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MATHEMATICS
FOR
AGRICULTURAL STUDENTS
McGraw-Hill BookGompany
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2j_
f
EDITED BY CHARLES S. SLIGHTER
MATHEMATICS
FOR
AGRICULTURAL STUDENTS
BY
HENRY C. WOLFF, PH. D.
ASSISTANT PROFESSOR OF MATHEMATICS
UNIVERSITY OP WISCONSIN
FIRST EDITION
McGRAW-HILL BOOK COMPANY, INC.
239 WEST 39TH STREET, NEW YORK
6 BOUVERIE STREET, LONDON, E. C.
1914
COPYRIGHT, 1914, BY THE
McGRAW-HiLL BOOK COMPANY, INC.
THE. MAPLE. PKES3. YORK-PA
£3
PREFACE
The present book was designed as the text for a working course
in elementary Mathematics given in the College of Agriculture of
the University of Wisconsin. It is not a text book of Advanced
Algebra, Trigonometry, Analytic Geometry, or "Practical Mathe-
matics." Students pursuing a scientific course, in which but a
year of college Mathematics is offered, receive little profit from a
formal course in Advanced Algebra, Trigonometry, or Analytic
Geometry. On the other hand, such students benefit little from a
so-called "practical" course of a type which does little more than
train them to substitute in given formulas, or to use formulas
merely as a means to an end. Such a course does little to develop
the habit or power of clear and logical thinking.
This book is the outgrowth of mimeograph notes used for the
past three or four years with undergraduate students of agricul-
ture. With the exception of the illustrations and exercises, the
book contains, nevertheless, nothing which is of exclusive in-
terest to agricultural students. It may be used equally well
with any class of scientific students who desire only a short
course in mathematics beyond elementary Algebra and Geometry.
The plan of the book is:
First, to select material primarily on the basis of its usefulness
to scientific students.
Second, to illustrate the principles with problems of interest to
the agricultural student, or with problems having direct applica-
tion to his work.
Third, to give minute and detailed explanations of all new
work, and to assume the minimum attainments in mathematical
preparation on the part of the student. On the other hand, details
are often omitted from work that is not new to the student, in
order that they may be supplied by the student himself.
Fourth, to give detailed directions for doing work "at home" so
vi PREFACE
that the student of moderate mathematical attainments may,
nevertheless, "learn by doing."
Fifth, to include certain material not elsewhere readily acces-
sible to the student of science, for the purpose of rendering the
book useful for reference throughout the four years of the college
course.
Sixth, to select the topics and the amount of material under
each topic so that either a half year or a full year may be
devoted to the work.
In writing this book the author had in mind the preparation
of a text which would :
First, train the student to do neat and careful work.
Second, encourage the student to make further use of elemen-
tary Algebra and Geometry.
Third, develop in the student the habit of careful and logical
thinking.
Fourth, train the student to study a problem with a view of
discovering the shortest and easiest method of handling it, rather
than attacking it by the "first thought-of " method.
Fifth, show the student, by illustrations and exercises, how
mathematics may be helpful in pursuing other subjects of study
••and useful in a "practical" way.
While this book includes enough material for a year's course of
study, it is believed that it does not contain too much material
for a shorter course. The inclusion of additional material in the
text, will at least let the student know of the existence of sub-
jects in mathematics not covered by him in the class room.
Even this superficial knowledge may be very helpful to him
later in connection with other scientific work, especially if he,
knows he can find brief discussions in a familiar book.
The author takes this opportunity to thank Professor C. S.
Slichter for assistance in the preparation of the entire manuscript;
and to acknowledge his indebtedness to Professor E. V. Hunting-
ton for valuable suggestions and criticisms. Acknowledgments
are due Mr. E. Taylor and Mr. T. C. Fry for suggestions based
upon their use of the preliminary notes in the class room.
A very brief review of elementary Algebra is given in the intro-
duction, besides a list of materials and instruments. In the
PREFACE vii
appendix is given, for reference, a list of common mathematical
symbols and a few formulas of mensuration.
Material in the book which may be omitted when less than a
full year is devoted to the course, is indicated by enclosing
exercise and section numbers, and chapter headings within
brackets.
HENRY C. WOLFF.
UNIVERSITY OP WISCONSIN
September, 18, 1914
CONTENTS
PAGE
PREFACE v
INTRODUCTION 1
CHAPTER
I GRAPHIC REPRESENTATION 23
II LOGARITHMS 60
III THE CIRCULAR FUNCTIONS: THE TRIANGLE 80
IV THE ELLIPSE 128
V THE SLIDE RULE 143
VI STATICS 155
VII PERMUTATIONS, COMBINATIONS, AND THE BINOMIAL EX-
PANSION. 189
VIII PROGRESSION 200
IX PROBABILITY 207
X SMALL ERRORS 231
XI POINT, PLANE, AND LINE IN SPACE 239
XII MAXIMA AND MINIMA 252
XIII EMPIRICAL EQUATIONS 264
APPENDIX 286
INDEX . 303
MATHEMATICS FOR
AGRICULTURAL STUDENTS
INTRODUCTION
1. Instruments and Materials. A large number of problems in
this book are to be solved graphically or solved both graphically
and analytically, and a large number of the
exercises are exercises in graphic representation
of observed or computed data. In order that
this work may be performed neatly and accur-
ately, it is necessary that the student have at
least a few simple drawing instruments. The
indispensable instruments, together with other
material required for the work, are here listed
and briefly described.
1. Two 4H Drawing Pencils. One, sharp-
ened to a fine point, is used for marking points
upon paper, or for sketching free-hand. The
other, sharpened to a chisel-point, is used for
drawing straight lines. Two views of a chisel-
pointed pencil are given in Fig. 1. When
drawing lines, a flat side of the chisel-point is
held against the straight edge of the guide,
thus causing the pencil always to move in a
direction with the chisel edge and to produce a
fine and distinct line. For fine and accurate
work, both pencil points must be kept sharp.
Neither point ought to become so rounded or
flat that any appreciable metallic luster will
appear at the wearing surface of the graphite.
FIG. 1.— Two
views of a chisel-
pointed pencil.
A good way of
keeping pencil points sharp is by occasionally whetting upon fine
1
2 MATHEMATICS [§1
sand-paper. Small pads of sand-paper for this purpose may be
procured from stationers or dealers in drawing materials. Hex-
agonal pencils should always be used. The student may prefer
to have one pencil sharpened at both ends; one end with a round
point, and the other end with a chisel-point.
2. One 3H Pencil, sharpened to a round point for lettering.
3. A small drawing board made of soft wood to which the draw-
ing paper may be fastened with thumb-tacks. A drawing board
should have at least one straight edge along which the T-square
may slide. A board 12 by 18 inches will be large enough.
FIG. 2. — T-square.
4. A small T-square about 15 inches long. A T-square, Fig.
2, consists simply of a thin, straight-edged blade, A, screwed to a
heavier block of wood, the head, B. The inside of the head has a
smooth, straight edge. The T-square is placed upon the drawing
board with its head held against the left-hand edge of the drawing
board. By sliding it up and down with the left hand, the draughts-
man is enabled to draw with it as guide a series of parallel lines
running from left to right. From this use of the T-square, it is
seen that the edge of the drawing board along which it slides
must be smooth and straight, otherwise the lines drawn along the
blade of the square would not be parallel. It is hot at all necessary
that the blade of a T-square be exactly at right angles to the inner
edge of the head.
5. A 60° and a 45° Transparent Celluloid Triangle. Drawing
triangles are thin triangular guides, usually made of wood or
celluloid, one angle of which is a right angle. (See Fig. 3.)
§1] INTRODUCTION 3
In a 60° triangle, the acute angles are 60° and 30°, and in a
45° triangle, the two acute angles are each 45°.
With these triangles straight lines may be drawn making angles
of 15°, 30°, 45°, 60°, 75°, and 90° with the lines drawn with the T-
square as a guide.
For the work in this book 6-inch triangles will be found large
enough, although an additional larger 60° triangle will be found
very convenient.
60 Triangle 45 Triangle
FIG. 3.— 60° and 45° triangles.
6. A protractor is an instrument for laying off or measuring
angles upon a drawing. One, consisting of angular graduations
upon a semicircular piece of transparent celluloid, about 6 inches
in diameter, is recommended for this work. (See Fig. 4.)
7. A scale divided decimally (the unit of length divided into
tenths) should be used rather than one in which the unit is divided
into eighths, twelfths, or sixteenths. It is advised that a triangular
box-wood scale about 4 inches long, or longer, be used in this work.
(See Fig. 5.)
8. A pair of 6-inch pencil compasses used for drawing circles
and arcs of circles. The lead in a cheap pair of compasses usually
is not found satisfactory and should be replaced by a good piece
of medium hardness, sharpened to a narrow chisel-point. The
lead should be so adjusted that when the two legs of the compasses
are brought together the two points will be together or directly
opposite each other.
4 MATHEMATICS [§1
9. A good rubber eraser for erasing lead pencil marks.
10. Twelve small thumb-tacks.
11. Fifty sheets of squared paper, form Ml, used for plotting
data.
FIG. 4. — Protractor.
12. Twelve sheets of paper, form M7, used for numerical
calculations and for tables.
13. Six sheets of polar coordinate paper, form M3.
14. Twenty-four sheets of heavy bond paper, letter size, 8£ by
11 inches; to be used for drawings.
Y
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\V
FIG. 5. — Triangular scale.
\
15. Two heavy paper note covers suitable for holding paper
85 by 11 inches without folding. These covers are to be used for
submitting daily work to the instructor.
16. One note cover large enough for filing 8J by 11-inch sheets
§2] INTRODUCTION 5
without folding. This cover is to be used for filing the written
work after having been examined and returned by the instructor.
17. A set of "Slichter's four-place logarithmic and trigono-
metric tables" used in connection with numerical computations.
There are other logarithmic and trigonometric tables published,
but in order to do efficient classroom work without waste of time,
there must be complete uniformity in the tables used by the mem-
bers of the class.
If the student wishes larger tables for his own private use, the
two following are among the best:
Albrecht's five-place, and
Bremiker's six-place.
While the above list of instruments and materials includes about
the minimum in number and smallest in size for efficient work,
more and better instruments may be purchased if the student
desires.
2. General Directions. All work must be done neatly. In
general, accuracy in graphic work can only be acquired by neat
work. At first, perhaps, much time will be sacrificed, but after
the habit of doing neat work has been established much time will
be saved both in avoiding errors and in the ease with which errors
may be located or the work checked. The student at the begin-
ning must not be discouraged if a large number of his drawings and
exercises are returned marked " redraw" or "rewrite," for neatness
of work will always be insisted upon as one of the means o* obtain-
ing accurate work.
Unless special directions are given, all work submitted to the
instructor is to be upon paper of letter size, 8£ by 11 inches, and
enclosed without folding in a paper folder. These exercises when
returned by the instructor are to be corrected in accordance with
instructions, and kept in order in a note cover. Exercises
marked "rewrite" or "redraw "are to be done over, and both
the original and the corrected sheets filed. This file of exercises
may be called for at any time by the instructor.
All drawings are to be done in pencil, unless the student has had
training in the use of the ruling pen, in which case he may, if he
desires, "ink in" his drawings with drawing ink.
6 MATHEMATICS [§3
All descriptive work upon a drawing is to be lettered, not
written.1 Each drawing or exercise handed in must be numbered
in the upper right-hand corner. It must have a complete yet
brief title, and contain a sufficient number of descriptive words
in the body of the drawing so that anyone familiar with the prob-
lem will be able to read the drawing; i.e., to understand the
meaning of every line or figure in the drawing.
ABCDEFGHIJK
a b cd efghijklmno
pqrstuv wxyz
1 234 567890
FIG. 6. — Set of letters and figures.
All exercises — not drawings — are to be done in ink. Use any
good writing fluid. It is suggested that the student acquire
the habit of working numerical problems, and the like, with
pen and ink.
3. To draw a straight line through a point with the T-square
or triangle as guide, place the triangle or T-square upon the paper,
so that the point is just visible at one of its edges; press the tri-
angle firmly against the paper with the fingers of the left hand and
with the pencil in the right hand hold a flat side of the chisel-point
against the straight edge and move the pencil from left to right.
The pencil should be held with its upper end tipped slightly away
1 A set of letters and figures is given in Fig. 6.
§3]
INTRODUCTION
from the triangle in order to bring the point of the pencil in contact
with the straight edge. (See Fig. 7.) As far as possible, the whole
right hand should be given a straight-line motion in order that the
pencil may be made to move parallel to itself.
(Pencil
Pencil
Triangle
Triangle
FIG. 7. — a, Incorrect way of holding a pencil; b, correct way of holding
a pencil.
Care must be taken that the triangle is well pressed down upon
the paper at the point where the line is being drawn. This is
partly accomplished by the pressure of the fingers of the left hand,
and partly by pressing down upon the triangle with the third and
fourth fingers of the right hand. This pressing down with the
fingers of the right hand not only helps to hold the triangle in
place, but, at the same time, gives a rest to the hand.
8 MATHEMATICS [§4
4. Parallel Lines. Horizontal parallel lines may be drawn with
the aid of the T-square.
Parallel lines making angles of 30°, 45°, 60° or 90° with the
horizontal may be drawn with the aid of triangles placed upon the
edge of the T-square.
Parallels other than those mentioned above may be drawn as
follows: Suppose that through the point P, Fig. 8, we wish to
draw a line parallel to the line AB. Place two triangles together,
as shown in the figure, one edge of No. 2 coinciding with the given
line. Hold triangle No. 1 fixed and slide No. 2 along it in the
direction of the arrow until its edge which originally coincided with
AB passes through the point P. Along this edge draw a line, PQ.
PQ is parallel to AB, for, by geometry, if two straight lines are cut
by a transversal, and if the interior exterior angles on the same
side of the transversal are equal, the lines are parallel.
Exercises
1. Fasten a sheet of paper upon the drawing board with its longer
dimension extending from left to right. About 2 inches from the
lower edge of the sheet draw, with the aid of the T-square, a horizontal
line 6 inches long. Divide this line into six equal parts. At the
points of division draw vertical lines with the aid of the T-square
and triangle. Beginning with the horizontal line mark off four equal
spaces, each 1 inch long, upon the left-hand vertical line. From the-
points of division just located draw horizontal lines, thus forming
a rectangle 4 by 6 inches, divided into twenty-four equal squares.
Through the lower left-hand corner draw lines making angles of
30°, 45°, and 60° with the horizontal. Divide the upper left-hand
square into 100 equal squares. A drawing (reduced) of this exercise
is given in Fig. 9. Complete the exercise by giving, to two decimal
places, the dimensions and quotients omitted in Fig. 9.
2. In a 45° triangle, prove by geometry that the hypotenuse is \/2
times as long as a leg.
3. In a 60° triangle, prove that the hypotenusejs 2 times as long
as the shorter leg. Prove that the longer leg is V3 times as long as
the shorter leg. The hypotenuse is how many times as long as the
longer leg?
4. Find, correct to three decimal places, V2; V3; V5.
6. If the hypotenuse of a 60° triangle is two units in length, what
§4]
INTRODUCTION
9
is the length of the shorter leg ? Of the longer leg ? If the hypotenuse
is four units in length, what is the length of each leg?
6. What is the length of the hypotenuse of a 45° triangle, if each
•C 5 w ^
Ss fc)
3 ^
fc. i: ^
^ ^ T^
fe] K|
K
leg is a unit in length? Two units in length? Three units in
length?
7. What is the length of each leg of a 45° triangle, if the hypotenuse
is one unit in length? Two units in length? Three units in length?
10
MATHEMATICS
[§4
8. Find the length in rods of a diagonal of a section of land, assum-
ing that the section is a true square 1 mile on a side.
9. A, B, C, and D are the corners of a section of land. Through
A, a line is drawn making an angle of 30° with the side AB, and inter-
secting EC at the point E. Find the length, in rods, of BE, and of
AE. Find the area in acres of the triangular plot ABE.
10. Continuing with exercise 9, let F be the mid-point of AB, and
G the mid-point of CD. Draw the quarter section line FG, cutting
the line AE at H. Find the length in rods of AH, and of FH. Find
the area in acres of the plot AFH.
FIG. 10.
11. The longer leg of a plot of land in the form of a 60° triangle is
80 rods. Find the area of the plot in acres.
12. A plot of land in the form of a 60° triangle contains 72 acres.
Find the length in rods of each side of the triangle.
HINT: Let x represent the number of rods in the length of the
shorter leg.
13. The shorter side of a rectangle is 100 feet, the diagonal is 200
feet. Find the length of the longer side.
§4]
INTRODUCTION
11
14. The diagonal of a square is 100 feet. Find the length of a
side.
16. A barn 36 feet wide has a gambrel roof. The lower rafters
make an angle of 60° with the horizontal. The upper rafters make an
angle of 60° with the vertical. The lower and upper rafters are equal
in length. Find the length of the upper edge of the rafters. See
Fig. 10.
HINT: Let x be the number of feet in the desired length.
16. If the rafters referred to in exercise 15 are 2 X 6's, and if
they run short 3/16 inch, find the dimensions of the block sawed
from each end. (See Fig. 11.) Find y and z.
./A
17. If the building referred to in exercise 15 is 84 feet long, find
the number of cubic yards 'space under the roof above the plates.
18. A circular concrete silo, 12 feet outside diameter, is 20 feet tall.
The wall is 8 inches thick. Find the number of cubic yards of con-
crete in the wall, no allowance being made for openings. After the
problem has been solved by the student, the instructor will show
how algebraic operations may be used in eliminating considerable
numerical work.
19. A man walks 3/4 mile up an incline making an angle of
30° with the horizontal. How many feet does he move in a hori-
zontal direction?
12 MATHEMATICS [§5
20. A ball rolls down a smooth inclined plane making an angle
of 45° with the horizontal. If the distance s measured in feet along
the incline is given by the formula
where t is time in seconds, find the vertical distance through which
the ball moves in one second; in two seconds; in three seconds.
6. Review of Elementary Algebra. The course for which this
book is intended presupposes a knowledge of elementary algebra,
and of plane and solid geometry, as usually covered in two years
of High School work. Many students, however, having had their
algebra and geometry some years since, may feel a need of review
in these subjects. When this is the case, the student should take
a hasty review touching only the essentials, and then study those
subjects or theorems which he wishes to apply, at the time he
wishes to apply them. For this purpose, a very brief review of
algebra, consisting mostly of examples and illustrations, is here
given.
6. Addition. To 3x + 4y - 6z
Add 2x - 5y + 2z
Sum 5x — y — 4z
Exercises
Find the sum of the following:
1. 3a - 66 + 7c and 5a - 2c - 76.
2. 3x2 + 6xy + yz and x2 — 5xy + y*.
3. 2xy - 3x2y + 3xy* + y2 and 2x2y - 3y* + 3xy - 7.
4. a + b + c - d and 2a - 7c - 2d + 66 - 7.
6. a - 26 + 6, 3a + 7c - 6 + 26, and 5a + 6 - 7c.
6. x2 + 2xy + y*, x2 - 2xy + y2, and x2 - 2j/2.
7. 3x3 + 6x2y + 6xy2 + y3, 5x3 — Qy3 + 6x2y, and 3x3 - y3
8. a + b + c + 3d, 2a - c + 76 - 5d + 6, and c - 7 + 2d - a.
9. a3 - 63, a26 + a&2, and - 4o26 + 2a62.
10. 25a3 - 2763, 5a26 - 6a62, and 2663 - 24a3 + a62.
7. Subtraction. From 3a - 66 + 7c - 6
Subtract 2a + 56 - 2c - 6
Difference a — 116 + 9c
§8] INTRODUCTION 13
Exercises
1. From 3o2 - 262 - Sab subtract a2 - 3a6 + 62.
2. From x2 - 2xy + y2 subtract x2 + 2xy + y2.
3. From x + 2xz + 3x3 - 2 subtract 2x + x2 + 7x3.
4. From x3 + 3x2y + 3xy2 + y3 substract Qx2y + 2y3.
5. From 3x2 - 2y + y2 subtract 2x2 - 2y + 6 + 7z2.
8. Multiplication. The product of two numbers having like
signs, either positive or negative, is positive. The product of two
numbers having unlike signs is negative. Thus,
(a) (6) = ab
(_a)(-6) = ab
(a)(- 6) = - ab
(-a) (6) = -ab
Multiply 3z2 - 2xy + y2
By 2x2 - sy + y2
- 2xt/3
___ _
Product 6a;4 - 7x37/ + 7x2y2 - 3xy3 + y*
Exercises
1. Multiply x3 + x?/ + y3 by x — y.
2. Multiply x3 — xy + y3 by x + y-
3. Multiply x2 + xy + y2 by x2 - xy + y2.
4. Multiply x — 7y by x — 2y.
5. Multiply x + y by x — y.
6. Multiply 3x + 7y by 3x + 8y.
7. Multiply a + b+c+dbya + b-c-d.
8. Multiply 2x - 3y + 6 by 2x + 7y + 2.
9. Multiply 3xy - 2x2 - 2x by 2xy - 2x - 3y - 6.
10. Multiply 9 - 3x + x2 by 3 + x.
A few simple multiplications may be performed mentally. We
see at once that (a + 6) (a — 6) = a2 — 62, where a and b are any
two numbers. From this we have
(3x - 2y)(3x + 2y) = 9x2 - 4?/2
14 MATHEMATICS
Exercises
Multiply the following mentally:
1. (3z -y)(3x+y).
2. (2x -y)(2x + y).
3. (a - 9) (a + 9).
4. (x*y -3a)(x2?/ + 3o).
6. (a +36) (a - 36).
6. (29) (31), or (30 - 1)(30 + 1) = 900 - 1 = 899.
7. (51)(49).
8. (52) (48), or (50 + 2) (50 - 2).
9. (103) (97).
10. (25) (35).
A few powers of binomials are:
(a + 6)2 = a2 + 2ab + b\
(a - 6)2 = a2 - 2ab + 62.
(a + 6)> = a3 + 3o26 + 3a62 + b3 .
(a - 6)3 = a3 - 3a26 + 3a&2 - b3.
(a + &)4 = a4 + 4a36 + 6a262 + 4a63 + 64.
(a - 6)4 = a4 - 4a36 + 6a262 - 4a63 4
Thus, (3 - a)3 = 27 - 27a + 9a2
and (x + ?/2)4 = x4
Exercises
Expand the following mentally:
1. (2o - a;)2. 4. (x - 3)4.
2. (x + 3y)2. 5. (1 - z)3.
3. (2z - I)3. 6. (2 + ?/)4.
7. (52)2, or (50 + 2)2, or 2500 + 200 + 4 = 2704.
8. (31)2, or (30 + I)2.
9. (29)2, or (30 - I)2.
10. (98)2.
The square of a polynomial is illustrated by the following:
(a + 6 + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc.
(a + 6 + c + d)2 = a2 + 62 + c2 + rf2 + 2o6 + 2ac + 2
+ 2bc + 26rf + 2cd.
(3 - x + ?/2)2 = 9 + xz + y* - 6x + 6i/2 - 2^.
§9] INTRODUCTION 15
Exercises
Expand the following mentally:
1. (o + b + 2)2. 4. (2o - x + 3)2.
2. (a + b - 2)2. 6. (x2 - 2?/2 + 4)2.
3. (a - b - c)2.
The product of two binomials having a common term may be
found by inspection. Illustrations:
(a + 6) (a + c) = a2 + (b + c)o + be.
(x + 7)(x + 2) = z2 + 9z + 14.
(x - 5)(z + 3) = x2 - 2x - 15.
(x2 - 2y)(x* - By) = x* - 5x2y + 6?/2.
Exercises
Find mentally the value of each of the following:
1. (x + 2)(x + 3). 6. (3x + 2y)(3x - 7y).
2. (x - 2)(x + 3). 7. (x2 - 3)(x2 - 4).
3. (x - 2)(x - 3). 8. (3xy - z)(3xy + 7z).
4. (x + 2)(x - 3). 9. (xV - 3)(xV - 10).
5. (x2 + 5y)(x2 - 5y). 10. (x - 2y)(2x - 2y).
9. Division. The quotient of two numbers having like signs
is positive. The quotient of two numbers having unlike signs is
negative.
Thus, (a) -(&)=-
f \ • f M a
(0) T ( _ 6) = _ _
(-a)-*- (6)= '-£
Divide a;3 - Go;2 - 19x + 84 by x - 7
x - 7(x* - 6z2 - 19x + 84) z2 + a; - 12 Quotient
x2 -19z + 84
a2 - 7x
- 12x + 84
-12x+ 84
16 MATHEMATICS [§10
Exercises
1. Divide 15x2 - llx - 14 by 3x + 2.
2. Divide 12a2 - 28a + 15 by 6a - 5.
3. Divide 8m5 - 14m2 - 18m + 21 by 4m3 + 6m - 7.
4. Divide 3x5 - 7x3 - 2x - 36 by x - 2.
5. Divide x8 + 2x6 - 3x4 + 7x3 - 2x + 69 by x - 3.
6. Divide x4 + x2y2 + y* by x2 + xy + y2.
7. Divide 8x4 - 22x3y + 43x2t/2 - 38xy3 + 24y* by
2x2 - 3xy + 4y2.
8. Divide x3 — y3 by x — y.
9. Divide x4 — y4 by x + y.
10. Divide x4 — y4 by x — y.
10. Symbols of Aggregation. If a sign of aggregation is pre-
ceded by the negative sign, change all signs within when the sign
of aggregation is removed. If the sign of aggregation is preceded
by the positive sign, all signs within remain the same when the
sign of aggregation is removed.
5x2 - [3y2 + {2x2 - (?/2 + 3x2) + 5y*} - x*]
= 5x2 - [3y2 + {2x2 - y2 - 3x2 + 5r/2} - x2]
= 5x* - [3y2+ {4t/2 -x"'} - x2]
= 5x2 - [3y2 + 4y2 - xz - x2]
= 5x2 - [7y2 - 2x2]
= 5x2 - 7y2 + 2a;2
= 7x2 - 7y2
Exercises
Simplify the following by removing the signs of aggregation :
1. ab - 462 - (2a2 - 62) - { - 5a2 + 2ab - 362} .
2. x - { y + z - [x - ( - x - y) + z}} + [z - (2x - y)].
3. a - { - a - [ - a - ( -a - 1)]}.
4. 3yz - [2yz + (9z - 2yz}].
5. - { - 1 -[-1 - ( -1)]}.
11. Factoring. Since (a + 6)2 = a2 ± 2ab + b2, any expres-
sion of the form of the right-hand side can be factored by inspec-
tion. Thus,
x2 - 6xy + Qy2 = (x - 3y)2
and
:.„ 4 + 4(o + 6) + (a + 6)2 - (2 + a + 6)2
§11] INTRODUCTION 17
Exercises
Factor the following by inspection :
1. 9x2 - 30xy + 25y2.
2. 4 + IQt + 16<2.
3. zV + lOxVz2 + 25z4.
4. 9 + 6(x2 + t/2) + (x2 + y2)2.
5. a4 + 4a262 + 464.
Since (a + 6) (a — 6) = a2 — 62, any expression of the form of
the right-hand side can be factored by inspection. Thus,
4a2 - 962 = (2a + 36) (2o - 36)
Exercises
Factor the following by inspection :
1. xV - 22. 4. 25 - 3x2.
2. (a + 6)2 - c2. 5. 81 - 625x4.
3. c2 - (a + 6)2.
Since (a + 6) (a + c) = a2 + (6 + c)a + &c, any expression of
the form of the right-hand side can be factored by inspection.
Thus,
x2 - 5x - 14 = (x - 7)(x + 2)
Exercises
Factor the following by inspection :
1. x2 + 7x + 10. 4. 9x2 - 18x - 27.
2. a2 + 4ay - 21y2. 5. 25 + 30a - 27a2.
3. 4x2 - ISxy + 18j/2.
Since (a + 6) (a2 — ab -f- 62) = a3 + &3, any expression of the
form of the right-hand side can be factored by inspection. Thus,
27 + 125x3 = (3 + 5x)(9 - I5x + 25z2)
Exercises
Factor the following by inspection:
1. zV + 1- 4- 125 + xy.
2. x6 + y\ 6. x3 + 8y6.
3. 8 + 27x3.
Since (a — 6) (a2 + 06 + 62) = a3 — b3, any expression of the
form of the right-hand side can be factored by inspection. Thus,
27 - 125z3 = (3 - 5s) (9 + 15* + 25z2)
18 MATHEMATICS [§12
Exercises
Factor the following by inspection :
1. x3y3 - 1. 4. 125 - x3y3.
2. x6 - y6, or (x3 + y3)(x3 - y3). 5. x3 - 8y*.
3. 8 - 9x3.
The following may be factored by grouping the terms. Thus,
a3m + a3n — m — n
= a3(m -\- n) — (m -\- n)
= (a3 - l)(m + n)
= (a - l)(a2 + a + l)(m + n)
Exercises
Factor the following :
1. ax — ay + bx — by. 4. x5 — xy* — x*y + yb.
2. x3 + 3x2 + 3x + 1. 6. x4 - xsy - xy3 + y*.
3. ax2 - 2axy + ay2 + 6x2 - 2bxy + by*.
12. Fractions. Multiplying or dividing both numerator and
denominator of a fraction by the same number, excepting zero,
does not change the value of the fraction. To reduce a fraction
to its lowest terms, factor both numerator and denominator and
then divide each by the common factors if there are any. Thus,
ax* + ax2y2 + ay*
a2z3 — a2y3
a(x* + xy + t/2)(a;2 - xy + yz]
a(x — y)(x2 -\-xy-\- y2)
x2 — xy + y2
(x - y)
Exercises
Reduce the following to lower terms:
^ ox2 — ay2 > ax + ay — x — y.
x3 — y3 ' x3 + y3
x2 + x - 6_ 27 - x3
2* x2 - 4 12- 7x + x2.
• x4 - y\
x6-6?/
§12] INTRODUCTION 19
To multiply fractions together, multiply the numerators
together for the numerator of the product, and multiply the
denominators together for the denominator of the product. To
simplify the result, divide both numerator and denominator by
common factors if there are any. Thus,
a3 - b3 b a26 - ab2
ab Xa2 + ab +¥ X (a -b)z
(a3 - 63)(6)(a26 - a&2)
" ab(a2 + ab + 62)(a - by
_ (a - 6) (a2 + ab + 62)(a62)(a - 6)
a&(a2 + ab + 62)(a - 6)2
= 6
In working this problem, the second expression should not be
written down, and possibly the third could be omitted.
Exercises
Simplify the following :
a4- y*
a2 - 2ay + y2
az(x + a)
x2 + lax + a
a2 + 2a.b + b2 - 4c2 (a - b)2 - 4c" (2a - b)2 - 4c2.
4a2 - (b + 2c)2 (a + 6)' - 4c2 a2 - (b - 2c)2
x3 - y3 x* — xy + y* x + y%
s\ . , , X
•7*3 _L_ 4/3 /7«2 _! TII —1— -i/2 />• _ _ 91
•*' \ y * \ -Ly i y * j/
_ oa; + ay bx — 36 x2 — xy + y2
b < x»+ j/3 X x2 - 5x +~6~'
To add fractions reduce each to the same denominator by multi-
plying numerator and denominator of each by a suitable number,
then add the numerators and place the sum over the common
denominator. Thus,
x -y _J_ xy
I -v I .. I _3 I
x2 - zt/ + 7/2 ' a: + y ' x3 + y3
x2 - y* x2 - xy + y2 xy
x3 + y3 "* a;3 + y3 +a;3 + y3
2a:2
20 MATHEMATICS [§13
Exercises
Simplify the following :
• • i n I o I o f*
2 i v-2
+ 3).
a; — 5 x — 1
. o
x- 5 x» - te + 5
y) (« + y)*'
y (*-»)«
13. Exponents,
Definitions: an = a a a . . . to n factors, if n is a positive
integer
ap/« = v/ap, if p and g are positive integers
a« = 1
a~8 = -, if s is positive
ft8
Laws of indices: aman = am+"
(am)« = amn
Exercises
1. Express the following using positive exponents only:
(a) x~2ab~3. ar^-'x-*.
(6) a-^b^. W c2«Sa;6
a~lbc~*
(c) (x-^y-^)«. (e) ^7c^i-
2. Perform the indicated operations:
(a) 25W. (6) 8?4. (c) (9/16)-*. (d) 27°.
3. Simplify:
§14] INTRODUCTION 21
4. Multiply Sa% - 4o6w + 2ow6 - &?* by 2ow - 36^.
6. Divide a* - 3a2 + a^ - 4 + 12ow - 4o% by a** - 4.
14. Radicals.
Simplify ^24 + 5^81 - 6^192 - 10^1029
+
= 2 i/3 + 15^3 - 24i/3 - 70^3
= - 77^/3
Rationalize the denominator of
(\/2 + \/3)(>/2 + \/3)
\/2 - \/3 (\/2 -
2-3
= 2 + 2\/6 +3
- 1
= - (5 + 2V6)
Exercises
1. Simplify Vl2 - 2\/27 + 9 A/48 - Vl08 - 6\/75.
2. Simplify V (a - b)*x - V(o + 6)2z - Va2x +
2
3. Rationalize the denominator of
4. Rationalize the denominator of
V3 - V2
2 - \/5
6. Rationalize the denominator of —7= 7='
V3 - V 5
16. Quadratic Equations. A quadratic equation may be solved
by completing the square, or by using the formula, or somtimes
by factoring.
Solve for x: x2 - 6x + 8 = 0.
By completing the square we have
x2 - 6z + 9 = 1,
or x - 3 = ± 1,
or x = 3 + 1,
or x = 4 or 2.
22 MATHEMATICS [§15
By factoring the left-hand side we have
(x - 4)(x - 2) = 0.
This equation is equivalent to the two equations.
x - 4 = 0
and z - 2 = 0
from which we have
x = 4
and x = 2
If axz + bx + c = 0 is the general equation of the second
degree, its solutions are
- b + V'b2 - 4ac
2a
In the above equation, a = 1, b = — 6, andc = 8. Substitut-
ing these values in the formula we have
6 + V31T-4-8
x — — '•
2.
= 3 ± 1,
or x = 4 or 2
Exercises
Solve the following equations for x:
1. x2 - 9x + 20 = 0.
2. 4z2 + 7x - 2 = 0.
3. x2 - 7z + 6 = 0.
4. 8x2 - 4x + 3 = 0.
6. 9x2 + 7x - 20 = 0.
CHAPTER I
GRAPHIC REPRESENTATION
The Straight Line. The Parabola. The Equilateral Hyperbola.
16. The Scale. In solving problems graphically the funda-
mental instrument used- is the scale. Unless otherwise stated,
the term scale, as used in this book, means what is commonly
understood by a scale, an instrument for measuring short linear
distances. The term scale, in a broader sense, may be defined as
any curve, including a straight line, the points along which are
supposed to correspond in order to the numbers of a sequence.
We are all familiar with various scales, such as the foot-rule,
divided into inches; the meter, divided into centimeters and milli-
meters; the thermometer scale, divided to correspond to degrees
and tenths of degrees of temperature; the scale on a burette,
divided to correspond to cubic centimeters and tenths of cubic
centimeters; and scores of other scales used for measuring different
quantities. If a scale is marked so that the subdivisions are all
of equal length, as the foot-rule, we say the scale is divided uni-
formly, or that we have a uniform scale; otherwise, the scale is a
non-uniform scale. For the solution of problems graphically, the
student should have for measuring lengths a scale divided into
centimeters and millimeters or into inches and tenths of inches.
Illustrations of non-uniform scales will be given later.
17. Drawing to Scale. A large number of graphic problems
consist in nothing more than drawing to scale, i.e., making a
picture of some object in which a definite length, as the inch, on
the drawing represents a definite length, as the foot, on the given
object. If the dimensions on such a drawing were to be given in
feet and inches, or if dimensions in feet and inches were to be
scaled from the drawing, a scale divided into tenths of inches
would apparently be very ill-adapted. For this kind of work,
scales may be procured on which the unit of length is sub-
23
24
MATHEMATICS
[§18
divided into twelve equal parts. If each unit throughout the
entire length of the scale is subdivided, the scale is called a full
_ divided or chain scale. If
only one unit, usually the
left-hand unit, is subdivided,
it is called an open divided
scale. These scales are made
with the unit J, \, f, £, \, 1,
2, 2, 3, etc., inches long, used,
respectively, when |, j, f , |, $,
1, 2, 2, 3, etc., inches on the
drawing are to represent 1
foot on the object.
Fig. 12 represents an 18-
^ inch, flat, open divided scale.
% The unit on one edge is f
^ inch, and on the other edge
]> it is j inch. This scale may
^ be used in drawing the plans
g, of a barn. The small scale
9 is used in drawing the plan
^ and elevations ? inch to the
foot, while the larger scale is
2 used in drawing, on a larger
scale, the details of the plank
frame truss bents. The
method of measuring lines is
illustrated in the figure. The
line AB is found to be 22 feet,
7 J inches long.
18. Positive and Negative
Directions. If upon a scale
some point is chosen as zero,
or origin, distances measured
in one direction, or the num-
bers represented by the points
on one side of it, are called positive, and are designated with the
positive, or plus, sign +. Distances measured in the other
§19]
GRAPHIC REPRESENTATION
25
direction, or numbers represented by points on the other side of
the origin, are called negative, and are designated by the nega-
tive, or minus, sign — .
19. Rectangular coordinate paper, or squared paper, is paper
ruled in both directions, horizontally and vertically. These
rulings are generally spaced uniformly, dividing the page into a
number of equal squares or rectangles. The size of these squares
or rectangles is entirely arbitrary. The length of a side may be
an inch, a tenth of an inch, a centimeter, a millimeter, etc.,
x-
Y'
FIG. 13. — Coordinate, or squared paper.
depending, to some extent, upon the use for which the paper is
designed. For convenience, every fifth, tenth, fourth, eighth, or
twelfth line is made somewhat heavier than the others. A sheet
of squared paper is represented in Fig. 13. Upon this sheet two
lines, mutually at right angles, are chosen as axes. The horizontal,
X'OX, is called the axis of abscissas; the vertical, Y'OY, the
axis of ordinates. The point of intersection, 0, of these axes is
called the origin.
A use of squared paper is illustrated by the following: In
Table I are given the maximum and minimum air temperatures
at Madison, Wisconsin, for each day of the month of October, 1910.
26
MATHEMATICS
[§19
TABLE I.— MAXIMUM AND MINIMUM TEMPERATURES AT MADISON
WISCONSIN, FOR OCTOBER, 1910
Date
Max. temp.,
°F.
Min. temp.,
°F.
Date
Max. temp.,
°F.
Min. temp.
op
1
66
55
17
81
53
2
71
48
18
81
57
3
75
58
19
69
45
4
68
55
20
45
40
5
62
53
21
49
41
6
58
45
22
52
34
7
66
43
23
60
37
8
68
47
24
60
49
9
60
44
25
52
44
10
67
42
26
60
40
11
75
49
27
42
32
12
61
46
28
35
30
13
69
45
29
38
26
14
73
52
30
60
31
15
76
50
31
63
39
16
80
56
These data are said to be represented graphically in Fig. 14, or the
data are said to be plotted upon squared paper. To do this: first,
points equidistant along the axis of abscissas are chosen to repre-
sent the days of the month. These points are numbered con-
secutively from 1 to 3 1.1 Distances from this horizontal line
represent temperatures. If the temperature is above zero, or
(+), it is represented by a distance above the horizontal axis;
if below zero, or ( — ), by a distance below. The distance above
or below is proportional to the magnitude of the temperature
reading. Since distances from the horizontal axis represent tem-
perature readings, points along the axis of ordinates are numbered
to correspond to degrees of temperature. In the figure, the verti-
cal distance between two adjacent horizontal lines represents
2°, but only every fifth line is 'actually numbered, since
numbering each line would not aid in any way in reading the
drawing. This vertical axis is in reality a uniform scale, for equal
1 Only the even numbers are placed in the drawing; otherwise the numbers would
be so near together as to be confusing. Omitting the numbers from every other
point does not make the drawing harder to read.
§19]
GRAPHIC REPRESENTATION
27
-^
<^
^
\D
V
v
1
<<;
>
>
>
H
3
p
<j
CS
?
PW ^^
S «
H %
o S
P Q °
S
o
O TJH
28 MATHEMATICS [§20
portions of it represent equal changes in temperature. Since the
scale is uniform, it is evident that it is immaterial whether we
think of the point or the vertical distance from the origin to that
point as representing degrees of temperature. Further, we may
look upon the scale as continuous, for every point upon it cor-
responds to some temperature.
The horizontal axis, as we are here using it, is not a continuous
scale for measuring time, for each numbered point along it, or
the distance of this point from the origin, represents a whole
interval of time from midnight to midnight; and the intermediate
points have no meaning whatever.
Now that we have chosen points to represent the days of the
month, a unit of length to represent a degree of temperature, and
have marked the axes, we are ready to transfer the readings from
the table to the squared paper. On the first of October, the maxi-
mum temperature was + 66°. Graphically, this is represented by
making a dot or small circle thirty-three divisions above the hori-
zontal axis and upon the vertical line passing through that point
of the horizontal axis representing the first day of October. This
reading is now said to be plotted. The maximum temperature
on the second of October was 71°, and is represented by placing
a point or circle thirty-five and one-half spaces above the hori-
zontal axis and upon a vertical line passing through that point
upon the horizontal axis marked 2. In a similar way, points
representing the maximum and minimum temperatures for each
day are plotted. To aid the eye in following the temperatures
from day to day the consecutive points are connected by short
straight lines. These broken lines will also aid one to see at a
glance how the maximum and minimum temperatures rose and
fell from day to day throughout the month; to see when the
temperature was high; when low; when the freezing point was
reached; etc. In short, Fig. 14 is a picture of all that is given
in Table I.
20. Smooth Curves Drawn Through Plotted Points. In Fig.
15 are plotted the data given in Table II, the hourly air tempera-
tures for May 14 and October 10, 1910, at Madison, Wisconsin.
It will be noticed that in this drawing the points are connected,
not by straight lines as in the previous illustration, but by means
§20]
GRAPHIC REPRESENTATION
29
of a smooth curve drawn through the points. The reason for
drawing a smooth curve is that the readings were taken so near
together that the points along the smooth curve are probably very
near the points that would have been obtained had additional
intermediate temperature readings been taken. Thus the curves
in Fig. 15 may be used for scaling off temperatures for any time of
the day.
•'£ 60
^ *-
S3
i
V
f •*
*
^>*
-^^
V,
is.
I
E
-°1
*v
^
S
i-^~
"^N
(X-1
k^
\
.a 50
fa
^
^
&
\—
X
1*>1111^^
| 40
if
i —
>.
''/
^
5
s
w 30
—
—
—
-
-
-
| 20
Hourly Air
Te
at
n V
r, l
an<
t.. I
•nperatures
Hm
Ma<
liso
Ma
Oe
Wisconsin
4.1910
i
0, 1910
0
i
2 4 6 8 10 12^ 2 4 ^6 8 10 1
< A.M; > | < F.M. *
Time of Day Hours
FIG. 15.
TABLE II.— HOURLY AIR TEMPERATURES AT MADISON, WISCONSIN.
MAY 14, 1910; OCTOBER, 10, 1910
Time
May 14, 1910,
temp., ° F.
Oct. 10, 1910,
temp., ° F.
Time
May 14,
1910,
temp., ° F.
Oct. 10,
1910,
Temp., ° F
1 a. m.
37
44
1 p. m.
58
65
2 a. m.
35
44
2 p. m.
61
66
3 a. m.
35
44
3 p. m.
63
67
4 a. m.
34
43
4 p. m.
62
67
5 a. m.
34
43
5 p. m.
62
65
6 a. m.
35
42
6 p. m.
61
62
7 a. m.
37
43
7 p. m.
59
57
8 a. m.
42
45
8 p. m.
56
55
9 a. m.
46
51
9 p. m.
53
55
10 a. m.
49
55
10 p. m.
50
55
11 a. m.
54
60
llp.m.
47
52
12 a. m.
56 ! 62 12p.m.
45
51
30
MATHEMATICS
[§21
21. Precipitation Charts. In Fig. 16 we have represented
diagrammatically, by the usual method, the mean monthly pre-
cipitation at Tacoma, Washington; Madison, Wisconsin; Tampa,
Florida; and Yuma, Arizona. In these charts the amount of pre-
cipitation is represented by the height of a shaded rectangle. This
method of representation is similar to that used in Fig. 14 and
enables one to see at a glance when the rainy and the dry periods
occur at these localities.
J FMAM J JASOND
Tacoma, Wash.
J FMAM J JASOND
Tampa, Fla.
. 6-
a>
-S5
a 4.
13
.2 3
3 2-
TIM Ml
I I I I I I I I IN
J F MAM J J A S O N D
Madison, Wis.
P-iU
J FMAM J JASOND
Yuma, Ariz.
FIG. 16. — Precipitation charts.
22. Marking Axes. Whenever data are represented upon
squared paper, both horizontal and vertical axes should be marked
in such a way as to make perfectly clear what quantities are
represented and what length represents a unit of measure of that
quantity. If two or more curves are drawn upon the same set of
axes, they should be so marked as to show definitely what each
curve represents.
Occasions may arise when it is desirable to draw upon the same
§22] GRAPHIC REPRESENTATION 31
sheet of paper, for comparison, two curves having in common only
one element, such as time. A scheme for doing this is illustrated
by exercise 12, below.
The length chosen to represent a unit is entirely arbitrary and
need not be the same along both axes. This is already illustrated
in Fig. 15. Further, the numbering in the horizontal and
vertical directions need not be placed upon the axes, as will be
illustrated by exercise 12, below.
Exercises
1. At what hour was the temperature a maximum on May 14,
1910? At what hour was the temperature a maximum on Oct. 10,
1910?
2. At what time was the temperature a minimum on May 14,
1910? At what time was the temperature a minimum on Oct. 10,
1910?
3. At what hour was the temperature increasing the fastest?
At what hour was the temperature decreasing the fastest?
4. During what hours of the day was the temperature decreasing?
During what hours of the day was the temperature increasing?
5. What was the temperature at 10:30 a. m.? 10:30 p. m.?
6. What was the temperature at 6: 15 a. m.? 6: 15 p. m.?
7. At what time was the temperature on May 14 the same as on
October 10?
8. For what portion of the day was it warmer on May 14 than on
October 10?
9. During what hours was the temperature below the freezing
point?
10. Plot data given in Table III.
11. Plot data given in Table IV.
12. Plot the data given in Table V upon squared paper. Use the,
same horizontal axis for all three curves. Put the temperature
curves above the discharge curve, using the same horizontal (time)
scale for both. Let 1 cm. (or 1/2 inch) on the vertical scale
represent 0.1 second-foot1 discharge, and 10° temperature. Start
the temperature scale with 60°, and place the 60, 6 cm. (or 3
inches) above the horizontal scale. Start the discharge scale with
3.4 placed on the horizontal scale.
1 Second-foot (or sec-ft.), when applied to the measurement of flow of water
means one cubic foot per second.
32
MATHEMATICS
[§22
TABLE III— MAXIMUM AND MINIMUM TEMPERATURES FOR
OCTOBER, 1911, AT MADISON, WISCONSIN
Date
Max. temp.,
°F.
Min. temp.,
°F.
Date
Max. temp.,
°F.
Min. temp.,
°F.
1
55
50
17
61
51
2
56
50
18
67
47
3
72
50
19
57
45
4
61
48
20
54
39
5
51
42
21
46
38
6
63
47
22
48
35
7
53
45
23
42
34
8
61
40
24
46
30
9
62
45
25
47
37
10
61
49
26
38
30
11
65
48
27
42
28
12
69
49
28
43
29
13
57
47
29
53
30
14
61
50
30
45
37
15
57
49
31
40
32
16
65
52
TABLE IV.— HOURLY TEMPERATURES AT MADISON, WISCONSIN,
FOR APRIL 11, 1911, AND NOVEMBER 10, 1911
Hour
Temp., ° F.,
Nov. 10,
1911
Temp., ° F.,
April 11,
1911
Hour
Temp., ° F.,
Nov. 10,
1911
Temp., ° F.,
April 11,
1911
a. m.
1:00
59
39
p. m.
1:00
62
59
2:00
59
40
2:00
60
61
3:00
59
40
3:00
59
59
4:00
59
40
4:00
51
58
5:00
60
40
5:00
44
56
6:00
60
40
6:00
39
53
7:00
61
41
7:00
35
53
8:00
62
44
8:00
31
52
9:00
64
47
9:00
28
51
10:00
66
50
10:00
25
47
11:00
68
55
11 :00
22
46
12:00 70 57
12 :00
20
45
§23]
GRAPHIC REPRESENTATION
33
13. What was the discharge at 7 : 00 p. m.? At 10 : 00 a. m.?
14. What was the minimum rate of discharge? At what time?
16. What was the water temperature when the discharge was a
minimum?
16. What was the maximum water temperature? At what time?
17. What was the maximum air temperature? At what time?
18. What was the greatest difference between the air temperature
and the water temperature?
19. At what hour did this "maximum" difference occur?
20. What was the minimum difference between the air temperature
and the water temperature? '
21. At what hour did this "minimum" difference occur?
22. Can you think of any cause which would produce the indenture
in the air temperature curve at 1:30 p. m., at which time a normal
temperature curve should have a maximum?
23. Can you think of any reason why the discharge curve should
have a minimum when the temperature curves have maxima?
TABLE V.— DISCHARGE OF A SEEPAGE DITCH
Time,
Aug. 24, 1905
Discharge of ditch,
sec.-ft.
Temp, of water,
°F.
Temp, of air,
0 F.
8 :00 a. m.
3.72
65
9:00 a. m.
11 :00 a. m.
1 :30p. m.
2 : 15 p. m.
3.70
3.66
3.52
3.49
70
79
83
74
84
85
3 : 30 p.m.
5: 30 p. m.
6 :00 p. m.
3.52
3.66
3.73
84
78
90
88
8: 00 p.m.
3.84
67
76
23. The Point. The location of a point upon a sheet of rec-
tangular coordinate paper is fixed by giving its distances from the
vertical and horizontal axe^. The distance from the vertical axis
is called the abscissa, and the distance from the horizontal axis is
called the ordinate of the point. These two distances are called
the coordinates of the point. The horizontal axis is commonly
called the X-axis, or the axis of x; and the vertical axis, the
Y-axis, or the axis of y. The abscissa and ordinate are then
3
MATHEMATICS
[§23
called, respectively, the x and y coordinates. When writing the
coordinates of a point, the value of the abscissa is placed before
the value of the ordinate, and separated from it by a comma.
Thus, "the point (3, 4)," (read, "three, four"), is a point three
units to the right of the F-axis, and four units above the X-axis.
The point (— 2, 6) is two units to the left of the F-axis and six
1
r
1
(-2,6
)
6
5
4
(3,4)
3
n
2
I
1
7 -
6 -
5 -
4 -
3 -
2 -
3 (-2,
1 O
1) -
1
l
2
3
4
5
6
7
.
2
(3,-:
S)
III
_
3
IV
4
_
5
.
6
7
FIG.
r'
17.
units above the X-axis. The point (—2, — 1) is two units to the
left of the F-axis and one unit below the X-axis. The point
(3, — 2) is three units to the right of the F-axis and two units
below the X-axis. These four points are shown on the sheet of
squared paper in Fig. 17.
The two axes divide the plane into four parts, called quadrants.
They are numbered from "one" to "four" as shown in Fig. 17.
§23] GRAPHIC REPRESENTATION 35
Exercises
1. Plot the following points: (a) (1, 1); (6) (2, 3); (c) ( - 3, 2);
(d) ( - 1, - 1); (e) ( - 2, - 3); (J) (3, - 4); (g) (V2, 1); (ft)
C - V3, V5).
2. Find the distance of each of the following points from the
origin: (a) (1, 1); (6) ( - V2, V2); (c) ( - 3, - 2).
3. Find the distances between the following pairs of points:
(a) (1, 1) and (2, 2); (&) ( - 2, 3) and ( - 5, - 6); (c) (3, - 5)
and (- 1, 0).
4. What is the y-coordinate of any point upon the axis of a;?
6. What is the x-coordinate of any point upon the axis of y?
6. What is the y-coordinate of any point upon a straight line
parallel to the axis of x, three units above it?
7. What is the ^-coordinate of any point upon a straight line
parallel to the F-axis, five units to its left?
8. What relation exists between the coordinates of any (every)
point of a line bisecting the angle between the positive directions
of the two axes? Between the negative directions of the two
axes?
9. What relation exists between the coordinates of any (every)
point of a line bisecting the angles having the positive direction of the
F-axis and the negative direction of the ^T-axis as sides?
10. What relation would exist between the x- and ^-coordinates of
any (every) point of the line in exercise 9, if it were raised five units
parallel to itself? If it were lowered six units ?
11. What relation would exist between the coordinates of any (every)
point of the line in exercise 9, if the line were moved up three units
parallel to itself? If it were lowered five units?
12. What are the coordinates of the origin?
13. Compute for every 10° temperature (between 0° and 200°)
read upon a Fahrenheit scale, the corresponding reading upon
a centigrade scale. Put your results in tabular form, using a sheet of
paper, form M7.
14. Compute for every 10° temperature (between — 20° and
100°) read upon a centigrade scale, the corresponding readings
upon the Fahrenheit scale. Put your results in tabular form, using a
sheet of paper, form Ml.
16. From the table constructed in exercise 13, convert the following
readings Fahrenheit into readings centigrade: (a) 12.6; (6) 21.7; (e)
28.55; (d) 116.75; (e) 151.86.
16. From the table constructed in exercise 14, convert the following
36 MATHEMATICS [§24
readings centigrade into readings Fahrenheit: (a) — 2; (6) 22.2;
(c) 61.3; (d) 62.6; (e) 75.4.
17. If x represents the number of degrees temperature, as read upon
a Fahrenheit scale, and if y represents the number of degrees of the same
temperature as read upon a centigrade scale, give the equation
connecting y and x.
18. Plot upon squared paper the numbers in the tables constructed
in exercises 13 and 14. Plot degrees Fahrenheit along the horizontal
axis and degrees centigrade along the vertical axis. Recompute
numbers, if any, whose plotted points do not fall upon a straight line
connecting the points (32, 0) and (212, 100).
19. Construct a table, for every sixty-fourth of an inch, con-
verting sixty-fourths into decimal equivalents. Compute to three
decimal places.
20. Plot the corresponding numbers found in exercise 19 upon
squared paper. Do all the points plotted fall upon a straight line
passing through the origin?
21. Construct a table converting centimeters, from 0 to 100, into
inches. One meter = 39.37079 inches. Compute to three decimal
places. Plot the numbers of the table upon squared paper. Do all
the points fall upon a straight line passing through the origin?
24. The Double Scale. From the preceding exercises we have
seen that relations between numbers may be represented by tables,
by equations, and by curves upon squared paper. A fourth
method of representing a relation between numbers will be ex-
plained by means of an illustration. If x represents a pressure
read in feet of water, and if y represents the same pressure read
in inches of mercury (since mercury is 13.55 times as heavy as
water), the relation between x and y is
12x
13.55
or
y = 0.886 x.
In Fig. 18, the scale on the under side of the horizontal line
represents pressure measured in feet of water, the scale on the upper
side represents pressure measured in inches of mercury. The
scales are so drawn that readings directly opposite the horizontal
line represent the same pressure, i.e., the length representing a unit
§25] GRAPHIC REPRESENTATION 37
on the upper scale is 1.13 times as long as the length represent-
ing a unit on the lower scale. From this double scale, one can easily
convert pressure readings in feet of water into inches of mercury ,
or vice versa.
Inches of Mercury, y
1 2 3 4 5
llllll
| l|ill||l.lj I 11111111.(1111 ,,|l I ll|lll,|lll.|l ,
123456789 10
Feet of Water, a;
FIG. 18. — Double scale showing relation between pressure measured
in inches of mercury and in feet of water.
Exercises
1. Draw a double scale connecting centimeters and inches.
2. Draw a double scale connecting inches and tenths of a foot.
3. Draw a double scale connecting pressure measurements in feet
of water and pounds per square inch. A cubic foot of water weighs
62.5 pounds.
25. The Equation of a Curve, the Graph of an Equation. Any
(every) point of a line parallel to the axis of x and three units above
it has an ordinate three, and no point whose ordinate is three is
found off the line. Since the equation y = 3 is satisfied by the
coordinates of every point of this line and by the coordinates of
no other point, it is said to be the equation of the line. The equa-
tion of a line parallel to the X-axis and ten units below it is
y = — 10. The equation of the X-axis is y = 0. The equation
of a line parallel to the F-axis, five units to its right, is x — 5.
The equation of a curve is an equation satisfied by the coordi-
nates of every point of the curve and by the coordinates of no other
point.
The graph of an equation is the locus of a point whose co-
ordinates satisfy the equation. To illustrate: if we have a
circle,1 radius 5 and center at the origin of coordinates, its equation
is xz -+- y2 = 52. For, if we draw from any point of this circle a line
to the origin and a perpendicular to the X-axis we have formed a
right triangle whose legs are x and y, and whose hypotenuse is 5;
1 The term "circle" as used in this book means the locus of a point at a constant
distance from a fixed point.
38 MATHEMATICS [§25
and if the point is chosen off the circle, the hypotenuse of the
right triangle is no longer 5. Thus, for points of the circle we have
x2 + y2 = 25, and for points off of the circle, x2 + y2 ** 25. *
The graph of the equation z2 + ?/2 = 25 is a circle, radius 5
and center at the origin.
The equation of a straight line passing through the origin and
making an angle of 45° what the positive direction of the axis of x is
y = x. The graph of the equation y = x is a straight line passing
through the origin, making an angle of 45° with the positive direc-
tion of the axis of x.
For brevity, instead of saying "the circle," or "the straight
line," or "the curve" — naming the equation — "representing
the equation" — we shall simply say the circle, or the straight
line, or the curve — naming the curve. Thus, the circle x2
+ y2 = 25; the straight line y = x, and the curve x3 — 3xy —
yz = 1.
Exercises
1. What is the equation of the axis of x?
2. What is the equation of the axis of y?
3. What is the equation of a line parallel to the X-axis, three
units above? Five units above? One hundred units below? Five
units below? Fifty units below?
4. What is the equation of a line parallel to the axis of y, five
units to the right? Ten units to the right? Six units to the left?
Eighty units to the left?
5. What is the equation of a right line passing through the origin
and bisecting the angle between the positive directions of the coordi-
nate axes?
6. What is the equation of a straight line passing through the origin
and bisecting the angle between the positive direction of the axis of
y and the negative direction of the axis of a;?
7. What is the graph of y = z? Of y = - xl Of y = x + 2?
Of y = x - 3?
8. What is the graph of y = - z? Of y = - x + 1? Of
y = - x + 3? Of y = - z - 6?
9. What is the equation of a straight line making an angle of 45°
with the positive direction of the axis of x and cutting the F-axis
five units above the origin? Ten units above the origin? Six units
below the origin? Twenty units below the origin?
i "^ " is read "is not equal to," or "is different from."
§25]
GRAPHIC REPRESENTATION
39
10. What is the equation of a straight line making an angle of 135°
with the positive direction of the axis of x and cutting the F-axis one
unit above the origin? Three units above the origin? Five units
below the origin?
11. What is the equation of the circle, center at the origin and
radius 1? Radius 2? Radius 3?
12. Show that the equation of a circle, radius 5 and center at the
point (2, 3), is (x - 2)2 + (y - 3)2 = 52, or z2 + y* - 4x - 6y = 12.
HINT: Take any point, P, of the circle. Its coordinates are x and
y. Draw PA perpendicular to the axis of x. Call the center of the*
circle C, and draw CB perpendicular to PA. Draw PC. PCB is a
right triangle, and
CB2 + BP2 = CP2,
or
. (x - 2)2 + (y - 3)2 = 52,
x2 - 4x + 4 + y2 - 6y + 9 = 25,
or
or
The student will show that the same equation is obtained when the
point P is to the right of, and below C', to the left of the F-axis ; below
the X-axis. When the point
P is to the left of the F-axis
the student must remember
that a; is a negative num-
ber; when below the x-axis
the y is a negative number.
13. Show that the equa-
tion of a circle, radius 5,
center at the point (3, 2), is
x2 + y2 - 6x - 4y = 12.
14. Show that the equa-
tion of a circle, radius 5,
center at the point ( — 2, 3).
is x2 + y2 + 4x - &y = 12.
15. Show that the equation of a circle, radius 5, center at the point
( - 3, 2), is x2 + y2 + 6z - 4y = 12.
16. Find the equation of a straight line passing through the origin
and making an angle of 30° with the positive direction of the X-axis.
Let T'OT, Fig. 19, be the line. Let P = (x, y) be any (every} point of
40
MATHEMATICS
[§26
the line. Let PA be the ordinate, y; and let OA be the abscissa, x.
OAP is a 30° triangle, and (see exercise 3, page 8)
is the equation of the line. For, the coordinates of every point of the
line satisfy the equation and the coordinates of any point not of the
line do not satisfy the equation. If the line weie raised three units,
x
its equation would be y = —7= + 3; if it were lowered ten units, its
V3
equation would be y =
- 10.
17. Show that the equation of a right line passing through the origin
and making an angle of 60° with the positive direction of the axis of
x is y = \3x. Show that if this line were raised six units its equation
would be y = V 3x + 6; and if it were lowered five units, its equation
would be y = \3x — 5.
18. Find the equation of a straight line cutting the F-axis one unit
above the origin and the X-axis one unit to the left of the origin;
cutting the F-axis one unit above the origin and the X-axis one unit
to the right of the origin;
cutting the F-axis one unit
below the origin and the
X-axis one unit to the right
of the origin; cutting the
F-axis one unit below the
origin and the X-axis one
unit to the left of the origin.
26. The Equation of the
Straight Line. Let T'OT,
Fig. 20, be any straight
line passing through the origin. If, in addition to stating that
the line passes through the origin, its direction is given, the
line is fixed. Its direction may be given by the angle XOT.
Let Q and R be any two points upon the line. Suppose a de-
scribing point slide along the line from Q to R, its ^-coordinate
FIG. 20.
§26] GRAPHIC REPRESENTATION 41
will change by the amount QS and its ^-coordinate will change
by the amount SR. If the line extends in the first and third
quadrants, both QS and SR will be positive, or both negative, de-
pending upon whether R is above and to the right of Q, or below
and to the left of Q. If the line extends in the second and fourth
quadrants, QS and SR will have opposite signs. If R is above and
to the left of Q, QS will be negative, and SR positive; if R is below
and to the right of Q, QS will be positive and SR negative. It is
SR
readily seen, however, that for any fixed line the ratio — is con-
Qb
stant (remains unchanged as the positions of Q and R are changed) .
SR
Let us abbreviate this ratio, — , by a. If the line extends into the
QS
first and third quadrants, a is positive; if into the second and
fourth quadrants, a is negative. If the line is very nearly hori-
zontal, a is numerically small. If the line is very nearly vertical,
a is numerically very large. For a line making an angle of 30°
with the positive direction of the axis of x, a is 1 / "V3. For a
line making an angle of 45° with the positive direction of the axis
of x, a is 1. For a line making an angle of 60° with the positive
direction of the axis of x, a is >3. For lines making angles of
120°, 135°, and 150° with the positive direction of the axis of x,
a is, respectively, — Vs, — 1, and r_-
Vs
If the line were to rotate counter-clockwise about the origin
from a horizontal position toward a vertical position, a' would in-
crease continuously from zero through all positive real values.
If the line were to rotate clockwise from a horizontal position to-
ward a vertical position, a would decrease continuously from zero
through all negative real values. From this, we see that every line
passing through the origin has an a, and for every value assigned
to a there corresponds one and only one straight line through the
origin.
The direction of a line is fixed by the algebraic value of a as well
as by the number of degrees in the angle between the line and the
positive direction of the axis of x. The numerical value of a is a
measure of the steepness of the line. It represents the number
42 MATHEMATICS [§26
of times faster a point moves in a vertical direction than in a
horizontal direction as it slides along the line, a is called the
slope of the line. The term slope is analogous to the term
percent grade in railroad work. A 1 percent grade means a
rise of 1 foot for every 100 feet of horizontal distance, or
a slope of 1 /1 00. Again we speak of the hydraulic gradient of a
stream as 2 feet per 1000, or as 8 feet per mile; these mean, re-
spectively, a slope equal to 2/1000 and 8/5280.
AP
Let P, Fig. 20, be any point on the line. Then -~ . = a, or
y/x = a,ory = ax. This, then, is the equation of a straight line
passing through the origin.1 If the line is raised two units, its
equation becomes y = ax + 2; if r.aised five units, y = ax + 5; if
lowered ten units, y = ax — 10. The equation of a line parallel
to the line y = ax, and cutting the F-axis b units from the origin, is
y = ax + b. If the line cuts the F-axis above the origin, b is
positive. If the line cuts the F-axis below the origin, b is negative.
y = ax + b is called the slope equation of the straight line, a is
called the slope, and b the y-intercept.
Exercises
1. Give the slope and y- intercept for each of the following lines:
(a) y = 2x + 1; (g) y = - Bx + 2; (ro) y = \x + 3;
(b) y = 2x - 3; (h) y = x + 1; (n) y = \x - 3;
(c) y = 3z - 6; (i) y = x - 1; (o) y = - %x;
(d) y = 2x - 1; (j) y =x+ 2; (p) y = - fx - 8;
(e) y = - 2z - 2; (fc) y = - a; + 6; (?) y = - fz;
(/) y = - 3x + 4; (I) y= -x + 2; (r) y = - O.Ola: - 6.
2. In exercise 1, what lines are parallel to (a)? To (c)? To (e)?
To (/)? To (A)? To (fc)? To (TO)? To (p)?
3. If a point were to move along the lines of exercise 1, so that its
x-coordinate is decreased by one unit, would the y-coordinate be
increased or decreased and how much?
4. Find the equation of a straight line passing through the points
P, (3, 2), and Q, (1, 5). To solve the problem is to find the value of
a and of 6 and substitute in the equation y = ax + b. As we pass
from the point P to the point Q, the x-coordinate decreases two units
1 The student must guard against getting the idea that the slope of a line is, in
general, y over x. This is only true when the line passes through the origin.
§27] GRAPHIC REPRESENTATION 43
and the ^-coordinate increases three units. Then the slope, a, of
the line is — 3/2. Let R be the point where the line through P and
Q cuts the F-axis. Draw PT perpendicular to the F-axis, meeting
it at the point T. Draw QS perpendicular to PT, meeting it at S.
PSQ and PTR are similar triangles, and we have
TP _ TR
SP ~ SQ*
or
3 _ TR
2 ~ 3 '
or
TB-\
and b = L23, since b = OT + TR. The equation of the line is
_ 3 13 ?
or
3z + 2y = 13.
This work can be checked, for, substituting for x and y the co-
ordinates of the points P and Q, the equation is satisfied in each case.
6. Does the line of exercise 4 pass through the point (2, 7/2)?
The answer is yes, for substituting 2 for x and 7/2 for y, in the equation
of the line, the equation is satisfied.
6. Find the equation of the straight line passing through the points
(a) (-3, 2) and (2, 1); (6) (1, 1) and (2, 2); (c) (-1, 3) and
(2, - 3); (rf) (- 3, -1) and (-1, 3); (e) (-1,2) and (6, - 1).
7. Which lines of exercise 6 pass through the point (— 2, 6)?
8. Find the equation of a straight line passing through the origin
and the point (3, 5); through the points (3, — 1) and (— 3, — 1);
through the points (— 3, 5) and (— 3, 2).
27. The Graph of an Equation of the First Degree between x
and y. In § 26 it was shown that every equation -of the form
y = ax represents a straight line passing through the origin; and
further, when the line was moved up or down its equation became
y = ax + b, where 6 represents the algebraic distance the line was
translated in the y direction. If the line were moved from a very
low position to a very high position, the b would take on all real
values between its very small initial value and its very large ter-
minal value. Thus, by moving the line over a greater and greater
44 MATHEMATICS [§27
Vertical distance, b can be made to take any real value whatever.
Any equation then of the form y = ax + b, for any values given
to a and b, represents a straight line. It is readily seen that this
form of the equation includes all lines excepting those parallel
to the F-axis. The equation of a line parallel to the F-axis is,
however, x = k, where the constant k is the number of units
the line is distant from the F-axis. The equation of any
straight line is then either of the form y = ax + 6 or of the
form x = k. Since any equation of the first degree between
x and y, Ax + By = C, may be put in one of these forms, it is
the equation of a straight line, and is called a linear equation.
To illustrate: 3x + % + 6 = 0 may be written y = — fx — 3,
the equation of a straight line having the slope — f, and a
y-intercept of — 3.
Exercises
Find the slope and intercept of each of the following lines:
1. 2z - 3y = 1. 4. 3x - 2y + 6 = 0.
2. 3x - 2y - 6 = 0. 5. y - 3x = 5.
3. x - y = 1. 6. y + x = 0.
7. Find the point of intersection of each line given above with the
axis of x. (The distance from the origin to the point of intersection
of a line with the X-axis is called the re-intercept.)
HINT: This is done by putt'ng y equal to zero, and solving the re-
sulting equation for x. Why does this give the rc-intercept?
8. Draw the straight lines of exercises 1 to 6 upon one sheet of
squared paper. Use 1 inch (or 2 cm.) to represent the unit and place
the origin near the center of the sheet.
9. Find the coordinates of all points of intersection of these lines
in so far as they intersect upon the sheet of squared paper.
The coordinates of the points of intersection are to be read to two
decimal places. To make this possible great care must be exercised
in drawing the lines. Tabulate your results. In the first column
place the numbers of the lines, as 2-5, meaning the line in exercise
2, and the line in exercise 5. In the second column place the
x-coordinate and in the third column place the y- coordinate of the
point of intersection of the two lines. A fourth and a fifth column
are to be filled in with the results obtained from the next exercise.
10. Consider all pairs of equations given in exercises 1 to 6 as
simultaneous equations and solve for x and y. Record the results in
the table of the preceding exercise. Compare the corresponding x
v
GRAPHIC REPRESENTATION 45
values and the corresponding y values found by the two methods.
Should the corresponding values be the same? Why?
28. Analytic Method of Finding a and b. In exercise 4, § 26,
there was described a geometric method for finding the a and the b
in the equation of a straight line passing through two points. An
analytic method of finding a and b will now be illustrated. Let
(3, 2) and (1, 5) be the points through which the line passes. The
coordinates of each point when substituted for x and y in the gen-
eral equation of the line, y = ax + b, must satisfy that equation.
Thus we have, upon substitution, 2 = 3a + b, and 5 = a + b.
These two equations are called conditional equations, for they are,
in mathematical language, statements of the two and only two
conditions imposed upon the line.1 The first equation is the
equation of condition which says that the point (3, 2) is upon
the line y = ax + b; the second equation is the equation of condi-
tion which says that the point (1, 5) is upon the line y = ax + b.
Solving these two conditional equations for a and b we ob-
tain fl = — f, and b = V. Substituting these values in
y = ax + b, the equation of the line through the two given
points becomes y = — lx -f V, or 3z + 2y = 13.
Exercises
1. Find, by the analytic method, the equation of the line through
each of the following pairs of points:
(a) (1, 2) and (- 3, 6); (c) (0, 0) and (- 1, - 1);
(b) (-2, 6) and (3, - 2); (d) (0, 0) and (- 2, - 3).
2. Plot upon squared paper the data given in Table VI. Plot
hydraulic gradient2 along the F-axis using 1 cm. (or 1/2 inch)
1 A conditional equation may be denned as an equation expressing some condition
given in the problem. To illustrate: If two numbers x and y are to be found such
that their sum is 9, and the sum of their squares is 25, we have
x + y = 9
and
x2 + j/2 = 25.
The first equation expresses the condition imposed upon x and y, that their sum
is 9. The second equation expresses the condition that the sum of the squares of
the numbers is 25.
2 By hydraulic gradient is meant the drop of the surface of water in moving down
stream a given distance. Thus in the table the hydraulic gradient of 10 feet per
thousand feet means a drop of the water plane (water surface) at the rate of 10 feet
for every 1000 feet measured along the horizontal in the direction of greatest
decline of the plane.
46
MATHEMATICS
[§28
to represent 10 feet. Plot velocity along the X-axis, using 2 cm.
(or 1 inch) to represent 10 feet. From the plotted points it
appears as if the points would fall (or would nearly fall) along a
straight line if there were no errors of observation, or if all the readings
were taken under the same conditions. The best that can be done in
this case is to assume a linear relation between hydraulic gradient
and velocity. Draw with the aid of a transparent triangle a straight
line among the points which in your judgment is the best line to repre-
sent this relation. Does the line pass through the origin? Compute
a by taking two points at or near the ends of the line. The student
must remember that a is the ratio of the number of units in the change
in the ^-coordinate to the number of units in the change in the
x-coordinate.
3. Plot upon squared paper the data for the 8-inch well given in
Table VII. Plot head1 along the F-axis using 1 cm. (or 1/2 inch) to
represent 1 foot. Plot yield along the X-axis, using 1 cm. (or 1/2
inch) to represent 10 gallons. As in the preceding exercise draw a
straight line among the points and determine the a.
TABLE VI. — VELOCITY OF WATER THROUGH SAND
Hydraulic
gradient,
feet per 1000
Velocity
of water,
feet per 24 hours
Hydraulic
gradient,
feet per 1000
Velocity
of water,
feet per 24 hours
31.9
16.9
21.0
10.1
20.8
11.4
119.1
58.8
54.1
10 0
24.5
4 3
55.8
22.7
TABLE VII.— YIELD OF AN 8-INCH AND OF A 14-INCH WELL
Size of
well,
inches
Head,
feet
Yield,
gallons
per minute
Size of
well,
inches
Head,
feet
Yield,
gallons
per minute
14
2.02
31.4
8
2.50
39.5
14
4.00
59.7
8
3.85
49.4
14
6.40
93.2
8
6.47
89.3
14
8.68
121.1
8
8.28
116.3
14
11 73
151.1
1 By "head" in this problem is meant the distance the water was lowered in the
well below its normal position. Thus, a head of 2.5 feet in the table means that when
39.5 gallons of water per minute were pumped from the well the elevation of the sur-
face of the water in the well was 2.5 feet below its normal elevation.
§29] GRAPHIC REPRESENTATION 47
4. Plot upon squared paper the data for the 14-inch well given in
Table VII. Draw a straight line among the points and find its a.1
29. Cost of Concrete. Suppose concrete mixture is made by
mixing cement, sand, and crushed rock. The expression " mixture
(ra, n, fe) " means the relative amounts, by volume, of the three
materials. Thus, a mixture (1, 2, 3) means that for each volume
of cement there are two volumes of sand and three volumes of
rock.
Curves will now be drawn from which the cost of a cubic yard
of concrete may be determined. The method of drawing the
curves will be illustrated by an example. The student is to con-
struct the drawing as explained. Suppose the percent voids of
rock is 45 and that of sand 33. Let us take the mixture (1, 2, 3).
For each cubic yard of cement, 2 cubic yards of sand and 3 cubic
yards of rock are used. In the 3 cubic yards of rock there are
3-0.45 = 1.35 cubic yards of voids. Then 1.35 cubic yards of the
2 cubic yards of sand will be taken up by the voids of the rock
leaving 0.65 cubic yard. Thus mixing 3 cubic yards of rock
and 2 cubic yards of sand will give but 3 + 0.65 = 3.65 cubic
yards of mixture. In the 2 cubic yards of sand there are
2-0.33 = 0.66 cubic yard of voids. Thus there is an excess of
1 — 0.66 = 0.34 cubic yard of cement above filling the voids of
the sand. Hence in mixing the rock, sand, and cement there will
be 3.65 + 0.34 = 3.99 cubic yards of mixture.
If Xi is the cost (say in. dollars) of a cubic yard of stone, the
cost, yi, of the stone in 1 cubic yard of concrete is
3
yi ~ 3^9 Xl
or
yi = 0.752 x, (1)
Upon a sheet of squared paper draw a straight line represent-
ing equation (1). Let 10 cm. represent $1. Then each centi-
meter will represent 10 cents. To draw this line join the points
(0, 0) and (2, 1.504).
1 The equations found in exercises 2, 3, and 4 very probably do not represent the
true relations among the measured numbers, and are at best only approximate
relations. Such equations are called empirical equations, or empirical laws.
48 MATHEMATICS [§29
If x2 represents the cost of 1 cubic yard of sand, the cost, y2,
of the sand in 1 cubic yard of concrete is
2/2 =
3799 2
or
?/2 = 0.501 x2 (2)
Plot equation (2) upon the same sheet of paper upon which
equation (1) was plotted, using the same scales.
A barrel (four sacks) of cement contains 3| cubic feet, or 5/36
cubic yard. If xs represents the cost of cement per barrel, the
cost, y3, of cement in 1 cubic yard of concrete is
36
z ~ 5-3.99 3
or
7/3 = 1.805 x3 (3)
Draw this line upon the same sheet of paper upon which equations
(1) and (2) were drawn.
Mark the first line drawn "rock;" the second "sand;" and the
third "cement." From these three lines the cost of material in
1 cubic yard of concrete can be scaled off, if the costs of stone,
sand, and cement are given. Thus, suppose stone worth $1.40 per
cubic yard, sand $1.25 per cubic yard and cement $1.10 per
barrel. From the point 1.4 on the X-axis go up to the rock line,
thence to the left to the F-axis and obtain the reading $1.05, the
cost of the stone in 1 cubic yard of concrete. From the point
1.25 on the X-axis go up to the sand line, thence to the left to
the F-axis and obtain the reading $0.63, the cost of the sand in
1 cubic yard of concrete. From the point 1.1 on the X-axis go
up to the cement line, thence to the left to the F-axis and obtain
$1.99, the cost' of the cement in 1 cubic yard of concrete.
The total cost of material in 1 cubic yard of concrete is then
$1.05 + $0.63 + $1.99 = $3.67.
The following table gives the number of cubic yards of rock,
cubic yards of sand, and barrels of cement in 1 cubic yard of
concrete based on 45 percent voids in rock and 33 percent voids
in sand.
§30]
GRAPHIC REPRESENTATION
49
Mixture
Rock,
cu. yds .
Sand,
cu. yds.
Cement,
barrels
1:2 :3
0.752
0.501
1.805
1:2 :4
0.881
0.446
1.586
1:2^:5
0.922
0.461
1.327
1:3 :5
0.951
0.475
1.125
Exercises
1. Verify the above table.
2. Construct curves from which to scale off the cost of material
used in a 1, 3, 5 mixture of concrete.
3. From the drawing of exercise 2, find the cost of the material in
1 cubic yard of concrete, if rock is worth $1.50 a cubic yard, sand
$1.10 a cubic yard and cement $1.30 per barrel.
30. The Graph of y = px2.
Exercises
1. Upon a sheet of paper, form M7, construct a table per the
following instructions. Head the first vertical column x, the second
x2, the third £x, the fourth |x2, and the fifth ?x2. In the horizontal
spaces in the first column place the numbers 0, 0.2, 0.4, 0.6,
0.8, 1.0, 1.2, . . ., 4.6, 4.8, 5.O.1 In the second column place
the squares of these numbers, as 0, 0.04, 0.16, 0.36, 0.64, 1.0, . . .,
21.16, 23.04, and 25.00. In the third, fourth and fifth columns
place, respectively, the halves, the thirds and the fourths of the num-
bers in the second column. Head this table "Table of Squares."
2. Plot upon squared paper a curve for y = x2. Let the X-axis
run the shorter and the F-axis the longer dimension of the sheet of
paper. Choose the origin near the lower center of the page. Plot
points for negative values of x as well as for positive values. Make
use of your "Table of Squares." Let 2 cm. (or 1 inch) on both
axes represent one unit. Use a very sharp pointed pencil in plot-
ting these points, and exercise great care in locating them as accurately
as possible. Do not draw a curve through the plotted points until
all exercises of this list have been completed.
3. Plot a curve for y = \x*. Follow directions given in exercise 2.
4. Plot a curve for y = $x2. Follow directions given in exercise 2.
6. Plot a curve for y = Jx2. Follow directions given in exercise 2.
6. Plot a curve for y = ix2. Follow directions given in exercise 2.
1 The marks " . . . " mean "and so on," indicating all corresponding inter-
mediate numbers.
4
50
MATHEMATICS
[§30
7. Prom the coordinates of the points of the curve y = z2 compute
the slope of lines passing through the consecutive plotted points. The
slope of the secant lines will be approximately the slope of the
tangent lines drawn to the curve midway between the points. Thus,
if we find a slope of 4.2 for the secant line drawn through the points
whose ^-coordinates are 2.0 and 2.2, we shall have the approximate
slope of the tangent line drawn to the curve at the point whose
x-coordinate is 2.1 Tabulate your results.
8. Same as exercise 7 but use curve y = £z2.
9. Same as exercise 7 but use curve y = |z2.
10. Same as exercise 7 but use curve y = |x2.
11. Same as exercise 7 but use curve y = £z2.
12. Plot upon squared paper the results of exercise 8. Plot x
along the horizontal axis, and slope, a, along the vertical axis. Draw
a straight line through the origin and among the plotted points.
Measure the slope of this line; call it A\.
13. Same as exercise 12 but use results of exercise 9. Call the
slope Az.
14. Same as exercise 12 but use results of exercise 10. Call the
slope A*.
15. Same as exercise 12 but use results of exercise 11. Call the
slope At.
16. Divide A 2 by f; As by |; and A^ by j. These quotients
and AI should all be equal to 2.
It can be shown by the method of the Calculus that the slope of a
tangent line drawn to the curve y = px2 (p being a constant) is 2px,
where x is the abscissa of the point of tangency.1
1 A formal proof is given
here. Let P, (x, y), Fig. 21,
be any point, and Q any second
point on the curve y = px*.
Let k, (PR), be the difference
of the two z-coordinates, and
let h, (RQ), be the difference of
the two ^-coordinates. The
coordinates of Q are then (x +
k, y + h) . For the point P we
have y = px*; and for the point
Q we have (y + h) = p(x +
k)t. Subtracting, we have h
= 2pkx + pk*. Dividing by
k we have h/k = 2px + pk.
This shows that the slope,
(h/k), of the secant line drawn
through the points P and Q is
§30] GRAPHIC REPRESENTATION 51
This is only saying that if a point move along the curve its upward
motion (downward motion if the point is to the left of the y-axis) is
at any instant 2px times its horizontal motion, or its y-coordinate is
changing 2px times as fast as its x-coordinate. Here x represents the
x-coordinate of the position of the moving point.
Experiment shows that a body falling freely from rest gives the law
s = 16. li2, where s represents the distance, in feet, passed over in
the time t, in seconds. If s be plotted along the vertical axis and t
along the horizontal axis the slope of the tangent line drawn to the
curve will be 32.2t. Distance s, then, is changing 32.2t times as fast
as t, i.e., the speed of a body falling freely from rest is 32.2J, when t
is the time, measured in seconds, elapsed from the instant the body
began to move.
If v represents speed, in feet per second, we have v = 32. 2t.
Plotting v along the vertical axis and t along the horizontal axis
we obtain a straight line representing the relation between speed
and time. The slope of his line is 32.2, i.e., v changes 32.2 times as
fast as t, or the acceleration of gravity is 32.2 feet per second per
second.
17. Plot a curve for y = x2 — 2x + 1 = (x — I)2. How is this
curve related to the curve y = x2?
18. Plot a curve for y = x* - 2x + 2 = (x - I)2 + 1. How is
this curve related to the one plotted in exercise 17?
19. Plot a curve for 3y = x* - 4x + 13 = (x - 2)2 + 9, or
y = 3(2 — 2)2 + 3. How is this curve related to the curve for
y = *x2?
20. Without plotting, compare the following curves:
(a) y = x2 and y = — x2.
(6) y = |x2 and y = - £x2.
(c) y = |x2 and y = — \xl.
(d) y = lx2 and y = — }x2.
(e) y = (x - I)2 and y = - (x - I)2.
2px + pk. Aa the point Q slides along the curve and approaches P, k, (PR), grows
smaller and smaller and approaches zero; the slope of the secant line then ap-
proaches 2px. But, as Q approaches P the secant line rotates about the point P
and approaches the tangent line PT. Therefore, the slope of the tangent line is
the limit of the slope of the secant line, i.e., 2px.
52 MATHEMATICS [§31
21. Have the curves considered in exercise 20 lines of symmetry?
Give the equation of the line of symmetry for each curve.1
22. Change each of the following equations to the form
y = p(x — q)z + r, where p, q, and r are constants, and find the coor-
dinates of the highest or lowest point on the curve:
(a) 4x2 + By = 3x + 5,
or 3y = - 4x2 + 3x + 5,
or - | y = x2 — f x — I ,
or - f y = x2 - f x + V* - f - &,
or - \y = (x - f)2 -ff,
or y = - f (x - f )2 + ||.
P = - I; 9 = 1; r = If.
The coordinates of the highest point are f and If.
(6) y = x* + 4x - 6. (f) x*+y = 6.
(c) 3y = x2 + 6x + 10. (0) x2 + 2z + y + 1 = 0.
(d) 3y = 4x2 + 6x + 10. (h) 2x2 - 5x - 6y + 10 = 0.
(e) 4x2 = y - 2x - 6. (i) 5x2 + 2x - y = 0.
31. The Parabola. All curves drawn and discussed in the
preceding section have equations reducible to the form
y = p(x — q)2 + r. They have a line of symmetry, x = q, and
change from an increasing to a decreasing curve or vice versa at
the point (q, r).2 Each curve is the curve y = px2 translated
horizontally a distance q, and vertically a distance r.3 Any curve
which is of the shape of y = px2 is called a parabola. The line of
symmetry of the curve is called the axis of the parabola, and the
point of intersection of the curve with its axis is called the vertex.4
1 A line such that line segments drawn perpendicular to it and terminated by
the curve, are bisected by it is called a line of symmetry of the curve. Thus any
line passing through the center of a circle is a line of symmetry for the circle.
A point such that line segments drawn through it atid terminated by the curve,
are bisected by it, is called a point of symmetry. Thus, the center of a circle is a
point of symmetry.
2 A curve is said to be increasing when its ordinate increases as its abscissa in-
creases; and decreasing when its ordinate decreases as its abscissa increases. The
increase in each case being understood to be the algebraic increase.
8 Distance here means algebraic distance. When q is positive the curve is moved
to the right, when negative to the left. When r is positive the curve is moved up,
when negative down.
4 It can be shown that the parabola is the locus of a point moving in a plane so
that its distance from a fixed line, the directrix, is always equal to its distance from
a fixed point, the focus. The parabola is also the curve of intersection of a right cir-
§31] GRAPHIC REPRESENTATION 53
Any equation of the form y = ax1 + ftx + 7, where a, /3 and
7 are constants, positive or negative, represents a parabola with
its axis parallel to the Y-axis; for
ax2 + 0x + y = a
which is of the form p(x — q)2 + r, where p = a, q = - — > and
2a
r = -- - ---- The expresson ax2 + fix + 7 is the most general
expression of the second degree in x, or it is a general quadratic.
Thus we see that the equation y = ax2 + 0x + 7 represents a
parabola with axis parallel to the Y-axis, and by completing the
square in the x terms we are able to locate its vertex. Problems
illustrating this point have already been given in exercise
22, § 30.
We have already seen that the slope of the tangent line drawn to
the curve y = px2 is 2px, i.e., 2p times the x-coordinate of the
point of contact of the tangent line. Another way of saying this is
that the y is changing 2px times as fast as the x. From this and
what was said in the preceding paragraph we see that the slope
of the tangent line drawn to the curve for y = ax2 + /3x -\- y is
2p(x — q), or 2p times the distance of the point of contact of
the tangent line from the axis of the parabola. Conversely, it
may be shown that if a number y changes 2p(x — q) times as fast
as the number x, x and y are connected by the parabolic law.
Exercises
1. Plot data given in Table VIII. Plot temperature along the
horizontal axis and weight along the vertical axis.
cular cone and a plane parallel to an element of the cone. The parabola has many
useful and interesting properties, one of which is: If a parabola be rotated about its
axis forming a surface of revolution, the paraboloid of revolution, rays of light passing
out from its focus are reflected by the surface into lines parallel to the axis of the
paraboloid. This property is made use of in the construction of reflectors.
54
MATHEMATICS
[§31
2. An empirical equation of the second degree in T for the data
given in Table VIII is W = 0.06T2 - 0.3677 + 0.47. Here W is
used in order to distinguish between weight computed by formula and
observed weight, W.
Copy Table VIII upon a sheet of paper, form M7. In a third
column headed W place weights computed by the empirical formula.
In a third column headed W — W place the differences between the
observed and computed weights. Plot W upon the drawing of
exercise 1. Mark the points representing W so that they are readily
distinguished from the points representing W.
TABLE VIII.— RELATION BETWEEN TEMPERATURE AND WEIGHT OF
10 LITERS OF WATER
T = temperature in degrees centigrade.
W = loss in weight, measured in grams, of 10 liters of water, as temperature
differs from 4°.
T
w
T
w
0
1.3
16
10.3
2
0.3
18
13.8
4
0.0
20
17.7
6
0.3
22
22.0
8
1.2
24
26.8
10
2.7
26
31.9
12
4.8
28
37.1
14
7.3
30
43.3
3. Plot data given in Table IX.1 Plot ^ along the horizontal
axis using 1 cm. to represent one-tenth unit. Plot velocities
along the vertical axis, using 1 cm. to represent two one-hun-
dred ths of 1 foot per second. Sketch a smooth curve among the
points. The curve may not pass through all the plotted points.2
Begin to number the vertical axis with 2.90. The true origin
will be far below the sheet of paper.
1 The velocity at any point of a moving stream is determined by a current meter
placed at that point.
2 This curve is called a vertical velocity curve. In practical work, however,
velocities are plotted along the horizontal axis and depths along the vertical axis, and
down from the origin. Your drawing gives the usual form of plotting if it is turned
90° in a clockwise direction. Vertical velocity curves are parabolic in shape, with
the axis of the parabola parallel to the surface of the water.
§32]
GRAPHIC REPRESENTATION
55
TABLE IX.— RELATION BETWEEN VELOCITY AND DEPTH AT A
POINT IN THE LOWER MISSISSIPPI
Depth at observed point = d; whole depth = D.
d
D
Velocity, feet per
second
d
D
Velocity, feet per
second
0.0
3.195
0.5
3.228
0.1
3.230
0.6
3.181
0.2
3.253
0.7
3.127
0.3
3.261
0.8 3.059
0.4
3.252
0.9
2.976
4. An empirical formula connecting velocity, V, and depth, x,
is V' = - 0.78(x - 0.296)2 + 3.261. Compute V from this formula
and tabulate your results following, in a way, directions given in
exercise 2.
6. A strip of tin L feet long and 40 inches wide is made into a trough
with rectangular cross section by bending up an equal portion of each
side. Find the width of each portion bent up such that the volume of
the trough is a maximum.
HINT: The volume is a maximum when the area of the cross sec-
tion is a maximum. Let y be the area of the cross section. Let x be
the width of the portion bent up. The dimensions of the cross section
are then x and 40 — 2z. Hence
y = x(40 - 2x)
y = - 2(z2 - 20x)
y = - 2(x2 - 2Qx + 100) + 200
y = - 2(» - 10)2 + 200
The equation is that of a parabola with its high point at (10, 200).
Thus the maximum area, 200, occurs when 10 inches of tin are turned
up.
6. Same as exercise 5, but with the width 60 inches and length
equal to 10 feet. Find the volume of the trough.
32. The Equilateral Hyperbola.
Exercise
1. Plot upon squared paper a curve for
123
(a) y = - (6) y = ~ (c) y - —
XXX
56 MATHEMATICS [§32
Choose the origin near the center of the sheet and plot for negative
as well as for positive values of x. Let 1 cm. (or 1/2 inch) rep-
resent the unit along each axis. When x is numerically small the
interval between successive values assigned to it must be small in
order to obtain points near enough together for sketching the curve.
Compare curves (a), (6), and (c).
These curves are equilateral hyperbolas. They are special
cases of the more general equilateral hyperbola, y = —' where a
X
is a constant. When one quantity, y, varies with another, x,
so that the relation connecting the two is represented by the equa-
tion y = — , y is said to vary inversely with x.
tC
If points plotted upon squared paper arrange themselves along
a curve which seems to be an equilateral hyperbola, y = -
X
may, as a first assumption, be taken as an empirical equation
connecting the variable numbers plotted. The problem now
becomes one of selecting a value for the constant a, and of deter-
mining the agreement between the values of the numbers plotted
and those called for by the equation. If the a cannot be selected
giving a fairly close agreement between observed and calculated
numbers, it means that the law we are dealing with in nature
is not that of one number varying inversely as the other.
An easy elementary method for determining the a, as well as
showing whether or not the law is that of the inverse power, will
be illustrated by means of an example.
In Table X are given the relative volumes of a mass of air
subjected to pressures ranging from 0 to 100 pounds per square
inch measured above atmospheric pressure. The temperature
of the air is constant, 60° Fahrenheit.
Plot the data upon a sheet of squared paper. Plot pressure
along the F-axis, using 2 cm. (or 1 inch) to represent 10 pounds.
Plot volume along the X-axis, using 10 cm. (or 5 inches) to rep-
resent the unit volume. This curve seems to be of the general
shape of y = — excepting that it crosses or touches the .X-axis
X
at unity instead of remaining above the X-axis and becoming
nearer and nearer to it.
§32]
GRAPHIC REPRESENTATION
57
To find an equation connecting x (volume) and y (pressure)
proceed as follows. Construct a third column of numbers for
Table X, by taking reciprocals of the numbers representing rela-
tive volumes given in the second column. Call these values the
x' values. (The prime is put upon the x to distinguish it from the
x which represents the true volume reading.) Plot a curve be-
100
90
•8
£ 70
#60
"
3 40
30
20
10
Reciprocal
I of Volume
A 8
0.2
0.4 0.6
Volume
0.8
1.0
FIG. 22.
tween pressure and x'. Let 2 cm. (or 1 inch) along the Y-
axis represent 10 pounds; let 2 cm. (or 1 inch) along the hori-
zontal axis represent one unit of x' '. Such a curve, reduced, is
given in Fig. 22. This curve is a straight line with slope equal
to 14.7 and y-intercept equal to — 14.7. The equation of the
58
MATHEMATICS
[§32
line is y = 14. 7o/ — 14.7. Since x' — — , the relation connect-
ing volume and pressure is represented by the equation
14.7
y = — - — 14.7. It is readily seen that this is the equation of
u/
14.7
the curve y = - - lowered 14.7 units; that the first curve
*c
plotted between pressure and volume is an equilateral hyperbola;
and that the pressure readings given in Table X, when increased
by 14.7, vary inversely as the volume.
TABLE X.— AIR COMPRESSION FROM ONE ATMOSPHERE AT
SEA-LEVEL
Pressure,
Ib. per sq. in.
Volume
Pressure,
Ib. per sq. in.
Volume
0
1,000
45
0.246
1
0.936
50
0.227
2
0.880
55
0.210
3
0.830
60
0.196
4
0.786
65
0.184
5
0.746
70
0.173
10
0.595
75 '
0.163
15
0.495
80
0.155
20
0.423
85
0.147
25
0.370
90
0.140
30
0.328
95
0.134
35
0.295
100
0.128
40
0.268
The pressure readings given in the table were taken by means of
a pressure gauge which registers the difference in pressure between
the outside and inside of the receptacle containing the air. If we
assume Boyle's law for the compressibility of a gas (the volume
varies inversely as the pressure) the y-intercept, 14.7, found
above, shows that when the experiment was conducted the pres-
sure gauge was registering 14.7 pounds less than the true pressure
exerted upon the air, or that the atmospheric pressure was
14.7 pounds per square inch.
If, in Fig. 22, the plotted points had not arranged themselves
along a straight line the method outlined in § 28 could have been
§32]
GRAPHIC REPRESENTATION
59
used in finding the slope and intercept. When the plotted points
arrange themselves -along a curve which is not a straight line, it
may mean that the original variables are not connected by the
equilateral hyperbolic law.
Exercise
Find the equation connecting I and W, using the data given in
Table XI,1 where 7 represents the indicated horse power of a
steam engine, and W the number of pounds of steam consumed
per hour.
TABLE XI
I
W
/
W
36.8
460.0
15.8
222.8
21.5
406.2
12.6
182.7
26.3
344.5
8.4
137.0
21 0
279 3
1 Taken from Elementary Practical Mathematics by John Perry.
CHAPTER II
LOGARITHMS
33. Logarithm Defined. The equation y = xz — 1 is solved
for y. If values are assigned to ' x, the corresponding values
of y may be calculated. The same equation may be written
x = ± \/y -f- 1. Here the equation is solved for x, and if values
are assigned to y, 1;he corresponding values ofz may be calculated.
The two equations represent the same law connecting x and y.
Exercises
1. Solve each of the following equations for x:
(a)y = x; (i) y = 3x2 - 2x + 1;
(6) y = x - 1; (j) y = xs;
(c)t/ = 2x+3; (k)y=x*-l;
(d)y = |x -6; (I) y = 3z» -3;
(e) y = x2; (m) y = x3 + 3x2 + 3x + 1;
(/) y = 2x^5 (n) y = X3 - 3x2 + 3x;
(0) y = x2 - 2x + 1; (o) y = ± Vx* - 25.
(h) y = 3x2 + 6x + 3;
2. Solve each of the following equations for x and also for y:
(a) x2 + y2 = 25; (i) xy = 2;
(6) x2 -0« = 25; (j) zy +y = 1;
(c) x2 + 2x + y2 = 25; (jfe) xy2 = 3;
(d) x2 - 2x + y2 = 25; (I) x*y = 5;
(e) x2 + y2 - 2y = 25; (m) x2y + xy2 = 3;
(/) x2 + 2x + y2 - 2y = 25; (n) x2?/ + xy + xy2 = 0;
(g) x2 + x + y2 - 2y = 16; (o) xy3 = 1;
(A) xy = 1; (p) x=y = - 1.
All equations of the preceding exercise can be solved for x by
means of the symbols and operations of elementary algebra. The
equation y = ax, where a is a known positive constant different
60
§33] LOGARITHMS 61
from zero, cannot, however, be solved for x by means of these
elementary operations. The best we can do is to say, x = the
exponent of the power to which a must be raised to produce y.
In this verbal sense, the equation is now solved for x. The
words "exponent of the power to which a must be raised to pro-
duce y" are abbreviated by "logarithm of y to the base a," and
in writing they are further abbreviated by "Iog0 y." Thus, the
equation y = ax, solved for x, gives x = loga y- Logo is then
a new symbol; considered as an operator, operating upon y, it
gives the number, x, such that when a has x for an exponent,
the power equals y. a is called the base. The definition then
of the logarithm of a number x to the base a is: "the exponent
of the power to which a must be raised to produce the number
in question."
Of the two equivalent equations, y = a* and x = logo y, the
first is called the exponential form, and the second the
logarithmic form.
Exercises
1. Write the following equations in logarithmic form :
(a) 100 = 102; (e) 25 = 52; (i) 1 = 10°;
(6)10 = 10'; (/) 9 = (V3)4; 0')2 = (V2)2;
\C) TC == *w j \Q ') -LD ^= ~ *j \fo) *5 ^~ \'\' *J ) j
(d) 8 = 23; (h) 125 = 53; (1) 8 = (V2)6;
(m) 2° = 1; (n) 5° = 1.
2. Write the following equations in exponential form:
(a) Iog24= 2; (/) logio 10 = 1; (ft) logs 9 = 2;
(b) logio 100 = 2; (0) logs 3 = 1; (I) logio 1000 = 3;
(c) logio 1 = 0; (h) Iog5 5 = 1; (m) logio 10,000= 4;
(d) Iog2 1 = 0; (i) logs 25 = 2; (n) logs 1 = 0;
(e) logs 1=0; (j) Iog2 32 =5; (o) Iog99 1 = 0.
3. Give the value of each of the following:
(a) Iog24; (h) log,0 100; (o) logio 0.1;
(6) logs 9; (i) logio 1000; (p) logio 0.01;
(c) logs 27; (j) logio 10,000; (a) logio 0.001 ;
(d) logt 25; (ft) logio 1; (r) logio 100,000,000;
(e)log44; (I) logs 1; («) logio 0.000000001.
(/) log« 6; (m) logs 1;
(o) logio 10; (n) logo 1;
62 MATHEMATICS [§34
4. Evaluate the following:
(a) Iog22 + Iog44 + logB5;
(6) logs 9 + Iog5 25 + logs 125 + logs 36;
(c) logio 1000 + Slogs 36-5 Iog2 16+2 logB 625;
(d) Iog2 1 - logs 1 + logs 3 - logio 10 + logs 27;
(e) \ logio 100 - | logio 10-| logio 0.1-2 log,0 0.01;
(/) logo a; (g) Iog0l
34. Theorem : The logarithm of the product of two numbers
is the sum of their separate logarithms.
Proof: Let
logo u = x and Iog0 v = y
Then
ax = u and a" = v
By multiplication
a*+v = uv,
or
Iog0 uv = x + y,
or
Iog0 uv = logo u + logo v
Thus the theorem : The logarithm of the product of two numbers
is the sum of their separate logarithms.
It is easily seen that the logarithm of the product of three or more
numbers is the sum of their separate logarithms.
If no base is indicated, the base 10 is understood.
Exercises
If log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990, and log 7 = 0.8451,
find the logarithms of the following:1
(a) 6 (g} 35 (m) 9
(6) 15 (h) 40 (n) 27
(c) 10 (t) 4 (o) 18
(d) 20 (j) 8 (p) 24
(e) 45 (k) 16 (q) 30
(/) 21 (i) 32 (r) 12.
1 The logarithms used in this book are taken from a four-place logarithmic table
and are correct to the nearest fourth decimal place.
§35] LOGARITHMS 63
35. Theorem : The logarithm of the quotient of two numbers is
the logarithm of the dividend diminished by the logarithm of the
divisor.
Proof: Let logo u = x and Iog0 v = y.
Then
a* = u and a» = v.
By division
_ u
v
or
logo (-) = x - y,
or
logo y-j = 10g0 U — logo V.
Thus the theroem: The logarithm of a quotient is equal to the
logarithm of the dividend diminished by the logarithm of the divisor.
Exercises
Using the logarithms given in preceding Exercise, find the loga-
rithms of the following: (a) f ; (6) f- ; (c) f ; (d) I; (e) J; (/) |.
36. Theorem : The logarithm of the power of a number is the
logarithm of the number multiplied by the index of the power.
Proof: Let logo u = x.
Then
u— a1,
Then
or
logo un = nx,
or
Iog0 un = n logo u.
Thus the theorem : The logarithm of the power of a number is
equal to the logarithm of the number multiplied by the index of
the power.
The theorem is true for fractional exponents as well as for integral
exponents, and from it then follows the law that the logarithm of the
64 MATHEMATICS [§37
root of a number is equal to the logarithm of the number divided by the
index of the root.
ILLUSTRATION :
If log 2_= 0.3010, log 4 = 2 log 2 = 2 X 0.3010 = 0.6020,
and log V2 = J log 2 = } X 0.3010 = 0.1505.
Exercises
1. Using the logarithms of the number given for the exercises)
§ 34, find the logarithms of the following: (a) 32; (6) 33; (c) 52; (d)27;
(e) 125; (f)V5; (g) 3?; (h) 7*; (i) 1251; (j) 625; (k)
2. If the log 2 is 0.3010, what is the (a) log 20? (6) log 200? (c)
log 2000? (d) log 20,000? (e) log 200,000? How do these loga-
rithms differ? What part of these logarithms is the same? Why?
3. If log 3 = 0.4771, what is the (a) log 30? (6) log 300? (c)
log 3000? (d) log 30,000? What part of these logarithms is the same?
Why?
4. If log 373 = 2.5717, what is the log 37.3? log 3.73? log 3730?
log 37,300? What part of these logarithms is the same? Why?
If a number is greater than unity, how does moving the decimal
point affect its logarithm?
37. Characteristic and Mantissa. If 10 is used as base, the
following equations are true:
log 10,000 = 4
log 1,000 = 3
log 100 = 2
log 10 = 1
log 1 = 0
log 0.1 = log 1 - log 10 = - 1 = 9 - 10
log 0.01 = log 1 - log 100 = - 2 = 8 - 10
log 0.001 = log 1 - log 1,000 = - 3 = 7 - 10
log 0.0001 = log 1 - log 10,000 = - 4 = 6 - 10
From the above table, which can be extended indefinitely in
both directions, we see that any number greater than 1000 and
less than 10,000 has a logarithm lying between 3 and 4, i.e., the
logarithm is 3 plus some (decimal) fraction. All numbers between
§37] LOGARITHMS 65
1000 and 10,000 have four digits preceding the decimal point, and
all numbers, excepting 1000, having four digits preceding the
decimal point lie between 1000 and 10,000. Therefore, the loga-
rithm of any number, excepting 1000, having four digits preceding
the decimal point is 3 plus a (decimal )fraction.
In a similar way it may be shown that the logarithm of any
number which is not an integral power of 10, consists of an integral
part plus a positive decimal fraction. For fractional numbers
the integral part of the logarithm is usually written in the form
of a binomial, as 9 — 10, 8 — 10, 7 — 10, etc.
The integral part of a logarithm is called the characteristic.
The positive decimal part is called the mantissa. The mantissa
of a positive integral power of 10 or of the reciprocal of a posi-
tive integral power of 10 is zero.
If the log 373 = 2.5717:
log 3.73 = log 373 - log 10 = 2.5717 - 1 = 1.5717
log 3.73 = log 37.3 - log 10 = 1.5717 - 1 = 0.5717
log 0.373 = log 3.73 - log 10 = 0.5717 - 1 = 9.5717 - 10
log 0.0373 = log 0.373 - log 10 = 9.5717 - 10 - 1 = 8.5717 - 10
log 0.00373 = log 0.0373- log 10 = 8.5717 - 10 - 1 = 7.5717 - 10
The logarithms of numbers between 1 and 0 are negative. It
will be noticed from the above illustration of the method of
writing negative logarithms that the mantissa is always the same
for the same sequence of digits, and is entirely independent of the
position of the decimal point.
The characteristic of the logarithm of a number depends only
upon the position of the decimal point in the number, and is entirely
independent of the sequence of the significant digits. A rule for
determining the characteristic of the logarithm of a number is:
the characteristic of the logarithm of a number, using 10 as a base,
equals the number of places the first significant figure is removed
from the units' place, and is positive if the first significant figure is
to the left of the units' place, and is negative if it is to the right of
units' place.
Thus the characteristic of the logarithm of 37.3 is + 1; of
3.73 is 0; of 0.373 is - 1, or 9 - 10; of 0.0373 is - 2, or 8 - 10,
etc.
5
66 MATHEMATICS [§37
Exercises
1. Give the characteristic of the logarithm of each the following
numbers :
(a) 375; (e) 87.36; (ft) 0.6127; (k) 17.31;
(6) 172.6; (/) 8.276; (t) 0.08217; (I) 0.00082;
(c) 37.62; (g) 0.561; (j) 0.00756; (m) 0.000002.
(d) 186.2;
2. Locate the decimal point in the numbers for which the follow-
ing are logarithms:
(a) 3.7283; (d) 1.8725; (g) 3.1786; (j) 7.6862 - 10;
(6) 3.6872; (e) 0.8271; (ft) 8.1268 - 10; (/c) 9.3621 - 10.
(c) 2.7826; (/) 9.1267- 10; (i) 0.3675;
3. Do you see any reason for using 10 as a base rather than any other
number?1
1 When 10 is used as base the system of logarithms is called the common system.
The system of logarithms employed in theoretical investigation has for its base the
sum of the infinite series
2 I 1 I 1 I 1 1 .
T 2 T 2-3 ^ 2-3-4
This number is represented by the letter e, and to seven decimal places is 2.7182818.
When the base e is used the logarithm is called the natural, the hyperbolic, or
the Naperian logarithm. The number e enters into a great many formulas rep-
resenting laws of nature. To illustrate: if the number of times faster y is changing
than x is proportional to y, it can be shown that the law connecting y and x is
y = cekx where candk are constants. This law is sometimes called the compound
interest law. The rate of growth of" bacteria is proportional to the number of
bacteria. The formula giving the number at any time is then the compound
interest formula.
It is an easy matter to find the logarithm of a number to the base e, if we know its
logarithm to the base 10. For, let logt N = x, where N is the number whose natural
logarithm is desired. Then
ex = N
i_
•,,x
e = N .
Taking the common logarithm, we obtain:
log e = ~ log N,
solving for x
log AT
=
~
log e '
whence
i w log ^
loge N = -. - .
log e
The log c = 0.434294. Then Ioge N = 2.302585 log N. This formula converts
the common logarithm of the number N into its natural logarithm. To find the
common logarithm of N if its natural logarithm is given we use the formula,
log AT = 0.434294 loge N.
§38]
LOGARITHMS
67
4. If the log 2 = 0.3010, log 3 = 0.4771, log 7 = 0.8451, and
log 11 = 1.0414, find the logarithms of the following:
. 121 . 70 . 84V24
(a) -v^-- (o) T-j-p '"*
' 15 144
007
I o\ I I I f\
21 1.25
(c)
16 / w' 0.064
5. Fill in the blanks of the following table:
55
/500\ i
I 6/ '
Number
Logarithm
Base
Number
Logarithm
Base
100
2
32
V2
125
5
81
5 0
81
4
16
2.0
5
2
17
5.0
27
3
V3
V2
5
10
3V3
1.5
625
5
38. Logarithmic Tables. A table of common logarithms con-
sists of the mantissas of the logarithms of numbers put in tabular
form. Since the characteristic of the common logarithm of a
numbed is dependent only upon the position of the decimal point,
it is not published in the tables, but is supplied by the computer;
and since the mantissa is independent of the position of the decimal
point, no decimal points are placed in the numbers in the table.
Thus, if one were to find the logarithm of 27.6 from a table of
common logarithms, he would enter the table with 276, take
out the corresponding mantissa, which is a decimal fraction, and
prefix the characteristic 1.
The tables commonly used are known as four-place, five-
place, and six-place; meaning by this that the mantissas are given,
respectively, to four, five, and six decimal places. In a four-
place table the number interval is usually 1 in the third place;
in a five-place table, 1 in the fourth place; and in a six-place
68 MATHEMATICS [§39
table, 1 in the fifth place. If one wishes to compute accurately
to four, five, or six places, he should use, respectively, four-, five-,
or six-place tables. Do not use a five- or six-place table if a
four-place table will suffice to give accurately the required
number of places.
The use of logarithms and logarithmic tables is primarily to
shorten the work of numerical computation, and to relieve the
computer of as much mental work as possible.
In the explanations of the use of logarithmic tables, it is as-
sumed that the student is using Slichter's four-place tables.1
All numerical work, unless otherwise stated, is to be done accu-
rately to four figures.
39. To Find the Logarithm of a Number from the Tables.
On the first page of the tables will be found the logarithms (i.e.,
the mantissas) of numbers from 1 to 1000. There are nine sets
of vertical columns. The first column of each set is headed No.
and contains numbers whose logarithms are given directly op-
posite in the second column, headed log. In the third column,
headed d, are found the differences (called tabular differences)
of the consecutive mantissas given in the second column. From
the table one finds the mantissa for the sequence of digits 376 to
be 0.5752. Hence
log 3760
= 3.5752
log 376
= 2.5752
log 37.6
= 1.5752
log 3.76
= 0.5752
log 0.376
= 9.5752 -
10
log 0.0376
= 8.5752 -
10
log 0.00376
= 7.5752 -
10
Exercises
Find the logarithm of each of the following numbers :
1. (a) 365. (6) 36.5. (c) 3.65. (d) 3650.
2. (a) 725. (6) 72.6. (c) 7.26. (d) 7260.
*" Slichter's logarithmic and trigonometric tables" are printed both in pamphlet
form and upon four leaves of cardboard bound together along the longer edge. For
desk use the cardboard form should be used.
If a student does any considerable amount of numerical work he cannot afford
to use poorly arranged and slow tables, such as are frequently found in text-books on
trigonometry and algebra.
§39] LOGARITHMS 69
3. (a) 816. (6) 81.6. (c) 8.16. (d) 8160.
4. (a) 261.
(b) 26.1.
(c) 2:61.
(d) 2610.
6. (a) 113.
(b) 1.13.
(c) 0.113.
(d) 0.0113.
6. (a) 216.
(6) 0.216.
(c) 0.00216.
(d) 216,000.
7. (a) 101.
(6) 1010.
(c) 0.0101.
(d) 0.000101
8. (a) 2.
(6) 0.2.
(c) 0.02.
(d) 0.00002.
9. (a) 9.99. (6) 99.9. (c) 0.999. (d) 9,990,000.
10. (a) 0.0136. (6) 0.000136. (c) 136,000. (d) 136,000,000.
In writing a decimal fraction without an integral part before
the decimal point, always place a zero before the decimal point;
as, 0.1276; 0.0281; 0.1201; 0.0026. In writing and reading the
mantissas of logarithms do not omit the last one or more digits,
even though they be zero. Thus, write:
log 399 as 2.6010 not 2.601
log 414 as 2.6170 not 2.617
log 455 as 2.6580 not 2.658
In reading the mantissas of the logarithms of numbers, read two
digits at a time, not one at a time. Thus, read 2.7162, "two,
(pause) seventy-one, sixty-two;" not, "two, point, seven, one,
six, two."
If the number consists of more than three digits the method of
finding the mantissa of its logarithm from the table is illustrated
by the following examples. Find the log 376.8. From the
table
log 376 = 2.5752
log 377 = 2.5763
The difference between these logarithms is 1 1 (disregarding the
decimal point). This difference is given in the table in the column
headed d. We see that a difference of 1 in the number pro-
duces a difference of 11 in the logarithm. A difference of
0.8 in the number will then produce (approximately) a difference
of 11 X 0.8, or 8.8 (use 9) in the logarithm. Then, since the
log 376 = 2.5752, the log 376.8 = 2.5752 + 0.0009 = 2.5761.
The 9 added to the logarithm of 376 is called the correction for
8.
Find log 521.7.
log 521 = 2.7168
70 MATHEMATICS [§40
Correction: 9 X 0.7 6
log 521.7 = 2.7174
Find log 8618.
log 8610 = 3.9350
Correction for 8 =4
log 8618 = 3.9354
Exercises
Find the logarithms of the following numbers :
1. (a) 172.6.
2. (a) 276.5.
3. (a) 365.8.
4. (a) 812.6.
6. (a) 8.887.
(6) 17.26.
(6) 27.65.
(6) 3.658.
(6) 0.8126.
(b) 8887.
(c) 1.726.
(c) 2.765.
(c) 0.3658.
(c) 0.08126
(c) 0.008887.
(d) 17,260.
(d) 27,650.
(d) 3,658,000.
(d) 812,600.
(d) 0.8887.
If there are five digits in the number the correction may be
found by multiplying the tabular difference by the last two digits of
the number. But, with the exception of some of the numbers oc-
curring near the beginning of the table where the tabular differ-
ences are large, no error will be introduced in the logarithm if the
number is fist reduced to four significant figures. Thus, to four
decimal places, the log 86326 is the same as log 86330.
40. Tables of Proportional Parts. In finding corrections the
tabular difference is multiplied by the digit in the fourth place
of the number, or by the last two digits of a five-place number.
This multiplication is performed by means of the tables of pro-
portional parts, the tables headed p. p. Thus, to find the correction
for the log 37,267, turn to the p. p. table headed 16, 16 being
the tabular difference. Opposite 6 we find 9.6, the correction for
6 in the fourth place; opposite 7 we find 11.2, the correction for
7 in the fourth place. But 7 is in the fifth place, therefore the
correction for it is 1.12. Add mentally 9.6 and 1.12, and obtain
for the complete correction 10.72 (use 11).
In taking out the mantissa from the tables, the addition of
corrections should be performed mentally. A good method to follow
is illustrated by example. To find log 13.78. First write down
the characteristic, 1. Then, with the table at your left, find
137 in the No. column and mark the corresponding mantissa
by placing the thumb nail above, or the finger nail below it.
§41]
LOGARITHMS
71
Do not read this mantissa, but read the tabular difference, 32.
From the p. p. tables find the correction, 26, for 8. Now, go back
to the mantissa marked by the finger nail, and read it increased
by 26, i.e., 1393. Then place 1393 after the characteristic 1
previously written down.
41. Arrangement of Work. All work should be arranged in a
vertical column and done with pen and ink. Make the digits
about 1/8 inch tall and space them horizontally about eight
to the inch. Study the formula in which you are substituting
and decide upon some arrangement of your work in the vertical
column which will make the additions, subtractions, etc., of
logarithms as systematic and easy as possible. Fill out the vertical
column as far as possible before turning to the table of logarithms.
This is called blocking out the work. The method of block-
ing out the work is illustrated by the following example: Find
the area, in acres, of a triangular piece of land, the sides a, b
and c being, respectively, 127.6, 183.7, and 201.3 rods. The
formula giving the number of acres, A, is1
A =
Vs(s — a)(s — b)(s — c)
160
1 The proof of this formula, as generally given, presupposes a knowledge of
trigonometry. A proof involving only algebra and geometry is given here.
C
FIG. 23.
Let ABC, Fig. 23, be the triangle whose sides are a, b, and c. Let CD be drawn
perpendicular to AB. Then
CD* - a* - (c - AD)* (1)
AD = \/b* - CD* (2)
Substituting in (1),
CD* = a* - c* + 2c\/b* - CD* - b* + CD*. (3)
and
72 MATHEMATICS [§41
where s is one-half the sum of the three sides. We may block out
the work as follows:
s = 256.3
a = 127.6
6 = 183.7
c = 201.3
2s = 512.6
s - a = 128.7
s - b = 72.6
s-c= 55.0
logs = 2.4087
log (s -a) = 2.1096
log (s -6) = 1.8609
log (s - c) = 1.7404
sum = 8.1196
l/2sum= 4.0598
log 160= 2.2041
log A = 1.8557
A = 71.73
In performing the work the expressions on the left of the
equality signs were all written down first. Then the numerical
<j2 _ C2 _ 62 = _ 2c&2 - CD*. (4)
Squaring (4),
(a* - C2 _ 62)2 = 4C2(62 _ CD*),
or
4c2&2 - (a2 - c* - b*)* = 4c2 CD*.
Factoring,
(2cb + a2 - C2 - 62) (2cfe - O2 + 62 + C2) = 4C2 (7^)2,
or
[a* - (c - 6)2][(c + 6)2 - 02] = 4c2 CD2.
Factoring,
(a + 6 - c)(o - 6 + c)(c + 6 + a)(c + 6 - a) = 4C2 CD*,
or
(2s - 2c)(2s - 26) (2s) (2s - 2o) - 4C2 CD*,
or
4«(* - o)(s - 6)(s - c) - c* CD*,
or
cCD
- o)(s - 6)(s - c)
= area of triangle.
§41] LOGARITHMS 73
values of a, b, and c were filled in and 2s, s, s — a, s — b, and
s — c computed. Then the logarithms of s, s — a, s — b, s — c,
and 160 were taken from the tables. Then the logarithms of s,
s — a, s — b, s — c were added, giving what is called the "sum."
Then one-half of this was taken giving what is called " 1 /2 sum,"
and from it the logarithm of 160 was subtracted giving log A.
From the tables of anti-logarithms1 the number 71.73 was taken
out corresponding to the mantissa 8557, the characteristic 1
fixing the decimal point after the second digit.
Exercises
Compute the areas, in acres, of the following triangles, a, b, and
c are expressed in rods. Record the time spent upon each exercise.
l.o= 93.6 2. a = 216.7
b = 101.7 b = 172.5
c = 127.3. c = 216.3.
3. a = 103.7 4. a = 101.73
b = 156.4 b = 127.27
c = 217.7. c = 131.38.
[5.] Plot a logarithmic curve. Plot numbers along the X-axis and
logarithms along the F-axis. Let x range from 0.1 to 10. Let 2
cm. (or 1 inch) on the X-axis and 10 cm. (or 5 inches) on the
F-axis represent one unit. Is this curve concave on the upward or
on the downward side? If the correction of a logarithm falls on a 0.5
shall we call it another one or drop the 0.5? Why? What would
be a similar rule in taking out the number from its logarithm?
Why?
[6.] With the aid of the T-square, triangle, compasses, and scale,
draw to scale a triangle with the sides given in exercise 1. Let 1/2
inch represent 10 rods. To do this use the scale marked 20 on the
triangular boxwood scale. Drop a perpendicular from any vertex
upon the opposite side. Measure the length of this perpendicular.
Multiply its length (expressed as rods) by the length of the base
(expressed as rods) and divide by 320. How could you get a more
accurate determination of the area?
[7.] Same as exercise 6, but use data given in exercise 2.
[8.] Same as exercise 6, but use data given in exercise 3.
1 A table of anti-logarithms is given on the last page of the Slichter table. From
it the number corresponding to a given mantissa is obtained. If the use of this
table is not apparent, the student should read the directions given on the first
1 cover.
74 MATHEMATICS [§42
[9.] Same as exercise 6, but use data given in exercise 4.
[10.] Upon a sheet of paper of good quality make a drawing accord-
ing to the following directions. Exercise great care in making this
drawing.
Draw a horizontal line 10 inches long. Divide it into ten equal
parts by drawing vertical lines 1/4 inch long on the lower side only.
Subdivide each of these ten intervals into ten equal parts by drawing
vertical lines 1/8 inch long on the lower side of the horizontal line.
We now have a unit of length (10 inches) divided into 100 equal parts.
In other words, the distance between any two consecutive vertical
lines represents one one-hundredth of one unit. Mark the longer
vertical lines beginning at the left-hand end 0.0, 0.1, 0.2, 0.3, . . .,
0.9, 1.0. Upon the upper side of the horizontal line and at each end
draw a vertical line 1/4 inch long. Mark the one at the left 1, and
the one at the right 10. Since the logarithm of 1 is 0, and of 10 is
1, it will be seen that the lines on the upper side of the horizontal
line representing the numbers 1 and 10 stand opposite their logarithms.
The logarithm of 2 is 0.3010. Find the location of this number
(estimating the position for the third place) on the lower scale, and
on the upper side draw a vertical line 1/4 inch long. Mark this line
2. In a similar way draw lines to be marked 3, 4, 5, . . . , 8, and 9,
standing opposite the points of the lower scale representing their
logarithms. In a similar manner draw vertical lines 1/S inch long
representing every tenth of a unit between 1 and 2, 2 and 3, and so
on. We now have a double scale. The lower is uniform, and the
upper, called a logarithmic scale, is non-uniform. Make a second
drawing by placing the uniform scale above and the non-uniform
scale, the logarithmic scale, below.
11. Find the values of the following, using logarithms :
(a) (27.83)3(1.621)5. (6) V(72.18)3 Vl.382.
(c) (0.6382)2 -=- 1.782. (d) Vl.278 -h 0.0183.
(e) (0.003782)^(1.7821)-. (/) 0.0007826 H- 0.007936.
(g) [3.782 X 26.78 -^ 37.92]l (h) [3.728 X 17.83 -f- (2.675)2]*.
(i) Vl.186 (0.02763)2 -r- (0.001726)3.
12. Find the amount of $10.00 in ten years at compound interest
at 6 percent; (a) interest compounded annually; (6) interest com-
pounded semi-annually.
42. Further Suggestions for Arrangement of Logarithmic
Work. If in a single formula several substitutions are to be made,
similar operations should be performed for each substitution before
§42]
LOGARITHMS
75
another is begun. This will be illustrated by the following
exercises :
1. Table XII indicates, in part, the amount of milk delivered
at the University of Wisconsin Creamery from Feb. 16, 1911,
to March 15, 1911, inclusive.
TABLE XII
Pat.
No.
Milk,
Ib.
Ave.
test
Price,
cents
Pat.
No.
Milk,
Ib.
Ave.
test
Price,
cents
4
4075
3.6
27
38
8228
3.9
27
5
749
3.5
27
39
2434
3.85
27
2
2395
3.8
27
63
6822
4.00
27
3
4046
3.6
27
80
1784
3.45
26
6
1503
3.6
26
65
1628
3.7
27
7
2020
4.15
27
102
1038
3.7
27
8
9012
3.6
26
105
705
4.15
27
9
5750
3.4
26
111
2646
3.9
27
10
1491
3.8
26
86
97
4.8
27
83
738
4.05
27
44
1603
4.6
27
36
2060
3.6
27
81
2291
3.65
27
37
2258
3.8
27
82
111
4.1
27
From these data the amount due each patron is to be computed
by multiplying each hundredweight of milk delivered, by the aver-
age test and then by the price. For the numerical work use a sheet
of computation paper, form M7. In the first column, the column
marked 0, on the first, second, third, fourth, fifth, and sixth lines
write, respectively, "pat. No.," "log of Ib.," "log of test," "log of
price," "log of amount," and "amount." Then on the first line of
the remaining columns put in from left to right the patron's num-
ber, as 4, 5, 2, 3, etc. Place one number in each vertical column.
If there are not enough vertical columns fasten two or more sheets
together. On the next line fill in the logarithms of the number of
hundredweights of milk delivered by the patrons. On the next
line fill in the logarithms of the numbers representing the average
test of the patron's milk. On the line below this fill in the loga-
rithms of the price paid each patron per pound of butter fat. On
the next line fill in the sums of the three logarithms above. Then
take out and place on the next line the anti-logarithms of the
76
MATHEMATICS
[§42
sums, and the numerical work is complete. By this arrangement
all numerical work pertaining to each patron is placed in a ver-
tical column headed by his number. The order of the work
should be as outlined above, so that all the operations of any
one kind are completed before those of another kind are begun.
As an illustration, the work for patron No. 65 will appear as
follows:
65
1.2117
0.5682
9.4314 - 10
1.2113
16.27
2. A solid of revolution1 is 10 feet long and has flat ends normal
to its axis of revolution. Its circumferences at various distances
from its larger end were measured and the results are given in
Table XIII. The numbers in the first column represent the dis-
TABLE XIII
Distance, feet
Circumference, feet
Distance, feet
Circumference, feet
0
10.00
5
6.28
1
9.00
7
5.47
2
8.15
8
5.00
3
7.40
9
4.68
4
6.80
10
4.40
tances, in feet, from the larger end to the points at which the
measurements were taken. The measurements are given in the
second column. Compute the area of the corresponding eleven
cross sections, arranging the numerical work upon a sheet of
paper, form M7, as follows: On the first horizontal line in the
first column of the sheet write "distance; " on the second line write
"log cir.;" on the next line write "2 log cir.;" on the next line
write "log area;" and on the line below this write "area." In the
columns headed I, II, III, etc., fill in from left to right, on the first
horizontal line, the distances, as 0, 1, 2, . . . , 9, and 10. On the
1 A solid of revolution is the volume bounded by a surface which may be generated
by rotating a curve 360° about a straight line.
§42] LOGARITHMS 77
line below this place the logarithms of the circumferences. Multi-
ply each of these logarithms by 2 and place the product in the
next horizontal line. Write the logarithm of -j- on the lower
edge of a slip of paper. (The logarithm of j- is to be computed
TiTT
separately, the work placed in systematic form on the lower
portion of the calculation sheet.) Place the slip of paper so that
the logarithm of -r- stands directly over the value of 2 log cir. in
1
column I. Add logarithm of j- and the value of 2 log cir.,
placing the sum on the fifth line. Slide the slip of paper to the
second column and add as before. Slide the paper to the third
column and add, then to the next, and so on for the eleven columns.
Take out the anti-logarithms corresponding to the logarithms on
the fifth line, and the numerical work is complete. Plot a curve
connecting area and distance. Plot distance along the X-axis,
using 2 cm. (or 1 inch) to represent 1 foot. Plot square
feet along the F-axis, using 2 cm. (or 1 inch) to represent
1 square foot. Find the area under this curve by counting the
squares. This area will be proportional to the volume of the
solid. Why? What is the volume expressed in cubic feet?
3. BCB'D, Fig. 47, represents the intersection of a right circular
cylinder with a plane perpendicular to the elements of the cylinder;
and BAB' A' the intersection with a plane inclined 30° to the first
plane. The two planes intersect in the line BE' cutting the axis
of the cylinder at the point 0. Our problem is to construct the
curve BPAB'. Take any point R upon the line B'OB. Let
OR = x. Draw RQ and RP, respectively in the normal and in-
clined planes, perpendicular to BOB'. P and Q, the points where
the two perpendiculars pierce the cylinder, are upon the same
element of the cylinder. PRQ is a 60° right triangle with the
angle R equal to 30°. Represent RP by y', and RQ by y. Then,
from the property of the 60° triangle,
V 2
y \/3
or
y' = 1.155y.
78 MATHEMATICS [§43
But, since BQCB' is a circle, y = \/az — xz, where a is the radius
of the cylinder. Then
y' = 1.155 Va2 -
or y' = 1.155A/(a — x)(a + x).
By giving to x, (OR), a series of values ranging from 0 to the
length of the radius, a, the corresponding perpendicular distances
RP, (y'}, may be computed and the curve BPAB' plotted.
Let a equal 2 feet. Using the above formula, compute y' for
the values x = 0, x = 0.1, x = 0.2, . . . , x — 1.8, x = 1.9,
and x = 2. Plot a curve between x and y', i.e., the curve BPAB'.1
4. A barn is provided with two 3^-foot diameter cupola venti-
lators connected with the basement. The intake ventilators in the
concrete basement wall are 7 inches in diameter, made by using
7-inch stovepipe and 7-inch elbows as forms in the construction of
the wall. Neglecting friction, and assuming no air leaks in around
the doors and windows, develop a formula giving n, the required
number of intakes, so that the velocity of the air in the intakes
shall be the same as the velocity in the flues of the cupolas.
Compute n without using logarithms.
43. To find the Anti -logarithm from the Table of Logarithms.
The use of the anti-logarithmic table is for instructional purposes
only. Since the method of taking out, from an anti-logarithmic
table, the number corresponding to a given logarithm is the same
as that for finding the logarithm if the number is given, the anti-
logarithmic table enables the student to become familiar with the
one method before he is to learn how to use a logarithmic table
backward, so to speak, in finding anti-logarithms.
An anti-logarithmic table has its advantages, but these are
more than offset by advantages gained by using the logarithmic
table backward. Five- and six-place tables are rarely, if ever,
provided with anti-logarithmic tables.
1 The curve thus plotted if drawn full sized would be a pattern (not allowing for
the lap in seam) for the saddle of a 4-foot circular ventilator, if the rafters of the
barn make an angle of 60° with the vertical.
Later an easier method will be given for drawing this curve as well as a method
for constructing a pattern for cutting the sheet iron for the flue of the ventilator, and
for the conical roof.
§43] LOGARITHMS 79
The method of taking out the number from the logarithmic
table which corresponds to a given logarithm will be illustrated by
example.
Find x if log x = 2.1786. By taking the mantissa, 1786, to
the table of logarithms, we see that the number x is between 150
and 151 (disregarding decimal point). Thus we can write down
at once the first three digits, 150, of the number and whatever
correction is to be made are digits suffixed. The given mantissa
is 25 greater than the mantissa for 150, and the tabular differ-
ence is 29. Hence the required number must be 25 /29 of one unit
greater than 150. Since 25 /29 = 0.86, the number corresponding
to the mantissa 1786 is 15086, or x = 150.86. The decimal point
is placed after the third digit to correspond to the characteristic 2.
The division of 25 by 29 may be effected by using the propor-
tional parts table. In the proportional part table headed 29
the nearest we can get to 25 on the smaller side is 23.2, which cor-
responds to 8. This means that in dividing 25 by 29 the first
digit in the quotient is 0.8, with a remainder of 25 — 23.2 = 1.8.
Again in the p.p. table we have 17.4 opposite 6, which means
that the nearest second digit of the quotient of 25 by 29 is 6.
Thus, at once from the proportional parts table we read that 25
divided by 29 is 0.86,
From the above illustration the student is not to infer that
with four-place logarithms one can compute accurately to five
places. This is, in general, not true. In general, only four sig-
nificant figures of the number should be found. If the number
lies near the beginning of the table where the tabular differ-
ences are large, the fifth digit may have significance and should
then be calculated. In this connection the student should read
section 72.
Exercises
Find, from the table of logarithms, the numbers corresponding to
the following logarithms. In finding corrections make use of the
p.p. tables.
1. 2.7864. 2. 3.1286. 3. 1.8152.
4. 9.6278 - 10. 6. 8.1278 - 10. 6. 6.1785 - 10.
CHAPTER III
THE CIRCULAR FUNCTIONS: THE TRIANGLE
44. Angular Magnitude. If a straight line, as AB, Fig. 24,
rotates about a fixed point 0 until it occupies some other (or the
same) position, as A'B', an angular magnitude is generated,
namely, the amount of rotation. All angles are or may be imag-
ined to have been generated in this way. The initial position of
the line is called the initial side, and the terminal position the
terminal side of the angle.
The point of intersection of
the terminal and the initial
sides is called the vertex of
the angle. This concept of
angular magnitude is more
general than that of elemen-
— B tary geometry in that the
angular magnitude here de-
fined is not only unlimited in
YIG. 24. size, but has a sense; i.e., the
angular magnitude represents
either counter-clockwise or clockwise rotation. If the rotation
is counter-clockwise the angle or the rotation is called positive;
if clockwise, negative.
If we fix our attention upon a particular spoke of a rotating fly
wheel and note its position at the beginning and at the end of a
definite interval of time, we say that this spoke has rotated through
an angle, the magnitude and sense of which is the amount and
direction of the rotation noted. If we had fixed our attention upon
a point of the wheel and had imagined it connected to the center of
rotation by means of a straight line, this imaginary line would have
rotated through the same angle as the spoke. We say that the
point has rotated through an angle. Since every point of the
80
§44] THE CIRCULAR .FUNCTIONS 81
wheel rotates through the same angle in any given interval of
time, we say the wheel has rotated through an angle.
The following is another illustration of a body rotating through
an angle. Suppose the telescope of a transit is first pointed at one
object and then at a second; we say that the vertical planes
through the two lines of sight form an angle with each other. This
same thought is expressed by saying that the line of sight of the
transit has been rotated through an angle. The magnitude and
sense of this angle is the amount and direction of rotation.
The practical unit for measuring angular magnitude is the one
three hundred and sixtieth part of one complete revolution. This
unit is called the degree. The degree is divided into sixty equal
parts called minutes. The minute is divided into sixty equal parts
called seconds.
Since the magnitude and sense of an angle is unchanged when it
is placed in different positions of its plane, we may move any given
angle until its vertex coincides with the origin of the system of
rectangular coordinates, and until its initial side coincides with
the positive part of the X-axis.
If the angle is positive and less than 90° its terminal side will
fall within the first quadrant, and the angle is said to be in the
first quadrant. If the angle is positive and greater than 90°, but
less than 180°, its terminal side will fall within the second quadrant
and the angle is said to be in the second quadrant. If the angle is
positive and greater than 180°, but less than 270°, its terminal side
will fall within the third quadrant and the angle is said to be in the
third quadrant. Similarly, a positive angle greater than 270° but
less than 360° is in the fourth quadrant; if greater than 360° but
less than 450°, in the first quadrant; and so on. If the angle is
negative and numerically less than 90°, it is hi the fourth quadrant ;
if numerically greater than 90°, but less than 180°, in the third quad-
rant; and so on. In general, an angle is said to be in that quad-
rant within which its terminal side would fall, if its vertex were
placed at the origin, and if its initial side were placed on the posi-
tive part of the axis of x.
Thus, in Fig. 25, a, 6, and c, represent angles in the first quad-
rant; d, e, and /, angles in the second quadrant; g, h, and i,
angles in the fourth quadrant; and j, k, and I, angles in the third
82
MATHEMATICS
[§44
quadrant. In each case the arrow indicates the sense and mag-
nitude of the rotation, a, b, d, g, h, i, and k are positive angles.
FIG. 25.
Exercises
1. In what quadrants are the following angles?
(a) 16°; (d) 217°; (g-) 87°; (j) - 76°; (m) - 276°;
(6) 291°; (e) 796°; (h) 102°; (ft) - 376°; (n) - 5°.
(c) 368°; (/) 560°; (i) 212°; (1) - 175°;
2. With the aid of the 77-square and the 60° and the 45° triangles,
THE CIRCULAR FUNCTIONS 83
draw lines through the origin, making positive angles of 15°, 30°
45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, and 180° with the positive
direction of the axis of x.
3. In a 60° triangle what is the ratio (a) of the length of the
hypotenuse to the length of the shorter leg? (6) of the length of the
hypotenuse to the length of the longer leg? (c) of the length of the
shorter leg to the length of the longer leg? (d) of the length of the
longer leg to the length of the hypotenuse?
4. In a 45° triangle, what is the ratio of the length of the hypo-
tenuse to the length of a leg?
5. Draw an equilateral triangle 6 inches on a side. From each
vertex draw a line to the middle of the opposite side. Mark the
numerical size of all the angles formed by these lines.
6. Draw a right triangle, one acute angle of which is 60°, and the
hypotenuse of which is 6 inches in length. What is the length of
the shorter leg of this triangle? What is the length of the longer leg?
What is the ratio of the length of the hypotenuse to the length of the
shorter leg? to the length of the longer leg? What is the ratio of
the length of the longer leg to the length of the shorter leg?
8. Same as exercise 7 but draw the hypotenuse 10 inches long.
9. What answers of exercises 7 and 8 are the same? Will these
answers remain the same for any length of hypotenuse? Why?
10. Draw a right triangle with hypotenuse equal to 6 inches in
length, and an acute angle equal to 45°. What is the length of the
legs? What is the ratio of the length of the hypotenuse to the length
of a leg? What is the ratio of the length of one leg to the length of
the other leg?
11. Same as exercise 10 but make the length of the hypotenuse
10 inches.
12. What answers in exercises 10 and 11 are the same? What
answers will remain the same for any length of the hypotenuse?
Why?
13. Draw a right triangle, one acute angle of which is 15°, and the
hypotenuse 6 inches in length. What is the length of the shorter
leg? of the longer leg? What is the ratio of the length of the hypo-
tenuse to the length of the shorter leg? of the length of the hypo-
tenuse to the length of the longer leg? of the length of the longer
leg to the length of the shorter leg?
14. Same as exercise 13, but make the hypotenuse 10 inches long.
16. What answers are the same in exercises 13 and 14? What
answers will remain the same for any length of the hypotenuse?
Why?
84 MATHEMATICS [§45
45. Trigonometric Functions Denned. Construct any angle in
the first quadrant. (This angle should be general, and not one
previously constructed in the preceding exercises.) Choose any
point, P, upon the terminal side, and from it drop a perpendicular
upon the initial side; i.e., upon the X-axis. We now have a right
triangle with the vertex of an acute angle (the original angle con-
structed) at the origin. Th's triangle is called the triangle of
reference for the constructed angle. The three sides, a, b, and c,
of this triangle may give rise to six ratios, viz. :
a b a b c c
-' ~> -' -» -' and -•
c c b a b a
From a property of similar triangles it is seen that these ratios are
entirely independent of the size of the triangle; i.e., independent of
the position of the point P upon the terminal side, provided, how-
ever, that the angle at the origin does not change. If, however,
the angle is made to increase or decrease by any amount, no
matter how small, each of the six ratios will change in value.1
Since these ratios depend for their values upon the size of the angle
and are entirely independent of the size of the triangle of reference,
they are functions2 of the angle, and are called the trigonometric,
or circular, functions.
In the triangle of reference the side opposite the given angle
(the angle whose vertex is at the origin) is called the opposite
(opp.) side. The side adjacent the angle is called the adjacent
(adj.) side.
The six trigonometric ratios are defined as follows, a is the
angle in question:
opp.
The ratio . — is called sine a; abbreviated by sin a.
hyp.
1 The increase or decrease in the size of the angle should not be so large that the
angle will not remain in the first quadrant. Later on it will be seen that the above
statement may not be true if the terminal side of the angle were to move to another
quadrant.
2 One quantity, y, is said to be a function of another quantity x, if, when x is known,
y is determined. Thus, the area of a circle is a function of its radius; for if the radius
is known, the area is determined. In this case, since we know the formula connecting
the radius and the area of a circle, we can actually compute the area. If an angle
in the first quadrant is known, the six ratios among the lengths of the three sides
of the triangle of reference are determined.
§45] THE CIRCULAR FUNCTIONS 85
The ratio , ' is called cosine a; abbreviated by cos a.
hyp.
opp.
The ratio .. is called tangent a; abbreviated by tan a.
adj.
The ratio , or the reciprocal of tan a, is called cotangent
opp.
a; abbreviated by cot a.
hi/D
The ratio ~TT, or the reciprocal of cos a, is called secant a;
duj .
abbreviated by sec a.
hwo .
The ratio ', or the reciprocal of sin a. is called cosecant a;
opp.
abbreviated by esc a.
It is to be remembered that in the order given above, the first
and last, the second and fifth, and the third and fourth ratios are
reciprocals.
The side opposite and the side adjacent are, respectively, the
^-coordinate and the z-coordinate of P, the point of the terminal
side from which the perpendicular was let fall upon the X-axis.
Since the point P is in the first quadrant, so that its x- and
^-coordinates are both positive, the side opposite and the side
adjacent of the triangle of reference are considered positive. The
length of the hypotenuse is the distance of the point P from the
origin, and is always considered positive.
If the angle is in the second quadrant, a triangle of reference is
formed as above by dropping a perpendicular from any point P
of the terminal side upon the X-axis (the initial side produced
through the vertex) . The six ratios of the lengths of the sides of
this triangle are called the trigonometric functions of the angle
and are named the same as if the angle were in the first quadrant.
In this case, however, the side adjacent, the z-coordinate of the
point P, is negative; hence the cosine, the tangent, the cotangent,
and the secant are negative.
In a similar manner, if the angle is in the third quadrant, we see
that both the adjacent and opposite sides of the triangle of refer-
ence are negative, and the sine, the cosine, the secant, and the
86 MATHEMATICS [§45
cosecant are negative. If the angle is in the fourth quadrant, the
sine, the tangent, the cotangent, and the cosecant are negative.
Exercises
1. In what quadrant or quadrants is the angle if:
(a) The sine negative? (6) The tangent positive?
(c) The secant negative? (d) The cosine positive?
(e) The cosecant positive?
(/) The sine negative and the tangent positive?
(g) The secant positive? (h) The sine positive?
(i) The tangent positive and the cosine negative?
(j) The cosine negative and the sine positive?
(fc) .The cosine negative?
(I) The sine negative and the cosecant negative?
(m) The cosine negative and the cosecant negative?
(ri) The cosine positive and the tangent negative?
(0) The secant negative and the cotangent positive? .
(p) The cotangent positive and the tangent positive?
(q) The sine negative and the tangent positive?
(r) The cotangent positive and the cosecant negative?
(s) The secant positive and the cosecant negative?
(t) The cosecant positive and the tangent positive.
(u) The cosine negative and the tangent positive?
(v) The cotangent negative and the secant positive?
(w) The sine negative and the secant positive?
(x) The sine positive and the cosecant negative?
(y) The tangent negative and the cotangent positive?
(z) The cosine negative and the secant positive?
2. Give the values of the six trigonometric functions of the
following angles. Tabulate your results. Do not reduce radicals
to approximate decimal fractions, (a) 30°; (6) 45°; (c) 60°; (d) 120°;
(e) 135°; (/) 150°; (g) 210°; (h) 225°; (t) 240°; (j) 300°; (k) 315°;
(1) 330°.
3. At 117 feet from the base of a flagstaff standing on level
ground, the angle of elevation1 of the top of the staff is 45°; what is
1 If a point A is at a higher elevation than a point B, the angle of elevation of the
point A at B, is the angle between the horizontal and a line drawn from B to A,
If a point A is at a lower elevation than a point B, the angle of depression of the
point A at B, is the angle between the horizontal and a line drawn from B to A.
§46] THE CIRCULAR FUNCTIONS 87
the height of the staff? What is the height if the angle of eleva-
tion is 60°? If 30°?
4. In exercise 15, page 11 how high above the plates is the
point of meeting of the upper edges of the upper and the lower
rafters? How far from the outside of the building will a plumb-line
dropped from this point strike the floor? How high above the
plates is the ridge of the barn?
6. The diagonal of a rectangular plot of ground is 82 rods. One
side is 41 rods. What is the length of the other side?
6. The sides of a rectangular plot of ground are 100 and 173.2
rods. What is the approximate length of the diagonal?
7. The sine a = f ; the cosine is positive. Find the values
of the remaining five functions.
HINT : Since the sine and cosine are both positive the angle is in
the first quadrant, and all functions are positive. A triangle whose
hypotenuse is five units in length and whose opposite side is three
units in length (hence the adjacent side four units in length) is a
triangle of reference for a. Hence we have, cos a = -5-, tan a = f ,
cot a = %, sec a = -f , and esc a = f .
8. Find the values of the trigonometric functions, construct to
scale a triangle of reference, and measure the angle with a protractor,
for each of the following:
(a) sin a = f , and the cosine is negative.
(6) cos a = -f , and the tangent is negative.
(c) tan a = %, and the sine is negative.
(d) sin a = — ^ and the tangent is negative.
(e) sec a. = 2, and the tangent is negative.
(/) sin a = |, and the cosine is negative.
(00 cos «=•§•, and the sine is negative.
(h) sec a = -f, and the sine is negative.
9. At two points, A and B, 100 feet apart and in line with the base
of a tower standing on level ground, the angles of elevation of the top
are 45° and 30°, respectively. Find the height of the tower.
10. In exercise 9, if the angles of elevation are 60° and 30°, what is
the height of the tower, and how far are the points A and B from its
base?
46. Functions of 0°, 90°, 180°, and 270°. The functions of 0°,
90°, 180° and 270° are not included in the above definition of
trigonometric functions, since for these angles there are no tri-
angles of reference. We shall define the functions of 0°, 90°, 180°,
88 MATHEMATICS [§46
and 270°, as the limits toward which the functions of any angle,
say a, approach as the angle a approaches 0°, 90°, 180°, or 270°.
Thus, if the angle a is in the first quadrant, and if the hypotenuse
of the triangle of reference is kept constant as the angle becomes
smaller and smaller, the opposite side will become smaller and
smaller and approach zero. The sine of this variable angle will
then approach zero as the angle approaches zero, or the sin 0° = 0.
The length of the adjacent side approaches the length of the
hypotenuse as the angle approaches zero, or the limit of the
cosine is 1, or cos 0° = 1. As the angle approaches zero, the
cotangent and the cosecant increase without limit.1
By considering a a small negative angle (hence in the fourth
quadrant) the cotangent and cosecant are both negative and
increase numerically without limit as a approaches zero. These
facts together with what was said above regarding cotangent and
cosecant are expressed by writing cot 0° — + °° and esc 0° = + oo .
Thus we have :
sin 0° = 0, tan 0° = 0, sec 0° = 1,
cos 0° = 1, cot 0° = ± oo esc 0° = ±
By allowing the angle to approach 90°, the student will show
that:
sin 90° = 1, tan 90° = ± oo sec 90° = ±00,
cos 90° = 0, cot 90° = 0, esc 90° = 1.
1 If 6 is a variable which approaches zero as its limit, and if a is a constant or a
a
variable whic approaches a limit c, the fraction - increases without limit. Thus,
6
if a = 1, and if b approaches zero by taking successively the values 1, 1, J, J, . . .,
the fraction has successively the values 1, 2, 3, 4, . . . By giving to 6 a value
sufficiently small, the value of the fraction, vt may be made larger, and for subse-
quent values of 6 remains larger, than any assigned number. We then say that
increases without limit as 6 approaches zero.
The expression, "r increases without limit," or "increases indefinitely," as 6
approaches zero is usually written
r = oo
The student is not to infer that °° (infinity) is a number or the limit of a number-
The expression, r= oo,is a symbol representing, "r increases without limit," or "r
increases indefinitely."
CO
§46] THE CIRCULAR FUNCTIONS 89
The student will further show that:
sin 180° =0, tan 180° = 0, sec 180° = -1,
cos 180° = -1, cot 180° = ± co, esc 180° = ± <»,
and that:
sin 270° = -1, tan 270° = + oo , sec 270° = ± °°
cos 270° = 0, cot 270° - 0, esc 270° = - 1
Exercises
1. Tabulate the functions of 0°, 90°, 180° and 270°.
2. Find the values of the functions of 360°; of 450°; of 540°; of
630°; of 720°; of 810°.
3. Is it possible for an angle to have a sine equal to 1/2? to 3/4?
to - 2?
4. Is it possible for an angle to have a cosine equal to 1/2?
to 3/4? to - 2?
5. Is it possible for an angle to have a tangent equal to 1/2?
to 3/4? to - 2?
6. Is it possible for an angle to have a cotangent equal to 1/2?
to 3/4? to - 2?
7. Is it possible for an angle to have a secant equal to 1/2? to 3/4?
to - 2?
8. Is it possible for an angle to have a cosecant equal to 1/2?
to 3/4? to - 2?
9. As the angle varies from 0° to 360° in a positive sense, how do
the values of the sine, the cosine, the tangent, the cotangent, the
secant and the cosecant change?
10. In the equation of the straight line, y = ax + b, a is what
trigonometric function of what angle?
11. Compare the sin 30° with sin ( — 30°) ; sin 45° with sin ( — 45°) ;
sin 60° with sin (- 60°); sin 135° with sin (- 135°); sin 300° with
sin (— 300°); sin a with sin (—a).
12. Compare cos 30° with cos (- 30°) ; cos 60° with cos (-60°);
cos 210° with cos (- 210°); cos a with cos (- a).
13. Compare tan 45° with tan ( - 45°) ; tan 150° with tan ( - 150°) ;
tan 330° with tan (- 330°); tan a with tan (-a).
14. Compare cos 30° with sin 60°; sin 30° with cos 60°; tan 30°
with cot 60°; sec 30° with esc 60°; sin 45° with cos 45°; tan 45° with
cot 45°; cos a with cos (90° — a); sin a with cos (90° — a); tan a
with cot (90° — a); sec a with esc (90° — a); esc a with
sec (90° — a), where a is an angle in the first quadrant.
90 MATHEMATICS [§47
16. Compare sin 135° with sin 45°; sin 120° with sin 60°; cos 120°
with cos 60°; tan 120° with tan 60°; sin 150° with sin 30°; cos 150°
with cos 30°; tan 150° with tan 30°; sin 240° with sin 60°; cos 240°
with cos 60°; tan 240° with tan 60°; sin 300° with sin 60°; cos 300°
with cos 60°; tan 300° with tan 60°; sin (180° - a) with sin «;
cos (180° - a) with cos a; tan (180° - a) with tan «; sin (180° + a)
with sin a; cos (180° + a) with cos a; tan (180° + a) with tan a;
sin (360° — a) with sin a; cos (360° — a) with cos a; tan (360° —a)
with tan a.
16. Express the following as functions of some positive angle less
than 90°: sin 112°; cos 127°; tan 171°; sin 92°; cos 170°; tan 131°;
cot 175°.
HINT: Construct an angle of 112°. Its terminal side falls in the
second quadrant. Draw a triangle of reference. Construct an angle
of 22°. Draw a triangle of reference for 22°. These two triangles of
reference are similar, and their homologous sides are in proportion.
Whence: sin 112° = + cos 22°; cos 112° = - sin 22° (because cosine
is positive in the first quadrant, and negative in the second), and
tan 112° = - cot 22°.
17. Show that
sin a
tan a —
COS a
and that
cos a
cot a = -s •
47. Even and Odd Functions. Let us consider the function
x2 (i.e., y = x2). If we give to x any two values numerically equal,
one positive and the other negative, the two corresponding values
of x2 (i.e., oiy) are equal. The curve for x2 (i.e., y = x2), plotted
upon squared paper is symmetrical with respect to the F-axis.
The same can be said of the functions x4, x6, x6, x2 + 3a;4, and
x2 + 5x4 — 6z6 — 7. These functions are said to be even. An
even function is one whose value remains unchanged when the
sign of the argument is changed.
Let us next consider the function x3 (i.e., y = x3). If we give to
x two values numerically equal but with opposite signs, the two
corresponding values of x3 (i.e., of y) are numerically equal but of
opposite signs. The curve for x3 (i.e., y = x3), plotted upon
squared paper is symmetrical with respect to the origin. The
§48] THE CIRCULAR FUNCTIONS 91
same can be said of the functions x5, re7, x, x + 3x3 — 7x&,
and x3 — 2z5 + 6z7. These functions are said to be odd. An
odd function is one whose value 'remains numerically the same but
changes sign when the sign of the argument is changed.
In previous exercises the fact was developed that cos a and sec a
are even functions, and that sin a, esc a, tan a and cot a are
odd functions. These facts are shown graphically in § 49.
48. Circular Measure. The circular unit of angular magnitude
is defined as an angle whose sides intercept an arc equal in length
to the radius of a circle when the vertex of the angle is placed at the
center of the circle. This unit is called the radian.1 This sys-
tem of angular measure is fundamental in mechanics, physics, and
pure mathematics. It must be thoroughly mastered by the
student.
Since the circumference of a circle is 2irr, where r is the radius of
the circle, there are 2ir radians in one complete revolution; or
360° = 2-ir radians
180° = TT radians
90° = K radians, etc.
6
To find the number of radians in an angle expressed in degrees,
divide the number of degrees by 180 (this gives the number of
straight angles in the given angle) and multiply the quotient by TT,
the number of radians in a straight angle. Thus, to change 176° to
1 7 ft
radians, we have T^TT = 3.07: or 176° = 3.07 radians.
loU
To find the number of degrees in an angle expressed in radians,
divide the number of radians by TT (this gives the number of
straight angles in the given angle) and multiply this quotient by
180 (the number of degrees in a straight angle). Thus, to change
4.6 radians to degrees we have — '- = 263. 6, or 4. 6 radians
7T
equals 263.6°.
1 If 0 is the angle measured in radians at the center of a circle of radius r, the
length of the arc intercepted by the angle is rO, and the area of the circular sector
is *r20.
92
MATHEMATICS
\
\
\0
y
7
Exercises
1 . Express in radians the
following: (a) 135°; (6) 45°; (c)
161°; (d) 275°.
2. Express in degrees the
following angles measured in
radians: (a) 7r/2; (6) ir/4; (c)
Tr/6; (d) 37T/2; (e) 1; (/) 2; fo)
3.6; (ft) 11.6.
[49.] Graphs of the Tri-
gonometric Functions. In
Fig. 26 let OPiPzPa be a
circle with unit radius. Let
OCPi be any angle at the
center of the circle. Then
PiAi represents the sine of
the angle 0CP,. Take OB
equal in length to the circum-
ference of the circle, i.e., equal
to 2-n-. Divide OB into thirty-
six equal parts, and erect per-
pendiculars at each point of
division. Each of these equal
distances of the X-axis, OB,
represents 10°, orT/18 radian.
Upon each vertical line plot a
point whose ordinate repre-
sents the sine of the corre-
sponding angle. To do this
divide the circumference of
the circle into thirty-six equal
parts, and join each to C, the
center of the circle. (Only a
few of these points, as PI, Pz,
PZ, PI, P&, are lettered in the
drawing.) Project these
points to the right upon the
vertical line corresponding to
the angle at the center of the
§49]
THE CIRCULAR FUNCTIONS
93
circle. Thus, the angle 50° determines the point P2. P2 pro-
jected upon the line passing through the point of the X-axis repre-
senting 50° gives P'2. ThenP'2A'2 = P2A2 = sin 50°.
The curve OKMNB (called the sine curve) is the graph of the
equation y = sin x, as x varies from 0 to 2?r. It is readily seen
that as x continues beyond this interval the curve is repeated for
every addition of 2ir to the X-axis.
The length of any portion of the X-axis represents an angle at the
center of the circle expressed in circular units.
The graph of y = 2 sin x may be obtained from the curve in Fig.
26 by making all ordinates twice as long; the curve y = 3 sin £ by
making all ordinates three times as long; the curve y = f sin x
by making all the ordinates one-half as long; the curve y •= % sinx
by making all ordinates one-third as long; etc.
FIG. 27.
If the sine curve in Fig. 27 were translated to the left a distance
ir/2, its equation would be y = sin (x + ir/2). Since the sin
(x + 7T/2) = cos x, the translated curve in Fig. 27 is the graph
for y = cos x.
To construct the graph y = tan x we proceed by a method
similar to that used in the construction of the sine curve.
Let C, Fig. 28, be the center of a circle of unit radius. Upon the
X-axis lay off the distance OD equal to 2ir, the circumference of the
circle. Divide OD into thirty-six equal parts, and at each point
of division erect a perpendicular. Let the F-axis, Y'OY, be
tangent to the circle. Beginning with the horizontal line draw
thirty-six radii at intervals of 10°, extending them to the tangent
94
MATHEMATICS
t§50
line Y'OY. Project the points of intersection with the tangent
line upon the vertical line whose distance from the origin repre-
sents the particular angle at the center of the circle. Thus, OCPi
is an angle of 30°. Project PI upon the third vertical line to the
right of the F-axis, and obtain P'i, a point upon the graph of
y = tan x. When the angle is in the second or third quadrants
the radius must be extended back through the center in order
to intersect with the tangent Y'OY. We see that as the angle
FIG. 28. — The tangent curve.
passes through 90° and 270° the tangent curve jumps from + °°
to — «» . The tangent curve extends to infinity in both y direc-
tions, and only a portion of it near the X-axis is represented in
the figure. Curves for y = cot x, y = sec x, and y = esc x may
be plotted from the tangent, cosine, and sine curves, respectively,
by taking reciprocals of the lengths of the ordinates.
[50.] The Line Representation of the Trigonometric Functions.
Let C, Fig. 29, be the center of a circle with unit radius. Let CO
be a horizontal line and CD a vertical line. Let OY and DQ be
[§50
THE CIRCULAR FUNCTIONS
95
tangents. Let OCP be an angle a, at the center of the circle.
Then, AP represents the sine of a; CA represents the cosine of a;
OP i represents the tangent of a; DP2 represents the cotangent of
a; CPi represents the secant of a; and CPz represents the cose-
cant of a.
AO, which is equal to 1 — cos a, is sometimes called the versed
sine of a (vers a).
FIG. 29.
Exercises
1. Draw a curve for y = sin x. Use 1.15 inches as the unit
of length. This gives" 2?r = 7.2 inches, approximately, or each 0.2
inch of the X-axis represents 10°.
2. Draw a curve for y = cos x.
3. Draw a curve for y = tan x.
4. Prove (sin a)2 + (cos a)2 = 1; or, as it is usually written,
sin2 a + cos2 a = 1, where a is any angle.
HINT: In a triangle of reference for any angle, excluding 0°, 90°,
180°, and 270°, the square of the adjacent side plus the square of the
opposite side equals the square of the hypotenuse. Note that this
is true even though we consider the sides negative. For the angles
96 MATHEMATICS [§51
excepted above, prove the truth of the formula by direct sub-
stitution.
6. Prove sec2 a = 1 + tan2 a.
6. Prove esc2 a. = 1 + cot2 «.
51. Trigonometric Tables. In preceding exercises we have
calculated the numerical values of the trigonometric functions of
a few angles; viz., 30°, 45°, 60°, etc. If the values of the functions
of other angles were desired they could be found by drawing the
angles with a protractor, constructing *a triangle of reference, and
from it scaling off the lengths of the sides and computing the ratios.
This method of finding the functions would not only be slow, but
the degree of accuracy would usually be unsatisfactory. A
method has been devised for calculating the trigonometric func-
tions to any number of decimal places, but it cannot be explained
in this place.
A table of natural trigonometric functions, for every ten minutes
from 0° to 90°, is given in Table XXII, pages 300-302. Since the
cosecant and secant are, respectively, reciprocals of the sine and
cosine, they need not be used in formulas, and their values are not
usually published in a trigonometric table. If the angle is less
than 45° the degrees and minutes of the angle are printed in the
left-hand column and the names of the functions are printed at the
top of the page. If the angle is greater than 45° the degrees and
minutes are printed in the right-hand column and the names of
the functions are printed at the bottom of the page. This
arrangement of the table is possible through the fact that the
cosine of an angle is the sine of its complement, and the cotan-
gent is the tangent of its complement.
52. Graphic Table of Trigonometric Functions. From a
sheet of polar coordinate paper, form MS, the sine and the cosine of
any angle, and the tangent of an angle less than 45°, may be scaled
off accurately to within two or three points in the third decimal
place.
Upon this sheet concentric circles, 2/100 of a unit apart, are
drawn. Every fifth circle is shown by a heavy line. Running
from their common center radiating lines for every degree are
drawn. Every tenth line is drawn heavy. Two additional circles
are drawn each passing through the origin, the center of the con-
§52]
THE CIRCULAR FUNCTIONS
97
centric circles, and each having a diameter equal to unity. The
circle with its center upon the F-axis is called the sine circle, the
other with its center upon the X-axis is called the cosine circle.
At the right is a vertical scale from which may be read the tan-
gent of any angle up to 45°. This scale is called the tangent scale,
and its use is apparent at once. In explaining the method of
scaling off the sines and cosines from the sine and cosine circles,
FIG. 30.
reference is made to Fig. 30. Let 0 be the center of the system
of concentric circles, and let CABD be the circle with radius equal
to unity, and center at 0. Let OEA be the cosine circle, and OFB
the sine circle. Let AOE be any angle, say a. Draw AE and
FB. OAE and OFB are right triangles. Then OE = OA cos a.
But OA = 1. Therefore OE = cos a. Upon form MS the
length OD may be read off at once by means of the system of
concentric circles.
Similarly, OF = OB sin OBF. But, since OB = 1 and the angle
OBF = angle a, we have OF = sin a.
7
98 MATHEMATICS [§53
63. The Graphic Solution of the Right Triangle. Let A, B,
and C be the vertices of any triangle; and let a, /3, and y be, re-
spectively, the interior angles of the triangle at these vertices. Let
the sides opposite the vertices A, B, and C be, respectively, a, b,
and c. In a right triangle the right angle is marked 7, and the
hypotenuse c.
The three sides together with the three angles are called the
elements of the triangle. If any three of the six elements, except-
ing the three angles, of a possible triangle are given, the triangle
may be constructed to scale, and the remaining three elements
measured. The triangle is then said to be solved graphically.
Exercises
Solve graphically the following right triangles; record the time spent
upon each problem.
l.o = 173, b = 216. 4. a = 175, a = 36°.
2. b = 136, c = 527. 5. c = 516, 0 = 73°.
3. a = 210, c = 728. 6. b = 172, a = 47°.
54. Solution of Right Triangles Analytically. Instead of solv-
ing a right triangle graphically we may solve it analytically.
This method will be illustrated by solving two right triangles.
ILLUSTRATION 1 : Given, a = 216 and a = 36°; to find b, c, and
/3. We see at once that /3 = 54°, the complement of 36°. To find c
a, a
use the relation c = — — , and to find b use the relation b = — —
sin a tan a
From the table of the natural trigonometric functions,
sin a = 0.5878, and tan a = 0.7265.
216 216
Substituting, c = --— = 367.5; and b = ~— - 297.3.
To check the result, make use of the relation
c2 - a2 = 62,
or \/(c — a)(c + a) = b.
c - a = 151.4.
c + a = 583.4.
V'(c^-~aXc +"a) = 297.3 = b
§54] THE CIRCULAR FUNCTIONS 99
This last value of 6 agrees with the value computed above.
This not only checks the value of b, but also the value of a, which
is used in computing the second value of 6.
ILLUSTRATION 2: Given c = 516.2 and a = 176.5; to find b, a,
andjS.
a 176.5
Solution: sin a = - = r< . _ = 0.3419
c 51o.2
From the table a = 19°.994
whence 0 = 70°.006
b = csin/3 = (516.2) (0.9397) = 485.1
Check: \/(c — a)(c + a) = b,
or \/235,310.19 = 6,
or 485.1 = 6,
which checks all work.
Exercises
Solve the following right triangles. Use natural values of trigono-
metric functions taken from Table XXII. Perform the necessary
multiplications and divisions by the arithmetical process, and hand
in all numerical work. Be sure to check your work. Keep a record of
time spent upon each problem.
[1.] a = 173, b = 216. [4.] a = 175, a = 36°.
[2.] b = 136, c = 527. [5.] c = 516, ft = 73°.
[3.] a = 210, c = 728. [6.] b = 172, a = 47°.
Instead of performing the multiplications and divisions by the
lengthy arithmetical method, logarithms may be used. In the
logarithmic solutions the natural values of the trigonometric func-
tions are not used, because their logarithms may be taken directly
from a logarithmic table. The two problems solved below illus-
trate the arrangement of the work. The student should go over
the solutions and checks of these two problems very carefully, in
order to become familiar with the use of a table of the logarithms
of the trigonometric functions. Especial study should be given
to the finding of the angles when the logarithms of the functions
are given.
100 MATHEMATICS [§54
ILLUSTRATION 1: Given a = 175.6 and 6 = 216.1; to find a
/?, and c.
a 6
Solution: tan a = ~ , /3 = 180 — a, c = — — -
6 sm /S
log a = 2.2443
log 6 = 2.3347
log tan a = 9.9096 - 10
a = 39° 4' .6
0 = 50° 55' .4
log sin ]8 = 9.8900 - 10
log c = 2 . 4447
c = 278.44
Check:
c - a = 102.84
c + a = 454.04
log (c - a) = 2.0122
log(c + a) = 2.6571
logfc2 = 4.6693
log 6 = 2.3347
The computed log 6 agrees with log b used above, which checks
all of the work.
ILLUSTRATION 2: Given, a = 176.3, a = 37° 17'.4; to find b, c,
and j9.
Solution: /3 (the complement of a) = 52° 42'. 6. b = a tan /3,
a
and c = — — -•
smp
log a = 2.2463
log tan /3 = 0.1183
logfe = 2.3646
6 = 231.53
log sin |8 = 9.7824 - 10
logc = 2.4639
c = 291 . 00
Check:
c- a = 114.7
c + a =467.3
log (c - a) = 2,0596
§54] THE CIRCULAR FUNCTIONS 101
log (c + o) = 2.6696
log&2 = 4.7292
log 6 = 2.3646
which agrees with log b above and thus checks the entire work.
This illustrative example was chosen so that the calculated value
of the logarithm of b would agree with the value found in the table.
This absolute agreement will, in general, not occur. For, in
finding the second logarithm of b, calculated numbers are used
which are correct only to four places, and the errors in these
numbers, especially in c-a will effect the calculated logarithm
of 6.
If the calculated logarithm differs from that taken from the
table by only one or two in the fourth decimal place, the student
may consider his work checked.
In checking the work it will be noticed that it is immaterial
whether the log a or log b is recomputed. If the hypotenuse and a
leg are given, the logarithm of the unknown leg should be recom-
puted. Otherwise the logarithm of the longer leg should be
recomputed by the check formula. The reason for this is that
it keeps the percent of error in the difference between the
hypotenuse and the leg as small as possible.
Exercises
Solve the following right triangles using logarithms. Keep a record
of the time spent upon each problem. Check the work.
l.o = 173, b = 216. 4. a = 175, a = 36°.
2. 6 = 136, c = 527. 5. c = 516, ft = 73°.
3. a = 210, c = 728. 6. b = 172, a = 47°.
7. Having measured a distance of 281.6 feet in a direct horizontal
line from the bottom of a tower, the angle of elevation of the top was
found to be 24° 16'.6. Find the height of the tower.
8. A person on top of a tower 75 feet high observes the angle of
depression of two objects on the horizontal plane, which are in line
with the base of the tower, to be 37° 27'.6 and 25° 17'.8. Find the
distance of each object from the base of the tower.
9. A tower stands by a river. A person on the opposite bank
finds the angle of elevation of its top to be 50°. He recedes 40 yards
102
MATHEMATICS
[§54
in a direct line from the tower, and finds the angle of elevation of the
top to be 40°. Find the breadth of the river and the height of the
tower.
10. A rope 100 feet long is fastened to the top of a building 40 feet
high. Find the angle the rope is inclined, if its lower end just touches
the ground.
11. From a balloon, which is directly over one town, is observed
the angle of depression of another town, 12° 16'. The towns are
8.5 miles apart. Find the
height of the balloon.
13. From a station A at
the base of a mountain, its
summit S is seen at an ele-
vation of 43° 17'. After
walking 5000 feet toward
the summit, up a plane
making an angle of 28° 13'
with the horizontal, to
another station, B, the angle
ABS was found to be 137°
17'. Find the height of
the mountain.
HINT: In solving any problem, first build up a formula giving
the unknown number. As far as possible put this formula in a form
suitable for logarithmic computation. This will be illustrated by
building up a formula for x, the height of the mountain. In Fig. 31,
SE = x is the required height. BA = 5000 feet; a = 28° 13' and
the angles /3, y, and 5 can be computed from the given data. BC is
perpendicular to AS.
BC = 5000 sin /3
BC 5000 sin ft
BS= .
sin 5 sin 8
DS = BS sin y
= 5000 sin 0 -7
sin 5
DE = BF = 5000 sin a
SUIT
x = DS + DE = 5000 sin /3 -
sin 5
5000
sin S
+ 5000 sin a
+ sin a
THE CIRCULAR FUNCTIONS
103
Compute /3, y, and 5, and substitute in the formula. Compute the
value of the first term within the parenthesis by using logarithms.
13. From a point in a window in the same horizontal plane with the
bottom of a steeple, the angle of elevation of the top of the steeple is
43° 16'. From another window, 18 feet below the first, the angle
of elevation of the top of the steeple is 52° 8'. Build up a formula
giving the height of the steeple. Find the height.
A c B A c B
FIG. 32.
55. The Law of Sines. Let ABC, Fig. 32, be any triangle. Let
p be a perpendicular let fall from C upon c. Then
p = a sin /3
p = b sin a.
a sin j8 = b sin a,
a _b_
sin a ~ sin £
In a similar way we may show that
or
a
Hence
sin a sin /3 sin 7
which is known as the law of sines.
Exercises
1. Suppose we are required to determine the distance from a
given point A to an inaccessible object B, Fig. 33. Let C be any other
104
MATHEMATICS
[§55
convenient accessible point. Set a transit over A and measure the
angle B AC. Set the transit over C and measure the angle BCA.
Measure the distance AC. Solve for AB, using the law of sines.
Data:
b = 500 feet
a = 101° 13' 6
7 = 42°28'.7
FIG. 33.
To find c = AB, use the formula
sin 7 sin /3
or
b sin 7
sin /3
The value of /3 may be found by subtracting the sum of a and 7
from 180°, for, from geometry, the sum of the three interior angles of
a triangle is equal to 180°.
2. Same as exercise 1, with the following data:
b = 375.6 feet
a = 27° 42'. 7
7 = 113° 17'.6.
HINT: Sin (113° 17'.6) = sin (66° 4.2' A.) = cos (23° 17'.6)
§56]
THE CIRCULAR FUNCTIONS
105
56. The Law of Cosines. Let ABC, Fig. 34, be any triangle.
Let p be a perpendicular let fall from B upon 6.
C2 = p2 + (6 - k)*
if 7 is acute (Fig. 34, a).
Since pz + #2 = a2, we have,
C2 = 02 + 52
and since k = a cos 7, we have
62 — 2a& cos 7.
FIG. 34.
If 7 is obtuse (Fig. 34, 6),
c2 = p2 + (6 + A;)2
c2 = p2 + 62 + 26fc + A;2
c2 = a2 + 62 + 2kb:
But, since k = a cos D(7# = — a cos 7, we obtain the same expres-
sion for c as above,
c2 = a2 + b2 - 2ab cos 7,
which is known as the law of cosines.
Exercises
1. Suppose we wish to determine the distance between two points
A and B on opposite sides of a building, Fig. 35. Choose any con-
venient point, C, and measure AC, CB, and the angle ACS.
Data :
a = 137.6 feet
b = 186.5 feet
7 = 76° 54'.6.
To find c use formula
C2 = a2 + b2 - 2abcos7.
106
MATHEMATICS
[§57
2. Same as exercise 1, with the following data:
a = 378.6 feet.
b = 276.5 feet.
7 = 112°27'.8.
57. The Law of Tangents. Let ABC, Fig. 36, be any triangle.
With C as center and with CA as radius describe an arc cutting BC
at E and BC produced at D. Draw AE and AD. BAD is a right
triangle with the angle AEC equal to
The angle EAB is
equal to
Draw EK perpendicular to AE.
K
FIG. 36.
The triangles BEK an4 BAD are similar. Hence
AD AE a +
- = tan —
AE EK 2
BD _ AD
BE ~ EK
Since BD = a + b, and BE = a — b,
a + b _ tan J (a +
a — b tan \ (a —
cot
(D
§57] THE CIRCULAR FUNCTIONS 107
which is known as the law of tangents. Since a + /3 + 7 = 180°,
2 2'
The law of tangents may be used instead of the law of cosines
for solving triangles in which two sides and the included angle are
given.
ILLUSTRATION : Given
a = 337. 6 feet
b = 213. 5 feet
7 = 27° 16'
to find c, a and |8.
Solution:
tan $(« - |8) = CI— ^- tan \(a + j8)
a + 6
a - 6 = 124.1
a + b = 551.1
J(a + j8) = 76° 22'
log (a - 6) = 2.0938
log tan |(a + 0) = 0.6152
log (o + 6) = 2.7413
log tan |(a - 0) = 9.9677 - 10
i(« _ 0) = 42° 52'
a = 119° 14'
/8 = 33° 30'
Find the value of c by using the law of sines,
b sin 7
c = —
sin |8
log 6 = 2.3294
log sin 7 = 9.6610 - 10
log sin /3 = 9.7419 - 10
log c = 2.2485
c = 177.2
108 MATHEMATICS [§58
Exercises
Solve exercises 1 and 2, § 66, by using the law of tangents.
58. The Addition Formulas. Let ABC, Fig. 37, be any tri-
angle. Let ACD be the exterior angle formed by producing the
side BC. Draw CK perpendicular to AB.
The law of sines gives,
AB sin a
smy = ~BC— (1>
Since a + ft = 180° — 7, sin (a + ft) = sin 7. Equation (1) may
then be written
. , _. AB sin a
sm (a + ft) = — —
or
sin (a + /3) = sin a cos /? + cos a sin ft. (2)
This formula gives the sine of the sum of two angles, provided the
sum is less than 180°. We shall show that this limitation may,
however, be removed. Let a + ft be greater than 180° but less
than 360°. We then have a + ft = 180° + € + ft, where e + ft
is less than 180°.
sin (a + ft) = sin [180° + ( e+ ft)] = -sin (e + ft)
By formula (2)
sin (e + ft) = sin e cos ft -f- cos e sin /3
§58] THE CIRCULAR FUNCTIONS 109
and since e = a — 180°,
sin (a + ft) = - [sin (a - 180°) cos/3 + cos (a - 180°) sin ft],
or
sin ( a + j8) = sin a cos /3 + cos a sin /3,
which shows that formula (2) is true if a. + ft is less than 360°.
Similarly it may be shown that formula (2) is true for any value
of the sum, a + ft-
To prove a similar formula for sin (8 — a), drop a perpen-
dicular, BM, Fig. 37, from B upon AC:
sin ft = sin (5 — a)
By the law of sines
s*n a
—
n/* =
BC
AM + MC .
_^_-Bin«
BMAM , MC .
or
sin (5 — a) = sin 7 cos a + cos 7 sin a (3)
Since sin 7 = sin 5 and cos 7 = — cos 5, equation (3) becomes
sin (d — a) = sin 5 cos a — cos 6 sin a (4)
Formula (4) gives the sine of the difference of two angles pro-
vided neither angle exceeds 180°. The student will show that this
limitation can be removed in a way similar to the way in which it
was removed from formula (2) above.
Formulas (2) and (4) are usually written together as,
sin (a ± ft) = sin a cos ft ± cos as in /3. (5)
Since cos (a ± ft) = sin [(90° - a) + ft],
formula (5) gives
cos (a ± ft) = sin (90° - a) cos ft + cos (90° - a) sin ft,
or
cos (a ± ft) = cos a cos ft + sin a sin /3. (6)
Formulas (5) and (6) are called the addition formulas for the
sine and cosine.
If ft = a, formula (6) gives
cos («+«) = cos a cos a — sin a sin a,
110 MATHEMATICS [§59
or
cos 2 a = cos2 a — sin2 a, (7)
or
cos 2a = 1 — 2 sin2 a, (8)
since cos2 a = 1 — sin2 a.
cos 2a = 2 cos2 a — 1 (9)
When /? = a, formula (2) becomes
sin (a + a) = sin a cos a + cos a sin a
sin 2a = 2 sin a cos a. (10)
Exercises
1. Find without using the tables, the exact value of sin 75°,
and of cos 75°.
HINT: 75° = 45° + 30°.
2. Find without using the tables, the exact value of sin 15°;
and of cos 15°.
HINT: 15° = 45° - 30°.
X I 1 — COS X
3. Show that sin— = -\ •
HINT: Let 2 a = x in formula (8).
4. Show that cos ~ = \ I —
Zi
HINT: Let 2 a = x in formula (9).
6. Show that
x 11 — cos x
tan - = \ —
2 \ 1 + cos x
1 — cos x
sin x
sin x
1 + cos x
HINT: Use results of exercises 4 and 5.
[59.] Area of Triangle by Drawing to Scale. In § 41, the
areas of triangles, with the three sides given, were found both
graphically and analytically.
§60] THE CIRCULAR FUNCTIONS 111
In the following exercises construct the triangle to scale. Drop
a perpendicular from any vertex upon the opposite side. Measure
the length of this perpendicular, and multiply this length by the
length of the side upon which it was dropped. One-half of this prod-
uct is the area of the triangle. In drawing the triangle use a
convenient scale, just small enough to admit drawing upon a sheet
of paper, 8J X 11 inches. Always mark in words upon the draw-
ing the scale used, as "scale 1 inch = 1 foot"; "scale 1/2 inch
= 10 rods"; "scale 1/4 inch = 20 rods," etc.
Exercises
Find the area of each of the following triangles. Record time spent
upon each problem. Place but one problem upon a sheet of paper.
l.o = 175.6 rods; b = 276.5 rods; /3 = 27° 35'.
2. a = 267.75 feet; ft = 37° 45'; 7 = 43° 17'.6.
3. b = 262.5 feet; a = 178.4 feet; /3 = 34° 47'.5.
4. b = 175.37 rods; c = 216.82 rods; a = 27° 47'.
6. b = 73.78 feet; c = 53.62 feet; a = 136° 43'.6.
6. b = 100.37 feet; a. = 27° 31'.6; ft = 73° 6'.7.
60. Area of Triangle by Formula. Case I. Two sides and the
included angle given.
Let ABC be any triangle with b, c and a given. From C draw
p perpendicular to the side AB. Then
area = %pc.
Since p = b sin a,
area = i&c sin a.
ILLUSTRATIONS: Find the area of the triangle having given
b = 172. 36 feet
c = 103. 27 feet
a = 27° 17'.6.
log b = 2.2364
logc = 2.0140
log sin a = 9.6614 - 10
log £ = 9.6990 - 10
log area = 3.6108
area = 4081 square feet
112 MATHEMATICS [§60
Exercises
Find the area of each of the following triangles:
l.o = 173.6 rods b = 263.8 rods y = 73° 47'.6.
2. a = 278.3 rods c = 172.7 rods 0 = 21° 21'.7.
3. 6 = 100.1 rods c = 216.5 rods a = 127° 26'.3.
Case II: One side and two angles given.
If two angles are given the third can be computed by sub-
tracting their sum from 180°. Let ABC be any triangle with
c, a and /3 given. From C drop a perpendicular, p, upon the
side c. Let M be the foot of this perpendicular.
Then
AM = p cot a
BM = p cot ft
and by adding
c = p(cot a + cot ft)
'cos a cos j8"j
. sin a sin 0 J
= P
= P
= P
sin ft cos a + cos ft sin
sin a sin ft
sin (a + 0)
in al
sin a sin /3
Then
_ c sin a sm p
P ~ sin (a + 0) '
Since area, A = %pc,
_ c2 sin a sin /
=
2sin(a:+p')
ILLUSTRATION: Find the area of the triangle having given
c = 263.75 feet, a = 33° 17'. Q, and ft = 73° 21'.6.
a + /3 = 106° 39'. 2
logc = 2.4212
logc2 = 4.8424
log sin a = 9.7395 - 10
log sin ft = 9.9814 - 10
log J = 9.6990 - 10
log sin (a + ft) = 9 . 9816 - 10
log area = 4 . 2807
area = 1909 square feet
§60] THE CIRCULAR FUNCTIONS 113
The log sin (a + /3), or the log cos 16° 39'.2, is subtracted
from the- sum of the four preceding logarithms. The subtrac-
tion is performed digit by digit, as the addition progresses
from right to left.
Exercises
Find the area of each of the following triangles. Record time
spent upon each exercise.
1. c = 216.3 feet; « = 24° 16'.7; ft = 61° 15'.6.
2. a = 372.6 feet; a = 37° 27'.7; /3 = 81° 21'.6.
3. b = 87.75 feet; ft = 16° 37'.6; 7 = 63° 21'.7.
4. c = 217.75 feet; a = 93° 17'.5; 0 = 21° 37'.6.
5. a = 126° 17'.6; 7 = 25° 57'.3; b = 200.37 feet.
[Case III:] Given two sides and an angle opposite one of them.
Let ABC be any triangle. Let a, c, and a be given. Drop a
perpendicular, p, from B upon AC. Let M be the foot of the
perpendicular.
MA = c cos a,
p = c sin a
MC = Va? - p2 = A/a2 - c2 sin2~a
AC = AM ± MC
The minus sign is used if 7 is obtuse. Substituting for AM and
MC, ^_____
AC = c cos a ± \/a2 — c2 sin2 a.
Then, since area = %pA C,
area = fc sin a [c cos a ± A/a2 — c2 sin2 a]
area = |c2 sin a [cos a ± \/(a/c — sin a)(a/c + sin a)]
If a /c < sin a, the quantity under the radical sign is negative,
which means that in this problem there is no real area. From
a figure it will be seen that this is the impossible case studied in
geometry.
If a/c = sin a we have a right triangle, and the formula reduces
to: area = ^c2 sin a cos a, which is as it should be.
If 7 is acute use the + sign before the radical; if obtuse use the
- sign.
If a > c, and if a is the interior angle of the triangle, there is but
one solution given by the formula with the + sign before the
radical.
114 MATHEMATICS [§61
ILLUSTKATION: Find the area of the'triangle if c = 216.7 feet,
a = 187.56 feet, and a = 37° 15'.6, and 7 obtuse,
log a = 2 . 2730
logc = 2.3359
loga/c = 9.9371 - 10
a/c = 0.8652
sin a = 0.6054
a/c — sin a = 0.2598
a /c + sin a = 1 . 4706
log(a/c - sin a) = 9.4147 - 10
log (a/c + sin a) =0. 1675
sum = 9.5822 - 10
£sum = 9.7911 - 10
\/(a/c)2 - sin2 a = 0.6181
cos a = 0.79581
cos a — \/(a/c)2 — sin2 a = 0. 1777
log 0.1777 = 9.2497- 10
log sin a = 9.7821 - 10
log c2 = 4.6718
log£ = 9.6990 - 10
log area = 3 . 4026
area = 2527 square feet.
Exercises
Find the areas in acres, of each of the following triangles. Record
time spent upon each problem.
1. c = 317.62 rods, a = 217.75 rods, a = 27° 41'.7, and y is obtuse.
2. c = 267.7 rods, a = 298.6 rods, a = 87° 8'. 6, and y is acute.
Case IV. Three sides given. This case has already been dis-
cussed in § 41.
61. Graphic Method of Constructing the Curve of Intersection
of a Right Circular Cylinder with a Plane. In exercise 3, § 42,
there was described a method of constructing the curve of inter-
section of a right circular cylinder and a plane making an angle of
60° with the axis of the cylinder. A graphic method will now be
1 In taking out the cosine or the logarithm of the cosine of an sngle, the correc-
tion, if any, must be subtracted, because, in the first quadrant, the cosine decreases
as the angle increases.
§61]
THE CIRCULAR FUNCTIONS
115
described for the construction of this curve, not only when the
angle is 60°, but for any given angle.
Let BQCB', Fig. 47, be the normal cross section of the cylinder,1
and let BPAB' be the section made by a plane making an angle
a with the normal cutting plane. From the explanation given in
exercise 3, § 42, it is readily seen that
RP =
RQ
cos a
= RQ sec a.
Draw the circle BQCB', Fig. 38, with radius OB equal to the
radius of the cylinder. Extend the diameter BOB' to some con-
venient point 0'. Consider O'B and O'Y' a set of rectangular
R B
FIG. 38.
coordinate axes. Through 0' draw the 45° line O'K and the line
O'M whose slope is equal to the value of sec a. Draw QN par-
allel to O'B intersecting O'K at N. Draw NS parallel to O'Y'
intersecting O'M at S. Draw SP parallel to O'B intersecting RQ,
produced, at P. P is a point on the curve to be drawn. For,
since O'E = EN = RQ, and since ES = RP = O'E sec a,
RP = RQ sec a.
In this manner any number of points may be located upon the
curve.
In the above construction the horizontal and vertical lines
need not actually be drawn; it is sufficient to locate only the points
N and S. If the drawing is done on squared paper the horizontal
1 A normal cross section is a section made by a cutting plane perpendicular to the
axis of the cylinder.
116
MATHEMATICS
[§62
and vertical lines of the paper will be of assistance; if the drawing
is done on plain paper the T-square and triangle should be used.
Exercises
Construct the curve called for in exercise 3, § 42, by the method
of this section. Construct the complete curve on a heavy sheet of
paper, cut out the inner portion and fold to form a model of the
saddle.
62. The Pattern for a Right Circular Cylindrical Surface Cut
by a Plane. In the preceding section a method was given for laying
out the saddle for a circular ventilator. A method will now be
FIG. 39.
described for laying out a pattern for cutting the sheet iron which
when rolled up into a cylindrical surface will give the flue fitting
upon this saddle. The length of this pattern will be equal to the
circumference of the cylinder.
Draw the line BQC, Fig. 39, equal in length to one-fourth the
circumference of the cylinder. This line corresponds to the arc
BQC, Fig. 47, i.e., if the sheet of paper upon which BC is drawn is
rolled up into a cylindrical surface such that lines perpendicular to
BC become the elements of the cylinder, BC becomes the arc
BQC of Fig. 47. From points of BC, such as Q, perpendiculars
must be laid off equal to QP, Fig. 47. From Fig. 47, it is appar-
ent that the distance
QP = RQ tan a
where a is the angle between the normal and inclined cutting
planes.
Extend CB to some convenient point 0. With 0 as center and
with a radius equal to the radius of the cylinder, describe the arc
§63] THE CIRCULAR FUNCTIONS 117
BQC. Through 0 draw the 45° line OK, and the line OM making
the angle MOC equal to a. By means of a protractor divide the
arc BQC into a number of equal parts (nine for convenience).
Divide the line BC into the same number of equal parts. From
each point of division of the arc, as Q, draw a horizontal line inter-
secting OK at N. Through N draw a vertical line intersecting
OM at S. Through S draw a horizontaliine intersecting at P the
perpendicular erected at Q of the line BC. P is a point
upon the desired curve. For, since QR = NE = OE, and since
PQ = SE = OE tan a,
PQ = RQ tan a.
Any number of points may be located in this way and a smooth
curve drawn through them. The pattern thus obtained may be
used four times, giving the complete pattern.
Exercises
1. Construct a pattern by the method explained above using the
dimensions in the exercise of the preceding section.
HINT: This drawing will require a sheet of paper larger than
8£ X 11 inches.
2. An elbow is formed by joining two right circular cylinders of
equal radii in such a way that their axes intersect at right angles. Lay
out a pattern for the surfaces.
63. Pattern for a Conical Roof. A pattern is to be drawn for a
conical roof, Fig. 40. It is apparent that the pattern may be
made by cutting out a cir-
cular sector from a circular
sheet of paper of radius r.
If R is given,
r = R sec a.
A >
The circumference of the
circular sheet of paper from FIG. 40.
which the circular sector is
cut, Fig. 41, is then 2irR sec a. The circumference of the lower
edge of the roof is 2irR. Hence the length of the arc
DE = 2irR sec a - 2-rrR
= 2irR(sec a - 1).
118 MATHEMATICS
Then the angle BOD expressed in radians is
_27rR(sec a - 1)
r
27rfi! (sec a — 1)
R sec a
= 2ir(l — cos a).
Exercises
1. Construct a pattern for a conical roof if a, Fig. 40, is 30°, and if
R is 20 percent greater than the radius used in exercise 1 of the
preceding section.
2. Build up suitable formulas for a pattern for the convex surface
of a tank in the form of a frustum of a right circular cone, if the radius
of the larger base is R, of the smaller base is r, and if the depth is H ,
Construct a pattern (reduced) if R = 5, r = 4, and H = 3.
Miscellaneous Exercises
1. The roof rises from the adjacent sides of a rectangular house at
an angle of 30°. Find the angle which the corner of the roof makes
with the horizontal. Find the length of this edge measured from the
plate to the ridge, if the building is 24 X 32 feet. What is the length
of the ridge of the roof?
2. The height of a butte is desired. At a point B, the angle of
§63] THE CIRCULAR FUNCTIONS 119
elevation of the base of a tree, A, standing on the top of the butte
was measured and found to be 63° 18'. 7. At a second point, C,
in the same vertical plane with AB, 'at the same elevation with, and
200 feet from, the point B, the angle of elevation of A was found to be
32° 7'. 8. Build up a formula giving x, the elevation of A above
B; and y, the horizontal distance of A from B.
3. Same as exercise 2, excepting' that instead of ABC being in the
same vertical plane the angle between the vertical planes through A
and B, and B and C is 163° 5'.6.
4. Same as exercise 2, excepting that the point C is 10 feet higher
than the point B.
5. Same as problem 3, excepting that instead of B and C being
on a level, the point C is 10 feet lower than the point B.
6. A 100-foot tape hangs down the face of a vertical cliff. The
zero mark is at A, and the 100 mark at B. At a point C in the
plane below the cliff, the angle of elevation of A is 43° 30'. 7, and
of B is 36° 20'. 7. Find the elevation of the point A above C.
7. A and B are two points on the face of a vertical cliff. At a
point C the vertical line AB subtends an angle of 7° 10'; i.e., the angle
BCA = 7° 10'. At a point D it subtends an angle of 6° 7'. If
A, B, C and D are in the same vertical plane and if C and D are
200 feet apart at the same elevation, find the elevation of A above C,
and the horizontal distance of A from C.
8. Same as exercise 7, but let D be 5 feet lower than C.
9. Same as exercise 7, but let the vertical plane through CD make
an angle of 137° 10'. 7 with the vertical plane through AC.
10. Same as exercise 9, but let D be 6 feet lower than C.
11. A, B, C, D, E, and F are the vertices of a six-sided field. Find
the area in acres if:
AB = 162.75 rods.
BC = 96.63 rods.
CD = 86.73 rods.
DE = 100.75 rods.
EF = 101.62 rods.
FA = 98.76 rods.
AC = 200.60 rods.
AD = 156.73 rods.
AE = 200.00 rods.
12. The line AB runs north and south. The line AC makes an
angle of 52° 8'. 6 with AB. Locate the line BC perpendicular to
AB 30 that the area ABC shall be 1 acre.
120 MATHEMATICS [§63
13. Same as exercise 12, excepting that the angle ABC is to be
29° 17'. 6 instead of 90°.
14. The area, in acres, of a four-sided plot of land with vertices
M, N, 0, and P, is desired. The side MP crosses a low swamp so
that it can not be chained. The sides MN, NO, and OP are
measured and found to be, respectively, 798.8 feet, 7103.7 feet and
2181.1 feet. The angles PMN and NOP are measured and found
to be, respectively, 76° 47'.6, and 126° 10'.2.
HINT: Draw the diagonal NP and find the area of each triangle.
15. Same as exercise 14, but instead of measuring the angle PMN the
angle MNP was measured, and found to be 63° 42'.6.
HINT: Compute the length of NP as a side of the triangle NOP.
16. Same as exercise 14, but angle ONM was measured instead of
angle, PMN and found to be 127° 0'.3.
HINT: Compute the size of the angle ONP as an angle of the
triangle NOP.
17. M, N, 0, and P are the vertices of a four-sided field.
NO = 3172.6 feet, Z MNP = 43° 21'.3, ZPNO = 51° 2'.1,
Z NOM = 55° 6'.5, and Z MOP = 47° 52'.6. Find the area of the
field in acres.
HINT: Calculate MN and angle MNO. Find the area of the
triangles NOP and MNP.
18. Find the volume, in cubic yards, of a right circular cylindrical
shell of uniform thickness. To make the problem specific, let us find
the number of yards of concrete in the vertical wall of a circular con-
crete silo, assuming, first, that there are no openings. Let R be the
outer radius, d the thickness of the wall and H the height, all
measured in feet. From the horizontal cross section of the silo, it
is seen that the area of the cross section of the wall is the difference of
the areas of the two circles, one with radius R and the other with ra-
dius R - d. This area A = irRz - ir(R - d)2 = TrlR2 - (R - d)2].
The quantity within the parentheses is the difference between two
squares, and may be resolved into the product of two factors, the one
the sum and the other the difference of the numbers. Then
A = ir(2R — d~)(d). This area multiplied by the height and divided
by 27 gives the volume, V, of the wall in cubic yards. Then
ir(2R -
27
a formula suitable for logarithmic computation.
ILLUSTRATION : Suppose the outside diameter of a silo is 13 feet,
§63] THE CIRCULAR FUNCTIONS 121
the thickness of the wall 7 inches (0.5833 feet, from Table XVII), and
heighl; 35 feet. Then
H = 35
d = 0.5833
2R - d = 12.4167
log H = 1.5441
log d = 9.7659 - 10
log (2R - d) = 1.0940
log TT = 0.4971
log 27 = 1.4314
log V = 1.4697
V = 29.49 cubic yards.
If only a single computation were to be made, the formula for
volume could be built up using the data directly. For the above
measurements :
7r(6.5)2 = area of larger circle
7r(6.5 — 0.5833)2 = area of inner circle
7r(6.52 - 5.91672) = 7r(6.5 + 5.9167) (6.5 - 5.9167)
= 7r(12.4167) (0.5833)
= area of cross section
Then
V = 35rr (12.4167) (0.5833) -h 27
= volume in cubic yards.
If several sets of measurements were given, and if volumes cor-
responding to all were to be computed, it would be desirable to derive
first the general literal formula and then substitute the given nu-
merical values. The example explained above illustrates how much
easier it is to effect the transformations when letters rather than
numerical values are used.
The student is reminded that in general it is advisable to represent
the known and unknown numbers by letters and build up, if possible,
a formula for each unknown in terms of known numbers before mak-
ing numerical substitutions.
19. Find the volume in cubic yards of a right circular cylindrical
shell, if the outer radius is 7.25 feet, the thickness 8 inches, and the
length 28.5 feet.
20. In exercise 18, the volume of the opening must be deducte
from the total volume of the wall. Let the sides of the opening be
two vertical planes passing, if extended, through the axis of the
122
MATHEMATICS
[§63
cylinder.1 Fig. 42 represents a cross section of the opening. Let S
be the width of the opening on the outside, and s the width on the
inside. These distances are to be measured along the arcs of circles
and not along the chords.
%RS = area of circular sector ABO.
%(R — d)s = area of circular sector CDO.
%[RS - (R - d)s] = area ACDB.
Before the last equation can be used we must find an expression for s,
for this is, as yet, unknown. From geometry,
s R - d
= ?
S R
or
_ S(R - d)
R ~~'
Substituting for s in the above formula for the area of ACDB we
have
K<»ACDB-i\SR _^I^~|
R J
= ~<2J2
ZK
1 Thii is not the true form of the opening in the silo; the exercise is only in-
iroduced here to illustrate how the formula is built up.
THE CIRCULAR FUNCTIONS
123
Multiplying this area by the height, h, of the opening, and dividing
by 27, the volume, v, in cubic yards is
54 R
ILLUSTRATION: Find v, if S = 38 inches (3.1667 feet) and h = 29.5
feet, for the cylinder whose volume was computed in exercise 18.
logd = 9.7659 - 10
log (2R - d) = 1.0940
logS = 0.5006
log h = 1.4698
log numerator = 2.8303
log R = 0.8129
log 54 = 1.7324
log denominator = 2 . 5453
log v = 0.2850
v = 1 . 928 cubic yards.
Find v for the set of dimensions given in exercise 19 if
S = 40 inches, and h = 21 feet.
21. Let Fig. 43 represent a vertical cross section of a conical con-
crete roof of a silo. Find the number of cubic yards of concrete in
the roof, assuming no open-
ing. Let k, r, and a be the
known dimensions as repre-
sented in the figure. The
volume sought is the differ-
ence between the volume
of AA'C'B'BDA and of
ACBDA. The first-named
volume, however, is the sum
of the volume of the right
circular cone A'C'B'D'A'
and the volume of the circular disc AA'D'B'BDA; and since
the volume of A'C'B'D'A' is equal to the volume of the right cir-
cular cone ACBDA, the required volume is equal to the volume of
the circular disc AA'D'B'BDA, or *r*(BB'). But, BB' = ^-^-
The final formula for the volume, u, then becomes
u =
FIG. 43.
COS a
Find u if r =6.25 feet, k = 3 inches, and a = 30°.
124
MATHEMATICS
22. Fig. 44 represents a water tank in the form of a frustum of a
right circular cone. Let R and r be, respectively, the radii of the
lower and upper bases. Let H be the altitude. Let x be the depth
of the water in the tank. AB is a vertical glass tube connected with
the tank as shown in the figure. The surface of the water in the
tube at C rises to a level with the surf ace of the water in the tank;
and as water is added to or taken from the tank, C will rise or fall.
FIG. 44.
If a uniform scale with its zero on a level with the bottom of the
tank be placed behind the tube, the reading upon it as indicated by C
will give the depth of the water within the tank, and will not give,
excepting indirectly, the volume of water within the tank.
The problem is to construct a non-uniform scale, such that, when
placed behind the tube, AB, it shall give the direct reading for the
volume of water in the tank. It is at once evident that the zero for
the scale must be on a level with the bottom of the tank, and that
the divisions corresponding to equal increments of volume of water
are farther and farther apart as the upper end of the scale is ap-
proached. This scale may be constructed by running known quan-
tities of water into the tank and marking the position of the point C;
or it may be constructed by computing, from a formula, values of x
THE CIRCULAR FUNCTIONS 125
for given values of the volume. Let V represent the volume of water
in the tank. Show that
ir7?3 Trjri rJ? ~\3
V =
_ «*f I"B _ I
~ 3 Ik ~X\
3fc
•where k = (R - r)/H.
Before solving for x, make the following abbreviations. Let
irR3 7T/C2 1 R
— = a, - — = — , and — = y. Then
or
x = y - /3(a - F),
where a, /3, and 7 are constants for a given tank. From this formula
we may compute x corresponding to any value or set of values for V.
These values of x are to be measured off from the zero upon the scale,
and the points thus located are to be marked, not with the x values,
but with the values given to V.
Compute x for every interval of 20 cubic feet in the value of V
ranging from 0 to 600 cubic feet, if R = 5 feet, r = 3 feet, and H = 10
feet. With the values of x computed construct a miniature scale on
which 1 inch represents 1 foot. Mark the divisions representing the
even hundreds of cubic feet much longer than the subdivisions rep-
resenting the even twenties. Number only the longer divisions, as
0, 100, 200, 300, . . .
In performing the numerical work, follow the outline given below.
Place the values of V, as 20, 40, 60, 80, 100, 120, 140, . . . , 600 upon
the first horizontal line of a sheet of paper form M7. Subtract these
values from the value of a, and place the differences on the second hori-
zontal line. The values of a, y, and log /3 should be computed at
some out-of-the-way place upon the lower part of the calculation
sheet. In subtracting V from a, write a at the lower edge of a slip
of paper and by sliding the slip along, the thirty subtractions may
be performed with the value of a written but once. Upon the third
horizontal line place the logarithms of a — V, to which add the
logarithm of /3 and place the sums upon the fourth line. Divide these
logarithms by 3, placing the quotients upon the fifth line. Take out
the numbers corresponding to the logarithms given on line 5, writing
them on line 6. Subtract these numbers from y, placing the differ-
ences on the seventh line, and the numerical work is complete.
In performing the numerical work, several sheets of computation
126 MATHEMATICS [§63
paper may be used, or by dropping down about ten lines, one sheet
may be crossed several times with values of V.
Before constructing the miniature scale, it is advisable first to plot
the values for V and x upon a sheet of squared paper. By so doing
any appreciable error in an x will be apparent by the corresponding
point not falling upon a smooth curve drawn through the plotted
points. The value of x for any point not falling upon the curve should
be recomputed.
The values for a, ft, and log 7 should be computed several times
to insure their correctness.
1000ft.
FIG. 45.
23. To find the distance from the flagstaff on University Hall to
the flagstaff on the Capitol building, a base line, AB, 1000 feet long,
Fig. 45, was laid off on Lakeside Ave., South Madison, and the
following angles measured:
CBA = 109° 53' 15".
DBA = 67° 35' 45".
DAB = 105° 1' 30".
CAB = 63° 2' 45".
§63]
THE CIRCULAR FUNCTIONS
127
From the triangle ADB compute the length of AD and of DB.
From the triangle ACB compute the length of AC and of CB.
Compute x from the triangle ACD, and also from the triangle DBC,
and take the mean of the two values.
24. To find the height of the ball on the flagstaff on University
Hall above the water table of North Hall, the following measurements
were taken with the transit placed at the points A and B, 100 feet
apart, Fig. 46.
Si Ball
100 ft.
FIG. 46.
The angle of elevation of S at A, Z SAS" = 27° 51' 30".
The angle of elevation of S at B, Z SBS' = 28° 35' 30".
The horizontal angle S"AB = 75° 31'.
The horizontal angle S'BA = 84° 21'.
At A the transit was 3.33 feet, and at B, 2.71 feet above the water
table of North Hall.
Using the law of sines, calculate the logarithm of the length BS',
and of AS". From the triangle BS'S calculate the length of SS',
and from the triangle AS"S the length of SS". To SS' add 2.71,
and to SS" add 3.33. Take the mean of the last two sums as the
height of S above the water table.
CHAPTER IV
[THE ELLIPSE]
64. The Ellipse Defined. We shall define the ellipse as the
curve of intersection of a plane with the surface of a right circular
cylinder. This definition includes the circle as a special case of
an ellipse.
FIG. 47.
In laying out the pattern for the saddle of a circular ventilator,
exercise 3, § 42, an ellipse was constructed by drawing a curve
through plotted points. In this exercise the equation of the curve
was derived for the case in which the cutting plane, the roof, made
an angle of 60° with the axis of the cylinder. The method of find-
ing the equation is, however, the same for all values of the angle.
128
§65] THE ELLIPSE 129
65. The Equation of the Ellipse. Let BCB'DB, Fig. 47, be a
normal cross section of a right circular cylinder. The curve
BCB'DB is then a circle. Let BAB'A'B be the intersection of
the cylinder with a plane inclined to the normal cutting plane at
an angle a. The curve BAB'A'B is then, by definition, an ellipse.
Let the normal and the inclined cutting planes intersect in the
diameter of the cylinder, BB'. Let the radius of the cylinder
be 6.
In the plane AB'A'BA let 0 be the origin, OB the positive direc-
tion of the X-axis, and OA the positive direction of the 7-axis,
of a rectangular coordinate system. If P be any point on the
curve AB'A'BA, OR is its z-coordinate and RP its ^/-coordinate.
Let Q be the point of intersection of the normal section with the
element passing through P.
Then OR2 + RQ2 = OQ2,
or
X2 + RQ2 = b2,
or, since
RQ = RP cos PRQ = y cos a,
x2 + y2 cos2 a = b2,
or
x2 y2 cos2 a _
b2 "b2 ~~ '
which is the equation of the ellipse.
When a is zero, this equation reduces to z2 + y2 = b2, the
equation of a circle, the normal cross section of the cylinder. As
a increases from 0° to 90°, cos a decreases from 1 to 0, and the
cos2 1
coefficient of y2, ,2 > decreases from ?2 to 0. When a = 90°,
the equation of the ellipse reduces to v^ = 1, or x2 = b2, or
z2 — 62 = 0, or (x — b) (x + b) = 0. The last equation is equiva-
lent to the two equations, z — 6 = 0, and z + 6 = 0, or x = b
and z = — b; for if the product of two expressions is equal to
9
130 MATHEMATICS [§66
zero, the equation is satisfied by equating either factor to zero.1
Each equation, x = b and x = — b, represents a straight line
parallel to the F-axis, 6 units from it, but upon opposite sides of
the origin. This is exactly what sliould be expected, for when
a = 90°, the cutting plane becomes parallel to the elements of the
cylinder and the curve of intersection reduces to two straight
lines.
66. General Shape of the Ellipse. Let OA, Fig. 47, be repre-
sented by a. Since OC = 6, and since / AOC = a, it follows
that - = cos a or - - = a. Substituting this value for
a cos a
in the equation of the ellipse., it reduces2 to
cos a
or, upon solving for y,
y = ±1 V&2 - x2
This equation shows that for every value assigned to x there cor-
respond two values for y, numerically equal but with opposite
1 It will be recalled that the method of replacing one equation by two equations
is that which may be used in solving quadratic equations. (See § 15).
Let x2 — x — 6 = 0 be the equation to be solved. By factoring the left-hand
side, this equation may be written (x + 2)(z — 3) = 0. To solve an equation for
x is to find all values such that when they are substituted for x in the equation, the
equation is satisfied, i.e., the left-hand side reduces identically equal to the
right-hand side. To solve then the above equation, considering it written in the
second form, means to find all values which when substituted for x reduce the left-
hand side to zero. To make the left-hand member of this equation zero in all
possible ways is to equate each factor to zero. Thus x + 2 = 0 and x — 3 = 0.
But, for x + 2 to be equal to zero, x must equal — 2, and for x — 3 to be equal to
zero, x must equal 3. Therefore, — 2 and 3 are the only values of x which when
substituted for x in the original equation satisfy that equation, i.e., — 2 and 3 is
the solution of the equation.
In general, if the right-hand member of an equation is zero, and if the left-hand
member breaks up into two or more factors, the equation is equivalent to the system
of equations formed by equating each factor to zero. Finding all solutions of the
system of equations gives all solutions of the original equation.
1 It must be remembered that Fig. 47 is only a perspective drawing of the
cylinder and the two cuttng planes, and that we are at present concerned with the
curve AB'A'BA as it appears upon the plane of the paper. We must imagine that
we are looking down upon the inclined cutting plane in such a way that OB extends
to our right and that OA is the upward vertical direction. The outline would then
appear to us as in Fig. 48.
§66] THE ELLIPSE 131
signs. Geometrically this means that the curve is symmetrical
with respect to the axis of x. For if the curve were plotted point
by point, the two numerically equal values of y, corresponding to
a single value of x, show that there are two points upon the curve
at the same distance from the .XT-axis; and the opposite signs of
the two values of y show that one point is above the X-axis while
the other is below. These two points, since they correspond to a
single value of x, are located upon the same side of, and at the
same distance from, the F-axis. The line joining the two points
is then perpendicular to, and bisected by, the X-axis. Thus the
two points are symmetrical with respect to that axis. Since all
points of the curve occur in pairs symmetrically situated with
respect to the X-axis, the curve is said to be symmetrical with
respect to that axis. As an illustration: let the radius, 6, of the
cylinder be 2 feet, let the angle between the normal and the in-
clined cutting planes, a, be 30.°
Then
__ _
" " ~° ~ ~ ' a
.— ,—
= ± V62 — x2 = ± — -- V 4 — x2.
When x = 1 the last equation gives
. 2V3 ,—
Thus, (1, 2) and (1, — 2) are two points on the curve, each, one
unit to the right of the 7-axis. One point is two units above the
X-axis, and the other is the same distance below.
From the equation y = ± -r V&2 — a;2 it is seen that y increases
in numerical value as the quantity under the radical increases, and
decreases as the quantity under the radical decreases. The
quantity under the radical, &2 — xz, increases as x decreases in
numerical value, and decreases as x increases. The value of
62 — x2, which is 62 when x is zero, decreases to zero as x increases
to 6, and becomes negative when the value of x exceeds that of b.
132
MATHEMATICS
[§66
The numerical value of y, which has a maximum when x equals
zero, decreases to zero as x increases to b. When the value of x
exceeds b, y becomes imaginary,1 since the number under the radical
sign is then negative. This says, geometrically that for all values
of x greater than 6, there is no real value of y and no point can
s A T be found whose coordi-
nates satisfy the given
equation; or the curve
does not extend to the right
of a vertical line, b units to
the right of the F-axis.
The curve is symmetrical
with respect to the F-axis.
This fact is shown by its
equation, since x occurs
only as the square. If
two values are assigned to
x, numerically equal but
with opposite signs, the
corresponding pairs of
values for y are the same.
Which says, geometrically
that for any point of the
curve there is another
point symmetrical to the
first point with respect to
the F-axis.2
From the preceding discussion it is seen that the curve does
not extend to the right of the line T'T, Fig. 48, nor to the left
of the line SS'; that the curve does not extend above the line ST
nor below the line S'T'. Therefore the curve is entirely within
the rectangle of dimensions 2a and 26. These conclusions are
geometrically apparent from Fig. 47.
1 The even root, as the square root, the fourth root, the sixth root, etc., of a nega-
tive quantity is called imaginary.
2 That the curve is symmetrical with respect to the F-axis may also be shown by
first solving the equation for x. For, in this form, it is readily seen that, for any
value of y, numerically less than a, these are two numerically equal values of x, one
positive and one negative.
S'
§67]
THE ELLIPSE
133
OE,( = b) and OA,( = a) are called the semi-axes of the ellipse.1
67. A Geometric Method of Locating Points upon an Ellipse.
With 0 as center, Fig. 49, and with OB, or b, and OA, or a, as
radii, draw two circles. From 0 draw any line making an angle 6
with the positive direction of the axis of x, cutting the smaller
FIG. 49.
circle at the point M and the larger circle at the point N. Through
M draw a line parallel to the F-axis, and through N a line parallel
to the .X"-axis. We shall show that P, their point of intersection,
is a point upon the ellipse. The x-coordinate, OS, of the point P
1 In text-books on Analytic Geometry it is customary to take the major semi-axis
of the ellipse coinciding with the axis of x. The ellipse would then appear as Fig.
48 rotated through 90°. Its equation is
where 6 > a.
134 MATHEMATICS [§68
is OM cos 6, or x = b cos 6. The T/- coordinate, SP, or TN, of
the point P is ON sin 6,ory = a sin 0. From these two equations ,
-T = cos 6
0
and
V
= sin
a
+ i - cos2 0 + sin2
J
or, since sin2 0 + cos2 0=1,
This is the equation of the ellipse. Therefore, P is a point upon
an ellipse. By the above method any number of points may be lo-
cated upon the ellipse and the curve drawn through them.
68. The Foci; the Eccentricity; the Sum of the Focal Radii.
At B, Fig. 49, draw a line tangent to the smaller circle cutting
the larger circle at the points E and E'. From E and E' drop
perpendiculars upon the F-axis meeting it at the points F and
F'. F and F' are called the foci of the ellipse, and the lines FP and
F'P, lines drawn from the foci to any point P of the ellipse, are
called the focal radii for the point P.
We shall show that FP + F'P = 2a (a constant for all points
upon the ellipse).
Let FP be abbreviated by n andP'Pby rt, and OF ( = OF')
by c. Then
or
ri» = x2 + y2 + c2 - 2cy,
or
ri2 = b2 cos2 0 + a2 sin2 0 + c2 - 2ca sin 0.
jt is easily seen from the rectangle OBEF that c2 = a2 — b2,
whence,
ri2 = b2 cos2 0 + a2 sin2 0 + a2 - b2 - 2ac sin 0,
or since cos2 0=1 — sin2 0,
ri2 = 62(1 - sin2 0) + a2 sin2 0 + a2 - b2 - 2ac sin 0,
§68] THE ELLIPSE 135
or ri2 = (a2 - 62) sin2 0 - 2ac sin 0 + a2
n2 = a2 - 2ac sin 0 -f c2 sin2 0
ri = a — c sin 0.
Similarly it can be shown that
TZ = a + c sin 0.
Upon adding,
ri + r2 = 2a,
or the sum of the lengths of the two focal radii is a constant.
This property of the ellipse furnishes an easy method for its
construction. From the relation c2 = a2 — 62, locate the foci,
F and F'. Fasten the ends of a string, whose length is 2a, at the
points F and f". Place a pencil point in the loop of this string
and draw taut. By moving the pencil, allowing the string to slip
around it, an ellipse will be described.
We shall prove the converse of the above theorem: The locus
of a point, moving in a plane such that the sum of its distances
from two fixed points remains constant, is an ellipse.
Let F and F' be the fixed points, and P any point upon the
curve. Let the F-axis pass through F and F'. Let 0, the mid-
point of FF', be the origin. Let the distance OF ( = OF') be rep-
resented by c. Let FP be represented by rt and F'P by r2.
n2 = xz + (c - 7/)2 (1)
r22 = z2 + (c + y)2. (2)
By adding,
n2 + r22 = 2(z2 + ?/2 + c2). (3)
By subtracting,
r22 — ri2 = 4cy. (4)
fi + r-i = 2a (constant sum) by hypothesis. (5)
Dividing (4) by (5),
By adding (5) and (6),
r2 = a+f- (7)
By subtracting (6) from (5),
* — * (8)
136 MATHEMATICS [§69
Squaring (7) and (8) and adding,
ri2 + r22 = 2(a2 + c2?/2/a2). (9)
Substituting in the first member of (3) and canceling the common
factor 2,
C2W2
a2 + — = x2 + y2 + c2.
Clearing of fractions,
or
a2z2 + (a2 - c%2 = a2(a2 - c2). (10)
Dividing (10) by the quantity a2(a2 - c2),
-Y.2 /j;2
x i y_ _ j
a2 — c2 a2
But, this equation is of the form
fc2 + a2 = L
which we have shown to be the equation of the ellipse.
The distance OF = OF' = c is called the focal distance. The
ratio, c/a, which is never greater than unity, is called the ec-
centricity of the ellipse. Referring to Fig. 47 the focal distance
is equal to AC, and the eccentricity is the sin a. A circle then
is an ellipse with zero eccentricity.
69. The Area of the Ellipse. Let ABCD and ABC'D', Fig. 50,
be two planes inclined to each other at an angle a. Let EFG'H'
be a rectangle in the plane ABC'D' with its base EF in (or parallel
to) the line of intersection of the two planes. Let EFGH be the
orthographic projection of EFG'H' upon the plane ABCD.1 Since
EH = EH' cos a, the area of EFGH = (area of EFG'H') cos a.
If, instead of a rectangle, an irregular area, as E'F'G'H',
1 If from a point a perpendicular is let fall upon a plane, the foot of the perpendicular
is called the orthographic projection of the point upon the plane. If from every
point of a curve a perpendicular is let fall upon a plane, the locus of the foot of this
perpendicular upon the plane is called the orthographic projection of the curve upon
the plane. If the bounding curve or curves of any surface be projected orthographic-
ally upon a plane, the area bounded by the projection is called the orthographic
projection of the area of the given surface upon the plane.
§69]
THE ELLIPSE
137
Fig. 51, on the plane ABC'D' be projected orthographically upon
the plane ABCD, divide the projected area into strips by equidis-
FIG. 50.
FIG. 51.
tant parallel lines drawn perpendicular to AB, and from the
points of intersection of each of these lines with the boundary of
138 MATHEMATICS [§70
the area draw lines to the right until they meet the next parallel.
In this way, on the plane ABC'D', is formed a series of rectangles
which project into another series of rectangles on the plane ABCD.
From what was shown above it is seen that the sum of the areas
of the projections of these rectangles upon the plane ABCD is
equal to the sum of the areas of the projected rectangles multi-
plied by the cos DAD'.
As the distance between the parallel lines is made smaller and
smaller, i.e., as the number of strips is increased indefinitely,
the sum of the areas of the rectangles in the plane ABC'D' ap-
proaches nearer and nearer the irregular area E'F'G'H' ; and the sum
of the areas of the rectangles in the plane ABCD approaches nearer
and nearer the area EFGH, the proj ection of E'F'G'H' . Therefore,
area EFGH = (area E'F'G'H') cos DAD'. Then (Fig. 47) the
area of the circle BCB'D = (area of the ellipse ABDB') cos a, or,
since cos a = b/a, and since area of the circle = irb2,
•jrb2 = (area of ellipse)—'
or
area of ellipse = Trba
70. The Orthographic Projection of a Circle. Let ABCD,
Fig. 52a, be a circle with radius a. Let AOC, Fig. 526, be an
end elevation of the same circle. Rotate this circle about BOD
as an axis through an angle, /3, to the position A"OC". Project
the rotated circle upon its original plane, into the curve A'BC'D.
We shall show that A'BC'D is an ellipse. Take any point P upon
the original circle. It rotates into the point P", and P" projects
intoP'. The equation of the circle is x* + y2 = a2, where x = MP.
To get the equation for the curve A'BC'D replace MP by its
equal MP' /cos 0. (See Fig. 526.) Whence,
MP'2
-— 4- v2 = a2
cos2 ,3 +y
Since MP' is the x-coordinate of P',
§71]
or
THE ELLIPSE
139
a2 cos2 /3 "" a2
Replacing a cos /3 by &( = OC"),
a;2 v2
the equation of an ellipse. The focal distance, Va2 — ft2, is equal
to C'C", and the eccentricity c/a = sin ft.1
FIG. 52.
[71.] Tangent to an Ellipse. Let PTQ, Fig. 47, be a tangent
plane to the cylinder with PQ the line of tangency. Let QT and
PT be, respectively, the lines of intersection of the tangent plane
with the horizontal and the inclined cutting planes. TQ is tangent
to the circle DBQCB' at the point Q, and TP is tangent to the el-
lipse BPAB'A' at the point P. These tangent lines meet at the
point T upon the shorter axis of the ellipse produced. This fact
will be made use of in drawing the tangent to an ellipse.
Let ABA'B', Fig. 53, be an ellipse to which a tangent line is to
be drawn at the point P. Draw the circle with B'B as diameter.
From P drop a perpendicular upon B'B cutting the circle at Q.
1 It can be shown that the section of a right circular cone made by a plane cutting
all of its elements on the same side of the vertex is an ellipse.
140
MATHEMATICS
[§71
Through Q draw QT tangent to the circle (perpendicular to OQ).
Draw TP, which will be tangent to the ellipse at the point P.
FIG. 53.
FIG. 54.
It is left for the student to show that the above statement is
correct.
We shall now show how to draw a tangent to an ellipse through
§71] THE ELLIPSE 141
a point S, not upon the curve. Let S, Fig. 47, be a point in
the plane of the ellipse, through which the tangent line is to be
drawn. Draw SS' parallel to PQ, intersecting the plane OCQ at
S'. Draw SK and S'K perpendicular to BB' From the similar
right triangles SS'K and AGO it follows that:
S'K/SK = CO/AO
Let ABA'B', Fig. 54, be the ellipse to which the tangent is to
be drawn from the point S. Draw ASW. Draw CW. Draw
SK perpendicular to BOW, cutting CW at the point S'. From S'
draw the two tangents to the circle BCB'D. Let Qi and Q2 be the
points of tangency. Draw RiQiPi and RiQ^Pz perpendicular to
B'OB cutting the ellipse in the points PI and P2. Draw PiS and
PzS. PiS and PyS are the tangents to the ellipse drawn from the
points S. PiS and QiS' intersect upon B'BW ; and P*S and Q2S'
intersect upon B'BW. It is left for the student to show that the
above statements are correct.
Exercises
1. Find the value of each semi-axis, the focal distance, and the
eccentricity for each of the following:
(a) + = 1- (c) lOOz2 + 4i/2 = 400.
4 y
(b) ^ + !^ = 1. (d) 7x* + 3*/2 = 11.
2. Construct an ellipse by the method given in § 67 if a =5
inches, and 6=4 inches.
3. Construct an ellipse by the method given in § 68 if a = 4
and a = 30°.
4. Find the weight of metal removed in cutting the opening in the
saddle for a circular ventilator 85 feet in diameter, if the rafters make
an angle of 30° with the horizontal, and if the ventilator is made of
20 gauge iron. Twenty gauge iron is 0.038 inch thick, U. S. standard.
The specific gravity of steel is 7.83. A cubic foot of water weighs
62.5 pounds.
[5.] Show that a right elliptical cylinder, has a circular cross section.
If,
x- + y- = i
16 ^ 9
142
MATHEMATICS
[§71
M
is the equation of its normal cross section, find the angle between the
axis of the cylinder and the plane giving the circular section.
[6.] Show that every section of a right elliptical cylinder is an
ellipse.
[7.] A circular window in the east wall of a building is 4.6 feet in
diameter. The wall runs true north and south. The rays of light
from the sun pass through the window and fall upon a horizontal
A floor. Find the area of the floor in the sunlight if the
angle of elevation of the sun is 57° 42'. 6, and if it is
37° 27'.7 east of the meridian.
[8.] Construct an ellipse with semi-axes equal to
1^ and 2^ inches. Draw a line bisecting the angle
between the axes. At the point of intersection of
this line with the ellipse, draw a tangent line.
Measure the distances from the center of the ellipse
to the points of intersection of the tangent line with
the axes of the ellipse produced.
[9.] Construct an ellipse
with semi-axes equal to 1\ and
3| inches. Through its center
draw a line making an angle of
30° with the minor axis. From
a point of this line 2 inches
from the center draw two
tangents to the ellipse.
[10.] Through the point of tangency of the tangent line drawn in
exercise 8 draw a line perpendicular to the tangent line. With the
protractor measure the angles between this perpendicular and the
focal radii.
[11.] Let OA, Fig. 55, be the vertical wall of a building; OB a hori-
zontal pavement; MN a ladder of length a + b. P is a point of the
ladder such that MP = b and NP = a. Show that as the foot of
the ladder is drawn out along the pavement, P moves along the arc
of an ellipse.
HINT: Consider OB as the X-axis and OA the F-axis of a rec-
tangular coordinate system. OT ( = x) and TP ( = y) are the co-
ordinates of P. - = sin a, and r = cos a. Square these equations
d 0
and add.
N
FIG. 55.
CHAPTER V
[ THE SLIDE RULE ]
72. Accuracy of Logarithmic Computation. The logarithms
given in tables are only approximations. In a four-place table,
they are correct to the nearest fourth decimal place; in a five-
place table to the nearest fifth decimal place; in a six-place
table to the nearest sixth decimal place; etc.
The log ir = 0.497 to three places,
0.4971 to four places,
0.49715 to five places,
0.497150 to six places,
0.4971499 to seven places, and
0.49714987 to eight places.
The log TT is given correct to three, four, five, six, seven and
eight places of decimals; but none of these values is absolutely
correct.
Since the logarithms of numbers taken from tables and used
in logarithmic calculation are not absolutely correct, the results
of the calculations are not absolutely correct. But in nearly all
numerical calculation, results are required accurate only to a
certain number of places; and the number of accurate places re-
quired is the fact that determines the table to be used. If a set
of logarithms are added (corresponding to multiplication) or
added algebraically (corresponding to multiplication and division)
the errors in the logarithms will be, in most cases, compensating
in nature. That is to say, the positive errors and negative errors
tend to balance, producing only a small error in the sum.
Below are written ten numbers selected at random. In the sec-
ond column are given their logarithms to six places; in the third
column their logarithms to four places; and in the fourth column
the errors in the four-place logarithms as compared with the
six-place logarithms.
143
144
MATHEMATICS
[§72
Number
Logarithm
Logarithm
Error
3.72
0.570543
0.5705
-0.000043
9.81
0.991669
0.9917
+0.000031
7.15
0.854306
0.8543
-0.000006
1.23
0.089905
0 . 0899
-0.000005
5.06
0.704151
0.7042
+0.000049
7.17
0.855519
0.8555
-0.000019
8.23
0.915400
0.9154
0.000000
3.56
0.551450
0.5514
-0.000050
1.17
0.068186
0 . 0682
+0.000014
8.75
0.942008
0.9420
-0.000008
Sum
6 4431
-0.000037
The sum of the errors is — 0.00004, which represents approximately
the error in the sum of the ten logarithms. The mantissa 4431
is found to correspond to the number 2774. Since the tabular dif-
ference at this point of the table is 15, the above error, 0.00004
makes only a difference of between 2 and 3 in the next (fifth)
place in the number. Thus by using a four-place table in multi-
plying these ten numbers together, the product is found correct to
the nearest fourth place.
If the sum of the ten logarithms had been 6.9872, the tabular
difference at this point of the table would have been 5, the error
of — 0.00004 would have produced an error in the corresponding
number of about 7 in the fifth place. Thus, the result of the prod-
uct of the ten numbers would have been one unit too small in the
fourth place.
We see from this illustration that an error in the logarithm,
the sum, may effect the fourth digit one, or possibly two points,
if the logarithm stands near the end of the table; while the fourth
digit is given correctly if the logarithm stands nearer the begin-
ning of the table.
Multiplying and dividing with a four-place table gives re-
sults correct to four places, or correct to three places while the
fourth place is in error one or two points only. The error in the
product or quotient will usually be less than 2/100 of 1 percent,
and in the majority of cases it will be very much less than this.
§72] THE SLIDE RULE 145
This is what is meant when we say a four-place table computes ac-
curately to four places.
In taking out the number corresponding to a logarithm, do not,
in general, write more than four significant digits if a four-place
table is used. If there are more than four digits preceding the
decimal point, determine the first four, and suffix ciphers between
the fourth digit and the decimal point. Thus, if log x = 6.1738,
x = 1,493,000. The fourth digit is determined by interpolation,
and three ciphers are suffixed. This number, however, occurring
near the beginning of the table, could have been corrected to
another place, in which case the fifth digit would have been correct
to within three or four points.
In numerical work, use the smallest table which will insure
the required degree of accuracy in the result. If a five-place table
will suffice, do not use a six-place table. If a four-place table
will suffice, do not use a five-place table. If the numerical work
of a problem is to be done with four-place logarithms, the student
who has had but little practice in logarithmic computation is
very apt to believe that by using a five-place table and discarding
the last digit he will save the time and trouble required in apply-
ing corrections. This method, however, does not gain the end
sought; for, by using a well-arranged four-place table, corrections
can be applied, after a very little practice, more expeditiously
and easily than the logarithms can be found from the larger ta-
ble containing ten times as many numbers.
Another oversight on the part of a large number of students,
is the attempt to compute more accurately than the observed
data warrant.
Thus, to illustrate, suppose a and b are two readings. They are
not absolutely correct. The first has a possible error ei and the
second a possible error e2. The two readings are to be multiplied
together. What is the possible error of the product? How many
places in the product are reliable?
(a ± ci) (b ± e2) = ab ± ae2 + 6ei + ci«2.
The sum of the last three terms in this product, taken with the
upper signs, is the largest possible error in the product due to
the errors ei and e2 in the factors a and b, respectively. eie2 is
10
146 MATHEMATICS [§73
small compared with ae2 + &ei, and may be neglected. The pos-
sible error in the product is then + (ac2 + &ei).
Suppose 325 and 52 1 are the readings. Suppose that in each case
1 in the third place is the least count of the instrument, i.e.,
there is a possible error of 0.5 in each reading. Then,
a = 325
b = 521
€i = 6z = 0.5
ab = 169,325
aei + 6e2 = 423
Thus we see that the product of the two numbers whose measure-
ments were attempted lies between 169,325 + 423 and 169,325 —
423. This shows that the product may be correct to three places
only, and should be written 169,000. The digits following the 9
in the product ab have no significance since all may be incorrect;
hence the digits should not be written, but ciphers substituted in
their places.
In the above illustration, even a four-place table carries
the accuracy of the work beyond what is required. In this prob-
lem the calculation could have been done with a three-place
logarithmic table, or with a slide rule.
73. The Principle of the Slide Ride. Let C and D, Fig. 56,
represent two uniform scales. The two scales are so placed that
the zero mark of C stands opposite the 3 mark of D. Any number
on D to the right of 3 is the sum of the number directly above it on
C, and 3. Thus, 5(read on scale D) is the sum of 2 and 3; 6 is the
sum of 3 and 3; 4| is the sum of 1J and 3; etc. In this way
any number may be added mechanically to the number 3. By
placing the 0 of C above any point of D, any other number may
be added mechanically to the number corresponding to this point
of the D scale.
It is immaterial whether the 0 of C or the 0 of D is set to cor-
respond to one of the numbers to be added. If the 0 of D is set,
the sum is read from the C scale.
By a similar process, subtraction may be performed mechanic-
ally. Thus, if 3 is to be subtracted from 6, move scale C so that
§73] THE SLIDE RULE 147
3 upon it stands above 6 of D, the number of D below the 0 of C,
or 3, is the difference.
In exercise 10, § 41, there were constructed two logarithmic
scales, each in juxtaposition with a uniform scale.1 These two
pairs of scales are represented (reduced) in Fig. 57. Bend scales
C and D along the horizontal lines; turn the lower part, the uniform
scale, of C back behind the non-uniform scale; and turn the
upper part of D back behind the lower part. Now place these two
scales together as shown in Fig. 58. The lines marked 1 and 10 of
these scales are called indices.
If we bring the left-hand index of C opposite 2 of D, any number
to the right of 2 on the D scale will be the product of the corre-
sponding number of C, and 2. Thus, in the figure, 6 of D stands
beneath 3 of C; 6 is the product of 3 and 2. To multiply 2 by
3 mechanically, bring the index of the upper scale over 2 of the
lower scale, and read the number on the lower scale which stands
directly under 3 of the upper scale. This number is the product.
For, upon the back of each scale, C and D is a uniform scale,
and by sliding C and D upon each other we are adding me-
chanically, as was shown above, the numbers of these uniform
scales. Thus, on the back of D is a number, which is the sum
of the number back of 2 on D and the number back of 3 on C.
But, the number back of 2 on D is the logarithm of 2, and the
number back of 3 on C is the logarithm of 3; then on the back
of D opposite 3 on C is the logarithm of 2 plus the logarithm of
3, or the logarithm of 6, since the sum of the logarithms of two
numbers is the logarithm of their product. On the face of D are
numbers whose logarithms are on the back. Therefore, opposite
the logarithm of 6 upon the back of D, i.e., opposite 3 of C, we
have 6, or the product of 2 and 3.
In constructing scales C and D, one unit of length only, of the
uniform scale was used. By extending these double scales indefi-
nitely in both directions, the uniform parts will include all al-
gebraic numbers, i.e., all numbers from — °° to + <» ; the non-uni-
form parts will include all positive arithmetical numbers, i.e., all
numbers from 0 to + °° . A port:on of one of these scales (the C
i If this exercise was not performed when the work on logarithms was taken up, it
should be performed now, or at least the description read carefully.
148
MATHEMATICS
[§73
O - : : -
§73] THE SLIDE RULE 149
scale) is represented in Fig. 59. The extended D scale is the same,
but inverted. For convenience of description, different segments
of these scales will be referred to as "block I," "block II," etc.
By block II is meant that portion of the double scale between 1 and
2 of the uniform part, or between 10 and 100 of the non-uniform
part; by block — I is meant that portion of the uniform part
between — 1 and 0, or between 0.1 and 1 of the non-uniform
part, etc.
Following directly from the facts that the mantissas of the com-
mon logarithms are independent of the position of the decimal
point in the number, and that by moving the decimal point one
place to the right or left increases or decreases, respectively,
the characteristic of the logarithm by 1, it is seen that all blocks
of the double scale, Fig. 59, are identical. To illustrate: the line
on the non-uniform part representing 20 is in the same position
relative to 10 as the line representing 2 is to 1. This must be so,
for the logarithm of 20 equals the logarithm of 10 plus the loga-
rithm of 2; and, since the logarithms are represented on the uniform
part, the logarithm of 20 is as far to the right of 1 on the uniform
part as the logarithm of 2 is to the right of 0. Since all blocks
are the same, one block will suffice for the entire indefinite scale.
In using scales C and D, Fig. 58, for mechanical multiplication
the operator may disregard the positions of the decimal points in
the factors, and point off the proper number, of places in the
product after the sequence of digits has been found.
Since the index at the right-hand end of C may be looked upon
as 1 as well as 10, the right-hand index as well as the left-hand
index may be set above one of the factors of the product. It
will be seen, however, that in multiplying two numbers together,
using only one block each of the C and D scales, only one of the
two indices of C can be set over the.mutiplicand, for if the other
index were set it would throw the product off the D scale into
the next block. By interchanging the multiplicand and multiplier
the other index of C may be set. To illustrate: let 3 be multi-
plied by 6. Set the left-hand index of C over 3, then 6 of C
falls beyond the D scale. In this case the product may be found
either by setting the right-hand index of C over 3 and reading
150
MATHEMATICS
[§74
cq O C)
under 6, or by setting the right-hand index
of C over 6 and reading under 3.
By means of scales C and D any two num-
bers may be multiplied together mechan-
ically. The principle involved, i.e., of
adding logarithms mechanically, is the un-
derlying principle of the construction of
the (Mannheim) slide rule.1 The opera-
tions performed by the mechanism may be
compared with logarithmic work as follows.
Locating the numbers on the scales may be
compared with taking out the logarithms of
the numbers and the anti-logarithm of their
product, while sliding the C scale above the
D scale may be compared to the addition
^ of the logarithms.
2 Since the logarithms of the factors and
3 the logarithm of their product are not read
"" they need not be printed upon the back of
^ the scales C and D.
| 74. The Slide Rule. The slide rule con-
§ sists of three parts: the body, the slide, and
e the indicator (Fig. 60). The body of the rule
£ consists of a wooden base about 10 inches
long,2 about 1 inch wide, and about 1/8
inch thick, upon one side of which are
fastened two parallel strips, each carrying
1 Many mechanical devices have been constructed for
performing special numerical operations. Often they are
in the form of one scale sliding over another, and for
• this reason are also termed, with a proper qualifying
adjective, -slide rules; as, stadia slide rule, sewer slide
rule, horse-power slide rule, etc.
The slide rule is also constructed by placing the
logarithmic scales upon two concentric circular arcs, one
rotating within the other.
' Slide rules are made in three sizes: viz., the 5-inch,
about 5 inches long; the 10-inch, about 10 inches long;
and the 20-inch, about 20 inches long. The 10-inch is
the size commonly used. The 5-inch is less accurate,
but is very convenient for carrying in the pocket and
for rough estimates.
§75] THE SLIDE RULE 151
a logarithmic scale engraved upon celluloid. These two scales are,
in Fig. 60a, marked A and D. A transverse section of the rule is
shown in Fig. 606. The two upper parallel strips of the body
are separated about 1/2 inch, and between them there is placed
the slide, a thin strip of wood carrying two logarithmic scales,
B and C. The scales C and D are the same as those explained
in the preceding section, and by means of them, multiplications
may be performed.
The indicator /, Fig. 60a, consists of a piece of glass, about an
inch square, mounted in an aluminum frame, which slides in
grooves lengthwise the rule. On the under side of the glass,
very nearly in contact with the surfaces of the four scales, is etched
a fine vertical line, or, "cross line" as it will be called.
Scales A and B are logarithmic scales exactly one-half as long
as scales C and D, i.e., upon A and B are two blocks of the in-
definite logarithmic scale. Scales A and B, then, have each
three indices, one at each end and one at the mid-point. When we
come to the explanation of the use of the A and B scales we shall
see that the end indices are considered to correspond to the num-
bers . . . 0.0001,0.01,1,100,10,000,. . . , while the central in-
dex corresponds to the numbers . . . 0.001, 0.1, 10, 1000,
100,000. . .
The scales A and B are placed upon the body of the rule
so that the left-hand index of A is directly over the left-hand
index of D, and so that the right-hand index of A is directly
over the right-hand index of D. When the cross line of the indi-
cator is set over the left-hand index of A it will also be over the
left-hand index of D; when set over the right-hand index of A it
will also be over the right-hand index of D.
75. The multiplication of two or more numbers will be il-
lustrated by an example. To multiply (37.5) (212) (11.2) (0.315)
proceed as follows: Set left-hand index of C over 375 of D. Set
cross line of indicator over 212 of C. (Directly under the cross line
on the D scale is the product of the first two factors, but do not
read it.) Set the left-hand index of C under cross line (i.e., over
the product of the first two factors on Z>). Set cross line over
112 on C. Bring the right-hand index of C under the cross line.
Move cross line over 315 of C, and under it on the D scale read
152 MATHEMATICS [§76
the product of the four factors. The digits read are 2805. To
locate the decimal point, find mentally a very rough estimate
of the product; thus, the product of the first two factors is about
i of 21,000, or about 7000; the next multiplier (about 10)
makes the product of the first three factors about 70,000; the
next multiplier is about ^, which makes the product of four
factors about 23,000. This shows that there are five digits be-
fore the decimal point. The product is then 28,050.
It is to be noticed that in the above operations no readings
are taken from the rule excepting the final product.
In a similar way any number of factors may be multiplied
together.
If several numbers are to be multiplied by the same number,
set the index of C over the common factor on D. Slide in turn the
cross line over the given numbers on C, and read off the various
products from D. In this way all the products can be read with
one, or at most two, settings of the slide.
76. Division. The division of one number by another may be
performed upon a slide rule in two ways. The principle of one
method is that subtracting the logarithm of the divisor from the
logarithm of the dividend gives the logarithm of the quotient.
The principle of the other method is that multiplying the quotient
by the divisor gives the dividend. Both methods will be illus-
trated by example. To divide 372 by 176, set the slide so that
176 of C is directly above 372 of D. (To do this easily, first set the
cross line over 372 of D and then bring 176 of C under it.) Read
the quotient from D, the number under the index of C. The
quotient is 2.113. By inspection it is seen that the decimal point
follows the 2.
This method of division corresponds to the subtraction of
logarithms. For, the distance from the index to 372 (measured
upon the uniform scale on the back of D) is the logarithm of 372.
The distance from the index of C to the number 176 is the logarithm
of 176. Then the distance from the index of D to the index of C
is the difference of these two logarithms, or the logarithm of the
quotient; and the number on the face of D is then the quotient.
To divide by the second method, set the index of C over the
divisor, 176, on D. Set the cross line over the dividend, 372, on
§77] THE SLIDE RULE 153
D. Under the cross line on the C scale read the quotient, 2. 1 13
It is easily seen that this method gives the quotient, for it is the
reverse of that used in multiplication.1
The second method of division is useful if several numbers are
to be divided by a common number; for example, to find the per-
centages which a set of numbers are of a given number, place the
index of C over the common divisor on D, bring the cross line over
the dividend on D and read the quotient from C under the cross
line. In this way a series of quotients may be found with one, or
at most two, settings of the slide.
77. The A and B Scales. Since the "block" on A is just one-
half as long as the "block" on D, the numbers (in imagination) on
the back of A are just double those directly opposite on the back
of D. The numbers on the face of A are then the squares of the
numbers directly opposite on the face of D. This follows from
the fact that the logarithm of the square of a number is two times
the logarithm of the number.
ILUSTRATIONS : To find the square of 3.77, either multiply
3.77 by 3.77 or set the cross line over 3.77 on the D scale and
read its square, 14.2, from the A scale. To extract the square
root of 7.13, set the cross line over 713 of the left-hand block
of the A scale, read off its square root, 2.67, from the D scale.
To extract the square root of 71.3, set the cross line over 713 of
the right-hand block of A, read off its square root, 8.44, from the
D scale.
In extracting square roots care must be exercised to enter the
number in the proper block of the A scale.
Scales A and B may be used for multiplication and division
in the same way that scales C and D are used. But on account of
the A and B scales being only half as long, the results are not as
accurate as when C and D are used.2
78. The Cube Root.3 The method of finding cube roots of
numbers will be illustrated by examples.
1 The first method may also be looked upon as the reverse of multiplication.
1 The Polyphase Slide Rule is one having two additional scales. One of these
scales gives cubes (and cube roots) directly; the other gives reciprocals of numbers
directly.
* Some rules are provided with a cube scale by means of which cubes and cube roots
may be obtained directly.
154 MATHEMATICS [§79
Example 1 : Find the cube root of 8. Set the cross line over
8 on the A scale (block I). Move the slide so that the number
on the B scale under the cross line is the same as the number on
the D scale under the left-hand index of the C scale. This is
found to be 2, the cube root of 8.
Example 2: Find the cube root of 80. Set the cross line over
8 on the A scale (block I). Move the slide so that the number on
the B scale under the cross line is the same as the number on the
D scale under the right-hand index of the C scale. This is found
to be 4.31, the cube root of 80.
Example 3: Find the cube root of 800. Set the cross line
over 8 on the A scale (block II). Move the slide so that the
number on the B scale under the cross line is the same as the
number on the D scale under the right-hand index of the C scale.
This is found to be 9.28, the cube root of 800.
The student will explain why these operations give the cube
roots.
79. Other Operations with the Slide Rule. It was the inten-
tion to give above only the underlying principle of the slide
rule; and to explain its use in connection with the simple opera-
tions of multipli cation, division, extracting square root, and
squaring. It is believed that the student will, after some prac-
tice with the slide rule, devise methods of procedure which will
involve the fewest number of settings of the slide, and the fewest
number of intermediate readings. Pamphlets and books may
be procured giving in considerable detail the methods of perform-
ing numerical calculation with the slide rule.
In performing numerical calculations with the slide rule, the
work should be blocked out as when calculating with logarithms.1
Exercises
The student will practice on simple multiplications, divisions,
squares, square roots, and cube roots.
1 Paper, form M 8, is ruled especially for slide rule calculations.
CHAPTER VI
STATICS
80. Vector and Scalar Quantities. In Physics and in Me-
chanics we have to deal with two kinds of quantities: vector
quantities and scalar quantities.
Quantities, such as velocity, force, acceleration, etc., which
require for their complete definition direction as well as magnitude,
are called vector quantities. When we say a body has a uniform
velocity of 10 feet per second, we mean that in every second of
time the body moves a distance of 10 feet and in some given
direction. Velocity, then, is a vector quantity.
Quantities, such as mass, time, area, volume, etc., having
only magnitude, are called scalar quantities. In order to convert
a gram of ice at zero degrees centigrade into water at the same
temperature, 79 units of heat must be added to the mass. This
quantity, 79 calories, is a scalar quantity.
81. Graphic Representation of Vector Quantities. In order
to represent a vector quantity graphically, choose some unit of
length, as the inch, the half-inch, the centimeter, to represent
the unit of magnitude; with this unit length, draw a line whose
length measures the magnitude of the quantity, and whose direc-
tion is that of the vector quantity. To illustrate: a body moves
FIG. 61.
from west to east with a speed of 3 miles per hour. Draw the line
AB, Fig. 61,3 cm. long and from left to right. The line drawn
3 cm. long represents the magnitude, the speed, 3 miles per hour,
and drawn from A to B represents the motion from left to right,
or, from west to east. An arrow is placed upon the line indicat-
ing its sense, i.e., from left to right.
A directed line, as AB, Fig. 61, is called a vector.
155
156 MATHEMATICS [§82
A vector quantity is also termed, for brevity, simply a vector.
In what follows it is assumed that all problems are of such a
nature that all vector quantities act in the same plane. The vec-
tors representing these vector quantities may then be drawn upon a
plane, the plane of the paper.
82. Vector Addition. In elementary algebra real numbers are
usually represented graphically by points along a straight line.
Thus, if 0, Fig. 62, be any point upon the straight line X'X,
FIG. 62.
points to the right of 0 represent the positive numbers, while
points to the left represent the negative numbers. To every real
number there corresponds some point upon the line, and for every
point of the line there exists a real number. If the points upon
the line are chosen such that distances between them are propor-
tional to the steps between the numbers which they represent,
we have what was previously called a uniform scale. In what
follows this uniform scale is assumed as the graphic representation
of the real number system. The points A,B,C, D, etc., or the
vectors OA, OB, OC, OD, etc., represent, respectively, the
numbers 2, 5, -3, -4, etc.
o A B c
X - 1 - 1 ____ i ___ >! ___ >! ____ i ___ J - X
FIG. 63.
The addition of two real numbers, say 2 and 3, may be looked
upon as consisting of the following operations: Move the vector
OA, Fig. 63, parallel to itself; place it so that its initial point,
which was originally at 0, coincides with the terminal point,
B, of the vector OB. Its new position is indicated by the dotted
line BC. Draw the vector OC. It is easily seen, since the scale
X'X is uniform, that the vector OC represents the number 5, the
sum of 3 and 2. It is readily seen that this process of addition
holds for any set of real numbers (negative as well as positive),
and that, in the process, either of the two vectors may be moved.
§82]
STATICS
157
In general, let OB and OA, Fig. 64, be any two vectors. Move
OA parallel to itself so that its initial point coincides with the
terminal point, B, of the vector OB. It then takes up the po-
sition BC. Draw the vector OC. The vector OC is defined as the
vector sum of the vectors OB and OA.
The negative of a vector is a vector of the same length, but
extending in the opposite direction (having opposite sense).
FIG. 64.
«
To subtract the vector OA from the vector OB means to find
a third vector OC, Fig. 65, such that the sum of the vectors
OC and OA is the vector OB. If OA is moved parallel to itself, so
that its terminal point, A, falls upon the point B, and its initial
point at C, the vector OC is such that the vector OC plus the vector
OA equals the vector OB. Therefore,
vector OC = vector OB — vector OA
FIG. 65.
From the figure it is readily seen that, to subtract the vector OA
from the vector OB is to add the negative of the vector OA , that
is, AO, to the vector OB.
To find graphically the vector sum of any three vectors, find
the sum of any two and to this sum add the third.
158
MATHEMATICS
[§83
A vector is designated by giving its length, and the angle
which it makes with the positive direction of the axis of x. Thus
(10, 30°) when referring to a vector means that the vector is ten
units long and its direction makes an angle of 30° with the positive
direction of the axis of x.
Exercises
Find graphically the sum of the following vectors:
1. (2, 10°) and (3, 20°). 3. (6, 70°) and (5, 217°).
2. (5, 30°) and (10, 135°). 4. (5, 110°) and (3, - 71°).
6. (3, 30°), (7, 121°) and (11, - 75°).
83. Resultant Velocity. Let a raft move along a straight
path with uniform speed of 2 feet per second; suppose a man
Direction of Raft
FlG. 66.
upon the raft walks at the rate of 3 feet per second in the direc-
tion indicated in Fig. 66. Let A be the point of the raft
occupied by the man at some particular instant. In one second
of time the point A "will be translated by the movement of the raft
to the point B. The length of the line AB represents 2 feet,
and the direction of AB is the direction of motion of the raft. If
§84] STATICS 159
the raft were not moving, the man would walk along the line A C,
and at the end of one second would be at the point C. The motion
of the raft, however, has carried the point C to the point D ; at the
end of one second of time the man is at the point D instead of at
the point C. ABDC is a parallelogram. The sides AB and AC
are vectors representing, respectively, the velocity of the raft
relative to the earth and the velocity of the man relative to the
raft. By the same reasoning it is seen that at the end of the
second second the man will be at the point G. The points A, D,
and G are upon a straight line. At the end of the first half
second the man was at the point K, the center of the parallelogram
ABDC, and at the end of any other fractional part of one second he
was at some point X on the line AD. From similar triangles, it
is easily shown that X is not only on the line AD, but at a position
such that AX is to A D as the time interval is to one secohd.
Thus, the man moves, relative to the earth, in the direction
AD, with a uniform speed, and over a distance AD in one second
of time. This is only another way of saying that the velocity
of the man may be represented by the vector sum of the
vector representing the velocity of the raft relative to the earth,
and the vector representing the velocity of the man relative to the
raft. In general, then, if a body is subjected to several velocities,
its resultant velocity may be found by taking the vector sum of
the vectors representing the component velocities.
84. Resultant of Concurrent Coplanar Forces Found Graphic-
ally. Experimentally, it is shown that the resultant of two
concurrent1 forces is such that it may be found in magnitude and
direction by considering it the vector sum of the two vectors
representing the two given forces. If a force of 10 pounds acts
horizontally to the right, and if a second force of 6 pounds acts in
a direction making an angle of 60° with the horizontal, the result-
ant of the two may be found graphically in the following manner.
Lay off a line AB 5 inches long, Fig. 67. The vector AB repre-
1 Forces acting upon a body are concurrent if they have the same point of applica-
tion; otherwise they are non-concurrent.
Forces acting upon a body are coplanar if their lines of action lie in a plane; other-
wise they are non-coplanar.
A single force, if it exists, which is equivalent to a system of forces is called the
resultant of the system.
160
MATHEMATICS
sents the 10-pound force. One inch in length represents a magni-
tude of two pounds. Draw the line AC 3 inches long making the
angle BAG 60°. The vector AC represents the 6-pound force.
Complete the parallelogram ABDC, or the triangle ABD, or the
Reduced B
FIG. 67.
triangle ACD. AD is a vector representing the resultant of the 10-
and the 6-pound forces. Measuring the length of AD it is found
to be 7 inches, which shows that the magnitude of the resultant
force is 14 pounds. Measuring the angle BAD it is found to be
approximately 21° 45', the direction angle of the resultant.
ABDC is called the parallelogram of forces.
Reduced
FlG. 68.
The resultant of three or more concurrent coplanar forces may be
found by repeated application of the parallelogram of forces.
Thus, to find the resultant of the five forces, PI, P%, Pz, Pt, PS,
whose magnitudes are, respectively, 3, 6, 2, 2, and 4 pounds, and
§85] STATICS 161
whose directions make, respectively, the angles 0°, 30°, 70°, 210°
and 170° with the positive direction of the axis of x, proceed as
follows: Draw the vector AB, Fig. 68, 3 inches long in a hori-
zontal direction. Draw BC 6 inches long making an angle
of 30° with the horizontal. AC is a vector representing the result-
ant of the forces PI and P2. Draw a vector CD representing the
force P3. AD is a vector representing the resultant of PI, P2 and
P3. Draw the vector DE representing the force P4. AE is the
vector representing the resultant of PI, P2, PS, and P4. Draw
EF, a vector representing the force P6. AF is the vector repre-
senting the resultant of PI, P2, PS, P4, and P5. AF is 5.59 inches
in length, which shows that the magnitude of the resultant force is
5.59 pounds. The angle BAF measured with a protractor is
found to be approximately 55°.
It is seen that in the construction, the lines AC, AD, and AE
are entirely unnecessary and should be omitted. The figure
ABCDEFA is called the polygon of forces. AF, the resultant, is
called the closing side. It will be noticed that the arrows on the
vectors representing the given forces all run in the same sense
around the polygon, while the arrow of the closing vector runs in
the opposite sense.
If the vectors representing the given concurrent forces of them-
selves form a closed polygon, the resultant is zero, or the system of
given forces is in equilibrium.
Exercises
Find graphically the resultant of the following concurrent, coplanar
forces :
1. (3, 30°), (5, 45°), and (7, 190°).
2. (2, 0°), (3, 40°), (5, 170°), and (7, 135°).
3. (2, - 10°), (6, 90°), (5, 0°), and (7, 135°).
4. (3, 180°), (2, 270°), (5, 45°), and (1, 150°).
5. (1, 270°), (5, - 30°), (7, 90°), and (2, 30°).
85. Resultant of Concurrent Coplanar Forces found Analytic-
ally. Let AD, Fig. 69, represent a force both in magnitude and
direction. If AB, be a line of any length, drawn in any direction,
and if ABiDCi be a parallelogram so constructed that ABi is a side
and AD a diagonal, AD may be considered the resultant of two
11
162
MATHEMATICS
[§85
forces represented by AE\ and ACi. Then in a process of analysis
the force represented by AD may be replaced by the two forces
represented by ABi and AC\. The force AD is said to be resolved
into the components AB\ and ACi. It is apparent that AD can
be resolved into an infinite number of pairs of components; and
if it were desirable,
either, or both, of
these components
could be resolved
further into two or
A BI B
•pIG gg more components.
If the resolution is
such that the components are parallel one to each axis of coordi-
nates, the force is said to be resolved into its x- and ^-com-
ponents, or into its horizontal and vertical components.
It is seldom desirable to resolve a force into other than its x-
and ^-components, and this form of resolution is very helpful at
times in finding the resultant force acting upon a body.
If OD, Fig. 70, represents a force acting in a direction making
an angle 6 with the positive direction of the axis of x, the magni-
tudes of the horizontal and vertical components are, respectively,
R cos 6 and R sin B, where R is v
the magnitude of OD. The
vector representing the force in
Fig. 70 is in the first quadrant;
but the student will easily see,
by sketching other figures, that
the algebraic sign of the sin 6 and
cos 6 will give to the component
force the proper algebraic sign,
i.e., positive when up or to the
right, and negative when down or to the left.
What is here called the analytic method of finding the resultant
of a set of given forces, consists of the following process: resolve
each force into its horizontal and vertical components, add alge-
braically the horizontal components, add algebraically the vertical
components, and from these two sums find the resultant as in-
dicated below.
FIG. 70.
§85]
STATICS
163
It is required to find analytically the resultant of the five forces
given in § 84. The work is put in tabular form:
Magnitude
Angle
Cosine
Sine
Jf-component
F-component
Pi
3
0°
1.000
0.000
3.000
0.000
P2
6
30°
0.866
0.500
5.196
3.000
P3
2
70°
0.342
0.940
0.684
1.880
p<
2
210°
- 0.866
- 0.500
- 1.732
- 1.000
Pi
4
170°
- 0.985
0.174
-3.940
0.696
Sum
3.208
4 576
In the first column are placed the letters indicating the forces.
In columns two and three are placed, respectively, the magnitudes
and angles of these forces. In the fourth and fifth columns are
placed, respectively, the cosines and sines of these angles. In
the sixth column are placed the products of the magnitudes of the
forces by the cosines of the angles. In the seventh column are
placed the products of the magnitudes of the forces by the sines of
the angles. In columns six and seven, then, are given, respectively,
the x- and ^-components of the five forces. Below are given the
algebraic sum of the z-components and the algebraic sum of the
^/-components. The five forces may be considered as being re-
placed by two; one of 3.208 pounds acting to the right along
the X-axis, the other of 4.576 pounds acting up along the 7-axis.
It now remains to unite these two forces into one resultant, the
resultant of the five forces. Since the two components of 3.208
and 4.576 pounds are mutually at right angles, the parallelogram
of forces (See Fig. 70) is a rectangle; hence, the magnitude
of the resultant R = ^(3.208) 2 + (4.576) 2 = 5.59. The direc-
tion of OC is given by tan 6 = ^~ = 1.426, or, 6 = 54° 58'.
Of the two values of the angle corresponding to tan 6, the
student must be careful to select the correct value for 6, by noting
the quadrant within which the resultant lies.
Exercises
1. Solve, analytically, placing work in tabular form, the exercises
given in § 84.
164
MATHEMATICS
86. The Simple Crane. To illustrate the application of the
polygon of forces to mechanical problems, let us consider a simple
crane. AB, Fig. 71, is the post, CE is the jib, and DE the rope
attached to the jib at the point E. At the end of the jib is sus-
pended a weight of 10 tons. It is required to find the tension in
the rope and the compression in the jib. The point E is acted upon
by three forces: one, of magnitude
10, acts down; one, of unknown JD
magnitude, acts in the direction
CE, and the third, of unknown
magnitude, acts in the direction
ED. Suppose CD = 12 feet,
\OT
FIG. 71.
FIG. 72.
CE = 8 feet, and DE = 7 feet. Draw a triangle to scale with
sides proportional to 12, 8, and 7, Fig. 72. Draw the side propor-
tional to 12 vertical. Draw the side proportional to 7 through the
upper end, and the side proportional to 8 through the lower end,
of the vertical side. The triangle CED is drawn to scale. A vec-
tor representing the 10- ton force will point in the direction DC;
a vector representing the compression (call it PI) in the jib will
STATICS 165
point in the direction CE, and a vector representing the tension
(call it P2) in the rope will point in the direction ED.
Draw the polygon of forces, DC'E', for the three forces, 10,
PI, and P2. This polygon must close if the point E is in equi-
librium. Scale off the length C'E', and the length E'D, which
give, respectively, the magnitude of the compression in the jib and
of the tension in the rope. If DC' is made 10 inches long, or 1
inch representing 1 ton, C'E' is found to be approximately 6.67
inches and DE' approximately 5.83 inches. This gives a compres-
sion in the jib of 6.67 tons, and a tension in the rope of 5.83 tons.
Exercises
Find graphically the compression in the jib, and the tension in the
rope, Fig. 71, if CD = 12 feet, CE = 8 feet, and ED = (a) 8 feet, (6)
9 feet, (c) 10 feet, (d) 11 feet, (e) 12 feet, (/) 13 feet, and (g) 14 feet.
Plot two curves upon a sheet of rectangular coordinate paper, one
representing the compression in the jib, the other representing the
tension in the rope. Plot forces as ordinates, and lengths of DE as
abscissas.1
87. Parallel Forces; Moment of Force. A force applied to a
body may act throughout the whole, or a part, of its mass; or it
may act over the whole, or a part, of its surface. A book lies
upon the table. Gravitational force acts throughout its entire
mass, whereas the force exerted upon it by the table acts only over
that portion of its surface in contact with the table. If the sur-
face over which the force acts is small compared with other dimen-
sions, the force is treated as though it acted at a single point of the
body. Thus, in Fig. 71, the post of the crane presses to the right,
1 This problem may be solved more easily by making uae of the theorem in ge-
ometry which says that homologous sides of similar triangles are in proportion. In
Fig. 72, the triangles DCE and DC'E' are similar, and
DC' C'E' DE'
DC " CE " DE'
or
10 C'E' DE'
12 * ~F " ~7~'
or
C'E' - 6J
DE' - 5.833
Solve the above exercises by this method.
166 MATHEMATICS
against the ceiling, and the ceiling presses against the crane with a
force of equal magnitude, but in the opposite direction. This
force, PS, is distributed, uniformly or non-uniformly, over the
entire surface of the post in contact with the ceiling; but this
bearing surface is so small compared with other dimensions of the
crane that the force P$ is considered acting, and is said to act, at a
point.
A weight of 10 tons is suspended from the end of the jib. Its
line of action is vertical and passes through the point E. If
the cable suspending the weight were two, three, or any number
of times as long, the weight would still exert the same influence
upon the crane. For the cable still pulls upon the crane with a
force of 10 tons and in a downward direction. The line of action
remains the same, i.e., vertical and through the point E. In
considering forces acting upon a body, the position of the line of
action, and not the exact point of application of the force, is
what enters into the solution of the problem.
It is readily seen that the line of action of the resultant of two forces
passes through the point of intersection of their lines of action.
In finding the resultant of two parallel forces it becomes neces-
sary to consider two cases: when the forces act in the same direc-
tion and when the forces act in opposite directions.
88. Two Parallel Forces Acting in the Same Direction. Let
AP, Fig. 73, be a vector representing a force, P, whose line of
action passes through the point A. Let BQ be a vector represent-
ing a force, Q, whose line of action passes through the point B.
P and Q are parallel forces acting upon a body in the same direc-
tion. The resultant of P and Q together with its line of action is
to be found. At the points A and B insert two forces, S and S',
having the same line of action, equal in magnitude but acting in
opposite directions. The resultant of these two forces is zero,
hence their introduction in no way changes the resultant force
acting upon the body. If AS and BS' represent the forces S and
iS', AP' represents P', the resultant of P and S; and BQ' represents
Q', the resultant of Q and S'. Since S and S' balance, the result-
ant of P' and Q' is the resultant of P and Q. The lines of ac-
tion of P' and Q' intersect at C. At the point C resolve the
force P' (represented by CP") into its original components P
STATICS
167
and S, represented by CP"' and CS", respectively. At the point
C resolve Q' (represented by CQ") into its original components Q
and Sf, represented by CQ'" and CS'", respectively. The forces
S and S' again balance, leaving the resultant of P and Q equal to
the vector sum of CQ'" and CP'". From similar triangles it is
seen that CQ'" is equal and parallel to BQ, and that CP'" is
-V-S'
equal and parallel to AP. Hence, the resultant of P and Q is
P + Q, whose line of action passes through the point C and is par-
allel to the lines of action of P and Q.
From similar triangles:
AP _ CK BQ _ CK
AS ~ AK a BS' ~ BK'
From these two equations, and since AS = BS', it follows that
_
BQ ~ AK'
If a is the distance of K from the line of action of P, and if 6 is
168 MATHEMATICS [§88
the distance of K from the line of action of Q, it follows from similar
triangles that
b BK
a~ AK'
Uniting this equation with the next preceding,
AP _b
BQ ~ a'
or
P _ b
Q ~ a'
or
Pa = Qb.
The product of the magnitude of a force by the distance of its
line of action from a fixed point is called the moment of the
force about the point. The point about which moments are taken
is called the origin of moments, and the perpendicular distance from
the origin to the line of action of the force is called the arm of the
force.
If the relation of the direction of the force to the origin of
moments is such as to suggest positive rotation about the origin, the
FIG. 74.
moment is considered positive. If the relation is such as to sug-
gest negative rotation, the moment is considered negative.
Thus, the moment of the force P about the point 0, Fig. 74, is
positive; the moment of the force Q about 0 is negative.
The moment of the force Q about the point K, Fig. 73, is Q 6;
the moment of the force P about the same point is — P a. The
(algebraic) sum of the moments is Q b — P a, but since in magni-
tude P a = Q b, this sum is zero. In other words, the sum of the
STATICS
169
moments of two parallel forces acting in the same direction about
any point in the line of action of their resultant is zero.
Let 0, Fig. 75, be any point in the plane. Let c be its perpen-
dicular distance from the line of action of R, the resultant of the
parallel forces P and Q; and d its perpendicular distance from the
FIG. 75.
line of action of the force Q; and e its perpendicular distance from
the line of action of the force P. Then
and since R = P + Q,
and since Pa = Qb,
c = e — a
Re = Re — Ra
Re = Pe + Qe- Pa - Qa
Re = Pe + Qe - Qb - Qa
Re = Pe + Q(e - b - a)
and since e — b — a = d,
Re = Pe + Qd.
This equation shows that the moment, about any point, of the
resultant of two parallel forces acting in the same direction is equal
to the sum of the moments of the two forces about the same point.
The proof given above considers the lines of action of the two given
forces, and the line of action of their resultant, to be upon one side
of the point 0. The student may easily show that the same law
holds when the point 0 is taken between the lines of action of the
two given forces, if by sum is meant algebraic sum.
170
MATHEMATICS
[§89
89. Two Parallel Forces Acting in Opposite Directions. Let
P and Q, Fig. 76, be two parallel forces acting in opposite direc-
tions. By analysis similar to that of the preceding section, the
student will show: that the resultant of the two forces is the alge-
braic sum of P and Q; that its line of action is parallel to that of P
and Q, and so placed that the moment of the resultant about any
point is the algebraic sum of the moments of P and Q about the
same point.
The above proof fails when P and Q are equal in magnitude, for
then AP' and BQ' are parallel and the point C is at infinity. In
FIG. 76.
this case the algebraic sum of the two forces is zero, and they can-
not be replaced by a single force whose effect upon the body is
equivalent to that of the two forces.
A system of two non-collinear parallel forces, equal in magnitude,
and acting in opposite directions, is called a couple. The product
of the magnitude of the forces by the perpendicular distance
between their lines of action is called the moment of the couple. If
the couple tends to produce positive rotation its moment is posi-
tive; if negative rotation, its moment is negative.
§90] STATICS 171
The student will show that the moment of a couple is equal to
the algebraic sum of the moments of the two forces of the couple
about any point. He will also show that the algebraic sum of the
moments of two equal parallel forces acting in opposite directions
is the same for all origins of the moments.
90. A System of Parallel Forces is Reducible Either to a Single
Force or to a Single Couple. In a system of three or more parallel
forces there must be at least two that do not form a couple.
These two forces may be replaced by their resultant force, as was
shown in §§ 88 and 89. The magnitude of the resultant is the
algebraic sum of the two components and its moment about any
point is the algebraic sum of the moments of the two forces about
the same point. In this way the system of forces is replaced by
another system having a number of forces less by one. The
algebraic sum of the forces and the algebraic sum of thier
moments about any point remain, however, the same.
By continuing this process of reducing the number of forces,
the system is eventually replaced by one consisting of only two
parallel forces. This system, if not a couple, can be replaced by a
single force, the resultant of the two forces. Thus, the resultant
of a system of parallel forces is either a couple or a single force.
The resultant can be a single couple only when the algebraic
sum of the given forces is zero, and when the algebraic sum of the
moments of all the given forces about any point is different from
zero. For, by the above method of reduction of the number of
forces of the system one at a tune, the algebraic sum of the magni-
tudes of the forces remains unchanged, and since this sum is zero
when a couple is obtained it must have been zero at the beginning;
the algebraic sum of the moments about any point remains un-
changed, and since the moment of the couple cannot be zero, the
sum of the moments of the given forces cannot be zero.
91. System of Parallel Forces in Equilibrium. A system of
forces applied to a body are said to be in equilibrium if the state of
motion of the body is unaltered by the application of the forces.
If the algebraic sum of a system of parallel forces is not zero, the
resultant of the system is a single force and the system is not in
equilibrium. One of the necessary conditions for equilibrium of a
system of parallel forces is, then, that the algebraic sum of the
172
MATHEMATICS
[§91
forces be zero. If the algebraic sum of the moments of the forces
about any point is not zero while the algebraic sum of the forces is
zero, the resultant is a couple and the body is not in equilibrium.
The necessary and sufficient conditions for a system of parallel
forces to be in equilibrium are: The algebraic sum of the forces must
be zero; and the algebraic sum of the moments of the forces about any
point must be zero.
The above law, it will be noticed, includes, as a special case, the
law of the lever.
Exercises
1. A lever 14 feet long is used for raising a weight of 500 pounds.
Where may the fulcrum be placed if a downward pressure of 100
pounds is available at the other end of the lever?
2. The two end clevis holes of a double-tree are 14 inches and 20
inches from the middle clevis hole. The team is pulling 300 pounds.
How much does each horse pull?
3. What must be the ratio of the longer arm to the shorter arm of a
double-tree: (a) if one horse is to pull 5 percent more than the other
horse ; (6) if one horse is to pull
5 percent less than the other
horse? If the team pulls 300
pounds, how much does each
horse pull in each of the above
cases ?
4. A horizontal bar, AB, 30
B feet long weighs 30 pounds per
linear foot. At the point A is
suspended a weight of 10 tons,
and at the point B is suspended
a weight of 3 5 tons. Find the
magnitude, the direction, and
the line of action of the resultant of the two weights and the gravi-
tational forces acting upon the bar.
HINT: The line of action of the gravitational forces acting upon the
particles of the bar passes through the geometric center of the bar.
5. Let AB, Fig. 77, be a plank hinged at the point A, and held in a
horizontal position by a vertical rope. The plank is 20 feet long and
weighs 110 pounds. The rope is fastened to the plank 15 feet from the
hinge. Find the tension in the rope when a man weighing 150 pounds
stands upon the plank x feet from the hinge.
FIG. 77.
§91] STATICS 173
HINT: If PI represents the tension in the rope, taking mo-
ments about A as origin gives 15Pi — (110)(10) — 150x = 0, or
PI = 10.r + 220/3. For, the system of parallel forces is in equilibrium,
and the algebraic sum of the moments about any point must be zero.
The above equation, which is linear, expresses the relation between
the tension in the rope and the distance of the man from the point A.
When the man is at the point A, the tension is 73J pounds. When he
is at the point B, the tension is 273 J pounds. P2 + PI — HO — 150 =
0, or P2 = 360 - Pi, or P2 = 286§ - 10z, which gives the reaction
at the hinge. For, by § 90 the sum of the parallel forces must
be zero. When the man is at the point B, P2 = 865 pounds.
200^
FIG. 78.
6. A weight of 200 pounds is to be raised by a lever x feet long,
Fig. 78. The distance from the weight to the fulcrum is 3 feet and the
lever weighs 4 pounds to the foot. If the lever is very short there is
but little advantage gained by its use, whereas if it is very long its
increased weight will more than compensate the gain in leverage.
Evidently there is some length which will make the force P a minimum.
2(300 + x2)
Show that P = — . Find the values of P when x equals
x
6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 17£, 18, 19, and 20. Plot
results upon a sheet of rectangular coordinate paper.
HINT: When substituting in the formula, place it in the form
»(?+•)
7. The end holes A and B of a double-tree, Fig. 79, are 2a inches
apart. The center hole C is placed d inches in front of the mid-point
of AB. If one horse is pulling ahead of the other as indicated in Fig.
79, find what percent more the rear horse is pulling than the forward
horse.
HINT:
Pi (a cos a — d sin a) = P2(a cos a + d sin a)
100(Pi - P2) _ 20Qd sin «
Pj a cos a — d sin at
174
MATHEMATICS
[§92
8. Same as exercise 7, but with a = 13 inches, d = 3 inches, and
= 30°.
9. With the statement of exercise 7, the difference between PI and
z is what percent of the load?
FIG. 80.
92. The Sum of the Moments of Two Non-parallel Forces.
Let AP and AQ, Fig. 80, represent two forces. Let AR represent
their resultant R. The moment of P about 0 is the magnitude of
§92] STATICS 175
P, (AP), multiplied by the perpendicular distance from 0 to the line
of action of P. This moment is then represented by twice the
area of the triangle OAP. Similarly, the moment of Q about 0 is
represented by twice the area of the triangle OAQ; and the mo-
ment of R by twice the area of the triangle OAR. Draw PP', RR',
and QQ' perpendicular to AO; and QB perpendicular to RR'.
The area of the triangle OAP = %(PP' -AO).
The area of the triangle OAQ = %(QQ' -AO).
The area of the triangle OAR = %(RR' -AO).
RR = RB -j- BR ,
or
RR' = PP' + QQ'.
By uniting these equations
2 (area of A OAP) + 2 (area of A OAQ) - 2 (area of A OAR) =
2(PP' + QQ' - RR') = 0,
which shows that the moment of the resultant, R, is equal to the
sum of the moments of P and Q.
In the above proof, the point 0 was so located that the algebraic
signs of the three moments were the same. The student will
prove the formula for the case in which 0 is taken between the
sides of the angle QAP. When this has been done it will be seen
that "sum" in the above law means algebraic sum.
From the above theorem and from what has been proved regard-
ing moments of parallel forces, it will be seen that the algebraic
sum of the moments of any number of forces is equal to the moment of
their resultant, and the algebraic sum of the moments of any number
of forces in equilibrium is zero.
A restatement of the above conclusions follows: The resultant
of a system of coplanar forces (not in equilibrium) is either a single
force or a couple. If the resultant is a single force, its magnitude
and direction may be found graphically by finding the sum of the
vectors representing the given forces; its line of action is such that
its moment about any point is the algebraic sum of the moments of
the several forces about the same point. If the resultant is a
couple, the vector sum of the vectors representing the forces
is zero, and the algebraic sum of the moments of the several
forces about any point is equal to the moment of the couple. For
176 MATHEMATICS [§93
a system of coplanar forces to be in equilibrium, the vector sum of
the vectors representing the several forces must be zero, and the alge-
braic sum of the moments of the several forces about any point must
be zero. The first condition throws out the possibility of the
resultant being a single force. The second condition throws
out the possibility of the resultant being a couple.
93. Equations of Equilibrium. In the preceding sections it was
shown that if a system of coplanar forces is in equilibrium, the
algebraic sum of the moments of the forces about any point must be
zero. This law is generally written in the form of an equation:
2Pa = 0
P represents the magnitude of force, a its moment arm, and 2
(the Greek letter Sigma) is a symbol meaning to take the alge-
braic sum. Another condition for equilibrium is that the vector
sum of the vectors representing the given forces must be zero.
Another way of stating this law is that if the given forces were con-
sidered concurrent, their resultant would be zero. It was shown,
however (§ 85), that one method of finding the resultant of a
system of concurrent coplanar forces is first to resolve each force
into its x- and ^-components, and from the algebraic sum (X) of
the z-components and the algebraic sum (F) of the ^/-components
compute the magnitude of the resultant (R) by the formula
R = \''X2 + F2. If R is zero, X and Y separately equal zero.
This law is generally written in the form of two equations:
2X = 0
and
27 = 0
The symbol 2 (the Greek letter Sigma) means to take the alge-
braic sum. X in the above equation means z-component. 2Z
then means the algebraic sum of the z-components. 2F means
the algebraic sum of the y-components.
The equations
2Z = 0
2F = 0
and
2Pa = 0
§93]
STATICS
177
are called the equations of equilibrium for a system of coplanar
forces.
The application of the equilibrium equations will be illustrated
by examples:
EXAMPLE 1: The crane, Fig. 71, considered as a whole, is acted
upon by four forces: viz., the downward force of 10 tons, the
upward push of the floor, P^, the push of the floor to the right,
PS; and the push of the ceiling to the left, P3. P4, Ps, and PS are
unknown in magnitude, and are to be found. Let AB = 12 feet
and FE = 4 feet. Then
2Z = P5 - P3 = 0
ZF =P4 - 10 = 0
and by taking moments about the point A,
ZPa = 12P3 - 10-4 = 0,
or PS = 10/3. The forces Pt and P5 contribute nothing to the
moment equation, for they act
through the point A about which
moments are taken; hence their
moment arms and moments are
zero.
The second equation gives P*
= 10 tons.
The first equation gives P3 =
PS = 10/3 tons. This means
that the post of the crane
pushes to the left against the
flange of the floor with a force
of 31 tons. The construction at
this point, as well as at the ceil-
ing, must be such as to with-
stand this pressure.
EXAMPLE. 2: Let AB, Fig. 81,
be a ladder 16 feet long, resting
against a smooth vertical wall,
CB. The lower end of the
ladder rests upon the horizontal ground, 6 feet from the building.
CB is 14.83 feet long. The ladder weighs 35 pounds, and a man
12
FIG. 81.
178
MATHEMATICS
[§93
who ascends weighs 150 pounds. Find the pressure of the upper
end of the ladder against the wall when the man stands on a rung
5 feet from the point A.
By taking moments about the point A,
SPa = +35-3 + 150-15/8 - 14.83 Pi = 0,
or PI = 26.04 pounds. (The weight of the ladder is considered
as a single force acting through the middle point.)
FIG. 82.
Taking moments about the point A, letting x be the distance
(measured along AB) the man stands from the point A,
SPa = 35-3 + 150?| - 14.83 Pi = 0,
or
Pi =
_ f225-
L 4
x + 105 -s- 14.83.
From this equation PI can be calculated for all values of x,
i.e., for any position of the man upon the ladder. The equation
is linear and hence its graphic representation upon a sheet of
squared paper is a straight line. In order to draw the straight
line all that is needed are the coordinates of two points. These
may be found by giving to x two values, as 0 and 16, and calcu-
lating the corresponding values of PI.
EXAMPLE 3 : Let all members of the bridge truss, Fig. 82, be
equal in length, k feet. The bridge sustains loads as indicated.
Find the reactions, PI and P2, at the points A and B. By taking
moments about the point A, it follows that
§94] STATICS 179
k 3k *>k
3k P2 - 1000 2 - 4000 k - 1000 y - 5000 2fc - 1000 ^
- 3000 -3k = 0,
or
3P2 - 500 - 4000 - 1500 - 10,000 - 2500 - 9000 = 0,
or
3P2- 27,500 = 0,
or
P2 = 9166f pounds.
SF = Pi + P2 - 2000 - 1000 - 4000 - 1000 - 5000 - 1000 -
3000 = 0,
or
P! + P2 = 17,000,
or
Pi + 9166| = 17,000,
or
PX = 7833i
Exercises
1. Find the magnitudes of the forces PI, P2, and Pa, Fig. 83, neglect-
ing the weight of the crane.
2. Find the magnitudes of the forces PI and P2, Fig. 84, also the
tension in the rope.
HINT: Draw ABC to scale, using 1/2 inch to represent a foot.
Neglect the weight of AC.
[94.] Tension and Compression. The tension or compression in
members of a construction may be easily computed, if the weights
of the members are neglected, if the loads are applied at the joints,
and if the members are discontinuous at the joints. By discon-
tinuous is meant that a single piece does not form a side of more
than one polygon. Thus, in the roof truss, Fig. 85, the tie AB con-
sists of three pieces, AD, DE, and EB, bolted together at the points
D and E.
If the loads are applied at the joints, the tension or compression
in every member is the same throughout the entire length of that
member, and the line of action of the compressive or tensile force
within the member is that of the longitudinal dimension of that
member.
180
MATHEMATICS
[§94
The method of finding tension and compression will be illus-
trated by solving the problem represented by Fig. 85.
Let AD = DE = EB = lOf feet
/\ 10 Tons
Let
and
FIG. 83.
CK = 8 feet
AC = BC = 17.88 feet
CD = CE = 9.61 feet
§94]
STATICS
181
To find the reactions, PI and Pt, of the supports use 2 Y = 0,
and 2Po = 0, taking moments about the point A.
FIG. 84.
/\ 10 Tons
or
FIG. 85.
27 = Pi + P2 - 500 - 2000 - 1000 - 1000 - 500 = 0,
P! + P2 = 5000.
182 MATHEMATICS [§94
SPa = 2000 lOf + 1000 16 + 1000-2H + 500-32 - P2-32 = 0,
or
P2-32 = 74,667,
or
P2 = 2333 pounds.
Then PI = 2667 pounds.
Imagine the members AC and AD cut by the section a and the
right-hand portion removed. In order to keep the left-hand por-
tion in equilibrium, Fig. 86, a, forces P3 andP 4, equal in magnitude
to the tension or compression in AC and AD, must be introduced
c'
A'
FIG. 86.
upon the cut ends. To express this in another way: suppose an
imaginary plane through the rafter AC divide it into an upper and
a lower part. The part below pushes against the part above, and
the part above pushes against the part below; the two compressive
forces are equal in magnitude and this magnitude is called the
compression within the rafter. If the upper part of the member
were removed, and if a force equal to the compression were
introduced acting upon the remaining cut end oi AC and in a direc-
tion toward A, the lower part of the rafter would be acted upon by
the same force as before the upper portion was removed.
The four forces acting upon the remaining portion of the truss,
Fig. 86, a, are in equilibrium. The two vertical forces acting
at the point A may be considered as one force acting up, of magni-
tude 2167 pounds. Draw the line C'K', Fig. 86, 6, to scale,
representing the upward 2167-pound force. C'K' is parallel to
CK, Fig. 85. Through C' draw a line parallel to AC and through
K' draw a line parallel to AK. These two lines intersect at A'.
§94] STATICS 183
K'C'A' is then the force polygon for the three forces, 2167,
Pa and P\. C'A' is the vector representing the force PS, and A'K'
is the vector representing the force Pi. The magnitudes of PS and
P* may be scaled off from this triangle. Since K'C'A' is a closed
polygon, the arrows run around the polygon in the same sense.
The arrow of C'A' then points from C" toward A', which says
that the arrow of PS, Fig. 86, a, was chosen to point in the proper
direction, or the member AC is in compression. Similarly, the
arrow of P\ was chosen in the proper direction, or the member AD
is in tension.
The magnitudes of PS and Pi may be found by another
method which does not require drawing the triangle K'C'A' to
scale. The triangles K'C'A' and KG A, Fig. 85, are similar, and
C'A' _ K'C't
CA == KG
or
C'A1 2167
17.88 " 8 '
or
C'A' = 4842,
PS = 4842 pounds.
Again :
A'K' _ K'C^
~AK ''= KG'
or
A'K^ _ 2167
16 8 '
or
A'K' = 4333,
P4 = 4333 pounds.
To find the tension or compression in the members CD and DE,
consider the portion about the point D removed by the cut 6,
Fig. 85.
184
MATHEMATICS
At the cut ends insert the forces PS and P6, Fig. 87, a, equivalent
to the tension or compression in the cut members. The force
polygon for the four forces is represented by the triangle K'C'D',
Fig. 87, b. This triangle is similar to triangle KCD, Fig. 85.
4333
Hence
or
or
Again:
or
43337)
2000^
FIG. 87.
D'C' C'K'
DC
D'C'
9.61
CK'
2000
D'
, ,2000^
K'
D'C' = 2402.5,
P5 = 2402.5 pounds tension.
K'D' 2000
KD
M'D' + 4333
8 '
2000
or
M'D' = - 3000 pounds.
The negative sign shows that the arrow of P& points in the wrong
direction.
P& = 3000 pounds tension
In a similar way by means of cuts c and d, Fig. 85, the tension or
compression in EB, BC, and EC may be found.
§95]
STATICS
Exercises
185
1. Find the tension or compression in the members BC, BE, EC,
and ED, of the truss given in Fig. 85.
2. Find the tension or compression in each member of the truss
given in Fig. 82.
95. Rope and Pulley. Let PI and P2, Fig. 88, be the tensions
in a rope upon opposite sides of a pulley over which it passes. If
FIG. 88.
there be no friction between the pulley and the axle, the line of
action of the force exerted by the axle upon the pulley will pass
through the center of the pulley. Taking moments about the
center of the wheel (calling a the radius of the wheel)
Pia-P2a = 0, orPi=P2
FIG. 89.
This says that if a flexible rope passes over a series of friction-
less pulleys, the tension at every point of the rope is the same.
In Fig. 89, the pull on each end of the rope is the same; and if the
186
MATHEMATICS
[§95
rope be cut at any point, K, a force equal to P must be introduced
upon each free end of the rope in order to hold the system in
equilibrium.
Fig. 90 represents a system of pulleys by which a pull, P, holds
a weight of 10 tons in equilibrium. To find the tension in the
rope passing around pulley E, imagine the rope cut by the section
/\ 10 Tons
FIG. 90.
a, and forces introduced upon the cut ends equal to the magnitude
of the tension in the rope. Represent these forces by PI. The
forces acting upon the part of the body below the section are in
equilibrium, and PI + PI — 10 = 0, or PI = 5. The tension in
the rope passing about pulley E is 5 tons.
If a second cut 6 be made, and if P2 represent the tension in the
rope passing about pulley F, P2 + P2 — P: = 0, or P2 + P2 —
5= 0, or P2 = 2J. The tension in the rope passing about pul-
ley F is 2£ tons.
§95]
STATICS
187
If a third cut c be made, and if P represents the tension in the
rope passing around pulleys G and H , P + P — P2 = 0, or
P + P - 2f = 0, or P = l£. The pull, P, necessary to hold
the weight 10 tons in equilibrium is lj tons.
Exercises
1. Find the pull P which will hold, by means of the system of
pulleys shown in Fig. 91, the weight of 10 tons in equilibrium. Find
the pull on the hooks A, B, and C.
10 Tons
FIG. 91.
2. Find the pull P which will hold, by means of the system of pulleys
shown in Fig. 92, the weight of 10 tons in equilibrium. Find the pull
on the hook A.
3. The differential wheel and axle shown in Fig. 93 consists of three
drums of radii a, b, and c. A force P is applied to a rope which unwinds
from the largest drum. As the drums rotate a rope unwinds from the
188
MATHEMATICS
[§95
smallest and winds up upon the medium sized drum, lifting a weight of
10 tons. Find P which will hold the weight in equilibrium. As P
moves down x feet, how far does the weight rise?
y' ~\10 Tons
FIG. 92.
/\ 10 Tons
FIG. 93.
CHAPTER VII
PERMUTATIONS, COMBINATIONS, AND THE BINOMIAL
EXPANSION
96. A Fundamental Principle. From a recitation room there
are two doors leading into the hall; from the hall there are three
exits. In how many ways can a person leave the building from
the room? Let the doors of the room be numbered 1 and 2, and
let the exits from the hall be marked A, B, and C. A person may
pass through door 1 and exit A; door 1 and exit B; door 1 and
exit C; door 2 and exit A ; door 2 and exit B; or door 2 and exit C.
After passing through door 1 there are three possible paths: one
through exit A; one through exit B; and one through exit C.
There are exactly the same number of paths passing through door
2. In all there are 2 X 3 or 6 ways of passing from the building.
If there were ra doors leading from the room into the hall, and
n exits from the hall, obviously there would be m X n ways of
leaving the building.
The following is a fundamental principle: // one thing can be
done in m ways, and if a second can be done in n ways, the two things
can be done in the order indicated in m X n ways.
It is obvious that the order in which the two things are done does
not affect the number of ways in which they may be done. They
may even be done simultaneously. To illustrate: suppose a
penny and a die are thrown simultaneously; the penny will, fall
in one of two ways; the die will fall in one of six ways; then, the
two will fall simultaneously in one of 2 X 6, or 12 ways. This
illustration shows that the fundamental principle applies not only
to the number of ways of doing two things but also to the number
of ways two events may occur.
It will be seen at once that if one thing can be done in m\
ways, a second in m* ways, a third in m$ ways, and so on, the num-
ber of ways they can be done together is m\ X m^ X m$ X . . .
189
190 MATHEMATICS [§97
97. Permutation and Combination Defined. Suppose we have
five cubes, one painted white, one black, one red, one blue, and
one green. From these five a selection of one is made. This can
be done in five ways. Each time a selection of one is made, a
selection of four is left consisting of a different combination of
colors.
A selection of two cubes is made. This may be done in ten
ways, as:
white and red, black and blue,
white and black, black and green,
white and blue, red and blue,
white and green, red and green,
black and red, blue and green.
Each time a selection of two cubes is made, a selection of three
cubes remains. Then a selection of three cubes from the five
cubes may be made in ten ways. A selection of four cubes may be
made in five ways, and a selection of five cubes may be made in one
way.
If from n things, a group of r (r ^ ri) are selected, the selection
is called a combination of the n things taken r at a time. The
symbol representing the number of combinations that can be made
of the n things taking r at a time is nCT. Thus, from the previous
illustrations, 5(7i = 5; £2 = 10; 5Ca — 10; £4 = 5, and 5^5 = 1.
Consider again the five colored cubes. Select two by first pick-
ing out one and then a second. This may be done in twenty ways
if the order of the selection is considered (the selection first black,
then white, is considered different from the selection first white,
then black). For, by the fundamental principle, the first selec-
tion may be made in five ways, and the second selection in four
ways; the total number of selections possible is 4 X 5 = 20.
Any selection of r (r ^ n) things taken from n things which
takes into consideration the order of selection is called a permu-
tation of n things taken r at a time.
The symbol representing the number of permutations of n
things taken r at a time is nPr.
It is to be remembered that with permutations we consider the
order of selection, or the arrangement, of the objects as well as the
PERMUTATIONS, COMBINATIONS, ETC. 191
individuals which go to make up the selection, while with combina-
tions we consider only the individuals of the selection without
reference to the order of their selection or arrangement. To
illustrate: If in a plane there are five points, no three of which are
upon one straight line, the problem of finding the number of
lines determined by the points is a problem in combinations.
For it is immaterial whether we think of the line drawn, for ex-
ample, from the point A to the point B, or from the point B to
the point A.
From the five digits 1,2, 3, 4, and 5, how many products may
be formed, using two and only two digits at a time? This is a
problem in combinations, for example, the product 3 X 2 is the
same as the product 2 X 3.
From the five digits 1, 2, 3, 4, and 5, how many two-figured
numbers may be formed? This is a problem in permutations;
for example, the number 32 is different from the number 23.
Exercises
1. Now many products can be formed from the digits 1, 2, 3, 4, and
5 taking (a) two and only two digits to form a product; (6) three and
only three digits to form a product; (c) four and only four digits to
form a product; and (d) five digits to form a product?
2. How many lines may be drawn through five points in a plane,
no three of which are upon one straight line?
3. How many planes are determined by five points in space, no
four of which are on one plane?
4. How many points of intersection are determined by five straight
lines on a plane, if no three intersect in the same point?
5. How many numbers can be formed from the digits 1, 2, 3, 4, and
5 (a) if two and only two digits are used in each number; (6) if three
and only three digits are used in each number; (c) if four and only four
digits are used; (d) if five and only five digits are used?
98. Formula for nPr. Let a, b, c, d, e, f be permuted two at a
time. The first letter may be selected in six ways, i.e., the first
letter selected may be the a, the b, the c, the d, the e, or the /.
When the first letter is selected, the second letter may be selected
in five ways from the remaining five letters. Then, by the funda-
mental principle, the number of ways two letters may be selected
from six letters is 6 X 5, or 30. Thus, ^ = 6 X 5 = 30.
192 MATHEMATICS [§98
If three letters are placed in each permutation, the first can be
selected in six ways, the second in five (6 — 1) ways, and the
third in four (6 - 2) ways. Thus, ePs = 6X5X4= 120.
If four letters are placed in each permutation, the first can be
selected in six ways, the second in five ways, the third in four ways,
and the fourth in three ways. Thus, ^4 = 6X5X4X3 = 360.
The student will show, by reasoning similar to the above, that
ePs = 6X5X4X3X2 = 720, and ^6 = 6X5X4X3X2
X 1 = 720.
If n things are permuted r at a time, the first selection is made
from the n things, or it may be made in n ways; the second
selection is made from the remaining n — 1 things, or it may be
made in n — 1 ways. Then, by the fundamental principle, the
first two things may be selected in n(n — 1) ways. The third
selection is made from the remaining n — 2 things, or it may be
made in n — 2 ways. Thus, the first three things may be selected
in n (n — !)(« — 2) ways. Continuing in this way the first four
things may be selected in n(n — l)(n — 2}(n — 3) ways, the
first five things in n(n — \)(n — 2)(n — 3)(n — 4) ways, the
first six things in n(n — l)(n — 2)(n — 3)(n — 4)(n — 5) ways,
and so on. The r things may be selected in
n(n - l)(n - 2)(n - 3) . • • [n - (r - 2)][n - (r - 1)]
ways. Therefore
nPr = n(n — l)(n — 2)(n — 3) • • -to r factors,
or nPr = n(n - l)(n - 2)(n - 3) • • • (n - r + 1) (1)
If r = n, formula (1) reduces to
nPn = n(n — l)(n — 3) • • • to n factors, or
nPn = n(n - l)(n - 2)(n - 3) • • • 3 X 2 X 1
nPn = n_ (2)
where J^ (read "factorialn") means
1X2X3X4 • • • (n - 2)(n - l)n, or the product of all the
positive integral numbers from 1 up to and including n.
The symbol nPn, the number of permutations of n things taken
all (n) at a time, will be written Pn.
From formula (1),
n-iPr-i = (n - l)(n - 2)(w - 3) • • • (n - r + 1)
§99] PERMUTATIONS, COMBINATIONS, ETC. 193
found by replacing n by n — 1, and r by r — 1. Then
nPr = n(n_1P,_1). (3)
By multiplying both numerator and denominator of the right-
hand side of formula (1) by n—r , it reduces to
,P,= -l^-. (4)
n — r
Exercises
1. Solve exercise 5 of the preceding section by formula (4).
2. How many different signals can be given with seven flags of
different colors, by hoisting them in a vertical line, any number at a
time?
3. How many different numbers can be formed from the digits
1, 2, 3, 4, 5, 6, if a digit cannot be used more than once in any one
number?
4. From twenty men how many nines can be formed, assuming
that every man can play any position? By nine is not meant merely
the aggregate of nine men, but has reference as well to the positions
played.
5. Same as exercise 4, excepting that only two can pitch and that
only three can catch.
6. Same as exercise 5, excepting that the pitchers and catchers
play no other positions.
99. Formula for nCr. Within each combination of n things
taken r at a time there are _£_ permutations. The number of
permutations of n things taken r at a time is equal to the product
of the number of combinations of n things taken r at a time by the
number of permutations within each combination taking r at a
time, or
npr =
or,
or
1:3
n — r
194 MATHEMATICS [§100
Every time a combination of r things is selected from n things, a
combination of n — r things remains. Thus, the number of com-
binations of n things taken r at a time is equal to the number of
combinations of n things taken n — r at a time. This fact can be
shown from formula (1) by replacing r by n — r, when it reduces to
nCn-r =
n — r
n
n—
r
n-n + r
a formula identical with (1).
Exercises
1. Solve exercises 1, 2, 3, and 4, § 97, by formula (1).
2. There are twenty points in a plane, no three of which are
upon the same straight line, excepting five, which are upon one
straight line. How many lines are determined by the points?
100. The Binomial Theorem. The product of the n binomials,
(fli + bi)(az + 62) (as + 63) • • • (a» + bn), is the sum of the
partial products formed by taking, in as many ways as possible,
one and only one letter from each of the parentheses.
When the binomials are all equal, the product becomes (a + &)
(a -f 6) . . . to n factors. By taking the first letter from each of
the parentheses the partial product a" is obtained. By taking a
b from one of the parentheses, and an a from the remaining n — 1
parentheses, a partial product, an~lb, is obtained. Several par-
tial products of this form may be obtained by selecting the b from
different parentheses. In fact, it may be selected in n ways,
giving n partial products of the form an~l b. By taking the b from
two of the parentheses, and the a from the remaining n — 2 par-
entheses, a partial product, an~z b-, is obtained. Several partial
products of this form may be obtained by selecting the b's from
different pairs of parentheses. The number of partial products
which may be formed in this way is equal to the number of ways in
which two b's can be selected from n b's, or, nCz. In the same way
it may be shown that the number of partial products of the form
an-s £3 js B(73- ^0 number of partial products of the form an~4, b*
is nCi, etc. Thus,
(o + b)n = an + nan-lb + nCzan~2b2 + nC3an~3b3
+nCtan-W + . . . +bn,
§101]
or
PERMUTATIONS, COMBINATIONS, ETC.
195
-(---JV^-C-lKn-Vv
n(n - l)(n - 2)(n - 3)
an-4b4
bn.
This is known as the binomial expansion, or binomial theorem.
The coefficients of the right-hand side of the equation are called
the binomial coefficients.
For n = 2, (a + 6)2 = a2 + 2ab + b\
For n = 3, (a + fe)3 = a3 + 3a26 + 3a62 + b3.
For n = 4, (a + 6)4 = a4 + 4a36 + 6a262 + 4a63 + 64.
For n = 5, (a + 6)5 = a5 + 5a46 + 10a362 + 10a263 + 5a64 + &5.
(2
6
X4
(2)3(3x2)
6X5X4
6X5X4X3X2Xl
(3x2)6>
or
(2 + 3x2)6 s 64
576x2
4320x6 + 4860a;8
If 6 is negative, it will be seen that every alternate term of the
expansion, beginning with the second, is negative.
Exercises
Expand the following :
1. (a - 6)*. 4. (x2 - 3)4.
2. (2 + 6)6. 5. (x* + 3z2)6.
3. (2 - 3x)6. 6. (x* + x-1)6.
101. The Binomial Expansion for Negative and Fractional
Exponents. In the preceding section the expansion was given
for a binomial with a positive integral exponent. It is shown in
higher mathematics that the expansion also holds for fractional
196 MATHEMATICS [§102
and for negative exponents, providing that the numerical value of b
is less than the numerical value of a.
For a positive integral exponent the binomial expansion consists
of a finite number of terms, n + 1. For fractional or negative
exponents the number of terms is infinite, or the expansion is said
to be an infinite series.
ILLUSTRATION 1: Expand (1 + a;)"1.
-
^ ^(_l)-4(a.)S +
or
ILLUSTRATION 2: Expand (1 + a;)* .
(1/2} (1/2 — 1}
(I _i_ xy = (1) + (1/2) (lr a; 4- (1)~^ (a;)2
2
or
Exercises
Expand the following to four terms:
1. (1 + x)U. 4. (1 - x)-1.
2. (1 - x)K. 5. (1 + x)~*.
3. (1 + x)H. 6. (1 - x)~*.
102. Approximation Formulas. If a; is numerically very small the
expansion
n(w — 1) n(n — l)(n — 2)
(1 + x)n ~ 1 + nx + ^—^ — - a;2 + - I 3 - x3 + • • •
is approximately equal to
1 + nx.
For, a;2, a;3, . . . , are higher powers of a small fraction and hence
§103] PERMUTATIONS, COMBINATIONS, ETC. 197
small compared with x. If the symbol = represents " approxi-
mately equal,"
(l + x)" = l + nx (1)
if x is numerically very small. Thus, to illustrate,
(1 + 0.06) w = 1 + M0.06) = 1.02.
The true value of this expression is, to five decimal places, 1.01961.
Again:
(1 - 0.06) * = 1 - 1(0.06) = 0.98.
If x, y and z are numerically very small, the student will show
by multiplication and division that the following are approxi-
mate formulas:
(2)
(3)
(1 + x)(l + y)(l + z) = 1 + x + y + z. (4)
It is to be remembered that the above formulas hold for nega-
tive values of x, y, and z as well as for positive values.
Thus, by formula (2)
(1 + 0.03) (1 - 0.05) = 1 + 0.03 - 0.05
= 0.98.
Exercises
Find the approximate value of each of the following:
1.03
2. (0.98) . '* 1.02'
H _ 1.03
3. (0.93) . 7- 0.97*
4. (l.Q3)(1.02). 8. (1.04) (1.06)(0.95).
103. The Compound Interest Law. If a dollars are loaned at r
percent per annum, compound interest,
at the end of one year the amount is a 1 + JQQ '
198 MATHEMATICS [§103
f r "12
at the end of two years the amount is a 1 + JQQ '
[r "is
1 + JQQ '
[r ~~\*
1 + TQQ
If the interest is compounded semi-annually instead of annually the
amount at the end of x years is
[r "12*
1 + 20oJ
if compounded monthly the amount at the end of the same period
is
[r "1 12*
f I200J
and if compounded n times a year the amount is
[
1
Now, if we find the limit of this expression as n increases indefi-
nitely, we find the amount if the interest were compounded con-
tinuously. For convenience let ,^,. be represented by — • Then
the amount, y, is1
lira TI ! 1^
y = a 1 + - 100
n = oo L u J ,
which may be written
u -
since a, r, and x are independent of n. By the binomial theorem,
n u(u — l)fln2
wJ
:m!
LwJ
lim
n = oo
. C7 means the limit of U as n increases indefinitely.
§103] PERMUTATIONS, COMBINATIONS, ETC. 199
Since u becomes oo as n becomes oo , equation (2) gives
i™ r1+i>=1 + 1+ i + i + i_ + i + - -(3)
u = oo L uj A A °
The sum of the infinite series on the right-hand side of (3) is
represented by e and is approximately equal to 2.71828. It is
the base of the natural or hyperbolic system of logarithms (see
footnote, page 66). Equation (1) then reduces to
TX
y = ae Too. (4)
Since r is any constant, r/100 is any constant, call it 6. Then
equation (4) may be written
y = ae>*, (5)
where a and b are constants. It is the equation then which
expresses the law: the rate of growth of a variable, y, is directly pro-
portional to the variable itself. The multiplier a is the value of the
variable when time, x, is zero, and 6 is the proportionality factor,
i.e., it expresses the ratio of the rate of growth of y to y.
The compound interest law is one of the very important laws of
nature.
ILLUSTRATION: Suppose it is found by experiment that 10 per-
cent of light is absorbed in passing through a pane of glass 1 cm.
thick. Then in passing through a second pane 1 cm. thick, 10
percent of the remaining 90 percent will be lost, and so on. The
intensity of light in passing into the glass obeys then the com-
pound interest law,
y = aebx
where y represents intensity of light and x distance into the glass.
When x = 0, y = a; thus, a is the intensity of the light as it falls
upon the glass surface. When x = 1, y = 0.9a, hence
0.9a = aeb,
or
0.9 = eb
or
b = log«0.9
Then with the constants a and 6 determined, the intensity of the
light at any depth below the surface is given by
aelog(0-9^ = a(0.9)x
CHAPTER VIII
PROGRESSIONS
104. Arithmetical Progression Defined. An arithmetical pro-
gression is a sequence of numbers such that any number of the
sequence except the first may be obtained from the preceding
by the addition (or subtraction) of a common number, called
the common difference. The following are arithmetical
progressions:
1. 2, 4, 6, 8, 10, 12. The common difference is 2.
2. 2, 5, 8, 11, 14. The common difference is 3.
3. — 7, — 2, 3, 8, 13. The common difference is 5.
4. — 4, —2, 0, 2, 4, 6. The common difference is 2.
5. 2^, 2, If, 1, f, 0, — f. The common difference is —1/2.
Exercises
From the following sequences select those which are arithmetical
progressions, and give the common difference:
1. 5, 7, 9, 11, 13. 5. x - y, x, x + y.
2. 2, 4, 8, 16. 6. x, xy, x + y.
3. 10, 8, 6, 4, 2. 7. 0, - 3, - 6, - 9.
4. 3, 2f, 2i 2, If. 8. a, a + d, a + 2d.
105. The General Term; the Sum of an Arithmetical Pro-
gression. Let a represent the first term, I the last term, d the
common difference, n the number of terms, and s the sum of all
the terms.
The general series is then
a, a + d, a + 2d, a + 3d, . . . , a + (w — l)d.
It is obvious that the coefficient of d in any term is one less than
the number of the term. Thus, in the fourth term it is 3, in the
twelfth term it is 11, and in the nth term it isn — 1.
A formula for the last term is then
1 = a + (n - l)d. (1)
200
§105]
PROGRESSIONS
201
The sum, s, is found by adding the n terms together, as
s = a + (a + d) + (a + 2d) + • • + [a + (n -
or
s = na+d[l + 2 + 3 + 4+ • . . +(ro-l)]. (2)
The sum of the n — 1 terms written within the brackets may be
represented graphically as follows: Let AoAn-i, Fig. 94, be a line
n —1 units long, divided into n — 1 equal parts by the points
A i, A2, As, etc. Upon the segments AoAi, A\Az, A2AZ, . . .,
An-zAn-i, construct rectangles with altitudes respectively 1, 2, 3,
Bo
An.2 An.\
FIG. 94.
4, . . . , n — 1 units long. The areas of these rectangles are,
respectively, 1, 2, 3, 4, . . . , n — 1 square units. The areas of
these rectangles may then be taken as representing the terms
within the brackets of equation (2) , and the sum of the areas of
the rectangles representing the sum of the terms within the
brackets.
Draw A 0C.
An-lC = AfiAn-1
The angle CA0An.l = 45°
The area of the triangle A^An^C = %(n — I)2
The area of each shaded triangle above the line A<>C is 1/2.
202 MATHEMATICS [§105
The sum of the n — 1 triangles is, then, %(n — 1). The sum
of the n — 1 rectangles is then
Kw- 1)2+ K»- 1) = *(«- l)n.
Therefore
«
Substituting in (2) ,
TO (TO — 1)
s = na-\ ---- - d,
or
s = °[2a+(n-l)d]- (3)
From formula (3) the sum of an arithmetical progression may
be calculated if the first term, the common- difference, and the
number of terms are given.
Formula (3) may be written.
* * 2 a+ o+ (n — l)d >
or, since a + (n — l)d = I,
' s= *(a+Z). (4)
ILLUSTRATION 1: Find the sum of 11, 22, 33, . . . to twenty
terms.
This is an arithmetical progression, for any term is eleven
greater than the next preceding term, o = 11, d = 11, TO = 20.
Substituting in formula (3),
s = jpJ2fl + (n - l)d\ ='~ [2-11 + (20 - l)ll] = 2310.
ILLUSTRATION 2 : Find the sum of \/2 + 2 \/2 + 3 \/2 4- . . .
to eleven terms.
a = V% d = \/2, and n = 11.
s = y[2A/2+ (11 - 1)V2J =66 V2.
ILLUSTRATION 3: Find the sum of 6 + 5i + 4f + 4 + . . .
to fifty terms.
a = 6, d = — f, and n = 50.
§107] PROGRESSIONS 203
Exercises
1. Find the sum of — 1 — 3£ — 6 . . .to twelve terms.
2. Find the sum of -£• + -5- + \ + . . .to eighteen terms.
3. Find the sum of the first 100 odd numbers.
4. Find the sum of the first 100 even numbers.
5. A man saves $200 each year, and at the end of each year places
this amount on simple interest at 6 percent. What is the amount of
his savings at the end of ten years?
6. For drilling a well a contractor is to receive 10 cents for the first
foot, and for each foot thereafter 1/2 cent more than for the pre-
ceding foot. What does he receive if the well is 600 feet deep?
106. Geometrical Progression Defined. A geometrical pro-
gression is a sequence of numbers such that any number of the
sequence, except the first, may be obtained from the preceding
term by multiplying by a common number, called the ratio.
The following are geometrical progressions:
1. 2, 4, 8, 16, ... The ratio is 2.
2. 2, - 4, 8, - 16, . . . The ratio is - 2.
3. 2, 2^2, 4, 4 V2~ . . . The ratio is \/2.
4. "s/3, - 3, 3 -v/3", - 9, ... The ratio is - \/3,
5. a, ar, ar2, ar3, . . . The ratio is r.
107. The Last Term; the Sum of a Geometrical Progresssion.
Let a represent the first term, I the last term, r the ratio, n the
number of terms, and s the sum of the terms of a geometrical
progression.
The general sequence is, then, a, ar, ar2, ar3, . . . , ar*"1. It
is obvious that the exponent of r in any term is one less than
the number of the term. Thus, in the fourth term it is 3, in the
twelfth term it is 11, and in the nth term it is n — 1.
A formula for the last term 's then
1 = ar"-1. (1)
The sum, s, is found by adding the n terms together, as,
s = a + ar + ar2 + ar3 + • • • + arn~l,
or
* = o(l + r + r2 + r3 + • '• • + r"-1). (2)
204 MATHEMATICS (§107
Since
1 + r + r2 + r3 + ' ' ' + rn~l = 1 ^r »
and equation (1) becomes
3.(.l rnj ff.\
s = t _- — (3)
ILLUSTRATIONS: 1. Sum the progression 2 + 4 + 8 + • • •
to ten terms.
a = 2, r = 2, and n = 10.
«d ~ r-) _ 2(1 - 2iQ) _
1-r 1-2
2. Sum the series \/2 + 2 + 2 \/2 + • • • to ten terms.
a = -\/2, r = A/2, and n = 10.
V2[l - ( V2)10] = 3lV2
1 _ ^/2 V2 - 1
3. Sum the series \/3 - 3 + 3 V3 - • • • to ten terms.
a = A/3~, ^ = — V3, and n = 10.
Exercises
1. Sum 3 + 9 + 27 + • • • to ten terms.
2. Sum — 3+6—9+ • • -to ten terms.
3. Sum 3 — 1 + 1/3 — • • • to six terms.
4. Sum — 3 + 6 + 15 + • • -to seven terms.
6. Sum \/3 + 2A/3 + 4 V3 + • • -to seven terms.
6. Sum \/3 + 2\/3~+ 3\/3 + • • • to seven terms.
7. Sum 2 — V2 + 1 — • • • to six terms.
8. Find the eleventh term in each of the above series.
1 An exception is made when r = 1, but in this case 1 + r + r2 + r3 +
r""1 = n.
§1()S] PROGRESSIONS 205
9. Show that the only right triangles whose sides are in arithmet-
ical progression are those whose sides are proportional to 3, 4, and 5.
HINT: Let x — k, x, and x + k represent the sides. Find k in terms
of x, and build up the progression.
10. A debt of $10,000 is to be paid in ten years. An equal amount
is to be paid at the end of each year. Find this amount if the indebted-
ness draws interest at 6 percent.
11. On a certain twenty-year life insurance policy $32 premium
is paid annually. What do the premiums amount to by the end of
the twentieth year, interest compounded annually at 3^ percent?
12. An equal amount of money is deposited at the end of each
year for twenty years as a sinking fund to replace a piece of machinery
valued at $10,000. How much must be deposited each year if the
deposits draw 4 percent compound interest?
108. The Infinite Geometrical Progression. From the formula
for the sum of a geometrical progression, s = — ^ , it is
evident that, if r is greater than unity in numerical value, the
numerical value of s increases without limit as r increases without
limit. For, as n grows very large, rn grows large without limit,
causing s to grow large without limit.
If r, however, is numerically less than unity, rn grows smaller
and smaller as n increases, and approaches zero as n increases
without limit. Thus, the quantity within the parentheses
approaches 1, or the limit of s is a/(l — r) as n increases without
limit. The sum, s, of a geometrical progression approaches a
definite finite limit as n increases without limit, when r is nu-
merically less than unity. This limit, a /(I — r), is called the
sum of the infinite geometrical progression.
ILLUSTRATIONS: 1. Find the sum of the infinite progression
1 + £ + i + i + ' • • to oo.
a = 1 and r = %.
* -o
• 1-r 1-t
2. Find the sum of the infinite progression \/3 + 1+1 \/3 —
' to 0°.
a = V3 and r = I/ \/3.
a \/3 3 = 3(V'3 + 1)<
\/3 - 1 = 2
206 MATHEMATICS [§108
3. Express the repeating decimal 0.237237237 ... in frac-
tional form.
237 237
The decimal fraction 0.237237237 . . . means ^^ + (1QOQ) 2
237
+ • • •, an infinite geometrical progression, in which
a = 237 /1000, and r = 1 /1000. Substituting in the formula for
the sum of an infinite geometrical progression,
237
1000 237 237
1 1000 - 1 == 999'
~ 1000
Exercises
1. Find the sum ofl— £ + i — |+ . . . to <» .
2. Find the sum of \/2 — ?\/6 + i\/2 — . . . to °° .
3. Find the sum of — 7= -^ + 1 — — 7=- — ;= + . . to oo .
\/2 + \/3 \/2 - \/3
4. Express 0.7261872618 ... in fractional form.
6. The middle points of the sides of a square, 10 inches on a side,
are joined forming a smaller square. The middle points of the sides
of this square are joined forming a third square. This process of form-
ing smaller squares is continued indefinitely. Find the sum of the
perimeters of the squares.
6. Same as exercise 5, but use an equilateral triangle 10 inches on
a side.
CHAPTER JX
'[PROBABILITY]
109. Probability Defined. When a coin is thrown upon the flat
surface of a table, it falls in one of two ways, heads or tails. If it
is given a whirling motion when thrown, the chances of its falling
heads are no greater than the chances of its falling tails, and the
chances of its falling tails are no greater than the chances of its
falling heads; that is, before the coin fell, we had no reason for
expecting one rather than the other. All that could be said was
that the coin must fall in one of two ways, both equally likely.
When a common die is thrown it falls in one of six possible ways,
all equally likely. One and only one of the six ways is that of the
ace up. The chances then of throwing an ace with one throw of a
die is one in favor and five against, or one to five, or one out of
six. This same thought is expressed by saying that the probability
of throwing an ace with one throw of a die is one to six, or 1 /6.
If a thing can happen in n ways, and can fail to happen in m
ways, it can happen or fail in just m -\- n ways. We say there
are m + n possible cases, m favorable cases, and n unfavorable
cases. If all these ways are equally likely to occur, — ^p — (the
ratio of the favorable cases to the possible cases) is called the
probability of the event happening. — j- — (the ratio of the
unfavorable cases to the possible cases) is called the probability
of it failing.
When a coin is thrown, the probability of heads turning is
1 /2; for the number of favorable cases is one, and the number of
possible cases is two. The probability of heads not turning is 1 /2.
When a die is thrown, the number of possible cases is six, the
number of favorable cases of the ace falling up is one. The proba-
bility of throwing an ace is then 1/6. The probability of not
throwing an ace is 5 /6.
207
208 MATHEMATICS [§109
In a set of twenty-eight dominoes there are seven doublets.
When one domino is selected at random the probability of its
being a doublet is 7/28, or 1/4. For, the selection of one
domino may be made is twenty-eight ways, or the number of
possible cases is twenty-eight; and since a doublet may be
selected in seven ways, the number of favorable cases is seven.
When two dice are thrown the number of ways in which they
may fall is 6 X 6 = 36. Of these thirty-six ways, one, and one
only, is a pair of aces. The probability of throwing a pair of aces
with two dice is then 1 /36. The probability of throwing a pair
is 6/36, or 1/6. The probability of throwing a sum less than
5 is 1 /6.
If an event is certain to happen, the number of favorable cases
is equal to the number of possible cases, and the probability of
its happening is one. If an event is certain to fail the number of
favorable cases is zero, and the probability of its happening is
zero. In all other cases the probability of an event happening
is a positive proper fraction.
If an event may happen in a ways, and fail in b ways, the
Q
probability of its happening is - — r» and the probability of its
a. ~T~ D
failing is — 7— r-« The sum of these two probabilities is 1.
Exercises
1. Two dice are thrown. What is the probability of throwing a
5 and a 3? A sum greater than 7? A sum equal to 10? Conse-
cutive numbers? One and only one 6?
2. A box contains three white, two black, and five red balls; if one
is drawn, what is the probability it is white? Black? Red? If two
are drawn, what is the probability that both are white? Both black?
Both red? One white and one black? One white and one red?
One black and one red?
3. Three coins are thrown. What is the probability that all turn
heads? That two turn heads? That only one turns heads?
4. In a box there are five red, two white, and seven blue balls.
If three are drawn one after the other, what is the probability of
their being drawn in the order red, white, and blue?
§110] PROBABILITY 209
HINT: The total number of ways, in order, in which three balls
can be drawn from fourteen balls is 14P 3 .
6. Four cards are drawn from a pack. What is the probability
that there is one from each suit?
110. Mutually Exclusive Events. Two or more events are said
to be mutually exclusive if the occurrence of one excludes the
occurrence of the others. Thus, a 5 and a 6 cannot be thrown
together with one die. The occurrence of one event excludes the
occurrence of the other. The two events are mutually exclusive.
Let A and B be two mutually exclusive events. There are three
possible cases:
(1) The event A may happen and the event B fail.
(2) The event B may happen and the event A fail.
(3) The event A and the event B may both faiJ.
If a, b, and c are, respectively, the number of equally likely
possible cases, the probability that either A or B happens is
— . For a + 6 + c is the number of possible cases, and
a + o + c
a + b is the number of favorable cases. The probability that A
happens is — — , and that B happens is — — . Hence
a + 6 + c' a + 6+c
the probability of the occurrence of one or the other of two mutually
exclusive events is the sum of the probabilities of the two separate
events.
It is easily seen that this law may be extended to any number
of mutually exclusive events. The above law applies only to
mutually exclusive events, and care must be exercised not to
apply it to events not so related. Thus, if A's probability of
winning a contest is 1 /3, and if B's probability of winning in the
same contest is 1 /4, the probability of A or B winning is 5- + j- =
o 4
7
^. For the two events are mutually exclusive. If A's probability
LZi
of solving a problem is 1 /3, and B's probability of solving the same
problem is 1 /4, the probability of the problem being solved is not
the sum of the separate probabilities, for the events are not
mutually exclusive. The solution of the problem by A does not
prevent B from solving it.
14
210 MATHEMATICS [§111
Exercises
1. The probability of throwing an ace with a common die is |;
the probability of throwing a 6 is f . What is the probability of
throwing an ace or a 6?
2. A ticket is drawn from a set numbered from 1 to 50. What
is the probability that it is a multiple of either 7 or 9?
3. A die is thrown. What is the probability that the throw will
be greater than 3?
4. Two dice are thrown. What is the probability that the throw
will be greater then 7?
5. A bag contains four red, five white, two blue, and seven green
balls. One is drawn. What is the probability that it is either red or
blue?
111. Independent Events. The occurrence of several single
events simultaneously or in succession constitutes a compound
event. Events are said to be dependent or independent, if the
occurrence of one does or does not affect the occurrence of the
others.
Thus, the drawing of a red and a white ball from a bag containing
red, white, and blue balls is a compound event.
Of two independent events, let a be the number of ways, all
equally likely, the first may happen, and let b be the number of
ways, all equally likely, it may fail; let ai be the number of ways,
all equally likely, the second may happen, and let 61 be the number
of ways, all equally likely, it may fail. The number of possible
cases for the first event is a + 6, and the number of possible cases
for the second event is a\ -\- b\. Each case of the first event may
be associated with each case of the second event. The number of
possible cases of the compound event is then (a + b)(a\ + &0,
all equally likely to occur. In aa\ cases both events happen;
in 661 cases both events fail; in 061 cases the first event happens and
the second event fails; and in a\b cases the first event fails and the
second event happens. Then the probability
of both events happening is -, —
of both events failing is -. — , ,., 1 , , . >
(a + 6)(ai + 61)
§111] PROBABILITY 211
of the first happening and _ 061 _ t
second failing is (a + &)(fli + &i)
of the first failing and second _ Oi& _ _
happening is (a
From the first probability given above it is seen that the
probability of the occurrence of two independent events is the product
of the probabilities of the occurrence of the separate events.
This law may be extended to include three or more single
events.
ILLUSTRATIONS: 1. A bag contains three white and four black
balls. The probability of drawing a white ball is 3 /7. If a
white ball is drawn and not replaced, the probability of drawing
another white ball is 2/6. The probability of drawing two white
u 11 • • • 3 ^ 1 l
balls in succession is ^ X ^ = ^'
lot
2. If A's probability of solving a problem is 1 /3, and if B's proba-
bility of solving it is l'/4, the probability that A does not solve it
is 2 /3, and the probability that B does not solve it is 3 /4. The
231
probability that both fail to solve the problem is ^ X 7- = «"•
Since 1 /2 is the probability the problem is not solved, 1 — 1/2, or
1 /2, is the probability that the problem will be solved by A or B.
3. Five coins are thrown. What is the probability that they
will all turn heads? The probability that a coin turns heads is 1 /2.
Hence the probability that the five turn heads is o'o'o'o'o =Q2'
Exercises
1. What is the probability that at least one ace will turn when
three dice are thrown?
2. A bag contains six white and five red balls, and a second bag
contains four white and three black balls. A ball is drawn from each,
what is the probability that both are white? That one is white and
one black? That one is white and one red?
3. What is the probability of holding four aces in a game of whist?
4. A sack contains three white and two black balls; a second sack
contains four white and seven black balls. A ball is drawn from the
first sack and placed in the second; then a ball is drawn from the
212 MATHEMATICS [§112
second sack. What is the probability that the second ball drawn is
white?
5. A bag contains three white and four black balls. Two persons
draw alternately one ball without replacing it. If A draws first,
what is the probability that he is the first to draw a white ball?
112. Expectation. If p denotes the probability of a person
winning a sum of money, m, the product mp is called the value
of his expectation, or his mathematical expectation. Thus,
if a man is to receive $10 if he draw a doublet from a set of
twenty-eight dominoes with one trial, his mathematical expec-
tation is $2.50, found by multiplying $10 by 1/4, the proba-
bility of drawing a doublet.
113. Successive Trials. On a single trial, let p be the proba-
bility of an event happening and let q be the probability of the
event not happening. Thus, if a die is thrown but once, the
probability of throwing an ace is 1/6 = p, and the probability
of not throwing an ace is 5/6 = q. If the die is thrown twice,
the probability of throwing an ace each time is p2, or (1 /6)2 = 1 /36.
The probability of not throwing an ace is q2 — 25 /36. The
probability of throwing an ace on the first trial and not throwing
155
an ace on the second trial is pq, or « X ^ = ™" The probability
o o oo
of not throwing an ace on the first trial and throwing an ace on
515
the second trial is qp, or _• X ^ = ~a' The probability of throw-
ing an ace once and once only with two trials is ™ + ™ = j^-
oo oo oo
The probability of throwing an ace at least once with two trials
' l°--l- — = —
18 36 + 36 ~ 36*
The probability of the event happening on each of the two
trials is p2. The probability of the event happening on the first
trial and not happening on the second trial is pq. The proba-
bility of the event not happening on the first trial and happening
on the second trial is qp. The probability of the event happen-
ing but once is 2pq. The probability of the event not happening
upon either of the trials is q2.
p* + 2pq + <? = 1
§113] PROBABILITY 213
In the illustration above,
P' + 2W + ««-Sg + SJ + |-!-»
The probability of the event happening on each of three trials
is p3. The probability of it happening on the first two trials
and not happening on the last trial is p'2q; of happening on the
first and last trials and not happening on the second trial is
p2q; and of happening on the last two trials and not happening
on the first trial is pzq. The probability of the event happening
twice and twice only with three trials is 3p2g. The probability
of the event happening once and once only with three trials is
3pg2. The probability of the event failing to happen with three
trials is q3. The probability that the event happens at least
once with three trials is p3 + 3pzq + 3pqz.
p3 + 3p2g + 3pq2 + q3 = 1
A sack contains three white balls and seven black balls.
Three successive draws of one ball are made, each ball being
returned to the sack before the next draw is made, p = 3 /10
and q = 7 /10. The probability of drawing a white ball each
time is (3/lQ)3 = 27/1000; of drawing two white balls and one
black ball is 3(3/10)2(7/10) = 189/1000; of drawing one white
ball and two black balls is 3(3/10) (7 /10)2 = 441/1000; and of
drawing a black ball each time is (7/10)3 = 343/1000. The
probability of drawing a white ball at least once is
27 189 441 657
I -| f\f\f\ I
1000 ' 1000 ' 1000 1000
If n trials are made the probability of the event happening
on each trial is pn. The probability of the event happening n — 1
times and failing to happen one time is npn-1q. For, the
probability of its failing to happen on the last trial only is pn~lq,
of its failing to happen on the next to the last trial only is pn~lq,
of its failing to happen on the second from the last trial only
is pn~lq, and so on for the n trials on which the event may fail
to happen; then the probability of the event failing to hap-
pen one time only in the n trials is the sum of these probabilities,
or np"-^. In a similar way it is shown that the probability
214
MATHEMATICS
[§113
of the event happening on n — 2 trials and failing to happen
on two trials is nC2pn~zqz (where nC2 is the number of combina-
tions of n things taken two at a time), or
n — 2
The probability that the event happens on n — 3 trials and fails
to happen on three trials is = pn-3qs. The probability
3 n-3
that the event happens on n — r (r
n
on r trials is
n) trials and fails to happen
n — r
The probability that the event
happens on each trial, on each trial but one, on each trial but two,
on each trial but three, etc., are, respectively, the first, second,
third, etc., terms of the binomial expansion,
(p + q)n = P" + np-'-'q +
=_ pn-2q2
2 n-2
3 I n-3
n — r
• +qn =
The probability that the event happens on k trials and on k
trials only is = p*qn~fc. The probability that the event
happens on at least k trials out of n trials is
npn-1q
2 n - 2
k-1 k+1
n -k k
Thus, if a die is thrown six times, the probability of throwing
an ace three times and three times only is
6
3 6-3
186,624
Exercises
1. What is the probability of throwing exactly three 6's in five
throws with a single die?
2. The odds are 2: 1 in favor of A winning a single game against B.
Find the probability of A winning at least two games out of three.
§114]
PROBABILITY
215
[114.] Graphic Representation of the Terms of (p + q)n. Let
A and B represent, respectively, the (r — l)st and rth terms of the
expansion of (p + q)n~l, and let C represent the rth term of the
expansion of (p + q)n. Then:
n- 1
A = n_iCr_ipn~i+V~2 = i — r~r" pn~r+1gr~2,
r — 2 n — r + 1 ^
n- 1
r- 1
C — r , T)n-r+l,,r— 1 —
»c'-iP 9 -
n — r
pn-rqr-l}
Y-
From these expressions it will be shown that
C = Aq + Bp.
Aq =
Bp =
n-l
r-2 n-r+1
n-l
r - 1
n — r
and
-l L -L-
— L r ~ 2 n-r + 1
-J_ n -r |:
or
Ap + Bq
= I71"1 I -r ~ 1
I T — 1 n — r+1
+
or
Ap + Bq =
r -I In-r+1
'-1 = C.
216
MATHEMATICS
[§114
The general term, the rth term, of the expansion of (p + q)n may
be found from the (r — list and rth terms of the expansion of
(p + g)"-1-
This property of the binomial expansion makes it possible to
construct graphically the terms of the expansion for the case
FIG. 95.
p + q = 1. Let AA' and BB', Fig. 95, whose lengths represent,
respectively, A and B, be drawn perpendicular to AB. CC' is a
third perpendicular drawn such that
AC = GE_
p ' q '
Draw AB' cutting CC' at K. From the similar triangles AA'B'
and KC'B',
KG' AA'
or
CB ' AB
AA' X CB
KG' =
AB
(1)
From the similar triangles BB' A and CKA,
CK _ BB'
AC~ AB'
§114] PROBABILITY 217
or
BB' X AC
AB '
By adding (1) and (2),
, BB'XAC + AA'XCB
AB ~
or
_ B(AC) + A(CB) _ B(AC) + A(CB)
AB AC + CB '
or
cc,
or, since p + q = 1, <7C' = JBp + Ag. Therefore <7C' = C,
the rth term of the expansion (p + 5)", when p + q = 1.
The first term of the expansion of (p + q)n equals the first
term of the expansion of (p + q)n~l multiplied by p, and may be
found graphically as in Fig. 95 by taking AA' zero. The last
term of the expansion of (p + q)n equals the last term of the
expansion of (p + g)""1 multiplied by q, and may be found
graphically as in Fig. 95 by taking BB' zero.
Let, 0, 1, 2, 3, 4, 5, 6, 7, . . . , Fig. 96, be points chosen
equidistant along a horizontal line. Let C\, C*, Ca, Ct, . . .
be intermediate points chosen such that
p OCi 1C2 2C3
q ~ Cil ~ C£ ~ C33 "
At these points of division of the line 07 erect perpendiculars
and let these perpendiculars be designated by the letter or numeral
at its lower end. We shall show how to lay off, by means of a
graphical construction, the successive terms of the expansion of
(p + q)n. With some convenient scale let IP = p and 2Q = q.
Then IP and 2Q represent the terms of the expansion when n = 1.
Draw OP, PQ, and 0.3. OP cuts Ci at ai, PQ cuts C2 at 0lf and
0,3 cuts C3 at 71. cti projected upon 1 gives 01, /Si projected upon
218
MATHEMATICS
[§114
2 gives 61, and 71 projected upon 3 gives Ci. By what was shown
in the first part of this section it will be seen that la\, 2b\, and 3ci
represent the terms of (p + q)z. In a similar manner it is shown
that laz, 262, 3c2, and 4d2 represent the terms of the expansion
(p + <?)3; 1«3, 263, 3c3, 4d3, and 5e3 represent the terms of (p + g)4;
Ia4, 264, 3c4, 4d4, 504, and 6/714 represent the terms of (p + <?)8.
Continuing in this way the terms of the expansion could be repre-
sented for any positive integral value of n.
c, \
ft
O Ci 1 C2 2
3 C4 4 Cs 5
FIG. 96.
C« 6
C, 7
If p + g is not 1, but k, the expression (p + q)» could
be written (kp' + /e?')re, or fcn(w' + q')n, where p' + q' = 1.
The terms of (p' + g')n may now be represented as above and
finally all ordinateo increased in the ratio of 1 : fc", or the vertical
scale changed so that all vertical distances represent quantities
kn times as great as originally.
If p — q, the configuration, Fig. 96, will be symmetrical with
respect to some vertical line. In this case (p = q = 1/2) the
ordinates represent the binomial coefficients. When p ^ q
the configuration is said to be asymmetrical or skew.
§114]
PROBABILITY
219
In Fig. 97 the ordinates represent the binomial coefficients of
the 999th power of (p + q). The drawing is due to Quetelet.1
FIG. 97.
From what was shown in the first part of this section the binomial
coefficients of any power may be calculated from those of the
expansion of a power one lower.
n
Binomial
Coefficients
I
1
1
2
1
2
1
3
1
3
3
1
4
1
4
6
4
1
5
1
5
10
10
5
1
6
1
6
15
20
15
6
1
7
1
7
21
35
35
21
7 1
Thus, in the above, any number is the sum of the number
directly above it and the number to the left of the latter.
Exercises
1. Represent graphically, as in Fig. 96, the terms of (p + q)n,
if p = \, q = -J, and n = 7.
1 The limit of the broken line at the top of the ordinates is known as the probability
curve; its equation is of the form
V - a«"kV
as is shown in treatises on the Theory of Probability.
220
MATHEMATICS
[§115
2. Represent graphically, as in Fig. 96, the terms of (p + q)n,
if p = q = 5 and n = 7.
116. Deviations from the Arithmetic Mean and the Probability
Curve. The readings of a set of direct measurements of a
magnitude (as the elevation of a bench mark, the length of a
base line, the angle subtended by a distant object, etc.) taken
by the same person or by several persons, will, in general, disagree
among themselves. This is especially true if the measuring
instrument is such as to enable the observer to read to a small
part of the unit; for example, as the reading of a level rod to the
nearest one-thousandth of a foot.
6.5 7.0 7.5 8.0 8.5
FIG. 98. — A frequency distribution polygon.
9.0
It is assumed that the measuring instruments are in adjust-
ment and that proper care is exercised in handling and reading
them. The disagreements of the readings are then due to what
are known as errors of observation.
In Table XIV are given readings of the area bounded by an
irregular curve drawn at random. The readings were taken with
an Amsler Polar Planimeter. In column I are given the readings
of the instrument and in column II are given the number of times
the corresponding reading was obtained. Fig. 98 gives a graphic
representation of the distribution of the frequency of the readings.
Areas are represented by abscissas and frequencies of readings by
ordinates. To represent more clearly to the eye the distribution
of the frequencies of the readings, the plotted points may be
joined, in order, by straight lines; or through each point a hori-
zontal line may be drawn giving a series of rectangles, as is
illustrated by the figure. The length of the base of each rectangle,
§115]
PROBABILITY
221
in this case, represents the least count of the instrument, while
its altitude represents the frequency of a reading.
TABLE XIV
I.
II.
ill
IV
V
VI
Area,
sq. cm.
No. of read-
ings
/
»
A1
126.4
0
1
-12.6
158.76
126.5
1
4
-11.6
538.24
126.6
126.7
3
4
8
13
-10.6
- 9.6
898.88
1198.08
8
126.8
6
17
- 8.6
1257.32
126.9
127.0
7
9
22
31
- 7.6
- 6.6
1270.72
1350.36
22
127.1
15
42
— 5 6
1317 12
127.2
127.3
18
22
55
69
- 4.6
- 3.6
1163.80
894.24
55
127.4
29
84
- 2.6
567.84
127.5
127.6
33
31
93
96
- 1.6
- 0.6
238.08
35.56
93
127.7
32
92
0.4
14.72
127.8
127.9
29
28
89
81
1.4
2 4
150.41
465.56
89
128.0
24
67
3.4
774.52
128.1
128 2
15
12
51
37
4.4
5 4
987.36
1078 92
51
128.3
10
31
6 4
1269.76
128.4
128 5
9
7
26
21
7.4
8 4
1423.76
1481 76
26
128.6
5
13
9.4
1148.68
128.7
128.8
1
2
8
5
10.4
11 4
865.28
649 80
8
128.9
2
4
12 4
615.04
129.0
0
2
13.4
359.12
2
Sums . .
354
1062
22,173.68
Mean area = 127.66 sq. cm.
h2 = 0.024
4= = 0.087
Vtr
<r = 4.3
E,.r = 3.08 or 0.308 sq. cm.
E = 0.1 or 0.01 sq. cm.
222 MATHEMATICS [§115
The graph obtained by either method of connecting the plotted
points is called a frequency distribution polygon.
From Fig. 98, as well as from the table of readings, one is lead
to believe that the true area sought lies somewhere between
127.3 and 128.0, probably near 127.6 square centimeters.
If it be granted that in the long run numerically small errors
occur more frequently than larger ones, and that positive and
negative errors, numerically equal, occur with equal frequency,
the frequency polygon for a very large number of readings would
present symmetry with respect to a vertical line, and the altitudes
of the rectangles would decrease with increasing distances from
this line of symmetry. In such a case the altitude of any rec-
tangle, or its area, would be proportional to the probability that
a single reading fall within the limits defined by the ends of its
base. For, the sum of the altitudes of all rectangles would
represent the total, or possible, number of cases, while the altitude
of a particular rectangle would represent the number of cases of
the corresponding reading. Since the bases of the rectangles are
equal, the above statement holds for the areas of the rectangles as
well as for their altitudes.
If the scales of the drawing are so chosen that the sum of
the areas of the rectangles is unity, the sum of the areas of any
number of adjacent rectangles is equal to the probability that a
single reading falls within the limits represented by the extremes
of the bases of the rectangles.
In a large number of cases it has been found that the plotted
points representing the frequency distribution of direct observa-
tions follow the curve representing the equation
y = ae-»*** (1)
where e is the base of the hyperbolic logarithmic system, approxi-
mately equal to 2.7183, a and h are positive constants, and
x the variable distance from the vertical line of symmetry, i.e.,
distance from the true value. Thus, x represents the error of
a reading. It is also found that the line of symmetry corre-
sponds to the arithmetic mean of the readings. (See Figs. 99
and 100. Also Figs, in Methods of Least Squares, by G. C.
Comstock.)
§115]
PROBABILITY
223
If the area under the curve is taken as unity, equation (1)
becomes
tf=-/£«^V (2)
-12-11-10-9 -8 -7 -6 -5 -4 -3-2 -1
A
0 1
234
A'
5 6 7 8 9 10 11 12
FIG. 99. — Graphic representation of frequency distribution of
readings given in Table XIV. The circles represent the distribution
of the readings, while the curve is the probability curve calculated
from them. The equation of the curve is y = Q.Q87e~0024x2. The
probable error of the mean is 0.01 square centimeters.
FIG. 100. — Graphic representation of frequency distribution of
readings of the elevation of a bench mark. The circles represent the
distribution of the readings, while the curve is the probability curve
calculated from them.
In what follows it is assumed that the units are chosen such that
the area under the curve is unity, so that the number of units of
area bounded by the curve (2), the X-axis, and any two vertical
lines is the probability that a single reading falls within the
limits corresponding to the positions of the vertical lines, or
224 MATHEMATICS [§115
the probability that a single reading has an error falling within
the values of x defining the vertical lines.1
The curve represented by equation (2) is known by various
names, as "the probability curve," "the error curve," "the normal
distribution curve."
While, probably, no set of readings which the student will
take, or will have occasion to study, will be large enough in
number to give a frequency distribution polygon following very
closely the probability curve, yet the laws and formulas derived
from this law will be used and applied to the data at hand. Thus,
to illustrate, the empirical law states that for a very, very large
number of observations the line of symmetry corresponds to the
arithmetic mean of the readings, and hence this mean gives the
1 For students who have a knowledge of the Calculus, and who may chance to
read this chapter, a few proofs of statements, which must of necessity be taken for
granted by the students for whom this book is intended, are inserted in footnotes.
The curve for
k-~*
y = ae~ h x
is symmetrical with respect to the Y-axis and approaches the X-axis asymptotically.
y has a maximum value when x = 0. If the area under the curve is unity, it follows
that
•J"
,
~h
dx
or, if
iru)
a = /=-•
vv
Hence
The probability that a single reading will have an error between a and ft is
h
'dx
§116]
PROBABILITY
225
true value of the measured magnitude. But for a very limited
number of readings the arithmetic mean gives, probably, only a
very close approximation to the true value. The empirical
law, however, does seem to indicate that for a set of repeated
readings the arithmetic mean is the best mean to use to represent
the true value.
116. Smoothing the Distributing Polygon. If only a few read-
ings are taken the frequency distribution polygon will in general be
very irregular. A more regular polygon may often be obtained by
grouping the readings within some larger interval. This method is
illustrated in column III, of Table XIV, and by Fig. 101. In column
6.3 6.6 6.9 7.2 7.5 7.8 8.1 8.4 8.7 9-0 9.3
FIG. 101. — Frequency distribution polygon.
Ill, to illustrate, 93 corresponding to area 127.5 is the sum of the
frequencies 29, 33 and 31 of column II, corresponding, respectively,
to the areas 127.4, 127.5 and 127.6. In Fig. 101 the numbers of
column III are plotted against areas.
Another method of smoothing the curve is to consider that in
a larger number of readings the frequency of any one would be
proportional to the mean of the observed frequencies of it and
its two adjacent readings. The frequencies given in column III
of Table XIV are obtained in this way from the observed frequen-
15
226 MATHEMATICS [§117
cies given in column II.1 In Fig. 99 frequencies, /, are plotted
against areas.
117. Measure of Precision; Standard Deviation. For a set
of readings it may not be enough to know merely the value of the
arithmetical mean. One may desire to know what degree of
confidence may be placed in this mean, or to know something of
the agreement or disagreement among the individual readings.
This information is given by the table of readings or by the fre-
quency distribution polygon, but it is desirable to have some definite
mathematical expression giving the precision of the set of readings.
If in a set of readings the number of small errors is relatively
large compared with the number of small errors of a second set of
readings, the first set of readings would, in general, command more-
confidence than the second set. Of the two corresponding fre-
quency distribution polygons the first would be tall and narrow as
compared with the second. Since the probability curve for the
first set of readings will have a greater maximum ordinate than
h
that for the second set, and since — j~- is the value of the maximum
VV
ordinate, h will be larger for the first set than for the second, h
is called the measure of precision. Hence of several sets of
readings, that set is considered the best for which h is a maximum.
Let x represent the distance of the center of the base of a
rectangle in a frequency distribution polygon from the point cor-
responding to the arithmetic mean. In Table XIV, these values
are given in column IV. It can then be shown that2
A _ 2/«2 m
2h* ~ n
1 In this case, instead of taking the mean, the sum of the three readings is taken.
This is done to avoid the division by 3, and at the same time to avoid fractions.
It will be noticed that by this process of smoothing, the arithmetic mean is not
changed.
2/i
" dv (approximately)
/""o
I
"Jo
e~ldt, where
is the area of a frequency rectangle (see next footnote). If the horizontal
§117] PROBABILITY 227
(where n represents the total number of readings) provided that
the unit of v is the interval corresponding to the base of the frequency
rectangle.1
In statistical work \/— — is usually denoted by <r (sigma) and is
called the standard deviation. Equation (1) may then be written
and the equation of the probability curve may be written:
In practice, instead of using the value of h as a basis of comparison
of the precision of sets of readings, the value of <r is used. Since a
varies inversely as h, that set of readings is considered the best
which gives <r the smallest value.2
scale is chosen so that At is unity, the sum of the frequency rectangles will be unity
if V = f/n, where n is the total number of readings. Thus:
1 y/**
2A2 ~ ^ n •
1 Frequency rectangle, as here used, means a rectangle of the frequency distribu-
tion polygon.
2 From equation (1) it is seen that for h to have a maximum value S/t2 must have
a minimum value. 2/t>2 is the sum of the squares of the deviations of the readings
from the arithmetic mean, the approximate value, or 2/c2 is the sum of the squares
of the residuals as they are called. Thus the sum of the squares of the residuals
must be a minimum. This principle is sometimes used in solving simultaneous
observation equations in which there are more equation? (corresponding to a
large number of readings) than unknown numbers. The method of uniting the
equations is called the method of Least Squares.
An observation equation, as here used, is an equation resulting from the substitu-
tion of readings in an equation assumed, or known, to be of the form connecting the
measured variables. In such an equation coefficients are the unknowns. Thus,
from Table VI, if x represents the hydraulic gradient and if y represents velocity, and
if it is assumed that the law connecting y and x is linear, i.e., y = ax + 6, there
result, by substituting the measured values of x and j/, seven equations, viz.:
16.9 = 31. 9o + 6
11.4 = 20.8o + 6
24.5 = 54.1o + 6
4.3 = 10. Oo + b
10.1 = 21. Oo + 6
58.8 = 119.1o + b
22.7 - 55.8o + 6
a and b are two unknown numbers to be found, having given seven equations.
The problem is not to find a set of values satisfying any two equations, but
a set of values which will represent the most probable line upon which the
points would fall if there were no errors of observation.
228 MATHEMATICS [§119
118. The Probable Error of a Single Reading. The probable
error of a single reading, Eg.r., is a number such that if it be sub-
tracted from and then added to the arithmetic mean, the proba-
bility of a reading falling within these limits is exactly equal to
1/2, i.e., the chance of a reading falling within these limits is
exactly equal to the chance of its falling without. Graphically,
the probable error determines two lines, as AB and A 'B', Fig. 99,
parallel to and equidistant from the line of symmetry of the
probability curve, such that the area between them, the curve,
and the X-axis is one-half unit. (The area under the entire
curve and the X-axis is unity.)
It is found that1
Es r = 0.6745 J— = 0.6745 <r
\ n
For the data given in Table XIV,
a = 4.56
Es.r. = 3.08
which corresponds to 0.308 square centimeters.
119. The Probable Error of the Arithmetic Mean. If several
sets of readings of the same magnitude were taken, the arithmetic
means of these sets of readings would, in general, disagree among
themselves. Thus, if one hundred students were each to take
an equal number of readings, say one hundred, of the area bounded
by a curve, the one hundred means of these sets would probably
not all agree. From these one hundred means a frequency dis-
tribution polygon could be constucted; a mean (i.e., a mean of the
means) and a probable error could be calculated. This probable
error, i.e., the probable error of a single mean of the one hundred
means, is called the probable error of the mean of any one of the
sets of readings. The probable error, E, of the mean of a set of
i The value of
is found by expanding e * x into an infinite series and integrating term by term
Tables for different values of the upper limit, or rather for x/<r, or hx, are usually
given in treatises on the Theory of Probability and in treatises on Statistical
Methods. From such a table the value of 0.6745 for x/a is found to correspond
to 1/4.
§120] PROBABILITY 229
readings, bears the same relation to a set of means as the probable
error of a single reading bears to the readings of a particular set.
A formula for the probable error of the mean is
The probable error of the mean of the readings of the area (Table
XIV) is
.,
\/1068
or expressed in square centimeters E = 0.01. The probable error
of a mean is usually indicated by affixing it with a double sign to
the value of the mean. Thus, in the above expression,
Area = 127.66 ± 0.01
This does not mean that the true area is somewhere between
127.65 and 127.67, but that with the data at hand the probability
of the mean of a second set of the same number of readings
falling within these limits is one-half (i.e., the chances are one to
one that it falls within these limits).
Exercises
1. Verify all numerical work done in connection with Table XIV.1
2. Using columns I and II, Table XIV, calculate h, <r, Es.r. and E.
Arrange your work on a sheet of paper form M8 or Ml
120. Biological Measurements. Frequently distributions of bio-
logical measurements are found to follow to a marked degree the
probability law. While the distribution polygons may be asym-
metric, or skew (See § 114), the formulas for standard deviation,
probable error, etc., are still applied as a method of comparison.2
Exercises
1. Construct the frequency polygon, find the mean, the standard
deviation, the probable error of the mean, for the readings given in
Table XV.
1 In treatises on the Theory of Probability and in treatises on Statistical Methods
are usually given tables of squares.
Crelle's or Peter's Multiplication Tables or the smaller table by Zimmermann
may also be found very helpful in numerical calculations. A collection of tables
and formulas is given in Statistical Methods by C. B. Davenport.
'At this point it is suggested that the student read pages 419—441, Principles of
Breeding, by E. Davenport.
230
MATHEMATICS
[§120
TABLE XV.— MEASUREMENTS OF THE LENGTH OF 327 EARS OF CORN'
Length
of ear,
inches
Number of
ears, or
frequency
Length
of ear,
inches
Number of
ears, or
frequency
Length
of ear,
inches
Number of
ears, or
frequency
V
/
V
/
t
/
3.0
1
6.5
8
9.5
63
3.5
0
7.0
12
10.0
38
4.0
1
7.5
19
10.5
21
4.5
0
8.0
32
11.0
8
5.0
2
8.5
40
11.5
2
5.5
3
9.0
67
12.0
1
6.0
9
1 These data are taken from Principles of Breeding, page 421, by E. Davenport.
CHAPTER X
[SMALL ERRORS]
121. Errors. Whenever an attempt is made to measure the
magnitude of a quantity with an observing instrument, the read-
ings thus obtained do not represent the true value of the given
magnitude. There is a difference between the reading of the in-
strument, the observation, and the true value of the magnitude.
This difference is called the error of the observation. Errors
of observation may be due to many causes, some of which are:
carelessness on the part of the observer or his inability to handle
the measuring instrument properly; inadjustment of the measuring
instrument; limitations due to the least count of the instrument,
etc. By the least count of the measuring instrument is meant
the smallest part of a unit which can be recorded with the in-
strument. To illustrate: when a volume of liquid is measured
with a burette graduated to cubic centimeters and tenths of
cubic centimeters, the readings are estimated to the nearest
tenth of one-tenth of a cubic centimeter; the least count of the
instrument is the one one-hundredth part of one cubic centimeter.
When direct measurements, which are in error, are used for
calculating other magnitudes, these calculated magnitudes will in
turn be in error. The larger the errors of the readings, the
larger the errors of the calculated quantities; the smaller the
errors of the readings, the smaller the errors of the calculated
quantities. When the errors of the observations are known
or admitted, it is at times desirable to know the corresponding
error in the calculated magnitude obtained by substituting
the readings in a formula. -If, on the other hand, the calculated
magnitude is desired only to a certain degree of accuracy, it is
usually desirable to know the degree of accuracy to which the
measurements must be made.
It will- be apparent that the error of the calculated magnitude
will depend upon the formula in which the readings are substituted.
231
232 MATHEMATICS [§122
Following are given the formulas for errors due to the simple
algebraic operations and then- applications to some of the formulas
previously derived in this book.
122. Error of Product. The derivation of the formula for
the error of the product of two numbers has been given in § 72,
but will be repeated here.
Let a and b be two readings. Let ci and 62 be, respectively,
their errors. The true values of the measured magnitudes are
thus assumed to be a ± ei and b ± €2. The product of the
two magnitudes is then
(a + €i)(6 + €2) = ab + aez + bei + eiez.
Since ab is the product obtained by multiplying the two readings
together, the sum of the last three terms of the right-hand side of
the above equation is the error of the product. Since e\ is small
compared with a and €2 is small compared with b, eicz is very
small compared with ae2 + bci and is negligible. Thus, if
ci and ez are, respectively, the errors of a and b, ocz + &d
is the possible error of the product ab. The true product may
fall anywhere between ab + (ac2 + &ci) and ab — (aez + bci).
All signs in the error are taken alike in order to obtain the greatest
possible error.
In what follows, expressions of the form a = 212 + 0.3 will
mean a reading of a equal to the first number (212), and a possible
error equal to the second number (0.3).
If a = 212 + 0.3 and b = 116 ± 0.5, the error in the product
of the readings is aez + bci = (212) (0.5) + (116) (0.3) = 150.8.
Thus, the true value of ab is between 24,592 +151 and
24,592 — 151. In calculating the product ab by multiplying the
two readings together there is no need of retaining more than
the first three digits, and the product should be written 24,600.
Exercises
1. Show that, if ei, e2, and ei are, respectively, the errors of a, b, and
c, the error of abc is a&es + bcei + ace2.
Calculate the possible error of the product for each of the following :
2. a = 217 ± 0.2; b = 117 + 0.3.
3. a = 1267 ± 0.5; 6 = 986 ± 0.4.
4. a - 3176 ± 0.3; b = 1012 ± 0.3.
5. a = 3176 ± 0.2; 6 = 1276 ± 0.3; c = 987 ± 0.4.
§124] SMALL ERRORS 233
123. Error of Quotient. Let a and 6 be the readings with
errors of ei and €2, respectively. The quotient of the true
magnitudes is
a
a ± *i _ a ~ b 2 '_^ _ a ±ac2±bfi
b ±e2 = b H 6 ± c2 ~ b H " ~b(b ± e2) '
Since -7 is the quotient of the two readings, the second term
of the right-hand side of the above equation is the error of the
quotient. Since €2 is small compared with 6, the quantity b + e2
may be replaced by &, and the possible error of the quotient
becomes
ae2 + tei
If a = 317 + 0.5 and 6 = 110 + 0.6, the error of the quotient
a .
(317X0.6) + (110) (0.5) ftn<?
(HO)2
Exercises
Find the possible error of the quotient for each of the following:
l.o= 1127 ± 0.5; b = 56 ± 0.2.
2. a = 2216 ± 0.6; b = 27 ± 0.1.
3. a = 2716 + 0.3; 6 = 25 ± 0.5.
124. Error of the Square. Let e be the error of the reading
a. The square of the true value is (a + e)2 = a2 + 2ae + e2.
Since a2 is the square of the reading, the sum of the last two terms
of the right-hand member of the above equation is the error of
the square. The possible error, 2ac + e2, may be written
(2a + e)e, and, since e is small compared with 2a, the expres-
sion within the parentheses reduces to 2a without appreciable
error. The possible error of the square is then
2ae.
If a = 112 ± 0.5, the error of a2 is (2) (112) (0.5) = 112.
Since a2 equals 12,544, the error of 0.5 in a, or about 1/2 of
1 percent, produces an error of about 1 percent in a2.
234 MATHEMATICS [§125
Exercises
Calculate the error and percent error of the square for each of the
following :
1. a = 111 + 0.1.
2. a = 316 + 0.2.
3. a = 416 ± 0.6.
4. Show that the percent error of the square of a reading is the
double of the percent error of the reading.
5. Show that the error of the cube of a + e is 3a2e.
6. Show that the percent error of the cube of a reading is the treble
of the percent error of the reading.
7. Show that the error of the fourth power of a + e is 4a3e.
125. Error in the Square Root. Let e be the error of the
reading a. The square root of the true value is then
(a ± e)1^ E= a* ± \a~\ + \a~3\z ± ia-**e« + • • •
Z o lO
by the binomial theorem. (See § 100.) The above may be
written
Since e is small compared with a, the terms within the parenthe-
ses, excepting the first, are small compared with 1, and may be ne-
glected. Hence
(a ± «)" = .« ±
approximately. Since a^ is the square root of the reading,
f
/== is the error of the square root of a.
If a = 121 + 0.5, the error of \/a is
0.5
(2) (11)
= 0.023.
Exercises
Find the error of the square root of each of the following :
l.o = 121 ± 0.1.
2. a = 146 + 0.3.
3. a = 216 + 0.6.
§126]
SMALL ERRORS
235
\c
4. Show that the percent error of the square root of a reading is
one-half of the percent error of the reading.
6. Show that the error of the cube root of a reading is s— sr'
OCt'3
6. Show that the percent error of the cube root of a reading is one-
third the percent error of the reading.
7. Show that the error of the fourth root of a reading a is T~57'
126. Error of Sine a. Let e, expressed in radians, be the
error of the measured angle a. The sine of the true angle is then
sin (a ± e). Let BOA, Fig. 102, be the angle a with its vertex
at the center, 0, of a circle
having a unit radius, OA. BE
is the sine of the reading a, CD
is the sine of the true angle, and
CF is the error of the sine due to
the error e of the reading a. The
angle FCB is very nearly equal
to the angle a, and approaches
it as e approaches zero. The
chord CB is very nearly equal
to the arc CB. FC is then very
nearly equal to the product of the chord CB by the cosine of
FCB, or the product of the arc CB by the cosine of a. But,
since e is expressed in radians, the arc CB equals the product
of e by the radius of the circle, which is unity. We thus have
CF = e cos a, approximately. The figure is drawn for e posi-
tive; the same conclusion is obtained if e be taken negative.
The sign of e regulates the sign of the error of the sine of a.
To illustrate the above formula, suppose a = 30° ±15'.
The error 15' expressed in radians is
Jg 4 - 0.00436.
The error then of the sine a is (0.00436) cos 30°,
or (0.00436) (0.866) = 0.0038.
This error (or the error of any trigonometric function of any
anclel may also be obtained from Table XXII. the table of
FIG. 102.
236 MATHEMATICS [§127
natural trigonometric functions. The sine of 30° is 0.5, the sine
of 30° + 15' is 0.5038, and the sine of 30° - 15' is 0.4962. In
each case the error in magnitude is 0.0038, the same as found
above by using the formula.
Exercises
1. Find by substituting in the formula the error of sin a for each
of the following:
(a) a = 27° ± 10'.
(6) a = 87° ± 20'.
2. Find by means of the table of natural trigonometric functions the
error of sine a for each angle given in exercise 1.
3. Find the error of cos a, if a = 26° + 25'.
4. Find the error of tan a, if a = 62° ± 17'.
127. Error of the Area of a Triangle Computed from Two Sides
and the Included Angle. The formula for the area, A, of a tri-
angle in terms of two sides, a and b, and the included angle, 7, is
A = \ ab sin 7.
Let ei €2, €3, be the errors of a, b, and 7, respectively. Let E
represent the error of sin 7. Then, by exercise 1, § 122, the
error of A is \ (abE + feei sin 7 + aez sin 7),1 where E — 63 cos 7.
In this formula, es must be expressed in radians. To illustrate:
let a = 111 + 0.2 rods; b = 72 ± 0.2 rods, and 7 = 37° + 15'.
From the table of natural functions, E is found to be 0.0035.
The error of A is then
i [(111) (72) (0.0035) + (72) (0.6018) (0.2) + (111) (0.6018) (0.2)],
or 50 square rods.
Exercises
Find the error of the area of the triangle for each of the following :
l.o = 126 ± 0.3 rods; b = 137 ± 0.3 rods; y = 47° + 10'.
2. a = 363 ± 0.8 rods; b = 216 + 0.5 rods; 7 = 26° ± 15'.
1 The value of E may be taken from a table of natural trigonometric functions, or
the expression given for it in the preceding section may be substituted, in which
case the expression for the error of A becomes
J(a&€scos-y + bei sin 7 + 0*2 sin 7).
§128] SMALL ERRORS 237
128. Error of the Area of a Triangle Computed from Three Sides.
The formula for the area, A, of a triangle in terms of three sides is
A = \/ s(s — a)(s — b)(s — c),
where s = |(a + b + c). This formula may be put in the form
A = J 2o»6» + 262c2 + 2 c2a2 - a4 - b* - c*
Let ei, e2, and €3 be the errors of a, b, and c, respectively. Then,
by § 124, the error of a2 is 2aei, of 62 is 26e2, of c2 is 2ce3, of
a4 is 4a3ei, of 64 is 4&3c2 and of c4 is 4c3e3. By § 122, the error
of a262 is 2a26e2 + 2ab2el} of 62c2 is 26c2e2 + 262c3, and of c2a2 is
2ca2e3 + 2c2oei. The error of the quantity under the radical sign
is then
4o26e2 + 4a62d + 462ce3 + 4&c2e2 + 4c2O€i + 4ca2e3 -
- 4c3c3,
or
4[a(62 + c2
- a2)Cl +
7 / 0 1 9 7 O\ 1
0(0^ + cr — 6z)e2 -|-
• c(a2 +
62 - c2)63].
By § 126 the
error of A
is found to be
1 Nf 4[a(62H
- c2 - a2)e
i + 6(c2 + a2 - 62)e2
+ c(a2
4-62-c2)e3],
4 X
2 V 2o262 -
f 262c2 + 2c2a2 -
a4 -64
-c4
or
a(b2 + c2
- a2)€l +
b(c2 + a2 - b2)€2 4
- c(a2 4
• b2 - c2)«3
84
If f\, (2, and c3 are each k percent of the length of the sides of
which they are errors, we have
ka
61 = i^
kb
e2 = -
100
^_
€3 ~ 100
Substituting these values in the above expression for the error
of A, it reduces to
(b2 + c2 - a2) + 6'(c* + a2 - 62) + c2(a2 + 62 - c2) ] ,
238 MATHEMATICS [§128
or
+ 26V + 2c2a2 - a4 - 64 - c4),
or
k 2kA
or
SWA™ zoo
This expression says that if the same percent of error, k, is made
in measuring the three sides of a triangle, the error of the calculated
area is 2k percent. It is to be noted that in order for this state-
ment to be true the three errors must have the same algebraic sign.
To illustrate: if a = 100 ±0.1 foot, b = 200 ± 0.2 foot, and
c = 150 + 0.1 foot, the posible error of A is
525,000 + 300,000 + 412,500
or 21 square feet or a possible error of about 3/10 of 1 per-
cent of the area.
Exercises
1. Find the possible error of the area of the triangle if a = 127 ± 0.2
rod, and b = 263 + 0.4 rod, and c = 290 ± 0.5 rod.
2. In a triangle a = 100 + 0.3, 0 = 37° + 15', and 7 = 52° ± 15'.
Calculate the possible error of b.
HINT: Use the law of sines and note that the possible error of a
is the sum of the errors of /3 and 7.
3. If the error of a is e, expressed in radians, show that the error
of cos a is e sin a.
4. If a = 100 + 0.2, b = 200 + 0.3, and « = 34° + 15', find the
possible error of
COS a
CHAPTER XI
[POINT, PLANE, AND LINE IN SPACE]
129. Rectangular Space Coordinates. Through any point 0,
Fig. 103, are drawn three right lines, X'X, Y'Y, and Z'Z, mutually
at right angles. The point 0 is called the origin of coordinates.
The three lines X'X, Y'Y, and Z'Z are called, respectively, the
X-axis, the F-axis, and the Z-axis of coordinates. The three
planes determined by the three axes taking them two by two are
called the coordinate planes. The coordinate planes are mutually
at right angles. The plane determined by the X- and F-axes is
called the XY- plane, by the F- and Z-axes the FZ-plane, and
by the X- and Z-axes the XZ-plane.
The three coordinate planes divide space into eight parts
called octants. The octants are numbered from one to eight.
The first octant is in front of the XY- plane, to the right of the
FZ-plane, and above the XZ-plane. The second octant is back
of the XF-plane, to the right of the YZ- plane, and above the
XZ-plane. The third octant is back of the XF-plane, to the left
of the FZ-plane, and above the XZ-plane. The fourth octant
is in front of the XY- plane, to the left of the YZ- plane, and
above the XZ-plane. The fifth, sixth, seventh, and eighth
octants are below the first, second, third and fourth octants,
respectively.
Distances measured along, or parallel to, the X-axis to the
right from 0 are called positive, to the left from 0 negative.
Distances measured along, or parallel to, the F-axis up from 0
are called positive, down from 0 negative. Distances measured
along, or parallel to, the Z-axis out from 0 are called positive,
back from 0 negative.
Any point, P, in space is located by giving its distance from
each of the three coordinate planes. FP, or OA, Fig. 103, its
distance from the FZ-plane, is called its x-coordinate; DP, or
OC, its distance from the XZ-plane, is called its y-coordinate;
239
240
MATHEMATICS
[§130
and EP, or OB, its distance from the .XT-plane, is called its
z-coordinate. The algebraic signs of the three coordinates of
points in the eight octants are given in the following table:
Octant
X
y
z
First
+
+
+
Second
+
+
Third
+
Fourth
+
+
Fifth
+
+
Sixth
+
Seventh
_
Eiehth..
4-
Since the axes of coordinates are taken mutually at right
angles, the three coordinates of a point P are called rectangular
coordinates. In designating a point its three rectangular co-
ordinates are written in the order: the x-coordinate first, then
the ^/-coordinate, and then the z-coordinate. Thus, the point
(1, 2, 3) is in the first octant, one unit from the FZ-plane, two
units from theXZ-plane, and three units from the .XT-plane. The
point ( — 1, — 2, 3) is in the eighth octant, one unit from the
FZ-plane, two units from the XZ-pl&ne, and three units from
the .XT-plane. The point ( — 1, — 1, — 2) is in the seventh
octant.
3. (-2, - 1, -5).
Exercises
Locate the following points:
1. (1, 1, 1). 2. (3, - 2, 1).
4. (3, -2, -3). 5. (-3,4,3).
130. Distance between Two Points. The distance of P, Fig.
103, from the origin is the length of the diagonal of the rectangular
parallelepiped whose edges are x, y, z, the coordinates of P.
The length of this diagonal, or the distance of P from the origin
is A/a;2 + i/2 + 22.
Let PI and Pz, Fig. 104, be any two points whose coordinates
§131]
POINT, PLANE, AND LINE IN SPACE
241
are, respectively, xi, y\, z\, and x2, yz, zz.
that the distance between PI and Pz is
The student will show
131. The Locus of an Equation.1 In geometry of two dimen-
oxons the equation x = 2 represents a straight line parallel to the
7-axis, two units to its right. (See § 26.) In geometry of
three dimensions the same equation, x = 2, represents a plane
sions
FIG. 104.
parallel to the 7Z-plane, two units to its right. For, the only
condition to be satisfied is that the y-coordinate of the point is
+ 2; the coordinates of all points of the plane satisfy this condition,
and the coordinates of a point not of the plane do not satisfy
the condition. The locus of all points satisfying the condition
x = 2 is a plane parallel to the FZ-plane two units to its right.
The locus of a point satisfying the condition x = 5 is a plane
parallel to the 7Z-plane five units to its right, x = — 2 represents
a plane parallel to the FZ-plane two units to its left, x = 0
represents the FZ-plane. x = k, where k is any constant, repre-
sents a plane parallel to the FZ-plane k units from it; to the right
if k is positive, to the left if k is negative. Similarly, y = k
represents a plane parallel to the XZ-plane, and z = k represents a
plane parallel to the XT-plane.
1 The locus of an equation is the locus of a point whose coordinates, x, y, and c,
satisfy the equation.
16
242 MATHEMATICS [§131
Any equation between x, y, and z, when solved for z, takes the
form z = f(x, y) . (The right-hand member of this equation,
f(x, y), is read "function of x and y," and is an abbreviation for
the right-hand side of the equation when solved for z. The f(x, y)
may contain x's and y's, but it does not contain z.) If, when to x
and y in the equation z = f(x, y) any two values are assigned,
there corresponds a real value of z, the three values, one for x, one
for y, and one for z, locate a point in space whose coordinates
satisfy the equation. If a second set of values is assigned to x
and y, and a second, or the same, value for z found, a second point
is located in space. If the difference between the two z-values
and the difference between the two ^/-values is made less and less,
the difference between the two z-values will in general grow less
and less. The locus of a point whose coordinates satisfy an equa-
tion is then seen to be a surface. Thus, in the equation
z = ± \/81 - (z2 + 7/2)
if any set of values is assigned to x and y such that x2 + y2 < 81,
there correspond two real values of z, numerically equal but with
opposite signs. If a second set of values is assigned to x and y,
a second (or the same) pair of values for z is obtained. If the
difference between the two x-values and the difference between
the two ?/-values are made smaller and smaller, the difference be-
tween the two positive values and between the two negative
values of z are made smaller and smaller. As x and y are made
to vary continuously, z varies continuously and the equation rep-
resents a surface in space. If the values assigned to x and y are
such that x2 + y2 > 81, the corresponding values of z are im-
aginary, which means that the surface does not extend to a
point whose x and y are the values assigned to x and y in the
equation. Thus, if x and y are each given the value 7 the equa-
tion reduces to z = ± V — 17, an imaginary. This says that the
surface represented by the equation contains no point whose x-
and y-coordinates are each equal to 7, or in other words, a line
parallel to the z-axis seven units each from the XZ- and YZ-co-
ordinate planes does not pierce nor touch the surface.
The equation z = ± V 81 — (z2 + ?/2) may be put in the form
x2 + yz + z2 = 81. The expression z2 + y2 + z2 is the square of
§132] POINT, PLANE, AND LINE IN SPACE 243
the distance from the origin to the point whose coordinates are
x, y and z. (See § 130.) This expression being a constant, 81,
shows that all points whose coordinates satisfy the equation must
be at a distance 9 from the origin. The locus of the equation is
then a sphere with center at the origin and radius equal to 9.
132. The Equation of a Locus. The equation of the locus of a
point is an equation satisfied by the coordinates of all points of the
locus and by the coordinates of no other point. Thus, the equa-
tion of a sphere, center at the origin and radius equal to 10, is
x2 + y* + z* = 100. The equation of the XZ-plane is y = 0.
The equation of the XY-p\a,ne is z = 0. The equation of the
YZ- plane is x = 0. The equation of the XY- and XZ-planes
considered as a single locus is yz = 0. The equation of the three
coordinate planes considered as a single locus is xyz = 0.
Exercises
1. Find the distance between the following pairs of points :
(a) (1, 1, 1) and (3, 2, 1). (6) (3, - 2, 6) and (- 2, 1, 7).
(c) (-1, -2, -3) and (2, 6, -1).
2. What loci are represented by the following equations :
(a) x = 3. (6) x = 0. (c) y = - 6.
(d) y = 0. (e) z = 0. (/) z = - 10.
(ff) x* + t/2 + z2 = 1. (h) x* +y* +z* = 25.
(i) xy = 0. (7) xyz = 0. (k) x = y.
(I) x = z. (TO) y = z.
3. Find the equations of the locus of (a) a point ten units from the
origin; (6) a point ten units from the point (3, 1, 2); (c) a point equi-
distant from the XZ- and the FZ-coordinate planes.
133. The Direction Cosines of a Line. Let P, Fig. 103,
be any point on the line OP passing through the origin. The
projections of OP upon the three coordinate axes are the co-
ordinates of the point P. Let a, ft, and 7 be the angles, called
the direction angles of the line OP, between OP and the positive
directions of the x-, y-, and z-axes, respectively. The cosines of
these angles are called the direction cosines of the line OP.
244
MATHEMATICS
[§134
The direction angles of a line through the origin are positive and
do not exceed 180°. From the figure it is seen that
x = OP cos a
y = OP cos (8
z = OP cos 7
Squaring and adding these three equations,
z2 + y2 + z2 = OP2(cos2 a + cos2 0 + cos2 7).
But, since
OP2 = x2 + ?/ + z\
it follows that
cos2 a + cos2/3 + cos2 7 = 1. (1)
The direction angles of a line not passing through the origin are
defined as the direction angles of a parallel line passing through
the origin.
The sum of the squares of the direction cosines of a line is equal
to unity.
134. The Projection of a Broken Line upon a Given Line.
Let ABODE, Fig. 105, be any broken line in the plane or in
A'
B' a'
FIG. 105.
E'
space. From the points A, B, C, D, and E perpendiculars are
dropped upon the line OP meeting it at the points A', B', C',
D', and E'. A'B', B'C', C'D', and D'E' are, respectively, the pro-
jections of AB, BC, CD, and DE. Let a, p, 7, and 5 be the
angles between the directed segments of AE and the line OP.
A'B' = AB cos a, B'C' = BC cos 0, C'D' = CD cos 7 (which is
negative, as 7 is obtuse) and D'E' = DE cos 6. Then A'E',
§135]
POINT, PLANE, AND LINE IN SPACE
245
the projection of a directed broken line upon any directed line,
is the sum of the products of the length of each segment of the broken
line by the cosine of the angle between it and the directed line upon
which it is projected.
135. The Normal Equation of a Plane. A plane is determined
if its normal distance, p, from the origin, and the direction angles
of this normal are known. Let ON, or p, Fig. 106, be the normal
FIG. 106.
drawn from the origin to the plane ABC. Let a, (3, and 7 be
the direction angles of this normal. Let P be any point in the
plane. Draw PN, which is perpendicular to ON. The sum of
the projections of x, y, z, and PN upon the normal is equal to
p. The projection of x upon the normal is x cos a, of y is y.cos @,
and of z is z cos 7. The projection of PN upon the normal is
zero, since ONP is a right angle. It then follows that
x cos a + y cos /3 + z cos 7 = p, (2)
which is called the normal equation of the plane.
In the equation Ax + By + Cz = D, let A, B, C, and D be any
real constants. The equation is then the general equation of the
246 MATHEMATICS [§135
first degree in x, y, and z. Upon dividing by ± V A2 + B* + C'2,
this equation reduces to
= ^x + - B
± VA2 + #2 + c2 " ± VA2 + £2 + c2
The sign before the radical is to be chosen so as to make the
right-hand side of the equation positive.
Since the numerical values of the coefficients of x, y, and z
in equation (3) cannot exceed unity, they are each the cosine of
some angle. Call these angles, respectively, a, /3, and 7. Then
cos2 « + cos2 ft + cos2 7 = A2 +^2 + c, + A.+%'+c.
,
h
A2 + 52
a, /3 and 7 are then, by § 133, the direction cosines of some line.
The left-hand side of (3) reduces to
x cos a + y cos /3 + z cos 7 ,
the left-hand side of the normal equation of a plane; and since
the right-hand side of (3) is a positive quantity, call it p, equation
(3) reduces to
x cos a + y cos j8 + z cos 7 = p,
the normal equation of a plane.
equation of a plane is of the first degree in x, y, and z;
and, conversely, every equation of the first degree in x, y, and z is
the equation of a plane. To reduce Ax + By + Cz = D to the
normal form, divide the equation through by ± V A2 + 52 + C2,
choosing that sign before the radical which mil make the right-hand
side positive.
To illustrate: let x + 2y — 3z = — 2 be the given equation.
To reduce it to the normal form divide by — \S\4, which gives
1232
§136] POINT, PLANE, AND LINE IN SPACE 247
2 1
The plane is — 1= units from the origin. The cos a, — —j=. is
V14 VI*
negative, which says that the normal drawn from the origin to the
plane extends into the third, fourth, seventh or eighth octant.
2
The cos /3, .^ is negative, which says that the normal from
the origin to the plane extends into the fifth, sixth, seventh or eighth
3
octant. The cos y, + ,— > is positive, which says that the nor-
mal drawn from the origin to the plane extends into the first,
fourth, fifth or eighth octant. The three direction cosines, then,
show that the normal drawn from the origin to the plane extends
into the eighth octant; or the plane and the three coordinate
planes form a tetrahedron in the eighth octant.
Exercises
Reduce each of the following equations of planes to the normal
form, find the distance of each plane from the origin, and locate the
octant in which the plane forms a tetrahedron with the coordinate
planes.
l.z+j/+z = l. 2. x - y + z = 6.
3. 3x - 2y + 5z - 6 = 0. 4. 2x + 7y - z + 7 = 0.
136. The Slope of a Plane, Given the Elevation of Three Points.
In this section is described a graphic method for determining the
slope, both in magnitude and direction, of a plane surface, if the
relative elevations of any three of its points are known. A
description and an illustration of the method will be given before
the proof.
Let A, B, and C, Fig. 107, be the three points whose elevations
are known. Suppose A the highest. Compute the gradient (in
feet per 100 feet, in feet per 1000 feet, in feet per mile, or in any
other desired unit) from A to B and from A to C. From A
toward B draw AB', whose length, drawn to scale, represents the
gradient from A to B. From A toward C draw AC', whose length,
to scale, represents the gradient from A to C. At the points B'
and C' draw lines, respectively, perpendicular to AB and AC.
248
MATHEMATICS
[§136
These perpendiculars intersect at the point X. Draw AX, which
represents in direction and magnitude the slope of the plane ABC.
To illustrate: suppose B and C are, respectively, 2.26 and 3.20
feet lower than A. Suppose AB = 838 feet, and AC = 866 feet
(measured along the horizontal). The gradient of the plane ABC
in the direction AB is 2.7 feet per 1000 feet; and in the direction AC,
3.7 feet per 1000 feet. Lay off AB' and AC', respectively equal
to 2.7 and 3.7 inches, thus making 1 inch represent a slope of 1 foot
per 1000 feet.1 AX is 3.79 inches long, which shows that any line
on the plane parallel to AX has a gradient of 3.79 feet per 1000
feet. The direction AX is that of greatest slope.
If A, B and C were points upon the plane surface of an
uneroded sedimentary rock, the length of AX would represent
the dip, and the direction perpendicular to AX would be the
strike of the stratum.
If A, B and C represent points upon the ground-water plane, AX
would represent the direction in which the water would be seeping.
1 The drawing here given is reduced. The student should redraw it to scale,
and check the results given here.
§137] POINT, PLANE, AND LINE IN SPACE 249
To lay out the direction of a line having a given slope, say 1 foot
per 1000 feet, proceed as follows: Upon AX as diameter describe
a circle. With A as center and with a radius equal to 1 inch
describe arcs cutting the circle at the points M and N. AM and
AN are then the directions having a drop of 1 foot per 1000 feet.
To find the elevation of any point P, draw AP intersecting the
circle AB'C'X at the point K. The length AK gives the slope of
the plane in the direction AP. With this slope and the distance
AP the elevation of P relative to A may be calculated. In our
specific illustration, AK is 3.25 inches, indicating a drop of 3.25
feet per 1000 feet in the direction AP. The length AP is 2.18
inches, making P 436 feet distant from A. (The scale of ABC
is 1 inch to 200 feet.) From these two measurements P is found
to be 1.42 feet lower than A.
Exercises
1. A, B and C are three points on a plane. AB = 500 feet, AC =
600 feet, BC = 400 feet, measured on the horizontal. The elevations
of the plane at the points A, B, and C are, respectively, 137.6 feet,
132.4 feet, and 135.7 feet. Find graphically the magnitude and
direction of the slope of the plane ABC. Find the elevations of two
points each 300 feet from B and 200 feet from C. Find two directions
having a slope of 3 feet per 1000 feet. Find the slope in the directions
from A to two points each 300 feet from A and 400 feet from B.
[137.] Proof of the Construction Given in the Preceding Section.
In the proof for the above construction let A, Fig. 108, be
taken as the origin of coordinates, and let the XY- plane pass
through the point B. The coordinates of A are then 0, 0, and 0.
Let the coordinates of B be a, — b, and 0; and of C be a', — b',
and c'. The general equation of a plane, by § 136, is
Ax + By + Cz = D. (4)
Since the plane passes through the origin, (0, 0, 0) must satisfy
its equation, which shows that D is zero. Since the plane passes
through the point B, its coordinates must satisfy the equation,
thus giving
aA-bB = 0. (5)
250
MATHEMATICS
[§137
Similarly, the coordinates of the point C must satisfy the equation,
giving
a' A - b'B + c'C = 0. (6)
Solving equations (5) and (6) for A and B we find
bc'C
C. (a,-b, c)
FlG. 108.
With these values, equation (4) reduces to
bc'x + ac'y + (ab1 - a'b)z = 0. (7)
The direction cosines of the upward drawn normal to this plane
are, by § 133,
be' ac' . ab' — a'b
—T=> — T=> and — — j=~
Vi Vi Vi
where
/ = 6V2 + aV2 + (ab'- a'6)2.
6 b'
The gradient of the line AB is — , and of A C is / ., , =j=-
a v a *• -f- c
The equations of the lines B'X and C'X, Fig. 107, on the XZ-
plane are, respectively, x = - and a'x + c'z = b'. The co-
(Z
ordinates of the point X are then x = — andz = - —,
§137] POINT, PLANE, AND LINE IN SPACE 251
Let AT, Fig. 108, be the upward drawn normal to the plane.
Take T such that AT = -y^-,. The coordinates of S, the
.AC
projection of T upon the .X"Z-plane, are —, -- — and
ac
V7 ab' - a'b & ab' - a'b , . ,
— r — 7^ — ' or, — and - — ; — , which are seen to be the
ac' \/I ' a ac'
coordinates of the point X. This shows that the slope of the
plane is in the direction AX. We shall now show that the length
of AX represents the magnitude of the slope of the plane.
The distance AX is ~ Vb*c'z+(ab'-a'b)*. By substituting
etc
the coordinates of X in the equation of the plane passing through
the points A, B, and C, we find the plane to be
units directly beneath X. Upon dividing this expression by the
length of AX we have
— V&V2 + (ab' - a'b)2,
which is equal to the length of AX found above. This shows that
AX represents the magnitude of the slope of the plane ABC.
Since any angle inscribed within a semicircle is a right angle,
a perpendicular drawn to AK, Fig. 107, at the point K will pass
through X. From what was proved above, AK must represent
the gradient of the plane ABC in the direction AP. For the same
reason, AM represents the slope (given) in the direction AM.
CHAPTER XII
[MAXIMA AND MINIMA]
138. Maxima and Minima Defined. Let y be a function of
x (See § 45). If, as x increases, y increases to a certain value,
M, and then begins to decrease, M is called a maximum value
of y. If, as x increases, y decreases to a certain value, m, and
then begins to increase, ra is called a minimum value of y.
A maximum value of a function is not necessarily the largest
value the function may have; and a minimum value is not neces-
sarily its smallest value. A function may have more than one
maximum, and more than one minimum, and a function may have
neither maxima nor minima.
£>'
FIG. 109.
Graphically the maxima and minima values of a function are,
respectively, the lengths of the ordinates of the high and low
points of the graph of the function.
Let ABODE, Fig. 109, be the graph of the function y = /(x).
As x increases from zero to OA', y increases to the value repre-
sented by the ordinate A' A. As x increases from OA' to OB',
y decreases. Then by definition, A'A represents a maximum
value of the function. Similarly C'C represents a second maxi-
252
§139]
MAXIMA AND MINIMA
253
mum value, and B'B and D'D represent two minimum values.
The minimum value DD' is larger than the maximum value A' A.
139. Maximum and Minimum Values by Plotting. When
observed data connecting two variables are plotted upon rec-
tangular coordinate paper, and a smooth curve is drawn through
the plotted points, the maximum and minimum values may be
found approximately from the graph. Thus, from Fig. 15.
the maximum air temperature was approximately 67.5° at 3 : 45,
13^'--
«*-
•< — --(13-2 *") >
FIG. 110.
If the equation of the curve be known it may be plotted point by
point and the approximate values of the maximum and minimum
values taken from the graph. To illustrate this method take the
following problem: From a square piece of tin 13 inches on a
side an equal square is cut from each corner. The edges are then
turned lip forming a box. Find the size of the square cut from
each corner such that the volume of the box shall be a maximum.
Let y be the volume of the box. Let x be the side of the square
cut out from each corner, Fig. 110. The volume of the box is
then y = (13 — 2x)2x. By assigning to x values and calculat-
ing the corresponding values of y the following table is con-
structed:
254
MATHEMATICS
[§139
X
2/ || z
y
0
0
7/2
126
1/2
72
8/2
100
2/2
121
9/2
72
3/2
150
10/2
45
4/2
162
11/2
22
5/2
160
12/2
6
6/2
147
13/2
0
From this table it is seen that a maximum value of y exists
when x lies somewhere between f and f . By giving to x, values
nearer together, the following table is constructed:
X
V
x
y
1.6
153.7
2.6
158.2
1.8
159.0
2.8
153.3
2.0
162.0
3.0
147.0
2.2
162.7
3.2
139.4
2.4
161.4
3.4
130.7
From this table it is seen that the maximum value of y occurs
for some value of x between 2.0 and 2.4 In Fig. Ill is given the
160
150
1.6
1.8
2.0
X
FIG. 111.
2.2
2.4
2.6
plot of the function for values of x ranging from f to f-. From
this graph it is seen that an approximate maximum value of y
is 162.7 A closer approximation may be obtained by assigning
to x a larger number of values between 2.0 and 2.4.
§140]
MAXIMA AND MINIMA
255
Exercises
1. Same as the illustrative problem above, but with the side of the
square of tin equal to, (a) 10 inches; (6) 15 inches; (c) 17 inches; (d)
20 inches.
2. If a represents the side of the square of tin in exercise 1, and x
the side of the square cut out, calculate the ratio x/a for each case.
3. Find the dimensions of the greatest rectangle that can be in-
scribed in a circle with radius 10.
140. The function ax2 + &x + c. In § 31 it was shown that
the function axz + bx + c has a maximum or a minimum value.
-i
-2
(••*.-*)
FIG. 112.
Thus, to illustrate, the function y = 4x2 + 8x — 2 may be put in
the form
y = 4z2 + Sx + 1 - 2,
or
y = (2x + I)2 - 2,
or
7/ = 4(x + l/2)2-2.
The last equation shows that the vertex of the parabola is at the
point ( — TJ-, — 2) , and that the axis of the parabola extends up
from the vertex, as is represented in Fig. 1 12. The function then
256 MATHEMATICS [§141
has a minimum value — 2, which is the value corresponding to
value x = — ^-.
Exercises
Find the maximum or minimum value for each of the following
functions :
1. x2 +2x + 1. 4. 9x2 + 3x - 2. 7. 2x - x2.
2. x2 + 3x - 6. 5. 7x2 _ 2x - 6. 8. 3 - 5x - 2x2.
3. 4x2 + 16x. 6. 8x2 - 3x - 7. 9. 3 + 5x - 2x2.
10. If a body is thrown vertically upward with an initial speed of a
feet per second, its height, h, in feet, at the end of t seconds is given
by the equation In = at — 16. li2. To what height will the body rise
if thrown with an initial speed of 32.2 feet per second? When will it
reach this height?
141. Maxima and Minima Found by Limits. In some problems
the maxima or minima values of a function may be found by a
method illustrated by the following examples :
EXAMPLE 1 : Find the maximum and minimum values, if any,
of the f uncton
2*4-1
To find the maximum and minimum values of the function is
to find the ordinates of the turning points of the graph of the
equation
a2 + 6
Upon solving for x the equation
.
x = y ±V(y-2)(y + 3) (2)
is obtained.
If, in equation (2), values are assigned to y (y = k, where k
is some constant) corresponding values of x are found; thus
locating points upon the graph of the function. These values
of x are given by
x = k ± V(k - 2)(fc + 3). (3)
The equation y = k represents a straight line parallel to the
x-axis, k units distant, and the values of x found from equation
§141]
MAXIMA AND MINIMA
257
(3) are the z-coordinates of the points of intersection of this
straight line with the graph of equation (2).
The second factor, k + 3, under the radical of equation (3)
is positive for all positive values of k, and for all negative values
of k numerically less than 3, or positive for all values of k greater
than — 3. It is zero when k equals — 3, and negative when
k is less than — 3. The first factor is positive when k is greater
than 2, and zero when k equals 2, and negative when k is less
than 2.
When both factors are positive, i.e., when k is greater than 2
or less than — 3, x is real, and the line y = k cuts the graph of
equation (2) in two real points. The z-coordinates of these
points are the values of x found from equation (3).1 When k
FIG. 113.
is greater than — 3 and less than 2, x is imaginary, for then the
quantity under the radical is negative. This shows that all
lines of the system y = k between the lines y = 2 and y = — 3,
fail to cut the curve. When k = — 3 or 2 there is in each case
but one value (or two equal values) of x. The lines y = 2 and
y = — 3 are tangent to the curve.
From the above discussion it is seen that the graph of equa-
tion (2) consists of two branches, one above and tangent to the
1 The student should plot the graph of equation (2). Give to y in (2), or to k in
(3), values ranging from — 10 to + 10.
17
258 MATHEMATICS l§141
line y = 2, the other below and tangent to the line y = — 3.
These two branches extend to infinity, one in the positive, the
other in the negative y direction, for x is real for all positive k
values greater than 2, and for all negative k values less than — 3.
The graph has two turning points, one with a ^-coordinate 2,
the other with a 7/-coordinate — 3. The first is a minimum value,
the second a maximum value of the given function. The mini-
mum value corresponds to the value 2 for x, the maximum value
corresponds to the value — 3 for x.
EXAMPLE 2: Find the maximum rectangle that can be in-
scribed in a circle with radius 10.
In Fig. 113, let x be one-half the length of the base, and z one-
half the length of the altitude. The area, y, is
y =.4z«. (1)
z = \/100 - z2. (2)
Substituting in equation (1)
y = 4z\/100 - x*. (3)
Squaring equation (3)
?/2 = 16z2(100 - z2). (4)
Solving (4) for xz
X2 = 800+ V(200-y)(200 + y) (fi)
16
In equation (5) the quantity under the radical is positive if y
is less than 200 and greater than - 200, (- 200 < y < 200),
and z2 is real. Beyond these values for y, x2 becomes imaginary.
200 then is the maximum value of y. When y equals 200, x2
equals 50, or x = \/50, and from equation (2), z = \/50. The
maximum rectangle is then a square with a side equal to 2 -x/50.
Exercises
1. A gutter with rectangular cross section is made from a strip of
tin 30 inches wide. Find the width of the portion of tin bent up upon
each side so that the carrying capacity of the gutter shall be a
maximum.
HINT : Let y be the area of the cross section, and let x be the width
of the tin bent up. Then y = x(3Q — 2x).
2. Find the length of the lever, exercise 6, § 91, which will make
the upward pull a minimum.
§142]
MAXIMA AND MINIMA
259
142. Another Method of Finding Maxima and Minima. The
method considered here of finding the maxima and minima of a
function depends upon a study of the slope of the tangent line drawn
to the curve representing the function. In Fig. 109, a tangent line
drawn to the curve at any point between P and A makes an acute
angle with the positive direction of the axis of x. The slope of any
such tangent line is then positive. Any tangent line drawn to
the curve between the points A and B makes an obtuse angle with
the positive direction of the axis of x. The slope of any such tan-
gent line is then negative. The slope of any tangent lines drawn
to the curve between the points B and C, and between the points
D and E, is positive. The slope of any tangent lines drawn to the
curve between the points C and D is negative. When the slope of
the tangent line is positive, the curve ascends; when it is negative
FIG. 114.
the curve descends. When the slope of the tangent line changes
from positive to negative the curve changes from ascending to
descending and possesses a high point. When the slope of the
tangent line changes from negative to positive, the curve changes
from descending to ascending and possesses a low point.
EXAMPLE 1 : In § 139 the volume, y, of a box formed from a
square piece of tin 13 inches on a side was found to be
y = (13 - 2z)2z = 169x - 52x2 + 4z3. (1)
Let us suppose that the curve in fig. 1 14 represents the graph of
the function. Let P be any point on the curve whose coordinates
260 MATHEMATICS [§142
are x and y. Let Q be any other point on the curve whose co-
ordinates are x + h and y + k. Since these points are on the
curve, their coordinates must satisfy equation (1), from which
we obtain
y = 169x - 52x2 + 4x3 (2)
and
y + k = 169(x + h) - 52(x + h)2 + 4(x + ft)3. (3)
Equation (3) reduces to
y + k = 169x + 169ft - 52x2 - 104ftx - 52ft2 + 4x3
+ 12x2ft + 12xft2 + 4ft3. (4)
Subtracting equation (2) from equation (4) there results
k = 169ft - 104ftx + 52ft2 + 12ftx2 + 12xft2 + 4ft3,
or
| = 169 - 104z + 52ft + 12.x2 + 12xft + 4ft2. (6)
k/h represents the slope of the secant line drawn through the
points P and Q. If the point Q is taken nearer and nearer the point
P, the secant line rotates about the point P and approaches the
tangent line drawn to the curve at that point. The slope of the
tangent line is then the limit of the slope of the secant line as the
point Q approaches the point P. But, as Q approaches P, h
approaches zero, and the right-hand side of equation (6) ap-
proaches 169 — 1042 + 12x2. Therefore, the slope of the tangent
line drawn to the curve representing the volume of the box is
r 1 04. 1 fiQ
12x2- 104x + 169, or 12 I x2 ~ jrf * + ^
or
slope =12(x-f)(x-f). (7)
Since the side of the square of tin is 13 inches, x, the side of the
square cut from each corner, must be less than 6j inches. Thus,
/ 13\
the last factor, ( x — ^-j , of equation (7) is always negative. The
second factor, (x — -g-j, is negative if x is less than -^-, and
13
positive if x is greater than^-- The expression representing the
13
slope is then positive for values of x ranging from 0 to - and
§142] MAXIMA AND MINIMA 261
for these values of x the curve representing the function rises.
13
For all values of x ranging from -~- to 13 the slope is negative,
and the curve descends. Thus, the function representing the
13
volume of the box has a maximum value when x = -^ • To find
o
the maximum volume substitute this value for x in equation (1),
which then gives
y = 162.74.
EXAMPLE 2: The area of a rectangle inscribed in a circle,
radius 10, is given by equation (3), example 2, § 141.
y = 4x\/100 - x2, (8)
or
y2 = 16x2(100 - a;2)
= IGOOx2 - 16x4. (9)
A value of x which makes y a maximum will at the same time make
y2 a maximum. For brevity call y1 = u, and equation (9)
becomes
u = 1600x2 - 16x4. (10)
As in the preceding example, let P and Q be any two points upon
the curve, Fig. 114, which represents the relation between u and x.
If x and u are the coordinates of the point P, and x + h and u + k
those of the point Q,
u = IGOOx2 - 16x4 (11)
and
u + k = 1600(x + hY - 16(x + hY
= IGOOx2 + 3200/ix + IGOO/i2 - 16x4 - 64/ix3
- 96/i2x2 - 64^3x - 16/i4. (12)
Upon subtracting equation (11) from equation (12) there results
k = 3200Ax + 1600/12 - 64/ix3 - 96/i2x2 - 64h3x - I6h*, (13)
or
kh = 3200* + 1600A - 64z3 - 96/ix2 - 64^2x - 16A3. (14)
As h approaches zero, the right-hand side of equation (14) ap-
proaches 3200x — 64x3, or
slope = 64z(50 - x2) = 64x( \/50 + x)( \/50 - x). (15)
262 MATHEMATICS [§142
From the right-hand side of equation (15) it is seen that the slope
is negative if £ is greater than \/5Q, and positive if x is positive
and less than -y/50- The curve ascends for all values of x from
0 to \/50, and descends for all values of x greater than \/50-
Thus, u has a maximum value when x = \/50- When u is a
maximum, y is a maximum; therefore the maximum rectangle is
that for which x = \/50. The remaining discussion of this
problem has been given in § 141.
EXAMPLE 3 : Find the most economical proportion of a closed
cylindrical can. Let z be the altitude, x the radius of the base,
v the volume, and y the surface of the can. The problem is to find
the ratio x/z which will make y a minimum for any constant
value of v.
V = TTZX2 (16)
y = 2irzx + 2irx* (17)
Upon eliminating z between equations (16) and (17)
y = ~ + 2irx*. (18)
U/
2v
y + k = — + 27r(a; + h}\ (19)
Subtracting equation (18) from equation (19)
+
- 2vh
or K
or
k
h
As h approaches zero, the right-hand side of equation (20) ap-
2v
preaches 4irx 2- Then
2v
slope = lirx -j
x2
or
A.™- / 11 \
(21)
§142] MAXIMA AND MINIMA 263
3 / V
If x is less than * /— the quantity within the parentheses of
3 / V
equation (21) is negative. If x is greater than -1/5— this quan-
3 / V
tity is positive. For all positive values of x less than \IJT- the
\ iir
slope is negative, and the curve representing y is descending; for
3 / V
values of x greater than \f^- the curve is ascending. Thus,
\ 2?r
8/7
the surface is a minimum when x = \l^- • From equation (16)
= \ — ,
\ 7T
whence -- = 4-- Since the v does not appear in
Z
this equation, the ratio of x to z is constant, ^, for all values of v.
The most economical proportions are such that the altitude is
equal to the diameter of the base.
The method used in the three above cases is applicable to any
problem where the functional relation between the two variable
quantities x and y is given by means of an equation.
Exercises
1. Find the most economical dimensions of a pan with square base
and vertical sides, if it is to hold 5 gallons (1 gallon = 231 cubic
inches).
2. Find the most economical proportions for a cylindrical can,
open at the top, if it is to hold 1 quart (1 quart = 57f cubic inches).
3. What is the ratio of the length of the legs of a right triangle
when its area is a maximum, if the hypotenuse is a constant, 10?
4. The strength, y, of a rectangular beam varies as the product of
its breadth, 2, by the square of its depth, x; find the dimensions of the
strongest beam which can be cut from a log 20 inches in diameter.
CHAPTER XIII
[EMPIRICAL EQUATIONS]
143. Empirical Equation Defined. A series of readings con-
necting two or more variable numbers should always be plotted
upon rectangular coordinate paper (squared paper), thus rep-
resenting to the eye the variation of one of the numbers with
respect to the other.
In Table VI, there are given seven readings of the velocity
of water through a sample of sand. These seven velocities
correspond each to a measured hydraulic gradient. This set of
readings when plotted upon squared paper is shown in Fig. 115.
Velocity- Feet 24 Hours
i-. to co *>. 01 o .j
OOOOOOOc
C
4
+
%
^
.'*
^
.,"
'"^
X
^
X1
'
x*
•
3
£X
•'
*r'
-'
^
r+
X-
*' L
10 20 30 40 50 60 70 80 90 100 110 120
Hydraulic Gradient-Feet per 1000 Feet
FIG. 115.
The plotted points do not lie upon a smooth curve. The reason
for this is that there are errors of observation — errors both in
the determinations of the velocities and in the hydraulic gradients.
If there had been no errors of observation and if all pairs of read-
ings h'ad been taken under exactly the same conditions, the plotted
264
§143]
MAXIMA AND MINIMA
265
points would have fallen upon a smooth curve. This smooth
curve which we are trying to locate by the plotted points represents
graphically the law of nature which we are trying to determine
or study, viz., the law governing the flow of water through this
particular sample of sand, relative to the hydraulic gradient.
From Fig. 115 it is apparent that the plotted points do not
justify the assumption of a locus other than a straight line along
which the points would fall if there were no errors of observation.
The true relation connecting the variable numbers measured
may not be the linear law, but with the data at hand this is the
only law which can be assumed as an approximation to the law
of nature. With the aid of a transparent triangle the line QA
Velocity- Feet per Second
to CO CO CO CO
05 0 J- JO OS
~-~ — '
•r1
>
X
f*
xt
I
)
^
\
>
I
\
\
\
1
\
\
\
\
.1 .2 .3 .4 .5 .6 7
Depth -Part of Total
FIG. 116.
.8 .9
is drawn among the plotted points. The position of the line is
located by the eye as the probable line upon which the points
would fall if the true law were linear and if there were no errors of
observation. The equation of this line is then y = ax -f- b,
where x represents the hydraulic gradient and y the velocity.
Since the line passes through the origin, b is zero, a is found to
be 0.51. The equation of the line becomes y = 0.51x, and is
called an empirical equation connecting velocity and hydraulic
gradient.
266
MATHEMATICS
[§143
Frequently it is desirable to express the relation existing be-
tween variable numbers in the form of an equation, even though
this equation represents only an approximation to the true
law. Such an approximate equation when found by means of
experimental readings is called an empirical equation.
As a second illustration of a method of finding the empirical
equation, consider the plotted points in Fig. 116. From this
figure it is seen that the law connecting velocity and depth is not
linear, but appears to be parabolic. The vertex of the parabola
is at, or nearly at, the point (0.29, 3.262)1. If the origin be moved
to this point the readings (which are given in Table IX) become:
X'
y \\ x1
y
- 0.29
- 0.067
0.21
- 0.034
- 0.19
- 0.032
0.31
- 0 . 082
- 0.09
- 0.009
0.41
- 0.135
0.01
- 0.001
0.51
- 0.203
0.11
- 0.010
0.61
- 0.286
and the equation connecting x' and y' is y' = a(x')2.
If (x')2 be replaced by X the equation becomes
y' = aX (1)
Hence if y"s are plotted against X's, the squares of the x"s,
a straight line will be obtained if the curve in Fig. 116 is a parabola
with its vertex at the point (0.29, 3.262). These plotted points
are represented in Fig. 117. The points indicated by circles cor-
respond to [positive values of x', while those indicated by dots
correspond to negative values. A straight line drawn among the
points with the aid of a transparent triangle gives the equation
(2)
(3)
Since X= (x'}'
Since
and
y' = - 0.77Z.
y' = - 0.77(z')2.
x' = x - 0.29
y' = y - 3.262,
1From Fig. 116 it may at first appear that the vertex of the parabola is on the
line x = 0.3; but by noting the vertical positions of the points plotted equidistant
from this line it is seen that the line of symmetry of the curve is slightly to its left.
§143]
EMPIRICAL EQUATIONS
267
equation (3) becomes
y - 3.262 = - 0.77(a; - 0.29)2,
or
y = - 0.77(z - 0.29)2 + 3.262,
or
y = - 0.77 xz + 0.447 x + 3.197,
an empirical equation connecting velocity with depth.1
(4)
(5)
y
-0.3
/°
/
,
/
'
.A
V
/
/
/
/
<
to
0
/
0.64 0.08 0.12 0.16 0.20 0.24 0.28 O.S2 0.36 X
FIG. 117.
In the first illustration above the empirical equation is of
the form y = ax + b, in the second illustration it is of the form
1 Compare equation (4) with the equation given in exercise 4, § 31.
268 MATHEMATICS [§144
y = ax2 + bx + c. These equations are each a special case of the
more general equation
y = a0x» + aiZ"-1^- azxn-z+ . . . + an_iz + an, (6)
where a0, 01, 02, . . . , an are coefficients to be determined and where
n is a positive integer. To determine the n + 1 coefficients of
equation (6), n + 1 equations are required. Hence the curve
corresponding to (6) may be made to pass through (n + 1)
points of the curve drawn among the plotted points. Empirical
equations of the form (6) are undesirable if more than three or
four terms are required.
Exercises
1. From the data given in Table VII, find empirical equations
connecting yield and head, for the 8-inch and 14-inch wells.
2. From the data given in Table VIII, find an empirical equation
connecting T and W.
144. Coefficients Determined by the Method of Least Squares.1
In Fig. 115 a linear law was assumed connecting velocity and
hydraulic gradient. A straight line was drawn among the points
and the values of the coefficients a and b were determined from this
line. If the plotted points had been so scattered that it would
have been difficult to estimate by eye the position of the line, the
coefficients a and 6 could have been determined by the method of
least squares. This method will be illustrated by using the two
illustrations of the preceding section. In the first illustration the
equation is assumed to be y — ax — b = 0. Upon substituting
the seven pairs of values given in Table VI there results:
16.9 - 31.9o -6 = 0
11.4 - 20. 8a -6 = 0
24.5 - 54. la - 6 = 0
4.3- 10.0a-6 = 0
10.1 - 21. Oa -6 = 0
58.8 - 119. la -6 = 0
22.7 - 55. 8a -6 = 0
1 The student will verify the numerical work of this section.
§144] EMPIRICAL EQUATIONS 269
Here are seven equations from which to solve for a and &. This
can, of course, not be done, since the number of equations is greater
than the number of unknowns. The values of a and b, which
are regarded as the best approximations, are found as follows:
First multiply each member of each equation by the coefficient of
a in that equation and add the resulting seven equations. In
the illustration this gives
10627.52 - 22216.51a - 312.76 = 0. (1)
Next multiply each member of each equation by the coefficient of
6 in that equation and add. In the illustration this gives
148.7 - 312.7a - 76 = 0. (2)
Next solve equations (1) and (2) for a and b. This gives in the
illustration
a = 0.48
and
6 = - 0.2
Thus, the equation of the straight line drawn among the points
determined by this method is
y = 0.48z - 0.2
The equation found by the method of the preceding section is
y = O.Slz
In the second illustration (See Fig. 116) the empirical equation
was assumed to be of the form y — ax1 — 0£ — 7 = 0. Upon
substituting the ten pairs of values given in Table IX there
results;
3.195 - 0.00 a- 0.0/3 -7 = 0
3.230 - 0.01 a - 0.1/3 - 7 = 0
3.253 - 0.04 a - 0.2/3 -7 = 0
3.261 - 0.09 a - 0.30 -7 = 0
3.252 - 0.16 a- 0.40 -7 = 0
3.228 - 0.25 a - 0.50 - 7 = 0
3.181 - 0.36 a- 0.60 -7 = 0
3.127 - 0.49 a- 0.7/3 -7 = 0
3.059 - 0.64 a - 0.80 -7 = 0
2.976 - 0.81 a - 0.90 - 7 = 0
270 MATHEMATICS [§145
To find the "best values" for a, /3 and 7 from these equations,
first multiply each member of each equation by the coefficient of
a in that equation and add the resulting ten equations. There
results
8.8289 - 1.5333a - 2.025/3 - 2.867 = 0. (3)
Next multiply each member of each equation by the coefficient of
/3 in that equation and add the resulting equations. There
results
14.09 - 2.025a - 2.85/3 - 4.57 = 0. (4)
Next multiply each member of each equation by the coefficient of
7 in that equation and add the resulting ten equations. There
results
31.762 - 2.85a - 4.5/3 - 107 = 0. (5)
Next solve equations (3), (4), and (5) for a, /3, and 7, and obtain
a = - 0.81
/3 = 0.48
7= 3.19
Hence an empirical equation found by this method is
y = - 0.81x2 + 0.48z + 3.19
This equation places the vertex of the parabola at the point
(0.296, 3.26).
These two illustrations show that a large amount of numerical
work is required to find the coefficients by the method of least
squares. If the plotted points fall upon, or nearly upon, a smooth
curve the methods of the preceding section should be used.1
145. Logarithmic Paper. The student will read § 73 again.
In Fig. 118 the square ADCB is one unit on a side. At the bottom
and at the left-hand side of this square are logarithmic scales, each
ranging from 1 to 10. Thus on the lower scale, AD, the numbers
are placed opposite their logarithms measured on the uniform
scale CB. For example, 1 is placed opposite 0; 2 opposite 0.301;
3 opposite 0.477; 4 opposite 0.602; 5 opposite 0.699; 6 opposite
0.778; 7 opposite 0.845; 8 opposite 0.903; 9 opposite 0.954,
and 10 opposite 1. The distance from A to the position of any
number on the scale AD is equal to the logarithm of that number.
1 The theory upon which the method of least squares is based may be found
in texts on "Least Squares."
§145]
271
Through the points of division of the AD and AC scales, vertical
and horizontal lines are drawn forming what is called logarithmic
coordinate paper.1 To illustrate the use of logarithmic coordinate
0 .1
1
.2 .3
I
.<
i
.5
i
.£
:
8
.9
LOg
, _ 9
X> p
^f
s"
^L
x^
*
**
7
^
^
-^
<L
^
n
t
~2L
-.5
,/'
4
3
-.2
-J.
. 0
Al 2 MZ 4 56789 10 D
FIG. 118. — A single square (reduced) of logarithmic paper.
paper for determining empirical equations, consider the data
given in the accompanying table. These data are plotted upon
squared paper in Fig. 119, and upon logarithmic paper in Fig. 118.
X
y
X
y
1.5
3.05
6.5
6.40
2.5
3.92
7.5
6.85
3.5
4.65
8.5
7.25
4.5
5.30
9.5
7.70
5.5
5.82
'For the work in this chapter the student should provide himself with Single
Logarithmic cross-section paper, form Af4, Semi-logarithmic coordinate paper,
form M5, and Multiple Logarithmic coordinate paper, form M 6.
272
MATHEMATICS
[§145
In Fig. 118 the plotted points lie upon a straight line EF. If we
imagine the logarithmic rulings to be removed and rulings corres-
ponding to the scales CB and BD to be drawn, the line EF would
then appear upon a sheet of squared paper with the origin at A
and the point (1, 1) at B. The equation of the line EF is then
y = aX + b, where X and Y are the rectangular coordinates of any
point, P, on the line.1 a, the slope of the line, is found by measure-
Y
^
^
^
^
^
X
s*
X
/
S
0123456789
FIG. 119.
ment,2 to be \. b, the ^-intercept AE, is found from the scale DB
to be equal to 0.398, which, from the scale AC, is seen to be equal
to log 2.5. The equation of the straight line EF is then
7 = iZ + log2.5 (1)
In this equation Y is the distance, PM, of any point, P, from the
line AD; but this distance is the logarithm of y. Similarly, X, the
distance PN, is the logarithm of x. Hence equation (1) becomes
log y = £log x + log 2.5,
1X and Y are used to represent the rectangular coordinates of a point in contra-
distinction to the x and y when the logarithmic scales are used.
2 The student is supposed to reproduce Fig. 118 on a sheet of logarithmic paper,
form Af4.
§145]
or
or
or
EMPIRICAL EQUATIONS
273
I
10
9
S
7
G
5
4
3
2
1
log y = log x^- + log 2.5,
log y = log (2.5x^2),
(
\
y
/
/
/
.
1
/
(2)
FIG. 120.
an empirical equation connecting the x and y values given in the
above table.
18
274
MATHEMATICS
[§145
As a second illustration, consider the data given in the accom-
panying table. These data are shown plotted upon squared paper
in Fig. 120, and upon logarithmic paper in Fig. 121. Upon the
E
X
: v \
1 *
y
1.2
2.15
2.0
5.90
1.3
2.50
2.3
7.80
1.5
3.35
2.5
9.30
1.7
4.30
CO .1
10| u
\
-.4
-0
A\ 2 3 ^ 5 6 1 8 9 10;;
FIG. 121.
logarithmic paper the curve is a straight line. Hence its equation
is Y = aX + b. The a is found by measurement to be 2. The b
is seen to be log 1.5. Then
or
Y = 2X + log 1.5,
; y = 2 logs + log 1.5,
§145]
or
or
or
EMPIRICAL EQUATIONS
logy = logz2 + log 1.5,
log y = log 1.5x2,
y = 1.5z2,
275
an empirical equation connecting the x- and y-values of the above
table.
As a third illustration consider the data given in the accom-
panying table. These data are shown plotted upon squared paper
in Fig. 122 and upon logarithmic paper in Fig. 121. The plotted
points in Fig. 121 lie upon the straight line GH, whose equation is
Y = aX + b. a is found to be — 1, and 6 the log 1.5. Hence
Y = -X+ log 1.5,
276
MATHEMATICS
or
or
or
or
X
. V
x y
1.5
10.0
4.5
3.30
2.0
7.5
5.0
2.98
2.5
6.0
6.0
2.49
3.0
5.0
7.0
2.12
3.5
4.25
8.0
1.87
4.0
3.73
9.0
1.65
log y - — log x + log 1.5,
log y = log - + log 1.5,
sc
log y = log -f- '
1.5
=
an empirical equation connecting the x and y values of the table.
From the above illustrations it is seen that if plotted points
upon logarithmic paper determine a straight line, an empirical
equation connecting the numbers plotted is of the form
y = Kz»,
where n is the slope of the line as it appears on the logarithmic
paper, and where K is the reading on the logarithmic scale of the
point of intersection of the line with the vertical line passing
through the point (1, 1). Conversely, any equation of the form
y = Kxn plotted upon logarithmic paper gives a straight line.
For, by taking the logarithm of each member of the equation there
results
log y = n log x + log K,
which is of the form Y = nX + b.
From the above illustrations it is seen that the uniform scales at
the top and right-hand edge of the logarithmic paper are not used
in the actual determination of the constants in the empirical
equation. These scales are placed there for instructional pur-
poses _only.
§146]
EMPIRICAL EQUATIONS
277
Exercises
1. Through the lower left-hand corner of a sheet of logarithmic
paper, form M 4, draw lines making angles of 30°, 45°, and 60° with the
horizontal. Find the equations connecting the x and y for each of
these lines.
2. Through the upper right-hand corner of a sheet of logarithmic
paper, form M 4, draw lines making angles of 30°, 45°, and 60° with the
horizontal. Find equations connecting the x and y for each of these
lines.
3. Through the upper left-hand corner of a sheet of logarithmic
paper, form M 4, draw lines making angles of 30°, 45°, and 60° with the
vertical. Find equations connecting the x and y for each of these
lines.
4. Through the lower right-hand corner of a sheet of logarithmic
paper, form Jl/4, draw lines making angles of 30°, 45°, and 60° with the
horizontal. Find equations connecting x and y for each of these lines.
5. From a sheet of logarithmic paper, form M 4, find to three deci-
mal places the logarithm of (a) 2.1; (6) 3.6; (c) 4.8; (d) 5.2; (e) 7.21;
(/) 9.3; (?) 9.8.
6. Find an empirical equation connecting x and y of the accom-
panying table.
X
y
1 x
y
1.5
3.55
4.0
1.59
2.0
2.80
5.0
1.22
2.5
2.34
6.0
1.14
3.0
2.02
7. Draw lines upon a sheet of logarithmic paper, form Af4, repre-
senting
(a) y = x2.
(6) y=x*.
(c) y = 2x*.
(d) y = 2x^.
(e) y = -.
(/) y = x
(g) y =
(h) y =
(t) y =
(7) y = 9.2z~?*.
146. Multiple Logarithmic Paper. In the preceding section
logarithmic paper was used (form M4) in which each of the two
278
MATHEMATICS
[§146
logarithmic scales extended from 1 to 10, or the corresponding
uniform scales extended from 0 to 1. It is apparent that if num-
bers falling beyond these limits were to be plotted, the coordinate
paper would have to be extended. Fig. 123 represents a sheet of
100
90
80
70
GO
GO
40
30
20
10
9
8
7
6
5
4
3
2
1
A
? H Q I
f
;
/
/
I
/
C
•/
/
F,
f
f
. /
/
/
/
1
I
/
/
/
/
D
F
1 2 3456 78910 20 30 40 C060 100
K P M
FIG. 123. — Four squares of multiple logarithmic paper (reduced).
coordinate paper in which each logarithmic scale extends from 1 to
100. The square ADBC is the same as the square ADBC of Fig.
118, or of form M 4.
Since
log 20 = log 10 + log 2
= 1 + log 2
= AD + AK
= AM
DM = AK
Thus the vertical line marked 20 is at the same distance from D
as the line marked 2 is from A. Similarly the vertical lines
§146]
EMPIRICAL EQUATIONS
279
marked 30, 40, . . . , 90, and 100 are at the same distances from
D, respectively, as the lines marked 3, 4, . . .,9, and 10 are
from the point A . Similarly the portion CG of the vertical scale is
the same as the portion AC excepting that the numbers attached
to the upper portion are ten times the numbers attached to the
corresponding points on the lower portion. By the addition
of squares like ADBC the logarithmic paper may be extended
over any range of values on the horizontal or vertical scales.
It is apparent from an examination of Fig. 123 that the point
(1, 1) (the origin) may be taken at any corner of a square. Thus,
if B is taken as origin, the point A is (0.1, 0.1) and / the point
(10, 10); if H is taken as the origin, A is the point (0.1, 0.01)
and I the point (10, 1). In this work it is supposed that the
student has a supply of multiple logarithmic coordinate paper,
form M6.
It is apparent that with logarithmic coordinate paper the origin,
i.e., the point (1, 1), may be taken only at the corners of the
squares, while with squared paper the origin, i.e., the point (0, 0),
may be taken at any convenient point.
To illustrate the use of multiple logarithmic paper consider the
data given in the accompanying table. The plotted points cor-
responding to these numbers are shown in Fig. 123. In plotting
these points the point B is taken as the origin. The slope of the
line PQ is 2 and the intercept is log 0.4. Hence the equation con-
necting the x- and ^/-values of the table is y = 0.4x2.
X
y II x
y
0.5
0.10
2.0
1.64
0.7
0.20
3.0
3.70
0.9
0.33
4.0
6.50
1.5
0.92
5.0
10.00
Exercises
1. Find an empirical equation between x' and y', using the numbers
given in the table on page 266.
2. Find an empirical equation connecting W and T, using the data
given in Table VIII.
HINT: First move the zero to correspond to 4° centigrade. Let T'
280
MATHEMATICS
[§147
be the temperature referred to this new zero. Find an equation be-
tween W and Tr, and then replace T' by T - 4.
3. Find an empirical equation connecting I and W, from the data
given in Table XI.
147. Translating the Curve on Logarithmic Paper. Consider
the data given in the accompanying table. These data plotted
upon multiple logarithmic paper are represented by small circles
in Fig. 124. The plotted points do not lie upon a straight
5
3
2
1
0.5
0.3
0.2
0.1
\
V
\
\
\
, \
V
\
\
\
\
*s
\
. V
\
\
\
°\
\
\
\
(
\
t
\
i
N
•
X
0.1
0.2 0.3
0.5 1
FIG. 124.
5 E 10
line, but if they are translated one unit to the right they fall upon
the straight line EF. Moving the points one unit to the right is
equivalent to plotting y against x + 1. Let x + 1 be represented
by x'. From the line EF, the equation connecting y and x' is
found to be
y = 1.05(x')-*. (1)
Since x'= x — 1, equation (1) gives
y = 1.05(x - 1)-*,
an empirical equation connecting x and y.
§148]
EMPIRICAL EQUATIONS
281
X
y i
X
y
0.1
8.00
2.0
0.40
0.2
4.75
2.5
0.31
0.3
3.30
3.0
0.25
0.5
1.95
4.0
0.17
0.7
1.3$
5.0
0.13
1.0
0.90
6.0
0.105
1.5
0 57
Frequently it may happen that by translating the curve in the
x direction or in the y direction the plotted points upon logarith-
mic paper will lie upon a straight line.
Exercise
1. By plotting upon logarithmic paper, form M6, find an empirical
equation connecting volume and pressure, from the data given in
Table X.
148. Equations Plotted upon Logarithmic Paper. If a large
number of readings, say x, are to be substituted in a formula giving
a second number, say y, it may be desirable to plot a curve repre-
senting the equation connecting x and y, and from this curve to
scale off the values of y corresponding to the readings of x. If this
equation plotted upon logarithmic paper gives a straight line it is
an easy matter to construct the curve. Thus, as an illustration,
consider the formula
Q = 3.37Ltf^, (1)
giving the discharge Q, in cubic feet per second (second-feet), over
a trapezoidal weir. L is the length, in feet, of the crest of the weir,
and H is the depth, in feet, of the water on the crest. This formula
for various values of L will give on logarithmic paper a series of
parallel lines of slope f .
Lines for L equal to 1, 2,. 3, 4, and 5 are given in Fig. 125.
If only one curve of the family is to be drawn, as the line for
L = 2, it is advisable to use a single square of logarithmic paper,
form M4. From Fig. 125 it is seen that the line L = 2 crosses six
squares. Instead of using these six squares a single square may be
traversed six times as shown in Fig. 126. The segments of the line
282
MATHEMATICS
[§148
in Fig. 126 are lettered the same as in Fig. 125. The segment AB
is not shown in Fig. 126 as it coincides with the segment F2.
The terms tenths, units, and tens written above the lines indicate
the denomination of the figures on the vertical scale. Similarly
L
5432 1
10
1.0
o.oi
lii\
//
s
M
Z
a
o.i i 10
FIG. 125. — Curves for Q = 3.37L.ff2£ plotted upon logarithmic paper.
the terms hundredths, tenths, etc., written below the lines indicate
the denomination of the figures on the horizontal scale. Thus, to
illustrate, if the reading of H is 0.7, the line segment DE is used,
and from it and the vertical scale the discharge, Q, is found to be
3.95 second feet.
§149]
EMPIRICAL EQUATIONS
Exercises
283
1. Upon a sheet of logarithmic paper, form M6, draw a series of
curves for
Q = 3.
for L = 1, 2, 3, 4, and 5.
z
7
^
1 8 9 10
FIG. 126.— Curve for Q=3.37-2HM plotted upon one square of log-
arithmic paper.
2. Upon a sheet of logarithmic paper, form M 4, draw a series of
line segments for
Q = 3.37LH*
if L =3, and if H ranges from 0.1 to 10.
149. Semi-logarithmic Paper. Semi-logarithmic paper, form
Mb, is coordinate paper ruled logarithmically in one direction and
uniformly in the other.
284
MATHEMATICS
[§149
The data given in the accompanying table, plotted upon semi-
logarithmic paper, is shown in Fig. 127.
X
y
0.2
3.18
0.4
3.96
0.6
5.00
0.8
6.30
or
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
FIG. 127. — Semi-logarithmic paper (reduced).
The equation of this straight line is
, Y = fcc + log 2.51,
log y = \x + log 2.51,
or
log
M x
2.51 2
§149]
or
or
EMPIRICAL EQUATIONS
y
2.51
285
y = 2.51(10)z/2,
an empirical equation connecting the x- and ^-values of the table.
Exercises
1. Find an empirical equation connecting the x- and the y-values
given in the accompanying table.
X
y
0.2
5.8
0.4
4.4
0.6
3.4
0.8
2.6
APPENDIX
GREEK ALPHABET
LETTERS
NAMES
LETTERS NAMES
LETTERS
NAMES
A
a
Alpha
I
4
Iota
P
P
Rho
B
ft
Beta
K
*
Kappa
2
ff, S
Sigma
r
y
Gamma
A
X
Lambda
T
T
Tau
A
d
Delta
M
n
Mu
T
V
Upsilon
E
6
Epsilon
N
t
Nu
4>
(f),<p
Phi
Z
r
Zeta
H
{
Xi
X
X
Chi
H
n
Eta
O
o
Omicron
^
t
Psi
e
6,&
Theta
n
*
Pi
n
0}
Omega
a X b, or a-b, or ab
a -T- b, or a : b, or -r
a^
a3
a4
a" = a a a . .a
a^ = Va
a1/* =\/a
a° = 1
SYMBOLS
plus.
minus.
plus or minus.
minus or plus.
equals.
a times b.
a divided by b.
parentheses, brackets, braces,
a square.
a cube
a to the fourth power,
a to the nth, or, a nth (n a's multi-
plied together, if n is a positive
integer),
square root of a.
cube root of a.
rth root of a.
286
APPENDIX 287
a', a", a'", . . ., a" (accents) read, respectively, "a
prime," "a second," "a third,"
etc.
Qo, 01, a2, . . . , a« (subscripts) read, respectively,
"a sub zero," "a sub one," "a
sub two," etc.
7* is not equal to.
> is greater than.
< is less than.
is equal to or greater than,
is equal to or less than.
.^ angle.
\n_ (read, "factorial n") means the
product of all the integral num-
bers from 1 up to and including
n. Thus: |5_ = 1-2-3-4-5.
n\ (1_^ is sometimes written n!).
|a| the numerical, or absolute, value
of a.
/Or); <t>(x); F(x)', ^(x), etc. function of x.
logo x the logarithm of x to the base a.
log x the logarithm of x to the base 10-
2 the sum of such terms as
i — ft the sum of such terms as, . . .,
2 asi varies from a to /3.
TT the ratio of the length of the cir-
cumference of a circle to the
length of its diameter. An ap-
proximate value of TT is 22/7. A
better approximation is 3.1416.
e the base of the natural (or hyper-
bolic) logarithmic system, e =
2.71828 approximately,
the number of permutations of
n things taken r at a time.
the number of permutations of n
things taken all at a time.
288
»CV =
00
lim
_l!L
In In
u.m Y
= (*
MATHEMATICS
the number of combinations of n
things taken r at a time.
infinite.
limit.
approaches.
the limit of Y as x approaches a.
FORMULAS
Algebra
5.
6.
7.
8.
9.
10.
~£~ a = d
\am]n = a
l"1
J =
a) (6) = - (06).
o)(- 6) = 06.
+ 6)2 = a2 + 2a6 + b2.
+ b)3 = a3 + 3a26 + 3a62 + b3.
+ 6)4 = a4 ± 4a36 + 6a262 ± 4a63
/ •• \
± -
= a" +
11. a2 + 2a6 + b2 = (a + 6)2.
12. a2 - 62 = (a + 6) (a - 6).
13. a3 - b3 = (a - 6)(a2 + ab + 62).
14. a3 +b3 = (a + b) (a2 - a& + 62).
64.
n(n - l)(n - 2)
-
15. If
c = 0, x =
16. logo (xy) = Iog0 x + logo y.
2C
17. logo - = logo x - logo y.
18. Iog0 xn = n logo x.
19. loga -- = - logo X.
£>
20. logo a; = (logb x)(loga 6).
22. log. x = log€ 10 logic x = 2.3026 logic x.
23. logic x = logic e log,, x = 0.4343 loge x.
APPENDIX
289
24. a + [a + d] + [a + 2d] + • • • + la + (n - l)d~]
25. a + ar + ar2 + ar3 +
26. a + ar + ar2 + ar3 +
27. 1+2+3+4 + • •
28. I2 + 22 + 32 + 42 + •
+ ar""1 =
to oo =
•(»-
- r")
1 - r
1 - r
> if |r|< 1.
+ » = 2 (n +
+ n2 = a(n + l)(2n + 1).
29. I3 + 23 + 33 + 4' + • • - + n3 = fj-(n +1)1*
to ~ = Hm r i + 1? =
» * «L n
so. i + p + + +
31. e* = 1 + x + + ~
= 2.71828.
to oo (for all values of x).
32. log, (1 + z) = + x - - + - + ... to oo (if - 1< x <+ 1).
33. e~x = 1 — x2 + -jTj — ^ + • • • to oo (for all values of x).
• (n - r + 1) =
34. nPr = n(n - l)(n - 2)
35. nPB = P» = n.
36. nPr = n(._iP,_i).
37. nCr = ~nPr =
n
In — *r
Trigonometry
38. sin a esc a = 1.
39. cos a sec a = 1.
40. tan a cot a = 1.
41. sin2 a + cos2 a = 1.
42. sec2 a = 1 + tan2 a.
43. esc2 a = 1 + cot2 a.
, .
44.
cos a
19
= tan a.
290 MATHEMATICS
. COS a
45. — - = cot a.
sin a
46. sin (90° ± a) = cos a.
47. cos (90° + a) = + sin a.
48. tan (90° ± a) = +_cot a.
49. sin (180° ± a) = + sin a.
50. cos (180° ± a) = - cos a.
51. tan (180° + «) = ± tan a.
52. sin (270° ± a) = - cos a.
63. cos (270° ± a) = ± sin a.
54. tan (270° ± o) = + cot a.
56. sin (a + )3) = sin a cos j3 + cos a sin /3.
56. cos (a ± /3) = cos a cos /3 + sin a sin /3.
tan a ± tan (3
67. tan («**)-
58. sin 2a = 2 sin a cos a.
59. cos 2a = cos2 a — sin2 a.
60. =1—2 sin2 a.
61. = 2 cos2 a - 1.
2 tan or
62. tan 2a =
z — — r
1 — tan2 a
DO. sin £
64. cos K
RK tor, a
1 \
/I + COS a
' \ 2
VI — COS a
bo. tan o
66.
CT
1 + COS a
1 — COS a
sin a
sin a
a + /3 a —
68. sin a + sin /? = 2 sin — s — cos — s-
^ ^
a + /3 a —
69. sin a — sin /3 = 2 cos — o — sm "
a + 0
70. cos a + cos /3 = 2 cos — o — cos
71. cos a — cos /3 = — 2 sin — ^ — sin — ^ — •
& &
72. — — = — — a = ~ ' Law of sines.
sm or sm /3 sm 7
APPENDIX 291
73. a2 = 62 + c2 — 26c cos a. Law of cosines.
a- j8
tan 2 a — b
74. -; — ?• Law of tangents.
tan^3 a + b
75. Area of triangle = A/S(S — a)(s — 6)(s — c),
where 2s = a + 6 + c.
cfe sin a
7b. = 2
c2 sin a sin 8
77.
78. Radius of inscribed circle = A/
79. sin2 a = |(1 — cos 2a).
80. cos2 a = 5(1 + cos 2a).
1 — cos 2a
1 + cos 2a
(1 - cos 2a)2
sin2 2 a
sin2 2 a
81. tan2 a =
82.
00
(1 + cos 2a)2
84. sin 3a = 3 sin a — 4 sin3 a.
85. cos 3a = 4 cos3 a — 3 cos a.
3 tan a — tan3 a
86. tan 6 a = — ^ o'+TTi —
1 — o tan2 a
a°
87. a > sin a > a — -„- if a is expressed in radians.
a3 a3 a&
88. a — -^ < sin a < a — -vr H — ^ if a is expressed in radians.
o o O
If a is small and expressed in radians:
89. sin a = a approximately, or
a3
90. sin a = a — r _ is a better approximation, or
o
91. sin a = a — „•- + ~=- is a still better approximation.
a2
92. 1 > cos a > 1 — ~~- if a is expressed in radians.
a2 a2 a4
93. 1 — ;-„• < cos a < 1 — rfc- + —r- if a is expressed in radians.
I « Le 4t
292 MATHEMATICS
a* *
94. 1 —
*r
if a is expressed
in radians.
If a is small and measured in radians:
95. cos a = 1 approximately, or
96. cos a = 1 — TTT is a better approximation, or
a2 a4
97. cos a = 1 — pr- + nr is a still better approximation.
\— L*.
Analytical Geometry
98. y = az + b, slope equation of straight line.
99. x cos a + y sin a = p, normal equation of straight line.
100. x2 + y2 = r2, equation of circle, center at origin, radius r.
101. (x — a)2 + (y — /3)2 = r2, equation of circle, center at the point
(a, /3), radius r.
102. y = px2, equation of parabola with axis coinciding with F-axis.
103. y = p(x — a)2, equation of parabola with axis parallel to the
F-axis, but a units to its right (to its left if a is
negative).
x2 v2
104. — j + rj- = 1, equation of ellipse, axes coinciding with the co-
ordinate axes, foci upon X-axis.
105. Area of the ellipse = irab.
2^2 4*2
106. — j • — j-j = 1, equation of hyperbola, axes coinciding with co-
ordinate axes.
107. x2 — y- = a2, equation of equilateral hyperbola, axes coinciding
with coordinate axes.
108. xy = k, equation of equilateral hyperbola referred to asymptotes
as axes.
109. Ax + By + Cz = D, equation of plane.
110. x cos a + y cos |8 + z cos y = p, normal equation of plane.
111. x2 + y2 + z2 = r2, equation of sphere, center at origin, radius r.
112. (x — a)2 + (y — b)2 + (z — c)2 = r2, equation of sphere, cen-
ter at the point (a, b, c).
113. cos2 a + cos2 ft + cos27 = 1, where a, 0, and 7 are the direction
angles of a line.
APPENDIX
293
114. cos 0 — cos «i cos at + cos 0i cos 02 + cos 71 cos 72, where 0
is the angle between two lines whose direction angles are,
respectively, a\, 0i, 71, and «2, 02, 72.
MENSURATION
General Triangle
116. Area = %bp.
116. a + 0 + 7 = 180°.
117. 5 = a + 0.
^T
Right Triangle
118. 62 + a2 = c2.
119. p2 = mn.
120. b2 = en.
Circle
r = radius.
121. Circumference = 2-irr.
122. Area = Trr2.
s = length of arc ab.
123. Area of sector = %rs.
Sphere
radius = r.
124. Volume = f*-r3.
126. Area of surface = 4irr2.
h = altitude of zone.
126. Area of zone abed = 2*rh.
a, b = radii of bases of
segment.
127. Volume of segment =
294
MATHEMATICS
A..
Right Circular Cone
r = radius of base.
h = alitude.
s = slant height.
s = r2 + h2.
128. Volume = \irr*h.
129. Area of convex
= wrs.
surface
Frustum of Right Circular Cone
R = radius of lower base.
r = radius of upper base
h = altitude.
s = slant height.
s = V/i2 + (R - r)\
130. Area of convex surface =
irs(R - r).
131. Volume
= ~(R^ + Rr + r*).
132.
3(R - r)
Anchor Ring
of
generating
r = radius
circle.
R = mean radius.
133. Volume = 2w*r*R.
134. Area of surface = 4v*rR.
APPENDIX
Theory of Probability
295
h
""
136. y = — —e"h * , probability curve.
136. <r = \/— , standard deviation.
137. E =
n
0.6745
_
\/2/t>2, probable error of mean.
TABLE XVI.— DECIMAL EQUIVALENTS OF COMMON FRACTIONS
8ths | 16ths
8ths | 16ths
1/8 =0.125
1/4 =0.250
3/8 = 0.375
1/2 =0.500
1/16 = 0.0625
3/16 = 0.1875
5/16 = 0.3125
7/16 = 0.4375
5/8 =0.625
3/4 =0.750
7/8 =0.875
9/16 = 0.5625
11/16 = 0.6875
13/16 = 0.8125
15/16 = 0.9375
TABLE XVII.— CONVERTING INCHES INTO FEET
Inches
0
1/8
1/4 | 3/8
1/2 | 5/8 | 3/4
7/8
0
0.000
0.010
0.021
0.031
0.042
0.052
0.062
0.073
1
0.083
0.094
0.104
0.115
0.125
0.135
0.146
0.156
2
0.167
0.177
0.188
0.198
0.208
0.219
0.229
0.239
3
0.250
0.260
0.271
0.281
0.292
0.302
0.312
0.323
4
0.333
0.344
0.354
0.365
0.375
0.385
0.396
0.406
5
0.417
0.427
0.437
0.448
0.458
0.469
0.479
0.490
6
0.500
0.510
0.521
0.531
0.542
0.552
0.562
0.573
7
0.583
0.594
0.604
0.615
0.625
0.635
0.646
0.656
8
0.667
0.677
0.688
0.698
0.708
0.719
0.729
0.740
9
0.750
0.760
0.771
0.781
0.792
V. 802
0.812
0.823
10
0.833
0.844
0.854
0.865
0.875
0.885
0.896
0.906
11
0.917
0.927
0.937
0.948
0.958
0.969
0.979
0.990
296
MATHEMATICS
TABLE XVIII— BOAED MEASURE
Table giving the number of board feet in timbers of different
dimensions; 1 board foot = 144 cubic inches.
Size in inches
Length in feet
8
10 | 12 | 14
16 18
20
1 X 4
2f
31
4
4!
5i
6
61
1 X 6
4
5
6
7
8
9
10
1 X 8
5|
61
8
9*
lOf
12
13|
1 X 10
6f
8|
10
ill
13|
15
16§
1 X 12
8
10 ;
12
14
16
18
20
2X4
51
6f
8
»*
10|
12
13*
2X6
8
10
12
14
16
18
20
2X8
10f
13*
16
18f
21|
24
26f
2 X 10
is*
18f
20
23i
26|
30
33£
2 X 12
16
20
24
28
32
36
40
3X4
8
10
12
14
16
18
20
3X6
12
15
18
21
24
27
30
3X8
16
20
24
28
32
36
40
3 X 10
20
25
30
35
40
45
50
3 X 12
24
30
36
42
48
54
60
4X4
10f
13*
16
18f
21|
24
26f
4X6
16
20
24
28
32
36
40
4X8
21*
26|
32
37|
42|
48
53i
4 X 10
26f
33|
40
46|
53|
60
66f
4 X 12
32
40
48
56
64
72
80
6X6
24
30
36
42
48
54
60
6X8
32
40
48
56
64
72
80
6 X 10
40
50
60
70
80
90
100
6 X 12
48
60
72
84
96
108
120
8X8
42f
53 *
64
74f
85|
96
106f
8 X 10
53£
66|
80
93§
106f
120
133^
8 X 12
64
80
96
112
128
144
160
10 X 10
66f
83J
100
116f
133|
150
166!
10 X 12
80
100
120
140
160
180
200
12 X 12
96
120
144
168
192
216
240
APPENDIX
297
TABLE XIX— WATER CONTAINED IN 1 FOOT LENGTH OF CIR-
CULAR PIPE OF d INCHES INSIDE DIAMETER
d
Gallons
Weight Iba. | d
Gallons
Weight Ibs.
H
0.010
0.085
7
1.999
16.683
1 0.041
0.340
8
2.611
21.790
1M 0.092
0.766
10
4.080
34 . 048
2 0.163
1.362
12
5.875
49 . 028
2H
0.255
2.128
14
7.997
66.733
3
0.367
3.064
16
10.44
87.162
SM
0.500
4.171
18
13.22
110.314
4
0.653
5.448 .
20
16.32
136.190
5 1 . 020
8.512
22
19.75
164.790
6
1.469
12.257
24
23.50
196.114
TABLE XX— TABLE OF HEADS OF WATER CORRESPONDING TO
GIVEN PRESSURE AND OF PRESSURE CORRESPONDING TO
GIVEN HEAD
Pressure, Ibs. per sq. in.| Head, feet || Head, feet
Pressure, Ibs. per sq. in.
1
2.307
1
0.434
2
4.614
2
0.867
3
6.920
3
1.301
4
9.227
4
1.734
5
11.534
5
2.168
6
13.841
6
2.601
7
16.147
7
3.035
8
18.454
8
3.468
9
20.761
9
3.902
10
23.068
10
4.335
20
46.135
20
8.670
30
69.203
30
13.005
40
92.271
40
17.340
50
115.338
50
21.675
60
138.406
60
26.010
70
161.474
70
30.346
80
184.541
80
34.681
90
207.609
90
39.016
100
230.677
100
43.351
298
MATHEMATICS
TABLE
XXI.— CONVERSION TABLES
Lengths
1 centimeter
= 0.3937
inch.
1 meter
= 39.37
inches.
1 kilometer
= 0.62137
mile.
1 inch
= 2.540
centimeters.
1 foot
= 30.480
centimeters.
1 yard
1 mile (5280 feet)
= 0.9144
= 1.609
meter,
kilometers.
Areas
1 square centimeter
1 square meter
1 square kilometer
1 square inch
1 square foot
1 square mile
= 0.1550
= 10.764
= 0.3861
= 6.4516
= 0.0929
= 2.590
square inch,
square feet,
square mile,
square centimeters,
square meter,
square kilometers.
Volume
1 cubic centimeter
= 0.0610
cubic inch.
1 cubic meter
1 liter
1 cubic inch
= 1.308
= 61.023
= 16.387
cubic yards,
cubic inches = 1 . 057 quarts
(liquid),
cubic centimeters.
1 cubic foot
= 7.480
gallons (liquid) = 28.317
liters.
1 cubic yard
1 pint (liquid)
1 gallon
= 0.7646
= 28.875
= 231
cubic meter,
cubic inches = 0.4732 liter,
cubic inches = 3. 785 liters.
Weights
1 gram
1 kilogram
1 grain
= 15.432
= 2.2046
= 0.0022857
grains = 0.03527 ounce
(Av.).
pounds (Av.).
ounce (Av.) = 0.064799
1 ounce (Av.)
1 pound (Av.)
= 437.5
= 0.45359
gram,
grains = 28 . 3495 grams,
kilogram = 7000 grains.
APPENDIX
299
Pressure
1 foot of water column = 0.4335
= 22.419
= 0.8826
= 0.0295
1 inch of mercury column = 1 . 133
= 0.4912
= 0.03342
1 pound per square inch
1 atmosphere
= 51.712
= 2.307
= 2.036
= 760
= 33,901
= 29.921
= 14.697
pounds per square inch.
millimeters of mercury.
inch of mercury.
atmosphere.
feet of water.
pound per square inch.
atmosphere.
millimeters of mercury.
feet of water.
inches of mercury.
millimeters of mercury.
feet of water.
inches of mercury.
pounds per square inch.
Weight and Volume of Water
1 cubic centimeter water
1 pint (liquid) water
1 cubic foot water
1 ounce (Av.)
= 1 gram.
= 1 . 043 pounds.
= 62.428 pounds (Av.).
= 28.35 cubic centimeters water.
= 1.730 cubic inches water.
1 cubic foot per second
1 acre foot per day
Flow of Water
= 450
= 0.5042
gallons per minute (approxi-
mately),
cubic foot per second. •
1 part per million
1 grain per gallon
Salts in Solution
= 0.05842 grain per gallon.
= 17.118 parts per million.
300
MATHEMATICS
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INDEX
Abscissa of point, 33
axis of, 25
Acceleration of falling body, 51
Accuracy of logarithmic com-
putation, 143
Addition, 12
formulas, 108
vector, 156
Aggregation, symbols of, 16
Algebra, formulas of, 288
review of, 12
Alphabet, Greek, 286
Analytic geometry, formulas of,
292
Angle of depression, 86
direction, 243
of elevation, 86
initial side of, 80
to measure an arc, 3
sense of, 80
terminal side of, 80
unit of, 91
vertex of, 80
Angular magnitude, 80
Anti-logarithmic tables, 73
Anti-logarithms found from loga-
rithmic tables, 78
Approximation formulas, 196
Area of ellipse, 136
of triangles, 71, 110, 111
Arrangement of logarithmic
work, 71, 74
Axes, coordinate, 25
marking, 30
of the ellipse, 133
Axis, X-, 33
Y-, 33
of abscissas, 25
of ordinates, 25
of parabola, 25
Bacteria, growth of, .66
Base of logarithmic system, 61
change of logarithmic, 66
Binomial coefficients, 195
expansion, 189, 195
for negative and fractional
exponents, 195
theorem, 194, 195
Biological measurements, 229
Board, drawing, 2
Boyle's law, 58
Charts, precipitation, 30
Characteristic of logarithm, 64
Circle, equation of, 37
orthographic projection of,
138
Circular measure, 91
Coefficients, binomial, 195
determined by the method
of least squares, 268
Combinations, 189, 190, 193
formula for number of, 193
Compasses, 3
Compound interest law, 66, 197
Compression, 179
of air, 58
Computation, accuracy of loga-
rithmic, 143
303
304
INDEX
Concrete, cost of, 47
mixture, 47
Cone, section of. 139
Coordinate axes, 25
paper, 25
planes, 239
x-, 34
V, 34
Coordinates of a point, 33, 239
Corrections to logarithms, 70
Cosines, direction, 243
law of, 105
Cost of concrete, 47
Couple, 170
moment of, 170, 171
Count, least, 231
Crane, simple, 164
Cube roots, found with the slide
rule, 153
Cross section, normal, of cylinder,
115
Current meter, 54
Curve, decreasing, 52
error, 224
equation of, 37
increasing, 52
of intersection of cylinder
and plane, 114, 118
normal distribution, 224
probability, 220, 223, 224
sine, 93
tangent, 94
line of symmetry of, 52
point of symmetry of, 52
Decreasing curve, 52
Degree, 81
Dependent events, 210
Depression, angle of, 86
Deviation, standard, 226, 227
Difference, tabular, 68
angles of line, 243
Difference, cosines of line, 243
negative, 24
positive, 24
Directions, general, for preparing
work, 5
Directrix of parabola, 52
Distance between two points, 240
focal, 136
Division, 15
with slide rule, 152
Eccentricity of the ellipse, 130
of the circle, 136
Elevation, angle of, 86
Ellipse, 128
axis of, 133
area of, 136
eccentricity of, 134, 136
equation of, 129
focal distance of, 136
foci of, 134
general shape of, 130
locating points upon, 133
sum of focal radii of, 134
tangent to, 139
Empirical, equation, 47, 54, 57
equations, 264, 266
Equation, of circle, 37
of curve, 37
of ellipse, 129
empirical, 47
graph of, 37
linear, 44
of locus, 241, 243
of plane, 245
solution of, by factoring, 130
of straight line, 40
of straight line through two
points, 45
of the first degree between
x and y, 43
Equations, conditional, 45
INDEX
305
Equafions, empirical, 264, 266
of equilibrium, 176
plotted upon logarithmic
paper, 281
quadratic, 21
Equilibrium, equations of, 176
of a system of parallel forces,
171
Error, probable, 228
Errors, 231
of area of triangle, 236, 237
of cube, 235
of cube root, 235
of observation, 220
of product, 232
of quotient, 233
of sine, 235
of square, 233
of square root, 234
small, 231
Events, dependent, 210
independent, 210
mutually exclusive, 209
Expansion, binomial, 189, 190,
195
Expectation, 212
Exponents, 20
Factoring, 16
Falling body, distance passed
over by, 51
velocity of, 51
acceleration of, 51
Functions, graph of trigonomet-
ric, 92
line representation of the
trigonometric, 94
Focal distance, 136 K
radii, sum of, 134
Foci of the ellipse, 134
Focus of the ellipse, 134
of the parabola, 52
20
Force, concurrent, 159
coplanar, 159
moment of, 165
Forces, in equilibrium, 170
polygon of, 161
resultant of, 161
Formulas, addition, 108
of algebra, 288
approximation, 196
of analytic geometry, 292
of theory of probability, 295
of trigonometry, 289
of mensuration, 293
Fractions, 18
Frequency distribution polygon,
222
Function, 84
Functions, circular, 80, 84
even, 90
odd, 90
trigonometric, 84
of 0°, 90°, 180°, and 270°, 87
Grade, percent, 42
Gradient, hydraulic, 42, 45
Graph, of an equation, 37
of an equation of the first
degree in x and y, 43
of y = px2, 49
Graphic tables of the trigono-
metric functions, 96
Graphs of the trigonometric
functions, 92
Greek alphabet, 286
Head of water in well, 46
Hyperbola, equilateral, 23, 55
Imaginary, 132
Increasing curve, 52
Independent events, 210
Infinity, 88
306
INDEX
Instruments, 1
Intercept, x-, 44
y-, 42
Law, compound interest, 66, 197
of cosines, 105
empirical, 47
of freely falling body, 51
of sines, 104
of tangents, 106
Least squares, 227
coefficients determined by,
268
Line to draw, 6
direction angles of, 244
cosines of, 243
equation of, 40, 45
projection of, 244
slope of, 42
straight, 23
of symmetry, 52
Lines, parallel, 8
Locus, equation of, 241, 243
Logarithm, 61
base of, 61
characteristic of, 64
common, 66
hyperbolic, 66
Naperian, 66
natural, 66
mantissa of, 64
of power, 64
of product, 62
of quotient, 63
of root, 64
Logarithmic curve, 73
paper, 270, 277
scales, 74, 147, 148
tables, 67
Materials, 1
Magnitude, angular, 80
Mantissa of logarithm, 64
Maxima amd minima, 252, 256,
259
Maximum or minimum of
ax2 + bx + c, 255
Mean, arithmetic, 220, 228
Measurements, biological, 229
Mensuration, formulas of, 293
Meter, current, 4
Minima (See maxima)
Minimum (See maximum)
Minute, 81
Moment, 168
of couple, 170
of force, 165
Moments, origin of, 168
sum of, 174
Multiple logarithmic paper, 277
Multiplication, 13
tables, 229
with slide rule, 151
Normal cross-section of cylinder,
115
equation of plane, 245
Octants, 239
Ordinate, 25, 33
Origin, 25, 239
of coordinates, 25
of moments, 168
Orthographic projection, 136
of circle, 138
Paper, coordinate, 25
logarithmic, 270
polar coordinate, 4
semi-logarithmic, 283
squared, 25
Parabola, 23, 52
axis of, 52
directrix of, 52
INDEX
307
Parabola, equation of, 52
focus of, 52
line of symmetry of, 52
property of, 52
vertex of, 52
Paraboloid of revolution, 53
Parallel forces, 165, 166, 170
Pattern for circular roof, 117
for right circular cylindrical
surface cut by a plane,
116
for saddle of ventilator, 78,
115
Pencil points, to sharpen, 1
Pencils, 4H, 1
3H, 2
chisel pointed, 1, 2
drawing, 1
hexagonal, 2
Permutations, 189
formula for number of, 192,
193
Plane, slope of, 247
normal equation of, 245
Planes, coordinate, 239
Plolling statistical data, 28
Point, 33, 239
abscissa of, 33
coordinates of, 33
ordinate of, 33
of symmetry, 52
Points, distance between, 240
locating, upon ellipse, 133
Polar coordinate paper, 4, 96
Polygon, force, 161
frequency distribution, 222,
225
Precipitation charts, 30
Precision, measure of, 226
Pressure, relation between vol-
ume of air and, 58
measured in inches of mer-
cury and feet of water,
36
Principle of slide rule, 146
Probability, 207
curve, 223, 224
theory of, formulas for, 295
Product, error of, 232
logarithm of, 62
Progression, arithmetical, 200
geometrical, 200, 203
infinite geometrical, 205
Projection, of line, 244
orthographic, 136
Proportional parts, tables of, 70
Protractor, 3
Pulley, rope and, 185
Quadrants, 34
Quadratic, general, 53
Quotient, error of, 233
logarithm of, 63
Radian, 91
Radicals, 21
Residual, 227
Resultant, force, 159, 161
velocity, 158
Revolution, solid of, 76
paraboloid of, 53
Rock, voids of, 47
Roof, pattern for conical, 117
volume of conical, 123
Root, logarithm of, 64
Rope and pulley, 185
Rotation, positive, 80
negative, 80
Saddle for ventilator, paitern for,
78
Sand-paper, 2
Sand voids of, 47
Scale, 3, 23
308
INDEX
Scale chain, 23
double, 36, 74
drawing to, 23
full divided, 23
logarithmic, 74
non-uniform, 23, 74, 124
open divided, 23
triangular, 4
Scales, double, 148
logarithmic, 147, 148
Second, 81
Section of cone, 139
Semi-logarithmic paper, 283
Sine curve, 93
Sines, law of, 104
Slide rule, 143, 150
cube roots found with, 153
division with, 152
multiplication with, 151
principle of, 146
square roots found with, 153
squares found with, 153
Slope, equation of straight line,
42
of line, 42
of plane, 247
Solid of revolution, 76
Solution of triangle, 98
Square, error of, 233
root, error of, 234
found with slide rule, 153
T-, 2
Squares, found with slide rule,
153
Statics, 155
Subtraction, 12
vector, 156
Sucessive trials, 212
Sum, of moments, 175
vector, 156
Symbols, 286
of aggregation, 16
Symmetry, line of, 52
point of, 52
Table of board measure, 295
converting inches into feet
295
conversion, 298
of decimal equivalents of
common fractions, 295
of head of water and pres-
sure, 297
of natural trigonometric
functions, 300
of water contained in cir-
cular pipe, 297
Tables, anti-logarithmic, 73
logarithmic, 5, 68
multiplication, 229
of proportional parts, 70
of trigonometric functions,
96, 300
Tangent to curve, 94
to ellipse, 139
to y = pxz, 50
Tangents, law of, 106
Temperature, relative between,
and density of water,
54
Tension, 179, 186
Terms of (p + q)n, 215
Translating curve on logarithmic
paper, 280
Triangle, 80
area of, 71, 110, 111
error of area of, 236, 237
drawing, 2
of reference, 84
solution of, 98
60°, 2
45°, 2
Trigonometry, formulas of, 289
Trials, successive, 210
INDEX
309
Unit, circular, 91
of angular measure, 81, 91
Variation, direct, 41
inverse, 56
Velocity of water through sand,
46
of falling body, 51
curve, 54
in lower Mississippi, 55
resultant, 158
Vector, 155
addition, 157
negative, 157
sum, 157
subtraction, 157
Ventilator, pattern for saddle for,
78
Versed sine, 95
Vertex of angle, 80
of parabola, 52
Voids of rock, 47
of sand, 47
Volume of air under pressure, 56
of conical roof, 123
of right circular cylindrical
shell, 120
of solid of revolution, 77
Water, relation between tempera-
ture and density of, 54
velocity of, through sand, 46
Well, yield of, 46
Work, arrangement of loga-
rithmic, 71, 74
blocking out of, 71
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