:ti(m
TO THE
MENTARY
JNCTIONS
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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