AN INTRODUCTION TO
MATHEMATICAL
ANALYSIS ,
By
FRANK LOXLEY GRIFFIN, PH.D.
Professor of Mathematics, Reed College, Portland, Oregon
HOUGHTON MIFFLIN COMPANY
BOSTON NEW YORK CHICAGO
(Cfce ttiterpiDe ptc??
I
COPYRIGHT, 1921, BY FRANK LOXLEY GRIFFIN
ALL RIGHTS RESERVED
CAMBRIDGE - MASSACHUSETTS
U • S • A
PREFACE
UNDER the traditional plan of studying 'trigonometry,
college algebra, analytic geometry, and calculus separately,
a student can form no conception of the character and possi-
bilities of modern mathematics, nor of the relations of its
several branches as parts of a unified whole, until he has taken
several successive courses. Nor can he, ea^y enough, get
the elementary working knowledge of mathematical analysis,
including integral calculus, which is rapidly becoming in-
dispensable for students of the natural and social sciences.
Moreover, he must deal with complicated technique in each
introductory course; and must study many topics apart
from their uses in other subjects, thus missing their full
significance and gaining little facility in drawing upon one
subject for help in another.
To avoid these disadvantages of the separate-subject plan
the unified course presented here has been evolved. This
enables even those students who can take only one semester's
work to get some idea of differential and integral calculus,
trigonometry, and logarithms. And specialist students, as
experience has shown, acquire an excellent command of
mathematical tools by first getting a bird's-eye view of the
field, and then proceeding to perfect their technique.
A regular course in calculus, following this, can proceed more rapidly
than usual, include more advanced topics, and give a fine grasp : the
principles and processes have become an old story. And the regular
course in analytic geometry can be devoted to a genuine study of the
geometrical properties of loci, since most of the type equations, basic
formulas, and calculus methods are already familiar.
iii
iv PREFACE
The materials presented here have been thoroughly tried out
with the freshman classes in Reed College during the past nine
years. Problems and methods which have proved unsatisfac-
tory have been eliminated. Care has been taken to make the
concepts tangible, relate them to the familiar ideas of daily
life, exhibit practical applications, and develop the attitude
of investigation. E.g., in many "leading problems," —in-
dicated in the text by bracketed numerals, — students are
asked to formulate for themselves methods not yet discussed.
The order of topics while unusual, — especially in starting
calculus before trigonometry, — is a natural one.
We begin with graphical methods because they afford a
simple and interesting means of introducing the function-
concept and the big central problems, — and also because
they tend to develop at the very outset the self-reliant habit
of attacking problems by " rough and ready " methods of
approximation when no better methods are known.
Refining the graphical methods leads naturally to the
calculus. After some work with this, the need for trigono-
metric functions is seen, and these are introduced. During
the work on trigonometry, analytic geometry, etc., the con-
tinuity of the course is preserved by frequent problems which
require calculus as well as these other subjects.
The intervals between the several parts of the calculus
are thus a gain rather than a loss : they give the big principles
a chance to emerge from the details. These principles are
kept before the student during almost the entire year, not-
withstanding the fact that systematic courses in trigonometry,
analytic geometry, etc., are worked in.
There is, by the way, considerable analytics in the course besides
what appears in Chapter VIII. (Cf. §§26-32, 40-41, 170-175, 244-
247, 258, 268, 274-278, 296, 298, ami the numerous plotting exercises
in Chapters II-IV, VII, IX-XII.) But the idea of coordinates proper
is not introduced until we are ready to use it in studying geometry. Up
PREFACE v
to that point the function-concept is the thing : we are interested simply
in the varying height of the graph, and do not need the more subtle con-
ception of a relation between the coordinates of every point along a
curve, — in other words, the idea of implicit functions.
The trigonometry also is continued for some time, — the analytical
portion being treated late, when needed. The transition to the general
definitions when we are ready to study periodic variations is made
smoothly.*
The natural difficulty in assimilating the many new ideas
in the course is largely overcome by close correlation, and by
the practice of assigning some review problem with advance
work. This last, with frequent rapid oral quizzes reviewing
recent studied material, enables us to work with each es-
sential topic long enough to fix it clearly in mind, although
proceeding rapidly. No effort is made to cover intricate
points of technique or to discuss subtle niceties of logic.
But we do insist upon clear ideas, grasp of the train of
thought running through the course, and ability to use the
processes accurately in simple cases. f Problem work in class
is prominently featured.
Certain topics, e.g., those treated in §§ 32, 72-73, 84, 101,
115-116, 118, 142-143, 151-154, 183, 219-221, 224, 241, 242,
281, 282, 284-286, 298-304, and Chapters XIV-XV, we have
usually dealt with very briefly, — but sufficiently to make
them clear at the time and enable the student to pick them
up again easily if he needs them later on. Still other topics,
such as those of §§ 82, 155-156, 228, 275-276, 318, 320-324,
* There seems to be a widespread idea that by giving the general defini-
tions of the functions at the outset all re-statements can be avoided. On the
contrary, to adapt such definitions to the solution of triangles, re-statement in
the form "(opposite side) -f- (hypotenuse)," etc., is necessary, and the two
statements must both be learned almost at the outset.
t Thus we introduce, for example, the easy " short method " of setting up
integrals, — but only after the exact methods are familiar. And the relation
of the various methods is constantly pointed out.
vi PREFACE
are merely touched upon in lectures for the sake of extending
the student's horizon and developing his imagination.
These latter topics coming mainly near the ends of chap-
ters are mentioned without appreciably reducing the time
spent upon essentials. E.g., students work investment prob-
lems, in part, while studying the rest of Chapter XIII.
Any of the more advanced topics can of course be omitted
if desired, and attention focused on drill work, — for which
an abundance of exercises is provided. Thus the presence
of these topics in the text merely makes for flexibility.
The course as given at Reed College takes four hours a
week through -the year, the number of lessons devoted to the
several chapters, when taken complete, having run about as
follows: 14, 4, 14, 8, 11, 12, 11, 16, 5, 7, 10, 6, 6, 5, 4. A
considerable shortening can be effected by omitting any of
the chapters IX, XII-XV, or §§ 181-184, or any work on
functions of functions, or many details in Chapter I.
The course is adapted to students of widely differing
preparations. A knowledge of plane and solid geometry
and of algebra through quadratics is the most suitable equip-
ment ; but a number of students who had had only two years
of secondary mathematics have carried the course very well.
On the other hand, students who have already taken trigo-
nometry and college algebra find in the present course very
little that merely duplicates their former work.
The problems of the course have been collected largely
from scientific, technical, and business sources. I am in-
debted to Miss Maurine Laber and to Miss Edna V. Johnston,
alumnse of Reed College, for drawing most of the figures in
Chapters I-V and VII-XIV, respectively. I am also under
great obligation to Professors C. S. Atchison, J. G. Hardy,
W. R. Longley, and W. A. Wilson for reading the galley
proofs and making valuable suggestions.
F. L. GRIFFIN
CONTENTS
PAGE
A PRELIMINARY WORD TO STUDENTS . . 1
CHAPTER I. FUNCTIONS AND GRAPHS , 3
!/
Some fundamental problems of variation : rates, mean val-
ues, extremes, zero values, formulas, etc.
• ?s CHAPTER II. SOME BASIC IDEAS ANALYZED . 58
Instantaneous rates, tangents, areas, etc., as limits.
V CHAPTER III. DIFFERENTIATION ... 76
Derivatives of polynomials and un. Rates, extremes, etc.
t CHAPTER IV. INTEGRATION 126
Cxndx. Area, volume, momentum, work, fluid pressure,
J falling bodies, etc.
CHAPTER V. TRIGONOMETRIC FUNCTIONS . . 156
Solution of right and oblique triangles. Applications.
CHAPTER VI. LOGARITHMS 189
Numerical calculations. Compound interest. Triangles.
^ CHAPTER VII. LOGARITHMIC AND EXPONENTIAL
FUNCTIONS 236
Compound Interest Law. Logarithmic and semi-logarith-
mic graphs. Laws discovered. Differentiation and
integration : log u, eu, uv, u/v.
CHAPTER VIII. RECTANGULAR COORDINATES 271
Mapping. Motion. Analytic geometry : line, circle, parab-
ola, ellipse, hyperbola ; translation ; intersections.
vii
viii CONTENTS
PAGE
CHAPTER IX. SOLUTION OF EQUATIONS . . 326
Quadratics: 6s— 4ac. Rational roots of higher equations.
Homer's and Newton's methods.
CHAPTER X. POLAR COORDINATES AND TRIGONO-
METRIC FUNCTIONS 343
Definitions. Radians. Periodic variations. Derivatives.
CHAPTER XI. TRIGONOMETRIC ANALYSIS . . 368
Basic identities. Equations. More calculus. Involute.
Cycloid. S.H.M. Damped oscillations. Addition
formulas. Sums and products, etc.
CHAPTER XII. DEFINITE INTEGRALS . . .392
Summation of "elements" : length, surface of revolution,
etc. Plotting a surface. Double integration. Partial
derivatives. Simpson's rule.
CHAPTER XIII. PROGRESSIONS AND SERIES . . 415
A.P. and G.P. Investment theory. Maclaurin series.
Calculation of functions. Binomial theorem.
CHAPTER XIV. PERMUTATIONS, COMBINATIONS, AND
PROBABILITY 440
Pn. r', Cn.r- Chance. Normal Probability Curve. Least
squares.
CHAPTER XV. COMPLEX NUMBER SYSTEM . . 460
Definition. Geometric representation. Operations. Roots
of unity. Application.
RETROSPECT AND PROSPECT .... 472
APPENDIX 485
Proofs for reference. Formulas. Integrals. Numerical
tables : roots, natural and common logarithms, trigo-
nometric functions for radians or degrees.
INDEX 509
ANSWERS . i
AN INTRODUCTION TO
MATHEMATICAL ANALYSIS
A PRELIMINARY WORD TO STUDENTS
(I) "What It is All About." In scientific ^work and in
daily affairs, we frequently observe that some two things
seem to be related, — that any change in the one produces
some corresponding change in the other. Often it is important
to ascertain precisely how the one will change with the other.
To illustrate : the speed of a locomotive depends in part on the
amount of fuel consumed. Just how will the speed vary with the con-
sumption of fuel? The blood-pressure in a healthy person is different
at different ages. Just how should the pressure vary with the age?
How should the price of corn vary with the size of the crop? Or the
cost of a reservoir with the capacity? Or the speed of development of
a photograph with the temperature of the developer? And so on.
Mathematical Analysis makes a systematic study of
different modes of variation, discovers the exact relations be-
tween the varying quantities, and devises easy methods of
making whatever calculations may be necessary. It has
played a leading part in the wonderful modern development
of the exact sciences and is being used more and more in
other fields of study, — in the social sciences, in medicine,
engineering, and business administration.
The subject is a large one, and could be studied for many
years without exhausting it. But the introduction given by
the present course will provide mathematical tools adequate
1
MATHEMATICAL ANALYSIS
for many kinds of scientific work. Also, — what is desirable
as a part of any liberal education, — it will give a clear
general idea of the nature, power, and uses of modern mathe-
mutics.
(II) Some Suggestions as to Methods of Study. No
subject can be mastered by merely receiving instruction.
One must study it actively for himself. Try, therefore, to
react on each new question, and to devise some " rough-
and-ready " method of your own for dealing with it.
Before studying each new lesson think over the recent
work. Recall it clearly. Then, after reading the assignment,
set down briefly in your own words just what each new process
is, what it does, and why it is valid. This will save you much
time in working the exercises. Study with care the ex-
amples solved in the text, as they often cover elusive points.
Now and then run rapidly over in your mind an outline
of the course to date, in order to see each topic in perspective.
Re-read occasionally the " summaries " of preceding chapters.
X amorous principles and processes will be covered, and you may find
it easy to forget them at first. But we shall return again and again to
the most important ones, so that, with a little persistence, you can make
them your own before the end of the course.
Practice quizzing yourself. That is, think of questions
which might come up in class, and see whether you can
answer them. If any point is not clear, make a note of it
in some place reserved for the purpose, and ask about it or
look it up soon. Note carefully the exact meaning of each
new technical term that is introduced. Make free use of
the index, pp. 509-512. In short, " get into the game,"
actively.
Some effort may be required for a thorough mastery of the
course, but the final achievement will be well worth it.
CHAPTER I
Maximum Capacity
FUNCTIONS AND GRAPHS
SOME FUNDAMENTAL PROBLEMS OF
VARIATION
(A) THE PROBLEM OF EXHIBITING VARIATION
§ 1. Graphs. One of the best ways of showing how a
quantity varies is by means of a graph.
What graphs are, and how widely they are used, will be
clear from the following typical examples. You will doubt-
less recall having seen many
others in your general reading.
Fig. 1 is reproduced from an ad-
vertisement explaining low charges
for transatlantic "cable letters"
sent during certain hours. The
height of the curve above the base
line at any hour represents the
rate at which messages are then being sent. Where the curve is high,
much business is being dispatched; where low, little business. The
fluctuations from hour to hour are
portrayed far more vividly than by
a statistical table.
Fig. 2 exhibits the growth of the
native and foreign born populations
in Portland (Ore.) from 1870 to
1910. It shows at a glance not only
the comparative sizes of the two
populations at any time, as repre-
sented by the heights of the two
curves, but also the comparative
rates of increase, and a peculiar fluctuation in the rate of increase of
the foreign-born.
3
A.M. 13579 11 13579 11 P.M
Time (N.Y.)
FIG. 1.
F - Foreign Born
N- Native ••
1870
1910
MATHEMATICAL ANALYSIS
[I, §1
Atmospheric pressure depends upon the elevation above sea-level.
Fig. 3 shows how the pressure decreases as the elevation increases.
Besides showing how a quantity
varies, — and calling attention to
peculiarities, as in Fig. 2, — a graph
is often helpful in explaining some
principle, or in studying some scien-
tific law. Figs. 4-6 illustrate this
use of graphs.
.
JO tO .
Me ration
(Thousan
FlG. .3.
.10 40
Selling Price
(Cents per qt.)
FIG. 4.
U.S.
Curves like those in Fig. 4 are used by
economists to show how the laws of Sup-
ply and Demand together fix the selling
price of a manufactured article, — e.g.,
ice-cream. Curve S shows how the
supply increases with the price : i.e., its
height at any point shows how much
would be made to sell at the price there
represented. Curve D shows the demand,
— i.e., the quantity that could be sold at each price. There is a price
where demand equals supply : this is the natural selling price. (Why ?)
Fig. 5 is often used in showing
how an iron rod acts under high
tension, stretching at an almost con-
stant rate until the "yield-point" is
reached, then lengthening rapidly, —
and breaking, if the point of "ulti-
mate strength" is reached. The
varying height shows how much
tension is necessary to produce vari-
ous elongations.
Diagrams like Fig. 6 are helpful in studying biological measure-
ments Here the height of each rectangle shows what percentage of
soldiers in certain Scotch regiments had the chest measure indicated
at tho base of the rectangle. (E.g., there were 18% whose measure
was between 40 and 41 in.)
The height of the curved line shows the relative fre<|ueney with
whifh any part imhr measure would be found in the long run.
The fact that the curve is low toward either extreme means that
.Of .04 .06 .03 .10 .12
Elongation (Inches)
FIG. 5.
I, §2]
FUNCTIONS AND GRAPHS
1*1
the chest-measure of very few men
departs widely from the average.
The same is true of many other ^'
physical measurements ; and prob- § lo~
ably also of mental ability. 1 5.
A third use of graphs, and ^ °-
the most important of all in
practical work, is in making
approximate calculations rapidly.
34 36
38 40 42 44
Chest Measure
(Inchea)
FIG. 6.
46 48
14,000
^12,000
I ^0,000
Q 8,000
| 6,000
O 4,000
2,000
>
/
jr
/
/
/
^
/
/
/
200 600 1000 1400 180
Capacity
(Thousand Gallons)
FIG. 7.
For example, a graph like Fig. 7
is used by a certain designer of large
concrete oil-tanks. He can read off
at a glance the approximate cost
of a proposed tank of any desired
size, and can submit a bid at once.
The graph is a "ready-computer"
which saves many hours of tedious
calculations.
We shall see various other
uses of graphs presently ; but
the basic principle underlying all of them is simply this :
Points along the base line represent values of one quantity,
while the varying height of the curve above the base line shows
how some other quantity varies with the first.
§ 2. Function Defined.* Whenever one quantity, y, varies
with another, x, in some definite way, y is called a function
of x.
E.g., the atmospheric pressure is a function of the elevation above
sea-level; for the pressure varies with the elevation in some definite
way, other things being equal.
Fig. 3, p. 4, shows how the pressure varies with the
elevation, — in other words, shows what sort of function
* The mathematical meaning of the word " function " is entirely different
from the ordinary meaning. Be sure to get it clearly in mind.
6
MATHEMATICAL ANALYSIS
[I, §3
the pressure is. Similarly Fig. 7 exhibits the cost of a tank
as a function of the capacity. The primary use of graphs
is to exhibit some quantity (y) as a function of some other
quantity (z).
The quantity x upon which y depends is called the Inde-
pendent Variable. It is regarded as running freely through
its range of values represented along the horizontal scale,
while y must vary with it in some definite way, as shown by
the changing height of the graph.
§ 3. How Graphs Are Drawn. The process of drawing
a graph will now be illustrated.
EXAMPLE. The amount of moisture, or weight of water
vapor, that a cubic meter of air can hold depends upon the
temperature. Table 1 shows the greatest amount possible
at various temperatures from —20° to +40°, Centigrade.
Plot a graph exhibiting the possible weight of vapor as a
function of the temperature.
TABLE 1
TEMPERATURE
(degrees)
._ WEIGHT
(grams) '
' TEMPERATURE
(degrees)
WEIGHT
(grams)
-20
1.
15
12.8
-15
1.5
20
17.2
-10
2.3
25
22.9
- 5
3.4
30
30.1
0
4.9
35
39.3
5
6.8
40
50.9
10
9.3
We first mark off on a horizontal line a series of points,
equally spaced, and label them as in Fig. 8 to represent the
temperatures shown in the table.
Now at 40° the weight is 50.9 gm. To show this we erect
50.5
I, § 4] FUNCTIONS AND GRAPHS 7
at the 40° point a vertical line 50.9 units tall. (The unit
may have any convenient size.) Similarly at the 35° point
we erect a line 39.3 units tall;
and represent likewise all other
weights given in the table.
These vertical lines or " ordi-
nates " show roughly how the weight ,.
. . .,, ,, -20 -10 0 10 20 SO 40
oi vapor varies with the tempera- Temperature
ture in saturated air. The varia- FlG- 8-
tion is shown better when we join the ends of the ordi-
nates by a smooth curve, as in
Fig. 9. This curve is the required
graph.
If, at any point on the hori-
zontal base line, representing any
-so n-io ^o +10 +20 +30 \lo temperature from -20° to 40°,
Temperature WQ erect an orcjinate
FIG. 9. . ., .„
up to the curve will represent the
maximum weight of vapor which 1 cubic meter of air can hold
at that temperature.
Indeed, one use for the graph is just thig : to ascertain by measuring
ordinates how much vapor can be held by air at other temperatures than
those given in the table.
For accurate measurement we plot on a large scale and use "graph
paper," ruled in squares. (See Fig. 7, p. 5.)
§ 4. Suggestions as to Details. — Before plotting always
mark off suitable scales of values along the base line and
some vertical line. Let the horizontal scale invariably
increase toward the right and the vertical scale upward. Never
mind whether this makes the curve higher at the right or at
the left. Run negative values toward the left or downward,
respectively.
Make the graph as smooth a curve as possible: free from
needless turns and abrupt changes of direction. Draw the
8 MATHEMATICAL ANALYSIS [I, § 4
curve lightly until it appears satisfactory. A help toward
smooth drawing is to turn the paper so that your hand is
on the inner or concave side of the curve.
If you find a hump in the curve, due to a value which does not fit
in smoothly with the other values, see whether you have plotted it
correctly.
Avoid drawing instruments, such as ruler, compasses, and
" French curves." To make the graph a series of straight
lines with different inclinations in successive intervals would
imply abrupt changes in the growth of the quantity, quite
unlike the smooth and gradual changes produced by forces
of nature.
Exceptions: In plotting statistics ab(out the fluctuations
of a quantity which we have no reason to suppose varies
regularly, or where no meaning can be attached to ordinate's
erected between those given, we join the ends of the given
ordinates by a series of straight lines. (Cf. Ex. 4 below.)
The graph then merely " carries the eye." Also, if the ends
of the ordinates happen to lie exactly in a straight line, we
of course use a ruler, and make the graph as straight as
possible.
In plotting a graph we use in reality only the ends of the ordinates.
But it is desirable at first to draw the entire ordinates, as in Fig. 9, to
fix in mind the important fact that the varying height is what we are
really studying.
EXERCISES*
1. (A) Plot carefully on graph paper the curve discussed in § 3,
using 1 large space to represent 10° horizontally, 10 gm. vertically.
(#) In your curve draw ordinates at temperatures of 34° and 18°,
and note their lengths. What weights of vapor can be held at those
* The most convenient paper is that having 10 small spaces to 1 large
space. For rapid practice this is more important than great accuracy of
ruling.
I, §4]
FUNCTIONS AND GRAPHS
temperatures ? (C) At what temperature will 1 cubic meter of saturated
air contain 3 gin,, of vapor? [For a list of answers, see p. 513.]
2. Table 2 shows the standard atmospheric pressure in inches of
mercury, at various elevations above sea-level. (A) Plot, using 1
large space for 5000 ft. horizontally, and for 5 in. vertically. (B) Find
the pressure at the summit of Mt. McKinley, 20,464 ft. above sea-
level. (C) How high is an airplane if the pressure shown by its ba-
rometer is 19.8 in.? (D) How much difference between the pressures
at 10,000 and 20,000 ft.?
TABLE 2
ELEVATION
PRESSURE
ELEVATION
PRESSURE
0
30.0
24,000
11.8
6,000
23.8
30,000
9.5
12,000
19.0
36,000
7.5
18,000
15.0
3. Table 3, used by life insurance companies, tells how many will
be living at various ages, out of an average group of 100,000 persons
at age 10. (A) Plot a graph showing the number of survivors as a
function of the age attained. (Use 1 small space for 1 year horizontally
and for 2000 persons vertically. Merely estimate odd hundreds in plot-
ting.) CB) How many survivors at 35 years? By what age will half
of the original company have died? (C) How many die between 45
and 55?
TABLE 3
AGE
LIVING
AGE
LIVING
10
100,000
70
38,569
20
92,637
75
26,237
30
85,441
80
14,474
40
78,106
85
5,485
50
69,804
90
847
60
57,917
95
3
65
49,341
10
MATHEMATICAL ANALYSIS
[I, §4
4. Exhibit graphically the world's yearly production of gold from
1872 to 1917. (In Table 4 the amounts are in millions of dollars.)
•
TABLE 4
YKAB
AMOUNT
YEAR
AMOUNT
1872
115.6
1897
236.1
1877
114.0
1902
296.7
1882
102.0
1907
413.0
1887
105.8
1912
466.1
1892
146.3
1917
419.4
6. Table 5 shows the number of kilograms equivalent to various
numbers of pounds. Plot, using 1 space vertically for 2 kg., hori-
zontally for 5 Ib. (What sort of graph? Why?) Read off the equiv-
alent of 28 Ib. ; of 1.44kg.
TABLE 5
TABLE 6
TABLE 7
La.
Ko.
0
0
5
2.268
10
4.536
15
6.804
20
9.072
25
11.340
30
13.608
A
p
25
19.90
30
22.70
35
26.30
40
31.00
45
37.40
50
46.20
55
58.30
60
75.00
h
D
10
3.9
50
8.7
100
12.3
160
15.6
200
17.4
300
21.3
400
24.5
6. The annual premium ($P) which a certain life insurance com-
pany charges for a $1000 policy taken out at various ages (A yr.) is
shown in Table 6. Plot, using 1 space for $10 vertically, for 10 yr.
horizontally. Find P at age 32.
7. The distance (D mi.) of the horizon at sea varies with the height
(h ft.) of the observer's eye above the water, as in Table 7. Plot D
as a function of h, using scales of 5 mi. and 50 ft. How much farther
can one see at a height of 360 ft. than at 80 ft.?
I, § 6] FUNCTIONS AND GRAPHS 11
8. Find three graphs in the Encyclopaedia Britannica, or other out-
side sources, and state what quantity each exhibits as a function of
what other. [See Aberration, Bacteriology, Bridges, Climate, Heat,
Influenza, Liquid Gases, Photography, etc.]
9. Which quantity would you plot vertically if given a table show-
ing the speed of a train at various times? The time of swing for pendu-
lums of different lengths? The cqst of running a locomotive at different
speeds ?
10. To practice using the index, pp. 509-512, -locate the pages on
which the following are mentioned: "probable error," "die-away
curve," "reciprocal."
§ 5. Interpolation. The operation of finding a value of a
variable quantity between those given in a table, and con-
sistent with them, is called interpolation. One way to inter-
polate roughly is to plot a graph and read off the required
intermediate values. This can be done rapidly if the scales
are well chosen.
§ 6. Choice of Scales. The most convenient scales are
those in which each space represents 1 unit, 10 units, or
100 units, etc. But if these would make the graph too large
for the paper or too small for accurate interpolating, we
let each space represent 2 or 20 units, etc., or J, 5, or 50 units,
etc.
Scales based upon 3, 4, or anything worse should be avoided. Even
if all the values in a table ran by 12's, we should employ scales of 10,
or 20, etc., in order to read off readily any required intermediate
values.
In short, the essentials are: (1) a convenient number
of units to each space ; (2) as large a graph as possible.
Always examine the table carefully at the1 outset with
these aims in view. Also decide, before marking off the
scales, which quantity is the function to be exhibited verti-
cally. Turning the paper afterward would make one of the
scales increase in the wrong direction.
12
MATHEMATICAL ANALYSIS
[I, §6
EXERCISES
1. The amount ($A) which a deposit of $1000 will yield after various
intervals of time (T yr.), drawing interest at 6%, compounded an-
nu:illv, is shown approximately in Table 1. Plot A as a function of T.
Find how much A will increase between T = 18 and 77 = 33. When
will the original sum have been quadrupled?
2. The number of years that an average person at any given age
will live is the expectancy for that age. This is shown in Table 2 for
various ages (A yr.) Plot E as a function of A. How much does a
man's expectancy decrease between the ages of 12 and 32? When is
E half as great as at age 20?
TABLE 1 TABLE 2 TABLE 3
TABLE 4
0
5
10
15
20
25
30
35
1000
1338
1791
2397
3207
4292
5743
A
E
10
48.7
20
42.2
30
35.3
40
28.2
50
20.9
60
14.1
70
8.5
80
4.4
0
10
20
30
40
50
60
75,000
70,000
55,000
30,000
13,000
6,000
3,000
DATE
ERROR
DATE
ERROR
Jan. 1
- 5.0
July 18
- 8.0
Feb. 1
-14.5
Aug. 1
- 7.2
Feb. 12
-15.6
Sept. 1
- 0.9
Mar. 1
-13.3
Oct. 1
9.0
Apr. 1
- 5.0
Nov. 1
15.2
May 1
2.0
Dec. 1
10.2
May 12
3.5
Dec. 15
3.5
June 1
2.0
Dec. 31
- 4.5
July 1
- 4.5
{3.1 The estimated value (V) of a certain piece of property at vari-
ous times (T yr. hence) is shown in Table 3. Plot, and find the
probable value 25 yr. hence. How much will V decrease from T = 5
to T = 10? What is the average rate of decrease per year for those
five years?
4. Table 4 shows the error of a certain sun-dial on various days of
the year, negative signs indicating when the dial is slow. (A) Plot,
treating months as equal, and drawing negative ordinates downward.
(7?) What error on April 10? (C) On what days, approximately,
should the dial be correct?
5. The probable error (E meters) of the U. S. army range-finder at
various ranges (R meters) is shown by Table 5. One of the values is
given incorrectly. Plot, and find which one; also what the value
should be.
I, §7]
FUNCTIONS AND GRAPHS
13
6. Table 6 shows the number of days of illness during a year for an
average person at various ages (A yr.). Plot. Find N for age 58.
Also the increase in N between the ages of 72 and 82.
7. Table 7 shows the number of people killed in 4th of July cele-
brations in various years. Show this graphically.
TABLE 5
TABLE 6
TABLE 7
400
1000
1500
2000
2500
2750
3000
.7
4.4
9.7
15.2
27.0
32.9
39.0
18
30
40
55
65
75
83
4.5
6.
8.3
19.
44.
105.5
171.5
YEAR
DEAD
YEAR
DEAD
1903
466
1910
131
1904
183
1911
57
1905
182
1912
41
1906
158
1913
32
1907
164
1914
40
1908
163
1915
30
1909
215
1916
30
8. Make a table of squares for the numbers 0, 2, 4, 6, 8, 10. Plot a
graph showing how the square varies with the number. Read off
(7.6)2 and VQ8, and check.
(B) THE RATE PROBLEM
§ 7. The Idea of a Rate. In studying a varying quantity
or function we often need to know how fast it is increasing
or decreasing, — in other words, the rate at which it is
changing.
The general idea of a rate is, of course, the amount of change
in the function per unit change in the independent variable.
Graphically this is represented by the distance the graph- of
the function rises or falls per horizontal unit. Thus a rate
is shown by the steepness of the graph, and not by the height
at any point.
If the graph is a straight line, rising by a fixed amount
in each and every horizontal unit, the function must be
increasing at a constant rate.
14
MATHEMATICAL ANALYSIS
[I, §8
6.000
6.000
Ti
g 4,000
« 3.000
'
/
/
/
/
/
1.000
S 10 15 20
Time (Yeart)
FlO. 10.
Thus in Fig. 10, representing the
growing value of a certain investment,
the value increases at the constant rate
of $200 per year. See also Fig. 5,
p. 4.
Conversely, if the rate is con-
stant, the graph must be straight,
since it must rise by the same
amount in every unit.
Most quantities, however, change
at a varying rate. In such cases
we distinguish between the average
rate of change during some interval,
and the instantaneous rate at some particular instant.
For instance, if the volume of a balloon increased by 1200 cu. ft.
in six hours, it increased at the average rate of 200 cu. ft. per hr. But
it may have been increasing more or less rapidly than this at any par-
ticular instant, — or even decreasing part of the time.
The distinction, and the relation, between the two kinds
of rates will be discussed more fully as we proceed. Both
kinds can be found approximately from a graph.
§ 8. Average Rates Found
Graphically. To find from a
graph the average rate of in-
crease of a varying quantity
or function in any interval,
we merely read off the amount
of "increase during the inter-
val, and divide by the length
of the interval or change in the
independent variable. Simi-
larly for a rate of decrease.
EXAMPLE. Find the average rate at which the weight of vapor in
saturated air increases with the temperature between 18° and 28°.
15 20 25 SO 35
Temperature (Deorce C.)
FlO. 11.
I, § 10] FUNCTIONS AND GRAPHS 15
The graph (Fig. 11) shows that the weight increases by 11.8 gm.
during this 10° interval. Hence the average rate is 1.18 gm. per deg.
§ 9. Instantaneous Rates Found Graphically. A straight
line tangent to a graph at any point will forever rise at the
same rate as the graph was rising at the point of tangency.*
Hence, when we wish to find how fast a given function
was increasing at a certain instant or point, rather than its
average rate of increase during some interval, we need merely
draw a tangent to the graph at the point in question, and
find the required instantaneous rate from it, using any con-
venient interval.
E.g., from the tangent line (Fig. 11) we see that if the weight con-
tinued to increase at the same rate as at 18°, it would increase by 9
gm. while the temperature rose 10°. Hence the instantaneous rate at
18° is .9 gm. per deg.
To draw a tangent line accurately by the eye, however,
requires great care. The ruler should have the direction of
the curve at the point of tangency and should run along the
curve closely in both directions near by.
In solving rate problems the respective increases should be clearly
labeled on the graph, as in Fig. 11. Also the answer should name
the units, — as grams per degree, etc.
Rate units are often written as fractions. Thus grams per degree
is abbreviated gm./deg., indicating that this rate is found by dividing
some number of grams by some number of degrees.
§ 10. Small Intervals. It is sometimes necessary to find
the average rate of increase of a quantity in an interval so
very short that the amount of increase cannot be read from
the graph with any accuracy, f
* This intuitive conception will be justified logically in § 41.
t Do not confuse the amount of increase with the rate. A function may
increase very little in a short interval and yet be increasing very fast, — just
as a train may run only an inch in a small fraction of a second and yet be
running at a very high speed.
16
MATHEMATICAL ANALYSIS
[I, § 10
In such a case we may reason as follows: The average
rate for the short interval would be very nearly the same as
the instantaneous rate at any instant during the interval, —
just as the average speed of a train during a small fraction
of a second would be nearly the same as the speed at any
instant during that short time.
Hence we may approximate an average rate in any short
interval by finding graphically the instantaneous rate, say
at the middle of the interval.
EXERCISES *
1. Table 1 shows the number of bacteria in a culture, i hr. after
first observing them. Plot. When had the number doubled? What
was the average rate of increase from / = 2.£ to J = 4.5 ? How fast
was N increasing at the instant t = 5.6 ?
2. The temperature of an object (T° Cent.) fell as in Table 2 after
various intervals (t min.). Plot. What was the temperature at
J = 3.2? How fast was the object then cooling? When was the tem-
perature 65°?
TABLE 1
t
N-
t
N
0
100
4
739
1
165
5
1220
2
272
6
2010
3
449
7
3310
3. Radium decomposes continually. Table 3 shows the quantity
(Q mg.) remaining after T yr., if the initial quantity was 1 gm. How
much will remain after 500 yr.? When will only half remain? What
is the average rate of change of Q during the first 1000 years? What
instantaneous rate at T = 1000?
* For further practice invent rate-problems using graphs already
plotted.
FUNCTIONS AND GRAPHS
TABLE 3 TABLE 4
17
TABLE 5
6
8
10
12
80
54.8
42.3
36.1
33
31.5
30.7
T
Q
0
1000
1000
681
2000
463
3000
315
4000
214
5000
146
6000
99
7000
68
t
X
0
0
1
17
2
128
3
405
4
896
5
1625
6
2592
7
3773
8
5120
9
6561
c
v
I
12
2
17.6
4
25
6
31
8
36
10
40
12
43.4
15
47.6
18
51
4. Table 4 shows the distance (x ft.) which a boat had traveled
t minutes after starting. Plot. How far did the boat go between
< = 3.5 and 2 = 8.5? What was the average speed during those five
minutes ? What speed at t = 4 ?
TABLE 6
TABLE 7
A
t
A
(Pre.)
(Ob8.)
(Pre.-)
(O6s.)
0
56.6
36
10.5
11.
4
49.1
44.3
40
8.3
9.2
8
42.
42.
44
6.0
6.5
12
35.4
33.6
48
4.3
4.2
16
29.7
30.2
52
2.9
3.1
20
24.5
24.2
56
1.8
1.8
24
20.1
22.1
60
.7
1.1
28
16.4
17.0
64
.2
.8
32
13.2
13.6
68
Healed
TIME
No.
TIME
No.
Oct.
65
Aug.
1293
Nov.
102
Sept.
1576
Dec.
129
Oct.
1843
Nov.
1971
Jan.
176
Dec.
1944
Feb.
225
Mar.
253
Jan.
1837
Apr.
320
Feb.
1710
May
424
Mar.
1562
June
722
Apr.
1376
July
996
May
1088
6. The speed of a locomotive (V mi./hr.) varied with the consumption
of coal (C tons/hr.), as in Table 5. Plot V as a function of C. What
V requires 3 tons hourly? How fast does V increase with C, at C=4?
What is the average rate of increase between C = 7 and C = 7.1?
18 MATHEMATICAL ANALYSIS [I, § 11
6. With Dr. Carrel's method of treating deep wounds, the date of
healing can be predicted accurately.* Table 6 shows the predicted
and the observed size (A sq. cm.) of a typical wound t days after the
first treatment. Plot together the theoretical and observed curves
of healing. Theoretically, at what rate should the wound have been
healing when J = 28?
7. Table 7 gives the number of soldiers in the A. E. F. (in thousands)
from October, 1917, to May, 1919. Show this graphically. At what
average rate were the A. E. F. increasing between May and October,
1918? Decreasing between February and May, 1919?
§ 11. Interpolation by Proportional Parts. If the inter-
vals between the values in a given table are small, we can
calculate intermediate values approximately without plotting
a graph. For we may regard the rate of increase as practi-
cally constant within a small interval.
Ex. I. Find from Table I the weight of vapor in sat-
urated air at 23° C.
TABLE I
TEMP.
WEIGHT
T°
W gm.
[20
17.21
6
1 26
[5.7
22.9 J
We simply have to find how muc'h W will increase while
T rises from 20° to 23°.
The 5° rise in T increases TF.by 5.7 gm.
/. The 3° rise in T increases W by 3/5 of 5.7 gm. (=3.4 gm.)
Adding this increase to 17.2, the weight at 20°, gives 20.6
gm. as the weight at 23°. This is evidently a reasonable
value.
* See Journal of Experimental Medicine, v. 24, pp. 429-460.
I, § 11]
FUNCTIONS AND GRAPHS
19
Observe in Fig. 12 that if the graph were straight, this calculation
would be strictly correct, as the increase in W between 20° and 23°
(denoted by the Greek letter A, "delta")
would be exactly three fifths of the whole
increase 5.7 gm.
To avoid blunders in more com-
plicated cases we may set the cal-
culation down in detail, as in the
following example.
Ex. II. Table II gives the "re-
ciprocals " of 4.42 and 4.43.* Find the
number whose reciprocal is .22591.
4
^S
t
^^r
5.7
^S^ A
1
^
3
< 5-t— >
17.2
10-
20° 23° 25"
Temperature
FlG. 12.
TABLE II
NUMBER (N)
RECIPROCAL (R)
.01
H-2
4.43
.226244
.225910
.225734
J334J
510
We first indicate the required value between the nearest
given values, as shown here. Then we form corresponding
differences in the two columns, using A for the difference
between the required value of N and 4>4%-
For a constant rate of change, these corresponding differ-
ences should be proportional :
_
.01 ~ 510*
That is, the partial difference in N is to the whole difference in N
as the partial difference in R is to the whole difference in R.
Multiplying through by .01 gives A = 3.34 -r- 510 = .0065.
Recalling what A stands for, we add it to 4.42, getting
4.4265.
* For a definition of "reciprocal," see § 30.
20
MATHEMATICAL ANALYSIS
[I, § 11
This result is reasonable, being between 4.42 and 4.43 and
nearer the latter, — as it evidently should be from a com-
parison of the given values.
Kf marks. (I) If we had used A to denote the difference between the
larger value (4.43) and the required value, we should have subtracted
the value of A finally from 4.43.
(II) How inaccurate the results obtained by this method of "Pro-
portional Parts" are, depends on how much the graph would deviate
from a straight line in the interval considered. In general, it is a waste
of time to calculate the value of A to many figures.
EXERCISES
(To be worked by Proportional Parts)
1. In Ex. II, solved above, show that if A were used to denote the
difference between the required value and '4.43, the final result would
be the same although the value of A came out differently.
2. If 1 cu. M. of saturated air contains 10 grams of vapor, what is
its temperature? (See Table 1, p. 6.)
3. How many "survivors" will there be at 43 years? (Table 3,
p. 9.)
4. What is the pressure at 4000 ft. elevation? (Table 2, p. 9.)
6. In how many years at 6% compound interest will any original
principal double itself? (Table 1, p. 12.)
6. (A) Draw a rough sketch similar to Fig. 12, to illustrate Ex. 2.
(B) The same for Ex. 3.
TABLE 1 TABLE 2
TABLE 3
TABLE 4
TABLE 5
p
T
T
D
N
5
N
R
760
787.7
816
845
100
101
102
103
26
28
30
13.532
13.528
13.523
890
891
892
792,100
793,881
795,664
1.40
1.41
1.42
.7143
.7092
.7042
•
7. Table 1 gives the boiling point of water (7'°) at
various pressures (P mm.). Find T when P = 800.
Also P when T = 102.8.
V
p
1200
.94
1500
1.06
1800
1.15
2100
1.15
24CO
1.07
2700
.94
3COO
.75
3300
.51
3600
.31
I, § 12]
FUNCTIONS AND GRAPHS
21
8. Table 2 gives the density of mercury (D gm. per cc.) at various
temperatures (T°). Find D when T = 27.2; also T when D = 13.526.
9. Table 3 gives the squares of 3 numbers. From it find approxi-
mately the square of 891.7. Check.
10. Table 4 shows the reciprocals of 3 numbers. Find the reciprocal
of 1.418.
11. In running a waterwheel the power obtained (p horsepower)
varied with the velocity (v revolutions per min.) as in Table 5. Find
graphically : (A) The value of p at v = 2500. (B) How fast p decreases
at v = 3000. (C) What velocity yields the maximum power. How
much power? Check (A) by Proportional Parts.
§ 12. Tables of Squares, etc. To save time in many
of the problems which follow, tarbles of squares and square
roots, etc., are given in the Appendix, pp. 500-501.* Some
sample lines are shown here in part.
N
AT*
<N
Vio*
5.0
25.00
2.2361
7.0711
5.1
26.01
2.2583
7.1414
This means, for instance, that
5.12 = 26.01 ; V5l = 2.2583; V51 = 7.1414, etc.
The numbers N given in the table all lie between 1 and
10. But the table can be used for larger or smaller values
because of these facts :
(A) Moving the decimal point one place in a number
will merely move it two places in the square.
(B) Moving the point two places in a number will merely
move it one place in the square root.
Ex. I
5.12 = 26.01 Since
512=2601 hence
.05 12 = .002601 also
etc.
Since
hence
also
Ex^II
Vy =2.2583
V5lO = 22.583
V.051 = . 22583
etc.
* More extensive tables of like character are included in the Macmillan
Logarithmic and Trigonometric Tables.
MATHEMATICAL ANALYSIS [I, § 12
The column VlO N gives the square root when the decimal
point has been moved one place, or 3 places, etc. But in
looking up a square root, the best way is not to think of
" N " and " 10 AT," but to decide in advance what the first
figure of the required root will be, and then look in the column
where that figure occurs.
If you were going to extract the root without a table,
the first step would be to point off " periods " of two places
each, starting from the decimal point: the required first
figure is found by extracting the square root of the leading
period.
Illustrations
VS'IO'OO'OO. starts with 2 ; has 4 digits before the decimal point.
starts with 7 ; has 2 digits before the decimal point.
starts with 2 ; in the second decimal place.
V.00'00'51 starts with 7 ; in the third decimal place.
In each case it is obvious which column of the table will give the required
root : the correct leading figure occurs in only one !
Cube-root tables are used similarly ; but the " periods "
consist of three figures each. For illustrations see p. 501,
Note (I).
Interpolation can be resorted to, when N has several
significant figures.
EXERCISES
1. Look up the squares of 4.7 ; 680 ; 25000 ; .72 ; .019. Try to
test each result roughly by common sense.
2. Look up the square roots of 3.3; 5600; 4200000; .028; .96.
Check by common sense.
3. Look up the cube roots of 5.2; 870; 43000; .38.
4. (A) Look up the square and square root of 87.5, interpolating
to take account of the third given figure. (#) The same for the number
.198. (C) The same for .000629.
5. Table 1 is taken from a table of "logarithms." Find log N
when # = 49056. Also N when log N = 69CXK )
I, § 12] FUNCTIONS AND GRAPHS
TABLE 1
23
AT
49040
49050
49060
logN
69055
69064
69073
6. Table 2 is from a table of " sines. " Find sin A when A = 12°35'.3.
TABLE 2
A
12° 35'
12° 36'
Sin A
.21786
.21814
7. Table 3 shows the temperature (T°) at several hours on a certain
day. About when was T = 0 ?
TABLE 3
H
8
10
12
2
T
-6
-2
6
9
8. For a cylindrical tank of a certain type and capacity, the cost
($C) depends upon the relation of the diameter and height, — varying
with the diameter (D ft.), as in Table '4. Plot, using 1 large space
TABLE 4
TABLE 5
D
c
80
8550
100
7200
120
6650
140
6600
160
6850
180
7250
200
7800
t
V
0
0
10
90
20
320
30
630
40
960
50
1250
60
1440
24
MATHEMATICAL ANALYSIS
[I, § 13
vertically for $500, and starting with $5000 at the base line. Find the
lowest possible cost, and what diameter gives it. How fast does C
increase with D per foot at D = 150?
9. Table 5 shows the speed (V ft./min.) of a boat at various times
(t min.) after starting. Find V when < = 22, graphically and by Propor-
tional Parts. Also find the acceleration when t = 3Q. [Acceleration
is the rate at which the speed is changing.]
[10.] In Ex. 9 can you devise some way to find the distance traveled
from < = 0 to < = 60? (Hint: About what speed did the boat average
during each 10 min. ?)
(C) THE MEAN-VALUE PROBLEM
§ 13. Average Value of a Varying Quantity. It is often
necessary to find the average value of a quantity which
is continually changing, — not the average of its values at
a certain few instants, but the average value maintained
throughout some interval of time.
If the varying quantity in question is the height of a
graph, the problem is simply to find the
average height in a specified interval or
strip.
This is usually not the same as the average
of the heights at the beginning and end of the
interval. For the latter average takes no account
of the way the height varies within the inter-
val,— whether the curve sags or arches upward.
NOT would the height at the middle of the inter-
val in general be the required average height.
By the average height throughout an
interval we mean simply: The height
which, multiplied by the base, would give
the area under the curve in that interval.
In other words, it is the height of a rec-
tangle equivalent to the area under the curve and having
the same base. (See MN in Fig. 13.)
The average height as thus defined is called the mean
M
Fio. 13.
I, § 14] FUNCTIONS AND GRAPHS 25
ordinate of the curve in the strip considered. To approxi-
mate it closely, simply draw a horizontal line across the
strip at about the right height, compare the triangular areas
thus formed (to the right and left of N), and move the line
up or down if either area appears to be the larger.
That this definition of average height is a suitable one will be seen
from the following applications.
§ 14. Distance Found from a Speed-time Graph. Sup-
pose that we have a graph, such as Fig. 14, showing how the
speed of a moving object varied, and that we wish to find the
distance traveled during some interval of time. The problem
is simply to find the average speed; for multiplying this by
the time would give the distance.
The varying speed is represented by the varying height
of the graph. Hence the average speed will be represented
by the average height, — provided the latter is properly
defined. That the definition in § 13 meets this requirement
is easily proved.
PROOF. Suppose the dotted line BC (Fig. 14) drawn at the exact
height A B which represents the average speed during the time AD,
Then ABXAD represents the product of the average speed by the time.
That is, the area of rectangle ABCD represents the distance traveled during
the interval.
Again, if AD were divided up into a billion smaller intervals, the
sum of the rectangular areas likewise formed for these tiny intervals
would still represent the distance traveled in the whole interval AD.
But these rectangles would be so narrow and their successive heights
differ by so little that they would virtually coincide with the area
under the curve. Hence, — very approximately at least, — the area
under the curve also represents the distance traveled. (That this represen-
tation is exact is proved in § 98, footnote.)
Therefore, since rectangle ABCD represents the same thing as the
area under the curve in this strip, it must be equivalent to the strip,
and its height AB must be the mean ordinate of the curve. I.e., the
average speed during the time AD is represented by the mean ordinate,
as defined in § 13 (Q. E. D.).
26
MATHEMATICAL ANALYSIS
[I, § 15
-ISO
100
15 20
me (Min.)
FIG. 14.
Hence in Fig. 14 the average speed during this interval
AD was, according to the vertical scale, about 140 ft./min.,
and the distance traveled
during this 5 min. was
about 140X5, or 700 feet.
Similarly the average speed
during the last 5 min. was
about 94 ft./ min. and the
distance about 470 ft.
jo is so 25 so In finding the total dis-
tance traveled during sev-
eral 5-min. intervals, time
would be saved by adding all the average speeds, as read
off from the mean ordinates, and multiplying the sum by 5
instead of multiplying each separately.
A rapid rise or fall of a graph may necessitate using very narrow
strips in some part of the curve.
§ 15. Further Physical Uses of Mean Ordinates.
(A) Finding Momentum from a Force-time Graph. A
force acting upon an obj ect imparts ' ' momentum " to it . The
amount of momentum equals the average force, multiplied
by the length of time. E.g., an average force of 35 Ib.
acting for 10 sec. imparts 350 Ib.-sec. of momentum. But if
the force is continually changing, how shall we know its
average value?
Suppose the force plotted as a function of the time, — i.e.,
represented by the varying height of a graph. Then the
average force in any interval is represented by the correspond-
ing average height or mean ordinate.
This can be proved in detail by reasoning just as in § 13 and slxwing
Iliat the areas of the rectangle and strip considered hnth represent the
momentum imparted.
I, §
FUNCTIONS AND GRAPHS
27
Using reasonably short intervals, we can estimate pretty
accurately the average force in each; and can find the
momentum.
(B) Finding Work from a Force-distance Graph. The
amount of work done in moving an object is found by multi-
plying the distance the object travels by the average force
used in moving it. E.g., an average force of 35 lb., moving
an object 10 ft., does 350 ft.-lb. of work.
If the force varies, and we have a graph exhibiting it as a
function of the distance, the average force during any dis-
tance will be represented by the average height or mean
ordinate of the curve. Hence the work can be found.
Remark. There are many more cases in scientific work in which mean
ordinates are useful, — in fact whenever the area under the graph has a
meaning. This will be so whenever the product of the two quantities
represented horizontally and vertically has one. E.g., in § 14, speed X
time = distance ; etc.
EXERCISES
1. The speed of an airplane (V ft./min.) after passing a certain
point varied with the time (T min.), as in Ta,ble 1. Plot, and find the
distance traveled during the hour covered by the table. How fast
was the speed changing at T = 30?
TABLE I
T
V
0
6080
10
6250
20
6120
30
5810
40
5440
50
5130
60
5000
TABLE 2
t
c
0
0
.002
379
.004
700
.006
915
.008
990
.010
915
.012
700
.014
379
.016
0
TABLE 3
t
V
0
1700
10
1575
20
1507
30
1507
40
1575
50
1700
28
MATHEMATICAL ANALYSIS
[I, § 15
2. As an auto traveled at a certain fixed speed, a point P on a
tire traveled with a varying speed (v ft./min.), shown in Table 2 at
various intervals of time (t min.) during one revolution. Find graph-
ically the speed of P at t = .001. Check by Proportional Parts. (Why
a discrepancy?) Find also the length of the path traveled by P during
a turn.
3. If a projectile were fired with an initial speed of 1700 ft. /sec.
at an elevation angle of 28°, and there were no air resistance, its speed
(y ft./sec.) would vary with the time (t sec.), as in Table 3. Plot and
find the length of the path traveled during the 50 sec.
4. The speed of a vertically falling body increases during each second
by 32.2 ft./sec. If thrown down with initial speed of 20 ft./sec., what
speed has it after 1, 2, 3, 4, 5 sec. ? Find graphically the total distance
fallen from t = Q to t = 5.
5. Table 4 shows the speed of Halley's Comet (V million mi./yr.)
at various times (T yrs.), since it was nearest the sun. (A) Find the
distance traveled until farthest away : 38. 5 'yrs. (B) Find how fast the
speed was changing at T=Q.
TABLE 4
T
0
1
3
6
9
12
15
21
30
38.5
V
1000
375
225
155
120
90
72
50
28
16
6. (o) Prove in detail that the area under a force-time graph
represents the momentum imparted, and hence that the mean ordinate
represents the true average force. (Use the idea of numerous tiny
rectangles, as in § 14.) (6) What meaning has the area under a force-
distance graph?
7. The pull exerted by a locomotive in starting a train exceeded
the resisting forces by F tons, varying with the time elapsed (t sec.), as
shown in Table 5. Find the total momentum given to the train in 8
seconds. Also the rate of increase of F at t = 1.
8. A weighing spring was stretched from a length of 6 inches to a
length of 6.8 inches, the pull used (/ Ib.) increasing with the length
(L in.), as in Table 6. Find the total work done in stretching the spring.
9. The resistance (R Ib.) offered by a tug-of-wur team after being
pulled x ft. drrro.isod as in Table 7. Find the total work done in
pulling the team 48 ft.
I, § 16]
FUNCTIONS AND GRAPHS
29
10. The force (/ Ib.) exerted by steam upon a piston varied with
the distance from one end of the cylinder (d ft.), as in Table 8. Find the
work done in moving the piston from d = 1.3 to d =3.2.
11. Table 9 shows the intensity (i amperes) of an electrical current
t sec. after the circuit was closed. Find the rate of increase of i at
t = .002 ; also the quantity of electricity passed in the first .005 sec.
(The quantity, Q coulombs, equals the average intensity X the
time.)
12. The electrical power (P kilowatts) used by a factory during a
half-day varied with the time (T hr.), as in Table 10. Find the total
amount of energy used. (The energy, E kilowatt-hours, equals the
average powerXthe time.)
TABLE 5 TABLE 6 TABLE 7 TABLE 8 TABLE 9 TABLE 10
L
/
6.0
0
6.1
5
6.2
10
6.3
15
6-4
20
6.5
25
6.6
30
6.7
35
6.8
40
X
R
d
/
0
1400
1.0
18400
9
1280
1.5
10250
18
1030
2.0
7300
24
810
2.5
5400
30
600
3.0
4250
36
410
3.5
3450
42
280
4.0
2900
48
200
t
t
0
0
.001
49.6
.002
71.8
.003
81.8
.004
86.3
.005
88.4
.006
89.3
.007
89.6
T
P
0
250
.5
500
1.0
650
1.5]
3.5 j
700
3.8
690
4.0
650
§ 16. Geometrical Uses of Mean Ordinates.
(A) To find the area within any dosed plane curve divide
the figure into narrow strips by parallel lines, and approxi-
mate each strip by a rectangle.
In this way, if the curve in Fig. 15
were a "contour line " running around
a hill at some given elevation, as de-
termined by a survey, we could find
the area of the horizontal cross-sec-
tion of the hill inclosed by that —
contour line. FlG- 15-
Engineers often use a " planimeter," which will measure
30
MATHEMATICAL ANALYSIS
[I, § 16
FIG. 16.
any small plane area approximately, however irregular the
boundary.
(B) To find the volume of an irregular solid, say a hill,
imagine it cut into thin slices by parallel planes, and approxi-
mate each slice by a cylinder.
Any slice, such as DE in Fig.
16, equals a cylinder of the same
height whose base area is some
average cross-section area within
the slice.
The areas of the various horizontal cross-sections A, B, C,
etc., can be found as in Fig. 15 above. Suppose this done,
and that we then plot a graph
showing how the sectional area
varies with the elevation. (Fig.
17.) The average height of
the graph in any strip will
represent the average area in
the corresponding slice of the
mound. If this is, say, 20,000
sq. ft. for the slice DE, whose thickness is 10 ft., the volume
of the slice is 200,000 cu. ft. Similarly for the other slices.
Reasoning as in § 14, we can show that the usual method of finding
a mean ordinate is valid here ; and that the area under the graph in
Fig. 17 represents the volume of the hill. That is, on the chosen scales,
the number of square units under the graph equals the number of cubic
units in the hill.
EXERCISES
1. A piece of land lies between a straight fence and a curved stream.
Table 1 shows the distance from the fence to the stream (y yds.) at
various points (x yds.) from one end of the fence. Map the land roughly
and find its approximate area. [Ans., 2700 sq. yds.]
2. A ship's deck has an axis of symmetry running lengthwise. The
semi-widths (w ft.) measured from this axis to one side of the deck
10 20
30 40 50
Elevation. (Ft.)
Fia. 17.
60
I, § 16]
FUNCTIONS AND GRAPHS
31
vary with the distance from the bow (d ft.), as in Table 2. Find the
approximate area of the deck.
3. The depth (D ft.) for a proposed railway "cut" for a level track
through a hill will vary with the distance (re ft.) from one end, as in
Table 3. Assuming the hill to slope smoothly, find the average depth
for each 100 ft. How much earth must be removed for a cut 20 ft.
wide with vertical sides ?
4. The area (A sq. ft.) of a horizontal section of a certain mound
varies with the elevation (E ft.) above the base, as in Table 4. Find
the volume of the mound, and also the rate at which A changes with
E at # = 20.
6. A plumb-bob is to be made with its horizontal cross-section area
(A sq. cm.) varying with the distance (x cm.) above the lowest point,
as shown in Table 5. What will its volume be?
6. According to a naval architect's drawings, the horizontal sections
of a certain ship at various heights (h ft.) above the keel will have the
areas (A sq. ft.) shown in Table 6. What will the volume of the ship
be, up to a height of 30 ft. ?
TABLE 1 TABLE 2 TABLE 3 TABLE 4 TABLE 5 TABLE 6 '
X
y
d
W X
D
E
A
0
0
0
0
0
0
0
82100
10
32
50
15
100
7
5
75000
20
49
100
24
200
20
10
67000
30
54
1501
300
25
15
58000
40
50
—
400
18
20
47400
50
40
—
27
500
12
25
33500
60
27
—
600
17
30
0
70
14
350 j
700
12
80
4
400
25
800
0
90
0
445
12
450
0
X
A
h
A
0
0
0
0
2
4.8
5
10900
4
16.
10
15000
6
28.8
15
15900
8
38.4
20}
10
40.
25
16000
12
28.8
30 1
14
0
35
15900
7. Can you suggest some way to find from the graph in Ex. 3,
p. 9, the average age attained by the 100,000 persons considered?
(If a horizontal line be drawn to the curve from any point on the ver-
tical scale at the left, what will it represent ?)
[8.] An uncovered rectangular tank is to have a square base and
contain 400 cu. ft. Materials for the base cost 30ji per sq. ft., and for
32
MATHEMATICAL ANALYSIS
[I, § 17
the sides 2Q£ per sq. ft. Can you suggest some way to figure out about
what dimensions would give the lowest cost ?
(D) THE EXTREME-VALUE PROBLEM
§ 17. Maximum and Minimum Values : Trial Method.
It is sometimes important to know the largest or smallest
value of a variable quantity, — e.g., the maximum power of
a motor for all different speeds, or the minimum cost of a
reservoir for different shapes.
Such " extreme values " can be found approximately by
experimenting repeatedly and comparing results. A graph is
often helpful in locating the highest
and lowest values.
Ex. I. Find 'the most economical
dimensions for an open, rectangular,
sheet-iron box, which is to have a
square base and contain 24 cu. ft.
(Fig. 18.)
Let us try various dimensions, and calculate each time the
number of square feet of material required.
If x=2: base area= 4 sq. ft.
To give a volume of 24 cu. ft., this requires a height of
6 ft. The total area of the four sides and base will then be
In like manner for x==3, 4, 5, etc., we find the other areas
shown in Table I.
TABLE I
X
A
X
A
2
52
5
44.2
3
41
6
62
4
40
I, §
FUNCTIONS AND GRAPHS
33
The smallest value here is A =40 when z=4 ; but a smooth
graph shows a still smaller value between 3 and 4' viz.
A = 39.6, at x = 3.6, approx. (Fig. 19.)
Thus the base should be about 3.6 ft.
square, which requires a height of 1.8 ft.,
approx. (How could the best values of x and
h be found still more closely ?)
In any such problem it is well to *46\
start experimenting with values near *i
those we think will be the best. But ^
we should go on a little beyond any ^
supposed maximum or minimum.
The table should have but two
columns : The quantity for which we
tried values and the quantity to be
made a maximum or minimum.
§ 18. Caution. A few questions
about maxima and minima can be
answered by elementary geometry, but we should never
jump at conclusions in such matters.
E.g., it would not do to argue that the box in § 17 should be
a cube to have the least area. We were not dealing with a
complete area : there was no top.
40}
38
345
Side x (In.)
FIG. 19.
EXERCISES
(Work these by experimenting and plotting results)
1. A rectangle is to have a perimeter of 30 in. For what dimensions
will its area be greatest?
25
Do these come out as you would expect ?
2. A long sheet of tin 25 in. wide
is to be made into a gutter, by turn-
ing strips up vertically along the two
sides. How many inches should be
turned up at each side to secure the
greatest carrying capacity, i.e., the greatest sectional area?
34 MATHEMATICAL ANALYSIS [I, § 18
3. A sheet of tin 20 in. square is to be made into an open box, by
cutting out equal squares at the four corners and turning up the result-
ing side strips. Find the size of the squares to cut out to give the
resulting box the maximum volume. What maximum? (Hint: What
if 1 in. squares are cut out? 2 in. squares, etc.?)
4. A rectangular sheep pen including 120 sq. yd. is to be built
against a long wall any part of which can be used as one side of the
pen. What lengths should the three new sides have to require the
smallest amount of new construction?
6. One ship (S) was 80 mi. straight north of another (£') at noon.
But S' was sailing east 12 mi. per hour and S was sailing south 16 mi.
per hour. When were they nearest, and how near? (Calculate their
distance apart at 1 P.M., at 2 P.M., etc., remembering that in any right
triangle the square of the hypotenuse equals the sum of the squares of the
two legs. Use tables of square roots if desired.)
6. In Ex. 2 above suppose the tin 30 in. wide and a 10-in. strip
turned up at each side, perhaps not vertically (Fig. 20). For what
depth will the gutter have the greatest capacity ?
(For any chosen value of y, the value to be used
for x must be : x = VlOO-?/2. Why ?)
7. The load (L Ib.) which a rectangular beam
of a certain length can carry is L = 10 xyz, where
x and y are the -width and depth of beam in
inches. Find what x and y give the strongest beam that can be cut
from a circular log 20 in. in diameter.
8. A printed page is to allow 60 sq. in. for printed matter and have
a margin of 1 in. at each side and 1£ in. at top and bottom. (Thus,
if the print lines are 10 in. long, the height of the print column must
be 6 in., making the page 12 in. wide and 9 in. high.) What shape of
page will require the least paper?
9. For a package to go by parcel post the sum of its length and
girth must not exceed 84 in. What are the dimensions and volume of
the largest rectangular package, with square ends, that can go?
10. A rectangular stockade is to contain 600 sq. yd. The fences
running one way will cost $3 per yd., the other two $2 per yd. What
dimensions will make the cost of fencing least, and how small?
11. Like Ex. I, p. 32, but containing 120 cu. ft.
12. A covered rectangular tank is to have a square base :md con-
tain 60 cu. yd. The base will cost $3 per sq. yd., I he sides and top
$2 per sq. yd. Find the most economical dimensions.
I, § 19] FUNCTIONS AND GRAPHS 35
13. Like Ex. 3 above for a sheet 20X30.
14. In Ex. 7 what x and y will give the largest beam that can be
cut from the given log? Do these come out as you would expect?
[15.] Suggest some way to solve the equation 4z3— 7x2+15 = 0.
(Hint : The question really is this : What values for x will make the
quantity 4z3 — 7z2 + 15 equal to zero?)
(E) THE ZERO-VALUE PROBLEM
§ 19. Elementary Equations Reviewed. In practical work
it is often necessary to solve an equation for some unknown
quantity. This can be done in any ordinary case, at least
approximately, by a simple graphical process. Before dis-
cussing this, however, it may be well to recall that certain
kinds of equations can be solved exactly by elementary
algebra, as indicated below. If these methods are not al-
ready familiar, they should be thoroughly mastered now.
Linear Equations
Any equation of the first degree (i.e., involving only the
first power of the unknown quantity) can be solved by a
simple transposition and division.
Ex. I. Solve 2x+5=0.
Evidently 2 x = — 5, whence x = — f .
Check : Substituting — f for x does make 2 x-{-5 equal to zero. Hence
— | is the required "root." *
Quadratic Equations
Any equation of the second degree can be solved by " com-
pleting the square." To understand this process observe
that, in the square of any binomial such as
* A root of an equation is a number which, substituted for the unknown
quantity, will make the two members of the equation equal.
36 MATHEMATICAL ANALYSIS [I, § 20
or, more generally,
the final term is the square of one half the coefficient of x.
Thus if the quantity z2-fl4 x appeared in an equation, we
could convert it into a perfect square by adding 72 or 49.
Ex. II. Solve 3z2-llz+7 = 0.
Transposing and dividing : x2 — ty x = — £.
Adding to both members the square of one half the coefficient of z:
The left member is now a perfect square. Extracting square roots :
= 11 + V37 11-V37
6 ~6^~
Check : Direct substitution shows these to be roots of the equation :
3/llW3JV _„ (Hi^lT) + 7, simplified, gives zero.
\ b / V b ' J
Remark. If we approximate V37 by decimals, the results cease to
be exact. But they will be more convenient for most purposes.
§ 20. Equations as Problems of Variation. Judged by the
foregoing methods, the problem of solving an equation does
not appear to be connected with our basic problem of deter-
mining how one quantity will vary with another. But it is.
For instance, we may think of the equation in Ex. II above
as follows : For every value of x, whether a root or not, the
quantity 3 z2 — llz+7 has a definite value, and it must there-
fore vary with £ in a definite way. The problem of solving
the equation
is simply the problem of finding where this varying quantity
t 3z2— llz+7 reaches the value zero.
The same idea will evidently apply to any other equation.
And this suggests the following general method of solution.
I, § 21]
FUNCTIONS AND GRAPHS
37
§ 21. Graphical Solution of Higher Equations. There is
no very elementary algebraic method of solving equations of
the third degree or above, unless factors can be seen by in-
spection. But the real roots of any equation of any degree
can be approximated graphically, if the equation contains
only one unknown quantity.
For instance, if we wish to solve the equation
2 xs- 15 z+10 = 0,
we simply plot a graph showing how the polynomial 2x3—l5x
+ 10 varies with x, and read off the values of x where the poly-
nomial becomes zero.
The first step is, of course, to calculate a table of values
of the polynomial by substituting various values for x :
x=4, poly. = 2(4)3-15(4) + 10 = 78;
z = 3, poly. = 2(3)3-15(3) + 10 = 19;
and similarly for the other values in the
adjacent table.
Plotting these values gives the curve
in Fig. 21, negative ordinates being drawn
downward as usual. The height of this
graph at any point represents the value
of the polynomial at the corresponding
value of x.
X
POLY.
4
78
3
19
2
- 4
1
- 3
0
10
i
23
-2
24
-3
1
-4
-58
A
\
x1
B
C
\
1 ,
W
8 20
J
* ^
\
\
\
\
/
U
«X
i
\/
V
19
-58
[/
r ->.
-.
Value
^V
tofx
^^
1
/
-JH/
,1 FIG. 21.
POLY.
19 }
23
38 MATHEMATICAL ANALYSIS [I, § 22
Evidently the polynomial becomes zero at the crossing points
A, B, and C, where x= -3.03, .71, and 2.31, approx. These
values are very close to the true roots of the given equation.
These roots could be found roughly from the table without plotting.
E.g., since the polynomial is negative
at x = 2, and positive at x = 3, we set
it equal to zero somewhere between, —
and use "Proportional Parts." This
gives z = 2.17, — a value less accurate 1 { A / —
than the graphical result above. (Why
should there be a discrepancy ?)
Further methods of solving equations will be discussed in Chapter IX.
§ 22. Number of Roots : Imaginaries. An equation of
the first degree has a single root ; one of the second degree
has two ; . and in general (as is proved in higher algebra) an
equation of the nth degree has n roots. These may be either
" real " or " imaginary."
By an imaginary number, you will recall, is meant a
number involving the square root (or any even root) of a
negative number, — e.g., 2+V — 5. Such a number cannot
be approximated by " real " numbers, either positive or
negative. Hence imaginary values cannot be represented
graphically, as long as we use scales consisting of real numbers.
If any of the roots of an equation are imaginary, they will
therefore not show in the graph.
So, if we find only a few roots for an equation of high degree,
it may mean that the others are imaginary. Or it may mean
that we have not carried the graph far enough. Even if the
curve has risen exceedingly high, possibly it will presently
fall and cross the base line again. Fortunately, a sure and
simple test as to this is available. (§ 23.)
The names "real" and "imaginary" are very misleading, as they
suggest that one kind of number actually exists and that the other
does not. The fact is, these are merely different kinds of numbers.
I, § 22] FUNCTIONS AND GRAPHS 39
Imaginary numbers can be given a perfectly concrete interpretation,
which makes them exceedingly useful in Electrical Engineering.
(Chapter XV.)
EXERCISES
1. Solve by completing the square :
(a) 2x2- 9x+ 3=0, (6) 3 z2-4 z-20 = 0,
(c) 12z2-10z+ 3 = 0, (d) 5z2+2z+.2=0,
(c) z2+ 4z+13=0, (f) lla;2_|_6a._ 9 = 0,
(flf) x4- 5x2+ 4 = 0; (A) 2 3^+5 z2- 11=0,
(i) 3z2+ 7z + c = 0, 0') 5z2-|- 6s- 7 = 0.
2. In Ex. 2, p. 33, how many inches should be turned up to give
the rectangle an area of 60 sq. in.?
3. How wide a margin (top, bottom, and sides alike) on a page
9 in. by 6 in. would leave 36 sq. in. for printed matter?
4. For a simple beam loaded and supported in a certain way, the
"bending moment" at any distance (x ft.) from one end is M =20 x— x2.
For what value of x will M = 80 exactly?
5. Solve graphically :
(a) x*-4x+ 2 = 0, (6) z2-z-7 = 0,
(c) z3-3z+ll=0, (d)
6. In Ex. 5 (6) check by completing the square. Also find the
roots directly from the table used in Ex. 5 (6). Why the discrepancies?
7. When a sphere of diameter 3 ft. and specific gravity 8/9 floats
in water, the depth of immersion (x ft.) is a root of the equation
2z3-9z2+24 = 0. Find that root. (Why must the other two roots
be excluded?)
8. In finding the maximum deflection of a 25-ft. beam loaded in a
Certain way it is necessary -to solve the equation
4x3-150x2+1500x-3125 = 0for x. Do this.
9. Determine the approximate location of the "real" roots of :
(a) z3+3z2-3z-18 = 0,
(b) x3 - 41 x2 +440 x- 495 = 0.
Apparently how many in each case?
10. In Ex. 9 (6) test the value of the given' polynomial at £ = 20.
Does this indicate further roots?
40 MATHEMATICAL ANALYSIS [I, § 23
11. Solve for x : x+2^5 = 8. Check your results.
12. What is the erroneous step in the following "proof" that 4 = 5?
Since (4)2-9(4) = (5)2-9(5),
.'. (4)2 -9(4) + y = (5)2-9(5)+^.
Hence, extracting square roots :
4-t=5-|.
/. 4 = 5.
§ 23. Synthetic Substitution. In solving an equation
graphically, the table of values is best calculated by the
following method.
Illustration. To substitute 2 for x in the polynomial
5z3+13z2-16z-20,
multiply the first coefficient (5) by 2, and add to the next
coefficient (13) ; multiply the sum by 2, and add to the next
coefficient (—16), etc.
5 +13 -16 -20 [2
10 +46 +60
+23 +30 +40
The final result, 40, is the value of the polynomial when
z = 2.
This can be verified by substituting directly :
+ 13(2)2-16(2) -20 = 40
The reason this process works is a simple one : Multiplying
Ihe 5 by any value of x and adding the 13 gives 5 z+13;
multiplying this sum by x and adding the -16 gives
5a:2+13x— 16; multiplying this sum by x and adding the
-20 gives 5x8+13x2-16o:-20, which is the value of the
polynomial. In other words, by multiplying by x at each
stage, before introducing the next coefficient, we have
multiplied each coefficient by x the proper number of times
in all.
-
I, § 23] FUNCTIONS AND GRAPHS 41
To illustrate further, let us substitute — 4 for x :
5 13 -16 -20 |-4
-20 28 -48
- 7 12 -68
The result is —68. Check this. Also examine these steps
carefully.
This process is quicker than direct substitution. Also it
shows with certainty when we have gone far enough to get
all the real roots.
E.g., in the present case, there can be no root above 2 nor any below
—4. For substituting —5 or —6, etc., instead of —4, would simply
make each successive product and sum numerically larger, while keeping
their signs alternately + and — , and could not produce a final zero.
Nor could the substitution of +3 or +4, etc., instead of +2. (Why
not?)
In any case, we have gone far enough in the negative
direction when the successive sums alternate in sign, be-
ginning with the leading coefficient ; and far enough in the
positive direction when the sums are all positive.
If any power of x is lacking in a given polynomial, its co-
efficient is zero ; and this zero must be inserted.
Thus, to substitute in 2 z5 — 15 z3+£2+7 x, use the coefficients
2+0-15+1+7+0,
since x* and the constant term are missing. (Observe the x2 term here.)
EXERCISES
1. Find approximately the real roots of
(a) z3-4z2- 3z+ll=0, (6) z3- 7 x+ 5=0,
(c) x*-2x*- 8z2 + 3=0, (d) x3- 6z-13=0,
(e) z4- x3-20x2- 5=0, (/) z5-
(0) z*-8 zM-15 x2- 6 = 0, (h) x9-
2. The length (L ft.) of the longest rectangular panel 1 ft. wide
which can be fitted diagonally across a door 4 ft. wide and 10 ft. long is
a root of the equation L4 - 1 18 L2 + 160 L - 1 15 = 0. Find it.
42 MATHEMATICAL ANALYSIS [I, § 24
3. The smallest safe diameter (d in.) for the bolts in a certain steel
shaft is a root of the equation d4+320 d2- 340 d— 4290 = 0. Find it.
4. Determine the approximate location of all the real roots of the
following equations :
(a) z4-39z3+354z2+640z-2500 = 0,
(6) z3+17z2+40z-290 = 0.
(Suggestion : Try values widely separated at first to get an idea of
how the polynomial runs.)
(F) THE PROBLEM OF EXACT REPRESENTATION
§ 24. Formulas. One way to show how any given quantity
varies with another is to draw a graph. Another way, also
very common, is to write a formula or equation which tells
the value of the varying quantity at any instant.
Illustration. If a bomb is dropped vertically, say from
an airplane, the distance (s ft.) through which it will have
fallen after t sec. is approximately
s=16*2. (1)
For instance, after 2 sec., s = 16(2)2 = 64; after 10 sec., s = 16(10)2 =
1600 ; etc. The bomb will have fallen 64 ft. or 1600 ft., respectively ; etc.
Evidently formula (1) is equivalent to a complete table of
values of s for all values of t} — until the bomb strikes, after
which the formula is no longer valid.
Formulas in general give very full information in a very
brief form ; and they can be carried around much more
easily than a graph, or can even be memorized. Moreover,
they can be used to make exact calculations, — even for rates
in very small intervals where a graph ceases to give reliable
results.
To illustrate in the case of the falling bomb, let us find
from formula (1) the average speed, or rate of motion, during
an interval of .01 sec. beginning at the instant t = 3.
At * = 3, «= 16(3)2 = 144.
At •'•- * = 3.01, s = 16(3.01)2= 144.9616. .
I, § 25] FUNCTIONS AND GRAPHS 43
During the .01 sec., therefore, the bomb falls .9616 ft.
Hence its average speed is
(ft./sec.) =96.16 (ft./sec.).*
Remarks. (I) Properly, the coefficient of t2 in formula (1) should
be 16 plus a small fraction, depending on the latitude of the place.
But even this would ignore the effect of air resistance! So, in illustrative
examples we shall always use the value 16 £2 for simplicity.
We shall also, as in Physics, use the notation s for the "space" or
distance traveled, and v for the "velocity" or speed. Do not mistake
s for speed.
Any algebraic expression involving x, whether it has any
concrete meaning or not, may be regarded as a function of x,
for it will vary with x in some definite way. E.g., the
quantity 2z3— 15z+10 varies with x} as shown in Fig. 21,
p. 37. Any formula, then, such as y = x2, expresses y alge-
braically as a function of x.
§ 25. Increment Notation. In calculating rates, etc., it
is desirable to have a short notation for the change in one
quantity produced by any change in another.
We have already used A (delta) to stand for a difference
or change in a quantity. Hereafter we shall affix the
name of the quantity to prevent any possible ambiguity.
Thus
Ax will denote a difference or change in x ;
Ay will denote the corresponding change in y.
Observe that Ax does not denote some quantity A times the
quantity x, but simply the change in z, as stated.
In this notation the average rate of increase in y per unit
change in x can be expressed simply as Ay /Ax.
* This, by the way, must be nearly the same as the speed at the instant
(=3.
44 MATHEMATICAL ANALYSIS [I, § 26
Similarly, in § 24, since s = 144 when J = 3, and 8 = 144.9616 when
f = 3.01, we may write
As = .9616, Ai = 01.
• As _Qg -IQ _ I average rate of increase of s per
AJ { unit change in t during this .01 sec.
§ 26. Plotting a Formula. From any given formula we
can calculate as extensive a table of values as we like, and
plot a graph, — to be used thereafter as a ready computer
in reading off further values. This is especially desirable
when the formula is very complicated.
For instance, the graph of the cost of oil-tanks, shown in Fig. 7,
p. 5, saves the designer several hours on each calculation.
Any graph which happens to be straight makes a good
computer, if drawn with a ruler on accurate paper.
EXERCISES
1. The assessed value ($T) of a certain house t years after construc-
tion will be F = 2000-30*. Plot this from t = Q to <=40, calculating
V every five years. What sort of graph? How do the original value
and the rate of change appear in the formula ? In the graph ?
2. In the formula y = a+bx what is the value of y when x = 1, 2, 3, 4 ;
k, k+1? How much does y increase every time that x increases by 1
unit? Hence what sort of graph must every formula of this type (first
degree) have? How many points are needed to plot it?
3. The amount which would accumulate on an original sum of $100
after t years with simple interest at 6% is ^ = 100+6*. (Why?)
Plot this from t = Q to t = 70. Answer the same questions as in Ex. 1.
Also read off A when < = 55.
4. The distance (s ft.) that an object will fall (from rest) in t sec. is
s = 16J2. 04) Calculate s at J = 3; also the average speed during .02
sec. beginning then. (B) Plot s as a function of t from t = 0 to t = 5',
and check the calculated speed.
6. The height of a ball t sec. after being thrown straight upward
was h = H2t — 16 Z2 feet. Plot. When was the ball highest and how
high? When was it 80 ft. high? Check the latter answer by putting
h =80 and solving the equation for t.
I, § 271 FUNCTIONS AND GRAPHS 45
6. In Ex. 5 calculate the average rate at which the ball rose from
£ = 2 to £ = 2.01. Check by the graph.
7. At any horizontal distance (x ft.) from the middle of a certain
suspension cable the height (y ft.) above the lowest point is given by
y = .002 x2. Plot the curve of the cable from x = 0 to x = 80. How fast
does the cable rise on the average, per horizontal foot, between z = 60
and x = 82 ? Check graphically.
8. Every horizontal section of a reservoir is a square, whose side
varies thus with the height (h ft.) above the bottom: s = 3Q+3h.
(A) Plot this from h = Q to h = W. Also plot the area of the section
as a function of h from 0 to 10.
9. In Ex. 8 find how much water must flow in to increase the depth
from 4 ft. to 10 ft.
10. The speed of an object (w ft. /sec.) after falling s ft. freely from
rest is, approximately, 0=8 Vs. Plot from s=0 to s = 36. (Hint:
Use s = 1, 4, •*£, 9, and other perfect squares.) Read off v when s = 10,
II, 12, • • -, to 20.
11. The time (T sec.) of a complete swing for a pendulum of length
I in. is: T = .32V7. By changing the vertical scale make the graph
in Ex. 10 serve for this formula. Read off T if Z = 6, 12, 18, 24, 30.
[12.] Plot a graph showing how the quantity y = QQ/(x— 3) varies
with x, from x= —2 to x = 7. Then find from the equation the value
of y when x = 3.01, and when z = 2.99. Is there any indication of the
upper and lower parts of the graph turning toward each other ?
§ 27. Varieties of Graphs. The character of the graph
of an algebraic function depends upon the type of formula.
E.g., if a formula is linear, — i.e., of the first degree, the
graph is straight. (See Ex. 2, p. 44.) To plot it we need
calculate but two points, well separated, and join these by
a straight line. A third point is desirable as a check.*
Again, if a function is irrational, substituting certain values
for x may give imaginary results, — which cannot be plotted.
Thus A/25 — x2 is real only between x= — 5 and x= +5, and
its graph does not go beyond these values.
Again, the graph of a polynomial, — i.e., the sum of integral
* A function like 20/(2 x — 5) which involves x in the denominator is not
called linear ; but only expressions of the form ax +6.
46
MATHEMATICAL ANALYSIS
U, 9 3
powers of x with given coefficients, — such as 2 xz— 15 z-f-10,
is a smooth curve, as in Fig. 21, p. 37. But the graph of a
function which has x in the denominator may be startlingly
different. (See § 28.)
§ 28. A Necessary Precaution, illustrated. Let us plot
the graph of the function
(2)
from
2-x
z=-4 toz = 8.
Substituting values for x gives the adjacent table.
X
y
X
y
-4
10
8
-60
-3
12
4
-30
-2
15
5
-20
-1
20
6
-15
0
30
7
-12
1
60
8
-10
2
?
(What happens when x = 2 will be discussed shortly.)
From —4 to 1 and from 3 to 8, the graph runs as indicated
by the curve in Fig. 22. Let us follow the left-hand branch
and see where it joins the other part.
Substituting again in the given formula :
60
at
at
y =
x = 1.99, y
2-1.99
600;
= 6000.
Evidently the curve is climbing faster and faster, and is
not approaching the other branch. Follow the latter back :
nrk
at
at
2-2.1
2.01, i/=-6000.
I, § 29]
FUNCTIONS AND GRAPHS
47
FIG. 22.
The farther we follow either branch toward x = 2, the
farther it goes from the other, — enormously ! What is
the explanation ? How about y when
x = 2 exactly?
The formula then reads : y = ^-.
But division by zero is impossible.
(§ 29, below.) Hence -%°- is a mean-
ingless symbol : it does not stand for
zero, nor for any other number. That
is, no value exists for y when x = 2.
This explains the peculiarity of the
graph. The curve nowhere crosses
the vertical line at x = 2 ; for if it did,
the ordinate at the crossing would
give a definite value for y when x = 2.
Hence the graph consists of two entirely separate branches.
There is a tremendous break at x = 2 : the function is
" discontinuous " there.
Remark. In general, whenever a function involves a fraction,
we must see whether any real value of x will reduce the denominator
to zero, and the numerator to some other value. If so, the curve will
break. To make sure^about this, simply set the denominator equal to
zero, solve for x, and then test near-by values.
§ 29. Operations with Zero. We can multiply by zero,
add or subtract zero, subtract any number from zero, or
divide zero by any other number, — in short, perform every
numerical operation with zero except one :
We cannot divide by zero.
For instance, to divide 60 by zero we should have to find
a quotient which, multiplied by zero, would give 60 :
if so- = Q, then60 = O.Q.
But no such quotient exists : any number whatever, multiplied
by zero, gives zero and not 60.
48 MATHEMATICAL ANALYSIS [I, § 30
Erroneous statements are often made in this connection about
"infinity," — whatever that may be. The correct statement is that
we cannot divide by zero.*
Observe further that such an expression as 7+Y- would
also be meaningless. We could no more add a number to \°-
than we could add a number to a color !
§ 30. Reciprocal Numbers. One number is called the
reciprocal of another if their product is +1.
Thus the reciprocal of 3 is £ ; the reciprocal of — f is — | ;
etc. To find the reciprocal of any number, simply divide
1 by that number.
Every number except zero has a reciprocal. But £ does not exist.
§ 31. Writing Formulas for Laws of Variation. To use
mathematics effectively in scientific work, we must know the
meaning of certain common statements concerning the
variation of quantities, and must be able to translate those
statements into equations. The following are particularly
important :
(I) When we say that one quantity is proportional to an-
other, or varies as that other, we mean that the ratio of the
two is constant, — that doubling the one will double the
other, etc.
For instance, if y varies as x*, then
L=k, Qiy = kx\ (3)
where k is some constant.
(II) To say that one quantity varies inversely as another
means that it varies as the reciprocal of that other.
Thus if y varies inversely as x4, then y varies as 1/z4, or
* An explanation of the proper technical use of the term "infinity"
may, of course, be given at this point, or postponed, at the discretion of the
instructor. See Appendix, p. 493.
-~
I, § 31] FUNCTIONS AND GRAPHS ^( ^49
(III) To say that y varies as u and as v means that it varies
as their product : y = k(uv). And so on.
N. B. Observe that the phrase " varies as x " has a very
definite technical meaning, but that y might vary with x in
any way whatever.
Observe also that we can find the constant k in any of the
cases above, if we know the value of y corresponding to any
value of x, — or u and v.
Thus, in equation (3), if y = 3Q when # = 2, we must have
36 = fc(22), /. k = 9.
And the definite formula for y would then be y = Q x2.
. 4).^ -,v ^-.i^cA
>y > EXERCISES
1. Draw the graphs of the following between a; =—30 and +40;
then read off the value of y at —13 and +27, and check.
(a) y = 2x-5, (6) y=-Ax+20, (c) ?/ = f z+10.
2. The voltage of a certain dynamo (Y volts) varies thus with the
speed (X rev./min.) : F = .817 X. Plot from X = 0 to Z = 900. Find
Y if X = 225; also X if 7 = 560.
3. Plot these formulas over the same base line: y — \x^ y = % x-\-7,
j/ = |x— 4. Similarly for this set: y = x+2, y = .3x-\-2, y=—x+2.
What graphical significance have a and 6 in the formula 7/ = a+6z?
4. Find by inspection whether the graphs of the following formulas
break ; and if so, where :
Plot the graph for (d), testing y near any supposed break.
6. When an object x in. away is photographed with a lens of "focal
length" F in., the plate should be at a distance y in. from the lens,
given by the formula y = Fx/(x-F). Taking F as 25, plot y from
x = Q to z = 90. Read off y for z = 12.5, and for a: = 75: (Values of x
less than 25 correspond to imaginary or "virtual" images.)
6. For quartz the "index of refraction" (n) varies thus with the
frequency of light vibration (/ trillion per sec.) :
50 MATHEMATICAL ANALYSIS [I, § 32
Plot n2 as a function of /, from /=10 to -90, substituting 10, 20, etc.,
also 35. Read off w2 when / = 25; 38; 77. (Imaginary values of n
correspond to an "absorption band," — as explained in Physics.)
7. The time of revolution (T yr.) for a planet at a distance of x
units from the sun is T = V^. [For the earth, x = I.] Plot from re = 0
to 35. Read off the value of T for z = 5.2 (Jupiter) ; also x if !T = 30
(Saturn).
In each of the following, obtain the formula expressing the law of varia-
tion, and calculate the further values asked for.
8. The elongation of a " spring-balance " varies as the weight applied.
If E = 4 when W = 50, what is the formula ? Find E when W = 35.
9. The volume of a gas under constant pressure varies as the "abso-
lute temperature." If 7 = 600 when T = 300, what is the formula?
What V for T = 347? What T gives 7 = 570?
10. The volume of a gas at constant temperature varies inversely
as the pressure applied. If F = 300 when p = 15, what is the formula?
Find V when p = 12 ; 20 ; 30.
11. The maximum range (R meters) of a projectile varies as the
square of the velocity (V m/sec.) with which it starts. If R = 18000
when 7 = 500, what formula? What R for 7 = 400? For 7 = 550?
12. The acceleration which gravity imparts to an object (A ft./sec.2)
varies inversely as the square of the distance (R mi.) from the center of
the earth. If A =32 when 12 = 4000, what formula? Find A when
fl = 8000; 200,000.
13. The consumption of coal (C tons/hr.) in a locomotive varies
as the square of the speed maintained. If C = 2 when v = 20, what
formula ? Find C if v = 30. What v if C = 20 ?
14. The speed of a falling object varies as the square root of the
distance fallen (from rest). If 7 = 80 (ft./sec.) when s = 100 (ft.), what
formula ? Find 7 when s = 900.
15. An electric current varies inversely as the resistance of the circuit.
If C = 3 (amperes) when R = 5 (ohms), find C when fl = 4; 15; 90.
§32. Discovering Linear Formulas. In working with
scientific formulas, the question naturally arises as to how
they are obtained in the first place.
Some are derived by reasoning from known principles, but
many others are discovered empirically. That is, experiments
are performed, or observations made, and the results noted
I, § 32]
FUNCTIONS AND GRAPHS
51
in the form of a table of values. The question then is purely
mathematical : To find a formula satisfied by all the values in
a given table.
There is no one process by which this can always be done.
But we shall from time to time see how all the more common
types of laws can be discovered. At present only the simplest
case will be considered.
THEOREM. If a quantity y increases at a constant rate
per unit change in another variable x, the formula for y in
terms of x must be linear, — i.e., of the first degree.
• PROOF. Let the value of y when z = 0 be denoted by a, and the
constant rate of increase by 6. Then, when x has increased from 0 to any
value X, y will have increased by bX (or b units per unit change in x).
Thus y will equal its original value a plus its increase bX :
y=a+bX.
That is, for every value of x, y is given by a linear formula. (Q. E. D.)
To find what values a and b should have in any particular
case, we may proceed as in Ex. I below.
Ex. I. Table I shows the amount of potassium iodide (W
grams) which will dissolve in 100 grams of water, at several
temperatures (T°). Find a formula for the amount which
will dissolve at any temperature.
TABLE I
T
W
T
W
10
136
40
160
20
144
50
168
30
152
60
176
By inspection of the table, W increases at a constant rate.
Hence the values all satisfy a formula of the type
(3)
52
MATHEMATICAL ANALYSIS
[I, § 32
For instance,
= a+106,
= a+206, etc.
To find a and 6, we simply take any two such equations
formed by substituting from the table, and solve simultane-
ously. Subtracting eliminates a and gives b = .S. Sub-
stituting this in either equation gives a= 128.
That is, if formula (3) is to fit the given table, we must have
a— 128 and b = .8. Thus the required formula is
TF = 128+.8T.
Remarks. (I) When a table runs at irregular intervals, we may not
be able to tell by inspection whether the rate is constant. But we can
tell by plotting a graph and seeing whether this is straight.
(II) If the rate is not constant, the formula is not linear, and we
cannot find it as yet. However, if the plotted values give an almost
straight line and apparently not a smooth curve, we assume that there
were slight experimental errors in making out the table, draw what
seems to be the most probable straight line among the points, and find
a formula to fit this line. A more reliable process will be explained
later. (§§ 342-343.)
EXERCISES
In each of the following exercises show that the formula is linear:
Find it, and check for some value in the table.
TABLE 1
TABLE 2
TABLE 3
TABLE 4
T
w
c
F
T
V
0
54
10
50
-33
160
20
64
60
140
- 6
178
40
74
100
212
12
190
60
84
160
320
27
200
80
94
200
392
42
210
13
18
23
11
18
25
32
I, § 33]
TABLE 5
FUNCTIONS AND GRAPHS
TABLE 6 TABLE 7
53
TABLE 8
70
85
100
115
130
557
546
536
526
515
h
w
M
61
118
124
63
126
132
65
134
140
67
142
148
69
150
157
71
158
166
No.-
COST
1000
152.50
2000
165.
3000
177.50
4000
190.
5000
202.50
W
POSTAGE
(HI)
(VI)
5
.14
.41
10
.24
.81
15
.34
1.21
20
.44
1.61
1. Table 1 gives the weight of KBr salt (W grams) which will dissolve
in 100 grams of water at various temperatures (T°). After finding and
checking the formula, find W for T = 13.
2. Table 2 shows several temperatures Centigrade with their Fahren-
heit equivalents. Find F when C = 190 ; also C when F = 100.
3. Table 3 gives the volume (V cc.) of a certain quantity of gas
at several temperatures (T°). What formula? Find V at IF = 50.
4. The rate at which the indicator of a certain gas-meter revolves
(AT rev. per sec.) for various velocities of the gas in the pipe (V ft./sec.)
is shown in Table 4. What formula? For what V is N = 0?
5. The latent heat of steam (L calories) is shown in Table 5 for
various temperatures (T°). Find an approximate formula by drawing
what seems to be the most probable straight line graph.
6. Table 6 gives the average weights (W Ib. and M Ib.) of women
and men of various heights (h in.). Express approximately the relation
between weight and height in each case.
7. The cost of publishing a certain pamphlet will vary with the
number (AT) to be printed, as shown in Table 7. Find C if AT = 3200
What is the meaning of each constant in the formula?
8. The postage required for packages of various weights (W Ib.) is
shown in Table 8 for "Zones" 3 and 6. What are the formulas?
§ 33. Summary of Chapter I. It is often necessary to study
the way in which some one quantity varies with another.
What value will the one quantity have for any specified
value of the other? What maximum and minimum values?
What mean value? Is it ever zero? What is the average
rate of increase in any interval? The rate at any instant?
54 MATHEMATICAL ANALYSIS [I, § 33
If a table of values has been obtained, — say experi-
mentally, — a graph can be plotted and used to answer such
questions approximately. Precise answers should not be
expected when our given information about the variable
quantities is merely a table of values.
If, however, we know a formula which expresses one of
the variable quantities as a function of the other, very ac-
curate calculations are possible. We can get any desired
value of the function exactly, also any average rate of in-
crease. By dealing with very small intervals, we can find
maximum and minimum values, mean values, and instan-
taneous rates as closely as may be desired, — though not
exactly, as yet.
Any algebraic function or formula .can be plotted, and
the graph thereafter used as a ready computer. If a de-
nominator becomes zero there will be a break in the graph.
The real roots of an equation of any degree can be ap-
proximated graphically, the synthetic method of substitu-
tion affording a sure test as to the sufficiency of the table.
Interpolation by proportional parts is too inaccurate here,
but is valuable when the intervals in a table are very small.
Any large error in a table of values can usually be dis-
covered from the resulting irregularity in the graph.
Remark. The exact calculation of an instantaneous rate may at
present appear almost hopeless, since we cannot select any interval,
not even a small one, for which the average rate will surely equal the
required instantaneous rate.
Nevertheless the calculation can be made easily as soon as we see
clearly just what an instantaneous rate is. This and similar questions
will be considered in the next chapter.
EXERCISES
1. Which quantity would you plot vertically and which hori-
zontally if you wished to show the relation hctwcon : The death rate of
bacteria and the amount of sunshine? The temperature in a mine and
I, § 33]
FUNCTIONS AND GRAPHS
55
the depth below the surface of the ground? The impulsive force of a
jet of water and the speed of flow?
2. Table 1 gives the pressure of steam at various high temperatures.
Plot the graph.
3. The area of a wound, A sq. cm., decreased with the time, t days,
as in Table 2. Plot the curve of healing. Find the rate at t = 8.
4. Table 3 shows the cash surrender value ($F) of a certain life in-
surance policy after T years. Find graphically and by proportional
parts what the value should be when T = 12. (Why. is there a dis-
crepancy?) What is the rate of increase per year at T = 20?
5. Based on the level of 1895 as 100%, the average price of com-
modities in general has increased as in Table 4. The cost of electric
power, however, has been reduced, as shown in the table. Exhibit
these facts graphically, — using a common base line.
TABLE 1
TABLE 2
TABLE 3
TABLE 4
240
259.2
274.3
286.9
297.8
307.4
316.0
10
20
30
40
50
60
70
0
4
8
12
16
20
24
16.2
10.7
6.5
3.8
2.1
1.0
.4
T
V
0
0
5
121
10
291
15
486
20
727
25
930
30
1000
^EAR
PRICE
ELEC.
1890
113
110
1894
96
101
1898
95
90
1902
113
75
1906
122
59
1910
126
27
1914
153
16
1918
321
16
6. The angle (A mils) at which to elevate a certain machine gun
for various ranges (R yd.) is shown in Table 5. Plot. What is the
range if A =80? Check by proportional parts. At what average rate
must A increase with R, from 72 = 1400 to # = 1600?
TABLE 5
R
A
R
A
200
3
2000
95
500
10
2500
157
1000
29
2800
207
1500
56
56
MATHEMATICAL ANALYSIS
[I, §33
7. Table 6 gives the depth of a river (D ft.) at various distances
x feet from one bank, going straight across. Find the approximate
area of the cross-section of the river.
8. The rate at which a flywheel was turning (R deg./min.) while get-
ting up speed was given by the formula: R = QQP(9— t), where t is
the number of minutes elapsed since starting. Plot and find the total
angle turned from t = Q to < = 6. When was the maximum rate of rota-
tion attained, and what maximum?
9. In Ex. 8, exactly how much did R increase from t = l to t = 3?
Hence, what average angular acceleration (or rate of increase of R)?
Check the latter result by the graph.
10. During the motion of a pendulum up and back, the speed
(v in. /sec.) varied with the time (t sec.), as in Table 7. (Negative values
indicate a reversed direction of motion.) (a) Find v at t = .3 ; also how
fast v was then changing. (6) How far did the pendulum travel in
reaching its highest point?
11. Solve graphically xz-\-x—7 = 0. Check by completing the
square. Also approximate one root by proportional parts. Why is
this inaccurate ?
TABLE 6
TABLE 7
X
D
0
0
200
5
400
21
600
30
800
26
1000
37
1200
25
1400
0
t
V
0
8.
.2
6.47
.4
2.47
.6
-2.47
.8
-6.47
1.0
-8.
12. Find approximately all the real roots of
(a) x3-12x+88=0,
(6) z*-z3-12z2+5 = 0.
13. In testing the insulation of certain telephone lines it is necessary
to find the resistance from the formula
15 000 000
It
-100000,
CHAPTER II
SOME BASIC IDEAS ANALYZED
THE EVER-RECURRING LIMIT-CONCEPT
§ 34. Instantaneous Speed. When we speak of the speed
of a moving object at a certain instant, precisely what do we
have in mind?
Not the average speed for the next hour, nor even for the
next minute or second. Nevertheless the average speed for
a very short interval would closely approximate the " in-
stantaneous speed " of which we are thinking. And by
making the interval shorter and shorter, we could bring the
average speed closer and closer to the instantaneous speed.
In other words, the speed at any instant is simply the
limiting value which the average speed would approach,
as closely as we please, if the interval were indefinitely
shortened, while always including the instant.
This statement will be taken as our definition of instantaneous
speed.
To calculate an instantaneous speed exactly, we must some-
how find the limiting value in question. To do this w.e shall
first find the average speed in an interval of arbitrary length,
— not a fixed interval such as .01 sec., but an elastic interval
of any length. Then we shall squeeze this interval down as
small as we please, and see what happens to the average rate.
Ex. I. Find the speed at which a ball was rising 3 sec. after it was
thrown straight upward, if the height (h ft.) after t sec. was
WP. (1)
58
I, § 33]
FUNCTIONS AND GRAPHS
57
for values of the voltage (7) running from 1 to 150. Plot the graph
for this interval. Would the complete graph break ?
14. A wire was stretched, its length (L in.) varying with the pull
(Plb.), as in Table 8. Find the formula for L in terms of P. Check.
Tell how you would find the work done in stretching the wire from
L=42.5 toL=43.
16. The quantity (Q gal.) of a certain mineral water which can be
sold at various prices (P£ per gal.) is shown in Table 9. The total
expense ($#) of marketing those quantities is also shown. What price
gives the greatest net profit?
16. Two burners A and B, 10 ft. apart, are of different power.
A gives 60 calories a second at a distance of 1 ft. ; B, 480 calories. The
intensity varies inversely as the square of the distance. What point
on the line A B receives the least heat from A and B combined ; and how
much heat?
17. Find the radius and length of the largest cylindrical package
which can go by parcel post. See Ex. 9, p. 34.
18. The deflection of a beam varies as the cube of the length. If
D = .002 when L = 10, find the formula giving D for any L. Find D
when L = 17.
TABLE 8
TABLE 9
P
L
100
42.5
200
42.7
300
42.9
400
43.1
500
43.3
p
Q
E
20
16200
2700
30
12800
2400
40
9800
2100
50
7200
1800
60
5000
1500
70
3200
1200
19. If the plates for printing a chart cost $40.00 and the cost of
printing is .6 cents per copy, what will be the total cost of x copies?
Plot from z = 1000 to x = 10,000. Read off the cost of 8750 copies.
How many copies for $75?
[20.1 When you speak of the speed of a projectile at some instant, as
distinguished from its average speed during some interval of time, what
do you have in mind?
II, § 35] SOME BASIC IDEAS ANALYZED 59
Consider any interval beginning at t = 3 ; and let At denote the length
of time in the interval, — which therefore ends at t = 3-\-At.
By (1) the heights at the beginning and end of the interval were
at t = 3, h = 100(3) -16(3)2 = 156;
att = 3+At, ft = 100(3 +AO- 16(3 +A£)2,
= 156 +4 At - 16 At2 (simplified) .
The difference of these heights, AA = 4 At — 16 At2, is the distance the
ball rose during the interval of At sec. (Fig. 23.) -Dividing by At :
— =4 -16 At = av. speed during A*. t=3+A.t O h=156+
M \4At-16At*
For instance, if At = .0l, we have 4 -16 (.01),
or 3.84, ft. /sec. as the average speed from t = 3
to t = 3.01. t=3 Qh=156
Now let At approach zero. The limiting
value approached by the average speed 4 — 16 At
is precisely 4. That is, the instantaneous speed
at £ = 3 is precisely 4 (ft. per sec.).
Remarks. (I) We do not say that the aver-
age speed 4 — 16 At will ever be exactly equal
to 4. Neither will it ever be equal to the in-
stantaneous speed. But the limiting value
which the average speed is approaching is
exactly 4 ; and this limiting value is precisely FIG. 23.
the instantaneous speed.
(II) If our resulting speed had come out negative, it would indicate
that the ball was falling, — i.e., that the height was decreasing. (For
it would show that the value of h at t = 3 -{-At was smaller than the value
at t = 3, which we subtracted, — at least, that this would be so when
At became small.)
§ 3€. The Limit Idea Is Essential to a satisfactory defini-
tion of an instantaneous speed.
The distinction between an interval of time and an instant
is like that between a line-segment and a point : an interval
has some length or extent, but an instant has none. No
distance whatever can be traveled " during an instant," for
an instant has no duration. Hence it would be meaningless
to define an instantaneous speed as " tfie distance traveled
60 MATHEMATICAL ANALYSIS [II, § 36
during the instant divided by the length of time in the in-
stant " (!).
Neither can we employ any such idea as " the speed during
the shortest possible interval of time." There is no such
thing : any interval, however short, has some definite extent,
and can be subdivided into billions of still shorter intervals.*
Again, it is useless to give any such vague definition of
instantaneous speed as " the rate of motion at the instant."
What is meant by the " rate at an instant " if the object
doesn't move at this rate for even a short interval? This is
precisely the thing to be defined.
Our definition of an instantaneous speed as the limiting
value of an average speed is, however, definite, and free from
logical objections. Many other familiar concepts can be
defined satisfactorily only by using a similar idea.
§ 36. Instantaneous Rates in General. If we say that
a balloon " is now expanding at the rate of 50 cu. ft. per
min.," precisely what do we mean? Simply this :
The average rate of expansion for any short interval be-
ginning now will be very approximately 50 cu. ft. per min. ;
and the limiting value which this average rate would approach,
if the interval were indefinitely shortened, is exactly 50 cu.
ft. per min.\
* Large and small are purely relative terms. Select any number which
you consider "small," and divide it by a billion. Repeat this division a
hundred times ; you then will have a number beside which the original
"small" number is enormously great. But even your new number is very
great compared with some others.
f Persons unfamiliar with this limit idea are generally unable to explain
the precise meaning of such a statement as the one above concerning the
balloon. For instance, they will often say: "This means that, if the
balloon kept on just as it is now growing, it would expand by 50 cu. ft. in
the next minute." (And so it would.) But precisely what is meant by
"just as it is now growing" ? What is meant by the way, or rate at which,
the balloon is expanding at the instant? This is precisely the thing to be
explained ! Such an explanation merely leads around a circle, and explains
nothing at all.
II, § 36] SOME BASIC IDEAS ANALYZED 61
Similarly in general, when we speak of the rate at which
any quantity is increasing " at a certain instant," we mean :
the limiting value approached by the average rate in an ad-
joining interval, as the interval is indefinitely shortened.
To calculate an instantaneous rate, then, we simply get
the average rate for an arbitrary adjoining interval, and see
what happens to this as the interval is indefinitely shortened.
Ex. I. The volume (V cu. in.) of a certain weight of a gas varies
with the pressure (p Ib. per sq. in.) thus :
F=?00.
P
Find the rate at which V changes per unit change in p, at the instant
when p reaches the value 17.
Solution :
Atp-iT, v.m.
The change in V due to the increase Ap is
AF= 200 200
17+Ap 17'
or, reducing to a common denominator and simplifying :
\y= -200AP
17(17+Ap)'
The average rate of increase per unit is, therefore,
Ay= -200
Ap 17(17+Ap)'
The instantaneous rate is the limit approached by this as Ap approaches
zero
Inst. rate= -^ = —.69, approx.
That is, the volume is at the instant in question decreasing at the rate of
.69 cu. in. per unit increase in p. (What shows that the volume is
decreasing? Should that be expected?)
62 MATHEMATICAL ANALYSIS [II, § 36
EXERCISES
1. Explain briefly the precise meaning of the following statements.
(Each necessarily involves the idea of a limiting value, approached as
an interval is indefinitely shortened.)
(a) The ice is melting more and more slowly ; just now it is melting
at the rate of 1.75 cc. per sec.
(6) The diameter of the balloon increased as the temperature rose :
the rate was 2 cu. ft. per degree when the temperature had just reached
60°.
(c) The water in a basin was flowing out at the rate of 20 cu. in.
per min. at the instant when it was 5 in. deep.
(d) The temperature of an iron bar was falling at the rate of 8° per
min. at the instant when it reached 150°.
(e) A wound was healing at the rate of .04 sq. cm. per hr. just four
days after the first treatment. ,
(/) The force exerted by steam against a piston was decreasing at
the rate of 2000 Ib. per ft. moved, when the piston had gone just 2 ft.
(0) Some salt was thrown into a pan of water ; 30 min. later it was
dissolving at the rate of .3 oz. per sec.
[2.] Precisely what is meant by the "area" or "number of square
feet" within a given curve? Such a space cannot be cut up exactly into
square feet.
3. The distance (s ft.) that a ball had rolled down an incline after
t sec. was s = 7 P. How far had it rolled when t = W? When
< = 104-A«? What average speed from t = lQ to £ = 10+A£? What if
AJ = .0001 ? Exactly what speed at the instant when t = 10?
4. A ball rolled up an incline, its distance from the starting point
after t sec. being s — SQt — 10 t2. What was its average speed from
* = 2toJ = 2+AJ? Its exact speed at t = 2?
6. A bomb was dropped from an airplane. Its height aboveground
t sec. later was h = 5000 - 16 t2. Find the rate of fall at t = 10.
6. A square metal plate is heated and expands. Find the rate at
which the area is increasing, per unit change in the edge x in., at the
instant when z = 5.
7. Like Ex. 6 for a circular plate. Find the rate at which A is
increasing, per unit change in the radius, when r = 6.
8. A volume of gas varied thus with the pressure: F = 500/p.
Find how fast V was changing per unit change in p at the instant when
p-21.
II, § 37] SOME BASIC IDEAS ANALYZED 63
9. The strength S of a beam is different for different thicknesses
(x in.). If £ = 9 x2, how fast does S change per unit increase in x at
the instant when x = 10?
10. As a rolling weight W moves along a beam, the "bending mo-
ment" B varies, being B = 6000+700 re -50 re2 when W is x ft. from one
end. Find how fast B is changing with x at the instant when x = 12.
§ 37. Length. Just what is meant by the " length " of
a curved line, or the " distance " along a curVe?
Consider how you would proceed to measure it approxi-
mately. Obviously you would measure only a small arc
at a time, — which would be nearly straight and hence
practically coincide with a part of your ruler. In reality,
however, you would be measuring
not the little arc itself but its chord.
The combined lengths of all the
chords (Fig. 24) would approximate
closely the thing which we call " the
length of the curve." The latter
we define, then, simply as the limit-
ing value approached by the total length of the chords
as their number is indefinitely increased and the length of
each approaches zero.
E.g., we define the circumference of a circle as the limiting value
approached by the perimeter of a regular inscribed polygon, when the
number of sides is indefinitely increased.
The limit idea is essential to a satisfactory definition. Thus
it would not do to say merely : " the length of a line is the
number of inches it contains." A curved line cannot be
cut up into parts each of which would coincide with an inch
rule or some fraction thereof. The phrase, " the number
of inches which it contains," is meaningless by itself. To
speak of " the number of inches to*which it is equivalent," would
be no better. For what is meant by their being " equiva-
64 MATHEMATICAL ANALYSIS [II, § 38
lent " other than that they " contain " the same number of
inches ?
Nor is it any definition to say that the "length of a curve is the
length which it would have if straightened out." A geometrical curve
must be taken as it is. So must the curved edge of a table, for instance.
Again, if we defined the length of a curve as the length of a rule
which would roll along the curve from one end to the other without
slipping, we should be presupposing an accurate definition of "rolling
without slipping," — which would itself be found to involve the limit.
idea.
In fact, the idea of a limiting value is needed to define fully
even the length of a straight line segment. E.g., the side
and diagonal of a unit square are incommensurable. The
diagonal " contains " the unit an irrational number of times
(viz. V2 times) ; but this needs explanation, and sooner or
later involves the idea of a limiting value.
§ 38. Area and Volume. By the " area " of a plane
figure bounded by a curved line we mean the limiting value
approached by the area of an inscribed polygon, as the
number of sides is indefinitely increased, each side being
indefinitely shortened.
The area of any plane figure can be approximated by
dividing it into narrow strips and replacing each strip by a
rectangle. (See Fig. 15, p. 29.) This same idea could be
used in defining the area, — say as the limiting value ap-
proached by the sum of the rectangular areas, as each strip
becomes indefinitely narrow.
The definitions of the area of a curved surface and the
volume of a solid are somewhat similar, but more complicated.
EXERCISES
1. Two runners, one with a long and the other with a short stride,
run a quarter mile on a curved track. If their footsteps follow the
same curved line, which steps the greater distance?
H, § 39] SOME BASIC IDEAS ANALYZED 65
2. Draw a circle of radius 3 in., and measure its circumference
approximately by rolling a ruler carefully along the circle. Also approx-
imate its area by using rectangles. Check.
3. (a) On some map measure the approximate length of the shore
of Lake Superior. (6) Likewise the length of the Mississippi River
from Minneapolis to the Gulf.
4. Criticize the following "Explanation." To say that the volume
of a sphere is 200 cu. ft. means that the sphere "contains" 200 cu.
ft., or that the "amount of space" is the same as in 200 foot
cubes.
[5.] What is meant by a tangent to a circle? Can you draw any sort
of curve to which that definition of a tangent line would not apply?
Can you suggest any definition which would always apply?
§ 39. Instantaneous Direction. A moving object usually
travels along a curved path, and thus does not move in
any one direction during even a small fraction of a second.
What then is meant when we
say that it "is now moving
in a certain direction "?
The short arc PQ (Fig. 25)
passed over in a short time
is nearly straight, and hence
nearly coincides with its
chord. The direction of the
FIG. 25.
chord or secant approximates
closely what we regard as the instantaneous direction of
motion, — the shorter the time, the better.
The instantaneous direction of motion is simply the
limiting direction of the secant, as the time-interval is in-
definitely shortened.
It is often said that the direction of motion at any instant,
or the direction of a curve at any point, is the direction of
the tangent line. But why? Simply because the tangent
line PT is the limiting position approached by the secant line
PQ, as Q approaches P along the curve.
66
MATHEMATICAL ANALYSIS
[II, § 40
§ 40. Definition of a Tangent Line. The last statement
above will be taken as our definition of a tangent to any
curve. Memorize it.
The limit idea is essential. Various definitions sometimes
given for a tangent to a circle are worthless for more com-
plicated curves.
(1) For example, the idea that a tangent is "a line perpendicular to
a radius" cannot be applied to either curve in Fig. 26, — unless we can
say what constitutes a " radius ' ' of such a curve. (Not even a short arc
of one of these could be a part of a circle, because of the continual change
of curvature.)
(2) The idea that a tangent is "a line meeting the curve at only one
point" is unsound. For evidently AA' in Fig. 26 is not what we mean
by a tangent, while BB' is clearly
tangent at 'B although meeting the
curve at several points.
(3) The idea that a tangent is "a
line touching the curve at a point
without crossing it there," is also un-
satisfactory. For instance CC' in
Fig. 26 meets the curve at C without
crossing it but is not what we mean
by a tangent, — its direction differing
from that of the curve at C, — while,
on the other hand, EE' has the di-
rection of the curve at E, and should
be regarded as tangent though it crosses. (Would you hesitate to call
EE' tangent if considering only the part of the curve to the right or
left of E alone?)
The important question is not whether a line crosses a
curve, but whether it has the same direction at the common
point. Our definition of a tangent as the limiting position
of a secant insures that the direction will be the same.
If different limiting positions aTe approached from the
right and left, as at D, there are two tangents. If because
of any peculiarity of the curve the secant fails to approach
a limiting position, there is no tangent.
II, § 41] SOME BASIC IDEAS ANALYZED 67
§ 41. Slope or Grade. To describe the rate at which a
line or curve rises, per horizontal unit, we speak of its slope
or grade.
In the case of a straight line, the slope is simply the number
of units the line rises in each and every horizontal unit. If
it rises 47 ft. in 100 ft. horizontally its slope is .47. Or its
grade is 47%.
The slope of a horizontal line is zero. There is no such
thing as the slope of a vertical line.
Of course a line may be very nearly vertical and yet have a slope.
The slope increases without limit if the line approaches a vertical posi-
tion indefinitely.
.In the case of a curve we speak o'f the average slope in any
interval, and also of the slope at any point.
By the average slope is meant the average rise per hori-
zontal unit during the interval. (This would equal Ay/ Ax in
Fig. 27.) By the slope at any
point P is meant the limiting
value approached by the average p Tis the limiting
slope Ay /Ax as Ax is indefinitely position of PQ
decreased.
Observe in Fig. 27 that Ay/ Ax
is also the slope of the secant
line PQ. Its limiting value is,
therefore, the slope of the tan-
gent line P T. Thus the average 3
i r i FIG. 27.
slope of a curve in any interval
is the same" as the slope of the secant; and the slope at
any point is the same as the slope of the tangent PT*
Ex. I. The height (y ft.) of a certain suspension cable above its
lowest point, at any horizontal distance (x ft.) from the center, is
Find the slope at the point where x = 100.
* See footnote, p. 15.
68 MATHEMATICAL ANALYSIS [II, § 42
Solution: We first calculate the average slope in any interval,
from x = 100 to x = 100 + Arc :
Atx = 100, T/ = . 002 (100)2 = 20;
At x = 100+Ax, y = .002 (100+ Ax)2 = 20 +.4 Ax + .002Ax2.
The difference of the two heights, At/ = .4 Ax +.002 Ax2, is the distance
the cable rises in the horizontal distance Ax.
.-. Al = .4+.002Ax = av. slope.
Ax
Next we let the interval Ax become smaller and smaller, and see what
limiting value is approached by this average slope. Evidently it is .4.
That is, the slope of the curve or of the tangent line at x — 100 is .4.
Or the "grade" is 40%.
Remark. When a calculated slope comes out negative, it means
simply that the curve is falling toward the right, inasmuch as the new
value of y after x has increased is smaller than the original value which
was subtracted from it. (Cf. Remark II, § 34.)
§ 42. Drawing a Tangent. Calculating the slope of a
tangent line furnishes a means of drawing the line exactly.
E.g., if the slope is .4, simply draw a line through the given
point of tangency, rising .4 units in each horizontal unit,
toward the right. If the slope is — f, draw a line which
falls toward the right £ of a unit in each horizontal unit, -
or, say, 4 units in any 3.
Whatever scales are used in plotting a curve are, of course,
to be employed also in drawing any tangent line.
EXERCISES
1. What is the slope of a line which rises 9 in. in each horizontal
foot? 12 ft. in 30 ft. horizontally?
2. What is the grade of a sidewalk which rises 8 ft. in 50 ft. hori-
zontally? Of a railroad which rises 132 ft. in 1 mi.?
3. Draw straight linos with those slopos : 2/3, —3/2, —3/5; also
lines with these grades: 30%, 100%, 80% down grade.
4. The height of a certain hill (y ft.) varies with the horizontal dis-
tance (x ft.) from the foot, as in the following table. Plot the curve,
II, § 43] SOME BASIC IDEAS ANALYZED
69
assuming it smooth. Measure the slope at z = 200 by drawing an
apparent tangent line.
X
y
X
y
0
0
400
80
100
8
500
100
200
28
600
108
300
54
6. Plot the graph of y=x* from x=— 3 to x = 3, using the same
scales horizontally and vertically. Draw an apparent tangent at
x = 1 and measure its slope. Also calculate exactly the rise of this curve
from x = \ to s = l-|-Aaj. What average slope? What slope at x = l?
How could an exact tangent be drawn to this curve?
6. In Ex. 5 calculate also the slope at z = 2. Draw a line having
that slope.
7. Calculate the slope of the suspension cable in Ex. 7, p. 45,
at the point where a: = 50. Show by drawing a line just how steep the
cable is at that point.
8. The height of a curve x in. from a certain point horizontally is
y = .8 x — .05 x*. Find the slope at x = 5.
9. Plot the graph of the function y = l/x from x = — 3 to x = 6, using
the same scale horizontally as vertically. Calculate the exact slope
of the tangent line at x = 3 ; and draw the line having that slope, — thus
checking your curve. (N.B. In getting the average slope from 3 to
3-f-Az it is necessary to subtract one fraction from another. How is
this always done?)
§ 43. A Gap in the Definitions. We have defined in-
stantaneous rates, slopes, lengths of curves, etc., as certain
limiting values.
But it might conceivably happen in some cases that no
limiting value would be approached. For example, a curve
might be so full of sudden turns that a secant line through a
given point would approach no definite limiting position while
the interval was indefinitely shortened. There would then be
no tangent line, nor any slope for the curve, at that point.
70 MATHEMATICAL ANALYSIS [II, § 44
Our definitions will, therefore, be complete only when we
can prove that the limiting values in question are actually
being approached. To deal systematically with such ques-
tions, we shall now define explicitly what is meant by a
" limiting value " or limit.
§ 44. Limit Defined. A variable v is said to approach a
constant c as its limit (written v-*-c), if the difference between
the variable and the constant will ultimately become and remain
numerically less than any specified positive number, no matter
how small.
To illustrate concretely, suppose that a weight suspended
by an elastic cord is pulled down and released : the cord will
contract, elongate, contract, etc. But the oscillations will
become smaller ; and the difference* between the varying
length of the cord and its original length will in time be-
come and remain less than any small distance which we
may specify. The variable length v, therefore, approaches
the original length L as a limit : v-*~L.
Again, more abstractly, suppose we have to deal with a
quantity s = 15+80 Ax, as Ax is indefinitely decreased.
Evidently s-»-15. For the difference between s and 15 is
80 Ax, which in time will become and remain as small as we
please.
Think of this as involving a contest. When I assert that s->-15, I
am saying that you may name any positive number you please, no
matter how small, and that I can then prove that the difference between
s and 15 will ultimately become and remain less than your specified
number. E.g., if you name .000 000 001, I must be ready to show that
the difference between s and 15 will after some stage of the variation
forever remain numerically less than this number. Otherwise I have
no right to assert that s-»-15.
N.B. Whether a variable reaches its limit or not has
nothing whatever to do with the question of its approaching
a limit. The sole essential is that the difference between v
II, § 45] SOME BASIC IDEAS ANALYZED 71
and c shall after some stage of the variation remain numeri-
cally less than any specified positive number, however small.
§ 45. A Doubtful Case. The limiting value approached
by any such ordinary expression as 15+80 Ax while Az-H)
seems obvious enough. But how about a quantity with very
large coefficients, such as
g = 27-2 000 000 Ax+2 000 000 000 Ax2?
Does this approach some limiting value ? Is the limit 27 ?
Let us try smaller and smaller values of Ax and see what
happens :
Az = .1, g = 27-200000+20000000 =
Az = .01, g = 27- 20000+ 200000= 180027;
Az = .001, 9 = 27- 2000+ 2000= 27;
Az = .0001, 9 = 27- 200+ 20= -153.
These results suggest that q does not approach 27 finally.
But let us go on.
For smaller values of Ax, q has the values in the following
table and approaches 27 closely. In fact
Ax = .000 000 000 001 would make
g = 27-.000 002+.000 000 000 000 002 = 26.999 998 +.
i .000 01
.000 001
.000 000 1
.000 000 01
7.2
25.002
26.800 02
26.980 000 2
Clearly by taking Ax small enough we could make both
2 000 000 Ax and 2 000 000 000 Ax2 (3)
as small as we please ; and thus make q differ from 27 by less
than any assigned positive number. That is, q+27.
72 MATHEMATICAL ANALYSIS [II, § 46
If you desire to prove this rigorously, observe that as soon as Ax < .001
numerically,* the second term in (3) is the smaller. Hence the sum or
difference of the two terms is surely less than twice the first, or
4 000 000 Ax. Hence if you assign any small positive number you please
(call it e), as soon as Ax is less than one four-millionth of e, the two
terms in (3), or the difference between q and 27, will be less than your
number e. (Q. E. D.)
§ 46. General Conclusions. From the foregoing example
it seems clear that any term containing Ax, Ax2, or Ax3, etc.,
as a factor, will approach zero as Ax-»-0, no matter how large
a coefficient it may have; also that the sum of any fixed
number of such terms must approach zero. That is :
(I) /bAxn-^0,
(II) =*= aAx± 6Ax2 . . . ± /vAx^O
as Ax->-0.
If a detailed proof of these two conclusions is desired, see
p. 486 of the Appendix. We shall assume them correct in
what follows.
§ 47. Further Limit Notation. To denote the limit which
a quantity approaches "as Ax-»-0, we shall write the symbol L
Aa->0
before the quantity. Thus L (17 Ax2) stands for the limit
Ace-X)
of 17 Ax2 as Ax->-0.
In this notation, statements (I) and (II) above may be
re-written thus :
(I) L (fcAx»)=0,
AarX)
(II) L (=taAx±6Ax2...=tfcAxn)=0.
A35-X)
Still another notation is to write lim in place of the L above.
(I) lim (fcAx)=0.
A05-X)
Also, the sign = is often used in place of ->-.
* < means "is less than."
II, § 48] SOME BASIC IDEAS ANALYZED 73
These notations should be studied until thoroughly fa-
miliar.
§48. Functional Notation. "Function" is often ab-
breviated/; and the variable on which it depends is written
after the /, in parentheses. Thus f(x) does not mean some
quantity / times x, but some function of x, — read briefly
"fof x." For instance, f(x) might stand for 4 z3+7, or for
10*, etc.
The value of f(x) when a: = 3 is denoted by /(3). Thus if
then /(3)=2(3)3-15(3)=9.
The statement that "when x = 3 the function = 9 " is
summed up in the brief equation /(3) =9.
To distinguish between different functions in the same
problem, we denote them by f(x) and F(x), or by /i(x) and
fz(x), etc.
EXERCISES
1. In what sense can we say that a ball thrown straight upward is
"instantaneously at rest" when it reaches the highest point, and not
similarly "at rest" at other instants?
2. "The speed of a pendulum is neither increasing nor decreasing
at the lowest point of the swing." Explain what this statement means,
in view of the fact that the speed is not constant during any interval.
3. Exactly what is meant by saying that: As x approaches 2,
(xsjrx — 1) approaches 9 as a limit?
4. Tell in detail exactly what these statements mean :
(a) .T4-»-16 as a»-2,
(6) L (*2
(c) lim (7+20Az+3Az2)=7.
5. What is the numerical value of L (z3-7 z2+60z-800)? Of
CB->-10
lim(*!±l? Of lim
\
74 MATHEMATICAL ANALYSIS [II, § 49
6. Point out precisely what is wrong with these notations :
(a) L (6+3z)-»-6, (6) As a»4), (z3+5)=5.
7. (a) If/(z)=*2-10:r, find/(l),/(20),/(a),/(2.5).
(6) If F(z)=z3+5, find F(0), F(k), F(2+Az), F(3+Az).
8. (a) Calculate the slope of the tangent to the graph of y = x3 at
x = 2.
(6) The angle (A°) turned by a wheel after t sec. was A=<3.
Calculate the average speed of rotation from t = 2 to t = 2+A£. Also the
instantaneous speed when t = 1.
(c) The edge of a cube is increasing. Calculate the instantaneous
rate of increase of its volume when the edge x = 2. Also the instan-
taneous rate when x = a, any value.
9. Find the slope of the curve y = 10 x — x2 at x =4 ; and draw a line
having that slope.
10. The repulsion (F dynes) between a certain pair of electrical
charges varies thus with the distance (x cm.) apart, F = 10/x2. Find how
fast F changes with x, at z = 2. (See Ex. 9, p 69, note.) .
11. The speed (V ft. /sec.) of an automobile t sec. after starting,
and until the full speed was reached, varied thus : V = 4 t — .1 12, Find
the rate at which the speed was increasing (i.e., the acceleration) at
§ 49. Summary of Chapter II. The idea of a limiting
value is essential to a satisfactory definition of an instantane-
ous speed, slope or rate, tangent line, direction or length of
a curve, area, or volume, etc. Moreover, each of these things
should be defined conditionally.
E.g., "if the average rate . . . approaches a limiting value . . .,
this limit is called the instantaneous rate."
By formulating these definitions accurately, we see how to
calculate the quantities in question. E.g., to calculate an
instantaneous rate, we must find what limit is approached
by an average rate in an adjoining interval which is being
indefinitely shortened.
II, § 49] SOME BASIC IDEAS ANALYZED 75
We discover also that several apparently distinct problems,
— such as finding the speed of a moving object, the slope of
a curve, and the rate of expansion of a metal cube, — may
in reality be one and the same problem. (Cf. Ex. 8, p. 74.)
In fact, a single process will suffice for calculating all in-
stantaneous slopes, speeds, and rates. And so, in the next
chapter, we shall reduce this process to a system, whereby
we can write formulas for these quantities at sight, and can use
the formulas readily for many purposes.
Quite apart from their immediate utility, the limit concepts which
we have been considering are a valuable possession -for any one. If
we realize their full import, and how constantly they occur in the affairs
of daily life, we shall develop a very helpful point of view and a fine
insight into the world about us.
Moreover, critical analysis such as we have attempted here, — the
effort to get at the inner meaning of terms, — is the very essence of all
accurate thinking. Much fruitless controversy would be avoided in
daily life by insisting upon clear ideas and accurate definitions.
CHAPTER III
DIFFERENTIATION
SOME IMPORTANT PHASES OF THE RATE
PROBLEM
§ 50. Rate-formulas. If we wish to know the speed of
a moving object at several different instants, we can save
time by deriving once for all a general formula for the speed
at any instant whatever. Similarly , for slopes and rates
in general.
Ex. I. The height (y ft.) of a vertically thrown ball after
t sec. was y= 100 t— 16 t2. Find the speed at any instant.
We proceed as when calculating the speed at t = 3 (§ 34),
but do not specify a particular value for t.
At any instant, y = 100 t- 16 t2.
M sec. later, « y=lQQ(t+At)-lQ(t+At)2.
Simplifying the latter value of y and subtracting the former
gives * Ay = 100 A* - 32 t At - 16 At2.
This difference in height is the distance traveled during the
At sec.
.'. ^= 100-32 £-16 A*.
At
This is the average speed during At sec. beginning at any
time t. The limit approached by this as AZ->-0 is
L (^ = 100-32*. (1)
At
* AJ will not combine with t as A/1, for it is not a product A • t but
simply the difference in t or "increment" of t. (§ 25.)
76
Ill, § 50] DIFFERENTIATION 77
This is the instantaneous speed at the beginning of our
interval At ; i.e., the speed at any time t sec. after the ball was
thrown.
For instance :
at t =0, speed = 100-32(0) = 100 (ft./sec.)
aU=3, speed = 100 -32 (3)= 4 (ft./sec.)
This last result was found in § 34. But now we can get the speed at
any number of instants, by merely substituting values for t in (1).
From this speed-formula we can also find exactly when the
ball was highest. For the speed was then exactly zero :
100-32 £ = 0. /. t
At that instant, y=100(¥)-16(^-)2 = 156i, — the great-
est height.
EXERCISES
1. A bomb was fired straight up, its height (y ft.) after t sec. being
y=3QQt — 16 P. Find the speed at any time. In particular what
speed at t = 10 ? At t = 20 ? When was the bomb highest ? How high ?
2. Plot the graph of y = x3 — Qx from x=— 4 to x = 4. Calculate
the exact slope at any point. In particular what slope at x = l and
x = 2 ? Find the exact values of x at which y has its maximum and mini-
mum (i.e., turning) values.
3. A volume of gas (V cu. in.) varied thus with the pressure
(P Ib./sq. in.) : F = 600/P. Find the rate of increase of V per unit
change in P at any instant. What rate at P = 15? At P=20?
4. As a ball rolled down an incline its distance (x ft.) from the top
varied thus with the time (t sec.) : x = &t2. Find the speed at any
instant. What speed at £ = 1, 2, 3, 4, 5?
5. Calculate the slope of a suspension-cable curve, whose height is
y = .003 re2 at any point. What slope at x = 5, 20, 100 ?
6. A spherical balloon expands as the temperature rises. Find
the rate of increase of the volume, per unit change in the radius, at any
time. * What rate when r = 20 ?
* Formulas for volumes, etc., are given in the Appendix, p. 492.
78 MATHEMATICAL ANALYSIS [III, § 51
§ 51. Derivative of a Function. Rate, slope, and speed
calculations present one and the same problem. Let us,
therefore, save time by formulating the problem abstractly
in such a way as to cover all these calculations at once, -
and perhaps others also.
Let y be any function of x. Then if x starts from any value
and increases by any amount Ax, y will increase by some A?/
(positive or negative). If Ax is made very small, Ay will
usually become very small also ; but the fraction Ay/ Ax will
ordinarily approach some definite limiting value.
The limit of Ay/ Ax as Ax approaches zero is called the
derivative of y with respect to x, and is denoted by dy/dx.
That is,
^= L f^V (2)
dx Ji-oW'
The notation dy/dx is read "d y over d x" ; but we are not using it
to stand for some quantity dy divided by some quantity dx. We are
using it simply as a notation for the limit approached by Ay /Ax.
Though a derivative is thus defined abstractly, yet it has
many possible concrete interpretations, — and hence every-
thing which we learn about derivatives will apply at once to
many different problems. The definition and the following
facts should therefore be learned with the greatest care.
§ 52. Interpretations of dy/dx. Some of the more funda-
mental interpretations of derivatives will now be mentioned.
(a) If y denotes the distance traveled by a moving object,
and x denotes the time, then AT/ = the additional distance
traveled in an additional time Ax ; and
^ = distance = a e d during Ax
Ax time
/. -^ = limit of average speed = instantaneous speed*
dx
* Why could we not find this speed by simply dividing the distance y
by the time x ?
Ill, § 53]
DIFFERENTIATION
79
X X+&X
(6) If y denotes the height of a graph and x the horizontal
distance from a fixed point, then Ay = the rise in a small hori-
zontal distance Ax ; and
— ^ = average slope of curve during
interval Ax ; (a)
= limit of average slope — slope at
point P.
(c) If y denotes any quantity which
varies with another, x (e.g., the volume
of a fluid, with the temperature x), Ay is
the increase (in volume) produced by a
rise of Ax (in temperature) ; and
— = average rate of increase (cubic
inches per degree).
/. -^ = limit of average rate = instantane-
ous rate.
Thus the derivative dy/dx means instantane- /c\
ous speed, slope, rate, etc., according to the
meaning of x and y.
Always, in fact, dy/dx gives the instantane-
ous rate at which y is changing per unit change
in x. Speed is simply the rate at which the
traveled distance is increasing per hour (or second, etc.). Slope is the
rate at which curve is rising per horizontal unit.
To find the rate at which any quantity is changing, simply get the deriva-
tive of the quantity.
§ 53. Differentiation. The process of calculating a deriva-
tive is called differentiation. It is the same as that of finding
a formula for an instantaneous rate, speed, or slope ; but is
usually condensed as follows.
Ex. I. Differentiate y=*x3.
FIG. 28.
80 MATHEMATICAL ANALYSIS [III, § 54
After x has increased, the new value of the function will
be the original value y plus some increment Ay.
i.e.,
Az
.. .
dx
Observe that we get the final value; dy/dx, not by putting
Az = 0 (which would make A?//Az meaningless), but rather
by seeing what limit Ay/Az approaches as Az-»-0.
The resulting derivative, just obtained, may be given various con-
crete interpretations :
1 . If y is the distance traveled by an object in any time x, and y = x3,
then the speed at any instant is 3 x2.
2. If y is the height of a graph at any horizontal distance x from
some fixed point, and y = x3, then the slope at any point is 3 x2.
3. If y and x are the volume and edge of a metal cube which is being
heated, then y = x3, and the rate of increase of the volume per unit
change in the edge is at every instant equal to 3 x2.
§54. Increasing or Decreasing? We shall often need to
test whether a given function y=f(x) is increasing or de-
creasing at a particular value of x. To be definite, let x itself
always increase.
Now the graph of a function is rising at any point where its
slope is positive, and falling where its slope is negative. (Cf .
Remark, §41.) That is,
y is increasing where dy/dx=+, ,„,
y is decreasing where dy/dx = — .
Remarks. (I) y may be increasing or decreasing even at points
where dy/dx = 0, — that is, where the slope is zero. (Discussed later :
564.)
Ill, § 54] m DIFFERENTIATION 81
(II) It is not a sure test of increasing to compare the value of y
at the given point with some near-by value. For y might be increasing
at the point, and yet have decreased before reaching the comparison
value.
But the test given in (3) above is sure. For if Ay /Ax is ultimately
positive, the neighboring point must ultimately be the higher, and
hence y must be increasing.
EXERCISES
1. Differentiate the following functions :
(a) i/ = z2, (6) y=x3, (c) y=x*,
(d)y = 3x*-9, (e) y=x*-9x+10, (f)y = 7/x, -
•(flf) 2/ = 7(H-5<2, (h) y = P+WQ/t, (i) y
Test whether each of these functions is increasing or decreasing at
x = l (or i = l).
2. A ball rolled up an incline so that its distance from the starting
point after t sec. was y = 7Qt — 5P. By Ex. 1 (g) what was the speed
at t = Q, 2, 8? When was it farthest up, and how far?
3. Plot y = x*-9x+W from z = 0 to z = 7. By Ex. 1 (e) what is
the exact slope at x =2? At x = 5? (Check.) Exactly, what z gives
the minimum y? How small a value?
4. If y denotes the speed of a moving object and -x the time, what
is the meaning of Ay/Ax, and of dy/dx?
5. If the speed of a car t sec. after starting was V = 4 t — .2 P, what was
the acceleration at any time? (Cf. Ex.9, p. 24, note.) What at f = 0,
5, 10, 15?
6. Along an arch of a bridge the height (y ft.) above the water at
any horizontal distance (x ft.) from the center is y = 7Q — .006 z2. Find
the slope at any point, and in particular at x = 50.
7. For the beam in Ex. 9, p. 63, find how fast the strength S will
vary with the thickness x in. at any value of x, and in particular
when x = 10.
8. The repulsion (R dynes) between a certain pair of electrical
charges varies thus with their distance apart (x cm.): R = 2Q/x2.
Find how fast R changes with x in general. In particular how fast at
x = 2.
[9.] Compare the results in Ex. 1 (a), (ft), (c), above. What would
you expect for dy/dx, if 7/ = z5? If y = xw? If y = xn?
82 MATHEMATICAL ANALYSIS [III, § 55
§ 55. Power Laws. In nature it is very common for one
quantity (y) to vary as some fixed power of another (x), say
y = kxn. (4)
Such quantities are said to " vary according to the Power
Law."
To study their rates of increase advantageously, we shall
now obtain a formula for differentiating any power at sight.
§56. Differentiating xn by Rule.
The adjacent table shows the deriva-
tives of three powers of x, as found in
Ex. 1, p. 81. These results suggest x*
that the derivative of any positive x* 8
integral power xn would be nxn~l. Lqt
us see.
If y = xn, then y+Ay=(x+Ax)n.
Multiplying out, (x-}-Ax)n would always give
(x+Ax)n=xn+nxn~1 Ax + (terms with Ax2, Ax3, etc.).*
Subtracting y=xn and dividing through by Ax gives
dydx
- = nxn~l — + (terms with Ax, Ax2, etc.).
Ax
The limit of this as Ax-»-0 is simply nxn~l. (§ 46, II.)
Thus *.«•-!. (5)
ax
In words : the derivative of any positive integral power of x
equals the exponent of the given power, times the next lower
integral power of x.
* If not familiar with the Binomial Theorem, you can check this expansion
as follows. (x+Ax)n means (z+Az) • (z+Az) • (z+Az) • (z+Az) ••• to n
factors. Multiplying together the z's of all the n factors gives zn. Multi-
plying the Ax in any one factor by the z's in all the others gives z*"1 A*
— and this term will occur n times in all when we use the Az of one factor
after another. Multiplying the Az's in two or more factors gives terms
containing Ax* or Axs, etc. — whose coefficients we do not need to know.
Ill, § 57] DIFFERENTIATION 83
By this remarkable rule, we may write certain derivatives
at sight. Thus
if y = x«, then ^ = 6 z5;
ax
iiy = xm, then ^ = 100z".
ax
Memorize carefully the verbal statement of rule (5), and
that of each similar rule that follows.
N.B. An important special case is the derivative of x
itself :
if y = x, ^=1 >x°=l*' *
dx
This result may also be obtained directly. Thus
if y = x, then
.'. !' = !, and -' = l.
Ax dx
This means that the rate at which y increases, per unit change in a:, is
1, — which is obvious, since y = x.
§ 57. Effect of a Constant Multiplier. What will be the
derivative of a power of x which is multiplied by some
constant fc?
If y = kxn, then y-\-Ay = k(x+Ax)n.
Every term in the expansion is multiplied once by the
factor k. Hence Ay/Az is precisely k times as large as if we
were differentiating xn alone, and must approach a limit, or
derivative, just k times as large.
A constant multiplier k simply multiplies the derivative by k.
Ex. I. If y = 10 x3, ^ = 10(3 x2} =30 z2.
dx
This means simply that 10 x3 increases just ten times as fast as x3.
Ex. II. Ify=-lxs, 5r=-i(8z7) = -fz7.
dx
Ex. III. Tfy = 17x, ^ = 17(1) = 17.
dx
* For the meaning of a zero exponent, if unfamiliar, see § 67.
84 MATHEMATICAL ANALYSIS [III, § ns
EXERCISES
1. Differentiate the following functions at sight, writing the functions
and their derivatives in parallel columns, like a table :
**>, 5x*, i*«, -8x, .002**, -.75x«, feri.| y Xf «|^ ^
i i —
2. The same as Ex. 1 for the following, differentiating with respect
to the variable named in each case :
.5/s, -9<2/64, 20 1, r4, -7 v\ irr\ f Trr*.
3. The range of a certain gun for various muzzle velocities is
R = .02 F2. How fast does R increase with V at V = 1200 ?
4. In Ex. 3 plot R from F = 800 to V = 1500, and check the calculated
rate.
6. Find how fast the volume of a cube increases with the edge x in.,
when x=20. .
6. The distance (D ft.) required for stopping an auto under normal
conditions varies as the square of the speed (V mi./hr.). If D=20.7
when F = 15, write the formula, giving D for any V. Also find how
fast D increases with V at 7 = 20.
7. Find the derivatives of the following quantities. Substitute some
numerical value and interpret each result as a statement about rates.
(a) The distance a ball had fallen after t sec. was s = 16 f2.
(b) The speed of a falling object after t sec. was v = 32 t.
(c) The kinetic energy of a moving car was E = 25 vz.
(d) The height of a suspension cable x ft. from the middle is
y = .QQ4x*.
(e) When a certain locomotive rounds a certain curve at a speed
of V mi./hr., the centrifugal force (F tons) is F = .09 V2.
(f) The consumption of coal (C tons/hr.) in a certain locomotive
varies thus with the speed (V mi./hr.) : C =
I § 58. Effect of an Added Constant. How does a constant
which is added to a function affect the derivative ?
If t/=/+/b, then y+Ay=(f+Af)+k.
Obviously k drops out in subtracting to get A?/. Thus
Ai//Ax, and hence dy/dx, has the same value as if we were
differentiating y=f alone. That is, a constant added to a
function contributes nothing to the derivative.
Ill, § 60] DIFFERENTIATION 85
Ex. I. Ify = zH-1000, = 4 z3.
dx
This means that (x4+1000) increases at the same rate as z4.
Graphically speaking, adding a constant to a function simply raises
or lowers the entire graph by a fixed amount, and. does not change the
slope at any point.
The derivative of an isolated constant is zero, since its
rate of change is zero. E.g.,
if t/ = 210 continually, ^ = 0.
dx
§59. Differentiating a Sum. If y is the " sum of two
functions of x, say y=f+F, and if x increases by Ax} then
y+Ay=f+Af+F+AF.
^ A|/ = Af +AF
Ax Ax Ax
.
dx dx dx'
That is, the derivative of the sum of two functions is simply
the sum of their separate derivatives.
The same is evidently true for the sum of any specified
number of terms.
Ex. I. If 2/ = z17+4z10, • ^=17z16+40z9.
dx
Ex. II. If y = f z8-15z+4, ^ = ^z7-15.
dx o
We can now write at sight the derivative of any "polynomial," and
use it to solve problems on rates, slopes, etc.
§ 60. Differentiating a Product or Fraction.
Ex. I. Differentiate 2/=(z2+6)(z3-4).
Multiplied out : y = x5 + 6 z3 - 4 x2 - 24.
Differentiated : ^ = 5 *4+18 x2-8 x.
dx
86 MATHEMATICAL ANALYSIS [III, § 60
Could this result be obtained by differentiating the factors (x2+6)
and (x'^-4) separately, and then multiplying the two derivatives
2 x and 3 x2 together?
Ex. II. Differentiate y=£±?2.
x
72
Divided out: y
x
Since we have as yet no rule for terms with x in the de-
nominator, we resort to the " A-process " (§ 53) ; and get
finally *f-9r 72
dx~ "?'
This result cannot be obtained by differentiating the numerator and
denominator separately.
These examples show that the derivative of the product or
quotient of two functions is NO T equal to the product or quotient
of the two derivatives. At present we can differentiate a
fraction only by the " A process " and a product only by
that process or by first multiplying out.
EXERCISES
1. Differentiate, writing the results in tabular form :
x*-3x, x*-2x*+x,
, -, -.
7 "5 2
2. The same as Ex. 1 for the following, in which a, 6, and c denote
constants :
' £;
36 43
3. Differentiate :
(a) 20z2(9-z); (6) Cr'-l)(z2+2 z+5) ; (c
4. The speed of a car ISGC. after starting was F = .09 F-.QQlt3.
Find the acceleration at £ = 40.
5. The height (y ft.) of a ship's deck, going forward from a certain
point A, varies as the square of the horizontal distance (x ft.) from A.
If y = 5.6 when x = 200, find the formula for y at any x. Find the slope
of the curve at x = 100.
Ill, § 60] DIFFERENTIATION 87
6. The volume (V cu. ft.) of a certain ship's hull up to a height of
x ft. above the keel is 7 = 1600 x2- 80 x3 +2 x4-. 02 z5. Find how
fast the volume of water displaced increases with the draught (i.e.,
how fast V increases with x) at z = 10; at z = 20.
7. The momentum of a locomotive t mm. after starting was
M = 2 P(t -20)2. Find how fast M was changing at t = 5 and at t = 15.
Increasing or decreasing?
8. Plot y = 30x — x3 from x =—3 to x = Q. Calculate the exact
rate of increase of y at x— — 2 and +2. Compare the graph. Is y in-
creasing or decreasing at x = 3? Make a sure test.
9. Plot y = x3+5 from x=— 3 to x = 3, calculating y at a? = |, — |.
Calculate the exact slope at x=0 and 1. Compare the graph.
10. (a) Plot y = .8x+3 from z = 0 to x = 10. Measure its slope;
and calculate the same by differentiation.
(6) Prove that the graph of any linear function y=ax+b has a
constant slope. Hence what sort of graph? [Cf. § 7, Fig. 10.]
11. (a) Plot y = xz and also y = xz-{-3, over the same base line, from
# = 0 to z = 4. Compare the slopes at any value of x by measurement.
Then find dy/dx in each case and compare.
(6) Likewise plot y = .7x2 over the same base line as in (a) and
compare its slope with that of the curve y=x*.
12. The following equations give the approximate percentage change
(T/) in the price of various farm products for any percentage change (x)
in the size of the crop.*
(a) y = .94-1.0899 Z+.02391 *2-.000234 x*. (Corn.)
(6) y = 8.22 -1.1904 x-. 00663 x*+. 000273 x3. (Oats.)
(c) y = 1.77 -1.5062 z + ,02489 z2-.000197 xs. (Potatoes.)
In each case find dy/dx when # = 10 and when z = 0. (The latter
value is related to the "coefficient of elasticity of demand," — dis-
cussed in Economics.)
13. Differentiate ?/ = (z3+5 z2+3)/z.
(Hint : Divide out ; then differentiate by the A-process, § 53.)
14. The same as Ex. 13 for y = (x*—7 z2+ll)/z3.
15. (a) — (e). Write at sight the rates, slopes, etc., called for in:
Exs. 9-10, p. 63 ; Ex. 8, p. 69 ; Exs. 5-6, p. 77.
[16.] The force (F Ib.) applied to an object varied thus with the time
(t mm.) : F=4Qt—t2. Find how fast F was increasing at t = lQ.
Hence, about how much did F increase in the next .02 min.?
17. In Ex. 16 was F increasing or decreasing at t =21 ?
* See H. L. Moore : Economic Cycles.
88 MATHEMATICAL ANALYSIS [III, § 61
§ 61. Note on Mensuration Formulas. A ready CCLYI-
mand of the elementary formulas for volumes, areas, etc.,
is essential in what follows. Recalling certain facts which
are proved in geometry may help to fix
those formulas in mind.
(I) An area is always the product of two
linear dimensions, while a volume is the
product of three. Thus a volume formula
could never be 2 Trrh, say.
(II) The same formulas apply to cylin-
ders as to prisms. The reason is that a
cylinder is the limiting form approached
by an inscribed prism, when the number of
sides of the base is indefinitely increased. For either solid :
Volume = (area of base) X (height).
N.B. "Height" means the perpendicular distance between bases.
(III) The same formulas apply to cones as to pyramids,
a cone being the limiting form approached by an inscribed
pyramid. For either solid :
Volume = I (area of base) X (height).
Lateral area — ^ (perimeter of base) X (slant height).
The term "slant height" is meaningless, however, unless the cone is
a right circular cone and the pyramid is regular.
(IV) The area and volume of a sphere are
A =4 Trr2, K^Trr3.
That is, the area exactly equals four "great circles" cut through
the center, — a very surprising fact.
The volume equals one third the area times the radius, — just as
if the sphere were composed of tiny pyramids with their vertices at
the center and their bases in the surface.
(V) The area and circumference of a circle are
4 = Trr2, and C=27rr.
Ill, § 62]
DIFFERENTIATION
89
If you tend to forget these, study Fig. 30 carefully until
you get the real significance of the formulas.
FIG. 30.
Obviously the area is less than
4 r2 and more than 2 r2 ; appar-
ently about 3 r2. Exact value :
irr2; or 3.1416 r2, very closely.
Each side of hexagon = r. Cir-
cumference is obviously a little
more than 6 r. Exact value:
2 vr ; or 6.2832 r, very closely.
EXERCISES*
1. Exactly how does the curved surface of a hemisphere compare
in area with the flat side?
2. If a square is circumscribed about a circle, approximately what
percentage of the square is contained in the circle?
3. The same as Ex. 2 for a cube circumscribed about a sphere.
4. If a circle is inscribed in each face of the cube in Ex. 3, except
the top and bottom, how does the combined area of these circles com-
pare with the area of the sphere?
6. If a cylinder and cone have the same base and height, how do
their volumes compare?
6. If a cylinder is circumscribed about a sphere (including the bases),
what fraction of the cylinder is contained in the sphere?
§ 62. Approximate Increments. For a small interval Az,
the fraction Ay /Ax and its limit dy/dx are nearly equal.
I.e., approximately,
,
Ax dx'
or
Ax.
(6)
Hence we can find the approximate change in y, due to a small
change in x, by simply multiplying the derivative by Ax.
* Hereafter when we speak of a "cylinder" or "cone " we shall mean a
right circular cylinder or cone, unless something is said to the contrary.
90
MATHEMATICAL ANALYSIS [III, § 62
Ex. I. If y = z10, how much will y increase when x changes
from 2 to 2.003?
Here = 10 x\
dx
FIG. 31.
5120 at x = 2.
(5120). 003 = 15.36, approx.
To find Ay exactly we should
have to calculate 2/+A?/ = (2.003)10
and subtract y = 210. This tedious
operation gives Ay = 15.464 •••,
nearly the same as the approxima-
tion obtained so easily above.
Ex. II. The edge of a cube
was measured as 20 in. but
was really 19.98 in. About
how much was the calculated
volume in error?
The question amounts to
this : By how much would the
volume change, if the edge
changed from 19.98 to 20 in.?
dV
:3ic2, = 1200atz = 20.
f^\ Ax = (1200) .02 = 24, approx.
The error m the calculated volume was about 24 cu. in.
•H
This result would be affected very little by taking the value of
dV/dx at x = 19.98 instead of x = 20. We chose the simpler value.
Compare the approximate formula AV = 3 z2Az with Fig. 31.
EXERCISES
1. If 2/=7x*, approximately how much does y increase while x
increases from 9.98 to 10.03? Exactly how much?
Ill, § 62] DIFFERENTIATION 91
2. If y = x6— 4 x3+ll x — 17, approximately what change in y
when x increases from 2 to 2.005?
3. In Ex. 2 about how much must x increase (starting from x = 2)
to change y by .04?
4. If a metal cube expands so that its edge x increases from 10 in.
to 10.002 in., approximately what change in its volume? In its total
area?
5. In Ex. 4 approximately what change in x would increase the
volume by .45 cu. in. ? The area by .3 sq. in. ?
6. Find the approximate change in :
(a) The area of a square if its edge increases from 8.47 in. to
8.52 in.
(6) The circumference and area of a circular plate if the radius in-
creases from 3.99 ft. to 4.01 ft.
(c) The volume and area of a ball-bearing if it wears down from a
radius of 5 mm. to one of 4.98 mm.
7. About what change would be necessary in :
(a) The radius of a spherical balloon, to increase the volume by
10 cu. ft., if the radius is approximately 20 ft.?
(6) The radius of a cylinder, to increase the volume by 4 cu. ft., if
the height is constantly 20 ft. and r is about 3 ft.?
8. Approximately what errors will there be in the calculated
(a) Area of a sphere if the radius is measured as 10 in. when its true
value is 10.03 in. ?
(6) Volume of a cube if the edge is measured as 8 in. instead of
7.985, the true value ?
(c) Area of a circle if the diameter is measured as 40 in. instead of
40.06 in.?
9. Approximately what errors would be allowable in the measured
(a) Side of a square (about 10 cm.), if the error in the calculated
area is not to exceed .25 sq. cm. ?
(6) Diameter of a sphere (about 80 cm.), if the calculated volume
is not to be in error by more than 100 cc. ?
10. In Exs. 8-9 what is the approximate percentage error in each
case?
11. In Ex. 3, p. 84, about how much greater a range is obtained
if V = 801.1 than if F=800?
12. In Ex. 7 (/), p. 84, about how much more coal is required per
hour if 7 = 20.05 than if F = 20? Exactly how much?
13. The maximum deflection of a beam varies as the cube of the
92 MATHEMATICAL ANALYSIS [III, § 63
length. If D = A when I = 10, write a formula for D. Approximately
how much larger is D for 1 = 12.08 than for 1 = 12?
14. The cost (y cents per hr.) of running a certain car at a speed of
V mi./hr. varies thus: y = .3 F2. Approximately how much will y
increase while V increases from 20 to 20.6? Exactly how much?
[15.] Criticize this "proof" that 1=2: When x = l, x*-x = xi-l.
Hence, factoring out x — 1 : x=x+l. That is, 1=2.
§ 63. Note on Zero Factors. A principle of algebra much
used in what follows is this: A product is zero if any one
of its factors is zero ; and, conversely, a product can be zero
only if some factor is zero.
Ex. I. 5z2(z-4)(z+2)=0.
This equation is satisfied if z2 = 0, or (x— 4) =0, or (x-f2) =
0, — that is, if 2 = 0, 4, or — 2. And it can be satisfied by no
other values.
This equation is the factored form of the equation
If we were solving the latter, and canceled out 5 x2, we should
lose the root x = 0, — one out of three possible values.
The indiscriminate canceling of factors may easily lead to an erro-
neous conclusion.
For instance, we might infer from the equation kx = 17 x that k = 17
necessarily, — which is not so ; k might be 1000, or any other number.
For if x happens to be zero, kx = 17 x, no matter what value k has.
We can, however, assert that, since kx — 17 x = Q,
/. (k- 17) * = 0.
And hence, either x=0, or else fc — 17 = 0 (making fc = 17).
Never cancel a factor out of an equation without considering
the possibility of its being zero.
§ 64. Horizontal Tangents. Any graph has a horizontal
tangent wherever its slope dy/dx = Q. Usually it is then
turning from rising to falling or vice versa, — as at A or B in
Fig. 32.
Ill, § 64]
DIFFERENTIATION
93
But not always. E.g., in the graph of y = x3 (Fig. 33), the
slope dy/dx( = 3x2) is zero at x = 0: but the curve rises
through C, being lower everywhere to the left and higher
everywhere to the right.
j A sure test as to whether there is a turning point can be
made by noting the sign of dy/dx near by. Thus if the sign
\
J 1 2
Graph of y=x3
FIG. 32.
FIG. 33.
runs H h as in Fig. 32, the curve must rise to A, fall to
Bj and then rise again.
N.B. This curve to the left of A is regarded as a rising rather
than a falling curve, because we always think, of going toward the right,
with x increasing.
EXERCISES
1. What sure conclusion as to the value of x can you draw from
each of the following equations :
(a) z2=20z? (6) 2^2 = 15^? (c) 7rz2(10-z) =0?
2. Plot each of the following curves, at least through the interval
indicated. Show clearly all horizontal tangents, and all intersections
with the base line.
(a) y = llx-x>,
(6) y = 5-x*,
(c) y=x*
(d) 7/=x4
-2 to +12;
-2 to +4;
-3 to +3 ;
-3 to +5.
94 MATHEMATICAL ANALYSIS [III, § 65
§ 65. Extreme Values. As any ordinary (rational alge-
braic) function y approaches a maximum or minimum value,
its graph rises or falls very slowly. At the turning point, the
slope is zero. »
By using this latter fact, we can locate the extreme values
(maxima and minima) of y without plotting. We have
simply to set dy/dx = Q, solve for x, and substitute in the y
equation. To determine whether each result.is a maximum
or minimum, we simply test the sign of dy/dx for values of x
just before and after each.
Ex. I. y = x*-4:X3+5.
Differentiating and factoring the derivative gives
^>=4*2(z-3).
dx
Hence dy/dx = 0 where x = 0, 3 ; and nowhere else.
Testing the slope dy/dx on both sides of x = 0 :
atx=-l,
atz=+l,
y is decreasing both before and after x = 0. No extreme here.
Testing dy/dx near x = 3 (signs only are important) :
at* = 2, dy/dx= (+)«(-) = -,
atz = 4, dy/dx= (+)*(+) = +.
y is decreasing before x — 3 ; then increasing. Hence it has a
minimum value at x = 3, viz.
2/=(3)4-4(3)3+5=-22.
Since x = 0 and x = 3 are the only values making dy/dx = 0,
y can have no maximum. The graph should, however, show
a horizontal tangent at z = 0. (Fig. 34.)
Ill, § 65]
DIFFERENTIATION
95
FIG. 34.
Remarks. (I) The value of y at C
is not a maximum, but merely the
largest reached so far. Inspection of
dy/dx shows the curve to be continu-
ally rising to the right of x = 3.
(II) The factored form of dy/dx is
the most convenient: both in seeing
that we have found all the points
where dy/dx = 0, and in seeing the sign
of dy/dx in the tests, without stopping
to calculate the value.
(III) Merely comparing the values
of y at z = 0 and x = 3, viz. 5 and —22,
might erroneously suggest that 5 is a maximum, and —22 a minimum ;
and would not show the true course of the curve at all.
(IV) Whether to substitute values of x in the ^/-equation or in
dy/dx depends simply on what information we are after: one tells the
height of the graph and the other the slope.
EXERCISES
1. Test each of the following functions for maximum and minimum
values. From this information draw the general shape of the graph.
(Get more points if in doubt.)
(a) ?/=z2-6z-f5, (6) y=x>+2x,
(c) ?/=a;3-
2. For a beam loaded in a certain way the deflection (y ft.) at any
horizontal distance (x ft.) from one end is y = .00001 (30 z2— z3).
Plot a graph showing how y varies with x from x = — 10 to +30. [This
graph will be the curve of the beam, exaggerated and inverted.l Calcu-
late the maximum y ; and check.
3. The work done by exploding a mixture of 1 cu. ft. of coal gas
with x cu. ft. of air is W=S3 2— 3.2 x2. Find the maximum value of
W. (Find dW/dx by the A process, for review.)
4. In Ex. 3, p. 34, express the volume of a box in terms of z, the
side of the square corners cut out. [Ans., simplified, F=4 x3— 80 #2 +
400 x.] Then find the maximum volume.
6. The speed of a point on a fly-wheel t sec. after starting was
F = 40f2— P. Was this increasing or decreasing at £=27, and how
fast? When was V greatest?
96 MATHEMATICAL ANALYSIS [III, § 66
6. A number x minus its square is to be made a maximum. Find x.
Check your result by calculating the specified difference for some
near-by values and plotting roughly.
7. Given a formula for any quantity, how would you find the rate
of increase at any instant ? The amount of increase in any very small
interval? The maximum and minimum values? Whether increasing
or decreasing at any point?
§ 66. Applied Maxima and Minima. Practical problems
concerning maximum and minimum values are usually stated
verbally rather than in terms of formulas or functions. In
such cases we must first set up a formula for the quantity
in question, expressing it in terms of some one variable, say x.
Then we can differentiate and proceed as formerly. (§ 65.)
Ex. I. If a rectangle is to have a perimeter of 40 in., what
is its greatest possible area?
(Perhaps you could prove geometrically that the rectangle
should be a square? But let us try out our new method.)
The area of any rectangle is
A = bh. (7)
And if the perimeter is to be 40 in., that is,
26+2/1 = 40,
then 6 = 20— h. Substituting this in (7) gives
h*. (8)
A is now expressed in terms of a single variable, h, and we
are to find what value of h will make A greatest. Differ-
entiating :
^ = 20-2 h.
ah
Equating this derivative to zero gives h = W. Testing the
signs of dA/dx at /i = 9 and h = 11, shows a maximum at 10 ;
viz., A = 20(10)-(10)2=100.
Ill, § 66]
DIFFERENTIATION
97
Remarks. (I) h = 10 requires also 6 = 10: a square.
(II) To keep the perimeter constant, b must change with h in a
definite way. Thus A, though expressed in (7) in terms of two variables
6 and h, is really a function of either alone ; and is not ready for differ-
entiation until so expressed, as in (8).
Ex. II. Find the volume of the largest right circular
cone which can be inscribed in a sphere of diameter 10 in.
For any cone, inscribed or not :
V=%irr*h. (9)
But in the present case we have, from the
right triangle in Fig. 35 :
whence
r2=lQh-h\
FIQ. 35.
Substituting this value for r2 in (9) above gives
3
(10)
dh 3^
— =0 when
dh
or
A test shows that h = 20/3 makes V a maximum ; viz.,
N.B. Similarly in any other problem, we first draw a figure (if
needed) and write some formula for the quantity which is to be maxi-
mized or minimized. Then by using the hypothesis or requirements of
the problem, we express everything in terms of one variable alone. When
in doubt, it is well to ask : What is to prevent our making all the quanti-
ties as large or small as we please. This will direct attention to the
limitation or specified relation among the several quantities.
EXERCISES
1. Find as in Ex. I above the largest possible area for a rectangle
of perimeter 80 in.
98 MATHEMATICAL ANALYSIS [III, § 67
2. Similarly show that a rectangle with any specified perimeter P will
have the largest area when it is square. «*
3. Find two numbers whose sum shall be 15 and whose product P
shall be as large as possible. Check your answer by calculating P for
some near-by values and plotting roughly.
4. The sum of a number and its square is to be a minimum. Find it.
Check by forming the sum for some near-by values.
6. In Ex. 2, p. 33, express the area in terms of x, and find its maxi-
mum value.
6. In Ex. 7, p. 34, express L as a function of x. [Ans.,
L=4000, x — Wx3.] Find what x and y will give the strongest
beam. \
7. What are the dimensions and volume of the largest rectangular
box with square ends which can go by parcel post? (See Ex. 9, p. 34.)
The specification with square ends is superfluous. Why?
8. In Ex. 5, p. 34, express the distance between the ships t hr.
after noon. [Ans., y = V400 t2 — 2560 t +6400.] Find the minimum
value of if ; and how large y was then.
9. In Ex. 15, p. 57, the tabulated values of Q and E satisfy these
equations : Q = 2(110-P)2, #=30(110-P). Express the net profit at
any price. [Ans., in dollars, AT = .02 P'-4.4 P2+272 P-3300.] Find
what P gives the maximum N.
10. In Ex. 4, p. 34, express the combined length of the three new
sides in terms of x. [Ans., L=x+24Q/x.] Find the minimum value
of L, getting dL/dx by the A process.
11. The total area of a rectangular box is to be 600 sq. in. Prove
that the volume will be greatest if the box is cubical.
12. A Norman window has vertical sides and a horizontal base,
but the top is a semicircle. If the perimeter is 30 ft., what dimensions
will give the largest possible area?
13. In locating a maximum or minimum value, why do we set the
derivative equal to zero?
§ 67. Negative and Fractional Exponents. To simplify
many differentiations, and other calculations which follow,
we must know the meaning of negative and fractional powers,
— such as x~*, x*, etc.
A positive integral power denotes simply the product of
several equal factors. /Thus, x5 stands for the product of
Ill, § 67] DIFFERENTIATION 99
five x's. But obviously x~5 can have no such meaning.
2
Likewise x* and x° are quite absurd from this standpoint.
The meanings assigned to such powers may already be
familiar to you from algebra. If not, master them thoroughly
now.
Definitions Illustrations
(1) x~p shall stand for - ; 10~3 = i = .001.
X J.U
(2) x»t« shall stand for -V& ; 10* = ^10* = 4.642-.
(3) x° shall equal 1, always. 10° = 1 ; (-3)°= 1, etc.
Remarks. (I) These definitions are the only ones possible
if the usual laws of exponents are to apply in all cases.
For instance, dividing z3 by x5 gives or2 by subtracting exponents ;
but gives 1/z2 by canceling. Thus x~2 must equal 1/z2, if the law of
subtracting exponents is to be valid here. Similarly z° would result
from dividing z3 by x3, and hence must equal 1.
Again, if the law of dividing exponents when extracting a root is
always to apply: ^x4 = x^, etc.
(II) These definitions suffice to determine the value of any
numerical expression consisting of rational powers. For
instance, y = 7(32)-*+8o+_i_
is the same thing as
Observe here that the negative exponent — 1 was applied to 4 and 5
separately, not to the denominator as a whole, — also that we may first
extract the fifth root of 32 and then cube the result.
EXERCISES
1. (a) Why do we take or4 as standing for 1/x4, and x° as denoting 1 ?
Illustrate by divisions, using definite powers. (6) Similarly, explain
why we take z* to mean ^x.
100 MATHEMATICAL ANALYSIS [III, § 68
2. When z = 10, what is the value of ar3? Of -x3? (Notice the
big difference in the meaning of the — signs. Also contrast x$ with 1/x3) .
3. What is the meaning of: x$, a*, x*, x~%, x~$?
4. Find the values of these expressions :
(a) 2-1-5~1 (6) 80+3-2, (c) 75X1Q-*,
(d) 30X20, («) 200-K2-1-3-1), (/)
07) 36*, (h) 2r*, (i)
6. Express hi a form free from negative and zero exponents, and find
the value of each quantity when x = 2 :
(a) 6x-*+x-*+.5x°, (6) 7 x~J
6. Why is 3 x~l not the same as — ? ]
3 x
7. Solve for x: 4 z-500 x~* = Q.
8. If y varies inversely as x3, and y = 5 wlien x = 2, write a formula
for y in terms of x. May this be regarded as a special case of the Power
Law (4), p. 82, for some values of k and n? Explain.
9. Express these equations in the form of the Power Law:
r_20 150 _VJ
y~x*> ~7x*' "'
[10.] Differentiate y = l/xz by the A-process. Also differentiate by
rewriting y asy=x~2 and using the formula for y = xn. Show that the
latter method gives a correct result on simplifying.
§ 68. Differentiating Negative Powers. Let us now see
about a rule for differentiating any negative power at sight.
Can the standard formula for any positive power y = xn be
used here?
This formula, applied for instance to y = x~l°j would give
since going down one unit from the exponent —10 would
bring us to —11.*
* The easiest way to be clear about negative numbers is to think of their
analogy to temperatures below zero.
Ill, § 68] \ DIFFERENTIATION 101
The correctness of this result can be tested by the original
increment method of differentiating. (§53.)
Since y = x~l° = 1/x10, we have here y+Ay = l/(z+Az)10,
1 1 xw-(x+Ax)l°
= -
' Ax (x-\-Ax)l°x10 Ax ,
The limit of the first fraction on the right as Ao»-0 is simply
1/x20. The limit of the other fraction is precisely the thing
we should have to find if we were differentiating z10, and hence
equals 10 x9. Hence the limit of Ay/ Ax in (11)" is
^=--kl03») = -10s-u.
dx x20
This is the same result as was obtained above from the
formula for y = xn. Hence that formula works correctly in
the case of y = x~10. ,
By precisely the same steps it can be shown to work in
the case of any negative power y = x~n. (Ex. 20 below.)
Hence we can differentiate many fractions without further
recourse to the increment method, — by simply regarding
the fractions as negative powers.
E.g., suppose that ?/ = 4/z100.
This is 4 times 1/x100, or 4x~100.
.'. ^ = -400*-'°', = -4™ [not -- L_l
dx xl0i I 400 zH
Notice how complicated this differentiation would be by the A-process.
Negative powers often arise in problems on maxima and
minima.
Ex. I. An open rectangular tank is to contain 500 cu. ft.
What is the least possible cost, if the base costs $3 per sq. ft.
and the sides $2 per sq. ft. ?
The base must be a square. (Cf. Ex. 7, p. 98.)
102 MATHEMATICAL ANALYSIS [III, § 68
The four sides contain 4 hx sq. ft. and cost 8 hx dollars.
The base costs 3 x~ dollars. Hence the total cost is
But as the volume is to be 500 cu. ft.,
A -5™ (12)
T is now expressed in terms of x alone. To minimize it :
.
dx x2
.'. x =^2000/3 = 8.74, approx.,
giving h = 6.55, and T = 687.5, approx.
Remark. To find the relative values of h and x in this tank, we may
best proceed thus :
A_500_500:c_ 500 x
xz x3 2000/3'
Simplified, this gives h = \ x; and T = 9 x\
EXERCISES
1. Differentiate, and write the results in tabular form :
(«)«-, 7^, -,*-, » ^ it, _'
, .
4 a: . 3 x6 x x2
2. For a certain weight of a gas the volume varied thus with the
pressure: F = 300/P. Find the rate of increase of V at P = 10. About
how much did V change while P increased from 10 to 10.137?
3. Gravitational acceleration (g ft./sec.2) varies thus with the
distance (r miles) from the center of the earth : g = 512000000/r2. Find
the rate of change per mile at r = 4000.
4. The intensity of light varies inversely as the square of the
distance from the source. If 7 = 120 when D = 10, write a formula for
HI, § 68] DIFFERENTIATION 103
7 at. any distance D. Also find the rate of change per unit distance
atl> = 20.
6. The electrical resistance of a wire varies inversely as the square
of the diameter x. If # = 20 when x=2, find the formula for R. Also
find the rate of change of R at re = 5.
6. The current in an electric circuit varies inversely as the resistance.
If c = 50 when R = 5, find the rate of change of c per unit change in R
at R = 10.
7. In Ex. 10, p. 34, express the total cost as 'a function of one
dimension alone, and find the minimum cost.
8. Find the most economical dimensions of an open rectangular box
which is to contain 400 cu. ft., if the base costs 5Q£ per sq. ft. and the
sides 30j£ per sq. ft.
9. The same as Ex. 8, if the base and sides both cost $2 per sq. ft.
10. An open cylindrical cup is to contain 125?r cu. in. What dimen-
sions will require the least material?
11. The same as Ex. 10, for a closed cylindrical can, to contain 250 IT
cu. in.
12. The same as Ex. 11, if the can is to have its top and bottom of
double thickness.
13. What number, added to twenty times its reciprocal, gives the
smallest sum? Check your result by calculating the sum for a few
near-by values and plotting roughly.
14. Find by using the derivative the smallest possible perimeter
for a rectangle whose area is to be 100 sq. in.
15. In Ex. 8, p. 34, express the area of the page as a function of the
length of the print lines. [Am., A =3 L+66+120/Z/.] Find the exact
minimum of A.
1C. Like Ex. 15, changing the area of the print column to 30 sq. in.
and the bottom margin to 2 in.
17. Like Ex. 10 above but containing 1 gallon (231 cu. in.).
18. A rectangular box with a square base is to contain 200 cu. in.
and have a cover cap which will slide on tight for 3 inches. Express
the total surface of the box and cap, — virtually two open boxes, —
and find its minimum value. [The resulting equation will need to
be solved by trial. § 21.]
19. Similar to Ex. 18 but with a volume of 800 cu. in., and a cap
which slides on to one fourth the way down.
20. Prove the differentiation formula (5) correct for any negative
integral power, y=x~n. [The steps are the same as for or10 above.]
104 MATHEMATICAL ANALYSIS [III, § 69
§ 69. Further Notation. The derivative of any function
f(x), being itself some function of x, is often denoted by
/'(z), read " / prime of x." Thus :
if/(x)=x4, then f(x)=4x*.
A derivative is often denoted also by writing — , or —, etc.,
ax dr
before the function differentiated. E.g.,
Thus, if y=f(x), the following notations are equivalent :
dx «f dx
§ 70. Repeated Differentiations. Suppose we have given
a formula for the distance (y ft.) traveled by an object in
t min., say
dist., 2/ = 20*3-*4. (13)
And suppose we wish to find the acceleration at any time, —
i.e., the rate at which the speed is changing.
Differentiating (13) gives the speed at any time :
speed, tf = 60£'-4J3.
But we wish to know how fast the speed is changing. Hence
we must differentiate the speed. (The fact that we have
already performed a differentiation in getting the speed
makes no difference.)
/. accel. = — = 120<~12Z2.
at
For instance, at t =*5, the acceleration is 120(5) -12(5)2 or 300 units.
That is, the speed is increasing at the rate of 300 ft. /min. gained per
min. This is abbreviated 300 ft./min.*
To make this calculation, we had to differentiate twice in
succession. If we had wished to know the rate at which the
Ill, § 71] DIFFERENTIATION 105
acceleration is increasing, we should have had to differentiate
a third time.
In general, to find the rate at which any quantity is changing,
we must find the derivative of that quantity, — no matter how
many differentiations may already have been performed in
getting that quantity.
There are many problems requiring repeated differentia-
tions. Another illustration follows.
(Ft,) 5 10
Ex. I. A steel beam loaded in
a certain way bends as in Fig. 37, ^
the deflection or ordinate at any
distance (x ft.) from one end being '0^
i/ = .00006(z3-15z2). M
Find how fast the slope is changing
per horizontal unit at x = 10.
Solution. Differentiating once gives the slope :
slope, I = .00006 (3 x2 - 30 x) . (14)
Differentiating again gives the rate at which the slope is changing :
rate, — = .00006 (6 x - 30) , = .0018 at x = 10.
That is, the slope is increasing at the rate of .0018 per horizontal foot.
To check this, let us calculate the slopes at x = 9 and a: = 11. By (14)
these are 1=-. 00162 and l = +. 00198. The increase is AZ = . 00360
in two horizontal units, making the rate .0018 per unit.
Definition. The rate at which the slope of a curve is changing at
any point is called the flexion.
§ 71. Derivatives of Any Order. The derivative of the
derivative dy/dx is called the " second derivative " of the
original function y=f(x), and is denoted * by -^ orf'(x).
Similarly the derivative of f"(x) is called the " third
derivative," and is denoted by f"'(x) or — ^. And so on.
ax*
* Observe where the indices 2 are written. We may think of these as
indicating that the operation denoted by — is to be performed twice upon y.
dx
106 MATHEMATICAL ANALYSIS [III, § 71
Ex. I. If y=xl°+l/x*, find d'y/dx3.
^ = 10 a* -2 or3
dx
^
dx2
^
dx3
EXERCISES
1. Find the second derivatives of the following :
y=x*+2x3+3x, t/ = 5/z10,
2. Find the fourth derivative of y = x5+6 z3-17 a: +2.
3. Find d7y/dx7 ify = .02 z10+zV3 + l/z.
4. For a ball thrown straight upward the height after t sec. was
y=8Qt — 1QP. Find the speed and acceleration at any instant. What
values at t = 1 ?
6. The same as Ex. 4 for the rolling ball in Ex. 4, p. 62.
6. In the curve y = x3 find the slope and flexion at z = 2.
7. The same as Ex. 6 for the suspension cable in Ex. 7, p. 45.
8. The same as Ex. 6 for the actual curve of the beam in Ex. 2, p. 95.
9. How fast is the slope of the curve y = x3— 2z-fl changing (per
horizontal unit) at 2 = 2? Check by finding the exact slopes at x =
1.99 and 2.01.
10. In t seconds after brakes were applied a train moved a distance
(s ft.) given by s=44< — 4 i2. Find how fast it was moving when
t =3. Also how fast its speed was then decreasing.
11. The distance traveled by an object in t minutes was y = 4t3 —
.1 V. Plot a graph showing y as a function of t from t = Q to 30, sub-
stituting 0, 5, 10, etc.
[12.] Plot a graph showing how the speed v increased with t in Ex. 11.
What were the maximum and minimum value of v, — by the graph, and
by calculation?
13. How would you proceed to find the rate at which the accelera-
tion of a moving object is increasing at any instant: (a) If given a
formula for the speed at any time? (6) If given a formula for the
distance traveled?
Ill, § 73]
DIFFERENTIATION
107
Graphs of
§ 72. Derived Curves. The slope of any given curve /
varies with x in some definite way. This variation is most
clearly shown by plotting another
curve /', whose height is every-
where equal to the slope of curve 'A»
/. (Fig. 38.)
Observe how the zero slope of
/at A, C and E shows in the
" derived curve " /' ; also the
maximum and minimum slopes
of /at B and D*
Further derived curves can be
drawn to show the variation of
/"(»), /'"(*)> etc., — and this is
often done in studying motion or
the bending of loaded beams. We may, then, interpret
}"(x) either "as the flexion of curve /, or as the slope of /',
or as the height of the second derived curve/".
§ 73. Points of Inflection. In drawing an accurate graph
it is helpful to locate the " points of inflection " (such as B
and D, in Fig. 38), where the curve has a maximum or
minimum slope. For near such points the curve is very
nearly straight, almost coinciding with its tangent line for
some distance.
Ex. I. Find the maximum and minimum slopes of curve/
in Fig. 38, if the height at any point is
(15)
FIG. 38.
y = x* — »zz-
The slope at any point is
ax
(16)
* Slope is not simply a measure of steepness. It has a sign, and at D
the slope of / has fallen to a negative value lower than that on either side
for some distance.
108 MATHEMATICAL ANALYSIS [III, § 74
To make this slope a maximum or minimum, we make its
derivative zero :
16 = 0. (17)
ax
This gives x = ± V4/3 = =*= 1 . 155, approx.
Testing x= — 2, —1, 1, 2, in (17) shows I to be a maximum at x =
— 1.155, and a minimum at x = 1.155. By (16) these maximum and
minimum slopes are 11.32 and —13.32; and by (15) they occur at
points B and D, having t/=8.19 and 5.96.
Observe that the derivative to be tested "before and after" is the
one which we set equal to zero.
§ 74. Maximum Rates. It is often important to know
when a quantity will be increasing most rapidly. This is very
different from the question as to when, the quantity will be
greatest.
For instance, in Fig. 38, the slope of curve / is greatest at
B, but is changing very slowly in that vicinity. Again, in
the same curve, the height is increasing most rapidly at B,
but is greatest at C. Another illustration follows.
Ex. I. The distance (?/ft.) traveled by an object in t rniu.
was
y = Qt*-t«.
Find when the acceleration was increasing most rapidly.
The question is not when the acceleration was greatest
but when its rate of increase was greatest.
6Z5, =speed,
at
-3(H4, =accel.
The rate of increase of the acceleration is
R = ^L = 360 t2 - 120 t3, to be max. (18)
at3
:. ^
at
Ill, § 75] DIFFERENTIATION 109
This gives t = 0 or 2. The latter value makes R a maximum,
for at *=land3: ^=+, -.
at
To find the maximum R, substitute t = 2 in (18).
Remark. It is helpful to label successive derivatives and introduce
a single letter (as R above) for the quantity which is to be a maximum
or minimum. Decide at the outset which quantity that is, and when
you reach it in differentiating set its derivative equal to zero.
EXERCISES
1. The distance traveled by an object in t min. was y = 6Qt*—t5.
Find when the speed was a maximum ; likewise the ^acceleration.
2. In Ex. 1 find when the acceleration was increasing most rapidly.
Check roughly by calculating the acceleration at several instants
before and after, and noting about how it increased.
3. Find where the slope of the curve ?/ = 18o;2— x* is a maximum.
Also where the slope is increasing most rapidly.
4. The height of a curve is y = 3Q x*—x5. Find where the slope is
a maximum, and where the flexion increases most rapidly.
6. The kinetic energy of a train t minutes after starting was E = 500 t4
+15 t5— t6. Find when E was increasing most rapidly and how rapidly
then.
6. A beam loaded in a certain way bends so as to form the curve
y = .00002 z4 - .003 x*. Plot from z=0 to x = 10. Show the exact
tangent at any point of inflection.
7. (A) Plot the curve y = x3 — 15 x, locating the maximum and
minimum y, and the point of inflection. What slope at the latter?
(5) Plot also the first and second derived curves.
[8.] Find dy/dx from y =((z7+l)2 after multiplying out. [Ans., 14 XB
(z7 + l).] Can you see any simple rule which would give this same
result without multiplying out?
§ 75. Indirect Dependence. Heretofore each function
that we have differentiated has been expressed directly in
terms of its independent variable, — say y in terms of x.
But there are cases in which y is given in terms of some other
quantity u, and u in terms of x. y is then in reality a function
of x, although expressed as such only indirectly.
110 MATHEMATICAL ANALYSIS [III, § 76
For instance, if y=u5 and u = x2, then y = xl°.
In this instance, it is easy to change from the indirect re-
lation between y and x in terms of u, to the direct relation
y = x10. And if we wish to know the rate of increase of y per
unit change in x, we simply differentiate this last equation,
getting dy/dx = 10 xg.
But it is sometimes very inconvenient to change to the
direct relation between y and x\ and we therefore need
some method of finding dy/dx even while y is expressed in-
directly in terms of u.
Notice the difference between dy/dx and dy/du. The latter would
be the rate of increase of y per unit change in u. We want dy/dx.
(To distinguish verbally between dy/dx and dy/du, we call one the
derivative with respect to x, and the other the derivative with respect
tow.)
§ 76. Differentiating a Function of a Function. If y is
given as a function of u, and u as a function of x, say
y = F(u), u=f(x), (19)
how can we find dy/dx immediately?
Increasing x by Ax would evidently increase u by some Au,
and hence y by some Ay. We seek dy/dx, the limit of Ay/ Ax.
But evidently
Ay = Ay,AUf (20)
Ax Au Ax
Taking the limits of these fractions (if the limits exist) gives
4y = dy , du (21)
dx du dx
That is, to find dy/dx in (19) above, we have merely to find
dy/du from the first equation, and du/dx from the second, and
then multiply the results.
Thus if y is increasing at the rate of 10 units per unit change in u,
and u at the rate of 6 units per unit of x, then y is increasing at the
rate of 60 units per unit of x.
Ill, § 77] DIFFERENTIATION 111
Ex. I. If y = u?+2 w4-3, and u = x*+l, find dy/dx.
Here ^ = 7u«+Su*, ^ = 5x*.
du dx
Hence by (21), = (7 ^6+8 w3)(5 s*).
It would be possible here to express y directly in terms of x, multiply
f out, and then differentiate. But this would be inconvenient.
Formula (21) will be used in deriving further differentia-
tion formulas ; and should be carefully memorized.
§ 77. Differentiating a Power of a Quantity. Let u de-
note any function of x, or quantity involving x. Then if
dy/du = nun~l, and hence by (21) above:
(22)
.
dx dx
That is, the derivative of any integral power of a quantity
equals the given exponent times the next lower power of that
same quantity, times the derivative of that quantity.
By this theorem we can now differentiate at sight any
integral power of a quantity, without first multiplying out
or using the A-process.
Ex. I. y=(xz-\-3x-7)m.
Here & = 100(z2+3 z-7)99 • (2 x+3).
ax
For any numerical value of x} this result can be calcu-
lated very quickly by logarithms. (Chap. VI.)
N.B. Here y is given directly in terms of x, but not as a power
of x. To differentiate by the short power-rule, we must regard y as
a power of a quantity u (=z2+3 x— 7) ; and thus from the practical
standpoint the case is one of indirect dependence, coming under (22)
or (21) above.
112 MATHEMATICAL ANALYSIS [III, § 78
Us.(M-*c)
Ex. II. y=— i.e., y
Here |-12(10-«)>4(-l), -^
(The factor — 1 comes in as the derivative of the quantity
in parentheses.)
Without formula (22) we should have had to resort to the A process.
EXERCISES
1. In the following cases of indirect dependence find dy/dx :
(a) j/=u3-6w+4, u=z2-l; (6) y=u*
(c) 7/ = u60-.002K10, u = 7x+3; (d) y = u™-75, u = x*.
2. In Ex. 1 (d) express y in terms of x, differentiate, and compare.
3. In the following find the derivative with respect to t :
(a) y=^f z = 3J4-17; (6)A=irr2, r = 25i?+t.
[4.] The edge of a cube (x in.) increased at the rate of .02 in./hr.
Can you find how fast the volume was increasing at the instant when
a: = 20? (Hint : We know dx/dt and wish to find dV/dt.)
6. Write at sight the derivatives of the following :
(a) T/ = (z*+25)100, (6) 2/ = (16-^)7,
(c) z = 5(2z+7)», (d) y = fr(l-x)8,
' 4(8-0' 4(9+*2)'
6. In Ex. 6, p. 49, find how fast n2 changes with /, at/ =30.
7. In Ex. 16, p. 57, express the total amount of heat as a function
of the distance x from A. Find the minimum value of H.
§ 78. Rate Problems Requiring Indirect Differentiation.
EXAMPLE : The edge (x in.) of an expanding metal cube is
increasing at the rate of .04 in./hr. How fast is the volume
increasing (per hour) at the instant when x= 10?
Ill, § 78] DIFFERENTIATION 113
This is in effect a problem of indirect dependence : We
know how V depends on x, and how x varies with the time t :
V = x*, and ^ = .04;
at
and we are to find how V varies with t at a certain instant.
V depends on t, — but through the medium of x, so to speak.
By § 76,
4V = dV _dx (23)
dt dx dt'
f-8*f- I
At the specified instant this becomes
dt
I.e., the volume is increasing at the rate of 12 cu. in./hr.
Notice particularly the factor dx/dt in (24). Evidently
this arises from the fact that we are differentiating with re-
spect to a different variable (t) than that in terms of which V
is expressed (viz. x). <
It is interesting to see how this extra factor enters if we differentiate
by the A process. During any interval At, x increases by some Ax and
V by some AV. Then
y-j-AF = (x+Ax)3=x3+3 x2Ax+3 x*x*+Ax*.
.'. A V = 3 x2Ax +3 xAx2 +Ax3.
Since we seek the rate per hour we must divide by At. This will not
cancel a factor, as dividing by Ax would ; but we can factor out a Ax :
At At
Now as At approaches zero, so does Ax. Hence the quantity in paren-
theses approaches j^r2 ; and the average rates Ax /At and A V /At approach
the instantaneous rates dx/dt and dV/dt, respectively.
.-. ^ = 3x2^.
dt dt
This shows the factor dx/dt in (24) coming in automatically. (Why
does it not enter when we differentiate V =x3 with respect to x?)
114 MATHEMATICAL ANALYSIS [III, § 79
This work suggests that in any case where we have one
quantity (y) expressed in terms of another (x), and we differ-
entiate with respect to the time, or any other third variable (t),
the result will be the ordinary derivative (dy/dx) multiplied by
an extra factor (dx/dt) :
if y =/(*), then f =/'(*) - g- (25)
This is, in fact, precisely what the general theorem on
indirect dependence shows, merely changing u and x in (21)
to x and t.
Hereafter, then, we may differentiate with respect to a
third variable t as just stated without explicitly employing
(21).
Ex.I. y=*«. Here ' !&=6z'^.
at at
Ex. II. S=4*r'. ^ Here ^ = 8 TTT^.
at at
N.B. Compare Ex. I carefully with this differentiation :
10 *».
In each case we are differentiating with respect to t a power of a quantity,
not simply t. In one case the quantity is (J10+l), in the other, x.
§ 79. Related Rates. It is profitable to look at the fore-
going problem of the expanding cube from another angle.
There we had given two related quantities V and x
V~af,
and the rate at which one of them was changing
dx/dt =.04,
and we had to find how fast the other was changing.
This is typical of many problems which arise in scientific
work. In any such case we simply differentiate the given
equation of relationship (like F = x3) with respect to t, and
substitute any given values.
Ill, § 79]
DIFFERENTIATION
115
Ex. I. From a conical filter whose height is three times the radius,
a fluid filtered out at the rate of .3 cu. in./min. How fast was the level
falling, when the fluid was 6 in. deep in the
middle? (Fig. 39.)
For the shrinking fluid cone we have
Given f=-.3;
To find — when h = 6.
dt
Now V = % trf-h.
But r = I h continually. (Why?) Substitut-
ing this :
. dV.
dt '
Substituting given values :
FIG. 39.
rvr» A
. . — = — .024, approx.
dt
The level was falling at the rate of .024 in./min.
EXERCISES
1. Find the derivative with respect to t of each of the following
quantities (assuming all the letters to vary with f) :
V=x3,
= 100/r3,
= 3.8s-s2,
2. The radius of a spherical balloon increased at the rate of .02
ft./min. Find how fast the volume was increasing at the instant when
r=20.
3. In Ex. 2 check your differentiation by the A process.
4. The volume of a cube was increasing at the rate of 600 cu. in./min.
at the instant when the edge was 20 inches. How fast was the edge
changing?
5. The height of a cone constantly equals the diameter of the base.
If the volume increases at the rate of 20 cu. in./hr., find the rate of
change of the radius when r = 2.
116
MATHEMATICAL ANALYSIS [III, § 80
6. Sand, falling at the rate of 2 cu. ft./min., forms a conical pile
whose radius always equals twice the height. How fast is the height
increasing when h = Wf [How do you account for the very small
answer?]
7. In § 79, Ex. I, take different dimensions and rates ; and solve.
8. A cylinder contracts so that its height always equals three times
its radius. If the volume is decreasing at the rate of 2 cu. in./hr., how
fast is r decreasing when r = 10?
9. A sphere is expanding at the rate of 12 cu. in./min. Find how
fast the radius and surface area are increasing when r = 10. About
how much will they increase in the next 6 sec. ?
10. The volume of a quantity of gas varied thus: F = 600/p. If
p increased at the rate of .2 Ib./min., how fast was V changing when
p = 20?
11. The volume (V cc.) of a kilogram of water varies with the tem-
perature (T° C.) thus: F = 1000-.05767:+.00756T2-.0000351 T3.
If T rises at the rate of .02 deg./min., how fast will V be increasing when
T = 50?
12. The volume of a balloon was increasing with the temperature
at the rate of 110 cu. ft./deg., when the radius r was 20 ft. How fast
was r then increasing, per degree?
§ 80. Differentiating Implicitly.
EXAMPLE: A ship A sailing east-
ward at the rate of 12 mi./hr. left a
certain point five hours before another
ship B arrived from the north coming
at 16 mi./hr. How fast was the dis-
tance AB changing two hours after
X A left?
Let Fig. 40 represent the positions
at any time. Then
12t
Fio. 40.
and we are to find dz/dt when a: = 24
and / = 48.
m, § 80] DIFFERENTIATION 117
At any time, t hr. after A left, we should have
x = 12t, y = SQ-Wt. (26)
.'. z= \/^H^= V6400-2560 J+400 t2.
We have as yet no formula for differentiating such a
function, this being a fractional power. But we can proceed
as follows.
Since z2 and x2+y2 are constantly equal, they must be
changing at the same rate. Hence their derivatives with
respect to t are equal :
But by § 77, differentiating z2 with respect to t gives 2 zdz/dt ;
etc.
Substituting the given values, with the corresponding value
of z, = V242-H82 or V2880 :
2V2880^ = 2(24)(12)+2(48)(-16).
at
. dz_ -480 _ OQ,
• • , ~ , _ : — o.»«j.
dt V2880
That is, the distance AB was decreasing at the rate of 8.9
mi./hr.
Remarks. (I) This method of finding dz/dt without first solving
explicitly for z is called implicit differentiation. Notice carefully the
reasoning involved ; also that the result would have been badly erro-
neous if we had overlooked the negative sign for dy/dt.
(II) When one side of a varying triangle remains fixed, its numerical
value should be used from the outset, rather than an unknown letter.
One term in the equation corresponding to (27) is then zero.
(III) The minimum value of z above was found in Ex. 8, p. 98,
by first finding the minimum of z2. We can now find it directly :
1 18 MATHEMATICAL ANALYSIS [III, § 80
Put dz/dt=*Q, and by equation (27) we must have
That is, 12 x — 16 t/ = 0. Or introducing t by (26) above :
12(12f)-16(80-160=0.
Solving this for t gives J=3.2, whence z = 48, — the minimum.
EXERCISES
1. In Ex. I above, just how does the equation 2 x dx/dt+2 y dy/dt =
2 z dz/dt follow from the one preceding ? What would the corresponding
differentiation give in case we had z2-{-2500 = z2 constantly? What if
we had z2+?/2 = 100 constantly?
2. A ladder 25 ft. long leans against a vertical wall. If its foot is
pulled away horizontally at the rate of .3 ft. /sec., how fast is the top
descending when 20 ft. high?
3. An airplane flying horizontally at the rate of 80 ft. /sec. passes
straight over a fort, at an elevation of 6000 ft. How fast is its distance
from the fort increasing 100 sec. later?
4. A launch is pulled upstream by a cable fastened to a bridge
60 ft. above. If the cable is pulled in at the rate of 6 ft./min., how fast
will the boat be advancing when 100 ft. of cable are out?
6. A balloon B was descending straight over a railroad track at the
rate of 60 ft./sec. An engine E was approaching at the rate of
80 ft./sec. ; but was 1800 ft. away from the point directly below B,
when B was 2600 ft. high. How fast was the distance BE changing
10 sec. later? When was BE least, and how small?
6. An auto running -constantly 60 ft./sec. passed directly under a
balloon just as a bomb was released. If the height of the bomb after t
sec. was /i = 800 — 16 i2, how fast was the distance between the bomb
and the auto changing when t — 5 ?
7. In Ex. 6 when were the bomb and the auto nearest?
8. The baseball "diamond" is a square 90 ft. on each side. A ball
was batted along the third-base line with a speed of 100 ft./sec. How
fast was its distance from first base changing .2 sec. after starting?
9. The same as Ex. 8, .5 sec. after starting, if the ball was moving
80 ft./sec.
10. A train running straight east at the constant rate of 40 mi./hr.
left a town 4 hours before another arrived from the north, coming ut the
Ill, § 81] DIFFERENTIATION 119
rate of 30 mi./hr. Find how fast the distance between the trains was
changing two hours after the first started.
11. In Ex. 10 when were the trains nearest? How near?
12, A rectangle is inscribed in a circle of diameter 20 in. If we
increase the base at the rate of .4 in./min., how fast will the altitude be
decreasing when equal to 12 in. ?
§ 81. Differentiating Fractional Powers. The formula for
differentiating un applies to fractional as well as integral
powers.
For instance, if
then, -*
dx dx
For here i/3 = w5; and differentiating this implicitly with
respect to x (§ 80) would give :
.
dx dx
. dy.-5 u^du
" dx~*y* dx
But by (25), y* = (uty = u**.
• <fy^s u* du^^idu
' dx ^-dx dx'
as stated above.
By precisely similar steps, the formula for un can be proved
to work in the case of any fractional power y = up/q.
Irrational quantities can therefore be differentiated by
first writing them as fractional powers.
Ex. I.
Ex. II.
120
MATHEMATICAL ANALYSIS [III, § 82
Ex. III. Problem 10, p. 121, involves finding the minimum value of
a function like the following :
180-s
Here
Equating this to zero (§ 65), transposing, squaring, etc., we find
x = 5, 2/=|+25=26f.
5
Testing dy/dx at 4 and 6 shows this value of y to be a minimum.
§ 82. Abrupt Extremes. The graph of
falls sharply to a minimum height of 1 at x = 0, and then rises
sharply. (Fig. 41.) It does not have a horizontal tangent at
the lowest point.
As always, the slope dy/dx is
negative just before the minimum
and positive just after it. But here
dy/dx changes from — to + by
" jumping," — not by going through
zero. The slope becomes indefi-
nitely great, positively on one side
and negatively on the other, as we
approach the point. The " derived
curve " (§72) would have a break
at x = 0, though there is no break in
the original graph.
The only powers of x which allow this combination at a
maximum or minimum are fractional powers.
\
Grai
\
ihof
near
\
x=o
/
•M
/
/
-1 0
Fio. 41.
Ill, § 83] DIFFERENTIATION 121
EXERCISES
1. What is the meaning of : z^, #2, x-^t x~z?
2. Differentiate each of the powers in Ex. 1, and express the results
in a form free from fractional powers.
3. Express as powers, and then differentiate :
4. In Ex. 11, p. 45, find how fast T increases with L at L = 9. About
how much longer does a 9.1-inch pendulum require to swing than a
9-inch pendulum?
6. In Ex. 7, p. 50, find about how much a planet's time of revolution
would increase if x increased from 1.52 to 1.54.
6. The volume of water (V gal./hr.) flowing through a pipe of
any radius (r in.) under a certain pressure is 7=630 r*. About how
much greater is V if r = 4.06 than if r = 4?
7. The distance of the horizon at sea (D mi.) varies thus with the
observer's elevation (E ft.) above the water: D = 1.22E*. Approx-
imately what change in D, if E increases from 100 to 102? s
8. In Ex. 10, p. 45, how fast does V increase with s at s = 100?
9. Differentiate the following fractional powers of quantities (§ 77) :
(a) y = (s» + D, (6) y = (*-), (c)
(d) z = 5ll6-z2, (e) s = 8175-z, (/) u = 16" xz -10 x +7,
(fl »~ 20°
10. Two towns A and B are 44 mi. apart on a, straight coast ; an
island C is 12 mi. out, directly off A. The trip from C to B is to be
made by a launch and auto, meeting somewhere along the shore, say
x mi. from A. If the launch goes 12 mi./hr. and the auto 20 mi./hr.,
what is the shortest possible time for the trip?
11. Plot y = 7-x$ from x= -8 to x = 8. (Take z=8, 2/, 1, fr, etc.)
Note the maximum.
12. Plot another graph showing how dy/dx varies with x in Ex. 11.
§ 83. Differential Notation. In some work it is con-
venient to be able to deal with a derivative dy/dx as a frac-
tion dy+dx. This can be done by giving suitable meanings
to dy and dx separately :
122
MATHEMATICAL ANALYSIS [III, i 84
Let dy and dx denote any two quantities, large or small,
whose ratio dy + dx equals the derivative f (x) or dy/dx. That is,
^derivative =, fraction.
dx (dx)
The quantities dy and dx are called differentials.
An equation like
dx
(28)
(29)
may now be written also in the form
dy = 3x*dx, (30)
by simply multiplying through by dx.
Treating a derivative as a fraction allows great freedom
of operation. For instance, a product like
dy du
du dx
may be simplified by merely canceling du. The value
(dy/dx) thus obtained is correct by the theorem on indirect
dependence, (21), p. 110.
Differentials are used very extensively in more advanced courses.
Here, however, we merely need to know their meaning and the fact
that a differential equation like (30) above is only another way of
expressing the value of a derivative, as in (29).
The following concrete interpreta-
tions of differentials may, nevertheless,
be of interest.
§ 84. Interpretations of Dif-
ferentials. In Fig. 42 the ratio
of HT to PH equals the rate at
P, — i.e., equals f'(x). Hence
HT and PH may be taken as dy
and dx, respectively.
Now HT is the amount that
y would increase while x in-
Ill, § 85] DIFFERENTIATION 123
creased by PH if the rate remained constant. Hence we
may say :
The differential of a function y is the amount that y would
increase while x increased by any amount (dx) if the rate re-
mained the same as at the instant considered.
Moreover, if dx is small, dy is nearly equal to Ay .(or HQ). But by
(28), dy =f'(x)dx. Hence Ay, the small change in y produced by a small
increase in x, is approximately equal to the derivative times the latter
increase. (Cf. § 62.)
EXERCISES
1. Differentiate the following, finding dy :
2. Write the differential of V, if V = x*. Verify that either dV/dx
or dV/dt is obtainable correctly from this by simply dividing.
3. Simplify by inspection: .
dV dx dQ dr dy_ du d/p dv
dx'dt' dr' dt' du dx dv ' It'
4. By what would you multiply ^ to get ^?
dx dt
^togetfr? ^toget-^? f^toget^?
ds dx dr dt dx da
6. Express in an equivalent form each of these statements :
dy=x*dx, dx = 3t*dt, dV = ±Trr*dr.
§ 85. Summary of Chapter III. To find the rate at which
any given quantity is increasing at any instant, we have
merely to find its derivative, — i.e., to differentiate.
The derivative is defined abstractly as a certain limit,
but it may have various concrete interpretations, such as
instantaneous rate, slope, speed, etc.
Derivatives can often be written at sight. In such cases
it is easier to find instantaneous rates than average rates.
(Cf. § 33, Remark.) Indirect differentiation with respect to
a third variable, say t, is possible. This introduces an extra
factor, dx/dt.
124 MATHEMATICAL ANALYSIS [III, § 85
The amount of change in y while x changes by a small
dx is approximately dy or f'(x)dx. The effect of an error
of measurement can be estimated in this way.
To locate the maximum and minimum values of any
quantity, we set its derivative equal to zero, and test the
same derivative before and after. (We also see whether
the derivative can change from + to — , or vice versa, by
"jumping.") In practical problems it is first necessary
to obtain a formula for the quantity to be maximized or
minimized, — and express this in terms of a single variable.
To find when a quantity is increasing most rapidly, — i.e.,
at the greatest rate, — requires repeated differentiations.
So do various problems on acceleration, flexion, etc.
Observe that although we can find the slope of a tangent line, we
cannot .as yet find its inclination, — i.e. the angle at which it rises
from the horizontal. This problem will be treated later. (§111.)
EXERCISES
1. What is meant by the derivative of y with respect to z? Explain
its significance as slope, rate, and speed.
2. Differentiate y = & — 10 x +5 by the A process.
3. Differentiate at sight :
(a) y=x5+|^-^/6+4; (6) z = 12^+2/v^;
» y = Vp +20* +85/30; (d) w = 5/x* -7/(10-z2)2.
4. A rectangular box is to contain 60 cu. ft. The materials for
the base cost 30ji per sq. ft., for the sides 10?f per sq. ft., and for the
top 20 jf per sq. ft. What is the smallest possible total cost, and what
dimensions will give it?
6. The space within a quarter-mile running track consists of a rec-
tangle with a semicircle at each end. To make the rectangle as large
as possible, how much of the quarter-mile (440 yd.) should be given to
the straight sides and how much to the curved ends?
6. The radius of a sphere is increasing at the rate of .04 in./hr.
How fast is the volume increasing at the instant when r = 30?
7. T|H> volume of a cube is increasing at the rate of 12 cu. in./min.
How fast is the total area increasing when the edge equals 20 in.?
Ill, § 85] DIFFERENTIATION 125
8. The edge of a cube is measured as 20 in. If this may be erroneous
by .02 in., about how inaccurate may the calculated volume be?
9. The radius of a sphere is measured as 5 in. About how accurate
must this measurement be, if the calculated volume is not to be erroneous
by more than 10 cu. in. ?
10. The height of a certain curve is y = 2Q x*—x5. Find where the
slope is a maximum and where the flexion is increasing most rapidly.
11. Plot the curve y=x4-\-l2 £3+2, showing any points of maximum
or minimum height or slope. What is the flexion at x = 1 ?
12. The speed of an object varied thus: v = 20ts — £4. Find when
the acceleration was increasing most rapidly, and how rapidly then.
13. A baseball is batted along the third-base line, going 60 ft. /sec.
How fast is its distance from first base changing one balf second after
it started? How fast when passing third base?
14. In Ex. 10, p. 118, change the rate of the eastbound train to
60 mi./hr., and of the other to 25 mi./hr., and solve.
15. In Ex. 14, find when the trains were nearest.
16. In each of the following, first write a formula for the variable
quantity, or function, in question ; then find the required rate.
(a) The velocity of flow of a gas escaping through a small hole
varies inversely as the square root of the diameter, D in. If F = 80
cu. in./hr. when D = .01, find how fast V changes, at D = .04.
(6) The "moment of inertia," /, of a flat circular disk varies as the
fpurth power of the radius. If 7 = 20 when r = 2, find how fast I in-
creases with r, at r = 5.
(c) In a railway curve the elevation of the outer rail should vary
inversely as the radius. If # = 2.5 (in.) when 72 = 3000 (ft.), find how
fast E changes, at 5 = 3600.
17. A bullet was fired straight upward, its height after t sec. being
2/ = 1600£— 16 <2. How fast did it start up? What was its greatest
height? When did it strike the ground, and with what speed?
[18.] The height of a ball t sec. after being thrown straight up was
y = 20 +802 — 16 £2. Show that the acceleration =-32 ft./sec.2 con-
stantly. What terms in this formula might have been different with-
out modifying this result? Write some other formulas to illustrate this.
[19.] A ball rolls up an incline, its distance from the bottom after
t sec. being x = a+bt — 5t2. Show that the acceleration = — 10 ft./sec2.
For what values of a, b, would x equal 10 and the speed 100 at t = 0?
[20.] The speed of an object after t seconds varied thus : v = 60 t2 — 4 i3.
Can you find the distance traveled at any time?
CHAPTER IV
INTEGRATION
THE RATE PROBLEM REVERSED
§ 86. Differentiation Reversed. We have seen how to find
the rate at which a given quantity is changing at any instant.
Consider now the reverse problem :
Given the rate at which a quantity is changing, to find how
large the quantity will be at any time.. (Of course, we must
also know how large it was at some particular time.)
If the given rate is constant, the problem is merely one of
arithmetic. But if the rate varies, we must in general pro-
ceed as follows :
The given rate is the derivative of the quantity whose
value is required. Hence we are given the derivative of a
function, and are to find the function itself. That is, we must
reverse the differentiation process.*
Ex. I. A bomb was dropped from an airplane 8000 ft.
high : t seconds later its height (h ft.) was decreasing at the
rate of 32 t (ft./sec.). Find the height at any instant.
The given rate or derivative is*
(Why-?)
at
To find h, then, we must think of some function which, if
differentiated, would give —32 t. One such function is
* Of course we might solve the problem approximately by some graphical
method. See Ex. 11, p. 131.
126
IV, § 87] INTEGRATION 127
But there are others. For instance, h — — 16 £2+500, and
h = — 16 t2 — 40, both have this same derivative — 32 1. So does
fe=-16«2+C, (1)
C being any constant whatever, positive or negative.
In other words, the given rate of change of h does not by
itself determine the value of h at any time. But we were
told also the height of the bomb at the start, viz. /i = 8000 at
£ = 0, when we began to count time. This fact requires the
constant C in (1) to have the value 8000 ; and (1) becomes
ft =- 16*2+8000. (2)
Check : This formula is a correct solution of our problem.
For at f = 0 it reduces to /i = 8000, and by differentiation
dh/dt = — 32 t, the specified rate.
It is instructive to compare the foregoing problem with one in
which the given rate is constant :
Ex. II. A captive balloon is being pulled down at the rate of
60 ft./min. How high will it be t min. hence?
Evidently, h = C — 50 t, where C denotes the present height, whatever
that may be. In other words, C is the value of h at t=Q, when we
begin to count time.
In each problem the value of h at any instant is completely determined
by the original value and the rate of change at all times.
§ 87. Integration. The process of reversing a differentia-
tion and finding the original function, when given its de-
rivative, is called integration. And the required original
function is called the integral of the given derivative.
When integrating, we must always add an arbitrary con-
stant C. For the given rate determines only the amount of
increase. And the total value of the function at any time
equals this increase plus the original value.
The value of this added " constant of integration " be-
comes definite if the value of the function is known at some
instant or point. For instance, in Ex. I of § 86, the value of
128
MATHEMATICAL ANALYSIS [IV, § 88
C was determined by the fact that h = 8000 at t = 0. Another
illustration follows.
Ex. I. The water in a rotating pail,
of radius 5 in., has its upper surface
hollowed out, forming a curve whose
slope at any distance x in. from the axis
of rotation is .2 x.* (Fig. 43.) If the
water is 8 in. deep at the highest point,
find its depth at any other point.
We are to find y as a function of x,
having given
PJQ 43 -p = .2 x, and y = 8 when x = 5.
We therefore seek a function which, if differentiated, would give .2 x.
One such function is .1 x2. But any constant might be added, making
7/ = .lz2+C. (3)
By the problem, however, y = S when z = 5. In (3) this gives
or C = 5.5. And hence the required depth at any distance x in. from
the center is
N.B. Even here C is the value of y at z = 0, as substitution would
show.
EXERCISE
[1.] Find by inspection a function which if differentiated will give
dy/dx = x3. (Check your answer.) Likewise for dy/dx = xl°, and xn.
Hence, integrating xn with respect to x will give what result?
§ 88. Integration Formula. To save labor let us systema-
tize the integration process.
Differentiating any power of x leads to the next lower
power. Hence integrating leads to the next higher. Thus
gives
(4)
* It is shown in Physics, by considering the forces involved, that for any
constant speed of rotation, the slope must equal some constant tinuAs x.
IV, § 88] INTEGRATION 129
The coefficient l/(n+l) is required to cancel the multiplier (n+1)
which would come from the exponent in differentiating.
By this formula, we can integrate various powers at sight
Thus
g-*» gives y-g+C;
=ar< gves 2/=
*-** gives if-^f
These last two results can be simplified, giving
-+c' and
These integrations should be checked by differentiating the results.
Formula (4) fails, however, if n= — 1. For then n-f 1=0
and cannot be used as a divisor. (It is clear anyhow that
x~l could not be obtained by differentiating the next higher
power x°.) The integral of a;"1 must be some other kind of
function, not a power of x. (Treated in § 178.)
Remark, The effect of a constant multiplier is simply to multiply
the resulting integral by the same factor. For instance :
^ = 20 x« gives y=2Q(%}+C.
But to integrate the product of two variable factors we must first mul-
tiply out. (Cf. § 60.)
To integrate the sum of several terms, we integrate term by term.
It is unnecessary to introduce a constant for each term integrated.
For a single C can have any value whatever, — say the sum of the
values which several C's might have.
130 MATHEMATICAL ANALYSIS [IV, § 89
§ 89. Uniqueness. If to an integral of a given derivative
any constant be added, the result will still be an integral.
For the constant would disappear on differentiating.
But may there not be some entirely different function
which would also be an integral, — perhaps a very compli-
cated function, whose derivative would simplify down to
the given quantity? No, this is impossible. In other
words:
Two functions which have the same derivative can differ only
by a constant.
For if the derivatives are equal, the functions must be changing at
the same rate ; and hence their difference is not changing but remains
constant. One function equals the other plus a constant.
Thus an integration can give only one result, aside from the possi-
bility of an added constant.
EXERCISES
1. Given each of the following values for dy/dx, find y itself, in-
cluding the arbitrary constant. Check each answer by differentiation.
(a) dy/dx=x*, x5, x, 5 x2, -§ z4, \x, 20,v -8,
(6) dy/dx =x$, z*, z*, 7z*, |Vz", -^x, 28v^+2,
(c) dy/dx = x~*, x-*, 1/z2, 5M -2/z3, fc/z», -7/2 z8+fc,
(d) dy/dx =x~*, x~l, or1-41, 1/Vx, -5/^z, 12/V^, fc/^z1".
2. Similarly find y if dy/dx = 5 z»-9 z+10vT-4+9/z2.
3. In the following, find y as a definite function of x, determining
the constant of integration :
(a) dy/dx=x*-7x+W, and ?/ = 100atz=0,
(6) dy/dx = (V^+25-Qa?)dx, and y= 50atz = l.
4. If dV=x*dx and V = 0 when z = 10, express V as a function of x.
6. The weight of a column of air whose cross section is 1 sq. ft.
and height fe h ft. increases with h approximately as follows : dW/dh =
.0805 — .00000268 h. Express W as a function of h, knowing that
TF = 0 when h =0. Find the weight of a column 10,000 ft. high.
6. The slope of a certain suspension cable at any horizontal distance
x ft. from the center is .004 x. Find the height (y ft.) at any point
IV, § 90]
INTEGRATION
131
if y = 20 at z=0. Plot the curve from x= -80 to +80, at intervals
of 20.
7. In Ex. I, § 87, change the slope to .3 x and the depth at the
cente to 4 in., and find y at any point. Plot from x = -3 to +3.
8. The speed of a car t sec. after starting was 60 t2- 4 t3. Find a
formula for the distance (y) traveled at any time. Calculate y when
t = W.
9. When a car had run for t sec., its momentum M was increasing
at a rate equal to (3 P+2, t). Find M at any time ;' at t - 10, 20, 30.
10. At a certain instant a quantity had the value 500, and t min.
later it was increasing at the rate of .6 Z2. Find its value at any time.
11. Table I shows the rate (R Ib. per hr.) at which a piece of ice
was melting t hours after being cut. Find the total amount which
melted during the five hours. (Hint: Plot a graph whose height
shall represent the rate of melting at any instant. What, then, will
represent the average rate during any hour?)j
TABLE I
t
R
t
72
0
0
3
21
1
9
4
24
2
16
5
25
12. In Ex. 1 1 the formula for R at any time is R = 10 t — t?. Calculate
the quantity melted from < = 0 to < = 5, and check your graphical result.
(Hint : What relation has the given rate R to the quantity f )
§ 90. Repeated Integrations. In some practical problems
it is necessary to integrate several times in succession.
To determine the value of the constant of integration
which enters at each step, we must know the numerical
value, at some instant or point, of the quantity represented
by the integral obtained at that step, — say, flexion, slope,
or height of a curve; or acceleration, speed, or distance
traveled by a moving object ; etc.
76
44
12
-20
-52
132 MATHEMATICAL ANALYSIS [IV, § 91
§ 91. Projectiles, Thrown Vertically. When an object
is thrown straight upward, its speed v decreases by 32 ft./sec.
in each second.* Thus if the speed is
140 ft./sec. at the start, after 1 sec. it will '
be 108 ft./sec.; etc. (See table.) After
5 sec. the object will be falling with a speed J
of 20 ft./sec., then 52 ft./sec., etc. Calling 2
these downward speeds negative, we can
say that v is still decreasing algebraically,
though increasing numerically. 6
That is, the acceleration or rate of change
of v, is —32 ft./sec2. whether the object is rising or falling.
To express this fact mathematically, recall that the speed
is the rate at which the height (y ft.) is changing : v = dy/dt,
whence the acceleration is
f=-32. (5)
By remembering this one simple equation, we can solve all
ordinary problems concerning vertically thrown projectiles,
if we understand the process of integrating. Separate for-
mulas for upward and downward motion, as commonly used
in elementary physics, are unnecessary.
Ex. I. A bomb was thrown straight down from a height of 2000
ft. with an initial speed of 80 ft./sec. Find its height t sec. later.
Integrating (5) gives the speed at any time :
at
But the speed was —80 at the start : i.e., dy/dt = — 80 at i =0. Hence
C=-80.
/. ^> = -322-80. (6)
at
Integrating again :
?/=
* See Remark I, § 24. Obliquely thrown projectiles are discussed in
§5 191, 222.
IV, § 91] INTEGRATION 133
But the height at the start was t/ = 2000. Hence fc = 2000. Thus,
finally,
y = 2000-80^-16i2. (7)
Observe the physical meaning of this result: The height at any
instant equals the original height (2000) minus the distance (80 0
the bomb would have fallen in t seconds if it had kept on at the original
speed "(80 ft. /sec.), and minus also the distance (16 22) that gravity
would pull it down in t seconds starting from rest. •
EXERCISES
1. Point out the physical meaning of formula (6) above, as has
just been done for formula (7).
In each of the exercises 2, 4, 5, 6, 10, 11, below, after getting the formulas
for dy/dt and y, point out the physical meanings of the terms in each.
2. A ball was thrown straight up from a roof 60 ft. high with an
initial speed of 80 ft. /sec. Find its height after t seconds, and also
when it struck the ground.
3. On a vertical line mark and label the positions of the ball in
Ex. 2 after 1 sec., 2 sec., etc., until it struck. Also plot a graph show-
ing how y varied with t during the flight.
4. A projectile was fired straight up from an airplane 2000 ft.
high with an initial speed of 1600 ft. /sec. Find when it was highest
and how high. When did it pass a balloon 10,000 feet high?
6. A bomb was dropped from an airplane 3000 ft. high. When
did it strike the ground and with what speed?
6. The same as Ex. 5 if the bomb was thrown down with an initial
speed of 50 ft. /sec.
7. In Ex. 6 mark on a vertical line the positions of the bomb after
1 sec., 2 sec., etc., until it struck.
8. A point on the rim of a flywheel was moving at the speed of
60 in. /sec. when the power was cut off. The speed thereafter decreased
at the rate of 4 in. /sec2. Find when the wheel stopped and how far the
point moved in stopping.
9. The same as Ex. 8, if the rate of decrease was —6 ft. /sec2., and
the original speed 75 ft. /sec.
10. A ball was thrown straight up from a window 96 ft. high, with
an initial speed of 60 ft. /sec. Express the height after t seconds.
When was the ball highest and how high? When did it reach the
ground, and with what speed?
134 MATHEMATICAL ANALYSIS [IV, § 92
11. A bomb dropped from an airplane struck the ground in 10 sec,
How high was the plane?
12. The acceleration of a point on the rim of a flywheel was <Py/dfr
™ .3 t — .01 J2, where y is in feet. Express y as a function of t if the wheel
starts from rest. Find the maximum speed, and the distance traveled
before reaching that speed.
13. For a beam loaded in a certain way, the flexion at any point
is dVdz*= -.00001 (z2 +21 a; -108). Express y in terms of z, if y =0,
and slope =.00288 at x=0.
14. Find the distance (y ft.) traveled by an object in t sec., if the
speed was 100 and the acceleration —30 at t = Q, and if d?y/dtz = 120 —
24 t at any time.
16. Find y if dy = (xl2-3 z»+7-l/Vs8+5/z4)<k and */=0 when
s = l.
[16.] Plot y = xz from x = Q to z = 4. Measure approximately the
area A under the curve from x = 0 to x = 2. t About how much larger
would A be if it reached to x = 2 + Arc, Ax being very small ? If the right
boundary is moving along, how fast is A increasing, at z = 2? At any
value of x? [Ans., dA/dx = xz.] From this result can you find the
value of A to any boundary (x) ?
§ 92. Integral Notation. The integral of any function
F(x), with respect to x, is denoted by the symbol
For instance, J xzdx denotes the integral of x2 with respect
to x.
This is usually read simply "The integral of x square — dx."
In this notation the integration formula (4), p. 128, reads:
rn+l
± -+C. (8)
71 + 1
This fails, however, if n=— 1 (p. 128). That case is treated later
(§ 178).
A constant multiplier simply multiplies the result. E.g.,
= 4
IV, § 93]
INTEGRATION
135
Thus a constant factor can be moved from one side of the
integral sign to the other. Variable factors, however, can-
not be so moved, for a product of two variable factors cannot
be differentiated by merely differentiating one factor.
Observe that the expression following the integral sign J is in the
differential form (§ 83). That is, the sign J stands for a quantity
whose differential is whatever follows.
Thus f x*dx stands for the quantity whose differential is~x*dx. But
of course this is the same thing as the quantity whose derivative with
respect to x is xz.
§ 93. A Growing Area. Many geometrical and physical
quantities can be calculated quickly and exactly by integra-
tion. The underlying idea is
much the same in all cases.
Let us consider first the typical
case of the area under any given
graph, — supposing of course
that the graph is free from
breaks, so that an area is actu-
ally bounded.
In Fig. 44 if CD is fixed and
PQ moves to the right, the area
A will vary with x in some def-
inite way. If we can determine
its rate of increase, dA/dx, we
can find A by integrating.
To get a vivid idea of the growing area A imagine a rubber sheet
with one end fastened at CD and the other end PQ being pulled along,
while the sides, attached to wires, constantly fit along the base line
and the curve. The area of the stretching sheet is the "growing area"
A, of which we are speaking.
While x increases by Az, A increases by
AA, =area of strip PP'Q'Q,
FlG-
136 MATHEMATICAL ANALYSIS [IV, § 94
where y is some average ordinate (read " y bar ").
/. ^-y.
Ax
This is the average rate of increase of A per z-unit. Its
limit is
^= L ®)=-pQ=y. (9)
OX Aaj-X)
That is, the rate at which the area A is increasing at any
instant is equal to the height of the curve at that point*
§ 94. Areas Found Exactly. Since the rate of growth of
the area A in Fig. 44 is dA/dx = y, we have simply
A=fydx. (10)
Thus we can find the area under any graph, if we know a
formula expressing the height y at any point in terms of the
horizontal distance x from some fixed point, and if we can
integrate the expression ydx.
To illustrate, let us find the area under the curve y = xs be-
tween a fixed ordinate erected at x= 1 and a moving ordinate.
Equation (10) becomes in this case
(11)
This is our growing area, and it is to start at x = 1 . That is,
A — 0 when # = 1.
Substituting these values in (11) gives
0=Kl)4+<?, or C=-i.
Hence the required area from x = I to any other value of x
is simply
* This result is reasonable. For suppose the ordinate PQ to move say
.001 inch to the right. Evidently the tiny strip added to the growing area A
Would be almost exactly .001 y (sq. in.). The average rate of growth during
this tiny interval would be practically y (sq. in. per horizontal inch). The
instantaneous rate is exactly *j.
IV, § 95] INTEGRATION 137
E.g., from x = l to x = 3, we have A =£(3)4— 1 = 20. Likewise,
from x = 1 to x = 10, A = £(10)4 — J = 2499f . And so on.
If we wished an area under this curve starting at some other
value than #= 1, however, we should have to redetermine the
constant C accordingly.
Observe, then, that we find an area between two fixed
ordinates by regarding it as the value to which a varying
area will grow, starting from one of the ordinates.
EXERCISES
1. Plot from x = 0 to 10 a line whose height at any point is y =2 x+5.
Find by elementary geometry the area under it. Calculate the same
area by integration.
2. Find by integration the area under the line 7/=3z+4, from
x=2 to 8. Can you check this result by geometry without plotting?
3. Find the area under the curve y=x* from x = 5 to any other xj
from x = 5 to x = 10.
4. Find the area under each of the following curves :
(a) 2/=z3-6z+8, from z = 2 to 4;
(6) 7/=z10--9z2 + ll, fromz=0to 1;
(c) y = Vx, 4 to 9; (d) y = #x, I to 8 ;
(e) ?/ = lM 2to5; (/) y = l/Vx, 1 to 4.
6. In a certain curve the height varies as the square of the hori-
zontal distance x from a certain point, and is 15 at x = 2. Find the
area under the curve from x = 1 to 10.
6. Plot that part of the curve y = l2 x—3 x2 in which y is. positive.
Calculate the area under it.
[7.] An object is moved against a force (F Ib.) which varies thus
with the distance x ft. : F = 8 x— x2. Plot F as a function of x from
x =0 to x = 8. Find graphically the work done from 0 to 8. Can you
calculate this work exactly?
§ 95. Momentum. Consider the momentum M imparted
to a moving object by a varying force in t seconds after
starting. (§ 15.)
In an additional interval AZ, further momentum AM is im-
parted. This equals the average force F acting during A£,
138 MATHEMATICAL ANALYSIS [IV, § 96
multiplied by the time A£ : AAf = F&t. Hence the average
rate of increase of M per sec. is
As AJ-M), F approaches the value of F at the instant con-
sidered :
dt
.: M= I Fdt. (12)
This agrees with our earlier statement (Ex. 6, p. 68) that M is rep-
resented by the area under a force-time graph. For y in that graph is
F and x is t, so that the area A = \ ydx becomes A = \ Fdt.
Ex. I. Find the momentum generated from t = 3 to any other
instant if the force varies thus: F = 16 — .6 t2.
; By (12) : M = J (16 - .6 P)dt ;
i.e., M = 16t-.2t*+C.
And, since M = 0 when t=3, we find on substituting: C= —42.6.
§ 96. Work. Consider the work done by a variable force
in moving an object any distance x.
In an additional distance Ax, additional work ATF is done.
This equals the average force F acting during Ax, multiplied
by the distance : ATF = FAz.
. ATF -^
" ^ = F*
The instantaneous rate at which W is increasing is therefore :
m=F
dx
/. W= fFdx. (13)
This indicates that W is represented by the area under a force-
distance graph. For there y = F and ( ydx becomes f Fdx. (Cf . Ex. 6,
p. 28.)
IV, § 97]
INTEGRATION
139
§ 97. Volumes. Let Fig: 45 represent any solid, and let
V be the volume between a fixed plane CD and a moving
plane PQ at a varying distance x
from some fixed point.
While x increases by Ax, V in- -y
creases by some A V :
= volume of slice
where A (" A bar ") is the aver-
age cross-section area in the slice.
Hence the average rate of in-
crease of V per x-unit is
FIG. 45.
Ax
L
Aa->0
area
= At.
(14)
That is, £/ie rate o/ increase of V at any instant equals the area
of the cross-section A8, at the point just reached.
7=
(15)
Hence we can find the volume of any solid, if we can do
two things : (1) Express the area of a moving cross-section
in terms of its distance x from some fixed point and (2) in-
tegrate the expression thus obtained.
Ex. I. Suppose the moving section in Fig. 45 to be a circle and its
radius to vary thus : r — .05 x2. Find the volume between two planes
CD and LK located at z = 3 and a; = 10, respectively.
A.=irr2=ir(.05 Z2)2 = .
V = f .0025 Ta*te
This is the growing volume between some fixed plane (to be taken
140 MATHEMATICAL ANALYSIS [IV, § 97
at CD) and the moving plane PQ. When PQ was just starting from
CD, the volume was zero: F=0 at x = 3.
.'. 0 = .00057r(3)5+C.
This requires C= — .1215 IT', and the growing volume is
V = . 0005 ITS*-. 1215 IT.
When PQ reaches LK, z = 10 and F=50r-. 1215 ir = 156.7 approx.
EXERCISES
1. The force (F Ib.) applied to an object varied thus : F = 120 t— 6 J2.
Find the momentum generated from t*=l to < = 4.
2. Like Ex. 1 for a force varying in each of these ways :
(a) F = 40J, < = 5to20; (b) F = 3.6*+20, < = OtolO;
(c) F = 125-*3, *=0to5; (d) F = l%Vt+l5, < = lto9.
3. The force (F Ib.) required to stretch a certain spring x inches is
F = 20 x. Find the work done in stretching it from its normal length
to an elongation of 5 in.
4. The force (F Ib.) exerted on a piston varied thus with the dis-
tance x in. from one end of the cylinder: F = 1 20000 /x*. Find the
work done from z = 27 to x = 64.
6. The force (F dynes) with which two spheres carrying certain
electrical charges will attract each other when their centers are x cm.
apart is F = 2Q/x2. Find the work done in moving them apart, from
3=5 toz = 10.
6. The force F Ib. with which the earth attracts a "1-lb. weight"
at a distance of x mi. from its center is F = 16000000 /x2. Find the
work necessary to drive such a weight from the earth's surface (x =
4000) to the distance of the moon (x = 240000). Ignore air resistance.
7. When an electron E is x cm. from a surface S, it is attracted
toward S with a force (F dynes) given by the formula : F = 5.25(10-2°) /z2.
Find the work necessary to draw E away from S, from x = 3tox = 15.
In the following problems draw a rough figure for yourself even where
one is shown.
8. Every section of a certain horn perpendicular to its axis is a
circle, whose radius varies thus with the distance x from one end:
r = .04 z2. Find the volume of the space within the horn from x = 10
tox = 20. (Cf. Fig. 45.)
IV, § 97]
INTEGRATION
141
9. Every horizontal section of a steeple x ft. from the top is a square,
whose side s ft. varies thus : s = .01 x2. Find the volume, if the
total height is 30 ft.
10. Every horizontal sec-
tion of a solid is a rec-
tangle, whose sides y and z
vary thus with the distance
(x in.) below the highest
point : y = 14 Vrc, z = x2/9.
Find the volume from x = 0
to z = 9.
11. Every horizontal sec-
tion of a solid is a ring between two concentric circles, whose radii
(R, r ft.) vary thus with the distance x ft. above the lowest point :
R = Vz, r = x*. Find the volume, x = 0 to 1 .
12. The base of a solid is a quarter circle of radius 10 in. (Fig. 46.)
Every section parallel to one face is a right triangle, whose altitude
equals 1.6 times its base. Find its volume.
(Hint : The area of the moving triangular section, .8 y2, must be
expressed in terms of x. This is easy, since x and y are legs of a right
triangle whose hypotenuse is the radius, 10 in.)
FIG. 46.
13. Find by integration the volume of a sphere of radius 10 in.
Check by geometry. (Hint: What sort of section is made by any
plane x in. from the center? What area, Aa?)
14. Find the volume of a segment cut from a sphere of radius 20 in.
by a plane 10 in. from the center.
15. The base of a solid is a circle of radius 10 in., but every section
perpendicular to one diameter is a triangle, whose height equals twice
its base. Find the volume. (Has this solid any relation to the type
of solid in Ex. 12? Cf. Fig. 46.)
142
MATHEMATICAL ANALYSIS [IV, § 98
§ 98. Setting up the Area-formula. In finding a volume
the area A, can often be expressed immediately in terms of
the dimensions of the moving section. But before integrat-
ing, this must be put in terms of x, the distance
of the section from some fixed point. The
transformation is often effected by the Pytha-
gorean theorem or a proportion.
For instance, suppose we wish to find the
volume of a wedge 7 in. high cut off from a
cylinder of radius 5 in. by a plane passed
through a diameter of the base. (Fig. 47.)
Any section perpendicular to that diameter is a right
triangle. (Why?) Its area is
As = \yz. (16)
But we must get this expressed in terms of x.
The radius, if drawn to the end of y, would form in the
base plane a right triangle with legs x and y, and hypotenuse 5.
"rf. (17)
v
FIG. 47.
Moreover, the vertical sectional triangle is similar to the
central right triangle whose sides are 5 in. and 7 in. (Why?)
whence
Substituting the values of y and z in terms of x in (16) above
gives
This is now ready to integrate :
If we consider half the wedge, starting the "growing volume" at
x = Q, then C=0. The moving plane finally comes to x = 5, making
V = .7(125 -4^) =i|A = 58J.
Doubling this gives the volume of the entire wedge.
IV, § 99] INTEGRATION 143
N.B. Observe once more that the area which we integrate is not
the area of some special fixed section (like the central triangle), but
rather the area of a general moving section, expressed in terms of its
distance x from some fixed point.
§ 99. Original Meaning of f . When we write A = Cydx,
we mean that A is a quantity whose derivative with respect to
x is equal to y.
But historically the sign j was originally an S, denoting
" sum of." The area under a curve was regarded as com-
posed of innumerable strips, — each having^a tiny base dx,
a practically constant height y, and an + \
area ydx. (Fig. 48.) The whole area
was the sum of these tiny " elements "
of area, or, as then written :
A = Cydx. (18)
" The integral " originally meant simply
" the whole," and integration was the
process of making whole. From this
point of view formula (18) has a very dx
••I * • FIG. 48.
tangible meaning.
Unfortunately, however, this reasoning is a bit crude. No
matter how narrow a strip may be, its area is not exactly ydx.
To get the exact value of A from (18) it is necessary to
use the sign j not in the old sense of the sum of elements
ydx, but as denoting the integral in the modern sense, — i.e.j
a quantity whose derivative with respect to x is y.
The old conception, nevertheless, can be modified slightly
so as to be free from logical objection. (This will be done in
Chap. XII.) And, when properly understood, this idea of
tiny elements will afford the simplest means of setting up
144 MATHEMATICAL ANALYSIS [IV, § 99
integral formulas. In fact, it is the method regularly used
by scientific men. Some further illustrations will make the
idea clearer.
(A) Volume of a Solid. According to the old conception, we may
consider the solid as composed of exceedingly thin slices, say like a
soap film, — so thin that the area of each face of the slice is the same.
The volume of the slice is this area A, (which depends on the distance
x from some fixed point) multiplied by the thickness of the slice dx.
The whole volume is the sum of these slices :
V =
This will give a strictly correct result, if we integrate instead of sum-
ming, as we know from (15) above.
(B) Distance Traveled at a Varying Speed. According to the old
conception, we may consider so short an interval of time dt that the
speed v remains constant. The distance traveled during this interval is
vdt ; and the whole distance is the sum of all these tiny distances :
s = f vdt.
If we integrate instead of summing, we get an exact value. For since
the speed is the derivative of the distance, the distance is the integral
of the speed.*
(C) Work Done by a Variable Force. According to the old concep-
tion, we may consider so short a distance dx that the force is constant.
The work done in this tiny distance is Fdx\ and the sum of all these
little bits of work is
This formula, too, as we know by (13), gives an exact value if we inte-
grate instead of summing, f
* Observe that this integral is also precisely the one which would have to
be calculated, if we wished to find the area under the speed-time graph, for
there -o takes the place of "y" and t the place of ''x." That is, the distance
is represented exactly by the area under the speed-time graph. Cf. § 14.
t Why this crude reasoning, despite its fallacy of considering certain
variable quantities as temporarily constant, lead to these formulas which are
strictly exact when we interpret the sign I in the modern sense, will become
clear in Chap. XII. We shall also see how to tell when this reasoning can
be relied upon.
IV, § 99]
INTEGRATION
145
EXERCISES
1. If the height of a cone is 10 in. and the radius of the base is 5 in.,
what is the area of a horizontal section x in. from the vertex? Calculate
the volume by integration, and check by geometry.
2. The same as Ex. 1, for a cone of height 24 and radius 7.
3. The same as Ex. 1, for a cone of any height h and radius r.
4. Draw the curve y = x2, roughly. Calculate the volume which
would be generated by revolving the area under this curve about its
base-line from x = Q to re = 10. (Hint:
What sort of figure will any section
perpendicular to the base-line be ? With
what radius?)
5. The same as Ex. 4, for the line
y = 2 rc+5 from rc = l to re = 4.
6. Find by integration the volume
of a sphere of radius 20 inches; also
the volume cut off from that sphere by
a plane 12 inches from the center.
7. Every horizontal section of a pier
50 ft. high is a square whose side increases uniformly from 10 ft. at
the top Jo 20 ft. at the bottom. Calculate the sectional area 10 ft.
from the top, 20 ft. from the top, etc., and find graphically the
volume of the pier.
8. Find the volume of the pier in Ex. 7 by integration.
9. Find the volume of a wedge 4 in. high cut from a cylinder of
radius 5 in. by a plane passing through a diameter of the base.
10. The same as Ex. 9, if the wedge has any height h, and the
cylinder any radius r.
11. The base of a solid is a quarter-circle of radius 6 in. Every
section parallel to one vertical face is a right triangle whose altitude
is twice its'base. Find the volume of the solid.
12. Find the volume common to two equal cylinders of radius
10 in. whose axes meet at right angles. (Hint : In the figure above,
show that every section one way is a square, whose area is 400—4 re2.)
13. Water is poured from a cylindrical cup 8 in. tall and 8 in. in
diameter. Find the volume remaining when the surface of the water
just bisects the bottom of the cup.
14. Two cylinders have a common upper base, and tangent lower
bases, — all circles of radius 10 in. Find the volume of the common
solid, if the height between bases is 20 in. (See the figure, p. 146.)
146 MATHEMATICAL ANALYSIS [IV, § 100
15. By using the idea of the summation of tiny "elements," set up
the integrals which express :
(a) The momentum generated by a variable force ;
(6) The volume of a sphere of radius 10 in., regarded as composed
of thin concentric shells at a varying distance r in. from the center.
[Hint : What are the area and thickness of
any shell?]
(c) The area of a circle of radius 20 in.,
regarded as composed of narrow concentric
rings, at a varying distance, r in. from the
center.
(d) The increase in the national wealth
during any period if the rate of increase
is some variable quantity R, — supposed known as a function of t.
[ 16. ] Could the quantity (x3+l)10 • 3 x2 be obtained by differentiating
some power of (z3+l), possibly multiplied by a numerical factor? If
so, find f (a^+l^oSaftiz.
§ 100. Water Pressure : Total Force. An important
engineering problem is this: To calculate the total force
with which water will press horizontally against a.vertical
wall or dam.
The pressure, that is to say, the number of pounds per
sq. ft., is different at different depths : 1 ft. below the surface
it is 62.5 Ib. per sq. ft. ; 2 ft. below, it is twice this ; and so
on, proportionally.
To find the total force against a dam, with the 'pressure
varying all the way down, we may proceed in either of two
ways.
(I) By using the old conception of " tiny elements."
According to this we may consider a very narrow strip
across the dam as being all at one depth, x ft. below the
surface of the water. The pressure against this strip is, then,
p = 62.5z (Ib. per sq. ft.).
The number of square feet in the strip is wdx, where w de-
notes the width of the dam at this depth. Multiplying the
IV, § 100] INTEGRATION 147
number of pounds per sq. ft. by the number of square feet,
we get the total force, or number of pounds, against the strip :
Force against strip = 62.5 xwdx (lb.).
The total force against the dam is the sum of all these little
forces :
* (19)
If we can get a formula for the width of the dam at any
depth x, the total force can be found quickly by integrating
(19).
If no such formula is obtainable for w, then F can merely be approxi-
mated by figuring out the force against many narrow strips, and adding.
(II) By reasoning exactly about a " growing force."
Let F denote the total force
against the dam down to any
depth x. Then while x in- w-^ I | -/"TAP
creases by Ax, F increases by w
some AF (the force against
the narrow strip in Fig. 49).
FIG. 49.
where x and w are some average depth and width in the strip.
The average rate of increase of F, per foot increase in x,
is, then,
^- = 62.5xw,
Ax
and the limit of this as Az-M) is
^ = 62.5^,
dx
whence we have (19) again, the J denoting " integral of." *
* For oil instead of water, the only change in (19) would be a different
numerical factor in place of 62.5. This is the weight of a cubic foot of water,
and would be replaced by the weight of a cubic foot of the oil.
148 MATHEMATICAL ANALYSIS [IV, § 101
Ex. I. The width of a dam x ft. below the surface of the water is
10 = 400— z2. Find the total force against it down to a depth of 20 ft.
Substituting in (19) the given value of w we have
F= 62.5 z(400-:r2)dz = 62.5 (400 x-a*)dx.
:. F = 62.5(200 x*-\ x*)+C.
But F = 0 at the surface, where x = 0. Hence C = 0. And when x = 20,
we have the whole force, F = 62.5 (80000 -40000), =2500000 (lb.).
§ 101. Integrating a Power of a Quantity. Can we ever
integrate an expression which involves a product, or a power
of a quantity, without first multiplying out? Yes, if the
given expression happens to be the derivative of a higher
power, — aside from a constant multiplier perhaps.
By (22), p. Ill, we know that differentiating any power
un gives nun~ldu/dx. Hence if we ar£ to integrate a given
expression and come out with un, the expression must consist
of a power of u, multiplied by the derivative of u, and possibly
also by a constant.
Ex. I. Integrate (x4-25)9
This is the 9th power of the quantity (x4— 25) times the derivative of
that quantity. Hence it would result from differentiating the 10th
power of that same quantity, — aside from a numerical factor 1/10.
25)10+C. (20)
Observe that the factor 4 x3 is used up in integrating the power (x4— 25) 9.
This will probably be clearer if you differentiate the result, and compare.
In the foregoing example, if we had been given 7 x3 instead
of 4 x3, the desired form 4 x3 could be found by taking the 7
outside, and multiplying and dividing by 4 :
7 x3dx = ? • - -)9 x3dx = ( • • •)' 4 x3dx.
The result would have been 7/4 times the result in (20) above.
If, however, we had been given a different power of x, say
x6 instead of x3, this could not be remedied by multiplying
IV, § 101] INTEGRATION 149
and dividing, for a variable cannot be moved from one side
of the integral sign to the other. [In fact, an x5 obviously
would not arise in differentiating (z4 — 25) 10.] We should
have to multiply out before integrating.
Ex. II. Find y= (^9-x2xdx.
We need —2x outside the radical. So we change the form thus :
Ex. III. Find
This is not yet possible, the factor xz being absent outside the radical.
EXERCISES
1. A gate of a canal "lock" has a constant width of 30 ft. and a
height of 40 ft. When the water level is 10 ft. from its top, what is
the total pressure against it? Calculate by integration, and also
without.
2. The width of a certain dam at a depth of x ft. is w = 500 — z2.
Calculate w at z = 0, 2, 4, 6, 8, 10; and make a rough drawing of the
dam down to that depth. Estimate its area and the total force of
water pressure against the part drawn.
3. Calculate the total force in Ex. 2 down to x = 10.
4. Find the total pressure down to a depth of 10 ft. if the width of a
dam varies thus : w = 500 — 4 xz.
5. Find the total force against one face of a triangular boar$ im-
mersed in water, if one vertex is at the surface, one side (3 ft. long) is
vertical, and the base (2 ft. long) is horizontal.
6. Find the following integrals and check each by differentiation :
(a) f(16+x4)'4x3^, (6)
(c) "V25=^afe, (d)
(e)
7. Find T(x2+3)2a:^ in two ways, and reconcile the answers.
8. Find the total pressure on the end of a cylindrical boiler of radius
4 ft., placed horizontally, and half full of water. (Hint: The depth
150 MATHEMATICAL ANALYSIS [IV, § 102
x and half -width w/2 are sides of a right triangle whose hypotenuse is
4 ft. Integrate the final expression as in Ex. 6 c.)
9. The same as Ex. 8, if the radius is 5 ft.
10. Explain precisely what is meant by the "number of Ib. per
sq. ft. at the depth of x ft." We cannot have a square foot of vertical
wall all at the same depth, x ft., no matter how narrow the strip.
11. A rectangular floor 30 ft. long and 20 ft. wide carries a load,
whose amount per sq. ft. (y Ib.) varies thus with the distance (x ft.)
from one end: y=4x. Express the total load as an integral: (a) By
using the old idea of elements ; (6) By a consideration of rates.
§ 102. Further Applications. Many physical quantities
calculated by integration are too complicated to discuss at
present. One more case will be cited, however, viz.,
The total attraction of a uniform rod upon an exterior
particle M, in its axis produced. By the law of gravitation :
M A B Every particle m of the rod
?__ attracts M with a force propor-
tional to the product of the
masses divided by the square
of the distance apart : F = GMm/xz, where G is a certain
gravitational constant. But there are particles at all dis-
tances, within certain limits. (Fig. 50.)
Using the old conception, consider a tiny piece of the rod
at any distance x from M. Its mass equals D, the mass per
unit* length, multiplied by dx, its length. The attraction of
this tiny mass Ddx upon M is GMDdx/x*. And the whole
force is
F= rGMDdx =-GMD_] c (21)
J x2 x
To check this, consider the growing attraction exerted by a varying
portion of the rod, over to a distance x from M. Increasing x by Ax
takes in additional force AF which equals GMDAx/2z, x denoting the
average distance from M to points of the added portion Ax. Then
AF/As - GMD/x*, and dF/dx = GMD/x*. Hence we have (21) again,
but with the sign f used in its modern sense, to denote a quantity
whose derivative is GMD/x2.
IV, § 103] INTEGRATION 151
§ 103. Summary of Chapter IV. Integration is the re-
verse of differentiation :
(x)dx = F(x) means dF(x)=f(x)dx. .
Its uses are of two sorts : (I) to derive a formula for some
varying quantity whose rate of change is known, — e.g., the
height of a projectile at any time ; and (II) to calculate some
fixed geometrical or physical magnitude, such as area, work,
etc.
But as each fixed area, etc., is the value to which some varying area,
etc., will grow, problems of type (II) in reality come under type (I).
The constant of integration is determined by the value
of x, or t, etc., at which the growing quantity starts.
In elementary geometry each new area and volume is
calculated by some new plan. But by integration we find
all volumes by one and the same process : Expressing the
sectional area At in terms of the distance x from some fixed
point, and then integrating A8dx* Similarly all plane areas
can be found by one process, all momenta by one process, etc.
Moreover many problems which we have not yet analyzed
yield to the same general method.
The science of calculating derivatives (or differentials) and integrals
is known as Differential and Integral Calculus. It was invented by
Sir Isaac Newton about 1670, and by Gottfried Leibnitz independently,
a little later. Each made many notable calculations with his new
invention — Newton's work in astronomy and physics being especially
remarkable, although his notation was much less convenient than that
of Leibnitz, which we are using.
In Differential Calculus the fundamental problem is to find the rate
at which a given quantity is changing ; in Integral Calculus it is the
reverse : Given the rate, to find the value of the varying quantity.
Our work up to this point is, of course, barely a start in this field.
* If no formula for Aa is known, we can often find one by a preliminary
integration. This will be discussed in Chapter XII.
152 MATHEMATICAL ANALYSIS [IV, § 103
EXERCISES
1. Express and calculate the attraction in Fig. 50 if M = .08, D = 2,
<7=.000 000 065, the rod is 20 cm. long, and the particle 10 cm. away.
2. A rod 10 cm. long has a mass of .4 gram per cm. Find its total
attraction on a particle of mass 5 grams, placed 6 cm. from the rod,
and in line. Take G as in Ex. 1.
3. Find the total weight W of a rod 10 ft. long if its weight per ft.
(w Ib.) varies thus with the distance (x ft.) from one end: w=.\x.
[Set up the integral in two ways.]
4. Find the total load on a beam 15 ft. long if the rate of loading
(y Ib. per ft.) varies thus with the distance (x ft.) from one end:
y=2Qx. [Set up the integral in two ways.]
6. Like Ex. 11, p. 150, for a floor 40 ft. by 25 ft., if y varies thus :
6. Find the area under the curve y = \/xi from x = 2 to x = 10.
7. The base of a solid is a circle of radius 10 inches, but every section
perpendicular to one diameter is an isosceles triangle whose height
equals half its base. Find the volume of the solid.
8. A triangular board is immersed vertically in water, its vertex
being at the surface and its horizontal base being 6 ft. below. If the
base is 8 ft. long, find the total force of water pressure against one face
of the board.
9. A point moved in such a way that d3y/dt3 = 12, y being the dis-
tance traveled. At t = Q the speed was 100 and the acceleration —6.
Find y at any time.
10. Which of these forms can be integrated by some method already
studied? (Do not work out.)
(a) sa(x4+8)dx, (6)
(c) (z2+l)20efo, (d)
11. A wedge 3 in. tall is cut off from a cylinder of radius 6 in. by a
plane passing through a diameter of the base. Find its volume.
12. In stretching a spring, the force (F Ib.) varied as the elongation
(x in.) and was 50 Ib. when x — \. Find the work done in producing
an elongation of 4 in., starting from normal length.
13. The force (F Ib.) with which steam drove a piston varied thus :
F = 12000 Ar*. Find the work done from z = 8 (in.) to a: = 27 (in.).
14. A bomb was thrown straight down from an airplane 8000 ft.
high with an initial speed of 120 ft./sec. Derive a formula for its
IV, § 104] INTEGRATION 153
height after t sec. How would you proceed to find the speed with
which the bomb struck the ground?
15. A bomb was dropped from an airplane 2200 ft. high when an
automobile running 80 ft. /sec. passed directly beneath. How far
apart were the bomb and automobile 10 sec. later, and how fast were
they separating?
16. In Ex. 15 when were the bomb and automobile nearest?
17. A bomb was fired straight up from an airplane 10000 ft. high,
with an initial speed of 400 ft. /sec. Find its height after t sec. When
did it strike the ground ? When was it highest ?
18. A beam loaded in a certain way has its curve defined by the
equation d*y/dx* = — .000036 x. Find y as function of x, if at x = 0
we have y = Q, slope = —.007, flexion = 0 and rate of increase of flexion
= .0006.
19. Find the volume within the surface generated by revolving the
curve y = xz— 4 about its base line, from x = 2 to x = 4.
[20.] Find by geometry the area of the smaller segment cut from a
circle of radius 10 in. by a chord which subtends an angle of 120° at
the center. Would it be more difficult to find the area if the central
angle were 100° instead of 120°?
§ 104. Unknown Integrals. We are not yet able to
integrate some expressions which arise in simple problems.
Ex. I. Find the work done by a force which varies thus,
from x = 10 to x — 20.
W= CFdx =
No integration formula has yet been given for J x~ldx.
(See § 88, p. 129.)
Ex. II. Find the area of the circular segment BCD.
(Fig. 51.)
A = 2J*ydx = 2JVlOO-a».
We cannot effect the integration as yet. (Cf . Ex. Ill,
§ 101.)
154
MATHEMATICAL ANALYSIS [IV, § 105
D
FIG. 51.
Later on we shall see how to per-
form both of these integrations. In
the meantime we can at least solve
such problems approximately by the
graphical methods of §§ 15, 16.
In fact, any unknown integral can
be approximated graphically. For
J F(x)dx = a,TQa, under graph of F(x),
and to find the value of such an integral we need merely plot
the function F(x) and measure the area under the graph.
Remark. The question naturally arises as to whether the area of
the segment in Ex. II above can be calculated by elementary geometry.
A = sector OBCD- triangle OBD.
The area of OBD is clearly 48 sq. in. The sectorial area is to the
entire area of the circle as Z BOD is to 360°.
The size of Z BOD is definitely fixed by the length of its chord BD
( = 16 in.). But we have as yet no means of finding just how many
degrees there are in the angle.
If we knew the precise relation of an angle of a triangle to the sides,
we could find the required area.
§ 105. Survey of Chapters I-TV. At the beginning of the
course we defined a function as a quantity which varies with
another in some definite way. And the central problem all
along has been to learn just how a function varies.
When given merely a table of values, we could only plot
the function and study it graphically. Average and in-
stantaneous rates could be approximated ; also extreme
values, mean values, and any quantity represented by the
area under a graph.
When the function was given by a formula we could make
some of these calculations exactly. But we did not at first
see how to find an instantaneous rate exactly, nor the area
under a graph.
IV, § 105] INTEGRATION 155
On defining an instantaneous rate accurately as a limit,
we saw that to calculate it we must find the limiting value
approached by an average rate, whose interval is being
indefinitely shortened. This brought us to differentiation,
the chief operation of the Differential Calculus. We wrote
derivatives of power functions at sight, and used them for
various purposes.
Finally, we saw how to calculate various quantities by
the Integral Calculus. But the usefulness of the integration
process is limited at present by our inability to integrate
many simple expressions ; e.g.,
id*.
x
Functions exist which have these differentials; but they
are of different sorts than any which we have studied. So
our next business will be to study some further kinds of
functions, which incidentally are useful of themselves.
As suggested by the problem of the circular segment
(§ 104) and that of the inclination of a curve (§ 85), we need
to know the relations between the sides and angles of a triangle.
To this matter we now turn our attention.
CHAPTER V
TRIGONOMETRIC FUNCTIONS
THE SOLUTION OF TRIANGLES
§ 106. Some Preliminary Ideas. The branch of mathe-
matics which deals with the relations between the angles
and sides of a triangle is called Trigonometry, — from two
Greek words meaning " to measure a triangle."
Trigonometry is the basis of all Surveying, and of many
calculations in engineering, physics, astronomy, and other
sciences ; — and yet it is perhaps
the simplest of all the branches of
mathematics.
Before proceeding with the subject
proper, however, let us note certain
very elementary methods of making
approximations by drawing and meas-
uring figures.
And, first, let us recall from geome-
try that an angle of any size can be
measured or drawn by using a pro-
tractor.
Fia. 52.
(I) Fig. 52 illustrates the measurement
of a given angle ABC ( = 73°). The radiat-
ing lines on the protractor should meet at the vertex B and the 0°
line fall directly along BC. If AB tends to cross any radial line, the
protractor is not placed correctly as to the vertex.
(II) In drawing an angle of 73°, we would first draw AB to serve as
one side. Then, placing the protractor so that its 73° line falls directly
over AB, we would draw BC along the 0° line of the protractor. (In
15G
V, § 108] TRIGONOMETRIC FUNCTIONS
157
allowing for the width of the pencil point, the protractor is slid along
in such a way as to keep its 73° line pointing directly along AB, con-
tinually.}
§ 107. Graphical Solution of Triangles. Any ordinary
surveying problem can be solved approximately by simply
drawing the figure to some chosen scale,
and reading off the required distances or
angles.
With good instruments and practice
the percentage of error can be kept very
low. Thus we can check roughly the
more refined methods developed pres-
ently, which use trigonometry proper.
Ex. I. Find the distance AB across a pond
(Fig. 53), if the distances CA and CB and Z C
have been measured as 900 ft., 700 ft., and pIG> 53
35°.*
We draw an angle of 35°, with a protractor, and lay off sides of 9 cm.
and 7 cm., to represent CA ( = 900 ft.) and CB ( = 700 ft.), respectively.
Joining the ends of these lines we get a triangle which must be similar
to the actual big triangle ABC. (Why?) The third side of the con-
structed triangle measures 5.4 cm. ; hence the distance AB is 540 ft.
§ 108. Force Problems. If two forces, acting in the
directions OA and OB (Fig. 54), are simultaneously applied
to an object at 0, the object will
move neither along OA nor along
OB, but in some intermediate
direction. More definitely, the
principle is this :
A (I) // two forces are repre-
FlG- 54< sented, in intensity and direction,
by two sides of a parallelogram OA and OB, drawn from a
common vertex, they are together equivalent to a single force
* The angle is measured by placing a surveyor's transit at C, sighting at
A., and then at B, and reading from the instrument the angle turned.
158 MATHEMATICAL ANALYSIS [V, § 108
represented on the same scale by the diagonal of the parallelo-
gram OR drawn from that vertex.
The resulting motion would be along that diagonal. The
single force represented by the diagonal is called the re-
sultant of the two given forces.
To find the resultant of two given forces: simply draw
the forces on some chosen scale, as in Fig. 54, complete the
parallelogram, and read off the force represented by the
diagonal. The direction of the resultant can be ascertained
by measuring the angle which it makes with one of the given
forces.
To find what third force would balance any two given forces
and maintain equilibrium: observe that the given forces
are together equivalent to a single force, —
their resultant. Hence the third force must
be equal to the resultant in intensity, but
oppositely directed. Thus it is easily drawn
and measured. In general :
(II) // three forces are in equilibrium, any
one of them must equal the resultant of the
other two, but with its direction reversed.
Ex. I. Find the two forces X and F in Fig.
55, which are just balanced by the upward force
of 200 Ib.
The resultant of F and X must be a force of
55 200 Ib. directed straight downward. Knowing the
diagonal of the parallelogram and the directions
of the sides, we easily construct the parallelogram, and read off the
intensities of F and X.
EXERCISES
1. Draw any triangle, measure its three angles, and check their
sum by geometry. Repeat for another triangle.
2. Draw a triangle with one side 10 cm. long, and with the adjacent
angles 40° and 80°. Measure the third angle and check.
V, § 108] TRIGONOMETRIC FUNCTIONS
159
200 v
3. Draw a parallelogram whose angles are 72° and 108°. Measure
opposite sides as a check.
4. To find the distance AB through a hill, lines AC = 748 ft., and
#C = 680 ft., were measured, also Z C = 50°. Find AB.
6. To find the distance between two points A and B, both across
a river, a line CD = 1000 ft. was laid off on this side, and angles were
measured as follows: ZCDB = 75°, ZCDA=40°,
Z DCB = 45°, Z DCA = 75°. Find AB.
6. Find the resultant R for two forces, X = 150 Ib. '
and F = 90 Ib., whose included angle is 60°. Also
find the angle which R makes with X.
7. The same as Ex. 6, but with X = 800 Ib., Y = 500
Ib., and their included angle 120°.
8. A horizontal force OH = 160 Ib. and a vertical
force OF = 120 Ib. are balanced by a single force F.
Find the intensity and inclination of F.
9. The same as Ex. 8, but with 0# = 330 Ib. and OF = 440 Ib.
10. Find the forces X and F in Fig. 55, after changing the angle
from 145° to 115°.
11. A 200 Ib. weight at the end of a rope swings around in a hori-
zontal circle, the rope making an angle of 30° with the vertical, as in the
figure above. What must be the
centrifugal force, X Ib., and the
pull in the rope, P Ib. ?
12. Find the pull (P Ib.) in
the cable, and the horizontal
thrust (T Ib.) exerted by the arm
of the adjacent crane at 0 to sup-
port the weight of 500 Ib.
13. The same as Ex. 12, but with the given angle 35° and the weight
7500 Ib.
14. Fig. 65, p. 171, shows a corner of a bridge-structure. What
forces, F and X, acting along the members meeting at 0 would just
be balanced by the vertical force of 2000 Ib.? (Hint : Draw all forces
away from O in their proper directions.)
15. The same as Ex. 14, but with the given angle 30° and the force
90,000 Ib.
[16.] At several points on one side of an acute angle, erect perpen-
diculars, and measure the sides of each right triangle so formed. Divide
the side opposite the given angle by the hypotenuse in each case, and
compare. Likewise the opposite side by the adjacent side.
30
o
}•
i
160 MATHEMATICAL ANALYSIS [V, § 109
§ 109. The Functions Sin A and Tan A. If we erect per-
pendiculars at various points of either side of any acute
angle (Fig. 56), the right triangles thus formed are all similar.
(Why?) Hence their corresponding sides are proportional.
E.g.,
a_a^_a^
b b' b'"
Thus the ratio a/b does not depend at all upon the size of
the triangle. But if we change the size of angle A, the tri-
angle changes shape, and a/b no
longer has the same value. In
fact, a/b varies with /. A in some
definite way, and is therefore some
function of 'A.
Likewise a/c, b/c, etc., are func-
tions of A. All of these are called
trigonometric functions, and each has a name of its own.
The ratios a/c and a/b are called respectively the " sine "
and " tangent " of Z A, written sin A and tan A. That is,
if a perpendicular is erected at any point in either side of an
acute angle A, forming a right triangle, then
sin,l=leg opposite/ A =a (1)
hypotenuse c
taa^ leg opposite/^ =q
leg adjacent to Z A b
(Memorize these definitions carefully. Also observe that
the hypotenuse does not appear in tan A at all.*)
The values of these ratios are given for various angles, correct to
three places, in the table on p. 161. For instance, sin 20° = .342.
This means that in any right triangle containing an angle of 20°, tho
opposite leg divided by the hypotenuse gives .342, no matter how
large or small the triangle may be. (How could we check this roughly ?)
* The reason for the name "tangent" appears in Ex. 27, p. 166 ; "sine"
is derived from a Hindu word.
V, § 110] TRIGONOMETRIC FUNCTIONS
SINES AND TANGENTS OF ACUTE ANGLES *
161
Ang
Sin
Tan
Ang
Sin
Tan
Ang
Sin
Tan
1°
.017
.017
31°
.515
.601
61°
.875
1.80
2
.035
.035
32
.530
.625
62
.883
1.88
3
.052
.052
33
.545
.649
63
.891
1.96
4
.070
.070
34
.559
.675
64"
.899
2.05
5
.087
.087
35
.574
.700
65
.906
2.14
6
.105
.105
36
.588
.727
66
.914
2.25
7
.122
.123
37
.602
.754
67
.921
2.36
8
.139
.141
38
.616
.781
68
.927
2.48
9
.156
.158
39
.629
.810
69
^.934
2.61
10
.174
.176
40
.643
.839
70
.940
2.75
11
.191
.194
41
.656
.869
71
.946
2.90
12
.208
.213
42
.669
.900
72
.951
3.08
13
.225
.231
43
.682
.933
73
.956
3.27
14
.242
.249
44
.695
.966
74
.961
3.49
15
.259
.268
45
.707
1.00
75
.966
3.73
16
.276
.287
43
.719
1.04
76
.970
4.01
17
.292
.306
47
.731
1.07
77
.974
4.33
18
.309
.325
48
.743
1.11
78
.978
4.70
19
.326
.344
49
.755
1.15
79
.982
5.14
20
.342
.364
50
.766
1.19
80
.985
5.67
21
.358
.384
51
.777
1.23
81
.988
6.31
22
.375
.404
52
.788
1.28
82
.990
7.12
23
.391
.424
53
.799
1.33
83
.993
8.14
24
.407
.445
54
.809
.38
84
.995
9.51
25
.423
.466
55
.819
.43
85
.996
11.43
26
.438
.488
56
.829
.48
86
.998
14.30
27
.454
.510
57
.839
.54
87
.999
19.08
28
.469
.532
58
.848
.60
88
.999
28.64
29
.485
.554
59
.857
1.66
89
1.00—
57.29
30
.500
.577
60
.866
1.73
Cos
Ctn
Cos
Ctn
Cos
Ctn
§ 110. Solution of Right Triangles. Finding the unknown
sides or angles of a definitely specified triangle is called
"solving the triangle." This can be done roughly by
* The bottom labels are for later use.
162
MATHEMATICAL ANALYSIS [V, § 110
90/
200
FIG. 67.
measuring a drawing, — as we have
— seen. For more accurate results
we calculate the unknown parts
of the triangle from given parts,
making use of the definition of
the sine or tangent, and consult-
ing the table on p. 161, or some
larger table. The following examples will illustrate this.
Ex. I. Find by tables the side a of the triangle in Fig. 57.
But
This result can be checked graphically.
Ex. II. Find the two forces X and Y in
Fig. 58, which are together equivalent to the
given force of 400 Ib.
X is represented by the side opposite the
25° angle, while Y is opposite the comple-
mentary angle -of 65°.
tan 33° = .649 (by table).
/. a =130 (nearly).
— = sin 25° = .423,
/. X = 169+,
N.B. We could get Y here after having found X, by using the
tangent. But any error in X would render Y incorrect also. It is
best, whenever convenient, to find each required part of a triangle
from given parts. To do this, simply choose whichever function (sine
or tangent) will bring in the unknown part, together with other parts
which are all given. When possible, get the unknown in the numerator,
and thus avoid division.
V, § 111] TRIGONOMETRIC FUNCTIONS
163
Ex. III. Given the hypotenuse of a right triangle, c = 615,
and one leg, a = 369, to find the angles and the other leg.
(Fig. 60, p. 169.)
whence A = 37° by the table ; and B = 90° - A = 53°. Also
b = V(615)2- (369)2 = 492, approx.
Or, since b is opposite /. B, we could get it by using
= sin # = .799, whence 6 = 491, approx.
615
The preceding method is better if a large table of squares
is available.
Since the table is accurate to three places only, so are the results.
Significant figures beyond the third should be dropped, and the nearest
figure in the third place taken.
The placing of a decimal point should always be checked by common
sense. Thus 615 X. 799 above clearly could not give 4915 nor 49.15,
as it must give a little less than 615.
Remark. At first thought there may not seem to be any
connection between this work of solving triangles and our
main problem of studying how one quantity varies with
another. But the solution of tri-
angles is possible only because the
sine and tangent vary with the
angle in a definite way, — studied
by mathematicians in the past and
recorded in the tables.
§ 111. Slope and Inclination.
One of the most important uses of
the tangent function is in finding FlG- 59-
the inclination (angle) of a line whose slope is known, or
vice versa.
164 MATHEMATICAL ANALYSIS [V, § ill
Since the slope I is the number of units the line rises in one
horizontal unit (Fig. 59), we evidently have tan 1 = 1/1, or
/ = tan/. (3)
For instance, if a hillside has an inclination of 15°, its slope is 1 =
tan 15° = .268. In other words, its grade is 26.8%.
Again, if a ship's deck has a slope of fa its inclination is given by
tan 7 = ^ = .083, /. 7 = 5°, nearly.
The slope of a curve, or of its tangent line, at any point is
easily found by differentiation: l = dy/dx. The inclination
can then be found by using formula (3).* Thus, in Fig. 59, 6,
tan 7 = -=1.800. /. 7 = 61°:
.5
The angle between two lines or curves in the same vertical
plane can be found by subtracting the inclination of the one
from that of the other.
Definition. When an object is viewed from any point, the inclina-
tion of the line of sight is called the angle of elevation, or angle of de-
pression, of the object. Literally, it is the amount we must elevate,
or depress, our gaze from horizontal to see the object.
EXERCISES
1. Draw a right triangle containing an angle of 40°, and verify by
measurement the values of sin 40° and tan 40° given in the table, p. 161.
2. Draw a right triangle whose legs are 9 and 4. Measure the
angle whose tangent is |, and check by the table.
3. Write out the values of the sine and tangent for each acute angle
in these right triangles.
* If I happens to be negative.we change its sign to + before using formula
(8). But we must remember that the line is falling at the angle 7 rather
than rising.
V, § ill] TRIGONOMETRIC FUNCTIONS 165
4. Given tan A =T\, find sin A exactly (without tables). Likewise
find tan B exactly if given sin B = $.
In Ex. 5-14, first solve by drawing to scale and measuring the required
part. Then solve by the table, directly from the given parts, without using
your measurements. Compare results.
5. Find the slope and the inclination of a hillside which rises uni-
formly 238 ft. in a horizontal distance of | mile.
6. Find the extreme height of Philadelphia City, Hall if the angle of
elevation, measured at a point 990 ft. from the base of the building on
a level street, is 29°.
7. The Pike's Peak Ry. rises 7800 ft. in a distance of 9| mi. along
the track. At what angle would the track have to ascend if the inclina-
tion were uniform?
8. In Ex. 11, p. 159, change the weight to 175 Ib. and the angle to
20°, and solve.
9. When light passes from one medium into another, the angles
of incidence and refraction (/ and R) are related thus : sin R = k sin I.
If k = | and 7 = 30°, find R.
10. How far would a swimmer be from the Statue of Liberty, if its
top (301 ft. above the water) had an elevation angle of 12°?
11. A boulevard runs in a direction 30° north of east. How far
east does it go to reach a street which is .8 mi. north of its starting
point?
12. Mt. Hood is 51 mi. from Portland, Ore., in a direction 14° south
of east. How far south is it, and how far east, from the city?
13. What direction is Mt. St. Helens from Portland, if 47 mi. north
and 24.5 mi. east?
14. (a) Find the slopes of lines whose inclinations are 3°, 20°, 45°,
78°, 85°. Express each slope also as a "grade."
(6) Find the inclination of a line if the slope is j, f , 4 ; also if the
grade is 6%, 20%, 180%.
15. On the Mt. Lowe (cable) railway the steepest grade is 67%.
What is the inclination at that point?
16. A ship's deck rises 1 inch in 1 ft. horizontally. What is its
inclination? What if the deck rises 2 in. in 27 in. horizontally?
17. Plot y = x2 from x = 0 to x = 7. Measure the inclination of the
tangent line at x = 2. Calculate the same without using the figure;
and check.
18. At what angle will a line whose slope is f cross one whose
slope is .3?
166 MATHEMATICAL ANALYSIS [V, § 111
Note : In the following exercises, the short-cuts given in the Appendix,
p. 491, would be useful.
19. In Fig. 65, p. 171, change the angle to 60° and the given force
to 12500 lb., and solve.
20. The rafters of a roof are inclined 40°. Find the height of the
ridge above the eaves if the distance between eaves is 25 ft.
21. Find the area of the segment cut off from a circle of radius
125 cm. by a chord whose length is 190 cm.
22. The hypotenuse and one leg of a right triangle are respectively
104 in. and 96 in. Find the other leg and the angles.
23. Using the tabulated values at 5°, 10°, etc., up to 85° plot a graph
showing how tan A varies with A. Does the tangent double when the
angle doubles?
24. Same as Ex. 23 for sin A.
25. In Fig. 58 change the given force to 650 lb. and the given angle
to 37° and solve for X and Y.
26. A circular filter paper, when folded for use, makes a circular cone
whose circumference is half the original circumference, and whose
slant height equals the original radius. What is the vertex angle of the
cone?
27. Draw an angle of 40° at the center of a circle whose radius is
1 unit. Where one side of the angle cuts the circle, draw a tangent,
prolonging it to meet the other side. How long is this tangent line?
This shows the origin of the name "tan A."
28. The eye sees colors incorrectly at certain angles from the center
of the field of vision, — different for different persons. In one experi-
ment red appeared as yellow when 14.3 cm. ahead and 8.5 cm. to the
right, and appeared as black when 15 cm. to the right. Find the angle
from the central line of vision in each case.
29. A ship is sailing 20° north of east at the rate of 14 mi./hr. How
fast is it going northward and how fast eastward ?
30. A man, running at the rate of 9 ft. /sec. in a shower of rain
falling vertically, holds his umbrella 20° from vertical for the best pro-
tection. How fast is the rain falling ?
31. In just what direction should a gun be aimed to fire at invisible
targets which are known to be : (a) 4 mi. north, 2.5 mi. east ; (6) 1200
yd. N., 3000yd. E.?
32. How far from the gun is each target in Ex. 31 ?
33. Seen from an airplane 15000 ft. high a town has a depression
angle of 32°. How far away is it, horizontally?
V, § 112] TRIGONOMETRIC FUNCTIONS 167
§ 112. Cosine and Cotangent. The ratio b/c (Fig. 60)
is called the " cosine " of angle A, written cos A. Also b/a
is called the " cotangent," written ctn A. That is,
hypotenuse c
ctn ^ adjacent leg = fr
opposite leg a
The ratio b/c is also the sine of Z B. That is, the cosine
of any angle is the sine of the complementary angle. Thus
cos 20° = sin 70° ; cos 50° = sin 40°, etc.
D
The name " cosine " is simply
a contraction of " complement's
sine." Similarly the cotangent is
the " complement's tangent."
Notice also that ctn A is the
reciprocal of tan A (i.e., I/tan A) ;
but cos A is not the reciprocal of b
sin A. FlG- 60-
Evidently a table of sines is also a table of cosines for the
complements of the angles listed. This fact is indicated by
the label cos at the bottom of each sine column, p. 161.*
By using cos A and ctn A we can solve some right triangles
more directly, and avoid introducing the second acute angle.
To fix the definitions of the four functions in mind, try repeating
them a few times :
sine = (opposite leg) -f- (hypotenuse) ; etc.
Then take a group of right triangles turned in various positions, and
practice picking off the functions as in Ex. 1 below. Do this fre-
quently at odd moments. It is exceedingly important to fix the definitions
permanently in mind, and to do so now.
* For convenience in looking up cosines and cotangents, you had best
write in, at the right of each group of columns, the angles complementary to
those printed on the left. (Every 5° or so will do.) This will call attention,
for example, to the fact that the value .643 labeled sin 40° is also cos 50°.
168 MATHEMATICAL ANALYSIS IV, § 113
EXERCISES
1. In Ex. 3, p. 163, read off the cosine and cotangent of each acute
angle.
2. Find the leg adjacent to an angle of 23° in a right triangle : (a) if
the opposite leg is 400 ft., (6) if the hypotenuse is '1000 ft.
3. Find an angle of a right triangle : (a) if the adjacent leg is 43
and the hypotenuse is 50; (6) if the adjacent leg is 273.6 and the
opposite leg is 300.
4. Given cos A = |^, find sin A, tan A, ctn A exactly, without tables.
6. Similarly, find sin A, cos A, ctn A, if given tan A =f.
6. Interpolate in the table to find cos 22° 24' and ctn 22° 24'.
7. A battleship 594 feet long, turned broadside toward us, subtends
an angle of 4°. How far away is it?
8. In Fig. 63, p. 169, let F be 2500 and Z A = 26°. Solve for X and Y.
9. Find the perimeter and area of a regular decagon inscribed in a
circle of radius 19.98 inches. »
10. Find the radius of the "Arctic circle," taking the earth's radius
as 3960 mi. (Latitude 66° 30'.)
11. Find similarly the radius of the "Tropic of Cancer."
12. Bisect one angle of an equilateral triangle whose sides are 10 in.,
and calculate from either right triangle so formed the sine, cosine, and
tangent for 30° and 60°. Compare the table.
13. Calculate geometrically the functions of 45°. Compare the table.
14. If the hypotenuse of a right triangle is c, and one acute angle
is A, what are the values of the two legs a and 6? Using this result,
draw several triangles with various angles and hypotenuses and write
at sight expressions for the legs of each.
15. Find the volume of water in a hori-
zontal cylindrical boiler of radius 28 in. and
length 5 ft. when the water is 10 in. deep in
the middle.
§ 113. Projections. We shall fre-
quently need to consider the projection
of a given line-segment s upon some
Flo 61 other line I, — i.e., the distance p be-
tween perpendiculars dropped from the
ends of s upon I. (Fig. 61 .) Much time will be saved by get-
ting a formula for the value of such a projection in any case.
V, § 114] TRIGONOMETRIC FUNCTIONS
169
Nowp/s = cos A, and therefore,
p = s cos A. (5)
That is, the projection equals the segment itself, multiplied
by the cosine of the included angle.
The same principle holds good for the projection of any
plane area upon another plane.
For suppose BQFB (Fig. 62) is
the projection of any area BPFB,
determined by dropping perpendic-
ulars from all points of the bound-
ing curve BPF. Then, regarding
each of the areas as the area under
a curve whose height above the
base-line at any point is y or Y,
respectively, the projection and the
original area are given by the integrals
Fm. 62.
=fydx,
But y = Y cos C by (5) above,
integral :
Substituting this in the first
= C(Y
cos
= cos C
CY dx =
= cos C - A.
(6)
That is, the projection equals the original area multiplied
by the cosine of the angle between the two planes.
Observe that the projection of a line
or area is always located at the feet
of the perpendiculars, and is smaller
than the original line or area.
§ 114. Components. Any two
forces X and F, which would together
be equivalent to a single force F, are
called components of F. If mutually
FIG. 63. perpendicular, — which is the way
170 MATHEMATICAL ANALYSIS [V, § 114
components are taken unless otherwise stated, — they are
easily found.
X = FcosA, Y = FcosB. (7)
Thus, the component of a force in any direction is equal to
the force itself, multiplied by the cosine of the included angle.
The value of Y in (7) is the same as Y = F sin A. (Why?)
Ex. I. Find the two components
N and T of the weight of the block
in Fig. 64, if the plane is inclined 20°.
The angles at the block are 20° and
70°. Hence
* ' N = 1000 cos 20° = 966,
T = 1000 cos 70° =342.
That is, the force with which the block presses against the plane is
966 lb., and the force tending to move the block down along the plane
is 342 lb. If there were no friction, a pull of 342 lb. up the plane
would just keep the block from sliding.
EXERCISES
1. Find the horizontal and vertical projections of a line 9.8 ft. long
inclined 20°.
2. What will be the apparent shortest diameter of a wheel of
radius 19.8 inches if its axle makes an angle of 40° with the line of sight?
3. The same as Ex. 2, for a diameter of 25 in. and an angle of 76°.
4. What area in a pipe of radius 10 inches would be obstructed by
a damper turned 70° from the position of complete obstruction?
5. What fractional part of the area would be obstructed in any
pipe with the damper turned 45°?
6. An erect cylinder of radius 5 in. is cut by a plane inclined 50°.
What are the area and the longest diameter of the sloping section?
7. Find the volume cut from a circular cylinder of radius 10 in.
by a plane through a diameter of the base inclined 40°.
8. In Ex. 7, find also the area of the sloping plane section.
9. In Fig. 64 change the angle to 18° and the weight of the block to
997 lb. and solve for the components T and N.
10. Find the horizontal and vertical components of a force of 1250
lb. inclined 33°.
V, § 115] TRIGONOMETRIC FUNCTIONS
171
11. A weight is suspended by two wires, each inclined 22°. If the
greatest straight pull which either wire could sustain is 450 lb., how
large a weight could the two support as specified?
12. If a ship is sailing 21° north of east at the rate of 15 mi. per hour,
what are its component speeds, northward and eastward?
13. Similar to Ex. 12, but sailing 17.3 mi./hr. 66° south of west.
14. If a wind is blowing 17.5 ft. /sec. and crosses the direction of
artillery fire at an angle of 38°, what are its component velocities along,
and directly across, the direction of fire ?
16. Same as Ex. 14, if the wind velocity is 9.8 m./s., crossing at an
angle of 72°.
16. A force F lb. inclined 23° has a horizontal component of 873 lb.
Find/''.
17. A sled is pulled on a level road by a cable inclined 18°. If
the pull in the cable is 200 lb. how much work is done in pulling the
sled 500 ft.? (Hint: Only the horizontal component does any work.)
18. How large a board held perpendicular to the sun's rays would
shade 1 sq. ft. of level ground when the sun is 60° above the horizon?
20° above? (What connection has this with the cause of the seasons?)
§ 115. Equilibrium of Forces : Component Method. The
unknown forces needed to balance a given force, or forces,
can be found without drawing a force
triangle or polygon.
To illustrate, let us find the forces
F and X acting along the two
members of a bridge-structure
(shown in Fig. 65) if the support-
ing force exerted by the pier is 200
tons. Fl°- 65-
We first tabulate the horizontal and vertical components
of all the forces :
[200
FORCE
HORIZ.
VERT.
F
F cos 40°
F cos 50°
X
X
0
200
0
200
172
MATHEMATICAL ANALYSIS [V, § 116
(Clearly the 200 ton force can have no effect horizontally,
nor X have any vertically.)
The component of a force along any direction measures
its tendency to produce motion in that direction. Hence,
the vertical components must balance one another.
F cos 50° = 200, .: F
200
311.
cos 50°
Similarly, the horizontal components must balance:
40 8
Mi,!
0
5 ^4
2/
I
In like manner, — as is shown in treatises on Statics, — it is possible
to go on to the other joints of a structure and find the forces acting
at each. Thus the forces along all the memb'ers can be found if the
supporting forces exerted by the piers are known. The principle by
which those forces are found is a very familiar one.
§ 116. Moment of a Force. As every one knows, a 50-lb.
boy can balance a 100-lb. boy on a " teeter " board by sitting
just twice as far from the sup-
porting rail or fulcrum. This
is because the " moment " of
each weight (i.e., its tendency
to produce rotation about the
point of support) is propor-
tional to its distance from that
point.
General Principle: The mo-
ment of any force about any
point equals the product of the
force by its arm, — that is,
by the perpendicular distance
from the point to the line of
FIG. CG. action of the force.
V, § 116] TRIGONOMETRIC FUNCTIONS 173
Thus, in Fig. 66 (a), we have about the point A :
Force Arm Moment
80 lb. 5ft. 400 Ib.-ft.
60 " 8 " 480 •"
100 " y " 100 y "
The first two forces tend to turn the beam about A in
one direction ; and the third in the opposite direction. If
the latter is just to balance the other two, its moment must
equal the sum of their moments :
/. 100^ = 880. .-. 0 = 8.8.
Likewise in Fig. 66 (6), if the force P is just large enough to prevent
the other forces from rotating the beam about B, its moment must
balance theirs :
P(25) = 100(3) +50(8) +200(18). /.P = 172.
And again, in Fig. 66 (c), if the force FD is to prevent rotation
about (7,
Fz>(40) = 1000(10) +3000(20) +2000(30). /. FD = 3250.
Similarly, to prevent rotation about D
Fc(40)= 1000(30) +3000(20) +2000(10). /. Fc=2750.
Check : FC+FD equals the sum of the three loads, 1000+3000+2000.
EXERCISES
1. A beam 30 ft. long weighing 10 lb. per ft. rests on a post at one
end A and is supported by a vertical cable at the other end B. It
carries a load of 500 lb. 12 ft. from A. Find the pull in the cable.
(Hint : Regard the weight of the beam as a single force acting at
its center.)
2. The same as Ex. 1 but with an additional load of 200 lb. 8 ft.
from B.
3. In Fig. 66 (c), what forces are exerted by the piers if the loads are
8000 lb., 15000 lb., and 12000 lb. at distances of 10 ft., 20 ft., and 30
ft. from one pier?
4. The same as Ex. 3, if the three loads are 1000 lb., 1000 lb., and
3000 lb.
6. The two piers at the ends of a bridge beam carry loads of 15000
lb. and 18000 lb. respectively, due to the weight of the bridge. If a car
174
MATHEMATICAL ANALYSIS
[V, § 117
weighing 4000 Ib. stands on the bridge one fourth way from the first
pier, what load will the other pier then carry ?
6. Solve Ex. 12, p. 159, by
the component method, chang-
ing the given angle to 50°.
7. In Ex. 11, p. 171, what is
w& the pull in each wire if the
Fxo. 67- supported weight is 100 Ib.?
8. The reaction R of the guide is vertical in Fig. 67. Find the force F.
9. In Fig. 65, change the supporting force to 25000 Ib., the angle
to 35°, and find F and X by considering components.
10. In Fig. 66 (£), add 50 Ib. to each force and find P.
§ 117. Larger Tables. The little three-place tables used
thus far are not accurate enough for much practical work.
Four places, however, will often suffice, and five places nearly
always, — though some scientific work requires even seven
or eight places. There is no point in using tables which
are much more accurate than the data of the problem
(measurements, etc.).
A few lines are reproduced here, from a typical five-place
table. The labels at the bottom are to be used with the
minutes at the right — as in the small tables. For example :
sin 72° 58' = .95613, tan 72° 59' = 3.2675.
17°
/
Sin
Tan
Ctn
Cos
0
1
2
.29237
265
293
.30573
605
637
3.2709
.2675
.2641
.95630
622
613
60
59
58
60
.30902
.32492
3.0777
.95106
0
Cos
On
Tan
Sin
/
72°
V, § 117] TRIGONOMETRIC FUNCTIONS 175
For intermediate values interpolate by proportional parts.
Check by common sense, noting whether your interpolated
valua lies between the tabulated values and nearer the right
one. If it does not, you may have overlooked the fact that
the cosine and cotangent grow smaller as the angle increases.
EXERCISES
1. Look up the five-place values of the sine, cosine, tangent, and
cotangent of the following angles :
25° 34', 3° 57', 88° 12', 47° 16'.
2. The same as Ex. 1 for the following angles, making the necessary
interpolations by proportional parts :
5° 9'.3, 28° 15'.6, 42° 58'.2.
3. In railroad construction a 6° curve is one in which a chord of
100 ft. subtends a central angle of 6° ; similarly for a 5° curve, etc.
(a) Find the radius of a 6° curve.
(6) If the radius is 2000 ft., -what is the degree of curvature?
4. As in Ex. 3, find the radius of a 4£° curve. Also the curvature
in a circle of radius 1 mi.
6. Find the area of the plane section common to two cylinders of
radius 5 in. whose axes cross at right angles.
6. The same as Ex. 5 for two cylinders of radius 10 in., crossing at
an angle of 60°.
7. A steel plate £ in. thick is to be bent 62° along a certain line.
How much longer will the outer surface be than the inner, if both
remain flat right up to the turn?
8. The same as Ex. 7 for a plate .75 in. thick if bent 71° 42'.
9. A triangular hole through a vertical dam is 6 ft. wide at the top,
and both sides are inclined 36° 40'. Find the total force of water pres-
sure against a gate closing the hole, if the surface of the water is level
with the top of the hole.
10. The top of a ladder rests against a vertical wall. The foot is
pulled away at the rate of 2 ft./min. How fast is the top descending
when the inclination is 50°? (Hint : Use any length.)
11. Solve Ex. I, p. 115, for a filter whose vertex angle is 100°.
176
MATHEMATICAL ANALYSIS
[V, § 118
§ 118. Oblique Triangles. The trigonometric functions
have been defined as ratios of the sides of right triangles.
They can, however, be used in solving oblique triangles as
well. We have merely to drop a perpendicular from some
vertex to the opposite side, and work with the right triangles
thus formed.
Any triangle whatever can be solved in this way if
enough parts are given to fix its size and shape, — in
other words, enough parts to let us draw the triangle.
By elementary geometry, this is possible if we know the
three sides, or two sides and any angle, or one side and any
two angles.*
NOTATION. In discussing these matters more fully we
shall use the following very convenient notation: Capital
letters will denote angles, and the
corresponding small letters the oppo-
site sides. Thus, for instance, A will
always stand for the angle opposite
side a, and hence included between
sides b and c.
•x \F Ex. I. Given a = 77, c= 40, £ = 121°.
— J Solve completely.
Plan: Let B' = 180° - B = 59°.
Solve ABCF for p and x. Then
2/ = 40+x. Knowing p and y} find b and A from A ACF.
Find C from A and B.
Results : x = 39.658, p = 66.002 ; A = 39°, b = 104.88, C = 20°.
Ex. II. Given a = 75, 6 = 65, c = 80. Find the angles.
Plan: Too few parts of either right triangle are known
to solve it alone. But by equating the values of p2 in the
* To have the three angles given would not suffice, as these alone do not
fix the size of the triangle. In fact, three angles are no better than two.
For if two are known, the third can be found from the fact that the sum of all
three is 180°.
—y—
FIG. 68.
V, § 1191 TRIGONOMETRIC FUNCTIONS
C
177
two triangles, we easily find x.
Thus
which, simplified, gives 160 x
= 5000, or £ = 31.25. Then
80 — x = 48.75 ; and angles A and B
are easily found, also C. [A= A
61° 16', 5 = 49° 27', C = 69° 17'.]
EXERCISES
1. (a) — (c) Carry out the method of solution outlined above for
the triangles in Figs. 68, 69.
2. Solve the following oblique triangles similarly :
(I) Given A = 17° 43', J5 = 82°55', c = 689;
(II) Given a = 735, 6 = 642, £ = 53° 17';
(III) Given a = 255, 6=388, 4=48° 65';
(IV) Given a = 850, 6 = 950, c = 1200.
§ 119. Cosine Law. To avoid the labor of dissecting
oblique triangles, let us now derive some formulas which will
show how the calculation must turn out in each case.
Proceeding as in Ex. II, § 118, for an acute-angled
triangle of any sides a, 6, c, we should have :
or solving this for a2,
a2 = 62+c2-2cz. (8)
But x is the projection of b on c, and by § 113 equals
b cos A. Substituting this value for x in (8), we find :
2bccosA. (9)
That is, the square of one side of a triangle is equal to the
sum of the squares of the other two sides, minus twice the
product of those sides by the cosine of their included angle.
178 MATHEMATICAL ANALYSIS [V, § 120
This theorem is called the " Cosine Law." It should be
carefully memorized. Applied to b and c it gives,
-2cacosB. (10)
2a6cosC. (11)
(Observe that in each case the formula begins and ends with
the same letter.)
What modifications, if any, must be made in these formulas to apply
them to obtuse angles will be discussed in § 121.
By using the Cosine Law we can solve an oblique triangle
very easily if given the three sides or two sides and their in-
cluded angle. We have merely to substitute the values of
the given parts in equation (9), (10), or (11) as required,
and solve for a required part. If some other combination
of parts is given, it is best to use a different formula.
(§ 120.)
Ex. I. Solve the triangle : a = 75, b = 65, c = 80.
By (9) : 752<=65*+802-2(65)(80) cos A.
.'. cos A* .48077, A => 61° 15.8',
Angles B and € are found likewise by starting \rith 65* or 80*.
Ex. II. Solve the triangle : b « 750, c = 860, A = 40°.
By (9) : a2 -7502-f 8602-2 (750) (860) cos 40a,
whence a is known. Angles B and C can be found as in
Ex. I, or as in § 123. Tables of squares and square roots
may be used.
Remark. If a given angle is 90°, the triangle should not be solved by
the Cosine Law but as a right triangle. // on unknown angle happens
to be 90°, this fact will soon be discovered, for the square of one side
will equal the sum of the squares of the other two.
§ 120. Sine Law. From the two right triangles in Fig. 70
we find
p = a sin B, p = b sin A.
Hence a sin B = b sin A. or - — = - — -.
sin A sin B
V, § 120] TRIGONOMETRIC FUNCTIONS
179
Similar equations can be derived likewise for sides a and c,
and for b and c.
a b c
.
sin A sin B sin C
(12)
That is, the three sides of a triangle are proportional to the
sines of the opposite angles. [Memorize.]
What, if any, modifications are necessary when 'the triangle contains
an obtuse angle will be discussed in § 121.
By using this " Sine Law,"
we can solve an oblique triangle
easily if given a side and two
angles, or two sides and the
angle opposite one of them. In
the latter case there are often
two possible triangles, one of
which involves an obtuse angle.
§123.
Ex. I. Solve the triangle: I
The third angle is known at once :
FIG. 70.
This case is postponed to
=750, 4
B = 60°.
0=80°.
One of these equations gives a and the other gives c.
Remarks. (I) When the two given angles are complementary, or
one of them is 90°, the triangle should be solved as a right triangle.
(II) The Sine Law will not solve a triangle in the cases covered by
the Cosine Law, though it may be helpful after some unknown part
has been found. The simplest rule as to which law to use in solving
any given triangle is this : Use the Cosine Law if given the three sides
or two sides and their included angle; and the Sine Law in all other cases.
EXERCISES
1. When should the cosine law be used to solve a triangle? The
sine law?
2. Given b=450, A =67° 23', (7 = 41° 34'. Find a, c, B.
180 MATHEMATICAL ANALYSIS [V, § 121
3. Given a = 600, 6 = 750, C = 40°. Find c, A, B.
4. Given a = 65, 6 = 75, c = 80. Find A, B, C, — independently
of one another. Check by adding.
6. To find the distance from a gun (G) to a target (T) a
line GO = 2375 yd. long was measured to an observation post, O, and
angles TOO and TOG were measured as 72° 15' and 80° 30'. Find GT.
6. To find the distance from a gun G to a target T beyond a hill
an observer at 0 found by a range finder GO =2037 yd., OT = 3258 yd.,
Z TOG = 69° 25'. Find GT.
7. On a certain day the distances of the earth and Venus from the
sun were 90,200,000 mi. and 66,200,000 mi. respectively. The angle
ESV between their directions was 69° 45'. Find EV, their distance
apart, at that time.
8. Two forces of 50 Ib. and 80 Ib. have an included angle of 120°.
Find their resultant force and its direction.
9. Solve Ex. 10, p. 118, if the first train runs 20° north of east.
§ 121. Sine and Cosine of an Obtuse Angle. The defi-
nitions of the sine, cosine, etc., which we have been using,
would be meaningless in the
case of an obtuse angle. For we
could not even get such an angle
into a right triangle, — much
less speak of the " opposite leg,"
" hypotenuse," etc.
Later on (§ 253), the defini-
tions will be restated in a form
applicable to angles of any size
whatever. But for present pur-
poses it will suffice to agree arbitrarily to let x, y, and 6
in Fig. 71 take the place of the adjacent leg, opposite leg,
and hypotenuse in the former definitions, thus making
smA=^ cosA=?-. (13)
0 0
We further agree to regard x as negative, as it runs in the
reverse direction from the actual side of Z A.
V, § 122] TRIGONOMETRIC FUNCTIONS 181
With these agreements we observe these facts :
Sine of an obtuse Z = Sine of the supplementary acute Z (14)
Cosine of an obtuse Z = — Cosine of supplementary acute Z .
E.g. sin 160° = sin 20°, cos 160° = - cos 20°,
sin 100° = sin 80°, cos 100° = - cos 80°, etc.
These agreements have been made arbitrarily, but there
is good reason for adopting them : They make the Sine Law
and Cosine Law valid for obtuse-angled triangles, as well as
acute.
Proof. In the large right triangle of Fig. 71
y = a sin B.
But by (13) :
y = b sin A.
Equating these we get finally the Sine Law :
_JL^= & ,etc.
sin A sin B
Also in Fig. 71, since x is negative, the base of the large right triangle
is c— x.
Or, since yz+x* = b2, and x = b cos A,
.'. a2 = 62+c2-2 be cos A.
Thus the Sine Law and Cosine Law are both valid.
§ 122. Solving Obtuse-angled Triangles. In solving any
triangle, then, whether acute-angled or obtuse-angled, we
use the same formulas. But in looking up the sine or cosine
of an obtuse angle, we must remember the relations (13)
above.
Ex. I. Solve the triangle : A = 110°,. 5 = 40°, 6 = 75.
a 75 c
sin 110° sin 40° sin 30C
(Sine Law.)
To look up the sine of 110°, simply look up the sine of the supplement
70°, — which it equals by (13). Then proceed as formerly.
182
MATHEMATICAL ANALYSIS [V, § 123
Ex. II. Solve the triangle : A = 110°, 6 = 75, c = 95.
a2 = 752+952-2(75)(95) cos 110°. (Cosine Law.)
To look up cos 110°, merely find cos 70° and prefix a negative sign.
(This will make the final term in the equation positive, and a2 greater
than 62+c8, — as it should be, by Fig. 71.)
WheofM£ known, find angles B and C by the Sine Law.
§ 123. Solving for an Angle. If we find from a triangle
that the cosine of an angle is negative, this means that the
angle is obtuse.
For instance, if cos A = — .76604, then A is obtuse, and its
supplement A ' has its cosine equal to .76604. By the table
A ' = 40°; hence A = 140°.
When we find from a triangle the, value of the sine of
an angle, the angle may be acute, or may be obtuse. For
instance, if sin B = .34202, this may
mean either that B = 20°, — by tables,
— or that B = 160°. Both values
should be tested out. (This is the
" ambiguous case " of elementary
geometry.)
Ex. I. Solve the triangle: a =
600, 6 = 800, A =40°.
600 800 c
sin 40° sin B sin C'
This gives sin B = .85705, whence B
is cither 58° 59', or the supplement
of this angle, viz., B'= 121° 1'.
There are two possible triangles, both having the given
parts a, 6, and .A.*
In one of these triangles, A =40°, J5 = 58°59', and hence
* When this case* arises in a practical problem, we have to decide by
means of additional information which triangle is the one we want.
V, § 123] TRIGONOMETRIC FUNCTIONS
183
C = 81° 1 ' ; etc. In the other, A = 40°, B = 121° 1 ', and hence
C = 18°59'; etc.
N.B. If the larger of the two given sides were opposite the given
angle, only one triangle would be possible. (Test this by construction.)
The fact would be discovered automatically in the process of solving :
the second value of B would be too large to go into a triangle with Z. A.
EXERCISES
1. Look up the sine and cosine of 108°; 165° 20'; 150° 12'.7;
128° 51 '.2.
2. Find the angles whose cosines are: —.92609, —.42683, and
— .10275. Also find the obtuse angles whose sines are : .39741, .81049,
and .22654.
3. Two given forces of 7 tons and 8 tons have an included angle of
60°. Find the magnitude of their resultant.
4. Two given forces of 200 Ib. and 300 lb., acting at a common point,
are balanced by a single force of 400 lb. Find the angle between the
given forces.
6. Mt. St. Helens and Mt. Jefferson are respectively 53 mi. and
74 mi. from Portland, Ore., the angle between then- directions being
115° 31'. Find their distance apart.
6. Find the distance AB through a hill if AC = 600 ft., £C = 700ft.,
and angle ^05 = 102° 17'.
7. Find the distance AB across a pond if AC = 495 ft., Z BAC = 3Q°
and Z^CB = 105°52'.
8. Solve the triangles some of whose parts are given below for the
missing parts :
a
6
c
A
B
c
i.
725
483
467
ii.
93.6
81.5
65.2
iii.
8.35
6.51
32° 17'
iv.
.927
1.035
138° 15'
V.
•
6845
5728
43d 12'.8
vi.
38.0
59.4
29° 48''
vii.
9806
'92° 13'
26° 39>.2
viii.
.0637
14° 57'
86° 23'
9. (i)-(viii). Find the area of each triangle in Ex. 8, (i)-(viii).
184 MATHEMATICAL ANALYSIS [V, § 124
§ 124. Successive and Simultaneous Triangles. To find
an unknown distance or angle, we often have to solve two
triangles in succession, or else obtain simultaneous equations
from two triangles.
Ex. I. Find x in Fig. 73; given A =20°, 5 = 30°,
2000 (ft.).
H
FIG. 73.
(1) Solution by successive triangles.
In A ABP, ZP = 10° (since Z5=ZA+ZP).
. y = sin 20° = . 34202*
2000 'sin 10° .17365'
whence y = 3939.2. Then in A BPH, x = y sin 30° = 1969.6.
(2) Solution by simultaneous right triangles.
In AAHP: 2000+2 = x ctn 20° = z(2.7475).
In A BHP: z = x ctn 30° = s(1.7321).
.'. 2000 =z(1.0154), [subtracting]
Observe here that ctn 30° — ctn 20° does not equal ctn 10° ; and that
sin 20° does not equal twice sin 10°.
EXERCISES
1. In Fig. 73 change the angles to 15° 3' and 25° 46' and the given
distance to 3000 ft., and find x.
2. Two houses in line with the base of a hill are 4000 ft. apart on
level ground. Observed from the hilltop they have depression angles
of 11° 2' and 18° 55'. Find the height of the hill.
3. Two boat landings 3000 ft. apart on the farther side of a river are
61° and 70° 40' downstream as seen from a landing on this side. How
wide is the river?
V, § 126] TRIGONOMETRIC FUNCTIONS 185
§ 125. Summary of Chapter V. In a right triangle the
ratios of the sides will vary in a definite way with either acute
angle ; and are consequently functions of either angle.
By using the tabulated values of these functions (sine,
tangent, etc.), we can solve right triangles, — and also
oblique triangles, by dropping perpendiculars or using the
Sine Law or Cosine Law.*
The sine and cosine of an obtuse angle have virtually
been denned in terms of the supplementary acute angle.
But no general definitions have been given for the functions
of any angle whatever.
By the principle of the parallelogram of forces, all prob-
lems on equilibrium of concurrent forces reduce to triangle
problems. They can be solved more easily by considering
components. A closely related idea is that of projections.
In solving triangles we use the theorem that the sum of the angles of
any triangle is 180°. You will recall from geometry that the proof of
this theorem rests upon the assumption that through a given point one
and only one line can be drawn parallel to a given line.
This assumption is not certainly known to be true. There are " Non-
Euclidean" geometries, perfectly logical, in which the assumption is
denied. According to these geometries, the angle-sum differs from
180°, — but imperceptibly in triangles of ordinary size. No one knows
which system of geometry is true of the space in which we live, but the
"Euclidean" geometry and trigonometry which you have studied are
simpler than the others, and are always used in practical work.
§ 126. Our Next Step. So much numerical calculation
is necessary in solving triangles, in finding the values of de-
rivatives and integrals, and in other scientific work, that it
is imperative to know the best methods of computing. This
matter will be considered in the next chapter.
Again, there are many scientific problems which use
* The tables were calculated approximately in ancient times by means of
certain formulas, but were greatly enlarged and improved in the sixteenth
century, chiefly by G. J. Rheticus, a German.
186 MATHEMATICAL ANALYSIS [V, § 126
trigonometry, not in solving triangles, but in studying how
some quantity varies. There we must know how fast the
functions sine, tangent, etc., change as the angle changes.
That is, we must know the derivative of each function. We
shall deal with this question in Chapters X-XI ; and shall
incidentally find formulas suitable for calculating the tables.
EXERCISES
1. The hypotenuse and one leg of a right triangle are respectively
104 in. and 96 in. Find without tables all four functions of the smaller
acute angle.
2. What is the slope of a line whose inclination is 17° 43'?
3. What is the inclination of a mountain trail if the grade is 8%?
What if 100%?
4. A projectile fell in such a way that the slope of its path on striking
was .8. If it struck rising ground whose grade was 15%, at what tingle
did its path meet the ground?
6. An invisible target is 12,000 yd. east and 1360 yd. north of a
gun. How far away is it, and in just what direction?
6. A color placed 40 cm. to one side and 90 cm. ahead of an observer
was seen incorrectly. What was the angle from the central line of
vision ?
7. If two cylindrical tunnels of diameter 20 ft. cross at an angle of
50°, what is the area of the common plane section? Also the longest
diameter of that section?
8. (a) Look up the sine and cosine of 110°, 169° 42', 128° 18\2.
(6) Find Z A if cos A = — .09875, and two possible values of /. B if
sin B = .10627.
9. A triangle ABC on level ground has 5 = 60°, C = 102°, a = 300 ft.
Find the height of a tree standing at A, if the elevation angle of its top,
observed from B, is 7° 35'.
10. If the sides of a triangle are 75, 80, and 95, find the smallest
angle.
11. Two forces of 50 Ib. and 80 Ib. have a resultant of 100 Ib. Find
their included angle.
12. At a certain time the distances of the earth and Mars from the
sun were respectively 91,100,000 mi. and 138,600,000 mi., the angle
7',',s'U I >H \vrni their directions being 152° 4'. Find EM, their dist.m:v
apart at that time.
V, § 126] TRIGONOMETRIC FUNCTIONS 187
13. A field is a parallelogram with sides 30 and 20 chains long, and
an area of 42 acres. Find the length of its shorter diagonal. (1 acre =
10 sq. ch.)
14. Find by measurement and by trigonometry side c of a triangle
if a =315, 6 = 521, and C = 40°.
15. Find the perimeter and area of a regular octagon circumscribed
about a circle of radius 25 ft.
16. A horizontal cylindrical oil tank has an inner diameter of 10 ft.
and length of 25 ft. Find the volume of oil in it when the oil is 8 ft.
deep in the middle.
17. A long horizontal rod weighing 4 Ib. per ft. is to carry a load of
100 Ib. two feet from one end which rests on a pier. The other end is
to be supported by a vertical cable. For what length of rod (x ft.)
will the pull in the cable (F Ib.) be least?
18. Draw to scale a triangle whose sides are 25, 30, and 40 units.
Measure the smallest angle. . A student tried to calculate this angle as
follows :
cos A = fft = .750, /. A =41° 25'.
What is wrong with this method?
19. An opening 6 feet square passes vertically through a ship's deck
at a place where the deck's slope is .07 going forward and — .04 going
outboard across-ship. How much higher is the highest corner of the
opening than tfre lowest?
20. An oblique cylinder has a vertical height of 3 in. and circular
bases of radius 2 in. The center of the upper base is directly over a point
on the circumference of the lower.
Find the longest and shortest
diameter of a plane section of the
cylinder passed at right angles to
the axis.
21. Given 6 = 10021, A =48°
59', 0 = 76° 3'; find the other
parts. L
22. Two forces OA = 16 Ib. and OB = 10 Ib. have an included angle of
120°. Find their resultant R, and the angle which it makes with OA.
Check by a protractor.
23. Like Ex. 22 but with OA = 125 Ib., OB = 99.8 Ib., and the included
an{$le equal to 50°.
24. Find TC (the height of the mountain top in the figure above),
if AB — l mi., and A, B, C are in a horizontal plane.
188 MATHEMATICAL ANALYSIS IV, § 126
25. Like Ex. 24, but making AB = 3875 ft., Z BAG = 82° 23', Z ABC
= 91° 40', and the elevation angle at A =4° 45'.
26. A 35-foot flag-pole on top of a building is observed from a point
P on a level street to have elevation angles of 50° and 45°, top and
bottom. How far is P from a point on the ground directly below the
pole?
27. A tree standing erect on a hillside whose inclination is 18° 12'
subtends at two points A and B directly in line down the incline angles
of 10° 25' and 15° 40'. If AB = 399.6 ft., find the height of the tree.
28. Find Z A from the equation (tan A)3 +3 (tan A) =47, first let-
ting tan A=x. (Equations like this often arise in finding the position
of a comet.)
[29.] Two stars A andB are at distances 3.5 XlO15 and 4Xl015miles
from the earth ; and the angle between their directions (from here) is
60°. Find their distance apart.
[30.] The sides of the triangular base of a prism are 7 X 10~4, 4X 10~4,
and 5 X 10~* cm. Find the largest angle.
31. A 20-foot ladder leans against a vertical wall. If its foot is
pulled away horizontally at the rate of .5 ft./sec., how fast is the top
descending when the inclination is 53° 7'.8?
32. The base of a solid is a quarter-circle of radius 10 in. Every
vertical section parallel to one side is a triangle whose base angles are
90° and 35°. Find the volume.
33. Find the total force of water pressure against a trapezoidal dam,
whose longer base ( = 20 ft.) is at the surface, whose acute angles are
both 70° and whose lower base is 8 ft. below the surface.
34. Find the slope and inclination of the curve y = .3x2 at x = 2.
35. Find the vertex angle of the largest cone which can be sent by
Parcel Post. (See Ex. 9, p. 34.)
36. Solve Ex. 6, p. 116, if the shape of the pile is changed so as to
make the vertex angle 110°.
37. In Ex. 22 above, if OB increases at the rate of .4 Ib./min., how
fast will R be changing when OB = 16 Ib. ?
CHAPTER VI
LOGARITHMS
NUMERICAL CALCULATION
§ 127. Estimating Results. In making a numerical cal-
culation it is important to estimate the result roughly in
advance. This will check any gross error, — such as mis-
placing the decimal point, etc.* Moreover, for some pur-
poses, rough estimates suffice in themselves, making accurate
calculations unnecessary. The following devices are fre-
quently useful.
(I) To estimate a product or quotient, use " round num-
bers," and cancel when convenient.
E 681.6X7 946 OOP OOP _700X8 OOP OOP OOP _300 aDDrox
20600000X(30.27)2 20000000X900
The actual value is about 273.6 ; but we are much better off to know
that it is somewhere near 300 than to have no idea at all whether it is
nearer, say, 5 or 5 000 000.
To make a closer estimate, notice whether each factor
has been increased or decreased, and by about what fractional
part. Make rough allowances accordingly.
(II) To estimate a root, use round numbers and group the
figures according to the index of the root, as in § 12. Treat
a fourth root as the square root of a square root, etc.
E.g., V265 800 = >/27 00 00 = 500+, = 520, say ;
also v/«QOOO 7989 = \/.000 080 = .04+, =.043, say.
* In using a Slide Rule, an estimate is the only simple means of pointing
off the result.
189
190 MATHEMATICAL ANALYSIS [VI, § 128
§ 128. Scientific Notation. In scientific work, numbers
which are very large or small are expressed briefly in such
a form as 3.67X1012, or 5.94 X10~8, instead of being written
out in full. This avoids operations with long rows of zeros
or decimal places.
To write out in the ordinary way any number given in this
" Scientific Notation," we simply perform the indicated
multiplication, — i.e., move the decimal point a number of
places equal to the exponent, supplying as many zeros as
may be needed.* E.g.,
7.69 X106 =7690000. (Point moved 6 places.)
4.27 X 10~5 = .000 0427. (Point moved 5 places.)
Conversely, to express in <{ Scientific Notation " any number
given in the ordinary way, we simply factor out the proper
power of 10. Placing the decimal point wherever we want
it, we note how many places it was moved to get it there.
Thus,
27 180 000 = 2.718 X107
.00000483 = 4.83 X10~6
For reasons which will soon appear, we usually place the
decimal point after the first significant figure. Thus, 2.718 X 107
is preferable to 27.18Xl06.or .2718 X108, — unless we have
to compare this number with other numbers carrying the
factor 106 or 108, or unless we need, say, 106 in order to
extract a cube root evenly, etc.
In calculating with numbers expressed in scientific no-
tation, we combine the various powers of 10 according to
the laws of exponents, viz.,
(A) Multiplying: 10* • 10" = 10^"
(B) Dividing: WX+WV = WX-"
* Since 10"" means l/10n, multiplying by a negative power of 10 is
the same as dividing by the corresponding positive power, — i.e., moving
the decimal point to the left.
VI, § 128] LOGARITHMS 191
(C) Finding powers : (10x)n =10nx
(D) Finding roots : VlO* = 10"
Ex. I. Calculate /- (1890° 000)
180 000 000 000
Rewritten j_ (1.89 X107)3X (6.15 X10"9) _1.893X6.15X1012
^34.18 XlO12 "v'MISXlO4
Estimate / = (7 X 6 -J- 3+) X 108 = 14 X 108.,
By calculation f = 12.794 X 10s.
Notice that the powers of 10 are combined much more easily than
their coefficients 1.89, 6.15, etc. If the latter could also be expressed
as powers of 10, the whole calculation would be very simple.
EXERCISES
1. Translate into ordinary notation: 6.54 XlO5; 3.91 X10~4;
diameter of an " H " molecule = 5.8 X 10~8 cm. ; weight = 4.6 X 10~24 gm.
2. Translate into "Scientific Notation," with the decimal point
after the first digit :
1 day =86 400 sec., 1 k.w. = 10 000 000 000 ergs,
1 mi. = 161 000 cm., 1 cm. = .000 006 21 mi.
1 gm. = .000 000 984 ton, 1 cc. = .000 0353 cu. ft.,
Air weighs .00129 gm./cc. ; hydrogen, .000 0896 gm./cc.
In Ex. 3-9 estimate the answer very roughly. Then calculate it, accu-
rate to three figures.
3. One wave of sodium light has a length of 5.89X10"5 cm. How
many waves to 1 mi. ?
4. The energy needed to raise the temperature of 1 gm. of water by
1° C. is 4.16 XlO7 ergs. How many tons of water can be raised 5° in
temperature by 9.62 XlO15 ergs?
6. How far does light travel in 1 hr. if its velocity is 2. 999 XlO8
meters /sec. ?
6. Evaporating at the rate of 1 cu. ft. /sec., how long would a cubic
mile of ice last ?
7. For each degree rise in temperature glass expands by .000025
of its volume at 0° C. How much expansion occurs between 300° and
425°?
8. The distance from the sun to the earth is 9.29 XlO7 mi.; to
Neptune, 2.79 XlO9 mi., and to the nearest fixed star 2.5 XlO13 mi.
192 MATHEMATICAL ANALYSIS [VI, § 129
Representing the distance to the earth by 1 inch, how far would it
be on this scale to Neptune? To the nearest star?
9. 1 ft.-lb. per sec. = 1.356 X 10~3 kilowatts. How many cubic feet of
water falling 1 ft. /sec. would supply the current needed for an 80-watt
lamp if only one tenth of the actual power can be converted into cur-
rent?
§ 129. Calculating by Combining Powers of 10. We come
now to the best system of numerical computation ever in-
vented, — by which we can make calculations in a few
minutes that would otherwise require days or even years.
Here is the idea : Every number is some power of 10, and
can be expressed as such by means of certain tables. Hence
to make a calculation we have merely to combine exponents.
For instance, suppose we wish to find ^/a3/b where a and
b denote some given numbers. And suppose the tables
show that
We build up the required quantity as follows :
a3= 102-73063 (Multiplying the exponent .91021 by 3.)
b =101-10054
/. ~ = 101-63009 (Subtracting the exponent 1.10054.)
o
= 10-i«i9 (Dividing the exponent 1.63009 by 11.)
And when we have seen from the tables what number this
final power of 10 equals, we shall have found the required
root.
Notice that in using these exponents the operation of cubing a is
replaced by the mere multiplication of an exponent by 3 ; a long division
(a3 -3-6), by the subtraction of an exponent; and the very difficult ex-
traction of an eleventh root, by the mere division of an exponent by 11.
VI, § 131] LOGARITHMS 193
Very little more work would be required to find even a 67th root or a
211th root.
The tables are easily used, but to understand them thoroughly we
must first note some further facts concerning powers of 10.
§ 130. Numbers as Powers of 10. The statement that
every number is some power of 10 should perhaps be ex-
plained briefly.
Consider, for instance, the number 75, which is clearly not
an integral power, being greater than 101 and less than 102.
Neither is it a fractional power. For a fractional power is
a root; and it can be shown that extracting -a root of any
integral power of 10 could never give 75 exactly.
When we say that 75 is some power of 10, we mean an
irrational power.
That is, fractional powers can be found which will approximate 75
as closely as we wish :
101-875 =74.99, 101-87506 = 74.9998, etc.
And the limit approached by a certain sequence of such fractional
powers, as the exponent approaches a certain irrational limiting value,
is exactly 75.
Similarly for other positive numbers. Negative numbers
will be considered later. (§§ 141, 321.)
§ 131. Logarithm Defined. In the equation
the exponent 1.87506 ••• is called the logarithm of 75, written
log 75. Thus,
log 75 = 1.87506 •••.
In general, the logarithm of any number is the exponent
of the power to which 10 must be raised to produce the number.
A logarithm usually consists of two parts : an integer and
a decimal. The decimal is found from a table, the integer
by inspection, — as will be explained shortly.
194
MATHEMATICAL ANALYSIS [VI, § 132
§ 132. Logarithms of Numbers between 1 and 10. A
principle which is very basic and will be used continually
is this :
// a number N lies between 1 and 10, its logarithm consists
of a decimal only.
For N lies between 10° and 101, and hence
N = I00+decimal, or log N = 0 + decimal.
Conversely, if log N is a positive decimal only, then N lies
between 1 and 10. (Proof?)
§133. Use of Tables. The logarithms of numbers between
1 and 10 can be read directly from a table. The logarithms
of other numbers are obtainable from them. (§ 134.)
Five-place tables are accurate enough for most purposes,
and their arrangement is like that of larger tables. A part
of a typical page is reproduced here.*
N
150
17609
638
667
696
725
754
782
811
840
869
51
898
926
955
984
*013
*041
*070
*099
*127
*156
52
18184
213
241
270
298
327
355
384
412
441
53
469
498
526
554
583
611
639
667
696
724
Explanation. The first three figures of the number are
shown in the N-column at the left, and the fourth figure in
the JV-line at the top. A decimal point is to be understood
after the first figure in the N-column, so that these numbers
are really 1.50, 1.51, etc.
The logarithms appear in the body of the table, and are
understood to be decimals only, and of five places throughout.
Their first two figures are printed (only occasionally) in
the 0 column at the left.
* The Macmillan Tables include a particularly good collection of auxiliary
tables.
VI, § 133] LOGARITHMS 195
Ex. I. To find the logarithm of 1.502, we look opposite 150 under
2. We read 667, with 17 at the left. Hence
log 1.502 = .17667, or 1.502 = 10-17667.
Ex. II. To find log 1.514, we look opposite 151 under 4 :
log 1.514 = .18013, or 1.514 = 10-18013.
(The asterisk indicates that the first two figures of the logarithm have
changed from 17 to 18, as the last three figures have changed from
984 to 013.)
Conversely, if we have given a logarithm and wish to
find the number, we simply locate the given logarithm in the
body of the table and see what number corresponds to it.
Ex. III. If given log AT = .18355, we locate this value opposite
152 and under 6.
/.AT = 1.526; i.e., 10-18355 = 1.526.
If a given logarithm does not appear exactly in the table,
we take the one nearest to it, or else interpolate by pro-
portional parts. (This is made easy in § 138.) Similarly
if a given number has more than four places. If it has
fewer than four, we mentally affix zeros; e.g., 1.5 = 1.500.
EXERCISES
1. Express in the language of logarithms the fact that 7 = 10-84510;
that 200 = 102-30103.
2. What is meant by saying that the logarithm of 80 is 1.90309?
That the logarithm of 1.1 is .04139? Express by an equation in each
case.
3. Express the following numbers as powers of 10 by inspection,
and state what the logarithm of each is : 1000; 10; 100; .1; .001.
4. Express as powers of 10 by means of tables: 2.718; 5.68;
7.945. Translate into logarithmic notation, as log 2.718= . . ., etc.
5. Given log a = .62459, log 6 = .78017, log c = .01442, log d = .96037,
express a, 5, c, d as powers of 10. Also look up their values, to the
nearest fourth figure.
6. What is the meaning of z?? Of x H? Of 101*? Of 101-41?
How could you find the approximate numerical values of 10- 5, 10 25,
10-125, and 10-75, without tables?
196 MATHEMATICAL ANALYSIS [VI, § 134
§ 134. Logarithms of Larger or Smaller Numbers. The
logarithm of a number greater than 10 or less than 1 can be
found by using the idea of Scientific Notation.
Ex. I. 1514000 would be 1.514X106.
And 1.514, being between 1 and 10, can be found in the table : 1.514 =
10.i8oi3 Multiplying this by 106 gives, on adding exponents :
1514000 = 106-18013, or log 1514000 = 6.18013.
Ex. II. .01514 would be 1.514X1Q-2.
/. .01514 = 10-18013 X 1C-2 = 10-18013-2.
For reasons to be explained presently (§ 136), it is customary in cases
like this not to combine the negative integer with the positive decimal,
but rather to keep the exponent expressed as a difference.
Observe that the decimal part of the logarithm is the same
for all these numbers: 1.514; 1514000; and .01514.
So would it be for any other number having these same
digits, 1, 5, 1, 4, in this same order. (Why?)
To find a number when given its logarithm, we simply
reverse the steps above, — as in the following examples.
Ex. III. Given log AT = 4. 18013, or N = 104-18013.
This is evidently the same as N = 104 X 10-18013. And the latter
exponent, being a decimal only, can be found among the logarithms
of the table : 10-18013 = 1.514.
.-. AT = 104X(1.514) = 15140.
Ex. IV. Given log N = .18184 -3, or N = 10-18184-'.
By tables this decimal power of 10, without the —3, would equal
1.52. The effect of the —3 is to multiply by 10~3, making
N = 1. 52 X10~3 = . 00152.
With practice all these operations may be abbreviated and
performed rapidly, — merely by inspection.
§ 135. Summary. (I) Every positive number is some
real power of 10 : the exponent of the power is the logarithm
of the number.
VI, § 135] LOGARITHMS 197
(II) The integral part of a logarithm (or characteristic,
as it is called) -is found by inspection :
For any number between 1 and 10, the characteristic is zero.
Thus, 1.52 = 1018184, or log 1.52 = 0.18184.
For any other number, think of its Scientific Notation.
Thus, 37200 = 3.72 Xl04=10dec-+4, or log 37200 = dec. +4;
and .00458 = 4.58 XlO-3=10dec-3, or log. .00458 = dec. -3.
Of course, we need not write out all these steps.*
(III) The decimal part of a logarithm (or mantissa, as it
is calldd) is read from a table. It is the same for all numbers
which differ only in the position of the decimal point.
(IV) In going from a logarithm back to the number, we locate
the mantissa in the body of the table, and read off the figures
in the number. If the characteristic is zero, the decimal
point falls in the standard position, after the first figure.
If there is a characteristic (±c), the point moves to the
right or left c places from the standard position.
(V) Calculations can be made by combining powers of 10.
The work should be so arranged that the exponents to be
combined will be near one another, and in a column.
Ex. I. Compute f-d51.4)'X6927
^735 000 000
151.4 = 102-18013 .'. (151.4)2 = 104-36026
6927 = 1Q3.84055
Product = 1Q8-2QQ81
735 000 000 = 108-86629 /. ^735 OOP OOP = 102-95543
Looking up the mantissa .24538 we find 1.759. The characteristic 5
moves the point five places. Hence /= 175 900.
* Simply point with the pencil at the standard position of the decimal
point after the first significant figure, and count up the power of 10 which
would factor out. Try this on the following :
6981 = 103+dec- 25 000 000 = 107+dec-
28.9=101+dec- 314.16=102+dec-
.657 = 10dec -1 .000 000 99 = 10dec--7
198 MATHEMATICAL ANALYSIS [VI, § 135
Remark. Many rates, maxima, areas, etc., as found by
differentiation or integration, must be calculated numerically
by means of logarithms. And that is possible only because a
logarithm varies with its number in a definite way. The
tables show this variation: in other words, they give the
values of a certain function called the " logarithm."
EXERCISES
1. Express these numbers as powers of 10 :
(a) 876.5 ; 7504 000. (First write the Scientific Notation.)
(6) 49.12 ; 582 000. (Think of the Scientific Notation.)
What is the logarithm of each of these numbers?
2. Write as powers of 10 the numbers whose logarithms are
3.495G8, 7.91219, 2.07734, 1.84510.
Look up each number, reading the fourth figure which is nearest.
3. Find the product of the numbers in 1 (a). Likewise in 1 (6).
Check each by actual multiplication.
4. Make each of the following calculations (to four figures) by
expressing the given numbers as powers of 10, in a vertical column, and
combining. Estimate each result roughly in advance as a check.
(a) 52.89X4791X3.809, (6) (7.058)3X20650,
(c) 986500-^287.4, (d) -086.17,
(e) (89.87)2X601.8^-4980, (/) Vl750-M2.5,
(g) (30.59)7-?-v/6031000, (h) 100000 -=-(419.9X9.083),
754-4
31.4X4.146' V98020000 '
5. Proceed as in Ex. 1 (a) for the numbers .00487 and .9216 ; also
as in 1 (6) for .0658, .000 097, and .000 5108.
6. Find what numbers the following powers of 10 equal :
1Q-81305-2 JQ.49136"7 1Q.73062-1
7. Calculate: (a) .0001683X246700; (6) .009875^-5.169.
8. Look up log 2.5486, interpolating by proportional parts.
9. Find the slope of the curve y = 1.73 xf at x = 20.
10. Plot a graph showing how log x varies with x from x = 1 to x = 10.
If the logarithms of two numbers were known exactly, would inter-
VI, § 137] LOGARITHMS 199
polation by proportional parts give too large or too small a value in
finding an intermediate logarithm?
§ 136. Avoiding Negative Mantissas. It would be in-
convenient at the end of a calculation to come out with
such a result as
for the tables give only positive mantissas. And if we used
the definition of a negative power, writing JV = 1-J-1039685,
we should have to look up the latter power and then perform
a long division to get N.
To avoid such difficulties we take care to .keep our man-
tissa positive at every step of a calculation. This can be
done, even when we have to subtract a larger logarithm from
a smaller, by using a simple device :
Increase the smaller logarithm by some integer, making it
now the larger, and at the same time indicate the subtraction
of a like integer, so as to keep the net value unchanged.
Ex. I. Calculate z = -^. [Estimate, x = . 000 35.]
By tables: 1.58 = 10-19866 and 4326 = 103-63609.
Increase the first exponent by 4, with - 4 affixed. 1 .58 = lO4-19866"4
Subtract the exponent 3.63609. 4326 = 103-63609
Look up the resulting positive mantissa, and point off
according to the characteristic —4. z = .000 3652.
§ 137. Operations with Negative Characteristics. When
looking up the logarithm of a small number, — as already
stated, — we do not combine the negative characteristic
with the positive decimal, but merely indicate the sub-
traction, in the form, say, .01514 = 10'18013-2. This pro-
cedure avoids negative mantissas, and also saves labor.
In working with such combination logarithms, there are
a few points to be looked out for, as shown in the following
examples.
200 MATHEMATICAL ANALYSIS [VI, § 137
(I) Raising to a power : say z = (.4074)5.
By tables : .4074 = 10-61002-1. .'. x = 103-0501°-5.
(Observe that we have multiplied the entire exponent by 5, — of course.)
The resulting exponent is clearly equal to .05010 — 2, simply dropping
3 —3, or zero. Looking up the mantissa .05010 and pointing off accord-
ing to the —2, we find z = .01122+.
(II) Extracting a root: say z =
By tables : .1998 = lO-30060'1.
Dividing this exponent by 3 would give .10020 — .33333, producing a
negative mantissa. To avoid this, we may add 2— 2 to the original
logarithm, making it 2.30060—3, still the same value. Then we can
divide evenly :
x = ^102.30060-3 = 10.76687-1 = >5846 (by tables).
To extract any other root we should likewise make the
negative integer exactly divisible by the index of the root.
Of course, we must do this without changing the value of
the combination, i.e., by adding zero in the form n — n.
(Ill) Dividing: say x = . [Est., .05+.]
By tables : Modified form (adding 1 — 1) :
.003166 = 10-50051-3 .003166 = lO1-5005'-"
.06314 =10-8003°-2 .06314 =10-8003°-2
The subtraction gives finally : x = 10-70021-2 = .05014.
EXERCISES
1. If we have to subtract —3 from —1 in making a calculation,
what will this give ?
2. Estimate the following values roughly, and then calculate to
four significant figures :
(a) 43.65-^917.8, (6) .5127 -i- .398, (c) .0683 -=-149.5,
(d) .002957 X. 6849, (e) (.287)8, (/) .0781 4 X. 00997,
(0) ^.000000007, (h) ^.007, (i) 2^.1,
0') ^. 00049728 X. 198, . (fc) (.4657)", (/) (.1624)15(l-642)30.
3. Likewise calculate the following :
(a) 9.875 X. 06543X21. 37, (b) .007968 X V^499 X ^6lU7,
/ , 6587 X. 04659 , « ^8725000
762.8 X. 561 ' 48.75 X. 548'
VI, § 138]
LOGARITHMS
201
4. Estimate and calculate Young's modulus for steel from the
formula Y =mgl/(-,rrzs), if ra = 1005, 0=980, 1=87, r = .0249, and
s = :02183.
5. The diameter (d in.) which a water pipe L ft. long must have
to discharge Q cu. ft. per sec. under a head of H ft. is d = 5.75 ^YLQ^/H
where F is a friction factor. If F = .0197, L = 250, Q = 7.5, and H = 1 1 .25,
find d. [If you can mentally find the product FL or replace Qz by its
value, the logarithmic work will be shortened.]
§ 138. Tables of Proportional Parts. In the margins
of logarithmic tables there are small auxili-
ary tables which make interpolation easy.
Ex. I. Find log 1.5146.
The required logarithm lies between
.18013 and .18041, which differ by 28
(units of the fifth place.) Select the
marginal table headed 28. This tells how
much to add to the smaller tabulated
logarithm [.18013] because of any fifth
figure in the given number. In our ex-
ample the fifth figure is. 6 : add 16.8 (i.e~y
17), making
log 1.5146 = . 18013+. 00017=. 18030.
Ex. II. Find N if log N = .18037.
The next smaller logarithm in the table is .18013, belonging
to 1.514. (This gives the first four figures of the required
number AT.)
Now log N exceeds .18013 by 24 units. Hence our
problem is this : What fifth figure in N would add 24 to the
logarithm, in a total difference of 28? The nearest to 24
shown in the marginal table is 25.2 and this is opposite
9, — the required fifth figure of N. Thus AT =1.5149.
Remarks. (I) In using these auxiliary tables, note carefully which
you are finding : how much to add to a logarithm, or what fifth figure
to affix to a number. With practice the operations can all be performed
mentally, and only the final result written.
STH
FIG.
ADD
TO LOG
1
2.8
2
5.6
3
8.4
4
11.2
5
14.0
6
16.8
7
19.6
8
22.4
9
25.2
202 MATHEMATICAL ANALYSIS [VI, § 138
(II) These auxiliary tables are based upon proportional parts. If
ten units in the fifth place of the number (one unit in fourth place)
make a difference of 28 in the logarithm, then 3 units make a difference
three tenths as large, or a difference of 8.4.
(III) What if a given number has six figures, say 1.51436? The
36 units of the sixth place will change the logarithm by .36X28. By
the little table, .3X28 = 8.4 and .06X28 = 1.68. Thus .36X28= 8.4 +
1.68 = 10.08. Simply add 10. (There is almost no chance of securing
greater accuracy by preserving figures beyond the fifth place.)
EXERCISES
Get all results in the following exercises accurate to the nearest
unit in the fifth place.
1. Look up the logarithms of :
387.26, 97193, 1.0487, .0056207.
2. Find the numbers whose logarithms are :
.30113, 8.47702, 8.69014-10, 4.96046-10.
3. Estimate and calculate :
, N (116.3)2(8.713) ,M ^3125
21739 9.8751X82.3'
4. The time of swing of a pendulum is T = 2 nVl/g. Estimate
and calculate T if 1 = 1 and g = 32.083.
6. The amount of $P with 5% compound interest after n years is
A =P(1.05)». Find A if P = 4750 and n = 30.
6. In Ex. 5 find the principal $P required to yield an amount of
$10000 after 20 years.
7. Find the radius of a steel sphere weighing 1 ton, if 1 cu. ft of
steel weighs 480 Ib.
8. Find the area under the curve y = .85 xl from x = 1 to 8.
[9.] Translate into logarithmic notation in a parallel column each
of the equations involving a power of 10 which was used in Ex. 3 (a).
Thus:
116.3 = 102-06658 1 log 116.3=2.06558,
etc. How was the "log" of the square obtained from the "log" of
the number ? The " log " of the product from the " logs " of the factors ?
The "log" of the fraction?
VI, § 139] LOGARITHMS 203
§ 139. Laws of Logarithms. Since logarithms are ex-
ponents, they combine according to the usual laws of ex-
ponents. These are already familiar ; but in what follows
it will be convenient to have them restated in logarithmic
form, as follows :
(I) The logarithm of a product equals the sum of the log-
arithms of the factors :
log (ac)=log a+log c.
(II) The logarithm of a quotient equals the difference of the
logarithms of the dividend and divisor:
log - = log a— log c.
c
(III) The logarithm of a power of a number equals the index
of the power times the logarithm of the number :
log an = n log a
(IV) The logarithm of a root of a number equals the logarithm
of the number , divided by the index of the root :
log\/a = -loga.
If a formal proof of these laws is desired, it can be given
as in the following illustrations :
PROOF OF (I) :
If log a = x and logc = y,
i.e., if 0 = 10" and c = 10«,
then ac = (10*)(10") = 10*+«,
which shows that logac = x+y = log a+log c.
PROOF OF (IV) :
If log a — x,
i.e., if a = 10*,
then
ic
The proofs of (ID and (III) are similar. (Ex. 1. :p. 205.)
which shows that logv/e^- =- log a.
n n
204
MATHEMATICAL ANALYSIS [VI, § 140
§ 140. Abbreviated Form. In calculating by combining
powers of 10, the actual operations arc performed upon the
exponents or logarithms. Hence it would suffice to set
down the logarithms alone, and work with them. This
should, however, be done in an orderly manner and labeled
clearly.
The following example shows a calculation worked out
with powers of 10 as heretofore, and the same calculation
in the abbreviated form.
Ex. L Calculate x
Exponential Form
J(25.89)3(.0125)
J 927
Logarithmic Form
25.89 = 101-41313
25.S93 = 104-23939
.0125 = 10-09691-2
product = 104-33630-2
927 = 102-96708
fraction = 101-36922-2
x = 1Q.68461-1
.*. x = . 48373.
No.
Loo.
•25.89
1.41313
25.89s
.0125
4.23939
.09691 -2
prod.
927
4.33630-2
2.96708
frac.
x
1.36922-2
.68461-1
= . 48373.
The latter form is the one which we shall use hereafter.*
§ 141. Arrangement of Work. Before looking up any
logarithms for a calculation, we should always plan the work
in full and lay out a " skeleton form," providing a place for
each step and labeling it. We can then devote our entire
attention to the tables and to the necessary arithmetic.
This will save time, eliminate many blunders, and keep
the calculation in a presentable form.
The following example shows an arrangement of work
pretty satisfactory for the more complicated calculations.
* In using this be careful not to write the = sign between a number and
its logarithm. The resulting confusion would be serious sometimes.
VI, § 141]
LOGARITHMS
205
Ex. I. Calculate x =
Plan : The logarithm of x will be obtained as follows :
log z = [J log a+| log 6]-| [5 log c+log d+% log e],
where a, b, etc., denote the given numbers, .5212, etc.
The following " skeleton form " (printed' in black type)
provides for these steps :
No.
Log
No.
Log
(a) .5212
19.71700-20 [2
9.85850-10
Va
bl
9.85850-10
1.90778
(b) 13.953
1.14467 X5
Nu.
De.
11.76628-10
2.94338
5.72335 |3
1.90778
(c) 8.2
0.91381 X5
z
4.56905
8.82290—10
(e) .0973
28.98811-30 (3
No.
Log
<*
d
<fe
4.56905
1.65501
9.66270-10
Prod.
5.88676 |2
De.
2.94338
9.66270-10
.'.X =.066512
Remarks. (I) The above form can of course be modified. Some
computers, for instance, would perform the simpler multiplications
and divisions mentally, and enter the results directly in the columns
where used. The essential thing is not the use of a particular form,
but the laying out of some good form in advance. It should provide
places for the logarithms as they come from the table, for the modified
logarithms, and for all necessary combinations.
(II) The negative characteristics above are so written that the sub-
tracted integer is 10 or a multiple of 10. This system is used by com-
puters, for reasons of uniformity and convenience in certain tables.
(§ 149.) Evidently -1 may be written either 9 ••• —10, or 19 ••• -20,
or 29 ••• -30, etc.
EXERCISES
1. Prove log - = log a— log c ; log an=n log a.
c
2. Prove log - = log a +2 log b-(log c+£ log d).
cvd
206 MATHEMATICAL ANALYSIS [VI, § 142
3. Estimate and compute to five figures :
(a) (3.6211)» (6) (9.2651)4
' (86.21)2 199870 »
,, (7.3) 5X. 06162 ,» 5.086 X(.OQ8769)3
V98 020 000 ' .98017 X (.019842K
I (.0
*V
058)3X421.61 X86
^.80008 *V50X. 045 X (200.15)*
4. The best elevation E in. for the outer rail of a railway curve
of radius R ft. is given by # = 12 GV*/(32.2 R) where G feet is the
gauge and V ft. per sec. the greatest speed used. Estimate and calcu-
late E if G=4.71, F = 66, and £ = 5730.
5. The volume of an oblate spheroid is V = $ irl&r. Estimate and
compute the volume of the earth if £ = 3963.3 and r = 3949.8.
6. An iron casting consists of a cone and hemisphere united, the
flat side of the latter coinciding with the base of the cone. If the
common radius is 9.8 ft. and the height of the conical part is 3.88 ft.,
find the total volume and surface area of the casting.
§ 142. Calculations with Negative Numbers. There is
no " real " value of x, positive or negative, for which 10* is
a negative number. That is, a negative number cannot
have a " real " logarithm.
But calculations involving negative numbers can be
made as follows: First decide by the elementary rules
of signs whether the final result should be positive or nega-
tive. Then find its numerical value by logarithms, treating
all the given numbers as positive.
Ex. I. Calculate s = \f- 3'14( 756'8)2 •
x(-17.5)5\/-100
The combined effect of all these negative signs is to make
x negative.
_ »| 3.14(56.8)2
17.5)
VI, § 143] LOGARITHMS 207
By logarithms the value of the radical itself is R = .14731.
.'. x=~. 14731.
N.B. It would be correct to write log R = $ [log a+ •••]; but not
to write any similar equation for log x, since x has no real logarithm.
§ 143. Sums and Differences. Suppose we have to make
a calculation which calls for the addition of. two quantities,
as in _
x = V(1.1825)20+ ^
The quantities can be expressed as powers of 10, giving say
a; = Vl01-45600+101-64743,
but they cannot be added by merely combining the ex-
ponents.
In what sort of calculation would you have to add the exponents or
logarithms ? •
We must evidently look up the numbers which these two
powers of 10 equal, and then add those numbers. And
similarly in any other calculation involving a sum or differ-
ence, we must go from logarithms back to numbers before
adding or subtracting.
In the example above the calculation could be arranged
conveniently as follows, using U and V to denote the two
quantities :
£7 = (1.1825)20, 7 = ^87556, x
No.
1.1825
(1.1825)20
Loo No.
.07280 87556
1.45600 ^87556
Loa No.
4.94229 U+V
1.64743
LOG
1.86321
.93160
(7 = 28.576 .'. V = 44.405 .'. x = 8.5428
72.981
Thus there are in reality three separate calculations: To find f/,
to find F, and to find x. The last cannot be started until we have
finished the first two and have added the numbers U and V.
208 MATHEMATICAL ANALYSIS [VI, § 144
§ 144. Short-cuts. Sometimes by making a preliminary
change in the form of the quantity to be computed we can
save considerable work.
(A) If some of the given numbers can be canceled or com-
bined, mentally, fewer logarithms will need to be handled.
E.g., A =?r(25)2, replace (25)2 by 625, and save one opera-
tion. Or, in V = %Trr2h, cancel 3 into the value of h; and
one fewer logarithms will be needed.*
(B) By factoring, a sum or difference can sometimes be
reduced to a product of known numbers.
E.g., the total area of a cylinder, A =2 irr-+2Trrh, may be
written A = 2 wr(r+h). If given r= 113.4 and /i = 246.6 we
have r+/i = 360; and multiplying by the factor 2:
A = (113.4) (720)7r'.
The two separate calculations needed to find A from the
first formula are thus replaced by one simple calculation.
Similarly, suppose we wish to find one leg a of a right triangle, having
given the hypotenuse and other leg, c = 983.5, 6 = 726.2.
Since a = V(983.5)2 — (72G.2)2, the calculation apparently involves
going from a logarithm back to the number three times in all. But
the difference of two squares is factorable: c2— 62 = (c+6) (c — 6).
Here
c+6 = 1709.7, c-6 = 257.3. .*. a = V(1709.7) (257.3).
Thus a is very readily computed.
EXERCISES
1. Estimate and compute to five significant figures :
(a) V485.72-321.42, (6) [V737l43-1.0252]i
2. (a) Estimate and compute by logarithms (Vs — VA)4.
(6) Calculate the same value by using tables of roots.
3. Find the minimum value of y = x*— 19z+8.
* log IT, being often used, should be memorized or inserted on the " 300
page " of the tables.
VI, § 145] LOGARITHMS 209
4. Find the total area of a cylinder whose base radius is 79.5 cm.
and whose height is 189.2 cm.
5. The hypotenuse and one leg of a right triangle are respectively
495.73 ft. and 312.45 ft. Find the other leg.
6. Find the angle of elevation of the sun when a tower 458.75 ft.
tall casts a shadow 1278.9 ft. long on horizontal ground. Check by
measurement.
7. Compute to five figures :
(a) «/-• 032963 ^ »/ (-268.94)3
* 7.9626 ' >7r(-.048167)2'
§ 145. Compound Interest Formula. When a sum of
money is left at compound interest for a long time, the
amount finally accumulated is rather tedious to calculate
by elementary arithmetic. Business men generally use
interest tables. But there are problems not readily solvable
by the tables. It is well, therefore, to know a general for-
mula, which can be used either to make ordinary calculations
quickly or to solve new types of problems.
For simplicity consider first some particular rate of in-
terest, say 6%. Then if the interest is figured annually,
the amount accumulated at the end of any year will be
106% of the sum at the beginning of the year. In other
words, the sum will be multiplied by 1.06 during each
year.
If the original principal is P, the amount after one year
will be P(1.06) ; after two years, P(1.06)2 ; after three years,
P(1.06)3; and so on. The final amount after n years will be
A = P(1.06)W. (1)
If the interest is compounded semi-annually, the sum will
gain 3% in each half-year, or be multiplied by 1.03. After
n years the original principal will have been multiplied by
this factor 2 n times in all, making
210 MATHEMATICAL ANALYSIS [VI, § 146
GENERAL FORMULA. From these special cases it appears
that the amount of any investment P, after n years, with
interest at any annual rate r (r being a fractional value,
as .06, say), compounded k times a year, will be
(2)
'('+0
This inference is easily proved correct.
Proof: Let S be the sum accumulated at the beginning
of any interest period. Then the interest gained during
the period (one fc-th of a year) will be rS/k ; and the amount
at the end of the period will be S+rS/k, or S(l+r/k). Thus
the sum will be multiplied by (l+r/fc) during each period;
and there are kn periods in n years. • Hence we have (2).
Remarks. (I) Formula (2) should be memorized carefully, as it
covers all cases. For instance, if the interest is compounded annually,
simply put k = l, getting A =P(l+r)n, like (1) above.
(II) The formula is strictly correct, however, only at the ends of
the interest periods, i.e., for integral values of kn. To find A after
10J periods, say, the exact method would be to find A after 10 periods,
and then add simple interest for three fourths of a period. But formula
(2) would give a very approximate result by simply putting n = 10.75.
(III) When compounding semi-annually, at the rate of 6%, the
amount after 1 yr. will be A =P(1.03)2 = P( 1.0609)'. Thus, due to
the frequent compounding, the effective rate of increase is 6.09%;
6% is merely the nominal rate used in figuring.
§ 146. Typical Problems. We can now solve various
typical problems in compound interest by merely substituting
the numerical values in formula (2), and using logarithms.
In each of the following examples, set up the formula
and logarithmic scheme for yourself. Then compare with
the work shown here in fine print.
Ex. I. What will be the amount after 20 years, on an
original investment of $2750 with interest at 5%, com-
pounded quarterly?
VI, § 146] LOGARITHMS 211
Here P = 2750, n = 20, fc = 4, r/k = . 05/4 = .0125.
/. A =2750(1. 0125) 80.
We have merely to add the logarithm of 2750 to 80 times the logarithm
of 1.0125, and look up the number A. [Ans., taking log 1.0125 as
.005395, A =7429.] Observe that it is best to reduce r/k to .0125
before substituting it in the formula.
Ex. II. How much must be invested now to yield $5000
thirty years hence, interest being at 3J%, compounded
annually ?
Here A =5000, n = 30, k = l, r/k = .035.
.'. 5000 = P(1.035)30, or P = 5000/(1.035)30.
We have merely to subtract 30 times the logarithm of 1.035 from
log 5000, and look up the number P. [Ans., P = 1781.40.]
Ex. III. At what rate of interest, compounded semi-
annually, would an investment of SI 750 yield $5000 after
20 years?
Here A =5000, P = 1750, n = 20, fc = 2.
/. 5000 = 1750 (l+-Y°.
Let the unknown quantity 1 +r/2 be denoted by x. Then
5000 = 1750 re40, /. x = .
1750
Subtracting log 1750 from log 5000 and dividing by 40, we find
log x = .01 140, whence x = 1 .0266.
This is l+r/2; that is, l+r/2 = 1.0266.
.-.- = .0266, r = .0532. [Ans., 5.32%.]
2
Ex. IV. In how many years would $983.50 amount to
$3875 with interest at 7% compounded semi-annually ?
Here A =3875, P = 983.5, fc = 2, r/k = .07/2 = .035.
/. 3875 = 983.5(1.035)2*.
There is no method in elementary algebra for solving an equation for
an unknown exponent. This problem will be discussed in § 147.
212 MATHEMATICAL ANALYSIS [VI, § 147
EXERCISES
In these exercises interest is to be compounded annually unless otherwise specified.
1. In 1626 the Dutch bought Manhattan Island for $24. To how
much would this amount in 1920 if it had been at 7% interest?
2. What sum, deposited now, would yield $17500 thirty years
hence, if interest is at 5%, compounded semi-annually ?
3. At what rate of interest will any sum be quadrupled in 25 years?
(Take any convenient sum, say $1.)
[4.] Compute by logarithms ' . Also see if you can solve the
.02531
equation (1.08)n = 3 for n. Can you think of any interest problem
which would require the solution of this equation?
6. Find how much must be invested now to yield $2500, fifteen
years hence, interest at 5%, compounded quarterly.
6. At what rate will $600 yield $2400 after 30 yr.?
7. At what rate will $6000 amount to $15000 in 20 years, com-
pounding semi-annually ?
8. At what rate will any sum double itself in 9 years?
9. What sum set aside when a boy is 1 year old would provide an
education fund of $2000 when he is 16, if 4% interest is obtained, com-
pounded semi-annually?
10. Find the amount of $100 after 1 year with interest at 8%, com-
pounded quarterly. What percentage is actually gamed during the
year, due to the frequent compounding, — i.e., what is the effective
rate?
11. A building costing $4000 must be rebuilt every 20 years. What
sum set aside when the building is erected will provide for its per-
petual replacement, if the cost remains constant and money is always
worth 4%? (Hint: The sum must produce itself plus $4000 in
20 yr.)
§ 147. Finding an Unknown Exponent. Suppose we
wish to solve the equation
2* = 25.
Since 2* and 25 are equal, their logarithms must be equal.
But the logarithm of 2* equals x times log 2 :
/. x log 2 = log 25, or z(.30103) = 1.39794.
VI, § 148] LOGARITHMS 213
That is, x multiplied by .30103 equals 1.39794; and hence
to find x we must divide 1.39794 by .30103 :
This result is evidently about right, since 24 = 16 and 25 = 32.
Notice then that this new problem of solving for an un-
known exponent calls for the division of a logarithm by a
logarithm, — not a mere subtraction of logarithms.
But we could of course avoid this long division by looking
up further logarithms, — just as if we had been given the
fraction to calculate in the first place. Subtracting the
logarithm of .30103 from the logarithm of 1.39794 would
give log x.
We can now return to Ex. IV, p. 211, and find n from the equation,
3875 = 983.5 (1.035)2".
Here log 3875 = log 983.5 +2 n log 1 .035.
By tables : 3.58827 = 2.99277+2 n(.01494).
Transposing 2.99277, and simplifying the coefficient of n :
.59550 = n (.02988).
By division: n= = 19.93.
(We could avoid division by looking up further logarithms.)
It would be useless to calculate n more accurately, since the interest
formula is exact only at the ends of interest periods.
§ 148. Depreciation. In any business it is necessary
to allow for depreciation in the value of buildings, machinery,
etc., due to wear which cannot be made good by current
repairs.
For simplicity it is commonly figured that the value will
decrease by a certain fixed sum during each year, until
finally reduced to the mere " scrap value." But for some
kinds of property it is more accurate to figure the loss during
each year as a certain constant fraction of the value at the
beginning of that year.
214 MATHEMATICAL ANALYSIS [VI, § 149
Ex. I. An automobile costing $2000 loses each year 30% of its
value at the beginning of that year. What will be its value after
5 years?
At the end of each year the value is 70%, or .7, of the value at the
beginning of the year. Multiplying by .7 each year, we get the final
value :
y = 2000(.7)5 = 721.40.
Remark. If 15% were deducted every half-year, this would be
"figuring the depreciation semi-annually at the yearly rate of 30%."
After each half-year the current value would be multiplied by .85,
and after each year by (.85)2.
In general, if depreciation is figured k times a year at any nominal
annual rate r, the value after n years would be
-rV" (3)
EXERCISES
1. Solve for z: 2* =5, 3* = 11, 4* = 100.
2. In how many years would
(a) $50,000 amount to $80,000 at 5%, compounded quarterly?
(6) Any sum quadruple itself, at 5%, compounded semi-annually?
(c) $1250 amount to $2250, at 4J%, compounded quarterly?
(d) $3250 amount to $8000, at 6%, compounded quarterly?
(e) Any sum be doubled, at 4%, compounded semi-annually?
3. An investment of $75,000 depreciates so as to lose in each year
3% of its current value. What will it be worth after 30 years?
4. The same as Ex. 3 if the rate is 5% and the original value $200,000.
6. If depreciation is figured semi-annually at the nominal annual
rate of 20%, what is the actual rate of depreciation per year? (See
Remark above; also (III), p. 210.)
§ 149. Logarithms of Trigonometric Functions. In solv-
ing a triangle, we can use logarithms to perform the
multiplications and divisions. To make this very con-
venient, there are special tables from which we can read
directly the logarithm of each sine, cosine, etc., which is
used, without first looking up the function itself. Part of a
typical page is shown here.
VI, § 149] LOGARITHMS
12° — LOGS OF TRIGONOMETRIC FUNCTIONS
215
t
L sin
d
L tan
cd
L ctn
L cos
d
0
1
9.31 788
9.31 847
59
9.32 747
9.32 810
63
0.67 253
0.67 190
9.99 040
9.99 038
2
60
59
18
19
9.32 844
9.32 902
58
9.33 853
9.33 913
60
0.66 147
0.66 087
9.98 991
9.98 989
2
42
41
59
60
9.35 154
9.35 209
55
9.36 279
9.36 336
57
0.63 721
0.63 664
9.98 875
9.98 872
3^
1
0
L cos
d
Lctn
cd
L tan
L sin
d
'
PRO. PTS.
60 58
2 12 11.6
3 18 17.4
4 24 23.2
5 30 29.0
6 36 34.8
V 42 40.6
8 48 46.4
9 54 52.2
LOGS OF TRIGONOMETRIC FUNCTIONS — 77°
Explanation. With every logarithm in the table —10
is to be understood, except in the third main column headed
L ctn. E.g., the first entry opposite 18' means that
log sin 12° 18' = 9.32344 -10 =.32844-1.*
The labels at the bottom and minutes at the right indicate
that this same value 9.32844-10 is log cos 77° 42'.
Interpolations can be made rapidly by using the marginal
tables of proportional parts, and the narrow columns marked
" d" or " cd " which give the differences between successive
logarithms.
Ex. I. Find log sin 12° 18'.7. Between 18' and 19', the value
increases by 58. By the marginal table seven tenths of 58 is 40.6.
Adding 41 to log sin 12° 18', we find 9.32885-10.
Ex. II. Find log ctn 12° 18'. 1. Here d = 60, and the value is de-
creasing. From log ctn 12° 18' we subtract one tenth of 60 (which being
obviously 6, is not shown in the marginal table), and get 0.66141. A
* Without these tables we should have to lookup sin 12° 18' ( = .21303),
and then look up log .21303 (.32844 — 1). Two interpolations would be
necessary in finding log sin 12° 18'.4.
216
MATHEMATICAL ANALYSIS [VI, § 150
common-sense check is that this result lies between the values given
for 18' and 19', and is much nearer the former.
Ex. III. Given log tan A =0.66130, clearly A = 77° 41'+ Here
d = 60, and the given logarithm exceeds log tan 77° 41' by 43. The
question is, how many tenths of a minute will make a difference of
43 in the logarithm? The marginal table headed 60 says .7 approxi-
mately. (More accurately, 43 is .72 X 60.) Thus A = 77° 41 '.7 approx.
Ex. IV. Given log cos 5 = 9.32852-10. Clearly 5 = 77° 41 '+.
Opposite 41' we read 9.32902, from which the given logarithm differs
by 50. The marginal table headed 58 shows 50 opposite .9. Hence
£ = 77°41'.9.
N.B. We always work from the value shown opposite the smaller
angle, whether this value is the smaller logarithm or not, — for the
simple reason that angles are written in the form 77° 41 '.9 rather than
77042'-.!'!
§ 150. Logarithmic Solution of Triangles. Typical ex-
amples.
Ex. I. Given 6 = 750, A =40°, C = 80°; find a, c, B.
(Estimate, using a protractor: a = 550, c = 850, 5 = 60°.)
Q. 1 a 750 c
s7n^~s7n^~sin~80^
We subtract log sin 60° from log 750, and then add log sin
40° or log sin 80°.
No.
Log
750
sin 60°
2.87506
9.93753
frac.
sin 40°
2.93753
9.80807
a
2.74560
No.
Log
frac.
sin 80°
2.93753
9.99335
c
a =
c =
2.93088
= 556.67
= 852.86
Ex. II. Given 6=21.75,
(Graphical estimate : B = 60°,
c =24.75, A =40°; find a, B, C.
C=80° a = 16.)
Cosine law: a' = (21.75)2 + (24.75)2-2(21.75)(24.75) cos 40°.
The addition and subtraction can bo performed only after going back
from logarithms to numbers. Thus the cosine law is inconvenient for
logarithmic work. Better formulas are derived in §§ 151-153.
VI, § 150]
LOGARITHMS
217
Ex. III. Given a = .8273, b = .9999, C = 90°; find B. This is a
right triangle and should be solved as such :
a .8273
Subtract log a from log b, and look up B directly. [Ans., # =50° 23'. 8.]
EXERCISES *
1. Look up the logarithms of sin 14° 27', sin 78° 22'.4, cos 56° 53'.7,
tan 23° 13'.8, ctn 84° 53'.7. (Check the first by looking up the sine
itself and then its logarithm.)
2. Look up Z A if log sin A =8.76966 - 10 ; log co§. A = 9.92379 - 10 ;
log tan A = 1.27960 ; log ctn A =0.46235.
3. The hypotenuse and one leg of a right triangle are 74.157 and
50.063 inches. Solve the triangle, finding the third side by means of
an angle.
4. In Ex. 3, check in part by finding the third side directly from the
given sides.
5. Solve graphically and by tables the triangle in which a = 738.1,
£ = 78° 14' 42", C = 54°26'.
6. The gravitational acceleration g (cm. /sec.2) is given for any lati-
tude L by the equation g = 977.989 [1 +.0052 (sin L)2]. Find g for the
latitude L= 45° 29'.
7. Solve each of the following oblique triangles for the missing parts :
a
6
c
A
B
C
i
368.42
62° 15'
93° 42'
ii
.038627
42 38
2 13
m
28.935
16 41
32 19
IV
1280.5
58 6.2
48°27'.5
v
47.198
75 12.8
5 8 .3
8. Find the radius of the circle inscribed in a regular decagon whose
perimeter is 286.5 ft.
9. What is the elevation angle of the sun when a pole 106.5 ft. high
casts a shadow 286.9 ft. long on level ground?
* For further triangles to solve see pp. 227, 231.
218 MATHEMATICAL ANALYSIS [VI, § 151
§ 151. Area of a Triangle. To find the area of a triangle
we may first solve for some one of its altitudes by dropping a
perpendicular, and then multiply by one half the correspond-
ing base.
Or if the three sides happen to be known, we can find the
area immediately by using a formula from geometry : *
S = Vh(h-a)(h-b)(h-c) , (4)
where S denotes the area, and h one half the perimeter, i.e.,
(5)
Remark. From (4) we can also derive a formula for the radius r
of the inscribed circle. For by Fig. 74 :
That is, S = hr, or r=S/h. Using here the value of S in (4) above
and putting our divisor h also under the radical, we get, on simplifying :
^
§ 152. Half -Angles. Since the center of the inscribed
circle lies on the bisector of each angle, we have from Fig. 74 :
tan (\ A) =7, tan (J B) =-, etc. (7)
/ m
But l-\-m-\-n is one half the perimeter, or h. (§ 151.)
.'. l = h-(m+ri)=h-a. (8)
* If this formula is unfamiliar, see p. 489 of the Appendix for its deriva-
tion.
VI, § 152]
LOGARITHMS
219
Similarly m = h—b,
Substituting in (7) above :
and
= h—c.
tan (J A)
h-a'
tan (i B)
h-b
, etc.
(9)
where R denotes the radical quantity in (6) above.
Formulas (9) can be used instead of the cosine law to solve
a triangle when the three sides are given. They are well
suited to logarithmic methods.
Ex. I. Find the angles and area of a triangle inVhich a =275.8,
6=361.4, c = 446.2.
The formulas are (9) above, together with
hr.
r-.J(A-o)(A-6)(ft-c)> ;
2 * h
(What steps are needed to find log r? log tan \ A ? log £?)
M>. -
Log No.
Log
£A=19°4' 22"
A=38 ,844
h = 541.7
r
'h-a
1.96348
2.42472
/i-c= 95.5
2.42472
2.25600
1.98000
tan (£ A)
9.53876 -10
r
h-b
1.96348
2.25600
B = 54 2 0
(Check*)
6.66072
2.73376
tan (£ B)
9.70748 -10
£ C = 43 5438
CQI-T .J Q -j f«
r
h
3.92696
r
h-c
1.96348
1.98000
— o< 4y ID
2.73376
••
tan (£ C)
9.98348 -10
S
4.69724
Area, S = 49801.
Final check:
The final check is satisfied closely enough for five-place tables if the
discrepancy between 180° and A+J5+C is less than 6".
To find C from A and B by the relation A +5+0 = 180° would be
undesirable, as it would leave us no simple check.
* A check on h— a, h—b, and h— c, is that their sum, 3/i — (a+&+c),
must equal A. (Why?)
220
MATHEMATICAL ANALYSIS [VI, § 153
EXERCISES
1. Find the angles of a triangle in which o = 63.89, 6 = 138.24,
c = 121.15. Find all independently, and check.
2. Derive the formula tan % B=r/(h—b).
3. In each of the triangles whose sides are given below find the three
angles, independently, and check. Also find the area.
a
6
c
a
b
c
i
289.6
462.5
378.1
iv
.9628
.4315
.6782
ii
514.7
625.8
981.4
V
.00681
.00419
.00745
iii
29.87
19.51
16.23
VI
12980
15642
18326
4. A ladder 36.45 ft. long is set 15.75 ft. from the foot of a sloping
buttress, and reaches 30.38 ft. up its face. Find the inclination of that
face.
5. If two forces of 638.9 Ib. and 1211.5 Ib. have a resultant of 1382.4
lb., what is the angle between them?
§ 153. Tangent Law. On any
side c of a given triangle, as
base, construct an isosceles tri-
angle ABF by extending the
shorter of the other two sides,
say 6, and making Z ABF =
ZA. (Fig. 75.)
Then in A BCF two of the
angles are A + B and A — B,
and the opposite sides are, say,
xandx-6. By § 152,
-A+B
FIG. 75.
h-(x-b) ~~~*v ' "h-x9 (10)
where r is the radius of the circle inscribed in A BCF, and
VI, § 153] LOGARITHMS 221
From (10), by dividing and substituting these values :
g) h-x a-b
h-(x-b) a+b'
Observe what this means for the original triangle A BC :
The tangent of half the difference of any two angles of a tri-
angle is to the tangent of half the sum as the -difference of the
opposite sides is to the sum of those sides.*
This " tangent law " is adapted to the logarithmic solu-
tion of a triangle, when two sides and their included angle
are given, say a, b, and C. For the sum %(A-£B) is known,
and by finding %(A — B) from (11), we can combine to obtain
A and B separately.!
Ex. I. If a = 37.485, 6=28.392, C = 40°, find A, B, c.
a-b = 9.093 A+B =180°-40°
a+6 = 65.877 J(A+*)=70" tan KA-J?) = 9.093
tan 70° 65.877
(How would tan \(A— B) be found from this last equation without
logarithms? How, therefore, when using logarithms?)
6
No.
Log
9.093
tan 70°
0.95871
0.43893
Product
65.877
1.39764
1.81873
tan %(A-B)
9.57891 —
A =90° 46' 6'
# = 49 13 54
c &
To find c use the Sine Law : -r™ = - — ~.
sin C sin B
* In Fig. 75 we took Z A as acute. If it happens to be obtuse, simply
produce b and BF backwards to meet at some point F'. Two angles in
ABCF' will be the supplements of A -B and A+B. Halving these angles
will give the complements of £ (A-B) and \ (A+B). The proof can then be
carried through as above, if we recall that the tangent of the complement of
£ (A -B) is ctn £ (A-B) or I/tan \ (A-B) ; etc.
t The method of § 118, Ex. I, may also be used.
222
MATHEMATICAL ANALYSIS [VI, § 154
Remark. Merely adding the three angles would give no check what-
ever upon the logarithmic work done in finding A and B. Suppose,
for example, that we had erroneously found in the case above :
£(4-.B) = 100, /. A =80°, £ = 60°.
Adding : A +B+C = 180°, which does not show the error.
Why does this fail to detect the error? [Where did we get the value
of | (A +B?)] What formula could be used as a real check upon A
and B?
EXERCISES
1. Given C = 124° 34', a = 52.8, 6 = 25.2. Find the other parts.
2. Given a=41.003, 6 = 48.718, C = 68° 33' 58". Find the other
parts.
3. (a), (6). Find the areas of the triangles in Ex. 1, 2.
4. Find the missing parts of the following triangles; and also the
areas.
a
b
c
A
B
C
i
1285.9
2684.5
42° 38'
11
.9248
.6983
98 15.2
111
62.875
39.487
20° 15.8'
iv
4.1635
5.2940
112 38
V
9.4683
5.6291
51° 16.3'
VI
96.285
112.34
106 28
§ 154. Other Bases. The logarithms which we have been
using are possible because of the fact that every number is
some power of 10. But it is equally true that every number
is some power of 2, or of 7, or of any other positive number,
except 1. Hence it is possible to have other systems of
logarithms, based upon powers of 2, or 7, etc.
For instance, if
5=22.32193,
the exponent 2.32193 is called "the logarithm of 5 to the base 2,"
written Iog2 5.
VI, § 155]
LOGARITHMS
223
And in general, the logarithm of any number to any base
is the exponent of the power to which the base must be raised
to produce the number.
The " common logarithms," to the base 10, which we have
been using, are by far the best for most numerical calcula-
tions, — because of the fact that moving a decimal point
in a number merely adds some integer to the characteristic.
Only one other base is very generally used for any purpose ;
this will be discussed in § 166. But it is well to be familiar
with the following general principles.
No matter what base B we may be using :
log 1 = 0,
For
and
and
log 5 =
(12)
The logarithm of any positive number to any base is easily found
with the help of common logarithms. For instance, suppose we want
Iog2 25. We simply let this equal x, and write the equivalent expo-
nential equation :
Iog225=z, 25 = 2».
Solving the latter equation as in § 147, we find x =4.6493.
§ 155. Slide Rule. Logarithmic calculations can be made
mechanically by means of a " slide rule." This has a fixed
scale F and a sliding scale S (roughly illustrated in Fig. 76),
each so ruled that the distance from 1 to any other number x
is equal to log x.
p log -1.75 •* wo ">
i
I
|j J"
"HITi'i'
I
Ij"
1.75 1 3
5 6 7 '. 9 10
F
i
FIG. 76.
When S is moved over to the position shown, its 1 being opposite
1.75 on F, every number (ri) on S will have moved a distance equal to
224
MATHEMATICAL ANALYSIS [VI, § 156
log 1.75 and hence will be opposite some number N on F whose loga-
rithm is the sum of log 1.75 and log n. This number N must be the
product of 1.75 and n.
Thus to multiply any number n by 1.75, we merely set the slide
as in Fig. 76 and then read off the number on F opposite n. (Observe
how this works for the simple product 2X1.75=3.5.)
Similarly for other multiplications : moving £ mechanically
adds logarithms. Divisions may also be performed, square
roots extracted, etc. Results
accurate to two or three places
can be obtained very fast. Full
directions are given in hand-
books supplied with the rule.
§ 156. Nomographic Charts.
In recent years much use has
been made of nomographs, —
i.e., charts of lines ruled with
number scales in such a way
that various calculations can
be made by merely laying a
straight-edge across the scales.
Fig. 77 illustrates this. The
cost of an automobile tire
per mile traveled can be read
off from scale B by laying a
ruler or stretching a thread
across from the original cost
of the tire on scale A to the
number of miles realized, as shown on scale C.
(A) (B) ,: (o
Cost
Price per Mile . Mileage
-100-
7 70-=
zl
— j.ooa
50^
l 62
?
70-
~****4 —
~
4^
-
- 1,500
60-
-
!>xx
5-
_ v^x
50-
^,
^ 2,000
L x/
2 -
Xx
40^
x.
- 2,500
35 •
-^"v.
J^
- 5,000
SO -
•«-z
^\
- 5,500
X5 -
-ff-E
E: ^
^4,000
to-
— .4-
- 5,000
.s —
— 0,000
15 -
.2-
— 7,000
— 5,000
— 0,000
10-
— .1-
— 10,000
FIG. 77.
Explanation. The scales here are logarithmic, the unit on B being
half as long as on A or C. Any line L through 1 on B passes through
equal numbers on A and C, — as it obviously should.
Raising L a distance equal to log 2 on A would bring it to a parallel
position L', passing through a number on A twice as large as formerly,
— just as with a slide rule. On C, L' will pass through a number half the
VI, § 157] LOGARITHMS 225
former value. Hence the ratio A/C will be four times as great as for
line L, — i.e., it equals 4. But the distance L' was raised, viz., log 2
on scale A equals log 4 on scale B; hence L' will cross B at 4, as it
should to give the value of A/C.
But every line which could be laid across the scales would be some
line L, raised or lowered ; and by a similar argument must cross B at
the right point.
§ 157. Summary of Chapter VI. In § 135 we have al-
ready summarized the definition and basic properties of
common logarithms. We have since observed that any
positive number (except 1) could serve as the base of a sys-
tem of logarithms.
Logarithms follow the laws of exponents, and are therefore
specially adapted to the calculation of products, quotients,
powers, roots, and unknown exponents. They are con-
tinually used in scientific work of many kinds, as is also
their mechanical substitute, the slide rule.
Sums and differences can sometimes be factored into
products. Otherwise we must go from logarithms back to
numbers before adding or subtracting. For this reason,
the cosine law is ill-adapted to the solution of triangles
where large numbers are involved. It may then be replaced
by the Half-angle Formulas or the Law of Tangents.
Calculations involving negative numbers can be made by
taking separate account of the combined effect of the minus
signs. (In Chapter XIII we shall define the logarithm of a
negative number.)
The tremendous power of the logarithmic method is hard to realize.
Notice how easily we could compute a root such as 20v/3.1416X10817,
and how fearfully complicated such a calculation would be by -pure
arithmetic.
Logarithms were invented by Lord Napier, a Scotchman, who
published the first tables in 1614. These were not to the base 10, but
to a base closely related to the one discussed later in § 166.
Our more convenient tables, to the base 10, were calculated soon
226
MATHEMATICAL ANALYSIS [VI, § 157
afterward by Henry Briggs, an Englishman, and Adrian Vlacq, a
Hollander, who unselfishly gave up several years to the tedious work.
EXERCISES
1. Calculate correct to five significant figures :
(a) ^047963
73.452
8.5125 '
( \ A 1.58642 (30.007)
U > .000099128 '
(e) V(74.157)2-(50.063)2,
, (.2685)20X(-47.168)^
.096416)2
.08754 (.4564) (4.6592)'
M) ,? 1 693.02 (-.04692)
V038412(-569.8)2'
(/)
\/
^
^
(.519)2 xVl!7.38
81 (.7)3 (3.4562) '
fi 13261(.78465)4 -^0834
2.7651X^^063524 '
(0 V489007r+(489)2.
2. Find the amount of $2000 after 20 yr. with interest at 5%, com-
pounded annually.
3. What principal will yield $15000 in 25 years, if interest is at 6%,
compounded annually?
4. At what rate of interest, compounded semi-annually, will $12250
yield $37500 in 25 years?
5. In how many years will $12250 amount to $37500 if interest is
at 5%, compounded quarterly?
6. Find H from the two equations :
if 7 = 516.38, T=6.4, r = 100, <*=6.07, Z=237, and o = .002.
Solve the following by trigonometry, and check by drawing to scale and
measuring the required distances or angles.
7. A tree is broken by the wind. Its top strikes the ground 45 ft.
from the foot of the tree, and makes an angle of 35° 56' with the ground.
Find the original height.
VI, §158] LOGARITHMS 227
8. From one bank of a river the angle of elevation of a tree on the
other bank directly opposite is 27°. From a point 129.5 ft. farther
away horizontally in a direct line its angle of elevation is 20° 40'.
Find the width of the river.
9. A boat, viewed from the top and from the bottom of a light-
house 92 ft. tall, had depression angles of 16° 20' and 10° 50' respectively.
Find the height of the rock on which the lighthouse stands.
10. A monument is 133 ft. high, and stands at the top of a hill.
At a point 276 ft. down the hill the monument subtends an angle of
12° 3'. Find the distance from this point to the top of the monument.
11. The sides of a triangle are 196.87, 281.45, and 358.16. Find the
length of the perpendicular from the largest angle upon the opposite
side.
12. A triangular lot has an area of 527.75 sq. yd., and two of its sides
measure 169.8 ft. and 67.4 ft. Find its perimeter. (Two solutions.)
13. A cliff rises vertically 250.92 ft. above sea-level. From its top
the angles of depression of two ships are 16° 21' and 14° 18'. At the
bottom of the cliff the angle subtended by the distance between the ships
is 127° 28'. How far apart are the ships?
§ 158. Looking Back. Let us now recall in brief outline
the work of the course up to this point.
We began by noting that the fundamental problem of
science, whether in studying the physical world or the social
and economic world, is to determine the relations between
varying quantities, — in other words, to ascertain precisely
how any one quantity mil vary with any other on which it de-
pends. And our aim all along has been to find methods of
dealing with this problem, — how to calculate rates of in-
crease, maximum and minimum values, etc. Incidentally
we have tried to get some idea of how these methods are
used in the practical affairs of daily life.
We first saw that approximate results can be obtained by
graphical methods, and that we can always fall back upon
these methods as a last resort.
Upon attempting to calculate instantaneous rates exactly,
we were led to differentiation. To reverse the rate-problem
228
MATHEMATICAL ANALYSIS [VI, § 158
and calculate the size of a growing quantity, we had to take
up integration.
Various integrations and numerical calculations which we
could not carry out showed the necessity of becoming familiar
with further types of functions, especially trigonometric
functions and logarithms.
We have now done this, in a measure, and are ready to
proceed with the main problem and make further applica-
tions of differentiation and integration in the study of vary-
ing quantities.
EXERCISES FOR REVIEW
Chapter I *
1. The average number of hours of artificial light used daily in a
certain city in the various months is shown in Table I. Plot the graph.
TABLE I
Mo.
HRS.
Mo.
HRS.
Mo.
HRS.
Jan.
6.53
May
2.95
Sept.
4.00
Feb.
5.38
June
2.55
Oct.
4.90
Mar.
4.10
Julv
2.60
Nov.
6.18
April
3.48
Aug.
3.15
Dec.
6.85
2. A concrete pedestal has horizontal sectional areas (A sq. ft.)
which vary with the distance (x ft.) above the ground, as in Table II.
Find the rate at which A changes with x at a: = 6.4; also the volume
from x=4 to a; = 16.
TABLE II
8
10
12
14
A
40
21.7
14.1
10.1
7.7
6.1
5.0
* Further graphical problems are included in the miscellaneous set
Allowing.
VI, § 158] LOGARITHMS 229
3. Solve for x exactly or approximately :
(a) 7x2-19x+4=0, (6) x3-5x + l=0.
4. If y varies as the square root of x, and y = 30 when x = 4, express
y in terms of x, and find y when x = 25.
6. Discover the formula satisfied by the values in this table :
2 8 17 29 44
y
4 8 14 22 32
Chapter II
1. State accurately and fully what is meant by : (a) Instantaneous
speed ; (6) a tangent to any curve ; (c) the area of a circle.
2. When is a variable v said to approach a constant c as its limit?
3. State accurately what is meant by saying: "This spiral bends
faster and faster; just at this point it is bending at the rate of exactly
2° per inch."
4. State clearly what is meant by the weight of a cubic foot of air
"at any height A ft." Evidently a cubic foot cannot all be at the same
height.
6. Can you see any definite interpretation that can be given to
this statement: "The amount of $1, with 10% interest for 10 years,
compounded continuously, would be $2.718"? How coul'd this amount
be verified approximately?
6. If y = x3+2x, and x increases by Ax, y will increase by some
amount Ay. If Ax— »0, so must Ay. Could you tell what limit the
fraction A?/ /Ax approaches, merely from the fact that Ax and A?/ are
both approaching zero? (Work out the value of Ay in detail; find
the limit of Ay /Ax.)
Chapter HI
1. Differentiate the function y=x2 by the A process. Explain the
meaning of each important step geometrically, and also from the
standpoint of rates.
2. Explain graphically how it would be possible for /'(x) to have a
large value (say 100) at some point, even if /(x) was everywhere small
(say never more than .5 nor less than zero).
3. The quantity of heat (Q cal.) which 1 cu. ft. of water holds is
a function of its temperature (T°). The instantaneous rate of increase
of Q per degree is called the "specific heat," at the temperature in
question. Write a familiar symbol expressing it.
230 MATHEMATICAL ANALYSIS [VI, § 158
4. Differentiate the function x*+3x — 12 by the A process. Show
from the derivative that the given function can have only one real root.
6. Differentiate at sight :
(a)
(6) ,
3
9 7 5(2 -x2)3'
6. Plot y = .l(x3 — 15z+2) from z=-4 to z = 4. Locate exactly
the maximum and minimum. Also find the exact slope and inclina-
tion of the tangent at the point of inflection.
7. An open rectangular tank is to contain 4000 cu. ft. Materials
for the base cost 80jf per sq. ft., and for the sides 60f£ per sq. ft. Find
the lowest possible cost.
8. The volume of a balloon (V cu. ft.) t hours after sunrise was
V = -(100 000 +80 t3 — 10 <4). When was V increasing most rapidly?
3 *
Approximately how much did V change from < = 3.98 to t = 4.02?
9. State briefly but clearly the principles used in Ex. 8.
10. If y = 10 x2, show by the A process that ^ = 20 x d-£.
at at
11. A cube grows. When are the volume and area increasing at
the same rate .numerically? Approximately how much must the edge
then increase to add .06 unit to the volume?
12. If the volume of a spherical balloon varies thus with the absolute
temperature: 7 = 40,000 T$, and if T is rising at the rate of 2° per
min., how fast is 7 increasing when T = 300? Also, how fast is the
radius increasing?
Chapter IV
1. Integrate, and check your answer by differentiation :
2. A bullet was fired straight up from an airplane 2000 ft. high
with an initial speed of 1600 ft. /sec. Find its height and speed 10
sec. later; also its greatest height; also when it struck the ground
and with what speed.
3. Find the area under the curve y = xi + l/x from x = \ to z = 10.
4. The force used in starting an object varied thus: f = 120 t2 — t3.
Find the momentum imparted in the first 10 seconds.
6. Find the pressure down to a depth of 10 ft. against a vertical
dam whose width varies thus : w> = 120— x2.
VI, § 158] LOGARITHMS 231
6. The "total utility" (M) and "marginal utility" (m) of any quan-
tity (x) of a commodity, — as denned in Economics, — have this
relation : du/dx = m. If we had a graph exhibiting m as a function of
x, what would represent u? Why?
7. A solid is hollowed out in the middle, so that every horizontal
cross-section is a ring between two concentric circles, whose radii
(r in. and R in.) vary thus with the distance (x in.) below the highest
point: r = V6x— z2 and R = ^/8x— x2. Calculate the volume, from
x=0tox=3.
8. In Ex. 7, if the weight of the material (w Ib.) per cu. in. varies
with x (say w = .QQx), devise some method for calculating exactly
the total weight of the solid from x=0 to a: = 3.
Chapter V
1. Given ctn A=T85, find the sine, cosine, and tangent, without
tables.
2. Find the slope and the horizontal and vertical projections of a
line 5 ft. long whose inclination is 27° 13'.
3. What is the angle of elevation of the sun when a vertical pole
55.7 ft. tall casts a shadow 125 ft. long on the ground?
4. The angle between the horizontal arm of a crane and the cable
is 42° 20'. What forces acting along the arm and cable would just
balance a suspended load of 12,375 Ib. ?
6. Look up the sine and cosine of 100° 18'. Also find (to the nearest
tenth of a minute) obtuse angles A and B for which sin A = .28691,
cos B=-. 94837.
6. Find the distance AB across a pond if AC =285 ft., J5(7=319 ft.,
andZAC£=58°43'.
7. In a triangle ABC, A£=48 in., BC = 12 in., Z<7 = 58°. How
large a force acting along AB would have a component of 10000 Ib.
along AC?
Chapter VI
1. Calculate to five significant figures :
(a) V(76.47)*- (21.38)', (6)
-38.893V. 078
(-261.17)2
2. If $1 had been at 6% interest, compounded annually, from the
beginning of the Christian era (say 1920 years), how large a gold ball
232 MATHEMATICAL ANALYSIS [VI, § 158
would be required to pay the amount due? Give the radius in miles.
(1 cu. ft. of gold is worth $362,620.)
3. If an investment depreciates, so as to lose in each year 5% of
its value at the beginning of that year, in how many years will it have
shrunk to one half the original value?
4. Ascending a hill by a straight path whose slope is .18, we see a
tree straight ahead. Observed from two points A and B of the path
115.3 ft. apart, the tree top has elevation angles of 15° 10' and 20° 55'
respectively. How far is the tree top from B?
6. The sides of a triangle measure .312 mi., .423 mi., .342 mi. Find
to five figures the length of the median upon the shortest side, using
logarithmic methods.
6. (a)-(c) Solve by logarithmic methods the following problems
on p. 187, (a) Ex. 21, (6) Ex. 22, (c) Ex. 26.
7. (a)-(c) The same as Ex. 6 for p. 183, Ex. 5, 6, 7.
MISCELLANEOUS AND COMBINATION
PROBLEMS
Chapters I-VI
1. To determine the height of a flag-pole standing at A, the angle
of elevation of its top viewed from a point B on the ground is measured
( = 9° 4'.7), also a line BC on the ground ( = 1158.7 ft.) and angles ABC
( = 19° 12') and ACB (=41° 450 are measured. Find the height, ac-
curate to five figures.
2. Sand, falling at the rate of 2 cu. ft. per min., forms a conical pile
whose vertex angle is constantly 140°. How fast is the base radius
changing at the iretant when the radius is 10 ft. ?
3. The speed (v ft. per min.) of a moving object t min. after starting
was v = 5tz(l2— t}. Find the distance traveled in the first 10 min.
Also find when the speed was increasing most rapidly, and the maximum
speed attained.
4. Plot a graph showing how the speed v in question (3) varied
from t = 0 to t = l2 ; and check your answers to (3).
6. A certain grade of oil exerts against the wall of its container a
pressure of 60 z Ib. per sq. ft., at a depth of x ft. below the surface.
(A) Explain precisely what this statement means, in view of the fact
that no square foot of wall could be at any one depth x ft. below the
surface. (B) Express by an integral the total force exerted by the oil
VI, § 158] LOGARITHMS 233
against the circular wall of a cylindrical tank of radius 20 ft., down
to any depth x ft.
6. An open reservoir has the shape of a hemisphere of radius 40 ft.
How much water will it contain when the water is 30 ft. deep in the
middle?
7. A bomb was thrown straight down with an initial speed of 30 ft.
per sec. from a balloon 3500 ft. high at the instant when an auto running
120 ft. per sec. passed straight under it. Find the distance of the auto
from the bomb 10 sec. later, and how fast that distance was then in-
creasing.
8. In Ex. 7 when was the bomb nearest to the auto?
9. Find the least possible weight for a cylindrical boiler which is
to contain 1375 cu. ft., figuring 11.2 Ib. to each sq. ft. of surface.
10. Starting with an initial velocity of 100 ft. per sec., a point
moves along a straight line ; its acceleration after t sec. varying as in
Table I. Discover the formula for the acceleration. Then find the
distance traveled at any time.
TABLE I
t
0.
1.2
2.7
4.5
6.
Ace.
2.5
3.3
4.3
5.5
6.5
11. The electromotive force (E volts) in a thermo-electric circuit
increases with the temperature (T°) of the hot junction at the rate
fl = .9+.013 T. If # = 1200 when T = 400, what should it be when
T = 500?
12. If y = xwl, find the numerical value of dy/dx when a: = .9875.
13. As a column of air (x in. long) was compressed m a cylinder,
the force F Ib. varied as in Table II. Find graphically the rate at
which F was changing when x = 20; and the work done while x
changed from 20 to 12.
TABLE II
24
20
16
12
8
F
90.6
117.1
160.4
240.7
426.3
234 MATHEMATICAL ANALYSIS [VI, § 158
14. The formula for Table II is F=8000/z1-41. Calculate to five
figures the rate and work in Ex. 13. (N.B. Here £ decreases as W
grows, making dW/dx negative. Change the sign in (13), p. 138.)
16. In Ex. 28, p. 188, change the 47 to 32, and solve.
16. A man bought a piece of property for $1000, and another piece
twenty years later for $2000. He used the annual income to pay
taxes and make improvements ; and ten years after the second purchase
sold both pieces for $17,000. To what rate of interest, compounded
annually, was this investment equivalent?
17. In how many years would $1000 with 12% interest, compounded
quarterly, amount to the same as $2000 with 6% interest, compounded
semi-annually, plus $3000 with 3% compounded annually, during the
same length of time ?
18. Find the inclination of the curve y = 2x3— x* at the point
where the slope is increasing most rapidly. Also state just what
steps would be needed in finding where the slope of this curve
equals —30.
19. Integrate: (a) (2-r»)2<fo; (6) (2-x*)*15xdx.
20. Water is poured from a cylindrical cup 4 in. in diameter until
the surface of the liquid bisects the bottom of the cup, the bottom
being then inclined 38°. Find the volume of water remaining and
the area of the surface of water exposed to the air.
21. An airplane leaves the ground with an initial speed of 80 ft. /sec.,
rising at a constant angle of 5°. If its acceleration after t sec. is 12 — .6 t,
how far will it be after 10 sec. from a point on the ground 800 ft. straight
behind the starting point?
22. Two forces, F Ib. and 125 lb., include an angle of 72° 15'
between their directions. If their resultant makes an angle of 31° 8'.6
with the 125 lb. force, find F. Solve by measurement and by trigo-
nometry.
23. A beam 40 ft. long and weighing 20 Ib./ft. rests on piers at its
ends A and B. A weight W = 6000 lb. moves from A to B at the rate
of 2 ft. /sec. Find the supporting force F at B when W has gone x ft.
How fast is F increasing when a: = 30?
24. A cylindrical tank is to contain 3000 cu. ft. The bottom in-
cluding foundations will cost $4 per sq. ft., the sides $2 per sq. ft., and
the top (in the form of a hemispherical dome) $1 per sq. ft. Find to the
nearest dime the least possible cost.
26. The force (F lb.) required to stretch a certain wire x inches
varied as in Table III below. Find graphically the total work done in
VI, § 158] LOGARITHMS 235
stretching the wire 1 inch. Also obtain a formula for F in terms of
x, and from this calculate the same work exactly.
TABLE III
X
.2
.3
.5
.7
.8
1.0
F
34
51
85
119
136
170
26. Find the pressure of water against a vertical dam, trapezoidal
in shape, which is 50 ft. wide at the bottom (10 ft. below the surface)
and whose sides are inclined 32°. Also state clearly how you could
proceed to find the depth below which half of all this pressure is sus-
tained.
27. The horse-power transmitted by a certain machine belt varies
thus with the speed : H = AS V- .000026 V3. Find the best speed.
28. The base of a solid is a circle of radius 25 in., and every vertical
section perpendicular to one diameter is an isosceles triangle whose base
angles are 80°. Find the volume.
29. A block of ice is drawn up an incline whose grade is 44% by
means of a rope passed over a pulley 10 ft. directly above the top of
the incline. If the block is to move at the rate of 2 ft./sec., how fast
must the rope be drawn in when the block is 20 ft. down the incline?
30. A safety-valve stopper is held down by a level rod x in. long,
weighing .1362 Ib. per in., and pivoted at one end 4 in. from the valve.
What force F Ib. would blow the stopper out, for any x ? About how
much larger is F if x = 20.015 than if x = 19.996?
CHAPTER VII
LOGARITHMIC AND EXPONENTIAL
FUNCTIONS
CONSTANT PERCENTAGE RATES OF GROWTH
§ 159. Our Aim. The only functions thus far differentiated
or integrated have been Power Functions, ^uch as y = xn or
y = un. The great practical importance of these lies in the
fact that it is very common in nature for one quantity to
vary as a power of another.
We now proceed to study an entirely different mode of
variation, which also is very common. A new type of
formula will be required to represent the varying quantities
studied. We shall see how to differentiate and integrate
the new functions, and shall solve further varieties of prob-
lems on rates, maxima, etc.
§ 160. Growing Like Compound Interest. Many quan-
tities in nature grow in the same way as a sum of money at
compound interest, — or rather, as such a sum would grow,
if the interest were compounded exceedingly often or continu-
ously.
That is to say : Money at interest grows faster and faster.
The percentage rate remains constant, as 6%, or 3J%, etc.,
but the total rate of growth (or number of dollars per year)
increases, — being proportional to the amount accumulated
at the beginning of the interest-period in question.
Thus, if we compound annually at 40%, the rate of growth at any
instant (as at P in Fig. 78) will be 40% of the value at the beginning
236
VII, § 161] EXPONENTIAL FUNCTIONS
237
of the year. If we compound semi-annually, the rate will be 40% of
the value at the beginning of the half-year in which P lies. And so on.
ACL
FIG. 78.
In each case AL represents 1 yr.
O L
If the interest were compounded exceedingly often, say
a trillion or more times a year, the periods would be so short
that the rate at any instant would be practically proportional
to the amount at that same instant. Just so for many quan-
tities in nature : the larger they become, the faster they in-
crease, — proportionately, — until stopped by modified physi-
cal conditions.
Many other quantities decrease in a similar way, — like an invest-
ment depreciating at a constant percentage rate, figured almost con-
tinuously.
§ 161. Effect of Compounding Continuously. To arrive
at a formula for quantities of the kind just mentioned, let
us see how the value of an investment will be affected if the
interest is compounded exceedingly often.
Compounding k times a year, the amount is (by § 145) :
kn
How will this be affected if k is indefinitely increased ?
Consider first a special case : The amount on $1 after 1
year with interest at 100%. Then P=l, n=l, r=l, and
(2)
238
MATHEMATICAL ANALYSIS [VII, § 162
Taking successively fc = l, fc = 10, fc = 100, etc., we find the
values of A shown in the following table. (Eight-place
logarithms are needed to get the last two values accurately.)
Notice that, although we are increasing k faster and faster,
A is increasing less and less rapidly, apparently approaching
some limiting value near 2.718. In higher analysis this is
definitely proved.
k
A
k
A
1
2
1000
2.717
10
2.594
10,000
2.718
100
2.704
Remark. This limiting value of A is called the result of compound-
ing continuously. The original dollar gains about $1.718 during the
year, or 171.8%. Hence compounding continuously at 100% is about
equivalent to compounding annually at 171.8%.
§ 162. The Number e. The limit approached by the
quantity (1 + 1/fc)* in (2) above, as k increases indefinitely
(written &->-<»), is denoted by e:
(3)
This number is very important in what follows. Approxi-
mately :
e = 2.7183, log e=. 43429.
Remark. $1 with 100% interest, compounded continuously for
one year, will amount to precisely e dollars.
§ 163. General Formula. Now consider the effect of
compounding interest continuously for any number of years
at, any rate.
Returning to the standard formula
\ kj '
VII, § 163] EXPONENTIAL FUNCTIONS 239
we are to let fc-»-oo without giving special values to P, r, and
n. The problem is, however, reducible to the special case
above by denoting r/k by 1/2. For then k = zr, and
Now, as k increases without limit, so must z. Hence the
bracketed quantity in (5) varies in the same way as the k-
quantity in (3), approaching e. Thus the limiting value of
A in (5) is
A = Pe™. (6)
' This is the amount of any principal P after n years with
interest compounded continuously at any annual rate r.*
Similarly, if a physical quantity Q grows at a constant
percentage rate r (per year, hour, or other unit of time), its
value after t units must be
Q = Pert. (7)
Ex. I. The number of bacteria in a culture increased at a rate
(per hr.) which was always 6% of the number then present. If the
original number was 1000, how many were there after t hr. ?
By (7), AT = 1000e-0«.
The value of N at any time is easily calculated by logarithms, since
log N=log 1000+ .06 t log e, and log e = .43429.
Ex. II. What rate of interest, compounded annually, will yield
the same amount as 6% interest compounded continuously?
By logarithms this gives 1 +r = 1.0618 ; whence r =6.18%.
Remark. Similarly hi Ex. I, the number of bacteria is multiplied
by e-06 or 1.0618, in each hour. Thus the gain is 6.18% of the number
at the beginning of the hour, though the rate of gain is but 6% of the
growing number at any instant.
* Since formula (4) is strictly correct at the ends of all interest periods,
(6) is correct at all times, and not merely for integral values of n.
240 MATHEMATICAL ANALYSIS [VII, § 164
Thus r in (7) is the instantaneous percentage rate, not the average
percentage rate during a whole unit. Similarly r in (6) is the nominal
rate used in compounding continuously, and not the effective rate
actually realized.
164. Depreciation. For negative values of r, formula (4)
above represents a depreciating investment. When r is
negative, so is z in (5). But by Ex. 14, p. 241, the quantity
(1-f 1/2)* still approaches e if k-*-vo .
Hence formula (6) gives the value of an investment which
depreciates at the nominal percentage rate r, figured con-
tinuously. If the rate is 8%, merely put r=— .08, getting
A = Pe-°*\
The corresponding formula (7) evidently holds for any
physical quantity Q which decreases in' an analogous manner.
EXERCISES
1. Find the amount of $1500 after t years with interest at 5%,
compounded continuously. How much after ten years?
2. At what rate, compounding annually, would the final amount
be the same as in Ex. 1 ?
3. In Ex. 1, when would the amount be $5000?
4. An investment depreciated at a constant percentage rate of 6%,
starting from an original value of $75,000. What was its value after
t years? After 10 yr.?
5. In Ex. 4, what percentage was actually lost per year?
6. In 1870 the population of a certain city (Portland, Ore.) was
8300, and this grew until 1910 almost like an investment with interest
at 8%, compounded continuously.
(a) Write a formula for the population at any time (t years after
1870), on this basis.
(6) Compare the population in 1900 and in 1910 (90,500 and
207,200) with the values given by your formula. [If the formula had
remained valid, what would the 1920 population have been?]
7. The number of bacteria in a culture increased at a rato (per
hour) always equal to 30% of the number. Find how many there
were at any time from an original 100. How many after 5 hours?
8. In Ex. 7, when had the original number doubled?
VII, § 166] EXPONENTIAL FUNCTIONS 241
9. Radium decomposes at a rate (per century) which at every
instant equals 3.8% of the quantity Q remaining. How much will be
left* after 1000 years from a present quantity of 100 mg.?
10. The speed (V) of a rotating wheel after the power was cut off
decreased at a rate (per sec.) which at every instant was 25% of V
itself. If V was originally 1000, what was its value after 10 sec.?
When was the speed reduced to one tenth its original value?
11. The speed (V) of a chemical reaction increases with the tem-
perature (T) at a rate constantly equal to 7% of V. If F = 30 when
T=0, wr^te a formula for V at any temperature. Find T for which
F = 60.
12. If the population of a state increases by 4% each year and is
now 2,000,000, write a formula for the population t years hence. To
what rate, figured continuously, would this be equivalent?
13. Calculate the value of the quantity (1+1/fc)* for k = l, 10, 100,
1000, 10000, — using 7 or 8 place tables if accessible.
14. The same as Ex. 13, for negative values of k.
§ 165. Equivalent Forms. Any such quantity as Q =
can be expressed also as a power of 10. For e = 10*43429
(§ 162) ;
^ Q = pe.02t _ p(lQ.43429)-02< _ p JQ.0086858^
That is, the forms e™ and 1O0086858' are equivalent. But the
2% rate which is clearly exhibited in the e form is entirely
hidden in the 10 form. Thus the e form is the more natural.
§ 166. Natural Logarithms. Common logarithms are
based upon the fact that every number is some power of 10.
(§ 130.) But it is equally true that every number is some
power of e. For instance
5 = e1-6094.
We call this exponent 1.6094 the logarithm of 5 "to the
base e."
Logarithms to the base e, being exponents, follow the
same four rules of combination as logarithms to the base 10.
(§ 139.) Tims, the logarithm of a product equals the sum
of the logarithms of the factors ; etc.
242
MATHEMATICAL ANALYSIS [VII, § 167
The base e is naturally suited to calculations concerning
the continuous compounding of interest.
•
E.g., in calculating q from g = 30 e-02' we should have, for any base:
log q= log 30 +.02 Hog e.
If the base is 10, log e = .43429 ; but if the base is e, log e = 1 simply.
Similarly to find t if q were given, the base e would be the simpler.
The chief reason for introducing the base e, and regarding it as the
"natural base," will, however, appear later.
§ 167. Use of Table. In the appendix (p. 502) there is a
table of natural or " Napierian " logarithms, referred to the
base e or 2.71828 — '. A few lines are reproduced here.
N
Q
j
2
3
5.0
5.1
1.6 094
292
114
312
134
332
154
351
174
371
194
390
214
409
233
429
253
448
273
467
10.0
2.3 026
036
046
056
066
076
086
096
106
115
This means, for example :
log 5. 14 = 1.6371; /. 5.14=e1-«71.
log 10 =2.3026; .'.10 =e*-z02«.
To find the logarithm of a number which lies beyond the
limits of the table, we use the idea of Scientific Notation.
For instance, 514 = 5. 14 X 102.
Hence, to get log 514, we would look up log 5.14 and add
twice log 10 :
log 514 = 1.6371+4.6052 = 6.2423.
We do this, in fact, with the base 10 ; but then twice log 10 is simply 2.
Conversely, if given log TV = 6.2423, we would subtract
log 10 twice, or 2 log 10, getting down to 1.6371. Then N
must be the number which corresponds to 1.6371, multiplied
by 102: # = 5.14X102 =
VII, § 168] EXPONENTIAL FUNCTIONS 243
Remark. In solving an equation like 749 = 135 e-06* for t, it would
simplify matters to divide through by 100, and have only the small
numbers 7.49 and 1.35 to deal with.
EXERCISES
1. Express as powers of 10 : e-02', e~Mt, e12', e~40t.
2. Look up the natural logarithms of :
4.85, 92.6, 913, 278000, .0681,. .00092.
Check each roughly by inspection, thinking of e as nearly 3.
3. The same as Ex. 2, interpolating for each given fourth figure :
6.283, 17.44, 60920, .005287.
4. Look up the numbers (to 3 figures) whose natural logarithms are :
2.0462, 5.3083, 9.3679, 9.2163-10, 6.9088-10.
6. The same as Ex. 4, interpolating to get a fourth figure, for
0.4682, 4.6928, 8.4179-10, 4.1263-10, 7.9182-10.
6. Calculate the following, using natural logarithms :
(a) A=75e-3-482*, (6) i/ = 1500 e1-8*6, (c) Q - .45 e2-65™.
7. Solve for the unknown n or r, after making any possible pre-
liminary simplifications :
(a) 985 = 159 e20--, (6) .0485 = .075 e--04".
(Hint : Both sides of any such equation may be multiplied or divided
by 10, 100, etc., without affecting the exponent.)
8. Express as powers of e : 2*, (1.06)4", 3-«, lO"3*.
9. By means of common logarithms calculate for yourself the natural
logarithm of 5, and compare with the table. [See §§ 154, 166.]
§ 168. Compound Interest Law. If any quantity y varies
with another, x, in such a way that its rate of increase or
decrease is constantly proportional to its value, it is strictly
analogous to an investment whose interest or depreciation is
figured continuously at a fixed percentage rate. (Cf. § 160.)
Such quantities are said to vary " according to the Com-
pound Interest Law " (abbreviated C. I. I/.).*
* Also called the Law of Organic Growth, or the Snowball Law.
244 MATHEMATICAL ANALYSIS [VII, § 169
Clearly, any such quantity is given by the formula
y = Pera> (8)
where P is the value of y at x = 0, and r is the fixed percentage
rate.
E.g., if y decreases at a rate always equal to 15 per cent of y, then
r=-.15, and y = Pe~-1**.
Conversely, any quantity given by a formula of type (8)
must vary according to the C. I. L. For instance, if given
•101
we would recognize this as the
formula for a quantity which in-
r
Graph of i=10 e** creases at & rate constantly equal
to 9% of its value.
Ex. I. An electric current does not
instantly vanish when the "EMF" is
cut off; but it falls off rapidly, as in
Fig. 79, decreasing at a constant per-
.002 .004 .006 .008 .01 i • i • T/j
... < ccntage rate wnicn is very great. It
this rate is, say, 30000% per sec.,
then r=— 300; and if the original
intensity is 10 amperes, then, after t sec., it will be
§ 169. Exponential Functions. The functions 2* and x2
grow in very different ways. (See the table.) The value of
2* is quadrupled every time we increase x by 2 ; whereas x2
is quadrupled every time we double the value of x. In fact,
these functions are of entirely different kinds. In one, we
raise a fixed number to a higher and higher power ; in the
other, we raise a larger and larger number to a fixed power.
X
1
2
3
4
5
' 6
2*
2
4
8
16
32
64
y?
1
4
9
10
25
36
VII, § 169] EXPONENTIAL FUNCTIONS
245
A variable raised to a fixed power is called a Power Func-
tion, as we have seen. (§ 55.) A constant raised to a
variable power is called an Exponential Function. (The
constant, however, is to be positive and not equal to 1.)
E.g., 2*, 1.06X, 1O3', e~-08x, and in general, &** are exponential
functions.
Any quantity represented by an exponential function, or
such a function times some constant, must vary according
to the C. I. L., — provided the exponent is of the first degree,
as in the illustrations just given. For the Constant base
which is raised to the variable power is itself some power of
e, and the given exponential function must therefore be re-
ducible to a power of e in the type form (8), p. 244. An
example will make this clear.
Ex. I. The speed v of a certain chemical reaction doubles with
every 10° rise in the temperature. Obtain a formula for v at any
temperature.
If v = P at !T=0, we have the following table.
But by the table of natural logarithms : 2 =
which falls under (7) or (8) and is a case of the C. I. L.
T
v
T
v
0
P
30
SP
10
2P
20
4P
10 x
2*P
Remark. Any quantity which doubles, or gains a fixed percentage,
at any regular intervals must increase according to the C. I. L.
It is therefore a characteristic feature of the Compound Interest
Law, and of the corresponding formula y = Perx, that adding a fixed
amount to x will multiply y by a fixed amount.
246
MATHEMATICAL ANALYSIS [VII, § 170
§ 170, Graphs. Exponential functions are so important
that we should be thoroughly familiar with their graphs.
Consider the following standard forms.
(I) .y=ex,
(III) y =
(II) y = ae*,
(IV) y =
Graph of ex
By § 169 these all vary according to the C. I. L., and
their rates of increase are constantly proportional to their
values.
The graph for (I) runs as in Fig. 80. The higher the curve,
the faster it rises. Toward the left the curve approaches
the base line indefinitely, never
reaching, it. (At x = — 100,
y = e~m. Whatdoese-100mean?)
The graph for (II) is the same,
except that every ordinate is a
times as great. With a change
of scale, Fig. 80 would do, pro-
vided a were positive. (What
if a were negative?)
In (III) the values of y at
x=l, 2, 3, •••, are the same as
those in (II) at x = k, 2k, 3k, •• .
If k is positive the graph is the same as for (II) , with a
change in the horizontal scale. If k is negative, the graph
is reversed as regards positive and negative values of x. It
then falls toward the right, and is the typical " die-away
curve." (Cf. Fig. 79, p. 244.)
The form (IV) is reducible to one of the preceding forms,
and thus has the same graph, to some scales.
The graph, however, is very different when the exponent involves
.•mything beyond the first power of x. E.g., the function y = c~s2,
which is very important in statistical studies, varies very nearly as
in Fig. 6, p. 5.
VII, § 171] EXPONENTIAL FUNCTIONS
247
EXERCISES
1. Plot the graph of y = e*, taking z=0, .5, etc., to x = 2.5. Over
the same base line plot y=log x from x = l to 10. Is there any simi-
larity between the two curves? For what reason?
2. Draw the graphs of the following functions roughly by inspection,
merely showing the general shape and location :
(a) y =
(c) 2/ =
(6)
(d)
3. The number of bacteria in a culture increased at a constant
percentage rate of 40%, the unit of time being 1 hr. ^If AT = 1000 at
t =0, write a formula for N at any time. When was N = 6000?
[4.] In Ex. 3 plot a graph showing how N increased, from t = Q to
t = 5. Also plot another graph showing how log N varied. Can you
explain the peculiarity of the latter?
6. Write a formula for a quantity Q which equals 50 at t — Q and
(a) increases at the constant percentage rate of 17%, 85%, 230%;
(6) decreases at the constant percentage rate of 35%, 1.6%, 3000%.
6. Express as a power of e : y = P 3-2*. What percentage rate ?
§ 171. Semi-logarithmic Graphs. In study-
ing the variation of a given quantity y, it is
sometimes desirable to plot a graph showing
how the logarithm of y varies with the inde-
pendent variable x. (Fig. 81.)
Such a " semi-logarithmic graph," as it is
called, will always be a straight line when y
varies according to the C. I. L.,
(9)
(10)
'
log y
X
FIG. 81.
For, taking logarithms to any base, we must have
log y = log- P +rx log e.
And as this equation is of the first degree in terms of log y and x, the
two quantities plotted, the graph must be straight. (Ex. 2, p. 44.)
Conversely, if that graph is straight, y must obey a C. L L. (Why?)
248
MATHEMATICAL ANALYSIS [VII, § 172
§ 172. Use in Statistical Problems. The characteristic
feature of the C. I. L. is that y varies at a constant percentage
rate. Hence we may say that whenever a quantity increases
or decreases at a constant percentage rate, its semi-logarith-
mic graph mil be straight. And conversely.
For this reason such graphs are much used by statisticians
in studying the growth of populations, bank clearings, bonded
indebtedness, etc. If the semi-
logarithmic graph of a popula-
tion is straight, the population
has increased at a constant per-
centage rate. If not, we can
see at a glance where the largest
percentage gains were made, by
simply noting where the graph
is steepest. By comparing such
graphs for many different popu-
lations, — e.g., for the native
and foreign-born populations of
Portland, Ore., in Fig. 82,-
we can see which made the largest percentage gain in any
interval.
This same idea is used by large business houses in comparing the
gains made by different departments of the business, or in comparing
the growth of their business with the volume of postal receipts or other
indications of general business conditions.
§ 173. Semi-logarithmic Paper. To make it easy to
plot semi-logarithmic graphs, a specially ruled paper has
been devised. Horizontally it has a uniform scale, to repre-
sent values of x ; vertically, a logarithmic scale (like a slide-
rule), to represent values of log y. (See Fig. 83.)
The number on the vertical scale at any point is a value,
of y, but the height of that point above the base line is 1<>^ ?/.
Thus the distance up to the point marked " 10 " is log 10, —
1870
.1910
VII, § 173] EXPONENTIAL FUNCTIONS
249
1000
100
10
or 1 unit, using the base 10. The distance up to " 100 " is
log 100, or 2 units. And so on. To erect an ordinate equal
to log 20, we have
merely to run it up to
the cross-line marked
" 20," etc.
Hence if we plot a
given table of values
(x and y), without
looking up any loga-
rithms, the paper will
automatically plot
log y as a function
of x, — i.e., will plot
the semi-logarithmic
graph. Similarly, if
given a formula, we
have simply to cal-
culate a table by substituting in the formula, and then plot.
In case the formula is a C. I. L. two points will be enough,
as the graph must be straight.
Ex. I. Plot the semi-logarithmic graph of i/ = 1000 e~M.
When t = 0, y = 1000. When t = 10, y = 100 e~* = 2.5, approx.
Plotting these two values of y, and joining by a straight line gives
the required graph. (Fig. 83, A.) Observe that further values of y
satisfying the given formula can now be read off directly. E.g., y = Ql
when t = 4c.
Ex. II. The native and foreign-born populations of Portland, Ore.,
have increased as shown in the table below. Plot the semi-logarithmic
graphs.
Taking the numbers of the vertical scale as thousands rather than
units, we get Fig. 83, F, N, — the same graphs as in Fig. 82, but ob-
tained now without looking up logarithms.
These graphs do not give intermediate values correctly, unless
the percentage rate of growth was constant during the interval
considered. .
800
600
400
300
200
SO
N
^^
x^
\
\
/
40
SO
20
8
— ->
Sy-
-/•
^f
J*
^^^
/
Sc
^
^
</
/
\
6
4
3
2
(
-
/r
X
C
A
^s
^y
9
^
> 2 k 6 8 10
1870 1880 1890 1900 1910
FIG. 83.
250
MATHEMATICAL ANALYSIS [VII, § 173
YEAR
NATIVE
FOREIGN
YEAR
NATIVE
FOREIGN
1870
5700
2600
1900
64600
25900
1880
11300
6300
1910
163400
43800
1890
29100
17300
EXERCISES*
^1. Table 1 shows the approximate number of telephones in use in
the United States at different times. Plot tho. ordinary and semi-
logarithmic graphs. (Does the latter graph show any facts which the
former does not?) How many telephones in 1902? In 1913?
TABLE 1
YEAH
No.
YEAR
No.
1895
280000
1910
5 140 000
1900
630000
1915
* 8 640 000
1905
2030000
2. Table 2 shows (in millions) the assets of a certain life insurance
company at various times, and also the amount of insurance in force.
Plot the semi-logarithmic graphs on the same sheet. Any peculiar
fluctuations?
TABLE 2
YEAB
A
INS.
YEAR
A
INS.
1867
3
37
1897
103
413
1877
18
64
1907
233
882
1887
29
148
1917
394
1604
* The scales printed on the paper usually run from 1 to 10 and repeat.
We re-label them to fit the problem. Thus the first " 1 " might bo taken as
100, the next as 1000, etc., making the first "2" mean 200, etc. Evidently
"200" should come just as far above " 100" on the scale as "2" above " 1."
For log 200-log 100 = loR (200/100) =log 2.
Accurately ruled paper can be obtained from dealers in scientific supj >''''*•
But paper good enough for rough practice cuu be run off on a mimeograph.
, § 173] EXPONENTIAL FUNCTIONS
251
3. Using the mortality table, p. 9, plot the semi-logarithmic
graph of the number of survivors. Does the percentage rate of decrease
become continually greater? (The absolute or total rate does not.)
4. The net profits of several firms in 1914 and 1916 are shown in
Table 3. Show these on semi-logarithmic paper. In which case was
the percentage increase greatest? Least?
TABLE 3
1914
1916
1914
1916
A
B
1 450 000
350000
10 992 000
5 983 000
C
D
5 590 000
36000
43 594 000
5 090 000
6. Plot the semi-logarithmia graph of the formula y = 2Q eAx.
6. The quantity of radium remaining after (f) years from an original
1000 mg. is given approximately by the formula Q = 1000 e~ °0038'. Plot
the semi-logarithmic graph, using t = Q and £ = 5000. Read off inter-
mediate values, and compare Table 3, p. 17.
7. Atmospheric pressure varies with the elevation according to the
C. I. L. At h = Q, p = 30, and at h = 30,000, p = 9.5. Plot the semi-
logarithmic graph and read off p when h = 6000 and 12,000.
8. The population of a state increased in 10 years from 518,000 to
1,142,000. Assuming the percentage rate to have been constant, what
was the population at the middle of the decade?
9. The population (in millions) of the United States has grown as
in Table 4. Plot a semi-logarithmic graph. Note how long the per-
centage rate was practically constant.
[10.] In the formula y = .2 x2 plot a graph showing how log y varies
with log x. Explain the peculiar result.
TABLE 4
YEAR
POP.
YEAR
POP.
1790
3.9
1860
31.4
1800
5.3
1870
38.6
1810
7.2
1880
50.2
1820
9.6
1890
62.9
1830
12.9
1900
76.0
1840
17.1
1910
92.0
1850
23.2
1920
105.7
252
MATHEMATICAL ANALYSIS [VII, § 174
§ 174. Logarithmic Graphs. In some statistical work
where it is necessary to handle very large and very small
values of y and z, it is customary to plot the logarithms of
both variables, — i.e., to plot a graph showing how log y varies
with log x. This greatly tones down the contrasts. E.g.,
log 100000 is only 5, and log .001 is -3.
Such " logarithmic graphs " are, however, mainly useful
in scientific work in studying Power Laws :
y = kxn. (11)
For any such law, the logarithmic graph is straight. For
log y = \og k+n log x, (12)
•
and this equation is of the first degree in terms of log y and
log x, the two quantities plotted.
Conversely, whenever the logarithmic graph is straight, y must
vary according to the Power Law. For by § 32 the relation between
log y and log x must be linear, — say
log y = a log x+b. And as this is of
the form (12), equation (11) must
hold, for some values of k and n.
Logarithmic graphs can be
plotted without looking up any
logarithms, by using a special
" logarithmic paper." This is
ruled with logarithmic scales
both horizontally and vertically.
(Fig. 84.) A point for which x = 20
and y = 8 has its actual distances
from the reference lines equal to
1000
-100
,
10
Values of &
FIG. 84.
loo log 20 and log 8.
Ex. I. Plot the logarithmic graph
of the Power Law, ?/ = 4 x1-13.
When x = l, y=4. When z = 100, y = 4(100) »•" = 728, approx.
Plotting these values and drawing a straight line through the
resulting points gives the required graph. (Fig. 84.)
VII, § 175] EXPONENTIAL FUNCTIONS 253
§ 175. Discovering Scientific Laws. As noted in § 32
many scientific laws are discovered experimentally. A
table of values (of x and y, say) is obtained by observation ;
and various mathematical tests are then applied to ascer-
tain what formula or formulas are satisfied by the tabulated
values.
A simple test for the three most common types of law
(§§ 32, 55, 168) can be made by plotting certain graphs :
(I) Ordinary graph straight : Linear Law, y = ax+b,
(II) Logarithmic graph straight : Power Law, y = kxn,
(III) Semi-logarithmic graph straight : C. I\ L., y = PeT*.
If none of these graphs is straight, the law is not one of these types.
Tests for some other types will be discussed in §§ 323-324*.
If the test shows a given table to satisfy one of these laws,
the required constants (k, n) or (P, r), etc., can be found by
merely substituting two pairs of values in the proper formula,
and solving algebraically. (Cf. § 32.)
Ex. I. Discover the formula for the following table.
X
y
i
2
4
6
8
10
3.3
5.4
14.8
40.2
109.2
296.8
The semi-logarithmic graph turns out to be straight.
Hence the required formula is a C. I. L. :
y = Per*.
Substituting the first and last values of the table :
296.8 =
Taking logarithms :
log 296.8 = log /b+10 r log e,
log 3.3 = log k+r log e. (13)
* In some cases it may be necessary to plot two semi-logarithmic graphs :
one with y plotted vertically, and the other with x. We may not know in
advance which to treat as the function varying according to the C. I. L.
254
MATHEMATICAL ANALYSIS [VII, § 175
Looking up the logarithms of 296.8 and 3.3 (either to the base
e, with loge = l, or to the base 10, with log e = . 43429), and
subtracting, we find r = .5, approx. Substituting this in
either equation of (13) gives k = 2. Hence the required
formula is finally
N.B. When log k happens to come out negative, we add 10 — 10.
Thus log k = -.175000 would be the same as log k =8.25000 — 10, which
could easily be looked up.
EXERCISES
1. The attraction (/ dynes) between two electric charges was found
to vary with the distance (x cm.) apart as in Table 1 below. Find the
formula and check.
2. Determine what kind of formula is satisfied by the values in
each of the Tables 2 and 3.
3. A cold plate was taken into a warm tunnel. The difference
between the temperatures of the air and the plate decreased as in
Table 4 after t hours. Find the formula for D at any time.
4. Plot the logarithmic graphs of the following :
(a) y = l/x from z = l to x = 100. Read off the reciprocals of 4.52;
25; 69.8.
(6) y = Vx from x = 1 to x = 100. Read off the square roots of 6
and 50.
6. The rate of rotation of a wheel under water decreased as in
Table 5 after the power was cut off. Find a formula for R at any time.
6. The speed of a certain chemical reaction doubles every time the
temperature is raised 10° C. Calling the speed 1 at 20°, make a table
of its values at several other temperatures and find a formula which
will represent the Table.
TABLE 1 TABLE 2 TABLE 3 TABLE 4 TABLE 5
X
/ x
V x
y t
D t
R
.5
144. 1
5. 1
5 0
29.4 0
3000
1.0
36. 20
5.6 8
14 .5
22.1 10
1348
5.0
1.44 400
54.8 50
35 1.0
16.6 20
606
20.0
.09 700
331.4 200
71 1.5
12.5 30
273
1000
2004. 450
106 2.0
9.4 40
122
VII,' §176] EXPONENTIAL FUNCTIONS
255
7. The number of bacteria in a culture increased as in Table 1,
p. 16. Discover the formula.
8. Discover the law satisfied by the following values of the pressure
exerted by an expanding volume of steam :
V
8
10
20
30
40
50
60
70
80
p
125
92.8
36.8
21.5
14.6
10.9
8.5
6.9
5.8
9. The distances of the planets from the sun and their periods of
revolution (T yr.) are given below. Discover the law. ("Kepler's
Third Law.")
D
Merc.
.387
Venus
.723
Earth
1.00
Mars
1.52
Jup.
5.20
Sat.
9.54
Ur.
19.2
Nep. I
30.1
T
.24
.615
1.00
1.88
11.9
29.5
84
165
j
§ 176. Derivative of log x. In many problems it is nec-
essary to know just how the logarithm of a number changes
with the number, — in other words, how the function log x
varies with x.
To find a formula for the derivative or rate of increase
at any instant we resort to the increment process. (§ 53.)
Let y — log x, to any base.
Then y+Ay = log (x+Ax).
. Aj/ log (x+Ax) -log x
The final step is to find the limit of this fraction as Az-M).
The numerator approaches log x— logo;, or zero. Thus we
have a quantity which is becoming very small, divided by
another, also becoming very small. Without more informa-
tion no one can tell what limit the fraction will approach.
But we can simplify the numerator. Subtracting one
logarithm from another gives the logarithm of a fraction :
256 MATHEMATICAL ANALYSIS [VII, § 176
_
AX AX
To simplify this further make the substitution :
^=i, or Az = 5. (16)
X Z Z
Then equation (14) becomes
A,
Ax x x
2
But multiplying a logarithm by z gives the logarithm of the
2th power.
A lo<1+1T
.-. ^=_V_£/
* Ax x
We can now see what happens as Ax->-0. By (16), 2 must
increase indefinitely. And by § 162, the quantity (1 + 1/2)*
approaches e, so that the limit approached by A?//Ax is
, *_•«£. 07,
This is the derivative of y = log x, no matter what base of
logarithms is used.
If the base is e, log e = 1 simply. Hence
if2/ = log.z, g = ^. (18)
But if the base is 10, log e= .43429, approx. (denoted by M).
Thus
= . (19)
VII, § 177] EXPONENTIAL FUNCTIONS 257
Remarks. (I) These formulas show that the rate at which a loga-
rithm increases with the number is inversely proportional to the size
of the number. Geometrically stated : The graph of log« x has a slope
of 1 at x = lj a slope of £ at z = 2, | at x — 3, etc.
(II) Formulas (17)-(19) rest ultimately on the existence of the
limit e, — proved in higher analysis, but assumed in this course. Com-
pare (16) above with the substitution used in § 163.
(III) Observe that e is the natural base to use in problems requiring
the differentiation of a logarithm, — because of the simplicity of formula
(18) as compared with (19). In calculus, when no base is specified,
e is always understood, not 10.
§ 177. Log u. To differentiate the logarithm^ a quantity,
say y=logw, (20)
we use the same principle as in differentiating a power of a
quantity, y = un. That is, we multiply dy/du by du/dx.
(§ 76.) Here
s.i.
du u
dx udx
Thus, the derivative of the logarithm of any quantity equals
one divided by that same quantity, times the derivative of that
quantity. (Memorize.)
Ex. I. 2/ = log(z3-l).
* 1
Remark. It is instructive to compare this with a power case, say
dx
Note the final factor 3 x2 in each case.
Ex. II.
dx x3 x
This result could have been foreseen, log x3 being the same as 3 log x.
258 MATHEMATICAL ANALYSIS [VII, § 178
This can be simplified greatly before differentiating. By
§ 139, the logarithm of the radical equals one half the log-
arithm of the fraction. And the latter equals what ?
/. 2/ = | [log (z4-l)-log (**+!)].
Each of these logarithms is easily differentiated.
(*-l)(x*+l)
N.B. We cannot yet differentiate the fraction (x* — 1 ) / (x* + 1 ) , with-
out resorting to the A process. (§53.) But, curiously, we have just
differentiated the logarithm of the square root of that fraction !
§178. C-dx. In Chapter IV we could not find this
J x
integral, because there is no power of x whose derivative is
1/x, or x~l. But we now see that
C-dx=log.x+C. (22)
•/ x
Hence we can now integrate every power of x : x~l by (22) and
any Bother power xn by (8), § 92.
Remark. It is also true that
(-dx=±-logwx+C. (23)
J x M
But this formula is not used if tables of natural logarithms are at hand.
Ex. I. The force (F Ib.) driving a piston varied thus with
the distance (x in.) : F = 6000/3. Find the work done from
x = W tox = 20.
Always W=(*Fdx. (§ 96.) In the present case,
f
J
x
.-. TF = 6000 log z-fC.
VII, § 178] EXPONENTIAL FUNCTIONS 259
Now the work starts (i.e., TF = 0) when x = 10.
.'. C= -6000 log 10,
and /. W = 6000 log x - 6000 log 10.
This is the work done from x = 10 to any other x. To x = 20 :
T7 = 6000 (log 20 -log 10) =6000 log 2.
By the table of natural logarithms, W = 4 16 (in.-lb.), approx.
EXERCISES
1. Find the instantaneous rate of increase of y = loge x at z = 2, 4,
and 8. Check the last value by finding from the tables the average
rate of increase in log x from a: = 7.99 to 8.01.
2. Approximately how much does loge x increase while x increases
from 2 to 2.0003 ? If log, 2 = .69315, find log 2.0003.
3. Plot y = loge x from x = 1 to x = 10, and check the results in Ex. 1.
4-6. Proceed as in Exs. 1, 2, 3, using the base 10, and log 2 = .30103.
7. Simplify and differentiate the following natural logarithms :
(a) y = log x20, y = log x7-6, y = logx&, y = log or4.
(6) ?/=log7z, ?/ = loglOOOz, y=\og5x3, y-k>g.lari
(c) y = logVx, y = \0g *x6, ?/ = log (5/x2), ^ = log (l/*x). •
(d) y = log z2+3 logVz+2 log(l/z), y = log z3+5 log^z+7 log(l/z2).
8. (a) Calculate the area under the curve y — \/x from x = 1 to x = 10.
(6) Check by plotting and measuring the area.
9. A rough table of logarithms could be constructed by measuring
areas under the curve in Ex. 8. Explain briefly.
10. The force used in driving a piston varied thus: F = 1200/2.
Find the work done from x = 20 to x — 3Q.
11. Find the volume generated by revolving about the base line
the area under the curve y = l/^/x from a: = 2 to z = 10.
12. The elevation (E ft.) above sea-level corresponding to any
atmospheric pressure (p in.) is given by the formula :
E = 88630 - 60000 log p. (Base 10)
Approximately what change in elevation corresponds to a decrease of
p from 30 to 29.7?
13. How fast is a balloon rising if the pressure recorded is decreasing
at the rate of .5 in./min. when p = 28? (See Ex. 12.)
260 MATHEMATICAL ANALYSIS [VII, § 179
14. Find dy/dx for each of the following functions :
(a) Basee: y = log (z4 + l), y=\og (z2-6z
(6) Base 10: y = log (z2-25), 2/=log (1-s6).
JV.B. 7n <te following differentiations the base is e.
15. Differentiate in two ways and check :
t/ = log x*», y = log (500 x\ 2/=log(20 z4).
16. Simplify each of the following and then differentiate :
(a) y-log-* (6) i/=log
(c) y=log (xz'-l), (d) i/=log
17. Differentiate and simplify the results :
(a) 7/ = z2-log (z2 + l), (6) 7/ = log (logx).
18. What is the derivative with respect to x of:
y=logu, 2/=logz, 9 = logr, w = log y.
§ 179. Differentiating Logarithmically. Many functions
which we cannot yet differentiate directly are easily handled
by introducing logarithms.
Taking the logarithms of both sides, and simplifying as in § 177 :
log y = * [log(:r<-l)-log(z<+l)].
Differentiating each term with respect to x gives :
. _ 4s3
ydx
Now dy/dx is what we are after ; so we multiply through by y, and then
substitute the value of y as originally given in terms of x.
" dx
or, simplified, ^ =
VII, § 180] EXPONENTIAL FUNCTIONS
261
N.B. It is desirable to simplify the logarithms as much as possible
before differentiating, also to combine the fractions obtained in differen-
tiating the right member, before multiplying across by the given function.
Compare this example with Ex. Ill, § 177.
EXERCISES
1. Differentiate logarithmically :
(c)
(d) q
2. Plot y = 2x and also y = x2, from 2 =—3 to x = 6, over the same
base line. Observe how differently y varies in the two cases, even at
the common points.
[3.] If y = 7, e3*3, find dy/dx by introducing logarithms. Similarly
prove that if y = eu, then dy/dx = evdu/dx.
§ 180. Derivative of e". Differentiating y — eu loga-
rithmically gives
%L=e.^ (24)
dx dx
That is, the derivative of any power of e is that very same
power, times the derivative of the exponent. (Memorize.)
Formula (24) bears no resemblance whatever to the formula
d(un)=nun~1du. The reason is that an exponential function
e* varies in an entirely different way than a power function
un. (§ 169.)
To differentiate an exponential function in any modified
form, as att, first express it as a power of e.
Ex. I.
Ex. II.
dx
y= 100 e*. Here =
This result is the same as .4 y. Thus the rate is proportional to y
itself. In fact the given formula is evidently a case of the C. I. L.
262 MATHEMATICAL ANALYSIS [VII, § 181
Ex. III. v = P-2-1T.
This is an e form in disguise. In fact by Ex. I, § 169 :
..
dL
That is, dv/dT = . 06931 v, which shows again that this y varies accord-
ing to the C. 7. L. Observe that the coefficient .06931 in the exponent
(fcT7) refers to the instantaneous rate of increase, rather than the average
rate during an entire unit.
§ 181. C. L L. by Integration. The general formula for
theC./.L., y = pe,,t (25)
was obtained in § 168 by considering the analogy to an in-
vestment whose interest is compounded continuously. The
formula can now be derived without any thought of that
analogy. The method is precisely the same as in the follow-
ing numerical illustration.
Ex. I. A quantity y increases with z at a rate constantly
equal to .04 y. If y = 100 at x = 0, find the formula.
Solution. We are given dy/dx = .04 y.
The problem, then, is simply to integrate this, and get y
in terms of x.
There is one difficulty: as the equation stands, the de-
rivative on the left side is taken with respect to x, whereas
the right member is expressed in terms of y. But let us
divide through by y :
*^ = .04.
y dx
The left member is now the derivative of log y, with respect
to x. (§ 177.) Integrating with respect to x gives
But y = 100 when z = 0, whence (7= log 100.
log i/ = . 04 z -flog 100.
VII, § 181] EXPONENTIAL FUNCTIONS 263
Transposing log 100, and remembering that the difference
log y — log 100 is the same as the logarithm of the fraction
y/100 (§ 139), we have
log ^ = .04*.
This means that .04 x is the exponent of the power to which
the base e must be raised to equal the fraction t//100.
/. -7^ = e^ or i/MOOe-04*.
JLUU
Remarks. (I) This formula is the same as would be obtained by
thinking of the analogy to compound interest, the given quantity y
growing at the rate of 4% compounded continuously.
(II) The formula can also be checked directly. Substituting z=0
gives y = 100 ; and differentiating gives
ax
as required.
EXERCISES
1. Differentiate each of the following functions :
(a) 7/=20e-0te, (6) i=4Qe~3<*,
(c) z = 10 e-*, (d) y =
(e) w = 7/e**\ (f) y =
2. (a) Differentiate y = 3*t by expressing y as a power of e; also
by taking logarithms. (6) Likewise differentiate w = 10*4 in two ways.
[3.] By differentiating logarithmically prove that if y = uv, where
u and v are functions of x, then dy/dx—u(dv/dx)-\-v(du/dx).
[4.] As in Ex. 3 show that if y=u/v, then dy/dx = [v(du/dx) —
5. To what interest problem is this equivalent: If ?/ = 1000 at t = 0
and grows at rate always equal to .15 y, what will be the value of y
at any time? Write the formula by inspection.
6. In Ex. 5 obtain the required formula also by integration.
7. The difference D between the temperature of a hot wire and that
of the air decreased thus : dD/dt=—.QD. To what sort of depreciation
is this analogous? What formula if D = 100 at t =0? Check.
8. In Ex. 7, derive the formula for D by integration.
264 MATHEMATICAL ANALYSIS [VII, § 181
N.B. In Exs. 9-15, obtain each required formula by integration.
9. An electrical current died out thus: di/dt= —60 i. Derive the
formula for i at any time if i = 30 when t=0.
10. In Ex. 9 plot a graph showing how i decreased, from t = 0 to .1
at intervals of .02.
11. The number of bacteria in a culture increased thus:
dN/dt = .3N. At the start, AT = 20. Derive the formula.
12. In Ex. 11 draw by inspection a rough graph showing how N
', increased from < = 0 to J = 9.
13. When an iron rod is heated its length increases thus :
dL/dT = . 00001 L.
Express L as a function of T, if L=60 (in.) when T = 0. At what
temperature will L have increased by 1% of its original length?
14. Passing through dark glass the intensity of light varied with the
distance (x in.) thus, di/dx=—.2i. If i was originally 60, derive a
formula for i at any distance.
15. Atmospheric pressure varies thus with the height above sea-
level : dp/dh = — .00004 p. Find a formula for p at any height, knowing
that p = 30 when h = 0. Calculate p at h = 6000.
16. Each of the quantities mentioned below varies at a rate con-
stantly proportional to the value of the quantity. In certain cases
the constant of proportionality has the value shown. Express these
facts in calculus notation, and write by inspection the result of each
integration. Check each result by differentiating.
(a) Rotary speed, with the time elapsed since the power was cut off :
fc= -.02, and 72 = 100 at t = 0.
(6) The length of a glass rod, with the temperature : k = .0000083 ;
L =20 when T = 0.
(c) Viscosity of olive oil, with temperature : k = — .023 and V = 3.265
when T = 0.
(d) An electric current, with time elapsed since the E.M.F. was cut
off: fc=-200andc = 10at J = 0.
(e) The quantity of sugar remaining t min. after being subjected to
a certain acid: fc = -.0014 and Q = 300 aU=0.
(/) Tension in a pulley belt, with the distance along the pulley:
A; = .04 and T = 30 when Z> = 0.
17. The number of negative "ions" passing between two charged
plates is given by the relation : dn = kndx, where k is the gas constant,
and x is the distance from the negative plate. Derive a formula for n
at any x, if n = N at x=0.
VII, § 184] EXPONENTIAL FUNCTIONS
265
§ 182. Derivative of a Product. Any product can be
differentiated logarithmically. But often it is more con-
veniently differentiated directly, by using the formula derived
in Ex. 3, p. 263, viz. :
dx
.
dx
(26)
That is, the derivative of the product of two variables is equal
to the first variable times the derivative of the second, plus the
second variable times the derivative of the first. (Memorize.)
Ex. I. Differentiate y=x* log x.
Here the first variable is x2 and the second is log x.
dx
simplified.
§ 183. A Typical Application. Suppose that, under
pressure, the height of a rectangular plate is decreasing at
the rate of .05 in./min. and the base increasing .02 in./min.
How fast is the area changing when h = 20 and 6 = 15?
A=bh.
. dA idh.idb ,n*7\
- -dT=bdi+hcu- (27)
Substituting given values for 6, dh/dt, etc.,
at
The area is decreasing at the rate of .35 sq. in. per min.
§ 184. Derivative of a Fraction. Differentiating y = u/v
logarithmically as in Ex. 4, p. 263, we find that
; r-
ay _ dx dx
dx~ ~~^~~
(28)
266 MATHEMATICAL ANALYSIS [VII, § 184
That is, the derivative of a fraction equals the denominator
times the derivative of the numerator, minus the numerator
times the derivative of the denominator, all divided by the
square of the denominator. (Memorize.*)
Ex. I. Find the maximum value of y= ^ x.
x*
if dy_l-4 log x
' j -- 1 - •
dx x*
To find the maximum value of y, we set dy/dx =0 :
1 -4 log re ^Q
z6
Multiplying through by x6 gives 1 -4 log z =»=0, or log x = \.
Since the base is e, this means that
x = e\ = ^ =1 .284 approx.
Substituting forz and log x in the original equation : 2/=|= T^-
Test : At x = 1 , dy/dx = + . At x = 2, dy/dx = — . Hence a maximum.
EXERCISES
1. Differentiate and simplify the results :
(a) 2/ =
(c) z = 2 x* log x— z2, (d) z=x2 • e4*,
(e) u=
(0) i/ = (log
(fc) Z=rlOO+r2, (Z) «-
2. Test for maxima and minima :
(a) y = Oogz)M (6) u=
(c) T/=rr/(x2+l), (d) 2=
* A help in remembering the order in (28) is to observe that the formula
begins and ends with v, the denominator. Notice also the negative sign
between the terms, as against the positive sign in the product formula.
VII, § 185] EXPONENTIAL FUNCTIONS 267
3. Differentiate 7/ = (x2 + I)3 logarithmically and compare (22), p.
111. The same for y = un.
4. A rectangular metal block has a square base whose edge is in-
creasing at the rate of .04 in./min. The height is increasing at the rate
of .06 in./min. How fast is the volume changing when a: = 10 and
JM*30?
5. If the radius of a cylinder is increasing at the rate of .2 in./min.
and the height is decreasing at the rate of .3 in./min., how fast is the
volume changing when r = 10 and h = 20 ?
6. For a certain quantity of a gas PF=500 T. If P increases at
the rate of .02 units per min., and V at the rate of .04 units per min.,
how fast will T increase when P=30 and F = 6000?
7. In Ex. 6, if P increases at the rate of .05 per mjn. and T at the
rate of .2 per min., how fast will V be changing when P = 40
and T = 300?
8. If one of two quantities is increasing at the rate of .02 per min.
and the other decreasing at the rate of .04 per min., how fast is their
product changing at the instant when
the first equals 15 and the second
equals 25?
9. How fast is the ratio of the first
to the second changing in Ex. 8 ?
10. Approximately how much will
be the area of a rectangle change if
the base and height increase slightly u
as in Fig. 85? Cf. § 182. FlQ 85>
11. The speed of signals through
an oceanic cable is proportional to the function S=*(log z)/:r2, where x
is the ratio of the thickness of the covering to the radius of the core.
Find the maximum value of S.
12. Find how high a wall-light L should be placed to secure the
maximum illumination I of a level surface at S 4 ft. from the wall, if 7
varies as the sine of Z S, and inversely as the square of distance LS.
[Show that the function to be tested is
§ 185. Summary of Chapter VII. A quantity which
varies according to the C. I. L. (§ 168) is strictly analogous
to an investment whose interest is compounded continu-
ously. Its value is expressible as an exponential function, —
i.e., as a varying power of e (§ 162) or of some other constant.
268 MATHEMATICAL ANALYSIS [VII, § 185
These formulas for the C. I. L. can be obtained either from
this analogy or by integration.
Logarithms to the base e are the simplest and most natural
in studying the C. /. L., and also in differentiating and
integrating.
The differentiation formulas for ev, uv, u/v, and un have all
been derived from the formula for log u, which assumes the
existence of the limiting value e. Thus all of our differentia-
tion and integration formulas to date rest upon this assump-
tion, — except that the formula for d(un) had already been
derived independently in the case of an integral or fractional
value of n. The (uv) and (u/v) formulas are also easily de-
rived independently by the A process.
The logarithmic method is the best for differentiating
complicated products, roots, etc.
Logarithmic and semi-logarithmic plotting are useful in
statistical work, and in discovering or studying Power Laws
and Compound Interest Laws.
Many quantities which we have not yet studied vary in
much the same way as some trigonometric function. To
deal with these effectively we need some further graphical
concepts, to the study of which we next turn.
EXERCISES
1. What differentiation formulas have been covered so far in the
course? Under which of these does each of the following forms come
primarily :
j/ = log(x10), 3/ = (logz)10, 2/ = (logz)/z, y = e*\ ?/ =
2. (a)-(e). Differentiate each of the functions in Ex. 1.
3. Differentiate each of the following functions :
(a) y = (log10z)», (6)
(c) i/ = (logz)/zB, (d) y = x/(\ogx),
(e) z =
(j) y-
VII, § 185] EXPONENTIAL FUNCTIONS
269
4. Approximately how much should log« x increase while x runs from
1 to 1.000275? Hence log 1.000275 = --?
5. Find where the slope of the curve y = e~xt is a maximum or
minimum.
6. For a certain gas, the pressure, volume, and temperature have
lis relation: pv = 5Q T. If T increases .3°/min. and p decreases .4
tits per min., how fast will V be changing when T = 320 and p = 40?
7. The speed of a point on a rotating wheel (v ft. /sec.) varied thus :
y = 400e~2'. Find a formula for the distance (x ft.) covered in any
length of time.
8. A sum of money, drawing interest compounded continuously,
doubles in 8 years. What is the nominal rate r?
9. In how many years will any sum double, drawing 5% interest,
compounded continuously ?
10. The speed v of a certain chemical reaction increases thus with
the temperature: dV/dT = .07 V. If 7 = 50 when T = 0, write by in-
spection the formula for V at any temperature. Derive this also by
integration. Find T when V = 500.
11. Draw by inspection a graph showing the general way in which v
increased in Ex. 10, and showing plainly the values v = 50, v = 500.
12. If a weight is to be suspended by a vertical rod, and the stress
on every horizontal section of the rod is to be the same, the sectional
area (A sq. in.) should vary thus with the elevation (x in.) above the
bottom: dA/dx = kA. If A =20 when x = 0, find the formula for A.
If k = .00002, will the change in A from x = 0 to x = 150 be appreciable?
13. The annual expenditure of the United States government (in
millions) has increased as in Table I. Plot together the semi-logarith-
mic graphs of this and of the population for the same period. [See
Table 4, p. 251.]
(I)
Yr.
1840
1850
1860
1870
1880
1890
1900
1910
Ex.
24
41
63
294
265
298
488
660
In which decade was the percentage rate of growth greatest for each?
Smallest? Which rate was the larger in the decade 1900 to 1910?
14. In a recent year there were in the United States 320 personal
incomes above $500,000; 230 above $1,000,000; and similarly for
other incomes, shown in Table II. Plot the ordinary and also the
logarithmic graph.
270
MATHEMATICAL ANALYSIS [VII, § 185
(II)
Income
3000
5000
10000
25000
50000
100 000
200000
Number
330 000
200000
85000
27000
10500
3600
1300
16. In Ex. 14 show that the tabulated values satisfy roughly the
power law : N = 7 300 000 000/7*.
16. In an experiment with light passed through a pinhole the inten-
sity was found to vary as in Table III with the distance (x in.) from the
hole. Find the formula for 7.
150
(HI)
10
(IV)
24
0
30
.67
50
.24
D
0
1
5
10
100
81.9
36.8
13.5
15
5.0
17. The intensity of light passing through a solution of copper
chloride varied with the depth (D cm.) as in Table IV. Find the law.
18. A wound treated by Dr. Carrel's method* decreased in size
(S cm.), after t days, as in Table V. Show that this follows roughly a
C. 7. L.
(V)
t
0
2
4
5
7
10
12
S
6.2
4.7
3.5
3.0
2.2
1.3
1.0
19. Discover the formula for Table 8, p. 29. Also find by calculus
the work asked for in Ex. 10, p. 29.
20. The following table shows the reading of a vacuum gauge t hours
after the pump broke down. (One value is grossly incorrect.) Dis-
cover the law. Also correct the error in the table.
t
0
1
2
3
4
5
R
29.2
15.6
9.4
4.48
2.4
1.28
* Cf. Ex. 6, p. 18.
CHAPTER VIII
RECTANGULAR COORDINATES
§ 186. Locating Points. A simple way to describe the
location of a point P is to tell its distances x and y from two
mutually perpendicular lines, XX' and YY1. (Fig. 86.)
To show on which side of each reference line or axis the
point P lies, we use a + or — sign, calling x negative for
points to the left of YY', v
and y negative for points
below XX'. Thus for
the point A, x = — 6,
y = 4 ; for 5, x = - 8,
y= —3 ; and for (7, z==4,
2/=-6.
The x and y of a point
are called its coordinates :
x the abscissa and y the
ordinate. 0 is called the
1 (
y
\(
i
6,
/>
r
B =
--
6'
1
y
-
I
:
j
*
y
Tl
0
1
P(
Ly-
C
(4
,-t
)
Fi
I
G.
/•'
8
G.
To designate a point we simply write its coordinates
within parentheses, x first. Thus ( — 6, 4) denotes the point
at which x= — 6 and 2/=4.
The idea of coordinates is used in daily life. E.g., we direct a man
to some point in the city or country by telling him that it is, say, five
blocks east and two blocks north ; or three miles west and four miles
south.
Coordinates afford the basis for the mapping of points,
for the scientific study of motion, and also, as we shall
presently see, for a very systematic and powerful method of
studying geometry.
271
272
MATHEMATICAL ANALYSIS [VIII, § 187
(A) THE STUDY OF MOTION
§ 187. Path of a Moving Point. The motion of a point
in any plane is conveniently studied by means of its varying
coordinates (x, y), referred to axes in the plane. If we have
a table giving the values of x and y at various instants, we
can map each successive position and draw the approximate
path.
Still better, if we have a pair of equations giving the values
of x and y at any time, we can calculate as many positions
as we please, and study the motion in detail. Such " equa-
tions of motion " are used in studying the motions of pro-
jectiles, airplanes, parts of machines, points on the vibrating
strings of musical instruments, etc.
Ex. I. The position of a projectile t sec. after firing was
x and y being in feet. Plot the path.
AtJ = 5; z = 5000, 2/ = 2500-400;
and similarly for the other values in the following table.
t
X
y
t
X
y
0
0
0
20
20000
3600
5
5000
2100
25
25000
2500
10
10000
3400
30
30000
600
15
15000
3900
Plotting these points (x, y}, we draw the path smoothly. (Fig. 87.)
IO.OOOT
0 5,000 10,000 15,000 20,000 25,000 30,000
FIG. 87.
VIII, § 187] RECTANGULAR COORDINATES ; 273
Remark. The projectile was highest when y was a maximum, or
500-32 t = Q, <=^i = l5f.
It struck the ground when its height y became zero :
500*-16f2=0 or t = 500/16 = 31^.*
The projectile struck at a point where x = 1000(31 J) =31,250.
EXERCISES
1. Plot the points (0, 0) and (6, 8), and join. , Calculate the slope of
the line ; also its inclination angle, and its length.
2. A city is laid out in squares, 10 to the mile. Map the following
points: A (0,0); 5(10,23); C(20, 15); D(22, -15)-E(-5, -12);
F(-24, -7); G(-18, 17). (The X-axis points east, and the F-axis
north.) How long a wire would be needed to reach from A to E?
Which plotted points are inside the 2£ mile circle about A ? Of points
on this circle, which is nearest A, via the streets? What direction is
F from A? How far is C from F by a straight subway? In what
direction ?
3. A gun is located at (4000, 5000) and a target at (16,000, 12,000)
meters. Find the distance and direction. (The X-axis points east
and the F-axis north.)
4. The same as Ex. 3 for a gun at (2000, —3000) and target at
(-1500, 9500).
5. Referred to X- and F-axes pointing east and north from Soissons,
the coordinates of several cities and towns (in miles) are: Rheims
(32, -8) ; Laon (13, 13) ; St. Quentin (-2, 32) ; Cantigny (-37, 19) ;
Paris (-44, -36); Chateau-Thierry (4, -23). Map these points.
Measure the direct distance of Paris from Cantigny ; from Chateau-
Thierry. Calculate the latter.
6. The curve of a ship's deck near the stern passes through the
points shown in Table I. Plot the curve on the scale of 1 inch to 2 ft.
each way. (Run the F-scale from 10 to 24 only.) By measurement,
how long is the plotted curve?
7. The cross-section of a ship's hull at one of the stern bulkheads is
a curve passing through the points shown in Table II. (All the abscissas
are to be taken both positive and negative.) Draw the curve. Find
the approximate area of the bulkhead, from the flat bottom up to the
30 ft.-level.
* Observe that t = 0 also satisfies the equation. What does this mean?
274
MATHEMATICAL ANALYSIS [VIII, § 188
8. The~samc as Ex. 7 for a bow bulkhead, using Table III. Find
the area from the curved bottom up to the 35 ft. level.
9. The positions (x, y) of a moving point after various intervals
(t sec.) are shown in Table IV. Draw the path.
TABLE I TABLE II TABLE III TABLE IV
0
2.25
4.5
6.75
9.
11.25
13.5
15.75
18.
19.85
20.50
21.10
21.65
22.15
22.60
23.00
23.35
23.65
±*
V
.6
0
2.0
5
3.2
10
5.2
15
8.7
20
13.8
25
18.3
30
20.1
33.5
±*
y
0
0
5
1.5
8.3
5
10.6
10
11.9
15
12.8
20
13.7
25
14.7
30
15.8
65
16.9
38.5
t
X
y
0
0
0
1
.6
.2
2
2.4
1.6
3
5.4
5.4
5
15.0
25.0
7
29.4
68.6
V10. A batted ball traveled thus: x = l2Qt, y = l6Qt-l6t, the X-
axis being horizontal and the Y-axis vertical. Calculate its position
at various instants from t = 0 to £ = 10. Plot the path, using the same
scale both ways. Measure the distance traveled through the air.
11. In Ex. 10 find when the ball was highest ; also where it struck
the ground.
§ 188. Speed and Direction of Motion. From a pair of
" equations of motion," we can find not only where the
moving point will be at any time, but also how fast it will be
moving, and in what direction.
Consider, for instance, the projectile P in Ex. I, § 187 :
The rate at which its height y is increasing at any time is the
rate at which P is then rising. That is,
vertical speed = ^ = 500 - 32 t.
at
E.g., at J = 10, P will be rising at the rate of 500-320, or
180, ft./sec.
VIII, §190] RECTANGULAR COORDINATES 275
Similarly, since dx/dt = 1000, P will be moving horizontally
at the rate of 1000 ft./sec. (Fig. 87.)
In reality the motion of P will be neither horizontal nor vertical.
But it is convenient to regard the actual motion as composed of two
independent motions, in the X and Y directions.
If we draw directed lines
or " vectors " to represent on
some scale these two rates -^ D T/
F Vx=1000
oi motion, or component
speeds," then the actual speed
and direction of motion will be represented by the diagonal
of the rectangle : *
.'. z; = V10002+1802=1016+, tan 4 = 1^ = . 18.
By tables, A = 10° 12'. Thus at £ = 10, P was moving with
a speed of 1016 ft./sec., in a direction 10° 12' above hori-
zontal.
If the two component speeds were maintained for one second, the
point P would move as just calculated, throughout the second.
§ 189. General Formulas. Let vx and vv denote the speeds
of any moving point (x, y) in the X and Y directions, re-
spectively, at any instant. Then, reasoning as in § 188,
the actual speed and direction of motion are given by
(1)
(2)
That is, the actual speed and direction are represented by
the diagonal of a rectangle whose sides, drawn from the same
vertex, represent vx and vv.*
§ 190. Distance Traveled. Knowing the speed v of a
moving object at every instant, we can find by integration
* A formal mathematical proof of formulas (1) and (2) is given in the
Appendix, p. 486.
276 MATHEMATICAL ANALYSIS [VIII, § 190
the distance s traveled during any interval of time. For
by (B), §99:
s= Cvdt.
To illustrate, suppose that an object moves thus :
x=t\ y=$t*-t.
Then v, = 2t, vv = P-l,
and by (1) above the speed at any time is
Hence the distance traveled is
•
s=£vdt= C(
The constant of integration is zero since s = 0 at t = 0.
N.B. The value of v which we must integrate in any case is the
general value in terms of t, and not the numerical value at some one
instant.
EXERCISES
1. A projectile moved thus: z = 1500J, ?/ = 400 t — 16 J2. Plot the
path. Measure the approximate distance traveled. Check the
direction of motion at t = 5 by calculating vx and vv and drawing the
corresponding vectors on the graph as in Fig. 88.
2. A bullet traveled thus: z = 800<, y = 600 t - 16 *2. Plot the
path. Find vx and vy at t = 5, and also at t = 0. Represent by vectors ;
and calculate the actual speed and direction in each case.
3. In Ex. 2 find when and where the bullet struck the ground, —
and with what speed and inclination.
4. A point moved thus: x = 3P, y = 3t — t3. Plot its path from
t=— 4 to £ = 4. Calculate vx, vv, at t = 2; also the actual speed and
direction then. Illustrate by vectors.
6. In Ex. 4 find v at any time. Also find the exact distance traveled
from t = Q to < = 4 ; and check by measurement.
6. A point moved thus : z = 9 <*, y = t3. Find the distance traveled
in the first t sec. (Hint: Show that v = 3t Vf2+36, and integrate as
in § 101.)
VIII, § 1911 RECTANGULAR COORDINATES 277
7. In each of the following, calculate the position of the moving
point at several instants during the specified interval, and plot the path.
Also find the speed and direction of motion at t = 2, and show this on the
plotted path.
(a) x = 3&, y = 2t3, from t = Q to £ = 4.
(6) x = P-2t, y = f$, from t = Q to 4.\
(c) z = 6(12-*2), y = t(l2-W, from t = -4 to +4.
(d) z = 20 W2-4), 0 = 20 W-4), from <= -2.5 to H-2.5, every .5.
(e) x = &(t-4), y = P(t-£), from t = -1 to +5.
8. (a), (6), (c). Calculate the lengths of the paths plotted in Ex. 7
(a), (6), (c). Check by measurement.
§ 191. Deriving the Equations of Motion. The foregoing
methods of studying motion exactly can be used only when
we know the " equations of motion," which give x and y in
terms of t. The question therefore arises as to how such
equations are obtained in the first place.
The method is different in different cases. Sometimes the
equations are discovered experimentally, the position of the
moving object being observed at various times, and a formula
being devised to fit the resulting table. But usually the
equations are deduced mathematically from some physical
or mechanical principle which governs the motion. A good
example is the motion of a projectile, or other object, fired
or thrown in any way.
If we ignore air resistance, there is no horizontal accelera-
tion, and the vertical acceleration is — 32 (ft. /sec.2) .* That is,
g-O, §=-32. (3)
Integrating both of these twice gives the desired equations,
the constants of integration being determined by the way
the projectile is fired.
Some further types of motion, equally important, will be discussed
later.
* See Remark I, p. 43.
278 MATHEMATICAL ANALYSIS [VIII, § 191
Ex. I. Find the equations of motion for a projectile fired
with a speed of 1000 ft./sec.- at an inclination of 30°.
Integrating (3) :
— = ^=—32*4- ' (4}
dt ~ ' dt
x = ct+k, y=—l6P+c't+kf. (5)
If we have chosen our axes so as to pass through the firing
point, then x — 0 and y = 0 at t = 0. Hence k = 0, k' = 0. To
•^ determine the values of c and
c', observe in (4) that these
constants are simply the values
HI ipTT^ "*** °f dx/dt and dy/dt (i.e., the
— a ^}S=10£) C08 30° component speeds vt and vv)
jto. 89. aU = 0. By Fig. 89 these are
simply :
v, = 1000 cos 30° = 866.03,
vv= 1000 sin 30° = 500.
Substituting these values of c and c' gives finally
x = 866.03 1, y = 5QQt-lQt2. (6)
Remarks. (I) These equations could now be used to study the
motion in detail. The general shape of the path could be seen by plot-
ting, as in § 187. Its precise geometrical character can be determined
later. (§223.)
(II) If the projectile were fired from an airplane 2000 ft. high,
directly above our origin (0, 0), we should still have fc = 0, but fc' = 2000.
EXERCISES
1. A projectile was fired with a speed of 1500 ft./sec. and an in-
clination of 21° 6'. Find its equations of motion, ignoring resistance.
2. When and where did the projectile in Ex. 1 strike the ground?
With what speed and inclination was it then moving ?
3. The same as Ex. 1 for a ball thrown with a speed of 100 ft./sec.
and an inclination of 30°. Also find when and where it was highest.
VIII, § 193] RECTANGULAR COORDINATES 279
4. A golf ball, driven with a speed of 125 ft. /sec. and an inclination
of 40°, over level ground, rolled 100 ft. after striking. Find the
equations of motion before striking, and the total length of the "drive."
[5.] Plot the points (3, 1) and (12, 13). Calculate their distance
apart. Can you derive a formula for the distance from (3, 1) to any
spther point (x, y}? From (x, y} to (x', y')?
[6.] Is the quadrilateral whose vertices are (1, 19), (23, 2), (88, 34),
(66, 52) a parallelogram? Plot, but also make a sure te.st by calculating
slopes.
(B) ANALYTIC GEOMETRY
§ 192. Formulas Needed. Coordinates are useful not
only in mapping points and studying motion, -but also in
studying geometry. The first step in this direction is to
derive certain standard formulas for distances, slopes, etc.,
by which those quantities can be calculated immediately
without the necessity of drawing a figure. These formulas
should be carefully memorized.
In deriving the formulas we shall denote any two given fixed points
by (xi, y\) and (x%, y^). Here Xz (read "x two") means simply the x of
the second point. Do not confuse it with re2.
§ 193. Distance Formula. The distance between any
two points (xif 2/1) and (x2, y2) is seen from Fig. 90 to be
"^ (7)
For d is the hypotenuse of a right triangle, whose legs are
the difference of the x's and the difference of the y's*
Ex. I. The distance between (2, 3) and (8, 15) is by (7) :
rf = V(8-2)2+(15-3)2=l80.
Remarks. (I) It makes no difference which point is considered as
Cd, 7/0 and which as (z2, 3/2). E.g., (2-8)2 equals (8-2)2.
(II) Formula (7) is correct even when some of the coordinates are
negative. E.g., for the points (-4, -10) and (-24, 6), (7) gives
* The formula may also be written c?=
280
MATHEMATICAL ANALYSIS [VIII, § 194
(-4.-10)
(*2.i/a> |
which agrees with Fig. 90
where the legs of the right
triangle are 20 and 16.
§ 194. Slope Formula.
The slope of the line
joining (x1} yi) and (z2,
2/2) is
(8)
FIG. 90.
For the line rises (2/2—2/1)
units in (x2— Xi) hori-
zontal units.
Ex. I. The slope of the line through (3, 4) and (8, 10) is
7 10-4^6*
8-3 5'
(This is simply the difference of the y's divided by the difference of
the z's, or Ay /Ax.)
Remarks. (I) The order of subtraction must not be reversed for
x or y alone.
(II) Formula (8) is correct even when the line descends toward the
right. For instance, for the line from ( -24, 6) to ( -4, - 10) in Fig. 90,
the formula gives l = ( — 10— G)T-( — 4+24) or — Jg-, which agrees
with the figure.
§ 195. Mid-point Formula. The
point (x, y) midway between (x\,
2/i) and (z2, 2/2) is : (x,
+y2). (9)
For the vertical line through (x, y)
bisects the base, and equals half
the height, of the right triangle
shown in Fig. 91. (Why?) That is,
FIG. 91.
which, simplified, reduce to the formulas in (9).
VIII, § 196] RECTANGULAR COORDINATES 281
The best way to remember (9) is to observe that x and y
are simply the averages of the x's and t/'s of the given mid-
points.
Ex. I. The point midway between (1, 14) and (9, 8) is
Ex. II. The point midway between (1, 14) and ( — 16, —6) is
§ 196. Direction. The direction of a line may be de-
scribed by telling its inclination measured from the positive
direction of the X-axis, upward or downward. - This can be
calculated from the slope, as in § 111.
The angle between two lines can be found from the two
inclinations.
A sure test whether two lines are parallel is to see whether their
slopes are equal. (Why?)
EXERCISES
'1. Calculate the distance from (8, 5) to (4, 2). From (13, -7) to
(-2, 1). From (0, 0) to (9, 12). From (-17, -31) to (95, 137).
2. In the triangle whose vertices are (6, 2), (9, —4), and (7, —1),
which is the longest side?
3. Show by the distance formula that the diagonals of the rectangle
whose vertices are (0, 0), (12, 0), (12, 5), (0, 5) are equal.
4. Which of the points (12, 12), (10, 13), (13, 11), (9, 15), (6, 17),
(8, 16) are on the circle with center at (5, 1) and radius 13 ; and which
are inside? Try to tell by plotting ; but check by exact calculation.
5. Find the slopes and inclinations of the lines joining : (8, 1) and
(13, -6), (-4, -12) and (9, 7), (0, 0) and (6, 8). Are any of these
lines parallel or perpendicular ?
6. Find the length and slope of each side of the quadrilateral whose
vertices are (2, -1), (12, 3), (-6, 5), (4, 9). What sort of figure?
7. Find the mid-point between (9, 12) and (3, 4). Between (4, -5)
and (-2, 9).
8. Find the mid-point of the hypotenuse of the triangle whose
vertices are (0, 0), (16,0), and (0, 12) ; and show that this point is equi-
distant from all three vertices.
282 MATHEMATICAL ANALYSIS [VIII, § 197
9. In the triangle whose vertices are (13, 4), (19, 12), and (7, -8)
show that the line joining the mid-points of any two sides is parallel
to the third side, and equal to half of it.
10. The same as Ex. 9 for the triangle (1, 8), (9, -2), (15, 6).
(N.B. Draw figures in the following exercises, but also use some sure
test.)
11. What sort of triangle is it whose vertices are (10, 35), (25, 15),
and (73, 51)? Can you find the area of this triangle?
12. What sort of quadrilateral has the vertices ( — 10, 40), (10, 10),
(43, 32), and (23, 62) ? Find the lengths of the diagonals.
13. A triangle has vertices (2, 48), (51, 61), and (38, 12). What is
its perimeter? Is it equilateral?
14. Find the length of the medians of the triangle whose vertices
are (1, 8), (9, -2), and (15, 6).
16. A triangle has vertices (4, 2), (8, 5), and (11, 1). Is it isosceles?
Equilateral? A right triangle? (Test by the lengths of the sides and
also by their inclinations.)
16. A quadrilateral has vertices (3, 1), (15, 9), (17, 6), and (5, -2).
Is it a parallelogram? A rectangle? (Test the diagonals.)
17. Certain buildings in a city are located as follows : City Hall, A
(0, 0) ; Post Office, B (3, 2) ; Court House, C ( - 1, 6) ; railroad stations,
D (15, 9) and E (3, 11). Plot. A garage is to be built midway
between D and E, and a law library midway between A and C. What
locations? How far is B from C, D, and E in air-lines?
18. A boulevard joining the points (5, 2) and (12, 7) is crossed by
a railroad from (15, 9) to (6, —12). What is the angle of crossing?
19. In a certain county there are several towns located as follows :
A (0, 0), B (8, 7), C (-3, 9), D (1, -11), the unit being 1 mi. and the
axes certain section lines, with OX pointing east. How far are B, C,
and D from A by airplane, and in precisely what directions? If
straight railways join AD and BC, at what angle will they cross?
20. A town located at (0, 0) gets its water from a reservoir at (14, 5).
How long is the water main, allowing 10% extra for grades?
§ 197. Test for Perpendicularity. Suppose that two
lines are perpendicular, and that one rises at an angle of
40°. Then the other must fall at an angle of 50°. (Fig. 92.)
Hence the two slopes are
Zi = tan40°, Z2=-tan50°.
VIII, § 198] RECTANGULAR COORDINATES
283
But tan 50° = ctn 40° ; hence k = - ctn 40°, = - I/tan 40°.
(§111.) That is,
/2=-p (10)
Hence, for these two perpendicular lines, the slope of one is
the negative reciprocal of the slope of the other.
Moreover, this is true for any pair of perpendicular lines.*
No matter at what angle AI one line ascends, the other must
descend at an angle A2 which is the com-
plement of AI. Thus the same reasoning
applies as above :
Ui lz= —tan Az= —ctn AI= — 1
Hence if one of two perpendicular lines has the
slope 3, the other must have the slope — $-. If one
has the slope — f , the other must have the slope +§-.
Conversely : if Z2 = — 1 A, the lines must
be perpendicular. For the perpendicular to
the first line at the common point would
have its slope equal to — 1/Zi, and would
have to coincide with the second line, since
there can be only one line through a given point with a
given slope.
§ 198. Equation and Locus. A sure test whether a point
(x, y) is on the circle with center (0, 0) and radius 10 (Fig. 93)
is to see whether
100. (11)
FIG. 92.
This equation is true for any point on the circle, no matter
where taken. But it is not true for points inside or outside
the circle.
Similarly, a sure test whether a point lies on any other
curve or line, say an ellipse or spiral or straight line, is to see
* Except a horizontal and a vertical line. The latter has no such thing
as a "slope," strictly speaking. See § 41.
284
MATHEMATICAL ANALYSIS [VIII, § 199
whether some other definite
equation is true for the x
and y of the point. Along
any graph, for instance,
there is some definite rela-
tion y=f(x).
Definition. An equation
which is true for the coor-
dinates of any point on a
curve, but not true for the
coordinates of any other
point, is called the equation
of the curve. " And the curve is , called the locus of the
equation.
E.g., (11) is the equation of the circle in Fig. 93. And that circle
is the locus: of equation (11). We have already plotted the loci of many
other equations, — such as y = x*, y = z3 — 12 x+5, etc.
§ 199. Descartes' Great Invention. The fact that any
one equation belongs exclusively to some particular curve makes
possible the solution of many v
geometrical problems by means
of coordinates.
ILLUSTRATION. If a point
(x, y) moves in such a way that
its distance from (20, 0) is
always twice its distance from
(5, 0), along what curve will it
move?
\
A(5,0) I
FIG. 94.
By hypothesis we have continually, in Fig. 94,
2AP.
Expressing BP and AP in terms of coordinates by the
standard distance formula (7), p. 279,
-0)2 = 2\/(*-5)2+(!/-0)2.
VIII, § 199] RECTANGULAR COORDINATES 285
Simplifying by squaring and collecting terms :
z2-40 z+400-i-2/2 = 40c2- 10
That is, P must move in such a way that z2-hf/2=100
continually. But by (11), p. 283, all points for which this is
true lie on a certain circle. (Fig. 93.) Hence P must move
along that circle.
This fact can also be proved by elementary geometry; but some
ingenuity is required to know what construction lines to introduce.
(Do you see what lines?)
Observe that in solving this problem by the coordinate
method, we had only to express by a standard formula the
given fact that BP = 2 AP, and then simplify according to
the standard rules of algebra. Much more difficult problems
can be handled easily by this method as soon as we are familiar
with the standard equations of certain curves, and can
recognize the equations at sight.
Coordinates were invented by Rene" Descartes, a Frenchman, who
published in 1637 a systematic treatment of geometry by means of
coordinates and equations. This was a great step in advance, for the
method is so systematic and powerful that it permitted a tremendous
extension of higher geometry. In particular, the problem at once arose
of finding the direction of any curve at any point; and this soon led
to the invention of calculus by Newton and Leibnitz. (Cf. § 103.)
Geometry thus studied is called "analytic" or "co6rdinate" or
"Cartesian" geometry, — Cartesius being the Latin form of Descartes'
name.
EXERCISES
1. Is the line joining (1, 4) and (9, 8) perpendicular to the line
joining (3, 14) and (13, -6)?
2. A triangle has vertices (1, 3), (15, -5), and (11, 7). Is the
median from the last vertex perpendicular to the side joining the first
two?
286 MATHEMATICAL ANALYSIS [VIII, § 200
3. The vertices of a quadrilateral are (4, 2), (16, 10), (6, —1), and
(18, 7). Is it a rectangle? Are the diagonals perpendicular?
4. The vertices of a quadrilateral are (4, 5), (10, 15), (40, 1), and
(8, —7). Show that the lines joining the mid-points of its four sides
form a parallelogram.
5. Where are all the points for which y = x? y = 3x? y=—x?
6. Which of the points (7, 24), (21, 15), (16, 20) are on the curve
xi -1-2/2 = 625 ? At what points on this curve is x = -20 ?
7. Which of the points (10, —10), (6, 5), and (50, 30) are on the
curve 2/2=20(a:— 5) ? Where does this curve meet the x-axis? (Hint :
What is the value of y at any such point ?)
8. At what points of the curve 4y2 = x3isy=x?
9. Where does the curve z2+?/2 = 25 cross the vertical line 3 units
to the left of the F-axis? Where does it cross the line bisecting the
angle between the negative X and F axes?
10. What value must r have if the curve xz+yz = r2 is to pass through
the point (3, -4)?
11. Is the point (8, 6) on the circle z2+?/2 = 100? Find the slopes
of the lines joining it to the ends of the horizontal diameter. What
theorem of geometry is illustrated ?
12. A square has its vertices at (5, 5), (5, —5), (—5, —5), and
( — 5, 5). A point (x, y) moves so that the sum of the squares of its
distances from these vertices is always 300. Find its path.
13. A point (x, y) moves so that its distance from (50, 0) is always
5 times its distance from (2, 0). Find its path. Check the points
where it crosses the X-axis.
14. A point (x, y) moves so that the sum of the squares of its dis-
tances from (—3, 0) and (3, 0) is always 50. Find its path. Check as
in Ex. 13.
15. A point (x, y} moves so that the lines joining it to (8, 0) and
(—8, 0) are always perpendicular. Find the path.
[16.] Express by an equation the fact that a point (x, r/) is on the
circle with center (9, 8) and radius 13.
§ 200. Linear Equations. If an equation is of the first
degree, — i.e., of the form
ax+by+c = 0, (12)
where a, 6, and c are constants, its locus is a straight line.
VIII, § 200] RECTANGULAR COORDINATES 287
PROOF. (I) If b is not zero, so that y is actually present,
we can solve for y, getting an equation of the form
y = lx+k, (13)
where I and k are some constants (viz., 1= —a/b, k = —c/b).
Differentiating this :
-r = Z, constant.
dx
That is, y increases at a constant rate, and its graph (or the
required locus) must be straight (§7). In other words the
locus has a constant slope ; and all points for which the
equation is true lie along a straight line. Conversely, by
§ 32 the coordinates of all points on the line satisfy one and
the same linear equation.
(II) If b happens to be zero, so that y is missing, equation
(12) gives simply x = —c/a. All points at which x has this
constant value lie on a certain straight line parallel to the
F-axis. And for all points on that line x has the value
— c/a.
Hence, whether 6 = 0 or 6 =£0, the locus of (12) is a straight
line.
To find the slope of a line whose equation is given, we
simply think of the equation as thrown into the form (13).
The coefficient of x will then be I, the slope.
To draw a line from its equation, we simply calculate two
points, well separated, and join them. A third point should
be calculated as a check.
Ex. I. 2z+32/+5 = 0. Here?/=-f z— |. /. Slope =-f
Ex. II. 4z-7&' = 8. Herej/ = fz-f /. Slope = f
EXERCISES
1. Can a point move and yet keep x = 5 constantly? How? Or
y= -3 constantly? Draw the lines x+11 =0, 2 x+7 = Q, ?/2 = 9.
2. Draw the following straight lines, checking each by calculating
a third point: x = y, x+y = 5, 2z+3?/-18 = 0, 4z+y+ll = 0.
288 MATHEMATICAL ANALYSIS [VIII, § 201
3. Draw the following lines with a single pair of axes, checking each :
Note how the constant 5 appears in the lines.
4. The same as Ex. 3 for the lines : y = 2 x+7, y=2 x — 1, y = 2 x+3,
y =2 x. Note the constant 2.
6. What is the geometrical significance of a and 6 in the equation
y = ax+b? Compare Exs. 3-4.
6. What slope has each of these lines :
2z+5i/ = 7, 3x-47/+9 = 0, x-y-5=Q?
Draw the lines.
7. Is the line 2 x—3 y+5 = Q perpendicular to the line joining (1, 9)
and (5, —3)? Does it pass through the point midway between these
two?
8. A point (x, ?/) moves so that its distances from (1, 3) and (9, —1)
are always equal. (A) Find the equation of its path, simplified.
OB) Show that the path is perpendicular to, and bisects, the line joining
the given points.
9. Show analytically that the locus of points equidistant from (3, 2)
and (7, —4) is the perpendicular bisector of the line joining these points.
10. A point moves so that the sum of its distances from the X and
y axes is constantly 10. Draw its path. (Is this properly an unlimited
line ? Discuss.)
11. How far is the point (13, 7) from the line along which x= — 3?
Likewise the point (6, 1) ? Likewise any point (x, y) ? How far is
(x, y) from the lines x= —8, y= —4, and y = 7?
§ 201. Type Equation of a Circle. Let (h, k) denote the
center and r the radius of any circle. Then for any point
(x, y) on the circle, and for no other points, we have by the
distance formula :
*)2 = r2. (14)
This is therefore the general equation for any circle. Ob-
serve that the coordinates of the center are subtracted from
x and y, — not added to them.
Ex. I. For a circle with center (5, 2) and radius 7, the
equation is
(z-5)2+(</-2)2 = 49.
VIII, § 202] RECTANGULAR COORDINATES
289
Ex. II. Find the equation of the circle having (8, 3) and
(4, —5) as ends of a diameter.
The center is the mid-point (6, — 1) ; and the radius is the
distance from (6, -1) to (8, 3) or (4, -5), viz., A/2a The
equation is
§ 202. Drawing a Circle from Its Equation. In case a
given equation represents a circle, we can easily recognize
that fact, and determine the center
and radius, by comparing the given
equation with the type equation
(14). The circle can then be
drawn with the compasses.
Ex. I. Draw the locus of
This is a case of
FlG
in which h = 8, k= — 6, r = 20. Hence, the locus is the circle
with center (8, —6) and radius 20. (Fig. 95.)
Ex. II. Find the locus of
2 z2+2 t/2+10 x+7 y- 10=0.
Dividing through by 2 and completing both squares gives
This represents a circle : center (— f, — -J-) ; radius, V229/4.
Ex. III. Is the locus of 2 z2+3 y*-5 x = 7 a circle?
No; for this equation cannot be reduced to the type equation (14),
— in which the coefficients of x* and y2 are both 1. (At present we
could plot the locus only by calculating points ; later we shall be able to
recognize precisely what curve it is.)
Remark. The only terms which can appear in the type equation (14)
when multiplied out are: x*+y2, with a common coefficient; x and y,
290 MATHEMATICAL ANALYSIS [VIII, § 202
with any coefficients, and a constant term. (The product xy cannot
occur, nor z2— t/2, nor higher powers.) Thus we can tell at a glance
whether any given equation represents some circle.
EXERCISES
1. Write the equations of the circles which have the following
centers and radii : (5, 12), r = 11; (-3, -4), r = 5; (2, 0), r = 3.
2. Find the equations of the following circles :
(a) With center (1, 2), and passing through the point (7, —6);
(6) With center midway between (2, 9) and (8, —1), and passing
through (1, 1) ;
(c) Having the line joining (4, 4) and (—8, 2) as a diameter.
3. What are the centers and radii of : (z-8)2 + (?/+3)2 = 9; (z+4)2-f
7/* = 25 ; x*+(y — I)2 = 1 ? Draw each circle by compasses.
4. Find the centers and radii of the circles x2+y2+lQx— 30?/ = 0;
z2 +yz — 12 x — 10 y + 12 = 0. Draw the circles.
5. The same as Ex. 4 for the circles 2 x2 +2 i/2- 12 x -15 ^ = 0, and
6. A point (x, y) moves so that the sum of the squares of its dis-
tances from (—6, 0) and (6, 0) is 200. Find the equation, and draw
the path. Select some special point on the curve and verify that it
fulfills the specified requirement.
7. A point moves so that the sum of the squares of its distances
from (3, 0) and (—3, 0) is any constant k. Find the character of its
path. Draw the path when fc = 22 and when /c = 34. In each case
check for some point.
8. A point moves so that its distance from (12, 0) is always twice
its distance from (0, 0). Find the equation of its path ; plot, and check
for some point.
9. In an "Addition" to a certain city a boulevard is to run east to
a point (12, 8), then swing around a quarter circle and run south from
(17, 3). What center and radius must the curved arc have? What
equation? Where will the curve meet a street on which x = 15?
10. The same as Ex. 9 if the curve starts at (10, 18) and ends at
(17, 11).
11. Draw circles with centers (0, 0) and (12, 16) and passing through
(30, 40). Find their equations. How much higher is the first at
x=0 than the second at x = 36? Are they tangent? Reason?
12. The curve of the under side of a bridge consists of three circular
arcs, as follows: (I) Center (0, 0), connecting (56, 192) and (-56,
VIII, § 203] RECTANGULAR COORDINATES 291
192); (II) Center (14, 48), running from (56, 192) to z = 104;
(III) Symmetrical with (II), on the left. Draw this compound curve.
Show that the intersection (56, 192) is on the line of centers, making the
arcs tangent.
13. In Ex. 12, calculate the radius of each circle and write each
equation. Calculate the height of the arch at the middle, above the
ends. Compare with your drawing.
14. Draw the circle x2 + (y — 8)2 = 25. How high above the
.XT-axis are the two points at which x = 4 ? If this circle were
revolved about the X-axis, what sort of surface would be generated?
How far would the center travel? Also the two points just men-
tioned?
15. Find by differentiation the slope of the tangent to the circle
z2+// = 100 at the point (8, 6). Also find independently the slope
of the radius drawn to (8, 6) and compare. What theorem is illus-
trated?
§ 203. Choosing Axes. In proving a geometrical theorem
analytically we must first introduce axes. We select these,
of course, in such a way as to
make the equations and coordinates
considered as simple as possible.
When we wish to be general we
use letters rather than special
numbers for the coordinates of the
given points. A couple of illustra- ^
tions follow.
(I) THEOREM: The perpendic-
ular dropped from any point of a
circle upon any diameter is a mean proportional between the
segments of the diameter.
PROOF. Choose the diameter in question as the X-axis,
and the center of the circle as origin. Then the equation of
the circle is simply x2+y2 = r*. The proposed perpendicular
is simply y. (Fig. 96.) Hence we are to prove y a mean
proportional between the segments of the diameter (r+z)
and (r— x).
292 MATHEMATICAL ANALYSIS [VIII, § 203
Now from the equation of the circle we have at once
or
~
= (r+x)(r-x).
y r+x
(Q. E. D.)
97.)
(II) PROBLEM : A point P (x, y)
moves in such a way that the sum
of the squares of its distances from
two fixed points F and F' is con-
stant. Along what curve does it
move ?
I (c'0) SOLUTION. Choose the line FF'
as the X-axis, and the mid-point
as the origin. Let c denote one
half the distance between F and
F', whose coordinates are then
simply (c, 0) and ( — c, 0). (Fig.
Let k denote the constant sum of the squares:
FIG. 97.
Then
The radicals disappear, and the equation reduces to
The path is a circle with center midway between F and F'.
EXERCISES
1. A point moves so that the sum of its distances from two mutually
perpendicular lines is constant. Prove analytically that its path is a
straight line. What if the difference of the distances were to be constant
instead of the sum?
2. A point moves so that its distance from one of two perpendicular
lines is twice its distance from the other. Find an equation for its
path. Plot and verify several points. What if the two distances have
any constant ratio ?
VIII, § 204] RECTANGULAR COORDINATES
293
3. Prove analytically that the middle point of the hypotenuse of
any right triangle is equidistant from the three vertices.
4. Prove analytically that the diagonals of any rectangle are equal,
but that they are perpendicular only if the rectangle is a square.
6. A point moves in such a way that the sum of the squares of its
distances from two fixed points is constant. Prove analytically that
it moves in a circle.
6. In Ex. 5 if the difference of the squares were constant instead of
the sum, what would the path be?
7. A point moves so that its distances from two fixed points have a
constant ratio. What sort of path? Any exception?
8. A point (a>, y) moves so that the sum of the squares of its distances
from the four sides of a square is constant. What sort of path?
[9.] A point (x, y) moves so that its distance from (3, 0) equals its
distance from the line x = — 3. (Cf. Ex. 11, p. 288.) Find the equation
of its path.
[10.] From the answer to Ex. 9 calculate several points and plot.
Then select several points on the curve and test by measurement
whether they meet the requirement stated in Ex. 9.
§204. The Parabola. We pro-
ceed now to study a few plane
curves other than circles, which are
used frequently in scientific work.
Definition: A parabola is the locus
of a point which is equidistant from
a fixed straight line and a fixed
point.
This means (Fig. 98) that any
point P on the parabola is equidis-
tant from the line DD' and the
point F ; and also that every point
thus equidistant is to be considered as part of the parabola.
The fixed line DD' is called the directrix; and the fixed
point F the focus.
Any number of points on a parabola can be found by simply
drawing lines parallel to DD', and cutting them by arcs
described from F with the proper radii.
FIG. 98.
294
MATHEMATICAL ANALYSIS [VIII, § 205
Or the parabola can be drawn by continuous motion. A triangular
ruler (Fig. 98) slides along DD', the edge DP being perpendicular to
DD'. A string just long enough to reach from A to D has one end
fastened to the ruler at A ; and
the other end fastened at F. The
pencil point P keeps the string taut,
-that is, keeps FP = DP, while the
ruler moves.. Hence P travels along
a parabola.
§205. Type Equation. De-
note by 2 p the -distance from
F to DD' ; and choose axes as
in Fig. 99, so that F has the
coordinates (p, 0) and DD' isp units to the left of the F-axis.
Then for any point P(x, y) :
= x+p,
FlG 99
At every point of the parabola, and at no others : DP = FP.
I.e.,
Simplified, this gives as the type equation of a parabola :
y* = *px. (15)
E.g., yz = l5x is the equation of a parabola in which p = *£. That
is, the distance from directrix to focus is J£.
§ 206. Nature of a Parabola. The general character of a
parabola can be seen from its equation
(1) It does not extend to the left of (0, 0). For x cannot be
negative in the equation. (2) It extends indefinitely far
toward the right. For y is real at all positive values of x.
(3) It is symmetrical with respect to the X-axis. For there
are always two values of y, ±, numerically equal. Consider,
then, only the upper half of the curve. (4) It has no highest
VIII, § 208] RECTANGULAR COORDINATES 295
point but rises continually. For y continually increases with
x. (5) Its slope, however, continually decreases. For, differ-
entiating gives
2 2/^ = 4 p, or & = 2£, (16)
ax as ?/
which shows that dy/dx grows smaller as y increases. That
is, although the curve continues to rise, it rises less and less
rapidly. (6) It makes no undulations. For the slope reaches
no maximum or minimum value.
The curve therefore appears as in Fig. 99. The axis of
symmetry is called the axis of the parabola ; and the point
where this axis crosses the curve, the vertex. The rapidity
with which the curve spreads apart depends upon the value
of p. When x=lj y,±vTji.
N.B. A curve may look very much like a parabola and yet not be
one. It may even have all six of the properties above; and still it
will not be a true parabola, unless all points on it are exactly equidistant
from some fixed point and fixed line.
§ 207. Applications. The parabola is a frequently en-
countered and much-used curve.
A steel girder, or the cable of a suspension bridge, if loaded
uniformly per horizontal foot, will hang in a parabola (the
axis of which is vertical). The arches of a bridge, or high
ceiling, are often made parabolic, — likewise the " crown "
of a pavement.
The hollow upper surface of a rotating fluid is parabolic.
(Fig. 43, p. 128.) So are the reflecting surfaces used in
searchlights and telescopes.
The orbits of some comets, the paths of projectiles in a
vacuum, and the graphs of many scientific formulas are
parabolas.
§ 208. Parabola with Axis Vertical. If the axis of a pa-
rabola is turned straight upward, the focus being at (0, p) on
296 MATHEMATICAL ANALYSIS [VIII, § 208
the F-axis, and the directrix p units below the X-axis, the
only change in the equation (15) will be that x and y will be
interchanged. [Draw a rough figure to illustrate this.]
The equation, then, will be
*2 = 4/.y. (17)
Ex. I. A parabolic suspension cable is to have its ends 200 ft.
apart and 40 ft. higher than the middle. Required, the equation of
the curve, and the height 50 ft. from the center.
With the origin taken at the lowest point the equation is of the form
(17). But we are to have y — 40 when x = 100, at the end. Substituting
gives 4p = 250.
/. z2 = 250y. (Check?)
At z = 50, this equation gives y = W. The cable will be 10 ft. above its
lowest point 50 ft. from the middle ; and a vertical strand to reach from
the cable to the bridge should be cut accordingly.
EXERCISES
1. Draw by inspection: 7/2 = 20z, y2 = l2x, y* = 2x. In each case,
where are the focus and directrix?
2. The same as Ex. 1 for : x* = y, x* = 10 y, x* = .04 y.
3. Test without plotting which of the following points are on the
parabola tf = 12x: (3, -6), (4, 7), (27, 18), (16, -14), (1/3, -2),
(75, 30), (0, 0), (-3, 6), (-12, -12).
4. For what value of p will the parabola y* = 4px pass through
(10, 20) ? Find the point on that parabola at which y = 8.
5. Find the equation of a parabola whose axis is vertical, whose
vertex is at the origin, and which passes through (10, 4).
6. The hollow upper surface of a rotating fluid is parabolic : 8 in.
deep and 40 in. across. Find the equation of the curve, taking the
lowest point as (0, 0). Find y at x = 10.
7. A ship's deck rises in a parabolic curve to a height of 9 ft. 3| in.
in a horizontal distance of 256 ft. 3 in. Find the equation of the curve ;
and the height at x = 50.
8. A suspension cable is so loaded as to hang in a parabola. Its
ends are 800 feet apart and 100 feet above the lowest point. Find
its simplest equation ; also its height at x = 80 and x = 200.
9. A level foot-bridge of span 200 ft. is supported by a parabolio
suspension cable from towers 30 ft. above the floor, the center of the
VIII, § 209] RECTANGULAR COORDINATES 297
cable being 5 ft. above the floor. How long a wire is necessary to
reach vertically from the cable to the floor 20 ft. from the center?
50 ft. from the center?
10. Write the equation of a parabola with focus (4, 0) and directrix
11. Finish deriving the equation y2 = 4 px in § 205.
12. A circle moves and changes size so as to be always tangent to a
fixed line and pass through a fixed point not on the line. Mark several
positions of the center. Apparently what locus? Proof?
13. The great reflecting telescope at Mt, Wilson is a parabolic mirror
5 ft. across. The distance from vertex to focus is 25 ft. Find the
equation of the parabola, with its axis vertically upward. How deep
is the mirror at the center?
14. Prove that the area under the parabola y = x* from z = 0 to
x =b, is precisely one third of the circumscribed rectangle having the
same base.
§ 209. Rotating a Curve 90°. It is important to know
how the equation of any curve will be modified when the
curve is moved to some new
position .without changing its \
shape or size. (xz,y*)
Let us first consider the
effect of merely turning a
curve 90° about the origin,
counter-clockwise. (Fig. 100.)
If (xi, 2/1) is a point on the
original curve, and (a^>, t/2) the
corresponding point on the
new curve, then
Xi
FIG. 100.
(18)
That is, each x in the original equation will be replaced by y
and each y by —x.
E.g., if the parabola i/2 = 4px is rotated 90°, its equation
will be
i.e., x* = 4py. [Cf. (17).]
298 MATHEMATICAL ANALYSIS [VIII, § 209
If rotated another 90°, so that its axis extends to the left,
along the negative X-axis, its equation changes to y2= — 4 px.
And, if rotated still another 90°, xz= — 4py. (Verify this.
Also see what effect a fourth rotation would have.)
Hence each of the equations
y'>=±±px, tf=±±py (19)
represents some parabola. We need not memorize how the
curve is turned for each of these equations ; but in any given
case simply make a substitution or two. Or notice which
variable, x or y, may have both positive and negative values.
Ex. I. Locate the parabola x2 = — 12 y.
Clearly x can be either positive or negative ; but y cannot be positive.
Hence no part of this parabola is above the X-axis ; and the curve must
extend along the negative F-axis. (Draw the figure.)
Of course the focus and directrix are carried along with a parabola
in its rotation. Hence the focus here is (0, —3) on the F-axis, and the
directrix is y = 3.
EXERCISES
1. Draw by inspection z2 = — 49 y and y2 = — 16 x. Mark each focus
and directrix, and test by measurement whether some of your points
are equidistant.
2. The same as Ex/1 for the curves 3 z2 = — 10 y and x+2 yz = 0.
3. Write the equation of a parabola with focus ( — 7, 0) and directrix
x = 7 ; likewise of a parabola with focus (0, —1) and directrix y = \.
4. Find the equation of a parabola through ( — 6, —9), with vertex
(0, 0) and axis vertically downward.
5. A roadway 40 ft. wide is 1 ft. lower at the sides than in the middle.
If the curve of the "crown" is a parabola, find its equation. What is
the drop in 10 ft. from the middle?
6. The curve of a ship's deck athwartship is a parabola which, in a
horizontal distance of 27 ft. from the center, falls 13.5 inches. Find its
equation. How much does the deck fall in the first 20 ft.? What is
the slope 15 ft. from the center?
7. The steel arch of a bridge is a parabola with axis downward.
The horizontal span is 400 ft., and the center is 120 ft. above the ends.
Find the equation of the arch with the origin at the vertex. Plot the
curve.
VIII, § 210] RECTANGULAR COORDINATES
299
8. In Ex. 7, a level road-bed, 20 ft. above the vertex, is supported
by vertical columns, from the arch, 40 ft. apart. Find the lengths of
the columns at a: =40 and x = 120.
[9.] A point (x, y) moves so that the sum of its distances from ( — 16,
0) and (16, 0) is always 40. Find the equation of its path, simplified.
(Transpose one radical before squaring.) Locate several points by
compasses and draw.
10. Prove this theorem : Any tangent to a parabola makes equal
angles with the axis (produced) and with the radius from the focus to
the point of tangency. [Hint : The slope of
PT is yi/TH, but by § 206 it is also 2p/yi.
Equating, and remembering that y? = 4 pxi,
we may show that TH = 2xi and hence that
TF = FP. Carry out the details of the proof.] ^
Remark. The reflection property of para- T
bolic surfaces depends on this theorem.
Rays of light entering parallel to the axis will
be focussed at F. Conversely, rays emanat-
ing from a source at F will emerge parallel
to TF as a non-scattering beam. FIG. 100 a.
§ 210. The Ellipse. Definition: An ellipse is the locus of
a point whose distances from two fixed points have a constant
sum. The two fixed points F and F' are called the foci.
Any number of points on an ellipse can be found by de-
scribing arcs from F and F' with various radii, FP and F'P,
whose sum is constant. (Fig. 101.)
Or the ellipse may be drawn by continuous motion. Take
a string longer than the distance FFf, and fasten its ends at
F and F'. Then, if a pencil point P keeps the string taut,
it will move in such a way that FP+F'P is constant, — i.e.,
along an ellipse. (A loop of string passed around two pins
at F and F', and drawn taut, accomplishes the same result
more conveniently.)
Evidently an ellipse must be a smooth symmetrical oval, as in
Fig. 101. But not every such oval is an ellipse. In a true ellipse, the
sum of FP and F'P must be absolutely constant for all points. (Cf.
Ex. II, § 203.)
300
MATHEMATICAL ANALYSIS [VIII, § 211
FlG 101
§ 211. Type Equation. De-
P(*,v) note by 2 c the distance between
the foci ; and by 2 a the con-
stant sum FP+F'P. (Fig. 101.)
Clearly 2 a must be greater than
2c.
Choose axes as in the figure,
making the coordinates of the
foci (c, 0) and (-c, 0). Then by the distance formula :
At every point of the ellipse, and at no others :
FP+F'P = 2a.
a.
Transposing one radical, squaring, and simplifying :
Squaring again and simplifying :
x2 y2
o2+a2-c2 = 1*
Since a>c, as noted above, a2 — c2 is positive and may be re-
garded as the square of some real number b. Substituting 62
fora*-*:
1.
(20)
This is the type equation of the ellipse, when the X and Y
axes are chosen as above.
Ex. I. What wffl the equation be if the ellipse is drawn with a string
10 in. long whose ends are 8 in. apart?
Here 2a = 10, 2c=8. Hence 6s = a2— cP-9, and the equation is
25 9
VIII, §212] RECTANGULAR COORDINATES 301
Remark. To draw this ellipse roughly by inspection of its equation,
simply find where it crosses the X and F axes :
^ = 0, x = ±5;' x=Q,y = ±3.
Then draw a smooth symmetrical oval through these four points.
§ 212. Axes and Foci. The constants a and b of equation
(20) appear very plainly in the ellipse. For the curve crosses
its axes of symmetry at the points
y = Q, x=±a; x = 0, y=±b.
Thus the diameters A A' and BB' are 2 a and 2 6.
Observe that the major axis 2 a equals the " constant
sum " mentioned in the definition, — i.e., equals the length
of string required to draw the ellipse.
Observe also in Fig. 102 that FB and F'B, whose sum is
2 a by the definition, must each equal a. That is, the dis-
tance from either end of the minor
axis to either focus is equal to half
the major axis. By using this fact,
we can construct the foci geomet-
rically, when a and b are known.
Knowing the foci, we can draw the
ellipse with a string, if desirable.
Moreover, if we get this tri-
angular relation clearly in mind,
we need not memorize the equa-
tion 62 = a2 — c2, for the triangle will supply it.
How flat an ellipse will be is determined by the relative
magnitude of the distance 2 c between the foci and the
" constant sum " 2 a. The ratio c/a is called the eccentricity.
1 Remark. If the ellipse is rotated through 90°, so that the major
axis is vertical and the foci are on the F-axis instead of the X-axis, the
equation (by § 209) will be simply
FIG. 102.
a2 '"&» 1-
302 MATHEMATICAL ANALYSIS [VIII, § 213
The only change is that y now has the larger denominator. But no rule
is necessary as Jo this; for inspection of the equation at x=Q and at
y = 0 will show the lengths of the two axes.
§ 213. Applications. Many ellipses are encountered in
nature: the orbits of the earth and other planets; the
meridians on the earth's surface; any oblique section of a
circular cylinder; the intersection of two equal circular
cylinders or tunnels.
Ellipses are also much used in practical work : In making
machine-gears, man-holes in ships' decks, the arches of many
artistic bridges, and, in general, wherever a shapely oval is
needed, as in a flower-bed, or an eye-glass, etc. The roofs
of " whispering galleries " are elliptical in shape : faint
sounds originating at one focus are reflected to the other,
and can be heard there, though inaudible between.
EXERCISES
1. Draw by inspection the ellipses
(c) 16 z2 +25 7/2 = 10000, (d) 9 z2+4 ?/ = 144.
Find the foci of each.
2. To draw accurately the ellipses in Ex. 1, how long a string would
be needed in each case, and how far apart should the ends be ?
3. Draw by inspection, showing the centers and foci, if any :
(a) fi+ir1- «*—*«.
(c) 25 x2 +4 7/2 = 100, (d) 3z2-17?y = 0,
(e)
4. Write the equation of an ellipse with center at (0, 0) and axes along
OX and O7, for which
(a) Longest diam. (horizontal) = 12 in., shortest diam. =6 in.,
(6) Longest diam. (vertical) = 7 in., shortest diam. =4 in.,
(c) Longest diam. = 10 in., foci are the points (—3, 0), (3, 0) ;
(d) Shortest diam. =26 in., foci are the points (Q, 5), (0, —5).
VIII, § 214] RECTANGULAR COORDINATES 303
5. An ellipse is drawn with a string 50 cm. long, whose ends are
30 cm. apart. Find its simplest equation. %
6. Using pins 12 in. apart, and a loop of string whose total length
is 32 in., what a and b will the resulting ellipse have? Verify by drawing.
7. The ellipse in which the earth travels around the sun has its
longest diameter = 186 000 000 mi., and the distance between foci =
3 000 000 mi. Draw it to some convenient scale, using a string. Also
write its equation.
8. (a) In the ellipse 16 z2+25 ?/2 = 400, inscribe a rectangle having
two sides in the lines x = 4 and x= — 4. Find its area. (6) Calculate
the area of the largest rectangle which can be inscribed in this ellipse.
9. The arches of London Bridge are semi-ellipses, the central one
having a span of 152 ft. and a height of 37:8 ft. Draw the arch to scale.
Also find its equation.
10. A bridge has an elliptical arch, of span 80 ft. and height 16 ft.
Find the equation and draw the curve. The level roadway is 5 ft.
above the vertex. How far is it above the arch at x = 10 and at x = 20?
11. If every ordinate of a circle of radius 10 inches is reduced to
half its value, show that the resulting curve is a true ellipse. (Hint:
If (x, y] is on the new curve, then (x, 2 y~) is on the circle.)
12. A point (x, y) moves so that its distance from ( — 16, 0) is always
$ of its distance from the line x= — 25. Derive the equation of the
path. Draw the figure, and check for some special point.
§ 214. Further Properties. An ellipse has numerous
interesting geometrical properties, two or three of which may
be mentioned here.
(I) Relation to the Major Circle. Let a circle be circum-
scribed about an ellipse, its diameter being the major axis.
Erect any ordinate y of the ellipse, and prolong it until it
meets the circle. Call its length up to the circle Y. Then
from the equations of the circle and ellipse,
a- o-
Hence y = (b/a) Y. That is, any ordinate of the ellipse equals
b/a times the corresponding ordinate of the major circle.
304
MATHEMATICAL ANALYSIS [VIII, § 214
E.g., in an ellipse having a = 10 and 6=6, every ordinate is three
fifths of the corresponding ordinate of the circumscribed circle.
(II) Converse of (I). If in any curve every ordinate y
equals some constant k times the corresponding ordinate Y
of a circle, the curve must be an ellipse.
For, calling the radius of the circle a,
we have
Y=±Va2-x2, y
Simplifying the latter equation by squar-
ing, transposing, etc., gives
1.
FIG. 103.
If k < 1, this represents an ellipse whose
vertical axis (ka) is the shorter; and if
k>l, an ellipse whose vertical axis is the longer. (Fig. 103.)
E.g., if we take a circle and reduce every ordinate to one half or two
thirds of its original length, we get a true ellipse. Or if we lengthen
every ordinate, say by 50%, we obtain an ellipse, with its major axis
vertical.
(Ill) Construction by Auxiliary Circles. From the common
center of two concentric circles, of radii a and b (a>6), draw
any radius meeting the outer circle
at Q and the inner at R. Drop an
ordinate QM from Q, and draw a
horizontal line from R meeting
QM at P. (Fig. 104.) Then the
ordinates of P and Q have the
same ratio as the radii. Hence
this construction reduces each or-
dinate of the larger circle in the
constant ratio b/a; and by (II)
the locus of P is an ellipse. We FIQ. 104.
VIII, § 214] RECTANGULAR COORDINATES 305
can construct accurately in this way as many points of an
ellipse as we wish, with any desired semi-axes a, b.
(IV) Projection of a Circle upon Another Plane. Let a semi-
circle of radius r be turned about its diameter until it makes
some angle C with its former plane. (Fig. 62, p. 169.) Its
projection on the former plane is some semi-oval, whose
precise shape we wish to know.
Every ordinate (y) of the semi-oval is the projection of
some ordinate (Y) of the semi-circle. By § 113, y= Y cos C.
Hence, by (II) above, the projection is a true ellipse, in which
a = r, b = r cos C.
Conversely, any ellipse, of semi-axes a and 6, is the projection of
some circle of radius a, inclined at an angle C whose cosine is 6/a.
(V) Area. Let A be the area of any ellipse, and A' the
area of the circle whose projection it is. Then, by § 113,
A = A'cosC, =A'(b-} = (7ra*)(
w
/. A=irab. (21)
Ex. I. The area of an ellipse of semi-axes 10 in. and 6 in. is 60 IT
sq. in.
Ex. II. A damper in a circular stove pipe turned 60° from the
position of complete obstruction cuts off an elliptic area in which a=r,
b = r cos 60° = .5 r, and hence A = .5 irr2.
EXERCISES
, 1. Every ordinate of a circle of diameter 50 in. was reduced in the
ratio 4:5. What axes had the resulting ellipse? What area?
2. The circular damper of a pipe is turned 35° from the position of
complete obstruction. What axes has the obstructed elliptical area ?
3. The same as Ex. 12, p. 303, but using the point (-32, 0) and
the line x= — 50.
4. Find the volume of a cone whose height is 20 in. and whose base
is an ellipse having longest and shortest diameters of 50 in. and 30 in.
6. The same as Ex. 4 for a height of 30 in. and diameters of 10 in.
and 20 in.
306
MATHEMATICAL ANALYSIS [VIII, § 215
6. (A) How long a loop of string should be used to lay out an
elliptical flower bed 10 ft. long and 6 ft. wide; and how far apart
should the fixed pins or stakes be? What area would the bed have?
(B) The same for a bed 17 ft. by 8 ft.
7. In a photograph, the circular rim of a cup appeared as an ellipse.
Explain this. Why do circular wheels, rings, lampshades, etc., appear
elliptical when viewed obliquely ? What shape, generally, is the shadow
of a circular plate ?
8. In a photograph of a new moon, the crescent is always bounded
by a semi-circle and semi-ellipse. Why?
9. Supposing you had measured a and 6 for the semi-ellipse in Ex. 8
(say 5 cm. and 4 cm., respectively), how could you proceed to calculate
the actual illuminated area of the moon shown in the crescent? (The
moon's radius is 2163 mi.)
10. The Colisseum at Rome was elliptical in shape : 620 ft. long and
510 ft. wide. What ground area did it cover? Draw the ellipse to scale.
11. Carry out the construction of Fig. ' 104, using two circles of
radii 5 cm. and 3 cm., approximately.
12. A straight line 80 cm. long moves with its ends on the X and
Y axes. Find the path of a point P 30 cm. from one end. (First mark
several positions of P as the line turns, and draw the path. Then
derive the equation.*)
[13.] A point moves so that the difference of its distances from
(—25, 0) and (25, 0) is always 40. Find the equation of the locus.
Construct enough points to deter-
Y • ^ mine the general character of the
curve.
§ 215. Hyperbola. The lo-
cus of a point whose distances
from two fixed points have a
constant difference is called an
hyperbola.
We can construct geometri-
cally as many of its points as
we wish, by describing arcs with centers at the fixed points
F, F', and with radii FP and F'P which differ by a fixed
* Large ellipses are often drawn with a "trammel," using the principle
of Ex. 12.
FIG. 105.
VIII, § 217] RECTANGULAR COORDINATES 307
amount. (Fig. 105.) Two separate curves are obtained,
according as we choose FP or F'P the larger. (The two
curves together are called the hyperbola.) These curves can
also be drawn by continuous motion.
An hyperbola is clearl}7" symmetrical with respect to the
line through the foci F, F'} and also with respect to the per-
pendicular bisector of FF'. A and A' are called the vertices
and the distance A A' the transverse axis.
§ 216. Type Equation. Let the foci F, F' be denoted by
(c, 0), ( — c, 0) ; and the constant difference between FP and
F'P by 2 a. (Clearly 2 a must be less than 2 c:)
Then at every point of the hyperbola, and at no others
V(x+c)*+y2-V(x-c}z+y* = 2 a, or -2 a.
Transposing, squaring, and finally putting c2 — a2 = 62 :
(Since a<c, c2 — a2 is positive and may properly be called 62.)
Ex. I. z29-
That is, FF' is 10 units and the "constant difference" 2 a is 6.
Ex. II. z2/16-i/2/9 = l: an hyperbola with a = 4, 6 = 3,
That is, FF' is 10 units, and the constant difference 2 a is 8.
Observe that in an hyperbola, a may be either larger or
smaller than 6. But c is greater than either.
§ 217. Nature of an Hyperbola. Because of the symmetry
of the curve, we need to discuss only that quarter in which
x and y are both positive.
Solving (22) for y gives
(23)
308
MATHEMATICAL ANALYSIS [VIII, § 217
When x<a, y is imaginary; when x = a, y = Q. "Thereafter
as x increases, so does y, and the curve continually rises.
But y is always less than
(b/a)Vx2. Hence the curve re-
mains always below the line
y—(b/d)x. This line is easily
drawn; and will help in sketch-
ing the curve if we can tell how
closely the latter will approach
the line. (Fig. 106.)
Fia- 106- The difference between the
ordinate of the line, Y, and that of the hyperbola, y, is
(24)
When x becomes large, so will Vz2 — a2. Will the difference
become large or small ? To find out, we multiply and divide
by the sum x+\/x2—a2, getting
Y-y = °--
a x+Vx2-a?
It is now clear that Y—y approaches zero, as x becomes very
great. Hence the hyperbola will come as close as we please
to the line if we draw it far enough. The lower half of the
curve must approach a similar descending line.
These two lines approached by the hyperbola, but not
reached by it, are called asymptotes. Their equations are
y=±-x.
a
(25)
Hence they pass through (0, 0) and through the points where
x = a and y= ±b.
Ex. I. Draw by inspection the hyperbola : ~7T7 = 1-
VIII, § 217] RECTANGULAR COORDINATES 309
When ?/ = 0, a; = ±3. (Vertices.) At each vertex we erect an
ordinate 6=4. Through (0, 0) and the ends of these ordinates we
draw straight lines, the asymptotes. Starting at (±3, 0) we draw
the curve, approaching an asymptote as it recedes. (Draw the
figure.)
Remarks. (I) Although the constant b of equation (22) does not
show itself in the curve, it appears in the height of the asymptotes
above the vertices.
(II) In Fig. 106, 0£ = vV+62. That is, by §216, OB is c, half
the distance between the foci F and F'. Hence to locate F and F'
geometrically, we need only describe a circle with center 0 and radius
OB. Fix this picture in mind, and you need not remember the formula
c2 = a2+62, nor tne equations y = ± (b/a)x.
EXERCISES
1. What are a, 6, and c for the hyperbolas z2/16 — i/2/9 = l, £2/4 —
y2/9 = l, andz2/3-i/2/5 = l?
2. Write the equation of an hyperbola whose foci are (±10, 0),
and whose constant difference 2 a is 12 ; also of another with foci
(±6, 0) and 2a = 10.
3. Draw by inspection, showing asymptotes and foci :
4. Draw by inspection, showing centers and foci, if any :
(a) *2+i/2 = 625, (6)
(c) z2-4i/ = 0, (d) 9z + 7/2 = 0.
6. A circle moves and changes size so as to remain always tangent
to two fixed unequal circles. Mark several positions of the center and
state what the locus is. Proof?
6. The same as Ex. 5 but using a fixed straight line and circle instead
of two fixed circles.
7. A point moves in such a way that its distance from (—25, 0)
is always £ of its distance from the line x = — 16. Find the equation
and draw the path. Check some particular point.
8. The same as Ex. 7 for (-50, 0) and the line x= -32.
310
MATHEMATICAL ANALYSIS [VIII, § 218
§ 218. Hyperbola with Axis Vertical. If the hyperbola
x2/a2—y2/b2=l is rotated 90°, its new equation, found by
replacing x by y and y by — x (§ 209), will be
4 O
(26)
b2
Thus the y2 term will be positive and the x2 term negative.
Either x2 or y2 may have the larger denominator.
No rule is needed to tell whether an hyperbola has a hori-
zontal or vertical position : simply try x = 0 and y = 0 in the
equation.
M=\
If z = 0, i/= ±3; if y = 0, x = imaginary. -
The curve meets the F-axis 3
units above and below (0, 0),
but does not meet the X-axis at
all. It is turned vertically.
The geometrical relation of
the foci, asymptotes, etc., to
the curve must be the same, no
matter how the curve may be
turned with respect to the X
and F axes. The entire figure of auxiliary lines is rotated
with the curve. (Fig. 107.)
Hence to draw this hyperbola, we start at (0, 3) and (0, —3), where
the curve meets the F-axis, lay off horizontal lines 5 units long, draw
the asymptotes, and fill in the curve.
§ 219. Sound Ranging. In warfare, an invisible enemy
gun can be located with the help of instruments called
" microphones," which record to .01 sec. the time when the
sound of the firing of the gun reaches various " listening
posts " along our front.
FIG. 107.
VIII, § 219] RECTANGULAR COORDINATES 311
To illustrate, suppose that a sound wave from a gun G
reaches post A .08 sec. sooner than post B. Then, according
to the velocity of sound, G is about 27 meters nearer A than B.
Now the locus of points 27 meters nearer A than B is one
branch of an hyperbola with foci at A and B, and having
2a = 27. G must lie somewhere along this hyperbola.
Similarly, considering other posts C, D, E, etc., G must lie
somewhere along certain other hyperbolas, and hence at the
common intersection of all.
The hyperbolas can be drawn ; for in each case the known distance
between posts (as AB, etc.) is 2 c ; and 2 a has been found by the
microphones. If, as is usual, the gun is far away, the hyperbolas will
practically coincide with their asymptotes in its vicinity ; and only
the asymptotes need be drawn. The value of 6, used with a in drawing
the asymptotes, (§ 217), is easily found, since 62 = c2— a2.
The microphones in like manner locate the bursting shells from our
own artillery and show whether we are shooting over or short, and
whether to the right or left.
EXERCISES
1. Draw these curves, showing the foci and asymptotes :
(c) 47/2-9rc2-36=0, (d) 9 z2-25 ?/2+900 = 0.
2. "Listening posts" are located at A (0, 2000), B (0, 1000), C (800,
400), and D (2000, -100). Microphones show a gun G to be located
506 m. nearer A than B, 280 m. nearer C than B, and 500 m. nearer
D than C. Plot ; and draw the required asymptotes to find G.
3. The same as Ex. 2 for posts located at A (0, 0), B (2000, 0),
C (3600, 0), and D (6000, 1000), and with the gun G 1414 m. farther
from A than from B, 954 m. farther from B than from C, and 1944 m.
farther from C than from D. '
'[4.] A point (x, y) moves so that the difference of its distances from
(10, 10) and (-10, —10) is always 20. Find the equation of its path.
Calculate a few points and plot.
[5.] Draw the ellipse 16 z2+25 y2 = 400 ; also the same curve moved
8 units to the right and 6 units upward, without rotation. What
should the new equation be ? [Hint : If (X, Y) is on the new curve,
312
MATHEMATICAL ANALYSIS [VIII, § 220
then (X— 8, Y— 6) was on the old one, and these values must satisfy
the old equation.] Compare the new equation with that of a circle
whose center has been moved from (0, 0) to (8, 6).
§ 220. Rectangular Hyperbolas. If an hyperbola of con-
stant difference 2 a has its foci at (a, a) and ( — a, —a), on
the line through (0, 0) inclined 45° to the axes, its equation
is, by Ex. 4, p. 314,
xy = a*/2.
By giving a different values, a set of such hyperbolas is
obtained whose equations are all of this form, or
xy = k. (27)
Moreover, for every value of the constant k, except k = 0, this
equation represents some hyperbola/, — viz., one in which
The asymptotes of all these hyperbolas are simply the
X and Y axes. For, by (27), when x becomes very great,
y-*-Q ; and vice versa.
Because of the fact that the asymptotes are mutually
perpendicular, these hyperbolas are called rectangular.
Such hyperbolas, with the axes
for asymptotes, are often en-
countered in scientific work.
They are also used in the busi-
ness world in making certain
_ calculations.
FIG. 108.
For instance, engravers use a chart
like Fig. 108 to read off the price of
a half-tone or zinc etching. Any
desired rectangular plate is siYnply
placed on the chart, with two of its edges along the X and Y axes. If
its fourth corner P falls, say, anywhere along the curve xy = 5Q, then
its area is 50 sq. in. The price to quote for a plate of that area is
marked on the hyperbola, and similar prices on the other curves, —
which come at frequent intervals, — so that no calculation is necessary.
VIII, § 221] RECTANGULAR COORDINATES 313
§ 221. Parabolic and Hyperbolic Formulas. As already
noted, it is very common for one quantity to vary as a power
of another: y = kxn (28)
Hence it is well to be familiar with the graphs of such for-
mulas.
When n= — 1, (28) becomes y = k/x or xy = k. The graph
is then a rectangular hyperbola. For any other negative
.-* -
632
FIG. 109.
FIG. 110.
value of n, the graph will be somewhat similar, approaching
the axes asymptotically. If n is odd, y will be negative when
x is. (Fig. 109.)
When n=+2, the graph is the parabola y = kx'2 with its
axis verti al. In fact, whenever n>l, the graph is very
similar (for positive values of x)} rising slowly at first and
then rapidly. (Fig. 110.)
When n=%, the graph is the parabola y1=(ky)x, with its
axis horizontal; and for other positive values of n less than 1,
and x positive the graph is very similar.
For any positive value of n, (28) is called a parabolic formula and the
graph a parabolic curve. Similarly for negative values of n the formula
and curve are called hyperbolic. For several illustrations of each,
see Ex. 3, p. 314; Ex. 7, p. 84; also pp. 102-103.
314
MATHEMATICAL ANALYSIS [VIII, § 222
EXERCISES
1. Verify the general shape of each curve in Fig. 109 by calculating
three scattered points on each.
2. The same as Ex. 1 for the curves in Fig. 110.
3. Show the general shape of the graph for each of the following
scientific laws. Here c, k, and g denote constants.
(a) Falling bodies: s = £ gP, v
(fe) Speed after falling h f t. :
(c) Time of a pendulum :
(d) Boyle's Law for gases : pv=k.
(e) Adiabatic expansion : pvlM=k.
(f) Magnetic repulsion : F = k/dz.
(g) Electric currents' intensity : i = k/R.
4. The point (x, y} moves so that the difference of its distances from
(k, k) and (— k, —k) is always 2 k. Find its path. From the definition,
what is the distance between vertices? Verify by the equation.
6. Draw the hyperbola xy = k for fc = 20, 30, 40. If zinc etchings
cost 20 i per sq. in., label each of your curves with the cost of any
plate which fits it as in Fig. 108.
6. Steel weighs 7.83 gm. per cc. State how you could make a chart
for reading off the weights of rectangular steel plates 1 cm. thick.
7. Find the area under the curve xy =50 from x=5 to x=20.
§ 222. Translating a Curve. Let us now see how the
equation of a curve will be affected if we move the curve
horizontally or vertically, with-
out rotating it or changing its
shape or size.
Let x, y be the coordinates
of any point on the original
curve ; and X, Y be the coor-
dinates of the same point after
the curve has been moved, say
h units to the right and k units
(X,Y)
FIG. 111.
upward. (Fig. 111.) Then
x-X-h.
(29)
VIII, § 222] RECTANGULAR COORDINATES. 315
That is, the old coordinates equal the new ones diminished
by h and k. Hence the equation of the curve in its new
position is obtainable by replacing each x in the original
equation by (x — h) and each y by (y—k).
Similarly, moving a curve h units to the left and k units
downward will replace x by x+h, and y by y+k. Such con-
stants h and k, which have the effect of sliding the curve
along bodily, we shall call " translaters " or " sliders."
Ex. I. If the circle z2-fz/2 = 100 is moved 4 units to the
right and 3 units upward, what will its new equation be?
Answer: (z-4)2+(i/-3)2=100.
(This agrees with § 201 for a circle with center (4, 3) and
radius 10.)
Ex. II. Recognize (2/-2)2 = 10(z-7).
This is the parabola f/2=10z, but moved 7 units to the
right and 2 units upward. To draw it, start from (7, 2) as
vertex instead of (0, 0) ; and run the axis horizontally to the
right, just as if you were drawing y2 = 10 x.
Ex. III. Recognize
9 z2+72 z-25 ?/2-100 i/+269 = 0.
Completing the squares for both the x and y terms :
9(z2+8 aj+16) -25(2/2+4 ^+4) = -269+144-100= -225.
. fa+2)2 Qr+4)2
9 25
This is the same curve as y2/9-x2/25 = l (Fig. 107, p. 310),
but moved 4 units to the left and 2 units downward. Hence
we merely draw the curve represented by the latter equation,
but starting from (-»4, —2) as the center instead of (0, 0).
Remark. Observe that in translating a curve we replace each x by
x±h and y by y±k; never by y*— k nor ax+h, nor any other expres-
316 MATHEMATICAL ANALYSIS [VIII, § 223
sion. Hence to recognize y-+I2 x— 7 =0 we must write it as yz = — 12
(x— -^j), rather than y* — 7 = 12 x or some other form.
§ 223. Path of a Projectile. By § 191, if we ignore air
resistance and take the firing point as (0, 0), the equations
of motion for a projectile fired in any way have the form
x = ct, y = kt-Wt2. (30)
We can now determine the precise geometrical character of
the path, — not by plotting, which merely shows the general
shape of the curve, but by considering the relation be-
tween x and y at all times.
EXAMPLE. Find the path if x = 80 t and y = 60 t - 16 P.
Here t = x/80, which gives in the ^/-equation :
Simplifying, and completing the square gives :
z2-300 z+1502= -400 7/+1502,
i.e., (z-150)8 = -400(y-56i).
This is the parabola x2= —400 y (with axis downward) but
moved 150 ft. to the right and 56£ ft. upward. Thus the
highest point is (150, 56£) ; and the focus is 100 ft. directly
below this, since 4 p = 400.
§ 224. Conies. It can be proved that the parabola, ellipse, and
hyperbola, though denned differently in what precedes, are all in reality
special cases of a single kind of curve called a conic, which is defined
thus:
A conic is the locus of a point whose distances from a fixed point and a
fixed line have a constant ratio.
Ex. 13 below and Ex. 12, p. 303, Ex. 7, p. 309, illustrate this.
(What must the "constant ratio " be for a parabola?)
VIII, § 224] RECTANGULAR COORDINATES 317
EXERCISES
1. Recognize the " translators " and draw the curves:
0, (d) x2+8 x
264, (/) z2+4 ?/2-24 ?/+ll =0.
2. Draw by inspection, showing the asymptotes and the foci :
9z2-90z-16?/2-320?/
3. An ellipse has its center at (4, 2), and is tangent to the X and
Y axes. Write its equation by inspection. What are the foci ?
4. An ellipse has the foci (4, 1) and (4, 7) and is tangent to the F-axis.
What is its equation?
5. A parabola has the point (8, —2) as focus and the line y = 4 as
directrix. Write its equation by inspection.
6. A point (x, y] moves so that the sum of its distances from (3, 2)
and (9, 2) is always 10. Derive the equation of its path and check by
your knowledge of the ellipse.
7. A point (x, y) moves so that its distance from (6, 0) is twice its
distance from (0, 0). Find the equation of the path. What curve?
8. A projectile moved along the curve y = 4 x/3 — z2/22500. Locate
the highest point by differentiation. Check by completing the square
and recognizing the sliders.
9. A bullet traveled thus: x = QWt, 7/ = 800 t-lGP. Show that
the path was a parabola ; and draw it by inspection.
10. A projectile was fired with an initial speed of 2080 ft. /sec. at
an elevation angle whose sine is T5j. Find its equations of motion, and
the equation of the path. Locate the vertex in two ways. Also find
the focus and directrix.
11. Draw by inspection : (z+100 000) ?/ = 15 000 000. [This formula
and curve are used by a certain telephone company in testing the insu-
lation of lines.]
12. A point moves so that its distance from (10, 0) equals a constant
e times its distance from the F-axis. Derive the equation of the path.
What is the nature of the curve if e = 1 ? If e = f ? If e = £ ?
318
MATHEMATICAL ANALYSIS [VIII, § 225
§ 225. Plotting the Locus of Any Equation. Certain curves
can be drawn by recognizing their equations. Many others
can be plotted by points.
Along any given curve y varies with x in some definite
way. If we can solve the equation of the curve for y in
terms of x, we have merely to plot the resulting function, -
just as we did frequently in Chapters I-III.
The amount of calculation required for such plotting can
often be greatly reduced by making a preliminary inspection
of the equation, and thus learning certain facts in advance.
ILLUSTRATION. Plot the locus of xy2 — 4 t/2+x3+4 z2 = 0.
Here
= ±x*
(I) Extent of the curve. For any value of x which makes
4+a; or 4— x negative, y is imaginary — i.e., if x is below
—4 or above +4. Hence the curve
exists only between x — — 4 and
z=+4.
(II) Symmetry. Wherever real,
y has two numerically equal but
opposite values ( ± ) . Thus the curve
is symmetrical with respect to the
X-axis. This is not so for the
Y-axis.
(III) Intersections with the axes.
When x = 0, y = 0. But when y = 0,
x may be 0 or —4. For one of the
factors x2 and (4+ re) must be zero,
and either may be. The curve
passes through the origin, and also
meets the X-axis four units to the left.
(IV) Vertical asymptote. At x = 4 the equation takes the
form t/2 =16(8/0), — which is entirely meaningless. (§ 29.)
Fio. 112.
VIII, § 226] RECTANGULAR COORDINATES 319
Trying a value near 4, say z = 3.999, makes the denominator
very small and hence y very large. If we let z->4, the
curve must rise indefinitely high, approaching the line # = 4
asymptotically.
(V) Table. Substituting values for x between —4 and 4,
we draw the curve. (Fig. 112.)
Remark. Even when the equation of a curve is too' complicated to
solve for y in terms o£ x, or vice versa, we may still be able to find its
intersections with numerous straight lines, suitably chosen, and by
plotting those intersections obtain enough points to draw the curve.
(See Ex. 5 below.)
EXERCISES
1. Draw the following curves roughly :
(a) y = (x-W)(x-7}(x-3)(x-l),
(6) 7/2 = Oc-
(c) 7/2 = (z
(First determine their intersections with the .X-axis, and see what
happens between each pair of such intersections.]
2. Plot y = x— 3 and y = \/(x— 3), using the same axes. Could
you have anticipated the character of the latter curve by careful inspec-
tion of the former?
3. Plot 2/2=-^-.
4— x
4. Plot yz = xt^4.
x— 4
What isolated point belongs to the complete locus of the equation?
6. Find where the curve x*-}-y3 = 6xy cuts the lines y = x, 2 x, ±x,
% x, \ x, 0, — \ x, etc. Plot these intersections and draw the curve.
[6.] Test by slopes whether (35, 39), (60, 60), and ( -20, -40) are on
the straight line through (20, 10) whose slope is £ . Can you express
by an equation the fact that a point (x, y) is on this line?
[7.] Draw 2x— 3y = ll and 2x+y = 7 and find the intersection.
How could you find this without plotting?
§ 226. Point-slope Equation of a Line. Various geo-
metrical properties of triangles relate to the intersections of
320 MATHEMATICAL ANALYSIS [VIII, § 226
certain straight lines. In studying sueh properties alge-
braically, the first step is to be able to write the equation of
any specified line. This is easy if we know the slope I and
some point (x\, y\) through which the line passes.
If (x, y) is any point whatever along the line, then
y-=y±=i. (3D
x-xi
For by § 194 this fraction is the slope of the line joining (x, y)
and (#1, y\) which is the line under consideration. Moreover,
(31) is not true if (x, y) is any point off this line* Hence (31)
is the equation of the line. Or, more simply,
y-yi = /(*-*i). (32)
E.g., the line through (4, 5) with slope 2 is
2/-5 = 2(z-4), i.e., 2x-y = 3.
Note the distinction between (x, y) and (xi, yi) ; also that
(32) cannot be applied to a vertical line, as there is then no
such thing as a " slope. " Along such a line, however, the
value of x must remain constant, and hence the equation
can be written at sight, in the form
# = some constant.
Ex. I. Find the equation of the perpendicular bisector of the line
joining (5, 6) and (11, 14).
Slope of given line : k = ^| = i
11— o o
Slope of required line : It = -f (§ 197)
The mid-point through which the required line passes is
*i = i (H+5) =8, </i = * (14+6) = 10.
Hence the equation of the required line through (8, 10) with slope -$ is
-10=-x-8 or
VIII, §227] RECTANGULAR COORDINATES 321
EXERCISES
1. Write the equations of the following straight lines :
(a) Passing through (8, 7) with the slope 3 ;
(6) Perpendicular to the line in (a) from (6, 10) ;
(c) Passing through (5, —1) parallel to the line in (a);
(d) Passing through (6, 9) and (12, 10) ;
(e) Through (6, -5) and bisecting the line from (4, -1) to (12, 15) ;
(/) Perpendicular to and bisecting the line from (—7, 6) to (1, 14) ;
(g) Through (0, 0) perpendicular to the line 2 x— 5 ?/ = 12;
(h) Through the mid-point of (8, 1) and (-18, 9) parallel to
2. The vertices of a triangle are (7, -2), (13, 10), -and (-1, 16).
Find the equations of the sides. Does any side pass through (0, 0) ?
3. Find the equations of the medians in Ex. 2. Is any one of them
perpendicular to the opposite side?
4. A point moves so as to be equidistant from (9, —4) and (17, 8).
Show analytically that its path is the perpendicular bisector of the line
joining those points.
5. The same as Ex. 4 for the points (11, 4) and (-1, 12).
6. At what point on the parabola ?/2 = 16 x is the slope equal to 4?
Write the equation of the tangent at that point.
§ 227. Intersections of Loci. If any point is common to
two lines or curves, its coordinates must satisfy both equa-
tions at once. Thus the
problem of finding the in-
tersection of two curves is
equivalent to the algebraic
problem of solving a pair
of simultaneous equations.
This is easy in the case of
straight lines, whose equa-
tions are always of the first
degree.
To see whether three
lines are concurrent (i.e., pass through a common point), we
solve for the intersection of two, and, by substituting in
(10,17)
(8.11)
FIG. 113.
322 MATHEMATICAL ANALYSIS [VIII, § 227
the third equation, test whether this intersection lies on the
third line.
Ex. I. Prove analytically that the medians of the triangle whose
vertices are (—4, 9), (6, 5), and (10, 17) are concurrent.
On each median we know a point, — viz., a vertex, — and we can
find the slope after getting the opposite mid-point.
Vertices: (-4, 9) (6, 5) (10, 17)
Opposite mid-pts. : (8, 11) (3, 13) (1, 7)
Medians' slopes : i — | J9ft
Simplified: x— 6y= — 58, 8z+3y = 63, 10s— 9 y= -53.
To find the intersection of the first two medians, we solve the first
two equations. Eliminating y gives x = 4; whence y = 10 \. Testing
(4, 10^) in the third equation shows that {his point lies on the third
median also. (Q.E.D.)
EXERCISES
1. Find the intersection of the lines 1x — 3y = 7 and z-f4z/ = 15.
Plot the lines and check your result.
2. The same as Ex. 1 for the lines 2 rc+i/ = 9 and 3x—2y = l2.
3. Find the intersection of two medians of the triangle whose vertices
are (8, 7), (4, -1), and (2, 11). Test whether the third median passes
through the same point.
4. The vertices of a triangle are (-3, 2), (3, -4), and (7, -2).
Find the equations of the perpendicular bisectors of the sides, and
show that these bisectors are concurrent.
6. In Ex. 4 prove that the three altitudes are concurrent.
6. Find the equation of the circle circumscribed about the triangle
whose vertices are (10, 5), (-4, -9), and (10, -7). (Hint: On what
lines must the center lie ?)
7. Find the equation of the circle through (0, 0), (8, 6), and ( -2, 10).
8. Show that the medians of the triangle whose vertices are (7, 5),
(3, —3), and ( — 13, 1) are concurrent.
9. The vertices of a trapezoid are (—20, 0), (20, 0), (-13, 6), and
(19, 6). Show that the non-parallel sides and the line joining the mid-
points of the parallel sides all pass through a common point.
10. The same as Ex. 9 for the vertices (-15, 0), (15, 0), (13, 2), and
(-11,2).
VIII, § 228] RECTANGULAR COORDINATES 323
11. Find the equation of a straight line passing through the inter-
section of 4x— 3y+5=0, and 3 x+2 y — 12 = 0, and also through the
point (2, -1).
12. If given the equations of the three sides of a triangle, how would
you proceed to find the equations (a) of the medians ? (6) Of the alti-
tudes ? (c) Of the circumscribed circle ?
§ 228. Summary of Chapter VIII. CoSrdinates are useful
in describing the location of a fixed point, or in following a
moving point. The speed and direction of motion at any
time can be found from vg and vt, which are merely the rates
of change of x and y. The distance traveled during any
interval can be found by integration. From the physical
law of acceleration, equations of motion can be found for
projectiles by repeated integrations. Also the precise
geometrical nature of the path can be found. Thus in the
study of motion, coordinates are almost indispensable.
Coordinates are also helpful in studying geometry. The
test as to whether a point lies on a given curve is to see
whether its coordinates satisfy a certain equation. This
connection between curves and equations permits the study
of geometrical properties of curves by means of their equa-
tions. (Cf . Ex. 10, p. 299.) Various theorems of Elementary
Geometry relating to loci and intersections of lines are also
easily proved analytically.
Perhaps this glimpse of Analytic Geometry, even if brief,
will suggest the possibilities of the method. In more ad-
vanced courses a vast number of properties of the foregoing
curves, and others, are worked out. One interesting fact is
that the ellipse, parabola, and hyperbola can all be obtained
by cutting a right circular cone by a plane.*
* The geometrical properties of these conic sections are of especial interest
historically since by utilizing them modern Astronomy has explained the
motions of the heavenly bodies and has froed mankind from the abject
terror formerly produced by every unusual celestial phenomenon such as
the apparition of a comet.
324 MATHEMATICAL ANALYSIS [VIII, § 228
Remark. The work of this chapter is closely connected with our
central problem of studying functions. For along any curve y varies
with x in some definite way and is therefore a function of x.
There is, however, an important new element in the recent work.
The equation of a curve is generally in the form of a relation between the
two variables x and y, rather than a formula giving y explicitly in terms
of x, as y=f(x}. The equation implies that y is a function of re, but it
defines y as such only implicitly. Thus we may be said to be studying
"implicit functions" now rather than " explicit functions."
We shall next see that the connection between equations
and curves is sometimes of value in solving equations.
EXERCISES
1. Draw by inspection : (a) 4 x -3 y = 12 ; (6) xz+yz = 16 ;
(c) (z-9)2 + (*/+4)2 = 25; (d) y* = l2x', (e} • x2 = 20 y.
2. A point moves so that its distance from (6, 0) is always twice
its distance from (0, 0). Find the equation of the path and draw it.
3. Write by inspection the equation of a parabola whose focus is
(0, 8) and whose directrix is 8 units below the X-axis.
4. The same as Ex. 3 if the focus is (3, 9) and the directrix is the
line x= — 5.
6. To draw an ellipse whose longest and shortest diameters are 20 in.
and 12 in., how long a string would you use and how far apart would
you fix the ends? How long a loop could be used?
6. If the foci of an ellipse are ( — 9, 4) and (—3, 4) and one vertex is
(—6, 8), what is the equation?
7. Derive the equation of the locus of a point (x, y) whose distance
From the 7-axis constantly equals its distance from (16, 0). Draw the
curve roughly by inspection.
8. The same as Ex. 7, if the distance from (16, 0) is always %
times the distance from the 7-axis.
9. A point moves so that its distance from (—34, 0) is always 30
units greater or less than its distance from (16, 0). Find its path.
10. An ellipse has foci (5, 18) and (5, —6), and is tangent to the
F-axis. Write its equation by inspection.
11. Find the center and foci, and draw each of the following curves :
16 x2 - 640 x + 25 1/2 = 1600; 9 z2-18z-257/2 + 100?/ = 116.
12. Recognize and draw the curve y2 — 12 y+12 x =0. What are the
vertex and focus?
VIII, §228] RECTANGULAR COORDINATES 325
13. An hyperbola has the foci (3, 4) and (13, 4), and one vertex is
(12, 4). Draw it roughly. Also write its equation by inspection.
14. A projectile was fired with an initial speed of 2000 ft./sec. at an
inclination whose tangent equals f . Find the equations of motion and
the equation of the path. Draw roughly. What focus ?
15. Hell Gate Bridge, New York City, has a parabolic arch with a
span of 977.5 ft. and a height of 220 ft. Draw the curve to scale.
TABLE I
H
B
0
-9000
5
-4500
10
+6000
25
13600
100
17200
150
18300
100
17250
25
14600
10
12500
5
11000
0
9000
- 5
4500
-10
-6000
etc., sym-
metrically.
TABLE II
X
V
3° 22'
49° 34'.9
20'
33'.5
, 10'
26'.5
3° 0'
19'.4
2° 50'
12'.2
40'
4'.9
30'
48° 57'.5
20'
SO'.O
16. As a magnetized piece of
steel was carried around the " mag-
netic cycle," the magnetizing force
H and the induction B ran through
the values shown in Table I. Plot
the curve. ("Hysteresis loop.")
Compare its area with that of
the rectangle with vertical and
horizontal sides, which just con-
tains the loop.
17. When the ' ' supergun ' ' bom-
barded Paris from St. Gobain, its projectiles traveled
a path along which the latitude y varied with the lon-
gitude x as in Table II. Plot this course, using 1 in.
for 10' (longitude), and 1 in. for 6.5' (latitude), —
the correct ratio for that vicinity. In what direction
did the course run?
18. A point P moves in such a way that the tangent to its path is always
perpendicular to a line joining P to a moving point on the F-axis 10
units higher than P. Find the equation of the path. Draw by inspection.
19. If a point P (x, y} moves on the curve yi = 16 z, and if Q (X, Y)
is the mid-point of the chord joining P to the origin, find the equation
of the curve in which Q moves. Draw the two curves roughly.
20. If a beam is embedded at one end in a wall, and carries a load at
the free end, its slope will vary as the distance (x ft.) from the wall.
What sort of a curve will it form?
21. The force (F Ib.) driving an object varied thus with the distance
(x in.) from a certain point: F = 5 +2Q/x. What sort of graph?
Find the work done from z=2 to 2 = 12.
22. The air resistance to an airplane (P Ib./sq. in.) varied as the square
of the speed (V mi./hr.), P being 180 when 7=60. Write the formula.
What sort of graph ? About what change in P between V = 49.9 and
7=50.1?
CHAPTER IX
SOLUTION OF EQUATIONS
§229. Summary of Earlier Methods. The methods of
solving an algebraic equation which should already be
familiar are as follows : *
(1) Completing the square for any equation in quadratic
form. (2) Factoring by inspection in simple cases, and
equating each factor to zero. (3) Making trial substitu-
tions synthetically to detect any integral roots. (4) Ap-
proximating any other real roots graphically.
We now proceed to refine some of these methods and obtain
others.
(A) EXACT METHODS
§ 230. Formula for the Roots of a Quadratic. The most
general equation of the second degree involving a single un-
known has the form
axz+bx+c = Q. (1)
By completing the square the roots of this are found to be
This result should be carefully memorized. It can be used
as a formula to write at sight the roots of any quadratic
equation.
Ex. I. Find the roots of 5 z2+ll z+3=0.
Here a = 5, 6 = 11, c=3;
_-ll±V6f
~10~
* See §§ 19, 21, 23 63.
326
IX, § 231] SOLUTION OF EQUATIONS 327
Ex. II. Solve x6 - 19 xz- 216 = 0.
This is really a quadratic in terms of x3. (We could let x3 = z, say.)
> .
There are six values for x. For, when rr3— 27 = 0, factoring gives
Hence one of these factors must be zero: x— 3=0, or z24-3 z+9=0.
Similarly z3= — 8 gives x= — 2, and two imaginary values.
§ 231. Plotting a Rotated Conic. In Analytic Geometry
it is proved that every equation of the second degree in x
and y represents some conic, — i.e., ellipse, parabola, hyper-
bola, or pair of straight lines, — unless the locus is imaginary.
If the equation contains no product term xy, it can be re-
duced to a type form with " translators." But if there is a
term xy with other terms, we must rely mainly on plotting
by points.
Ex. I. Plot the locus of
First we solve for y (say) in terms of x, noting that
We may now calculate points by substituting values for x.
The curve is clearly real, no matter how large x becomes, either
positively or negatively. Hence, as it is some conic, it must be an
hyperbola. Its axis is tilted about 42°.
EXERCISES
1. Solve axz+ &z+c = 0 by completing the square. [Cf. (2) above.]
Can the two values of x ever be equal? If so, how? What part of the
results determines whether the roots will be real or imaginary?
328 MATHEMATICAL ANALYSIS [IX, § 232
2. Using formula (2) write by inspection the roots of :
(a) 7x*+13x+12 = 0, (6) 8 x2-13 x+5 = 0,
(c) 11 ^-6 z+5 = 0, (d) 4.8 x2+. 75 x -.0125 = 0.
3. Solve each of these equations for y :
(a) y*-10y2+9 = 0, (6) i/6-26 y3-27 = 0.
4. If a wooden column (x in. square) is to carry a certain load, the
smallest safe value of x is a root of x* — 125 x2- 10368 = 0. Find that
root.
6. The deflection of a loaded beam x ft. from one end is, under
certain conditions, y = k (3 x5 - 4000 x3 + 1 120000 x) . At what value of
x is y a maximum?
6. The same as Ex. 5 if y = k(3 x5 — 10 Px3+7 Z4x) where I is the
length of the beam.
7. Plot each of the following equations by calculating a table of
points. Apparently what sort of curve is each locus?
(a) 2x2-2xi/+2/2 = 2, (6) x2+*/2-2x7/-8x-8i/+16 = 0,
(c) x2+5x2/+4i/2 = 9, (d) 6x2-x7/-27/2+7x+77/-5 = 0.
§ 232. The Discriminant, bz— 4 ac. The nature of the
roots of the equation az2+6z+c = 0 is determined by the
quantity 62— 4 ac which appears under the radical in (2).
The roots are imaginary if b2 — 4 ac is negative. Otherwise
they are real.
The roots are rational if 62— 4 ac is a perfect square.
[They are then free from radicals.] Otherwise they are
irrational.
The roots are equal if 62— 4 ac is zero. [For ( — 6+0)/2 a
is the same as ( — b — 0)/2 a.] Otherwise they are unequal.
E.g., in the equation 121 x2 — 176 x+64=0 we have
62-4ac = (176)2-4(121)(64)=0.
Hence the roots are real, rational, and equal.
These criteria are often useful in determining quickly
whether two given lines or curves intersect. E.g., to find
whether the line y = 2 z+12 meets the circle z2+y2 = 25,
IX, § 233] SOLUTION OF EQUATIONS 329
we need only see whether the two equations have a real
simultaneous solution. Eliminating. y gives
i.e., 5 z2+48 x+ 119 = 0 simplified.
Here 62-4 ac = 482-4(5)(119) = -76.
The values of x are imaginary : the loci do not meet.
§ 233. Factorability of a Quadratic. If we subtract from
x each of the roots in (2), multiply the resulting expressions,
and simplify, we find
That is, x minus each root is a factor of ax2+bx-\-c, the left
member.
E.g., if a quadratic has the roots 3+V5, 3 — Vs, it has the factors
(x-3-VB), (z-3+V5).
Thus e#en/ quadratic is factorable into linear factors of
some kind. These will be rational if the roots are, but not
otherwise. Hence a sure test whether any quadratic is
rationally factorable is to see whether (62— 4 ac) is a perfect
square.
Ex. I. Test 99 x* -42 x -16.
Here 62— 4ac = 8100. Hence the factors are rational. They can
be found by inspection, or by writing x minus each root, by formula.
EXERCISES
1. Determine the nature of the roots of the following :
(a) 9z2 + 17z+8 = 0, (b) 12 x2 +31 x +24 = 0,
(c) x*-x-I=Q, (d) z2
(e) 18 z2+30z+ 12.5 = 0, (/) 72 z2- 179 * -216 = 0.
2. Without actually solving for the intersections determine whether
the circle z2+7/2— 3 x+7 y+Q = 0 meets the X and Y axes.
330 MATHEMATICAL ANALYSIS [IX, § 231
3. Test the rational factorability of the following quadratic expres-
sions. (The actual factors, if any, are not required.)
(a) llx*-lSx-8, (6) 12 x*- 19 x -136,
(c) 388 z2- 1164z+813, (d) 12 z2-42 z+27.
4. For what value or values of fc would the equation fcz2— 6 z+4 = 0
have equal roots? What are those roots?
5. The same as Ex. 4 for the equation 2 x*—2 fcz+4-fc=0.
6. The circle z2+y2-8 z-14 y+1 =0 is cut by the line 2/ = Z.r.
Find the two intersections (in terms of I). For what value of I would
the two points come together, making the line tangent?
7. For what k will 4 y = 3 x+k be tangent to z2+y2 = 100?
[8.] Divide 2 z<-3 z*+7 z2-110 z + 13 by rc-4. Also substitute
z = 4 synthetically in the given quantity. Compare the synthetic
substitution with your quotient and remainder.
i
§ 234. Synthetic Division. Before proceeding to solve
equations of the third degree and above, we need to become
familiar with a certain easy method of dividing out factors.
To see the underlying principle, divide the polynomial
f(x) = 4 z3-z2- 19z+10
by (x — 3), and compare with the synthetic substitution of 3
for x in the same f(x).
4 -1 -19 +10
4 s»-12 x2 12 33 42
Ilz2-19z 4 +11 +14 +52
Ils2-33rc
14 a; -42
52
By § 23 we know that the final sum in the substitution
process (+52) is the value of f(x) when z = 3. But observe
that it is also the remainder resulting from the division by
(x — 3). Further, the other sums in the substitution (4, 11,
14) are precisely the coefficients in the quotient. Hence this
I£, § 235] SOLUTION OF EQUATIONS 331
synthetic substitution could have been used as a quick and
easy method of performing the division by (x — 3) .
The reason the process works is simply this : In the sub-
stitution, we at each step multiply by 3, and add; whereas
in the division, we multiply by —3 and subtract.*
Synthetic Substitution is also called " Synthetic Division,"
since it builds up the result of a division. Observe above
that the leading coefficient (4) was brought down to give the
complete quotient.
§ 235. Integral Roots by Trial. It will now be easy to
find all the integral roots of an equation of any degree what-
ever. We have merely to test a few numbers as roots, and
at the same time factor the given polynomial by Synthetic
Division.
Ex. I. Solve4z3-z2-19z+10 = 0.
Substituting 2 synthetically 4 -1 -19 +10 [2
gives zero as the final sum. 8 14—10
Hence 2 is a root. 4 IpT _ 5
This substitution also shows that the remainder after
dividing out (x — 2) would be zero; and that the quotient
would be 4z2+7z— 5. Hence the original equation, fac-
tored, is
Setting the factor 4 x2+7 x— 5 equal to zero gives two more
roots :
_-7±Vl29
There can be no further roots ; for any value of x that reduces the
original polynomial to zero must make one of the factors zero.
* For a formal proof that a similar process will always work, see the
Appendix, p. 487.
332 MATHEMATICAL ANALYSIS [IX, § 236
In this example we could have told in advance that any
integer, such as 3, which is not a factor of 10, could not be a
root of this equation. For multiplying the next to the last
sum (an integer) by 3 could not furnish the —10 necessary
to produce the final zero.
Likewise, in any other case, the only possibilities for integral
roots will be the divisors of the final term, — providing the
equation has been cleared of fractions.
Ex. II. Factor x4-17 x2-34 z-30.
The only possible integral roots are ±1, ±2, ±3, ±5, ±6, ±10, ±15,
±30. And the test of 5 shows there can be no root above 5. (§ 23.)
Since 5 is a root, we divide out 1 0 -17 -34 -30 [5_
(x-6), and test the quotient. 5 25 40 30
Similarly, we divide out (x +3) and 1 +5 +8 +6
solve the remaining quadratic. —3 —6 —6
1
z2 +2 x +2=0 gives x
Roots: 5, -3,
Factors: (x-5), (z+3), (z-fl-V-1),
Lowest rational factors: (x-5), (z+3), (x2+2 a; +2).
§ 236. Fractional Roots by Trial. Some equations have
fractional roots. It is easy to tell in any such case what
fractions need be tested.
To get the idea let us see, for instance, under what condi-
tions f might be a root of any equation
axn+bxn~1+ ." +fc = 0,
in which the coefficients a, b, •••, k, are all integers.
Substituting synthetically, let S denote the next to the
last sum :
a +b ... ( ) +k If
+fa ( ) +f£
) .-. S+(k+%S)
IX, § 236] SOLUTION OF EQUATIONS 333
For f to be a root, we must have
Now, S cannot be a fraction with the denominator 3 ; for at
no step could this denominator be introduced. Hence 3
must be a divisor of k.
Again, the first multiplication introduces a fraction, —
which will persist and prevent a final zero, — ' unless 2 is a
divisor of a.
Thus f can be a root of an equation only if the numerator
is a divisor of the constant term (k), and the denominator is a
divisor of the leading coefficient (a). Similarly for any other
fraction p/q.
Ex. I. 1526-19z5+723-ll22+4=0.
The only possible fractional roots are those whose numerators are
factors of 4, and whose denominators are factors of 15 :
vnd their negatives. A test shows f .to be a root :
15 -19 +0 +7 -11 +0 +4 L|
10 -6 -4 2 -6 -4
15 -9 -6 +3 -9 -6
The remaining equation is, after canceling a factor 3,
5 25-3 24-2 23+22-3 2-2=0.
The only possibilities now are ±£, ±§, ±1, ±2.
None of these is a root. Hence any further roots must be imaginary
or irrational.
Ex. II. 23+1722+6 2-24 = 0.
The leading coefficient is 1. The only possible "fractional" roots
have a denominator 1, and must be integers.
EXERCISES
1. Find all the roots of the following equations :
(a) 23-13 2 + 12=0, (6) 23-222-2+2 = 0,
(c) 24 +2 2*- 26 22 +27 2+18 = 0, (d) 24 -35 22 -90 2— 56 = 0,
(e) 25-1223-4622-85 2-50=0, (/) z5-20 22-441 2-420 = 0,
334 MATHEMATICAL ANALYSIS [IX, § 237
3x-2 = 0, (h) 6z3 + 13rc2-14z+3=0,
(i) 4:r<+8:r3-9z2-24z-9 = 0, (» 4 z<-ll z2+65 z-30=0.
2. Find all the roots and rational factors of
(a) 2z3-21z2+74z-85 = 0, (6) 3 r'+ll z+14 = 0,
(c) 36a^-x2-2x-l=0, (d) 4 a^-4 z3+z2-4 z-3=0,
(e) 6x<+5a:3+28a;2-6a:-5=0, (/) x6-3 z4+4z3-4z2+3 x-l =0.
3. Find the lowest rational factors of 16 a^+24 z'+8 a;2 +2 x+1.
4. The same as Ex. 3 for 4 z5 +8 z4— 41 z3 + 10 z2+20 z-8.
6. The deflection of a loaded beam x feet from one end was y = k
(2 a^-30 2?+ 1000 x). At what value of x was y a maximum?
6. The same as Ex. 5 if y = k(x*-4Q z3+480 z2-1600 x).
7. The rate of rotation of a flywheel (ft deg./sec.) t sec. after the
power was cut off was R = t3-75 t +250. Find when the wheel stopped.
[8.] Plot y = x3+3x*-3 x-lS, from x = -4 to +5. What root has
the polynomial? What would be the equation of this curve if trans-
lated two units to the left? (Multiply out.) What would the former
root then become?
§ 237. Further Roots. When we have found by trial all
the rational roots of an equation, any further roots must be
imaginary or else real and irrational.
If imaginary, we cannot find them as yet, unless the equa-
tion is in quadratic form or easily factorable into quadratic
forms. But if merely irrational, we can at least approximate
them, — roughly by a graph, and then more closely by suc-
cessive substitutions near the supposed root.
In treatises on the Theory of Equations it is proved that every
polynomial of degree n has precisely n linear factors, real or imaginary,
— and hence n roots. (Some of the factors may be equal; likewise
the roots.)
Also it is proved that, if the given coefficients are real, any imaginary
roots must occur in pairs, like 3 + V— 2 and 3 — V^2, etc.
(B) METHODS OF APPROXIMATE SOLUTION
§ 238. Diminishing a Root. The labor involved in ap-
proximating an irrational root closely can be minimized by a
simple device.
IX, § 239] SOLUTION OF EQUATIONS 335
Suppose, for example, that the unknown root is 2.1768 •••,
and that we have located it between 2 and 3. If we move
the graph 2 units to the left, the root will be reduced to
.1768. (Fig. 114.) We can
easily locate it between .1 and
.2 by testing these values. If
we move the graph .1 more,
the root will be .0768 .... We
can locate it by testing .07 and
.08; and the multiplications
involved will be far simpler than FIG
if we were testing 2.17 and 2.18.
By continuing to move the graph we can make each succes-
sive test by using a multiplier of a single digit rather than
several digits.
To make this device the more effective we shall now find
a very quick method of getting the new equation of the graph
after each successive translation.
§ 239. Translating a Graph Synthetically.
THEOKEM : If the graph of any polynomial y =f(x) is moved
h units to the left, the coefficients in the new equation will be
simply the remainders which would result from dividing f(x)
synthetically by (x — h), the quotient by (x — h), the new quotient
by (x — h), and so on.
o
PROOF : Let the new equation be
y = xn+axn-i>.- +bx*+cx+d.
Now, whatever values the new coefficients a, ••• 6, c, and d may have,
they must be such that, if the graph were moved back h units to the
right, replacing each x by (x—h), we should get back the original
equation y=f(x). Hence
f(x) = (x-h)n+a(x-h)n~1 — +b(x-h)+c(x-h)+d.
336 MATHEMATICAL ANALYSIS [IX, § 239
Clearly, then, dividing the original polynomial /(x) by (x—h) would
give a remainder equal to d, the quotient being
-h)"~2 - +b(x-h)+c.
Again, dividing Q by (x—h) would give c as the remainder, with
h)»-*..- +b
as the next quotient. Dividing this by (x—h) would give 6 as the
remainder. And so on for the other coefficients. (Q.E.D.)
To illustrate the actual working of the process, let us move
the graph of 2/ = z3+3 z2 — 3 x— 18 two units to the left :
Dividing f(x) by (x- 2): 1+3 - 3 -18 j_2
quotient, Q = xz+5x+7 +2 +10 +14
remainder, —4 1+5+7 -4
Dividing Q by (x-2) : 1 +5 + 7 [2_
quotient, Q' = x+7 +2 +14
remainder, 21
Dividing Q' by (z-2) :
quotient, = 1
remainder, 9
These remainders, 9, 21, —4, are the coefficients in the new
equation which results from translating the graph, — viz.,
y = x3+9x*+21 x-4.
Remarks. (I) In practice the cal- 1 +3 —3 —18 12
culation can be condensed and made +2 4-10 +14
rapidly, as shown here. 1 +5 + 7 —4
New function : +2 +14
1 +7 +21
*3+9x2+21x-4. +2
1 +9
(II) Another statement of this process is that the new coefficients
are found by substituting 2 for x in the original / (x} and in the quotients
resulting from successive divisions by (z— 2).
IX, § 240] SOLUTION OF EQUATIONS 337
^ 240. Horner's Method, Complete. Let us now see how
the foregoing process works in solving an equation.
ILLUSTRATION: z3+3 x2 — 3 x— 18 = 0.
I (A) Plot the graph : (B) Move graph 2 units (left) :
1+3 - 3 -18 [2
4 units below at x = 2, +2 +10 +14
27 units above at z = 3. 1 +5 +7 — 4
[ence, a root near 2.1.* +2 +14
After moving the graph, 1 +7 +21
the root will be near . 1 . +2
+9
z3+9 z2+21 z-4.
II (A) Test new f(x) at (5) Move graph .1 unit (left) :
.1, etc., by substitution :
1 +9 +21 -4 |_1
1.809 below at x = .1. + .1 + .91 +2.191
.568 above at z = . 2. 1 +9.1 +21.91 -1.809
Hence, a root near .17. + .1 + .92
After moving the graph, 1 +9.2 +22.83
the root will be near .07. + .1
1 +9.3
III (A) New f(x) =z3+9.3 z2+22.83 x- 1.809. Testing
shows a root between .07 and .08, — near .077.
(B) Moving the graph .07 to the left, the root should
be near .007.
IV (A) New /(z)=z3+9.51z2+24.1467z-.164987.
Testing shows a root between .007 and .006, —
near .0068.
(B) Moving the graph .006 to the left, the root should
be near .0008.
V New /(a;) =z3+9.528 z2+24.260928 z-,019764224.
* This estimate is made roughly by inspection : comparing the values 4 and
27 indicates that the crossing is several times as far from x = 3 as from x = 2.
338 MATHEMATICAL ANALYSIS [IX, § 240
Instead of continuing as above we can now get some
further figures in the root as follows. Since x is now very
small, the terms z3 and 9.528 x2 are practically negligible.
Ignoring them, our equation is approximately :
24.260928 x - .019764224 = 0.
By ordinary division this gives
x =.0008146+.
Recalling the several translations, the original root is
x = 2.1768146+.
Remarks. (I) To test both at x = 2 and at rr = 3 before translating
the graph 2 units was very important. This not only insured us against
moving the curve a wrong amount, but alsb showed about how many
tenths to test in the next stage. For similar reasons, at every stage,
we make tests until the root is definitely located.
Plotting the graph is not essential ; but we should determine whether
it would rise or fall near the root sought.
(II) To approximate negative roots, slide the curve to the right,
To do this, use negative substitutions instead of positive.
Another method is to change the sign of x throughout the given
equation, and then seek positive roots of the new equation.
(III) After n figures of a root have been found by testing, and the
next f(x) has been obtained, approximately n more figures can be
obtained by a simple division, as in the last step above.
(IV) This method of approximating irrational roots was invented
about 1820 by W. G. Horner, an Englishman. It applies only to
equations in the standard polynomial form; but is the best known
method for such equations, and is much used.
EXERCISES
1. Move the curve y = z2— 6 x+7 three units to the left, using the
synthetic process. Check by "translaters."
2. Find to 6 decimals the root of z3-20 z+8 =0 which lies between
4 and 5. (Hint : Get the last three places by division.)
3. In Ex. 2 locate the other roots and approximate each to 4 deci-
mals.
IX, § 241] SOLUTION OF EQUATIONS 339
4. Locate graphically the real roots of xz — 5 z+l=0, and approxi-
mate one of them to 4 decimals.
6. In Ex. 4 approximate the other roots to two decimals.
6. Show graphically that the equation z3+2 z2-23 x— 70 = 0 has
only one real root. Approximate this to six decimals.
7. What are the possibilities as regards the number of real and
imaginary roots for an equation of degree two? three? four? If you
had found one real unrepeated root of a quartic equation could you
draw any conclusion as to the other roots?
8. Approximate to four decimals every real root of x* +x3 +x — 1=0.
9. Solve x3 — 12=0 by Horner's method to four decimals and check
directly.
10. The "index of correlation" between the eye-colors of a certain
group of people and of their great-grandparents is approximately a root,
between 0 and 1, of the equation : .024 o?*+. 137 0?+. 035 x*+x-. 225 = 0.
Find x to two decimals. [C. B. Davenport.]
11. The diameter (d in.) of the bolts needed in certain cylindrical
shafts is a root of the equation cZ4+800 d2 — 18 d— 360 = 0. Find d to
two decimal places.
12. At what point on the curve y — x*— 8 z2+3 z+10 is the slope
equal to +35?
13. Where should an ordinate be erected to the parabola y=xz-\-10
to make the area under the curve between the F-axis and the ordinate
100 square units?
14. The greatest and least distances of Jupiter's Fifth Satellite from
the center of the planet are approximately roots of the equation
z3-5z2-t-6.27396z -.060385=0, the unit of distance being Jupiter's
radius, 45090 miles. Find those roots to 3 decimal places.
15. A magnet placed with its ends in a "magnetic meridian" will
neutralize the earth's magnetism at certain points. To calculate the
position of these points in a certain case, it was necessary to solve the
equation :
20000 x
(z2-100)2
= .2.
Simplify and solve. (There are two values, — one large and one
small.)
§ 241. Newton's Method. Another excellent method of
approximating irrational roots, which can be used even for
equations that involve trigonometric and exponential func-
340 MATHEMATICAL ANALYSIS [IX, § 242
tions, etc., was invented by Sir Isaac Newton. It does not
move the graph, but works throughout with the original
f(x) and the derivative }'(x). An example
will show the idea.
Ex. I. x3+3x2-3rc-18 = 0.
The graph shows a root near 2.2. (Fig. 114, p.
'if is ' 335.) Substituting this value in the given func-
tion and in the derivative 3x*+Qx— 3 gives as
the height and slope at that point :
y = .568, slope = 24.72.
To reach the crossing, we must evidently go
back to the left some horizontal distance Ax.
Assuming the graph practically straight that far, the slope is approxi-
mately .568/Az. *
/. = 24.72, whence Ax = .023, approx.
Ax
Subtracting Ax from 2.2 gives 2.177 as the root.
Repeating the operation with this result as a starting point instead
of 2.2 would give a very fine approximation.
§ 242. Isolating the Roots. We can approximate closely
the irrational roots of an equation, — provided we can first
locate them roughly.
Systematic substitutions will usually show any change
of sign in f(x) . But suppose there were two roots between
2 and 3, so that the graph should cross and recross in this
interval, leaving f(x) positive both at x = 2 and at x = 3. We
might not discover that the graph ought to cross at all.
Such double crossings can usually be detected by calculus.
For between the crossings there will be a maximum or
minimum height, — at which ff(x) = 0. // we can solve this
" derived equation/' /'(#) =0, we can find all the maxima and
minima, and thus discover that the curve has crossed and
recrossed.
IX, § 243] SOLUTION OF EQUATIONS 341
A sure test in all cases, — which was invented about 1830 by
J. C. F. Sturm, a Swiss, — is given in treatises on the Theory of Equa-
tions. The methods suggested above suffice, however, for almost all
practical problems.
EXERCISES
1. Starting from the result of Ex. I, § 241, find the crossing still
more closely by the same method. (Cf. the value in § 240, p. 338.)
2. Solve x7 — 144 = 0 approximately by Newton's method, — in
one step, starting from z = 2. Check by logarithmic calculation.
3. Find by Newton's method the largest root of 2 x3— 15 re +10=0.
(Cf . § 21.)
4. Find graphically the real roots of 3 z3-27 z+3i =0. Make a
sure test whether the minimum y is positive or negative.
§ 243. Summary of Chapter IX. The new methods of
solving equations which have been developed in this chapter
are of two kinds.
(A) Exact solutions. Finding the roots of a quadratic
by formula, and the rational roots of any equation by trial
substitutions and removal of factors.
(B) Approximate solutions. Finding the irrational roots
of an equation by Horner's or Newton's method.
Remark. We have taken up the solution of equations immediately
after Cartesian geometry because of the close relationship between these
two topics. E.g., the idea of translating a curve underlies Horner's
method, and conversely, the formula for the roots of a quadratic is very
useful in analytical geometry in studying tangent lines, etc.
We shall now return to the field of analytic geometry and consider
another system of coordinates which is very useful in many practical
problems.
EXERCISES
1. Approximate by Newton's method the largest root of
s»- 7 3+5 = 0.
2. Check the answer to Ex. 1 by Horner's method.
3. The diameter of a pipe (x in.) which will discharge water at a
342 MATHEMATICAL ANALYSIS [IX, § 243
certain rate under a certain pressure is a root of the equation:
z5 -38 x — 101=0. [M. Merriman.] Find that root to two decimals.
4. Find any rational roots exactly, and approximate any irrational
real roots to two decimals :
(a) 16z*-24:r3+8z2-2z+l=0,
(6) 4z6-8z*-41z3-102;2+20z+8 = 0.
5. Factor into their lowest rational factors :
(a) 8z4+12rJ-10a?-29z-15,
(6) 16 z6-40 ^-104 z3 + 194z2-101z+35,
(c) z8-10z6+16z5-16z3+10z2-l.
6. Solve graphically the simultaneous equations z2+t/ = ll,
7. Solve the equations in Ex. 6 algebraically, finding the irrational
roots to 1 decimal.
8. In how many years would $1000 with 8% interest compounded
quarterly amount to the same as $2000 with 4% interest compounded
semi-annually plus $3000 with 2% interest compounded annually?
9. The rate of interest which a bond will net if purchased at a
certain price is given by 900 x21— 925 x20 -1000 x +1025 = 0, x being
l+r/2. Solve this equation approximately by Newton's method.
10. What is the nature of the roots of 4z2— 7 x— 2 = 0?
11. Solve f or x: e*+10e-* = 7. (Hint: Lete* = y.)
12. For what value of r would the circle xz-\-y2 = r2 be tangent to the
line i/ = 3 x-10? (Cf. p. 330, Ex. 6, 7.)
13. Solve by Newton's method : 2 x— log« x = 5.
14. Find the area under the curve y=l+l/x, from x = l tox = X.
For what value of X is A = 7 ?
15. Where is the slope of the curve y = x*+ e* equal to 10?
CHAPTER X
POLAR COORDINATES AND TRIGONO-
METRIC FUNCTIONS
PERIODIC VARIATION
§ 244. Locating Points. In daily affairs we frequently de-
scribe the location of a point by telling how far it is from
some known point, and in what direction.
For instance, we say that a certain town is "20 miles from here in
a direction 12° north of east." . •
The" same idea is used extensively in Mathematics. A
fixed point 0 is chosen as origin or " pole " ; and a fixed line
OA as a direction axis. Any
point P is then definitely lo-
cated, as soon as we know /
the distance OP and the angle VN
AOP. (Fig. 115.)
OP is called the radius vec-
tor of P ; and is denoted by r.
Z AOP is called the longitude
or vectorial angle of P, and is denoted by 9 (Greek letter
theta).
The two values (r, 6) are called the polar coordinates
of P.
If P moves around 0 continually, 6 will increase up to
360°, — and beyond, if we consider the whole angle turned.
Thus angles of any size whatever may arise in considering
rotary motion.
343
344 MATHEMATICAL ANALYSIS [X, § 245
§ 245. Positive and Negative Angles. 6 is always meas-
ured from the " polar axis " OA to the radius vector OP, —
usually in the counterclockwise direction. When measured
clockwise, 6 is regarded as negative. Thus in Fig. 115,
P' has 0 = 330° or -30°, and thus has the coordinates (160,
330°) or (160, -30°).
Merely adding 360° to 6 will not change the position of a
point. Thus any given point has innumerable sets of co-
ordinates, the 6 values differing by multiples of 360°, but r
having a single positive value, the distance OP.*
To plot a point when given its polar coordinates, simply construct
6 by protractor, and lay off r. Or use "polar plotting paper." (Fig.
127, p. 366.)
Notice the resemblance of this paper to, a circumpolar map of the
earth, with the meridians radiating from the pole and cut by the parallel
circles of latitude. •
§ 246. Path of a Moving Point. If we know the values
of r and 6 for a moving point at various times, we can plot
the corresponding positions, and obtain the approximate
path.
The distance traveled during any interval can be estimated
by rolling a ruler along the path.
EXERCISES
1. The polar coordinates of five snow-peaks visible from Portland,
Ore., are (104, 66°), (53, 62° 30'), (75, 39° 30'), (51, -14°), and
(74, — 53°), r being in miles and 6 being measured from the east (toward
the north, if plus). Plot these points. Calculate the distance between
the last two ; and check by measurement.
2. A golf ball driven from (0, 0°) for a flag at (400, 0°) came to
rest at (210, 10°). How far was it from flag?
3. Halley's comet was nearest the sun in April, 1910. Its position
then and after various intervals (t yrs.) is shown in Table 1. (The sun
* Negative values of r can be given an interpretation as in Ex. 7
below, which is useful in certain kinds of work.
X, § 247]
POLAR COORDINATES
345
t
r
1
0
.59
0
&
.87
70° 30'
£
3.08
130°
1
5.01
142° 38'
3
10.4
156° 43'
8
19.0
166° 8'
18'
28.9
172° 54'
28
33.8
176° 48'
38
35.4
180°
is at the origin ; and the unit distance is the TABLE 1
mean distance from the earth to the sun.) Plot
these positions and draw the path, — a half-
'' ellipse. About what r and 6 has the comet now,
apparently? Estimate the distance traveled in
the ten-year period, £ = 28 to 38; also in the
first year.
4. On Jan. 1, 1913, the coordinates of the
planets were: Merc. (.41, 195° 13'), Venus
1 (.72, 30° 45'), Earth (.98, 100° 30' ), Mars (1.49,
252° 34'), Jup. (5.26, 267° 42'), Sat. (9.07,
62° 6'), Ur. (19.8, 303° 14') Nep. (30, 114° 34').
Plot these positions. Estimate and calculate the distance from Jupiter
to Uranus.
6. A point P moved so that r = 50 always and 0 = 3 & after t sec.
Mark the positions of P at £ = 0, 1, 2, •--, 10. Find the rate of rotation
(in degrees per sec.) at £ = 5.
6. A point moved so that r = 2 t and 0=60 t after t sec. Plot the
path from t = Q to £ = 10.
7. Let negative values of r be laid off from the pole in the reverse
of the direction indicated by the value of 0. Then the point 0 = 17°,
r= -10 is the same as 0 = 197°, r = +10. Plot this and the following
points: (-10,80°), (-5,160°), (-8,240°), (-20,340°). How could
each point be designated with r positive ?
§ 247. Circular Motion. Polar coordinates are especially
suited to the study of circular motion : r remains constant,
and we have only to consider how 6 varies.
The rate at which 9 is increasing, — say
the number of degrees per sec., — is the
rate at which the radius OP is turning.
(Fig. 116.) This is called the angular
speed of P ; and is denoted by the Greek
letter a? (omega).
From co we can find how fast P is moving,
— say in feet per sec., — in other words, the linear speed of P,
Ex. I. A point P moved in a circle of radius 5 ft. so that
346 MATHEMATICAL ANALYSIS [X, § 248
6 being in degrees, and t in seconds. Find the position and
speed of P when Z = 20.
0 = . 1 (203) - .002(204) = 480 at t = 20.
That is, the radius OP had then turned 480° since starting.
=56aU = 20.
.
dt
That is, OP was then turning at the rate of 56 deg./sec.
Hence P was moving at the rate of -//$ of a circumference per
sec., i.e., with the speed
0 = -jftfr(10ir) /i./sec., =4.89 ft./sec.
§ 248. Angular and Tangential Acceleration. The rate at
which the angular speed co is increasing is called the angular
acceleration. It is denoted by the Greek letter a. (alpha).
= dco = ^0
dt dt2 '
The rate at which the linear speed v is changing is called
the linear acceleration in the direction of motion, or the tan-
gential acceleration, at*
Like v, at refers to the actual motion of P along the path ;
whereas a, like a>, refers only to the rotation of the radius
OP, though called the angular acceleration of P.
For instance, in Ex. I of § 247, we should have
.., =2.4 at t =20.
at2
That is, the rotational speed of the line OP is increasing at the rate of
2.4 deg./sec2. Hence the linear speed of the point P is increasing at
the rate of 2.4/360 of a circumference per sec2., or
* This is usually not the total acceleration, as denned in Physics. For
in curved motion there is an acceleration perpendicular to the path, —
which changes the direction of motion, — as well as the acceleration a< along
the path, which changes the speed.
X, § 249] POLAR COORDINATES 347
EXERCISES
1. A point moved in a circle so that 0 = .2 tz (degrees). Find «
and a when t = 5 (seconds) .
2. A point moved in a circle of radius 20 in. so that 0 = 2 t2 — . 05 t3
(degrees). Find the speed and the distance traveled at Z = 20.
3. For 30 seconds a wheel turned so that 0 = .03 t3 — .0005 t4 (degrees) ,
after which the speed remained constant. Find the angular speed and
acceleration when t = 10.
4. A wheel of radius 5 ft. started from rest in such a way that
cP0/c^2 = 24 t— 6 t2, where 6 is in revolutions and t in minutes. Find 0
and u and the distance traveled by a point on the rim when t = 5.
6. A point moves in a circle of radius 10 in. in sucH a way that
0 = 90 P—P (degrees). Find the maximum speed of the point.
[6.] How long an arc is intercepted in any circle by a central angle
of 1°? How large a central angle will intercept an arc equal to the
radius?
, § 249. Radians. An angle which if placed at the center of
a circle would intercept an arc exactly equal to the radius is
called a radian, — written l(r). (Fig. 117.)
' Since the radius is contained precisely
IT times in a semicircumference, there
are TT radians in a central angle of 180° :
irC*) = 180°. (1)
Dividing both sides of this by TT, or
3.1416 approx.,
l(r) = 57° 17'44".6. FlG< 117*
Since a radian is an angle of perfectly definite size, we can
measure any other angle by the number of radians it contains,
just as well as by the number of degrees. This greatly
simplifies the study of circular motion.
E.g., for every radian through which line OP turns (Fig. 116), the
point P travels a distance equal to the radius. If the angular speed is
4 radians per sec., P is moving with a speed of 4 radii per sec.*
* Do not confuse a radian with a radi-us. A radian is not an arc or line
but an angle. (Exactly what angle?)
348 MATHEMATICAL ANALYSIS [X, § 250
§ 250. Tables of Equivalents. By the tables on p. 504
of the Appendix, we can quickly convert any number of
radians into degrees, or vice versa. The following examples
show the reduction of 2.16(r) to degrees ; and of 7° 12' 45" to
radians.
2.(r) 114° 35' 30" 7°= 7X.017453(r) = .12217(r)
.1
.06
2.16
5 43 46 12'=12X.000291 =.00349
3 26 16 45 " = 45 X. 000005 =.00022
123° 45' 32" 7° 12' 45" =.12588(r)
Any simple fraction of 7r<r> or 180° is best transformed without the
tables. E.g.,
7T (r) 7T (r)
5 =30°, t 90° = | '
EXERCISES
1. In a circle of radius 30 in., find the lengths of the arcs intercepted
by central angles of 3«, 1.85W, and .08651 W.
2. A wheel of radius 10 in. is turning with an angular speed of 3
rad./min. What is the speed of a point on its rim?
3. If the same wheel turns so that 6 = .2 P (radians), find the speed
and tangential acceleration of a point on the rim, when i = 5.
4. A wheel of radius 20 in. turned in such a way that, after t sec.,
0 = .0006 t3 — .00001 t* (radians.) Find the maximum speed attained
by a point on the rim.
6. In Ex. 4 find the maximum tangential acceleration attained.
6. A wheel of radius 2 ft. had, after t sec., an angular speed
w = .009 £2 — .0002 P (rad./sec.). Find the angle turned and the dis-
tance traveled by a point on the rim from t = 0 until the maximum speed
was reached.
7. Without tables find the number of degrees in 3<r>, .2<r>, £(f .
8. How many radians in 90°? In 1°? In 17°?
9. Verify without tables the equivalents in § 250.
10. Using tables, find the equivalents of 5W, 2.37<r>, 6.2832W.
11. The same as Ex. 10 for 7T/6 W. Check without tables.
12. If a watch keeps correct time, what is the angular speed of the
second hand in rad./sec.?
13. What is the angular speed of rotation of the earth in rad./sec.?
X, § 252] POLAR COORDINATES 349
14. What is the linear speed of a point on the earth's equator, due
only to the rotation? (Take the earth's radius as 3960 mi.)
15. The same as Ex. 14 for a point in latitude 45° 30'.
[16.] In any circle what is the length of an arc intercepted by a central
angle of 1"?
§ 251. Arc and Central Angle. If a central angle in a
circle contains 0 radians, its intercepted arc equals 0 times
the radius :
s = r0. (2)
This relation is much simpler than if the angle were expressed
in degrees.
A central angle of 1° intercepts an arc equal to ^-^(2 TIT),
or .01745329 r, approx. And for an angle of 0° the arc is
s=?01745329r0. (3)
Whatever the units of measure may be, the relation be-;
tween the arc and the angle is always of this form :
s = ko, (4)
k being the length of arc intercepted by a unit angle, — what-
ever that may be.
§ 252. Estimates Involving Very Small Angles. By Ex.
16 above, a central angle of 0" intercepts an arc whose
length is
s=4.85XlO~6r0. (5)
This formula is useful in making approximations involving
very small angles, — as illustrated in the following example.
Ex. L A comet subtends at the
earth an angle of 2" at a time when
it is known to be one billion miles
away. Find its approximate diam- T=JOS
eter- FIG. 118.
Imagine a circle to be drawn with
a radius of one billion ( = 109) miles, and with its center at the earth.
(Fig. 118.) The part of its circumference intercepted by a central
850 MATHEMATICAL ANALYSIS [X, § 253
angle of 2" is relatively so short an arc as to be practically straight.
That is, the arc would approximate the required diameter of the comet.
Substituting 0 = 2 and r = 109 in (5) gives
s = (4.85 X 10-«) X 109 X 2 = 9700.
The diameter of the comet is approximately 9700 miles.
N.B. The value of s in (5) will not approximate closely the distance
between two points, unless the line joining them is practically perpen-
dicular to the bisector of the angle.
EXERCISES
1. Find the diameter of a sun-spot which, if viewed perpendicularly,
would subtend at the earth (92,500,000 mi. away) an angle of 40".
2. Estimate the diameter of a crater on the moon, at the center of
the disc, if it subtends at the earth (240,000, mi. away) an angle of 100".
3. How far away is Jupiter when its diameter (90,000 mi.) subtends
at the earth an angle of 40"?
4. The distance of the earth from the sun subtends at the nearest
fixed star an angle of 0".6, approx. Find the distance to the star.
5. How large an angle would the diameter of a dollar (1.5 in.)
subtend at a distance of 8 miles? (Compare the angle in Ex. 4.)
6. Two ships crossing the same meridian have a difference of latitude
of 16' 40". How far apart are they? (Take the earth's radius as
3960 mi.)
7. A mountain 10 mi. away has an elevation angle of .2<r>. About
how high is its summit above the observer?
8. A mountain rising 7500 ft. above an observer has an elevation
angle of .15<r>. How far away is it, roughly?
9. Find the angle which the earth's diameter would subtend if seen
from the moon.
10. The same as Ex. 9, if seen from Mars when 80,000,000 mi. away.
11. Verify the values of k in (3) and (5) above for angles measured
in degrees or seconds. Also find k for angles measured in minutes.
12. A flywheel of radius 5 ft. is turning at the rate of 8 rad./min.
How fast is a point on the rim moving?
TRIGONOMETRIC FUNCTIONS
§ 253. General Definitions. The functions sine, cosine,
tangent, etc., were defined in Chapter V for any acute angle,
X, § 253]
POLAR COORDINATES
351
— as certain ratios of the sides of a right triangle containing
the angle.
Evidently these definitions would be meaningless in the
case of very large angles : we could not even get the angle
into a right triangle, — much less speak of
" the opposite leg," etc. The definitions,
however, can be restated now in such a
way as to make them applicable to angles
of any size, — and yet leave their meaning
unaltered for acute angles.
In the case of an acute angle (Fig. 119),
the " adjacent leg " and " opposite leg "
are simply x and y, — the rectangular co-
ordinates of the point P referred to the
axes shown. Hence the former definitions might now be
stated as follows :
Let the initial side of any angle 0 be taken as the X-axis, and
the vertex as the origin; and let P(x, y) be any point on the
terminal side of the angle. Then
FIG. 119.
. A y ordmate of P
sin u = - = — -— ,
r radius vector of P
Q _ x _ abscissa of P
r radius vector of P'
*«« n V ordinate of P
Tan w == == *
x abscissa of P
(6)
FIG. 120.
I These statements have a mean-
ing even for very large angles,
such as 6 and 0' in Fig. 120.
For no matter how large an angle
IV is, any point P on its terminal
line has some definite coordinates
x, y, and r.
352
MATHEMATICAL ANALYSIS [X, § 254
The "quadrants" into which the X and Y axes divide the piano are
numbered for convenience, starting from the positive Jf-axis and going
counterclockwise. Angles between 90° and 180° are called ' ' angles
of the second quadrant." And so on.
Each function denned above is positive in certain quadrants and
negative in others. No rules are necessary as to this if you memorize
thoroughly the definitions (6). Simply picture to yourself the position
of P for any given angle 6, and notice whether x and y are positive or
negative : r is always positive.
§ 254. Reciprocal Functions. The reciprocals of the three
functions above, in reverse order, are called the cotangent,
secant, and cosecant, written ctn, sec, esc :
cos0=-,
tan 0 = %
CSC 0 = -,
y
sec 6 = ?,]
x
ctn0=-«
y
(7)
To remember easily this pairing of reciprocals, practice
reading the order down the first column and up the other.
Also observe that a co-function is
never the reciprocal of a co-function.
§255. Finding All from One. Given
any one function of an angle, you can
find all the others, — without tables.
Simply draw the angle and read off
the desired values. Except for certain
extreme cases, there are two possible
sets of values, due to the fact that
there are two angles in different quad-
rants, for each of which the given
function has the specified value.
Ex. I. cos 0 = —3/5. Find the other functions.
Here x/r = — 3/5 for every point on the terminal line. Taking r = 5
Fio. 121.
X, § 255] POLAR COORDINATES 353
requires x= — 3. This is satisfied both at P and at P', in Fig. 121.
Thus 0 may be in either the second or third quadrant.
Quadrant II Quadrant III
sin 0 = 4/5, esc 0 = 5/4, sin B = -4/5, esc 0 = - 5/4,
cos 0 = - 3/5, sec 0 = - 5/3, cos 6 = - 3/5, sec 6 = - 5/3,
tan 0= -4/3, ctn 0= -3/4. tan 0=4/3, ctn 0 = 3/4.
Ex. II. tan 0 = 3/2, but 0 >90° and <360°. Find sin 0, cos 0.
Here y/x = 3/2 for every point on the terminal line. Since 0 is not
in Quadrant I, both x and y must be negative. Take x=—29
=-3. Then r = V(^2)2-(-3)2
Q
sin 0 = -- ~. cos 0 =
Vl3*
EXERCISES
1. Draw angles of 160°, 200°, and 340° ; and by measuring lines cal-
culate the approximate values of the six functions for each. List
separately, noting + and — signs carefully.
2. Prove that for any obtuse angle (i.e., between 90° and 180°) :
sme of obtuse angle = sine of supplementary acute angle;
cosine of obtuse angle = —cosine of supplementary acute angle.
(Compare § 121.) How about the tangent of an obtuse angle?
3. Draw and measure the following angles :
(a) between 90° and 270°, having ctn 0= —1.50,
(6) between 90° and 270°, having sin 0= -.90,
(c) between 180° and 360°, having sec 0 = 1.25.
4. Given each of the following functions, draw each possible angle
< 360°, and write by inspection the values of the five remaining func-
tions, — listed separately :
(a) cos 0=-5/13, (&) esc 0=-5/4,
(c) tan 0= -3/4, (d) sin 0 = 3/4,
(e) sec 0 = V5/2, (f) ctn 0=4.
6. Find approximate values of the six functions of —80°. How
do the sine and cosine of —80° compare with the same functions of
+80°, numerically and as to sign? Is the same thing true for —130°
and +130°? For any -0 and 0?
354 MATHEMATICAL ANALYSIS [X, § 256
^§:256. Special Angles. The functions of any angle can
be found approximately by measurement. For certain
special angles they can also be calculated exactly by ele-
mentary geometry, e.g., for angles which differ from 180°
or 360° by 30°, 45°, or 60°.
Ex. I. Functions of 300°.
Taking r=10 gives z = 5, y=— 5\/3; for the right tri-
angle formed by x, y, and r is half of an equilateral triangle.
(Draw the figure.)
10 2 '
Ex. II. Functions of 225°.
Taking r=10 gives x = y= - V50 = -$V2. (Draw the
figure.)
cos 225° =
§ 257. Quadrantal Angles. The functions of 0°, 90°, 180°,
etc., can be read off directly from a figure. This should be
done often, until their values are fixed in mind.
Ex. I. Functions of 270°, or f TT radians.
Taking r = 10 gives x = 0, y = — 10. Hence
sin 270° = y/r = - 1, esc 270° = r/y=-l,
cos 270° = x/r = 0, sec 270° = -,
tan 270° = -, ctn 270° = x/y = 0.
The tangent and secant do not exist for 270° ; for it is im-
possible to divide by x when x = 0. They exist, however, for
angles as near 270° as we please. (See § 258.)
X, § 258] POLAR COORDINATES 355
EXERCISES
1. Without tables find the sine and cosine of 90° and 180°, noting
the sign in each case.
2. By measurement or inspection find the sine and cosine of 0°,
30°, 60°, 90°, and other such angles in other quadrants. Plot a graph
of sin 0 and cos 0 from 0 = 0° to 360°.
3. Find the exact values of all the functions of 90° and 360° which
exist.
4. Find the exact values of the functions of :
(a) 225°, (6) 330°, (c) 315°, (d) 150°.
6. What can you say about the value of tan 0 when 04s just a little
less than 90°? A little more than 90° ?
6. What are all the angles less than 360° for which sin 0 = 0? Cos 0
= -1? Tan0=-l?
7. The same as Ex. 6, if ctn 0 = 0; sec 0 = 1 ; esc 0= —1.
8. Express in radians all angles less than 2 *-W for which cos 0=0;
ctn 0 = 1; esc 0 = 1.
§ 258. Graphs. Let us now see how the functions vary
as the angle 6 increases from 0° to 360° and beyond.
If we keep r fixed, we need
only consider what happens to
x and y. With r fixed, the point
P (Fig. 119, p. 351) moves in a
circle. Hence y starts from zero, I \ /
increases to +r, decreases to — r FIG. 122.
and -increases again. Thus sin 9( = y/r) takes the values:
8
0
90
180
270
360
sin 6
0
1
0
i
0
The graph is the wavy curve in Fig. 122, reaching a maxi-
mum height of 1 unit at 90° and a minimum of — 1 at 270°.
(Note the radian equivalents.)
Similarly the graph of cos B( = x/r) is seen to run as in
Fig. 123.
356 MATHEMATICAL ANALYSIS [X, § 259
The graph of tan 6( = y/x) is less simple. As 6 approaches
90°, y becomes nearly equal to r. Dividing by x, which is
almost zero, makes tan 6 exceed-
, ingly large. As soon as 6 passes
Li
i sn° I 7
rSoosTrco ' 90°, there is a startling change :
FTQ 123 tan 6 jumps to an exceedingly large
negative value, — for x is now nega-
tive. The graph has, therefore, a tremendous break. It
approaches asymptotically the vertical line drawn at 6 = 90°.
Similarly at 0 = 270°.* (Fig. 124.)
The graphs of the cotangent,
secant, and cosecant can be drawn _
by inspection of the foregoing /
graphs. That of esc 6 is shown • /
dotted in Fig. 122.
The sine, cosine, and tangent
curves are extremely important. They should be thor-
oughly fixed in mind, — together with the radian equivalents
of the angles.
§ 259. Some Important Observations.
(I) Limitations on size. Since sin 0 and cos 9 are restricted to values
numerically less than 1, their reciprocals esc 0 and sec 0 are always
numerically greater than 1. There is, however, no limitation upon
tan 0 and ctn 0. These may have any value whatever, positive or
negative.
(II) Periodicity. Adding 360° to 0 leaves the values of x, y, r
unchanged, and hence also the values of the functions. Thus all the
graphs repeat themselves every 360° or 2 u-W . In fact, the graphs of
tan 0 and ctn 0 repeat every 180° or *-<r>. For changing 0 by 180° affects
x and y only by changing their signs, and does not affect y/x.
* It is customary to say that the tangent of 90° is infinite, written
tan 90° = cc. But this is intended merely as a short way of stating that
while tan 90° does not exist, tan 0 increases without limit as 0-»-900.
Similarly for 270°. (Cf. Appendix, p. 493.)
X, § 260] POLAR COORDINATES 357
(III) Relation of the sine and cosine curves. Rotating the line OP
(Fig. 119, p. 351) through 90° would replace x by y and y by — x.
(§ 209.) That is,
x for Z0 = y for Z (0+90°)
x/r for Z d = y/r for Z (0+90°)
/. cos0 = sin (0+90°). (8)
Hence the graph of cos 6 is the same as that of sin (0+90°), — i.e., the
same as the graph of sin 0, but moved 90° to the left. •
§ 260. Reducing to Acute Angles. Trigonometric tables
run only to 90° ; but they can be used to find the functions
of any angle whatever.
For instance, as is clear from Fig. 125, cos 160° is nu-
merically equal to cos 20°, since 160° is just as far from 180°
as 20° is from 0°. Similarly for
cos 200°, - and for cos 340°, with 'urve
°
respect to 360°. Hence cos 160
cos 200°, and cos 340° can all be >
found by looking up cos 20°, and
f, . , . FIG. 125.
prefixing the proper sign, -f- or — .
In general, to find any function of any large angle,* we
have simply to take the difference between the given angle and
180° or 360°, whichever gives an acute angle, look up the re-
quired function and prefix the + or — sign, according to the
quadrant. No rule is necessary as to signs. Simply visual-
ize the angle and note whether x and y are positive or nega-
tive.
This method is easily seen to be correct by examining the
graphs of sin 6, cos 0, tan 6, etc. Or it can be proved in
detail by geometry, using the definitions in §§ 253-4.
Ex. I. Find sin 190°.
This angle differs from 180° by 10°. By tables,
sin 10° = .17365.
* Of course, an angle larger than 360° is first reduced by a multiple of
360°, — or a negative angle is similarly raised, — until between 0° and 36QP.
358 MATHEMATICAL ANALYSIS [X, § 261
But for 190°, y/r is negative. Hence,
sin 190° =-.17365.
Ex. II. Find ctn 275°.
This angle differs from 360° by 85°. By tables,
ctn 85° = .08749.
But for 275°, x/y is negative. Hence,
ctn 275° =-.08749.
§ 261. The Reverse Operation. To find an angle of any
size when given one of its functions, we have simply to look
up the acute angle which has the same function numerically ;
and combine this acute angle with 180° or 360°, according to
the given sign.
Ex. I. sin 6=-. 17365: find 6.
By tables: .17365 = sin 10°.
Now sin 0 is negative when y is. Hence we must combine 10° with
180° or 360° in such a way as to get an angle 6 in Quadrant III or IV :
.'. 0 = 180° +10° = 190°, or 0 = 360° -10° =350°.
Ex. II. tan 0= -5.6713. ( = y/x)
By tables : 5.7613 = tan 80°.
Since tan 0 is negative when x and y have opposite signs, we combine
80° with 180° or 360° so as to get into Quadrants II and IV :
0 = 180° -80° = 100°, or 0 = 360° - 80° = 280°.
Ex. III. sec 0= -1.30541. (=r/x)
The reciprocal is cos 0 = —.76604 ; and 0 must be in Quadrant II or III,
where x is negative.
By tables :' .76604 = cos 40°.
/. 0 = 180° -40° = 140°, or 0 = 180° +40° = 220°.
EXERCISES
1. Find from tables the sine, cosine, and tangent of 216° ; 304° ; 92°.
2. The same as Ex. 1 for 158° 12', and for 260° 15'.4.
3. How would you look up the cotangent and the secant of 340°?
4. Find both values of 0<360° for which sin 0= -.38725.
X, § 262] POLAR COORDINATES 359
6. The same as Ex. 4 if cos 0 = . 25601 ; also if tan 6= —3.4874.
6. Find all the angles < 360° for which (a) sin A = .88712;
(6) cos B=-. 42893; (c) ctn C= 2.8375.
7. Find the sine, cosine, and tangent of 1.5 W. (See table, p. 504.)
8. The same as Ex. 7 for 4.137<r>. (Change to degrees.)
9. Find hi radians both angles < 2 7r<r> for which (a) sin A =
-.67880; (6) cos B = . 71995; (c) tan C= -.06301.
10. (a) By inspection of the graph of cos 6, draw a. rough graph for
sec 0. (6) The same for sin 6 and esc 0. (c) The same for tan 6 and
ctn 6.
11. According to the graphs in Figs. 122-4 what are the angles
<2 TJ-M, for which sin 6=0? cos 0=0? tan 0 = 1?
12. How often does the sine curve repeat? The tangent curve?
§ 262. Oscillating Physical Quantities. Many physical
quantities vary periodically, in much the same way as the sine
of an angle.
E.g., an alternating electric current rises to a maximum
intensity in one direction, sinks to zero and on down to a
minimum (i.e., a maximum in the opposite or negative
direction), rises again, etc. The varying intensity is repre-
sented by some such formula as
i = 10 sin (2000, (9)
where t is the number of seconds elapsed, and 200 t is the
number of radians in the " phase angle." *
The graph, on some scale, is the sine curve. (Fig. 122,
p. 355.) The greatest value of a sine being 1, the maximum
i in (9) is 10 units, represented by the greatest height of the
sine curve.
Of course, the oscillations occur very rapidly. Thus in
(9) i completes an oscillation when the angle 200 t reaches
the value 2 TT :
* That the formula involves a trigonometric function is not strange
— inasmuch as the current is generated by circular revolutions of an "arma-
ture."
360
MATHEMATICAL ANALYSIS [X, § 263
That is, a complete oscillation takes about -£$ sec., or the
current alternates about 60 times a second. The base of each
arch of the sine curve here represents .0157 sec. of time.
The rate at which i is increasing at any instant can be found approx-
imately from the graph. To find it exactly we must be able to differen-
tiate a sine function. (§ 263.)
EXERCISES
The angles here are in radians.
1. At what values of t is the graph of ?/ = 30 sin 100 t just starting
up from j/ = 0? Down from y = Q? Draw the graph roughly by
inspection.
2. The same as Ex. 1 for each of the following :
(a) t/ = .25 sin 4 t, (6) y = 20 cos 10 /, (c) y = 5 cos rt.
3. As a tuning fork vibrated, its displacement (x cm.) from the
position of rest varied thus with the time (t sec.) : x = .06 sin 800 t.
Draw the graph by inspection, showing
the maximum displacement and the time
elapsed during a vibration.
4. The same as Ex. 3 for an oscillating
mechanism whose displacement (y in.)
varies thus : y = 10 cos 20 t.
6. The same as Ex. 3 for a pendulum
whose angular displacement varies thus:
0 = .l sin 2 irt.
FIG. 126.
§ 263. Derivatives of sin 0 and
cos 0 : Radian Measure. To dif-
ferentiate sin 6 by the original pro-
cess (§53), we first let 6 increase by A0 and see how much
the sine increases.
By definition, v
sin 0=",
r
in which y and r may be taken for any point on the terminal
line of 0. But by choosing that point P whose r=l, we
have simply „.„ a_y> (1Q)
sin 9
X, § 263] POLAR COORDINATES 361
and the change in sin 6 due to any change in 6 is simply
Ay.
But when we increase 9, keeping r=l, P must travel in a
circle, and Ay is easily calculated from the small right tri-
angle in Fig. 126. Z Q = 6+% A0, these angles having sides
which are mutually perpendicular.
.'. Ay = chord PQ - cos Q = chord PQ • cos (6+% A0).
If A0 is in radians, arc PQ' = rA0 (§ 251), =A0 simply.
Now as A0->-0, the ratio of the chord to the arc ap-
proaches 1.
/. ^ = cos 0, or 4(sin e) =c°s e- (12)
ad au
I.e., the derivative of sin 0 is cos 0, if the angle is in radians.
In like manner, noting that x = cos 0 in Fig. 126, and that
the change in cos 0 due to A 0 is simply Ax, a negative quantity,
we find
A arc
.-. ^= -sin 0, or 4 (cos 0) = -sin 0- (13)
a0 c9
Physically, formulas (12) and (13) mean that the rates
at which sin 0 and cos 0 change, per radian increase in 0, are
constantly equal to cos 0 and —sin 0, respectively.
Ex. I. Find the rate of change of sin 6, per radian, at 0 = .5<r>.
Answer : — (sin 0) = cos .5 = .87758.
ad
Check. By tables, sin .49*= .47063, sin .51 = .48818. Thus sin 9 in-
creases by .01755 while 6 increases by .02<r>, — or at an average rate of
.8775 per radian for this interval.
362 MATHEMATICAL ANALYSIS [X, § 264
§ 264. Modification for Degree Measure. If 6 is in
degrees instead of radians, then in Fig. 126, arc PQ =
.017453 A0. (§ 251.) Hence dividing Ay by A0 ( = arc
PQ/.017453) gives instead of (11) :
du
.017453
. 017453 cos 6. (15)
Likewise ^ = - .017453 sin 6. (16)
ad
That is, the rate of change of sin 6 and cos 0, per degree change in 9,
is only a small fraction (.017453 . . .) of the1 rate of change per radian,
— which is evidently reasonable.
The great simplicity of (12) and (13) as compared with (15) and (16)
is the reason for using radian measure in practically all problems requir-
ing the differentiation of a sine or cosine.
§ 265. Sin u and cos u. If u is any function of 6, and
y ~= sin u
then, by (21), p. 110,
dy = dy du n_
dd du ' dO'
*-«-«• g- (18)
In words : the derivative of the sine of any angle equals
the cosine of that same angle times the derivative of the angle,
— expressed in radians.
Similarly for the derivative of a cosine.
Ex. I. Differentiate i = cos (100J) •
Answer : ^= -sin (100 <)J- (100 0 = -100 sin (100 0. (10)
X, § 266] POLAR COORDINATES 363
EXERCISES
1. Find the instantaneous rate of increase of sin 0, per radian, at
0 = .5<r>. Check by finding from tables the average rate from 0 = .59<r>
to .61 W.
2. The same as Ex. 1 for cos 0.
3. Find the instantaneous rate of increase of sin 0, per degree, at
0 = 60°. Check by comparing the actual increase from 59° 30' to 60° 30'.
4. The same as Ex. 3 for cos 0.
5. Differentiate the following, the angles being in radians :
(a) y = 75 sin 10 t, (6) r = .6 cos 8 t,
(c) z = .12 sin (2 £+5), (d) z = 20 cos (f-.3),
(e) s = 40 sin tt\ (f) Z = .07 cos tt3)^
6. Differentiate the following, the angles being in degrees :
(a) 2/ = 100 sin (3 $+20), (6) z = 3 cos (.02 t-S).
7. The centrifugal acceleration of points on the earth's surface,
due to the rotation, varies with the latitude thus: A = .11 05 cosL,
approximately. Find the rate at which A changes with L, per degree,
atL = 30°.
8. The intensity of a certain alternating current varies thus:
i = 200 sin 500 t. (t is the number of seconds elapsed, and 500 Z the
number of radians in "the phase-angle.") Find i and di/dt when
< = .002.
9. In Ex. 8 plot i from <=0 to .01, and check your results graphi-
cally. (In plotting take t = Q, .002, .004, •••, besides noting maxima,
etc.)
10. In Ex. 3, p. 360, find the displacement and speed of the fork at
< = .002.
11. The same as Ex. 10 for the mechanism in Ex. 4, p. 360.
12. The same for the pendulum in Ex. 5, p. 360.
§ 266. Notation for an Angle. The symbol sin-1 is
commonly used to denote an angle whose sine is . . . (what-
ever number follows) .
Thus sin"1. 5 denotes an angle whose sine is .5. This angle
might be 30°, or 150°, or 390°, etc.
Likewise 0 = tan"1 2.88 means simply that 0 is an angle
whose tangent is 2.88, or tan 0 = 2.88.
364 MATHEMATICAL ANALYSIS [X, § 2G7
Observe that the —1 is not an exponent but simply a part of the
new symbol for an angle. It does, however, have a significance some-
what analogous to that of a negative exponent, in this way: Looking
up 0 = sin~1 .5 is the reverse of the operation of looking up sin 0; just
as multiplying by 10"1 is the reverse of multiplying by 101.
§ 267. Notation for Powers. To indicate a power of a
trigonometric function, it is customary to apply the exponent
directly to the function, rather than to write it after the angle.
Thus
sin2 6 means (sin 0)2,
sec3 6 means (sec 0)3,
etc., while sin 02 would mean the sine of an angle whose num-
ber of units is the square of the number in 6.
An exception occurs, however, in the case of the — 1
power. We cannot write (sin u) ~l in the abbreviated form
sin"1 u, having adopted the latter symbol to denote an angle
whose sine is u*
Ex. I. Differentiate ?/=sin100.
Here y is not primarily a trigonometric function but rather a power,
viz., y = (sin 0)10. This is of the form y = un.
.'. ^ = 10(sin 0)9 cos 0 = 10 sin9 0 cos 0.
U0
[Where does the factor cos 0 come from ? If in doubt, think how you
would differentiate the form y = (x8+7)10. Note the 3 z2. (§ 77.)]
EXERCISES
1. Look up, in radians and in degrees, the smallest positive angles
for which :
(a) sin-1 .86742, (6) cos-1 (-.02920), (c) tan-^-l).
2. The area of a segment cut from a circle of radius r ft. by a chord
x ft. from the center is A =rz cos-1 (x/r} —x\/r*-x* sq. ft., the angle
being in radians. Calculate this if r = 10 and z = 6.
3. Calculate sin3 1.2(r> and cos~2 35°.
* To avoid this confusion the symbol arcsin is often used for an angle in
I>l:u- • of sin"1.
X, § 268] POLAR COORDINATES 365
4. Differentiate y = 3 sin4 0, and x =7 cos3 (5 f), the angles being
in radians.
§ 268. Curves in Polar Coordinates. As a point (r, 6)
moves along any curve, r varies with 0 in some definite way.
That is, r=/(0). Conversely, all the points whose polar
coordinates satisfy a given equation lie along some definite
curve.
Ex. I. Plot the curve r = 10 sin 0. (20) e
Substituting values for 0 gives the adjacent table. o
Plotting the positive values of r, we get the 30
60
curve in Fig. 127. (Plotting the negative values Q0
of r, according to the system mentioned in Ex. 7, 120
p. 345, would merely retrace the same curve.)
ISO
0
5
8.66
10
8.66
5
0
-5
This curve is a true circle. For sin 6=y/r, 2io
which gives in (20) :
r2=10i/,
i.e., £2+2/2=10 y.
EXERCISES
1. A point moves so that r = .05 0 always. Make a table of values
of r for values of 0 at intervals of 60°, and draw the path from 0=0 to
360°.
2. What sort of a curve is r = 0/3? r = k0t (If in doubt make a
table.)
3. The same as Ex. 1 for r0 = 60; but try also small values of 0,
say 1°. .1°.
4. Plot the curves r = cos 0 and r = sin 2 0, from 0 = 0 to 901
(a) r=sin2 0, (6) r = cos2 0,
(c) r = 10 sin 3 0, (d) r = .5 cos 3 0.
5. Draw roughly by inspection r = 2 0, from 0 = 0(r) to 0=2 7r(r).
6. The same as Ex. 5 for r = e™.
7. In Ex. 6 find the rate of change of r, at 0 = .8<r) .
8. For each curve in Ex. 4 find the rate of change of r, per degree,
at 0=20°.
9. The equation of the path of Halley's Comet is r = 1.158/(l +
,9673 cos 0) . Calculate r when 0 = 0° and when 0 = 180°. (Cf . Table 1,
p. 345.)
366
MATHEMATICAL ANALYSIS [X, § 269
180s-
i'JO*
350'
§ 269. Summary of Chapter X. Polar coordinates are
useful in locating points, studying motion, and defining the
trigonometric functions of angles in general. In higher
courses they are used in study-
ing curves analytically.
The simplest unit angle in
differentiations, and in motion
problems generally, is the ra-
dian. A circular arc, equals
its radius times the number of
radians in its central angle.
An angle may have any size
whatever, positive or negative.
Very small angles are best
expressed in seconds, especially
in making estimates which regard an arc as equal to its
chord, etc.
The graphs of the trigonometric functions are important
in many scientific problems, as well as in showing how to
look up any given function.
When one function of an angle is known, all the others
can be calculated exactly, without tables. This shows in-
cidentally that there must be some definite relations among
the several functions. We turn now to the consideration of
such relations and their uses.
EXERCISES
1. With minor fluctuations, the pressure of steam in a boiler varied
one afternoon as shown in Table I :
(I)
TIME
12
1
2
3
4
5
6
P (LBS.)
96
105
116
77
95
120
10
Plot the polar graph, using 15° for 1 hour. (In the automatic recording
charts the radial lines curve slightly backward, but the graph appears
much as here.)
X, § 2691 POLAR COORDINATES 367
2. Table II shows several positions of the planet Mercury during
one of its revolutions about the sun. Plot the path.
(II)
r
.31
.36
.44
.47
.43
.35
.31
e
73°
158°
215°
255°
308°
360°
73°
3. Find by measurement and by trigonometry the rectangular co-
ordinates of the points whose polar coordinates are (10, 70°), (10, 155°),
and (20, 200°). f
4. Vice versa find the polar coordinates of the points whose rec-
tangular coordinates are (5, 12), (-8, 15), (20, -15).
5. Write formulas for the rectangular coordinates (x, y) of any
point whose polar coordinates are (r, 6) ; and vice versa.
6. When an object travels in a circle of radius r ft. with an angular
speed of co rad./sec., its centrifugal acceleration is A = co2r (ft. /sec2.).
Find A for points on the earth's equator, taking r = 3960 (mi.). Also
show that for points in any latitude L : A = .1105 cos L, approx. [Cf.
Ex. 7, p. 363.1
7. What are the values, in degrees < 360, and in radians < 27r, of
(a) sin^O, (6) cos"1 0, (c) tan'1 (-1), (d) ctn^O?
8. Given ctn A = —4/3, find without tables the other five functions
for both possible angles <360°.
9. Find the diameter of a sun spot in the center of the disc if it
subtends at the earth, when 91,000,000 mi. away, an angle of 8".
10. A point moved in a circle of radius 5 in. so that after t min.
0 = 60Z2— I3 (radians). Find its maximum speed; also its tangential
acceleration when £ = 10.
[11.] Divide both members of the equation xz+y*=r* by r2, and
express the resulting equation in terms of trigonometric functions.
The same, dividing by #2 ; by yz.
12. Find the slope and flexion of the sine curve ?/ = sm 6 at 0=.4W.
13. Show without plotting that the curves r = We^ andr = 51og0
are spirals of some kind. In each find how fast r increases, per radian,
at 6 = 2.
14. Solve Kepler's equation 6— e sin0=M f or 6 when e = .3 and
M=.75. (See §241.)
CHAPTER XI
TRIGONOMETRIC ANALYSIS
FUNDAMENTAL RELATIONS AMONG THE
FUNCTIONS
§ 270. The Basic Identities. The coordinates x, y, and r,
used in defining the trigonometric functions (§ 253), always
have this relation :
r\ (1)
Dividing by r2 gives (z/r)2+ (y/r)2 = 1. That is,
(cos0)2+(sin0)2=l,
or rearranging and using the notation of § 267 :
sin26+cos26 = l. (2)
Equation (2) is true for every angle large or small : for
300° or for .02 ". For this reason sin2 0+cos2 6 is said to be
identically equal to 1, and equation (2) is called an Identity.
Other identities result from dividing (1) by x2 or y2 ; viz.,
l+tan26 = sec2e, (3)
l+ctn26 = csc26. (4)
Still others come from tan 6 = y/x and ctn 0 = x/y, by divicU
ing numerator and denominator by r :
=
x/r cos 6
y/r sin 6
368
XI, § 271] TRIGONOMETRIC ANALYSIS 369
These identities (2) -(6) and the reciprocal relations
esc 6 = ^—, (7)
-— , --, —,
tan 6 cos 0 sin 0
will be used frequently, and should be memorized very care-
fully.
Doing this thoroughly now will save much time and trouble. Prac-
tice writing the list (2)-(7) from memory. Then, if you get any wrong,
study those especially. Notice the similarity of (3) and (4), and of
(5) and (6) ; also that (3) involves no co-functions, and (4) only co-
functions.
§ 271. Some Applications. The foregoing identities are
useful in solving equations, and in simplifying complicated
expressions before differentiating or integrating.
Ex. I. The angle at which a certain gun should be ele-
vated to shoot a certain distance is to be found from the
equation
sin 0cos 0 = .25. (8)
This involves two unknown quantities, sin 0 and cos 0.
But always
sin20+cos20=l: (9)
and, by combining these two equations, we can solve for
sin 0 or cos 0. The simplest way is to add twice (8) to (9),
to get a perfect square, viz., (sin 0-fcos 0)2 = 1.5, — or
similarly to subtract twice (8). Thus
sin 0+cos 0 = VT5 = 1.2247, (10)
sin 0-cos 0= ± Vj5~= ±.7071.
Adding these gives sin 0 = .9659 or .2588. Subtracting gives
cos 0 = .2588 or .9659. There are two angles, 0 = 75° or 15°.
How is this possible physically? Also why could we not take sin 0
-f-cos0=-Vr5in (10)?
370 MATHEMATICAL ANALYSIS [XI, § 272
Ex. II. Simplify and differentiate y = (esc 6 — sin 0)/ctn 6.
By (6), (7) :
sin 6
_sm 6 l-sm2
cos 6 ' cos 6
sin 6
By (2) this reduces to t/ = cos2 0/cos 0 = cos 0.
/ *=-sin0.
dB
N.B. It is usually best to express all the given functions
in terms of the sine and cosine, as this can always be done
without radicals. But it may be better to put ctn 6 = I/tan 0,
if only these two functions appear ; or to use (3) when only
even powers of tan 0 and sec 0 appear ; etc.
§ 272. Trigonometric Equations. In solving a trigono-
metric equation there are usually three steps : (a) Expressing
all the given functions in terms of a single function ; (6) solv-
ing algebraically for the value of that function ; and (c) find-
ing all possible angles.
Ex.I. Solve 5 sin 0-10 cos 0+11 =0. (11)
Replacing cos 9 by =±= Vl — sin2 0, transposing and squaring gives:
125 sin2 0+110 sin 0+21 =0.
This is a quadratic equation for sin 0. By formula (2), p. 326,
. Q -110±Vll02-4(125)(21) 7 3
-250- ~25' ~5
0 = sm-1(-1&),orsin-1(-|)
Substituting these values of sin 0 in (11) gives cos 0 = |f or f.
Since 0 has a negative sine and positive cosine, it lies in the fourth
quadrant ; and is found by subtracting from 360° (or 2 7r<r>) the acute
angle whose sine is ^ or f .
Ex. II. For a projectile fired with a speed of 4000 ft. /sec., and at
any inclination 0, the path is
2
y=x tan ^~
XI, § 272] TRIGONOMETRIC ANALYSIS 371
For what 0 will this strike a balloon at (20000, 8000) ? We must have
2/ = 8000 when x = 20000.
.'. 8000 = 20000 tan 0-^-.
COS20
The easiest way to solve this equation is to observe that
«* ,-l+tan.*
.*. 8000 = 20000 tan 0-400 (1+tan2 0)
or tan2 0-50 tan 0+21 =0,
/. 0=88° 51', or 22° 47'.
EXERCISES
1. Look up the sine, cosine, and tangent of 40° ; and verify formulas
(2) and (5) arithmetically for this angle.
2. Express each of the following in terms of sin 0 and cos 0 and
simplify. Then find the derivative of each (in radian measure).
(a) cos 0 tan 0, (&) ctn 0/csc 0,
• (c) esc 0 — ctn 0 cos 0, (d) sec 0— tan 0 sin 0,
f •. sec2 0— tan2 0 m esc2 0 — ctn2 0
\C) ' - - - "» VJ/ " - „ - >
csc 0 sec 0
(0) , Sln ^ +t C°S e , W cos 0Vsec20-l, [0<90°].*
1— ctn 0 1— tan 0
3. Establish each of the following identities by reducing the left
member to the form on the right side :
(a) cos20-sin20 = sin e cos
x x
ctn 0— tan 0 1+cos 0 sin 0
tan 0 sin 0 tan 0— sin 0
tan 0+sin 0 tan 0 sin 0 '
tan0csc0-ctn0sec0 = sec ^ cgc ^
sm 0— cos 0
= csc 0-ctn 0, 0<90°.
1+cos 0
(Hint : Multiply above and below by 1 —cos 9.)
* This fact is given to determine the sign of the function which will
replace the given positive radical in simplifying.
372 MATHEMATICAL ANALYSIS [XI, § 273
4. Reduce to simpler forms :
(a) sin3 0 cos 0+cos3 0 sin 0,
1+sin 0' ^' sec0-l
(e) In Ex. 2 (e), (f), (h), simplify in part without changing to sin 0
and cos 0.
(/) Simplify and differentiate : y = (Vl+sin 0 + Vl-sin 0)2.
6. Simplify and integrate each of the following expressions :
(a) (tan 0+ctn 0) sin 0 cos 0,
(6) (1+ctn 0-csc 0) (1+tan 0+sec 0).
6. (a) Differentiate i/ = tan 0. [Use (5), p. 368, and § 184.]
(6) In similar fashion differentiate y = esc 0.
7. Solve for some function of 0, and Jook up each possible angle
<360°:
(a) 5 ctn 0 = 3 esc 0, (6) tan 0+3 ctn 0 = 4, (c) sin 0 cos 0 = .15.
8. The same as Ex. 7 for each angle, in radians, <2 u-W, if;
(a) 5 sec2 0-8 tan2 0 = 4, (6) 2 sin2 0+3 cos 0 = 0.
9. For a projectile fired with a certain speed the range is R = 100000
sin 0 cos 0. For what 0 will 72 = 4800?
10. In Ex. 9 find the maximum value of R as 0 varies.
11. When a block of weight W Ib. rests on a plane of inclination 0,
and with a "coefficient of friction" /, the pull up the plane necessary
to prevent sliding is P = W (sin 0— / cos 0). If / = .15, what 0 will just
make P = 0?
12. Find for what inclination of a line of fixed length the sum of the
horizontal and vertical projections will be greatest.
13. Two tangents to a circle are 20 in. long. The chord joining
their points of tangency is 4 in. from the center. Find the angle be-
tween the tangents.
14. If the path of a projectile is y=x tan 0— z2/250000 cos2 0, for
what (smaller) value of 0 will it strike the point (50000, 1875) ?
§ 273. Further Derivatives. Differentiation formulas for
the tangent, cotangent, etc., can be obtained by expressing
these functions in terms of the sine and cosine, and using the
fraction formula (§ 184).
XI, § 273] TRIGONOMETRIC ANALYSIS 373
Ex. I. ctn 0 = 52*1.
sm 6
sin 0 j- (cos 6) - cos 6 — (sin 6)
dO sin2 6
_sin2 0+cos2 0 = 1
sin2 0 sin
Ex. II. sec0 = -i—
cos 0
:. 4- (sec 0) = - 1 (cos 0)~2 • ~ (cos 0)
cos2 0 cos 0 cos 0 ,
#.£. For degree measure multiply by .017453 • • ; and for deriva-
tives with respect to t, multiply by dd/dt.
Tan 0 and esc 0 are differentiated similarly. All the for-
mulas are listed in the Appendix, p. 493. They should be
memorized if time permits their use at any length.
EXERCISES
1. Using the formulas (E), p. 493, differentiate:
(a) y = tan 4 t, (&) z = sec 5 0,
(c) w> = ctn (5f2), (d) u = csc (3/0,
(0) r = log (sin 0), (h) s = 3 0+3 ctn 0— ctn3 0,
(1) x = e3t sin t, 0') y = e* sec t.
2. Differentiate, after making any helpful simplifications:
ctn20
(c) tr = log ysecx"t"i, (d) Z = ctn0-sec0csc0(l-2sin20).
3. (a)-(e). Differentiate the right members of Ex. 3 (a)-(e), p. 371.
4. When a comet moves in a parabolic orbit, the equation of its
path has the form r = p sec2 (0/2), p being the distance of nearest ap-
374 MATHEMATICAL ANALYSIS [XI, § 274
proach to the sun. If p = .8, find the rate at which r is changing, per
radian, when 0 = 7T/2<r>.
5. What is the slope of the tangent curve (Fig. 124, p. 356) at
Of the curve y = ctn 6, at 6 = 7T/4 fr> ?
§ 274. Further Motion Problems. In Chapter VIII we
saw how 'to study the motion of a point (x, y) when its equa-
tions of motion are known, — i.e.,
equations giving x and y in terms
of t. (§§ 187-191.) Some further
types of motion may now be con-
sidered, in whose equations trigo-
nometric functions are involved.
Ex. I. ^ taut string is unwound
from a circle of radius a. Find how
its free end P travels.
The sides of the right triangle
Fio. 128. in pig. 128 are (x-X) and (Y-y).
The hypotenuse is the length s unwound, =ad by § 251.
-
aB
whence x = X+adsm6, = a (cos 6+6 sin 0),
y=Y-adcos d, =a (sin 6-8 cos 6).
If the string is unwound at a constant angular rate of k radians
per sec., then 6 = kt. Substituting this in (12) gives the equa-
tions of motion for P. (See Ex. 3-4, below.)
EXERCISES
N. B. In the following problems, t denotes the number of radians in
the angles concerned, and also the number of seconds of time elapsed.*
1. A point (x, y) moves so that a; = 40 cos t, t/ = 40 sin t. Plot the
path from £ = 0 to J = 6.5. Could you have anticipated the result by
eliminating I? Explain.
* Suggestion : Keep a few of these problems to work from day to day.
XI, § 276] TRIGONOMETRIC ANALYSIS 375
2. In Ex. 1 find the speed at any time ; also the distance traveled
from 2=0 to 2 = 2 ir. (See §§ 187-191.)
3. A point moves so that x = 10 (cos t +2 sin 2), y = 10 (sin t— t cos 2).
Plot the path from 2 = 0 to 2=8, taking t at intervals of 1W. Two
figures in the value of each x or y will suffice.
4. In Ex. 3 show that the speed at any instant equals 10 t; and
find the length of the path plotted. [Note the products t sin t, etc.
Cf. § 182.]
6. A point moves thus: x = W (t— sin t), y = 10(1— cos 2). Plot
the path, taking 2 = 0, 1», 2W, to 8<r>. Exactly when does the first
arch end?
6. In Ex. 6 find the speed at any instant.
7. A point Q moves in a circle of radius 10 with an angular speed of
2 rad./sec. Meanwhile its projection P on the horizontal diameter
oscillates. Derive a formula for the distance of P from the center
at any time.
8. A point P moves so that x = 10 cos3 t ; y = 10 sin3 t. Plot the
path from 2 = 0 to 2 = 3.5, at intervals of .5. For exactly what value
of t is y greatest, and when is y = Q again?
9. In Ex. 8 find the speed at any time, and in particular, when P
is nearest the origin. Find also the length of the path from 2 = 0 to
t = 7T/2.
10. A point moves so that a: = 10 cos 2; y=Q sin 2. Plot the path,
2 = 0 to 2 = ?r, and measure its length. What sort of curve is it?
§ 275. Involutes. The path of any point on a string which
is being unwound from a given curve is called an involute
of that curve.
The spiral in Ex. I, § 274, is an involute of a circle. Arcs
of this involute are much used in designing gears, cams, etc.,
because of the excellent rolling contact obtainable with
such arcs. (This is explained fully in books on machine
design.)
§ 276. The Cycloid. When a circle rolls along a straight
line" without slipping, any point on it traces out some
definite curve, — a series of arches. This curve is called a
cycloid.
376 MATHEMATICAL ANALYSIS [XI, § 277
To study it, choose axes through A where P starts up.
Then
x=AQ-u, y=QC-v. (13)
But AQ equals the arc PQ which rolled along it. That is,
AQ = a6. Also QC = a, u = asmd, v = acosO. Hence (13)
becomes
x = a(e-sin 0), y = a(l-cos 6). (14)
f ^
P/£
)'~""N ;
r^ ^L^V
v~t TO
/ ^^
¥
A
uQ
FIG. 129.
If the circle rolls at a constant angular speed, k radians per
sec., then 9 = kt, and substituting this in (14), we have the
equations of motion.
The cycloid has various interesting geometrical and physical proper-
ties, established in higher courses, — of which the following may be
mentioned here :
(1) The length of one arch is exactly four times the diameter of the
rolling circle, and the area under the arch is exactly three times that of
the circle.
(2) The curve down which a particle would slide from one given point
to another in the shortest possible time is a cycloid, — inverted. (3) The
time of sliding to the lowest point is the same, no matter where the point
starts on the cycloid. (4) The involute of a cycloid, starting to unwind
at the middle of an arch, is an equal cycloid. Hence a pendulum swing-
ing between two inverted cycloids, with its string unwinding alternately
from each, travels along a cycloid. (5) The time of swing of such a
"cycloidal pendulum " is strictly independent of the angle through which
it swings. This is only approximately true of an ordinary pendulum.
§ 277. Simple Harmonic Motion. If a point Q moves in
a circle with a constant speed, its projection P upon any
XI, § 277] TRIGONOMETRIC ANALYSIS
377
diameter will oscillate back and forth in a certain way.
(Fig. 130.) This type of oscillating motion is called Simple
Harmonic Motion, — abbreviated
S. H. M.
To study the nature of an S. H. M.,
we need an equation giving the dis-
tance x from the center of oscilla-
tion at any time.
Now
x = r cos 6.
And by hypothesis the angular FlG' 130'
speed of Q is some constant dB/dt = k, whence
6 = kt+C,
C being the value of 6 when we begin to count time.
(15)
This is a general formula, true for any S.H.M., and giving
x, the " displacement " from the center, at any time.
Differentiating twice gives
dx/dt= -kr sin (kt+C), =speed of P.
d2x/dt2 = - Wr cos (kl+C), = accel. of P.
.\ d2x/dt2=-k2x. (16)
That is, the acceleration is constantly proportional to the dis-
placement x, — negatively proportional, — which is the
characteristic feature of every S.H.M.
The acceleration is zero at the center, where the speed is
a maximum. It is greatest at the left extreme, when x= — r,
though the speed is then passing through zero from negative
to positive.
S. H. M.'s occur frequently in machinery. Also many motions which
are not simple harmonic may be regarded as the result of combining
several or many such motions. The oscillations of an alternating electric
current and of waves of light and sound are of this general character.
378 MATHEMATICAL ANALYSIS [XI, § 278
EXERCISES
1. What are the equations of motion along a cycloid if the generating
circle has a radius of 50 in. and an angular speed of .2 rad./sec.? What
2. A circle of radius 20 inches rolls along a straight line with an
angular speed of .3 rad./sec. Find the speed of a point of the
circumference when at half its greatest height, and also when
highest.
3. If Q (Fig. 130) moves with an angular speed of .4 rad./sec., and
r = 5, find the speed of P when passing through the center. Also find
the position, speed, and acceleration of P three seconds after starting, if
4. WThat is the equation for an S. H. M. if the time of a complete
oscillation is 2 sec., and the maximum displacement of 15 in. occurs
at t = 0 ? Find the speed at t = .4.
6. Like Ex. 4 if the period is TT sec. and the maximum displacement
is 20 in. occurring at t = ir/Q sec.
[6.] Which of the following is the derivative of some one of the six
trigonometric functions aside from sign : sin 0, tan 0, sec 0, sin* 0, tanz 0,
sec2 0?
§ 278. Damped Oscillations. The exponential curve
(Fig. 79, p. 244) shows how a direct electric current will
" die away " after the E.M. F. is cut off. But an alternat-
ing current continues to alternate while dying out, the
Graph of intensity at any instant
i=e~™isin200t being given by some such
equation as
(17)
t (seconds) The " damping factor " e~wt
rapidly decreases, and makes
the successive waves of the graph smaller and smaller. But,
by (17), the waves all come at the same intervals of time.
For i = 0 only when sin 200 £ = 0, — i.e., when 200 J = 0, IT,
27r, etc., or * = 0, .0157, .0314, etc. (Fig. 131.)
XI, § 2791 TRIGONOMETRIC ANALYSIS 379
To locate the maxima, differentiate (as in § 182) :
di/dt = e-™ cos 200 t (200) +e~m sin 200 t (-50).
Putting di/dt=Q gives, since e~60' cannot be zero,
200 cos (200 0-50 sin (200 0 = 0.
To solve this equation easily, divide by 50 cos 200 t :
tan (200 0=4.
In radians, the acute angle whose tangent is 4 is 1.33.
The next, in Quadrant III, is larger by7r( = 3.14) radians, etc.
/. 200 *=1.33, 4.47,
.'. Z=.0066+, .0223+.
The first of these values makes i a maximum, the next a minimum,
and so on. (Verify by testing di/dt at 200 Z = 0, 7r/2, IT, etc.)
§ 279. Trigonometric Integrals.* In practical work it is
often necessary to integrate a trigonometric expression.
In some cases this is merely a matter of reversing a standard
differentiation formula. Thus by (12), (13), p. 361,
fcos 0d0 = sin 0+C,
(18)
Observe that the first of these results is not —sin 0+C. We are
not differentiating cos 0, but are finding a function which, when dif-
ferentiated, will yield cos 0.
Some other forms are immediately reducible to known
derivatives. For instance, tan? 0 may be written (sec2 0—1)
and then integrated. Thus
tan 0-0+C. (19)
* If time is lacking, the integrations may be confined to the sine and
Cosine.
380 MATHEMATICAL ANALYSIS [XI, § 279
Observe that the integral of tan2 9 is not } tan3 9. Differentiating
the latter would give tan2 9 but multiplied by the derivative of tan 9,
viz. sec2 9.
Similarly the integral of sin2 9 would not be | sin3 9 (nor cos2 9, etc.),
but a very different form, found later (Ex. I, p. 385).
By §§ 77, 101, a power of a quantity (whether a trigo-
nometric function or something else) can be integrated, with-
out changing its form, only if we have present besides the
power the derivative of the quantity which is raised to the
power. Any power of tan 6 can be integrated if it is multiplied
by sec26d6] likewise any power of sin 6 if multiplied by
cos 6d6. And so on. [See Ex. II below.]
The systematic integration of trigonometric expressions
' of all sorts is treated in texts on Calculus. Tables of integrals
are available, covering many forms. (§ 286.)
Ex. I. An alternating electric current varied thus : i = 17 sin 200 t.
Find the quantity of electricity transmitted in any time.
1 The intensity of the current is the rate of flow : i=dq/dt.
.'. q = ( i dt = 17 j*sin 200 t dt ==H cos 200 t+C.
Since g=0 when t=Q, C = — cos 0= — .
.'. g = — (1 -cos 2000-
Check this result by differentiating; also by substituting t—0.
Ex.. II. Find f sin3 9 cos 9 d9.
Since cos 9 d9 is the differential of sin 9, this is like having u* du to
integrate, which would give i*4/4, — the u being sin 9.
'sin3 9 cos 9 d9 = J sin4 9+C. (Check ?)
EXERCISES
1. Draw by inspection a graph for each of the following quantities,
showing a few waves, the value at t—0, and the values of t where the
function is zero. Also find at what time the first maximum and mini-
mum are reached.
XI, § 280] TRIGONOMETRIC ANALYSIS
381'
(a) An alternating current died out thus : i = 10 e~zm cos 400 t ;
(6) Another died out thus : i = 30 e~m sin 200 t ;
(c) The displacement of a pendulum thus : 0 = .2 e~-05' sin 2 vt ;
(d) The elevation of a wave in water thus : y = 3 e~2t sin 3 t ;
(e) The displacement of a tuning-fork thus : x = .08 e~* cos 800 1.
2. Integrate: sin 10 tdt; cos 5 £ eft; 7 sin 802 eft; .4 cos .U eft. Check
each result by differentiation.
3. An alternating current varied thus under steady conditions:
t = 10 sin 400 1. Find i and di/dt when t = .003. Also find the quantity
of electricity passed from t = Q to £ = .003.
4. The angular speed of a pendulum varied thus: w = .l TT cos irt.
Find the angle swung through, from £ = 0 to any time.
6. Show that the following integrals are special cases of f undu,
and find each :
(a) f sin7 0 cos
(c) f cos 0 de
J sir>4 fl '
(6) f tan4 0 sec2 0 d0,
'sec2 0 de
tan 0
J sin4 0
6. Simplify and integrate :
, x r(2 sin 0 cos 0 — cos 0)
•J 1 —sin 0+sin2 0 — cos2
7. (a)-(/i). Integrate the expressions in Ex. 2 («)-(/*), p. 371.
§ 280. Addition Formulas . For various purposes we
need to know how the sine of the sum of two angles (A -\-B)
is related to the functions of
the two separate angles, A
and B.
Fig. 132 illustrates the case
where (A -\-B) is an acute
angle. From any point P on
the terminal line of (A-\-B),
perpendiculars are dropped to
the initial line and to the
terminal line of Z A ; and,
from Q, the foot of the latter,
382 MATHEMATICAL ANALYSIS [XI, § 280
perpendiculars are dropped to the initial line and to the
first perpendicular. Then, by definition,
sin(A+5)=£- (20)
r
The construction divides y or HP into two parts : HF( = GQ)
and FP. Also it makes Z FPQ= Z A. (Why?) Solve
A GQO for GQ, and A FPQ for FP, — using the sine or
cosine of A, — and you will find that
GQ = OQ sin A, FP = PQ cos A.
Substituting the sum of GQ and FP for y in (20) gives
sin (A+B^^^^A±PQ^A.
But OQ/r = cos B, PQ/r = sin B, whence
sin (A+B) = sin A cos 5+cos A sin B. (21)
In words, the sine of the sum of two angles equals the sine of
the first times the cosine of the second, plus the cosine of the first
times the sine of the second.
E.g., sin (45° +30°) = sin 45° cos 30°+cos 45° sin 30°.
Thus sin 75° can be calculated from the functions of 45° and 30°, which
are known from simple right triangles.
In like manner, starting with cos (A + B) = OH/r, and re-
placing OH by OG-FQ, we find
cos (A+B) = cos A cos B — sin A sin B. (22)
Verbal statements of (21) and (22) should be memorized
very carefully.
Remark. So far (21) and (22) have been established only when
(A +B) is acute. In the Appendix, p. 488, they are proved valid for
any angles A and B whatever. Incidentally it is shown that
sin (A — B) = sin A cos B — cos A sin B, (23)
cos (A— B) =cos A cos B+sin A sin B. (24)
XI, § 282] TRIGONOMETRIC ANALYSIS
383
Observe that these formulas are just like (21) and (22) except for the
sign in the middle on both sides. If we remember this fact, it will
suffice to memorize (21) and (22) alone.
§ 281. Some Applications. The addition formulas (21)-
(24), above, are useful in making simplifications, in solving
equations, in studying simple harmonic motion, in calcu-
lating tables (§318) and in other ways.
Ex. I Expand x = 10 cos (kt - -} .
By (24), re = 10 (cos kt. cos 7r/3+sin kt sin 7r/3).
But by tables cos Tr/3 = .5 and sin x/3 = .86603.
.'. x = 5 cos kt +8.6603 sin kt.
That is, the S.H.M.,x = 10 cos (to - *-/3), is equivalent to two S. H.
x = 5 cos kt, and a: = 8.6603 sin kt, combined.
Ex. II. In Fig. 133, X and ,
Y are components of F. As these
are physically equivalent to F,
their combined effect along any
other line OP should equal the
effect of F along that line. Let
us see whether this checks by
actual calculation.
Effects of X and Y along OP : FIG. 133.
XcosB, Y cos (90° -B).
Or, since X = F cos A and Y = F sin A, and since cos (90°— B) =sin B,
these amount to *
/** cos A cos B-t-F sin A sin B.
But the effect of F along OP by Fig. 133 is F cos (A — B) ; and expanding
the latter cosine by (24), we get the same result as from the effects of
X and Y.
§282. Multiple Angles. Expanding sin (6+6) and
cos (6+6) by the Addition Formulas (21) and (22) we find
sin 2 6 = 2 sin 6 cos 9, (25)
cos 2 0 = cos2 6 - sin2 8. (26)
384 MATHEMATICAL ANALYSIS [XI, § 282
In like manner we can apply the Addition Formulas to reduce
functions of any multiple of 0 to functions of 6.
Such formulas are frequently used in scientific work, par-
ticularly (25) and (26).
Ex. I. The range of a certain gun varies with the elevation 6 thus :
R =40000 sin 0 cos 6.
What elevation gives a range of 10000 ft. ? The maximum range?
By (25) the formula may be re-written
R = 20000 sin 26.
Hence to make R = 10000, we need merely make sin 2 6 = .5 ;
2 0 = 30° or 150°,
/. 0 = 15° or 75°. , (Cf. Ex. I, § 271.)
R will be greatest when sin 2 6 is. This will be when the angle 2 0 is
90°, or 6 = 45°. Then R = 20000.
EXERCISES
1. Carry out in detail all the steps of the derivations of (21) and
(22) above.
2. Using formulas (21) and (22) and the values of the sine and cosine
of 1° and 2° as given in the tables, calculate sin 3° and cos 3°.
3. From the functions of 1' calculate sin 2' and cos 2'.
4. Knowing sin 45° = ^2" and sin 30° = £, calculate sin 75°, also
sin 15°.
6. Derive formulas (25) and (26) in detail. Write verbal statements
for these, — as "The sine of twice any angle equals . . .," etc.
6. Show that the area cut from a circle of radius r by a chord which
subtends an angle 0W at the center is : A = £ rz(d — sin 0).
7. Derive a formula for cos 3 6, in terms of cos 6.
8. Expand and simplify the following :
(a) sin (90°+A), (6) cos (180°-A),
(c) sin (45°+A), (d) cos (60°-A).
9. Given sin A=|, cos B = A> A and B acute, find sin (A +B)
and cos (A -B).
10. The same as Ex. 9, but with neither A nor B in Quadrant I.
XI, § 283] TRIGONOMETRIC ANALYSIS 385
11. Simplify: (a) sin (A+B)— sin (A— B),
(6) Bin7g+sin3g. [Hint: 7 0 = (5 0+2 0) 5 3 0 = (5 0-2 0).]
cos 7 9 -{-cos 3 6
(c) cos (A+B) cos £-{-sin (A +B) sin B.
12. Expand and simplify : s = 10 sin(3 £+30°).
13. The same as Ex. 12 for the following :
(a) z = 7 cos (5 £+80°), (6) x = C sin (400 «-A).
14. For what values of c and A would x=ccos (20 t— A) give the
expanded form x =8 cos 20 t -J-6 sin 20 t?
§283. Half-angle Formulas. Combining (26) with the
identity l = cos2 0+sin2 6, we find
sin20 = |(l-cos20),
(27)
cos20 = |(l+cos20).
These two formulas should be carefully memorized as
they are used in many calculations, reductions, and inte-
grations.
Ex. I. Integrate sin2 0 dO.
By (27) :
sin2 Od8 = (l -cos 2 0)d0 = £ 0-^sin 2 0+C.
Js
Ex. II. Knowing cos 30°= .86603, calculate cos 15°.
When 0 = 15°, formula (27) gives cos 15°== V|(l+cos 30°) ;
i.e., cos 15° = V53302 - .96593.
The tables were originally calculated partly by this method. (How
could it be continued?)
Formulas (27) give the sine and cosine for half of any
known angle 2 0, and are called " Half-angle Formulas."
Often they are written
A)=j(l-coB A), cos2 ft A) =i(l-f-cosA),
. (28)
2
386 MATHEMATICAL ANALYSIS [XI, § 284
§ 284. Angle between Two Curves. The tangent of the
difference of two angles is easily found from the sine and
cosine :
tan (A — B) = s*n (A- — B)__ sin A cos B — cos A sin B
cos (A — B) cos A cos B— sin A sin B
Dividing both numerator and denominator by cos A cos B
gives
which expresses tan (A — B) in terms of tangents only.
This is useful in finding the angle (K)
between two lines or curves. For K is the
difference between the inclination angles 72
and 1 1 at the point where the curves cross.
(30)
l-ftan/2tan/i
y Now tan 7 is the slope of a curve,
FIG. 134.
= I, =dy/dx.
Hence the angle K can be found by (30) directly from the slopes or
derivatives without first looking up the two separate angles.
EXERCISES
1. (a) Show how to integrate cos2 0 and sin2 0. (6) What would
you suggest for sin2 (3 0) ? For cos2 (7 0) ?
2. In a right triangle one leg is 6=999.95 and the hypotenuse is
1000. Try to find /.A directly, but also solve by using the formula
sin1 (A /2) = HI -cos A).
3. Knowing cos45° = |V2, =.70711, calculate cos 22° 30' and cos
11° 15'. Explain how you could also calculate cos 67° 30', and from it
cos 33° 45', etc.
4. From formulas (28) derive the formulas: (A) tan | 0 = sin 0/
(H-cos 0) ; OB) tan \ 0 = (1— cos 0)/sin 0.
6. Obtain formulas for ctn % 0, similar to those in Ex. 4.
6. Derive a formula for tan (A +B) in terms of tan A and tan B.
XI, § 285] TRIGONOMETRIC ANALYSIS 387
7. What is the angle between two straight lines if their slopes are :
(a) 3 and 2; (&) 7 and -5; (c) 4 and -5?
8. Where does y = xi meet z2+7/2 = 20, and at what angle do they
cross each other? Draw a rough figure by inspection.
9. The same as Ex. 8 for xy = 12 and 2 x-3 y = 6.
§ 285. Sums, Differences, and Products. With the help
of the Addition Formulas, it is easy to change sums and
differences of certain functions to product forms which are
more convenient for some purposes.
E.g., suppose we wish to change* sin 40° — sin 28° to a
product.
Let A and B be two angles whose sum is 40° and whose
difference is 28° :
Then
sin 40° = sin (A-\-B) =sin A cos 5+ cos A sin B,
sin 28° = sin ( A — B) = sin A cos B — cos A sin B.
/. sin 40° -sin 28° = 2 cos A sin B = 2 cos 34° sin 6°.
(Q. E. F.)
The same method can be used to prove the following for-
mulas : f
sinw+sinv =2 sin % (u+v) cos | (u— v) ; (31)
sin u — sin v =2 cos ^ (u+v) sin -J (u— v) ; (32)
cos w+cos v =2 cos \ (u+v) cos -J (u— v) ; (33)
cos u — cos v— —2 sin -J (u+v) sin -J (u—v). (34)
* Observe that this is not sin (40° -28°).
t If you expect to work in higher mathematics or engineering, memorize
statements of these formulas in words : "The sum of two sines equals twice
the sine of half the sum (of the angles) times the cosine of half the difference.'^
Etc.
388 MATHEMATICAL ANALYSIS [XI, § 286
A method of going from the product form to a sum or differ-
ence is shown in Ex. Ill below.
Ex. I. Solve for x : sin 3 x+sin x = sin 2 x.
By (31), sin 3 z+sin x = 2 sin 2 x cos x.
.'. 2 sin 2 x cos x = sin 2 x.
This gives cos x = .5, or else sin 2 x =0. (§ 63.)
Hence x = 60°, 300°, etc. Or else 2 z = 0°, 180°, etc. ; i.e., a; =0, 90°, 180°
etc.
*"•"• «"-££»
By (32) and (33): sin 60° -sin 10° = 2 cos 35° sin 25°; and cos 60°
-f cos 10° = 2 cos 35° cos 25°. Thus y =tan 25° simply.
Ex. III. Integrate cos 4 x cos x dx.
We first change the product to a sum or difference. As the product
involves only cosines, it comes under (33), 4 x being half the sum, and
x half the difference of the angles. Hence u+v = 8 x, u—v = 2x, giving
.'. cos 4 x cos x = \ (cos 5 x+cos 3 x).
We can now integrate each term, getting
£ [isin5z-Hsm3z]+C.
EXERCISES
1. Derive in detail formulas (31)-(34).
2. Transform into products and simplify :
, ^ sin 5 0— sin 3 0 /.\ sin 8 0-fsin 4 0
cos 5 0 — cos 3 0' cos 8 0+cos 4 0'
fc) sin 50° -sin 10° ,~ cos 30° - cos 20°
sin 50°+sin 10°' cos 30°+cos 20°*
3. Transform into sums or differences :
(a) sin 6 0 cos 4 0, (6) cos 8 0 sin 6 0.
4. Show that f cos 7 0 cos 3 0 d6 = -fa sin 10 0-f-J sin 4 0-fc.
§ 286. Tables of Integrals. By various methods numerous
integrals have been worked out and tabulated. Thus many
integrals needed in practical work can be looked up in a table
without working them out for ourselves.
XI, § 287] TRIGONOMETRIC ANALYSIS 389
There is a small table in the Appendix, pages 494-497.
Larger ones are given in texts on Calculus ; also separately.*
Ex. I. Find ( sec3 x dx.
This comes under (39), p. 497, with n =3 and a = l :
.'. f sec3 x dx = \ sin x sec2 z-f f j sec x dx.
The latter integral is given in (9) of the table; viz., log (sec z+tan x}.
(Formulas like (39) are called Reduction-Formulas.)
r xdx
Ex-n- Find Jv^ri'
Clearly x dx suggests x2, as does also the even power, x*.
Putting z2 = t, gives x dx = | dt, and
r x dx i /• dt
J Vx4+9 2 J
This comes under (23), with a = 3.
• f x dx = | log («+V£+9) = i log
J Vx4+9
EXERCISES
1. Find from tables the integrals of the following :
(d) sin? 8 cos Od8, (e) tan6 ^ d8, (/) sec5 0 d0.
(g) sin 3 0 cos 0 d0, (A) x2 log x dx, (i) e^ sin 3 t dt.
2. Find by integration the area of the ellipse : 4 z2+9 y* =36.
3. Find the volume generated by revolving about the X-axis the
area under one arch of the sine curve. (Radian measure.)
§ 287. Summary of Chapter XI. The basic formulas
derived in this chapter may be classified as follows : (I) For-
mulas involving a single angle only ; (II) Addition formulas ;
(III) Half -angle formulas; (IV) Conversion formulas for
sums and products ; (V) Differentiation formulas.
* B. O. Peirce, A Short Table of Integrals, gives several hundred forma
and is sufficiently complete for all ordinary purposes.
390 MATHEMATICAL ANALYSIS [XI, § 287
All of these should be carefully memorized if you expect
to use mathematics extensively as a working tool. In any
case it will pay to make a full list of the uses of each set of
formulas, so far as shown. The integrals of the several func-
tions, so far as obtained, need not be memorized ; but the
methods by which they were obtained should be familiar.
We shall next turn to a further study of the uses of inte-
gration, and methods of setting up integrals.
EXERCISES
1. Prove the identities :
(a) sin* 9+tan> » = sec> 0-cos* 6, (b) tan2M-tan».gnl»
tan 20 — tan 6 sin 0
2. Given cscA = —^ and secJ5 = ^5, with A and B in the same
quadrant. Find all the functions of A — B.
3. A "synmotor" plows a spiral furrow, being drawn steadily in-
ward by a cable which winds up on a cylindrical "drum." What kind
of curve is this spiral, by definition?
4. A point moved in such a way that x = lO cos t, ?/ = 8 sin t. Find
the speed at any instant ; also what kind of curve the path was.
6. From the formulas for sin (A+B) and cos (A+B) derive a
formula for sin 3 9 in terms of sin 0 alone.
6. Telephone wires, h feet above the ground, exert a horizontal
pull of P pounds on each pole. The last pole is strengthened by a guy
wire I feet long inclined 0°, not reaching the top. Show by § 116 that
for a perfect balance Ph = Fl sin 0 cos 0, where F is the force along the
guy. Find what value of 0 will make F least, I being constant.
7. The time required for a small force (F Ib.) to raise a 100-pound
weight 30 feet by pulling it up a smooth plane, of inclination 0, is
T = V60/D, where D = sin 0 (F — 100 sin 0). What value of 0 will make
T a minimum? (Hint: Simply make the denominator D a maximum.)
8. Along the "great circle" from San Francisco to Manila, the lati-
tude L varies with the longitude 0 thus: tan L = .24 sin 0 — .95 cos 0,
approx. Does the circle pass north or south of Honolulu (0 = 157° 50',
L = 21° 20') ? How much does L change with 0, per degree, at 0 = 150°?
(Differentiate implicitly. § 80.)
9. Find an equation of the form tan L = c sin (0 — A), which is equiv-
alent to the equation given in Ex. 8. [Hint: Expand sin (0— A) and
compare.]
XI, § 287] TRIGONOMETRIC ANALYSIS
391
10. Two halls 17.28 feet wide and 10 feet wide meet at right
angles. Find how long a pole can be carried from one into the other,
while kept horizontal. (Hint: Show that the shortest distance across
at the turn is expressible in the form 17.28 esc 0+10 sec 0.)
11. In just what direction should
the wind blow to make the sum of its
eastward and northward velocities a
maximum, — the actual velocity of
the wind being constant?
12. When a ship sails "into the
wind," the driving force is approxi-
mately proportional to sin 6 sin (A —6),
where A is the angle between the
course of the ship and wind's direction
(reversed), and 6 is the angle between
the wind and the sail. Find what
value of 0 gives the maximum force.
13. A right triangle has a horizontal hypotenuse 20 ft. long. Show
that its area is A =200 sin 7 cos 7, where 7 is the inclination of either
leg. For what 7 will A be greatest?
14. Along every great circle of the earth the latitude (L) varies with
the longitude (6) according to the formula
tan L=a sin 0+6 cos 0,
where a and 6 depend on the course of the circle in question. Find
a and 6 if the circle is to pass through St. Johns, N. F. (0 = 52° 40',
L =47° 40'), and the Azores (0=29°, L = 38°). If an airplane is to fly over
this course, what should its latitude be when in longitude 36°?
15. The same as Ex. 14 for a great circle from St. Johns to Queens-
town (0 = 8° 20', L = 51°50/).
16. Where would the circle in Ex. 8 above cross the equator?
Where does it go farthest north?
FIG. 135.
W, wind ; 8, sail ; K, keel.
CHAPTER XH
DEFINITE INTEGRALS
THE SUMMATION OF MINUTE ELEMENTS
§ 288. Constant of Integration. We proceed now to note
certain facts about integration which will enable us to apply
it more easily to practical problems. And first we shall
shorten somewhat the calculation of areas, volumes, etc., by
observing that the constant of integration and the final result
must always assume a certain form.
Consider for example the area under the curve y = \/x from x = 2
toz=7.
A = \ y dx = f - dx =log *+C.
Since A =0 when x = 2, we have 0 = log 2+C, or
C=-log2. (1)
Hence the area from z=2 to any other value of a; is
A=log x— log 2.
In particular, the area from x =2 to x = 7 is
A = log 7 -log 2.
Similarly the area from x=3iox = ll would be log 11 —log 3.
The final result is simply the difference between the values
of the integral function at the beginning and end of the interval.
Clearly the same thing must be true for the area under
any other graph. For the constant of integration must equal
the value of the integral at the starting point, — with the
sign changed.
392
XII, § 290] DEFINITE INTEGRALS 393
§ 289. Definite Integrals. The symbol
is used to denote the difference of the values of the integral
function at x = b and at x = a. It is called "the definite
integral from a to b of f(x) dx " ; and a, b are called the
limits of integration*
This difference is also denoted by writing the symbol
after the integral function. Thus
X7 I -17
-dx = \ogx =log7-log2.
_ X _\2
§ 290. Calculations Abridged. By § 288 the area under a
curve may be expressed in the form
Cydx.
Ja
(2)
Similar reasoning shows that the volume of a solid, the work
done by a varying force, etc., are simply :
V= {*A.dx, W=Tfdx,etc. (3)
«/a Jn
That is, in finding any such quantity we need not consider
the constant of integration, but merely form the difference
of two values of the integral.
Ex. I. Find the momentum generated from t = 2 to t = 5 by a force
varying thus : / = 100 t - 12 Z2. «
M = f dt = 5 (100 t- 12 &}dt
= 50 Z2-4 *3 =750-168 = 582.
* An ordinary integral function involving an arbitrary constant is some-
times called an indefinite integral.
394 MATHEMATICAL ANALYSIS [XII, § 291
Remark. The constant of integration is present here in disguise,
already determined, in the term -168. [Cf. (1), § 288.]
EXERCISES
w
1. Evaluate: §*6xzdx, j*_*z4 dx, j** e3* dx, j^2 cos x dx.
2. Find the area under one arch of the sine curve, plotted in radian
measure (Fig. 122, p. 355).
3. Find the area under the curve y = I/ Vz from x = 4 to x = 9.
4. Find the work done by a gas in driving a piston from x = 20 to
x = 60 if F = 1200/z continually.
6. The force applied to a car varied thus : F = 6Qe~-P. Find the
momentum generated between £ = 10 and £ = 20.
6. How much water must flow into a hemispherical cistern of radius
7 ft. to raise the depth (at the middle) from 3 ft. to 5 ft. ?
7. Find the volume generated by revolving about the X-axis the
area under the curve 7/ = tanz (Fig. 124, p. 356) from z = p to x = ir/4.
Note. In Exs. 8-9, find the integrals from tables.
8. A "conoid" has circular base of radius 10 in. Every section
perpendicular to one diameter is an isosceles triangle, whose height is
15 in. Find the volume.
9. A cylindrical tank of radius 5 ft., lying horizontally, is half full of
oil which weighs 60 Ib. per cu. ft. Find the pressure against one end.
[10.] Plot the curve y=x<*-\-\ from x=0 to x=2, drawing the ordinatos
at 2=0, .5, 1.0, etc. Calculate the total area of the four rectangles
inscribed under the curve in these strips. Likewise find the area for a
similar set of 10 inscribed rectangles with bases x=0 to .2, .2 to .4, etc.
Compare the results with the area under the curve from x=0 to 2, as
found by integration.
A
§ 291. Fundamental Theorem. We proceed now to es-
tablish a theorem of »the very greatest importance.
Let y be any quantity which varies continuously with x,
and let j/i, y%, ••• yn, be its values taken at equal intervals Ax
from x = a to x = b. Multiply each of these values by Ax,
and consider the sum of the products :
2/iAx+7/2Ax ••• +2/Ax.
XII, § 292]
DEFINITE INTEGRALS
395
If Az-M), this sum mil approach a limit, — that limit being
Xb
ydx] i.e.,
_
L (y i Ax+y2 A* • - • +yn Ax) = Cy dx. (4)
*/<z
Fio. 136.
PROOF. The several products yi Arc, ?/2 Ax, etc., are equal to the areas
of rectangles inscribed in the graph of the varying quantity y (Fig. 136).
And the limit of the sum of those rectangles as their number is in-
definitely increased is precisely the area
under the graph, or the definite integral
in (4).
This theorem has been stated ab-
stractly but is of great practical im-
portance, for it shows that
Any quantity whatever, which is ex-
pressible as the limit of a sum, of the
type in (4), is therefore, without further
argument, equal to a definite integral, as
in (4). .
By means of this " Fundamental Theorem " we can easily
set up many integrals. Consider, for instance, the volume
of a solid. If cut into n slices of thickness Ax by parallel
planes, the volume of each slice will be approximately its
face-area A times Ax, and the entire volume approximately
AiAx+AzAx ••• -\-AnAx,
or, exactly, the limit of this sum as Ax->-0. Hence without
further proof :
V=j*Adx. (5)
This integral can be set up still more quickly by reasoning freely as
in § 99. But the present argument is logically sound and exact.
§ 292. Infinitesimal Analysis. Integration as defined
nowadays is the process of finding a quantity when given
396 MATHEMATICAL ANALYSIS [XII, § 292
its derivative or rate of increase. This was in essence the
conception of Newton, who first devised the process.
We have now seen that integration is also a method of
calculating the limit of a sum of a certain type. This is
almost the point of view of Leibnitz, who shares with Newton
the honor of having invented the calculus, and who con-
sidered integration simply as a method of summing.
For instance, he regarded the area under a curve as composed of
exceedingly many " infinitesimal " strips, — so narrow that the height
y does not change within a strip ( !) Calling the base of each strip dx
and the area of each y dx, the whole area was the sum of all these areas :
A=("ydx, (6)
Ja
the sign f being simply an S, standing for "sum of."
We have already touched upon this conception (§ 99), and
have seen that although logically defective it seems to work
as a method of setting up integrals. We are now in a posi-
tion to understand the matter more fully.
No matter how narrow a strip is, its height is not a fixed
value y. Thus y dx is not the area of the strip, but of a rec-
tangle inscribed in the strip. (Cf. Fig. 48, p. 143.)
The desired area is not the sum of the rectangles y dx, but
the limit of that sum. And the limit of the sum is the true
integral as we now define it. Thus Leibnitz set up a formula
which is strictly correct in the sense in which we now use the
sign.
Moreover he obtained strictly correct results by making
another error which he saw would compensate for the first.
Although he said that he was going to " sum up," actually
he did not do so, but instead found the limit of the sum, -
which is the thing he should have been seeking. In other
words, he used rules for " summing " which were the same
as our rules for integrating.
XII, § 292] DEFINITE INTEGRALS 397
The relation between Leibnitz' method of setting up integrals and
the strictly logical procedure may become clearer if we refer again to
the case of the volume of a solid.
By the old conception, as noted in § 99, the slices are regarded as so
thin that the volume of each equals its face-area A times its thickness
dx, making the whole volume the sum of these elements Adx :
§Adx.
What we should say, reasoning exactly as in § 291, is that the vol-
umes of the slices are approximately
A2Az,
that the entire volume V is approximately the sum of these, and ex-
actly the limit of that sum, or by (4) the integral
r=)Adx. (7)
If the old method is regarded merely as a short way of
stating the correct argument about the limit of a sum, and
the person using it understands what he is doing, the old
conception may properly be used in setting up integrals.
Indeed it is the method regularly used by scientific men.
The method, even if not " rigorous," is very " vigorous."
Of course the question arises as to when we can rely upon
this older method to give a correct integral. The answer is
simple : Whenever the quantity under consideration is the limit
of a sum of the type in (4), above.
If for convenience we regard a quantity as a certain sum,
when in reality it is the limit of that sum, and if we then
actually work out the limit of the sum (by integrating),
clearly we shall get an exact result.
EXERCISES
1. Tell how an exact argument in terms of limits and theorem (4)
above would run in setting up integrals (B) and (C) of § 99 .
2. The same as Ex. 1 for the integrals of : (a) § 100; (6) § 102.
398 MATHEMATICAL ANALYSIS [XII, § 292
3. By considering "infinitesimal elements," and also by an exact
argument, set up the integral in each of the following cases :
(a) The momentum generated by a variable force F ;
(6) Radium decomposes at a variable rate R, s&yR=f(t'). Write
an expression for the total amount lost from t = t\ to t = t2.
(c) A wound is healing at a variable rate, =F(t). Express the total
area healed in any time.
4. The density of the earth (D Ib. per cu. mi.) varies with the
distance (x mi.) from the center: D = F(x). Express the total weight
of the earth from x = 0 to 3960. [Hint : Regard a spherical shell in
the interior as "so thin that it is all the same distance x ft. from the
center" !] Also give an exact argument.
5. Plot the parabola 7/ = x2 and the line 7/ = 3.T+4, and measure
the area bounded by them. Calculate the area by a single integral.
[Hint: If the area be divided into "infinitesimal rectangles," running
parallel to the F-axis, what will be the length of any one in general ?)
6. The same as Ex. 5 for the curves y =xz and y2 = x.
7. A beam 30 feet long carries a load (L Ib. per ft.) which varies
thus with the distance (x ft.) from one end : L = 120 x —4 x2. Find the
total load.
8. In Ex. 7 find the total moment of the load about the end men-
tioned. (§ 116.)
9. A horizontal semicircular plate of radius 10 ft. weighs 2 Ib. per
sq. ft. What is the total moment of its weight about its straight side?
10. Find the total weight of a circular plate of radius 5 in., if the
weight per sq. in. varies thus with the distance (x in.) from the center :
w =4 — .2 x. [Consider a narrow ring x in. from the center.]
11. Find the total weight of a rectangular plate 20 in. long and
4 in. wide if the weight per sq. in. varies thus with the distance (x in.)
from one end : w = 6 + .4 x.
12. A hemispherical cistern of radius 10 ft. is full of water. Find
the work required to pump the water to a level 2 ft. above the top.
[Hint : Consider a thin sheet of water at any distance (x ft.) below the
top. Water weighs 62.5 Ib./cu. ft. roughly.]
13. The speed of a car varies thus with the time : t> = 10 t2. , Calcu-
late the speed every 2 seconds from t =0 to 10 inclusive, and by averaging
the values numerically for each interval, calculate the approximate
distance traveled. Repeat the calculation, getting the speed every half
second. Compare the results with the exact distance found by inte-
gration.
XII, § 293] DEFINITE INTEGRALS
399
§ 293. Length of a Curve. If PQ is a chord joining any
two points of a smooth curve, then
But A^/Az, the slope of PQ, must
equal dy/dx, the slope of the tangent,
at some point on the arc PQ.
.x. (8)
Now the length of the curve, s, is
the limit of the sum of the lengths
of its chords PQ, as each approaches zero:
s= L
Hence, by (4) above, s equals a definite integral.
(9)
E.g., for the curve y = x3, we have dy/dx =3 z2,
a
^_^
1
FIG. ]
dx
[37.
Remark. Using the brief " vigorous "
method, we would consider any very short
arc ds as straight, and as forming with
dx and dy a right triangle. (Fig. 137.)
dx\
Hence as in (9) above the entire length
of the curve is
400
MATHEMATICAL ANALYSIS [XII, § 294
d0
§294. Length in Polar Coor-
dinates. Let us also express the
* length of a curve whose equation
P is given in polar coordinates.
(I) Using the short " vigorous
method." When dO is infinitesi-
. mal, the circular arc PN (Fig.
138) is regarded as straight, and
being perpendicular to its radius
ON, it forms with PQ and NQ a right triangle (!), in which
PQ = ds, NQ = dr, and PN = rdd, if 6 is in radians. (§ 251.)
(10)
(II) Using an exact method. In Fig. 138 let a straight line PN' be
drawn perpendicular to OQ. Then
(chord PQ
This may also be written
or
or
Now the length of the curve is the limit of the sum of the lengths of
the chords, as A0-^0. But the fractions PJV'/arc PN and N'Q/NQ both
approach 1. Hence s is the integral
which agrees with (11).
§ 295. Surfaces of Revolution. When any curve y=f(x)
is revolved about the X-axis, it generates some curved sur-
face. Let us find the area S of this surface.
XII, § 295] DEFINITE INTEGRALS
401
Using the short method, any tiny
arc of the curve, ds, is regarded as
generating a narrow band running
around the surface, — of length 2 Try
and width ds. The area of this
tiny band of surface is then 2 wy ds.
Or, substituting for ds its value
(§ 293), we have as the surface :
TIG. 139.
S-.
1+
(13)
This integral can also be set up by the " rigorous method "
of limits ; but far less simply.
To calculate the area of a general curved surface, which is
not obtainable by revolving a plane curve, is a more difficult
problem, which will not be treated in this course.
Ex. I. Find the area generated by revolving the parabola ?/2=4 x
about the z-axis, from x = 0 to x = 8.
:. S
dx
EXERCISES
N.B. When necessary find the required integrals from tables.
1. Find the lengths of the following curves :
(a) y=iCc — l)* from x = l to x=9.
(6) The entire circle xz+y* = 25. (Check.)
(c) The parabola 12 y =z2 from x = 0 to x = 8.
2. Find by integration the areas generated by revolving the follow-
ing curves about the X-axis :
(a) The cubical parabola y =x3, from x = 0 to x = 1.
(6) The parabola ?/2 = 4 x, from x =3 to 15.
(c) The line y = %x, from x = 0 to 10. (Check by elementary
geometry.)
402
MATHEMATICAL ANALYSIS [XII, § 296
(d) The line y = 2x, from x = 1 to. x = 5. (Check.)
(e) The circle x2 +2/2 = 100. (Check.)
3. Express as an integral the area of the surface generated by
revolving any given curve about the F-axis.
4. Find the lengths of the following curves, each from 0 = 0 to 0 = TT^ :
30,
(6) r =
(a)
(c) r = e0, (d) r = 10sin0.
5. (a)-(d) Plot each curve in Ex. 4. Measure each desired length.
6. Like Ex. 9, p. 398, for r = 20 and a weight of .4 Ib. per sq. ft.
7. Like Ex. 12, p. 398, for a radius of 8 ft. and a level 3 ft. above
the top.
§ 296. How to Plot a Surface. To understand the calcu-
lation of volumes bounded by curved surfaces in general, we
need to know how surfaces can be represented by equations.
Any plane curve is represented by some equation y=f(x),
which tells exactly how high the curve is above the X-axis
at every point.
Similarly for a surface. We first select a horizontal
reference plane, and in it choose X and Y axes. The height z
of the surface above this plane
will vary from point to point
in some definite way. The
surface will be definitely de-
scribed, if we tell by an equa-
tion z =f(x, y) how high it is
(2,2) above every point (x, y) in
FIG. 140.
V(0,2)
(4,2)
the reference plane.
To plot a surface from its equation we proceed as in the
following example.
Ex. I. Draw that part of the surface
z = x*+y*+4y (14)
which stands directly above a rectangle in the .XT-plane
bounded by the axes x = 0 and 2/ = 0, and the lines z = 4 and
= 2.
XII, § 297] DEFINITE INTEGRALS 403
First draw the specified base in perspective. (Fig. 140.)
Its corners are the points (4, 2), (0, 2), (0, 0), (4, 0).
Then calculate the height of the surface above each corner,
and at various other points, using (14) :
x = 4, y = 2, give 2 = 16+4+8 = 28,
x = 0, y = 2, give z = 0+4+8 = 12; etc.
Thus the height of the surface above the corners ^
A (4, 2) and B (0, 2) is 28 units and 12 units, re- 0
spectively. Similarly for the other values of z in
the table.
0
5
V2
9
16
21
28
Erect perpendiculars to represent the height of 2
the surface above the base plane at these several 2^
points, and join the ends of the vertical lines by 4
smooth curves. These are curves on the required
surface, and show its general shape.
The surface forms a sort of tent-like roof over the space
between it and the base plane.
§ 297. Volumes by Double Integration. Let us find the
volume of the solid drawn in Fig. 140 above.
Consider a section perpendicular to the X-axis at any dis-
tance x from the origin. If we can express its area (At) in
terms of x, an integration will give the required volume. But
as this section is not one of the figures of elementary geometry,
we have no formula for its area, and must perform a pre-
liminary integration to find its area.
Throughout this section, x has a constant value ; but the
height of the surface (z) varies with y. The element of area
in this section is, then, zdy ; * and
/. Aa = Czdy = f
c/O */0
* If this is not clear, make a rough drawing showing how the section
would appear, if seen at right angles, looking into the end of the required
volume.
404 MATHEMATICAL ANALYSIS [XII, § 297
Since x is a constant during this integration, we find (§ 88),
\ 2/3+2 2/2T=2 *2+f .
6 Jo o
We now have the sectional area in terms of x, and can find
the volume as in earlier cases :
This calculation would have been slightly modified, if the base of
the solid, instead of being a rectangle, had been bounded, say, by the
X-axis and the curve y =x3, from z =0 to z = 2.
As above, we first draw the
part of the X F-plane which is
to be the base of the solid.
(Fig. 141.)
The curve y = x3 need not be
drawn accurately to scale for
this purpose. But in calculat-
ing the height of the surface
FIG. 141.
COM)
above any point on this curve,
we must use the proper value
of y as well as x at that point, — the value of y being first found
from the equation of the curve y = x*. E.g., at C (Fig. 141).
x = 2, y = S', :. 2 = 100.
A, is again found by integrating z dy ; but the values between which y
runs are not the same for all sections. The upper limit for y is a value
depending upon the x of that section, viz., it is x3. Hence *
* If it looks peculiar to have z8 as a limit of integration, remember that
in this first integration x is a constant.
XII, § 298] DEFINITE INTEGRALS 405
Remark. It is interesting to interpret this process from
the standpoint of " infinitesimal elements." Consider a vol-
ume to be composed of tiny columns, of height z and bases
dy dx. The whole volume is then the sum of these, or
V=§§ zdydx.
Summing first with respect to y all columns having the
same x gives a slice A dx. Summing these slices, as to x,
gives V.
EXERCISES
Draw that part of each of the folio wing, surf aces which stands over
the specified portion of the X F-plane ; and calculate the inclosed
volume :
Surface Base
1. z = x2+y2, 2/ = Oto3, z = lto4.
2. z=x2+y2, y = 0 to y=xz, x=0 to 2.
3. z = xy, y = Qtoy = x, z = lto5.
4. z=xy, y = Qtoy = Vx, z = 0to4.
6. z = x2+Qy, y = Qtoy=x2, z = 0to4.
6. z = 12x+y2, y = ltoy=x, x = 2tox=4.
7. z = xy+y3, y = ltoy = Vx, * = ltox = 4.
8. z = 4x+5y, bounded by X -axis and y = 4 x — x2.
§ 298. Special Plane Sections. In studying the shape of a
surface, it is helpful to know the nature of the cross-sections
made by various planes. Sections perpendicular to an axis,
— in which x or y is constant, — are the most easily studied.
Ex. I. What sort of curve is cut from the surface
by the plane y = 2 ? What is its slope at any point ?
Putting y = 2 makes z = #2+26. (Here z is the height of the
curve and x the horizontal distance.) The section is there-
fore a parabola, extending in the positive Z direction, and
406 MATHEMATICAL ANALYSIS [XII, § 299
raised 26 units. Differentiating gives dz/dx = 2x, the slope
at any point.
This slope could be found directly from the equation of the surface
by simply treating y as constant while differentiating. Similarly, in any
section perpendicular to the Ar-axis, x would be a constant; and the
slope would be dz/dy = 4 y3. (In such a section y would be the horizontal
coordinate and z the vertical.)
§ 299. Partial Derivatives. Whenever, as in § 298, we
differentiate a function z=f(x, y), treating y as a constant,
we are said to find the " partial derivative " of z with respect
to x, written dz/dx. Similarly for dz/dy.
Thus in the case 2 = z2+2/4+10 above,
^ = 2*, £-4*.
dx dy
Similarly, if z =x6+5 sy+8 y2,
-^ = 6 z6+10 xtf, ^=15 sV+16 y.
dx dy
Geometrically interpreted : dz/dx is the slope of a section
of the surface z —f(x, y) made by a plane y = c. Physically, it
is the rate at which z changes per unit change in x, if y re-
mains constant.
In general, if 2 is a function of several variables x, y, u,v, . . . , then
dz/dx will give the rate at which z will change with x, while y, u, v,
. . . , remain fixed, — or, as we say in daily life : "Other things being
equal."
§ 300. Extreme Values. It is sometimes necessary to
find the maximum value of a function of two variables which
can change independently of each other, say z =f(x, y) .
This amounts to finding the highest point on a surface,
2 =f(x, y) • Such a point must be the highest on each of the
two special sections (x = Ci, y =» c2) passing through it. Hence,
XII, § 300] DEFINITE INTEGRALS 407
unless the surface rises sharply to the point, the slope of each
section must be zero :
a*=0, -^ = 0. (15)
dx dy
And similarly for a minimum.
More generally, let z be a function of several variables,
z =f(x, y,u, . . .) . Its maximum value must be the greatest
obtainable by varying x and keeping y, u, . . . constant.
Hence dz/dx = 0. Similarly dz/dy = 0, etc.
Ex. I. Test z=z2-6 z+?/2-4 2/+30.
<^ = 2z-6 = 0, -^ = 2i/-4 = 0.
dx dy
This gives x=3, y = 2, whence 2 = 17.
Testing each derivative on both sides shows this point to be the
lo wesson each sectional curve. This suggests that it is the lowest
point on the surface ; but is not a sure test, since the surface might go
lower somewhere between the sections. Systematic methods of making
sure tests are discussed in more advanced courses.
EXERCISES
1. Plot and find the volume under the surface z = x3-\-y2 from y = Q
to y = x2 and from x = Q to z = 2.
2. In Ex. 1 what sort of section is made by the plane x = 2? Find
its slope at the point where y = 3.
3. What is the character of the sections of each of the following
surfaces made by the specified planes, and what is the slope of each
section at their common point?
Surface Cutting Planes
(a) z = z2+4i/+5, z = 2, y = l.
(6) z = 100-x*-y2, x = &, y = Q.
(c) z=xy, x = 5, 2/ = 10.
(d) z = 6x-3y, x = 3, y = Q.
4. (a)-(e). The same as Ex. 3 for the surfaces in Ex. 1, 4, 5, 6, 8,
p. 405, and the cutting planes x = 2, y = l.
5. In Ex. 3 (6) find the maximum z. Also test Ex. 3 (a).
408 MATHEMATICAL ANALYSIS [XII, § 301
6. Locate any possible maxima and minima for the following func-
tions, and make a sure test if you can :
(a) z=^r2+4i/2-6z+82/+30, (6) z = 40-x*-y*,
(c) z=a:3+32/2-6</+20, (d) z = 10z2 + 122/2-z4-i/<.
7. The temperature at any point (x, y) of a square metal plate
varied thus: :T-z2+2/2-12:c-10?/+300. Where was the coolest point ?
§ 301. The Mean Value Problem Resumed. We saw in
§ 96 that the work done by a variable force F in moving an
object any distance, say from x = a to x = b, is
r
W= I Fdx.
The average force acting during that distance is by definition
the total work divided by the distance ; i.e.,
/ft
Fdx
r = _
6 — 0
Let us compare this definition of average force with the
usual idea of an average as a value obtained by averaging up
a number of distinct values.
Consider the actual average of n values of the force F
selected at equal intervals of distance, Ax, from x = a to x = b :
av. of n values
The larger we take n, the nearer this will come to what we
would call the average of all the force-values between a and 6.
To see what the limiting value is, which this approaches,
multiply numerator and denominator by Ax :
av. of n
Now as n becomes indefinitely great, Ax-»-0, and by the Fun-
damental Theorem of § 292, the limiting value of the numera-
XII, § 302] DEFINITE INTEGRALS 409
Xi
F dx. The denominator nAx is simply the whole
_
distance from x = a to x = b,i.e., (b—a). Hence we consider
the average of all the force-values between a and b to be
C
\
—a
Fdx
6-0 '
the same value as that obtained in (16).
Evidently the same argument would lead to the following
definition of the average value of any varying quantity
V =f(x) from x = a to x = b :
- (17)
; b-a
Geometrically interpreted, the numerator of (17) is the area under
the curve y=f(x). So that the average height y is simply the area
divided by the length of base considered, — agreeing with our definition
of mean ordinate in § 13.
Ex. I. If y =xz, find the average value of y between x = 1 and x = 5.
i
§ 302. Simpson's Rule. Many of- the quantities con-
sidered in the natural and social sciences are representable
by the area under some graph, and can be approximated by
measuring that area. It is desirable, however, to be able
to approximate such an area without plotting and without
integrating.
The problem is virtually to find the average height of the
graph, which, multiplied by the base, would give the area.
The following rule, devised about 1750 by Thomas Simp-
son, an Englishman, gives excellent results for many
curves :
410 MATHEMATICAL ANALYSIS [XII, § 303
The average height yfrom x = a to x = b is found by averaging
the heights at a and b with four times the height at the middle :
2/ =(</«+ 2/6+4 i/mH 6. (18)
And the area under the graph is this y times the base (b — a) :
-(b-a). (19)
Ex. I. Find the area under the parabola y = x2 from x = 2 to x = 10.
The first height, at x = 2, is y = 4. The final height, at z = 10, is
I/ = 100. The middle height, at z = 6, is i/ = 36. Hence by (18) :
£ = [4+ 100+4(36)] +6 = 41f
Multiplying y by the base, from z=2 to 10, gives the area: A =330f.
Remarks. (I) In this case we can check by integration :
= f y dx = f 10z2 dx = \ (992) = 330f .
But the rule is most useful when integration is impracticable.
(II) The rule gives exact results whenever the formula for the height
of the graph is of the first, second, or third degree. (This is proved in
the Appendix, p. 490.) Generally it gives only an approximation, — but
a close one if the interval is small. A large interval may be split into
smaller sub-intervals, and the rule applied to each. In fact, Simpson's
rule is usually stated in a more general form than (19), giving the total
result for n intervals.
p
§ 303. General Applicability. Any definite integral, what-
ever its original physical meaning, can be thought of as the
area under some curve, and hence can be approximated by
Simpson's Rule.
E.g., suppose the work done by a force was to be found from
This same integral would give the area under the curve
2/ = Vx2-fl from x=l to 5. So to approximate the integral
we would simply calculate the value of Vl-j-x2 at 1, 3, and 5,
average according to. the rule, and multiply by the length of
the interval, 4.
XII, § 304] DEFINITE INTEGRALS 411
In general, we would proceed similarly with any function f(x) to be
integrated, — getting its values at the beginning, middle, and end, etc.
Thus, approximately,
(*) dx = itfla) +/(&) +4 /(*»)] (b -a) , (20)
where xm is the value of x midway between a and b.
EXERCISES
1. Find the approximate area under a curve from # = 10 to a: = 20
if the height of the curve is 8 at x = 10, 12 at x = 15, and 6 at x = 20.
2. Find by Simpson's Rule the area under y=x2 from x =3 to 5.
Check by integration. What is the average height?
3. The same as Ex. 2 for the curve y=x3 from re = 2 to 6.
4. How much error is there in the area under y = l/x from x = 1 to 5,
as given by Simpson's Rule without subdividing?
6. What area problem would call for the calculation of the integral,
r16 /-
J4 Vx dx? How accurately does Simpson's Rule give this area?
6. Find by Simpson's Rule the integrals :
(a) (g\^dx, (b) ( 8(x*+Gx + l)dx, (c)
7. Express as a definite integral, and then approximate by (20) :
(a) The work done from x = 2 to z = 6 by a force varying thus:
F = .2z2. (Check.)
(6) The volume from a; = 0 to x = 4 of a solid whose cross-section area
varies thus : A = Va£+l.
8. Find the exact mean value of each of the following :
(a) Of x5 from x = 0 to x = 4 ; (6) Of 1/rc2 from x = 1 to 3 ;
(c) Of Vx~3 from x = 0 to x =4 ; (d) Of sin x from x = 0 to v/2.
9. For an electric current : i = 20 sin 400 t. Find the mean current
flowing during a half-period.
10. For a pendulum : w = .2 cos irt. Find the mean angular velocity,
w, during an upward swing.
§ 304. Prismoid Formula. Applied to volumes, Simp-
son's rule may be stated in a simple form, called the Pris-
moid Formula.
412
MATHEMATICAL ANALYSIS [XII, § 304
The volume of a solid between two parallel planes is
/(a)
B=f(b)
(§97)
The values of the area-func-
tion A at the beginning and
end are simply the areas
of the bases B and B'.
(Fig. 142.)
where M is the area midway
between bases, and h (or
6— a) is the height. That is, the volume of a solid equals its
height times the average cross-section area, found by averaging
the bases with four litres the mid-section.
Ex. I. A uniformly tapering timber 20 ft. tall has square bases
2X2 and 8 X8 inches. Find its volume.
The mid-section is a square, 5X5. Hence
5 = 4, £'=64, M = 25, A -20(12) =240.
/. 7=1(4+64 + 100)240 = 6720.
This value is exact. For the cross-section area varies as the square
of the distance x from a certain point. (What point?)
EXERCISES
1. Find by the Prismoid Formula the volume of a cone of height 12 in.
and base radius 4 in. Check by geometry.
2. The same as Ex. 1 for a sphere of radius 20 in.
3. A uniformly tapering timber 30 ft. long has ends 1 ft. square and
2 ft. square. Find its volume.
4. Find the approximate volume of a barrel 30 in. long with a radius
of 12 in. at each end and 15 in. at the middle, — inside dimensions.
6. A solid has three mutually perpendicular elliptic cross-sections,
with semi-axes of 12 in., 10 in., and 8 in. Find its volume.
6. The frustum of a cone is 18 in. tall and has base radii of 4 in.
and 8 in. Find its volume.
XII, § 305] DEFINITE INTEGRALS 413 '
7. The base of a granite column is 3 ft. high and has a radius of 2 ft.
at the top, 3 ft. at the middle, and 5 ft. at the bottom. Find its approxi-
mate volume.
8. Every horizontal section of a certain solid z in. above the lowest
point is an ellipse : z2/36 z+y2/l6 z = 1. (What axes have the sections
at 3 = 10, 1, 0?) Draw the solid from 2 = 0 to z = 10. Also find its
volume by integration, and by the Prismoid Formula.
9. A sphere of radius 10 in. is cut by two parallel planes 6 in. and 8 in.
from the center, on the same side. Find the included volume by the
Prismoid Formula.
10. Find the volume in Ex. 9 by integration.
§ 305. Summary of Chapter XII. Any quantity ex-
pressible as the limit of a sum, of a certain form (§291), is
equal to a certain definite integral, i.e., the difference of two
values of an indefinite integral.
E.g., the area under a curve y =f(x) from x = a to x = b is
A= L \_yi&x+ ••• +ynkx]= | ydx,
Az-X) «/a
i.e., A=F(b)-F(a),
where F is the integral function, whose derivative equals y
or f(x) . Further cases are listed in the Appendix, p. 498.
Originally, a definite integral was regarded as the sum of
numerous infinitesimal elements. But the formulas then
used in " summing " did not give the sum. They were the
same as our formulas for integrating, and thus gave the limit
of the sum, — the value really required.
In finding the volume of a solid two successive integrations
may be required : one to find the area of a cross-section, and
the other to get the volume itself.
To show graphically how a quantity z varies with two
other quantities x and y on which it depends, we must draw
a surface. The variation of z with x or y alone is shown by a
section of the surface perpendicular to the Y or X axis. The
rate of change is then dz/dx or dz/dy.
414 MATHEMATICAL ANALYSIS [XII, § 305
The mean value of a function is conveniently defined in
terms of definite integrals. Its value may often be ap-
proximated by Simpson's Rule.
EXERCISES
(Integrate may be found from the Table, p. 494, if necessary.}
1. Find the area bounded by the curve y* = x3, the X-axis, and the
ordinate at x = 5.
2. Find the length of the curved arc considered in Ex. 1.
3. Find the area bounded by the X-axis and the curve 7/ = l— x2.
(What kind of curve is this, and how located?)
4. Find the volume under the surface z=xy, above the part of the
.XT-plane from y = Q to y=xz and from z = 0 to rr = 2. Plot.
6. Find the volume from z = 0toz = 3o£a solid whose every hori-
zontal section is an ellipse, with one of these equations :
, . xz i ?/2 i /t\ x2 i y* i
1 lOOz 642~ ' 100(1 +22) 36(1 +z2)"
6. The thickness (T in.) of certain pavement x ft. from the middle
line is : T = 3 + .003 x2. Find the average thickness, x = 0 to x = 10.
7. In Ex. 7, p. 398, change the length of the beam to 20 ft., and
let the load vary thus : L = 6 x. Find the total moment.
8. In Ex. 10, p. 398, change the radius to 20 in., and let w vary
thus : w = 6— .4 x. Find the total weight.
9. Find by Simpson's Rule the area under the curve 2/=sin x from
0 to 7T/2, — cutting this into three parts. Compare the exact value.
10. If a circle of radius 10 in. be revolved about a line 12 in. from
its center, what sort of surface will be generated ? What sort of section
will be made by any plane perpendicular to the axis of rotation ? Find
the volume.
11. A tunnel has a level floor 12 ft. wide and arched walls forming a
parabola with the vertex 16 ft. high. If it is full of water, what will
be the pressure against a vertical end of the tunnel ?
12. The capstone of a monument has square horizontal sections,
whose vertices all lie on two vertical semicircles of radius 3 ft. Find
the volume of the cap, if 2 ft. high.
13. In a certain type of tank the weight of steel used in the wall
varies thus with the diameter x and height y : W= .01 zV- Plot a sur-
face showing this variation, from x = 20 to 100 and y=0 to 10.
CHAPTER XIII
PROGRESSIONS AND SERIES
INVESTMENT THEORY — CALCULATION OF
FUNCTIONS
§ 306. Problems to be Considered. We have now defined
and studied briefly the following kinds of functions : loga-
rithmic, exponential, trigonometric and power functions,
derivatives and integrals. To round out our knowledge of
these various functions, — as far as practicable in an intro-
ductory course, — we shall now consider a very general
method of finding their values, by which we can calculate
tables and discover relations among the different functions.
As a preliminary, however, we must recall certain formulas
of elementary algebra. Incidentally we shall see how the
fundamental problems of the Theory of Investment are
solved.*
§ 307. Arithmetical Progressions. A series of numbers
like
3, 7, 11, 15, 19, etc.,
which have a constant difference is called an Arithmetical
Progression, — abbreviated as A. P.
If a denotes the first term and d the constant difference,
any A. P. may be written
a, a+d, a+2d, a+3 d, . . .
* In business such problems are usually solved by merely consulting
tables of interest, discount, annuities, etc. But our methods will handle
cases not covered by the tables (cf. Ex. 2, p. 438 and Ex. 18, p. 482), and
will show also how the standard tables were first obtained.
415
416 MATHEMATICAL ANALYSIS [XIII, § 308
The fourth term is a +3 d. What would the 10th be? The 17th?
If there are n terms in all, the last one is evidently
l=a+(n-l)d. (1)
A formula for the sum of all the terms is easily found.
Write the terms in the order above and also reversed :
Add, and observe that each sum on the right reduces to
(a+l) :
.: 2S=(a+l)+(a+l) ••• +(a+l)=n(a+l).
.: S=%a+l). (2)
A i
This formula has a simple interpretation which makes it easy to
remember: The average value of the terms is Ka~H)> and the sum
equals this average value multiplied by the number of terms.
Ex. I. Find the last term and sum of the A. P., 2, 5, 8, 11, . . . to
20 terms.
Here a = 2, d=3, n = 20.
Z = 2+19(3)=59, S = 2^(2+59)=610.
Ex. II. A debt is to be paid off in 30 payments running as follows :
$200, $195, $190, etc. Find the total amount to be paid.
The payments form an A. P. with a = 200, d= — 5, n = 30. Hence
the last term is J = 200+29(-5) =55; and
S=Y(200+55)=3825.
§ 308. Geometrical Progressions. A- series of numbers
like
2, 6, 18, 54, 162, etc.,
each of which equals the preceding multiplied by some con-
stant, is called a Geometrical Progression, — abbreviated G. P.
If a denotes the first term of &G. P., and r the constant
multiplier or ratio, the progression may be written :
a, ar, ar2, ar3, ...
XIII, § 308] PROGRESSIONS AND SERIES 417
Here the 4th term is ar3. What would the 10th term be ? The 17th ?
If there are n terms in all the last one is evidently
l=ar"-1. (3)
A formula for the sum of all the terms,
S=a+ar+ar*+ ••• -far-1
is easily found. Multiplying by r and subtracting S from
rS gives
.'. rS-S=-a+ar*.
Factoring both sides and solving for S, we have,
S=?^l. (4)
Ex. I. Find the last term and the sum of the G. P., 2, 6, 18, etc., to
20 terms :
Here a»2, r-=3, ra*=20, whence J = 2(3)13 and
Approximate values of I and <S can be found quickly by using loga-
rithms.
EXERCISES
1. What is the charge for ten postage stamps, one of each denomina-
tion from 1^ to 10^? Check by addition.
2. Find the sum of the first 30 odd integers. Of all odd integers
<100.
3. Find the sum of the first 20 even integers. Of all even integers
<175.
4. Find the sum of all integers between 100 and 800 which end in 3.
Of all integers between 1 and 999 which are divisible by 5.
5. 500 raffle tickets are sold, at all prices from 10 to $5.00. What
are the total receipts ?
6. In a contest there are to be 12 prizes: $500, $475, $450, etc.
What will the total amount be ? Check.
7. If we make 10 monthly deposits of $60 each, beginning now, and
are allowed simple interest at 4%, how much will there be to our credit
one year hence?
8. In the G. P., 8, 12, 18, . . ., what are a and r? Express by a
formula the 12th term ; also the sum of 30 terms.
418 MATHEMATICAL ANALYSIS [XIII, § 309
9. The same as Ex. 8 for the G. P., 900, 600, 400, . . ..
10. Each stroke of a pump removes one fourth of the air remaining
in a vessel. What fraction of the original weight will remain after 20
strokes?
11. In the G. P., 3, 6, 12, . . ., find by formula the 10th term and the
sum of the first 6 terms. Check by direct calculation.
12. In the G. P., 100(1.06)6, 100(1.06)«, . . ., 100(1.06)20, what are
a, r, and n? Find the sum.
13. If we make 10 annual deposits of $500 each, beginning now,
and these draw interest at 3 5%, compounded annually, how much will
there be to our credit 30 years hence? (Before trying to sum up,
express the amount accumulated by the 1st deposit, 2d deposit, and last
deposit.)
14. A present debt of $600 on a piano is to be paid in monthly install-
ments of $30, plus accrued interest at 8%. How much will the first
payment be, 1 mo, hence? The last payment? The total amount
paid?
15. The same as Ex. 14 for a debt of $1000 on an automobile, to be
paid off in monthly installments of $100 with 6% interest.
16. A sewer assessment of $225 is payable $22.50 per year, beginning
now, plus interest at 6%. How much will be paid in all?
17. The problem of dividing an octave into 12 equal semitones
is mathematically the same as the problem of inserting 11 "geometric
means" between 1 and 2, — i.e., 1 and 2 are to be the first and last terms
of a G. P. of 13 terms. Find the third and seventh terms.
18. Carry out in detail the derivation of formula (4), as given in
§ 308, in a special case, say n = 7, writing out all the terms. Then
check by dividing out (r7 — l)-?-(r — 1) and getting the original G.P.
§ 309. Investment Problems : Accumulation. In business
it is often necessary to find how large a fund will be accumu-
lated at a specified date by making certain deposits or pay-
ments at stated intervals. Or, conversely, how large the
payments must be to yield a certain sum finally.
To solve such problems, simply express the amount ac-
cumulated on each payment at the end of the time, using
the interest formula :
A =
XIII, § 309] PROGRESSIONS AND SERIES 419
Then sum up. This is quickly done when the amounts form
B.G.P.
Ex. I. If we make 20 annual deposits of $100 beginning
now, and are allowed 4% interest compounded annually, how
much will there be to our credit 30 years hence ?
1st deposit, with 30 yrs.' int. will amount to 100(1.04)30
2nd deposit, with 29 yrs.' int. will amount to 100(1.04)"
20th deposit, with 11 yrs.' int. will amount to 100(1.04)"
The 20th deposit, being made at the beginning of the 20th year, or
19 years hence, and running until 30 years from now, will be at interest
for 11 years.
These amounts form a G. P. For, starting from the
bottom, if we multiply any of them by 1.04, we obtain the
next above.
/. o = 100(1.04)u, r=1.04, n = 2Q.
The total amount to our credit 30 years hence will be the
sum:
Remarks. (I) In calculating S by logarithms, we must go from
the logarithm of 1.0420 back to the number before subtracting the 1.
The denominator, .04, may be canceled into the 100 to save work.
(II) The amounts above also form a G. P. starting from the top.
But then the constant multiplier r is ^, which is inconvenient.
(III) In the G. P. formula n denotes the number of terms to be added,
i.e., the number of payments, — not some number of years. Also r
denotes the ratio for the G. P., — not the interest rate.
Before trying to sum, always write out a few terms of the G.P., to
recognize a and r ; and be especially careful as to the time that each
payment draws interest. The number of years elapsing between the
first and last payment is one less than the number of payments.
* Five-place tables will not give this very accurately.
420 MATHEMATICAL ANALYSIS [XIII, § 310
§ 310. Life Insurance. To determine what " annual
premium " to charge for insuring a man's life, an insurance
company must ascertain what sum, invested annually at a
certain estimated rate of interest, would yield the amount of
policy at the expected time of death. (It is known from
Tables of Mortality that an average group of men at the age
of the insured have a certain number of years to live.)
The company plays safe by insuring no one whose health
is poor, and by figuring interest at a lower rate than is
actually earned. The premium charged is thus larger than
necessary; and the excess charge is returned annually as a
" dividend." The annual premium, by the way, includes a
charge to cover administrative expens.es. But we seek here
only the "net premium" which goes to provide the face of
the policy.
Ex. I. Fifteen annual premiums of $P each, beginning now, are to
provide a $10,000 policy 40 years hence. Find P, figuring interest
at 3i%.
The last premium payable 14 years hence will draw interest 26 years.
1st premium, with 40 yrs.' int., will yield P (1.035)40
2d premium, with 39 yrs.' int., will yield P (1.035)39
last premium, with 26 yrs.' int., will yield P (1.035)28
These amounts, starting from the bottom, form a G. P. in which
a = P(1.035)2«, r = 1.035, n = 15.
The sum of these accumulated amounts must be $10,000 :
35°
1.03526[1.03516-1]
By logarithms we find P =211.89, — the net annual premium.*
* In actual practice premiums are not calculated in this way. It ia
necessary to figure on the cost of a large group of policies rather than a
single " average " policy. The actual procedure is explained in treatises
oil life insurance.
XIII, § 310] PROGRESSIONS AND SERIES 421
EXERCISES
Here and in what follows all interest is to be compounded annually
unless otherwise specified.
1. Twenty annual deposits of $600 each are to be made beginning
now. With 4% interest, what will the total accumulation be 25 years
hence ?
2. What sum deposited now at 4% would yield the same final amount
as in Ex. 1 ?
3. How much should a firm set aside annually beginning one year
hence and ending 30 years hence to replace a $100,000 building at the
time of the last deposit if interest is at 4£%?
4. When a boy was 1 year old, his father began depositing $50 a
year in a bank which paid 4% interest, compounded semiannually.
What sum had been accumulated when the boy reached the age of 17?
(Include the final $50.)
5. A city has just issued bonds for $500,000 to construct an audi-
torium. What sum raised annually by taxation, beginning 1 year
hence, and set aside at 6% interest, will meet the face of the bonds
when they mature 10 years hence?
6. A bridge costing $35,000 must be replaced every 20 years. What
sum set aside annually for 10 years, beginning 1 year after the construc-
tion of the bridge and drawing 5% interest compounded quarterly,
would yield enough for the first renewal ?
7. In a certain society the annual dues are $5, payable in advance.
A life membership costs $50. If a member lives 15 years after joining,
which arrangement would have been the more economical, figuring
interest at 6% on any dues paid in either way?
8. What net annual premium, payajble at the beginning of each year,
would provide for. a life insurance policy of $1000, 38 years hence, if the
company earns 5% net?
9. As in Ex. 8, find the premium in each of the following cases :
Time to run Interest Rate Face of Policy
(a) 40 yrs. 5.2% $ 2,000
(6) 35 yrs. 6% $ 1,000
(c) 35 yrs. 3% $ 1,000
(d) 42 yrs. 4.8% $10,000
(e) 25 yrs. 4.5% $ 1,000
10. A balance of $1125.60 owing on a mortgage Sept. 1, 1916, was
paid in monthly installments of $37.50, including accrued interest at
422 MATHEMATICAL ANALYSIS [XIII, § 311
7% and beginning Oct. 1^ 1916. How much was still due after the
payment of Jan. 1, 1919? (Figure two separate funds: the amount
which would have been due if nothing had been paid ; and the amount
which would have been accumulated by the payments if not applied
on the debt. Subtract.)
§ 311. Present Value. Money payable at some future
date is not worth the same amount now. Its " present
value " is only so much as would have to be invested now
(at the prevailing rate of interest), to yield the specified
payment at the time when due.
E.g., if money will now earn 6%, compounded annually,
the present value of $1000 payable 10 years hence is only
$558.40. For calculations show that $558.40 invested now
at 6% would yield $1000 in 10 years. '
A formula for the present value of any amount A , payable
n years hence, is obtained immediately from the interest
formula:
A =
:. P=
p. v.
For the present value is simply the principal which will yield
A. This " P. V" formula is useful in many investment
problems.
Observe that, the more remote the payment of a ^_
sum, the smaller its P. V. Thus the P. V. of $1000 o
after various intervals (n years) decreases approxi- 5
mately as in the adjacent table, — if interest is 15
at 6%, compounded annually. 20
1000
747
558
417
312
§ 312. Investment Problems : Disbursement. A very
common problem in business is this : To determine how
large a sum would have to be deposited now to provide for
XIII, § 312] PROGRESSIONS AND SERIES 423
certain stipulated payments at specified future dates. Or,
conversely, how much must be paid at stated intervals to
use up a certain original sum, or pay off an original debt, in
a certain length of time.
All such problems, where an original sum is to be dis-
bursed, are conveniently solved by using the P. V. formula.
For problems in which a final amount is to be accumulated,
the interest formula is best.
Ex. I. How much must we deposit now to get back 20 annual
installments of $600 each, beginning 15 years hence, if interest is at 4%,
compounded semiannually ?
The last $600 is due 19 years after the first, or 34 years hence.
P.V. of first $600, due 15 years hence, is 600
(1.02)3°*
P.V. of second $600, due 16 years hence, is 600
P.V. of last $600, due 34 years hence, is 600
(1.02)68*
The sum of these present values is the total amount we must deposit
now to get back all 20 installments. In this G. P.,
60°
..
(1.02)68 1.022-1
By logarithms/ S = $4667, the total amount to be deposited now.
Setting aside the sum is called "capitalizing the annuity."
Ex. II. $2000 now due on a house is to be paid off with interest
at 8% hi 60 equal monthly installments beginning 3 months hence.
How large must the installments be?
The last payment will be due 59 months after the first, or 62 months
hence. The interest is figured monthly, making k = l2 and
r/k = .08/12 = .0066667. If each payment is $A, then
P. V. of 1st payment, due 3 months hence, is
^ J. .
P.V. of last payment, due 62 months hence, is
.
424 MATHEMATICAL ANALYSIS [XIII, § 312
The sum is the total present value of all the payments and should equal
the present debt, viz., $2000. Or since
°=a^7p r=LOOG7' n==60'
. s= 4- a.QQ676°-i)_2000
(1.0067)62 .0066667
. A ^2000(.0066667)(1.0067)<g
1.006760-!
By logarithms, A =41.09, — the monthly installment.
Observe that to round off the factor .0066667 as .0067 would produce
a considerable error in A. But to round off 1.0066667 as 1.0067 does
not.
Observe, too, that we have not credited against the present debt
of $2000 the full amount ($A each) of the coming payments, but only
the present values of those payments. The total of the sixty installments
is 60 A, or $2465.40; so that the debtor pays'interest amounting in all
to $465.40.
EXERCISES
1. Find the present value of $15,000 payable 20 years hence, if
interest is at 3£%, compounded semiannually.
2. Verify the third and the last values shown in the table of P.V.'s
on page 432.
3. How much, deposited to-day, would provide for 20 annual pay-
ments of $800 beginning 25 years hence, if interest is at 4£%?
4. The same as Ex. 3, for 10 annual installments of $600 beginning
10 yr. hence, if interest is at 4%.
6. If we invest $10,000 now at 6%, how much can we get back each
year, 20 times, beginning 15 years hence?
6. A balance of $2500 now due on a house is to be paid off in 50
monthly installments beginning one month hence, with interest at 8%.
Find the installment.
7. The same as Ex. 6, for a balance of $2000 to be paid in 40 monthly
installments beginning 3 months hence, with interest at 6%.
8. How much deposited now would yield forty quarterly installments
of $300 each, beginning 10 years hence, if interest is at 4%, compounded
quarterly?
9. Express by a formula the amount of the installment if :
(a) A balance of $500 on an auto is to be paid off in 8 monthly
installments beginning one month hence, with interest at 6%.
XIII, § 313] PROGRESSIONS AND SERIES 425
(6) A balance of $3500 on a house is to be paid off in 9 annual install-
ments beginning one year hence, with interest at 8%.
(c) An Insurance Company which earns 5% is to make 60 quarterly
payments beginning now, in lieu of paying $10,000 now.
10. (a)-(c) . Calculate the values of the installments in Ex. 9, (a)-(c).
11. How expensive a house can be purchased by paying $1500 down,
plus 96 monthly installments of $60 each, if interest is at 6%, computed
monthly ?
12. A bond calls for the payment of $1000 ten years hence. If
money is worth 4%, compounded semiannually, what is the present
value of the future payment?
§ 313. Valuation of Bonds. When a corporation or a
government issues a bond, it promises to pay on a certain
date the sum specified in the bond, and meanwhile to pay at
stated intervals a certain amount of interest.
The market value of a bond, prior to the date of maturity,
is usually different from the face value, — due to the fact that
the rate of interest prevailing in the money market rarely
happens to be the same as the rate named in the bond.*
Normally, the market value of a bond is simply the sum
of the present value of the principal payable at maturity,
plus the present values of the several interest payments to
be made on specified dates.
Ex. I. A $1000 Liberty Bond maturing 10 years hence
carries interest at 4^%, payable semiannually. Calculate
its present value, assuming money now worth 5%, com-
pounded semiannually.
The government is to pay $1000 ten years hence ; and also
pay $21.25 interest every half-year, beginning 6 months
hence, until maturity. The P. T.'s of the 20 interest pay-
ments, with money now at 5%, are
21.25 21.25 21.25
(1.025)' (1.025)2' '(1-025)20'
* The value is affected also by the nature of the security, privileges of
conversion or tax-exemption, etc., — factory which cannot be discussed here.
426 MATHEMATICAL ANALYSIS [XIII, § 313
or in all :
s= 21.25 = 001 27
(1.025)20 .025
The present value of the principal, $1000, payable ten years
hence, is
The total, S+P = 941.54, should be the present price of the
bond.
N.B. The rate of interest named in the bond merely fixes tho
amount of the interest installments. All present values are determined
by the rate which money is now worth.
Observe, too, that the government will pay on the bond $1000 +
20(21.25), or $1425 in all, but the P. V ofi these payments now is only
$941.54.
EXERCISES
1. A $1000 bond maturing 20 years hence carries 4^% interest,
payable semiannually. If money is now worth 5%, compounded semi-
annually, what is the present value of the bond ?
2. Express by formulas the present values of the following bonds :
Description of Bond Maturity Market Int.
(a) $1000, 4-|%, semi. 15 yrs. hence 6%, semi.
(b) $500, 5%, quart. 20 yrs. hence 4.8%, quart.
(c) $5000, 4|%, semi. 10 yrs. hence 5%, semi.
(d) $10,000, 6%, quart. 8 yrs. hence 5.8%, quart.
(e) $50, 4%, semi. 12 yrs. hence 8%, semi.
3. (a)-(e) Calculate the values in Ex. 2 (a)-(e).
4. How much must we deposit in a bank annually for 20 years,
beginning now, in order to draw out $600 a year for 15 years beginning
30 years from now, if interest is at3|%? (Hint: Equate the total
present value of all the deposits to the total present value of all the
withdrawals.)
6. (a) Express by a formula the amount which we could draw out
quarterly in 60 installments beginning 20 years hence, if we deposit
$300 semiannually 30 times beginning now, and interest is at 4%.
(6) Calculate the amount in (a).
XIII, § 315] PROGRESSIONS AND SERIES 427
§ 314. Infinite Series. It is sometimes necessary to deal
with aG.P.OT other series of terms which runs on indefinitely,
- never ending. Such an " infinite series " cannot in any
literal sense be said to have a sum. But we may need to
find the sum of any number of terms, and see what happens
as more and more terms are added on.
To illustrate, consider the simple series
1, —x, +z2, —xs} -fz4, ... (unending).
The sum of the first n terms (a G. P. having a = 1, r = — x) is
_l.[(-x)"-l]_ 1 .. ,
(_X)-1 -i+x L-
If x is numerically less than 1, then xn->Q as n-^oo. And the
limit of Sn is simply l/(l+x).
This limit is called the sum of the series to infinity. We
write simply
.. (6)
We do not mean by this that the fraction equals the sum of several
terms, but that it is the limit approached by the sum as more and more
terms are taken. The idea is the same as when we write
The fact that the fraction is the limit approached, ^§jjjsa expressed by
saying that the series " converges toward the value of the fraction."
Remember, however, that it does so in the example above only when
x is numerically less than 1.
§ 315. An Application. In more advanced courses it is
proved legitimate to differentiate or integrate an infinite
power series term -by term : the resulting series will equal
the derivative or integral of the function represented by the
original series. Let us integrate both sides of (6) above :
(7)
When x = 0, this gives log 1 = C. .'. C = 0.
428 MATHEMATICAL ANALYSIS [XIII, § 316
This equation is valid if x is numerically less than 1. E.g.,
letting z = .l,
log 1.1 = .1-^+^ ••• =1-.005+.0003 .-.,
'_ o
.-. log 1 . 1 = .9953, approx. (Base e.)
In this way we can calculate the logarithms of numbers near
1. The logarithms of larger numbers are found by a com-
bination of series.
Remark. In obtaining the differentiation formula for log u (§ 177),
— used above in integrating (6), — we employed tables of logarithms
of numbers near 1. But that formula can be obtained otherwise.
Hence (7) could have been used to calculate the tables originally.
§ 316. Maclaurin's Series. Equation (7) above expresses
the function log (1+z) as the "sum" of an infinite series
of powers of x. This suggests that perhaps many other
functions, such as sin x, ex, etc., might be similarly expressed.
This is indeed the case ; and it is easy to determine pre-
cisely what the series should be for any ordinary function.
An example will make the process clear.
Ex. I. Assuming that cos x equals some series of the form
cosx = A + 5x+Cx2+Dx3+fe4+ •••, (8)
find what the coefficients A, B, C, etc., must be.
The method is simply to differentiate several times, and
then substitute x = 0 in each of the resulting equations, -
and in the original equation.
Substituting x = 0 in (8) gives at once cos 0 = A, or A — 1.
Differentiating (8) repeatedly :
4
-cosz= 2C +6D
+sino; = 6D +24 Ex +
24 # +
Putting x = 0 :
.'.-sin 0 = 5, = 0
.'. -cosO = 2C, =-1
/.sin 0 = 6 D, = 0
XIII, § 317] PROGRESSIONS AND SERIES 429
Continuing thus we have A = l, 5 = 0, C=— ^, Z> = 0,
E=2l$, etc. Substituting these values in (8), that series
becomes
or simply
/y-2 /r4
l-+-.. (9)
That is, if cos x is the " sum " of any series of the form (8),
this must be the series. It can also be proved that this
series (9) does actually approach cos x as its limiting value,
no matter how large x may be.
Remarks. (I) In (9) x is necessarily the number of radians in the
angle. (Why so?)
(II) The possibility of expanding many functions into series of the
form (8) was discovered by C. Maclaurin, a Scotchman, about 1740;
and series of this type are called Maclaurin series. These are, however,
only a special case of a more general type of power series expansion
discovered by B. Taylor, an Englishman, about 1715.
(III) Maclaurin series are useful not only in calculating values of a
function (§§ 315, 318), but also in performing integrations otherwise
difficult or impossible. For instance, from (9) we could find :
x \2 24 4 96
an integral not otherwise obtainable by elementary methods.
§ 317. Factorial Notation. The product of the integers
from 1 to n inclusive is called "factorial n" and is denoted
by n I Thus 5 ! = 1 • 2 • 3 • 4 • 5 = 120. And so on.
By means of this notation the series for cos x in (9)
above can be written more conveniently. Thus the de-
nominator 24, which arose as 4 • 3 • 2 • 1 in the repeated
differentiations, is simply 4 !. Likewise the denominator 2
may be written 2 !, and series (9) becomes
l-'+ M... (10)
430 MATHEMATICAL ANALYSIS [XIII, § 318
According to this beginning, how should the series continue? Cam
you tell from the derivation of (9) whether your inference is correct ?
EXERCISES
1. Find Maclaurin's series for e* as far as x5. By inspection write
three more terms. What would be the term containing x20? xn?
2. In the series of Ex. 1 substitute x = 1 and thus calculate e approx-
imately. (The calculation can quickly be carried to many decimals;
for dividing the third term by 3 gives the fourth, dividing this by 4
gives the fifth, and so on.)
3. Differentiate the series for e* and note the result carefully.
/• _£ _ -I
4. Approximate \ -- dx as far as x4.
6. What does the series for e* become if we replace x throughout by
—x2? Use the resulting series to find f \e~x<i dx to 5 decimals.
«/o
6. Find Maclaurin's series for sin x, as far as x5. Write three more
terms by inspection.
7. Calculate sin .2W and compare the tables.
8. The same as Ex. 7 for sin 1°. (Express 1° in radians.)
9. By differentiating the series for sin x obtain a scries for cos x,
and compare (10) above.
10. Derive the cosine series also by integrating the sine series.
Note the constant of integration.
11. (a) Findf* *™*dx. (6) Find f ' l~cosxdx.
Jo x Jo x2
12. Expand (a+z)6 into a Maclaurin series as far as possible, [a is
constant. 1
13. Expand (a+z)n as far as z5.
[14.] In the series for e* in Ex. 1 let x = V — 1 u, and simplify.
Collect separately the terms free from V — 1 and those involving it,
and compare the cosine and sine series.
§ 318. Calculation of Trigonometric Tables. The sine and
cosine of 1° are easily calculated by substituting the radian equivalent
of 1°, viz., x = . 017453, in the Maclaurin series:
(U)
XIII, §319] PROGRESSIONS AND SERIES
431
The Addition Formulas, § 280, will then give the sine and cosine for
(1°+1°) or 2°; then for 3°, 4°, etc., to 45°. Beyond 45° we need not
go. (Why not?) Tangents can be found from the sines and cosines.
(How?)
The Half -Angle Formulas, § 283,. can be used for certain fractions of
1°. Or sin 1' and cos I7 can be calculated from the series (11), making
a fresh start.
Still other methods were used in calculating the tables originally.
§ 319. Binomial Theorem. The standard formula for
expanding (a+u)n is obtainable by Maclaurin's method :
{A=an
Cu+3 Du2+4 Eu?
= 2C+6Du+12Euz
B = nan~l
C = n(n- !)„„_,
2!
n-2
This formula is called the Binomial Theorem. Notice
how the terms run. The exponent of a is n at first and de-
creases by 1 at each step. The exponent of u increases si-
multaneously, keeping the sum of the two exponents always
equal to n. In the coefficients, notice the factorial de-
nominators ; and also that each new factor in the numerator
is less by 1 than the preceding. (Judging by the fourth
term, what would the sixth be?)
From these facts you can see that the rth term will involve
(r — 1) ! in the denominator, ur~l, an~(r~l\ and factors from
n down to (n—r—2) in the numerator :
rth
(13)
(r— 1)1
If n is an integer, the series (18) will end presently. (When?)
432 MATHEMATICAL ANALYSIS [XIII, § 319
If n is a fraction the series never ends ; but, as is proved
in higher algebra, the series converges, and the sum of
r terms approaches the value of (a+it)n, provided u<a
numerically.
Certain facts as to the Binomial Theorem were pointed out in obtain-
ing the differentiation formula for xn. (§56.) We have now gone
farther, and have used that differentiation formula to get fuller infor-
mation about the Binomial Theorem. An independent and purely
algebraic proof of this theorem can also be given.
The Binomial Theorem has many uses, both in making
numerical calculations and in integrating radical forms which
will yield to no other treatment.
/r2 _
Ex. I. Find Vl-x3 •
^
dx
The radical is a case of (a-\-u)n, where a = l, u= —x3, n = $.
Clearly every power of " a " equals 1 here and need not be
written. Thus :
.*(-*)(-*)/ ^Y\
3!
= .2 -i (.0016) -••• = . 1998.
This definite integral could be approximated by Simpson's Rule;
but here we have found the indefinite integral also, valid if x< 1.
EXERCISES
1. Write the following expansions, and simplify the terms :
(a) (2+z)10, as far as x6, (6) (3 -a:)4, complete.
(c) (1+z4)8, as far as z12, (d) (2 +z2)5, complete.
(e) (1 -z2)*, as far as x6, (/) (1 -z4)-*, to x*.
(0) (l-x)-», as far as x4, (/i) (3+x*)~2, to x12.
XIII, § 320] PROGRESSIONS AND SERIES 433
2. Find 1/(1— x} by division as far as x* and compare Ex. 1 (0).
From this series find by integration a series for log (1 — x).
3. From the series in Ex. 2 for log (1 — x) calculate :
(a) Gog .95), (6) Jo3 log (1 -x) dx.
4. Find I v 1+a:3 dx approximately by expanding to four terms.
6. The same as Ex. 4, for f ' dx/^/l— x*
Jo
6. Find "v'l.OG approximately by regarding this as (1+.06)^ and
expanding to three terms.
7. The same as Ex. 6 for V^982 [ = (1 -.018) i].
§ 320. Relation of Exponential to Trigonometric Func-
tions. There is a very close connection between the ex-
ponential series
and the sine and cosine series
'(15>
as will be obvious if we let x have an imaginary value, say
x=V — l u, in (14).
Or, separating the real terms from the imaginaries,
434 MATHEMATICAL ANALYSIS [XIII, § 321
The two sets of terms in parentheses are seen to be simply
the cosine and sine series (15). Hence
=cos u+ sin u. (16)
In this equation u is the number of radians in the angle.
(Why?)
N.B. To speak of an imaginary power of e is meaningless in the
ordinary sense of a power. But since (14) is valid for all real values of
x, this series is commonly taken as the definition of what is to be under-
stood by an imaginary power.
(What is the definition of 10~3? Why is that definition adopted?)
§ 321. Imaginary Logarithms. When U = TT, equation (16)
above becomes
e*V=i = CQS T + vnTj- sin T< (17)
But in radian measure, cos w= —1 and sin TT = O.
Thus — 1 equals e raised to this imaginary power. Or, with e as base :
log (-l) = aV=l,=3.1416V=l.
It is equally true, however, letting u = 3 TT, 5 TT, etc., that
c3irV^I=_l> e6*V=I= _1? etc
Hence —1, though it has no real logarithm, has an unlimited number of
imaginary logarithms,
n-V^T, 3 a-V^I, 5 TT v^l, etc.
Similarly any positive number, though it has only one real logarithm,
has infinitely many imaginary logarithms, likewise differing by multiples
Of 2 *V-1. (Cf. Ex. 1 below.)
The theory of'imaginary logarithms, developed fully in higher courses,
is basic for advanced studies in electrical engineering.
§ 322. Fourier Series. If we plot the function
i/ = sin rc+J sin 3 x— -fa sin 5 x— -fa sin 7 x •••,
the pjaph turns out to be a set of connected straight lines,
as in Fig. 143.
XIII, §322] PROGRESSIONS AND SERIES
435
i.o
y
Graph of
y= sin x+^ sin 3x-
•55- sin 5x— ^9 sin 7x + • • •
Conversely (as was shown about 1820 by J. B. J. Fourier,
a Frenchman), if we have given this broken-line graph, or any
other graph which has a
definite height at every
point and a limited
number of discontinu-
ities and maxima and
minima, the function
represented by it can be
expressed as the " sum "
of an infinite series of
sines (or
both).
cosines, or
0 80 100 120 HO 160
X (degrees)
FIG. 143.
180
Such "Fourier series" are treated in detail in higher courses, being
admirably adapted to the study of the vibrations of a string of a musical
instrument, other types of wave motion, the flow of heat, etc.
The Fourier series for any given graph can be found as far as the first
80 terms by a machine, called an "harmonic analyzer," invented by
Lord Kelvin and improved by A. A. Michelson. For instance, it will
give an approximate equation for a human profile!,
EXERCISES
1. In formula (16) substitute the following values fora:, also each
value plus 2 TT or 4 TT, and interpret each result in terms of logarithms :
(a) x =
(d) x =
(6) s-r/3,
(e).x = l,
(c)
2. Formula (16) is valid only if x denotes the number of radians in
the angle considered. Why so?
3. Make a table of values for z = 0°, 30°, 60°, etc., to 360°, and
plot:
(a) ?/ = sin rc+J sin 3 z+£ sin 5 x ...,
(6) ?/=sin x— \ sin 3 z-f 33 sin 5 x—
sin 7 x ...,
(c) y = cos x-l cos 3
(d) y = cos x+ cos
cos 5x— \ cos 7 x ...,
cos 5 x
cos 7 x ....
436 MATHEMATICAL ANALYSIS [XIII, § 323
§ 323. Discovery of Laws Resumed. In §§ 32 and 175 we
saw how to discover certain types of scientific laws, viz.,
Linear, Power, and Compound Interest Laws. We may now
consider two further types : Trigonometric and Polynomial
Laws.
By using an harmonic analyzer any law can be approxi-
mated in the form of a Fourier series. But simpler types of
trigonometric laws can often be discovered by drawing the
graph and recognizing it as a sine curve, cosine curve, or
some simple combination of these.*
§ 324. Polynomial Laws. A quantity y sometimes varies
according to some polynomial formula of the type
y = A + Bx+Cx2+>-. (18)
Whenever this is the case, it is easy to discover the fact
from a given table of values, and to find the proper values
for the coefficients A, B, C, etc.
Suppose that the values of x in the table run at constant
intervals, Ax = &.• (If they do not, we can plot a graph and
read off values which do.)
Form the differences (A?/) between successive y values.
If these vary, form their successive differences, and denote
these " second-order differences " by A?y. If these also vary)
form their differences (A3?/). And so on.
If the first-order differences (A?/) are constant, y increases
at a constant rate, and the tabulated values satisfy a first
degree formula. More generally :
THEOREM. If the differences (An*/) of order n are constant,
the tabulated values of x and y satisfy a formula of degree n.
And conversely.
* This has been done with considerable accuracy in studying apparent
cycles of rainfall in the Ohio valley during some decades past. See
H. L. Moore, Economic Cycles, Their Law and Cause.
XIII, § 324] PROGRESSIONS AND SERIES 437
This theorem, proved in more advanced courses, makes it
possible to discover any Polynomial Law.
Ex. I. Discover the law for the following table.
1st
2d
Table :
Diffs.,
Diffs.,
x
7
17
27
37
47
57
y
A?/
60
136
76
192
56
228
36
244
16 -4
240
tfy
-20
-20
-20
-20
These second-order differences being constant (= —20), the required
formula is of the second degree :
To find A, B, C, we substitute values of x and y from the" table :
240 = A +57 B +3249 C
192 = A+275+ 729(7
60 = A+ 7 B-\- 49 C
Subtracting the third equation from each of the others gives two
equations free from A. Solving these for B and C, and substituting
back to get A, we obtain finally A = -5.1, 5 = 10, C=— .1. Hence
the required formula is
This is satisfied by all the tabulated values, as direct substitution would
show.
Remark. If the differences never become exactly constant, but are
very nearly so at some stage, the discrepancies may be due to slight
experimental errors in the table. Anyhow, an approximate polynomial
formula can be obtained by substituting as above. (This will be a sort
of Maclaurin series for the function.) The most probable values of
A, B, C, etc., can be found as in §§ 342-343, later.
EXERCISES
1. From the formula y=x2— 7 z+3 calculate a table of values of y
when x — 0, 10, 20, 30, 40, and 50. Verify that the second difference,
A?y are constant in your table.
2. Proceed as in Ex. 1 with the formula y = .1 x3+.2 z+5, and show
that your third differences A3?/ are constant.
3. Discover a formula satisfied by the values in each of the following
tables :
438
MATHEMATICAL ANALYSIS [XIII, § 325
X
0
2
4
6
8
10
y
X
0
0
32
5
88
10
168
15
272
20
400
25
y
X
-7
10
-2
20
13
30
38
40
73
50
118
60
y
X
-15
0
-5
2
25
4
75
6
145
8
235
10
y
X
11
10
13
20
11
30
-7
40
-53
50
-139
60
y
3.10
3.76
4.15
3.97
2.92
.70
(Observe that the
X
0
24
62
110
218
290
y
X
125
0
113
5
94
8
•70
13
16
16
-20
20
y
0
75
96
91
64
0
(a)
(c)
(d)
(e)
(f) Table 5, p. 31. Solve Ex. 5, p. 31, exactly.
4. The same as Ex. 3 for the following tables,
intervals Az are unequal.)
(a)
(6)
§ 325. Summary of Chapter XIII. Formulas relating to
A. P.'s and G. P.'s can be used to solve problems on invest-
ments for which no tables would serve. Also they may be
used to calculate further tables.
Many functions can be expressed as the " sum " of an
infinite series of powers of x [Maclaurin] or sines and cosines
of multiples of an angle [Fourier]. By the " sum " of such
a series in general is meant the limit approached by the sum
of n terms as n->~ao. If no limit is approached, we cannot
speak of the " sum " of the series.
Various useful approximations are obtained by taking the
first few terms of a series. Logarithmic and trigonometric
tables may be so calculated. Many indefinite integrals can
be obtained as infinite series — and in no simpler form.
Functions of imaginary variables are usually defined by
means of series which are valid when the variables are real.
In this way we can give a meaning to imaginary logarithms.
When the values in a table satisfy a trigonometric or poly-
XIII, § 325] PROGRESSIONS AND SERIES 439
nomial law, the formula can usually be discovered. Formulas
are called empirical when found to fit a table.
We shall next consider some further topics of algebra, closely related
to series and the binomial theorem, which are basic in the scientific
study of statistics.
EXERCISES
Express the answers to Exs. 1-5 by formulas ready for calculation.
1. If we make 40 semiannual deposits of $300 each', beginning now,
and interest is at 4% compounded semiannually, how much will there
be to our credit 30 years hence ?
2. What sum set aside annually, 15 times, beginning 1 year hence,
would provide a sinking fund amounting to $50,000 twenty years hence,
if interest is at 5%, compounded quarterly?
3. What sum set aside now by the state would provide an 'accident
indemnity of $600 a year, for 30 years, beginning 1 year hence, if interest
isat6%?
4. How much should an insurance company pay annually for 20
years, beginning 1 year hence, in lieu of paying $20,000 now, if it earns
5. What is the present value of a $1000 bond bearing 4£% interest
payable semiannually, and maturing 15 years hence, if money is now
worth 4|%, compounded semiannually?
6. Approximate Vl.012 by the Binomial Theorem.
7. As in Ex. 6 find an approximate formula for Vl±# when x
is very small. Use your formula to find Vl.0008, V 1.0032, V^98.
8. In finding the ratio of the lever arms of a fine balance, it was
necessary to approximate Vl.OOO 000 023. Do this by inspection.
9. Discover a formula satisfied by the values in this table :
10 20 30 40 50 60
y 3 28 103 228 403 628
10. A debt of $1000 is to be paid off, beginning now, by paying
$100 monthly on the principal, plus interest at 6%. How much will
be paid in all?
11. Find Maclaurin's series for log (1+z) as far as x4. Use it to
calculate log 1.2, and compare tables.
[12.] How many different "chords" could be sounded by striking
three of the four keys A, C, E, G? Write out the combinations.
[13.] How many code "words" could be spelled by using any three
of the four letters A, C, E, G?
CHAPTER XIV
PERMUTATIONS, COMBINATIONS, AND
PROBABILITY
FUNDAMENTALS OF STATISTICAL METHOD
§ 326. The Problem of Arrangements. It is sometimes
useful to know in how many different orders a given set of
objects can be arranged, — using all or a part of the set at
a time. Such questions may be reasoned out as follows :
EXAMPLE. Three flags are to be placed in a vertical row as
a signal. If we have seven different flags, how many signals
are possible?
The top place can be filled by any one of the seven flags, —
that is, in 7 ways. The middle place can then be filled in 6
ways; and the lowest place in 5 ways. For each way of
filling the first place, there are 6 ways of filling the second :
hence 7X6 ways of filling the first two places. Similarly
there are 7X6X5 ways of filling the three places. I.e., there
are 7X6X5 possible signals.
§ 327. Formula. The foregoing example shows that the
number of possible orders of 7 objects taken 3 at a time is
7X6X5. Can you see from this how many orders are possible
for 9 objects taken 4 at a time? For 13 objects taken 6 at a
time?
An order, or arrangement in sequence, is also called a
" permutation." The number of permutations of n objects
taken r at a time is denoted by Pn,r. Thus the result in
the example above may be written
P1 3 = 7 x 6 X 5. (3 factors, from 7 down.)
440
XIV, § 328] PROBABILITY 441
This suggests that Pn> r is the product of r factors, viz.,
/V = n(n-l)(n-2) .-» (n-r+1). (1)
To prove this, simply reason as in the example above.
If all n objects are used every time, n = r ; and hence
Pn.n = n(n-l)(n-2) ••• 1, or P/»,/, = n!. (2)
E.g., the number of possible orders for 6 books on a shelf is
6! ( = 720).
Remark. Factorial notation can be used to express Pn>r
also. For multiplying Pntr by (n— r) ! would give all the
factors from n down to 1, — or n ! That is,
Ex. I. How many batting orders are possible for a ball nine to be
selected at random from 12 men?
12J
3!
Ans. Pi2.9=-^7 =12 • 11 - 10 - 9 - 8 - 7 • 6 - 5 - 4.
§ 328. Restricted Arrangements. Whenever an arrange-
ment is to be made, subject to some restriction, it is im-
portant to consider the restricted groups first.
Ex. I. On each side of a car are 30 seats. In how many ways can
60 persons be seated, 20 of whom insist upon sitting on the sunny side ?
We first take 20 of the 30 sunny seats and assign them to the 20
restricted persons in some order. This can be done in P3o,2o( = 30!-:-
10 !) ways. The other 40 persons may sit anywhere in the remaining
40 seats, in P40, «[ = 40 !] ways.
For each seating of the 20 persons, there are 40! seatings of the
company as a whole. Hence, in all
P VP nr30! 40!
"», 2oX/4o, w or
10!
is the total number of arrangements possible. These factorials are
easily computed by logarithmetic tables with a special first page.
EXERCISES
1. (a) How many numbers can be formed, using any three of the
digits, 1, 2, 3, 4, 5, and 6, without duplication in any one number?
442 MATHEMATICAL ANALYSIS [XIV, § 329
(6) How many code "words" are possible, using any three letters
of the alphabet without duplication ?
2. Ten men compete in a race hi which the first four places score.
In how many ways may the scoring turn out, barring ties?
3. Five persons enter a car in which 8 seats are vacant. In how
many ways can they be seated?
4. A class of 15 men meets in- a room which has 19 blackboards.
In how many ways may the men be assigned one board each?
6. If 10,000 persons register for a drawing of 500 pieces of land of
different grades, in how many ways may the allotment result ?
6. How many numbers can be formed from 1, 2, 3, and 4, using all
the digits each time ? Three each time? In all possible ways?
7. In how many numbers between 100 and 1000 is there neither a
repeated digit nor a zero?
Note: If the location of any objects is restricted, consider the arrange-
ment of those objects first.
8. A car has 20 seats on each side. In how many ways may 40
people be seated if 12 of them insist on sitting on the sunny side?
9. In how many ways may a basketball team line-up if one of the
men can play only as a forward, and two of the others only as guards?
10. In how many ways can eleven men line up as a football team,
if three of the men can play in the line only, and two others in the
backfield only?
11. How many six-place numbers can be formed from the digits
1, 2, 3, 4, 5, and 6, if 3 and 4 are always to occupy the middle two places?
12. In renumbering a city's streets, house numbers were traded
extensively. How many numbers could be formed, using all or part of
the figures 1, 3, 5, and 7? How many from 8, 6, 2, and 0?
§329. The Problem of Combinations. It is often im-
portant to know how many different sets of r objects can be
chosen from n objects. This is not a question of the number
of orders or permutations, but rather the number of groups
or combinations. A set, however, is regarded as different if
even a single individual is changed.
EXAMPLE. How many triangles can be drawn with vertices
chosen from among five points A, B, C, D, E, no three of
which are in the same straight line ?
XIV, § 330] PROBABILITY 443
There will be as many triangles as there are sets of three
letters.
Each set of 3 letters could be arranged in 3 ! ( = 6) different
orders. Hence the number of sets is only one sixth as large
as the total number of possible orders, — which is PB. 3.
Hence the number of sets, or triangles, is
P6.3_5-4-3
3! 1-2-3
10.
§ 330. Formula. The number of sets or combinations of r
objects that can be chosen from n objects is denoted by Cn§f.
Evidently
c-=7f (4)
For each set of r objects has r ! possible orders ; and hence the
total number of orders must be r I times the number of sets.
Or the number of sets = — rX the number of orders. This
r !
gives finally
Ex. I. How many speaking tubes would be needed to connect
each of 5 rooms privately with every other room?
There must be as many tubes as there are pairs of rooms, viz.,
2!3! 1-2
Ex. II. A pack of 52 cards contains "spades," '/clubs," "dia-
monds," and "hearts" in equal numbers. In how many ways can a
hand of 12 cards be drawn, so as to contain precisely 5 spades?
Any 5 of the 13 spades might be drawn, which can be done in CIS.B
ways. By hypothesis, the other 7 cards may be any 7 of the 39 clubs,
diamonds, and hearts. These can be drawn in £39,7 ways. Each set
of 5 spades can go with any set of the 7 other cards. The total number
of hands possible is then
13! 39!
444 MATHEMATICAL ANALYSIS [XIV, § 331
Remark. Formula (5) would give in the case of C9t 9
Q f i
sy <J i __ JL f(\\
9>fl~9!0!~oT
Now 0 ! is meaningless : we cannot speak of the product of
the integers starting at 1 and running up to zero. But if we
arbitrarily assign the value 1 to 0 ! (just as we assigned the
value 1 to z° in § 67) equation (6) will then give C9t 9=1 ;
which is clearly correct.
§ 331. Selecting and Arranging. When a problem in-
volves both the selection and arrangement of objects, with
a limitation upon either, it is best to consider the two steps
separately. That is, ask (a) In how many ways can a
suitable set of objects be chosen, and (6) In how many ways
may each chosen set be arranged.
Ex. I. How many line-ups are possible, choosing a foot-
ball eleven of 6 seniors and 5 juniors, from a squad contain-
ing 10 seniors and 15 juniors?
(a) The 6 seniors may be chosen in Cio.c ways, the 5
juniors in Cnt B ways. Hence the set of players may be chosen
in Cio.eXCiB, 5 ways.
(6) Any one set of 11 men can line up in 11 ! ways. Hence
the total number of possible line-ups is
Ci0,eXCi5.5Xll! =^r,X-i^
[Would it be correct here to reason that the number of orders for
the seniors would be Pio,6, and for the juniors Pis.sl and hence, in all,
PIO. « X PU, 6 ? No, for this allows only for shifts of the seniors and juniors
among themselves and not of seniors with juniors. J
§ 332. Cn, r in the Binomial Theorem. The general term
in the expansion of (a+u)n is by § 319 :
n(n-l) - (n
r!
XIV, § 332] PROBABILITY 445
If n is a positive integer, this is the same thing as Cn>rar un~r.
Hence the binomial theorem may be rewritten :
(a+u)n = an+Cn, Ki^u+Cn, 2a"-2u2+--.. (8)
In fact, another proof of the theorem for positive integral
values of n is easily given from the standpoint of combina-
tions.
To illustrate the idea, think of (a+w)10 as
(a+w)10 = (a+w)(a+w) ... (a+u), 10 factors.
Multiplying the a's in three factors and the u's in the other seven would
give a3u7. This particular term will arise as many times as there are
sets of 3 factors, — i.e., CWi3 times. Hence the expansion will contain
Cw.3a*u7.
You may see from this illustration how the proof would run that the
coefficient of arun~r in (a+u)n must be Cn,r«
EXERCISES
1. In how many ways could :
(a) A committee of five be named from a group of 10 men?
(6) A bodyguard of six be chosen from 15 secret service men?
(c) Six trees be selected from 20 for cutting ?
(d) A bowling team of four be chosen from a club of 25 men?
(e) A jury of 12 men be drawn in a town where 500 are eligible?
(/) Five sprinters draw "lanes" if the track has six in all?
(g) A president, a secretary, and a treasurer be elected from a club
membership of 32?
(h) A dramatic club of 10 be chosen from a public-speaking class
of 25?
(i) Six speakers find seats on a platform containing 8 chairs?
(j) Thirty out of 100 passengers be admitted to the "diner" on the
"first call"?
(k) A clown make up a three-piece " suit," if he has 8 coats, 4 vests,
and 10 pairs of trousers?
(Z) A committee of six Republicans and five Democrats be chosen
from a legislature of 47 Republicans and 41 Democrats?
(m) A collection of 3 gold coins and 4 silver coins be selected from 8
gold and 11 silver pieces, all different?
(n) An arbitration board consisting of 3 employers, 3 laborers, and
3 outsiders be chosen from 12 employers, 20 laborers, and 7 outsiders?
446 MATHEMATICAL ANALYSIS [XIV, § 332
2. If there are 8 "jitneys" in one town and 3 'busses in another, in
how many ways could the round trip be made, going by "jitney" and
returning by 'bus?
3. In a group of 10 houses, how many speaking tubes are needed
to connect each house privately with every other one ?
4. How many possible committees of 5 could be selected from 50
senators, of whom only 8 are able to serve as chairman, and these 8
are unwilling to serve otherwise?
6. In how many ways can a university fill vacancies consisting of
3 instructorships of equal rank in physics, 2 in chemistry, and 4 in
English, from 5 available candidates in physics, 6 in chemistry, and 20
in English?
6. How many batting orders are possible for a ball nine if the out-
fielders must bat before any of the other players, and catcher and
pitcher bat after all the others?
7. From 10 seniors and 8 juniors how many basketball teams can
be chosen, if each includes 3 seniors and 2 juniors? How many "line-
ups" are obtainable thus?
8. How many line-ups are possible for a ball nine, if only 3 men
can pitch or catch, but every man can play any other position?
9. How many football line-ups are obtainable from a squad of 30,
if only 5 can play as ends and only 7 as backs, and these 12 men can
play nowhere else?
10. If we draw 5 balls at random from a bag containing 8 red and
12 white balls, in how many ways may we get 3 red and 2 white?
11. In how many ways can 20 books be shelved in a row, without
separating one set of 5 volumes or another of 4 volumes?
12. If we draw 9 cards from a pile containing 10 "spades" and
8 "hearts," in how many ways may we get 5 spades and 3 hearts?
13. Ten men enter a Marathon race in which various prizes are
offered for winner, second, third, and fourth men. In how many ways
may the prizes fall ?
14. How many different recitation schedules could a student get,
in registering for 3 out of 8 electives, which come at different hours ?
16. How many straight lines are determined by 12 points on a given
circle? How many triangles have three of those points as vertices?
16. Write the expansions of the following binomials, expressing their
coefficients in the Cn, r notation :
(a) (a+xY; (6) (a2 -y2)10, to four terms,
(c) (1-H)7. [Note the expression thus obtained for 27.]
XIV, § 334] PROBABILITY 447
§333. The Idea of Chance. The theory of chance is
fundamental in many lines of scientific work — e.g., in
statistical studies of physical, biological, and social phe-
nomena, in the theory of errors of measurement, etc. The
basic ideas of chance underlying these studies are familiar to
every one, as the following illustrations will show.
(1) If we toss up a coin, we say that the chance of its
falling " heads " te 1/2. We mean that it may fall in either
of two ways, which are regarded as equally probable. In
several thousand trials we should expect " heads " just about
half the time, and " tails " about half.*
(2) If we name a date at random, as May 12, 1987, the
chance that it will be a Tuesday is 1/7. For there is no
reason to suppose that we are any more likely or less likely
to hit upon Tuesday than any other day of the week.
(3) If we draw a ball at random from a bag containing
3 red and 7 black balls, the chance or probability that it
will be red is 3/10.
§ 334. Probability Defined. If an event can occur in x
ways, and can fail in y ways all apparently equally likely,
we say that the probability of success (p) and the probability
of failure (q) are respectively :
(9)
-~, -
x+y' x+y
That is, the probability of an event equals the number of ways
it can occur, divided by the total number of ways it can occur or
fail. ^
?
* When we speak of the turn of a coin as a "chance event," we do not
mean to imply that there is nothing which determines how it will fall. But
the determining factors are so complex, — and so far beyond our knowledge
when the coin is honestly flipped, — that we are quite unable to predict
the fall. And we sum all this up in calling the turn of the coin a "chance
event."
448 MATHEMATICAL ANALYSIS [XIV, § 335
Observe that p-\-q=l. That is, the chance of success
+the chance of failure equals 1. If an event is sure to occur,
0=0.
.-. P=i.
Thus 1 is the symbol of certainty in questions of chance.*
Ex. I. If 5 balls are drawn at random from a bag containing 7 red
and 8 black balls, what is the chance for 3 red and 2 black balls?
Any three of the 7 red balls might be drawn : this could happen in
CT, 3 ways. Likewise two black balls could be drawn in Cg, 2 ways ;
and hence the required combination can be obtained in C7. 3 X C8, 2 ways.
But the total number of ways of drawing some 5 balls is Cu,6. Hence
the probability of succeeding is
7! 8!
CV.3XC8.2__3!4! 2!6!^140
Ci5.6 15! 429*
5T10T
There is approximately "one chance in 3."
§ 335. Compound Probability. What is the probability
that two events, independent of one another, will both occur?
Suppose the first can occur in x\ ways and fail in yi ways ;
and that the second can occur in xz ways and fail in yz ways.
Any one result for the first event might be associated with
any result for the second. Hence the two events can both
occur in x\ x2 ways ; and can occur or fail in (xi+?/i) • (#2+2/2)
ways. Thus the chance that both will occur is
z _6 ^
where p\ and p2 are the probabilities of the two events con-
sidered separately.
Ex. I. If a ball is drawn from a bag containing 3 red balls and
7 black ones, and a card is drawn from a pack, what is the probability
that the ball will be red and the card a spade?
* The mathematical study of games of chance will show in general how
hopeless it is to try to " beat a game " in the long run.
XIV, § 335] PROBABILITY 449
The chance for a red ball is pi = 3/10. The chance for a spade is
p2 = 13/52 = 1/4. Hence the required chance is
That is, in the long run, a red ball would be drawn in three tenths of all
the trials; and, in one fourth of these cases, a "spade" also would be
drawn.
Ex. II. What is the chance that a coin tossed 3 times will fall
"heads" every time?'
The chance for a "head" on any throw is obviously 1/2. Hence
for three throws :
[Why would it be incorrect to argue similarly that the chance of
drawing a "spade" from a pack is 1/4 ; and hence the chance that two
cards drawn in succession will both be spades is p = (l/4)2? Would
this argument be correct if the first card were replaced before drawing
the second?]
EXERCISES
1. What is the probability: (a) of throwing a "six," when rolling
a single die; (6) of getting a "spade" when drawing one card from a
"pack,"-— without the joker; (c) of hitting upon a Monday, when
naming a date at random ; (d) of choosing a man, when selecting one
person by lot from a group of 20 men and 10 women?
2. If 5 balls are drawn from a bag containing 6 red and 9 white balls,
what is the probability: (a) that all will be red? (6) that 3 will
be red and 2 white ?
3. Selecting 5 persons from the group in Ex. 1 (d) what is the
probability for 4 men and one woman?
4. Drawing 7 cards out of a pack, what is the chance : (a) that all
will be black ? (6) that 5 will be black and 2 red ?
5. Drawing 7 cards as in Ex. 4 what is the chance : (a) that all
will be hearts? (6) that 5 will be hearts and 2 diamonds? (c) that
5 will be hearts and the other two something else?
6. Naming a date at random, and simultaneously selecting 3 persons
as in Ex. 1 (d), what is the chance of getting a Friday and 3 men?
7. Drawing 3 balls from the bag in Ex. 2 and simultaneously throw-
ing a die, what is the chance of getting two red balls and a "five" on
the die?
450 MATHEMATICAL ANALYSIS [XIV, § 336
8. Drawing three cards from a pack, and simultaneously tossing a
coin, what is the probability of 3 aces and a head?
9. (a) If 4 coins are tossed, what is the chance that all will fall
"tails " ? (6) The same question for 10 coins.
10. If 3 dates are named at random what is the chance : (a) that
all will be Sundays ? (6) that the first will be a Sunday, and the other
two something else ?
11. In a certain locality 90% of July days are clear. If three
successive days are selected in advance, what is the probability:
(a) that all will be clear? (6) that the first two will be clear and the
third not?
§336. An Illustrative Problem.
If five dice are thrown at random, what is the chance that
some pair will fall " aces " (i.e., one-spots) and the other three
dice fall in some other way?
First consider the probability that a certain pair will fall
aces, etc.
For each ace the chance is 1/6 ; for the pair it is (1/6)2.
For any other die to fall in some other way than ace, the
probability is 5/6; for all three dice to do so it is (5/6)3.
Hence the chance that a certain pair will fall aces, and the
others not aces is (1/6)2 (5/6)3.
The chance for some pair to fall aces, etc., is much larger,
for there are €&t 2 ( = 10) possible pairs, and any one of these
might be the pair to fall aces. Hence the total chance, as re-
quired in the question above, is
••• p-cviwct)1. (ID
This is also the chance that a single die, thrown 5 tunes, would fall
an ace twice and only twice.
§ 337. r Successes in n Trials. The result in (11) merely
illustrates the following theorem : If p is the chance of success
in one trial, and q( = l—p) is the chance of failure, then the
probability that the event will occur precisely r times in n
trials is
pr=Cn.rpr<T^. (12)
XIV, § 337] PROBABILITY 451
PROOF. The probability that a certain r trials will all
succeed is pr ; the probability that all the others will fail is
qn~~r ', and the r successful trials could be selected from n
trials in Cn>r ways.
Ex. I. If 20 dates are named at random, what is the chance that
precisely 5 will be Sundays?
The chance that a single date would fall on Sunday is 1/7. Thus
p = 1/7, q = 6/7 ; and hence
, approx.
EXERCISES
1. What is the probability of getting precisely
(a) Two heads, in flipping 5 coins?
(6) Four aces, in throwing 10 dice?
(c) Three Sundays, in naming 25 dates at random?
2. In a certain locality 40% of the days in April are rainy. If 3
days are selected in advance, what is the chance that precisely 2 of them
will be rainy days?
3. 80% of the days in June are fair. What is the chance for good
weather on 3 consecutive days chosen at random ?
4. In a certain class there are 30 men and 25 women. If 5 names
are drawn at random from a box containing all 55 names, what is the
chance that 3 men arid 2 women will be drawn?
5. If 9 coins are tossed up, and simultaneously 2 dice are thrown,
what is the chance that 6 of the coins will fall heads and both dice fall
fives?
6. If 7 cards are drawn from a pack and 3 dice are thrown, what is
the probability for 3 aces among the cards and 2 aces on the dice ?
7. If 10 dates are named at random, and 5 coins are tossed, what is
the probability that precisely 4 dates will be Mondays and 3 coins fall
"heads"?
[8.] If 5 coins are tossed, what is the probability of five heads? Of
four heads? (And so on, to no heads.) Represent these several
probabilities graphically by ordinates, equally spaced.
[9.] Show that the probability in Ex. 1 (c) equals one term in the
expansion of the binomial
452
MATHEMATICAL ANALYSIS [XIV, § 338
§ 338. Normal Binomial Distribution. If we toss up 10
coins, there are 11 possible results: 10 heads, 9 heads, ...,
1 head, no heads. The chances for these several results
are represented by the heights,
— or preferably we may say
by the areas, — of the several
rectangles in Fig. 144. In
other words, this " staircase "
shows the relative frequency
with which the various num-
7 ,8 9 10 bers of heads would occur in
a very large number of trials,
tossing 10 coins each time.
(Fig. 145, p. 453, shows the same thing for 100 coins. Ob-
serve how slight the chances are for more than 65 heads
or fewer than 35.)
The probability of r heads when tossing n coins once is
Number of Heads
FIG. 144.
Since the factor (l/2)n occurs in every pr, the several proba-
bilities pr are proportional to the coefficients Cn>r in the ex-
pansion of (p+q)n. For this reason a distribution of fre-
quencies like that in Fig. 144 is called a Normal Binomial
Distribution.* Such distributions are very common in
statistical studies. Some illustrations follow.
(I) The "staircase" in Fig. 6, p. 5, which shows the relative
commonness of various chest measures among a number of soldiers,
resembles a Normal Binomial Distribution. So would a similar chart
for statures, and many other biological measurements.
(II) A series bf rectangles similarly drawn to show the relative
frequency of marriages at various ages also resembles the normal
binomial staircase. Likewise for the relative commonness of life
* The distribution of frequencies for the various possible numbers of
"aces" when throwing n dice is a different type of Normal Binomial Distri-
bution, not symmetrical with respect to the vertical center Hue, or any
other line, — p and q being unequal in thia case, £ and f .
XIV, § 340]
PROBABILITY
453
insurance policies taken out at various ages, or of various wages in
some industries ; etc.
§ 339. Errors of Measurement. If we measure the length
of a room many times, say with a yardstick, our results will
disagree by small fractions of an inch. But the values will
tend to cluster closely around their average, — which we
would regard as the true value. Errors of 1/8 in. or 1/4 in.
will be far more common than errors of an inch or more. A
series of rectangles representing the relative frequency of
errors of various sizes will closely resemble the normal bi-
nomial " staircase."
Half of all the errors will be less than a certain small
amount, which is called the " probable error." The other
half of the errors will exceed this. With more accurate
methods of measuring, the staircase would be condensed
from either extreme toward the center, and the " probable
error " would be reduced.
§ 340. Normal Probability Curve. If instead of 10 coins
we toss up 10,000, and draw a " staircase " representing the
chances for no head, 1 head, 2
heads, etc., to 10,000 heads, making
the bases of the rectangles much
smaller, the " staircase " will ascend
by such tiny steps as to be prac-
tically a smooth curve. (Cf. Fig. -
145 for 100 coins.) Indeed, if the
bases of the rectangles are suitably
decreased while the number of steps
is indefinitely increased, the limiting form approached by the
staircase can be proved to be the " normal probability curve " :
y = ae-nx* (13)
whose shape is shown in Fig. 146. (See also Fig. 6, p. 5.)
Similarly, in measuring the length of a room, the finer
40 50 60
Number of Heads
FIG. 145.
454
MATHEMATICAL ANALYSIS [XIV, § 341
the units which we distinguish, the more closely will the
staircase distribution of errors approach the bell-shaped
curve (13). The area of any rectangle represents the proba-
bility of making an error within the limits of the base of the
rectangle. The area of an " infinitesimal strip " y dx under
the curve represents the probability of an error falling within
the tiny base dx, when exceedingly fine units are dis-
tinguished.
Thus the probabilities of making any two errors x\ and #2
are to each other as
canceling the common factor dx. Here n is a constant, de-
pending upon the degree of precision attainable with the
instruments used.
§ 341. Errors of Artillery Fire. If a gun is fired many times,
under as nearly the same conditions as possible, the shots will not all
/T\
SB
2E -E Q E
FIG. 146.
HE
strike at any one point, but because of slight discrepancies in aim, pow-
der, atmosphere, etc., they will scatter somewhat. They will, however,
cluster about a central point, — "the center of impact," — large errors
being few and small errors numerous, in accordance with the Normal
Probability Curve (Fig. 146).
The best half of the shots will fall within a certain distance E ft.
of the center, over or short. (E is called the "probable error" for
the gun at the range in question.) All the shots will fall within a
distance of 4 E from the center, and will in the Jong run be distributed
as in the following table :
XIV, § 341]
PROBABILITY
50% ZONE
455
—4E-3E-2E -E 0
E 2E 3E 4E
2
7
16
25 1 25
to
16
7
2
PERCENTAGES OF SHOTS FALLING IN EACH ZONE.
E, 2E, ETC., ARE DISTANCES FROM CENTER OF IMPACT.
N. B. If the center is not correctly laid on the target, more than half the shots will
be *' overs " or " shorts." >
With this information an artillery officer can correct his fire. For
example, if only 9% of the shots are seen to strike beyond the target,
the distance to the target from the center of impact of the shots must
be about 2 E, which is pretty accurately known for the gun in question,
— and the gun is re-aimed accordingly. Similarly if 15% of the shots
fall short, the target must be about one third of the way from the
— 2 E mark to the — E mark on the scale above ; etc.
Remark. These percentages, 25, 16, 7, 2, give the distribution not
only for errors of artillery fire, but for all sorts of errors of measurement,
and also the distribution of many physical characters. This fact is
very fundamental in the science of statistics.
EXERCISES
1. If a gun has a P. E. of 70 m. and the center of impact has been
erroneously placed 140 m. beyond the target, find from the table above
what percentage of shots will fall short and what percentage will be
"overs" in the long run.
2. For the gun in Ex. 1, if 25% of the shots fall short, how far is the
target from the center of impact, and in which direction ? What if
40% of the shots are shorts?
3. The table shows that the probability of a positive error greater
than 3 P. E. on a single shot is about .02. What is the probability that
2 consecutive shots will both have such an error?
4. What is the probability that 3 successive shots will fall short
and by less than 1 P. E.?
5. Suppose that the distribution of male statures in a nation follows
the normal probability curve, 68 in. being the average stature, and 2 in.
the deviation which is exceeded in half the cases. If a man's height is
70 in., what percentage of his compatriots are taller? What if his
height is 72 in. ? 74 in. ? 62 in. ?
456 MATHEMATICAL ANALYSIS [XIV, § 342
6. In certain measurements the probability of an error of x mm.
is proportional to e-&. How does the chance for an error of 1 mm.
compare numerically with the chance for an error of .5 mm. or .2 mm. ?
7. In Ex. 6 what is the probability that two successive measure-
ments will have errors of .5 mm. and .2 mm., as compared with the
probability of an error of 1 mm. each time?
§ 342. Method of Least Squares. If we measure an
object several times, with different results, how shall we
know the most probable size of the object?
For example, suppose there are to be three measurements,
— with some unknown errors x\, xz, #3. The chance that
this particular set of errors will be made is the product of
the chances that each error will be made separately, and
hence is proportional to
I.e., adding exponents,
D = e-n(xi*+xf+xf)
The most probable set of errors is that for which this p is
largest ; and hence that for which
is least.
Similarly, in general, the most probable value of the thing
measured is that which would make the sum of the squares of
the errors least, — provided, of course, that the distribution
of errors follows the normal probability curve.
Ex. I. The force (/ Ib.) required to stretch a wire x thousandths of
an inch varied as in the adjacent table. These values should x
satisfy an equation of the form
3
f=mx, 5
11
/
20
73
whose graph is a straight line through the origin. But there
are slight discrepancies, due to errors of measurement. Find the most
probable value of m,
XIV, § 343] PROBABILITY 457
When z = 3, 5, 11;
the true / is : / = 3 ra, 5 m, 11 m.
The errors in the tabulated values of / are therefore
3m-20, 5m-33, 11 m-73.
The most probable value of m makes the sum of the squares,
S = (3m-20)2+(5m-33)2+(llra-73)2,
least. Differentiating and equating dS/dm to zero gives :
4^- =2(3w-20)(3)+2 (5m-33)(5)+2(llm-730)(ll)=0.
dm
..'. 155 m = 1028, ra = 6.63+.
And the most probable values of / for x = 3, 5, 11, are :
/=3w = 20.-, / = 5m = 33.16, /=llm = 72.95.
§ 343. Several Coefficients. In trying to find the most
probable " Polynomial Law " f or a given table, say
y = a+bx+cx2+--, (14)
the sum of the squares of the errors S is a function of several
unknown coefficients, a, b, etc. For a minimum of S we set
each of its partial derivatives equal to zero (§ 300) :
10
20
30
0, etc.
era 00
v Ex. I. Find the most probable linear formula y=a+bx
.. r. for the adjacent table.
4.2 For each pair of tabulated values, a +bx should equal y. Any
6.6 difference is an error. That is, the errors here are :
a + 10 6-1.6, a+20 6-4.2, etc.
The sum of the squares of the errors is, then,
-6.6)2.
— = 2(a+lti6-1.6)+2(a+206-4.2)-h2(a+30 6-6.6).
dtt
-1.6)+40(a+206-4.2)+60(a+30 6-6.6).
db
458
MATHEMATICAL ANALYSIS [XIV, § 344
Equating these to zero, canceling 2 or 20, and collecting :
3a+ 606-12.4 = 0
6 a + 140 6-29.8 = 0
Solving these equations gives a = — £f , 6 = £.
Hence the most probable law is
EXERCISES
1. Find the most probable value of m in the formula y=mx for each
of the following tables :
(a)
X
5
10
y
4
8
20
17
(6) _JL UJ
1
3
rj
7.5
22.4
37.8
2. The following elongations of a rod should, if strictly accurate,
vary with the stretching force according to the formula e = kF.
(a) Find the most probable value of k.
5000
.231
10000
.465
15000
.691
10
20
30
5.1
6.9
9.1 '
2
5
10
.59
3.01
7.00
T
0
10
30
50
100
W
35.7
35.8
36.3
37.0
39.8
(6) Using your value of k, calculate e from the formula when F = 5000,
etc., and compare the table.
3. Find the most probable values for a and 6 in the formula y = ax -\-b
for each of these tables :
x_
y
4. The weight of common salt which will dissolve in 100 g. of water at
various temperatures is shown in
the following table. Find the most
probable linear formula, W = aT+b.
6. The height of a desk was measured, with the same care each time,
as 20 in., 20.2 in., 20.1 in., and 19.9 in. What is the most probable
height, x in.? [Show that your result is the arithmetical average.]
§ 344. Summary of Chapter XIV. The formula for the
number of combinations of n things taken, r at a time is
derived from the formula for the number of arrangements.
Besides its more important uses, it enables us to write the
binomial theorem for integral exponents in a new way.
XIV, § 344]
PROBABILITY
459
The foregoing problems in permutations and combinations
are confined to the simplest cases. And we have had only a
glimpse of probabilities and the method of least squares,
both of which are very fundamental in the study of statistics.
But even this glimpse may serve as an introduction, and
familiarize us with the idea of Normal Binomial Distributions
and the Normal Probability Curve for chance events.
To round out bur knowledge of the number system of
algebra, and of the use of " imaginaries " in studying certain
kinds of variation, we shall conclude the course with a brief
study of " Complex Numbers."
EXERCISES
1. How many baseball batting-orders are obtainable in choosing a
nine from 15 sailors and 20 soldiers so as to contain 3 sailors and 6
soldiers, if the sailors are to bat before the soldiers'?
2. If 10 cards are drawn at random from a pack, what is the chance
for precisely 3 spades? Show that your result is reasonable.
3. What is the probability that the first shot from a gun will go
over the target by more than 2 P. E. and the second fall short by more
than 1 P. E. if the center of impact is correctly placed on the target?
4. The weight of silver nitrate which will dissolve in 100 g. of water
at various temperatures is shown
here. Find the most probable
linear formula.
5. By Weber's law in psychology we should have in the adjacent
table: A i=mi. [Here A i is
the smallest increase in the inten-
sity i of a light which a certain
observer could detect.] Find the most probable value of m.
6. In studying the relation of a firm's advertising expenditure and
volume of business, it was necessary to find the most probable linear for-
mulas for y and x in these tables :
T
0
20
50
80
W
122
222
455
669
i
25
60
120
300
Ai
.2
.5
.8
2.5
8
84
The index of correlation is (dy/dx) • (dx/dy). Find this.
X
y
2
4
6
60
68
75
11
60
80
100
X
2
3.6
5.3
120
6.8
CHAPTER XV
COMPLEX NUMBERS
OPERATIONS WITH DIRECTED QUANTITIES
§ 345. The Real Number System. Elementary arithmetic
deals only with positive numbers, — running from zero up-
ward. In algebra we invent another set of numbers, -
" negative numbers," — running from zero downward.
In arithmetic it is impossible to subtract 7 from 5, or 9
from 0, or any other number from a smaller one. In algebra
this is possible : 5 — 7 gives simply — 2 ; etc. The introduc-
tion of negative numbers makes subtraction possible in all
cases.
Still more important, the complete set of positive and
negative numbers is adapted to the study of opposite quantities,
such as temperatures above and below zero, elevations above
and below sea-level, latitudes north and south, gains and
losses, etc.
The complete set of positive and negative numbers is called
the Real Number System.
The positive numbers can be represented by the points
of a line in one direction from a chosen point or origin ; the
complete set of positive and negative numbers by all the
points of a line, in both directions from the origin or zero
point.
Students just beginning algebra sometimes wonder how negative
numbers are possible. Can there be any number lower than zero?
Not if we are thinking of numbers as in arithmetic. But this is just
the point : In algebra we are talking about a new kind of number.
But are not such numbers purely fictitious or abstract? Yes, until
we exhibit some concrete interpretation for them, — some definite set
460
XV, § 346] COMPLEX NUMBERS 461
of objects to which they can be applied, such as temperatures below
zero, etc.
To deal with negative numbers, certain rules of operation are agreed
upon, such as (-)X ( — ) = (+), etc. These, while arbitrary, are
justified by the useful way in which the rules work. The same will
be true of what we shall say about "imaginary numbers."
(Of course, in inventing any new kind of number, we have a right
to prescribe the rules of combination, — just as the inventor of any
game, such as chess^, had the right to specify how the " pieces" should
move.)
§ 346. " Imaginary " Numbers. In elementary algebra,
as long as we know only the real number system, it is im-
possible to solve the equation x2=— 4. For the square of
any real number, positive or negative, is positive, and hence
never —4.
But we can solve this and similar equations by inventing
a still different set of numbers called imaginary or complex
numbers. We- may do this as follows :
Let i denote a number whose square is —1, that is,
i2=-l, ori = V~^l. (1)
And let —i denote the result of subtracting i from zero;
that is, (— t)=0 — (i). Squaring shows that (— i)2 = i2= — 1.
Thus there are two numbers, i and — i, whose square equals
— 1. So —1 has two square roots, i and —i. We shall de-
note either by v — 1.
Observe that i and — i are not "real" numbers; and are not to be
regarded as positive or negative, greater or less than zero. They are,
however, opposites.
With the introduction of the " imaginary unit " i, and
multiples of i, we can extract the square root of any negative
number. E.g.,
=2i, or -2i.
or -
462 MATHEMATICAL ANALYSIS [XV, § 347
In fact, by using combinations of real and imaginary units,
such as 3+2 i, 7 — Qi, etc., it is possible to extract the square
root, or any other root, of any real or complex number.
For instance : \/7-24i = 4-3 i or -4+3 i. This may be verified
by squaring either result.
Indeed, the number system composed of all possible com-
binations, such as a+bi, where a and b are any real numbers,
suffices for all the purposes of ordinary algebra.
§ 347. Operations. We agree that complex numbers
shall be combined according to the usual rules of algebra,
multiplying sums term by term, adding like terms, etc.
For instance :
(2+3 i) + (5+8i) =7 + 11 i
(2+3 i)(5+8 i) = 10+15 z + 16 i+24 V
Since &= — 1, this last reduces to —14+31 i.
To perform a division, we first indicate the result as a
fraction, and then rationalize the denominator. If the de-
nominator is a -\-bi we have simply to multiply above and
below by a—bi.
For instance :
(2+3 i) -h (5+8 i) -?+3» = (2+3 i)(5-8 i)
5+8 i (5+8i)(5-8t)
52-(8t)2 89 89
In this way, the quotient of any two complex numbers is
reducible to the standard form of a complex number, c+di.
(How could any division be checked?)
THEOREM. Two complex numbers can be equal only if their
real parts are equal, and also their imaginary parts.
For instance, if x+t/i = 3+4i, then z = 3 and y = 4. For x — 3 =
(4— y)i, and if both sides were not zero, we should have a real number
equal to an imaginary.
XV, § 348]
COMPLEX NUMBERS
463
To extract a square root, denote the required root by
a-\-bi. Then square, and compare with the given number.
For instance, let us find V?— 24i, =a-rbi, say.
Squaring : Vy — 24 i = a2 +2 abi+bH*. Equating real parts, and also
imaginary parts : a2 — 62 = 7, 2 ab = —24. Solving gives a = =*= 4, 6 = =F 3.
.'. V7-24i=4-3z, or -4+3 i.
EXERCISES
1. Perform these additions and subtractions, and simplify :
(a) (4+3 i) + (9+0, (6) (7+60 -(2+80,
(c) (_6-80 + (4-30, (d) (-3+20 -(-9-lli).
2. Perform these multiplications and divisions, in each case reducing
the result to the standard form x-\-yi :
(a) (7+40(12-20, (&) (5-0 ^ (3-2 t),
(c) (-1 + 130*, (d) (-1 + 13 0-K
(e) (3+5 Otf3-*2), (/) 7 -*- (8+50,
(0) (2-03, (A) t-K6*-4).
3. Find two square roots for each of the following :
(a) 4+3 1, (6) 4-3 i, (c) -4+3 i, (d) -3-4 i,
(e) 5-12 i, (/) -12-5 i, (0) 15+8 i, (h) -8-15 i,
(0 40+9 i, .0') 16-63t, (fc) -21+20i, (I) -ll-60i.
§ 348. Geometrical Representation of Complex Numbers.
In any given plane let a pair of rectangular coordinate axes
OX and OY be selected.
And let any complex num-
ber x+yi be represented
by the directed line, or
" vector," OP drawn from
the origin 0 to the point P
whose coordinates are x
and y.
E.g., the number — 3 +4 i
is represented by the vec-
tor from 0 to the point
(—3, 4). Observe that it FIG. 147.
Y.
(-5,4)
464
MATHEMATICAL ANALYSIS [XV, § 348
is not the length of the vector (viz., 5) which represents
the complex number, but rather the vector itself, — i.e., the
directed line.
The length of the vector OP is called the absolute value or
numerical value of the number x-\-yi, written | x +yi \. Thus
Num. value of x-\-yi — \ x+yi \ =
(2)
Observe that when we say that the numerical value of — 3+4 i is
V(3)2 + (4)2 = 5, we are not saying that -3+4i = 5. The numerical
value of — 10 is 10, but — 10 does not equal 10.
With this representation of a complex number, let us see
what will represent the sum of two such numbers. We have
agreed that
(x+yi) + (x'+y'i) = (x+xf) + (y'+y)i. (3)
Hence the vector representing the sum should run from 0
to the point whose coordinates are (x+x't y+y').
In other words, to add two
O(z-jVi/-H/) complex numbers graphically,
proceed just as in finding the
combined effect of two forces
which act from a common
point. Make a parallelogram.
Remark. Such vectors, combined
in this way, have all the properties
of complex numbers. Hence to
picture to yourself what sort of
thing an "imaginary" number is,
simply think of these directed lines.
Thus a complex number is not
something vague and impossible of
existence in a real world. It is
simply a more general kind of
number than the "real" numbers whose vectors all lie along the
.X-axis.
P(x,y)
Fio. 148.
XV, § 349]
COMPLEX NUMBERS
465
§ 349. Polar Form of a
Complex Number. In the
form x + yi, a complex num-
ber is expressed in terms of
the rectangular coordinates
x and y of the end of its
vector. For many purposes
it is better expressed in terms
of the polar coordinates (r, 6)
of that point. (Fig. 149.)
Clearly z = rcos0, y = r
sin 6, hence
r(cos8+isin8). (4)
FIG. 149.
This latter is called the " polar form " of the complex
number. As before, r is called the absolute value of the
number ; 6 is called the argument
or simply the angle.
Cos 6+i sin 6 is often abbrevi-
ated cis 6. Thus
x+yi=rcisQ. (5)
To change easily from the rec-
tangular to the polar form, or
vice versa, simply draw the vec-
tor; and calculate the required
values of (r, 6) or (x, y) from the
figure. This is exceedingly im-
portant and should be practiced
FIG. iso. freely.
Ex. I. Find the polar forms for 8+6 i; -2-7i; -6; 10 i.
Fig. 150 shows the vectors representing these numbers.
, 0 = 36° 52'
(1) Herer = 8 = 10; tan 0 =
/. 8+6 i = 10(cos 36° 52'+i sin 36° 52') =10 cis 36° 52'.
466 MATHEMATICAL ANALYSIS [XV, § 350
(2) Here r = V(-2)*+(-7)2 = V53, tan0=-7/-2, 0=180°+74°3'.
/. -2-7 i= V53 cis 254° 3'.
(3) Here, by inspection, r = 6 ; 0=180°.
/. -6=6(cos 180° +i sin 180°) =6 cis 180°.
(4) Here, by inspection, r = 10 ; 6 = 90°.
/. 10 i = 10(cos 90° +i sin 90°) = 10 cis 90°.
EXERCISES
• 1. Mark the points and draw the vectors representing these numbers :
-5i, 4+6 i, -3-7 i, -9,
3.5 i, 9-2 i, -l+i, +2.
2. Perform the following additions algebraically, and also geo-
metrically by means of the vectors representing the given numbers :
(a) (4 +2 i) + (2 +7i), (6) (8+5 i) + (1-9 0,
(c) (6+0 + (-7+40, (d} (3 + 110 + C-3-20,
(«) (9-5 i) + (3i), (/) (-24) + (-60.
3. Draw the vector representing each of the following numbers,
find its r and 6, and re-write the number in polar form :
(a) 2+2 i, (6) -2-2 i, (c) 3 i,
(d) 3-4i, (e) -4\/3+4i, (/) -lOi,
(g) -3, (A) -4A/3-4t, (0 +12.
What are some of the different values of 6 which can be chosen in
c, /, g, i if we do not limit the size of 0?
4. Draw the vector for each of the following, find the x and y belong-
ing to the number, and re-write the number in the form x+yi.
(a) 2 cis 70°, (6) 5 cis 310°, (c) .3 cis 180°, (d) 7 cis 270°,
(c) 12 cis 240°, (/) 8 cis 495°, (g} 9 cis 0°, (A) .4 cis 720°.
6. Plot, and find the standard polar form for these numbers :
(a) cos 80° -i sin 80°, (6) -8 cis 40°, (c) -10(cos 50° +i sin 20°).
§ 350. Multiplication and Division, in Polar Form. Any
two complex numbers can be expressed in the form :
r (cos e+i sin 0), r' (cos B'+i sin 6').
XV, § 350] COMPLEX NUMBERS 467
Multiplying these together will give
rr' [(cos 0 cos 0'-sin 6 sin 0')+t (sin 6 cos 0'+cos 6 sin 6')].
But by the Addition Formulas of Trigonometry (§281), the
first parenthesis is cos (0+0'), and the second is sin (0+0').
Hence the product above reduces to
rr' [cos (0+0') +i sin (0+0')]-
r cis,0 • r' cis Q' = rr' cis (0+60. (6)
That is, to multiply two complex numbers, multiply their
absolute values and add their angles. To divide, simply re-
verse this process. (If the numbers are given in the form of
x+yi, first put them into the polar form.)
Ex. I. Multiplying 5 cis 300° by 7 cis 40° gives 35 cis 340%
i.e., (35 cos 340°) + (35 sin 340°)i, or 11.97-32.89 i.
Ex. II. Fmdz=(2cisl5°)10.
For this repeated multiplication, we keep on multiplying the
r's and adding the 0's, until we finally get
z = 210 cis 150° = 1024(-.866+.500 i) = -867+512 t.
Remark. By § 320, cos 6+i sin 0 = e™, (0 in radians).
Hence equation (6) above may be written :
reiO . r'ei6'=rr' . et'(0+0')
In other words, this shows that the usual law for exponents in multi-
plying holds good even when the exponents are pure imaginaries, id
and id'.
EXERCISES
1. Find the following products and quotients, expressing the results
in both the polar and rectangular forms, and drawing the various vectors
involved :
(a) 10 cis 30° X 3 cis 20°, (6) 10 cis 30° -j- 3 cis 20°,
(c) 2 cis 1 10° X 5 cis 250°, (d) .6 cis 110° -5-. 02 cis 250°,
(e) (2 cis 50°)2, (/) -7 i+2 cis 100°,
(gr) (5cis20°)3X2i, (h) (3 cis 200)2-J-(2 cis 62°)6.
2. Calculate a: = 3 (cos 20° -i sin 20°) -?-4 cis 70°'.
(The dividend is not in standard form : consider its vector.)
468 MATHEMATICAL ANALYSIS [XV, § 351
3. Calculate z = 20/(G cis 50°).
(The vector for 20 has 0 = 0° or 360°, etc. Which is best here?)
[4.] Can you find a number whose cube is 8 cis 30°?
§ 351. Powers and Roots. From (6) it follows that for
any positive integral value of n :
(r cis 0)n = rn cis nO.
Thus we can very quickly find any high power of a complex
number which is given in the polar form.
This same idea furnishes a means of extracting any root of a
complex number.
ILLUSTRATION. Find x= \/7 cis 300°-
Let us denote any possible value of x by r cis 0 :
r cis 0 = ^7 cis 300°.
Cubing : r3 cis3 6 = 7 cis 300°.
This equation is satisfied by r = *Vl and 0= 100°.
/. -\/l cis 300 = v'Tcis 100° = - .332+1.884 i, approximately.
This, however, is not the only possible cube root of the
given number. For adding any multiple of 360° to the given
angle would not change the value of its sine or cosine ; and
thus the given number could have been written in any of the
forms :
7 cis 300°, 7 cis 660°, 7 cis 1020°, 7 cis 1380°, ...
The cube roots obtained from these would be
vY cis 100°, ^7 cis 220°, ^7 cis 340°, ^7^8460°, ...
The last of these equals the first, however. And further
forms would only repeat some of the first three.
Thus, 7 cis 300° has three cube roots which are distinct, —
and no more.
In getting r there is no ambiguity. For r is real, and 7 has only one
real cube root.
XV, § 352] COMPLEX NUMBERS 469
In general there are n distinct nth roots of any number,
real or imaginary. They can be found by expressing the
given number in the several polar forms :
r cis 6, r cis ((9+360°), r cis ((9+720°), ...,
and then extracting the nth root of r and dividing each angle
by n.
Further illustrations follow.
Ex. I. Find the fourth roots of z = 81 cis 20°.
The given number z may also be written :
_z=81 cis 380°, 81 cis 740°, 81 cis 1100°, ... .
.'. *z = 3 cis 5°, 3 cis 95°, 3 cis 185°, 3 cis 255°.
Ex. II. Find the square roots of i.
Proceeding as in § 349, we express i in the form
i = cis90°, or cis 450°,
/. Vi = cis45°, or cis 225°,
That is, since sin 45° and cos 45° are both V2/2 (by geometry), )
or^/?(l-i). (Check?)
EXERCISES
1. Verify that squaring either answer to Ex. II above will give i.
2. Find the following roots (all values of each) :
(«) \/27 cis 120°, (&) ^16 cis -60°, (c) A/32 cis 280°,
(d) ^/_8i, (e) V9(cos80°-isin80°), (/) -N/COS 10°- i sin 10°.
3. Find the five fifth-roots of — 1 ; and express numerically in the
x +yi form by trigonometric tables. (Hint: What is the polar form
for -1?)
4. Find the three cube roots of +1, and plot roughly the points
representing them. Where do those points lie?
6. The same as Ex. 4 for the four fourth-roots of +1.
§ 352. nth Roots of Unity. The nth roots of the number
+ 1 are interesting, and, in Higher Algebra, very important.
Let us consider first the sixth roots.
470
MATHEMATICAL ANALYSIS [XV, § 353
Fia. 151.
Expressed in the polar form,
l = cisO°, cis 360°, cis 720°, cis 1080°, etc.,
/. ^l = cisO°, cis 60°, cis 120°, cis 180°, etc.
The vectors representing these sixth roots all have r=l,
and their successive angles differ by 60°. Hence their ends
lie on a circle of unit radius,
and are vertices of a regular in-
scribed hexagon.
Similarly the nth roots of
+ 1 are represented by vectors
drawn to the vertices of a regu-
lar inscribed n-gon.
The nth roots of any number
other than 1 would be repre-
sented by vectors to the vertices
of a regular n-gon inscribed in
some circle perhaps of a different radius, and the first vertex
being perhaps at some point off the real axis.
EXERCISES
1. Find the following nth roots and mark the n points representing
them in each case :
(a) VI, (6) x/3, (c)
(0) V±i, (h} V=7, (i) VT.
2. The same as Ex. 1 for the following :
(a) V27 cis 60°, (6) VQ cis 140°, (c) \/4 cis 320°.
3. In Ex. 1 (a), (6) show graphically and also by calculation that
the sum of all the n roots is zero.
4. In what further parts of Exs. 1-2 is the sum of the roots zero ?
§ 353. Application to Electricity. In studying electricity, it
is customary to represent a simple alternating current by a vector in
the "complex plane." The length of the vector represents the uuixi-
XV, § 354] COMPLEX NUMBERS 471
mum intensity of the current and the angle of the vector represents
the time during the periodic oscillation when that maximum is reached.
It can be proved that if a circle be drawn on the said vector as
diameter, and a varying vector be drawn from the origin to the circle,
the latter vector as it turns will represent the varying intensity of the
current during the entire period of oscillation.
Moreover, if another current, represented by any other vector at
right angles to the first, is also impressed upon the same circuit, the
resulting combination » current will be represented by the resultant
(or sum) of the two vectors.
These facts and others, together with the laws of imaginary numbers,
make such numbers exceedingly useful in the study of alternating
currents, and they are discussed at length in treatises on that subject.
§ 354. Summary of Chapter XV. The number system of
elementary arithmetic consists solely of real positive numbers,
representable by points on a line in one direction from a
chosen origin. The complete number system of algebra
consists of: (I) All real numbers, representable by all the
points along a line or all vectors drawn from the " origin "
to those points ; (II) All pure imaginaries, representable by
vectors to all points on another line perpendicular to the
first ; (III) All complex numbers (combinations of reals and
pure imaginaries) and representable by vectors to all points
in the plane.*
The so-called " imaginary " numbers are susceptible of
concrete interpretation, and become very useful in elec-
trical engineering and elsewhere.
The rectangular form x+yi is best for addition and sub-
traction ; the polar form r (cos B-\-i sin 6) for finding powers
and roots. Either is good for simple multiplications or
divisions.
There are n distinct nth roots of any number. The two
square roots (+ or — ) as found in elementary algebra are
merely a special case of this.
* Strictly speaking, (III) includes (I) and (II).
472 MATHEMATICAL ANALYSIS [XV, § 354
EXERCISES
1. Carry out these calculations, marking the points and getting the
results in both polar and rectangular form :
(a) (3+6i) + (5-4t), (6) (3+6 t)-s- (5-4 1),
(c) 2 cis 60° X 5 cis 90°, (<f) 8 cis 120° -^3 cis 40°,
(e) (7cis28°)3, (/) *16 cis 320°.
Retrospect and Prospect
We have now considered some of the more important prob-
lems relating to variation and the mutual dependence of
quantities ; and have developed methods of dealing with
these problems, at least in the simpler cases.
Most of the mathematical processes covered here and in
more elementary courses originated in the effort to salve
certain practical problems. For instance, geometry de-
veloped out of problems of mensuration, trigonometry fromi
problems of surveying, and algebra from the attempt to
systematize certain kinds of calculations. The basic prob-
lem of differential calculus is to find the rate at which some]
quantity will vary with some other quantity on which it de-
pends. Integral calculus seeks to determine how large a
varying quantity will be at any time, knowing its presena
size and its rate of increase at all times. Or, what amounts!
to the same thing, to find the sum of all infinitesimal cle-|
ments of a quantity, knowing how each element that we
add will vary in size. It would be hard to think of any
problem of science or business which is more common or more
important than these. Moreover, the calculus methods of
analysis, — once they are thoroughly understood and linve
become a habit of thought, — are invaluable in analyzing!
new problems on variation or summation.
As yet, however, we are only on the threshold of Mathe-
matics. We have differentiated only the most elementary!)
XV, § 354] COMPLEX NUMBERS 473
functions, — trigonometric, logarithmic, exponential, and
power functions, and simple combinations of these. There
are many other types of functions, relating to important
kinds of variation, not discussed in this course. The study
of their differentiation, integration, and application makes
calculus a vastly bigger subject than the very brief intro-
duction given in this course might suggest.*
Analytic geometry also is an exceedingly extensive sub-
ject. Besides rectangular and polar coordinates, numerous
other systems of coordinates have been invented, especially
for studying geometry upon various kinds of surfaces. There
is scarcely any limit to the variety of curves and surfaces
whose geometrical properties have been, and are being, inves-
tigated analytically. Then, too, there are the non-euclidean
geometries, and the geometry of hyperspace or n dimensions.
And in modern times new methods of investigation, of a
purely geometrical character (" projective " methods), have
led to many beautiful theorems concerning triangles, circles,
and other figures.
Algebra, too, has many higher branches, — dealing with
number relations, the solution of equations, the simplifica-
tion of expressions by algebraic substitutions, infinite series,
etc. Further kinds of numbers, which combine according to
different laws even than the " imaginary numbers," have also
been invented and studied.
In fact, the science of Mathematics has in modern times
* There is in fact no limit to the possible variety of functions, for a
quantity may vary with another in any manner whatever. Hence if we state
simply that y is a function of x, all we are saying is that To every value of x
there corresponds a value (or values) of y, according to some definite law or
system or agreement.
E.g., the postage on a letter is a function of the weight. For any weight
up to 1 oz., the postage is 2^. For any weight from 1 oz. up to 2 oz., the
postage is 4j£, jumping instantly from 2ff to 4^ as soon as the weight passes
an exact ounce. And so on. This is a very peculiar type of function,
represented graphically by a series of horizontal lines, entirely separated.
474 MATHEMATICAL ANALYSIS [XV, § 354
grown to such vast proportions that no one can now hope to
have a detailed knowledge of the whole field. Moreover it
is still growing, and more rapidly than ever. Hundreds of
research papers are published each year developing new
processes and announcing theorems previously unknown.
Much of this higher mathematics is very abstract. But it
is not therefore valueless, even from the standpoint of ap-
plications. Several subjects which originally developed in a
theoretical way with no thought of a practical application
have later been taken over bodily by some practical science,
e.g., electrical engineering, crystallography, etc. Several
others have contributed powerful methods to the solution of
particular problems. The theoretical mathematics of to-day
may be practical mathematics to-morrow.
But this is only one aspect of the matter. The intellectual
values obtainable from a contemplation of the power, elegance,
and absolute precision of mathematical reasoning, and of the
perfect harmony existing among the various branches, are
very great indeed. In elementary courses, — such as this,
and those immediately following, — which are designed for
all classes of students, the practical aspects deserve particular
emphasis. But men and women who have time to get an
understanding of the more advanced branches find nothing
finer and more inspiring than the wonderfully abstruse in-
vestigations of pure mathematics, — achievements of the
reason which far transcend the realms of physical sense, and
time and space.
GENERAL REVIEW
Exercises on Chapters I-VI
1. In a series of experiments the yield of various crops
(Y bushels per acre) was found to vary with the amounts
80
160
(F Ib.) of nitrate fertilizer used as shown in the table.
Plot the graph. Apparently, what F gives the maximum Y ? 040
320
2. Differentiate: ?/ = z4- — -6v^H — — + (x*+l)§ +7VH.
3 6— x2
16
24
28
24
21
Integrate: x*-24x+l9-—+-: dx.
3. The load (L Ib.) per sq. ft. on a certain rectangular floor varies,
being L = (x — 15)2 at a distance of x ft. from one end. (a) Find the
total load on the floor, if the length is 30 ft. and the width is 20 ft.
(6) Explain in what sense we can say that "the load per sq. ft. is 100 Ib.
at a distance of 5 ft. from one end," — in view of the fact that a square
foot of floor cannot all be at exactly that distance.
4. A cable car weighing 13975 Ib. stands on an incline whose grade
is 60 per cent. How hard does it press against the rails, and what is
the pull in the cable, ignoring friction?
5. A plane inclined 29° 41' passes through a diameter of the base
of a cylinder of radius 9 in. (a) Find the area and greatest length of
the sloping section. (6) Find the volume of the wedge cut off.
6. The force applied to an object varied thus : F=4522 — t3. Find
the momentum imparted from t = Q to £ = 40. When was the force
increasing most rapidly? What was the maximum force?
7. Plot the force in Ex. 6 as a function of t, and check your an-
swers graphically.
8. A cylindrical tank is to contain 1125 cu. ft. and is to have a hemi-
spherical screen for a roof. If the bottom costs 40 cents per sq. ft.,
the curved wall 30 cents per sq. ft., and the roof 20 cents per sq. ft., find
the least possible total cost.
9. For a quantity of gas: y = 6000/p. If p is increasing at the
rate of .2 unit per min., how fast is v changing, when p = 2Q?
10. To find the distance from a gun G to a target T a line
GP = 428.5 yd. was laid off and angles measured as follows : TGP = 68°
15'.6, TPG = Wl° 12'. Find GT.
475
476 MATHEMATICAL ANALYSIS
11. A triangular farm lies at the junction of two rivers having a
frontage of 780 yd. on one and 930 yd. on the other. Its area is 291000
sq. yd. Tell precisely what steps you would take to calculate the angle
between the rivers, and the length of fence required for the third side
of the farm. Mention any formulas needed.
12. Calculate: M,j38764.(-.09627)«;fr-.2
\ -198.98
13. A bullet was fired straight up from an airplane 6000 ft. high,
with an initial velocity of 2000 ft. /sec. Find its height at any time.
When was it highest and when down to earth ?
14. A cylindrical tank of radius 4 ft. lies horizontally,
0
30
60
and is half full of oil weighing 60 Ib. per cu. ft. Find the
pressure on one of its circular ends.
15. The adjacent table shows the angular velocity
(w deg./sec.) of a flywheel of radius 10 ft. at various times 120
(t sec.). Find the angular acceleration when £ = 100, and 150
0
240
840
1620
2400
3000
the total angle turned from t = 0 to t = 180. 180 3240
16. What measurements of lines and angles made on one side of a
river would suffice to let us find the distance between two objects on the
other side ? Explain.
Chapter VII
1. Differentiate: y = %a(x2 + l), y = log^ y = e~**, y
2. Integrate: 10 e2' dt, (20 /x) dx, (e*-e~*}zdx.
3. Find the maximum value of y = Gog x)/x*.
4. What rate of interest, compounded annually, gives the same
result as 20%, compounded continuously?
5. The speed of a certain chemical reaction increases thus with the
temperature: dV/dT = M9 V. If F = 20 when T = 0, YR. VALUE
write by inspection a formula for V at any temperature.
1850
1880
3.2
6.8
7.3
10.7
13.2
16.8
rate of increase greatest? Least? 1910 34.9
7. By the Weber-Fechner law in psychology, the amount of sensation
E produced by any stimulus R varies thus : dE/dR = C/R, where C is a
constant. Integrate. Does this come under the C. I. L. ?
Derive the same formula by integration. „„
6. The value of farm lands and buildings in the
United States at various times is shown in billions in
the adjacent table. Plot the ordinary and semi-loga-
rithmic graphs. In which decade was the percentage 1900
GENERAL REVIEW 477
8. The values in the adjacent table satisfy a Power gg 12 40
law. State precisely how you could discover that fact j QQ 7 44
for yourself. Also find the exact law. 2.25 4.96
576 3.10
Chapter VIII
1. Calculate the length, slope, and inclination of the line joining
(2, 7) and (5, 12).
2. Is the line joining (0, 0) and (3, —4) perpendicular to that joining
(0, 0) and the mid-point between (6, 2) and (2, 4) ?
3. Draw the lines : 2 x— y = 7, x-{-y = 5, x = 9.
4. Simplify the following equations and draw the curves by inspec-
tion. Show the foci and asymptotes, if any.
9 *2+4 ?/2+90 z+16 y+177 = 0, z2+20 y = 40,
5. A projectile was fired with an initial speed of 1600 ft. /sec. and
an inclination angle = tan"1 (|). Find the equation of the path. Locate
the vertex by two methods.
6. What sort of curve is the graph of Boyle's law : pv = k?
7. (a) How would you draw an ellipse whose longest and shortest
diameters are 50 in. and 30 in.? What area would it have? (6) A
horizontal beam casts a curved shadow on the wall of a cylindrical gas
tank. Precisely what sort of curve is it, and why?
8. A suspension cable, loaded uniformly per horizontal foot, has a
horizontal span of 400 ft., and its ends are 60 ft. higher than the middle.
Find the equation of its curve, the height 50 ft. from the center, and the
position of the focus and directrix. (Cf . § 208.)
9. A point moves in such a way that the lines joining it to (0, 0)
and (6, 8) are always perpendicular. Find the equation of the path.
Draw the path. Check your result by elementary geometry.
10. The vertices of a triangle are (8, 7), (6, 3), and (0, 9). Show
analytically that the three medians are concurrent.
Chapter IX
1. Solve forz: 3 z4-17 z2+5 = 0.
2. Is 12 x2-51 z+45 rationally factorable?
3. Find the lowest rational factors of 2 x*— 3 x3 — 7 x2— 5 x—3.
4. Find a root of z3 — 4 z — 2 = 0, accurate to four decimals.
478
MATHEMATICAL ANALYSIS
Chapter X
1. Given 0 = ctn — 12/5, find without tables all the other functions
for both possible values of 0<360°.
2. Find the radian equivalent of 52° 23', and the degree equivalent
of 2.182<r>.
3. Find from tables the sine, cosine, and tangent of :
105°, 200°, 348° 10'.
4. Find both possible angles <360° for which :
sin A = - .28765, cos B = - .42859, tan C = 3.6962.
6. A point P moved in a circle of radius 10 in. so that 6 = .2P
(radians). Find its speed after 10 sec.
6. An alternating current varied thus: i = 10 sin (60 •*£), the angle
60 irt being in radians. Find i and di/dt at the instant when t = .02.
When did i reach its first maximum and when did it next become zero ?
7. The following table shows the intensity of illumination i at various
inclinations 0° from an arc light. Plot in polar coordinates. What is
the maximum i ? For what 0?
0
60
20
0
-20
-30
-40
-50
-70
i
170
250
370
760
1050
1230
1200
700
(a)
Chapter XI
1. Differentiate : y = (tan 0+ctn 0) -^-sec 0 esc2 0.
2. Integrate : 10 sin 6 t dt, cos 0 d 0, tan 0 d 0.
3. Find from the table, pp. 494-497, the following integrals :
C dd /tx C dx
J sin3 0'
(c)
J sin 4 0 cos 2 0 dd.
4. Find all values of 0<360° for which sin 0 cos 0 = .18.
6. Given sin A = % and cos B = — }| (A and B in the same quadrant) ;
find the functions of A +# and A — B.
6. If cos 0 = f, and 0 is acute, find sin 0/2 and sin 2 0.
7. A point moved thus along a straight line : x = S cos 10 rt. What
type of motion is this? With what period and amplitude? Find the
speed .01 sec. after the point passed through the center.
8. What is a cycloid? An involute?
9. An alternating current died out thus: i = 20e~40< sin 400 1.
Find i, and the rate at which i was changing, at t = .0015. When was
the first maximum reached? How frequently did i become zero?
GENERAL REVIEW 479
Chapter XII
1. Find the value of f 6 x3 dx'by integration and by the prismoid
formula.
2. By considering concentric shells of "infinitesimal" thickness
express the volume of a sphere of radius 10 in. as a definite integral.
Work out and check.
3. Find the area generated by revolving the curve y = .5xz about the
X-axis from x = 0 to z = 4.
4. Draw that part of the surface z = x*+yz+2 which stands above
the XF-plane from y = 0 to y = x, and x = 0 to 2. Calculate the volume
shown in the drawing.
6. Find the slope of the section of the surface in Ex. 4 made by
the plane x = 3 at the point y = 2. What sort of curve is that section?
6. What is the lowest point on the surface in Ex. 4?
Chapter XIII
1. Find the 10th term and the sum of the first 8 terms for each of
the progressions :
(a) 3, 12, 21, ..., (6) 3, 12, 48, ...,
(c) 36, 24, 16, ..., (d) 36, 24, 12, ....
2. What is the present value of a bond for $100 bearing 5% interest,
payable semi-annually and maturing 10 years hence, if money is now
worth 4£% compounded semi-annually?
3. What net annual premium would provide for an insurance policy
of $2000, if the company earns 5% and the man lives for 37 years?
(No installment 37 years hence.)
4. A balance of $3250 now due on a house is to be paid off in 60
equal monthly payments beginning 1 month hence. How large should
each installment be if the interest is at 8% ?
5. Because of the accidental death of a workman, the state is to
pay his widow 200 monthly installments of $30 for herself and 100
monthly installments of $5 for a minor child, beginning 1 mo. hence.
What sum set aside to-day and drawing 6% interest, compounded
monthly, would suffice to meet these payments?
6. Find the Maclaurin series for e~x as far as x4.
480 MATHEMATICAL ANALYSIS
7. Expand (2+z)10 by the Binomial Theorem, as far as x3.
8. How is it possible to give a meaning to an imaginary exponent,
or logarithm ? Find log ( — 1) .
9. How can you ascertain whether the values in a given table satisfy
a polynomial formula: y = a+bx+cx2 ...?
Chapter XIV
1. How many "words" (spellings) of 4 letters can be made from the
26 letters without repetition in any word?
2. How many committees of 5 could be selected from a class of 20?
3. How many basketball line-ups are possible when organizing a
team of 2 seniors and 3 juniors, chosen from 7 seniors and 10 juniors?
4. If 8 cards are drawn from a pack, what is the probability that
all will be spades?
6. If we name 5 dates at random, what is the chance that precisely
two will be Wednesdays?
6. If we name two dates at random, throw three dice, and draw four
cards from a pack, what is the probability that both dates will be
Fridays, two dice "fives," and three cards aces? (Express the answer
numerically, without calculating it.)
7. If the probable error of a gun is 60 m., and 65% of the shots are
"overs," how far is the target from the center of impact and in which
direction ?
8. Find the most probable formula of
20 30 40
the type y = ax for the adjacent table: y 17.9 26.9 36.2
9. The following tables relate to the statures of a certain group of
fathers and sons. Find the most probable linear formulas for y and x\
also the index of correlation. (See Ex. 6, p. 459.)*
64
y 166.7
66
67.6
68
69.0
70 ?/ I 65 I 67
69.7 x 165.4166.5
69 I 71
67.9 I 69
Chapter XV
1. Calculate: (3+2 i) • (4-7 i), and 8i-s-(3-5i).
2. Express the following numbers in polar form and find the fifth
root of each: i, -32, 3+4 i.
3. Calculate (cos 40°+ i sin 220°) .6
* Also cf. G. U- Yule, Intro, to Theory of Statistics.
GENERAL REVIEW
Miscellaneous Exercises
INCLUDING SOME PROBLEMS INVOLVING COMBINATIONS OF
PRINCIPLES
1. The speed V of a chemical reaction is 36 units at a temperature
of 20°, and doubles with every rise of 10°. Obtain a formula for the
speed at any temperature T°. Find V when T = 25.
2. A point moves so that its distance from (3, 0) is always twice
its distance from (0, 0). Find the equation of the path and draw it.
3. Express by formulas the answers to these questions :
(a) How much deposited now would provide for 18 annual install-
ments of $900 each, beginning 24 years hence, if interest is at 6%?
(6) How much must we pay quarterly, 20 times, beginning 3 months
hence, to pay off $3000 now due on a house, with interest at 8% com-
pounded quarterly?
4. Simplify esc 0-s-(tan 0-fctn 0) and find its value when 0 = 290°.
6. Find all possible values of 0<360° for which
cos 0(2 sin 0+l)(sin 0-cos 0) =0.
6. The same as Ex. 5 for 27 sec2 0-54 tan 0-35 ctn 0+9=0.
7. Given a table of bank clearings each year for 50 years past, what
is the best way to plot, to exhibit percentage gains or losses in various
intervals? Why?
8. Given any table of experimental values, how would you proceed
to discover the law?
9. Find all the rational roots of 3 x5+5 x*-17 x*-22 x*+15x-2
and approximate one irrational root to 2 decimals.
10. A string is unwound from a circle of radius 10 in. at an angular
rate of .1 (rad./sec.). Find how fast its end P is moving when t = 2.
Also find how far P travels in the first 10 sec.
11. A hemispherical cistern of radius 10 ft. is full of water. Calculate
the volume of water by elementary geometry, by the prismoid formula,
and by integration.
12. (a) Calculate the wet area of the cistern in Ex. 11 by integration,
and check. (6) Knowing that the pressure x ft. below the surface is
62.5 x Ib. per sq. ft., find the total force with which the water presses
against the cistern.
481
482 MATHEMATICAL ANALYSIS
13. In Ex. 11 find the work required to pump all the water up to a
level 4 ft. above the top.
14. Differentiate and simplify :
(e) tog' ® e-20' sin 400*.
15. Integrate, using tables if necessary :
(a) cscHctntdt, (6) Vx2 - 100 x dx,
( \ dx / 7\ dx
} (100 -a*)*' U *V^25'
(e) 25 e-* cos 3 t dt, (f) 30 sin 200 t dt.
16. Find the numerical values of
f'
)2 dx, ctn 6dd'
17. A point moves so that its distance from the point (0, —16) is
always f of its distance from the X-axis. Find the equation of its path.
Also draw the path roughly, but so as to show clearly its character.
18. A man bought a piece of property for $1000, and ten years later
bought another for $2000. After five years more, he sold the two for
$5000. The income had meantime just paid for taxes and repairs.
To what rate of interest, compounded annually, would this profit be
equivalent? (Give the answer correct to the nearest tenth of 1%.)
19. What is the curve 10 x2— 2 xy+y2 = 3Q; and what if any hori-
zontal or vertical boundaries has it ?
20. Derive Maclaurin's series for ex, to several terms. Calculate e
to four decimals; also • (ex — 1) dx/x.
21. Would $50, deposited now in a bank which pays 4% interest,
yield enough for 30 annual payments of $3 beginning 1 year hence?
What balance or deficit at the time of the last payment ?
22. Prove analytically* The perpendicular bisector of the line
joining the points A(l, —9) and B(5, 3) is the locus of points equidistant
from A and B.
• 23. At what inclination should a projectile be fired from (0, 0),
with an initial speed of 2000ft./sec., to strike the point (40,000, 20,000),
ignoring air resistance?
GENERAL REVIEW 483
24. If a ball nine containing 3 seniors, 2 juniors, 4 sophomores draw
batting positions at random, what is the chance that the seniors will
bat before all others and the juniors after all others?
25. If the chance for a certain event to occur twice in three trials
is .15, what is the chance, p, that it will occur in a single trial? (Find
p correct to 2 decimals.) t i
An electric current died out as in the adja-
.002
100
36.79
•JO CO
time. Find the rate of decrease at £ = .004 and the '^X^ ^'gg
1.83
.67
cent table. Find a formula for the intensity at any
time. Find the rate of decrease at £ = .004 and the 'QQQ
quantity of electricity passed from t = 0 to < = .01. 008
.010
27. (a) Prove that the volume of a segment cut from a sphere of
radius 10 ft. by a plane h ft. from the center is V = ^ (2000—300 h+hz).
o
(6) Using this formula, find how deep the water would have to be
in a hemispherical cistern of radius 10 ft. if the cistern were half full.
28. If 9 coins are tossed up, what is the chance for no head, 1 head,
2 heads, .etc., up to 5 heads? Plot the " staircase " distribution.
29. How many triangles could be drawn having vertices at points
A,B,C, D, E, F, G, H, 7, J, and K, no three of which are in one straight
line?
30. On a certain day two planets had the positions U (19.8, 303° 14')
and N (30, 114° 34'). Find their distance apart and their rectangular
coordinates, at that time.
31. A point (x, y} moved thus: z = 10 cos t, y = W sec t. Find the
speed and direction of motion at t = ir/4. Also find precisely what kind
of curve the path was, and the area under that curve from x = .5 to x = 1.
32. Write by inspection the product of the two complex numbers
7 (cos 80° +i sin 80°) and 2 (cos 5° +i sin 5°) . Verify your result by
multiplying out and comparing.
33. The "hyperbolic sine" and "hyperbolic cosine" of x are two
higher functions defined as follows :
sink x = \ (ex—e~x), cosh x = \ (e?+e~x).
Find the value of each when x = 0 and when x = 1 . What is the deriva-
tive of each? Find Maclaurin's series for cosh x as far as x3.
34. A flywheel of radius 5 ft. was turning with an angular speed
of 16 rad./min. when the power was cut off, after which the acceleration
was d2 6 /dp = 12 t2— 24 t. Find how far a point on the rim traveled
while the wheel was coming to rest.
APPENDIX
I. PROOFS OF CERTAIN THEOREMS.
II. NUMERICAL SHORT-CUTS.
III. FORMULAS.
IV. THE IDEA OF INFINITY.
V. TABLE OF INTEGRALS.
VI. NUMERICAL TABLES.
INDEX.
ANSWERS.
PROOFS OF CERTAIN THEOREMS
(A) SOME THEOREMS ON LIMITS
(I) // Ao»-0 and k is any fixed number, then k Az->-0.
For k Ax will become and remain numerically less than any number
e that you may name, as soon as Ax becomes and remains less than e/k.
(II) // Az-M) and n is any positive integer, then Axn->-0.
For as soon as the numerical value of Ax is less than 1, any power of
Ax will be smaller than Ax itself. Hence Ax will become and remain
numerically less than any small positive number e, as soon as Ax becomes
and remains less than e.
(III) // Az-M) and a, b, c, ..., k, are any positive numbers,
then the quantity (=*=a Ax ±6 Ax2 ±c Az3 ••• ±/b Azn)-M).
For each power of Ax approaches zero by Theorem II and hence
can be made as small as we please. Let the sum of the numerical
values of a, b, c, ... fc, be denoted by m, some fixed number. Then as
soon as Ax is small enough to make each power less than e/m, the given
quantity will be less than f ±a— ±&— ±c— ••• ±fc— V and hence less
\ m m m mj
than (a+b+c ... +k)e/m, or m e/m, or e.
(B) INSTANTANEOUS SPEED AND DIRECTION OF MOTION
(I)
(2)
Proof of (1) : Let As be the length of arc PQ traveled during a short
time A£, just following the instant. Then the required speed t; is the
limit of the average speed As/AJ.
There is no simple relation between As, Ax, and Ay. But
chord PQ2=Ax2+A?/2,
/ AsV /chord PQV /AxV /Aj/ \« (3)
\U] \ As / \AtJ \At)
486
APPENDIX
487
Let A«->-0. Then (As/AZ)-^, (&x/M)-*~vf, (Ay/Af)->-vy. Also the
ratio of chord PQ to arc As approaches 1 as the arc becomes more nearly
straight. Taking limits in (3) we have (1).
FIG. 152,
Proof of (2) : The direction of motion is the direction of the tangent
line.
tanA=^.
dx
Dividing numerator and denominator by dt, we have (2).
(C) DIVISION BY SYNTHETIC SUBSTITUTION
Theorem: The quotient and remainder which result from
dividing any given polynomial
by (x— k) can be found by substituting k for x synthetically :
••• +an -fan+i [fc
a\ Si 02 >S»n_i Sn
[Here Si denotes the first sum ; £2, the second sum ; etc., i.e.,
Sl = kal+az, S^kSi+at, etc.] (4)
In other words : the remainder is Sn and the quotient is
488 MATHEMATICAL ANALYSIS
Proof: To verify this (or any other) division we need merely show
that multiplying the supposed quotient by the divisor (x — k) and
adding the supposed remainder [Sn] will give the original quantity.
x-k
[+Sn]
But, by (4) above, <S»i— fciai = oj, Sz — kSi = aS) etc.
Hence the multiplication and addition give the original quantity
aixn+aixn-l+a3xn~* ... +aBB-f-an+i, which verifies the division.
(D) ADDITION FORMULAS FOR THE SINE AND COSINE:
ANGLES OF ANY SIZE
sin (A+B) = sin A cos l?-f cos A sin B, .
cos (-4 -ffl) = cos A cos 5- sin 4 sin B.
These formulas are established in § 281 for the case where
A , B and A-\-B are all acute angles. The generalization of
this proof will be effected in three steps :
(I) The formulas are valid when A and B are acute angles
out (A + B) is obtuse.
The proof is again geometrical, the construction and steps being
identically the same as in § 281, — with the single exception that since
cos (A -\-B} is now negative, we must equate it to a difference of lines so
chosen as to be negative.
That is, formulas (5) are valid for all acute angles A, B,
whether A+B is acute or obtuse.
(II) The formulas are valid if either A or B is increased by
90°, — that is, if one angle is obtuse and the other acute.
For suppose A =90° +A', where A' and also B are acute. Then
sin (A+£)=sin(90°-M'+£)=cos (A'+B).*
* If two angles differ by 90°, the sine of the larger equals the cosine of the
smaller; and the cosine of the larger equals minus the sine of the smaller.
(See § 259.)
APPENDIX 489
But as A' and B are acute, we already know that
cos (A'+B) =cos A' cos B — sin A' sin B.
And A' being 90° smaller than A, we may substitute
cosA' = sinA, — sin A' = cos A.*
.'. sin (A+B) =cos (A'+B) =sin A cos £+cos A sin J5,
which is the first of formulas (5) valid for A obtuse and B acute. The
same argument can be applied to cos (A +B).
Evidently the above argument can be repeated if either angle is
again increased by 90° ; and so on indefinitely. Hence (5) are true
for all positive angles, no matter how large.
(Ill) Finally, (5) are true when A and B are either or both
negative.
For the functions of a negative angle are the same as those of a
positive angle, larger by a multiple of 360°. And the formulas are
known to be valid for all positive angles.
Remark. When B is negative, =—B' say, (4) becomes
sin [A + (-£')] = sin A cos (-B')+cos A sin (-£')•
But, as can be seen from a figure by using the definitions in
§254:
cos (-B')=cos5', sin (-£') = -sin B'. (6)
.'. sin (4 — B') =sin A cos B' — cos A sin B'.
Likewise
cos (4 — 5') = cos A cos B+ sin A sin B. (7)
These may be considered as Subtraction Formulas.
(E) AREA OF A TRIANGLE. (Cf. § 151)
S=*Vh(h-a)(h-b)(h-c).
Proof: Let p be the altitude of the triangle drawn to the base 6,
— cutting b into segments x and b—x. (Draw a figure.)
Then p2 = os-^ = cs-(6-x)2,
Whence x = (a5 +6* - c2) -^ 2 b.
490 MATHEMATICAL ANALYSIS
Substituting this value of x in p- = (a+x)(a— x), gives:
[(a+6)2-c2][c*-(a-6)2H462
462
Now the quantities in these parentheses are seen to be
2 A, 2fc-2c, 2A-26, 2A-2a
. z^lQh(h-c)(h-b)(h-a)
46*
And since the area of the triangle is ? bp, we have
which simplifies at once to the formula above.
(F) SIMPSON'S RULE
Theorem : Simpson's rule gives the exact value of the inte-
gral from a to b of any cubic function : y = k+lx+mxz+nx3*
Proof : At x = a, yi = k+la+ma?-)-na3,
at .T = 6, 7/2 = /c-K6
and mid-way between, at x= (a+6)/2,
Multiplied out and used in the rule, this gives
. i(?/i+2/2+4j/m)(6-a)=fc(6-a)+|(62-a2)-|.|(
But this is precisely the value of the integral :
Q.E.D.
* For n = 0 this covers also quadratics. And so on.
APPENDIX 491
NUMERICAL SHORT-CUTS
(A) In Squaring Numbers Mentally
(I) To square any number ending in ^, say (n-\-%), simply multiply
the integer n by the next higher integer, and add \.
Thus (6£)2 = (6X7)+i=42|; (10|)2 = 110i
Similarly, 652= 4225; 1052 = 11025; etc.
(II) To square a number near 100, add to the number its excess over
100, consider the result as " hundreds," and add the square of the
excess.
E.g., 1072 = (107+7) hundreds +72 = 11449.
K N is less than 100, its excess is negative. Thus for 93, e = —7:
/. 932 = [93 + (-7)] hundreds + (-7)2 = 8649.
(III) To square a number near 50, find its excess over 50, add this to
25 to get the "hundreds" ; then add the square of the excess.
E.g., 542 = (25 +4) hundreds +42 = 2916.
(B) In Multiplying and Dividing
(I) Multiplying by 25 : multiply by 100 and divide by 4.
Dividing: reverse this. E.g., 124375 -r- 125 = 124.375X8 =995.
(II) Multipliers slightly less than an even thousand or hundred. E.g.,
to multiply by 297, multiply by 300 and subtract 3 times the number.
(III) To multiply together two nearly equal numbers which differ by
an even integer, use the idea that (a+b) (a—b) = a2— 62.
E.g., 43X37 = (40 +3) (40 -3) = 1600 -9.
(C) Some Simple Approximations
(When 2 is a small fraction.)
Formulas Illustrations
(l±a;)« = l±nx (1.002)4 = 1.008; (.999)3 = 1-.003
l ±x/n Vl. 006 = 1.001% V.9996 = .9998.
log.(l ±z) = ±s-| x2 loge (1.0025) = .0025
sin x& = tan z<r> = x. sin .004 W = .004
cos oj« = 1-| x2. cos .006<'> = 1 - .000018.
492 MATHEMATICAL ANALYSIS
SOME STANDARD FORMULAS
(A) MENSURATION. (Cf. § 61)
Circle: C = 2irr, A=Trr\
Sphere: S = 4irr2, V = %irr*.
Cylinder : S = 2 irrh, V =Trr*h.
Cone: S=irrs, V = $irr2h.
Frustum : S = ir(R+r)s, V = $Trh(R+Rr+r2).
Segment, of height h, cut from sphere of radius r :
(B) ALGEBRA
Roots of Quadratic, az2+6z-fc = 0: o;=
Interest: 4=P(l+r/fc)*", P = A/(l+r/fc)*».
Geom. Progression : l = arn~l, S = a(rn— l)-i-(r — 1).
Arith. : Z = a+(n-l)d, S = $(a+l).
Relation of base e to base 10 : log. N = 2.30259 Iog10 N.
(C) TRIGONOMETRY
Definitions :
sin 0 = y/r, ( = ordinate-i-rad. vector), etc. [§ 253.]
Basic Identities :
sin2 0-f cos2 0=1, tan 0 = sin 0/cos 0, etc. [§ 270.]
Addition Formulas :
sin (A + B) = sin A cos B-f cos A sin B, etc. [§ 280.]
Double-angles :
sin 2 0 = 2 sin 0 cos 0, cos 2 0 = cos2 0— sin2 0.
Half -angles :
sin (4 0) = V$(l-cos0), cos (£ 0) = V£(l+cos 0).
Triangle-laws: Sines, Cosines, Tangents, Half-angles,
Area : see text references in Index.
Projections: p = s cos A, P = S cos A. [§ 113.]
APPENDIX
493
(D) ANALYTIC GEOMETRY
- Distance : ^d=^ix^—x^^. (MZ— 2/0 2-
Mid-point : x = %(xi +z2) , y = i (2/1 +2/2) .
Slope : l=(yz- 2/1) -5- (x2 - zi) .
Inclination : tan 6 = I.
Perpendiculars : l-J^ — — 1, or Z2= —\/l\.
For equations of lines and curves, see Index under name of that curve.
(E) DERIVATIVES
2/ =
dy/dx =
un
xn
nun~l du/dx
nxn~l
* log u
\ogx
-i du/dx
u
J.
X
eu
ex
a*
e* du/dx
e
a* log. a du/dx
y=
dy/dx =
f sin u
cos u
cos u du/dx
— sin u du/dx
tan u
ctn u
sec2 u du/dx
— esc2 u du/dx
sec u
CSC U
sec u tan u du/dx
— esc u ctn u du/dx
uv
u
u dv/dx+v du/dx
v du/dx— u dv/dx
V
V*
THE IDEA OF "INFINITY"
ILLUSTRATION. If y = 60/ (2 — x) , then
(a) There is no possible value for y when x = 2 ;
(b) As x approaches 2, y exists and increases without limit.
These facts are often stated briefly by saying that
or
" y approaches infinity (y-^oo) as x =
"y = infinity (y — oo ) , when x = 2."
These statements, however, must not be misunderstood as saying
that there is some enormous number (oo ) which y equals when x = 2.
They are to be used only in a technical sense, as a brief way of stating
the two facts (a) and (b) above.
* Base e. t Radian measure.
TABLE OF INTEGRALS
GENERAL HINTS
I. Sums of several terms : integrate term by term.
II. Products or powers: multiply out if necessary and feasible.
III. Fractions: often simplified by dividing out, or by writing as
negative powers.
IV. Radicals: may be regarded as fractional powers.
V. High powers : use reduction formulas (32)-(44).
VI. Quadmtic expressions like ax*-\-bx-{-c can be reduced to binomial
forms like a(t-+k) by completing the square :
VII. A constant should be added to each integral below.
INTEGRALS
. Cundu = -^-
J n+1
Here n may have any positive or negative value except —1. E.g.,
fx\ dx = \x\ ; /or1'71 dx = -^ aT'7*.
(2)— (4) are special cases of (1).
2. f (axm+b)nxm~l dx=— --- — • (axm+b)n+l.
J n-f-1 ma
This includes forms like Vx*+25 x3 dx, V4-x2 xdx, (3 x2+7)10 x dxt
x dx/(x* — 16)5, etc. Simply use in (2) the values of a, 6, ra, n, which appear
in each of these forms.
n + 1
3. lsinnz cos a; dx= sinn+1
J
4. f cosnx sin x dx= -- cosn+1 x.
J n + 1
These include forms like sin6 x cos x dx, sin x dx/coa x, etc.
494
APPENDIX 495
6. J — = log u.
(6)-(10) are special cases of (5).
6. (xm~l d* = — log (ax™+6).
•/ axm+6 ma
This includes forms like x*dx/(2x*— 5), Vz dz/(7 z*+9), etc.
JTcos ax 1
ctn ax dx— I — — dx = — log sin ax.
•/ sui ax a
8. (tan ax dx = —- log cos ax, =- log sec ax.
J a a
9. fsecaxdx = f (se°2 a*+sec a* tan a*)^ = 1 log (sec ax+tan ax).
J J sec ax+tan ax a
10. f esc ax dx= — log (esc ax+tan ax).
J a
11. i sin ax dx = — cos ax : f cos ax dx = - sin ax.
J a J a
12. f . dx ^ = * log tan £±^, where k = tan'1 6/0.
Jasmx+6cosx Va2+62 2
/• /• i
13. \ eu du = eu', \ e^dx = - &*.
/* fcj;
14. \ e^ sin axdx = — ~— -(k sin ax — a cos ax).
JxJfcX
ete cos ax dx= (fc cos ax + a sin ax).
&2+a2
^«» i * sin (cz — u)X i sin (Cf ~i&)jC
16. jcosaxcos&xdx^-^^.+^J^^.
17. |sin ax sin 6x dx=sin g-g*-S|g±&j*
18. f sin ax cos 6x dx= -cos («-&)«-COB (a+fc)^
J 2 (a -6) 2(a+6)
19. f cos2 axdx = -=- [ax+£ sin 2 ax].
J 2 a
20. f sin2 ax dx = — [ax - £ sin 2 ax] .
«/ 2 a
496 MATHEMATICAL ANALYSIS
21 ( du - 1 log u~a
J u2-a? 2 a u+a'
22. r_^L_ = ltan-^, ==1 ctn'^.
J u-+a2 a a a a
23. f_ ^_. = l0g (W + V^^^)T
J Vu*±a*
24.
. f Va2-u2 dw = ^V^^+^ sin"1 -.
•^ 2 2 a
. f
J
r
uVu*—a? a a a
27
'
28. f — ^== f^T, where r = vW+&; Use (21) or (22).
J uVau+b Jr*-b
29. J sin-1 x dx
30. Jtan-1 x dx = x tan"1 a; - i log (1 -f x») .
81. V log a; dx = o:'»+1[(n+l) log x -l]-f-(n+l)».
REDUCTION FORMULAS
32. (xne**dx=\xnek*-1lCxn-'i<**dx. Leads to (14).
J K kJ
33. Cx" cosaxdx = — sin ax-- ( xn~l sin ax dx. i
J a aJ
Lead to (11).
r yn
34. i xn sin ax dx= — — cos ax+~ (xn~l
* a aJ
35. f sipn qx dx = ~sinn"1 gx cos «*+*=! fsin"-2 ox dx. dD,
J na n J (19))
or
na n J (20).
36. f cos* ax dx = cosn"1 ax sin flj^— 1 cos"-8 ax d». (11) or
J
APPENDIX 497
37. f tan" ax dx = *§5TlJ* _ (tan"-2 ax dx.
J (n-l)a J
38. f ctn» ax dx = -ctpn~1 ax - ("ctn»-« a* <fe.
J (n — l)o J
39. f secn ax dx = - - — - sin ax sec""1 ax+^—^ \ sec71"2 ax <fx.
J (n-l)a n— lJ
40. i cscn ax dx = - cos ax cscn-1 ax-\ — ^— ( cscn~2 ax dx.
J ,(»— l)a n — JV
In (41) — (44) below, u denotes axn-\-b, and each formula is valid as long
as its denominator is not zero. When a denominator is zero, the expression
is integrable by some other formula, such as (2), (6), (21), etc., or by sub-
stituting axn-\-b = t or xnt.
41. (xm(axn+b)?dx= l (xm+lu*>+npb
J mnl\
42.
bn(p+l)
+ (m+n+np+l) §xmu?+l dx\
43. m
(m+l)b
(
44. (xm(axn+b)Pdx = - - -
a(m+np+l)
N. B. By (41) — (44) the power of the binomial can be raised or lowered
by one unit at each step, or the power outside be increased or decreased by
the power inside the parentheses.
These formulas cover such types as :
/V(2 z3+5)f dx, [Use (44) twice] ; f dx } [m=0, (42) twice] ;
^(x+16)3
dx, [(41) once, then (24)] ; f f * ' [(44) once].
498 MATHEMATICAL ANALYSIS
APPLICATIONS
45. Area under a curve y=f(x), A= (ydx.
46. Volume of a solid, sectional area A,, V= \ Aadx
47. Length of curve, y=f(x), s
48. Surface of revolution about X-axis, S
49. Length of curve, r=/(0), s= (Vr*+(dr/d8)zdd.
50. Work of a force, W=(Fdx.
51. Momentum generated, M-
52. Total water pressure, F
53. Total attraction of rod, F--
54. Quantity of electricity flowing, Q= \idt.
55. Amount of increase at rate R, 1
56. Average value of a varying quan- f Qdx
tity Qi from x=a to x = b: v= a^_a '
SOME IMPORTANT CONSTANTS
T = 3.14159265, Iog10 7r = .49714987.
e = 2.71828183, Iog10 e = .43429488.
loge 10 = -^— = 2.30258509.
log,0 e
1">=57°.2957795
1W = 206264/'.806
1° = .017453290
1" = . 4848137^X10-'
1ft. = 30.48 cm.
1 cu. ft. = 2831 cc.
lib. = .4536 kg.
1 gal. = 231 cu. in.
1 acre = 10sq. chains
1 cm. = .0328 ft.
1 cc. = .0000353 cc.
1 kg. = 2.2046 Ib.
1 cu. ft. water weighs 62.4 Ib.
1 ch. = 66ft.
Equatorial, 3963 mi.
APPENDIX
The Earth's Radius
I Polar, 3950 mi.
499
Gravitational Acceleration (sea-level, lat. 45°)
0 = 980.53 cm./sec.2 0 = 32.17 ft./sec.2
For any other latitude L° multiply by (1-.0026 cos 2 L).
Newtonian Gravitational Constant (Cf . § 102)
In C. G. S. system, G = 6.66X10~8 llnft.-lb.-sec. system, £=1.08X10-9
Distribution of Errors in Normal Case
(E = Probable error)
0 to E, 25% ; EtoZE, 16.1% ; 2 E to 3 E, 6.9%.
More such values are given in the following table.
Zero to ...
Errors be
\E %E
'ween
\E
Zero
E
and
$E
Various Values
\E \E 2 E
3E
4E
49.7
Percentage
6.7
13.2
19.4
25.0
30.0
34.4
38.1
41.2
47.8
The standard deviation <r (or square root of the mean of the squares
of all the deviations of n values from their average) should in the long
run approach 1.4826 E; or
#=.6745 ff.
Binomial Coefficients
The coefficients in the expansion of (a+6)n run j 1
as in Pascal's triangle: e.g., for n = 3, they are 121
1,3,3,1, etc. Each row is formed easily from that 1331
above. 14641
1 5 10 10 5 1
500
MATHEMATICAL ANALYSIS
SQUARES AND SQ. ROOTS — CUBES AND CUBE ROOTS
N
N*
VN-
Vio~N
N'
&r
V/10N
V/100N
1.0
1.00
1.0000
3.1623
1.000
1.0000 2.1544
4.6416
1.1
1.2
1.3
1.21
1.44
1.69
1.0488
1.0954
1.1402
3.3166
3.4641
3.6056
1.331
1.728
2.197
1.0323
1.0627
1.0914
2.2240
2.2894
2.3513
4.7914
4.9324
5.0658
1.4
1.5
1.6
1.96
2.25
2.56
1.1832
1.2247
1.2649
3.7417
3.8730
4.0000
2.744
3.375
4.096
1.1187
1.1447
1.1696
2.4101
2.4662
2.5198
5.1925
5.3133
5.4288
1.7
1.8
1.9
2.89
3.24
3.61
1.3038
1.3416
1.3784
4.1231
4.2426
4.3589
4.913
5.832
6.859
1.1935
1.2164
1.2386
2.5713
2.6207
2.6684
5.5397
5.6462
5.74SU
2.0
4.00
1.4142
4.4721
8.000
1.2599
2.7144
5.8480
2.1
2.2
2.3
4.41
4.84
5.29
1.4491
1.4832
1.5166
4.5826
4.6904
4.7958
9.261
10.648
12.167
1.2806
1.3006
1.3200
2.7589
2.8020
2.8439
5.9439
6.0368
6.1269
2.4
2.5
2.6
5.76
6.25
6.76
1.5492
1.5811
1.6125
4.8990
5.0000
5.0990
13.824
15.625
17.576
1.3389
1.3572
1.3751
2.8845
2.9240
2.9625
6.2145
6.2996
6.3825
2.7
2.8
2.9
7.29
7.84
8.41
1.6432
1.6733
1.7029
5.1962
5.2915
5.3852
19.683
21.952
24.389
1.3925
1.4095
1.4260
3.0000
3.0366
3.0723
6.4633
6.5421
6.6191
3.0
9.00
1.7321
5.4772
27.000
1.4422
3.1072
6.6943
3.1
3.2
3.3
9.61
10.24
10.89
1.7607
1.7889
1.8166
5.5678
5.6569
5.7446
29.791
32.768
35.937
1.4581
1.4736
1.4888
3.1414
3.1748
3.2075
6.7679
6.8399
6.9104
3.4
3.5
3.6
11.56
12.25
12.96
1.8439
1.8708
1.8974
5.8310
5.9161
6.0000
39.304
42.875
46.656
1.5037
1.5183
1.5326
3.2396
3.2711
3.3019
6.9795
7.0473
7.1138
3.7
3.8
3.9
13.69
14.44
15.21
1.9235
1.9494
1.9748
6.0828
6.1644
6.2450
50.653
54.872
59.319
1.5467
1.5605
1.5741
3.3322
3.3620
3.3912
7.1791
7.2432
7.3061
4.0
16.00
2.0000
6.3246
64.000
1.5874
3.4200
7.3681
4.1
4.2
4.3
16.81
17.64
18.49
2.0248
2.0494
2.0736
6.4031
6.4807
6.5574
68.921
74.088
79.507
1.6005
1.6134
1.6261
3.44S2
3.4760
3.5034
7.4290
7.4889
7.5478
4.4
4.5
4.6
19.36
20.25
21.16
2.0976
2.1213
2.1448
6.6333
6.7082
6.7823
85.184
91.125
97.336
1.6386
1.6510
1.6631
3.5303
3.5569
3.5830
7.6059
7.6631
7.7194
4.7
4.8
4.9
22.09
23.04
24.01
2.1679
2.1909
2.21:10
6.8557
6.9282
7.0000
103.823
110.592
117.649
1.6751
1.6869
l.GOSo
3.6088
3.6342
3.6593
7.7750
7.8297
7.8837
5.0
25.00
2.2361
7.0711
125.000
1.7100
1.7213
1.7325
1.7435
3.6840
7.9370
5.1
5.2
5.3
26.01
27.04
28.09
2.25X3
2.2804
2.3022
7.1414
7.2111
7.2801
132.651
140.608
148.877
3.70S4
3.7325
3.7563
7.9S9C.
8.0415
8.0927
5.4
5.5
5.6
29.16
30.25
81.36
2.3238
2.3452
2.3664
7.3485
7.4162
7.4833
157.464
166.375
175.616
1.7544
1.7652
1.7752
3.7798
3.8030
3.8259
8.1433
8.1932
8.2426
5.7
5.8
r,.o
32.49
33.64
34.81
2.3875
2.4083
2.4290
7.5498
7.6158
7.r,sn
185.193
195.112
205.379
1.7863
1.7967
1.8070
3.8485
3.8709
H.8MO
8.2913
8.3396
8.8872
6.0
36.00
2.4495
7.7460
210.000
1.8171
3.9149
8.4343
APPENDIX
501
SQUARES AND SQ. ROOTS — CUBES AND CUBE ROOTS
N
N2
VrT
NS
#r
^/10N
VlON
^/lOON
6.0
36.00
2.4495
7.7460
216.000
1.8171
3.9149
8.4343
6.1
6.2
6.3
37.21
38.44
39.69
2.4698
2.4900
2.5100
7.8102
7.8740
7.9372
226.981
238.328
250.047
1.8272
1.8371
1.8469
3.9365
3.9579
3.9791
8.4809
8.5270
8.5726
6.4
6.5
6.6
40.96
42.25
43.56
2.5298
2.5495
2.5690
8.0000
8.0623
8.1240
262.144
274.625
287.496
1.8566
1.8663
1.8758
4.0000
4.0207
4.0412
8.6177
8.6624
8.7066
6.7
6.8
6.9
44.89
46.24
47.61
2.5884
2.6077
2.6268
8.1854
8.2462
8.3066
300.763
314.432
328.509
1.8852
1.8945
1.9038
4.0615
4.0817
4.1016
8.7503
8.7937
8.8366
7.0
49.00
2.6458
8.3666
343.000
1.9129
4.1213
8.8790
7.1
7.2
7.3
50.41
51.84
53.29
2.6646
2.6833
2.7019
8.4261
8.4853
8.5440
357.911
373.248
389.017
1.9220
1.9310
1.9399
4.1408
4.1602
4.1793
8.9211
8.9628
9.0041
7.4
7.5
7.6
54.76
56.25
57.76
2.7203
2.7386
2.7568
8.6023
8.6603
8.7178
405.224
421.875
438.976
1.9487
1.9574
1.9661
4.1983
4.2172
4.2358
9.0450
9.0856
9.1258
7.7
7.8
7.9
59.29
60.84
62.41
2.7749
2.7928
2.8107
8.7750
8.8318
8.8882
456.533
474.552
493.039
1.9747
1.9832
1.9916
4.2543
4.2727
4.2908
9.1657
9.2052
9.2443
8.0
64.00
2.8284
8.9443
512.000
2.0000
4.3089
9.2832
8.1
8.2
8.3
65.61
67.24
68.89
2.8460
2.8636
2.8810
9.0000
9.0554
9.1104
531.441
551.368
571.787
2.0083
2.0165
2.0247
4.3267
4.3445
4.3621
9.3217
9.3599
9.3978
8.4
8.5
8.6
70.56
72.25
73.96
2.8983
2.9155
2.9326
9.1652
9.2195
9.2736
592.704
614,125
636.056
2.0328
2.(k08
2.0488
4.3795
4.3968
4.4140
9.4354
9.4727
9.5097
8.7
8.8
8.9
75.69
77.44
79.21
2.9496
2.9665
2.9833
9.3274
9.3808
9.4340
658.503
681.472
704.969
2.0567
2.0646
2.0723
4.4310
4.4480
4.4647
9.5464
9.5828
9.6190
9.0
81.00
3.0000
9.4868
729.000
2.0801
4.4814
9.6549
9.1
9.2
9.3
82.81
84.64
86.49
3.0166
3.0332
3.0496
9.5394
9.5917
9.6436
753.571
778.688
804.357
2.0878
2.0954
2.1029
4.4979
4.5144
4.5307
9.6905
9.7259
9.7610
9.4
9.5
9.6
88.36
90.25
92.16
3.0659
3.0822
3.0984
9.6954
9.7468
9.7980
830.584
857.375
884.736
2.1105
2.1179
2.1253
4.5468
4.5629
4.5789
9.7959
9.8305
9.8648
9.7
9.8
9.9
94.09
96.04
98.01
3.1145
3.1305
3.1464
9.8489
9.8995
9.9499
912.673
941.192
970.299
2.1327
2.1400
2.1472
4.5947
4.610
4.6261
9.8990
9.9329
9.9666
10.0
100.00
3.1623
10.0000
1000.000
2.1544
4.6416
10.0000
Notes: (I) To determine in which column to find a required root, use
the pointing-off method, as in § 12. Thus
V.0076 =V/- 00-76- ; starts with 8 hence .087178.
-y^.076 =V/-Q76-OQQ"; starts with 4 hence .42358.
V/7600000 = V/7- 600- 000- ; starts with 1 hence 196.61.
(II) For any third figure / in N, add / tenths of the difference between
tabulated values; e.g., ^7\64 = 1.9661+T4ff (1.9747-1.9661) = 1.9695.
502 MATHEMATICAL ANALYSIS
NATURAL LOGARITHMS (Base e)
N
Hj
0
123
456
789
0.0 000
100 198 296
392 488 5S3
676 770 862
1.1
1.2
1.3
953
0.1 823
0.2 624
*044 *133 *222
906 989 *070
700 776 852
*310 *398 *484
*151 *231 *311
927 *001 *075
*570 *655 *740
*390 *469 *546
*148 *221 *293
1.4
1.0
1.6
0.3 365
0.4 055
700
436 507 577
121 187 253
762 824 886
646 716 784
318 382 447
947 *008 *068
853 920 988
511 574 637
*128 *188 *247
1.7
1.8
1.9
0.5 306
878
0.6 419
365 423 481
933 988 *043
471 523 575
539 596 653
*098 *152 *216
627 678 729
710 766 822
*259 *313 *366
780 831 881
2.0
2A
2.2
2.3
931
981 *031 *080
*129 *178 *227
*275 *324 *372
0.7 419
885
0.8 329
467 514 561
930 975 *020
372 416 459
608 655 701
*065 *109 *154
502 544 587
747 793 839
*198 *242 *286
629 671 713
2.4
2.5
2.0
755
0.9 163
555
796 838 879
203 243 282
594 632 670
920 961 *002
322 361 400
708 746 783
*042 *083 *123
439 478 517
821 858 895
2.7
2.S
2.9
3.0
3.1
3.2
3.3
933
1.0 296
647
969 *006 *043
332 367 403
682 716 750
*080 *116 *152
438 473 508
784 818 852
*188 *225 *260
543 578 613
886 919 953
986
*019 *053 *086
*119 *l5l *184
*217 *249 *282
1.1 314
632
939
346 378 410
663 694 725
969 *000 *030
442 474 506
756 787 817
*060 *090 *119
537 569 600
848 578 909
*149 *179 *208
3.4
3.5
3.6
1.2 238
528
809
267 296 326
556 585 613
837 865 892
355 384 413
641 669 698
920 947 975
441 470 499
726 754 782
*002 *029 *056
3.7
3.S
3.9
1.3 083
350
619
110 137 164
376 402 429
635 660 686
191 218 244
455 481 507
712 737 762
271 297 324
533 558 584
788 813 838
4.0
4TT
4.2
4.3
863
888 913 938
962 987 *012
*036 *061 *085
1.4 110
351
586
l.'U 159 183
375 398 422
609 633 656
207 231 255
446 469 493
679 702 725
279 303 327
516 540 563
748 770 793
4.4
4.5
4.6
816
1.5 041
261
839 861 884
063 085 107
282 304 326
907 929 951
129 151 173
347 369 390
974 996 *019
195 217 239
412 433 454
J.7
4.8
4.9
476
686
892
497 518 539
707 728 748
913 933 953
560 581 602
769 790 810
974 994 *014
623 644 655
831 851 872
*034 *054 *074
5.01.6 094
114 134 154
174 194 214
233 253.. 273
X
e*
.05
1.051
.10
1.105
.15
1.162
.20
1.221
.25
1.284
.30
1.350
.35
1.419
.40
1.492
.45
1.568
.50
1.649
.6
1.822
.7
2.014
.8
2.225
.9
2.460
1.0
2.718
1.1
3.004
1.2
3.320
1.3
3.670
1.4
4.055
1,5
4.482
1.6
4.953
1.7
5.474
1.8
6.050
1.9
6.686
2.0
7.389
2.1
8.166
2.2
9.025
2.3
9.974
2.4
11.023
2.5
12.182
3.0
20.0S5
3.5
33.115
4.0
54.600
4.5
90.017
5.0
148.413
5.5
244.692
6.0
403.429
6.5
665.13
7.0
1096.6
7.5
1808.0
8.0
29S1.0
Notes : When given a larger or smaller value of N, express it
In Scientific Notation (j 167).
Thus 1720 = 1.72 X 103. .'. log 1720 = log 1.72 + 3 log 10.
When given a logarithm outside the table, reverse this operation.
MULTIPLES OF Loge 10
log 10 - 2.3026 4 log 10 - 9.2103 - log 10 = .6974 - 3
2 log 10 - 4.6052 5 log 10 -11.5129 - 2 fog 10 - .3948 - 5
3 log 10 - 6.9078 6 log 10 = 13.8155 - 3 log 10 = .0922 - 7
Note: Don't, interpo-
late in this small t;il>l<>.
Locate further values
of x among Ions in tho
main table, and read e*
from N-column.
APPENDIX
503
NATURAL LOGARITHMS (Base e)
In q
5.0
1.6 094
114 134 154
174 194 214
233 253 273
5.1
5.2
5.3
292
487
677
312 332 351
506 525 544
696 715 734
371 390 409
563 582 601
752 771 790
429 448 467
620 639 658
808 827 845
5.4
5.5
5.6
864
1.7 047
228
882 901 919
066 084 102
246 263 281
938 956 974
120 138 156
299 317 334
993 *011 *029
174 192 210
352 370 387
5.7
5.8
5.9
405
579
750
422 440 ' 457
596 613 630
766 783 800
475 492 510
647 664 681
817 834 851
527 544 561
699 716 733
867 884 901
6.0
918
934 951 967
984 *001 *017
*034 *050 *066
6.1
6.2
6.3
1.8 083
245
405
099 116 132
262 278 294
421 437 453
148 165 181
310 326 342
469 485 500
197 213 229
358 374 390
516 532 547
6.4
6.5
6.6
563
718
871
579 594 610
733 749 764
886 901 916
625 641 656
779 795 810
931 946 961
672 687 703
825 840 856
976 991 *006
6.7
6.8
6.9
1.9 021
169
315
036 051 066
184 199 213
330 344 359
081 095 110
228 242 257
373 387 402
125 140 155
272 286 301
416 430 445
7.0
459
473 488 502
516 530 544
559 573 587
7.1
7.2
7.3
601
741
879
615 629 643
755 769 782
892 906 920
657 671 685
796 810 823
933 947 961
699 713 727
838 851 865
974 988 *001
7.4
7.5
7.6
2.0 015
149
281
028 042 055
162 176 189
295 308 321
069 082 096
202 215 229
334 347 360
109 122 136
242 255 268
373 386 399
7.7
7.8
7.9
412
541
669
425 438 451
554 567 580
681 694 707
464 477 490
592 605 618
719 732 744
503 516 528
631 643 656
757 769 782
8.0
794
807 819 832
844 857 869
882 894 906
8.1
8.2
8.3
919
2.1 041
163
931 943 956
054 066 080
175 187 199
968 980 992
090 102 114
211 223 235
*005 *017 *029
126 138 150
247 258 270
8.4
8.5
8.6
282
401
518
294 306 318
412 424 436
529 541 552
330 342 353
448 460 471
564 576 587
365 377 389
483 494 506
599 610 622
8.7
8.8
8.9
633
748
861
645 656 668
759 770 782
872 883 894
679 691 702
793 804 815
905 917 928
713 725 736
827 838 849
939 950 961
9.0
972
983 994 *006
*017 *028 *039
*050 *061 *072
9.1
9.2
9.3
2.2 083
192
300
094 105 116
203 214 225
311 322 332
127 137 148
235 246 257
343 354 364
159 170 181
268 279 289
375 386 396
9.4
9.5
9.6
407
513
618
418 428 439
523 534 544
628 638 649
450 460 471
555 565 576
659 670 680
481 492 502
586 597 607
690 701 711
9.7
9.8
9.9
721
824
925
732 742 752
834 844 854
935 946 956
762 773 783
865 875 885
966 976 986
793 803 814
895 905 915
996 *006 *016
10.
2.3 026
036 046 056
066 076 086
096 106 115
z
e-*
.05
.951
.10
.15
.20
.25
.30
.35
.40
.45
.50
.905
.861
.819
.779
.741
.705
.670
.638
.606
.6
.549
.7
.8
.9
1.0
1.1
1.2
1.3
1.4
1.5
.496
.449
.406
.368
.333
.301
.272
.247
.223
1.6
.202
1.7
1.8
1.9
2,0
2.1
2.2
2.3
2.4
2.5
.183
.165
.149
.135
.122
.111
.100
.091
.082
3.0
.050
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
.030
.018 •
.011
.007
.004
.0024
.0015
.0009
.0005
8.0
.0003
Note: Further
values : e~x = — •
See N column for
e? values, 5 being
in body of Table.
504 MATHEMATICAL ANALYSIS
Trigonometric Functions (Radian Measure)
0(r)
sin 6
COS0
tane
8<r>
sinO
COS0
tan0
.00
.000
1.000
.000
1.0
.841
.540
1.557
.05
.10
.15
.20
.25
.30
.35
.40
.45
.60
.050
.100
.149
.199
.247
.296
.343
.389
.435
.479
.999
.995
.989
.980
.969
.955
.939
.921
.900
.878
.050
.100
.151
.203
.255
.309
.365
.423
.483
.546
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
.997
.909
.598
.141
-.351
-.757
-.978
-.959
-.706
-.279
.071
-.416
-.801
-.990
-.936
-.654
-.211
.284 .
.709
.960
14.101
-2.185
-.747
-.143
.375
1.158
4.637
-3.379
-.996
-.291
.60
.70
.80
.90
.565
.644
.717
.783
.825
.765
.697
.622
.684
.842
1.030
1.260
7.0
8.0
9.0
10.
.657
.989
.412
-.544
.754
-.146
-.911
-.839
.871
-6.800
-.452
.648
Radians to Degrees
100 = 57° 17' 44". 806
RADIANS
TENTHS
HUN-
DREDTH8
THOU-
SANDTHS
TEN-THOU-
SANDTHS
HUNDRED-
THOU-
SANDTHS
1
2
3
57 17 45
114 35 30
171 53 14
5 43 46
11 27 33
17 11 19
0 34 23
1 8 45
1 43 08
0 3 26
0 6 53
0 10 19
0 21
0 41
1 02
2
4
6
4
5
6
229 10 59
286 28 44
343 46 29
22 55 06
28 38 52
34 22 39
2 17 31
2 51 53
3 26 16
0 13 45
0 17 11
0 20 38
1 22
1 43
2 04
8
10
12
7
8
9
401 4 14
458 21 58
515 39 43
40 6 25
45 50 12
51 33 58
4 0 38
4 35 01
5 9 24
0 24 04
0 27 30
0 30 56
2 24
2 45
3 06
14
16
19
Degrees to Radians
This change is not often necessary. To make it, multiply out, using
1° -.01 7453293 W, l'». 000290888(0, 1 " = .00000484 8 fr>
APPENDIX
505
TRIGONOMETRIC FUNCTIONS
and their common logarithms
An-
gle
SINE
Value log
TANGENT
Value log
COTANGENT
Value log
COSINE
Value log
0°
.0000 .
.0000
1.0000 0.0000
90°
1°
2°
3°
4°
.0175 8.2419
.0349 8.542S
.0523 8.7188
.0698 8.8436
.0175 8.2419
.0349 8.5431
.0524 8.7194
.0699 8.8446
57.290 1.7581
28.636 1.4569
19.081 1.2806
14.301 1.1554
.9998 9.9999
.9994 9.9997
.9986 9.9994
.9976 9.9989
89°
88°
87°
86°
5°
6°
8°
9°
.0872 8.9403
.1045 9.0192
.1219 9.0859
.1392 9.1436
.1564 9.1943
.0875 8.9420
.1051 9.0216
.1228 9.0891
.1405 9.1478
.1584 9.1997
11.430 1.0580
9.5144 0.9784
8.1443 0.9109
7.1154 0.8522
6.3138 0.8003
.9962 9.9983
.9945 9.9976
.9925 9.9968
.9903 9.9958
.9877 9.9946
85°
84°
83°
82°
81°
10°
11°
12°
13°
14°
.1736 9.2397
.1908 9.2X01}
.2079 9.3179
.2250 9.3521
.2419 9.3837
.1763 9.2463
.1944 9.2887
.2126 9.3275
.2309 9.3634
.2493 9.3968
5.6713 0.7537
5.1446 0.7113
4.7046 0.6725
4.3315 0.6366
4.0108 0.6032
.9848 9.9934
.9816 9.9919
.9781 9.9904
.9744 9.9887
.9703 9.9869
80°
79°
78°
77°
76°
15°
16°
17°
18°
19°
.2588 9.4130
.2756 9.4403
.2924 9.4659
.3090 9.4900
.3256 9.5126
.2679 9.4281
.2867 9.4575
.3057 9.4853
.3249 9.5118
.3443 9.5370
3.7321 0.5719
3.4874 0.5425
3.2709 0.5147
3.0777 0.4882
2.9042 0.4630
.9659 9.9849
.9613 9.9828
.9563 9.9806
.9511 9.9782
.9455 9.9757
75°
74°
73°
72°
71°
20°
21°
22°
23°
24°
.3420 9.5341
.3584 9.5543
.3746 9.5736
.3907 9.5919
.4067 9.6093
.3640 9.5611
.3839 9.5842
.4040 9.6064
.4245 9.6279
.4452 9.6486
2.7475 0.4389
2.6051 0.4158
2.4751 0.3936
2.3559 0.3721
2.2460 0.3514
.9397 9.9730
.9336 9.9702
' .9272 9.9672
.9205 9.9640
.9135 9.9607
70°
69°
68°
67°
66°
25°
26°
27°
28°
29°
.4226 9.6259
.4384 9.6418
.4540 9.6570
.4695 9.6716
.4848 9.6856
.4663 9.6687
.4877 9.6882
.5095 9.7072
.5317 9.7257
.5543 9.7438
2.1445 0.3313
2.0503 0.3118
1.9626 0.2928
1.8807 0.2743
1.8040 0.2562
.9063 9.9573
.8988 9.9537
.8910 9.9499
.8829 9.9459
.8746 9.9418
65°
64°
63°
62°
61°
30°
31°
32°
33°
34°
.5000 9.6990
15150 9.7118
.5299 9.7242
.5446 9.7361
.5592 9.7476
.5774 9.7614
.6009 9.7788
.6249 9.7958
.6494 9.8125
.6745 9.8290
1.7321 0.2386
1.6643 0.2212
1.6003 0.2042
1.5399 0.1875
1.4826 0.1710
.8660 9.9375
.8572 9.9331
.8480 9.9284
.8387 9.9236
.8290 9.9186
60°
59°
58°
57°
50°
35°
36°
37°
38°
39°
.5736 9.7586
.5878 9.7692
.6018 9.7795
.6157 9.7893
.6293 9.7989
.7002 9.8452
.7265 9.8613
.7536 9.8771
.7813 9.8928
.8098 9.9084
1.4281 0.1548
1.3764 0.1387
1.3270 0.1229
1.2799 0.1072
1.2349 0.0916
.8192 9.9134
.8090 9.9080
.7986 9.9023
.7880 9.8965
.7771 9.8905
55°
54°
53°
52°
51°
40°
41°
42°
43°
44°
45°
.6428 9.8081
.6561 9.8169
.6691 9.8255
.6820 9.8338
.6947 9.8418
.7071 9.8495
.8391 9.9238
.8693 9.9392
.9004 9.9544
.9325 9.9697
.9657 9.9848
1.0000 0.0000
1.1918 0.0762
1.1504 0.0608
1.1106 0.0456
1.0724 0.0303
1.0355 0.0152
1.0000 0.0000
.7660 9.8843
.7547 9.8778
.7431 9.8711
.7314 9.8641
.7193 9.8569
.7071 9.8495
50°
49°
48°
47°
46°
45°
Value log
COSINE
Value log
COTANGENT
Value log
TANGENT
Value log
SINE
An-
gle
Note : log sec x = —log cos x,
log esc x = —log sin x.
506
MATHEMATICAL ANALYSIS
COMMON LOGARITHMS (Base 10)
N
01234
56789
u. d.
10
11
12
13
14
0000 0043 0086 0128 0170
0414 0453 0492 0531 0569
0792 0828 0864 0899 0934
1139 11/3 1206 1239 1271
1461 1492 1523 1553 1584
0212 0253 0294 0334 0374
0607 0645 0682 0719 0755
0969 1004 1038 1072 1106
1303 1335 1367 1399 1430
1614 1644 1673 1703 1732
4.2
3.8
3.5
3.2
3.0
15
16
17
18
19
1761 1790 1818 1847 1875
2041 2068 2095 2122 2148
2304 2330 2355 2380 2405
2553 2577 2601 2625 2648
2788 2810 2833 2856 2&78
1903 1931 1959 1987 2014
2175 2201 2227 2253 2279
2430 2455 2480 2504 2529
2672 2695 2718 2742 2765
2900 2923 2945 2967 2989
2.8
2.6
2.5
2.4
2.2
20
21
22
23
24
3010 3032 3054 3075 3096
3222 3243 3263 3284 3304
3424 3444 3464 3483 3502
3617 3636 3655 3674 3692
3802 3820 3838 3856 3874
3118 3139 3160 3181 3201
3324 3345 3365 3385 3404
3522 3541 3560 3579 3598
3711 3729 3747 3766 3784
3892 3909 3927 3945 3962
2.1
2.0
1.9
1.8
1.8
25
26
27
28
29
3979 3997 4014 4031 4048
4150 4166 4483 4200 4216
4314 4330 4346 4362 4378
4472 4487 4502 4518 4533
4624 4639 4654 4669 4683
4065 4082 4099 4116 4133
li'.!!' 4249 4265 4281 4298
4393 4409 4425 4440 4456
4548 4564 4579 4594 4609
4698 4713 4728 4742 4757
1.7
. 1.6
1.6
1.8
1.:,
30
31
32
33
34
4771 4786 4800 4814 4829
4914 4928 4942 4955 4969
5051 5065 5079 5092 5105
5185 5198 5211 5224 5237
5315 5328 5340 5353 5366
4843 4857 4871 4886 4900
•19X3 4997 5011 5024 503S
5119 5132 5145 5159 5172
5250 5263 5276 5289 5302
5378 5391 5403 5416 5428
1.4
1.1
1.3
1.3
1.3
35
36
37
38
39
5441 5453 5465 5478 5490
5563 5575 5587 5599 5611
5682 5694 5705 5717 5729
5798 5809 5821 5832 5843
5911 5922 5933 5944 5955
5502 5514 5527 5539 5551
5623 5635 5647 5658 5670
5740 5752 5763 5775 5786
5855 5866 5877 5888 5899
5966 5977 5988 5999 6010
1.2
1.2
1.2
1.1
1.1
40
41
42
43
44
6021 6031 6042 6053 6064
6128 6138 6149 6160 6170
6232 6243 6253 6263 6274
6335 6345 6355 6365 6375
6435 6444 6454 6464 6474
6075 6085 6096 6107 6117
6180 6191 6201 6212 6222
6284 6294 6304 6314 6325
6385 6395 6405 6415 6425
6484 6493 6503 6513 6522
1.1
1.0
1.0
1.0
1.0
45
46
47
48
49
6532 6542 6551 6561 6571
6628 6637 6646 6656 6665
6721 6730 6739 6749 6758
6812 6821 6830 6839 6848
6902 6911 6920 6928 6937
6580 6590 6599 6609 6618
6675 6684 6693 6702 6712
6767 6776 67S5 6794 6X03
<iXf,7 CiSlifi (5X75 r.xsl r.x«i3
6946 6955 6964 6972 6981
1.0
.9
.'.I
.'.)
.'.)
50
51
52
53
54
6990 6998 7007 7016 7024
7076 7084 7093 7101 7110
7160 7168 7177 71x5 7 1'.»3
7243 7251 7259 7267 7275
7324 7332 7340 7348 7356
7033 7042 7050 7059 7067
7 MX 7126 7135 7143 7152
7202 7210 7218 7226 T2\K
7J.SI 7292 7300 730X 7316
7364 7372 7380 7388 7396
• 1
.8
.s
.8
.8
Note: The column u. d. ( = unit difference) may be used in interpolating. Mul-
tiply the u. d. value by fimm- in 4th place of given number and add to logarithm read
from table for first 3 figures of number.
APPENDIX
COMMON LOGARITHMS
507
55
56
57
58
59
7404 7412 7419 7427 7435
7482 7490 7497 7505 7513
7559 7566 7574 7582 7589
7634 7642 7649 7657 7664
7709 7716 7723 7731 7738
7443 7451 7459 7466 7474
7520 7528 7536 7543 7551
7597 7604 7612 7619 7627
7672 7679 7686 7694 7701
7745 7752 7760 7767 7774
.8
.8
.8
.7
.7
60
61
62
63
64
7782 7789 7796 7803 7810
7853 7860 7868 7875 7882
7924 7931 7938 7945 7952
7993 8000 8007 8014 8021
8062 8069 8075 8082 8089
7818 7825 7832 7839 7846
7889 7896 7903 7910 7917
7959 7966 7973 7980 7987
8028 8035 8041 8048 8055
8096 8102 8109 8116 8122
.7
.7
.7
.7
.7
65
66
67
68
69
8129 8136 8142 8149 8156
8195 8202 8209 8215 8222
8261 8267 8274 8280 8287
8325 8331 8338 8344 8351
8388 8395 8401 8407 8414
8162 8169 8176 8182 8189
8228 8235 8241 8248 8254
8293 8299 8306 8312 8319
8357 8363 8370 8376 8382
8420 8426 8432 8439 8445
.7
.7
.6
.6
.6
70
71
72
73
74
8451 8457 8463 8470 8476
8513 8519 8525 8531 8537
8573 8579 8585 8591 8597
8633 8639 8645 8651 8657
8692 8698 8704 8710 8716
8482 8488 8494 8500 8506
8543 8549 8555 8561 8567
8603 8609 8615 8621 8627
8663 8669 8675 8681 8686
8722 8727 8733 8739 8745
.6
.6
.6
.6
.6
75
76
77
78
79
8751 8756 8762 8768 8774
8808 8814 8820 8825 8831
8865 8871 8876 8882 8887
8921 8927 8932 8938 8943
8976 8982 8987 8993 8998
8779 8785 8791 8797 8802
8837 8842 8848 8854 8859
8893 8899 8904 8910 8915
8949 8954 8960 8965 8971
9004 9009 9015 9020 9025
.6
.6
.6
.6
.5
80
81
82
83
84
9031 9036 9042 9047 9053
9085 9090 9096 9101 9106
9138 9143 9149 9154 9159
9191 9196 9201 9206 9212
9243 9248 9253 9258 9263
9058 9063 9069 9074 9079
9112 9117 9122 9128 9133
9165 9170 9175 9180 9186
9217 9222 9227 9232 9238
9269 9274 9279 9284 9289
.5
.5
.5
.5
.5
85
86
87
88
89
9294 9299 9304 9309 9315
9345 9350 9355 9360 9365
9395 9400 9405 9410 9415
9445 9450 9455 9460 9465
9494 9499 9504 9509 9513
9320 9325 9330 9335 9340
9370 9375 9380 9385 9390
9420 9425 9430 9435 9440
9469 9474 9479 9484 9489
9518 9523 9528 9533 9538
.5
.5
.5
.5
.5
90
91
92
93
94
9542 9547 9552 9557 9562
9590 9595 9600 9605 9609
9638 9643 9647 9652 9657
9685 9689 9694 9699 9703
9731 9736 9741 9745 9750
9566 9571 9576 9581 9586
9614 9619 9624 9628 9633
9661 9666 9671 9675 9680
9708 9713 9717 9722 9727
9754 9759 9763 9768 9773
.5
.5
.5
.5
.5
95
96
97
98
99
9777 9782 9786 9791 9795
9823 9827 9832 9S36 9841
9868 9872 9877 9881 9886
9912 9917 9921 9926 9930
9956 9961 9965 9969 9974
9800 9805 9809 9814 9818
9845 9850 9854 9859 9863
9890 9894 9S99 9903 9908
9934 9939 9943 9948 9952
9978 9983 9987 9991 9996
.5
.5
.4
.4
.4
ABBREVIATIONS AND SYMBOLS
A., amount, area, attraction.
A. P., arithmetical progression.
C. I. L., compound interest law.
G. P., geometrical progression.
M., mass, moment, momentum.
P., pressure, principal, probability.
P. V., present value.
R. P. M., revolutions per minute.
S. H. M., simple harmonic motion.
V., value, volume, speed (velocity).
-T-. derivative (as to x) .
ax
logb, logarithm of • • • , base b.
sin, cos, sine, cosine,
tan, ctn, tangent, cotangent,
sec, esc, secant, cosecant.
Cn,r,Pn,r. (See §§ 327, 330.)
Cis, cosine +'i sine. (See § 350.)
A, triangle.
A (delta), increment.
9 (theta), polar angle.
w (omega), angular speed.
a (alpha), angular acceleration.
n \ factorial n.
l(r>, radian.
i, imaginary unit
| •••!. absolute value of
, integral.
— >, = , approaches the limit
L , limit of • • • , as Ax— > 0.
A.v— >0.
oo, infinity.
— »oo , increases without limit.
<, is less than (algebraically).
>, is greater than (algebrnirjilly).
INDEX
Abbreviations, List of, 508.
Abrupt Extremes, 120.
Abscissa, 271, 351.
Absolute value, 464.
Acceleration, 24, 74, 81, 104, 132,
277, 346, 377.
Addition formulas for sine and co-
sine, 381, 488.
Amortization of debts, 422 ff.
Analytic geometry, 279-325, 343,
365, 375, 386, 402-406.
Angle between curves, 164, 386.
Angular speed, 56, 345 ff., 377.
Annuities, 422-426.
Arc and angle, 349; chord, 63, 361.
Area: of a surface, 88, 169, 400; of a
triangle, 218, 489; under a curve,
29, 64, 136, 393, 410.
Arithmetical progression, 415.
Arrangements, 440.
Artillery, 454. See Projectile.
Asymptotes, 308, 312, 319, 356.
Atmospheric pressure, 4, 251, 264.
Attraction of a rod, 150.
Average value, 24-31, 408-412.
Axes of coordinates, 271, 291, 343.
Axis of a curve, 294, 301, 307.
Bacterial growth, 16, 247, 255, 264.
Beams, 63, 105, 134, 173, 325, 328.
Binomial Distribution, Normal, 452.
Binomial theorem, 431, 444.
Bonds, Valuation of, 425.
Bridges, 171 ff., 290, 298, 303, 325.
Bullet. See Projectile.
Calculation of tables, 185, 225, 430.
Calculus, 151; 76-155, 188, 229 ff.,
255-267, 360-362, 372-385, 388,
392-414, 428-433, 456-458, 473.
Cartesian geometry. See Analytic G.
Chance. See Probability.
Characteristic, 193-200.
Circle: area, 89; equation, 288.
Circular motion, 345-348.
Co-function, 167, 351 ff.
Combinations, 442-445.
Completing the square, 35, 289, 326.
Complex numbers, 460-471.
Component, 169-172, 275, 383.
Compound interest, 209-213, 236 ff.,
418-426. C. /. Law, 243-251, 262.
Computation, Logarithmic, 192-223.
Concurrent lines, 321.
Cone, 88, 323.
Conic, 316, 323, 327.
Constants, 49, 83, 84, 127, 392, 498.
Contour lines, 29.
Coordinates, 271-325, 343-350, 365,
463-465.
Cosecant, 352, 368.
Cosine, 167-171, 177-182, 351-362,
368, 382. C. Law, 177-182.
Cotangent, 167, 352, 368.
Cubes, roots, etc., 21, 500.
Curvature, 175.
Cycloid, 3?5.
Cylinder, 88.
Damped oscillations, 378.
Definite integrals, 392-414.
Degree-measure (calculus), 362, 373.
Delta (A), 19, 43, 59, 78 ff., 255,
361.
Dependent variable. See Function.
Depreciation, 213, 240.
Depression, Angle of, 164.
Derivatives, 78-125, 255-267, 360,
372, 493; higher, 105; partial, 406.
Derived curves, 107.
Descartes, 284.
Die-away curve, 244, 246.
Differential, 121-123, 151.
Differentiation, 76-125, 255-267,
360, 372, 406, 472.
Diminishing a root, 334-338
Direction, 65, 274, 281, 343, 486.
Directrix, 293.
Discovering laws, 50, 253, 436.
Discriminant, 328.
509
510
INDEX
Distance: between points, 279, 344;
traveled, 25, 144, 275, 375.
Division: by zero, 47; of complex
numbers, 462, 466; synthetic, 330,
487.
Double integration, 403.
e (Napierian base), 238, 241.
Electric current, 29, 50, 233, 244,
264, 359, 378, 380, 470.
Elevation, 259; Angle of, 164.
Ellipse, 299-306.
Empirical laws, 50, 253, 436.
Engravers' charts, 312.
Equation of curve, 283 ff.
Equations, Solution of, 35-41, 212,
326-342, 370.
Equilibrium, 158, 171 ff.
Errors, 91, 453-459.
Estimates, 189, 349. See Graphical
solution; Graphs.
Exponential equations, 212, 253.
Exponential functions, 244 ff., 378;
relation to sine and cosine, 433.
Exponents, 98, 190, 212, 433, 434.
Extremes. See Maxima and Minima.
Factorability of a quadratic, 329.
Factorials, 429, 440-444.
Factors of polynomials, 92, 329-338.
Falling bodies, 42, 45, 50, 126, 153.
Flexion, 105.
Focus (foci), 293, 301, 307.
Force problems, 157, 169-173.
Formulas, 42, 44, 48, 50, 76, 88, 128,
142, 209, 313, 492; for roots of a
quadratic, 326; Addition, 381.
Fourier series, 434-436.
Fraction, Derivative of, 85, 265,
372.
Fractional exponents, 98, 119.
Function: definition, 5; notation, 73;
kinds, 45, 78, 82, 104, 110, 127,
HU), 163, 167, 198, 236, 244, 324,
350, 413, 433, 473, 483.
Fundamental theorem of Integral
Calculus, 394.
Gas laws, 50, 61, 102, 116, 233, 258.
Geometrical principles, 34, 63-66,
88, 141 jf., 160, 185, 279 jf.
Geometrical progression, 416-426.
Geometrical representation of com-
plex numbers, 463.
Grade. See Slope.
Graphical representation: of forces,
157; of functions, see Graphs.
Graphical solution: of equations, 37;
of triangles, 157.
Graphs, 3-57, 79, 107, 246, 318, 355,
365; logarithmic, etc., 247-255;
of function of two variables, 402.
Half-angles, 218, 385.
Horizontal tangents, 92, 120, 407.
Horner's method, 337.
Hyperbola, 306-312.
Hyperbolic formulas, 313.
Identities, Trigonometric, 368-371,
381-385, 387, 433.
Imaginaries, 38, 318, 329, 434, 460 jf.
Implicit functions, 116., 295, 324.
Inclination, 163, 281.
Increasing test, 80, 94.
Increment, 43, 80, 89.
Independent variables, 6, 406. •
Indirect differentiation, 110, 111,
257, 362.
Infinite, 48, 356, 493; series, 427 jf.
Infinitesimal analysis, 144, 395-401.
Inflection, Points of, 107.
Instantaneous. See Direction, Hate,
Slope, Speed.
Insurance. See Life Insurance.
Integral, integration, 126 155. LV.S.
276, 379, 388-414, 427, 49 1 ff.
Interpolate, 11, 18, 201.
Intersections, 318, 321, 329,
Investment, 210-214, 342, 418-426.
Involute, 375, 376.
Irrational roots, 35-41, 334, 337 jf.
Isolating a root, 340.
Laws. See Compound Interest, Co-
sine, Discovery, Power, Sine.
Least squares, 456-458.
Leibnitz, 151, 396.
Length of a curve, 63 ff., 399, 400.
Life Insurance, 420.
Limit, 58-75, 78, 193, 395, 427, 486.
Limit of (1+1/*X 238.
INDEX
511
Linear equations, 35, 45, 50, 87, 253,
286, 319.
Locus (loci), 283 Jf.
Logarithm, 193-226, 241, 247-260,
427, 434; to any base, 222.
Logarithmic differentiation, 260 ff.
Logarithmic plotting, 252.
Logarithmic solution of triangles,
216-222.
Maclaurin series, 428 ff.
Mantissa, 193-200.
Mass, 150, 398.
Mathematical analysis, 1.
Maxima and Minima, 32, 94-98,
108, 120; for two variables, 406.
Mean ordinate, 24-31, 409.
Mean value, 24-31, 408-412.
Measurement, Errors of, 91, 453 jf.
Mensuration formulas, 88, 492.
Midpoint, 280.
Moment of a force, 172, 398.
Momentum, 26, 137, 393.
Motion, 58, 62, 78, 104, 132, 272-278,
344-346, 374-377.
Multiple-angle formulas, 383.
n-th roots, 469.
Napier, 225.
Napierian base (e), 238, 241.
Natural logarithms, 241-243.
Negative rates, etc., 59, 61, 68, 80.
Newton, 151, 339, 396.
Nomographic charts, 224, 252.
Normal probability curve, 453-459.
Notation, 43, 72-73, 104, 121, 134,
190, 193, 363, 508.
Oblique triangles, 176-183, 216-222.
Obtuse angles, 180-182, 351, 357 ff.
Operations, with complex numbers,
462, 466-470; with logarithms,
199, 203.
Ordinate, 7, 25, 271.
Origin, 271, 343.
Original meaning of /, 143, 396.
Oscillations, 359, 376-378, 435.
Parabola, 293-299, 313, 316.
Parabolic formulas, 313.
Partial derivatives, 406.
Path of motion, 272, 316, 344.
Pendulum, 45, 56, 73, 360, 411.
Percentage, 91, 209, 236-251, 455.
Periodic oscillations, 359.
Permutations, 440.
Perpendicularity test, 282.
Pistons, 29, 174, 233, 258.
Planets, 50, 121, 180, 255, 302.
Planimeter, 30.
Point-slope equation, 319.
Polar axis, 343.
Polar coordinates, 343-350, 365, 465.
Polynomials, 37 ff., 85, 436.
Power laws, 82, 100, 252-253, 313.
Power series, 427 ff.
Present value, 422.
Pressure, Fluid, 146, 175, 188, 232.
Prismoid formula, 411.
Probability, 447-458.
Probable error, 453.
Product, Derivative of a, 85, 265,
375, 379.
Products and sums, Trig'metric, 387.
Progressions, 415-426.
Projectile, 42, 50, 58, 76, 125, 132,
277, 316, 370, 454.
Projection, 168-170, 305.
Proportional parts, 18, 201.
Protractor, 156.0".
Quadrants, 352, 354.
Quadratic equation, 35, 326-329.
Radian, 347, 356, 360, 400, 429, 434.
Radium, Decomposition of, 16, 241.
Radius vector, 343, 365.
Rates, 13-16, 42, 60, 76, 79, 105, 108:
of growing areas, etc., 135-150;
percentage, 236-251; related, 114-
118; reversed, 126-151.
Rational roots, 331-333.
Real number system, 38, 460.
Reciprocals, 48, 99, 283, 352.
Rectangular coordinates, 271-325,
374, 463.
Rectangular hyperbola, 312 ff.
Reducing to acute angles, 357-358.
Related rates, 114.
Repeated differentiation, 105-108;
integration, 131 ff., 277, 403.
Resultant force, 158 ff.', speed, 275.
512
INDEX
Roots, of an equation, 35-41, 326-
341, 370; Tables of, 21, 500.
Rotating a curve 90°, 297, 327.
Rotating fluid, 128, 295, 296.
Scales, 11, 224, 248-254.
Scientific Notation, 190, 242.
Secant, 65 #; 352, 368, 373.
Sections of a surface, 405.
Segments, 141, 153, 166, 492.
Semi-logarithmic plotting, 247 ff.
Scries, 427-435.
Simple harmonic motion, 376.
Simpson's rule, 409, 490.
Simultaneous equations, 52, 321, 437,
458.
Simultaneous triangles, 181.
Sine, 160, 180-182, 214, 217, 351-
362, 382. S. law, 178-182, 216.
Slide rule, 223, 247-252.
Sliders. See Translate s.
Slope, 67, 68, 79, 107, 163, 280, 405.
Solution. See Equation, Triangle.
Sound ranging, 310.
Speed, : -5, 58, 76 ff., 274, 486.
Sphere, 38, 141, 146.
Squares, roots, etc., 21, 500.
Straight lines, 45, 253, 286, 319.
Successive triangles, 184.
Summaries, 53, 74, 123, 151, 154,
185, 225, 227, 267, 323, 341, 366,
389, 413, 438, 458, 471.
Sums and products, Trig' metric, 387.
Surface, Plotting a, 402, 405.
Surface area, 88, 169, 400.
Suspension cable, 45, 67, 130, 295 if
Symbols, List of, 508.
Synthetic substitution, or division
40, 330, 487.
Tables, 161, 500-507.
Tangent line, 15, 65-68, 92, 107, 275.
Tangent (of an angle), 160-164
218-222, 351, 356, 368, 373, 386.
Tanks, 5, 45, 101, 187, 233, 394. .
Taylor series, 428 ff.
Telescope, Reflecting, 295, 297 ff.
Time rates, 112j7.
Translaters of a curve, 314, 335.
Triangles, 116, 156-188, 216-222.
Trigonometric equations, 370.
Trigonometric functions, 160-188,
214-222, 350-391, 428-431, 433.
Trigonometry, 156. See also above.
Variables, 1-6, 70, 324, 413.
Varies as, 48.
Vectors, 158, 275, 463^.
Vectorial angle, 343.
Vibrations, 359, 378, 435.
Volume, 30, 64, 139, 397, 403, 412.
Water pressure, 146, 175, 188.
Work, 27, 95, 138, 398.
Zero, 35, 47, 92, 99, 354, 444.
Some Further Applications to Various Fields> mainly in
Exercises
Aeronautics, 10, 27, 166, 325, 391.
Agriculture, 34, 87, 390, 436, 475.
Architecture, 41, 98, 166, 228, 306, 413.
Astronomy, 12, 28, 188, 306, 339, 350.
Biology, 5, 9, 31, 53, 339, 455, 4so.
Business, 12, 55, 57, 224, 342, 459.
( h. -mistry, 53, 166, 191, 254, 264, 458.
i ivil Eng'r'g, 63, 145, 150, 159, 269.
Economics, 4, 55, 87, 231, 269, 476, 481.
Heat, 16, 20, 53, 57, 191, 263, 408,
Hydraulics, 39, 121, 192, 201. 398.
Light and photography, 49, 102. 112,
165, 191, 264, 207. 270. 30(5. 478.
Machinery, 42, 56, 84, 254, 264, 270.
Magnetism, 81, 140, 226, 325. 389,
Map-making, 29. 30. 273, 282, 344.
Medicine, 1, 11, 13, 18, 54, 270.
Metal work, 34, 91, 17."), 1S6. 20<i. ::or,.
X;ivi-;ition, 24, 118, 171, :;'.»<). : 91,
Physiography, 121, 168, 171, 217, 363.
Psychology, 5, 166, 186, 459, -I so.
Hail ways, 28, 31, 84, 87, 106, Ki.',, 20(1.
Shipbuilding, 30, 87, 187, 27."., 2<)<;/.
Sociology, 3, 13, 250 ff., 270, 452, 479.
Sound, music, 302, 360, 381, -I is, 435.
Sport. 118, 124, 274, 279, 344, 442 ff.
'IVlrphony, etc., 3, 56, 2f,0, L'ii7. :;'.«).
Warfare, 55, 118, 171, 180, 273, 325.
MAY 14 1954
QA Griffin, Prank Loxley
300 An introduction to mathe-
G74. matical analysis
Physical &
Applied Sci.
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