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Full text of "Introduction to the elementary functions"

:ti(m 



TO THE 



MENTARY 
JNCTIONS 



fcGLENON AND RUi 






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MI6 



Gfarnell Untocrattg Slibrarg 



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Cornell University Library 
QA 331.M16 

Introduction to the elementary functions 



3 1924 003 988 148 



u 



,f 



INTRODUCTION 

TO THE 

ELEMENTARY FUNCTIONS 



BY 

RAYMOND BENEDICT McCLENON 

WITH THE EDITORIAL COOPERATION OF 

WILLIAM JAMES EUSK 



GINN AND COMPANY 

BOSTON • NEW YORK • CHICAGO • LONDON 
ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO 













COPYRIGHT, 1918, BY 
GINN AND COMPANY 



ALL HIGHTS RESERVED 
118.8 



Vjt gtjitnguiii grend 

GINN AND COMPANY • PRO- 
PRIETORS ■ BOSTON • U.S.A. 



PREFACE 

This book is an attempt to solve the problem of the first-year 
collegiate course in mathematics. That the problem is a very 
real one is attested by the many discussions constantly taking 
place among teachers of mathematics and others interested in 
education. The traditional Freshman course, consisting of " college 
algebra," trigonometry, and solid geometry or elementary analytic 
geometry, is very generally regarded as unsatisfactory. 

There are three main objections to this traditional course : first, 
it is not unified, so that it sacrifices time and fails to hold the 
student's interest; secondly, much of the subject matter should 
come after a first course in calculus, when it would gain vastly 
in significance; thirdly, the usual plan has deprived the large 
majority of college students of any introduction to the calculus, 
which is the heart and soul of modern mathematics and natural 
science. Only that small number electing to go beyond the first 
year of collegiate mathematics have the opportunity to become 
acquainted with the subject, which unquestionably represents one 
of the most important lines of development of human thought 
during the past two centuries. 

Accordingly, we decided to construct a course with the funda- 
mental idea of functionality as its unifying principle, and leading 
up to some elementary work in calculus as its culmination. The 
advantages of this arrangement are that it not only meets the 
objections stated in the preceding paragraph, but saves time 
by avoiding the repetition inevitable in the triple arrangement 
of subjects ; and, what is more important, it leads to a deeper 
understanding of the significance of mathematical principles 
and relations than the student is likely to gain through the tra- 
ditional course. Thus, whether he goes farther into the study of 



iv THE ELEMENTARY FUNCTIONS 

mathematics or has but one year for the study, we believe that 
he will be the gainer by taking the unified course. 

The course presented in this book is the result of our expe- 
rience in Grinnell College, mimeographed copies having been used 
and revised during five successive years. The material included 
comprises the simpler and more important parts of plane trigo- 
nometry and analytic geometry, followed by an introduction to 
the differential calculus, including differentiation of the simpler 
algebraic functions and applications to problems of rates and 
maxima and minima. The conic sections are not studied as 
extensively as in most textbooks of analytic geometry, but 
enough has been given to make the student feel familiar with 
these important curves. 

The trigonometric functions are introduced early, and the gen- 
eral definitions are given at once, instead of those valid only for 
the acute angle. Thus the connection with the coordinate system 
is established from the first, and a clearer idea of the meaning of 
the trigonometric functions is obtained than if the student's atten- 
tion is for some time confined to the case of the acute angle. The 
applications of the trigonometric functions to the solution of right 
triangles and problems depending upon them is made without the 
use of logarithms, as experience shows that the early introduction 
of logarithms may easily lead to mechanical methods of work. 

The number of numerical exercises is not large, as the teacher 
can easily supplement those in the text by as many others as he 
wishes. "We feel that the purely computational work can easily 
be overdone in the first-year course. Four-place tables may be 
used in this work ; the Wentworth-Smith tables, or others of like 
nature, are very satisfactory. 

The arrangement is almost exclusively inductive, and the style 
direct and informal throughout. The explanations are not always 
as detailed as in many texts, the object being to lead the stu- 
dent to supply the connecting links for himseK, where they 
are not explicitly given in the text. Moreover, many of the 
important results are altogether left to the student as exercises. 
In such cases, as in the case of all the most important formulas 



PEEFACE V 

throughout the book, attention is called to their importance by 
the use of black type. 

The course will be found suitable for an advanced course in 
the secondary school, as well as for the first year in college, and 
in this case it might very well be made a five-hour course. For 
the student who has somewhat lost touch with his previous work 
in mathematics, a small amount of review matter has been placed 
in appendixes. If this work has to be taken, it will probably be 
unwise to attempt to cover all of the text, and certain paragraphs 
have accordingly been starred, to indicate that they may be omitted 
without interfering with the unity of the course. 

We have not gone so far in the way of radical changes in 
subject matter as our personal feelings would lead us, because 
we believe that progress in the teaching of mathematics, as in 
everything else, should be in the nature of evolution rather than 
revolution. For instance, no work in the integral calculus is 
given, although we firmly believe that this topic should eventually 
be included in the first-year course; but it seemed to require 
a greater departure from the traditional course than is as yet 
practicable. We hope that the present book may prove a contri- 
bution to the solution of the problem presented by the first year 
of college mathematics, and that experience will indicate what 

further steps may advantageously be taken. 

R. B. M. 

W. J. R. 




»1 



Cornell University 
Library 



The original of tiiis bool< is in 
tine Cornell University Library. 

There are no known copyright restrictions in 
the United States on the use of the text. 



http://www.archive.org/details/cu31924003988148 



CONTENTS 

PAGE 

CHAPTER I. THE GRAPHICAL REPRESENTATION 1 

Measurement. Construction of segments of given length. Directed 
segments. Correspondence between numbers and points on a straight 
line. The coordinate system. Application to some problems of ele- 
mentary geometry. Point that divides a segment in a given ratio. 

CHAPTER II. FUNCTIONS AND THEIR GRAPHS 19 

Variables and the equation of a locus. Graphical representation of 
functions Importance of functional relation. Graphical representation 
of statistical data. 

CHAPTER III. APPLICATION OF GRAPHICAL REPRESENTA- 
TION TO ELEMENTARY ALGEBRA 28 

Graphs of linear equations. Graphical solution of simultaneous linear 
equations. Determinants ; applications to solution of simultaneous 
linear equations with two or three unknowns. The quadratic func- 
tion. Graphical and algebraic solutions of quadratic equation. Alge- 
braic solution by completing the square, by formula, and by factoring. 
Test for solvability of quadratic equation. Graphical interpretation of 
discriminant test. Construction of tangent to a parabola. Sum and 
product of the roots of a quadratic equation. Factor theorem for 
quadratic equation. Maximum and minimum values of a quadratic 
function. 

CHAPTER IV. INTRODUCTION TO THE TRIGONOMETRIC 

FUNCTIONS 61 

Definition and measurement of angles. Definition of the trigonomet- 
ric functions. Simple relations among the trigonometric functions. 
Applications of the trigonometric functions to the solution of right 
triangles, to problems in velocities and forces, and to the slope of a 
straight line. 

CHAPTER V. SOME SIMPLE FRACTIONAL AND IRRATIONAL 

FUNCTIONS. THE LOCUS PROBLEM ... 85 

Graphs of rational functions ; asymptotes. Graphs of irrational func- 
tions. Definition of locus. Simple examples of locus problems. 

vii 



viii THE ELEMENTAEY FUNCTIONS 

PAGE 

CHAPTER VI. THE STRAIGHT LINE AND THE CIRCLE ... 97 
Equation of straight line through a given point and having a given 
slope. Proof that the equation of every straight line is of the first 
degree in x and y, and conversely. Normal form of equation of 
straight line. Distance from a line to a point. General equation of 
the circle. 



CHAPTER VII. THE PARABOLA, ELLIPSE, AND HYPERBOLA 114 

Equation of parabola. Construction of parabola. Definition and 
equation of ellipse. Definition and equation of hyperbola. Equation 
of its asymptotes. 

CHAPTER VIII. SIMULTANEOUS EQUATIONS 141 

Intersection of straight line and conic. Tangent to a conic. Alge- 
braic solution of certain types of simultaneous quadratic equations. 

CHAPTER IX. FURTHER STUDY OF THE TRIGONOMETRIC 

FUNCTIONS. POLAR COORDINATES 154 

The Law of Sines. The Law of the Projections. The Law of Co- 
sines. Solution of the oblique triangle. The half-angle formulas. 
Functions of the sum and difference of two angles. Sum and differ- 
ence of two sines or cosines. Graphical representation of the trigo- 
nometric functions. Polar coordinates. 

CHAPTER X. THE EXPONENTIAL AND LOGARITHMIC FUNC- 
TIONS 188 

Laws of operation with exponents. Definition of logarithms. Laws 
of logarithms. Use of logarithms in computation. Derivation of the 
half-angle formulas for the triangle. MoUweide's Formulas and the 
Law of Tangents. 

CHAPTER XI. INTRODUCTION TO THE DIFFERENTIAL CAL- 
CULUS 203 

Average rate of change, and instantaneous rate of change, of a func- 
tion. Limits. Theorems on limits. The derivative. Its use in find- 
ing the slope of the tangent to a curve. Rules for finding derivatives. 
Application of the derivative to drawing graphs, and to maximum 
and minimum values. Differentiation of irrational and implicit func- 
tions. Various applications. 



CONTENTS ix 

PAGE 

APPENDIX A. PROOF THAT THE DIAGONAL OF A SQUARE 

IS INCOMMENSURABLE WITH ITS SIDE 233 

APPENDIX B. LAWS OF OPERATION WITH RADICALS ... 235 

APPENDIX C. to CONSTRUCT A SEGMENT HAVING A 
RATIONAL LENGTH 236 

APPENDIX D. TO CONSTRUCT THE SQUARE ROOT OF ANY 
GIVEN SEGMENT 237 

APPENDIX E. SIMULTANEOUS LINEAR EQUATIONS IN TWO 
UNKNOWN QUANTITIES 238 

APPENDIX F. THE QUADRATIC EQUATION IN ONE UNKNOWN 
QUANTITY 240 

INDEX 243 



THE ELEMENTARY FUNCTIONS 

CHAPTER I 

THE GRAPHICAL REPRESENTATION 

1. Measurement. The study of algebra is concerned witli nuTn- 
bers ; that of geometry, with space. The object of this chapter is 
to bring into closer relation these two portions of elementary 
mathematics, in order that each may be used to help the other. 
The simplest process in which number and space are combined is 
measurement. 

2. The important r61e which measuring plays in the work of 
the world is so apparent to everyone that it needs no mention. 
The nature of this important process, 

as applied to distances, is made clear ' ' ' ' ' ' 

by the following considerations : If c D 

we wish to measure a line segment 
(that is, a limited portion of a straight 

line) AB, we begin by choosing any convenient line segment CD, 
which we call a unit ; we then apply CD to AB so that C coin- 
cides with A, when D will fall upon H, a point between A and B, 
in case AB is > CD. Thus AE=CD. We next apply CD to EB, 
C coinciding now with E, and D falling at F. Thus EF=CD, 
and AF= 2 • CD. By continuing this process again and again 
we may eventually find that D falls upon B, let us say, the mth 
time we applied CD successively. Then AB = n • CD, and we say 
the length of AB is n units. We also say the unit is contained 
exactly in the segment AB. Whenever this happens, the length 
of the segment is an integer. 

But in practice this does not usually happen. More often the 
unit will be contained exactly, let us say n times, in a segment 
AB', less than AB, whereas the remainder B'B of AB is less than 

■1 



Fig. 2 



2 THE ELEMENTARY EUNCTIONS 

the unit CD (Fig. 2). We can then assert that the length of AB 
is greater than n units but less than n + 1. That is, we have 
measured the segment AB to the nearest integer less than its 
length. If we wish to get a closer approximation, we may divide 
the unit CD into any number of parts, ^, ^ 

as k, and proceed to measure the seg- ' ' ' ' ' ^~^ 

ment B'B, using one of these parts of ^ ? 

CD as a new unit. If this part of the 
unit is exactly contained in B'B, say 
m times, then B'B = m kt\\s of a unit, and the length of AB 

equals w + — • This number can be written ; > that is, in 

k k 

the form of the quotient of two integers. 

If it happens that the ^th part of CD is not contained exactly 

in B'B, m of these parts being less than B' B, while w + 1 of them 

exceed B'B, we can at any rate assert that the length of AB is 

yyt fyry I "1 

greater than « + — units but less than n -\ ; — : that is, we 

k k 

have measured AB to the nearest kt\\ part of the unit less than its 

length. If we wish a stUl closer approximation, we have only to 

repeat the same process, taking a smaller fractional part of CD than 

we did before, that is, taking a larger value of k. How large a 

value of k we take (that is, how small a fractional part of CD we 

use) is a question that purely practical considerations answer. If 

the unit is the foot, often k = 12 would be sufficient ; that is, we 

should be satisfied with measurement which is carried out to the 

nearest inch. We might, however, wish to know the length of 

our segment to the nearest tenth of an inch, in which case we 

should of course take ^ = 120, the unit being a foot. 

The careful thinking through of this process will prepare the 
student to realize the truth of the following fact : The length of 
any line segment (measured by any unit whatever) is always given, 
either exactly or approximately, as an integer or as the quotient of 
two integers (that is, either as an integer or as a fraction). 

The actual measurement of several concrete, objects should be 
carried out by the student with special reference to the degree 



-5 la S 



Fig. 3 



THE GEAPHICAL EEPEESENTATIOK 3 

of accuracy attained in cases where the segment measured does 
not seem to contain exactly the unit or the part of the unit 
chosen. Evidently a segment might happen to have such a length 
that some fractional part of the unit would be contained exactly 
in it, even though repeated trials might not show just what part 
is so contained. Thus, if a segment were exactly 3^3^ units long, 
a measurement that gave the length to the nearest third of a unit 
would seem nearly exact, as 3| is only ^L too small ; and if we 
used tenths of a unit, we should get 3J^ as the length, — a result 
that differs from the actual length by only -^^^. For most pur- 
poses this would be an entirely sufficient degree of accuracy, 
but we should never find an 

exact measure of the segment ' 

unless we divided the unit into 
precisely 13 parts (or some mul- 
tiple of 13). Considerations of 
such examples of measurement as this one (and each student 
may easUy work out several for himself) may very well suggest 
the conclusion that any segment could be measured in terms of 
a given unit and fractional parts of it, if we could only dis- 
cover the correct fractional part to try. This conclusion, plausible 
though it seems, is, however, not correct; there exist segments 
which can never he measured hy a particular unit nor ly any 
fractional part of it. One example of such a segment is the 
diagonal of a square whose side is the unit.^ Such segments 
are said to be incommensurable with the unit. The student will 
no doubt recall numerous examples. Of course these incom- 
mensurable segments can be measured to any desired degree of 
approximation, exactly as is done in the case of such a seg- 
ment as the one mentioned just above, that was assumed Sj-^^ 
units long. 

3. Summarizing the results thus far obtained, we see that in 
measuring a line segment, after choosing a unit length, one of 
three things can happen : (1) the unit is contained exactly in the 

1 See Appendix A for a proof that the above statement is true of the diagonal 
of a square. 



4 THE ELEMENTARY FUNCTIONS 

segment to be measured; in this case the length of the segment 
is an integer; or (2) the unit itself is not contained exactly in 
the segment, but some fractional part of the unit is ; in this case 
the result of the measurement is a fraction, that is, the quotient 
of two integers ; or, (3), neither the unit nor any fractional part of 
it is contained exactly in the segment ; in this case the segment 
is incommensurable, and the length can be only approximately 
given as an integer or fraction. This length is, however, still 
spoken of as a number, but is called an irrational number, while 
the lengths of commensurable segments (that is, either integers 
or fractions) are called rational numbers. 

The laws of elementary algebra include rules for working with 
irrational numbers, and with these rules the student is assumed 
to be familiar.! 

4. Construction of segments of given length. The converse 
problem to measurement is constriction of segments having a given 
ratio to the unit segment, that is, having a given length. The 
constructed segment will be said to correspond to the number 
given as its length, so that a segment of length 4 units will be 
said to correspond to the number 4. 

(a) Rational numbers. Any segment whose length is a rational 
number can be constructed at once, because elementary geometry 
provides us with a method of constructing any fractional part of 
a unit segment. If the student has forgotten how to do this, let 
him review the method carefully. A brief statement of it is found 
in Appendix C. 

(b) Irrational numbers. One would naturally take it for granted 
that, inasmuch as the segments whose lengths are irrational num- 
bers are incommensurable, they would also be inconstructible ; 
and in general this is true. But many such segments can be very 
easily constructed, namely, all those depending on square roots 
alone. The Pythagorean Theorem ^ gives us the means of doing 

1 See Appendix B for statement of these rules, with exercises. 

2 " The square on the hypotenuse of a right triangle is equal to the sum of 
the squares on the two legs." This very important theorem was discovered 
by the famous Greek philosopher and mathematician Pythagoras, in the sixth 
century b.c. 



THE GEAPHICAL REPRESENTATION 5 

this for numbers like ^2, Vd, y/}, Vf , etc. For iostance, V5 is the 
length of the hypotenuse of a right triangle whose legs are 1 and 2 
in length ; Vf is the length of one leg of a right triangle in which 
the hypotenuse is f and the other leg f ; Vs is the hypotenuse 
of a right triangle in which the legs are 1 and V2. To construct 
a segment whose length is the square root of the length of any 
given segment, see Appendix D, where the method is explained 
and illustrated. Finally, any segment whose length involves only 
rational combiuations of square roots, that is, only additions, sub- 
tractions, multiplications, or divisions of square roots, can be con- 
structed by combinations of the above-mentioned constructions, — 

for instance, such lengths as 1-1-^2,^ — ^^, \^(=\/V9) -■ 

3_V5 ^ ^'2-V3 



EXERCISES 

Construct accurately segments of the following lengths (using 
the same unit for all): 

^- i- 6 ^~^ . 8- "^^ 11- 3-9- 2- V3 

2-ff ' ^ 9.^. 12. 3-V3. '^■2 + V3' 

V7. 6.1.7. 3^ ^ 15.4^. 

10. — ^- 13. 



3 



4. 1-1- V2. 7. Vs. ' VE' ' 5-V6 

5. This is as far as we can go in the construction of seg- 
ments of given length, with the means of elementary geometry, 
— namely, straightedge and compass ; for no cube root, fifth 
root, or other irrational num- 
ber not reducible to square 4 S. £ f 

roots, can be so constructed.^ -pia. i 

But we are none the less 

instinctively certain that there exists a definite segment corre- 
sponding to any length, even to these inconstructible lengths. 
For example, if AB = BC=1, we are convinced that there exists 

1 This statement cannot be proved here, as it involves more advanced 
considerations than the student is yet prepared for. 



6 THE ELEMENTAEY FUNCTIONS 

a point X between B and C such that AX='^; and similarly 
for other such numbers. We can thus say that to any number, 
rational or irrational, corresponds a definite length, when once a 
unit has been chosen. 

Negative Numbees 

6. Directed segments. If we start from any point on a straight 
Une to lay off a distance, it often makes a great difference in 
which direction we proceed to take the distance in question. It is 
accordingly useful to have some way of distinguishing, in such 
cases, which direction is to be taken. This is done by prefixing 
to each number used a label, or sign, + if the distance is to be 
taken in the one direction, and — if in the opposite direction. 
For instance, segments directed toward the right on a horizontal 
line may be given the sign + and called positive segments, while 
those directed toward the left 

will then be given the sign — > 1 1 

and called negative segments. Fig. 5 

In Fig. 5, AB, AC, and BC 

are positive segments, while BA, CB, and CA are negative. The 
numbers expressing the lengths of such directed segments are then 
given the same signs as the segments themselves, and are spoken 
of as positive or negative numbers, as the case may be. Thus, if 
the measure of the segment AB is 5 units (Fig. 5), and if that of 
BC is Z units, AB = + 5, while BA = - 5 and CB = - 3. A line, 
such as ABC in this illustration, on which every segment has a 
direction as well as a length, is called a directed line. 

7. Theorems on directed segments. If A, B, and C are any 
points on a directed line, then 

AB = -BA (1) 

and AC + CB = AB. (2) 

The first theorem results from the definition of a directed seg- 
ment. The second is self-evident when C is between A and B, 
and it is true whatever be the relative positions of the three 
points. Thus, in Fig. 5, ^C = -f- 8, C5 = - 3, and ^5 = -(- 5 ; and 
it is true that -|- 8 + (— 3) = + 5. This theorem (2) is of great 



THE GRAPHICAL EEPEESENTATION 7 

importance in our further work, and hence the student should 
verify it by numerous examples showing the various possible 
relative positions of the points A, B, and C Indeed, the truth 
of (2) is obvious geometrically, since to pass from A to C, and 
then from C to B, will give the same displacement to the right 
(or left) as the single amount AB. 

8. The student must be careful to notice that whenever we 
speak of negative numbers as the lengths of segments, the word 
"negative" refers merely to the direction in which the corre- 
sponding distance is measured ; it is a qualitative, not a quanti- 
tative, adjective. It wiU be remembered that this is the case with 
all the illustrations of negative numbers that are given in text- 
books of elementary algebra ; for example, if we call temperature 
above zero +, then temperature below zero is called — ; if an 
amount of capital (asset) is considered +, then a liability is 
called — ; and so on. In fact, whenever two sets of numbers can 
be associated with each other in such a way that one set gives 
one kind of number and the other set the opposite hind, we may 
call the numbers of the one set + and those of the other set — . 
Whatever we agree shall be the meaning of the sign +, the 
sign — means the exact opposite. 

9. Correspondence between numbers and points on a line. 
Let us take any point as starting-point, and (with any con- 
venient unit) construct on the same straight line the segments 
0A= + 1, 0B=+2, 0C= + 3, 0A' = -1,0C'=-S, etc. As B is 

the end-point of the seg- 

.. w ■ J v. i-u c' B' A' o A B c , 
ment determmed by the _j 1 ■ ■ > 1 > > > — 



length + 2, we may say no. 6 

that the number + 2 de- 
termines the point B, and, similarly, that the number + 3 deter- 
mines the point C, the number - 1 the point A', the number - 3 
the point C", and so on. We see that in the same way any number 
determines some one definite point on the straight line (and, con- 
versely, any point on the straight line determines one definite 
number), provided only that a starting-point, as 0, and a unit 
segment, as OA, are assumed. This gives us a definite one-to-one 



8 THE ELEMENTARY FUNCTIONS 

correspondence between numbers and points on a (directed) straight 
line. Note that the number ^ corresponds to the point 0. 

10. The phrases " greater than " and " less than." Since, 
as we have seen, the word "negative" appKed to numbers or 
distances is a qualitative, not a quantitative, adjective, it is evi- 
dently impossible to use the words " greater than " or " less than " 
in their ordinary quantitative sense when referring to negative 
numbers. A distance — 3, for instance, is neither less than nor 
greater than a distance + 3 in the ordinary sense, but, rather, 
an equal distance in the opposite direction. A meaning of these 
phrases that will be useful when applied to negative numbers is 
developed by the f oUowing considerations : 

When all numbers are represented as points on a horizontal 
straight line, in the way that we have just illustrated, we find 
that, of two positive numbers, the point corresponding to the 
lesser is always to the left of that corresponding to the greater. 
Thus, the statement " 3 < 4 " is equivalent to the statement " the 
point 3 lies to the left of the point 4 " ; and so for all positive 
numbers. Now it is equally true that — 1 lies to the left of 0, 
— 2 to the left of — 1, and so on ; and so it is natural to express 
these facts also ly the same words that are used in the case of 
the positive numbers: " — 1 is less than 0," "—.2 is less than —1," 
etc. In every case it should be remembered that the expression 
has no longer a quantitative, but a directional, meaning. 

EXERCISES 

1. Restate the last paragraph for the case where the numbers are 
represented as points on a vertical line ; take the upward direction 
as positive. 

2. Arrange in order of magnitude : 2,-3, Vs, 4, — 1.1, — VlO, 
V3, - 2, - 2.8, 3.9, Vi5, - VH, -v^, ^. 

3. State the laws of addition, subtraction, multiplication, and 
division with negative numbers. 

1 Many students are in the habit of thinking of "zero" as if it were not a 
number at all — sometimes even reading it " nothing." This is a mistake ; zero 
is a number, in some respects indeed the most Important number there is. 



THE GEAPHICAL REPRESENTATION 



C' 



11. The system of coordinates. We have now seen that every 
number determines a certain point on a directed straight line, and 
that, conversely, to every point on a directed straight line corre- 
sponds a number, a starting-point (or zero-point) and a unit segment 
having been assumed. We now extend this correspondence be- 
tween points and numbers to include all the points of a plane. 

Let XX' and YY' be two per- 
pendicular lines intersecting at 0, 
and suppose for convenience that B r \ A 

XX' is horizontal. These lines will 
be referred to as the X-axis and the 

y-axis respectively. Their intersec- xA 1 — ^i — ^^^1 1^^ — i hX 

tion is called the origin. Then any 

point in the plane can be located 

exactly by giAring its distances from 

the X-axis and from the F-axis. 

Thus, the point ^ is -I- 1 from the 

r-axis and -|- 2 from the X-axis, 

while the point 2f is — 1 from the 

r-axis and -|- 2 from the X-axis. 

Similarly for any point in the plane. 

To locate the point it is sufficient 

to mention these two distances, imt 

forgetting the proper sign. It wiU 

also save words if we agree once 

for aU that the horizontal distance, 

or distance from the T-axis, shall 

always be given first. Thus, merely 

mentioning the pair of numbers ^^^^ g 

(1, 2) is sufficient to locate the 

point A, (- 1, 2) locates the point B (Fig. 7), while (2, 1) locates the 

point E (Fig. 8), and (2, - 1) the point F (Fig. 8). The student 

will be able to decide for himself what pair of numbers locates the 

point C, or the point D, in Fig. 7, or the point E' in Fig. 8. 

It is thus clear that a pair of numbers can be considered as 
determining a definite point, and, conversely, that a given point 



r 

Fig. 7 
Y 



E 



E 



H \-X 



»P 



10 THE ELEMENTARY FUNCTIONS 

determines a definite pair of numbers, namely, its distances from 
the X- and Y-axes. The horizontal distance, or (directed) distance 
from the Y-axis to the point, is called the abscissa of the point ; 
and the vertical distance, or (directed) distance from the X-axis 
to the point, is called the ordinate of the point. Thus, the abscissa 
of the point B (Fig. 7) is — 1 (the directed segment BB or OS), 
and its ordinate is 2 (the directed segment SB or OR). Give the 
abscissa and the ordinate of the points E and F (Fig. 8). The 
abscissa and ordinate of a point are called its coordinates. 

This system of connecting points with pairs of numbers was 
devised by Descartes, a great French mathematician and philoso- 
pher (1596-1650), and published by him in 1637 in a work called 
" La Geometrie,'' which is justly regarded as a milestone in the 
history of human thought. In honor of him the system of coor- 
dinates which has just been outlined is called the Cartesian 
coordinate system. 

Locating points in a drawing by means of their coordinates' is 
called plotting the points. For the sake of convenience in plot- 
ting points, squared paper is used, as it enables us to lay off any 
required distance both rapidly and accurately. In making all 
drawings a hard lead pencil with a fine point should be used, ink 
being reserved for certain lines which are meant to stand out 
prominently. In this first set of drawing exercises, draw the X- and 
y-axes in ink. It is scarcely necessary to say that the greatest 
neatness and accuracy are absolutely indispensable in all problems 
involving drawing, as indeed in all mathematical work. 

EXERCISES 

1. Plot the points (2, 1), (2, - 1), (- 2, 1), and (- 2, - 1). What 
kind of figure do they form ? 

2. Plot the points (2, 0), (-1, 0), (V2, O), (5, 0), (- 4^, 0), 
(- V2, O), (3f, 0). On what line are they all found? 

3. On what line are all the following points located? (1, 1), 
(-2, -2), (3, 3), (-i, -i), (5, 5), (2.1, 2.1), (-V2, -V2), 
(H-V2, l-i-V2),(-^,-J). 



THE GEAPHICAL EEPEESENTATION 11 

4. Plot the points (- 4, - 3), (3, - 4), (4, - 3), (3, 4), (4, 3), 
(— 3, 4). Prove that they all lie on a circle, and give the coordinates 
of other points on the same circle. 

5. Plot the points (0, 0), (y2, V2), (-V2, V2). What kind of 
figure do they form ? 

6. An equilateral triangle whose side is 4 units is placed with 
one vertex at the origin and with the opposite side perpendicular 
to the A"-axis. Find the coordinates of the other two vertices. 
(Two solutions.) 

7. A square whose side is 2 units is placed with one vertex at 
the origin and with a diagonal lying along the A' -axis. Find the 
coordinates of the other three vertices and of the center. (Two 
solutions.) 

8. A regular hexagon whose side is the unit is placed with one 
vertex at the origin and with its center on the A'-axis. Find the 
coordinates of the other 5 vertices, and the area of the hexagon. 
(Two solutions for the vertices.) 

9. Plot the points (3, 1), (0, 4), and (3, 4). What kind of 
figure do they form ? Find the lengths of its sides, its area, and the 
coordinates of the center of the circumcircle.^ 

10. How far is the point (3, 4) from the origin in a straight line ? 
the point (1, 5)? the point (-1, - f)? the point (a, b)? 

11. Plot the points (3|, 3f), (h 3f), and (3*, -1%), and find the 
center and radius of the circle through the three. 

12. What are the coordinates of the points P, R, and S in Fig. 7 ? 
What are the coordinates of the origin ? 

13. What are the coordinates of the point halfway between the 
origin and the point (4, 4)? of the point halfway between the origin 
and the point (3, 6) ? 

14. Plot the points (2, 1) and (4, 5). What are the coordinates of 
the point midway between them ? 

15. Answer the same question for the points (1, 3) and (— 3, 1); 
for th,e points (1, 0) and (0, 1); for the points (3, - 2) and (- 5, 4). 

1 Circumcirole : a circle passing through the vertices of a triangle or of 
any polygon. 



12 



THE ELEMENTARY FUNCTIONS 



B 
C 


\ 


A 


Y 






X 


D E 


O 



16. Let (cCj, ^j)^ and (x^, y^ be any two points. Find the coordi- 
nates of the mid-point of the segment joining them. 

2 2 

17. Find the distance between the points (—1, 2) and (— 4, 6). 

Solution. If A (Fig. 9) represents the point (— 1, 2) and B the point 
(— 4, 6), and if we draw BD and AE perpendicular to the X-axis, and AC 
perpendicular to BD, we have formed a 
right triangle ABC whose hypotenuse 
is the required distance AB. But the 
lengths of .i4C and CB can easily be 
found, as follows : 

0E = -1 
and Oi» = - 4. 

Therefore ED=AC=-Z. 

Likewise DB = 6 Fig. 9 

and DC = 2. 

Therefore CB = 4. 

By the Pythagorean Theorem, AB^ = AC^ + CB\ 

Therefore ABfi = 9 -(- 16 = 25. 

Hence AB = 5. 

18. Find the distance between the points (3, —1) and (4, — \); 
between the points (6, 1) and (— 6, 6). 

19. Find the distance from (0, 3) to (4, 2); from (1, 1) to 
(- 2, - 3); from (3, 5) to (- 4, 1). 

20. Find the distance between the points (x^, y^ and (x^ yX 
Am. ^{x^ — x^y + (y^ — yj'. This result is easily established 

when the given points (x^, y^ and (x^ y^ both have positive coordi- 
nates, but the student should show that it is true for all positions 
of the two points. The theorems in § 7, p. 6, will be found useful 
for this purpose. 

12. Application of coordinates to some problems of elementary 
geometry. This method of representing points by pairs of num- 
bers is very useful because it gives us a means of handling many 

' Read "z-one, j/-one," meaning "the first s," and so on. Such subscripts 
are often very convenient, and should not be confused with exponents. 



THE GEAPHICAL EEPKESENTATION 



13 



geometric problems in an algebraic way. The significance of this 
statement will become much more apparent as we proceed farther 
in our work, but the simple examples which follow will provide 
at least a basis for appreciation of it. 

Example 1. Let us take the well-known problem of elementary geometry, 
Prove that the diagonals of a rectangle are equal. In such problems, where 
the object is to prove geometric theorems by the aid of coordinates, the 
secret of a simple solution lies in a wise 
choice of the coordinate axes. We are at 
liberty to choose them in any position 
we please with reference to any figure 
already given. In this case we choose 
as X- and F-axes two adjacent sides of 
the given rectangle, and suppose the 
length of OP (Fig. 10) is a and that 
of OR is b. Then the coordinates of 

O are (0, 0), those of P are (a, 0), those of R are (0, 5), and those of 
Q are (a, b). We now apply Ex. 20, above, which gives us 




Fig. 10 



RP = V(a - 0)2 + (0 - by = Va^ + i 



and 

Therefore 



OQ = V(a - Of + (6 - 0)2 = Va2 + ( 
RP = OQ. 



Q.E.I>. 



Example 2. Another well-known theorem of elementary geometry which 
can easily be proved by the use of coordinates is. The diagonals of a paral- 
lelogram bisect each other. 

Proof. Let OPQR (Fig. 11) be the parallelogram. Choose one side 
OP as the X-axis, and as the origin. The coordinates of P, the other 
vertex on the X-axis, may then be repre- 
sented by (a, 0), those of R by (b, c), and 
those of Q bj (a + b, c). (For MR = b 
and RQ= 0P = a. .: MQ = a + b, the 
abscissa of Q.) 

Then the coordinates of the mid-points 
of the diagonals can be found by Ex. 16, 
above. For the diagonal RP the mid- 
point is thus I 1 -|i and for the 

diagonal OQ it is (^-^^ |)' that is, the same point. 

points coincide, the diagonals bisect each other. q.e.d. 



Y 










M 


R 






Q 


r 


\ ^ 




-J 




U 


^-^ 


v/ 


1 
X 


O 




1 
Fig. 11 


p 





Since their mid- 



14 



THE ELEMENTARY FUNCTIONS 



EXERCISES 

Prove the following theorems by means of coordinates : 

1. The line joining the vertex of any right triangle to the mid- 
point of the hypotenuse is equal to half the hypotenuse. 

2. The line joining the middle points of two sides of a triangle 
is equal to half the third side. 

3. The distance between the middle points of the nonparallel 
sides of a trapezoid is equal to half the sum of the parallel sides. 

4. In any quadrilateral the lines joining the middle points of 
the opposite sides and the line joining the middle points of the 
diagonals meet in a point and bisect each other. 

5. If the lines joining two vertices of a triangle to the middle 
points of the opposite sides are equal, the triangle is isosceles. 

13. Point that divides a segment in a given ratio. Ex. 16, 

p. 12, enables us to find the coordinates of the mid-point of any 





Y 






I^(X2 


y,> 




p^-.y/^^^ 


"^ 


^ 








B 


X 


o 


c 


7 1 


3 


J 


i 



Pi^^'Vi) 




Fig. 12 

segment; we can do more, and get the coordinates of the point 
that divides a given segment in a»y given ratio. 

Let ^ = («!, 2/j) ^ and ^ = {x^, y^ be the end-points of the given 
segment, and let P be the required point dividing the segment 
ij-^ in the ratio m : n, that is, so that P^P : PI^ = m:n. Let {x, y) 
be the coordinates of P. Then, if P^G, PD, and I^E are per- 
pendiculars from Jl, P, and J^ respectively, upon the X-axis, 
00 equals the abscissa of ij (that is, x-^, OE=x^, and OD = x. 
Form the right triangle I^AI^ by drawing through ij the line 

1 Read "Pj, which is the point (x^ y^" or "P^, whose coordinates are 



THE GRAPHICAL REPRESENTATION 



15 



I^A parallel to the X-axis, meeting FD in B and J^U in. A. 
Then IIA=C£J= x^~ x^ (notice that this is true even if ^ is 
to the left of i^, as in Fig. 12 6, since x^- Xj^ is then negative), 
^j5= CD= X — ajj, and BA = DE= x^ - x. 

By hypothesis ^P:PP^ = m:n, and since the triangles I^BP 
and P^A^ are similar (why?), we have P^P:PP^=I[B:BA. But 
this last ratio is equal to 



Solviag (1) for x, 



X- 

) — 


- ■''! 


hence 


«2 


— X 




X- 


- X^ _ 


m 


X^ 


— X 


n 


x = 


TTUTj 


+ nx^ 



(1) 
(1') 



m + n 

In exactly the same way, drawing PF perpendicular to ^A (not 
shown ia Fig. 12), we find AF=BP= y— y^, FP^ =y<i—y, and thus 

y - yi _ m 



Solving (2) for y, 



y-i-y n 



(2) 
(2') 



m + n 

Notice that these results (!') and (2') reduce to the results of 
Ex. 16, p. 12, when m = n. Observe also that m and n are not 
necessarily the lengths of the segments P^P and Pi^, but are 
merely numbers in the same ratio as those lengths, m correspond- 
ing to the segment JJP (that is, to the segment nearest i^), while n 
corresponds to the segment F^ (that is, to the segment nearest ^). 

Example 1. Find the coordinates of the point which divides the segment 
from (1, 3) to (6, 2) in the ratio 2 : 3. 

Solution. Here 

m = 2, n = 3, 
^1 = 1> yi = 3, 
^2 = 6, 2^2 = 2. 
Substituting these values in (1'), 



2-6 + 3-1 
2 + 3 



= 3, 



Pi(i5r 



Pi."!^ 



fi(6,2) 



Fig. 13 



and from (2'), 
Therefore 



_ 2 ■ 2 + 3 ■ 3 
^~ 2 + 3 
P = (S,^^). 



13 
5 ■ 



16 



THE ELEMENTARY EUNCTIONS 



Example 2. Find the coordinates of the point ^ of the way from 
(-4, 2) to (2, -1). 

Solution. The point P being ^ of the way from P^ to P^, the segment 
PjP = JPiPj, and hence PP^ = § PjPj, so that PjP : PP^ = 1:2. (A rather 
natural although careless mistake that is often made in problems of this 
kind is to say m = 1, re = 3, since the point P is to be ^ of the way from 
Pi to Pj.) The student can now finish the work for himself. The result 
isPs(-2, 1). 

Example 3. Prove that the medians of any triangle meet in a point f of 
the way from a vertex to the mid-point of the opposite side. 

Proof. Let A = (Xj, y^), B = (x^, y^), and C= (xg, j/g) be the vertices of 
the triangle. Then the coordinates of M, the mid-point of AB, are 

(Ei±Ii, yi + yi \ ; those of A', the mid-point of BC, are (^s±^, ^^^) ; 
and those of Q, the mid-point of 



CA, are (^^, 2^)- 

Next, the coordinates of the point 
f of the way from C to M are as 
follows (m = 2, n = 1) : 

2 • ^L±^ + 1 • Xg 

3 ' 



2' = 



2-H 
2 . a±l2 + 1.3, 



'.+ y> 




Fig. 14 



2-1-1 



Working out the coordinates of the point § of the way from A to iV, 
also of the point § of the way from B to Q, we get in each case the same 
pair of coordinates. This proves the theorem. 



EXERCISES 

1. Find the coordinates of the point which divides the segment 
from (1, 3) to (— 2, 6) in the ratio 1:2; in the ratio 2:1. 

2. Find the coordinates of the point f of the way from (— 2, — 3) 
to (7, 3) ; § of the way from (^, 3§) to (- 1, - ^). 

3. Prove that in any parallelogram A BCD, if M is the middle 
point of the side AB, the line MD and the diagonal AC trisect 
each other. 



THE GRAPHICAL EEPEESENTATION 17 

4. rind the coordinates of the point that divides the segment 
from (0, 3) to (2, 5) externally in the ratio 3 : 2. 

Hint. Here the point P is on the line P^P^ produced, and since F-^P : PP^ 
equals numerically |, P^P must be greater than PP^ ; that is, P must be nearer 
to Pj than to Pj, or, in other words, it is beyond the point P^. P^P and PP^ now 
being measured in opposite directions, we may think of the ratio m : n as being 
negative. With this understanding it will be found that the reasoning by which 
the formulas (1) and (2) were obtained is still valid, and they may be used in 
these cases also. Here m = 3, n = — 2 (or else m = — 3, m = 2), so that we have 

3-2-2.0 „ 3-5-2-3 . 

x = = 6, y = = 9. 

3-2 ' " 3-2 

Hence the required point is (6, 9). 

5. Find the coordinates of the point that divides the segment 
from (— 2, — 3) to (3, — 5) in the ratio —2:3; in the ratio —1:2; 
in the ratio —2:1. 

6. A line AB is produced to C so that AC = SBC. Find the 
coordinates of C if ^ s (2, 0) and B=(— 3, 1). 

7. Find the coordinates of the point that divides in the ratio \ 
the segment from (1, 2) to (— 1, 3) ; from (aj^, y^) to (x^ y^. 

8. Find the point of intersection of the medians of the triangle 
whose vertices are (2, 3), (4, - 5), and (3, - 6). (Do not use the 
result of Ex. 3, p. 16, but follow the same method, using the special 
numbers of this problem.) 

9. In what ratio does the point (2, 3) divide the segment from 
(-2, -3) to (4, 6)? 

Hint. Use equations (1) and (2), p. 15. 

10. In what ratio is the segment from (—2, 1) to (3, —9) divided 
by the point (1, — 5) ? 

\\. li A = (2, —1) and B = (5, \), in what ratio is AB divided 
by the point C = (4, 0)? 

12. Using the same segment AB as in the preceding problem, 

show that, for the point Z)=(-5, -4), |^=-io' ^^""^^ 
y ~y\ = _ ? , a different ratio. What conclusion can you draw 

y^-y ^ 

from this peculiarity? 

13. Test as in the preceding problem the three points A = (3, 2), 
B = {-1, -10), and C= (0, - 8); the points (5, -1), (2, 2), and 
(-1, 5). 



18 THE ELEMENTARY FUNCTIONS 

14. Formulate the results of Exs. 12 and 13 in the form of a 
theorem that gives a test as to whether three points A, B, and C are 
in the same straight line or not. 

15. Given the three points A = (2,1), B = (-1,3), and C = (1,-2), 
find the coordinates of D, the fourth vertex of the parallelogram 
determined by ^, B, and C. (Three solutions.) 

16. Proceed as in Ex. 16 for the three points (0, 5), (7, 3), and 
(- 2, - 3). 

17. Prove that the sum of the squares of the medians of any 
triangle equals three fourths the sum of the squares of the sides. 



CHAPTER II 

FUNCTIONS AND THEIR GRAPHS 

14. In the preceding chapter we have seen how to associate 
with every segment a number, — its length; and by means of the 
system of coordinates we have associated with every point in 
the plane a pair of numbers. We apphed this to the proof of a 
few theorems of elementary geometry. In this chapter we shall 
consider not merely fixed points, as heretofore, but variable points, 
that is, points that are free to 
occupy any number of positions, 
subject to certain conditions. ^ p, 

For example, let us say that a ' *~ 
point is free to move so as to be x 



always at the distance +2 from ^'(^ ^ 

the X-axis. There are evidently Pm ^5 

an unlimited number of such 

points, but they will all be found on one straight line, the line 
ABC, parallel to the X-axis and 2 units above it. We use the 
expression. The locus of points at the distance -|- 2 from the 
X-axis is the line parallel to the X-axis and 2 units above it. 
Furthermore, to specify that the point shall be at the distance 2 
from the X-axis is the same thing as saying that its ordinate shaU 
equal 2 ; or, writing y for the ordinate of any one of the points 
in question, the original condition is equivalent to the statement 
that y = 2 (or y — 2 = 0). We may say, then, that the locus of 
the equation y=2(ovy—2 = 0) is the straight line parallel to the 
X-axis and 2 units above it. We call the line also the graphical 
representation, or simply the graph, of the equation. 

15. As another example, let us make the following condition: 
A point moves so as to be always twice as far from the X-axis as 
from the Y-axis. Here again we know from elementary geometry 

19 



20 



THE ELEMENTARY EUJSTCTIONS 




P(x,y) 



Q 



Fig. 16 



that the locus of these points is a straight line P'OP, for it P is 

twice as far from the X-axis as from the Y-axis (that is, if 

QP= 2 • BP= 2 • OQ), then the same thing is true for any point 

on the line OP, and for no other 

points.^ Since the given condition 

is equivalent to the condition that 

the ordinate of the point shall equal 

twice its abscissa, we can express this 

condition in the form of an equation, 

thus : y=2x, where x represents the 

abscissa of the moving poiat, and y 

its ordinate. 

16. This example illustrates the 
use of the general numbers («, y) to 
represent the coordinates of a variable, or moving, point, — a 
notation which is very convenient and which will be frequently 
employed throughout this book. The quantities that we have to 
deal with in physics, engineering, astronomy, and indeed in aU 
applications of mathematics, are largely variable, and hence it is 
necessary to learn to use and understand symbols that represent 
such variable quantities. To represent constant quantities we shall 
use the earlier letters of the alphabet or 
letters with subscripts, as x-y, y-^, etc. 

17. As a third example let us make 
the following condition : A point moves 
so that its ordinate is always equal to the 
square of its abscissa. If the coordinates 
of this variable point are represented by 
{x, y), then evidently the equation of the 
locus is y = o?; but elementary geometry 
gives us no information as to just wliat 
this locus is, that is, just how the point 

can be located if it satisfies this condition. We can, however, 
get a good idea of the locus by starting with the equation y = o^ 

' For the present this assertion may be accepted without proof. The proof 
is given on page 92, where this example is taken up again more in detail. 




Fig. 17 



PUNCTIONS AND THEIR GRAPHS 21 

itself and finding a number of pairs of values of x and y which 
satisfy the equation. If we then plot the points having such 
values {x, y) as their coordinates, we shall get a series of points 
satisfying the given condition; that is, we shaH get points on the 
locus. Thus, when x = l,y = l- when x=2,y = 4:; when a; = 1 
y = i ; when a; = 0, y = ; etc. Hence (1, 1), (2, 4), (i, i), and 
(0, 0) are points on the locus. We may conveniently arrange 
these pairs of values in the form of a table, thus : 



X 





h 


1 


1 


2 


-1 


-2 


-h 


_ 3 


y 





\ 


1 


1 


4 


1 


4 


i 


9 



etc. 



We now plot the points carefully, and when enough points 
have been located, it will be possible to see that they suggest 
a curved line, concave upward, symmetrical with respect to the 
r-axis, and with its lowest point at the origin. This curve should 
now be carefully drawn on a large scale, and we can then infer 
that we have a fairly accurate drawing of the locus of the equa- 
tion y = a?. The curve is called a parabola. (It must not be over- 
looked, however, that we cannot prove that all points on the 
curve satisfy the condition y = a^ ; but we asswme this to be the 
case, and if the drawing has been made accurately, this assump- 
tion will be in fact approximately correct.) 

18. As a fourth example, let us suppose that a point (x, y) 
moves so that the following relation between x and y always 
holds : y = a?—29i? — x + 4=. As before, we proceed to compute 
a table of values of x and y that satisfy this equation. The result 
may be exhibited thus : 



X 





1 


2 


3 


-1 


- 2 


- 3 


i 


1 


-i 


y 


4 


2 


2 


10 


2 


-10 


-38 


H 


If 


H 



Plotting the points given by this table, and connecting them 
by a smooth curve, as in Fig. 18, we get approximately the 
graphical representation of the equation given. 



22 



THE ELEMENTARY FUNCTIONS 



la the following exercises a point is to be understood as 
varying subject to the condition given each time in the form 
of an equation between x and y. 
Draw the graph of the equation 
carefully, determining the exact 
locus in Exs. 1-10, and a very 
close approximation in the others, 
as was done in the last two ex- 
amples above. Also, in the first 
12 exercises state in words the 
condition which the equation gives 
in symbols : for example, in No. 1, 
" If a point moves so that its ab- 
scissa equals 2, what is its locus ? " Fig. 18 




1. a; = 2. 

2. y=-3. 

3. a; = 0. 

4. y — 5 = 0. 

5 . 2/ = 3 a;. 

6. x=— 2. 

7. y = 0. 

8. y=—2x. 

9. 2/ = — a;, and y = — x + 2, 

in the same figure. 

10. y = i X, and y = ^ x — 3, 

in the same figure. 

11. 2/ = 2a; —1. 

12. y = 3x + i. 
13- y=-ix+l. 

14. y = 2x'^-2. 

15. y=-Sx^-5. 



EXERCISES 

16. y = x^ — X. 

17. y = x''—2x. 

18. y =— x^ — x. 



= — a;^ — a; + 5. 
= -2x' + x +1. 



19. y^=—x + 3x. 

20. y = 2 x^ — 3 a;. 

21. y = 2cc^ — a;-|-3. 

22. y =—x 

23. y 

24. y = x^. 

25. y = a;'— 1. 

26. y = 7? — X. 

27. y = x^ -^x. 

28. y =zx^ — x^. 

29. y = - x' + 3 xl 

30. y=:x''-3x2-|-l. 

31. y =— 2x''— 6a;. 



FUNCTIONS AND THEIR GRAPHS 23 

32. y = x^ + x\+x+l. 35. y=-x^ + 2x' + 3. 

33. y = x''-x^' + x-l. 36. y==-2x' + x'-x + 2. 
Si. y = x'' — x^—x—l. 37. 2/ = - a;= - a;2 - a; - 3. 

19. Definition of function. In all of these later exercises, pairs 
of values (x, y) that would satisfy the equation given were deter- 
mined by taking a set of values of x and then using the equation 
to compute the corresponding values of y. Whenever a pair of 
(variable) numbers x and y are related in this way, that is, in. 
such a way that to a value of the one corresponds a definite 
value of the other, the second is said to be a function of the first, 
or the relation is called a functional relation. The symbol for 
this relation between x and y is y—f{x), read "3/ is a function 
of X " or « y =f of x." For example, if y = 2 «, y is a function of 
X, because to a given value of x corresponds a definite value of y ; 
or, again, ily — a?—2x'^ — x + 4:,y\sa. function of x. In these 
examples x is called the independent variable, and y, or the 
function of x, is called the dependent variable, because its value 
depends upon that of x. 

20. Importance of functional relation. The idea of this 
functional relation between two variable numbers is of very far- 
reaching importance, because it can be applied to any pair of 
quantities that are representable by numbers if one of these 
numbers is determined by the value of the other. For instance, 
the temperature at any place is a function of the time, be- 
cause at any definite time there is a definite temperature at that 
place ; or, again, if a man walks at the rate of 3 miles per 
hour, the distance he has gone is a function of the time he has 
walked ; or, again, the population of a city is a function of the 
time (date) when the estimate or count is made. Considera- 
tion of the wide variety of such possible illustrations wiU make 
clear the very great importance of a study of various kinds of 
functional relations. 

21 . For such a study the graphical representation just illustrated 
is a most valuable aid, because it gives us a vivid ("graphic") picture 
of the functional relation, thus assisting materially in forming a clear 



24 



THE ELEMENTARY FUNCTIONS 



idea of the nature of that relation. This may be seen in the ex- 
ample mentioned above of the variation of temperature at a certain 
place ; suppose the weather observer had made the following record : 



Time 


Tempebatuke 


7 a.m. 


51° 


8 a.m. 


55° 


9 a.m. 


60° 


10 a.m. 


62° 


11 A.M. 


70° 


12 m. 


72° 


1 P.M. 


72° 


2 p.m. 


73° 


3 p.m. 


71° 


4 p.m. 


66° 


5 p.m. 


61° 


6 p.m. 


58° 



80 1 

70° 
60" 

SO'-f 
40'- 
30°' ■ 
20" 
10° 




7A.1 



-+- 



-+- 



::It 



-t- 



-4- 



H 1- 



10 1112M.1F.M. 2 3 4 6 

Fig. 19 



Since the first observation was taken at 7 a.m., we represent 
that time as the origin and construct the times as abscissas and 
the temperatures as ordinates, thus getting the above figure. A 
single glance at the figure evidently gives us a clearer and more 
comprehensive idea of the variation of the temperature on that 
particular day than the table of values does. 

22. To take a more elaborate illustration of this use of the 
graphical representation, we find in the United States Statistical 
Abstract for 1915 the following table showing the progress of 
shipbuilding in the United States from 1900 to 1915: 



Year 


Tonnage Built 


Yeak 


Tonnage Built 


1900 


393,790 


1908 


614,216 


1901 


483,489 


1909 


238,090 


1902 


468,831 


1910 


842,068 


1903 


436,152 


1911 


291,162 


1904 


378,542 


1912 


232,669 


1905 


330,316 


1913 


346,155 


1906 


418,745 


1914 


316,250 


1907 


471,332 


1915 


225,122 



FUNCTIONS AND THEIE GRAPHS 25 

Here the tonnage built is a function of the time. Plotting the 

number of the year as abscissa (starting of course with 1900 

at the origin) and the corresponding tonnage as ordinate, we 

get a figure like the one 

adjoining. Let the student 

complete the figure, using 

as large a scale as possible. 400,000 

Now a glance at the broken 

hne joinmg these points 100,000 



600,000- ■ 
600,000 



gives a much clearer idea iLix izoiui, is i, isi, io ii L iz I 



.X 



Fig. 20 



of the variation in the ship- 
building during these years 
than does the mere table from which the diagram was constructed. 
23. In drawing these graphical representations of statistical 
tables it is not necessary to try to draw a smooth curve joining 
the points located, since we have no possible way of telling how 
the curve should look, between the located points ; hence we draw 
only a broken line, that is, join each point to the next following 
by a straight line. In the graphical representation of equations, 
however, as in the exercises on page 22, it would not be right to 
do this, because we can locate points just as close together as 
we need, thus determining the shape of the curve to any desired 
degree of accuracy. For instance, in the example of § 18 we 
had (among others) the points. (0, 4) and (—1, 2), which are so 
close together that it would be natural to connect them by a 
straight line (just as we should do if there were no way of tell- 
ing how the curve reaUy lies between these points) ; but by taking 
x = — ^ we found y = 3|, which shows that the straight line 
joining (0, 4) and (—1, 2) would be decidedly incorrect. Between 
x = l and a; = 2 it would be natural to make the same mistake 
again, and join (1, 2) and (2, 2) by a horizontal Hne; but the 
point (|, 1|) corrects this. The detailed study of the graphical 
representation of functions in this way is made much simpler and 
more practical by the use of a chapter in mathematical analysis 
which is called the Differential Calculus. Until the use of that 
very powerful method has been learned, however, the student 



26 



THE ELEMENTARY FUNCTIONS 



must be content to plot accurately a sufficient number of points 
so that the form of the curve is evident. 

Summarizing, the graphical representation of functions given 
by mathematical equations can be carried out to any required 
degree of accuracy, whUe in the case of functions given by 
tables of observations or statistics we cannot, of course, locate 
more than the number of points given in the tables. 



EXERCISES 

1. The Statistical Abstract for 1915 gives the following figures 
for the values of exports and imports of merchandise for the years 
1900-1915 : 



Year 


Exports 


Imports 


Year 


Exports 


Imports 


1900 


1,394,483,082 


849,941,184 


1908 


1,860,773,346 


1,194,341,792 


1901 


1,487,764,991 


823,172,165 


1909 


1,663,011,104 


1,311,920,224 


1902 


1,381,719,401 


903,320,948 


1910 


1,744,984,720 


1,556,947,430 


1903 


1,420,141,679 


1,025,719,237 


1911 


2,049,320,199 


1,527,226,105 


1904 


1,460,827,271 


991,087,371 


1912 


2,204,322,409 


1,653,264,934 


1905 


1,518,561,666 


1,117,513,071 


1913 


2,465,884,149 


1,813,008,234 


1906 


1,743,864,500 


1,226,562,446 


1914 


2,364,579,148 


1,893,925,657 


1907 


1,880,851,078 


1,434,421,425 


1915 


2,768,589,340 


1,674,169,740 



Make a graphical representation of these statistics. 

2. We find in the Statistical Abstract that the number of immi- 
grants admitted to the United States during each of the years from 
1892 to 1915 was as follows : 



Year 


Number 


Year 


Number 




Number 




Admitted 




Admitted 




Admitted 


1892 


623,084 


1900 


448,572 


1908 


782,870 


1893 


502,917 


1901 


487,918 


1909 


751,786 


1894 


314,467 


1902 


648,743 


1910 


1,041,570 


1895 


279,948 


1903 


857,046 


1911 


878,587 


1896 


343,267 


1904 


812,870 


1912 


838,172 


1897 


230,832 


1905 


1,026,499 


1913 


1,197,892 


1898 


229,299 


1906 


1,100,735 


1914 


1,218,480 


1899 


311,715 


1907 


1,285,349 


1915 


326,700 



Draw a graphical representation of these facts. 



rtJNCTIONS AND THEIR GRAPHS 



27 



3. The number of persons killed in railway accidents in the United 
States during each of the years from 1892 to 1910 was as follows : 



Yeak 


Number of 
Persons 


Year 


Number of 
Persons 


1802 


2554 


1902 


2969 


1893 


2727 


1903 


3606 


1894 


1823 


1904 


3632 


1895 


1811 


1905 


3361 


1896 


1861 


1906 


3929 


1897 


1693 


1907 


4534 


1898 


1958 


1908 


3405 


1899 


2210 


1909 


2610 


1900 


2550 


1910 


3382 


1901 


2675 







Make a graphical representation of these statistics. 

4. Find statistics for the growth of the population of the United 
States since 1790, and draw the graphical representation of these 
figures. 

5. Answer the same question for three or four of the states, from 
the time they were admitted to the Union to the present. 

6. Make a graphical representation of the amount of flOOO at 
simple interest for one year, as a function of the rate per cent 
(from 1% to 10%). 

7. Answer the same question for the amount of flOOO at 6% as 

a function of the time (from 1 yr. to 10 yr.). 

Note. Ample material for further work in statistical graphs may be found in 
the government census reports, crop reports, and in the financial journals, etc. 



CHAPTER III 

APPLICATION OF GRAPHICAL REPRESENTATION 
TO ELEMENTARY ALGEBRA 

24. In the preceding chapter we saw how to use the coordinate 
system to draw the graphs of many simple functional relations be- 
tween X and y ; and we observed that such graphical representation 
of equations in the form y=f{x) is of very wide application iu 
various fields. In this chapter we shall use this geometrical work 
of drawing graphs for the purpose of throwing new light upon 
certain problems of elementary algebra. As a preliminary step it 
will be foimd useful to make a simple classification of functions 
according to their degree. 

25, Degree of a function. The expressions 2a;— 3, —?>x+2, 

1 X 

Z X, -x — 1, — + 7 are all functions of x, according to the defi- 

Ji 

nition of the word " function," for the value of any one of these 
quantities is determined by the value of x. In each of them the 
variable number x occurs to no higher power than the first. 
Functions in which this is the case are said to be "of the first 
degree " or " linear " in x. A general form of such functions of x 
would be ax+h, where a and h are general numbers, that is, may 
have any value we please. 

A function which contains x^ but no higher power of x is said 
to be " of the second degree " or " quadratic " in x. A general form 
for it would be aa^+hx+c. A function which contains a? but no 
higher power of x is said to be " of the third degree " or " cubic " 
in 03, a general form being aoc? + hx^ + cx + d. In the list of prob- 
lems on page 22 we had examples of all these three kinds of 
functions, — linear, quadratic, and ciibic. In the same way it is 
possible to write down functions of the fourth, fifth, • • • degree, but 
we shall seldom need to go beyond the third degree in this book. 

28 



If this 


equa- 


c 


a 


y = b- 


T" 



ELEMENTARY APPLICATIONS 29 

26. Linear equations. An equation of the form 

ax + hy = c (1) 

is called a linear equation in the variables x and y. 

tion be solved for y, a linear function is obtained : 

For example, if we had 2x + 3y= — 1, then y= — ^ — f a^, which 
is a linear function of x. Let the student write down, at random, 
three or four equations of the form (1), choosing any numbers 
whatever for a, h, and c, and then make a careful graph of the 
function in each case. Each graph will turn out to be a straight 
Hne. Later on we shall be able to prove that this is necessarily 
the case, but for the present we merely assume it to be a fact, — 
that is, we make, as yet without prx)of , the following assumption : 

The graph of every equation of the first degree in x and y is a 
straight line. 

This is the reason why such an equation or functional relation 
is called "linear." Accordingly, in making the graph of such 
an equation it is sufficient to plot two points whose coordinates 
satisfy the equation, and then the straight line joining these two 
points wiU be the graph of the equation. (In practice a third 
point shoiild also be plotted, as a check; if it is not exactly on 
the graph, a mistake has been made.) For the two points it is 
often most convenient to choose the points where the hne crosses 
the X-axis and the T-axis. For the point of intersection with the 
X-axis the ordinate equals (since the ordinate of any point on 
the X-axis equals 0) ; hence let y = in the equation of the hne, 
and solve for the value of x. Similarly, to get the point of inter- 
section with the F-axis, let a; = in the equation. The (directed) 
distances from the origin to these points of intersection with 
the X- and F-axes are called the X- and F-intercepts of the line. 
In general, if in the equation of any locus we let a; = 0, we obtam 
the value of the ordinate of the point where the locus meets the 
F-axis ; and if we let y = 0, we get the abscissa of the point 
where the locus meets the X-axis. These two points are especially 
useful in the practical work of drawing graphs. 



30 



THE ELEMENTARY FUNCTIONS 



27. Simultaneous linear equations. It is a famUiar and simple 
problem of elementary algebra to find a pair of values {x, y) that 
will satisfy each of two linear equations in the variables x and y. 
We can now solve this problem graphically, that is, geometrically, 
in a very simple manner. Let the equations be 

ax + hy = c (1) 

and a'x + h'y = c'. (2) 

We draw in the same figure the graphs of the two equations; 
then the graph of equation (1) contains all points whose coordi- 
nates {x, y) satisfy that equation (for this is the definition of the 
graph of an equation), and the graph of equation (2) contains 
all points whose coordinates (x, y) satisfy that equation. Hence 
any point whose coordinates {x, y) satisfy both equations at once 
must be found on both graphs, that is, must be their point of 
intersection. The solution of the problem can accordingly be 
obtained, at least approximately, by a glance at the figure; and 
since the graphs of linear equations are straight lines, there will 
be one and only one such point of intersection, unless the lines 
are parallel, when there will be none. The equations (1) and (2) 
have, therefore, in general one and only one solution, or, in case 
the graphs are parallel lines, 
they have no solution. 

Example 1. Solve graphically the 
simultaneous equations 

\2x-%y = T, (1) 

L3 a; + !/ = 5. (2) 

Solution. To make the graphs, 
we may determine the X- and Y- 
intercepts of each of the lines (1) 
and (2) 1; in (1), when ^ = 0, a; = J, 
and when a; = 0, y = — J ; in (2), 
when y = Q, X = ^, and when x = 0, 
^ = 7. As a check point take any 

value of X at random and compute the corresponding value of y\ the 
point thus determined should lie upon the graph. If correctly drawn, 

1 The expression "line (1)" is used for brevity, instead of the complete 
expression "line whose equation is (1)." 



G 




Fig. 21 



ELEMENTARY APPLICATIONS 



31 



the lines (1) and (2) will result as in Fig. 21. The coordinates of the 
point of intersection P are evidently about (2, — 1), which fact gives 
(approximately) the solution of the problem, namely, x = 2, y = —X. In 
this case the graphical method has given us the exact solution, as we can 
easily verify by substituting a; = 2, t/ = - 1 in the equations (1) and (2). 

Example 2. Solve graphically the simultaneous equations 

p + 4)/ = 10, (1) 

1.5 a; -8 2^ = 1. (2) 

Solution. Making tables of values for x and y in order to determine 
points on the graph, we have in tabulated form : 

Check. 

Y 



(1) 


X 





10 


2 


y 


n 





2 










Check. 


(2) 


X 





i 


2 


y 


-4 





1 




Fig. 22 



The coordinates of the point of intersection P are about (3, 2), which is 
in fact very nearly the correct solution, although not exactly so (as was 
the case in Ex. 1). Algebraic solution ^ shows that x = 3, y = 1^ is the 
exact solution. 

Example 8. Solve graphically the simultaneous equations 

■4y = 5, 



c 



(1) 

x-Sy = 7. (2) 

The graphs result as in Fig. 23, and apparently the lines are parallel. 

That they actually are so is shown by attempting an algebraic solution. 

If we mviltiply equation (1) by 2, 

we have 

6 a; — 8 y = 10, 

whereas (2) requires that 

6 a; - 8 2^ = 7. 




Fig. 23 



Both cannot be true at once; hence 
there is no pair of values (x, y) that 
will satisfy both equations; that 
is, the lines (1) and (2) do not intersect, as was indicated by the figure. 

1 See Appendix E for the method, in case it has been forgotten. 



32 



THE ELEMENTARY FUNCTIONS 



EXERCISES 

Solve the following pairs of simultaneous equations both graphi- 
cally and algebraically : 



f3x + t 


,y=n, , f4x -7^=11 


, ^ rSx-l- 9^ = 93, 


■U + y = 3. "■ l3x + 2y=l. " l7x-5y = n. 


fx-4y = l, f3x-z/ = 3, r|x-|y=7, 
■\2x + y=-7. \6x-3y = 2. ' Ux-5y = 20. 


(x + 2y = l, fl5x+7y=ll, r7.2 x + 1.5 y= 0.42, 
■ l2x + 8y = 5. 'l^x = 2y. * U.8x- 2.5 y= 0.98. 


10. 


r3x 4y_ 


fx-l 2/4-2 11 
5 6 30' 
2x + 3 3y-l , 
[7 ' 5 -^• 


12. ■ 


3^4 4 


4x 2?/ 3a; 19y 




3 6 - 4 ' 40 




■x + 32/ 2x + y x + y 3x — 5y 




8 5 2 7 ' 


13. 


3x + 2,v 1-32/3 5 

[7 ' 4 "*"4 


x+7 5X + 82/ 
6 ' 2 ' ^'" 



14. Prove algebraically that the simultaneous equations ax + by 
+ c = and a'x + b'y + c' = will have one and only one solution 
unless ab' = ba'. 

Determinants 

28. Before passing on to another application of the graphical 
representation to elementary algebra, it will be useful to take up 
a new method of solving simultaneous linear equations. This 
new method is indeed a very valuable aid in many more advanced 
mathematical studies. It is the method of determinants, which 
are defined as follows : 

A determinant is a symbolic expression in the form 

a b 
c d ' 
which is understood to represent the quantity a • d — b • c. Thus, 



ELEMENTARY APPLICATIONS 



33 



-4. 2 = 15-8=7; 



is a determinant and equals 3 ■ 5 

is a determinant and represents x^ — y^. Before reading farther 
the student should write down some other determinants and 
give their values, so that this symboKo expression may become 
somewhat familiar. 

29. Solution of simultaneous linear equations by use of deter- 
minants. If we solve by the ordinai'y elementary algebraic method 

the pair of equations 

rax + ly = c, (1) 

\a'x + Vy = c', (2) 

we obtain the values 



cV - he' 
aV — ha' 



y- 



ac' 



ca 



ah' — ha' 



Now both numerator and denominator of each of these fractions 
are differences of products, and hence can be written in the 
determinant notation, as follows : 



x = 



c 


I 




a 


c 


c' 


h' 


y = 


a' 


c' 


a 


h 


a 


h 


a' 


V 




a' 


h' 



These fractions can be written down by inspection of the equa- 
tions (1) and (2) if we observe the followiag directions : The 
denominator is the same for x and y, and is formed by copy- 
ing the coefficients of x and y ia the equations, in the exact order 



in which they occur. 



V 



The numerator for x is formed by 



writing, instead of the coefficients of x, the numbers on the right- 
hand side of the equations (c and c'), thus : , , , > the second 

colunm of the determinant being the coefficients of y {h and h'). 
Finally, the numerator for the value of y is formed by writing in 
the first column the coefficients of « in the equations (a and a'), and 
in the second column, instead of the coefficients of y, the numbers 

on the right-hand side of the equations (c and c'), thus : 



34 



THE ELEMENTARY FUNCTIONS 



Example 1. Solve in this way the equatidns 



{: 



4 a; + 5?/ = 17. 



(X) 

(2) 



By the preceding rule the values of x and y are written down directly, thus : 

10-(-119)_129_g 



2 -7 
17 5 



,15 -(-28) 



43 



3 


2 


4 


17 


3 


-7 


4 


5 



51-8 
43 



= 1. 



The solution is accordingly (3, 1), which pair, in fact, satisfies both 
equations. 



Example 2. Solve by determinants : 



if- 



1, 



I- 
4 

• a: + 2 = 0. 



(1) 

(2) 



Here the second equation is not in the form a'x + Vy = c', so we first write 
it in that form : — x -\- y =—2. The solution can now be written down : 



1 


-i 


_ 2 


1 


i 


-i 


-1 


1 



i- 



f = 6, 



i 


1 


-1 


-2 


4 


-i 


-1 


1 



=s=*- 



Therefore (6, 4) is the solution, and this pair of values satisfies both 
equations. 

EXERCISES 

Solve by the method of determinants the problems in the exercises 
on page 32. 

*30.i Determinants of the third order. This method of deter- 
minants can be used to solve a system of threfe linear equations 
in three unknown quantities. The determinants thus far used 

1 Starred paragraphs may be omitted if desired, without interfering with 
the unity of the course. 



ELEMENTARY APPLICATIONS 



35 



are called determinants of the second order, and we now define 
determinants of the third order as being symbolic expressions 
of the form 

«2 \ "2 ' (1) 

a, hr, c. 



*3 



which is imderstood to represent the quantity 



(2) 



This rather long expression can be written down without bur- 
dening the memory at all, by observing that it contains three 
products with + sign and three products with — sign, and that 
the first three products can be read off by following the directions 
of the arrowheads lq the following scheme : 



(I) (in (III) 

2 2 ^2 




(3) 



jug^s ; from the 



From the line marked (I) we get the product a 
line marked (II), the product liC^a^ ; and from the line marked 
(III), the product c-fi^a^. These are the three products that have 
the + sign in the value of the whole determinant. The other 
three products are read off in a similar way by following the 
direction of the arrowheads in this figure: 



(4) 



(IV) 




The lines marked (IV), (V), and (VI) give us the products a^\cy, 
l^c^a^, and c^^"-^, which, except that the negative sign must be 
prefixed to each, are the last three terms in the expression (2) 
for the whole determinant. 



36 



THE ELEMENTAEY FUNCTIONS 



*31. Two illustrative examples will make this procedure 
entirely clear, and will show that it is in reality very simple. 

Example 1. Write the value of the determinaiit 

3 4 1 
2 3 4 
6 6 8 

It is better not to draw the guide lines marked (I), (II), (III), etc. in 
(3) and (4) above ; we merely think them into the figure. Thus, the value 
of the determinant is 

3 -3 -3 + 4-4-6 +1-6-2-6-3-1-6-4-3-3-4-2 
= 27 + 96 + 12 - 18 - 72 - 24 = 21. 

Example 2. Write the value of the determinant 



■1-6 
2 -3 
1 -2 



Here some of the numbers a, b, etc. are negative, but of course this fact 
introduces no new difficulty, except that care must be taken about the 
sign of each product. The result is 

(- 1) (- 3) (- 1) + (- 6) - 5 - 1 + 4(- 2) ■ 2 - 1 (- 3) . 4 - (- 2) - 5 (- 1) 
- (- 1) (- 6) - 2 = - 3 - 30 - 16 + 12 - 10 - 12 = - 59. 

EXERCISES 



Write the value of each of the following seven determinants : 



3 6 7 
2 13 

4 3 7 

12 3 

2 3 4 

3 4 6 



4. 



7. 



5 
-3 
-2 

3 -1 

7 -6 
6 -3 

X y 

1 3 

2 -1 



2 -1 

5 -6 

6 1 

4 

-2 

1 

z 
4 
1 



1 a a^ 
1 b P 



a h 


9 


h b 


f 


9 f 






ELEMENTARY APPLICATIONS 



37 



8. Prove that interchanging two adjacent rows (or two adjacent 
columns) of the terms of a determinant changes the sign of the 
determinant; that is, 



h 


«i 




«1 


h 


"i 




h 


«2 


= — 


% 


h 


"8 


= + 


K 


«8 




«2 


h 


"2 





«3 

"1 



etc. 



(Note that the same statement can be made about a determinant of 
the second order as well.) 

9. Prove that if any two rows (or two columns) in a determinant 
are identical, the determinant equals 0. 

10. Prove that 



ttjW 


b^n 


CjU 


«2 


K 


^ 


% 


h 


"s 



that is, if a number is a factor of every term in a single row (or 
column) of a determinant, it is a factor of the determinant. 

11. Prove that 



a^ b„ G„ 



12. A minor determinant is a determinant obtained from a given 
one by suppressing all the terms in any one single row and also all 



those in any one single column. Thus, 
obtained from 



is a minor determinant 



by suppressing all the terms in3<the first row and all those in the 
first column. This minor determinant will be symbolized by A^ and is 
said to " correspond " to the term a^ at the intersection of the row and 

the column struck out. Similarly, to 5j corresponds Bj= ^ ^ and 



soon. Prove that aj^, — 6j5j + c,Cj=a.j^j — a242+a3^3= 



and also that a^B^ — a^^ + aji^ 



0. 



o «8 
"1 \ 
«2 h 



38 



THE ELEMENTARY FUNCTIONS 



y- 



*32. Solution of simultaneous linear equations in three vari- 
ables. If we solve, by the method of elementary algebra, the 

system of equations 

'' a^x + \y + c^z = (ij, (1) 

a^x + \y + c^z = d^, (2) 

^a^x + l^y + c^z = d^, (3) 

we obtain the values (after dividing out a common factor in 
numerator and denominator) 

^ ^ d^h^c^ + \c^dg + Cjbgd^ - dgb^c^ - b^c^dj^ — c^\d^ ^ 
a-fi^c^ + \c^a^ + HhH — HK^i — hV\ — ^3*1*2 

(Al-tQ/nCn ~\~ C(/-iOnUin —\~ l/-< C&q wa ttottnCl CX/oCnt^-t CnLt-tiA/n 

and a similar fraction for z. 

Each numerator and denominator can be written in the form 
of a determinant, as follows: 



x = 



where the denominator is in each case the determinant formed 
by copying the coefficients of x, y, and z in the equations (1), 
(2), and (3); and the numerators are formed in an exactly analo- 
gous way to that used in solving two equations in two variables. 

Example. Solve the set of equations 

'2x-Zy + 5z = -\0, (1) 

x-^2y-z = 9, (2) 

lbx-y + iz=7. (3) 

Writing down the values of x, y, and z according to the method just explained, 



d. 


h 


Cl 


d. 


h 


h 


d. 


h 


h 


«1 


h 


h 


«2 


b. 


h 


«3 


bz 


h 



«i 


d. 


«1 


«2 


d. 


^2 


«3 


ds 


«8 


«1 


\ 


«1 


«2 


h 


C2 


«3 


bs 


«3 



«1 


*1 


d, 


^2 


\ 


d. 


«3 


bs 


ds 


«1 


b^ 


«i 


«2 


b. 


«2 


«3 


bs 


«3 





-10 


-3 


5 




9 


2 


-1 




7 


-1 


3 




2 


-3 


5 




1 


2 


-1 




5 


-1 


3 



60 + 21 - 45 - 70 + 10 + 81 _ - 63 
12 + 15-5-50-2 + 9 ~ - 21 



ELEMENTARY APPLICATIONS 



39 



-21 



2 


-10 


5 




1 


9 


-1 




5 


7 


3 






-21 






2 


- 3 


-10 




1 


2 


9 




5 


- 1 


7 





54 + 50 + 35 - 225 + 14 + 30 _ - 42 
- 21 ■ ~ - 21 ' 



^2, 



28 - 135 + 10 + 100 + 18 + 21 _ 42 
- 21 ~ - 21 



:-2. 



Therefore x = d, y = 2, z = — 2 is the solution of the problem, and in fact 
this set of values satisfies all three equations. 



EXERCISES 



Solve by the method of determinants : 



3a; -2/ -2s =-7, 

[x + y + 3z = i. 



3. 



[ix + iy + ip = 38. 



. + 1 + 1 = 20, 
- + __«_8. 



( ax + by — cz = 2 ah, 

4. ■i.by + cz — ax = 2 be, 
\^cz + ax — by = 2 ac. 

Ua + b)x+(a — 6)s = 2Sc, 

5. \(b + c)y+(b — c)x = 2ac, 
[ (c + a) » + (c — a) y = 2 a&. 

*33. Linear equations in three variables do not lend themselves 
to graphical representation in the same way as linear equations 
in two variables, because a point is determined by a pair of 
values, — its abscissa and ordinate. In three-dimensional geome- 
try, however, a set of three coordinates may be introduced, corre- 
sponding to the three variables x, y, and z, and then any equation 
containing these variables may be given a graphical interpretation. 
This is found very useful in all higher study of geometry, but 
we shall not make any use of it in this book. 

The Quadratic Function 

34, The next application of the graphical representation to ele- 
mentary algebra arises when we consider the quadratic function. 
We have drawn the graphs of a number of such functions (see 



40 



THE ELEMENTARY FUNCTIONS 



Exs. 14-23, p. 22), and in every case the locus was a curve of a 
particular form, which we called a parabola. We assume that 
this is always the case, just as we assumed that the graph of 
every hnear function is a straight hne, without being able as yet 
to prove the correctness of the assumption.^ Stated in the form 
of a theorem the assumption is as follows: 

The graph of every function of the second degree, that is, of the 
form aa? + 'bx + c (or, what is the same thing, the graph of every 
equation of the form y = ay? + 6a; + c), is a parabola. 

Knowledge of this fact will be useful in checking the accuracy 
with which the coordinates of points on the curve have been com- 
puted ; for if the points, when plotted, do not lie on a curve of the 
general shape that we have learned to recognize as a parabola, a 
mistake has been made, either in the computing or in the plotting. 

35. Intersection of parabola and straight line. Let us draw 
in the same figure the graphs of the following equations : 
'y = a;^ — a; — 4, (1) 

Vy = 1x. (2) 

The tables of values for determining 
points on the graphs are 

(1) (2) 



I' 



X 


y = x^ — X 


-4 





-4 




1 


-4 




2 


-2 




3 


2 




4 


8 




-1 


-2 




-2 


2 




\ 


-4i 




-i 


-31 





Check. 




Fig. 24 



Now the graph of (1) contains all points whose coordinates 
{x, y) satisfy that equation, and the graph of (2) contains aU 

1 It is to be hoped that the student has not failed to notice that the word 
" parabola" has as yet no precise meaning at all ; it is, so far, only a word used 
in order to give a name to a more or less vaguely defined curved line. For 
this reason, if for no other, it would be impossible at this point to prove the 
assumption above. 



ELEMENTARY APPLICATIONS 41 

points whose coordinates satisfy the second equation. Hence any 
point whose coordinates satisfy both, equations must be found 
on both graphs, that is, must be one of their points of intersec- 
tion. By inspection of the figure we see that these points are 
approximately (—1, — 2) and (4, 8) ; and in fact these are the 
exact points of intersection, for these pairs of values wUl satisfy 
both equations. 

EXERCISES 

Find in this way the points of intersection of the following : 
f 2/ = x^ + X - 3, ^ry = 2.x^ + 2x-l, 

'\y= — x. ■ 1?/ = — 3x +1. 

f2/ = 2x^-5x-|-l, ^ r2/=6cc2-8x-12, 

^' ly = X - 3. ■ ty = - X + 8. 

36. Algebraic solution. Let us return to the example above 
and consider how it could be solved algebraically. To this end. we 
may substitute the value of y, obtained from one equation, in the 
other equation ; thus, putting 2 x, the value of y obtained from 
the second equation, in place of 2/ in the first equation, we have 

2x = 3?— X— 4:; 
that is, x^ — Zx = 4. 

This equation is an example of what is called in elementary 
algebra a " quadratic equation in one unknown quantity," and the 
student should be able to solve such an equation.^ The result is 
a; = 4, or a; = - 1. Since y=2x,y=B, when a; = 4, and y = - 2 
when a; = — 1. Hence the solutions are (4, 8) and (— 1, — 2), 
exactly as we had already discovered by the graphic method. 

EXERCISES 

Solve the above four problems algebraically. 

37. The general quadratic equation. Any quadratic equation 
in the one unknown quantity x can be reduced to the form 

ax^ + hx + e = 0, (1) 

1 If not, sufacient time should be taken to review this process thoroughly ; 
for this purpose Appendix F gives a full explanation, with examples. 



42 THE ELEMENTARY FUNCTIONS 

which is accordingly a general quadratic equation. In this equa- 
tion a, b, and c may be any numbers whatever, either rational or 
irrational (except that a cannot equal zero). Usually, however, 
these numbers will be rational, in the problems with which we 
shall have to deal. When we have solved the equation (1), we 
shall have a solution for every possible quadratic equation in one 
unknown quantity, since (1) represents any such equation. To 
solve (1), we use the method of elementary algebra known as 
" completing the square " (see Appendix F). 
Dividing (1) by a and transposing, we have 

a;2 + -« = -£. (2) 

a a 

To " complete the square," that is, to make the left side of (2) 

a perfect square, we must add \-z—] : 

a \2a) \2a) a 



! a) \2 a) 
that is, {a; + -— )=-—--- = 



J2 — 4 ac 
4a2 



(3) 



m, ( , ^ ^^"^ — Aac 

Therefore x -\ = ± . 

2a 2a 



-6±V6'-4ac 
or x = (4) 

rn, . 1 4 4-1, — h + y/H^ — 4: ac , —h — y/V^—Aac 

Ihese two values of x, then, and , 

2a 2a 

wUl satisfy the equation (1). The values of the unknown quan- 
tity which will satisfy a given equation are known as the 
roots of that equation. So the results (4) are the roots of the 
quadratic equation ax^ + hx + c^ 0. Since these are the roots 
of a general quadratic equation, the form (4) may be used as 
a symbolic representation, or formula, for the roots of any 
particular quadratic equation whatsoever, as the next paragraph 
illustrates. 



ELEMENTAEY APPLICATIONS 43 

38. Solution of quadratic equation by formula. If we take 
the equation that we have solved before (§ 36), 

x^ —3x = i, 
we can write it in the form aa^ + bx + c = by mere transposi- 
tion, thus: x'-Sx-A=0, 

so that a = 1, 6 = — 3, and c = — 4. Now the roots of the equation 
ax^ + bx + c= are, as was just found, 

— b ±V62 _ 4 ac 
2a 

hence, using the values of a, b, and c that apply to our special 
equation. 



-(-3)±V(-3)^-4-l(-4 ) 3±V9 + 16 . 
X = -^ '- "^-^ ^ '- = = 4, or - 1, 

which are of course the same results that we obtained before (§ 36). 
Again, take the equation 

x + -=3. 

X 

Here the equation is easily reduced to the form ax^ + bx + c = 
by multiplying both sides by x and transposing, thus : 

a^-3a: + l=0. 

By the formula the roots are 

3±V9-4 3 ±V5 
" = 2 = ^— 

(These results should be checked.) 

EXERCISES 

Solve by the formula the exercises on page 41, and also the 
following : 

1. 3a;2-7x + 2 = 0. 5. x^ - 6x - 775 = 0. 

2.7.^ + 6.-1 = 0. e.^ + ^ = ^^. 

cc — 1 X — 2 X + 1 

3. 4a;^-a;-3 = 0. ^ ^_ 2 2-x 9 

7. — = -• 

4. 10x'-13a; + | = 0. .-1 x + 1 4 



44 THE ELEMENTAEY FUNCTIONS 

39. Solution by factoring. Suppose the quadratic function 
aoP' + bx + c can be resolved into two rational factors, thus : 

ax^ + bx + c = a{x — a)(x — ^). (1) 

Now a product can equal zero only if one of its factors equals 

zero ; hence, if aa? + bx + c=^0, that is, if a{x — a) (x — 0)= 0, 

then either „ 

X — a = 

or a; — /3 = ; 

that is, a; = a and x = ^ are the only values of x that can satisfy 
the equation ax^ -\-lx + c = 0. This means that a and /3 are 
the roots of the equation 

ax^ + 6a; + c = 0. 

Example 1. Let us take once more the equation 
a;2 - 3 a; - 4 = 0, 
which we have already used several times. The factors of a:'' — 3 a; — 4 are 

(X - 4) (a: + 1). 

Hence a; — 4 = and a; + 1 = are the only possible ways in which our 
equation can be satisfied. Thus the roots are 4 and — 1, agreeing of course 
with the results already found. 

Example 2. 5 a:^ - 2 a; - 3 = 0. 

The factors of 5 x^-2 a;-3 are (5 a: + 3) (x-1); hence, if 5x^-2 a; -3 = 0, 
then either 5a; + 3 = 0ora: — 1 = 0. Hence a; = — | or x = 1. 

This method is evidently the simplest way that we have met of finding 
the roots of a quadratic equation. The only difiiculty is that it is often 
impracticable to find the factors of the given quadratic function. In that 
case the previous method, that of the formula, should be used. 



EXERCISES 

Solve the following quadratic equations by factoring : 

1. x'' + 3a; + 2 = 0. 5. 2x2 - 5x + 2 = 0. 

2. x^ + i a; - I = 0. 6. 3 a:^ - a; - 2 = 0. 

3. a:''-6r + 6 = 0. 7. 3 «'' + 2a; - 1 = 0. 

4. a;' -5a-- 6=1 0. 8. 4x2 - 203- + 9 = 0. 



ELEMENTAKY APPLICATIONS 45 

40. Summarizing the results thus far obtained, we have seen 
that a quadratic equation iu one imknown quantity can be solved 
(1) by " completing the square "; (2) by the formula ; (3) by factor- 
ing. Of these only the third or the second is suited to practical 
use. Of course the imknown quantity need not be x, as it was 
in the examples we have considered ; on the contrary, the student 
shoiild practice solving quadratic equations in various letters, 
until the process becomes thoroughly familiar. A number of prob- 
lems are added, as an aid in this work and for review purposes. 

EXERCISES 

Solve and check (algebraically) : 

1. 2t' + t-2 = 0. 6. s^-10s-M6 = 0. 

2. 5y^-3y-2 = 0. 7. 3 _ 2 ^^ 

r — <o r — o 

3. 7)5^ -8!! + 8 = 0. y 21 1 

4. 92/^-ll2/-12 = 0. ^' im'^2E^j^~^' 

4 111, 

5. r-|-3 = T- 9. q i + z. + q-— = 4. 

r — 1 1 — s^l — sl + s 

Solve the following pairs of simultaneous equations both alge- 
braically and graphically : 



10 



ly = 6. ■ I: 



^-3a;, 



ry = x'-3x + 4:, nj = x'-3x, 

^'■{2/ =2. ^^•\2x + y = 2. 

ry=.x% ^g ry = 2x^-6x + l, 

12- •[y_a;-2=0. \x-h2y = 3. 

41. Has every quadratic equation a solution? Let us try to 
solve this quadratic equation : 

a;^ - 2 a; + 2 = 0. (1) 

Applying the formula, we get 



2 ± V4 - 8 2 ± V-4 
x = = t: ■• \'^) 



46 THE ELEMENTARY FUNCTIONS 

Now (2) gives an expression for the roots of the equation (1) ; but 
what is the actual value of that expression? V4 is equal to 2 
or to - 2, since 2^ = (- 2)^* = 4 ; but what is the value of V— 4 ? 
Not — V4, as is sometimes hastily concluded, for the simple 
reason that — V4 = — 2, and (— 2)* is + 4, not — 4. As a matter 
of fact there is no positive and no negative number whose square 
is — 4. Hence the " solution '' (2) for x is merely a formal ex- 
pression, without any value among aU the numbers that we are 
acquainted with. The same thing is evidently true of any indi- 
cated square root of a negative number. We shall call such 
expressions " complex expressions." They are not, for us, subject 
to laws of operation, like numbers ; but whenever they enter into 
a problem, we shall consider that the problem has no solution. 

It is indeed found useful in higher mathematics to introduce 
new numbers such that their squares are negative ; these new 
numbers are called " complex numbers," and simple, practical 
rules are adopted for performing arithmetical operations upon 
them. We shall not need to use them, however, and so we shall, 
as stated, consider as unsolvable any problem which leads to 
complex expressions. 

We accordingly conclude that our quadratic equation 

a;2-2« + 2 = 
is unsolvable, because its roots are complex expressions. 

EXERCISES 

Show which of the following equations are unsolvable, that is, 
have complex roots : 

1. x'' - a; + 1 = 0. 4. a;' + 3 a; + 2 = 0. 

2. 2a;^-3a;-3 = 0. 5. 5a;^ + 3a; - 1 = 0. 

3. 2x''-5x + 6 = 0. 6. 10a;2-7a; + 2 = 0. 

42. Test for solvability of a quadratic equation. The situa- 
tion described in the last paragraph can only happen when in 

— b± V&2 — 4 ac 
the formula , for the roots of the equation, a 



ELEMENTARY APPLICATIONS 47 

negative number appears under the square-root sign. But the 
quantity under the square-root sign is b^ — 4:ac; hence the roots 
will be complex when h^ — iac is negative. On the other hand, 
if 6^ — 4 ac is positive, the formula gives two different values for 
the roots of the equation, these values being rational if 6^ — 4 ac 
equals a perfect square, and irrational if 6^ — 4 ac is not a per- 
fect square (on the assumption that a, h, and c are all rational 
numbers). One .)ther possibility must not be overlooked, — that 
is, J^ — 4 ac may equal zero ; in this event the formula takes the 

very simple form > and thus there is only one root, 

^ ^ 2a " 2a 

We describe this situation by saying, " The roots of the equation 

ax^ + hx + c = Q are equal when 6^ — 4 ac equals zero." 

Summarizing, we have the following facts: 

' >0, the equation has two unequal roots. 
If 6"— 4ac J =0, the equation has equal roots. 

< 0, the equation is unsolvable (has complex roots). 

The quantity 6" — 4 ac is called the discriminant of the quadratic 
equation ax^ + 'bx + c = (i, since it enables us to determine the 
nature of the roots of the equation. 

EXERCISES 

Determine the nature of the roots of the following equations 
(without actually finding the roots) : 

1. a;2_3a;+l=0. 6. 4x=' -12a; -f- 9 = 0. 

2. a;^ — 4x4-4 = 0. T-Sas^-I-Sx — 4 = 0. 

3. 2x^-x-l=0. 8. ^x'' -I- 5a; -f 3 = 0. 

4. a;'*-f-x=l. 9. 8a;''- 8x -I- 2 = 0. 

5. 3x^- 5a; -I- 6 = 0. 10. §x' - |a; -f 1= 0. 

43. Graphical interpretation of discriminant test. Let us return 
to the problems concerning the intersection of a parabola and a 
straight line, from which we were led to the study of quadratic 
equations. Take again the example on page 40, § 35. 



48 



THE ELEMENTARY FUNCTIONS 



Example 1. 






X — 4:, 



(1) 

(2) 



We saw that by eliminating y between the two equations there results the 

quadratic equation in x, 

a;2-3x-4 = 0, (3) 

which must be satisfied by the abscissas of the points of intersection 
of the graphs of (1) and (2). Now the discriminant of this equation is 
equal to 9 — 4(— 4) = 25, a positive number, and accordingly the equa- 
tion has two unequal roots. Therefore there exist two abscissas (the roots 
of (3)) which determine points on both of the graphs ; that is, the graphs 
intersect in two different points, as of course we proved by actually 
finding the points (4, 8) and (— 1, — 2). But the method of this section 
enables us to prove that the graphs really intersect, without actually find- 
ing the coordinates of their point of intersection, by merely noting the 
value of the discriminant 6"— 4 ac. 



{ 




Example 2. Let us consider the 
graphs of the two equations 

fj/ = 2 x2 _ 3 a: _ 4, (l) 
x-y-Q = 0. (2) 

Substituting the value of y 
from (2) in (1), we have 

2a;2_42; + 2 = 0, 
that is, z^ - 2 X -1- 1 = 0, (3) 

which must be satisfied by the 
abscissas of the points of inter- 
section of (1) and (2). In (3), 
ft^ — 4 ac = ; hence there is only 
one root of (3), which means that j-jq 25 

there is only one point common 

to the parabola (1) and the straight line (2). In other words, the line 
is tangent to the parabdla. The figure, of course, verifies this conclusion. 

Example 3. Let us find the points of intersection of 
2/ = 2a;2-3x-4 
and y = 2 a; — 8. 

Substituting the value of y from (2) in (1), we obtain 
2x2-5x-l-4 = 0, 



(1) 
(2) 

(3) 



which must be satisfied by the abscissas of the points of intersection 
of (1) and (2). But in (3), i^ _ 4 „£ = 25 - 82 = -7; hence there is no 



ELEMENTAEY APPLICATIONS 49 

solution to (3), which means that there is no point common to the parabola 
(1) and the straight line (2). Moreover, a drawing of the two graphs verifies 
this conclusion, as of course it must. This verification is left to the student. 

SuMMAEY. Given two equations, representing a parabola and a 
straight line, after eliminatiag one of the variables from the two 
given equations we obtain a quadratic equation in one unknown 
quantity (the equation (3) in each of the above examples), whose 
roots are the abscissas (or ordinates) of the points common to the 
two graphs. If the discriminant of this equation is positive, the 
graphs iatersect in two distinct points ; if the discriminant equals 
zero, they have only one poiut in common, which means that 
the liue is tangent to the parabola; and if the discriminant is 
negative, the line does not meet the parabola at all. 

EXERCISES 

Determine by the discriminant test, before drawing the graphs, 
whether each of the following pairs of lines will intersect, be tangent, 
or fail to meet. Verify by drawing the graphs. 

ry = 2a^-x-3, ^ Cy = o^-2x+l, 

'\y=Zx. ■l2/ = 0. 

ry = a? + x+l, ry = 3x' + Bx + 3, 

^•\y = 3. '"•lcc + y = 0. 

, ry = x'-5x-3, ry = x' + x+l, 

^■\y=-3i. ■Uy-3 = 0. 

ry = x^-ix + 2, , ^2 ry = ar' + 5aj+l, 

\2x + y+l=0. '1^ + 3 = 0. 

ry = of + 3x+l, ry = x^ + x, 

^■iy = ix-3. '^•ly-2 = 0. 

^- [y = 0. 12/ + 2 = 0. 

ry=:x^-2x-3, jg ry=3x^-4:X + 2, 
■^^ |y + 4 = 0. ■ \x + y + 3=0. 

ry=2x'-3x + 8, ig ry=ix'-12x + 9, 
^•\y = 5x. ■l2/ = 0. 



50 



THE ELEMENTAEY FUNCTIONS 



17. 



18. 



19. 



fy = 4:x' -12 X + 9, 

Cy = ix'-12x + 9, 
12/=- 2. 



\y = x^ 



■{ 
■{ 



2a; -3. 



y = x', 
X + y = 0. 

y = 3?-l, 
X + y + 3 = 0. 



23 



24 



25 



{ 



2/ = 2x^-3x + 3, 
a; + 2/=l. 

= 2a:'=-3x + 3, 
-y+l=0. 

2/ = 6a;^- 3a; -2, 
2/ + X = 6. 

3/ = 4x2-3x-2, 
2/ = X - 3. 

y = x'' + X + 1, 
1 



44. Construction of a tangent to a parabola. This discriminant 
test is especially useful for discovering tangents to a given 'parab- 
ola (and also to certain other kinds of curves, as we shall see 
later). Thus, let us draw the parabola 

y = a;2 + a; + 2 (1) 

and the straight lines 

y = 3 X + c (2) 

for several values of c. Fig. 26 shows 
the lines for c = 0, c = 1, c = 2, and 
c = — 1. We observe that the line for 
c = 1 seems to be tangent to the curve ; 
to prove that it actually is so, we take 
equations (1) and (2) simultaneously, 
and, substituting «/ = 3 x + c in the 
equation y = x^ + x + 2, we get 

x2 - 2 a; + 2 - c = 0, (3) 
which has for its roots the abscissas 
of the points of intersection of (1) 

and (2). If (2) is tangent to (1), the roots of (3) will be equal, that is, 
62-4acwill equal zero. But &2_4«c = (- 2)^— 4(2 — c) = 4c-4, 
which will equal zero when c = l. Hence the value c = l does in fact 
make the line (2) tangent to the parabola (1), just as was indicated 
by the graphical solution. We have thus found a line, namely, the 
line y = 3 X + 1, which is tangent to the parabola y = 3?--\-x-\-1. 




Fi6. 26 



ELEMENTARY APPLICATIONS 



51 



Next, let us try the same parabola (1) with the lines 

y = 2x + c. (2') 

On drawing the lines (2') for several values of c it will be seen 

that we have, as before, a set of parallel lines ; but these make 

a smaller acute angle with the X-axis. Fig. 27 shows the lines 

for c = 0, c = 1, c = 2, and c = 3. 

The line for c = 2 seems to be 

tangent to the parabola, but we 

apply the discriminant test as 

above, to verify this conclusion. 

Eliminating y by substituting 

its value from (2') in (1), we get 

ic2_ic + 2-c=0. (3') 

The discriminant of (3') is 4 c — 7, 
which will equal zero when and 
only when c = |. This shows 
that the Mne y = 2x + ^ is tan- 
gent to the parabola, while the -^la. 27 
line y = 2 a; -I- 2 is not actually 

tangent. This illustrates again the fact that the graphical method 
cannot always be relied upon to give accurate results. Of course 
this is an unavoidable defect of graphic methods in general, 
whereas the algebraic method is precise. 




EXERCISES 

1. Eind for what value of c the line y = — a; -|- c is tangent to the 
parabola y = x^ + x + 2. 

2. Answer the same question for the parabola y = ar' — 4a; — 3 
and the line y = ^x + c; for the line y =— ^x + o; for the line 
y = c. 

3. The same question for the parabola y = 2x^—Zx + 2 and the 
line y = mx. ^ns- m = 1 or - 7. 

4. For what value of k will the line y = mx + k(m being a fixed 
number) be tangent to the curve y = x^ ? 



62 



THE ELEMENTARY FUNCTIONS 



5. Find the coordinates of the point where the tangent line of 



Ex. 4 meets the parabola. 



Ans. 



2'TJ- 



6. Find the coordinates of the point where the tangent line of 

Ex. 4 meets the F-axis. , /. rrv'\ 

Ans. [0, - — j- 

Draw a conclusion from the results to Exs. 5 and 6. 

*45. Sets of curves. In the equation y = 3 a; + c of the example 
in § 44 the number c was understood to be capable of assuming 
any value we chose, thus giving rise to a set of straight Hues. 
Such a general number as c in this equation, which can be given 
any particular value desired, is called a parameter. The presence 
of a parameter in an equation, then, indicates that the graphical 
representation of the equation wiU consist of an unlimited set of 
lines, straight or curved according to the nature of the equation. 
In § 44 we considered the prob- 
lem of discovering which one of 
a set of straight lines is tangent 
to a given parabola, but the dis- 
criminant test can equally weU 
be used to discover which one of 
a set of parabolas is tangent to a 
given straight line. 



Example 1. 



Cy = x^ ■ 



■ ix + c, 



(1) 

(2) 




Fig. 28 



Fig. 28 shows the parabolas for the 

values 2, 3, 4, and 5 of the parameter c. 

The curve for c = 4 seems to be tangent 

to the line (2), that is, to the X-axis ; and the discriminant test verifies 

the result indicated by the figure. For, substituting in (1) the value y = 



from (2), we get 



• 4 X + c = 0. 



(3) 



The discriminant of this is 16 — 4 c, which evidently equals for c = 4, 
and only for that value. 



ry = 2x^ + kx + 8, 
Example 2. < 

Putting y = in (1), we have 2 x^ + ifcx + 8 = 0. 



(1) 
(2) 

(3) 



ELEMENTARY APPLICATIONS 



53 



If (1) is tangent to (2), the discriminant of (3), which is k^ - 64, must 
equal 0. This will happen when ^ = ± 8, giving two curves of the set (1) 
that are tangent to (2), that is, to the X-axis. (If k is any number > 8, 
the discriminant of (3) will 
be positive, and the cor- 
responding parabola will 
meet the X-axis in two dis- 
tinct points ; while if k is 
any number between — 8 
and 4- 8, the discriminant 
will be negative, and hence 
the corresponding parabola 
will not meet the X-axis at 
all ; if k<- 8, there will 
again be two intersections.) 
Draw the figure on a 
large scale. Fig. 29 




EXERCISES 

Find, by means of the discriminant test, for what values of the 
parameter k the curve and straight line will be tangent in each of 
the following exercises. Draw the graphs for these and also for 
several other values of k. 

"^y = a? + 5x + k, fy = x' — .lx-\-k, 

0. ' \y = 2. 






2. 



3. 



5. 



6. 



5, 



' y = a^ — kx + i, 

i = o. 

y = lj(^ — kx + 12, 
Ly = 3. "'\y^2x-3. 

*46. Sum and product of roots of a quadratic equation. The 

roots of the quadratic equation 

am? + l)x + c = 



iy = 

Cy = kx^ — 3 a; — 
\y = x + l. 

{y = kx^ — a; -f I, 
2/ = 



are 



— h -f V62 _ 4 «c , 

X-, = and 



2a 

The sum of the two is 
and their product is 



h - V&a -4tac 
2a 
26 -b 
a 



X^Xo — 



62. 



2a a 

■ (h^ — A ac) _4: ac _ c 
4a2 ""4a2~a' 



(1) 
(2) 



54 THE ELEMENTARY FUNCTIONS 

The results (1) and (2) enable us to write down hy inspection the 
sum and the product of the roots of a quadratic equation. 

Example 1. 2 a;" _ 5 a; + 3 = 0. 

Here = - , which is accordingly the sum of the roots ; and - = - , 

which is the product of the roots. We can verify this by actually finding 
the roots 2 x" — 5 a; + 3 = (2 a; — 3) (a: — 1) = 0. Hence the roots are x = 1 
and a; = 5. Their sum is in fact J, and their product ^, as just found. 

Example 2. Use the results (1) and (2) to find the values of x^ + x^ and 



Solution. 


in terms of a, 
We have 


h, 


and 1 


+ 


^2 




-b 

a 


and 










X, 


x„ 


_ 


c 



(1) 
(^> 

To get Xj2 + x|, begin by squaring (1) : 
Xi^ + 2x1X2 + x| = -. 

Multiply (2) by 2, 2 x,x„ = — . 

a 

Subtracting, x ^ + x^ = ^ . (•3) 



To get — + — > which equals ^ ^ ^^ , we need only to divide (3) by 

Xj Xg Xj Xg 

x,^xi, that is. by '-^ . This gives ^^tfl = Ijzl^ ^f.^ '^-^ac 

EXERCISES 

1. Write down the sum and the product of the roots of each of 
the following equations, and verify the results by actually finding 
the roots : 

(1) x''- 335-4 = 0. (5) 3cc^-lla: + 6 = 0. 

(2) a;^ + 10a; + 9 = 0. (6) 2a;^ - 3a; = 0. 

(3) 2a5»-3x-2 = 0. (7) x'' + jsx + ? = 0. 

(4) 5a;2 - 6a; + 1= 0. (8) aiV + 2ma; + 1 = 0. 

2. Write down the sum and the product of the roots of each of 
the following equations : 

(1) mV + 2 (m - 2 a)a; + 1 = 0. 

(2) {h^ + aV)x^ + 2 a?mx + a\l - b") = 0. 

(3) (a + b + c)a^+(a + b- c)x + a' + 6^ - c^ + 2 aj = 0. 



ELEMENTAEY APPLICATIONS 55 

3. Find two numbers whose sum is 22 and whose product is 117. 

4. Find two numbers whose sum is 47 and whose product is 420. 

5. Find the value of each of the following expressions in terms 
of a, b, and c (it being understood that x^ and x^ are the roots of the 
equation ax^ + ftx + c = 0) : 

(1) xl + xl (4) xixl + xlxl. (x, , X, 

(2)xt + xl .5. J_,^ ^H^/^ 

(3) x',x?, + xlxl ^"^ x-^x^ "^ x^xl ■ 

*47. Factor theorem for quadratic equatibn. On page 44 we 
learned that any factorable quadratic equation can be solved by- 
inspection. We used the principle that " a product of two factors 
wiU equal zero when and only when one of the factors equals 
zero." The result of that investigation may be stated as follows : 

If x — a is a factor of the function aoP- + 6a; + c, then a is a 
root of the equation ax^ + 6a; + c = 0. (I) 

The object of this paragraph is to show that the converse^ of 
this theorem is also true, that is, to prove : 

If a is a root of the equation ax^ + 6a; + c = 0, then x — a is a 
factor of the function ax^ + bx + c. (II) 

The statements (I) and (II) are called the factor theorem for 
the quadratic equation. 

Proof ot (11). ax^ + bx + c = a(x^ + -x + -\ 

\ a a/ 

If a and P are the two roots" of the equation ax'^ + hx + c = 0, then, by 

the preceding paragraph, 

a 

and aR = -• 

a 

Therefore ax^ + bx + c = a^x'^ - (a + /3)x + ar/3] = a (a; - a) (x - /3). This 
proves the theorem. 

1 The converse of a theorem is a new theorem in which the hypothesis of the 
original theorem becomes the conclusion of the new one, and vice versa. Thus, 
the theorem "If the sides of a triangle are equal, its angles are equal" has 
for its converse the theorem " If the angles of a triangle are equal, its sides 
are equal." Here both the original theorem and its converse are true, but 
that is not always thfe case. The student should think of examples in which a 
theorem is true but its converse false. 

" p will equal a in case 6^^ — 4 oc = 0, but the proof is valid in this case also. 
The case b^ — iac<0 cannot happen under the hypothesis we have made. 



56 THE ELEMENTARY FUNCTIONS 

A second proof. Another method of proof of this theorem can be worked 
out by the student as follows : Divide ax^ + bx + chj x- a, using the ordi- 
nary process of long division ; the quotient will be found to be a:c + aar + b, 
with the remainder aa^ + ba + c. Now, by the hypothesis, a is a root of 
the equation ax^ + 6x + c = ; hence aa^ + ba + c = 0; that is, the remainder 
when ax^ + bx + cis divided by x — a is 0. This proves the theorem. 

A third proof. This theorem can also be proved in the following 
manner, which is especially instructive, and which, moreover, leads us to 
another result of importance. 

Begin with ax^ + bx + c = a(x^ + -x + -j a,a before. This time we will 

" complete the square " of the first two terms in the parenthesis, that is, 
add ( — ) , so as to make a perfect square. We then have 

\ a a/ L a i a" i a^ aA 

_ r _ -b-W-iac l r _ -b + Vb^-iac l 

= a (x — a) (x — /3), 

- 6 - VJ2 -iac , o -b + Vi2 - 4 ac , , 4. - ^i. 

where a = and p = are the roots of the 

2a 2a 

equation ax^ + bx — c = 0. 

This completes the proof of the theorem. But it does more 
than that. The form (A) shows that the function aso^ + bx + c 
can be written as the difference of two squares whenever 6^— 4ac 
is positive ; while if S^ — 4 ac is negative, the form (A) will be 
the sum of two squares; and if 6^— 4ac equals zero, the function 
aoa^ + bx-\-c is shown to be itself a perfect square. It follows 
from this that in each of the last two cases (6^ — 4 ac ^ 0) the 
sign of the function aa^ + bx + c wUl be the same as the sign of 
a for all values of x (for neither the sum of two squares nor a 
single perfect square can be negative). This is often a useful fact 
to know concerning such a function^. For example, the function 
a? — x + 1 will be positive for all values of x, since J^ — 4 ac< 
and a is positive; and the function — 2 a? + 10 x — 13 will be 



ELEMENTARY APPLICATIONS 



5T 



negative for all values of x, since 6^ — 4 ac < and a is negative. 
In the case where 6^—4 ac>0 the result is almost equally simple, 
for then (A) gives us asi? + bx + c = a(x -- a) (x — /3), so that the 
sign of the function will be the same as that of a if both the 
factors x — a and » — /3 are positive or if both are negative, 
that is, if x>^ or <a; but the sign of the function will be 
opposite to that of a if a; — /3 is negative, while as — a is positive, 
that is, if X is between a and /3 (a < a; < /8). 

Example. — x"^- 5x - 6 = -l(x + B)(x + 2). 

Here a = — 3, /3 = — 2, and evidently the sign of the function will be 
negative (the same as that of a) except when x is between — 3 and — 2 
(— S<x<— 2), in which case the last factor a; + 2 is negative; the factor 
X + S, however, is positive, so that their product is negative, and hence the 
function — 1 (a; + 3) (a; + 2) is positive. This result is well illustrated by 
the graphic representation of the function. If we take as usual x for the 
abscissa and y =f(x') = — x^ — 5x ^ 6 for the ordinate, the graphical 
representation will give the parabola shown in Fig. 30. Of course ?/ = 
when a; = — 3 or — 2, that is, at the points A and B (because a; + 3 and 
a; + 2 are factors of /(a;)) ; and we see that the function is positive (the 
curve is above the J^T-axis) for all values of x between — 3 and — 2, but 
that it is negative (the curve is below the .X-axis) for all other values of x. 

Y 
Table of values 



X 


y 





-6 


1 


-12 


-1 


-2 


-2 





-3 





-4 


-2 


-H 


+ i 


-^ 


+ feto. 




Fig. 30 



EXERCISES 

1. Test both algebraically and graphically the signs of the 
following functions : 

(1) ar'-Saj + S. (3) a;^ - 10 a; - 7. (5)3x^-x+l. 

(2) x^-3a; + 2. (4) 2a;^ - 3x - 4. (6) 6a;' - 6x +1. 



58 THE ELEMENTARY FUNCTIONS 

2. Test algebraically in a similar manner the product 

and apply the information gained to make a rough sketch of the graph 
of the function, without computing any pairs of values exactly. 

3. Test in the same way the product 2(x—iy(x — i); the product 
-3(x +l)\x - Sy-, the product -x{x+l)(x + 3f. 

*i8. Maximum and minimum values of a quadratic function. 

The expression (A) on page 56 enables us to draw still another 
conclusion. Since we have 



ax^ + bx + c-- 



7 bV h^-4: acl 

\"+2^j — 4^r 



and since the term I x + ^r— 1 is always positive, or at least zero, 

the least possible value of the whole expression in the square 

/ i V 
brackets will be obtained when ( x + — — ) is equal to zero : that 

\ 2 a/ 

is, x = — —- will give the least possible value of the expression 

which multiplies a, and therefore also the least possible value of 
ax^ + bx + c it a is positive. On the other hand, this value of x 
will obviously give the greatest possible value of aa^ -\-hx + G 

if a is negative, because the term I x + -— 1 is then a subtractive 

term, so that its least value will give the greatest value of the 
whole function. 

Example. -2x^ + i,x -1 = -2{x'^ - ^x + \) 

= -2[(x-i)2-ff+H 

Since the least value that (x — J)^ can have is zero, x = f will give the 
least value of that term, which, being subtractive, gives then the greatest 
value of the whole function. Thus, in this example the greatest value of 
the function is -V-, the value obtained when x = f . This agrees with the 

result just proved in general; namely, that x = — —- will give the greatest 

_ J _ 5 5 2 a 
value of the function, for here = = -. 

2a -4 4 



ELEMENTAEY APPLICATIONS 59 

It is always better to go through with the complete process of reducing 
the given function to the form (A) for each special case, as was done in 

the above example, rather than to use the value x = ^— as a formula. 

2 a 



EXERCISES 

Find the maximum or minimum values of each of the following 
functions, and verify your result by making a graph of the function : 

(1) x'-Sx + G. (3) -x^-6a;+7. (5) Sx'' - 6x + 5. 

(2) 2a;2-3x-4. (4) - 3 a;'' + a; - 1. (6) - a;^ - 10 a- + 1. 

Note. An easier method of finding maximum and minimum values of 
functions is developed in the Differential Calculus. 

MISCELLANEOUS PROBLEMS INVOLVING QUADRATIC EQUATIONS 

A few concrete problems leading to quadratic equations are here 
given. In solving such problems the most important thing is to be 
sure of understanding the problem itself thoroughly, then to trans- 
late the concrete language of the problem into the more abstract 
language of algebra. 

1. The length of a rectangular field exceeds its breadth by 2 rd. 
If the length and the breadth were each increased by 4 rd., the area 
would be 80 sq. rd. Find the dimensions of the field. 

2. The area of a certain square may be doubled by increasing its 
length by 10 ft. and its breadth by 3 ft. Find the length of its side. 

3. A rectangular grass plot 12 yd. long and 9 yd. wide has a path 
of uniform width around it. The area of the path is f of the area 
of the plot. Find the width of the path. 

4. A farmer sold a number of sheep for |120. If he had sold 5 
less for the same money, he would have received $2 more a head. 
How much did he receive per head ? 

5. A man agrees to do a piece of work for $48. It takes him 4 days 
longer than he planned, and he finds that he has earned $1 less per 
day than he expected. In how long a time did he plan to do it ? 

6. A circular grass plot is surrounded by a path of a uniform 
width of 3 ft. The area of the path is J the area of the plot. Find 
the radius of the plot. 



60 . THE ELEMENTAEY FUNCTIONS 

7. A boat steams down a river 12 mi. and back in 2 hr. 8 min., 
and its rate in still water is 4 times the rate of the current. Find 
the rate of the steamer and that of the current. 

8. A straight line AB, of length I, is divided at a point X in such 
a way that AX is the mean proportional between AB and XB (that is, 
AB:AX=AX: XB). Find the lengths 

V5-1 ^ 5 ^ 

AX and XB. Am. AX = — l. 

2 Fig. 31 

AB is said to be divided in extreme 
and mean ratio at X. Prove that if BA is produced beyond A 
by a length AX' = AX, then X'B is divided in extreme and mean 
ratio at A. 

9. In a right triangle ABC whose hypotenuse AB equals 20, a 
perpendicular CD is drawn from C to AB. If BC = f AD, find the 
lengths of BC, AD, and AC. 

10. Divide 100 into two such parts that their product shall be a 
maximum. 

11. Find the dimensions of the largest rectangle that can be 

inclosed by a perimeter of 60 ft. 

Hint. If x represents the number of feet In the length, then 30 — x will 
represent the number of feet in the breadth, and x (30 — x) is to be made a 
maximum. 

12. An isosceles triangle has the dimensions 10, 13, 13. Find 
the dimensions of the largest rectangle that can be inscribed in this 
triangle, one side of the rectangle lying in the base of the triangle. 

13. A window is to be constructed in the shape of a rectangle 
surmounted by a semicircle. Find the dimensions that will admit 
the maximum amount of light, if its perimeter is to be 48 ft. 




CHAPTER IV 

INTRODUCTION TO THE TRIGONOMETRIC FUNCTIONS 

49. Definition and measurement of angles. We have considered 
in some detail the question of measurement as applied to distances ; 
we have now to consider its application to angles, — a question 
which is of equally fundamental importance. We shall find it 
convenient to think of an angle as resulting from a rotation 
of a straight line from one position to another, the vertex of 
the angle being the center of rota- 
tion. Thus, the angle BAG (Pig. 32) 
may be considered as generated by 
the rotation of a line about A, from 
the initial position AB to the ter- fig, 32 

minal position ^C; and the numerical 

measure of the angle BAC gives the amount of this rotation 
in terms of a conveniently chosen unit. The most common unit 
of measure is the degree, which is g^^ of a complete rotation 
about the vertex of the angle. Thus, an angle of 90°, or a right 
angle, is produced by one fourth of a complete revolution about 
the vertex ; an angle of 180°, or a straight angle, is produced by 
one half of a complete revolution about the vertex; an angle 
of 60° is produced by one sixth of a complete revolution ; and 
so on. Any angle can thus be measured by the amount of rota- 
tion from its initial line to its terminal line; and it is evident 
that an angle can have a numerical measure greater than 180° 
or even greater than 360°, for we can easily have an amount of 
rotation that is greater than 360°, that is, greater than one com- 
plete revolution. Thus, an angle of 450° would mean one complete 
revolution (360°) and 90° more. 

50. Moreover, as rotation can take place in either of two opposite 
directions, we are led to make a distinction between positive and 

61 



62 



THE ELEMENTARY FUNCTIONS 



negative angles. It is usual to regard angles that are produced by 

a counterclockwise direction of rotation as positive, and those that 

are produced by a clockwise direction of 

rotation as negative. Thus, in Fig. 33, 

if the rotation is from AB to AC, the 

angle is positive ; if the rotation is 

frovi AC to AB, the angle is negative. Fig. 33 




EXERCISES 

Construct (with a protractor) angles equal to 250°, — 30°, 390°, 
- 430°, - 400°, 1000°, - 180°, - 350°, 490°. 

51. Relation to coordinate system. For convenience of reference 
the four parts into vs^hich the plane is divided by the X- and Y-axes 
are called quadrants and are numbered in the counterclockwise 
order, as shown in Fig. 34. If we 
now consider an angle to be formed 
by a rotation about the origin, the 
initial position being the positive 
halt of the X-axis (OX), then the 
terminal Unes of all angles between ~ 
0° and 90° will lie in the first 
quadrant; those of all angles be- 
tween 90° and 180° will lie in 
the second quadrant ; and so on. 
Thus, the terminal line of the angle 
225° wUl be in the third quadrant, 
bisecting the angle between the 
X- and r-axes (Fig. 35). We shall 
say, " The angle 225° is in the third 
quadrant," and similarly for any 
angle, meaning that the terminal 
line of the angle is in that quad- 
rant when the initial line coincides 
with the positive half of the X-axis. 
Likewise, an angle of — 225° is 
in the second quadrant, and so on. Fig. 35 



III 



IV 



Fig. 34 



TEIGONOMETKIC FUNCTIONS 



63 



EXERCISES 

1. In what quadrant is each of the following angles : 150°, 300°, 
550°, - 20°, - 150°, - 200°, 1040°, - 500°, - 36,280° ? Using a 
protractor, construct each terminal line, and mention the number 
of complete revolutions in case it is more than one. 

2. If a wheel makes 80 revolutions per minute, through how 
large an angle does it turn in 15 sec. ? in ^ sec. ? in 5 min. ? in 
1 min. 60 sec. ? 

3. Through what angle does each hand of a clock rotate in 1 hr. ? 
in24hr.? 

52. Definition of the trigonometric functions. Having become 
familiar with the idea of an angle of any magnitude, including 



A 

Ke X 





Fig. 36 



the distinction between positive and negative angles, we are now 
prepared for the study of some important numerical values that 
depend upon angles, or are determined by angles. 



64 



THE ELEMENTARY FUNCTIONS 



Let ^ ^ be any angle, and suppose it has been formed by a rotation 
from the initial position OX to the terminal position OA (Fig. 36). 
Let P be any point on the terminal line of the angle Q, and con- 
struct line segments representing the ordinate and the abscissa 
of P(Fig. 37). These segments are QP and OQ in each of the 



^v 



fp^ 




Q'Q 




\P' 



QQ'O 





Fig. 37 



four parts of Fig. 37. Now the ratio of QP to OQ will be the 
same for any one angle 0, whatever point P is taken on the ter- 
minal line OA. For if P' is any other point on OA, the triangle 
OQ'P' is similar to the triangle OQP (why ?), and hence 



Q'P' 
OQ' 



QP 
OQ 



1 See Greek alphabet at end of book. 



TRIGONOMETRIC FUNCTIONS 



65 



In other words, the ratio — • that is, the ratio — ; — : > 

OQ abscissa of P 

depends for its value only upon the magnitude of the angle 6, and 
not upon the particular point F that we take. This ratio is accord- 
ingly a function of the angle 0. The same thing can evidently 

be said of the ratio ^— > which is the ratio — ;; > 

OP distance from O to P 

and also of the ratio -— > which is the ratio — ;; — - — - ■ 

OP distance from to P 

(The distance from to P is called the radius vector of P.) 

All these ratios are thus functions of the angle 0. They are 

called the trigonometric functions and are named as follows (see 

A 





(a) 
Y 



Q X 




A-^P 



6 



(b) 




y 






c 


^ X Q 




k 


^y~^ 


•^A 



(<=) 



(d) 



Fig. 38 



Fig. 38, where x, y, and r symbolize the values of the abscissa, 
ordmate, and radius vector of the point P) : 

The ratio ordinate of P ^ gf ^ y i^ ^^^^^ tl^e tangent of the 
abscissa of P 0(^ X 
angle and is written tan 6. 



66 THE ELEMENTARY FUNCTIONS 

_, ^. ordinate of P QP y . n i ii . c t.\. 

The ratio = ■^— = - is called the sine or the 

radius vector of P OP r 

angle 6 and is written sin 6. 

mi i- abscissa of P OQ x . ,, j ,, . « , i 

The ratio = — — = - is called the cosine or the 

radius vector of P OP r 

angle 6 and is written cos 6. 

The reciprocals of these ratios are named as follows : 

1 abscissa oi P X . ■,■,■,.■, ^ ., in 
= = - is called the cotangent of the angle 

tan ordinate of P y 

and is written cot 6. 

1 radius vector of P r . ,i j ,, j, ,, , /, 

= = - IS called the cosecant or the angle 

sin ordinate of P y 

and is written esc 0. 

1 radius vector of P r . n , ,, » i, , /, 

= = - IS called the secant of the angle 

cos abscissa of P X 

and is written sec 0. 

These names must be thoroughly memorized and the definitions 

made famUiar both in terms of the words used and in terms of 

the corresponding lines in the figure. Angles should be constructed 

in various positions, and the trigonometric functions of each one 

obtained. In doing this it must never be forgotten that both the 

ordinate and the abscissa of any point are directed distances, so 

that particular attention has always to be paid to the question of 

sign. It will be noticed that some of the ratios are negative on 

this account. The radius vector is always to be taken as positive 

in determining the signs of the ratios. For example, in Fig. 38, (c), 

oc 
cos = -> and x is negative ; hence, r being positive, the ratio is 
r 

negative, which means that the cosine of an angle in the third 
quadrant will be negative. 

EXERCISES 

1. Which of the ratios are negative when 6 is in the second 
quadrant? when $ is in the third quadrant? in the fourth? 

2. With the aid of a protractor, construct the following angles 
and determine as accurately as possible, by careful measurement, 
the values of the six trigonometric functions of each : 20°, 60°, 



100°, 230°, 280°, 350°, - 30°, 
results in decimal form. 



TEIGONOMETEIC FUNCTIONS 67 

300°, - 1300°, 2000°. State the 



3. Show that sin (-20°) = -sin 20°; sin (- 70°) = - sin 70° • 
sin (-100°) = -sin 100°; sin (-210°) = -sin 210°. Generalize these 
results to apply to any angle a ; that is, show that sin (-«) = - sin a. 

4. Show that cos (- 20°) = cos 20° ; cos (- 70°) = cos 70° • 
cos (-100°)= cos 100°; cos (- 210°) = cos 210° ; and generalize 
these results to apply to any angle a; that is, show that 
cos (— a) = cos a. 

5 . Prove that sin^ o + cos'' o = 1 . (We usually write sin'' a, cos^ a, 
etc. for (sin ay, (cos ay, etc.) 

XI MP 

Solution, sin a = - = , 

r OP 

X OM 

cos a: = - = 

r OP 
Therefore 

V^ x^ 

sin^ a = ^ and cos- a = —\ 

1/2 . 




hence 



sin'' a + cos^ a = 



Fig. 



But, by the Pythagorean Theorem, 
Therefore 



y'^ + x^ = r^. 
sin^o: + cos" a = 1. 



The student should draw figures with a in different quadrants, 
and show that the proof holds in these cases also. 



6. Prove that tan^a + 1 = sec" a. 

7. Prove that cot" a + 1 = esc" a. 



„ T. . , , sin a 

8. Prove that = tano. 



Note. The results of Exs. 5, 6, 7, and 8 are such important relations among 
the trigonometric functions that they should be memorized. 

9. Construct an angle whose tangent is 2 ; find its value in 
degrees, with the aid of a protractor. What are the values of the 
other five trigonometric functions of this angle ? 

10. Construct an angle whose sine is J ; find its value in degrees. 
Give the values of the other trigonometric functions of this angle. 

11. Proceed as in Ex. 10, for an angle whose sine is — f ; for an 
angle whose cosine is — ^ ; for an angle whose tangent is — |. 



THE ELEMENTARY FUNCTIONS 



53. Functions of 45°, 30°, and 60°. (I) If we construct an angle 
of 45°, we observe that the ordinate and the abscissa of any 
point on the terminal line are equal {0Q= QP, since the triangle 
OQP is isosceles); that is, y = a;. Therefore 

tan 45° = 1 and also cot 45° = 1. 

Now r^z=a? + y^=2y'^; 

hence i' = y • ^• 

Therefore 



sin 45° = -^ = 



and hence 
Since 

Therefore 
Hence 



yV2 V2 
CSC 45° = V2. 



^H 



Y 




» 


/ 


/ 






A 






r/ 


/ 


y 






Xb' 






X 


O 


X 


Q 





Fig. 40 



x = y, 
cos 45° = sin 45°. 
J__V2 
V2~ 2 

= V2. 



cos45° = — = = — -■ 



sec 45° 

(II) If we construct an angle of 30°, we find that r = 2 y, 
as is easily seen by producing PQ its own length beyond OX 
(giving the point P') and joining 
OP', thus completing an equi- 
lateral triangle OP'P; whence 
0P=P'P=2y. 

But x^ — r^ — y\ 

Therefore oi? = 4:y^ — y^=3y^, 
and hence x = y • V3. 

Now the trigonometric func- 
tions of the angle 30° can be 
written down as follows : 




Pig. 41 



tan30°-^- y.- L = 

» yv3 V3 


4^. 


sin30° = ^ = ,2/ ^1, 
r 2y 2 




cos30°-:^-^^-^l 
r 2y 2 





TRIGONOMETRIC FUNCTIONS 



69 



The other three ratios can then be written down as reciprocals of 
these values respectively. 

The values of the trigonometric functions of 60° can be obtained 
in a similar way. The work is left as a problem for the student. 

54. Functions of 180° -ff etc. (I) A very simple relation 
exists between the trigonometric functions of an angle 6 and 
those of its supplement 180° — ^. Thus, let d be any angle XOA', 
and let ZXO^ = 180°-^. 
Denoting BA by y, OB by 
X, and OA by r, as usual, 
we have, from the defini- 
tions of the trigonometric 
functions, 

tan(180°-6l) = ^- (1) 

Now take A' on the ter- 
minal line of 6 so that 
0A!= OA, and draw A'B' perpendicular to OX. Then the triangles 
OAB and OA'B' are equal (being right triangles with the hypote- 
nuse and an acute angle of one equal respectively to the hypotenuse 
and an acute angle of the other), so that B'A' = BA, that is, y' = y. 
Further, B0 = OB'; and, since x=OB (not BO), therefore x = —x'; 
and r = r', since every radius vector is positive. Hence 




Fig. 42 



y_. 

X 



y 



- — tan 0. 



— X' 



Therefore, from (1), 

tan(180°-6l) = - 



tan 0. 



Again, 
that is, 
and 
that is, 



sin (180° -6')=^ = ^ 
r r 



sin (180° - 
cos,(180°- 
cos(180°- 



•^)=sin(9 
r 



sin ^ : 



• = — cos ; 



0) = - cos 0. 

In the figure, was an angle in the first quadrant; but the 
proof holds, word for word, if ^ is in any other quadrant. (If 



70 



THE ELEMENTARY FUNCTIONS 



is in the third quadrant, for example, 180° — ^ is constructed by 
rotating first through 180°, then through an amount equal to —6, 
which will yield a terminal 
line in the fourth quadrant, 
as Fig. 43 shows. The student 
should draw figures illustrat- 
iag the various possible values 
of e and 180° -0.) 

(II) An equally simple rela- 
tion exists between the trigono- 
metric functions of an angle 6 j-ig 43 
and those of the angle 90° -t- 6. 

Let e be any angle, and let Z XOA= 90° -|- 6. Then (Fig. 44) 




tan(90°-f-^) = ^ = — • 
^ ' X OB 



(1) 



Now take A' on the terminal line of 6 so that OA' = OA, and 
draw A'B' perpendicular to OX. Then the triangles OAB and 
OA'B' are equal (why?), and hence 
BA = OB', that is, y = x'; also 
BO=B'A', that is, x = -y' {iov 
BO = —x and B'A'=y). Hence 

^ = - cot 61. (2) 



y. 

X 



y 



Comparing (1) and (2), 

tan (90° -h 61) = -cot 6'. 
Similarly, 




Fig. 44 



sia(90°-h6l) = ^ = ^ = cos6l 



and 



cos(90° + 6')=' 



= — sin ^. 



Fig. 44 shows in this case also an angle in the first quad- 
rant, but, as in (I), the student should construct figures showing 
that the proof stiU holds in case 6 is in the second or any other 
quadrant. 



TRIGONOMETRIC FUNCTIONS 71 

EXERCISES 

1. Prove that sin (180° + 6) = - sin ^, cos (180° + «) = - cos 6, 
tan(18O° + e)=tan0. 

2 . Prove that tan (270° - «) = cot $, sin (270° - ^) = - cos 6 
eos(270°-e) = -sin^. 

3. Give the exact values of the following : sin 120°, tan 150°, 
sin (-120°),. tan (-150°), tan 225°, cos 240°, sin 300°, tan 300°, 
cos 330°, cos (-300°). 

4. Show that sin 150° + cos 240° = 0. 

5. Show that tan 60° + sin 240° + cos 150° = 0. 

6. Show that sin 150° • cos 60° + sin 60° ■ cos 150° = sin 210°. 

7. Show that cos 330° ■ cos 210° + sin 330° • sin 210° = cos 120° 

8. Show that -, f"^'''^!^" = tan 60°. 

1 + cos 120 

9. Show that 2 • sin 120° ■ cos 120° - sin 240° = 0. 

10. Show that sin 0° = 0, cos 0° = 1, tan 0° = 0. 

11. Show that sin 90° = 1, cos 90° = 0, cot 90° = 0. 

Applications of the Trigonometric Functions 

55. Having learned the meaning of the trigonometric functions, 
we now take up the question of their applications. These are very 
numerous, large parts of physics, surveying, and astronomy being 
indeed based entirely on the use of these functions. We shall here 
consider only a few of the simplest applications. 

56. Solution of the right triangle. To "solve" a triangle means 
to find the unknown parts (angles, sides, etc.) from sufficient data. 
But what are "sufficient data"? Elementary geometry answers 
this question for us by showing what combinations of sides and 
angles are sufficient to determine a triangle. In the case of the 
right triangle it is proved that any of the following combinations 

is sufficient: 

1. A leg and an acute angle. 

2. A leg and the hypotenuse. 

3. The hypotenuse and an acute angle. 

4. The two legs. 



72 THE ELEMENTAEY EUNCTIONS 

EXERCISES 

1. Construct accurately, witli rule and compass, a right triangle 
corresponding to each of the four problems indicated in § 57. 

2. State what would be "sufficient data" for constructing an 
isosceles triangle ; a scalene triangle not right-angled. Make con- 
structions for each case. 

57. Notation. The notation of Fig. 45 will be found con- 
venient and will be used throughout this section. The angles 
a, y8, and 7 are opposite the sides 
a, h, and c respectively, and 7 is the 
right angle. 

58. The way in which right tri- 
angles may be solved by the help of the 
trigonometric functions may be best 
explained by giving some examples. 

Example 1. In the right triangle ABC, given b = 150, a = 75°, to find 
the unknown parts. 

Solution. Since the value of the angle a is given, all of its trigonometric 
functions are determined, and can be computed after the manner of Ex. 2, 
p. 66 ; or their values may be read out of a printed " table of trigono- 
metric functions." From now on we shall use the latter method.^ We find 




tan 75° = 3.7321. But tan a = 7 > and hence we have 







tan 75° 


= 3.7321 


a 
150 






Therefore 


u 

jhe value of 


= 150 ■ 3.7321 = 

c : 


= 559.8. 




Next, to find 1 








cos a 


h 
= - = cos 
c 


75° = 


0.2588. 




Therefore 


150 
c 


= 0.2588, 


c = 


150 

.2588 


579.6 



(1) 



(2) 



Finally, as a check on the accuracy of these results, 

a 559.8 ...„ 

sin a = - = = .9658. 

c 579.6 

1 Any one of the printed tables on the market may be used. Four decimal 
places give ample accuracy for this work. 



TRIGONOMETRIC FUNCTIONS 73 

Referring to the table, we find sin 75° = .9659. Thus the results check. 
(A discrepancy of 1 or 2 in the fourth place may be expected when 
four-figure tables are used.) 

Another method of checking would be to use the Pythagorean Theorem, 
c^ = a^+b^; this also verifies the correctness of the results obtained above. 

59. We observe that the method of solving such problems 
consists in writing down the value of one of the trigonometric 
functions of the given acute angle, being careful to choose a ratio 
that contains the given side. Thus, in the above example we 

started with tan a = — rather than with sin a = - > because neither 
c 

a nor c was given. In case two sides are given, the modification 

of the method to be used is nearly self-evident and is illustrated 

by the following example: 

Example 2. In a right triangle ABC, given a = 15, c = 20, to solve the 
triangle. 

Since the two sides a and c are known, we write down a trigonometric 
function which contains both a and c : 

- = sin a. (1) 

c 

Therefore sin a = ^^ = .75. 

Using the tables, we find that the angle whose sine is .75 is 48° 35' ; hence 

a = 48° 35' . 

Therefore /8 = 41° 25' . 

To find 6, - = cos a = cos 48° 35' = .6615. 

c 

Therefore 6 = 20 • .6615 = 13.23 . 

Check. tana = J = ^ = 1.134. 

And as 1.134 is in fact the value of tan 48° 35', the above results are 
checked. 

60. In dealing with right triangles in practice, it will not always 
be convenient to turn the figure about so that the vertex of the 
acute angle we are using shall fall on the origin, and one side 



74 THE ELEMENTAEY FUNCTIONS 

along the X-axis. Thus, in Fig. 46, to write down the ratios tliat 

form the trigonometric functions of the angle a we should consider 

A to be the origin and ^C to be the X-axis, when of course AC 

would be the abscissa, and CB the ordinate, of the point B. Then 

CB a . CB a 

tan a = — = - > sm a = = - 1 etc. 

AC h AB c 

This is not altogether an easy process, however, especially for 

such an angle as /3 in Fig. 46, and we shall find it more practical 

to think of the definitions of the trigonometric functions in terms 

of the sides of the triangle themselves. 

We shall then regard a as " the side 

opposite the acute angle a," b as " the 

side adjacent to the angle a," and 

c as " the hypotenuse of the right 

triangle." Using these terms, we can ^^^ ^g 

easily restate the definitions of the 

trigonometric functions for an acute angle in a right triangle in 

such a way that we shall not need to think of an X- or I'-axis at 

all. Moreover, since the angles concerned are necessarily acute 

angles, we shall have positive values for all the trigonometric 

functions, and so may think of each side as undirected; that is, 

we may take its length as positive. Careful consideration wiU 

show that, in whatever position the angle may appear, the values 

of the trigonometric functions may be stated as follows: 

side opposite a _a 




tana = 



: — > 



side adjacent a b 

side opposite a a 

sm a = £i = - , 

hypotenuse c 

side adjacent a: b 

cos a = = - • 

hypotenuse c 

And the values of cot a, esc a, and sec a are the reciprocals of 
these three respectively. This way of thinking of the trigonometric 
functions wUl be found useful in problems that involve right 
triangles, but it must not be forgotten that it applies only to 
right triangles. 



TRIGONOMETRIC FUNCTIONS 



75 



EXERCISES 

1. Construct a right triangle with the sides 3, 4, and 5, and give 
the values of the six trigonometric functions of each acute angle. 

2. Solve the same problem, the given sides being 5, 12, and 13 ; 
8, 15, and 17. 

3. In Fig. 46, read oE the values of the functions of a and of jS. 

4. Compare the values of the functions of a with those of ^, and 
thus show that for any acute angle a the following relations hold : 

sin (90° — a) = cos a, 
cos (90° — a) = sin a, 
tan (90° — a) = cot a. 

5. Prove that the results of Ex. 4 are true for any angle a. 

6. In Fig. 47, ZACB = 90° and ZADC = 90°; show that the 
three right triangles formed are similar, and hence write down the 
trigonometric functions of the angles 
a and yS, each in three ways. 

„,, . h a a 

Thus, sin a = 7 = -^ = — ; 

' b a p + q 

In each of the following prob- 
lems, through Ex. 22, construct the 
triangle determined by the given 

data, and make an estimate of the values of the unknown parts; 
then compute these values and check the results arithmetically. 




7. a = 6 in., a = 30°. 

8.b = 75 ft., /3 = 15°. 

9. c = 1.3 ft., a = .9 ft. 

10. a = 1 ft., J = 2 ft. 

11. i = 2fin., c = 5in. 

12. a = 67°, a = 356 ft. 
13.^=88°, c = 110ft. 
14. a = 15.8 mi., b = 6.3 mi,. 



15. a = 72 ft., /3 = 25°. 

16. c = 10 in., a = 70°. 

17. c = 40 in., a = 6 in. 



18. a = 1 ft., 

19. a = 40°, 

20. p = 16°, 

21. a = 73°, 

22. « = 42°, 



c = 2 ft. 
a = 1 mi. 
b = 23.4 ft. 
b = 17.3 ft. 
= 3950 mi. 




76 THE ELEMENTAEY FUNCTIONS 

Each of the following problems depends for its solution upon 
the solution of a right triangle. In most cases merely drawing the 
figure will give the clue to the method to 
be employed. One technical term needs 
explanation; the "angle of elevation" of 
an object means the angle formed by the 
line of sight to the object, and the hori- 
zontal line, — the angle CBA in Eig. 48. Fig. 48 

23. Determine the height of a tree if its angle of elevation, from 
a point 200 ft. away, is 20°- 

24. A standpipe 100 ft. high stands on the bank of a river. Erom 
a point directly opposite, on the other bank, the angle of elevation 
of the standpipe is 29°. How wide is the river there ? 

25. A rectangle is 40 in. x 17 in. What angle does the diagonal 
make with the longer side ? 

26. In a circle of radius 5 in., how long is a chord that subtends 
an angle of 20° at the center ? 

27. How long is the chord that subtends an angle of 1° at the 
center of a circle of radius 100 ft. ? 

28. Eind the side and the area of a regular nonagon (9-sided 
polygon) inscribed in a circle of radius 16 ft. ; circumscribed about 
the same circle. 

29. Solve the same problem for the regular dodecagon (12-sided 
polygon). 

30. Solve the same problem for the regular 15-sided polygon. 

31. Eind the radius of the fortieth parallel of latitude; of the 
eighty-fifth. (Assume the earth a sphere of radius 3960 mi.) 

32. Eind the length of the perimeter of a regular inscribed poly- 
gon of 24 sides when the diameter of the circle equals 1. Solve the 
same problem for the regular circumscribed polygon of 24 sides. 

33. Solve the problem of Ex. 32 for the regular inscribed and 
circumscribed polygons of 48 sides. 

34. Solve the problems of Exs. 32, 33 for the regular inscribed 
and circumscribed polygons of 96 sides (given sin 1^° = .032719 and 
tan 1|° = .032737). 

35. Use the results of Ex. 34 to determine an approximate value 
of TT. Ans. IT is between 3.1410 and 3.1428. 




TRIGONOMETRIC FUNCTIONS 77 

61. Velocities and forces. A second useful application of the 
trigonometric functions is to physical problems involving velocities 
or forces. These are indeed, as we shall see, merely special cases of 
the solution of right triangles, but as they involve certain difficul- 
ties of their own, it is better to consider them in a separate section. 

62. Suppose a body is moving with known velocity v in the 
direction AB, making an angle 6 with the northerly direction, as 
in Fig. 49. Then the problem arises, How fast is the body moving 
toward the east ? If AB represents the distance traveled in a unit 
of time, then AB = v. If A]^ repre- 
sents the direction north from A, 
the angle NAB will equal the given 
angle d. Drawing BC perpendicular 
to AN, we have a right triangle ABC 
in which CB = x represents the dis- 
tance the body has traveled toward 
the east in unit time ; that is, x is the '**' 
numerical value of the required velocity toward the east. Hence 
the solution of the problem is the same thing as the solution of 
the right triangle ABC. The side AC = y will give the numerical 
value of the velocity toward the north, x and y are called the 
eastern and northern components of the given velocity v; v ia called 
the resultant of the velocities x and y. Note that 3? + y^ = v^. 

Directions are usually given with reference to north or south as 
standards ; thus, the direction 20° east of north is written N. 20° E. 
(read " North, 20° east "), and the direction 20° south of east, 
S. 70° E. (not E. 20° S.). 

EXERCISES 

1. A train is running at a speed of 40 mi. per hour in a direction 
10° north of east (N. 80° E.). How fast is it moving eastward, and 
how fast northward ? 

2. A man walks, at the rate of 3^ mi. per hour in the direction 
S. 36° E. How far south has he gone in 3 hr.? 

3. A boat is steaming in a direction N. 70° E. at the rate of 20 
knots per hour. What are the eastern and northern components of 
this velocity ? 



78 THE ELEMENTAEY FUNCTIONS 

4. A point moves in a vertical plane at the rate of 20 ft. per 
second in a direction inclined 53° with the horizontal. Find the 
horizontal and vertical components of this velocity. 

5. The horizontal and vertical components of a velocity are re- 
spectively 30 ft. per second and 40 ft. per second. Find the resultant 
velocity and its direction. 

6. A balloon is rising vertically at the rate of 660 ft. per minute 
and encounters a wind blowing horizontally at the rate of 15 mi. per 
hour. In what direction will the balloon continue to rise and with 
what velocity ? 

7. From the platform of a train going at the rate of 40 mi. per 
hour a boy throws a stone in the direction at right angles to the 
train's motion with a velocity of 50 ft. per second. In what direction 
will the stone go, and how fast ? 

8. A projectile from an 8-in. gun on a warship has a velocity 
of 2000 ft. per second, and the ship is moving 22 knots per hour 
(1 knot = 6080 ft.). If the gun is fired in a direction perpendic- 
ular to the ship's motion, in what direction will the projectile 
actually go ? 

9. A man rows at the rate of 4 mi. per hour, and the rate of the 
current in a river is 3 mi. per hour. If he starts to row straight 
across at a point where the river is 350 ft. wide, how far down will 
he reach the other bank ? 

10. If in Ex. 9 the man wishes to land directly opposite his 
starting-point, in what direction must he row ? 

11. A hunter is traveling straight north in an auto at the rate of 
20 mi. per hour, when he notices a rabbit in a field about 100 ft. 
away. He fires when he is due east of the rabbit. If the velocity of 
the shot is 1000 ft. per second, in what direction must he aim if he 
is not to miss ? 

63. Problems involving component forces are identical mathe- 
matically with these problems in velocities. For example, if two 
forces, one of 40 lb. and the other of 30 lb., act at right angles 
to each other upon a point P, the effect is equivalent to that 
of a single force F acting upon the point P in the direction 
PQ, the diagonal of the rectangle PAQB, and with an intensity 



TRIGONOMETEIC FUNCTIONS 



79 



numericaUy equal to the length of PQ. F is called the resultant 
of the component forces PA and PB. Here, evidently, F=50 lb., 
and by solving the right triangle PAQ 
we find e = 36° 52'. 

This relation between the component 
and resultant forces is familiar to students 
of physics under the name " parallelogram 
of forces." It may be stated as follows : 
If a point P is acted upon by two forces, 
represented ia magnitude and direction 
by PA and PB, then the diagonal PQ of the parallelogram 
PAQB represents, in magnitude and direction, the resultant force. 




Fig. 50 



EXERCISES 

1. Two forces, of 45 lb. and 75 lb., act at right angles to each other 
upon a point. Find the direction and intensity of the resultant force. 

2. The resultant of two forces at right angles to each other is 
100 lb., and it makes an angle of 30° with the horizontal force. 
Find the horizontal and vertical 
components. 

3. A weight of 250 lb. lies on 
an inclined plane whose angle 
is 20°. With what force does 
it press against the side of 
the plane, and with what force 
does it tend to slide down the 
plane ? (In Fig. 61, ii WW rep- 
resents 250 lb., then WR and WS represent the required forces.) 

4. Two forces, one of 100 lb. and the other of 75 lb., act on a 
body, one force pulling N. 20°E., the other N. 40° E. Find the 
residtant force. 

Hint. Find the eastern and nortjiern components of each force. 

64. Slope of a straight line. Another important application 
of the trigonometric functions is in the study of the linear equa- 
tion in ,« and y. We saw on page 29 that the graph of such an 
equation is a straight line, although we have not yet proved this to 




Fig. 51 



80 



THE ELEMENTARY FUNCTIONS 




Fig. 52 



be true. The question we shall now consider is the determination 

of the angle Q which a straight line makes with the X-axis. (By 

"the angle which one line (1) 

makes with another line (2)" we 

mean the positive angle through 

which (2) must rotate to come 

into coincidence with (1). Avoid 

the expression, "the angle he- 

tween (1) and (2)," because that 

is ambiguous, there being always two angles between any two 

intersecting lines, unless the lines are perpendicular.) 

65. Now, to find the angle which a given straight line^ makes 
with the X-axis, take any two points F and Q on that Hne, and 
suppose their coordinates to be 
(x^, 2/i) and (x^, y^. Draw PM 
and QN parallel to the F-axis, 
and PB parallel to the X-axis, 
thus obtaining the right tri- 
angle PBQ, in which the angle 
BPQ = 6. Using this triangle, 

tan0 = ^ 
PB 

and BQ=NQ-NB-. 




Fig. 53 



■■ y% - Vv 



while 
that is. 



PB = MN= ON- 0M= x^-x^; 



tan^ = ^?-^l- 



(A) 



This simple result is of great importance, as we shall see. The 
truth of the equation (A) is not dependent upon which particular 
points P and Q we choose, on the line ABPQ. Thus, if we choose 
P and Q as in Fig. 54, the reasoning is as follows : 

In the triangle PBQ, tan = ^ 
^ PB 

and BQ=:NQ-NB = y^-y^, 

while PB = MO+ON = -x^ + x^ = :r^-x^, 



1 The given line is assumed not to be parallel to either coordinate axis. 



TRIGONOMETRIC FUNCTIONS 



81 



the only difference being that PR is the sum of the absolute lengths 
of MO and ON, but x^, the abscissa of P, is not MO, but OM'^ ; 



that is. 


MO 


= - 


Therefore 




BQ 


Vi- 


Vi 


PR 


x^- 


^1 




Fig. 54 



SO that formula (A) is 
true in this case also. 
The student should ex- 
periment further with 
points P and Q in various other positions on the line, and satisfy 
himself that in every case formula (A) is true. 

When the angle 9 is obtuse, the same result will again be 
found to hold. Thus, in Tig. 55, let P and Q be any two points on 
the given straight line, 
and let their coordi- 
nates be (a;^, y-^ and 
{x^, 2/2) respectively. 
Draw perpendiculars 
to the JT-axis from P 
and Q, and the perpen- 
dicular to the F-axis 
from Q, forming the triangle PRQ. Eegarding Q as the vertex of 
the obtuse angle 6, the definition of tan 6 is 



R 


P ^ 


B 






Q 




R 


I 


J 


/ vl---^..^^^ 



Pig. 55 



Now 
and 



RP I RP' 

tan 6 = ( observe that it is not — — ■ 

QB\ RQ 

RP=MP--MR = y^- 2/2, 

QR = NO + 0M= —x^ + x^ (since NO = — x^). 



Therefore tan. 6 = ^ — ^ . which equals ^ — ^ ■ 
x^ — ^2 *2 ^ *i 

Thus, formula (A) is correct in this case also. 

If the line is parallel to the X-axis, we shall say 0=0, and 

formula (A) stUl holds true. 

1 A very common mistake in work of this kind is to write OM = - x^, on 
account of the abscissa of P (or of M) being negative. By de/imtion the abscissa 
is OM, and hence MO = — Xj. 



82 THE ELEMENTARY FUNCTIONS 

In work of this kind it will be noticed that the all-important 
thing is to keep in mind the fact that every segment parallel to 
the X- or the Y-axis is a directed segment, and that abscissas are 
always measured from ilie Y-axis, ordinates from the X-axis. 

66. The number tan 6 is called the slope of the line PQ. 
In words, the slope of a straight line is the tangent of the angle 
which it makes with the X-axis. We shall hereafter usually 
designate the slope by m. Formula (A) is then 

1/ — y 
tan ff = m = slope of PQ — _ , 

or, in words, the slope of the straight line through two points is 
the difference of their ordinates divided by the difference of their 
abscissas (both differences being taken in the same order). 



EXERCISES 

Draw careful figures in all cases. 

1. Find the slope of the line through the points (2, 3) and (5, 6); 
through the points (3, 1) and (— 3, — 1). 

2. Find the slope of the line through the origin and the point 
(4, 3); through the points (- 2, - 3) and (1, 0). 

3. Find the slope of the line through (2, 0) and (0, — 3); through 
(a, 0) and (0, b). 

4. Write down the values of sin 6 and cos in each of the 
Exs. 1-3 above. 

5. What is the slope of the line x — y = 2? (In drawing the 
graph you necessarily get two points on the line; hence its slope 
can be found.) 

6. What is the slope of the line 2x — 3y = 5? of the line 
a! + 22/ = 3? 

7. What is the slope of the line x — ni/ = 5? of the line 
2/ = mx + k? ot the line ax + by + c — 0? 

8. Find the angle which the line ix-i-Gy = 7 makes with 
the X-axis. 



TRIGONOMETRIC FUNCTIONS 83 

9. Prove that if two straight lines are perpendicular to each 
other, and if the slope of one is J, the slope of the other is — |. 

10. Prove that if two straight lines are perpendicular to each 

other, and if the slope of one is m, that of the other is 



REVIEW PROBLEMS ON THE TRIGONOMETRIC FUNCTIONS 

1 . Prove that tan + cot 6 = sec $ • esc 0, being any angle. 

1 



Solution. By definition, cot = 



Therefore tan 5 + cot 6 = tan 6 + 



tand 

1 tan^e + l 



tan tan ■, 



sec^'fl „ sec0 . cos 5 

= - — ^ = sec e • - — ^ = sec • -r-^ 

tan 6 tan sin tr 

I cos 5 

= sec 6 • y: = sec ^ • CSC 0. Q.E.D. 

sin 6 
In problems 2-14, and a represent any angle. Prove : 
2. sec fl — cos = sin ^ • tan 0. 

1 — sin cos 

3 • = • 

cos fl 1 + sin 

4. sin^e — cos''^ = sin''^ — cos^fl. 

5. (sin^a — cos'^a)^ =1 — 4 sin'^a eos'^a. 

6 ^ + ^ =1. 

1 + tan^a 1 + cot^a 

7. (sin a + cos a)^ + (sin a — cos af = 2. 
sec e — esc 5 tan ^ — 1 

Q ^ • 

■ sec ^ + CSC tan e + 1 
9. (cot e + 2) (2 cot ^ + 1) = 2 csc''^ + 6 cot 0. 
tan ^ — sin g _ secg 
^®" sin»e ~l + cos(9' 
tan a — cot a _ 2 _ . 
' tan a + cot a csc^a 

12. sec'^e csc^'e - 2 = tan^e + cotf^^. 

13. cot e - sec esc 6(1 - 2 sin'^^) = tan 6. 

14. (3 sin a - 4 sin'a)'' + (4 cos'a - 3 cos af = 1. 



84 



THE ELEMENTARY FUNCTIONS 



In each of the following problems, a and (3 are the acute 
angles of the right triangle ABC, and a, b, and c the sides. 

15. Prove that tan - = ^ 

Z a 

Solution. Bisect ^a and draw the perpen- 
dicular from B upon this bisector, producing it 
to meet AC (produced) at F. Then ABCF is 
similar to AAEF, and 



ta,n^ = ta,nZFBC = — = - — - ■ 
2 EC a 

Prove : 

16. sin 2 a; = cos (/8— a). 

17. cos 2a = -^^ 'hr -• 



Q.E.D. 




{P>a) 



18. tan 2 a = 



0' 

2 ah 



s 2 a6 
19. cos(j8 — a)= — ;^- 



(b + a)(b — a) 



20. tan- 



« _ io — b 
2~yc + b 



+ b 
4y=12 makes with 



21. Find the angle which the line 3x 
the A'-axis. 

22. Find the angle which the line x + i/ = 3 makes with the line 
3 X — 1/ = 5. Find also the coordinates of the point of intersection 
of the two lines. 

23. The equations of the sides of a triangle are 2x — 3y = 5, 
X + 5y = 9, and 3 a; + 2 // = 1 ; find (a) the coordinates of the 
vertices, and (b) the angles of the triangle. 



CHAPTER V 

SOME SIMPLE FRACTIONAL AND IRRATIONAL FUNCTIONS; 
THE LOCUS PROBLEM 

67. Review questions. What is the meaniag of the statement 
" y is a function of a; " ? What is meant by the " graph of a 
function " ? Classify functions according to degree. What is the 
nature of the graph of a function of the first degree ? of the 
second degree ? How can a quadratic equation be solved graphi- 
cally ? How can the nature of the roots of a quadratic equation 
be determined without solving it? Define the six trigonometric 
functions of an angle. State the most important relations among 
these functions. Define "slope of a straight line." When wiU a 
line have a negative slope ? 

68. Rational and irrational functions. The functions whose 
graphs we have constructed (Exercises, p. 22) involved only the 
operations of addition, subtraction, and multiplication, so far as the 
independent variable x was concerned. Such functions are caUed 
integral rational functions. The most general form of such a func- 
tion of X is /(«)= aga;"-f aia;"~i-Fa2*"~^+ • • • +*«-i* + *»- 
As in Exs. 10 and 13, p. 22, the coefficients of certain terms 
may contain fractions, but the function is still called integral. 
Thus, the quadratic function y = ax^ + bx + c is an integral 
rational function, whatever may be the values of a, b, and c. 

If the independent variable occurs in the denominator of a frac- 
tion in its lowest terms (that is, if the operation of division has to 
be performed with the independent variable in the divisor), then the 
function is called a fractional rational function, — for example, 

2 B + X X^ r> ii, 4-1, I, ^ * J ^^ + ^ 

y = -, y = — ■ — , y = s • t)n the other hand, - and — - — 

are integral rational functions of x. 

85 



86 



THE ELEMENTARY FUNCTIONS 



Either of these two kinds of function is called a rational function, 
so that a rational function may be defined as one that involves any 
or aU of the four fundamental operations, — addition, subtraction, 
multiplication, and division. 

If the functional relation is such that a root must be extracted 
in order to arrive at the value of the function, it is said to be 
an irrational function, — for example, y = v a;, y = v 2 x^ + 1, 

y 

69 



■ for example, y =^ ■ 

' = -vl , y = Vuj^ — 5 X, y = \x + 'Vx. 

Graphs of rational functions. 

Example 1. Draw the graph of the function - 

1 ^ 

sponding values of x and y, when »/ = - > is as follows : 



The table of corre- 



X 


1 


2 


3 


-^ i 


i 


-1 


- 2 


-3 


-i 


y=f(x) 


1 


i 


i 


-i 


2 


3 


-1 


_ 1 

■J 


-i 


-2 



etc. 



The value x = 0, it should be observed, gives no value of y at all, since 
division by zero is impossible. This means that the graph does not 
meet the I'-axis, because, for points 
on the y-axia, ^ = 0. But for very 
small positive values of x, y is very 
large, so that the graph is very far 
above the ^-axis when near the 
y-axis in the first quadrant. For 
negative values of x near to the 
value of y is very large numerically, 
but negative, so that the graph 
is very far helow the A'-axis when 
near the F-axis in the third quad- 
rant. Thus the curve is separated 
into two parts, or " branches," each 
of which approaches the F-axis 
more and more closely as the 
numerical value of y increases.^ 



Fig. 57 



When a curve continually approaches a straight line in such a way that 
as either x or y increases without limit the distance between the curve and 

1 What was just stated was'of course the converse of this, namely, that when 
X decreases, y increases; but the converse statement is equally obvious from 

the equation y =-• 

X 



FRACTIONAL AND IRRATIOKAL FUNCTIONS 87 



the line eventually becomes and remains less than any assignable value, the 
line is called an asymptote to the curve. Thus, the I'-axis is an asymptote 

to the graph of the function - ■ It can easily be seen by writing the equa- 

1 ^ 

tion in the form a; = - that the A'-axis is also an asymptote to this curve, 

since, as x increases without limit, y decreases and becomes less than any 

assignable value. The curve is called a hyperbola. 

a: + 1 

Example 2. Draw the graph of the function 

X —% 

X + 1 

The table of corresponding values of x and y, when y = , is as follows : 

X — 2 



X 





1 


2 


3 


4 


5 


6 


-1 


-2 


-3 


-4 


1 


3 


V 


-\ 


-2 


No value 


4 


S 


2 


I 





i 


f 


i 


-5 


7 



Here the value x = gives a definite value for y (jj = ^), but x = 2 makes 
the denominator zero and hence gives no value of y. As in the case of the 
I'-axis in the preceding example, 
so here the line x = 2 is easily seen 
to be an asymptote to the graph, 
because, for very large values of 
y, X is very near the value 2. 
There is also another asymptote, 
which can be found by letting x in- 
crease without limit. To do this it 
will be found convenient to change 
the form of the fraction as fol- 
lows : divide both numerator and 
denominator by x, thus obtaining 

1 

1 -I- 

X -I- 1 



H-- 



Now as X in- 




--2 i_2 

X 12 •^'''- ^^ 

creases without limit, both - and - 

1 X 

diminish and become eventually less than any assignable value. Thus, 

1 + i 

^ + 1 i approaches 1, that is, y approaches 1. This means that y = \ 

x-2 i_2 

X 

is an asymptote. This curve is also a hyperbola. 

(The fact that ?/ = 1 is an asymptote could also have been discovered by 

solving the equation y = ^^^ for x as a function of y, which gives x = -j--^ , 

and the form of this fraction shows that !/ = 1 is an asymptote to the curve.) 



88 



THE ELEMENTARY FUNCTIONS 



EXERCISES 



Draw the graphs of the following, showing the asymptotes if 
there are any : 



1. xy = 4. 

2. xij = — 10. 

1 

3. y = 

4. 2/ = 

5. y = 

6. y = 

'•y-^x-1 



X 


-4 




1 


X 


+ 4 


X 


+ 3 


X 


+ 5 


X 


-2 


X 


-1 


2 


x-1 



8. y = 

9. y = 

10. y = 

11. y = 

12. 2/ = 

13. y = 



2x + S 
3 X — 6 

— X — 1 

2^- + 5 

a; 



1-a; 

3j-4 
5-2x' 

7 « — 1 



a;-l 




20. Prove that the graph of the function y 

of , a 

asymptotes x = — - and. y = -• 

C G 



ax + b 



-d 



x' + X - 

will have as 



70. Irrational functions. We shall consider only such irra- 
tional functions as involve square roots. Several examples of 
irrational functions will be given, in order to bring out all the 
details that must be attended to in the study of this class of 
functions. 



Example 1. /(x)=±Vl + : 



Write, as usual, y =f(x), that is,y=± Vl + x. For any value of x that 
is < — 1, y will be complex ; hence no point to the left of x = — 1 can be 
found upon the graph. For x =—1, y = 0; and for any value of x which 
is > — 1, y will have two values that are equal numerically but of opposite 
sign, as in the following table : 



etc. 



Plotting these points and joining them by a smooth curve, we have the 
graph of the function (Fig. 59). The curve is symmetrical with respect to 



X 


-1 





1 


2 


3 


4 


5 


6 


7 


8 


y 





±1 


±V2 


±V3 


±2 


±V5 


±V6 


±v^ 


±V8 


±3 



FRACTIONAL AND IRRATIONAL FUNCTIONS 89 



the X-axis because of the fact just noted, that for every vahie of a; (> — 1) 
y has two values which are numerically equal but of opposite sign, thus 
giving two points symmetrically 
located with regard to the X-axis. 
The curve is a parabola. 

The functional relation of this 
example can be written in the 
form y^ = \-\- X, or y^ — x = 1, or 
y^ — x — \ = (i. In any of these 
forms y is said to be an implicit 
function of x, because the func- 
tional relation is definitely im- 
plied by the equation ; when the equation is solved for y, y is said to be 
an explicit function of x. 

Example 2. f{x) ■ 




Fig. 59 



±vr 



Writing y =f(x), we see that y is complex if x^ > 1, that is, if x > -1- 1 
or < — 1 . Hence the only values of x that will give points on the graph are 
those between — 1 and -I- 1 (inclusive). The table of values is as follows : 



X 


-1 





i 


1 


i 


1 


(any negative value > — 1) 


y 





±1 


±Vi| 


± VI 


iVxV 





(same as for corresponding -|- value) 



Plotting these points and joining them by a smooth curve, we find that 
the graph seems to be a circle (Fig. 60). This is in fact the case, as wiU 
be shown later, y is given as an 
implicit function of x by the equa- 
tion y^ = 1 — x^, or x"^ + y^ = 1. 



Example 3. f{x) = ± V4 - 4 x\ 

If, as usual, f(x) is represented 
by y, then the equation can be 
written in the implicit form 
4 a;2 -I- / = 4. 

Within what limits must x lie in 
order that y may have real values? 
Having answered this question, the 
table of corresponding values of x 
and y should be drawn up, as follows : 




Fig. 60 



X 





±v 


±i 


±1 


±1 


y 


±2 


± Vi 


±V^ 


±vi 






90 THE ELEMENTARY FUNCTIONS 

Joining the points by a smooth curve, we have the graph of this 
function as in Fig. 61. It is a closed curve, symmetrical with respect 
to each of the coordinate axes. The curve is called an ellipse. 




Fig. 61 



Fig. 62 



Example 4. f{x) = ± Vx^ — 1, or x^ — y^ = 1, when written in the im- 
plicit form. What are the limitations on the values of x here ? The table 
of values is as follows : 



X 


±1 


±2 


±3 


±4 


±9 


V 





±V3 


±V8 


±VT5 


± VI 



The graph of this function consists of two parts, symmetrically located 
with respect to the F-axis, each part being symmetrical with respect to 
the X-axis. The curve is a hyperbola ; its similarity in form to the graph 

of y = - is evident, the only difference being in the position of the curve. 
A characteristic feature of the hyperbola is the presence of two asymp- 
totes. In the case of ^ = - they were the A^-axis and the F-axis, while here 

X 

they are the lines through the origin with slopes -f 1 and — 1 respectively. 
The distances from points on the curve to either line become and remain 
less than any positive distance that can be mentioned, as the values of x 
or y increase without limit. This assertion will be proved later (see p. 134). 
Fig. 62 shows the asymptotes as well as the curve. 



FEACTIONAL AND IREATIONAL FUNCTIONS 91 

EXERCISES 

Make a careful graph of each of the functions of x given by the 
following equations, and state in every case for what values of x, 
if any, the function fails to have a value. If the graph is a hyper- 
bola, draw the asymptotes, as near as you can get them. 

l.x' + 3y' = 12. S. 2x''+5y^=5. 15. 23^- 3y'+5 = 0. 

2. ar" - 2 2/2 = 4. 9. f + x = i. 16. 4x2- 9 j^'' = 36. 

3.x^ + 4:f = i. 10. 4y2_a; = 8. IT. ix" + 9t/ = 36. 

i.i/-ix = 4:. 11. 7a:2 + y2=28. 18. 4a! - 9/ = 36. 

5. x^ + y^ = 9. 12. 4y_a;2 = 4. 19. 4a; + 9?/* = 36. 

&.3x^+f=6. 13. 4 a;" -2/2 = 4. 20. 4 x" - 9 y'' ^ - 36. 

7. f -0^ = 1. 14. x^ + 3y''=5. 21. 2x' + 2y^ = 11. 

The Locus Peoblem 

71. One of the most important uses of the graphical repre- 
sentation of functions is in the study of geometric loci. The word 
"locus" has already been used (p. 19), but before going farther it 
will be well to review carefully a few examples which will aid 
in making the precise meaning of the word clear. 

1. The locus of points equidistant from two fixed points is the 
perpendicular bisector of the line segment joining the two points. 

2. The locus of points equidistant from two fixed intersecting 
straight lines is the bisectors of the angles formed by the lines. 

3. The locus of points at a constant distance from a fixed 
point is a circle about the fixed point as center, and with the 
constant distance as radius. 

72. Definition of locus. As these illustrations remind us, the 
locus of points that satisfy a certain condition means the totality 
of points satisfying that condition. The locus must, first, contain 
aU the points that satisfy the condition, and, secondly, must not 
contain any point that fails to satisfy the condition. No statement 
about a " locus of points '' can be justified unless it can be shown 
that both these things are true of the alleged locus. Each of the 
three examples given above should be carefully tested to see if 



92 



THE ELEMENTARY FUNCTIONS 



it fulfills both requirements of a locus. Notice that 2 is not true 
if the italicized word bisectors be changed to bisector. Notice also 
that 1, 2, and 3 are not true if we consider points outside of the 
plane. What is the corresponding locus in space in each case ? 

73. It was stated above that the graphic representation can 
be used in solving problems concerning loci. This fact has been 
partly brought out in the work at the beginning of Chap. II, 
p. 19, and those examples wiU now be considered again. 

Example 1. The locus of points at the distance 2 from the A'-axis is 
evidently the straight line parallel to the A'-axis and 2 units distant from 
it, because, first, all points on the line ABC (Fig. 15, p. 19) are at the 
distance 2 from the A'-axis, and, secondly, every point that is at the dis- 
tance 2 from the A'-axis lies on this line, since a point above the line ABC 
will be at a distance greater than 2 from the A'-axis and a, point below 
the line ABCynW be at a distance less than 2 from the A'-axis. (Of course, 
if we were not dealing with directed distances, the locus would consist of 
two straight lines, one above and one below the A'-axis ; but the latter is the 
locus of points whose distance from the A'-axis is — 2.) 

Example 2. The locus of points that are twice as far from the A'-axis as 
from the F-axis is a straight line through the origin, with slope 2. 

The result of this problem was stated on page 20, and indeed it is nearly 
self-evident that this locus is a straight line ; but it is important even in 
these simple cases to make sure that the 
essential nature of the locus problem is 
not lost sight of. To prove that the line 
P'OP (Fig. 16, p. 20) is the locus of points 
that are twice as far from the A'-axis as 
from the F-axis, it is necessary to show, 
first, that every point on this line satisfies 
the condition, and, secondly, that every 
point that satisfies the condition is on the 
line. First, if P is any point on the line, 
and if OM is the abscissa and MP the 

MP 

OM 
hypothesis ; that is, MP = 2 OM, which 
means that P is in fact twice as far 
from the A'-axis as from the I'^axis. Secondly, let P' (Fig. 63) be any 
point that is twice as far from the A-axis as from the F-axis. Then 
M'P' = 2 OM' ; hence triangle OM'P' is similar to triangle OMP ; and 



ordinate of P, then 



: tan 9 = 2, by 




M' M 



Fig. 63 



THE LOCUS PROBLEM 



93 




therefore /.M'OP' = ZMOP. Therefore I" is on the line OP. This 
completes the proof of the theorem as stated. 

It was seen on page 20 that the equation of this locus is !/ = 2 z ; hence 
we have now proved that the graph of the equation y = 2 a; is the straight 
line through the origin with slope 2. 

Example 3. What is the locus of 
points at a constant distance of 3 
units from a fixed point? 

Let the fixed point be chosen as 
origin, and let (x, y) be the coordi- 
nates of any point P on the locus. 
Then the geometric condition of the 
problem is Qp _ g ^-i -, 

Now, by the Pythagorean Theorem, 

OP=V^^Tf. (2) 

Hence Vx^ + i/ = 3. (3) 

This algebraic condition is thus p,„ qa 

equivalent to the geometric condi- 
tion of the problem, so that the graph of equation (8) is the locus required. 
(In this case the locus is a circle, from the very definition of that curve. 
The same reasoning proves that the graph of Example 2, p. 89 (Fig. 60), 
was a circle, as was there asserted. The student should state the locus 
problem corresponding to that example.) 

74. These examples illustrate the process of obtaining equations 
corresponding to geometric locus problems. This process may be 
summarized thus : First, represent by the variables (x, y) the coor- 
dinates of any point on the required locus (the X- and F-axes 
being chosen in any convenient position). Secondly, translate the 
geometric condition of the problem into an algebraic condition 
or equation involving x and y, or either one of them alone. 
The graph of this equation will then be the locus required. The 
equation is called " the equation of the locus." 

To be sure, no advantage of this process is evident in cases 
like those just given, where the locus was easily obtained by 
the apphcation of well-known theorems of elementary geometry. 
But in many cases it is difficult, perhaps even impossible, to gain 
definite knowledge of the locus directly ; whereas the equation of 
the locus can be obtained, and then the graph of this equation. 



94 



THE ELEMENTARY FUNCTIONS 



found by plotting points whose coordinates satisfy the equation, 
will be the desired locus. In later chapters we shall see that 
many important facts about a locus can often be established by 
drawing conclusions from the equation of the locus. For the 
present, however, the derivation of the equation is the main thing. 



EXERCISES 

Draw a figure for each. 

1. Find the equation of the locus of points at a distance 2 from the 
F-axis ; from the .Y-axis. 

2. Find the equation of the locus of points at a distance — 5 from 
the -Y-axis. 

3. What is the equation of the -Y-axis ? of the F-axis ? 

4. Find the equation of the locus of points that are 3 times as far 
from the F-axis as from the A'-axis. What is the loous ? Prove the fact. 

5. Find the equation of the locus of points that are — 2 times as far 
from the -Y-axis as from the F-axis. What is the locus ? Give proof. 

6. Find the equation of the locus of points at a distance c from 
the F-axis. 

7. Find the equation of- the locus of points that are at a constant 
distance equal to 6 from the origin. 

8. Find the equation of the locus of a point that moves so as to 
be always equidistant from the points (2, 1) and (3, 5). 

Solution. Let A = (2, 1) and B = (3, 5) ; also let P= (x, y) be any point on 
the required locus. Then the condition of the problem is PA = PB. Each 
of these distances can be expressed algebraically by means of the result of 
Ex. 20, p. 12, which proved that the 
distance between two points (xj, j/j) 
and (xj, 2/2) is 

Therefore 



PA =^(x-2y-\-{y-\y 

and PB=^(x-Zf+{y-by. 

Since PA = PB, the correspond- 
ing algebraic equation is 



y 


\ 




~~-~-\p(^,y) 




A 

X 









Fig. 65 



V(x-2)2+(y-l)2 = V(x-3)■^-t-(2/-5)^ 
or, simplifying, 2 x -t- 8 y = 29. 



THE LOCUS PROBLEM 



95 



This is accordingly the equation of the locus. The graph is found to be 
a |traight line, which, as we know, is the perpendicular bisector of the line 
segment AB. (Of course the mid-point of the segment AB must lie on the 
locus ; its coordinates are (—- — > —~\, that is, (5 ' 3 ) , and these coordi- 
nates must therefore satisfy the equation of the locus, 2x + Sy = 29; in 
fact, 2 • § -I- 8 • 3 = 5 -I- 24 = 29, checking partially the accuracy of the 
result obtained.) 

9. Find the equation of the locus of a point that moves so as 
to be equally distant from the points (2, 3) and (6, 1). 

10. Find the equation of the locus of points equidistant 

(a) from the points (1, 3) and (— 1, 4); 

(b) from the points (3, 2) and (— 5, — 1) ; 

(c) from the points (— 1, — 2) and (3, — 5) ; 

(d) from the points (2, 2) and (0, 0) ; 

(e) from the points (2, 0) and (0, 2); 

(f) from the points (0, — 4) and (— 6, 0); 

(g) from the points (a, 0) and (0, b). 

11. A point moves so that its distance from the origin is con- 
stantly equal to 10. Find the equation of its locus. 

12. A point moves so that its distance from the point (2, 4) is 
always equal to 6. Find the equation of its locus. 

13. Find the equation of the path of a point that moves so as 
to be equally distant from the points (— 2, — 3) and (2, — 5). 

14. A point moves so thatits distance from the point (2, 0) is always 
equal to its distance from the F-axis. Find the equation of its locus. 

Solution. Let P = (x, ?/) be any position of the moving point (that is, any 
point on the required locus). Then the condition of the problem is AP=BP 
(Fig. 66). The algebraic equivalents of 
these geometrical quantities are 

AP=V(x-2y+f 
and BP = x. 

Hence the required equation is 
y/(x - 2)2 + y'^ = x. 
That is, x'^-i:X + i:-\-y'^= x\ 
or ^2 = 4 1 — 4, 

„ _ y' + 4 





Y 

P(x.v) 




/ X 





A (2,0) 

Fig. 66 



96 THE ELEMENTARY FUNCTIONS 

This is therefore the equation of the required locus. Here we have an 
illustration of a problem whose solution could not be obtained by using 
any of the theorems of. elementary geometry merely, but by drawing a 
careful graph of the equation of the locus we can gain a very good idea of 
the appearance of the locus itself. It will be found to be a parabola. 

15. A point moves so that its distance from the X-axis is always 
equal to its distance from the point (0, 4). Find the equation of its locus. 

16. Find the equation of the locus of a point which moves so that 
its distance from the point ( 6, 1) is always equal to its distance from 
the line x = 2. 

In each of the following problems, find the equation of the locus 
of a point that moves according to the condition given, and draw 
the graph of the equation : 

17. Its distance from the point (— 2, 5) is always equal to its 
distance from the line y = — 3. 

18. Its distance from the point (4, 0) is always twice its distance 
from the line x — 2. 

19. Its distance from the point (5, 3) is always one half its 
distance from the line y = 1. 

20. Its distance from the point (—1, 4) is always one half its 
distance from the F-axis. 

21. Its distance from the point (3, 0) is always two thirds its 
distance from the line a; = 4. 

22. The sum of its distances from the points (2, 0) and (— 2, 0) 
equals 6. 

23. The difference of its distances from the points (2, 0) and 
(- 2, 0) equals 3. 

24. Its distance from (3, 0) is constantly equal to twice its 
distance from (— 3, 0). 

25. The sum of the squares of its distances from the points 
(3, — 2) and (— 3, — 4) is always equal to 70. 

26. The difference of the squares of its distances from the points 
(2, 1) and (— 3, — 6) is always equal to 19. ' 

27. The square of its distance from the origin is always equal to 
the sum of its distances from the X-axis and the F-axis. 



CHAPTER VI 



THE STRAIGHT LINE AND THE CIRCLE 

75. Having become somewhat familiar with the locus ^oblem 
in a general way, we now take up the study of the simpler 
types of loci in detail, in order to learn the great power of the 
methods of algebra as applied to geometric problems. We shall 
see that to certain types of equations between x and y correspond 
certain loci, the connection being so simple that we can tell at a 
glance what the locus of a given equation will be. 

Example 2, p. 92, proved that the locus of points twice as far 
from the X-axis as from the Y-axis corresponds to the equation 
y=1x, and that the graph is the straight line through the origin 
with slope 2. In the same way we are led to the general theorem : 

The equation of the straight line through the origin with slope 
m is y = mx. 

The proof wiU be given, although it is practically a repetition of 
the work of the problem just referred to. We must show, first, 
that if P = (x, y) is any point on the straight line AOB (the 
straight line through the 
origin with slope m), then 
the coordinates (x, y) of P 
satisfy the equation y = mx ; 
and, secondly, that if the 
coordinates of any point 
P'= (of, y') satisfy the equa- 
tion y = mx, then the point 
P' lies on the line AOB. 

Proof. First, if P is any point on the line AOB (Fig. 67), 

=^ = tan /.XOA = m; 

X 

that is, y = mx. 

97 




Fig. 67 



then 



98 



THE ELEMENTAEY FUNCTIONS 



Secondly, if the coordinates of P' (x' and y') satisfy the equation y = mx, 
then ,/ 



that is, 



X 

tan Z XOP' =tanAXOA. 



Thereioie ZXOP' = ZXOA or else 180° + ZXOA, andin either case the 
point P' is on the straight line A OB. This completes the proof that the 
equation y = mx is the equation of the straight line through the origin 
with sjcpe m. 

EXERCISES 

Write down the equations of the straight lines through the origin 
with each of the following slopes : 1,-2, §, ^, — 5, 7, — 1, — J, 
V3 3 



V2, 



Draw a figure for each. 



76. As our next problem, let us find the equation of the straight 
line through the point (1, 3) with the slope 1. 

If P s (aj, y) is any point on the straight line (Fig. 68), then, 
by theorem (A) on page 80, the slope P(x,y) 

of the line is „ 

y-3 



x-l 








Since the slope is equal to ^, 




y-3. 

x-l 


1 

"2" 






That is, 2 y - 6 = 


= X- 


-1, 




or as — 22/ + 5 = 


-0, 






which is accordingly 


the 


equation 


of 




the line AP. 



Fig. I 



EXERCISES 

1. Find the equation of the line 

(a) through the point (2, - 3) with slope 1; 

(b) through the point (0, 3) with slope 2 ; 

(c) through the point (-1, - 2) with slope |; 

(d) through the point (7, 1) with slope - 3 ; 

(e) through the point (- 2, - 3) with slope - f 



THE STRAIGHT LINE AND THE CIRCLE 99 

2. Find the equation of the straight line through the point (0, k) 
with slope m. 

Am. y = mx+k. This equation is called the slope-intercept form 
of equation of the straight line, because it shows at a glance the 
slope (m) and the F-intercept (/c) of the line (cf. p. 29). Any equa- 
tion of a straight line can be reduced to this form (if y is present 
in the equation) by solving for y. Thus, 3 a; — y = 2 is equivalent 
to 2/ = 3 a; — 2, in which form m = 3 and k = — 2. 

3. By reducing each of the following equations to the slope- 
intercept form, read off the value of the slope and the F-intercept. 
Verify by the graph, as usual: x-\-y = 2, 3a5-|-4y = 5, 7a; — 3y = l, 
^x + ly = Z, Ix + y = b, X — ny = q. 

4. Find the equation of the line through the point (— 3, — 5) 
with slope m. 

5. Find the equation of the line through the point (x^, y^ with 

slope m. ^^g y_ZLJL = ^ or y-y, = m(x-xX 

x — x^ " "'■ ^ " 

This is called the slope-point form of equation of a line. 

6. Write down the equation of the line through the point (5, 3) 
with slope 2. 

7. Write down the equation of the line through the point (— 1, 2) 
with slope — 1. 

8. Write down the equation of the line through the point (2, — 3), 
parallel to the line y=^x-\-&. 

9. Find the equation of the line through the origin, parallel to 
the line x -|- 2 y = 3. 

10. Find the equation of the line through the point (1, —1), parallel 
to the line ^ x — 3 y = 6. 

1 1 . Find the equation of the line through the point (- 5, 1), perpen- 
dicular to the line 2 x — 3 y = 4. 

Hint. The slope of the required line can be foujid from that of the given 
line by means of Ex. 10, p. 83. 

13. Find the equation of the line through the point (1, 4), perpen- 
dicular to the line x — 3 y = 10. 

13. Find the equation of the line through the origin, perpendicular 
to the line x -f- y = 4. 



100 THE ELEMENTARY FUNCTIONS 

14. Find the equation of the straight line through the points 

(3, 4) and (4, 1). 

Hint. Find the slope of the line by the theorem on page 80, then follow 
the method of the preceding exercises. 

15. Find the equation of the line through the points 

(a) (4, 1) and (- 1, - 6) ; (c) (4, 3) and (4, - 1) ; 

(b) (4, - 4) and (3, 2); (d) (0, 0) and (- 6,-3); 

(e) (3, 4) and (-2, 4); 

(f) i'^v yd ^"^^ (^2' yd- 

Ans. '^^^ = y^^^y^, or x„ y„ 1 = 0. 



x^ 


2/i 


1 


^1 


y. 


1 


X 


y 


1 



16. Find the equation of the line whose X- and F-intercepts are 

a and b respectively. j^^^g ^ ^y. — i 

ah 

17. Find the equation of the line through the point (— 2, 3) and 
making an angle of 136° with the A'-axis. 

18. Find the equation of the line through the point (1, 2) and 
making an angle of 60° with the X-axis. 

77. These exercises illustrate the fact that the equation of a 
straight line can be found if its slope and a point on it are given. 
Ex. 5 brought this out definitely, and it is of fundamental impor- 
tance in the study of straight lines. We can now state as a theorem 
what we have hitherto tacitly assumed : The equation of any straight 
line is of the first degree in x and y. 

For every straight line has a definite slope (unless it is perpen- 
dicular to the X-axis, in which case its equation is x = k, which 
is of the first degree), and hence its equation is y — y^ — m (x — x-^, 
where (x^, y{) is any fixed point on the line. But this equation is 
of the first degree ; hence the theorem is proved. 

78. Conversely, any equation of the first degree in x and y has 
for its graph a straight line. 

Proof. The general equation of the first degree in x and y can be written 

ax + by + c = 0. 
If J ;!t 0, we can divide by b, getting 

that is, y = ^ ~ 7' 

b b 



THE STRAIGHT LINE AND THE CIECLE 



101 



This is now in the form y = mx -V k, where m = and k= But 

b b 

Ex. 2 above proved that this is the equation of the straight line whose 

CL c 

slope is m, that is, — - 1 and whose F-intercept is k, that is, Hence 

• ft 

the equation ax + by + c = represents a straight line whose slope is 

c * 

and whose i'-intercept is , provided ft ^ 0. 

But if 6 = 0, the equation is simply 

ax + c = 0; 



that is. 



(Naturally a cannot equal 0, else the equation would not have been of the 

first degree.) This is the equation of a line parallel to the F-axis, so that 

the theorem is proved 

also for the case ft = 0. 

Note that if o = 0, 

the line is parallel to 

the .X-axis. Moreover, 

a A 

since m = — - , m = U 
ft 

in this case, so that 
lines parallel to the 
X-axis have the slope 
zero. This of course 
results also from the 
definition of " slope,'' 
as we observed on page 81. 

79. Normal form of the 
equation of a straight line. 

Let AB be any straight line 
not passing through the ori- 
gin, and OiV the perpendic- 
ular from the origin upon 
the line. Let the length of 
OiV be denoted by p, and the 
angle XON by a. We shall, 
moreover, consider ON as 
being a directed line, the 
positive direction being from 




102 



THE ELEMENTAEY FUNCTIONS 



to N, in whatever position AB may lie. The angle a may have 
any value from 0° to 360° (0 ^ (u < 360). The line ON is called 
the normal to the line AB. 

In case we have to do with 
a line AB through the ori- 
gin, the perpendicular to AB 
through {OC in Fig. 70) is 
still called the normal to AB, 
and it is directed so that the 
angle m is between 0° and 
180°(0Sa)<180). 

Evidently, under these stipu- 
lations, any line AB determines 
a pair of values o and jp ; and, 
conversely, any pair of values of m and j> determines a straight line. 




Fig. 70 



EXERCISES 

Construct the lines for which 

(1) <u = 30°, _?J = 3 ; (4) o) = 200°, ^ = f 

(2) 0, = 165°, i? = 1 ; (5) a. = 90°, p = 5 

(3) 0. = 330°, p = 2; (6) 0. = 100°, ^ = 



(7) o. = 180°, J? = 4 ; 

(8) «, = 0°, j9 = 0. 

(9) «) = 90°, ^ = 0. 



Since <» and p determine the straight line AB completely, it 
must be possible to express the equation of AB in terms of 
these values. This is easily 
done by considering that the 
line AB passes through the 
point iV (Fig. 71), and that its 
slope is the negative recipro- 
cal of the slope of ON; that 

is, it equals — : > which 



tan a) 



equals 



cos ft) 



Hence the 



sm o) 
equation of AB is 

COSO) 

y-y\ = -~ — 

sm<u 



{x-x^). (1) 




Fig. 71 



THE STRAIGHT LINE AND THE CIRCLE 103 

But 2/1 = MN (Fig. ll) = psmm, 

and ajj = OM=p cos tu. 

(1) is accordingly the same as 

cos ft) , 

y — « sin o) = -, (x — p cos eo) : 

sino) ' ' 

that is, X cos a> -\-y sin ft) — j? (sin^ft) + cos^ft)) = 0. 

Since sin2ft) + cos2&) = 1, 

the equation takes the simpler forin 

Of cosw + j/sinw— />=0, (2) 

which is known as the "normal form of the equation of a straight 
line." Write in this form the equation of each of the nine lines 
in the above exercises. 

80. Reduction of general linear equation to normal form. Any 
equation of a straight line can be reduced to the form (2). For 
example, suppose we have the equation 3 x — 4 ?/ = 5, and that we 
wish to reduce it to the form (2). The only change that we can 
make in an equation that will not change its graph is to add the 
same quantity to both sides of the equation or to multiply both sides 
by the same constant quantity. Since the right-hand side of (2) is 
0, we first write our equation so that its right-hand side becomes : 
3a;-4y-5 = 0. (3) 

The only other change that can be made is to multiply both sides 
by the same number, say k ; 1 that is, to write (3) in the form 

3kx-4:ky-5k=0. (4) 

1 A precise statement of the fact used here is, If 

a^x + 6j2/ + Cj = (1) 

represents the same line as a^x + b^y + Cj = 0, (2) 

then O2 = tei, \ = k\, and c^ = kc^ ; or (2) can be obtained from (1) by mul- 
tiplying it by a constant k. This theorem may be proved as follows: The 

slope of (1) is — ^, and that of (2) is ^ . Hence, if the lines are the same, 

^ = ^ ; that is, ^ = ^ . Also, the F-intercept of (1) is - ^ , while that of (2) 

is _ ^. Hence, if the Mnes are the same, — = -^; that is, -^ = -?• Therefore 

^ = 2? _ ^ . Call the value of this common ratio fe, and we see that a^ = fca„ 

flj 6j Cj 

62 = A;6i, and c^ = kCj . Q'E-d. 



104 THE ELEMENTARY FUNCTIONS 

If (2) is the same line as (4), h must have such a value as to make 
each term of (4) equal to the correspondiug term of (2) ; that is, 

cos ft) = 3 ^, sia o) = — 4 ^, p = Zlc. 
Therefore cos^w + sin^w = 9 ^ + 16 Z:;^ = 25 A;^. 

But cos^a) + sw?m = 1. 

Therefore Ic^ = ^, 

and p = 5k = + l, 

the sign being determined by the fact that p is always positive. 

Hence only the + sign may be chosen for k. 

Therefore cos <» = f , sin (u = — l^, p = l, 

and the normal form of the equation Sa? — 4^ = 5 is accordingly 
|x-|2/-l=0. (5) 

By inspection of equation (5) we see that the length of the normal 
is 1, and that it makes with the X-axis an angle whose sine is — |^ 
and whose cosine is |-; that is, an angle in the fourth quadrant. 

General case. The general case is treated in a similar way. 
Let the equation of a straight line be ax + by + c= 0, and let it 
be required to reduce this equation to the form (2) of § 79. 

If (2) and ax + by + c=0 represent the same line, we must 
be able to get the one equation from the other by multiplication 
by a constant k. The equation 

akx + bky + ck = 
must then be exactly the same as 

X cos Q) + ysma>—p = 0. 

Therefore cos to = ak, sin a) = bk, p=— ck. 

Therefore cos^ea + sin^co = (a^ 4. 52^ p. 

Therefore ky^ = 

and k = 



a2 + 62 
1 



±Va2 + j2 

Since p is positive and equals — ck, the sign of k must be 
opposite to that of c. If c = 0, the angle <o is between 0° and 



THE STRAIGHT LINE AND THE CIECLE 



105 



180° (§ 79); hence sin to is positive, so that the sign of k must 
be taken the same as that of b. 

81. This result may be summarized as follows: To reduce the 
equation of a straight line ax + hy + c = to the normal form, 
divide each term by ± v a^ _j_ 52^ choosing the sign of the radical op- 
posite to that of c, or, if c= 0, choosing the same sign as that of b. 



EXERCISES 

1. Reduce each of the following equations to the normal form : 
5x + 12y = l, x + j2/ + 5 = 0, x + y + l = 0, 2x — 3y — 8 = 0, 
3 X + y = 4. Draw a figure for each. 

2. Find the distance between the parallel lines x + 3y — 3 = 
and 2a; + 62/ +1=0. 

3. Find the equation of a line parallel to the line 3x — Ay = 5 
and at a distance 3 from it. (Two solutions.) 

82. Distance from a line to a point. One of the most impor- 
tant uses of the normal form of the equation of a straight line 




Fig. 72 



is to determine the distance from a given line to a given point. 
Let the equation of the given line AB be 

a; cos a) + y sin Q) — J? = 0. (1) 

Let P= («!, y■^) be the given point, and d its distance from AB; 
draw through P the line PP' parallel to AB; the distance from 



106 



THE ELEMENTARY FUNCTIONS 



the origin to PP' will then equal p + d. The equation of PP' 
is accordingly xcos<o + y Bm(i>-{'p + d)= 0. (2) 

Since P is on tliis line, its coordinates {x.^, y^) must satisfy the 
equation (2); that is, 

ajj cos a) + 2/i sin oj — p — d={). 
Therefore d = Jf^ cos <o + j/i sin « — />. (3) 

The result (3) may be stated in words thus : In the normal form 
of the equation of the given line, substitute for x and y the 
coordinates x-^ and y-^ of the given point; the result will he the 
distance from the given line to the given point. 

If the equation of the given line is in the form aa; + &?/ + c = 0, 
it must be reduced to the norfiial form 
ax hy 



: + - 



: + 



:=0. 



Then, substituting the coordinates of the given point (ajj, y^) for 

{X, y), we have ax^+by^ + c ... 

a= , » W 

iVa' + b" 

the sign of the radical being opposite to that of c. 

83. In Fig. 72 the point P was on the opposite side of the 
line from the origin, so that the direction from the line AB 
to the point P was the same 
as the positive direction of the 
normal ON. Hence d, that is, 
QP, may also be considered as 
positive, while in case P were 
on the same side of AB as 
the origin (as in Fig. 73), the 
direction from the line AB to 
the point P would be opposite 
to the positive direction of the 
normal ON. Hence d may in 
this case be considered negative. 

Thus, for all points P on the opposite side of AB from the 
origin, d would be positive, and for all poiuts P on the same side 




Fio. 73 



THE STRAIGHT LINE AND THE CIRCLE 107 

of AB as the origin, d would be negative. The student should 
now verify the fact that the formula (3) above gives the sign 
of d correctly accordiug to this agreement. Every straight line 
thus divides the plane into two parts, which may be called the 
positive and negative sides of the line. The origin is, then, 
always on the negative side of any line (unless the line passes 
through the origin, in which case the upper side of the line is 
the positive side) (see Fig. 70). 

EXERCISES 

1 . How far is the point (1, 4) from the line x + y + l = 0? 
Solution. Using the formula (4), 

± Va2 + /,2 

1 + 4 + 1 6 r- 

we have in this case d = == = — = — .3 V2 = — 4.24 + . 

- Vl + 1 V2 

Draw the figure and note that the result thus obtained by the 
formula is correct in sign as well as numerically. 

2. Find the distance from the line 3x + iy + 1 = to each of 
the points (- 1, - 3), (2, 5), and (0, - 4). 

3. Find the distance from the line y — 2x = to each of the 
points (3, - 2), (1, 4), and (- 2, 2). 

4. Find the equation of the locus of a point that moves so as 
to be equally distant from the lines 

3a; -42/ + 5 = (1) 

and 5a; + 122/-6 = 0. (2) 

Solution. Let P = (x, y) be any point on the locus, and d^ and d^ the 
distances from the given lines to the point P (Fig. 74). 

„, , 3 1 - 4 ?/ + 5 
Then d^ = —^ 

, 5 X + 12 ?/ — G 
and «2 = z^ 



108 



THE ELEMENTAEY FUNCTIONS 



Since the point P is equally distant from the lines (1) and (2), we must 
have either 



or 

that is, either 







d,-- 


--d. 










d,-- 


-— d^\ 






3a; 


-42, 
-5 


+ 5 


5 a; + 12 2/ - 
13 


6 




3a; 


-42/ 
-5 


+ 5 


5a; + 12y 
13 


— 


6 



(A) 
(B) 



(A) contains all points equally distant from (1) and (2) for which the 
distances are liotli positive or both negative ; (B) contains all points equally 
distant from (1) and (2) for which one of the distances is positive and 




Fig. 74 

the other negative. These loci are of course the bisectors of the angles 
formed by the lines (1) and (2), and (A) is the one passing through the 
angle containing the origin. 

The simplified forms of the equations are 



and 



64a; + 82/ + 35 = 
14 a; - 112 y + 95 = 0. 



(A) 
(B) 



5. Eind the equations of the bisectors of the angles formed by the 
lines 3 a; + 4 2/ = 6 and 3 a; — 4 3/ = 10 ; by the lines x — 2y + Z = Q 
and 2a;— 2/— 7 = 0. 

6. Find the equations of the bisectors of the angles formed by 
the lines 3 a; — y = and a; + 3y — 4 = 0. 



THE STRAIGHT LINE AND THE CIRCLE 109 

7. Find the equations of the interior bisectors of the angles of 
the triangle formed by the lines 6 x — 12 y = 0, 6 a: + 12 y + 20 = 0, 
and 12 a; — 5 y — 30 = 0. Show that they meet in a point. 

8. Find the equations of the interior bisectors of the angles of 
the triangle formed by the lines 4x — 3y=5, 5a; — 12^ = 10, and 
4 a; + 3 2/ = 12. Show that they meet in a point. 

9. Answer the same question for the triangle formed by the 
lines y = 4, y = 2x, and y =—2x. 

MISCELLANEOUS PROBLEMS ON THE STRAIGHT LINE 

1. Find the equations of the medians of the triangle whose 
vertices are the points (2, 3), (4, — 5), and (— 2, — 1). Show that 
they meet in a point. 

2. Find the equations of the perpendicular bisectors of the sides 
of the same triangle, and show that they meet in a point. 

3. Find the equations of the altitudes of the above triangle, and 
show that they meet in a point. 

4. Show that the three points of intersection obtained in 
Problems 1-3 lie on a straight line. 

5. Find the center and radius of the circumcircle of the triangle 
whose vertices are the points (0, 6), (7, — 3), and (— 2, 2). 

6. Find the area of the triangle whose vertices are the points 
(0,0), (-2,3), and (-4,-1). 

7. Prove that the area of the triangle whose vertices are the 
points (0, 0), (Xj, y^, and (a-^, y^) is h (x^i/^ - x^^- 

8. Find the center and radius of the inscribed circle of the tri- 
angle whose sides are the lines 3x + 4y — 8 = 0, 4a; + 32/ + 6 = 0, 
and 5 X - 12 y - 13 = 0. 

9. Find the center and radius of each of the escribed circles ^ of 
the triangle in Problem 8. 

10. Prove that the medians of any triangle meet in a point. (Take 
the origin at one vertex and let the A'-axis coincide with one side ; 
that is, let the vertices have the coordinates (0, 0), (a, 0), and {b, c).) 

1 An escribed circle of a triangle is a circle which is tangent to one side of 
the triangle and to the other two sides produced. 



110 THE ELEMENTARY FUNCTIONS 

11. Prove that the perpendicular bisectors of the sides of any 
triangle meet in a point. 

12. Prove that the altitudes of any triangle meet in a point. 

13. Prove that the three points of intersection found in Problems 
10, 11, and 12 lie on a straight line. 

14. Show that the locus of the equation 

(ax + bi/ + c) (a'x + b'lj + e') = 
consists of the two lines ax + Sy + c = and a'x + b'y + c' = 0. 

15. Prove that the locus of the equation ax^ + bxij + cy'' = is a 
pail- of intersecting lines if i^ — 4 ac > 0, that it is one straight line 
if ^2 _ 4 oc = 0, and that it contains no real point except the origin 
if 6^ - 4 ac < 0. 

ie. Prove that the two straight lines bx^ — cxy + ay" = are 
respectively perpendicular to the lines as? + cxy + by'^ = 0. 

17. Taking the vertices of a triangle as (cCj, y^), (x^, y^), and (x^, y^), 
find the equations of the perpendicular bisectors of the sides. Show 
that they meet in a point. 

18. Prove that the locus of the equation 

ax ^- by -\- c ■\- k {a'x + b'y + c') = 

is a straight line through the points of intersection of the lines 
ax + by + c = and a'x + b'y + c' = 0. 

19. Prove that the area of the triangle whose vertices are the points 
(»i> yi), (a:^, y^, and {x^, y^) is \ (x^y^ - x^^ + x^^ - x^^ + x^y^ - x^y^, 
that is, 



1 


^x Vx 


1 


2 


% 2/2 


1 




"=s Vz 


1 



The Circle 

84. Several of the locus problems in the preceding chapter 
(for example, Exs. 11, 12, p. 95) led to circles as their result. 
Indeed, it is easy to determine the equation of any circle in terms 
of its radius and the coordinates of the center. 



THE STRAIGHT LINE AND THE CIRCLE 111 



Let C={cc,^} be the center and r the radius (Fig. 75). If 
P = {x, y) is any point on the 
circle, then 



CP. 



P(x,y) 



But CP=yf{x- af + (y - ^f. 

Therefore 

{x-ay + {y-pf = r\ (1) 

which is accordingly the equa- 
tion of the circle. 

CoEOLLAEY. The equation of 
the circle whose center is the origin and whose radius is r is 

x-^-\-y^ = r\ (2) 




Fig. 75 



EXERCISES 

1. Draw the graphs of the following equations : le' -\- i^ = 1, 
x^ + f = 2, x' + f=l x' + i/' = 3.24. 

2. Write the equation of the circle whose center is 

(a) (4, 6) and whose radius is 5 ; 

(b) (— 1, — 2) and whose radius is 3 ; 

(c) (2, 0) and whose radius is 2 ; 

(d) (a, 0) and whose radius is a. 

3. Show, by reducing to the form (1), that the locus of each of 
the following equations is a circle, and construct it. 

(a) x'' + f-ix + 2j/+l=0. 

Solution. To get the standard form (1), complete the square of the 
terms in x and also of those in y, thus : 

(i2 - 4 a; + 4) + (^2 + 2 ,y + 1) = 4. 
That is, (x - 2)2 + (y + 1)^ = 4. 

This is in the form (1), with a = 2, ^ = — 1, r^ = i ; hence the graph is 
a circle whose center is (2, — 1) and whose radius is 2. The student may 
now draw the figure. 

(h) x^ + 2f-x-iy=2. (f) x'' + f=3(x + 3). 

(c) a;'^ + y^-|-16x-122/ + 84 = 0. (g) 3x^+32/'-5a;+ 24?/= 0. 

(d) x^ + y^ + Ax + Sy-il = 0. <h) x^-|-y^-|-16y + 36 = 0. 

(e) x' + f — x — 'i/ = 0. 



112 THE ELEMENTARY FUNCTIONS 

4. Find the equation of the circle whose center is (3, 0) and 
which passes through the origin. 

5. Find the equation of the circle whose center is (2, — 1) and 
which passes through the point (— 1, — 5). 

6. Find the equation of the circle whose center is (3, — 2) and 
which is tangent to the F-axis. 

7. Find the equation of the circle which has the line joining the 
points (3, 4) and (— 1, — 2) as diameter. 

8. Find the equation of the circle whose center is on the 
A'-axis and which is tangent to the lines y = 3 and x = l. (Two 
solutions.) 

9. Find the equation of the circle whose center is on the F-axis 
and which is tangent to the lines // = 2 a; — 1 and y =— 2x; of 
the circle whose center is on the A'-axis and which is tangent to the 
same two lines. 

10. Find the equation of the cii-cle whose center is on the line 
y = 1 and which is tangent to each of the lines 3x — y = 6 and 
X — 3y = 3. 

11. Find the equation of the circle that is tangent to the three 
lines X = 0, y = 0, and x = 3. 

12. Find the equation of the circle that is tangent to the three 
lines X = 5, y = X, and y = — x. (Four solutions.) 

13. Find the equation of the circle that is tangent to the three 
lines 3x-4y — 5 = 0, ix + 3y-8 = 0, and 4a; - 32/ -|- 12 = 0. 

14. Find the equation of the circle whose center is on the F-axis 
and which passes through the points (3, 4) and (2, — 1). 

15. Find the equation of the circle which passes through the three 
points (4, - 1), (- 2, - 3), and (- 1, 3). 

16. Find the equation of the circle which passes through the three 
points (0, 6), (3, 0), and (- 2, - 4). 

17. Find the equation of the circle which passes through the 
points (— 1, 9) and (6, 8) and is tangent to the A'-axis. 

Am. x" + y^ — ix — 10 1/ + i = 

or x'^ + y^- 244 ,r - 1690 y -|- 14,884 = 0. 



THE STRAIGHT LINE AND THE CIRCLE 113 

18. Find the equation of the circle which passes through the 
points (8, 8) and (1, 7) and is tangent to the line 3 as + 4 y = 6. 

Am. aj^ + j/'- lOz- 8y + 16 = 

OTx^ + y^-2x-64:i/ + 400 = 0. 

19. Find the equation of the circle which passes through the point 
(2, 4) and is tangent to the X- and F-axes. (Two solutions.) 

Prove that each of the following loci (Exs. 20-22) is a circle, by 
showing that its equation assumes the form (1) (p. 111). To this 
end, choose the X- and F-axes in a convenient position with refer- 
ence to the given points or lines. 

20. A point moves so that the sum of the squares of its distances 
from two iixed points is constant. 

21. A point moves so that the ratio of its distances from two iixed 
points is a constant (not equal to 1). 

22. A point moves so that the square of its distance from a fixed 
point, divided by its distance from a fixed line, is constant. 

23. Find the equation of the circle passing through the mid-points 
of the sides of the triangle whose vertices are (4, 0), (— 2, 0), and 
(0, 6). 

24. Prove that the circle of Ex. 23 passes through the feet of the 
altitudes of the triangle, and also through the points halfway 
between the vertices and the orthocenter of the triangle. This circle 
is called the Nine Points Circle of the triangle. 

25. Do as in Exs. 23, 24 for the triangle whose vertices are (a, 0), 
(b, 0), and (0, c). ^^^ ^j^^ equation of the Nine Points Circle is 



CHAPTER VII 



THE PARABOLA, ELLIPSE, AND HYPERBOLA 

85. The parabola. A particularly important locus is that of a 
point which moves so as to be equally distant from a fixed point 
and from a fixed line. The student should construct carefully 
such a locus, taking the fixed point and the fixed line quite at 
random, and locating a sufficient number of points on the locus 
to make the shape of the curve clear (cf. Ex. 14, p. 95). This 
locus is called a parabola, and thus we have for the first time 
a definition of that word, which we have often used heretofore 
merely for the sake of giving a name to a curve of a certain 
shape. We have now the definition : 

A parabola is the locus of a point wJiose distance from a 
fixed point is always equal to its distance from a fixed line. The 
fixed point is called the focus, and the fixed line the directrix, of 
the parabola. 

86. Equation of the parabola. The loci of Exs. 14-17, pp. 95, 
96, were parabolas, and accordingly the equations obtained were 
equations of parabolas. In order 
to get the simplest possible form 
of equation, proceed as follows : 

Let F be the focus and DD' 
the directrix. Draw the per- 
pendicular from F to DD', and 
choose this line as the X-axis, 
with the direction from DD' 
to F as the positive direction. 
Choose as origin the point mid- 
way between F and DD', and 
represent the distance OF by a 
(a, 0), and the equation of DD' is x 

114 



D 


Q 


Y 






\ , 


N 
D' 





F 



Fig. 76 

Then the coordinates of F are 
a. Now if P s (x, y) 



THE PARABOLA, ELLIPSE, AND HYPERBOLA 115 

is any point on the locus, the two distances MP and FP must 
be equal, by the definition of the parabola. 

But MP=x-\'a 

and FP = y/(x-af+y\ 

Therefore V(a; -af+if=x + a (1) 

Simplifying, we get y'^ = ^ax. (2) 

This equation must then be satisfied by the coordinates of all 
points on the parabola; and, conversely, all points whose coordi- 
nates satisfy (2) must lie on the parabola, because we can reason 
back from (2) to (1). The possible double sign in (1) does not 
make any difference here, as a; + a cannot be negative. 

Equation (2) is therefore the equation of the parabola, and it 
may be considered as a standard form for the equation of the 
curve. By analyzing the equation several important properties 
of the curve may easily be obtained. 

87. (1) Solving (2) for y, we have 

y =±V4aa;. 

Hence, for any positive value of x, y has two corresponding values, 
which are equal numerically but of opposite sign. This means 
that the curve is symmetrical with respect to the X-axis. The 
line through the focus, perpendicular to the directrix, of a parab- 
ola is accordingly called the axis of symmetry, or simply the axis, 
of the curve. The point where it intersects the parabola is called 
the vertex of the curve. Here the vertex is the origin. 

(2) The form of the equation y^=Aax shows that x and a 
must have the same sign, since y"^ cannot be negative; that is, 
X must be positive, since we have taken a as positive. Hence the 
curve lies entirely on the positive side of the F-axis. 

(3) As x increases, the positive value of y increases also, but 
less rapidly (since y is proportional to the square root of x), so 
that for very large values of a; a small change in x wUl produce 
scarcely any change at all in y. This means that the curve 
becomes more and more nearly horizontal as it recedes from 
the vertex. 



116 



THE ELEMENTARY FUNCTIONS 



(4) The chord of the parabola drawn through the focus perpen- 
dicular to the axis is called the lahis rectum. The student may 
show that the coordinates of B (Fig. 77) are {a, 2 a) and that 
those of C are (a, — 2 a), and hence that the length of the latus 
rectum is 4 a. 

Summarizing, the equation y^ = 4 ax has for its locus the 
parabola with focus at (a, 0), directrix x = — a, axis the X-axis, 
vertex the origin, and latus rectum 4 a. 

If we take a as a negative quantity, the parabola will be on 
the negative side of the F-axis, and the other inferences that 
have been drawn will require cor- 
responding modifications. Also, if 
we should choose the line through 
the focus, perpendicular to the 
directrix (that is, the axis of 
the parabola), as Z-axis instead 
of as X-axis, the effect would be 
merely to interchange x and y 
in the equation (2), giving 

jc=' = 4 ay. (3) 

The focus is now the point (0, a), 

the vertex is the origin, and the 

curve is concave upward if a is 

positive, concave downward if a is negative. Draw the figure 

for this choice of coordinate axes, taking the F-axis vertical 

as usual. What is the equation of the directrix ? 




Fig. 77 



EXERCISES 

Draw the loci of the following equations, locate vertex, focus, 
directrix, and axis of symmetry, and find the length of the latus 
rectum : 

1. / = 8x. 5. 2/2^1 K. 9. x'^-by. 

2. 2/' = 10 a;. &.y'^ = -i,x. \0.y = A,x^. 
%.f=x. I.f = -Vax. 11. x^ + 32/ = 0. 
^.f = \x. 8. a;' = 4y. 12. y^-|-ix = 0. 



THE PAEABOLA, ELLIPSE, AND HYPERBOLA 117 



*88. Generalized standard equation of the parabola. We often 
have to deal with problems in which the X- and Y-axes cannot be 
chosen at pleasure but are already prescribed in a less favor- 
able position than that which we chose in deriving the standard 
equation (2), § 86. Exs. 14-17, pp. 95, 96, are examples of such 
problems. In problems of this kind the equation of the curve 
wUl not take so simple a form as (2) or (3), §§ 86 and 87. The 
simplest generalization, and the only one which we shall here 
consider, is the case where the axis of the parabola is parallel to 
either the X- or the F-axis and the vertex and the latus rectum 
are given. In this case the equa- 
tion of the curve can be found by 
using a certain f imdamental prop- 
erty of the parabola, as follows : 

If P s (x, y) is any point on 
the parabola indicated in Fig. 78, 
we have seen that the coordinates 
of P must satisfy the equation 

y'^ = 4l ax. 

But y = MP and x = VM, so that the 
equation y'^ = 4^ax is the same as 

MP^ = 4 a • VM. 

Fig. 78 

This relation must then hold true 

for all points P on the parabola whose vertex is V and whose 
latus rectum is 4 a. It accordingly states an intrinsic property 
of the curve, which, expressed in words, is the following 

Theoeem. If from any point on a parabola a perpendicular 
is drawn to the axis of the curve, then the square of this perpen- 
dicular (MP^) is equal to the product of the latus rectum {4 a) by 
the distance from the vertex to the foot of the perpendicular (VM). 

This theorem is now stated in such a way as to be inde- 
pendent of the position of the coordinate axes. We can use it 
to write down at once the equation of a parabola whose axis 
is parallel to either the X- or the F-axis. Let the vertex be 



y 


E 

f 


f 




V 





F M 



118 



THE ELEMENTARY FUNCTIONS 



Y 


1 B 

A 


( 


f 


R 


k 

1 ^ 


F M 





Q 


S 



Fig. 79 



V={a, /8) (Fig. 79), and let the latus rectum be 4 a, the axis 
being parallel to the X-axis. Then ii P = {x, y) is any point on 
the curve, we have, by the 
above theorem, 

Jfp2 = 4 a . VM. 

But MP = SP-S3f 

= 5P - QF 

= 2/-/3, 
and VM=BM-Rr 
= OS-OQ 
= X — a. 
Therefore 
(y-^)^ = ^a{x-a). (4) 

Since this equation is true 
for all points {x, y) on the 
parabola, and for no other points, it is the equation of the curve. 
If the axis of the curve had been parallel to the F-axis instead 
of to the X-axis, the equation would have been 

(a;-a)2 = 4a(j/-/8). (5) 

The student should draw the figure and give the proof for this 
case also. 

EXERCISE 

Give the coordinates of F, B, and C, and the equation of the 
directrix, both for Eig. 79 and for the case where the axis of the 
parabola is parallel to the F-axis. 

*89. Summary. The equation (y — /3)2 = 4 a (a; — a) represents a 
parabola with vertex at the point (a, /3), but otherwise the same 
as the locus of the equation 2/^ = 4 ax. Its locus can be thought 
of, indeed, as the same curve as y^ = 4 ax, merely moved along so 
that its vertex takes the position (a, /8), the axis of the curve re- 
maining horizontal. Likewise, the equation (x — af — ^ a(jj — /8) 
represents the same parabola as a;^ = 4 ay, moved so that its vertex 
takes the position (a, /8), the axis of the curve remaining vertical. 
A movement of this sort is called a translation. 



THE PAEABOLA, ELLIPSE, AND HYPERBOLA 119 



*90. By using these results we can also recognize and draw 
the graph of any equation of a parabola whose axis is parallel to 
the X-axis or to the J-axis. 

Example. Draw the graph oi 1/ + 2 x ~ i y = 6. This equation can be 
reduced to the form (4), (y-.pf = ia(x- a), 
by completing the square of the y-teims. 

Thus, y^~4:y = -2x + 6. 

Adding 4 to complete the square, 

y^-iy + i = -2x + 10; 
that is, (!/ - 2)^ = - 2 (2: - 5). 

This is now in the form (4) with o: = 5, ;8 = 2, 4 a = — 2 (that is, a = — I). 
Hence the graph is the parabola whose vertex is the point (5, 2), whose axis 
is parallel to the X-axis, and whose 
latus rectum is 2. The focus is to 
the left of the vertex, because a is 
negative. Hence the coordinates 
of the focus are (4i, 2), and the 
equation of the directrix is i = 5 J. 
With this information it is easy to 
draw the graph without computing 
the coordinates of any points on 
the curve. As a check, however, 
it is always well to determine the 
intercepts on the A'- and F-axes. 
Thus, when y = 0, 2 x = Q, x = & ; 
hence the point (3, 0) is on the 
curve. When x = 0, 3/''— 4 i/ — 6 = 0, 

2/=2±VlO = 5.16+or-1.16+. Hence the points (0,5.16+) and (0,-1.16+) 
are on the curve. These facts agree with Fig. 80. 

EXERCISES 

Draw the graphs of the following equations and locate vertex, 
focus, and directrix. Check by finding the intercepts of the curve 
on the X- and F-axes. 

1. y''-2x-2y = l. e. 2 1/ + ^x - tj + ^ = 0. 

2. 2/=-5a;-82/ + l = 0. 7. x^ - Sx + Sy = 0. 
i. f - X - y = 0. 8. ai2 - a; + 2 2/ = 3. 

4. 42/^ — 6x— -2/ = 5. 9. 4a;^— 9a; +-2/ — 1 = 0. 

5. 32/^ + 4a;-52/ + 2 = 0. 10. lOx^ - 15a; + 8 // = 20. 




Fig. 80 



120 



THE ELEMENTAEY FUNCTIONS 



D 




P 












X 


A 


F 


Q 






D 




. 


P' 

s 





Fig. 81 



91. Construction of parabola. (1) By points. Since any point 
on the parabola is equally distant from the focus and from the 
directrix, we can locate any 
number of points on the curve 
as follows : 

Let F (Fig. 81) be the focus 
and Djy the directrix. Draw 
the axis AFX, and through Q, 
any point on the axis, draw 
BS perpendicular to AX. From 
F as center, with radius AQ, 
describe an arc meeting RS 

in the points P and P'- These will both be points on the 
parabola whose focus is F and whose directrix is DD'. (Why ?) 

(2) By continuous motion. 
Place a right triangle JKH 
(Fig. 82) with KH along the 
axis AX and with KJ along 
DD'. One end of a string of 
length KH is fastened at H, the 
other end at F. If now a pencil 
point P be pressed against the 
string, keeping it taut as the tri- 
angle JKH is slid along the 
directrix, then P will trace an arc of a parabola. Prove this. 



J 

K 


D 


\ \ 


A 


F 

D' 

Fig. 82 



PROBLEMS ON THE PARABOLA 

1. The expression y^ — 4: ax, which equals zero for all points 
(x, y) on the parabola whose equation is 2/* = 4 ax, is positive for all 
points outside the curve and negative for all points inside. 

2. Every point outside the parabola is nearer to the directrix 
than to the focus ; every point inside the parabola is nearer to the 
focus than to the directrix. 

3. If PP' is any chord of a parabola passing through the focus, 
then the circle on PP' as diameter is tangent to the directrix. 



THE PAEABOLA, ELLIPSE, AND HYPERBOLA 121 

4. If FP is the line segment joining the focus to any point on 
the parabola y^ = 4 ax, then the circle on FP as diameter is tangent 
to the F-axis. 

5. Given the directrix of a parabola, and two points on the curve, 
to find the focus. 

6. Given the focus of a parabola and two points on the curve, to 
find the directrix. 

7. The straight line passing through the mid-points of any two 
parallel chords of a parabola is parallel to the axis of the curve and 
bisects all other chords that are parallel to the original ones. 

8. Given a parabola, to find its axis, focus, and directrix. 

9. The vertex of a parabola is 0, and P is any other point on the 
curve. Through P two lines are drawn, one perpendicular to the axis 
and the other perpendicular to OP. These lines meet the axis in Q 
and R respectively. Prove that QR is equal to the latus rectum. 

10. Two perpendicular chords are drawn through the vertex of a 
parabola. Show that; the line joining their other intersections with 
the parabola meets the axis in a fixed point. 

11. Prove that the following is a correct construction for a parab- 
ola : A being the vertex and F the focus, produce AF to B, making 
FB=AF, and draw the circle with B as center and BA as radius. At Q, 
any point on the axis, erect the perpendicular QR to AB (ij being on 
the circle), and lay off QP (and QP') equal to AR. Then P and P' are 
points on the parabola. In this way any number of points can be found. 

12. Given A and F, the vertex and focus respectively, draw FA 
and produce it to C, so that ^C = iFA. Let Q be any point on the 
axis FA, and draw the circle having CQ as diameter. Draw the chord 
RAR'± CQ, and the tangent QP. The line RP W AQ meets QP in P, 
which is a point of the parabola. Prove this.^ 

The Ellipse 

92. Definition of the ellipse. Another important locus is that 
of a point which moves so as to be always half as far from a 
fixed point as from a fixed line. The student should construct 
this locus carefully, taking a point and a line entirely at random 

1 The constructions of Exs. 11 and 12 are of Arabian origin. They are found 
in a work written by an Arabian mathematician named Abti'I WSfS, who lived 
in the tenth century. 



122 



THE ELEMENTARY FUNCTIONS 



and locating enough points on the locus to make the shape of the 
curve evident (of. Exs. 20, 21, p. 96). This locus is called an 
ellipse. The locus would also be an ellipse if we changed the 
word "half" in the first sentence of this paragraph to "one third," 
"two thirds," or any other positive number less than one. The 
complete definition of an ellipse is accordingly: 

An ellipse is the locus of a point whose distance from a fixed 
point is to its distance from a fixed line in a constant ratio less 
than one. The fixed point is called the focus, the fixed line the 
directrix, and the constant ratio the eccentricity. 

The student should now construct carefully several ellipses, 
using different values of the eccentricity, — • for instance, ^, |, |, ^, 
|, ^^. Notice how the shape of the curve is affected by increasing 
or diminishing the eccentricity. 

93. Equation of the ellipse. The simplest form of the equation 
of an ellipse can be obtained as follows: Let F be the focus, 
D-D' the directrix, and e the eccentricity, a positive number less 
than one. Choose as A-axis the line through the focus perpen- 
dicular to the directrix, with the direction from F to DD' as the 
positive direction. Two points on the curve can be located at once, 
namely, the points A and A' where 
the curve meets the X-axis ; for 



FA 



= e 



and 



(1) 
(2) 



r 




D 

X 


A' 


F A. 


B 
D' 



Fig. 83 



AB 
A'F 
A'B' 

so that A and A' can be found 
by a simple construction of ele- 
mentary geometry. Take now as origin the mid-point of the segment 
AA, and represent the distance OA by a. Then A'O also equals a. 
In order to get the equation of the ellipse in the simplest possible 
form, we shall need to obtain the lengths OF and OB in terms of 
a and e. To accomplish this, rewrite the equations (1) and (2) above : 

FA = e- AB, (1) 

A'F==e-A'B. (2) 



THE PARABOLA, ELLIPSE, AND HYPERBOLA 123 

(3) 



Adding, we get 
But 
and 

since 

Therefore (3) gives 

that is, 



A'F + FA = e{A'B+AB). 
A'F + FA=A'A = 2a, 
A'B + AB={A'0 + OB) + {0B- OA) 
= 2 OB, 
A'0 = OA. 
2a = e.2 0B; 

a 
0B=-- 
e 

In order to get the length of OF, subtract (1) from (2): 

A'F -FA = e {A'B - AB) 

= e ■ A' A = 2 ae. 

But A'F - FA = {A' + OF)- {OA - OF) 

= 2 OF, 

since A'0 = OA. 

Therefore (5) gives 20F=2ae; 

that is, OF=ae. 



(4) 
(5) 



(6) 



The important results of equations (4) and (6) may be stated thus : 
the coordinates of F are {ae, 0), and the equation of DB' is a; = - • 

Now, to get the 
equation of the ellipse, 
let P = {x, y) be any 
point on the curve; 
then, by the*definition 
of the ellipse, 

PF 
'FQ 
or PF=e-PQ. (7) 
But 



: = «' 



Y 
R 


P(x,y) 


D 

Q 


\ 


A' 


F A 


B 
D' 



Fig. 84 



and 



or 



PF = y/{x — ae)^-iry^ 

PQ = RQ-RP = --x. 

e 

Therefore V(a; - ae)2 -\- y^ = e(^ - x\= a - ex, 
{x — aef + y'^ = {a — ex)\ 



(8) 



124 THE ELEMENTARY FUNCTIONS 

Multiplying out and rearranging terms, 

x^{l-e2) + f = a^{l-e% (9) 

Equation (9) is the equation of the ellipse, because it is the equa- 
tion which must be satisfied by the coordinates (x, y) of aU points 
on the curve; and it is not satisfied by the coordinates of any 
other points, because, if (9) is true, we can reason backwards 
through (8) to (7), the possible change of sign in (8) affecting 
only the direction of PQ, which here makes no difference, because 
e is the ratio of the numerical lengths of PF and PQ. 

94. It is usual to make, in equation (9), the abbreviation 
Ifi^d^^V — ^, which is allowable because e< 1 and hence a2(l— e^) 
is necessarily positive. Then (9) becomes 

— a^ + y2 = J2. 
a'' 

that is. ^"+§=1- (10) 

This is the simplest form of the equation of an ellipse, and it 
may be considered as a standard form for the equation of the 
curve. By analyzing the equation, several important properties of 
the curve may be obtained. 

95. (1) Solving (10) for y as an explicit function of x, we ob- 



tain « = ±-Va2_a32_ rpjjjg siiows that a? cannot be>a2; for if 
a 

it were, y would be complex, thus giving no point on the curve. 
This means that no point of the curve lies to the right of a; = a 
(the point A) or to the left of x=— a (the point A'). For any 
value of X between — a and + a, y has two values, numerically 
equal but opposite in sign, so that the curve is symmetrical with 
respect to the JT-axis. 

(2) Solving (10) for x as an explicit function of y, and reason- 
ing as in the preceding paragraph, we find that the ellipse is 
symmetrical to the T-axis, and that the points B s (0, b) and 
P' = (0, — h) are respectively the highest and the lowest point on 
the curve. The segment B'P is called the minor axis of the eUipse; 



THE PARABOLA, ELLIPSE, AND HYPERBOLA 125 



A' A is called the major axis. The end-points of the axes, A, A', 
B, and B', are called the vertices of the ellipse. The intersection 
of the axes, 0, is called the center and is the mid-point of any 
chord drawn through it.^ 

(3) The relation among the three quantities a, b, and e (namely, 



■ = a^{l-e^)-. 



aV) may be easily remembered by noting 



that in a right triangle with 
b and ae as the two legs, 
a will be the hypotenuse. 
Thus, in Pig. 85, the triangle 
BOF has OF =ae,OB = b; 
hence BF = a. This fact 
enables us to find the focus 
of an ellipse when the major 
and minor axes are given. 
(4) As in the parabola, 
the chord through the focus, 
perpendicular to the major 
axis, is called the latus rectum. The student may show that the 
coordinates of its end-points are 




Fig. 85 



(a.,^')and(ae,-^') 



62\ 2 &2 

> so that its length is — •• 
a/ a 



r 



96. Summarizing, the equation — -|- f- = 1 has for its locus 

the ellipse with major axis 2 a, minor axis 2 h, center at the 
origin, vertices (± a, 0) and (0, ± b), focus {ae, 0), and directrix 

x = — > where the value of e may be obtained from the fact that 

e jg2 

a2g2 ^a2_ffl_ Thus, the equation — 4- y^ = 1 has for its locus 

the ellipse with center at the origin, major axis 2>/5, minor 
axis 2, vertices (±V5, O) and (0, ±1), focus (2, 0), and directrix 
aj = f (Fig. 86). 

1 Because if (ij, y-^ is a poiirt satisfying the equation of tlie ellipse, 
E_ 4- IL = 1, then (— x,, — y^) will also satisfy the equation ; and the origin is 
the mid-point of the segment joining the points (x^, y^ and (- Xj, - y.^). 



126 



THE ELEMBNTAEY FUNCTIONS 



In deriving the equation of the ellipse, if we had chosen the 
line through the focus, perpendicular to the directrix, as F-axis 
instead of as X-axis, 
the effect would have 
been merely to in- 
terchange X and y 
in the equation (10) 
(§ 94), giving 

1+^-1- (11) 

Thus, the equation 

— H =1 has for its 

3 2 Fig. 86 

graph the ellipse with 

center at the origin, major axis sVs, minor axis 2V2, vertices 

(O, ±V3) and (±V2, O), focus (0, 1), and directrix y = 3. It is 

the same curve as the ellipse — -f- ^ = 1, rotated about the origin 
through an angle of 90°. Draw the figure. 



EXERCISES 




Draw the graphs of the following equations, locating focus and 
directrix and finding the eccentricity : 



'1-, 



8. ■lx^ + Zf= 21. 

9. a? + ^>f = :^. 
10. 2x' + Zif=l. 
H. 4x^-1-52/'= 6. 

12. Prove that the point (— ae, 0) and the line x= are also 

X V 

a focus and directrix of the ellipse -; -|- t;; = 1. 



'•9+-i=^- 



2.j + f=l. 



3 ^+l!-l 
^- 25 + 16"^- 



i ^ + r 

4^2 



5.| + y^=l. 



6. x^ + -^=l. 



13. Find the equation of the ellipse whose center is the origin 
and whose foci are on the A'-axis, if a =10 and e = i. 



THE PAEABOLA, ELLIPSE, AND HYPERBOLA 127 



*97. Generalized standard equation of the ellipse. As in the 
case of the parabola, the simple standard equation of the ellipse 
gives a geometric property common to all points of the curve, 
and this property will enable 
us to obtain a more gen- 
eral form for the equation of 



the curve. 



Thus, ^ + § = 1 




is equivalent (see Fig. 87) to 

1 = 1, this relation 

a^ J2 

being expressed in words by 

the following 

Theorem. If from any •point Fig. 87 

on an ellipse perpendiculars are 

drawn to the rnajor and minor axes, the square of the perpen- 
dicular upon the minor axis (ATP^), divided by the square of the 
semi-major axis (a^, plus the square of the perpendicular upon 
the major axis (MP^), divided by the square of the semir-minor 
axis (JP), equals unity. 

Using this theorem, we can write down the equation of an 
ellipse whose axes are parallel to the X- and Z-axes, and whose 
center is any point (a, /S). Let the major axis be parallel to the 
X-axis and of length 2 a, y 
the minor axis having the 
length 2b. Then, by the 
theorem just stated, 

NF^ MP^_ 

But NP=TP-TX 
= OS-OQ 
= X— a, 
and MP=SP-SM 




Fig. 88 



128 THE ELEMENTARY FUNCTIONS 

Therefore the equation of the ellipse is 

If the major axis had been parallel to the F-axis, we should 
have had , ., , .^^ 

(fi^+(LLa.i. (13) 

The figure should be drawn and the proof given for this 
case also. 

EXERCISE 

Give the coordinates of F, F', A, A', B, B', and the equation of 
each directrix (both for equation (12) and for equation (13)). 

*98. Summary. The equation ^^ — 7r-^+ ,„ =1 represents 

an ellipse with center at {a, /8), but otherwise the same curve as 

the ellipse -5 + 75 = 1- It can be thought of, indeed, as the same 

curve, merely translated from the position ia which the origin is 
the center to the position in which the point («, /S) is the center. 
By using these results we can recognize and hence draw the 
graph of any equation that is reducible to either of the forms 
(12) or (13). 

Example. Draw the graph of the equation x^ + 4 v^ — 6 a; + 4 j/ — 6 = 0. 

We see that by completing the square of the x-terms and of the 2/-terms 

it will be possible to reduce this to the form (12) or (13). Hence we write 

it first in the form 

(a:2-6x) + 4(),2 + y) = 6. 

To complete the square we must add 9 inside the first parenthesis and \ 
inside the second (which amounts to adding 9 + 1 to the left member 
of the equation): 

(x2 _ 6 X + 9) + 4 (y2 + 2, + i) = 6 + 9 + 1 = 16 I 
that is, (x - 3)'' + 4 (j, + \y = 16, 

or (^ - 3)' , iy+hy _. 

16 4 

which is in the form (12) with a: = 3, |8 = - ^, a = 4, ft = 2. Hence ths 
graph is the ellipse whose center is the point (3, — J), whose major axis 



THE PARABOLA, ELLIPSE, AND HYPEEBOLA 129 

is 8, and whose minor axis is 4. The vertices are accordingly the points 
(7, - i), (- 1, - i), (3, li), and (3, - 2^). To find the focus, 

aV = a2 - «2 = 16 - 4 = 12. 

Therefore ae = Vl2 = 2 Vs = 3.46+. 

TV. * 2V3 V3 -_ 

Therefore e = = = .87-. 

4 2 

The foci are therefore at the distance 2v3 from the center, and the 

directrices are at the distance - = — — = from the center. With the 

help of these data the graph is easily drawn. As a check the X- and . Y- 

intercepts should be found. When a; = 0, 4y^ + 42/ — 6 = 0, 23^^ + 2)/ — 3 = 0, 

— 1±V7 
y = = .8 or — 1.8, approximately. When y = 0, x^ — 6 x — 6 = 0, 

X = S ± Vl5 = 6.87 or — .87, approximately. These intercepts should agree 
with the figure. 

EXERCISES 

Draw the graphs of the following equations, and locate vertices, 
foci, and directrices. Check in each ease by finding the intercepts 
of the curve on the X- and F-axes. 

1. 4a;^ 4- 9 2/" -16a; +18 3/ -11 = 0. 

2. 3x' + 9f-6x-27y + 2 = 0. 

3. Ax' + y^-Sx + 2y + l = 0. 

4. a;'' + 152/' + 4a; + 602/ + 15 = 0. 

5. x'^ + 2f + 3x + y=0. 

6. 2x^ + iy^ + x-8y = 0. 



PROBLEMS ON THE ELLIPSE 

1. Eind the equation of an ellipse, given 

(a) foci at (3, 0) and (— 3, 0), one directrix a; = 4 ; 

(b) foci at (1, 1) and (— 1, 1), one directrix x = 2; 

(c) major axis = 8, foci (4, 3) and (— 2, 3); 

(d) major axis = 2, foci (0, ^) and (0, — J). 

2. Find the equation of the locus of a point which moves so that 
the sum of its distances from the points (3, 0) and (— 3, 0) is con- 
stantly equal to 10. What kind of curve is this locus ? Draw it. 



130 THE ELEMENTAEY FUNCTIONS 

3. Find the equation of the locus of a point which moves so 
that the sum of its distances from the points (c, 0) and (— c, 0) is 
constantly equal to 2 a. (a > c) . x'^ y^ _ . 

A.71S, — r -| ^ T — 1. 

This result, being the equation of an ellipse, establishes the 
following 

Theorem. The locus of a point which moves so that the sum of its 
distances from two fixed points is a constant greater than the dis- 
tance between the points is an ellipse having the fixed points for 
foci and the constant distance for its major axis (of. Ex. 22, p. 96). 

4. Prove the theorem of Ex. 3 by showing that the distances 

from any point (x^, y^ on the ellipse ^ + ji = 1 to the foci are 
a + eXj and a — ex^, so that their sum is the constant 2 a. 

5. Use the theorem of Ex. 3 to construct an ellipse by continuous 
motion, with the help of a piece of string and two thumbtacks. 

x^ If^ 

6. Given an ellipse "i + f^ = 1- (a>b-) Let b be gradually in- 
creased until it finally equals a. How will the ellipse change ? What 
will happen to the foci ? to the eccentricity ? to the directrices ? 

7. The lines joining a point on an ellipse with the ends of the 
minor axis meet the major axis in 5 and T. Prove that the semi- 
major axis is the mean proportional between OS and OT, being 
the center of the ellipse. 

8. A and A' are the ends of the major axis of an ellipse; P is 
any point on the curve, and PM and PN are perpendiculars to PA 
and PA' respectively, M and N being on the major axis. Prove that 
MN is equal to the latus rectum. 

9. The same construction as in Ex. 8, except that perpendiculars 
are drawn to PA and PA ' at A and A ' respectively. Let these per- 
pendiculars intersect in Q. Prove that the locus of Q is an ellipse 

whose semi-axes are a and f 



10. If a point P moves around an ellipse, starting from one end 
of the major axis, its distance from the center will decrease until 
it reaches the end of the minor axis. 

11. The circle on any focal radius as diameter is tangent to the 
circle on the major axis as diameter. 



THE PARABOLA, ELLIPSE, AND HYPERBOLA 131 

The Hyperbola 

99. A third important locus is that of a poii>t which moves 
so as to be always twice as far from a fixed point as from a fixed 
line (cf. Ex. 18, p. 96). This locus should now be constructed 
carefully, enough points being located so that the shape of the 
curve is clear. The locus is called a hyperbola. It would also be 
a hyperbola if we changed the word " twice," in the first sentence 
of this paragraph, to " three times," or " 1 J times," or any other 
number greater than 1. The complete definition of a hyperbola 
is accordingly : 

A hyperhola is the locus of a point whose distance from a fixed 
point is to its distance from a fixed line in a constant ratio greater 
than one. The fixed point is called the focus, tlie fixed line the 
directrix, and the constant ratio the eccentricity. 

The student should now construct carefully several hyperbolas, 
using different values of the eccentricity, — for instance, e= 3, e = 4, 
e = 11, e = l^,e = 5. Notice how the shape of the curve is affected 
by increasing or diminishing the eccentricity'. (In drawing these 
curves be sure to take account of the whole of the locus, and 
not only of part of it.) 

100. Equation of the hyperbola. Since the definition of the 
hyperbola differs from that of the ellipse only in the fact that the 
eccentricity is greater than one instead of being less than one, 
the equation of the hyperbola can be obtained in exactly the same 
way. The details are left to the student to carry out, and care 
should be taken that the figure drawn corresponds to the facts. 
For instance, the points A and A' will now be on opposite sides 
of the directrix; but we obtain, exactly as for the ellipse, the 

facts that 

0B = -, OF=ae 
e 

(using the same letterings as for the case of the ellipse. Fig. 83). 
That is, the coordinates of the focus are (ae, 0), and the equation 

of the directrix is a; = - (do not overlook the bearing of the fact 
that e>l). ^ 



132 THE ELEMENTARY FUNCTIONS 

The equation of the hyperbola takes accordingly the same form 
as that of the ellipse, in (9), p. 124, namely, 
a,'2(l-e2) + 2^2 = a2(l-«2). 

If we wish to abbreviate this, however, as we did in the case of 
the ellipse, we cannot do as we did before and put V^ = a^{l— ^), 
because 1— e^ is negative in the. case of the hyperbola. We may, 
however, let 6^ = a^ (g2 _ \\^^ \^ which case our equation takes the 
form _ w 

_^a^ + y2^_J2. 

a'' 
that is. ^"fe^"^" ^^^ 

This is the simplest form of the equation of a hyperbola, and it 
may be considered as a standard form for the equation of the 
curve. By analyzing the equation we can obtain several impor- 
tant properties of the hyperbola. 

101. (1) Solving (1) for y as an explicit function of x, we obtain 

y = ±- V'a;2 _ (i\ This shows that x^ cannot be < a^ ; for if it 
a 

were, y would be complex, thus giving no point on the curve. This 
means that no point on the curve lies between the points where 
x = a and where x = — a (that is, between the points A and A'). 
The points A and A' are called the vertices of the hyperbola. For 
any value of x> a or <— a there are two values of y, numeri- 
cally equal but of opposite sign, so that the curve is symmetrical 
with respect to the X-axis. Since this axis crosses the curve (at A 
and A'), it is called the transverse axis. As x increases indefinitely, 
the positive value of y also increases without limit; hence the 
curve extends indefinitely far from both the X- and the Y-axis. 

(2) Solving equation (1) (§ 100) for x as an explicit function 
of y, we get ^ , « ,/ a . m 

h ^ 
This shows that for every value of y there are two values of x, 
equal numerically but of opposite sign ; hence the curve is sym- 
metrical with respect to the F-axis. This axis of symmetry is 
called the conjugate axis of the hyperbola, and it does not meet 



THE PARABOLA, ELLIPSE, AND HYPERBOLA 133 



the curve at all, since x cannot equal 0. The length B'B = 2 6 is 
called the length of the conjugate axis, while A' A = 2 a is called 
the length of the transverse axis. The point of intersection of the 
transverse and conjugate axes is called the center and is the mid- 
point of every chord drawn through it (cf. footnote, p. 125). 

(3) The relation among the three quantities a, b, and e, namely, 
62 = a2 ^g2 — 1) = a2g2 _ ^2^ jjjj^y 133 easily remembered by noting 
that in a right triangle with a and b as the two legs, ae will be the 
hypotenuse. Thus, in 
Fig. 89 the triangle 
OAC has OA=a, 
AC=b, and hence 
OC=ae. This fact 
enables us to find the 
focus of a hyperbola 
when the transverse 
and conjugate axes 
are known. 

(4) The chord pass- 
ing through the focus, 

perpendicular to the transverse axis, is called the latus rectum. 

2 &2 
The student may show that its length is 

(5) As we have already noticed (pp. 87, 90) in the preliminary 
study of functions whose graphs were hyperbolas, we may expect 
to find that a hyperbola has two asymptotes intersecting at the 
center of the curve. We have not yet seen how to locate them 
exactly, however. This is accomplished by the following 

Theorem. The equations of the two asymptotes of the hyperbola 
— = i are obtained by replacing the 1 by in the equation 




Fig. 89 



rfi 



of the curve, thus : 

^~&2 

Proof. The graph of the equation ^ - tj = {- + r I ( " ~ i ) = ^ consists 
of the two straight lines 



- + ^ = and 
a b 



y 



= 0. 



(Ex. 14, p. 110) 



134 THE ELEMENTARY FUNCTIONS 

These are the lines through the origin with slopes and - respectively. 

(Note that the latter line is in fact the line OC of Fig. 89, of course pro- 
duced indefinitely.) To prove that the line - + j = (that is, bx + ay = 0) 
is an asymptote to the hyperbola — — ^ = 1, it is sufficient to show that 

as the point (x^, y,) recedes indefinitely along the curve, the distance from 
the line bx + ay = to the point (ar^, j/j) approaches more and more 
closely. Now the distance from the line bx + ay = ix> the point (Xj, y^) is 

^ ^ b^ i + aVi . (Formula (4), p. 106) 

By the equation of the hyperbola, 

a2J2 
that is, ftij + ayi - 



a^b"^ 
Replacing bx-^ + ay^ in d by 



d = 



aW 



(6xi - ay^) Va2 + J^ 

Since the line Ja; + a^ = passes through the second and fourth quad- 
rants, we are concerned with points {x^, ^j) for which either x^ < and 
y^ > 0, or else Xj > and y^< 0. In either case, as Xj and y.^ increase in- 
definitely in numerical value, bx^ — ay^ increases indefinitely also, and 
hence d approaches 0. This proves that the line bx + ay = Q is an asymp- 

tote to the hyperbola ~; ~ f^ = 1- The proof for the line bx— ay = is 

similar and should be carried through by the student. 

102. Summarizing the facts obtained from the standard form 

of the equation of the hyperbola, — — ^ = 1, the locus of this 

equation is the hyperbola whose center is the origin, with trans- 
verse axis 2 a, conjugate axis 2 I, vertices (± a, 0), focus {ae, 0), 

directrix a; = - (where aV = a^ + P), and asymptotes — = 0. 

Thus, the equation -- — ^ = 1 has as its locus the hyperbola 

whose center is the origin, transverse axis 4, conjugate axis 2V5, 
and vertices (± 2, 0). aV = ^2 ^ j2 ^ 4 ^ 5 Therefore ae = 3, 



THE PARABOLA, ELLIPSE, AND HYPERBOLA 135 

3 a 4 
e = -, — = -• Hence the focus is the point (3, 0), and the direc- 

2 e 6 . 

4 
trix is the line « = o " '^^^ asymptotes are given by the equation 

— — ^ = 0, and are the lines through the origin with slopes ± -— ■ 

If in deriving the equation of the hyperbola we had chosen the 
line through the focus, perpendicular to the directrix, as F-axis 
instead of as X-axis, the effect would have been merely to inter- 
change X and y m the equation (1), giving 

^-^ = L (2) 

Thus, the equation = 1 has for its locus the hyperbola 

with vertices (0, ± 2), focus (0, 3), directrix y = ^, and asymp- 

totes =0. It is in fact the same hyperbola as that of the 

example above, merely rotated about the origin through an angle 
of 90°. Draw the figure. 

EXERCISES 

Draw the graphs of the following equations, and locate focus, 
directrix, and asymptotes : 

1. -- — ^ = 1. 



4 9 



X' X _ ■ 



2. — -^ = 1. 

16 9 7. 

,2 



6. x' 

x^ 



144 



y^ = 2. 


9. 


4 12 ~ • 


-S-- 


10. 


y'-x'=2. 


11. 


32/"-4a;= = l. 


f-- 


12. 


9 4~-'- 



3. ^ - 2/= = 1. 

* 8 ^ 

4. a;=-92/= = 4. " 4 

Compare the graph of Ex. 12 carefully with that of Ex. 1, and note 
that they have the same asymptotes, the transverse axis of the one 
coinciding with the conjugate axi'S of the other. Two such hyperbolas 
are called conjugate hyperbolas. Draw them both in the same figure. 

13. — — — = 1. What is the equation of the conjugate hyperbola ? 
9 7 

Draw them both in the same figure. 

14. 4:x^ — ^y^ = 12. What is the equation of the conjugate 
hyperbola? Draw both curves. 



136 THE ELEMEKTARY EUKCTIONS 

15. If e and e' are the eccentricities of two conjugate hyperbolas, 
prove that - + — = 1. 

16. Prove that the point (— ae, 0) is also a focus, and the line 
X = is a directrix, of the hyperbola ~^ — f^ = 1- 

*103. Generalized standard equation of the hyperbola. As in 

the case of the parabola and the ellipse, so the simple standard 
form of the equation of a hyperbola can be generalized' to apply 
to any hyperbola with axes parallel to the X- and Y-axes. The 
student should work through the details and obtain the result 
that the equation of the hyperbola whose center is the point 
(a, /3), whose semi-transverse axis is a, and whose semi-conjugate 
axis is & is (^ (y-)g)' , .„. 

if the transverse axis is parallel to the X-axis, and 

if the transverse axis is parallel to the Z-axis. 

The graphs of equations (3) and (4) can be thought of as the 
same curves as (1) and (2) respectively, merely translated from the 
position in which the origin is the center to the position in which 
the point (a, /3) is the center. 

By using (3) and (4) as standard forms the graphs of many 
equations can be recognized and drawn very easily. 

Example. 9 a:^ - 4 j/^ - 18 a; + 24 y - 63 = 0. 

This equation can be reduced to one of the forms (3) or (4) by com- 
pleting the square of the x-terms and of the ^-terms. Hence we write the 
equation 9(^2 _ 2:,) _ 4(^2 - Gy) = 63 

and note that to complete the square of the x-terms we must add 1 inside 
the first parenthesis, and that to complete the square of the ^-terms we 
must add 9 inside the second parenthesis. This gives 

9 (x2 - 2 X + 1) - 4 (2^2 _ e y + 9-) = 63 4- 9 - 36 = 36 ; 
that is, 9 (s _ 1)2 _ 4 (2, _ 3)2 _ 36^ 

or (x-1)^ (y-3)2_, 

4 9 ~ ' 



THE PARABOLA, ELLIPSE, AND HYPEKBOLA 137 

■which is in the form (3), with or = 1, ^ = 3, a = 2, 5 = 3. Its graph is 
accordingly the hyperbola whose center is the point (1, 3), whose transverse 
axis is parallel to the X-axis and of length 4, and whose conjugate axis is 
of length 6. The vertices are accordingly the points (3, 3) and (—1, 3). 
Moreover, c?e^ = a^ + b^ = 13. Therefore ae = v 13. Hence the distance 
from the center to either focus is VlS. Further, 

Vl3 
' = -2-' 

hence - = -4= = ^ ^^IS = 1.11-, 

e Vl3 13 

the distance from the center to either directrix. The asymptotes are the 
lines through the center, that is, through the point (1, 3), with slopes | 
and — |. What are their equations ? With these data, the graph is readily 
drawn. As a check the intercepts should be found. When a; = 0, 4 j/^ — 24 )/ 
+ 63 = 0, an equation which has complex roots (b^ — iac being negative) ; 
hence the graph does not meet the K-axis. When ^ = 0, 9 x'' — 18 a; — 63 = ; 
that is, x2 — 2 X - 7 = 0, I = 1 ± VS = 3.83, - 1.83, approximately. These 
intercepts agree with the figure if it is correctly drawn. 

EXERCISES 

Draw the graphs of the following equations, and locate vertices, 
foci, directrices, and asymptotes. 

1. x'-2/''-2x + 8y-3 = 0. 

2. x'- 2 if +10^ = 0. 

3. ix^-f + 8x-2i/-l = 0. 
i. x' - 5if + 6x -lOy =: 0. 

6. 3x^-2/^+12a; + 2y+14 = 0. 
6. 2x^-3f + x + ij+10 = 0. 

PROBLEMS ON THE HYPERBOLA 

1. Find the equation of a hyperbola, given 

(a) foci at (3, 0) and (— 3, 0) and directrix x=l. 

(b) foci at (0, 2) and (0, — 2) and directrix y = i- 

(c) transverse axis = 3, foci (2, 3), and (— 2, 3). 

(d) vertex (4, 0), asymptotes y = 3x, and y = — 3x. 

2. Find the equation of the locus of a point which moves so that 
the difference of its distances from the points (6, 0) and (— 5, 0) is 
equal to 6. What kind of curve is this locus ? 



138 THE ELEMENTARY FUNCTIONS 

3. Find the equation of the locus of a point which moves so that 

the difference of its distances from the points (c, 0) and (— c, 0) is 

constantly equal to 2 a. (a <. c) x' 1/^ _i 

Arts, —z ^ 5 — 1. 

a^ r — a' 

This result, being the equation of a hyperbola, establishes the 
following 

Theorem. 27i6 locus of a point which moves so that the difference 
of its distances from two fixed points is a positive constant less 
than the distance between the points is a hyperbola having the fixed 
points for foci and the constant distance for its transverse axis 
(cf. Ex. 23, p. 96). 

4. Prove the theorem of Ex. 3 by showing that the distances 

from any point (x^, y^ on the hyperbola -;; — ^ = 1 to the foci are 

eXj + a and ex^ — a, so that their difference is the constant 2 a. 

This theorem suggests a mechanical construction of a hyper- 
bola by continuous motion, thus : fasten pegs or thumb tacks at 
the foci, then pass around both a string whose ends are held 
together. If now a pencil 
point be fastened at P and 
both ends of the string 
be pulled down together, 
the point P will move 
along an arc of a hyper- 
bola, because PF'-PF 
will remain" constant. Fie. go 

5. What is the eccentricity of a hyperbola in which a = b? Such 
a hyperbola is called equilateral. 

6. What is the angle between the asymptotes of an equilateral 
hyperbola ? 

X^ 11^ 

7. Show that the foci of the hyperbola -i; — 7; =1 and those of 

a^ b' 

its conjugate all lie on the circle whose equation is x^ + y^=a' + h^. 

8. Show that the circle of Ex. 7 meets either of the two hyper- 
bolas on the directrix of the other. 




THE PAEABOLA, ELLIPSE, AND HYPEEBOLA 139 

9. The latus rectum of the hyperbola -^ — ^ =1 is extended by 

the amount k so that it just reaches the asymptote. Prove that k is 
equal to the radius of the circle inscribed in the triangle formed by 
the asymptotes and the line x = a. 

10. Prove that the foot of the perpendicular from a focus of a 
hyperbola upon an asymptote lies on the directrix corresponding to 
that focus and also upon the circle described upon the transverse axis 
as diameter, and that its length is equal to the semi-conjugate axis. 

11. The distance of any point of an equilateral hyperbola from 
the center is the mean proportional to its distances from the foci. 

12. Through any two points P and Q of a hyperbola, lines are 
drawn parallel to the asymptotes, forming the parallelogram PRQS. 
Prove that the diagonal RS passes through the center. 

13. If a straight line cuts a hyperbola at the points P and P', and 
its asymptotes at R and R', prove that the mid-point of PP' will also 
be the mid-point of RR'. 

14. If a point moves along a hyperbola, the product of its distances 
from the two asymptotes remains constant. 

104. The curves that have been studied in this chapter and 
the preceding — namely, parabola, ellipse, hyperbola, and circle — 
are called conic sections, or simply conies, because they can all be 
obtained as plane sections of a circular cone. They were origi- 
nally studied from that point of view, and nearly all their ele- 
mentary properties that are known to-day were proved by the 
Greek geometers more than two thousand years ago.^ The conic 
sections are of especial interest because of the fact that the 
paths of all the heavenly bodies are curves of this kind. This 
fact was first established by the great German astronomer and 
mathematician Johannes Kepler, in 1609. He showed that the 
planet Mars moves in an ellipse; the other planets, including 
of course the earth, do the same, while many comets move in 
parabolas or hyperbolas. 

1 The most complete study of the conic sections among the Greeks was made 
by Apollonius of Perga, about 200 b.c. 



140 THE ELEMENTARY FUNCTIONS 

105. From the point of view of their equations it will be noted 
that all these curves have one thing in common : the equation of 
every conic section is of the second degree in x and y ; that is, when 
cleared of fractions and radicals and reduced to its simplest form, 
each of our standard equations has been of the second degree. We 
cannot as yet prove that every conic section must have as its 
equation one of the second degree, but this is a fact, and the proof 
of it will be possible at a later stage of the mathematical course. 
The converse statement, that every equation of the second degree 
in X and y has for its locus a conic section, is not true, because 
many equations of the second degree have no locus at all.- If there 
is a locus, however, it must be either a conic section or something 
simpler — a pair of straight lines (as a^— 3/^= 0), a single straight 
line (as o?—1xy + y'^ = ^), or a point (as a^ + 2/^^=0). This asser- 
tion also will not be proved here. 

The most general form of equation of the second degree in x 

^ ao^ + hxy + ey"^ + dx + ey +/ = 0, 

where at least one of the coefficients a, h, c is different from 0. 
The student should show how to get each one of the standard 
forms of equation of the conies by giving special values to a, i, 
c, d, e, and /. For example, the circle a^ + y^ = r^ is obtained if 
we put a = l, h = 0, c = l, d = e = 0,f = —7^. 



CHAPTER VIII 

SIMULTANEOUS EQUATIONS 

106. Review questions. Give the standard forms of the equa- 
tions of straight line, circle, parabola, ellipse, and hyperbola. How 
are the asymptotes of a hyperbola determined? How can the 
coordinates of the point of intersection of two straight lines be 
found ? of a straight line and a parabola ? 

107. This last question brings us to the problem of this chap- 
ter, which is, to find the points of intersection of a straight line 
and any conic section, or the points of intersection of two conies. 
Some important geometric results can then be obtained, based on 
the solutions of these problems. As a preliminary step the prob- 
lems on page 45 should be reviewed, that there may be no difficulty 
in the algebraic solution of a quadratic equation in one unknown 
quantity. 

108. Intersection of straight line and conic. The same method 
which was used in Chapter III, to find the intersections of a straight 
line and a parabola, can be used in the case of a straight line 
and any conic section. An example will make this clear. 

Example. Find the points of intersection of the straight line 

X + 2/ = 2, (1) 

and the conic i x^ + y'' = i. (2) 

Making the graphs of (1) and (2), we see that their intersections are the 

points A and C (Fig. 91). A is evidently (0, 2), and C is not far from ( j, 1 J). 

To determine algebraically the exact values of the coordinates of A and C 

(A we have found exactly, because (0, 2) satisfies both equations (1) and (2), 

but (|, j) does not satisfy the equations) we obtain from (1) the value of y 

as a linear function of x, thus : „ 

y = '2-x. 

Substituting this value of y in (2), 

4 x2 + (2 - rf = 4. (3) 

141 



142 



THE ELEMENTARY FUNCTIONS 



Since (3) has been obtained by using hoth (1) and (2), the values of x which 
it determines will be the abscissas of points lying both on (1) and on (2) ; 
in other words, the roots of equa- 
tion (3) are the abscissas of A 
and C, the points of intersection 
of (1) and (2). 
Simplifying (3), 



5 a;2 - 4 a; = 0. 
3;(5z — 4) = 0. 
Therefore a; = or 4- 



(4) 




Hence the abscissas of A and C, 

the points of intersection of the 

straight line and the ellipse, are 

and I respectively. Sincex + !^ = 2, ^^^ g^ 

X = corresponds to y = 2, and 

I = I to y = f . Therefore (0, 2) and (5, f) are the points of intersection. 

Evidently this method can be used in any case where it is 
desired to find the points of intersection of a straight line and a 
conic. We may solve the linear equation for y as a function of x 
(or for a; as a function of y) and substitute in the equation of 
the conic, getting a quadratic equation in the one variable x (or y), 
whose roots will be the abscissas (or ordinates) of the required 
points of intersection. The other coordinate is then found by sub- 
stituting in the linear equation the value of the coordinate that 
has been found. (The equation of the conic should not be used 
for this last substitution, because it would usually give two values 
where only one is correct. Thus, in the example above, if we had 
substituted a; = in the equation of the ellipse, we should have 
found y = + 2 or — 2, whereas y = + 2 is the correct value.) 



EXERCISES 

Solve both graphically and algebraically the following pairs of 
simultaneous equations : 

ra;^ + 2/' = 25, ^ fy^^Sx-Sx' + T, 

0. 



rx' + f = : 
\x — y=l. 



y = 2x'~3x-i, 



ry = ^x-~ 
\y-x = 2. 



ry — ax — ox-\- 
' i3a;-2y + 5 = 



4. 



25, 



l2a- -y = 4. 



SIMULTANEOUS EQUATIONS 


1-3x^+22/^=11, 
" U-3y=7. 


10. 


36 20 ' 


1. ] 25 ^ 9 ' 
.2x-2/=14. 


11. 


.a; — y = 4. 

rx-3y=l, 
a;y + 7/2 = 5. 


r3x''+16y2==192, 
\x + 2y=lQ. 


12. 


•4y = 5x+l, 
l2xy = 33-a;l 


■ \x-2y+l=0. 


13. 


r7a;'^-8jci/=159, 
\hx-\-2y=l. 


rx'^-2tf = i, 
' 3x-2y=10. 


14. 


ra;2/ + 3 2/^ = 42, 
l2x + 2/=13. 



143 



8. 



109. Tangent to a Conic. Just as in the case of the parabola 
(p. 49), so here for any conic, the sign of the discriminant of the 
quadratic equation corresponding to (3), p. 141, enables us to tell 
whether the straight line (1) and the conic (2) have two points 
in common, or only one, or none. In case they have only one 
common point (which happens when the discriminant equals 
zero), the line will, in general, be tangent to the conic. 

Example. For what values of c will the line 

Sx + 'i:y = c (1) 

be tangent to the curve x^ + y'^ = 251 (2) 

c — S X 
Solving (1) for y, y = — 

Substituting this value of y in (2), 

that is, 25x2-6cx + c2-400 = 0. . (3) 

The roots of (3) are the abscissas of the points where (1) meets (2). If 
the line is tangent, these abscissas must be equal, and this requires that the 
discriminant of (3) shall equal zero. The discriminant of (3) is 

D = (- 6 c)2 - 4 • 25 (c2 - 400) 

= 36 c" - 100 6-2 + 40,000 = 40,000 - 64 c\ 
If 2) = 0, 64 c" = 40,000, 

c' = 625. 
Therefore c = ± 25. 



144 THE ELEMENTARY FUNCTIONS 

Hence the line Sx + iy = 25 or 3x + 4y = — 25 will be tangent to the 
circle (2). (Verify this by a figure.) If c^ > 625, that is, if c > 25 or < - 25, 
the value of the discriminant will be neyatioe, and the straight line will not 
meet the circle at all. If c^ < 625, that is, if — 25 < c < 25, the value of 
the discriminant will be positive, and the straight line will meet the circle 
in two distinct points. Each of these possibilities should be illustrated 
by a figure. 

EXERCISES 

Determine for what values of k the following lines and conies will 
be tangent, and discuss the values of k that will give intersection or 
nonintersection of the two graphs : 

' [3y = ix + k. ' \x + 2i/ = k. ' \kx + y = 3. 

rx'-,f=9, \^,m1 = , 8 (2x'-3y'=5, 

l5x-42/=A-. 5. 36"^25 ' Ux + h/=5. 

3. 16 "'"25 ' jy^=4x, 9. ■! 9 4~' 

[kx + 4:y=20. ^'\t/ = 2x + k. [6x-kif=9. 

10 r2/' = 4x + 8, ^^ r4:a^ + 9f=36, 

' \x + ky = 2. ' \tj=ix + k. 

12/ = mx + k (th a, fixed number). 

ry^ = 4 ax (a a fixed number), a 

\y = mx + k (m a. fixed number). ' m 

This result means that the line ?/ = mx -\ is tangent to the 

m 

parabola y" = 4 ax for any given value of m (not 0). 

14. Eind the coordinates of the point of contact of the tangent 

line of Ex. 13. / a 2 a\ 

Am. ( — , — I- 
\nr m I 

15. Find the F-intercept of the tangent line of Ex. 13. Draw a 
conclusion from the results of the preceding exercise and this one 
(cf. Exs. 6, 6, p. 52). 

16. Find the Z-intercept of the tangent line of Ex. 13. Compare 
this with the abscissa of the point of contact. 

17. Use the results of the preceding problems to find the equation 
of the tangent to the parabola y'^ = 6x at the point (f , 4) ; at the 
point (6, 6); at the point (x^, y^. 



SIMULTANEOUS EQUATIONS 



145 



18. Prove that the tangents to the parabola y^ = 4 ax at the ends 
of the latus rectum are perpendicular to each other and intersect on 
the directrix. 

19. Give a geometric construction for the tangent to a parabola 
at any given point. 

20. Discover a construction for the two tangents to a parabola 
from an external point. 

21. The normal to a curve means the perpendicular to the tan- 
gent, drawn through the point of contact. Find the equation of the 
normal to the parabola y" = 4: ax, at the point (x^, y^). 

22. Prove that the subnormal is constant for all points on the 
parabola ■i^ = iax. (The subnormal is the distance from M, the foot 

Y 




Fig. 92 



of the perpendicular from the point of contact to the Z-axis, to N, 
the intersection of the normal with the X-axis. See Fig. 92.) 

23. Prove that the tangent to a parabola bisects the angle between 
the focal radius at the point of contact and the line parallel to the 
axis through the same point. 

24. For what values of k will the line y = mx + k (m a fixed 
number) be tangent to the ellipse 4 a;^ -|- 9 y^ = 36 ? 

x^ y^ 

25. Answer the same question for the ellipse -^ + Ti = 1 and the 

line y = mx + k{ma. fixed number). {Ans. k = ± Vs^ + aW.) This 
result shows that the \mey = mx ± VPT^Wis tangent to the ellipse 

2 2 * 

?- + ^ = 1 for any given value of m. The double sign shows that 
a^ b^ 

there are two such tangents, that is, that there are two tangents with 
any given slope m, — a fact which is geometrically self-evident. 

26. Find the coordinates of the point of contact of the tangent 
line of Ex. 25. 



146 



THE ELEMENTAEY FUNCTIONS 



110. Construction of a tangent to an ellipse. The results of 
Exs. 25 and 26 enable us to construct the tangent to an ellipse, thus : 

The tangent line FT 
(Fig. 93) has the equation 

y^mx+^h^ + a^m?. (1) 
This meets the X-axis at 
the point T, whose ab- 
scissa is the X-intercept 
of (1). This is found by 
setting y = in (1), giving 

= mx + y/j)^ -h ahn^, 



or x = — - 



Vz,2 + ahiv^ 




m 



Fig. 9.3 



0T = - 



that is, 

But, by Ex. 26, 0M= 

Therefore OT 



V&2-I-, 



m 

mAtn 



V62. 
0M= a^ 



or 0T-. 



a'' 

X, 



where x-^ = OM = the abscissa of the point of contact. 

This is a very important and remarkable result, because it shows 
that the distance OT is independent of b, that is, does not depend 
upon the minor axis 
of the ellipse. In 
other words, if we 
have a set of ellipses, 
aU having the same 
major axis A' A but 
different minor axes, 
the tangent to any 
one of the ellipses, 
at the point whose 
abscissa is x-^, wUl 
pass through T. If 
b = a, the ellipse be- 
comes the circle with Pig. 94 




SIMULTANEOUS EQUATIONS 147 

A' A as diameter, and hence its tangent (at P', the point whose 
abscissa is x^) is easily constructed. It is the line P'T, perpen- 
dicular to OP' (Fig. 94). The intersection of this tangent with 
the axis A' A produced is the common point T where the tangents 
to all the ellipses meet the axis. Hence we only need to join 
this point T with the point P on the ellipse to get the required 
tangent TP. 

PROBLEMS 

1. In Eig. 93, prove that F'T: FT = F'P : PF, and hence that PT 
bisects the exterior angle of the triangle FPF' This theorem gives 
another construction for the tangent to an ellipse at any point. 

2. Obtain the equation of the tangent to the ellipse -^ + y^ = 1 

SC OC 1/7/ 

at the point (Xj, yj in the form -\ + ^ = 1. 



a 



Hint. It passes through (Xj, y^) and ( — , Oj 



3. Find the equation of the normal to the ellipse -^ + y^ =1, at 
the point (x^, y^. 

4. Find the ■ JT-intercept of the normal of the preceding problem. 

5. Prove that OF, the distance from the center to the focus of the 

a.2 ,,2 
ellipse -^ + y^ = 1, is the mean proportional between the -Y-intercepts 



a' 



V" 



of the tangent and the normal. 

6. State and prove the corresponding theorem for the F-intercepts 
of the tangent and the normal. 

7. Construct the two tangents from an external point to an ellipse. 

A similar set of problems can be worked out for the hyperbola, 
but the results do not show enough difference from those for the 
ellipse to justify stating them explicitly here. This whole subject 
is considered from a higher point of view in the chapter, Introduc- 
tion to the Differential Calculus. 

111. Intersection of two conies. To obtain graphically the 
coordinates of the points of intersection of two conies is a sim- 
ple matter; unfortunately the algebraic method, by which alone 



148 THE ELEMENTARY FUNCTIONS 

we could be certain of obtaining exact results, is not practicable ^ 
except in a comparatively few special types of problem. In each 
of the following exercises the graphical solution will be found 
simple, in some cases even leading to exact results. 

EXERCISES 

Find (graphically) the coordinates of the points of intersection of 
each of the following pairs of conies : 

la 



{ 



ix + if=ll. 
a?-tj= 17, 




112. Solvable type; both equations homogeneous. If we have 
two equations of conies in which every term containing the vari- 
ables is of the second degree, then algebraic solution is always 
possible. Tlie most general form of such an equation is 

ax^ + hxy + cy'^ = d, 

which is called a homogeneous equation of the second degree in 
X and y. When each of the given equations is of this type, we 
can eliminate the constant term and. get an equation of the form 

a' a? + h'xy + c'y^ = 0, 

which can be factored if the points of intersection have rational 
coordinates, and often even when they do not. 

Example. f 2 a;^ - 3 a;y + 5 / = 14, (1) 

[ x2 + 4 a;2, _ 2 2/2 = 19. (2) 

To eliminate the constant term, multiply (1) by 19, and (2) by 14, and 
subtract. The result is 

24 x^ - 113 xy + 123 y'^ = 0. (3) 

1 That is, it would involve more advanced algebraic work than we are yet 
prepared for. 



SIMULTAKEOIJS EQUATIOi^S 149 

Solving (3) for a; as a function, of y, we get 



_ 113 y ± Vl2769 y^ - 11808 y^ 

48 
_ 113y ^VQGly" 

48 
_ 113yi:31y 
48 

= 3y or |l y. 

(Equation (3) could of course have been factored, thus : 

24 x' - 113 x^ + 123 ;/•- = (x - 3 ?/) (24 a: - 41 y) = 0. 
Therefore a: = 3 2^ or |1 y.) 

(a) Using a; = 3 y, we get, from (1), 

2(3 3,)2-3j,(3y) + 52/^ = 14, 
14 / = 14, 

2/^ = 1, 
y = ±l. 

Since a; = 3 y, when y = 1, x'=%, and when 3^ = — 1, x = — 3. Check these 
pairs of values by substituting them in both (1) and (2). 

(b) Using x = %^y, we get, from (1), 

2(ti3^)''-3y(li3r)+5 2/>' = 14, 



V235 23o 

Since ^ = ti ^> 

x = ±^\V2m. 

The graphical solution is not so simple as the algebraic one in 
this case, because (on account of the xy term) the equations can- 
not be reduced to any of the standard forms of the equations of 
conies. To draw the graphs we shall therefore have to compute 
the coordinates of enough points so that the form of each curve 
becomes clear. Solving (1) tov y as, a, function of x, 



_3a;±V9a^-20(2a!2-14)_3a;±V280-31a;2 
^~ 10 ~ 10 ' 



150 THE ELEMENTAEY FUNCTIONS 

Using this, we get the table of values as follows : 



X 





1 


2 


3 


±2.6+ 


y 


±1.7 


1.8 or - 1.2 


1.8+ or - 6+ 


1 or .8 






Negative values of x give the same results but with opposite sign. 
It is evident from the value of y above that x cannot be > S'*'- 
The graph is an ellipse. 

r 




Fig. 95 



Similarly, from (2), 



y = 



4a;±Vl6a^-8(19-a^)_2a:±V6a^-38 



The table of values is as follows: 



X 





1 


2 


3 


4 


±4.3+ 


y 


complex 


1 or 5 


.2 or 7.8 






The graph is a hyperbola. Fig. 95 shows the graphs of (1) and (2), 
their four intersections corresponding to the algebraic solution. 
The straight lines given by (3) (p. 148) are also shown. 



SIMULTANEOUS EQUATIONS 161 



EXERCISES 


Solve for x and y, checking 


graphically where practicable : 


^_ (x^-f=l, 

\j)i? — xy + y^ =1. 


f 2 2/^ -4x2/ + 3x^=17, 


Ixy + y' = A. 


Cx^-xy-f = 5, 
■ 12x^ + 3x2/ + 2/^ =28. 


Cx' + 2xy-if=li, 


rx^-x2/ = 35, 
• 1x2/ + 2/^ =18. 


f x^ + xy + 2 y2 = 44, 


rx^-x2/ + 2/^ = 21, 
• V- 2x2/ =-15. 



*113. Algebraic solution of some equations of higher degree. 

In the case of simultaneous equations, one or both of which is of 
higher degree than the second, it is only in special cases that alge- 
braic solution by elementary methods is possible. In some cases, 
however, it is so very simple that it is worthy of consideration. 
As for the graphical solution, we shall neglect it entirely, because 
it involves in nearly all cases too difficult a process of computing 
tables of corresponding values of x and y. 

*114. The following illustrative examples should be carefuUy 
studied, until the general methods used are well understood. It 
will be noticed that in each case the given equations are combined 
in such a way as to lead to a linear and a quadratic equation, 
which pair can then always be solved by the methods with which 
we are already familiar. 

Example 1. x' + y^ = 133, (1) 

x + y = 7. (2) 
Here we notice that if (1) is divided by (2), we obtain a quadratic equa- 
tion. In fact, division gives ^^ - xy -V y"^ = 19. (3) 
We may now solve (3) with (2) as usual, or we may proceed thus : 
Squaring (2), x' + 2xy + y'^ = 49. (4) 
Subtracting (3) from (4), Zxy = 30. 

xy = 10. (5) 

Multiplying (5) by 4 and subtracting from (4), 
x^ — 2xy + y^ = 9. 

X - 2/ = ± 3. (6) 



152 THE ELEMENTAEY FUNCTIONS 

Adding (6) and (2), 2 a; = 7 ± 3 = 10 or 4. 

Therefore a: = 5 or 2. 

Therefore 2/ = 2 or 5 (since x -\- y = 7). 

Hence the solutions are (5, 2) and (2, 5). 

Another method is to cube (2) and subtract (1) from the result. This 
gives ^xy(x + y) = 210, and hence 3 xy = 30, xy = 10, since x + y = 7. From 
here we can proceed as usual, substituting the value of x (or y) from (2) in 
the equation xy =\0\ or we can continue as from equation (5) above. 

Example 2. x^ + y* = 706, (1) 

a- + y = 8. (2) 

Raising (2) to the fourth power, 

a;* + 4 x^i/ + 6 xh/ + 4 a;y^ + j/* = 4096. 

Subtracting (1), ^x^y + Q xV ^^xy^ = 3390. (3) 

Squaring (2) and multiplying by 4 xy, 

4 x^y + 8 xhf + 4 x!/S = 256 xy. (4) 

Subtracting (3) from (4), 2 xV = 256 xy - 3390. 

xy - 128 xy + 1695 = 0. 

{xy - 15) {xy - 113) = 0. (5) 

We can now use (5) and (2) together, getting (5, 3) and (3, 5) as the only 
real solutions. 

Example 3. x< + x^ + y* = 481, (1) 

x''-\-xy + y^ = 37. (2) 

Dividing (1) by (2), x'' - xy + y^ = 13. (3) 

Subtracting (3) from (2), 2 xy = 24. 

xy = 12. (4) 

We can now obtain (x + yy and (x — yy by using (4) with (2) and (3) ; 
the remainder of the work is left to the student. 

EXERCISES 

Solve for x and y, and check : 

' \x + y=5. ■15x2 + 6 2/^=50. 

2 [»' + / = 72, ra;^ + y2 + ^ + 3/=36, 

' \x + y=&. • Xxy = 10. 

\a;2/(a; + 2/) = 30. ' \xy — 4.. 



SIMULTANEOUS EQUATIONS 153 

la; + 3/ = 4. ' Ixy (x — y) = 30. 

g f a;s - y5 = 3093, ^^ f x^ - 2/^ = 14 - X, 

■ \x-y = 3. ' \f=^2x + 6. 
p-a;y + / = 7, f a;^ + ^^ + 2^^ = 19, 

■ Ix* + x'y' + if = 133. • la: + xy + y = 1. 

13. The sum of two numbers is 28, and their product is 147. 
Find the numbers. (Of. also Exs. 3 and 4, p. 55.) 

14. The product of two numbers is 180, and their quotient is #. 
A^'hat are they ? 

15. The diagonal of a field is 89 rd. long, and another field which 
is 3 rd. less both in length and in breadth has a diagonal 85 rd. long. 
What are the dimensions of each field ? 

16. The diagonal of a rectangle is 68 cm., and if the length were 
increased by 2 cm. and the breadth diminished by the same amount,' 
the area would be diminished by 60 sq. cm. Find the dimensions. 

17. A sum of money and its interest for one year amount to 
fl3,520. If the sum is increased by |200 and the rate of interest 
by ^%, the amount will be |13,794 for one year. Find principal 
and interest. 

18. The fore wheel of a carriage turns in a mile 132 times more 
than the rear wheel, but if the circumferences were each increased 
by 2 ft., it would turn only 88 times more. Find the circumference 
of each. 

19. A merchant buys a certain amount of wheat for |322. The 
market goes up 3^, and he can then for |323 get 10 bushels less than 
he could before for |322. What was the price per bushel ? 



CHAPTER IX 

FURTHER STUDY OF THE TRIGONOMETRIC FUNCTIONS. 
POLAR COORDINATES 

115. Review questions. Define the trigonometric functions. 
Give their signs in each of the four quadrants. State four rela- 
tions among the functions of an angle. What are the values of 
the functions of 180° - 6 and of 90° + ^ in terms of functions 
of 6 1 Hov? can a right triangle be solved ? How is the resultant 
of two forces found ? Give the slope of the straight line joining 
the points {x^, y^) and {x,^, y^. What is the slope of the line 
^a; + my + m = ? State the relation between the slopes of two 
perpendicular lines ; of two parallel lines. 

116. Chapter IV included a few of the simplest applications of 
the trigonometric functions. In this chapter we shall take up some 
other problems m which they are used, and also study their graphs. 

117. Solution of the oblique triangle. We saw in Gliapter IV 
how to solve any right triangle. Inasmuch as any oblique-angled 
triangle can be divided into two right triangles by drawing an 
altitude, the unknown parts can be found without the use of any 
new principles. A more practical method, however, is obtained if 
we discover the relations that exist among the sides and angles 
of the oblique triangle itself, thus avoiding the necessity of solving 
the two right triangles separately. Moreover, the study of these 
relations and of many others involving the trigonometric functions 
has very great value for its own sake and on account of its wide 
application in mathematical work of a more advanced character. 

I. The Law of Sines 

118. The following discussion applies only to Fig. 96, (a), whicli 
represents an acute-angled triangle. 

Draw the altitude CD, thus forming the right triangles BCD 
and ACD. 

154 



THE TRIGONOMETfilC FUNCTIONS 



155 



h 



In the triangle A CD, 

Therefore 

In the triangle BCD, 

Therefore 

From (1) and (2) it follows that 

a sin /S = & sin a, 

or, dividing by sin a sin/3, 

a h 



— = sm a. 



h = b sin a. 

h . a 

- = sm p. 
a 

h = a sin yS. 



(1) 

(2) 
(3) 
(4) 



sin. a sin /3 

If we draw the altitude from B, we shall get, in the same way 

a c 



Therefore 



sm a sm 7 
a b c 



(5) 
(6) 



sin a sin j3 sin y 

This extremely important relation is known as the Law of 





(a) 



Pig. 



(6) 



Sines. It may be stated in words as follows : In any triangle 
the sides are pro20ortional to the sines of the opposite angles. 

119. We have proved only that this law is true for acute- 
angled triangles ; in Fig. 96,(&), where the triangle ABC is obtuse- 
angled (a being the obtuse angle), we see that the altitude CD falls 
outside the triangle. Accordingly, 

- = sm ZJ»^C = sin (180°- a) = sin a (p. 69). 
b 

"With this hint the student may complete the proof for himself, thus 
establishing the truth of the Law of Sines for any oblique triangle. 



156 



THE ELEMENTARY FUNCTIONS 



II. The Law of the Peojections 

120. Let the projections ^ of the sides a and h of the triangle 
ABC upon the side c be ^ and q respectively ; then (Fig. 97, (a)) 

p = a cos /S and q = h cos a. 
Since p -\-q = c, 

c = a cos ^ + & cos o. (7) 

By drawing the altitude upon the side I we get, in the same way, 
6 = a cos y + c cos a, (8) 

and by drawing the altitude upon the side a, 

a=bcosY + ccosfl. (9) 

In the obtuse-angled triangle (Fig. 97, (6)) the side c is not the 
sum, but the difference, of the projections p and q ; but equation (7) 

c c 





(a) 



Fig. 97 



(6) 



is true in this case also, because q=h cos (180°— a) = —h cos a (p. 69). 
The details should be worked out by the student. 



III. The Law of Cosines 

121. If we apply the Pythagorean Theorem to the right triangle 
BCD (Fig. 97, (a)), we have 

a^=p^ + h\ (10) Ar 

Eeplaciug p by its value 

a^ = (c- qf + W 

= c'-2cq + q^ + h\ (11) p,^.98 

1 The projection of a line segment AB upon a line CD is the segment A'B, 
between the feet of the perpendiculars from A and B upon CD (Fig. 98). 



THE TEIGONOMETRIC FUNCTIONS 157 

But <f'+'h? = &2 

Therefore a? = c^ - 2 c^ + J^. (12) 

And, finally, g' = J cos a. 

Therefore tf = 6^ + c" - 2 6c cos a. (13) 

By starting with the triangle ACD instead of the triangle BCD 
we obtain, in exactly the same way, 

6^ = a'' + c" - 2 ac cos ^. (14) 

And by drawing the altitude from ^ or £ we get, likewise,^ 
c" = a" + 6" - 2 a& cos y. (15) 

Equations (13), (14), and (15) are known as the Law of Cosines. 
This law may be stated in words as follows : 

The square of any side of a triangle is equal to the sum, of the 
squares of the other two sides, minus twice their product times the 
cosine of the included angle. 

As in the case of the other theorems of this section, the student 
should carry through the proof for Fig. .9 7, (b), and show that the 
same relation holds true in an obtuse-angled triangle. 

122. By the aid of Theorems I-III it is possible to solve any 
oblique triangle without first drawing an altitude and solving the 
auxUiary right triangles thus formed. The student should state 
what Laws I, II, and III give when apphed to a right triangle. 

Example 1. Given a = 36°, ji = 69°, a = 35 ft., to find y, h, and c. 
y = 180° - (a + /3) = 75°. 
h a 



Using the Law of Sines, 



sinjS sin a' 



XI. ^ . 35 -sin 89° 35-0.9.836 ,, _. 

therefore b = . = ^ .„„„ = 55.59 . 

sindD O.ooYO ^=^= 



Again, 



c a 



sm y sm a 
a sin 7 35-0.9659 



^, , asmy do - u.aooa ^ ^. 

therefore c = —. 1- = — t-^t^t;^— = 57.54. 

sm a O.OOYO =^ 

Check = — - — '■ — = '- Hence the results obtained 

■ sin 13 sin y sin 69° sin 75° 

are verified. 

1 Equations (14) and (15) are the same formula as (13), only expressed in 
different letters ; it is logically correct to derive them from the latter by merely 
changing the letters suitably. 



158 THE ELEMEISITARY FUNCTIONS 

Example 8. Given a = 15, b = 20, y = 51°, to find c, a, and ^. 
Using the Law of Cosines, c'^ = a^ + V^ — 2 ab cos y 

= 225 + 400 - 600 cos 51° 

= 247.42. 
Therefore c = 15.73. 



To find a, use the Law of Sines, thus : 

sin a _ sin y 

n c 

mv, jr 15 sin 51° „„„ 

Therefore sin a = — — = .7410. 

Hence a = 47° 50'. 



SUuilarly, sin^ ^ siiiy 

Therefore sin 3 = ^ V° '"'^° = -SSSO- 

15.73 

Hence ^ = 81° 10' . 

Check. a + l3 + y = 180°. 

EXERCISES 
Solve the triangles in Exs. 1-16, drawing an accurate figure for 
each and checking both by measurement and by computation : 



1. 


a = 7 in., 


13 = 68°, 


y = 54°. 


2. 


a =16 ft., 


a = 80°, 


/? = 37°. 


3. 


a = 62 ft.. 


b = 55 ft., 


y = 42°. 


4. 


a = 8.6 ft., 


c = 11.3 ft.. 


;8 = 70°. 


5. 


b = 21.2 ft., 


y = 71°, 


/3 = 60°. 


6. 


a = 5 in.. 


6 = 6 in.. 


c = 7 in. (Use the Law of Cosines.) 


7. 


a = 13, 


6 = 15, 


0=16. 


8. 


a = .82, 


6 = .59, 


c = .71. 


9. 


y = 38°, 


6 = 14, 


c = 20. 


10. 


/3=19°, 


y = 83°, 


6 = 35.2 ft. 


11. 


a = 15, 


a = 150°, 


/3 = 12°. 


12. 


6 = 26, 


/3 = 16°, 


y = 125°. 


13. 


a = 21, 


6 = 35, 


y = 110°. 


14. 


6 = 17, 


r = 19, 


.; = 115° 


15. 


« = 12, 


6 = 14, 


r = 22. 


16. 


a = 1.6, 


6 = .3, 


c = 2. 



THE TEIGONOMETEIC FUNCTIONS 



159 



17. Prove that the area of any triangle is equal to ^ab sin y, or 
^ ao sin j8, or ^ be sin a. 

18. Derive the Law of Sines by circumscribing a circle about a 

a b c , 

where 



triangle ABC and showing that 2 R 
R is the radius of the circle. 



sin a sin fi sin y 



Hint. In Fig. 99, 



ZI) = a. (Why ?) 
Z CB2> = 90°. 



Using the right triangle GBD will lead to the required result. Or use Fig. 100, prov- 
ing that Z BOD = a and then using the right triangle BOD to establish the result. 





Fig. 



Fig. 100 



19. Use the Law of the Projections to derive the Law of Cosines 
algebraically. 

Hint. Multiply the equations (7), (8), and (9) by a, b, and c respectively, 
and combine the three resulting equations by addition and subtraction. 

123. In general three independent data are sufficient to deter- 
mine, and hence to solve, any triangle (but observe that the three 
angles are not three independent data). There are four possible 
combinations of sides and angles that are essentially different, 
and it may be found convenient to regard them as four " cases," 
or groups of data, for the solution of a triangle. The four are 
as follows : 

Case I. Given two angles and one side. 

Case II. Given two sides and the included angle. 

Case III. Given the three sides. 

Case IV. Given two sides and the angle opposite one of them. 



160 THE ELEMENTARY FUNCTIONS 

It will be noted that examples of each of these possible com- 
binations have occurred in the exercises on page 158. Before read- 
ing farther the student should make constructions of triangles 
from given parts, according to each of the first three cases. In 
each case, at least one example of an acute-angled triangle and 
one of an obtuse-angled triangle should be taken. 

124. The "ambiguous case" in solution of a triangle. In 
Case IV appears a slight difference from the other three cases. 
The construction is, to be sure, equally simple. For instance, if we 
are given a, h, and a, we construct first the given angle a and lay 
off on one side the length AC =h; then, from 
C as center, with radius equal to the given 
opposite side a, we describe an arc which may 
cut the side AX in two points B^ and B^. 
Then either of the triangles 
ACBj^ or ACB^ is a correct 
solution, for either contaius 
the two given sides and the 
angle a opposite the side a. 

Case IV is accordingly 
often called the " ambiguous 
case" ia the construction 
and computation of trian- 
gles, because, when the sides and angles are as in Fig. 101, 
there are two equally correct solutions or constructions. 

125. This ambiguity may, however, not occur if the relations 
of the given sides and angles are different from what they were 
in the figure above. If, for example, the side a is equal to or 
greater than the side b, there will be only one intersection with 
the line AX, and hence only one triangle can be constructed. 
Or, again, the side a may be exactly equal to the perpendicular 
distance CD from C to the opposite side AX, and in this case the 
arc with radius a will be tangent to AX, thus determining but one 
point BD. The right triangle ABC is then the only construc- 
tion. Note that in case this happens, a = CD = 6 sin a. Finally, 
the side a may be shorter than the perpendicular distance from 




THE TRIGONOMETRIC FUNCTIONS 161 

C to AX, and then the arc with radius a will not meet AX at 
all, so that no construction is possible. In this case a < 6 sin a. 
A careful figure should be drawn for each of these possibilities. 

126. Summarizing, in Case IV, when a, h, and a are given, a 
being an acute angle, there wiU be two solutions when and only 
when a is less than h and at the same tune greater than h sin a. If 
a=i there will be only one solution ; if a = & sin a there wUl also be 
one solution, in this case a right triangle ; and, finally, if a < & sin a 
there wiU be no solution. 

EXERCISES 

1. Show that in Case IV, if the given angle is obtuse, there 
cannot be two solutions. When will there be none ? 

2. Determine the number of possible constructions (solutions) in 
each of the following cases : 



(1) 


a = 30°, 


6 = 10, 


a =12. 


(2) 


a = 30°, 


6 = 5, 


a = 3. 


(3) 


a = 30°, 


6 = 100, 


a = 50. 


(4) 


P = 30°, 


6 = 25, 


a = 50. 


(5) 


y = 30°, 


6 = 25, 


c = 30. 


(6) 


^ = 30°, 


6 = 15, 


c = 25. 


(7) 


^=60°, 


6 = 7, 


a = 10. 


(8) 


y = 20°, 


a = 20, 


c = 10. 


(9) 


y = 110°, 


6 = 60, 


c = 100. 


(10) 


y = 150°, 


c = 25. 


a = 30. 



3. In Kg. 101, writing Cj for AB^, and c^ for AB^, prove that 
Cj + Cj = 2 6 ■ cos a. 

4. Show that (in the same figure) Cj — c^ = 2 a • cos j8^. 

5. Solve the following triangles, checking the results both by 
measurement and by computation. In the two-solution case, check 
by the formula of Ex. 3. 



(1) 


a = 39, 


6 = 65, 


a = 25°. 


(2) 


6=23, 


c = 4.1, 


fi = 31°. 


(3) 


a = 16, 


c = 24. 


y = 49°. 


(4) 


6 = 78, 


c = 50. 


y = 62°. 


(5) 


a = 153, 


6 = 136, 


P = 40°. 



162 THE ELEMENTARY FUNCTIONS 

MISCELLAHEOUS PROBLEMS 

In the following list of miscellaneous problems, draw an accurate 
figure whenever possible, and in every case check the result by 
computation : 

1. In order to find the distance between two objects A and B, sepa- 
rated by a swamj), a station C is chosen, and the distances CA = 355 ft., 
CB = 418 ft., and Z A CB = 36° are measured. Find the distance from 
A toB. 

2. Two objects, A and B, were observed from a ship to be in a 
line bearing N. 15° E. The ship then sailed N. W. 5 miles, when it 
was found that A bore due east and B bore N. E. Find the distance 
from A to B. 

3. In a circle with radius 3 find the area of the part included 
between two parallel chords (on the same side of the center) whose 
lengths are 4 and 5. 

4. The angle of elevation of a tower is at one point 63° 30'; at a 
point 600 ft. farther from the tower, in a straight line, it is 32° 16'. 
Find the height of the tower. 

5. A tower makes an angle of 113° 12' with the hillside on which 
it stands ; and at a distance of 89 ft. from the base, measured down 
the hill, the angle subtended by the tower is 23° 27'. Find the height 
of the tower. 

6. To determine the distance between two points A and B, a base- 
line CD is measured, 500 ft. long, and the following angles are ob- 
served: ACB = 58''20', ACD = 95° 20', ADB = 53° SO', BDC = 98° iS'. 
Find the length AB. 

7. Two inaccessible points A and B are visible from JO, but no 
other point can be found from which both are visible. A point C 
is taken, from which A and D can be seen, and CD is found to be 
200 ft., while Z.ADC = 89° and Z 4 CD = 50° 30'. Then a point 
E is taken, from which D and B are visible, and DE is found 
to be 200 ft., ZBDE = 64° 30', ZBED= 88° 30'. At D, Z.ADB is 
observed to be 72° 30'. Compute the distance AB. 

8. A pole 10 ft. high stands vertically, and from its foot the angle 
of elevation of the top of a tree is 32° 27'. The angle of elevation 
of the top of the pole from the foot of the tree is 14° 48'. Find the 
distance between the tree and the pole, and the height of the tree. 



THE TRIGONOMETRIC FUNCTIONS 



163 




Fig. 102 



127. The half-angle formulas. We now turn from the appli- 
cations of the trigonometric functions to the study of further 
theorems concerning the functions themselves. 
Problems 5-8 on page 67 gave four important 
relations among the trigonometric functions of 
an angle. We shall now discover the rela- 
tions that exist between the functions of an 
angle and those of an angle twice as large. 

128. Let ABC (Fig. 102) be an isosceles 
triangle, and let ZBAC = ZBCA = a. Let 
each of the equal sides be represented by 
a, and the base by h. If AB be produced, 
the exterior angle CBE = 2 a. (Why ?) 

If we now apply the Law of Sines to the triangle ABC, we get 

a _ 5 6 h 

sin a ~ sia /.ABC ~ sin (180°- 2a) ~ sin 2 a ' ^ ^ 

But & = 2DC=2acosa;, (2) 

since DC = a cos a. 

Putting this value of b into equation (1), 

a 2 a cos a 
sin a sin 2 a 
that is, sin 2 a = 2 sin a cos a. (3) 

Next, let us apply the Law of Projections (p. 156) to the 
triangle ABC: 

a = b cos a + a cos (180° — 2a)=2a cos^a — a cos 2 a, 

since 6 = 2 a cos a, by (2) ; that is, 

cos 2 a = 2 cos^ a — 1. (4) 

129. This relation can be written in other forms by making 
use of the fact that (Ex. 5, p. 67) 

sin^a;-f-cos^a; = 1. 
Therefore cos 2a = 2(1— sin^a) — 1 ; 

that is, cos 2 a = 1 — 2 sin'* a. (5) 



164 THE ELEMENTAEY FUNCTIONS 

Replacing the term 1 in equation (5) by cos^a + sin^a;, 

cos 2 a = cos' a — sin" a. (6) 

It will be noted that these formulas have been proved for any- 
acute angle a ; they are true for any angle whatever, but that 
fact must for the present be left without proof. 

EXERCISES 

1. Given sin 10° = .1736, cos 10° = .9848, find the values of 
sin 20°, cos 20°, and tan 20° to four decimal places. 

-r^ , 1 . . r> 2 tan a 

2. Prove that tan 2a = : — 5— • 

1 — tan-'a 

3. In formulas (4) and (6), change 2 a: to a, which requires that 
a be changed to - > thus getting 

cos a =: 2 cos^ 5 ~ 1 ^nd cos a = 1 — 2 sin" - > 

and thence derive the formulas 



a 11 + cos a , .a jl 



. — cos o 
cos- 



4. Find the values of sin 16°, cos 15°, and tan 15° from the fact 

that cos 30° = — ^ (p. 68). _ _ 

' ^n.. sinl5° = iV^3^=^^^^, 

Ii 4: 

cosl5° = |V^Wl= -^ + ^ , 

tan 15° = \ r ~ \^ = 2 - Vs- 
>2+V3 

5. Find the values of sin 22 i°, cos 22^°, and tan 22^° from the 
known values of the functions of 45°. 



Ans. sin 22^° = ^ V2 — V2, 



cos22i° = i\/2 + V2, 
tan 22^° = -y 



;^ = V2-I. 



2 + V2 

«T> i.i_i..^ — I-' — COS a sin a 1 — cos a 

6. Prove that tan - = x| = = :. . 

+ cos o 1 + cos a sin a 



'f=^ 



THE TRIGONOMETRIC FUNCTIONS 165 

7. Derive the values of the functions of 15° geometrically by 
constructing a regular dodecagon and computing the exact values 
of the ratios of the apothem^ and side to the radius. Compare with 
the results of Ex. 4. 

8. As in Ex. 7, construct a regular octagon and thereby find the 
exact values of the functions of 22^°. 

9. As in Ex. 7, use the regular decagon to find the values of the 
functions of 18° (cf. Ex. 8, p. 60). 

V6— 1 

Ans. sin 18° = ; ; 

4 

cos 18° = \ VlO + 2 V5, 

tanl8°=Jl^l4 = JVi^^^I^. 
>10 + 2V5 5 

10. As in Ex. 7, use the regular pentagon to find the values of 

the functions of 36°. , 

Ans. sin 36° = i V 10 - 2 VS, 

cos 3d = - — ) 



tan 36° = V5 - 2 V5. 
11. Prove that tan 7^° = V6 - Vs + V2 - 2. 

130. The problems which follow are designed to aid the stu- 
dent in fixing in mind the relations among the trigonometric 
functions which we have thus far considered, and also to develop 
the power of discovering new relations. The illustrative examples 
should be studied, as indicating the general method to be followed, 
the details, however, varying from problem to problem. 

Example 1. Prove that cos*0 — sin^O = cos 2 0. 

Proof. cos«e - sin^e = (cos^S - sin^^) (cos^g + ain^e) 

= cos^e - sin^e, 
since cos^S + sin^^ = 1. 

But eos^^ - si-D^O = cos 2 ; 

hence the theorem is proved. 

1 Apothem : the distance from the center to a side of a regular polygon. 



166 THE ELEMENTARY FUNCTIONS 



Example 2. Prove that 



2 tan - 

9 



l + tan^" 

Proof. l + tan2? = sec2^. 

2 tan - 2 tan - 

Therefore = 1 = 2 tan - ■ cos^ - • 

1 + tan- - sec^ - 

. X 
sm- 

1 2 
Now tan - = 

2 X 
cos- 

2 

ar 3r XX 

Therefore 2 tan- • cos^— = 2 sin — cos— = sinx. q.e.d. 



Another method. This formula can also be proved as follows : 



and 

Therefore 

Therefore 



tan- 


sma; 


1 + cos X 


tan^ - 


1 — cos X 
1 + cos X 


1 + tan^l 


_ ^ 1 1 — cosx 2 


1 + cos X 1 + cos X 


2tan| 


2 sin a; 
1 + cos a; 


1 J- (-.5.T,2*' 


2 -«'^^- 



2 1 + cos X 



Q.E.D. 



EXERCISES 
Prove each of the following formulas : 

1. tan X + cot a; = 2 CSC 2 x. 4. cot a; — tan a; = 2 cot 2 x: 

. 6 e I r 

2. sm- + cos-= Vl+sin0. 1— tan^^ 

5. = cosa;. 

3. tan(45° + |)=seca; + tanx. l+tan^^ 

Hint, tan (45° + ^) = tan ^J^l±l. 6. ?' '^"^"'^ = 2 sin »■+ sin 2 x. 

\ 2/ 2 1 — cos a; 



THE TRIGONOMETRIC FUNCTIONS 167 

7. sin - — cos ^ = Vl — sin x. 

8. sin 2 a; sin x = (1 — cos 2 x) cos x. 

^ . 1+sinfl— cose 

9. tan - = : — . 

2 l+sme+eose 

10. cos^a; + sin^a; = 1 — ^ sin^2 x. 

11. CSC X — 2 cos X cot 2 x = 2 sin x. 

lo 6 -6 o /h sin''2a;\ 

12. cos°a! — sin^a; = cos 2 a; ( 1 — 1 • 

13. cos^x + sin^a; =1—3 sin^x cos^'a;. 

14. sin 4x = 4 sin a; cos a;— 8 sin'a; cosx=8 cos'a; sin a;— 4 cos x sin x. 

15. cos 4 X = 1 — 8 cos^x + 8 cos*x =1—8 sin^x + 8 sin*x. 

, „ . „ . _ cos X + sin X 

16. tan 2 X + sec 2 x = ■ — -. 

cos X — sin X 

sec^x 

17. sec2x = jr 5--. 

2 — sec^x 

18. tanfl- tan 2^= sec 2^ — 1. 

^ g _ 2 sin e — sin 2 e 

19. tan2-2si^^^sij^2e' 

20. l+sec2e+tan2e 

sin a + sin 2 a a 

21. ^ = cot-- 

cos a — cos 2 a 2 



l-tan^ 



22. (1+ cot'' |) sin a- tan I = 2. 

2 cot a 

23. tan 2a = 



25. 



eot^a —1 

sin X. 
cos X — sin X 1 — sin 2 x cos 2 x 



24. sin'' I f cot I — ij = 1 — I 



cos X + sin x cos 2 x 1 + sin 2 x 



tan a tan a 1 1 

26. tan 2a = 3 1 ^- rTT = ^i i^ m: ^' 

1 — tan a 1 + tan a 1 — tan a 1 + tan a 

11 1 I . o._ (l+ tan g)^ ^ (l + cot 0)^ ^ (1 + tang) (1 + cote) 
"^ l + tan^e l + cot^0 tane + cote 



168 THE ELEMENTAEY FUNCTIONS 

sin 2x 2 tan x 2 cot x 



28. 



l+sin2x (1 + tanic)'' (1+cotx)^ (l + tanx) (l+cotcc) 
29. 2cos^=\2 + V 2 + y/2-\ + V2 (w + 1 radical signs). 



Of\0 I . — 

30. 2 COS -— = y 2 + V 2 H + VS (w + 1 radical signs). 

131. Functions of the sum or difference of two angles. For the 

further study of the trigonometric functions no theorems are more 
important than those now to be proved, which enable us to find 
any function of the sum or difference of two angles from the func- 
tions of each angle. Two proofs will be given, one algebraic and 
the other geometric. 

132. First proof. This proof uses the Law of Sines, and also 
the formula (7) on page 156 : c = a cos /3 + J cos a. By the Law 
of Sines (in the form given in Ex. 18, p. 159), c = 2iJsin7, 
J = 2 -B sin /3, and a = 2 i? sin a. 

Therefore 2 ^ sin 7 = 2 iJ sin a cos /3 + 2 -B sin /8 cos a. 

Now, since a + /S + 7 = 180°, 

7 = 180°-(a + /3); 
therefore sin 7 = sin [180° — (a + /8)] = sin (a + /8). 

Therefore sin (o + ^) = sin a cos )ff + cos a sin p. (I) 

To find the value of cos (a + /S) : Begin with equations (8) and 
(9) on page 156, ' « = j oos7 + ccos/3, 
6 = a cos 7 + c cos a. 
Multiplying these together, we get 

ab = ab cos^7 + c cos 7 (6 cos a -\- a cos yS) + c^ cos a cos /3. 
But, by (7) on page 156, h cos a + a cos/8 = c. 
Therefore 

a6 (1 — cos^7) = c^ cos 7 + c^ cos a cos ;8 = c^ (cos 7 + cos a cos /8). 
By the Law of Sines, as above, a = 2 i? sin a, J = 2 .B sin (S, and 
c = 2 ^ sin 7 ; also 1 — COS27 = w?'^. 
Therefore 

4 if 2 sin a; sin ^ . sin27 = 4 iJ^ gin2^ ^^Qg ,y ^ cos a cos /3). 



THE TRIGONOMETRIC FUNCTIONS 169 

Dividing both sides of this equation by ^B^sin^y, 

sin a sin /3 = cos 7 + cos a cos /8. 
And, since a+/S + 7 = 180°, 

cos 7 = cos [180° - (a 4- /3)] = - cos (a + /8). 
Therefore co6 (o + )ff ) = cos a cos ^ — sin a sin ^. (II) 

133. Second proof. Let a and /8 be any two acute angles. Place 
them with a common vertex and a common side OiV^ as in 
Fig. 103, so that the angle MOP = a +y8. From any point A in 
OM draw AB perpendicular to ON at B, and produce it to meet 
OB at C Notice that this assumes that 
a + yS is an acute angle. Then we can 
express the areas of the triangles formed, 
as follows (Ex. 17, p. 159) : 

Area. AAOC = ^OA • OC sin(a:+/3). 

Area AAOB = ^OA- OB . sin a. 

Area A^OC =10-8. OC • sin/8. 

Since AAOC = AAOB +ABOC, 

therefore 

^OA. DC- sin(a + /3) = J0^. 0.B ■ sina + 105 • OC- sinA 

Therefore 

^^ OB . OB . „ 

sm(a; + /3) = --sma + — smA c/ 




Fig. 103 



But 

and 

Therefore 



OB 

OC' 

qB_ 

OA 



■ coS/S 



• = COS a. 




Fig. 104 



sin (a + jff) = sin a cos jff + cos a sin p. (I) 

This same method can be used to find the value of sin (a - /8), 
as follows: 

Let ZAOC=a (Fig. 104) and ZBOC=^, AC being perpen- 
dicular to OC. Then ZAOB=a-0, and, just as before, 

lOJ. OC-sma = ^OA-OB-sm{a-^) + ^OB- OC- sin^ 



170 THE ELEMENTARY FUNCTIONS 

Therefore sin (a — /3 ) = — sin a sin B, 

^ ' OB OA 

or sin (a — ^) = sin a cos ^ — cos a sin p. (HI) 

From (I) and (III) the values of cos {a + /3) and cos (a — /S) can 

be found. For ^ . „.o a\ 

cos 6 = sm(90 — &). 

Therefore cos (a -f /3) = sin [90° - (a + y8)] 

= sin[(90°-a)-;S]. 

This last expression is in the form of the sine of the difference 
of two angles, both of which are acute, since 90°— a is acute if 
a is. Therefore we can apply (III), gettrug 

sin[(90°-a)-;S] = sm(90°-a)cos;8-cos(90°-a:)sin/S 

= cos a cos /8 — sin a sin /3 ; 

that is, cos (a+/3)= cos a cos ^ — sin a sin /S. (II) 

To obtain the value of cos (a — /S) : 

cos (a - /3) = sin [90° -{a- 0)] = sin [(90° -a) + 13]. 

Since the angles (90° — a) and /S are both acute, we can apply (I) : 
sin [(90° - a) + ^] = sin (90° - a) cos /3 + cos (90° - a) sin /3 

= cos a cos /3 + sin a sin /8. 
Therefore cos (a — ^) = cos acos ^ + sin a sin p. (IV) 

134. The first method of proof (§ 132) assumes that a and /3 
can be taken as two of the angles of a triangle ; that is, that 
their sum is less than 180°. The second proof (§ 133) assumes 
that a, /8, and a +13 are acute. We can, however, remove these 
restrictions and establish the fact that formulas I-IV hold for 
any angles whatever. 

To accomplish this, take a =90°+ a', where a' is an acute 
angle, and suppose /8 to be acute, as before. Then 

sin(a;+^)=sm[90°+a' + /8]=cos(a;' + iS). 

Since a' and /3 are acute angles, (II) can be applied, giving 

cos {a' + /3) = cos a' cos jS — sin a' sin /3. 



THE TRIGONOMETEIC FUNCTIONS 171 

But cos a' = cos (a — 90°)= sin a, 

and sin a' = sin (a — 90°) = — cos a. 

Therefore cos (a' + /3) = sin a cos y8 + cos a sin /S. 

Therefore sin (a + /3) = sin a; cos /3 + cos a sin /8. 

Hence formula (I) is true if one angle is obtuse and the other 
acute, and by exactly the same process we can establish its truth 
if we add 90° to any angle a' for which the formula has already 
been proved. Thus, we can add 90° to /3, proving the formula 
true when hoth angles are obtuse; then we can add 90° to a, 
proving its truth when one angle is between 180° and 270°, the 
other being either acute or obtuse ; and so on. The same reason- 
ing holds for successive subtractions of 90°. Hence (I) is true for 
angles of any magnitude whatever. The theorem is, as was said 
before, of the very greatest importance. It is called the Addition 
Theorem for the trigonometric functions. 

Similar reasoning can be used to establish the truth of II, III, 
and IV for angles of any magnitude. 

EXERCISES 

1. Work through the same reasoning as above to prove (II) 
in general. 

2. Prove (III) by changing ;8 to — /3 in (I). (This is allowable, 
since (I) has been proved true for all angles.) 

3. Prove (IV) in- a similar way. 

4. Eind the values of sin 76° and cos 76° from the known values 
of the functions of 45° and 30°. 

5. Find the values of sin 16° and cos 15° from the values of the 
functions of 45° and 30°. Compare with the results of Ex. 4 and 
also of Ex. 4, p. 164. 

6. Derive formulas for tan (a +/3) and tan (a — ;8) by using I-IV. 

tan a + tan S , . .. tan « — tan/? 
Ans. tan(« + ^)= j3^^^^^; *^°(^-^)= 1 + tana tan^" 

_ , . , „, cot a cot fl — 1 

7 . Prove that cot (a+B) = — , ,, , , > 

^ "^^ cot ;8 + cot a 

cot a cot /8 + 1 
and that cot(a - ^) = ^^^J^^^ ' 



172 



THE ELEMENTARY FUNCTIONS 



8. Prove formulas (I) and (II) by using Fig. 105. Show that 

„, DE +FB 
sm (^a + P) = 



OB 








DE 
OE 


OE 
OB 


+ 


FB 
EB 


EB 
OB 




= sin a cos ^ + cos a sin fi ; 
and, similarly, derive (II) by starting 

with 

„, OC OD—FE 
cos(a+^)=— = — ^^— 

9. In (I) and (II) let ^ = a, thus 
getting formulas for sin 2 a and cos 2 a. 

Compare with the results of §§ 127, 128. Notice that we are now 
for the first time able to assert that these results are true for 
all angles whatsoever. The " half-angle formulas " of Ex. 3, p. 164, 
are thereby also assured universal validity, because they were 
derived by purely algebraic processes from the values for cos 2 a. 

10. Find the values of the functions 
of 75° from those of 150°, and compare 
with the results of Ex. 4 above. 

11. Prove that 
sin (a — 13) — sin a cos ^ — cos a sin /3 

by using Fig. 106. ABC is any triangle, 

CD is the altitude from C, DE=AD; 

show that Z.ECB = a —fi, and then apply the Law of Sines to the 

triangle CBE. 

12. Prove the Addition Theorem from the Law of Sines, using 

a 
b' 




the fact that if - = - > then ; 

b d b -\-d 



Prove each of the following formulas (13-30): 

sin (x + y) 
13. ^^ — -^^ = tana^ + tany. 

sin 2 a- + sin 2 y 



cos X cos y 
14. tan(x -f- 2/) = 



cos 2 x + cos 2 y 

15. cos (x + 30°) + cos (x - 30°) = V3 • cos x. 

16. sin (a; + 60°) + sin (a; - 60°) = sin x. 

17 . sin e + sin (6 - 120°) + sin (60° - 0) = 0. 



THE TRIGONOMETRIC FUNCTIONS 173 

18. sin 3 aj = 3 sin a; — 4 sin'a;. 

19. cos 3 a; = 4 cos'a; — 3 cos x. 

20. sec (a + 46°) sec (a - 46°) = 2 sec 2 a. 

21. tan (45° + a) - tan(45° - a) = 2 tan 2 a. 

22. tan (5 + 46°) + tan (45° - ^) = 2 sec 2 0. 

23. sin {a + P) + sin (a — |8) = 2 sin a cos ;8. 

24. sin (or 4- /3) — sin (a — ^) = 2 cos a sin /8. 

25. cos (a + y8) + cos (a — /?) = 2 cos a cos /?. 

26. cos(a: + y8) — cos (a — y8) = — 2 sin « sin /3. 

27. sin (a + ^8) sin (a- p)= sin'a - sin'jS. 

28. cos (a + ;8) cos (a — j8) = cos^a + cos^/3 — 1 = cos^a — sin'^/? 

= cos'^yS — sin^a. 

29. sin (a -\- p) cos (a — p)= sin a cos a + sin /3 cos p. 

30. sin (a — P) cos («+/?) = sin a cos a — sin p cos /?. 

31. Obtain the values of the functions of 18° by the follow- 
ing method : sin 36° = 2 sin 18° cos 18°, and also sin 36° = cos 64°. 
But, by Ex. 19, above, cos 54° = 4 cos' 18° - 3 cos 18°. Therefore 
2 sin 18° cos 18° = 4 cos» 18° - 3 cos 18°. Solve this for sin 18° and 
compare the result with that of Ex. 9, p. 165. 

32. Prove that sin (a + /?) cos p — cos (a + p) sin p = sin a. 
Solution. This can be proved by writing out the values of sin (o: + /3) 

and cos (a +/8), but it is simpler to observe carefully the combination of 
terms that we have here, namely, sin x cosp — cosa: sin /?, x representing 
a + p. This is the value of sin (x — /3), that is, of sin (or + ^ — ^), or sin a. 

33. Prove: sin(a — ;8)cos/3 + cos(a: — /3)sin/3 = sina. 

34. Prove : sin (a + P) sin /3 + cos (a + P) cos /3 = cos a. 

„^ .„ tan(a: — y8)+ tan;8 , 

35. Prove: q 1 — / o\^ '"- = tana:. 

1 — tan (a — P) tan p 

_ cos a —• cos 6 cos (Q + a) . . 

36. Prove: : — 77; — ^^ ^ = sinSseca. 

cos a sm (6* + a) 

37. Prove that the angle Q, made by a line whose slope is m^ with 

m — TO. 
a line whose slope is m^, is given by tan B = -r-j^ 

38. Find the angle that the line 3x — 2 ij + 1 = makes with 
the line x + y — 3 = 0. 



174 



THE ELEMENTARY FUNCTIONS 



39. Find the angles of the triangle whose vertices are the points 

(3, 6), (7, - 1), and (- 2, 3). 

40. The theorem known as the Ptolemaic Theorem ^ is as follows : 
In any inscribed quadrilateral the sum of the products of the 
opposite sides is equal to the product of the diagonals. 

(a) Prove this theorem by elementary geometry. 

(b) Assuming its truth, prove the Addition Theorem. 

Hints. For (a) draw from B a line BE (not shown in Fig. 107), so that 
/.ABE = p, E being the point where this line meets AG; then the triangles 
ABE and BDC are similar, as are the triangles BEG and ABD. From these 
facts the theorem can be proved. 





Fio. 107 



Fig. 108 



For (b) construct the given angles a and /3 on opposite sides of the diameter 
through A (Fig. 108), and complete the inscribed quadrilateral ABGJD. Then 
apply the fact that a = 2fisina etc. to the right triangles ABC and AGD; 
also the Law of Sines to the triangle ABD. 

135. Sum and difference of two sines or two cosines. The 

equations that were proved in Exs. 23-26, p. 173, were as follows : 

sin (a + y8) + sin (a — /3) = 2 sin a cos /3, (1) 

sin (a + /8) — sin (a — /S) = 2 cos a sin y8, (2) 

cos (a + /3) + cos (a — /3) = 2 cos a cos /8, (3) 

cos (a + /3) — cos (a — /3) = — 2 sin a sin /3. (4) 

These equations give the sum or difference of two sines, or of 
two cosines, in the form of a product, a form which we shall often 

1 From Ptolemy, the famous Greek astronomer of the second century. This 
theorem is included in his great work called the "Almagest," which explained 
the astronomic system which, under the name of the "Ptolemaic system" was 
universally accepted for 1400 years. It was only overthrown after a long 
struggle between its adherents and those of the system of Copernicus. See 
the articles "Ptolemy" and "Copernicus" in the Encyclopedia Britannica. 



THE TRIGONOMETRIC FUNCTIONS 175 

find useful. These equations are written in a more practical form 
by setting a + ^ = x and a — ^ = y. 

Therefore 2a = x + y and 2^ = x— y. 

x + y , ^ X — y 

a = " and j8 = —-^■ 
2 • 2 

Thus the equations (1) — (4) are equivalent to 

x+y x—y 

sin Jr + sin y = 2 sin cos » (5) 

sin Jf — sin tf = 2 cos sin » (b) 

2 2 

x + y x-y 

C0SX+C0SU=2 cos COB- » (/) 

2 2 ^ 

. x + y . jr-y 

COS j: — cos w = — 2 sin sin • (») 

2 2 



EXERCISES 

^^ ^ sin 33° + sin 3° , , _„ 

1. Prove that 5^5- 53 = tan 18° 

cos 33 + cos 3 

Proof. By (5) and (7), 



. . 36° 30° 
2 sin — — cos — — 
sin 33° + sin 3° _ 2 2 _ sin 18° 

cos 33°+ cos 3° „ 36° 30° cos 18° 
2 cos -^ cos -— - 

„ sin 3 a + sin 5 a , . 

2. Prove: 5 ; ^ = tan 4 a. 

cos Sa + cos o a 

sin 75°+ sin 16° , „„» 

3. Prove : . „.,o r-^^ = tan 60 . 

sin 75 — sm 16 

^ tan X + tan y , , 

4. Prove : —r , r^ = tan x tan y. 

cot X + cot y 

x + y 

tan— r— 

„ sin r + sm y ^ 

5. Prove: -. : — - = 



= tan 18°. 



Sin x - sin y . x-y 



176 



THE ELEMENTARY FUNCTIONS 



6. Prove the formula (3), p. 174, by means of Fig. 109. This 
method of proof is found in an astronomical work of the tenth 
century, by an Arabian scholar, Ibn Junos. The formulas (1), (2), 
and (4) can also be proved in a similar way. 




Hints. Letting 
and 

it follows that 
and 
Therefore 



ZNCP = /.Z'CE'= a, 
ZE'CD" = ZECD = p, 

JD" = cos (a + /3) (if r = 1) 
D'D = cos(a-/3). 
J'D = cos (a: + (3) + cos (a — /3) . 
Next, show LD = ^ J'D = cos a ■ cos /3 hy showing that LD = BD ■ cos a. 

7. cos X + cos 3 a; + cos 5x + cos 7x = 4 cos x • cos 2 x ■ cos 4 x. 
Proof. cos X + cos 3 x = 2 cos 2 a; cos x, 

cos5x + cos7:r = 2cos6a;cosx. 

Therefore cos x + cos 3 x + cos 5 x + cos 7x = 2 cos x (cos 6 x + cos 2 x) 

= 2 cos X (2 cos 4 X cos 2 x) 
= 4 cos X (cos 2 X cos 4 x). q.e.d. 

8. sin X + sin 3 a! + sin 5x + sin 7a; = 4 cos x • cos 2 x ■ sin 4 x. 

tan a; + tan /8 _ sin (a + ;8) 
tan a — tan fi sin (a — j3) 



THE TRIGONOMETEIC FUNCTIONS 177 

sin a + sin 2 a + sin 3 a 

10. ~ — ; = tan 2 a. 

cos a + cos 2 a + cos 3 a 

, , sin a -}- sin 2 a + sin 3 a + sin 4 a , 5 a 

11. ^ -. — = tan-;7-- 

cos a + cos 2 a + cos 3 a + cos 4 a 2 

If a, j3, and y are the angles of a triangle, prove that each of the 
following relations is true : 

12. sin a + sin yg + sin y = 4 cos ^ • cos ^ • cos ^ ■ 

Proof. sin a + sin /3 = 2 sin — ;— ^ cos — ^-^ • 

siny = sin [180°- (a + )3)] = sin (a + ;8) 

= 2 sin ^4^ cos 2±^. ((3), p. 163) 

Therefore sin a + sin/3 + siny = 2 sin " ^ cos " + cos — —^ 

- . 180° -yr„ a /yi 
= 2 sm ;r — t 2 COS "5 cos c: 

a B y 
= 4C0S- COS^COS^- Q.K.D. 

, . a . B . y 

13. cos a + cos j8 + cos y = 1 + 4 sm - sin o s^"' o ' 

14. tan a + tan j8 + tan y = tan a tan j8 tan y. 

15. sina + sin/3 — siny = 4 sin - sin - cos -• 

. . «■ P ■ y 

16. cos a + cos /3 — cosy = — 1+4 cos -cos 2 sin ^• 

17. cot I + cot I + cot I = cot I cot I cot |- 

a B By y "^ t 

18. tan - tan I + tan | tan ^ + tan ^ tan - =1. 

19. cos^a + cos^yS + cos^y = 1 — 2 cos cr cos /3 cos y. 

20. sin^a + sin^/S + sin'y = 2 (1 + cos a cos /S cos y). 

21. 1 - sin^l _ sin^l - sin^| = 2 sin ^ sin | sin |- 

22. sin2« + sin 2^8 + sin 2y = 4 sin a sin ^8 siny. 

23. sin 2 a + sin 2 yS — sin 2 y = 4 cos a cos j8 sin y. 



178 



THE ELEMENTARY FUNCTIONS 



sin 2 a + sin 2 fi + sin 2 y „ . a . B . y 

24. -. ■ -. — r^ = 8 sin - sin ^ sin ^ • 

sin a + sin yS + sin y ^ 2 z 

25. cos 2 a + cos 2 j8 + cos 2 y = — 1 — 4 cos a cos ^ cos y. 

26. cos 2a + cos 2/3 — cos 2y =1— 4 sin a sin /3 cos y. 

27 . cot a + cot 13 + cot y = cot a cot ^ cot y + esc a esc y3 esc y. 



28. sin-"^ = 

LI 

a 



•J/ _ (sin j8 + sin y — sin a) (sin y + sin a — sin j3) 



,/8 



29. cos ;- + cos ^ + COS 
Z u z 



4 sin a sin j8 
4 



-t = 4 COS . ' cos T^ COS — - 



4 
3n; 3;3 3y 

4 cos -^r- COS -h- COS -r^ ■ 

L Ji Z 

.2 1 _ (sin tt + sin /3 — sin y) (sin a + sin ^8 + sin y) 

31. COS „ — A • • n 

2 4 sm a sin /3 



30. sin 3 a + sin 3 j8 + sin 3 y : 



Changes in the Trigonometric Functions as the 
Angle changes 

136. Our experience with the use of the trigonometric func- 
tions has given us considerable information as to the way in 
which they change as the 
angle changes. Let us now 
follow out these variations 
more carefully. 

137. The sine. Since 
the sine of an angle equals 
the quotient of the ordi- 
nate by the radius vector, 
we shall find it easy to 
follow the changes in the 
sine function if we choose 
points on the terminal line 
of the angle such that the 
radius vector is constant. 
This will be accomplished 

if the points are taken on the circumference of a circle, as in Fig. 110. 

As the angle a increases, the point P will move aroimd the 

circumference of the circle, from a position on the X-axis, when 




Pig. 110 



THE TRIGONOMETRIC FUNCTIONS 179 

the angle a = 0°, through the four quadrants, till it reaches the 
same position again, when a = 360°. The question is. What hap- 
pens to sin a as a; passes through this series of values ? When 

a = 0°, the ordinate of P is 0, and hence the ratio — = ; 

radius vector 

that is, sin 0° = 0. Now as a increases, the ordinate increases 

also ; and since the radius vector is constant, sin a must increase. 

This increase of the ordinate, and therefore of the sine function, 

continues through the first quadrant, until, when the angle is 90°, 

the ordinate equals the radius vector, and therefore sin 90°= 1. 

As the angle increases from 90° to 180°, the sine of the angle 

evidently decreases from 1 to ; as the angle passes through the 

third quadrant, sin a decreases from to — 1 ; and, finally, as a 

increases from 270° to 360°, sin a increases from — 1 to 0. 

138. This review of the variation in the sine function should 
be completed by drawing up a table of the values of the function 
for all angles at 15° inter- 
vals, from 0° to 360°. Do 
not use the printed tables for 
this purpose, but take the re- 
sults obtained from our pre- 
vious study (cf. pp. 68, 164). 

When this table of values has Fto. Ill 

been written down, it will be 

found easy to construct the graph of the function y = smx from 
a;=0toa;=360by taking the values of the angle {x) as abscissas 
and the corresponding values of sin x as ordinates. Plotting the 
points located by the table in this way, and joming them by a 
smooth curve, we shall have a part of the graph of the function. 
Fig. Ill gives a smaU-scale representation of the curve obtained. 
The student should make a careful drawing on a larger scale. 
(The scale on the T-axis will need to be much larger than that 
on the X-axis, for obvious practical reasons.) 

139. Since the increase of an angle beyond 360° means merely 
increasing the amount of rotation beyond one complete revolution, 
the sine function will change in precisely the same way while the 




180 



THE ELEMENTARY FUNCTIONS 



angle changes from 360° to 450° as it does while the angle changes 
from 0° to 90°; in other words, beyond 360° the same cycle of 
changes is repeated, because the terminal Une of the angle merely 
occupies the same positions again, in the same order. Thus the 
sine function repeats itself periodically at intervals of 360°. It 
is accordingly called a periodic function, with the period 360°. 
Anticipating the next paragraphs, we can see that all the trigo- 
nometric functions are periodic, with a period 360°. For negative 
values of the angle we have the fact that sin (—«)=— sin x, which 
enables us to extend the drawing of the sine graph to the left of the 
F-axis. This should be done in the final construction of the curve. 
140. The cosine. By similar reasoning to that for the sine, and 
by referring to Fig. 110 again, the student may show that the 
cosine varies as follows : 



Angle 


0° to 90° 


90° to 180° 


180° to 270° 


270° to 360° 


Cosine 


1 toO 


to -1 


-1 to 


to 1 



And, as in the case of the sine, a table of values of the cosine func- 
tion should be drawn up for angles at 15° intervals, from 0° to 360°. 
From this table points can 

be plotted and the graph of \q 

y = cos X drawn. y 

141. The tangent. Since 
the tangent of an angle 

, ^, ^. ordinate 
equals the ratio 




abscissa 
we see at once that tan 0°=0 
and that, as the angle in- 
creases, the tangent also 
increases through the first 
quadrant. For it we keep 
the abscissa constant, as in 
Fig. 112, the ordinate will 
evidently increase as the 
angle increases. As the angle a approaches 90°, the ordinate of the 
point Q, where the terminal line of the angle meets the line A^^, 




Fig. 112 



THE TRIGONOMETEIC FUNCTIONS 181 

grows larger and larger. This ordinate will eventually exceed any 
value that can be named if the angle a be taken sufficiently 
near to 90°; and thus tana will exceed any assignable value if 
a is near enough to 90°. If a= 90°, however, tan a does not exist, 

since the ratio — — -. — would then take the form (the 

abscissa 

abscissa of any point on the F-axis being 0), and a fraction whose 

denominator equals zero has no value whatever.^ The fact that 

the tangent function will exceed any assignable value, for angles 

in the vicinity of 90°, is expressed by saying, "tana becomes 

infinite as a approaches 90°," or, in symbols, tan 90° = co. (This 

may be read "equals infinity," but it is not to be understood 

as giving a valii,e to tan 90°, which, as we have just seen, is 

impossible ; on the contrary, it only expresses in brief symbolic 

form the fact that the tangent of an angle will exceed any 

assignable value if the angle is near enough to 90°.) 

As soon as the angle a passes into the second quadrant, the 
tangent is negative and numerically very large. Thus there is an 
enormous jump in the value of the function as the angle passes 
from a value a little less than 90° to one a little more than 90°. 
Contrast this behavior of the tangent function with that of the 
sine, which changes gradually or continuously as the angle changes 
from a value a little less than 90° to one a little greater than 90°. 
This is an example of the important distinction between a func- 
tion which is continuous at a point (as the sine at 90°) and one 
which is discontinuous at a point (as the tangent at 90°). Of 
course it is obvious that both functions are continuous for any 
value of the angle between 0° and 90°. 

Eetuming to the variation of the tangent: as the angle a 
increases from 90° through the second quadrant. Fig. 112 shows 
4;hat the length of the ordinate decreases (the abscissa being kept 
constant), and hence tan a decreases numerically ; but as its value 
is negative, it actually increases from — co to 0. (The symbol — co 

1 Notice that it is a very different tiling to say that a certain expression 
"has no value" and to say it "equals zero." Zero is a perfectly definite value 
(cf. note on page 8). 



182 



THE ELEMENTARY FUNCTIONS 



is to be interpreted in accordance with what was said above con- 
cerning the symbol oo.) As the angle increases from 180° to 270°, 
the tangent becomes positive and increases from to oo. Is it con- 
tinuous or discontinuous at 180 ? at 270 ? FiuaUy, as the angle 
increases from 270° to 360°, the tangent increases from — co to 0. 

142. Remembering that tan (180°+ 6)= tan0, we see that this 
function has a period of 180°, unlike the functions sine and 
cosine, which do 
not repeat their 
cycle of values for 
a smaller interval 
than 360°. The 
student should 
now make a table 
of values of tan x 
for all values of 
X at intervals of 
15°, from 0° to 
360°, and make 
the graph of the function y = tana;. Fig. 113 gives a smaU-scale 
representation of a part of this curve, which should be drawn on 
a large scale, with considerable accuracy. 

143. The graphical representation of these functions, as in the 
case of all others that we have studied, is a great help in forming 
definite and accurate ideas of the way in which the functional 
values change. To gain more detailed knowledge of these changes 
one must study that branch of mathematics which is known as 
Differential Calculus (see Chapter XI). That study enables us to 
answer questions about the rate of increase or decrease of func- 
tional values, whereas for the present we must be satisfied with 
the general information which the graph gives. 




Tig. 113 



EXERCISE 



Study in the same way the variations in the other three trigo- 
nometric functions, and draw their graphs. 



THE TRIGONOMETRIC FUNCTIONS 



183 




Fig. 114 



Polar Coordinates 

* 144. We have seen how any point can be located by means of 
its distances from two perpendicular straight lines; it is also 
possible to locate a point in various other ways, which are found 
useful in solving a great many prob- 
lems. The most important of these 
other ways is by means of _the radius 
vector of the point and the angle which 
the radius vector makes with a fixed 
line. Thus, the point P is located by 
the radius vector r and the angle 6, which is called the vectorial 
angle of P. The line OA is called the initial line, and the angle 6 
may have any value, positive or negative. The values (r, d) are 
called the polar coordinates of the 
point P. 

Thus, in Fig. 115 the point P 
is completely located by the radius 
vector 2 and the vectorial angle 25°, 
which values are its polar coordi- 
nates. The radius vector is always 

mentioned first, so that the statement P = (2, 25°) locates definitely 
the point P. 

Similarly, Q = (i 200°) locates the point Q in Fig. 116, and 
Q={^, — 160°) locates the same point, as does also Q= (1 560°). 
It is thus clear that a point can have an indefinite number of pairs 
of polar coordinates (but a 
single pair of coordinates 
determines only one point). 
Moreover, it is customary 
to consider that a negative 
radius vector gives a point 

at the same distance from the origin as the corresponding posi- 
tive radius vector, but measured in exactly the opposite direction. 
Thus, the point P in Fig. 115 is not only (2, 25°) but also 
<^2, 205°), and Q in Fig. 116 is (-.^,20°) or (-1, 380°). 




Fig. 115 




184 



THE ELEMENTAEY FUNCTIONS 



EXERCISES 

1. Locate each of the following points : (1, 45°); (5, 90°); (2, 0°); 
(^, 270°); (- 1, 100°); (2, 320°). Choose also other pairs of coordi- 
nates at random and locate the corresponding points. 

2. Where can a point be if its radius vector is 2 ? if its vectorial 
angle is 180° ? 0° ? 

3. Prove that the line joining the points (1, 45°) and (1, 135°) is 
parallel to the initial line. 

4. Show that the points (0, 0), (2, 30°), and (2, 90°) form an 
equilateral triangle, and find the (polar) coordinates of the mid-points 
of its sides. 

5 . The rectangular coordinates of a point (that is, its abscissa and or- 
dinate) are (2 V2, 2 V2). Find its polar coordinates. Answer the same 
question for the point (— 2 V2, 2 V2) ; for the point (2 V2, — 2 V2). 

* 145. If (r, d) represent variable coordinates, r being a function 
of expressed by the equation r=f{0), then the corresponding point 
will take various 
positions, the total- 
ity of which form 
the graph of the 
function 

The following ex- 
amples wlU illus- 
trate the way in 
which such graphs 
are studied. 

Example 1. r—\0. 

Making a table of 
corresponding values 
of r and 6, we have Fig. 117 



etc. 



Evidently the graph is a spiral, as illustrated in Fig. 117. It is known as 
the Spiral of Archimedes. 




e 





30° 


45° 


60° 


75° 


90° 


135° 


180° 


T 





n 


Hi 


15 


18| 


22^ 


33f 


45 



THE TRIGONOMETRIC FUNCTIONS 

Example 2. '• = sin 6. 

Making a table of values of r and 6, 



185 



r 





h 


.87 


1 


.87 


i 





-h 


e 





30° 


60° 


90° 


120° 


150° 


180° 


210° 



etc. 



Evidently all values of 6 in the third or fourth quadrant will give negative 
values of r, and hence the corresponding points will be in the first or second 




Fig. 118 

quadrant. Furthermore, sin (90°+^) =sin (90°-^), so that the graph is sym- 
metrical with respect to the line ^ = 90° The curve is a circle (Fig. 118). 

Example 3. r = sin 2 d. 

Here the table of values is as follows : 



r 





.5 


.87 


1 


,87 


.5 





-.5 


e 





15° 


30° 


45° 


60° 


■ 76° 


90° 


105° 



r 


-.87 


-1 


-.87 


-.5 





.5 


.87 


6 


120° 


135° 


150° 


165° 


180° 


196° 


210° 



etc. 



in the same order as for 6 in the first and second quadrants. 

We do not need to pay attention to the values of beyond 180°, because 

increasing B by 180° increases 2 « by 360°, and hence gives the same value 



186 



THE ELEMENTARY FUNCTIONS 



of sin 2 5 as if we had used $ itself. Thus, sin (2 • 195°) = sin (2 • 15"), 
sin (2 • 210°) = sin 60°, etc. Of course the values of r are negative for all 
values of 6 in the second quadrant (for 2 ^ is then in the third or fourth 




quadrant); hence the points on the curve are in the fourth quadrant 
(see Fig. 119). When 6 is in the third quadrant, however, 2 6 is in the 
first or second ; hence r is positive, and the points on the curve are in the 
third quadrant. The graph is the curve known as the four-leaved rose. 



EXERCISES 



Draw the graphs of each of the following equations in polar 
coordinates : 



1. r = e. 

2. r = cos 6. 

3. r = l — coBO. 
i. r = cos 2 d. 
B. r = tan 6. 

6. r'' = cos 2 e. 



■ 6 

7. r = sin-- 

8. r = tan-' 

o 

9. ?• = tan 5 • sin 5. 
10. . = 1 



11. r = sin- +1. 

Z 

12. /• = sec ± 1. 

13. r = 1 + sinf fl. 
6 



14. r = 



1 — 2 cos e 



THE TRIGONOMETRIC FUNCTIONS 187 

8 3 

1 + 2 cos e 1 + cos e 3 

8^ ,„ 8 22. r = acsc4- 

16. »' = acos'-r- 19. r = - -• 2 

o 3 — cos 6 a 

17. ?-cos2fl = a''. 20. »- = 4(l-2cose). 23. r = asm''-- 

24. ?• = atan^^secfl. . 6 d 

29. r = sin- + C0S7^• 
25. r=asm3e. , "^ ^ 



26. r = a cos 3 6. an j.2 = _ 

a'sm^d + b^cos^e 

27. »-2 = cos4e. 

31. r = a tan 3 d. 

28. r^ cos e = a' sin 3 e. 



CHAPTER X 

THE EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

146. In elementary algebra, the symbol a" was defined, where 
n represents a positive integer, as meaning a • a • a ■ a • • • to n 
factors. The number a is called the base, and n the exponent. 
The laws of operation are as follows : 
I. a*. a* = a" + * 

II. ^ = a''-*if A>;fc 
= ^ if k>h. 

III. («*)* = a**. 
. IV. («&)* = a* . bK 

EXERCISES 

The following easy exercises in review of these important laws 
should be worked over as often as is necessary in order to insure 
perfect mastery of them : 

a;y ' a + b ' (x — yy ' 
oio 
2 . 2" . 2 = ? y = ? 2"' + 2 = ? (Do not multiply out.) 

3 (x' + /)'■ _. [(a + bY-ir _^ 

• (x'-yy ■ (a + b-iy ■ 

i. of ■x^" = ? w" ■ av = ? 

7. What is the value of (»y? of a''? (Note that (a"')" is wo< the 
same as a™".) 

8. What is the difference in meaning between 2^ and (2'')^? 

188 



EXPONENTIAL AND LOGARITHMIC FUNCTIONS 189 

147. It is evident that our definition of a* has no meaning 
unless w is a positive integer, because we cannot multiply together 
a fractional number of factors or a negative number of them ; 
but it is equally evident, after a little thought, that the laws I-IV 
do Tiot lead to any absurdity if we suppose that the exponent 
can become fractional or negative. For instance. Law I gives for 
}i=zh = \, a* • a^ = «i J that is, a' is a number which, when mul- 
tiplied by itself, gives a. 

But " a number which, when multiplied by itself, gives a " is 
the square root of a. Hence Law I will be absolutely true for 
h = k = i if we simply regard a^ as meaning y/a. We shaU accord- 
ingly take this as the definition of a^. (Notice that Law III leads 
to the same resvdt, by taking h = ^, k=2.) By using the same 
process of reasoning with three factors we have a^ • a^ ■ a^ = a, 
and hence we must define a^ as meaning v a ; and by extending 

the method farther we find, in general, a« = "v^. (Or, still better, 

we may get this result by using Law III with h = - and k = n.) 

EXERCISES 

1. Show (from Law III) that a^ ={-^T = ^ ; that a* = Vo" ; 

m 

that a* = -v^; and, in general, that a" =Var={VaJ'. 

2. Write down the values of 9*, 8^, 4^, 16*. 

148. Now let us proceed in the same way to find a meaning 

for a negative exponent by testing the laws I-IV. For instance, 

what should be the meaning of a-^ ? If Law I is to hold, a-^ • a^ 

= a2-i — a. 

. a 1 

Therefore «" = ~2 ^ 37 ' 

a^ a 

Or h can be kept general : 

Therefore "'^""^"a" 



Similarly, it can be shown that 



«-» = ■ 
a" 



190 THE ELEMENTARY FUNCTIONS 

EXERCISES 

1. Show that the last statement is true whether w is an integer or 
a fraction. 

2. Write down the values of i'', 4"*, 4"^, IQ-i, 10-", 9"^, 8~^, 
8-* 8"*, 16-*. 

3. Show that the first part of Law II holds even if A<A;. What 
does it become if h = k? What is therefore the meaning that we 
must give to a"? 

4. Make a table of the powers of the base 4, using as exponents all 
values that give rational results, from 4"° to 4*, inclusive. The table 
should then have twenty-five entries. By its aid many problems of 
multiplication and division can be solved without any computation. 

Thus, 32 X 16 = ? 

Solution. From the table, 32 = ii, 

16 = 42. 

Therefore 32 x 16 = 4^ . 42 = 4! = 512. 

Again, 4096 -=-128 = ? 

Solution. From the table, 4096 = 4^ 

128 = 4^ 
Therefore 4096 h- 128 = i^-i = ii = 32. 

Work the following in a similar way : 

(1) 8 ^ 128 = ? (5) -v''4048 = ? | 32.64 ^ ^ 

(2) 256 • 8 = ? (6) VIO24 = ? ^ ^ N2.256 ■ 

(3) ^V • 2048 = ? (7) ■v'1024 = ? (10) (^ ■ 128)^ = ? 

(4) 5I5 . 64 = ? (8) V256 • -v/612 = ? 

> 
149. Such a table as that of Ex. 4 is called a table of loga- 
rithms. For instance, in the equation 4^ = 64, 3 is called the 
logarithm of 64 to the base 4. In general, if a^=b, x is the loga- 
rithm of h to the base a. This is abbreviated j: = log„&. The 
number h in this equation is called the antilogarithm. Thus a 
logarithm is the exponent of the power to which the base must 
be raised in order to get the antilogarithm. "8~i=^" means 



EXPONENTIAL AND LOGARITHMIC FUNCTIONS 191 

exactly the same as "loggi =-^." 8 is the base, - 1 the loga- 
rithm, and I the antilogarithm. The student should now practice 
translating equations from the exponential form to the logarithmic 
and back again. 

EXERCISES 

8' = 2 is equivalent to log, 2 = ^. 
logj^lOO = 2 is equivalent to lO'' = 100, 

1. Work the following in a similar way : 

(1) log,9 = ? (6) log,3 = ? (11) logj,8 = ? 

(2) log^oT^TT = ? (7) log,,a) = ? (12) log^81 = ? 

(3) log^^lO = ? (8) log^^lOO = ? (13) log^^lO . log,„100 = ? 

(4) log327 = ? (9) log^3 = ? (14) log,27 • log^3 = ? 

(5) log,16 . log,,8 =? (10) log,125 • log^S = ? 

2. Translate into the logarithmic form, 

(1) 10' = 1000. (2) 16- J = f (3) 32-^ = ^. 

3. What is the value of log^^l? of log„l? 

150. Having thus fixed in mind the fundamental fact that 
logarithms are exponents, we see that the laws of operation with 
exponents must naturally give corresponding laws of operation 
with logarithms. Law I, »*.«*= «'' + * tells us that the exponent 
of a product is equal to the sum of the exponents of the factors, 
the base being the same. Hence, using the word "logarithm" for 
its equivalent, "exponent," we have, as the First Law of Loga- 
rithms, The logarithm of a product equals the sum of the logarithms 
of thg factors, all to the same base. 

a* 

151. Similarly, Law II, — = a''-'', tells us that, the base beiug 

a* 

the same, the exponent in a quotient equals the exponent in the 
dividend minus the exponent in the divisor. Hence, using the 
word "logarithm" for "exponent," we have, as the Second Law of 
Logarithms, The logarithm of a quotient equals the logarithm 
of the dividend minus the logarithm of the divisor. 



192 THE ELEMENTARY FUNCTIONS 

152. Similarly, Law III, («'')* = «''*, gives as the Third Law 
of Logarithms, The logarithm of a power of a number equals the 
logarithm of the number, multiplied by the exponent of the power. 

For, calling 

^ a^ = N, (1) 

TV* = a**. (Law III) (2) 

Translating (1) into logarithmic form, 

h = log^N. 

Translating (2) into logarithmic form, 

M = log„(iV*). 
Therefore log„(iV*) = h • log^N. 

Example, logg (4') = 3 log8 4. 

We can verify this by writing down the values of the logarithms on 
both sides of this equation, thus : 

4" = 64, 
log,64 = 2; 
and log8 4 = |, 

31og34 = 3-§=2 = log,64, 
as was to be shown. 

153. Since the nth root is the same as the — th power, this law 

n 

also includes as a corollary the following : The logarithm of a 
root of a number equals the logarithm of the number, divided by 
the index of the root. 

154. If we now consider the application of these laws of loga- 
rithms (taking for base some number > 1), we see at once that 
the tables we can make are very restricted in their scope. Thus, 
with the base 4, or the base 8, we can easUy give the values of 
log 2, log 4, log ^, etc., but not of log 3 or of log 5 ; and wiflh the 
base 9 we can give the value of log 3, but not that of log 2. We 
do not even know that there is any power of 4 that will give 3, 
or of 9 that will give 2. Thus, no matter what base we take, there 
will be large gaps in our table. 

How can this incompleteness be overcome ? Only by intro- 
ducing an assumption, which, although it cannot now be proved, 



EXPONENTIAL AND LOGAEITHMIC FUNCTIONS 193 

yet leads to consistent results and is of the very highest utiUty. 
This assumption is, With any base greater than 0, except 1, every 
positive number has a logarithm ; that is, there exists a number x 
such that a^ = N, where iVand a are any positive numbers (a ¥= 1). 
It turns out that this logarithm x is in most cases an irrational 
number ; thus, log^3 and log9 2 are irrational. To find their approxi- 
mate value as rational numbers is a problem of no very great 
difficulty ; in fact, it is not much harder than extracting the cube 
root of numbers by the arithmetical method. This computation of 
logarithms, however, is best taken up at a later point in the mathe- 
matical course ; for the present we shall use the results obtained 
by other computers and pubhshed in tables of logarithms. (For 
further information see the articles "Tables'' and "Logarithms" 
in the Encyclopedia Britannica.) 

For practical purposes the base 10 is the most convenient. We 
shall therefore use that base altogether, so that, unless otherwise 
stated, logiVis hereafter to be understood as meaning logjgiV: 

EXERCISE 

Make a table of the positive and negative integral powers of 10, 
from 10" to 10- ^ and translate each item into its equivalent loga^ 
rithmic equation, thus : 10^ = 10, log 10 = 1, etc. 

155. Using this elementary table, it will be seen that the loga- 
rithm of any (positive) number can be determined to the nearest 
integer at a glance. Thus, log 11, log 12, and the logarithm of 
any number between 10 and 100 will be between 1 and 2, that is, 
wiU equal 1 plus a decimal ; the logarithm of any number between 
100 ^d 1000 will be between 2 and 3, that is, wiU equal 2 plus a 
dechnal ; and so on. The decimal part of the logarithm is called the 
mantissa ; the integral part, the characteristic. Using these words, 
we have at once the First Rule : TJie characteristic of the logarithm 
of a number betvjeen 1 and 10 is 0, of a number between 10 and 
100 is 1, of a number between 100 and 1000 is 2, and, in gen- 
eral, is one less than the number of digits to the left of the decimal point. 
But if there are no significant figures to the left of the decimal 



194 THE ELEMENTARY FUNCTIONS 

point (that is, if the antilogarithm is less than 1), another rule 
must be formulated. Noting that the characteristic of the logarithm, 
of a number between .1 and 1 is — 1 (since the logarithm is then 
— 1 plus a decimal), that the characteristic of the logarithm of a 
number between .01 and .1 is — 2, and so on, we have the Second 
Eule : When a number is less than 1, the characteristic of its loga- 
rithm is negative and numerically one more than the number of zeros 
after the decimal point before the first significant figure. 

EXERCISE 

Determine the characteristic of the logarithm of each of the 
followLQg numbers : 531, 24, .62, 7.1, .006, 4.06, .05001, 56.3, .403, 
.9, 45,000,000. 

156. For the mantissa we use printed tables of logarithms, 
as has been said ; but their use is facilitated by the follow- 
ing principle : The mantissa of a logarithm is not changed hy 
moving the decimal point in the corresponding number ; that 
is, log 2, log .2, log .0002, log 200, log 20,000,000, all have the 
same mantissa. This important fact is very easily proved, thus : 
Given logiV, then we have, from §§ 150 and 151, 

log (10 i\^) = log ^ + log 10 = (log N) + l; 

and • log(^^ = logiV^-loglO=(logiV^)-l; 

in general, log (iV . 10*) = (log N) + k, 

k being an integer (positive or negative). 

Example. Log 2 = 0.3010, log 20 = 1.3010, log 2000 = 3.3010, log .2 = - 1 
+ .3010 (written 1.3010), log .00002 = - 5 + .3010 = 5.3010. 

157. "We have now the ability to find, with the help of the 
table, the (approximate) values of the logarithms of aU positive 
numbers. The student should practice doing this untU the use of 
the printed tables becomes very easy. Then it is time to apply 
the laws of logarithms to problems. A few examples wUl first be 
worked out. 



EXPONENTIAL AND LOGARITHMIC FUNCTIONS 195 

Example 1. Find the product of 36 by 124. 

Solution. By the First Law, log (36 • 124) = log 36 + log 124. 

From the table, log 36 =i 1.5563 that is, IQi-^^ea = gg 

Also log 124 = 2.0934 that is, loa-0884 = 124 

log (36 • 124) = 3.6497 ioi-6668 + 2.0934 - gg . 124 

Therefore 36 ■ 124 = 4464 By the table, lO'**" = 4464 

Example 2. Find the quotient of .0031 by .0925. 

iir -4- -0031 

Solution. Write x = : 

.0925 

then logx = log .0031 - log .0925. (Second Law) 

From the table, log .0031 = 3.4914 

log .0925 = 2.9661 '' 

log X = 2.5253 Therefore x = .03352 

Translate each equation also into the exponential form, as was done 
in Example 1. 

Example 3. Find the value of v3 to three decimal places. 

Solution. Let x = "v3 ; 

then log X = J log 3, 

by the Corollary to Law III (§ 153). 

From the table, log 3 = 0.4771. 

Therefore log x = 0.2385, 

and X = 1.732 , 

correct to four figures, that is, to three decimal places. 

, 213 X 3.23 



Example 4. Find the value of • 



/594 



Solution. If X is the required value, log x = log (dividend) - log (divisor) 
: log 213 + log 3.23 - i log 594. 

We find log 213 = 2.3284 

log 3.23 = 0-5092 

2.8376 

Log 594 = 2.7738, ^ log 594 = 0-9246 

log X = 1.9130 Therefore x = 81.84 + 



196 THE ELEMENTARY FUNCTIONS 



Find (by 
expressions : 

1. 3.14 X 


using 
.216. 


logs 

3. 
4. 
5. 


EX] 

irithms) 
V21.5. 


ERCISES 

the value of each of the 

9.003 X .3874 
■ 25 X .0067 


following 


„ 293 


</l.331. 
(l.Ol)i^- 




"• .4772 


7. Vl4x 


21 X 35 X 112 X 3.42. 



8. Find the area of a circle of radius 2^ in. (Take log 7r = 0.4971.) 

9. Find the volume of a sphere of radius 3 ft. 3 in. ; find also 
its weight if its specific gravity is 7.2 and a cubic foot of water 
weighs 62.5 lb. 

10. Find the amount of $1 at compound interest for 10 yr. at 
6% ; for 20 p. ; for 100 yr. 

Hint. The amount at the end of 1 yr. is $1.06; at the end of 2 yr., 
|1.06 . 1.06 = f 1.062 ; and so on. 

11. Use logarithms to find the unknown parts and the area of 
the A ABC, given a =15.63, a = 26° 10', y = 90°. 

b 



Solution. 




a 


: cot a. 








Therefore 




b = 


-■ a cot a. 








Therefore 




log 6 = 


= log a + 


log cot a. 






Also 




c = 


a 

sin a 








Therefore 




logc = 


: log a - 


log sin a. 






log 


loga 
cot a 
log 6 


= 1.1940 
= 0.3086 
= 1.5026 




logo 

log sin a 

logc 


= 1.1940 
= 9.6444 - 
= 1.5496 


-10 




b 


= 31.81 




c 


= 35.45 




Check. c2- 


-62 = 


a^ = {c + b){c- 
c + 6 = 
c-b = 

log(c + 6) = 

log(c-6) = 

log a^ = 


-6). 
67.26. 
3.64. 

1.8278 

0.5611 

: 2.3889 












loga = 


: 1.1944, 


which checks the above results. 



EXPONENTIAL AND LOGARITHMIC FUNCTIONS 197 

For the area, S =—■ 

2 

log a = 1.1940 

log b = 1.5026 

log I = 9.6990-10 

log S = 2.3956 

Therefore the area is 248.6+. 

Note. The student will observe that the logarithms of the trigonometric 
functions are given in a separate table. Negative characteristics are iisually 
written " 9 - 10" for " - 1," and so on. 

12. Use logarithms to solve, check, and find the area, given 

(1) a = 421, ;8=54°35', y = 90°. 

(2) a = 37° 15', |8 = 7r54', a = 4.263. 

(3) j8 = 115°, y = 30°30', a = 1.00L 

(4) a =10° 40', y=150°10', c = .124. 

(5) a = 41° 34', a = 46.70, b = 60.03. 

(6) a = .698, c = .615, y = 38° 15'. 

(7) a = 3.16, S = 5.09, /8=147°59'. 

158. These last problems make it clear that the use of loga- 
rithms is practical in solving triangles by the use of the Law of 
Sines; but it is evidently impossible to use them to so good 
advantage in cases where the Law of Cosines applies, because 
the Law of Cosines contains additions as well as multiplications, 
and these cannot be performed by logarithms. So it is necessary, 
in order to gain the advantage of the use of logarithms in the 
computations, to obtain modified formulas that shall include no 
additions or subtractions. We begin with three sides given. 

By the Law of Cbsines, 



cos a = 



2 6c 



(Z (X 

but we have also cos a = 1 - 2 sin^ - = 2 cos^ - - 1. (Ex. 3, p. 1 1 9) 

^ 7i2 _L rt2 /v2 

Therefore 2 sin^ - =1— cosa = l r-7 

2 2 oc 

2lc-W--c^+a^ _ d'--(b- ef 

" 2 be ~ 2bc 

{a + b — c)(a — b + c) 

~ 2 he 



198 THE ELEMENTAEY FUNCTIONS 

The numbers a + b — c and a — b + c are very easily found, and 
so we have a practical formula for logarithmic computation : 

. 1 l(a+b -c){a-b + c) 

'^r=\- ibc ■ (') 

It is usual to abbreviate by writing 2s = a + b + c, or 

a + b + c 
s= ~ 

Therefore 2s — 2a = 6 + c — a, 

2s—2b = a — b + c, 
2s — 2c = a + b — c. 



Therefore (1) becomes sin - a = -\ ^ 'z^ -■ (1 ) 

2 y be 

Similarly, by starting with 2 cos^ 1 a = 1 + cos a, let the student 

show that 1 rz r 

1 \s(s — a) „, 



Dividing (1') by (2), tan^a=^<i-^J^^. (3) 

Using the other angles of the triangle instead of a, 



tani;8=-x|(^H«HiEi). 
2^ \ s(s-6) 

tani,=JiH«)(£3. 
2' \ s{s-c) 

The formulas for the tangent are generally the best to use, be- 
cause all three angles can be found with very little more work 
than would otherwise be needed to find one angle. 
One further abbreviation is very useful: 



Since 
and 


tanj«= 1 is-a)is-bns-c) 
2 s —a y s 

tan^^- ^ {s-a)(s-b)is-c)^ 
2 s — 6 \ s 


and 


tan^- 1 (s-a)is-b)(s-c) 
2 s — cy s ' 



EXPONENTIAL AND LOGARITHMIC FUNCTIONS 199 
therefore, if we write 

^_ j(^:r^(s-b){s-c) ^ 

we have tan-a^ 



2 s— a 
2^^ s-b 



taii-y = ; 

2 s — c 

or, using logarithms, 

log tan 1^ a = log r — log (s — a), 
and log r- = 1 [log {s-a)+ log {s-h)+ log (s-c)- log s]. 

Example. Given a = 46.34, 6 = 31.27, c = 55.03, to find the angles. 

a = 46.34 s-a = 19.98 

6= 31.27 s-b =35.05 

c = 55.03 s-c= 11.29 

2s = 132.64 66.32 = s 

s = 66.32 (Notice that (s - a) + (s - i) + 

(s — c) = s, necessarily. Why so?) 

To find log r: 

logr = i[log(s - a) + log(s - b) + log(s - c) - logs]. 

log (s- a) = 1.3006 

log (s- 6) = 1.5447 

log (s-c) = 1^0527 

3.8980 

logs = 1.8216 

log r2 = 2.0764 

logr =1.0382 
Therefore log tan ^ a: = 9.7376 - 10 

log tan i/3 = 9.4935 -10 
log tan iy =9.9855-10 

Therefore ia = 28=39' 

i;3=17n8' 
^y = 44°3' 

90°00' Check 

Therefore a = 57° 18', j8 = 34°36', 7 = 88^- 



200 



THE ELEMENTARY FUNCTIONS 



159. If the area is to be found, we may use the formula already 
proved (p. 159): Sz 



Now 



sma : 



1 be sm a. 
2sm-cos-; 



or, using (1') and (2) (p. 198), 



sin a = —y/s(s — a){s — b){s — c). 

DC 

Therefore S = ^s{s - a){s - h){s - c). 

This important result is called Hero's Formula} Notice that 
S = r • s also. 

EXERCISES 

1. Solve and check, and find the area: 

(1) a = 13, b = li, c = 15. 

(2) a = .00365, b = .00846, c = .00697. 

(3) a = 100.1, b = 102.1, c = 104.1. 

(4) a = 3.194, b = 5.235, c = 6.118. 
J (s-a)(.s-b)(s- 

B 



2. Prove that r 
inscribed circle. 



is the radius of the 



160. The only remaining case of the solu- 
tion of a triangle that requires new formulas ^ 
for logarithmic work is that in which two 
sides and the included angle are given. Sup- 
pose the given parts are a, b, y, and a >b. 
Construct the triangle ABC (Fig. 120), and 
locate the point D on CB, so that CD = b ; 
then BD= a~b. Produce BC to JS, making (^ 
CE= b ; then 5^= a + b. 

The student may show that 



ZAEC = ZCAE = i, 



and that 

Therefore 
Also 



ABAD = a- 
ZDAE=W 



-^ 



a+^ a—^ 




Fio. 120 



1 From Hero of Alexandria, who was the first to prove it (about the first 
century a.d.). His method was a purely geometric one. 



EXPONENTIAL AND LOGARITHMIC FUNCTIONS 201 

Therefore /.BAE= 90° + ^^ • 

Now apply the Law of Sines, first to ABDA and then to ABU A. 

. a — B . a — B 
sm — - — sm — - — 

We get = ^ = , (1) 

sm — - — cos ~ 
2 2 

sml90 H -— ) cos '^ 

a+6 \ 2 / 2 

and — !— = = (2) 

c -7 .7 ^ ' 

sin 4- sm - 

2 2 

Formulas (1) and (2) (leaving out the second fraction in each line) 
are called MoUweide's Formulas.^ By dividing (1) by (2) we get 

a-B a-B 

tan — tan C 

^Z^ = ?_^ ?_. (3) 

a+b a+fi V 

tan — cot — 

2 2 

This important result is known as the Law of Tangents ; it pro- 
vides an easy way of solving by logarithms in case two sides and 
the included angle are given. 

Example. Given a = 26.71, b = 20.89, y = 61° 32', to find a, p, and c. 

a + 6 = 47.60, 

a-h = 5.82, 

9i±A = 90° - 3^ = 90° - 30° 46' = 59° 14'. 
2 2 

From the Law of Tangents, 

log tan ^^^ = log (a - ft) + log tan ^^^ - log (a + 6). 

log (a- h) = 0.7649 

, ^ a + S 0.2252 
logtan^ = ^;^g^ 

log(a + 6) = 1.6776 

Therefore log tan ^^^ = 9.3125 - 10 



and 



a - 



j8. 



: 11° 36' 



1 After C. B. MoUweide (1774-1825), who, however, was not the first to dis- 
cover them. Formula (2) was first given by Newton in 1707, and Formula (1) 
by Friedrich Wilhelm von Oppel in 1746. 



202 THE ELEMENTARY FUNCTIONS 

Therefore a = ^^ + ^^ = 70° 50^ 



^ = 4^-^ = 4^38; 

Check. a + /3 + y = 180°. 

To find c, use either of MoUweide's Formulas ; thus (1) gives 

log c = log(a -b) + log cos ^ - log sin ^y^- 

log(a-i)= 0.7649 (a + J)sin5: 

log cos I = 9.9341-10 Check. c = ^ 

10.6990 - 10 cos "-P 

.'^~P— a 5n94 _ 1 n 2 



log sin ^^-2 = 9.3034 - 10 
rc= 1.3956 
"= ?M5 """ ^ 11.3865 - 10 



log (a + 6)= 1.6776 
logc= 1.3956 log sin ^= 9.7089-10 



log cos ^^ = 9.9910-10 
logc= 1.3955 
EXERCISES 

Solve, check, and find the area : 

(1) a = 75, b = 70, y= 56° 18'. 

(2) a = 463, b = 499, y = 28° 6'. 

(3) b = 2.68, c = 1.69, a = 80° 54'. 

(4) a = 41.3, b = 28.7, y = 136° 28'. 

Note. Inasmuch as many of the formulas used have not been proved for 
the case a > 90°, the student should supply this omission here. 

161. These examples illustrate but' a few of the very large 
number of applications of logarithmic computations. The subjects 
of surveying, navigation, and astronomy make constant use of th js 
labor-saving device. It may, indeed, be said to have revolutionized 
numerical work in every field. 

We now return to the study of the logarithmic function itself. 
The best way to get a general idea of this function (as of any other) 
is by drawing a careful graph of it. Accordingly, writing y = logi()a;, 
we lay off a> values as abscissas and y-values as ordinates (choosing 
a suitable scale, as usual). As we have had no definition of a loga- 
rithm of a negative number, no points can be located to the left of 
the y-axis. The details of forming the graph are left to the student. 



CHAPTER XI 

INTRODUCTION TO THE DIFFERENTIAL CALCULUS 

162. The preceding chapters have been devoted to a study of 
various types of functions, — their fundamental properties, their 
graphical representation, and some algebraic and geometric results 
derived therefrom. The linear function, the quadratic function, 
the fractional function, the irrational function, the trigonometric 
functions, the logarithmic and exponential functions, all have 
their individual characteristics, which are in each case intimately 
connected with their graphs, and which must be well understood 
as a preliminary to the solving of many important problems. A 
complete vmderstanding of a function would involve a complete 
knowledge of the way in which the function changes as the inde- 
pendent variable changes. The object of this chapter is to lay the 
foundation for such an understanding. The branch of mathematics 
which investigates the rate of change of a function is called the 
differential calculus. 

163. Although this branch of mathematics is of vast complexity 
and difficulty in its advanced aspects (indeed, no one can even 
faintly imagine the extent which future discoveries may give to 
its applications), yet in its simpler aspects it is exceedingly simple. 
This win be seen by examining one of the simple functional rela- 
tions, which has already been mentioned (p. 23), — the case of a 
man walking at the constant rate of 3 mi. per hour. If t represents 
the number of hours he has walked, and s the number of miles he 
has gone, then s is a function of t[s=f{t)]. Evidently the func- 
tional relation is the linear one s = 3 t. If we consider two different 
values of t, say t = 2 and t = 2|, the corresponding values of s 
are 6 and 6| ; that is, the additional distance he has gone in the 
extra 1 hr. is | mi. Again, if we take t = 3 and t = 3^, we find 
that the additional distance he has gone in the extra ^ hr. is 

203 



204 THE ELEMENTARY FUNCTIONS 

again | mi. In fact, it is evident that any ^-hr. increase in the 
time spent will produce an increase of | mi. in the distance cov- 
ered ; and that is exactly what is involved in the statement that 
his rate is constant. 

164. Contrast with this the functional relation between dis- 
tance and time that exists in the case of a body falling freely 
near the surface of the earth. The Law of Falling Bodies ^ 
gives this functional relation to be (neglecting the resistance of 
the air) s = 16 <2, where t is the number of seconds the body has 
been falUng and s the number of feet it has fallen. If we give 
t two different values, say t=2. and t = 2\, we find the corre- 
sponding values of s to be 64 and 81 (that is, the additional fall 
during the extra \ sec. is 17 ft.) ; but if we take t=2> and t = 3|, 
we find the corresponding values of s to be 144 and 169, the 
additional fall during the extra \ sec. being thus 25 ft. In fact, 
we shall find that an increase of ^ in i will produce a different 
amount of change in s, according to the value of t chosen to 
begin with. This is what is involved in the statement that the 
rate of fall is not constant. 

165. These two simple examples contain the essential material 
out of which the structure of the differential calculus is built; 
namely, the ideas of functional relation and rate of change of a 
function. A number of examples of a similar kind will now be 
considered, and although the computations involved are, as was 
said above, extremely simple, still they bring out the essential 
thing, which is, the effect upon a function of changing the inde- 
pendent variable by a certain amount. This " amount of change " 
is called an incre^nent, and is to be considered positive if the 
change is an increase, negative if the change is a decrease. Incre- 
ments are usually symbolized by the Greek letter A prefixed to the 
letter standing for the variable in question. Thus, " the increment 
of t " means " the amount of change in t " and is symbolized by At. 
In the examples above, A< = ^ (that is, ^ hr. in the first example and 
^ sec. in the second example). As a further abbreviation the symbol 

1 Discovered and proved by Galileo (1564-1642) ; published in his " Dialogues 

on Two New Sciences," in 10o8. 



INTRODUCTION TO DIFFERENTIAL CALCULUS 205 



/(2) is used for « the value of f{t) when t = 2" or « the value 
of /(«) when x = 2." Thus, itf{t) = 16 <2, then /(2) = 16 • 2^= 64, 
/(O) = 16 . 02 = 0, /(a) = 16 . a2, /(J + 1) = 16 (J + 1^2^ g^^ 

Example 1. Take the function s = 5 < + 6, and find the effect upon the 
value of s of an increase of 1 in the value of i} (a) when t = 2, (b) when 
t = 4, (c) when t = J, (d) when t = t^. 

Solution, s =f(t) = 5^ + 6. 

(a) When < = 2, s =/(2) = 5 • 2 + 6 = 16. 
When < = 2 +1, s=/(8) = 5-3 + 6 = 21, 

Therefore the change in s, As, is equal to/(3) — /(2) = 21 — 16 = 5. 

(b) When < = 4, s =/(4) = 5 • 4 + 6 = 26. 
When / = 4 +1, s=/(5) = 5-5 + 6 =31. 

Therefore the change in s, As, is equal to/(5) — /(4) = 31 — 26 = 5. 



(c) When 
When 

Therefore 

(d) When 
When 

Therefore 



t=h> =/(*) = 5 ■ i + 6 = 8|. 
( = J + 1, s =/(f) = 5 • I + 6 = 13^ 
^s=/(i)-/(i) = 13i-8i = .5. 



; = «i + 1, s =/(«! + 1) = 5 ft + 1) + 6 = 5 <i + 11. 
As=/ft + l)-/ft) = 5. 

This result shows, since <j is any value of t, that As = 5 for aH values of 
when A< = 1 ; that is, a change of 1 y 

in « produces a change of 5 in s, no 
matter what value of t we start with. 
In the same way it can be shown that 
a change of any amount A; in t pro- 
duces a change of 5 A< in s, whatever 
value of t be chosen to begin with. 
This means that the rate of change 
of s, compared with that of t, is con- 
stant, s changing five times as much 
as t does (As = 5 A(). 

The graphical representation of 
the function s will throw more light 
upon these results. If we use x and y 
to stand for the variables, instead of 
J and t, the functional relation will 
be y = 5 a: -I- 6. The graph of this is the straight line with slope 5 and 
F-interoept 6. The work above proved that for any value of x, Ai/ = 5 

^ In symbols this would be stated, Find the value of As when Ai = 1. 




206 



THE ELEMENTARY FUNCTIONS 



so that the rate 



when Aa; = 1. This means that, whatever point on the straight line we 
start with, whether A,B, C, D, or any other, an increase of 1 in a; produces 
an increase of 5 in ;/. Geometrically this is evident, because of the fact 
that the graph is a straight line. (The student may prove by elementary 
geometry that this is true.) Show further that the function y = rnx + k 
gives always Ay = m- Ax, whatever value of x be chosen to begin with. 

Example 2. Let us now consider a quadratic function, choosing y = x^ 
as the simplest one. Starting with any value of x we please, say x = x^, let 
us find out what change in the value of y will be produced by a change Ax 
in the value of x. When x = x^, y =f(x^ = x^. 

When X = Xj^+ \x, y =f{Xi + Ai) = (x^ + Aa;)^. 

Therefore £i.y =f{x^ + Ax) -/(a^i) = {x^ + Aa:)^ - x ^ = 2 ijAa; + (Aa;)'''. 

This is evidently not the same for all values of 
of change of this function is \ y 

not constant. For instance, if 
Xj = 1, A?/ = 2 Aa; + (Ax)^; while 
if Xj = J, A^ = Ax + (Ax)^, and if 
Xj = 3, A?^ = 6 Ax + (Ax)''. Refer- 
ence to the graph will make this 
fact still clearer. Thus,inFig.l22, 
if Ps(xi, ^j) and if PR = Ax, 
then RQ = Ay, SP=r y^ =A^i), 
TQ=f(ix^ + Ax) = y^ + Ay. It _ 
is geometrically evident that for 
the same value of PR the length 
of -RQ wiU be different accord- 
ing to where P is located upon 
the curve. This of course agrees 
with the result of the algebraic 
work above, which showed that RQ = 2x^Ax + (Ax)^, a length which is 
different according to the value of Xj chosen. 




Fig. 122 



EXERCISES 

Work through the same study as in Examples 1 and 2 for each 
of the following functions, illustrating by the graph : 

1. s=At + 2. 4. s = i!!-|. 7.y = 2x'-l. 

Z. s=-2t+l. 5. y=-6a; -f- 3. 8. 2/=l-a;». 

3. y = 10x-f 25. 6.y = x' + 3. 9. y = 2-3x\ 

In particular, find the value of Ay (or As) when x (or ^) = 1 and 
Aa; (or A^) = l, .1, and .01 respectively. 



nsTTRODUCTIOlSr TO DIFFERENTIAL CALCULUS 207 

166. Instantaneous rate of change of a function at a point. 

Keturning to the function y = x^oi Example 2, § 165, we had 
established the fact that for any given value of x, such as a; = a;,, 
Ay = 2 ajj . Aa; + (Aa;)^. This is the total' change in y produced by 
increasing x by the amount Aa;. Hence the average change in y 
per unit of Aa; will be found by dividing Ay by Aa;,i the average 

Av 
rate of change — ^ . or 2 a;j + Aa;. If now we take Aa; smaller and 

smaller, we get the average rate of change in y throughout an ever- 
diminishing interval, and this average rate will in this case (and 
generally) approach a deiinite value, here 2 a;j, as Aa; approaches 0. 

Av 
This value, which — ^ approaches as Aa; diminishes, is called the 

instantaneous rate of change of y relative to x at the point x — x^. 
This instantaneous rate of change is what we mean when we speak 
of the rate of change of any variable at a definite point or at a 
definite time. Thus, to state that a train is at a certain moment 
moving at the rate of 30 mi. per hour does not necessarily mean 
that it continues to move at that rate for an hour, or even for 
a single second ; it means, rather, that the instantaneous rate 
(namely, the limiting value of the average rate throughout an in- 
terval of time or of space, as the interval is conceived to diminish 
toward zero) is 30 mi. per hour. Of course it would be impossible 
by any process of observation or experiment to determiue what 
this limiting value is ; it is determined by a process of reasoning, 
which may be very simple or very difficult, according to the 
nature of the functional relation involved. Only in case the 
variable is changing at a constant rate, as in Ex. 1, p. 205, can 
we avoid this consideration of a limitiug value, because only then 
can we take any fixed interval we please and be sure that the 
rate of change for this interval is the same as that for any other 
interval. In that example (p. 205) we saw that As = 5 A<; that is, 

— = 5, a constant, so that the rate of change of s is always 5 times 

A^ 

the rate of change of t. But if the rate is not constant, we cannot 

1 If the total change in y were 10 for Ax = 3, the average change in y per 
unit of Ai would be J^. 



208 THE ELEMENTARY FUNCTIONS 

As / AwN 
do this, but must consider what happens to the ratio — I or —I 

as the interval At (or Aa;) becomes smaller and smaller. It is thus 
evident that a study of the rate of change of a variable will nec- 
essarily involve a study of limiting values, or, more briefly, limits. 
167. Limits. The student has already met with a few examples 
of limits ; for instance, in elementary geometry it is stated that 
the limit of the perimeter of a regular polygon inscribed in a 
circle, as the number of sides is indefinitely increased, is the 
circumference of the circle, and that the limit of the apothem 
of the polygon is the radius of the circle. The following are 
other examples of limits. 

1. If X represents a variable which assumes the series of values 
hhhh ' ' '• ^^^^ ^ approaches the limit ; in symbols, lim x=0, 
or x = Q. 

2. If X takes on the series of values .9, .99, .999, • • •, then 
lima; = 1, or x = l. 

3. If y==- and a; = 10, then y = -— ; this statement is abbrevi- 
ated thus : 



limy = 777' or lim (-) = -—• 



4. If 2/ = 2 +h and h = 0, then y = 2; in symbols, lim (2 + h) = 2. 

5. If y = — and x is positive and diminishes toward as limit, y 

^ ■ ■ ■ /1\ 

will increase without limit ; in symbols, lim y = oo, or lim ( — 1 = oo. 

6. If y = - and x is negative and approaches as limit, y 

wUl decrease without limit (since it is negative and its numerical 
value becomes larger and larger); in symbols, limy = — oo, or 



ii„(l)=. 

x=0-\x/ 



00. 



'^- I^ y = oS ' "^liere m = 1, 2, 3, 4, • • •, y approaches as limit ; 



in symbols, lim ( — 1=0. 

n = Qo \^ / 

8. If y = ^' where 
without limit ; in symbols, lim (— ) 



^- ^^ 2^ =^' where ?!, = — 1, —2, —3, —4, ••-, y increases 



INTRODUCTION TO DIFFEEENTIAL CALCULUS 209 
9. If a; = l + (-i)» where 71 = 1, 2,3,4, • . ., lima; = L 

10. If a; = and z increases without limit, x approaches 

— 1 as limit, that is, lim a; = — 1 ; if a = 0, lim a; = |. 

11. If y = &m.x and a; = 90°, y = l; that is, limy = l. 

1 a: = 90 .. 

12. If y = sin - and x^O, y has tw limit, because - increases 

iC . X 

without limit, and hence sin- will take on all values between 

X 

— 1 and + 1 over and over again ; that is, the values of y will 
oscillate as x diminishes. 

13. li y = log X, lim y = 0. 

x = l 

14. If P^ is the straight line joining any two points F and Q on 
a curve, and if Q approaches P along the curve, the straight line 
PQ will usually approach a limiting position PT, which is called 
the tangent to the curve at P. Also, ZBCQ=ZBDT (Fig. 123). 





Fig. 123 

15. If two circles C and C, of equal radii, intersect, and if 
C approaches C, its center 0' remaining on the straight line O'O, 
then the points of intersection ^ and -^ will approach Q and S 
respectively, where Q and B are the ends of the diameter through 
O, perpendicular to O'O (Fig. 124). 

168. Definition of limit. These examples make the idea of 
limit sufficiently clear so that we are prepared for a definition 
of the word: A variable v approaches a constant I as a limit if 



210 THE ELEMENTARY FUNCTIONS 

\v — l\ becomes^ and remains less than any assignable positive 
number e (epsilon). The above thirteen algebraic examples should 
be thought through carefuUy to verify that this definition is satis- 
fied in each case. In the geometric examples (Exs. 14 and 15) 
it is either an angle or a length that is the variable, so that here 
also the definition will be seen to be satisfied. 

EXERCISES 

Show that the following limit equations are true : 

1. If y =1+ -) where /i =1, 2, 3, 4, • • •, then v =1. 

n 

„ 1 -f-a; 1 

2 . i± V = ) hm V = -■ 

3. lim-^=l. 
x=oi- — x 

4. lim = 0. 

X = cO J- -^ 

5. lim-^=l. 

x = at J- ~r iC 

6. lim [tan^(a; —■ ff)— cos a;] = sec^^. 

x = 180- 

3-^ — 1 r'' — 1 

7. lim f = 3 ; lim ~ = 2. 

x = 2X—i- x^lX—1 
^8 -j^ ^n '\_ 

8. lim — = 3 ; lim —■ = n,n being a positive integer. 

x=\ ^ 1 x=i ^ — 1 

9 lim ^(^+^ ) 1 

■ „^oo(»i + 2)(n + 3). -"• 

169. Theorems on limits. The following theorems are easily 
seen to result from the definition of a limit : 

I. If lim v = l, then lim (v + c)=l + c, c being any constant. 
II. If lim V = 1, then lim (cv) = cl, c being any constant. 

III. If lim v■^ = Zj, and lim v^ = l^, then lim (v■^ ± v^) =l^± l^ 

IV. If lim'yi=Zj, and \imv^ = l^, then \\vsx.{v^v^=l.^l^. 
V. If lim v^ = \, and lim v^ = l^, then 



lim/^ j= -1 {provided l^ + 0). 



1 |b — i| means " the numerical value of u — T'; that Is, i) — i itself if »>/, 
but I — n ii v <l. 



INTRODUCTION TO DIFFERENTIAL CALCULUS 211 

170. The derivative. The most important case of a limiting 
value with which we have to do is that of the instantaneous 
rate of change of a function, described in § 166. It is the limit 

of the quotient -^ as Ax approaches 0. In the case of the 

function y = x^ we found that -^ = 2 ajj + Ax. Hence for this 
function we have the fact that 

Hm--^ = 2x.. 

This limiting value of the quotient -^ is called the derivative 

of y with respect to x at the point x = x^. In words, the 
derivative is the limit of the ratio 

increment of the function 



increment of the independent variable 

as the increment of the independent variable approaches 0. The 
symbol used is I>a:y]x^xi> ^^^^ "*''^® derivative of y with respect 
to X at the point x = Xj." 

Thus DM.-. = lim^ = ^■^/(^. + A^)-/(^,) 

If the function is s =/(t), the corresponding statement will be 
/,,.],_, =lim^*=lim>^^l±^?z£(i). 

Other symbols often used for the derivative of f(x) with 
respect to x are Dxf{x) and /'(«). The last form is especially 
convenient if we wish to show clearly for what value of x the 
derivative is to be understood ; for example, f'(Xj), /' (0), etc. 
mean " the derivative at the point x = x-^, x= 0," etc. 

171. The derivative as the slope of the tangent to a curve. We 
have seen that the derivative of a function gives the value of the 
instantaneous rate of change of the function in relation to that of 
the independent variable. It can also be interpreted in another way, 
as can be seen by referring to Fig. 122, p. 206, which is the graph 



212 THE ELEMENTARY FUNCTIONS 

of the function y = x^. In this figure EQ —Ay =/(x^ + Ax) — /(xj) 

= 2 x, Aa; + (Ax)\ and PH = Ax. The quotient — ^ = — is the 
^ ' ^ " ^ Ax PB 

value of tan Z.BPQ ; that is, it is the slope of the line PQ. If we 

now let Ax diminish toward zero, the point Q will approach the 

point P along the curve, and the chord PQ wUl approach as 

limiting position the line PP', which is the tangent to the curve 

at P. Also, the angle EPQ will approach as limit the angle 

RPP' = 8, and therefore tan ZBPQ = tand; that is, 

Ay 
lim _^ = tan fl. 
Ax=o Ax 

Ay 
In words, the limiting value of — ^ as Ax approaches is the 

slope of the tangent to the curve at the point x = x-^; or, the 
derivative of y with respect to x gives the slope of the tangent 
to the curve at the point for which the derivative was formed. 

Thus, the derivative of the function y = ^ heing 2 iCj at any 
point a; = a;j (§ 170) we can say that the slope of the tangent to 
the curve y = a^ at the point x = x.^ is 2x^. 

172. We shall now consider by means of some examples the 
method of finding the derivatives of a few functions. 

Example 1. Find the derivative of the function y = 2x''' — x at any 
point X = Xy 

Solution. Let y^ be the value of y when x = Xj. Then y^ =fixj) — 2x? — x^, 

and 2/i + Ay =f(x^ + Aj:) = 2 (Xj + \xf - (ij + Ax). 

Hence, by subtraction, 

A)/ = 4 XiAz + 2 (Ax)2 - Ax 



and 


^ = 4x,-l+2Ax. 
Ax ^ 


Therefore 


lim ^ = 4x^-1; 


that is, 


D.y]x=..= 4xi-L 



Thus the slope of the tangent to the curve y = 2x^ — x at the point 
(Xj, y^) is 4 Xj — 1. Draw a figure. 



INTRODUCTION TO DIFFERENTIAL CALCULUS 213 
Example 2. Find the derivative of the function y = -■ 

X 

Solution. From now on we shall drop the subscripts and write, for the 
point at which the derivative is to be found, (x, y) instead of (x^, y^. 

Then 2,=/(z) = i. 

X j^ 

y ■¥ /^y=f{xJr iix)-- 



Therefore Ay = 



X + Aa; 
1 1 -Ax 



x + Ax X x(x+Ax) 

Ay^ -1 
Ax X (x + Ax) 



and 

Hence lim ^ = - i (Theorem V, § 169), provided x 5^ ; 

a3;=oAx ^ x2 

that is, Dj, (- ) = ^ > or the slope of the tangent to the curve y = - at 

the point (x, y) is -■ Draw a figure. 

X — 1 

Example 3. Find the derivative of at any point (x, y). 

'^ 2x + o 

X — 1 
Solution. y = 



2x + 3 

, . (x + Ax)-l 

^ + ^^ = 2(x + Ax) + 3- 
Therefore 

X + Ax — 1 X — 1 

^2' = 2(x + Ax) + 3~2x + 3 

(2x + 3)(x-l) + (2x + 3)Ax-(2x + 3)(x-l)-2Ax(x-l) 
(2 X + 3) (2 X + 3 + 2 Ax) 

5 Ax 

'^(2x + 3)(2x + 3 + 2Ai;)' 



Ay 5_ 

aI~(2x + 3)(2x + 3 + 2Ax)' 

1- ^y - 5 

Therefore, ^lim — - ^^^_^_^y - 

/x-l\_ 5 
that is, ^xy^ ^ ^ 3; (2 X + 3)2 

The student should verify this result in a figure, using several special 
values for x, such as x = 1, x = 0, x = — 1. 



214 THE ELEMENTARY FUNCTIONS 



EXERCISES 



Find the derivatives of each of the following functions, and draw 
figures : 

l.y = x' + 5x. ^ 2x + Z _ 

X 



'■y = ^r^r- ^^•^ = 5^^-6- 



2. y = x". 

X ,_ 3 



3.y = 3x^-4.x. ^-y^T^' ^^■y^\-x' 

4. y = x^-ix+l. ^ ^ 2x 

5. y = 2x^ + 3x-2. ^- y = y:^' "• y ~ T+T^ 

1 cc — 2 X 

6. y = -— r- 10. 2/ = — ^- 14.2/ 



x+1 " x-3 •' (1 + x^) 

15. Find the derivatives of x^ -\-\, x' —1, x' — 3, and x^ + k. 
Compare with the value of Dja?). 

173. Rules for finding derivatives. The derivatives of all the 
algebraic functions that we have studied in this book can be found ^ 
by methods like those illustrated by the above examples. In prac- 
tice these computations can be greatly shortened by breaking up 
more complicated functions into simple parts, according to the 
following rules : 

I. The derivative of a constant is zero. D^c = 0. 

For a constant does not change its value at all ; hence if y = c, 

Ay = for all values of A«. Therefore lim -^ = 0, which proves 
the theorem. 

II. The derivative of a variable with respect to itself is unity. 

For if y = x,Ay = Ax. Therefore — ^ = 1 for all values of Ax, 

, Av ^* 

and hence lim — ^ = 1. 

Axio Aa; 

III. If u and V are two functions of x, each having derivatives 
D^u and D^v, then the function u + v has as its derivative 
D^u + D,v. D, {u + v) = D^u + D^v. 

^ That is, they can be found at every point for which they exist. A function 
inay have a value for some values of the independent variable for which never- 
theless no derivative exists, — for example, V* for a; = 0. 



INTRODUCTION TO DIFFERENTIAL CALCULUS 215 

For if y = u-\-v, and if the increments of y, u, and v, caused 
by giving x the increment Aa;, are represented by Ay Aw, and 
Av respectively, then 

y+Ay = (u+Au) + (v JtAv) ; 

therefore Ay = Au+ Av, 



Ay Am Av 
Ax Ax Ax 



By hypothesis. 



1- ^w ^ , ,. Av ^ 

iim — - = B^u and hm — - = B^v. 
Aj;ioAa; AiioAx 



Hence, by Theorem III, § 169, 

Iim -^ = B^u + B^v. Q.E.D. 

Ax = o Ax 

Evidently this result can be extended to the sum of any number 
of functions. Stated in words it is as follows : 

The derivative of the sum of any number of functions whose 
derivatives exist equals the sum of their derivatives. 

Example 1. We have found that D^ (x'^) = 2x, and that D^ (x^) = 3 x^ 
(Exercise 2, § 172). Therefore D^.{x^ + x^) = 3x^ + 2x, so that it is not 
necessary to work out the computation for the function x^ + x^ anew. 

Example 2. D^ (x") = 3 x% (Exercise 2, § 172) 

and D^(x''_3H-l) = 2x>-3. (Exercise 4, § 172) 

Therefore i)^(x3 + x^ - 3 x + 1) = 3 x''' + 2 x - 3. 

IV. If u and v are two functions of x, having derivatives B^u 
and B^v, then the function u • v has as its derivative uB^v + vB^u ; 
that is, B^{uv) = uB^v + vB^u. 

Proof. liy = ««, and Ay, Au, and Au have the usual meaning, then 
y + Ay = (« + Au) (v + Ay) = «u + uAv + v\u + A«At>. 

Therefore Ay = «Ak + «A« + AuAv, 

Ay Aw , A« j^ A« . 
^-^ -i = ''-Kx^''^x^^x-^"- 



216 THE ELEMENTARY FUNCTIONS 

AjT A?/ 

By hypothesis, lim —~=Dj.v, lim -— =Z)a?">^i"i) by Theorem rv,§ 169, 

lim — • At! = 0, since the limit of the first factor is D^u and the limit of 

the second factor is 0. 

A?/ * 

Therefore lim — ^ = uD^v + vD^u. q.e.d. 

Ax=0 Aa; 

In words, TAe derivative of the product of two functions is 
equal to the first times the derivative of the second, plus the 
second times the derivative of the first. 

Evidently this rule can be extended to a product of any 
number of functions ; thus, ii y = uvw, By = u ■ D (vw) + vw ■ Bu 
= uv • Bw + uw - Bv + vw • Bu, and so on for any number of 
functions. 

Special cases. (1) If v = c, any constant, the rule becomes 

B^ (c ' u) = cB^u + iiB^c. 
But, by Eule I, B^c = 0. 

Therefore I^xi'^'") = cB^u. 

Example. Since D^ (x^) = 2x, D^ (5 x") = 10 x, Da;(50 x^) = 100 x, etc. 

(2) Another special case is y = m", n being any positive integer. 
Then, using the extended form of the rule (for n factors), we have 

Z)^y = M"-ii)^M + M''-iZ)^M+M«-iX)^MH {n such terms all 

precisely the same). 

Therefore B^y = n . u'"-'^B^u. 

Example 1. Find the derivative of x^". 

D^ (xi») = 10 • a;» • D^x. 
By Rule H, D^ = 1. 

Therefore D^ (x^") = 10 x*. 

In general, we can write down the result for x" just as easily : 
Dx (x") = n • X" - ^D^x = n • x" - 1 (n any positive integer). 

Example 2. Find the derivative of (2 x^ + 3 x - 2)^. 
Let 3^ = (2 x" + 3 X - 2)= = u^, 

where « = 2 x^^ + 3 x - 2. 

Then R^y = 5 ■ u*D^u. 



INTRODUCTION TO DIFFERENTIAL CALCULUS 217 

By Ex. 5, § 172, D^u = JD^ (2 x^ + 3 ^ _ 2) = 4 j; + 3. (This can also be 
obtained by direct application of Rule IV (1), Rule III, and Rule I.) 
Therefore Dx{2x'' + 3a; ^ 2)^ = 5(2a;2 + 3a; - 2)4(4:X + 3). 



-«(:)= 



vDm — uD^v 7 7 • „ . 

— ^ ^^) u and V being any two functions 

having derivatives D^u and D^v. 

In words. The derivative of a fractional function is equal to 
the denominator times the derivati've of the numerator, minus the 
numerator times the derivative of the denominator, divided by 
the square of the denominator (this excludes of course points at 
which the denominator equals zero). 

Proof. Let y = -• 

Then y + ^y ' 



Therefore Ay = 



« + Au 

M + A« M _ uA« — uAi) 
V + Ay V v(v + Aw) 



and 



Au All 
Ay Ax Ax 
Ax v(v + Ay) 



Therefore lim ^ = f^^^L^f^V. (Theorems III and V, § 169.) 

i,x=o Ax y2 

Q.E.D. 

Special case. If u = c, any constant, this rule takes the form 



•ifi v^ 

since D^c = 0. 

The case d = c is not to be taken as a special case of this rule, 

because - is not really a fractional function at all (cf. § 68, p. 85). 

" 1 . . 

Eather it is to be treated as a product - • u, givmg 



c 



for example, D^\^j = ^ D^ = -g- 



218 THE ELEMENTARY FUNCTIONS 

174. By the help of the foregoing rules we can work out, 
with the minimum expenditure of time and effort, the derivative 
of any rational function. (The process of finding derivatives is 
called differentiation.) 

Example 1. Dififerentiate the function 3 :t:* — 6 a;' + 5 1'^ — 1. 
Solution. By Rule III, 
i)^(3 x*-%x^ + 0x^-1) = D^(Z X*) - D,(6 x') + D^(o x^) - D^ (1) 

= 3 D^ (x*) - 6 jDj; (xS) + 5 Z>a; (x^) - (Rules IV (1) and I) 

= 3-4x'-6-3x2 + 5.2x (Rule IV (2)) 

= 12 x' - is x2 + 10 X. 

Of course it is possible to take the intervening steps mentally, and the 
student should observe that we can vrrite down at once 

£>j:(3 X* - 6 x« + 5 x2 - 1) = 12 x3 - 18 x' + 10 x.. 

Example 2. Differentiate the function (4 x'' — 3 x)^ 
Solution. By Rule IV (2), 

Z>x(4: x2 - 3 x)' = 3 (4 x" - 3 x)^ Z»^(4 x" - 8 x) 
= 3(4x2-3x)2(8x-3). 
Here also the intervening step can be taken mentally, enabling us to 
write at once ^^ (4 x^^ - 3 x)' = 3 (8 x - 3) (4 x= - 3 x)K 

X" 

Example 3. Differentiate the function : 



l-x2 

c, ■ 7^ / ^M (l-x2)2x-x2(-2x) 2x 

Solufon. D^ (^ = i L__J ) = __ 



EXERCISES 

Differentiate each of the following functions (doing as much of 
the work as possible mentally): 

ax + h 2x'-3a:'+l 





l+a;' 


2. 


©' 


3. 


1 


4. 


10 

x'' 



5. 



ex -\- d X 

■ (^ • 
\l+xl 



6. (.^V. 10.1+^ + ^' 



+ xl 1+ X — x^ 

1+x 11. (.T* - 5 x^ 4- 1)'. 



'''• l+x" 



12. 



8. x'-5x« + x'. i^ + c^x^ 

13. Verify the results of Exs. 1-14, § 172, using rules I-V. 




Fig. 125 



Fig. 126 



INTRODUCTION TO DIFFERENTIAL CALCULUS 219 

175. Use of the derivative in drawing graphs. Since the value 
of the derivative of y with respect to x (D^y) gives the slope of 
the tangent to the curve y=f{x) at any point, the derivative 
shows the direction 
of the curve at that 
point, the direction 
of the curve being 
that of the tangent 
line. In particular, 
if the derivative is 
positive, the tan- 
gent line makes an 
acute angle with the JPaxis, and the curve is rising as we go along 
it toward the right (Fig. 125). If the derivative is negative, the 
tangent line makes an ob- 
tuse angle with the X-axis, 
so that the curve is falling 
toward the right (Fig. 126). 
Finally, if the derivative 
is equal to zero, the tan- 
gent line is parallel to the 
X-axis (or coincides with it). 
Fig. 127 shows various forms 
of curve where this happens. 
Tlius, in drawing the 
graph of a given function 
it will often be foimd useful to compute the value of the derivative, 
and especially to note at what points it is positive, negative, or zero. 

(positive, the curve is rising, 
negative, the curve is falling, 
zero, the curve has a horizontal tangent. 
Example. Draw the graph of y = 2 a;' + 2^ — 8. 
Making a table of Gorresponding values of x and y, we find 




Fig. 127 



X 





1 


2 


- 1 


-2 


y 


-8 


-5 


12 


-9 


-20 



220 



THE ELEMENTARY FUNCTIONS 



These points give a general idea of the shape of the curve, but we can 
learn more about it by making use of the derivative of y with respect to x. 

This expression has the value for a; = and for x = — \, and is positive 
for all values of x except those in the interval from — ^ to inclusive, 
where it is negative. 

Thus the curve is rising except for values of x in the interval from 
a;=— Jtoa; = 0. Atar=0 (where y= — 8) the tangent is horizontal, also 
at a: = — ^ (where y — — 7f f ). Only between these points (A and B in 
Fig. 128) is the curve falling. 
This short interval would 
in all probability have been 
overlooked if we had not 
had the help of the deriva- 
tive in locating it. At all 
events, we could not have 
been certain what were the 
highest and lowest points on 
the little wave in the curve. 
AVe now know that A is the 
highest point in its imme- 
diate vicinity (because the 
curve is rising on the left of 
A and falling on the right 
of A), and that B is the 
lowest point in its imme- 
diate vicinity (because the 
curve is falling on the left 
of B and rising on the right of B). Such a point as A, that is, a point 
which is the highest point in its immediate vicinity, is called a maximum 
point ; while a point (such as B) which is the lowest point in its immediate 
vicinity is called a minimum point. 

176. Maximum and minimum values. Summary. Thus we see 
that, in general, the value x = a will give a maximum point on 
the curve y =f{x) when D^fix) is zero at the point x = a, and 
is positive just to the left, and negative just to the right, of that 
point ; that is, f'{a -h)>0, f{a) = 0, /'(a + A) < 0, h being a 
small positive number. For in that case f{x) is increasing as x 
approaches a from the left, and is decreasing as x passes beyond 
a toward the right. Hence at x = a, f[x) must have the greatest 




Fig. 128 



INTRODUCTION TO DIFFERENTIAL CALCULUS 221 



value that it can have for any point in that immediate vicinity. 
For similar reasons the value x — a will give a minimum on the 
curve y =f{x) when f'{a -h)<0, f'{a) = 0, and f'{a + h)>0. 
These results are summarized in the following table: 





f(a - h) 


/'(«) 


/'(a + h) 


Maximum . . . 


>0 





<0 


Minimum . . . 


<0 





>0 



Example. Draw the graph oi y = 



1- 



As we saw on page 87, the graph of any fractional function will 
have a vertical asymptote corresponding to any value of x that gives 
the denominator the value zero. 
Hence we have here the asymp- 
totes X = 1 and X = —1. The 
value of the derivative will give 
the direction of the curve at 
any point : 

l- x^-x(-2x) 



Dxy = 



(1 - x^y 



(1 -x^y 



Since this is always positive, the 
curve is always rising; hence 
there can be no maximum or 
minimum point. 



Fig. 129 



EXERCISES 

Draw the graphs of the following functions, making use of all 
the information which the derivative gives, especially with regard 
to maximum and minimum points : 



1. y = x'— 4a; 4- 4. 

2. 2/ = 3a;'-a;^+l. 

3. y = x\ 

4. y = x'-a;'+l. 



h. y = 



6. y 



x+_2 

l-x' 

x + 2' 






8. y- 



1+x 



\-irX-\-x'- 



9. Discuss the function ax^ + te + c for maximum and mmimum 
values. Compare the results with those obtained by more elementary 
methods in § 48 (p. 58). 



222 THE ELEMENTARY FUNCTIONS 

177. Differentiation of irrational functions. The only irrational 
functions that we have considered have been those involving square 
roots. Any such functions can be differentiated as follows : 

If y = Vm, where u is any function of x that has a derivative, 

then 

y + Ay = V w + Am. 



Therefore Ay = y/u+Au — ^ni 

Ay _ Vm + Am — Vm 



and 



Aa; Aa; 



Am 
In this form it is not easy to see what limit the quotient — ^ 

will approach as A« — ; but by multiplying both terms of the 
fraction by Vm + Am + Vm we get 



^y _ (^^ + Am — Vm)(Vm + Am + 'v^m) 
A* Aa; ( Vm + Am + Vm ) 

_ (m + Am) — w _ 1 Am 

Aa;(VM + Am + Vm) Vm + Am + Vm ^* 

Now, as Aa; = 0, — = Bji and Vm + Am = Vm, since Am = 0. 

Aa; 

Therefore lim —^ = — ;= Bjw : 



Ax=oAa; 2V 



M 



that is, B^y = — -= • B^u. (1) 

2VM 

In words this result is, The derivative of the square root of 
a function is equal to 1 divided hy twice the square root of the 
function, times the derivative of the function. 

Notice that (1) can be written 

i)^(M*) = lM-*D,M, 

which is exactly the form we should get by using Rule IV (2) 
(p. 216) with n = \; that is, the rule B^{u^)=n • u^-^ ■ B^u 
holds when n has this fractional value, as well as for the posi- 
tive integral values for which the rule was proved. As a matter 
of fact, the rule is still true if n has any fractional value, but 



INTRODUCTION TO DIFFERENTIAL CALCULUS 223 

we do not need to use that general fact in our work. For our 
purposes it wHl usuaUy be found simpler to use the form of 
equation (1) than to change to the form of a fractional exponent. 

Example 1. Find the derivative of V2 a; — 3. 

Solution. D^ V2^-3 = ^ D^(2x-S) = . 



2V2X-3 ' V27:r3 

Example 2. Find the derivative of x Vl — x^. 



Solution. D^{x Vl - x^) = Vl - x" Z»^x + xD^ VT 



= Vl - a;2 + : 



x" 
-2x l_2a 



2Vl-a;2 Vl-a;2 
EXERCISES 

Find the derivatives of each of the following functions : 
1. V5x. 4. Vx^ - 4. 7. VlOO - 25 a?. ,„ b 



10. - Va^ - 3?. 



a 



2. Va;2 + 1. Ia;+1 8. ■ 

^- \^Z^- Va;+1 11. a:Wx+l. 

-I a; „ 3a; 1 

Draw the graphs of the functions in Exs. 1-9, locating any 
maximum or minimum values. 

178. Differentiation of implicit functions. When we have an 
equation in the variables x and y which determines y as an im- 
plicit function of x, it is easy to write down the value of B^y 
without first solving the equation for y. Eules I-V (pp. 214-217) 
are all that is needed. 

Example. a;^ + 4 / = 4 (1) 

(1) being true, the derivative of the left-hand side must equal the derivative 

of the right-hand side, „ , „ „ . 

^ Z)^(3:2-f-4y2) = Z)^(4); 

that is, D^ (^2) + D^ (4 y^ = 0. 

Therefore 2 z -I- 4 • 2 ^ D^y = 0. 

Therefore D^y^ — -. — 

42/ 

(The above work assumes that the function y has a derivative. This 
assumption is justified in every case with which we shall have to do.) 



224 THE ELEMENTARY FUNCTIONS 

EXERCISES 

Find D^y from each of the following equations : 

1. 0^ + %/= 9. Ans. D^ij=--- 

2. 4x^ + 2/' = 4. ^ 

3. 9x2 -42/= =36. 

4. xy -\- y'^= 4. Ans. D^y = — _^ , ^ ^ ■ 
6. x''-2xy + y'' + ix-Sy=0. 

6. x' + / = 65. 

7. x"y + a;?/" = 26. 

8. (x-«f +(y_^/ = rl 

9. 6V + ay = a^'Sl 

a? ^ b' ~ 

12. ax" 4- 6x2/ + cy" + c?x + ey +/= 0. 

13. Prove that the equation of the tangent to the parabola 
2/" = 4 ax at the point (Xj, y^) is yy^= 2 a(x + x^). 

14. Obtain the equation of the tangent to the hyperbola xy =1 
at the point (Xj, y^) in the form y^x + x^y = 2. 

15. Show that the hyperbola of Ex. 14 and the circle x' + y'' = 2 
have the same tangent at their common points ; that is, that they are 
tangent to each other. 

16. At what point on the ellipse ix^ + 9 y^ = 36 is the tangent 
parallel to the line y =— x? 

17. Verify the results of Ex. 17, § 109, and Ex. 2, § 110, by 
finding the value of D^y in each case. 

18. Prove that the tangents to a parabola at the ends of the 
latus' rectum intersect at right angles on the directrix. 

179. Summary. We have now seen how to find the derivative 
of any rational function, and of irrational functions of degree not 
higher than the second. Since the derivative of y with respect to 
X gives the slope of the tangent at a point, we can find the equa- 
tion of the tangent to any curve that is the graph of one of these 
functions. 



INTEODUCTION TO DIFFERENTIAL CALCULUS 225 

The derivative gives also the rate of change of the function 
per unit change of the independent variable, so that this impor- 
tant problem is solved for all the algebraic functions mentioned. 
We have accordingly reached the point where we can consider 
the applications of this work to problems from various fields. 

Example 1. The Law of Falling Bodies. 

If s = 16 fi (1) 

states the relation between distance and time in the case of a falling body, 

then the instantaneous rate of change of the distance with respect to the 

time, that is, the velocity of falling, is given by Dts. Since DfS = 32 t, we have 

at once the formula ti on < /ox 

V = Dts = 32 t, (2) 

which is well known from elementary physios, where it appears as a second 
formula, as if it were independent of (1). But we now see that it is not, 
since, if (1) is true, (2) must be true also. 

Example 2. If a body moves according to the law s = <^ — 10 < + 2, find 
its velocity at any instant t. When is it moving in the positive direction, 
and when in the negative ? 

Solution. The velocity is given by D(S = 2t — 10, which is negative until 
t — 5, when it equals zero ; for <> 5 it is positive. Hence the body is moving 
in the negative direction until t = 5, when it has zero velocity, after which 
it moves in the positive direction. 

Example 8. A ladder 40 ft. long rests against the side of a house. If 
its foot A is pulled away from the house at the rate of 5 ft. per second, 
how fast is the top of the ladder descending 
when the bottom is 10 ft. from the house? 

Solution. Let x = AC, the distance from the 
house to the foot of the ladder ; and let y = CB, 
the height of the top of the ladder. Then y is a 
function of x, and the problem is to find the rate 
of change of y, knowing that of x. From the 
right triangle ABC ^2 ^ igoo - x^ 

Therefore 2 yD^y = -2x, 

X 

or DxV = ~ " ' 

that is, y is changing at any instant times as 

fast as x is. We are to find the rate of change of 2/ when :r =K). When x= 10, 

y = VI6OO - 100 = Vl500 = 10 Vlo ; 

W - J_ 
hence ^^^ = " I^VW ~ VlS' 




226 THE ELEMENTAEY FUNCTIONS 

Therefore y is at that instant changing ■r= times as fast as x. But, by 

vl5 



the conditions of the problem, x is changing at the rate of 5 ft. per second, 

and accordingly the rate of change of y is p= • 5 = — - vl5 ft. per 

Vl5 ^ 

second. The negative sign shows that y is diminishing. 



EXERCISES 

1. Two persons start from the same point and walk in directions 
at right angles to each other at the rates of 3 mi. per hour and 4 mi. 
per hour respectively. How fast are they separating after 15 min. 
has elapsed ? 

2. A light is 10 ft. above a street crossing, and a man 6 ft. tall 
walks away from it in a straight line at the rate of 4 mi. per hour. 
How fast is the tip of his shadow moving after 10 sec? after 1 min.? 
How fast is the shadow lengthening in each case ? 

3. A point moves along the parabola j/" = 8 a;. How does the rate 
of change of the ordinate at the point (^, 2) compare with that of 
the abscissa ? 

4. At what point on the parabola of Ex. 3 are the ordinate and 
the abscissa changing at the same rate ? 

5. When a stone is thrown into still water, how fast does the 
area of the circle formed by the ripples change in comparison with 
the radius ? 

6. If the radius of a spherical soap bubble is 3 in. and is increasing 
at the rate of ^ in. per second, at what rate is the volume increasing ? 

7. If a point moves along the curve a? + 2y^ = 2, how does the 
rate of change of the ordinate compare with that of the abscissa at 
the point (I, ^)? 

8. An automobile track is in the form of an ellipse whose major 
axis is ^ mi. long and whose minor axis is \ mi. long. If a car moves 
around the track at the rate of 40 mi. per hour, how fast is it mov- 
ing toward the north, and how fast toward the east when it is at 
the end of the latus rectum ? (Suppose the length of the track to 
be from north to south.) 

9. A man on a wharf is pulling in his boat by means of a rope 
fastened to its prow. If his hand is 10 ft. above the boat and he 
pulls in the rope at the rate of 3 ft. per second, how fast is the boat 
coming in when 20 ft. away ? when 6 ft. away ? 



INTRODUCTION TO DIFFERENTIAL CALCULUS 227 



10. A railroad crosses a road at right angles by an overhead 
crossing 30 ft. high. If a train and an auto cross at the same time, 
the one at the rate of 30 mi. per hour, the other at the rate of 15 mi. 
per hour, how fast will they be separating after 1 min.? 

11. One ship sails east at the rate of 10 mi. per hour, another 
north at the rate of 12 mi. per hour. If the second crosses the track 
of the first at noon, the first having passed the same point 2 hr. before, 
how is the distance between the ships changing at 1 p.m.? How was 
it at 11a.m.? When was the distance between them not changing? 

180. Problems involving maxima and minima. As we have 
seen, the use of the derivative of a function enables us to discover 
for what values of the independent variable the function will have 
a maximum or a minimum value. 
This fact is of use iu solving a great 
variety of problems. 

Example 1. Find the rectangle of 
greatest area which can be inscribed in 
a given circle. 

Solution. Let r be the radius of the 
given circle, and 2 x and 2 y the dimen- 
sions of the rectangle (Fig. 131). Then 
^2 _ ^ _ J.2 or y = Vr" — x^. The area of 
the rectangle is 4 xy, which is to be made 
the greatest possible by choosing the 
proper value of x (and hence of y, wh ich 

is a function of x). Now i xy = i x Vc^ — x- =f{x), so that the question is, 
For what value of x will f(x) have its maximum value ? For that value 
of X, as we have seen (p. 220), we may expect to find D^f^x) = ; that is, 




Fig. 131 



Therefore 

But Da;[x^r^-x^] 

Therefore 






: Vr2 - !» - ■ 



■2x^ 



■■-2x^ = 0, 



Vr" - x'' 



V2 

When X < -^ > Dxf(x) > 0, and the function is increasing as x increases ; 
V2 

when a;>-7=' Dxf(^)<0> ^.i^ t^® function is decreasing as x decreases. 

V 2 



228 



THE ELEMENTARY FUNCTIONS 



That is, the function is increasing before x reaches the value 



V2 



and 



decreasing after x passes the value 



V2 



Hence that value of x gives a 



maximum value of /(x) ; that is, of the area of the rectangle. Geometrically it 
is also evident that this value is a maximum rather than a minimum, because 
as X increases from very small values the area of the rectangle also increases, 
but beyond a certain point it will diminish again, approaching as i 
approaches r. The locating of this "certain point" as being the value 

is made possible by the use of the derivative. 



r 



Notice that when x = — = , 

V2 
a square. 



: — -= , so that the maximum rectangle is 
V2 



Example 2. What would be the dimensions of a cylindrical steel tank, 
open at the top, to require the least material possible 
and still contain 50 gallons ? 

Solution. The quantity of material used will be least 
when the area of the surface is least. The surface con- 
sists of the side, whose area is 2 wxy, and the bottom, 
whose area is ■itx'^ ; total, S = 2 ■n'xy + ■rrx^ (x represent- 
ing the radius of the base and y the altitude, as in 
Fig. 132). In this expression y is a function of j, 
whose value is determined by the fact that the volume 
is to be 50 gal.; that is, 6.684 cu. ft. The volume of the cylinder = irx^y. 




Fig. 132 



Therefore 



nx^y = 6.684, 
6.684 

2' = -Tir- 



Therefore 



.. „ ■ 6.684 , 2 13-368 ^ , 



This function of x is to be made a minimum. 

Therefore Da:S = 0. 

T^ C-, 13.368 . „ 

DxS = ;— -I- 2 w = 0. 



13.368 



3/6.684 
X = -»/ = 1.286, approximately. 



From the nature of the problem this value makes the function S a 
minimum, not a maximum (as can also be verified by noting that Dq.S 



INTRODUCTION TO DIFFERENTIAL CALCULUS 229 

changes sign from — to + as i increases through the value 1.286), so that 
the value x — 1.286 gives the radius of the base that will require the least 
amount of material for the tank. 

When X = 1.286, 

y — — — -- = 1.286, approximately ; 
that is, y = X. 



The exact value of x is -* / — > which equals V^, so that x = Vifiy- 

Therefore x = y exactly, in fact. 

Thus the height of the tank should equal the diameter of the base in 
order to require the least amount of material. 



EXERCISES 

1. Work through Ex. 2 above, using V as the volume of the tank, 
instead of 50 gal. Show that y = x will give the proportions to 
require the least material. 

2. Divide a length AB into two parts so that the product of 

the lengths of the segments shall be the __;; __^_ 

greatest possible (Fig. 133.) 

Fig. 133 

3. Divide a length AB into two parts 

so that the sum of the squares of the lengths of the segments 
shall be the least possible. 

4. A piece of cardboard 10 in. square has a small square cut out 
at each corner, and the remainder is folded up so as to form a box. 
How large a square should be cut out so that the box may be the 
largest possible ? Ans. | in. is the side of the square cut out. 

5. A rectangular piece of cardboard, 15 in. x 7 in., is to be made 
into a box by cutting out a square of equal size from each corner. 
How large should this square be in order ^ 
that the box may have the largest possible 
contents ? 

6. A person in a boat, 6 miles from 



6 miles 

10 milea Q 



Beach 

the nearest point of the beach, wishes to i-m. 134 

reach in the shortest possible time a place 

10 miles from that point along the shore; if he can walk 5 mi. per 
hour, and row 4 mi. per hour, where should he land ? (Fig. 134.) 




230 THE ELEMENTARY EUNCTIONS 

7. A rectangular box, open at the top, is to be constructed 
to contain a certain volume V. What must be its dimensions 
in order to require the least amount of material ? 

8. Find the largest rectangle that can be inscribed 
in an isosceles triangle whose base is 4 in. and whose 
slant height is 10 in. (Fig. 135). 

9. Find the largest cylinder that can be inscribed 
in a given sphere. 

10. Find the largest cone that can be inscribed in a 
given sphere. 

11. What should be the dimensions of a cylindrical tin can (both 
ends being closed) in order to require the least material ? 

12. The lower corner of a sheet of paper whose width is a is 
folded over so as just to reach the other edge of the paper. How 
wide should the part folded over be in order that the length of the 
crease may be the minimum ? 

13. Find the shortest and the longest distance from the point 
(4, 6) to the circle x^+f= 25. 

14. Find the shortest distance from the point (0, 1) to the 
hyperbola xy = 1. Show that this distance is measured along the 
normal to the curve through the given point. 

15. If the cost of running a steamboat is proportional to the 
cube of the velocity generated, what is the most economical rate of 
steaming against a 3 mi. per hour current ? 



PORTION OF GREEK ALPHABET 



Alpha 


a 


Lambda 


Beta 


i8 


Pi 


Gamma 


y 


Eho 


Delta 


A 8 


Tau 


Epsilon 


e 


Phi 


Theta 


6 


Omega 



<^ 



231 



APPENDIX A 



PROOF THAT THE DIAGONAL OF A SQUARE IS 
IISrCOMMENSURABLE WITH ITS SIDE 

This theorem proves the fact asserted in the text (p. 3), that there 
exist segments which cannot be measured in terms of a specified 
unit. The fact to be proved may be stated in the following words : 

If the side of a square be taken as unit, there exists no number — 

n 
(m and n being integers) such that the length of the diagonal of 

the square is — 
n 



Proof. By the Pythagorean Theorem, 

d^ = 1^ + 12 = 2. 



(1) 



We can prove that the assumption rf = — , where m and n are integers, 

leads to a contradiction. Suppose — to be in its lowest terms; that is, 

n 

suppose m and n to have no common divisor. 

Then, since d^ = 2, 

?n2 _ 

Therefore m^ = 2 n\ (2) 

Therefore rrfi is divisible by 2. This 
is only possible if m, is an even integer 
(since the square of an odd integer is 

odd) ; and since m is even and — is in 
' n 

its lowest terms, n must be odd. 

Further, since m is even, we can write m = 'ik, where k is some integer. 

Therefore nfi = 4:B. , (3) 

Comparing (3) with (2), 2 ra^ = 4 P ; 

that is, n2 = 2F. (4) 




Hence n^, and therefore n, is even. 



233 



234 THE ELEMENTARY FUNCTIONS 

But we proved above that n is odd ; hence a contradiction results from 

the assumption that d = —■ Therefore the assumption is false, and there 
n 

wi 
does not exist any number — representing the diagonal of a square whose 

side is the unit. 

Note. This proof is of ancient origin, being given in almost exactly the 
above form by Euclid, in his famous " Elements of Geometry," in the third 
century b.c, and being referred to by Aristotle, as well known, much earlier. 
It is very probable that it was discovered by Pythagoras himself in connection 
with the study of his famous theorem (see p. 4, footnote). 



APPENDIX B 
LAWS OF OPERATION WITH RADICALS 

The following are the laws of operation with irrational numbers 
that are in the form of radicals : 

I. Multiplication. Va • a/* = -Vab. 
Special case. Va ■ Vfi = Va6. 

Examples. V2 • Vs = Vg ; 712 = Vi • VI = 2 Vs. 

II. Division. —= = ->M-- ; but in no ease may the denominator 
equal 0, since division by is an impossible operation. 

Examples. J| = ^; ./! = Je^lVG; Jl = JI = 1 Vs. 
V4 2'\3 \9 3 '\5 \25 5 

III. Addition and subtraction. Only similar radicals can be added 
or subtracted; that is, radicals that involve the same root of the 
sarne numher. 

Examples. V2 + Vs cannot be simplified, nor can V2 + V^, but 
V2 + Vl8 =V2 + 3V2 = 4V2= V32. 

(It is a very common error to write v a + ft = V a + Vj, whicli violates 
this law.) 

EXERCISES 

Give the value of each of the following in the simplest form : 
1. VI. 4. (2-V3)(2 + VI). ^ V6 + V2 

o 2-V3 ^ V^^_V^3-3 ■ 2 + V3 

2 + Va _ ■ V^TjTi" + Va;-3 „ 4 - V6 - V2 
3. 2-3V2 e. ^^ ■ V6-V2 

3 + 2 V2 1 - Vl - 3; Y 

Hint. A fraction containing Va + Vft in the denominator is simplified by 
multiplying both numerator and denominator by Vo — V6. Accordingly, in 
Ex. 3, we should multiply both terms of the fraction by 2 — V3. 

9. Prove: (a) -^ = ^. (b)-;^— = 1+V2. (c) ^^^E^li^ = 9 - 4 Vs. 
■v^ 2 ^ W2-I 5V5+IO 

(d) V3 - 2 V2 = (V2 - 1). 

235 



APPENDIX C 

TO CONSTRUCT A SEGMENT HAVING A 
RATIONAL LENGTH 

1. To divide a given segment AB into any number of equal parts : 
Draw any line AE making an angle with AB. On AE lay off any 
length, as AD, n times, and let the last division point be E. Draw 




EB, and from all the other division points draw lines parallel to EB. 
These parallels intersect AB in the points of division required, and 

one of the resulting segments is - oi AB. (Why ?) 



EXERCISES 

Take a random segment as unit, and construct J, J, J, \, J, J, If, 2J, 
0.6, 3.1, 5.3. Take two random .segments, a and b. Construct Ja + J 6, 
3a + 4i, l|a + 2ii. 

2. To divide a given segm,ent AB into two segments having any 
given ratio, as m : n. 

This problem is solved by a method very similar to that above, 
and it is accordingly left to the student as an exercise. 



236 



APPENDIX D 

TO CONSTRUCT THE SQUARE ROOT OF ANY 
GIVEN SEGMENT 

This important problem of construetion is solved by means of 
the following 

Theorem. In a right triangle the altitude drawn from the vertex 
of the right angle is the mean proportional between the segments into 
which it divides the hypotenuse. 

The proof is found in any textbook of elementary geometry. 

By using this theorem we can construct the mean proportional 
between any two given segments, as follows : 

Let a and b be the two segments. Draw AB = a, and produce it 
to C so that BC = b. Draw a semi-circle on 4 C as diameter. At B 




erect the perpendicular to A C, meeting the semi-circle in D. Then 
BD is the mean proportional between AB and BC ; that is, between 
the given segments a and b. 

The proof follows at once from the fact that the angle ADC is a 
right angle. 

Now, to construct the square root of any given segment m, construct 
as above the mean pro portio nal to m and 1. This mean proportional 
then has the length Vm ■ 1 ; that is, it is the required Vro. 

EXERCISE 

Take any random segment and construct its square root. 



237 



APPENDIX E 

SIMULTANEOUS LINEAR EQUATIONS IN TWO 
UNKNOWN QUANTITIES 

If we have the two equations 

r3x-72/ = l, (1) 



{; 



I2x+ 2/ = 12, (2) 

which are both to be satisfied by a certain pair of values (x, y), we 
proceed to combine the two equations in such a way as to obtain a 
new equation which shall contain only one unknown quantity. To 
do this we may add (1) and (2) as they stand or after they have 
been multiplied by any number we please. If we multiply (2) by 7, 
and add the result to (1), it is clear that we shall get rid of the 
y-term altogether; we call this "eliminating y" from (1) and (2). 
Thus, (2) multiplied by 7 gives 

14a; + 7?/ = 84. 
Adding this to (1), 17 x = 85. 

Therefore x = 5. 

If we substitute this value of x in (1), we get 

15-7y = l. 

7y = 14. 

y = 2. 

Therefore the pair of values (6, 2) will satisfy both equations. 

We could also have eliminated x from (1) and (2) by multiplying 
(1) by 2 and (2) by 3 and subtracting : 

6a;-14y = 2, 

6a; + 3 2/ = 36. 

Therefore 17?/ = 34, 

y=2. 
238 



APPENDIX E 239 

Substituting ?/ = 2 iu (1), 

3x-14 = l. 
3x = 15. 
X = 5. 

Therefore (5, 2) is the solution. This result should be checked 
by substituting x = 5, y = 2 in (1) and (2). 

It is clear that this procedure can always be adopted, giving a 
new equation in which one or the other of the two unknowns fails 
to appear. 



APPENDIX F 

THE QUADRATIC EQUATION IN ONE UNKNOWN 
QUANTITY 

Suppose we wish to solve the equation x'' + 6 r = 7. The method 
of solution most commonly employed in elementary algebra is that 
known as " completing the square," because it consists in adding to 
the left side of the equation such a number that that side becomes a 
perfect square. Success in using this method therefore depends upon 
familiarity with the form that an expression must have if it is a per- 
fect square. We know that (x + /.-y = x' + 2 kx + k", which shows 
that when the terms of a complete square are written in this order, 
the middle term (2 kx) is tirice the product of the square roots of the 
other two terms. As one of these square roots is x (the first term 
being x^), the other one must be half the coefficient of x in the 
middle term. Thus, in the equation x^ + &x = 7 above, a?-\-&x-\-k^ 
will be a perfect square if 6 = 2 A, that is, if /c = 3 ; hence the 
quantity to be added to x'' + 6 x to make it a perfect square is 2>^ ; 
x^+6x + 9=(x + 3)l 

Similarly, to x^ + 8 x we must add 4^ to complete the square, 

to x'^ -\- 5x we must add ( x ) , and to x^ -\- mx we must add ( -x- ) • 

In words, The quantity to be added is the square of half the coeffi- 
cient of X, when the coefficient of\c'^ is 1. 

EXERCISES 

Complete the square of each of the following : 

x^ + 10 X. z^ + (m + n) X. x^ + ^ x. 

x^ + 15 X. x^ + ax. 2 ab 

x^ + ^x. x^ + ^x. a + b ' 

Solve each of the following equations : 
1. a;2 + 6 3; = 7. 

Solution. To complete the square of the left side we must add 9 : 

x2 + 6a; + 9 = 16. 
Extracting the square root, x + 3 = ± 4; 

that is, X = 1 or — 7. 

240 



APPENDIX F 241 

2. x^ + bx = 6. 4. x'' + 7x = - 12. 6. x^ + 12 x = V- 

3. x2-5x = 6. 5. x2 + 12x = -20. T. x^ - (a + l)x = - ab'. 

8. x2 - 2 mx + m2 - «« = 0. 

9. a6x2 = (a'-'-i2)a; + ah. 

10. x2 - (a + i)x = 2 a2 + 2 i2 - 5 oi. 

11. x-* — (a + c)x + = 0. 



INDEX 



Abscissa, defined, 10 

Ambiguous case, 160 

Angles, general definition, 61 

ApoUonius, 139 n. 

Asymptotes, 87 

Axis, of parabola, 115; of ellipse 
(major, minor), 124, 125 ; of hyper- 
bola (transverse, conjugate), 132 

Center,of ellipse,125; of hyperbola, 138 

Characteristic of logarithm, 193, 194 

Circle, equation of, 110, 111 

Component forces and velocities, 77 

Conic sections, 139 

Conjugate hyperbolas, 135 

Constants, 20 

Construction, of rational numbers, 4 ; 
of irrational numbers, 4-6 ; of pa- 
rabola, 120 ; of hyperbola, 138 

Continuity of trigonometric functions, 
181 

Coordinate system, 9 

Correspondence between numbers and 
points, 7 

Cosines, Law ^f, 156, 157 

Degree of function, 28 

Derivative, 211 ; as slope of tangent, 
212 ; rules for finding, 214-218 ; used 
in drawing graphs, 219 

Descartes, 10 

Determinants, 32-39 ; minors, 37 ; 
applied to solution of simultaneous 
linear equations, 33, 34, 38, 39 

Diiference of two angles, trigonometric 
functions of, 170, 171 

Difference of two sines or cosines, 
174 

Differentiation, of rational functions, 
214-218 ; of irrational functions, 
222 ; of implicit functions, 223 

Directed segments, 6, 7 

Directrix, of parabola, 114 ; of ellipse, 
122 ; of hyperbola, 131 

Discontinuity of trigonometric func- 
tions, 181 

Discriminant of quadratic equation, 47 



Discriminant test graphically inter- 
preted, 44-53 

Discriminant test used in obtaining 
tangents, 50 

Distance between two points, 12 ; from 
a line to a point, 105-107 

Division of segment in given ratio, 
14-16 

Eccentricity, of ellipse, 122 ; of hyper- 
bola, 131 

Ellipse, 90 ; defined, 121 ; equation of, 
122-128 

Equation, linear, 29 ; quadratic, 41-47 ; 
of straight line, 97-109 ; of circle, 
110, 111 ; of parabola, 114-118 ; of 
ellipse, 122-128 ; of hyperbola, 131- 
136 ; of asymptotes to hyperbola, 
133 

Exponents, laws of, 188 ; fractional 
and negative, 189 

Factor theorem, 55 

Eocus, of parabola, 114 ; of ellipse, 
122; of hyperbola, 131 

Forces, problems on, 77 

Functions, defined, 23; linear, 28; quad- 
ratic, 39 ; sign of quadratic, 56, 57 ; 
maximum and minimum values of 
quadratic, 58, 59 ; trigonometric, de- 
fined, 63-66 ; relations among trigo- 
nometric, 67, 69; trigonometric, of 
30°, 45°, 60°, 68 ; fractional, 85-88 ; 
irrational, 88-90 ; logarithmic, 202 

Galileo, 204 n. 

Graph, of linear function, 29 ; of quad- 
ratic function, 40 ; of fractional func- 
tions, 86-88 ; of irrational functions, 
88-91 ; of trigonometric functions, 
178-182 ; of logarithmic function, 
202 

Graphical representation, of number 
pairs, 9, 10 ; of equations, 19-22 ; of 
statistical data, 24-26 

Graphical solution, of simultaneous 
linear equations, 30-32 ; of quad- 



243 



244 



THE ELEMENTARY FUNCTIONS 



ratio equation, 40 ; of simultaneous 
quadratic equations, 141-150 
"Greater than," 8 

Half-angle formulas, 163 ; applied to 

plane triangles, 198, 199 
Hyperbola, 87 ; asymptotes of, 87, 133, 

134 ; defined, 131 ; equation of, 131- 

136 I construction of, 138 

Incommensurable segments, 3 
Increment, 204 
Independent variable, 23 
Integer, 4 
Intercepts, 29 
Irrational functions, 85 
Irrational numbers, 4 

Kepler, 139 

Latus rectum, of parabola, 116 ; of 
ellipse, 125 ; of hyperbola, 133 

" Less than," 8 

Limits, 208-210 ; theorems on, 210 

Linear equation, graph of, 29 

Locus, 19, 91-94 

Logarithms, definition, 190 ; laws of, 
191, 192 ; characteristic and man- 
tissa, 193, 194 

Mantissa of logarithm, 193, 194 
Maximum and minimum values, of 

quadratic function, 58, 59 ; with 

help of derivative, 220 ; problems 

involving, 227 
Measurement, 1 
Mid-point of segment, coordinates of, 

12 
Mollv?eide's Formulas, 201 

Negative numbers, 6-8 
Newton, 201 n. 

Normal form of equation of straight 
line, 101-105 

Oblique triangle, solution of, 154, 157, 

159 
Ordinate, defined, 10 



Parabola, 21 ; defined, 114 ; equation 
of, 114-118 ; construction of, 120 

Periodicity of trigonometric functions, 
180, 182 

Polar coordinates, 183-187 

Projections, Law of, 156 

Ptolemy, 174 n. 

Pythagoras, 4 n. 

Pythagorean Theorem, 4 n. 

Quadratic equation, 41-47 ; formula 

for, 43 ; solution by factoring, 

44 
Quadratic function, 39 ; sign of, 56, 

57 ; maximum and minimum values 

of, 58, 69 

Radius vector, 65, 183 
Rate of change, 203, 207 
Rational functions, 185 
Rational numbers, 4 
Resultant forces or velocities, 77 
Right triangle, solution of, 71-76 
Roots of quadratic equation, 42 ; sum 
and product of, 53-55 

Simultaneous linear equations, 30-39 

Simultaneous quadratic equations, 
141-152 

Sines, Law of, 155 

Slope of straight line, 79-82 

Straight line, equation of, 97 ; normal 
form of, 101-105 

Sum of two angles, trigonometric func- 
tions of, 168-171 

Sum of two sines or cosines, 174 

Tangent, to parabola, 50 ; to any 
conic, 143 ; to ellipse, 146 ; slope 
of, found by means of derivative, 
211-214 

Tangents, Law of, 201 

Variables, 20 ; independent, 23 ; de- 
pendent, 23 

Velocities, problems on, 77 

Vertex, of parabola, 115; of ellipse, 
125; of hyperbola, 132