of
elimination. It consists in multiplying the given equations by
successive powers of x until we have one more equation than we
have powers of x. The eliminant is then found as in 1.
The method may also be used to eliminate one of the unknown
quantities from two equations containing two unknown quantities.
0
«i
\
^l
0
«2
K
^2
«1
\
^1
0
O-c,
K
Co
0
PROBLEMS
25
PROBLEMS
Find the value of each of the following determinants i
1.
4.
14 51
3 6
9 \^
^' \y
n IX
^- X2
I 1 o|
[10 4|"
II -71
|2 3|
12 3
2 3 1
3 12
7.
10.
1
0 1
3
2 0
1
1 0
0
a b
a
0 c
b
c 0
a
h g
h
b f
9
f c
1
1 1
a
b c
a2
62
C2|
11.
12.
13.
X y 1
12-1
4 3 2
1
1
1
0
0
1
1
1
1
0
1
1
1
1
0
1
0 ai bi Ci
0 ag 62 C2
«! bi ci 0
02 62 C2 0
Prove the following relations ;
14.
15.
16.
20.
21.
22.
23.
4 2 1
1
-2
-3
1 2 3
3 4 2
= 0.
17.
-2
1
3
=
2 1 3
6 6 3
-3
1
2
3 1 2
5 4 1
3 2 1
= 0.
X y
z
1 1 1
2 1 1
18.
x2 y2
Z2
= xyz
xyz
X8 w3
Z8
x2 2/2 z'^
a 0 0 6 0
0 a 0 0 b
X y I z w
= (ad-
-6c)2
1 1
1
c 0 0 d 0
19.
X y
Z
= {x-y){y-z){i
0 c 0 0 d
X2 2/2
22
ai 61 0 0
a2 62 0 0
_ ffli &i
Cl
di.
0 0 ci di
02 ^2
C2
da
0 0 C2 ^2
1 4-3
6
5
0
2 1
0 6-8
- 1
1 -
-6
-3 4
2-3 4
2
—
3
8
4 -3
5 1
3
4
4
1
2
5
«! + di 6i
a2 + d2 bo
as + dz 63
«! 61
a2 62
as 63
ao 3ai 3a2 as 0
0 ao 3ai 3a2 as
ao 2ai a2 0 0
0 ao 2ai 02 0
0
0
ao 2 oi a2
ao
di 61
Cl
d2 62
ca
ds 63
cs
ai 2a2 as 0
0 tti 2a2 as
ao 2 ai a2 0
0 ao 2ai ao
ao {a^a^ - G aoajaoas + 4 aoa| + 4 a^as - 3 af aa^}.
26 ELIMINATION
Solve the following equations :
24.
|4-x 3
3 d-x
= 0.
25.
1-x 2
2 3-x
3 5
3
6
— X
= 0.
Write the following equations in their expanded forms :
26.
27.
X y \
1
X2 + 2/2 X
y
1
2-3 1
=
0.
28.
5 1
2
1
1 6 1
13 2
-3
1
2 -1
-1
1
x^ + y^ X y
1
1 0 1
1 1 0
0 0 0
1
1
1
= 0.
29.
a — X h
h b — X
= 0.
a
— X
h
9
30.
h
b-x
f
= 0.
9
f
C — X
= 0.
Solve the following equations ;
31. 4x-5y + 0 = 0,
7x-9y + 11 = 0.
32. X + 2 y - 2 + 3 = 0,
2x- y -5 = 0,
X +22-8 = 0.
y z
1 1 .
- + - = 4.
z X
34. 2x+ 4y + 32 - 2 = 0,
X- 5j/+ 2+1 = 0,
3x+ lOy + 52 - 5 = 0.
35. 2x+ y+ 2 + 2 = 0,
6x + 2y + 32 + 6 = 0,
2x + 3y-22 + 2 = 0.
36. X + 2/ + 92 -7 = 0,
5x - y + fi2 - 5 = 0,
3x- j/ + 3a-2 = 0.
37. 10x-3y + 122- 5 = 0,
4x- y+ 62-3 = 0,
5x-2y + 32 =0.
38. X + y + z - a,
y + z + w = b,
2 + 10 + X = C,
w + X + y = d.
39. lOxi + 4x2 + 6x3 = 0,
3xi+ X2 + 2x3 = 0.
40. xi + 5x2 + 3x3 = 0,
3 xi + 3 X2 + X3 = 0.
41. 2Xi + 4X2 + X3 = 0,
3xi + 6x2 -X3 = 0.
42. 2xi+ X2-5X3+ X4 = 0,
3xi -2x2 - 4x3 - 2x4 = 0,
«i + X2 + 2 X3 - X4 = 0.
43. 2xi - 3x2 + 2x3 - 3x4 = 0,
4xi + 5x2 + 4x3 - 6x4 = 0,
3 xi - 7 X2 - 2 X3 + 3 X4 = 0.
44. 7xi - 5x2 + 3x3 - 4x4 = 0,
3xi+ 2x2- 6x3 + 9X4 = 0,
6X1 - 16 xo + 21 X3 - 35 X4 = 0.
PROBLEMS
27
Find whether or not the equations in each of the following examples have
a common solution :
45. 2x - 2/ + 3 = 0,
3a; + y- 1 = 0,
3x-4y + 10 = 0.
46.
6X-2 2/ + 7 = 0,
3a;- y + 6 = 0,
x + 32/-l = 0.
47. X- 2y+ 3z- 1 = 0,
2x+ y- 2 + 1 = 0,
X- 3?/+ 2z + 2 = 0,
X - 19 y + 22 2 - 4 = 0.
48. X-22/+ 1 = 0,
2/-2z + 2 = 0,
2-2x + 3 = 0,
X + ?/ + 2 = 0.
49. For what values of a are the
following equations consistent ?
X + a'h) + a = 0,
ox + y + a2 = 0,
a^x + ay + 1 = 0.
50. Eliminate x from the equations
X2/ + 3 X + 1 = 0,
2x2/ -4?/ + 2 = 0.
51. Eliminate x from the equations
xy2 + 22/ + 3 = 0,
xj/ + 4 X + 1 = 0.
52. Eliminate x and 2 from the
equations
xy + 2/z — X + 2 + 2 = 0,
X2/-2x + y + 2 + 2 = 0,
X + 3Z-2 =0.
53. Find the condition that
ax^ + 6x + c = 0,
and x2 = 1,
have a common root.
54. Show that the condition that
ax2 + 6x + c = 0,
and x^ = 1,
have a common root is
= 0.
55, Show that if
flix + hiy + ci = 0,
a^x + hiv + C2 = 0,
asx + ftsy + C3 = 0,
have a common solution, there can
always be found three numbere Z, A:,
?n, such that
«li + ^2^ + «3"i = 0»
61/ + 62^ + ftsWi- = 0,
Cii + C2^ + Cz>n = 0.
a
6
c
6
c
a
c
a
b
CHAPTEE II
GRAPHICAL REPRESENTATION
10. Real number. The science of mathematics deals with vari-
ous kinds of numbers, each of which has arisen through the desire
to perform, without restriction, some one of the fundamental oper-
ations. The simplest numbers are the positive integers, or whole
numbers. If one restricts himself to the use of these, he may add
or multiply together any two of them without obtaining a new
kind of number ; but he may not divide one number by another not
exactly contained in it, nor subtract a larger number from a smaller.
In order that division may always be performed, the common frac-
tions, which are the quotients of one integer divided by another,
are necessary. In order that subtraction may always be possible,
the idea of a negative number must be introduced. The integers
and fractions, both positive and negative, together form the class
of rational numbers. On these numbers the operations of addition,
subtraction, multiplication, and division may always be performed
without leading to a new kind of number.
The operation of evolution, however, leads to two new kinds of
numbers, — the irrational, exemplified by V2 ; and the complex, of
wliich V— 2 is an example. The complex numbers will be noticed
in § 12 ; we shall here speak only of the irrational numbers. An
irrational number is defined as one wliich cannot be expressed
exactly as an integer or a common fraction, but which may be so
expressed approximately to any required degree of accuracy. The
simplest examples are the roots of rational numbers ; for example,
V? may be approximately expressed as f f^^ ro-Uh ^^^■' ^^^^ c^^"
not be expressed exactly. There are also irrational numbers which
are not the roots of numbers and cannot be expressed by means of
radical signs. A familiar exj^mple is the number 7r = 3.14159---.
An irrational number may be either positive or negative. The
28
ZEKO AND INFINITY 29
rational and the irrational numbers together form the class of real
numbers.
A rigorous investigation of the nature and properties of these
numbers, especially of the irrational numbers, is too advanced for
this book. An elementary discussion, however, is given in any
course in algebra, and is here assumed as known.
The real numbers may be represented graphically on a number
scale, constructed as follows :
On any straight line assume a — j — i — i — i — i — i — i — i-i-f
•^ *^ -U -3 -2 -1 0 1 2 S U
fixed point 0 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. 1, it is usual, but not necessary, to lay off
the positive numbers to the right of 0 and the negative numbers
to the left. 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.
11. Zero and infinity. There are two mathematical concepts
usually included in the number series, for which special rules of
operation are needed. These are zero, represented by the symbol 0,
and infinity, represented by the symbol oo.
Zero arises in the first place by subtracting a quantity from an
equal quantity ; thus, a — a = 0. It signifies in this sense the
absence of quantity — nothing. It cannot, then, either operate
upon a quantity or be operated upon; for all operations imply
the existence of the quantities concerned. Literally, then, the
n n
expressions a x 0, - > - > are meaningless. However, it is possible
to put into these symbols conventional meanings, as follows:
Take the three expressions ax, -> -> and consider what hap-
a X
pens when x is taken smaller and smaller, constantly nearer to
zero but never equal to it. It requires only elementary arith-
metic to see that ax and - may each be made as small as we
a
30 GRAPHICAL KEPllESENTATlOX
please by taking a; sufficiently small, while - becomes indefinitely
great as x decreases, and may be made larger than any quantity
we may choose to name. We may express the first two results
concisely by the formulas
a X 0 = 0, - = 0.
a
We can express the last result in a formula, however, only by
introducing the concept infinity. Wlien the value of a quantity
is indefinite, but the quantity is increasing or decreasing in such
a way that its numerical value is greater than any assigned quan-
tity, however great, it is said to become infinite. It is then denoted
by the symbol oo, called infinity. We can accordingly express our
third result by the formula
a
0 = "'
which means that when the denominator of a fraction decreases, be-
coming constantly nearer to zero, the value of the fraction increases
and becomes greater than any quantity which can be named.
The symbols a ^
a X 00, — > —
00 a
are also literally meaningless. We can, however, give a conven-
tional meaning to them by writing ax, -> -, and studying the
X a
effect of increasing x indefinitely. Elementary arithmetic leads
to the results expressed by the formulas
a X 00 = oo, — = 0, — = 00.
CO a
Two other forms also occur in practice, namely, - and — • These
0 00
arise when we have a fraction - in which the numerator and
the denominator either approach zero together or increase indefi-
nitely together. The value of the fraction cannot be determined
unless we know a law to govern x and y. These fractions are
consequently called indeterminate forms, and will be considered
later in the course.
COMPLEX NUMBERS 31
Neither zero nor infinity can be said to have an intrinsic alge-
braic sign. In some cases a quantity may increase in value,
remaining always positive. It is then said to be + co. At other
times it may increase numerically, remaining always negative.
It is then said to be — co. Often, however, the quantity is iadefi-
nitely great in such a way that the sign is ambiguous. An
example is tan 90°. If an acute angle is made nearer and nearer
to 90°, its tangent increases indefinitely, remaining positive. But
if an obtuse angle is made nearer and nearer to 90°, its tan-
gent increases indefinitely, remaining negative. Hence we say
tan 90° = 00, and no algebraic sign can be attached to it.
Similar considerations hold for the sign of zero.
12. Complex numbers. If one restricts himself to the use of
the real numbers, named in § 10, it is impossible to perform the
operation of evolution without exception ; for the even root of a
negative number is not a real number. It is therefore necessary,
if the generality of all algebraic operations is to be maintained, to
introduce a new kind of number, called a complex number. These
numbers will be used very little in this volume, and the following
resume of the matter usually contained in algebra is sufficient for
our present purposes. A further discussion will be given in the
second volume.
The imaginary unit is V— 1, and is denoted by i. Then
e=-i.
By multiplying this equation successively by i, we find
i^=— i, i* = 1, i^ = i, i^= — l, • • • ;
and, in general, — —
i'' = l, 'i" + ' = t, i"'^' = -l, i^^+'" = -i,
where k is zero or any integer.
If h is any real number, the product M is called a pure imagi-
nary number. The square root of any negative number is pure
imaginary ; thus,
32 GRAPHICAL REPRESENTATION
If a and h are any two real numbers, the combination a + bi
is caUed a complex imaginary number, or, more simply, a complex
number. A complex number reduces to a pure imaginary number
when a = 0, and to a real number when 5 = 0. If a = 0 and 5 = 0,
the complex number a + bi = 0; and conversely, if a + bi = 0,
then a = 0 and 5 = 0.
All operations with complex numbers are carried out by using
the ordinary laws of algebra and replacing all powers of i by their
values just determined.
Ex. 1. V-3 X V32 z= iVs X i V2 = ?:2 Ve = - Ve.
3j-V34_3 + 2i 2 + 2i_6 + 10i + 4i2 2 + lOi l+sV^
Ex. 2.
4 2-2i 2 + 2i 4-4 i^
Two complex numbers such as a + 5* and a — bi, where a and b
have the same values in each, are called conjugate complex numbers.
Their product is a real number ; thus,
(a-{-bi)(a-bi) = a^-\.b\
It is clear that the complex numbers have no place on the num-
ber scale of § 10.
13. Addition of segments 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 consid-
ered, and consequently it is immaterial whether the line be called
AB or BA ; but in the work to follow it is often important to con-
sider the direction of the line as well. Accordingly, if the direction
of the line is considered as from A to B, it is called ^j5; but if
the direction is considered from B to A, it is called BA. It will
^e seen later that the distinction
-4 B c between AB and BA is the same
Fig. 2 ^^ t-hat 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 begin-
ning of the second. The segment AC is called the sum of AB and
BC, and is expressed by the equation
AB+BC=AC. (1)
SEGMENTS OF A STEAIGHT LINE 33
This is clearly true if the points are in the position of fig. 2, but
it is equally true when the points are in the position of fig. 3,
Here the line BC, being opposite in
direction to AB, cancels part of it, j[ J ^
leaving ^C. F^,, 3
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)
If the point C is at the left of A, as in fig. 4, we still have
4B + BC = AC, where AC = -CAhy (2). ,
It is evident that this addition
— ^ '^ ^ is analogous to algebraic addition,
■pj^ ^ and that this sum may be an arith-
metical difference.
From (1) we may obtain by transposition a formula for sub-
traction, namely,
BC = AC-AB. (3)
This is universally true since (1) is universally true.
This result is particularly important when applied to segments
of the number scale of § 10. For if x is any number corresponding
to the point M, we may always place x = OM, since both x and OM
are positive when M is at the right of 0, and both x and OM are
negative when M is at the left of 0. Now let 31^ and M^ be any
two points, and let x^ = OM^ and x^ = OM^. Then
3I^M^ = OM^ - OM^ = «2 - ^r
On the other hand,
M^M^ = OM^ — 031^ = x^—x,_= — M^M^.
It is clear that the segment 31^31^ is positive when 31^ is at the
right of Jtfp and is negative when M„ is at the left of 31^
Hence, the length and the sign of any segment of the number
scale is found ly 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 ser^m^ent.
34
GRAPHICAL REPKESENTATION
14. Projection. Let AB and MN (figs. 5, 6) be any two straight
lines in the same plane, the positive directions of which are respec-
tively AB and MN. From A and B draw straight lines perpendicu-
lar to 3IJSf, intersecting it at points A' and B' respectively. Then A' B'
is the projection oi AB on MN, and is positive if it has the direction
MN (fig. 5), and is negative if it has the direction NM (fig. 6).
Denote the angle between MN and AB by ^, and draw A C par-
allel to 3IN. Then in both cases, by trigonometry,
AC=ABcos(f>.
But AC=A'B', and therefore
A'B' = AB cos </).
Hence, to find the projection of one straight line upon a second,
multiply the length of the first hy the cosine of the angle hetiveen the
2)ositive .directions of the two lines.
Ex. It is customary in meclianics to represent a foi'ce by a straight line,
tlie lengtli and tlie direction of wliich denote respectively the magnitude and the
direction of the force. Then the component of the force in any direction is the
projection upon that direction of the line which represents the force.
—N
In particular, let Fi and F2, represented respectively by AB and AC (figs.
7, 8), be two forces acting at A along the same line, and let MN be a line
which makes an angle <f> with AB.
COORDINATE AXES
35
The respective components of Fi and F2 are represented by A'B' and A'C,
and the resultant component is represented by A'B' + A'C".
But A'B'=Fi cos 0, and ^'(7'= F2 cos (j> ; hence, by substitution, tlie resultant
component is F^ cos 0 + jFs cos cj). It is to be noted that in fig. 8 Fi and F2 have
opposite signs.
15. The projection of a broken line upon a straight line is defined
as the algebraic sum of the projections of its segments.
Let ABCDE (fig. 9) be a broken line, MN o. straight liue in the
same plane, and AE the straight line
joining the ends of the broken line.
Draw AA', BB', CC, BD\ and
EE' perpendicular to MN', then ^^_
A!B\ B'C, CD', J)'E\ and A'E' ^
are the respective projections on
MNoi AB, BC, CD, DE, and AE.
But
A'B' + B'C + CD' + D'E' = A'E'.
(^T § 13)
Hence, the projection of a broken line upon a straight line is
equal to the projection of the straight line joining its extremities.
Ex. If ABCDE (fig. 9) represents a polygon of forces, we have the result:
the component of the resultant in any direction is the sum of the components
of the forces in that direction.
X-
M
16. Coordinate axes. Let X'X and Y' Y be two number scales
at right angles to each other, with their zero points coincident at 0,
as in fig. 10.
Let P be any point in the
plane, and through P draw
straight Imes perpendicular to
X'X and Y'Y respectively,
intersectmg them at M and iV".
If now, as in § 13, 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.
Y'
Fig. 10
X
36 GRAPHICAL REPRESENTATION
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 per-
pendicular as MP. Then MP = OiV^= y and 0M= x.
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 perpendiculars
to X'X and Y' Y respectively through M and N. These perpen-
diculars intersect at the required point P. Or, as is often more
convenient, a point M corresponding to x may be located on its
number scale, and a perpendicular to X'X may be drawn through
M, and on this perpendicular the value of y laid off. In fig. 10,
for example, M corresponding to x may be found on the scale X'X,
and on the perpendicular to X'X at M, 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 per-
formed when the paper is ruled in squares, as in fig. 10.
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 distanpes from Y' Y and X'X
to the point, the direction as weU as the magnitude of the distances
being taken into account. Instead of designating a point by writing
x = a and y = — b,itis customary to write P(a, — h), the abscissa
always being written first in the parenthesis and separated from
the ordinate by a comma. X'X and Y' Y are called the axes of
coordinates, but are often referred to as the axes of x and y
respectively.
17. Distance between two points. Let P^{x^, y^) and P,{x^,y^
be two points, and at first assume that P^P^ is parallel to one of
the coordinate axes, as OX (fig. 11). Then y^^y^ Now M^M^,
the projection of P^P^ on OX, is evidently equal to P^P^. But
Y M^M^ = x^-x^{^ 13). Hence
I\T^=i X^ Xy (1)
In like manner, if x^ = x^ P^P^ is parallel
to OY, and
Fio.ii P.n=^y-yy (2)
^.
R
Ml 0
M, ^
DISTANCE BETWEEN TWO POINTS
37
If x^ 4-- x^ and y^ ^ y^, P^P^ is not parallel to either axis. Let
the points be situated as in fig. 12, and through ij and P^ draw
straight lines parallel respectively to OX and OY. They will meet
at a point R, the coordinates of which are readily seen to be
{x„ y,). By (1) and (2),
P^B = x^— x^, RP^ = ^2 - Vv
But in the right triangle P^RIl,
PP =
whence, by substitution, we have
Fig. 12
PxP. = ^(^2 -*'i)'+ (2/2-2/1)'- (3)
It is to be noted that there is an ambiguity of algebraic sign on
account of the radical sign. But since P^Il is 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 P^P^ 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
Pi(l, 2), P2(- 1, - 2), and Pz(2, - 5).
Let P (x, y) be the required point. Then
PiP= P^P and P2P = P3P.
But
PiP = V(x - 1)2 + {y- 2)2,
P2P = V(a; + l)2 + (2/ + 2)2,
P3P = V(x - 2)2 + (y+ 5)2.
V(x - 1)2 + {y- 2)2 = V(x + 1)2 + (y + 2)2,
V(a; + 1)2 + {y + 2f = V(x - 2)2 + (y + 5)2,
ehence, by sohition, x = f and y - - ^. Therefore the required point is
38
GRAPHICAL EEPRESENTATION
18. CoUinear points. Let P{x, y) be a point on the straight line
determined hy P^{x^, y^) and Bi{x^, y^, so situated that P^P = lil^P^).
There are three cases to consider according to the position of
the point P. If P is between the points P^ and P (fig. 13), the
3
X
P/
R/
/ 0
M,
M
M, '
Fig. 13
Fig. 14
segments P^P and P^P^ have the same direction, and P^P<P^P^;
accordingly Hs a positive number less than unity. If P is beyond
P^ from i^ (fig. 14), P^P and P^P^ still have the same direction, but
P^P > P^P^ ; therefore / is a positive number greater than unity.
Finally, if P is beyond P^ from P^ (fig. 15),
PyP and P^P, have opposite directions, and
/ is a negative number, its numerical value
ranging all the way from 0 to oc.
In the first case P is called a point of
internal division, and in the last two cases
it is called a point of external division.
In all three figures draw P^M^, PM,
and I^M^ perpendicular to OX. In each
figure OM=OM^ + M^M; and since P^P = I (P^P,), M^M= 1{M^M^,
by geometry.
. • . OM = OM^ + / {M^M^),
I
Fig. 15
whence, by substitution,
x = x^-\- l(x2—x^).
(1)
By drawing lines perpendicular to OY we can prove, in the
same way,
y = y.+ i{y.-yx)- (2)
COLLINEAE POINTS
39
In particular, if P bisects the line P^P^, I = ^, and these formulas
become
2-^2
X =
Ex. 1. Find the coordinates of a point J of the distance from Pi (2, 3) to
P2(3, -3).
If the required point is P(x, y),
X = 2 + f (8 - 2) = 22,
2/ = 3 + -?(-3-3) = f.
Ex. 2. Prove analytically that the straight line dividing two sides of a tri-
angle in the same ratio is parallel to the third side.
Let one side of the triangle coincide with OX, one vertex being at O. Then
the vertices of the triangle are 0(0, 0), A{xi, 0),
B{X2, 2/2) (fig. 16). Let CD divide the sides OB
and AB so that OC = 1{0B) and AD = 1{AB).
If the coordinates of C are denoted by (X3, 1/3)
and those of D by (X4, 2/4), then, by the above
formulas,
Xa = 1x2, 2/3 = ly2,
and X4 = Xi + Z(xo — xi), 2/4 = l)/2-
Since 7/3 = 2/4, CD is parallel to OA. Fig. 16
19. Let us now see what happens as different real values are
assigned to I. Wlien I = 0, P coincides with P^ (fig. 17). As /
increases in value, the
point P moves along the
line toward P^ till, when
/ = 1, it coincides with P,.
As the value of I con-
tinues to increase, the
point P continues to move
along the line away from
^ and in the same direc-
tion as before.
If negative values are assigned to /, in ascending order of numer-
ical magnitude, the point P moves along the line, away from P^, in
the opposite direction from P,.
Fig. 17
40 GRAPHICAL REPRESENTATION
It follows that
^1 + ^ («2 - ^i) and y^ +l{y^— 2/1)
may be made to represent the coordinates of any point of the
straight line determmed by the pomts P^ and i^ by assigning the
appropriate value to I, the range of values for each segment of
the line being indicated in fig. 17.
Ex. Consider the straight line determined by the two points Pi(— 1, — 4)
and P2(5, 6). Any other point P on tliis line has the coordinates
x = -l + 6Z, ^ = -44-10^.
When Z < 0, it is clear that a;< — 1, y < — 4k; hence P lies at the left
of Pi. When 0 < i < 1, it is clear that — l<x<5, — 4<2/<6; hence P
lies between Pi and P2. When l>\, it is clear that a; > 5, y >Q; hence
P lies at the right of P2.
20. 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. Wlien 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 fvmction of the area.
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.
EXAMPLES OF FUNCTIONS
41
Ex. 1, An important use of the graph of a function is in statistical work.
The following table shows the price of standard steel rails per ton in the
respective years:
1895 $24.33
1896 28.00
1897 18.75
1898 17.62
1899 28.12
1900 $32.29
1901 27.33
1902 28.00
1903 28.00
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 doUare, we shall locate the points Pi, P^, . . ., Pio in fig. 18. In
order to study the variation in price, we join these points in succession by straight
Fig. 18
lines. The resulting broken line serves merely to guide the eye from point to
point, and no jwint 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 for isolated
values of x.
Ex. 2. As a second example we take the law that the postage on each ounce or
fraction of an ounce of fii-st-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 x, and each two y
cents of postage by one unit of y, we have the
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 elementary mathe-
matical .symbols. A peculiarity of the graph is
the series of breaks. The lines are not connected ,
but all points of each line represent correspond-
ing values of x and y. Fig. 19
O
42
GKAPHICAL KEPRESENTATION
Ex. 3. As a. third example, differing in type from eacli of the preceding, let
us take the following. While it is known that there i.s some physical law con-
necting the pressure of saturated steam with its temperature, so that to every
temperature there is somo corresponding pressure, this law has not yet been
formulated mathematically. ^'everthele.ss, knowing some corresponding values
of temperature and prassure, we can construct
a curve that is of considerable value. In the
table* below, the temperatures are in degrees
Centigrade and the pressures are in millimeters
of mercury.
Y
J
7
r
j
7
t
zr
1
t
^
i
.1
r
1
J
r
r
t
t.
1 _
t -
J
t _
- 4
J.
J
r
ol
1 1 X
Temperature
Pressub
100
760
105
906
110
1074.7
115
1268.7
120
1490.5
125
1743.3
130
2029.8
135
2353.7
140
2717.9
146
3126.1
150
3581.9
Let 100 represent the zero point of tempera-
ture, and let each unit of x represent 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 pre.ssure given in the above
table. Through the points thus located draw a
smooth curve (fig. 20) i.e. one which has no sudden
changes of direction. AVhile only the eleven points
located are exact, all other points are approxi-
mately accurate, and the curve may be used for
approximate computation as follows : A-ssume any
temperature, and, laying it oif as an abscissa,
measure the corresponding ordinate of the curve.
While not exact, it will, nevertheless, give an approximate value of the corre-
sponding pressure. Similarly, a pre.ssure may be assumed 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.
Fig. 20
»From C. H. Peabody's " Steam Tables," computed for sea level at a latitude of
45 degrees.
CLASSES OF FUNCTIONS
43
Ex. 4. As a final example, we will take the law of Boyle and Mariotte for per-
fect 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 repre-
sent the pressure by x and the corre-
k
spending volume by y. then y = -,
X
where fc is a constant and x and y are
positive variables. A curve (fig. 21) in
the first quadrant, the coordinates
of every point of which satisfy this
equation, represents the comparative
changes in pressure and volume, show-
ing that as the pressure increases by a
certain amount the volume is decreased
more or less, according to the amount
of pressure previously exerted.
This example differs from the pre-
ceding 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 con-
struct the curve to any required degree of accuracy.
Fig. 21
21. Classes of functions. We shall consider in this book only
those functions of one variable which can be expressed by means
of elementary mathematical symbols. The simplest kind of such
functions is the algebraic polyTiomial, expressed by
ao«"-f ftj^t-"-^ + • • • + «„_!« + «„,
where all the exponents are positive integers and the coefficients
a^, «!, •••,«„_ J, «„ are real or complex numbers or zero. The
number n is the degree of the polynomial. These functions are
discussed in Chaps. Ill and IV.
The quotient of two algebraic polynomials is a rational algebraic
fraction, expressed by
«„»" -f ajiz;"""^ +••• + «„_]''» + «„
Examples of functions of this kind are discussed in Chap. VL
44 GKAPHICAL REPRESENTATION
If a function requires for its expression the use of radical signs
combined with algebraic polynomials, it is an example of an irra-
tional algebraic function ; for example.
Ab + JIZI.
Examples of such functions are found in Chap. VI.
The general definition of an algebraic function is given in
Chap. IX, and examples of non-algebraic, or transcendental func-
tions, are given in Chap. XIII.
22. Functional notation. When y is a function of x 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 f{a). For example, if
tben /(2) = 2«+3-22-f-l=21,
/(0)=0«+3-0^+l = l,
/(-3) = (-3)H3(-3)^+l = l,
/(a) = a'+3a2-f 1.
If more than one function occurs in a problem, one may be
expressed asf(x), another as F{x), another as <l>(x), and so on. It
is also often convenient in practice to represent different functions
by the symbols f(x), f^(x), f^{x), etc.
If /(a;) is any function, and we place
y =/('«),
we may, a^ 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.
PROBLEMS 45
The curve is said to be represented by the equation, and the equa-
tion 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.
PROBLEMS
1. Find the perimeter of the triangle the vertices of which are (2, 3),
(- 3, 3), (1, 1).
2. Prove that tlie triangle the vertices of which are (— 4, — 3), (2, 1),
( — 5, 5) is isosceles. ,
3. Prove that (6, 2), (- 2, - 4), (5, - 5), (-1, 3) are points of a circle the
center of which is (2, — 1). What is its radius ?
4. Prove that the quadrilateral of which the vertices are (2, 2), (4, 5),
(—1, 4), (— 3, 1) is a parallelogram.
5. Find a point equidistant from the points (—3, 4), (5, 3), and (2, 0).
6. Find the center of a circle passing through the points (0, 0), (—3, 3),
and (5, 4).
7. Find a point on the axis of x which is equidistant from (0, 4) and
(-3,-3).
8. A point is equally distant from the points (1, 1) and (— 2, 3), and its
distance from OY is twice its distance from OX. Find its coordinates.
9. Find the points which are 4 units distant from (2, 3) and 5 units distant
from the axis of y.
10. A point of the straight line joining the points (—4, — 2) and (4, — 6)
divides it into segments which are in the ratio 3 : 6. What are its coordinates ?
11. Find the coordinates of a point P on the straight line determined by
Pi (2, - 1) and P2 (- 4, 5), when |^ = ^ •
12. On the straight line determined by the points Pi (2, 4) and P2( — 1, — 3)
find the point three fourths of the distance from Pi to P2.
13. If P (x, y) is a point on the straight line determined by Pi (xi, 2/1) and
■P2(X2, 2/2), such that — ^ = -, prove
PP2 ^2
hxi + hxi ?i2/2 + hVi
x = 1 w = •
h + I2 k + k
46 GKAPHICAL REPRESEKTATION
14. The middle point of a certain line is (1, 2) and one end is the point
(—3, 5). Find the coordinates of the other end.
15. 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 ?
16. One end of a line is at (2, — 6) and a point one fourth of the distance
to the other end is (— 1, 4). Find the coordinates of the other end of the line.
17. Find the points of trisection of the line joining Pi(0, 3) and P2(6, — 3).
18. Find the lengths of the medians of the triangle (2, 1), (0, —3), (— 4, 0).
19. Given the three points A{- 3, 3), B{S, 1), and C{6, 0) upon a straight
A.D AB
line. Find a fourth point D such that =
DC BC
20. Given four points Pj, P2, P3, P4. Find the point halfway between Pi
and P2, then the point one third of the distance from this point to P3, 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.
21. 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.
22. Prove analytically that the straight line drawn between two sides of a
triangle so as to cut off the same proportional parts measured from their com-
mon vertex is the same proportional part of the third side.
23. Prove analytically that if two medians of a triangle are equal the tri-
angle is iso.sceles.
24. Prove analytically that in any right triangle the straight line drawn
from the vertex to the middle point of the hypotenuse is equal to one half the
hjrpotenuse.
25. Prove analytically that the lines joining the middle points of the opposite
sides of a quadrilateral bisect each other.
26. Show that the sum of the squares on the four sides of any quadrilateral
is equal to the sum of the squares on the diagonals, together with four times the
square on the lin6 joining the middle points of the diagonals.
27. Prove analytically that the diagonals of a parallelogram bisect each
other.
28. Prove analytically that the line joining the middle points of the non-
parallel sides of a trapezoid is one half the sum of the parallel sides.
29. OABC is a trapezoid of which the parallel sides OA and CB are per-
pendicular to OC. D is the middle point of AB. Prove analytically that
OD = CD.
PROBLEMS
47
30. The following table gives the price of a bushel of wheat in the New
York market from 1890 to 1904. Construct the graph.
1890
.983
1895
.669
1900
.804
1891
1.094
1896
.781
1901
.803
1892
.908
1897
.954
1902
.836
1893
.739
1898
.952
1903
.863
1894
.011
1899
.794
1904
1.107
31. The following table shows hourly barometric readings at a United States
weather bureau station. Construct the graph.
1 A.M.
28.85
9 A.M.
29.04
6 P.M.
29.13
2
28.87
10
29.05
6
29.18
3
28.90
11
29.05
7
29.21
4
28.92
12 m
29.05
8
29.24
5
28.94
1 P.M.
29.05
9
29.25
6
28.97
2
29.06
10
29.29
7
28.98
3
29.08
11
29.29
8
29.02
4
29.10
12
29.29
32. The following table shows the number of inches of rainfall in Boston
during the years 1880-1891. Construct the graph.
1880
38.89
1886
46.47
1881
49.22
1887
41.91
1882
48.42
1888
60.27
1883
35.56
1889
54.79
1884
53.86
1890
50.21
1885
44.07
1891
49.63
33. The following is a portion of a railway time-table. The letters indicate
stations, and the adjacent number gives the distance fi'om A to each of the othei'
stations. The second and the third columns give the times at which two trains
running in opposite directions leave each of the stations. Make a graph showing
the motion of each train and thus determine the time and place of their passing.
A
10.45 AM.
2.00
F99
1.06 P.M.
10.48
B 21
1.30
G 126
9.53
C44
11.50
12.56
H 151
2.59
8.56
D64
12.11 P.M.
1177
7.48
E 84
11.30 A.M.
K200
4.15
7.00 A.M.
48
GRAPHICAL REPRESE:N^TATI0X
34. The following table shows the amount of SI. 00 put at interest at 4%
compounded annually. Construct the graph.
6yr.
1.217
30 yr.
3.242
10
1.480
35
3.946
16
1.801
40
4.801
20
2.191
45
5.841
25
2.666
50
7.116
35. Make a graph showing the relation between the side and the area of a
square.
36. Make a graph showing the relation between the radius and the area
of a circle.
37. Make a graph showing the relation between the radius and the volume
of a sphere.
38. The space s through which a body falls from rest in t seconds is given
by the formula s = I gt^. Assuming g = 32, construct the graph.
39. The velocity acquired by a body thrown towards the earth's surface
with a velocity Vq is given at the end of t seconds by the formula v = Vo + gt.
Construct the graph.
40. Two particles of mass mi and mz at a distance d from eacK other attract
each other with a force F, given by the formula
F =
cP
Assuming mi = 5 and m^ = 20, construct the graph of F.
41. Ohm's law for an electric current is
^ ^ Electromotive force
Current = ..
Resistance
Assuming the electromotive force to be constant, plot the curve showing the
relation between the resistance and the current.
42. n/(x) = X* - 3x2 + 7 X - 1, find/(3), /(O), f{a), f{a + h).
43. If /(x) = x8 + 1, show that/(2) - 4/(1) =/(0).
44. If /(x) = x< + 2x2 + 3, prove that/(- x) =/(x).
45. If /(x) = x6 + 3x8 - 7x, prove that/(-'x) = -/(x).
46. If /(x) = x2 - o2, prove that/(a) =/(- a).
4 7. If /i (x) = x2 + o2, and /2 (x) = 2 x, prove that /i (a) - af. (a) = 0.
48. If /(x) = (x - I) (x2 - 1) (.3 _ i^^ , prove that / (a) = -/(I) .
PROBLEMS 49
49. If /(x) = ^-^, prove that /(a) • /(- a) = 1.
50. If fix) = ^' + 2a:8 + 2x + l^ p^^^^ ^j^^^ ^ Jl\ ^
X2 \X/
51. If/(x)=.^p±l,find/(3),/(0),/(-3),/(a),/Q).
52. If /(x) = |x, prove that (x + l)/(x) =/(x + 1).
53. If /i(x) =-v/- + \/|' and f^iz) = \/- - \/^' P^O'^'^ ^^^^^^
[/i(a=)?- [/2(a;)P =[/!(«)?•
54. If /(x) = ^^ , prove that /[/(x)] = x.
CHAPTEE III
THE POLYNOMIAL OF THE FIRST DEGREE
23. Graphical representation. An algebraic polynomial of the
first degree is of the form mx + 1, where m and h are numbers,
which may be positive or negative, integral or fractional, rational
or irrational. We shall restrict the values of m and h, however, to
real numbers. In particular cases h may be zero, when the poly-
nomial becomes the monomial mx.
To obtain the graph of the polynomial, we write
y = mx + h,
(1)
and proceed as in the examples of the previous chapter. We assign
to X any number of values assumed at pleasure, say x^, x^, x^, x^,
etc. ; compute the
corresponding values
of y, namely.
y^ = mx^ + h,
y^ = m,x^+h,
y^ = mx^+h,
y^ = mx^+h,
(2)
and plot the points
m^z> 2/3). P^i' Vi)
(fig. 22). We then
Fig. 22 draw the straight
lines p,p„ nn, P,P„
each connecting two successive points, and shall prove that these
lines form one and the same straight line. For that purpose draw
no
THE STllAIGHT LI:NE 51
through each point Imes parallel to the coordinate axes, forming
the triangles shown in fig. 22. Then, by § 13,
T[li.^ = x,^ x^, J^A^ = x^ x^, J^A^ = x^ x^,
^2^2 = ^2 - Vv ^^Z^l = ^3 - 2/2 . ^-^4^ = 2/4-2/3-
By subtracting each equation in (2) from the one below it, we
have
2/3-2/2 = ^'K-«3-^"2)'
y4-2/3 = ^^('''4-'^3).
Whence |Z|l ^ |^ ^ |Z| = ,,. (4)
or, by (3),
jpi, liR, IIR^
Hence the triangles of the figure are similar, and the angles
R^P^P^, R,P,Ps, RiPPi are equal. Therefore the line P^P^P^P^ is a
straight line.
Again, let us take on this line any other point, such as ^,
which has not been used in constructing the graph, and draw
JIR^ and J?^^ parallel to OX and OY respectively. Then, since
the triangles PJi^P^ and I^R^I^ are similar,
R,P, _R.^J^,
p,r,~i(k'
that is, '.l^^ =. '!^~^ = m. (by (4))
0 4 2 1
Therefore y^ = mx^— mx^ + y^,
whence, by substituting the value of y^ given in (2),
2/5 = '^^5 + ^'
Hence the coordinates of P-, satisfy the equation (1).
We have now shown that all points the coordinates of which
satisfy eqiiation (1) lie on a straight line, and that any point on
52 THE POLYNOMIAL OF THE FIRST DEGREE
the line has coordinates which satisfy (1). We have accordingly
proved the foUowijig proposition : The equation y = mx + h always
represents a straight line.
24. The general equation of the first degree. The equation
Ax + By + C={),
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
equation of the first degree. We shall prove : The general equa-
tion of the first degree vnth real coefficients always represents a
straight line.
1. Suppose A ^ {^ and B ^ 0. If any value of x is assumed,
the value of y is determined. Therefore y is a, function of x,
which may be expressed by solving the equation for y ; thus,
A C
y = X
^ B B
This equation is of the form y = mx + h, and therefore repre-
sents a straight line by §23.
2. Suppose yl = 0, 5 T^ 0. The equation is then
By + C=0, or 2/=- 1'
B
All points the coordinates of which satisfy this equation lie
on a straight line parallel to OX at a distance units from it ;
B
and, conversely, any point on tliis line has coordinates which
satisfy the equation. Hence the equation represents this line.
3. Suppose A^O,B = 0. The equation is then
Ax+C=0, or x = --,
A
and represents a straight line parallel to 0 F at a distance — —
units from it. -^
Therefore the equation Ax + By + C=0 always represents a
straight line.
THE STRAIGHT LINE
53
25. In order to plot a straight line it is, in general, convenient
to find the points L and K (fig. 23), in which it cuts OX and OY
respectively. If the coordinates of L are (a, 0) and those of K are
(0, 5), these coordinates will satisfy the equation Ax + By+ C= 0.
By substitution we find F
C
a = >
A
B
Fig. 23
The quantities a and h, which
are equal in magnitude and
sign to OL and OK respectively,
are called the intercepts of the
straight line. It is evident that
the h found here is the same as
in y = tnx + h.
If C=0, i.e. if the equation is Ax-\-By=Q, then a = 0 and
5 = 0, and the straight line passes through the origin. To plot
the line, we must find by trial the coordinates of another point
which satisfy the equation, plot this point, and draw a straight
line through it and the origin.
Ex. 1. Plot the line 3a; — 52/ + 12 = 0. Placing y = 0, we find a = — 4.
Placing X = 0, we find b = 2f . We lay off Oi = - 4, OK - 2f , and draw a
straight line through L and K.
Ex. 2. Plot the line 3x — 5 i/ = 0. Here a = 0 and 6=0. If we place x = 1,
we find 2/ = 4. The line is drawn through (0, 0) and (1, ^).
26. Any straight line may he represented hy an equation of the
first degree.
The proof consists in showing that the coefficients A, B, C,
in the general equation of the first degree, may be so chosen that
the equation may represent any straight line given in advance.
Let (x^, y^) and (x^, y^) be any two points on a given straight
line. The coordinates of these points will satisfy
Ax-\-By + C=0, (1)
provided A, B, C have such values that
Ax, + By,+ C = 0,
Ax^ + By^+C=0,
54 THE POLYNOMIAL OF THE FIRST DEGREE
Solving these equations for the ratios oiA,B, C, we have (by § 8)
A:B:C =
|y, 1
X, 1 .
^'i yi
\y. 1
X, 1
^2 2/2
(2)
If these values are used in (1), that equation represents a
straight line which has two points in common with the given
lin^, and therefore coincides with it throughout. Hence the
theorem is proved.
The result of substituting from (2) in (1) is
= 0,
which is the equation of a line through two given points.
27. Slope. Let P^{x^, y^) and P^{x^, y^) (figs. 24, 25) be two
points upon a straight line. If we imagine that a point moves
along the line from ij to i^, the change in x caused by this
motion is measured in magnitude and sign by x^—x^, and the
X y
1
^1 Vx
I
«2 ^2
1
Fig. 24
Fig. 25
change in y is measured by y^—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,
m =
2^2-^1
ANGLES 55
It appears from equations (4) (§23) that the letter m in the
equation y = mx + J has the meaning just defined. It follows that
if the equation of a straight line is in the form Ax+By + C = 0,
its slope may he foimd by solving the equation for y and taking the
coefficient of x, thus,
AC A
y = X — —•> whence m =
"^ B B B
A geometric interpretation of the slope is readily given. For if
we draw through I^ a line parallel to OX, and through ^ a line
parallel to 0 Y, and call E the point in which these two lines inter-
BB
sect, then x^— x^ =I\B, and y^ — y^ —BB^ ; and hence m = — ^ •
It is clear from the figures, as well as from equations (4) (§ 23),
that the value of m is independent of the two points ij and B^
and depends only on the given line. We may therefore choose B^
and B^ (as in figs. 24 and 25) so that I^B is positive. There are
then two essentially different cases, according as the line runs up
or down toward the right hand. In the former case BB^ and m
are positive (fig, 24) ; in the latter case BJF^ and m are negative
(fig. 25). We may state this as foUows :
The slope of a straight line is positive when an increase in x
causes an increase in y, and is negative vjhen an increase in x
causes a decrease in y.
When the line is parallel to OX, y„ = y^, and consequently m = 0,
as explained in § 11. If the line is parallel to OY, x^=. x^, and
therefore m = oo in the sense of § 11.
28. 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
56 THE POLYNOMIAL OF THE FIRST DEGREE
the line, and to consider LX as the initial line of this angle. We
shall denote this angle by <^. Then if we take P any poiut on
Fig. 26
Fig. 27
the terminal line of ^ and drop the perpendicular MP, we have,
in the two cases represented by figs. 26 and 27,
tan^ =
MP
LM
MP
But ^^^^^ is equal to the slope of the line. Therefore
LM ^ ^
tan 0 = m.
If the straight line is parallel to OF, <^ = 90° and tan (f> = cc.
If the line is parallel to OX, no angle ^ is formed ; but since 7n = 0,
we may say tan <^ = 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
and Ax +By + C„_ = Q,
represent parallel lines.
More generally, two straight lines,
and A„x + B„y + C, = 0,
are parallel if
^1
J5„
= 0.
ANGLES
57
3. Perpendicular lines. Let AB and CD (fig. 28) be two lines inter-
secting at right angles. Through P draw PR parallel to OX and
let RPD = <^^ and RPB = (f)^. Then tan <f>^ = m^ and tan ^^ = ^2.
where m^ and ^/ig ^^"^ the slopes
of the lines. But by hypothesis,
(^,= </,,+ 90°,
whence
1
tan ^2 = — cot (f)^ =
which is the same as
tan </)j
Fig. 28
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 product of their slopes is minus unity.
It follows that two straight lines whose equations are of the type
and
Ax+Pi/+C^ = 0
Bx — Ai/+C^ = 0
are perpendicular.
4. Angle hetvjeen two lines. Let AB and CD (fig. 29) inter-
sect at the point P, making the angle BPD, which we shall
call /3. Draw the line PR
.D
parallel to OX and place
RPB = (f>^ and RPD = (f}^.
Then
Fig. 29
and hence
tan yS = tan {<^„ — </>j)
_ tan <^„ — tan ^^
1 -|- tan 0., tan <^^
58 THE POLYN^OMIAL OF THE FIRST DEGREE
But tan <f>^ = 7n^ and tan (f)^ = m^, where m^ is the slope of CD
and m^ is the slope of AB. Therefore
„ m„—m,
tan p =
1 + m^mj
If <^2 is always taken greater than ^^, tan yS will be positive or
negative according as /3 is acute or obtuse.
29. Problems on straight lines. We shall solve in this article
certain important problems which depend on the equation
y = mx + h.
The essential problem is, in every case, to determine m and h so
that the line will fulfill certain conditions. Since two quantities
are to be determined, two conditions are necessary and sufficient ;
hence, in general, one and only one straight line can be found to
satisfy two given conditions.
1. To find the equation of a straight line which has a known
slope and passes through a known point. Let m^ be the known
slope and P^{Xy, y^ be the known point. Tlie equation of the line
will be of the form y = m^x + h, where h, however, is unknown.
But the line contains the point Py Therefore
y^ = 7n^x^+h,
whence h = y^— m^Xy
The required equation is, therefore,
y = m^x + y^-my)c^\
or, more symmetrically,
y-y^ = m^{x-x^).
Ex. Find the equation of a straight line with the slope - § passing tlirough
the point (5, 7).
Fir^ method. We have y = — § x + 6 ;
then 7 = -§(5) + &,
whence 6 = s^L.
Therefore the required equation is
or, finally, 2x + 32/-31 = 0.
PROBLEMS OX STRAIGHT LINES 59
Second method. By substituting in tlie formula we have
y_7 = -|-(a;-5),
whence 2x + 3y — 31 = 0, as before.
2. To find the equation of a straight line passing through a
known jpoint and parallel to a known line.
The slope of the required line is the same as that of the given
line, which can be found by § 27. Hence the problem is the same
as the preceding.
Ex. Find the equation of a straight line passing through ( — 2, 3) and parallel
to3x-5y + 6 = 0.
First method. The slope of the given line is ^. Therefore the required line is
2/-3=f(x + 2), or 3x-5y + 21 = 0.
Second method. As explained in § 28, 2, we know that the required equation
is of the form
Sx-Sy + O^O,
where C is unknown. Since the line passes through (—2, 3),
3(- 2) -5(3) + 0 = 0,
whence C = 21. Therefore the required equation is
3x-5y + 21 = 0,
3. To find the equation of a straight line passing through a
known point and perpendicular to a known line.
The slope of the required line may be found from the slope of
the given line, as in § 28, 3. The problem is then the same as
problem 1.
Ex. 1. Find a straight line through (5, 3) perpendicular to7x + 9y + l = 0.
First method. The slope of the given line is — ^. Therefore the slope of the
required line is |. By problem 1, the required line is
2/-3 = f(x-6), or 9x- 72/ -24 = 0.
Second method. As shown in § 28, 3, 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 9x — 7j/ — 24 = 0.
60 THE POLYNOMIAL OF THE FIEST DEGKEE
Ex. 2. Find the equation of the perpendicular bisector of the line joining (0, 5)
and (5, — 11). The point midway between the given points Ls (^, — 3), by § 18.
The slope of the line joining the given points is — ^-, by § 27. Hence the required
line passes through (§, — 3), with the slope ^^. Its equation is
y + 3 = /5(x-|), or lOx - 32y - 121 = 0.
4. To find the equation of a straight line through two known
points.
This problem has already been solved in § 26, and the result
given in the form
which is the same as
X y I
^1 2^1 1
a^2 ^2 1
X
-•»! y -Vx
X,
— X
2 ^1-^21
= 0,
= 0.
(Ex. 2, § 3)
Or, by § 27, the slope of the required line is
x^ x^
Hence, by problem 1, the equation of the required line is
Ex. Find a straight line through (1, 2) and (- 3, 6),
By the formula,
5 — 2
y - ^ = _ 3 _ ^ (X - 1), or 3x + 4y-ll = 0.
5. To find the condition that three known points should lie on
the same straight line. If the three points are (x^, y^), (x^, y^),
and («3, y,), the condition that they should lie on the same straight
line is - -
as is evident from 4.
x^
Vl
1
^2
y^
1
«8
yz
1
= 0,
INTEKSECTION OF STRAIGHT LINES
61
30. Intersection of straight lines.
Let A^x + B^y + C; = 0
and A^x + B^y + Cg = 0
(1)
be the equations of two straight lines. It is required to find their
point of intersection. Since the coordinates of any point on one
of the lines satisfy the equation of that line, the coordinates of a
point on both lines must satisfy both equations simultaneously.
Hence the coordinates of the point of intersection of the lines is
found by solving the two equations.
There are three cases.
1.
A A
^0.
A ^2
The solutions are then
X =
c,
c.
^1
A
A
^1
^2
y =
A c,
A ^2
A A
A ^2
The two straight lines intersect in the corresponding point.
^1
B.
= 0, but at least one of the determinants.
B„
and
not equal to zero.
The equations are then contradictory and the straight lines do
not intersect. In fact, § 28 shows that the straight Imes are
parallel.
This case may be brought into connection with case 1 as
follows : In case 1 suppose that A^B^ — A„B^ is very small, but
not zero. The values of x and y are then very large, assuming
that the numerators are not small, and the point of intersection
is then very remote.
62 THE POLYNOMIAL OF THE FIRST DEGREE
Let now the lines be changed in such a manner that ^1^2—^2-^1
approaches zero. The values of x and y increase indefinitely, the
point of intersection recedes indefinitely, and the lines approach
parallelism.
= 0,
C„ B„
= 0,
A.. C,
= 0.
The equations are then not independent but represent the same
straight line.
In this case the attempt to use the solutions as given in 1
leads to the indeterminate form ^ (§ 11).
31. If the three straight lines
A^oi+B^y + C^=Q,
A„x + B^y + 0.-,= 0,
(1)
(2)
(3)
pass through the same point, the three equations have a common
solution, and therefore
A
A
Cx
A
B,
a
A
A
cl
= 0.
(4)
Also, if the three straight lines are parallel, the determinant (4)
is zero. For if (1), (2), and (3) are parallel, A^B^ — A^B^=^0,
0, A^Bj^—A^Bg = 0, and therefore
A^B,-A,B.
Conversely, if
A
A
Ci
A
A
c.
A
A
c.
A
A
Cr
A
A
c.
A
A
C3
= 0.
= 0,
the Imes (1), (2), and (3) either pass through the same point or are
parallel. For, by § 7, if two oi' the lines intersect, the coordinates
of the point of intersection satisfy the other.
DISTANCE FROM A STRAIGHT LINE
63
32. Distance of a point from a straight line.
Take the equation of any straight line, written in the form
y — mx — b = 0, (1)
and consider the polynomial
y — mx — h, (2)
which stands upon the left-hand side. We may substitute in (2)
the coordinates (x^, y^ of any point P^, and thus obtain a value
of (2) which is zero when P^ lies
on the line (1), but not other-
wise. We wish now to obtain
the meaning of
y^— mx^ — h
when P^ is not on (1). For that
purpose, let LK (fig. 30) be the
line (1), and let MP^, the ordinate j^,g 30
of ij, cut LK in Q. Then the
abscissa of Q is x^ and its ordinate is MQ. From (1)
MQ = mx^-\- h.
Hence y^ — ma;^ — ^ = 2/i ~ (^^i + ^)
= MPi-MQ = QP^.
It is clear that y^ — inx^ — & is a positive quantity when {x^, y^
lies above the hne LK, and is a negative quantity when {x^, y^ lies
below LK. It is also evident from the triangle P^QR, and from
a like triangle in other cases, that the length of P^R is numeric-
ally equal to P^Q cos <^. But tan <^ = m, and hence
We have, then.
cos<^
P^R
mx^ — h
±ViT
m
64 THE POLYNOMIAL OF THE FIRST DEGREE
We may, if we wish, always choose the + sign in the denomi-
nator. Then P^R is positive when ij is above y = mx + h, and
negative when i^ is below.
If the equation of the straight line is in the form
Ax + By+C=Q,
A C
m= and & = Therefore
B B
and
Ax^ + By^ +C=B{y^— mx^ — 5),
Ax, + By,+ C
y/A' + B^
It appears, then, that the polynomial Ax^ + By^ + C and the per-
pendicular PyR are 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.
33. Normal equation of a straight line. Let LK (fig. 31) be any
straight line and let OD be the normal (or perpendicular) drawn
from the origin. Let the length of OD be p and let the angle
XOD be a. Take P any point on LK. The projection of OP on
OD is equal to the sum of the
projections of OM and MP (§ 15).
But the projection of OP on OD
is p, since ODP is a right angle.
The projection of OM on OD is
xcoBa (§ 14), and that of 3IP is
y cos (a — 90°) = 2/ sin a. Hence
Fig. 31
or
p = x cos a + y sin a,
X cos a-\- y sin a — j? = 0.
This equation, being true for the coordinates of any point on
LK and for those of no other point, is the equation of LK, It is
called the normal equation of a .straight line.
NORMAL EQUATION 65
Since sin'^a; + cos^a = 1, it follows from § 32 that
Xj^ cos a + 2/j sin a — p
is numerically equal to the distance of {x^, y^ from
X cos a + 2/ sin a; — ^ = 0.
It is sometimes desirable to change an equation
Ax +By + C = 0
into X cos a + 2/sina — jt? = 0.
For that purpose it is enough to notice that since any value of
{x, y) which satisfies one equation must satisfy the other, the one
is a multiple of the other. Hence
J A = k cos a, B = k sin a, C = — ly,
where k is an unknown factor. But from these last equations we
have
A' + B'' = k^.
Therefore cos a =
sma =
p =
±^A' + B''
B
-c
±y/A' + B-
Since p is to be positive, the sign of the radical must be oppo-
site to that of C.
PROBLEMS
Plot the graphs of the following equations :
1. 5a; -3?/ + 10 = 0. 3. a; + 3?/ - 7 = 0. 5. 3x + 52/ = 0.
2. 4x + 6?/ + 12 = 0. 4. 2x-9y = 0. 6. 4a; + 7 = 0.
7. .5y-8 = 0.
8. Two numbers are to be found such that one half of one plus one third
of the other is eciual to unity. Show how one number may be graphically
found when the other is known.
66 THE POLYNOMIAL OF THE FIRST DEGREE
9. A plane figure is in the form of a square, 3 ft. on one side, surmounted
by a triangle constructed on one of its sides as a base. Express the area of the
above figure in tenns of the altitude of the triangle, and plot the graph of the
function.
10. Express the number of inches in any length as a function of the number
of centimeters, and express the same as a graph.
11. A uniform elastic string of length I is subjected to a stretching force/.
If V is the new length, l'=l(l + mf), where m is a constant. Plot the graph,
showing the relation between I' and /.
12. K t represents the boiling point in degrees Centigrade at a height h in
meters above sea level, then approximately h = 295 (100 — t). Plot the graph.
13. The pressure on a square unit of horizontal surface immersed in a liquid
is equal to the weight of the column of liquid above it. Express the pressure at
a depth x below the surface of a body of water, the density of the water being
taken as unity. Express also the pressure x units below the surface of a body
of water over which is a body of oil of density .9 and of depth 8 units. Plot
the graphs.
14. A road starts at an elevation of 100 ft. above sea level and has a uniform
up grade of 15 per cent ; i.e. it rises 15 ft. in every 100 ft. of horizontal length.
Express the distance above sea level on the road as a function of the horizontal
distance from the point of departure, and construct the graph.
15. A tank of water contains 100 gal. A tap is opened, causing the water
to flow out at a uniform rate of § gal. per minute. Express the amount of
water in the tank as a function of the time, and construct the graph.
16. Find the equation of the straight line of which the slope is 7 and the
intercept on OY is — 3.
17. Find the equation of the straight line passing through the point (0, — 3)
and making an angle of 135° with OX.
18. y ind the equation of a straight line making an angle of 60° with OX and
cutting off an intercept — 5 on OY.
19. A straight line making a zero intercept on OY makes an angle of 120°
with OX. Find its equation.
20. A straight line making a zero angle with OX cuts OY at a point 5 units
from the origin. Find its equation.
21. Find the acute angle between the lines 2x-3y+5 = 0 an^ x+2y + 2 = 0.
22. Find the acute angle between the lines 2x + Sy - 0 = 0 and 2x + y+l = 0.
23. Find the acute angle between the lines 4x + y — 2 = 0 and 3x + oy + S = 0.
24. Show that 2x + 14i/ — 17 = 0 bisects one of the angles between the lines
8z4-6y-ll = 0 andSx -4y + 3 = 0.
25. Find the equation of the straight line through the point (- 4, 5) parallel
to the line 6x — iy+l = 0.
PROBLEMS 67
26. Find the equation of the straight line through (3, — 1) parallel to the line
X - 2/ = 8.
27. Find the equation of the straight line through the point (2, — 11) per-
pendicular to the line 9x — 8y + 6 = 0.
28. Find the equation of the straight line through the origin perpendicular
to the line 6x + 5?/— 3 = 0.
29. Find the equation of the straight line through the points (— 2, — 3)
and (0, 4).
30. Find the equation of the straight line through the points (2, — 1)
and (3, 2).
31. Find the equation of the straight line through the points (—1, 3)
and (—1, 5).
32. Find the angle between the straight lines drawn from the origin to the
points of trisection of that part of the line 6x + iy = 24: which is included
between the coordinate axes,
33. Find the equation of the perpendicular bisector of the line joining
(-3, 6) and (-4, 1).
34. A straight line is perpendicular to the line joining the points (—4, — 2)
and (2, — 6) at a point one third of the distance from the first to the second
point. AVhat is its equation ?
35. Find the equation of the straight line through (3, 5) parallel to the
straight line joining (2, 5) and (— 6, —2).
36. Find the equation of the straight line parallel to the line 2z — 3^ + 5 = 0
and bisecting the straight line joining (— 1, 2) and (4, 5).
37. Find the equation of the straight line perpendicular to 3x — 5^ = 9 and
bisecting that portion of it which is included between the coordinate axes.
38. What is the equation of a straight line the intercepts of which on the
axes of X and y are 2 and — 5 respectively ?
39. What is the equation of the straight line the intercepts of which on the
axes of X and y are — 4 and — 7 respectively ?
40. In the triangle A {- 2, - 2), B {1, - 8), C (0, - 7), a straight line is
drawn bisecting the adjacent sides AB&nd BC. Prove that it is parallel to ^C
and half as long.
41. Find the equation of a straight line through (4, ^) and the point of
intersection of the lines 3x — 4y — 2 = 0 and 12x — loy — 8 = 0.
42. Find the equation of the straight line passing through the point of inter-
section of X — 22/ — 5=0 and 2x — 3y — 8 — 0 and parallel to3x — 2?/ + 2 = 0.
43. Find the equation of the straight line through the point of intersection
of6x — 2y — 11 = 0 and 4x — Cy— 5 = 0 and perpendicular to4x— 2/ + l = 0.
68 THE POLYNOMIAL OF THE FIRST DEGREE
44. Find the equation of the straight line joining the point of intersection
of the lines 2x — y + o = 0 and x+ y + 1 — 0 and the point of intersection of
the lines x — y — 7 = 0 and 2x + y — o = 0.
45. Determine the value of m so that the line y = mx + 3 shall pass through
the point of intersection of the lines y = 2x + 1 and y = x + o.
46. Find the vertices and the angles of the triangle formed by the lines
x = 0,x-y + 2 = 0, and 2x + 32/-21 = 0.
47. Find the distance of (3, 5) from the line y = 4 x - 8. On which side of
the line is the point ?
48. How far distant from the line 2x + Sy + 8 = qis the point (7, - 4), and
on which side of the line is it ?
49. Find the distance from the point (6, — a) to the line - + - = 1
a b
50. The base of a triangle is the straight line joining the points (- 1, 3) and
(5, - 1). How far is the third vertex (6, - 2) from the base ?
51. The vertex of a triangle is the point (6, - 2) and the base is the straight
line joining (- 3, 2) and (4, 3). Find the lengths of the base and the altitude.
52. Find the distance between the two parallel lines 4x + Sy -10 = 0 and
4x + 3 2/ - 8 = 0.
■ 53. A straight line is 7 units distant from the origin and its normal makes
an angle of 30° with OX. What is its equation ?
54. The normal to a straight line which is 5 units distant from the origin
makes the acute angle tan-i J with OX. What is the equation of the line ?
55. A straight line 4 units distant from the origin makes an angle of 45°
with OX. What is its equation ?
56. The normal to a straight line makes an angle tan-i|with OX. The
line passes through the origin. What is its equation ?
57. The normal to a straight line makes an angle of 90° with OX. The
line is 7 units distant from the origin. What is its equation ?
58. Find a point on the line 4 x + 3 y = 12 equidistant from the point"
(-1, -2) and (1,4).
59. Find the equation of the perpendicular bisector of the base of an
isosceles triangle having its vertices at the points (3 2) (-2 -3^ and
(2, - 5). V ' /' V . /,
60. A point is equally distant from (2, 1) and (- 4, 3), and the slope of the
straight line joining it to the origin is §. Where is the point ?
61. A point is 7 unite distant from the origin, and the slope of the straight
hue joining it to the origin is §. What are ite coordinates ?
62. Perpendiculars are let fall from the point (.5, 0) upon the sides of the
triangle tlie vertices of which are at the points (4, 3), (- 4, 3) and (0 - 5)
Show that the feet of the three perpendiculai-s lie on a straight line
PEOBLEMS 69
63. Find a point on the line x + 2 y — 3 = 0, the distance of which from the
axis of X equals its distance from tlie axis of y.
64. One diagonal of a parallelogram joins the points (4, — 2) and (— 4, — 4).
One end of the other diagonal is (1, 2). Find its equation and length.
65. Find the equations of the straight lines through the point (—2, 0)
making an angle tan-i | with the line 3x + 4y + 6=:0.
66. Find the equations of the straight lines through (2, 2) making an angle
of 45° with the line 3 x - 2 ?/ = 0.
67. Find the equations of the straight lines through the point (2, 1) making
an angle tan-i ^ with the line 2x — y — 3=0.
68. Derive the equation of the straight line making the intercepts a and h
on the axes of x and y respectively.
69. Prove analytically that the locus of points equally distant from two
points is the perpendicular bisector of the straight line joining them.
70. Prove analytically that the medians of a triangle meet in a point.
71. Prove analytically that the perpendicular bisectors of the sides of a tri-
angle meet in a point.
72. Prove analytically that the perpendiculars from the vertices of a tri-
angle to the opposite sides meet in a point.
73. Prove analytically that the perpendiculars from any two vertices of a
triangle to the median from the third vertex are equal.
74. Prove analytically that the straight lines joining the middle points of
the adjacent sides of any quadrilateral form a parallelogram.
75. Prove analytically that the straight lines drawn from a vertex of a paral-
lelogram to the middle points of the opposite sides trisect a diagonal.
CHAPTER IV
THE POLYNOMIAL OF THE JNTth DEGREE
34. Graph of the polynomial of the second degree.
The polynomial of the second degree is aa^+hx + c. Its
graph may be plotted by equating it to y and proceeding as in
§§ 20 and 23.
Ex. 1. a;2 + 2x + 2.
Place y = x^+2x + 2 and assume integral
values of x. The corresponding values of x and
y are given in tlie following table :
y
X
y
- 1
1
-2
2
-3
5
-4
10
-5
17
As in § 20, we plot these points (fig. 32), and
are then to draw a smooth curve through them.
But we notice that these points are nearer
together in some places than in others. It follows
that in some parts the curve would be more
accurate than in others. To obviate this diffi-
culty we assume such fractional values of x
as will locate points between the more widely
separated points already plotted.
We thus form the table :
x
y
1.5
7.3
2.5
13.3
3.5
21.3
-2.5
3.3
X
y
-3.5
7.3
-4.5
13.3
-5.5
21.3
Fig. 32
Plotting these points also, and drawing the curve, we have (fig. 32) the graph of
the given polynomial, a;^ + 2x + 2. The graph lies entirely above the axis of x,
and recedes constantly from it as x increases numerically, since the polynomial
is positive for all values of x, and increases in value as x increases.
70
POLYNOMIAL OF THE SECOND DEGREE
71
Ex. 2. 2x2 + x -6.
Place y = 2x^ + X — Q and assume integral values of x
Hence the table :
y
-6
-3
4
15
X
- 1
-2
-3
y
On plotting these points (fig. 33) we see that it
is desirable to assume fractional values of x.
Hence the table :
X
y
1.5
0
2.3
6.9
2.6
10.1
-1.5
-3
-2.5
4
X
y
-3.3
12.5
-3.7
17.7
- .5
-6
- .3
-6.1
- .7
-5.7
Fio. 33
In obtaining this new set of points we have
assumed — .5, — .3, — .7 as values for x, with the
aim of locating as closely as possible the turning
point, or vertex, as it will be called, of the curve.
Plotting these points also, we draw the curve (fig. 33).
It is especially to be noted that the curve cuts the
axis of X when x = — 2 and when x = 1.5. But these two values of x, since they
make 2 x'^ + x — 0 equal to zero, are the roots of the equation 2 x^ + x — 6 = 0.
As the graph of the polynomial in Ex. 1 did not intersect the axis of x, we
conclude that the equation formed by placing it equal to zero has no real roots.
Solving that equation we find that, in fact, the roots are — 1 ± V— 1.
35. Let us now consider the general polynomial of the second
degree, ax^+hx + c, of which the two polynomials just plotted
are special cases.
If we place y = ax^ + hx + c, we can write
7/ = a
0 b cl
3y'+-x+-\
a aj
h y c
^- + — + -
2 a/ a
¥— 4ac
4 a'
+ x-h
h
72 THE POLYNOMIAL OF THE Nth DEGREE
-4
4: a'
The expression in brackets is the constant, -— ^ — > plus
a function of x, {x + -^-]> which is always positive except for
h V 2 a/
x = — ——, when it is zero.
2 a
At first we shall regard a as positive. It follows that y has its
least value when x = Therefore the lowest point, the vertex,
2 a
of the curve will be ( > ) • As values greater and
\ 2 a 4a /
less than are assigned to x, x + - — increases numerically,
2 a 2 a
y increases, and the corresponding point of the curve rises in
the plane. Moreover, if x is assigned the values — 1- k and
h
A;, A; being any assumed constant value, the corresponding
2 a
values of y are the same. Hence the curve is symmetrical with
respect to the straight line x = — - — > which line passes through
the vertex of the curve parallel to the axis of y.
If a is negative, it can be proved in the same way that the curve
has an axis of symmetry, a; = — - — > which passes through its
vertex, which is in this case the highest point of the curve.
36. Now that we have proved that the graphs of all quadratic
polynomials in x are alike, having a vertex and an axis of sym-
metry passing through it, we can plot them more easily than was
possible before, as is shown by the two following examples.
Ex. 1. 4z2_4x + l.
I y = 4a;2_4x + l = 4(x2_x + i) = 4(a;-^)2.
I Therefore the vertex of the graph is (^, 0), and the
/ axis of symmetry is the line x= ^. Beginning with
/ the value ^, we assign to x values greater and less
/ than ^, thereby locating points on both sides of the
axis of symmetry, and plot the graph which is repre-
sented in fig. 34.
V
We see that the equation 4x2_4x + l = 0 has two
equal real roots, the graph being tangent (§ 37) to the
Fig. 34 axis of x at the point x = \.
DISCRIMINANT
73
Ex.2. -2x2 + 3x.
y = -2x2 + 3x
= -2(x2-|x)
= -2[{x- 1)2-^9^].
Therefore the vertex of the graph is (|, |) and
the axis of symmetry is the line x = |. The graph
is represented in fig. 35. We see that it crosses
the axis of z at two different points. Hence the
equation — 2x2 + 3x=0 has two unequal real
roots, which are found to be 0 and f .
Fid. 35
37. Discriminant of the quadratic equation. Turning now to
the constant — in the equation
4a^
y = a
h^—4ac
4 a'
+ x +
2a
of § 35, we have three cases to consider.
1. If &'— 4 ac > 0, the vertex of the graph is below the axis of
X when a > 0, and above the axis of x when a < 0, and accord-
ingly the graph intersects the axis of x in two points.
2. If &'— 4 ac = 0, the vertex of the graph is on the axis of x,
and hence the graph intersects the axis of a; in a single point.
3. If b'^—4:ac < 0, the vertex of the graph is above the axis
of X when a > 0, and below the axis of x when a < 0, and the
graph does not intersect the axis of x at all.
Now let us suppose that different values are assigned to the
constants a, h, and c, in such a way as to make &^— 4 ac decrease,
beginning with a positive value. Then the vertex of the graph
rises or falls in the plane until, when J^— 4 ac = 0, it lies on the
axis of x. At the same time, the points in which the graph inter-
sects the axis of x have been approacliing each other, and finally
coincide, when the graph is said to be tangent to the axis of x.
Eecalling that the abscissas of the points of the graph on the
axis of X are the real roots of the equation formed by placing the
expression equal to zero, we can tabulate the following results.
74
THE POLYNOMIAL OF THE iS^TH DEGKEE
1. If &^ — 4 ac > 0, the graph of a^^ +hx + c intersects the axis
of X at two points, and the equation aa^ -\-bx-i- c = 0 has two real
roots, which are unequal.
2. If 6^— 4 ac = 0, the graph of aa^+ hx-\- c is tangent to the
axis of X, and the equation aa?+hx + c = 0 has two real roots,
which are equaL
3. If &^— 4 ac < 0, the graph of ax^-\- hx + c is entirely on one
side of the axis of x, and the equation ai? + Jx + c = 0 has only
imaginary roots.
The expression J^ — 4 ac is called the discriminant of the quad-
ratic equation, as its sign indicates the nature of the roots of the
equation.
Y 38. Graph of the polynomial of the nth degree.
Let the polynomial be
a^x"-\- a^af-'+ a^3(f-'^-\ 1- a^_^x+a^.
In general this polynomial contains n + 1 terms.
If any term is lacking, we may consider that its
coefficient lias become zero.
We will begin by plotting the graphs of some
5^ special numerical cases.
Ex. 1. z3.
Place y = x^ and assume values of x. Hence the table ;
Fio. 36
X
y
0
0
1
1
2
8
- 1
- 1
- 2
- 8
.5
.1
- .5
- .1
z
y
1.5
3.4
- 1.5
- 3.4
2.3
12.2
- 2.3
- 12.2
2.7
19.7
- 2.7
- 19.7
Drawing a smooth curve through these points, we have the curve of fig. 36.
It is called a cubical parabola.
EXAMPLES OF GRAPHS
76
Ex. 2. X*.
Place y = X* and assume values of x. Hence the table ;
X
y
0
0
1
1
2
16
-1
1
-2
16
.5
.1
.7
.2
.8
.4
.9
.7
1.1
1.6
1.3
2.9
1.6
6.1
X
y
1.7
8.4
1.9
13.0
- .5
.1
- .7
.2
- .8
.4
- .9
.7
-1.1
1.6
-1.3
2.9
-1.5
5.1
-1.7
8.4
-1.9
13.0
The curve is represented in fig. 37.
Ex. 3. x5.
Place 7/ = x^ and assume values of x.
Hence the table :
Fig. 37
-X
X
y
0
0
1
1
2
32
- 1
- 1
-2
-32
.7
.2
.9
.6
1.2
2.6
1.4
5.4
1.6
10.6
1.7
14.2
X
y
1.8
18.9
1.9
24.8
- .7
- .2
- .9
- .6
-1.2
- 2.5
-1.4
- 5.4
-1.6
-10.5
-1.7
- 14.2
- 1.8
- 18.9
- 1.9
-24.8
Fig. 38
The curve is represented in fig. 38.
In each of the three examples above, the curve crossed
the axis of x at the origin, and the corresponding equation
had the root zero.
76
THE POLYNOMIAL OF THE Nth DEGREE
Ex. 4. x^ -2x- + ox -6.
Place y = x^ — 2 x^ + Sx — a and assume values of x.
Hence the table :
x
y
0
- 6
1
- 4
2
0
3
12
-1
-12
Fig. 39
- 2 i - 28
The curve is represented in fig. 39.
This curve crosses the axis of x at
the point x = 2, and hence the equation
x^ — 2x2 + 3x-6 = 0 has 2 for a real root.
Its other roots are imaginary, i.e. ± V— 3.
Ex. 5. 4x8 + 4x2- 9x - 9.
tlace y = 4x' + 4x2 _ 9x — 9 and assume
values of x. Hence the table :
X
y
1.5
- 2.6
2.5
4.6
2.7
7.2
-1.5
-18.4
-1.7
-21.8
X
y
0
- 9
1
-10
2
21
-1
0
-2
- 7
X
y
1.6
0
1.3
-5.2
1.7
6.9
- .5
-4.0
-1.5
0
-1.3
.7
This curve is represented in fig. 40. It crosses the axis of
X at three points, — when x = 1.5, when x = — 1.5, and when
x = — l. Hence ±1.5 and — 1 are real roots of the equation
4x8 + 4x2 -9x -9 = 0.
Without discussing any more numerical examples
we can see that, in general, the abscissas of the points
on the axis of x of the graph of the polynomial
Fig. 40
are real roots of the equation
a^of+ a^x"-'+a^oif-'+ ... + a„_,x+ a„ = 0.
SOLUTION BY FACTORING 77
Conversely, the real roots of the equation
a(,af + a^af~^+ a^o(f~^-\- • • • + a„_ia;+ «„ = 0
are the abscissas of the points at which the graph of
ftpaf + a^x''~^+ a^3(f*~^+ • • • + a„_jiz; + a„
intersects the axis of x, for they make y = 0.
Moreover, if the graph of the polynomial does not intersect
the axis of x, the corresponding equation has no real roots; and
conversely, if the equation has no real roots, the graph of the
polynomial does not intersect the axis of x.
39. Solution of equations by factoring. Let f{x) be a poly-
nomial which can be separated into factors fy{x), fj^x), f^{x), • • • ,
each of which is necessarily of lower degree than /(a?). Then the
equation
/(^)=0 (1)
may be written in the form
A{^)-U^)-U^)'-- = ^- (2)
It is evident that any value of x which makes one of the fac-
tors f-^{x), f^(x), fJx), ■ • • zero, satisfies equation (2), and hence
equation (1), i.e. is a root of equation (1). But such a value of x
is evidently a root of some one of the equations
Conversely, any root of equation (1) must satisfy equation (2),
and hence must make some one of the factors /^{x), f^(x), fj^x), • • •
zero ; for if no one of these factors is zero, their product cannot be
zero. Hence the solution of the equation f{x) = 0 is reduced to
the solution of the separate equations
/i(^)=0, /.(^)=o, /3(^)=0,
In applying this method it is usually desirable to have no fac-
tor of higher degree than the second ; but there is no advantage
in carrying the factoring any further, as any quadratic equation
can be readily solved.
78 THE POLYNOMIAL OF THE Ath DEGREE
Ex. 1. Solve the equation x^ = 8.
By transposition, x'^ — 8 = 0 ;
whence, by factoring, (x — 2) (x^ + 2 a; 4- 4) = 0.
.-. x-2 = 0 or x2 + 2x + 4 = 0;
whence x = 2 or — 1 ± V— 3.
Since the original equation might have been written x = '^S, we see that the
three values of x which have been found are each a cube root of 8. In fact,
every number has three cube roots, which may be found by solving the equation
formed by placing x^ equal to the number.
Ex. 2. Solve the equation x* + 9 = 0.
This equation may be written
(x* + 6x2 + 9) -6x2 = 0;
whence, by factoring, (x2 + Vo x + 3) (x2 — V6 x + 3) = 0.
.-. x2 + Vox + 3 = 0, or x2 - V6x + 3 = 0 ;
- V6 ± V3^ Ve + V^Te
whence X — — = or —= - .
2 2
It is to be noted that every number has four fourth roots, which may be found
by a method similar to that suggested above for linding its three cube roots.
40. Factors and roots. It follows immediately from the pre^
ceding article that if x — r is a factor of f{x), then r is a root of
the equation f{x) = 0.
Conversely, if r is a root of the equation f(x)=0, then the
polynomial f {x) is divisible hy x — r.
Let f{x) = a^x^^ a^af -' + • • • + a,^_^x + a„,
and let r be a root of f{x) = 0. Then
f{r) = a.,r^ + a^r'-i + . . . + a^_^r + a„ = 0.
r,f{x)=f{x)~f(r)
_ = (a^aj" + a^a;"-' + • • • + rr.,,,, a? + aj
- («o^" + «i^"~' H + «„_ir + a„)
= a„(af - r") + a^(a--' - r'-^) + • • • + a„_>(.t^ - r).
As f{x) is expressed as a series of terms each of which, being
the difference of the same positive integral powers of x and r, is
divisible by x-r, it follows that /(a-) is divisible by a;-r.
FACT0E8 AInD ROOTS 79
Ex. By inspection — 1 is a root of the equation
a;* + x3 + 2x2 + 3x + l = 0. (1)
Hence x + 1 is a factor of tlie left-hand member of tlie equation, which may
accordingly be written
(x + l)(x3 + 2x + l) = 0. (2)
Additional roots of equation (1) may now be found by solving the equation
x3 + 2x + l = 0by methods given in §§ 62 and 63.
It is to be noted that the solution of the original equation has been simplified
by making it depend upon the solution of a depressed equation, i.e. one of degree
lower than the degree of the original equation.
41. By means of the second theorem we can form an equation
which shall have any given quantities, r^, r.,, • • •, r^ as roots. For
if r^, r^, • • • are the roots of the equation, its left-hand member
must contain the factors x — r^, x — r^, • ■ • , the right-hand mem-
ber being zero. Therefore the equation
(x — r^) (x — r^) ■ ' • (x — rj = 0
has the required quantities as roots. Moreover, this equation can
have no other roots, since any other value of x will make no fac-
tor equal to zero, and hence the product will not be zero. There-
fore the required equation is
(x — Vj) {x — r„)---(x — r„) = 0.
Ex. 1. Form the equation having as roots 2 -f- 3 V — 1, 2 — 3 V— 1, — -J.
The required equation is
(a; _ 2 - 3 V^) (x - 2 + 3 V^) (x + -J) = 0,
or [(X - 2)2 + 9] [3x + 1] = 0,
or 3x3-11x2 4- 35x + 13 = 0.
This method of forming an equation suggests a method of factor-
ing a quadratic expression. For if r^ and r^ are the roots of the
quadratic equation ax^+ hx+ e= 0, then aaf-i- hx + c i& divisible
by x — r^ and x — r,, ; and hence
ax^ + bx + c = a (x — r^) (x — r^).
80 THE POLYNOMIAL OF THE .Vth DEGREE
Ex. 2. Factor 6 x^ + x - 1.
The roots of the equation 6x2 + x — 1 = 0 are _ ^ and ^.
,-. 6x2 + X - 1 = 6(x + ^) (X - ^)
= 2(x + ^).3(x-i)
= (2x + l)(3x-l).
Ex. 3. Factor 4 x^ + 4 x - 2.
The roots of the equation 4x2 + 4x — 2 = 0 are
... 4x2 + 4x - 2 = 4(x - ^if^^) (x - :^if^)
= (2x + 1 - V3) (2x + 1 + V3).
Ex. 4. Factor x2 + 4 x + 6.
The roots of the equation x2 + 4x + 6 = 0 are — 2 ± V— 2.
.-. x2 + 4x + 6 = (x + 2 - V3^) (x + 2 + V^).
42. Number of roots of an equation. The fundamental propo-
sition concerning the roots of an equation is that every equation
formed by placing a polynomial equal to zero has at least one root.
The proof of this proposition, however, depends upon methods
too advanced to be used here. We shall therefore assume it as
proved, and proceed to prove, as a consequence of it, that every
equation of the nth degree has n roots, and only n roots.
Let the given equation
a^af + ajOf-^H \.a^_^x+ a^ = 0
be denoted by f{x) =0. (1)
Since this equation must have at least one root, let r^ be that
root. Then f{x) is divisible hy x~r^ (§ 40) and therefore
f{x) = {x-r,)f,{x), (2)
/j(ic) being the other factor, and necessarily of degree n — 1.
Equation (1) can now be written
{^-r,)f,{x)=0, (3)
and any root of f^{x) = 0 (4)
is a root oif{x) = 0 (§ 39).
NUMBER OF ROOTS 81
But equation (4) must have at least one root ; and if we let r^
be that root, and reason as before, we may write
U(x) = (x-T^f,{x), (5)
f^ip^ being of degree n — 1.
By substitution in (2) we shall have
f{x) = {x-r^(x-T^f^{xy s (6)
After separating n linear factors in this way, the last quotient
will be (Xq. Therefore we shall have
f{x) = aj^a - r^) {x-r^)---{x- rj, (7)
the polynomial being expressed as the product of n linear factors.
Then the equation f(x) = 0 may be written
af,{x — r^) {x — r^---{x — r„) = 0, (8)
whence it is seen to have n roots (§ 39), i.e. r^, r^, • • •, r„.
It can have no other roots ; for if we let x have any value other
than r^ 1\- ■ -, ov r„, no factor of the first member of (8) is zero,
and hence the product in the first member is not equal to zero.
Therefore the equation of the nth. degree has n, and no more
than n, roots, and the polynomial of the nth. degree can always
be separated into n linear factors. In general, however, it is not
possible to determine these factors where n > 4.
It is to be noted that the roots may all be different, or some of
them may occur more than once. In the latter case the equation
is said to have multiple roots.
43. If now the left-hand member of equation (8) of § 42 is
expanded, the equation appears in the original form
a^af + a^af' ^ -j h a^,_iX + «„ = 0,
and it is evident that
(-r,) + (-r,) + (-r3)-|-...+(-r„)=%
"■0
(1)
and that (- r,) (- r,) (- r^) ■ - ■ (- O = ^ '
(2)
82 THE POLYNOMIAL OF THE Ntu DEGKEE
Equations (1) and (2) express respectively the following theorems:
1. The sum of the roots of an equation with their signs changed
is the coefficient of x^'^ divided hy that of «".
2. The product of the roots of an equation vnth their signs
changed is the constant term, divided hy the coefficient of af.
Other theorems of this type are given in works on the theory
of equations, but only these two have been stated here, since they
are of special service in finding the remaining root of an equation
after all the others have been determined.
Ex. 1. Three roots of the equation 2 x* + Tx^ + 8 a;2 + 2 a; - 4 = 0 are - 2,
_ 1 _ V— 1, and — 1 + V— 1. Find the fourth root.
The sum of all the roots is — ^, and the sum of the three roots known is — 4.
Therefore the fourth root is — | — (— 4), or ^.
Ex. 2. Two roots of the equation 36 x^ — 7 x + 1 = 0 are ^ and — J. Find
the third root.
Tlae sum of the two roots known is — ^, and the sum of all the roots is 0,
since the coefficient of x^ is 0; therefore the third root is 0 — (— ^), or ^.
Or the product of the roots known is — ^, and the product of all the roots
is — 3^5 ; therefore the third root is (— ^^) -;- (— ^), or ^.
44. Conjugate complex roots. Nothing was said in § 42 as
to the nature of the roots r^, r„, • • • , r„. But if the coefficients
«o» «i> • • • » ^n ^re all real, and if a + hi is one of the roots, then
a — hiSs, also a root.
For if a + Z>t is a root of f{:£) = 0, then f{a + hi) = 0. When
f{a + hi) is expanded the terms can be separated into two sets,
— those containing a alone or involving only even powers of hi
as a factor, and those involving only odd powers of hi as a factor.
By § 12 the terms of the first set are all real and their sum may
be denoted by A ; and the terms of the second set contain i to the
first power as a factor, and their sum may be denoted by Bi (B,
of course, being real). Then f{a + hi) = 0 may be written
A + Bi = 0,
whence (§ 12), ^ = 0 and 5 = 0.
If, in the above, we replace hi by — hi, it is evident that the
terms in the first set are not affected, as they involve only even
GRAPHS OF PRODUCTS 83
powers of li as a factor, and those in the second set, involving
only odd powers of hi as a factor, are changed in algebraic sign
only. Therefore we have f\a + (— hiy] = A — Bi. But we have
seen that A = Q and J? = 0 ; therefore f\ct + (— hi)] = 0. Since
f\a + (— 11)]= f [a — hi), however, it follows that f{a — hi) = 0,
and a — &i is a root of the given equation f{x) = 0.
This fact is usually stated by saying that complex roots occur
in pairs.
It follows that an equation of even degree may not have any
real roots, and that an equation of odd degree must have an odd
number of real roots, and thus at least one real root.
45. It was proved in § 42 that every polynomial is equivalent
to the product of n linear factors, i.e.
«o(^ - ^i) {x-r^---{x- r„),
where r^, 7\, • • • , r„ are the roots of the corresponding equation.
Now if any one of these roots is complex, there will be a corre-
sponding conjugate complex root. Let a + hi and a — hi be two
such roots. Then the corresponding factors are (x — a — hi) and
(x — a + hi), which combine into (x—a)^-i-h^, a real quadratic
factor.
Therefore every polynomial with real coefficients is equivalent
to the product of real linear and quadratic factors.
46. Graphs of products of real linear and quadratic factors.
1. All the factors linear and none repeated, as
a^{x - r,) {x-r^)---{x- r„).
Placing y equal to this expression, we have
y = a^{x - r,) {x-r,y--{x- r„).
It is evident that the graph intersects the axis of a; at ?i dis-
tinct points for which x =■ r^, x = r^, ■ • ■ , x = r^, and at no other
points, as no other values of x make y zero. Now let the quan-
tities T^, r^, ■ • • , r^ be arranged in the order of their magnitude,
r^ being the least. Then if at first x < r^, all the factors are nega-
tive ; and if x changes so that r^< x < r^, the first factor becomes
positive while all the others remain negative. Therefore y changes
84 THE POLYNOMIAL OF THE Ath DEGREE
sign when x changes from being less than r^ to being greater
than rj, and the curve crosses the axis of x at the point x = r^.
Again, if x changes so that at first 7\< x < i\ and then
r^<x< rg, the second factor changes sign from minus to plus,
the others retaining their original signs. Hence y again changes
sign, and the curve crosses the axis of x again at the point x = r^.
Continuing in this manner, we can show that the curve crosses
the axis oi x n times as it is traced from left to right.
2. All the factors linear, some being repeated, as, for example,
a'o{^-ri){x-r^Y{x-r^)\
the corresponding equation being
y = a,{x- r,) (x - r^Y {x - r^)\
If the r's are arranged in ascending order of magnitude, it may
be proved, as in the previous case, that the graph crosses the axis
of X at the points x = r^, and x = r^, but not at the point x = r^.
For if at first r^< x <r^ and then r^<x < r^, it is seen that no
factor changes sign. But since ^ = 0 when x = r„, the graph has
a point on the axis of x when x = r^; in fact, it is tangent to the
axis of X. And it can be proved in general that, if a linear factor
occurs an even number of times, the graph does not cross the axis
of X at the corresponding point.
3. Some of the factors quadratic, as, for example,
«o (^ - ^i) (^ - ^2? [(^ - «)' + ^']'
the corresponding equation being
y = a,{x- r^) (x - r„f [{x - af + 5=^.
The only new type of factor is [x — af + W, and this is positive
for all values of x. Hence there is no new point to be discussed
in regard to the intersection of the graph with the axis of x.
In general, the graph has as many points on the axis of x as
the polynomial has different linear factors; it does not cross the axis
at any point corresponding to a factor occurring an even number
of times ; and it crosses the axis of x at any point corresponding
to a factor occurring an odd number of times.
EXAMPLES OF GRAPHS
85
Ex.1. 2/ = .5(x + 2)(x+.5)(a;-2).
1. If X = — 2 or — . 5 or 2, y = 0,
and there are three points of the
curve on the axis of x.
2. If X < — 2, all three factors are
negative ; therefore 2/ < 0, and the
corresponding part of the curve lies
below the axis of x. If — 2 < x < — . 6,
the first factor is positive and the
other two are negative ; therefore
2/ > 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 nega-
tive ; therefore y <0, and the corre-
sponding part of the curve lies
below the axis of x. Finally, if
X > 2, all the factors are positive ;
therefore y>0, and the correspond-
ing part of the curve lies above the
axis of X.
3. Assuming values of x and
finding the corresponding values
of 2/, we plot the curve, as repre-
sented in fig. 41.
Fig. 41
Fig. 42
Ex.2. y = .5(x + 2.5)(x-l)2.
1. If X = — 2.5 or 1, 2/ = 0, and there
are two points of the curve on the
axis of X.
2. If X < — 2.5, the first factor is
negative and the second factor is posi-
tive ; therefore y <0, and the corre-
sponding part of the curve lies below
the axis of x. If — 2.5<x<l, both
■X f Victors are positive ; therefore y>0,
and the corresponding part of the curve
lies above the axis of x. Finally, if
« > 1, we have the same result as when
— 2.6<x<l, and the curve does not
cross the axis of x at the point x = 1,
but is tangent to it.
3. Assuming values of x, and finding
the corresponding values of y, we plot
the curve as represented in fig. 42.
86
THE POLYNOMIAL OF THE Nth DEGREE
Ex. 3. 2/ = .5(x + 3)
(x2- 2.5 a; + 3.5).
■ 1. If X = — 3, y = 0, and this
curve has but one point on the
axis of X.
2. If X < — 3, the first factor is
negative and the second factor is
positive, as it always is, since it is
equivalent to (x — 1.25)2 + 1.9375;
therefore y <0, and the corre-
sponding part of the curve is
below the axis of x. If x > — 3,
the first factor is positive ; there-
fore 2/ > 0, and the corresponding
part of tlie curve is above the
axis of X.
3. Assuming values of x, and
finding the corresponding values
of y, we plot the curve as repre-
sented in fig. 43.
47. Location of roots.
From the work of the last
article it is evident that the
real roots of the equation f(x)=0 determine points on the axis
of X at which the graph of f{x) crosses or touches that axis.
Moreover, if x^ and x^ (x^ < x^) are two values of x, such that
/(Xj) and f{x^) are of opposite algebraic sign, the graph is on one
side of the axis when x = x^, and on the other side when x = x^.
Therefore (§56) it must have crossed the axis an odd number of
times between the points x — x^ and x = x^. Of course it may-
have touched the axis at any number of intermediate points.
Since a point of crossing corresponds to an odd number of roots
of an equation, and a point of touching corresponds to an even
number of roots, it follows that the equation f(x) = 0 has an odd
number of real roots between x^^ and x^.
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 x^, 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
Fig. 43
DESCARTES' RULE OF SIGNS 87
nearly do 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.
48. Descartes' rule of signs. When in a polynomial a term
with one sign is immediately followed by one with the opposite
sign, there is said to be a variation of sign. For example, in the
polynomial 3 aj* + 2 a;^ — 3 ic^ + ic — 2 there are three variations.
The variations of sign in the left-hand member of an equation
are often of value in locating the real roots of the equation, for
the number of positive roots of the equation f (x) = 0 cannot exceed
the number of variations of sign in its left-hand member. This
rule is known as Descartes' rule of signs.
For example, the equation 3x^-i-2x^ — 3x^ + x— 2 = 0 cannot
have more than three positive roots, as there are three variations
of sign in its left-hand member.
To determine the greatest possible number of negative roots,
replace ic by — x'. The roots of the resulting equation will be
those of the original equation with their signs changed. Accord-
ingly the original equation can have no more negative roots than
this new equation has positive roots.
If, in the equation 3 x* -\- 2 x^ — 3 x'^ + x — 2 = 0, x is replaced
by — x', tlie new equation is 3 x'* — 2 x'^ — 3 x'^ — x' — 2 = 0. As
this equation cannot have more than one positive root, the original
equation cannot have more than one negative root.
Sometimes, l)y Descartes' rule, we can prove that an equation
has imaginary roots. For example, the equation 3 a;^ -|- «^ -f- 2 = 0
can have no positive root, and not more than one negative root.
Being of odd degree, it has at least one real root (§ 44); therefore
it has one negative root and two imaginary roots.
In order to prove Descartes' rule we will first prove that if any
polynomial f(x) is multiplied by x — r, ivhere r is a positive quan-
tity, the product has at least one more variation than has f{x).
Assuming the first term of f{x) to l)e positive, we will inclose
all the terms preceding the first minus §ign in a parenthesis. In
a second parenthesis we will inclose all the terms with a minus
sign before a positive sign occurs again, and so on. Suppose,
then, that the first minus sign appears in the term containing
88 THE POLYNOMIAL OF THE Nth DEGREE
x""*, the next plus sign occurs in the term containing af-\ etc.,
and that all the terms after that containing x""" have the same
sign as that term. We can now write
— {a^^af-^ H \- a,_iaf -'+')
±K^"'" + ••• + ««)> (1)
where all the terms within each parenthesis are of the same sign,
i.e. plus. Therefore each parenthesis marks a variation.
To multiply f(x) hj x — r we shall multiply first by x, then by
— r, and add the partial products.
The result is an equation of the following form : '
+ (&^-' + i±...)
±(&„af-"' + ^±...)HFa„r, (2)
where h^. = a^ + ra^._i, hi = ai + rai_^, etc., and accordingly are
positive.
The signs before each parenthesis of (2) are the same as in (1),
but the signs within the parenthesis are not necessarily all plus.
But however the signs may occur within any parenthesis, there
is at least one variation between tlie first term of one parenthesis
and the first term of the following parenthesis. Hence, if we con-
sider the parentheses only, the number of variations in the prod-
uct is not less than the number of variations in f{x).
But, in addition, we have the last term of the product, i.e. :f a„r,
the sign of which differs from the sign of the first term in the
last parenthesis. Hence there is at least one more variation in
{x — r) f{x) than in f{x), as we set out to prove.
Now the equation having the roots r^, rg, • • • , r„ is (§ 41)
{X- T^) {X- r^) ■ ■ ■ {X - t;)= {).
In expanding the left-hand member every time we multiply
by a factor corresponding to a positive root, we add at least one
variation of sign. Hence the number of positive roots cannot
exceed the number of variations, as stated in Descartes' rule.
EATIONAL BOOTS 89
49. Rational roots. The real roots of any equation are either
rational or irrational (§ 10), and the rational roots must be
either integral or fractional. We will now derive methods of
finding the rational roots, beginning with the integral roots.
An easy method of determining the integral roots depends upon
the following theorem : If the equation is written in the forni
a^a^ + a^a""-^ H h «„_!«; + a„ = 0, (1)
ivhere all the coefficients are integers, any integral root r must he
a factor of a„.
It has been proved in § 40 that the left-hand member of (1) is
divisible hy x — r. Since the coefficient of x is unity, and all the
coefficients in the dividend are integers, all the coefficients in the
quotient are integers. But the last coefficient in the quotient
multiplied by r must be «^„, since there is no remainder. Hence
the theorem is proved.
Accordingly, to find the integral roots of any equation with
integral coefficients, we have merely to try the integral factors of
a„. When an integral root has been found, we depress the degree
of the equation as in § 40, and apply the process to the new
equation. In this way all the integral roots may be found. In
case no integral factor of a„ proves to be a root, it follows that
the equation can have no integral root.
Ex. Find the integral roots of the equation
4 X* - 4 x3 - 26x2 + X + 6 = 0.
The integral roots of this equation must be factors of 6, so that we have to
tiy ± 1, ± 2, ± 3, ± 6. By trial it is found that — 2 is a root, and the degree
of the equation is depressed by dividing the left-hand member by x -1- 2, the
depressed equation being 4 x^ — 12 x^ — x -H 3 = 0. The only possible values of
integral roots of this equation are db 1, ± 3, and 3 is found to be a root. Dividing
the left-hand member by x — 3, we have, as the depressed equation, 4x2— 1 = 0,
the roots of which are ± ^.
Therefore the roots of the original equation are — 2, 3, ± J.
While all the integral roots of an equation may be found by
tliis method, it is evident that it fails for fractional roots, as there
is no way of determining what fractions ought to be tried. This
difficulty is obviated by the two theorems in the next article.
90 THE POLYNOMIAL OF THE 3th DEGREE
50. If a^ is unity and all the other coejgHcients are integers,
the equation cannot have a rational fraction in its lowest terms
as a root.
Let the equation be
af + a^af-'^+a^af-^-\ ha„_^x + a„=0,
p
and, if possible, let the rational fraction - , wliich is in its lowest
terms, be a root. Then
(f)"-'(f)""-'(f)""--»-(f)----
Multiply through by q"'^, and transpose to the second member
all terms but the first. Then
^ = - a^p"-'- aj)''-\ "n-ii??""' - «„2""'-
By hypothesis p and q have no common factor, and therefore —
is a rational fraction in its lowest terms, while the right-hand
member of the equation is an integral expression. But two such
p
expressions cannot be equal, and hence — , the rational fraction
in its lowest terms, cannot be a root of the equation.
Moreover, every equation in the form
ttoX" + ttjO^-' + a^cff"-- -\ h a^_^x + a„ = 0,
in which a^ is not unity, can he tra-nsformed into an equation with
integral coefficients in tvhich the coefficient of the highest power of
the unknow7i quantity shall he unity.
For, dividing tlirough by «„, we have
a. a„ a , a
ic"+-iiC»-i + -^af'-2-|-...-f--^^a;-}-- = 0. (1)
If « is a root of this equation, let a? = — > 7n. being an integer, and
substitute in (1). Then
w" ffoWi a.m'^ «„ m a,, ^ '
RATIONAL ROOTS 91
Multiplying (2) by wi", we have
x"'-\-[-^m x'"-' + {-^m^]x'"-' +
In-l,l 2 2l^./n-2
We can now determine m by inspection in such, a way that
all the coefficients of (3) shall be integers. The roots of this new
equation are m times the roots of the original equation.
Ex. Transform equation 12x3 + 16a;2 _ 6x — 3 = 0 to an equation having
integral coeflBcients, the coefficient of the highest power of x being unity.
Dividing by 12, we have
Multiplying the roots of this equation by an integer m, we insert in each
term a power of to such that the sum of its exponent and that of x' shall be
equal to the degree of the equation, thus obtaining
X'8 + (J to) X'2 - (^\ to2) X' - (4 7n,8) = 0.
For ^ m to be an integer, to must equal 3 k where k is an integer. Then ^'^ m2
becomes j''^ (9 k^), and this is an integer only when k = 21; i.e. in = 61, I being
an integer. Finally, ^ to^, or ^ (6 Z)^, is an integer if 1 = 1, the least value of to
being the one desired.
Therefore we let to = 6, and our required equation is
x'3 + 8x'2 - 15x' - 64 = 0,
the roots of which are six times the roots of the original equation.
The roots of this equation are found by the method of § 49 to be — 2, 3,
and — 9. Hence the roots of the original equation are — ^, ^, and — ^.
We are thus in a position to determine the rational fractional
roots of any equation with rational coefficients.
51. We now see that to find all the rational roots of any equa-
tion, we first find all its integral roots and then all its fractional
roots, as indicated in the following example.
92 THE POLYNOMIAL OF THE .Vth DEGREE
Ex. Find all the rational roots of the equation
2x*-5x3_2a:2-7x + 30 = 0. (1)
By Descartes' rule of signs this equation cannot have more than two posi-
tive roots, and not more than two negative roots. If any of the roots are inte-
gral, they will be among the factors of 30, i.e. ± 1, ± 2, ± 3, ± 5, ± 6, ± 10,
± 16, ± 30. By trial we find + 2 to be a root, and the depressed equation is
2x8 -x2-4x- 15 = 0. (2)
By trial we find that this new equation has no integral roots, no factor of 15
being a root. Accordingly we proceed to find fractional roots.
Dividing equation (2) through by 2 and then multiplying the rpots by m, we
have x'3 - (^ m) x'2 - (2 m^) z' - (V- rnS) = 0. (3)
To make the coefficients of (3) integral we take wi = 2, and the equation becomes
a;'8_x'2_8x'-60 = 0. (4)
By trial we find an integral root of this equation to be 5, and the depressed
equation is „ . ,„ „
^ x2-|-4x + 12 = 0, (5)
the roots of which are — 2 ± 2 V— 2.
Therefore the three roots of the transformed equation (4) are 5 and
— 2 ± 2 V— 2, and the roots of the first depressed equation (2) are ^ and
— 1 ± V— 2, so that the roots of the given equation are 2, ^, and — 1 ± V— 2.
It is to be noted that in this example, after having found all the rational
roots, we were able to find the remaining roots also, since the last depressed
equation was of no higher degree than the second.
52. Irrational roots. It should be borne in mind that rational
roots occur only for special values or systems of values of the
coefficients. Hence, after removing the rational roots, if any, by
the previous methods, we have, in general, to determine irrational
roots in order to have all the real roots of the equation. But
from the definition of an irrational quantity (§ 10) it is evident
that we cannot find an irrational root exactly. We may, however,
find an approximate value to any required degree of accuracy.
There are various methods of approximation, one of which imme-
diately follows. A more rapid method is given in § 63.*
* A method of solving algebraic equations, known as Horner's method, is found
in most treatises on the theory of equations. It is convenient in arrangement of
work and speedy in the hands of an expert. It may therefore be recommended to
one who has often to solve equations. On the other hand, the methods of §§ 52, 6.3 of
this book have two advantages. They may be applied to other than algebraic equa-
tions (see § 162), and depend upon principles which, if once mastered, are not easily
forgotten.
IRRATIONAL ROOTS
93
Fig. 44
Let the given equation 'bef(x)=0, and the graph of the left-
hand member be as in fig. 44, where OM^ = x^ and OM^ = x^.
Then M^F^ =f{x^) and M^P^=f{x^, and since f{x^ and f{x^ are
of opposite sign, the curve crosses the axis of x between M^ and
M„, and there is at least one
real root of f{x)=Q between
x^ and x^ (§ 47).
Not only does the curve cross
the axis of x at some point be-
tween Jfj and M^, but it is
evident from fig. 44 that the
straight line P^P^ also intersects
the axis of x at some point
between M^ and M^, as M^. If
the points M^ and M^ are near
together, i.e. if x^ and x^ differ
only by a small amount, the curve in most cases differs only slightly
from the straight line P^P^. Hence, if we replace the curve by
the straight line, the abscissa of the point at which P^P^ intersects
the axis of x will be approximately the root of the equation.
If OJ/3 is denoted by x^, it is evident (fig. 44) that there is a
root of f(x)=0 between x^ and x^, a smaller interval than that
between x^ and x^, in which the root was first located.
If, however, the graph oi f{x) had
been as in fig. 45, the root would
have been between x^ and x^, an
interval smaller, of course, than that
between x^ and x^.
If f{x^ has the same sign as f{x^,
we have the first case (fig. 44) ; and
if f{x^ has the same sign as f{x.^,
we have the second case (fig. 45).
In the first case, repeating the proc-
ess, using iCg in place of x^, we can
find an x^ between which and x^
the root must lie ; and in the second case, using x^ in place of x^,
we can find an x^ between which and j\ the root must he.
94 THE POLYNOMIAL OF THE Nth DEGREE
Moreover, it is evident that the successive values of x, ie. x^,
«4) ^6> • • •> found in this way are each nearer to the true value of
the root of f(x)=0 than the one preceding.
Ex. Find the root of the equation x3 + 2x — 17 = 0 between 2 and 3.
Here Xi = 2 and X2 = 3 ; al8o/(2) = - 5 and/(3) = 16. The equation of the
straight line determined by the points (2, - 5) and (3, 16) is (§ 29)
, - 5 - 16 ,
y + 5= ^_3 (X - 2).
Its intercept on OX, found by letting y = 0, is 2.2 +, and /(2.2) = — 1.952.
Since /(2.2) has the same sign as/(2), the second straight line is determined
by the points (2.2, - 1.952) and (3, 16). Its intercept on OX is 2.28 +,and
/(2.28) = - 0.587648.
Since /(2.28) and /(2.2) have the same sign, the third straight line is
determined by the points (2.28, -0.587648) and (3, 16). Its intercept on
OX is 2.3+, and /(2.3) = — 0.233. The fourth straight line is determined
by the points (2.3, -0.233) and (3,16). Its intercept on OX is 2.31 +, and
/(2.31) = - 0.053609. The fifth straight line is determined by (2.31, -0.053609)
and (3, 16). Its intercept on OX is 2.312.
Hence the irrational root of x^ + 2 x — 17 = 0, accurate to two places of
decimals, is 2.31.
By continuing this process we can find any desired number of decimal places
of the root. It is to be noted that we are obliged to find one more decimal place
than the number of decimal places to which the root is to be accurate. The
approximation is more rapid if the first decimal place is found by the method
of § 47.
PROBLEMS
Plot the graphs of the following quadratic expressions, in each case locating
the vertex of the graph and determining the nature of the roots of the corre-
sponding equation :
1. 2x2 + 3x-2. 4. -3x2 + 5x.
2. 9x2-3x-2. - 5. -9x2 + 12x-7.
3. 4x2 + 4x + 3. 6. 4x2-4x-l.
7. For what values of a are the roots ofax2 + 3x + 7 = 0 equal ? What are
the roots ?
8. Prove that the roots of (ex + —J - 8 ax = 0 are equal for all values
of a and c, and find them. ^ ^
9. Prove that there is no real value of m for which the roots of
x2 + (mx + 3)2 - 16 = 0 are equal.
PKOBLEMS 95
For what values of k are the roots of the following quadratic equations (1)
equal ? (2) real and unequal ? (3) imaginary ?
10. 2x2 + 3a; + 2 = A;. 11. x2 + (2 - A;)x + 1 = 0.
12. {k + l)x2 + (A; - l)x + (^ + 1) = 0.
Plot the graphs of the following polynomials :
13. x3 - ax. (a > 0.) 19. x^ - 12 x + 3.
14. x3 - 4x2 + X + 1. 20. 2x* + x3 - 4x2 - lOx - 4.
15. x8- 3x2 + 1. 21. 4x* + 12x3 + 7x2-28x-6.
16. x3 + x2 + 2x + 5. ' 22. 3x* - 10x3 - 5x2 ^ 2x.
17. x3 - x2 + X - 4. 23. X* + 6 x8 + 10x2.
18. x3 + 6x-6. 24. 2x5 + 2x*-7x3-8x2-4x.
Find all the roots of the following equations :
25. 8x3 = 27. 28. 5x6 + 27x2 = 2x«-64x*.
26. 8x6 - 63x8 - 8 = 0. 29. {2x - a)* - (3x + a)* = 0.
27. x6-5x3 + 12x = 2x3 + 3x. 30. x*- 2(a2+l)x2 + (a2 -1)2 = 0.
Form the equations having the following values for their roots :
31.0,2,3. 32. a + Vb,a-Vb, -a.
33. 0, 0, 2 a ± 6, ± V2b.
34. Form a quadratic equation with real coefficients having 2 + 3i for one
of its roots.
Factor the following quadratic expressions :
35. 4x2 + 8x- 7. 38. x2 + 2ax- a + a2.
36. 4x2 + 12 X + 11, 39, ^2x2 + 2 a6x - a.
37. 4 a2x2 + 2 ax + 1. 40. a2x2 + 2abx + b + b'l
If ri and r^ are the roots of the equation x2 + px + g = 0, find the values of
the following expressions in terms of p and q without solving the equation :
41. r2 + r|. 42. rf + r.i'. 43.1 + 1. 44. 1 + i. 45.^ + ^.
If ^ii ''25 ra are the roots of the equation x^ + _px2 + gx + r = 0, find the val-
ues of the following expressions in terms of the coefficients without solving the
equation : , . '
46. (rl + r| + r.2) + 2 (nrz + rgrs + nn) + 3 rirgrs-
47. r{r2r3 + rhaTi + rhir^. 48. h 1
■^ nrz r2r3 rsn
49. Show that if a + VS is a root of an equation with rational coefficients,
then a — Vft is also a root.
96 THE POLYNOMIAL OF THE Nth DEGREE
Plot the graphs of the following expressions, and find all the roots of the
corresponding equations :
50. (X + 1) (a; - 2) (z - 4). 56. (2z + 5)(x2 + 2x + 3).
51. (x-2)(x-4)(2x + 3). 57. (x-5)(2x2 + 3x + 2).
52. (X - 4) (2x + 1) (3x + 5). 58. (x + 2) (x - 3) (x - 2)2.
53. (X + 3) (X - 1)2. 59. (X - 2) (X + 2) (x2 + 2).
54. (2x - l)(x - 3)2. 60. (X - 2)2(2x2 + 2x + 1).
55. (x-2)(2x + 3)2. 61. (x + l)(2x-l)(3x2 + 2x + 3).
Find all the roots of the following equations :
62. x» - 4x2 - 2x + 6 = 0. 67. 8x3 - 28x2 + 30x - 9 = 0.
63. x8 - 3 x2 + 4 = 0. 68. 12x8 - 44 x2 + 5x + 7 = 0.
64. 3x3 _ 7x2 - 8x + 20 = 0. 69. 3x3 + 10x2 + iqx - 12 = 0.
65. 4x3-8x2-35x4-75 = 0. 70. 3x3 + 10x2 + 2x - 8 = 0.
66. x8 + 4x2 + 4x + 3 = 0. 71. 4x< + 8x3 + 3x2 - 2x - 1 = 0.
72. 6x* - 11x8 - 37x2 + 36X + 36 = 0.
73. 3x< - 17x3 4- 41x2 - 53x + 30 = 0.
74. 2x* - 9x3 - 9x2 + o7x - 20 = 0.
75. 18x< - 27x3 + 10x2 + I2x - 8 = 0.
76. 16x* + 16x8 - 72x2 - 20x + 25 = 0.
77. x8 - 2x< - 4x3 - 4x2 + 15x + 18 = 0.
78. 4x5 + 12x* + 11x3 + 5x2 - 3x - 2 = 0.
79. 12x5 + 44x* - 55x3 - 95x2 + 63x - 9 = 0.
80. 2x6 - 6x* - 13x3 + 13x2 + 5x - 2 = 0.
Determine by Descartes' rule of signs the nature of the roots of the follow-
ing equations :
81. x3 + 6x-7 = 0. 84. 3x* + 4x8 + 4x + 3 = 0.
82. x3 + 2x + 3 = 0. 85. X* + x2 - X - 6 = 0.
83. x8 + 2x2 + 5 = 0. 86. X* - 4x2 + 1 = 0.
Find the real roots of the following equations, accurate to two decimal places :
87. a^ + 3x - 7 ^ 0. 89. X* - 12x + 7 = 0..
88. i8 + X + 5 = 0. 90. X* - 3x3 + 3 = 0.
91. x3-x2-6x + l = 0.
f
I
CHAPTEK V
THE DERIVATIVE OF A POLYNOMIAL
53. Limits. A variable is said to approach a constant as a
limit, when, under the law ivhich 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 narned,
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 x is given in succession the values
2" — 1
■^1 ~" 2"' "^2 ~" 1» "^3 — t' » "^n ~ On
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 suffi-
ciently great ; and for all larger ^ ^
values of n the difference be- ? | f T ¥ /
tween x and 1 is still smaller. * 234
This may be made evident
graphically by marking off on a number scale the successive values
of X (fig. 46), when it will be seen that 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
«i = J, ^2= — ^, «^3 = -4, ^4 = ~ 5' ■ ■ ■ ' ^n — \~^) n + \ '
X approaches 0 as a limit (fig. 47).
~s~7 0 s e i 2 f
-t— ) 1 H 1 + 1
Fig. 47
If the variable is not independent but is a function of x, the
values which it assumes as it approaches a limit depend upon
97
98
THE DERIVATIVE OF A POLYNOMIAL
For example, let y =f{x),
the values arbitrarily assigned to x.
and let x be given a set of values
approaching a limit a. Let the corresponding values of y be
Vv 2/2' 2/3. 2/4' •'•> Vny ••••
Tlien 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 ^ as a limit
as X approaches a in the man-
ner indicated. This may be seen
graphically in fig. 48, where the
values of x approacliing a are
seen on the axis of abscissas apd
the values of y approaching A
are seen on the axis of ordinates.
The curve of the function is con-
tinually nearer to the line y =A.
In the most common cases, the
limit of the function depends only
upon the limit a of. the inde-
pendent 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. 48.
Ex. 1. Consider the function
x2 + 3 X - 4
y = z '
X — 1
and let x approach 1 by passing through the succession of values
x = l.l, x = 1.01, x = 1.001, x = 1.0001, ••..
Then y takes in succession the values
2/ = 5.1, 77=5.01, y = 5-001, 2/ = 5.0001.
It appears as if y were approaching the limit 5. To verify this, we place x = 1 + A,
where h is not zero. By substituting and dividing by h we find 2/ = 6 + A.
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 6
is the limit of j/ as x approaches 1. Moreover, it appears that this limit is inde-
pendent of the .succe.ssion of values which x assumes in approaching 1.
SLOPE OF A CURVE
99
Ex. 2. Consider y =
as X approaches zero.
1-vT-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
VT
by 1 + Vl — X we find
y = 1 + vT^ X for all values of x except zero.
Hence it appears that y approaches 2 as x 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
\Am.f{x) = A
is read " the limit oi f{x), as x approaches a, is A."
54. Slope of a curve. By means of the conception of a limit
we may extend the definition of " slope," given in § 27 for a
straight line, so that it may be
applied to any curve. For let i^
and ^ be any two points upon a
curve (fig. 49). If ij and P^ are
connected by a straight line, the
slope of this line is — ^ • If P„
and i^ are close enough together,
the straight line PyP^ 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 P^ Now this
approximation is closer, the nearer the point P, is to Py Hence we
are led naturally to the following definition :
The slope of a curve at a point P^ix^, y^) is the limit approached
Fig. 49
by the fraction
«//(> ^"~ W-i
where x„ and y., are the coordinates of a
second point P^ on the curve, and where the limit is taken as P^
moves toward P^ along the curve.
100
THE DERIVATIVE OF A POLYNOMIAL
Ex. 1. Consider the curve y = x^ and the point (5, 25) upon it, and let
Xi = 5, yi = 25.
We take in succession various values for x^ and y^ corresponding to points
on the curve which are nearer and nearer to (xi, j/i), and arrange our results in
a table as follows :
X2
2/2
Xo, - Xi
2/2 - 2/1
2/2-2/1
Xg-Xi
6
36
1
11
11
5.1
26.01
.1
1.01
10.1
6.01
25.1001-
.01
.1001
10.01
5.001
25.010001
.001
.010001
10.001
The arithmetical work suggests the limit 10. To verify this, place X2 = 5 + /;.
Vi — ^1
Then ya = 25 + 10 A + A^ Consequently = 10 + A, and as Xa approaches
Xo — Xi
Vn — Vi
Xi, h approaches 0 and approaches 10. Hence the slope of the curve
X2 — Xi
y = X* at the point (5, 25) is 10.
Ex. 2. Find the slope of the curve y = - at the point (3, ^).
We have here
We place
xi = 3,
2/1
Z2 = 3 + A, 2/2 =
Then X2 — Xi = A, j/2 — 2/1 =
-h
9 + 3A
, and
3 + A
X2 — Xi
9 + 3A
As P2 approaches Pi along the curve, h approaches 0, and the limit of
^2 — 2/1 ■ 1
— - — IS — - ; hence the slope of the curve at the point (3, ^) is — J.
In a similar manner we may find the slope of any curve the
equation of which is not too complicated ; but when the equation
is complicated there is need of a more powerful method for find-
ing the limit of ^ _^- This method is furnished by the opera-
tion known as differentiation, the first principles of which are
explained in the following articles.
55. 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 x changes from 5 to ^\, its
CONTINUITY
101
increment is ^. If it changes from 5 to 4|, the increment
is — ^. So, in general, if x changes from x^ to x^, the increment
«„— a.%. It is customary to denote an increment by the
I
is
symbol A (Greek delta), so that
Aa = x^— x^, and x^ = x^ + Ax.
If 2/ is a function of x, any mcremeut added to x will cause
a corresponding increment of y. Thus, let y =f(x), and let x
change from x^ to x^. Then y changes from y^^ to y^, where
2/i =/K) and y„ =f{x.;).
Hence Ay =f(x^) -f(x^).
But, as shown above, x^ = x^ + Ax,
so that Ay = f{x^ + Aa:;) — f{x^.
56. Continuity. ^ 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 C, lying
between A and B, for at least
one value of x between a
and h (fig. 50).
In particular, if f{a) is posi-
tive and f{h) is negative, f{x) = 0 for at least one value of x
between a and 5.
An algebraic polynomial is a continuous function, but we shall
omit the proof. The postage function (§20) is an example of a
function which is discontinuous at certain points. Other examples
are found in §§ 149, 154.
x=a
Fig. .50
102 THE DERIVATIVE OF A POLYNOMIAL
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.
57. Derivative. When y is a continuous function ofx,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.
dy
The derivative is expressed by the symbol -~ \ or, if y is expressed
by f{x), the derivative may be expressed by f'{x).
Th.ViS,iiy=f{x),
^ =f'(x) = Lim ^ = Lim /(^ + ^/)-/(^) .
dx Ax=oA£c Ax=o Ax
The process of finding the derivative is called differentiation,
and in carrying out the process we are said to differentiate y with
respect to x.
The process of differentiation involves, according to the defini-
tion, 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 hmit approached by this quotient,
as the increment of x approaches zero.
Ex. 1. Find the derivative oi y = z^.
(1) Assume Ax = h.
(2) Compute Ay = {z + h)^ -x^ = Sx^h + Sxh^- + h^
(3) Find :^ = 3 x2 + 3 xA + K\
Ax
(4) The limit is evidently 3x2. Hence — = 3x2.
dz
Ex. 2. Find the derivative of - •
X
(1) Place y = - and assume Ax = h.
X
(2) Compute Ay ^ ^ ^
X + h X x2 + xA
(8)Find^ = I
Ax x2 + xA
(4) The limit is clearly , and therefore ^ = - i.
x2 dx x2
I
FOKMULAS OF DIFFERENTIATION 103
It appears that the operations of finding the derivative of f{x)
are exactly those which are used in finding the slope of the curve
y =zf{x). Hence the derivative is a function wliich gives the slope
of the curve at each point of it.
58. Formulas of differentiation. The obtaining of a derivative
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. — = naaf'^, where n is a positive integer and a any
dx
constant.
Let y = ax^.
(1) Assume Ax = h.
(2) Then Ay = a(x+hy— aaf
= afnaf-'h + ""^"j ^K''-'7i'+ • • ■+h\
(3) ^ = a(naf-'+'^^''r'^K-'h + • • • + h"-').
^ Ax \ [2 /
(4) Taking the limit, we have -^ = naaf \
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
SiDce c is a constant, Ac is always 0, no matter what the
value of X. Hence — =0, and consequently the limit -j- = ^^
Ax <^^
104 THE DERIVATIVE OF A POLYNOMIAL
4, The derivative of a polynomial is found hy adding the
derivatives of the terms in order.
Let y = a^af'-\-a^af-'+--' + a^_,x + a„.
(!) Assume Ax=h.
(2) Then
Ay = a,{x + A)"+ a^{x + /i)"-'+ • • • + a„_,{x + h)-{-a„
- [a^^af+ a^3f-^+ • • • + a„_iX + aj
= h [naf^af-^ + {n — 1) a^af "^-j h a„_^]
+ ^[n(n-l)a,s(f-' + {n-l){n-2)a,af-'-h--' + a,^_,]
H + h^'a^.
(3) ^ = «aoic"-i + (?i - 1) a,af -2+ • • ■ + a„_,
Aa;
+!()+. ..+A
n-l.
"0*
2
(4) Taking the limit, we have
-^ = waoaf-i + (w — l)a,af-^H h «„_!•
Ex. Find the derivative of
/(x) = 6x5 - 3a;* + 5x3 - 7 a;2 + 8x - 2.
Applying formulas 1, 2, or 3 to each term in order, we have
/'(x) = 30x* - 12x3 + 15x2 - 14x + 8.
59. Tangent line. A tangent to a curve is the straight line
approached as a limit hy a secant line as two points of intersection
of the secant and the curve are made to approach coincidence.
It is immaterial in what manner the two points of intersection
are made to approach coincidence. In § 37 this was done by
considering the curve as moved in the plane. In § 88 the secant
is considered as moving parallel to itself until it becomes a
tangent. In this article we are especially interested in determin-
ing a tangent at a known point of the curve. Let us call this
TANGENT LINE 105
point JJ and a second point on the curve P,. Then if a secant is
drawn through ij and i^ of a curve (fig. 51), and the point F^ is
made to move along the curve toward i^, which
is kept fixed in position, the secant will turn on
^ as a pivot, and wUl approach as a limit the
tangent F^T. The point F^ 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 § 54.
The equation of the tangent is readily written by means of
§ 29, when the point of contact is known. For, let (x^, y^ be the
point of contact, and let [-t-] denote the value of -7- when x = x^
\"'Vi ^^ /dy\
and y = y^. Then {x^, y^ is a point on the tangent and ( -p j is
its slope. Therefore its equation is ^ '^
The equation of the tangent may also be written in terms of
the abscissa of the j)oint of contact. Let a be the abscissa of the
point of contact of a tangent to a curve y =f{x), and let f{x)
represent as usual the derivative of f{x). Then the ordinate of
the point of contact is f{a) and the slope of the tangent is f (a),
in accordance with § 22. Hence the equation of the tangent is
y--f(a) = (x-a)fia). (2)
Ex. 1. Find the equation of the tangent to the curve ^^ = x^ at the point
(Xi, Vi) on it.
Using formula (1), we have
y -yi = Sxl{x-Xi).
But since (xi, j/i) is on tlie curve, we have 7/1 — xf. Tlierefore the equation
can be written
J/ = 3 xf X — 2 Xi^
106
THE DERIVATIVE OF A POLYNOMIAL
Fig. 52
Ex. 2. Find the equation of the tangent to
y = x^ + Sx
at the point the abscissa of which is 2.
We will use equation (2). Then
/(x) = z2 + 3x,
/(x) = 2x + 3.
/(2) = 10, /(2) = 7.
Therefore the equation is
y — 10 = 7(x — 2), or y = 7x — 4.
If PT (fig. 52) is a tangent line and cf> the angle it m^kes with
dy
OX,'ita slope equals tan <f), Ijy § 28. Hence tan ^ = — •
60. Sign of the derivative. A function of x 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 increas-
ing, and runs down toward the
right hand when the function is
decreasmg. Thus x'—x— 6 (fig. 53)
is decreasing when « < |-, and in-
creasing when x>'^.
The sign of the derivative enables
us to determine whether a func-
tion is increasing or decreasing
in accordance with the following
theorem :
When the derivative of a func-
tion is positive the function is in-
creasing ; when the derivative is
negative the function is decreasing.
To prove this, consider y =/(«), and let us suppose that
dy . . dy Av Av
-r- is positive. Then, since ~ is the limit of -.^ , it follows that -r^
C'X cix Ax Ax
SIGN OF THE DERIVATIVE
107
is positive for sufficiently small values of Ax; that is, if Ax is
assumed positive, Ay is also positive, and the function is increas-
ing. Similarly, if -r- is negative. Ay and Ax have opposite signs
for sufficiently small values of Ax, and the function is decreasing
by definition.
dy '
Ex. 1. If y =: a;2 _ a; _ 6, — = 2« — 1, which is negative when x<^ and
positive when x > ^. Hence the function is decreasing when x<^ and increas-
ing when x>^, as is sliown in fig. 53.
Ex.2. If y = |(x3-3x2-9x+27), Y
dx ^ *
= |(x + l)(x-3).
Now — is positive wlien x < — 1,
negative when — 1 < x < 3, and positive
when X > 3. Hence tlie function is
increasing wlien x < — 1, decreasing
when X is between — 1 and 3, and
increasing when x > 3 (fig. 54).
It remains to examine the
cases in which -^ = 0.
ax
Eefer-
FiG. 54
ring to the two examples just
given, we see that in each the
values of x which make the
derivative 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 f'{x) is a continuous function of x, the
values of x for which the derivative is positive are separated from
those for which it is negative by values of x for which it is zero
(§ 56 ). Now in most cases which occur in elementary work
f'{x) is 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.
108 THE DERIVATIVE OF A POLYNOMIAL
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
\{x^-^^+21x-\%
I Its derivative is ar^— 6 a? 4-9 = (a?— 3)^
/ which is always positive. The func-
/ tion is therefore always increasing.
^ Wlien a; = 3 the derivative is zero
and the corresponding shape of tlie
graph is shown in fig. 55.
61. Maxima and minima. The
X turning points of the graph of a
function correspond to the maxi-
mum and the minimum values of
the function. These terms are more
precisely defined as follows :
/(a) %8 a maximum value of the function f [x) when f [a ±h) <i f(a)
for all values of h sujficiently small, i.e. for all values of h nu-
merically less than some finite quantity.
f{a) is a minimum value of the function f {x) when f {a ± h) >f{a)
for all values of h sufficiently small.
In passing through a maximum value the function changes
from an increasing to a decreasing function, and in passing
through a minimum value the 'function changes from a decreas-
ing to an increasing function. From the work of the previous
article we may accordingly frame the following rule for finding
the maxima and the minima values of a function :
Find the derivative of the function, place it equal to zero, and
solve the resulting equation. Take each root thus found and see
if the derivative has opposite signs as x is taken first a little
smaller and then a little larger than the root. If the sign of the
derivative changes from plus to mimis, the root substituted in the
fwnetion gives a maximum value of the function. If the sign of
the derivative changes from minus to plus, the root suhstituted in
the function gives a minimum value of the function.
MAXIMA AND MINIMA
109
This rule is most readily applied when the derivative can be
factored. The change of sign is then determined as in § 46.
In § 62 will be given a method of distinguishing between a maxi-
mum and a minimum, which may be used when the factoring of
the derivative is not convenient. In practical problems the ques-
tion as to whether a value of x for which the derivative is zero
corresponds to a maximum or a minimum can often be deter-
mined by the nature of the problem.
Ex. 1. Find the maximum and the minimum values of
/(x) = xs _ 5x* -f- 5x3 + 10a;2 - 20a; + 5.
We find /(x) = 5 x* - 20x3 + 15x2 -|- 20x - 20
= 5(x2-l)(x2-4x + 4)
= 5(x + l)(x-l)(x-2)2.
Tlie roots of /' (x) = 0 are — 1, 1, and 2. As x passes through — 1, /'(x)
changes from + to — . Hence x = — 1 gives /(x) a maximum value, namely 24.
As X passes through + 1, /"(x) changes from — to +. Hence x = + 1 gives /(x)
a minimum value, namely — 4. As x passes through 2, /' (x) does not change
sign. Hence x = 2 gives /(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 inches, required the
largest box which can be found.
Let X be the side of the square cut out. Then if the cardboard is bent along
the dotted lines of fig. 56, the dimensions of the box are 30 — 2 x, 20 — 2 x, z.
Let y be the volume of the box. Then
y = X (20 - 2 x) (30 - 2 x)
= 600x- 100x2 + 4x3.
dy
dx
= 600 - 200 x + 12x2.
Equating this to zero, we have
3x2 _60x + 150 = 0,
25 ± 5 V?
Hence
dy
dx
3.9 or 12.7.
12(x-3.9)(x-12.7).
a-
X
[ 30-2X
Ti
?!
1
Fig. 56
dx
changes from + to — as x passes through 3.9. Hence x = 3.9 gives the
maximum value 1056+ for the capacity of the box. x = 12.7 gives a mini-
mum value of ?/, but this has no meaning in the problem for which x must
lie between 0 and 10.
110 THE DERIVATIVE OF A POLYNOMIAL
Ex. 3. The deflection of a girder resting on three equally distant supports
and loaded uniformly is given by the equation
V = C (- l^ + Slx^ - 2x*),
where C is a constant, I the distance between the supports, and x the distance
from the end support. Required the point of maximum deflection.
dz
Equating this to zero, we have
8x3-9^2 + 18 = 0.
It is clear that in the practical problem x<l. We find by trial that a root
lies between x = Al and x = .51. We will place
y -Sx^ -dlx^ + P,
and apply the method of § 62. The straight line connecting (.41, .072^3) and
(.5L -.25i3)is
y - .072 Z3 =: _ 3.22 i2 (X - .4 0
and this cuts the axis of x when
-(■^-Si)'=-
This is approximately the root of the equation. As a check we note that when
X = .42 i, y = .006104 1^ ; and when x = .43 i, y = - .028044 1\ Hence the root
lies between .42 1 and .43 J.
If more accuracy is required, the straight line connecting (.42 Z, .006104^3)
and {A31, — .028044 i^) may be found. Its intercept on OX is
x=.4215Z.
As shown in § 63, Ex. 2, this is correct to four decimal places.
62. The second derivative. Since -^ is in general a function of
ax
X, it 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 -r- (-r-)' which is commonly abbreviated into — ^ •
'' dx\dx/ ^ dod^
When a function is denoted by f{x) and its derivative by f'{x),
its second derivative is denoted by f"{x) ; thus, if
^= /'(»)= 3a^- 6 a; +6,
ax
0=/"(.)=6.-6.
THE SECOND DEKIVATIVE
111
Again, by differentiating — ^ or f"{x), we may obtain an expres-
sion called the third derivative, denoted by — ^ or f"'{x). By dif-
ferentiating this we obtain the fourth derivative, and so on. To
distinguish -^ from these higher derivatives it is sometimes called
ax
the first derivative. ^
The significance of — 4 for the graph is obtained from the fact
dy d^y
that -^ is equal to the slope ; hence — ^ is the derivative of the
CtX 72 clx
slope. Therefore, by § 60, if — ^ is positive, the slope is increas-
d^y
ing ; if — ^ is negative, the slope is decreasing. We may have,
accordingly, the following four cases :
1. ^ is +,
dx
dy?
IS +.
The graph runs up toward the right with
increasing slope (fig. 57).
2. "f. is +,
dx
d'y,
da?
IS — .
The graph runs up toward the right with
decreasing slope (fig. 58).
3. "f is -,
dx
da?
IS +.
The graph runs down toward the right.
The slope which is negative is increasing
algebraically and hence is decreasing
numerically (fig. 59).
4.^ is
dx
da?
IS
Tlie graph runs down toward the right
and the slope is decreasing algebraically
(fig. 60).
Fig. 60
112
THE DERIVATIVE OF A POLYNOMIAL
The consideration of these types leads to the following con-
clusion '■ If ^^ positive, the graph is concave upward ; if — ^
d3?
da?
is negative, the graph is concave dovmward.
From this we may deduce the following rule to distinguish
maxima and minima in that we take accoimt of the fact that
the graph is concave upward when y is a minimum and concave
rj. dy . -, d?y .
downward when y is a maxunum. Ij -f- u zero and — ^ %s posv-
tive, y has a minimum value ; ij — is zero and -77^ is negative, y
has a m,aximum value. 1
This rule cannot be applied to the case in wliich ~r~^ ^^^
— = 0, and hence it is not so complete as the rule in § 61, but it
da?
is sometimes more convenient in application, and especially when
the first derivative cannot be factored.
When the curve changes from concavity in one direction to con-
cavity in the other, — ^ = 0. Tlie corresponding point is called a
point of inflection. Hence to find the points of inflection we
must solve the equation — -^ = 0, and see if the second derivative
changes sign as x passes through
each root.
Ex. 1. y^^^ix^-Qx"^),
dx
x = -x{x-4),
dy
Fig. 61
d'^y
d^i/ 1 1
^^ = -x-l = -(x-2).
dx^ 2 2^ '
The curve (fig. 61) is concave down-
ward when X < 2, is concave upward
when a; > 2, and has a point of inflec-
tion when x = 2. When x = 0, — = 0
fPy dx
and -— < 0 ; the corresponding value
of y is therefore a maximum. When
« = 4, -— = 0 and " ^ > 0 ; the corresponding value of y is therefore a minimum.
EXAMPLES
113
3 x2 + o,
Ex. 2. y = x^ + ax = x(x^ + a),
dy
dx
dx2
The curve is concave downward when x < 0, is con-
cave upward wlien x > 0, and has a point of inflection
when X = 0. In addition we distinguish two cases :
(1) a positive.
dy
dx
is always positive, and the curve
cuts OX only at the origin (fig. 62).
(2) a negative. The curve has a maximum ordinate
when X =
I
and has a mini-
mum ordinate when
H-
y = +
2a
Fig. 62
0, or -H V — a
It cuts OX when x
(fig. 63).
Ex. 3. y = x^ + ax + b.
The graph of this function may be obtained
by moving the graph of Ex. 2 through the dis-
tance b up or down, according to the sign of b.
Our interest is especially with the intei'cepts
on OX. The curve obtained from (1) of Ex. 2
cuts the axis of x in one and only one point.
The curve obtained from (2) of Ex. 2 will
intersect OX in three points, will intersect
OX in one point and be tangent in another,
or will intei'sect OX in one point only, accord-
ing as the numerical value of b is less than,
equal to, or greater than the distance of the
tui'ning point of the curve from OX ; that is,
according as
b^ =
2a r^a\
Y\ 3/
This condition reduces to
62
Fig. 03
4 27
0.
114
THE DERIVATIVE OF A POLYNOMIAL
{,2 qS
It is to be noticed that when a > 0, - + — > 0. Hence we may cover all
4 27
cases by the statement :
The eauation x^ + ax + b = 0 has three unequal reed roots, two equal real roots
52 a' <
and one other real root, or one real and two complex roots, according aa - + — = 0.
63. Newton's method of solving numerical equations. The
results of this chapter may be applied to finding approximately
the irrational roots of a numerical equation. We first find, by the
method of § 47, two numbers x^ and x^, between which a root
of /(a:)=0 is known to lie. It is necessary to take care that
neither /'(a;) norf"{x) is zero for any value of x between x^^ and x^.
Then f{x) is always increasing or decreasing between x^ and x^
and hence only one root of f{x) = 0 lies between x^ and x^. Also
N
Xi D/ /C
U
(1)
D X2
(S)
N
Fig. 64
the curve y= /(a?) is always concave upward or concave down-
ward between aj^ and x^. Hence the curve has one of the four
shapes of fig. 64.
It appears that in each case a tangent at one of the points
M ox N will intersect the axis of ic in a point C which lies
between x^ and x^. In practice it is most convenient to sketch
the curve with attention to the signs of the first and the second
derivative, and to find the tangent at that end at which it lies
between the curve and the ordinate of the point of contact.
The intersection of the tangent with OX is then nearer to the
J
I
NEWTON'S METHOD 115
intersection of the curve, i.e. to the required root of the equation,
than is the abscissa of the point of contact. For example, in
fig. 64, (1) and (4), the equation of the tangent is
fix )
and its point of intersection with OX is x — •^ , ^ • Hence the
root which was at first known to lie between x^ and x^ is now
fix )
known to lie between x. and x„ — ^; ^' •
It is well in practice to combine this method with the method
of § 52. For, if we draw the secant MN, it will intersect the
axis of ic in a point. D, and the root of the equation lies between
C and D. But C and D are closer together than are x^ and x^, so
that we have narrowed down the interval within which the root lies.
Ex. 1. Find the root of a;^ — 6 x — 13 = 0, which lies between 3 and 4.
Here /(a;) = x^ - 6 x - 13,
/'(x) = 3x2-6,
/"(x)=6x.
When X = 3,/(x) = — 4 ; and when x = 4,/(x) = 27 ; while between x = 3 and
X = 4, f'{x) and f"{x) are positive. Hence the graph is as in fig. 64, (1), where M
is (3, - 4) and N is (4, 27). The tangent at N is
y -21 = 42(x-4).
Hence, for C, x = 4-2j = 3.36.
The equation of MN is y - 27 = 31 (x - 4).
Hence, for D, x = 4 - §^ = 3. 13.
Therefore the root lies between 3.13 and 3.36.
As this does not fix the first decimal figure of the root, it is advisable to apply
§ 47 again. We find /(3. 1) = - 1.809 and /(3.2) = + .568. Hence the root lies
between 3.1 and 3.2. Accordingly, the point M is now (3.1, — 1.809), and the
point N is (3.2, .568). The equation of the tangent at N is
y- .568 = 24.72(x-3.2),
and for the new point C x — 3.17702.
The secant MN is y - .568 = 23.77 (x - 3.2)
and for D x = 3.176.
The root of the equation therefore lies between 3.176 and 3.177. This result
is close enough for most practical purposes, but if the operations are carried
out once more it is found that the root lies between 3.1768148 and 3.1768144.
116 THE DERIVATIVE OF A POLYNOMIAL
Ex.2. In §61, Ex.3, the root of 8x^ -Olx"^ + 1^ = 0 was found to lie
between A21 and .43 Z,
Placing /(a;) = 8x8-9ix2 + Z3,
we have /'(x) = 24 a;- - 18 Ix,
/"(x) = 48x-18Z,
so that/'(x) is negative and /"(x) positive, when x is between A21 and AS I.
Hence the curve has the shape of fig. 64, (3). The tangent at (.42 Z, .005104^3)
meets OX where x = .42153 1. The chord connecting (.42 1, .005104 P) and (.43 1,
- ,028044 1^) meets OX where x = .42154 Z. The root is therefore determined to
four decimal places.
64. Multiple roots of an equation.
If f{x) = aQ3if+a^xr-^+a^x"-^-] \- a„_^os^-{- a„_iX + a^,
f'{x) = na^af'-'^ + {n - l)a^c(f'-'' + {n - 2)a^af-'^-\
+ 2 a„_^x + a„_^,
f\x) = n(n- l)a^ar-'+{n -l){n- 2)a^o^-^
+ {n-2){n- 3) a^af -*+ • • • + 2 a„_j,
f"'{x) = n{n-l){n-2) a^-"
+ {n-l){n,-2){n — S)a^af-*-] ,
and so on. Now let /(a), /'(a), f"{a), /'"{a), etc., denote the result
of placing x = a va. these functions, and f{a + h) denote the result
of placing x = a + hin f{x). One readily computes that
f{a + ^) = f{a) + hf'(a) + ^ f"{a) + ^ f"'(a) + . . . + «^». (1)
In (1) place h = x — a and it becomes
f{x) =:f(a) + {x- a)f'{a) + ^^^V"(«)
+ ^^^/'»+--- + «o(^-<. (2)
If now a is a double root oif{x) = Q,f{x) is divisible by {x—af,
by § 42, and therefore, by (2), f{a) = 0, f{a) = 0. If a is a triple
root of f{x) = 0, /(.>:) is divisible by (x— a)\ and therefore /(a) = 0,
f'(a) = 0, /" (a) = 0. Similar statements may be made for multiple
roots of higher order.
MULTIPLE KOOTS
117
Conversely, if f{a) = 0 and f'{ci) = 0, (2) shows that f(x) is
certainly divisible by {x — aY and perhaps by a higher power of
X — a. Therefore a is a multiple root of f(x) = 0. We have then
the result:
A multiple root of/{x) = Ois also a root off'{x)=0, and conversely.
Hence we may find the multiple roots of f{x) = 0 by equating
to zero the highest common factor of f{x) and f'{x) and solving
the resulting equation.
The condition that an equation f{x) = 0 should have multiple
roots is the vanishing of the discriminant of the equation, which
is the eliminant of the equations /{x) = 0 and f'{x) = 0, and may
be found by the method of § 9.
Ex. 1. Find the discriminant of ax^ + 6x + c = 0.
We have to find the condition that the two equations
ax2 + 6x + c = 0
and 2 ax + 6 = 0
should have a common root. Multiplying the last equation by x, we have
2 ax2 + 6x = 0,
and the determinant of the coefficients and the absolute terms of the three
equations is
a b c
0 2 a 6=0,
2a & 0
62 _ 4 ac = 0.
Ex. 2. Find the discriminant of x^ + ox + 6 = 0.
We must find the eliminant of this and
3x2 + a = 0.
Multiplying the first equation by x, and the second by x and x^, we have the
five equations
X* + ax2 + 6x = 0,
x3 + dx + 6 = 0,
3x* + ax2 =0,
3x8 +ax =0,
3x2 +a = 0,
1 0 a 6 0
0 1 0 a 6
3 0 a 0 0=0,
0 3 0 a 0
0 0 3 0a
4a3 + 2762 = 0. (See § 62, Ex. 3.)
■and their eliminant is
118 • THE DERIVATIVE OF A POLYNOMIAL
PROBLEMS
Find the respective slopes of the following curves at the points noted:
(1) by an approximate numerical calculation, as in § 54 ; (2) by placing x equal
to the abscissa of the given point, plus A, and allowing h to approach zero :
1. y = a;3 at (2, 8).
2. y = a;2 - 3 X at (0, 0).
3. y = a;3_3x + l at (1, -1).
4. Find the derivative of x^ — a; by using the definition but not the formulas.
5. Find the derivative of 3 x* + 2 x by using the definition but not the
formulas.
Find the derivative of each of the following expressions by the fonnulas :
6. |X6 -|X5 + X.
7. 4x8-6x2 + 5x-8.
8. 6x9-6x8 + 7x«-4x4-2x2 + 3x-9.
9. By expanding and differentiating show that the derivative of (3 x + 2)*
isl2(3x + 2)8.
10. By expanding and differentiating show that the derivative of (x + a)" is
n(x + a)»-i.
11. Find the equation of the tangent to the curve i/ = x* + 3 at the point the
abscissa of which is — 2.
12. Show that the equation of the tangent to the curve y = x^ + ax + 6 at
the point (Xi, i/i) is y = (3x,'' + a)x - 2xf + 6.
13. Show that the equation of the tangent to the curve y = ax- + 2 6x + c at
the point (xi, yi) is y = 2(axi + b)x — axf + c.
14. Determine the point of intersection of the tangents to the curve y =
x^ — 6x + 7 at the points the abscissas of which are — 2 and 3 respectively.
15. Find the angle between the tangents to the curve y = 2x2 — 3x + lat
the points the abscissas of which are — 1 and 2 respectively.
16. Find the area of the triangle included between the coordinate axes and
the tangent to the curve y = x^ at the point (2, 8).
17. Find the points on the curve y = x* — 3x + 7at which the tangents are
parallel to the line y = dx + S.
18. How many tangents has the curve y = x8— 4x2 + x — 4 which are
parallel to the line y + 4x + 7=0? Find their equations.
19. Find the points on the curve y = x^ + x^ — 6 at which it makes an angle
of 45° with OX.
PROBLEMS 119
Find the values of x for which the following expressions are respectively
increasing and decreasing:
20. x2 + 4a;-7. ' 22. a;* + 8 x - 10.
21. x3-2x2 + 8. 23. X*- 2x2 + 6.
24. Find the lowest point of the curve y = 3x^ — 8x + l,
25. Find the turning points of the curve ?/ = i x* — 2 x2 + i.
Find the maximum and the minimum values of the following expressions :
26. 3x8 _ 2x2 _ 5a; + 1. 27. 3x5 - 25x3 + 60x - 50.
28. Prove that the largest rectangle with a given perimeter is a square.
29. A rectangular piece of cardboard a in. long and b in. broad has a square
cut out of each corner. Find the length of a side of this square when the box
formed from the remainder has its greatest volume.
30. Find the dimensions of the greatest rectangle which can be inscribed in
a given isosceles triangle with base b and altitude h.
3 1 . Find the right circular cylinder of greatest volume which can be inscribed
in a sphere of radius a.
32. Find the right circular cylinder of greatest volume which can be cut from
a given right circular cone.
• 33, Find the point of the line 3x + y = Q such that the sum of the squares
of its distances from the two points (5, 1) and (7, 3) may be a minimum.
34. Among all circular sectors with a given perimeter find the one which
has the greatest area.
35. 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 thickness of
material or waste in construction, what are the dimensions of the largest box
that can be made ?
36. 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 ir : 4.
37. A piece of galvanized iron b ft. long and a ft. wide is to be bent into a
U-shaped water pipe b ft. long. If we assume that the cross section of the pipe is
exactly represented by a rectangle on top of a semicircle, what are the dimensions
of the rectangle and the semicircle that the pipe may have the greatest capacity ;
(1) when the pipe is closed on top ? (2) when it is open on top ?
38. A stream flowing with the velocity a strikes an undershot water wheel,
giving it the velocity x. Assuming that the efficiency of the wheel is propor-
tional to the velocity x of the wheel and the loss of velocity a — x of the
water, what is the velocity of the wheel when it has its greatest eflBciency ?
120 THE DERIVATIVE OF A POLYXOMIAL
39. A gardener has a certain length of wire fencing w4th which to fence
three sides of a rectangular plot of laud, the fourth side being made by a wall
already constnicted. Required the dimensions of the plot which contains the
maximum area.
40. For a continuous girder of uniform section, uniformly loaded, and con-
sisting of three equal spans, the deflection in the middle span is given by the
equation v — C {l^x — 6 l^x- + lOlx^ — 5x*), where C is constant, I the length of
the span, and x the distance from a point of support. Find the greatest
deflection.
41. If p is the density of water and t the temperature between 0° and 30° C,
p = po(l + lt + mt^ + vi^), where po is the density when t = 0, and / = .000052939,
m = - .0000065322, n = .00000001445. Show that the maximum density occure
when « = 4.108°.
42. Show that the curve y = ax^ + bx + c is concave upward or downward
according as a is positive or negative.
43. Show that the curve y = x^ + ax + b is concave upward when x is posi-
tive and concave downward when x is negative.
Determine the values of x for which the foUowhig curves are concave
upward or downward :
44. y = x8-3x2-24. 45. y = a;5_5a; + 6.
Find the points of inflection of the following curves :
46. Gy = x^-Gx^-\-Gx + 1. 47. 12y = x* - 6x^ + 12x^ - 2x+ 1.
48. y = 3x^- lOx* + 10x8 + 6x - 8.
49. y = 3x5 - 5x* 4- 20x8 - 60x2 + 20x - 5.
50. Prove that the curve y = ax^ + bx^ + ex + d always has one and only
one point of inflection.
Find the real roots of the following equations accurate to two decimal
places :
51. x8 - x2 - 2x + 1 =0. 54. x^ - 3x2 - 2x + 5 = 0.
52. x8 + 3x2 + 4x + 5 = 0. 55. x* - x^ - x2 + x - 1 == 0.
53. x8-2x -5 = 0.
Show that each of the following equations has equal roots and solve it :
56. x8-x2-8x+ 12 = 0. 57. x* - 2(l-a)x8 + (l_ 3a)x2 + a = 0.
Find the condition that each of the following equations should have equal
roots :
58. x8 + 3 ax2 + 6 = 0. 60. x* + 4 ax + 6 = 0.
59. X* + 4ax8 + 6 = 0. 61. aox3 + 3aix2 + Soox + as = 0.
CHAPTEE VI
CERTAIN ALGEBRAIC FUNCTIONS AND THEIR GRAPHS
65. Square roots of polynomials. In the previous chapters the
discussion has been restricted to the polynomial. We will next
study the square root of the polynomial.
At first let us assume that the polynomial can be separated
into 71 linear real factors, as in § 42. We have, then,
y
= ± Va, (X - r,) {x-r^)---{x- r„), (1)
and the graph of this function can readily be constructed by con-
sidering the graph of
y = «o (^ - ^i) {-^ - ^^2) •••(•»- r„), (2)
as given in § 46.
In the first place, the graph of (1) will intersect the axis of ic in the
same points as the graph of (2), i.e. in the points x = 1\, x = r^, • • •,
as for these values of x the product under the radical sign is zero.
In the second place, wherever the graph of (2) is below the
axis of X, the expression under the radical sign in (1) is negative,
tlie value of the radical is imaginary, and hence there is no cor-
responding point of the graph. If, however, the graph of (2) is
above the axis of x, there are two values of y in (1), equal in
magnitude and opposite in sign, and correspondingly there are two
points of the graph situated symmetrically with respect to OX.
Therefore OX is an axis of symmetry.
As the negative values of the expression under the radical sign
are separated from the positive values by zero, it follows that the
values of x which make the expression zero, i.e. r^,r„,---, r„, are of
the utmost importance in plotting these graphs. In fact, the lines
X = 1\, x=.r^,---,x = r^ divide the plane into sections bounded by
straight lines parallel to 0 F, in which there will be no part of the
121
122
CEKTAIN ALGEBRAIC FUNCTIONS
graph if the corresponding values of x make the expression negative,
and in which there will be a part of the graph if the corresponding
values of x make the expression positive. Hence the first step in
plotting the graph is the drawing of these lines and the determi-
nation of which sections of the plane should be considered.
Ex. \. y = ± V(x + 2) (X - 1) (X - 5).
K X = — 2, 1, or 5, y = 0, and the graph intersects the axis of x at three points.
The lines x = — 2, x = l, x=5 divide the plane (fig. 65) into four sections.
If a; < — 2, all three factors of the product are negative ; hence the radical
is imaginary and there can be no part of the graph in the corresponding section
of the plane.
Fig. 65
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 1 < X < 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.
GRAPHS
123
Finally, if x > 5, all three factors are positive ; hence the radical is real and
there is a part of the graph in the corresponding section of the plane.
Therefore the graph consists of two separate parts, and is seen (fig. 65) to
consist of a closed loop and a branch of infinite length.
Ex. 2. y = ± V(x + 4) (X + 2) (x - 1) (x - 4).
If X = — 4, — 2, 1, or 4, y = 0, and the graph intersects the axis of x at
four points.
The lines x = — 4, x= — 2, x = l, and x = 4 divide the plane (fig. 66) into
five sections.
I
-X
Fig. 66
If X < - 4, all four factors are negative ; hence the radical is real and tlaere
is a part of the graph in the first section.
If - 4 < X < - 2, the first factor is positive and the others are negative ;
hence the radical is imaginary and there can be no part of the graph in the
second section.
If - 2 < X < 1, the first two factors are positive and the other two are nega-
tive ; hence the radical is real and there is a part of the graph in the third section.
124
CERTAIN ALGEBRAIC FUNCTIONS
If 1 < X < 4, the fii-st three factors are positive and the last is negative ;
hence the radical is imaginary and there can be no part of the graph in the
fourth section.
Finally, if x > 4, all the factors are positive ; hence the i-adical is real and
there is a part of the graph in the fifth section.
In this example we see that the graph consists of three separate parts, and
is seen (fig. 66) to consist of a closed loop and two infinite branches.
Ex.3, y = ± V- (X + 4) (X + 2) (X - 1) (X - 4),
The plane is divided into five sections (fig. 67) by the lines x =
X = 1, and X = 4.
- 4, X = - 2,
Fig. 67
Proceeding as in the previous two examples, we find y to be real if
-4<x<-2, or l<x<4, and to be imaginaiy for all other values of x.
Therefore the graph consists of two separate parts, and is seen (fig. 67) to
consist of two closed loops.
GKAPHS
125
66. In the examples of the last article no two factors were
alike, i.e. no factor occurred more than once. If any factor does
occur more than once, only its first power will be left under the
radical sign, or, to put it more generally, no perfect square will
be left as a factor under the radical sign. As a result, there will
be before the radical a factor involving x, and the presence of
this factor will of necessity change the course of the reasoning to
some extent, as is shown in the following examples.
Ex. 1. y = ± V(x + 2) (x - 1)2.
This will be written as
y = ± (X - 1) Vx + 2.
The line x = — 2 divides the plane
(fig. 68) into two sections.
Proceeding as in the previous ex-
amples, 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
such value as to make the coeflicient 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. 68.
Comparing this example with Ex. 1
of § 65, we see that by changing the
factor X — 5 to x — 1 we have joined
the infinite branch and the loop,
making a single curve crossing itself at the point
Ex. 2. y = ± V(x + 2)2(x - 1) = ± (x + 2) Vx - 1.
The line x = 1 divides the plane (fig. 69) into two sections.
If X > 1, the radical is real and there is a part of the graph in the
corresponding section of the plane. If x < 1, the radical is imaginary and
there will be no points of the graph except for such values of x as make
the coefficient of the radical zero. There is but one such value, i.e. - 2,
and therefore there is but one point of the graph, i.e. (—2, 0), to the
126
CERTAIN ALGEBRAIC FUNCTIONS
left of the line x = 1. The graph consists, then (fig. 69), of the isolated
point A and the infinite branch.
Comparing this example also with Ex. 1 of § 65, we see that by changing
the factor x — 5 to x + 2 we have reduced the loop to a single point, leaving
the infinite branch as such.
Fig. 70
Ex.3, y = ± V_ (X + 4) (X + 2)2(x - 4) = ± (X + 2) V- (X + 4) (X - 4).
The lines x = - 4 and x = 4 divide the plane (fig. 70) into three sections.
If - 4 < X < 4, the radical is real and there is a part of the graph in the cor-
responding portion of the plane. If x < - 4 or x > 4, the radical is imaginary ;
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. 70.
Comparing this example with Ex. 3 of § 65, we see that the changingof x - 1
to X + 2 has brought the two loops together, forming a single closed curve cross-
ing itself at the point ( - 2, 0).
FUNCTIONS DEFINED BY EQUATIONS
127
67. Functions defined by equations of the second degree in y.
If we have given an algebraic equation involving both y and x,
y is thereby defined as a function of x. For if x is assigned any
value, the corresponding values of y are determined by means of
the equation. In particular, if the equation involves no power of
y higher than the second, it may be readily solved for y, and the
work of finding the graph is similar to that already done.
In many important cases the solution of the equation is of
the form
y = c± y/(x — Vj) (x — r^)--'.
Comparing this case with the previous one, we see that y = c
is an axis of symmetry instead of y — 0, and that in all other
respects the work is similar.
Ex. 2x2 + 2/2 + 3x -42/- 5 = 0.
Solving for y, we have
y = 2± V-2x2_ 3x + 9,
or, after the expression under the radical
sign has been factored,
y = 2± V-2(x-|)(x + 3).
The lines x = — 3 and x = | divide
the plane (fig. 71) into three sections, and,
proceeding as before, we find that the
curve is entirely in the middle section,
i.e. when — 3 < x < ^, and that the line
2/ = 2 is an axis of symmetry.
Fig. 71
In case the given equation is
of higher degree in y than the
second, but of the first or the second degree in x, it is evident
that we can solve for x in terms of y and proceed as above,
working from the y axis instead of the x axis.
It should be added that given any equation in x and y, since
either may be regarded as the independent variable and the other
as the function, we have perfect freedom of choice to solve for y in
terms of x, or for x in terms of y, according to convenience.
128
CERTAIN ALGEBRAIC FUNCTIONS
68. Functions involving fractions. If the expression defining a
function contains fractions, the function is not defined for a value of
X which makes the denominator of any fraction zero (§ 11). But if
a? = a is a value which makes the denominator zero, but not the
numerator, and x is allowed to approach a as a limit, the value of
the function increases indefinitely and is said to become infinite. The
graph of a fimction then runs up or down indefinitely, approaching
the line x = a indefinitely near, but never reaching it. We have
thus a graphical representation of the discussion of infinity in § 11.
When a fimction becomes infinite it is discontinuous (§ 56).
In fact, this is the only kind of discontinuity which can occur in
an algebraic function.
Ex. 1. y
Fig. 72
a;-2
It is evident that y is
real for all values of x;
also if a; < 2, y is negative,
and if a; > 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,
that value being excluded
by § 11. The curve is repre-
sented in fig. 72.
It is seen that the nearer
to 2 the value assigned to x,
the nearer the correspond-
ing 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 indefinitely 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.
FUNCTIONS INVOLVING FRACTIONS
129
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 a; = 2 is determined by the value of x which makes the function infinite.
It is clear that all equations of the type
1
y =
X — a
represent curves of the same gen-
eral shape as that plotted in fig. 72.
Ex.2. y =
+
x + 2
If X = — 2 or if X = 2, y is
infinite ; hence these two values
may not be assigned to x, all
other values, however, being
possible. The curve is repre-
sented in fig. 73.
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
y
1
4-
X — a X — b
and it is not difficult to see how
the curve represented by
1 1 1
y
+
+
+
X — a X — b X — c
will look for any number of terms.
1
Ex. 3. y =
(X - 2)2
All values of x may be assumed
except 2. The curve is represented
in fig. 74. It is evident that the lines
x = 2 and y = 0 are asymptotes.
This curve is a special case of
that represented by
_ 1
^~(x-a)2'
which is itself a special case of
1 1
Fig. 74
(X - a)2 (X - 6)
+
130
CERTAIN ALGEBRAIC FUNCTIONS
Ex. 4. y* =
x-3
As in § 67, we solve for y, forming tlie equation
ion y = ±yj^
The line
X = 3 (fig. 75) divides the plane into two sections, and it is evident that there
can be no part of the curve in that section for which a; < 3. Moreover, this
Fig. 75
line X = 3 is an asymptote, as in the preceding examples. The curve, which
is a special case of that represented by
Is represented in fig. 75. It is to be noted that the axis of x also is an asymptote.
Ex.5. y = ?i±i.
X
To plot this curve we write the equation in the equivalent form
y = x + -.
(1)
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 = x,
(2)
a straight line passing through the origin and bisecting the first and the third
quadrants.
SPECIAL IRRATIONAL FUNCTIONS
13i
Comparing equations (1) and
(2), we see that if any value xj,
is assigned to x, the corre-
sponding ordinates of (1) and
(2) are respectively xi -\ and
Xi ■,
Xi, and that they differ by — .
Moreover, the numerical value
of this difference decreases as
greater numerical values are
assigned to Xi, and it can be
made less than any assigned
quantity however small by tak-
ing Xi sufficiently great. It
follows that the line y = x isan
asymptote of the curve. It is
also evident that the line x = 0,
determined by the value of x
which makes the function infi-
nite, is an asymptote. The curve
is represented in fig. 76.
(£)A
Fig. 70
69. Special irrational functions.
Ex. 1. 2/2 = x3.
Writing this equation in the form y = ±x Vx,
we see that y is an irrational function of x, and
that its graph is symmetrical with respect to
OX and lies entirely to the right of the axis y.
It is represented in fig. 77, and is called the
semicubical parabola.
In general, if the equation expressing the
_y function is of the form
y = kx",
the function is rational or irrational according
as n is integral or fractional. In § 38 we have
plotted the graphs of some of the rational func-
tions of this type for the special case when k= 1
and n has the values 3, 4, and 5 respectively.
Above we have just plotted the graph of one of
the irrational functions, i.e. when n = ^.
The grajDhs of the irrational functions y = x^,
■pio, 77 y = X*, and y = x^ may be obtained by assuming
values for x and plotting as above, or by rewriting
the equations in the forms x = y^, x = y*, and x = y^, when it is immediately
evident that their graphs are respectively the same in shape as those of the
132
CEKTAIN ALCiEBRAIC FUNCTIONS
Fig. 78
rational functions y = x^, y = x*, and
y = x^ already plotted, the axes of
X and y, however, being changed in
position.
It is to be noted that the graphs of
all the functions expressed by the
equation y = x" pass through the points
(0, 0) and (1, 1).
Ex. 2. a;' + 2/5 = a'.
If y is defined as a function of x
the equation x' + y^ = a\ it is
evident that its graph will lie entirely
in the first quadrant, since both x
and y must be positive, and that its relative positions with respect to the
two axes of coordinates are the same
(fig. 78). The curve is a parabola
(§ 79). If the equation is put in the
form 2/ = (a* — x^)-, it is .seen that y
is an irrational function of x.
Ex. 3. xi + y^ = al
Writing this equation in the form
y = ± (a* — x^)-, we see that 2/ is an
irrational function of x, and that its
graph is symmetrical with respect to
OX and bounded by the lines x = — o
and X = a. In the same way we may
show that the graph is symmetrical
with respect to OY and bounded by the
lines y = ~ a and y= a. It is repre-
sented in fig. 79, and is a four-cu.sped hypocycloid.
Ex. 4. x3 + y^ -3axy = 0.
The graph of this equation,
by which y is defined as an
irrational function of x, is repre-
sented in fig. 80, and is known
as the Folium of Descartes.
It is symmetrical with respect
to the line y = x and has the
line X + y + a = 0 as an
asymptote. While it may be
plotted by assuming values for
X and solving the corresponding
cubic equations for y, it is more
easily plotted when different
axes of coordinates are chosen
(see Ex. 38, Chap. X).
PROBLEMS
.00
PROBLEMS
Plot the graphs of the following equations :
1. y2 = (X - 1) {x'^ - 4). 29. 2/3 = a;2(x + 2).
2. y2 ^ (X + 2) (8x - x2 _ 15). 30. {y + 2)3 = (x - 1) (x2 - 4).
3. 4 2/2 = (x + 3)(2x-3)2. 31. X2/ = 7.
4. 42/2 = x2{x + l). • 32. xy = -1.
5. 2/2 = (X- 3)2 (.5-2 X). __ 16
66. y =
6. ?/2 = (3x + 2)(9x2-4). ^-*
7. 2/2 = (X - 2)2(4x2 - 4x - 15). 34. ?/ = ^- L_.
(X - 1)2 (X + 3)2
8. y2^(4x2-l){x2-4).
9. 2/2 = {2x+5)2(6 + x-x2). ^^- 22/ = 3x4---
10. 2/2 = -x2(x + 3)2(x + l). 36. 2/-2 = 2(x-l)+ ^ .
11. 2/2 = x2{x-2)2(x-3). „„ 1
37. (2/ -2)2 = ^.
12. 2/2 = (1 - x2) (x2 - 9). a^ + 1
13. 2/2 = (2x-5)(x2 + 2). 38. y2 = 5i^±^.
14. 2/2=(x-2)2(x2-f-2).
15. 2/- = (x-2)(2x-3)2(x2+x+l). ^^ y^^^Zr^-
16. 10^2- 42-4 _a;6. 3
40. 2/2 = ?
17. x2-2/2-4x + 6y-l = 0. x2-6x + 8
18. 4x2 + 92/2+ 4x- 122/ -31=0. *^- a;22/2 + 36 = 4 2/2.
19. x2 - 2/3 + 32/2 + 2/ - 3 = 0. ^2- l^'^'^/' = 62a;2(a2 - 2ax).
20. x2-2/4(4 + 2/) = 0. 43. 2/2 = ^!^^^±^.
21. [x2 + .S(2/-l)][x2-3(2/-l)]=0.
44. 2/(x2 + a2) = a2(a-x).
22. (2/-l)2 = (x-l)2(x-4).
45. 2/2(x2 + a2)_a2a;2.
23. (2/ _ x)2 = 9 - x2.
46. a*2/2 + 62a;4 = a262x2.
24. (X + ?/)2=: 2/2(2/ + 1).
^ ^ •^Vi'^; 47. 2/2(a2 + x2) = x2(a2-x2).
25. x2 - 4 X2/ + 8 2/2 - 2/* = 0.
48. xy2 = 4a2(2a-x).
^^■^h^h'- 49.. = x= + i.
27. 2/3 = x4. J
28. 2/3 = X (x2 - 4). • ^ ^ ^ + ^"
CHAPTEE VII
CERTAIN CURVES AND THEIR EQUATIONS
70. The circle. Wheu a curve has been defined by a geometric
property it is often possible to find the equation of the curve by
expressing the definition in algebraic symbols. This equation
serves, then, as a means for plotting the curve and also as a basis
for examiniug its other properties. In this chapter we shall derive
the equation of certain important elementary curves, beginning
with 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 {d, e) (fig. 81) be the coordi-
nates of the center C, and r the
radius of the circle. Then if P (x, y)
is a point on the circle, x and y
must satisfy the equation
(x-df^{y-ef = r', (1)
by § 17.
Conversely, if x and y satisfy
the equation (1), the point {x, y)
is at a distance r from {d, e) and
therefore lies on the circle.
Therefore (1) is the equation of the circle (§ 22).
Equation (1) expanded gives
x-+f~ 2dx-2ey + d'+e^-r^=0;
and if this is multiplied by any quantity A, it becomes
Ax' + Af+2Gx+2Fy + C=0,
where
Fig. 81
(2)
d=--
A
e=--, d?Jre^-r' = -'
A A
134
THE CIRCLE 135
Ex. The equation of a circle with the center (^, — J) and the radius § is
(a;-J)2 + (y + J)2 = 4,
which reduces to 12 a;^ + 12 ys _ 12 x + 8 y - 1 = 0.
71. Conversely, the equation
where A^ 0, represents a circle, if it represents any curve at all.
To prove this, we will transform the equation as follows :
a?+2-x + y^ + 2-y=--,
A ^ A"^ A
^,0^ , <^\ 2,0^ , ^' G^+F^-AC
4/ y A I A'
There are then three possible cases :
1. G^ + F^-AOO. The equation is then of the type (1), § 70,
1, ^ ^ ^' -2 G^+F^~AC ,^, ,
where a = 7'^ = ,r=^ — , and therefore represents
A A A^ .
a circle with the center ( , | and the radius \
\ A AJ > A^
2. G'' + F--AC=Q. The equation is then
which can be satisfied by real values of x and y only when
G F
x = and y— Hence the equation represents the point
— -> I . This may be called a circle of zero radius, regarding
it as the limit of a circle as the radius approaches zero.
3. G^-\-F^ — AC<0. The equation can then be satisfied by no
real values of x and y, since the sum of two positive quantities
cannot be negative. Hence the equation represents no curve.
136 CERTAIN CURVES AND THEIR EQUATIONS
Ex. 1. The equation x2 + j/2-2x + 4y4-l = 0 may be written
(X - 1)2 + (2/ + 2)2 = 4,
and represents a circle with center (1, - 2) and radius 2.
Kx. 2. The equation x^ + y- - 2 x + iy + o = 0 may be written
(x'- 1)2 + (y + 2)2 = 0,
and is satisfied only by the point (1, - 2).
Ex. 3. The equation x^ + y^ -2x + 4y + 7 = 0 may be written
(X - 1)2 + (y + 2)2 = - 2,
and represents no curve.
72. To find the equation of a circle which will satisfy given
conditions, it is necessary and sufficient to determine the three
quantities d, e, r, or the ratios of the four quantities A,G,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, m 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 {x^, y^ must satisfy the equation
of the circle ; therefore d, e, and r must satisfy the condition
{x,-df+{y-ef = r\
2. Let the condition be imposed upon the circle to be tangent
to the known straight line Ax + By + C=Q. Then the distance
from the center of the circle to this line must equal the radius ;
therefore, by § 32, d, e, and r must satisfy the condition
Ad + Be + C
- = ± r.
y/A' + B^
The sign will be ambiguous, unless from other conditions of the
problem it is known on which side of the line the center lies.
THE CIRCLE 137
3. Let it be required that the center of the circle should lie on
the line Ax + i?y + C= 0, Then d and e must satisfy the condition
Ad + Be + C=Q.
Ex. 1. Find the equation of the circle through the three points (2, — 2),
(7, 3), and (6, 0).
The quantities d, e, and r must satisfy the three conditions
(2-d)2 + (-2-e)2 = r2,
(7 - d)2 + (3 - e)2 = r2,
(6 - d)2 + (0 - e)2 = r2.
Solving these we have d = 2, e = 3, and r = 5. Therefore the required
equation is
(x-2)2 + (y._3)2 = 25,
or x2 + 2/2 - 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 Sy + x — IS = 0.
The quantities d, e, and r must satisfy the conditions
(2-d)2+ (-3-e)2 = r2,
(_4_d)2 + (_l_ e)2 = r-2,
3 e + d - 18 = 0.
Solving these equations we find d = |, e = y-, r2=-l-|-S.. Therefore the
required equation is
or a;2 + ^2 _ 3a; _ 11 2/ - 40 = 0.
Ex. 3. Find the equation of a circle which is tangent to the lines
17 x + 2/ -35 = 0 and 13x + II2/ + 50 = 0,
and has its center on the line 88 x + 70 2/ + 15 = 0.
The quantities d, e, and r must satisfy the conditions
17d + e-35
■■±r.
V290
- 13d -lie -50
= ±r,
V290
88d + 70e + 15 = 0.
138 CERTAIN CURVES AND THEIR EQUATIONS
These equations have the two solutions
and d = 6, e = — ^^ ,
6 '
3V29O
20
Hence each of the two circles
3a;2 + 32/2+ 5x- 5j/-20 = 0
and 40x2 + 402/2 _400x + 620y + 2429 = 0
satisfies the conditions of the problem.
Ex. 4. The equation of a circle through three given points is most readily
found by means of the equation
^x2 + .42/2 + 2Gx + 2Fy-\-C=0.
If (a^i, yi), {3^1 1/2)1 and (X3, 2/3) are the three given points, the quantities
A, G, F, C must satisfy the equations
^x 2 + Ay^ + 2Gxi + 2Fyi + C=0,
Ax^ + Ayl + 2 Gx2 + 2 Fyo + C = 0,
Ax^ + Ay^ + 2 Gx3 + 2 Fys + C = 0.
There are here four homogeneous equations in the unknowns .4, G, jP, C,
and the result of eliminating the unknowns is, by § 9,
x2 + ?/2 X y 1
^X+Vl *! Vi 1
xi + yi X2 2/2 1
^s+v! X3 Vs 1
0,
(1)
which is the required equation of the circle.
It is to be noticed that the coefficient of x2 4- 2/2 in (1) is
xi Vi 11
X2 2/2 1 1 .
xs Vz 11
When this is zero, equation (1) is of the first degree and represents a straight
line. But when
Xi
2/1 1
X2
Vi 1
X8
2/8 1
= 0,
the points (Xi, 2/1), (X2, 2/2), and (X3, 2/3) are on the same straight line (§ 29, 5) and
cannot determine a circle.
THE ELLIPSE
139
73. The ellipse. A71 ellipse is the locus of a point the sum of
the distances of which from two fixed points is constant.
The two fixed points are called the foci. Let them be denoted
by F and F' (fig. 82) and let the axis of x 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
F'P + FP=2a. (1)
From the triangle F'PF it follows that
F'F<2a.
Hence there is a point A on the axis of x and to the right
of F which satisfies the definition. We have then
F'A + FA=2a,
or {F'O + OA) + {OA- OF) =2 a,
whence OA = a.
Let us now place
OF
OA
= e, where e < 1.
Then the coordinates of F and
F' are {±ae, 0). Computing
the values of F'P and FP by § 17, and substituting in (1), we have
V(a?+ aeY+ y^ + V(^ — ae)'^ + y^ = 2>
(2)
By transposing the second radical to the right-hand side of the
equation, squaring, and reducing, we have
a — ex= ^{x—aeY+f = FP. (3)
Similarly, by transposing the first radical in (2), we have
a + ex = y/{x + aef +f = F'P. (4)
140 CERTAIN CURVES AND THEIR EQUATIONS
Either (3) or (4) leads to the equation
or -,+ ./ , =1. (6)
Since e <1, the denominator of the second fraction is positive
and we place „ , ,o
thus obtaining ~2 + ^2 ~ -^- C^)
We have now shown that any point which satisfies (1) has co-
ordinates which satisfy (7).
To show, conversely, that any point whose coordinates satisfy (7)
is such as to satisfy (1), let us assume (7) as given. We can then
obtain (6) and (5), and (5) may be put in each of the two forms
ar^ + 2 aex + a^c^ + if = «- + 2 aex + ^o?,
ar^ — 2 aex + cj^c^ + if = ar—1 aex + e^x?,
the square roots of which are respectively
F'F=±{a + ex),
FP = ±{a — ex).
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 nega-
tive numbers 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
(7) satisfies (1), and therefore (7) is the equation of the ellipse.
THE ELLIPSE
141
74. Placing y = 0 in (7), § 73, we find x = ± a. Placing x = 0,
we find y = ±h. Hence the ellipse intersects OX in the two points
A{a, 0) and A' {— a, 0), and intersects OF in two points B{Q, h)
and -B'(0, — h). The points A and ^' are called the vertices of the
ellipse. The line AA', which is equal to 2 a, is called the mayor
axis, and the line ^^', which is equal to 2 h, is called the wtTior
axis of the ellipse.
Solving (7) first for y and
then for x, we have
M
y = ±-^a'-ci»
A'
and
a; = ± - V&-^ - y"^
B'
Fig. 83
iV B' K
These equations show (1) that
the elHpse is symmetrical with
respect to both OX and OY, (2) that x can have no value numer-
ically greater than a, (3) that y can have no value numerically
greater than b. If we construct the rectangle KLMN (fig. 83),
which has 0 for a center and sides equal to 2 a and 2 h respec-
tively, 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. 82 and 83.
75. Any equation of the form (7), § 73, in which a > b,
represents an ellipse with the foci on OX. For if we place,
as in §73, &"' = a'(I-e'), we find
Vc^^'
and may fix i^and F', which in § 73 were arbitrary in position, by
the relation 0F = — OF' = ae.
The foci may be found gi'aphically by placing the point of a com-
pass on B and describing an arc with the radius a. This arc will
intersect AA' in the foci ; for since OB = h and OF = Va^ — b\
BF=a.
142 CERTAIN CURVES AND THEIR EQUATIONS
Similarly an equation of the form (7), § 73, in wliich h>a,
represents an ellipse in which the foci lie on BB' at a distance
VV^—a^ from 0. In this case BB' = 2 & is the major axis and
AA' = 2 a is the minor axis.
It may be noted that the nearer the foci are taken together, the
smaDer is e and the more nearly h = a. Hence a circle may he
considered as an ellipse with coincident foci and equal axes.
76. The hyperbola. An hyperhola is the locus of a point the
difference of the distances of which from two fixed points is constant.
Fig. 84
The two fixed points are called the foci. Let them be F and
F' (fig. 84) and let FF' be taken as the axis Of x, the origin being
lialfway 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 F', we have either
or
F'P-FP = 2a,
FP-F'P = 2a.
(1)
(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 O and F which
satisfies the definition.
THE HYPERBOLA 143
Then F'A—AF=2a,
or {F'0 + 0A)-{0F-0A)=2a;
whence OA = a.
We may therefore place
— - = e, where e > 1.
OA
Then the coordinates of F and F' are (± ae, 0) and equations
(1) and (2) become
V(aj + aef + / — V(a; — ae)"-^ +f = 2a (3)
and V(a; — aef + 2/"^ — V(a; + aef + y- = 2 a. (4)
By transposing one of the radicals to the right-hand side of
these equations, squaring, and reducing, we obtain from (3) either
+ a= y/{x + aef + f = F'P,
— a = y/{x — aef -\-f=FP',
and from (4) we obtain either
ex
or ex
■ {ex + a) = y/{x + aef+y'' = F'P,
or — {ex — a) = ■>J{x — aef + / = FP.
Any one of the last four equations gives
{l-e')x'+f = d\l-e\ (5)
^ , f
a^ a\l-e^)
^.+ ..f ..=!• (6)
But since e >l,a^{l — e'^) is a negative quantity and we may
write a^{l — e^) = — b^, thus obtaining
^-C = l. (7)
b
144 CERTAIN CURVES AND THEIR EQUATIONS
Then any point wliich satisfies (1) or (2) satisfies (7). Conversely,
by retracing our steps, we find that if the coordinates of a point
P satisfy (7), then
F'P = ±(ex + a)
and FP = ±{ex — a).
Hence we must have either
F'P — FP = 2a,
— F'P+FP = 2a,
F'P + FP = 2a,
or -F'P — FP = 2a.
The equation F'P + FP = 2 a is impossible, for F'P + FP> F'F,
and 2a < F'F. The equation — i^'P — i^P = 2 a is also clearly
impossible. Hence any point which satisfies (7) satisfies either
(1) or (2). Therefore (7) is the equation of the hyperbola.
77. If we place y = 0 in (7), § 76, we have x = ±a. Hence
the curve intersects OX in two points, A and A', called the vertices.
It x= 0, y is imaginary. Hence the curve does not intersect OY.
Solving (7), § 76, for y and x respectively, we have
a
and x=±- Vy+i^.
These show (1) that the curve is symmetrical with respect to
both OX and OY, (2) that x can have no value numerically less
than a, and (3) that y can have all values.
Moreover, the equation for y can be written
THE HYPERBOLA
145
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 asymptotes of the hyperbola. They are the
diagonals of the rectangle constructed as in fig. 85, and are used
F
Fig. 85 ,
conveniently as guides in drawing the curve. The line AA' is
called the transverse axis and the hne BB' the conjugate axis
of the hyperbola. The shape of the curve is shown in figs.
84 and 85.
78. Any equation of the form (7), § 76, where a and h are any
positive real values, represents an hyperbola with the foci on AA'.
VaF+l?
For if we place —h^ = a'^(l — e'), we find e = ■ and may
146 CERTAIN CURVES AND THEIR EQUATIONS
find the position of the foci from the equations OF = — OF' = ae.
Similarly any equation of the form
represents an hyperbola mth the foci on BB'.
\ih = a, the hyperbola is called an equilateral hyperhola and its
equation is either ar^— z/^ = a^ or — ar^+ y^ = a^
79. 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 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. 86 let i^ be the
focus, BS the directrix intersecting
OX at D, and let P be any point on
the curve. Then the coordinates of
-X F are (p, 0), those of D are {—p, 0),
and the equation of BS is « = —p.
Draw from P a line parallel to OX
intersecting BS in N. If F is on the
right of BS, P must also lie on the
right of BS, and by the definition
FP = NP.
If, on the other hand, F is on the left of BS, P is also on the
left of BS and
FP = PN=-NP.
In either case FP^ = NP^.
But FP = (x-pY-\- f, and NP = x + p;
hence {x~pf+f = (x + p)',
which reduces to -f = i^px.
(by § 17)
(1)
THE PARABOLA
147
Any point on the parabola then satisfies this equation.
Conversely, it is easy to show that if a point satisfies this
equation, it must so lie that FP = ± NP, and hence lies on
the parabola.
Equation (1) shows (1) that the curve is symmetrical with
respect to OX, (2) that x must have the same sign as p, and (3)
that y increases as x increases numerically. The position of the
curve is as shown in fig. 86 when jp is positive. When p is neg-
ative F lies at the left of 0 and the curve extends toward the
negative end of the axis of x.
Similarly the equation a? = 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.
80. If Pi{x^, 2/i) and Po{x^, y„) are two points on the parabola
y^ = 4:px (fig. 87), then
yl = ^px^;
hence
Fig. 87
That is, tlie squares of the ordinates
of a parabola are to each other as the
abscissas. Conversely, if in any curve the squares of the ordinates
are to each other as the abscissas, the curve is a parabola.
For let i^ be a known point and P any point on the curve.
Then, by hypothesis.
which may be written
y^ = ^x.
But this is the same as y^ = 4:px, where p =
148 CERTAIN CURVES JlSD THEIR EQUATIONS
81. The conic. A conic is the locus of a point the distance
of which from a Jixed 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, and a line through
the focus F as the axis of x, and shall caU the coordinates of the
focus (c, 0), where c represents OF and
is positive or negative according as F
lies to the right or the left of O.
Let P be any point on the conic;
connect P and F, and draw PN per-
pendicular to OY. Then by definition
FP = ±eXP, (1)
according as P is on the right or the
left of OT. In both cases
FP* = e'- XP\
But FP* = (a: — ef + y», by § 17, and
NP = z. Therefore for any point on
the conic
r
/
/
/•I
/
0
\
\
\
\
\
\
\
\
\
\
\
\
Fio. 88
{x-c)'+f = t^Ji*.
(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 « of § 73 and < 1.
For if P (fig. 89) is a point on the ellipse -i + ^ = 1> we found
in § 73 that a 6-
FP = a- ex, F'P = a + ex,
or FP = e(^-x\ F'P = e(- + x\
THE CONIC
149
If now we take the point D so that OD — - » and Z)' so that
OD' = » draw the lines DS and Z>'*Sf' perpendicular to OX, the
line N'FN perpendicular to I>S, and the ordinate MF, we have
- — x = OI)— OM = MD = PN,
e
- + x = D'0-{- OM = D'M = N'P.
Fig. 89
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.
In § 114, Ex. 3, we shall prove that the conic is always either
an ellipse, a parabola, or an hyperbola.
82. The witch. Let OBA (fig. 90) be a circle, OA a diameter,
and LK the tangent to the circle at A. From 0 draw any line
intersecting the circle at B and LK at C. From B draw a line
parallel to LK and from C a line perpendicular to LK, and call
the intersection of these two Hnes P. The locus of P is a curve
called the witch.
150 CERTAIN CURVES AND THEIR EQUATIONS
To obtain its equation we wiU 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=x, MP = y,
MP OB
0A=MC=2a.
OBOC
In the triangle OMC, — — = — - =
° MC OC OC
If AB is drawn, OB A is a right angle and consequently
0B0C = 0A\ also OC' = OM^'+MC^
(1)
Fig. 90
Therefore
that is,
and finally.
JifP ^ OA
^C~ OM^ + MC^
y __ 4 a'^ .
2a
y
x'+4.a'
(2)
(3)
(4)
Conversely, if equation (4) is satisfied by any point, we can
deduce equations (3), (2), and (1) in order, and hence show that
the point is on the witch.
Solving (4) for x, we have
. ^ \2a-y
THE CISSOID
151
This shows (1) that the curve is symmetrical with respect to OY,
(2) that y cannot be negative nor greater than 2 a, and (3) that
?/ = 0 is an asymptote.
83. The cissoid. Let ODA (fig. 91) be a circle with the diam-
eter OA, and let LK be the tangent to the circle at A. Through
0 draw any Ime intersecting the circle in D and LK in E. On
OE lay off a distance OP 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 coordinates
and OA as the axis of x, and will
call the diameter of the circle 2 a.
Draw MP perpendicular to OA.
Then if ^ and Z> are connected, a
triangle ADE is formed similar
to OMP ; whence
OP _AE
MP~ DE
By hypothesis DE = OP.
Therefore
W^MPAE. (2)
Also, in the similar triangles
OAE and 0PM,
AE _ MP .
0A~~ OM'
whence, from (2),
^2 _ OA . MP'
or X -\- y =
whence y^ =
OM
2af,
9.n.—
This equation is satisfied by
the coordinates of any point
upon the cissoid.
Fig. $)1
152 CERTAIN CURVES AND THEIR EQUATIONS
Conversely, if we assume equation (6), we may deduce (5) and
(4), aud then by aid of (1) and (3) we have OP = DE.
Therefore (6) is the equation of the cissoid. It may be written
y = ±x
1 X
\2a-x
From this it appears
(1) that the curve is
symmetrical with re-
spect to OX, (2) that
no value of x can be
greater than 2 a or
less than 0, and (3) that
the line x = 2a is an
asymptote.
84. The strophoid.
Let ZA' and i?^' (fig. 92)
be two straight lines
intersecting at right
A angles at 0, and let A
be a fixed point on LK.
Through A draw any
straight line intersecting
RS in. D, and lay off on
AD in either direction a
distance DP equal to OD.
The locus of P is a curve
called the strophoid.
To find its equation,
take LK as the axis of
X and BS as the axis of
y, and call the coordi-
nates of A (a, 0). By
the definition the point
P may fall in any one
of the four quadrants.
THE STROPHOID 153
If we take the positive direction on AD as measured from A
towards D, we have
OD=PI>
when F is in the first quadrant,
OD = -PD
when F is in the second quadrant,
-OD = -FD
when F is in the third quadrant, and
-OD = FD
when F is in the fourth quadrant.
These four equations are equivalent to the single equation
od'=fi>\ (1)
From the similar triangles OAD and APM,
OD MF y
AD AF V(ic-
■cif+f
FD MO OM
AD AO OA
X
a
f
{x-af+y^ a"
Hence ^^_af+f-d^ ^^
is an equation satisfied by any point on the curve. Conversely,
if (2) is given, (1) may be deduced. Therefore (2) is the equation
of the strophoid.
It may be written
= ±Xy^
\a — x
^ ^a-\- X
This shows (1) that the curve is symmetrical with respect to
OX, (2) that no value of x can be less than - a nor greater than
+ a, and (3) that « - - « is an asymptote.
154 CEKTAIN CURVES AND THEIR EQUATIONS
85. Examples. The use of the equation of a curve in solving
problems connected with the curve will be constantly illustrated
throughout the book. The following examples depend upon prin-
ciples already given.
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. 93)
A'Mi ■ M2A '
Let the coordinates of Pi be
X (p^ii Vi) ^-nd let those of P^ be
(X2, 2/2)- Then
(g + xi){a - xi)
Fig. 93 y\ a'^ - x^ (« + ^^H'^ - ^)
But yi = MiPi, a + Xi = A'O + OMi = A'Mi, a-Xi=OA - OMi = MiA,
y-i = M2P2, a + X2 = A'Mz, a — X2 = M2A. Hence the proposition is proved.
Ex. 2. If MiPi is the ordinate of a point Pi of the parabola, y2 _ 4 px^ and
a straight line di-av?n through the middle point of ilfiPi parallel to the axis of x
cuts the curve at Q; prove that the intercept of the line MiQ on the axis of y
equals § MiPi.
Let the coordinates of Pi (fig. 94) be
y?
(Xi, yi). Then xi = — from the equation of
the parabola.
4p
By construction, the ordinate of Q is
Vi
Since Q is on the parabola its abscissa is
found by placing y = ~ in y'^ = 4px. The
/ 2 \
coordinates of Q are then ( — - , —)• The
\16p 2/
cobrdinates of M^ are (xi, 0), which are the
same as ( — » Ol. Hence the equation of
MQ is, by § 29,
Fig. 94
8px + 3yiy-2y^ = 0.
The intercept of this line on OF is § yi = 2 ilfiPi, which was to be proved.
PROBLEMS 155
PROBLEMS
1. Find the equation of the circle having the center (2, —4) and the radius 3.
2. Find the equation of the circle having the center (— §? ^) and the
radius 6.
3. Find the equations of the circles having the line joining (2, 3) and ( — 3, 1)
as a radius.
4. Find the equation of the circle having the line joining (a, — h) and
( — a, 6) as a diameter.
5. Find the equations of the circles of radius a which are tangent to the
axis of y at the origin.
6. Find tiie equations of the circles of radius a which are tangent to both
coordinate axes.
7. Find the equation of the circle having as a diameter that part of the line
2x — Sy + 6 — 0 which is included between the coordinate axes.
8. Find the center and the radius of the circle x^ + ?/2 + 4 a; — 10 y — 36 = 0.
9. Find the center and the radius of the circle x^+y^-\-4x-Qy-\-l = 0.
10. Find the center and the radius of the circle 3x^ + 3y^ — 9x + 6y — 2 = 0.
1 1. Find the center and the radius of the circle 5x^+ 5y^ + 2x — 4ty + 1 = 0.
12. Prove that two circles are concentric if their equations differ only in the
absolute term.
13. Show that the circles x^ + y^ + 2Gx + 2 Fy + C = 0 and x^ + y^ + 2 G'x
+ 2 Yy + C" = 0 are tangent to each other if
V((?-G')2 + (F-i?")2 = Vg2 + F2 _ c- ± VG'2 + 2f"2 _ c".
14. Find the equation of the circle which passes through the points (0, 3),
(3, 0), (0, 0).
15. Find the equation of the circle circumscribing the triangle with the
vertices (0, 2), (- 1, 0), (0, - 2).
16. Find the equation of the circle circumscribed about the triangle the sides
of which are x + y — 2 = 0, 9x + oj/ — 2 = 0, 2/ + 2x — 1 = 0.
17. Find the equation of the circle passing through the point (— 2, 4) and
concentric with the circle x2 + ?/2— 5x + 4?/ — 1 = 0.
18. A circle which is tangent to both coordinate axes passes through (4, 2).
Find its equation.
19. The center of a circle which is tangent to the axes of x and y is on the
line 2x — 3y + 6 = 0. What is its equation ?
20. A circle of radius 5 passes through the points (2, — 1) and (3, — 2).'
What is its equation ?
21. The center of a circle which passes through the points (1, — 2) and
(- 2, 2) is on the line 8x-42/ + 9 = 0. What is its equation ?
156 CERTAIN CURVES AND THEIR EQUATIONS
22. A circle which is tangent to OX passes through (-3, 2) and (4, 9).
What is its equation ?
23. The center of a circle which is tangent to the two parallel lines x — 3 = 0
and a; - 7 = 0 is on the line y = 2 x + 4. What is its equation ?
24. The center of a circle is on the line 2 x + y = 0. The circle passes
through the point (4, 2) and is tangent to the line 4x-3?/-15 = 0. What is
its equation ?
25. Find the equation of the circle circumscribing the isosceles triangle of
which the altitude is 4 and the base is the line joining the points (— 3, 0) and
(3, 0).
26. Find the equation of the ellipse the foci of which are (±3, 0) and
the major axis of which is 8.
27. Find the equation of the ellipse the foci of which are (0, ± 2) and the
major axis of which is 6.
28. Find the equation of an ellipse when the vertices are (±6, 0) and one
focus is (4, 0).
29. Determine the semiaxes o and h in the ellipse — H = 1, so that it will
pass through (1, 4) and (2, — 3).
30. If the vertices of an ellipse are (± 5, 0) and its foci are (±3, 0), find
its equation.
31. The center of an ellipse is at the origin and its major axis lies along OX.
If its major axis is 8 and its eccentricity is ^, find its equation.
32. Find the equation of an ellipse when its center is at the origin, one focus
at the point (— 3, 0), and the minor axis equal to 8.
33. Find the equation of an ellipse the eccentricity of which is ^ and the
foci of which are (0, ± 6).
34. Given the ellipse 9x2 + 16y2 = 144. Find its semiaxes, eccentricity, and
foci.
35. Find the eccentricity and the equation of an ellipse, if the foci lie half-
way between the center and the vertices, the major axis lying on OX.
36. Find the equation of an ellipse the eccentricity of which is f and the
ordinate at the focus is 5, the center being at the origin and the major axis
lying along OX.
37. Find the equation and the eccentricity of the ellipse if the ordinate
at the focus is one fourth the minor axis.
38. Find the eccentricity of an ellipse if the line connecting the positive ends
of the axes is parallel to the line joining the center to the upper end of the
onlinate at the left-hand focus.
39. Find the equation of an ellipse when the foci are (±2,0) and the
directrices are x = ± 5.
40. Given the ellipse 2x^-\-Zy^ = \. Find its semiaxes, foci, and directrices.
PKOBLEMS 157
41. Find the equation of an hyperbola if the foci are (± 3, 0) and the trans-
verse axis is 4.
42. Find the equation of an hyperbola if the foci are (0, ± 4) and the trans-
verse axis is 4.
43. An hyperbola has its center at the origin and its transverse axis along
OX. If its eccentricity is ^ and its transverse axis is 5, find its equation.
44. Find the equation of an hyperbola when the vertices are (± 4, 0) and
the eccentricity is I.
45. Show that the eccentricity of an equilateral hyperbola is equal to the
ratio of a diagonal of a square to its side.
46. Find the equation of an hyperbola the vertices of which are halfway
between the center and the foci, the transverse axis lying on OX.
47. Find the equation of the hyperbola with eccentricity 3 which passes
through the point (2, 4), its axes lying on OX and OY.
48. Find the equation of an equilateral hyperbola which passes through
(5, - 2).
49. Find the equation of the hyperbola which has the points (0, ± f V2) for
foci and passes through the point (2, — 1).
50. The sum of the semiaxes of an hyperbola is 17 and its eccentricity is
\^. Find its equation, if its axes lie on OX and OY.
51. Find the equation of the hyperbola which has the asymptotes y = ± ^x
and passes through the point (1, 1).
52. Express the angle between the asymptotes in terms of the eccentricity
of the hyperbola.
53. If the vertex of an hyperbola lies two thirds of the distance from the
center to the focus, find the slopes of the asymptotes.
54. Given the hyperbola 4x^ — 25]/^= 100. Find its eccentricity, foci, and
asymptotes.
55. Find the equation of the hyperbola which has the lines y = ± § x for
its asymptotes and the points ( ± 4, 0) for its foci.
56. Show that 1 = 1, where k is an arbitrary quantity,
a^ — k^ 62 _ j(2
x2 v^
represents an ellipse confocal to — I- — = l,when ^2 <; 52 j and represents an
a;2 y2 a^ t>'^
hyperbola confocal to — + ^= 1, when k'^ >62 but< a"^, a^ being considered
greater than &2. «^ ^^
57. Find the equation of an hyperbola when the foci are (± 7, 0) and the
directrices are x = ± 4.
X2 ?/2
58. Given the hyperbola — = 1. Find its eccentricity, foci, directrices,
and asymptotes.
158 CERTAIN CURVES AND THEIR EQUATIONS
59. A perpendicular is drawn from a focus of an hyperbola to an asymptote.
Show that its foot is at distances a and b from the center and the focus
respectively.
60. Show that in an equilateral hyperbola the distance of a point from the
center is a mean proportional "between its focal distances.
61. Determine p so that the parabola y'^ = ipx shall pass through the point
(2, - 3).
62. An arch in the form of a parabolic curve is 29 ft, across the bottom and
the highest point is 8 ft. above the horizontal. What is the length of a beam
placed horizontally across the arch 4 ft. from the top ?
63. The cable of a suspension bridge hangs in the form of a parabola. The
roadway, which is horizonUl and 240 ft. long, is supported by vertical wires
attached to the cable, the longest being 80 ft. and the shortest being 30 ft.
Find the length of a supporting wire attached to the roadway 50 ft. from the
middle.
64. Find the equation of a circle through the vertex and the ends of the
double ordinate through the focus of the parabola y^ = A px.
65. Find the equation of the circle through the vertex, the focus, and the
upper end of the ordinate at the focus, of the parabola y^ + 12x = 0.
66. Find the equation of the locus of a point the distances of which from
(8, - 2) and (-4, 1) are equal.
67. Find the equations of the locus of a point the distance of which from the
axis of X equals five times the distance from the axis of y.
68. Find the equation of the locus of a point the distance of which from the
axis of X is one third its disUnce from (0, 3).
69. 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).
70. What is the locus of a point the distance of which from the line
3x + 4y — 6 = 0 is twice its distance from (2, 1) ?
71. 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.
72. A point moves so that the square of its distance from the point (0, 2)
equals the cube of its distance from the axis of y. Find its locus.
73. Find the locus of the points at a constant distance 6 from the line
4x + Sy -6 = 0.
74. Find the locus of points equally distant from the lines 2x + 3w-6 = 0
and Sx — 2y + 1 = 0.
75. Show that the locus of a point which moves so that the sum of its dis-
tances from two fixed straight lines is constant is a straight line.
76. Find the equations of the locus of a point equally distant from two fixed
straight lines.
PROBLEMS 159
77. A point moves so that its distances from two fixed points are in a con-
stant ratio k. Sliow tliat tlie locus is a circle except when k — \.
78. A point moves so that the sum of the squares of its distances from the
sides of an equilateral triangle is constant. Show that the locus is a circle and
find its center.
79. 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.
80. A point moves so that the sum of the squares of its distances from the
four sides of a square is constant. Find its locus.
81. A point moves so that the sum of the squares of its distances from any
number of fixed points is constant. Find its locus.
82v 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.
83. Find the locus of a point such that the lengths of the tangents from it
to two concentric circles are inversely as the radii of the circles.
84. 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.
85. Find the equation of the locus of a point the tangents from wliich to
two fixed circles are of equal length.
86. 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.
87. 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 inter-
section of these lines.
88. A point moves so that the product of the slopes of the straight lines
joining it to A {—a, 0) and B («, 0) is constant. Prove that the locus is an
ellipse or an hyperbola.
89. If, in the triangle ABC, taw A tan^B = 2 and AB is fixed, show that
the locus of C is a parabola with its vertex at A and focus at B.
90. Given the base 2 6 of a triangle and the sum s of the tangents of the
angles at the base. Find the locus of the vertex.
91. Find the locus of the center of a circle which is tangent to a fixed circle
and a fixed straight line.
92. 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.
93. A point moves so that its shortest distance from a fixed circle is equal to
it8 distance from a fixed diameter of that circle. Find its locus.
160 CERTAIN CURVES AXD THEIR EQUATIONS
94. 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
OPOQ = k^. Find the locus of P.
95. 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.
96. 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.
97. AB and CD are perpendicular diameters of a circle and M is any point
on the circle. Through Jf, AM and BM are drawn. AM intersects CD in N,
and from J\r a line is drawn parallel to AB meeting BM in P. Find the locus of P.
98. Given a fixed 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.
99. Let OA be the diameter of a fixed circle. From J5, any point on the
circle, draw a line perpendicular to OA and meeting it in D. Prolong the line
Z>B to P, so that OD:DB=OA: DP. Find the locus of P.
100. Two straight lines are drawn through the vertex of a parabola 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.
101. In the parabola y^ = ^px an equilateral triangle is so inscribed that
one vertex is at the origin. What is the length of one of its sides ?
102. Prove that in the ellipse half of the minor axis is a mean proportional
between AF and FA'.
103. Prove that in the ellipse or the hyperbola the ordinate at the focus
is an harmonic mean between AF and AF'.
104. If from any point P of an hyperbola PK is drawn parallel to the
transverse axis, cutting the asymptotes in Q and R, then PQ- PR- a^. If PK
is drawn parallel to the conjugate axis, then PQ ■ PR = - b-.
105. Show that the focal disUnce 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.
106. Prove that the product of the distances of any point of the hyperbola
from the asymptotes is constant.
107. 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.
108. 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.
109. Pi is any point of the parabola y^ = 4px, and P^Q, which is perpen-
dicular to OPi, intersects the axis of the parabola in Q. Prove that the pro-
jection of PiQ on the axis of the parabola is always 4 p.
CHAPTEE VIII
INTERSECTION OF CURVES
86. General principle. If f„J(x, y) is an expression involving
"^"'^- /..(»=. y) = o (1)
is the equation of a curve containing all points the coordinates
of which satisfy (1), and containing no other points. Similarly if
/„(ie, y) is any second expression in x and y,
/„(^,^) = 0 (2)
is the equation of a second curve. It follows that if we consider
these two equations, any point common to the two corresponding
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 joints of intersection of two curves, solve their equa-
tions simultaneously.
We have already discussed in § 30 the simplest case of this
problem, i.e. the intersection of two straight lines. We shall now
discuss some more complex cases.
87. /,(x, t/) = Oand/,(;c, (/) = 0. Let
fM,y)=^ (1)
be a linear equation, and f^ {x, y)=0 (2)
be a quadratic equation. Since a linear equation always represents
a straight line, this problem is to find the points of intersection of a
straight Ime and a curve. Solving (1) for either x or y, and substitut-
ing the result in (2), we obtain in general a quadratic equation, as,
for example, aa^ + hx + c=:0,
if (1) has been solved for y. We shall call this equation the
resultant equation {§ 9). If the roots of this equation are denoted
161
162
INTERSECTION OF CURVES
by x^ and x^, x^ and x^ are the abscissas of the required points of
intersection. The corresponding ordinates are found by substi-
tuting x^ and x^ in succession in (1).
But according to § 37 there are three cases to be considered in
the solution of the resultant equation. (1) The roots x^ and x^
may be real and unequal, in which case there are two points of
intersection. (2) The roots x^ and x^ may be real and equal, in
which case the corresponding ordinates are equal and the two
points coincide. As in § 37, we may regard this case as a limit-
ing case when the position of the curves is changed so as to make
x^ and x^ approach each other, i.e. so as to make the points of inter-
section of the straight line and the curve approach each other along
•the curve. Accordingly, the straight line represented by equation (1)
is tangent to the curve represented by equation (2). (3) Finally, the
roots x^ and x^ may be imaginary, in which case no real points of
intersection can be found, and the curves do not intersect.
Ex.
and
1. Find the points of intersection of
3x-2y-4 = 0
r
(1)
(2)
Fig. 96
Solving (1) for y and substituting tlie result in (2), we have x^ - 6x + 8 = 0,
the roots of which are 2 and 4. Substituting these values of x in (1), we find the
STRAIGHT LIKE AND COKIC
163
corresponding values of y to be 1 and 4. Therefore the points of intersection
are (2, 1) and (4, 4) (fig. 95).
Ex. 2. Find the points of
intersection of
and
6x-4y -9 = 0
a;2 - 4 2/ = 0.
(1)
(2)
Solving (1) for y and substi-
tuting the result in (2), we have
x2-6x + 9 = 0. The roots of
this equation are equal, each
being 3. Hence the straight line
is tangent to the curve. Substi-
tuting 3 for X in (1), we find
2/ = I ; hence the point of tan-
gency is (3, f ) (fig. 96).
Ex. 3. Find the points of
intersection of
3x-2y-5 = 0 (1)
and x2 - 4 y = 0. (2)
Proceeding as in the two previous examples, we obtain x^ — 6 x + 10 = 0,
the roots of which are 3 ± V — 1. Hence the straight line does not intersect the
curve (fig. 97). The correspond-
ing values of y are 2 ± | V— 1.
It is to be noted that the
straight lines of these three
examples all have the same
direction, differing only in the
intercept on the axis of y.
88. The work of the last
article suggests a method
'^ of finding the tangent to
any curve represented by
an equation of the second
degree, the slope of the
tangent being given. For
if m of the required tan-
gent is known, its equation may be written y = mx + b, where
h is not known. According to the definition of a tangent, how-
ever, b must have such value that the points of intersection of
Fig. 97
164
INTERSECTION OF CURVES
straight line and curve shall be coincident. This condition enables
us to determine b, as shown in the following examples.
Ex. 1. Find the equation of tlie
tangent to the parabola 3 x^ + 2 7/ =0,
the slope of the tangent being 2.
Since the slope of the tangent
is 2, its equation may be written
y = 2x + b. Substituting this value
of y in the equation of the parabola,
•we have the equation
3a;2 + 4x + 26 = 0.
Since the line is to intersect the
curve in two coincident points, this
equation must have equal roots.
The condition for equal roots, by
§ .37, is (4)2 - 4 (.3) (2 b) = 0, whence
we find 6 = |-.
= 2x + |-, or 6x-Sy + 2 = 0 (fig.
Fig. 98
Therefore the required tangent is
Ex. 2. Find the equation
of the tangent to the ellipse
x2 -I- 4 y2 = 4, the slope of the
tangent being ^.
The equation of the tangent
is y = ^z + b. Substituting this
value of y in the equation of the
ellipse, we have
x2 + 2 6a; + (2 62 - 2) = 0.
Fig. 99
The condition that this equation
shall have equal roots is (2 6)2 - 4 (2 62 - 2) = 0, whence 6 = ± V2.
In this case there are two tangents having the required slope -^ (fig. 99), the
equations of which are respectively j^ = i x + Vi and y = ^x-V2
or x-2y±2V2 = 0.
By this same method the following formulas for a tangent with
known slope m may be derived :
1. The tangent to the parabola y- = Apx is
y = mx +
EXCEPTIONAL CASES
165
2. The tangents to the ellipse — + ^ = 1 are
= mx ± Va^m^ + b^.
X IT
3. The tangents to the hyperbola — — 4 = 1 are
a- ■ l)-
y
= mx ± ^ d^m^— If.
89. It was stated in § 87 that the result of the substitution is
in general a quadratic equation. In exceptional cases, however, the
resultant equation may be linear,
as in the first of the following
examples, or even impossible, as
in the second example.
Ex. 1. Consider
2a; -5?/ -10 = 0 (1)
and 4x2-252/2 = 100. (2)
Fig. 100
Substituting in (2) tlie value of y
from (1), we have tlie equation 40 x — 200 = 0, wlience x = 5. .-. y = 0, and
tlie straight line and the curve intersect in a single point (o, 0) (tig. 100).
E.\. 2. Consider
and
2x-5y + 4 = 0
4x2 -25?/2 + i6x- 84 = 0.
(1)
(2)
Fig. 101
Substituting in (2) the value of y from (1), we have — 100 = 0. But this
equation is impossible. Hence the given equations are contradictory, and the
straight line and the curve do not intersect (fig. 101).
166 INTERSECTION OF CURVES
These exceptional cases, of which the above are illustrative
examples, may be regarded as limiting cases as follows :
If x^ and x^ are the roots of the resultant equation aaP + bx + c = 0,
2c
X, =
_5 + v^
- 4ac
2a
-&-V&^-
-4ac
-6-V&--4ac
2c
2a —h -f Vft^— 4
ac
Now as a == 0, the resultant equation approaches the linear equa-
tion &« + c = 0. At the same time x.= and a;„ = cx) . There-
l ^
fore, if a is made to approach zero by changing the position of
either the straight line or the curve in the plane, the case in which
only one solution of the linear and the quadratic equations is
found is the limiting case of intersection of the straight line and
the curve as one point of intersection recedes indefinitely from
the origin.
If a = 0 and & = 0, both x^ and x^ increase indefinitely. Hence
the case in which the linear and the quadratic equations are con-
tradictory is the limitmg case of intersection, as both points of
intersection recede indefinitely from the origin.
90- /i(^, 1/) = 0 and /„(x, y) = 0. Let
/i(-^> y) = 0 (1)
be a linear equation, and fj^x, y) = 0 (2)
be an equation of the wth degree where w > 2. The degree of a
curve is defined as equal to the degree of its equation. Accord-
ingly, this problem is to find the points of intersection of a straight
line and a curve of the w,th degree where %->% and the method
is the same as that of § 87. The resultant equation, after sub-
stitution from the linear equation, is, in general, of the wth degree,
and its solution is found by the methods of Chaps. IV and V.
The number of points of intersection will be the same as the
number of real roots of the resultant equation. Hence a straight
STRAIGHT LINE AND CURVE OF A^th DEGREE 167
line can intersect a curve of the nth degree in n points at most.
If the resultant equation has multiple roots, they correspond, in
general, to points of tangency of the straight line and the curve,
as in § 88 ; and if the resultant
equation is of degree less than n, it
can be shown that the straight line is
the limiting position of one in which
one or more points of intersection
have been made to recede indefinitely.
Ex. 1. Find the points of intersection of
y = 2x (1)
and y2 = x(x-3)2. (2)
The resultant equation is
a;[(x-3)2-4x] = 0,
or a;[x2-10x + 9] = 0.
Its roots (§ 39) are the roots of x = 0 and
x2-10x + 9 = 0, which 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. 102).
Ex. 2. Find the
points of intersection
of
y = 3x + 2{l)
and y = x^. (2)
Fig. 102
The resultant equa-
tion isx^ — 3x — 2 = 0.
One root is found (§ 49) to be 2, and the depressed
equation isx2 + 2x + 1 = 0. Its roots are equal, both
being - 1. The corresponding values of y, found
from (1), are 8 and - 1. Therefore the points of
intersection are (2, 8) and (- 1, - 1), the latter being
a point of tangeucy (fig. 103).
Fig. 103
168
INTERSECTION OF CURVES
Ex. 3. Find the points of intersection of
2x + 2/-4 = 0 (1)
and y2 = x(x2-12). (2)
The resultant equation is x^ - 4x2 + 4x - 16 = 0, or (x - 4) {x^ -f 4) = 0,
the roots of which are 4 and ± 2 -Z^. The corresponding values of 2/, found
Fig. 104
from (1), are — 4 and 4^4 V— 1. The only real solution of equations (1) and
(2) being x = 4 and i/ = — 4, the straight line and the curve intersect in the single
point (4, - 4) (fig. 104).
91. /„(x, i/)=Oand/„(;t, i/)=0. Let
LM^y) = ^ (1)
be an equation of the wth degree and
/„(•«, 2/) =0 (2)
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 cases, i.e.
the elimination of either x or y, the solution of the resultant equa-
tion, and the determination of the corresponding values of the
unknown quantity eliminated. The equation resulting from the
CURVES OF Mth AND Nth DEGREE
169
elimination is, in general, of degree mn, and the number of simul-
taneous solutions of the original equations is mn. If all these
solutions are real, the corresponding curves intersect at m7i points.
If, however, any of these solutions are imaginary, or are alike, if
real, the corresponding curves will intersect at a number of points
less than 'tnn. Hence, two curves of degrees m and n respectively
can intersect at mn points and no more.
No attempt at a complete discussion will be made, on account
of the unlimited number of cases which are possible. We shall
merely solve a few illustrative examples, noting any interesting
geometrical facts that may occur in the course of the sokition:
Ex. 1. Find the points of intersection
of
y2_2x = 0 (1)
and x2 + 2/2_8 = o. (2)
Subtracting (1) from (2), we elimi-
nate y, thereby obtaining the resultant
equation a;^ + 2 a; — 8 = 0, the roots of
which are — 4 and 2. Substituting 2
and — 4 in either (1) or (2), we find the
corresponding values of ?/ 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. 105).
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.
Y °
Ex. 2. Find the points of inter-
section of
x2 - 3 2/ = 0 (1)
and 2/2_3x = 0. (2)
Substituting in (2) the value of y
from (1), we have x* — 27x = 0.
This equation may be written
x(x-3)(x2 + 3x + 0) = 0,
the roots of which are 0, 3, and
_ 3 4- 3 V — 3
— Substituting these
values of x in (1). we find the corre-
Fio. lOG spending values of y to be 0, 3,
Fig. 105
-X
170
INTEKSECTION OF CUEVES
and
- 3^8^^
Therefore the real solutions of these equations are a; = 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 — — S when x = 3.
But — 3 for 2/ 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 corre-
sponding curves intersect at the two points (0, 0) and (3, 3) (fig. 106).
We 8ee, moreover, that any results found must be tested by substitution in both
of the original equations.
The remaining two solutions of these equations found by letting x =
_3±3Vr3
are imaginary.
Ex. 3. Find the points of intersection of
2x2 + 32/2 = 35
and xy = 6.
(1)
(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 x^ + 3 m'h? = 35 and
mx2
= 6, whence
x2 =
35
(4)
2 + 3m2
and
x2 =
6
m
(6)
35
6
5
m
(6)
2 + 3 jft2
from which we
find m =
i
or 4.
Fig. 107
If m = ^, from (5) x = ± 2 ; and
from (3) the corresponding values
of y are ± 3.
If m = |, in like manner we find
X = ± '^ Vg and y — ±^ V6.
Therefore the ellipse and the hyperbola intersect at the four points (2, 3),
(- 2, - 3), (I V6, f V6), (- I V6, - I V6) (fig. 107).
It should be noted that (3) is the equation of a straight line through the
origin, so that when we solve (6) for m we determine the slopes of the straight
lines passing through the origin and intersecting the two given curves at their
common points.
SYSTEMS OF CURVES
171
Ex. 4. Find the points of inter-
section of
2 2/2 = x-2 (1)
and x2 - 4 2/2 = 4. (2)
Eliminating y, we have
a;'^
2a; = 0,
Fig. 108
the roots of which are 0 and 2.
Wlien X = 0 we find from either
(1) or (2) y = ± V— 1, and when
X = 2 eitlier (1) or (2) reduces to y'^ = 0, whence y = 0.
two curves are taijgent at the point (2, 0) (fig. 108).
Therefore these
Ex. 5. Find the points of intersec-
uuxiux x2 = 2y
(1)
and x^ — 3 XT/ + 2/3 = 0.
(2)
Eliminating j/, we have
x6- 4x3 = 0,
which may be written x^ (x^ — 4) = 0.
X The real roots of this equation are 4^
and 0, the latter being a triple root,
and the two imaginary roots are
4*(-l±V-3) ^ ,.
— i -• Corresponding
2
values of y are found to be 2^, 0, and
2-^(- 1 T ^^^^)- Therefore the
Fig. 109 curves intersect at the two points
(4*, 2i) and (0, 0) (fig. 109).
At the point (0, 0) the parabola (1) is tangent to one part of (2) and passes
through another part of (2), and for this reason the point is to be regarded as a
triple point of intersection.
^^' Ifmi^t y)~^ kfn{x, i/)=0. If we have two expressions
f^{x, y) and f,^{x, y), we have seen in § 86 that we can form the
equations of two curves by placing each of them separately equal
to zero, i.e.
A(^,2/) = 0, (1)
and /„(^,2/) = 0. (2)
Let us now form the equation of a tliird curve by multiplying
fjx, y) and fjx, y) by I and k respectively, where I and k are
172
INTERSECTION OF CURVES
any two quantities which are independent of both x and y, and
placing the sum of the products equal to zero, i.e.
ifj<^>y)+¥n{^>y) = ^-
(3)
This third curve has the following two important properties :
1. It passes through all points common to curves (1) and (2).
For the coordinates of any such points make /„,(a?, y) = ^ and
fj^x, y) = 0, since they satisfy (1) and (2). Hence they will
satisfy (3), ie. be coordinates of a point of curve (3).
If either / or A; is placed equal to zero, (3) reduces to either (2)
or (1) as a special case.
2. If neither I nor k is zero, it intersects curves (1) and (2)
at no other points than their common points. For the coordi-
nates of any point on (1), for example, but not on (2), make
/^{x, y) = 0 and /„(«, y) different from zero. Hence they will not
satisfy (3), and the corresponding point cannot be a point of (3).
It follows that if (1) and (2) have no points in common, (3)
intersects neither (1) nor (2). If we treat (1) and (2) apart from
possible geometrical interpretation, however, it is evident that the
imaginary solutions of (1) and (2) are solutions of (3).
By assigning different values to I and k we may make (3) satisfy
another condition, as will be illustrated in the following examples :
Ex. 1. Find the equation of the straight line passing through the point of
intersection of the lines
Y
2x + y-l = 0 (1)
and x + 2y -3 = 0 (2)
and the point (1, 2).
l{2x+7j-l) + k{x + 27j-3) = 0 (3)
passes through the point of intersection of
(1) and (2), and is the equation of a straight
line, since it is an equation of the first
degree. Since (3) is to pass through the point
(1, 2), (1, 2) must satisfy (.3). Therefore
i (2 + 2 - 1) + fc (1 + 4 - 3) = 0, or 3 ; + 2 /<: = 0. Therefore, if we substitute
k = - ^lin (3) and simplify, we shall have the equation of the required line.
It is found tobex-4y + 7 = 0 (fig. 110).
SYSTEMS OF CUKVES
173
Ex. 2. Find the equation of a straiglit line passing through the point of
intersection of the lines
and
and parallel to the line
x-2y + 1 = 0
x-3y + 3 = 0,
2x + 3y + 8 = 0.
As in Ex. 1, the required line is
lix - 2y + 1) + k{x - 3y + 3) = 0,
which may be written
{I + k)x + (- 21 - Sk)y + {I + 3k) = 0.
Since this line is to be parallel to (3),
. l + k
2l-3k
(1)
(2)
(3)
(4)
(§ 28, 2), whence
2 3
* = — I ^ Substituting this value of k in (4) and simplifying, we have as our
required line 2a; + 3y-12 = 0 (fig. 111).
Fig. Ill
Both of these examples could also have been solved by finding the point of
intersection of the given lines and then, by the methods of Chap. Ill, passing
the line through the point subject to the given condition.
93. In the two examples of the last article both equations were
of the first degree. In this article we will solve some examples
in which one or both equations are of the second degree.
174
INTERSECTION OF CURVES
Ex. 1. Find the equation of a circle determined by the points of intersec-
tion of the straiglit line
2x-y-6 = 0 (1)
and the circle
x2 + y2-6x-6y -7 = 0 (2)
and the point (1, - 1) (fig. 112).
The equation
l{2x-y-6)
+A;(x2+2/2-6x-6y-7) = 0 (3)
is the equation of a circle, since the
coefficients of x^ and y^ are equal ;
and since it passes through the points
of intersection of (1) and (2), it only
-piQ 112 remains to choose I and k so that it
shall pass through the point (1, —1).
3 I
Substituting (1, — 1) in (3), we have 3 Z + 5 fc = 0, whence k = Accordingly,
the equation (3) of the required circle, in simplified form, is
3 x2 + 3 2/2 - 28 X - 13 ?/ + 9 = 0.
Ex. 2. Find an equation representing the system of circles passing through
the points of intersection of the circles
x2 + 2/2 _ 9 = 0
and x2 + j/2 - 4x - 2 2/ - 11 = 0
(fig. 113).
The equation
Z(x2 + 2/2- 9)
+ A:(x2+2/2-4x-2 2/- 11)=0
is the required equation, for by its form it is
the equation of a circle, and passes through
the two points common to (1) and (2). By
assigning different values to I and k we can
make (3) represent any, and hence every,
circle passing through the common points of
(1) and (2). In other words, it represents the required system of circles.
In particular, if I and k are assigned such values as to make the coefficients
of x' and y^ vanish, i.e. k = — I in this example, the equation reduces to
Fig. 113
2x + 2/ + l = 0.
(4)
But this is the equation of a straight line, and since it must, from the way in
which it was formed, pass through the points common to the two circles, it
must be the equation of their common chord.
PROBLEMS 175
In general, if x^ + y^ + 2 GiX + 2Fiy + Ci = 0, (6)
x2 + y2 ^. 2 Ggx + 2 F^y + Ca = 0, (6)
are the equations of any two circles, we derive the third equation
2{Gi-G2)x + 2{Fi-F2)y + {Ci-C2) = 0 ' (7)
by assuming k = — I.
If the circles intersect, (7) is the equation of their common chord ; but if
they do not intersect, (7) is called their radical axis. It may easily be proved
to be perpendicular to the line of centers and is the locus of points from
which equal tangents, one to each circle, may be drawn.
PROBLEMS
Find where and how the straight line intersects the curve of the second
degree in each of the following cases :
1. 2x + 3y = 5, 4x2 + 92/2 + 16x-182/-ll = 0.
2. X - y + 1 = 0, (X + 2)2 _ 4 2/ = 0.
3. x-22/ + 4 = 0, 2x2-?/2 + 8x + 22/ + 13 = 0.
4. 2/ - 2x = 0, x2 + ^2 _ a; + 3y = 0.
5. X - 2 2/ + 4 = 0, 5x2 - 4 2/2 + 20 = 0.
6. 2/ = 8x-5, 2x2 + X2/-3y2 + 6x4-42/ + 4 = 0.
7. 2x + 32/-6 = 0, x2 + 4 2/2-4 = 0.
8. X + 2/ - 4 = 0, x2 - 2 X2/ + 2/2 - 20 = 0.
9. Find the length of the chord of the circle x2 + 2/2 + 8x — 42/ + 10 = 0
cut from the line 2x — Sy + S = 0.
10. Find the tangent to the curve x2 + 6x-22/ + 5 = 0 with slope 2.
11. For what value of p will the parabola y^ = ipx be tangent to the line
2/-3x + l = 0?
12. Find the tangents to the ellipse 4 x2 + 9 7/2 = 36 which are parallel to the
line joining the positive ends of the axes.
13. Find the tangent to the curve b^x^ + a^y^ + 2 a62x = 0 perpendicular to
the line ax + by = ab.
14. Prove that the line y = -mx + 2c Vm. is always tangent to the hyper-
bola xy = c2, and that the point of contact is ( —^ ' c Vmj •
15. Find the point of contact of tlie tangent to the curve
x2_4y2 + 2x2/-2x + 42/ = 0 with slope ^.
16. Find the points of intersection of the line 82/ - 26x = 0 and the curve
x22/2 + 36 = 4 2/2.
17. Find the points of intersection of the line 2/ = 2x - 3 and the curve
42/2 = (x + 3)(2x- 3)2.
176 INTEKSECTION OF CURVES
18. Find the points of intersection of the line x — 2y + 2 — 0 and the cissoid
X (x2 + 2/2) = 4 2/2.
19. Find the points of intersection of the line x = 2y and the curve
16 2/2 = 4 X* - a*.
20. Find the points of intersection of the line y = 2 x — 2 and the cissoid
a;(x2 + 2/2) = 4 2/2.
21. Find the points of intersection of the line y = mx and the cissoid
x(x2 + 2/2) = 2a2/2.
22. Find the points of intersection of the line x — y — 1 = 0 and the witch
y
x2 + 4
Find tlie points of intersection of the following pairs of curves :
23. 42/2 = x2(x + 1), 2/2 = x(x + 1)2.
24. 2/2 = 12 x, ^2 = (a; + 2) (X - 3)2.
25. x2 = y^{y + 2), x2 = (2/ - 1)2(2/ + 1).
26. Find the points of intersection of the parabolas 2/2 = 4 ax + 4 a2 and
2/2=-46x + 4 62.
27. Find the points of intersection of the parabola x2 = 4a2/ and the witch
8a3
y =
X2 + 4 o2
28. Find the points of intersection of the cissoid y^ = and the
parabola 2/2 = 4 ax. a — x
29. Find the poiaits of intersection of the cissoid y^ = and the circle
x2 + 2/2-4ax = 0. 2a-x
30. Find the points of intersection of the hyperbola xy = 2aP- and the witch
_ %d?
^~x2 + 4a2"
31. Find the points of intersection of the witch y = and the cissoid
4«3 *^ + 4a2
X2 =
5a — 42/
32. Find the points of intersection of the circle x2 + 2/2=5 a2 and the witch
x2 + 4 a2
33. Find the equation of a straight line through the point of intersection
of 7x - 2/ - 18 = 0 (1) and x - 3 2/ - 14 = 0 (2) and the point (- 2, 1), without
finding the point of interaection of (1) and (2).
34. Find the equation of a straight line through the point of intei-section of
2 X - y + 5 = 0 (1) and x - 42/ + 13 = 0 (2) and parallel to the line 2x + 62/ + 2 = 0,
without finding the point of intersection of (1) and (2).
PROBLEMS 177
35. Find the equation of a straight line through the point of intersection
of 4x - 6?/ - 5 = 0 (1) and G a; - 4?/ - 5 = 0 (2) and perpendicular to the line
X — Sy + 1 = 0, without finding the point of intersection of (1) and (2).
36. A circle passes through the origin of coordinates and the points of
intersection of the circle x^ + y^ = 14 and the line 2x + Sy + b = 0. Find
its equation.
37. Prove that (1, 1) is a point of the common chord of the two circles
a;2 + 2/^ — 4 X = 0 and x^ + |/2 _ 4 y = 0.
38. Find the circle passing through (1, — 3) and the points of intersection
of the two circles x2 + 2/2 — 4x — 4^/ — 8 = 0 and x^ + y^ + x + y — 4 = 0.
39. Find a curve of the second degree passing through (1, 1) and the points
of intersection of the curves Sx^ + 5y'^ — 15 = 0 and 2x^ — 3y^ — G = 0, and
tell what kind of a curve it is.
40. Prove that a parabola can be passed through the points of intersection
of the curves x^ - 2y^ + x + 2y + 1 = 0 and 3x2 + 4 2/2_2x-2 = 0.
41. The center of a circle is at the vertex ^ of a parabola y^ = 4px, and its
diameter is 3 J.F, i^ being the focus of the parabola. Prove that their common
chord bisects AF.
42. Show that the circle described on any focal radius of a parabola as
diameter is tangent to the tangent at the vertex of the parabola.
43. Show that the circle described on any focal chord of a parabola as
a diameter is tangent to the directrix of the parabola.
44. If a circle is described from a focus of an hyperbola as center, with its
radius equal to half the conjugate axis, prove that it will touch the asymptotes
at the points where they intersect the corresponding directrix.
CHAPTER IX
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
94. 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, Y, and Z be three variables, such that LimX = J,
Lim Y = B, JAm Z = C. From the definition of Hmit (§ 53) we
may write X = A+a, Y = B + b, Z=C + c, where a, b, 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
X+Y-\-Z = A+B + C+a + b + c.
Now if e is any assigned quantity, however small, we may
make a, b, and c each numerically less than - > so that a + b + c is
numerically less than e. Then the difference between X+Y+Z
and A+B + C becomes and remains less than e, that is,
lim {X+ Y + Z) = A + B+C= Lim X+ Lim Y+ 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 Y such that Lim X = A and
Lim Y = B. As before, we have X = A + a and Y = B -i-b. Hence
XY = AB + bA + aB + ab.
178
THEOREMS ON LIMITS 179
Now we may make a and h so small that hA, aB, and ah are
e
3
Lim XY = AB = (Lim X) (Lim Y).
each less than - . where e is any assigned quantity, no matter how
small. Hence
Consider now three variables X, Y, Z. Place XY= U. Then,
as iust proved,
UmUZ = {UmU){lAmZ);
that is, lim XYZ = (Lim XY) (Lim Z)
= (Lim X) (Lim Y) (Lim Z).
Similarly the theorem may be proved for any finite number of
variables.
3. The limit of a constant times a variable is equal to the con-
stant times 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 quo-
tient of the limits of the variables, provided the limit of the divisor
is not zero.
Let X and Y be two variables such that LimX = ^ and
limY^B. Then, as before, X = A + a, Y = B-{-b.
„ X A + a , X A A + a A aB — bA
Hence — = > and = r- = — :; — ; — •
Y B + b Y B B + b B B^ + bB
Now the fraction on the right of this equation may be made
less than any assigned quantity by taking a and b sufficiently
small.
TT ^. X A Lim X
Hence Lim — = — =
Y B LimF
The proof assumes that B is not zero.
95. Theorems on derivatives. The definitions of increment,
continuit} , and derivative given in Chap. V are perfectly general,
although they are there applied only to algebraic polynomials.
180 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
In order to extend the process of differentiation to other func-
tions, we shaU. need the following theorems :
1. The derivative of a function plus a constant is equal to the
derivative of the function.
Let w be a function of x which can be differentiated, let c be a
constant, and place y — ^jl. c
Then if x is increased by an increment Ax, u is increased by
an increment Au, and c is unchanged. Hence the value of y
becomes u + Au + c.
Whence Ay = (w + Au + c) — {u + c) = Au.
Therefore ^ = ^'
Ax Ax
and, taking the limit of each side of this equation,
dy _ du
dx dx
Ex. 2/ = 4x3 + 3,
^ = 1(4x3) = 12x2.
dx dx
2. The derivative of a constant times a function is equal to the
constant times the derivative of the function.
Let -JA 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 correspond-
ing increments of u and y. Then
Ay = c(u-{- Au) — cu = cAu.
(by theorem 3, § 94)
Hence
Ay _ Au
Ax Ax
and
, . Ay -r • Aw
Lim — ^ = c Lim -—
Ax Ax
Therefore
dy du
-^ = c — >
dx dx
by the definition of a derivative.
THEOREMS ON DERIVATIVES 181
Ex. 2/ = 5(x3 + 3x2 + l),
dv d
-^ = 5--(x3 + 3x2 + 1) = 5(3x2 + 6x) = lo(x2 + 2x).
uX uX
3. The derivative of the sum of a fnite 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, V, w, and y be Aw, Av, Aw, and Ay. Then
Ay = [u + A'2<- + V + Av + w + A'Z^;) — [u + v + w)
= Ait + A^ + Avj ;
, Ay Aw Av Aw
whence t^ = 7 — ^-r~ + -i;—'
Ax Ax Ax Ax
Now let Ax approach zero. By theorem 1, § 94,
-r . Ay _ . Aw -r . A«7 -r • Aw .
Lim — ^ = Lim f- Lim 1- Lim --— >
Ax Ax Ax Ax
that is, by the definition of a derivative,
dy _ du dv dw
(A/*A/ tl/t/j- (A/t/y Ct'X'
The proof is evidently applicable to any finite number of
functions.
Ex. ?/ = x*- 3x3 + 2x2-7 X,
^ = 4x3-9x2 + 4x-7.
dx
4. The derivative of the product of a. finite member 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,
^^^let y=^uv.
182 DIFFERE^^TIATION OF ALGEBRAIC FUNCTIONS
Give X an increment Aa;, and let the corresponding increments
of u, V, and y be Aw, Av, and Ay.
Then Ly = (w + Aw) (v + At-) — wv
= w Av + ■?; Aw + Azt • Av
, Aw Av Aw . Aw .
and —^ = w-— +?;—- + -— -Av.
A;r ^x Ax Aa;
If now Aic approaches zero, we have
,. Aw ^. Av . -r- Aw , -r- Aw _. .
Lim — ^ = w Lim \- v Lim h Lim -— • Lim Av.
A« Aa; o^x Ax ,(. q -^,
T^ ^ T • Aw i^w ^ . Aw du T . Av dv , -r • a a
But Lim —^ = -f-, Lim — = — - > Lim -— = -- > and Lim A?; = 0 ;
Ax dx Ax dx Ax dx
^, .„ dy dv , du
thereiore -^ = w — — I- ^ -7- •
dx dx dx
Again, let y = uvw.
Eegarding uv as one function and applying the result already
obtained, we have
dy dw d (uv)
-^ = uv — — I- w —^ — -
ax dx dx
dw .
= uv — — I- w
dx
[dv dul
dx dxj
dw dv du
= uv— — f- uw -— + vw-—-
dx dx dx
The proof is clearly applicable to any finite number of factors.
Ex. y = (3z-5)(x2 + l)ic3,
^ = (3x - 5) (x^ + D^-i^ + (3x - 6).3^(?!_±i) + (,. + i),3l(^^^
dx ^ '^ ' dx ^ ' dx ^ ' dx
= (3x - 5)(x2 + l)(3x2) + (3x - 5)x3(2x) + (x^ + l)x3(3)
= (18x3 - 25x2 + 12x - 15)x2.
5. The derivative of a fraction is equal to the denominator
times the derivative of the numerator minus the numerator times
the derivative of the denominator, all divided by the square of
the denominator.
THEOREMS ON DERIVATIVES 183
l^et y where u and v are two functions of x which can be
v
differentiated. Let Aic, Ai*, Av, and Ay be as usual. Then
. _u-\- Au ti _v Ail — u Av
V + Av V v^-{- V Av
and
Au Av
Ay Ax Ax
Ax v^+ V Av
Now let Ax approach zero. By § 94,
-r . Au -, . Av
V Lim u Lim —
Ay Ax Ax
Lim-^— = 7-, T"- — 1
Ax V + V Lim Av
dy
du dv
dx dx
whence
dx
v'
Ex. y — ,
" x2 + 1
dy (a;2 + l)(2x)-
-(x2-l)2a;
1)2
4x
dx (x2 +
(X2 + 1)2
^. 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
Ax 1
Ay^Ay'
Ax
whence lim -r— =
Ax
dx 1
that is, ~r = t" '
dy dy
dx
7. If y is a function of u and ic is a function of x, then y is
a fu7iction 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.
184 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
An increment Ax determines an increment Au, and this in turn
determines an increment Ay. Then evidently
Ay _ Ay Au
Ax Au Ax
whence
Lim — ^ = lim — ^ •
Ax Au
Lim -— >
Ax
that is,
dy _ dy du
dx du dx
Ex. y =
= W2
+ 3tt
+ 1
, where u = —,
dy _
dx~
= (2
u + 3)
'(-
2\ 2 + 3x2 2 _
X3/ ~ X2 X8 ~
4 + 6x2
X5
The same result is obtained by substituting in tlie expression for y the value
of u in terms of x, and then differentiating.
96. Formulas. The formulas proved in the previous article are :
d(u + c) _du
(2)
(3)
(4)
(5)
(6)
(7)
(8)
dx dx
d(cu) du
dx dx
d(u + v) du dv
dx dx dx
d{uv) dv 1 du
dx dx dx
, /u\ du dv
\v/ dx dx
dx ^ v^
dx 1
dy~ dy'
dx
dy _ dy du
dx du dx
dy
dy_du
dx~ dx'
du
brmula (8) is
a combination of (6) and (7),
DERIVATIVE OF U"" 185
97. Derivative of u". If tt is any function of x wliich can be
differentiated and n is any real constant, then
—- — - = nu — •
To prove this formula we shall distinguish four cases :
1. When ri is a positive integer.
d{u'') _ divJ') du
dx du dx
„_, du
(by (7), § 96)
m.»-^-. (by(l), §58)
2. When % is a positive rational fraction.
P
Let n = — where p and q are positive integers, and place
p
y = u\
By raising both sides of this equation to the g-th power we have
?/« = w".
Here we have two functions of x which are equal for all values
of X. If we give x an increment Aic, we have
A(y'') = A(wa
A(y^)_A(^0.
Hx Ax
and therefore - "^ — \ >
dx dx
whence QV^'^ — = pW'^ — - »
dx dx
since ^ and q are positive integers. Substituting the value of i)
and dividing, we have
dx q dx •
Hence in this case also
d hi") „ \du
dx dx
186 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
3, When w is a negative rational number.
Let n = — m, where tw is a positive number, and place
y =z U "" =
d{u'^)
Then -^ =
dx
dx
m 1 du
dx
—
^2m
=
-m-1 <^^'
— mu -—
dx
Hence in this case also
d{u^) _
dx
n 1 du
dx
(by (5), § 96)
(by 1 and 2)
4. When n is an irrational number.
The formula is true in this case also, but the proof will not be given.
As a particular case of this formula, it appears that 1, § 58, is
true for all real values of n.
Ex.1. y = (x3 + 4x2- 5a; + 7)3,
-^ = 3(a;3 + 4x2 - 5a; + 7)2— (x^ + 4x2 - 5x + 7)
dx dx
= 3(3x2 + 8x _ 5)(x3 + 4x2 _ 5x + 7)2. (by § 58)
#
Ex. 2. y = ^/x2 + — = x^ + x-3,
x3
^ =!«■*- 3 x-" (by (3), §96)
_^ _3
Ex. ^. y-{x + l)Vx2 + 1,
= (a; + i)[i(x2 + i)-'.2x] + (x2 + i)i
= ^(^ + (x2 + l)i
_2x2 + x + l
Vz2 + 1
HIGHER DERIVATIVES 187
dy_l / a; \- ^ d^ / x \
cto "" 3 \x8 + 1/ da; U* + 1/
1 /x8 + i\3 1-2x8
_ 1-2x8
~3x3(x3 + l)4'
98. Higher derivatives. It has been noted already (§ 62) that
the derivative of the derivative of a function is called the second
derivative of the function. Similarly the derivative of the second
derivative is called the third derivative, and so on. The succes-
sive derivatives are commonly indicated by the following notation.
y =^f(x), the original function,
dx
-^ = f'{x), the first derivative,
-— I — ) = — =^ =f"(x), the second derivative,
ax \dxj dx
— ( — 4 ) = -4 =f"'(x), the third derivative,
\dx / dx'^
dx \dx
dx'
d"v
•L = /^"^(a;), the wth derivative.
It is noted in § 22 that /(a) denotes the value of f{x) when
x — a. Similarly f{a), f"{a), f"'{a), are used to denote the values
of f'{x), f"{x), f"'{x) respectively when x = a. It is to be empha-
sized that the differentiation is to be carried out before the sub-
stitution of the value of x.
Ex. If/(x) = ^-^^, find/"(0).
,,, , -x2 + 2x + l.
^^''^- (x-^ + l)2 '
^„, , 2x8-Gx2-6x + 2,
^^^^ = WT^'
.-. /"(O) = 2.
188 DIFFERENTIATIOX OF ALGEBRAIC FUNCTIONS
99. Differentiation of implicit algebraic functions. Consider
any equation of the form
P^ + I>xf~'' + I>2t~'' + I>zy'"^+ ■ • • +Pn-iy+K = ^> (1)
where w is a positive integer, and where some or all of the coeffi-
cients Pffl Pi, • • • , jt>„, are polynomials in x. By means of this equa-
tion, if a value of x is given, values of y are determined. For if a
numerical value is given to x, the coefficients become numerical
and the equation is of the kind discussed in Chap. IV, which has
been shown always to have n roots. Hence (1) defines y as a
function of x. This is the most general form of an algebraic
function. When (1) is solved for y, so that y is expressed in
terms of x, y is an explicit algebraic function. When (1) is not
solved for y, y is an implicit algebraic function.
For example,
^ a? — ^ xy + h y^ — ^ x + 1 y — ^ = ^,
which may be written
5 2/' + (7 - 4 ic) y + (3 a;' - 6 a: - 8) = 0,
defines y as an implicit function of x.
If the equation is solved for y, giving
-7+4a^±V209-}-64x-44ar^
^ = lo '
y is expressed as an explicit function of x.
It may be shown by advanced methods that y defined by (1)
is a continuous function of x and has a derivative with respect
to X. Assuming tliis, it is possible to find the derivative 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 article, as shown
in the examples.
Ex.1. Given x2 + y2 = 5.
Then djz^ + y^)^
dx '
dy
that IS, 2x + 2y— =0',
dx
whence — = — .
dx y
IMPLICIT functio:n^s 189
The derivative may also be found by solving the equation for y. Then
y = ±y/b- iC'^,
dy _ ^ -X _
x
dx V5 - x2
y
Ex. 2.
Given
y^ -xy -1 = 0.
Then
d(2/3) d(xy)_^
dx dx
Hence
Sy^^/-xf--y = 0
dx dx
and
dy y
dx Sy^ — X
The second derivative may be found by differentiating the result thus
obtained.
Ex. 3. If a;2 + y2 = 5, we have found ^ = - -.
dx y
Therefore
^ _ _ d^ /x\
c2 dx \y/
d?y
dx^ dx \y,
dy
dx
Ex. 4. If 7/3 — xy — 1 = 0, we have found
y2
y^ + x^
yS
dy y
dx 3 2/2 _ a;
,dy di^y'^-x)
{Sy^-x)-f-y ' ' V ,
_,, • d^y dx dx
Then —^ =
dx2 (3 2/2-x)2
(3 2/2 _ x)2
(3z/2_x)-^ yi^ 1)
^ \S 2/2 - X \3 y2 - a: /
(3 2/2-a;)2
_ -2x2/
■" (32/2 -x)3'
190 DIFFERENTIATION OF ALGEBKAIC FUNCTIONS
100. Tangents. It has been shown in § 59 that the tangent
to a curve ?/ =/(a^) at a point (x^, y^ is
where ( — | denotes the value of ^ at Ix, yX We wiU apply this
\dxji dx
to some of the curves of Chap. VII, obtaining results for future
reference.
Ex. 1. Consider the circle Ax"^ + Ay'^ ^r 2Gx + 2Fy + C = 0.
Differentiating, we have
2Ax + 2Ay^ + 2G + 2F— = Q;
dx dx
dy_
Ax + G
L6I1C6
dx
Ay + F
Hence the
equation
of the tangent
. is
it is,
y-
-yi = -
Axi + G
Ayi + F
■^yi -t r
Axix - Ax^ + Ayiy - Ay^ + Gx - Gxi + Fy - Fyi = 0.
This equation may be simplified by adding to it the identity
Ax^ + Ay^ + 2 Gxi + 2 Fi/i + C = 0,
which follows from the fact that (xi, y{) is on the circle. There results
^xix + Ayxy + G (x + Xi) + F{y + ^i) + C = 0.
This result is easily remembered from its resemblance to the equation of the
circle.
The proofs of the next three examples are left to the student.
Ex. 2. The tangent to the ellipse - + — = 1 is — + — = 1,
Ex. 3. The tangent to the hyperbola ^-^ = lis?l?-^ = l.
a2 62 (j2 ft2
Ex. 4. The tangent to the parabola j/2 = 4px is y^y = 2p(x + Xi),
NORMALS
Ex. 5. Consider the witch x'h/ + 4 a^y — 8 a^ = 0
Differentiating, we have
191
ox dz
Hence the equation of the tangent is
y -Vi
2X1^1
; (« - «i) ;
that is,
But
Xi'' + 4 o2
zly + 4 a2y - 4 a^y^ + 2 Xi^ix - 3 xfyi = 0.
XjVi + 4 o2yi - 8 a3 = 0.
Hence the equation of the tangent may be written
2 xi^ix + (X 2 + 4 a2) 2/ + 8 a^yi - 24 a^ = 0.
Ex. 6. In the same manner the tangent to the cissoid x^ + xy- — 2 ay^ = 0 at
the point (xi, y{) is found to be
(3 x^ + y^) X + (2 xxyi - 4 ayi) y-2 ay^ = 0.
101. Normals. The normal to a curve at any point is the
straight line perpendicular to the tangent at that point. To find
its equation first find the slope of the tangent and then apply
problem 3, § 29.
X^ w2
Ex. 1. For the ellipse h — = 1 the slope of the tangent at (Xi, yi) is
62x, "' *'
Hence the equation of the normal at (xi, yi) is
a-'yi
ah/i
y-yi = ^{x- xi),
which is
a%ix - b^iy - (a2 - b!')xiyi = 0.
If y = 0,
a2 - 62
X = Xi = C'^Xi.
a2
Hence in fig. 114
NF=OF-ON-ae- e'^xx,
F'N = F'O + ON=ae + e^Xi.
Then
F'N _ a + exi _ F'P-i
'nF ~ a-exi ~ FPi '
(§73)
Fig. 114
and therefore, by plane geometry, the angle F'PiF is bisected by NPi ; that is,
in an ellipse the normal bisects the angle between the focal radii drawn to the
point of contact.
192 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
102. Maxima and minima. The discussiou of § 61 applies here
without change.
Ex. 1. A lever with the fulcrum at one end A (fig. 115) is to be used to lift a
jp weight w applied at a distance a from the
fulcrum by means of a force applied at
the other end B. The lever weighing n
units per unit of length, required the
length of the lever that the force required
may be a minimum.
Let X = AB, the length of the lever, 6
the angle it makes with the horizontal,
and F the force applied at B. Then the
weight of the lever is nx, and may be
considered as applied at C, the middle point of AB. By the law of the lever.
Fig. 115
Fx cos e = wa cos 6 + nxl-) cos 6,
Then
and
Fz=
wa
X
nx
dF_
dx
wa
-I
d^F
2wa
dx2
X3
2ioa dF ^ , d^F .
\l , — = 0 and >0.
\ n dx da;2
When X
Therefore this is the required length.
Ex. 2. Light travels from a point A in one medium to a point B in another,
the two media being separated by a plane
first medium is Vi and in the second v-z,
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. 116, where j\fiV represents
the intersection of the plane of the motion
and the plane separating the two media, and
ACB represents the path.
If the velocity in the
Fig. 110
MAXIMA AND MINIMA 193
Let MA = a, NB = b, MN = c, and MC = x. Then AC = VaF+1^^ and
CB = V(c — a;)2 + 62, xhe time of propagation from J. to 5 is therefore
_ Va2 + a-2 V(c - a;)2 + 62
, dt X c — X
whence — = ,
^ Vi Va2 + X2 1)2 V(C - X)2 + 62
d2< a2 62
and — = +
dx^ t)i(a2 + x2) J »2 [(c - a;)2 + 6^]^
dH
Since — is always positive, tlie time is a minimum wlien
Vi VcC^ + X2 V2 V(C - X)2 +
(1)
This equation may be solved for x, but it is more instructive to proceed as
follows :
X MC
Va2 + x2 --l^'
= sin ^.
c - X _ C'-ZV' _
: sin \j/.
V(c - X)2 + 62 <^B
sin 0 Vi
sin f V2
Then equation (1) is
Now 0 is the angle made by J. C with the normal at C and is called the angle
of incidence, and xp 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.
A case of a maximum or a minimum value sometimes occurs
when the derivative is infinite and consequently discontinuous.
Therefore the case is not included in the previous discussions.
In practice the infinite values of the derivative may be examined
by the rule of § 61.
Ex. 3. 2/ = </{x - 1) (X - 2)2 = (x - 1)^ (x - 2)^
^ = \ix-irHx-2)^ + lix-l)Hx-2)-i
ox 3 o
= 1 (X - ir?(x - 2)-i[(x - 2) + 2(x - 1)]
3x-4
Sy/{x- l)2(x-2)
194 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
•^ = 0 when x = |, and changes from + to — as a; passes through ^•
Therefore « = * gives a maxi-
mum value to the function.
dy
— = 00 when a; = 1 or 2. When
dx ^
X = 1 — • does not change sign.
dz ^y
When X = 2 — changes from —
dx, °
to + . Then x = 2 gives a mini-
mum value of the function. Its
graph is in fig. 117.
for it is that , ,
32- doif
103. Point of inflection.
A point of inflection was
defined in § 62 as a point
at which the curve clianges
from being concave upward
to concave downward, or
vice versa; and the condition
changes sign. Hence only those values of x which
d'y
make — ^ zero or infinity need be considered in the examination of
a curve for points of inflection.
Ex. 1. Find the points of inflection of the witch y =
8a3
dy
By differentiation, — =
16a5x
dx (x-^ + 4 a2)2
x2 + 4 a^
d^y _ 16a3(3x2-4a2)
dx^~ (x2 + 4 a2)3
It is evident that — - = 0 if x = ± — ^ 1 and that no real finite value of x
makes — ^ infinite.
dx2 9.
We have, then, to consider only the points for which x
48
Writing — in the form
«'(-^.)(
-^,)
Vs'
, we see that if x <
(x2 + 4 a'-^)3
2a d^y ^. ,.. ^ 2a d^y ^
— -, < 0 ; and if x > — ~, ~=- > 0.
V3 dx.'^ Vs dx:^
2a
dx2
d^y n •* 2 a
_|>0; If -— : <x
dx2 V3
Hence the curve is concave downward between tlie two points for which
X = and X = respectively, and concave upward at all other points.
Then there are two points of inflection (fig. 90, § 82) for which x = ± — - • The
ordinates are found from the equation to be
LIMIT OF RATIO OF ARC TO CHORD
Ex. 2. Examine the curve y = (x — 2)^ for points of inflection.
dy _ 1 d^y 2
195
By differentiation,
dx 3(x-2)^ dx^ 9(x-2)S
It is evident that — = oo if a; = 2, and that no value of x makes — = 0.
d^y '^'^ d^v ^""^
If X < 2, — ^ > 0 ; and if x > 2, — ^ < 0. Hence the point for which x = 2 is a
dx^ dx2
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.
118). The ordinate of this point isO.
Fig. 118
104. Limit of ratio of arc
to chord. The student is fa-
miliar 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. 119. 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 there-
fore increases without limit. The
.B manner in which this limit is ob-
tained is a question of the Integral
Calculus, and will not be taken up
here.
We may use the definition, how-
ever, 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, i.e. as the ends of the
arc approach each other along the curve.
Accordingly, let P^ and P. (fig. 120) be any two points of a curve,
i^T^ the chord joining them, and F^T and P^T the tangents to the
curve at those points respectively. We assume that the arc P^I^
lies entirely on one side of the chord ^^, and is concave toward
Fig. 119
196 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
the chord. These conditions can in general be met by taking
the points J^ and i^ near enough together. Then it foUows from
the definition that
whence
P^T+TP^ arc P^J^
PP
>
PP
>1.
Fig. 120
If TH is the perpendicular from
T to P^P^, and if the angles P^P^T and
-^i^T are denoted by a and yS respectively, then P^T = P^R sec a,
and TP^ = RP^ sec ^ = (^^ - P^R) sec /S.
.-. i^T + T^ = iji^ sec a; + (iji^-iji^) sec /3
= P^P^ sec yS + i^^ (sec a — sec /3),
and
P^T+TP^_ P,P^ sec ^ + P^R {seca— sec /3)
PP ~
PP
12
Pi?
= sec yS + -'— (sec a — sec )Q).
Now, as Pi and JP approach each other along the curve, a and ^
both approach zero as a limit, whence seca and secyS approach
PR
unity as a limit ; and since -^~ is always less than unity, it fol-
-^■'2
PT+ IP
lows that the limit of — r^ is unity.
PP
arc PP ^
Hence ^—^ lies between unity and a quantity approaching
?^. arc PP
unity as a limit, and therefore the limit of — -, — ^ is unity, i.e.
J\P,
the limit of the ratio of an arc to its chord as the arc approaches
zero as a limit is unity.
105. The derivatives -r- and -y- • On any given curve let the
as as
distance from some fixed initial point measured along the curve to
any point P be denoted by s, where s is positive if P lies in one
DIRECTION OF A CLTRVE
197
direction from the initial point, and negative if P lies in the
opposite direction. The choice of the positive direction is purely
arbitrary. 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 direction of OX
and the positive direction of the tan-
gent by 4>.
Now for a fixed curve and a fixed
initial point, the position of a point P
is determmed if s is given. Hence x
and y, the coordinates of P, are func-
tions of s, which in general are con-
tinuous and may be differentiated. We will now show that
dx
ds
= cos <f).
-^ = sin 6.
ds
Let arc PQ = As (fig. 121), where P and Q are so chosen that As
is positive. Then PE = Ax and RQ = Ay, and
PR chord PQ PR
Ax
As hiqPQ arc PQ chord P©
chord PQ
cos RPQ.
arc PQ
RQ chord PQ
RQ
Ay_
As arcPQ arc PQ chord P^
chord PQ
arc PQ
sin RPQ.
Taking the limit, we have, since Lim "_ _ = 1 and
Lim RPQ = 4>,
arc PQ
dx
ds
COS(f),
dy . ,
-^ = sin 9.
ds
(1)
198 DIFFERENTIATIOK OF ALGEBRAIC FUNCTIONS
From (1) we obtain by division
dy
ds dy
tan <p = -7- = -7- ;
ax dx
ds
(2)
by (8), § 96. This agrees with § 59.
Again from (1), by squaring each equation and adding them,
we have
ds
H^h-
/dsV /dsV
By multiplying (3) by | — | and again by I — | and applying
(7), § 96, we have \^-^/ W/
1 +
\dx
and
(%h'-
(4)
(5)
These last are the familiar trigonometric formulas
1-f tan'^^ = sec'^^,
cot^<^ + 1 = cosec^<^.
For convenience we have used
a figure in which <f) is acute. But
as s increases (f) may be in any
V , quadrant. This may be seen on
the circle of fig. 122.
The student may verify that
formulas (l)-(5) are true in all
cases.
106. Velocity. An important application of the conception of
a derivative is found in the definitions of the velocity of a moving
body.
Fig. 122
VELOCITY 199
If a body moves so that the space traversed is proportional to
the time, the motion is said to be uniform, and the velocity is the
quotient of the space divided by the time, and is therefore con-
stant. \i t represents time, s the space traversed in the time t,
s
and V the velocity, then for uniform motion v = -• When the
space is not proportional to the time but is some other function
of it, the quotient of the space divided by the time is called the
mean or average velocity during the time. Thus if a railroad
train goes 200 miles in 5 hours, the mean velocity is 40 miles
an hour. So, in general, if a body traverses a small increment of
As
space As in a small increment of time A^, the quotient ■— is the
mean velocity in the time A^. The mean velocity depends upon
the value of A^. To obtain a definition of the velocity at the
beginning of the interval A^, we think of A^, and consequently of
. . As
As, as approaching zero as a limit, and take the limit of — as the
velocity v ; that is,
T . As rfs
V = Lim — - = — .
A^ dt
We note that if •?; > 0, an increase of time corresponds to an
increase of s ; while if v < 0, an increase of time causes a decrease
of s. Consequently, the velocity is positive when the body moves
in the direction in which s is measured, and negative if it moves
in the opposite direction.
Ex. 1. If a body is thrown up from tlie eartli with an initial velocity of 100 ft.
per second, the space traversed, measured upward, is given by the equation
s = 100t- IC) t'.
Then • t> = ^ = 100 - 32«.
, at
When « < 3^, u >0 and when e > 3^, u <0. Hence the body rises for 3^ seconds,
and then falls. The highest point reached is 100 (3^) - 16(3|)2 = 156^.
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.
200 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
Let A (fig. 123) be the position of tlie man and C that of the boat. Let
^B = /i = 20, AC= s, and BC = x.
dx
We wisli to find
dt
Now
therefore
X = Vs'^ - 400 ;
X s ds
ds
dt
dt Vs-' - 400 '^t
But, by hypothesis, s is decreasing at the rate of 3 ft. per second ; therefore
= — 3, and tlie required expression for the velocity of the boat is
3s
dx
dt y/g2 _ 4y(J
To express this in terms of the time we need to know the value of s when
t = 0. Suppose this to be Sq ; then
and
s
dx
s,i — 3 1.
3s„ + 9«
'^ Vs^j'-400-O.So« + 9«2
107. Components of velocity. When a body moves along its
path, straight or curved, from P to ^ (fig. 124), where PQ = AS,
X changes by an amount PR = A«, and y changes by an amount
RQ = Ay. We now have
Lim 'T~ — 'T-='^' = velocity of the body
in its path.
Ax dx
Lim — - = -J- = v^ = component of ve-
locity parallel to OA'.
Lim — = J = ^y = component of ve-
locity parallel to O Y.
Fig. 124
Otherwise expressed, v represents the velocity of P, v^ the
velocity of the projection of P upon OX, and v^ the velocity of
the projection of P on 0 Y.
COMPONENTS OF VELOCITY
201
dx
dt
By (7), § 96, and (3), § 105,
dx ds
ds dt
dsY
dt)'
V^ = V COS (f),
dy _ dy ds
dt ds dt
dxV /dy'' "^
dt) \dt
whence
v^ = V sin ^,
V- = V- + V'
Ex. A man walks across the diameter of a circular courtyard at a uniform
rate. A lamp, at one extremity of the diameter perpendicular to the one on
which he walks, throws his shadow
on the wall. Required the velocity
of the shadow along the wall.
In fig. 125 let L be the lamp,
M the man, and S the shadow. Let
a be the radius of the courtyard
and c the uniform velocity of the
man. Let the variable OM = Xi
dX]
where — = c. Then the equation
dt
of the line LS is
ax — Xiij — axi = 0,
and that of the circle is
a;2 + ?/2 =: (• ".
Solving these equations, we have,
for the coordinates of S,
2a^x
a-"
Henc^-
a2 + a;f
y =
a- + x{
a- — Xy
and
and
dt
dy
di
(a2 + x{f
2a^x? dx, ^ „
— = 2 a-c - ^ „. .,
dt (a2 + x^y
4:a\ dxi ,, ,
-— = - 2a-c-
2axi
v'i= { — ) +
\dt
(a^ + xf)- dt " ~ (a;- + x{r
fc-» + 2 a2xf + x^
m-*^'
(a2 + X,-)*
2 a-c
The requii-ed velocity is ^
a -r X|
The above solution can be simplified by the use of trigonometric functions.
See Ex. 2, § 1<>3.
202 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
108. Acceleration and force. When the motion of a body is
not uniform, the velocity at the end of an interval of time is not
the same as at the beginning. Let v be the velocity at the begin-
ning of the interval A<, and ?; + Ai; the velocity at the end. Then
the limit of the ratio of the change in the velocity to the change
in time, as the latter approaches zero as a limit, is called the
acceleration in the path ; that is, if a denotes this acceleration,
_dv _ d (ds\ _ d?s
"'~dt~di\di/~df'
When a is positive an increase of t corres'ponds to an increase
of V. This happens when the body moves with increasing velocity
in the direction in which s is measured, or with a decreasing
velocity in the direction opposite to that in which s 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 s is measured, or with increasing velocity
in the direction opposite to that in which s is measured.
The force F which acts in the direction of the path of a moving
body is measured by the product of the mass rii and the accelera-
tiona. Thus ^^ ^^
. F — ma = m — = w — r •
dt df
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 s is measured, and negative when it acts in the opposite
direction.
Ex. Let s = A + Bt+\ Ct\
Then v = B+Ct,
a = C,
and F= mC.
If So and Vo denote the values of s and v when t = 0, -we have, from the last
equations,
So = ^, Vo = B,
and the original equation may be vsritten
S=so + vot+ ]^ai?.
ILLUSTRATIONIS OF THE DEEIVATIVE 203
As a special case, suppose a body of mass m thrown vertically upward from
a point h ft. above the surface of the earth with an initial velocity of Vq ft. per
second. Then, if s is measured upward from the surface of the earth, we have
80 = A, F = — mg, a = — g,
where g is the acceleration due to gravity. Then the expression becomes
109. Other illustrations of the derivative.
1. Bate of change. \i y = f{x), a change of Aa; units in the
value of X causes a change of Ay units in the value of y. Then
A?/
— ^ is the change in y per unit of change in x ; that is, the change
in y which would be caused by the change of a unit in x, if A2/
were proportional to Aic. Passing to the limit, we have
dv
-— = rate of change of y with respect to x.
For example, the velocity of a moving body is the rate of change
of the space with respect to the time, and the acceleration is the
rate of change of the velocity with respect to the time.
2. Momentum. The momentum of a moving body is the product
of the mass and the velocity ; that is, if M is the momentum,
M = mv.
Now, from S 108, F = m -— = , ' = — — •
^ dt dt dt
The force is therefore the derivative of the momentum with
respect to the time, or, in other words, the rate of change of the
momentum with respect to the time.
3. Kinetic energy. The kinetic energy of a moving body is
equal to half the product of the mass into the square of the
velocity; that is, if E is the kinetic energy,
E = \ mv^.
^, dE d(lmv^) dv ds dv dv _,
Then -^- = -^ ~ = mv-- = m — -— =m-- = F;
ds ds ds dt as at
that is, the force is the derivative of the kinetic energ}' with
respect to the space traversed, or, in other words, the rate of
change of the kinetic energy with respect to the space.
204 DIFFEKENTIATION OF ALGEBEAIC FUNCTIONS
4. Coefficient of expansion. Let a substance of volume ?? be at a
temperature t. If the temperature is increased by A^, the pressure
remaining constant, the vohnne is increased by A v. The change
A?;
per unit of volume is then — » and the ratio of this change per
unit of volume to the change in the temperature is
1 Av
V A^'
The
limit of this ratio is called the coefficient of expansion ; that is,
the coefficient of expansion equals - -y- • In other words, the
coefficient of expansion is the rate of change of a unit of volume
with respect to the temperature.
5. Elasticity. Let a substance of volume v be under a pressure
p. If the pressure is increased by A^, the volume is increased by
Av
— £^v. The change in volume per unit of volume is then
V
The ratio of this change per unit of volume to the change in the
1 Av
pressure is — j and the limit of this' is called the compres-
V Lp
sibility ; that is, the compressibility is the rate of change of a unit
volume with respect to the pressure.
The reciprocal of the compressibihty is called the elasticity,
wliich is therefore equal to ~ v — •
dv
Ex. For a perfect gas at constant
temperature,
P = --
Therefore the elasticity is
dp I k\ k
-v-- = -v(- ) = -=p;
dv \ vV V
that is, the elasticity of a perfect gas is
equal to the pressure.
G. Areas. Let2/=/(^)(fig.l26)
Ite any curve, C a fixed point, and
F{x, y) a variable point upon it.
We shall assume that P lies at the right of C and that the por-
tion of the curve between C and P lies above the axis of x.
M
¥u.. 120
INTEGRATION
205
Draw the ordinates BC and MP and let A denote the area
BMFC. Then ^ is a function of x, since it is determined when
OM = a; is given. Give x an increment A^ = MN, and draw the
ordinate NQ and the Unes BR and QS parallel to OX. Then
BQ = Ai/, MNQP = A A,
MNBP = MB • MN = y Ax, MNQS = NQ • MN = (y + Ay) Ax.
But, from the figure,
MNRB < MNQB < MNQS * ;
that is, yAx<AA<{y + Ay) Ax,
whence
y^-A-x^y^^y-
Now as Ax approaches zero as a limit, — approaches — — j 3/ is
AA
unchanged, and y + Ay approaches y. Hence -- — > which lies be-
tween y and y + Ay, also approaches y ; that is,
dA^
dx
If the curve lies below the axis of x (fig. 127), and we place,
as before, MNRB = y Ax and
MNQS= {y + Ay) Ax, these areas
are negative. We shall then
have, as before, —
dA
dx
= y,
but the area is now considered
as negative.
110. Integration. In many
applications of the calculus the
derivative is known, and the
problem presents itself to find
M N
R
Q
Fio. 127
* If the curve runs down toward the right, the inequality signs will be reversed.
206 DIFFEEENTIATION OF ALGEBRAIC FUNCTI0:J^S
the function which has that derivative. For example, it may be
required to find a curve when its slope is known, or to find the
space traversed by a particle with known velocity or acceleration,
or to find the area bounded partly by a known curve, or to find a
function which has a known rate of change.
The process by which a function is found from its derivative
is called iiitejration. Differentiation and integration are then
inverse processes, as are addition and subtraction, multiplication
and division, involution and evolution. The methods of integra-
tion are in general complex and must be studied later in the
integral calculus. At this time we shall give some simple exam-
ples where the integration can be carried out by reversing the
formulas of differentiation.
In the first place, however, we must notice that the integration
of a given function does not lead to a unique result. For, as we
have seen already (§95),
d {u + c) __ du
dx dx
where c is any constant whatever ; that is, two functions which
differ hy an additive constant have the same derivative.
Conversely, if two functions have the same derivative, they differ
hy an additive constant.
_- , , dv du
For let -r = -r'
dx dx
Then
dv du _ ^
dx dx
or
d{v-u) _Q
dx
Hence*
V — M = c, where c = constant ;
that is,
V =ic + c.
The constant c cannot be determined by integration, but must
be fixed by the special conditions of the problem in which it
occurs.
* A proof of this conclusion will be given in the second volume.
INTEGRATION
207
Ex. 1. Required the curve the slope of which at any point is twice the
abscissa of the point.
By hypothesis,
dy
dx
Therefore y = x^ + c. (1)
= 2x.
Any curve whose equation can be
derived from (1) by giving c a defi-
nite value satisfies the condition of
the problem. If it is required that
the curve should pass through the
point (2, 3), we have, from (1),
3 = 4 + c ; whence c = — 1,
and therefore the equation of the
curve is „, ^o .
y = x^ — I.
But if it is required that the curve
should pass through (—3, 10), we
have, from (1),
10 = 9 + c ; whence c — 1,
and the equation is
y = x^ + -i.
Ex. 2. Required the space traversed by a particle if its velocity is equal to
the square of the time.
By hypothesis.
Therefore
V = — — t-.
dt
s- 1 1^ + c.
The constant c can be determined if we Itnow the position of the particle at a
given time. For instance, if when t — 0 the particle is at the point from which
8 is measured, we must have c = 0. On the other hand, if wlien t = 0 the particle
is two units from the point at which s = 0, we have c = 2.
Ex. 3. Required the space traversed by a body if the acceleration is propor-
tional to the time.
-nr 1- dv d^s ,.
We have o = — = ■ — = kt,
dt *2
where A; is a known constant. Then v =
ds_l
di~2
kt^ + ci.
and
S = -M^ + Cit + C2.
The constants Ci and C2 can be determined if we know the position and the
velocity of the body at a given time. If, for examole, we know that when < = 0,
a = 0, and v = 4, we have cj = 0, ci = 4.
208 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
Ex. 4. Find the area bounded by the curve y = ^ (x^ — 3x'^ — Ox + 27), the
axis of X, and the ordinates x = 4 and x = 5.
If A is the area C DP M (fig. 129), where OC = 4 and OM = x, then (§ 109, 6)
d A 1
— = -(x3-3x2-9x + 27).
whence
(1)
Therefore
1 /x* 9
8\4 2
x)
/ 2
Fig. 129
If X = 4, MP coincides with CD and therefore J. = 0. Substituting in (1)
the corresponding values x = 4, yl = 0, we find c = — |.
9
2'
If X = 5, ^ = CDEF. Hence
CZ>J?F = ^ (fi-l A _ 125 - 2f A + 135) - I = 2/j.
Ex. 6. Find the area bounded by the axis ofx and tlie portion of the curve
y = ^ (x^ — 3 x2 — 9 X + 27) between x = — 3 and x = 3.
We now let A = the area 6?iVQ (fig. 129).
Then, as before.
dA_l
dx ~8
(x3 -3x2- 9x + 27).
1 /x* 9 \
^ = - I x3 - - x2 + 27 X ) +
8\4 2 /
When X = - 3, ^ = 0 ; therefore c = 2^^ ,
and A = -(-~x^-~x'^ + 21x] + ^^'^
8\4 2 '
32
Placing X = 3, we have area GQH — 13 J^.
PROBLEMS 209
PROBLEMS
dy
Find -^ in each of the following cases :
dx
1. 2/ = (3a; + l)(x2 + 2x + 1).
16.
y
2. y = (3x2 + 6x + 1) (5x2 + lOx + 5). x^ + x2 + 1
x + a Vx2 + 1
4. y:
5. y
6. y
x + a
x3 + l
X3-1
2x2-
4x + 3
3x2-
6x + 1
x^ — x-
2 + X-
1
18. 2/ = (2x -3)2(x + l)3.
19. 2/ = (3x - 5)2(x2 - 5x + 1).
20. 2/ = (x + l)Va;2- 1.
21. 1/ = (x2 - 4x + 3)'^(x3 + 1)3.
2.2. y = Vx + 1 + Vx - 1.
a;'*-l 23. 2/ = x + Vx2 + 1.
2
7. 2/ = 2xi + 8x^-- + -.. 2L y=W^^Tl + -^.
V3x2 + 1
8. z/ = 4x2-0x + ^-|. 25. 2/=^(|-J,
9. 7/ = V^-J-. „„ Vx2 + 1
Vx 26. y = — •
^ X X — 1
10. y = ^X2 — -^X + = =:• „„ (X2 + 1)5
Vx ^x2 27. y = ^^ ^,
^X vx (X3 + 1)*
11. y = (3x2 -ox + 6)2. ^
12. 2/ = (x2 + l)3. 28. 2/
13. ?/ = V4x2 + 5x-6. -Q
3/ — ^~ /
14. y = Vx2 + X - 1. ^02 - x2
15. 2/^ -J—. 30.2/
X + Vl + X2
X
x2 + 1 * x + Va2 + x2
Find — from each of the following equations :
dx
31. X* - 4x22/2 + 2/3 = 0. 34. x5 + 2/* - x' - 2/ = 0.
32. x5 - 2/5 - x3 + 2/ = 0. 35. (X + 2/)^ + (X - 2/)^ = a.
33. x«2/* + (X - yY = 0. 36. y"- = ^-^ •
X y
Find ^ and —^ from each of the following equations :
dx dx2
37. 5 x2 + 2 2/2 = 10. 40. 2/' = ct^ (X + 2/).
38. x^ + y' = a^. 41_ 2/' + 2/ = x^.
39. t^ + ^ = 1
?! 4. tL_
a2 62 "• 42. 2/' -2x3 + 4x2/ = 0.
210 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
43. Find the tangent and the normal to the parabola y^ — 4y — 6x — 9 = 0
at a point the abscissa of which is — 2.
44. Find the equations of the tangent and the normal to the circle
x^ + y^-iz + 6y-2i = 0
at the point (1, 3).
45. Find the equation of the tangent.to the witch y = at the point
for which x = 1.
46. Find the tangent to the curve x^ — y^ + x^ — y = 0 at the point the
abscissa of which is 1.
47. Find the tangent to the curve x^y + x^ — x^ + y = 0 at the point the
abscissa of which is 1.
48. Find the equation of the tangent to the curve y^ — xy — a = 0 at the
point {xi, Vi).
49. Find the equation of the tangent to the curve x = y^ + 1 &t the point
(xi, yi).
50. Find the equation of the tangent to the curve y- = x^ at the point (xi, yi).
51. Find the equations of the tangent and the normal to the curve y — x -i —
at the point (xi, yi).
52. Find the equation of the tangent to the curve Vx + Vy = Va at the
point (Xi, yi).
53. Find the equation of the tangent to the curve x^ + y^ — a* at the point
(xi, yi)-
54. Find the tangent and the normal to the ellipse 3x^ + 5y^ = lo at the
upper end of the ordinate through the right-hand focus.
55. Find the equations of the tangent and the normal to the hyperbola
4 x^ — 2/2 — 12 at a point the abscissa of v/hich is equal to its ordinate.
56. Find in terms of x, y, and — the projections upon OX of the portions
dx
of the tangent and the normal between the point of contact and OX. These
are called the suhtangent and the subnormal.
dy
57. Find in terms of x, y, and — the lengths of the portions of the tangent
dx
included between the point of contact and the coordinate axes.
58. Prove that a normal to an hyjjerbola makes equal angles with the focal
radii drawn to the point where the normal intersects the hyperbola.
59. Prove that a normal to a parabola makes equal angles with the axis of
the parabola and the line drawn from the focus to the point where the normal
intersects the parabola.
60. 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 a^ : b"^.
61. Find the point at which the tangent to the curve x^ — xy — 1 = 0 has
the slope 2.
PROBLEMS 211
62. Find the coordinates of a point on the ellipse 1- — = 1 such that the
tangent there is parallel to the line joining the positive extremities of the major
a,nd the minor axes.
63. Find a point on the ellipse -^ + ^ = 1 such that the tangent there is
equally inclined to the two axes.
64. Prove that the portion of a tangent to an hyperbola included by the
asymptotes is bisected by the point of tangency.
65. 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 oh the transverse axis.
66. If a tangent to an hyperbola is intersected by the tangents at the verti-
ces in the points Q and B, show that the circle described on QR as a diameter
passes through the foci.
67. Prove that the ordinate of the point of intersection of two tangents to a
parabola is the arithmetical mean between the ordinates of the points of con-
tact of the tangents.
68. If P, Q, and R are three points on a parabola, the ordinates of which
are in geometrical progression, show that the tangents at P and R meet on the
ordinate of Q.
69. Show that the tangents at the extremities of the latus rectum* of a
parabola are perpendicular to each other.
70. Prove that the tangents at the ends of the latus rectum of a parabola
intersect on the directrix.
71. Prove analytically that if the normals at all points of an ellipse pass
through the center, the ellipse is a circle.
72. Prove that the tangent at any point of the parabola y'^='ipx will meet the
directrix and the latus rectum produced in two points equidistant from the focus.
73. Find the length of the perpendicular from the focus of the parabola
y2 = 4px to the tangent at any point (xi, j/i), in terms of Xi and p.
74. If perpendiculars are let fall on any tangent to a parabola from two given
points on the axis which are equidistant from the focus, prove that the difference
of their squares is constant.
75. Show that the product of the perpendiculars from the foci of an ellipse
upon any tangent equals the square of half the minor axis.
76. Find the equation and the length of the perpendicular from the center
to any tangent to the ellipse 1- — = 1.
77. At what angles t do the loci 2/2-4a;-|-4=Oand2/-a; + l = 0 intersect ?
*The latus rectum of a conic is the chord through the focus perpendicular to
the axis.
fThe angle between two curves is the angle between their tangents at their
point of intersection.
212 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
78. Find the angle between the straight line y — 2x — 2 and the cissoid
X {x^ + y''^) — i y'^ at each of their points of intersection.
79. At what angle do the circles x^ -hy- -Q = 0, x^ + y^ — iix-6y + 9 = 0
intersect ?
80. Prove that the center of each of the circles
x^ + y^ = a2 and x^ + y^-2ax = 0
is a point of the other, and find the angle at which they intersect.
81. At what angle do the circle x'^ + y^ = 21 and the parabola y'^ = ix inter-
sect each other ?
82. Show that the curves \- — = 1 and ^ = 1 cut each other at
right angles and are confocal. '
83. Prove that an ellipse and an hyperbola with the same foci cut each
other at right angles.
84. If two concentric equilateral hyperbolas are described, the axes of one
being the asymptotes of the other, show that they intei-sect at right angles.
85. Find the angle between the parabolas y"^ — iax and x^ = 4ay at each
of their points of intersection.
86. Find the angle between the parabola x" = iay and the witch y = —
at each of their points of intersection. "^
87. Prove that the cissoid y^ = and the parabola y^ = 4ax intersect
at right angles at the origin. ""
88. Find the angles of intersection of the cissoid y- = and the circle
x2 + y2_4ax = 0. 2a -X
89. Find the angle of intersection of the witch
8 a^ 4 7/8
y = — ^ and the -cissoid x^ —
x'^ + ia^ 5 a — 4 y
90. Find the angles of intersection of the circle x^ + y- = a a^ and the witch
8a3
y
«2 + 4 a^
91. Find the angle between the strophoid y = ±x\ and the circle
0,09 \ a + X
92. Find the angles of intersection of the curves
2/2 = 2 ax and x^ + y^ — S axy = 0.
93. 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. Required
the dimensions of the rectangle which will require the least amount of fencing.
94. A man on one side of a river, the banks of which are assumed to be
parallel straight lines i mi. apart, wishes to reach a point on the opposite side
of the river and 3 mi. further along the bank. If he can walk 4 mi. an hour
and swim 2 mi. an hour, find the route he should take to make the trip in the
least time.
PKOBLEMS 213
95. A rectangular piece of land to contain 96 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 fence
may be required ?
96. What are the dimensions of the rectangular beam of greatest volume
that can be cut from a log a ft. in diameter and b ft. long, assuming the log to
be a circular cylinder ?
97. The hypotenuse of a right triangle is given. How shall the sides be
chosen so that the area shall be a maximum ?
98. Two towns A and B are situated respectively 2 mi. and 3 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 neai'est
points of the river to A and B respectively are 10 mi. apart?
99. AB and CD are two parallel lines distant b units apart. A transversal
BF is drawn, intersecting the transversal AD at E. For what position of F is
the sum of the ai'eas of the two triangles AEB and FED a minimum ?
100. A right cone is generated by revolving an isosceles triangle of constant
perimeter about its altitude. Show that the cone of greatest volume will be
obtained when the length of the side of the triangle is three fourths the length
of the base.
101. Into a full conical wine glass whose depth is a and angle at the base is
2 a there is carefully dropped a spherical ball of such size as to cause the greatest
overflow. Show that the radius of the ball is
a sin a
sin a + cos 2 a
102. Two ships are sailing uniformly with velocities w, v along lines inclined
at an- angle 6. Given that at a certain time the ships are distant respectively
a and 6 from the point of intersection of their courses, show that the least dis-
tance between the ships is
(av — 6(t)sin 0
(u2 + v"^ -2uv cos 6)^
103. Find the least ellipse which can be described about a given rectangle,
assuming that the area of an ellipse with semiaxes a and b is -n-ab.
104. Find what sector must be taken out of a given circle in order that it
may form the curved surface of a cone of maximum volume.
105. The stiffness of a rectangular beam varies as the product of the breadth
and the cube of the depth. Find the dimensions of the stiffest rectangular beam
that can be cut from a circular cylindrical log of radius a in.
106. Tlie strength of a rectangular beam varies as the product of its breadth
and the .s<juare of its depth. Find the dimensions of the strongest rectangular
beam that can be cut from a circular cylindrical log of radius a in.
214 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
107. The fuel consumed by a steamship 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 fixed distance against a current flowing a mi. an hour, find the
most economical rate.
108. A cistern in the form of a circular cylinder open at the top is to
be constructed to contain a given amount. Required the dimensions that the
amount of material expended may be the least.
109. Required the right circular cone of greatest volume which can be
inscribed in a given sphere.
110. A power house stands upon one side of a river of width b mi. and a
manufacturing plant stands upon the opposite side a mi. 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.
111. Find the isosceles triangle of greatest area which can be cut from a semi-
circular board, the vertex of the triangle being at the center of the diameter.
112. Find the isosceles triangle of greatest area which can be placed in a
figure bounded by a portion of a parabola and a straight line perpendicular to
the axis of the parabola, assuming that the vertex of the triangle lies in the
straight line and that the base is parallel to the straight line.
113. Find the point of inflection of the curve y = a + {b — x)^.
114. Find the points of inflection of the curve y =
x2 + 1
115. Examine the curve y = (x — l)^{x + 1)^ for maxima and minima and
points of inflection.
116. Find the maximum and the minimum ordinates and the points of
inflection of the curve y^ = x(x^ — a^).
a
117. Find the points of inflection of the curve y =
X2-1-4
118. Show that the strophoid y = ± x \ has no point of inflection.
\ a + x
119. Find the points of inflection of the curve a*y'^ = a^x* — x^.
120. Find the points of inflection of the curve |-| + f-J = 1.
121. Find where the rate of change of the ordinate of the curve
j/ = a;3 -6a;2 + 3x + 5
is equal to the rate of change of the slope of the tangent.
122. A body moves in a straight line according to the law 8= ^t* — At^ + 16t^.
Find its velocity and acceleration. When is it stationary ? When is its velocity
a maximum ? During what interval is it moving backward ?
123. A particle is moving along the curve y^ — 4x, and when a; = 4 its ordi-
nate 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 along the curve ? Where
will the abscissa be changing ten times as fast as the ordinate ?
PROBLEMS 215
124. Two points, having always the same abscissa, move in such a manner
that each generates one of the curves y = x^ — 12x^ + 4x and y = x^ — 8x^ — 8.
Wlien are the points moving with equal speed in the direction of the axis of y?
What will be true of the tangent lines to the curves at these points ?
12a. The top of a ladder a units long slides down the side of a vertical wall
which rests on horizontal land. Find the ratio of the velocities of its top and
bottom.
126. The altitude of a variable cylinder is constantly equal to the diameter
of its base. If when the altitude is 6 ft. it is increasing at the rate of 2 ft. an
hour, how fast is the volume increasing at the same instant ?
127. A boat moving 8 mi. an hour is laying a submarine cable. Assuming
that the water is 100 ft. deep, that the cable is attached to the bottom of the
sea and stretches in a straight line to the stern of the boat, at what rate is the
cable leaving the boat when 120 ft. have been paid out ?
128. A ball is swung in a circle at the end of a cord 4 ft. long so as to make
100 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
10 sec. after the cord breaks, if no allowance is made for any new force acting?
129. A body slides down an inclined plane at such a rate that the distance
traversed at the end of t sec. from the time it begins to move is 5 1^. If the plane
is inclined to the horizon at an angle of 30°, what is the vertical velocity of the
body at the end of 3 sec. ?
130. A roll of boJt leather is unrolled on a horizontal surface at the rate of
5 ft. per second. If the leather is ^ in. thick and at the start the roll was 2 ft.
in diameter, at what rate is the radius decreasing at the end of 3 sec, if the roll
is assumed to be a true circle ?
131. An elevated car running at a constant elevation of 40 ft. above the
street pas.ses directly over a surface car, the tracks of the two cars crossing at
right angles. If the speed of the elevated car is IG mi. per hour and the speed
of the surface car 8 mi. per hour, at what rate are the cars separating 5 min.
after they meet ?
132. Find the curve the slope of phich at any point is 3 more than the
square of the abscissa of that point and which passes through the point (1, — 3).
133. x'^ind the curve the slope of which at any point is equal to the square
of the reciprocal of the abscissa of the point and which passes through (2, 1).
134. Find the curve the slope of which at any point is equal to the square
root of the absci.ssa of the point and which pas.ses through (4, 9).
135. Prove that any curve the slope of which at any point is proportional
to the abscissa of the point is a parabola.
136. Find the curve the slope of which at any point is proportional to the
square of the ordinate of the point and which passes through (1, 1).
137. Find the area of each arch of the curve ?/ = 150a; - 25x2 _ ^s.
216 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS
138. Find the area of the arch of the curve y = x^ — Sx^ ~ dx + 27.
139. Show that the area bounded by any parabola y" — 4px, the axis of x,
and tlie ordinate tlirougli any point of tlie curve is two thirds the area of a rec-
tangle the sides of which are the coordinates of the point.
140. Expre.ss the area between the curve y = x", the axis of x, and the ordi-
nate through the point (h, k) of the curve as a rational function of h and k.
141. Find the area of the three-sided figure bounded by the coordinate axes
and the curve x^ -f- 2/^ = a' (§ G9).
142. Find tlie area between the parabola y- = ^x and the straight line
2 y - x = 0.
143. Find the area between the parabolas y'^ = 4 ax and x^ = 4 ay.
144. Find the area of the crescent-shaped figure between the curves
y = x"^ -h 5 and ?/ = 2 x- -|- 1.
145. Find the area of the closed figure bounded by the curves y'^ — \(jx
and 2/2 ;= x^.
CHAPTEE X
CHANGE OF COORDINATE AXES
111. Introduction. So far we have dealt with the coordinates
of any point in tlie 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, i.e. 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 transformation
of coordinates never alters the position of tlie point in the plane,
the coordinates alone being changed because of the new standards
of reference adopted.
112. Change of origin without change of direction of axes.
In this case a new origin is chosen, but tlie new axes are respec-
tively parallel to the original axes. ,
Let OX and or (fig. 130) 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
Xq and ^o- Let P be any point
in the plane, its coordinates being
X and y with respect to OX and
OY, and x' and y' with respect to
O'X' and O'Y'. Draw P3IM' paral-
lel to OY, intersecting OX and O'X' at M and M' respectively.
217
0'
M
N
Fig. 130
M'
218 CHANGE OF COORDINATE AXES
Then 0M= x, MP = y, 0'M'= x', M'P = y', 0N= y^, and N0'= x^
But OM = NM' = NO' + O'M',
and MP = MM' + M'P = 0N+ M'P.
.-. x = x^+ x', y = yo + y\
whicli 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 respec-
tively to the first set and intersecting at (1, — 1) with respect to OX and OY ?
Here a;o = 1, yo = — 1, x = 3, and y = — 2. Therefore 3 = 1 + x' and
— 2 = — 1 + y', whence x' = 2 and y' — —\.
Ex. 2. Transform the equation i/^ — 2?/ — 3x — 5 = 0toa 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 = \-\- y". Therefore
the equation becomes
(1 + y')2 _ 2(1 -I- 2/0 - 3(- 2 + X') - 5 = 0,
or 2/'2-3x' = 0.
Ai 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 y^ — Sx = 0,
or y^ = S X, the new axes being now the only ones considered.
113. 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 y^ = 3 x is simpler than the
original equation, and from its form we recognize the curve as a
parabola. It is obvious, however, that the position of the new
origin is of fundamental importance in thus simplifying the equa-
tion, and we shall now solve an example illustrating a method of
determining the new origin to advantage.
Ex. Transform the equation y^ — iy -x^ — Sx^ — 3x-|-3 = 0to 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.
CHANGE OF ORIGIN 219
The formulas of transformation are
X = jco + x' and y = yo + y',
where suitable values of Xq and yo are to be determined. The equation becomes
(2/0 + VT - 4 (yo + yl - {xo + a;')^ - 3 (xq + x'f - 3 {xo + x') + 3 = 0,
or, after expanding and collecting like terms,
y'^ + (2 2/0 - 4)2/' - x'3 - (3xo + 3)x'2 - (Sx^ + 6xo + 3)x'
+ (2/o - 42/0 - x^ - 3x2 _ 3a.^ + 3) = 0.
By the conditions of the problem we are to choose Xo and yo so that ,
2 yo - 4 = 0, 3x2 + 6xo + 3 = 0,
two equations from which we find Xo = — 1 and r/o = 2.
Therefore (—1, 2) should be chosen as the new origin of axes, and the new
equation is y'^ — x'^ — 0, or y'^ — x^, after the primes are dropped.
114. In particular, this method of simplifying an equation is
of considerable importance in studying the conies defined in
Chap. VII. For consider the equation
a" '^ V ' ^ '
If we place x = Xq + x', y = y^ + y', (1) becomes
^ + |f = 1^ (2)
a 0
which is the equation of an ellipse with its center at x' = 0, y' = 0,
and its axes along O'X' and O'Y'. Therefore (1) is an ellipse'
with its center at x = x^, y = y^, and its axes . parallel to OX
and OY. V^^ITft^
Furthermore, if a > &, e = ■ ; and the foci of the ellipse
a
are at (x-' = ± ae, y' = 0), or, what is the same thing, (a; = ± ae + x^,
y = y^). The directrices are x' = ± - > or x = x^ ::r -■
In a similar manner
220 CHANGE OF COORDINATE AXES
is the equation of an hyperbola with its center at {x^^ y^ and its
axes parallel to OAT and OY; and
represents a parabola with its vertex at {x^, y^ and its axis parallel
to OX.
Any equation which can be reduced to a form similar to one
of these can be discussed in a similar manner. A general treat-
ment of such equations will be found in Chap. XI. We shall
give here some examples.
Ex.1. 16x2 + 252/2 + 64a; -1502/ -111 = 0.
Rewriting, we have
16(a;2 + 4a;)+ 25(2/2 _ 6 y) = 111,
whence 16(x2 + 4x + 4) + 25(2/2 - 62/ + 9) = 400,
or (x + 2)2 (2/-3)2^^
25 16
Placing now x = — 2 + x', y — Z -\- y',
we have \-^— = \.
25 16
This is an ellipse with semiaxes 5 and 4, and eccentricity |. Its center is at
(x' = 0, 2/' = 0), its foci are at (x' = ± 3, y' = 0), and its directrices are x' = ± '^^-
= ±83-
Hence the original equation represents an ellipse with semiaxes 5, 4, and
eccentricity 3. Its center is at (— 2, 3), its foci are (—5, 3) and (1, 3), and its
directrices are x = — 10^ and x = 6^.
Ex.2. 5^2 _ 102/ - 4x -7 = 0.
Rewriting, we have 6 (2/2 — 2 ?/) = 4 (x + |),
5(2/2_2y + l) = 4(x + |+|),
or (2/ - 1)2 = 4 (X + 3).
Placing now x = — 3 + x',
2/ = 1 + 2/',
we have 7/2 = 4 x'.
which represents a parabola with vertex (x'= 0, 2/'= 0). Its axis is along (YX';
its focus is (x' = ^, 2/' = 0), and its directrix is x' = - 1.
Therefore the original equation represents a parabola with its vertex at
(—3, 1) and its axis parallel to OX. Its focus is (—24, 1) and its directrix is
a; = - 3.L
CHANGE OF DIRECTION OF AXES 221
Ex. 3. (X - C)2 + 2/2 - e2a;-2.
This is the equation of tlie conic, as found in § 81. We may write it as
(1 - e2)a;2 _ 2 ex + 2/2 = - A
Then if e ^i 1, we may proceed as follows:
(1 - e2) (x^ - -l^x + ] + 2/2 ^ _ c2 + _£!_ ,
(1 - e2) (x —Y+ 2/2 = -^ ,
^V l-e2/^^ l-e2
2/-*
+ — ^ — = 1.
c2e2
(1 - e2)2 1 _ e2
(<2g2
We may now place = a^,
(1 - e2)2
= a2(l-e2) = ±62^
1 - e2
and
1 - e2 e
the sign of 6 being ± 1 according as e < 1. The equation is then
a2 62
The equation accordingly represents an ellipse or an hyperbola with center
at 0,0).
If e = 1, the equation (x — c)2 + y^ = ^x^
becomes y^ = 2cx - c* = 2 c (x 1 ,
which represents a parabola with the vertex at /- , 0| .
115. Change of direction of axes without change of origin.
Case I. Rotation of axes. Let OX and OY (fig. 131) be the
original axes, and OX' and 0 Y' be the new axes, making Z </>
with OX and Oy respectively. Then ZXOY' = 90° + <f>, and
ZYOX' ^90° -(f>.
222
CHANGE OF COOKDl^ATE AXE«
Let P be any point in the plane, its coordinates being x and y
with respect to OX and OY, and x' and y' with respect to OA"'
and 0 F'. Then by construction OM = x,
0N= y, OM' = x', and J/'P = y'. Draw
OP.
The projection of OP on OX is OM, and
the projection of the broken line OM'P
on OX is OM' cos(f> + i»/'P cos (90° + <f>)
-X or OM' cos ^ — 3f'Psin^.
.-. 0Jf=01f' cos (^-J/'P sin (^, (1)
by § 15.
In like manner the projection of OF on OY is ON, and the
projection of the broken line OM'P on OF is Olf' cos (90° — <^)
+ J/'Pcos</).
.'. 0N= OM' sin ^+M'P cos 0, (2)
by § 15.
Replacing OM, 02i, OM', • •• by their values, we have
X = x' cos (f>~ y' sin ^,
y = a;' sin^ + y' cos^.
Ex. 1. Transform the equation xy = 5 to new axes, having the same origin
and making an angle of 45° with the original axes.
x'
Here <p = 45°, and the formulas of transformation are x =
w' x' + y'
V2 '" V2
Substituting and simplifying, we have as the new equation x? — y"^ = 10,
from which we recognize the curve to be an equilateral hyperbola.
Ex. 2. Transform the equation 34x2 j^ \\yi — 24x2/ = 100 to new axes with
the same origin, so choosing the angle 0 that the new equation shall have no
term in xy.
The formulas of transformation are
X = x' cos <t> — y' sin 0,
y = x' sin 0 + y' cos 0,
where 0 is to be determined.
Substituting in the equation and collecting like terms, we have
(34 cos2 0 + 4 1 sin2 0 — 24 sin 0 cos 0) x^
+ (34 sin2 0 + 41 cos2 0 + 24 sin 0 cos 0) t/2
+ (24 sin2 0 + 14 sin 0 cos 0 - 24 cos2 0)xj/ = 100.
By the conditions of the problem we are to choose 0 so that
24 sin2 0 + 14 sin 0 cos 0 - 24 cos2 0 = 0.
OBLIQUE COOKDINATES 223
One value of 0 satisfying this equation is tan-i |. Accordingly we substitute
sin^ = I and cos^ = |, when the equation reduces to x^ + 2y^ = 4, which is
the equation of an ellipse.
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 imder consideration in this article. The
formulas for such a transformation are evidently x = y', y = x'.
Case III. Rotation and interchange of axes. Finally, if the
axes are rotated through an angle <^ and then interchanged, the
formulas, being merely a combination of the two already found, are
x = y' cos<f> — x' sin0, y = y' sin^ + x' cos^.
A special case of some importance occurs when cf)= 270°. We
have then x = x', y = — y'.
Cases II and III, it should be added, occur much less frequently
than Case I.
In case 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.
116. 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. 132, let OX and OY intersect at
some angle &> other than 90°.
We now define x for any point in the
plane as the distance from OY 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 according to the same rules as
were adopted in § 16.
It is immediately evident that the rectangular coordinates are
but a special case of this new type of coordinates, called oUique
224
CHANGE OF COORDINATE AXES
coordinates, since the new definitions of x and y include those
previously given. In fact, the term Cartesian or rectilinear co-
ordinates includes both the rectangular and the oblique.
Oblique coordinates are usually less convenient than the rectan-
gular, and are very little used in this 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 risht angle.
117. Change from rectangular to oblique axes without change
of origin. Let OTand OY (fig. 133) be the original axes at right
angles to each other, and
OX^ and OY' the new axes,
making angles <^ and (/>'
, respectively with OX.
n Then co = cf)' — ^. Let P
be any point in the plane,
X its rectangular coordinates
being x and y, and its ob-
lique coordinates being x'
and y'. Draw PM parallel to OY, PM' parallel to OY', M'N
parallel to OY, and RM'N' parallel to OX. Then Z RM'P = <\>'.
But OM =0N + NM =0N+ M'N' = 031' cos <b+3I'P cos (f>',
MP = MN' + N'P = NM' + N'P = OM' sm <^ -F M'P sin <^'.
.*• x= x' cos<f>-{- y' cos(f>',
y = x' sin <f> + y' sin (f>'.
Ex. Transfonn the hyperbola
x-*
a2 62
= 1 to its asymptotes as axes.
Since the equations of the asymptotes are 2/ = ± - x, 0 = tan- ' / — 1 , and
(f/ = tan-i - , if we choose to have the hyperbola lie in the first and the third
quadrants with respect to the new axes. The formulas of transformation become
b
Va-2 + 62
{X' + y'), y =
Va- + 62
i-x' + y').
Substituting and simplifying, we have as the new equation xy
Unless b — a, the axes are obli(iue and w = 2 taii^' - •
a
a2 + 62
PROBLEMS 225
118. Degree of the transformed equation. In reviewing this
chapter we see that the expressions fur the original coordinates in
terms of the new are all of the first degree. Hence the result of
an}' transformation cannot be of higher degree than that of the
origmal equation. On the other hand, the result cannot be of
lower degree than that of 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
if 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 succes-
sive transformations. In particular, the proposition that any equa-
tion 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 (2, 3), {— 4, 5), and (5, — 7)
if the origin is transferred to the point (.3, — 2), the new axes being parallel to
the old ?
2. Transform the equation x^ + Ay^ — 2x + 8y + l = 0to new axes parallel
to the old axes and meeting at the point (1, — 1) with respect to the old axes.
3. Transform the equation ?/3 - 6?/2 + 3a;2 + 12 ?/ - 18x + 35 = 0 to new
axes parallel to the original axes and meeting at (2, — 3) with respect to the
original axes.
4. Find the equation of the ellipse when the origin is taken at the lower
extremity of the minor axis, and the minor axis is the axis of y.
5. Find the equation of the ellipse when the origin is at the left-hand vertex,
the major axis lying along OX.
6. Find the equation of the hyperbola when the origin is at the left-hand
vertex, the transverse axis lying along OX.
7. Find the equation of the strophoid when the origin is at A (fig. 92), the
axes being parallel to those of § 84.
8. Find the equation of the strophoid when the asymptote is the axis of y,
the axis of x being as in § 84.
9. Find the equation of the witch (fig. 90) when LK is the axis of x and
OA the axis of y.
226 CHANGE OF COORDINATE AXES
10. Find the equation of tlie witch when the origin is taken at the center of
the circle used in constructing it, the axes being parallel to those of § 82.
11. Find the equation of the cissoid when its asymptote is the axis of y and
its axis is the axis of x.
12. 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 § 83.
13. Find the equation of the parabola when the origin is at t'.io fo -as and
the axis of x is the axis of the curve.
14. Find the equation of the parabola when the axis of the curve and the
directrix are taken as the axes of x and y respectively.
15. Transform ?/2 — 8x — 10 2/ + 1 = 0 to new axes parallel to the old, so
choosing the origin that the new equation shall contain only terms in y^ and x.
16. Transform the equation 12 x^ + 18 y^ _ 12 x + 12 ?/ — 31 = 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.
17. Show that any equation of the form xy + ax + 6j/ + c = 0 can always
be reduced to the form xy = k hj choosing new axes parallel to the old, and
determine the value of A;.
18. Show that the equation ax^ + by^ + ex + dy + e = 0 (a ?i 0, hjtiQi) can
always be put in the form ax^ + hy'^ = fc by choosing new axes parallel to the
old, and determine the value of k.
19. Show that the equation y^ + ay + bx + c = Q (^ ?^ 0) can always be
reduced to the form y^ + 6x = 0 by choosing new axes parallel to the given ones.
20. Find the equation of an ellipse if its axes are 6 and 2, its center is at
(—3, 2), and its major axis is parallel to OX.
21. Find the equation of an ellip.se if its axes are a and ^, its center is at
(— 2, — 3), and its major axis is parallel to OX.
22. Find the equation of an hyperbola if its transverse axis is 4, its conju-
gate axis 2, its center at (1, — 2), and its transverse axis parallel to OX.
23. Find the equation of an hyperbola if its transverse axis is v2, its con-
jugate axis V §, its center at (2, 3), and its transverse axis parallel to OX.
24. The vertex of a parabola is at (3, — 2) and its focus is at (5, — 2). Find
its equation.
25. The vertex of a parabola is at (4, 5) and its focus is at (4, 1). Find its
equation.
26. The center of an ellipse is at the point (2, 3), its eccentricity is ^, and
the length of its major axis, which is parallel to the axis of x, is 10. What is
the equation of the ellipse ?
27. Find the equation of an ellipse when the vertices are (—2, 0), (4, 0), and
one focus is at the origin.
PROBLEMS 227
28. The center of an hyperbola is at (- 1, - 2), its eccentricity is 1^, and its
transverse axis, which is parallel to OX, is 4. Find its equation.
29. The vertex of a parabola is at the point (- 4, - 2), and it passes through
the origin of coordinates. Find its equation, its axis being parallel to OX.
30. Given the ellipse 4x^ + 9y^ + 8x - S6y + i -0; find its eccentricity,
center, vertices, foci, and directrices.
31. Given the ellipse 3a;2 + 5y2 + i8 x - 20y + 32 = 0 ; find its eccentricity,
center, vertices, foci, and directrices.
32. Given the hyperbola 9 x^ — 4 y^ — 6ix — 32 y — 19 = 0 ; find its eccen-
tricity, center, vertices, foci, directrices, and asymptotes.
33. Given the hyperbola 3x2 — 2 2/2 ^ 6x + 8y— 11 = 0; find its eccentricity,
center, vertices, foci, directrices, and asymptotes.
34. Given the parabola 72x2 + 48x + 180y - 37 = 0; find its vertex, focus,
axis, and directrix.
35. Given the parabola y^ — 5x + 6y — 1 — 0; find its vertex, focus, axis,
and directrix.
36. What are the coordinates of the points (0, 1), (1, 0), (1, 1) if the axes
are rotated through an angle of 60° ?
37. Transform the equation 3x2 + 3?/2 — lOxy + 8 = 0 to a new set of axes
by rotating the original axes through an angle of 45°, the origin not being
changed.
38. Find the equation of the folium x^ + y^ — 3 axy = 0 after the axes have
been rotated through an angle of 46°.
39. By rotating the axes through an angle of 45° and changing the origin,
prove that the curve x' + y^ = «^ is a parabola.
40. Transform 6x2 — 12xy + 10?/2 _ 14 = o to a new set of axes, making
an angle tan-i | with the origiual'set.
41. Show that the equation x2 + ?/2 = a^ will be unchanged by transforma^
tion to any pair of rectangular ^.xes, if the origin is unchanged.
42. Transform the equation x"^ - y^ = 36 to new axes bisecting the angles
between the original axes.
43. Transform the equation ix'^ - 3xy + 8y^ = I to one which has no
xy-term, by rotating the axes through the proper angle.
44. By rotating the axes through the proper angle transform the equation
3 x2 + 2 Vs xy + ?/■- + 2 X - 2 V3 ?/ = 0 to another which shall have no term in xy.
45. Transform the equation
x2 _ 6 2/2 _ 6 V3 xy + [2 + 12 V3] X + [20 - 6 V3] y - 15 + 12 V3 = 0
to a new set of rectangular axes making an angle of 60° with the original axes
and intersecting at the point (-1, 2) with respect to the original axes.
228 CHANGE OF COORDINATE AXES
46. Transform the equation ix- + 9y^ = 36 from rectangular axes to oblique
axes with the same origin, making angles tan-i^ and tan-i(— J) respectively
with OX.
47. Find the equation of the hyperbola Sx^ — 4y^ = 12 referred to its asymp-
totes as coordinate axes.
48. 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.
49. Transform the equation 2x^ — Sy^ = Q from rectangular axes to oblique
axes having the same origin and making the angles tan-' ^ and tan-' ^ respec-
tively with OX.
50. Prove that the formulas for transposing 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' + y' cos w,
y = y' sin w,
where w is the angle between the oblique axes.
51. By transforming the equation y = mx + 6 by the formulas of example
60, show that the equation of a straight line in oblique coordinates is
sind>
sin(« — <f>)
where u is the angle between OX and OY, <f> the angle between the line and OX,
and c the intercept on OY.
52. Derive the result of example 51 directly by use of the trigonometric
formulas connecting the sides and the angles of an oblique triangle.
53. By use of the transformation of example 60, prove that the equation of
a circle in oblique coordinates is
(X - d)2 + (y _ e)2 4- 2 (X - d) (y - e)cos u = r^,
where w is the angle between the axes, and (d, e) is the center.
54. Obtain the result of example 53 directly by use of the trigonometric
relations connecting the sides and the angles -of an oblique triangle.
CHAPTEE XI
THE GENERAL EQUATION OF THE SECOND DEGREE
119. Introduction. The most general equation of the second
degree is of the form
Ax'+ 2 Hxi/ + Bf-{-2Gx+2Fy+C=0,
where the coefficients may have any values, including zero, except
that A, B, and H cannot be zero together.
We shall proceed to show that this equation always represents
an ellipse, an hyperbola, a parabola, or a limiting case of one of
these, if it represents any curve, and shall derive criteria by which
the nature of tlie curve can be readily determined.
120. Removal of the xy-t&rm. Let us make a transformation
of coordinates to new rectangular axes, making an angle ^ with
the original ones, the origin being unchanged. The formulas of
transformation are (§ 115)
x = x' cos 4> — l/' sin (f),
y = x' sincf) + y' cos (f>.
Substituting, we have
A'x''' + 2 H'x'y' + B'y'^ + 2 G'x' + 2 F'y' + C" = 0,
where A' = A cos^ ^ + 2H sin (f> cos<f) -\-B sin^ </>,
H' = (B — A) sin (}> cos (f)-\-H (cos^<^ — sin^</)),
J5' = ^ sin*^ 0 — 2 ^ sin ^ cos ^ + -B cos^ 0,
(T' = G^cos<^ + i^sint/);
i^' = ii^ cos </>-(? sine/),
and C'=C.
229
230 GENERAL EQUATION OF SECOND DEGREE
We may now determine <f) so that H' shall vanish; that is,
so that
2{B — A) cos <^ sin <^ + 2 H(cos'(f> - sm^cfy) = 0.
This equation is equivalent to
2 ^ cos 2 <^ + (5 - ^) sin 2 <^ = 0,
2ff
whence tan 2(f> =
A-B
2H
or ^ = J tan '
A-B
To compute the values of A^ and B\ we have
A! =A cos^<f>+ 2Hsia.<l) cos(f> +B sia^<f>
.l+cos2<^ . 1— cos2<^
= A -+B:sin2(t>+B
= ^[^+5 + (^-i?)cos2<^4- 2^sin20].
2H
But, since tan 2 ^ =
A-B
sin 2 <^ = ± "' > cos 2 <^ = ±
and therefore A! ^Wa-^B± (^-^f+4^1
2L V(^-J5f+4^'J
= 1 [^ + j5 ± V(^-^f+4^'].
Sunilarly, 5' = ^ [^ 4- -S :f ^ {A-Bf^- A.H''\
From these results it follows that
A'B' = AB-H^.
Hence if AB — H^ is positive, A' and ^' have the same sign ; if
AB — H^ is negative, ^' and B' have opposite signs ; if AB — ^^
is zero, either A' or B' is zero.
EQUATION WITHOUT THE xy-TERM 231
The discussion of the general equation is then reduced to that
of the simpler equation
This equation we will consider in the next two articles, dropping
the primes for convenience.
121. The equation Ax"" + By" -\- 2 Gx + 2 Fy-\- C = 0. We
shall prove the theorem : The equation
Aaf + Bif+2Gx+2Fy + C=0,
where the coefficients are such that
AF^' + BG^-ABC^^ 0,
represents a conic, if it represents any curve at all. In particular,
(1) when A and B have the same sign, it represents an ellipse *
or no curve ;
(2) when A and B have opposite signs, it represents an hyperhola ;
(3) when either A or B is zero, it represents a parabola.
Suppose first that neither ^ nor 5 is zero. Then the equation
may be rearranged as follows :
A(cc'+2~x\+B(f+2^y) = -C.
We may then complete the squares of the expressions in the
parentheses; thus,
Aj Y BJ AB
* The circle is considered a special case of an ellipse (see § 75).
232 GENERAL EQUATION OF SECOND DEGREE
Since AF'^ + BG^ — ABC is not zero, we may divide by the right-
hand member of the equation, obtaining
M N
where, for convenience, we place
AF' + BG'^ — ABC
M =
N =
A'B
AF^ + BG--ABC
AB-
We may now transfer the origin of coordinates to the point
(C F\
> — - b the new axes remaining parallel to the old, by the
formulas ^ ^
A ' ^ B ^
The equation is then \-^ = \.
^ M N
Now if A and B have the same sign, M and N wUl have the
same sign. If this sign is positive, we may place M= a^, N= V^,
and the equation is
d' ^ h' '
which represents an ellipse.
The axes of the ellipse are parallel to the original coordinate
(C ft'X
J — - I referred to the
original axes, li A=B, the ellipse is a circle.
If M and N are both negative, the equation
^ + ^' = 1
M N
can be satisfied by no real values of x and y.
EQUATION WITHOUT THE xy-TERM 233
If A and B have opposite signs, M and N have opposite signs,
and we may place either M= a^, iV = — h'\ or J/ = — a~, JV = V^,
thus obtaining either ,., ,.,
^- _ /; ^
a' J)' '
-^ + 1^ = 1'
either of which represents an hyperbola.
The axes of the hyperbola are parallel to the original coordinate
axes, and its center is at the point ( > ^ | referred to the
• 1 V ^ ^/
origmai axes. ^ ^
The first and the second parts of the theorem are therefore proved.
Consider now the case in wliich either tI or ^ is zero. If, for
example, A = 0, B ^ 0, the equation is
Bf+2Gx-\-2Fij + C=0,
and the condition to be fulfilled by the coefficients is BG^ ^ 0,
which is equivalent to G "^ 0, since B cannot be zero.
We may arrange the equation as follows :
1^+2— y=— 2— X
^ B^ B B
Completing the square, we have
F\^ 2GI . C F^
y + -^)=--^{^ + ^^-
Bj B\ 2G 2 GB
If now we transform to a new origin by placing
X— 1 \- x', y = — — + y>
2G 2GB ^ B ^'
we have ■y'^ = x',
which is the equation of a parabola.
Similarly, if ^ = 0 but A^ <d, the equation may be reduced to
the form
which is also a parabola.
2F
a?'2 = - — y'.
234 GENERAL EQUATION OF SECOND DEGREE
In each case the axis of the parabola is parallel to one of the
original coordinate axes.
Hence the third part of the theorem is proved.
122. The limiting cases. AVe shall consider now the equation
Ax' + Bf+2Gx+2Fy + C=0
when the coefficients are such that
AF'' + BG''-ABC=0.
The figures represented are limiting cases of a conic, since the
equation of this article may be obtained from that of the previous
article by allowing the coefficients to change in such a way that
AF^ + BG^—ABC approaches zero. We have three cases:
1. A and B have the same sign.
By proceeding as in § 121, we may put the equation in the form
and if, as before, we place
X = -— + X', y = - — + x',
A B
we have Ax"^ + By'"" = 0.
Since A and B have the same sign, we may consider them as
positive, and factor the equation as follows :
(Vja:' + IVBiJ) {y/lix' — iVBy') = 0,
which is satisfied by real values of x' and y' only when x' = 0,
G F
y = 0, or in the old coordinates x = > ?/= ■•
^ A "^ B
Hence in this case the equation represents a point. This may be
considered the limiting case of the ellipse.
2. A and B have opposite signs. We may put the equation in
the form , ^2 / r^\2
^(. + _') + 5(,+ _) = 0,
or Ax"' + By"' = 0.
THE DETERMINANT AB - H^ 235
Since A and B have opposite signs, we will consider A as posi-
tive and B as negative. The equation can then be separated into
two real factors
{VAx' + y/'^y') {VAx' ~ V^y') = 0.
Consequently the equation represents the two straight lines inter-
sectmg in the point ic' = 0, y' = 0, or a? = > 2/ =
This may be considered the limiting case of the hyperbola,
3. One of the coefficients ^ or -B is zero. For example, let
^ = 0, ^ ^ 0. Then the condition AF'' + BG^-ABC=0 becomes
G = 0. Hence the equation is
Bf+2Fi/-{-e=0.
This may be factored into
B{y-y^{y-y^) = ^,
and accordingly represents either two parallel straight lines, two
coincident straight lines, or no real locus, according as y^ and y^
are real and unequal, real and equal, or imaginary.
This is considered a limiting case of the parabola.
123. The determinant AB — //'. Returning now to the gen-
eral equation of the second degree,
Ax^+2Hxy-{-Bf+2Gx+2Fy + C=0,
and remembering that if it is reduced to the form
^ V + B'y"" + 2 G'x' + 2 F'y' + C' = Oy
we have AB-H^'^A' B',
we may state the foUo^snng theorem :
The equation
Ac(^+2JH'xy+By^+2Gx+2Fy-\-C=0
always represents a conic or one of the limiting cases, if it repre-
sents any curve at all.
286 GENERAL EQUATION OF SECOND DEGREE
1. If AB—H' > 0, the equation represents an ellipse, a point,
or no curiae.
2. If AB — H^ < 0, the equation represents an hyperbola or two
intersecting straight lines.
3. If AB —H^ = 0, the equation represents a parabola, two par-
allel lines, two coincident lines, or no curve.
124. The discriminant of the general equation. We have seen
in § 122 that
A'x''' + B'y'^ + 2 G'x' + 2 F'y' + C = 0 (1)
represents one of the limiting cases of the conic sections when
A'F'^ + B'G''-A'B'C' = 0.
It is useful to have this condition in terms of the coefhcients of
the general equation
Ax'-^2Hxy + By''^-2Gx + 2Fy-\-C=0. (2)
This might be done by substituting for A', B', G' , F', and C the
values given in § 120, but this method is tedious. We may obtam
the result by noticing that the first member of (1) can be factored
rationally in x and y when it represents a limiting case, and not
otherwise. The same must be true of equation (2). We shall pro-
ceed then to find the condition under which (2) can be factored.
1. Assume ^ ^ 0. (2) may now be considered as a quadratic
equation in x, and factored by the method of § 41. Solving (2)
for X, we have
^-{Hy+ G) ±-^/JlP-AB)y'+ 2 y{HG-AF) + {G'-CA)
It is necessary, however, that y should not appear under the radi-
cal sign, and for this it is necessary and sufficient that the quantity
under the radical sign must be a perfect square. The necessary
and sufficient condition for this is (§ 37)
{HG-AF)--{H''-AB){G''-CA)=Q,
that is, ylBC+2FGH-AF^-BG^-Cff^ = 0. (3)
CLASSIFICATION OF CONICS 237
2. Assume ^ = 0, but B ^^ 0. The equation may then be con-
sidered as a quadratic equation in y, and handled in the same
manner as before with the same result.
3. Assume A = (), B = 0. Then If cannot equal zero. The
equation can consequently be written
r* V c
The factors of this, if they exist at all, are clearly of the form
{x+a){y + h)^Q,
whence a = — > 5 = — » ah =
H H 2H
The necessary and sufficient condition that two quantities a and h
can be found satisfying these equations is
But this is just what (3) becomes when ^ = 0, ^ = 0. Hence,
the necessary and sicfflcient condition that
Ax^+2Hxy + By^-\-2Gx+2Fy + C=Q
represents a limiting case of a conic is
ABC +2 FGH -AF^- B G^ - 6'//' = 0.
The expression (3) is called tlie discriminant of (1) and is
denoted by A. In determinant form
A =
125. Classification of curves of the second degree. The results
of the previous articles are exhibited in the table on the following
page, which gives the simplest forms to which the general equation
Ax}+ 2Hxy+Bf+ 2Gx+2Fy + C=0
can be reduced under the various hypotheses, where
D = AB-H\
A =ABC+ 2FGH-AF^-BG--CIP.
A
H G
H
B F
G
F C
238 GENERAL EQUATION OF SECOND DEGREE
At^O
A = 0
D>0
l2 7/2
or no curve
a;2 7/2
D<0
x2 ?/2
Hyperbola- -^ = 1,
X2 w2
-^2 + ^ = 1
Two Intersecting straight lines
02 62
D = 0
Parabola 2/2 = 4 pa;,
or x^ = 4py
Two parallel straight lines
{y - 2/i) {y - 2/2) = 0,
or (x - a;i) (x - X2) = 0,
or no locus
126. Center of a conic. It is frequently desirable to find the
center of a conic represented by the general equation. Now, if
the origin of coordinates is taken at the center of the curve, the
equation can contain no terras of the first degree in x and y ; for
if it is satisfied by any point (x^, y^, it must also be satisfied by
the symmetrically placed point (— x^, — y^. We will accordingly
take the center £is (ic^, y^ and make the transformation
The general equation then becomes
Ax'^ + 2 Hx'y' + By" + 2 {Ax, + Hy, + G)x'+2 {Hx, + By, + F) y'
+ ^<+ 2Hx^, + Byl+ 2 Gx,+ 2Fy,+ C=0,
where, by the condition for the center,
Ax, + ffy,+ G = 0,
irx,-{-By, + F=0.
By multiplying each of these by a properly chosen factor and add
ing, we obtain the equivalent equations
(1)
(AB - H"") x^ = HF- BG,
(AB - H') y^=HG- AF.
(2)
CENTEK OF A CONIC 239
Three cases then occur:
1. AB — H^'^O. Equations (2) have then a single solution
and the curve has a center. This occurs for the ellipse, the
hyperbola, and their limiting cases.
2. AB — IT^ = 0, but not each of the expressions HF—BG
and HG—AF equal to zero. At least one of equations (2) ex-
presses an absurdity, and hence equations (1) have no solution
and the curve has no center. This occurs in the case of the
parabola.
3. AB-IT'' = 0, JTF-BG=0, ITG-AF=0. Equations (2)
are each 0 = 0. Equations (1) are identical, and any point on the
line expressed by each of them is a center of the curve. In this
case one easily calculates that A = 0. The curve then consists of
two parallel straight lines (§ 125), and the line of centers is the
line halfway between the two parallel lines.
127. If for the equation
Ax'+2 Hxy + Bf+1Gx+2Fy + C=0
the origin is transferred to the center of the curve, when such
exists, the equation becomes
Ax'^+2Hx'y' + By'''+C' = Q,
where C = Ax^ + 2 Hx,y, + Byl +2Gx^+2 Fy^ + C.
This quantity C may be expressed in terms of the original coeffi-
• cients as follows. Take the equations (1) of § 126, multiply the
first one by x^, the second by y^, and add them. There results
Ax^ + 2 Hx^, + By^ + Gx, + Fy, = (i.
Subtracting this from the value of C', as given above, we have
C' = Gx, + Fy,+ C,
whence, by substituting the values of x^ and y^, as given by (2)
(§ 126), we have
ABC+2FGH-AF^-BG^-Cff^ A
C' =
AB-H' D
240 GENERAL EQUATION OF SECOND DEGREE
128. Directions for handling numerical equations. In case it
is necessary to reduce a uumerical equation to its simplest form,
the procedure, based on the foregoing discussion, is as follows :
First compute AB — H^ and determine the type of the curve
(§ 123). A may also be computed if wished, but it is not necessary.
If AB — H^ ^ 0, find the center, as in § 126, and transfer the
origin to it. Then, as in § 120, turn the axes through an angle
(p = w tan = tan
A-B ■s/(^A-Bf+4:H''±{A-B)
computing A' and B' by the formulas of § 120. The two values
of tan^ are the slopes of the axes of the curve.
If AB — H'^ = 0, write the equation in the form
{y/~Ax + ^yf+2Gx + 2Fy + C=Q,
Vb being taken with the same sign as H, and let
, -y/Ax + V^y , Vbx — VAv
y' = — 7=^' «' = — , ^-
■■Va + b ^a + b
Solve these equations for x and y and substitute in the given
equation.
The equation is now in the form y'^-i- 2 G'x' + 2F'y' + C" = 0,
and the further reduction is made by the method of § 121.
Ex.1. 8x2 _ 4xy + 52/'2-36x + 18?/ + 9 = 0.
Here AB — 11^ = 30, and the curve is an ellipse or a limiting case of an ellipse.
The center, found by § 126, is (2, — 1), and the equation transferred to the
center as origin becomes
8 x'2 - 4 xY + 5 ?/'2 - 36 = 0.
We now turn the axes through (p = ^ tan-i(— |) = tan-i2 or tan-i(— ^),
and find, from § 120, ^' = 9 or 4, B' =4 or 9.
The ambiguity is removed by noticing that if we take tan (f> — 2, the formulas
of transformation (§ 115) are
, z" — 2 ij" , 2 x" + y"
2/ =
which give A' = 4, B' = 9.
The simplest equation is then
4 x"2 + 9 ?/"2 = 36.
The slopes of the axes are 2 and — J^,
CONIC THROUGH FIVE POINTS
241
Ex. 2. 36a;2-48x2/ + Wy^ + 52x - 260y - 39 = 0.
(6 X - 4 r/)2 4- 52 a; - 260 y - 39 = 0,
6x — 4?/ Sx — 2y
Here
We write
and place
y
V52 Vl3
4x — Gy — 2x
8y
V52 Vl3
Solving for x and y and substituting, we have
y'S 4. VlSj/' + Vl3x' -3=0,
or y''2 = — VTs x",
Vl3
^~
The curve is a parabola, the axis of which is y" = 0 or 6 x
where
Vl3
y =
¥
4 2/ + 13 = 0.
129. Equation of a conic through five points. The general
equation of the second degree
1x^+2 Hxy + By^+2Gx + 2Fy + C=Q
contains six constants, the ratios of which are alone essential. Five
independent equations are sufficient to determine these ratios.
Therefore a conic is, in general, determined by five conditions. The
simplest conditions are that the conic should pass through the five
points (a?„ y^, [x^, y^, (x^, y^), {x^, y^, and {x.^, y,). The five equa-
tions to determine the ratios of A, If, B, G, F, and C are then
Axl + 2 Hx^y^ + Byf + 2Gx^+ 2Fy^+ C = 0,
Axl+2Hx^y^ + By^+ 2 Gx^+2Fy^+C = 0,
Axl+ 2Hx,y^ + By^+ 2 Gx^+2Fy,+ C ={),
Axl+2Hx^^ + Byl+ 2 Gx^+ 2Fy, + C=0,
Ax^ + 2 Hx^y^ + Byl -^ 2Gx^+ 2Fy^+C = Q.
Eliminating the coefficients between these and the general equa-
tion, we have
= 0,
x-"
xy
f
X
y
xl
^iVx
yl
^1
2/1
^!
^^y^
yl
x^
y.
^^8
^^z
yl
x^
2/3
x!
^43/4
yl
X,
y^
<
^6^5
yl
X,
y.
242 GENERAL EQUATION OF SECOND DEGREE
which is the required equation of a conic through five given points.
The equation of a conic through five points may also be found
in the following manner :
Let us take any four of the given points and connect them by
straight lines so as to form a quadrilateral (fig. 134).
Let the equation of I^J^ be A^x + B^y + C^ = 0, or, more shortly,
f^{x, y) — 0. Similarly, let the equation of P^l^ be fj^x, y) = 0, that
of ^^ be f.J^x, y) = 0, and that of F^J^ be f^{x, y) = 0.
Form now the equation
Vi{x, y) ■ Ai-^, y) + ¥2(«. y) ■ A{^> y) = o. (i)
where I and k are undetermined factors. This equation is of the
second degree in x and y ; therefore
it represents a conic section. More-
over, this conic section passes through
Il\ for the coordinates of P^ make
f^{x, y) = 0 and f^{x, y) = 0, and
therefore satisfy equation (1). Simi-
larly, this conic passes through ^,
^, and j^. If now we substitute in
(I) the coordinates of j^, we de-
termine values of I and k, which we
must assume in order that the conic
may pass through i^. We thus de-
termine the equation of a conic through the five given points.
Ex. Let it be required to pass a conic through the points Pi(2, 3), P^i — 1, 2),
P3(-3, -1), P4(0, -4), P5(l, 1).
The equation of P1P2 isa;-3y + 7 = 0, that of P2P3 is3x-22/ + 7=:0,
that of PsP4 is X + y + i =: 0, and that of P4P1 is7a;-2y-8 = 0.
We form the equation
l{x -Sy + '!)(x + y + 4) + k(Sx - 2y + 'J){7 x - 2y - 8) = 0,
and, substituting the coordinates of P5, find k = ^l.
Hence the required conic is
(X - 3 2/ + 7) (a; + 2/ + 4) + I (3 a; - 2 y + 7) (7 X - 2 y - 8) = 0,
or 109x2-108x2/ + 82/2 + 169x -10?/ -168 = 0.
If three of the points lie in a straight line, the method is appli-
cable, but it is evident that the conic must be one of the limiting
CONIC THKOUGH FIVE POINTS 243
cases, for it must consist of the straight line in which the three
points lie, and the straight line connecting the other two points.
If four or five of the points lie in a straight line, the method is
not applicable. It is geometrically evident that in this case the
problem is indeterminate ; for the conic may consist of the straight
line in which the four points lie, together with any line through the
fifth point, if that is not on the line with the four, or any line what-
ever if the fifth point lies on a straight line with the four others.
If it is required to determine a parabola, only four points are
necessary. This follows from the fact that one relation connecting
the coefficients is always given, namely, AB — H^ = 0. We form,
as before, the equation
^/i(«'. y) ■ M^> y) + ^M^> y) ■ f^i-^^ y) = o-
We form, then, the equation AB — H'^ = 0 out of the coefficients
of this equation. The result is a quadratic equation in -> and
K
hence we will have two, one, or no real parabolas, according as the
values of j are real, equal, or imaginary. It should be noticed that
fC
in this connection " parabola " may mean two parallel straight lines.
Ex. Let it be required to pass a parabola tlirough the points Pi(l, — 1),
P2(2, 3), Ps(2, - 6), P4(5, 7).
We find the equations of the following lines : P1P2, 4x — y — 6 = 0; P2P3,
X - 2 = 0 ; P3P4, 4x-2/-13 = 0; P4P1, 2a;-y-3 = 0. The equation of
the conic is then
Z(4x - 2/ - 5) (4x - 2/ - 13) + fc(a; - 2)(2x - 2/ - 3) = 0,
or (16Z + 2A:)x2 + (-8Z- A:)x2/ + i2/2
+ ( _ 72 J - 7 fc) X + (18 i + 2 i) 2/ 4- «5 Z + 6 fc = 0,
and the condition AB — R"^ = d'xs
(16Z + 2A:)Z-(4Z + ^A:)2 = 0,
whence k-Oov—%1.
There are accordingly two parabolas,
16x2 - 8 X2/ + 2/2 - 72 X + 18 2/ + 65 = 0,
and 2/=^-16x + 2 2/ + 17 = 0.
The first equation, however, represents a limiting case of a parabola, since
it factors into 4x_2/-5 = 0 and 4x-2/-13 = 0,
which represent two parallel straight lines.
244 GENEEAL EQUATION OF SECOND DEGEEE
130. Oblique coordinates. We have assumed, thus far, that the
general equation is referred to rectangular coordinates. If, how-
ever, the equation
Ax" + 2 Hxy + Bf+2Gx-j-2Fy + C=0
has reference to oblique coordinates, it may be transformed to any
conveniently chosen pair of rectangular coordinates. Formulas for
this purpose are given in § 117, and it has been proved in § 118
that such a transformation does not alter the degree of the equa-
tion. Therefore the new equation is of the form
A'x'^ + 2 H'x'y' 4- B'y"" + 2 G'x' + 2 F'y' + C'=0.
This equation may now be investigated by the methods of this
chapter.
Hence we have the result :
Any equation of the second degree, whether referred to rectangu-
lar or to oblique coordinates, represents a conic.
PROBLEMS
Determine the nature and the position of the following conies :
1. 4xj/ + .3 2/2_8x + 16?/ + 19 = 0.
2. a;2 _ 6xy + 9 2/2 -280«- 20 = 0.
3. Ilx2_4xy + 14i/2_26x + 32y + 59 = 0.
4. 5x2 - 26xy + 51/2 + 10x-2Gy + 71 = 0.
5. 4:xy + 6x-8y + 1 = 0.
6. x2-2x2/ + y2 + 2x-22/ + l = 0.
7. 13x2 + 10 xy -}- 13 2/2 + 6x - 42y - 27 = 0.
8. x2 -4xy - 2 2/2- 14x + 4 2/ + 25 = 0.
9. 6x2 - 5x2/ - 62/2 -46x- 9?/ + 60 = 0.
10. 4 x2 - 8 x?/ + 4 2/2 + 6 X - 8 2/ + 1 = 0.
11. x2 + 6x2/ + 92/2 -6x -18 2/ + 5 = 0.
12. 41x2 - 24x2/ + 342/2 - 188x + 116?/ + 196 = 0.
13. 31 x2 - 24 X2/ + 21 2/2 + 48x - 84 2/ + 84 = 0.
14. Show that, if A and B in the general equation have opposite signs, the
conic is an hyperbola.
15. Show that the conic represented by the general equation is an equilateral
hyperbola when A = ~ B.
PEOBLEMS 245
16. Prove that the necessary and sufficient conditions that the general
equation should represent a circle are A = B, H — 0, provided the axes are
rectangular.
17. Show that, if the general equation contains the term in xy and not more
than one of the terms containing x^ or y^, the conic is an hyperbola.
18. Show that xy + ax + by + c = 0 is the general equation of the hyperbola
when the axes of coordinates are parallel to the asymptotes. ,_. ^ ... -
19. Prove that any homogeneous equation in x and y represents a system of
straight lines passing through the origin.
20. Find the angle between the two straight lines represented by the equation
Ax^ + 2 Hxy + By^ = 0.
21. Show that the asymptotes of the hyperbola are parallel to the two
straight lines Ax^ + 2 Hxy + By^ = 0,
22. Show that, if the focus is taken upon the directrix, the conic becomes
one of the limiting cases.
Find the equations of the conies through the following points :
23. (3, 2), (- 2, - 3), (J, - 3), (2, - 2), (|, - f).
24. (1, 2), (6, 3), (3, 2), (2, 1), (9, 2).
25. (0, a), (a, 0), (0, -«),{- a, 0), (a, a).
26. (1, 1), (-1,6), (2,4), (0,3), (3,1).
27. Find the equation of a parabola through the four points (4, — 4), (9, 4),
(6, - 1), (5, - 2).
28. A point moves so that the sum of the squares of its distances from two
intersecting straight lines is constant. Prove that the locus is an ellipse, and find
its eccentricity in terms of the angle between the lines.
CHAPTER XII
TANGENT, POLAR, AND DIAMETER FOR CURVES OF THE
SECOND DEGREE
131. Equation of a tangent. It has been shown in § 59 that
the tangent to a curve at a point {x^, y^ is
where ( — ) denotes the value of -^ at (x., yX
\dx/i ax
Applying this theorem to the conic
Aa?+2 Hxy + Bf+ 2Gx+2Fy + C = 0,
we first find, by differentiation,
2Ax-h2Hy+2Hx^ + 2By^ + 2G+2F^=0,
ax ax ax
. dy Ax + Hy + G
whence , = — -77 tt 7; '
dx Hx+By + F
Therefore the equation of the tangent at the point {x^, y^ is
Ax^ + Hy^+G. ,
y — y, = ^^ ix — X.),
^ ^' Bx^ + By^ + F^ '''
that is, Ax^x — Ax^ + Hxy^ + Hx^y — 2 Hx^y^ + ^ViV ~ ^Vl
+ Gx — Gx^ + Fy — Fy^ = 0.
This equation may be simplified by adding to it the identity
Ax*-^ 2Hx^y^ + Byl+ 2Gx^+2Fy^ + C= 0,
which follows from the fact that {x^, y^ is on the conic. There results
Ax^x + H{x^y + xy;) + By^y + G{x + x^) + F{y + y;)+ C =Q.
This result is easily remembered from its resemblance to the equa-
tion of the conic.
246
POLAR 247
132. Definition and equation of a polar. We have just seen in
§ 131 that the equation
Ax^x + H(x,y + xy^) + By^y +G{x+x;) + F(y + y,) + C = 0 (1)
represents the tangent line to the conic
As(^-\-2Hxy+By^+2Gx + 2Fy + C=0, (2)
provided the point (x^, y^ is on the conic. But no matter what is
the position of the point {x^, y^, (1), being of the first degree, repre-
sents some straight line which from the form of the equation must
in some way be related to the conic (2) and the point {x^, y^.
This line is called the polar of the point {x^, y^ with respect to
the conic, and the point is called the pole of the line.
The tangent line now appears as only the special case of the
polar which occurs when the pole is on the conic.
Ex. 1. The polar of the point (3, — 2) with respect to the ellipse
4a;2 + 5 2/2_2x + 3y-l = 0
is 12x-10y-(x + 3) + §(2/-2)-l = 0,
or 22x-17y-14 = 0.
Ex. 2. Find the pole of the line 2 x — 3^ + 6 = 0 with respect to the hyper-
bola 4x2-5T/2 + 4x-2i/ + 3 = 0.
The polar of (Xj, yi) is
4xix - 5yiy + 2{x + Xi) - (2/ + 2/i) + 3 = 0,
or (4xi + 2)x + (- 5yi -l)y + 2xi - yi + S = 0.
This will be the same as the given line if
4xi + 2 ^ 5?/i + 1 _ 2xi - yi + 3
2 3 6 '
These reduce to the two equations for Xi and yi,
12x1-102/1 + 4 = 0,
2x1-11^1 + 1=0;
whence • Xi = - ^|, ?/i = ^V-
133. Fundamental theorem on polars. When the equation
Ax'+2Hxy+By^+2Gx+2Fy + C = 0 (1)
represents one of the limiting cases of the conies, the polar has
little importance. We shall therefore assume that the conic is
248
TANGENT, POLAE, AND DIAMETER
either an ellipse (including the circle), a parabola, or an hyperbola.
The properties of its poles and polars are then conveniently found
by use of the proposition :
If P^is any 'point on the polar of another point ij, the polar of
II passes through Py
For the polar of Il(x^, y^ with respect to (1) is
Ax^x + H{x^y + xy^) + By^y +G{x + x^) + F{y + y^) + C = 0, (2)
and if P^ix^, y^ is on (2), we must have
Ax^x^ + II{x^y^-\- x^y^ + By^y^_ + G (^,+ x^ + F{y^-^ y,) + C= 0. (3)
Again, the polar of P^ with respect to (1) is
Ax^x + H{x^y + xy^) + By^y j^G{x + x^} + F{y + 7/,) + C = 0, (4)
and this passes through {x^, y^) because of (3).
134. Chord of contact. An inspection of the figures of the conies
shows that a point not on a conic must lie so that in general either
two tangents or no tangent can be drawn from it to the conic. In
the former case the point is said to be outside the conic ; in the
latter case, inside. Let us take
now a point 7^ outside the conic,
and let the two tangents drawn
from it to the conic touch the
conic in L and K (fig, 135). Now
the polar of a point on a conic
is the tangent to the conic at
that point (§ 132). Hence I^L is
the polar of L, and I^K is the
polar of K. Therefore, by the fundamental theorem (§ 133), the
polar of 7J must pass through L and K. Hence the polar is
the straight line LK, which is called the chord of contact of
tangents from 7J.
Conversely, if a straight line intersects a conic, its pole is the
point of intersection of the tangents at the points of intersection.
The proof of this is left to the student.
POLAR
249
The chord of contact may be used to find the equations of the
tangents through a point not on the conic.
Ex. Find the tangents to the conic x^ + 2xy + y^ + 2x + 6y + l = 0 which
pass tlirougli the point (4, — 2).
Since this point is not on the conic, its coordinates not satisfying the equa-
tion of the conic, we form the equation of its polar, i.e. 3x + 5?/ — 1 = 0, which
will be the chord of contact of the tangents drawn from the point to the conic,
provided any can be drawn. Solving the equations of the polar and the conic
simultaneously, we find that they intersect at the points (7, — 4) and (2, — 1).
Hence there are two tangents which are respectively 2x + 3y — 2 — 0 and
x + 2y = 0. •■■■'■ ■ ■•- ■■■ -■•■■'•': - , ■
135. CottstiUction of a. ^lar. Whethfer a point lies inside or
outside a conic, the polar may be obtained by the following con-
struction. Draw through
i^ (fig. 136) two straight
lines, one intersecting the
conic in L and K, and
the other intersecting the / / ^y ^^^^s
conic in M and N. Let
the tangents at L and K
intersect in H and the
tangents at M and iV in-
tersect in S. Then B is the pole of LK and S is the pole of UN,
by § 134 Since i^ lies on both LK and 3IN, its polar passes
through B and S by the fundamental theorem. Therefore ES is
the required polar.
This construction may also be used when ^ is outside the conic.
136. The harmonic property of polars. An important property
of poles and polars is stated in the theorem : Any secant passing
through I^ is divided har-
monically hy the conic and
the polar of P^.
Let -^iNT (fig. 137) be any
secant through P^, M and
N be the points in which
P^N cuts the conic, and Q
the point in which it cuts
the polar of P^. We are to prove that the line MN is divided
Fig. 137
250 TANGENT, POLAR, AND DIAMETER
harmonically, i.e. that it is divided externally and internally in
the same ratio. We are to prove, then, that
P^M _ MQ
F^N~ Qjsr'
whence, by placing MQ =F^Q—P^M, QN=P^N—P^Q, and solving
for P^Q, we have 2 PM ■ PN
^^ P^M+P^N
Let the point P^ be {x^, y^, the equation of the conic be
Aa?-\-2Hxy + Bf+2Gx + 2Fy + C=Q, (1)
and that of the polar of P^ be
Ax^x + H{x^y + xy^ + By^y + G{x + x;) + F{y + ^/j) + C = 0. (2)
Let {x, y) be a variable point on F^N, r the variable distance JJP,
and d the angle made by F^N and OX. Then
. ^-«^i . ^ y-Vx
cos V = , sm a = ,
r r
that is, x = r cos 0 + x^, y = r sin. 0 ■{■ y^ (3)
Now if P coincides with either M or iV, the values of x and y
given by (3) satisfy (1). Substitution gives
r^ [A cos' 0+2II sin 0cos0+B sin' 0]
+ 2r [Ax^ cos 0 + -^'(iCj sin 0 + y^ cos ^)
+ %, sin ^ + G! cos ^ + i^ sin 6'] + C" = 0,
where C = Ax'- + 2 Hx,y, + J5yf + 2 (?a;i + 2 Fy^ + C.
The roots of this equation are P^M and P^N. Hence, by § 43,
2 [J«,cos ^ +//(«! sin ^ + yj cos0)-i-By^ sin ^ + (?cos^ ^-i^sin ^]
F^M-F^N
whence
A cos' (9 +2 7/ sin 0cos0+B sin'^
^ cos'^ + 2^sin ^ cos ^ + 5 sin^^
2F,M-F,N
FJf+P,N ,
(4)
^a;jCos^+-H'(ajiSin^ + 2/jCOS^)+%iSin^ + <9cos^+i^sin^ ^
EECIPROCAL POLAES 251
Also, if the point P coincides with Q, the values of x and y
given by (3) satisfy (2). Substitution gives
r [Ax^ cos 6 + H{x^ smd + y^ cos d)+By^ siD.d + Gco80 + Fsin. 6]
+ C" = 0.
The root of this is F^Q. Therefore F^Q
C
(5)
Ax^ cos 6-\-H(x^ sin 6 + y^ cos 6) + By^ mid+G cos 6 +i>''sin 6
Comparing (4) and (5), we have
_ 2F,MF^N
^^~ f^m+f^n'
which was to be proved.
The theorem of this article is often made the basis of the
definition of the polar.
137. Reciprocal polars. Consider a given conic and a rectilinear
figure, such as the triangle ABC with sides a, b, c (fig. 138). Con-
struct the lines a', b', c', the polars of A, B, C, respectively with
respect to the conic. The lines a', b', c' form a new triangle A'B' C.
The fundamental theorem shows that A', B', C' are the poles of
a, b, c respectively. Hence the two triangles are so related that the
vertices of one are the poles of the sides of the other. They are
called reciprocal polars. A similar construction holds for any
figure composed of straight lines.
Consider next any curve K and a tangent line a (fig. 139). Let
A be the pole of a with respect to a conic C. As the tangent rolls
252
TANGENT, POLAR, AND DIAMETER
around the curve K, the point A describes another curve h Let
a and h be two tangents to K, and M their point of intersection,
and let A and B be the two corresponding points of k, and m the
chord AB. Then, by the fundamental theorem, m is the polar of M.
Now let a and h approach
coincidence. Then M ap-
proaches a point on K, B
and A approach coinci-
dence, and m approaclies a
tangent to h. Hence tlie
points of K are the poles
of the tangents to k.
We have then two curves such that the points of either are the
poles of the tangents of the other. These curves are called reciprocal
polars.
The study of reciprocal polars forms an important part of geom-
etry, but lies outside the limits of this work.
138. Definition and equation of a diameter. A diameter of a
conic is the locus of the iniddle points of a system of parallel chords.
I, i^t
Asg'+2 Hxy+ Bf-^ 2 Gx
+ 2Fy + C=Q (1)
be any conic (fig. 140), RS an^^
chord which makes the angle 6
with OX, and -?J(«i, y-^ the mid-
dle point of this chord. Take
P{x,y) any point on the chord,
and let P^P = r, where r is posi-
tive if iJP has the direction of
as, and negative if I^P has the direction SB. Then for any
position of P we have
-X
Fig. 140
= cos 6,
r '
2/ -2^1
sin^:
whence
X = x^+ r cos d, y = y^+ r sin 0.
(2)
DIAMETER 253
Now if P coincides with either B or S, the values of x and y in
(2) satisfy (1). Substituting, we have
r" [^ cos' (9 + 2 iT sin 6^ cos 6" + ^ sm='^]
+ 2 r [Ax^ cos 6 + Hx^ sin ^ + Hy^ cos 6
-\-By^&ind + Gco^d + F&md]
+ [Axl + 2 Hx^y^ + %,2 ^ 2 Gx^ + 2 i^'^/i + C] = 0, (3)
the roots of which are P^S and P^R. But, by hypothesis, P^R
= — -^*S^. Hence the roots of equation (3) are equal in magnitude
and opposite in sign. Therefore the coefficient of r in (3) must be
zero, that is,
Ax^ cos 6 +Hx^ sin 6 + Hy^ cos ^ + %j sin 6* + G^ cos ^ + i<" sin ^ = 0. (4)
If, for convenience, we assume that cos 6^0, and this will gen-
erally be the case, we may divide by cos 0 and replace tan 0 by the
usual symbol for the slope m, thus obtaining
Ax^ + Hy^ +G + m {Hx^ + By^ + F) = 0. (5)
If we allow RS to move parallel to itself, so that m remains
fixed but P^ changes, (5) always holds true, and in fact shows that
7J is always a point of the straight line
Ax + Hy + G + m{Hx + By+F)=Q. (6)
Conversely, any point P^{Xy, y^) on line (6) makes the values of
r in (3) equal in magnitude but opposite in sign, and if i^ lies so
that these roots are real, it will be the middle point of a chord
with slope m. ,-. g lo i?i.^s;r> iii"
The straight Ime (6) is of infinite length, and it is customary to
regard the entire line as the diameter, though it is evident that
not all of its points correspond to chords of the system which
intersect the conic in real points. i^w .-iuj ui.na
139. The last statement of the previous article may be explained as follows :
The equation y = ??ix + 6 may be made to represent any line of slope m by
assigning an appropriate value to b. For some values of b the corresponding
line intersects the conic (1) of § 138 in real points, and is one of the chords
bisected by the diameter (6).
254 TANGENT, POLAK, AND DIAMETER
For other values of 6, however, the line does not intersect the conic in real
points, the simultaneous values of x and y satisfying their equations being
imaginary. But if these imaginary values of x and y are substituted for
Xi, X2 and yi, yi respectively in the formulas x = — ^ , y = ^ — — of
§ 18, the resulting values of x and y are real, and furthermore they satisfy the
equation of the diameter.
This fact is sometimes expressed by saying that the line is a choi'd of the
conic which intersects it in imaginary points, and that its middle point is a
real point of the diameter. It is from this point of view that the entire line is
regarded as the diameter, since every point of it is the middle point of some
chord of the system.
140. If the conic has a center, every diameter passes through the
center. For, by § 126, the center satisfies the equations
Ax + Hi/ + G = 0, Hx+By + F=0,
and hence satisfies (6) of § 138 for any and all values of m.
In the parabola, hovjever, all diameters arc parallel to each other
and to the axis ; for the slope of the diameter is, from (6), § 138,
A. + Hm I
But for the parabola H = ^ AB, so that the slope of
^ + ^^ A -J~A~ \l~
the diameter becomes --^= » which reduces to ;= •
SAB + Bm y/B
This is independent of m, and equal to the slope of the axis (§ 128).
It is evident that the axes of a conic are diameters, for from the
symmetry of the curves they contain the middle points of all
chords which are perpendicular to them. In fact, they are the
only diameters which are perpendicular to the chords which they
bisect, as will be proved later on.
141. Diameter of a parabola. If the equation of the parabola
is written in its simplest form, 'if = 4:px, the equation of the
diameter becomes y = — •
m
From this equation it is evident that the only diameter perpen-
dicular to the chords which it bisects is the axis of the parabola.
Ex. 1. Find the equation of the diameter of the parabola 2y2^,3x = 0
bisecting chords with slope 2.
Since m = 2 and p = — ^, the equation of the diameter is, y = ^ » or
2 2/ + 3 = 0. ^
DIAMETER OF A PARABOLA
255
Ex. 2. A diameter of the parabola y^ = 2x passes through the point (2, — 1).
What is its equation, and what is the slope of the chords bisected by it ?
If m is the slope of the chords bisected, the equation of the diameter is
y — —. But (2, — 1) is a point of this diameter.
m 1 J
.-. — 1 = — , whence rn = — 1 ; also the diameter is y = , ov y = — \.
tn — 1
This equation of the diameter could have been written down immediately,
for the diameter is parallel to OX, so that if one of its points is distant — 1 from
OX, all its points are distant — 1 from OX, and its equation isy = — 1.
If we solve the equations of the diameter and the parabola
simultaneously, we find the coordinates of 0' (fig. 141), their point
p 2p^
of intersection, to be
The equation of the tangent at 0' is found to be 3/ = mx +
whence it is seen that its slope is m.
Calling 0' the end of the diameter, we express the above theo-
rem as follows : The tangent at the end of a diameter is parallel to
the chords bisected hy the diameter.
If we consider the tangent as the limiting position of a chord
which is moved, yet retains its original slope, the above theorem
seems almost immediately evident.
142. Parabola referred to a diameter and a tangent as axes. Let
O'X' (fig. 141) be a diameter of parabola
f = ^px, (1)
bisecting chords of slope m, and O'Y' be the tangent at 0'. Then
p 1p^
^m^ m
and the slope of O'Y' is m.
First transposing (1) to O'X' and
O'Y", where O'Y" is parallel to OY,
we have the formulas of transfor-
mation
the coordinates of 0' are
x = ^, + x".
m
y = ^ + y".
The new form of the equation is
y"^+^y"=4:px".
m
Fig, 141
256 TANGENT, POLAR, AND DIAMETER
Using now the 'formulas of transformation of § 117, which become
a?" =x'+ ^ > y" = /^^ >
Vl + mt: c:. . , ' -.:..: . . V 1 + tlV^
since <^ = 0 and <^' = tan" ' m, we have, finally,
,, . ; : By § 1 7, however, FO' = P(^-^/'^) .
Therefore if we denote FO' by p', after dropping the primes from
« and 2/, the equation becomes
y^ = 4 j?'x.
^^ ""It is to be noted that an equation in the form y"^ = 4:]px always
represents a parabola, the x axis being a diameter, the y axis a
tangent, and the distance of the focus from the origin being one
fourth the coefficient of x.
143. Diameters of an ellipse and an hyperbola. If the equation
+ -. ~ 2 2
of the ellipse is written in its simplest form, — + ^ = 1, and the
a^ If
common slope of the chords is denoted by m^ the equation of the
diameter becomes
V
V = 7, — X.
If the slope of the diameter is denoted by m,, m^= — >
whence m,w„ = 5 •
' . -If b =^ a^ m^ni^ csLnnot in general be —1, cmd the diameter of an
ellipse cannot in general he perpendictdar to the chords which it
bisects. The single exception is when the chords are parallel to
either axis, in which case the diameter is the other axis and is
perpendicular to the chords which it bisects, as noted above.
If Z> = a, the ellipse becomes a circle, and m^ni^ is always equal
to — 1. Hence the diameter of a circle is always perpendicidar to
the chords which it bisects.
DIAMETERS OF CENTRAL CONICS
257
62
jtti being
4
becomes —
a^mi 9 mi
Ex. 1. Find the equation of a diameter of the ellipse ix^ + 9y^ = 3(j bisoft-
ing chords parallel to the line a;-|-2y + l = 0.
Here a^ = 9, h"^ = 4, and in\ = — ^. .-. the diameter is y = —
or 9 2/- 8x = 0.
Ex. 2. 2y + 3x = 0isa diameter of the ellipse 4x2 + 9 ^2 = 3^
the slope of the chords which it bisects ?
The slope of the diameter is — |, and by the formula is —
the slope of the chords bisected. As a^ = 9 and 62 = 4, —
3 4, 8
.-. = , whence mi = —
2 9 77H 27
Ex.3. Find the diameter of the circle 4x2 + 4 2/2 + 43; — By — 11 = 0
bisecting chords of slope 2.
The center of the circle is (— -^, 1), so that the required diameter will be
2/ - 1 = - |(x + I), or 2 X + 4 ?/ - 3 = 0.
Ex. 4. Find the diameter of the circle 4x2 + 42/2 + 4x — By — 11=0, which
passes through the point (2, — 1).
The center of the circle is (— ^, 1), and the straight line determined by the
two points (2, — 1) and (— i, 1), i.e. 4x + oy — 3 = 0, is the required diameter.
In the case of the hyperbola —, — —,
1 it is to be noticed that
the parallel chords may be drawn in two ways. They may join
points on the same branch of the hyperbola, or points of one
branch to points of the
other branch, as repre-
sented in fig. 142.
In whichever way
the chords are drawn, if
their common slope is
denoted by m^, the equa-
tion of the diameter is
y =
Fig. 142
This equation differs from that for the diameter of the ellipse
only in the sign of the right-hand member. .^
If W2 is the slope of the diameter, m^m^ = — > and, as in the
case of the ellipse, a diameter of an hyperbola cannot be perpendic-
ular to the chords it bisects, except in the two special cases of the
transverse axis and the conjugate axis.
258 TANGENT, POLAR, AND DIAMETER
144. Conjugate diameters. In § 143 we have seen that if the
slope of the chords of the ellipse — + ^ = 1 is denoted by m^, and
the slope of the diameter is denoted by m^,
w„ = —
am.
whence m^m^ = —
(1)
Similarly, if the slope of the chords is m^, the slope of the diam-
eter bisecting them must be — -^ — > which, by (1), must be m,.
am\
Hence the proposition: If m^ and m^ are the slojjes of two
diameters of an ellipse, and
rthj)n„ = >
then each diameter bisects all
chords parallel to the other.
Such diameters are called
conjugate diameters. As the
major and the minor axis
each bisects chords parallel
Fig. 143
to the other, they are conjugate diameters.
It follows that :
1. The two axes are the only pair of conjugate diameters which
are perpendicular to each other.
2. If one of two conjugate diameters of an ellipse makes an
acute angle with the axis of x, the other makes an obtuse angle
h^
with the axis of x. For if m^ > 0, m^< 0, since m^m^= r^-
But a positive slope corresponds to an acute angle, and a negative
slope to an obtuse angle. Hence the upper portions of conjugate
diameters always lie on opposite sides of the minor axis, as OA^
and OB^ in fig. 143, A^A^ and B^B^ being conjugate diameters.
In similar manner for the hyperbola —
of two diameters m^ and m„ are such that
m,m„ = — >
y _
= l,if the slopes
CONJUGATE DIAMETERS
259
the corresponding diameters are conjugate, and each bisects all
chords parallel to the other. The transverse and the conjugate
axes are conjugate diameters, each of which bisects chords parallel
to the other.
It follows that :
1. The two axes are the only pair of conjugate diameters that
are perpendicular to each other.
2. Two conjugate diameters make either both acute or both
obtuse angles with the transverse axis ; for m^m^ being always
positive, vi^ and m^ have the same sign.
3. Two conjugate diameters lie on opposite sides of either asymp-
tote ; for since m.m„ = — > if ?», < - > then m^> - > and the corre-
^ ^ a^ a 'a
sponding conjugate diameters are on opposite sides of the asymptote
2/ = ^x(fig.l46).
145. Ellipse and hyperbola referred to conjugate diameters as
axes. Let the conjugate diameters OA^ and OB^ of the ellipse
^ + 1 = 1
a'^ b'
(1)
(fig. 144) be chosen as new
axes OX' and OY', and let
them make angles (f> and ^'
respectively with OX.
Then the formulas of trans-
formation are
x = x' cos <f> + y' cos <f>',
y = x' sin 4> + y' sin ^',
(2)
where
Le.
tan <^ tan <^
I _
Fig. 144
b^
sin <^ sin </>' cos <^ cos <f>' _ n
b' + a' "'
(3)
since OX' and OY' are conjugate diameters.
260
TANGENT, POLAR, AND DIAMETER
Substituting in (1) and collecting like terms, we have
cos^(f> sin^^X ,2 , o/cos ^ cos (f>' sin (f> sin <f>'\ , ,
v
+ ,'2^' + !«^V=l.
x'y
(4)
But the coefficient of x'y' is zero, by virtue of (3) ; and if the
intercepts on OX' and OY' are denoted by a' and b' respectively,
i.e. OA^ = a' and OB^ = b', (4) becomes
4- =^ = 1
(5)
where a' =
and &' =
Icos^
(f) sin^<f)
^l
cos-(f)' sm^<f)'
The equation of an ellipse can assume the form (5) only when
the axes chosen are a pair of conjugate diameters, as only then will
the coefficient of xy be zero. Conversely, any equation in form (5)
is an elhpse referred to a pair of conjugate diameters as axes.
In similar manner, the equation of the hyperbola referred to a
pair of conjugate diameters as axes is — ^ — ^ = 1, where at pres-
ent no geometrical meaning will be assigned to b'.
146. Properties of conjugate diameters.
1. The tangent at the end of a diameter is parallel to the conjugate diameter.
We shall prove the theorem for the ellipse, the same form of proof being appli-
cable to the hyperbola.
In fig. 145 let Ai have
coordinates (a;i,2/i). Then
the slope of OAi is — ,
and if the slope of OBx is
7/i2 , m2 — = SI whence
Xi
a2'
b^Xi
a^yi'
m^ = s — The equa-
tion of the tangent .4 iTi
a. ^, «f + ?g^ = l,
the slope of which is
Hence this tangent is parallel to the conjugate diameter 0B\.
CONJUGATE DIAMETERS 261
2. The sum of the squares of the halves of two conjugate diameters of an ellipse
is constant and equal to the sum of the squares of the halves of the major and the
minor axes, i.e. a"^ + b"^ = a^ + h"^.
b^Xi
We have just seen that the slope of OBi (fig. 146) is , so that its
equation is i,2r
y = -^x. (1)
Solving this equation simultaneously with the equation of the ellipse, in order
to find coordinates of Bi, we substitute the value of y from (1) in h — = 1-
Ct V
As a result x^ = — - — s- But Ai(xi, yi) is a point of the ellipse, so that
2 2 ""2/1 + ^^^1
^ + ?^ = 1, or 62x2 + a2y 2 = a2ft2.
«' ^' 2 = ^ = ^'
• • * a2^2 62 '
and X =± —— '
b
bxi
By substitution in (1), 2/ = T
Therefore the coordinates of J5i are ( -^ , — ^ ) •
\ b a /
If, as in § 145, we denote OAi by a' and OBi by b', by § 17,
a'2 = a; 2 + 7/2^
a2y2 62x2
and ^" = -^+^'
and hence a'2 + 6'2 = ^^1+^ x ^ + ^^^±^ y f
a2 o2
= a^ + f>^,
X V
since — + — = 1, as noted above,
a2 62
3. The area of the parallelogram formed by drawing tangents to an ellipse at
the ends of conjugate diameters is constant and equal to 4 ab. Let Ti T^ T-s T4
(fig. 145) be a parallelogram formed by the tangents at the ends of the con-
jugate diameters A1A2 and B1B2. Now the area of this parallelogram is evi-
dently four times the area of the parallelogram ^lOlJiTi. But^iTi= Oi'i
\a*v^ 4- 6*x^ iCiX
= 6' = ~ i , from work above ; and since the equation of AiTi is — -
a^ 22
, ViV _ 1 the perpendicular distance from 0 to ^ 1 Ti is, by § 32, — ; -_ •
62 "^ / VaV + 6V
/^a*y^ + ¥x^\/ a^b-^ \ , , ,,
Hence the area of AiOBi Tj = ( \ ^- I , = a&, and the
\ a& /\^a*yl + b*xl/
area of the large parallelogram is 4 ab, as was to be proved.
262
TANGENT, POLAR, AND DIAMETER
147. It was noted in § 144 that conjugate diameters of the hyperbola
= 1 lie on opposite sides of the asymptotes, whence it follows that if
one of two conjugate diameters intersects the hyperbola, the other cannot inter-
sect it. In order, then, to state for the hyperbola propositions analogous to 2 and
3 of the last article, it is customary to consider, in connection with the above
hyperbola, the hyperbola 1 = 1. These two hyperbolas are called con-
jugate hyperbolas, either one being considered the primary and the other being
called the conjugate.
It may readily be proved that if the slopes of two diameters are such that
62
rmm^ = — , they are conjugate diameters of both the above hyperbolas. More-
over it is evident (fig. 146)
that if one diameter in-
tersects one hyperbola, the
other intersects the conju-
gate hyperbola.
Now if OAi and 0B\ are
conjugate diameters, and
OAx is called o', as in
§ 145, and we apply the
same method as was ap-
plied to the ellipse, we shall
find OBi = 6' of § 145.
With this value of 6',
theorem 2, § 146, becomes
for the hyperbola a'2 — 6'2 = a2 — 62, while theorem 3 is the same for the
hyperbola as for the ellipse.
The proofs of these last statements are left to the student, the work being
exactly like that for the ellipse.
Fig. 146
PROBLEMS
Find the polars of each of the following points with respect to the given
conic, and find the points in which the polar intersects the conic :
1. (1, 2), 23x2 -\\xy + iy^ + 36x - 9y + 9 = 0.
2. (-1, -2), 3x2-3xy + 4x + y-3 = 0.
3. (0, 0), 2x2-22/2-2x + 22/-l = 0.
4. (4, -2), 5^/2 + \%y 4-4x4. 5 = 0.
Find the poles of each of the following polars with Vespect to the given conic :
5. 2x-y = 0, x2 + 8xy- 22/2 -12x4- 62/ -9 = 0.
6. x-32/4-2 = 0, x2-f 1/2 _ 2x4-42/ = 0.
7. x-f 2 2/-13 = 0, 3x2 + 82/2 -26x- 762/ -I- 231 = 0.
8. 3x-22/-9 = 0, 3x2-4j/24-6x-242/-45 = 0.
PEOBLEMS 263
Find the equations of the tangents from each of the following points to the
given conic :
9. (2, 3), 4x2 - 5X2/ + 2 2/2 + 3a; - 22/ = 0.
10. (0, 1), 3x2 - 42/2 + 12x = 0.
11. (1, -2), 2x2 -22/2- 6x- 62/ -1 = 0.
12. (2, 4), x2 + 2/2 - 6x - 2 2/ + 5 = 0.
13. (2, 0), 5 2/2 + 4x - 2 2/ - 8 = 0.
14. (-1, -1), 3x2 + 8 2/2- 8X-122/ + 4 = 0.
15. Prove that the polar of a given point with respect to any one of the
circles x2 + 2/^ — 2 fcx + c2 = 0, when k is variable, always passes through a fixed
point whatever the value of k.
16. T is the pole of a chord PQ of the parabola y^ = 4px. Prove that the
perpendiculars from P, T, and Q upon any tangent to the parabola are in
geometric progression.
17. If P is any point, LM its polar with respect to any central conic, C the
center of the conic, R the point in which the perpendicular from C to LM meets
LM, and S the point in which the perpendicular from P to LM meets the axis
of the conic, prove CR ■ PS = b'^.
18. Prove that the perpendicular from any point (xi, 2/1) to its polar with
respect to any central conic intersects the axis of the conic at a distance e2xi
from the center of the conic.
19. Prove that if in any conic the pole of the normal at P lies on the normal
at Q, then the pole of the normal at Q lies on the normal at P.
20. If Pi and P2 are any two points, and C the center of a conic, show that
the perpendiculars from Pi and C to the polar of P2 are to each other as the
perpendiculars from P2 and C to the polar of Pi.
21. If mi is the slope of the polar of a point Pi with respect to the ellipse
x2 w2
1 = 1, and mg is the slope of the line joining Pi to the center, show that
o* f>^ yi
mirrii = Find the similar relation for the hyperbola.
a2
22. Prove that the portion of the axis included between the polars of two
points with respect to a parabola equals the projection on the axis of the line
joining the points.
23. Show that for any conic section the polar of the focus is the directrix.
24. Where is the polar of the center of an ellipse or hyperbola with respect
to that curve ?
x2 ifl
25. In the ellipse 1- — = 1 find the equations of two conjugate diameters,
o2 62
one of which bisects the chord deteimined by the upper end of the minor axis
and the right-hand focus.
264 TANGENT, POLAR, AND DIAMETER
26. If Pi and P2 are the extremities of any two conjugate diameters of the
ellipse 1- ~ = 1, prove that the sum of the squares of the perpendiculars drawn
from Pi and P2 to the major axis of the ellipse is equal to 6^.
27. Show that there can be only one pair of equal conjugate diameters of the
ellipse h ^ = 1, namely y — -x^y — x.
a^ 6'^ a a
28. Show that the equation of any ellipse referred to its equal conjugate
diameters as axes is x^ + y^ =
2
29. In any ellipse show that the diameters parallel to the lines joining the
extremities of the axes are conjugate.
30. One diameter of the ellipse \- — = 1 passes through the upper end of
the right-hand latus rectum. What is the slope of the conjugate diameter ?
31. What must be the relation between the semiaxes a and b of an ellipse
when the diameters passing through the upper extremities of the left-hand latus
rectum and the right-hand latus rectum are conjugate ?
32. Show that the polar of any point on a diameter of a central conic is
parallel to the conjugate diameter.
33. Show that if an ellipse and an hyperbola have the same axes in magni-
tude and position, then the asymptotes of the hyperbola coincide with the equal
conjugate diameters of the ellipse.
34. Prove that tangents at the ends of conjugate diameters of an hyperbola
intersect on the asymptotes.
35. Prove that the straight line joining the ends of a pair of conjugate diam-
eters of an hyperbola is parallel to one asymptote and bisected by the other.
36. If an hyperbola has a pair of equal conjugate diameters, prove that it is
an equilateral hyperbola.
37. Show that in an equilateral hyperbola conjugate diameters are equally
inclined to the asymptotes.
38. Show that in an equilateral hyperbola all diameters at right angles to
each other are equal.
39. Show that every diameter of an equilateral hyperbola is equal to its
conjugate.
40. Prove that the tangents at the ends of any chord of a conic intersect on
the diameter which bisects the chord.
41. The chords which join the ends of any diameter to any point of the
curve are called supplemental chords. Prove that two diameters which are
parallel to any pair of supplemental chords are conjugate.
42. If the tangent at the vertex A of an ellipse cuts any two conjugate
diameters produced in T and t, show that AT ■ At = — tfi.
PROBLEMS . 265
43. Show that if any tangent meets any two conjugate diameters, the prod-
uct of its segments is equal to the square of the lialf of tlie parallel diameter.
44. If from the focus of an ellipse a pei-pendicular is drawn to a diameter,
show that it will meet the conjugate diameter on the corresponding directrix.
45. The tangent at any point Pi of an ellipse cuts the equal conjugate
diameters in T and T^. Show that the triangles TCP\ and TiCPi are in the
ratio CT'^ : CTi\
46. Show that the product of the focal distances of any point of a central
conic is equal to the square of half the corresponding conjugate diameter.
47. Find where the tangents from the foot of the directrix will meet the
hyperbola, and what angles they will make with the transverse axis.
48. Show that the perpendicular from the focus upon a polar with respect to
an ellipse or an hyperbola meets the line drawn from the center to the pole on
the corresponding directrix.
CHAPTER XIII
ELEMENTARY TRANSCENDENTAL FUNCTIONS
148. Definition. Any function of x 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 graphs and the derivatives of these functions.
149. Graphs of trigonometric functions.
Ex. \. y = sinx. » ■
The values of y are found from a table of trigonometric functions. In plot-
ting it is desirable to express x in circular measure ; e.g. for the angle 180° we
lay off X = TT = 3.1416. When x is a multiple of tt, y = 0; when x is an odd
TT
multiple of — , y = ± 1 ; for other values of x, y is numerically less than 1. The
graph consists of an indefinite number of congruent arches alternately above
and below the axis of x, the width of each arch being tt and the height being 1
(fig. 147).
Y
Fig. 147
Tlie curve y = sin x may be con.structed without the use of tables by a method
illustrated in fig. 148.
Let Pi be any point on the circumference of a circle of radius 1 with its
center at C, and let .40 be a diameter of the circle extended indefinitely. With
a pair of dividers lay off on A 0 produced a distance OiVj equal to the arc 0P\.
This may be done by considering the arc OPi as composed of a number of straight
lines each of which differs unappreciably from its arc. From ^i draw a line
266
TRIGONOMETRIC FUNCTIONS
267
perpendicular to A 0, and from Pi draw a line parallel to A 0. Let these lines
intersect in Qi. Then iViQi = JfiPi = CPi sin OCPi. But CPi = 1, and the cir-
cular measure of OCPi is OPi= ONi. If, then, we take ONi = x, NiQi= y.
Fig. 148
Qi is a point of the curve y = sin x. By varying the position of the point Pi we
may construct as many points of the curve as we wish. The figure shows the
construction of another point Q2.
Ex. 2. y = a sin bx.
TT TT
When X is a multiple oi ~, y = 0; when x is an odd multiple of — ? 2/ = ± o ;
b 26
for all other values of x, y is numerically less than a. The curve is similar in
its general shape to that of Ex. 1, but the width of each arch is now - , and its
height is a. Fig. 149 shows the curve when a = 3 and 6 = 2.
Fig. 149
Ex. 3. y = a sin {6x + c).
c
Place X =
+ x', y = y'.
The equation then becomes y' = a sin 6x'.
The graph is consequently the same as in Ex. 2, the effect of the term + c
being merely to shift the origin.
268 ELEMENTARY TRANSCENDENTAL FUNCTIONS
Ex. 4. y = a cos hx.
This may be written y = asmlbx -\ — |,
which is a cur\'e of Ex. 3. Hence the graph of the cosine function differs from
that of the sine function only in its position.
Ex. 5. y = sin X + ^ sin 2 x.
The graph is found by adding the ordinates of the two curves y = sin x and
y = ^ sin 2 X, as shown in fig. 150.
,-7/=^ sin 2x
^ -^y^sin x-t-jsin 231
-s y=sin X
Fig. 150
Ex. 6. y = sin
y = 0 when - = Attt, i.e. when x = -, where k is any integer. Hence the
X k "
graph crosses the axis of x at the points 1, ^, ^, \, \, etc. Between any con-
secutive two of these points y varies continuously from 0 to ± 1 and back to
Fig. 151
zero. It follows that as x approaches 0, the corresponding point on the graph
oscillates an infinite number of times back and forth between the straight lines
INVERSE TRIGONOMETRIC FUNCTIONS
269
y = ±1. It is therefore physically impossible to construct the graph in the
neighborhood of the origin. This is shown in fig. 151 by the break in the cur%e.
It should be borne in mind, however, that the value of y can be calculated
for any value of x no matter how small. E.g. let x = — - ; then - = = 10 x
125 X 12
+ — ir, and y = sin — = sin 75° = .9659.
12 ' 12
The value of 2/ is not defined for x = 0, and the function is discontinuous
at that point.
Ex. 7. 2/ = tanx.
When X is a multiple of tt, y = 0; when x is an odd multiple of - , y is
infinite, in the sense of §§ 11 and 68. The curve has therefore an iinlimited
number of asymptotes perpendicular to OX, namely x = ± — » x = ± — » • • • ,
which divide the plane into an infinite number of sections, in each of which
is a distinct branch of the curve, as shown in fig. 152.
Fig. 162
150. Graphs of 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.
'270 ELEMEIs^TARY TliANSCEXDENTAL FUoS'CTIONS
Ex. 1. 2/ = sin-la;.
From this x = sin y, and we may plot the graph
by assuming values of y and computing those of x
(fig. 153).
Fig. 1o3
151. Limits of
Fig. 154
Ex. 2. y = tan-ix.
Then x = tan j/, and the graph is as in fig. 154.
These curves show clearly that to any value of x
corresponds an infinite number of values of y.
sin^
and
1 — cos A
In order to apply the
h
h
methods of the differential calculus to the trigonometric functions, it
is necessary to know the limits approached by — : — and
as h approaches zero as a limit, it
being assumed that h is expressed
in circular measure.
1. Let AOB (fig. 155) be the
angle h, r the radius of the arc AB
described from 0 as a center, a the ^ '
length of AB,p the length of the per-
pendicular BC from B to OA, and
t the length of the tangent drawn
from B to meet OA produced in D. Fig. 155
^^
p
V'
^^^^^1^^
c
^1 V
-^
1 /
^^^
1 /
■v^
/ /
■^^
/ /
'^^v
//
^v
,.
/
^^.
/
B'
CERTAIN LIMITS 271
Eevolve the figure on OA as an axis until B takes the position
B'. Then BCB' =2p, BAB' = 2 a, B'D = BD. Evidently
BD + DB' > BAB' > BCB',
whence t> a> p.
Dividing through by r, we have
tap
T r r
that is, tan h> h> sin h.
Dividing by sin h, we have
cos h sin h
or, by inverting, cos h < — - — < 1,
Now as h approaches zero, cos h approaches 1. Hence >
which lies between cos h and 1, must also approach 1 ; that is,
-r . sin h .
Lim —— = 1.
1 cos Jb
2. To find the limit of — — - — - » as h approaches 0, we proceed
as follows :
, ; 2sin^J sin'^^ , /sin^^
1 — cos h ^ 2 ^ 2 ^ h 2
h h h ^\ Ik
2 \ 2
. h
sm-
Now as h approaches zero as a limit, — -^ approaches 1, as
just shown, and therefore - ■ j approaches zero, by 2, § 94.
Therefore Lim = 0.
A±0 h
272 ELEMENTARY TRANSCENDENTAL FUNCTIONS
152. 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
—- sm u = cos u—-} . (1)
dx dx
d . du ,-,
— cos u= — sm u-—> (2)
dx dx
d . 2 <^^ /ov
— tan -i^ = sec M — - > (3)
dx dx
d . ^ du ..^
-— COtW = — CSC M-—> (4)
dx dx
d du
— sec u = sec u tan u-—> (5)
dx dx
d . du _^
-— cscw =— cscw cotM — -• (6)
dx dx
1. By ( 0> S 96, -- sm % = -— sm w • -- •
dx du dx
To find — - sin u, we place y = sin u.
du
Then if u receives an increment Aw, y receives an increment A?/,
where / * \ »
. . , * s • o / Aw\ . Am
Ay = sm {u + Azt) — sm u^ 1 cos i u + -— I sm — >
the last reduction being made by the trigonometric formula
. , „ a -\-h . a — b
sm a — sm o = 2 cos — - — sm — - —
2 2
Then we have * a
. Au . Aw
V sm — — , sm -—
Ay ^^ / Au\ 2 / , Au\ 2
— ^ = 2 cos I u + — — = cos u + — —
Au \2/Aw V2/Aw
2
Let Au approach zero. By 2, § 94,
. Aw
sm— -
Lim — = Lim cos ( w + — - ) Lim — r '
Aw \ 2 / Aw
DIFFERENTIATION . 273
But Lini — ^ = -^ , Lim cos ( m + — - ) = cos u, and Lim
Alt du \ 2 /
= 1(§151). 2
Hence -— sin w = cos u
du
d . du
and —- sm u = cos u - —
dx dx
2. To find — cos u, we write
dx
■ i-^
cos w = sin I — — t*
Then -— cos w = -7- sin \— — ^t
dx dx \l
/tt \ d (it
=°°%2-'7rf:;i2-''' <''y<'»
"K \du
du
= — sm w ^- •
dx
3. To find — tan u, we write
sinw
tan u =
cos w
^, d ^ d smu
Then -r tan -26 = -^
aa; dx cos -it
cos u — sin u — sm u —- cos u
ax ax
(hvi^) ^^^)
cos^u
\"j \ /> 0 /
. 0 K du
(cos^u + siw'u) —
cos^u
(by (1) and (2))
0 du
= sec w -7--
dx
274 ELEMENTARY TRANSCENDENTAL FUNCTIONS .
4. To find -T- cot u, we write
ax
cos u
cot u —
sin w
^, d ^ d cosu
Then --cotu = : —
ax dx sinw
d d .
sin u — cos u — cos ti — sin u
= 3i^.^ (by (5). § 96)
^-sin^^-cos^.^ (by (1) and (2))
= — CSC M • — -•
dx
5. To find — - sec u, we write
dx
sec u = = (cos u) \
cosu
Then — sec u= — (cos ti)~ ^ -— cos u (by § 97)
dx dx
_ sin u du
cos^w dx
du
= sec it tan u — — ■
(by (2))
6. To find — CSC u, we write
dx
CSC w = -; = (sm u) .
suiti
Then — esc ?t = — (sin u)~^ — - sin u (by § 97)
aa? dx
= — CSC M cot w — • (by (1))
ax
Ex. 1. y = tan 2 x — tan^ x = tan 2 x — (tan x)^.
— = sec2 2 X (2 x) — 2 (tan x) — tan x
dx dx dx
= 2 sec2 2 X — 2 tan x sec^ x.
DIFFERENTIATION
275
Ex. 2. A particle moves in a straight line so that
s = A; sin bt,
where t = time, s = space, and b and k are constants. Then
velocity = w = — = bk cos bt,
dt
acceleration = a = — '- = — b^k sin bt = — 6%,
dt^ '
force = F= ma = — mb^s.
Let 0 be the position of the particle when ^ = 0, and let OA = + k and
OB = — 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
_ 2 77"
OA in the time — , and moves from B to A and back to B in the time
26 -6
The formula F = — mb^s shows
that the particle is acted on by a
force directed toward O and pro-
portional to the distance of the
particle from 0.
The motion of the particle is
called simple harmonic motion.
Ex. 3. A wall is to be braced
by means of a beam which must
pass over a lower wall 6 units
high and standing a units in front
of the first wall. Required the
shortest beam which can be used.
Let AB = I (fig. 156) be the beam, and C the top of the lower wall.
Draw the line CD parallel to OB and let EEC = 0. Then
that is, when
1 = BC + CA
= EC esc 0 + DC sec 0
= bcsc0 + a sec 9
— = — 6 CSC ^ cot ^ + a sec tf tan 0
do
_ asin^e - bco^^0
Si\vfi 0 cos"^ 6
— = 0, when a sin^ 0=b cos^ 5,
de
bi
tan 0 = — •
270 ELEMENTARY TRAKSCENDEKTAL FUNCTIONS
When e has a smaller value than this, a sin^d < h cos^^, and when d has a
larger value, a sin^tf > b cos^tf. Hence Hs a minimum when tan d ■= — Then
o*
i = 6 CSC ^ + a sec tf
6 Vg^ + 6^ g VaJ +
6* g*
= (a^ + b^)^.
153. Differentiation of inverse trigonometric functions. The
formulas for the differentiation of the inverse trigonometric func-
tions are as follows :
1. —&\n~^u= rr^ — > when sin"'w is in the first or the
VI — w fourth quadrant;
= —> when sin~'w is in the second or
V 1 — w ^j^ third quadrant.
2. --T- cos~*w = — > when cos"'w is in the first or the
^ ~ ^ second quadrant ;
> when cos~'w is in the third or the
fourth quadrant.
Vl — tt'^ dx
ax 1 + u ax
. d ^ , 1 du
4 -- cot-^u = — -— •
ax 1 + u ax
5. — -sec"^^ = - — — > when sec" 'w is in the first or the
dx . uVu' - 1 dx ^j^.^^ quadrant ;
— > when sec~^ic is in the second
wVit IX ^^ ^-^^ fourth quadrant.
6, — csc~'w= y^^z^ — > when esc ^w is in the first or
dx uy/u' - 1 dx ^-^^ ^j^.^^^ quadrant ;
— , when csc"'?^ is in the second or
y/u^ — l dx
uy/u
the fourth quadrant.
DIFFERENTIATION 277
The proofs of these formulas are as follows :
1. If y = sin"^w,
then sin y = u.
Hence ^^^^'^~'J~' (^"7 § 152)
dy 1 du
or • _£ =
dx cosy dx
But cos y = Vl — u^, when y is in the first or the fourth quad-
rant, and cosy =— vl — w^ when y is in the second or the third
quadrant.
2. If y = cos'^w,
then cos y = u;
dy du
whence —siny-f- =
or
lA/Jb (a/Ju
dy _ 1 du
dx sin y dx
But sin 3/ = + Vl— u^ when y is in the first or the second
quadrant, and siuy = — Vl — ti"-^ when y is in the third or the
fourth quadrant.
3. If y = tsinr^u,
then tan y = u.
Hence
whence
4. If
then
Hence
whence
2 dy
du .
dx
dy _
1
du
dx
1+'
u^ dx
y =
■■ cot"
'u.
coty =
u.
2 <^V
du
dx
dy _
1 (
dx
1
+ w' (
du
278 ELEMENTARY TRANSCENDENTAL FUNCTIONS
5. If
y = sec"'tf,
then
sec y = u.
Hence
dy du
secytany-f- = — ,
dx dx
whence
dy 1
du
dx sec y tan y dx
But sec y = u and tan y = + ^u^ — 1 when y is in the first or
the third quadrant, and tan y = — ^li^ — 1 when y is in the second
or the fourth quadrant.
b. if
y = csc"'w,
CSC y = u.
Hence
dy du
— CSC y cot y -T- = -rr-y
dx dx
whence
dy 1 du
dx CSC y cot y dx
But CSC y = u and cot y = + v ?t^ — 1 when y is in the first or
the third quadrant, and cot y = — vw^ — 1 when y is in the second
or the fourth quadrant.
Ex. 1. y = sin-i Vl — x^, where y is an acute angle.
^ Vl - (1 - a;2) <i^ Vl-x2
This may also be clone by noticing that sin-^ v 1 — x^ = cos-^x.
Ex. 2. The example of § 107 may be solved by drawing a straight line from
S to 0 (fig. 125), denoting the angle YOS by 6 and the subtended arc by s.
Then s = ad,
rL.S = 2tan-i
OL a
and 6 = 1- YLS = 2 tan-i = 2 tan-i^
Hence
and
s = 2 a tan-' — ,
a
1
ds - a
V = — = 2a
dt. x'{
dxi 2 a'^c
dt a2 + xl
EXPONENTIAL AND LOGARITHMIC FUNCTIONS 279
154. The exponential and the logarithmic functions. The
equation
2/ = 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. li x = n, an integer, y is determined by raising a to
P
the nth. power by multiplication. If a is a positive fraction — > y is
the g-th root of the ^th power of a. If ic is a positive irrational num-
ber, the approximate value of y may be obtained by expressing x
approximately as a rational number. If a? = 0, 3/ = a" = 1. Finally,
if x = — m, where m is any positive number, 3/ = a,""* = — •
Practically, however, the value of a^ is most readily obtained by
means of the inverse function, the logarithm ; for if
then
a; = log„y.
When a = 10, tables of log-
arithms are readily accessible.
Suppose a is not 10, and let h
be such a number that
10* = a,
I.e.
and
h = log,oa.
Then we have
y = a^={lQy = W.
Hence hx = log^^^y,
^ log,„y log,o2/
logio^
Fig. 167
Ex. 1. The graph oi y = log(i.6)X is shown in fig. 157.
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 z.
280 ELEMENTARY TRAKSCENDEXTAL FUXCTIONS
Ex. 2. The graph oi y = (1.5)* is shown in fig. 158.
Fig. 158
155. The number e. In
the theory and the use of the
exponential and the logarith-
mic functions, an important
part is played by a certain
irrational number, commonly
denoted by the letter e. This
number is defined by an
X infinite series, thus :
It will be shown in the second volume that this series converges;
i.e. 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^,x is called the natural or Napierian logarithm of y.
The student will discover as he proceeds with his study that the
use of Napierian logarithms in theoretical work causes simpler
formulas than would arise with the use of the common logarithm.
Hence, in theoretical discussions, the expression logic usually means
the Napierian logarithm. On the other hand, when the chief inter-
est is in calculation of numerical values, as in the solution of tri-
angles, logic usually means logjoX In this hook we shall use log x
for log^x.
Tables of values of log^a: and (f 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 § 154.
We give the graphs of certain other functions which involve e
and present other points of interest.
GRAPHS
281
Ex. 1. y = e-=^.
The curve (fig. 159) is symmetrical with respect to OY and is always above
OX. When x = 0,y = 1. As x increases numerically y decreases, approaching
zero. Hence OX is an asymptote.
Y
Fig. 159
Ex. 2. y
-(ea + e »).
This is the curve (fig. 160) made by a string held at the ends and allowed to
hang freely. It is called the catenary.
Ex. 3. y = e~°^sin6x.
The values of y may be computed by multiplyi:ig the ordinates of the curve
y _ e-ax by the value of sin bx for the corresponding abscissas. Since the values
of sin6x oscillate between ± 1, the value of e-'"sin6x cannot exceed those of
e-°^. Hence the graph lies in the portion of the plane between the curves
y = e-«^ and ?/ = - e-"^. When x is a multiple of - , y is zero. The graph
282 ELEMENTAEY TIIANSCEKDE:N^TAL FUNCTIOI^S
therefore crosses the axis of x an infinite number of times. Fig. 161 shows the
graph when a = 1, 6 = 2 7r.
1
\
1 \ ^\
/\^^ ~}^--<z~
0
i \. y2 :::
-1
/""
Fig. 161
Ex. 4. y = e^.
When X approaches zero, being positive, y increases without limit. When x
approaches zero, being negative, y approaches zero; e.g. when x= yoV(J'
y = el"*'", and when « = — y^Vo' y = er'^'**^ ~ "wm' ^^® function is therefore
discontinuous for x = 0.
3
T
;
L
\
0
Fig. 162
The line y = 1 is an asymptote (fig. 162), for as x increases without limit,
being positive or negative, - approaches 0 and y approaches 1.
CERTAIN LIMITS
283
10
Ex. 5. y =
L
1 + e^
As X approaches zero positively, y approaches zero. As x approaches zero
negatively, y approaches 10. As z increases indefinitely, y approaches 6.
The curve (fig, 163) is discontinuous when x = 0.
10
Fig. 163
156. Limits of (l + hf and
e^'-l
In obtaining the formulas
for the differentiation of the exponential and the logarithmic func-
tions it is necessary to know certain limits, 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.
1. We require the limit of (1 + A)* as h approaches zero. We
1.
begin by expanding (1+ ^)* by the binomial theorem and making
certain simple transformations ; thus :
a+.)i=i+i..+i6:
h' +
¥
A« +
_ 1 (1-;^) (i-;^)(i-2A)
■-^^i-*-|^+ii +
-hB,
284 ELEMENTAEY TRANSCENDENTAL FUNCTIONS
where 7^ represents the sum of all terms involving h, h^, h^, etc.
Now it may be shown by advanced methods that as h approaches
zero R also approaches zero, and at the same time
i + T + g + @ + --
approaches e. Hence i
Lim(l+A)A = e.
A= 0
e*— 1
2. We require the limit of — - — as h approaches zero.
h
Let us place e* — 1 = k,
where evidently k is a number approaching zero as h approaches
zero. Then
e* = 1 + ^, whence A = log (1 + k).
Then we have
g"-l_ k _ 1 _ 1
h -log(l4-^^)-Log(i+Z:)"log(l + >t)^'
K
1
Now as h approaches zero k approaches 0, and (1 + ^)* approaches
e by the previous proof. Hence log(l + ky approaches log e, which
is 1. Therefore e*— 1
Lim — - — = 1.
»=o h
157. Differentiation of exponential and logarithmic functions.
The formulas for the differentiation of the exponential and logarith-
mic functions are as follows, where, as usual, it represents any func-
tion which can be differentiated with respect to x, log means the
Napierian logarithm, and a is any constant:
dx dx
d 1 I du
-7- log W = ;- • (2)
dx u dx
— a-^a-loga^^. (3)
dx "" dx ^ '
d , _ log„g du .
dx u dx
DIFFERENTIATION 285
dx du dx
To find — e", place y = e". Then if w receives an increment Aw,
du
y receives an increment Ay, where
Ay = e" + ''" — e" = e" (e^" — 1).
Then
Aw "" Aw
Now let Aw
approach zero.
By § 94,
Lim — ^ = 6" Lim — -^
Aw Aw
But
-..Ay ^y c? „
Lim -^ = -/■ = — e",
Lu du du
and
Lim — - — = 1.
Aw
Therefore
— e" = e",
ait
and
d „ ^du
2. If
y = logw,
then
e>' = u.
Hence
ydy _du
dx dx
whence
dy 1 du _ldu
dx e" dx u dx
by 2, § 156
by(l)
3. Let y=a".
Then it is always possible to find a quantity b such that
a = e\
whence ^ = log a.
286 ELEMENTARY TRANSCENDENTAL FUNCTIONS
Then
y = {e")" = e"\
and
dx dx
dx
= (log«)».|.
4 If
y = log^-?^,
then
a« = u,
and
„ T dy du .
a^ log a -^ = — - >
dx dx
whence
dy 1 1 du
dx log a u dx
But if
log a = h,
a = e^,
1
whence
(^ = e,
and therefore
\ = log.*.
Hence ■ ^^\og^dv.
dx u dx
Ex.1, y = log(x2_4x + 6).
iZj/ 2x — 4
by(i)
by (3)
dx x2 _ 4 x + 6
Ex. 2. y = e-=^. _
: ■ da; ■
Ex. 3. 2/ = e-'^cosftx.
dy .d,, d,,, ,, .,
-^ = cosox — (e-'«^) + e-"^ — (cosox) = — ae-°^cos6x — oe-»*sinc>x.
dx dx dz ,
EXPONENTIAL FUNCTION 287
158. An important property of the exponential functions is
expressed in the following theorem : If the rate of change of a
function is proportional to the value of the function, the function
is an exponential function.
Let -^ = ay.
Then l^ = a.
y dx
Hence log y = ax + c^,
or y = g^^^+'^i = e'^'e'^ = ce'^.
Ex. Let p be the atmospheric pressure at the distance h above the surface
of the earth and p the density of the air. We will assume that the density is
proportional to the pressure. Then if po and po are the density and the pressure
respectively at the surface of the earth,
Po Po'
whence p = — • p.
'^ Po
Let now the height h be increased by a distance Ah. The pressure will
be increased by an amount Ap, where — Ap is equal to the weight of a column
of air standing on a base of unit area and having a height Ah. If p is the density
at the height h and p — Ap the density at the height h + Ah, it is evident that
the weight of this column of air lies between (p — Ap) Ah and pAh ; that is,
(p — Ap) Ah< — Ap < pAh,
Ap
whence p — Ap < r < P-
Ah
Taking the limit, we have
dp ... .^Ap PO
■— z=Limit— =- p = - ^p.
dh Ah Po
-Bih ... .
Therefore p = ce po . .•..•■...
_?o A .....
•Since when it = 0, e i'o =1 and p = po, it follows that c = p^.' "' •
Hence p = poe I'o ,
, Po, Po
/. = -log^.
288 ELEMENTARY TRANSCENDENTAL FUNCTIONS
159. Sometimes the work of differentiating a function is sim-
plified by first taking the logarithm of the function and then
applying the formulas of this article.
Ex. 1. Let
= Vr
X2
+ X2
-
Hence
and
+ X'''
= ^l0g(l-x2)-^.l0g{l+«2).
1 dy _ X X
y dx~ I - x^ 1 + X*
-2x
~ (1 - x-2) (1 + x2)
dy _ —2xy
dx~ (1- cc2) (1 + a;2)
-2x l-x2
x'^) \1
(1 - x2) (1 + x-i) \l + x3
-2x
(l + x2)V'l-x<
This method is especially useful for functions of the form u",
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«'°^.
Then log y = log (x"" ^)
= sin X • log X.
Therefore — - = (sin x) - + cos x ■ log z,
y dx X
dy
and -^ = x"° ^- J • sin X + x*'°^ cos x • log x.
dx
160. Hyperbolic functions. Certain combinations of exponential
functions are called hyperbolic functions. In their names and
properties they are analogous to the trigonometric functions, but
the reason for this cannot be shown at present. The fundamen-
tal hyperbolic functions are the hyperbolic sine (sinh), the hyper-
bolic cosine (cosh), the hyperbolic tangent (tanh), the hyperbolic
HYPERBOLIC FUNCTIONS
289
cotangent (coth), the hyperbolic secant (sech), and the hyperbolic
cosecant (cosech), defined by the equations
sinha; =
cosh X =
tanh X =
coth a; =
sech X =
cosech X =
e^—e-^
2 '
e^+e-^
2 '
sinha;
e^—e
X
cosh a:
e^-f-e"
X
cosh X
e^^+e-
-X
sinh X
e* — e~
X
1
2
cosh a?
e^'+e-
X
1
2
sinh a? e^ — e"
Fig. 164
Fio. 105
The graph of sinh x is given in fig. 164, that of cosh x in
fig. 165, and that of tanh x in fig. 166.
290 ELEMENTARY TRANSCENDENTAL FUNCTIONS
Relations between hyperbolic functions may be derived by
expressing each in terms of the exponential functions. The
student may in this way prove the following relations:
cosh'^a; — sinh^a; = 1,
tanh^ic + sech^a? = 1,
coth'* X — cosech^ x = \,
sinh (a? ± y) = sinha? coshy ± cosh a; sinhy,
cosh {x ± y) = cosh X cosh y ± sinh x sinh y,
^ 1 / , X tanhaj±tanhy
tanh(a;±y)— ^
Fig, 166
The derivatives of the hyperbolic functions are readily obtained
by differentiating the equations which define them. We have in
this way : , ,
•' a . , , an
-T— smh u = cosh u-—>
ax ax
d , . , du
-r- cosh u = smh u — - ,
dx dx
d ^ , 1 2 du
-— tanh u = sech w -— ,
dx dx
d ^, , ^ du
-r- coth u=— cosech u—-,
dx ax
-T- sech u =— sech u tanh u — - ,
dx dx
—— cosech u = — cosech u coth u — -•
dx dx
INVERSE HYPERBOLIC FUNCTIONS 291
161. Inverse hyperbolic functions. If
X = sinli y,
then ' y = smh"^a;,
called the inverse hyperbolic sine of x.
This function may be expressed as a logarithm as follows :
We have y = sinh~ ^ x,
and X = sinh y = —
Placing e~^= — and clearing of fractions, we have
e^''—2xe''=l.
Treating this as a quadratic equation in «", we have
e^=x±Va^+l;
but since we know that for any real value of y, e" is positive, we
discard the minus sign before the radical and have
y = sinh"' ic = log {x + vV+T).
In the same manner, the student may prove the following :
cosh~'a; = log {x ±Vic^— 1)
= ±log
tanh~'a; = | log
= ± log (x + ^y? — 1)'
coth"'a;= | log
\—x
x + 1
x — 1
, l±Vl-a;*-^ _^, l + Vl-ar'
sech"' a? = log = ± log
^ X X
cosech"^' ic = log •
292 ELEMENTARY TRANSCENDENTAL FUNCTIONS
The derivative of the inverse hyperbolic functions can be
obtained by differentiating the expressions just obtained, or by
proceeding in the same manner as in § 153. In either way
we find: , . ,
a . , , 1 du
— sinh u =■—= — J
dx V M^ -\-l dx
d , _, , 1 du
— cosh '■u = ±—: —-,
dx ^u'-l dx
d ^ , _, 1 du
-— tanh ^u = - r-— ,
dx 1 — u dx
d ^. _, 1 du
--coth u=- --—,
dx 1 — udx
d , , 1 du
— sech ^u = zf — — ,
dx u\l— u^ dx
d , _, 1 du
— cosech u =
dx uV 1+ u^ dx
Ex. Consider the motion of a particle of unit mass falling from rest, and
impeded by a force proportional to the square of its velocity. The total force
acting on the particle is then g — kv'^, where g is the acceleration due to gravity,
and A; is a constant. Hence
— = g-kv^;
at
1 dv
whence = 1,
g - kv^ dt
1 1 dv ,
or = 1.
ff ^ k . dt
9
To bring this under one of the known formulas of differentiation we will
place
whence
dv _ jg du
We have, therefore, — - — = 1 ;
whence — — =tanh-i u z=t + c,
Vkg i-u^ dt
1
{k
tanh-i-%/-i' = < + f.
TRANSCENDENTAL EQUATIONS
293
But since the body falls from rest, when t = 0, v = 0; therefore c = 0.
The equation may be written
that is,
Hence
V = ■%- tanh t Vkg^
^ sinh t ^</kg
* COShtV/i-gr
8 = - log cosh t y/kg + c.
k
162. Transcendental equations. Equations involving transcen-
dental functions can often be solved by methods similar to those
used for algebraic equations. Graphical methods can often be used
to advantage.
Ex. 1. sin a; = a.
The solutions of this equation are the abscissas of the points of intersection
of the curve y = sinx and the straight line y = a (fig. 167). If a > 1 or a < — 1,
there are no real solutions ; otherwise there are an infinite number of solutions.
TT
Let us call the smallest positive root Xj, where 0 < xi < - if a is positive, and
]
r
-Ztt/ Nc'T 0
/ \7r 2w/
\ y
\ „.» / -
Fig. 167
TT < Xi < 2 TT if a is negative. The value of Xx must be found from a table or
approximately from the graph. The next largest positive root is then tt — Xi
when a is positive, and 3 tt — Xi when a is negative ; and all other roots, positive
or negative, are found by adding or subtracting multiples of 2 tt. Hence the
general solution is 2 A;7r + Xi and (2 A; + 1) rr - Xi, or, more compactly written,
/c7r + (-l)'^^Xi,
where k is any positive or negative integer or zero.
Ex. 2. cosx = a.
The general solution is 2 ^tt ± xi, where Xi is the smallest positive solution
and k is an integer or zero. The proof is left to the student.
Ex. 3. tanx = a.
The general solution is kiz + Xi. The proof is left to the student.
294 ELEMENTARY TRANSCENDENTAL FUNCTIONS
Ex.4, cos 2 a; = 2 cos X.
When an equation involves two or more trigonometric functions it is well to
write it in terms of one. The above equation may be written
2 cos^x — 1=2 cosx,
which is a quadratic equation in cos x. Solving, we have in the first place
cos x = -J- ± -^ V3,
but the plus sign may be disregarded, since for real angles cos x is not greater
tlian 1 numerically. The equation
cos X = J - 1 Vs
is now to be solved as in Ex. 2. There results x = 2 Arir i 1.946.
Ex. 5. tanx = kx.
The roots of this equation are the abscissas of the points of intersection of
the curve y — tanx and the straight line y = kx (fig. 168).
Fig. 168
The two intersect at the origin, but the other intersections depend upon the
value of k. Since the slope of the curs'e y — tan x is 1 when x = 0, and > 1 when
0<x<-, we need to distinguish three cases, according as A; >1, 0<fc < 1, or A;<0.
TRANSCENDENTAL EQUATIONS 295
The graph shows that if fc > 1, the smallest positive root lies between 0 and
— ; if 0 < A; < 1, the smallest positive root lies between tt and — ; and if fc < 0,
the smallest positive root lies between — and tt.
We shall now find the smallest positive root in the special case
tan X = 2 X.
We must first locate the root (§ 47), either by the graph or by means of a table.
If a table is used, it must be one in which angles are given in radians. We shall
use the table on page 132 of Professor B. O. Peirce's "Short Table of Integrals."
We find, by looking for a place in the tables where the tangent of an angle is
approximately equal to twice the angle, that when x = 1.1G36 (G6° 40'), tanx
= 2.3183, and when x = 1. 1665 (66° 50'), tan x = 2.3369. Consider now the curve
y = tan x — 2 x.
When xi = 1.1636, yi = - .0089, and when Xg = 1.1665, yz = .0039.
Hence the curve intersects OX between xi and X2, and a root of the equation
tan X — 2 X = 0
is therefore located to two decimal places. To locate the root more closely we
will use the method of § 63. We have
^ = sec2x - 2,
dx
and — ^ = 2 tan x sec^ x,
both of which are positive when x is between Xi and Xg. Hence that portion of
^i^^onrye y = tanx-2x
appears as in fig. 64, (1), and its intersection with OX lies between the tangent
at (X2, 2/2) and the chord connecting {xi, yi) and {X2, 2/2)- The tangent at (X2, 2/2) is
y - .0039 = 4.461 (x - 1. 1665) ;
0128
the chord is y -. 0039 = '-^— (x - 1 . 1665) ,
and the point of intersection of each with OX is found to be
x = 1.1656
to four places of decimals. This is therefore the root of the equation to four
decimal places.
Ex.6. e^-4x2-2x + 3 = 0.
The roots of this equation are the abscissas of the points of intersections of
the curves y = e^ and j/ = 4x2 + 2x - 3, and may be found graphically or by
means of tables to lie between - 1 and - 2 and between 0 and 1. To determine
the root between 0 and 1, we place y = e^ - 4x2 - 2x + 3. When Xi = 0,
2/1 = 4, and when X2 = 1, 2/2 = — •282.
296 ELEMEXTAKY TKAXSCENDENTAL FUNCTIONS
Also ^ = e^-8x-2,
dx
and — = e^ - 8,
which are both negative when x is between 0 and 1. Hence the portion of the
curve in question has the shape of fig. 64, (4), and its intersection with OX lies
between that of the tangent at {x^, y^) and that of the chord connecting (xi, yi)
and (Xj, yo). The tangent is
y + . 282 = - 7.282 (x-1),
which intersects OX when x = .97 — . The chord is
2/ + .282 = - 4.282 (x-1),
which intersects OX when x = .93 + .
If we now place Xi = .93, yi = .2149, and if Xj = .97, 2/2 = — -0657, the tan-
gent at (x^, y,) is ^ ^ ^.„ ^ _ ^^^gi (^ _ .97),
which intersects OX where x = .9608— ; and the chord between (Xi, yi) and
(«2, Vi) is _ oaof?
y + .0657 = (X - .97),
.04 ^ ^'
which intersects OX where x = .9606+.
Hence a root of the equation lies between .9606 and .9608.
PROBLEMS
Plot the graphs of the following equations :
1. y = ctnx, 11. y = xsin-.
x
2. y = secx. j
_ 12. y = x^sin--
6. y — cscx. x
4. y = versx. . 13. y = el^.
5. y = ^ sin 3 x. L-^x
is ,10 14. y = xe ^ .
6. y = sm X + ^ sin 3 X. ^
7. y = sin X + sin 2 x. 15. y = xe^.
8. y = 2 sin X - sin 2 X. 16. y = log (sin x).
9. y = cosx+^cos3x. 17. 2/ = tan- 1 (ox + 6).
X — 1
10. y = 1 - A cosx - * cos 2 x. 18. y = log .
■* X + 1
19. Plot the graph of the equation y = - sin x, and determine what relation
it has to the hyperbolas xy = ± 1.
20. Plot the graph of the equation y = sin x', and show that the distance
between two consecutive intercepts on OX approaches zero as a limit.
PROBLEMS 297
dt/
Find — in each of the following cases :
dx
21. 2/ = sin (ox + 6) cos (ax — 6). 45. y = sin- 1 (2 x Vl — x^).
22. y = tan (ax + 6) ctn (ox + c). ._ , 2x — 1
46. y = csc-i
23.
8ec4x
24.
ctn 2 X + 2
y — •
csc2x
25.
sec3x
y = - — ^ — -^-
2 V^
X
47. y = sec- 1 V4 x2 + 4 X + 2.
48. y = CSC- • - I x2 + — I .
49. y = e^ + 2ar.
lauox-t-i L2 _ a2
26. y = csc2x-ctn2x. 50. y = \ogy^^^ ^ ^^-
27. y = sec"'nxcsc»mx. ^
28. 2/ = sec22x + tan2x. 51. y = a^i-=^e^i-^.
29. 1/= ctn4xcsc2x. 52. y = log(2x + 1 + 2 Vx2 + x).
30. y = sin(xcosx). 53. y = e»: -v^a.
31. y = (cos2x + f)sin8x. 54. y = o*""^.
32. 2/ = (2 sec^x + 3 sec2x) sinx. 55. y = x^ logx2.
sin2x 56. y = otan^^ec^'^.
33. y =
Vl — cos2x 57. y = tan 2 x • a"*= '- "'.
34. y = cos Vl — x2. 58. y = e(<' + *)''sinmx.
__ sin2x cos2x 59. w = csc-i(sec2x).
35. y = _
sinx COSX CA / , V tan-iV^
SO. y — {x + a)e ^ <-.
36. y = sin-i :• , e"' - e-^
Vi + x2 61- 2' = ta"-'^^rqr^x-
37. y = tan-i-^=. g2 ,, - ip.. ^ ^^" '^ + ^
Vrr^ '*''• ^~ "^^ tanx + 3
38. y = cos-i:l-^- . 63. y = sec-i^ —
1 + x ^
(J 64. y = e^=<'««cos(xsina).
39. y = sec- 1 — --• 65. 2/ = log tan (x2 + a2).
a-x 66. y = x cos->x - Vl - x2.
40. y = sin-i ^
" + * 67. 2/ = - log (sec ax + tan ox).
41. y = sec-il(|Vx + ^y gg ,^^(a; + oVr3^)e«--'-.
42. 2/ = CSC- 1 (x2 + 2 X). 69. y = log Vl + x2 + x ctn- 1 x.
43. 2/ = ctn-i^^-2tan-:l 70. y = ^^ - log VT^.
^ x2 - a2 a Vl-x2
.. 1 ,6 + acosx _. e<^(o sin mx — m cos mx)
44. y = — cos-^ ; 71. v = — ^ ; ;, "
Va2 - 62 a + 6cosx " m^ + a^
298 ELEMENTARY TRAJSTSCENDENTAL FUNCTIONS
72. y = a;2 ctn-i _ -|- a^ tan-i ax.
X a
-g , , * X sin-ix
X
X
y = \og( ^ )-
\1 + Vl-X2/
74. y = log(x + Vx2 - a2) + cos-i
75. y = log(x + Vx2 - a2) _ csc-i-.
76. y = tan-i Vx^ - 2x - Ig^jg^lj).
Vx2 — 2 X
,/ /a -6^ x\
=^ Un- 1( \ tan - I .
b2 \\a + 6 2/
77.2/ = -^
Va2
78. 2/ = log tan (2 x + 1) + esc (4 x + 2).
mn r^ ', 1 "^2 OX — X2
79. y = v2 ax — x^ + a cos-i
80. 2/ = X Va2 + x2 + a21og(x + VoM^).
o, , 2 v^ - 1 2V^sec-'2Vx
81. y = log ^ —
\2Vx+l V4x-1
ort X — a ,/r r , a*^ . ,x — o
82. y = V2ax — x2-) sin-i
83. y = tan- 1 (x - Vx2 - 1) + log
Vx* - 1
84. y = -M^±IL + csc-i V^^^T^.
Ve2^+ 2e*- 1
85. y = ^ Vl - x2 _ A _ ?!\ tan-i Vl _ x2.
86. 2/ = -^ log , ^ + log(x + Vl + x2).
2V2 V2 + 2x2 + x
87. y = (sin V^)'*°^. 90. y = (x)««.
88. y = </^nx. 91- ^ = (e)=^.
89. y=x(^). 92. y = (o + x)*^* (»+*>.
Find — in each of the following cases :
dx
93. x!' + secxy = 0. 97. e^ sin y - ei' cos x = 0.
94. ytan-ix-2/2 + x2 = 0. ^8. y sin x + x cos 2/ = x2^.
95. ysinx — cos (x — y) = 0,
99. 7/ logx = xsiny.
ni> 100. X2/ = tan-i-.
96. ye"v = ox™. '^ y
PROBLEMS 299
1-x , 1+x
„. , dy d'hj dhj
Find -^, -4, -4 m each of the following
dz dx^ dx^ ^
101. log (x2 + 2/2) - 2 tan-i ?^ = 0. 103. log ^— ^ - log i^t^ = 1.
X l+y l-y
102. e- + ev = l. 104. x-2/ = log(a; + y).
105, &^+y = x«.
106. At what points is the curve j/ = sin x + sin 2 x parallel to the axis of x ?
107. What value must be assigned to m that the curve y = h tan-'(x+m)
bmx
may be parallel to OX at the point the abscissa of which is 1 ?
108. Find the angle of intersection of the curves 3/ = sinx and y = cosx.
109. Find the angle of intersection of the curves y = sin x and y = sin (x + a).
110. Find the angle of intersection of the curves y = sinx and y = sin2x.
111. Show that the portion of the tangent to the curve
o, a + Va2 - x2
y = - log ===: — V a^ — x2
2 a - Va2 - x2
included between the point of contact and the axis of y is constant. (From this
property the curve is called the tractrix.)
112. Find the points of inflection of the curve y = 2 sin x — ^ sin 2 x.
113. Find the points of inflection of the curve xy = a^log--
114. Find the points of inflection of the curve y = e-^
115. Prove that the curve
y = ^x — ^sinx + j^ 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.
116. Plot the curve y = sin^x, finding maxima and minima, and points of
inflection.
117. Plot the curve y = e-«^cos6z, and prove that it is tangent to the curve
y = g-oa: wherever they have a point in common. Find maxima and minima and
points of inflection of this curve when a = 6 = 1.
118. Plot the curve y = x"e-^ (n > 0), finding maxima and minima and points
of inflection.
119. A body moves in a plane so that x = a cost + b, y = a sint + c, where
t denotes time and a, 6, and c are constants. Find the path of the body, and
show that its velocity is constant.
120. A rectilinear motion is expressed by the equation s = 5 — 2 cos' t. Show
that the motion is a simple harmonic motion, and express the velocity and the
acceleration at any point in terms of s.
300 ELEMENTARY TRANSCEKDE:N^TAL FUNCTIOXS
121. A, the center of one circle, is on a second circle witli center at B.
A moving straiglit line, AMN, intersecting the two circles at M and N respec-
tively, has constant angular velocity about A. Prove that BN has constant
angular velocity about B.
122. Two particles are moving on the same straight line, and their dis-
tances from the fixed point O on the line at any time t are respectively
X = a cos ut and x' = a cos lut-] — j, w and a being constants. Find the greatest
distance between them. ^ '
123. A ladder b ft. long leans against a side of a house. Its foot is drawn
away in the horizontal direction at the rate of a ft. per second. How fast is its
center moving ?
124. If a particle moves so that
s = e- 2 «^' (a sin ht + b cos hi),
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.
125. If s = ae*"' -|- be-'-', 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.
126. BC is a rod a ft. long, connected with a piston rod at C, and at B with
a crank AB b ft. long, revolving about A. Find Cs velocity in terms of AB's
angular velocity.
127. A man walks along the diameter, 200 ft, in length, of a semicircular
courtyai-d 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 ? .
128. How fast is the shadow in the preceding problem moving if the sun's
rays make an angle a with the diameter ?
129. Given that two sides and the included angle of a triangle have at a
certain moment the values 6 ft., 10 ft., and 30° respectively, and that these
quantities are changing at the rates of 3 ft., — 2 ft., and 10° per second respec-
tively, what is the area of the triangle at the given moment, and how fast is it
changing ?
130. One side of a triangle is I ft., and the opposite angle is a. Find the
other angles of the triangle when its area is a maximum.
131. A tablet 8 ft. high is placed on a wall .so that the bottom of the tablet
is 20 ft. from the ground. How far from the wall should a person stand in order
that he may see the tablet at the best advantage, i.e. in order that the angle
between the lines from the observer's standpoint to the top and the bottom of
the tablet may be the greatest ?
132. 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 ?
PROBLEMS 301
133. 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 4 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 ?
134. 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
that the edge of the table may be most brilliantly lighted, given that the illumi-
nation varies invereely as the square of the distance and directly as the cosine
of the angle of incidence '?
135. 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 hori-
zontal width of the girder is neglected, how wide must the passageway be in
order that the girder may go around the corner ?
136. Find the area of an arch of the curve tj = sin x.
137. Find the area bounded by the axis of y and the portion of the curves
?/ = sin X, y = cos x, lying between x = 0 and x = tt.
138. Find the area bounded by the portions of the curves y =: ^ sin 2 x and
?/ = sin X + ^ sin 2 x that extend between x =^ 0 and x — tt.
139. Find the area between the curve y = e^, the axis of x, and the ordinates
X = 0 and x = 1.
140. Find the area bounded by the axis of x, the catenary, and the ordinates
X = ± a.
141. Find the area bounded by the axis of x, the curve y = -, and the
ordinates x = 1 and x = 2. ^
142. Find where the ordinate of the witch should be drawn in order that
the area between that ordinate, the witch, the axis of y and the axis of x should
be equal to the area of the circle used in the definition.
143. Show that for the catenary — = -(e" + e~«), and thence find an
expression for the length of s.
144. Find the curve the slope of which at any point is k times the reciprocal
of the abscissa of the point, and which passes through (2, 3).
145.. Find the curve the slope of which at any point is k times the ordinate
of the point, and which passes through the point {a, h).
146. Find the space traversed by a moving body in the time t if its velocity
is proportional to the distance traveled.
Solve the following equations :
147. tanx = cosx. 152. tanx = x.
148. cos2x = Jcosx. 153. tanx=^x.
149. sin 2 tf cos 2 fl 4- 2 sin ^ = 0. 154. x - ^ .sin x = ■^^.
150. sin4x - 2sinxcos2x = 0. 155. e^=«2.
151. sin*x+3cos*x-4sin2xcos2x = 0. 156. logx = ^x.
CHAPTEE XIV
PARAMETRIC REPRESENTATION OF CURVES
163. Definition. Thus far we have considered a curve as
represented by a single equation connecting x and y. Another
useful method is to express x and y each as a function of a
third independent variable ; thus :
where t is an independent variable and f^{f) and f^{t) are
continuous functions of t. As t varies, x and y also vary,
and the point {x, y) traces out a curve. By eliminating t be-
tween the two equations the curve may often be expressed
by a single equation between
X and y.
164. The straight line.
Let Pi{x^, 2/1) (fig- 169) be a
fixed point on a straight line
and <^ be the angle which
the line makes with a line
P^R parallel to OX. Let
P{x, y) be any point on
the line, and r the distance
from P^ to P, where r is
positive when P is on the
terminal line of ^, and negative when P^ is on the backward exten-
sion of the terminal line. Then, for all possible positions of P
Fig. 169
x — x^ y — y^
— -— = cos</), — -^ = sin<^;
whence x = x^-\-r cos ^, y = y^-\-r sin <^.
302
THE CIRCLE AND THE ELLIPSE
303
This is a parametric representation of the straight line, where
r is the arbitrary parameter. Illustrations of the use of these
equations have been given in §§136 and 138.
Another parametric representation of a straight line is furnished
by the equations of § 19,
X = X^ -f- fr \^^ ~~ '^i/i
where I is the parameter and {x^, y^ and {x^, y^ are fixed points.
More generally, the equations
x = a-\-ht, y=zf + gt,
where a, b, f, g, are constants, and t is an arbitrary parameter,
represent a straight line. For these equations are equivalent to
165. The circle. Let
P{x, y) (fig. 170) be any
point on a circle with its
center at the origin 0, and
its radius equal to a. Let ^
be the angle made by OP
and OX. Then from the defi-
nition of the sine and cosine
x = a cos<f>,
y — d sin^,
Fig. 170
the parametric equations of the circle with <^ as the arbitrary
parameter.
166. The ellipse. Take the ellipse
^ + ^ = 1
(a,>h)
and on its major axis as a diameter construct a circle. Take
F{x, y) (fig. 171) any point on the ellipse, draw the ordinate MF
304 PARAMETRIC REPRESENTATION OF CURVES
and prolong it imtil it meets the circle in Q. Call the coordinates
of Q {x, y'). Then from the equation of the circle
y
'=V^^
and from the equation of
the ellipse
Fig. 171
^ a
^ Hence y == — y'.
Draw the line OQ, mak-
ing the angle XOQ = <f).
Then, as in § 165,
x = a cos <^,
y' = a sin cf).
By substituting for y' its value in terms of y, we have
X — a cos (f>, y = ^ sii^ ^>
the parametric equations of the ellipse.
^ is called the eccentric angle of a point on the ellipse, and
the circle x^+'i^=o?' is called the auxiliary circle.
Ex. The parametric equations of an ellipse may be used to find its area.
For if A is the area bounded by the ellipse, the axis of y, the axis of x, and
any ordinate MP (fig. 171), then (6, § 109)
dA
— = V-
dx
dA dA
dA _d^_ d<t>
dx dx — asinc^
But
(1)
and y = 6 sin <^,
Therefore (1) is equivalent to
dA
d(j)
Hence A
= abf
ah sin2 (p = ab
sin 2 0 <f>
cos 2 (^ — 1
+ c.
THE CYCLOID
305
When <6 = - , ^ = 0 ; hence c = .
2' 4
Therefore
A = ab(^^
2 4/
When (f> = 0, A is one fourth the area of the ellipse. Therefore the whole
area of the ellipse equals Trab.
167. 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. 172) be its center at any time of its motion, N its point of
contact with OX, and P the point on its circumference which
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
OJ:i= arc FK
Draw 3fP and CJSF each perpendicular to OX, PR parallel to
OX, and connect C and P. Let
NOP = 0.
Then x = OM=ON- MN
= arc NP - PR
= a(j) — a siu </>.
y = MP = NC — RC
= a — a cos (f).
Hence the parametric representation of the cycloid is
x= a(<j) — sin <f)),
y = a(l— cos<f>).
306 PARAMETRIC REPRESENTATION OF CURVES
By eliminating <j> the equation of the cycloid may be written
x== a cos
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 2 rra.
168. The trochoid. When a circle rolls upon a straight Une,
any poiat upon a radius, or upon a radius produced, describes a
curve called a trochoid.
M ^
Fig. 173
Let the circle roll upon the axis of x, and let C (figs, 173 and
174) be its center at any time, N its point of contact with the
axis of X, F(x, y) the point which describes the trochoid, and
Fig. 174
K the point in which the liae CP meets the circle. Take as
the origin 0 the point found by rolHng the circle toward the
left until K is on the axis of x. Then
ON^iixcNK.
THE EPICYCLOID
307
Draw PM and CN perpendicular to OX, and through P a line
parallel to OX, meeting CN or CN produced, in R. Let the radius
of the circle be a, CP be h, and NCP be <^. Then
x=^OM = ON-MN
= sivc NK-PB
= a^ — h sin <f).
y = MP = NC-BC
= a — h cos <^.
169. 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.
r
Fig. 175
Let 0 (fig. 175) be the center of the fixed circle, C the center of
the rolling circle, N its point of contact with the fixed circle, and
308 PAKAMETRIC REPRESENTATION OF CURVES
P{x, 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 KN = arc NP.
Take 0 as the origin of coordinates, and OK as the axis of x.
Draw PM and CL perpendicular to OX, PS parallel to OX, meet-
ing CL in R, and connect O and C. Let the radius of the rolling
circle be a, that of the fixed circle h, and denote the angle OOP
by e, the angle KOC by 4>. Then
whence
arc KN = h(f), arc NP = aO :
We now have x = Oil = OL-h LM
= OC cos KOC - CP cos SPC
= (a -i-h) cos <f> — a cos (^ + ^)
= (a + 6) cos 9 — a cos 9.
y = MP = LC-RC
= OC sin KOC - CP sin ^PC
= (a + &) sin ^ — a sin (^ + ^)
/ , 7,\ • ji . a + b
= (a + 0) sm 9 — a sin 9.
The curve consists of a number of congruent arches the first of
which corresponds to values of 0 between 0 and 2 ir, tliat is, to-
2(177
values of ^ between 0 and
Similarly the Xth arch corre-
sponds to values of <j> between — — -— ^ — and — Hence
0 b
the curve is a closed curve when, and only when, for some value
of k, ^^ — is a multiple of 2 tt. If a and b are incommensurable,
0
CL 7) 10
this is impossible, but if - = — > where — is a rational fraction in
b q q
its lowest terms, the smallest value of Tc = q. The curve then con-
sists of q arches and wiuds p times around the fixed circle.
THE HYPOCYCLOID
309
170. 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 equations of the hypocycloid are
x = (b — a) cos ^-\- a cos rf>,
a
y = (b — a) sin <f)— a sin (f>.
The proof is left to the student. The curve is shown in fig. 176.
Fig. 176
171. Epitrochoid and hypotrochoid. The epitrochoid and
hypotrochoid are generated by the motion of any point on
the radius of a circle which rolls upon the outside or the
310 PARAMETRIC REPRESENTATION OF CURVES
inside of a fixed circle. If h is the distance of the generating
point from the center of the moving circle, and the notation
is otherwise the same as in the previous articles, the equations
of the epitrochoid are
x = {a-\-h) cos (f> — h cos (f>,
y = (a + b) sin.(f) — h sm c^i,
a
Fig. 177
and of the hypotrochoid are
T
x = (b~ a) cos <f) + h cos (f>,
y = (h — a) sin d> — h sin 6.
a
THE EPITROCHOID
311
The proofs are left to the student. The curves are shown in
figs. 177, 178, and 179, 180 respectively.
Fig. 178
172. The involute of the circle. If a string, kept taut, is
unwound from the circumference of a circle, its extremity
describes a curve called the involute of the circle. Let 0 (fig. 181),
be the center of the circle, a its radius, and A the point at which
the extremity of the string is on the circle. Take 0 as the origin
of coordinates and OA as the axis of x. Let P {x, y) be a point on
the involute, PK the line drawn from P tangent to the circle at
K, and ^ the angle XOK. Then PK represents a portion of the
unwinding string, and hence
KP = Q.VC AK = a<^.
THE INVOLUTE OF THE CIRCLE
313
Now it is clear that for all positions of the point K, OK makes
TT
an angle </> — -^ with 0 Y. Hence the projection of OK on OX is
always OK cos^= a cos^, and its projection on OY is OK cos
7r\ . TT
^ — — I = « sin<^. Also KP always makes an angle <f>~ ^ with
(
Fig. 181
OX and ir — (j> with 0 Y. Hence the projection of KP on OX
is KP cos(<^ — — I = a<^ sin^, and its projection on OF is KP
cos (tt — (/>) = — a^ cos<f>. The projection of OP on OX is x, and
upon OY is 2/. Hence, by the law of projections, § 15,
X = a cos^ + a<f> sin(f>,
y = a sin ^ — ci<^ cos ^.
173. Time as the arbitrary parameter. An important use of the
parametric representation of curves occurs in mechanics in find-
ing the path of a moving point acted on by known forces. Here
the independent parameter is usually the time.
314 PARAMETRIC REPRESENTATION OF CURVES
Ex. 1. A particle moves "in a circle with uniform velocity, k. Then, if s
represents the arc traversed, and a the radius of the circle,
s = kt and d> = - = — .
a a
Therefore the equations of the circle are (§ 165),
U
x = a cos — ,
a
. kt
w = a sin — .
a
This shows that the projections of P on the coordinate axes have simple
harmonic motions of the same amplitude.
Ex. 2. A particle Q moves with uniform velocity along the auxiliary circle
of an ellipse (§ 166) ; required the motion of its accompanying point, P.
kt
As in Ex. 1, 0 = — . Hence the equations of the path are
kt
X = a cos — ,
a
^ . kt
y = 0 sm — ,
a
showing that the projections of P upon OX and OY have simple harmonic
motion of amplitudes a and b respectively.
Ex. 3. A projectile is shot with an initial velocity vq in an initial direction
which makes an angle a with the horizontal direction. Then the initial com-
ponent of velocity in the horizontal direction is vo cos a and in the vertical
direction is Vq sin a. If the resistance of the air is neglected, the only force
acting on the projectile is that of gravity. Hence if we take the origin at the
initial position of the projectile, and the axis of x horizontal, we have
¥'="'
d^y _
dt^ ^'
which give
X = Cit -r C2,
But when t =
= 0, we have
X = 0,
y =
= 0.
y = -yt^ + Cst + €4.
dx , dy
, — = Vo cos a, and -^ :
dt dt
Vo sm a.
Hence the parametric equations of the path of the projectile are
X = Vot cos a,
y — Vot sin or — J gt^.
Eliminating t from these equations, we have
y = X tan a —
2 Vq cos2 a
or 2 Vq y cos^ a = 2 Vo^x sin a cos a — gx^
which shows that the curve is a parabola.
THE DERIVATIVES 315
174. The derivatives. When a curve is defined by the equations
dy
(1)
dt
Ex. For the cycloid q
X = a{<(> — sin^),
we have, by (8). §96. | = |
y = a(l — cos^y,
dy
dy d4> a sin <t>
-cot*^
dx dx a(l— COS0)
2
d<p
Now -p is the tangent of the angle
made by the tangent with the axis of x.
Therefore this angle is •
* 2 2
From this follows a simple construction of the tangent and normal. For if
the line NC (fig. 182) is prolonged until it cuts the circle in Q, and PQ and PN
are drawn, the angle CQP — - • Hence PQ makes the angle with OX
and is therefore the tangent. Pl>i^ being perpendicular to PQ, is the normal.
If it is required to find -r-^' we may proceed as follows :
d (dy
d?y d {dy\ _ dt \dx
da? dx \dx/ dx
di
(2)
Ex. For the cycloid
dy _
dx~
cot-,
2
dx
d(p ~
a(l — cos(^) =
2 0 sin2 -
2
d-^y
— cosec2 -
2 2
2asin2-
2
1
dx2
4 a sin*
2
31G PARAMETRIC REPRESENTATION OF CURVES
Formula (2) may be expanded as follows:
d (dy
d^y dx dj^x dy
~dt~~d¥'di
dxV
dt)
dy\_d Idt \ _ df
dt \dx) dt\dx\
\dil
^y dx dj^x dy
cPy 'de'di~~dell
dj? /dxV
\dt}
/dsV
By multiplying equation (3), § 105, by ( — ) > we have
\dt/
dsY_/dxY /dyV
dt) ~ \dt/ \dt
(3)
(4)
175. Application to locus problems. In finding the Cartesian
equation of the locus of a point which satisfies a given condition,
it is often convenient to employ the principles of parametric rep-
resentation ; for by fixing the attention upon a single point of
the required locus, it is frequently possible to express its coordi-
nates in terms of a single parameter. The required equation is
then found by eliminating the parameter.
Ex. 1. Locus of the point of inter-
section of perpendicular tangents to a
parabola.
Let the parabola be y^ = i px
(fig. 183), and let the equation of
any tangent to it be written (§ 88)
y = mx +
P
(1)
If m is replaced by
110
y=( )x + _L.
\ mj 1
, we have
Fig. 183
X
mm,
m
(2)
as the equation of a tangent perpendicular to (1). Therefore, if P(x, y) is the
point of intersection of (1) and (2), P is any point of the locus.
LOCUS PROBLEMS
317
Solving (1) and (2), we find
and
X = — p
(S)
(4)
which are the parametric representations of the locus, the parameter evidently
being m. But for all points of the locus x = — p, and (3) is the Cartesian equation
of the locus. It is to be noted that in this example the elimination of the param-
eter is unnecessary, since one of the equations does not contain it.
As (3) is the equation of the directrix, we have the proposition : Perpendicular
tangents to a parabola meet on the directrix.
Ex. 2. Locus of the point of
intersection of perpendicular tan-
gents to an ellipse.
Let the ellipse be 1 —
(fig. 184), and let the equation of
any tangent to it be written (§ 88)
y = mx ± Va^m'^ + 6^.
Then the equation of a tangent
perpendicular to (1) will be
y = --±
rn
Ia2_
\m2
+ 62,
and P{x, y), the point of inter-
section of (1) and (2), will be any
point of the locus.
Solving (1) and (2), we find
Fig. 184
± m r-v/^ + 62 _ Va2m2 + 62!
y =
»n2 + l
\m^
-t
+ 62 + Va2»i-
+ 62
m2-M
(8)
(4)
as the parametric representations of the locus in terms of the parameter m.
To eliminate m, we square the respective values of x and y and add, the
result being
x2 + 2/2 = a^ + 62. (5)
The locus is seen to be a circle concentric with the ellipse and having its radius
equal to the chord joining the ends of the major and the minor axes of the ellipse.
While (3) and (4) form the explicit parametric representation of the locus, x
and y being expressed explicitly in terms of the parameter m, <{1) and (2) may
be regarded as the implicit parametric representation of the locus, for x and y,
the coordinates of any point of the locus, are expressed implicitly in terms of m.
318 PARAMETRIC REPRESENTATION OF CURVES
From this point of view it is evident that we may eliminate m directly from
(1) and (2) to find the Cartesian equation of the locus. Accordingly we write
(1) and (2) in the forms
y — mx = ± Va%i2 + 62,
my + x — ± Va2 + b^m^,
and square and add, the result being
(1 + m2) (X2 + 2/2) = (1 + ^2) (o2 + 62)^
or (1 + m2) (x2 + ^2 _ a2 _ 52) = 0.
As 1 + m' cannot be zero, since by hypothesis m must be real, we may
cancel out this factor. The result,
x2 + 2/2 _ a2 _ 62 = 0,
is the same as that found by the previous method.
Ex. 3. Lociis of the foot of the perpendicular from the focus of a parabola to
any tangent.
Let the parabola be 2/2 = 4px (fig. 185), and let
P
y = mx +
(1)
be any tangent. Then the perpendic-
ular to the tangent from the focus is
y = -l^{x-p). (2)
Their point of intersection, P(x, y),
is any point of the locus.
Solving (1) and (2), we find
Fig. 185
and
x = 0
y
(8)
(4)
The locus is therefore x = 0, the tangent at the vertex of the parabola.
If we proceed from the implicit parametric representation, we may elimi-
nate the parameter m by substituting in (1) its value found from ('2). The
result is x[2/2 + (p — x)2] = 0, which breaks up into two equations, i.e.
X = 0, y^ + (x — p)2 =0. As the last equation represents a single pointj it is
evident by the geometry of the problem that the required equation is x = 0,
as was found by the other method.
We see then that when we eliminate the parameter from the equations
expressing x and y in terms of it, we must examine our result carefully to
be sure that no extraneous factor is left in it.
LOCUS PKOBLEMS
319
Ex. 4. Locus of the foot of the perpendicular from the vertex of a parabola
to any tangent.
Let the parabola be y'^ = 'ipx (fig. 186), and
y = 7nx +
(1)
be any tangent. Then the perpen-
dicular to (1) from the vertex is
y = X.
m
Solving (1) and (2), we find
-P
y =
m2 + l
P
(2)
(3)
(4)
m (m2 + 1)
as the explicit parametric represen-
tation of the locus. The Cartesian
equation of the locus is most readily-
found by substituting in (1) the value of m from (2), and reducing. The result is
x8
Fig. 186
y^ = -
(6)
p + x
which is the equation of a cissoid (§ 83) situated on the negative axis of x.
The last two loci are special examples of pedal curves, i.e. loci of the feet of
perpendiculars drawn from any chosen fixed point to tangents to a given curve.
176. In the examples of the last article the parametric repre-
sentation of the locus was in terms of a single parameter. In
the examples of this article the parametric representation, whether
implicit or explicit, is in terms of two parameters, which are not
independent, however, since they are connected by a single equa-
tion. The problem of finding the Cartesian equation of the locus
is, then, the elimination of two parameters from three equations.
Ex. 1. Through the vertex of a parabola a line is drawn perpendicular to
any tangent. Required the locus of the intersection of this line and the ordinate
through the point of contact of the tangent.
Let Pi(xi, 2/i) be any point of the parabola y^ = ipx (fig. 187), PiT the
tangent at Pi, and OT the perpendicular to PiT from the vertex 0. Then the
equation of PiT is
yiy = 2p{x + xi), (1)
2/1
and the equation of 0 T is
y = - —X.
2p
The equation of the ordinate MiPi through Pi is
X = Xi.
(2)
(8)
320 PARAMETRIC REPRESENTATION OF CURVES
If P(x, y) is the point of intersection of (2) and (3), P is any point of the
locus, and (2) and (3) form the implicit parametric representation of the locus
in terms of the parameters Xi and yi. Since Pi (Xi, yi) is by hypothesis any
point of the parabola, its coordinates satisfy the equation of the parabola, and
the parameters Xi and yi satisfy the equation
y^ = 4pxi. (4)
Fig. 187
Solving (2) and (3) for Xi and yi and substituting their values in (4), we
thereby eliminate them and have, as the Cartesian equation of the locus.
2/2 = _ x8.
P
(5)
From the form of the equation tlie locus is seen to be a semicubical parabola.
It may be added that the explicit parametric representation of the locus is
- Vi^i
2p
readily found to be x = Xj and y =
, where y^ - 4pxi.
LOCUS PROBLEMS
321
Ex. 2. Locus of the middle points of chords of an ellipse, drawn through one
end of its major axis.
Let the ellipse be 1 = 1 (fig. 188), and Pi(a;i, yi) be any point of the
a^ b^
ellipse. Then APi is any chord through A, and P(x, y), its middle point, is
any point of the required locus. Since the coordinates of A are (a, 0), by § 18
xi + a
and
y =
y\
Then (1) and (2) are the explicit para-
metric representations of the locus in
terms of the parametere Xi and y^
which satisfy the equation
(3)
— + — = 1,
02 62
Fig, 188
since Pi is any point of the ellipse.
To find the Cartesian equation of the locus, we substitute in (3) the values of
Xi and yi from (1) and (2). The result is
Accordingly the locus is an ellipse with its center at ( -, 0 ) and its semiaxes
equal respectively to - and - ■
Ex. 3. Locus of the point of intersection of tangents at the ends of conjugate
diameters of an ellipse.
"~^
^/-"""T^:^--
^
\(i)
y^ ^icC"^
^
^
>^^r^\
0-
^
J J
\s,^^ Jir-——— __
-^
-^^2 ^y
^--^
X
Fig. 189
X2 V^
Let the ellipse be 1-^ = 1 (fig. 189), and OAi and OBi be any two con-
a2 62 / 6 \
jugate diameters. If Ax is (xi, yj), -Bi is ( ^, — ^-\ by ICx. 2, § 146.
\ 6 a J
322 PARAMETRIC REPRESENTATION OF CURVES
Then the tangents at Ai and Bx will be respectively
and
where
Solving (1) and (2), we find z =
y =
62
ab ^ ab~ '
a2 62
6xi — ayi
bxi + ayi
(1)
(2)
(3)
(4)
(6)
as the explicit parametric representations of the locus.
If we write (4) and (5) in forms bz — bxi — ayi and ay = bxi + ayi respec-
tively, and square and add, we have
62a;2 + a22/2 = 2 (62x2 + a^^),
or 62x2 + a2y2 = 2 a262, (6)
by virtue of (3).
y^
As (6) may be written
+
= 1, we see that the required locus is
(a V2)2 (6 V2)2
an ellipse, concentric with the given ellipse and with the semiaxes a V2 and 6 V2.
Ex. 4. P1P2 is any chord of an ellipse perpendicular to its major axis A\Ai.
Find the locus of point of intersection of AiPi and A2P2.
Fig. 190
x2 y^
Let the ellipse be 1- — = 1 (fig- 190), and the coordinates of Pi and P2 be
a2 62
respectively (xi, yi) and (xi, — yi). Then the equation of AiPi and A2P2 are
respectively „
y
xi + a
Vi
a — xi
{x + a),
(z - a).
(1)
(2)
PROBLEMS 323
which are accordingly the implicit parametric representation of the locus. The
parameters Xi and yx satisfy the equation
^ + g = l. (8)
Taking the product of (1) and (2), we have
y^ = -lL^{x^-a\ (4)
which may be written y"^ — — (x2 — dF), (6)
a2
by virtue of (3).
As (6) may
hyperbola concentric with the ellipse and having the same semiaxes.
As (6) may be written = 1, we see that the required locus is an
a2 62
PROBLEMS
1. Show that X = pt^, y = 2pt are parametric equations of the parabola.
2. Find the equations of the tangent and the normal to the parabola when
the equations of the parabola are as in problem 1.
3. Find the parametric equations of the parabola when the parameter is
the slope of a line through the vertex.
4. Find the equations of the tangent and the normal to a parabola when
the equations of the curve are as in problem 3.
5. Find the parametric equations of the ellipse when the parameter is the
slope of a straight line through the center.
6. Find the parametric equations of the ellipse when the parameter is the
slope of a straight line through the left-hand vertex.
7. Find the parametric equations of the cissoid when the parameter is the
angle AOP (fig. 91).
8. Show that x = t, y — are parametric equations of the witch.
a2 + <2
2 (I 2 (I
9. Show that x = , y — — are parametric equations of the
\^t^ <(l + «2) ^ ^
cissoid. What is the geometric significance of i ?
10. Find the equation of the tangent to the cissoid if the equations of the
curve are as in problem 9.
324 PARAMETRIC REPRESENTATION^ OF CURVES
Find the Cartesian equations of each of the following curves :
. , o <2 + 2 a t^ + 2t
11. X = - ■ , y = ~ ■ — .
2 P+\ 2 i^ + 1
12. X — , y =
\-^t^ 1 + «3
,o *:^ , kH
l«j. x = a-\ -^, y = at +
a(l + <2) a(l + «2)
,. l + 2« l + <
14. x = , y =
15. X = -, y
t-1 t^-1
.„ e' + e-' e' — C-'
16, X = , y = .
{e'-2e-'y^ {e'-2e-'f
17. x = t^ + St + 2^ y = t^-l.
, o ct ct^
^°- ^ = '. \ ; ;r' y
(a + U) (1 + «2) " (a + 6t) (1 + «2)
19. Eliminate < from
X _ cos < — sin < y _ cos < + sin ^
a e' a e'
and prove that the curve represented is a logarithmic .spiral (§ 178).
20. Let 0 be the center of a circle with radius a, ^ a fixed point, 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 para-
metric form, using the angle AOB as the arbitrary parameter, OA as the axis
of a;, and 0 as the origin. Also find the Cartesian equation of the locus.
21. OBCD is a rectangle with OB = a and BC = c. Any line is drawn
through C, meeting OB in E, and the triangle EPO is constructed so that the
angles CEP and EPO are right angles. Find the parametric equations of the
locus of P, using the angle DOP as the parameter, OB as the axis of x, and 0
as the origin. Find also the Cartesian equation of the locus.
22. Let AB be a given line, 0 a given point, a units from AB, and k a
given constant. On any line through 0, meeting AB in M, take P so that
OM ■ MP = A:2. Find the parametric equations of the locus of P, using 0 as the
origin, the perpendicular from O to ^Z? as the axis of x, and the angle between
OX and OP as the parameter. Also find the Cartesian equation.
23. A and B are two points on the axis of ?/ at a distance — a and 4- a
respectively from the origin. AH ifi any line through A meeting the axis of x
at H. BK is the perpendicular from B on AH, meeting it at K. Tiirough K a
line is drawn parallel to the axis of x and through H a line is drawn parallel to
the axis of y. These lines meet in P. P"ind the parametric equations of the locua
of P, using the angle BAK as the parameter. Also find the Cartesian equation.
PROBLEMS 325
24. Let OA be the diameter of a fixed circle and LK tlie tangent at A.
From 0 draw any line intersecting the circle at B and LK at C, and let P be
the middle point of BC. Find the parametric equations of the locus of P,
using the angle A OP as the parameter, OA as the axis of y, and O as the origin.
Find also the Cartesian equation.
25. Show that the tangent to the ellipse at any point and the tangent to the
auxiliary cii'cle at the corresponding point pass through the same point of the
major axis.
26. Prove that the eccentric angles of the ends of a pair of conjugate
diameters of an ellipse differ by — •
27. Show that the perpendicular from either focus upon the tangent at any
point of the auxiliary circle of an ellipse equals the focal distance of the corre-
sponding point of the ellipse.
28. Q is the point on the auxiliary circle of the ellipse, corresponding to
the point P of the ellipse. The straight line through P parallel to OQ meets
OX at L and OY at M. Prove PL = b, and PM = a.
29. Find the equation of the tangent at any point of an ellipse in terms of
the eccentric angle at that point.
30. What elevation must be given to a gun to obtain the maximum range
on a horizontal line passing through the muzzle of the gun ? (In this and the
following examples the resistance of the air and the effect of all forces except
gravity are neglected.)
31. What elevation must be given to a gun to obtain a maximum range on
an oblique line passing through the muzzle of the gun and making an angle ^
with the horizontal ?
32. What elevation must be given to a gun that the projectile should pass
through a point in the horizontal line passing through the muzzle and 6 units
from it ?
33. A gun stands on a cliff h units above the water. What elevation must
be given to the gun that the projectile may strike a point Tn the water b units
from the base of the cliff ?
34. Find the parametric equations of the curve described by any point in
the connecting rod of a steam engine.
35. 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 ?
36. Show that the hypocycloid generated when the rolling circle has J the
radius of the fixed circle has the Cartesian equation x* -f- y^ = 6'.
37. 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.
326 PARAMETRIC REPRESENTATION OF CURVES
38. If a wheel rolls with constant angular velocity on the circumference
of a fixed wheel, find the velocity of any point on its circumference and on its
spoke.
39. 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 dis-
tance from the ground is lialf the radius of the wheel.
40. If a string is unwound from a circle with constant velocity, find the
velocity of the end in the path described.
41. AB and CD are perpendicular diameters of a circle of radius E.
AM is a chord of the circle, rotating about A so that the angle BAM varies
uniformly. AM is extended to N so that J!f^=the chord MB. Find the
path of N, the velocity of N in its path, and the components of the velocity
respectively parallel to AB and CD.
42. 0, CK, 0" are three points on a straight line and 0"(y = ^0(7. LK is
drawn through C pei-pendicular to 00" ^ and any point M is taken on LK.
From M a straight line is drawn perpendicular to 0"Jf, and through 0 a
straight line is drawn parallel to 0"M. These lines intersect in P. Required
the locus of P.
43. 0 is a fixed point and LK a fixed straight line. Any point M is taken
on LK, and the line OM is drawn and prolonged to P so that OM ■ OP = k^,
where A; is a constant. Find the locus of P.
44. Show that the locus of points symmetrical to the vertex of a parabola
with respect to its tangent lines is a cissoid.
45. Let OA be the diameter of any circle and LK the tangent at A,
Through 0 draw any line intersecting the circle in D and LK in E. Lay off on
OE produced the distance EP = OD, and find the locus of P.
46. Let a circle with center at 0 intersect the axis of y at J. and the axis
of X at C. Take two points G and E on the circle equidistant from A. If the
ordinate of G intersects the line CE in P, prove that the locus of P is a cissoid.
47. From a point a units from the axis of x lines are drawn to OX, and
from the point where each line meets the axis a line of tlie same length is
drawn at right angles to the first line. Find the equation of the locus of the
end of this last line.
48. OA is a diameter of a circle and LK the tangent at A. Through 0 any
line is drawn meeting the circle in B and LK in C. Through B a line is drawn
perpendicular to OA and meeting it in M. Finally MB is prolonged to P so
that MP = AC. Find the locus of P.
49. Find the path described by any point of a tangent line which rolls upon
a circle without slipping.
50. CD is perpendicular to OX and distant a units from O. Through A , any
point on CD, a straight line OA is drawn, and from A a perpendicular is drawn
PEOBLEMS 327
to OA, intersecting OX at B. From B a straight line is drawn parallel to OY,
intersecting OA at P. If m denotes the slope of OA, find the parametric and
the Cartesian equations of the locus of P,
51. Prove that the pedal of a parabola with respect to any point is a cubic
curve which passes through that point.
52. Prove that the pedal of the ellipse \- — = \ with respect to the center
is the curve (x2 + y2)2 = a'^n^p. + 52^2. <^^ ^
53. A line of constant length k moves with its extremities on the two axes
of coordinates. Find the locus described by any point of the line.
54. A straight line has its extremities on the coordinate axes and passes
through a fixed point. Find the locus of its middle point.
55. If the ordinate NP of an hyperbola be produced to Q, so that NQ, = FP,
find the locus of Q.
56. Find the locus of the points of intersection of normals at corresponding
points of the ellipse and the auxiliary circle.
57. P is any point of a parabola, A the vertex, and through A a straight
line is drawn perpendicular to the tangent at P. Find the locus of the point of
intersection of this line with the diameter through P, and also the locus of the
point of intersection of this line with the ordinate through P.
58. Two equal parabolas have their axes parallel and a common tangent at
their vertices, and straight lines are drawn parallel to the axes. Show that the
locus of the middle points of the parts of the lines intercepted between the
curves is an equal parabola.
59. Find the locus of the intersection of the ordinate, produced if necessary,
of any point on an ellipse with the perpendicular from the center upon the
tangent at that point.
60. Two parabolas have the same axis, and tangents are drawn from points
on the first to the second. Prove that the middle points of the chords of con-
tact with the second lie on a parabola.
61. Chords of an ellipse are passed through a fixed point. Find the locus of
their middle points.
62. From a point P on an ellipse straight lines are drawn to the vertices A
and A\ and from A and A' straight lines are drawn perpendicular to AP and
A'P. Show that the locus of their point of intersection is an ellipse.
63. Show that the locus of the point of intersection of two tangents to a
parabola, the ordinates of the points of contact of which are in a constant
ratio, is a parabola.
64. If the tangent to the parabola y^ = 4px meets the axis at T and the
tangent at the vertex A at J5, and the rectangle TABQ is completed, show
that the locus of Q is the parabola y^ + px = 0.
328 PARAMETRIC REPRESENTATION OF CURVES
65. Find the locus of the feet of the perpendiculars from the focus to the
normals of the parabola y^ = 4px.
66. Show that perpendicular normals to the pai'abola y- = 4px intersect on
the curve y'^ = px — Sp^.
67. Find the locus of the intersection of a pair of perpendicular tangents to
an hyperbola.
68. Two tangents to an ellipse are so drawn that the product of their slopes
is constant. Show that the locus of their point of intersection is an ellipse or
an hyperbola according as the product is negative or positive.
69. Prove that the locus of the point of intersection of two tangents to a
parabola is a straight line if the product of their slopes is constant.
70. Find the locus of the foot of the perpendicular from either focus of an
hyperbola to any tangent.
71. Let AB be the diameter of a circle and O its center. Let NQ be the
ordinate of a point Q on the circle and P another point of the circle, so related
to Q tjiat OP revolves uniformly from OA through a right angle in the same
time that QN travels at a unifoi-m rate from A to 0. If OP and QN intersect
in R, find the locus of R.
72. Find the equations of the cycloid when the tangent at its highest point
is the axis of x, the normal at the vertex is the axis of y, and the angle 0 is
the angle through which the radius has rotated after passing through the
highest point.
73. Prove that the area of an arch of the cycloid above the axis of x is
three times the area of the rolling circle.
74. Prove that for a cycloid — = 2 a sin - , and thence find its length from
. dd> 2
cusp to cusp. ^
75. Show that for an epicycloid — = 2(a + 6)sin — , and thence find its
length from cusp to cusp, ^
CHAPTEE XV
POLAR COORDINATES
177. 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 direction, as follows :
Let 0 (fig. 191), called the origin or foh, be a fixed point, and
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 Q are called the 'polar coordinates of the point P(r, 6),
and when given will completely determine P.
Fig. 191
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 raditis 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
O along the backward extension of the terminal line. It follows
that the same point may have more than one pair of coordinates.
320
330
POLAR COORDINATES
Thus (2, 195°), (2, - 165°), (- 2, 15°), and (- 2, - 345°) refer to
the same point. In practice it is usually convenient to restrict 6
to positive values.
Plotting in polar coordinates is facilitated by using paper ruled
as in figs. 192 and 193. The angle 6 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 num-
ber which indicates 6, according as r is positive or negative.
When an equation is given in polar coordinates the correspond-
ing curve may be plotted by giving to 6 convenient values, com-
puting the corresponding values of r, plotting the resulting points,
and drawing a curve through them.
Ex. 1. r = a costf.
a is a constant which may be given any convenient value. We may then find
from a table of natural cosines the value of r which corresponds to any value of 0.
165)
180
195
By plotting the points corresponding to values of 6 from 0° to 90° we obtain the
9,TcABC0 (fig. 192). Values of 6 from 90° to 180° give the arc ODEA. Values of
GRAPHS
331
e from 180° to 270° give again the arc ABCO, and those from 270° to 360° give the
arc ODEA. Values of d greater than 300° can clearly give no points not already-
found. The curve is a circle (§ 184).
Ex. 2. r = o sin 3 6.
As 0 increases from 0° to 30°, r increases from 0 to a ; as ^ increases from 30°
to 60°, r decreases from a to 0; the point P{r, 6) traces out the loop OAO (fig.
193). As e increases from 60° to 90°, r is negative and decreases from 0 to — a •,
5 if
as 6 increases from 90° to 120°, r increases from — a to 0 ; the point (r, 0) traces
out the loop OBO. As 6 increases from 120° to 180°, the point (r, 0) traces out the
loop OCO. Larger values of 0 give points already found, since sin 3 (180° + 0)
= — sin 3 $. The three loops are congruent because sin 3 (60° + 0) = — sin 3 0.
This curve is called a rose of three leaves.
178. The spirals. Polar coordinates are particularly well
adapted to represent certain curves called spirals, of which the
more important follow.
332 POLAR COORDINATES
Ex. 1. The spiral of Archimedes,
r = ae.
In plotting 6 is usually considered in circular measure. When $ =z 0, r = 0,
and as 6 increases r increases, so that the curve winds infinitely often around
Fig. Voi
the origin while receding from it (fig. 194). In the figure the heavy line repre-
sents the portion of the spiral corresponding to positive values of 0, and the
dotted line the portion corresponding to negative values of 6.
Ex. 2. The hyperbolic spiral,
rd = a,
a
L-
K
Fig. 19.J
As 0 increases indefinitely r approaches zero. Hence the spiral winds infi-
nitely often around the origin, continually approaching it but never reaching it
THE SPIRALS 333
(fig. 195). As 0 approaches zero r increases without limit. If P is a point on
the spiral and NP the perpendicular to the initial line,
■»TT, • « sin^
NP = rsmd = a
e
Hence as 6 approaches zero as a limit, NP approaches a (§ 161). Therefore
the curve comes constantly 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. 3. The logarithmic spiral,
r = e"*.
When 6 = 0, r = 1. As ^ increases r increases, and the ciirve winds around
the origin at increasing distances from it (fig. 196). When 0 is negative and
increasing numerically without limit, r approaches zero. Hence the curve
winds infinitely often around the origin, continually approaching it. The
dotted line in the figure corresponds to negative values of 0.
Fig. 196
A property of this spiral is that it cuts the radii vectors at a constant angle.
The student may prove this after reading § 187.
We shall now give examples of the derivation of the polar equa-
tion of a curve from the definition of the curve.
334
POLAR COORDINATES
179. The conchoid. Take a fixed point 0 (fig. 197) and a fixed
straight line BC. Through 0 draw any line OR intersecting BC
in D, and on OB lay off a constant distance DP or DQ, measured
from D in either direction. The locus of P and ^ is a curve called
the conchoid.
From the definition the conchoid consists of two parts, one
generated by P, the other by Q. We may obtain the whole curve,
Fig. 198
however, by allowing the line OR to revolve in the positive direc-
tion through an angle of 360° and always laying off the distance h,
measured from D in the direction of the terminal line of the angle
AOR. Then if AOR is in the first quadrant, we obtaia the upper
half of the curve described by P ; if ^ OR is in the second quad-
rant, we have the lower half of the curve described by Q; HA OR
is in the third quadrant, we have the upper half of the curve
THE CONCHOID
335
described by Q ; and if A OB is in the fourth quadrant, we have
the lower half of the curve described by P.
To find its polar equation, take 0 as the origin and the line OA
perpendicular to BC as the initial line. Let OA = a and the con-
stant distance DP = h.
Call the coordinates of P (r, 6), where 6 =A OR. When ^ is in
the first or the fourth quadrant, r = OD + DP = OD + h ; when 6
is in the second or the third quadrant, r = — OD -\-DQ = — OD + h.
Fig. 199
But OD = a sec d when ^ is in the first or the fourth quadrant ;
and OD = — a sec 6 when ^ is in the second or the third quadrant.
Hence for all points on the conchoid
r = a sec ^ + &.
The conchoid has three shapes according as a > i (fig. 197),
a = 6 (fig. 198), a<l (fig. 199). If & = 0, the conchoid becomes
the straight line BC and its equation becomes r = a sec ^, the
equation of the straight line (§ 183).
336
POLAR COORDINATES
180. The limacon. Through any fixed point 0 (fig. 200) 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 ^ thus
found is a curve called the limagon.
Take 0 as the pole, the diameter OA as the initial line of a sys-
tem of polar coordinates, and call the diameter of the circle a and
Fig. 200
the constant length h. Then it is clear that the entire locus can
be found by causing OD to revolve through an angle of 360° and
laying off DP = h always in the direction of the terminal line
of ADD.
Let the coordinates of P be (r, 6), where 0=AOD. Then
r = OD + DP when 6 is in the first or the fourth quadrant,
and r = — OD + DP when 6 is in tlie second or the third
quadrant. But it appears from the figure that OD = OA cos 6
when 6 is in the first or the fourth quadrant, and OD = — OA cos d
THE LIMAgON
337
when 6 is in the second or the third quadrant. Hence for any
point on the limagon
r = a cos 6 + 1).
In studying the shape of the curve there are three cases to be
distinguished.
O^cos\-i)
Fig. 201
1. h>a. r is always positive and the curve appears as in fig. 200.
2. b < a. r is positive when cos d > ' negative when
7 h ^
cos 6 < ' and zero when cos 0 = • The curve appears as
a a ^'^
in fig. 201.
3. h = a. The equation now becomes
r = a (cos ^ + 1) = 2 a cos^ - ■
r is positive except when 6 = 180°, when it is zero. The curve
appears as in fig. 202 and is called the cardioid.
338
POLAE COORDmATES
The cardioid is an epicycloid for wMch the radii of the fixed and
the rolling circles are the same. The proof of this is left to the
student.
Fig. 202
181. The ovals of Cassini. If a point moves so that the product
of its distances from two fixed points is constant, it generates a
r
Fig. 203
curve called an oval of Cassini. Let F^ and F^ (fig. 203) be the
two fixed points, called the foci, and If the constant product of
THE OVALS OF CASSINI 339
the distances of a point of the curve from F^ and F^. Take F^F^
as the initial line and the point .0, halfway between F^ and F^, as
the pole of a system of polar coordinates, and let P be a point on
the curve. Then, by definition,
F^P . F^P = l\ (1)
By trigonometry,
:^' = OB'^j^OF^- 2 OP • OF^ cosF^OP = r'+ €0"+ 2 ra cos(9,
where (r, 6) are the coordinates of P and 2a= F^F^. Also
TJ^'' = op'^j^OF^- 2 OP • OF^ QosF^OP = r'+ a"- 2 ra cos^.
Substituting in (1), we have
(r2+ay_4aV'cos'(9 = J*,
which is the same as
/_ 2 aV cos 2 (9 + a*- 5* = 0. (2)
To determine the form of the curve, it is convenient to solve (2)
for r^, obtaining
r" = a^cos2e± ^a* cos'2 0 -{a*-b*). (3)
We have, then, three cases to consider
1. a^ < If. The quantity under the radical sign in (3) is posi-
tive and greater than a* cos^ 2 6 for all values of 6. Therefore r^
in (3) has two real values, one positive and one negative. Conse-
quently r has two, and only two, real values equal in magnitude
and opposite in sign. The curve therefore consists of a single oval,
symmetric with respect to the origin (fig. 203).
a*— h*
2. a^ > If. When cos*^ 2 0 > — the quantity under the rad-
ical sign in (3) is positive and less than a* cos^ 2 6. Hence for
these values of 6 there are two real positive values of r^ and there-
fore four real values of r, two positive and two negative. When
a^—h* . .
cos^ 2 0 < — the quantity under the radical sign in (3) is
340 POLAE COOKDINATES
negative, and hence all values of r are imaginary. When cos*^ 2 Q
a* — b* Va* — b*
= — there are two real values of ?', namely r = ±
The curve consists of two distinct ovals (fig. 204).
-M
Fig. 204
3. a^ = b^. Equation (2) then factors into the two equations
r^ = 0 and r^— 2 a^ cos 2 ^ = 0. But r^ = 0 is satisfied only by
the origin, which is also a point on the second equation.
Fig. 20.3
Hence
r' = 2 a' cos 2 d
(4)
is the full equation of the locus in this case. From (4) it appears
that r has two real values equal in magnitude but opposite lq sign
when 0<^<->or— <6'<— ,or— <6'<2 7r. Further,
4 4 4 4
r = 0 when 6 = — , —^ > —7- , or — — ; and r is imaginary when
4 4
5 TT
4 4 4 4
4 4
The curve appears as in fig. 205
and is given the special name of the lemniscate.
CHANCxE OF cooedi:n:ates
341
182. Relation between rectangular and polar coordinates. Let
the pole O and the initial line OM of a system of polar coordinates
be at the same time the origin and the axis of a; of a system of
rectangular coordinates. Let P (fig. 206) be any point of the plane,
{x, y) its rectangular coordinates, and {r, 6) its polar coordinates.
Then, by the definition of the
trigonometric functions,
X
cos 6 = -)
sm a = - >
r
whence follows, on the one hanC,
X = r cos 6,
y = r sin Q,
and, on the other hand.
Fig. 206
r = -Vaf + y , sin ^ =
y
\^x^ + y^
cos tr =
Va;^
r
(2)
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 tlie cissoid (§ 83) is
2/2 =
2a — X
Substituting from (1) and making simple reductions, we have the polar
equation
2asin2 6i
r =
cos^
Ex. 2. The polar equation of the lemniscate is
r2 = 2 a2 cos 2 e.
Placing cos2^ = cos2 0 — sin^e and substituting from (2), we have the rec-
tangular equation
(x2 + ?/2)2=:2a2(x2-2/2).
342 POLAR COORDINATES
183. The straight line. Take the equation of the straight hne
in the normal form x cos a-{-y sin a—p = 0 and substitute the
values of x and y from (1), § 182. There results
r (cos 0 cos a + sin 0 sin a) — j? = 0;
whence r cos {0 — a) = p.
A reference to § 33 shows that {p, a) are the polar coordinates
of the point in which the normal from the origin meets the straight
line. If a = 0 and p = a, we have the special equation
r cos 0 = a,
or r = a sec 0,
as found in § 179.
If the straight liue passes through the origin, ^ = 0. The equa-
tion of the line then becomes
cos {6 — a)= 0,
or simply 0 = — + a,
which is of the form 0 = c.
184. The circle. If (d, e) are the rectangular coordinates of the
center of the circle and a its radius, its equation is
If (h, a) are the polar coordinates of the center and (r, 0) those
of any point, the pole and the initial line of the polar coordinates
being the origin and the axis of x, respectively, of the rectangular
system, we have, by (1), § 182,
x = r cos 0, y = rsm0,
d = b cos a, e = h sin a.
We obtain, by substitution,
r'—2rb (cos 0 cos a + sin ^ sin a)+h^— a^ = 0,
or r^-2rbcos(0-a) + b^-a^==O. (1)
P{r,e)
THE CONIC 343
This result may also be directly obtained from fig. 207 by noticing
CF"" ='0C^ +0F'' - 2 OP • OC cos Foa
"When the origin is at the
center of the circle, b = 0,
and (1) becomes simply
r = a. (2)
When the origin is on
the circle, b = a, and (1)
becomes
r—2a cos (^ — a) = 0, ^^
which may be written ^'^- -^^
r = aQ cos 6 + a^ sin 6, (3)
where a^ and a^ are the intercepts on the lines ^ = 0 and 6 = —
respectively.
Wlien the origin is on the circle and the initial line is a
diameter, (3) becomes
r = «() cos 6. (4)
When the origin is on the circle and the initial line is tangent
to the circle, (3) becomes
r = a^ sin 6. (5)
185. The conic, the focus being the pole. From § 81 the equa-
tion of a conic, when the axis of x is an axis of the conic and the
axis of 3/ is a directrix, is
{x — cf + y^ = e^x^.
We may transfer to new axes having the origin as the focus and
the axis of x as the axis of the conic by placing
x = c + x', y = y',
thus obtaining x'^-\- y'^ = e'^{x' + cf.
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
x' = r cos 6, y' — "f sin 6.
344
POLAR COORDINATES
The equation then becomes
r^ = e^ {r cos 6 + cf,
which is equivalent to the two equations
. r =
1 — e cos 0
ce
1 + e cos 6
Either of these two equations alone will give the entire conic.
To see this, place 6 = 6^ in the second equation, obtaining
T, =
— ce
l-\- e cos 6^
Now place 6 = 'Tr -\- 6^ in the first equation, obtaining r = — r^.
The points {9^, rj and (tt + 6^, — r^ are the same. Hence any
point which can be found from the second equation can be found
from the first.
ce
Therefore
r =
e cos
d
is the required polar equation.
186. Examples. We shall now give examples of the use
of polar coordinates in solving
problems.
Ex. 1. Prove that if a secant is
drawn through the focus of a conic,
the sum of the reciprocals of the seg-
ments made by the focus is constant.
Let PiPo (fig- 208) be any secant
through the focus F, and let FPi = ri
and FPi = r^, and the angle MFP = 0.
Then the polar coordinates of Pi are
(ri, 0) and those of Pa are (ra, d + ir).
From the polar equation of the conic
we have
Fig. 208
ri =
r2
1 — ecos^
ce
Hence
1-
Ti r^ ce
e cos {d + tt)
ce
1 + e cos 0
DIRECTION OF A CUEVE
345
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. 209) and let 0 be any point in the
plane. If O is taken for the pole and OC for the initial line of a system of polar
coordinates, the equation of the ^^^^ p
circle is
r2-2r6cose + 62_a2 = 0. (1)
Let P1P2 be any chord through
0 and let OPi = ri, OP2 = rg.
Then ri and r% are the two roots
of equation (1) which correspond
to the same value of d. Hence
n + r2 = 2 6 cos 6.
If Q is the middle point of
P1P2, and we now place OQ = r,
we have
ri + r2 , .
r = = ocos^. ,, „^^
2 iiG. 209
But this is the polar equation of a circle through the points 0 and C.
187. Direction of a curve. 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 F(r, 6) (fig. 210)
be any point on the curve,
VQ /'■ PT the tangent at P, and
i/r the angle made by PT
and the radius vector OP.
Give 6 an increment
A6=P0Q, expressed in
circular measure, thus fix-
ing a second point of the
curve Q(r + /^r, 6 + A6).
To determine Ar describe
a circle with center 0 and radius OQ, intersecting OP produced
in R. Then
OB = OQ = r + Ar,
PE = Ar,
and SiTG PQ = As,
Fig. 210
s being measured from some initial point A.
346 POLAR COORDINATES
Draw also the chord PQ and the straight line QS perpendicular
to OP and meeting it in S. Then
^^ = (r + Ar) sin A^,
OS = {r + At) cos A6,
SB = OE-OS
= (r + Ar)(l— cosA^),
and PS = PE-SB
= Ar — {r + Ar) (1 — cos A^).
As A0 approaches zero, the chord PQ approaches the limiting
position PT and the angle EPQ approaches i/r. But in the
triangle SPQ
SQ
tanEPQ = ^^
(r + Ar) sin A^
Ar — (r + A?') (1 — cos A^)
sin A^
(r + Ar)
A6>
Ar , . , 1 — cos A^
-— — (r + Ar) -
(1)
A^ ' ' Ad
Now as.:A^ approaches zero
T- / A \ T- sinA^ .
lam (r + A?-) = r, Lim — —r- = 1,
A0
,. Ar ch , ,. 1— cosA^ a/ciki\
Hence, by" taking the limit of (1),
■tan '«ir = 4^. (2)
ar
•To
If it is desired to find the angle MNP = (f), it may be done by
the evident relation
DERIVATIVES WITH RESPECT TO THE ARC 347
188. Derivatives with respect to the arc. In the triangle FQS
(fig. 210)
SQ
BmSPQ-
chord P^
SQ QxcPQ
arcFQ chord P^
_(r + Ar) sin A^ arc PQ
As chord P^
^ ,sinA(9 Ad arc PQ
= (r + Ar)
A0 As chord Pg ,
As A^ approaches zero, SPQ approaches yjr, Lim — —^ = 1, and
Lim f^'^f^ = 1 (§ 104) ; hence
chord P^ ^^ ^
sin '\lr = r—-- (1)
as
By dividing (1) just obtained by (2) of the previous article,
dr
ds
cos ■\(r = — — (2)
From (1) and (2) we obtain
/dsY
By multiplying (3) by | -r^ I we obtain
d0j ^\ddi ' ^ '
/dsV
and by multiplying (3) by ( ^r- ) we obtain
CI
drj \ dr
'J.yjr'^)'+l. (6)
348
POLAR COORDINATES
189. Area. Let C (fig. 211) be a fixed point and P {r,6) a
variable point on the curve r =f(6), and let A denote the area
of the figure OCF, bounded by
the arc of the curve CP and
the radii OC and OP. Then A
is a function of 6, since the
value of 0 fixes the position of
the point P. If 6 is increased
by A^ = angle POQ, A is in-
creased by A A = area POQ.
From 0 describe arcs of circles
PS and QR with radii OP = r
and OQ = r + Ar respectively.
Yio. 211 Then in the figure
aresi POS <AA< aresi EOQ.
But the area of tlie sector of the circle POS is
lOP-PS=lr'Ae,
and SiTeaBOQ=^^OQ-BQ==^^{r + AryAe.
We have then ^ r^AO <AA<l(r + ArfAd ;
AA
whence ^t^ < -rz <\{^ + Ar)l
Taking now the limit as A^ approaches zero, we have
Ex. Find the area of a loop of the lemniscate r^ = 2 a* cos 2^.
We will take C as the point for which ^ = 0, and P as any point for which
0<»<--
4
Then
dA
= o2 cos 2 » ;
de
whence
0^
^ = — sin 2 ^ + c.
But when e = 0, A = 0; therefore c = 0. Also when 6 = -, A- I area of
^ 4
the loop. Hence the area of the loop is a^ sin - = a^.
PROBLEMS 349
PROBLEMS
Plot the following curves :
1. r=asin2^. 13. r = a(l + cos2^).
2. r=acos3tf. 14. r = a(l + 2 cos2e).
3. r = atan^. 15. r = a(l— cos2fl).
4. r = a(l + sintf). 16. r = a(l + cosSfl).
5. r = a(2 + sin«). 17. r = a(l + 2 cos3tf).
6. r = a(l + 2 sintf). 18. r = 4 + 5 cos 5^.
7. r = a0~i. 19. r = 2 + sin S 0.
8.
r =
asec2-.
2
9.
r =
a
0-b
10.
r =
a — he.
11.
r =
. e
a sin - .
2
12.
r =
e
■ a cos - .
3
20.
r = a tan - •
2
21.
r = a sin8 - .
■ 3
22.
r2 = a2 sin 0.
23.
r2 = a^sinStf.
24.
r cos 5 = a cos 2 tf.
25.
a a
r = h ^
cos 0 sm fl
Find the points of intersection of the following pairs of curves :
26. r cos ( 0 ]= a, r cos 10 i = a.
27. r cos / ^ 1 = — , r = asin0.
\ 2; 4
28. r2 = a2 sin 0, r^ = a^ sin 3 0.
29. r = a sin 20, r = a (1 — cos 2 0). [(ri, ^1) and ( — n, ^i + tt) are the same
points.]
30. 0 is a fixed point and LK a fixed straight line. If any line through O
intersects LK in Q and a point P is taken on this line so that OP • OQ = k^,
find the locus of P.
31. A straight line OA of constant length revolves about 0. From 4
a pei-pendicular is drawn to a fixed straight line through 0, intersecting it
in B. From B a perpendicular is drawn to OA intersecting it in P. Find the
locus of P.
32. MN is a straight line perpendicular to the initial line at a distance a
from 0. From 0 a straight line is drawn to any point B of MN. From B a
straight line is drawn perpendicular to OB. intersecting the initial line at C.
From C a line is drawn perpendicular to BC, intersecting MN at D. Finally,
from I> a straight line is drawn perpendicular to CD, intersecting OB at P.
Find the locus of P.
350 POLAR COORDINATES
Transform the following equations to polar coordinates :
33, y2 = 4px, 36. x2 + 2/2 - 8 ox - 8 ay = 0.
34. xy = 7. 37. x* + x2y2 _ a2y2 = q.
35 ^ + ?^ = 1 ^^- ^'^^ + ^')' = ''' (•^' ~ ^'>-
' a2 62 • 39. x^ + y3-3axy^0.
40. Find the polar equation of the cissoid when the pole is A and the initial
line is OA (fig. 91).
41. Find the polar equation of the strophoid (1) when the pole is 0 and
the initial line OA (fig. 92); (2) when the pole is A and the initial line is OA.
42. In the strophoid (fig. 92) show that
112
AP.APi = a:^, and -t^ + -!— = -^,
AP APi AN
where AN is the projection of ^0 on AD.
Transform the following equations to rectangular coordinates :
43. rcosM-- J + rcosM+-| = 12. 46. r = atan«.
44. r = asin«. 47. r2 = a2sin«.
g
45. r = a(cos2^ + sin2«). 48. r2 = a2sin-.
49. Find the Cartesian equation of the rose of four petals »• = a sin 2 6.
50. Find the Cartesian equation of the cardioid r = a(l — cos 5).
51. Find the Cartesian equation of the ovals of Cassini
r* - 2a2r2cos2^+ a* - 6* = 0.
52. Find the Cartesian equation of the limagon r = a cosO + b.
53. Find the Cartesian equation of the conchoid r = a sec 5 + 6.
54. Find the Cartesian equation of the logarithmic spiral r = e"^.
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 perpen-
dicular 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 the 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
■ Boinfs of its path, its focal distances being 5 and 15 million miles and the angle
between them being 90°. What is its distance from the sun when it is nearest it ?
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.
PROBLEMS 351
59. The focal radii of a parabola are extended beyond the curve until their
lengths are doubled. Find the equation of the locus of their extremities.
60. If Pi and P2 are the points of intersection of a straight line drawn from
any point O to a circle, prove that OPi ■ OP2 is constant.
61. If Pi and P^ 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 straight line such that
00 = ^ ^^^ ' ^^^ , find the locus of Q.
^ OP1 + 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 0 a straight line is drawn intersecting a fixed circle at P,
and on this line a point Q is taken so that OP ■ OQ = k^. Find the locus of Q.
64. Find the polar equation of a conic if the pole is a vertex and the initial
line an axis.
65. Find the locus of the middle points of the focal chords of a conic.
66. Find the locus of the middle points of the focal radii of a conic.
67. If P1PP2 and QiFQ-2 are two perpendicular focal chords of a conic,
prove that 1 is constant.
PiF-FPi. Q1F.FQ2
68. 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.
69. 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.
70. If a straight line drawn through the pole 0 perpendicular to a radius
t"
vector OP meets the tangent in A and the normal in B, show that OA = —
dr ^^
and OB = — v;
de , de
These are called the polar subtangent and the polar subnormal respectively.
71. If p is the pei-pendicular distance of a tangent from the pole, prove
that p =
nRI
72. When a point traverses the curve r=f{d) with a uniform angular
velocity, find the rate at which r is changing and the rate of the point along
the curve.
73. When a point moves along the curve r=/{S) at a uniform rate, find
the rates at which r and 6 are changing.
352 POLAR COOEDINATES
74. Find the velocity of a point moving in a lima^on when B changes
uniformly.
75. A point moves along the radius vector with a constant velocity a, while
the radius vector revolves about 0 with a constant velocity w. Find the path of
the point.
76. Find the total area bounded by the curve r^ = a^ sin 0.
11. Find the area of a loop of the curve r^ = a^ sin 3 d.
78. Find the area swept over by the radius vector of the spiral of Archimedes
as 0 changes from 0 to ir.
79. Find the area swept over by the radius vector of the logarithmic spiral
as 0 changes from 0 to tt.
Q
80. Find the area swept over by the I'adius v-ector of the curve r = asin-
as 0 changes from 0 to 2 tt.
81. Find the area swept over by the radius vector of the curve r = atanfl
TT
as 0 changes from 0 to — •
4
82. Find the total area of the limagon (6 > a).
83. Find the total length of the cardioid.
84. Prove that the length of an arc of the logarithmic spiral is proportional
to the difference of the radii vectores drawn to its ends.
85. Show that if the angle between the tangent to a curve and the radius
vector to the point of contact is one half the vectorial angle, the curve is a
cardioid.
CHAPTER XVI
CURVATURE
190. Definition of curvature. If a point describes a curve the
change of direction of its motion may be measured by the change
of the angle <f> (§ 59).
For example, in the curve
of fig. 212, a AP^ = s and
F^I^ = As, and if ^^ and (f)^
are the values of <f) for the
points i^ and i^ respectively,
then ^2 ~" ^1 i^ the total
change of direction of the
curve between ij and I^.
If ^2 — ^^ = A<^, expressed
in circular measure, the
A^
As
ratio
is the average
Fig. 212
change of direction per linear unit of the arc -^^. Regarding
^ as a function of s and taking the limit of — -^ as As approaches
zero as a limit, we have -^ > which
as
is called the curvature of the
curve at the point Py 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 (§ 109).
If -~ is constant, the curvature
as
is constant or uniform; other-
Applying this definition to the
Fig. 213
wise the curvature is variable.
circle of fig. 213 of which the center is C and the radius
353
354
CURVATUKE
Hence
-J- = -, and the circle is a curve of constant
ds a
is a, we have A^ = -?^Ci^ ; and hence As = aA<j). Therefore
A^_l
As a
curvature equal to the reciprocal of its radius.
191. 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, %vith 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 curva-
ture 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.
Since the curvature is -^>
ds
_}__ds_
^^ d4~ d4>'
ds
(by (6), § 96)
K the equation of the curve is in rectangular coordinates.
and
whence
dx N \dxl
(by § 105)
\dx/
(by § 59)
A
d<^ dx^
dx ^Jdy^
\dx)
ds
ds dx
' ^^d4)~'df
(by (8), § 96)
dx
Mm
i
d'y
cb?
RADIUS OF CURVATURE 355
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.
Ex, Find the radius of curvature of the ellipse h — = 1.
Here dy_^_^
dx a?y
and — ^ =
dx2 a^y^
{a*y'^ + 6*x2)*
•■• P =
Another formula for p, i.e,
r /dx^^-^^
(Tx
df
may be found by defining ^ as the angle between OY and the
tangent, and interchanging x and y in the above derivation.
ds
192. According to the definition, i.e. p = -tt' it is evident that
d<p
p is positive when s is measured so that s and </> increase at the
same time, and is negative when one increases as the other decreases.
For convenience we shall assume in the following work that s
always increases from left to right * along the curve (figs. 214-217).
Then ^ is always in the first or the fourth quadrant, and hence
sec ^ is always positive.
But sec </> = Vl+tan^^= \l 1 + ( -^^ ) • Therefore in the formula
NIII .
da?
the sign of p is the same as the sign of -—• Hence p is positive
when the curve is concave upward, and negative when the curve
is concave downward.
•The results and the proof are the same if s is measured from right to left along
the curve ; hence the proof is left to the student.
356
CURVATURE
193. Coordinates of center of curvature. The center of the
circle described in § 191 is called the center of curvature corre-
sponding to the point. Let C{a, /3) (fig. 214) be the center of
curvature corresponding to the point P{x, y) of the curve. Draw
CL and FM parallel to OY, and NR through P parallel to OX.
Then
OL = 0M+ ML = OM+ PiV;
Now
and
LC = LN+NC = MP + NC.
ZBPC = <f>+ 90°,
PC = p,
Fig. 214
since p>0, the curve being concave upward. Therefore, by the
definition of the trigonometric functions,
PiV = PC cos i^PC = /? cos (<^ + 90°) = - /3 sin </>,
NC = PC sin PPC = p sin(<^ + 90°) = p cos<^.
.'. a = X — p sin<f>,
^ = y i- pcos<f).
There are three other cases represented in figs. 215, 216, 217
respectively. The construction in all these figures is the same
as in fig. 214, and the proof from fig. 215 is the same as that
EVOLUTE AND INVOLUTE
357
just given. The proof from figs. 216 and 217 differs only in that
RFC = 0—90°, and PC ~ — p, since /a < 0, the curve being con-
cave downward. Hence the
above expressions for a and
/3 are universally true.
Since cos ^ =
and
sin0 =
dx
Fig. 210
the formulas for a and /S may be written
a = x
dx\_ \dx/ J
^ = y +
v<
c£y_
dx^
dy^^'
dx,
dx^
Fio. 217
In the example of § 191 we found
dx a'hj
Ex. Find the coordinates of
tlie center of curvature for any
x^ w2
point of the ellipse — | = 1.
a'2 62
and -^— =
Substituting in the above formulas and simplifying, we have
/a2 - 62\ /a2 _ 62N,
" = (-^r-)-^ ^ = -(-6^)^'-
194. Evolute and involute. With the single exception when
d^v
— ^ = 0, in which case p becomes infinite, there will be a center
d3(r
of curvature corresponding to each point of the curve. The locus
358
CURVATURE
b*
y",
X2 y2
— + ~ = 1.
a-2 62
of these centers of curvature is a curve called the evolute of
the given curve, and the given curve is called the involute. In
fig. 218 (1) is the involute
and (2) is the evolute.
To find the evolute we
find the coordinates of
the center of curvature
in terms of x and y, and
then eliminate x and y
from these two equations
by the aid of the equa-
tion of the curve.
Ex. To find the evolute of
the ellipse, we have then,
in the example of the last
article, to eliminate x and y from the three equations
a2-62 - „ a2-62
a = -— a:% /3 = -
a*
From the first two equations
X _ / aa \
o ~ W - 62/ '
b W - by
Substituting these values in the
third equation and simplifying, we
have
{aa)i + (6^)^ = (o2 - 62)?
as the equation of the evolute. The
ellipse and its evolute are shown in
fig. 219.
It may be noted that equa-
tions expressing a and y3 are,
in fact, the parametric repre-
sentation of the evolute, x and
y being two independent param-
eters connected by the equation
of the given curve.
Fig. 219
EVOLUTE AND INVOLUTE 359
195. Properties of evolute and involute. From the equations
a = x — p sill 4>, /3 = y + /) cos </>, we may find the slope of the
evolute at any point by assuming a, /3, x,y,p, and </> as functions
of s, the length of arc along the involute. Then
da _dx d^ • M^P
ds ds ds ds
= cos (f) — p COS <^ ( - ) — sin <^
= — sinrf)-f--
ds
dfi dy . . d(b , , dp
-7- = -/^— /^sm</>-^ + cos</>-p
as ds ds ds
dp
ds
= sbi(f) — p sin (f> {-)-{- cos ^ -~
\p/ ds
= cos<i— •
as
^ d£
. ds ^ , , ^ ds dB
..-— = — ctn d> ; but -—- = -—>
da da da
ds ds
by (8), § 96 ; and if tan^' is the slope of the evolute at the assumed
dfi
point, — = tan </>', and hence tan <f)' = — ctn (f). Hence <f>' and </>
differ by 90°, and the tangent to the evolute at any point is
perpendicular to the tangent to the involute at the corresponding
point (fig. 218).
If we square and add the above equations, we have
/da\
\ds)
daV (d^Y_(dp
ds/ \ds/ \ds
But if we denote the length of arc along the evolute by s', we
have _= - i_|_/z^\ ; and if we regard s', a, ^, as expressed in
da N \da)
terms of s, the length of arc along the involute, we have
360 CURVATURE
ds I \ds/ .
da ~~ I /daV '
ds n| W
whence
Hence
as ^\ds/ \ds/
c?s/ \ds/
and -y- = ± ,
as as
ds' _ C?/3
6?S
•'• s' = ±p + c. (by § 110)
/if follows, then, that as the center of curvature moves along the
evolute the radius of curvature increases or decreases hy exactly the
distance traversed hy the center (fig. 218).
From these two properties we see that an involute may be
described by a pencil attached to the end of a string which is
unwound from the evolute, the free portion being kept taut and
tangent to the evolute. From any one evolute any number of
involutes may be described by changing the length of the string.
196. Radius of curvature in parametric representation. If x
and y are expressed in terms of any parameter t, the radius of
curvature may be fovmd as follows:
ds dt
But
and
cix
dt
p =
d(f) d<f)
dt
ds
dt'
-4m<tT
dy
<!>-
= tan-^-/ = tan-^--;
dx ax
(by (8), § 96)
RADIUS OF CURVATURE
361
whence
dt
dx\ dj^y (dy\ d?x
di/d^~\di)df
ldy\
\dti
(dxY
\dtl
dx d^y dy d^x
Tt"df~^t'~df
(dxV /dy
\dt) \dt
Therefore, by substitution,
P =
dx d^y dy d?x
'dt'~de~~dt"de
Ex. Pind the radius of curvature of the cycloid
Here the parameter is 0
x = a<t>
— a sin 0,
y = a-
acos<f>.
dx
d<t>
acoscft,
d^x
— = asinc^,
dy
-^ = asin^,
d(p
d"'y
Hence, by substitution, p
= a cos<t>.
_ [a2(l-cos0)2 + a2.sinV]^
a (1 — cos <p) ■ a cos <p — a sin <p {a sin <p)
= -2^a(l-cos0)^
= -2^a-(2sin^^\
— 4a sin
197. Radius of curvature in polar coordinates. The equation
of any curve in polar coordinates may always, theoretically, be
expressed in the form r=f{d). Then, since r may be regarded
362 CURVATURE
as a function of 6, the equations x = r cos B, y — r sin Q, are the
parametric equations of a curve. From them we may accordingly
derive the formula for p in polar coordinates by substituting in
the formula of § 196 as follows:
dx dr n • /I
dy dr . a , n
J = -^smfl + .cos«,
dr V drT dv
_| = _siu^ + 2^cos(>-.sin«.
Substitutmg these values and simplifying, we have, as the
required formula,
\ddl d&'
Ex. Find the radius of curvature of the cardioid r = a(l — cos^).
Here — = a sin d and — — a cos 6.
de dff^
_ [a2(l - cos^)2 + a? sin2fl]i
' a2(l- cos^)2 + 2a2sin2tf _ a(l - cose)acos9
[2a2(l-cos^)]3 2?a„ „,i
= ^^ -^ = (1 - costf)*,
02(3-3 00861) 3 ^
p = f(2ar)i.
PROBLEMS
fj - _ -
1. Find the radius of curvature of tlie catenary y — ~ {e" + e ").
2. Find the radius of curvature of the cissoid y^ —
2a — X
3. Find the radius of curvature of the four-cusped hypocycloid x^ + y^ = a*.
4. Find the radii of curvature of the curve a*y^ = a^* — x^ at the points
(0, 0) and (a, 0).
5. Find tlie radius of curvature of the curve (-) +(-) =1 at the
point (0, b). ^"^ ^^^
PROBLEMS 363
6. Find the radii of curvature of tlie curve y^ = ax{x — Sa) at the points
where it crosses the axis of x.
7. Find tlie radius of curvature of the curve e'^ = sin y at the point (xi, yi).
8. Find the slope and the radius of curvature of the curve y + log(l — x^) = 0
at the origin of coordinates.
9. Show that the radius of curvature of the curve r = a (sin 6 + cos 0) is
constant.
10. Find the radius of cui^ature of tlie curve r = a (2 cos^ — 1).
11. Find tlie radius of curvature of the curve r = a sin^ - . Find the greatest
and the least values of the radius of curvature.
12. Find the radius of curvature of the lemniscate r^ = 2a^ cos 2 8.
13. Given the curve x = 2 cos t — cos 2t, y = 2 sint — sm2 1. Find the radius
of curvature in terms of t, and show that it will be greatest when t = Tr.
14. Find the e volute of the parabola y^ = ipx.
15. Find the radius of curvature of the tractrix
a, a + Va2-x2 _
y — -log — ■ Va2 - x2.
2 a _ Va2 - x2
16. Prove that the evolute of the tractrix is the catenary.
17. Prove that the evolute of a cycloid is an equal cycloid.
18. Find the evolute of the four-cusped hypocycloid x = a cos^4>, y = a sin^^.
19. Find the evolute of the ellipse from the parametric equations x = a cos^,
y = bsm (f>.
20. Prove that the center of curvature of any point of the logarithmic spiral
is the point of intersection of the normal with the perpendicular to the radius
vector.
21. Find the circle of curvature of the curve y = er'^ when x = 0.
22. Show that the catenary ?/ = ^(e^ + e-^) and the parabola 2/ = 1 + ^x^
have the .same tangent and the same circle of curvature at their point of
intersection.
23. Find the point of minimum curvature on the curve y — log x.
24. Find the points of greatest and of least curvature of the sine curve
y = sinx.
25. Find the points on the ellipse for which the curvature is a maximum or
a minimum.
26. Show that the curvature of the parabola y = ax? + 6x + c is a maximum
at the vertex.
364 CURVATURE
27. Find the condition for a maximum or a minimum of the curvature k,
dx2
where A; =
Mm'
28. At what points on tlie curve y = log sin x is the radius of curvature
unity, and in what direction from the point on tlie curve is the center of
curvature ?
29. Show that the product of the radii of curvature of the curve y — ae~^
at the two points for which x = ±ais a'^(e + e-i)^.
30. If the angle between the radius vector to the point of contact and the
straight line drawn from the pole perpendicular to the tangent is either a maxi-
ANSWEKS
(The answers to some problems are intentionally omitted.)
CHAPTER I
Page 25
1. 9. 3. x-x2. 5. 17. 7. 1.
2. x-y. 4. 4. 6. -18. 8. 2aJc
9. ahc + 2fgh-aP-hg'^-ch'^. 11. Ix-Qy-b.
10. a62 + hc^ + ca^ - ac^ - ba^ - cb^. 12. 3.
13. 2aia2CiC2 + 01616202 + 02616301 — a^c^ — aScf — ai6,Jci — 0261%.
Page 26
2^ 13 zfcVei 33. a;= 2, 2/=-2, z = f.
2 ' 34. a; = - 5, 2/ = 0, z = 4.
25. 0, 6 ± V39. 35. X = - 1, 2/ = 0, z = 0.
26. 8x + 2/-13 = 0. 38. x = i(a-26 + c + d),
27. x2 + 2/2 - X - ?7 = 0. 7/ = i (a + 6 - 2 c + d),
28. x2 + 2/2-3x + i/- 4 = 0. z = i(« + ^ + c-2d),
29. x2 - (a + 6) X + tt6 - /i- = 0. lo = { (- 2 a + 6 + c + d).
30. x3 - (a + 6 + c) x2 + (a6 + 6c 39. Xi : X2 : X3 = 3 : - 5 : - 2.
+ ca — /2 — </2 — /i"-) X — a6c 40. xi : X2 : X3 = 1 : — 2 : 3.
- 2fgh + a/2 + 6^2 + cA2 = 0. 41. X3 = 0, Xi : X2 = - 2 : 1.
31. X = 1, y = 2. 42. xi : X2 : X3 : X4 = 4 : — 3 : 2 : 5.
32. X = 2, y = - 1, z = 3. 43. Xi = 0, X2 = 0, X3 : X4 = 3 : 2.
51. 2/2 + 11 2/ + 12 = 1
2 52. 2/ - 4 = 0.
50. 2 2/2 + 6 2/ - 3 = 0. 53. 62 - (a + c)2 = 0.
CHAPTER II
Page 45
1. 5 + 3 Vi. 7. (- I, 0). 10. (- 1, - 3i), (1, - 41).
8. (lA, 3H). 8. (- 2|, - 1|), (- If, \l). 11. (- I, If).
6. (H, Hi)- 9- (-5, 3 ± V7). 12. (- i, - 11).
Page 46
14. (5, -1). 16. (-10,31). 18. ^VSO, 1V53, V26.
15. (-14, 17). 17. (2, 1), (4, - 1). 19. (15, -3).
366
366
A2TSWERS
CHAPTER III
Page
66
21.
tan-i|. 22.
tau-if 28. -.
4
Page
67
26.
x-y -4=0.
32. tan-iT-'j.
27.
8x + dy + 83 = 0.
33. 2x + 8y-17 = 0.
28.
5x -6y = 0.
34. 9x-Gy -2 = 0.
29.
7x-2y + S = 0.
35. x-y + 2 = 0.
30v
3z-y-7 = 0.
36. 4x-62/ + 16 = 0.
81.
X + 1 = 0.
37. 25x + loy- 24 =
Page
68
44.
2x + Sy + 1 = 0.
48. IfVlS.
45.
i-
^^ 62 _ a2 _ aft
46.
(0, 2), (0, 7), (3, 5)
; V52 + a2
25. 5x - 4j/ + 40 = 0.
38. 5x-2y-10 = 0.
39. 7x + 4y + 28 = 0.
41. 12x-1.3y-8 = 0.
42. 3x-2y-7 = 0.
43. x+iy -4 = 0.
0.
— , tan-13, tan-15.
4 ^'
50.
Vl3.
47. jJyVn,
Page 69
51. 6V2, HV2.
52. |.
58. (3, 0).
59. 2x - y- 4 = 0.
60. (-21, -li).
61. (±fiVl3,±ffVi3).
63. (1, 1), (- 3, 3). _ 66. 5x + 2/-12 = 0, x-5?/ + 8 = 0.
64. 5x-j^-3 = 0; 2 V26. 67. 3x-4y -2 = 0, x - 2 = 0.
65. 17x + 6j/ + 34 = 0, x + 18y + 2 = 0.
CHAPTER IV
Page
94
7.
a =
Page
95
10.
(1)
11.
(1)
(3)
12.
(1)
(2)
(3)
_ 4 2
8.
2a
26. f,
fc=|; (2)A:>|; (3) A:<|.
A; = 0or4; (2) fc<0orA;>4;
0<A;<4.
A; = - 3 or - ^ ;
-3<A;<-1;
A;< - 3 or k> — |.
|(-3±3 V-3).
26. 2, -i, -1±V^, {{1±-
27. 0, H±l±Vl3).
0, 6, ± V^ ± V^.
28.
29.
30.
31.
0, -2ffl, — (-l±5i).
32. x3 - ax2 - (a2 + 6)x + a^ - ab
±{a + 1), ± (a -
6x3- 13x2 + 6x
= 0.
= 0.
33. x6 _ 4aj;5 + {4a2 - 62 - 26)x* + 8 aSx^ - (8 a26 - 263)x2 - q.
34. x2 - 4 X + 13^ 0.
35. (2x + 2-Vll)(2x+2 +VlT).
36. (2 X + 3 + V^) (2 X + 3 - V- 2).
37. (2 ax + J +4 V- 3) (2 ax + J - ^ V^).
38. (X + a + Va) (x + a - Va).
ANSWERS 367
89. {ax + b + Va + 62) (ox + 6 - Va + ^).
40. (ax + b + i Vb) (ax + b - i Vft).
41.p2_29. ^^ p^-2q 46. p2-3r.
42. S^jg — p3. ^2 ' 47. j)r.
43. -P. 46 ^'^-^g. 48.^.
9 Q r
Page 96
62. 1, ^(3±v^). 66. -3, H-l±^^^)- 70--f,-l±'^.
63. 2,2, -1. 67. ^, §, f. 71. - 1, _ 1, ± i.
64- 2, 2, - |. 68. i, I, - f 72. 3, - 2, f , -j.
65. -3, f, f. _ 69. I, -2dhV-2. 73. 2, |, 1±V^,
74. 4, - f , 1(3 ± V5). 80. 1, -2, -^,2 ± Vs.
75. ±f, J(3±V^>. 87. 1.41.
76. i, -5, 1(1 J.V6). 88. -1.52.
77. 2, 3, - 1, - 1 ± vCr2. 89. 2.05, .59.
78. -2, ±J, |(_l±Vr3). 90. 1.18,2.87.
79. I, ± # , _ 2 ± Vs. 91. .16, 2.93, - 2.09.
CHAPTER V
Page 118
1. 12. 3. 0. 14. (4|, 55f). 16. lOf.
2. -3. 11. 32x + j/ + 45 = 0. 15. tan-i^V 17. (2, 9), (- 2, 5).
18. 4x + 2/ + 2 = 0, lOSx + 27y + 58 = 0.
19. (-1, -6), (1, -512).
Page 119
20. Increasing if x > — 2 ; decreasing if x < — 2.
21. Increasing if x < 0 or^ > ^ ; decreasing if 0 <x < f.
22. Increasing if x > — V2 ; decreasing if x < — V2.
23. Increasing ifx>lor— l<x<0; decreasing if 0 < x < 1 or x < — 1.
24- (I, I)- 26. (0, i), (± 2, - 3|).
26. Maximum value, f f f ; minimum value, — 3.
27. Maximum values, — 12, — 66 ; minimum values, — 34, — 88.
29. 1 (a + 6 - Va2 + 6^ - ab).
30. Altitude, -; base, -•
2 2
31. Altitude, | a V3 ; radius of base, ^ a Ve.
32. Altitude is one third the altitude of the cone. 83. (1^, f).
84. The one in which the radius of the circle from which it is cut is one
fourth the perimeter of the sector.
35. Altitude is one half .a side of the base.
37. (1) Height of the rectangle is equal to the radius of the semicircle.
(2) Semicircle of radius—. 88. J a.
368
ANSWERS
Page
120
39.
Length is twice the breadth.
52.
- 2.21.
40.
.06 CIK
53.
2.09.
44.
Upward if x> 1;
, 54.
1.20, 3.13, -1.38.
downward if x < 1.
55.
1.51, -1.18.
45.
Upward if x > 0,
downward if x < 0.
56.
57.
2, 2, - 3.
1, 1, - a ± Va-^ - a.
46.
(2, - h).
58.
b{b + 4a^) = 0.
48.
(0, - 8).
59.
62(27a< -6) = 0.
49.
(1, - 27).
60.
¥ - 27 a* = 0.
51.
0.45, 1.80, - 1.25.
61.
See Ex. 23, Chap. I.
CHAPTER VII
Page 155
5. x2 + 2/2 ± 2ax = 0.
6. x2 + 2/2 -t 2 ax ± 2 ay + a2 = 0.
7. x2 + 2/2 + 3 x^ 2 2/ = 0.
8. (- 2, 5); V6o.
9. (-2, 3); 2 V3.
10. (3., - 1) ; 1 Vlil.
11- {~hi)> 0.
14. x2 + 2/2-3x-32/ = 0.
15. X2 + 2/2- 3x -4 = 0.
16. x2 + 2/2 + 26 X + 16 2/ - 32 = 0.
17. x2 + 2/^ - 5x + 4 y - 46 = 0.
18. x2 + 2/2- 20x- 202/ + 100 = 0,
x2 + 2/2-4x-42/ + 4 = 0.
19. x2 + 2/2 - 12x - 12 2/ + 30 = 0,
25x2+25 2/2 -l-60x-602/ + 3G = 0.
20. x2 + 2/2 + 2x + IO2/+ 1 = 0,
a;2 + y2 _ 12 X - 4 2/ + 15 = 0.
21. 2x2 + 22/2 + 6x + 3 2/ -10=0.
Page 156
22.
x2 + 2/2 + 22 X - 34 2/ + 121 = 0,
30.
x2 + 2/2-2x- 102/ + 1 = 0.
31.
23.
X2 + 2/2- 10x-28 7/ + 217 = 0.
32.
24.
x2 + 2/2 + 22 X - 44 y - 20 = 0,
33.
x2 + 2/2 + 2 X - 4 2/ - 20 = 0.
34.
25.
4x2 + 42/2 ±7y- 36 = 0.
35.
26.
7x2+ I6y2_ii2 = 0.
36.
27.
9x2 + 5y2_ 45 = 0.
37.
28.
5x2 + 92/2-180 = 0.
38.
29.
I V385, i V105.
39.
40. ^ V2, i V3; (±
iAo)
age
157
41.
5x2-42/2-20 = 0.
48.
42. 3 2/2 -x2- 12 = 0.
43. 28x2-36 2/2- 175 = 0.
44. 24x2-252/2-384 = 0.
46. 3x2 - 2/2 - 3a2 = 0.
47. 8x2-2/2-16 = 0,
8y2_a;2_ 124 = 0.
16 x2 + 25 2/2 - 400 = 0.
5x2 4.9y2_80 = 0.
16 x2 + 25^2 _ 400 = 0.
196x2 + 132 2/2 - 14553 = 0.
4,3; J V7; (±V7, 0).
J; 3x2 + 4^2 _ 3^2= 0.
6x2 + 9^2 _ 405 = 0.
a;2 + 4 2/2 _ a2 ^ 0 ; }, V3.
\ V2.
3x2 + 52/2-30 = 0.
; 2 X ± Ve = 0.
48. x2 - 2/2 - 21 = 0.
49. x2 -82/2 + 4 = 0.
50. 25x2 _ 144 2/2 _ 3500 = 0.
51. 252/2- 9x2- 16 = 0
52. cos-
53. ^ V5, i V5.
ANSWERS
360
54.
58.
Page
61.
62.
63.
71.
72.
Page
80.
81.
82.
Page
94.
95.
i -v^; (±v^, 0);
2x±5y = 0.
iVl3; (±VT3, 0); 13x±9 Vl3
158
55. 52x2 _ll7y2_ 576 = 0.
57. 3x2-41/2-84 = 0.
0; 2x±3y = 0.
2^? V2.
67. y ± 5x = 0.
68. x2 _ 82/2 _6y + 9 = 0.
69. 2/2 + 4?/-2x + 11 = 0.
64. x2 + y2 _ 5px = 0.
65. x2 + 2/2 + 3x-6y = 0.
66. 7x- 32/ + 2 = 0.
70. 91x2 + 842/2 - 24X2/ - 364x - lb2y + 464 = 0
2/2 -10x + 25=0. 73. 4x + 32/-31=0, 4x+ 32/ + 19 = 0
(2/ - 2)2 ± x8 + x2 = 0. 74. 5 X + y - 5 = 0, X - 5 2/ + 7 = 0.
159
Circle. 83. Concentric circle.
Circle. 84. Straight line.
Circle. 85. Straight line.
91. Parabola.
86. Circle.
87. Two straight lines.
90. Parabola.
93. Two parabolas.
160
Circle.
Parabola.
96. Hyperbola.
97. Parabola.
98. Hyperbola,
99. Witch.
101. 8i)V3.
CHAPTER VIII
Page 175
1. (1, 1), (- 2, 3).
2. (0, 1).
4. (0,0), (-1, -2).
5. (1, 21).
6. (1,3), (I, -1).
8. (2 ± VB, 2 1^ V6).
9. ^6, Vl3.
Page 176
18. (2, 2).
19. (0,0), (±V2, ±i>^)
20. (2, 2), (f, - I).
21. (0, 0), 1^-
22. (2, 1).
. + »l2 1 + ,„2
10. 2x-?/ + 2 = 0.
11. -3.
12. 2x + 32/±6V2 = 0._
13. bx-ay + ab± ab V2 = 0.
16. (0, 0), (4, li).
16. (±lf, ±5), (±li, ±3J).
17. (I, 0), (1, - 1).
23. (0, 0), (-1, 0).
24. (1, ±2V3), (6, ±6V2).
25. (± I- V48 T 3 Vl3, Q )•
26. {b - a, ± 2 Va6).
27. (±2a, a).
28.
29.
30.
31.
Page
35.
36.
(0,0), (fa, ±faV2).
(2 a, a).
(±2a, a).
177
3x + y -I = 0.
5 a;2 + 5 2/2 + 28 X + 42 2/
(0, 0), (- 2 a + 2 a A ± 2 a V2 Vi - 2)
32. (±2 a, a).
33. 5x + 42/ + 6 = 0.
34. 2x + 5y-13 = 0.
38. 3x2 +3^2 + 13a; +13?/ -4 = 0.
39. x2 + 82/2 -9 = 0.
370
ANSWERS
CHAPTER IX
Page 209
1. 9x2 + 14x + 6.
2. 20(x + l)(3x2 + 6x + 2).
2a
(X + o)2
6x2
(X8 - 1)2
14 (1 - X)
(3x2-6x + l)a'
1
^^^^V|-^>
(X + 1)2
3Vxi
8. 2(4x
9
3) + _(6_7x).
x + 1
2xVx
10. l/A_JL_J_ + _^>
A^x <^2 a;^x x^x2y
11. 2(3x2_5x+6)(6x- 5).
12. 6x(x2 + l)2.
IS «* + ^
14.
2 V4x2 + 6x-6
2x + 1
16. -
16. -
17. -
3 v^(x2 + X - 1)»
2x
(X2 + 1)2
6(3x2 ^-2x)
(X8 + X2 + 1)2
5x
\^(X2 + 1)6
18. 5(2x-l)(2x-3)(x + l)2.
19. (3x- 6) (12x2 -55 X +31).
20 2^' + ^ + l
21.
Vx2 + 1
3x*-10x3 + 6x2 + x-2
(x2-4x + 3)*(x3 + l)l
22. V_L=-f-i=V
2VVx + l Vx-1/
23. 1 + ^ .
Vx2 + 1
24. 2x/
25.
1
1
30.
31.
32.
33.
34.
41.
42.
.\^(3X2+1)2 \/ (3x2+1)
1
.)
(X + l)Vx2-l
X + 1
27.
28.
29.
(X - 1)2 Vx2 + 1
X-X2
(X2 + l)i(x8 + 1)*
2x2 + 1
Vl + X2
-2x.
a-'
(a2 - x2)^
X — Va2 + x2
cfi Va2 + x2
4 X (2 y2 _ x2)
2/(3 2/ -8x2)'
5x^-3x2
52/* -1
3 [x2y* + (X - 2/)2]
3 (X - 2/)2 - 4 x^ys '
3x2- 5 x<
4y8_i
35.
36.
37.
38.
X + Vx2 - 2/2
y
V
X - 2/(x - 2/)2
_6x. 25
2 y ' 2 2/3
x6 6 a^x^
y
18
6*
39. ; -
40
6a*y
3 2/2 - a2 (a2 - 3 y^f
3x2 6x(l-32/2)
32/2 + r (1 + 32/2)3 '
2(3x2-2 2/). 128x2/
3 j/2 + 4 X (3 2/^ + 4 x)3
ANSWERS
371
Page 210
43. tan., 3x + y + 6 = 0;
nor. ,x — 3y-\-b = Q:
tan. , 3x — 2/4-9 = 0;
nor. ,x + 3y — 7 = 0.
44. X - 6 y + 17 = 0, 6 X + 2/ - 9 = 0.
45. x + 2y - 2 = 0.
46. 4x-3?/-l = 0.
47. x + 2y -1 = 0.
48. (Sy^-xi)y-yiX-Xiyi-Sa=0.
49. X - 3 y^y + 2 Xi - 3 = 0.
50. 2yiy -3x^x + xl = 0.
51. (2 - xf)x + xfy - 3xi = 0.
52. xi~*x + yi~^?/ = a*.
53. Xi~^x + 2/i~i?/ = a^.
54. 52/ + Vl0x-6 VH = 0,
lOy- 5V1OX + 4 V5 = 0.
tan., 4x — 2/ — 6 = 0;
nor., x + 4y — 10 = 0:
tan., 4x — 2/ + 6 = 0;
nor. , X + 4 2/ + 10 = 0.
X
65
66
dy
ax
61,
dy
dx
(-0.57,2.08)
Page 211
62.
(±ia V2, ±
(1 '' ,
lbV2]
76.
62 N^
a22/ix - 62xi2/ = 0 ;
o262
63.
^6^x2
TT .
-, tan
4
73.
V Va2 + 62
Vp(p + Xi).
2+62/
77.
+ a*y{
age
212
78.
79.
80.
0, tan-i |.
TT
2'
TT
3'
86. -, tan-i|.
2 *
86. tan-13.
88. -, tan-i V2.
2
90. tan-i|.
91. 0, tan-isVs.
92. 0, -, tan-ijv^.
93. The length is twice
81.
tan-1 5 Vs.
89. tan-1/5.
the breadth.
94. He walks 2.86 mi.
Page 213
95. 8rd., 12 rd.
96. Cross section is a square.
97. Of equal length.
98. 4 mi. from nearest point on
bank to A.
99. jFD = (V2 -\)AB.
103. Area of ellipse is - area of
rectangle.
104. Central angle of sector is
|7r VO.
105. Breadth a, depth a Vs.
106. Breadth, f a V3; depth, fa V6.
Page 214
107. Velocity in still water \a mi.
per hour.
108. Radius of base equals alti-
tude.
109. Altitude is | radius of sphere.
110. a-
6m
Vn2 - m2
6n
mi. on land,
Vna"
mi. in water.
372
ANSWERS
111. Altitude is J '^ radius of
semicircle.
112. Altitude is jr distance between
vertex of pai-abola and bound-
ing straight line.
113. (&, a).
114. (±|V3, I).
115. maximum value when x = i,
minimum value when x — \;
points of inflection when x= — 1
1±V6
or
5
122. Stationary when < = 0, 4, or
moving backward when 4 <
123. 20, 10 V5; (100,20).
116. minimum ordinate, x = — -\
V3
maximum oixlinate, x— —
points of inflection
(± a, 0).
117. (±i V3, I).
/± - V27 -3 V33,
\ 6
V3
(0, 0),
119
± :^ V5 V33
12
120.(±|V^,±^>^).
121. (1, 3), (5, - 5).
8 ; maximum velocity when t
t<8.
rii)
= 1.
Page 215
124. Whenx= i; parallel.
126. Velocity of top: velocity of bottom = distance of bottom from wall
distance of top from ground,
126. 54 TT cu. ft. per hour. 132. 3 y = x» + 9z - 19.
127. 389 ft. per minute. 133. Sx-2xy -2 = 0.
128. 41.9 ft. per second. 134. 3y = 2x^ + U.
129. 15 ft. per second.
130. 0.2 hi. per .second.
131. 17.9 mi. per hour.
136. - -1 = A;(l-x).
y
137. 90,000, 677iV-
Page 216
138. 108.
140.
hk
n + 1
141. J «2. 142. 851. 143. K^ a^. 144. 10|. 145. 17 1
CHAPTER X.
Page 225
1. {- 1, 5), (- 7, 7), (2, - 5). 2. a;2 + 4 2/2 - 4 = 0.
3. 2/^ - 15 ?/2 -f 3 x2 + 75 y - 6 X - 106 = 0.
4. 62a;2 ^ a2y2 _ 2 a^by = 0.
5. 62x2 4. a^yi _ 2 ab^-x = 0.
6. 62a;2 _ a2y2 - 2 ab^ = 0.
X (a + x)2
7. 2/2 = -
Page 226
2a + X
10. y =
4 a3 - ox"
x2 + 4 a2 '
11.
y.^ (2a + x)8
8.
2/'' =
9. 2/ =
(X - a)2(2 a-x)
2ax2
xa + 4 o2
12. 2/^ = (^±^.
a-x
13. y^ = 4px -f 4p*
14. 2/2 = 4px — 4j)'.
ANSWEES 373
16. 2x2 + 32/2-6 = 0. ■ 4a 4 6
n.ah-c. 20. x2 + 9^2 + 6a;_36y + 36 = 0.
21. 196 x2 + 900 2/2 + 784 X + 5400 y + 8875 = 0.
22. x2 - 42/2 _ 2x- 162/- 19 = 0. 25. x2 - 8x + I62/ - 64 = 0.
23. 2x2-62/2-8x + 36y-47=0. 26. 3x2 + 42/2- 12x - 242/ - 27=0.
24. 2/2 + 42/-8x + 28 = 0. 27. 8x2 + 92/2 - 16x - 64 = 0.
Page 227
28. 5x2 - 42/2 + 10x- 162/ -31 = 0.
29. 2/2 + 4 2/ - X = 0.
30. i V5j (- 1, 2) ; (2, 2), (- 4, 2) ; (- 1 ± V5, 2) ; 6x + 5 ± 9 V5_= 0. "
31. I VlO; (-3, 2); (-3±V5, 2); (-3±V2, 2) ; 2x + 6 ± 5 -^2 =_0.
32. i Vl3; (3, -4); (5, -4), (1, -4); (3±vT3,-4); 13x-39±4 Vl3 = r, ;
3x-2y- 17 = 0, 3x + 22/- 1 = 0.
33. 1 VlO; (-1, 2)_; (-1 ±V2, 2); (- 1 +V5, 2); 5x + 5i:2 V5 = 0;
V3(x + l)±V2(2/-2) = 0.
34. (-1, ^); (-^, -I); 3x + l=0; 82/-7=0.
35. {- 2, - 3) ; (- I, - 3) ; 2/ + 3 = 0 ; 4x + 13 = 0.
36. (i V3, i), (i, - 1 V3), 40. x2 + 14 2/2 - 14 = 0.
/I + V3 1 - V3\ *2. xy = - 18, or xy = 18.
y ^' — 2 — ;■ ^^- 17x2 + 7 2/2-2 = 0,
37. x2- 42/2 -4=0. or 7x2+ 172/2-2 = 0.
,„ , x2(3aV2-2x) 44. x2±2/ = 0, or2/2±x = 0.
^^.y^ = -^ ^ : 45.2x2-2/2-1 = 0.
3 a V2 + 6x
Page 228
46.5x2 + 8 2/2 = 40. 48. 1^ + VV ^(^^^ - ^ + V
47. 4 X2/ = 7. ■ \x - 2/7 aV^ + x-y
49. 5x2-62/2 = 30.
CHAPTER XI
Page 244
1. Hyperbola; center, (—7, 2); slopes of axes, 2 and — \.
2. Parabola; slope of axis, \; vertex, (||, — 4if).
3. No curve.
4. Hyperbola ; center, (—1,0); slopes of axes, 1 and — 1.
5. Hyperbola; center, (2, — |); slopes of axes, 1 and — 1,
6. The line x — 2/ + 1 = 0 taken twice.
7. Elliijse; center, (— 1, 2); slopes of axes, 1 and — 1.
8. A pair of straight lines intersecting at (3, — 2), and having the slopes
- 1 ± i Ve.
9. A pair of straight lines intersecting at (3, — 2), and liaving tlie slopes §
and — |.
10. Parabola; slope of axis, 1; vertex, (— 4^, — \\).
374 ANSWERS
11. The parallel straight lines x + 3y-5 = 0, x + 3y-l = 0.
12. Ellipse; center, (2, - 1); slopes of axes, f and - |.
13. Point, (0, 2).
Page 245
20. tan-i :
A + B
23. 2x2 + 3xy + 2/2 + 12x-13?/-50 = 0.
24. xy - 2r/2 _ 2x + 4y = 0. 25. x^ - xy + y"^ - a"^ = 0.
26. 6x2 + 5x2/ + 2/* - 29x - 13?/ + 30 = 0.
27. 9x2-12xy + 4y2_ii7x + 78y + 380 = 0,
or 49x2 _ 66 xy + 16i/2 -621x + 354y + 1964 = 0.
V. -Jtan2^-1
1 - tan2^ if tan ^ < 1, if tan ^ > 1 (/3 the angle between
2 2 ^ 2
28
tan-
the lines). 2
CHAPTER XII
Page 262
1. 6x-y + 3 = 0; (0,3), (-1,-2). 5.(0,3).
2. X + y - 3 = 0 ; (0, 3), (1, 2). 6. (|, - \).
3. X- 2/ + 1 = 0; (-1,1). 7.(2,3).
4. 2/-2x + 6 = 0;(l,-3),(2,-l). 8. (1, -2).
Page 263
9. 8x-2y = 0,x-2/ + l=0. 14. x - 3?/- 2 = 0, 2x - 2/ + l = 0.
10. x = 0, x-2/ + l=0. 21 -.
11. 3x + 2/-l = 0. ■ a2
12. 2 X - 2/ = 0, X + 2 2/ - 10 = 0. 24. At infinity.
18. X + 2 2/ - 2 = 0, X - 3 y - 2 = 0. 25. 6ex - oj/ = 0, bx + aey = 0.
Page 264
30. -e. 81- 0^^ = 2 62.
Page 265
47. (ae, ± ^) ; tan-J(±e).
CHAPTER XIII
Page 297
21. a cos 2 ax.
22. o [sec* (ax + h) ctn (ax + c) - tan (ax + &) esc* (ox + c)].
28. -8csc24x. 26. sec2x.
2(2ctn2x — 1) 27. mn sec™ nx esc" wx (tan ?ix — ctn mx).
csc2x 28. 2sec2 2x(2tan2x + l).
3sec3x(tan3x-l) 29. - 2 csc2x(2 csc24x + ctn 4xctn 2x).
(tan3x + l)2 30. cos(xcosx) (cosx — xsinx).
ANSWERS 375
31. 5sin2xcos3x. 49. 2(x + l)e^ + 2»:.
32. 8 sec^x — 3 sec x. 2 a^
33. — — cosx.
V2
X* — a*
51. e^T^^^cF^^i-^^^ ^—\
V(l - X2)* Vl - xV
34. — == sin Vl-x2. V(l - x2)^
36. secxtanx. Vx^Tx
^■^,- .,...VH(i-!5?).
37. . 64. a'ana^loga- sec^x.
Vl-x2 55. 2x{l + 21ogx).
gg 1 66. loga-sec2x(sec2x + 2tan2x)at''n3-sec»a:.
(1 + x) Vx 67. 2sec2x(sec2x + loga- tan22x)a«e«''^
1 68. [2{a + x)sinmx + m cosmx]e(« + =')^
69. -2.
39
Va2 - x2
40.
a + X \x ^ ^ '
-, 1 ox.
41. — • e2a: + e-2x
(x + 1) vx
1 61.
62.
63.
42. -• 3 + 5sm2x
(x2 + 2 X) Vx2 + 2 X - 1 ^^ 2
43. 0.
44. ^^ 64. e^<:os«cos(a + xsina).
65.
45. -• sm(2x2 + 2a2)
66. cos-^x.
46.
47.
48. -
a + b cos X
2
Vl-X2
1
(2x-l) Vx2-x
1
2x2 + 2x + l
4x
CO
4x
67. sec ax.
68. (a2 + l)e««i""*'".
69. ctn-ix. .
cos-^x
70.
V(l - X2)8
X* + 1 71. ef^sinmx.
Page 298
72. 2xctn-i«. ^g_(x-l)log(x_D.
* (x2-2x)3
sin-ix 1
— :;i 77.
78.
a + b cos X
74. ^ _ 1 78. 4csc(4x + 2)[l-ctn(4x + 2)].
Vx2 - Ofi Va2 - X2 Yg ^ .
76
1 jx + a
X \x — a'
V2 ax - a;2
80. 2 Va2 + x2.
376 ANSWERS
sec- 12 Vx 91. 2/x^(l + loga;).
• Vx(4x-1)3' 92 yrtan-i(a + x) ^ log(a + x)-|
82- V2ax-x2. L « + ^ l+(a + x)d'
83.
2 - Vx2- 1 2/ (tan xy )
2x(x2-l) ■ 93
84. -
, log X — X tan xy
y
(e^ + 2) Ve2^+ 2e^-l ^^^fTxa
(e2x+e-)log(e^ + 2)^ 94. ^^^^^^_,^
V(e2a: + 2ex_i)3 ^^ y C08X + sin (x - y)
85. X tan-i Vl — x^. ' sin {x — y) — sinx
86. ^^i±^'. ' 96. '"^
2 + x2 {ny + \)x
V r r nH e2' sin X + e^ sin w
87. -^(sec2 Vxlogsin Vx + l). 97. ^.
2 Vx ^ ° 01 cos X — e* cos y
y a» y(l -cos x)- cosy
88. i;^(xctnx-logsinx). M- ^i^^ _ ^(i + ,5^^)' .
89. yx^fl + logx + (logx)2l. 99. ^^^^V -V
L* J X log X — x2 cos y
90. ye^(l + logxV IQQ. Vd-^^-y^)
\a^ / x(l+x2 + y2)
Page 299
101 ^ + ^, 2(x2+y2) 4(x + 2y)(x2 + y2)
■ X - y (X - y)3 (x - y)5
102. - e^-y, - e*-J'(H- e^-p), - e^-2'(l + e^-») (1 + 2e*-!').
103 ^-y' 2(x + y)(y2-l) 6 (x + y)2(l - y^)
■ X2 - 1 ' (X2 - 1)2 ' (pfl - 1)3
104 ^ + y -^ 4 (X + 2/) 8(x + y)(l-2x-2y)
■ x + y + l' (x + y + l)3' (x + y + l)6
jQg y-g y(l+logx)-2x 2y[l + logx + (logx)2]-3x(l+logg)
■ x(l-logx)' [x(l-logx)]2 ' [x(l-logx)]3
- 1 ± V33
106. X
8
107. 1 or 2. ,,„ , o ; , '^
112. X = «7r, X = 2 KTT ± - •
108. tan-i2V^. 3
113. (ae^ fae-^).
109. tan-i(2tan-sec-V
\ 2 2/
110. tan-4, tan-13.
114.
(^^--*)
116. Maxima when x = (2 A; + 1)-, minima when x — kir] points of inflection
-when X = (2fc + !)-•
117. Maxima when x =(2fc + |)7r, minima when x =(2*; + |)7r; points of in-
flection when X = kit.
ANSWERS
377
118. Maximum when x_= n, minimum wlien x = 0, (n = 2fc): points of
inflection when x = n ± Vn, (n ?i 1) ; x = 2, (n = 1); x = 0, (n = 2 k + 1).
119. Circle.
120. 2 V(s - 3) (5 - s), 4(4-8).
Page 300
122. a.
123.
ab
128.
500 sin a
; , where x is
2 V62_a2t2•
126. -bsind
62 sin 5 cos ^
times
Vl0,000-x2sin2a
the distance from the center.
1 29 . 1 5 sq. ft. ; 9. 03 sq. ft. per second.
Va2-
62 sin2 e
130. each.
velocity of AB.
1 where
2 2
e = CAB.
131. 23.7.
500
132. At an angle tan-^fc with the
127.
, wl
lere x is the
Vl0,000-x2
ground.
distance from the center.
Page 301
-
133.
134.
Sin.
1:V2.
145. log- = ft(x-a).
6
135.
5V5ft.
146. s = ce^'.
136.
2.
147. A7r + (-l)«^.6661.
137.
V2-I.
148. 2fc7r ± .567, 2kTr ± 2.206.
138.
2.
149. ktr.
139.
e-1.
150. fcTT, (2A: + l)j, 2kir±1-
4 0
140.
e
151. A;7r ±^, ^^'ri -•
4 0
141.
log 2.
142.
X = 2 a.
152. 4.4934.
a j '^ *\
163. 4.275.
143.
- (ga _ e «) -f c.
154. 0.199.
«
155. -0.7085.
144.
2/ - 3 = & log - .
2
166. 1.857, 4.54.
CHAPTER XIV
Page 323
2. tan., X - iy + p«2 = 0 ; nor., tx + y -2pt- pt^ = 0.
4p 4p "
3. X = — ^» 2/ =
<2 ' ^ i
4. tan., <2x _ 2 «y + 4p = 0 ; nor., 2 t^x + t^y - ipt^ -Sp = 0.
ab , aht
6. r.=z± —— y = J:— —
y6-2 + a2«2 V62 + a2«2
_ a(62- OT2a2) _ 2a62m ^
■ ^ - 62 + m2a2 ' ^ ~ 62 + m2a2 '
. „ 2asin3(^
7. x = 2asm2<^, y =
COS0
10. (l + 3«2)x-2i3y-2a=:0.
378 ANSWERS
Page 324
■ ^ 2x-a 3i2 + l
12. a;3 + 1/3 _ 3 oxy = 0. 16. (3 y - x)2 = 2 Vx2 - 2/2.
13. 2/2 (ax - a2) = x2 (a2 + A;2 - ax). 17. 9 y = (x - ?/)2 - 6 (x - 2/).
14. 3 2/2 - 4 xy + 2 2/ - 1 = 0. 18. (x2 + y^) (ox + 62^) = cxy.
/— - — tan'
■ll
19. Vx2 + 2/2 = a V2 e^
o^ a/.„^^\. 2x + a ja-x
20. X = a cos2 - , 2/ = - I sin ^ + tan - ; ?/ = \ ~
22\ 2/ 2\x
21. x = (a — c tan^) sin2^, 2/ = (acosfl — csin^)sin^; 2/(x2 + 2/2) = x(a2/ - ex).
22. X = - (a2 + A;2 cos2 ^), y = - {a^ tan » + fc2 gin 0 cos fl) j
a(x- a) (x2 + 2/2) = i2a;2.
a{a2 — x2)
23. X — a tan ^, 2/ = a cos 2 ^ ; 2/ = -^; r-^ •
a2 + x2
Page 325
24. X — a sin d (cos ^ + sec 6), y = a cos ^ (cos d + sec tf) ;
2/(x2 + 2/2) = a(x2 + 2 2/2).
29. 6x cos 0 + a2/ sin <^ — a6 = 0.
30. !^. 31. !r + ^. 32. ^sin-i?-^
4 4 2 2 «„^
8,. ,,„.,'l±:^<±lMs:^\
gb
34. X = a cos ^ + i V62 _ a2 sin"^ S, y = {I — I) a sin ^, where the center of
the driving wheel is the origin, a the length of the radius of the driv-
ing wheel, b the length of the connecting rod, and lb the distance along
the rod from the wheel to the point.
35. Straight line.
37. 2 aw sin - ; w Va^ — 2 aAcos 0 + ^2^ where u is the constant angular
velocity.
Page 326
38. 2 au sin - , w v a'*' — 2 ah cos 6 + h^. ■
2
40. adw, where a is the radius of the circle, ad the distance through which
the point of the string in contact with the wheel has moved along the
rim of the wheel, and w the constant angle of velocity.
41. X = a(cos2e + sin2tf), 2/ = a(l + sin2^ - cos2d) :
2 V2 aw, 2 a (cos 2 ^ — sin 2 ^) w, 2 a (sin 2 ^ + cos 2 ^) u.
^„ „ x2(3a + x) 48. The witch.
43. 1/2 =: ^ •
a — x 49. X = a (cos 6 + 0sm 0),
43. Circle. y = a (sin 0 — 0 cos 0).
46. ?/ (x2 + 2/2) = 2 a (x2 + 2 2/2). 50. x = a(l 4- m2), ?/ = ma (1 + ?n2) ;
47. y = x — a. ay^ = x2(x — a).
ANSWERS
379
Page 327
53. Ellipse.
54. Hyperbola.
55. Straight line.
56. Concentric circle.
57. z + 2p-0, py^ = afi.
59. Concentric ellipse.
61. Ellipse.
Page 328
65. Parabola.
67. Concentric circle.
70. Concentric circle.
TTX
71. y = xctn
2a
72. x = a{4) + sm<t>),
y = a{-l + cos^).
74. 8 a.
8a{a + b)
b
75
Page
26.
27.
28.
349
CHAPTER XV
(l 0353 a, ^y
(l^-l)'(i^'T)-
29. (0,0),(a,±^),(a,±^)
80. Circle.
31. r = acos2^.
a cos 2 0
32. r
coa^9
Page 350
40. r^ sin2(? cos ^ + (2 a + r cos 6)^ = 0.
41. r cos 61 = a cos2 (?, (r^ + a^) cos^ + 2 ar = 0
49. (x2 + 2/2)3 _ 4 a-^x^y^ = 0.
60. (x2 + y^ + ax)2 - a2(x2 + y"-) = 0.
61. (a;2 4- 2/2)2 _ 2 a2 (x2 _ 2/2) + a* - 6* = 0.
82. {X2 + 2/2 _ ax)2 = 62(a;2 + 2/2).
63. (X - a)2 (x2 + 2/2) = &2x2.
54.
log(x2 + y2)-2atan-i^-
56. 25,000,000.
57. 1,200,000, or 4,800,000.
Page
59.
61.
62.
63.
2c
64. r =
351
r = -
1 — costf
Straight line.
Circle.
Circle.
2 ce cos ^
65. r =
66. 2r =
ce2 cos d
— eP'CO&^B
ce
1 — e cos B
72. «/'(«), « Vt/(«)]2 + [/'(fl)]2.
1 - e2 cos2 ^
78
^r(e)
V[/(e)]2 + [ne)f V[/(fl)]2 + [/'(<?)]«
380 ANSWERS
Page 352
74. w Va2 + 62 ^. 2 ab cos 6. ^9 J_(e2.a_i\
75. Spiral of Archimedes. 4 a
76. 2a2. 80. \ira^.
tt. ia2. 81. ia2 (4 -■»■).
78. i7r%2. 82. i,r(a2 + 2 62).
83. 8 a.
CHAPTER XVI
Page 362
a
2 ox' (8 g - 3 x)^
4. \ a, a.
a?
3(2a-x)2 ■ *• 36"
Page 363
6. fa2, |a2. ,o 8 ■ «
-X, 13. -sm
7. e '^. 3 2
8. 0, \. ,4 „,__4
10.
14. y2_ (a; _ 2r>)s,
a(6-4co8g)8 27p^
9-6cos« i« a Va2 - .r^
3 fl ^
11. -asin2-; greatest value, - a ;
15.
-U.0111--; t-iettLest value, -a ,o , .j ,i .-i
4 3^ '4 ' 18. (a; + 2/)J + (j-- ?/)1^ 2.0
least value, 0. 21. «2 ^ ^2 _ ^ _ q.
12. 2-^.
3r
Page 364
23. Minimum curvature when x =
24. Maximum curvature when x = (2 A; + 1) — ;
minimum curvature when x = krr.
25. Maximum curvature at ends of major axis;
minimum curvature at ends of minor axis.
L \dx/ J dx3 dx \dxV
28. (2fe + l)|.
INDEX
[The numbers refer to the pages.]
Abscissa, 36
Acceleration, 202
Addition of segments of a straight line,
32
Algebraic functions, 43, 121
differentiation of, 178
implicit, 188
Angle between two lines, 57
between two curves, 211
eccentric, 304
vectorial, 329
Angles, 65
Arc
length of, 195
limit of ratio to chord, 195
derivatives with respect to, 197, 347
Archimedes, spiral of, 332
Area, 204
of an ellipse, 304
in polar coordinates, 348
Asymptote, 128
of an hyperbola, 145
Auxiliary circle of ellipse, 304
Axes
of an ellipse, 141
of an hyperbola, 145
Axis, of symmetry, 121
of a parabola, 147
radical, 175
Bisection of a line, 39
Cardioid, 337
Cassini, ovals of, 338
Catenary, 281
Center of a conic, 238
Change of origin without change of
direction of axes, 217
of direction of axes without change
of origin, 221
from rectangular to oblique axes
without change of origin, 224
from rectangular to polar coordi-
nates, 341
Chord of contact, 248
Circle, 134
through a known point, 136
tangent to a known line, 136
with center on known line, 137
through three known points, 138
parametric equation of, 303
involute of, 311
polar equation of, 342
of curvature, 354
Cissoid, 151
Classes of functions, 43
Coefficient of an element of a determi-
nant, 8
Collinear points, 38
Complex numbers, 31
roots of an equation, 82
Components of velocity, 200
Concavity of a curve, 112
Conchoid, 334
Conic, l48, 229
classification, 237
through five points, 241
polar equation, 343
Conjugate complex numbers, 32
axis of an hyperbola, 145
diameters, 258
hyperbolas, 262
381
382
INDEX
Constant, 40
of integration, 206
Contact, point of, 105
chord of, 248
Continuity, 101
Coordinate axes, 35
cliange of, 217
Coordinates
rectangular, 35
transformation of, 217
oblique, 223
Cartesian, 224
polar, 329
relation between rectangular and
polar, 341
Curvature, 353
radius of, 354, 360, 361
circle of, 354
center of, 356
Curve, Cartesian equation of, 44
slope of, 99
degree of, 166
parametric equations of, 302
polar equation of, 330
Curves, intersection of, 161
Curves of second degree, 229
Cycloid, 305
Degree of a curve, 166
Depressed equation, 79
Derivative, 102
of a polynomial, 97, 103
sign of, 106, 111
second, 110
higher. 111, 187
theorems on, 179
of w, 185
illustrations of, 203
■with respect to an arc, 196, 347
Descartes' rule of signs, 87
folium of, 132
Determinants, 1
elements of, 4
minors of, 4
properties of, 6
expansion of, 8
Diameters, of a conic, 252
of a parabola, 254
of an ellipse, 256
of an hyperbola, 257
conjugate, 258
Differentiation, 102
of a polynomial, 103
of algebraic functions, 178
successive, 187
of implicit functions, 188
Differentiation, formulas of
for a polynomial, 103
general, 184
for M», 185
for trigonometric functions, 272
for invei-se trigonometric func-
tions, 276
for exponential functions, 284
for logarithmic functions, 284
for hyperbolic functions, 290
for inverse hyperbolic functions,
291
Direction of a curve, 197
in polar coordinates, 345
Directrix
of a parabola, 146
of a conic, 148
of an ellipse, 149
of an hyperbola, 149
Discontinuity, 101
examples of, 41, 128, 268, 282
Discriminant, 117
of a quadratic equation, 73, 117
of a cubic equation, 114, 117
of the general equation of the sec-
ond degree, 236
Distance between two points, 36
of a point from a straight line, 63
e, the number, 280
Eccentric angle of ellipse, 304
Eccentricity of a conic, 148
Elasticity, 204
Elements of a determinant, 4
Eliminants, 23
Elimination, 1
INDEX
383
Ellipse, 139
referred to conjugate diameters as
axes, 269
parametric representation of, 303
Energy, kinetic, 203
Epicycloid, 307
Epitrochoid, 309
Equation in one variable
solution by factoring, 77
with given roots, 79
depressed, 79
number of roots of, 80
sum and product of roots of, 82
complex roots of, 82
solution of, 89
Newton's method of solution of,
114
multiple roots of, 116
resultant, 161
Equation of a curve, 44
Equations in several variables,
linear, 1
systems of, 12
homogeneous, 21
Equations in two variables
of first degree, 52
of second degree, 229
Equations, transcendental, 293
Evolute, 357
Expansion of a determinant, 8
coefficient of, 204
Explicit algebraic function, 188
Exponential functions, 279
differentiation of, 284
Factoring, solution of equations by, 77
of quadratic expressions, 79
Factors and roots of an equation, 78
of a polynomial, 81, 83
Foci of an ellipse, 139
of an hyperbola, 142
Focus of a parabola, 146
of a conic, 148
Folium of Descartes, 132
Force, 202
Function, 40
Functional notation, 44
Functions, classes of, 43
algebraic, 43, 121
irrational, 44, 131
transcendental, 44, 266
defined by equations of the second
degree, 127
involving fractions, 128
trigonometric, 266
inverse trigonometric, 269
exponential, 279
logarithmic, 279
hyperbolic, 288
inverse hyperbolic, 291
Graph, 40
Graphical representation, 28
Harmonic property of polars, 249
division of a line, 250
motion, 275
Homogeneous equations, 21
Horner's method, 92
Hyperbola, 142
equilateral, 146
referred to asymptotes as axes, 224
referred to conjugate diameters as
axes, 259
conjugate, 262
Hyperbolic functions, 288
inverse, 291
differentiation of, 290, 292
Hyperbolic spiral, 332
Hypocycloid, 309
four-cusped, 132
Hypotrochoid, 309
Imaginary numbers
(see Number, Complex)
Implicit algebraic function, 188
Increment, 100
Infinity, 29, 128
Inflection, points of, 112, 194
Initial line, 329
Integration, 205
Intercepts, 53
384
INDEX
Interchange of axes, 223
Intersection of curves, 161
number»of points of, 169
Involute, 357
of circle, 311
Irrational number, 28
algebraic functions, 44, 131
roots of an equation, 92, 114
Isolated point, 126
Kinetic energy, 203
Latus rectum, 211
Limit of ratio of arc to chord, 195
. sin ft , 1 — cos h _^-
of and —, 270
h h
of (1 + h)'' and
1
283
Limiting cases of a conic, 234
Limits, 97
theorems on, 178
Locus, 45
Locus problems, 316
Logarithm, Napierian, 280
Logarithmic function, 279
differentiation of, 284
spiral, 333
Lemniscate, 340
Lima^on, 336
Maxima and minima, 108, 112, 192
Minors of a determinant, 4
Momentum, 203
Motion, uniform, 199
harmonic, 275
Multiple roots of an equation, 116
Napierian logarithm, 280
Newton ■;S method of solving numerical
equations, 114
Normal, 64, 191
Normal equation of straight line, 64 '
Number, real, 28
complex, 31
Oblique coordinates, 223
Ordinate, 36
Ovals of Cassini, 338
Parabola, 146
referred to tangent at ends of
latus rectum, 132
referred to a diameter and a tan-
gent as axes, 255
cubical, 74
semi-cubical, 131
Parallel lines, 56, 59
Parametric representation of curves,
302
Pedal curves, 319
Perpendicular lines, 57, 59
Plotting, 36, 329
Point of division, 38
Polar of a point, 247
Polar coordinates, 329
Polars, reciprocal, 251
Pole of a straight line, 247
of a system of polar coordinates,
329
Polynomial, 43
of first degree, 50
of second degree, 70
of nth degree, 74
factors of, 81, 83
derivative of, 97
\ square root of, 121
Problems on straight lines, 58
Products, graphs of, 83
Projection, 34
Radical axis, 175
Radius vector, 329
of curvature, 354, 360, 361
Rate of change, 203
Rational algebraic function, 43
number, 28
roots, 89
Reciprocal polars, 251
Resultant, 23
equation, 161
Roots of an equation
and factors, 78
INDEX
385
Roots of an equation
number of, 80
sum and product of, 82
complex, 82
location of, 86
rational, 89
irrational, 92, 114
multiple, 116
Rose of three leaves, 331
Rotation of axes, 221
Slope of a straight line, 64
of a curve, 99
Solution
of simultaneous equations, 12
of algebraic equations, 77, 89, 114
of transcendental equations, 293
Spiral
of Archimedes, 332
hyperbolic, 332
logarithmic, 333
Straight line, 50
satisfying two conditions, 58
normal equation of, 64
parametric equations of, 302
polar equation of, 342
Strophoid, 152
Subnormal, 210
polar, 351
Sub tangent, 210
polar, 351
Supplemental chords, 264
Sylvester's method of elimination, 24
Systems of curves with common points
of intersection, 171
Tangent, 73, 84, 104, 190
to a conic with given slope, 163
to a conic at a given point, 246
Tractrix, 299
Transcendental functions, 44, 266
equations, 293
Transformation of coordinates, 217,
341
Transverse axis of hyperbola, 145
Trigonometric functions, 266
differentiation of, 272
inverse, 269
differentiation of, 276
Trochoid, 306
Turning points of a graph, 107
Variable, 40
Variation of sign, 87
Vector, radius, 329
Vectorial angle, 329
Velocity, 198
components of, 200
Vertex, 71
of a parabola, 147
Vertices
of an ellipse, 141
of an hyperbola, 144
Witch, 149
Zero, 29
Date Due
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